MATHEMATICAL ANALYSIS

MATHEMATICAL ANALYSIS
SECOND EDIT/ON
TOM M. APOSTOL
California Institute of Teclmologj
ADDISON- PUBLISHING COMPANY
Read.g, Maadausetts
Amsterdam • London • Manila • Singapore. Sydney • Tokyo

WORLD STUDENT SERIES EDITION
FIFTH PRINTING 1981
To my parents
A complete nd unabridged reprint of the original American textbook, this World
Student Series edition may be sold only in thoae countries to which it is o3nIgned
by Addison-Wedey or its authorized tade ditrlbutorx It may not be re-exported
from the country to which it has been consigned, and it may not be old in the
United States of America or its poaeions.
Consulting Editor: Lyn Loomis
Copyright © 1974 by Addison-Wesley Publishing Company, Inc. Philippines copy-
right 1974 by Addison-Wesley Publishing Company, Inc.
All right reserved. No part of  publication may be reproduced, tored in a
trieval system, oz transmitted in any form o by aay means, electronic, mechanical,
photooopying, recording or otberwie, without the prior written permission of tbe
publisher. Original edition ptnted in the United States of Am©rica. Publixbed ximul.
taneously in Canada. Library of Congre Catalog Card No. 72-11473.

PREFACE
A glance at the table of contents will reveal that this textbook treats topics in
analysis at the "Advanced Calculus" Ievel. The aim has been to provide a develop-
ment of the subject .which is honest, rigorous, up to date, and, at the same time,
not too pedantic. The book provides a transition from elementary calculus to
advanced courses in real and complex function theory, and it introduces the reader
to some of the abstract thinking that pervades modem analysis.
The scfond edition differs from the first in many respects. Point se topology
is developed in the setting of general metric spaces as well as in Euclidean n-space,
and two new chapters have been added on Lebesguc integration. The material on
line integrals, vector analysis, and surface integraIs has been deleted. The order of
some chapters has been rearranged, many sections have been completely rewritten,
and several new exerc/ses have been added.
The development of Lehesgue integration follows the Riesz-Nagy approach
which focuses directly on functions and their integrals and does not depend on
measure theory. The treatment here is simplified, spread out, and somewhat
rearranged for presentation at the undergraduate level.
The first edition has been used in mathematics courses at a variety of levels,
from first-year undergraduate to first-year graduate, both as a text and as supple-
mentary reference. The second edition preserves this flexibility. For example,
Chapters 1 through 5, 12, and 13 prosdde a course in differential calculus of func-
tions of one or more variables. Chapters 6 through lt, 14, nd 15 provide a course
in integration theory. Many other combinations are possible; individual instractors
can choose topics to suit their needs by cosulting the diagram on the next page,
which displays the logical interdependence of the chapters.
I would like to express my gratitude to the many people who have taken the
trouble to write me about the first edition, Their comments and suggestions
influenced the preparation of the second edition. Special thanks are due
Charalambos Aliprantis who carefully read the entire manuscript and made
numerous helpful suggestions. He also provided some of ¢ new exercises.
Finally, I would like to acknowledge my debt to the undergraduate students of
CaItech whose enthusiasm for mathematics provided the original incentive for this
work.
Paxadena T.M.A.
September 1973

LOGICAL I/qTERDEPENDENCE OF THE CHAPTERS
THE REAL AND COM-
PLEX NUMBF__R SYSTF_S
SOIVI BASIC NOTIONS
OF SET TItEORY
ELEMENTS OF POINT
8ET TOPOLOGY
1
CONTI17UITY
DERSVATIVE8
FTINCTIONS OF BOUNDED
VARIATION AND REC-
TIFIABLE CURVF_
s I 12 /
INFINITE SERIES AND [ MULT[VARIABLE DIF-
[ "'HE RIMANN-I I IM,L,CIT CTIONS ]
/ I 8'rlELTJE$ INTF_RAL I I AND EXTREMUM
! I " -
S 1. UNCES OF
10 "
 B
I ,1
FOURIER SERIES AND
FOURIER INTEDRALS
, 16 
CAUCHY S THEOREM AND
[ THE RESIDUE CALCULUS
.MULTIPLE LEBF_,UE
INTEGRALS
CONTENTS
1
1.1
1.2
1.4
1.5
1.6
1.7
1.8
1.9
1.10
1.11
1.12
1.13
!.14
!.15
1,16
1.17
1A8
1,19
1.20
1.21
1.22
] .23
1.24
1.25
1,26
1,27
1.28
1,29
1.30
1.31
1.32
] .33
The  and Complex Nmnl Sstems
Introduction ................. l
The field axioms ................ I
The order axioms ............... 2
Geometric representation of real numbers ........ 3
Intervals ................. 3
Inte/xs ................. 4
Tim unique factorization theorem for integm's ....... 4
Rational numbers ............... 6
Irrational numbers ............... 7
Upper bounds, maximum element, least upper bound
(supmmum) ................ 8
The complcCmcss axiom ............. 9
Some propgtti¢ of the suprcmum .......... 9
Properties of tile integer deduced from the completeness axiom , 10
Tim Archimedean property of tho real-numtmr system ..... 10
Rational numbers with finite decimal ropresentation ..... 11
Finite decimal approximations to real mmabers ....... 11
Infinite dccimaI representatitm of real nmnbers ....... ! 2
Absolute values and the triangle inexlualily ........ 12
The Caue.hy-Schwarz inequality ........... 13
Plus and mus infinity and the extended real number system R* 14
Complex numbm's ............. 15
Gometric reprcstation of complex numbers ....... 17
Tim imaginary unit ............... 18
Absolute value of a complex number .......... 18
Impossibility of ordering th, complex numbers ....... 19
Complex exponentials ............. 19
Furth properties of complex ¢aporntials ........ 20
The argument of a cmplcx number .......... 20
ntcgral powers and roots of complex numbers ....... 21
Complex logarithms .............. 22
Complex powers ............... 2
Complex sines and cosines ............. 24
Infinity and the ¢xtmaded ¢omplt plan¢ C* ........ 24
Excrdr .................. 2S

2,2
2`3
2.4
2.5
2.6
2,7
2.8
2.9
2.10
2.12
2.13
2.i4
2.15
Introducti ................. 32
Notions ................. 32
Ordered pairs .............. 33
 product of two  ............ 33
Relations and functions ............. 34
Fuzthcr terminology concerning functions ........ 35
On-to-onc functions and invccscs .......... 36
ComIxite functions .............. 37
Sequences ................ 37
Similar (equinomerous) s6ts ............ 38
Finite and infinite  .............. 38
Cotmtsbl and uncountable sets .......... 39
Uncountability of the real-number systn ........ 39
Set algebra ............... 40
Countable collectio of countable sets ......... 42
Ex.eises .................. 43
 3
3,1
3.2
3.3
3,4
3,5
3.6
3.7
3.8
3,9
3.10
3.11
3.12
3-t3
3.14
3,15
3,16
Introduction ................ 47
Euclidean space R m ............... 47
Open balls and open sets in R" ........... 49
Th sttre of  sets in R 1 ........... 5{)
Cleed se[s ............... 52
Adherent points, Accumulation poinla .......... 52
Clol   adherent points ........... 53
The ]lzmzo-Weiecstras  .......... 54
Tim Cantr intexsecfioa theonm ........... 56
The Lindel6f covexin8 theon=n ........... 56
The l-leine-lkrel covering theorem .......... 58
Compactne in R" ............... 59
Metric spao ................ 60
Point se topolo" in metric spa3es .......... 61
Compact stabbers of a ractric SltCe .......... 63
Boundary of a sol ............... 64
4.1
4.2
4.3
4.4
4.5
4.6
Introduction ................ 70
Convergent sequences in a metric  ......... 70
Cauchy scqucnccs ............... 72
Complete metric spaces ............. 74
Limit of a function .............. 74
Limits of comple-valu:l functions .......... 76
4.7 Limits of vector-valued functions ........... 77
4.8 Continuous functions .............. 78
4,9 Continuity of composite functions ........... 79
4.10 Continuous complex-valued and vector-valued functions .... 80
4.11 Examples of continuous functions  ...... 80
4.12 Continuity and inverse imag of op¢ or closl sets ..... 81
4,13 Functions continuous on ctnpact ses ....... 82
4.14 Topological mappings (homeomorphisms) ........ 84
4.15 Bolzaxzo's tb_corcm ..... 84
4.16 Conntdncs .....
4.17 Components of • metric spa.
4. I 8 Arcwi connctcdness ....
d,.19 Uniform continuily ..... 90
4.20 Uniform continuiiy and compact se
4.21 Fized-point tlotea for contractions ......... 92
4.22 Disc, ontinuitics of rcal-valucxl functions ......... 92
4.23 Monotonic functions .............. 94
ses .................. 95
5,2
5.3
5A
5.5
5.6
5.7
5.8
5.9
5.10
5,11
5,12
5,13
5.14
5.15
5.16
¢a0te 6
6.1
6.2
6.3
6.4
6,5
Introduction ................. t
Definition of derivatiw .............. 1{)4
Derivatives and continuity ............. 105
Algebra of derivatives ............. 106
The chain rule ................ 106
One-sided derivative and infinite derivatives ....... 107
Functions with nonzero d'ivative .......... 108
Zero der[vatiws and local extrema .......... 109
Rolle's theorem ................ 110
The Mean,Value Theorem for dexivativ ........ 110
Intermediale-valu thorn for derivativ ........ I I 1
Taylor's formula with remainder ........... 113
Derivatives of veor-valued functions ......... 114
Partial dexivatiws ............... 115
Diff¢i-¢ntiation of functions of a complex variable ...... 116
Tl Cauchy-Ricmarm cquatiom ........... 118
Exercises . . . ............... 121
Ihaios of Bounded Variation and Rectifiable Cta'ves
Introduction ................. 127
Properlies of monotonic functions .......... 127
Functions of bounded variation ........... 128
Toter variation . • ............. 129
Additive property of total variation .......... t 30

6,6 Total variation on [a, x] as a function of x ........ 131
6.7 Fi of  v]ation p  e d of
ing funi ............... 132
6.8 Continus fui of  a .......  32
6.9
6A0 ible t
6A I Additi  loily p of  lh ...... 135
6A2 uiv of pa    ....... 136
 ................. 137
Chaler 7
7.2
7,3
7.4
7.5
7.6
7.7
7.8
7.9
7,10
7,11
7,12
7.13
7.14
7.15
7,I6
7.17
7.18
7.19
7.20
7,21
7,22
7.23
7.24
7.25
7.26
7.27
8,1
&2
8.3
&4
8.5
Introduction ................. 140
Nolation .................. 141
The definition of the Riemann-Slieltjes integral ....... 141
Linear properties ............... 142
Integration by parts ............... 144
Change of variable in a Riemann-Stieltjes integral ...... 144
Reductioa to a Riemaan integral ........... 145
Step functions as integrators ............ 147
Reduction of a RieanannStiettjes integral to a finite sum . 148
Euler's summation formula ............ 149
Monotonically increasing integrators. Upper and lower integrals. 150
Additive and linearity propertic of upper and lower integrals 153
Ricanann'$ condition .............. 153
Comparison theorems ..............
Integrators of hounded variation ...........
Sufficient conditions for existence of gicmann-Stieltjes integnds . 159
Necessary condilions for ey, istertce of Riemann-$tieltjes integrals. 160
Mean Value Theorems fdr Riemann-Sfieltjes integrals ..... 160
The integral as a function of the interval ......... 161
Second fundamental theorem of integral calculus ..... 162
Change of variable in a Riemann integral ........ 163
Second Mean-Value Theorem for Rieanann integrals ..... 165
Rk:mann-Stieltjes integrals depending on a ly, trameter ..... 166
Differentiation under the integral sign ......... 167
Interchanging the order of inte, grat[on ......... 167
Lebesgue's criterion for existence of Riemann integrals . . ' . . 169
Complex-val0ed Riemann-Stieltjes integrals ........ 173
Exercises .................. 174
Infinite Secies and Infinite Prodncts
Introduction ...............
Convergent and divergent sequences of complex numbers . 183
Limit superior and Iimit ifcrior of a real-valued sequence 184
Monotonic sequences of real numbers ......... 185
Infinite series ................ 185
8.6 Inserting and removing parentheses . 187
8.7 Alternating series ......... 188
8.8 Aisolute and conditional convergence . 189
8,9 Real and imaginary parts of a complex series . 189
8,10 Tests for convergence of series with positive terms 190
8,11 The geometric scics ........ 190
8.12 The integral test .......... 191
8.13 The big oh and little oh notation ..... 192
8.14 The ratio test and the root test ..... 193
8,15 Dirichl's est and Abel's test ............ 193
8,16 Partialsurnsoftlgeometrieserics £zwontlunitcirclelz[ = i , 195
8.17 Rearrangements ofsexies ............. 196
8,18 Ricmann's theorem on conditionally convergt series ..... 197
8.19 Subseries ................. 197
8.20 Double sequences .......... " ..... I99
8,21 Double series ................ 200
8,32 Reaxrmagement theorem for double scrim ....... 201
8.23 A sufficient condition for equality of iterated series ...... 202
8,24 Multiplication of scxies .............. 203
8,25  summability ............... 205
8.26 Infinite poducs ................ 206
8.27 Euler's product for the Riemann zm function ....... 209
Exercises .............. 210
 9
9.2
9.3
9.4
9.5
9.6
9.7
9.8
9.9
9.10
9,11
9,12
9.13
9.14
9,15
9.16
9.17
9.18
9.19
9.20
9.21
Sequ of Fanetims
Pointwise convergence of scquerg of functions ...... 2t8
Examples of sequences of re, at-valued functions ....... 219
Definition of uniform convergence .......... 220
Uniform concergenc¢ and continuity .......... 22I
The Cauchy condition for uniform convergen ...... 222
Uniform convergence of infinite series of functions ...... 223
A space-filling curee ............. 224
Uniform convergence and Riemann-Stieltjes integration 225
Nonuniformly convergent sequences that can be integratod term by
term ................... 226
Uniform convernce and differentiation ........ 228
Sufficient conditions for uniform convergence of a series 230
Uniform convergence and double S:lUCnCeS ...... 23t
Mean convergnc ............. 232
Power se, xis ................. 234
Multiplication of power reties ......... 237
The substitution theorem .......... 238
Reciprocal of a power series ......... 239
Real IOWer scris ............ 240
The Taylor's series generalcd by a function ........ 241
Bcrnstein's theorem ............... 242
The binomial series ............... 244

9.22 Abel's limit theorem .............. 244
9.23 Tauber's theorem ............... 246
Exercises .................. 247
10,!
I0.2
I0.3
10,4
10.5
10.6
10.7
I0.8
10.9
10.10
10,11
10,12
10.13
10.14
10,15
10.16
10.17
10.18
10.19
10220
10.21
10.22
10.23
10.24
10.25
Introduction ................. 252
Th integral of a st function ........... 253
Monotonic sequences of step functiom ....... .÷ . 2.91
Upper fimctiom and tlmir integrals ....... 256
Riemann-integrable functions as ey.amples of uPlX:r functions 259
The class of Lebesgue-integrable function on a 8tetal interval . 260
Basic propertie of the Lebesgue integral ......... 261
Letsgu¢ integration and sets of measure zexo ....... 264
Th L monoton convergen tho0nm ........ 265
Applicatiom of Lebesgn's dominatod convexmc¢ theorem 272
Lebesgue integrals on unboumld inttzvals as limit, of integra on
hotmded in ............... 274
Impropex Ritm'mnn inttaI$ ............ 276
Mmu-abi¢ functions .............. 279
Continuity of funetiot defined by Lcbc,tm inttal$ ..... 281
Differentiation under the intral sign ......... 283
Interchanging the ordta" of integration ......... 287
Measurable sts on tim real line .......... 289
Th Lebesgue integral over arbitrary sabsts of ! ..... 291
I_,besgue integrals of complex-valued functions ....... 292
Inner products and norms ............ 293
The set L 2( I) of squam-intcgrable functions ....... 294
The set L(I) as a semimetri¢ space ......... 295
A on theor,m for sics of fmiom in L2(/) .... 295
The Ri0sz-Fisehtr t ............ 297
Omler 11
I!.1
11.2
11,3
!1.4
11.5
11.6
11.7
11.8
11.9
ll.10
ll.ll
Introduction ................. 6
Orthogonal systems of futfiorm ........... 306
The theorem on best approximation .......... 307
The Fourier series of a function relative to an orthonorma! system . 309
Properties of the Fourier coeffents . . • ........ 309
The Ricsz-Fischer theorem . , ' ........ . . 311
The convergenc and representation problems for trigonometric series 312
The Ri¢mann-Lobtzm lemma ........... 313
Tlm Dirichlet integrals ............ ' . 314
An integral reprtmtation for the partial sums ofa Fourior seri 317
Riemann's localization theorem ........... 318
I1.12
11.13
11.14
11.15
11.16
11.17
11.18
11,19
11.20
11.21
! 1.22
Clmpl 12
12.1
12.2
12.3
12.4
12.5
[2.6
12.7
12,8
12.9
12.10
12.11
12.12
t2+13
12.14
Clmpler 13
13.1
13.2
133
13.5
13.6
13.7
Otaer 14
t4.1
Suffichmt cnditions for contence of a Fourie, x sics at a particular
point ...................
Cesro summability of Fourier series ..........
Consequence, of Fej,r's theorem ...........
The Weierstras approximation theorem ........
Other forms of Fourier series ...........
The Fourier inteal theorem ...........
The exponential form of the Fourier integral theorem ....
Integral transforms . . . . - ..........
Convolutions ...............
The convolution theotam for Fourier transforms .....
The Poisson summation formula .......
Exercises .................
Multivariable Differottial Calculus
319
319
321
322
322
323
325
326
327
329
332
335
Multiple Riemann Integrals
Introduction ................. 388
The measure of a bounded interval in R" . ....... 388
lmldicit Functits and Exlremum Problems
Introduction ................. 367
Functions with nonzero Jacobian determinant ....... 368
The inverse function theorem ............ 372
The.implicit function theorem ......... 373
Extrema of real-valued functions of one variable . ' ..... 375
Extrema of real-valued functions of several variables ..... 376
Extremum problems with side conditions ........ 380
Exercises .................. 384
Introduction ................. 344
The directional derivative ............. 346
Directional derivatives and continuity ..... 345
The total derivative ............... 346
The total derivative expressed in terms of partial derivatives 347
Art application to complex-valued functions ...... 348
The matrix of a linear function ........... 349
The Jacobian matrix .............. 351
The chain rule ................ 352
Matrix form of the chain rnle ............ 353
The Mean-Value Theorem for differentiable functions ..... 355
A sutficient condition for differentiability ........ 357
A sufficient condition for equality of mixed partial derivatives 358
Taytor's formula for functions from li" to R x ....... 361
Exercises .................. 362

14.3
14.4
14.6
14.7
14.8
14.9
14.10
The Riemann integral of a bounded function defined on a compact
interval in IP ................. 389
Sets of measure zero and Lcbesgac's criterion for istc¢ of a multiple
R icmann integral ............... 391
Evaluation of a multiple integral by iterated integration .... 391
Jordan-measurable sets in R" ............ 396
Multiple integration over Jordan-measurable sets ...... 39"/
Additive property of the Riemaxm integral ........ 399
Mean-Value Theorem for multiple integrals ........ 40
Chalter 15
15.1
15.2
15,3
15.4
i55
15,6
15,7
15,8
15,9
15.10
15.11
15.12
I5.I3
Introduction ................. 405
Step futio and tleit intcgmh ........... 40
Upper functions end Lehesgue-intezble functiom ..... 406
Measurable ftmctions and meastmtble sets in R e ...... 407
Fubini's reduction thcoztmi for ttm double inlegml of a step funation 409
Som¢ properties of sets of nmasum zro ......... 411
Fubini's reduction theorem for double integrals ...... 413
The Tone, ifi-Hobson test for inteability ........ 415
Coord/nate transformations ............ 4I 6
The transformation formula for mutdpl© integrals ...... 421
Proof of the transformation formula for linear coordinate tmnsforma-
tioas ............... . .... 421
Proof of the transformation formula for the ch function of a
compact cube ................ 423
Completion of the proof of the transformation formula .... 429
Exercises .................. 430
Olalma" 16
16.1
16.2
16.3
16,4
16,5
16,6
16,7
16.8
16.9
16.10
16.11
16.12
16.13
16.14
16.15
Canchy's Theorem and the Residue Caleu
Analytic functions ............... 434
Paths and cttrves in the complea plane ......... 435
Contour integrals ............... 436
The integral along a circular path as a function of the radius. 438
Cauchy's integral thcortm for a c'mzl¢ ......... 439
Homotopic curves ............... 439
Invariance of contour integrals under homolopy ...... 442
General form of Cauchy's inVral theorem ........ 443
Cauchy's integral formula ............. 443
The winding numlr of a circuit with respect to a point .... 444
• The, unbouaOcdnoss of the, sot of points with winding number ztro 446
Analytic functi0as dofa 1 contour integrals ....... 44"/
Pow'-scrics ¢xpam/ons for analytic functions ....... 449
Cauchy's incqties. Liouville's tbtorcm ........ 450
Isolation of the zros of an analytic futon ....... 451
16,16
16.17
16.18
16.19
16.20
16.21
16.22
16,23
16,24
16.25
16.26
16.27
Cemtents zvB
The identity theorem for analytic functions ........ 452
The maximum and minimum modulus of an analytic function 453
The op0n mapping theorem ............ 454
Laurent expansions for functions analytic in an anndus . • . . . 455
Isolated singularities  ............. 457
The residue of a function at an isolated singular point ..... 459
The Cauchy residue theorem ............ 460
Counting zeros and poles in a region .......... 461
Evaluation of real-valued integrals by means of residues 462
Evaluation of Gauss's sum by residue calculus ....... 464
Application of the residue theorem to the inversion formula for Laplac
transforms ............. 468
Conformal mappings .............. 470
Exercises .................. 472
ladex of Speeial Symbels ............ 481

CHAPTER
THE REAL AND
COMPLEX NUMBER SYSTEMS
1.1 IN'IIIODUL--TION
Mathematical analysis studies concepts related in some way to real numbers, so
we begin our study of analysis with a discussion of the real-number system.
Several methods arc used to introduce real numhtm. One method starts with
the positive inters I, 2, 3,... as undefined concepts and uses them to build a
larger system, the positive rational numbers (quotients of positive integers), their
negatives, and zero. The rational numbers, in turn, axe then used to constt-uct the
irrational numbers, real numbers like  and vhich are not rational. The rational
and irrational numbers together constitute the real-number system.
Although these matters are an important part of the foundations of math-
ematics, they will not be described in detail here. As a matter of fact, in most
phases of analysis it is only the properties of real numbers that concern us, rather
than the methods used to construct them. Therefore, we [halt take the real numbers
themselves as undefined objects satisfying certain axioms from which further
properties will be derived. Since the reader is probably familiar with most of the
properties of real numbers discussed in the next few pages, the presentation will
be rather brief. Its purpose is to review the important features and persuade the
reader that, if it were necessary to do so, all the properties could be traced back
to the axioms. More detailed treatments can be found in the references at the end
of this chapter.
For convenience we use some elementary set notation and terminology. Let
S denote a set (a collection of objects). The notation x ¢ S means that the object x
is in the set S, and we write x ¢ S to indicate that x is not in S.
A set S is said to be a subset of T, and we write S _ T, if every object in S is
also in T. A set is called nonempty if it contains at least one object.
We assume there exists a nonempty set R of objects, called real numbers,
which satisfy the ten axioms listed below. The axioms fall in a natural way into
three groups which we refer to as the field axioms, the order axioms, and the
completenes, axiom (also called the least-upper-bound axiom or the axiom of
continuity).
1.2 THE FIELD AXIOIviS
Along with the set R of real numbers we assume the eistence of two operations,
called addition and multiplication, such that for every pair of tea! numbers x and y

the sum x + y and the prodtct xy are real numbers uniquely determined, by x
and y satisfying the following axioms. (In the axioms that appear below, x, y,
z represent arbitrary real numbers unless something is said to the contrary.)
Axiom 1. x + y -- y + x, xy -- yx (commutative taws).
Axiom 2. x + (y + z) = (x + y) + z, x(yz)  (xy)z (associative laws).
Axiom J. x(y + z) = xy + xz (distributive taw).
Axiom 4. Given any two reM numbers x and y, there exists areal number z such that
x + z = y. This z is denoted by y -"x; the number x - x i denoted by O. (It
can be prtmed that 0 is independent o/ x.) We write -x for 0 - x and call --x the
negative of x.
Axiom 5, There exists at least one real number x ¢; O. If x and y are two real
numbers with x  O, then there exists a real number z such that xz --- y. This z is
denoted by y/x ; the number x/x is denoted by 1 and can be shown to be independent of
x. We write x-l for 1/x if x ¢: 0 and call x - t the reeiprocai of x.
From these axioms all the usual laws of arithmetic can be derived; for example,
-(-x) = x,(x-l) -1 = x, -(x-y) =y- x, x-y = x + (-y),etc. (For
a more detailed explanation, see Reference I.l)
13 THE ORDER AXIOMS
We also assume the existence of a relation < which establishes an ordering among
the real numbers and which satisfies the following axioms:
Axiom 6. Exactly one of the relations x = y, x < y, x > y holds.
NotrE. x > y means the same as y < x.
Axiom 7. If x < y, then for every z we have x + z < y + z.
Axiom g. lf x > 0 and y > O, then xy > O.
Axiom 9. lf x > y and y > z, then x > z.
worE. A real number x is called positive if x > 0, and negative if x < O. We
denote by R + the set of all positive real numbers, and by R- the set of all negative
real numbers.
From these axioms we can derive the usual rules for operating with inequalities.
For example, if we have x < y, then xz < yz if z is positive, whereas xz > yz if
z is negative. Also, if x > y and z > w where both y and w are positive, then
xz > yw. (For a complete discussion of these rules see Reference l.l.)
No'rE. The symbolism x < y is used as an abbreviation for the statement:
"X < )¢ or X =
Thus we have 2_< 3 since 2 < 3; and 2 < 2 siuc¢ 2 - 2. The symbol > is
similarly used. A real number x is called nonnegat&e if x > 0. A pair of ;imut-
taneous inequalities such as x < y, y < z is usually written more bri fly as
x<y<.
The following theorem, which is a simple consequence of Ihe foregoing axioms,
is often used in proofs in analysis.
Tleorem 1.1. Given real numbers a and b such that
a < b +  for every e > O. (1)
Then a < b.
Proof If b < a, then inequality (1) is violated for e = (a - b)[2 because
a-b a+b a+a
b+--b+------<-- =a.
2 2 2
Therefore, by Axiom 6 we must have a _< b.
Axiom 10, the completeness axiom, will be described in Section I.I I.
1.4 GEOMETRIC REPRESENTATION OF P-,FL NUMBERS
The real numbers are often represented geometrically as points on a line (called
the real line or the real axis). A point is selected to represent 0 and another to
represent !, as shown in Fig. 1.1. This choice determines the scale. Under an
appropriate set of axioms for Euclidean geometry, each point on the real line
corresponds to one and only one real number and, conversely, each real number
is represented by one and only one point on the line. It is customary to refer to
the point x rather than the point representing the real number x.
t-- : = : Figm'e 1.1
O l x ,*I
The order relation has a simple geometric interpretation. If x < y, the point
x lies to the left of the point y, as shown in Fig. 1.1. Positive numbers lie to the
right of 0, and negative numbers to the left of 0. If a < b, a point x satisfies the
inequalities a < x < b if and only if x is between a and b.
1.5 INTERVALS
The set of all points between a and b is called an interval. Sometimes it is important
to distinguish between intervals which include their endpoints and intervals which
do not.
NOTATION, The notation {x" x satisfies P} will be used to designate the set of
all real numbers x which satisfy property P.

Defmltlon 1.2. 3aaume a < b. The open interoal (a, b) is defined to be the set
(a,b) = {x:a < x <
The closed interval [a. b] is the set {x : a. < x < b}. The half-open interoals
(a, b'[ and [a, b) are Mmilarly defined, using the inequalities a < x  b and
a _  < b, respectively. Infinite interoals are definedas.foltows:
(a, = {x:x > a}, [a, = {x:x >
a) = (x:x < a}, = {x:x <_
The real line R is sometimes referred to as the open interval (-oo, + ). A
single point is considered as a "degenerate" closed interval.
NOa'E. The symbols +  and -  ate used here purely for convenience in notation
and are not to be considered a being real numbers. Later we shall extend the
real-number system to include these two symbols, but until this is done, the reader
should understand that all real numbers are 'Tmite.'"
1.6 INW_F.RS
This section describes the integer#, a special subset of R. Before we define the
integers it is convenient to introduce first the notion of an inductive set.
Defmiffoa 1.3. A set of real numbers is called an inductwe set if it ha# the foffowing
two properties:
a) The number 1 is in the set.
b) For eoery x in the set, the number x + 1 ia also in the aet.
For example, R is an inductive set. So is the set R . Now we shall define the
positive integers to be those real numbers which belong to every inductive set.
DeflMtio 1.4, A real number is called a positive integer f it belongs to eery
inductive set. The set of positive integers is denoted by Z +.
The set Z + is itself an inductive set. It contains the number I, the number
I + I (denoted by 2), the number 2 + ! (denoted by 3), and so on. Since Z + is a
subset of every inductive set, we refer to Z + as the smallest inductive set. This
property of Z ÷ is sometimes called the principle of utuction. We assume the
reader is familiar with proofs by induction which arc based on this principle.
(See Refercaace 1.1.) Examples of such proofs are given in the next section.
The negatives of the positive integers arc called the negative integers. The
positive integers, together with the negative integers and 0 (zero), form a set Z
which wc call simply the set of nte@er#.
1,7 THE UNIQUE FACTORIZATION THEOREM FOR INTF_,GERfi
Ifn and dare integers and ifn --- cdfor some integer c, we say dis a divisor of n,
or n is a multiple of d, and we write din (read: d divides n). An integer n is called
a prime ff n > 1 and if the only positive divisors of n are 1 and n. If n > 1 and n
is not prime, then n is called composite. The integer 1 is neither prime norcomposite.
This section derives some elementary results on factorization of integers,
culminating in the unique factorization theorem, also called the fundamental theorem
of arithmetic.
The fundamental theorem states that (1) every integer n > 1 can be represented
as a product of prime factors, and (2) this factorization can be done in only one
way, apart from the order of the factors, li is easy to prove part (1).
Tkeorem 1.5. Eoery integer n > | is ether a prime or a product of primes.
Proof We use induction on n. The th¢orem holds trivially for n = 2. Assume
it is true for every integer k with 1 < k < n. If n is not prime it has a positive
divisor d with 1 < d < n. Hence n = cd, where 1 < c < n. Since both c and
d are <n, each is a prime or a product of primes; hence n is a product of primes.
Before proving part (2), uniqueness of the factorization, we introduce some
further concel  t.s.
If dla and dlb we say d is a common divisor of a and b. The next theorem
shows that every pair of integers a and b has a common divisor which is a fineax
combination of a and b.
l'lorem 1,6, Every pair of integers a and b has a common diLcor d of the,form
d=ax+by
where x .and y are integers. Moreover, ewry common divisor of a and b divides
this d.
Proof. First assume that a _> 0, b  0 and use induction on n -- a + b. If
n--- 0thena---- b = 0, and we can take d -- 0withx-y = 0. Assume, then,
that the theorem has been proved for 0, 1, 2, ..., n - 1. By symmetry, we can
assumea:> b. Ifb=0taked= a,x: I,y-- 0. If b_> 1 we can apply the
induction hypothesis to a - b and b, since their sum is a -- n - b <_ n - 1.
Hence there is a common divisor d of a - b and b of the form d -- (a - b) + by.
This d also divides (a - b) + b = a, so d is a common divisor of a and b and
wc have d = ax + (y - x)b, a linear combination f a and b. To complete the
proof we need to "show that every common divisor divides d. Since a common
divisor divides a and b, it also divides the linear combination ax + (y - x)b = d.
This completes the proof ira > 0 and b > 0. Ifone orboth of a and b is negative,
apply the result just proved to ta[ and Ibl.
OT_ If dis a common divisor of a and b of the form d = ax + by, rhea -dis
also a divisor of the same form, -d = a(-x) + b(-y). Of these two common
divisors, the nonncgative one is called the #reatest common divisor of a and b,
and is denoted by gcd(a, b) or, simply by (a, b). If (a, b) = I then a and b
said to be relatively prime.
Theorem 1.7 (Euclid's Lemma), lf albc and (a, b)  1, then

Proof. Since (a, b) --- I we car write t -- ax + by. Therefore c -- acx + bey.
But alacx and albcy,
Theorem 1.8. If a prime p divides ab, then pin or plb. More generally, {f a prime
divides a product a I ... ak, then p divides at least one of the fac, tors.
Proof. Assume plab and that p does not divide a. If we prove that (p, a) = 1,
then Euclid's l.emma implies plb. Let d -= (p, a). Then dip so d = 1 or d =- p.
We cannot have d -- p because dla but p does not divide a. Hence d -- I. To
prove the more general statement we use induction on k, the number of factors•
Details are left to the reader.
l'lmrm 1.9(Unique faclorization theorem). Every integer n > I can be repre-
sented a, a product of prime factors in only one way, apart from the order of the
factor
Proof We use induction on n. The theorem is true for n = 2. Assume, then,
that it is true for all integers greater than 1 and less than n. If n is prime there is
nothing more to prove. Therefore assume that n is composite and that n has two
factorizations into prime factors, say
n =plp.-.p= qjq-.-q.
We wish to show that s = t and that each p equals some q. Since Pl divides the
product qlq2 """ q,, it divides at least one factor. Relabe] the q's if necessary so
that Pl [q 1- Then Pl = q i since both p and q  are primes. In (2) we cancel Pl
on both sides to obtain
n
Since n is composite, 1 < n/p I < n; so by the induction hypothesis the two
factorizations of n/pl are identical, apart from the order of the factors. Therefore
the same is true in (2) and the proof is complete.
1.8 RATIONAL NUMBERS
Quotients of integers a/b (where b # 0) are called rational numbers. For example,
1/2, -7/5, and 6 are rational numbers. The set of rational numbers, which we
denote by Q, contains z as a subset. The reader should note that all the field
axioms and the order axioms are satisfied by Q.
We assume that the reader is familiar with certain elementary properties of
rational numbers. For example, if a and b are rational, their average (a + b)/2 is
also rational'and lies between a and b. Therefore between any two rationaI numbers
there are infinitely many rational numbers, which implies that if we are given a
certain rational number we cannot speak of the "next largest" rational number.
1.9 IRllATIONAL NUillEII
Real numbers that are not rational are called irrational. For example, the numbers
r, e,  and e  are irrational.
Ordinarily it is not too easy to prove that me pactieular number is irrational.
The is no simple proof, for example, of the irrationality of e'. However, the
irrationaIity of certain numbers such as ,/ and x/ is not too difficult to establish
and, in fact, we easily prove the following:
Tkem,em 1.10. If n is a positive integer which s not a perfect square, then x/-n is
irrational
Proof Suppose first that n contains no square factor > I. We assume that %n is
rational and obtain a contradiction. Let n = a]b, where a and b are integers
having no factor in common. Then nb z = a z and, since the left side of this equation
is a multipfe of n, so too is a z. However, ifa3 is a multiple of n, a itself must be a
multiple of n, since n has no square factors > t. (This is easily seen by examining
the factorization of a into its prime factors.) This means that a = cn where c is
some integer. Then the equation nb z = a 2 becomes nb 2 = c2n 2, or b z = nc z.
The same argument shows that b must also be a multiple of n. Thus a and b are
both multiples of n, which contradicts the fact that they have no factor in common.
This completes the proof if n has no square factor > !.
If n has a square factor, we can write n -- rnZk, where k > 1 and k has no
square factor > 1. Then /g = mv/i; and if ,fg were rational, the number
would also be rational, contradicting that which was just proved.
A different type of argument is needed to prove that the number e is irrational.
0Ve assume familiarity with the exponentiaI e  from elementary calculus and its
representation as an infinite series.)
Tlworem 1.11, If e  = 1 + x + x2/2[ + x13! + "" + x/n! + "", then the
number e is irrational.
Proofi We shall prove that e-1 is irrational. The series for e  is an alternating
series with terms which decrease steadily in absolute value. In such an alternating
series the error made by stopping at the nth term has the algebraic sign of the first
neglected term and is less in absolute value than the first neglected term. Hence,
ifs, = = (-1)/k[, we have the inequality
from which we obtain
1
0 <: e -t -- s2_ l <:--,
(2k)!
1 1
0 < (2 - 1) (e '
• - sz  -1) <- < (3)
for anq integer k > 1. Now (2k - I)r sk_ is always an integer. If e -t were
rational, then we could.choose k so large that (2k - 1)! e -1 would also be an

integer. Because of (3) the difference of these two intents would be a number
between 0 and ½, which is impossible. Thus e-1 cannot be rational, and hence e
cannot be rational.
proof that • is irrational, see Exercise 7.33.
The ancient Greeks were aware of the existence of irrational numbers as early
as 500 ]t.c. However, a satisfactory theory of such numbers was not developed
until late in the nineteenth century, at which time three t theories were
introduced by Cantor, Dedekind and Weierstrass. For an account of.the theories
of Dedekiud and Cantor and their equivalence, see Reference 1
L10 UPPER BOUIW, ]cJMUM ELEMENT, LEAST UPPER BO
Irrational numbers arise in algebra when we try to solve cean quadratic equa-
tions. For example, it is desirable to have a real number x such that x a ffi 2. From
the nine axioms l/sled above we cannot prove that such an x exists in R because
these nine ax/oms are also satisfied by Q and we have shown that the[e is no
rational number whose squar is 2. The completeness axiom aUows us to introduc
irrational ambers in the real-number system, and it gives the real-number system
a property of continuity that is fundamental to many theorems in analysis.
Before we describe the completeness axiom, it is convenient to introduce
additional terminology and notation.
13¢ftaizim, ld2. Let S be a set of real numbers. If there is a real manber b such
that x < b for every x in $, then b is called an upper bound for S and we say that
S is bounded above by b.
We say an upper bound because every number greater than b will also be an
upper bound. If an upper hound b is also a member of S, then b is called the
largest member or the max/mum element of S. There can be at most one such b.
If it exists, we write
b = max S.
A set with no upper bound is said to be unbounded above.
l)efinitions of the terms lower bound, bounded below, smallest member (or
minimum element) can be similarly formulated. If S has a minimum dement we
denote it by rain S.
I. The set R + = (0, + ) is tmbonded above. It has no upper bounds and no max.
imum element. It is bounded below by 0 but ha no minimum ©lemet.
7.. The closed interval S = [0, 1 ] is bounded above by ! and is boanded below by 0.
In fact, max $ = I md rain $ = 0.
3. The half-open interval  = [0, 1) is botmded above by 1 but it has no maximum
element. Its minimum ekanent is 0.
For sets fik¢ the one in Example 3, which are bounded above but have no
maximum element, there is a concept which takes the place of the maximum
ment. It is called the least upper bound or supremum of the set and is defined as
follows:
Definition l.lJ. Let S be a set of real numbers bounded abo. ,4 real number b is
called a 1east upper bound for S if it has the following two propertie :
a) b is an upper bound for S.
b) No number le than b is an upper bound for S.
 If S = [0, 1 ] the maximum element I is also a least upper bound for S. If
S = [0, 1) the number I is a leastupper bound for S, even though S has no maximum
elemt.
It is an easy exercise to prove that a set cannot have two different least upper
hounds. Therefore, if there is a least upper bound for S, there is only one and we
can speak of the least upper bound.
It is common practice to refer to the least upper bound of a set by the mote
concise term upremum, abbreviated sup. We shall adopt this convention and write
b = sups
to indicate that b is the supremum of S. If S has a maximum element, then
max S = sup $.
The greatest lower bound, or infimum of S, denoted by inf S, is defmed in an
analogous fashion.
1.11 THE COMPIA-'IY#iIS8 AXIOM
Our final axiom for the real number system involves the notion of supremuin.
Axiom 10. F2ery nonempty set S of real numbers which is bounded abo has a
aupremum; that is, there is a real number b such that b = sup S.
As a consequence of this axiom it foltows that every nonempty set of real
numbers which is bounded below has an infimum.
1.12 SOMI PROPERTE OF THE SUPREMU]91
This section discusses some fundamental properties of the supremum that will be
useful in this text. There is a corresponding set of properties of the infimum flaat
the reader should formulate for himself.
The first property shows that a set with a supremum conaim numbers arbi-
trarily close to its supremum.
Titeorem 1.14 (Appro property). Let S be a nor, empty set of real numbers
with a supremum, say b - sup S. Then for eery a < b there is some x in S such
that
a<x<__b.

10
ld m Cp Nlmllm- $¢m
1.15
Proof. First of all, x < b for all x in S. If we had x  a for every x in S, then a
would be an upper bound for S smaller than the least upper bound. Therefore
x > a for at least one x in S.
Theorem ld5 (Addiase praperty). Given nonempty subsets A and B of R, le C
denote the set
C= {x+y:xeet, yB}.
If each of A and B has a supremum, then C has a supremum and
sup C = sup ,4 + sup B.
Proof. Let a= supA, b = supB. If zeC then z = x +y, where x•
yeB, soz =x +y < a + b. Henca + b is an upper bound for C, soChasa
supremum, say c= supC, and c a + b. We show next that a + b < c.-
Choose any  > 0. By Theorem 1.14 there is an x in A and a y in B such that
a- n < x and b.- e <y.
Adding these inequalities we find
a+b - 2e <x+yc.
Thus, a + b < c +  for every e > 0 so, by Theorem ].1, a + b ..<_ ¢.
The proof of the next theorem is left as an exercise for the reader.
Tleorem 1.16 (Comlmdaon property). Given nonempty subsets S and T of R such
that s <_ t for eery s in S and t in T. ]f T has a supreVaurn then S has a aupremum
and
supS< supT.
1.13 PROPERTIES OF TIlE INTEGERS DEDUCED FROM THE
COMPLETENESS AXIOM
TAeerem ldT. The set Z + of positive integers I, 2, 3, ... is unbounded above.
Proof. If Z ÷ were bounded above then Z + would have a suprcmum, say a =-
sup Z +. By Theorem 1.14 we wouM have a - 1 < n for some n'in Z ÷. Then
n + 1 > a for this n. Since n + 1 • Z ÷ this contradicts the fact that a - sup Z +.
leorem 1.18, For etyery real x there is a poxitive integer n such that n > x.
Proof. If this were not true, some x would be an upper bound for Z +, contra-
dicting Theorem I. ! 7.
1.I4 THE ARCHIMEDEAN PROPERTY OF TIlE REAL NUMBER SYSTEM
The next theorem describes the Archimedean property of the real number system.
Geometrically, it tells us that any line segment, no matter how long, can be
covered by a finite number of line segments of a given positive length, no matter
how small.
Theorem 1.19. If x > 0 and if y is an arbitrary real number, there is a positive
integer n such that nx > y,
Proof. Apply Theorem 1.1.8 with x replaced by y/x.
1.15 RATIONAL NUMBERS WITH FINITE DECIMAL REPRESENI'ATION
A lea/number of the form
r = a 0 + al + a2 an
1--0  +"" + 1---'
where a o is a nonnegative integer and at,..., n are integers satisfying 0 < a < 9,
is usually written more briefly as follows:
r = a O.ala 2-''a a.
This is said to be a finite decimal representation of r.
For example,
1 = 5 0.5, 1 2 0.02, 29 7 + 2 + 5 = 7.25.
2 10 50 10  4 10 10 z
Real numbers like these arc nccessarily rational and, in fact, they all have .the form
r = silO', where a is an integer. However, not all rational numbers can be ex-
pressed with finite decimal representations. For example, if- could be so expressed,
then we would have  = a/10* or 3a -- I for some integer a. But this is im-
possible since 3 does not divide any power of 10.
1.16 FINITE DECIMAL APPROXIMATIONS TO REAL NUMBERS
This section uses the completeness axiom to show that real numbers can be
approximated to any desired degree of accuracy by rational numbers with finite
decimal representations.
Tlworem 1.20. Assume x > O. Then for every integer n > I there is a finite
decimal r. = a o . aa z ... a, such that
r. -<x< r.+ 10 .
Proof. Let S be the set of all nonnegativc integers <x. Then S is nonempty,
since 0 • S, and S is bounded above by x. Therefore S has a supremum, say
ao = sup S. It is easily verified that a0 • S, so a0 is a nonnegativc integer. W¢
call ao the greatest integer in x, and we write a o = I'x]. Clearly, we have
ao < x < ao + 1.

Now le! at ffi [10x- 10ao], the greatest integer in 10x " 10a 0. Since
0 10x- 10a o = 10(x- no) < ]0, we have 0 _< at gand
al < 10x- 10no < al + ].
In other words, a is the largest integer satisfying the inequafities
a a + ]
ao+_<x<a o+ 10
Move generally, having chosen a t .... , a._ 1 with 0 < a s _< 9, let g, be the
largest integer satisfying the inequalities
ao + al + ... + an a 1 a + 1
1-6 x <a° +T6+"'+ (4)
Than0 < a. < 9 andwehave
1
r < x < r + IO '
where r. = ao .aaz - • • a.. This completes the proof. It is easy to verify that x is
actuall the suprmum oftlae set of rational numbers r, r,
I.I7 INFINITE DEC1MAL REPRFENYATIONS OF REAL NUMBERS
The integers no, a, a,.., obtained in the proof of Theorem 1.20 can be used to
define an infinite decimal representation of x. We write
X  Io,ala  "'"
tO mean that a is the larger integer safiffying (4). For example, if x = f we fmd
ao ffi O, a t = l,a = 2, a = 5, andg, : 0 for all n >._ 4. Therefore we can
write
 = 0.125000...
If we interchange the inequality signs < and. < in (4), we obtain a slightly
differ t dfmition of decimal ¢xpaasiom. The finite decimals r. satisfy r, < x <
r, + l0 -" although the digits a, al, a,.., need not be the same as those in (4).
For example, if we apply this seoond definition to x --  we fiad the infinite decimal
representation
= 0.  2  9  9 .:.
The fact that.a real number might have two different decimal representations is
merely a reeetion of the thct that two different sets of real numbers can have the
same supremum,
1.18 ABSOLUTE VALUES AND THE TRIANGLE INEQUAL/TY
Calculations with inequalities arise quite frequently in analysis. They arc of
particular importance in dealing with the notion of absolute value. Ifx is any real
1.22
number, the absolute value of x, denoted by Ixl, is defined as follows:
Ixl = ,1" x, if x > 0,
-x, if x _< 0.
A fundamental inequality concerning absolute values is given in the following:
Tlteorem 1.21. If a > O, then we have the inequaliO, Ixl -<- a if, and only if,
-a<_x<a.
Proof. From the definition of Ixl, we have the inequality -Ixl <- x <_ kxl, since
x = [xl orx = -Ixl. lfwe assume that Ixl --- a, then wecan write -a _< -Ixl _<
x  Ixl <- a and thus half of the theorem is proved. Conversely, let us assume
-a <: x < a. Then ifx _> 0, wehavetx[ =x < a, whereasifx < 0, wehave
Ixl = -x < a. In either case we have Ixl < a and the theorem is proved.
We can use this theorem to prove the triangle inequality.
Theoeem 1.22, For arbitrary teal x and  we have
Ix + Yl < Ixl + lYl (the triangle inequality).
Proof. We have -Ixl < x  Ixl and -lyl< y < lYl. Additioh gives us
-(Ix[ + lY[) -< x + y < Ix[ + lYl, and from Theorem 1.21 we conclude that
Ix + y[ <: Ixl + l Yl- This proves the theorem,
The triangle inequality is often used in other forms. Fo example, if we take
x -= a - c andy = c - b in Theorem 1,22 we find
In-hi < In- ct +lc-bl.
Also, from Theorem 1.22 we have Ixl > ix + y[ - lYl- Taking x = a + b,
y = -b, we obtain
la+bl > lat-
Interchanging a and b we also find la ÷ bl >- Ibl - lal -- (lal - Ibl), and
hence
+ II  1 - I  1t.
• By induction we can also prove the generalizations
Ix + x + --" + x,I --< Ix,i + IxJ ÷."' ÷ Ix, I
and
Ixt + xz + "'" + x,I _> ixll - Ixzl ..... Ix,.
1.19 THE CAUCHY--SCHWARZ INEQUALITY
We shall now derive another inequality which is often used in analysis.

14
 an Cemptex N 
Theorem 1,23(Cauchy-Sclcarz inequality). If aj, . . . , a and b .... , b. are
arbitrary real numbers, we have
Moreover, if some ai ¢: 0 equality holds if and only if there is a real x such that
ax + bk = O for each k = I, 2, ..., n.
Proof A sum of squares can never be negative. Hence we have
 (ax + t,0  > 0
for every real x, with equality if and only if each term is zero. This inequality can
be written in the form
Ax z + 2Bx + C >_ 0,
whel-
A = , a, B : a,b,, C = b[,
IrA > 0, putx = --B[A to obtain B' - AC < 0, which is the desired inequality.
If A = 0, the proof is trivial.
NOTE. in vector notation the Cauchy-Schwarz inequality takes the form
(a'b) z _ liall211blf z,
where a = (al, ..., a), b --- (b t, • • • , b,) are two mdimensional vectors,.
a. b = k atbi"
is their dot product, and Ilall = (a-a) t:2 is the length ofa.
1.29 PLUS AND MINUS INFINITY AND THE EXTENDED REAL NUMBER
SYSTEM R*
Next we extend the real number system by adjoining two ideal points" denoted
by the symbols + o and - m ("plus infinity" and "minus infinity").
Definition 1.24. By the extended real number system R* we shall mean the set oj"
real numbers R t. ether with two symbols + o and - oo which satisfy the followin.q
properties:
a) If x  R, then we have
x + (+m) = +co, x + () -
x- (+co) = -co, x - (-:o) = +o.
x/(+ m) = x/(- ) = O.
per. t.2s 
15
b) l f x > O, then we haue
x(+o) = +o, x(-o)
c) l f x < O, then we have
x(+o) = -m,
d) {'+co) + (+oo) = (+m)(+oo) --- (-m)(-co) = +o,
(-o) + (-m) = (+o)C-m) - -.
e) If x • R, then we have -
NOT^nON. We denote R by ( - o0, + m) and R* by [- m, + o]. The points irt R
are called "finite" to distinguish them from the "infinite" points + oo and -m.
The principal reason for introdudng the symbols + m and -m is one of
convenience. For example, if we define + m to be the sup of a set of real numbers
which is not bounded above, then every nonempty subset of R has a supremum
in R*. The sup is finite if the set is bounded above and infinit if it is not bounded
above. Similarly, we define - co to be the inf of any set of real numbers which is
not bounded below. Then every nonempty subset of R has an inf in It*.
For some of the later work concerned with limits, it is also convenient to
introduce the following terminology.
Dejfnition 1,2& Every open interval (a, + co) is called a neighborhood of + m or
a ball with center + m. Every open interval (- co, a) is called a neighborhood of
- or a ball with center -.
1.21 COMPLEX NUMBERS
It follows from the axioms governing the relation < that the square of a real
number is never negative. Thus, for example, the elementary quadratic equation
x  = - I has no solution among the real numbers. New types of numbers, called
complex numbers, have been introduced to provide solutions to such equations. It
turns out that the introduction of complex numbers provides, at the same time,
solutions to general algebraic equations of the form
a + a,x + ... + a,#¢" = O,
where the coefficients a0, at,... , a. are arbitrary real numbers. (This fact is
known as the Fundamental Theorem of Algebra.)
We shall now define complex numbers and discuss them in further detail.
Defuu'rion 1.26. By a complex number we shatl mean an ordered pair of reaf numbers
which we denbte by (x, xz). The first member, x, is called the real part of the
complex number; the second member, xz, is called the imaginary part. Two complex
numbers x = (x, xz) and y = (y,))z) are called equal, and we write x - y, if,

and only if, x t = Yt and x2 = y2. We define the sum x 4- y and the product xy by
the equations
x + y = (xj + )q, xz + Yz), xy = (xyt - xzYz, x)'z +
rr. The set of all complex numbers will be denoted by C.
Zlumeem 1.27. The operations of addition and multiplication lust defined satisfy
the commutative, associative, and distributive laws.
Proof. We prove only the distributive law; proofs of the others are impler. If
x = (x, x2) , y = (),t,)'2), and z : (z I, zz), then we Imve
x(y + z) :- (xt, x2)(J + z, , + zz)
= (x,y, - x,y, x,y + xzy,) + (x,z - xz, xz + xaz)
7r,om 1.28.
(x,. x) + (0, 0) -- (x,, x), (x,, x2X0, 0) = (0, 0),
(x,, x)(I, O) -- (x,, x), (x. x) + (-x,, -xz) = (0. 0).
Proof. The proofs here are immediate from the definition, as are the proofs of
Theorems 1.29, t .30, 1.32, and 1.33.
Tkeorem 1.2. Given two complex numbers x-- (xt, xz) and y = (,,yz), there
exi$1$ a complex number z such that x 4- z = j. In fact, z = (Y.l - x, Yz -
This z is denoled by ¥ - x. The complex number ( -x,, -x) i denoted by
TIeorem 1.30. For an two complex number x and y, we have
(- x)y : x( - y) : - (xr) = (- I, OXxy).
 1.31. If x = (x, x) # (0, O) and ), are complex numbers, we define
eom 132. If x and j are complex number: with x # (0, 0), there exists a
complex number z such that xz ---- y, namely, z = yx-. .
Of special interest are operations with complex numbers whose imaginary
part is 0.
Tkeorcm 1213. (xt, 0) + (y, 0) - (x, + y,
(x,, 0Xy, 0) - (xty,, 0),
(x, 0)/(y, O) = (x,/J,, 0), if y,  O.
NorE. It is evident from Theorem 1.33 that we can perform arithmge operations
on complex numbers with zero imaginary part by performing the usual real-num-
ber operations on the real parts alone. Hence the complex numbers of the form
(x, 0) have the same arithmetic properties as the real numbers. For this reason it is
convenient to think of the real number system as being a special case of the complex
number system, and we agree to identify the complex number (x, 0) and the real
number x. Therefore, we write x = (x, 0). In particular, 0 = (0, 0) and 1
1.22. GEOMETRIC REPUTATION OF COMPLEX NUMBFRS
Just as real numbers are represented geometrically by points on a line, so complex
numbers are represented by points in a plane. The complex number x = (x j, x)
can be thoue, ht of as the "'point" with coordinates (x, x). When this is done, the
definition ofaddilion amounts to addition by the parallelogram law. (See Fig. 1.2.)
0  (0, 0) x - (xt, o)  1
The idea of expressing complex numbers geometrically as points on a plane
was formulated by Gauss in his dissertation in 1799 and, independently, by Argand
in 1806. Gauss later coined the somewhat unfortunate phrase "complex number."
Other geometric interpretations of complex numbers a_re possible. Instead of
using points on a plane, we can use points on other surfaces. Riemann found the
sphere particularly convenient for this purpose. Points of the sphere are projected
from the North Pole onto the tangent plane at the South Pole and thus tb, ere
corresponds to each poim of the plane a definite point of the sphere. With the
exception of the North Pole itself, each point of the sphere corresponds to exactly
one point of the plane. This correspondence is called a stereographic projection.
(See Fig. 1.3.)

1, THE IMAGINARY
It is often convenient to think of the complex number (xt, x) as a two-dimensional
vector with components xl and x2. Adding two complex numbers by means of
Definition 1.26 is then the same as adding two vectors component by component.
The complex number ! = (!, 0) plays the same role as a unit vector in the hori-
zontal direction. The analog of a unit vector in the vertical direction will now be
introduced.
LM 1.34. The complex number (0, !) is denoted by i and ia called the imag.
inary unit.
T&eorem 1.35. Every complex number  = (xt, x2) can be represented in the form
x= x + ixz.
Proof. x, -- (x,, 0),
x, + ix2 = (x,. 0) + (0. x2) = (x,, x)..
The xt theorem tells us that the complex number i provides us with a solution
to the equation x a = - l.
Tkeerem 1.36, i z = - I.
Proof. i  = (0, IX0, I) = (-I, 0) = -l.
1.74 ABSOLUTE VALUE OF A COMPI.F NUMBER
We now extend the concept of absolute value to the complex number system.
 13. If x = (x, x), we define the modulus, or absolute value, of x to
be the nonnegative real number I xl ive n b
i) I0, 0)1 = 0, ad Ixl > 0/fx  0.
iii) lx/yl - Ixl/lyl,/fy  0.
ii) Ixyl = txl lyl.
iv) I(x, 0)1 = Ix, I.
Proof. Statements (i) and (iv) are immediate. To prove (ii), we write x = x + /xa,
y = y + iy:, so that xy = xy - x2y + i(xy + x:y). Statement (ii)
follows from the relation
Equation (iii) can be derived from (ii) by writing it in the form ix[
Geometrically, IM represents the length of the segment joining the ori#n to
the point x. More generally, Ix - .vl is the distance between the points x and y.
Using this geometric interpretation, the following theorem states that one side of
a triangle is less than the sum of the other two sides.
Tkeaeem 1.39. lf x and y are complex numbers, then we have
Ix ÷ yl < Ixl ÷ iyt (triangle inequality).
The proof is left as an exercise for the reader.
1.2 IMPOSSIBILITY OF ORDERING THE COMPLEX NUMBERS
As yet we have not defined a relation of the form x < y if x and y are arbitrary
complex numbers, for the reason that it is impossible to give a definition of < for
complex numbers which will have all the properties in Axioms 6 through 8. To
illustrate, suppose we were able to define an order relation < satisfying Axioms
6, 7,and 8. Then, since i 4, 0, we must have either i > 0 or i < 0, by Axiom 6.
Let us assume i > 0. Then taking, x --- y =  in Axiom 8, we get i z > 0, or
- I > 0. Adding 1 to both sid¢ (Axiom 7), we get 0 > !. On the other hand,
applying Axiom 8 to -1 > 0 we find I > 0. Thuswehave both 0 > l.and
1 > 0, which, by Axiom 6, is impossible. Hence the assumption i > 0 leads us
to a contradiction. [Why was the inequality - 1 > 0 not already a contradietion?]
A similar argument shows that we cannot have i < 0. Hence the complex numbers
cannot be ordered in such a way that Axioms 6, 7, and 8 will be satisfied.
1.26 COMPLEX EXPONENTIALS
The exponential e  (x real) was mentioned earlier. We now wish to define e  when
z is a complex number in such a way that the principal properties of the real
exponential function will be preserved.. The main properties of e  for x real are
the law of exponents, ¢'e  :- e'+'L and the equation e ° = I. We shall give a
definition of e  for complex z wMch preserves these properties and reduces to the
ordinary exponential when z is real.
If we write z = x + iy (x, y real), then for the law of exponents to hold we
want e +' = de °. It remains, therefore, to define what we shall mean by
Defmiaoa 1.40, If z = x + iy, we define e  = e + to be the complex number
e a -- e (cosy + i siny).
This definition* agrees with the real exponential function when z is real (that
is, y = 0). We prove next that the law of exponents still holds.
* Several arguments can be given to motivate the equation e ¢ - cos y + i sift .v. For
example, let us write • t" = f(y) + ig(y) anc try to determine the real.valued functions f
and g so that the usual rules of operating with real[ exponentials will also apply to com¢lex
exponentiaIs. Formal differentiation yields e  = fl'(y) - if(y), if we assume that
(e)" = ie v, Comparing the two expressions for e . we see that fand # must satisfy the
equations f(y) = d(y), f'(y)  -y). Elimination of g yields f(y) = -if(y). Since
wewante ° = I, wemust hayer(0) = ! andf'(0) = 0. It follows thatf(y) = cosyand
a(Y) = -f(Y) = sin y. Of course, this argument proee$ nothing, but it strongly suggests
that the definition e i = cos y + i-sin y is reasonable.

lm,e 1.41.
then we hae
If z 1  x 1 + iy I and zz : x: + iy 2 are two complex number,
e, e =
eJe = ele[cos
+ i sin Yl), e = e( c°s. Y2 + i sin
Yl cos Y2 - sin Yl sin y2
sin y + sin y cos
Now ee " : e  +, since x 1 and x are both real. Also,
cos y cos Yz - sin y sin y = cos (yt -+
and
eo y sin Y2 + sin y cos y =
and hence
e'e' = e'*[ cos (Yl + Yz) + /sin (y +
1.27 FURTHE PROPERTIES OF COMPLEX EXPONENTIALS
In the following theorems, z, zt z2 denote complex numbers.
Tlorem 1.42. e  is never zero.
Proof. ee - -- e  -- 1. Hence e  cannot be zero.
Tlorem 1.43. If x is real, then ]el = I.
Proof. [e"t  = cos  x + sin  x = l, and ]el > 0.
Theorem 1.44. e  -- I if, and only if, z is an integral multiple of 2rti.
Proof. If z = 2nin, where n is an integer, then
e  = cos (2nn) + i sin (2nni = 1.
Conversely, suppose that e  = I. This means that e  cos y = ! and e  sin y = 0. _
Since e   0, we must have siny = 0, y = kx, where k is an integer. But
co(kx)= (-l). Hence e  =-1) , since e cos(/n) = 1. Since e > 0,
k must be even. Therefore e  : 1 and hence x =- 0. This proves the theorem.
1Theorem 1.45. e ' = e  if, and only ij e I - z = 2xin (where n is an integer).
Proof. e ' = e  if, and only if, e '- = 1,
1.28 THE ARGUMENT OF A COMPLEX NUMBER
If the point z -= (x, y) = x + iy is represented by polar coordinates r and 0, we
can write x ffi rcos0 and y rsin0, so that z = rcos0 + irsinO = re 
The two numbers  and 0 uniquely determine z, Conversely, the positive number
r is uniquely determined by z; in fact, r = Izl. However, z determines the angle 0
only up to multiples of 2t. There are infinitely many values of 0 which satisfy the
equations x -- Izl cos 0, y :- Izl sin 0 but, of course, any two of them differ by
some multiple of 2n. Each such 0 is called an argument ofz but ore of these values
is singled out and is called the principal argument of z.
Del"bdtlon 1,46. Let z = x + iy be a nonzero complex number. The unique real
number 0 which satisfies the conditions
x -- Izl cos 0, y -- Izl sin 0, -n < 0 : +n
is called the principal argument of z denoted by 0 = at8 (z).
The above discussion immediately yields the following theorem:
Tleaeem 1.47. Every complex number z  0 can be represented in the form
z = re , where r = Izt and 0 -- arg (g) + 2nn, n being any integer.
No. This method of representing complex numbers is particularly useful in
connection with multiplication and division, since we have
(rleXre) = rtre ii÷ and re 1 = .r_
r e m r
Theorem 1.48. If ztz2  0, then ag (ztz2) = ag (zl) + arg (z) +
where
0, /f -rr < arg(z) + arg(z2) < +n,
n(z,z2) = +l, /f-2rr < arg(z) + arg(z) <
Proof Write z I = Iz, le e', z2 -- Iz2le , where 01 = arg (z,) and 0
Then zlz2 = Izlz2te '÷'. Since -n < 01 -<: +n and -r < 02  +r, we
have-2n < 01 + 02 < 2n. Hence there is an integernsuchthat-n < 0 +
02 + 2mz  x. This n is the same as the integer n(z 1, z2) given in the theorem,
and for this n we have arg (ztz2) = 01 + #2 + 2nn. This proves the theorem.
1.29 INTEGRAL POWERS AND ROOTS OF COMPLEX NUMBERS
Given a complex number z and an integer n, we define the nth power
z ° = l, z "+1 ---= zz, tf n > 0,
z-" = (z-ly, if Z # O and n > O.
Theorem 1.50, which states that the usual laws of exponents hold, can be proved
by mathematical induction. The proof" is left as an exercise.

Tkeorem 1.50. Gipen two integers m and n, we hae, for z  O,
z'z" = z "+- and (z,zz)" =
ct mptex mrs Zo, z,, ..., z._, (cMt the ,th roos o/z), ch that
 : z, foreak = 0, 1,2,..,n - 1.
Furthermore, these rts are given by tbe formu
,=g(z+2 (=0,,,...,,-).
.  n ,th o ofz a qually s on  ¢i1 of rius R =
 at the
Proof. ¢ n mple num Re , 0  k  n - I, a disnct d each
  rt of z,
(Rey = R"  = Izl "'''+'a = z.
We mt now show t e a no oer n  of z. Suppo w = Ae ;= is
a complen namer such that  = z. Th Iwl" = Izl, and hen A
 = I=l ''. o,   z   writ e " = e '*"o, which impli
 - arg (z) = 2nk for some .inr k.
H a = [ O) + 2]/,. But when k runs through all integral values,
mk only e distinct vues z0,. -., z,_ . (. Fig. I
F, 1.4
1.30 O:IMPLEX LOGARITHMS
By Theorem 1.42, e* is never zero. It is naturat to ask if there are other values
that e  cannot assume. The next theorem shows that zero s the only exceptional
value.
 1_$ Cx Powers
Tlworem 1.52. If z is a complex number # O, then there exist complex numbers w
such that e  = z. One such w i, the complex number
log Izl + i arg.(z).
and any other such w must bae the form
log Izl + i arg (z) 4- 2ni,
where n is an integer.
Proof. Since egZt+'"")= elzl *r( = Izl earg(z) - z, we S¢¢ that w =
log [z[ + i arg (z) is a solution of the equation e" = z. But if w I is any other
solution, then e" - e  and hence w - w = 2ni.
o 1.5. Let z  0 be a given complex number. If w is a complex number
such that e" = z, then w i called a logarithm of z. The particular wlue of w given
w = log Izl + i arg (z)
called the principal logarithm of z, and for this w we write
1, Since
2, Since
w = Logz.
Ill = ] and arg (i) = =/2, Log (i) =
I-il -- ! and ar (-i) = -=/2, Log (Ci) = -tr/2.
3. Since t-1[ = 1 and arg(-I) = =, Log(-1)  tci.
4. If x > 0, Log (x) = log r, since Ix[ = x and arg (x) = 0.
S. Sinc I1 + il = / and ar ( + ) = x/4, Log'(i + i) = log 4 +
Theorem 1.54. lf z,z # O, then
Log (z,z) = Log za + Log z 2 +' 2in(zl, zx),
where n(z,, zz) is the integer defined in Theorem 1.48.
Proof.
Log (zaz) = log zz2[ + i arg (zz)
-- log z,[ + log Iz=l + i [arg (z0 + arg (z2) + 2nn(z,, z2)].
1.31 COMPLEX POWERS
Using ¢ompkx logarithms, we can now give a definition of complex powers of
complex numbers.
Defmit 1.55. If z  0 and if w is any complex ,umber, we define
Z w _-- ewL.

1..32 COMPLEX SINI AND COSINES
DcflOo'  1.58. Give  complex number z, e defie
 + e -  -
N. When z is al, t uations ag wi finition 1 ..
r i9, If z = x + iy then we 
cosz= sxcoshy- inxsinhy,
sinz ffi sinxshy + icosxy.
Prof.
2COSZ = e iz + e
= e-'(s x +  sn x) + d(cm  -  sin x)
= 2 cos x cosh y  2i sin x sink y.
 woof for sn  is simi.
Purr WO of ses and cosines arc vcn in the cxc.
1.33 INFIN_rI'Y AND THE EXTENDED COMPLEX PLANE C*
Next we extend the complex number system by adjoining an ideal point denoted by
the symbol ,
DefutitioR 1.60. By the extended complex number system C* we shall mean the
complex plane C along with a ymboi m which satisfies the fottowi 9 properties:
a) If z E C, then we have z + m --- z - m = m, z/m = O.
Def. 1.61
b) /fz  C, but z # 0, then z(oo) = m and zlO =
c) o +  = (X03)--
DeflMff 1.6I. Every set in C of the form {z : Iz  > r > 0} /s called a neighbor-
hood of oo, or a ball with center at m.
The reader may wonder why two symbols, + m and -o, are adjoined to R
but only one symbol, 03, is adjoined to C. The answer ties in the fact that there is
an ordering relation < among the real numbers, but no such relation occurs
among the complex numbers. In order that certain properties of real numbers
involving the relation < hold without exception, we need two symbols, + m and
-m, as defined above. We have already mentioned that in R* every nonempty
set has a sup, for example.
in C it turns out to be more convenient to have just one ideal point. By way
of illustration, let us recall the stcrcographie projection which establishes a one-
to-one correspondence between the points of the complex plane and those points
on the surface of the sphere distinct from the North Pole. The apparent exception
at the North Pole can be removed by regarding it as the geometric representative
of the ideal point oo. We then get a one-to-one correspondence between the
extended complex plane C* and the total surface of the sphere. :It is geometrically
evident that if the South Pole is placed on the origin of the complex plane, the
exterior of a "large" circle in the plane will correspond, by stereographie projection,
to a "small" spherical cap about the North Pole. This illustrates vividly why we
have defined a neighborhood of 03 by an inequality of the form Izl >
1.1 Prove that there is no largest prime. (A proof was known to Euclid.)
!.2 If n is a positive integer, prove the algebraic identity
d' - b" = (a b) ,= d/¢ '-t-'.
!.3 If 2 n - I is prime, prove that, is prime. A prime of the form T' - 1, where p is
prime, is called a Mersenne prime.
1.4 If 2" + 1 is prime, prove that n is a power of 2. A prime of the form 2
called a Fermat prime. Hint. Use Exercise 1.2.
1.5 The Fibonacci numbers l, l, 2, 3, 5, 8, 13,... are defined by the fecursion formula
x+] = xn + x,_, with xj = x = 1. Prove that (x,,x,+l) -- I and that Xa =
(a  -- b')/(a -- b), where a and b are the fools of the quadratic equation x 2 - x -
1.6 Prove that every nonempty set of positive inteiers contains a smallest member.
This is called the well-ordering principle.

Ratioml and irratkmal mmdrs
1.7 Find the rational numl>er whose decimal expansion is 0.33.44444...
1,8 Prove thai the decimat ¢xlmnsion of x will end in ze.ro (or in nines) if, and only if,
x is a rational number whose denominator is of the form 25 ', where m and n are non-
negative integer.
1,9 Prove that -,/. + -f is irrational.
1.10 Ira, b, c, dare rat/onat and ifx is irrational, Orovethat (ax + b)/(ex + d) is usually
irrational. When do exceptions occur?
1.11 Given any real x > 0, prove that tlr¢ is an irrational number tween 0 and x.
1.12 If a/b < e/d with b > 0, d > 0, prove that (o + e)/(b + d) lies between
and c/d.
1.13 /.t a artd b be positive integer,, Prove that x/ always lies between the two fractiohs
a/b and (a + 2b)/(a + b). Which fraction is closer to /?
1.14 Prove that x/n - I + x/ + 1 is irrational for ever integer n >_ 1.
1.12 Given a real x and an integer N > 1, prove that there exist integers h and/c with
0 < k _< N such that I/cx - hi < I/N. Hint. Consider the N + 1 numbers tx -[tx]
for t = 0, 1, 2,..., N and show that some pair differs by at most I/N.
1.I6 If x is irrational prove that there are infinitely many rational numbers h/l with
k > 0 such that Ix - Mk] < I/k z. Hnt. Assume there are only. a finite number
ht[kj,. ÷, h,/k, and obtain a contradiction by applying Exercise 1,15 with N > I[#,
where # is the smallest of the numbers Ix -
1.17 Let x be a positive rational numher of the form
t= k!'
where each a is a nonnegative integer with a < k - I for k > 2:an6 an > 0, Let Ix]
denote the greatest integerin x. Prove that a -- [x], that a = [k! x] - k[(k - l)! x]
for k = 2 ..... n, and that n is the smallest integer such that n! x is an integer. Con-
versely, show that every positive rational number x can be expressed in this form in one
and only one way.
Ulcer bolmaS
1.111 Show that the sup and inf of a set are uniquely determined whenever they exist.
1.19 Find the sup and inf of each of the following sets of real numbers:
a) All numbers of the form 2 - + 3 - + 5 -r, wherep0 q, and r take on all positive
integer values.
b) S {x:3x  - 10 + 3 <
e) S= {x:(x  aXx- bXx - cXx- d) < 0}, wherea < b < c < d.
1,20 Prove the comparison property for suprema (Theorem 1.16).
1,21 Let A and B be two sets of positive numbers bounded above, and tet a -- sup A,
b - supB. Let Cbetheset of all products of the form xy, wherexA andyeB
Prove that ab = sup C+
10.2 Given x > 0 and an integer k >_ 2, Let ao denote the largest integer _<x and,
assuming that no, a,..., a._ have been defined, let a denote the largest integc[ such
that
a o -I- a ÷
a) Prove that 0 _< a < k - t for each
b) Let r, = ao + ak - + ak - + --. + al:-" and show that x is hesupof the
set of rational numbers r, rz
NOT. When k = 10 the integers no, ax, az .... are the digits in a decimal representation
of x. For general k they provide a representation in the scale of k.
1.23 Prove Laglange's identity for real numbers:
ab, =
Note that this identity implies the Cauchychwar'z inequality.
1,! Prove that for arbitrary real a, b, c we have
1.?.,5 Piove Minkowski's ineqoality;
This is the triangle inequality I1 ÷ bll -< Jail ÷ IIll for n-dimensional vectors, where
a -- (a ..... an), b = (h,..., b.) and
1.26 If a >_ a2 >- "'" >- a and b > bz >- • "" -> b,, prove that
a b _< n ab.
= =
Hint. ., ..:,,<. (o - aXt,, .- f,) >_ O.
Cmplex
1.27 Express the following complex numbers in the form a + hi.
aj (I ÷ i) z, b) (2 + fj/(3 - 40,
c) i  + i , d) ;(1 + i(I + i).
1.2/! In ech case, determine all real x and y which satisfy ,,he given relation.-
t00

RI md  Numm" Sym
i.29 If a = x + iy, x and y real, the comphx conjugate of z is the complex number
g = x -- iy. o t:
d) z +  = t
0 (z  )/ = i the i  of z.
I ri ottilly e  of mpl num r which mtisfi h of th
following nitions:
11 Giv th mpx num zt, za, z such that Iz=l Iz  [ : Iz  [ = i and
z + z + z = 0. Show lt th num are vcrli of an uilatgml triangle
inked in the unit civic wih nl at t origin.
12 Ira and b a mpx numbs, o hat:
a) la - l z  (t + lall + Ibl%
b) If a # 0, thn la + bl : [al + Ibl iL and only if, b/a is l d nonnativ¢.
1 Ira and b a mplcx numbs, po
if, and only if, Il
1 If a and c a 1 conslanls, b mple, show It the ualion
az + b + gz + c= 0
top--ms a ci in t xpla.
I R the dnition of th in nnt: given a [ num t, n - (t) is the
uniq ! humor 0 which ties lhe wo conditiom
-- < O< + - , tanO= t,
2 2
If z -- x + iy, show that
a) arg (z) = tan- 1 (-Y,), ifx > o,
d) arg (:) -- - if x -- 0, y > 0; arg (z) = -  if x = o, y < 0.
2 2
1.36 IN.'fine the following "'psoudo*ordexlng*' of the complex numbs:
i) Izl < Iz=l or ii) Iz, I = Iz=l a afz) < (za).
h of io 6, 7, 8, 9  ti  is lation?
17 Whi ofoms 6, 7, 8, 9  tis ift r is n  foito?
We y (xt, Yx) < (x, y) •  ve eit --
i) x < x or it) xt =
1 State d  a t ous to ¢om I., ping  (z/z) in
ts of a (z) and g (z).
1 State and v¢ a  anflo o  1., p  (zdzz) in
r of  (z) d 
1. o t lh¢ nth  of 1 (
•  ..... , w • = cz=t., and show tt    1 mt  ti
i + x+2 +.,-+-t =0.
1,41 a) P that [zq < e  for l mplcx a # O,
b) vc tt th is o snt M > 0 such tt [ z[ < M for fll mplex z.
1. a)  that g (z +) = w g z + i., whcm. is  in.
( 0 +  sin 0 = s () + i  (),
ii) Sh tt, in , the clion -x <   + is n in (i) b ing
0= -=.a= .
iii) If+  m inter, ow thai
In this  it is kn+m 
sin30= 3+OsinO- sin +0,
30= +0+ 3OsinZO,
va]id fox l 0. Am  valid w
l, nc t z = (sin )j(c z) and
sin  + i sinh 2y
s2x + ch2y
1.47 l w  a v complex num. If w   1, show that tr¢ t two val of
z = x + iy mtis/ying the conditions c z = w and - • < x  + g, nd th val
whenw= iandwnw= 2.

1.48 Prove Lagrangc's identity for complex
U his to du a h inty for compl num,
1. a)  uating in  in Mo[v's fula v tt
b) IfO < 0 < ]2, po t
sin ( + 1)0 = sia'+OP{[ z 0)
w P. is t lynomal of d m n by
Use this to show that P,, ha zero at the m distinct points x = cot 2 {k/(2m + i)}
c) Show tt the sum of h¢  of P is  by
 2m+ 1 3
and that t sum of their squ  gin 
 colt k = 2m - l)(4m 2 + lore - 9)
 2m + 1
( Ex 8.46 and 8.47.)
1 ove that z" - 1 = I= (z - e '*) for all mplex z. U this to derive the
formals
sin = for n  2,
SUGGESTED REFERENCES FOR IRTHER STUDY
l.i Aposto], T. M., Calculus, Vo[. 1, 2nd ed. Xerox, Waltham, 1967.
1.2 Birkhoff, G., and MacLane, S., A Survey of Modern Aiyebra, 3rd ed.
New York, 1965.
1.3 Cohen, L, and Ehrlich, G., The Structttre of the Real-Number System.
trand,.Princeton, 1963.
i .4 G[eason, A., Fundamentals of Abstract Aualysis. Addison-Wesley, Reading, I
1.5 Hardy, G. H., A Course of Pure Mathematics, 10th ed. Cambridge University Press,
t952.
Macmillan,
Van Nos-
Refet'eces 31
! .6 Hobon, E. W., The TheoryofFunetior of a Real Variable dnd tlur Theory of Fourier's
Series, Vol. t, 3rd ed. Cambrid Univca-sity Pxess, 1927.
1,7 Landau, E., Foundations afAnalysis, 2nd od. Chelsea, New York, 1960.
1.8 lobinson, A., Nan-standard Analysis. North-Holland, Amsterdam, 1966.
1.9 Thurston, H. A., The Number System. Blackie, London, 1956.
1A0 Wilder, It. L., Introductian to the Foundationz of.Mathematics, 2nd ed. Wiley, New
York, 1965.

CHAPTER
SOME BASIC NOTIONS
OF SET THEORY
2.1 INTRODUCTION
In discussing any branch of mathematics it is helpful to use the notation and
terminology of set theory. This subject, which was developed by Boole and Cantor
in the latter part. of the 19th century, has had a profound influence on the develop-
ment of mathematics in the 20th century. It has unified many seemingly discon-
nected ideas and has helped reduce may mathematical concepts to their logical
foundations in an elegant and systematic way.
We shall not at-tempt a systematic treatment of the theory of sets but shall
confine ourselves to a discussion of some of the more basic concepts. The reader
who wishes to explore the subject further can consult the references at the end of
this chspter.
A collection of objects viewed as a single entity will be referred to as a set.
The objects in the collection will be called elements or members of the set, and they
will be said to belong to or to be contained in the set. The set, in turn, will be said
to contain or to be composed of its elements. For the most part we shall be inter-
ested in sets of mathematical objects; that is, sets of numbers, points, functions,
curves, etc. However, since much of the theory of sets does not depend on the
nature of the individual objects in the collection, we gain a great economy of
thought by discussing sets whose elements maybe objects of any kind. It is because
of this quality of generality that the theory of sets has had such a strong effect in
furthering the development of mathematics.
2,2 NOTATIONS
Sets will usually be denoted by capital letters:
and elements by lower-case letters: a, b, c,..., x, y, z. We write x ¢ S to mean
"x is an element of S," or "x belongs to S.'" If x does not belong to S, we write
x  S. We sometimes designate sets by displaying the elements in braces; for
example, the set of positive even integers less than 10 is denoted by {2, 4, 6, 8}.
If S is the collection of all x which satisfy a property P, we indicate this briefly by
writing S = {x: x satisfies P}.
From a given set we can form new sets, called subsets of the given set. For
example, the set consisting of all positive integers less than 10 which are divisible
32
by 4, namely, {4, 8}, isa subset of the set of eve intesm less than 10. In general,
we say that a set A is a subset of B, and we write A  B whenever every dement
of A also belongs to B. The statement ,4 _= B does not rule out the possibility
that B  A. IJa  we have both A _ B and B  A if, and only if, A argl B lutve
the same elements. In this case w droll call the sets A and B equal and we write
A -- B. If A and B are not equal, we write A # B. If A _ B but A  B, then
we say that A is a proper subset of B.
It is convenient to ecmsider the possibility of a set which contains no elements
whatever; this .set is called the empty set and w agree to catl it a subsist of every
set. The reader may fred it helpful to picture a set as a box containing.certain
objects, its elements. The empty set is then an empty box. We denote the empty
set by the symbol g.
Suppose we have a set eonsing of two dements a and b; that is, the ¢t {a, b}.
By our definition of equality this set is the same as the set {b, a}, ince no question
of order is involved. However, it is also necessary to consider sets of two elements
in which order is important. For example, in analytic geometry of the plane, the
coordinates (x, F) of a point represent an ordered pair of numbers. The point (3, 4)
is different from the point (4, 3), whereas the set {3, 4} is the same as the set {4, 3}.
When we wish to consider a set of two elements a and b as being ordered, we shall
enclose the elements in paxentheses: (a, b). Then a is called the first element and
b thesecond. It is possible to give a purely set-tbeoretie definition of the concept
of an ordered pair of objects (a, b). One such definition is the following:
2.1. (a, t,) = HaL {a, b}}.
This definition states that (a, b) is a set containing two elements, {a} and
{a, b}. Using this definition, we can prove the following theorem:
Tleorem 2.2, (a, b) :- (c, d) if, and only if, a = c and b = d.
This theorem shows that Definition 2. ! is a "reasonable" definition of an
ordered pair, in the sense that the object a has been distinguished from the object
b. The proof of Theorem 2.2 will be an instructive exercise for the reader. (See
Exercise 2. In)
2,4 CARTESIAN PRODUCT OF TWO SETS
Defmitiott 2.3, Given two sets A and B, the set of all ordered pair (a, b) such that
a  A andb  B is calledthe cartesianproduct ofA andB, andis denotedby A × B.
Example. If R denolcs the set of all real numbers, then R × R is the t of all complex
numbers.

2.5 RELATIONS .AND FUNCTIONS
Let x and y denote real numbers, so .a the ordere pair (x, y) can  ot
as representing the ctangular cordnates of a point  t xy-pIane (or a
plex humor). We freqafly encounr sach expressions as
xy = 1, x  + y = 1, x z + yz  , x < y. (a)
Each of these expressions defines a rtain t ef ordered pais (x, 0 of real
numrs namely, he t of all pairs (x, y) for which the expression is
Such a t of ordered, pai i.s called a pNne relation The co.nding set of
points plott in the xy-plane is called the #raph of the relation, e graphs of
the relations descried n (a) are shown in Fig. 2.1.
x<y
Fe
The concept of relation can be formulated quite generally so that the objects
x and y in the pairs (x, y) need not be numbers but may be objects of any kind.
Definition 7.4. Any set of ordered pairs is called a relation.
If S is a relation, the set of all elements x that occur as first members of pairs
(x, y) in S is called the domain of 5", denoted by N(S), The set of second members
y is called the range of S, deno by (S).
The first example shown in Fig., 2.1 is a special ldnd of relation known as a
function,
Deflnlaon 2.$, A junction F is a set of ordered pairs (x, y), no wo cf which have
the same first member. That is, if(x, y)  F and (x, z)  F, then y
The definition of nction requires that for every x ir the domain of F here is
exactly one y such that (x, y) s F. It is customary to call y the vane off at x arid
to write
y =
instead of" (x, y)  F to indicate that the pair (x, y) is in the set Fo
As an alternative o describing a function Fby specifying the pair, it convains.
it is usualty preferable to describe the domain of T and then, for each x in the
domain, to describe how the function value F(x) is obtained. In this connection,
we have the .feowing heorem whose proof is left as an exercise for the reader
Theorem 2,6. 7:v functions F and G are equa f and on@ if
a) (F) = (G) (F.and G have he same domain), and
b) F(x) = G(x) /be every x in (F),
2.6 FURTHER RMINOLOGY CONCERNING .FVNCTIONS
When the domain (F) is a subse of R, then F is called a fianction ef one rea
variable, If .@(F) is a subset of C, the comNex number system then F is called a
function of a complex variable,
if (F) is a subset of a cartesian product A × B, then F is called a function
of two variables. Ir this case we denote tlae function values by F(a, b) instead of
F((a, b)). A fuller.ion of two real variables is one whose domain is a subset of
R×R.
If S is a subset of N(F), we say that F is defined on S. In this case, the
ofF(x) such that x  S is caIled the ima.qe ors under Fand is denoted by F(S).. If
T is any set which contains F(S), ghen F is Mse called a mapping from S to T.
This is often denoted by wr/ting
F:S T.
If F(S) = T the mapping is said to be onto T A mapping of S into itself s some-
times called a tranfformation
Consider, for example, the function of a complex variable defined by the equa-
tion F(z) = z ". This function maps every sector S of the form 0 < arg (z)
a < n/2 of the complex z-plane onto a sector S) described by the ineqaalities
0  arg IF(z)] .._<_ 2o (See Fig. 2.2.)
Figure
If two functions F and G satisfy the inclusion relation G _ F, we say that G
is a restriction of F or that F is an extension of G, In particular, if S is a subset of
(F) and if G s defined by the equation
G(x) = F(x) for all x in S,
then we cMl G the restriction of F to S. The function G consists of those pairs
(x, F(x)) such that x e & ls domain is S and its range is F(S)o

Z.70NFTO-ONE FUNCTIONS AND 
De.anitDa 2.7. t F be a c d  . We y F
d ly  for  x  y  S,
x) = F(y) plies x= 7-
s is e e as mg at a fuon which is ontne on
• s fon valu to dis m of & Such
jt. ey m nt u, as we shall pfly
Mses. Hor, fo s  nifion of the inve of a
nent to u a mo ! notion, tt of  nr of a Iafion.
 2. Gin a ret , t w relation  ded by
 lled the verse of S.
us  od  ( b) 1on   i. d only if,
¢lemen , longs 1o S.  S Js a  relati  mply s
tt e ph of  is the fltion of  ph of S   to  line
 2. Supse that the relati F  a fct. Coir the conrse
reon , which y or may not  a ti. If  
led t rse of F  is not by F- .
Fi 2,3(a) IIustt an exp ofa on Ffor wh :is not a funon.
In Fig. 2.3)  F d i nve  funsal.
 next m tells us at a funon wch is onne  its don
ways  an
(b)
• lorem 2.10. If the function Fis one-to-one on its domain, then  is also a function.
Proof To show that #isa function, we must show that if(x, y)  and (x, z)  ,
then y = z. But (x, y) / means that (y, x)  F; that is, x - F(y). Similarly,
(x, z)   means that x = F(z). Thus F(y) = F(z) and, since we are assuming
that F is one-to-one, this implies y ---= z. Hence,  is a function.
lqtrre. The same argument shows that if F is one-to-one on a subset S of (F),
then the restriction of F to S has an inverse.
2.8 COMPOSITE FUNCTIONS
lgefmiaon 2.11. Gituen two functions F and G such thai I(F) _ (G), we  form
a wnction, the msite G FofG and F, deedfollow:for ry x 
of  , =
Sin (  (, the element F(x) is  e down of G, and tbefo it
k n  n G[F(x)]. In gen¢l, it is not t tt G  F = F  G.
In fa, F  G y  mningless units ¢ ran o G i nin in t down
of  Hoover, the assiativ¢ Mw,
Ho(6o = (HoG) oF,
always holds whenever each side of the equation has a meaning. (Verification will
be an interesting exercise for the reader. See Exercise 2.4.)
2.9 SEQUENCES
Among the important examples of functions are those defined on subsets of the
integers.
Deflll01l 2112. 2y a finite sequence of n terms we shall understand a function F
whose domain is the set of numbers { I, 2, ...,
The range of F is the set {F(I), F(2), F(3),..., F(n)}, customarily written
{FI, F, F3, .... F,}. The elements of the range are called terms of the sequence
and, of course, they may be arbitrary objects of any kind.
Definition 2.13. By an infinite aequence we shall mean a function F whose domain
is the set { I, 2, 3,... } of all positive integers. The range of F, that is, the set
{F(t), F(2), F(3),... }, is also written {F, F2, F,... }, and the function oalue F
is called the nth term of the sequence.
For brevity, we shall occasionally use the notation {F,} to denote the infinite
sequence whose nth term is
Let s = {s,} be an infinite sequence, and let k be a function whose domain is
the set of positive integers and whose range is a subset of the positive integers.

Assume that k is "order-preservi" that is, assume tha
k(m) < kCn), if m < n.
Then the comlxsite function s o k is defined for all integers n > 1, and for every
such n we have
Such a composite fuuctioa is said to be a subequence of . Again, for brevity,
w ofn use the notation {;c,} or {,.} to denote the subsequcnc¢ of {s,} whose
nth term is s,.
Km  s -- {l/n} and let k be defined by k(n) = 2". Tben s o k  {1/2"}.
2.1 SlkfllA (EQOUb3 SETS
De 2.14. Two sets A and B re called similar, or equinumerous, and we write
A  8, d y the ext a otiun F wh   t : A
 wh re  t t B.
We also y t F esbfies a ot corresn twn  
A and B. ly, e t  is  to if (ke Fto  the "idenfi *" fon
for wch Ffx)  x f  x in ). Fu, if A  B en B  , u
if F is a ontone function wch  A mil o B, en F-  wffi  B
sitoA. Al,if BandifB  C,A  C. hefisleto
2.11 FINITE AND INFINITE SETS
A set S is ealledfinite and is said to contain n elements if
S-, {l, 2, ..., n}.
The integer n is called the catch'ha! monber of S. It is an easy exe:ise to prove
that if {1, 2,..., n} N {1, 2,..., m} then m = n. Therefore, the c'dinal
number of a finite set is well defined. The empty set is also considered finite. Its
cardinal number is defined to be O.
Sets which are not finite are called infinite et, The chief differenee between
the two is that an infinite set must be similar to some proper subset, of itself,
whereas a finite set cannot be similex to any proper subset of itself. (See Exercise
2,13,) For example, the set Z + oral! positive integers is similar to the proper subset
{2, 4, 8, 16.. } consisting of powers of 2. The one-to-one function F which
makes them similar is defined by F(x) = 2" for each x in Z ÷.
A set S is said to be coutably infinite if it is equJnumerous with the set of all
positive integers; that is, ff
s~ 0,2,
In this case there is a function f which establishes a one-to-one correspondence
between the positive integers and the elements of S; hence the set S can be dis-
played as follows:
S  {f(1),f(2),f(3),... ).
Often we use subscripts and denote f(k) by a (or by a similar notation) and we
write S = {a, a, a_ .... }. The important thing here is that the correspondence
enables us to use the positive integers as "labels" for the elements of S. A count-
ably infinite set is said to have cardinal number R0 (read: aleph nought).
 2.15. A mt S i called countable if it is eitherfinite or counmbly infinite.
A et which is not countable is called uncountable.
The words denumerable and nondenumerable are sometimes uso¢! in place of
countable and uncountable.
llteorem 2.16. Every subset of a countable set is countable.
Proof  S be the given countable set and assume A  S. If A is finite, there
nothing to prove, so we can assume that .4 is infinite (which means S is also in-
finite). Let s = {s,} be an infinite sequence of distinct term such that
Defln© a funetion on the positive integers as follows:
Let k(1) be the smallest positive integer m such that s,, • A. Assuming that
k(l), k(2),..., k(n- 1) have been defined, let k(n) be the smallest positive
integer m > k(n - i) such that s. • A. Then k is order-preserving: m > n
implies k(m) > k(n). Form the composite function s, k. The domain ors o k is
the set of positive integers and the range of s o k is A. Furthermore, s o k is one-
to-one, since
implies
which implies k(n) = k(m), and this implies n = m. This proves the theorem.
2.13 UNCOUNT OF .THE  NUMBER SYSTEM
The next theorem shows that there are infinite sets which are not countable.
Tkeorem 2.I7. The set of all real numbers is uncountable.

Proof. It suffice to show that the set of x satisfying 0 < x < I is uncountable.
If the re numbers in this interval were countable, there would be a sequence
s -- {s.} whose terms would constitute the whole interval. We shall show that this
is impossib|e by constructing, in the interval, a real number which is not a term
of this so:lucnce. Write each s. as an infinite decimal:
where each u.,t is 0, I ..... or 9. Consider the tea! number , which has the decinmt
expansion
y  O.VlVZV 3...,
where
Thea no term of the sequence {s,} can be ¢lUal to y, since ), differs from st in the
first decline1 place, differs from s 2 in the second decimal place, ..., from s, in
the nth decimal place. (A situation like s, = 0.1999... and y -- 0.2000...
cannot oour here because of the vay the t, are chosen.) Since 0 < y < l, the
theorem is proved.
TItcerem 2,18. Let Z + denote the act of all positioe integers. Then the cartesfin
product g + x Z + / countable.
Proof. Define a function f on Z + x Z + as follows:
f(m, n) : 2"3', if(m, n) E Z + x Z +.
Thenfis one-to-one on Z + × Z + and the range off is a subset of Z +.
2.14 SET ALGEBRA
Given tivo sets A and A2, we define a new set, called the union of A 1 arid A2,
denoted by/11 w .4z, as follows:
Defaftien 239. The union A,  /12 is the set of those elements which belong
either to A 1 or Io A z or to both.
This is the same as saying that .4 t  12 consists of those elements which belong
to at least one of the sets A, A2. Since there is no question of order involved in
this definition, the anion At  A is the same as Az  A ; that i, set addition is
commutative. The definition is also phrased in such a way that set addition is
associative:
.4 w (az o As) = (A
2.22
The definition of union can be extended to any finite or infinite collection of
sets:
Deftm'riaa 2.20. If F is an arbitrary collecffon of sets, then the union of all the sets
in F is defined to be the set of hose elements which belong to at least one of the sets
in F, and is denoted by
If F is a finite collection of sets,
AF  1
If F is a countable collection, F
UA.
F = {At,... , A}, we write
--A A'.'
= {A,, A,... }, we write
Dftm'tim 2,21. If F   bitrary coltecfion of sets, t tcion of 1 sets in
F is defined to be the set of those elements which belong to ry one of t sets in F,
and  denoted by
The interaction of two s A and Az is deno by At  Az d consists
of tho elements common to both . Ifd  and dz  no clemm  mmon,
 A  Az is  empty t d At and d a id to  disjot. If F is a
ite llion (as above), we wtc
If the  in e lIfion ve no NemenB in common, r interfion is e
empty t. Our definiom of union and intuition apply, of , even when
F is not countable, use of e way we have defin mons and inons,
the mmutiv¢ d iafive h m automatically tisfied.
 Z22. The complement of A relat to B, denoted by B - A, ia defined
to  the
B--  = {x : x 6 B, but x ¢ A}.
Note that B- (B- A) = d whenever A G B. Also note that B- X = Bif
B  A h empty.
e notions of union, interon, and complement a illustted in Fig. 2.4.

A B
B-A
Figure 2.,$
Theorem Z23.
and
Let F be a cMIectMn ( sets. Then for any se B, we have
Prae;£ Le S= ,eA T= (,(B A). If xB- & thenxeB, but
x 6 S. Hece, R is not true that x logs to at Iea.s one A in F; therefore x
b.longs to no A in F. Hen for every A in F x , B - A. But this implies
x  7 so that B - S  Z Reversing the steps, we obtain T  B - & and this
proves hat B - S = T To prove the cond satement, se a similar argument,
2.15 COUNTabLE COLLECTIONS OF COUNTABLE SETS
Definition 2,24, If F is a collection qf sets such that every two distinct sets in Fare
dLoint, then F is said w be a collection of a'{joint .sets.
Theorem 2,25, UF is a countable collection of di#/oint sets, say F = {A, A>.,, },
such that each set A is coumab&, ¢:hen the union i  A is also
Proof Let A = {at,., a2,,,a,....}, n = 1,2,.., and l.et S = k. A..
Ten every element x ef S s in at 1east one of the s n F and hence x = a,. for
some pair of integers (m n). The pair (m, n) is uniquely determined by x, sin
F is a collec6en ef doint ses. Hence the Nnction fdefined byfix) = (m,
x = %:,,, x  N has domain & The ranger(s) is a subt ofZ + x Z + (where Z +
i the set of positive ntegers) atd hence is countaNe Bm.fs one-to-one and there-
fore S .. JS), which mans that S is .also coumaNe
Torem2,26, If F= {A, A2,.,. } is a countable collection qf
G = {B., B2 .... }, where B = A and, for n > 1,
Then G is a collection of di/oim sets and we have
TG. 2.27 E.xercis 43
Pro Each set , is constructed so that it has no eiements ia common with the
earlier sets B,, Ba,.o, &_. Hence G is a ¢oItecti.on of dis)oint sets. Let
A =  Aand B- U%B> Wesha.t.l show that A - & First of alL
xeA, then xA [br some k. If n i the smMl:est such k, then x
x &, which means that x e B., md thereffre x e B. Hence A g B.
Conversely, if x  & then x e B.. for some n, and herefl:re x e A, for tNs same
Thus x e A and this proves that B  A.
Using Theorems 225 and 2.26, we imm.ediatey obtair
.Theorem 2.27. If b is a countable cotlecdon qf countable ses, h#n de
vet. in F is also a countable
Exam#e 1. The t Q of all rational umr is a coumaNe L
Proof t A, denme the set of all posi6ve ra6onal numrs h.avmg denominator
The set of all positive rationa mimrs is a o Uk, .4. From this it follows hat
Q is countable, since each A is countable.
Exampie 2. 2he set S of intervals with rational endoms is a ceuntaNe set.
ProOf t {x, x z .... } denote *he set of ratiopal numbrs aad et -,G  the st of all
intervals who left endpoi is :v and whoa right dint is rational Then
• countable sad S = tJ  A,.
EXERCISES
2.1 Prove Theorem 22. HInL (.a,, b) (.G d) meas {{a:/, {a, b)} = {{c), {c.. d '
Now apral to the definition of  uaii:y.
2,2 { S  a relafio and let (S.)  its domain The relation S is :said te 
i) reflexhe if a e (S) imp*i (a, a)  &
ii) symmetric if (a, b) e S implies (b, a) 
iii) transitive if (.a, b)  S and (.b, c)  S impties (a c) e S..
A relation which is symmetric, reflexive, and ransitive is called an equiveteece relation,
Determine which of he 9rores is .possd by S. f S is e se of all pairs 7f
humors (x, y) such
a} x y. b.) x < y, c} x <
d) x  +;v  - 1, e.) x a + y2 < 0, f).:v z + x = y: +
2.3 The following func:ios F and G are defined for ait reM x by the uations given.
 eac ca where th e comite function G  F can  form.., gve he domMr of
G :. F and a formula (or formu1) for (G : F){xL
a) F(x) =  -.. x, G(x) = x 2 + 2x.
b} Y(x} = x + 5, G(x) = }x/x,. ffx # O, G(O} =

Find F(x) if G(x) and G IF(x)] are given as follows;
e) G(x) = 3 +  + x , G[F(x)]
2.4 Given th funclions  G, H, wt iefis mt
so hat t folling four comsi funions
Assuming at H  (G  F) and [Ho G)  Fn  n, o t ialive law:
2.5 Pro thc following t*lheotic identit for union d inteion :
g)(A - B)B= Aif, andontyif, BG
6 Let f : S  T  a funion. If  and B a arbitry subts of S, pm that
f(  B) = f() f(O) and f(
Genetize to arbit unions and iat¢ions.
2.7 Letf : S  Ta fion. If Y   we dote
whichf into Y. Tt
f-(Y) = {x : x  S andf(x)
The set/-(Y) is 11¢ the inverse imee of F underf Pro the fotlowing fo bitra
sub,is X of S and Y of T.
f) Generali (c} and (d) to arbitra unions d intemions.
2.8 Refer to Exercise 2.7. ove thatf-(Y)] = y for ¢ sut Y of T if, and
only if, T =
2.9 tf:S  T a function, ove that the following gate,hiS am uivalenL
a) fis ontone on S.
b) ]  B) = f(A) f(B) for all sub,is M, BofS.
c) - ()] =  forew sut  orS.
d) For al 6isjoint suts  and B of S, the imaffd) and/(B) a disjoint.
e) For all subets A and/ of S with B __q A, we have
f(A -- B)
2.10 ove that ira  B d B  C,
2.11 If {I, 2 ..... n}  {i, 2,..., m},
2.12 If S i$ an ini L P tat S ntains a unb]y infinite su. Hint. Ch
 elent a in S and ider S - {o
2.13 ove thal eve nfini   contains a pror sut similar
2.14 If A i$ a ulable  and B  ununble
ZI5 A I humor is I] a[bic ff it is a    abic lion f(x) = 0,
wre f(x)  an + otx + -.- + o is a ]yno] with inr ients, o
t the t of I lomials with int t$
t of atbraic n  a unmhle.
2.16 t Sa finite t nsistg of n ts d t T  ltioa of all suts
of S. Show t T is a finite t d find the n of elects in
2.17 t R note e  of r num and let S dote the t of all l-v fun
lions wh do,in is R. Show that S d R a no1 uinuus. Hnt. Au
S  R d f a ote fui sh thatf)
the i-lued fcfion in S wh ns o t 1 nem a. Now e h by
he tion x)  1 + Ox(x) if x  R, and
2.18 t S  t ]{ion of all n who t a the inte 0 and 1. Show
that  is unnable.
2,19 Show that t following ls  counble:
a) Ihe t of ¢i in t p pl¢ having rational ii and nle with
tion inat,
b) any lllion of diinl inteals of tti nh
 t f  a l-val functio fin for e x in the intca1 0  x s 1.
Supp th is a iti humor M vi t fo]ling oy: for  choi of
a finite humor of ints x, xz,..., xa in t inteal 0  x  I, the sum
t S  t 1 ofth x in 0  x   for whichf(x)  0. ov¢ that S is counlable.
1 Find the fatla in t following "proof ' that t t of all inta]s of positi
t {x, xz,... } dote t uaabl¢  of rational num and I¢ 1  any
intel of itive knglh.  I ntains infinitely many tiol poin x but among
t t will  e wit s/test ix n.  a function F  ns of the uation
F(i) = n, if x is t rational humor with smalll Jx in Ihe inteal 1. is function
lablis a ontn¢ ¢ornd  the  of all interls and a sut of the
poslive int. H te  of all inels is
2. t S note the coll¢ion of all sus of a gi¢ t  tf : S  R  e real-
 function defin on S. e functionfis ] i iff(A  B) = f(A) + f(B)
wr A and Ba disjoint su of T. If f is aditive, p that for any two sub,is

A and B we have
f(A v B) = f(A) ÷ f(B - A) and f(A  )  f() + f() - f(  .
 R t E 2.. A f is addifi d u al tt t following
o hold for  icu su A d B of T:
f(A  B) = f(A + f(B -- f(Af(B')
f(A  B)  f(A)f(B), f(A) + f(B) 
w A" = T - , B" = - B.    ations if(, d m-
pule  
2.1 Boas, R. P., A Primer of Real Ftmetians. Carus Mo No. 13. Wi, N
YorL t.
2.2 F, A,, At t , 3 . NoHod. A 15.
3 G, A., Fnt of t y. An-W, Rd 1
2A  P. R., Nai t , V N N Yo, !.
2.5 K, E., ry of. F. , tor. , N Yo, 19.
2.6 gap, i., t  a Metric Ss. t and , ston, 1 2.
2.7 Ro, B., d  G. T.,  y of tz a TrfiMte Nurs.
, N Y, 1.
CHAPTER 3
ELEMENTS OF
POINT SET TOPOLOGY '
3.1 INTRODUCTION
A large part of the previous chapter dealt with "abstract" sets, that is, sets of
arbitrary objects. In this chapter we specialize our sets to be sets of real numbers,
sets of complex numbers, and more generally, sets in higher-dimensional spaces.
In this area of study it is convenient and helpful In use geometric terminology.
Thus, we speak about sets of points on the real line, sets of points in the plane, or
sets of points in some higher-dimensional space. Later in this book we will study
functions defined on point sets, and it is desirable to become acquainted with
certain fundamental types of point sets, such as open sets, closed sets, and compact
sets, before beginning the study of functions. The study of these sets is called
point set topology.
3.2 EUCLIDEAN SPACE R"
A point in two-dimensional space is an ordered pair of real numbers (xt, x).
Similarly, a point in three-dimensional space is an ordered triple of real numbers
(x t, x z, x3). It is just as easy to consider an ordered n-tuple of real numbers
(xt, x2,. • •, x.) and to refer to this as a point in n-dimensional space.
Definition 3,1. Let n > 0 be an integer. AIn ordered set of n real numbers
(Xl, x2,..., xj) is called an n-dimensional paint or a vector with n components.
Points or vectors wilt usually be denoted by single bold-face letters," for example,
x = (x,, x 2, ... , x,) or y = (y,, yz .... ,
The number x it called the kth coordinate of the point x or the kth component of
the vector x, The set of all n-dimensional points is called a-dimensional Euclidean
space or simply n-space, and is denoted by
.The reader may wonder whether there is any advantage in discussing spaces of
dimension greater than three. Actually, the language of n-space makes many
complicated situations much easier to comprehend. The reader is probably familiar
enough with three-dimensional vector analysis to realize the advantage of writing
the equations of motion of a system having three degrees of freedom as a single
vector equation rather than as three scalar equations. There is a similar advantage
if the system has n degrees of freedom.
47

4tl Elcme of Put Set Tolmq Def. 3.2
Another advantage in studying n-space ['or a general n is that we are able to
deal in one stroke with many properties common to l-space, 2-space, 3-space,
etc., that is, properties independent of the dimensionality of the space.
Higher-dimensional spaces arise quite naturally in such fields as relativity, and
statistical and quantum mechanics. In fact, even infinite-dimensional spaces are
quite common in quantum mechanics.
Algebraic operations on n-dimensionai points are defined as follows:
a) Equality:
b) Sum:
t X : (Xl, 1 1 I , X andy = (Yl .... , ) be in R% We define:
x = y  and only  xl =
x + y = (x: + Ya,.-., x, +
¢) Multiplication by real numbers ('scalars): -.
ax = (axa,... , axe) (a real).
d) Difference:
e) Zero vector or origin:
f) Inner product or dot product:
x- y =x + (-I)y.
o = (0,..., 0).
x'y =  xy t.
g) Norm or length:
The norm [Ix - Yl] is called the distance between x and y.
uoa In the terminology of linear algebra, R* is an example of a linear space.
Tkeorem $3. Let x and y denote points in R". Then we haee :
a) llxli > 0, and IIxll -- ]/f, and on6' if, x : O.
b) ]laxl = la¢ I[xlf for every real a.
c) IIx - yll = fly -
d) Ix- Yl < IIxll I[Y[I (Cauch-Sehwarz inequality).
e) IIx + Y[I  llxll + llylt (triangle'inequality).
Proof. Statements (a), (b) and (c) are immediate from the definition, and the
Cauchy-Schwarz inequality was .proved in Theorem 1.23. Statement (e) follows
Def. 36 Olma Balh and 01 Se in R' 49
from d) because
/=1 kxl
: Ilxll 2  2x-y + Ilyl[ x  Ilx[[  + 2llxli Ilykl + Ilyli 2  (llxl[ + [lyl]) ,
OT. Sometimes the tfangle inequality is written in the form
This follows from {e) by replacing x by x - y and y by y - z. We al
fiMtn 3. The un# crnate ctor  in R  is t ctor whose h com-
nent is I and whose remaking mnents are zero. Thus,
e  (!,0 ..... 0), u2 = (0, 1,0,..., 0), ..., = (0, 0,...,0, 1).
if x = (x,...,xD then x = xa +-'-+
x,e2 .... , x = x'a,. The vto u,..., u, are  l sis ctors.
3.3 OPEN BALLS AND OPEN SETS IN R •
Let a be a given point in R  and let r be a given positive number. The set of
points x in R" such that
Itx - aH < r,
is called an open n-ball of radius r and center a, We denote this set by B(t) or
by B(a; r).
The ball B(a; r) consists of all points whose distance from a is tess than r.
In R j this is simply an open haterval with center at a. In R 2 it is a circular disk,
and in R  it is a spherical solid with center at a and radius r.
3.5 Dejiaon of n ierior poi. Let S be a subset of R , and assume that a  S.
Then • is called an interior point of S if there is an open n-ball with center at a, all of
whose points belong to S.
In other words, every interior point a of S can be surrounded by an n-ball
B(a) _ S. The set of all interior point of S is called the interior of S and is
denoted by int S. Any set containing a ball with center a is sometimes called a
neiyhborhood of a.
3.6 Definition of an olden set. A set Sin R" is ealledopen if all its points are interior
points.
tOrE. A set S is open if and only if S = int S. (See Exercise 3.9.)

50 Eletmmts o Point  TOlmio TI 3.7
Emnple, In R j the simplest type of noncmpty open set is an open intexval. The union
of two or more open intervals is also open. A closed interval [o, b] is not an open set
because the endpoints a and b are not inter/or points of the interval.
Examples of oln sets in the plane are: h¢ interior of a d/sk; he cartesian producl of
two one-dimensional open intervals. Th© reader should be cautioned that an open interval
in R 1 is no longer an open set when it is considered as a subset of the plane. In fact, no
subset ofR x (except the empty set) can be open in R z, because such a set cannot contain
a 2-.ball.
In R" the empty set is open (Why?) as is the whore space R'. Every open n-ball
is an open set in R'. The cartesian product
(o,, b,) ×-..×
ofn one-dimensional open intervals (a, bt) .... , (a., b.) is an open set in 14" called
an n-dimensional open interval. We denote it by (a, h), where a -- (al,. l l , a and
b =
The next two 1heorems show how additional open sets in R" can be constructed
from given open sets,
The union of any collection of open sets is an open set.
Proof. Let F be a collection of open sets and let S denote their union, S = l,J,r A.
Assume x 6 S. Then x must belong to at least one of the sets in F, say x e A.
Since A is open, there exists an open n-ball B(x) _= A. But A ___ S, so B(x)  S
and hence x is an interior point of S. Since every point of S is an interior point,
S is open.
Theorem 3.& The interection of a finite collection of open set is open.
Proof Let S = 1"]'=1 A where each A is open. Assume x e S. (lfS is empty,
there is nothing to prove.) Then x e A for every k =- l, 2, .., m, and hence
there is an open n-ball B(x; e) __. A. Let r be the smallest of the positive numbers
rt, rz .... ,r,. Then xeB(x;r) _ S. That is, x is an interior point, so Sis
open.
Thus we see that from given open sets, new open sets can be formed by taking
arbitrary unions or finite intersections. Arbitrary intersections, on the other hand,
will not always lead to open sets. For example, the intersection of all open intervals
of the form (-l/n, I/n), where n = I, 2, 3 .... , is the set consisting of 0 alone.
3A THE STRUCTURE OF OPEN SETS IN [I t
In R  the union of a countable collection of disjoint open intervals is an open set
and, remarkably enough, every nonempty open set in R 1 can be obtained in thig
way. This sect/on is devoted to a proof of this statement.
First we introduce the concept of a component interval.
Th. 3,11 TI S'ucIu of Oln Sels ill R  1
3,9 Definition of comtmnem interd. Let S   o bset of R . An on
intel 1 (which may  finite or inite)  lted a compont ter of S if
!  S d  there is no open inter! J  1 ch that I  J  S.
In other words, a eomncnt interval ors is not a pror subt of any other
on inal conn in S.
Trem 320. Every int of a nonempty open set S belays to one and only one
compont interval of S.
Proof. Aume x e S.  x is nned in some on instal I with I  S.
Thc are many such laterals but the "largest" of the will  tbe desi com-
ponent inlervaL We leave it to the rder to verify that this largest lateral is
L = (x), x)), where
a(x) = inf {a : (a, x)  S}, b(x) = sup {b : (x, b) U S}.
He a(x) might  -  and b(x) might  + . Clearly, there is no on interval
J such that I  Y  S, so 1 is a eomnent instal of S containing x. If J
is o¢r mnent instal of S nting x, en the union 1 w J is an
on inteal contain in S and containing both 1 and J. Hence, by e defi-
nition of component inteal, it follows that I w J = I and I w J = J, so
Torem 3.11 (Represeion tom for on sets on t re ). Every non-
empty on set S in R  is the union of a countable collection of disjoint comnent
intea of S.
Proof. lfx  S, let 1 denote the comnent lateral ors ntaining x. The union
of all such inteals 1 is clrly S. If two of them, 1 and I, have a point in
common, then their union I w I is an on intern[ contained in S and containing
both 1 and Is. Hence 1  lr = I and I,  i = I so 1, = I r Thefo the
inteals 1 form a disjoint call,ion.
It mains to show that they form a countable call,ion. For is purpose,
let {x, xz, xn, ... } denote the countable t of rational numrs. In each com-
ponent ina] 1 there will  infinitely many x,, but ong th there will 
exactly one with smallest index n. We then. define a function F by mns of the
equation F(I) = n, if x, is the rational humor in I with smallest index n. This
function F is oneqone since F(i) = F(Ir) = n implies that I and I have x, in
common and this implies I = I r Therefore F establishes a one-to-one corre-
spondence twn the inteals I and a subt of the sifive integers. This
completes the proof.
OT+ This pntation of S is unique. In fact if S is a union of disjoint on
intervals, then the inteais must  the component inteals of S. is is an
immediate conquenee of Theorem 3.10.
if S is an on interval, then the rprntation contains only one comnent
interval, namely S it]f. efo an on inlerval in R 1 cannot  expres as

the union of two nonempty disjoint open sets. This property is also described by
saying that an open interval is connected. The concept of connectedness for sets
in R  will be discussed further in Section 4.16.
3.5 CLOSED SETS
3.12 Deflaitlom of a closed set. /I set S in R  is called closed (It" its complement
R  - S is open.
Examlfles. A closed interval [a0 b] in R t is a closed set. The carlesian l:n'oduct
[., b,] × -.. × [.,
of n one-dimensional closed intervals is a closed st in R" cal]c.A an ndinumMonal closed
interval [a, b].
The next theorem, a consequence of Theorems 3.7 and 3,8, shows how to
construct further closed sets from given ones.
Tkeoeem 3.13. The union of a finite collection of closed sets is closed, and the
intersection of an arbitrary collection of closed sets is closed.
A further relation between open and closed sets is described by the following
theorem,
Tluoeem 3.I4. If A is open and B is closed, then A - B is open and B - A is
closed.
Proof. We simply note that A - B = A c (R" -- O), the intersection of.two
open sets, and that B - A ---- B ta (R" - A), the intersection of two closed sets.
3.6 ADHERENT POINTS. ACCUMULATION POINTS
Closed sets can also be described in terms of adherent points and accumulation
points.
3.15 Defmitiom of an adlterent point. Let S be a subset of R , attd x a point in R',
x not necessarily in S, Then x is said to be adherent to S if every n-ball B(x) contains
at least one point of S.
1. If x  S, then x adheres to S for the trivial reason that every n-ball B(x) contains x.
2. If S is a subset of R which is bounded above, then sup S is adherent to S.
Some points adhere to S because every ball B(x) contains points of S distinct
from x. These are called accumulation points.
3.16 Definition of an accumulation paine. If S _ R  and x  R , then x is called
an accumulation point of S if every n-bali B(x) contains at least one point of S
distinct from x.
In other words, x is an accumulation point of S if, and only if, x adheres to
S  {x}. If x e S but x is not an accumulation point of S, rhea x is called an
isolated point of S.
1. The s of numbers of the form I/n, n  1, 2, 3,..., has0 asan accumulation point.
2. The set of rational numbers has every real number as an accumulation point.
3. Every point of the closed interval [a, b ] is an accumulation point of the set of num-
 in the open interval (a, b).
Theorem 27.17. If x is an accumulation point of S, then eery n-ball B(x) contains
infinitely many points of S.
Proof Assume the contrary; that is, suppose an n-ball B(x) exists which contains
only a finite number of points of S distinct from x, say a, 2, -. -, ,. lfr denotes
the smallest of the positive numbers
then B(x; r12) will be an n-ball about x which contains no points of S distinct
from x. This is a contradiction.
This theorem implies, in particular, that a set cannot have an accumulation
point unless it contains infinitely many points to begin with. The converse, how-
ever, is not true in general. For example, the set of integers {!, 2, 3, ... } is an
infinite set with no accumulation points. In a later section we will show that
infinite sets contained in some n-ball always have an accumulation point. This is
an important result known as the Bolzano-Weicrstrass theorem.
3.7 CLOSED SETS AND ADT POINTS
A closed set was defined to be the complement of an open set. The next theorem
describes closed sets in another way.
Tkeorem 3,1& d set S in R"/s closed if, and only if, it contains all its adherent
points.
Proof Assume S is closed and let x be adherent to S. We wish to prove that x  S.
We assume x ¢ S and obtain a contradiction. If x ¢ S then x  R  - S and, since
R"  S is open, some n-ball B(x) lies in R" - S. Thus B(x) contains no points of
S, contradicting the fact that x adheres to S.
To prove the converse, we assume S contains all its adherent points and show
that S is dosed. Assume x $ R  - S. Then x ¢ S, so x does not adhere to S.
Hence some ball B(x) does not intersect S, so B(x)  R" - S. Therefore R" -- S
is open, and hence S is closed.
3,19 Dcfudtioa of clore. The set of all adherent points of a set S is called the
closure of S and is denoted by .

54 Elements of Point Se{ Telmlog3 r Th. 3.20
For any set we have S _ S since every point ors adheres to S. Theorem 3.18
shows that Ihe opposite inclusion .  S holds if and only ifS is closed. Therefore
we have:
Theorem 3.20. A set S is closed if and only if S = 3.
3.21 De.fmiffol of deriedset. The set of all accumulation points of a set S is
called the derived set ors and is denoted by S'.
Clearly, we have ; = S L) S' for any set S. Hence Theorem 3.20 impl!es that
S is closed if and only ifS'  S. In other words, we have:
Theorem 3.22, A set S in R" is closed if, and only if, it contains all its accumulation
points.
3.8 THE BOLZNO-WEIERSTRASS THEO -
3.23 /)e/it/ott of a boumled set. A set Sin R" is said to be bounded if it lies entirely
within an n-ball B(a; r) for some r > 0 and some a in R*.
Tltearem 3.24 (Bolzano-IVeieestras). If a bounded set S in R" contains infinitet),
many points, then there is at least one point in R" which is an accumulaton point of S.
Proof TO help fix the ideas we give the proof first for R t. Since S is bounded,
it lies in some interval f-a, a]. At least one of the subintervals l-a, O] or [0, a]
contains an infinite subset of S. Call one such subinterval [at, bl]. Bisect [al,
and obtain a subinterval [a, bz] containing an infinite subset of S, and continue
this process. In this way a countable collection of intervals is obtained, the nth
interval [a,, bj] being of length b, - aj = a[2"-t. Clearly, the sup of the left
endpoints a, and the inf of the right ¢ndpoints b must lye equal, say to x. [Why
are they equal?] The point x will be an accumulation point of S because, if r is
any positive number, the interval [a,, b,] will be contained in B(x; r) as soon as n
is large enough so that b, - a. < r/2. The interval B(x; r) contains a point of S
distinct from x and hence x is an accumulation point ors. This proves the theorem
for R !. (Observe that the accumulatioff point x may or may not belong to S.)
Next we give a proof for R", n > l, by an extension of the ideas used in treating
R t. (The reader may find it helpful to visualize the proof iff R 2 by referring to
Fig. 3.1.)
Since S is bounded, S ties in some n-ball B(O; a), a > O, and therefore within
the n-dimensional interva J defined by the inequalities
--a<_x <_a (k - 1, 2, . .. , n).
Here J, denotes the cartesian product
that is, the set of points (xt,..., x), where x  I ) and wher each I lr is a
one-dimensional interval -a < x _< a. Each interval I 1 can be bisected to
Figure 3.1
form two subintervals I} and -,.2,r°) defined by the inequalities
Next, we consider all possible cartesian products of the form
1(1) x r¢1 x --- x I ' (a)
t,t I a2J
where each kl = | or 2..There are exactly 2 * such products and, f course, each
such product is an n-dimensional interval. The union of these 2  intervals is the
original interval J,, which contains S; and hence at least one of the 2 J intervals in
(a) must contain infinitely many points of S. One of these we denote by J2, which
can then be expressed as
J = rl 2 × 1( ) x .--× 1. (,
where each I () is one of the subintervals of I, (1 of length a. We now proceed
with J2 as we did with Jr, bisecting each interval i (2 and arriving at an n-dimen-
sional interval J3 containing an infinite subset of S. If we continue the process,
we obtain a countable collection of n-dimensional intervals J, Jz, J3, --., where
the ruth interval J,. has the property that it contains an infinite subset of S and
can be expressed in the form
J,, = I " × I ") x ,., x I, ('), where I
Writing
we have
(k = 1,2,'", n),
For each fixed k, the sup of all left endpoints a[ "q, (m - I, 2,, ,, ), must therefore
be equal to the inf of all right endpoints b[ "), (m = 1, 2, ... ), and their common
value we denote by t. We now assert that the point t = (t,, tz ..., t,) is an

accumulation point of S. To sec this, take any n-ball B(t; r). The point t, of
course, belongs to each of the intervals Jr, Jz .... constructed above, and when
m is such that al2 "- < r]2, this neighborhood will include J,. But since J=
contains infinitely many points of S, so will B(t; r), which proves that t is indeed
an accumulation point of S.
3.9 THE CAN'IR INTERSECTION THEOREM
As an application of the Bolzano-Weierstrass theorem we prove the Cantor
intersection theorem.
 3.25. Let {QI, Qz, • • • } be a countable collection ofnonempty sets in R 
such that:
i Each set Q is cloed and Q  is
Th the intersection   Q i closed d no,reply.
Prf tS =  Q. en Sis cl u of 3.I3. To ow
that S is no, we hit a int x in S. We n  tt ch Qa con-
ns infiteIy y in; oe  pfis . Now fo a Ron
ofdt inA = {x,x,...}, wh xQ. s A is  te
n     Qj, it   u nt, y x. We'
ow at x  S by vefifyg at x  Q for  k. It w s to w t
is an ution int of h Q, sin t a  clo . But
neighrh of x ns initdy many ints of , d si alf
(ibIy) a fini nr of e ints of A long to Q, ts ih al
con iniIdy y n of Q efo x is an muhtion int of
Q d  t is ov.
In this section we introduce the Concept of a covering of a set and prove the
Lindelbf covering theorem. The usefulness of this concept will become apparent
in some of the later wociL
3.26 Deftl¢ jfm ¢¢ri. A collection F of .ets L¢ said to be a covering of a
given set 3; if S _ Ur A. The collection F is also aid to cover S. If F i a
collection of open sets then F is called an open coering of S.
Exam.s
i. The collection of all imerva|s of th form l[n < x < 2/n, (n = 2, 3, 4,... ), is an
open covering of the interval 0 < x < I This is an example of a countable covering.
2. The real liner  is covcrca:l by thecoltcctioo of all open intcxvals (a. b). This covering
is not coumablc. Howcver it contains a countable covering ofR t, namely, all inter-
vals of tl form (n, n + 2), where n runs through the inlegcrs
3. Let  = {(x, y) : x > 0, y > 0). The ¢ollcfion F of all circular disks with
at Oc, x) and wih radius x, whcf¢ x > 0, is a coving of $. This ¢owring is not
countable. Hovcever, it contains a countable covering of S, narnly, all those disks
in which x is rational (So0 Exercise
The Lindel6f covering theorem states that every open covering of a set $ in
contains a countable subcollection which also covers S. The proof makes use of
the following preliminary result:
Tlmaeem3,27 Let G = {A, A, ...} denote tAe countable collection of all n-
balls tuing rational radii and centers at point with rational coordinates. Assume
x  R  and let S be an open set in R" which contains x. Then at least one of the
n-balls in G containz x and is contained in S. That is, we have
x  A _ S for some A in G
Proof. The collection G is countable because of Theorem 2.2. Ifx  R " and ifS
is an open set containing x, then there is an n-hall B(x; r)  S. We shall find a
point y in S with. rational coordinates that is "near" x and, using this point as
nter, will then find a neighborhood in G which ties within B(x; r) and which
contains x. Write
and let y be a rational
k :- 1,2,...,n. Then
X - (xt,.x2,-,-,
number such that IY - xz] < r/(4n) for each
Next, |et q be a rational number such that r/4 < q < el2. Then x  B(y; q) and
B(y; q) c:_ B(x; r) ___ S. But B(yg q)  G and hence the theorean is provod.
(See Fig. 3.2 for the situatio in
Theorem 3.2 (l.lavlelf coerimj theorem). As.none d _ " and let F be an open
coeriag of A. 'hen there is a countable subcollection of F which also covers A.
Proof. Let G = {J, A .... } dcnot the countable ¢oflection of all n-balls
having rational centers and rational radii. "[his st G will he ased to help as extract
a countable subeollcction of F which covers f.

Assume x ¢ A. Then there is an open set S in F such that x • . By Theorem
3.27 there is an n-ball A ia G such that x ¢ A  S. Tbee ar, of 0ourse, infinitely
many such A corresponding to each S, but we choose only 6he of these, for ex-
ample, the one of smallest index, say m - re(x). Then we have x e A,t ) _ S.
The set of all n-balls A,,x ) obtained as x varies ovr all elements of d is a countable
collection of open sets which covers A. To get a countable subcoltection of F_
which covers .4, we simpty correlate to each s¢ AI) one of the sets S of F which
contained Ax ). This completes the proof.
3.11 TH HKINE--BOP..EL COVERING THEOREM
The Lindel6f covering theorem states that from any open covering of an arbitrary
set A in W we can extract a countable covering. The Heinc-Borel theorem tells
us that if, in addition, we know that `4 is closed and bounded, we can reduce the
cover/rig to a finite covering.. -Tile proof makes use of the Cantor intersection
theorem.
em :L29(Heine-Borel). Let F be an open covering of a closed and bounded
set A in R'. Then a finite subcoilection ofF also covers A.
Proof. A countable subcollcction of F, say {I, I2, ... }, covers A, by Theorem
3.28. Consider, for m >_ 1, the finite union
This is open, since it is the union of open sets. We shall show that for some value
of m the un/on S,, covers A.
For this purpose we consider the complement R  - S.,, which is closed.
Define a countable coll¢:tion of sets {Q, Q .... } as follows: Q = 4, and for
That is, Q, consists of those points of A which lie outside of S,,. If we can show that
for some vMu¢ afro the set Q,, is empty, then we will have shown that for this.m
no point of d lies outside S,; in other words, we will have shown that some S,
covers A.
Observe the following properties of the sets Q,: Each set Q,, is close(i, since
it is the intersection of the closed set A and the closed set R" - S,. The sets Q,
are decreasing, since the S,, are increasing; that is, Q,,+ _ Q,.. The sets
being subsets of-4, are all bounded. Therefore, if no set Q,, is empty, we can apply
the Cantor intersect/on theorem to conclude that the intersection 1; Q is
also not empty. This means that there is some point in `4 which is in all the sets
Q,, or, what is the same thing, outside all the sets S,,. But this is impossible, since
A _q [J'_ Sk. Therefore some Q, must he empty, and this completes the proof.
3.31 CompaetlaS ia R  59
3.12 COMPACTNESS IN
We have just seen that if a set S in R  is closed and bounded, then any open
covering of S can be reduced to a finite covering. It is natural to inquire whether
there might be sets other than closed and bounded sets which also have this
property. Such sets wilt be called compact.
.30 DefiMtio of  compact set. A set S in R  is said to be compact if, and only if,
every open covering ors contains a finite subcover, that is, a finite subcotfection which
ato cover S.
The Heine-Borel theorem states that every closed and bounded set in
compact. Now we prove tbe converse result.
Theorem 3.211. Let S be a subset of R'. Then the following three stateraen1 are
equivalent:
a) S is compact.
b) S is closed and bounded.
c) Every infinite subset of S has an accumulation point in S.
Proof. As noted above, (b) implies (a). If we prove that (a) implies (b), that (b)
implies (c) and hat (c) implies (b), lhis will establish the equivalence of all three
statements.
Assume (a) holds. We shall prove first that S is bounded. Choose a point
in S. The collection of n-balls
By compactness a finite subcollection also covers S and hence S is bounded.
Next we prove that S is closed, Suppose S is not closed, Then there is an
accumulation point y of S such that y
is positive since y ¢ S and the collection {B(x; r) : x • S} is an open covering of
S. By compactness, a finite number of these neighborhoods cover S, say
=
Le! r denote the smallest of the radii r, r,..., r. Then it is easy to prove that
the ball B(y; r) has no points in common with any of the baits B(x; r), In fact,
if x • B(; r), then IIx - yll < r _< r, and by the triangle inequality we have
IlY - xdl -< IlY - xll ÷ Ilx - xl[, so
[Ix - xll > fly - xll - IIx - Ylt = 2r - IIx - Yll > r,
Hence x ¢ B(x; r D. Therefore B(y; r) c S is empty, contradicting the fact that
y is an accumulation point orS. This contradiclion shows that S is closed and hence
(a) implies (b).
Assume (b) holds, In this case the proof of (c) is immediate, because if T is
an infinite subset of S then T is bounded (since S is bounded), and hence by the
Bolzano-Wcierstrass theorem T has an accumulation point x, say. Now x is also

an accumulation point of S and hence x ¢ S, since S is closed. Therefore (b)
implies (c).
Assume (c) holds. We shall prove (b). If S is unbounded, then for every
m > 0therccxitsapointx=in Swith x=ll > m. Thecollvction T  {xl, x2,...}
is an infinite subset of S and hence, by (c), T has an accumulation point y in S.
But for m > I + I[yll we have
IIx=- yll > I[x.II- Ilyll > m- yff > i,
contradicting the fact that y is an accumulation point of T. This proves that S is
bounded.
To complete the proof we must show that S is dosed. Let x [x an accumulation
point of S. Sine, every neighborhood of x contains infinitely many points of S,
we can consider the neighborhoods B(x; Ilk), wher¢ k -- I, 2 .... , and obtain a
countable set of distinct points, say T = {x,, xa, ... }, contained in S, such that
x  B(x; Ilk). The point x is also an accumulation point of T. Since T is an
infinite subset of S, part (c) of the theorem tells us that T mu,t haw an accumula-
tion point in S. The theorem will then IX proved if we show that x is the only
accumulation point of T.
To do this, suppose that y # x. Then by the triangle inequality we have
Iry - xll < Ily - x,ii ÷ iix - xll <tly - x,I[ ÷ Ilk, ifx  T.
If k o is taken so large that lt'k < ½11Y. - xtl whene,er k .>_ ko, th last inequality
leads to ½1IY - xll < ]IY - xtll. This shows that xt ¢ B(y; r) when k > k o, if
r -- ½[y - xll. Hence y cannot [x an accumulation point of T This completes
the proof that (c) implies (b).
3.13 METRIC SPACES
The proofs of some of the theorems Of this chapter depend onIy on a few properties
ofthe distance between points and not on the fact that the points are in R'. When
these properties of distance are studied abstractly they lead to the concept of a
metric space.
3212 Defugfie of a mcffrk: paee. A metric pace is a nonempt set M of objects
(called points) together with a function d from M × M to R (called the metric of
the space) satixfying the following four properties for all points x, y, z in M:
2. d(x, y) > O/fx # y.
3. d(x, y) : d(y, x).
. d(x, y) < d(x, z) + d(z, ).
The nonnetive number d(x, y) is to be thought of as the distance from x to
y. In these terms the intuitive meaning of properties 1 through 4 is dear. Property
4 is called the triangle inequality.
Poi Set Tetralogy la Met¢ Slm¢
61
We sometimes denote a metric space by (M, d) to emphasize that both the set
M and the metric d play a role in the definition of a metric space.
1. M = Its; d(x, y) = I[x - y[. This i called the Euclidean metric. Whenev¢r we refer
to Euclidean space R', it will be understood that the metric is the Euclidean metric
unl¢ anothex metric is specifically mcntioncxf.
2. M = , the cmpkm plane; d(z, zz) -- Iz - zxl. As a metric space, C is indistin-
guishabl from Euclidean space R x because it has the same points and the tme metric.
3. M any noncmpty t; d(x, y) = O if x = y, d(x. ) = 1 ifx # y. Thisiscallod the
diacrete metric, and (M, d) i callvd a diaerete metric apace.
4. If (M. d) is a metric space and if S is any non©m0W tub,t of M, thtm (S, d) is also a
metric space with the same metric or, more precisely, with the restriction of d to
,g × ,g as metric. This is sorrmtirmm called the relative metric induced by d on S, and
8 is called a metric abspace of M. For ¢xampl the rational numbea O with th
m¢t d(x, y) = [x - Yl form a nric subspm of 
$. M= R; d(x,)-  - y,) + 4(xl -a) 2, wher x = (x,,x2) and .v =
(yl, y2). The metric space (M, d) is not a metric supace of Euclidean space it 
because the metric is different.
6. M = {(xt, x):xi 2 + x22 -- 1), the unit circle in R2; d(x,) = the length of the
smaller rc joining the two points x mad  on the unit circle.
7. M = {(xt, x, x) : xl + x22 + x = 1 }. the unit sphere in R; d(x, y) = the length
of tim smaller arc along the great circle joining the two points x and y.
& M = R'; d(x. y) = Ix. - Y,[ +'" + Ix. - Y.I-
9. M = R; d(x, y) = max {Ix, - YI ..... ix, - y, IL
3.14 POINT SET TOPOLOGY IN ME'IIC SPACES
The basic notions vf point set topology can lx etended to an arbitrary metric
space (M, d).
If a  M, the ball B(a; r) with center a and radius r > 0 is defined to be the
set of all x in M such that
d|x,a) < r.
Sometimes we denote this ball by B(a; r) to emphasize the fact that it points
come from M. IfS is a metric subspace of M, the ball Bs(a; r) is the intersection
of S with the ball Bu(a; r).
EXamltlt In Euclidean space R 1 the ball B(0; !) is the open interval (- I, I). In the
rrric subspa S = [0, 1] th¢ ball Bs(0; I) is the half-open interal [0, 1L
sor. The geometric appearance of a ball in R" need not be "spherical" if the
metric is not the Euclidean metric. (See Exercise 3.27÷)
If S _ M, a point a in S is called an interior point of S if some ball Ba; r)
lies entirely in S. The interior, int S, is the set of interior points ors. A set S is

called open in M if all it, points ar interior points; it is called elosvd in M if M - S
is open in M.
1. Every hall B,(a; r) in a metric Slmee M is open in M.
2 In a discrete metric stac¢ Mecry subset Sis open. In fact, ifx  S, the ball B(x; ½)
¢ousists entirely of points of S (since it oontains only x), so S is open. Thefore every
subset of M is also dosed !
3. In the metric subspace S -- [0, ] ] of Euclidean space R a, every interval of ghe form
f0, x) or (x, 1 ], where 0 < x < .t, is an open set in S. These   not open in R t.
Example 3 shows that if S is a metric subspace of M the open sets in S
not b¢ open in M. The next theorem describes the relation between open sets in
M and those in S.
Ylg, om 3..L Let (S, d) be a metric mbspace of (M, d), and let X be a ubset of
S. Then X is openin S if, and only
for some $et ,4 which ia open in M.
Proof Attme `4 is open in M and let X = `4 ca S. If x  X, then x ¢ A so
Bu(x;r) _ Aforsomer> 0. HenceBs(x;r) =Ba(x;r)SG A fS= X
so X is open in S.
Convel-sly, assume X is open in S. We will show that X = ,4  S for some
open set A in M. For every x in X there is a ball B(x; r,,) contained in X. Now
Bs(x; r=) = Bu(x; r  S, so if we let
A-- 0
then A is open in M and it is easy to verify that `4  S = X.
Theorem 34. Let (S, d) be a metric subspace of (M, d) and let Y be a subset of
S. Then Y is closed in S if, and only  Y -- B  S for some set B which is closed
inM.
Proof. If Y = B ca S, where B is closed in M, then B ffi M - ,4 where ,4 is open
inMso Y= SrBffi S(M-A) ffi S-,4;hencc Y is closed in
Conversely, if Y is closed in S, let X -- S - Y Then X is open in S so
A  S, where ,4 is open in M and
where B = M - A is closed in M. This compJetes the proof.
If 3" _ M, a point x in M is called an adherent point of Sifevery ball Bu(x:
contains at least one point of S. If x adheres to S - {x} then x is called an
ccumulation point of S. The closure  of S is he set of all adherent points of S,
and the derivedset S' is the set ofa]l accumulation points ors. Thus, $
The following theorems arc valid in every metric space (M, d) and are proved
exactly as they were for Euclidean space R'. In the proofs, the Euclidean distance
IIx - yli need only be replaced by the metric d(x, y).
Tlworem 3,35. a) The union of any collection of open sets is open, and the inter-
section of a finite collection of open sets is open.
b) The union of a finite collection of ctosed sets is closed, and he intersection of any
collection of closed sets is closed.
If ,4 is open and B is closed, then A - B is open and B  A is
Theorem 3:17. For any subset S of M the following tatements are equfvalent :
a) S is closed in M.
b) S contains all its adherent points.
c) S contains all its accumulation points,
d) sf-g.
Examlie. Let M = Q, the set of rational numbers, with the Euclidean raettic of R x.
Let S consist of all rational numbers in the open interval (a, b), where both a and b are
irrational. Then S is a closed subset of Q.
Our proofs of the Bolzano-Weierstrass theorem, the Cantor intersection
theorem, and the covering theorems o/" Linde]6f and Heine--Bore! used not only the
metric properties of Euclidean space R" but also special proper :ies of R  not gen-
eraIIy valid in an arbitrary metric space (M, d). Further restrictions on M arc
required to extend these theorems to metric spaces. One of these extensions is
outlined in Exercise 3.34.
The next section describes compactness in an arbitrary metric space.
3.15 COMPACT SUBSETS OF A METRIC SPACE
Let (M, d) be a metric space and let S be a subset of M. A collection F of open
subsets of M is said to be an open covering of S if S _ [je ,4.
A subset S of M is called compact if every open covering of S contains a finile
subcover. S is called bounded if S  B(a; r) for some r > 0 and some a in M.
Tkeorem 3.38. Let S be a compact subset of a metric space M. Then:
i) S is closed and bounded.
ii) Every infinite subset ors has an accumulation point in S.
Proof. To prove (i) we refer to the proof of Theorem 3.31 and use that part of the
argument which showed that (a) implies (b). The only change is that the Euclidean
distance IIx - yll is to be replaced throughout by the metric d(x, y).

To prove (ii) we argue by contradiction. Let T be an infinite subset of S and
assume that no point of S is an accumuration point of T. Then for each 10int x in
S there is a bali B(x) which contains no point of T(ifx  T) or exactly one point
of T (X itself, if x  T). As x runs through S, the union of these balls B(x) is an
open covering of S. Since S is compact, a finite subcol]ection covers S and hence
also covez T. But this is a contradiction because T is an infinite set and each ball
contains at most one point of T.
Nor In Euclidean spaze R , each of properties (i) and (ii) is equivalent to com-
pactness (Theorem 3.31). In a general mewic space, property (ii) is cquivatent to
compactness (for a proof see Reference 3:4), but property (i) is not. Exercise 3.42
gives an example of a metric space M in which certain closed and bounded subsets
arc not compact.
Tleoeem 3d9, Let X be a closed subset of a compact metric stace M. Then X is
compact.
Proof. Let F be an open covering of X, say X _ Uar ,4, We will show that a
finite number of the sets A cover X. Since X is closed its complement M - X is
open, so F u {(M - ,g)} is an open covering of M. But M is compact, so this
covering contains a finite subcovcr which we can assume includes M - X. There-
fom
M  A I -'-u A,u (M- X).
This subcovcr also covers X and, since M - X contains no points of X, we can
delete the set M - Xfrom the subcover and still.cover X. Thus X _ A ...  A,
so X is compact.
3.16 BOUNDARY OF A SET
Deletion .40, Let S be a subset era metric space M, A point x in M is called a
boundary point of S if every ball Bu(x; r) contains at leas one point of S and at
least one point elM - S. The et of att boundary point ors is called the boundary
of S and  denoted by
The reader can easily verify that
S.= gerM- S.
This formula shows that S is closed in M
Example InR', the boundary of a ball B(a ; r) is the set of points x such that Ix -  = r.
In R , the boundary of the set of rational numbers is all of R .'
Further properties of metric spaces arc developed in the Exercises and also in
Chapter 4.
3.1 Prove that an open interval in R  is an open set and that a closed interval is  closed.
Set.
3-2 Imine all tl accumulation points of the following sets in R t and decide whether
the sets are open or closed (or neither).
a) All integczs,
b) The interval (a, b].
c) All numbers of the form l[n, (n - 1,2,  .... ).
d) All rational numbers.
¢) All numbers of rig form 2 - + -=, (m, n = 1, F., ,,. ),
f) All numbers of the form (- IF' + (l/m), (m, n = I, 2,... ).
g) All numbers of the form (l/n) + 0/m), (m,  = 1, 2,... ).
h) Allnumbets ofth© form (- Iy/[l + (l/n)], (n = 1, 2,...).
3.3 The same as Exercise 3.2 for th¢ following sets in R z :
a) All complex z such that Iz[ > I.
b) All comp z such that Izl -> i.
c) All complex numbers of tht form (l[n) + (i/m), (m, n = 1, 2 , .  ).
d) All points (x, y) such that x z - yz < i.
e) All points (x, y) such that x > 0.
f) Air points (x, y) sm:h that x > 0.
3.4 Prove that every noncmpty open set S in R  contains both rational and irrational
numbers.
3.5 Prove lhat the only sets in R  which arc both open and dosed are the ¢npty set and
R  itself. Is a similar statement true for Rz?
3,6 1Wove hat every closed set in IR  is the intersection of a countable collection of open
3.7 Prove tlt a nonernpty, bounded dosed set $ in R  is either a dosed interval, or that
S can be obtained from a closed interval by removing a countable disjoint collection of
open intexvals whose endpoints belong to S.
Olmm aml clo scs tn R =
3-8 Prove that open n-halls and n-dL, nemional open intervals arc open sets in R'
3.9 Prove that the interior of a set in R" is open in
3.10 IfS' _ R', prove tlt int S is the union of all open subsets err" which ate contained
in S. This is described by saying that int S is  laxst open subset of
3.11 ff A' and T are subsets of R , pt'ove that
(int S)  (int T)  im (S t T), and (int )  (int T)  int (S t T).

Elements 0[ Point Set Topology
3.12 L,t S' denote the derived set and S the closure of a set S in RL Prove tt :
a) S" is cl in R"; that is, (S)' 
b) lfS  T, thS' T'. c) (S ' = S'w
d) ($)' = S'. e) N is cl in R .
0 $ is the inttion of all cl sub,Is og R" ntaining S. That i  is t
alt cl t containing S.
3.13 t Sand TsutsofRL ovethat S  T    and that S   G   T
if S is
. e statts in i 3.9 thou# 3.13 a e in any tric
3.14 A t S in R   d noex if, for eve i r of in x and y in S and e al
0 tiffying0 < 0 < i, weve + (I - 0D  & Intem this statet metric-
ally (in R 2 and R 3) d 0 that:
a) Ev wll in R  is convex.
b) E nMiional on inteal is
O e t of a convex t is nvex.
d)  ctu of a ¢on t is n.
3.1 t F a co]ltion of ts M R a, and let S  0a,r A and T = ay M. For
ch of t following statts, ith # a 0rf or eahibit a
a) If is  aumeMtion int of T, th x is an amutation int erich t
AMF.
b) Ifx is an aumuMtion point of & then x is an aumelation int of at Ist one
AinF.
3,16 o that ¢ t S of tiol num in t intM (0, l) ot  exp
as  Mion era unmb colfion aon . Hint. Wfim
au S = =t St, where eh St is o, d nsct a q {,} of cloud
intals su that ** G Q G Sa d such that xa ¢ =.  u the ntor inter-
ioa t to obin a comiion.
3.17 If S _c R', prove that the collection of isolated points of S is countable.
3.18 Prove that the set of open disks in the xy-plane with center at (x, x) and radius
x > 0, x rational, is a countable covering of th¢ set {(x, y) :
3.19 The collection For open intervals of the form (l/n, 2/n), where n = 2, 3 ..... is an
open covering of the open interval (0, 1). Prove (without using Theorem 3.31) that no
finite subcollection of Fcovers (0, I).
3.20 Give an example era set S which is closed but not bounded and exhibit a countable
open coveriug F such that no finite subset of F covers S.
3.21 Given a set S in R" with the property that for every x in S there is an n-ball B(x)
such that B(x)  S is countable. Prove that S is countable.
3.Z2 Prove that a colk:ction of disjoint open sets in R * is ne/x.sarily countable. Give an
eample of a collection of disjoint closed sets which is not countable.
3.23 Assume that S _ R', A point x in R" is said to be a condensation point of S if every
n-ball B(x) has the property that B(x)  S is not countable. Prove that if 8 is nc t cotmt-
able, then there exists a point x in S such that x is a condematiou point of
3.2t Assume that S G R" and assume that S is not countable. Let T denote the set of
condensation points of S. Prove that;
a) S - T is countable, b) S  T is not countable,
c) 7'is a closed set, d) T contains no isolated points.
Note that Exercise 3.23 is a special case of Co).
3J.5 A set in R" is called perfect if S = S', that is, if S is a closed set which contains no
isolated points, prove that every uncountable closed set F in R a can be expressed in the
form F = .4  B, where ,4 is perfect and B is countable (Cantor-Bendixon theorem).
Hint. Use Exercise 3.24.
Metric slaaces
3.26 In any metric space (M, d), prove that the empty set 0 and the whote spaoe M are
both open and closed.
3,27 Consider the following two metric in R":
dr(x, r)= max txa - Yd,
In each of the following metric spaces prove that the ball B(a; r) has the geometric
appearance indicated:
a) In (R 2, at), a sqtJal with sides parallel to the coordinate axe,
b) In (R z, d2), a square with diagonals parallel to the axes.
c) A cube in (R a, di).
d) An oetahedron in (il s,
3.28 Let d, and d 2 be the metrics of Exercise 3.27 and let IIx - wll denote the usual
Euclidean metric+ Proc¢ the following inequalities for all x and y in R':
3.29 If (M, d) is a metric space, define
d'(x, y) = -
t
d(x, y)
Prove that d' is also a metric for M. Note that 0 <_ d'(x, y) < t for all x, y in M.
3.30 Prove that every finite sub of a metric space is closed.
3,31 In a mettle space (M, d) the closed ball of radius r >. 0 about a point- a in M is the
set B(a; r) = {x : d(x, a) <_ r ).
a) Prove that B(a; r) is a dosed set.
b) Give an example of a metric space in which B(a; r) is not the closure of the open
ball B(a; r).

3.3: In a rrmrric space M, if subsets satisfy A _  :_ ,], where  is the cloure of
A is said to be dense in $. For example, the st Q of rational numbers is dense in R. If
A is dense in S and if S is dense in T, prove that ,4 i$ dense in T.
3.33 R©fer to Exercise 3.32. A mcAric space M is said to be aelaarable if there i$ a coantab/e
subset A which is dense in M. For example, R is separable because the set Q of rational
numbers is a ¢x)untabi¢ dense subset. Prove that every Euclidean space R  is separable.
3.34 Refer to Exercise 3.33. Prove that the Lindel6f covexing theorem (Theorem 3.28)
is valid in any sparable metric space.
3.5 Refer to Exercise 3.32. ffA is densein S and if B is open in S, prvethat B _ A & B+
Hit. F.ztercise 3.13;
3.36 Refer to Exercise 3.3:2. If each of A and B is dense in 5,and ifBis open in S, prove
that A  B is dense in S.
3..T7 Givm two metric spaces (S, dj) and (Sz, dz), a metric p for the Cartesian product
5"1 x z can be constructed from dl and d in many ways. For exampl©, ifx = (xa, xa)
and .v = (Yz, Y) are in S, x S±. IL-t p(. y) - d,(x,. Yt) + dz(x, Yz)- Prove that , is
a metric for Sz × Sz and constru further examples.
Prow each of t foBowing s ing  i r s (M, d) d
sus S, T of M.
3 AS T M. Shin(M,if,donlyif, Sism[n
  su (T, 
3 If S is d d T  , t S  T is m.
3     i i of    M  .
3.41  ion of a i num     M  m,
t[o.  S is a cl d 0n  ofQ   not 
If ,4 and B denote arbitrary subsets of a metric  M, prove that:
3A3imA = M- M- A.
3.44inI(M-A)=M-.
3.45 int (int A) = int A,
3.46 a) int (N'=  A,)= (== (int A), whcmeach a, _ M.
b) int (['a,- A) _ (, (int A), ffFis an infinite ¢ou:tgct of subsets of M.
c) Give an example where equality does not hold in (b).
3.47 a) U.,t- oat A) _ int (Ur A).
b) Give an example of a finil¢ collection Fin  equality does not hold in (a).
3.48 a) int (¢A) = 0 if A is open or if A i dos in M.
b) Give an example in which int (A) = M.
3.49 ff int A = int B --- B and if A is cloeed in M, then int (A  B) - 0.
3.0 Give an example in which int A = int B =  but im (A  B) = M.
3..b'l A =   M - ,4 and A  M - ,4).
3_2 ]fA= 0, thenAuB) = AuB.

CHAPTER 4
LIMITS AND
CONTINUITY
&l INTRODUCTION
The readcx is already familiar with the limit concept as introduced in elementary
cakmlns wh in fact, sevofl kinds of limits arc usually prcs¢ntcd. For ¢xampI,
the limit ofa aequence of real numbers {x,}. denoted symbolically by writing
lim x
means that for every number  > 0 there is an integex N such that
Ix. - A I <  whemcver n > N.
This limit process conveys tim intuitive idea that x, can be made arbitrarily clos
to A provided that n is sufflcienfly larg. Thcm is also the limit of a fum:tion,
indicated by notation such as
lira f(x) = A,
which means that for every, > 0 th¢ is another number & > 0 such that
If(x) - fl <  whenever 0 < Ix - Pl < &
This conveys tim idea that f(x) can be made arbitrarily close to A by taking x
sufficiently close to p.
Applications of calculus to gometrical and physical problems in 3-space
and to functions of several variables make it necessary to extend thc concepts
to R'. It is just as easy to go one step further and introduce fimits in tbe more
general setting of mric space. This achieves a simplification in the th¢ory by
stripping it of unnecessary rstrictions and at the same time covers nearly all the
imlmxtant aspcOa needed in analysis.
First w0 discuss limits of sequences of points in a metric space, then w discuss
fimits of functions and the concept of continuity.
CONVERGENT SEQUENCES IN A ME'IC SPACE
 4J. A aequence {x,} of points in a metric paee (S, d) is aid to converge
f there is a point p in , with the foilowing property :
For every  > 0 there is an intejer N such tlmt
d(x,, p) < • whenever n > N.
We also ay that {x,} converges to p and we write xo - p as n - o, or simplj
x, - p. If there ia no such p in S, the aequenee {x,} is said to diverge.
uo3. The definition of convergnc, impfies that
x, , p if and only if d(x,,, p) - O.
The convergence of tim sequence {d(x, p)} to 0 takes plac, in the Euclidean metric
spa¢ R t.
1. In EuIidcan spao¢ R1, a sequence {x,} is called/ncrea.t/n ifx _< x,+ t for all n. If
m  tClUtm¢ is bounded above (that is, ff x, < M for ome M > 0 and
all n)0 titan (x=} onvergm to tl suprexnum of its rans, sup {xt, x2 .... }. Similarly,
{X.}  a]I deere,t if x,+ 1  x, for all /. Evt'y  ::[uix: whic]l is
botmdcd llow convta-, to tl infanum of its .rm. For ¢xample {I/n} converg
toO.
2. If{a,}and {b,}arartmlstxlucncconvtxging to0, then {at + ha} also converl to 0.
If 0 _< e,  at for all n md if {a,} con to 0, thn {e,} also conq to 0.
These  pro of soqu in R t can ! u! to simplify some of the
lmms ¢xmceming limits in a
3. In the complex plane C, It z, = 1 + n-  + (2  l]n),'. Thtm {=} convexg to
1 + 2" bcautm
1 1
d(,, 1 + ) =
so d(zt¢ 1 + 20 -¢ O.
'heore d.2.  sequence {x=} in a metric space (S, d) can eonvere o at mot one
point in .
Proof Assume that x,  p and xo --, q. We will prove that p = q. By the
triangle inequality we have
Since d(p, x.) --, 0 and d(x.. q) , 0 this implies at d(p, q) = O, so p = q.
If a quence {x.} converge, the unique point to which it converges is ealled
the limit of the sequence and is denoted by lira x, or by lira,.® x.
Exam#e, In Euclidean space R t we have lir_ / = 0. The same seqtte in the
metric subspae T = (0, I ] does not convene because the only candidate for the limit is
0 and 0  T, This eymple shows that the convergence or divergs:e of a sluence detencl
on the underlying space as well as on the metric.
T/revere 4.3. In a metric space (S, d), assume
Be the range of {x,}. Then."
a) T is bounded.
b) p is an adherent point of T.

Proof. a) Let V be the ieteser nding to e = 1 in the defiultion ofcon-
vergence. Then everyx, withn 2 N/iesinthehallp; l), so every point in T
lies in the bali B(p; r), where
r ffi 1 + max {d(p, x),..., d(p, xs-:)}.
Therefore T is bounded.
b) Since every hall B(p; ) contains a po/at of T, p is an adhert point of T.
 If T is intimate, every ball B(p; n) contains infinitely many points of T, so
p is an accumulation point of T.
Irltem,em 4.4. Gitcn a metric ace (S, d) al a mbt T _ 8. If a point p in S i
an adlrent point of T, then tilers is a equence {x} of points in T wMh ¢onoerge$
top.
Proof. For every integer n : I there is a point x. in T with d(p, xD < l/n.
Hence d(p, x -, 0, so x.  p.
Proof. Assume x, - p and considex any suhsequence {x¢.}. For every e > 0
there is an N such that n > N implies d(x, p) < e. Since {xt.} is a subsequence,
there is an /neger M such that k(n)  N for n > M. Hence n > M implies
d(xt,3, p) < , which proves that xt¢l - p. The converse statement holds trivially
since {x.} is itself a subseqeence.
If a sequence {x.} converses to a limit p, its terms must ultimately become dose to
p and hence dose to each other. This property is stated more formally in the next
theorem.
 4.  t/mr {x,} ¢ontcrge8 in a metric apace (S, d). Then for every
e > 0 there/ an n 31uh that
Proof. Let p = lira x a. Given s > 0, let N be such that d(x, p) < .el2 wheaever
n  31. Then d(x.,,e) < /2 if m > N. If both n > N and m > N the, triangle
4.7  of a Camel Sequence. A sequence {x,} in a metric space ($, d) is
called a Cauchy sequence if it satisfies the following condition (called the Cauchy
condition):
For every e > 0 there is an integer N such that
d(x, x) <  wheneer n > Nand m >_ N.
Theorem 46 states that every convergent sequence is a Cauchy sequence. The
converse is not true in a general metric space. For example, the sequence {I/n} is
a Cauehy sequence in the Euclidean subspac¢ T ---- (0, I'[ of Rt but this sequence
does not converge in T. However, the cenvers¢ of Theorem 4.6 is true in every
Euclidean space R t,
Tlteoeem 4.& In Euclidean space R  every Cauchy equence i converTent.
Proof. Let {x} be a Cauchy sequence in R t and let T = {xi, x2,... } be the range
of the sequence. If Tis finite, . then all except a finite number of the terms {x} are
equal and hence {x,} converges o this common value.
Now suppose T is infinite. We use the Bolzano-Weierstrass theorem to show
that T has an accumulation point p, and then we show that {x,} converges to p.
First we need o know that Tis bounded, This follows from the Cauchy condition.
In fact, when  = 1 there is an N such that n > N implies IIx. - x  l[ < L This
means that all points x, with n >_ N lie inside a ball of radius 1 about xs as center,
so T lies inside a ball of radius 1 ÷ M about 0, where M is the largest of the
numbers x[], ..., IIxll- Therefore, since T is a boaaded infinite set it has an
accumulation point p in R  (by the Bolzano-Weierslras theorem). We show next
that {x,} converges to p.
Given  > 0 there is an N such that Itx, - x,l! < /2 whenever n > N and
The bail B(p; /2) contains a point x= with m > N. Hence if n > N we
so lira x,, - p,
IIx. - Pll -< Itx, - x, Jl + IIx,, - Plt<  +  = ,
This completes the proof.
1. Theorem 4.8 is often used for proving the convergence of a sequence when the limit
is not known in advance. For ¢xarnpl¢ consider the sequence in R  defined by
If m > n _> N, w¢ find (by taking siv¢ crrns in pairs) that
n 1
Ix.-- xl =  1 n+ 2
+...± <_<

so Ix, - xx= i <  as oon as N > I], Therefore (ab) s a Cauchy nce and
it converts o some limit. ]t n  shown ( Exerci 8A8) tt this !irnlt is log
a fact wh.ch is not immiateiy obvious.
 Given a reM ence {a) such that a.+ - a+,  {a+ - a for all n
We n prove that {a} co.verbs without kowing its iim* t b  ]a+   a.
Tn0  b+x . b.i2so, byind]on.,b.+l  b[2 . Hb.  O. Also, fire >
h¢tce
]a m- an!  b _< b 1 +- +,-.+ < 2b.
This imp|ins that {..an). is a Cauchy sequence, so {a,) econverges=
4M COMPLETE METRIC SPACES
Definition 49, A metric space (S, d) is catled complete if every Cauchy sequence
in S converges in S. A .bset T of S is called complete if the metric subspace (T,. d)
i complete.
Exam#e 1. Every Euclidean space R * is comple,e (Theorem 4.8), In particular, R  is
com.plee, but the subs.pace T = (0. 1 ] is not complete°
Exam#e 2. e space R  with the metric d(x, y) = maxx ¢  Ix,  YI is comNete.
The next theorem relates completeness with compactn6ss
Theorem 4.10. In any metric space (S, d) every compact subset T is complete,
Proof Le {x,} be a Cauchy sequence in T and let A = {x, xz,.... } denote the
range of {x}. If A is finite, then {.x} converges to one of the elements of A, hence
(x,} converges in T.
If A is irtfinie, Theorem 338 tells as that A has an accumulation point p in
T since T is compact. We show next that x.., p. Given  > 0, choose N so hat
n 2 N and m > N implies .d(x, x,.) < 2. The ball B(p; e/2) contains a point
x,, with m > N Therefore if n  N he triangle inequality gives us
g(x., p) <_ d(x,, x + d(x,,,, p) <  + ,. = ,
so x,  p. Therefore every Cauchy sequence in Thas a limit in T, .so Tis complete.
4,5 LI?vlIT OF A FUNCWION
in this section we consider two metric spaces (S, d) and (T, dr), where ds and dr
denote the respective metrics, Let A be a subset of S and let f : A - T be a
function from A to T
TIL 4,12 :Hmi of a F 75
Definition 4.11..Ifp is an accumutaUon poitt of A and if b _: T, the notation
is" defined to mean the followirta. :
iimf(x) = b, (1)
F'or every  > 0 there is a .8 > 0 such that
dr(f (x), b) <  whenever x  A, x # p, and ds(x,p) < 6.
The symbol in (I) is read "the limit off(x), as x tends o p, is b." or "f(x)
approaches b as x approaches p.Y We sometimes indicate, this by writingf(x) - b
The. definition conveys the intuitive idea that f(x) can be made arbitrarily
close to b by taking x suffidently close to p, (See Fig. 4.1..) We require that p be
an aumulation point of A to make certai that there will be points x in A
su.cienty close to p with x # p. However, p eecl not be in the domain off,
and b need no.t be in the range off.
Figure
?qorv.. The definition can also be formulated in terms of balls. Thus, (1) holds if
and only if, for every bali Br(b), there is a bali Bs(p) such that .Bs(p)  A is not
empty and such that
f(x) e Br{b) whenever x s B(p) ¢',, A, x ,. p..
When formulated this way, the definition is meaningful when p or b (or both) are
in the extended real number system R* or in the exte:ded compbx number system
C*. However, in what follows, i is to be understood that p and b are finite unless
it is explicitly stated that they can be infinite,
The next theorem relates limits of functions to limits of convergent sequences°
Theorem 4.12. As,vume p is an accumulation point ef A and assume b e T;. Then
lim f(x) = b, (2)

lira .fix.) -- b, (3)
for ever), ece {x) of im in J - p} hich conge to p.
oo If (2) holds, tn for  ¢  0  is a  > 0 sh t
Now e any q x,} in A - {p} w n to p. For   in (4),
• e is an intent N sh t n  N impli dx p) < . Thfo (4)
d(x, b) < e for n  N, d hen {f(x)} nve to b. ¢fo (2)
impli (3).
To pm  conve we ume tt (3) holds d t (2) h fal d ve
at a naon. If (2) is f,  for me • > 0 and eve  > 0  is a
int x in A (wh¢ x y dnd on ) sh t
o < dAx. p) <  but dx). b)  . ()
Taking   l/n, n = 1, 2,..., this means there is a conding uen of
in {x,} in A - (p} such that
0 < d(x,, r) < /n but dr(x,), b)  e.
rly, s  {x,} nver to p but the qu¢ {f(x} d not con-
v to b, commdiig (3).
N s 4.12 d 4.2 thcr show t a fustian not v¢ two
difft ts as   p.
4.6 LIMrIS OF COMPLEX-VALUED FUNCTIONS
Let (S, d) be a metric space, let A be a subset of S, and consider two complex-
vatued ftmctionsfand g defined on `4,
f : A - C, g : A - C.
The sum f + g is defined to be the function whose value at each point x of A is
the complex number f(x) + g(x). The difference f - 9, the product f- g, and the
quotientflg arc similarly defined. It is understocl that the quotient is defined only
at those points x for which g(x) # 0.
The usual rules for calculaling with limits are given in the next theorem.
Tlaeoeem 4.13. Le f and g be complex-valued function. defined on a subset `4 of a
metric space (S, d). Let p be an accumulation point of A, and assume that
lira
lii
b.
x = a, gx ffi
4,14   Veetof-Vth Fumetiom 77
Then we also have:
a) lim.., f(x) _+. g(x)'[ - a +_ b,
b) lim.a,f(x)g(x ) = ab,
) lim+,f(x)lg(.x)  afb fib  O.
Proof. We prove (b), leaving the other parts as exercises. Given  with 0 < . < 1,
let ' be a second number satisfying 0 < ' < 1, which will be made to depend on
e in a way to be described later. There is a  > 0 such that ifx  A and d(x, p) < 6,
then
If(x) - al < .' and Io(x) - bl < '.
Then
If(x)l = la + (f(x) - a)l < lal ÷ e' < lal ÷ !.
Writingf(x)a(x) - ab = f(x)a(x) - bf(x) + ¥x) - , we have
< (11 ÷ )' ÷ Ibis' -- e'(lal + ]hi ÷ O-
If we choose e" = el([al + Ibl + 1), we see that .f(x)g(x) - ab[ <  whenever
x  A and d(x p) < 6, and this proves (b).
4.7 LIMITS OF.VF_XTrOR-VALUED FUNCTIONS
Again, let (S, d) be a metric space and let .4 be a subset of S. Consider two vector
vaiued functions f and g.deflned on .4, each with values in R ,
f:AR t, g:AR .
Quotients of vector-valued functions are not defined (if k > 2), but we can define
the sum f + g, the product M (if). is real) and the inner product f" g by the respec-
tive formulas
(f + g)(x) = f(x) + g(x), (f)(x) = ).f(x), (f-g)(x) ffi f(x)-g(x)
for each x in A. We then have the following rules for calculating with limits of
vector-valued functions.
Theoeem 4.14. Let p be an accumulation point of A and assume that
lim f(x) ffi a, Iim g(x) = b.
Then we also have:

78
LmRs  CotiauRy
Proo/ We prove only parts (c) and (d). To prove (c} we write
The rarg.Ie ieqality and he Ca'chy-Schwarz nequalhy giwe us
Each tem on the rit ends to 0 a x  ?,  f(x).g(x) -., a-b. This proves
(c)... To prove {d) ote hat :(x) - a   'f(x)" 
o-r. Le f .... ,./ be n real-vMued funct.ios defined on A, and let f" A  R. 
 the vector-valed func6oe defined by ,he equation
f(x) --= (f(x),j(x) ..... £(x)) i.f.x s A,
Then ./ ..... £ are lted he components of f, and we also write f
{o denote tNs relationship.
If a = (a ..... , a)., then lr each r = 1, 2, .., a we have
These inequalities show that im. fix) = aiL and only iC lim.._.j;(x) =
for eac r.
4.8 CONTINUOUS FUNCTIONS
The definition of continuity presented in eIementary calculus can be extended o
functions from one metric space to another
Defirdtion4.LL Lee {S:, ds. ) ad T, d.) be metric spaces and et f: S .- T be a
function .#on ,.5" ;o T. The Juncio. / is seM ;o be comn;,ous a a point p in S 
./br e,ery : > 0 there ) a 5 > 0 sue}} dmt
d6,.fx), f(p)) < *: wl, eeever ds(x, p} < 6.
(/'im continuous a e every poim o.f a subset A of S, we say f is continuous on A.
This definito reflects the ntnhive idea ha.t points dose to p are mapped by
.finto points close tof(p). It can also be stated m terms of balls: A .nction fis
continuous at p if and only iL fLr every a > 0, there s a 6 > 0 such that
Here Bs-(p; fi) is a bali in S; i8 image under f mast e contained in the bail
Be
r,J(P); e) : T. (See Fig. 4.2.)
Ifp is an accumula6on poini of S, h.e definition of continuity implies ihat
lira f(x) = f(
Th. 4.i7
Ifp is an isolated point of S (a point of S which is not an accumulation point of
S), then every fdefined at p will be continuous at p becau fr sufficiently small.
there is only one x satisfying ds(x, p) < 6, namely x = p, and dr(f(p), f(pJ) = O.
Theo'em 4,I6, Let f : S --, The a function from one metric space (S, ds) to another
(.T, dr.), and assume p  S. Then f is continuous at p if, and only"  for every sequence
{xo} in S convergent to p, the sequence {f(x.)} in T conver.qes to f(p); i: symbols,
lira. f(x.)= f(2i x.).
The proof of this theorem is similar to that of Theorem 4.12 and is left .as an
exercise for the reader. (The result can also be deduced from 4.12 but. there is a
minor co.mpiication in the argument due to t.be fact that some terms of the sequence
{x,} could be equal to p.}
The theorem is often described by saying that for continuous functions the
limit symbol can be interchanged with the functiota symbol, Some care is needed
in interchanging these symbols because, sometimes {f(x,)} Converges when {x}
diverges.
Examl!e If x, .-- x and y . y in a metric spa.c, (& d), then d(x,,
(Exercise 4.7). The reader can verify daat d is continuous on tb, e met.rc space (S × S,. p),
where p is the metric of Exercise 3.37 with S. = .Sa = S.
OTE. Conthuity of a function fat a point p is called a tocaiproperty .off because
it depends on the behavmr of foamy in the immediate vicinity of p. A property of
fwhich concerns the whole domain off is calted a 910bal property. Thus, continuity
off on its domain is a global:: property.
4.9 CONTINUFIY OF" COMSITE FUNCTIONS
Theorem 4,17. Let (& ds), (Z dr.) and (U, 4) be metric spaces. Let f .: S --, T
and g :f(S) - U be Jknctions, and et h be the composite function dfined on S by
the equation
h(x) = g(f(x}) for x in S.
tf f ts continuous at p and g./'g is cominuous air(p), then h is cont#mous a.t f7.

Pro@ Le ,5 = f(p). Given r. > O, there is a 6 > 0 sch tha
For this 6 thee i:s a 6' such h.at
d{f(x)f(p)) < 6 whenever d.x, p) < '
Combining hese tw statements and taking y = f(:), we find that
du(h(x) h(p)) <  whenever ds(x, p) < 6',
so h is cntinuous at p.
4.10 CO ITINUOUS COMPLEX-%LLUED AND VECTOR-VALUED F3,SNCTIONS
Theorem 4J& Let f and # be complex-valued functions continuous at a pint p in
a metric space (S, d)o Then f÷ .q, f- 9, andf.y are each continuous at p.. The
quotient JTg is alo continuous at p if g(p)  O.
Proof The restlt is trivial if p is an isolated point of S. If p is an aceumulation
point of S, we obtain the result from Theorem 4A3.
There. is, of coure., a corresponding theorem for vector-valued functions, which
is proved in the same way, umg Theorem 4.14.
Theorem 4d9. Let f and g be fhnctions continuous at a point p in a metric space
(;S, d), and assume that f and g have values in RL Then each of the fetlowirgl is
continuous at p: the sum f + g, the product :f for every real 2, the nner product
f" g, and the norm I1
Theorem .20. Let f  .... , f be n reaMalued functions defined on a subset A of a
metric pace (S, ds), and let f = (./],  .., f,)o Then f is eomfnuous at a point p
of A (f and on6' if each of the functions f , . . , f, .is cominuous at p
Proof tfp is an isolated point of A there is nothing to prove. Ifp is an accumula-
tion point, we note that fix) -, f(p) as x , p if and only ify.(x) - q(p.) for each
k = 1,2,.o.,n.
4A1 EXAMPLES OF CONTINUOUS FUNCTIONS
Let S = C; the complex .plare It is a trivial exercise to show that the following
complex-valued functions are continuous on C:
a) constant functions, defined by f(z) = c for every z in C;
b) the identity function defined byf(z) = z for every z in C.
Repeaed application of Theorem 4.18 estaNishes the continuity of every poly-
nomial:
f(z) = a e + a.z + az  + "" + Gz",
the a being complex numbers°
If S is a subt oft C on which the polynomial fdoes not vanish, then ldfis
continuous on S Therefore a ratioral function, g/d; where y and fare polynomials,
is cominueus at these its of C at which b.e denominator d:s not vanish.
The familiar r at-valued functions of elementary calculus, sc.h as he ex-
nential, trig.ee.etri.c, ad 1e,fithmic functions, are al continuers wherever
they aN defi.ned. Tb:e continaity of thee elememary functions justifies the common
.practice of evadng certain Iimit by substituting the limiting vae of the
'qndendent variable"; for examNe,
lira e:" =. e ° = .
The continui V of the complex exponential and trigonometric functions is a
coqen of the continity of the corresponding realwalued fncten.s and
eorem
4.12 CONTINUITY AND N¥.EItSE IIAGE$ OF OPEN ON CL(TSED SE
The conpt of inver image can  used to gve. two im,ant gobal descriptio
of contieoas functions.
4.21 Deflation of i,rse ige. t f : S  T be a jnetion from e set S o a
.et Z If Y s a subset of T, the inverse image of Y under f noted
defined to be the [arges ubset of S which f maps into } that
f-(Y)= .{x:xS and f(x) Y}.
NOTZ ffhas an nver fnctJenf-  the inverse image of Y under fin the same
as the mage of Y nder f -', and in this ca there is no am.biguky in the ntatin
f-(Y). Note aso thatf-(A)  f-(B) if A  B  Z
Torem 4d2. Let f : S - T be a function from S to Z [f Y
then we have:
a) X = f-(Y) im:4ies f(X}  Y.
e proof of Theorem 4,22 5 a dr ranslakn of the definition or" the sym-
bos f-(Y) and J'(X), and  left to the reader., k shud  o.b that;
gezeraL we camot conclude that Y = f(.Y) hp[Jes X = f-(Y), (See he example
Fig. 4.3..)
Figure 4,3

Note that the statements in Theorem ,22 can also be expressed as follows:
f[f-l(y)] _: r, x f-'[f{x)].
Note also thatf-(A  B) = f-t(A) f-l(B) for all subsets A and B of T.
re 4.23. Let f : S  T be a function from one metn'e pace (S, ds) to another
(T, dr). Then f is continuou on S if, and only if, for every open set Y in T, the
inverse image f-(1 r} is open in S.
Proof. Let f be continuous on S, let Y be open in T, and let p be any point of
f-(Y). We will prove thatp is an interior point off-l(Y). Let y ffi f{p). Since
¥is open we have Br(y; -) - Yfor some e > 0. Sincef is continuous atp, there
is a 6 > 0 such thatf(Bs(p; 6)) _= Br(y; ). Hence,
Bs(p; )  f-[f(Bs(p: ))] _ f- [Br(y; -)] 
so p is an interior point off-(Y).
Convcxsely, assume that f- (Y) is open in S for every open subset Y in T.
Choose p in S and let y ffi f(p). We will prove thatfis continuous atp. For every
8 > 0, the ball Br(y; ) is open in T, so f-(Br(y; e)) is open in S. Now,
P efl(Br(y; 0) so there is a  > 0 such that Bs(p; ) _ ff'(Br(y; e)). There
fore, f(B(p; b)) _ Br(y; 0 sofis continuous at p.
Theorem 4.24. Let f: S  T be a function from one metric space (S, ds) to another
(T, dr). Then f i continuous on S if, and only if, for every closed set Y in T, the
inverse imoe f- t(y) is closed in S.
Proof If k" is dosed in T, then T - Y is open in T and
f-  (r
" Now apply Theorem 4.23.
Examples. The ima of an open set under a continuous mapping is not ncssarily open.
A simple counterexample is a conslant function which maps all of S onto a single point
in R. Similarly. he image of a dosed set under a continuous mapping need not be closed.
For example, the reaI*valued function f(x) =  x maps R I onto the open intetnnd
(- n/Z,
4.13 FUNCTIONS CONT/NUOUS ON COMPACT SETS
The next theorem shows that the continuous image of a compact set is compact.
This is another global property of continuous functions.
Tlorc.m 4.25, Let f : S - The a function from one metric space (S, ds) to another
(T, dr). lf f's continuous on a compact .rubset X of S, then the image f(X) is a
compact subset of T; in particular, f(X) is closed and bounded in T.
Proof. Let F be an open covering off(X), so thatf(X) _ v A. We will show
that a finite number of the sets .4 coverf(X). Sineefis continuous on the metric
subspace (X, ds) we can apply Theorem 4.23 to conclude that each ¢,et f- (1) is
open in (X, ds). The sets f- (A) form an open covering of X and, since X is
compact, a finite number of them cover X, ay Z - f-(.40  "'" f-t(A,) •
Hence
sol(X) is compact, As a corollary of Theorem 3.38, we ee thatf(X) is closed and
bounded.
Dean 4,2. .4 function " S  R  is called bounded on S if there  a positive
number M such that f(x)ll < M for all x in $.
Since 1" is bounded on S if and only if f(S) is a bondl subset of R
the following corollary of Theorem 4.25.
'leoem .27. Let [; S - R  be a function feom a metric space S to Euclidean
space R . If  is eontinuer on a compact subset X of S, then  is bounded on X.
Thi theorem has important implieation for real-valued functions. If f is
real-valued and bounded on X, then f(X) is a bounded subset of II, so it has a
supremum, supf(), and an inlimum, inff(X). Moreor,
inI f(X)  f(x) < sup f(X) for every x in X.
The next theorem show that  continuous f actually takes on the valuta sup fiX)
and inff(X) if X is compact.
Theoeem 4;20. Let f: S  R be a real-tmlued function from a met¢ie sttce S to
Euclidean space II. Arne that fig continuous on a compact ubset X of S. Then
thee exist points p and q in X such that
= inrf(x) and f(q) = s  pf(X).
rox. Since f(p) < f(x)  f(q) for all x in X, the numbers f(p) and f(q) are
called, respectively, the absolu or global minimum and maximum values of
]on X.
Proof. Theorem 4.25 shows that f(X) is a closed and bounded sulmet of R. Let
m = inff(). Then m is adherent to f(X) and, since f(X) is closed, m el(X).
Therefore m = f(p) for somep in X. Similarly, f (q) = spf(X) for some q in X.
111eoeem 4.2, Let f: S - The a function from one metric sttce (S, dO to another
(T, dr). Assume that f is one-to-one on S, so that the inverse function f-a exists.
If S is compact and if f is continuous on 9,. then f - is continuou oaf(S),
Proof By Theorem 4.24 (applied to f- t) we need only show that for every dosed
set X in S the image f(X) is closed in T. (Note that f(X) is the inverse image of

Xunderf-l.) $inc Xis closed and S is compact, Xis compact (by Theorem
sol(X) is compact (by Theorem 4.25) and henccqX) is closed (by Theorem
This completes rbe proof.
Examl Tiffs example shows that oomgectness of  is an essenlial part of Theorem
4.29. Let S = [0, ])withtheusual metricofR z and consider thecomplex-valued function
£dee by
f(x) = • ' for 0 _< x < 1.
This is a one-to-one continuous mapping of the half-open interval [0, 1) onto the unit
circle [21 = 1 in the complex plane. Hower, f- t is not continuous at the point f(0).
For ¢xample if x = 1 - l/n, the sequence {ffx.)} converges tot(0) but {X} does not
conwre in S.
.14 TOPOLOGICAL MAPPINGS (HOMEOMORPHISMS)
Defadn 430. Let f: S - T be a function from one metric space (S, ds) to
another (T, d?). Assume also that f is one-to-one on S, $o that the inverse function
f- t exists. If f is continuous on S and if f- 1 is continuous on f(s), then f is called
a topological mapping or a homeomorphism, and the metric spaces (S, ds) and
(f(S), dr) are said to be homeomorphic.
Iff is a homeomorphism, then so is f- 1. Theorem 4.23 shows that a homeo-
morphism maps open subsets of S onto open subsets off(S). It also maps closed
subsets of S onto dosed subsets off(S).
A property of a set which remains invariant under every topological mapping
is called a topological propert?. Thus the properties of being open, closed, or
compact are topological properties.
An important example ofa homeomorphism is an isometry. This is a function
f : S -, T which is one-to-one on S and which preserves the metric; that is,
atf(x), f( y)) -- gs(x, y)
for nil points x and ? in S. If there is an isometry from (S, d s) to (f(S), dr) the
two metr/c spaces are caled ixometric.
Topological mappings are particularly important in the theory of space curves.
For example, a simple arc is the topolog/cal image of an interval, and a simple
closed curve is the topoIogical image of a circle.
THEOREM
This section is dcvoted to a famous theorem of Bolzano which concerns a global
property of real-valued functions continuous on compact intervals [a, b] in R.
If the graph off lies above the x-axis at a and below the x-axis at b, Bolzano's
theorem asserts that the graph must cross the axis somewhere in between.. Our
proof will be based on a kwal property of continuous functions known as the
sign-preserving property.
rem 4.31.. Let f be defined on an interval S in R. Assume that f is continuous
at a point c in S and that f(c)  O. Then there is a I-ball B(c;  such that f(x)
has the same sign as f(e) in B(c; b')  S.
Proof Assume f(c) > 0. For every t > 0 there is a 6 > 0 such that
f(c) -  < f(x) < f(e) +  whenever x  B(c; ) c S.
Take the # corresponding to  = f(c)12 (this e is positive). Then we have
f(c) < f(x) < f(c) whenever x v B(c: 6)  S,
sot(x) has the same sign, asf(c) in B(c; 6)  S. The proof is similar iff(c) < 0,
except that w take  = --½f(c).
Tlarem 4.32 (gr.aa). It f be real.valued and continuous on a compact interd
[a, b] in R, and suppose that f(a) and f(b) have opposite .igns: that i, as.aline
f(a)f(b) < O. Then there is at least one point c in the open interal (a, b) such that
f(c) -- O.
Proof. For definiteness, assume f(a) > 0 and f(b) < O. Let
A = {x:x[a,b] and f(x)  0}.
Then A is nonempty since a  A, and A is bounded above by b. Let c = sup A.
Then a < e < b. We will prove that f(c) = O.
Iff(c) # 0, there is a 1-ball B(c; 6) in which f has the same sign as f(c). If
f(c) > 0, there .are points x > c at whiehf(x) > 0, contradicting the definition
of c. Ill(c) < 0, then c -- 6/2 is an upper bound for A, again contradicting the
definition of c. Therefore we must hayer(c) = 0.
From Bolzano's theorem we can easily deduce the intermediate tw2ue theorem
for continuous functions.
T&,orem 4213. Asaume f is real-valued and continuous on a compact interval S in
R. Suppose there are two points  <  in S such that f(,)  f(fl). Then f takes
ener alue beteen f() and f() in the interrl (, ).
Proof. Let k be a number between f() and f(.fl) and apply Bolzano's theorem to
the function g defined on [, ] by the equation g(x) = f(x) - k.
The intermediate value theorem, together with Theorem 4.28, implies that the
continuous image of a compact interval S under a real-valued function is another
compact interval, namely,
[inff(S), sup/(S)].
(If f is constant on S, this will be a degenerate interval,) The next section extends
this property to the more general setting of metric spaces.

4.16 CONNF.L'TEDNESS
This section describes the concept of conneetedness and its relation to continuity.
/D¢ 4.34. A metric space $ is called dgrconnected if $ - A  i where A
and B ate disjoint nonempty open sets in $. We call S connected if ft g¢ not d
connected.
qorF.. A subset X of a metric SlmCe ,Y is called connected if, when resaxded as a
metric suhspace of ', it i a connected metric space.
1. Themetrkspace$ = R -- {O} with themmal Etraidean metricis  since
it is the union of two disjoint nonernpty open se tim ImSidvc real nttmbers and the
7,. Every open iterval in R ts connected. Thiswas Imbed in Section 3.4asacome
qu¢¢ of 3.11.
& The mt O of mtional numbers, regarded as a metricsubspamofEudidean spaoeR 1,
is d/sconnecte In fact, O = ,q L7 R, where A consists of all rational numbers
< ",/ ant Baall rational numtm, > -,/]. Similarly, everyba in O is disconnected.
4. Every metric space $ contains nonempty connected subsets. In fact, for each p in S
To relate connectedness with continuity we introduce the concept of a two-valued
function.
 4.35. A real-valuedfunctionf which is cont on a metric ce S is
mid to be two-valued on ; if f() _ {0, 1}.
In other words, a two-valued function is a continuous function whose only
possible values are 0 and 1. This can be regarded as a continuous function from S
to the metric spaoe T = {0,  }, where T has the discrete metric. We recall that
every subset of a discrete metric space T is both open and closed in T.
Tlg, m,em 4,36 A metric space S is connected if, and only if, eery two-valued
function on S is cormtant.
Proof Assam¢ S is connect and letfbe a two-valu function on S. We must
show that fis constant. Let A ---f-l({0})and B-- f-({1}) be tbe invcr
images of the subets {0} and {1}. Since {0] and {1} are open subs of tbe
discrete metric SlmCe {0, 1 }, both A and B arc open in S. Hence., S =  u B,
where 1 and B are disjoint open sets. But sinc $ is connected, either  is empty
and B  S, or else B is empty and A -- S. In either  f is constant on S.
Conversely, assume that S is disconnected, so that S = A u B, where A and
B are disjoint nonempty open subsets of S. We will exhibit a two-valued function
on S which is not constant. Let
f(x)={ ifxcB.ifx¢4"
Colpmm of a Mettle Sim
Since/1 and B are nonempty, f takes both values 0 and 1, so f is not constant.
Also, f is continuous on S because the inverse image of every open subset of {0, !}
is open in S.
Next we show that the continuous image of a connected set is connected.
Theorem 4.37. Let f: S " M be a funcaon flora a metric space S to anotler
metric space M. Let X be a connected subset of S. lf f is continuous on X, then
f(X) is a connected subset of M.
Proof Let g be a two-valued function on f(X). We will show that g is constant.
Consider the composite function h defined on X by the equation h(x) = g(f(x)).
Then h is continuous on X and can only take the values 0 and i, so h is two-valued
on X. Since X is connected, h is constant on X and this implies that g is constant
on f(X). Therefore f(X) is cormected.
Example Since an interval X in R  is connected, every continuous image f(X) is con-
netted. If f has real values, the image f(X) is another interval. Iffhas values in R', the
imagef(X) is  a curve in R . Thus, every curve in R = is connected.
As a corollary of Theorem 4.37 we have the following extension of Bolzano's
theorem.
Teorem 4.38 (Itermiale-ml tk¢orcm for renl  f). Let f be
real-valued and continuous on a connected subxet S of R . If f take on two ¢fi'fferent
values in S, say a and b, then for each real c between a and b there exists a point x
in S such that f(x) = c.
Proof The image f(S) is a connected subset of R t. Hence, f(S) is an interval
containing a and b (see Exercis 4.38). If some value c between a and b were not
in f(S), then f(S) would be disconn¢:L
4.17 COMPONENTS OF A METRIC SPACE
This section shows that every metric space S can be expressed in a unique way as
a union of connected "pieces'" called components. First we prove the following:
l'lteorem 4.39. Let F be a collection of connected subet8 of a metric space S such
that the intersection T =   is not empty. Then the union U = a, A is
connected.
Proof Since T ¢ 0, there is some t in T. Let f be a two-valued function on U.
We will show tlmtfis constant on U by showing that f(x) = f(t) for all x in U.
If x  U, then x  A for some A in F. Since A is connected, f is constant on A
and, since t 6 A,f(x) = f(t).
Every point x in a metric space S belongs to at least one connected subset of
S, namely {x}. By Theorem 4.39, th union of all the connoted subsets which
contain x is also connected. We call th/s union a component of S, and we denote it
by U(x). Thus, U(x) is the maximal connected subset of S which contains x.

lmem 4.40. Eoery point of a meteic space S be[on Io a uniquely determined
component orS. In other wovd, the eompanents of S form a collection of disjoint
.ets whose union s S.
Proof. Two distinct components cannot contain a point x; othcrwi (by Theorem
4.39) their union would b a larger conctcd set containing x.
4.18
This section describes a special
 by  t n all)    E  r.
 b in N there is a  cti
, Shafhathfromatob. IfO) lof
[0, 1]
n   lthcted. Ift) =tb+ (! - t f
0  t
ofa. In ,     con.
   in F 4,4 (a i  two t  )  i .
4.4
3. The set in Fig. 4.5 consists of those points on the cure described by y = sin (I/x),
0 < x _< |, along with the points on the horizontal segment - l _< x _< 0. This set
is connected but not arewis¢ connected (Eaercis¢ 4.46).
4.5
The next theorem relates arcwise connectedness with connectedness.
Tleorem 4.42. Every arcwise connected set S in R  is connected.
Proof. Let g be two-valued on S. We will prove that g is constant on S. Choose
a point # in S. If x e S, join a to x by an arc F lying in S. Since F is connected,
g is constant on F so g(x) = g(). But since x is an arbitrary point of S, this shows
that g is consrant on S, so S is connected.
We have already noted that there are connected sets which are not arcwise
connected. However, the concepts are equivalent for open sets.
]'keoretn 4.43o Every open connected set in R  is arcwise connected.
Proof. Let S be an open connected set in R" and assume x e S. We wi|[ show that
x can be joined to every point 3' in Shy an arc lying in S. Let A denote that subset
of S which can be so joined to x, and let B = S - A. Then S = A u B, where
A and B are disjoint. We will show that A and B are both open in
Assume that a E A and join # to x by an arc, say F, lying in S. Since a  S
and S is open, there is an n-hall B(*) __- S. Every y in B(t) can be joined to a by
a line segment (in S) and thence to x by F. Thus y  A if y  B(a). That is,
B(a)  A, and hence A is open.
To  that B is also open, assume that b  B. Then there is an n-ball B(b)  S,
since S is open. But if a point y in B(b) could be joined to x by an arc, say F',
lying in S, the point b itself could atso be so joined by first joining b to y (by a
line segment in B(b)) and then using V'. But since b  A, no point of B(b) can be
in A. That is, B(b) _= B, so B is open.
Therefore we have a decomposition S = A  B, where A and B are disjoint
open sets in R . Moreover, A is not empty since x E A. Since S is connected, it
fotlows that B must be empty, so S = A. Now A is clearly arcwie connected,
because any two of its points can be suitably joined by first joining each of them to
x. Therefore, S is arcwise connected and the proof is complete.
oa' A path f : ['0, 1]  S is said to be polygonal if the image of I'0,
is the union of a finite number of line segments. The same argument used to prove
Theorem 4.43 also shows that every open connected set in R" is polyonalfy con-
nected. That is, every pair of points in the set can be joined by a polygonal are
tying in the
71"lcorca .44. Eery open set S in R" can be expressed in one and only one way as a
countable disjoint union of open connected sets.
Proof. By Theorem 4,40, the components of S form a collection of disjoint sets
whoe union is S, Each component T of S is open, because if x  T then there is
an n-ball B(x) contained in S. Since B(x) is connected, B(x) _ T, so T is open,
By the Lindelfif theorem fTheorexn 3.28), the components of S form a countable
collection, and by Theorem 4.40 the decomposition into components is unique.
 4.45. A set in R"/s called a region fit is the union of an open connected
se with some, none, or all its boundary points. If none of the boundary points are

included, the region is called an open region If all the boundary points are included,
the region is called a closed region.
N(:XE. Some authors use the term domain instead ol" open region, especially in the
complex plane.
4.19 UNIFORM CONTINUITY
Supposefis defined on a metric space (S, ds) , with values.in another metric space
(T, dr) , and assume thatfis continuous on a subset '4 of S. Then, given any point
p in A and any s > 0, there is a  > 0 (depending on p and on t) such that, if
x  A, then
dr(f (x), f{v)) < t whenever ds(x, p) < &
In general we cannot expect that for a fixed t the same value of will serve equally
well for eery point p in ,4. This might happen, however. When it does, the
function is called uniformly continuous on ,4.
Defun'tian 4.46. Let f : S  The a funetion frorn one metric space ( S, ds) to another
(T, dr). Then f is said to be uniformly continuous on a subset A of S if the following
condition holds:
For every r > 0 there exists a 5 > 0 (dependin only on ) such that if x  A
and p  A then
dr(f(x),f(p)) <  whenever ds(x , p) < 5. (6)
To emphasize the difference between continuity on A and uniform continuity
on fl we consider the following examples of real-valued functions.
1. Let f(x) = 1Ix for x > 0 and take A -- (0, 1 ]. This function is continuous on A
but not uniformly continuou on .4. To prove this, lea • = 10, and suppose we could
find a ,, 0 < 3 < 1, to satisfy the condition of the definition. Taking x = , p -- d/I 1,
we obtain Ix - Pl < - and
- = - > 10.
Hence, for these two points v would always have If(x) - f(P)l > I0, contradicting
the definition of uniform continuity.
2. Let f(x) = x  if x e R 1 and lake A = (0, 1 ] as above. This function is uniformly
continuous on A. To prove this, observe that
If Ix - Pl < a, then If(x) - f)l < 2& Hence, if t is given, we need only take
 = t]2 to guarantee that If(x) - f(p)] < e for every pair x, p with Ix - PI < &
This shows that f is uniformly continuous on A.
Tit. 4.47
Unif ContilmltV aml Cmnln S
91
An instructive exercise is to show that the function in Example 2 is not uni-
formly continuous on R t.
4,7,0 UNIFORM CONTINUITY AND COMPACT SET
Uniform continuity on a set A implies continuity on ,4. (The reader should verify
this.) The convere is also true if "4 is compact.
Tleorem 4.47 (Heine). Let f : S  T be a function from one metric space (S, ds)
to another (T, d). Let A be a compact subset ors and assume that f is continuous
on A. Then f is uniformly continuous on A.
Proof. Let  > 0 bc given. Then each point a in A has associated with it a ball
B(a; r), with r depending on a, such that
dr(f(x), f(a)) <
2
whenever x  Bs(a; r) ta A.
Consider the collection of balls Bs(a; r[2) each with radius rt2. These cover
and, since A is compact, a finite number of them also cover ,4, say
In any ball of twice the radius, B(a; rt), we have
/(a,)) < -
2
whenever x e Bs(at; rt)  A.
Let  be the smallest of the numbers rt/2, ..., r=[2. We shall show that this
works in the definition of uniform continuity.
For this purpose, consider two points of A, say x and p with ds(x, p) <
By the above discussion there is some bail Bs(a; rd2) containing x, so
dr(f (x), f(a)) < .
By the triangle inequality we have
ds(p, a) < ds(p, x) + ds(x, a t ) <  + r < rA + r = r.
-
Hence, p  BAa; r,) m S, so we also have dr(f(p), f(at)) < e/2. Using the
triangle inequality once more we find
dr(f(x),f(p)) < dr(f(x),f(ai)) + dr(f(a),f(I))) < 5 + e
 2 2
This completes the proof.

41 FIXEB-POINT THEOREM FOR CONTRACTIONS
Let f : S  S be a function from a metric space (S, d) into itseIf. A point p in
S is called a fixed point offiff(p)  p. The function f is called a contraction of
S if there is a positive number 0 < I (called a contraction constant), such that
d(f(x), fO')) -< dfx, y) for all x, y in S. (7)
ClearIy a contraction of any metric space is uniformly continuous on 5".
Tkeorem 4.48 (Fixed-point tlteorem). A contraction f of a complete metric space S
ha. a unique fixed poin p.
Proof. If p and p' are two fixed points, (7) implies d(p, p')  ,td(p, p'), so
so d(p, p') - 0 and p -- p'. Hence f has at most one fixed point.
To prove it has one, take any point x in S and consider the sequence of iterates:
x, f00, f0"(x)) ....
That is, define a sequence {pj} inductively as follows:
Po  x, P,+ a -f(P), n = O, I, 2 ....
We will prove that {p,} converges to a fixed point off First we show that {p,} is
a Cauchy sequence. From (7) we have
p,) = a(f(p),y(p._,)) ;,,_,),
so, by induction, we find
a(p,+t, p, <_ " d(p,, Po) =
where c = d(pl, Po)- Using the triangle inequality we find, for m > n,
at--I
-a ' c
Since ¢t" -) 0 as n  o, this inequality shows that {p.} is a touchy sequence. But
S is complete so there is a point p in S such that p.  p. By continuity off,
f(p) = f(lim p = lira f(p,)= lira
P,
so p is a fixed point off. This completes the proof.
Many important eistence theorems in analysis are easy consequences 3f the
fixed point theorem. Examples are gfven in Exercises 7.36 and 7.37. Refererme
4.4 gives applications to numerical analysis.
422 BISCONTINUITIES OF REAL-VALUEU FUNCTIONS
The rest of this chapter is devoted to special properties of real-vMued functions
defined on subintervals of R.
Disamtitm#ies of Real-Vali Ftmctiotts
Let f be defined on an interval (a, b). Assume c e [a, b). If [(x) - a as
x - c through values greater than c, we say that A is the riyhthand limit of fat c
and we indicate this by writing
lira f(x) = A.
The righthand limit A is also denoted byf(c+ ). In the , 6 terminology this means
that for every : > 0 there is a 6 > 0 such that
If(x) -f(c+)l < e wenever c < x < ¢ + 6 < b.
Note that f need not be defined at the point c itself. Iff is defined at c and if
f(c + ) = f(c), we say thatfis continuous from the rieht at c.
Lefthand limits and continuity from the left at c off similarly defined if
c
Ifa < c < b, thenfis continuous at c if, and only if,
f(c) :- f(c+) =/(c-).
We say c is a discontinuity of/iffis not continuous at c. In this case one of
the following conditions is satisfied:
a) FAtherf(c+) orf(c-) does not exist.
b) Bothf(c+) and f(c-) exist but bare different values.
c) Both f(c+ ) and f(c-) exist andf(c+)  .f(c-) # f(c).
In case (c), the point c is called a removable diacontinuity, since the discontinuity
could be removed by redefining fat c to have the valuer(c+) = f(c--), in case
(a) and (b), we coil c an irremovable discontinuity because the discontinuity cannot
be removed by redefining f at c.
Deftaitlo 4.49. Let f be defined on a cioed interval [a. b]. lf f(c + ) andj'c-)
both exist at some interior point c, then:
a) f(c)  f(c-) ls called the lefthandjump off a* e,
b) f(c + ) -- f(c) ia called the righthtmd jump of fat c,
c) f(c+ ) - f(c-) is called the jump off at e.
If any one of these three number is different from O, then c is called a jump din-
continuity off.
For the endpoints a and b, only one-sided jumps arc considered, the righthand
jump at a,f(a+) --f(a), and the lefthand jump at b,f(b) -f(b-).
1. The function fckfined byf(x) = x/[x[ ifx  0,f(0) -- ,4, has a jump discontinuity
at 0, regardless of the value ofA. Herg/(0+) = I and f(0-) = - I. (See Fig.
2. The function [ defined by [(x) = i if" x # O, f(0) - O, has a removable jump dis-
continuity at 0. In this case f(0+) = f(0--) = 1.

Fig.re
3. The function ./defined by f(x) = ]iX if x ¢ 0, f(0) = A, has an irremovable dis-
cont[nuhy at 0. In this case neither f(0+) norf(O-)exists. (See Fig. 4.7.)
4. The functionfdefined byf(x) =. sin (1Ix)ifx =# 0,f(0) = A, has an irremovable dis-
continuity a! 0 since neilherf(0+) nor f(0-) exists. (S¢ Fig. 4.8.)
5. The function f defined by f(x) = x sin (t/x) if x  0,/(0) = I, Ires a removable
jump discontinuity at 0, sinc/(0+) = f(0-) = 0. (See Fig. 4.9.)
Figure 4.8
4.9
4.2,3 MONOTONIC FUNIONS
Definition 4.50. Let f be a real-valued function defined on a subset S of R. Then
f is said to be increasing (or nondecreasing) on S if far every pair of points x and y
inS,
x < y implies f(x) <<_ f(y).
lf x < y irnpliesf(x) < f(y), then fis said to be strictly increasing on S. (Decreasint
functions are similarly defined.) A function is ealled moaotonic on S if it is increasing
on S or decreasing on S.
lffis art increasing function, then -f is a decretsing function. Because of this
simple fact, in many situations involving monotonic functions it suffices to consider
only the case of increasing functions.
We shall prove that functions which are monotonic on compact intervals
always have finite right- and lefthand limits, Hence their discontinuities (if any)
must be jump discontinuities.
Tleorem 4,51. l/f is increasing on [a, b], then f(c + ) and f(c-) both exist for each
c in (a, b) and we have
f(c-)  f(c) < f(c+).
the endpoints we have
f(a) < f(a +) and f(b -)  f(b).
Proof. Let A = {f(x):a < x < c}. Since f is increasing, this set is bounded
above by f(c). Let z = sup 4. Then • < f(c) and we shall prove that f(c-)
exists and equals
To do this we must show that for every, > 0 there is a 6 > 0 such that
c - fi < x < c implies If(x)
But since ,:x - sup A, there is an element f(x0 of A such that  -  < f(xt)
Since f is increasing, for every x in (x,, c) we also have  -  < f(x) < t, and
hence If(x)  '1 < . Therefore the number  = c- xt has the required
property. (The proof that f(c+) exists and is >f(c) is similar, and only trivial
modifications are needed for the endpoints.)
There is, of course, a corresponding theorem for decreasing functions which
the reader can formulate for himself.
Tleorem 4,52. Let f be strictly increasing on a set S in R. Then f-  exists and is
strictly increasing on f(S).
Proof. Since f is strictly increasing it is one-to-one on S, so f- exists. To see
that f- t is strictly increasing, let y, < y2 be two points inf(S) and let x =f- (y),
x2 = f-[y2). We cannot have x,  x:, for then we would also have yt >_
The only alternative is
X I < X2
and this means that f-* is strictly increasing.
Theorem 4.52, together with Theorem 4.29, now gives us:
Tleorem 4.53. Let f be strictly increasing and continuous on a compact interval
[a, b]. Then f-  is continuous and strictly increasing on the internal If(a), f(b)].
NOTE. Theorem 4,53 tells us that a eontinuous, stricdy increasing function is a
topological mapping. Conversely, every topological mapping of an interval [a, b]
onto an interval [c, d] must be a strictly monotonic function. The verification of
this fact will be an instructive exercise for the reader (Exercise 4.62).

EXERCISF_,S
Limits of 
4.1 Prove each of the following statements abou[ sequences in C.
a) :-,0 iflzl < l; {z')diverges iflz I > 1.
b) if z. --, 0 and if {c.} is bounded, then (ctz.} -, O.
c) z=/n[  0 for every complex z.
,1.2 Ifa,+z = (a+ 1 + a.)/2 forall n  I, showthat am- (a, + 2az)/3. l'lintaM+z -
a.+l = ½fa. -
4.3 If 0 < xt < 1 and if xm+  = t -  - x, for all n _> 1, prove that [Xa} iS a
decreasing sequence with limit 0. Prove also that xa+l/x  ½.
4.4 Two sequences of positive integexs {a,} and {b.} are defined recurgively by taking
a = bt = 1 and equating rational and irrational parts in the equation
Prove that am -- _
and that 2b,/a  "2 through values < V'2.
,1.5 A real sequence {x,} satisfies 7x,+ = x, + 6 for n _> 1. If xt = ½, prove that
sequence increases arid find its limit. What happens if x, =  or if x, =
t 2
4.6 if la.t < 2 and la.+ - a.+.l -< ta.+ - at for all n  1, prove that
converges.
4.7 In a metric space (S, d), assume that x. , x and .v, -, y. Prove that d(x, y,)
dfx, y).
4.11 Prove that in a compact metric spae ($, d), every sequence in $ has a subsoquenee
which converges in S. This propexty also impl/es that S is compact but you are.not re-
quired 1o prove this. (For a proof s¢¢ either gefcrence 4.2 or 4.3.)
4.9 Let Ab¢ a subset of a metric space S. IrA is complete, prove that A is closed. Prove
that the converse also holds if $ is complete.
Limils o[ functions
?qOT. In Exercises 4.10 through 4.28, all functions are real-vaIued.
4.10 Let f dned on an on intal (a, b) and sum¢ x  (a, b). Consider the lwo
statements
a) lira f(x + h) -f(x)[ = O; b) ]im If(x + ) -f(x - h)l = O.
hO
Prove thai (a) alwa impli (b), and  an example in whi (b) hol& but (a)d not.
4.11 tf d¢fin on R . If
lim f(x, y) = L
and if he one-dimensional limits limxaf(x, y) and I[mrsf(x, y) th ¢aist, p that
tim [im/(, ?)1 = tim [im f(, y)] ffi L.
b) f(x, y) =
d)/(x, y) =
functions fdefined on R z as follows:
if (x, y) ¢= (0, 0), [(0, O) = O.
x 2 + yz
(xy)  + (x -
1
- sin
|(x + y)sin (I/x)sin
io
if (x, y) ¢ (0, 0), f(o, 0) = 0,
tfx  O,f(O, y) ---- y.
ifx ¢ Oandy  O,
ifx = 0ory : 0.
if tan x  tan y,
[_si.n. _x- sin .v
e) f(x, y) = tan x - tan ,
/ cos 3 x if tan x  tan y.
In each of the prcgling examples, determine whether the following limits exist and
evaluate those limits that do tmist :
lira film f(x, y)] ; lira [lim f(x, y)] ; lim [(x, y).
xO ],.=.0 y-O x-.O (r,.r)--,-(O.O)
4.12 If x  [0. l ] prove that the following limit exists,
lira film cos " (m! :rx)],
and that its value is 0 or !. according to whether x is irrational or rational.
Coafinuity of real-nalaed functions
4,13 Let f be continuous on [a, b] and let f(x)  0 when x is rational. Prove that
.f(x) = 0 for every x in [a, b ].
4.14 Let f b¢ continuous at the point a = (at, a,..., an) in R". Keep az, a ..... a,
fixed and define a new function g of one real variable by the equation
Prove that .O is continuous al the point x = at. (This is sometimes stated as follows:
A eontlnuous [unction of n variable is continuous in each variable sepaeately.)
4.15 Show by an example that the converse of the statement in Exewise 4.14 is not true
in gn©ral.
4.16 Let f, 9, and
f(x)
X)
(0)
h be defined on [0, I ] as follows:
= g(x) = h(x) = 0, whenever x is irrational;
= t and gC¢) = x, whenever x is rational;
= I/n, if x is lhe rational number rain (in lowest terms);
Prove that[is not conlinuous anywhere in [0, 1 ], that g is continuous only at x = 0, and
that h is continuous only at the irrational points in [0, I ].
4.17 For each x in [0, 1], let f(x) = x if x Ls rational, and let [(x) = I - x if x is
irrational. Prove that:
a) f(f(x)) = x for all x in [0, 1 ]. b) f(x) + f(1 - ) = I for all x in [0, 1 ].

LLtn uad Contiuity
c)/is continuous only at the point x = J.
d) f assumes every value between 0 and I.
e) f(x + y) - f(x) - f(y) is rational for alt x and y in [0, 1 ].
4.18 Letfbe defined on R and assume t_hat there eists at least one point xo in !1 at which
fis continuous. Suppose also that, for every x and y in It,/satisfies the equation
Prove that there exists a constant a such that f(x) = ax for all x.
4.19 Letfheconlinuous on in, b] and define# as foltoves:g(a) = ]'(a) and, for a < x < b.
let g(x) be the maximum value of fin the sttbinteal in, x]. Show that t is continuous on
1,,, ,].
4.2 Let fl .... , f, be m real-valued functions dned on a et S in R . Assume that each
f is continuous at the point a of S. Define a new function/as follows: For each x in S,
f(x) is the largest of the m numbersfl(x) .... ,/'x). Discu the continuity of fat
4.21 Let f: S  It be continuous on an open set S in R , assume that 1  S, and assume
that/'f) > 0. Prove that there is an n-ball BO; r) such that f(x) > 0 for eve x in the
ball.
,I.27, Letf be defined and continuous on a closed set S in R. Let
A = {x:xeS and f(x)=
Prove that A is a closed subset of R.
423 Given a function f: R - R, define two sets A and Bin Itx as follows:
 = {(x,y):v < S`(x)}, o =
Prove thal S`is continuous on R if, and only if, both A and Bare open subsets ofR z.
4.2,1 LetS'be defined and bounded on a compact interval S in It, If T  S, the number
Q¢{T) = sup {f(x) - f{?)
is e-ailed the oscillation (or span) of/on T. Ifx  S, the oscillation of fat x is defined to
tm the number
ox) - lira fl.(B(x ih)  S).
k-O+
Prove that this limit a]ways exists and that ox) = 0 if, and only if, [is ¢ontimtous at x.
•  tf ntinuou on a compact intml a, hi. SuO thatfh a I1 m-
imam at x and a I] imum at x. Show that the must  a third point twn
x t and x whfhas a tl minimum.
N. To y that fh a I1 mimum at xt mns that the s a t-ball B(x} such
thatf(x)  f(xt) for all x in B(x)  in, b]. LI minimum is similarly defined.
4. t f a l-vald funetlon, ntinuous on [0, 1 ], with the following prorty:
For e 1 y, citer t is no x in [0, I ] for which [(x) = y or the is exactly one
such x. Prove that f is strictly monotonic on [0, I ].
7 t f  a function dn on [0, I] with the fotlowing prorly: For every [
numr y, either te is no x in [0, I ] for ichf(x) = y or the a exaOly two.valu
of x in 10, 1 ] for whichf(x) = y.
a)
b)
c)
f(s) =
a) s = (0, D, r = (0, t].
b) S = (0, 1), r = (0, 1)  (1, 2).
c) S = R t, T = the set of rational numbers.
d) S = [0, 1, 12, 3], T= 0, 1},
e) S=[0,]× [0,1], T= .
f) s- [o,] × [O,l], r= (o, 0 × (o,I).
g) S=(0,1) x (0,1), T=R 2.
Pro that fcannot be continuous on [0, I]1.
Construct a function fwhich has the above property.
Prow that any function with this property has infinitely many discontirtuiti on
[0, l 1.
each case, give an example of a function .f, continuous on S and such that
T, or else explain why there cn be no surga f:
Ctinuity in metric
In Exercises 429 through 4.33, we a.ume that f : S --, T is a function from one metric
space (S, d s) to another (T, dr).
4.29 Prove that fis continuous on S if, and only if,
f--t(int B) __- int f-t(B) for every subset B of T.
4.30 Prove thatfis continuous on S if, and only if,
f() / for every subset A of S.
4.31 Prove that f is continuous on S if, and only if, f is continuous on every compact
subset of S. Hint. If x  p in S, the set {p, x,, x2, • • • ) is compact.
4.32 A function f: S  Tis called a ctosedmappina on S if the imagf(A) is dosed in T
for every closed subset A of S. Prove thatfis continuous and closed on $ if, and only
if, f(.) =  for ery subrt A of S.
4.33 Give an example of a continuous/'and a Cauchy sequence {x,} in some metric
space S for which {f(x.)} is not a Cauehy .luence in T.
4.34 Prove that the interval (-1, l) in lli t is homeomorphie to R , This shows that
neilher boundedness nor completeness is a topologica| prolrty.
4.2 Section 9.7 contains an example of a function  continuous on [0, l l, with
/([0, I]) ffi [0, t] x [0, I]. Prove that no suchfcan be one-to-one on [0,
4, Prove that a metric space S is disconnected if, and only if, there is a nonempty subset
A of S, A # S, which is both open and closed in S.
4,3/ Prove that a metric spae S is connected if, and only if, the only subets of S which
are both open and closed in S are the emlty set and S itselF.
4.38 Prove that the only connected subet$ of R are (a) the empty set, (b) sets consisting
of a single point, and (c) intervals (olxn, closed, half-open, or infinite).

4.39 Let X be a connected suUset of a mec space 5. Let Y b¢a subset of , such that
X  Y _ 3[', where X is the closure of X. Prove that Y is also connected. In particular,
this shows that X is connected.
4.40 If x is a point in a metric space S, let U(x) he the component of S containing x.
Prove that U(x) is closed in A'.
&41 Let 5 be an open subset of R+ By Theorem 3.11, Sis the union of a countable dis-
joint colkion of open intervals in R. Prove that each of these open intervals is a com-
ponent of the metric subspace S, Explain why this does not contradict Exercise 4.40.
4,42 Given a compact set 5 in R = with the following properly: For eveT pair of points
a and h in S and for every • > 0 there exists a finite set of points {xa, xt,.
with x o = a and x = b such that
IIx-xt_1] <e fork = 1,2 .... ,n.
Prove or disprove: S is connected.
4.43 Prove that a metric space 5 is connected if, and only if, every nonempty proper
subset of S has a nonempty boundary.
44 Prove lhat every convex subset of R'* is conitected,
4.45 Given a function f: R" - R" which is one-to-one and continuous on
open and diso3nnectvd in R', prove that f(A) is open and disconnected in
4.46LetA-- {(x,y):0<x< 1, y=sinl]x}, Bffi {(x,y):y:O, -t_<x<0},
and let S = A  B. Prove lhal S is cormected bul not arcwise connected. (See Fig. 4.5,
Section 4. I
4.47 Let F = {F1, F z .... } be a countable vollection of connected compact sets in It*
such that Ft ÷ x _ F for each k > 1. Prove that the intersection N a Ft is connected
and dosed.
4A8 Let Sbc an open connected set in R'. Let Tbea component of R" - S. Prove that
It" - T[s connected.
&49 Let (S, d) be a conrecled metric space which is not hounded. Prove that for every
a in S and every r > 0, the set {x : d(x, a) -- r } is nonempty.
Uniform continuity
4 Prove that a function which is uniformly continuous on S is also continuous on 5.
4.51 If f (x) - x 2 for x in R, prove that fis not uniformly continuous on IL
4.52 Assume thatfis un[forrnly continuoos on a botmded set A' [n R'. Prove that fmust
be bounded on S.
4.53 Let f be  function defined on a set S in It" and assume that f(S) _ R'. Let g be
defined on f(3) with value in R t, and let b denol© the compite furion defin¢! by
h(x) - g if(x) ] if x  S. If f is uniformly continuous on  and ifg is oformly continuous
on f(S), show that h is uniformly continuous on S.
4.q4 Assume f: S  Tis uniformly conlinuous on S, whe 5and Tar mric slaces.
If {x,} is any Cauchy soquenc in S, prove that {f(x,)} is a Cauchy sequence in T. (Com-
pare with Exercise 4.33.)
F.J.erelses 10I
4, Let f : 5 - T be a function from a metric space 5" to another metric space T.
Asstunef is uniformly continuous on a subset 4 of 5 and that T is complete. Prove that
there is a unique extension of/to , which is uniformly continuous on
4.56 In a metric space (S, d), let A be a nonempty subset of S. Define a function
f : S  R by the equation
Z,(x)  inf {d(x, y) : y  .4 )
for each x in $. The numbetf(x) is calhsd the distance from x to
a) Prove thal fa is uniformly continuous on 5.
b) Prove thal  = {x : x  , mul fa(x) =
4.5"7 In a metric ¢ace (5, d), le1.4 and B be disjoint closed subsets of 5. Pove that there
exist disjoint op,m sub, s Uand VofSsuch thal A _ U and B  Y. Hint. Let
g(x) = A(x) -- fs(x), in the notation of Exercise 4.56, and consider g- 1(_ , 0) and
Diseemlnettes
4,58 Loeam and classify the discontinuities of the functions fdefined on R t hot the follow-
ifx  0,f(0) = O.
if x  0, f(O) = O.
if x ¢ o,/¢o) = o.
if x  0,f(0) = O.
Locate t¢ points in R z at wh/¢h  of the functions in Exercise 4.11 is not on-
tinuous.
Manooni¢ fams
4.60 I_¢t[be defined in the open interwal (a, b) and assume that for each interior point x
of (a, b) the. e.xisls a l-ball B(x) in which f [s increasing. Prove that f is an increasing
function throughout (a,
4.61 Lgtfbg continuous on a compact interval [a, b] and assume thatf does not have a
local maximum or a local minimum at any interior point. (Se the so're following
Exercise 4.25,) Prove thatfmust be monotonic on [a, b].
4.62 Iffis one-to-one and continuous on [a, b], pro that fmust bc strictly monotonic
on [a, b]. That is, prove that every topological mapping of [o, b] onto an interval [c, d]
must be strictly monotonic.
4.63 Let f be an increasing function defined on [a, b] and let x] ..... x, be a points in
the interior such that a < x I < x < -÷, < x, < b.
a) Show that,_,= If(x,+) - f(xs-)] _ f(b-) - f(a+).
b) Deduce from part (a) that the set of discontinuities off is countable.
c) Prove that f has points of continuity in every open subinterval of
4.64 Give an ¢gample of a function f, defined and strictly increasing on a set 5 in R, such
that f-  is not continuous on f(S).

10"  d Cguiy
4.6 Let f be strictly increasing on a subset S of R, Assume that the image1-(3) has one
of the following properties: (a)f(3) is open; Co)1-($) is connected; (c)f(S) is closed. Prove
that fmust be continuous on S.
Metric spae¢ a! Ied pols
4.66 t B(S) otc the  of all [-ld run.ions whi  d and
o a no--ply  $, [ff B(S), let
/11 = sap
e num f is lled the "sup no'" of
a)  tt the foula d( ) = I[f - 0   a mc d on
b) Pm at e 'ric s (B(S), d) is ple. Hint. If {} is a uchy
quen in B(S), ow that {(x)) is a Cauchy qu¢ ofl num for ch x
4.67 Refer to  4.66 and let C(S) denote the subt of B(S) consisting of all func-
tions ntinus and on on S, we now S is a metric
a) ov¢ tt C(S) is a cloud sut of B(S).
b) Pro tt the ric su C(S) is complete.
4. Ref to the proof of the fid-t theom ffh 4,) fr notation.
This inequality, whh is uful in numl work, provi an limate for t stan
from p. to t  int p, An p is ven in (b).
b) T¢(x) = (x + 2ix), S = [I, +). lhatfis antraction of Swith
confection nstant  :  and fia int p = . Form the uen
stating with x = Po = ! and show that IP.  /  2 -*.
4 Show  untepl that the fix-int  f ntraions d not
hold if eilh (a) the unrly[ng metric s is not cple, or (b) the contraion
•  tf: S  S  a run.ion from a complete metric spa (S, d)into it.ll. Aume
th is a 1 u {} which con to 0 s lhat d(fn(x), f(y))  (x,
for all n  ! and all x, y in S, whf s is the h iterate off; t ,
fa(x) = f(x), f'+X{x) = £(f'(x)) for n  ,
 tf a iq  L HI, A  -t  tof  for a
sb 
4.71 f: S  S  a function from a tfie s (S, d) into iif such that .
wher x  y.
a) ov¢ tt/has at mt o  int, and give an ¢mpl¢ of such an with no
fix int.
b) If S is mpt,  that f s extly one fix int. HinL Sh thai
x) : d(x, f(x)) attains its minimum on S.
c) Give an ample with S compact in which fis not a ntraction.
103
472 Assume that f satisfies the condition in Exercise 4.71. If x  S, let Po = x,
pj÷, = f(p,,), and cj  d(p, p+ 1) for n > 0.
a) Prove that {era} is a decreasing sequence, and let c =Iim c,.
b) Assume there is a subsequence {Pt.} which converges to a point q in S. Prove
that
¢ = d(q,f(q)) -- d(f(q),fLf(q)]).
Deduce that q is a fixed point off and that p, , q.
SUGG REFERFCES FOR FURTHER STUDY
4.1 Boas, R. P., A ',imer ot"bat Functiottt. Cants Monograph No. 13. Wiley,
York, !.
4.2 Glen, A., Fnta ofAt Aysls. AdwW, Ring, 1.
4.3 So G. F., lntn to Toy  M Ais. Mw-Hiil,
N Y 13.
4.4 Td, J., S ofNur Asis. Mw-HI, N Y 1

CHAPTER 5
DERIVATIVES
5.1 INTRODUCTION
This chaplet trots Ihe derivative, the central concept of differential calculus. Two
different types of problem--the physical problem of finding the instantaneous
velocity of a moving particle, and the geometricaI probIem of finding the tangent
line to a curve at a given point--both lead quite naturally to the notion of deriva-
tive, .Here, we shall not b concerned with applications to mechanics and geometry,
but instead will confine our study to general properties of derivatives.
This chapter deals primarily with derivatives of functions of one real variable,
specifically, reabvalued functions defined on intervals in R. It also discusses
briefly derivatives of vector-valued functions of one real variable, and partial
derivatives, s/ace these topics involve no new ideas. Much of this material should
be famitiar to the reader from elementary calculus. A more detailed treatment of
derivative theory for functions of several variables involves significant changes
and is dealt with in Chapter 12.
The last part of the chapter discusses derivatives of complex-valued functions
of a complex variable.
5,2 DEFINITION OF DATIVI
Iff is defined on an open interval (a, b), then for |wo dislinct points x and c in
(a, b) we can form the difference quotient
f(x) - f(c)
We keep c fixed and study the behavior of this quotient as x  c.
Defudtion 5ol. Let f be defined on an open interval (a, b), and assume that c  (a, b).
Then f is said to be differentioble at c whenever the limit
lira f(x) - f(c)
exixrs. The/imit, deno/ed by, f'(c), is called the derivative of fat c.
This Iimit process defines a new function f', whose domain consists of those
points in (a, b) at which f is differcntiable. The funclion f' is called the first
104
derivative off Similarly, the nth derivative off, denoted byft*, is defined to be
the first derivative off '-1, for n -- 2, 3,.... (By our definition, we do not
consider f ' unless f t*- is defined on an open interval.) Other notations with
which the reader may be familiar are
df dy I [where y = f(x)],
or similar notations. The function f itself is sometimes written ft0). The process
which produces f' from f is called differentiation.
5.3 DERIVATIVE AND CONTINUITY
The next theorem makes it possible to xiucc some of the theorems on derivatives
to theorems on continuity.
Theorem 5,2. If f is defined on (a, b) and differentiable at a point c in (a, b), then
there is a function f* (depending one and on c) which is continuous at c and which
satisfies the equation
f(x)  f(c) = (x -- e)f*(x), (1)
for all x in (a, b), with f*(c) : f'(c). Conversely, if tlere is a functio f*, con-
tinuous at c which satisfies (l), then f is differentiable at c and f'(c)  f*(c).
Proof
Iff'( 0 exists, let f* be defined on (a, b) as follows:
f*(x) = f(x) - f(.¢_) if x  c, f*(c) = if(c).
X -- C
Then f* is continuous at c and (I) holds for all x in (a, b).
Conversely, if(l) holds for some f* continuous at c, then by dividing by x - c
and letting x  c we see thatf'(c) exists and equatsf*(c).
As an immediate consequence of (1) we obtain:
TIeorem 5.. lefts differentiabte at c, then f gs continuoua at c.
Proof Let x - c in
or_. Equation (1) has a geometric interpretation which helps us gain insight
into its meaning. Since f* is continuous at c,f*(x) is nearly equal tof*(c) =
ifx is near e. Replaeingf*(x) by f'(c) in (1) we obtain the equation
f(x) =- f(c) + f'(e)(x - c),
which should be approximately correct when x - c is small. In other words, if f is
differentiable at c, thenfis approximately a linear function near c. (See Fig. 5. I),
Differential calculus continually exploits this geometric property of functions,

106
Flgur
5.4 ALGEBRA OF DERIVATIVES
The next theorem describes the usual formulas for differentiating the sum, differ-
ence, product and quotient of two functions.
Teorem 5,4, Assume f and # are defined on (a, b) and differentiable at c. Then
f + g,f -- g, andf. g are alo differentiable at c. This is also true off/ ifg(c)  O.
The derivatives at c are 9iven by the foltowin.q formulas"
a) (f + g)'(c) = f'(c) + g'(c),
b) (f.a)'(c) = f(c)#(e) +
c) (f/o)'(c) = -
provided g(c) =A O.
Proof. We shall prove (b). Using Theorem 5.2 we write
Thus,
= f(c) + (x - c)f*¢x),
g(x) = g(c) + (x - c)g*(x).
f(x)gfx) --f(c)g(c) = (x - c)[f(e)*{x) + f*(x)g(c)] + (x -- c}f*(x*(x).
Dividing by x - c and letting x  c we obtain {b). Proofs of the other statements
are similar.
From the definition we see at once that iff is constant on (a, b) then f' = 0
on (a, b). Also ill(x) = x, then f'(x) = 1 for all x. Repeated application of
Theorem 5.4 tells us that if f (x)  x" (n a positive integer), then if(x) = nx*-x
for all x Applying Theorem 5.4 again, we see lhat every polynomial has a deriva-
tive everywhere in R and every rational 'unction has a derivative wherever it is
defined.
5.5 THE CHAIN RUI,E
A much deeper result is the so-called chain rule for differentiating composite func-
tions.
DO', 5,6 Onvide Dedvtlives ad kfftC l)eriatives 107
Theorem 5.5 (C&ffn rul). Let f be defined on an open interval S, let g be defined on
f(S), and consider the composite function O of defined on S by the equation
(0 of)(x) :
Assume there is a point c in S such that f(c) is an interior point off(S), lf f is
differentiabIe at c and if g is differentiable at tic) then 0 °f is differentiabfe at e
and we have
(g o f)'(c) =
Proof. Using Theorem 5.2 w.e can write
fix) - f(c) - (x - c)f*(x) for all x in S,
where/* is continuous at c and f*(c) = f'(c). Similarly,
a(y) - = [y -
for all y in some open subinterval T off(S) containingf(c). Here 9" is continuous
at f(c) and g*[f(c)] = g'[f(c)].
Choosing x in S so that y = f(x)  T, we then have
gDc(x)] -off(c)] = if(x) -f(c)]o*[f(x)] - (x - c)f*(x)o*[f(x)]. (Z)
By the continuity theorem for composite functions,
9*[/(x)] -* g*[f(¢)] = g'[f{c)] as x - c.
Therefore, if we divide by x - c in (2) and let x -o c, we obtain
lim off(x)] - air(c)] _ #'[f(c)]f'(c),
x,e X -- C
as required.
5.6 ONE-SIDED DERIVATIVES AND INFINITE DERIVATIVES
Up to this point, the statement that f has a derivative at c has meant that c was
interior to an interval in whichfwas defined and that the limit definingf'(c) was
finite. [t is convenient to extend the scope of our ideas somewhat in order to discuss
derivatives at cndpoints of intervals. It is also desirable to introduce infinite
derivatives, so that the usual geometric interpretation of a derivative as the slope
of a tangent line will still be valid in case the tangent line happens to be vertical.
In such a case we cannot prove that fis continuous at c, Therefore, we explicitly
require it to be so.
Defim'tion 5.6. Let fbe defined on a closed interval S and assume that f is continuous
at the point c in S. Then f is said to have a ri.qhthad dericative at c if the righthand
limit
lira f(x) - f(c)
x-,c+ X  e

1o Dcdralh In..%7
exists as a finite value or if the limit is + o or - ¢o. This limit will be denoted by
f (c). Lefthand derivatives, denoted by fL(c), are similarly defined. In addition,
if c is an interior point of S, then we say that f has the derivatie f'(c) = + o if
both the riyht- and lefthand derivatives at c are + c. (The derivative f'(c) = -to
is similarly defined.)
It is clear thatfhas a derivative (finite or infinite) at an interior point c if, and
only if, f.(c) = f_ (c), in which casef(c) = f'_(c) -f'(c).
Xl  xl z 5 •
Figure 5.2
Figure 5.2 illustrates some of these concepts. At the poim x we havef_(x) =
- oo. At x 2 the lefthand derivative is 0 and the righthand derivative is - 1. Also,
f'(x) = - c, f'_ (x,) = - 1, .f; (x4) = + i, f'(xt) = + o, and f'_ (x7) = 2.
There is no derivative (one-sided or otherwise) at xs,. since f is not continuous
there.
5.7 FUNCrIONS WITH NONZERO DERIVATIVE
Tlteorem 5.7. Let f be defined on an open interval (a, b) and assume that for some
c in (a, b) we havef'(c) > 0 orf'(c) = + oo. Then there is a I-halt B(c) c_ (a, b)
in which
f(x) > f(c) if x > c, and f(x) < f(c) if x < c.
Proof. lff'(c) is finite and positive we can write
f(x) --f(c)  (x - c)f*(x),
where f* is continuous at c and/*(c) = f'(c) > 0. By lhe sign preserving property
of continuous functions there is a 1-ball B(c) c_ (a, b) in whichf*(x) has the same
sign asf*(c), and this means that f(x) - f(c) has the same sign as x - c.
lff'(c) : + o, there is a l-bail B(c) in which
f(x) - f(c)
> I whemver x # e.
In this ball the quotient is again positive and the conclusion follows as before.
A result similar to Theorem f7 holds, ofcurse, iff'(c) < 0 or iff'(c) = -
at some interior point c of (a, b).
5.8 ZERO DERIVATIVES AND LOCAL Ex'rREMA
DeflMti $,8, Let f be a real.valued function defined on a ubset S of a metric
space M, and assume a ¢ S. Then f is said to have a local maximum at a if there is
a bail B(a) such that
f(x) < f(a) for all x in B(a)  S.
If f (x) ; f(a) for all x ire B(a)  S, then f is said to have a local minimum at a.
No'r. A local maximum at a is the absolute maximum off on the subset B(a) c S.
Iffhas an absolute maximum at a, then a is also a local maximum. However, f
can have local maxima at several points in S without having an absolute maximum
on the whole set S. -
The next theorem shows a connection between zero derivatives and local
extre.ma (maxima or minima) at interior points.
lrltweem 5.9. Let f be defined on an open interval (a, b) and assttrne that f has a
local maximum or a local minimum at an interior point c of(a, b). lf f has a derivative
(finite or infinite) at c, then f'(c) must be O.
Proof Iff'(c) is positive or +o, then f cannot have a local extremum at c
because of Theorem 5.7. Similarly, f'(c) cannot be negative or -. However,
because there is a derivative at c, the only other possibility isf'(c) = 0.
The converse of Theorem 5.9 is not true. In general, knowing that f'(c) ffi 0
is not enough to determine whether f has an extremum at c. I, fact, it may have
neither, as can be verified by the example f(x) = x J and c = 0. In this case,
f'(O) = 0 but f is increasing in every neighborhood of 0.
Furthermore, it should be emphasized that/can have a local extremum at c
withoutf'(c) being zero. The examptef(x) = txl has a minimum at x = 0 but,
of course, there is no derivative at 0. Theorem 5.9 assumes thatfhas a derivative
(finite or infinite) at c. The theorem also assumes that c is an interior point of
(a, b). In the example f(x) = x, where a < x < b, f takes on its maximum and
minimum at the endpoints but f'(x) is never zero in [a, hi.

5.9 ROLLE'S TI-IgOREM
It iS geometrically evident that a sufficiently "smooth" curve which crosses the
x-axis at both cndpoints of an interval [a, b] must have a "turning point" some-
where between a and b. The precise statement of this fact is known as Rolle's
theorem.
Theorem 5.I0 (Rolle). Assume f h a derivative (finite or infinite) at each point of
an open interval (a, b), and assume that f is continuous at both endpoints a and b.
If f (a) = f(b) there is at least one interior point e at which f'(c) - O
Proof. We assume [' is never 0 in (a, b) and obtain a contradiction. Since f is
continuous on a compact set, it attains its maximum M and its m/nimum m some-
where in [a, hi. Neither extreme value is attained at an interior point (otherwise
f' would vanish there) so both are attained at the endpolnts. Since f(a) = f(b),
then m = M, and hencefis constant on [a, b]. This contradicts the assumption
thatf' is never 0 on (a, b). Thereforef'(c) = 0 for some c in (a, b).
5,10. THE MEAN-VALUE THEOREM FOR DERIVATIVES
Theorem 5.11 (Mean-lYMue Tl, eorem). Assume that f has a derivative (finite or
infinite) at each point of an open interval (a, b), and assume also that f is continuous
at both endpoints a and b. Then there is a point c in (a, b) such that
f(b) -f(a) =f'(c)(b - a).
Geometrically, this states that a sufficienlly smooth curve joining two points
A and B has a tangent line with the same slope as the chord AB. We will deduce
Theorem 5. ] 1 from a more general version which involves two functionsfaad , in
a symmetric fashion.
Theorem 5.12 (GeneraRr.ed Mean. Value Tlutorem). Let f and # be two functions,
each having a derivative (finite or infinite) at each point of an open interval (a, b)
and each continuous at the .endpoints a and b. Assume also that there is no interior
point x at which botk f'(x) and g'(x) are infinite. Then for some interior point c we
have
f'(c)[g(b) - g(a)] =g'(c)[f(b) -f(a)].
No'rE. When g(x) = x, this gives Theorem 5.1 I.
Proof. Let h(x) = f(x)[y(b) - g(a)] - #(x)[f(b) - f(a)]. Then h'(x) is finiie if
bothf'(x) and .q'(x) are finite, and h'(x) is infinite if exactly one off'(x) or #'(x) is
infinite. (The hypothesis excludes the cas/: of both f'(x) and '(x) being infinite.)
Also, h is continuous at the endpoints, and h(a) = h(b) --f(a)y(b) - .q(a)f(b).
By Rolle's theorem we have h'(c) = 0 for some interior point, and this proves the
assertion.
Co. .15 ltct'mcdc-Valu  111
N00rE. The reader should interpret Theorem 5.12 geometrically by referring to the
curve in the x.v-plane described by the parametric equations x = #(t), y = f(t),
a<_t<_b.
There is also an extension which does not require continuity at the endpoints.
m 5.13. Let f and 9 be two functions, each having a derivative (finite. or
infinite) at each point of (a b). At the endpoints assume that the limits f(a+ ),
t?(a+), f(b-) and tl(b-) exist as finite values. Assume further that there is no
interior point x at .which both if(x) and y'(x) are infinite. Then for some interior
point c we have
f'(c)[a(b - ) - #(a + )] - l'(e)[f(b - ) - flf )].
Proof. Define new functions F and G on [a, b] as follows:
F(x) = f(x) and G(x) = g(x) if x  (a, b);
F(a) =f(a+), G(a) = e(a+), F(b) = f(b-),
Then F and G are continuous on I'a, b] and we can apply Theorem 5.12 to F and
G to obtain the desired conclusion.
The next result is an immediate consequence of the Mean-Value Theorem.
]Tg, orcm 5d4. Assume f has a derivative (finite or infinite) at each point of an open
interval (a, b) and that f is continuous at the endpoints a and b.
a) If f' takes only positive tues (finite or infinite) in (a, b), then f is strictly
increasing on [a, b],
b) If f' takes only negative values (finite or infinite) in (a, b), then f is strictly
decreasing on [a, b].
¢) lf f' is zero everywhere in (a, b) then f is constant on [a,
Proof. Choose x < y and apply the Mean-Value Theorem to the subinterval
[x, y] of [a, ] to obtain
f(y) - f(x) = f'(cXy - x) where c  (x, y).
All the statements of the theorem follow at once from this equation.
By applying Theorem 5.t4 (c) to the difference f -  we obtain:
Corollary LIS. lf f and g are continuous on [a, b] and have equat finite derivatives
in (a, b), then f - g is constant on [a, hi.
5.11 INTERMEDIATE-VALUE THEOREM FOR DERIVATIVES
In Theorem 4.33 we proved that a function f which is continuous on a compact
interval [a, b] assumes every value between its maximum and its minimum on

112 Deritallv Tb. 5.16
the interval. In particular,(assumes every value betweenf(a) andf(b). A similar
result will now be proved for functions which are derivatives,
rkeorem 5.16 (lntrmediate-lbe tlumm foe dees). Assume that f is de-
fined on a compact intert [a, b l and that(has a derivative (finite or infinite) at each
interior point. Assume also that(has finite one-sided derivatives f (a) and f'__(b) at
the endpoints, with f(a)  fL(b). Then, if c is a real number between f(a) and
f'_(b), there exists at least one interior point x such that f'(x) - c.
Proof. Define a new function g as follows:
a(x) = f(x) - f(a) if x  a, g(a) --- f(a).
Then g is continuous on the closed interval [a, bl. By the intermediate-value
theorem for continuous functions, g takes on every value betweer f(a) and
If(b) - f(a)]l(b - a) in the interior (a, b). By the Mean-Value Theorem, we have
g(x) = f'(k) for some k in (a, x) whenever x  (a, b). Therefore f' takes on every
value between f(a) and If(b) - f(a)]/(b - a) in the interior (a, b). A similar
argument applied to the function h, defined by
h(x) ffi f(x) -f(b) ifx ¢: b, h(b) = f_(" b),
shoves that(' takes on every value between If(b) - f(a)]f(b - a) andf'_(b) in the
interior (a, b). Combining these results, we see that(' takes on every value between
f,(a) andf'_(b) in the interior (a, b), and this proves the theorem.
rOl. Theorem 5.16 is still valid if one or both of the one-sided derivatives
f(a), f'_(b), is infinite. The proof in this case can be given by considering the
auxiliary function # defined by the equation g(x) = f(x) - cx, if x  [a, b].
Details are leR to the reader.
The intermediate-value theorem shows that a derivative cannot change sign
in an interval without taking the value 0. Therefore, we have the following
slrengthening of Theorem 5.14(a) and (b).
Theorem 5,1L /Issume f has a derivative (finite or infinite) on (a, b) and is con-
tinuous at the endpoints a and b. l/f '(x) # 0 for art x in (a, b) then f is strictly
monotonic on [a, bl.
The intermediate-value theorem also shows that monotonic derivatives are
necessarily continuous.
Theorem 5.18. Assume(' exists and is monotonic on an open internal (a, b). Then
f' is continuous n (a, b).
Proof. We assume(' has a discontinuity at some point c in (a, b) and arrive at a
contradiction. Choose a closed subinterval [,/1] of (a, b) which contains c in its
interior. Since/' is monotonic on [,/] the discontinuity at c must be a jump
To. £0 Taylor's Fcmula with llmaibr 113
discontinuity (by Theorem 4.51). Hence f' omits some value between f'(a) and
f'(, contraAieting the intermediate-value theorem.
,12 TAYLOR'S FORMULA WITH REMAINDER
As noted earlier, if/is diffcrentiable at c, then/is approximately a linear function
near c. That is, the equation
f(x) =/(c) + f'(cXx- c),
is approximately correct when x - c is small. Taylor's theorem tells us that, more
generally,(can be approximated by a polynomial of degree n - ! if f has a deriva-
tive of order n. Moreover, Taylor's theorem gives a useful expression for the eor
made by this approximation.
Tkeoem 5.19 (Taylor). Let f be a function havin finite nth derioatie f'} ever),-
where in an open intercal (a, b) and assume that( <'- is continuous on the closed
interval [a, bl. Assume that c  [a, b]. Then, for eery x in [a, b], x # c, there
exists a point x t interior to the internal joining x and c such that
f(x) = f(c) + E f')(c), c)  + -- (x --
Taylor's theorem will he obtained as a consequence of a more general result
that is a direct extension of the generalized Mean-Value Theorem.
Theorem 520, Let f and # be two functions having finite nth dericatioes f' and
g{'} in an open interal (a, b) and continuous (n - 1)st derivatives in the closed
interval [a, bl. Assume that c • [a, b]. Then, for eery x in [a, b], x # c, there
exists a point x #nerior to the interat joiningx and c such that
.- ] [
qor. For the special case in which if(x) = (x - c) ", we have gS(c) = 0 for
0 < k < n - 1 andg')(x) = n!. This theorem then reduces to Taylor's theorem.
2roofi For simplicity, assume that c < b and that x > c. Keep x fixed and define
new functions F and G as follows:
r(t) = f(t) +
-|
for each t in [c, x]. Then F and G are continuous on the closed interval [c, x]
and have finite derivatives in the open interval (c, ), Therefore, Theorem 5.|2 is

114 IMqmtit
appficable and we can write .-
F'(xt)[G(x) - G(c)] = G'(x,)[F(x) - F(c)], where x,  (c, x).
This reduces to the equation
'(x,)[#(x)- (e)] = '(xl)Lt'(x) -(c)],
since G(x) = g(x) and F(x) = f(x). If, now, we compute the derivative of the sum
defining F(t), keeping in mind that each term of the sum is a product, we find that
all terms cancel but one, and we are left with
F'(t) - (x - t) b-'
(n - 1)!
Similarly. we obtain
¢7'(0 (x -
(n -
If we put t = x and substitute into (a), we obtain the formula of the theorem.
5.13 DERIVATIVES OF VECTOR-VALUF_ FUNC'IaONS
Let f: (a, b)  R" be a vector-xalued function defined on an open interval (a, b)
in R. "Fhen f = (f,..., f) where each component f is a real-valued function
defined on (a, b). We say that f is differentiable at a point c in (a, b) if each com-
ponentf is different/able at c and we define
f'(O = (f;(),,,,,
In other words, the derivative f'(e) is obtained by differentiating each component
of f at c. In view of this definition, it is not surprising to find that many of the
theorems on differentiation are also valid for vector-valued functions. For example,
if f and g are vector-valued functions differentiable at c and if l is a .real-valued
function differentiable at c, then the sum f + g, the product )of, and the dot product
f' g are differentiable at e and we have
(f + g)'(c) = r'c) +
(r. g)'(c) = f'()- g(c) + re)- g'(e).
The proofs follow easily by considering components. There is also a chain rule for
differentiating composite functions which is proved in the same way. If f is vector-
valued and if u is real-valued, then the composite funclion g given by g(x) -
flu(x)] is vector-valued. The chain rule states that
g'(c) =
if the domain of f contains a neighborhood of u(c) and if u'(c} and f'[u(c)] both
exist.
The Mean-Value Theorem, as stated in Theorem 5.1 I, does not hem for vector-
valued functions. For example, if f(t) =- (cos t, sin t) for all real t, then
f(2t) - f(0) = 0,
but f'(t) is never zero. In fact, llf'(t) = t for all t. A modified version of" the
Mean-Value Theorem for vector-valued functions is given in Chapter t2 (Theorem
12.8).
5.14 PARTIAL DERIVATIVES
Let S be an open set in Euclidean space R ", and let f : S - R be a real-valued
function defined on S. If x = (xx,..., x) and e = (c .... , c, are two points
of S having corresponding coordinates equal except for the kth, thin is, if x, ---- e
for i  k and if x  ct, then we can consider the limit
lim f(x) - f(c)
-.q x - ct
When this limit exists, it is called the partial deritmtioe off with respect to the kth
coordinate and is denoted by
or by a similar expression. We shall adhere to the notation Dtf(e).
This process produces n further functions Df, Dz..., D.f defined at those
points in S where the corresponding limits exist.
Partial differentiation is not really a new concept. We are merely treaing
f(xt, ..., xn) as a function of one variable at a time, homing the others fixed.
That is, if we introduce a funetion# defined by
g(x) = f(c, ..., c,-t, xt, q+,, ..., c.),
then the partial derivative Dot(©) is exactly the same as the ordinary deriva_tive
9'(c). This is usually described by saying that we differentiate f with respect to
the kth variable, holding the others fixed.
In generalizing a concept from R * to R , we seek to preserve the important
properties in the one-dimensional case. For example, in the one-dimensional c,ase,
the existence of the derivative at c implies continuity at c, Therefore it seems
desirable to have a concept of derivative for functions of several variables which
will imply continuity. Partial derivatives do not do this. A function ofn variables
can have partial derivatives at a point with respect to each of the variables and yet
not be continuous at the point. We illustrate with the following example of a
function of two variables:
x + y, ifx = 0ory = 0,
f ( x, ,) = I, otherwise.

The partial derivatives Df(O, O) and Dz]'(0, 0) both exist. In fact,
Dtf(O , O) = |ira f(x, 0) - f(0, 0) = lim _ x =
-0 X -- 0 .x,,-.o X
and, similarly, D2f(O, 0) = I. On the other hand, it is clear that this function is
not continuous at (0, 0).
The existence of the partial derivatives with respect to each variable separately
implies continuity in each variable separately; but, as we have just scan, this does
not necessarily imply continuity in all the variables simultaneously. The difficulty
with partial derivatives is that by their very definition we are forced to consider
only one variable at a time. Portia[ derivatives give us the rate of change of a
function in the direction of each coordinate axis. There isa more general concept of
derivative which does not restrict our considerations to the special directions of
the coordinate axes. This will be studied in detail in Chapter 12.
The purpose of this section is merely to introduce the notation for partial
derivatives, since we shall use them occasionally before we reach Chapter 12.
lff has partial derivatives Dlf, ..., Df on an open set S, then we can also
consider their partiat derivatives. These are called second-order partial derivatives.
We write D,,,f for the partial derivative of D,f with respect to the rth variable.
Thus,
9,.d =
Higher-order partial derivatives are similarly defined. OIher notations are
D,,f -- Ox Oxl ' DP'vf - Ox cx dx, "
5.15 DIFFERFJqTIATION OF FUNCTIONS OF A COMPLEX VARIABLE
In this ,'-tion we shall discuss briefly derivatives of complex-valued functions
defined on subsets of the complex plane. Such functions are, of course, vector-
valued functions whose domain and range are subsets of R z. All the considerations
of Chapter 4 concerning limits and continuity of vector-valued functions apply,
in particular, to functions of a complex variable. There is, however, one essential
difference between the set of complex numbers C and the set of n-dimensional
vectors R  (when n > 2) that plays an important role here. In the complex number
syste m we have the four algebraic operations of addition, subtraction, multiplica-
ti on, and division, and these operation s satisfy most of the "usual" laws of algebra
that hold for the real number system. In particular, they satisfy the first five
axioms for real numbers listed in Chapter l. (Axioms 6 through 10 involve the
ordering re]ati0n <, which cannot exist among the complex numbers.) Any
algebraic system which satisfies Axioms I through 5 is called a field. (For a
thorough discussion of fields, see Reference 1.4.) Multiplication and division, it
turns out, cannot be introduced in R  (for n > 2) in such a way that R" will
117
become a field]" which includes C. Since division is possible in C, however, we can
form the fundamental difference quotient if(z) - f(c)[(e -- c) which was used
to define the derivative in R, and it now becomes clear how the derivative should be
defined in C.
Defudtion 5.21. Let f be a complex-valued function defined on an open set S in C,
and asume c  S. Then f is said to be differentiable at c f the limit
lim f(_z.) - f(c_) .... f'(c)
ets.
By means of this limit process, a new complex-valued function f' is defined at
those points z of S where f'(z) exists. Higher-order derivatives f', f', ... arc,
of course, similarly defined.
The following statements can now be proved for complex-valued functions
defined on an open set S by exactly the same proofs used in the real case:
a) f is differentiable at c if, and only if, there i a function f*, continuous at c, such
that
f(z) -- f(c) = (z - e)f*(z),
for all z inS, with f*(c) = f'(c).
rOr. If we let 9(z)  f*(z) - f'(c) the equation in (a) can be put in the form
f(z) : f(c) + f'(c    z - c) + - c),
where (z)  0 as z  c. This is called afirst-order Taylor formula forf
b) If f is differentiable at c, then f is continuous at c.
c) If two functions fond  have derivatives at c, then their sum, difference, product,
and quotient also have derivatives at c and are given by the uauai formula:
Theorem 5.4). In the case o f fig, we must assume y(c)  O.
d) The chain rule is valid; that is to say, we have
of)'(c) =
if the domain of a contains a neighborhood of'f(c) and if f'(c) and g'[f(c)]-
exist.
Whenf(z) = z, we findf'(z) = 1 for all z in C, Using (c) repeatedly, we find
thatf'(z) = nz "- wbenf(z) = z" (n is a positive integer). This also holds when
" For example, if it were possible to define multiplication in R a so as to make R a a field
including C, we could argueas follows: For every x in R  the veelors !, x, x 2, x a would
be linearly dependent (see Refetcnc 5.1, p. 558). Hence for each x in R , a relation of
the form ao + ax. + a2 x + aax a = 0 would hold, where no, at, a, aa are real
numbers. But every polynomial of degree Ihro with real coeflScients is a product of a
linear polynomial and a quadratic polynomial with real cool, tents, The only roots such
polynomials can hav are either real numbers or complex numbers.

I11 Derfvtti Tit. .22
n is a negative integer, provided z # 0. Therefore, we may compute derivatives
of complex polynomials and complex rational functions by the same techniques
used in elementary differential calculus.
5.1K THE CAUCI-IY-N EQUATIONS
lffis a complex-valued funon of a complex v'iable, we can write each function
value in the form
f(z) = u(z) + i(z),
where u and v are real-valued functions of a complex variable. We can, of course,
also consider u and v to be reai-valued fitnctons of two real variables and then
we write
f(z) = u(x, y) + (x, y),
In either case, we writer = u + iv and we refer to u and v as the real and ima
inary parts off, For example, in the case of the complex exponential funct/on f,
defined by
f(z) = e: = eZ cos y + ie siny,
the real and imaginary parts are #yen by
u(x, y) =
Similarly, whenf(z) ffi z z ffi (x ÷ iy), we find
u(x, y) ffi x 2 - y2, (x, y) = 2zy.
In the next theorem we shall see that the existence of the derivativef' places a
rather severe restriction on the tea] and imaginary parts
Tkeorcm 5.22. Let f = u + iv be defined on an open set S in C. If f'(c) exists for
some c in S, then the partial derivatives D,u(c), D2u(c), D jr(c) and D2v(c ) also
exist and we ham
if(c) ffi Olu(c ) ÷ / D(c), (3)
and
f'(c) -- D:(c) - i D2u(c). (4)
This implies, in particular,
Du(c) ---- D2(c) ad Dv(c) = -- D2u(c).
NOT. These last two equations are known as the Cauchy-Riemann equations.
They are usually seen in the form
Proof. Sincef'(c) exists tere is a function f* defined on S such that
f(z) : ffc) = ( - c).f*(z),
(S)
Cauchy-Rienumn Equati
119
where f* is continuous at c andf*(c) = f'(¢). Write
z = x + iy, c = a + ib, and f*(z) =
where (z) and B(z) are real. Note that :f(z) -, A(c) and B(z) -., B(c) as z -, c.
By considering only those z in S with y -- b and taking real and imaginary parts
of (5), we find
u(x, b) - u(a, b) = (x - a)A(x + ib), v(x, b) - v(a, b)
Dividing by x - a and Jetting x -, a we find
Dtu(c) = A(c) and Dry(c) -- B(c).
Sincef'(c) = A(c) + iB(c), this proves (3).
Similarly, by considering only those z in S with x = a we find
o:(e) - (c) and n:(c) = -
which proves (4).
Applications of the Cauchy-Riemann equations are given in the next theorem.
Theorem 5.2J. Let f---- u + ic be a function with a derivative eer,vhere in aa
open disk D cettered al (a, b). If any one of u, , or ]ft is constant* on D, the
f is constan o D. 41so, f is constant if f "(z) - O for all z in D.
Proof. Suppose u is constant on D. The Cauchy-Riemann equations show that
Dv = Dv -- 0 on D. Applying the one-dimensional Mean-Value Theorem twice
we find, for some y' between b and y,
v(X, y) -- V(X, b) = (y - b)D2v(x, ") -- O,
and, for some x"between o and x,
v(x, b) - v(a, b) = (x - a)D,vx', b) = O.
Therefore v(x, y)  v(a, b) for all (x, y) in D, so v is constant on D. A similar
argument shows that if v is constant then u is cc;nstant.
Now suppose Ill is constant on D. Then lfl z -- u 2 + ,2 is constant on D.
Taking partial derivatives we find
uD,u + vD:v = O, uDzu + vD,v = O.
By the Cauchy-Riemann equations the second equation can be written as
rDlu - uDl, = O.
Combining this with the first to eiiminate Dv we find
u 2 + v 2  0, thenu-- v = 0, sol= 0, lfu ; + v z :# 0then Du = 0;hence
u is constanl, saris constant.
" Hem ]f[ denotes the function whose value at z is

Finally, iff' = 0 on D, both partial derivatives Die and Day are zero on D.
Again, as in the first part of the proof, we flndfi constant on D.
Theorem 5.22 tells us that a necessary condition for the functionf : u + iv to
have a derivative at c is that the four partials Dlu, Dxu , Dry , Dxe , exist at c and
satisfy the Cauchy-Rieanann equations. This condition, however, is not sufficient,
as we s¢ by considering the following example.
Example. Let u and v be defined a follows:
x 3 _ y3
u(x, y) = if (x, y) # (0, 0), u(0, 0) = 0,
x 2 + yZ
x  + y
x,y) - if(x,y) # (0,0L o(0,0) = 0.
x  + yZ
It is easily sn that D,u(O, 0) - Dry(0, 0) = ! and that Du(0, 0) = - Dz0, 0) = - 1,
so that the C_.auchy-Riemann quations hol0 at (0, 0). Nevertheless, the function f =
u + iv cannot have a derivative at z -- 0. In fact, for x = 0, the difference quotient
/(z)-f(0) -y+y
-- =1+i,
z-O iy
whereas for x  y, it becomes
'(z) - f(o) _
z--0
and hence f'(O) cannot exist.
xi I+i
x+ix 2
in Chapter 12 we shall prove that the Cauchy-Riemann equations do suffice to
establish existence of the derivative off = u + iv at e if the partial derivatives of
u and v are continuous in some neighborhood of c. To illustrate how this result is
used in practice, we shall obtain the derivative of the exponential function. Let
f(z) = e x = u + iv. Then
and hence
u(x, y) = e  cos ,,
v(x, y) = e  sin y,
Dtu(x, y ) = ecosy = Dzv(x, y), Dzu(x, y) = -esiny = -Dtv(x, y ).
Since these partial derivatives are continuous everywhere in R 2 and satisfy the
Cauchy-Riemann equations, the derivativef'(z) exists for all z. To compute it we
use Theorem 5.22 to obtain
f'(z) =  cos y + ie  sin y = f(z).
Thus, the exponential function is its own derivative (as in the real case).
121
R el.valned fuactits
In the following exercises assume, where necessary, a knowledge of the formulas for
differentiating the elementary trigonometric, exponential, and logarithmic functions.
5.1 A funclion f is said to satisfy a Lipschitz condition of order 0¢ at c if thece exists a
positive number M (which may depend on c) and a i-ball B(c) such that
If(x) -f(c)[ < Mix - el"
whenever x • B(c), x ¢ c.
a) Show that a function which satisfi¢ a Lipsehitz condition of order  is oonfinuous
at c if a > 0, and has a derivative at c if , > l.
b) Give an example of a function satisfying a Lipschitz condition of order ! at c for
which f'(c) does not emist,
..2 In each of the following cas, determine the inttawals in which the function f is
increasing or decreasing and find the maxima and minima (if any) in the set where eachf
is defined.
a) t'(x) . x  + ax + b,
b) f(x) - log (x 2 - 9),
c) f(x) = xt(x - IP,
d) f(x) = (sin x)]x if x  O,.f(O) = ,
x6R.
lxl > 3.
0_<x_<l.
O < x <_ n/2.
5.3 Find a polynomial f of lowest possible degree such that
f(xl) =- a,, f(x2) = a, f'(x,) = b, f'(xz) =
where x ¢ xz and a t, az, b, b 2 are given real lumbers.
,4 I)efinefasfollows:.f(x) = e -ttx ifx  0,f(0) = O. Show that
a) f is continuous for all x.
b) f') is continuom for all x, and that f')(0) = 0, (n = 1, 2,... ).
Dfinef, o, and h as follows:f(0) = 0(0) -- 0)  Oand, ifx ¢ O,f(x) = sin (I/x),
= x sin (l/x), h(x) = x 2 sin (l/x). Show that
if) f'(x) = - 1/x 2 cos (I Ix), if x  0;
b) O'(x) = sin (t/x) - llx cos (l/x), if x  0;
c) h'(x) = 2x sin (fix) - cos (t/x), ifx ¢ 0;
f'(0) does not exist.
#'(0) does not exist.
t,'(0) = 0;
limx o h'(x) does not exist.
5,6 Derive Leibnitzs formula for the nth derivative of the product h of two functions
fund g:
(:)=
gO(x ) = 1"o'(x)a"-°(x), where k ! (n - /,')!
&=O
5,7 Let land g be two functions defined and having finite third-order derivatives f'(x)
and g'(x) for all x in R. ffffx)g(x) = | for all x, show that the relations in (a), (b), (¢),

anti (d) hold at those points where the denominators are not zero:
a)/-(x)//(x) + v'(x)/(x)  O.
b) f'(x)ff'(x) - 2f'(x)l/fx) - '(x)l'(x) = O.
c) f'(x) 3 f'tx)o"tx) f"(x) (x) _ o.
f'(x) f(xD'(x) f(x)
,  pion which a on th l si of (d)
i off at x.
g) how lhatfaad 0 w lh¢ e han iti if
ex) = tar(x) + b[cf(x) + d],   - bc ¢ O.
Hint. If c # O, it¢ ( + b)l(cf + d) : (a[c) + ( - )/[¢f+ d)], d aply
 ().
s] A,A, an,   four fan,long v divaliv in (a, b). n¢ F s of
the
odx) o(x ! "
a) Show lhat F'(x) ¢s for  x  (a, b) and
F'(x) = f( )
b) State and  a mo  ult for h  deti.
$, Given n funio ..... f,  vig nth or dili in (a, b). A run, ion
W, i e Waski of .... ,f i fin  foi[o: For
e value of the lint of orfler n who element in the kth
- )(x) wbe k  I, 2,..., n and m = 1,  .... n. e expion)&) is writt
rot&(x).]
a) Show that W'(x) n  obtn by lang the [i w of the deteint
0efining W(x) by the nth deritivs)(x),...,
b) Agsumin$ the exist of n consists ct .... , %, not all o. such. that
cA(x) + ..- + c,(x) = 0 for ¢ x in (a, b}, show
x in (a, b).
n.   of function tisfying such a iation is id to  a finearl dennt set
on (a, b).
c) The vanishing of the Wronskian throughout (a, b) is n, bu not sucit,
for lir ded off .... , L. Sh that in the  of two futions, if the
Wronskian vanlsh ouout (a, b) and if o of the functions ds not vanish
in (a, b), t th fo a tin,fly dedem t in (a, b).
 123
Mtn-Val 
$.10 Given a funclion f definexl and having a finite ivative in (a, b) and such
limx _ f(x) = + . Prove tt limxo_ f'(x) either fails to exist or is inite.
5.11 Show that lhe formula in t Mn-Vale ¢om n  writ as foil--s:
f(x + h) -f(x)= f'(x + Oh),
h
whe 0 < 0 < 1. teine 0  a fusion of x and h when
a) f(x) = x , b) f(x) = x ,
c) f(x) = e , d) f(x) = log x, x > O.
K x ¢ 0 , d find li o 0 in  .
5.12 Tef(x) = 3x  - 2x  - x  + I dx)  4x s - 3x  -  in  5.
Show lhat/'(x)lv'(x) is no l to the quoter if(l)- f(O)]/[#(1)- #(0)] if
0 < x  I, How do you I this th 1h¢
ffb)  f(a) _ if(x,)
obinabl¢ from  5.20 when n = 1 ?
5.13 In  of t followi sial  of ¢om ., t¢ n = I, c = a, x = b,
and show lt x t = (a +
a) [(x) = sin x, x)   x; b) f(x) = e , (x) = e -x.
n you find a 1 cis ofsu i of funclions fandfl for which x will always
(a + b)[2 d such t lh mpl (a) and )  in thi
5.14 Gin a funion fdn d having a [ defivatif' in the .lbofl int¢al
0 < x  I d su that Ifx)l < 1. n a, = f(lln) for n  l, 2, 3,..., and show
tt [im  at ist. Hint. uchy
5.15 Au ltfh a finite divafiv¢ al h int of ¢ on lateral (a, b). Assu
a tt lt, tf'(x) ls d is i f  interior t c. o that t v
of this limit must  if(c).
$.16 t f ntinuous on (a, b) th a finite rivati f' evwh in (a, b),
ibIy at c. ff iimxtf'(x)  d h t vM A ow tt f'(c) musl al exist
and ha ¢ vM A.
5,17 tf ntinus on [0, 1 ], f(0) = O,f'(x) fini for ch x in (0 1).  that
iff' is an ining function on (0, 1),  so t i ! fi  d by the ua-
tion () = f(x)]x.
5.1g Assume fh a i divative in (a, b) and is ntinuo on [a, b] with f(a)
f(b) = O. ov¢ thai for  r!  Ih¢ is so c in (a, b) such that f'(c) =
Hint. Apply RoH¢ Ih to O(x)f(x) for a ible  dewing on .
$.19 Assumef is ntinuous on [a, b] d  a finite ond derivatiw ff in the
itl (a, b). Aum¢ that the line m joining Ih¢ ints A  (a,f(a))
B = (b,f(b)) inters the ph of fin a third int P differ from A and B.
lhatf'(c) = 0 for m¢ c in (a,

5.20 If/has a finite third ckrivativef" in [a, b] and if
f(a) ffi f(a) = f(b) -- f'(b) ffi O,
prove thal f'(¢) ffi 0 for some c in (a, b).
5.1 Assurr f is ntive and has a finite third dm'ivative f" in the open intea'val
(0, l). If f (x) = 0 for at least two values of x in (0, I), prove that f'(c) = 0 for some c
in (0, l).
 Assume fhas a finite dcrlvative in some intetwal (a, -t o).
a) If f (x) -* I and f'(x)  c as x - + co, prove that c -- 0.
b) If f '(x) -, 1 as x  + , prove tlmt f(x)/x -* I as x - + .
c) If f '(x)  0 as x  + oo, prove that f(x)/x -} 0 as x - + o.
£3 Let k be a fixed positive nunxber. Show that th¢ is no function f satisfying the
following thr conditions: f(x) exists for x >- O, f'(O) = O, f'(x) > h for x > 0.
K3A If/ > 0 mad if f "(x) exists (and is firfitc) for every x in (a - /t, a + h), and fffis
continuous on [a - /, a + hi, show that ,,ve have:
a) f(a + ) - f(a - h) = f(a + Oh) + $'(a - 0k), 0 < # < 1 ;
b) ]'(a + 4) - 2.f(a) + .f(a - h) _-p(a + ih) -P(a - ), 0 < A < 1.
c) Ill(a) exists, show that
f(a) = Iim f(a + h) - 2f(a) + f(a- h)
/tO h 2
d) Give an exaznpIc where the limit of the quotient in (c) exists but wheref'(a) does
not exist.
5.2 Let f have a finite derivative n (a, b) and assume that c  (a, b). Consider the
following condition: For every t > 0 there exists a l-bail B(c; d), whose radius  depends
only on • and not on c, such that if x  B(c; ), and  # c theft
t
Show lhat" i conliuous o (, ) if thi coditb holds throughout (a, ).
 Assume f has a finil¢ dvative in (a, b) and is continuous on [a, hi, with a
f(x) _< b for all x in [a, b ] and If'(x)l -< " < 1 for all x in (a, b). Prove that f has a
unique fied point in [a. b].
$.27 Give an emimple of a pair of functions/and g having finite dedivativc, in (0, 1),
such that
lira f(x) = 0,
but such thai [imof'(x)lg'(x ) does not exist, choosing g so that g'(x) is never zero.
125
5.28 Prove the following theom:
Lez f nd g be two functionz haeing finite nth derivatives in (a, b). For ome iterior point c
in (a, b), sura¢ that f(c) = f'(c) ..... f('-)(c) = O, and that g(c) = g'(c) ....
= g¢'-)(c) - O, but thatg((x) is never zero in (, b). Show
wx; is intto tinljoinigxandc.  1 -  = (x - Xl)/(x - c), Shw
tt 0 <  < 1 d  t fong fo of the ainder m (d  uchy):
(1
c)')[ + 0 - )c.
(n- I)Z
Hint. T t)  g(t) = t in  f of 
Veet'-valeml
$.30 If a vector.valued function f is differentiable at c, prove that
f'(c) ffi lira t_ [[(¢ + h) - f(e)].
Conversely, if this limit exists, prove that f is differentiable at c.
.31 A vector-cabled function f is differentiable at each point of (a, b) and has constant
norm I[f[- Prove ,hat f(0"
532 A vector-valued function f is never zero and has a derivative
continuous on R. If there is a real function ,1 such that f'(t) -- ,.(t)f(t) for all t, prove
that there is  positive tea| funclion  and a constant vector e such that f(t)  u(t)c
for all t.
Partial derivatlv¢
£33 Consider the function fdcfined on R 2 by the following fotrnulas:
f(x, y) -- xy if (x, y) # (0, 0) f(O, 0) = 0.
xz+y 2
Prove that the partial derivatives Dlf(x, y) and D2f(x., y) exist for every (x, y) in R  and
evaluate these derivatives explicitly in terms of x and y. AIso, show that f is not con-
tinuous at (0, 0).

5,34 Letfbe defined on R 2 a follows:
f(x, ¥) -- y if (x, y)  (0, 0), f(0, 0) -- 0.
Compute the first- and second.order partial dm-ivatives of fat the origin, when they exist.
-valuetl fmcfims
-- Let S be an open set in C ad let $* be the et of complex eongates z-, where z e S.
If f is defined on S, defincg on S* as follows: tT(zT) -- )r, tle complex conjugate off(z).
lffis differentiable at c prove that  is differenlhble at ff and tlat g'() = f--r.
5.36 i) In ¢ac.h of the following tarnples write f = u +  and find explicit formulas
for u(x, .v) and v(x, y):
a) f(z) :- sin z,
c) fO) = Izl,
e) f(z) = arg z (z ¢: 0).
g)/0) = e:,
b) £(z)= cc.z,
f) £(z) = r_g z (z  o),
h) f() = z" (at complex, z # 0).
(These functions are to 1 defined as indieattl in Chapter 1.)
ii) Show that a and  mtisfy the hy-Rnn tions for t following l
of z: 1 z in (ak ), ); no z in (¢), (d), (¢); all z pt 1 z  0 in (f), (h).
(In  (h), the hy-Rann uati hold for all z if • is a nonne@ti
Mt, d t hold for  z # 0 if a is a ti in.)
iii) Compute e diti/z) M (a), (b), (0, ), (h), uming it exist.
 Wtef  u + ieand uttfh a ti at eh int of an o di D
 at (0, 0). If au 2 + be 2 is co.ant on D for  i a and b, not th O, pro
tt/is nsmnt on D.
SUGG REFERFACES FOR FURTHER STUDY
5.1 Apostol, T. M., CaicMus, Voi. 1, 2nd ed. Xerox, Waltham, i967.
5.2 Chaundy, T. W., The Differential Calculus. Clarendon Press, Oxford, 1935.
CHAPTER 6
FUNCTIONS OF
BOUNDED VARIATION AND
RECTIFIABLE CURVES
6.1 INTRODUCTION
Some of the basic properties of monotonic functions were derived in Chapter 4.
This brief chapter discusses functions of bounded variation, a class of functions
closely related to monotonic functions. We shall find that these functions are
intimately connected with curves having finite arc length (rectifiable curves). They
also play a role in the theory of Riemann-Stieltjes integration which is developed
in the next chapter.
6.2 PROPERTIES OF MONOTONIC FUNCTIONS
Theorem 6.1. Let f be an increasing function defined on [a, b] and let xo, x,, . . . , x.
be n + 1 points such that
a = x < x I < x < "'" < x n = b.
Then we have the inequality
 If(x,+) -- f(x,--)] _ f(b)  f(a).
/--1
Proof. Assume that y (x. x+). For 1  k  n - I, wehavef(x+) <f(YD
and f(yt_)  f(x-), so that f(x+) - f(x-) <_ f(y) - f(Yt-t). If we add
these inequalities, the sum on the right telescopes to f(y,_) -f(Yo). Since
f(Y- ) - f(Yo) < f(b) - f(a), this completes the proof.
The differencef(x+) - f(x-) is, of course, the jump of fat x. The fore-
going theorem tells us that for every finite collection of points x in (a, b), the sum
of the jumps at these points is always bounded byf(b) - f(a). This result can be
used to prove the following theorem.
Theorem 6,2. lf f is monotonic on [a, b], then the set of discontinuities off is
countable.
Proof. AsSume thatfis increasing and let S, be the set of points in (a, b) t which
the jump off exceeds I/m, m > 0. Ifx < xz < "'" < x,_t re in S,, Theorem
6.1 tells us that
.... <_ f(b)  f(a).
127

This means that S. must be a finite set. But the set of discontinuities of fin (a, b)
is a subset of the union U==t S. and hence is countable. (lffis decreasing, the
argument carl b¢ applied to -f)
FUNCTIONS OF BOUNDED VARIATION
De 6.t. If Is, b] i a compact intert, a set of points
P = {xo, x .... ,
satisfying the inqualities
s called a partition of In, b'L The ik-termi Ix,_ ,, x,] is called the kth subinterval
of P and we write Ax t = xt -- x_ ,, so that , Ax = b -- a. The collection
of all poffble partitions of In, b] will be denoted by [a, I,].
Defudtion6.4o Let f be defined on In, hi. lf P = {xo, x, ..., x,} is a partgtion
of [a, b'], write AA = f(xO  f(x_ ), for k = 1, 2,..., n. If there exists a
positive number M such that
 IAftl  M
for all partitions of [a, b], then f i said to be of bounded variation on In, hi.
Examples of functions of bounded variation are provided by the next two
Theorem 6.5. If f is monotonic on In, b'L then f is of bounded riation on In, b].
Proof Let f be inereasing. Then for every partition of In, b] we have A A > 0
and hence
t-I /t=l
Theorem 6.6. lf f is continuous on Is, b] and if f' exists and is bounded in the
interior, ay If'(x)l < A for all x in (a, b), then f is of bounded variation on
Proof. Applying the Mean-Value Theorem, we have
Af = f(x) - f(x_ ) = f'(tt)(xt - x_ ,),
This implies
where t • (x_ , x),
 IAfl =  l/'(tt)l Ax, <_ A  Ax = A(b - a).
I=1 k-----I g=l
Theorem 6.7. If f is of bounded variation on [a, hi, say , taft] <_ M for alt par-
titions of[a, b], then f is bounded on [a, hi. In fact,
If(x)l -< I.f(a)l + M for aft x in In, hi.
6.11 Ttal Variation
129
Proof. Assume that x e (a, b). Using the special partition P ffi {a, x, b}, we find
if(x)- f(a)I + If(b)- f(x)l -< M.
This implies If(x) -f(a)l-< M, If(x)t _< If(a)l + M. The same inequality holds
ifx = aorx  b.
1. It is easy to construct a continuous function which is not of bounded variation. For
example, letf(x) = x cos {/(7.x)}ifx # tl, f(0) = 0. Then /is condnuous on [0. !],
but if we consider the partition into 2n subintervals
1 [ ! 1 1}
P= 0'2n' 2n- 1"""3' 2'
an easy calculation shows that we haw
±n 1 1 1 1 1 t 1 1
This is not boullde6 for all n, since the series ,.,.=, (l/n) diverges. In this example
the dexivativef" exist in (0, I) buff' is not bounded on (0, 1). Howeved, f' is bounded
on any compact interval not containing tl origin and henc, f will be of boundod
variation on such an intexval.
An exampi similar to the first is give by f(x) = x  cos (1]x if x # 0, f(0) = O.
This f is of bounded variation on [O, 1 ], sino f' i bounded on [O, 1 ]. In fact,
f'(0) -- O and, for x # O, f'(x) = sin (1]x) + 2x ¢xs (l]x), o that lf'(x)[ : 3 for
a|l x in [0, I ].
Bounde.dness of/" is not neceary for fro be of bounded variation. For example, let
f(x) = x ta. This function is monotonic t'md 1¢ of bounded variation) on every
finite interval. However, f'(x) -* + m as x  O.
6.4 TOTAL VARIATION
Defudtio 6,& Let f be of bounded tmriation on I-a, hi, and let . (P) denote lhe xurn
X-- IAAI corresponding to the partition P ffi {xo, xt, .., x,} of Is, hi. The
number
VAn, b) = sup {Z (P): e
is called the total variation off on the interval In, b].
orts. When there is no danger of misunderstanding, we will write V- instead of
V,(a, b).
Sincefis of bounded variation on [a, b'], the number V, is finite. Also, V! > O,
since each sum  (P) _> O. Moreover, V.e(a, b) -- 0 if, and only if, f is constant
on In, b].

Tkem, em 6.9. Asume that f and g are each of bounded oariation on (.a, b]. Then
so are their sum, difference, and product. Also, we ha
v,  v + v,  v.  v s+
where
A  sup {Ig(x)l : x  [a, b]}, B : sup {If(x)l : x  [a, b]}.
Prf  x) = f(x)g(x). For eve paition P of [a, b], we have
Ts impli that h is ofund varation and tt V  AV t + B. The proofs
for the sum d diffence are simpler and will  omitS.
. Quotients we not includ in t forgoing theom u e rprl
ofa funon of bounded fiation n not  o und vation. For emple,
ill(x)  0  x  x0, then l/f will not  und on any inteaI nning xo
d (by m 6. 1 ]fcnot  of bounded variation on such an instal. To
exd eom 6.9 to quorts, it suffi to exclude functiom whose values
ome arbitrly clo to ro.
Torem 6dO. t f be of bounded riation on [a, b] d me that fix unded
away om zero; that is, uppase that there exists a sit nr m such that
0 < m  If(x) for aft x in [a, b]. Tn 0 : l/f is alm of ud variation on
[, ], a   Vlm .
Proof.
6.5 ADDITIVE PROPERTY OF TOTAL VARIATION
In the last two theorems the interval [a, b] was kept fixed and V.(a, b) was con-
sidered as a function off If we keep f fixed and study the totaI variation as a
function of the interval [a, b], we can prove_ the following additive property.
Theorem 6.11. Let f be of bounded variation on [a, b], and assume that c • (a, b).
Then f is of bounded variation on [a, c] and on [c, b] and we have
Vefa, b) = vAa, c) + vAe, b).
Proof. We first prove thatfis of bounded variation on [a, c] and on [e, b]. Let
P1 be a partition of [a, c] and let P2 be a parlition of [c, b]. Then P0 - Pt u P2
is a partition of [a, b]. If Y (P) denotes the sere Y. IAff corresponding to the
partition P (of the appropriate interval}, we can write
Th, tZ Telal Vtdllt  [ x] l • lttaleo f x 131
This shows that each sum  (P) and  (P2) is bounded by Vy(a, b) and this means
thatfis of bounded variation on [a, el and on [c, b]. From (1)we also ob-.ain the
inequality
V(a, e) + VAe, b )  V(a, b),
because of Theorem I.I 5.
To obtain the reverse inequafity, let P  {xo, x,..., x,} • '[a, b] and let
Po -- P u {c} be the (possibly new) partition obtained by adjoining the point c.
If c  [xt_ , xt], then we have
lf(x) --f(xt-0l < If(x) - .f(c)l + If(c) -
and hence E (P) < E (Po). Now  poims of Po in [a, C] determine a partition
Pt of [a, c] and those in [e, b] determine a partition Pz of [c, b]. The corre-
sponding sums for all these partitions are connoted by the relation
Z (e) < E (Yo) = E (P,) + Z (e9 <- VAa, c) +VAc, b).
Therefore, V/(a, c) + V/(c, b) is an upper bound for every sum Z (P). Since this
cannot be smaller than the least upper bound, we must have
v(a, )  vt(a, c) +vAc, b),
and this completes the proof.
6 TOTAL VARIATION ON la, x] AS A FUNCTION OF x
Now we keep the function land the left endpoint of the interval fixed and study
the total variation as a function of the right endpoint. The additive property
implies important consequences for this function.
l"keorem 6.12. Lei f be of bounded variation on [a, b]. Let V be defined on [a, b]
as follows: V(x) = Vl(a , x)ira < x <_ b, V(a) = O. Then:
i) V is an increasing function on [a, b],
it) V-f is an increasing function on [a, b]-
Proof. If a < x < y < b, we can write Vy(a, y) = Vy(a, x) + Vy(x, y). This
implies Y(y) - V(x) = V(x, y) >_ 0. Hence V(x)  V(y), and (i) holds.
To prove (il'), let D(x) = It'(x) - f(x) if x • [a, b]. Then, if a < x < y _< b,
we have
D(y) - D(x)  V(y) - V(x) - If(y) -/(x)] = Vi(x, y) -- If(y) -
But from the definition of V(x, y) it follows that we have
f(y) - f(x) < V:(x, y).
This means that D(y)  D(x) >_ O, and (it) holds.
OTE. For some functions f, the total variation V/( a, x) can be expressed as an
integral. (See Exercise 7.20.)

6.7 FUNCTIONS OF BOUNDED VARIATION EXPRF_,,ED AS THE
DIFFERENCE OF INCREASING FUNCI'IONS
The following simple and elegant characterization of functions of bounded varia-
tion is a consequence of Theorem 6.12.
]Twerem 6.13. Let f be defined on [a, b]. Then f is of bounded r¢n'iation on [a, b]
if, and only if, f can be expressed as the difference of two increasing functions.
Proof lff is of bounded variation on [a, hi, we can write f = V - D, where
Vis the function of Theorem 6.12 and D = V --f Both Vand D are increasing
functions on [a, hi.
The converse follows at once from Theorems 6.5 and 6.9.
The representation of a function of bounded variation as a difference of two
increasing functions is by no means unique, lff = ft - f2, where f, and f2 are
increasing, we also have f = (f, + 9')  ffz + g), where g is an arbitrary in-
creasing function, and we get a new representation off if # is strictly increasing,
the same will be true offt + g andfz + 9. Therefore, Theorem 6.13 also holds
if "'increasing" is replaced by "strictly increasing."
6,8 CONTINUOUS FLTNCTIONS OF BOUNDED VARIATION
Tlorem 6.14. Let f be of bounded variation on [a, hi. If x  (a, hi, let V(x)
Vr(a, x) and put V(a) --- O. Then every point of continuity off is also a point of
continuity of V. The converse is also true.
Proof. Since V is monotonic, the right- and lefthand limits V(x+) and
exist for each point x in (a, b). Because of Theorem 6.I3, the same is true of
f(x + ) and f(x- ).
If a < x < y < b, then we have [by definition of V/(x, y)]
0 _< If(y) -f(x)] < V(y) - V(x).
Letting y --, x, we find
0 ._< If(x+) -f(x)l _< V(x+)- v(x).
Similarly, 0 < if(x) -f(x-)l < V(x) - V(x-). These inequalities imply that
a point of corttinuity of V is also a point of continuity off.
To prove the converse, letfbe continuous at the point c in (a, b). Then, given
e > 0, there exists a 6 > 0 such that 0 < Ix - cl < 6 implies If(x) - f(c)l <
For Ibis same e, there also exists a partition P of [c, hi, say
P= {xo, xt,.-.,x}, xo = c, x, = b,
such that
VAc, b) -  < IAAI,
=1
'rh. 6.11 Carves and Patlts 133
Adding more points to P can only increase the sum  IAAI and hence we can assume
that 0 < xt - xo < 6. This means that
IAft = If(xa) -f(e)l <
and the foregoing inequality now becomes
since {x, xz,..., x,} is a partition of [xt, hi. We therefore have
But
V.Ac, b) - b) <
0 <_ V(x,)- V(c)= VAa, xO- VAa, e)
= V.#, = v.,(,:, VA  ,. <
Hence we have shown that
0<x
implies 0 <_ V(x) - V(c) < ..
This proves that V(c + ) = V(c). A similar argument yields V(c-) = V(c). The
theorem is therefore proved for all interior points of [a, b]. (Trivial modifications
are needed for the endpoints.)
Combining Theorem 6.14 with 6.13, we can state
/'/werem 6.15, Let f be continuous on I-a, b]. Then f is of bounded variation on
[a, b] if, and only if, f can be expressed as the difference of two increasin 9 continuous
functions.
rcrm The theorem also holds if "increasing" is replaced by "strictly increasing."
Of course, discontinuities (if any) of a function of bounded variation must
be jump discontinuities because of Theorem 6.13. Moreover, Theorem 6.2 tells us
that they form a countable set.
6.9 CURVES AND PATHS
Let f: I-a, b] - R  be a vector-valued function, continuous on a compact interval
[a, b in R. As t runs through [a, b], the function values fit) trace out a set of
points in R" called the graph of f or the curve described by f. A curve is a compact
and connected subset of R" since it is the continuous image of a compact interval.
The function f itself is called a path.
It is often helpful to imagine a curve as being traced out by a moving particle.
The intercM ['a, b] is thought of as a time interval and the vector f(t) specifies the
position of the particle at time t. In this interpretation, the function f itself is
called a motion.

Different paths can trace out the same curve. For example, the two eomplex-
Valued functions
f(t) = e 2"", g(t) = e -z*', 0  t < 1,
eaJzh trae out the unit circle x a + y2 - l, but the points are visited in opposite
directions. The same circle is traced out five times by the function h(t) - e t°=",
0_<t<I.
6.10 RECTIFIABI PATHS AND ARC LENGTH
Next we introduce the concept of arc length of a urve. The idea is to approximate
the curve by inscribed polygons, a technique learned from ancient geometers. Our
intuition tells us that the length of any inscribed polygon should not exceed that
of the urve (since a straight line is the shortest path between two points), so the
length of a curve should be an upper bound to the lengtim of all in$cribed polygons.
Therefore, it seems natural to define the length of a curve to be the leat upper
bound of the lengths of all possible inscribed polygons.
For most cun that ari in practice, this gives a useful definition of arc
length. However, as we will  prently, there are curves for which there is no
upper bound to the lengths of the in-ribed polygons. Therefore, it becomes
necessary to classify c,,rves into two categories: those which have a length, and
those which do not. The former are called rectable, the latter nonrectable.
We now turn to a formal description of these ideas.
Let f: In, b] , R" be a path in R'. For any partition of In, b] given by
? = {to, q,..., t.},
the points f(to) , f(ti),... , f(t=) are the vertices of an inscribed polygon. (An
example is shown in Fig. 6. i .) The length of this polygon is denoted by At(P) and
is defined to be the sum
At(P) =  [[fftD - f(tt-011.
t=l
Delta/t/oR 6,16. If the set all, umbers A(P) is bounded for all partffiona P of In, b],
then the path f is said to be rectifiable and it, arc length, denoted by At(a, b), is
 fire)
defined by the equation
At(a, b) = sup {At(P): e
If the set of numbers At(P ) is unbounded, f is called nonrectifiable.
It is an easy matte to characterize all rectifiable curves.
Theorem 6.17, Consider a path f: In, b]  R" with components f = (J',... ,f).
Then f is rectifiable if, and only if, each component  is of bounded tariation on
[a, b]. If f is rectifiable, we have the inequalities
V,(a, b) < Aa, b) < V,(a, b) +... + V,(a, b), (k = l, 2,..., n), (2)
where V(a, b) denotes the total pariation of A on [a, hi.
Proof If P - {to, t .... , t=} is a partition of [a, b] we have
- _< at(v) _< -
i----I Iml
for each k. All assertions of the theorem follow easily from (3).
1. As noted earlier, the fun-lion given by f(x) = x cos {](2x)} for x # 0, f(0) -- 0,
is condnuou but not of bounded variation on [0, I ]. Therefore its graph is a non*
rectiabte curve.
Z It an be shown (Exerc.is 7.21) that if f' is ontinuous on In, hi, then f is rectifiable
and its arc length can be eXleSsed a an integral,
At(a, b) ffi f' Jlf'(t) ll dr.
6.11 ADDITIVE AND CONTINUITY. PROPERTIES OF ARC LENGTH
Let f = (ft,-.., fo) be a rectifiable path defined on In, hi. Then each component
f is of bounded variation on every subinterval [x, y] of In, b]. In this section we
keep f fixed and study the are length At(x, y) as a function of the.interval Ix,
First we prove an additive property.
Theorem 6,1&. If c ¢ (a, b) we have
A  a, b) = At(a, e) + b).
Proof. Adjoining the point c to a partition P of [a, b], we get a partition P of
[a, c] and a partition Pz of [c, b] such that
AKe) <_ At(e,) + At(e,) _< At(a, e) + &(c,
Tkis implies /if(a, b)  At(a, c) + At(c , b). To obtain the revm' inequality, let
Pt and P be arbitrary partitions of t'a, c] and It, b], respectively. Then
P= P uPz,

is a partition of [a, b for which we have
+ At(e2) = At(e) _< N(a, b).
Since the supremum of all sums A(Pt) + At(P2) is the sum A,(a, c) + At(c, b)
(see Theorem 1.15), the theorem follows.
Tlor.m 6.19. Consider a rectifiable path f defined on In, b-[. If x • (a, b], let
s(x) = At(a, x) and let s(a) = O. Then we have:
i) The function s so defined is increasing and continuous on In, b].
ii) If there is no subinterval of [a, b] on which [ is constant, then s is strictly
creasing on In, b].
Proofi If a < x < y  b, Theorem 6.18 implies s(y) - s(x) = At(x, y) > O.
This proves that s is increasing on In, b]. Furthermore, we have s(y) - s(x) > 0
unless Adx, y) = 0. But, by inequality (2), At(x, y) -- 0 implies V(x, y) ffi 0 for
each k and tls, in turn, implies that f is constant on Ix, y]. Hence (ii) holds.
To prove that s is continuous, we use inequality (2) again to write
o s(y) - s(x) ffi a,(x, y) <_ v,(x,
If we let),  x, we find each term V(x, y) - 0 and hence s(x) ffi s(x+). Similarly,
s(x) = s(x-) and the proof is complete.
6.12 EQUIVALENCE OF PATHS. CHANGE OF PARAMETER
This section describes a class of paths having the same graph. Let f : [a, b]  R"
be a path in R*. Let u: [c, d]  In, b] be a reaLvalued function, continuous and
strictly monotonic on [c, d] with range In, b]. Then the composite function
g ffi fougivenby
g(t) ffi flu(Z)] for c _%< t < d,
i a path having the same graph as f. Two paths [ and g so related are called
equivalent. They are aid to provide different parametric representalions of the
same curve. The function u is said to define a chatje of parameter.
Let C denote the common graph of two equivalent paths f and g. If u is
strictly increasing, we say that f and g trace out C in the same direction. If u is
strictly decreasing, we say that f and g trace out C in opposite directions. In the
first case, u is said to be orientation-preserving; in the second case, orientation-
reversing.
Tkeorem 6.20, Let f : [a, b] , R" and g : [c, d] -, R" be two paths in R*, each
of which is one-to-one on its domain. Then f and g are equivalent , and oni), if, they
hae the same graph.
Proof. Equivalent paths necessarily have the same graph. To prove the converse,
assume that land g have the same graph. Since f is one-to-one and continuous on
the compact set I'd, b], Theorem 4.29 tells us that f- 1 exists and is continuous on
its graph. Define u(t) = f-x[g(t)] if t • It, all. Then u is continuous oa l-c, d]
and g(t) = fin(t)]. The reader can verify that u is strictly monotonic, and hence
f and g are equivalent paths.
EXERCISES
Ftmc/ioas of buded varialion
6.1 Determine which of the following functions arc of bounded vexiation on [0, 1 ].
a) f(x) = x 2 sin (I/x) if x # 0,f(0) = 0.
b) f(x) = / sin (I/x) ifx  0,f(0) = 0.
6.2 A function f, defined on [a, b ], is said to satisfy a uniform Lipschitz condition of
order a > 0 on In, b] if there erdsta a constant M > 0 such that If(x) - f(.v)] <
M]x- YI" for all x and y in [a, b]. (Compare with Exercise 5.1.)
a) fffis such a function, show that a > t impliefis cot-tant on In, b], whereas
t -- 1 impliesfis of bounded variation In, b].
b) Give an example of a function fsatisfying a uniform Lipschilz condition of order
 < 1 on [a, b] such that fis not of bounded variation on In, b].
¢) Give an example of a function f which is of bounded variation on [a, b] but
which satisfies no uniform Lipschitz condition on [a, b].
6.3 Show that a polynomial fis of bounded variation on evelT compact intetwal In, b]
Describe a method for finding the total variation off on In, b ] if the zeros of the derivative
f' are known.
6.4 A nonempty st S of real-valued functions defined on an interval In, b] is called a
linear space offunction if it has the following two properties:
a) Iff S.., then ere S for every real number c.
b) Iff Sandg S, thenf + g S.
Theorem 6.9 shows that the set Vof alI functions of bounded variation on In, b] is a linear
space. If S is any linear space which contains all monolonie functions on In, b], prove
that V _ S, This can be described by saying that the functions of boundext variation
form the smallest linear space containing all monotonic functions.
6 Let fbe a real-valued function defined on [0, 1] such that f(0) > O,[(x)  x for
all x, andf(x) _< f(y) whenever x _< y. Let A = {x :f(x) > x}. Prove that sup .4  A
and that f(1) > 1.
6.6 If f is defined everywhere in R , then f is said to be of bounded variation on
(- o, + a¢) if f is of bounded varialion on evexy finite interval and if there ¢xisls a positive
number Msuch that V.r(a, b) < Mfor all compact intervals In, b]. The total varialion of
/'on (- o, + o0) is then defined to be the sup of all numbers V.r(a, b), o < a < b <
+ o% and is denoted by V.e{- co, + c¢). Similar definitions apply to half-open infinite
intervals In, + co) and ( co, b].
a) Stale and prove theorems for the infinite interval (-co, + 0o) analogous to
Theorems 6.7, 6.9, 6.10, 6.11, and 6.12.

b) Show lhat Theo 6.5 i true for (- co, ÷ co) if "'monotonic" is replaced by
"bounded and monotonic." State and prov a milar modLrgtion of Theor¢
6.13.
6.7 Assume that fis of bounded variation on [a, b| and let
As usual, wTite
The numbers
are calIed respectively, the positive and negative variations of fan In, hi. For each x in
(a, hi, let V(x)
n(a) = 0. Show that we have:
b) 0 <_ p(x) < V(x) and 0 < n(x) <_ V(x).
c) p and n are increasing on In, b ].
d) f{x) -- f(a) + p(x) - n(x). Part (d) gives an akernative proof of Theorem 6.13.
e) 2(x)
f) Every poinl of continuity off is also a point of continuity alp and ofn.
6.8 Lctfand g be complex-valued functions defined as fol[ows:
f(t) = e  ift [0 1], (t)-- e 2 ff [0,2]
a)  tt fd g ha e se aph but e ot cquivt aording to the
dnition  ioa 6.12.
b) Pro that e Ih ofg is twi tt off
6.9 Let f  a ible path of Ih L deed on In, b], and s¢ that f  not
nsnt on any subinal of [, b], t s denote  arlch function v by
s(z) = a,x) ifa < x S b,s() = 0.
a) Prove that s- k and is continuous on [0, L].
b) fine ) = f[s-(t)] if t  [0, L] ad show that g is qu[[t [o f. S[n
f(t) = g[s(0], the funion g is said to pvi a rrenta1[on of the graph off
wi c tgth as .
6"10 t fd g  two l-vMu contuo functions of bod vMtion
on In, b], with 0 < f(z) < g(x) for  x in (a, b}, f(a) = g(a),
the mplex-u run,ion dned on the instal [, 2b - a]  folios:
139
a) Show that h ¢leua'ibe a rectiftable curve
b) Explain, by means of a sketch, the geometric relationship between.f,
c) Show that the set of points
s -- {(x, y) : a <_ x <_
is a resion in R 2 whose boundary is the 6trv F.
d) Let/'/be the complex-valued function defined on [a, 2b - a] as follows:
H(t) = t - ½Jig(t) - f(t)], ira _< t _< b,
n(t) = t + ½i[a(2t, - t) -fOb - 0], ifb  r  2 -_ a.
Show that H describes a rectifiable curve Fe which is the boundary of the region
so = {(x,y):a <_ x _< b, '(x) - a(x)
e) Show that So has the x-axis as a line of symmetry. Crhe region o is ealle the
symmtrization of S with repe to the
f) Show that the length of V doe not exceed the length of r.
Absokltely ¢omhmees fenetimm
A real-valued function fdefined on [a, b] is said to be absolutely continuous on [a, b] if
for evca-ye > 0thereisa$ > 0 suchthat
 If(b) -f(a)] < 
for every n disjoia open subintervals (at bt) of In, hi, n = l, 2,..., the sum of whose
lengths ,-_ x (b - a) is less Ihan $,
Absolutely continuous fttnctiots occur in the Lcbesgue theory of integration and
differentiation. The following exercises give some of their elementary properties.
6.11 Prove that every absolutely continuous function on [a b] is continuous and of
boundexl variation on In, b].
OT. There e×ist functions which are continuous and of bounded variation but not
absolutely continuous.
6.12 Prove that fis absolutely contituous if it salisfies a uniform Lipsch[tz condition of
order i on [a, hi. (See Exercise 6,2.)
6.13 If f and g are ablutely continuous on In, hi, prove that each of the following is
also: If[, cf(cconstant),f + g,f-fl; also fig ifgis bounded away from zero,
SUGGESTED CES FOR FURTltF_. STUDY
6.1 Apostot, T. M., Calculus, Vol. , 7.rid ed. Xerox, Walthasn, 1967.
6.2 Natan, I. P,, Theory of Functions o]" a Real Variable, VoL
Boron, translator. Ungax, New York, 1961.

CHAPTER 7
THE RIEMANN-STIELTJES INTEGRAL
7.1 IN'I'RODUCTION
Calculus deals principally with two geometric problems:finding the tangent line
to a curve, and finding the area of a region under a curve. The first is studied by a
limit proce known as differentiation; the second by another limit process
integration--to which we turn now.
The reader will recall from elementary ¢ateulus that to find the area of the
region under the gaph of a positive function f defined on ['a, b], we subdivide
the interval I'a, b] into a fmit number of subintervals, say n, the kth subinterval
having lengtk Axt, and we consider sums of the form z.= 1 f(t) Ax b where t t is
some point in tim kth subinterval. Such a sum is an approximation to tim area by
means of rectangles. If f is sufficiently well behaved in ['a, b]r, ontinuous, for
example--then there is some hope that these sums will tend to a limit as we let
n - co, making the successive subdivisions finer and finer. This, roughly speaking,
is what is involved in Riemann's definition of the definit¢ integral j f(x) dx. (A
procise danitionis given below,)
The two concepts, derivative and integral, arise in entirely differtmt ways and
it is a remarkable fact indeed that the two are intimately connected. If we consider
the definite integral of a continuous function f as a function of its uplg limit,
say we write
F(x) = f(t)
then F has a derivative and F'(x) = f(x). This important result shows that
differentiation and integration are, in a sense, inverse operations.
In this chapter we study the process of integration in some detail. Actually
we conskier a mor¢ general concept than that of Riemann: this is the Riemann-
Stieltjes integral, which invotvcs two functions f and t. Tim symbol for such an
integral is f(x)do(x), or something similar, and the usual Riemann integral
occurs as the spodal case in which ,(x) -- x. When t has a continuous derivative,
the definition is such that the Stieltjcs integral  f(x) d(x) becomes the Ricmann
integral _f(x) '(x) dx. However, the Sti¢ltjes integral still makes sense when t
is not differentiable or even when  is discontinuous. In fact, it is in dealing with
di, conanuous ,, that the importance of the Stieltjes integral becomes apparent. By
a suitable choice of a discontinuous t, any finite or infinite sum can he expressed
as a Stieltjes integral, and summation and ordinary Riemann integration then
140
become special cases of this more general process. Problems in physics which
involve mass distributions that arc partly discrete and partly confnuous cam also
be trgated by using Stieltjes integrals. In the mathematical theory of probability
this integral is a very useful tool that makes possible the simultaneous treatment
of continuous and discrete random variables.
In Omptcr 10 we discuss another generalization of the Ricmann integral
known as the Lebeague integral.
7.2 NOTATION
For brevity we make certain stipulations concerning notation and terminology to
b¢ uscgt in this chaptex. We shall b¢ working with a compact interval Ia, b] and,
unless otherwise stated, all functions denoted by f, g, a, fl, etc., will be assumed to
b¢ real-valued functions defined and bounded on In, b]. Complex-valued functions
axe dealt with in Section 7.27, and extensions to unbounded functions and infinite
intervals will be discussed in Chapter I0.
As in Chapter 6. a partition P of In, hi is a finite set of points, say
P = {xo, xt,..., x,},
such that a = x0 < xa <,"" < x,_ t < x= b. A partition P' of In, b.'] is said
to befiner than P (or a refinement of P) if P _ P', which we also write P'  P.
The symbol A denotes the difference &ui = (xt) - u(xt-t), so that
The set of all posdble partitions of [a, b] is denoted by [a, b].
The norm of a partition P is the Iength of the largest subinterval of P and is
denoted by liPI[. Note that
P'  P implies ItP'II < ilPII.
That is, refinement of a partition decreases its norm, but the converse does not
necessarily hold.
7.3 THE DEFINITION OF THE RIEMANN.-STIELTJES INTEGRAL
Defudtion 7.1. Let P =. {xo, x, . . . , x,} be a partition of In, b] and let t, be a
point in the subinterval [xt_ t, xi]. ,4 sum of the form
s(P, f, =
is called a Riemann-Stiettjes sum of f with respect to a. We say fia Riemann-
interable with respect to • on In, hi, and we write "re R(r 0 on In, b]'" /f there
exists a number A having the following property: For every g > 0, there exists a
partition P of In, b] such that for eoery partitien P finer than P and for every
choice of the point$ tt in [Xt_ l, X], we have [S(P,f, ,t) - Al < .

Wlgn suc a number A exists, it is uniquely determined and is denoted by
_fd or by ,f(x) d(x), We also say that the Ricnann-Stieljes ia'ra _fd
exists. The functions f and  are referred to as the/ntegrand and the integrator,
pectively. In the special ease when (x)ffi x, we write S(P,f) instead of
S(P,f, ), andf R instead off R(). Te integral is then called a Riemann
integral and is denoted by fdx or by f(x)dx. The numerical value of
[_f(x) d(x) depends only on f, , n, and b, and does not depend on the symbol x.
The letter x is a "dummy variable" and may be replaced by any other convenient
symbol.
Tr. This is one of several accepted definitions of the Riemann-Stieltjes integral.
An alternative (but not equivalent) definition is stated in Exercise 7.3.
7.4  PROPF_,RTII
It is an easy matter to prove that the integra! operates in a linear fashion on both
the integrand and the integrator. This is the context of the next two theorems.
Tkeoeem 7.2. lf f  R() and f g  R() on [a,b], then ctf + c2g e R( ) on
[a, b]. (for any wo constants ct and cD and we
Proof. Let h = cf + c,g. Given a partition P of [a, b], we can write
= cxS(P,f, z) + c,S(P, g,
Given n > O, choose P-" so that P  P implies [S(P,f, ) - J.fda[ .< , and
choo P" o that P  P[ im#i IS(P, a, a) -  g dl < *, If we take
P, = P' .P, then, for P finer than P. we have
and this proves the theorem.
Tkeorem 7.3. lf f  R(a) and f  R() on [a, hi, the f  R(c, + cz[Y) on [a, b]
(for any two constants c and cz) and we have
The poof is similar to that of Theorem 7.2 and is left as an exereise.
A result somewhat analogous to the prrvious two theorems tells us hat the
integral is also additive with respect to the interval of integration.
Tlorem 7.4. Assume that c  (a, b). If two of the three integrals in (1) exist, then
the third also exists and we have
fda+ffd=ffd. (I)
Proof. If P is a partition of [a, b] such that c e P, let
P' -- P ta [a, c] and P" = P  [c, b],
denote the corresponding partitions of [a, c] and It, b], respectively. The Rie-
mann-Sfieltjes sums for these partitions are connected by the equation
s(P,f, ) = s(P',f, ,) + s(e',f, ).
Assume that fd, and j',Sfd,, exist. Then, given e > 0, there is a partition
P', of [a, e] such that
Is(P"f" - 21d  l 2 whenever P" is ftner than P:,
and a partition P: of [e, b] such that
S(P"f'')-fdal<' 2 whenever P" is fmer than p:.
Then P, = P  P7 is a partition of [a, b] such that P finer than P, implies
P' _= P; and P _ P]. Hence, ifP is finer than P,, we can combine the foregoing
results to obtain the inequality
This proves that jfdt exists and eqeals fd + fd. The reader can easily
verify that a similar argument Voves the theorem in the remaining cases.
Using mathematical induction, we can prove a similar result for a decomposi-
tion of [a, b] into a finite number of subintervals.
ro'r. The preceding type of argument cannot be used to prove that the integral
j'_fda exists whenever _fd exists. The conclusion is correct, however. For
integrators a of bounded variation, this fact will later be proved in Theorem 7.25.
Definition 7.5. If a < b, we define jf d -- -.f d= whenever J,f d exists.
We also define  f du. = O.
• The equation in Theorem 7A can now be written as follows:
ff dt + ff du + f du - O.

7.5 1NTE, GRATION BY PARTS
A remarkable cormeetion exists between the integrand and the integrator in a
Riemanntieltjes integral. The existence of fdz implies the existence of
j'. • dr, and the converse is also true. Moreover, a very simple relation holds
between the two integrals.
Tlworem 7.6. If re g() on in, b], tlen   R(f) on in, b] and we hae
f2 f(x) dw(x) ÷ f2 (x) df(x) f f(b)(b) - f(a)a(a).
rcrr. This equation, which provides a kind of reciprocity law for the integral, is
known as the formula for integration by larts.
Proof. Let  > 0 be given. Since . f d exists, there is a partition P, of [a, b]
such that for every P' finer than P,, we have
Consider an arbitrary Riemann-Stieltjes sum for the integral j a dr, say
Subtracting the last two displayed equations, we find
The two sums on the right can be combined into a Single sum of the form S(P',f,
where P' is that partition of in, b] obtained by taking the points x and h together.
Then P' is finer than P and hence finer than P. Therefore the inequality (2) is
valid and Ibis means that we have
whenever P is finer than P. B,at this is exactly the statement that S  df exists
and equals A - J f d.
7,6 CHANGE OF VARIABLE IN A RIEMANN-STIELTJES INTEGRAL
Theorem 7.7. Let f  R(ot) on [a, b] and let g be a strictly monotonic continuous
function defined on an interval S having endpoints c and d. Assume that a
7.7 Reduction to a Riemam Intcllral 145
b = 9(d). Let h and fl be the composite functions defined as follows:
h(x) = f[.q(x)], ,6'(x) = a[a(x)], tf x e S.
Then h  R(ff) on Smd we haoe  f &t -  h dfl. That is,
=
Proof. For definiteness, assume that g is strictly increasing on S. (This implies
c < d.) Then g is one-to-one and has a strictly increasing, continuous inverseg -
deane on [a, Therefore, for every partition P = {Yo,---, Y,} of [c, d],
there corresponds one and only one partition P = {x0, ..., x} of [a, b] with
x, = g(Y0. In fact, we car write
P' = g(P) and P =
Furthermore, a refinement of P produees a corresponding refinement of P', and
the converse also holds.
If e > 0 is given, there is a partition P of [a, b] such that P' finer than
implies IS(P',f, ) - ,fdul < e. Let P = g-(P) be the corresponding par-
tition of re, d-J, and let P = {Yo,..., Y,) be a partition of re, arJ finer than P,.
Form a Riemann-Stieltjes sum
where u t[yt_,yt] and A/ t = fl(y)-- fl(Yt-t). If we put t t =#(us) and
x = #(Yt), then P' = {x0, ..., x.} is a partition of [a, b'] finer than P. Moreover,
we then have
S(P, h, [3) = E f[g(u,)]{cz[g(y,)] - a[O(Y,-,)]}
sinc t, • [x,_t, xL]- Therefore, IS(P, h, fl) -- fda I < e and the theorem is
proved.
o]. Thig theorem applies, in particular, to Riemann integrals, that is, when
g(x) = x. Aaother lheorem of this type, in which g is not required to b¢ mono-
tonic, will later b¢ proved for Riemann integrals. (S¢ Theorem 7.36.)
7.7 REDUCTION TO A NN INTEGRAL
The next theorem tells us that we are permitted to replace the symbol d(x) by
t'(x) dx in the integral f(x) &z(x) whenever u has a continuous derivative '.

l"lteorcm 7.8. Assume f  R() on in, b] and assume that  ha a continuous
derivative ¢," on [a, b]. Then the Pdema, n integral _ f(x)a'(x) dx exists and we have
Proof Let g(x) -- f(x)a'(x) and consider a Riemann sum
The same rfition P and the e choi of e t n  used to fo e
RiemanSfiettjes sum
Atn8 e Mon-Vue eomm, we  te
d hence
Sincefis bounded, we have If(x)l _< M for all x in in, b], where M > 0. Con-
tinuity of t' on in, b] implies uniform continuity on in, b-]. Hence, if ¢ > 0 is
given, there exists a # > 0 (depending only on ) such that
0  Ix --Yl <  implies lac'(x) - '(.v)l <
2M(b - a)"
If we take a partition P,' with norm [I P'¢]I < , then for any finer partition P we
will have la'(vt)  g'(tt)l < /[2M(b - a)] in the preceding equation. For such
P we therefore have
ls(P,f, - s(e, a)l < -.
2
On the other hand, since fe R(r 0 on ['a, hi, there exists a partition P;' such that
P finer than P7 .implies
Combining the last two inequalities, we see that when P is finer than P, = P: w P',
we will have IS(P, ) - fdl < ,, and this proves the theorem.
NOTE. A stronger result not requiring continuity of ct' is proved in Theorem 7.35.
7.8 STEP FUNC"IONS AS INTEGRATOIL
If  is constant throughout in, b'l, the integral ,fd exists and has the value 0,
since each sum S(P,f, t) = 0. However, if • is constant except for a jump dis-
continuity at one point, the integral ,fda need not exist and, if it does exist, its
value need not be zero. The situation is described more fully in the following
theorem:
Theorem 7.9. Given a < c < b. Define ct on in, b] as follows: The values
a(c), a(b) are arbitrary;
• (x) -- (a) if a < x < .e,
and
a(x) = a(b) if c < x < b.
Let f be defined on [a, b] in such a way that at least one of the functions f or
continuous frora the left at c and at least one is continuous frora the right at c. Then
f  R() on [a, b] and we haee
f &, = f(c)[(c + ) -- a(e- )].
NOTE. The result also holds if c = a, provided that we write ,(e) for (c--), and
it holds for c = b if we write a(c) for r,(c+). We will prove later (Theorem 7.29)
that the integral does not exist if both faad  are discontinuous from the right or
from the left at c.
Proof Ifc • P, every term in the sum S(P, f, ) is zero except the two terms arising
from the subinterval separated by c, say
S(?,f ) = f(tt_t)[c(c) - (c-)] +/(it)in(c+) - a(c)],
where t_  < c < tv This equation can also be written as follows:
= if(t,_,) - + [f0O -
where A = S(P,f. ) - f(c)[a(c+) - (c-)]. Hence we have
IAI s: If(t-)  f(c)[ I  (c) - + If(t) - f(c)i -
lffis continuous at c, for every e > 0 there is a fi > 0 such that [IPJt < ,5 impfies
If(h- ) - f(c)l <  and If(tO - f(c)l < .
In this case, we obtain the inequality
IAI < ela(c) - + ela(c+) -
But this inequality holds whether or notfis continuous at c. For example, if f is
discontinuous both from the right and from the left at c, then (c) = ,,(c-) and
t(c) = t(c+) and we get A = 0. On the other hand, iffis continuous from the
left and discontinuous from the right at c, we must have -(c) = 0t(c+) and we get

14 The Riemmm-ieRks lntegr4d Def. 7.10
IAI -< elf(c) - (c-)l. Similarly, iff is continuous from the right and discon-
tinuous from the left at c, we have c) = a(c-) and IAI -< lm(c+) -
Hence the last displayed inequality holds in every case. This proves the theorem.
F_mmg. Theorem 7.9 tells us that the valu ofa Ri©mann-$tieltjes integral can be altered
by changing the value of f at a single point. The following example show that the
existence of the integral can also be affected by such a cheng¢. Let
• (x) -- O, if x ¢: O, (0) = -1,
f(x)= 1, • if-I _< x_< +i.
In this case Theorem 7.9 implies J'-i f&t = 0. But if we finefso thatf(0) = 2 and
f(x) = 1 ifx # 0, we can easily see that t t fak will not exist. In fact, when P is a par-
tition which includes 0 as a point of subdivision, we find
= f(t,) -
where x_2 -< t_l -< 0 _< t _< xt. The value of Ihis sum isO, 1,or -1, depending on
the choice of tt and t_ x. Hence, jq_ f&t does not exist in this case. However, in a
Riemann integral j' f(x) dx, the values of f can be changl at a finite number of points
without affecting either the existence or the value of the integral. To prove lhis, it sutfcs
to consider the case where f(x) = 0 for all x in in, b] except for one point, say x = c.
Bet for such a function it is obvious that [S(P,f) I < I/<c)l WII. Singe I1  $ can be made
arbitrarily small, it follows that j f(x) dx  O.
REDUCTION OF A RIEMANN-STIELTJES INTEGRAL TO A FINITE SUM
The integrator  in Theorem 7.9 is a special case of an important class of functions
known as step functions. These arc functions which are constant throughout an
• interval except for a finite number of jump discontinuities.
Definition 7.10 (Step function). A funclion • defined on in, b) is caHed a step fimction
if there is o partition
such that ct is constant on each open sub#tervai (xt_ , x). The member =(x +) -
ct(x-) is called the jump at x if I < g < n. The jump at x is (x x +) - (x).
and the jump at x. is cffx.) - (x.-).
Step functions provide the connecting link between Riemann-Stieltjes integrals
artd finite sums:
Theorem 7.11 (Redwtion of a Riemana-Stieltjes inte! to a finite sum). Let  be
a step function definedon in, b] with jump  at xs, where xt,..., xn are as described
in Definition 7.10. Let f be defined on [a, b] in such a way that not both fond  are
discontinuous from the right or from the left at each x. Then , f da exists and we
have
f(x) da(x) = 
Proof. By Theorem 7.4, fd can be written as a sum of integrals of the type
considered in Theorem 7.9.
One of the simplest step functions is the greatest-integer function. Its valt at
x is the greatest integer which is less than or equal to x and is denoted by ix].
Thus, ix] is the unique integer satisfying the inequalities ix] < x < ix] + 1.
Theorem 7,12. Every finite sum can be written as a IO'emann-Stieltjea integral. In
fact, gven a sum .-_ a t, define fan [0, n] as folfows:
f(x) = a /fk-- 1 <x< k (k = 1,2..,.,n), f(0) =0.
where ix] is the greatest integer <_ x.
Proof. The greatest-integer function is a step function, continuous from the right
and having jump 1 at each integer. The function f is continuous from the left at
1, 2 .... , n. Now apply Theorem 7.
7.10 E--qJLERS SUMMATION FORMULA
We shall illustrate the use of Riemann-Stieltjes integrals by deriving a remarkable
formula known as Eulbr's summation formula, which relates the integral of a
function over an interval [a, b] with the sum of the function values at the integers
in [a, b]. It ean sometimes be used to approximate integrals by sums or, conversely,
to estimate the values of certain sums by means of integrals.
l"keort, m 7.13 (Euler's summation formula). If f has a continuous derivatite f' on
[a, b], then we have
 f(n)= fbf(x)dx+ f'f'(x)((x))dx+ f(aX(a))-/(bX(b)) ,
where ((x)) = x - Ix]. When a and b inteoers, this becomes
NOT. V__.<. means the sum from n =

150 "Ehre Rte'rnStielljea i[n|egat lh. 7A3
Proq£ Ap.ptyirg Theorem 7.6 (integration by parts), we have
Since tl:te grea{est*iteger function ha_ unit jumps at {he integers
[a + 2,..., [b]., we can: write
If we combine tJhis with the previous equa(io, fhe theorem %flows a. once.
%11 MONOTONICALLY tNCASING INTIEGRATORS. UPPER AND
LOWER ILN'[EGRALS
The further theory of Riemann-Stieltjes inegra6or will now be developed fm
moaotonically increasing inte.graors, and we shaII see later (in Teorem 7.24) hat
f;ar many p*arposes this is )u:st as general as studying he theory for integrawrs which
are of bounded variation.
When  is increasing, the difli:rences A which appear i the Riemann-
Stieltjes sms are aH nonnegative This simple fact plays a vital role in the develop-
ment of the teory. For brevity, we shall use the abbreviation
mean. thin ' is increasing on
As stated earlier, to find {he area of the region under the graph of a function
fwe consider Riemane sums f(q) Ax as approximations to lhe area by means
of reca.ng]es. Such sums also arise qie naturally in certain physica] problems
requiring the use of integration for their solution. Another approach o these
problems is by means ofuRper and low,or Riemann sums. For example, in the case
of areas, we can consider approximations from "°above" and from "below'" by
means of the surf, s M, kx ad m kx> where M: and m,: denote, respectively,
the sup and iM of he thnction vales in the kth subinterval Our geometric
intuition elIs us *haf the upper sums are a least as b{g as the area we seek, whereas
• {he Iower s,ams canm exceed this area, (See. Fg. 7.t.) Therefore it seems natura
Th. %15
15i
o ask: What is the smalesg possiNe value o.f the upper sums? Th:is leads s to
consider the inf of all upper .stms. a number called the upper ig<gra[ ofji The
hwer itegrat is similarly defined to >e the sup of at/lower sams. For reasorabk
funcfion. (for example, continuous hmctions) both these in.tegrals will be equal to
.[f(.x) dx. However, in general, these integrals wiII be different ad it becomes an
important proNem to :find conditio.es on t1e function .hich will. ensure ihat the
upper and Iower integrals will be the same. We now diuss this ,ype of problem
br Riemann-Stiettjes imegrals.
Definition 7.14.
Let P be a partition of [a b] and &t
M(f} -.-.. sup {.f(x) :x e [x._.,, x?},
and .L(P...g r*). = _ mdf )
are called, re,Epectively, the upper and lower S*ieltjes sums off with respect w
the partition P
oT. We always have mdf}  M(J) If   on [G hi, then Ae  0 ad we
ca also write m.(f) A  .M(f) Ae, from whicg it tIlows that the ower sums
do not exceed the upr sums. Furthermore, f t s [&_ , x], then
m(f) < f(tD 
Therefore, when , ¢, we have the inequalities
relating the upr and lower s.ms to the Riema--gtie]0es sums. The in.equali-
ties, which are frequently used in the material that fNows, do not necessarily hold
when  is not an increasing function.
e next theorem shows that, tr icreasigg  refinement of te partition
increases he lower sums and deereas the upr sums,
Tkeorem L15. Assume ghag   on [a, b].
i) [f P' is finer than P, we have
ii) [br any two partitions P, and P, we hm.e

Proof. It suffices to prove (i) when P' contains exactly one more point than P,
say the point c. If c is in the ith suhintereal of P, we can write
u(e',f, e) = , M,(f) a + M'[(c)- <x-l)] + r°[(x) --
where M' and M" denote the sup off in ix.._1, c] and ['c, xi]. But, since
M'  M,(f) and M" ___ M(f),
we have U(P', f, ) < U(P, f, ct). (The incqtudity for lower sums is proved in a
similar fashion.)
To prove (ii), let P = P u P2- Then we have
.(P,f, ) < .(P,f, )  U(P,f, ) < u(P,f, a).
OT. It follows from this theorem that we also have (for increasing n)
role(b) - e(a)]  Z(P,,f, ) _< V(P,f, e)  M[e(b) - (a)],
where M and rn denote the sup and inf of fan in, b].
Definition 7.16. Assunu that • , on [a, b]. The upper Stieltjes integral off with
respect to , is defined as follows:
The lower Stieltjes integral is similarly defined:
f de : sup {t_(e,y, ): P  [, ]}.
lqOT. We sometimes write l(_f, e) and _/(f, (z) for the upper and lower integrals.
In the special case where e(x) = x, the upper and lower sums are denotod by
U(P,f) and L(P,f) and are ealIed upper and lowcr Riemann sums. The corre-
sponding integra denoted by f(x) dx and by f(x) dx, are called upper and
lower Riemann integrals. They were first introduced by J. G. Darboux (1875).
Theorem 7.17. ¢ssume that  on in, b]. Then !(f, ') < i(f, ).
Proof. If ¢ > 0 is given, there exists a partition P such that
u(P,, f, ) < l(f, ) + e.
By Theorem 7.15, it follows that i(f, e) + e is an upper bound to at! lower sums
L(P,f, a). Hence, _/(f, e)  l(f, e) + , and, since e is arbitrary, this implies
!(f, a) <
ErmlPl¢. It is easy to give an example in which !(f, ) < i(f, ). Let t() = x and
define f on [0, I ] as follows:
f(x) = !, if x is rational, f(x) = 0, if x is irrational.
7.19 Rlemann's Condiliem 153
Then for every partition P of [0, 1 ], we have M(j') = 1 and m(f) = 0, since every
subinterval contains both rational and irrational numbers. Therefore, U(P, f) = 1 and
L(P,f') = 0 for all P. It follows that we have, for [a,b] = [0, 1],
f dx = ! and [ ) f dx = o.
Observe that the same result holds if f (x) = 0 when x is rational, andf(x) = 1 when x is
irrational.
/.12 ADDITIVE AND L1NEARITY PROPER]ES OF UPPER AND
LOWER INTEGRALS
Upper and lower integrals share many of the properties of the integral. For
ample, we have
if a < c < b, and the same equation holds for lower integrals. However, certain
equations which hold for integrals must be replaced by inequalities when they are
stated for upper and lower integrals. For example, we hve
y] (f + #)de <_ Ida + ode,
and
These remarks ear[ be easily verified by the reader. (See Exercise 7.11.)
7,13 RIF_.MANN'S CONDITION
If we are to expect equality of the upper and lower integrals, then we must also
expect the upper sums to become arbitrarily dose to the lower sums Hence it
seems reasonable to seek those functions f for which the difference U(P, f, e) -
L(P,f, e) can be made arbitrarily small.
Definition 7.18. We say that f satisfies Rieraann's condition with respect to  on
in, b] if, for every e > O, there exists a parlition P¢ such that P finer than P, implies
o <_ v(e,f, ) - z(e,f, ) < .
Tlarem 7.19. Assume that •  on [a, hi. Then the followiny three statements are
equivalent:
i) f s e() o.[, ].

ii) f atisfies lO'emann's condition with respect to a on In, hi.
iii) l(_f, ¢t) = i(f, ).
Proof. We will prove that part 0) implies [ii), part (ii) implies (iii), and part (iii)
implies (i). Assume that (i) holds. If z(b) = a), then (ii) holds trivially, so we
can assume that ct(a) < (b). Given t > 0, choose P, so that for any finer P and
all choices of t and t in [x_ t, x, we have
t0aa-,4 <3 5'
.where "4 =  f dt. Combining these inequalities, we fred
Sinc Mdf) - ink(f) = sup {/'(x) -]'(x') : x, x' in [Xk-1, X)}, it follows that
for every h > 0 we can choose t and t so that
Making a choice corresponding to h = /[g(b) - t(a)], we can write
< If(t,) -fog)] A, + h  a < s.
Hence, (i) implies (ii).
Next, assume that (ii) holds. If e > 0 is given, there exists a partition P, such
that P finer than P, implies U(P,f, ,,) < L(P,f, ) + . Hence, for such P we
have
lff,) <_ V{P,f,a) < L(P,f,) +  < !tf,) + .
Tat is, iG';, ,,) < If_f, ) +  for every  > 0. Therefore, l(f, ) < !Cf, ). But,
by Theorem %17, we also have the opposite inequaIity. Hence (ii) implies (iii).
Finally, assume that l0 r, at) = IG, a0 and Let A denote their common value.
We wilI prove that , f d0t exists and equals "4. Given  > 0, choose P so that
U(P,f, ) < l(f, ) + e for all P finer than P[. Also choose P: such that
Lff', f, a) > !(£ ,) - *
for all P finer than P:. If P, = P;  P:, we can write
! a) -  < L(l",f, ,) <_ S(l',f, ,) < V(,,f, ) < I(£ ) + 
for every P finer than P. But, since I(f, ') = if, f, ) = A, this means that
tS(P,f, t)  AI <  whenever P is finer than P,. This proves that fak exists
and equals A, and the proof of the theorem is now complete.
gx)  x) for all x  [ ], t we 
Pr For ev¢ fion e,  onSng aeltj
n,  on [a, hi. From   m foBo y.
 #1, s  p   x) x)  0 w e(x) z 0
In, b]    the ity
M -- m = sup {x) - ) : x, y in [x_
Sine  iq ]/(x)[ - Iy)l] a If(x) - )l  hol, it fono that
Mr(If[) - m(Ifl)  Mt -
Mulfipl#ng by t d suing on k, we obn
U(P, Ill, ) - P, Ill, )  U(P, ) - P,£ ),
for  ifion P of In, hi. By appI Riemann's nfio we d t
Ill  R() on In, b]. e inu M  m follo  m g = Ill in
m 7..
 e nr ofm 7.21  not ¢. ( Ex 7.12.)
rem L22. Asme tMt •  on In, b]. lff
Prof. Using e nofion of Defion 7.14, we
M z) = [n(Ifl)]  d (f2)
MffO - mff 2) = [MO£1) + mlfl)][M(lfl) -

where M is an upper bound for Ill on [a, b'i. By applying Riemann's condition,
the conclusion follows.
Theorem 7.23. Aume that  , on In, b]. lf f  R(¢) and g  R() on In, b], then
the product f.g
Proof We use Theorem 7.22 along with the identity
.fx)a<x) : f(x) + a(x)]  - [fO03  - [gO0] .
7.15 INTEGRATORS OF BOUNDF_. VARIATION
In Theorem 6.13 we found that every function ,, of bounded variation on ['a, b]
can be expressed as the difference of two increasing functions, lf --- 1 -- =2 is
such a decomposition and iff¢ R(=i) andf R(%) on In, b], it follows by linearity
thatfe R() on In, b]. However, the converse is not always true. If f6 R(=) on
In, b'i, it is quite possible to choose increasing functions =t and ¢z such that
ct - =t - 0¢2, but such that neither integral fdg, =fd= exists. The difficuIty,
of course, is due to the nonuniqueness of the decomposition = = =1 - 2- How-
ever, we can prove that fi)ere is at least one decomposition for which the converse
Ls true, namely, when d t is the total variation of g and a2 = gl - . (Recall
Definition 6.8.)
Titeorem 7.24. Assume that a is of bounded variation on In, b]. Let V(x) denote the
total variation of = on In, x] ifa< x  b, and let V(a) ffi O. Let f be defined and
bounded on In, b]. lf f  R(=) on [a, b], then f  R(IO on In,
Proof If V(b) - 0, then V is contant and the result is trivial. Suppose therefore,
that V(b) > 0. Suppose also that If(x)[ _< M if x ¢ In, b]. Since V is increasing,
we need onIy verify thatfsatisfics Riemann's condition with respect to V on [a, b].
Given  > 0, choose P, so that for any finer P and all choices of points t and
g, in x_ i, x] we have
and
V(b) < 2 [A%I +
*= 4M
For P finer than P, we will establish the two inequalities
and
/--1 2 
E [M(/) - m/(f)] It,=.l < -,
which, by addition, yield U(P,f, V) - L(P,f, V) < e.
To prove the first inequality, we note that AV - IA=I > 0 and hence
, [M,O r) - mdf)](^ -IA,I) _< 2M 2 (AV -
To prove the second inequality, let
and lt h = /V(b). ff k  A(P), choose t and t;, so that
f(td -f(tl) > M,(f) - m,0') - h;
but, if k  B(P), choose tt and tl so that f(6) - fff,) > M(f) -- m,(.f) -- h.
 [M,U) - mJ_f)] IzS=l <
,k  1 k---1
  _=
<  + V(b) =  + 4
It follows thatfe It(V) on In,
lqO. This theorem (together with Theorem 6.12) enables us to reduce the theory
of Riemann-Stieltjes integration for imegrators of bounded variation to the case
of increasing integrators. Riemann's condition then becomes available and it
turns out to be a particularly useful tool in this work. As a first application we shall
obtain a resuR which is closcIy related to Theorem 7.4.
Tleorem 7.25. Let =.be of bounded variation on In, b and asaume that f  a(=) on
In, b]. Then f  R(,) on every subinterval It, d of In,
Proof Let V(x) denote the total variation of a on [a, xl, with V(a) = O. Then
a = V - (V - =), where both V and V -  are increasing on ['a, b] (Theorem
6.12). By Theorem 7.24,f a(I0, and hencvf R(V  ) on [a, b. Therefore,
if the theorem is true for increasing integrators, it follows thatf R(I0 on [c, d]
and/ R(V - ) on [c, d, soft R(=) on ['c, d].

Hence, it suffrces to prove the theorem when cz/ on in, b]. By Theorem 7.4
it suffices to prove that each integral j'fd0t and Jfd, exists. Assume that
a < c < b. If P is a partition of in, x], let A(P, x) denote the difference
of the upper and lower sums associated with the interval in, x]. Since fE R()
on [-a, b'], Riemarm's condition holds. Hence, if e > 0 is given, there exists a
partition P, of [-a, b] such that A(P, b) <  if P is finer than P. We can assume
that c  P,. The points of P, in [-a, c] form a partition P' of in, c]. if P' is a
partition of [-a, c] finer than P[, then P = P'  P, is a partition of i-a, b]
posed of the points of P" along with thor points of P, in [c, hi. Now the sum
defining A(P', c) contains only part of the terms in the sum defining A(P, b). Since
each term is _> 0 and since P is finer than P, we have
A(P', c) _< A(P, b) < .
That is, P' finer than P implies A(P', c) < e. Hence, f satisfies Riemann's con-
dition on [a, c] and Jf dz exists. The same argument, of course, shows that
Jfdot exists, and by Theorem 7.4 it follows that fd0t exists.
The next theorem is an.application of Theorems 7.23, 7.21, and 7.25.
Torem 726. Assume fe R(a) and # • R(cO on in, hi, where ct on in, hi.
Define
2°
and
6(x) =. f: o(t) /: x s [a, b].
Then f ¢ R(, g  R(, and the prct f'g • R() on [a, hi, and we have
 f(x)g(x) d(x) = f f(x) dG(x)
Proofl The integral **f'9  exists by Theorem 7.23. For every partition P of
in, b] we have
and
Therefore, if M = sup {Ig(x)l " x  in, hi}, we have
IS(P, f, G)--f f'o a  l= {f(t)--f(t)}O(t)d*)l
_< M, If(tD -f(t)l d(t) <_ M, iMp(f) - m(f)]
Sin f e R(), for eveW e > 0 th is a position P, su tt P r than P,
impfies U(P. g) - P,  ) < g- Ts prov tt fe R( on in, b] and
that Jf.g rig= f. A simir agument shows t 9  R( on in, b] d
s. Theorem 7.26 is also valid if  is of unded variation on in, hi.
"I.16 SUFFICIENT CONDITIONS FOR EXIS'rF_CE OF RIEMANN-STIELTffE
INTEGRALS
In most of the previous theorems we have assumed that ceatain integrals existed
and then studied their properties. It is quite natural to ask: When does the integral
exist? Two useful sufficicut conditions will b obtained.
Tl,orem 7,27. If f is continuous on in, b] and if  is of bounded variation on
then f e R() on in, b].
o. By Theorem 7.6, a second sufficient condition can be obtained by inter-
changing f and :, in the hypothesis.
Proofi It suffices to prove the theorem when :, :" with 0(a) < b). Continuity
off on [a, b] implies uniform continuity, so that if • > 0 is given, we can find
 > 0 (depending only on :) such that Ix - Yl < # implies If(x) -
where 4 = 2let(b) - a(a)]. IfP, is a partition with norm lIP, It < , then for P
finer than P, we must have
since Mf) - mCO = sup {f(x) - f(y) : x, y in rxt_ x, xt]}. Multiplying the
inequality by Ag and summing, we find
U(P, f, or) -- L(P, f, =) <_  ,- , 2
and we see that .Riemann's condition holds. Henc¢,f R() on ['a, hi.
For the special case in which (x) = x, Theorems 7.27 and 7.6 give the foltowing
corollary:
Theorem 7,28. Each of he followin9 conditions is uflcient for de existence of the
Riemann integral o f(x)
a) f is com:nuous on [a, b. b) f is of bounded vriaaon on [a,

7.17 Y CONDITIONS FOR EXISTENCE OF RIEMANN--STIELTJES
INTEGRALS
When u is of bounded viation on a, hi, continuity off is sufllcicnt for the exis-
tence of Jfd. Continuity off throughout [a, b] is by no means necessary,
however. For example, in Theorem 7.9 we found that when ,, is a step function,
thenfcan be defined quite arbitrarily in ['a, b] provided only thatfis continuous
at the discontlnuitics of . The next theorem mils us that common discontinuities
from the right or from th left must be avoidod if the integrbJ is to exist.
Tlor¢ 7J9, Assume that ¢t  on [a, b] and let a < c < b. dssume further
that both  and fare disconthou from the right at x = c; that is, assume that there
exits an  > 0 such that for every 5 > 0 there are values of x and y in the interval
(c, e + b') for which
If(x) -f(e)[ _> ana I  (y) - "(c)l .>
Then the integral ,f(x) dot(x) cannot exist. The integrat also faib to exist if  and
fare discontinuous from the left at c.
Proof. Let P t a l:'fition of [¢, b'] containing c as a point of subdivision and
form the difference
:----1
If the ith subinterval ha c as its left endpoint, then
u(P,f, - z  e,f, [M,(D - -
since each term of the sum is > 0. if c is a common discontinuity from the right,
e can assume that the point X is chosen so that a(x) - (c) > e. Furthermore,
the hypothesis of the theorem implies Mf) - m,(f) > . Hence,
c'(P,f, - L(e,f, >
and Riemaan's condition cannot be satisfied. (If c is a common discontinuity
from the left, the argument is similar.)
7.18 MEAN-VALUE THEOREMS FOR RIF_,MANN-STIELTJES INTEGRALS
Although integrals occur in a wide variety of problems, there are relatively few
cases in which the explicit value of the integral can be obtained. However, it
often suffices to have an estimate for the integral rather thax its exact value. The
Mean Value Theorems of this section are espccially useful in making such estimates.
Tltcorem 7.30(First Mean-Value Theorem for Riemmm-Stidtjes istegrMs). Assume
that  7 and let f  R(¢) on a, hi. Let M and m denote, respectively, the sup and
inf of the set {f(x) : x ¢ [a, b]}. Then there exists a real number c satifyin 9
rh. 7.32 Integral as a Functi.o of the lateal 161
m <_ c .< M such that
f f(x) dct(x) ffi e f d(x) -- c[(b) - (a)].
In particular, if f is continuous on [a, b], then c = f(xo) for some x o in [a, bJ.
Proof. If a(u) = (b), the theorem holds tfivialty, both sid being 0. Hcn we
n assume that (a) < =(b). Sin all upr and lower sums mtisfy
the integral Jfdu must lie twn e se bounds. Therefore, the quotient
c = (fda)/( da) ties n m d M. Whenfis ntinuous on D, bJ, the
intermm vue m yields c = f(x) for some Xo in [a, hi.
A nd theorem of this t n  obtained from the fit by using inmgra-
tion by
Torem 731 ( MV orem for RSties ir).
Ame that  is contuous and tt f  on [a, b]. T there exits a t x o
 [a, b] ch that
/(x) dx) = f(a) dx) + f(b) dx).
Proof. By Theo 7.6, we have
AppLying Theorem 7.30 to the in on the right, we find
wh xo  [a, b], wffich is the st we set out tO prove.
7,19 THE INTEGRAL AS A FUNCTION OF THE INTERVAL
If re R(u) on [a, b] and if u is of bounded variation, then (by Theorem 7.25) the
integral fdn exists for each x in [a, b] and can be studied as a function of x.
Some properties of this function will now b¢ obtained.
Tleorem ZJ2. Let t be of bounded vatiation on [a, b] and assume that f  R(ct) on
[a, b]. Define F by the equation

Then ,e have:
i) F i of bounded variation on [a,
ii) Every paint of continuity of  is also a point of continuity ofF.
iii) If  : on [a, hi, the derivative F'(x) exists at each point x in (a, b) where a'(x)
exists and where f is continuoux. Foe such x, we have
F'(x) = f(x)a'(x).
It suffices to assume that g: on Is, hi. If x  y, Theorem 7.30 implies
Praof .
that
where m < e  M (in the notation of eorem 7.30). Smtemen 0) and 0i)
follow at on from this fion; To pve (iii), we vide by y - x and obese
tc x) asy  x.
When m 7.32 is  in njunon with m 7.26,  ob the
foowing  which n a Riem in of a pries f.g into a
Rintidtj inI Jf wi a nfinuous intetor of bod
vaaon.
Torem 73S. lff  g d g e R on Is, b], let
F(x) = ff f(t) dt, = ff g(,) a, if x  [a, b].
Tn F  G e continuous ctiona of ud iation on Is, hi. Also,
Proof- Pa (i) d (i 0 of m 7.32 how that F d G a nuous func-
tioas of und variation on Is, hi. e sn of t  d e two
rol ro  rono by ruing (x) = x in  7.26.
. When (x) = x, rt 0i0 o m 7.32 is sometim lled the first
funental theorem of inte#ral eal. It state that F'(x) = x) at h point
of conlinuity off A mpion nlt, called the sendftal theorem, is
v in the ext ion.
7.211 SE, CO1RD FUNDAMENTAL THEOREM OF INTFaGRAL CALCULUS
The next theorem tells how to integrate a derivative.
Tleorem 7=14 (Seeand fndamentel rlrem of integral ). Assume that f  R
on Is, b]. Let g be a function defined on Is, b] such that the derivative g' exists in
(a b) and has the value
g'(x) = f(x) for every x in (a, b).
At the endpaints assume that O(a +) and g(b--) exist and satisfy
Tlen we hm
g(a) - a(a+) = g(tO - g(b-).
For every partition of Is, b] we n
163
where t, is a point in (x-t, xD determined by the Mean-Value Tbeore, of
differential calculus. But, for a given  > O, the partition can be taken so fine that
o(b) - g(a) - f(x) dx = f(t) Axt -- f(x) dx < e,
and this proves the theorem.
The second fundamental theorem can be combined with Theorem 7.33 to give
the following strengthening of Theorem 7.8.
17earem 7.J5. Assume f e R on Is, hi. Let , be a function which is continuous on
Is, b] and whose derivative a' is Riemann integrable on Is, hi. Then the following
integrais exist and are equal:
y(x) ,t(x) =  y(xy'(x) .
Proof. By the second fundamental theorem we have, for each x in Is, hi,
(x) - a) =  '(t)
Tki  = '  m 7.33  ob  7.35.
. A re ult is dcfi in ci 7.M.
7,1 CHANGE OF VARIABLE IN A RIEMANN INTEGRAL
The f f  =  h dff of eomm 7.7 for ¢hng the variable in an
intel msum c fo
f'"
f(x) dx = f[(¢)]O'(/) at,

when a(x) = x and when g is a strictly monotonic function with a continuous
derivative g'. R is valid fff R on In, hi, When f is continuous, we can use
Theorem 7,32 to remove the restriction that g be monotonic. In fact, we have the
following theorem:
Trem 7,36 ( C&mge of variable i a Riemmm inrd). Aamme that g has a
continuoua derivative g' on an interval [c, d]. Let f be continum on g([c, d]) trod
define F by the equation
F(x) = f(t ) dt if x e g([c, d']).
Then, for each x in [c, d] the inte9rai f[e(t)]e'(t) dt exists and has the value
F[(x)]. In particular, we haoe
f(x) dx -- f[g(t)]g'(t) dr.
Proof. Since both g' and the composite function fo g are continuous on [c, d]
the integral in question exists. Define G on [c, d] as follows:
¢(x) = f:Y[o(t)]a'ft)
We are to show that G(x) = Fig(x)]. By Theorem 7.32, we have
a'(x) =
and, by the chain rule, the derivative of Fig(x)] is also f[9(x)]g'(x), sino¢ F'(x) =
f(x). Hence, G(x) - Fig(x)] is constant. But, when x : c, we gct G(C) = 0 and
Fig(c)] = 0, so this constant must be 0. Hence, G(x) = F-g(x)] for all x in
[c, d]. In particular, when x -- d, we get G(d) = F[-g(d)] and this is the last
equation in tim theorem.
NOTE. Some texts prove the preceding theorem under the added hypothesis that
g' is never zero on [c, d], which, of course, implies monotonicity ofg. The above
proof shows that this is not needed. It should be noted that g is continuous on
[c, d], so g(l-c, d]) is an interval which conta/ns the interval joining 9(c) and g(d).
i ,
Figure 7.2
X'n. 737 Second Man-Valu¢ l"heorem for Rknmnn Integrals 165
In particular, the result is valid ify(c) -- g(d). This makes the theorem especially
useful in the applications, (Sec Fig. 7.2 for a permissible
Actually, them is a more general version of Theorem 7.36 which does not
require continuity offer ely', but the proof is considerably more difficult. Assume
that h  R on I-c, d] and, if x e [c, d], let g(x) = S h(t) dr, where a is a fixed
point in [c, d]. Then iff R on g([c, d]) the intcgra! Sa, f[g(t)] hO) dt exists and
we have
/(x) dx = f[a(t)]h(O dr.
This appears to be the most general theorem on change of variable in a Riemann
integral. (For a proof, see the article by H. Kestelman, Mathematical Gazette,
45 (1961), pp. 17-23.) Theorem 7.36 is the special case in which h is continuous on
[c, d] andfis continuous on g([c, d]).
7.22 SECOND MEAN-VALUE THEOREM FOR RIEMANN INTEGRALS
11eorem 7d7. Let g be continuous and assume that f/ on [a, hi. Let A and B be
two real numbers satisfying the inequalities
,4 _.< ffa +) and B >_ f(b -).
Then there exists a point x o in [a, b] such that
i) ; f(x)g(x) dx= A f? a(x) dx + B g(x) dx,
ln particular, if f(x) > O for all x in [a, hi, we have
ii) f(x)O(x) dx = B
ro'r. Part (ii) is known as
Proof. If (x) J, g( ) dr,
O(x) dx, where xo
Bonnet's theorem.
then ' = , Theorem 7.31 is applicable, and we get
f(x)o(x) dx = f(a) e(x) dx + f(b) 9(x) dx.
This proves (i) whenever A = f(a) and B = f(b). Now if A and B are any two
real numbers satisfying A < f(a+) and B > f(b-), we can redefine fat the end-
points a and b to have the values f(a) = ,4 andf(b) = B. The- modified f is siill
ir,.xeasing on In, b] and, as we have remarked before, changing the value of fat
a finite number of points does not affect the value of a Riemann integral. (Of
course, the point xo in (13 will depend on the choice of A and B.) By taking A = 0,
part (ii) follows from part (i).

7. "gJl  DING ON A
Theorem 7d#. t f  mo  ch t x, y) of a
Q = [(x,y):axb cy d
e tt a  of d ation  [a, b] d let F  t ction
uous on Q. en, ven  > 0,  es a  > 0 (dng
s t for e pair ofin z = (x, y) and z" = (x', ) in Q
IF() - F(y')]  f If(x, y)  f(, Y')I d()  <b) - (a).
s tabfi e nfinuiW of F on [c, .
Of coup, when (x) = x, this omes a connty eorem for giemann
instals involg a er. However, we  deve a muc mo ful
ult for a in--Is n at obmin by mply tting x) = x  we
ploy  7.26.
m 739. lf f is cti  the rectIe [a, b]  [c, d],  if g  R on
.[a, b], tn the fction F dn by l equation
a x
Proo If a{x) =  t) dr,  7.26 shows that F(y)
Now apply com 7.38.
7.24 DIFFFdRF2qTIATION UNDER THE INTEGRAL SIGN
Theorem 7.40. Let Q = {(x, y) : a < x _< b, c _< y _< d}. lssume that , is of
bounded variation on [a, b] and, for each fixed y in [c, d], assume that the integral
Fly) -- I f(x, y) d(x),
exists. If the partial derivative Dzf i continuou on Q, the derivative F'(y) exiats
for each y in (c, d) and is iven by
F'(y) ffi  Df(x, y) d(x).
In paraear, when g  R on [a, b] d (x) ffi J: U(t) dr, we get
F(y) =  g(x)f(x, y) dx and
Proof.
F'(y) = ; g(x) Dff(x, y)
If Yo  (e, d) and y  Yo, we have
F(Y) - F(Yo) --- ff f(x" Y) - f(x" Y°) dx) = f D'f(x'y-) dx(x)'y yo y
where ? is between y and Yo- Sinee Dz/" is continuous on Q, we obtain the con-
clusion by arguing as in the proof of Theorem 7.38.
725 INTERCHANGING THE ORDER OF INTEGRATION
leorem 7.41. Let Q - {(x, ¥) : a < x < b, c < y  d}. Assume that t i of
bounded variation on [a, b], fl is of bounded variation on [c, dr], and f is continuous
on Q. If (x, y) e Q, define
F(y) = f ffx, y) d(x),
In other worda, we may interchange the order of integration aa foflowa :
Proof. By Theorem 7.38, Fis continuous on [, d] and hence F e R( on [c, d].
Similarly,   () on [a, b]. To prove the eluality of the to integrals, it suffices
to consider the ease i which ,, on [a, b] ad #, on [c, d].

By uniform continuity, given e > 0 there is a 6 > 0 such that for every pair of
points z  (x, y) and z' = (x', 7') in Q, with Iz - z'l < 6, we have
If(x, y) - f(x', y')l < e.
Let us now subdivide Q into n 2 equal rectangles by subdividing In, b'] and [e, d]
each into n equal parts, where n is chosen so that
Writing
(b-a)< 6 and (d-c) < 
xk =a+ k(b--a) and y-- c+ k(d--c)
for k = 0, 1, 2, ..., n, we have
f(x, y) d#(j/) dx) --   y(x, y) d  d(x).
We apply m 7.30 t on the fight. The double s
whc (x, y) is in thv glv  having (z, y) and (x+, y+)
v. Similarly, wc find
k=O
where (x,, y) e Od" But lf(xg, .v) - .f(x,, Y.)I < e and hence
IfG(x) d(x)-fF(y)dfl(Y)]
: - -
Since e is arbkrary, this imptics equality of the two integrals.
Theorem 7.41 together with Theorem 7.26 gives the following result for Rie-
mann integrals.
7.43
Theorem 7.42. Let f be continuous on the rectangle In, b] x [c, d]. If 9  R on
[a, b] and if h • R on [c, d], then we have
Proof Let a(x) = ] g(u) du and let (y) =  h(v) dr, and apply Thrums 7.26
and 7.41.
7.2i LEBESGUE'S CRITERION FOR EXISTENCE OF RIEMANN INTEGRALS
Every continuous function is Riemann integrable. However, continuity is certainty
not necessary, for we have seen thatfa R whenfis of bounded variation on [a, b].
In particular, f can be a monotonic function with a countable set of discontinuities
and yet the integral Jf(.¢) dr will exist. Actually, there are Riemann-integrable
functions whose discontinuities form a noncountable set. (See Exercise 7.32.)
Therefore, it is natural to ask "'how many" discontinuities a function can have and
still be Riemann integrable. The definitive theorem on this question was dis.
covered by Lebesgue and is proved in th/s section. The idea behind Lebesgue's
theorem is revealed by examining Riemann's condition to see the kind of restriction
it puts on the set of discontinuities off.
The difference belween the upper and lower Riemann sums is given by
k=l
and, roughly speaking, f will be integrabl¢ if, and only if, this sum can be made
arbitrarily small. Split this sum into two parts, say S + Sz, where _ comes from
subintervals containing only points of continuity off, and S contains the re-
maining terms. In S, each difference Mr(f) - m(.f)is small because of continuity
and hence a large number of such terms can occur and still keep S small. In
however, the differencesM(f) - mz(f) need not be small; but because they are
bounded (say by M), we have ISz[  M Y.Ax so that Sz will be small if the sum
of the lengths of the subintervals corresponding to Sz is small. Hence we may
expect that the set of discontinuities of an integrable function can be covered by
tervals whose total length is small.
This is the central idea in Lebesgue's theorem. To formulate it moe precisely
we introduce sets of measure zero.
Definition 7.41. A set S of real numbers is naid to hae raeatre zero if, for every
 > O, there is a countable, covering of S by open interval, the  ofwhoe ientht
i legs than .
If the intervals are denoted by (a, b), the definition requires that
S  O (ak, b) and  (b - a) < e. (3)

If the.collection of intervals is finite, the index k in (3) runs over a finite set. If the
collection is countably infinite, then k goes from I to oo, and the sum of the lengths
is the sum of an infinite series given by
Besides the definition, we need one more result about sets of measure zero.
Theorem 7,44, Let F be a countable collection of sets in R, say
F= {F, Fz, • • • },
each of which has measure zero. Then their union
s: 13 F,,
also has a measure zero.
Proof Given  > 0, there is a countable coveting of F by open intervals, the sum
of whose lengths is less than g/2 . The union of all these coverings is itself a
countable covering of 5' by open intervals and the sum of the lengths of all the
intervals is less than
Faamltles. Since a set consisting of just one point has nasare zero, it follows that every
countabl© subset of R has nmasure zero. In particular, the set of rational numbers has
measure zero. However, there are tmeountabte sets which have measure zero. (See Excx-
cise 7.32.)
Next we introduce the concept of oscillation.
DeflMtian 7.,15, Let f be defined and bounded on an interval 8. If T _ ,, the
number
fs(T) = sup {f(x) - f(y) : x • T, y • T},
i called the oscillation off on T. The oscillation off at x is defined to be the number
= lim nAe¢x; h) n S).
OTE. This limit always exists, since f-(B(x; h) r S) is a decreasing function of
h. In fact, T t G Tz implies f)y(Tj)  F/(T2). Also, o.(x) -- 0 if, and only if,
fis continuous at x (Exercise 4.24).
The next theorem tells us that if my(x) <  at each point of a compact interval
[a, b], then f(T) <  for all sufficiently small subintervals T.
Theorem 7.46. Let f be defined and bounded on [a b], and let  > 0 be given.
Assume that e)(x) <  for every x in [a, b]. Then there exists a 5 > 0 {depending
1 7,48 Lebesgue's Crilet 171
only on ) such that for every closed subinterval T _ [a, b], we have (T) < e
whenever the length of T is tess than &
Proof For each x in [a, b] there exists a l-ball B,, = B(x; 5) such that
The set of all halfsize balls B(x; 3/2) forms an open covering of [a, b']. By
compactness, a finite number (say k) of these cover [a, b']. Let their radii be
5/2, ..., d2 and let  be the smallest of these k numbers, When the interval
T has length <, then T is partly covered by at least one of these balls, say by
B(x,; d2). However, the hall B(x; ¢5) completely covers T (since fi,  23).
Moreover, in B(xn; On) r [a, b] the oscillation off is tess than e. This implies
that fl/(T) <  and the theorem is proved.
T/worem 7d7, Let f be defined and bounded on [a, b]. For each  > 0 define the
set J, as follows:
{x : x [a, . Ax) e}.
Then d, ia a closed set.
Proof. Let x be an accumulation point of d,. If x  J,, we have Ax) < .
Hence there is a 1-ball B(x) such that
Thus no points of B(x) can belong to , contradicting the statement that x is an
accumulation point of J,. Therefore, x • d, and d, is closed.
Tlorcm 7d$ (Lebesge's ,rRerian for Renm.,meauddlity). Let f be defined
and bounded on [a, b] and let 1)denote the :wt of discontinuities of fin [a, b]. Then
f• R on [a, b] if, and only if, D has mea.,ure zero.
Proof. (Necessity). • First we assume that D does not have measure zero and show
thatfis not integrable. We can write D as a countable union of" sets
where
If x • D, then ,.(x) > 0, so D is the union of the sets D, for r = 1, 2,
Now if D does not have measure zero, then some set D, does not (by Theorem
7.44). Therefore, there is some e > 0 such that every countable collection of Open
intervals covering D has a sum of lengths >4. For any partition P of In, b] we
have
u(e,f) - L(ef) = , [Mt(.f) -- mt(/)] axt = S t + S

where St contains those terms coming from subintervals containing points of D
in their interior, and $2 contains the remaining terms. The open intervals from S t
cover D, except possibly for a finite subset of D,, which has measure 0, so the sum
of their lengths is at least t. Moreover, in these intervals w¢ have
1 
M(f) - ms(f) >_ - and hence S 1 > -.
¥ r
This means that
U(P,f) - L(P, f) .. -,
r
for every partition P, o Riemann's condition cannot be satisfied. Therefore f is
not integrable. In other words, fife R, then D has measure zero.
(Suciency). Now we assume that D has measure zero and show that the
Riemann condition is satisfied. Again we write D = 13= t, D,, where D, is the set of
points x at which ot(x ) > lit. Since D, _ D, each D, has measure 0, so D r can
be covered by open intervals, the sum of whose lengths is < lit. Since D, is compact
(lheorem 7.47), a finite number of these intervals cover D,. The union of these
intervals is an open set which we denotf by A,. The complement B r - [a, b] - A,
is the union of a finite number of closed subintervals of Is, b]. Let I be a typical
subinterval orB,. Ifx e 1, then (x) < 1Jr so, by Theorem 7.46, there is a 6 > 0
(depending only on r) such that Ion b¢ farther subdivided into a finite number of
subintervals T of length <6 in which flT) <l]r. The endpoints of all these
subintervals determine a partition P, of [a, b]. If P is finer than P, we can write
vff,, f) - L(,, 19 =  [uM) - m(D] ax, = Sl + s,.
=!
where $1 contains those terms coming from ,ubintervals containing points of
D,, and S x contains the remaining terms. In the kth term of S x we have
1 b-a
Mr(J0 - mr(f) < - and hence S z < --
Since A, covers air the intervals contributing to S, we have
S t _< --,
r
where M and m are the sup and inf of fun Is, b'l. Therefore
M-m+b-o
O(', f) - L(P, f) <
r
Since this Ixolds for every r _> !, we see that Riemann's condition holds, sof R
NOTE. A property is said to hold almost everywhere on a subset S ofR  if it holds
everywhere on S except for a set of measure 0. Thus, Lebesgue's theorem states
that a bounded function fun a compact inteawal Is, b] is Riemann integrable on
Is, b] if, and only if, fis continuous almost everywhere on Is, hi.
The following statements (some of which were proved earlier in the chapter)
are immediate consequences of Lebesgue's theorem.
Theatre 7.49. a) lf f is of bounded variation on Is, hi, then f e R on [a, hi.
b) lf f e R on Is, b], then f , R on [c, d] for every subinterval [c,
tfl • R and f2e R on [a, b]. ,4tso, f. e R on Is, b] whenetr # • R on
e) lf f $ R and a e R on Is, hi, then fig  R on Is, b] whenever g is bounded away
o,n O.
d) lf f and e are bounded functions htwin e the same dscontinuities on a, b, then
f e a on [a, b] if, t only if,  e ¢ on [a, ].
e) Let g e R on [a, b] tmd aasume that m  g(x) _< M for alt x in [a, b]. lf f is
contuos on Ira, m], the co,eoite funcaon  dealed y (x) = f[(x)] is
Riemann-integrable on Is, b].
Hoax. Statement (e) need not hold if we assume only that f• R on I'm, M].
(See Exercise 7.29.)
77 COMPLEX-VALUED RIEMANN-STIELTJE$ INTEGRALS
Riemanntieltjes integrals of the form $.fd, in which f and u are complex-
valued functions defined and bounded on an interval Is, hi, are of fundamental
importance in the theory of functions of a complex variable. They an be intro-
duced by exactly the same definition we have used in the rest ease. In fact,
Definition 7.1 is meaningful when f and a axe complex-valued. The sums of the
product f(tt)[x(xt) - xt_)] which are used to form Riemann-Stieitjes sums
need only be interpreted as sums of products of complex numbers. Since complex
numbers satisfy the commutative, associative, and distributive laws which hold
for real numbers, it is not surprising that complex-valued integrals share many of
the properties of real-valued integrals. In particular, Theorems 7.2, 7.3, 7.4, 7.6,
and 7.7 (as well as their proofs) are all valid (word for word) when f and t are
complex-valued functions. (In Theorems 7.2 and 7.3, the constants e and c2 may
now be complex numbers.) In addition, we have the following theorem which, in
effect, reduces the theory of complex Stieltjes integrals to the real ease.
'lteoeem 7.$0. Let f = ft + if2 and  = t + iz be complex-valued functions
defined on an interval Is, b]. Then we have
whenever all four integrals on the rieht exist.

174
The proof of Theorem 7.50 ts immediate from the definition and is left to the
reader.
The use of this theorem permits us to extend most of the important properties
of real integrals to the complex ca. For example, the connection between
differentiation and integration established in Theorem 7.32 remains valid for
complex integrals if wc simply define such notions as contnulty, diffcrentJability
and bounded variation by components, as with vector-valued functions. Thus, wc
say that the complex-valued funcon • = i + -2 is of bounded variation on
[a, b] if each component sq and % is of bounded variation on [a, b]. Similarly,
the dcrivativc u'(t) is defined by the equation '(t) = (t) ÷ i(t) whenever the
derivatives (t) and (t) exist. (One-sided derivatives are defined in the same
way.) With this understanding, Theorems 7.32 and 7,34 (the fundamental theorems
of integral calculus) both remain valid when f and • are complex-valued. The
proofs follow from the real case by using Theorem 7.50 in a straightforward
manner.
We shall return to complex-valued integraIs in Chapter 16; when we study
functions of a complex variable in more detail.
EXERCISES
%1 Prove that S dax) = (b) -- a), directly from Definition 7.1. -
7-Z If re/a 0 on [a, b] nd if f  0 f  fwh  motonic  [a, b],
 tt • must  constt on [a, b].
7 c following tion of a t intl  ofl  in t I]tt:
We yfis ]nteble th s to • ifth i a 1 nr A   pro
that for   > 0, t m a  > 0 sh at for  fion P of [a, b] with
no I[ell <  d for w o or  in [xt_t, xt],   ]S(P, ) - al <
a) Show at if  [ exists g to this dn]tion, t it  sts ardi
to ition 7.1 and t o int  ual.
b) (x) = x) = 0 f a   < c, (x}  ¢(x) = I for c < x  b,[(c) = O,
c) = ]. Sh that [f ists ng to niti 7.1 but d not
by th d dition.
%4 Iffe R [ng to n]tion 7.1, pr t [f(x) dx  is aing to
dition of ei 7.3. [nst with i 7.3(b).] Hint. t I  f(x)
M = p {]f(x)l : x e to, hi}. G  > 0,  P  that U(Pg,  < I +
(notion of tion 7.11). t N  t aum of suion in in Pe d let
 = I(2MN). If lell < . write
whe St is the sum of ter aring fm  subintewals of P contng no [n of
P, d S2 is t sum of t ining te. en
1  u(eo f) < I + el2 and Sz  NM]IPI[ < wga =
175
and hencg U(P,f) < I + e. Similarly,
L(P, f) > I - e if P < ' for some
HCrlC IS(P,f) - Zl < • if PJI < rain (a,
Let {a,} be a sequence of real numbers. Forx  0, define
where [x] is the greatest integel" in x and empty sums are interpreled a zero. I.t fhe.ve
a continuous derivative in the interval 1 _< x _< a. Use StieRjes integrals to derive the
followiag formula:
7.6 Use Euler's summation formula, or integration by parts in a Stieltje integral, to
derive the following identities:
b) --Iogn- -" dx+ 1,
7.7 Assume f" is continuous on [I, 2n] and use Eule's summation formula or integra-
tion by parts to prove that
= (-l)V(k) -- f'{xX[x] - 2[x/2])
7.8 Lt Pl() = . -- IX] -- " if x  integer, and let ¢1(x) = 0 if x = integer. ALso,
let ¢(x) =  ¢(t)dr. If f" is continuous on [1, n] prove that Eulet's summation
formula implies that
fO) +f(n)
7.9 Take f(x) = log x in .Exerc 7.8 and prove that
Iogn[= (n + ½) logn n + 1 + dr.
31 t
7.10 If x _> I, let agx) denote the number of primes < x, that is,
x) =  I,
where the sum Jo extCzld  all primes p < x. The prime number theorem states that
l/m (x) Iog__x_x = 1.

176
whc again the sum is extended ovex all pr/mes p < x Both functions  and  am step
functions with jumps at Om prins, This exorcise shows how the RJcammn-Stieltj¢
integral can b used to re.lat these two fnactioas.
a) If x > 2, lxov¢ that a() and x) can b¢ ¢pressed as th following Riemann-
Stieltjcs integrals:
(x) = log : t), a(x) =  d(t).
zo. Th lowr 5mit ca  replaced y  numlr in  open imerL (1, 2).
5) If  z 2, m integration L pa_2s to shw that
(x) = agx) log x - ( 0.) d,
log x .] t log z t
These ¢quatiom can be used to trove that the  number theonma is equ/vat
to the relation lim_.. O(x)lx  !.
7.11 If ¢/ on [, hi, prove that we have
a) 2. /d = /d + /d
b) (f + )d,, : fda + o
7o12 Give an ommlle of a bounded function f and an increasing funotion a defined on
[a, b] such that If[  R(a) but for which t.fd do not exist.
7.13 Let a be a continuous function of bounded variatioa on [a, hi. Assume ,
on [a, b] and define ,8(x) -- (t) da(t) ifx  [a, b]. Show that:
a) [ff, on [a, hi,  xists a point Xo in [a, b] such that
; f:"
fd ffi ](a)  ,, + f{) g
b) If, in addition, fis continuous on [a, hi, we also haw
f(x)a(x) d(x} = f(a)  a,, + f(b)  da. "
17"!
7.14 Assurer f R(a) on [a, b], whero a is of boundod variation on [a, b]. Lt F(x)
dcnotvthc total variation ofaon [a,x]forcachxin(a, b], andlet V(a) = 0. Show that
where M is an upper bound for Ill on[a, b]. In partiar, whea (x) - x, t inequality
7.15 Let {,} be a sequemc of functions of bounded variation on [a, b]. Suppose thole
exists a function • defined on [a, b] such that the total variation ofa - ¢h on [a, b] tends
to 0 as n --, o. Assume also that a(a) = a(a)  0 for .h n = 1, 2,... If fis con-
tinuous on , b ], prove that
7.16 lff
2
When =/ on [a, b ], deduce the CauchySchwarz inequality
(Compare with Exercise 1.23.)
7.17 Assume thatfe R(a),   R(), andf-O e R(g) on [o, b]. Show that
 ; [ff (/(y) - f(x)Xo(y) - (x)) da(y) ] da(x)
If a/ on [a, b ], deduce the inequality
when bolhfand are incre.asing (or both am decreasing) on [a, hi. Show that the reverse
inequality holds if f increases and t decreases on lag hi.

%18 Aum©f R on [, b]. Ue  7.4 to prove hat he limit
- a+
 d  t vai yx) .  tt
I = k n n z- n, lim(nz+ kZ)_,,=,(1 + ).
7,19 
f(x) e   , x)= t z+ 1
a) Show that #'() + fx)  0 f 1  d u t (x) + f(x)  hi4.
Hm [ e-' dt = _1 .
7  g ¢ R on [a, b ] d e f(x) = ff (:)  if x  In, b]. ove that
in  (:)l   t to  f [ x].
721  f  (f, ..., f  a -v i wi a continuous ti f"
[ b].  that t    f  lh
A: b) = s if'0)II -
a) Show tlt
k! '
b) Use (a) to p e n  Tayl's foula  5.19)  an inl.
7  f nfinuous on [0, a].  x  [0, a], A(x) = f(x) and let
+,(x) =  (x - O"f(t) dr, n = O, I, 2 ....
a) Sh at t nth iti A ists d
b) P  followg  of M. F&:  n   in si of f
in [ a] is not  than  humor of ch  si in  o  of
/(a), A(a),...,
Hint. Use nmthexnatical induction.
¢) U (b) to  th following thoom of L, Fejr: The number .¢"  in
aign ¢'fin [0, a] is not ie thnn tl  of in sign in the  sel
f(O)" ffof(t)dt' ffo tf(t)dt" '"' o t'f(')'.
7,2 Letfbe a poeifivecontinuosftmctionin [a, b]. 1. Mdenotethemaxumvalu
off on [a, hi, show that
lira ffxp - M.
7.2 A function f¢" tw real vaxiab$ is defined for each pcfint (x, y) in the trait squaxe
0_<x- 1,0_<y < I asfollows:
I, fix is rational,
f(" Y) = 2y, if x i irrational.
a) C.am0ute io  f(x, y)  d I f(, Y) dx in term of y.
b) Shiny that  f(, y)  ¢Asts fvr each fed x and campute  f(, y) dy in terms
ofxaIt for0 _< x __< 1,0 - t < I.
c) Let (x)  I f(x, y) d. Show that o  ¥() dx exim and fred ts value.
7.26 Letfbe defined on [0, I ] as follows:f(0)ffi 0;if2 -'- < x < 2-',theaf(x) 2 -,
forn ffi 0, ,2....
a) Give two reasom why ' f(x) d
b) Let F(x) =  f(t) dr. Show that fat 0 < x < I we have
where A(x) = 2 -[-'=;'] and where [y] is th gatet integer in y.
7..-/Assume f has a derivative which is monotonic decreasing and atisfies f'(x) >_
m > 0forallxin [a,b]. Provethat
Hint. Multiply and divkl the integraml by f'(a) and use  7.37(ii).
7., Given a da'asiag sequen¢ of real numbers {G(n)} such that C,(n)
De,an, a functionfon [0, I ] in tems of {G(n)} as follows:/(0) = I; ifx is irrational,
f(x)  0; if x is tl rational m/n ('m lowt trmn), thenf(m]n) = G(n). Compute the
ormation a(x) at eada x in [0, I ] and show that fE R o [0, 1 ].
7.9 l.-tfbcdcfined as in  7.28 with G(n) -- fin. I.t #(x) = 1 if0 < x < I,
t(0) = 0.  that the ccmspoe, it function h &flncxl byh(x) = gLf(x)] is not -
intcgrabl on [0, lalthoughbothfRamlUeRon [0, I].
7.30 U¢ Le', thcore to prove  7.49.
7.31 Use Lebe,u's theorem to In'ov that if re R and g  R on [u, b] and if f (x)
m > 0fotallxin In, b],thn the ftmctionh denedby
x) = f(xy 
i eiemazn-integrabk on [a, 1,1.

7. Let l -- [O, ] ] and let A -- l - (, ) l that suhset of l obtain by removing
vots wh:h in t open middk third, o;thaX ,A _- [0.] L, ]. Let
Aa be that  of A o by    d i  [0, ] d of
a) 0<f(x)< lln O < x < 1.
b)  kth f(0) df(l) is  in.
c) F(O) and F(I) tr integefg
d {F'(x) sin  - Ffx)  }.
 U(a) (e)O < 1) + F(O) < 1 if hiS Sufly. 
 Gi a vu fc6oa , finuous on  in [ b] d v[ng a
un   on (a, b).   d und  [, b] d  t
th
vf(x) x)  'f(x) (x) 
7  the follog  whi p 1 a fon wlh a fi int
mu  iti  in.  tfe R [ b] d tl 0  f(x) 
that 1 > O.  e  T  {x : f(x)  h} s a fini num of in, 
s of wh ls i a I & Hint  P  a fi of [a, b ] sh that 
R s P,f) =  f(tD t fi P, > I. Sit P,f) into o
If k ¢ A, u e ii f(t D  M; • k ¢  ch t t  t f(tD < h. u at
The following exercises illustrate how the fixed.point theorem for contractions
4.48) is ttsed to prove existence theorems for solutions of certain integral and differential
equations. We denote by C [a, b] the metric space of all real continuous functiom on
aG e, O) = If- ] = max If(x) - (x)l,
7.36 Given a function 0 in C [a, b], and a function g continuous on the ctangle
 = [a, b] × [a, b], oimder the function 2"defined on C[a, b] by the equation
where J. is a vea constant.
a) Prove that T maps C [a b  into itself.
b) If [K(x,y)[ _< Mort Q, where M > 0, and if Il < M-'(b - ,)-, poe that
Tis a contraction of C[a, b] and hence has a fixed point  which is a solution of
th illtegral equation ¢(x) .= O(x) ÷   K(x, t)C(t) dr.
7.w Assume f is continuous on a rec  : [a - S, a + hi × [ - t,
where h > 0, k > 0.
a) Let ¢ be a runctinn, ontinaous on [a - S, a + ], such that (x, ¢(x))  0 for
all x in [a - h, a + h]. If 0 < e <_ h, prove that ¢ satis the diffenmtial
equation y' = f(x, y) on (a  c, a + c) and the initial oondition ¢(a) -- b if,
and only if, ¢ satis the integral equation
b) Assume that. If(x, Y)I -< M on Q, where M > O, and let e = rain {h, k/M}.
Let S denote the metric subspace of C [a - c, a + c] consisting of all  uch
that It(x) - bl < Mc on [a - c, a + e]. Prove Jat $ is a dosed subspace of
C [a  c, a + c] and hence that ,. is itself a complete metric space.
¢) PRrve that the function T defined on S by the equation
maps S into itself.
d) How assume that f satiies a Lipschitz condition of the form
If(x, y) - f{x, )1  Aly - zt
for every pair of points (xo y) and (x. ) in Q. where A > O. Prove that T is
contraction of $ if h < 1]A. Deduce that for 1, < I/A the differential equation
" = f(x, y) has exactly one solution y = ¢(x) on (a - ¢, a + ¢) such that
¢(a) = t,.

SUGGEbWED REI:ERENCI FOR FURTHER STUDY
7.1 I-Iildebrandt, T. H., Introduction to the 7lteory of ltte#ration. Academic Pre New
York, 1963.
7.2 Kestelman, H., Modern Ttteories of lntefrratfon. Oxford University Pre Oxfd,
1937.
7.3 Rankin, R. A., An ltroduction W MathmatimlAnMysis. Pergamon Pre Oxford,
1963.
7.4 R, W. W., Volume and Integral. Wiley, New York, 1952.
7.5 Shilov, G. I., and Gurevich, B. L., In,Orcd, Measure and Derimt: A Uned
Apprm R. Silvetman. translator. Prentice-Hall, Englewood Gifts, 1966.
CHAPTER
INFINITE SERIES
AND INFINITE PRODUCTS
g.l INTRODUCTION
This ctpter 8es a brief development of the theory of infinite series and infinite
prdus. These are nrely spec infinite sequences whose terms e red or
complex numbers. Convergent sequenc were  in Chapter 4 in the setting
of general metr SliCes. We recxH some of the concepts of Chapter 4 as they apply
to sequences in C with the usual Euclidean meuic.
If{a,} converges top, wewrite lin..® a, = p and call p the iimit of the sequene.
A sequence is tailed dergent if it is not convergent.
A sequence in C is called a Cauchy sequence if it satises the Cauchy condit/on;
that is, for every, > 0 there is an integer V such that
la .-
Since C is s complete metri space, we know from Chapter 4 that
is convergent if, and only if, it is a Cauchy sequence.
The Cauchy condition is particularly useful in establishing convergence when
we do not know the actual value to which the sequence converges.
Every convergent sequence is bounded (Theorem 4.3) and hence an unbounded
sequence necessarily diverges.
If a sequence {a,} converges to p, then every subsequence {at,} also converges
to, (Theorem 4.5).
A sequence {a,} whose terms are real numbers is said to diverge to + co if,
for every M > 0, there is an integer ?/(depending oa M) such that
a, > M whenever n > N.
In this case we write lira._.® a
If lim,.., (-a,) = +00, we write lira,_.® a, = ro and say that {} diverges
to -o. Of course, there are divergent real-valued sequences which do not diverge

to + orto -o. For example, the sequence {(-I)'(! + l/n)} divces bet does
not diverge to +or or to -cv.
K3 LIMIT SUPEIJO Al   OF A IAIVA/_LTED SEQUENCE
 &Z Let {a.} e a uowe of r  Su t s a r
<U+.
U = tim sup
If t set  d  t t  low   {} h no finite
riar, th we  l sup, a = - . T limit inferior (or Ioer lt) of
]ima, = -  . where b = -a. for n = 1,2 ....
. St (i)  t uy 1  of   lie to the le
of U + e. SmUt (i  t indy y tes  to ¢ t of U -
It is cr t     n o U which fies th ) and
Eve  u s a limit suEor d a limit infeor in e td al
   y  s of e foHong s:
b) e nce  if  y i lira sup, a  tim inf.   a, e both
Nffr. A sequence for which lira inf,_®  # lira sup_. a is tid to oscillate.
Tleorem &4. Assume that a, < b, for each n  1, 2,... Then w have:
lira inf a, < lira inf b. and lira sup a, _< lira sup b,.
. a, -- (-ff'O + ln),
Z.a,= (- Y',
3. a, = (-1)n,
4. a. = n  sin  (½w),
lira inf._.  a. = -- 1.
lira inf,.® , - -1,
tim inf, a, = o,
lira supa. = +1.
lim sup.- a. = + 1,
lira sup a. =
8.4 MONOTONIC SEQUENCES OF REAL NUMBERS
De_fm//hm
increasing and we write a,' if a < a,+ z for n = 1,2, ... If a >_ a.÷ lfor all n
we say the sequence is decreasing and we write  " . A sequence is called monotonic
if it is increa.ffng or if it is decreasing.
The convergence or divergence of a monotonic u:quence is paxticularly easy
to detemfine. In fact, we have
Ttomm 8.6. A monotonic sequence corwerge if, and only if, it is bounded.
Proof If a,',
inf{a,:n = 1,2....}.
Let {a,} be a given sequence of real or complex numb and form a new sequence
{s,} as follows:
s. ffia,+-.-+a, ffi    ,
oa &7. The oedered pair of sequences ({a}, {s,}) is called an infinite series.
The number s, i called the nth partial sum of the eries. The series is aid to co-
verge or to diverge according as {s,} is convergent or divergent. The followin
symbols are used to dote the series defined by (1):
a, ÷ a 2 +'"+ a.+"', a + a 2 + a 3 +
 The letter k used-in Yk  a is a "dummy variable" and raay b¢ replaced
by any other convenient symbol If p is an integer >0, a symbol of the, form
,-, b, is interpreted to mean 3-'_1 a where a, = b.+_. When there is no
danger of misunderstanding, we write ,.b. instead of Y'_, b..

If the sequence {s.} defined by (1) onvers to s. the number s is called the sum
of the series and we write
Thus, for convc¢gnt series the symbol  is used to denote both the sexis and
its sum.
EXaml lfxhas the infinite decimal ¢xpazmion x - ao.axa2 -.. (see Section 1.17), then
the series V_ao a,10 -t convcrg to x.
7&,orem 8,8. Let a =  ond b = _.b, be comergent serif. Then, for ecery
pair of constants a and , the .eries (a, 4- b) comerye to the sum aa 4- b.
That is,
Tiarcm 8.9. ,,lxsume that a.  Of of each n = I, 2,... Thin  ¢onccrge 
and only if, the quence of partial  is bognded abcme.
Proof. Lets. = a + ... + a. Then s., andwe can apply Theorem 8.6.
m.S.10 (yets,a,i ). Let {a.} aad {b.} be two  =uch t
exists, in which case we
Tlerem 8.11 (Caaeky  fc, r ¢rles). The serie  cnzes if, and only
if, for eery  > 0 there exists an integer N 'uch that n > iV implies
I.+ + "'" + a.+,l < e for eachp = I, 2,...
(2)
Proof Let s. :-- ,V_.__: at, write s.+ - s, -= a.+ + --- + a+, and apply
Theorem 4.8 and Theorem 4.6.
Taking p - 1 in (2) we find that lira,.® a, = 0 is a necesxm'y condition for
convergence of Y'.a.. That this condition is not uffieient is seen by considering
example in which a, = l/n. When n = 2" and p = 2" in (2), w¢ fred
I 1 2" 1
a+ + -'- + a,+, 2 = + 1 4- "" + 2" ÷ 2" -> 2"+2" - 2'
and hence the Cauchy condition cannot be satisfied when a < ½. Therefore the
series Y',__  i[n diverges. This series is called the harmonic series.
8.6 INSERTING AND REMOVING pAI2'HESE
Dfmitlm¢ 832. Let p be a fanction whose domain is the set of positive integers and
whose range is a subet of the poMtie integers tch that
 e(n) < e(m), if n < m.
Let  and b, be two eries related a follows:
Then we  dmt .b. is obtained from a. by inserting porenthese, wi tlmt , is
obtained from .b. by removing parentheses.
71¢eom 8.13. If  converges to s, ever ,ries .l,. obtained from . by 
f   d .   by (i and t¢ s. =  :  t. =  bt.
ea {t.} is a su of {s.}. In f t. = s.. efo, r of
{,} to s impfi n of {t,} to s.
Ro  y oy n. To   der 
 .  w h   0 (oboy nvt).  n) :  d 
 = (- 1.  (0 d 00 hold but  .
P   ov if we fuer    p.
Proof. If . cOnverges, the result follows from Theorem 8.13. The whole
dilficulty lies in the converse deduction. Let
t.-b +'"+b.,
Let e > 0 be given and choose N so that n > N implies
i.- tl <- and la.I < .
2 2M
If n > p(N), we can find m  N so that N _< p(m) < n < p(m + I). [Why?]
For such n, we have
s, : a z + "" + ag=+t) - (a,.+: + a.+ z + --- +
=- :.+ - (a.+ + a.+z + "'" +a=+)),

and hence
2 2
This proves that lim, s. - t.
g.7 ALTERNATING SERIES
Dcfmilion 8.I. If a. > 0 for each n, the rie -=1 (-1) "+1
alternating sers.
17keorem 8.16. If {a.} is a decreaMl sequence conging to O, the alternating
seres ,(-- ly '+1 a converges, lf s denotes its um cmdr. its nth partial sum, we have
the inequality
0 < (- l)(s - s o < a.+l, forn = 1,2 .... (3)
O', Inequality (3) tells us that when we "'approximate" x by s the error made
has the same sign as the first neglected term and is less than the absolute value of
this term.
Proof. We insert parentheses in (- 1)  +  a=, grouping together two terms at a
time. That is, we take p(n) = 2n and form a new series _b. according to Definition
8.12, with
Since a. - 0 and p(n + I) - p(n) = 2, Theorem 8.14 tells us that 2(- I) "+ a.
converges if Y_b. converges. But Y_b. is a series of nonnegative terms (since a. ,),
and its partial sums are bounded above, since
/¢ 1
Therefore b. converges, so Y(-I) "+t a. also converges.
Inequality (3) is a consequence of the following relations:
and
8,8 ABSOLU AND CONDITIONAL CONVERGENCE
Deyfni" n 8.17. A series .a. is called olutely convergent if,la.[ converges It
i called conditionally convergent if .a. converges but Y.la[ diverges.
Tlworem 8.18. /tbsolute convergence of a implie convgence.
Proof Apply the Cauchy condition to the inequality
la.+ + "" + a.+,[ < la.+l + "'" +
To see that the converse is not true, consider the exampte
This alternating series converges, b Thcorem $36, b it does not convmge
absolutely.
Theorem 8.19. Let  be a given serie with real-valued terms and define
P. la.[ + a. la.I - a.
-- , q. = (n = 1, 2 .... ). (4)
2 2
i) If .a. is conditionally convergent, both ,p. and .q. diverge.
ii) lf Nla.I converges, both p. and ,q. converge and we have
oe. p = a,, and q. -- 0 if a. > 0, whereas q. -- -a and p. = 0 if a. : 0.
Proof. W have a = p. - q=, lal  p. ÷ q, To prove (i), assume that 
converges and Y.lg.I diverges. If Y.q. convergeg then p. also converges (by
Theorem 8.8), since g. = a,. + q.. Similarly, if p. conrges, then qo also
converges. Hence, if either Y.p. or V_,,q. converges, both must converge and w¢
deduce that Zla.I converges, since la.I = p, + q.. This contradiction proves (i)
To prove (ii), we simply use (4) in conjunction with Theorem g.8.
8.9 REAL AND IMAGINARY PARTS OF A COMPLEX SERIF
Let .c. be a series with complex terms and write c. - a. + /b., where a and b.
are real. The series  and :..b, arc called, respevely, the real and imaginary
parts of c., in situations involving complex series, it is often convenient to treat
the  and imaginary parts separately. Of course, convergence of both a. and
N'b= implies convergence of _. Conversely. convergence of c. implies cow
vcrgence of both a, and b.. The same remarks hold for absolute ccmvergence.

However, when e. i comh'tiomdly convergent, one (but not both) of  and
b might bc absolutely convergent. (See Exercise 8.9)
If c= converges absolutely, we can apply pat ( of Theorem 8.19 to the real
and imaginary pa g1aratdy, to obtain hc decomposition.
U.c, = U.(p, + zuj- F.(q, +
where Y'.p., ¢., Yu., v. are convergent series of nonnegative terms.
g.10 TF_STS I¢OR CONVERGENCE OF
r 80 (C te). If
if there exist siti cotts c  N ch
a. < eb. fo, n "N,
then comergence of _.b. implies convergence
Proof. The partial sums ofa. are boundcxl if the pardal sums of,b, are bounded.
By Theorem 8.9, this completes the proof.
Theorem 8.21 (Limit ¢omporison tet). Assume that a > 0 and b. > 0 for
n = 1, 2, ..., and szppose that
Then  comerges if, and only if, _b, converges.
Proof There exists N such that n >_ N implies ½ < a./b. < z . The theorem fol-
lows by apply/ng Theorem 8.20 twi.
NO. Theorem 8.21 also holds if lim...® ,.lb. = c, provided that c ¢: 0. If
lim,. ® a/b. = 0. we can only conclude that convergence of ,k. implies con-
GEOMETRIC SERIES
To use comparison tests effectively, we must have at our disposal some examples of
series of known behavior. One of the most important series for comparison
purposes is the geometric series.
Tlgrem 8.22. If Ixl < 1, the series I + x + x z + "" converges and has sum
1/(1 - x). If[xl > 1, the series dizroes.
Proof. (1 - x)-o x* = '-=o ( x' -- x+t) = I -- X"* '. When Ixl < l, we
find Iim.... x '+j = 0. If Ixl > 1, the -n  ral term does not tend to zero and the
series cannot converge.
T'b. &23 The  Test 191
Further examples of series of known behavior can be obtained very simply by
applying the integral test.
Tkeorem 8.23 (lteral text), let f be a positive decreasing function defined on
[], +oe) such that fimx+®f(x) = O. For n = 1, 2,..., define
 f(k), &= I'f(x)dx, d. = s.-
Then we have:
i) 0 <f(n + 1) < d.+, <_d. gf(l), forn= 1,2,...
ii) lim. d. ex#u.
iii) _% t f(n) converges if, and only if, the sequence {t.) converges.
iv) 0 _< d - lim,_ d. <= f(k), for k = I, 2,...
Proof To prove (i), write
t.., = f(x) dx --- f(x) dx < f(k) dx
This implies thatf(n + I) = &+t - s < &+ - t,+ = d,+t, and we obtain
0 <f(n + 1) < d.÷,.
But we also have •
d. - d.+ = t.. - t. - (&+a - sJ = f(x) dx -f(n + 1)
f(. + )dx -f(. + o,
and hen d.+, g d.  dt = 1). s proves (i).
implies Oi) and at (ii) imp (i).
To prove  (iv), we u (5) a@in to write
0  d. - d=+  f(n) dx -f(, + 1) = f()  f( + D-
Sming on n, we get
0 _<  (d. - d..,) _<  (f(n) - f(n + 1)),
But now it is dear that (i)
When we evaluate he sums of these telescoping series, we gel (iv).
(5)
ifk > I.

NOT Let D = lim,_.,o d,. Then (0 implies 0 < D _ f(l), whexeas (iv) gives us
0 f(t) - f(x) dx- o <_ f(n).
(6)
This inequality is extremely useful for approximating certain finite sums by
integrals.
8.13 THE BIG OH AND LrrrLE OH NOTATION
De &.24. Givet two ,quences {a.} and {b,} such that b  0 for all n. We
write
a. = O(b.) (read: "a. is big oh of b.'),
if there exists a conxtant M > 0 such that la.I _< Mb.for all n. We write
a. = o(b.) as n -, o (read: "% is little oh
NOX. An equation of the form a. = c. + O(b.) means a. - c. = O(b.). Sim-
ilarly, a. = c. + o(b.) means a. - c. = o(b.). The advantage of this notation
is that it allows us to replace certain inequalities by equations. For example, (6)
implies
+
(7)
Exam 1.  f(x) = llx in Theorem 8.23. We find t. = log n and hence .l]n
diverges. However. (i estbIishes the existence of the limit
a famous number known as Fader's constant, usdIy denoted by C (or by 7)- Equation (7)
=tog.+c+o . (S)
Examl 2. Let f(x) -- x-'. s  1. in Theorem 8.23. We find that ,n - mverg if
s > I and diverges if s < 1. For s > l, this series defines an iml3ortant function known
as the Riemann zeta function:
For s > 0. s ¢ ]. we can apply (7) Io write
- + c() + O
tft I -S '
8A THE RATIO TEST AND ROOT TESW
Tkeorem 825 (Ratto test). Given a series __, of nonzero complex terms, let
r = lira inf la.÷,[, R -- lira sup
a) The series ,a. converges absolutely if R < I.
b) The series _,, diverges if r > t.
c) The test is inconclusive if r  I < R,
Proof Assume that R < I and choose x so that R < x < 1. The deflation of R
implies the existence of N such that la.+,la.I < x ifn > N. Since x =
this means that
]a,+l < la.l  lasl ifn _> N,
and hence la.I < cx" if n _> N, where c = [aslx -. Statement (a) now follows by
applying the comparison test.
To prove (b), we simply observe that r > 1 implies [a.+ t] > la.] fr all n _> N
for some N and hence we cannot have lira... ® a. = 0.
To prove (c), consider the two examples .-  and ,,n -. In both eases,
r = R -- I but "n-  diverges, whereas n-  conve.
71teorem &26 (Root esO. Given a series  of complex ferric, let
p = lira sup /l.l.
a) The series ,a, converges absolutely if p < I,
b) The series a, diverges if p > l.
c) The test is inconclusive if p = 1.
Proof. Assume that g < 1 and choose x so that p < x < 1, The definition of
implies the existence of N such that la.I < x" for n > N. Hence, Z[a.I converges
by the comparison lest. This proves (a).
To prove (b), we observe that p > I implies la.I > 1 infinitely often and
hence we cannot have lim, ® a = O.
Finally, (c) is proved by using the same examptes as in Theorem 8.25.
ffr. The root test is more "powerful" than the ratio test. That is, whenever the
root test is inconclusive, so is the ratio test, But there are examples where the ratio
test fails and the root test is conclusive, (See Exercise
8.15 DIRICHLET'S TEST AND ABEL'S TEST.
All the tests in the previous section help us to determine absolute convergence of a
series with complex terms. It is also important to have tests foe determining

convergence when the series might not converge absolutely. The tets in this
section are particularly useful for this purpose. They all depend on the part//
summation formula of Abel (equation (9) in the next theorem).
Thearem 8.27. If {a} and {b,} are two sequeaces of complex numbers, define
n   t intity
(9)
Therefore, -1 atb converges if both the serie ,t At(b+l - b) and the
sequence {Ab, + I } converge.
Proof Writing ,4 o -- 0, we have
(At- At-1)b ffi  Atbt-  Atbt+l + Ab+t.
tml tml tml k----1
The second assertion follows at once from this identity.
qo'r. Formula (9) is analogous to tim formula for integration by parts in a
Ricmann-Sticltjes integral.
Theorem 8.28 (Dirlcklet'a test). Let .a be a series of compfex terms whose partial
auras form a bounded zequence. Let {b} be a decreasing sequence width converges
to O. -Then Y.ab converges.
Proof Let A, -- a I + '-- + a, and assume that IA,I _< M for all n. Then
lira A.b. + _ = O.
Therefore, to establish 3nvergcnce of,ab, we need only show that .At(bt   - bt)
is convergent. Since b ",, we have
But the series '.(b ÷  - hi) is a convergent telescoping series. Hence the com-
parison tet implies absolute convergence of At(bi +1 - b).
em S9(.4bei's test). The series ,ab, converges if _.a converges and if
{b} is a. monotonic convergent #equence.
Proof. Convergence of Y.a. and of {b} establishes the existence of the limit
Iim._,. Ab.+ 1, where A. = a + ... + a.. Also, {A.} is a bounded sequence.
The remainder of the proof is similar to hat of Theorem 8.28. (Two farther tests,
similar to the above, are given in Exercise 8.27.)
PARTIAL SUMS OF THE GEOME'rRIC SERIES z" ON THE
u  wr Izl =
To use Dirichlet's test effectively, we must be acquainted with a few series having
bounded paxta[ sums. Of course, all convergent series have this property. The next
theorem gives Rn exampl© of a divergent series whose partial sums arc bounded.
• This is the geometric series  with Izl -- 1, that is, withz = e  where x is real.
The formula for the partial sums of this series is of fundamental importance in the
theory of Fourier series.
Theorem 830. For ever), real x  2ran (m is an integer), we have
 ea  = e  I - ' _ sin (nx/2)e,.+)/. (10)
t-t t -- e t. sin (x/2)
oz. This identity yields the following estimate,"
ea -< [sin (xt2)l
(11)
Proof. (1 - e Y,= e a = ,V_=z (e a - e'er+t) 0 = e ' - e< "+z)=. Thisestab
lishes the first equality in (10). The sond follows from the identity
1 e _ e - e -''a
I - e t e /z - e -t12
NOT, By considering the real and imaginary parts of (10), we obtain
-, cos kx sin ax _
= -- cos (n + I) sin x
=t 2 2
=  _1+ _tsin(2n + l)/sin x_,
2 2 2
(12)
sin kx = sin a__x sin (n + 1) sin _x. (13)
2 2
Using (10), we can also write
eU_ ) = e_C, ea:o _- sin nx
= tt SiO x
(14)
an identity valid for every x # mt (m is an integer). Taking real and imaginarY

parts of (14) we obtain
gin x
sin (2k - l)x --  sinz ax (16)
Formulas (12) and (16) occur in the theory of Fourier series.
8.17 RGEMF_,NTS OF SERIES
We recall that Z ÷ denotes the set of positive integers, + = {1, 2, 3,... }.
Definition 8.tl. Let f be a function whose domain is Z + and whose range is Z +,
and assume that f is one-to-one on g +. Let .a, and .b, be two series such that
b, = arc,) for n = !, 2,... (17)
Then b, is said to be a rearrangement of
NO'F. Equation (17) implies a. = b;-1¢.) and hence a, is also a rearrangement
Theorem &32. Let .a, be an absolutely cotwergent series having sum s. Then
every rearrangement of a, also converges absolutely and has sum s.
Proof. Let {b,} be defined by (17). Then
Ibll +--- ÷ Ih.I ffi la;¢,  l + "' + tat  .)l lad.
so  Ib, i has bounded partial sums. Hence b, converges absolutely.
To show that .b, ffi s, bet t,  b +... + b. s, = a +... + a,. Given
e > 0, choose Nso that ]ss - sl < /2 and so that y__%a la,,d " e/2. Then
It.-sl-< It.-s  t+ls  -sl<tt.-s  ,l
2
Choose Mso that {t, 2,..., N} _ {f(I),f(2),..., f(M)}. Then n > Mimplies
f(n) > N, and for such n we have
It,  sl ffi bl +--- + b, - (a +--- + as)l
: la,o) + -'- + a.(. - (a + --- + a)i
sin all the terms a,..., a s cancel out in the subtraction. Hence, n > M im-
plies It. -- sl <  and this means b. = ,.
&l$ .Nn  ON CONDITIONALLY CONVERGENT SERIES
The hypothesis of absolute coevergtnaoe is essential in Theorem 8.37-. Riemann
disooveved that any comh'tionally convergent series of real terms can be rearranged
to yield a series which converges to any prca'ibed sum. This remarkable fact is a
¢onstluem of" the folknvivg tin:
lflcacem &... Let ¢t be a cmtditiomdly convergent series with real.valued terms.
Let x and y be given  in the cloud interval [- , + c], with x < y Then
there exists a rearrangement .b, of a, uch that
lira inf , -- x and lira sup t, = y,
where t, -- b + .-. + b.
Proof. Discarding those terms of a serk:s which are zero does not affect its con-
vergeno¢ or divergence; Hem we might as well assume that no terms of  are
zero. Let p, denote the nth lmsith tetra of,V_.)g and let - q, denote its nth negative
term. Then d, and q, are both dit series of positive terms. [Why?]
Next, construct two sequences ofrea2 numbers, say {x,} and {y.}, such that
lira x, = x, limy, =y, withx, <y,, y > 0.
The idea of the proof is no quite simple_ We take just enough (say k) positive
terms so that
" followed by just enough (say rt) neatiw terms so that
Next, we take just czough further positive terms so that
followed by just enough furtlzer zjth terms to satisfy the inequality
+ Ptc= -- q + 1 ..... qra <
These steps are possible since ,ae, ad Yq. are both divergent series of positive
terms, if the process is cged it this way, we obviously obtain g rearrangement
o[ .a,. We leave it to the rzder |o sbowtht the partial sums of this rearrangement
have limit superior y and limit izffmSor x_
8.I9 SUBSERIIS
Defim'tio 8J4. Let f be a fuio whose  is Z + and whose range is 
infinite bset of Z +, d  t f b to-one on Z +. Let a. and , 

two series such that
Ten Y_b, ix said to be a bseries of Y.a:
Tllorm &35. If a, converges absolutely, every suberies  also converges
absolutely. Moreover, we hae
Proof. Given n, le! N be Zhe largest integer in the set {f(1),..., An)}. Then
The inequality :1 Ibkl < Y.E- lal implies absolute cnvergence oE_b..
m 8.36. Let {f, f, ... } be a countable collection of functim, each defined
on Z +, ha/ng the followin properties:
a) Each(, is _one-to-one on Z +.
b) The rage f,(Z +) is a bet Q. of Z +.
¢) {Qi, Qz, . . . }  a collection of disjoint sets whoe mion is Z +.
Let Y.a be an absolutely convergent serie aeut define
b(n) = %,, ifn  Z +, k  Z +.
i) For each k, Y-% x bk(n) is an absolutely con,ergent derie of Y.a r
ii) If s = y_.  b(n), .the ries y__%   converges absolutely and has the sume
Proof. Theorem 8.35 impli (i). To prove (ii), let t = Isll ÷ "'" ÷ Ixal. Then
t _<  Ib(n)l ÷ "'" ÷  IbRn31 = Y. (Ib,(n)l ÷ "'" ÷ IbRn)l)
But Y_,-t (la.a.>t + --- + la,,)l) _< T_..%, It. This proves that ls, I has
bounded paxtiai sums and hence ,.t converges absolutely.
To find the sum of .st, v proceed as follows: Let  > 0 be giw- and choose
N so that n  N implies
 .  (18)
Choose enough functions(a,... ,f, so that each term al, az, ..., a will appear
somewher in the sum
The number r depends on N and hence on . If n > r and n > N, we have
]sa + sz +."'+ s,- .o] <_ [a+] + [a÷] ÷... < -
because the terms at, a .... , a cancel in the subtraction. ow (18) implies
When this is combined with 09) we find
(19)
ifn > r,n > N.
$1 ÷ "'" ÷ $ -- 
This empletes e proof of (ii).
.20 DOUBLE SEQUF2CCES
DefMo 837. A function (whose domain is Z + × Z ÷  called a double sequence.
OT. We shall be interested only in real- or cornplex-valucl double .quences.
DeJtionJS. If a C, we write lim,..®f(p,q)= a and we say lat the
dubte equence f converge to , provided that the following condition is satisfied:
For every e > O, there exists an N sch that If(P, q) - a[ <  whenever both
p > Nandq > N.
Theorem #219. Assume that lim,.. f(p, q) = a. For each fixed p, aume t/at
the limit lim.. f(p, q) exists. Then the lmit lim,® (limq_ f(p, q)) also exists
and ha the alue a.
o__ To distinguish lim,..f(p, q) from limr. = (lim_®ffp, q)), the first is
called a double limit, the second an iterated limit.
Proof. Let F(p) = lim,_ f(p, q). Given  > O, choose N, so lhat
 ifp > N, and q > N. (20)
If(P, q) - al < ,
For each p we can choose N so that
IF(p) -- f(p, q)l <', if q > Na. (21)

(Note that N2 depends onp as well as on e.) For eachp > Na choose N2, and then
choose a fixed q greater than both NI and N 2. Then both (20) d (21) hold and
hence
Therefore, limr,® F(p) -- a.
so. A similar result holds if we interchange the roles alp and q.
Thus the existenve of the double limit tim.® f(p, q) and of lim,.® f(p, q)
implies the existence of the iterated limit
Iim (lim f(p, q)).
f(,q)- P (p= ,2,..., q= 1,2,...).
Then limef(p,q) = 0 and hen lim.® (lim..,f(p, q)) = 0. But f(p,q) -- ½
when p - q and f(p, q) = ] when p = 7.¢, and lumce i! is dear that the double limit
cannot exist in this case.
A suitable converse of Theorem 8.39 can be established by introducing the
notion of uniform convergence. (This is done in the next chapter in Theorem 9.16.)
Further examples il|ustrating the behavior of double sequences are given in
Exercise 8.28.
8.21 DOUBLE SERIF
Def'tion 8.40.
by the equation
Let f be a double sequence and let s be the double sequence defined
fm---- I
The pair (f, s) is called a double series and is denoted by the symbol ,., f(m, n) or,
more briefly, by .J'(m, n). The double ser¢es is said to converge to the sum a if
]im (p, q) ffi o.
Each number f(m, n) is called a term of the double series and each s(p, q) is
a partial sum. If r(m, n) has only positive terms, it is easy to show that it con-
verges if, and only if, the set of partial sums is bounded (See Exercise 8.29.) We
say _ff(m, n) converges absolutely if Y_[f(m, n)[ converges. Theorem 8. t8 is valid
for double series. (See Exercise 8.29.)
8.?.2 REARRANGEMENT THEOREM FO] DOUBLE SERIES
n 8.41. Let f be a double sequence and let g be a one-to-one function
defined on Z + with range Z + × Z +. Let G be the sequence defined by
G(n) - fig(n)] (f . e Z +.
Then g is said to be an arrangement of the double sequence f into the sequence G.
lleerem &42. Let ,W__j'(m, n) be a given double series and let g be an arrangement
of the double sequence f inw a sequence G. Then
a) ,G(n) converges absolutely if, and only if, ,.ff(m, n) converges absolutely.
,,luming that f(m, n) does converge absolutely, with sum S, we have further:
b) y_,.%, a(n) = s.
c) _..  f(m, n) and .= a f(m, n) both converge ubsolutety.
d) If A, = ,_ff=  f(m, n) and B, = Z.  f(m, n), both series ./t. and ZB,
converge absolutely and both hatw sum S. That is,
m--| nl =1
Proof. Let T - IGO)[ + --" + IG(k)l and let
s(e, q) = If(,,,,
Then, for each k, there exists a pair (p, q) such that T < S(p, q) and, conversely,
for each pair (p, q) there exists an integer r such that S(p, q) < T,. These
equalities tell us that IG(n)I has bounded partial sums if, and only if, Elf(m,
has bounded partial sums. This proves (a).
Now assume that .lf(ra, n)[ oonverges. Before we prove (b), we will show that
the sum of the series .G(n) is independenl of the function .q used to construct G
fromfi To see this, Iet h be another arrangement of the double sequencefinto a
sequence H. Then we have
G(.) = lEg(n)] and H(n) = fib(n)].
But this means that G(n) = H[Mn)], where k(n) = h-a[e(n)]. Since k is a one-
to-one mapping of Z + onto Z +, the series 57H(n) is a rearrangement of G(n),
and hence has the same sum. Let us denote this common sum by S'. We will
show later that S' = S.
Now observe that each series in (c) is a subseries of" 57G(n). Hence (c) follows
from (a). Applying Theorem 8.36, we conclude that YM,, converges absolulely
and has sum S'. The same thing is Irue of Y'B,. ]1 rerains to prove that S' -- S.

For this purpose let T = lim,,® S(2, q). Given  > 0, choose N so that
0 < T - S(p, q) < 8/2 whenever p > N and q > N. Now write
Choose M so that t¢ includes ai terms f(m, n) with
I <m<<_N+ ], 1 <nAN+ l..
Then ta - s(N + l, N + 1) is a sum of terms f(m, n) with either m > N or
n > N. Therefore, ff n. _> M, we have
Similarly,
it.- s(N+ 1, N + I)l < T- S(N + I,N + 1) < ,.
IS- s(N + 1, N + 1)1 < T- S(N + 1, N + 1) < -.
- 2
Thus, given  > 0, we can always find M so that I t. - SI <  whenever n _> M.
Since lira...® t. -- S', it follows that S' = S.
Ior. The series _ t V_..%_  f(m, n) and ,_,, 1 -.,-  f(m, n) are called "iterated
series". Convergence of both iterated series does not imply their equality. For
example, suppose
f(m, n) = - I,
O,
Then
ifm =n + I, nffi 1,2,...,
ifm =n- l,n= 1,2,...,
otherwise.
but  :S(m,n)= L
lK23 A SUFFICIENT CONDITION FOR EQUALITY OF ITERATED SERIES
Tbear¢m &45. Lel f be a complex-mlued double sequence. Assume that ..%_  f(m, n)
eonverea absolutely for each fixed m and that
converges. Then:
a) The double serie, ,., f(m, n) converges absolutely.
b) The series Y  fm, n) converges absolutely for each n.
c) Both iterated eries _%  _,, f(m, n) and _._  _,%  f(m, n} converge
absolutely and we have
Proof Let g be an arrangement of the double sequencefinto a sequence G. Then
G(n) is absolutely convergent since all tim partial sums of l G(n)]_ are bounded
by _._- ,V_. If(m, n)l. By Theorem 8.42(a), the double series .,.f(m, n)
conv absolutely, and statements (b) and (c) also follow from Theorem 8.42.
As art application of Theorem 8.43 we prove the following theorem concerning
double series y_,.f(m, n) whose terms can be factored into a function of m times
a function of n.
Tkerem 8.44. Let , and .b. be two abxolutely conrgent seres with sums
A and B, respecaely. Letfbe the double equence defined by the equation
f(m, n) = ad  ,, if (m, n) Z + × z +.
Then x..,f(m, n) converges absolutely and has the sum AR
Proof. We have
Therefore, by Theorem 8.43, the double series ,V_,,, a=b. converges absolutely and
has sum AB.
MULTIPLICATION OF 
Given two series  and ,V_.b., we can always form the double series d"(m, n),
wheref(m, n) = ab.. For every arrangement g off into a sequence G_, we are led
to a further series C(n). By analogy with finite sums, it seems natural to refer to
]'f(m, n) or to G(n) as the "product" ofa. and y_.b,, and Theorem 8.44 justifies
this terminology when the two given series  and ,,b, are absolutely convergent.
However, if either  or ,,b. is conditionally convergent, we have no guarantee
that either 'Ef(m, n) or a(n) will convere. Moreover, if one of them does
converge, its sum need not be 4B. The convergence and the sum will depend on
tim arrangement g. Different choices ofg may yield different values of the product.
There is one very important case in which the terms f(m, n) are arranged "'diag-
onally'" to produce YC_r(n), and then parentheses are inserted by grouping together
those terms a,b= for which m + n has a fixed value. This product is called the
Cauchy product and is defined as follows:

two seres T_.o , and T-.,%o b,, define
e. =  atb._, if n = O, 1, 2,... (22)
kiO
The series 7.%o c= is called the Cauchy product of j and T_.h,.
NOlX The Cauchy product arises in a natural way when we multiply two power
series. (Se¢ Exercise 8.33.)
Because of Theorems 8.44 and 8.13, absolute convergence of both a= and
b. implies convergence of the Cauchy product to the value
This equation may fall to hold if both .a= and T_, are conditionally convergent.
(S¢ Exercise 8.32.) Howccr, we can prove that (23) is valid if at least one of
a, b, is absolutely convergent.
Tieerem &46 (MerU). .4ne that -%0 a. commrges ablutely and tu m
A, and auppose ,T_% e b converges with aura B. Then the Cauchy product of these
two series convergeJ and has sum ,4B.
Proof. Define A. = T_:,= = =
0 at, T_.I=0 c. X  -o wh  re is #yen by
(22). Let d. ---- B -- B. and e= =  =o atd-t- Then
where
Then (24) bocomcs
Jfn :> k,
ffn <k.
k=O nlO k=O =--k tlO m=O
=  at(B - de_k) = AeB - %.
To complct the proof, it suilices to show that %  0 as p -, co. The sequence
{d.} converges to 0, since B = T_,,b=. Choose M > 0 so that Idol < M for all n,
andletK = T_%o la=l, Given, > 0. choose Nso that > Nimplies Id.[ < t/(2K)
and also so that
--s+l 2M
Then, for p > 2N, we can write
la, I < + =,.
- 2K --0 =+=  
This proves that e, - Oasp  co, and hence Ce  ABaspo co,
A la m (due to AI), in wch no aIu nr is sd,
will  prov in e next chapter. ( eom 9.32.)
Anoer pr, known 
in te Theo of Num. We take a o = b o = 0 and, ins of dning c. by
(), we u the fou
=
wh 1. ms a sum n over M] siti disors of n (including I and
). For cmp]c, c = aab e + ab 3 + abz + ab t, and c =atb + aTbt.
The a]og of Mecns' theom holds al for this pruct. The Difichl p
as in a nature] way when we multiply Didch]et fies. ( Exe 8..)
Defudtion 8.47. Let s denote the nth partial sum of the Series _j%, and let {tv=} be
the ,equence of arithmetic meana defined by
s + "'" + s.
= , = ....
The series .a, i$ aid to be Cesdwo summable (or (C, I) summable) if {o.} cmwerge$.
If lim, ® r, - S, then $ is called the CeVo sum (or (C, 1) sum) of ,, and we
write
Za. = S (C, ).
Example 1. 1 a. = z'. ]zl = 1, z # 1. Then
1 z" ! ! z(! - z")
.= and ta=
1 - z I - z ! - z n (1 - z)z "
in particular,
 z "- = -- (C, !).
Jtl

Eaton#e2. I_t a. = (-l)+tn. In this case,
lira sup r. = ½, lira inf r, = 0,
and hence (- ly'+'n is not (C, 1) gumnmble.
Theorem 8.#& lf a series is comrgent with sum S. then it is also (C, 1) suramable
with Cesdro sum S.
Proof Let ,, denote the nth partial sum of the series, define r, by (26), and
introduce t, = s, - S, % = r, - S. Then we have
"t, -- , (27)
and we must prove that r, --, 0 as n --, o. Choose A > 0 so that each It.] _< .4.
Given, > 0, choose. N so that n > N implies It,[ < . Taking n > N in (27),
we obtain
I%1 < Itl +-'- + t:1 + It+,l + "'" + ]t.I < NA
Hence, lira sup,= [t,l < s. Sinc • is arbitrary, it follows that Iim,® I*.l = 0.
NO. We have really proved that if a sequence {s,} converges, then the SlUence
{cry} of arithmetic means also converges and, infact, to the same limit.
Cos&to summability is just one of a large cla of "summability methods"
which can b¢ used to assign a "sum" to an infinite  l  8.48 and
Example I (following Definition 8.47) show us that 's method has a wider
scope than ordinm'y oonvergence. The theory of summability methods is an
important and fascinating subject, but one which we cannot enter into here. For
an ecellent account of the subject the reader is referred to Hardy's Divergent
5'er/e (Reference 8.1). We sludl e later that (C, 1) summability plays an impor-
tant role in the theory of Fourier series. (See Theorem 11.15.)
88 E PitODUCTS
This section gives a brief introdion to the theory of infinite products.
 &.49, Gizn a sequence {u,} of real or complex numbers, let
Pl = u,. p = uu. p, = utu2""u, = [ u. (28)
The ordered pair of sequences ({u}, {p,}) is called an infinite product (or simply,
a product). The number p, i called the nth partialproduct and u, is called the nth
factor of the product. The fottowin symbols are used to denote the product dgned
y (25):
uiuz " " u. - " , ] u.. (29)
O. The symbol .=+  u. means I .  u+.. We also write I-[u when the
is no danger of misunderstanding.
By anaIogy with infinite series, it would seem natural to call the product (229)
convergent if {p} converges. However this definition would be inconvenient
since every product having one factor equal to zero would converge, regardless of
the behavior of the remaining factors. The following definition turns out to be
more useful:
Definition &.50. Gitwn an infinite pro&,et I'I%  u,, &t p, = =  u,.
a) If infinitely my factors u, are zo, we y the proct dirges to zero.
b) lf no factor u. is zero, we say the prct conrges  tre exits a number
p # 0  tt {p,} corges W p. In th ce, p is fed 1he lue of the
prct  we write p = =  u,. lf {p} crges to zero, we say t pruct
dioerges to zero.
c) If there exits an N ch that n > N p[ies u  0, we say   u conrges,
prid tt =s+ u rge  descrd  (b). In th e, the lue
of the procl %
d)  t u is 1 rgent if it do not e  scrd in )  (c).
Nora that e vMue of a nt i pruct n  ro. But  pns
if, and only if, a fite nmr of fao m o. The nvergen of an infinite
pr is not  by  or mong a finiM humor of factor, zero or
not. h is s fa wMch mak finifion 8. v¢ convenient.
e. = (l + l/n) d
p = n + I, and in the nd
TIteorem 8.51 (Caucky emuliaon f presets). The infinite product H u, con-
verges if, and only if,.for ever), e > 0 there eMsts an N such that n > N implies
lu=+,u.+2"'" u+ - 11 < n, fork = 1, 2, 3,... (30)
Proof. Assume that file product IIu, converges. We can assume that no u, is
zero (discarding a few terms if necessary). Let/7, = u ... u. and p -- lint,_,® p.
Then p # 0 and hence there exists an- M > 0 such that Ip=l > M. Now
satisfies the Cauchy condition for sequences. Hence, given  > 0, there is an N
such that n > N implies IP.+t - P,I < ,M for k = t, 2,... Dividing by
we obtain (30).

How seme that condition (30) holds. Then n > N implies u,  0. [Why?]
Take -- ½ in (30), let N O be the corresponding N, and let
ifn > No. Then (30) implies ½ < lq.l < ½. Therefore, if {q.} converges, itcannot
converge to zero. To show that {q,} does converge, let  > 0 bo arbitrary and
write (30) as follows:
I
I
Tkcorem 8,54. dbsolule convergence of II(l + a,) implies convergence.
Proof. Use the Cauchy condition along with the inequality
I(I + a.+j)(l + a.÷.z).-.(l 4- a.+) - I[
Nor. Theorem $.52 tells us that II(l + a} converges absolutely if, and only if,
, converges absolutely. In Exercise 8.43 w¢ give an example in which II(I + a)
convrls but .a. diverges.
A sult analogous to Theorem 8.52 is the following:
lkearcm 855. ,aaume that each a > O. Then the product rio - a0 converges
if, and only if, the series ,a a com:erges.
Proof. Convergence of  impfies absolute convergenc (and hence convergence)
of II(1 -
To prove the converse, as,ume that  diverges. If {a,} does not converge to
zero, then II(l  a,) also diverges. Therefor¢ we can assum that a, -, 0 as
n - o. Discarding a few terms if necessary, we can assume that each a, ..< ½.
Then each factor !  , > ½ (and hence # 0). Let
p.-- - aDO -
Since we have
(1- a)(1 + at)-- 1- a< 1,
we can write p, < 1/q,. But in the proof of Theorem 8.52 we observed that
q, -, +  if :a diverges. Therefore, p,,  0 as n  o and, by part (b) of
Dgfinition 8..50, it follows that I(l -- a) divgrges to 0.
8.27 E'S PRODUCT FOR THE RIEMANN ZETA FUNCI'ION
We conclude this chapter with a theorem of Euler which expresses the Riemann
zeta function (s) - =: n - as an infinite product extende over all primes.
Ttworem 8.56. Let p denote the kth prime number. Then if s > ! we have
= - = -
The product converges absolutely.
Proof. We consider the partial product P. = 1-I'--t (l - p-')-t and show that
P, - (s) as m - o. Writing each factor as a geometric series we have
a product of a finite number of absolutely convergent series. When we multiply
these series together azd rearrange the terms according to increasing denominators,
we-get another absolutely convergent series, a typical term of which is
1
= - where n = p ".-

210
and each a 20 Therefore we have
P - ns •
where 't is summed over those n having all their prime factors <p.. By the
unique factorization theorem (Theorem 1,9), each such n oozurs once and only
once in '-1. Subtracting P. from (s) we get
1
I'  t B s 2 I s 
where 2 is summed over those n having at least one prime factor >p... Sncc these
n occur among the integers >p., we have
P.I-<
As m   the last sum tends to 0 becaus _)-" converges, so
To prove that the product cnverges absolutely we use Theorem 8.52. The
product has the form I](l + a,), where
1 1
at:---I----l-..-
The series ,.a t converges absolutely since it is dominated by n-*. Therefore
[(1 + at) also converges absolutely.
8.1 a) Given a teal-valued sequence {a} bounded above, let u, = sup {a t : k > n}.
The us'- and hence U ffi lira,_,® u, is either finite or -. Prove that
U : lira sup a m -- Ih31 (sup {at : k >_
b) Similarly, if {a.} is bounded beIow, prove that
V - lim hff a= -- lira (inf {a : k > n}).
If U and V are finite, show that:
c) There exists a subsequece of {a.} which converses to U and a subsequence
which cOnverges to V.
d) If U = V, every sulCluene of {a.J convergg to
8.2 Given two real-valued sequences {as} and {bs} bounded below. Prove that
a) lira sups_ (a. + b.) < lirn sup..® a. + lira up.®
211
b) lira sup..  (a,,b.) _< (lira sups_  a.)(lim sups_  b) if a. > 0, b. > 0 for
and if both lira sups_  a. and lira sup,_ b. are finite or both are infinite.
8.3 Prove Theorems 8.3 and 8.4.
8.4 If each a > 0, prove that
lim inf as+t  lira inf / <: lira sup / _< lira sup
g.  a = nlnL Show that fim,_,o a+la,  e and use Exercise 8.4 to deduce that
,..,® (n!)/s
8.6 Let {as} be a real-valued sequence and let #. = (a + ... + a,,)ln. Show that
lira inf a, _< lira inf r _< lira sup  : llm sup
8.7 Find lira sup._  a. and lira inf._,  a. if a, i given by
a) csn, b)(l +-ln) conn , c) nsin nn
3'
sin - cos -,
 In (f), Ix] dcaotes the greatest inte4 _<x.
8.8 Let a. = 2/ - _= I/x/. Prove that the sequence {a, converges to a limit p
in the interval 1 < p < 2.
In each of Exercises 8.9 through 8.14, show that the real-valued sequence {as} is con-
vetgent. The given conditions arc asumed to hold for all n > 1. In Exercises 8.10
through 8.14, slw that {a.} has the limit L indicated.
8.10, at > ,0,, a, > O, a,,,+ - (a,,a.,,+) t:, L --- (aa) ;,
a2a2.- 
8.11 a = 2, a = 8, aasst = (a2, + azs+), a2,+z -
--,L=4.
L = 1. Modify a t to makeL = -2.
8.13
3+a,
8.14 a, - +, wh bt
Hint.  tt b,.ab. - b+ = (-IT  and u t Is.  a..[ < n -, if
n>4.

Test for conversence (p and q denote fixed reaI numbers).
8.16 Let S = {n, n,... ] denote the coilectio of thoe positive integers that do not
involve the digit 0 in their dccin representation. (For example, 7  S but 10l ¢ $.)
Show that __ t 1/ converges and hasa sum less than 90.
8.17 Given intege at, a2 .... such that 1 <__ a, _< n - I, n = 2, 3 .... Show that the
sum of the series .,%1 a./n! is rational if, and only if, there exists an intcg Nsuch that
a,, -- n - 1 for aIia >_ N. Hint. For suiciency, show that ,-z (n - I)/n! is a el-
moping sexies with sum I.
.18 Let p and q be fixed integers, 9  q -> !. and let
a) Use formula (8) to prove that lira.... + x, = log (p/q).
b) When q = I, p --  show that sz, -- x. and deduce that
 (- 1)'+ -
c) Rearrange te 'fies i (b), ,tig alternately  liti ler foilM b q
negative terms and use (a) to show that this rearrangement has sum
log 2 +
d) Find the sum offf= (-1) + ](I/On - 2) - 1/(3a -. I)).
&19 Le c. = a, + ib., whera. = (- l)a/x/, b. = !/ . Show thatc.iscondltionally
convergent.
b)
/-" i + a.
80.6 Determine all real values of x for which the following series converges:
(if no a, =  1),
8.27 Prove the following statem(mts:
a) .aOb converges if .g, ecmverges and if (b, - b,+ ) converges absolutely.
b) b, convers if a. has bounded partial sums ad ff(b. - b,+ ) converges
absolutely, provided that ba - 0 as n - 0o.
8.28 Investigate the existence of the two itcrat limits and the double fimit of the double
a) f(p, q) = 1 b) f(p, e) -- P
p+q P+q

2,14
¢) f(p. q) - (- I)np d) f(p,q)--(-1)n+a(+ )
e) f(p, q) - ( l)n , f) f(p, q) = ( 1) n+,
q
g) f(p, q)  h) f(p. q) ---- -" sin
q .-_: P
Answer. Double limit exists in (a), (d), (e), (g). lth iterated limits exist in (a), (b), (h).
Only one iterated limit exists in (c), (e). Neither iterated limit exists in (d), (f).
8.29 Prove the following slaternents:
a) A double series of positive tenm conver if, and only if, the set of partial sums
is bounded.
b) A double series convexg if it conveq absolutely.
c) _ .,,e
830 Assume thai the double series ,,., na)x " converses abolulely for Ixf < 1. Call
its sum S(x). Show that each of the following series also converses absolutely for Ixl < 1
and has sum
consists of p complex numbers of which one has absolute value , thr satisfy
I1 + 2t't < lm + inl < 2,], five satisfy I1 ÷ 311 -< Im+ inl _< 3, etc. Verify this
gmqrkally and deduce e inequality
= ,,,. (n  +
2 a) $how that the Cauchy product of=o (- IP+z/ + l with itselfisa divergent
) Show that the Cauchy product of_ff=o (-l):+z/(n + I) with itself is the
+ i+'"+ "
Does this converge? Why?
 Given two absolute converSnt power  say ,.o a' and _ b -', having
sums A(x) and B(x), respcliwly, show that :,0 c x= - A(x)B(x) where
834 A series of the form Y% 1 aa[t/ is called a Dirietdet *r/es. Given two ahsolutely
converlt Dirichlet series. ay _.= 1 a,/n and _.. z b n, having suras A0) and B(s),
respectively, show that :,  cln*  A(a)B(s) where e.
aSS If I(#) -- % /,,  > , lOW that fz(s) -- y.,%, a(n)/,, where '(n) is the
number of positive divisors of n (including 1 and n).
Cemo ntbillq
8.3 Show that each of the following  ha (C, I) sum 0:
a) 1- 1- +1 + 1--1- 1 + 1 + 1--++..-.
c)  x +  3x +  5x + .--(x ,x  ).
37 Gi a  , t
at, t, = kat,  n =
a) , = ½a. + (-
c),_ff.a(-ly'(l +½ +,..+ l/n) = -iog (C,I).
8..39 Determine whether or not the following infinite products converge. Find the value
of each cont product.
2 b) (1 n-±),
a) 1 nin + I) "
I0 If each partial sum a, of the convergt seal,s  is not z=ro and if the sum itself
is not zero, show that the infinite product a I17= (I + ads._ ) converges and has the
value :  a.

• 41 Find tim values of the following p:luc.s   t [ng i d
:
a) 1 + =2. 2-*. b) 1 + =2 I
&42 Determine all real x for which the produc = 1 cos (x/2 converges and lind the
value of the product when it does converge.
8.43 a) Let a, = (- IY'/n for n = I, 2 .... Show that I-[(1 + a,) divert but that
'O a
b) Let az,_ = -1/x/n,%. = |/xn + l/aforn - 1,2,... Showthal l-I(! +
converges but that ,r, diverges,
&44 Aume 1hat a,  0 for each n = !, 2 .... Assume further that
Show tat II= (a + (-;)%) to.verbs if, and ora ][, X,%, (-])% converges.
tlA A complex-valued sequemce {f()} is called multiplicatioe Elf(l) :- 1 and if f (ran) =
f(m)f(n) whenever m and n are relatively prime. (See Section 1.7.) It is calIed cony
pletely mult71iatire if
f{l) = I and f(tnn) = f(m)f(n) for all m and n.
a) If {,'(n)} is multipticative and if the series ,f(n) converges absolutely, prove that
E f(n) -- H {1 + f(p + f(p:) + --. },
where p denotes the kth prime, the produc being absolutely convergent.
b) If, in addition, if(a)} is completely mutfip]icat/ve, prove that the forraula in (a)
Note that Euler's product for () (Theorem 8.56) is lh¢ special case in whifh
8.46 This exercise outlines a simple proof of the formula (2) = g2/6. Slarl: with the
inequality sin x < x < tan x, valid for 0 < x < n/2, ake reciprocals, and square each
member to obtain
cotZx <--I < 1 +cot x.
X 2
Now put x -- kzr/(2m + I), where k and m are integers, with I <_ k .g m, and sum on k
to obtain
217
Use the formula of Em:rcise 1.49(c) to deduce the inequality
m(2m- 1)= a<* i < 2re(m+ 1)n =
30n + 1) 2  .. 3(2m + I) =
Now  m --, o to obtain (2) = x16.
 Um an argument  to thax outlined in Excrcis &46 to prove that '(4) -- :/90.
SUGGESTED REIcEiIENCES FOR  SI%Y
8.1 Hardy, G. H., Dlternt Ser/es. (htford University Press, Oxford, 1949.
8.2 I-lincltmana, I. I., iFm#e er/es. Holt, Rimlmrt and Winston, New York, 1962.
8.3 Knol K,, T&-oey and dp/d/cat/oa of lnfuffte Scri 2rid o R. C. Young, tran
latex. Hafer, lqew Yot 1948.

CHAPTER
SEQUENCES
OF FUNCTIONS
9.t I'OINTWl CONVGF_.NCE OF  OF ONS
  dls wi  } who s a 1- or mex-val
fons  a on domMn on e 1 lira R or in e mpl pMne C.
For  x in the domn   fo o  {f,(x)} wh tes
 nding function values.  S denote  t of x for wMch this nd
u nve. e funionfde by the fion
f(x) = lira f,(x), if x  S,
is   litctn of  n {}, d we y at {} crges
twise fon e t S.
r chief int in this ¢hapr is e following  of qufion : If h
function of a n {A} h a n pmy, such  nfinuity, diffen-
bili, or inequality, to wt exit is ts pmy sf to t t
funon? For emple, ifch funon  is afinuous at c, is e lit fuaion
f nfinuous at c7 We sI  tMt, in nerai, it is not. In f we fl
d t intwi conveen is lly not strong ough to ner any of the
pes mention ave from the indifidual s A to the limit funon f
Thefo we a l to study stronger methods of nvern that do p
e proi. e mt implant of the is e notion of orm nven.
fo  tu unifo convergent, let  foulate one of our
questions in another way When we k whether ntinm ofh at c implies
continui of the limit function fat c, we a Ity king whher the equation
implies the equation
lira f(x) = f(c). (1)
But (I)can also be written as follows:
lira lira f,(x) -- lira lira f(x). (2)
Therefore our question about continuity amounts to this: Can we interchange the
timit symbols in (2)? We shall see that, in general, we cannot. First of nil, the
limit in (I) may not exist. Secondly, even if it does exist, it need not be equal to
f(c). We encountered a similar situation in Chapter 8 in connection with iterated
'ies when we found that ,-.=1 5".f(m, n) is not necessarily equal to
Thz general question of whether we can reverse the order of two limit pro.
cesses arises again and again in mathematival aaa]ysis. We shall find that uniform
convergence is a far-reaching sucient condition for the validity of interchanging
certain limits, but it does not provide the complete answer to the question. We
shall encounter example in which the order of two limits can be interchanged
ahhough the sequence is not uniformly convergent,
9.,2 EXAMPLES OF SEQUENCES OF REAL-VALUED FUNCTIONS
The following examples illustrate ome of the possibilitie that might arise when
we form the limit function of a sequence of teat-valtted furions.
Irure 9.1
Exam 1. A equence of continuous fimctlons with a
(x) = xl(l + x ) if x  R, n ffi 1,2,... pofaftsin
Fig. 9.1. In is  1. fx)
{ iflxl<',
f(x) = if [xl = I,
if Ix[ >
  is ntinuo on R, t f  ditinuous at x = ! and x = - 1.
 2, Aqoffctiofichl
fx)=l - ifxR,n-1,,+. If0x 1 ellmif(x) = lim,(x)
ists and uals 0. ( Fig. 9,Z) H yf(x) dx = O,
= n  (t - t)t'dt- -
n+ 1 n+2 (n+ )(n+ 2)

mtegl of the limit fmion. "l'nefefore the  of "'limit'* and
catmm always be intetdmnged.
Let.f,,(x) = (sin nx)lifxR, n = 1.2,... Then lim._,®,f/fx) = 0 forevetyx. But
f(x) = / o0s x, so lim..®.f(x) does not exist for any x. (See Fig,
9.3
9.3 DEFINITION OF UNIFORM CONVEItGENCE
Let {f} be a sequence of functions which converges pointwise on a set S to a
limit function f This means that for each point x in S and for each • > 0, there
exists an N (depending on both x and ) such that
implies [f.(x) -f(x)l < t.
Tll. 9, Uaifetm Cemvmm
If the same N works equally well for every point in S, the convergence is said to be
un/form on S. That is, we have
Deftahfon 9.1. A sequence o f functions {f-} is said to cormerge uniformly to f on a
set S if, for ever),  > O, there exists an N (depending only on ) uch that n > N
implies
[f.(x) - f(x)l < ¢, for every X in S.
We denote this symbolically by writincj
£  f uniformly on S.
When each term of the sequence {f,} is real-valued, there is a useful geometric
interpretation of uniform convergence. Th© inequality If,,{x) - f(x)l < t is then
equivalent to the two inequalities
.f(x) -- ts < ..f,(x) < f(x) + t. (3)
If (3) is to hold for all n > N and for all x in S, this means that the entire graph
off. (that is, the set {(x, y) : y = f,(x), x e S}) lies within a "band" of height 2t
situated symmetrically about the graph off. (See Fig. 9.4.)
Flgllre 9.4
A sequence {.f.} is said to be uniformly bounded on S if there exists a constaat
M > 0 such that If,(x)l < M for all x in S and all n. The number M is called a
uniform bound for {f.}. If each individual function is bounded and if f. - f
uniformly on S, then it is easy to prove that {f.} is uniformly bounded on S. (See
Exercise 9.1.) This observation often enables us to conclude that a sequence is
not uniformly convergent. For instance, a glance at Fig. 9.2 tells us at once that
the sequence of Example 2 cannot converg uniformly on any subset containing a
neighborhood of the origin. However, the convergence in this example/.¢ uniform
on ¢¢ery compact subinterval not containing the origir.
9.4 UNIFORM CONVERGENCE AND CONTINUITY
Theorem 9.2. Assume that f. -, f uniformly on S. If each f. i continuous at a
point c of S, then the limit funciion f is alo continuous at c.
qoTls. If c is an accumulation point of S, the conclusion implies that
lira lira f.(x) - lira limf,(x).

Proof. If c is an isolated point of S, then f is automatically continuous at c.
Suppose, then, that c is an acatmulation point of S. By hypothesis, for every
 > 0 there is an M such tlmt n > M implies
If(x)
for every x in S.
Since fu is continuous at c, there is a neighborhood B(c) such that x  B(c) c S
implies
3
But
Iffx) -/(c)I <_ If(x) -f(x)l + Iffx) -A,(c)l + IA,(c) -/(c)l.
Ira  B(c) ¢ S, each term on the fight is less than t/3 and hence If(x) - f(c)l < .
This proves the theorem.
or Uniform convergence of {f,} is sutticient but not necessary to transmit
continuity from the individual terms to the limit function. In Example 2 (Section
9.2), we have a nonuniformly convergent sequence of continuous functions with
a continuous limit function.
THE CAUCHY CONDITION FOR UNIFORM CONVERGENCE
T/amrem 9.3. Let {f.} be a sequence of functiord defined on a set S. There exists a
function f such that f. - f uniformly on S if, and only if, the followiny condition
(called the Cauchy condition) i sat£tfied: For every e > 0 there exists an N such
that m > N and n > N implies
]f(x) - fe(x)l < e, for eoery x in S.
Proof Assume thatf. --, funiformly on S. Then, given z > 0, we can find N so
that n > N impfies [f.(x) - f(x)l < e[2 for all x in S. Taking m > N, we also
have [fx) - f(x}l < e[2, and hence If.(x) -/,(x)l <  for every x in S.
Conversely, suppose the Cauehy condition is satisfied. Then, for each x in S,
the sequence {f(x)} converges. Letf(x) = Faa,..®./'.(x) ifx S. We must show
that f. - f uniformly on S. If  > 0 is given, we can choose N so that n > N
implies If.(x) - f.+i(x)l < g/2 for every k = 1, 2, ..., and every x in S. Ther¢
fore, in® IL(x) -L.(x)] = If.(x) -f(x)l < 12. Hence, n > N implies
lf.(x) - f(x)[ <  for every x in '. This proves that f. - f uniformly on S,
ro'rE. Pointwise and uniform convergence can be formulated in the more general
setting ofmetr/e spaces. Iff, and fare functions from a nonempty set S to a metric
space (T, dr), we say that f, -, funiformly on S, if, for every  > 0, there is an
N (depending only on e) such that n > N implies
d(f,(x), f(x)) < e for all x in S.
9.7 UMfeum  el' laflnite  2
Theorem 9.3 is valid in tkis more general setting and, if S is a metric space, Theorem
9.2 is also valid. The same proofs go through, with the appropriate replacement
of the Euclidean metric bythe metrics ds and dr. Since we are primarily interested
in real- or complex*valued functions defined on subsets of R or of C, we will not
pursue this extension any further except to mention the following example.
gXaml Consider the metric space (B(S), d) of all bounded real-valued functions on a
nonempty set s, with metric drY, #) -- If- gll, where ]tfll = sup. [f(x)l is the sap
norm. (See Exerc/se 4.66.) Then.f, --,/'in the metric space (B(S), a r) if and only iff,  f
uniformly on S. In other words, uniform convergence on S is the same as ordinary con-
vergeax in the metric slxsce (B(S), dr).
9.6 UNIFORM CONVERGENCE OF INFINITE SERIES OF FUNCTIONS
Def¢niti 9.4. Giren a sequence {f} of functions defined on a set S. For each x in
S, tet
so(x) =  A(x) (n = , 2,...). (4)
If there exists a function f such that s. - f uniformly on S, we say the seres r.(x)
converges uniformly on S and we write
 f(x) = f(x) (uniformly on
,=l
Tlleorem 9$ (Calelly condition for gclifofm comrgenee of serles). The infinite series
_,f(x) converges uniformly on S if, and only if, for every  > 0 there is an N uch
that n > N implies
[  ft(x)t< , foreachp= 1,2,...,andeoeryxinS.
Proof Define s. by (4) and apply Theorem 9.3.
Tlieorem 9.6 (Weierrct, M-tet). Let {M.} be a sequence of notmegatitw numbers
such that
0 _< lf.(x)l < M., for n = 1, 2,..., undfor every x in S.
Then Lf(x) conoerges uniformly on S if 1!. eotwerges.
Proof Apply Theorems 8.t I and 9.5 in conjunction with the inequality
Tlg, orem 9.7, ,ssume that f,(x) = f(x) (uniformly on S). lf each f. is continuous
at a point xo of S, then f is also continuous at x o.

Proof. Define s, by (4). Continuity of each f, at x o implies continuity of s, at
xo, and the conclusion follows at once from Theorem 9.2.
No. If x o is an accumulation point of S, this theorem permits us to interchang
limits and infinite sums, as follows:
• e,-.x 0 .tl  '1 ,--1
9.7 A SPACE-FILLING CURVE
We can apply Theorem 9.7 to construct a space-//ll/g curve. This is a continuous
curve in R 2 that passes through every poitt of the unit square [0, I] × [0, I].
Pcano (1890) was the first to give an example of such a curve. The example to be
presented here is due to I. J. Schoenberg (Bulleti of the ,merican Mathematioff
Society, 1938) and can be described as follows:
Let  be defined on the interval [0, 2"] by the following formulas:
f 0, if0< t < ,orif < t_<2,
t(t) = 3t - !, if'½ < t < ,
!, if.< t<,
-3t + 5, if< t  .
Extend the definition of  to all of R by the equation
¢(t + 2) --
This makes b periodic with period 2. (The graph of b is shown in Fig. 9.5.)
-2 -1 0 !  3 4
Figar¢
Now define two functions f and f2 by the foIiowing luations:
• =1 :2" ' , 1
Both series converg absolutely for eae. real t and tlmy conver¢ uniformly on
R. In fa  t, since i  (t)l 1 for all t, the Weierstrass M-tt is applicable with
M - 2-". Since # is contiuous on R, Theorem 9.7 tells us that f] and fa are
also continuous on R. Let.f = (f],fa) and let F denote the image of the unit
interval [0, 1] under f. We will show that F "fills" the unit square, i.e., that
First, it is clear that 0 <f,(t) < ! and 0<A(t)_< 1 for each t, since
_.% ] 2-  = I. Hence, F is a subset of the unit square. Nt, we must show that
.a, b) ¢ F whenever (a, b) ¢ [0, I] × [0, 1]. For this purpose we write a and b
in the binary system. That is, w¢ wri
where each  d h b, s ee 0 or I. (  I..) Now I
c = 2 _c. wc._ = a. andc. = b.,n = 1,2
im 3 g 
Oly, 0  c  1 fin 2 t 3-" = 1. We I1 show thatA(c) = a and that
A(c) = b.
 we n pve t
3tc) = c,+ t, for h k = 0, I, 2, .... (5)
then we will  .(3'-c) = c._ =  d (3-c) = c2. = b., and this
wiII Ove usA(c) = a,A(c) : b. To pro (, we te
3c = 2 .= +2 c" ,=t+l 3 = ( even integer) +
¢(3'c) =
If c,+z = 0, n  h 0    2% 3-" = , d hen  = 0.
Buttn   I(= I. fo,(3*e)=
fll  d s pv tft(e) = a,f(c) = b. H, F fills  unit .
9. UNIFORM CONVERGENCE AND RIEMANN.-b-WIELTJES INTEGRATION
Tbeorem 9.& Let  be of bound varimion on [a, b]. Assume that each term of
the equence {.f.} is a real function such that f ¢ R() on [a, b] for each
n -- 1, 2 .... Aasume thatf  funiformly on [a. b] anddefineg.(x) :  f.t0 d(t)
f x e [a, b], n = 1,2,... Then we have:
b) --, # [a, hi, #(x) --
sorry. The conclusion implies that, for each x in [a, b], we can write
lim " f(t) d(t) -- [" lira f,(t) do(t),
This property is often dcscril::l by saying that a uniformly collvergnt sequence
can be integrated term by term.

Proof We can assume that  is increasing with a) < (b). To prove (a), we
will show that f satisfies Riemann's condition with respect to • on In, bl. (See
Theorem 7.19.)
Given  > 0, choose N so that
If(x) - f  (x)l <
- '
Then, for every partition P of In, b-l, we have
for all x in I-a, hi.
IU(P,I  A, )1 -< y and IL(P,f - f, )I -< _e,
(using the notatiot of Definition 7.I4). For this N, choose P so that P finer than
P, implies U(P, fmct) - HP, fs, a) < /3. Then for such Pwe have
U(P, f, ) - L(P, f, ) <_ U(P, f -- fs, ) - L(P, y - f, )
+ fs, - c(P, A,
This proves (). To prove (b), let  > 0 be given and choe N so that
If,(r) - f(01 <
for all n > N and every t in [a, hi. If x  In, b], we hae
- 2 - 2
s proves that .
 9.9. t •  ofund tion on [a, b] dme that (x)  f(x)
(oiy on a, b]), wre each f is a reallved function ch that   R() on
Prf Apply eorem 9.8 to the quen of rtial sums.
. is eorem is d by ying tt a unifory nvernt des
on  ina term by te.
9.9 NONUN]WORMLY CONVERGENT SEQUENCES THAT CAN BE
INTEGRATED TERM BY ]qM
Uniform convergence is a sufficient but not a necessary condition for term-by-
term integration, as is seen by the following example.
Tit. 9.11 Ncmmifm'm  S,mcea 2,27
or
 9.6
Example. Letf,(x) = ifO < x_ 1. (SeeFig. 9.6.) Thelimit fm'tionfhas thevau©
0 in [0 1) and f(l) = l Since this is a sequence of continuous functions with discon-
tinuous limit, the convezgece is not uniform on [0, 1 ]. Nevertheless, term-by-term
integration on [0, ! ] leads to a correct result in this case. In fact, we have
-- -- 0 aS 1¢ -- ,
so tim_,  J fn(x) dx = ]' f(x) dx --- O.
The sequence in the foregoing example, although not uniformly convergent
on [0, I]. is uniformly eonrgent on every dosed subinterval of [0, 1] not con-
raining 1. The next theorem is a gneral resfilt which permits term-by-term inte-
gration in examples of this type. The added ingredient is that we assume that {f.}
is uniformly bounded on In, b] and that the limit function y'is integrable.
Defmit 9.10. A equence of functiog¢ {f,}/x said to be boundedl, convertent on
T if {f}/s po/ntw/,e congruent and uniformly bounded on T.
Theorem 9.1L Let {f,} be a boundedly conuer#ent equence on [a, b]. Assume that
each f, R on In, b], and that the timit function f  R on [a b], Assume also that
there is a partition P of In, hi, say
P -- {x0, x,..., x,},
such that, on etery subintenaff [c, d] not containing any oJ-the points x the sequence
{f} converges uniformly to f Then we have
Proof. Siuc f is bounded and {f,} is uniformly bounded, there is a positive
numbm M such at Iy'(x) _< M and If,(x)l <- M for all x in l-a, b] and all
n > I. C, ivea s > 0 sudh that 2 < IIPII, let h ffi #{2m), when: m is the number
of subintervals of P, and consider a new partition P' of I-a, b] given by
Since If - fl is integable on [a, b] and boundd by 2M, the sum of the inlegrals

of If -- f.f taken over the intervals
Do, x0+h], D,-h, Xl+h], ...,
is at most 2M(2mh) = 2Me. The remaining portion of In, b] (call it S) is the
union of a finite number of closed intervals, in each of which {f} is uniformly
conwrgent tof. Therefore, there is an integer N (depending only on t) such that
for all x in S we have
If(x) -_f.(x)t <  whenever n  N.
Hence the sum of the integrals of If - f.I o,mr the intervals of Sis at most e(b - a),
flf(x) -f(x)l dx <_ (2M + b - a) whenever N.
This proves that f.(x) dx f(x) dx as n --, m.
There is a stronger theorem due to Atzelt which makes no reference whatever
to uniform convergence.
Tlttortm 9.12 (AtztI). Assume that {f.} is boundedly convergent on In,hi and sup-
pose each f. is Riernann-integrable on In, b]. Assume also that the limit function
f is Riemann-integtable on In, hi. Then
The proof of Arz.el.'s theorem is considerably more difficult than that of
Theorem 9.11 and will not be given here. In the next chapter we shall prove a
theorem on Lebesgue integrals which includes Arzel's theorem as a special case.
(See Theorem 10.29).
so're_ It is easy to give an example a boundedly convergent sequence
of Riemann-integrable functions whose limit f is not gi6man-integrabIe. If
{r, rz .... ) denotes the set of rational numbers in [0, !], definef.(x) to have the
value I if x -- r, for all k = !, 2,..., n, and put_f,(x) = 0 otherwise. Then the
integral ofo(x) dx - 0 for each a, but the pointwise limit function f is not
Riemann-integrable on [0, 1].
9.10 UNIFORM CONVERGENCE AND DIFFERENTIATION
By analogy with Theorems 9.2 and 9.8, one might expect the fo]lowing result to
hold: lff. - f uniformly on In, b] and if f[, exists for each n, then f' exists and
f - f' uniformly on In, hi. However, Example 3 of Section 9.2 shows that this
cannot be true. Although the sequen {f,} of Example 3 converges uniformly on
R, the sequence {f} does not even converge pointwise on R. For example,
{f,'(0)} diverges since f(0)-= x/- Therefore the analog of Theorems 9.2 and
9.8 for differentiation must take a different form.
9.13 Unifm'm Cmp,gm and Differentiation 229
Theorem 9.1:L Assume that each term of {f.} is a real-valued.function having a
finite derivative at each point of an open interval (a, b). 4ssume that for at least one
point xo in (a, b) the sequence {f(xo) } converges. Assume further that there exit
a function g such that f - @ uniformly on (a, b). Then:
a) There exists a function f such that f,  f uMforraly on (a, b).
b) For each x in (a, b) the derivati if(x) exists and equals
Proof. Assume that c  (a, b) and define a new sequence {y.} as follows:
g,(x)
.f(c) ifx = c.
The sequence {9} so formccl depends on the choice of m Convergence of
follows from the hypothesis, since g(c) ffi f(c). We will prove next that
converges uniformly on (a b). If   c, we have
- ffi (9)
where h(x) --- f(x) - f(x). Now h'(x) exists for each x in ('a, b) and has the value
f(x) -f(x). Applying the Mean-Value Theorem in (9), we get
where x lies between x and c. Since {f converges uniformly on (, b) (by hy
pothesis), we can use (10), together with the Cauchy condition (Theorem 9.3),
to leduce that {9} converges uniformly on (a, b).
Now we can show that {f converges uniformly on (a, b). Let us form the
particular sequence {g} corresponding to the special point c  xo for which
{f(xo)} is assumed to converge. From (8) we can write
f.(x) = f.(x0) + (x -
an equation which holds for every x in (a, b}. Hence we have
This equation, with the help of the Cauchy condition, establishes the uniform
convergence of {/,1 on (a, by. This proves (a).
To prove b), return to the sequenco {g.) defined by (81 for an arbitrary, point
c in (a, b) and let G(x) = lim,_,. ,q.(x). The hypothesis that .f; exists means that
lim_¢ g.(x) = g,(c). In other words, each # is continuous at c. Sinoe g.  G
uniformly on (a, by, the limit funC:tion G is also continuous at c. This means that
G(c) = lira G(x), (I

the existence of the limit being part of the conclusion. But, for x  c, we have
C,(x) = lim g,(x) = lira f(x) - f.(c) = f(x) - f(c)
Hence, (11) states that the derivat/vef'(c) exists and equals G(c). But
C-(c) = lim g(c) = lira ft.(c) : /(c);
hencef'(c) -- g(e). Since c is an arbitrary point of(a, b), this proves
When we reformulate Theorem 9.i3 in terms of series, we obtain
Tlteorem 9.14. /lssume that each f is a reaI-votued function defined on (a, b) such
that the derivative f(x) exists for each x in (a, b). Assume that, for at leat one
point x o in (a, b), the series af(xo) corwerges. Assume further that there exists a
function y such that _3:(x) = y(x) (uniformly on (a, b)). Tlen:
a) There exists a function f such that f(x) - f(x) (uniformly on (a, b)).
b) If x e (a, b), the derivatie f'(x) exists and equals f(x).
9.11 SUFFICIENT CONDfflONS FOR UNIFORM CONVERGENCE OF
A SERIES
The importance of uniformly convergent series has been amply illustrated in some
of the preceding theorems. Therefore it seems natural to seek some simple ways of
testing a series for uniform convergence without resorting to the definition in each
case. One such test, the Weierstrass M-te,t, was described in Theorem 9.6. There
are other tests that may be useful when the M-test is not applicable. One of these
is the analog of Theorem 8.28.
Theorem 9.15 (Dirld'a te for uniform cmmergence). Let F,(x) denote the nth
partial sum of the aeries _,f(x), where each f is a complex-valued function defined
on a set S. gsume that {F}/a uniformly bounded on S. Let {g} be a sequence of
real-valued functions such that g+ (x)  g.(x) for each x n S and for eoery
n : 1, 2 .... , and assume that g  0 uniformly on N. Then the series .f(x)(x)
converges uniformly on S.
Proof. Let s,(x) : Y_.  ft(xt(x). By partia! summation we have
and hence if n > m, we can write
k =
T. 9.16 Unlfecm Coevergeace ami Dooie Sequences 231
Therefore, ff M is a uniform bound for {Fo}, we have
r 
= a.+ :(x).
Sin 9,  0 unffoly on S, this inuality (loather  he uchy ndion)
c  shonld  no lff in exnding m 8.29 (Al's t)
in a similar way so at it yieIds a t for uform converge. (Exe 9.13.)
  ()    e a. In  lt pt (  8.),   
ii ](x)l  ltl (x/2), vid f  ml x   (m is  int). 
IF,(x)]  l/sin (/2) if # g x   - .
lions of  9.15,   nclu t   Ax   unifoy
on [,  - ]. In , if   9x)  1In, s bfish  ifo n-
 of e 
on [6, 2a - #]if0 < , < x. Note that the Weietslrass M-let carmot be  to estab-
lish uniform convergence in this case, since [el : 1.
As a different type of application of uniform convergence, we dcdace the following
theorem on double sequences which can be viewed as a converse to Theorem 8.39.
17tearem 9.16. Let f be a double sequence and let Z + denote the ,et of positive
integers. For each n = I, 2 . . . , define a function , on Z + as follow."
a.(m) - f(m, n), if m  Z +.
Asume that g, -, g uniformly on Z +, where g(m) : Hm,..®f(m, n). If the iterated
limit linh,_ (lim,..f(m, n)) exista, then the double limit lim,,,.®f(m, n) also
exists and has the same
Proof Given  > O, choose N, so that n > N implies
• If(m, n) - g(m)l <

231 Sequul¢ of lhmctiolls Def, 9.17
Tk 9.t9 Ms  
Let a = lim.® (lim._.f(m, n)) - lim._. g(m). For the same e, choo N so
that m > N implies [g(m) - a[ < e/2. Then, if N is the larger of N 1 and N, wc
have ]f(m,n) - al < e whenever both m > N and n > N. in other words,
lira ....  f(m, n) = a.
9.13 MFak_N CONVERGENCE
The functions in this section may be real- or complex-valued.
De,'an 9.17 Let {f.} be a sequence of Riemann-integrabte functions defined on
In, hi. Assume that f ¢ R on In, b]. The sequence {f.} is said to converge in the
mean to f on Ea, hi, and we write
l.i.m.f.
lira f If.(x) - .f(x)i : ax = 0.
If the inequality If(x) -f.(x)l < e holds for every x in I-a, hi, then we have
.[. If(x) - f.(x)l  dx < r.(b - a). Therefore, uniform convergence of {f.} to f
on [a, b] implies mean convergence, provided that each f. is Riemann-integrable
on [a, hi. A rather surprising fazt is that convergence in the mean need not imply
pointwise convergence at any point of the interval. This can be seen as follows'
For each integer n > 0, subdivide the interval I-0, 1] into 2" equal subintervals
and let I2.+, denote that subinterval whose right endpoint is (k + I)[2 , where
k = 0, t, 2, ... , 2   1. This yields a collection {I Iz,... } of subintervals of
[0, 1], of which the first few are:
l = [0, 1], I = [0, ½"], I '= [-, 1],
', = [0, ¼], r, _- [t, ½], t, = [½, t],
and so forth. Definef. on [0, 1] as follows:
Then {f.} eon,ergcs in the mean to 0, since o t [f.(x)t z dx is the length of 1,, and
this approaches 0 as n --, o. On the other hand, for each x in [0, .1] we have
lira supS(x) = 1 and lim inff(x) = 0.
[Why?] Hence, {f(x)} does not converge for any x in [0, 1].
The next theorem il]ustrates the importance of mean convergence.
wh/eh is a direct appfication of the Cauchy-Schwarz inequality for integrals. (See
Exercise 7.16 for the statement of the Cauchy-Schwarz inequafity and a sketch of
its proof.) Given • > 0, we can choose N so that n > N implies
If(t) - f(t)l  dt < --, (13)
A
where A = 1 +  I#(t)l z dr. Substituting (13) in (12), we find that n > Nimplies
0 _< [h(x) - h.(x)l <  for every x in In, b_'].
This theorem is particularly useful in the theory of Fourier series. (See Theorem
11.16.) The following generalization is also of interest.
leoem 9.19. Assume that I.i.m.,_®f. =f and l.i.m.._,g. -- g on
Define
h(x)= f f(t)f(t) dt,
if x  in, b]. Then h. - h uniformly on In, b].
Proof We have
h(x) - (x)=
The proof is now an easy consequence of Theorem 9.18-
O < (I [f - f.l lg - g.l dt) < (I lf - f.lZ dtf lg - g.lZ dt)"
Applying the Cauchy-Sehwarz inequality, we can write

9.14 POWER 
An infinite series of the form
written more briefly as
 a.(z - zo)', (14)
RED
is callod a power seris in z - z0. Here z, zo, and g, (n = O, l, 2 ... ) are complex
numbers. With every power series (14) there is associated at disk, called the d/ak
of convergence, such that the series converges absolutely for every z interior to
this disk and diverges for every z outside this disk. The center of the disk is at z0
and its radius is called the radiu of comrgenee of the power series. (The radius
may be 0 or + t in extreme Cases.) The next theorem establishes the existence of
the disk of convergence and provides us with a way of calculating its radius.
T&,arem 9.20. Given a power ries ,% o a=(z - zoy, let
 1
,l ffi lira sup r = -,
(where r = 0 if it ffi + oo and r = + oo if 2 = 0). Then the series converes
abaolutely if [z - zol < r aml diverges if Iz - zol. > r. Furthermore, the series
converges uniformly on every compact subaet interior to the disk of convergence.
Proof. Applying the root test (lheorem 8.26), we have
tim sup a(z - zo)'l- Iz - zo[
and hence ,a,(z - zo) = converges absolutely if Iz - zo[ < r and diverges if
Iz - zd > r.
To prove the sccoad assertion, we simply observe that if T is a compact subset
of lhe disk ofconvergence, there is a point p in T such that z • T implies
Iz-zol lp-zJ <r.
Hence, la.(z - zo)'l  la.fp - zo)'[ for each z in T, and the Weierstrass M-test
is appfieable.
or. If the limit lim,,® la./a.+tl exists (or if this limit is +), its value is also
equa/to the radius of convergence of (I4). (See Exercise 9.30.)
Examlde 1. The two sexies -ffio z and . t z'jn  have the same radius of eonvergem,
narndy, r = 1. On the boundary of the disk of convergera the first converges nowhere,
the second converg ¢vcrShere.
7.. The scris#V_.at z'ln has radius of convergcc r - 1, but it does not con-
a'gc at z = 1. However, it does conver ¢¢crywhcrc else on the boundary because of
Dirichkt's tet frheorem 8.28).
These examples illustrat why Theorem 9.20 makes no assertion about the be-
havior of a power series on the boundary of the disk of convergence.
.4ssurae that the power series -%o a.(z - zoY' converges for each
Then the function f defined by the equation
lrlecrem 9.22.
the equation
where
is known to be oalid for each z in some open subset S of B(ze ; r). Then, for each
point z t in S, there exists a neighborhood B(z ; R) _ S in which f haa a power
series expansion of the form
If z c S, we have
= o (z - z,f(z -
=;0 k--0
06)
(17)
is continuous on B(zo; r).
Proof Sitlee each point in B(zo; r) belongs to some compact subset of B(zo; r),
the conclusion follows at once from Theorem 9.7.
NOT. The series in (15) is said to represent fin B(zo; r). It is also callod a power
eries expansion off about z 0. Functions having power series expansions are
continuous inside the disk of convergence. Much more than this is true, however.
We will later prove that such functions have derivatives of every order inside the
disk of convergence. The proof wilt make use of the following theorem:
e that .a,(z - zo)  converges if z • B(zo; r). Suppose that

where
Now choose R so that B(z, ; R) ___ S and assume that x
iterated series ,V_.% 0 -.%0 cAk) conwrgcs absolutely, since
Then the
(18)
and henc the series in (18) converges. Therefore, by Theorem 8.43, we can inter-
change the order of summation to obtain
where b, is given by (17). This compIetes the proo£
NCrr. In the course of the proof we have shown that we may use any R > 0 that
satisfies the condition
8(z,; R)  S.
Theorem 9.23. Assume that .a,(z - zoy' converges for each z in B(zo;
the function f defined by the equation
ha a derivative f'lz) for each :z in 8(zo; r), given by
NOT£.
Proof.
indicated in (16)..
/'(z) =  na.(z - Zo) -'.
(19)
Then
(2O)
(20
n=l
The series in (20)and (21)have the same radius of convergence.
Assume that z e B(zo; r) and expand f in a power series about z, as
Then, if z e B(z ; R), z  z,, we have
./()-](z,) = b, +  b.+,(- z,)'. (22)
Z  z I
]8'y continuity, the right member of (22) reads to b as  - zt. Hence,/'(z) exists
and equals b. Using (17) to compute b, we find
b, :  na.(z, - zor-'.
Since z is an arbitrary point ofB(z0; r), this pro-s (21). The two sm'is have the
same radius of convergence because / - 1 as n - co.
NCn By ge.atcd application of (2I), w¢ fred that for each k = l, 2, ..., the
derivative f)(z) exists in B(zo; r) and is given by the series
o n!
/a)(z) =   (. - a.(z -
If  put z = zo in (23), we obtain tl important formula
f*zo) = kla, (k = 1, 2,...).
()
This equation tells us that if two power series .(z - zoY' and ,V_./,.(z - zo)" both
represent the same function in a neighborhood B(ze; r), then a. = b. for every n.
That is, the power scri¢ expansion of a function labour a given point zo is uniquely
determined (if it exists at all), and it is given by the formula
f(-) = 22/'°) (z -
valid for each z in the disk of convctgnc.
9,1
MULTIPLICATION OF POWER SERIF
Given two power series expansions about the orWin, say
f(z) = 
if z • B(O;
V z /0; R).
Then the product f(z)90) is #iven b.y the power series
where
/.fz • B(O; r) r s(o;
(n = 0, I,2,..,}.

Proof, The Cauchy product of the two given series is
and the conclusion follows from Theorem 8.46 (Mertens" Theorem).
NOTe. If the two series are identic, we get
ftz)  = 2
where c. = o ata.---,1+-=. . The symboI __j+.__. indicates
that the summation is to be extended over all nonnegative integers mt and m2
whose sum is n. Similarly, for any integer p > 0, we have
where
9,16 THE SON THEOREM
9.25. Given two power series expan about the origin, ,my
a--0
f z e B(O; r),
o(z) =  b.e, if   0; s).
If, for a fixed z in B(0; R), we have .*=--o Ib'[ < r, then for this z we con write
where the coefficients ct are obtaed a follows: De.fine the number bk(n ) by the
equaffon
9.26
By hypothesis, we can choose z so that T_.o Ib'l < r. For this zwe have
< r and hence we can write
If we are allowed m interetu,e t. order of summation, we obtain
Now each number b(n) is a finite sum of the form
and hence Ibd)l -< E...,..-. -..-,, Ib..I "'" I..I- On the other hand, wehw
(25)
where Ba(n) = y_.+ ... +.. [b.l "'" lb-.I. Returning to (25), we have
" " " " " I-
and this establishes the convergence of (25).
9,17 RECROCAL OF A POWER SERIES
As an application of the substitution theorem, we will show that the reciprocal of
a power series in z is again a power series in z, provided that the constant term is
not O.
Theorem 9.26. Assem that we have
a(z )" = b = bt(n)z, t. p(z ) = pz °,
Then et = -%o abt(n) for k -- O, 1, 2,...
 The series _,% oct zt is the power series which .arises formally by substituting
the series for g(z) in place of r in the expansion off and then rearranging terms in
increasing powers of z.
ff z  B(0; h),
where p(O)  O. Then there exists a neighborhood B(0; 6) in whi the reciprocal of
p a a power series expansion of the form
1 _  q,z'.
p(z)
Furthermore, qo = l fpo.

Proof. Without loss in generality we can assume that Po -- 1, [Why?] Then
p(0) -- 1. Let P(z) -- I + Y-,t Ip.z'l ifz ¢B(0; h). By continuity, there exists
a neighborhood B(0; 6) such that IP(z) - It < 1 if z  B(0; . The conclusion
follows by applying Theorem 9,25 with
f(z) - l _ z .o
and
9.18 RFAL POWER SERIES
If x, x e, and a, are real numbers, the series a.(x - xoY is called a real power
• er/es. Its disk ofconvergen intersects the real axis in an interrai (xo - r, xo + r)
called the interval of convergence.
Each real power series defines a real-valued sum function whose value at each
x in the interval of convergence is given by
The series is said to represemf in the interval of convergence, and it is called a
power-series expmn off about
Two problems concern us here:
1) Given the series, to find properties of the sum functionf
2) Give a function f, to tind whether or not it can be represented by a power
It turns out that only rather special functions possess power-ries expansions.
Nevertheless, the class of such functions includes a large number of examples that
arise in practice, so their study is of great importance.
Question (1) is answered by the theorems we have already proved for complex
power series. A power series converges absolutely for each x in the open subinterval
(xo - r, x o + r) of convergence, and it converges uniformly on every compact
subet of this interval. Since each term of the power series is continuous on R, the
sum functionfis continuous on every compact subset of the interval of convergence
and hencefis continuous on (xo - r, x e + r).
Because of uniform convergence. Theorem 9.9 tells us that we can integrate a
power series term by term on every compact subinterval inside the integral of con-
vergence. Thus, for every x in (x0 - r, Xo + r) we have
The integrated series has the same radius of convergence.
Th sum ftmction has derivatives of very order in the interval of convergence
and they can be obtained by differentiating the series term by term. Moreover,
D 9.27
f')(xo) ----- n  so the sum function is represented by the power series
= - :o)".
-o ri!
(26)
We turn now to question (2). Suppose we arc given a real-valued function f
defined on mine open interval (x o -- r, xo + r), and suppose f has derivatives of
every order in this interval. Then we can rtainly form the power ies on the
right of (26). Does this sedes converge for any x besides x = xo ? If so, is its sum
equa/to f(x)? In general, the answer to both questions is "'No." (See Exercise
9.33 for a counter example.) A necessary and sufficient condition for answering
both questions in the affirmative is given in the next section with the help of
Taylor's formula (Theorem 5.19.)
9.19 THE TAYLOR'S SFAtlF GENERATED BY A FUNCTION
DeftMtion 9.27. Let f be a real-valued function defined on an interval l in R. If f has
derivatives of every order at each point of l, we write f  C  on I.
Iffe C ® on some neighborhood of a point c, the power series
-f('(e) (x c)',
n--0 irl!
is called the Taylor's series about c generated by f To indicate that f generates
this series, we write
f(x) -- (x - c)".
.=o
The question we are interested in is this: When can we replace the symbol -.- by
the symbol = ? Taylor's formula states that iffe C ® on the closed interval I'a, b]
and if c  in, b'], then, for every x in Ea, bJ and for every n, we have
f(x) = =o *'f()(c)-i (x - c)' + f(")(X)n[ (x - c)*, (27)
where x is some point between x and c. ]'he point x t depends on x, c, and on n.
Hence a necessary and sufficienl condition for the Taylor's series to converge to
f(x) is that
lira f(")(x,) (x  c)  = 0. (28)
in practice it may be quite difficult to dent with this limit because of the unknown
position of x. In some cases, however, a suitable upper bound can be obtained
forja°(x) and the limit can be shown to be zero. Since A'/n! , 0 as n  o for

all A, equation (28) will certainly hold il' there is a positicv constant M such that
for all x in In, hi. In other words, th Taylor's series of a function fconverges if
the nth derivativef ') grows no faster than the nth power of some positive number.
Tiffs is stated more formally in the next theorem.
r¢m 9,28. Agntme that f e C  on [a, b] and tet e  In, hi. Assume that there
is a neighborhood B(e) and a constant M (which might depend on c) such that
[f*)(x)[ < M" for every x in B(c) ca [a. b] and ever), n = l, 2,,.. Then, for
each x in B(e)  [a, hi, we e
9.20 BERNSTEIN'S THEORFAI
Another sufficient condition for convergence of the Taylor's series off, formulated
by S. 8ernstein, will be proved in this section. To simplify the proof we first obtain
another form of Taylor's formula in which the error term is expressed, as an
integral.
Tlwoeem 9.29. Assume f has a continuous derivative of order n + 1 in some open
interval t containing c, and define E,(x) for x in I by the equation
f(x) ffi  yt)(c) (x - c)  + Ed(x). (29)
--o k!
Then E(x) is also giten by tlie integral
E,(x) =  (x - t)" ft.+ t)(t ) dr.
Proof. The proof is by induction on n. For n = I we have
(3o)
where u(t) = f(O - f'(c) and v(t) =  - x. Integration by parts gives
This proves ¢30) for a = I. Now we assume () is true for n and prove it for
n + t. From (29) we have
(n + )
We write (x) as a_n integral and note that (x - cy '+ - (n + !) j'[ (x - )" dt
to obtain
1 (x - t)ft'+t) dt (x - t)" dt
1.+ (x) = 
= l f; (x - ')" [f'+'(t) - f'+"(c)] dt = l-- f? u(t) tgt)'n!
where u(t) = f'÷ t(t) -f'÷t)(c) and v(t) = -(x - t)'+ : /(n + 1). Integration
by parts gives as
l f; 1
.,+(x) -- - n v(t) du(t) = ( + )!
This proves (30).
o'nz. The change of variable t ffi x + (c - x)u raasforms the integral in (30)
to the form
n! u'f'+[x + (c - x)u] du. (31)
17tearem 9.0 (BernstebO. Assume f and all its deritatines are nonnegative on a
compact interat[b,b + r]. Then, fib < x < b + r, the Taylor's serie$
converges to f(x).
Proof. By a translation we can assume b ffi 0. The result is trivial if x = 0 so
we assume 0 < x < r. We use Taylor's formula with remainder and write
f(x) =  f()(O)  + (x).
We will prove that the error term satisfies the inequalities
0 < E,(x) < f(r).
This implies that (x)  0 as a --, o since (xlr) "*t --, 0 if0 < x < r.
To prove (33) we use (31) with c = 0 and find
E,(x) -  u'/'÷ (x - xu) du,
for each x in [0, r]. If x  0, let
(32)
(33)

9.30
The functionfl "+ is monotonic increasing on [0, r] since it derivative is non*
negative. Therefore we have
if 0 : u : l, and this implies F,(x)  F,(r) if 0 < x <_ r. In other words,
E.(X)IX'* ' .< F(r)/r .+ l, or
E,(x) _< 04)
Putting x = r in 02), we see that E,(r) < f(r) since each term in the sum is
nonnegafive. Using this in (34), we obtain (33) which, ia turn, completes th© proof.
92,1 THE BINOMIAL SERIES
As an example illustrating the use of Bcrnstein's theorem, we will obtain the fol-
lowing expansion, known as the b/nora/a/eries:
where a is an arbitrary real number and (,0 : a(a- 1)-'-(a - n + l)lnl.
Bernstein's theorem is not directly applicable in this case. However we can argue
as follows: Letf(x) = (l - x)-', where c > 0 and x < 1. Then
= c(c + 0" (c + n - l)(l - x)-'-',
and hence f')(x)  0 for each n, provided that x < 1. Applying Bernstein's
theorem with b = - 1 and r = 2 we fred that f(x) has a power series expansion
about the point b ---- --1, convergent for -I < x < 1. Therefore, by Theorem
9.22, f(x) also has a power series expansion about 0, f(x) = _a%of(i(O)x/k!,
convergent for -I < x < 1. Butfi(0) = ([')(-1) t k!, so
(1 ; --lyx , if--I < x < 1. (300
Replacing c by -a and x by -x in 06) we find that 35) is valid for each a < 0.
But now (35) can be extended to all real a by successive integration.
Of course, if a is a positive integer, say a = m, then ()  0 for n > m, and
(35) reduces to a finite sum (the Binomial Theorem).
9.22 A.IlF'S LIMIT THEOREM
If - I < x < 1, integration of the geometric series
1 --X --0
11t. 9.31
gives us the series expansion
log(l-x)=-
(37)
also valid for - 1 < x < 1. If we put x = - 1 in the righthand side of (37), we
obtain a convergent alternating series, namely, (-l)'+X/n. Can we also put
x -- - 1 in the lefthand side of (37)2 The next theorem answera this question in
the affirmative.
Tkeorem 9.31 (Abel's limit tkeoem). Assume that we haoe
lf the series also conperges at x  r, then the limit lim_,,_ f(x) exists and we hwae
lira f(x) : 
Proef For simplicity, assume that r = 1 (this amounts to a change in scale).
Then we are given that f(x) = ,X" for -1 < x < 1 and that za, converges.
Let us write f(l) = ,W_.,,% o a. We are to prove that lim_,_ f(x) = f(l), or, in
other words, that fig continuous from the left at x = 1.
If we multiply the series for f(x) by the geometric series and use Theorem
9.24, we find
f(x) =  c,x", where c, =  at.
1 - x .--0 t=o
Hence we have
By hyesis, lim,. c, = f(l). efo, #yen  > 0, we n find N such t
n 2 N implies Ic= - t)[ < /2. If we split the sum (39) into two as, we get
f(x)- f(l)= (1 -- x)  [¢.- f(1)] + (1 - x)  [¢.- f(l)]x'. ()
Let M deno e largest of the N n Ic, -- 1)1, n  0, 1, 2, ,.., N - 1.
If0 < x < I, ()gives us
 x  8
21-x 2

Now let 6 = ,[2NM. Then 0 < 1 - x < 6 implies If(x)-f(l)l < , which
means lim._ f(x) -- f(l). This completes the proof.
Example, We may put x ffi - I in (37) to obtain
log2 =  (-1)÷'
(Se Exercise 8.18 for anoth derivation of this formula.)
As an application of Abel's theorem we can derive the following result on
multiplication of series:
T,erem 9.32. Let -=o a= and --o b. be two convergent series and let .,=o c.
denote their Cauchy product, lf _Y= o c. converges, we have
orr.. This result is similar to Theorem 8.46 except that we do not assume absolute
convergence of either of the two given series. However, we do assume convergence
of their Cauehy product.
Proof The two power series .ac" and _,b' both converge for x = !, and hence
ty converge in the neighborhood B(0; 1). Keep Ixl < 1 and write
using Theorem 9.24. Now let x --, 1 - and apply Abel's theorem.
9223 TAUBER'S THEOREM
The converse of Abel's limit theorem is false in general. That is, iff is given by
(38), the limit f(r-) may exist but yet the series a.r" may faiI to converge. For
example, take a. = (-i)". Thenf(x) :- i/(I + x)if -1 < x < ] and f(x)  az
as x --, 1-. However, Y(- l)t' diverges. A. Tauber (I 897) discovered that by
placing further restrictions on the coefficients an, one can obtain a converse to
Abel's theorem. A large number of such results are now known and they are
referred to as Tauberian theorems. The simplest of these, sometimes called Tauber's
first theorem, is the following:
Theorem 9.33 (Tadmr). Let f(x) ffi '=0 a=x fo r -I < x < 1, and assume that
lim. no. = O. 1f f(x)  S as x -, 1 -, then -=a a. converges and has sum S.
Proof Let no. = ,--o k[as]. Then a.  0 as n --, c¢. (See Note foilowing
Theorem 8.48.) Also, lim,_.f(x.) -- S if x. = ! - i]n. Hence, given e > 0,
247
we can choose N so that n _> N implies
If(x,) - Sl < ,< 3 nt  ,t < -3
Now let s. = _.o a. Then, for - 1 < x < 1, we can write
s. S = f(x) - S +  %(1- x )- 
k=O
Now keep x in (0, 1). Then
for h k. efo, if n  N d 0 < x < 1. we have
Is. - Sl <_ If(x) - Sl + (l - x) 2", kl.,I + •
--o 3n(1 -- x)
Taking x = x. = 1 - l/n, we find Is, - SI < e[3 + ]3 + e/3 = :.
pletes the proof.
OT. See Exercise 9.37 for another Tauberian theorem.
EXICISES
Uniform ctmvergece
9.1 Assume that f,  f unifolmly on S and that each f is hounded on S. Prove that
{f,) is uniformly bounded on S.
9.2 Define two sequences {/.} and {g,,]- as follows:
a.(x) - 1
+-
Let h.x) = L(x)a.(x).
if x  ll, n= 1,2,....
ifx = 0 or ifxis irrational,
a
if x is rational, say x = ,
b>0.
a) Prove that both {f.f and |o,} converge uniformly on every bounded interval.
b) Prove that {h.} does not converge uniformly on any hounded interval.
9.3 Assume that f. -/uniformly on S. #. --, g uniformly on S.
a) Prove that f. + #, - [ + # uniformly on S.
b) Let hn(x) = fd(x)o(x), h(x) = f(x)a(x), if x  S. Exercise 9.2 shows that the
assertiott h, --, h uniformly on S is, in general incorrect. Prove thai it is correct
if each fn and each gn is boundel on S.

9,4 Assume that f, - f uniformly on S and suppose there is a constant M > 0 such
that jf(x)j _<  f   Jn S d all n.    nfinuous  the ¢u of
ifoy on .
 but not uno  (0, )
folly  (0, 1).
9  fx)  . ¢  {f} n int but not ify  [0, 1 ].
   uous  [0, 1 ] wlthe(1) = 0.  tt t  {x} cons
oy  [0, ].
9.7  t   f uoly on S, d t    nuo  S. If x
 {xe}  a  of in in S su that x=  x.  tfx.)  ffx).
.8  }  a  of nfinuous fis  on a mt  S and
fy  S if,  ly if, the foflong o tions hold:
i)  limit fonfjs nfinuom on S.
fi) For > 0, msm> 0da4>'0shtt n > md
Hint. To  t scie of ( d (fi), ow t f  xo in S t is a
 B(xo) d   k (g  Xo) sh t
Jfx) - f(x)[ <  if x  a(xo).
 , a te  of in, y A = {k .... , k,),  t o t, for
 x M S,  k in A ti IA(x) - f(x)J < . Onfo  is 
  this fact.
9& a) U  9.8 to pro t followi th of : If {f}.is a uq of
ald tin feio i intw to a  limit
b) U ¢  in i 9.5(a) to ow tt m of S is ti
i's t.
 [0, 1 ] for   c. ine  c for whi t n is ff on
[0, ! ] d th for which t--t intt  [0 ! ] !  a co ult.
9,11 M at (I  x) conve int but not unifoly on [0, I ], whets
(- 11 - x) n ifoly on [0, 1 ]. is illustra tt ifo
of(x) Mong wi intw n ofl fx)[ d t Iy  reply unif
 of ZL()I.
9.12 Au at e+l(X)  Ox) for  x in Td ch n = !, ..., O sup
te  0 ifly on T. vo at (- 1 + tx) n ffo on Z
9.13 ve Al's tl for unifo n:  {}  a u of 1-
functions such tt g=+t(x)  a(x) for cb x in Tand for n = I, 2 .... If
249
is uniformly bounded on T and it :fx) ¢nvePgs nliforly on T, tn f.(x)#(x)
 nvcr uniformly on T.
9,14 t (x) = x/(1 + nx ) if x
 {f.} and e lit fiong of
a) ove that f'(x) exgts for ev¢ x but that f'(0)  a(0). For what vu of x is
if(x) =
b) In wt subint ofR dA  funifoly?
)  what sublls ofR df  g unifoly?
9.15 t A(x)  (l/n -'' if
tf;  0 intwim R, but t the  of ([g} is not uniform on y intal
nng  o.
9.16 t [}  a u of l-vMu eontinuo fclions de on [0, i ] and
sume lhatf, /unifoiy on [0, t ]. ve or dpm
9.17 Mas from Slobv did tt e em inte
I so they pl it  the Stoi integral, d as fol[ow: If f is a function
d on the t Q of mtiol n in [0, 1 ], e Slobvian intl of denoted
by S), is defin to  the limit
whenever th limit sts. t {}  a  of run.ions such tt S.) ists for
 n  su t X -- f unffoy on Q.
ists, d that S)  3(
9.18 t fx)  1/(1 + nZx z
intwi but not ifoly on [0. 1 ]. Is term-by-t inttion missible?
9.19  tt . x](l + ) n unifoly on eve finite ini in R
if  > . I the conven
. tt = ((- Ip/) sin
m sut of R.
9.11 ov¢ that the fi %0 (xz*/ + 1) - +/( + 2)) nr pointwi
but not ufoly on [0, I ].
9, ov¢ that t an sin nx and =  a. s nx a unifoly convnt on R if
9. t a,}  a dsing uen of silive ter. ove that the r[ . sin nx
n unifoly on R if, and only if,
9. Gin a vt fi = a.. ove that the Diricet  %t a -
nv unholy on the atf-infinit¢ instal
llm,o+ E, ° - = 

9.25 Prove that the series (s) -- _.,t n-" converges uniformly on every half-infinite
interval 1 + h _< s < + , whom h > 0. Show that the equation
]log n
is valid for each s > I and obtain a similar formula for the kth derivative
Meml
90.6 Let f(x) = n3/2xe -m. Prove that {.f. } converges pointwise to 0 on [- 1, 1 ] but
hat I.i.m.m_,mf, = 0on i-I, 1].
9"/ Assume that f,} convcfg pointwis¢ to f on [a, b] and that l.i.m.._, f = g on
[a, b]. Prove that f =  if both land .q ar continuous on [a, b].
9.28 Let f(x) -- co@x if 0 < x _< n.
a) Prove that l.i.m.,_,f m = 0on [0, n] hut that (f,()} does not converge.
b) lTove that {f,} converges pointwise but not uniformly on [0,
9.29 LetA(x ) = 0if0 _< x _< l/norif2/n <_ x <_ 1, andletf,(x)
Prove that {fa} converges poinlwis¢ to 0 on [0, 1] but that I.i.m.,_f, # 0on [0, 1].
Poa" series
9.30 Ifr is the radius of convergence of .a(z - zo)", where each a m # 0, show that
liminfI+ z r <__,, timsup[,__,
931 Given that the power series _,% 0 az" has radius of onvergence 2. Find th radius
of convergence of each of the following series:
In (a) and (b), k is a fi.xod positive integex.
9.30. Given a por series ,%0 a c= whose ooefficients are telau:d by an equation of the
form
a + a_i + am   0 (n -- 2,3,.,.).
Show that for any x for which the scales converges, its sum is
ao + fat + Aao)x
! + x+Bx 
9.33 Letf(x) = e -u ifx = 0,f(0) -- 0.
a) Show thatf¢(0) exists for all   1.
b) Show that the Taylor's series about 0 generated by f converges everywhcxe on R
but that it eprcscts f only at th origin.
751
havior at the points x = _+ 1.
a) If x = - 1, the series converges for  >_ 0 and diverges for • < 0.
b) If x = 1, the series diverges for • _< -1, converges conditionally for  in the
interval - 1 < , < 0, and converges absolutely for a _> 0.
9.3 Show that "_ax  converg uniformly on t0, 1] if'_a converges, Us¢ this fact to
give another proof of Abel's limit theorem.
9.36 ff each am >- 0 and if a, diverges, show that ac = - + oo as x  t -, (Assume
,,,x  converges for ]xl < 1.)
937 If e.aeh a _> 0 and if iim__ a.x = exists and ¢¢luds A, prove that am converges
and has sum A. (Compan with Theorem 9.3.)
9,38 For each foal t, define (x) ffi xe/(e  - I) if x 6 R, x  0, (0) = !,
a) Show that tbex is a disk B(0; o') in whichf is represcnt.cd by a power se_a'ics in x.
b) Defin¢ t'o(t), P,(O, P(t) ..... by the equation
and se  identity
polynomial. These are the Bernoulli polynomials. The numbers B = P(0)
( : O, 1, 2,... ) are caIled the Bernoulli monbers. Derive the following further
h)
SUGGFSTED REF£LFCS FOR FURTHER STUDY
9.1 Hardy, G. H., Diroent Series. Oxford Univ. Press, Oxford, 1949.
9.2 Hirschmann, L I.. Infinite Series. Holt, Rinehart and Winston, New York, 1962.
9.3 Knopp, K,, 77Jeory andApptication oflfinite 3eries, 2rid  R; C. Young, lran$-
htor. Harrier, Nvw York, 19.

CHAPTER !0
THE LEBESGUE INTEGRAL
10.1 IN'IIlODUCTION
The Riemann integral .f(x)dx, as developed in Chapter 7, is well motivated,
simple to describe, and serves all the needs of elementary calculus. However, this
integral does not meet all the requirements of advanced analysis. An extension,
called the Lebesgue integral, is discussed in this chapter. It permits more general
functions as integrands, it treats bounded and unbounded functions simultaneously,
and it enab]es us to replace the interval [a, b] by more general sets.
The Lebesgue integral also gives more satisfying convergence theorems. If a
sequence of functions {f,} converges pointwise to a limit funetionfon In, hi, it
is desirable to conclude that
lira f(x) dx = x) dx
with a minimum of additional hypotheses, The definitive result of this type is
Lebesgue's dominated convergence theorem, which permits term-by-term integra-
tion if each {f,} is Lebesgue-intcgrable and if the sequence is dominated by a
Lebesgue-integrable function. (See Theorem 10.27.) Here Lebesgue integrals are
essential. The theorem is false for Riemann integrals.
In Riemann's approach the interval of integration is subdivided into a finite
number of subintervals. In Lebesgue's approach the interval is subdivided into
more general types of sets called measurable sets. In a classic memoir, Integrate,
longueur, dire, published in 1902, Lebesgue gave a definition of measure for point
sets and applied this to develop his new integral.
Since Lebesgue's early work, both measure theory and integration theory have
undergone many generalizations and modifications. The work of Young, Daniell,
Riesz, Stone, and others has shown that the Lebesgue integral can be introduced
by a method which does not depend on measure theory but which focuses directly
on functions and their integrals. This chapter follows this approach, as outlined
in Reference 10.10. The only concept required from measure theory is sets of
measure zero, a simple idea introduced in Chapter 7. Later, we indicate briefly
how measure theory can be developed with the help of the Lebesgue integral.
10.2 THE INTEGRAL OF A STEP FUNCTION
The approach used here is to define the integral first for step functions, then for a
larger class (called upper functions) which contains limits of certain increasing
sequences of step functions, and finally for an even larger class, the Lebesgue-
integrable functions.
We recall that a function s, defined on a compact interval In, hi, is called a
step function if there is a partition P -- {x0, x,..., xn} of I'd, b] such that s is
constant on every open subinterval, say
s(x) = c if x  (x_ 1, x).
A step function is Riemann-integrable on each subinlerval [x_t, x] and its
integral over this subinterval is given by
s(x) dx -- c,(x, -
regardless of the values of s at the endpoinls. The Riemann integral of s over
t-a, b] is therefore equal to the sum
/or. Lebesgue theory can lie develol without prior knowledge of Riemann
iategration by using equation (1) as the definition of the integral of a step function.
It should be noted that the sum in (1) is independent of the choice of P as long as s
is constant on the open subintervals of
It is convenienl to remove the restriction that the domain of a step function be
compact.
DeflMffon 1.1. Let I denote a enerai interval (bounded, unbounded, open, closed,
or half-open). 4 function s is called a step function on I if there 8 a compact
subinterval [a, b] of I such that s is a step function on [a, b] and s(x) = 0
ifx i' [a, The integrat of s o  er 1, de,,ot  d by I, s(x) dx or J, is de, inca
to be the integral of s ooer [a, b]. as aiven by
There are, of course, many compact intervals [a, b] outside of which s vanishes,
but the integral of s is independent of the choice of [a, b].
The sum and product of two step functions is also a step function. The follow-
ing properties of the integral for step functions are easily deduced from the fore-
going definition :
for CVel'y COl$tanl

Also, if I is expressed as the union of a finite set of subintervals, say
I = U,=  ['a,, b,], where no two subintervals have interior points in common, then
s(x) dx : s(x)
10.3 MONOTONIC SEQUENCF_. OF STEP .JNCTIONS
A sequenc of eal-valued functions {f.} defined on a set S is said to be increasing
on S if
f.(x) _< f+ l(X) for all x in S and all n.
A decreasing sequence is one satisfying the reverse inequality.
NOTE. We remind the reader that a subset T of R is said to he of measure 0 if,
for every t > 0, T cn b¢ ¢ovcrm:l by a countable collection of intervals, the sum
o whose lengths is less than . A property is said to hold almost everywhere on a
set S (written: a.e. on S) if it holds everywhere on S except for a set of measure 0.
I<orATIOI<. If {f.} iS an increasing sequence of functions on S such that f, - f
almost everywhere on S, we indicate this by writing
f, , f a.e. onS.
Similarly, the notation f x, fa.e. on S means that {f,} is a decreasing sequence
on S which converges tofalmost everywhere on S.
The next theorem is concerned with decreasing sequences of step functions on
a general interval L
Theorem 10.2. Let {s,} be a decreasing sequence of nonnecjative step functions such
that s, " 0 a.e. on an interval L Then
lira f s, -- 0.
Proofi The idea of the proof is to write
where each of A and B is a finite union of intervals. The set A is chosen so that
in its intervals the integrand is small if n is sufficiently large. In B the integrand
need not be small but the sum of the lengths of its intervals wilt be small. To carry
out this idea we proceed as follows.
There is a compact interval In, b] outside of which s vanishes. Since
0 < s,(x) < s (x) for all x in L
each s vanishes outside In, hi. Now s, is constant on each open subinterval of
lb. 10.2 M(motonl¢  o# Ste0 F.m:tiu 7
some partition of In, hi. Let D, denote the set of endpoints of these subintervals,
and let D = U:= t D,. Since each D, is a finite set, the union D is countable and
therefore has measure 0. Let E denote the set of points in In, b] at which the
sequence {s,} does not converge to 0. By hypothesis, E has measure 0 so the set
F=DwE
aim has measure 0. Therefore, if e > 0 is given we can cover F by a countable
collection of open intervals F,, F,..., the sum of whose lengths is less than e.
Now suppose x  In, b] - F. Then x ¢ E, so s,(x) --, 0 as n  . Therefore
there is an integer N = Nix) such that ss(x) < e. Also, x 6 D so x is interior to
some interval of constancy of ss. Hence there is an open interval B(x) such that
s(t) < e for all t in B(x). Since {s,} is decreasing, we also have
s,(t) < e for all n >_ N and all t in B(x). (2)
The set of all intervals B(x) obtained as x ranges through I-a, b] - F, together
with the intervals Fx, F2, ..., form an open covering of In, hi. Since I-a, b] is
compact there is a finite subcover, say
$= ] r= 1
Let N O denote the largest of the integers N(x), ..., N(xe). From (2) we see lhat
s,(t) < e for all n >_ No and all I in  B(x). (3)
Now define A and B as follows:
Then ,4 is a finite union of disjoint intervals and we have
First we es6mate the integral over B. Let M be an ur bound for st on In, hi.
Sin {s,} is decreasing, we have s,(x)  s(x)  M for all x in In, hi. The sum
of/he lengths of the inteals in B is less than e. so we have
n. % 
Next we estimate the inteal over A. Sn A  g t B(x), the inequality
in (3) shows that s(x) < e if x  d and n  No. The sum of the ]englhs of the
intervals ia A does nol exd b - a, so we have the estimate
so < (b - a) ifn >_ No.

The two estimates together give us t s, _< (M 4- b - a)¢ if n _> No, and this
shows that ]irta._  r s, = 0.
Tleam 10.3. Let {t.} be a sequence of step functions on an interval l such that:
a) There is a function f such that t,  f a.e. on I,
b) the sequence { t.} converges.
Then for any tep fuction t such that t(x) _ f(x) a,e, on I we hae
Proof. Define a new sequence of nonnegative step functions {s.} on las follows:
s'(x)={t(OX)-t"(x) ifift(x)>t'(x)'t(x) < t.(x),
Note/3aat s.(x) = max {t(x) - t.(x), 0}. Now {s.} is decreasing on I since {t.} is
increasing, and s.(x)  max {t(x) - f(x), 0} a.e. on jr. But t(x) < f(x) a.e. on L
and therefore s, 0 a,e. on L Hence, by Theorem 10.2, lim._., .[ s. = 0. But
s.(x)  t(x) - t.(x) for all x in 1, so
Now let n  0o to obtain (4).
10.4 UPPER FUNCTIONS AND THEIR llIGRALS
Let S(/) denote the set of all step funetions on an interval/. The integral has been
defined for all functions in S(/). Now we shall extend he definition to a larger
class U(/) which contains limits of certain increasing sequences of step functions.
The functions in this class are catled upper functions and they are defined as follows:
Deftain lOd. A real-alued function f defined on an iraerual I is called an upper
function on I, and we write f e U(I), f there exists an increasing sequence of step
funct/ons {.} such that
a) s,,, f a.e. onI,
b) lim._ t s /s finite.
Thesequence {s.} is said to generate f. The integral off oer I is defined by the
equation
f = lira fl s.. (5)
ore. Since { s.} is an increasing sequence of real numbers, condition (b) is
equivalent to saying that {X s.} is bounded above.
The next theorem shows that the definition of the integral in (5) is unambiguous.
 lO-f. lssume f¢ U(I) and let {s.} and {t.} be two sequences wnerat/n9
Then.
Proof The sequence {G,} satisfies hypotheses (a) and (b) of Theorem 10.3. Also,
for twery n we have
s(x) < f(x} a.e. on 1,
so (4) gives us
Since this holds for every n, w have
The same argument, with the sequences {s,} ad {t,} interchanged, gives the reverse
inequality and completes the proof.
It is easy to see that every step function is an upper function and that its
integral, as given by (51, is the same as that given by the e.adiet definition in
Section 10.2. Further properties of the integral for upper functions are described
in the next theorem.
Tlteom 10.6. As.uane f  U(I) and g  U(I). Then:
a) (f+ a)e
b) cf e U(I) for every constant e > O, and
c) It f < I, g gf(x ) < if(x) a.e. on I.
sOa'. In part (b) the requirement c _> 0 is essential. There are example for
whiehf U(/) but -f¢ U(/). (See Exercise 10.4.) However, fife U(F) and if
s  S(/), thenf - s e U(/) sincef - s = f + (-s).
Proof. Parts (a) and (b) are easy consequences of the corresponding properties
for step functions. To prove (c), let {s,} be a sequence which generates f, and let

10.7
{t,} be a sequence which generates g. Then s,, ' fand t,  ga.e. on/, and
But for each m we have
s.(x) < f(x) <_ O(x) = lim t.(x) a.e. on I.
Hence, by Theorem 10.3,
Now, let m -, o to obtain (c).
The next th¢or¢m describes an important onsequence of part (c).
Tkeotem 10.7, If f  U(I) and 0  U(I), and if f(x) = g(x) almost everywhere on L
Proof. We have both inequalitiesf(x) < g(x) and O(x) < f(x) almost everywhere
on I, so Theorem 10.6 (c) gives ',f < I, ,q and j',
Defudao 10.8. Let f and g be reaI-alued funetion defined on L We define
max (f, 9) and min (f, g) to be the functions whose values at each x in I are equal to
max {f(x), g(x)} and rain {f(x), 9(x)}, respectively.
The reader can easily verify the following properties of max and rain:
a) max U;, + rain (/; a) =f+ g,
b) max (f + h, g + h) = max (f a) + h, and min (f + h, g + h) :- min (f, g) + h.
[f/* fa.e. on/, and ifg ,, 9 a.e. on/, then
c) max (f.. g.)  max (f, g) a.e. on/, and rain (f., .q,)  rain (f, g) a.e. on I.
T&rem 10.9. lff  U(I) andg  V(l), then max (f, g)  V(I) andmin (f, g)  U(1).
Proof Lt {s} and {t} 1 glnc¢ of step functions which generate f and g,
reslzCtively, and let t¢ = max (s, t), v. = rain (, Q. Then u. anti v. are stp
functions such that u. )' max (f, g) and v. / rain (f, g) a.e. on L
To prove that rain (f, g)  U(/), it suffices to show that the sequence { v} is
bounded above. But v = min (s, t,3 < fa.e. on/, so  v, < .)'x f Therefore the
sequence { v,} converges. Bu the sequence {j u} aIso converges since, by
property (a), u. = s. + t, - v. and henc
The next theorem describes an additive property of the integral with respect
to the interval of integration
l?erem 10.10. Let I be an interval which is the union of two subimemals, say
I = I a  I, where I and lz have no interior points in common.
a) If re O(1) and if f  0 a.e. can I, then f  U(l,),f  U(l), and
f,f=f,.f+f,f (6,
b) de f s U(l),fz  U(I), and let f be defined o lag follows:
Then f
f(x) =
Proof. If {s,} is an increasing raquence of step functions which generates fort L
let s,+(x) = max {s,(x). O} for each x in L Then {s, + } is an increasing sequence of
nonneoative step functions which generates f on I (since f > 0). Moreover. for
every subinterval Joflwe have ]' s, +  ]'r s, + < Jx fso {s. + } generatesfon J. Also
so we let n  oo to obtain (a). The proof of (b) is left as an exercise.
o. There is a corresponding theorem (which can be proved by induction) for
an interval ,hich is ex as the union of a finite number of subintervals, no
two of which have Jut, riot points in common.
10,5 RIEMANN-'E--* IJNCTIONS AS EXAMPLES OF UPPER
XJNC'rloNs
The next theorem shows that the class of upper functions includes all the Riemann-
integrable functions.
Theorem lO.ll. Let f be defined and bounded on a compact internal In, hi, and
assume that f is continuous almost everywhere on [a, hi. Then f  O([a hi) and the
integral off, as a function m U([a, hi),/s equal to the Riemann integral , f(x) dx.
Proof Let P, -- {x, xa,... , x,} be a partition of [a, b] into 2" equal sub-
intervals of length (b - a)/2*. The subintervals of P.+ are obtained by bisecting
those of P,. Let
mt= inf {f(x) : x e [xt_ t. x]} for 1 _< k <

and define a step function  on [a, b'] as follows:
s.(x) = rat if 1 < x < s.(a) = ms.
Then x,(x) : f(x) for all x in [a, hi. Also, {s,} is increasing because the inf off
in a subinterval of [xt- , art] cannot be less than that in ['xtr
Next, we prove that s,(x)  f(x) at each interior point of continuity off. Since
the set of discontinuities off on Ea, b'l has measure 0, this will show that s. -r f
almost ehere on [a, b'[. If f is continuous at x, then for every e > 0 there is
a 6 (depending on x and on e) such that f(x) - e < f(y) < f(x) +  whenever
x-  <y < x + & Let m(b') ffi inf{f0,):y(x- ,x + 6)}. Then
f(x)   < m(), so f(x) < m(b') + ,, Some partition Ps has a subinterval
[xt-s, xt].containing x and lying within the interval (x -- 6, x + ). Therefore
s  (x) -- m, _< f(x) < + e < + =
But sa(x)  f(x) for all n and ss(x) - s,(x) for all n  N. Hence
$.(x) <_ f(x) <_ s,(x) +  if n _>
which shows that s.(x) f(x) as n -,
The sequence of integrals { s,} converges because it is an increasing sequence,
bounded above by M(b - a), where M = sup {f(x) : x • ['a, b]}, Moreover,
where L(P,f) is a lower Riemann sum. Since the limit of an increasing sequence
is equal to its supremum, the sequence {j x.} converges to the Riemann integral
off over I-a, b']. (The Riemann integral .f(x) dx .exists because of Lebesgue*s
criterion, Theorem 7.48.)
o As already mentioned, there exist functions fin U(/) such that -f¢ U(/).
Therefore the class U(/) is actually targer than the class of Riemann-integrab|e
functions on I, since -f R on Iiffe R on L
10.6 THE CLASS OF UE- FUNCTIONS ON A
GF_,NERAL INTERVAL
If u and v are upper functions, the difference u - v is not necessarily an upper
function. We eliminate this undesirable property by enlarging the class of inte-
grable functions.
Definition 10.12, l,Ve denote by L(1) the set of all functions f of the form f = u - v,
where u  U(I) and v  U(1). Each function f in L(1) is said to be Lebesyue-
iteorable on I, and its inteoral is defined by the equation
fife L(I) it is possible to writefas a difference of two upper functions u - v
in more than one way. The next theorem shows that the integral off is independent
of the choice of u and v.
T1tcorem lOd3. Let u, v, u, and v he functions in U(1) such that u - r = u s - ,
Then
Proof. The function u + v s and u s + v are in U(I) and u + v = u + .
Hnce, by Theorem 10.6(a), we have fi u + fi  - I, us + J v, which proves (8).
roa. If the interval lhas endpoints a and b in the extended real number system R ,
where a < b, we also write
f or f;f(x) dx
for the Lebesgue integral rf. We als define ,f - - .[f
If [a, b] is a compact interval, every function which is Riemann-integrable on
[a, b] is in U([a, b]) and therefore also in L([a, hi).
10.7 BASIC PROPEliTII OF THE LEllESGUE INTEGRAL
Tleon,n 10.14. Assume f  L(I) and g • L(1).
a) (af + by)  L(1) for every real a and b, and
b) I,f > 0 iff  x)
Then we alye:
+ = a
>_ 0 a.e. on L
= g(x) a.e. on I.
Proof. Part (a) follows easily from Theorem I0,6. To prove 03) we write
f = u - p, where u  U(1) and v  U(/), Then u(x) >_ x) almost everywhere
on I so, by Theorem 10.6(c), we have 't u >  v and hence
Part (c) follows by applying (b) to f- y, and part (d) foIIows by applying
twice.
Deftm'tion 10.15. If f is a real-valued function, its positive part, denoted by f+, and
its negative part, denoted by f-, are defined by the equations
f+ -- max (f, 0), /- = max (-f, 0),

 lO.1
Note thatf + and f- are nonnegative functions and that
/ = f+ -- f-, l/I = .f+ +/-.
Examples are shown in Fig. 10.1.
7eom 10.16. lf f and g are in L(1), then so are the functions f+, f-, Ill,
max (f, 9) and min (f, g). Moreover, we have
Proof Write f ffi u - v, where u e U(/) and v ¢ U(/). Then
f+ ffi max(u - v, 0) ffi max(u, v) -
But max (u, v)  U(1), by Theorem 10.9, and v  U(/), so f+ ¢ L(/). Since
f- = f÷ - f, we see that f- v L(/). Finally, I/I --
since -I/(x)l  f(x) < I/(x)l for all x in I we have
which proves (9). To complete the proof we use the relations
max (f, ,) = ½(f + 9 + If - gl), rain (f, g) = Rf + g - If - gl)-
The next theorem describes the behavior ofa Lebesgue integral when the inter-
vat of integration is translated, expanded or contracted, or reflected through the
origin, We use the following notation, where c denotes any real number:
I + c = {x + c'xel), el=
Tiprem 10d7. Assume f e L(1). Then we hae:
a) lnvariance under translation, lf g(x) = f(x - c) for x in I + c, then g  L(I + c),
b) Behavior under epansion or contraction.
c > O, then g 6 L(cl) and
f
If g(x) = f(x/c) for x in cL Where
c) lnvariance under reflection. If g(x ) = f(-x) for x in -I, then   L(-I) and
. If I has endpoints a < b, whe a and b are i the eeded real nur
sm N*, the fN in {) n N  tten  folle:
fix - c) x  f(x) ax.
os ) d (c) n  mbin into a single form which incla 
positive and netive values of c:
ff f(x/c) dx = lcl ff f(x) dx ff c  O.
Prfi In pvg a m of s , e prdu is ys the me. Ft,
we vfy e eom for step functions, en for upr funons, and filly for
guinteble funons. At eh sp the argument is sttfa, 
 omit e .
Tlore 10.18. Let ! be an interval which is the union of two subinteroMs, say
I = I  lz, where I and I have no interior points in common.
a) If re L(1), then f  L(Ix),f  L(Iz), and
) ,4ssume A  L(&), A  L(&), and let.f be
f(x)
A(x)
Then f ¢ L(1) and jf = It, f, +
Proof. Write f-- u -  where u  U(I) and v ¢ U(I), Then u -- u ÷ - u- and
 = v + - v-, so f-- u* + v- - (u- + v÷). How apply Theorem 10.10 to
each of the nonnegative functions u + + v- and u- + o ÷ to deduce part (a). The
proof of part (b) is left to the reader.
o, There is an extension of Theorem 10.18 for an interval which canbe
expressed as the union of a finite number of subintervals, no two of which have
interior points in common. The reader can formulate this for himself.
We conclude tkis section with two approximation properties that will be
needed later. The first tells us that every Lebesgue-integrable function f is equal
to an upper function u minus a nonnegative upper function v with a small integral,
The second tells us that fis 'eqraal to a step function s plus an integrable function

g with a small integral. More precisely, we have:
Tlcorcm 10.19, Assume f _ L(I) and let e > 0 be given. Then:
a) 7tere exist funclions u and v in U(1) such that f ---- u - v, where v i non-
n(gatiae a.e. on I and [ v < .
b) There exists a step function s and a function g in L(I) such that f -- s + g,
where  Igl < e,
Proof Since f L(/), we can write f = u s - va where u t and vj ate in U(/).
I. {t,} be a sequence which gcncra.t va. Since J' t   v, w can choose N so
thatO_<=[(v-- t)< . Now letv----t -- tandu---u - t. Then both
u and  are in U(1) and u - v = u - v = f. Also, v is nonnegtive a.e. on I
and   < e. This proves (a).
To prov© (b) we u¢ (a) to chos u and v in U(/) so that v > 0 ae. on I,
f-u-  and 0_<  < 2
How choose a step function s such that 0 _<  (u - s) < ]2. Then
wllere g = (u - s) - v. Hence g  L(/) and
2- 2
1@. LEBESGUE INI_.RAION AND SETS OF MEASURE ZERO
The theorems in this section show that the behavior of a Lebesgueintegrable
function on a set of measure zero does not affect its integral.
l'heorem 10.20. Let f be defined on L lf f : 0 almost et¢rywhere on I, then
f  .(0 and f = O.
Proof. Lt s.(x) :-: 0 for all x in I. Then {s,} is an incasing squence of step
functions which conveq to 0 everywhere on L Hence {s} converg©s o falmost
cvcrywh¢ on L $in [, s. = 0 the SlUCnce {j', s.} converges. Therefore f is
an upper function, sole L(/) and ,f -- lim,®  ** = 0.
Tiear¢m 10.21. Let f cmd g be defined an L If f  L(I) and if f = g almosl every,
where on l, then a  L(1) analyst=
Proof Apply Theorem 10.20 tof- g. Thcnf - g 6 L(I) and , (f - g)  0.
ml)e, Define/on tl)e ioumal [0, I]  fows:
{; ifx isrational
,f(x)  if x is irrational.
Then f = 0 aknost ¢wrwhea on [0, ] so f is tct,aitcra on [0,
 1 is 0. As not in  7, is fuon is not
 [o, H,
.  10.21 suts a finition of  in for run, iota at arc
d t ¢wh¢ on L Ire is such a fion d ifg(x) = f(x) almt
h on I, whef  0,  Y t   L(I) aM that
10.9 THE LEVI MONOTONE CONVERGENCE THEORF2dS
We turn next to convergence theorems concerning term-by-term integration of
monotonic sequences of functions. We begin with three versions of a famous
theorem of Beppo Lvi. The first concerns sequences of step functions, the second
seqnces of upper functions, and the third sequences of Lcbcsgue-integrable
functions. Although the th©orems are stated for increasing sequences, th©re are
corresponding results for decreasing sequences.
Tlearem 10.22 (Leol tkeorem for ace9 ftug'tiaas). Let {,} be a sequence of step
functions such that
a) {s,} increases on an interval I, and
b) lim, j s. exists.
Then {s} converges almost everywhere on I to a limit function f i U(I), and
Proof. We can assume, withom lss of generality, that the step functio , are
nonnegative. (If not, consider instcad the sequence {s, - sj}. If the theore is
true for {s. - }, then it is also tre for {/-) Let D be the el fx in/for hich
{s,()} diver, and let  > 0 be gve. We will pme that D has measure 0 by
showing that D can be covered by a countaNe collection of intervals, the sum of
whose lengths is < e.
Since the sequence { s.} converges it is bounded by some positive constant
M. Let
= s.x if x  I,
where [Y] denotes the greatest integer <y. Then {t,} is an increasing sequence of
step functions and eaeh funelion value t.(x) is a nonnegative integer,
If {s,(x)} converges, then {s,(x)} is bounded so {t,(x)} is bounded and hence
t,+ ,(x) - t.(x) for all suciently large n, since each t,(x) is an integer.
If {s,(x)} diverges, then {t,{x)} also diverges and t.+ ,(x) - t,{x) >__ I for
infinitely many values of n. Let
D. = {x:xel and t,+,(x)-t.(x)> 1}.

Then D. is the union of a finite number of intervals, the sum of whose lengths we
denote by tD.I. Now
so if we prove that ,,1 ID.I < e, this will show that D has measure 0.
To do this we integrate the nonnegative step function t.+ t - t ocer I and
obtain the inequalities
Hence for every m > I we have
Therefore .=l ID.I < e/2 < e, so D has measure 0.
This proves that {s.} converges almost everywhere on/. Let
f(x)={ m's"(x) ifxD.ifxl-D'
Then fis defined everywhere on ! and s, --,, falmost everywhere on L Therefore,
fe U(1) and S,f = lim.o St s..
Theorem 10.23 (Le tlovem for apper fmtions). Let {f} be a sequence of upper
functions such that
a) {f.} increases almost everywhere on an interval I,
and
b) lim._., }'f. exists.
Then {f} converges almost everywhere on 1 to a limit function fin U(I). and
Proof. For each k there is an increasing sequence of step functions (s.} which
generatesf. Define a new step function t. on I by the ectuation
t.(x) = max {s..t(x). s.,z(x), ... , s..(x)}.
Then {t.} is increasing on I I:'cause
t.+,(x) -- max {s.÷ t.,(x) .... , s.+ t,.-t(x)} _> max {s,.t(x ) .... , s,,,.÷ ,(x)}
_> max {s., (x) ..... s...(x)} =
10,2,1
But s.(x) < f,(x) and {2} increases almost everywhere on/, so we have
t.(x) _< max {A(x) .... ,y.(x)} = L(x)
(10)
almost everywhere on L Therefore, by Theorem t0.6(c) we obtain
,t. <: f,y.. 01)
But, by (b), {jx f.} is bounded above so the increasing sequence {: t.} is also
bounded above and hence converges. By the Levi theorem for step functions,
{h} converges almost everywhere on I to a limit function fin U(/), and rf =
fim._ r t.. We prove next thatf. -, faimost everywhere on I.
The definition of t.(x) implies s..i(x) < h(x) for all k < n and aB x in I.
Letting n   we find
(x) < f(x) almost everywhere on I. (12)
Therefore the increasing sequence {(x)} is bounded above by f(x) almost every-
where on L so it converges almost everywhere on I to a fimit function 9 satisfying
g(x) < f(x) almost everywhere on L But (10) states that t.(x) < f.(x) almost
everywhere on I so, Ietting n  , we find f(x) < g(x)almost everywhere on I.
In other words, we have
lira f,(x) - f(x) almost everywhere on I.
Finally, we show that ]'f -- fim.® Sf.. Letting n  o in (11) we obtain
f < lim ff.. (13)
Now integrate (12), using Theorem 10.6(c) again, to get ft g j'f Letting
k - oo we obtain lim_. ]'f < f which, together with (13), completes the
proof.
ro. The class U(1) of upper functions was constructed from the class S(I) of
step functions by a certain process which we can call P. Beppo Levi's theorem
shows that when process ' is applied to U(I) it again gives functions in U(O. The
next theorem shows that when P is applied to L(I) it again gives functions in
L(1).
TIgeem 10.24 (I.ol t&.orem for .qattme# of Lebesgaitegctzble fmg.tio).
{f} be a sequence of functions in L(I) such that
a) {f} increa.ces almost everywhere on I,
Let
b) fim._. L exists.

Then {f.} comserges almoa't eoerywhere on I to a I t f  L(, 
We all d s  from an v¢nt st s for series of
funons
a) eh g¢  at mt rywre on L
d
en the series  g crge almost rhere on I to a  ctn g 
L(1),  we 
Proof Since g, ¢ L(I), Theorem 10.19 tetIs us that for every t > 0 we can write
where u.  U(I), v. ¢ U(I), v. >_ 0 a.e. on l, and
corresponding to • = (y'. Then
= ¢. + v., where j
The inequality on js v¢ assures us that the series ,%t i v. converges. Now
u. > 0 almost everywhere on I, so the partial sums
form a sequence of upper functions {U.} which increases almost everywhere on I.
Since
the sequence of integrals { U¢} converges because both series ,_.  i g and
_a=  v converge. Therefore, by the Levi theorem for upper functions, the
sequence {U.} converges almost everywhere on ! to a limit function U in U(/),
and  U ffi lirn._  U.. But
The Levi Mmetm¢  
SO
Similarly, the sequence of partial sums {V,} given by
converges almost everywhere on I to a limit function V in U(1) and
Thcrcfor U- V e L(I) and the sequence {Y,= g} = {U." V.} converges
almost everywhere on Ito U- V. Letg = U- V. Theno L(/) and
This completes the proof of Theorem 10.25.
Proof of Theorem 10.24. Assume {f=} satisfies the hypotheses of Theorem 10.24.
Let gt -- f and let g, = f - f_ x for n >. 2, so that
Applying Theorem 10.25 to {g.}, we find that W_  go converges almost everywhere
on 1 to a sum function 9 in L(1), and Equation
almost everywhere on land  g =
In the following version of the I_,vi theorem for series, the terms of the series
are not assumed to be nonnegative.
Theorem 10,26. Let {ft.} be a sequence of functions in L (1) such that the series
is convergent. Then the series .%  g= coverges almost everywhere on I to a sum
function g in L(1) and we haoe
n----i lt---I
Proof Write g. =- g.+ - g- and apply Theorem 10.25 to the sequences {g.+}
and (gf } separately.
The following examples illustrate the use of the Levi theorem for sequences.

 L Let f(z) -- z" for x > O, f(O) = O. Prove that the Lebesgue integral
.ff(x) dx etists and ires the valu !/($ • 1) ifs > -1.
8olut/on. Ifs > O, thenfis bounded and Riemmm-integrableon [0, I] andits Riemann
intogral is oqual to 1/( + 1).
If s < 0, then f is not bounded and hcno not Riemann-intcgrable on [0, 1 ]. Define
a sequence of functions {f=} a follows:
/,(x) = {0x" ifx>lln,
if O < x< lln.
Thetl {fl} i in.easing afiafa f oll [0, 1]. Each f. is Riemmm-integrable
mid hence Leintegrable n [0. 1 ] and
Z,( x ) c =   - I -
lfs + 1 > 0, the seq {.[ } convttl to l/(a + 1). ore, the Levi thoorem
for soqu shows that !0 ftmists and equals l/{s + 1).
Kvam#e Z. The same  of argument shows that the l.besgue inmgral .IS e-'x "-  dx
e.ists for  real y > 0. This ilatcgral will be used  in discussing the Gamma
function.
Levi's theorems have many impotent eonsoquen, The first is Lebesgue's
dominated convergence theorem, the cornerstone of Lcbesgue's theory of inte-
graft.on.
'iaatrem 10.27 (L&   ammm,). Let {f,} be a sequence
of Lebexgue-terable fuctions on ¢m interval I. Agytm that
a) {f,} converges almost eerywhere on I to a limit function f,
b) there is a nonnegaHve function g n L(1) such that, for olin > 1,
If,(x)l < g(x) a.e. oa 1.
Then the limit function f• L(I), the sequence {Sf,} converges and
f,f= (15)
crr. Property (b) is describ by saying that the sequence {f,} is dominated by
g almost everywhere on L
Proof. The idea of the proof is to obtain upper and lower bounds of the form
g.(x) f.(x) _< G.(x) 06)
where {go} increases and {G} decreases almost everywhere on I to the limit function
f. Then we use the Levi theorem to show that fe L(I) and that
lim...o y, = lim...  G, from which we obtain
To construct (g,} and {G.}, we make repeated use of the Levi theorem for
sequences in L(I). First we define a sequence {Gj.1} as follows:
G..(x) = max {ft(x),fx(x),... ,f,(x)}.
Each function G.,t • L(I), by Theorem 10.16, and the sequence {G,.s} is in-
creasing on I. Since ItT.,,(x)l < (x) almost everywhere on I, we have
Therefore the increasing sequence of numbers { G.,x} is bounded above by
S g, so lim,_.  G,. exists. By the Levi theorem, the sequence {G. } converges
almost everywhere on I o a function G, in L(I), and
Because of (17) we also have the inequality - g _< J G v Note that if x is a
point in/for which G,.t(x ) - G(x), then we also have
G,(x) = sup {A(x), A(x),... }.
In the same way, for each fixed r  1 we let
= max {f,(x),f.+ ,(x) ..... f,(x))
for n _> r. Then the sequence {G,.,} increases and converges almost everywhere
on I to a limit function G, in L(/) with
Also, at those points for which G,,(x)  G(x) we have
G,(x) = sup {f,(x), f,+ l(X), . . . },
SO
f,(x) <_ G,(x) a.e. on L
Now w¢ examine properties of the sequence {G,(x)}. Since , _ B implies
sup A < sup B, the sequence {G,(x)} decreases almost everywhere and hence
converges almost everywhere on L We show next that Go(x)  f(x) whenever
lira f,(x) = f(x). (18)
If (f 8) holds, then for every  > 0'there is an integer N such that
f(x) - s < f,,(x) < f(x) + e. for all n >_ N.

Hence, if m > N we have
f(x) - e : sup {f.(x),f,+x(x),... }  f(x) + .
In other words,
m > N implies
and this implies that
f(x) - 6.(x) f(x) +
lim G=(x) = f(x) almost everywhere on I. (19)
On the other hand, the decreasing sequence of numbers {1 G.} is bounded below
by -x g, so it converges. By ([9) and the Levi theoretn, we see that f• L(/) and
timrG.=ff
By applying the same type of argument to the sequence
g..,(x) = min {fx),+(x),.,.,
for n  r, we find that {g.} dec and nv almt eye, here to a
limit function g, in L(I), where
.q,(x) = inf {f.(x),)r, +, (x),... } a.e. on L
Also, almost everywhere on I we have g.(x) < f,(x), {g,} increases, lim,_ g,(x) =
f(x), and
Since (16) holds almost everywhere on I we have jx go _< j,L < a., Letting
n --, oo we find that {'t f,} converges and that
10.11 APPLICATIONS OF LEESGUE'S DOMINATF__ CONVERGENCE
THEOREM
The first application concerns term-by-term integration of series and is a companion
result to Levi's theorem on series.
Irheoem 10.28. Let {g.} be a sequence of functions in L(1) such that:
a) each g. is nonnegative almost everywhere on 1,
and
) tlw series ,'-= g, com,erges almost everywhere on I to a function g which is
bouned above bj a function in L(I).
Then O e L(/), the series .%   . coverges, and we have
Let
f,,(x) =  O,(x) if x • I.
Then f  g almost everywhere on/, and {f.} is dominated almost everywhere
on I by the function in L(/) which bounds g from above. Therefore, by the Le-
besgue dominated convergence theorem, g 6 L(/), the sequence f} converges,
and  g ffi lim,..® Jf,. This proves the theorem.
The next application, sometimes called the Lebesgue bounded convergence
theorem, refers to a bounded interval
Theorem 10.29. Let I be a bounded interval. 4ssume {f,} is a sequence of functions
in L(1) which is boundedly convergent almost everywhere on L That is, assume there
is a limit function f and a positive constant M such that
lira f(x) = f(x) and If(x)[ < M, almost everywhere on 1.
Proof Apply Theorem 10.27 with g(x) = M for all x in L Then g • L(/), since
1 is a bounded interval,
NOT. A special case of Theorem 10.29 is Arzelt's theorem stated earlier CFheorem
9.12). If {f,} is a boundedly convergent sequence of Riemann-integrable functions
on a compact interval [a, b], then each f, eL([a, b]), the limit function
f• L([a, b]), and we have
If the limit function f is Riemann-integrable (as assumed in Arzelh's theorem),
then the Lebesgue integral fis the same as the Riemann integral .f(x) dx.
The next theorem is often used to show that functions are Lebesgue-integrable.
Theoeem 10.30. Let {f,} be a sequence of functions in L (1) which converges almost
everywhere on I to a iimit function f Assume that there is a nonnegative function g
in L (D such that
If(x)l < g(x) a.e. on 1.
Then f  L(I).
Proof Define a new sequence of functions {9,} on I as follows:
g. :- max {rain (f., g), -g}.

10.2
Geometrically, the function g. is obtained from f. by cutting off the graph off.
from above by g and from-below by -9, as shown by the example in Fig. 10.2.
Then lg.(x)i _< (x) almost evhcre on I, and it is easy to verify that g. - f
almost everywhere on L Therefore, by the Lebesgue dominated cxnvergence
theorem, f  r(/).
1.12 LEllESUE IN'IEGRAI ON UNIR)URDErl IlgIEilVALS AS LIMITS
OF INTEGRAI ON BOUNDED INTERVAUS
lmm 1031. Let f be defined on the half-infinite interval I = [a, + oo). lsavne
that f is Lebexgue-interable on the compact interval [a, b] for each b > a, and
that there is a poMtie constant M such that
 [f/ < M for all b
Then f ¢ L(1), the limit lim.+ . f existx, and
f lira f (21)
Proof. Let {b.} be any increasing sequence of real numbers with b, > a such that
lira,_,® b, = + oo. Define a sequence {f,} on ! as follows:
f"(x) = {fo(x) otherwis.ifa<x<b"
Each f. e L(/)(by Theorem 10.I8) andf.  fon L Herte¢, If.I -, Ill on/. But
If.I is incxrasing and, by (20), the sequence t If.I} is bounded above by M.
Ther¢fore Iim..** j' If.I exists. By the Levi theorem, the limit function Ill e L(/).
Now each IfJ < Ill andf.  fort L so by the l.¢besgue dominated eonvergenoe
.theorem, re L(D and lirn., j'1f. = fir- Therefore
lira f = f
for all luence {b.} which increase to + o. This completes the proof.
There is, of course, a corresponding theolx:m for the interval (-m, a] which
concludes that
provided that  Ifl < M for all c < a, If ,s lfl < M for all real c and b with
c . b, the two theorems together show thatfe L(R) and that
f= lira f+ lira f.
Example 1. Le.tf(x) = 11(1 + x 2) for all x in R. We shall prove that/e Ltll) and that
a f = n. Now fis nonnega|ive, and if c _< b w¢ have
fc  f dx - argtarl b arctan c < .
f= 1 +x 2 -
"rherffom, f  L(R) and
_ c 1 + x z + lira +
,--, +  1 + x z 2 2
Examlfle 2. In this example the limit on the right of (21) ¢ists but f6/.I). Let
I = [0, + ) and define f on I as follows:
f(x)  (-1)" if n- 1 : x < n, forn= 1,2,...
If b > 0, let m = [b], the greatest integer <_ b. Then
ff= ffff + f: f= k(-l)'+.=-n (b- m)-l)m+lm + I
As b  + c the last term  0, and we find
fir k
lira .... log 2.
Now we assumef¢ L(I) and obtain a contradiction. I_¢t f. be defined by
for 0 < x _< n,
forx> ,
Th {} i anti A(x)  If(x)l everywhere on L Since fe L(I) we also have
Ifl  L(I). But If.(x)[ -< I/'(x)l everywhere on r so by the Lebesgue dominated con-
vergnce theorem the sequence {j' f,} convert. But this is a contradiction since
= Ill = g-' + 
as-4

10.13 IMPROPER RIEMANN 
In Example 2 of the foregoing section the improper Rivraann integral
 © f(x) dx exists but f is not Lebesguintegrabte on [0, + ). That example
should be contrasted with the following theorem.
Tlworem 10.$3. Assume f is Riemann-integrable on [a, b] for every b  a, and
there is a poMtive constant M such tlm¢
If(x)l dx < M
for every b  a.
Then both f and Ifl are improper Riemann-integrable on [a, +
Lebesgue-integrable on [a, + co) and the Lebe.e integral off is equal to
proper Rdemann integral off
(22)
Also, f is
Proof Let F(b) = , [f(x)l dx. Then Fisan increasing function whiehis bounded
above 5y M, so lim_+ ® F(b) ey£sts. Therefo If  is improper Rieraann-integrable
on [a, + oo). Since
0 If(x)l -f(x) <_ 21f(x)l,
the limit
tim [ {lf(x)l -f{x)} #x
also exists; hence the imit lims ÷ ® .,f(x) dx exists. This proves thatfis improper
Riemann-integrable on [a, + co). Now we use inequality (22), along with Theorem
10.31, to deduce thatfis Lebesgu.integrable on [a, + o) and that the Lebesgue
integral off is equal Io the improper Riemann integral off.
There are corresponding resutts for improper Riemann integrals of the
' f(x) dx = lira f[ f(x) dx,
f(x) dx = lira f(x) dx,
ff f(x) dx = lira ff f(x) dx,
which the reader can formulate for himself.
If both integrals JL  f(x) dx and +  f(x) dx exist, we ay that the imegrat
+_  f(x) dx exists, and its value is defined to he their sura,
f(x) dx = f(x) ax + f(x) ax.
If the integral +_,f(x) dx exists, its value is also equal to the symraetric lirait
However, it is important to realize that the symmetric limit might exist even when
_ f(x) dx does not exist (for example, taker(x) = x for all x). In this case the
symmetric limit is calIe.x[ the Cauchyprincipal tlue of_+ f(x) dx. Thus [+_, x dx
has Cauchy principal value 0, but the integrat does not exist.
Examlllel. I.tf(x) = e-Xxv-l, where y is afixed real number. Sie e-t2x -I -* 0
as x -, +, there is a constant M such that e-X/zxv-: < M for all x _> 1. Thea
e-.:¢- t <_ Me-X/z, so
 ,f(x)l dx <_ M e e-x/" dx = 2M(l - e'') < 2M.
Here the integral j z e-X.-, dx mdsts fr every real y, both as an iml:a'Olf 
integral and as a  integral.
 2. The Gamma fam:tlo ./niece"a/. Adding the integral of Example 1 to
inttgrat .[ e-'x -t dx of Exampte 2 of Section 10.9, v find that the Lclxsgu integral
= e-  x '-'
exists for ach  y > O. The ftmction r so  is  th Gamma fuactRm.
Example 4 below shows its relation to th¢ P.ieamum zeta function.
r, rorr_ Many of the theorems in Chapter 7 concerning Riemann integrals can be
converted into theoreras on improcer giemann integrals. To illustrate the straight-
forward manner in which some of these extensions can be made, consider the
formula for integration by parts:
, j f(x)$'(X) dx = f(b)g(b) -- f(.)g(a) --  g(x)f'(x)
H
Since b appears in three terras of this equation, there are three limits to consider

as b  + oo. If two of thee Hmits exist, the third also exists and we get the
formula
f(x)a'(x) dx = l f(b)g(b) - ffa)g(a) - g(x)f'(x) dx.
+
Or thrums on Rn inls n  ext in much ¢ e way
to imr Rienn intls. Hr, it is not ns to deop the
of tensions any luther, sin in any cr examplg it sus to aly
the ui eom to a com inal In, b] and
 3.  futio eqtion y  11 = yF(y). If 0 < a < b, tion by
a  O+ db +,y + 1) = y).
 & lntel recantation for t nn zeta fu.
fion { is n for s > 1  e ti
= 7"
This example shows how the Levi con theorem for series can be used to derive an
integral reprmentation,
The intedlxal exists as a l.besgue integral.
In the integral for F(a) we make th¢ ehango of variable t = nx, n > O, to obtain
He.nee, if s > O, we ha-e
n-q's) = _(: e-"Xx - dx.
If s > 1, the seri¢ ..% n - converg o we have
the  on the right ing t. Sin
ven th  10.25) [ls  that e i  e - - n
almt h to a sum fui which s intb on [0, +
(a)r(s) = e-: -  = e--
But ifx > O, wehaveO < e -x < land
n- 1 -e -x -1'
t  g a  . f  ha
 --1
.-1 ex -- ]
   [0, + ), in f h cxpt
(s)U(s) = --
. - 1
10.14  ILrNCTIONS
lv¢T function fwhieh is Lebcsgue-integrable on an interval I is the limit, almost
everywhere on I, of a ertain sequence of step functions. However, the converse
is not true. For examp|e, the constant function f = 1 is a limit of step functions
on the real line R, but this function is not in L(R). Therefore, the class of functions
which are limits of step functions is larger than the class of Lebesgue-integrable
fune6ons. The functions in this larger class are called measurable functions.
Defmim 10.34. A function f defined on 1 is called measurable on L and we write
f • M(I), if there exists a sequence of step functions {s} on I .ruch that
lira s(x) = f(x) almost everywhere on I.
No. Iff is measurable on I then f is measurable on every subinterval of L
As already noted, every function in L(I) is measurable on I, but the converse
is not tree. The next theorem provides a partial converse.
Theorem 10.35. lf f 6 M(I) and if If(x)l ¢: #(x) almost everywhere on l for some
nonnegative  in L(1), then y e L(I).
Proof. There is a sequence of step fanct/ons {s,} such that s,(x) - f(x) almost
everywhere on L Now apply Theorem 10.30 to deduce that f• L(I).
Corollarj I. If f• M(I) and Ifl  L(I), then f e L(I),
C.ordy 2. lf f is measurable and bounded on a bounded interval 1, then f • L(I).
Further properties of measurable functions are given in the next theorem.
l"heottm 1036. Let q be a real+valued function eontgnuoas on R . lf f • M(I) and
 • M(1), define h on I by the equation

10.37
Then h  M(]). In particular, f + g, f . g, Ill, max  g), and rain (f, g) are in
M(I). Also, life M(1) if f(x) ¢ 0 almost everywhere on L
Proof. Let {s,} and {t,} denote sequences of step functions such that s, -, land
t, - g almost everywhere on L Then. the function u, = 9(s,, t) is a step function
such that u, - h almost everywhere on L Hence h • M(/).
The next theorem shows that the class M(/) cannot be enlarged by taking
limits of functions in M(/).
Tl,arem 10.37. Let f be defined on I and assume that {f.} is a sequence of measur-
able functions on I such that f(x)  f(x) almost everywhere on L Then f is measur-
able on L
Preef. Choose any positive function g in L(I), for example, g(x) = 1/(1 + x 2)
for all x in L Let
= y.(x) for x in t.
i + If,(x)l
F.(x) -.
I +
almost everywhere on 1.
Let F(x) : g(x)f(x)]{I + If(x)f}. Since each F, is measurable on 1 and since
IF,(x)] < g(x) for all x, Theorem 10.35 shows that each F, e L(l)h Also, IF(x)l <
#(x) for all x in Iso, by Theorem 10.30, F L(I) and hence Fe M(1). Now we
have
f(x){g(x) - /F(x)I} = f(x)g(x) t 1
for all x in/, so
If(.x)l _
1 + If(x)tJ I + If(x)l
f(x) -
Therefore f M(I) ince each of F, g, and IF t is in M(I) and g(x) - IF(x)l > 0
for all x in 1.
NOT.. There exist nonmeasarable functions, but the foregoing theorems show that
it is not easy to construct an example. The usual operations of analysis, applied to
measurable functions, produce measurable functions, Therefore, every function
which occurs in practice is likely to be measurable. (See Exercise 10.37 for an
example of a nonmeasurable function.)
10.15 CONTIUITY OF FUNCTIONS DEFINED BY LI,ESGUE INTERAL._q
Let f be a real-valued function of two variables definc.d on a subset of R 2 of the
form X × Y, where each ofXand Yis a general subinterval of R. Many functions
in analysis appear as integrals of the form
F(y) = fxf(x, y) dx.
We shall discuss three theorems which transmit continuity, ditterentiability, and
integrability from the integrand f to the function F, The first theorem concerns
continuity.
Tlwaeem lO3& Let X and Y be two subintertcds of R, and tet f be a function defined
on X x Y and $atisfying the following conditions:
a) For each fixed y in Y, the functDn f defined on X by the equathm
f,(x) = f(x, y)
is measurable on X.
b) There exists a nonnegative function g in L(X) such that, for each y in Y,
If(x, Y)I <-- g(x) a.e. on X.
c) For each fixed y in Y,
lira f(x, ) = f(x, y) a.e. on X.
Then the Lebesgue integral f(x, y) dx exists for each y in Y, and the function F
defined by the equation
e(y) -:- f  f(x, y) dx
is continuous on Y. That is, if y  Y we have
,-,lim fxf(x' t) dx = fxlimf(x' ')dx"
Proofi Sincef is measurable on X and dominated almost everywhere on X by a
nonnegative function  in L(X), Theorem 10.35 shows thatf  L(X). In other
words, the Lebesgue integral xf(X, y) dx exists for each y in Y.
Now choose a fixed y in Y and let { y} be any sequence of points in Y such that
lira y, = y. We wilt prove that lira F(y,) = F(y). Let G,(x) = f(x, y,). Each
G • L(X) and (c) shows that G,(x)  f(x, y) almost everywhere on X. Note that
F(y,) = r G,(x)dx. Since (b) holds, the Lebesgue dominated convergence

theorem shows that the sequence {F(y)} converges and that
F(y.) -- ff(x, y) dx =
Jx
Exalttlfle L Continuity of the Gamma function F(y) - j'o +® e-Xx -z dxfor.v > 0. We
apply Th©orem 10.38with X = [0, +o0), y= (0, +0o). Foreachy > 0theint¢grand,
as a function of x, is continuous (hone, tricasorable) almost everywtmre on X, so (a) holds.
For each fi×od x > 0, the integrand, as a function of y, is continuous on Y, so (c) holds.
Finally,  verify (b), not on Y but on eyeD' compact subinterval [a, b]. whel 0 < a < b.
For ¢aeh y in [a, b ] the integrand is dominated by tl function
{x - if0 < x .-< 1,
g(x) -- Me -#t2 ifx > l,
wher M is some posffiw constant. This a' is Legu-inttable on X, by Theorem
10.18, so Theorem 10.38 mils us that F is continuous on In, b]. But since this is ta-ue
for e'ry subintcxval [o, b], it follows that F is continuous on Y
  sin X
F(V) = • - dx
x
for y > 0 In this example it is uriC(nod that tim quotient (sin x)lx is to tm rolacod
by 1 when x = 0. Let X = [0, + co), Y  (0, 4 co). Conditions (a) and (c) of Thcorean
t0.38 ai satisfied. As in Exampl I, w¢ verify (b) on each subinterval
a > 0. Since [(sin x)lx t <_ 1, the intogrand is dominated on Ya by the font(ion
g(x) - e - for x >_ 0.
Since O is I.*bcsgue-integrable on , F is continuous on Ya for evt a > 0; Irene, F is
ontinuous on Y = (0, + o)
To illustrate another ue of the Lebcsgue dominated convergence theorem we
hall prove that F(y) - 0 as y --, +
Let {y,} h¢ any increasing sequenc of real numbers such that y, > 1 and
y, - + 0o as n , co. We will prove that F(y) - 0 as n -, oo. Let
sin x
f.(x) = e -" for x _> 0.
x
Then lim,.., f(x) - 0 almost everywhere on [0, +), in fact, for all x exc,pt 0.
Now
y. _> 1 implies tf(x)[ < e - for all x > 0.
Also. each f= is Riemanmintegrable on [0, b] for every b > 0 and
Therefore, by Theorem 10.33. f. is Ldxgucintegrable on [0, +oo). Since the
sequence {f} is dominated by the function y(x) = e - which is Lebesgue-inte-
grable on [0, + m), the Lcbesgu dominated convergence theorem shows that the
sequence
0 f.} converges and that
lira f. : lim f. = 0.
But   L = F(yo), so F(y) - 0 as n -* co. Hence, Fry)  0 as y - + co.
OTE. In much of the material that follows, we ghall have occasion to deal with
integrals involving the quotient (sin x)/x. It will be understood that this quolient
is to be replaced by l when x = O+ Similarly, a quotient of the form (sin xy)/x is
to be replaced by y, its limit as x -, O. More generally, if we are dealing with an
integrand which has removable discontinuities at certain isolated points within
the interval of integration, we wilt agree that these discontinuities are to be "re-
moved" by redefining the integrand suitably at these exceptional points, At points
where the integrand is not defined, we assign the value 0 to the integrand.
10.16 DIFFERENTIATION UNDER THE INTEGRAL SIGN
Theorem 10.39. Let X and Y be two subintervals of R, and to(f be a function defied
on X × Y and satisfying the following conditions:
a) For each fixed y in Y, the function f defined on X by the equation f(x) = f(x, y)
is measurable on X, and f  L(X) for some a in ¥.
b) The partial derivative Df(x, y) exists for each interior point (x, y) of X x Y.
c) There is a nonnegatwe function G in L (X) such that
IOzf(x, y)l  G(x) for att interior points of X x Y.
Then the Lebesgue integral :t f(x, y) dx exists for every y in Y, and the function F
defined by
e(y) = ]xf(x' y) dx
is differentiabte at each interior point of Y. Moreover, its derivative is given by the
formula
F'(y) = Jx Df(x, y) dx.
NOTE. The derivative F'(y) is said to be obtained by differentiation under ttte
integral sign..
Proof. First we establish the inequality
IL(x)[-< If.(x)l + ly - al G(x), (23)

for all interior points (x, y) of X × Y. Th© Mean-Value Theorem gives us
f(x, y) - f(x, a) - (y - a) D:tf(x, c),
where c lies between a and );. Since lDzf(x , c)l <_ G(x), this implies
If(x. Y)I -< If(x, a)l +
which proves (23). Sincef is measurable on X and dominated almost everywhere
on X by a nonnegafive function in L(X), Theorem 10.35 shows that f, c L(X).
In other words, the integral Sxf(x, Y) dx exists for each y in Y.
blow choose any sequence {y.} of points in Y such that each y, # y but
lira y. ffi y. Define a sequence of functions {q,} on X by the equation
q.(x) = f(x, r.) - y(x, y)
y.- y
Then q.  L(X) and q.(x)  Df(x, y) at each interior point of X. By the Mean-
Value Theorem we have q(x) = D2f(x , e, wher c. lies between y. and y. Hence,
by (c) we have Iq.(x)l  G(x) almost everywhere on X'. Lebesgue's dominated
convergence theorem shows that the sequence {Sx q.} converges, the integral
r Dzf(x, .v) dx exists, and
l  zf(x, y) dx.
q, =  {f(x, Ys)  f(x, y)} dx ----
y. -- y y  y
Since this last quotient tends to a limit for all sequences {y.}, it follows that F'(y)
exists and that
Example I. iti ft G/anetion. e divalive F'(y) exists for ch y > 0
and is n by the iatl
(Y) = e- -  log x
obta/ncd by differentiating the integral for F(.V) under the inttgral sign. This is a conse-
qugnc¢ of Theorem 10.39 because for each y in [a, b], 0 < a < b the partial deriva-
tive D(e-x r-t) is dominated a.e. by a function# which is integrable on [0,
fact,
-y(e -xr-t) = e-xX 1 log X ifx >
so if y > a tlm partiaJ derivative is dominated (pt at 0) by tlm function
a-t]log xl if0 < x < l,
g(x) ffiie-¢ ifxif x > 1,= O,
where M is sort positive constant, The reader can easily verify that g is Lebesgue-
intqlrable on [0, + o).
Exam#e 2. F_aluaian o/ the inte#ral
Appng  10.39,  find
F') = -
fo +m sn x
oe-'Kmxdx ify > 0.
(As in Example 1,  prove the result on every interval Y. = [a, + ), a > 0.) In this
example, the Riemann integral I e- sin x dx can be calculated by the mhods of
elen'gmtary calculus (using integration by parts twice). This gives us
fe -" sin x dx = e-Y(-'v sin b - cos b) + __1. (24)
I + I +yZ
for all real y. l.tting b  + o we find
f+ I
e -'sinxdr= 1+ yz if y> 0.
Therefor F'(y) = -1/(1 + yz)ify > 0+ Inmgration of th/s equation gives us
F(y) - F(b) = - 1 + t 2 alkali b arctan y, for y > 0, b > 0.
Now let b - +0. Then arctan b --, /2 and F(b) --, 0 (see Example 2, Section I0.15).
so F(y) = n/2 -- artan y. In other words, we have
fa © sin x n
e -u -- dx = - - arctan y ify > 0+ (25)
x 2
This equation is also valid if y = O. That is, we have the formula
sin x dx =  (26)
x 2
However, we cannot deduce this by putting y - 0 in (25) because we have not shown that
F is continuous at 0. In fact. the integral in {26) exists as an improper giemann integral.
It does not exist as a Lebesgue integral. (See Exercise 10.9.)

 3. Proofoftlformula
 sin x dx lira
t }  the u of fio n for aH 1 y by  uation
g.¢_v) = f e -' sin _ x x dx.
First we not that g.(n) - 0 a n  c¢ since
[e,(n)l < f*o e-,, dx = l f 1
(27)
Now we differentiate (27) and use (24) to obtain
g(y) =  (" e - sin x dx : - e-v(- ysinn- cash)+ 1
l+yZ '
an equation valid for all real y. This shows that e(y) - - 11(1 + j) for all .v and that
I +3 ,2
for alJ y _> 0.
Therefore the function f= defined by
if0_< y< n,
if)' > n,
is Lebesgue-integrable on [0, + ) and is dominated by the nonnegatie funiou
a(.v) = e-('v + 1) + I
l+y 2
Also, g is Lebesgu-integrable on [0, +oo). Sinef(y) -, - I/(1 + y2) on [0, + =), the
Lebesgu¢ dominatl eoavergence theorem implies
Jim f,, = _ dy 
"-' 1 + y2 2
But we have
f = a;,(y) dy = a,(,) - g,(o).
Letting n  oc, we find g(0)  /2.
Now if b > 0 and if n = [hi, we have
sin x dx sin x dx + dx
X X X
0_< tdn x dx < dx- <--0
lim fo u sin x d lira #0) = - _
This formula Will he ttvedvd in C-2mptm" 11 in the study of Fourier
10.17 INTERCHANGING THE ORDER OF INTEGILATiON
Tktartm lOd#. t X  Y  mo te of R,  let k  a fction whi
is fi, cot,  ed  X x Y, say
Ix, )[  M far all (x, y) in X x Y.
a) For eh y  Y, t e tral x f(x)k(x, y)  exits,  t cti
F d  Y by t ion
e(y) =  f(x)k(x, y) dx
 t on Y.
b) For  x  X, the  trd r g(y)x, y)  et,  t funetion
G d on X by the eion
 ntuo on X.
c) T o e informs r g(y)F(y) @ d rf(x)G(x)  exit  e
eaL Thal ,
Pmof For h  y in Y, let(x) = f(x)k(x, y). en is measuble on X
and tisfi e inty
Ifx)[  Ix(x, Y)I  Mlx)l for all x in X.
so, si k is nfinaous on X x Y  
l f(x)k(x, t) = f(x)x, y) for
e,  (a) fallows fm m 10.38. A sitar

10.40
Now the product f- G is measurable on • and satisfies the inequality
Iffx)¢7001 < If(x)[ r I(y)l Ik(x, Y)I dy  g' If(x)l,
whex M' = M Jr I(Y)I dy. By Theorem I0.35 we see that f. G • L(X). A
similar argunnt shows that g-F e L(Y).
Next we prove (28). First we note that (28) is true if each off and # is a step
function. In this case, each off and g vanishes outide a compact interval, so each
is Riemann-integrable on that interval and (28) is an immediate conse.zluence of
Theorem 7.42.
Now we use Theorem 10.19(b) to approximate each off and g by step functions.
If e > 0 is given, there are step functions s and t such that
Therefore we haw
where
lf- sl <  and ;r la - Xt < e,
f,,f "  = fx s ' G + al, (29)
Also, we have
where
Therefore
where
I#(y)[ lk(x, y)[ dy < tM r
G(x)= .fr fl(y)k(x, y)dy= fr t(y)k(x, y)dy+A,
#s'G = xS(X)[rt(Y)k(x, y)dyldx + A3,
 M yx {Is  fl + lfl}
Igl.
YI s(x)[t(y)k(x,, y)dy] dx + At + A..
and
But the iterated integrals on the right of (30) and (31) are equal, so we have
Since this holds for every  > 0 we have-}'xf. G = ]' e" F, a required.
mrrr. A more general version of Theorem 10.40 will M proved in Chapter 15
using double integrals. (See Theoro
111.18  SETS ON THE ILEAL LINE
 101. Gioen any nonempty subset S of  The cti s  by
{10 if x S,
x) = x e R - S,
is iled the charactstic ction ors. If S  empty   x) = O f l x.
m 10.42. t R = (- , + ). Tn  :
a) IfShmereO, th   L) $t  = 0.
b) If   L(R)    = O, t S  mre O.
f. P (a) follo by raking f = s in om 10.20. To ove (b), t
f, =  for all n. en ]f,I =  so
= s =0.

 10.43
By the Levi theorem for absolutely convergent series, it follows that the series
-.,,% t f,(x) converges everywhere on R except for a set T of measure O. If x e S,
the series cannot converge sine each term is 1. If x ¢ S, the series converges
because each term is 0. Hence T = S, so S ! mcasu 0.
Defmit 10.43, A subet S of It is called measurable if its characteristic function
Zs is measurable. If, in addRion,  is Lebesgue-integrable on R, then the measure
 S) of the set S is defined by the equation
v(s) =
If Z is measurable but not Lebestjue-integrable on R, we define ($) = + oo. The
function 1' so defined i,t called Lebeudue measure.
1. Theo:m 10.42 shows that a set S of mrasure zexo is measurable and that/(S) = 0.
2. Every interval I (bounded or unbounded) is measurable. If I is a bounded intcawal
with endpoints a _< b, then MI) -- b - a. If I is an unbounded interval, theal
aft)=
3. If A and B are measurable and A _ B, then t*(f) </(B).
Tlteorem 10-44, a) If S and T are measurable, so s S - T.
b) If St, 52, .., , are measurable, so are U=I s and (=
Proof To prove (a) we note that the characteristic function of S - Tis
To prove (b), let
v. = s. v. = &, v = U s,, = (I s,.
Then we have
Zv. = max (Xa,,. •., Xa.) and Zv. = rain (Zs,,---, Zs.),
so each of U. and V, is measurable. Also, Xv = lim_. Zv. and Zv = lim,_.
so U and V are measurable.
Theorem 10.45, If A at, d B are disjoint measurable sets, then
: va) +
Proof Let S = A  B. Sinoe 4 and B are disjoint we have
(32)
Suppose that Zs is integrable. Sinc both Xa and Zs are measurable and satisfy
0 < X(x) < Z(XL 0 _< X(x) < /a(x) for all x,
Theorem 10.35 shows that both ga and  are integrable. Therefore
In this ca¢ 02) holds with both members fmit.
If Xa is nt integrable then at least one of Xa or X i not integrabl¢, in which
case 02) holds with both mmbers infinite.
The following extension of Theorem 10.45 can be proved by induction.
Tlteorem 10,46. If {X .... , A.} is a finite disjoint collection of measurable ets,
NO'm. This property is der, cribed by saying that Lebesgue measure is finitely
addffoe. In the next theorem we prove that Lebgue measure is eotmtably additive
Tlteorem 10.47, If {, A2,... } is a countable disjoint collection of measurable
sets, then
t A = A9. 03)
we have
Since t is finitely additive,
We are to prove that/(T.) - #( a8 n  . No tt V(T  +) so
{T} is an inca quen.
We consid two . V(  finite,  Xr and ch X is intcgmble. ,
• c qn {(T} is und avc by  so it n.  the 
dona convgcn thcom, v(T  (T).
If p( = + , n z  not b. m 10.24 plics at 
some X. is not ble or el e , is able.but (T  + . In er
ca 03) holds  both m i.
For a fuer study of measu theo and i lafion to inteation, the der
n nst the mfe at the end of this cpter.
10.19 THE UE INTEGRAL OVErt ARBITILAR¥ SUBSETS OF R
Definition 10.48. Let f be defined on a raea,mrable subset S of R. Define a new
fum:tion  on R as follows:
j.(x) = {of(x) /fx e S,
ifxeR - S.

If f is Lebesgue-integrable on R we say that f is Lebeague-integrable on Samt we
write f  L(S). The integral of foyer S is defined by the equation
This definitior/immttiatly gives the foltowing properties."
If f  L(S), t.hen f  L(T) for every ubset oft orS.
If S ha finite measure, then i(S) = s 1.
The following theorem de'rlbc, a countably additive prolx:rty of the Lebcsgue
integral Its proof is left as an exercise for the reader.
leorem 10.49. Let {A x, Az, ... } be a countable disjoint collection offers in R,
andlet S = J'=l A. Let fbe defined on S.
a) If f c L(8), then f ¢ L(A3 for each i and
b) lf f ¢ L(A,) for each i and if the series in (a) commrges, then f c L(S) and the
equation in (a) holds.
10.20 UE INTEGR&LS OF COMPLEX-VALUED FUNCTIONS
Iff is a complex-valued function dfined on an interval I, then f = u + iv, where
u and v are real We sayfis Imbcsgue-intgrabie on 1if both u and v are Lebesgue-
integmbl¢ on I, and we define
Similarly, f is called measurable on/if both u and v are in M(I).
It is easy to verify that sums and products of complex-valued measurable
functions are also measurable. Moreover, sinc
Ifl = (u z +
Theor¢m 10.36 shows that Ill is measurable if ]'is.
Many of the theorems concerning I.,besgu integrals of real-valued functions
can be extended to complex-valued functions. However, we do not discuss these
extensions since, in any particular case, it usually suffices to write f = u + 
and apply the theorems to u and v. The only result that needs to be formulated
explicitly is the following.
Tluorem 1050. If a complex-valued function f is Lebeue-integrable on I, then
ill ¢ L(I) and we have
Since f is me  s  ab  e tfl < I.I ÷ I  l, Theorem
Proof. Write f = u + it,.
10.35 shows that Ill  L(/).
L¢t a = Jf. Thor a = re , where r  ]a[. We wish to prove that r <: r If I-
b={:-" ifr > O,
ifr = 0.
Ttmn tbl = 1 and r = ba = b [f = Jr bf Now write b/= U + iV, where
U and V ar real. Then bf = .v, since bfis real. Hence
10.21 1NNEll PRODUC'q AND NORMS
This section introduces inner products and norms, concepts which play an im-
portant rolt in the theory of Fourier series, to b¢ discussed in Chapter 11.
n 10.51, Let f and g be two reat-oalued functions in L(I) whose product
f . g ia in L(I). Then the integral
f('O0('O d, (30
is called the inner product off and g, and is denoted by (f, g). If f   L(I), the
narmegatioe number (f,f)t/, denoted by ]lfll, is called the L-norm off.
orE. The integral in 04) resembles the sum =  xy which defines the dot
product of two vectors x = (x,..., x) and y : (yi .... , y,). The function
valuesf(x) and g(x) in (34) play the role of the components x, and y, and integra-
tion takes the place of summation. The LZ-norm of f is analogous to the length of
a vector.
The first theorem gives a sufficient condition for a function in/_I) to have an
L2-norm.
Tlorem 10.52. [f f  L(1) and if f is bounded almost everywhere on  then
Proof. Sincef L(I),fis measurable and hencef  is measurable on I and satisfies
the inequality If(x)[  _< MIf(x)[ almost everywhere on 1, where M is an upper
bound for [f]. By Theorem 10.35, f  L(1).

10.22 THE SET £(/) OF SQUARFINTEGRABLE FI.CTIONS
Definition I0. We de, re by L(I) the se of all real-lu mrable fctions
f on l ch tt f   . The ncti in LZ(l) e aid o  sqintegrle.
N ¢ t L( is nther  tn nor sm than L(. For emple,
the function v by
f(x) = x - for 0 < x  1, f(0) = 0,
is in [0, 1]) but not in La([0, 1]). Similly, the funcon (x) = I/x for x  1
is in L2([I, + )) but not in [1, + )).
m 104. If f e L(I)  g  Lz(, th f " g e 1) d ( + )  LZ(1)
for ry al a  b.
Pof Bothfd g a mumble f.g  M(1). Sin
2
com 10.35 shows thatf.g  . , ( + ) $ M( d
( + ) = af + f.g +
us, the inner pu  g) is n for ev r of funs f d g in
Le(I).   p of inner p d nos   in e xt
m.
TKemm 10.5.
a) (];, g) = (g,f)
b) if+ g,h) = h) +
c) (cf, ,) = ,)
d) flcfll ---- IcJ II/11
e) I(f,g)l < IIfll
0 Ill + ll < IIf +
lf f, 9", and h are in L(I) and f c is real we have:
(commutativity).
(linearity).
(assoe/aa/ty).
(homogeneity).
( Cauchy-Schwarz /nequa/ity).
( triamjle inequality).
Proof Parts (a) through (d) are immediate consequences of the definition. Part (e)
follows at once from the inequality
ftzJf(x)g(Y)-g(x)f(Y)'Zdyldx>O-
To prove (f) we use (e) along with the relation
Ill÷ gll = -- (f+g,f+g) = (f,f) + 2(f,g) + (g,g) = I[frj = ÷ r]ar]= + 2(f,g).
Tit. 105 Serlet af lhateflem ta L=(/) Z95
NOT. The notion of inner product can be extended to complex-valued functions
fsuch that I/I  L2(/). In this case, (f, g) is defined by the equation
(f, 0) = f, f(x)--(3 dx,
where the bar denotes the complex conjugate. The conjugate is introduced so that
the inner product of f with itself will be a nonnegative quantity, namely,
(f,f) = jr Ill 2. The LZ.norm off is, as before, Ill41 = Lf, f) lj=.
Theorem 10.55 is also valid for complex functions, except hat prt (a) must be
modified by writing
(f, ) = (,f). 35) _
This implies the following companion result to part (b:
In parts (c) and (d) the constant e can be complex. From (c) and (35) we obtain
The Cauchy-Schwarz inequality and the triangle inequality are also valid for
complex functions.
10.23 THE SET L±(/) AS A SE,MIMETRIC SPACE
We recall (Definition 3.32) that a metric space is a set Ttogether with a non negative
function don T x Tsatisfying the following properties for ail points x, y, z in T:
!. d(x,x) = O. 2. d(x, y) > 0 ifx #y.
3. d(x, y) = d(y, x). 4. a(x, y)  d(x, z) + d(z, y).
We try to convert L2(I) into a metric space by defining the distance d(f, 9) between
any two complex-valued functions in L(1) by the equation
d(f, O) = Ill- 011 = If - l/I 2 •
This function satisfies properties l, 3, and 4, but not 2. Iffand 9" are functions in
LZ(/) which differ on a nonempty set of measure zero, then f # g butf - 9' = 0
almost everywhere on/, so d(f, q) = 0.
A function d which satisfies !, 3, and 4, but not 2, is called a semimetric. The
set L2(I), together with the semimetric d, is called a semimetric spate.
10.24 A CONVERGENCE THEOREM  SERI] OF FUNCTIONS I L±(/)
The following convergence theorem is analogous to the Levi theorem for series
(Theorem I0.26).

10.56. Let {g,) be a sequence of functions in L(1) such that the series
converges. Then the eries of functions ,1 g converges almost everywhere on I
to a function g in LZ(l), and we hae
/'roof. Let M -- y_,,l II0'.1t. The triangle inequality, extended to finite sums,
gives us
! •
< Irg, II < M.
=1
This implies
:If x 6/ let
Ig(x dx = Ig, < M .
(37)
The sequ©nce {f,} is increasing, each f, e Lff)" (since each gi  L2(/)), and (37)
shows that ,f  M . Therefore the sequence {,f,} converges. By the Levi
theorem for sequences (Theorem 10.24), there is a function f in L(1) such that
f,  f almost everywhere on I, and
f = lira ff. < M 
Therefor the series , 9i(x) converges absolutely almost everywhere on L Let
g(x) = lira  g(x)
at th ints where the timit eisU, and let
Then each G  L(I) and G.(x)  tg(x)l z almost everywhere on L Also,
G.(x) <_ f(x) _<_ f(x) a.e. on I.
Therefore, by the Lebesgue domina'ted convergence theorem, Igl  ¢ L(I) and
Since g is measurable, this shows that y  L(1).
so (38) implies
ilg[I : -- ]im g,,r
and this, in turn, implies (36).
<M 
10.25 THE RIESZ-FISCHER THEOREM
The convergence theorem which we have just proved can be used to prove that
every Cauchy sequence in the semimetric space L(I) converges to a function in
L(1). In other words, the semimetfic space LZ(1) is complete. This result, called
the Riesz-Fischer theorem, plays an important role in the theory of Fourier series.
T&eorem I0.57. Let {f} be a Cauchy sequence of complex-valued function in
L(I). That is, assume that for every e > 0 there is an integer N such that
IlL, -f.ll < . whenever m >_ n >_ N. (39)
Then there exists a function f in L2(/) such thai
lim IIf. - fit = 0. (40)
Proof. By applying (39) repeatedly we can find an increasing sequence of integers
n(l) < n(2) <... such that
1
whenever m > n(k).
Let gl = fn(ll, and letg, = f¢ - f.¢,_,) for k > 2. Then the series
converges, since it is dominated by
Each g, is in L(1). Hence, by Theorem 10.56, the series _..%, g. converges almost
everywhere on/to a function fin LZ(l). To complete the proof we will show that
Itf. - fl! "-' 0 as m - co.
For this purpose we use the triangle inequality to write
Ill,,- fll-< [If,- fml[ + [[f,t,)- fll. (41)
Ifm > n(k), the first term on the right is < ]/2 *. To estimate the second term wc
note that

298
Since nk) --r  as k - oo, this shows that llf, -- fll - 0 as m  co.
NOTE. In the course of the proof we have shown that every Cauchy sequence of
functions in Lx(/) has a subsequence which converges pointwis¢ almost everywhere
on I to a limit function fin L2(/). However, it does not follow that the sequence
{f} itself converges pointwise almost everywhere tofon L (A counterexampIe is
described in Section 9.13.) Although {f,} converges to fin the semimetri space
L2(1), this convergence is not the same as pointwise convergence.
EXERCISES
UPlIer functions
10,1 Prove that max (f, g) + rain (f, g) = f + g, and that
max(f+ h, + h) = max (f,) + /,, min (f+ t,f + /t) = rain 0',) +
10.2 Let {f,} and {g,}  ing u  functions   int £ t u, =
 (, d . =
a) o at (u.) d (v,)  inng on I.
b) ffAflf a.e. onIdff, g a.e. on I,  that u  g) and
I  hat  u { .} di.
10.4 i i   p of an up funmion f on the ial I = [0, I ]
 t -f U(1). t (r, r,...  d the t of tional num in [0, I ] and
@  A(x) = I if   I., f) = 0 if   I., and let s =   ..... . $h
t {.}   ing u of st functions which  f Ts
o tt f ¢ U(1).
b)  that f f 
c)    fan.ion  li ()  -() on A show lht s(x)
d) A thal -/¢ (I)  u (b) d (c) to oblain a
'l
299
ocr, In the following excrciscs, the intcgrand is to he assigned the value 0 at points
wbre it is undefined.
10.6 Justify the following equations:
a) log I x = n . n
h) f x-i-I - x log () dx = [i 0- (n +1 p)2 (p > 0).
10.7 Prove Tarmta-y's cnvegnc theorem for RJcmatm integrals: Given a sequence of
functions {f} and an increasin sequence {p.} of real numbers such that p. --> +
n , o. Assume that
a) f - f uniformly on Is, b ] for eeery b >_ a.
b) f/* Riermmn-interable on Is, b ] for every b > a.
e) IA(x)l <- a(x) almost eerywhere on Is, + ¢o), where  is nonneaetie and im-
proper Rz'emonnqntetlrable on Is, + o ).
Then both f and lfl are improper l'emonn-intearabfe on Is, + o), the sequence {
colltteres, and
f(x) dx = lira /(x) dx,
d) Use Tarmery's theorem to prove that
f:
lira | - x  dx = e-x  dr, if g > - 1.
10.8 Prove Fatou's ltma: Given a sequence {f,} af nonnegalie flciona
tt (a) {A) nrs st ryw  I to a fit fctien   )
A > Odalln  1. nthelimitfctlonfe l)tf  A.
. It is not  tt {tA) r. (m with T
Hint. tax)  i {fx), + l(X) .... }.  a=  f a.e.  land
A  ]im. t g. ¢xhts d is aA. Now apply  10..
ImllrOce¢ Rknlmm Integrals
10,9 a) lfp > 1, prove that the integrM la +® x -' sin x dr exists both as an improper
Riemann intral and as a I_¢besgue integral Hint. Integration by parts,

b) If 0 < p -< l, fn'ove that the integral in (a) exists as an improper Riemann
integraI but not as a Lebmgue integral. Ht. Le
10,10 a) Ue the o ifity   = 2 sin x  x, alg wi the fou
 sin x/x   n to ow tt
f sinxx  
x 4
b) U iation   in (a) to   f
f  x  
 2
c) U  itity sin 2 x + 2 X = l, o with (b), to o
sin*x dx -- -.
x  4
d) Use the result of (c) to obtain
sn 4 x
10.11 ff ,, > 1, prove that the integral .[¢+ r v (log xYCdx exists, both as an improper
Rienmn integralandasaLcbesipaeintegralforal|qifp < - I.orforq < - 1 ifp --
10.12 Prove that each of the following integrals exists, both as an impropca" Riemann
integral and as a Lebesgu integral.
10.13 Determine whether or not each of the following integrals exists, ©ithcr as an
improper Ricmann integrdl or as a Lebesgue integral.
a) f:e-*2+-dt, b) f: c°sx
c) x(x _ 1) z dx, d) e - sin xl
¢) f: t°g x si° l- dx'x f) f: e- l°g (c°s: x) dx"
301
10.14 Determine those values ofp and q for which the foIIowing Lebesgue integrals exist.
10.15 Prove that the fo|lowing imppex Riemann integrals have the values indicated
(m and n denote positive integers).
f:sin 2"+' t(2n)!
a) x X dx  2z,+t(n!)z, b) - dx -- n -2,
J0
(m + n)!
10.16 Given that [ is Riea-nann-integrabIe on [0, 1 ], thai f is periodic with period 1, and
that I ffx} d.r = 0. Prove that the improper Riernasm integral ff  x-" f(x) dx ¢sts
ifs > 0. B/nt. Let O(x) = ff £(t)  and write ff x-" £(x) ax = ff x-"
10.17 Assmne that f R )n [a, b] fc¢ every b > o > 0, Define  by the equation
xg(x) = ff f(t) dt if x > 0, assume that the limit lim_,, ® g(x) exists, and denote this
limit by B. If a and b are fixed positive numbers, proe that
f
j x
b) im = Bog b.
c) f(ax) I(bx) #x = s og  + J. T "
d) Assum that th limit lirno+ x ff(t)t - dt exists, demote this limit by
and lOve that
©) Combine (c) and (d) to deduce
f(ax) - f(bx) dx -_ (B -- A) log 
x
and use this result to evaluate the following integrals:

I0,I$ Pro',-* that ach of th, followin exists as a ll.sg inteal.
10.19 Assume that f is continuous on [0, !],/(0) = O, f*(O) exists. Prove that the
Ltg,a integral ; f(,-)-  ¢im.
t0.20 Prov that the integrals in (a) atd (¢) exist as Lebesgue integrals lmat that those in
(b) and (d) do not.
c)  d)
I ÷ xinZ x '
.ge -ru=:t dx,
I + x 2 sin 2 x"
Hint. Obtain uppm" and lower bounds for the integrals over suitably chosen
hoods of the points lt (a = 1, 2, 3,... ).
10.21 Determine tim set S of thos¢ real values of y for which each of tim following
integnds exists as a Lesgu¢ integral.
a) f/ cosxydr,! +x z b) foo(x2+Y) -:dx,
c) f sin2 xy dx,xz d) ff e- cos 2xy dx.
10 -- Let F(y) = [ e - cos 2xy dx if y 6 R. ow that
ti {y)+ 2y F)= 0 d  at F(y)= -,
 e -  = ,  in  7,19.)
t0 t F(y) =   xy/x 2 + 1)  ify > 0.  tt
to   follog fis,  for y > 0 d a > 0:
  )  =  0 - e-%
x =
x2 + a  = - e -°- you may  ---
2 ' x
(x + yp" (x
10.25 Show that the order of integration cannot beinterchangcd in the following integrals:
lfl.2fi Letf(x,y) -- dt/[(I + xzt2)(1 + yzt2)] if(x.y) # (0, 0), Show (by methods
of elcnmntary calculus) that f(x, y) - ½n(x + y)-'. Evakmte the iterated integral
1o [ f(x, y) dx] dy to derive the formula:
o a (arctan x)2 dr = g log 2.
10.2"/ Let fO') = 1 ° sin x cos xy/x dr if y > 0. Show fl¢ methods of elementary
calculus) thatf(y) = r/2if0 _< y < ! and thatf(y) = 0ify > I, Evaluate the integral
.f f(y) d to derive the formula
10.2 a) If s > 0 and a > 0, show that the se.ri
lff  m2  x _n
con,rges and prove that
 I f" sin 2mrx
tim ,. - j, dx -- O.
b) Letf(x) -- _,,% sin (2nx)/n. Show that
ff f(x) dx fo
xS = (2a)-(2 - ) t  d:, if 0 < s < 1,
where  denotes the R.iemann zeta function.
10.29 a) Derive the following formula for the nth deriva/ve of the Gamma function:
r°)(x) = f eP - (log t)" dt (x > 0).
b) When x = I, show that this can be written as follows:
I"°°(t) = f2 P + (-l)'e'-tt')e': -z (log t)" dr.
c) Use (b) to mow tim r*o(1) has the same sign as
In Exercises 10.30 and 10.31, I" denotes the Gamnm function,
103e ose the result j' e-" dr  ½",/ to prove that r'(½) = V-. Prove that r(, + 1)=
n! and that Y'(n + ½) = (2n)! x[/4"n! ifn = 0 1, 2,...

1031 a) Show that for x > 0 wv have the series re--ration
r(x) = (- j)" 1 +
where c = (l/n!)j t-te-t(Iogt)dt. Hint: Write J' = j + j allduse
an appropriate pvw series expansion in vach integral.
b) Show that the power ¢rie _,o e- zt converes for every complex z and that
the series ffo i(-lY'/n[]l(n + z) converges for every complex z 7 0, -1,
10.32 Assume that f is of bounded varhttion on [0, b] for every b > 0, and that
lim_+ f(x)e_Asts. Denote this limit byf() and prove that
v--O 4-
Hint. Use integration by paris.
10.33 Assume that fig of bounded variation on [0, l]. Prove that
lira y x'-tf(x) dx =/(0+).
10.34 |ffis Lebesgue-integrable onan open interval I and if f'(x) exists almost every-
where on L prove that f" is measth-ebl¢ on L
10.35 a) Let {st} be a sequ¢oc¢ of step run,ions such that ,  f everywhere on
Prove that, for eveq, real a.
f-((, +))-- /-  +-, +oo ,
b) If f is measurable on R, prove that for every open subset A of R the set f- (A)
is measurable.
10.36 This exercise describes an example of a nonmeasurable set in R. If x and y are real
numbers in the interval [0, 1 ], we say that x and y are equivalent, wriJten x ~ y, whenever
x - y is rational. The relation  is an equivakmc relation, and the interval [0, 1 ] can
be expressed as a disjoint union of subsets (called equivalence classes) in each of which
no two distinct points are equivalent. Choose a point from each equivalence class and
let E be the set of points so chosen. We assume that E is mea;urable and obtain a contra-
diction. Let A = {r, rz .... } denote the set of rational numbers in [- 1, 1 ] and -let
E = ro + x:xE}.
a) Prove that each F_ is meurable and that/z(E) = #(E).
b) Prove that {E, E2,... } is a disjoint collection of sets whose union contains
[0, ] l and is cotine in [- 1, 2]. _
¢) Use parts (a) and (b) along with the countable addit[vity of Lebesgue measure
to obtain a contradiction.
10.37 Refer to Exercise t0.36 and prove that the characteristic function Z is not measur-
able. Let f -- :r - t-z where I = [0, 1 ]. Prove that [fl  L(/) but that f
(Compare with Corollary ! of Thea3rem 10.35.)
In Exercise 10.38 through 10.42 aH fcfions a aum o  in L(I). e L-no
[f is  by lhe fula, Ilffl = ( Ill2) .
10 If li.= f. = f = 0 and if lim.=A(x) = g(x)almost eyeshot on L prove
10. f.  fifoly on a mp t L d ifh is cominuous on L prove
I1 If lim. II --fll = 0. prove th [i fi f. : fi . or   in
(I).
if-.
SUGGESTED CES FOR FURTHER STUDY
10.1
10.2
10.3
10.4
10.5
10.6
10.7
10.8
10.9
IO.lO
10.11
I0.12
10.13
10.14
Asplund, E., and Btmgart, L., A First Course in Integration. Holt, Rinehar and
Winston, Nc,v York, 1966.
Baxtle, R., The Elements of Integration. Wiley, New York, 1966.
Burkill, J. C., The Lebesgue lntegraL Cambridge University Press, 1951.
Halmos, P., Measure Theory. Van Nostrand, New York, 1950.
Hawkins, T., Lebesgue*s Theory of Integration: Its Origin and Development.
University of Wisconsin Press, Madison, 1970.
Hfldebrandt, T, H,, Imroduetion to the Theory of Integration. Academic Press,
New York, 1963.
Kestelman, H,, Modern Theories of Integration. Oxford Univea'sity Press, 1937.
Korevaar, J., Matmatieal Methods, Vol. 1. Academic Press, New York, 1968.
Munroe, M, E., Measure andlntegration, 2nd ed÷ Addison-Wesley, Reading, 1971.
Riesz, F., and Sz.-Nagy, B.. Functional Analys. L. Boron, translator. Ungar,
New York, 1955.
Rudin, W., Principles of Mathematical Aalyis, 2nd exi. McGraw-Hill, New York,
1964.
Shilov, G. E., and Gutovich, B, L., Integral, Measure and Derivative: A Unified
Approach. Prentice-Hall, Eng]ewood Clffs, 1966.
Taylor, A. E., General Theory ofFunction and lntegratian. Blaisdell, New York,
1965.
Zaanen, A. C., Integration, North-Holland, Amsterdam, 1967.

CHAPTER II
FOURIER SERIES
AND FOURIER INTEGRALS
11.1 INTRODUCTION
In 1807. Fourier astounded some of his contemporaries by asserting that an
"'arbitrary" function could be expressed as a linear combination of sines and co-
sines. These linear combinations, now called Fourier series, have become an
indispensable tool in the ana]ysis of certain periodic phenomena (such as vibra-
tions, and planetary and wave motion) which are studied in physics and engineering.
Many important mathematical questions have also arisen in the study of Fourier
series, and it is a remarkable historical fact that much of the development of
modern mathematical analysis has been profoundly influenced by the search for
answers to these questions. For a brief but excellent account of the history of this
subject and its impact on the development of mathematics see Reference I I.i.
11.20RTHOGONAL SYSTEMS OF FUNCTIONS
The basic problems in the theory of Fourier series are best described in the setting
of a more general discipline known as the theory of orthogonal functions. There-
fore we begin by introducing some terminology concerning orthogonal functions.
or_. As in the previous chapter, we shall consider functions defined on a general
subinterval I of It. The interval may be boundeM, unbounded, open, closed, or
half-open. We denote by L2(/) the set of all complex-valued functionsfwhich are
measurable on land are such that If[ 2 6 L(1). The inner product (f g) of two such
functions, defined by
(f, = f, f0,)g(x) dx,
always exists. The nonnegative number Jill] = (f, f)l/2 is the LX-norm off.
Definition lid. Let S -- {¢Po, ¢P, q2, . . . } be a collection of functions in L2(I). If
(a,, q,) ffi 0 whenever ra q: n,
the collection S is said to be an orthogonat system on L If, it* addition .each q has
norm I, then S is said to be orthonormai on I.
UOT. Every orthogonal system for which each Ilq,ll # 0 can be converted into
an orthonormal system by dividing each p, by its norm.
We shall be particularly intmxsted in the special trigonometric system
S = {0, Pt, P2 .... }, where
I cos nx sin nx
for n : !, 2 .... It is a simple matter to verify that S is orthonormat on any
interval of length 2. (See Exercise 11.1.) The system in (1)consists of real-valued
functions. An orthonormal system of complex-valued functions on every interval
of length 2 is given by
coS lX + i sin llX
, 0, 1,2,...
11.3 THE THEOREM ON BEST APPROXIMATION
One of the basic problems in the theory of orthogonal functions is to approximate
a given function fin L2(/) as closely as possible by linear combinations of elements
of an orthonormal system. More precisely, let S = {0o, 91, 92, -.. } be ortho-
normal on I and let
where b0, b,..., b, are arbitrary complex numbers. We use the norm IIf - t, ll
as a measure of the error made in approximatingfby t.. The first task is to choose
the constants b0, ..., b. so that this error will be as small as possible. The next
theorem shows that there is a unique choice of the constants that minimizes this
error.
To motivate the results in the theorem we consider the most favorable case.
fffis already a linear combination of qo, ql, ---, ¢P,, say
then the choice t, = fwill make [If  t,[I = O. We can determine the constants
co, .... c, as follows. Form the inner product (f, =), where 0 < m <2 n. Using
the properties of inner products we have
since (, ¢p) = 0 if k :# m and (¢p=, ¢) = 1. In other words, in the most
favorable case we have c, :- (f, ¢p=) for m -- 0, 1, ..., n, The next theorem shows
that this choice of constants is best f6r all functions in L2(1).

11.2
Tlorem 11.2. Let {oo, qh, Po---} be orthonormal on I, and ame that
f ¢ Lz(I). Define two sequences of functions {s,} and {t} on I a follow$:
where
k =0
and ha, bt, b2, ... , are arbitrary complex numbers.
Then for each n we kave
(2)
Ill- s,,ll < [If- t, ll.
Moreover, equality holds in (3) if, and only if, b = c for k = 0, 1,..., n,
Proof We shall deduce (3) from the equation
ItS- ,AI ' = liSP -  I<.1  + . 1. - .1 , (4)
I--0
It is clear at (4) implies (3)  the right memr of (4) h i smMlt value
wh b = c for ch k. To prove (4), write
[If- t.ll x : (f - t.,f - t = (f) - (L t.) - (G, + (G, t.
Using the proies of inner pructs we find
k:O m=O
and
Also, (t.,f) = (/, ,) : y_;o b?,, and hence
tl/-t.II 2= tlfll 2- : c  : bt +  Ibd 
k=O
11,4 TIlE FOURIER SF..RIF_ OF A FUNCYEION RELATIVE TO AN
OIITHONORMAL SYSTEM
 lid. Let S = {qo, IPl, P .... } be orthonormaf on l and assume that
f  L(I). The notation
f(x) ...  cjp.(x)
will mean that the numbers Co, cl, c2 .... are given by lhe formulas."
(,, : o, I, 2,...). (6)
The serier in (5) is called the Fourier series off relative to S, and the numbers
co, c, c2, ... are called the Fourier coecients of f relative to S.
or When I = [0, 2x] and S is the system of trigonometric functions described .
in (1), the series is called simply the Fourier series generated byf. We thett write (5)
in the form
f(x) .-. a_o + y. (a. cos nx + b. sin nx),
the coefficients being given by the following formulas:
a. = - f(t) cos nt dr, b . : -- f ( t ) sin ,t tit.
In this case the integrals for a. and b. exist fife L([O, 2n]).
(7)
11.5 PROPERTIES OF THE FOURIFA! COEFFICIENTS
Theocem 11.4. Let {Po, i, 2 .... } be orthonormal on I, assume that f E LZ(l).
and suppose that
Then
a) The series  leJ  coneerges and satisfies the inequality
 lc.I 2 g Ilfll  (Bessel's inequality). (8)
b) The equation
(Parseval"s formula)

11.4
holds if, and onty if, we al hcee
lira
where {s.} is the sequence of lraa!
sums defined by
Proof. We take b : e in (4) and observe that the left member is nonnegative.
Therefore
kd 2 -< llffl =
This establishes (a). To prove Co), we again put bE = cs in (4) to obtain
Ill- s, lf e = llff[ z - Ic  l 2.
Part (b) follows at once from this equation.
As a further consequence of part (a) of Theorem 11.4 we observe that the
Fourier coefficients c, tend to 0 as n --, o (since Y Ic, l z converges). In particular,
when q(x) = e/x/n and I = [0, 2rt] we find
lira f(x)e -" dx = O,
formulas
f(x) sin nx dx = O.
(9)
from which we obtain the important
lira f(x) cos nx dx
These formulas are also special cases of the Riemann-Lebesgue lemma (Theorem
I1.6).
lqo'r.. The Parse.val formula
is analogous to the formula
2
tix]l  = x + x2 2 +-.-+ x,
for the lenglh of a vector x = (x .... , x,) in RL Each of these can be regarded
as a generalization of Ihe Pythagorean theorem for right triangles.
'1 11.5
311
11.6 THE  THEOREbt
The converse to part (a) of Theorem 11.4 is called the Riesz-Fischer theorem.
llteorem 113. Assume {¢0, ¢1, "'"  S orthonormal on L Let {c,} be any sequence
of complex numbers such that . Icd z converges. Then there is a function fin L(I)
such that
a) (f, ¢) = c for each k >_ O,
b) [tfl[ z =  led 2.
Proof. Let
We will prove that them is a function fin Lx(1) such that (f, 0r) : c and such that
lim Ils.  f = O.
Part (b) of Theon i ! .4 then implies  (b) of Theorem 11.5.
First we note that Is,} is a Cauchy sequence in the semimetric space L(1)
because, if m > n we have
Ils. - .11 =
t--'n+ 1
and the last sum can be made less than e if m and n are sufficiently large. By
Theorem 10.57 there is a functionf in Lx(I) such that
tim IIs, - fll -- 0.
To show that (f, #) = c we note that (s., ¢) = c ifn _> k, and use the Cauchy-
Schwarz inequality to obtain
Since IIs  f[] "-" 0 as n --, oo this proves (a).
Norris. The proof of this theorem depends on the fact that the semimetric space
L2(/) is complete. There is no corresponding theorem for functions whose Sclaares
are Riemann-integrable.

312
Foar Series an Foariet Intrah
11.7 THE CONVERGENCE AND REPRESENTATION PROBLEMS
TRIGONOMETRIC SERIF.
Consider the trigonometric Fourier series generated by a function f which is
Lebesgue-integrable on the interval I =- [0, 2r], say
.f(x)  a__o + _, (a, cos nx + b sin nx).
2 n=l
TWO questions arise. Does the series converge at some point x in 17 If it does
converge at x, is its sumf(x)? The firgt question is called the convergence problem;
the second, the representation problem. In genera], the answer to both questions
is "No." In fact, there exist Lebesgue-integrabte functions whose Fourier series
diverge everywhere, and there exist continuous functions whose Fourier series
diverge on an uncountable set.
Ever since Fourier's time, an enormous fiterature has been published on these
problems. The object of much of the research has been to find sufficient conditions
to be satisfied by fin order that its Fourier series may converge, either throughout
the interval or at particular points. We shall prove later that the convergence or
divergence of the series at a particular point depends only on the behavior of the
function in arbitrarily small neighborhoods of the point. (See Theorem 11.11,
Riemann's localization theorem.)
The efforts of Fourier and Dirichlet in the early nineteenth century, followed
by the contributions of Riemann, Lipschitz, Heine, Cantor, Du Bois-Reyrnond,
Dial, Jordan, and de la Vallde-Poussin in the latter part of the certury, led to the
discovery-of sufficient conditions of a wide scope for establishing convergence of
the series, either at particular points, or generally, throughout the interval.
After the discovery by Lebesgue, in 1902, of his general theory of measure and
integration, the field of investigation wag considerably widened and the names
chiefly associated with the subject since then are those of Fejr, Hobson, W. H.
Young, Hardy, and Littlewood. Fejr showed, in 1903, that divergent Fourier
series may be utilized by consid.ering, instead of the sequence of partial sums
the sequence of arithmetic means {r,}, where
so(x) + s,O,) +
n
He established the remarkable theorem that
"'" + s,_,(x)
the sequence {o(x)} is convergent
and its limit is ½If(x+) + f(x-)] at every point in [0, 2az] where f(x+) and
f(x-) exist, the only restriction on f being that it be Lebesgue-integrable on
[0, 2n] (Theorem 11.15.). Fejr also proved tat every Fourier miles, whether it
converges or not, can be integrated term-by-term (Theorem 1 t.16.) The most
striking result on Fourier series proved in recent times is that of Lennart Carleson,
a Swedish mathematician, who proved that the Fourier series of a function in
L(I) converges almost everywhere on L (Acta Mathematica, 116 (1966), pp.
135-157.)
1X I1.
313
In this chapter we shall dduce some of the suffmient conditions for convergence
of a Fourier series at a particular point. Then we shall prove Fejr's theorems.
The discussion rests on two fundamental fimit formulas which will be discussed
first. These limit formulas, which are also used in the theory of Fourier integrals,
deal with integrals depending on a real parameter =, and we are interested in the
behavior of them integrals as  -, + m. The first of these is a generalization of (9)
and is known as the Riemann-Lebesgue lemma.
11 THE PdF_,MANN-LIEGUE LEMMA
Tkeoeem ll.fi. Assume tha f  L(I). Then, for each real [J, we have
lim ff(t) sin (¢¢t + if) dt =- O. (10)
Peoof. Iff is the characteristic function of a compact interval In, b] the result is
obvious since we have
The result N holds iff is constxt on the open iterval (e, b) and ro outside
[a, hi, regardless of how e define f(,) andf(b). Therefore (10) i vNid iffis a
step function. But now it is easy to prove (10) for every l.,besgue-integrable
functionf
If, > 0 is given, there exists a step function s such that J' If -- sl < /2 (by
Theorem 10.19(b)). Since (10) holds for step functions, there is a positive M such
that
If, s(t)sin(at+a)dt<-if>-M2
Therefore, if  >- M we have
If, f(t)sin( + ,)at I < lf, (f(t)- s(O)sin(  t +
_ < If(t) - s(t)l at +
This completes the proof of the RiemannLebesgue temma.
F_amni. Taking # = 0 and B = nl2, we lind, if re/..i),
lira ff(t)singt dt = lira ftf(r)cosgt dt = O.

314
11.7
As an application of the Riemann-Lebesgue lemma we derive a result that will
be needed in our discussion of Fourier integrals.
Tieorem 11,7. lf f  L( - o, + oo), tee have
• --' + t
wleever the Lebesgue integral on the right exists.
Proof For each fixed u, the integral on the left of (l l) exists as a Lebesgue
integral since the quotient (I- cos rt)/t is continuou and bounded on
(-oo, + o). (At t  0 the quotient is to be replaced by 0, its limit as t  0.)
Hence we can write
f-f(t) l-coStdt:f:f(t)l-c°s'dt+f:t t  f(ol-coSUtdt - t -
= f® If(t) f(-0] - cos
dt
Jo
t
When z. + o, the last integral tends to O. by the Rimann-Lebesgue lemma.
11.9 THE DIRICHLET INTEGRALS
Integrals of the form ' (tXsin t)/t df (called Dirichlet integrals) play an im-
portant role in the theory of Fourier series and also in the theory of Fourier
integrals. The function g in the integrand is assumed to have a finite right-hand
limit 9(0+) = limv.0+ g(t) and we are interested in formulating further con-
ditions on 7 which will guarantee the validity of the following equation:
tim g(t)--- dt = g(0+). (12)
To get an idea why we might expect a formula like (12) to hold, let us first consider
the case when g is constant (g(t) -- g(0+)) on [0, 6]. Then (12) is a trivial con-
sequence of the equation  (sin t)/t dt = x/2 (see Example 3, Section 10.16),
since
_sin .__  t
More generally, if   L([0, 3]), and if 0 < t < 6, we have
lira - 2
0,
Th. llJ "Fee lkkhlet  315
by the Riemanrt-Lebesgue lemma. Hence the validity of (12) is governed entirely
by the iocalbehavlor ofg near 0. Since 9(0 is nearly g(0+) when t is near 0, there
is some hope of proving (12) without placing too many additional restrictions on g.
It would seem that continuity of g at 0 should certainly be enough to insure the
existence of the fimit in (12). Dirichlet showed that continuity of g on [0, 6] is
sufficient to prove (I2), if, in addition, g has only a finite number of maxima or
minima on [0, 6]. Jordan later proved (12) under the less restrictive condition
thatg be of botmded variation on [0, b. However, all attempts to prove 02) under
the sole hypothesis that g is continuous on [-0, 6] have resulted in failure. ]n fact,
Du Bois-Reymond discovered an example of a continuous function  for which the
limit in 02) fails to exist. Jordan's result, and a related theorem due to Dini, will
be discussed here.
Tlteerem 11.8 (rordan), lf g is of bounded variation on [0, 6], then
lim ...... 2O(t)sintdt -- g(0+). (13)
Proof. It suffices to consider the case in which g is increasing on [0, . If a > 0
and if0 < h < 6, wehave
(14)
let us say. We can apply the Riemann-Lebesgue Iemma to I(, h) (since the
integral  O(t)/t dt exists) and we find l(t, h) --, 0 as  - + co. Also,
-- 0(0+)  in2 de --, - (0+)
 2
aSet --, +C.
Next, choose M > 0 so that Ij (sin t)/t dtl < M for every b _> a _> 0. It follows
that I(sin=t)/tdr[ < M for every b > a  0 if = > 0. Now let e > ¢) be
given and choose h in (0, 6) so that Ig(h) - g(0+)l < /(3M). Since
90) -- .q(O+ ) >_ 0 ifO <: t < h,
we can apply Bonnet's theorem (Theorem 7.37) in 1(,, h) to write
m dr,

316 Fomr Series sd Fourier hsfer TL 11,9 Th. ll,10  Seals of Fmer Series
where c • [0, h I. The definition of h gives us
Ill(a, h), : tg(h)- g(O+)l lf sint dt] < t 3M
For the same h we can choose .4 so that g > A impfies
Ila(, h)l <  and II,(, h) -  I j 
Then, for • > .4, we can combine 04), (15), and (16) to get
[f a(t) sin "- dt - - °(°+) < *" t 2
-Ibis proves (l 3).
3
< -. (16)
3
A different kind of condition for the validity of (13) was found by Dini and
can be described as follows:
Theorem ll.9(Diai). Assume that g(0+) exists and suppose thatjbr some 5 > 0
the Lebesgue integral
f a(t) - e(0+) dt
exists. Then we have
lira -2 f g(t) __sin at dt = g(0+).
Proof Write
¢n   + , the fit term on the fight tends  0 (by the Riemann-sgue
lemma) and the ond term tends to
N. If e  Ma, ]) for eve sifive a < fi, it is easy to show that Dini's
condition is satisfied whenever g satisfies a "right-handed" Liphi condition at
0; that is, whenever the exist two positive constants M and p such that
[g0) - g(0+)l < Mt , for cvc t in
( Exeise 11.21.) In rticar, the Liphitz condition holds with p = 1
whenever  has a fighthand derivative at 0. It is of intrust to note that there exist
funons which fisfy Dini's condition but which do not satisfy Jordan's con-
dition. Similarly, there a functions which satisfy Jordan's condition but not
Dini's. ( Referen 11.10.)
11.10 AN INTEGRAL REPRESENTATION FOR THE PARTIAL SUMS OF A
FOURIER SE'RIF
A function fis said to be periodic with period p # 0 if f is defined on R and if
f(x + 9) = f(x) for all x. The next theorem expresses the partial sums of a
Fourier series in terms of the function
This formula was discussed in Section 8.16 in connection with the partial sums of
the geometric series. The function D, is called Dirichlet's kernel.
l'lg, aeem 11.10. As, n'e that f • L([0, 2z)) and suppose that f is periodic with
per/m/2g. Let {s} denote the sequence of paraal sums of the Fourier series generated
by.f, say
s,(x) = a__o +  (a cos kx + b sin kx), (n -- 1, 2 .... ). (18)
Then we haoe the integral representation
s.(x) = 2_f*cf(x + t) + f(x - t)O.(t)dr.
x Jo 2
(19)
Proof The Fourier coefficients of fare given by the ingrals in (7). Substituting
these integrals in (18) we find
s,(x) = -
1 f(t)
+  (oos kt cos kx + sin kt sin kx) dt
/=1
+ cosk(t- x dt =-1 t)D,(l - x) dL
Since both f and D. arc periodic with period 2n, we ean replace the interval of
integration by Ix - r, x + r] and then make a translation u = t - x to get
s.(x) l f:: f(t)D,(*
= - - x) dt
Using the equation D,(-u) = D(u), we obtain (19).

Fom Set4 ad Fom lateM
T]L ltJl
11.11 RIEMANN'S LOCAIJZATION THIOREM
Formula (I9) tells us that the Fourier series generated byfwill converge at a point
x if, and only if, the following limit exists:
lira _2 f*/(x + t) +/(x - t) sin (n + ½)t
dr,
(20)
,-. x Jo 2 2 sin ½t
in which cas¢ the value of this limit will be the sum of the series. This integral is
sentially a Dirichlet integral of the tyl discussed in the previous soction, except
that 2 sin ½t appears in the denominator rather than t. Ho,ver, the giemann-
Lcbsgu¢ Icmma allows us to replace 2 sin t by t in (20) without affecting either
the existence or the value of the limit. More proisely, tl R/cmann-Lebesgu
lcmma implies
lim  o; (
.- 2 sin ½ 2
because the function Fdefined by the equation
i ! if 0 < t <
F0) = sin ½t
ift ---- O,
is continuous on [0, hi. Therefore the convergence problem for Fourier series
amouats to finding conditions on f which w/l[ guarantee the existence of the
following limit:
lira _2 f*f(x + t) + f(x -- t) sn (n + )t dr.
Using the Ricmann-Lebesgue lemma once more, we need only consider the limit
in (21) when the integral ' is replaced by j'g, where , is any positive number <
because the integral jl tends to 0 as n -* oo. Therefore we can sum up the rsults
of the previous section in the following theorem:
Tiorem 11.11. Assume that f e /.([0, 2hi) and suppose f las period 2. Then
the Fourger eries eaerated by f will converge for a 9ien calue of x if, and only if,
for ome positive ,$ < n the followia 9 limit exists:
lira _2 ff(x + t) + f(x - t) sin (n + ½)t
dr,
• ;z .Jo 2 t
in which case the adu¢ of this limit is the sum of the Fourier series.
This theorem is known as Riemaans tocalization theorem. It tells us that the
convergcnc¢ or divergence of a Fourier series at a particular point is governed
entirely by the behavior off in an arbitrarily small neighborhood of the point.
This is rather surprising in view of the fact 1hat the coefficients of the Fourier
depend on the values which the function assumes throughout the entire
int,rv EO, 2  3,
11.12 SUFFICIENT CONDrrlONS FOR CONVERGENCE OF A FOURIER
 AT A PARTIC]JI,Aig POINT
Assume_ that fe L([0, 2]) and suppose that fhas period 2g. Consider a fixed x
in [0, 2a] and a positi, 6 < . Let
a0) = f(x + t) +/(x - t) if [o,
2
and let
s(x) =
whettevg't this limit exists. Note that s(x) = f(x) iffis contintmes at x.
By comb/ning "Dmorem ll.ll with Thcoct:tt ii.8 and 11.9, respectively, we
obtain the foBowing sufficient ¢amditions for convergence of a Fourier series.
Tkem, em 11d2 ( tct), lf f i of  variation oa the compact interval
Ix - b, x + ] for   < ,  the tirm'l x) exists and the Fourier series
generated by f converges to
Tkfm,em 11.13 (Dis tt)o If the linu't .x) exits and if the Lebesgue integral
- s(x) dt
ex2sts for same  < , lhen the Fourier series #eaerated by f converges o s(x).
11.13 CESRO IT OF FOURIER SERIES
Continuity of a f f is not a ver fruitful hypothesis when it comes to
studying conve of the Fo'ier series getmrated by f In 173, Du Boi
Reymond gave an example of a function, continuous throughout the interval
0, 2hi, whose Four s¢m fmls to couverge on an uncountable subset of[0,
On tl other hand, ¢ontfrmity do suffe to establish Ceso summability of the
ses. This result (due to F¢.) and some of itg consequences will be discussed
Our first tak is to obtain an fntcgral representation for the arithmetic means
of the partial sums of a Fourier so'tics,
 ILl4. A'ae tha f L([O, 2:]) and suppose that f is periodic with
period 2n. Let s denote the nth partial sum of the Fourier series 9enerated by f and
let
a  (x) = so._(_x) + s,(x) + .-. + s,_,(x) (n = t, 2,...). (23)

11214
Then we htto¢ the integral representation
¢r,(x) = I__ f'f(x + t) + f(x - t) sin  _t
tit.
nz J0 2 sin  ½t
Proof ff we use the integral representation for s,fx) given in (19) and form the
sum defining ¢r,(x), we immediately obtain the required result because of formula
(t6), Section 8.16.
;oTa. If we apply Theorem l1.14 to the constant function whose value is t at each
point we find r,(x) = s,(x) -- 1 for each n and hence
1 in2
nn,v sin 2 ½t dt = 1. (25)
Therefore, given any number $, we can combine (25) with (24) to wrte
1 f:{f(x+t)+f(x-t) }sin½nt
,(x) -- s - -- - s dr. (26)
nn 2 sin 2 ½t
ff we can choose a value of s such that the integral on the right of {26-) tends to 0
as n  m, it will foRow that a,(x) - s as n --* m. The next tho0rem shows that it
suffices to take s = If(x+) + f(x-)]/2.
Yl'htos 11.1.$ (Year). asmane that f • L([0, 2n]) and suppose that f is periodic
with period 2ft. Define a.function s by the following equation:
s(x) = lira f(x + t) + f(x - t)
,-+ 2 ' (27)
whenever the limit exists. Then, for each x for which s(x) is defined, the Fourier
series generated by f is Cesro summabte and has (C, 1) sum s(x). That is, we have
lira a,(x) = s(x),
where {o,} is the sequence of arithmetic means defined by (23). If, in addition, f is
continuous on [0, 2n], then the sequence {a.} converges uniformly to fan [0, 2a].
eroof. Let at(t) - [f(x + t) + f(x - t)]/2 - s(x), whenever s(x) is defined.
Then g(t) --* 0 as t - 0+. Therefore, given t >
• such that Ig(t)l < #2 when¢ver 0 < t < & Note that a depends on x as ,all as
on . However, iffis continuous on [0, 2], then fis uniformly continuous on
[0, 2n], and there eists a a which serves equally well for every x in [0, 2n]. Now
we use (26) and divide the interval of integration into two subintervals [0, ] and
sin - ½- - 2nn sin  ½t dt = 2 - '
Th. 11.16  of Fei Tncot'n J21
because of (25). On I/i, n]] we have
-n g(t) inZ ½t dt <<. laat)l dt < ,
. nn sin z  nz sin z ½6
where l(x) =  I9(t)l dr. Now choose N so that l(x)/(N n sin  ½) < ]2. Then
n > N implies
In other words, ¢,(x) --, s(x) as n -+ c¢.
sin  ½nt
sin: ½t dt < e.
If f is continuous on [0, 2n], then, by periodicity, f is bounded on R and there
is an M such that [g(t)l  M for all x and t, and we may replace l(x) by M in
the above argument. The resulting N is then independent of: and hence or. - s = f
u, iformly on [-0,
11.14 CONSEQUENC OF FEJIIi THEOREM
Torem 11.16. t f be continuous on [0, 2hi d periodic with period 2. t
{s} denote the sequence of liat sums of the Fourier series generated by  say
2 ant
en we ha :
a) l.i.m., s, = fan [0, 2x].
b) -1 tf(x)l  dx = __a +  (a + b) (Pareval'sformula).
 2 nl
c) The Fourier series c be inteyrated term by term. That is, for all x we have
ff(t) dt =a° + I;(asn'+ b*sinnt) dt'2
the integrated series bein# unormly conrgent on ery intenl, etn  the
Fourier series in (28) diveryes.
d) If the Fourier series in (28) converges for same x, then it converges
Proof Applying formula (3) of Theorem 1 t.2, with t.(x) = a(x) = (I {n) S
we obtain the inequality
If(x) - s,(x)l z dx  If(x) - a.fx)l z dx. (29
But, since a.  funiformly on [0, 2z], it follows that l.i.m._ ¢, = fan [0,
and (29) implies (a). Part (b) follows from (a) cau of Theatre I1A. Part

also follows from (a), by Theorem 9.18. Finally, if {s,(x)} converges for some x,
then {a,(x)} must converge to the me limit. But since a,(x) - f(x) it follows
that s,(x)  f(x), which proves (d).
11.15 THE rElllSlo$8 APPROXIMATION
Fejr's theorem n al   to pro a fo eom of Weietss wch
stat at ev ntinuom function on a compact inteal n  uniformly
approximated by a lom. Mo ply, we have'
Tm 1L17. t f be reaI.ued d etuo on a coct tei [a, hi.
Th for e  > 0 there  a iomi p (whh y nd  )  tt
If(x) - x)l < ¢ for ery x m [a, hi. ()
Proof If t  [0, g), let g(t) = f[a + t(b - a)/]; if 
f[a + (2n - tb - a)/] and define g ouide [0, 2hi so t g h ri
For the e  in the theom,  n aly Fejr's theom to find a funon
defined by an uation of e form
t) = A 0 + (Ackt + B, sin
such at I#(t) - t)l < e/2 for eve t in [0, 2hi. (No tt
dends on e.) Sin o is a fini sum of tonomee functions, it s a
w  exertion aut e o#n which ng uniformly on eve fini
MM. e M sums of is wer fi non nsfitu a uen of
lomial y {p,}, such t p,  a uormly on [0, 2]. Hen, for the
mine , the exits  m sh that
 for every t in [0, 2hi.
If.(0 - o(t)l
Therefore we have
Ip,(t) - g(t)l < , for every t in [0, 2r]. (31)
Now define the l:mlynomial p by the formula iv(x) = p=[x  a)j(b - a)]. Then
inequality (31) becomes (30) when we put t = x -- a)/(b L a).
11.16 OTIIER FORMS OF FOURIER SERIES
Using the formulas
2 cos nx = e ' + e -" and 2/sin nx = e  - e -'x,
the Fourier series generated byfcan be expressed in terms of complex exponentials
as follows:
f(x) .. a_? +  (a, cos nx + b. sin nx) = a_ +  (%e,  + [j,e_,,) '
2 ,= 2 ,=t
whete = (a - /b,)/2 and ], ffi (a. + /b,)/2. Ifweputao = a0/2and'_, = 1,,
we can write the exlnential form more briefly as follows:
f(x).  e .
The formtdas (7) for the coefficients now become
t f()e -dt (-- 0, +L +2 ,,).
if f has period 2r, the interval of integration can be replaced by any other interval
of length
Mote generally, fife/..([0, p]) and if f has period p, we write
f(x)  a° + a, cos + b. in x
to mean that the coefficients are given by the formulas
:2 .fir(0 cos 2_n.n_ dr,
an= - .- P
2 I 2nt
b, ffi  f(t) sin--p dt (n = O, L 2 .... ).
In exponential form we can write
where
f(x) ~  olne 2nfI,
" = -1) f(t)e - dr, if n = 0, + 1, ___ 2,....
All the convergence theorems for Fourier series of period 2rt can also be applied
to the case of a general period p by making a suitable change of scale.
11.17  FOURIF. IIAL THEOREM
The hypothesis of periodicity, which appears in all the convergence theorems
dealing with Fourier series, is not as serious a restriction as it may appear to be at
first sight. If a function is initially defined on a finite interval, say [a0 hi0 we can
always extend the definition off outside [a, b] by imposing some sort of periodicity
condition. For example, iff(a) = f(b), we can definefeverywhare on (-
by requiting the equation f(x + p) = f(x) to h61d for every x, where p = b  a.
(The condition f(a) --f(b) can always be brought about by changing the value
of fat one of the endpoints if necessary. This does not affect the existence or the
values of the integrals which are used to compute the Fourier coefficients off)
However, if the given function is already defined everywhere on ( - oo, + o) and

37.4 Fea'ier Sles ai Fetf Integrals T. 11.18
is not periodic, then there is no hope of obtaining a Fourier series which represents
the function everywhere on (-- o, + o). Nevertheless, in such a case the Function
can sometimes be represented by an infinite integral rather than by an infinite series.
These integrals, which are in many ways analogous to Fourier series, are known as
Fourier integrals, and the theorem which gives sufficient conditions for representing
a function by such an integral is known as the Fourier integral theorem. The basic
tools used in the theory are, as in the case of Fourier series, the Dirichlet integrals
and the Riemann-Lebesgue lemma.
Tlrem 11.18 (Fourkr intelral theorem). Assume that f  L(-o, + oo). Suppose
there is a point x in R and an interval Ix - 6, x + 6] about x such that either
a) f is of bounded ,,ariation on Ix - 6, x +
or else
b) both timits f(x + ) and f(x- ) exist and both Lebeszue integraIs
2 f(x+t)-f(x+)dtt and f2f(x-t)-f(x-)at
exist.
Then we have the formula
f(x+) + ](x-) = - fo" [L f(u)c°sv(u - x)du]d'2 n (32)
the inte.qral  being an improper Riemann integral.
Proof. The first step in the proof is to establish the following formula:
lira -1 f f(x + t) sintt dt =f(x+) +f(x-)
(33)
For this purpose we write
f(x + t) sin a___t dt = + + + .
When a - + oo, the first and fourth integrals on the right tend to 0, because of
the Riemann-Lebesgue iemma. In the third integral, we can apply either Theorem
11.8 or Theorem 1t.9 (depending on whether (a) or (b) is satisfied) to get
lim
Similarly, we have
; f(x + t) sina--t d' = f2f(x - t) sint dt -*f(X-) as"  +' tt nt 2
Thus we have established (33). If we make a translation, we get
fo f(x + t)_sin ____t dr= f f(u) sin ,t__(u - x)du,
and if we use the elementary formula
sin .x(u - x) f
= cos 0(u - x)
U -- X
the limit relation in (33) becomes
lira -tfLf(u)[fleos(u-x)avldu = f (x + ) + f (x - )
.+  _ 2
(34)
But the formula we seek to prove is (34) with only the order of integration reversed.
By Theorem 10.40 we have
f i [ f c°s au - x) du] dv = f [f f(u) c°s - x) d°] du
for every g > 0, since the cosine function is everywhere continuous and bounded.
Since the limit in (34) exists, this proves that
lira i- [[;,f(u) cos o(u - x)du] dv =f-(x+) + f(x-)
-.+  2
By Theorem 10.40, the integral  f(u) cos v(u - x) du is a continuous function
of v on [0, 0], so the integral  in (32) exists as an improper Riemann integral.
It need not exist as a Lebesgue integral.
11.18 THE EXPONENTIAL FORM OF THE FOURIEI INIEGRAL THEOREM
Theorem lldg If f satisfies the hypotheses of the Fourier integral theorem, then
we have
Proof. Let F(v)= J_f(u) cosv(u-x)du. Then F is continuous on
(-oo, +oo), F(v) -- F(-v) and hence o,, F(v) dt, = J F(-v)dv = j F(v) dv.
Therefore (32) becomes
f(x+)+f(x-)-lim2 ,,-,. ÷  n-1 fl F(v) 'av = =÷lim 2,,-tf , F(v)dv.
(36)

Now define G on (- co, + co) by the equation
= f:® f(u)sin o(u - x)a..
Then G is everywhere continuous and G(v) ffi -G(-v). Hence J', G(v) a = 0
for every t, so lim,..+® , G(v) ab = 0. Combining this with (36) we find
f(x+) +f(x-)= lira 1 f
2 ... +® 2n {F(o) + iG(v)}
This is formula (35).
II.LO INTEGRAL TRANSFORMS
Many functions in analysis can be expressed as Lebesgue integrals or improper
Riemann integrals of the form
g(y) ffi f g(x, y)f(x) dx.
(37)
A function g defined by an equation of this sort (in which y may be either real or
complex) is c.ailed an integral transform off. The function K which appears in the
integrand is referred to as the kernel of the transform.
integral transforms are employed very extensively in both pure and applied
mathematics. They are especiaIly usefuI in solving certain boundary vaIue prob-
lems and certain types of integral equations. Some of the more commonly used
transforms are listed below:
ExponentiaI Fourier transform :
Fourier cosine transform:
Fourier sine Iransform:
Laplace transform:
Mellin transform :
_ e-i'f(x) dx.
:cos xy_f(x) dx.
:sin xy f(x)
f : e- 7(x) dx.
x'- 'f(x)
Since e -:' = cos xy - i sin xy, the sine and cosine transforms are merely
special cases &the exponential Fourier transform in which the functionfvanishes
on the negative real axis. The Laplace transform is also related to the exponential
Fourier transform. If we consider a complex value of y, say y = u + iv, where
u and v are real, we can write
e-'f(x) dr = dx = e-"o¢,(x)
whe ,(x) = e-'f(x). erefo the p tnsfo n also  rerd
as a sl  of the ¢xnentl Fourier tfform.
No. An tion such as (3 ig mimes written mo briefly in the form
#   or  = : whe  deno the "'otor'" which convesfinto
Sin integration is involved in this eqtion, the omtor  is fe to as an
tegr orator. It is dear thai  is also a linear otor. That
if a and a z  constants. e otor defined by the Foufi transfo is
deno by  and at defin by the place tnsfo is denoted by .
The exponentl form of the Fouer intent theom can  expres in
terms of Fourier transforms as follow. 1 # denote the Fourier tnsform of
so that
(38)
en, at points of confin of formula (35) mes
f(x)= lira --1 f"
• + 2a d- g(u)e " du, (39)
and this is called the inrsion fmula for Fourier tnsfos. It tells us that a
continuous function f tisfyi e conditions of the Fourier integral m is
uniquely detein by its Fouler transform g.
se. If  denotes the otor defined by (38), it is customa to denote by
the orator dned by (39). uarons 08) and (39) can  expres symboally
by writing  = ffandf  f-g. e inveion foula tells us how to
the equation 9 = fforfin terms
Befo we pursue the study of Fourier tnsfos any further, we in a
new notion, the convolution of two run,ons. is n  intet as a sial
kind ofintegl transform in which the kernel K(x, ) dends only on the difference
x-- y.
ll.iO CONVOLUTIONS
Definition 11.20. Given two functions f and 9, both Lebesgue integrable on
( - oo, + co), let S denote/he set of x for which the Lebesgue inle¢ral
h(x) = f f(t)O(x - t) dt
(4O)

328 Farler ¢¢ies and Fmu Integrals Ti 111
exists. This integral defines a function h on S called the convolution off and g. We
alao write h = f , 9 to denote this function.
O. It is easy to see (by a translation) that f, g --- g-fwhenever the integral
exists.
An important special care occurs when bothfand g vanish on the negative real
axis. In this case, g(x - t) = 0 if t > x, and (40) becomes
h(x) = f f(t)g(x - t) dr. (41)
It is clear that, in this case, the convolution will be defined at each point of an
interva| [a, b] if both f and g are Riemann-integrable on [a, b]. However, this
need not be so if we assume only that fond g are Lebesgue integrable on [a, b].
For example, let
1 1
-- , if0<t<l,
f(t) = x/ and g(t) = -,/1 - t
and letf(t) = g(t) = 0 if t < 0 or if t  1. Thenfhas an infinite discontinuity at
t =-0. Nevertheless, the Lebesgue integral _f(t)dt = j t-t/2dt exists.
Similarly, the Lebesgue integral j'T, g(t) dt = 'o t (1 - t) -l/z dt exists, although
9 has an infinite discontinuity at t = 1. However, when we form. the convolution
integral in (40) corresponding to x :-- 1, we find
f(t)g(!  t) dt =  t-  dr.
Observe that the two discontinuities off and g have "coalesced" into one dis-
continuity of such nature that the convolution integral does not exist.
This example shows that there may be certain points on the real axis at which
the integral in (40) fails to exist, even though both fond 9 are Lebesgue-integrable
on (- oo, + o0). Let us refer to such points as "singularities" of h. It is easy to
show that such singularities cannot occur unless both f and g have infinite dis-
continuities. More precisely, we have the following theorem:
Theorem ll.MI Let R = (- oo, + co). Assume that f  L(R), g *: L(R), and that
either f or 9 is bounded on R. Then the convolution integral
i(x) = - t) dt
(42)
exists for every x in R, and the function h so defined is bounded on R. If, in addition,
the bounded function f or # is continuous on R, then h is also continuous on R and
Th. 11.2.3 Ctmvolufiou TItcoecm foe Fomic¢ lX, attsfoem$ 3
Proof. Sincef, g = g *f, it suffices to consider the case in which g is bounded.
Suppose ]el -< M. Then
If(t)g(x - t)l < Mlf(t)[. (43)
The reader can verify that for each x, the product f(t)g(x - t) is a measurable
function of t on R, so Theorem 10.35 shows that the integral for h(x) exists. The
inequality (43) also shows that Ih(x)l < M ]" tfl, so h is bounded on R.
Now if is also continuous on R, then Theorem 10.40 shows that h is continuous
on R. Now for every compact interval [a, b] we have
f]'h(x)ldx<--f;?olf(t)lla(x-t)ldt]dx
=  'f(t)l [f2 la(x - t)l dx] dt
= If(t)[ ta(y)ldy at
so, by Theorem 10.31, h  L(R).
Theorem 11,22. Let R = (--oo, + co). Assume tlmt f e LZ(R) and g • L(R)..
Then the convolution integral (42) exists for each x in R and the function h is bounded
on R.
Proof. For fixed x, let ,(t) = g(x - t). Then g, i measurable on R and
g e L(R), so Theorem 10.54implies that the productf,g • L(R). In other words,
the convolution integral h(x) exists. Now h(x) is an inner product, h(x)
hence the Cauchy-Schwarz inequality shows that
Ih(x)l  Ilffl IIgl[ -- Ilfll
so h is bounded on R.
11.21 THE CONVOLUTION THEOREM FOR FOURIER TRANSFORMS
The next theorem shows that the Fourier transform of a convolution f* g is the
product of the Fourier transforms off and ofy. In operator notation,
-(f* ,q) = :(y). (g).
Theorem ll.2J. Let R = (-co, + oo)÷ Assume tftat f  L(R), g  L(R), and that
at least one off or .¢ is continuous and bounded on R. Let h denote the convolution,

• - ]eter  a! Foorl  Th. 11.23 Convoietit "llte(m for Foorlet" TtnmMmms 331
h = f* g. Then foe every real u we have
f_ b(x)e-' dx = (_f(t)e-" dtl (_ 3(y)e-" dy I . (44)
e integral  the It exists th  a Lee intral d   impror
Rie intrat.
oof A¢ at 9 is nfinua and bounded on R. t {} and {b.}  two
in--sing u of posie ! n such t a=  +  and b.  + .
ne a qu of functions {} on R as follows:
Since
f.(t) = e-"" O(x - t) dx.
 le -' O(x - t)l dt < J_
for all compact intervals [a, b'[, Theorem 10.31 shows that
The translation y = x - t gives us
e= ' g(x -- t) dx = e- t, e-  g(y) dy,
and (45) shows that
lim f(t)f.(,)=f(t)e-"(oe-",(y)dy )
for all t. Nowf. is continuous on  (by Theorem 10,35), so the productf-f, is
measurable on R. Since
If(t)f.(t)l <_ If(')I f®
the product f.f is Lbcgue-integtabl¢ on R, and the Lebesgue dominated con-
vergence theorem shows that
lira ,it
But
Since the function k definod by k(x, t) = O(x - t) is continuous and bounded
on R 2 and since the integral  e -' dx exists for every compact interval In, hi,
Theorem I0.40 permits us to reverse the or6er of integration and we obtain
L ;' [L ]
f(t)f.(t) d = e -t' f(t)a(x - t) dt dx
 e-'h(x) dx.
Therefore, (46) shows that
;' (L )(L )
lim h(x)e ' dx = f(t)e - dl (y)e -° dy ,
which proves (44). The integral on the left also exists as an improper
Riemann integral because the integrand is continuous and bounded on R and
 ]h(x)e -i'll dx < _ ]hi for every compact interval In, b].
As an application of the convolution theorem we shall derive the following
property of the Gamma function.
Extmlfle. lfp > 0 and q > 0, we have the formula
f F(p)F(q) (47)
x-a(I - x) °Èt dx -- r'(p + )
The inlegral on the left is called the Betafumtion and is usuafly donot by B(p, q). To
0rove (47) we let
f(t)= { -xe-' ift<ift >O,o.
Tln f • L(R) and _® f,(t) dt  I t-e=t dt = F(g). Let h deole the convolution,
h = f, • f. Taking u -- 0 in the convolution formula (44) we find, if p > 1 or  > !,
Now we calculate lho integral on the left in another way. Sinee both  and f vani-h on
the ative teal axis, we have
The change of variable t -- ux gives us, for x > 0,

Th. 11.74
11.22 THE POISSON SUMMATION FORMULA
We conclude this chapter with a discussion of an important formula, called
Poisson's summation formula, which has many applications. The formula
can be expressed in different ways. For the applications we have in mind, the
following form is convenient.
lteorem 11.24. Let f be a nonnegative function such that tle integral _ f(x) dx
exists as an improper Riemann integral. Assume also that f increases on (- co, O]
and decreases on [0, + oo). Then we have
÷. "f(__m+) + f(m-)= ÷,. f?f(,)e_2.., d,. (49)
each series being absolutely convergent.
Proof. The proof makes use of the Fourier expansion of the function F defined
by the series
F(x)---- ,, f(m + x). (50)
First we show that this series converges absolutely for each real x and that the
convergence is uniform on the interval [0,
Sincefdecreases on [0, +co) we have, for x > 0,
E f(m + x) <_ f(O) + f(m) ; f(O) + ff(t)(It.
Therefore, by the Weierstrass M-test (Theorem 9.6), the series .=of(m + x)
converges uniformly on [0, +co). A similar argument shows that the series
y_..2 _® f(m + x) converges uniformIy on (- co, I]. Therefore the series in (50)
converges for all x and the convergence is uniform on the intersection
(-co, l] [0, +co) =- [o, l]r
The sum function F is periodic with period 1. In fact, we have F(x + 1) ---
Y+,,=_f(m + x + I), and this series is merely a rearrangement of that in (50),
Since all its terms are nonnegative, it converges to the same sum. Hence
F(x + = F  x).
Next we show that F is of bounded variation on every compact interval, if
0 < x _< , then f(m + x) is a decreasing function of x if m _> 0, and an in-
creasing function of x if m < 0. Therefore we have
so F is the difference of two decreasing functions, Therefore F is of bounded
variation on [0, ½]. A similar argument shows that F is also ofhounded variation
on l'--½, 0]. By periodicity, F is of bounded variation on every compact interval.
Now consider the Fourier series (in exponential form) generated by F, say
F(X) "- E °h
Since F is of bounded variation on [0, 1] it is Riemann-integrable on [0, 1]. and
the Fourier coefficients are given by the formula
=  F(x)e -2* dx. (51)
Also, since F is of bounded variation on every compact interval, Jordan's test
shows that the Fourier series converges for every x and that
+ V(x.-) =
' =_
To obtain the Poisson stmamation formula we express the coefficients . in
another form. We use (50) in (51) and integrate term by term (justified by uniform
convergence) to obtain
The change of variabIe t = m + x gives as
since e )== = I. Using this in (52) we obtain
When x -- 0 this reduces to {49).
)o. In Theorem I 1.24 there are no continuity requirements on f. However, if
fis continuous at each integer, then each termf(m + x) in the series (50) is con-
tinuous at x = 0 and hence, because of uniform convergence, the sum function F
is also continuous at 0. In this case, (49) becomes
f(m) = f(t)e- ""' at. (54)
The monotooicity requirements on fcan be relaxed. For example) since each
member of (49) depen.ds linearly on f, if the theorem is true forfl and forfz then
it is also true for any linear combination atft + a2f:, In particular, the formula
holds for a complex-valued function f = u + it) if it holds for u and ,, separately.

EXaml 1, Tramform/onfor,/afor/he the/a funcion. The dleta function 0 is defined
t'or all x > 0 by the equation
We shall use Poisson's formula to derive the transformation equation
O(x) =  0 " for x > O. (55)
For fixed  > 0, let f(x) = e - for all real x. This function satisfies all the hypoth-
esis of Theorem 11.24 and is continuous everywhere. Thexffote, Poisson's formuIa
The left member is 0(]). The integral on the right is equal to
whem
F(y) = f: e-X2cos2xy dx.
But F(y) = ½ 4 e-
Ug s  (56) d g
  Partl-frti
is n for x # 0  t uafion
e  + 1
x
e z 1
We 1  Poin's fo to i the  ial-fi omsifion
()
fx > O. F > O,let
f(x) = ifx  O,
ifx <
 fdy tis
pt at O, whef(O+)  1 andf(O-)  O. o, the Polka foula pl
(8)
The um on lh¢ left is a geometric series with sum l/(e" - I), and lh¢ integral on the right
is equal to 1/( + 2n/n). Thercfot (58) becomes
 + e  1  + 2 + '
  -2in
and this gives (57) when a is replaced by 2x..
Ollegeml s3am
11.1 Verify .that the trigonometric system in (1) is orthonormal on [0, 2].
11.2 A finite collection of functions {'0, 9,, ..... g} is said to he linearly independent
on In, b] if the equation
__octct(x) = 0 for all in b]
hnplies co -- c ..... c = 0. An infinite collection is called tinearly independent on
In, b] if every finile subset is linearly independent on [o, hi. Prove that every orthonrmal
system on [a, b] is linearly independent on [a, hi.
113 This exercise describes the Gram-chmidtproces for converting any linearly inde-
pendent system to an orthogonal system. Let {re, f: .... } be a linearly independent
system on [a, b] (as defined in Exercise 11=2). Define a new system {go, 0%... } recur-
dvdy as follows:
k----l.
where a, = zr II.  d # o, and = 0 if lt,o',,J] = 0. Prove that is
orthogona] to each of go , 9:,..., 9, for every n  0.
11.4 Refer to Exercise 11.3. Let Or, g) = [f(t)g(t)dt. Apply the Gram-Schmidt
procms to the system of polynomials {I, t, t 2 .... } on the interval [-I, I ] and show that
g(t) = t, gz(t) : t  - , grit) - t  It, g(t) = t'  t  + .
11. a) Assuincf R on [0, 2x], whercfis real and has lOd 2, Povc that for
c > 0 thcr is a continuous function g of pcziod 2x such that Ill - gll
Hint. Choose a paxtition P= of {0, 2x] for whichfsatisfies Ricrnann's cndition
U(P, f) - £(P, f) <  and construct a pieccw[sc linear g which agrees with f
at the points of
b) Ut¢ paxt (a) to show that Theorem l 1.16(a), (b) and (c) holds iffis Ricmamz
intcgrab]¢ on [0, ].
11.6 In this cxcxcise all functions are amumcd to be continuous on a compact interval
[a, hi. Let {o, Wz .... } b¢ an orthonormat system o In, b].
a) Prove that the following three statements arc equivalent.

Foatr S¢ie ad Fmaet Integrs
1) (f, )  (0,  for all n implies f = 0- wo dfi ntuous functions
not ha   Foufi .) - •
2)  9,) = 0 for   plf  0. , ly continuom fion orthogonal
to   is   functMn.)
3) If T is  hono t  [a,b] s t {o,,--.)  th
{o, ¢t,- • • ) ffi T. (We nnot i c oonoml .) s pmy is
 by ng that {o,  .... } is il or coyote.
b)  x)  e] for n  t, d y that t  {, : n q Z} is -
pe on  inl of  .
II.7 ffxRdn = 1,2 ..... let(x) = (x 2 - 1 and 
It is  that  Js a Iomi. is is   re ial of  n. 
t few 
s(x) = x  - ix, ,(x) = M - Mx 2 + {.
  fotling proi of  ]omia:
x 2- 1
.(x) = x4,_t(x) + 
$, t C difftial uati [(1 - xZ)] ' + n + l)y  O.
[0 - x z) x)]" + [m + ) - n + )].(x).(x) = 0,
t {o, t, 2,.-- } k onal  [- , I].
a)
b)
c)
d)
e)
f)
,I-
1 2n+ 1
The polynomials
2*(n!) 2
ari by applying the Gram-Schmidt proc¢ to the system (I, t, t 2, ... } on the interval
[- 1, 11. (See Exercise 11.4.)
11.8 Assume that/e L([-n, nl} mad that/ha I,iod 2n. Show that the Fourier seris
generated by fassumes the following Slcial forms under the conditiom stated:
a) Iff(-x) = f(x) when 0 < x < n, then
ao  2
[(x) ~  + . a. cos nx, where a. = - [(t) co m dr.
b) If f(-x) = -f(x)when 0 < x < n, then
/'(x) ,,,  b, sin
whe b, = _ n 2 fff(t) sin m dr.
In Exexc.is 11.9 through 11.15, show that each of the ¢pansions is valid in the range
indicated. Suaeestien. Use .ercie 11.8 and Thtorem 11.16(c) when possible.
11.9 a) x  2 
-- - , ifO < x < 2n.
X2 t 2 'm 
-- = rx- -- + 2 -- ifO < x <
b) 2 3  n  ' - -
When x = 0 this gives (2) = 216.
n  sin (2n - 1)x
11.10 a)  = 2n - 1
II=l
cos (2n - 1)x
"_ )2
I1.11 a) x = 2 .. (-1)- sin nx 
if0<x<n.
if0_<x_< n.
, if-n<x<n.
n 2  (-1) cos nx
b) x2=. -+4 n2 , if-nxn.
3
11.12 x2= 3-4rrz+4n_-t  ,(e°s---nXn  _nsin nx_) , if O<x< 2n.
8  n sin 2nx
11.13 a) cosx =-2. --, if0 < x < n,
 =t 4n - 1
2 4 =-,  cos 2nx
b) sin x - -
r, r,'= 4n - t
I)_"n_ _sin_
11.14 a) x cos x  -½ sin x + 2 2_
b) x sin x = 1 -  cos x - 2  -- (- I.)" cos nx
ifO<x<n.
----, if-_< x_< r.

11.15 a) log sin -- -log 2 -- cos nx,
" 41-----1 n
if x  2kn (k an integer).
if x  (2k +
c) log tan = -2 nZ-1 ' ifx/ca.
11.16 a) Find a continuous fun--tion ou
,  ( ln-  s nx.
n6/945.
b) U an aofiate Fou6er 6 n conjunion kh Paval's foula to
show that if(4) =
11.17 Assume that f has a continuous derivative on [0, 2n] thatf(O) = f(2n), and that
fo'f(t)dt - O. Prove that Ilfql -> ]fll, with equalily if and only if f (x) -- a eosx +
b sin x. Hit. Use Parseval's formula
11.18 A sequence {B,} of 10¢ri0dic fonctions (of period t) is defined on R as follows:
B2,(x) = (-I +1 2(2n)!   cos__2_nkx (n - 1, 2, .L
(2n) z -t k2" '"
2(2n + 1)!  sin 2rkx
Bz.+(x) = (-lY'+ • (2,z)2,+ 
=1
(n = 0, 1, 20...).
(/, is called the Bernoulli function of order n.) Show that:
a) B,(x) -- x - [x] - ½ ifx is not an inleger. ([xl is the greatest integer <_x.)
b) [(x) dx = 0ifn >_ t and B(x) -- n_(x) ifn >_ 2.
c) (x) = P,(x) if 0 < x < I, where P, is the nth Bernoulli polynomial. (See
Exercise 9.38 for the definition of P,.)
- k" (n  I, 2,...).
11.19 Lel)"be the function of period 2n whose values on [-r. zc] are
f(x)= 1 if0 < x < n, /'(x)= -1 ifr < x < O,
f(x) = 0 ifx = 0orx = n.
a) Show that
sin (2n - l)x
for
X.
This is one example of a class of Fourier series which have a curio property known as
Gibbs'phenomenon. This exercise is design! to illustrate this phenomenon. In that which
follows, s¢(x) denotes lhe nlh partial sum of the series in part
h) Show that
• ,(x) - _ 2 fx sin 2nt dr.
 .]o sin t
c) Show that, in (0. ), s has local maxima at x,, x ..... xz,_  and 1peal minima
at xz, x,,..., x._z, where x. = ½mn/n (m = 1, 2, .... 2n - I).
d) Show that an(g/n) is the largest of the numbers
e} Into'pint a{!t/n} as a Ricraann sum ad prov that
The value of the limit in (e) is about 1.179. Thus, although fhas a jump equal to 2 at the
origin, the gaphs of the approximaling curves sa ted to approximate a vrtical segment
of length 2.3-58 in the vicinity of the origin. This is the Gibbs phenomenon.
11:?,0 Iff(x) ~ no/2 + y__% (aa cos nx + b sin nx) and iffis of bounded variation on
[0, 2ar], showthat a, = O(I/n) andb = O(I/n). Hint. Writer-- a  h, wheregandh
are increasing on [0, 2n]. Then
If"
ann g(x) d(sin nx) - --m h(x) d(sin ux).
Now apply Theorem 7.31.
1121 Suppose g e L([a, e]) for every a in (0,.b') and assume that g satisfies a "right-
handed" Lipschitz condition at 0. (See the Note following Theorem I .9.) Show that the
Lebesgue .integral j' Iv(t) - o(O+)l/t dt exists.
11.22 Use Exercise 11.2f to po'ce that diffea'ntiability of fat a point implies
of its Fourier series al the tint.
11.X3 Let# be continuous on [0, 1] and assume that J' t'(t) dt= 0 forn -- 0, 1, 2,....
Show that :
a S u(tV at   z(tXO} - eO)) dt for ever pornomla '.
b)  #(t)  dt = O. Hint. Use Theorem I] .17.
e) t(t) = 0 for every t in [0, 1].
11.24 Use the WeierstraSs approximation theorem to pove each of the following state-
a) If/is continuous on [I, +0o) and ill(x)  a s x  +c, thenfcan be uni-
formly approximated on It, + ce) by a function # of the form (x) =
where p is a polynomial.
b) If  is continuous on [0, + c) and it" f(x) --, a as x --, + , then/- can be
uniformly approximated on [0, + ) by a function  of the form (x) =
where p is a polynomial.
11.2.5 Assume that f(x) ~ ad2 + ,_.,, (a cos nx + b, sin x) nd let {%} be the
sequence of arithmetic means of .the partial sums of this series, as it was given in (23).

Show that:
a) r,(x) =  + 1 -- (a cos kx + b sin kx).
£.
b) If(x) - a,,(x)l : d.  If(x)l z ak
Assume also that the derivative/'  R on [0, 2].
a) Prove that the sexics __'_® aalt z converges; then use the CauchySchwarz
inoqualiy to dduc¢ that ,ff_  I.l
b) From (a), deduoe that the series +ff_  e m converges uniformly to a con
tinuous sum/traction  on [0, 2a]. Then prove that/=
11,2"/ Iffsa the hypothoscs oftho Fourier integral theorem, show that:
a) If/is even, that is, if/(- t) = f(t) forcvery t, then
cos x /(u) cos vu du d.
2 --t4-
b) If f is odd that/s, if f(- t) = -f(t) for every t, then
f(x+) + f(x-) = 2_ lira sin vx
Us the Fourier integral t to evaluate the improp integrals in Exercises 11.28
through 11.0. Sgestio. Us ExewAs¢ 11.27 when possibl
11. 2 f sin v cos vx & = if Ixl > 1,
nJo v if]xl = 1.
f; COS  : Ilb
11.2 b + xd=2be , if b> 0
Hnt, Apply Fnse 11.27 with/(u) = e-l't.
341
11.30 f; x__i sin+ x ax dx -- --14a 2_. e-'' if a  O.
11.31 a) Prove that
F(p)r(p) _ 2 x-t(l - x) p-' dx.
r(2p)
b) Make a suitable change of variable in (a) and derive the duplication formula for
the Gamma function:
F(2,)F(½) = 2zt-F(P)F(p ÷ ½).
sore. In Exercise 10.30 it is shown that F(½) =
113: If f (x) = e -x:/2 and g(x) = xf(x) for all x, prove that
sin xy dr.
11.33 Th/s exercise describes another form of Poisson's summation formula. Asaun'
that/is nonnegative, docrcasing, and continuous o [0, '+ oo) and that j' f(x) dx exists
as an improper Riemarm integral. Let
g(y)= (f;f(x)cosxydx.
ff a and ,8 axe positive numbers such that /] = 2n, prove that
1134 Prove that the transformation formula (55) for 0(x) can be put in the form
where aB = 2n,t > 0.
11.35 If s > 1, prove that
and derive tb_¢ formula

where 29(x) ffi lx) - 1. Use this and the transformation formula for x) to prove that
Laplace traasferms
Lel c be a positive number such that the integra| S' e-lf(t)l dt exists as an improper
Rkmann integral. I_¢t z = x + iy, where x > e. It is asy to show that the integral
F(z) -- f: e-" f(t) dt
exists both as an impfo Riemann integral and as a Lebesgue integral. The function F
so defined is called the Laplace transform off, denoted by .L'(f). The following execcises
describe some prOlXXt of Lapla transforms.
11,36 Verify the entries in the following table of Laplace transforms.
z=x+iy
(z- a) -t (x> a)
zl(z+ + ,,2) (x > 0)
I'(e + ly(z - mF +' (x > a, p > O)
1137 Show that the convolution h --- f, e assuas tM form
v/non both f and g vanish on the negative ttal axis. Us¢ the convolution theorem for
Fourier transforms to prove that .¢(f, a) = GO.
11.38 Assumefis continuous on(0, + ) and let F(z) = S e-f(t) dt for z
x > ¢ > 0. If > vanda > 0 provethat:
h) IfF(+ a) = 0forn = 0,1,2,..., thenf(t) = 0for t > 0. Hiat. Use
Fercis¢ I 1.23.
c) If'h is continuo on (0, +o) and if fend h haste the sam Laolac¢ transform,
tlamf(t) = gt)for  t > 0.
I1_9 Let F(z) = .[ e-'f(t)dtforz =- x + iT, x > e > 0. Let t be a point at whichf
satisfies on, of th¢ "local" conditions (a) or Co) of the Fourier intclai theorem (Theorem
ll.lg). Provcthat foreacha >
/(t+) +/'(t-) _  lirn fr e'+'tF(a + ) de.
2 2 r-+®
This is called the nverslon formula for Lap/ace transforms. Th¢ limit on th© right is usuaIly
evaluated with the help of residue calculus, as descrilxut in Section 16.26. Hint, Let
g(t) = e-f(t) for t _> 0, 9(t) -- 0 for t < 0, and apply Theorem I 1.19 toe.
SUGGESTED REFFJIFCES FOR FURTHF STUDY
11.3
I 1.4
11.$
11.6
II.7
11.8
11.9
11.10
11.1 C.arslaw, H, S., Introduction to the Theory of Fourier's Serie ami lntegrals, 3rd ed..
Macmillan, Londoth 1930.
11.2 Edwards, R. F, Fourier Series, A Modern Introduction, Vol. 1. Holt, Rinehart and
Winston, New York, 1967.
Hardy, G. H., and Rogosinski, W. W., Fourier Series. Cambridge University Press,
1950.
Hobson, P÷ W., T/w Theory of Functions of a Real Variable and tl¢ Tlory of
Fourier's Series, Vol. I, 3rd ed. Cambridge University Press, 192"/.
Indritz, J., Methods in Analysis. Macmillan, New York, 1963.
Jackson, D., Fourie# Series and Orthogonal Plynomiafs. Carus Monograph No. 6,
Open Court, New York, 1941.
Rogosinski, W. W., Fourier Series. H. Cohn and F. Steinhardt, translators.
Chelsea, New York, 1950.
Titchmarsh, E. C. Theory of Fourier Inte#rals. Oxford University Press, ! 937.
Wiener, ]q., The Fourier InteaeeL Cambridge University Press, 1933.
Zygmund, A., Trigonometrical Series, 2nd ed. Cambridge University Press, 1968.

CHAPTER 12
MULTIVARIABLE DIFFERENTIAL
CALCULUS
17..1 INTRODUCTION
Partial derivatives of functions from R  to It  were discussed briefly in Chapter 5.
We also introduced derivatives of vector-valued functions from R 1 to k . This
chapter extends derivative theory to functions from R ' to R'.
As noted in Section 5.14, the partiaI derivative is a somewhat unsatisfactory
generaIization of the usmd derivative because existence of_all the partial derivatives
Dr f,..., Dt-at a particular point does not necessarily imply continuity off at
that point. The trouble with partial derivatives is that they.treat a function of
several variables as a function of one variable at a time. The partial derivative
describes e rate of change of a function in the direction of each coordinate axis.
There is a slight generalization, called the directionol derivative, which studies the
rate of change of a function in an arbitrary direction. It applies to both real- and
vector-valued functions.
12.2,  DIRECTIONAL DFJIIVATIVE
Let S be a subset of R', and let f : S --, R = be a function defined on S with values
in R". We wish to study how f changes as we move from a point ¢ in S along a
line segment to a nearby point ¢ + a, where a  0. Each point on the segment
can be expressed as ¢ + /m, where h is real. The vector a describes the direction
of the line segment. We assume that ¢ is an interior point of S. Then there is an
n-hall B(©; r) lying in 8, and, if h is small enough, the line segment joining c to
c + hn will lie in B(c; r) and hence in S.
Definition 12,1, The directional derivative of f at ¢ in the direction u, denoted by
the symbol f'(c; a), is defined by the equation
f'(c; tt) = lira f(¢ +, ha) - f(¢) , (1)
whenever the limit on the right exists.
NOTE. Some authors require that glaIJ -- l, but this is not assumed here.
1. The definition in (I) is meaningful ifu = 0. In this case f'(c; 0) exists and equals 0
for every e in S.
g. tf u = . the kth unit coordinate vector, then f'(c; uk) is called a partial derivative
and is denoted by Dkf(e). When f is real-valued this agrees with the definition given
in Chapter 5.
3. If f - Oct ..... ft), then f'(e; u) exists if and only if f(e; u) ediists for each k :
1, 2 .... , in, in which case
f'(e; u) = (/;(c; n),...,/(e;
In particular, when e = u we find
Pal{c) = (odt{c),-.-,
4. IfF(t) -- f(© + tu), then F'(0) = f'(¢; u). More generally, F'(t)  f'(c + tu; u) if
either derivativ exists.
0) --/(e + tu) = (c + m).(c +
so F'0) = 2c-u + 2tuH2; hence F(0) = f'(©; u) = 2¢- u.
6, Linezrfunctios. A function f: R  - R m is called iineor ift'(ox + by) = of(x) + bf(y)
for every x and y in R " and every ptr of scalars  aad & If f is linear, the quotient
on the right of(D simplifies to f(u), so f(e; u) = flu) for every ¢ and every u.
12.3 DIRECTIONAL DERIVATIVES AND CONTINUITY
If f'(e; u) exists in every direction u, then in particular all the partial derivatives
Df(e),..., D,f(c) exist. However, the converse is not true. For example,
consider the real-valued function f: R z - R I given by
f(x, Y) = { + y
i fx = 0ory = O,
otherwise.
Then Df(0, 0) :-- Df(O, 0) = 1. Nevertheless, if we consider any other direction
a : (a, a), where a t d: 0 and a :# 0, then
f(0 + hu) - f(0) f(hu) _ 1
t h h
and this does not tend to a limit as h  0.
A rather surprising fact is that a function can have a finite directional derivative
f'(¢; a) for etrery  but may fail to be continuous at ¢. For example, let
f(x, y): { ya/(x2 + 3'') ifxifX =# 0.0'
Let u - (a,, az) be any vector in R a. Then we have
f(O +_ n) - f(O) _ f.(_h_a,, haz) _ a'.__
h h a 2, +

ad hence
f,(O;u)={2/at if slits' =:/:0'0,
Thus, f'(0; u) exists for all u. On the other hand, the functionftakes the value ½
at each point of the parabola x = yZ (except at the origin), soils not continuous
at (0, 0), since f(0, 0) = 0.
Thus we see that even the existence of all directional derivatives at a point fails
to imply continuity at that point. For this reason, directional derivatives, like
partial derivatives, are a somewhat unsatisfactory extension of the one-dimensional
concept of derivative. We turn now to a more suitable generalization which implies
continuity and, at the same time, extends the principal theorems of one-dimensional
derivative theory to functions of several variables. This is called the totalderivative.
12.4 THE TOTAL DERIVATIVE
In the one-dimensional case, a functionfwith a derivative at C can be approximated
near e by a linear polynomial. In fact, iff'(c) exists, let Ec(h ) denote the difference
Eh) - f(c + h)  :(c) :'(c) if h  0,
and let E,(0) ffi 0. Then we have
f(c + h) = f(c) + f'(c)h + hEo(h), (4)
an equation which holds aiso for h - 0. This is called the first-order Taylor
formula for approximating f(c + h) -f(c) by f'(c)h. The error committed is
hE,(h). From (3) we see that E,(h) -, 0 as h --, 0. The error hEc(h ) is said to be
of smaller order than h as h - 0.
We focus attention on two properties of formula (4). First, the quantity
f'(c)h is a linear function of h. That is, if we write T,(h) = f:(c)h, then
T,(ah + bhx) -- aT(ht) + bT(h2).
Second, the error term hE,(h) is of smaller order than h as h  0. The total
derivative of a function f from R" to R" will now be defined in such a way that it
preserves these two properties.
Let f : S --, R = be a function defined on a set S in R" with values in R =. Let ¢
be an interior point of S, and let B(c; r) be an n-ball lying in S. Let v be a point
in R  with Ilvll < r, so that e + v  B(e; r).
Defmide 12.2. The function f is said to be differentiabte at e  there exists a linear
function T" R  --, R" such that
f(e + v) : f(e) + T(v) + Ilvll E Jr),
(5)
where E(v) -, 0 as v -, 0.
Th, 12 The Total Dexivaliv¢ 347
NOTE. Eqoation (5) is called afirst-order Taylor formula. It is to hold for all v in
it" with Ilvll < r. The linear function T, is called the total derivative of f at c We
also write (5) in the fo[m
ffc + v) = f(e) + T,(v) + of[IvlD as v-, o.
The next theorem shows that if the total derivative exists, it is unique. It also
relates the total derivative to directional derivatives.
Theorem 12.3, Assume f is differentiabte at c with total derivative T¢. Then the
directional derivative f'(e; u) exists for eery u in R  and we have
Proof
v#0.
= f'(e; (6)
If ¢ ---- 0 then f'(e; 0) = 0 and T(0) : 0. Therefore we can assume that
Take v = ha in Taylor's formula (5), with h # 0, to get
f(c + ha) - f(e) = T,(hu) + llhull E  (v) : hT0,) ÷ lhl Italt E(v).
Now divide by h and let h  0 to obtain (6).
Tlteorem 12.4. If f is differentiable at c, then f is continuous at e.
Proof Let v --, 0 in the Taylor formula (5). The error term llvll E.(v)  0; the
linear term T(v) also tends to 0 because if v = vtut + .-' + v,a, where
ut ..... u. are the unit coordinate vectors, then by tinearity we have
T,(u) = vtT(nt) + "'" +
and each term on the right tends to 0 as v -- 0.
NOT. The total derivative T is also written as f'(©) to resemble he notation used
in the one-dimensional theory. With this notation, the Taylor formula (5) takes
the form
f(e ÷ -- f(c) ÷ f'(c)fv) ÷ Ilvtl E¢(v), (7)
where E,(v)  0 as v -, 0. However, it should be realized that f'(e) is a linear
function, not a number. It is defined everywhere on R*; the vector f'(e)(v) is the
value of f'(c) at v,
Example. If f is itself a linear function, then f(c + v) = f(e) + f(v), so the derivative
f'(c) eists for every c and equals f. In other words, the total derivative of a linear function
is the function itself.
12.5 THE TOTAL DERIVATIVE EXPRESSED IN TERMS OF PARTIAL
DEITJVATIVES
The next theorem shows that the vector f'(c)(v) is a linear combination of the partial
derivatives of f.
Theorem 12.5. Let f : S --, R  be differentiable at an interior point c of S, where
S _ R*. If v = vtu + "" + vu=, where ut, -.., u, are the unit coordinate

Maltivarleble Dilrealgsl Ca]talus
In particular, if f is real-valued (m = l) we have
f'(c)(v) =
the dot produc¢ of v witft the vecto Vf(c) - (Df(e), ..  ,
Proof. We use the linnarity of f'(e) to write
f'(cXv) =  f'(cXvuD- ]
k=l k--1
NOTE.
at each point where the partials Dif,,.., Dff exis|.
real-valued/"now takes the form
f(e + v) =f(e) + V/(c).v ÷0(tlvlt)
k=l kl
The Vector Vf(e) in (8) is called the gradient vector, off at c, it is defined
The Taylor formula for
12.6 AN APPLICATION TO COMPLEX-VALUED FUNCTIONS
Let f ffi u + iv be a complex-valued function of" a complex variable. Theorem
5.22 showed thata necessary condition forfto have a derivative at a point c is that
the four partials Dtu , Dzu , Dlv , D2v exist at c and satisfy the Cauchy-Riemann
equations:
DIu(c) = D2v(e), Jl/)(c) ffi -- D2Ig(c ).
Also, an example showed that the equations by themselves are not sufficient for
existence off'(c). The next theorem shows that the Cauehy-Riemann equations,
along with differentiability of u and t, imply existence off(c).
Tl,orem 12.6. Let u and v be two real-valued functions defined on a subset S of the
complex plane. -Assume also that u and v are differentiable at an interior point c
of S and that the partiaI derivatives satisfy the Cauchy-Riemann equations at c.
Then the function f = u + iv has a derivative at c. Moreover,
if(c) = DIu(c) + iDlY(C).
Proof. We have f(z) -f(c) = u(z) - u(c) + i{v(z) - v(c)} for each z in S.
Since each of u and v is differentiablc at c, for z sufficiently near to c we have
and
Here we use vector notation and consider complet numbers as vectors in R 2. We
then have
Writing z = x + iy and c = a + /b, we find
{ Vu(e) + i Vv(e)}. (z - c)
ffi Otu(c)(x -- a) + D2u(c)(y - b) + i {Dl(e)(x -- a) ÷ Dzv(c)(y - b)}
-- o,u(e)((x - a) + i(y - b)} + io  v(c){(x - a) + i(y - b)},
because of the Cauchy-Riemann equations. Hence
f(z) -f(c) = {Du(c) + iDle)} (g -- C) + O([Ig -- ell)-
Dividing by z - c and letting z  c we see thatf'(e) exists and is equal to
Du(c) + iDly(c).
12.7 THE MATRIX OF A LINEAR FUNCTION
In this section we digress briefly to record some elementary facts from linear
aigebra that are useful in certain calculations with derivatives.
Let T:R" - R" b¢a linear function. (In our applications, T will be the
total derivative of a function f.) We will show that T determines an m × n matrix
of scalars (see (9) below) which is obtained as follows:
Let ua, ..., u. denote the unit coordinate vectors in R'. If x e R" we have
x -- xtut + " • " + x.u, so, by linearity,
T(x) ffi  x,'l'().
Therefore T is completely determined by its action on the coordinate vectors
Ul, . . . ,
NOW let el, • • • , e. denote the unit coordinate vectors in R". Since T(a0
we can write T(a) as a linear combination of e,..., e., say
$----1
The scalars ttt,...-, t are the coordinates of T(u), We display these scalars
vertically as follows:

Malivariable Dilt'erentia! Calculus
This array is called a column ,,ector.
T(#), L.., T(u,} and place lhem side by side to obtain the rectangular array
We form the column vector for each of
(9)
Suppose that S and T have matrices (so) and (to), respectively.
S(et) =  xv for k = I, 2,..., m
and
This is called the matrix* oft and is denoted by re(T). It consists of m rows and
n columns. The numbers going down the kth column are the components of
T(a). We also use the notation
re(T) -- rt
L ili.-- t Or re(T) = (t,)
to denote the matrix in (9).
Now let T : R* --, R = and S : R = -, R e be two linear functions, with the domain
of S containing the range of T. Then we can form the composition S o T defined by
(So T)(x) : SiT(x)] for all x in RL
The composition S o T is also linear and it maps R" into R e.
Let us calculate the matrix m(S o T). Denote the unit coordinate vectors
R*, R =, and R e respectively, by
an, • • •, an, el, • • •, e,., and WL, . . .
This means that
T(u_) =  tye forj = 1, 2,..., n.
,=l k=l
Then
so
In other words, m(S  T) is a p × n matrix whose entry-in the ith row and jth
* More precisely, the matrix of T relative to the given bases u .... , u. of R n and
¢ ..... ¢ ofR .
The Jobian Matrht 351
column is
 ,.i..j 
=1
the dot product of the ith row of re(S)with thejth column of re(T). This matrix
is also called the product m(S)m(T). Thus, m(S  T) = m(S)m(T).
12,8 THE JACOBIAN MATRIX
Next we show how matrices arise in connection with total derivatives.
Let f be a function wi'h values in R" which is diffcrentiable at a point ¢ it R',
and let T = f'(c) be the total derivative of f at e. To find the matrix of T wc
consider its action on the unit coordinate vectors a .... , ar By Theorem 12.3
we have
T(u = l'(c; u) = Df(c).
To express this as a linear combination of the unit coordinate vectors e, _ _., e, of
R" we write f = (f,... ,f,,) so thal Df = (Df ..... Df,,), and hcce
T(u) = D,f(c) =  Df,-fe)e e
Therefore the matrix of T is re(T) = (Df,(e)). This is called the Jacobian matrix
of f at c and is denoted by Dr(e). That is,
Dr(c) = Dt.f(c ) Dz,fz(c ) ,,. D.f(c)
D,j2(c) O:/i(c) ""
The entry in the ith row and kth column is Dtf;(c ). Thus, to get the carries in the
kth column, differentiate the componems of f wih respect to the kth coordinate
vector. The jacobian matrix Dr(c) is defined a each poitt c in R" where all the
partial derivatives Df(c) exist
The kth row of the Jacobian matrix (10) is a vector in R" called lhe #radic, m
rector off, denoted by Vf(c). That is,
vA(e) = (t),A(e),...,
In the special case whenfis rcalwalued (m = I), the Jacobian malrix consists
of only one row. In this case Df(c) = Vf(c), and Equation (8) of Theorem 12.5
shows thor the directional derivative f'(c; v) is the dot product of the gradient
vector V/c) with the direction v.
For a vector-valued function f = (f ..... f,,) we have
h=l k:l

32 Mnlfivarla lffentil CAdculu Th. 12.7
so the vector f'(¢Xv) has components
(v/l(c). v, ..., v/.(c), v).
Thus, the components of |'(c)(v) are obtained by taking the dot product of the
successive rows of the Jacobian matrix with the vector v. If we regard f'(cXv) as
an m × t matrix, or column rotator, then f'(c)(v) is equal to the matrix product
Eff(c)v, where Df(c) is the m x n Jacobian matrix and v is regarded as an n x 1
matrix, or column vector.
NOT Equation-(11), used in conjunction with the triangle inequality and the
Cauchy-Schwarz inequality, gives us
Thor, fore we have
Ilf'(c)(v) _< Mlfvll,
where M = _ t 11Vf(c)[[ This inequafity will be used in the proof of the chain
rule. It also shows that f'(¢Xv) -, 0 as v --. 0.
12.9 THE CHAIN RULE
Let f and g be functions such that the composition h = f o g is defined in a
neighborhood of a point at. The chain rule tells us how to compute the total
derivative of h in terms of total derivatives of f and of g.
Theorem 12.7. lssume that g is differemiable at , with total derivative g(a). Let
b = g(a) and a.rsurae that f is differentiable at b, with total derivative f'(b). Then
the composite function h = f o g is differentiable at a, and the total derivative h'(a)
is given by
W(a) = f'() o
the composition of the linear functions f'(b) and g'(a).
Proof. We consider the difference a + y) - h(a) for small [lyll, and show that
we have a first-order Taylor formula. We have
h(a + y) -- h(a) = f[g(a + y) -- fig(a)] ---- f(b + v) - f(b), (13)
where b = g(a) and v = g(a + y) - b. The Taylor formula for got + y) implies
v -- g'(a)(y) + Ilyl[ E,(y), where E,(y) - 0 as y  0. (14)
The Taylor formula for f(b + v) implies
f0} ÷ v) - fib) f'(t0(v) ÷ I]vtl where Eb(v ) --, 0 as v  0. (I 5)
Using (14) in (15) we find
f(b + v) - f(b) = f'(b)[g'(a)(y)] +
-- f'(b)[g'(a)(y)] + liYl[ Y), (16)
where E(0) = 0 and
y)  f'(b)[E(y)] +
To complete the proof we need to show that E(y) * 0 as y  0.
e fit term on the fight of(l tends to0asy
ond rm, the factor (v)  0 cause v  0 as y  0. Now we show that
the quotient [[vli/liyl[ remains unded a y  0. Using (14) and (12) to estimate
• ¢ numetor we find
Ilvlt IIg'(a)(y)ll + Ilyll II  ff)ll
llvl  M
 ILvil/Dil mains bounded as y  0. Using (13) and (16) we obn the Taylor
form
h(a + y) -- h(a) = f'(b)[g'(a)(y)] + IlYll
whc y)  0 as y  0. This proves that k is diffentiabl¢ at a and that its
total dcrivativ at a is the composition f'{b)
12.10 MATRIX FORM OF THE CHAIN RULE
The chain rule states that
h'(n) = U(b) o g'(a),
where h = f o g and b = g(a). Since the matrix of a composition is the product
of the corresponding matrices, 08)implies the following relation for lacobian
matrices:
Dh(a) = Df(b)Dg(a). (19)
This is called the matrix form of the chain rule. It can also be written as a set of
scalar equations by expressing each matrix irt terms of its entries.
SpecificMly, suppose that a 6 R p, b = g(a) E R*, and f(b) e R". Then h(a) • R =
and we can wdt¢
g -- (g,,:-.,a.), f : (f,,...,f,), h : (h,,...,
Then Dh(a) is an rn x p matrix, Dr(b) is an m x n matrix, and Dg(a) is an n x p

1 Mulhrariable Differential Clculm T 12
matrix, given by
Dia) = [D.ho(a)],l, Dr(b) = [Dfi(b)'j, , (a) = [D(a)]:= .
e ix uafion (19) is equivent to the mp al equations
D2h(a) =  Dtfgb)Dat(a), for i = 1, 2,..., m and j = 1, 2,..., p. (20)
 equations press ¢ partial divafiv of ¢ components of h  t¢ of
the pmial derivatives of e componts of f and g.
e uafions in (20)   put in a form t is der to rememr. Write
y = f(x) d x = t). en y = fg(0] = ), d (20) comes
=  y Ox (2)
t  x ty '
where
Oy__ = Dh, -dY-5 = DtJ, and .x_ = Dsg.
E, Supm = 1. ¢nlhfdk = fgl-vudep
tions in (), one for  of the i derivati of h:
Dh(a) :  D)Dg(a), ]  1, 2 .....
¢ ridt mr is the dot oduct of the two o Vf(b) and Dya). In this 
Eqtioa (21) mk ¢ f
In rticul, ifp = I we get only o uation,
Wh the Jabian matrix ) is a c61mn re.or.
e chain rule can  used to give a simple proof of the foltowing theorem for
differentiating an integl with st to a parameler which apa both in the
integrd and in the limits of integration.
Totem 12.& t f and Dzf  contgnuous on a rectyte [a, b] x [c d]. t p
and q be derentile on [e, d], where p(y)  [a, b] d q(y) e [a, b] for each y in
[c, d]. Oefine F by the equation
F(y) = f(x, y) dx, if y  [c, d].
12.9 Me,m-Value Themem fer Digeneatlable Fanctioas 35
Then F'(y) exists for each y in (c, d) and is #iven by
F'(y) = Dzf(x; y) dx + f(q(y), y)q'(y) - f(p(y), y)p'(y).
Proof Let G(x,, xz, x) = j f(t, xa) at whenever , and x are in Ca, b] and
x e It, d]. Then F is the compote fetion given by F(y) = G(p(y), q(y), y).
The chain rule implies
By Theorem 7.32, we have OaG(x , x z, xj) = -f(x,, x) and OzG(x,, x, x) =
f(x,, x3). By eom 7., we also have
Using the results in the formula fr F'() we obtain the theorem.
12.11 THE MEAN-VALUE THEOREM FOR DIFFERENTIABLE FUNCTIONS
The Mean-Value Theorem for functions from R a to R i states that
f(y) - f(x) = f'(z)(y - x),
(22)
where z lies between x and y. This equation is false, in general, for vector-valued
functions from R" to R', when ra > 1. (See Exercise 12.19.) However, we will
show lhat a correct equation is obtained by taking the dot product of each member
of (22) with any vector in R', provided z is suitably chosen. This gives a useful
generalization of the Mean-Value Theorem for. vectorvalucd functions.
in the statement of the theorem we use the notation L(x, y) to denote the line
segment joining two points x and y in R". That is,
L(x,y) = {tx +(I - t)y:0 < t < 1}.
Theorem 12.9 (Mean-Val Theorem.) Let S be an open subset of R  and assume
that f : S --, R = is differentiable at each point of S. Let x and y be two points in S
such that L(x, y)  S. Then for every vector a in R" there is a point z/n/.(x, y)
such that
a' {f(y) - f(x)} -- a" [f'(z)(y - x)}. (23)
Proof. Letu = y- x. Since S is open and L(x, y) __. S, thereisa6 > 0such
that x + tu e Sfor al| real t in the interval (-6, 1 + of). Let a be a fixed vector in
R  and let F be the real-valued function defined on (-5, I + 6) by the equation
F(t) = a. f(x + tu).
Then F is differentiable on (- 6, 't + ) and its derivative is given by
F'(t) :- a'f'(x + /u; u) = a'{f'(x + tu)(u)},

By the usual Mean-Value Theorem we have
F(l) -- F(0) = F'(O), where 0 < 0 < 1.
F'(0) = a- {f'(x + = a- (f'(z)(r - x)},
where z = x + 0 ¢ L(x, y). But F(I) -- F(0) = a, {f(y) - f(x)}, so we obtain
(23). Of course, the point z depends on F, and hence on a.
NOTE, If S is convex, then L(x, y) _ S for all x, y in S so (23) holds for all x and
yin S.
1. Iffis real-valued (m = 1) we can take a = I in (23) to obtain
[(y) - f(x) = ['(zXy - x) = vftz)- (y - x). (24)
2. If f is ctor-valued and ira is a unit vector in R =, a = I, Eq. (23) and the Cauchy-
Sehwarz inequality giv us
Using (12) we obtain the inequality
Ill(Y) - f(x) <. rally - 11,
M -- Xro, ] Note that M depends on z and hence on x and y.
3. If S is convex imd if all the imrtial derivatives D.f are bounded on S, then there is a
constant .4 > 0 such that
fiX(y) - f(x)l[ -< Ally - xll-
In other word, f satisfies a Lipschitz condition on S.
The Mean-Value Theorem gives a simple proof of the following result concern-
ing functions with zero total derivative.
Tlcorem 12.10. Let S be an open connected subset of R , and let f: S , R ' be
differentiable at each point orS. tff'(¢) -- O for each e in S, then f is constant on S.
Proof Since S is open and connected, it is polygonally connected. (See Section
4.I8.) Therefore, every pair of points x and y in S can be joined by a tlygonal
arc lying in S, Denote the vertices of this arc by Pt, .--, Powhere Ih = x and
Pr = Y. Since each gment L(p+, lh) - S, the Mean-Value Theorem shows that
for every vector a. Adding these equations for i = 1, 2,..., r - l, we find
• " - f(x)l 0,
for every a. Taking a = f(y) - f(x) we find f() = f(y), so f is constant on S.
Th. ll.tl Suaei¢ Cdlthm for Diffentlabillty
12.12 A SUFFICIENT CONDITION FOR DIFFERENTIABILITY
Up to now we have been deriving consequences of the hypothesis that a function
is differentiable. We have also seen that neither the existence of all partial deriv-
atives nor the existence of all directional derivatives suffices to establish differ-
entiabJlity (since neither implies continuity). The next tt¢nrem shows that
continuity of all but one of the partials does imply differentiability.
Theorem 12,11. Assume that one of the partial derivatives DL .-., D,f exists at c
and that the remaininy n - 1 partial derivatives exist in some n-ball B(c) and are
continuous at e. Then f is differentiable at e.
Proof, First we note thata vector-valued function f = (f,... ,f=) is differentiable
at c if, and otly if, each componemA is differcntiable at e. (The proof of this iz an
easy exercise.) Therefore, it suffices to prove the theorem when f is real-valued.
For the proof we suppose Ihat l).,f(e> exists and that the continuous partials
are Dzf, -..,
The only candidate forf'(e) is lhe gradient wetter Vf(e). We will prove that
f(¢ + v) - f(e) = Vf(c)-.v + o(Ikvil) as v  0,
and this will prove the theorem. The idea is to express the differencef(c + v) - f(c)
as a sum of n terms, where the kth term is an approximation to Dsf(e)vt.
For this purpose we write v = 2y, where [[Y[I = 1 and.)° = I[v[I. We keep
small enough so that c + v lies in the ball B(e) in which the partial derivatives
Dzf, ..., DC'exist. Expressing y in terms of its components we have
y = yn + "-- + y,.,
where u is the kth unit coordioate vector. Now we write the differencef(¢ + v) -
f(c) as a telescoping sum,
f(c + v) -f(c) =f(c + )oy)--f(c) =  {f(e + .v) -f(c + ,:.v_,)}, (25)
where
v 0 = 0, v ---- yiUt, v  yllll -1- y2112 ..... V n = yu t + " " " + ynll.
The first term in the sum is f(c + 2yu) -f(c). Since the two points c and
c + slyu differ only in their first component, and since Dtf(c ) exists, we can
write
f(¢+ ,tytt) -f(e) = ,tytOff(e) +
where E(,t) , 0 as 2 , 0.
For k > 2, the kth term in the sum is
f(c + 2v_ t + 2yug) --f(c + 2v,_ ) -f(h, + 2y,u,) -f(b),
where b = e + 2v_:. The two points b and bt + 2yu differ only in their
component, and we can apply the one-dimensional Mean-Value Theorem for

evafives to write
(20
wher at lies on the line sgrnent joining b to bt + Aytu t. Note that bt - c and
hence  - ¢ as ,I, -, 0. Since each Dd'is continuous at ¢ for k > 2 we can write
DJ() --- Dd'(e) + Et(.),- where E(.)  0 as , - 0.
Using this in (26) we find that (25) becomes
where
This ¢ompletts the proof.
Noax Continuity of at least n - 1 of the partials D,f,..., Df at e, although
sufficient, is by no means necessa for diffctcntiabillty of f at ¢. (See Exercises
12.5 and 12.6.)
12.13 A SUFFICIENT CONDITION FOR EQUALITY OF MIXED PARTIAL
DERIVATIVES
The .partial derivatives D,f,..., D,,f of a function from R" to R '  themselves.
functions from R" to R" and they, in turn, can have partial derivatives. These are
called second-order partial derivatives. We use the notation introduced in Chapter
5 for real-valued functions:
Ozf •
Higher.order pardal derivatives are similarly defined.
The example
f(x, y) = x y(xz - YZ)l(x" + yZ) if (x, y) * (0, 0),
(o
if(x, y) = (o, o),
SHOWS that D,.2f(x , y) is not necessarily the same as D2,,f(x , y). In fact, Jn this
example we have
y) -- + 4x'y' - y*) if(x, y) ¢ (o, o),
(x z + y)2 ,
Th. IX12 Sult Cemlit for Eomllt ef Mite¢l Partial
and Dtf(O, 0)  O. Hence, Dr f(0, y) -- -y for all y and therefore
D,tft0,y) - - 1, Dz,tf(0, 0) : - 1.
On the other hand, we have
D2f(x, y) = x(x* - 4x2y 2 -- y4) if(x, y) # (0, 0),
(XZ + y2)2 '
and Dzf(O, O) = O, so that Dzf(x, O) = x for all x. Therefore, Dt .f(x, O) = I,
Dt.f(0, 0) = 1, and we see that O.tf(0, 0) # Dt.zf(0, 0).
The next theorem gives us a criterion for determining when the two mixed
partials D ,2f and D.tf will be equal.
Tkeorem 12.12. If both partial derivatives Df and Df exist in at, n-ball B(¢; ) and
if both are differentable at c, then
D,,tf(c) = Dtff(e)- (27)
Proof. if f -- (ft,.. ,f,), then Dtf = (Dtft,..., Dor,). Therefore it suffices
to prove the theorem for real-valued f Also, since only two components are
involved in (27), it suffices to consider the case n = 2. For simplicity, we assume
that ¢ = (0, 0). We shall prove that
D,,2f(O, 0) ---- Dx.,f(O, 0).
• Choose h # 0 so that the square with vertices (0, 0), (h, 0), (h, h), and (0, h)
lies in the 2-ball B(0; ). Consider the quantity
a(h) = f(h, h) - f(h, O) - f(O, h) + f(O, 0).
We will show that A(h)/h  tends to both D2.tf(0, 0) and Dt.xf(O, O) as h  O.
Let G(x) = f(x, h) - f(x, 0) and note that
= - a(o). (28)
By the one-dimensional Mean-Value Theorem we have
G(h) - G(0) = hG'(x) = h{Dtf(xt, h) - Dtf(x,, 0)}, (29)
where x, lies betwn 0 and t. Since Dfis differentiable at (0, 0), we have the
first-order Taylor formulas
D,f(x,, h) - D,f(O, O) + D,.tf(O, O)x, + Ox,tf(0, 0)h + (X + h2)tfZEl(h ),
and
Otf(xx, O) = D,f(O, 0) + Dr,t f(0, 0)x t + txtl E2(]I),
where Et(h) and Ex(h) --} 0 as h - 0. Using these in (29) and {28) we find
= + E(h),

Multivartable Dhgcmuti 
Tit, lg.13
where E(h)
SO
Since Ixll - Ihl, we have
0 <_ IE(h)l __- / h z IEl(h)l + h 2 IE(h)l,
D2 ,f(0, 0).
lira h2 = .
/t-c0
Applying the same proodure to the function H(y)= f(h, y) -f(O, y) in
ptace of G(x), we find that
(h) = Dl.2/(0,
lim h
which completes the proof.
As a consequence of Theorems 12.11 and 12.12 we have:
Theorem 12.1,L If both partial derivatives O,f and Dd exist in an n-baH B(e) and
if both D,.f and Dt.,f are continuous at c, then
NOTE. We mention (without proof) another result which states that if D,f, Dtf and
Dtfl are continuous in an n-ball B(c), then D,.tf(c) exists and equals Dt,,f(e).
If f is a real-valued function of two variables, there are four second-order
partial derivatives to consider; namely, DtAf, Dt.2f, Dz.tf, and Dz,zf. We have
just shown that only three of these are distinct if f is suitably restricted.
The number of partial derivatives of order k which can be formed is2 . If all
these derivatives are continuous in a neighborhood of the point (x, y), then
certain of the mixed partials will be equal. Each mixed partial is of the form
D,,, ..., ,f, where each r. is either I or 2. If we have two such mixed partials,
D,,,... , rkfand D,,, ..., ,f, where the k-tuple (rt,... , rE) is a permutation of
the k-tuple (Pt ..... pt), then the two partials will be equal at (x, y) if all 2 t partials
are continuous in a neighborhood of (x, y). This statement can be easily proved
by mathematical induction, using Theorem 12.13 (which is the case k -- 2). We
omit the proof for general k. From this it follows that among the  partial
derivatives of order k, there are only k + 1 distinct partials in general, namely,
those of the form D,, . ..., ,f, where the k-tuple (rt, ..., rt) assumes the following
k + 1 forms:
(2, 2,..., 2), (1,2,2 ..... 2), (l, I, 2,..., 2) .... ,
1,..., 2), (1, ..... t).
Similar statements hold, of course, for functions of n variables, in this case,
there are n  partial derivatives o( order k that can be formed. Continuity of all
these partials at a point x implies that D, ..... ,if(x) is unchanged when the
indic rt,..., r are permuted, Each r is now a positive integer <n.
TIk 12.14 Tayk' Fot'mula 
12.14 TAYLOIS FORMULA FOR FUNCTIONS FROM R  TO R 
Taylor's formula (Theorem 5.19) can be extended to real-valued funetionsfdefined
on subsets of R". In order to state the general theorem in a form which resembles
the one-dimensional case, we introduce special symbols
for certain sums that arise in Taytor's formula. These play the role of higher-
order directional derivatives, and they are defined as follows:
If x is a point in R" where all second-order partial derivatis off exist, and if
t -- (t,..., t,) is an arbitrary point in R', we write
f'(x; t): 1 "
also define
if all third-order partial derivatives exist at x.
defined if all ruth-order partials exist.
These stuns are analogous to the formula
i---1 j=l It-t
The symbol fl'l(x; t) /s aimilarly
f'(x; t) =
for the directional derivative of a function which is differentiable at x.
Theorem 12,14 (Ta¥1or's formul). Assume that fund all its partial derivatives of
order <m are differentiable at each point of an open set S in R . Ira andb are two
points of S such that L(a, b) =_ S, then there ia a point z on the line segment I.(a, b)
such that
' m-I I 1
f(b) -- f(a) t--t:  fO'(a; b - a) +  f°(z; b - a).
Proof. Since S is open, the is a 6 > 0 such that a + t(b - a) e S for all real
t in the interval -6 < t < 1 + 3. Define g on (-, 1 + 6) by the equation
= flu + tfb -
Then f(b) - f(a) = 9(1) - g(0). We will prove the theorem by applying the
one--dimensional Taylor formula to g, writing
Now g is a composite function given by g(t) = f[p(t)], where p(t) = • +t(b - •).
The kth component of p has derivative p[(t) = b  as. Applying the chain rule,

inltlvarildl Dllllmmtl Csulm
we see that g'(t) exists in the interval (- 6, 1 + 6) and is given by the formula
e'(t) =  Oif[_p(l)](b i - ai) = f'(p(t); b --
.r=!
Again applying the chain rule, we obtain
;=1 =1
Similarly, we find that ,(')(t) = .ta')(p(t); b - t). When these are used in (30)
we obtain the theorem, since the point z = a + 0(b - a)  L(a, b).
12.1 Let S M an OlXm subset of R', and let f: S  R  a l-valu function
finite iM vafi Dt .... O . Iffh a lo1 ximum or a lt nimum
at a Mt ¢ in S, pm tt Dff(e) = 0 f  k.
1 CakuMtc MI fitrr Rial deriti d the diional vati f'(x;
for h of t i-vM functions n on R =  follo:
. a) f(x) = a" x, whe a is a ed or  R =.
b) f(x) = IxP.
0 f(x) = x' x), where L : "  R  is a Inear run, ion.
d) f(x)=   a,x,x s, wh a,= a.
1 t f d 8  fiom with vatues in R  such that the ditional divativ
f'(e; e) d #(e; e)ist. ove thin the sum f + 8 and dot pr f. g have diionM
divativ gin by
(f + g)'(e; u) = f'(c; u) + g'(c; eJ
and
(f- g)'(c; u) = (c)- g'(c; u) + c. f'(c;
12M IfS  R =, let f: $  R =  a run, ion with valu in R , and ite f =(,... ,f).
 that f is diffetiab at an interior int c of S iL nd only iL ch  is diffemntiable
12-5 Gin n l-lucd functions ft ..... , h diffentiabic on an on intewal
(a, b) in R. F eh x = (x,..., xn) in t nMinsial o inleat
S = {(x ..... x.):a < x < b, k = 1,2,...,
nef(x) = f(x) + ... + (x,). o thatfis diffemntiable at ch int of S and
that
Exm:ises 3(d
12.6 Given n real.valued funclions fl .... , fs defined on an open set 3 in R". For each x
in S, define f(x) = fl(x) + --. + f(x). Assume that for each k = ], 2,..., n, the
following limit exists:
lira fa(Y) -- f{z)
Call this limit a(x) Prove that f is differentiate at x an that
f'(x)(u'l =  at,(x)u, ifu = (ul .... , u).
12.7 Let f and g be Functions from R  to R". Assume that f is differentiable at c, that
f(c) = 0, and that g is continuous at c. Let h(x)  g(x) - f(x). Prove that h is differen-
tiable at © and that
h'c)(u) -- g(e) • {f'(c)(u)} if u •
12-8 Let f: R z --, R 3 be defined by the equation
f(x, y) = (sin x cos y, sin x sin y, cos x cos y).
Determine the Jacobian matrix De(x,
12.9 Prove that there is no real-valued functionfsuch that f'(c; u) > 0 for a fixed point
¢ in R" and every nonzero vector  in R'. Give an exampte such lhst f'(e; a) > 0 for a
fixed direction a and every e in RL
12.10 Let f = u + iv be a complex-valued function such that the derivative f'(c) exists
for some complex c. Write z -- c + re " (where u is real and fixed) and lel r -- 0 in the
difference quotient [/-(z) - f{c)]/(z - c) to obtain
f'(c) = e-[d(c; a) + iv're;
where a = (cos 0 sin ), and u'(C; a} an6 d(c; a) are directional derivatives. Let b ---
(cos/L sin fl), where// =  + -I-n, and show by e similar argument that
f'(c) = e-t['(c; b) - iu'c; b)].
Deduce that u'(c; a) = dc; b) and d(c; a) = -d(¢; bL The Cauchy.-Riemann
tions (Theorem 5.22) are a special case.
Gradients aud the chain ruIe
12.11 Let f be real-valued and differentiable at a point c in R", and assume that
IlVf(e)lt  0. Prove that there is one and only one unit vector u in R" such that
If're; ")1 = IlVf{OII, and that this is the unit vector for which If'4e; u)[ has its maximum
value.
12.12 Compute the gradient vector Vf(x, y) at those points (x, y) in R e where it exists:
a) f(x, y) = x2y  log (x 2 + y2) if (x, y) # (0, 0), f(O, 0) = 0.
1
b) f(x, y) = xy sin xZ + y if (x, y) =/- (0, 0), /(0, 0) = 0.

Mlivartabk Diffmtlal CJdculm
12.13 Let fand # be reaJ.valued functions defined on g  with continuous second de.riva-
tires f" and ', Define
F(x, y) = fix + #(y)] for each (x, y) in R z.
Find formulas for all partials of For first and second order in terms of the derivatives of
land a'. Verify the relation
(DtFXD=.2F) :
17_1=1 Given a function f defined in R 2. Let
F(r, O) = f(r cos 0, r shl 8).
a) Assume appropriate differentiability properties of f and show that
DF(r, O) = cos 0 Dxf(x, y) + sin 0 D2f(x, y),
Dt.tF(r, 0) -- cos z ODj .if(x, Y) + 2 sin 0 cos ¢ D ,zf(x, ¥) + sin  OD.zf(x, y),
where x = r cos 0, y = r sin 0,
h) Find similar formulas for D2F, DLF, and
c) Verify the formula
t [D2F(r ' 0)]2 ,
IlVf(r cos 0, "sin 0) 2 : [DF(r, 0)]
17-15 If land g have gradient vectors Vf(x) and Vg(x) at a point x in R  show that the
product function h defined by k(x) = f(x)g(x) also has a gradient vector at x and that
vx) = f(x)va(x) + (x)vf(x).
State and prove a similar result for the quotient fig.
12,16 Let f be a function having a derivative f" at each point in R  and [et # be defined
on R  by the equation
g(x,y,z) = x ± +yz + z .
tf h denot¢ the composite function h = f • #, show that
12.1 Assumefis differentiable at each point (x, y) in R 2. Let g and #z be defined on
R 3 by the equation
g(x,y,z) = x 2 + y2 + z z, g2(x,y,z) = x + y + z,
and ]¢t g be the Vector-valued function whose values (in R 2) are given by
g(x, y, z) = (g,(x, y, z, gz(x, y. z)).
Let h he the composite function h = f g and show that
12.t8 Let fbe defined onan open set S in R n. We say thatfis homogeneous of degree p
over S iff(gx) = ,'f(x) for every real 2 and for every x in S for which .x e S. If such a
function is diffexentiable at x, show that
x. Vffx) = p]'(X).
OT. This is known as Euler's fheorem fo homogeneous functions. Hint. For fixed x,
define fl0.) = f(2x) and compute
Also prove the converse That is, show that if x. Vf(x) = pf(x) for all x in an open
set S, thenfmust be homogeneous of degree p over S.
Meam-Valee theorems
1/,.19 Let f: R  R 2 be defined by the equation f(t) = (cos t, sin t). Then f'(t)(u) =
u( sin t, cos t) for every real u. The Mean-Value formula
f(y) - f(x) = f'(zXy - x)
cannot hold when x = o, y = 2r, since the left member is zero and the right member is a
vector of tength 2n. Nevea'theless, Theorem 12.9 states that for every vector a in R  there
is a z in the interval (0, 2n) such that
a. {fly) - f(x)} : a. (f'(z)fy- x)).
Determine z in terms of a when x  0 and y = 2zt.
12.20 Let ]'be a real-valued function differentiable on a 2-ball B(x) By eonsdetirg the
function
prove that
where z= • L(x, y) and z z  L(xz, yz).
12i State and prove a generalization of the result in Exercise 12.20 for a real-valued
function differentiable on an n-ball
12.22 Letfhe real-valued and assume that the di'ectional derivativef'(e + tu; u) exists
for each t in the interval 0 _< t __. 1. Prove that for some  in the open interval
have
/(c + u -/(¢) =/'(c + Ou; u).
12.23 a) If/is real-vatued and if the directional derivative.f'(x', u) = 0 for every x in an
n-ball B() and every direction u, prove that fis constant on
b) What can you conclude about fiff'(x; u = 0 for a fixed direction [] and every
x in B(e)?
Derivatives of higher order and 'aylor's formula
12.24 For each of the following funelions, verify thal the mixed partial derivatives D .2f
and D2. j f are equal
a) f(x,y) = x 4" + y'* - 4xZy .
b) f(x, y) - log (x 2 + y2), (x, y) : (0, 0).
c) f(x, y) -- tan (x/y), y  O.

 Mitlvarible Dffmtial
ll. Let/be a function of two variables. Use induction and Theorem 12.13 to prove
that if the 2 k partial derivatives off of order k are continuous in a neighborhood of a point
(x, y), then all mixed partiMs of the form Dp, ..... ,fand Dp, ..... ,fwil! be equal at (x, y)
if the k-tuple (r I ..... r) contains the same number of ones as the k-tuple (Pl, • - • •/r).
1126 Iff is a function of two variable having continuous partials of order /c on some
open set S in R 2, show that
where in the rthtermwe havcp ..... p, = ! and,r+t ='2' =Pt  2. Use this
result to ive an alternarivg expression for Taylor's tormula (Theorem t2.14) in the case
when n -- 2. The symbol (,i) is the binomial eoe, ffteient k!/[r! (k - r)!].
127 Use Taylor's formula to express the following in powers of (x - t) and (y - 2):
a) f(x,y) = x 3 + y3 + xy, b) f(x,y) = x 2 + xy +
SUGGESTED REFERENCES FOR FURTHER STUDY
12.1 Apostol, T M. Calculus, Vol. 2, 2nd ed. Xerox, Waltham, 1969.
12.2 Chaundy, T. W., The Differential Calculus. Clarendon Press, Otford, 1935.
12.3 Wail, J. W., Functions of Several Variables. Harcourt Brace and World, New York,
-I
CHAPTER 13
IMPLICIT FUNCTIONS
AND EXTREMUM PROBLEMS
13,1 INTRODUCTION
This chapter consists of two principal parts. The first part discusses an important
theorem of analysis called the implicit function theorem; the second part treats
extremum problems. Both parts ue the theorems developed in Chapter 12.
The implicit function theorem in its simplest form deals with an equation of
the form
f(x, t) = 0. (1)
The problem is to decide whether this equation determines x as a function of t.
If so, we have
x =
for some function #. We say that  is defined "implicitly" by (1).
The problem assumes a more general form when we have a system of several
equations involving several variables and we ask whether we can solve these
equations for some of the variables in terrns of the remaining variables. This is
the same type of problem as above, except that x and t are replaced by vectors,
and f and e are replaced by vector-valued functions. Under rather general con-
ditions, a solution always exists. The implicit function theorem gives a description
of these conditions and some conclusions about lhe solution.
An important special case is the familiar problem in algebra of solving n linear
equations of the form
asx s . t (i = !, 2 ..... u), (2)
where the a and t are considered as given numbers and x, ..., x, represent
unknowns, in linear algebra it is shown that such a system has a unique solution
if, and only it', the determinant of the coefficient matrix A = [as ] is nonzero,
UOT. The determinant of a square matrix A -- [a] is denoted by det/1 or
det [as ]. if det [as ]  0, the solution of (2) can be obtained by Cramer's rule
which expresses each xt as a quotient of two determinants, y xt = AfD, where
D = det I'a,.] and A is the determinant of the matrix obtained by replacing the
kth column of [a,s] by tt,..., .. (For a proof of Cramer's rule, see Reference
13.1, Theorem 3.14.) In particular, if each tz = 0, then each xs --- 0.
367

Next we show that the system (2) can he written in the form (1). Each equation
in (2) has the form
f,x, t) = o
and
wherex= (xx,...,x,), t =(t,...,t,),
f(x, t) =  atxj - t.
Therefore the system in (2) can he expressed as one vector equation f(x, t) = 0,
where f = Cfl, ... • f). If Djf, denotes the partial derivative of 3 with respect to
thejth coordinate xl, then D2fi(x , t) = aj. Thus the coefficient matrix A = [a,j]
in (2) is a Jaeobian matrix. Linear algebra tells us that (2) has a unique solution if
the determinant of this Jacobian matrix is nonzero.
In the general implicit function theorem, the non-canishing of the determinant
of a Jaeobian matrix also plays a role. This comes about by approximating f by
a linear function. The equation f(x, t) = 0 gets replaced by a system of linear
equations whose coefficient matrix is the Jacobian matrix of f.
NOTATION. If f= (f,, ... ,.f,) and x = (xt,... ,x,), the lacobian matrix
Dr(x) = [Djfx)] is an n × n matrix. Its determinant is called a Jacobian
determinant and is denoted by 2"f(x). Thus,
The notation
Jr(x) = det Df(x) = det I]Dff(x).'].
is also used to denote the Jacobian determinant Jr(x).
The next theorem relates the Jacobian determinant of a complex-valuec[
function with its derivative.
Theorem 1J.1. If f - u + to is a coraplex-oalued funetion with a deritative at a
point z in C, then Jr(z) = if'(z)l z.
Proof We havef'(z) = Du + idly, so If'(z)l  = (Dtu)  + (Dv) z. Also,
d(z) = det [_Dt t Dvl -- Dxu Dzv  Oto Du = (Du) z + (D,v) z,
by the Cauchy-Riemann equations.
13.2 FUNCTIONS WrI'H NONZERO JACOBIAN DETERMINANT
This section gives some properties of functions with nonzero Jacobian determinant
at certain points. These results will be used later in the proof of the implicit function
theorem.
f
Theorem 1J,2. Let B = B(a; r) be an n-bali in R , let B denote its boundary,
s = {x fix - all = r},
and let B = B  3B denote its closure. Let f = (f, ,.., f.) be continuous on B,
and assume that all the partial derioatioes Df(x) exist if x  B. Assume further
that f(x) # f(a) if x  B and that the Jacobian determinant Jr(x) # 0 for each
x in B. Then f(B), the image of B under f, contains an n-ball with center at f(a).
Proof Define a real-valued function g on OB as follows
g(x) -- Uf(x) - f(a)ll if x s
Theng(x) > 0 for each x in Bbecause f(x) # fin) ifx  dB. Also, g is continuous
on ¢3B since f is continuous on B. Since 3B is compact, y takes on its absolute
minimum (call it m) somewhere on dB. Note that rn > 0 since fl is positive on
Let T denote the n-bait
We will prove that T  f(B) and this will prove the theorem. (See Fig. 13.1.)
To do this we show that y  T implies y  f(B). Choose a point y in T, keep
y fixed, and define a new real-valued function h on B as follows:
h(x) : Ill(x) - y if x e B.
Then h is continuous on the compact set B and hence attains its absolute minimum
on B. We wi|! show that h attains its minimum somewhere in the open n-ball B.
At the center we have h(a) = [If(a) - Yll < rn/2 since y c T. Hence the minimum
value of h in B must also be <m/2. But at each point x on the boundary dB we
h(x) = Ill(x) - yli = Ill(x) - f(a) - (y -
=> Ill(x) - f()ll - Ilk(a) - yll > 0(x) -  => m,
2 2
so the minimum ofh cannot occur on the boundary OB. Hence there is an interior
point ¢ in B at which h attains its minimum. At this point the square ofh also has

 Implit Fio  Ezmnm l,olm T 13.3
a minimum, Since
h2(x) = if(x) - Yll  =  f(x) - y]2,
r=l
and since each partial derivative D(h 2) must be zero at c, we must have
 [f(e) - y,]Dfr(e) = 0 for k -- 1, 2,..., n.
But this is a system of linear equations whose determinant Jr(e) is not zero, since
¢ • B. Therefore f(e) = y for each r, or fie) = y. That is, y f(B). Hencc
T _ f(B) and the proof is complete.
A function f: S  T from one metric space (S, ds) to another (T, dr) is
called an open mappiny if, for every open set A in S, the image f(A) is open in T.
The next theorem gives a sufficient condition for a mapping to carry open sets
onto oln sets. (See also Theorem 13.5.)
Theorem 13.J. Let A be an open subset of R" and assume that f : A - R" is con-
tinuous and has finite partial derivatives Djf on A. lf f is one-to-one on ,4 and if
Jr(x) :# O for each x in A, then f(A) is open.
Proof. If b • f(A), then b = f(a) for some a in A. There is an n-bait B(a; r) _ A
on which f satisfies the hypotheses of Theorem 13.2, so f(B) contains an n-ball
with center at b. Therefore, b is an interior point off(A), so f(A) is open.
The next theorem shows that a function with continuous partial derivatives is
locally one-to-one near a point where the Jacobian determinant does not vanish.
Theorem IL4. Assume that f--(f,.. ,f.) has continuous partial derivatives
Df¢ on an open set S in R*, and that the dacobian determinant Jr(a) q 0 for some
point a in S. Then there is an n-ball B(a} on which f is one-to-one.
Proof Let Za,...,Z, benpoints in Sandlet Z = (ZI;...;Z,) denote that
point in R "" whose first n components are the components of Z, whose next n
components are the components of Z2, and so on. Define a real-valued function
h as follows:
h(Z) = det [Df(Z,)].
This function is continuous at those points Z in R " where h(Z) is defined because
each Df is continuous on S and a determinant is a polynomial in its n z entries.
Let Z be the special point in R * obtained by putting
Z = Z2 ='" = Z= a.
Then h(Z) -- Jr(a) #= 0 and hence, by continuity, there is some n-ball B(a) such
that det [Ds, f(Zi) ] q* 0 if each Z  a(a). Wc will prove that f is one-to-one on
B(a).
"1"5. 13. Ncmz¢o Jeeeim Detertuinnt 371
Assume the contrary. That is, assume that f(x) = f(y)' for some pair of points
x :# y in B(a). Since B(a) is convex, the line segment L(x, y) _ B(a) and we can
apply the Mean-Value Theorem to each component of f to write
0 = f(y) - f(x) = Vf(Z) "(y - x) for i = I, 2 ..... n,
where each El •/.,(x, y) and hence Z e B(a). (The Mean-Value Theorem is
applicable because f is differentiable on S.) But this is a system of linear equations
of the form
-(y - x)a = 0 with a,  D(Z;).
The determinant of this system is not zero, since Z  B(a). Hence y - x = 0
for each k, and this contradicts the assumption that x # y. We have shown,
therefore , that x # y implies f(x)  f(y) and hence that f is one-to-one on
crr. The reader should be cautioned that Theorem 13'.4 is a local theorem and
not a global theorem. The nonvanishing of Jr(a) guarantees that f is one-to-one
on a neighborhood of a. It does not follow that f is one-to-one on S even when
df(x) :/: 0 for every x in S. The following example illustrates this point. Letfbe
the complex-valued function defined byf(z) = g ifz • C. Ifz = x + iy we have
Jy(z) = If'(z)l 2 = tel 2 - e 2.
Thus dz) q: 0 for every z in C. However, f is not one-to-one on C because
f(z) = f(z) for every pair of points z and z which differ by 2hi.
The next theorem gives a global property of functions with nonzero Jacobian.
determinant.
Tleorem 13.5. Let A be an open subset of R* and assume that f :  , R" has
continuous partial den'tratives Df on A. If Jr(x) :# 0 for all x in A, then f is an
open mapping.
Proof. Let S be any open subset of A. If x  S there is an n-ball B(x) in which f
is one-to-one (by Theorem 13.4). Therefore, by Theorem 13.3, the image f(B(x))
is open in R". But we can write S = 3s B(x). Applying f we find
[3,s f(B(x)), so f{S) is open
tOT., ifa function f = f, ... ,f) has continuous partial derivatives on a set S,
we say that f is continuously differentiable on S, and we write f e C' on S. In view
of Theorem 12.11, continuous differentiabitity at a point implies differentiability
at that point.
Theorem 13.4 shows that a continuously differentiable function with a non-
vanishing Jacobian at a point a has a local inverse in a neighborhood of a. The
next theorem gives some local differentiabihty properties of this local inverse
function.

13.3 THE  VUNCTION THEOREM
leorem 13.6. Assume f = ,... ,)  C" on  op et S in R ,  let
T = f(. If the Jobi termnt Jr(a) # O for some nt a in S then here
are two o sets X  S d Y  T  a ily determed fution g ch tt
a) a  X  f(a)  Y,
c) f  ontne on X,
Proof. c non J¢ is continuous on $ and, sin J(a) # O, there is an n-ball
B(a) such at J(x) # 0 fox all x in Bt(a). By ¢orem 13A, there is an n-ball
B(a)  B(a) on wch f is onton¢ t B  an n-ll  nr at a and
dius smaller an at ofa). cn, by com 13.2, B) nmins an n-!
with  at f(a). note s by Y and let X : f-x(  B. en X is on
sin th ff(Y) and B arc on. (See Fig. 13)
f
lnre 13,2
The set B (the closure of B) is compact and f is one-to-one and continuous on
B. Hence, by Theorem 4.29,. there exists a function g (the inverse function f- t of
Theorem 4.29) defined on f(/) such that g[f(x)] = x for all x in J. Moreover, g
is continuous on f(B). Since X _= B and Y  f(B), this proves parts (a), (b), (c)
and (d). The uniqueness of g follows from (d).
Next we prove (e). For this purpose, define a real-valued function h by the
equation h(Z): det [D,f(Z)], where Zt,... , Z are n points in S, and
Z = (Zt ; • • • ; Z,) is the corresponding point in R "2. Then, arguing as in the proof
of Theorem 13.4, there is an n-ball Bz(a) such that h(Z) # 0 if each Z,. • B2(a ).
We can now assume that, in the earlier part of the proof, the n-ball B(a) was chosen
so that B(a) G B2(a ). ThenB _c Ba(a) and h(Z) # 0if each Z• B.
To prove (e), write g = (gl, -- -, g,). We will show that each , • C' on Y.
To prove that D, gt exists on Y, assume y • Y and consider the difference quotient
['',(y + ttb) - gt(y)]/t, where u, is the rth unit coordinate vector. (Since Y is
373
open, y + t  Y if t is sufficiently small.) Let x = g(y) and let x' = g(y + tu).
Then both x and x' are in X and f(x') - f(x) -- in,. Hence f,(x') - f(x) is 0 if
i =# r, and is t if i = r. By the Mean-Value Theorem we have
fx') -/,(x) _ Vf,(Z) • x' - x for i = 1, 2,..., n,
! t
where each Z, is on the line segment joining x and x'; hence Zi e B. The expression
on the left is 1 or 0, according to whether i = r or i • r. This is a system of n
linear equations in n unknowns (. - xt)]t and has a unique solution, since
det [Ojf,(Zb] = • 0,
Solving for the kth unknown by Cromer's rule, we obtain an expression for
[ai(Y + t) - gt(y)]/t as a quotient ofdetexminants. As t - 0, the point x - x,
since g is continuous, and hence ¢ach Z - x, since Zt is on the segment joining
x to x'. The detcrmiuant which appears in the denominator has for its limit the
number det [Dff(x)] = d(x), and this is nonzero, since x e X. - Therefore, the
following limit exists:
lira fAY + m,) - e,(y) =
t--'O
This establishes the existence of D,e(Y) for each  in Y and each r = I, 2,..., n.
Moreover, this limit is a quotient of two determinants involving the derivatives
Dff(x). Continuity of the Dft impfies continuity of each partial DWt. This
completes the proof of (¢).
Noax The foregoing proof also provides a method for computing Oddt(y ). In
practice, the derivatives D,, can be obtained more easily (without recourse to a
limiting process) by using the fact that, if y = f(x), the product of the two Jacobian
matrices Dr(x) and Dg(y) is the identity matrix. When this is written out in detail
it gives the following system of n z equations:
D,{)D() -- if i  j.
Fo each fied i, we obtain  linear equations as ] rims through the values
l, l .... , ÷ These n tn be sol"t for t
by Crame's rle, or by e oth ethod.
13.4 'rile IMPLICIT FUNCTION THEOREM
The reader knows that the equation of a curve in the xy-plane can be expressed
either in an "'explicit" form, such as ), -- f(x), or in an "implicit" form, such as
F(x, y) = 0. However, if we are given an equation of the form F(x, y) = O, tiffs
does not nccesmrily represent a function. (Take, for example, x
The equation F(x, y)  0 does always represent a relation, namely, that set of all

pairs (x, y) which satisfy the equation. Thd following question therefore presents
itself quite naturally: When is the relation defined by F(x, y) ffi 0 also a function?
In other words, when can the equation F(x, y) ---- 0 he solved explic/tly for y in
terms of x, yielding a unique solution? The implicit function theorem deals with
this question loealty. It tells us that, give a point (x0, Yo) such that F(xo, Yo) = 0,
under certain conditions there will be.a neighborhood of (Xo, Yo) such that in th/s
neighborhood the relation defined by F(x, y) - 0 is also a function. The conditions
are that F and DF he continuous in some neighborhocai of (x0, Yo) and that
DzF(xo, Yo)  O. In its more general form, the theorem treats, instead of
equat/on in two varlables, a system of n equations in n + k variables:
f,(xt,... , x,; tl,... , t,) ---- 0 (r ---- 1, 2,..., n).
This system can be solved for xt,..., x, in terms of tl .... , ts, provided that
certain partial derivatives are continuous and provided that the n x n Jacobian
determinant d(f,... ,f,)/d(xt ..... x,) is not zero.
For brevity, we shall adopt the following notation in this theorem: Points in
(n + k)-dimensional space R "+t will be written in the form (x; t), where
x =-- (x,, ...,x,) • R" and t = (t, .... , t)  R t.
Tkeaeem 13.7 (Implicit fimetiatt tit•or•m). Let f = (ft, • .-, f,) be a vector-valued
functgon defined on an open set S in R +  with values gn R . Suppose f • C' on
Let (xo; to) be a point in S for winch f(xo; to) = O and for which then x ndetermi-
nant dot [-Dft(xo; to)] # 0. Then there exists a k-dimemional open set T O con-
tainim d t o and one, and only one, vector-valued function g, defined on T o and having
values in R', such that
a) g e C' on To,
b)
c) f(g(t); t) -- OforeveeytinTo.
Proof. We shall apply the inverm function theorem to a certain vector-valued
function F = (F, ..., F,; F,÷,..., F+ defined on S and having values in
R +. The function F is defined as follows: For I < m  n, let F,,(x; t) - f,,(x; t),
and for 1 < m < k, let F,+,(x; t) = t,,. We can then write F = (f; I), where
f ffi (f,, ..., f) and where I is the identity function defined by l(t) ffi t for each t
in R . The Jacobian Jr(x; t) then has the same value as the n × n determinant
dot [Dff(x; t)] because the terms which appear in the last k rows and also in the
hast k columns of Js(x; t) form a k x k determinant with ones along the main
diagonal and zeros elsewhere; the intersection of the first n rows and n columns
consists of the determinant dot [Dff(x; t)'], and
.DF,+#(x;t)= 0 for! <i< n, I <j.<;k.
Hence the Jacobian dr(x; to) ¢ O. Also, F(xo; to) = (0; to). Therefore, by
Theorem 13.6, there exist open sets X and Y containing (x,; to) and (0; to),
respectively, such that F is one-to-one on X, and X" = F-(Y). Also, there exists
 of RcaI-Valal
a local inverse funetion G, defined on Y and having values in X', such that
t)] : (x; 0,
and such that G • C' on Y.
Now G can be reduced to components as follows: G ffi (v; w) where
v = (v, .... v.) is a vector-valued function defined on Y with values in R' and
w - (wt ..... w) is also defined on Y but has values in 11 , We can now determine
v and w explicitly. The equation G[F(x; t)] = (x; t), when written in terms of the
components v and w, gives us the two equations
v[F(x; t)] ffi x and w[F(x; t)] = t
But now, eveeypoint (x; t) in Ycan be written uniquely in the form (x; t) -- F(x';
for some (x'; t') in X, because F is one-to-one on X and the inverse image F- (Y)
contains X'. Furthermore, by "the manner in which F was defined, when we write
(x; t) --= F(x'; t'), we must have t' -- t. Therefore,
v(x; t) = v[F(x'; t)-[ = x' and w(x; t) - w[F(x'; t)] = t.
Hence the function G can he described as fotlows: Given a point (x; t) in Y, we
have G(x; t) -- (x'; t), where x' is that point in R" such that (x; t) = F(x';
This statement implies that
Fly(x; t); t] -- (x; t) for every (x; t) in Y.
Now we are ready to define the set T o and the function g in the theorem. Let
7"0 = {t:teR , (0;t) Y},
and for each t in T O define g(t) = v(0; t). The set To is open in R *. Moreover,
g • C' on T O because G • C' on Y and the components of g are taken from the
components of G. Also,
g(to) ffi v(O; to) = x o
because (0; to) = F(xo; to). Finally, the equation FI-v(x; t); t] = (x; t), which
holds for every (x; t) in Y, yields (by considering the components in R*) the
eouation t'l-(x; t); t] = x. Taking x = 0, we see that for every t in To, we have
fl'g(t); t] ffi 0, and this completes the proof of statements (a), (b), and (c). It
remains to prove that there is only one such function g. But this follows at ooce
from the one-to-one character of f. If there were another function, say h, which
satisfied (c), then we would have fig(t); t] = fib(t); I], and this would imply
(g(t); t) = (h(t); t), or g(t) ffi h(t) for every t in T 0.
13.5 EXTREMA OF REAL-VALUED FUNCTIONS OF ONE VARIABLE
In the remainder of this chapter we shall consider reabvalued functions f with a
view toward determining those points (if any) at which f has a local extremum,
that is, either a local maximum or a local minimum.

We have already obtained one realt in this connection for functions of one
variable/Theorem 5.9). In that theorem we found that a necessary condition for
function f to have a ]oca! extremum at an interior point c of an interval is that
if(c) -- 0, provided thatf'(c) exists. This condition, however, is not st, as
we can see by takingf(x) = x a, c = 0. We now derive a sufficient condition.
kcorem l&& For some integer n = 1,/elf have a continuous nth derivative in the
open interval (a, b). Suppose q[so that for some interior point c in (a, b) we have
f'(e) = if(c) ..... = o, but o.
Then for n en, f ha a local min,num at c if *)(e) > O, and a local maximum at
c if fear(c) < O. If n is odd, there is neither a local maximum nor a local minimum
Proof Sincef')(c) # 0, there exists an/ntcrval B(c) such that for every x in B(e),
the derivative J''(x) will have the same sign as f'(c). Now by Taylor's formula
(Theorem 5.19), for every x in B(c) we lmve
f(x) - f(c) = (x - c)', where xl •
If n is even, this equation implies f(x) > f(c) when f'(c) > 0, and f(x)  f(c)
when f(e)  0. If n is odd and f*)(c) > 0, then f(x) > f(c) when x > c, but
f(x) < f(c) when x < c, and there can be no extremum at c, A similar statement
holds ifn is odd andf*)(c) < 0. This-proves the theorem.
I3.6 EXTREMA OF REAL-VALUED FUNC'rlONS OF SEVERAL VARIABLES
We turn now to functions of several variables: Exercise 12.1 gives a necessary
condition for a function to have a local maximum or a local minimum at an interior
point a of an open set. The condition is that each partial derivative Dtf(a) must
he zero at that point. We can also state this in terms of directional derivatives by
saying that if(a; u) must be zero for every direction e.
The converse of this statement is not true, however. Consider the following
example of a function of two real variables:
f(x, y) = (y - x  )(y - 2z2).
Here we have Df(O, O) = Dzf(O , 0) ---- 0. Now f(0, 0) = 0, but the function
assumes both positive and negative values in every neighborhood of (0 0), so
there is neither a local maximum nor a local minimum at (0 0). (See Fig. 13.3.)
This example illustrates another interesting phenomenon, if we take a fixl
straight tine through the origin and restrict the point (x, y) to move along this line
toward (0, 0), then the point will finally enter the region above the parabola
y - 2x 2 (or helow the parabola y = x 2) in which f(x, y) becomes and stays
positive for every (x, y) # (0, 0). Therefore, along every such line,f has a minimum
at (0, 0), but the or/gin is not a local minimum in any two-dimensional neighbor-
hood of (0,
" unshaded r,ions
Ftgttre 13.3
Defudtw"  13,9 If f is differentiabfe at a and if Vf(a) = 0, he point a is called a
stationary point off. A stationary point fs called a saddle point if every n-ball B(a)
contains points x such that f(x) > f(a) and other points such that f(x) < f(a).
In the foregoing example, the origin is a saddle point of the function.
To determine whether a function of n variables has a local maximum, a local
minimum, or a saddle point at a stationary point a, we mut determine the algebraic
sign off(x) - f(a) for all x in a neighborhood of a. As in the one-dimensional
case, this is done with the help of Taylor's formula (Theorem 12.14). Take m = 2
and y = a + t in Theorem 12.14. if the partial derivatives of fare differenfiable
on an n-ball B(a) then
f(a + t) -f(a) = Vf(a)" t + ½ff(z t), (3)
where z lies on the line segment joining a and a + t, and
if(z; t)=
At a stationary point we have Vf(a) --- 0 so (3) becomes
f(a + t) -f(a) = ½if(z; t).
Therefore, as a + t ranges over B(a), the algebraic sign of f(a + t) -f(a) is
determined by that off"z; t). We can write (3) in the form
f(a + t) -f(a) -- ½if(a; t) + IltllZE(t), (4)
where
The inequality
Iltll2E(0 = -  ff(z; t) - J  f"(a; t).
shows that E(t) - 0 as t  0 if lhe second-order partial derivatives of f are
continuous at a. Since ]lt]lZE(t) tends to zero faster han I]Itl , it seems reasonable
to expect that the algebraic sign off(a + 1) - f(a) should be determined by that
of f "(a; 0. This is what is proved in the next theorem.

3TR Implicit Functions and Extrmnn Problems T'n. 13.10
lorem 13.10 (Second-derivative test for extrema). Assume that the second-order
partial derivatives D r,jf exixt in an n-ball B(a) and are continuous at a, where a is a
stationary point off. Let
q(t) = t) = D,,;f(*)t,tj.
i=l
a) lf Q(t) > O for all t  O, f has a relative minimum at a.
b) If Q(O < 0 for all t e O, f has a relative maximum at a.
c) If Q(t) takes both positive and negative values, then f has a saddle point at a.
Proof The function Q is continuous at each point t in RL Let S -- {t : Iltll -- 1}
denote the boundary of the n-ball B(0; I). If Q(t) > 0 for all t  0, then Q(t) is
positive on S. Since S is compact, Q has a minimum on S (call it m), and m > 0.
Now Q(ct) =- c2Q(t) for every real c. Taking c = 1/llt]l where t # 0 we see that
de Sand hence cZQ(t) > m, so Q(t) >_ mlltlJ . Using this in (4) we find
f(a + t) -f(a) = Q(t) + IltllE(t) _> m Iltll  + I[t[IZE(t).
Since E(0  0 as t  0, there is a positive number r such that IE(t)[ < ½m
whenever 0 < Iltll < r. For such t we have 0 < Iltll z IE(t)l < ½mlltll , 
f(a + t) -f(a) > ml[tll  - ½mlltll  = mlltll  > 0.
Therefore f has a relative minimum at a, which proves (a). To prove (b) we use a
similar argument, or simply apply part (a) to fi
Finally, we prove (c). For each 2 > 0 we have, from (4),
f(a + 2t) -f(a) = QQ.t) + ).lltll2E(;t) = A{Q(t) +
Suppose Q(|) ¢ 0 for some t. Since E(y)  0 as y --, 0, tttere is a positive r such
that
Ittll  E(,  t) < ½1Q(01 if0 < a < r.
Therefore, for each such 2 the quantity 2{Q(I) + IltllE(2t)} has the same sign as
Q(t). Therefore, if0 < 2 < r, the differencef(a + 2t) -f(a) has the same sign
as Q(0. Hence, if Q(t) takes both positive and negative values, it follows that f
has a saddle point at #.
rffr. A real-valued function Q defined on R" by an equation of the type
iffiL j=l
where x = (x,..., x.) and the a o are real is called a quadratic form. The form is
called symmetric if ai = a: for all i and j, psitive definite if x  0 implies
Q(x) > 0, and negtnive definite if x # 0 implies Q(x) < 0.
In general, it is not easy to determine whether a quadratic form is positive or
negative definite. One criterion, involving eigervaIues, is described in Reference
"I'll, 1].11
£xtrcm of Fm-'tkm of Sevet'al Variabi
13.1, Theorem 9.5. Another, involving derminants, can be described as follows.
Let A = det[ao- ] and let A denote the determinant of the k × k matrix obtained
by deleting the last (n - k) rows and columns of [a.]. Also, put A o -=- 1. From
the theory of quadratic forms it is known that a necessary and sufficient condition
for a symmetric form to be positive definite is that the n + 1 numbers
A0, A, ..., A be positive. The form is negative definite if, and only if, the same
n + t numbers are alternately positive and negative. (See Reference 13.2, pp,
304-308.) The quadratic form which appears in (5) is symmetric because the
mix¢l partials Dijf(a ) and D#.(a) are equal. Therefore, under the conditions of
Theorem 13.10, we see thatf has a local minimum at • if the (n + 1)numbers
A 0, A,..., A are all positive, and a local maximum if these numbers are
alternately positive and negative. The case n = 2 can be handled directly and gives
the following criterion.
Tleorem 13all. Let f be a reat-valued function ith continuous second-order partial
deriuatipes at a stationary point a in R . Let
 -: D.if(), B = Dt.f(a), C -- D,f(a),
and let
Then we have:
a) IrA > 0 and A > O, f has a relative minimum at a.
b) If A > 0 and A < O, f has a relative maximum at a.
c) If A < O, f has a saddle point at a.
Proofi In the two-dimensional case we can write the quadratic form in (5) as
follows:
Q(x, y) = ½{Ax  + 2axy + Cy}.
If A ¢ 0, this can also be written as
1 {fAx + By z + Aya}.
If A > 0, the expression in brackets is the sum of two squares, so Q(x, y) has the
same sign as /. Therefore, statements (a) and (b) follow at once from parts (a)
and (b) of Theorem 13.10.
If  < 0, the quadratic form is the product of two Inear factors. Therefore,
the set of points (x, y) such that Q(x, y) -- 0 consists of two lines in the xy-plane
intersecting at (0, 0). These lines divide the plane into four regions; Q(x, y) is
positive in two of these regions and negative in the other two. Therefore fhas a
saddle point at a.
NOT. If A ---- O, there may be a local maximum, a local minimum, or a saddle
point at a.

13.7  PROll].JS  SIDE CONDITIONS
Consider the following type of extremum problem. Suplse that
represent the temperature at the point (x, y, z) in space and we ask for the maxi-
mum or minimum value of the temperature on a certain surface. If the equation of
the surface is given expficitly in the form  = h(x, y), then in the expression
f(x, y, z) we can replace z by h(x, y) to obtain the temperature on the surface as a
function of x and y alone, say F(x, y) = fix, y, h(x, y)]. The problem is then
reduced to finding the extreme values ofF. However, in practice, certain ditficulties
arise. The equation of the surface might be given in an implicit form, say
.q(x, y, z) = 0, and it may be impossible, in practice, to solve this equation
explicitly for z in terms of x and y, or even for x or y in terms of the remaining
vadable. The problem might be further complicated by asking for the extreme
values of the temperature at those points which lie on a given curt,e in space. Such
a eure is the intersection of two surfaces, say 9,1(x , y, z) = 0 and ,(x, y, z)
If we could solve these two equations simultaneously, say for x and y in terms of z,
then we could introduce these expressions into f and obtain a new function of
z alone, whose extrema we would then seek. In general, however, this procedure
cannot be carded out and a more practicable method must be sought. A very
elegant and useful method for attacking such problems was developed by Lagrange.
Lagraage's method provides a necessary condition for an extremum and can be
described as follows. Letf(xl, ..., x) be an expression whose extreme values are
sought when the variables are restricted by a certain number of side conditions,
say #1(xl,-.-, x,)= 0,. ..... g,(x ..... x,)---0. We then form the linear
combination
(xl ..... x.) = f(x, .... , x.) + al(xl, .-., x.) + --- + z.a.(x,,..., x.),
where 21, ..., 2, are m constants. We thendifferentiate b with respect to each
coordinate and consider the following system of n + m uations:
O,eb(xl ..... x.) = O, r = I, 2,..., n,
t(x,...,x,) =- 0, k = l, 2,..., m.
Lagrange discovered that if the point (xx, ..., x,) is a solution of the extremum
problem, then it will also satisfy this system of n + m equations. In practice, one
attempts to solve this system for the n + m "unknowns," 2, ..... 2,,, and
x,..., x,. The points (x,..., x,) so obtained must then be tested to determine
whether they yield a maximum, a minimum, or neither. The numbers 21,
which are introduced only to help solve the system for x .... , x, are known as
l_range's multipliers. One multiplier is introduced for each side condition.
A complicated analytic criterion exists for distinguishing between maxima and
minima in such problems. (See, for example, Reference 13.3.) However, this
criterion is not very useful in practice and in any particular prolem it is usually
easier to rely on some other means (for example, physical or geometrical consider-
ations) to make this distinction.
The following theorem establishes the validity of Lagrange's method:
T&,ore.m 13d2. Let f be a real.valued function such that f • C' on an open set S
in R'. Let gl, -.-, , be rn reat-alued functions such that g = (gl, • .., g,)  C'
n S, and assume that rn < n. Let Xo be that subset of Son which g vanishes, that s,
Xo = {x" x S, g(x) =
Assume that xo e X o and assume that there exists an n-ball B(xo) such that f(x) <
f(x) for all x in X o c B(xe) or such that f(x) > f(xe) for all x in X c B(xo).
Assume also that the m-rowed determinant (let [D,g(Xo)]  0. Then there exist
ra real numbers 21, ....  such that the following n equations are satisfied:
rOT. The n equations in (6) are equivalent to the following vector equation:
W(xo) + ,q vax(Xo) + '" + = 0.
Proof. Consi(ler the following system of rn linear equations in the m unknowns
• ...,
This system has a unique solution since, by hypothesis, the determinant of the
system is not zero. Therefore, the first m equations in (6) are satisfied. We must
now verify that for this choice of 21, --., 2,, the remaining n - m equations in
(6) are also satisfied.
To do this, we apply the implicit function theorem, Since rn < n, every point
x in S can be Written in the form x = (x'; t), say. where x' e R'* and t ¢ R*-"
In the remainder of this proof we will write x' for (xl,.-., x,) and t for
(x,,+t .... , x), so that t--x,,÷. In terms of the vector-valued function
g = (,,...., .q®), we can now write
g(xh; to) = 0 if x0 = (x; to).
Since g  C' on S, and since the determinant det ['Dy(x; to)] # O, all the
conditions of the implicit function theorem_ are satisfied. Therefore, there exists
an (n- m)-dimensional neighborhood T o of t o and a unique vector-valued
function h = (hi,..., h,,), defined on T o and having values irt R" such that
h • C' on To, h(to) = x,, and for every t in To, we have g[h(t); t] = 0. This
amounts to saying that the system of m equations
t(xl, . . . , x,) = 0,..., g,,(x ..... x) =- O,
can be solved for x,, .. -, x, in terms ofxs,+t, -.. , x,, giving the solutions in the
form x -- h,(x,÷,,..., x,), r = 1, 2,,.., m. We shall now substitute these
expressions for x,,..., Xm into the expressiort f(x t, .., xn) and also into each

expression O,(xl, ..., xO. That is to say, we define a new function Fas follows:
F(x.÷l,..., x.) ffi J-[  il(x.÷l ..... x.) ..... ..., x.): x.÷, .... , x.];
and we define m new functions G ..... G, as follows:
..... x.) = a,[hl(x.÷,, . . .. xO ..... h.(x.÷,, . . . , x.); x.,1 ..... x.].
More briefly, we can write F(0 = f[H(t)] and G(t) = gv[H(t)], where H(t) =
(k(0; t). Here t is restricted to lie in the set To.
Each function G 0 so defined is identically zero on the ¢t To by the implicit
function theorem. Therefore, each derivative D, Gp is also identically zero on To
and, in particular, D,G(to) - O. But by the chain rule (Eq. 12.20), we can com-
pute these derivatives as follows:
(r = i,2 .... ,n-m).
But H(t) = h(t)if 1  k ._< m, and H(t) = xt if m + I < k < n. Therefore,
when m + 1 < k < n, we have D/(t) -= 0 if m + r # k and D/,+,(t) = 1
for every t. Hence the above set of equations becomes
-- = l, 2, ,, - m. (7)
By continuity of It, there is an (n - rn)-balt B(%) _ To such that t  B(to)
implies Ca(0; t)  B(xe), where B(xo) is the n-ball in the statement of the theorem.
Hence,  • B(t0) implies (h(t); t) • Xo c B(xo) and therefore, by hypothesis, we
have either F(t) < F(_to) for all t in BOo) or else we have F(t) > F(to) for all I in
B(t0). That is, Fhas a local maximum or a local minimum at the interior point 1o.
Each partial derivative DeU(to) must therefore be zero. If we ue the chain rule to
compute these derivatives, we find
D,F(to) =  Df(xo)OrH(lo) . (r -- 1 .... , n  m),
k---[
and hence we can write
(r = 1,..., n - m). (8)
If we now multiply (7) by 2v, sum or p, and add the result to (8), we find
k--I .el p=l
for r = I .... , n - m. In the sum over k, the expression in square brackets
vanishes because of the way 21, ..., 2= were defined. Thus we are left with
D,+,f(xo) +  AD,+,#(xo) = 0 (r = 1, 2
and these are exactly the equations needed to complete the proof.
UOT. In attempting the solution of a particular extremum problem by Lagrange's
method, it is usually very easy to determine the system of equations (6) but, in
general, it is not a simple matter to actually solve the system. Special devices can
often be employed to obtain the extreme values offdirectly from (6) without first
finding the particular points where these extremes are taken on. The following
example illustrates some of these devices:
Example. A quadrlc surface with center at the origin has the equation
Ax  + ly z + Cz z + 2Dyz + 2Ezx + 2Fxy ffi 1.
Find the lengths of it semi-axes.
Solution. Let us write (xl, xz, xa) instead of (x, y, z), and introduce the quadratic form
J--I '--1
where x = (xl, x2, x3) and the a o = a are chosen so that the equation of the surface
becomes q(x) = 1. (Hence the quadratic form is symmetric and positive definite.) The
problem is equivalent to finding the extreme alues of f(x) --
subject to the side condition #(x) -- 0, where#(x) -- q(x) - 1. Using Lagrange's mehod,
we introduce one multiplier and consider the vector equation
(since V# -- Vq). In this particular case, both f and  are homogeneous ['unctions of
degree 2 and we can apply Euter's theorem (see Exerci 12.18) in (10) to obtain
x- Vf(x) + .J.x- V,(x) = 2/(x) + 2M(x) = O.
Since q(x) -- I on the stJrface we find 2 -- -f(x), and (0) becomes
where t -- lff(x). (We cannot bawl(x) - 0 in this problem.) The vector tlUation ( I)
then lead to the following three equations for x,, x2,
(al| -- l)x I ÷ al2X 2 +
Since x : 0 cannot yield a solution to our problem, the determinant of this system must

13.1 Let f be the complex-valued function defined for each complex z # 0 by the
equationf(z) = liP. Show that l(z) = -[zl-*. Show that fis one-to-one and compute
f- J explicitly.
13.1 Lt f = (f.f2,f3) be the vector-valuod run.ion defined (for  point (x;, x. xa)
in 113 for which xt + x2 + x # - 1) as follows:
A(xt, x2, x) = x (k = 1, 2, 3).
i + x t + x z +
Show that dt(xi, xz, x0 = (1 + xj + xa + x)-'. Show that f is oae-to-or and
comput f-  ©xplicifly.
13.3 Let f = (fx,..., f,) be a ,aaar-valucd ftmctiotl dcfinod in R", suppose f
on R =, aad let di(x) dtmot the Jacobian domninaat. Lea at ..... g, be n real-valued
fimctions 6aed on R l and having continuous d-ivativs gi,..., g=. Let /(x) =
f[at(xt),..., g,(x)], k = 1, 2, .... n. and put h = (h, ..... h). Show that
13.4 a) If x(r, 0) = r cos 0, r, 0) = t sin 0, show that
O(r. 0)
b) If xr, 0, #,) = r cos 0 sin t, r 8, ,) = r sin 0 sia , z -- r cos ,, show that
0(x,y,z)_ r :sine.
]t3.$ a) State oanditiom on f and g which will ensur that the equations x
y = g(u, v) can be solved for u and o in a righborhood of(xa, Yo). If the
tioas aro  = F(x y), v = G(x, y). and if d --  g)/(u, e), show that
OF 10g OF t f oG 10a G i f
b) Compute J and the partial derivative of F and G at (Xo, Yo) : (l, !) when
f(u. v) = u  - 2, a(u, ) = 2av.
13.6 L,t fond g lag related as in Theorem 13.6. Consider the cas n = 3 and show that
we have
Jt(X)Dlg,(Y) = ;,2 D2fz(x) D2fa(x) (i = i, 2, 3),
where y = f(x) and 6t. = 0 or 1 according as i # j or i  j. Use this to deduce the
formula
There are similar expressions for the other eight derivatives
13.7 Let[-- u + i be a complex-valued function satisfying the following conditions:
u  C" and v • C'on the open disk A = { : Izk < i };fis continuou on the closed disk
. -- {z: lgl <_ 1); u(x,v) -- x and o(x,y) = y whenever x z + yZ  1; the Jacobian
Jr(z) > 0 ifz 6 A. I.¢tB = f(A) chmote the imageof A under fond prove that:
a) -If Xis an open subset of A, tbenf(X) is an open sul:tset orB.
b) B is an open disk of radius I.
c) For each point u o + i o in B, there is only a finite number of points z in A such
thatf(z) -- u o +
Extremum problems
13.8 Find and classify theextremevalues (if any) of the functions defined by the following
equations:
a) f(x,,) = y + x , + x 4,
b) f(x,y) ffi x  + y2 + x + y +
d) [(x, y) = yz _ x .
13.9 Find the shortest distance from the point (0, b) on the y-axis to the parabola
x z  4y = O. Solve this problem using Lagrange's method and also without using
Lagrange's method.
13,10 Solve the following geometric problems by Lagrange's method:
a) Find the shortest distance from the point (a, az as) in R a to the plane whose
equation is btx + bzx  + bsxa + bo = O.
b) Find the point on he line of intersection of the two planes
axt + a2x  + aa.vs + ao = 0
and
bx + bx2 + baxz + bo = 0
which is nearest the origin.

13.11 Find the maximum value of I,,-i a,l, if,'=l x, - !, by us
a) the uy-hwa inlity.
b) 's th.
13.1Z Find Ihc maximum of (xx • -- x)  d the rlriion
x + .-- + x 
the result to dive tl folIowing inequalhy, valid for positiv real numbers al,. •., an:
13.13 fir(x) = x + -,. + x,, x = (xt ..... x), show thai a local extreme off, subject
to the condition x l + o,, + Xa = a, is an l-t.
13.14 Show that all poinls (xl, x2, xa, x) whew x + x has a local extremum subjec
to the two sidg conditions x + x + x = 4, x] + 2xa  + 3x[ -- 9, are found among
(0,0, _+4, _+), (0, ±1, +2,0), (_+,o,o, +_x/), (_+2, _+,0, o).
Which of the-so yield a local maximum and which yiotd a local minimum? Giv© reasons
for your conclusions.
13.15 Show that the ¢xtrgmc values off(x, xz, xz) = x + x + x], subject to the two
side conditions
and
bxx + bzx 2 + b3g 3 ffi O, (hi, b2, b3)  (0, O, 0),
arc t[ , tz'-z, whr¢ tt and t are the roots of th equation
Show that this is a quadratic equation in t and give a geometric argument to explain why
the roots q, tz are real and positive,
13.16 L¢1 A = det [xt] and I©t Xt  (x,..., x.). A famous theorem of Hadamard
states that [&l -< d,---d,, if d ..... d arc n p0sitiv¢ constants such that IIx, l? -- d/
(i = 1, 2,..., a). Prove this by treating A as a function of #z variables subject to n
constraints, using Lagrange's method to show that, when A has an extrcm¢ und©r these
conditions w mst hav¢
A 2 _
SUGGESTF REFERENCES FOR FURTHER STUDY
13.1 Aiaostol, T, M., Calculus, Vol. 2, 2nd ¢d. Xe¢ox, Waltham, 1969.
13.2 Gantmacher, F. R., The Theory of Matrices, Vol. 1. K. A. Hirsch, traaslator.
Chelsea, New York, 1959.
13.3 Hanecw. H., Theory of Maxima andMlnima. Ginn, Boston, 1917.

CHAPTER 14
MULTIPLE RIEMANN INTEGRALS
14.1 INTRODUCTION
The Riemann irtegral J, f(x) dx can be generalized by replacing the interval [a, b]
by an n-dimensional region in which f is defined and bounded. The simplest
regions in R" suitable for this purpose are n-dimensional intervals. For example,
in R 2 we take a rectangle I partitioned into subreetangles 1 and consider Riemann
sums of the form Yf(x, y)A(l), where (xt, Yt) e I and A(I,) denotes the area of
I÷ This leads us to the concept of a double integral. Similarly, in R a we use
rectangular parallelepipeds subdivided into smaller parallelepipeds 1 t and, by
considering sums of the form Yf(xt, y, zv)V(ID, where (x, y, zt)  It and V(I)
is the volume of It, we are led to the concept of a triple integral. It is just as easy
to discuss multiple integrals in R , provided that we have a suitable generalization
of the notions of area and volume. This "'generalized volume" is called measure or
content and is defined in the next section.
14 THE MEASURE OF A BOUNDED INTERVAL IN
Let A,..÷, A denote n general intervals in R I ; that is, each A may be bounded,
unbounded, open, closed, or half-open in R t. A set A in R" of the form
A =- A × -'- × A, - {(x,...,x):xeAt fork
is called a general n-dimensional interval. We also allow the degenerate case in
which one or more of the intervals A consists of a single point.
If each At is open, closed, or bounded in R , then A has the corresponding
property in
If each A is bounded, the n-dimensional measure (or n-measure) of .4, dettoted
by #(A), is .defined by the equation
=
where g(dt) is the one-dimensional measure (length) ol- Ak. When n =
called the area of A, and when n = 3, it is called the volume of A. Note that
/(A)  0 if#(A) = 0 for some k.
We turn next to a discussion of Riemann integration in R". The only essential
difference between the case n = 1 and the case n > I is that the quantity
Ax = x - x_ t which was used to measure the length of the subinterval
I-xt- , xv..] is replaced by the measure/,(1 of an n-dimensional subinterval. Since
the work proceeds on exactly the same lines as the one-dimensional case, we shall
omit many of the details in the discussions that follow.
14.3  RIF_£MANN IN-'IGRAL OF A BOUNDED FUNCTION DEFINI}
ON A COMPACT INTEIIVAL IN R 
Defmiffan 14d. Let A = A  × " • x A, be a compact intereal in R'. ff P is a
partition of ,4t, the cartesian product
P= P, ×""  P,,
is said to be a partition of A. If Pt divides A into-m t one-dimensional subintervals,
then P determines a decomposition of A as a union of mt'"ra, n-dimensional
intervals (called subintervals of P). A partition
P _ P'. The set ofallpartitions of A wilt be denotedby
Figure 14. i illustrates partitions of intervals in R
Definition 14.2. Let f be defined and bounded on a compact interval I in R . If P
s a partition of l into m subintemals 11, ..., 1,. and if tt lt, a sum of the form
k-l
is called a Riemann sum. We say f is Riemann-inteyrable on I and we write f e R on
I, whenever there exists a real number A havin9 the following propertv: For every
e > 0 there exists a partition P ofl such that Pfiner than P implies
Is(*', f) - a l <
for all Riemann sums S(P, f). When ,tch a number A exists, it is uniquely

determined and is denoted by
, x,) d(x .....
No. For n > 1 the integral is called a multiple or n-fold integral. When n = 2
and 3, the terms double and triple integral are used. As in R 1, the symbol x
j', f(x) dx is a "dummy variable" and may be replaced by any other convenient
symbol The notation ]'xf(xl, .-., x,,)dxt-.-dx, is also used instead of
rf(xl, ..., x) d(xt,..., x. Double integrals are sometimes written with two
integral signs and triple integrals with three such signs, thus:
Dcfinitim 14.3. Let f be defined and bounded on a compact interval I in R . If P
is a partition of l into m subintervals I, ..., I=, let
m(f) = inf {ffx) : x e Ik}, M(f) = sup {f(x)" x elk}.
The numbers
U(P,f) = E M(f)p(l) and
L(V,f) = k
are called upper and lower Riemann aunts. The upper and lower Riemann integrals
off over I are defined as follows:
ftfdx inf{U(P,f): (I)},
P
f dx = sup {LP,f) . P e '(I)}.
The function f is said to satisfy Riemann's condition on I if, for every  > 0, there
exists a partition P, of l such that P finer than P implies U(P,f) -- L(P,f) < t.
tor. As in the one-dimensional case, upper and lower integrals have the following
properties"
a)
b) if an interval 1 is decomposed into a union of two nonoverlapping intervals
11, lz, then we have
and
The proof of the following theorem is essentia|ly the same as that of Theorem
7.19 and will be omitted.
Tkeoeem 14.4. Let f be defined and bounded on a compact interval I in
the following statements are equivalent:
i) f RonL
ii) f satisfies Riemann's condition on L
iii) _ f dx = 1f dx.
The/1
14.4 SETS OF MEASURE ZERO AND LEBE.SGUEn CRITERION FOR
EXISTENCE OF A MULTIPLE RIEMANN INTEGRAL
A subset T of R" is said to be of n-measure zero if, for every . > 0, T can be
covered by a countable collection of n-dimensional intervals, the sum of whose
n-measures is <.
As in the one-dimensional case, the union of a countable collection of sets of
n-measure 0 is itself of n-measure 0. Ifm < n, every subset of R', when considered
as a subset of R ", has n-measure 0.
A property is said to hold almost everywhere on a set S in R" if it holds every-
where on S except ['or a subset of n-measure 0.
Lcbesgue's criterion for the exis|ence of a Riemann integral in R  has a
straightforward extension to multiple.integrals. The proof is analogous to that of
Theorem 7A8.
rlworem 14.5. Let f be defined and bounded on a compact intertal I in R'. Then
f  R on ! if, and only if, the set of discontinuities off in I has n-measure zero.
14- EVALUATION OF A MULTIPLE INTEGRAL BY ITERATED
INTEGRATION
From elementary calculus the reader has learned to evaluate certain double and
triple integrals by successive integration with respect to each variable. For
example, if f is a function of two variables continuous on a compact rctangle Q
in the xy-plane, say Q = {(x, y) : a ___ x _< b, c  y < d}, then for each fixed y
in [c, d] the function F defined by the equation F(x) = f(x, y) is continuous (and
hence integrable) on [a, hi. The value of the integral  F(x) dx depends on y and

de.fins a new function G, where G(y) = f(x, y)dx. This functioa G is con-
tinuous (by Theorem 7.38), and hence integrable, on [c, d]. The integral S G(y) dy
turns out to have the same value as the double integral J(f(x, y) d(x, y). That is,
we have the equation
(This formula will be proved later.) The question now arises as to whether a
similar result holds whenfis merely jntegrabl¢ (and not necessarily continuous) oft
Q. We can see at once that certain difficulties are inevitable. For example, the
inner integral S,f(x, y) dx may not exist for certain values or" y even though the
double integral exists. In fact, iff is discontinuous at every point of the line
segment y = Yo, a  x _< b, then jof(x, Yo)dx will fail to exisl. However, this
line segment is a set whose 2-measure is zero and therefore does not affect the
integrability off on the whole rectangle Q. in a case of this kind we must use
upper and lower integrals to obtain a suitable generalization of (1).
Theorem 14.6. Let f be defined and bounded on a compact rectangle
Q = [a,b] × [c,d]
Then we have:
ii) Statement (i) holds with  replaced by  throughout.
iv) Statement (iii) hold with , replaced by  throughout.
v) When  f(x, y) d(x, y) exists, we have
Proof. To prove (i), define F by the equation
F(x) = jf f(x, y) dy, if x • [a,
Then IF(x)l < M(d - c), where M = sup {If(x, Y)I: (x, y)  Q}, and we can
consider
Similarly, we define
Let P, = {xo, xt,-.-, x.} be a partition of [a, b] and let
be a partition of [c, d],, Then P  P, x P is a partition of Q into mn sub-
reetangle Qo and we define
L ] f; -If,'
I o = f(x, y) dy ax, 1o = f(x, y) dy ax.
Since we have
f(x, y) ty = f(x, y)
1=I
dx < f(x, y) dy dx
ffi f(x, y) d
Jml  t j-I'
That is, we have the inequality
Similarly, we find
If we write
and
then from the inequality m < f(x, y) < M], (x, y) ¢ Qo, we obtain
m = inf {f(x, y) : (x, y) e
Mj = sup {f(x, y) : (x, y)  Q,},

 f(:, y) dy dx 
Summing e i anj and using the above inequalities, we get
L(P,f) N ! N  N U(P,f),
Since this holds for all partitions P of Q, we must have
Ths preve tatement (i).
It is clear that the preceding proof could also  carried out if te function F
were originally defined, by the formula
f(x) = I f(x, y)
and hence (ii} follows by the same argument,
Statements (iii) and (iv) can  similarly proved by interchanging the roles of
x and y. Finally, statement (v) is an immediate consequence of statements {0
through (iv).
As a corollary, we have the formula mentioned earlier:
f(.:, y) d(x, y) = fix, y) dy dx = .f(, y) dx dy,
wNcb is valid when f is continuous on Q, is is often called Fubini's theorem.
SOTE. The existence of the iterated integrals
• j'(x, y) dy dx and f(x. y) dx dy,
does oI jmply the existence of fef{x, f) d{x, y). A counger example is Wen in
Eercise 14.7.
Beff>re commemng on the analog of Theorem 14.6 in ', we flst introduce
some I)rther notatio and terminolo,. If k S n, the set of x in R" for which
x = 0 is called the cooedinau/oerplaee . Given a t S in R', the proieaion
& of S on  i:s defined to be the image of S under that mapping whose vabe at
each point (x, x .... , x.) i S is (x .... , x_, 0, x.,..., x.). It is easy to
Figure i4.2
./ ....
show thal such a mapping is continuous on S. It fotlows that Jf N is compact, :each
projection & is compact, Also, f S is connected, each S is conncvtedo Projections
in 1R  are illstraed in Fig. 14.Z
A theorem entirely analogous to Theorem 14,6. holds for nfotd i.tegraM It
will suffice to indicate how the extension goes wb.e e = 3 in this case,f is defined
and bounded on a compact interval
and statement (i) of Theorem 14.6 is rephced by
j fdx -': fd(xz.,, x) dx
S %.. fd{x2. , ;%) dx 
where Q is the projection of Q onhe coordinate plane ]]. When 'ef(x) dx
exists, the analog of part (v) of Theorem 4.6 is the formula
fix) dx =. f d(x> %) dx,. ,'." f dx. d(x> %).,
As in Theorem 14.6, similar statements hold with appropriate replacements of
upper integrals by lower integrals, and there are also analogous formulas for the
prections Q and Q>
The reader should have no difficulty in stating analogots results for n-fold
integrals (they can be proved by the method used in Theorem 4.6).. The special
case in which the n-fold integral fef(x) dx eists is of particular importance and

canbc stafed a,s fb]lows:
Theorem 1.47 Let f be d<fimd and boum/ed on a comac 9ter.,;M
# RL Asume g.hat , JX) dx ex.v.s The
Similar fbrmus hoh ih zer i¢qcd' rep&e'd by A:wer imegraL' ad ,'i Q.
rcp&ced by Q ghe ;rojedion GF Q on e
14.6 JORDAN-?ASURABLE SETS IN R"
Up to his point the multiple mtegrM . f(x).dx .has bee defin:ed o.  .tbr i.ntervaE
L TNs, of corse, is too restrictive fbr the application of integration  is ot
dNcul.t to extend he definition to encompass more gemral se cMed
measurable sets.. These arc di:ussed in this section. T1e definitio mak ue of
the boundary of a set S in RL We recaI{ that a point x m R"is calted a boundary
point of S if every n-ball B(x) cem.ains a point in S and aso a point not in S, The
set of all boundary points of S is caiJed tge boundary of S and is denoted by
(See Section 3.t6.)
Definition t4.& Let S be a subset :5e'a compact #ervag i in R . N?r every partition
P of I d@;ne (P, S) to be the s'um of he measur<s ( those subingerva/s qf P which
contab en(v interior points R[ S nd ]e JP, S) be he sum of the men.lures
C(S) = sup {g(.e, S) : P 
i:(S) = inf {Y(.P, s) : p e
a:e cat:e& re.syecive@, the (nd#nen:;ionaJ) )ner and outer Jordan ,:'onen f S.
The se S is said o be dorda>measurabIe fo(S} = E(S), in :hich <Yz'e H#s common
cane is cai&d He do.rdan cement qfi S, denoted by
t.¢ is easy to verify that f(S) a.d 8(S) dend oMy on S and nof o.e he interval
I which contains N. Also, 0 . g(S)  (S).
If S has content zero, hen £{S) = (5) = 0. Hence, fi: every c > 0, S car be
covered bya finite c:ollecion of {mervals, the sum of whose measures is <g. Note
hat content zero is deribed in terms offiRe coverings, wereas measure zero
described in terms of .counlabD coverings. Any set wkh contert :zero a&o has
measure zero, bin the converse is no, ecessarfly
Every compact rl{erva O is Jorda-measurab{e and is uoenL c{@), is equal
its measure, g{O}- fk < n/ffe ndhrensional coierJ. of every bounded set in R 
J.ordan-measnrable ses S in R z am aso said o have area .c(S). h:t hs case, the
sums d(P, S) sad Y(P, S) represent approximatioes .) the area from the "inside"
Figure
and the "outside" of& respectively. This is illustrated i Fig. 14,3, where the lightly
shaded rectangles are counted in ](P, S), the heavily saded recta.nNes in J(P
For sets m R , c(S) is also called, the volume of &
The next theorem shows that a bounded set has Jordan connt ifi and only
its boundary snt too "thick."
• heorem 14.9. Let S be a bounded set in R" d let ?S denote is bundao, Then
we have
Hence, S is Jordan-me,utah# b4 and only (f, 3S has cement zero
Pro.@ Let I  a compact nterval conmini.ng S and d& Then for every partition
P of I we have
J(P, aS) = J(P, S) - J(P S)
Therefore, 3(P, S)  (S) - (S) and hen (dS)  P(S) - g(S). To obtain
[be reverse inequality, .le .e > 0 . givem choose P so that J(P, S) < g(S) +
and choose P2 so tat J(P> S) > ::(S) - e/2. Let P  P u Pz. Since refine*
men{ increases the inner sums d and dreases the outer sums J, we find
< a(S) - dS) + ..
Since  is arN{rary this means that (S)  d(,5) - f(S). erefoe, (gS} 
a(S) - f(S) ad the proof i complete
14.7 MULTIPLE IN,.GRATION OVER JORDAN-MEASU.I1ABLE :SETS
Definition t4dO. Let f be defined and bounded on a bounded gordanmeasurable set
S in RL Let I be a compact interval containing S and define y on I asJbtlows:
g(x.) = {;(x) fxeS,

Then f is said to be Riemann-integrable on S and we wrile f e R on S, whenever the
integral x g(x) dx exists. Fee also write
fs.ffx) ax -- fx U(x) dx-
The upper and tower integrals sf(x) dx and _sf(x) dx are similarly define£
soa. By considering the Riemann sums which approximate j'r g(x) dx, it is easy
to see that the integral sf(x) dx does not depend on the choice of the interval 1
used to enclose S.
A necessary and sufficient condition for the existence of sf(x) dx can now be
given.
Thereto 14.11. Let S be a Jordan-measurable set in R', and let f be defined and
bounded on S. Then f e R on S if, and only if, the discontinuitie off in S form a
set of measure zero.
Proof. Let I b¢ a compact interval containing S and let g(-x) = f(x) when x • S,
/7(x) -- 0 when x  I - S. The discontinuities off will be discontinuities of g÷
However, g may also have discontinuities at some or all of the boundary points of
S. Since Sis Jordan measurable, Theorem 14.9 tells us that c(3S) = 0. Therefore,
/7 e R on I if, and only if, the discontinuities off form a set of measure zero.
..1'1.8 JORDAN CONTENT EXP AS A RIEIMN'N INTEGRAL
Theorem 14.12. Let S be a compact Jordan-measurable set in R*. Then the integral
Js 1 exists and we have
dS) =  1.
Proof. Let I be a compact interval containing S and let s denote the characteristic
function of S. That is,
{: ifxS,
Zs(x) = ifxl - S.
The discontinuities of 2(s in I are the boundary points of S and these form a set
of content zero, so the integral 't ;fs exists, and hence 's 1 exists.
Let P be a partition of I into subintervals I,..., I,, and let
A = {k : 1 o S is nonempty}.
Ifk  A, we have
Ms) = sup {s(x) :x  I,} ---- 1.
and M,(xs) = 0 if k ¢ A, so
Since this holds for all partitions, v have f s = ?:(S) = c(S). But
14.9 ADDnTVE PROPERTY OF THE RIEMANN IN'IEGRAL
The next theorem shows that the integral is additive with respect to sets having
Jordan content.
T&mrem 14.13. Assume f e R on a Jordan-meamable set S in R'. Suppose
S = A  B, where A and B are Jordan-measurable but hae no Otteror point.¢ in
common. Then f ¢ R on .4, f • R on B, and we hape
isf(x) dx = If(x) dx + I,f(x) dx. (4).
Proof Let I be a compact interval containing S and define # as follows:
,(x) = {0f(x) ifx•S,
ifxl - $.
The existence of ]', f(x)dx and nf(x)dx is an easy consequence of Tlxcorem
14.11. To prove (4), let P be a partition of/into m subintervals I,, ..., I., and
form a Riemann sum
s(P, =
k--1
If S, denotes that part of the sum arising from those subintervals containing
point of A, and if Sn is dmilarly defined, we can write
whre Sc contains thos terms coming from subintervals which eotain both points
of A and points of B. In pattieuIax, all points common to the two boundaries A
and 8B will fall in this third class. But now S, is a Riemann sum approximating
the integral xf(x) dx, and S is a Riemann sum approximating ,f(x) dx. Since
c($A o gB) = 0, it follows that I$cl can be made arbitrarily small when P is
sufficiently fine. The uqution in the thexrem is an easy consequenc of th¢
1o1 Formula (4) also holds for upper and lower integrals.

400 Mtllk Riana ]lntr Th, 14.14
For sets S whose structure is relatively simple, Theorem 14.6 can be used to
obtain formulas for evaluating double integrals by iterated integration, These
formulas are given in the next theorem.
Theorem 14,14, Let opt and dp be two continuous functions defined on in, b] such
that ?l(x) <_ dp2(x) for each x in in, b]. Let S be the compact set in R z given by
s = {(x,y):a _< x _< b, l(x) _< y
If f  R on % we have
= f(x, y)
rcrr. The set S is Jordan-measurable because its boundary has content zero,
(See Exercise 14.9.)
Analogous statements hold for n-fold integrals. The extensions are too obvious
to require further comment.
Figure 14.4
Figure 14,4 illuslrates the type of region described in the heorem. For sets
which can be decomposed into a finite number of Jordan-measurable regions of
this type, we can apply iterated integration to each separate part and add the results
in accordance with Theorem 14.13.
14.10 MEAN-VALUE THEOREM FOR MULTIPLE INTEGRALS
As in the one-dimensional case, multiple integrals satisfy a mean value property.
This can be obtained as an easy consequence of the following theorem, the proof
of which is left as an exercise_
Theorem 14.I5, Assume f • R and #  R on a Jordan-measurable set S in R*. If
f(x) _< #(x)for each x in S, then we have
14.17 Mean.Value Theoeem foe Multiplo Integrals 401
Theoeem 14.16 (Mean-Value Theorem for mltlple integral). Assume that g • R
and f• R on a Jordan-measurable set S in R  and suppose that oo(x) _> O for each
x in S. Let m = inff(S), M = supf(S). Then there exists a real number 3. in the
interval m <_ 2 <_ Af such that
In particular, we have
f(x)o(x ) dx = ,l s #(x) dx. (5)
me(S) <_ sf(X),ix <_ Me(S). (6)
No'rE. If, in addition, S is connected andfis continuous on S, then 2 = fxo) for
some xo in S (by Theorem 4.38.) and (5) becomes
In partieular, (7) implies sf(x) dx = f(xo)c(S), where x0 • S.
Proof. Since (x) _> 0, we have torT(x) < f(x) g(x) < M(x) for each x in S By
Theorem 14.15, we can write
If s.q(x)dx ffi 0, (5) holds for every 2. If sy(x) dx > 0, (5) holds with
2 = sf(X)t(x) dx/s g(x) dx. Taking g(x) --- 1, we oblain (6).
We can use (6) to prove that the integrand fcan be disturbed on a set of content
zero without affecting the value of the integral In fact, we have the following
theorem:
Theorem 14.17. Assume that f e R on a Jordan-measurable set S in R . Let The a
subset of S having n-dimensional Jordan content zero. Let  be a function, defined
and bounded on S, such that y(x) = f(x) when x • S - T. Then g e R on S and
fsf(x) dx = .q(x) dx.
Proof. Let h = f - q. Then [- h(x) dx = r h(x) dx + Is-r h(x) dx. However,
r h(x)dx = 0 because of (6), arid s-r h(x)dx = 0 since h(x) = 0 for each
xinS- T.
oxE. This theorem suggests a way of extending the definition of the Riemann
integral .(sf(x)dx for functions which may not be defined and bounded on the
whole of S. In fact, let S be a bounded set in R  having Jordan content and let T
be a subset of S having content zero. Iff is defined and bounded on S - T and

if s_rf(x) dx exists, we agree to write
fs.f(x) dx  fs f(x) dx'_r 
and to say thatfis Riemann-integrable on S, In view of the theorem just proved,
this is essentially the same as extending the domain of definition off to the whole
ors by dofiningf on Tin such a way that it remains bounded.
fsfl(xl)"'fi(.In) d(,l, . .-, x) - (f|ilfl(xl_) d.xI) .- . (fg fl('Xl) n ) •
whSffi [,b]] -.- [a. ].
14 tf  d un on a m   ffi [, b] x [e, d]  R z.
 at f   y  It, d], f(x, y)   g fion of x, d t for
x In, b ], f(x, y) is  g fion ofy. o tf Ron
1 Evflg  of  foHog ubl¢ ts.
t  t.
14.4  fl = [0, 1] x [0, 1] d ht¢ hf(x,y)dy in
a) f(x,y) = 1 -- x-- y ifx +y  t, f(x,y) = Ooi.
b)f(x,y)f x  + yz gx  +   !, f(x,y)= Oo.
c) f(x,y)= x + y  x   y  , f(x,y) = O o.
14 fontu= [0,1] x [O.I]foHo:
,  I if x is rationS,
f(x, y)  , 2y if x is
0  tt  f(x, y)  s for 0  t  ! and
and
This shows that I ['f(x, y) dy] dx exists and equals 1.
b) Prove that f [gf(x, y) dr] dy exists and find its value,
c) Prove that the double integral J'o f(x, .v) d(x, y) does not exist.
14.6 Dn f on the square Q = [0, 1 ] x [0, 1 ] as follows:
0 if at least one of x, 3' is irrational,
f(x, y) = 1/n ify is rational and x =
where m and n ara relatively primo integers, n > 0. Prove that
f(x, y) dr = f(x, ,) ,ix ay = f(x, y) d(x, y) = o
but that I f(x, y) dy does not exist for ratiomd x.
14.7 Ifpt denotes the kth prime number, let
-- , :---- 1,2 ..... /-- 1, m= l,,...,p--
a) Prove that S is dense in Q (that is, the closure of S contains Q) but that any line
parallel to the coordinate axes contains at most a finite subset of
b) Define f on Q as follows:
f(x,y) = 0 if(x,y)S, f(x,y) = ! if(x,y)eQ - S.
Prove that $o  iIf(x, y) dy]  = I IIf(x, y) ] de = L but that the
double integral J'o f(x, y) d(x, y) does not exist.
Jordan contain
14`11 Let $ be a bounded set in R  having at most a finite amber of accumulation poims.
Prove that e(S) -- 0.
1,1.9 Let f be a continuous real-valued function defined on In, b]. Let S denote the
graph off lhatis, S -- {(x, y):y = f(x), a <_ x < b }. Provethat Shastwa-dimcnsiona
Jordan content ero.
14.10 Let F be a rectifiable curve in R'. Prove that F has n-dimensional Jorda content
zed'o.
14`11 Let fbe a normegative function defined on a set 6; in R'. The ordinate set of foyer
S is defined to be the following subset ol'R'+ :
{(x ..... x,,x÷):(x,,...,x,) S. 0 <_ x,+, < f(x,...,x,)}.
If S is a Jordan-measurable region in R n and ill is continuous on S, prove that the ordinate
set of foyer S has (n + l)-dimensiorud Jordan cotttenl whose value is
Interpret this problem geometrically when n = I and n = 2.

14.12 Assua thatf R on S and suppose J'sf(x) dI  0. (S is a subset of Ra). Let
A = (x : x  S,/(x) < 0} and assume that c(A) = 0. Prove that there exists a set B of
measure zero such that f(x) = 0 for ¢ac.h x in S - B.
14.13 Assume thatf¢ £ on S. where Sis a region in R a andfis continuous on S. Prove
that there exists an interior point x o of S such that
s f(x) dx =
14.14 Letfb¢ continuous on a rectangle Q  [a, b] x [c, all. For each interior mint
(xi, xa) in Q, detlae
Prove that D.zF(x, x2) = Dz.F(xL, x) = f(xl, x:).
14.15 Let Tdenote the following triangular region in th© plane:
T= { (x'y):O<- x+ y  1}a- b- , warren >0, b> O.
Assume thatfhas a cotinuous second-order partial derivative D.2fon T. Prove that
there is a point (xm Yo) on the segment joining (a, 0) and (0, b) such that
fr th.xf(x, y) dfx, y)  (0, O) -/'(a, 0 )+ aD.f(x0,
SUGGESTED REFERENCF_,S I¢OR FURTHER STUDY
14-.1 Apostol, T, M., Calculus, Vo]. 2, 2rid ¢d, Xerox, Waltham, 1969.
14.2 Kesteiman, H., Modern lheorie f Integration, Oxford University Press, 1937.
14.3 Rogosinski, W. W., Volume and Integral. Wiley, New York. 1952.
CHAPTER
MULTIPLE LEBESGUE INTEGRALS
15.1 INTRODUCTION
The l.¢besgue integral was described in Chapter tO for functiom defined on subsets
of Rt. The method used there can be neralized to provide a theory of Lebesgue
integration for functions defined on subsets of -dimensional space R'. The
remidng integrals are called multiple interalr. When n = 2 they are called double
integrals, and when n = 3 they are called triple iate.rala,
As in the one-dimensional case, multiple Lelxgu¢ iategration is an extension
of multiple Riemann integration. It permits mor general functions as integrands,
it treats unbounded as well as bounded functions, and it encompasses more
general sets as regions of integration.
The basic definitions and the principal convergence thcotgms are camplctely
analogous to the one-dimensional case. However, there is one new featur that
does not appear in R . A multiple integral in R  can be evaluated by calculating
a succession of n one-dimensional integrals. This result, called Fubini',¢ Theorem,
is one of the principal concerns of this chapter.
As in the one-dimensional case we define the integral first for step functions,
then for a larger class (called upper functions) which contains limits of certain
increasing sequences of step functions, and finally for an even larger class, the
Lebesgue-integrabte functions, Since the development proceeds on exactly the
same lines as in the one-dimensional case, we shall omit most of the details of
the proofs.
We recall some of the concepts introduced in Chapter 14. If I = It ×
is a bounded irterval in R , the n-measure of I is defined by the equation
_-
where g(l) is the one-dimensional measure, or length, of
A subset T of It" is said to be of n-measure 0 if, for every
covered by a countable collection of n-dimensional intervals, the sum of whose
n-measures is <
A property is said to hold almost everywhere on a set 5" in R  if it holds every-
where on S except for a subset of n-measure 0. For example, if {f,} is a sequence
of functions, we sayf - falmost everywhere on S if lira,_,, f(x) = f(x) for all
x in S except for those x in a subset of n-measure 0.

STEP FUNCTIONS AND  INTEGRALS
Let I be a compact interval in R', say
wher each I is a compact subinterval of R t. IfPt is a partition of/, the cartian
product P ffi Pj × --- × P. is called a .partition of I. If Pt decomposes Ik into
m t one-dimensional subintervals, then P decomposes I into m = mt...mt
n-d/mensionai subintervals, say Jr, ..., J=.
A function s defined on I is called a step function if a partition P of ! exists such
- that s is constant on the interior of each subinterval J, say
s(x) = c ifx eint J.
The integral of s over I is defined by the equation
blow let G b¢ a general n.dimensional interval, that is, an interval in R  Which
ncod not be compact. A function s is called a step function on G if there is a
compact n-dimensional subinterval I of G such that s is a step function on I and
s(x) ffi 0 if x e G - I. The integral of s over G is defined by the formula
where the integral over / is given by (1). As in the one-dimensional case the integral
is independent of the choice of/.
15.3 UPPER FUNCTIONS AND LEBE.SGUE-INTEGRABLE FUNCTIONS
Upper functions and Lebesgue-integrabl¢ functions are defined exfictly as in the
one.dimensional
A real-valued functionfdefined on an interva ! in R  is called an upper function
on L and we writefe U(/), if there exists an increasing sequence of step functions
{s} such that
a) ,%  falmost everywhere on
and
b) lira,.  , s, exists.
The sequence {s,} is said to generater The integral of foyer I is defined by the
equation
f = lim f: s.. (2)
 Ftmede aml Metl'al Sets in R"
We denote by L(I) the st of all functions f of the form f- u - v, where
u e U(F) and v  U(/). Each function f in L(/) is said to be Lebesgue-integrabl¢
on I, and its integral is defined by the equation
Since these definitions are completely analogous to the one-dimensional case,
it is not surprising to learn that many ofthe theorems derived from these definitions
arc also valid. In particular, Theorems 10.5, 10.6, t0.7, 10.9, 10.10, I0.I, 10.13,
10.14, 10.16, 10.17(a) and (c), 10.18, and 10.19 arc al[ valid for multiple integrals.
Theorem 10.17(b), which describes the behavior of an integral under expansion or
contraction of the interval of integration, needs to be modified as follows:
Iff L(F) and if(x) = f(x/c), where c > O, then g  L(c/-) and
In other words, expansion of the interval by a positive factor c has the effect of
multiplying the integral by c , where n is the dimension of the space.
The Levi convergence theorems (Theorems 10.22 through 10.26), and the
Lebesgue dominated convergence theorem (Theorem 10.27) and its consequences
(Theorems [0.28, 10.29, and 10.30) are also valid for multiple integrals.
rOTATIO. The integral  fis also denoted by
The notation f(xt,... , xa)dxi-., dx, Js also used. Double integrals are
sometimes written with two integral signs, and triple integrals with three such signs,
thus:
f f f(x, y) dx dy, f f f f(x, y, z) dx dy dz.
!
I-A MEASUR.4LE FUNCTIONS AND MEASURABLE SETS IN R"
A real-valued function f defined on an interval I in R" is called measurable on 1,
and we write fe M(1), if there exists a sequence of step functions {&} on I such
that
lira s,(x) = f(x) a,e, on I.
The properties of measurable functions described in Theorems 10.35, 10.36, and
10.37 are also valid in this more general setting.

A abset S of R ' s cM|ed measurable i.f hs characeisi.c uct]:or b; is
{.he set S s defied by the equatio
if L s mea.srable bt a.o {n L,:(R) we define g(S) = -? .. Th.e funclo  so
The propertie of measure descri n Theorems ]0 hroagh t0.47 are also
valid for mdimenaioaI Leagne measure. Aso, the L@sgue integrat cat be
defined for arNrary .sbse:s of R ; by the mettled md b. S,tio t0.I%
We crop.size m parficuNr the coumably additive prorB: of Lekesge
measure descried ia Theorem 10AT:
K {A, A,.. } is a ceumable dioint colbc:ien of measraNe ses i R ,
I:: is meaaab?e and
 A. = g( )
The nex*: te.orem shows that every on sbset of R  is measvrabb.
Tkeorem 1£1. Ever:: op set S in R  ca be expressed as e um q/ a couta.b/e
dAgoie: collection (f bo,m'd cubes wtm,se c:os'ure is contained b S, ThereJb:e S
mouse,rob&. Moreoce< ( S is boumA, d., tke: g(S) isfin#e.
Pro@ Fix ar meger m 2 I and consider 1 haK-on intervaIs in R  of {be form
.:;;, 2  for k  0,.  l, .'h 2
A{ he imervals are of tegth 2-% ned hey %rm a coamable dioim
whose union is R) The cartesian product of  such intervals is an
cube of edge4esgh 2 % Let : denote the co/he,ion of nit these cbes. Then
is a coumaNe .di]oinveol-tion whose ueion is R< Note hat the cus in f+
are obtained by NsectMg the edges of those i 6:'. There,)am, if Q is a cabe in
and if . is a cube i +,, then either +  . or ,+. and  are
Now we extr-ac, a subcoecfioa G= from Y= as {blIows If m =. {, G cosists
of a cubes h F. whose do:sre ies in S. If m = 2. G cosists oF.a1 cs n 
whez closure lies {n S but net in any of te cubes n G, If m = 3, G
of all cubes i F:: whose closure lies in S but not in any of the cbes i G or G,
ned so on. The coastr.w.io is ]liust}at.ed in Fig, 5 i where S is a qu.arer of an
open disk  R< Te blank sqare is i G, the ighty shaded ones are in
and the rker ones are in
Now
4
That is, T is he uion of all the cubes in G, G2 .... We wilt prove that S -= T
and' thi.,; wit{. prove the theorem caue T is a coumabie dsjon colbcio, of
c:us whose closure {its i. S. Now T
Hence we need oly show that S
Let p = (p...,p) apointin S, Since Sison, there :s a cbe with.
center  and edge:-.legB  > 0 which ies
The= for each ¢ we have
Now choose k. so
k k + 1
..... < P.e :'24 ...... : ..............
and let Q be the Cartesian prodm:{ of the intervals (ke2"% (k + t)2 '] %r
' = , 2 ..... m Then p  Q r some cu O a/;, If m is the smaiies integer
with this property, then Q  G, so p e T; t--{eace S  K The sat.emems about
he measurabiky of S *ltow a once from he countaNy addMve property of
Lebesgue measure.
dosed subse of R  is measurable.
FUBINYS REDUCfION THEOREM FOR TILE. DOUBLE INTEGRAL OF
STEP lvUN6TION
Up to this poinL Lesgue theory in R ' is completely analogous to the one
dimenbsat case. New ideas are toque.red wen we come to Fubini's deorem
catcuIatieg a m1fipb nteg.fa{ in R  by ierated towe>.dmens{ona imegrals. To
b:ter aderstand what js needed we consider flrs the two-dimensiona1 case.
Let :us re:atl he correspodbg resuh for muitipb Riemarm imegraIs,
] .... In, b] x [c, d] is a compact merval in R  arid iff is

on I, then we have the folIowing reduction formula (from part (v) of Theorem
I4.6):
f(x, y) d(x, y) = f(x, y) dx dy, (3)
There is a companion formula with the lower integral }' replaced by the upper
integral , and there are two similar formulas with the-order of integration re-
versed. The upper and lower integrals are needed here because the hypothesis of
Riemann-integrability on I is not strong enougtt to ensure the existence of the
one-dimensional Riemann integral f(x, y) dx. This difficulty does not arise in
the Lebesgue theory. Fubini's theorem for double Lebesgue integrals gives us the
reduction formulas
ff(x" y) d(x' Y) = f [f f(x, y) dx] dy -- f tff f(x, y) dy] dx,
under the sole hypothesis thatfis Lebesgue-integrable on L We will show that the
inner integrals always exist as Lcbesgue integrals. This is another example illus-
trating how Lebesgue theory overcomes difliculti inherent in the Riemann theory.
In this section we prove Fubini's theorem for step functions, and in a later
section we extend it to arbitrary Lebesgue integrable functions.
Theoeem 15.2. (Fttbini's tlteorem for step fanetio). Let s be a step function on
R 2. Then for eachfixedy in R 1 the integral E, s(x, y) dx exists and, as a function
of),, is Lebesgue4ntegrable on R , Moreover, we have
(4)
Similarly, for each fixed x in R  the integral J,, s(x, y) dy exists and, as a function
of x, is Lebesgue4ntegrable on R . Also, we have
(5)
Proof. This theorem can be derived from the reduction formtOa O) for Riemann
integrals, but we prefer to give a direct proof independent of the Riemann theory.
There is a compact interval I :- Is, b] × [c, d] such that s is a step function
on I and s(x, y) = 0 ff (x, y) e R z - L There is a partition of I into mn sub-
rectangles lq = ]'x_ 1, x] x [Yt-, Yj] such that s is coastant on the interior of
s(x, y) = e u ff (x, y) e int 1.
Then
s(x, y) d(x, y) -- ct(x i - x,_Xy - yi_) = s(x, y) dx
J-1 t-t
Summing on i andj we find
Since 8 vanishes outside I, this proves (4), and a similar argument proves (5).
To extend Fubini's theorem to Lcbesgue-integrable functions we need some
further results concerning sets of measure zero. These arc discussed in the next
section.
1.6 SOME PROPERTIES OF SETS OF MEASURE ZIRO
Theorem 15.3. Let S be a subset of R'. Then S has n.measure O if, and only if, there
exists a countable collection of n-dimensional intervals {J, J2, . }, the sum of
whose n.measures is finite, such that each point in S belongs to Jt for infinitely
many k.
Proof. Assume first that S has n-measure 0. Then, for every m > I, S coo be
covered by a countable collection of n-dimensional intervals {I,,, 1.,2 .... }, the
sum of whose n-measures is <2-". The set .4 consisting of all intervals I,a for
m = 1, 2,..., and k = !, 2 .... , is a countable collection which covers S, and
the sum of the n-measures of all these intervals is < ,:  2-" =- 1. Moreover,
if a¢ S then, for each m, a•l,, for some k. Therefore if we write
A = {J, Jz, --. }, we see that a belongs to Jt for infinitely many k.
Conversely, assume that there is a countable collection of n-dimensional
intervals {Jr, J, ... } such that the series ,_ /z(Jt) converges and such that each
point in S belongs to Jt for infinitely many k. Given  > 0, there is an integer N
such that
Each point of S lies in the set U J*, so S _ U-- J*" Thus, S has been
covered by a countable collection of intervals, the sum of whose n-measures is
<e, so S has n-measure O.
Defmin 1.4. If S is an arbitrar), subset of R z, and/f (x, y) • R z, we denote by
Sv and S  the followin# subsets of R  :
S,--{x:x•R and (x,y)•S),
S ' = {y:yeR  and (x,y)•S}.

412
Examples are .shown in Fig.
x-axis of a horizomaI coss sectoa of S; and S - is the prqjeaion on the .wax.is of a
vertical cross section of S.
Tosem 1£5, .fS t a subse qfR  w# 2-measure O, then S, has" 4seasure Ojbr
aImost a# y in R , and S  has t-measure
Proof We wi1.1 prove that @ has 1-measure 0 for almost al y n R . The proof
makes use of Theorem  5,3.
Sir S has 2-measure 0, by Theorem
rectangges {I:) such hat the seri
 U) converges (6)
and such. tha every point .(x, y) of S klosgs o 1. 2)r infinitdy mar G k. Write
2 = X x }, where X and Y are subintervals of N  The
Th (6) mpes that the series
Now {g} s a seqgec.e of nennegaive fuse,.ions m L.{R ) such tha xhc series;
2-, i J' &. ceweees. Therefore, by the Levi theorem (Theorem 0.25}. the :series
 y cow, verges amos everywhere on R In oher words ere is a sbset
T of R  of lmeasre 0 ach tha*; the series
 4,);O( Y comerges %r all v is R  -- 71 (7}
Take a poi y is. R  - T; keep y fixed arid consider the set So We wl prove
@ has Imeasre zero.
We can assm.e tha S is nonempy; otherwise he res is rivial Let
The A(y) is a com.abte collection of one-dimensioeai imervals wh{ch we reiabe
as {Y., Ja .... }. The um of the legs of ai te inteva:s & converges because of
(7) If x s @, the (x, y) s S so (x, y)  1, = X x y for ilnkely many k, and
hence x e J i:r infinitely ma W k. By the one-dimensional version of Theorem
I5.3 it foltows that 5. has l-measu zero. This shows that S, has --measare zero
for Mm.os. al y in R *, and a similar argument proves that S" has 1--measure zero
fbr a:most all x i R  ,
15.7 FUBINI'S REDUCrlON THEOIed£;M FOR DOUIgLE INTEGRALS
.Toem I££ Assume f is
a) There is a see T f l-measure 0 such thae ghe Lebesgue mg<qra]  f{x, y) dx
exsis Jbr aY y m R  ...,
b) The fimction G defined on R
G(y) =
is Lebesue-imqrabe on "R
f{.x,
NOT5:, There is a corresponding zesut which concludes ha
Proof We have aready proved he theorem for sep Ngctions, We prove it ext
%r upper Nnctoes. Kfe U(R ) there s an increasing seqece of step Panctions
{.,  such tha s,,(x, y) ./L. y)
measure 0;

Now (x, y) e R 2 - S if, and only if, x e R t - S,. Hence
s.(x, y)  f(x, y) if x • R i -- S r ($)
t t.(y) = a s.(x, y) .  tel es for eh l y d is  inbl¢
fon ofy. Moover, by vom 15.2 
ts. fo, by e  theom m 10.24)  is a fon t in
t) such at t  t ost ev¢wh on R
T of l-m 0 su at t(y)  t(y) ify  R  - Tv Moov,
gy) dy -- lim
Again, since {t,} is increasing, wc have
t,(y) = f s,(x, y) dx <_ t(y) if y  R  - Tv
J!
Applying the Levi theorem to {s,} we find that if y  R  - T1 there is a function,
g in L(R ) such that s,(x, y) - g(x, y) for x in R  - A, where A is a set of l-
measure 0. (The set d depends on y.) Comparing this with (8) we see that if
)'R 1 - Tt then
(, y) =/(x, y) if x  R  - (  s,). (9)
But A has l-measure 0 and S, has t-measure 0 for almost all y, say for all y in
R  - T2, whcr T± has t-measur 0. Lc't T = T  Tv Then T has l-measure 0.
If y  R  - T, the set .t  Sy has l-measure 0 and (9) holds. Since the integral
. g(x, y) dx exists if y  R t - T it follows that the integnfl l, f(x, y) dx also
exists if y  R  -- T. This proves (a). Also, if ), e R  - T w¢ have
Sinc t /_.(R), this proves (b). Finally, w¢ have
Th. 15 Tonclll-Hcisoa Test fo¢ lntegrai 415
Comparing this with (10) we obtain (c). This proves Fubini's theorem for upper
functions.
To prove it for Lcbcsgue-integrable functions we write f :- u - v, where
u  L(R z) and v 6 L(R z) and we obtain
As an immediate corollary of Theorem 15.6 and the two-dimensional analog
of Theorem 10.11 we obtain:
Tleorem 15.7. Assume that f is defined and bounded on a compact rectangle
I = [a, b] x [c, d-J, and thatfis continuou almost everywhere on I. Thenf6 L(I)
and we hae
NOTIL Tke one-dimensional imcgral ,[. f(x. y) dx .¢ts for almost all y in [e, d]
as it Lebesgue iIcgral, It need not exist as a Riemann integral. A similar remark
applies to the i,tegral jf(, y) dy. I th Riemann theory, the ingr integrals
in the reduction formula must be replaced by upper or lower integrals.
Theorem 14.6, part (v).)
There is, of course, an extension of Fubini's theorem to higherimensional
integrals. If f is Lebesguc-integrabfe on R "+ the analog of Theorem t5÷6
concludes that
Here we have written a point in R =+* as (x; y), where x  R = and y • R . This
can be proved by an extension of the method used to prove the two-dimensional
case, but we shall omit the details.
15.8 THE TONELLI-HOBSON TEST FOR INTEGRABILITY
Which functions are Lebesgue-integrabte on R27 The next theorem gives a uscfu[
sufficient condition .for integrability. Its proof makes use of Fubini's theorem.
1"theorem 15.8. ,4ssume that f is measurable on R 2 atld assume that at least one of
the two iterated integrals

exists. Then we have:
a) fe L(R2).
Proof. Part (b) follows from part (a) because of Fubini's theorem. We will also
use Fubini's theorem to prove part (a). Assume that the iterated integral
Jr, [_J=, If(x, Y)I dx] av exists. Let {s,} denote the increasing sequence of nonneg-
alive step functions defined as follows:
s,(xy)- n_ if lx, < n and , y, < n,
otherwise.
Let f.(x, y) = min {a,(x, y), If(x, y)l}- Both s, and Ifi are measurable so f is
measurable. Also. we have 0 < f(x, y) < a,(x. y), so f. is dominated by a
Lebesgue-lntegrable function, Therefore, f  L(R2). Hence we can apply Fubini's
theorem tof, along with the inequality 0 <_ f(x, y) < If(x, Y)I to obtain
Since {f,} is inereasing, this shows that the limit lim,_. $$: f. exists. By the Levi
theorem (the two-dimensional analog of Theorem 10.24), {f=} converges almost
eve .rT*'here on 112 to a limit function in LOIS). Butf,(x, y)  If(x, Y)I as n --,
so Ifl • L(R2)- Siucefis measurable, it follows thatfe L(R2). This proves
The proof is similar if the other iterated integral exist,.
15.9 COORDINATE TRANSFORMATIONS
One of the most important results in the theory of multiple integration is the
formula for making a change of variables. This is an extension of the formula
f(x) dx =
which was proved in Theorem 7.36 for Riemann integrals under the assumption
that 7 has a continuous derivative e' on an interval T
continuous on the image
Consider the special ease in which #' is never zero (hence of constant sign) on
T. If7' is positive on T, then t/is increasing, so
and the above formula can be written as follows:
Th. 15.10 Coortlim 'I'mmfomutikms 417
On the other hand, if g' is negative on T. then g(T) = [g(d), g(c)] and the above
formula becomes
Both cases are included, therefore, in the sin#ie formula
f f(x)dx = f fig(t)] ]t/'(t)I dr. (I I)
d
.Jr
Equation (t 1) is also valid when c > d, and it is in this form that the result will be
generalized to multiple integrals. The function r which transforms the variables
must be replaced by a vector-valued function called a coordinate transformation
which is defined as follows.
De)iiao 15.9. Let T be an open subset of R'. A ector-alued function g : T  R"
is called a coordinate transformation on T if it has the fottowin three properties:
a) gC' onT.
b) g is one-to-one on T.
e) The dacobian determinant Jr(t) = det Dg(t) # 0 for all | in T.
o, A coordinate transformation is sometimes called a diffeomorphism.
Property (a) states that g is continuously differenliable on T. From Theorem
13.4 we know that a continuously differentiable function is locally one-to-one
near each point where its Jacobian determinant does not vanish. Property (b)
assumes that g is tobatly one-to-one on T. This guarantees the existence of a
global inverse g- t which is defined and one-to-ote on the image g(/'). Properties
(a) and (c) together imply that g is an open mapping (by Theorem 13.5). Also, g- t
is continuously differentiable on g(T) (by Theorem 13.6).
Further properties of coordinate transformations will be deduced from the
following multiplicative property of Jacobian determinants.
Theorem 15.10 ( Multiplieatio tlteorem for Jacobian determinants). Assume that g
is differentiable on an open set T in R" and that h is differentiabfe on the ima#e g(T).
Then the composition k = h o g is differentiable on T, and for everj t in T we have
J(t) = Jt[g(t)]Jt(t). (12)
Proof The chain rule (Theorem 12.7) tells us that the composition k is differen-
tiable on T, and the matrix form of the chain rule tetls us that the corresponding
Jacobian matrices are related as follows:
Dk(t) = Dh['g(t)] Dg(t). (I 3)
From the theory of determinants we know that det (AB) = det A det B, so (!3)
implies (12).

418
This theorem shows that if g is a coordinate transformation on T and if h is a
coordinate transformation on g(T), then the composition k is a coordinate trans-
formation on T. Also, if h = g-l, then
k(t) = t for all t in T, and Jk(t)  1,
SO Ju1g(t)]J(0 ---- 1 and g- ' is a coordinate transformation on
A coordinate transformation g and its inverse g-' set up a one-to-one corre-
spondence between the open subsets of Tand the open subsets of g(T), and also
between the compact subsets of T and the compact subsets of g(T). The following
examples are commonly used coordinate transformations.
Examine 1. Potr ¢ordlne in R'r  t]]  1A
T= {(tt, h):t > 0,0< tz <
and we let g - (el, #2) be the function defined on Tas follos:
at(t)  tl cos t, g(t)  tt sin
It is customary to denote the components of t by (r, 0) rather than (tl,
ordinate transformation g maps each point (r, 0) in T onto the point (x, y) in g(T) given
by the familiar formulas
x = rcosO, y = r sin 0.
The image g(T) is the set R 2 - (x, 0) : x >_ 0L and the Jacobian determinant is
cos 0 sin 0
Jt(t) = _rsin0 teas0 = r.
Emtple 2 Cylindrical coordinates in Ra÷ Here we write t = (r, O, z) and we take
T= ((r,O,z):r> O, 0 < 0< 2a, -o < z< +o0).
The coordinate transformation g maps each point (r. 0, z) in T onto the point
in g(T) gi'cn by the equations
x = rcosO, y = r sin 0, z = z.
Figure 15.3
419
The image g(T) is the set R  - {(x, 0, O):x >_ 0}, and the Jacobian determinant is
iven 
cos 0 sin 0 0
Jr(t) = -rsin0 teas0 j = r.
0 0
The geometric significance oft, 0, and z is shown in Fig. 15.3.
Europe 3, Slhericat coordinates in R 3. In this case w¢ write t = (p, 0, p) and we lake
The coordinate transformation g maps each point (p, 0, ¢) in 7" onto the point (x, y, z)
in g(T) given by the equations
x  pcos0sin , y = psin0sin ¢, z-- pcos#,.
The image g(.T) is the set R  - [{(x, 0,0):x > 0} {(0,0, z):zR}], and the
Jacobian determinant is
I cos0sin¢ . sin0sin ¢ cos¢ ]
d(t) = [-psinlsin #cos0sin ¢ 0
I
lPc°s0c°s9 psin0cos  -psin ¢
 _p2 sin ,
The geometric significance of p, 0, and ¢ is shown in Fig. 15.4.
Figure 15.4
Example 4. Linear transformations in R . Let g : R"  R" be a linear transformation
reprcnted by a matrix (a2) = re(g), so that
g(t) = az,, .., . aofl ,
j=l
en g = (g] .... , g) wh¢ g(t) = :, nat j, and the abian matrix is
us the Jacobian determinant Jr(t) is constant, and aals det (aty), the determinant of
the mato (az). We a[ 11 this the deteinant of g and we ite
det g = det

A linear transformation g which is on,-to-one on R" is called nonsingutar.
We shall use the following elementary facts concerning nonsingular transforma-
tions from R" to R'. (Proofs can lx fr, und in any text on linear algebra; so, also
Reference 14.1.)
A linear transformation g is nonsingular if', and only if, its matrix .4  re(g)
has an inverse A- t such that .4.4 -  = I, where I is the identity matrix (the matrix
ot" the identity transformation), in which case A is also calIed nonsingular. An
n × n matrix A is nonsingular if, and only if, det A # 0. Thus, a linear function
g is a coordinate transformation if, and only if, det g # 0.
Every nonsingular g can be expressed as a composition of three special types
of" nonsingular transformations called elementary transformations, which we refer
to as types a, b, and c. They are defined as follows:
Type a: g(t .... , it, ..., t.) = (tl, ..., ).tt, ..., t), where - # 0. in other
words, g, multiplies one component of t by a nonzero scalar A. In particular, g,
maps the unit coordinate vectors a follows:
g(u) -- ,.u tor some k, g(u3 = u for all i # k.
The matrix ot" g can be obtained by multiplying the entries in the kth row of the
identity matrix by ,L ALso, del g¢ = )..
Type b: go(t1,..., t .... , t.) = (tl ..... t + ti, . . . , t,), where j # k, Thus,
g replaces one component oft by itself plus another, in particular, g maps the
coordinate vectors as follows:
g(a) = tl + u for some fixed k and j, k #j,
g(u¢) = u for alli  k,
The matrix g can be obtained from the identity matrix by replacing the kth row
of I by the kth row of/plus thejth row of L Also, det g, = l.
Type c: g(tt, ..., tl ..... t,..., t) ---- (t t ..... t3, ..., t, ..., t), where i # j.
That is, g¢ interchanges the ith and jth components of t for some i and j with
i # j. in particular, g(u) = u., g(u) = u, and g(u)  ul for all k # i, k # j.
The matrix of g¢ is the identity matrix with the ith and jth rows interchanged. In
this case det g -- - I.
The inverse of an elementary transformation is another of the same type. The
matrix of an elementary transformation is called an elementarj matrix. Every
nonsingular matrix /I can be transformed to the identity matrix I by multiplying
A on the left by a succession of elementary matrices. (This is the familiar Gauss--
Jordan process of linear algebra.) Thus,
where each T is an elementary matrix Hence,
a = T, .-.
Th. 15,12 Tramffocmatimi Fotmmla for Linear Coordinate Transformations 421
If .4 ---- re(g), this gives a corresponding factorization of g as a composition of
elementary transformations,
15.10 THE TRANSFORMATION FORMULA FOR MULTIPLE INTEGRALS
The rt of this chapter is devoted to a proof of the following transformation
formula rot multiple integrals.
Theorem I.%11. Let T be an open subset of R  and let g be a coordinate transfor-
mation on T. Let f be a real-valued function defined on the image g(T) and assume
that the Lebesgue interat .[rf(x) dx exists, Then the Lebesgue inteyral
zf[g(t)'l IJt(t)l dt also exists and we have
,,r,,f(x) dx = r f[g(t)'l IJs(t)] dr. (14)
The proof of Theorem t 5. t t is divided into three parts. Part ] shows that the
formula holds for every linear coordinate transformation =. As a corollary we
obtain the relation
tz[m(et)] = Idet tl/(g),
for every subset A of R* with finite Lebesgue measure, in part 2 we consider a
general coordinate transformation.g and show that ([4) holds when f is the
characteristic function of a compact cube. This gives us
t(K) : | IJt(t)l dr, (15)
for every compact cube K in g(T). This is the lengthiest part of the proof. In part
3 we use Equation (I 5) to deduce (14) in its general form.
15.t[1 PROOF OF THE "IRANSFORMATION FORMULA FOR LINEAR
RDA ANSTIONS
Tem 15.I2. Let  : R   R  be a linear coordinate trfortion. If the
Lebesgue integral t.f(x) dx exists, then the besgue integral l.f[m(t)] IJ.(t)l dt
also exits, and the two inteyrals e equal.
Prof. First we note that if e theatre is true for • and , then it is also te for
the comsition  =   a au
ft f(x'dx=f, f[a(t)][d'(t)ldt= ft f('t)])lJ't)]JIJt'[dt. .
= f /[t)]/J,(t)l dr,

Therefore, since every nonsingular linear transformation  is a composition
of elementary transformations, it suffices to prove the theorem for every elemen-
tary transformation, It also suffices to assumef .> 0.
Suppose  is of" type a. For simplicity, assume that t multiplies the last
component of I by a nonzero scalar 2, say
Then t J=(0[ =[det "1 = 121. We apply Fubini's theorem to write the integral off
over R  as the iteration of an (n - 1)dimensional integral over R*-
dimensional integral over R 1. For the integral over R 1 we use Theorem 1O.17(b)
and (c), and we obtain
where in the last step we use the Tone]]i-Hobson theorem. This proves the theorem
if", is of type a. If  is of type b, the proof is similar excep! that we use Theorem
10.17(a) in the one-dimensional integralk In this case IJ.(01 -- 1. Finally, if $ is
of type c we simply use Fubini's theorem to interchange the order of integration
over the ith and jab coordinates. Again, 1J=(t)l = I in this case.
As an immediate corollary we have"
'heotem 15.13. If " R  - R" is a linear coordinate transformation and if A is
any subset of R* such that the Lebes.que integral S,f(x)dx exists, then the
Lebesgue integral ja f[t(0] IJ=(t)l dt also exists, and the two are equal.
Proof. Let](x) =f(x) ifx  (A), and lel](x) = 0 otherwise. Then
As a corollary of Theorem 15.13 we have the following relation between the
measure of 4 and the measure of (A).
15.15 "luformatit Formula for a Compact  423
l"lg, m,tm 15.14. Let  : It"  It" be a linear coordinate transformation. If A is a
subset of R" with finite Lebesgue measure I(A), then (A) also has finite Lebesyue
measure and
,[:(a)] = Idet [/(A). (16)
Proof. Write A = at-(B), where B -- (A), Since m- is also a coordinate
transformation, we find
This proves (16) since B = (A) and det (-) = (det at) -1.
7rlmarem 15.1. If .4 is a compact Jordan-measurable subset of It , then for any
linear coordinate transformation  : R*  R* the image ,,(A) is a compact Jordan-
measurable set and its content is given by
c[(A)] = Idea tt c(A).
Proof The set (A) is compact because t is continuous on A. To prove the
theorem we argue as in the proof of Theorem 15.14. In this case, however, all the
integrals exist both as Lebesgue integrals and as Riemann integrals.
15.1Z PROOF OF THE TRANSFORMATION FORMULA FOR THE
CHARAC FUNCTION OF A COMPACT CUBE
This section contains part 2 of the proof of Theorem 15.11. Throughout the '
section we assume that g is a coordinate transformation on an open set T in
Our prpose is to prove that
-- at,
t  (K)
for every compact cube K in T. The auxiliary results needed to prove this formula
are labelled as Iemmas.
To help simplify the details, we introduce some convenient notation. Instead
of the usual Euclidean metric for R* we shall use the metric d given by
d(x,y)= max lxz-
This metric was introduced in Example 9, Section 3.13. In this section only we
shall write .fix - Yll for d(x, y).
With this metric, a ball B(a; r) with center a and radius r is an n-dimensional
cube with center a and edge-length 2e ; that is, B(a; r) is the cartesian product of
n one-dimensional intervals, each of length 2r. The measure of such a cube is
(2r)', the product of the edge-tengths.

then
If a R"  R  is a linear transformation represented by a matrix (ate), so that
We also define
It[ = max  lat2l. (18)
This defines a mec II - ll oa the sp of all liner transfoafions fm
R" to W. The fit 1emma v m¢ proi of this
Le 1. t • and  denote linear transformations fiom R" to R'. Then we ha:
d) IlIl[ = I, where I i the identiy lrortion.
Proof Sup that ms . 1 alj ig alain for i = p. Te
a  0, x = - 1 i a2 < 0, and x 3. = 0 ifj  p. en sl = I and IIlF
la(x)i!_, which proves (a).
Pa (b) follows at on from (17) and (18). To prove (c) we use (h) to
Tang x with tlxll = I  that I[(a o fx)l[ = j[ o 1[, we obtain (c).
Finally, if I is the idtity transformation, then eh sum  1 laj[ = 1 in
(18) so ill[l_ = 1.
The coordinate transformation g is diffentiable on T, so for h t in T the
 rivative g'(t) is a liner transformation fm R' to R" pnt by the
Jobian mai t) = (DgI)). Thefo, ting  = g'(t) in (18), we find
= max 
1 [a  1
We note that IIg'0)ll is a continuous function of I sin all the pai derivaiv
Dyz are continuous on Z
If Q is a compact sut of T, eh function Dg is und on Q; hen
I[g'(011 is also unded on Q and we define
l..mmm 3 Trmmformali Fomulla for a Compact Cul 42
The next lemma states that the imag¢ g(Q) of a cube Q of edge-length 2r lies
in another cube of edge-length
/m 2. Let Q = {x" IIx - all  r} be a comzr cbe of edge-length 2r
lying in Z Then for each x in Q we hiroe
IIg(x) - g(a)II  rAn(Q).
(20)
Therefore g(Q) ties in a cube of edge-length
Proof. By the Mean-Value tlmorem for real-valued functions we have
where z lies on the line segment joining x and a. Therefore
fx) - a)l -<  Dz,)l Ix - a-f  IIx - all
./-1 j--!
and this implies (20).
so'r. Inequafity (20) shows that g(Q) lies inside a cube of content
(2rt,(Q))  = {
Lemm . lf A is any compact Jordan-measurable subset of T, then g(A) is a com-
pact Jordan-measurable subset of g(T).
Proof. The compactness of g(A) follows from the continuity of g. Since A is
Jordan-measurable, its boundary A has content zero. Also, 8(g(,4)) = g(tSA),
since g is one-to-one and continuous. Therefore, to complete the proof, it suffices
to show that g(tA) has content zero.
Given e > 0, there is a finite number of open intervals A,..., A, lying in
T, the sum of whose measures is < e, such that cA  t,.J'=l A,. By Theorem 15.1,
this union can also be expressed as a union U() of a countable disjoint collection
of cubes, the sum of whose measures is < . If  < I we can assume that each
cube in U(e)is contained in U(I). (If not, intersect the cubes in U(e) with U(I) and
apply Theorem 15.1 again.) Since d.,l is compact, a finite subcotlection of the cubes
in" U(,) covers &4, say Q1, ---. Q. By Lernma 2, the image g(t) lies in a cube of
measure {&(Q}c(Qt). Let . = ,(U(-'0Y). Then Ad(t) <: 3. since , _ tT0j.
Thus g(cq) is covered by a finite number of cubes, the sum of whose measures
does not excee6 ) __  c() < ÷ Since this holds for every  < 1, it follows
that g(&4) has Jordan content 0, so g(d) is Jordan-measurable.
The neat lemma relates the content of a cube Q with that of its image g(Q).

lmma 4. Let Q be a compact cube in T and let h -- ++  g, where ¢+ : R" - R" is
an) nonsingutar linear transformation. Then
c[g(Q)] < Idet 'l- {3,(Q)}"c(Q). (21)
Proof, From Lemma 2 we have c[-g(Q)] _< {2/(Q)}'c(Q). Applying this inequality
to the coordinate transformation h, we find
c[h(Q)] <
But by Theorem 15.15 we have c[h(Q)] = c[m(g(Q))] = Idet al c[g(Q)], so
c[g(Q)] = Idet l-t c[h(Q)] < Idet l-
Lemm 5, Let Q be a compact cube in T Then for every  > 0, there is a < > 0
such that if t  Q and a e Q we hae
IIg'(a)+" o g'(t)ll < 1 +  wheneer lit - a < & (22)
Proof. The function [Ig'(0-'l[ is continuous and hence bounded on Q, say
Ilg'(t)-ll < M for ail t in Q where M > 0. By the continuity of llg'(t)ll, the is
a  > 0 such that
IJ#Ct) - g'(a)l[ < -- whenever lit - all < &
M
If I denotes tle identity transformation, then
g'(s)- o g'(t) - (t) -- '(a) - o {f(0 - so if lit - all <  we have
Hg'()-' g'(t) l(011 -< IIg'(a)'ll IIg'(t) g'(a)ll < M e
M
The triangle inequality givms us IIll <- U/ill + II -/hi- Taking
s : g'(a)| o g'(t) and / -- (0,
we obtain (22).
Lemm 6. Let Q be a compact cube in T. Then we hae
cl-g(Q)] _< o IJ+(t)l dr.
Proof. The integral on th fight exists as a Riemann integral becaus the inte-
grand is continuous and bounded on Q. Therefore, given e > 0, there is a partition
P of Q such that forevery Rlemana sum S(P, [J=t) with P finer than P, we have
Is(P, IJl) - a lJi(t)] dt[ < .
Take sh a partition P into a finite number of cubes Qa, .... Q=, each of which
Luna 7 Transformafion Formula for a Compact Cube 427
has edge-length <& where  is the number (depending on ) given by Lemma
Let a denote the center of Q and apply Lemma 4 to Q with t - g'(a+)-
obtain the inequality
c[g(Q,)] _< [dct g'(a,)l {2(Q,)}" c(Q,), (23)
where h = a o g. By the chain rule we have h'(t) = t'(x)  g'(t), where x
But ,'(x) = • since ,, is a linear function, so
h'(t) =  o g'(t) - g'(a,)- o g'(t).
But by Lemma 5 we have IIh'(t)l[ < 1 ÷ e ill  Q, so
(Q) = sup ItW(t)ll -<  ÷ .
Thus (23) gives us
c[g(Qi)] .< Idct g'(a)l (I + e)" c(Q),
Summing over all i, we find
e[g(Q)] _< (I + t)   Idet g'(a31
Sin dct g'(a) = J,(a), c sum on the right is a Riemann sum S(P, ]Jtl), and
since S(P, J) < @ IJ(t)] dt+ e, wc find
c[g(Q)]  (i +{f [J,(t)tdt+}.
But e is bitr, so this impli c[g(Q)]   I J,(01 dr.
Le 7. t K be a comct cu in g(T). Then
(24)
Proof. e imeal exists as a Remann inteat ause the integrand is con-
finuous o the compt t g- t(K). Al, by mma 3, the inteal over g- (K)
is equal to at over the interior of g-(). By Th¢om 15.1 we can
;=l
wh¢ {A , A2,... } is a countable disjoint colllion of cus who closure ties
in the interior of g-(K)+ us, int g-(K) = = Q where each Q is the
closure of A. Sin the integral in (24) is also a sgue inteM, we can use
countable additivity along with mma 6 to write
[Jt(t)l dt = IJt(t)l dt 2 rig(Q,)] = v Q, = v(K).
-IR ) =l + i=l

Lemma 8, Let K be a compact cube in g(T). Then for any nonnegatime upper
function f which is bounded on K, the integral ,-qf[g(t)] IJ.(t)l dt exists, and
we have the inequality
 f(x) dx < f, fig(t)] IJ.(t)l dr. (25)
- qg)
Proof Let s be any nonnegative step function on/C Then there is a partition of
K into a finite number of cubes K t .... , K, such that s is constant on the interior
of each K, say s(x) = a, _> 0 if x ¢ int Kv Apply (24) to each cube K,., multiply
by al and add, to obtain
f, ,(x) ax <_ s[g(t)] lJ,(t), dt. (26)
Now let {s} be an increasing sequence of nonnegative step functions which
converges almost everywhere on K to the upper function f. Then (26) holds for
each st. and we let k -, c io obtain (25). The existence of the integral on the
right follows from the Lebesgue bounded convergence theorem since both
f[g(0] and IJ(t)l are bounded on the compact set g-t(K).
Threm 15.16, Let K be a compact cube in g(T). Then we haoe
, Id(t)l dr.
#(K) = -,¢)
(27)
Proof. In view of Lemma 7, it suffices to prove the inequality
f, fJ,(OI at _< ,(r).
As in the proof of Lemma 7, we write
(28)
intg-l(K)= 0 A, = 0 Q:,
where {A , A,... } is a countable disjoint collection of cubes and Q.. is the closure
of A. Then
f,_ ',g, IJt(t)l dt = ,= j'e, [J,(|)l dr. (29)
Now we apply Lemma 8 to each integral JO, IJ,(t)l dt, taking f = IJ,I and using
the coordinate transformation h - g-. This gives us the inequality
which, when used in (29) gives (28).
Th. 15,17 Completioll of the Proof of the TranMmmath Formula 429
15.13 COMPLETION OF THE PROOF OF THE TRANSFORMATION
I)II_MULA
Now it is relatively easy to complete the proof of the formula
f,(r) f(x) dx = frf[g(t)] 'J'(t)l d|" (30)
under the conditions stated in Theorem 15.1 I. That is, we assume that T is aa
open subset of R', that g is a coordinate transformation on T, and that the integral
on the left of(30) exists. We are to prove that the integral on the right also exists
and that the two are equal. This will be deduced from the special case in which the
integral on the left is extended over a cube K.
Tloeem 1.17. Let K be a compact cube in g(T) and assume the Lebesgue inte.qral
Jxf(x) dx exists. Then the Lebesyue inte#ral -,,af[g(t)] I J,(01 dt also exists,
and the two are equal.
Proof. It suffices to prove the theorem whenfis an upper function on K. Then
there is an increasing sequence of step functions {s} saeh that s  f almost
everywhere on K. By Theorem 15.16 we have
for each step function &. When k  c, we have jr s(x) dx   f(x) dx. Now
let
if re R  - g-'(K).
so
By the Levi theorem (the analog of Theorem 10.24), the sequence {f} converges
almost everywhere on R ' to a function in L(R'). Since we have
ift  g-'(K),
ifte R" - g-'(K),
almost everywhere on R", it follows that the integral J,-,t)f[g(t)] jY,(t)l .dr exists
and equals  f(x) dx. This completes the proof of Theorem 15.17.

Proof of Theorem 15.11. Now assume that the integral J'go)f(x) dx exists. Since
g(T) is open, w¢ can write
(T)   2J  A,,
where {A, Az, ... } is a countable disjoint colkon of.cubes whose closure lies
in g(T). Let K dgnote the closure of ,4 v Using countable additivity and Theorem
15.17 w have
15.1 fife/T), where T is the triangular region in R 2 with yea-rices at (0, 0), (!, 0),
and (0, 1), prow that
15.2 For fi c, 0 < c < 1, f R 2  follv.
f(x,y) = [l-yr/(x-yr ifoy<x,O < x< !,
ve at f 2) d  the double  fez f(x, y) d(x, y).
1  S  a mb su ofR  wi ite m 8). U  nation of
iti I5.4,  tt
lE4f(x,y)=e -xy if x 0, y20. d I](x,y)= 0 o.
ist d  ,  tt  u in offo R  d not ist. AI, =plain
why ts d not tmdict
431
1K Let f(x, y) = (x  - y:)/(x  + y2)z for 0 _< x _ I, 0 < y _< t, and let f(0, 0) =
O.Prov¢ that both iterated integrals
exist but are not equal. This shows thatfis not Lebsgu-integrabI¢ on [0, 1 ] × [0, 1 ].
LS. Let r = [0, l] x I0, 1], et /(x, y) = (x - Y)tCx + y) if (x, y)• ,r, (x, y) ¢
(0, 0), and let f(0, 0) = 0. Prove thatf¢ L(I) by considering the iterated integrals
15.7 Let I = [0, I ] x [1, + o) and let f(x, y) = e -; - 2e -2 if (x, .v) ÷ 1. Prove
that f ¢ L(I) by considering the iterateA integrals
15.# The following formulas for transforming double and triple integrals occur in ele-
me.ntary calculus. Obtain thtan as consequences of Theorem 15.11 and give restrictions
on T and T  for validity of these formulas,
a) fff(x,y)dxdy=fff(rcosO, rsinO)rd, dO.
b)
yffl(x,y,z) dxdydz=fff/(rcosO, rsmO, z)rdrdedz.
15.9 a) Prove that
coordinates.
b) Us pan (a) to
c) Us part (b) to
d) Use part (h) to
15.10 Let V.(a) denote the n-measur¢ of th© n-ball B(0; a) of radius
outlines a proof of the formula
e -(x+') d(x,y)= n by transforming the integral to polar
prove that j'_ e -a2 dx = x/r.
prove that J't e-II'll d(xt,..., xa) = t a/.
calculate ' e -t dx and (._ x  e - dx, t > O.
• %(a) = ---
a) Use a linear change of variable to prove that//.(a) = aV(l).

b) Aume n  3, expr¢ Ihe integral for V,(I) as tl',e itexation of  (n - 2)-fold
in and a double al, d u  (a) for  (n - 21 Io obin t
fula
(1) = _(1) (1 -
c) Fr the on foa in ) d thai
e) -
r({ + 1)"
15.1 ¢f to Exi 1.t0 d p tt
f, x[ (,,..., .) - e)
;i)
1i2 Refer to Exe 1.10 and
(n - Ifold int g a oiot in#, to obn t uon foula
(l)
t x = c t in the instal, d u e fuM of 15.10 to du that
¢ t dt -
the n-mu of $a), is
a) U a liar ang¢ of variable t v¢ thal V.(a)  (I).
b) Au n  2, p  intaI for Vl)  an ittion ofa ondimensi@nal
inl¢l d  (n - lfold intl,  (a) to show that
1
and dedu that VI)
15,14 Ifa > 0 d n 2 2, let S(a) denote h¢ following ml in R':
Let (a) noe t e of $,(a). Um • th sut by ei 15.13 o
.15  Q,(a) Oeno t 't quaint" of the  0:a) given by
t f(x) = x--- x and pm
SUGGESTED REFERENCES FOR FURTHER STUDY
15.1 AsDlund, E., and Bungart, L., A First Course in lnteflration. Holt, Rinehart, and
Wirtston, New York, 1966.
15.2 Battle, R.,  Elements of lntegration. Wiley, New York, 1966.
15.3 Kestelman, H., Modern Thearies of lntegratlon. Oxford Univexsity Press, 1937.
15.4 Korcvam', J., MatMmaticalMethods, Vol. !. Academic Press, New York, 1968.
15.5 Riesz, F., and Sz.-Nagy, B, Functional .4nalys/s. L. Boron, translator. Ungar,
New York, 1955.

CHAPTER 16
CAUCHY'S THEOREM
AND THE
RF  IDUE CALCULUS
16.1 ANALYTIC FUNCTIONS
The concept of derivative for functions of a complex variable was introduced in
Chapter 5 (Section 5.15). The most important functions in complex variable theory
are those which possess a continuous derivative at each point of an open set.
These are called analytic functions.
Defmitia 16.1. Let f = u + iv be a complex-valued function defined on an open
set S in the complex plane 12. Then f is said to be analytic on S if the derivative f'
exists and is continuous* at every point of S.
NOT. If T is an arbitrary subset of C (not necessarily open), the terminology
"fis analytic on T" is used to mean thatfis analytic on some open set containing
T. In particular, f is analytic at a point z if there is an open disk about z on which
fis analytic.
It is possible for a function to have a derivative at a point without being
analytic at the point. For example, iff(z) - [z[ z, thenfhas a derivative at 0 but
at no other point of C.
Examples of analytic functions were encountered in Chapter 5. If f (z) = z"
(where n is a positive integer), then fis analytic everywhere in C and its derivative
is f'(z) = n-1, When n is a negative integer, the equation f(z) -- z" if z # 0
defines a function analytic everywhere except at 0. Polynomials are analytic
everywhere in C, and rational functions ate analytic everywhere except at points
where the denominator vanishes. The exponential function, defined by the formula
e  -- e(cos y + i sin y), where z = x + iy, is analytic everywhere in C and is
equal to its derivative. The complex sine and cosine functions (being linear
combinations of exponentials) are also analytic everywhere in C.
Let f(z) = Log z if z  0, where Log z denotes the principal logarithm of
z (see Definition 1.53) Thenfis analytic everywhere in C except at those points
z - x + iy for which x < 0 and y = 0. At these poims, the principal logarithm
fails to be continuous. Analyticity at the other points is easily shown by verifying
* It can be shown that the existence off' on ,S automaticalIy implies continuity off" on
S (a fact discovered by Goutsa! in 1900). Hence an analytic function can be defined as
one which merely possesses a derivative everywhere on ,S'. However, we shall incIude
cOIltinuity off" a part of the definition of analyticity, since this allows some of the proofs
to run more smoothly.
that the real and imaginary parts off satisfy the Cauchy-Riemann equations
{Theorem 12.6).
We shalt see later that analyticity at a point z puts severe restrictions on a
function, it implies the existence of all higher derivatives in a neighborhood of z
and also guarantees the existence of a convergent power series which represents
the function in a neighborhood oft. This is in marked contrast to the behavior of
real-valued functions, where it is possible to have existence and continuity of the
first derivative without existence of the second derivative.
162 PATHS AND CURVES IN THE COMPLEX PLANE
Many fundamental properties of analytic functions are most easily deduced with
the help of integrals taken along curves in the complex plane. These are called
contour integrals (or complex line integrals) and they arc discugsed in the next
section. This section lists some terminology used for different types of curves,
such as those in Fig. 16.1.
axe Jordan ate  curw Jonhm curv
 16,1
We recall that a path in the complex plane is a complex=valued function y,
continuous on a compact interval [a, hi. The image of [a, b'l under  (the graph
ofT) is sa/d to be a cure described by y and it is said to join the points a)
and y(b).
If /(a) # ?(b), the curve is called an arc with endpoints l'(a) and (b).
If , is one-to-one on [a, hi, the curve is called a simple arc or a Jordan arc.
If "t(a) = "t(b), the curve is called a closed curve. If y(a) = ?(b) and if 5' is
one-to-one on the half-open interval [a, b), the curve is called a simple closedeume,
or a Jordan curve.
The path 7 is called rectifiable if it has finite arc length, as defined in Section
6.10. We recall that 5' is rectifiable if, and only if, 5' is of bounded variation on
[a, b]. (See Section 7.27 and Theorem 6.I%)
A path 5' is called piecewise smooth if it has a bounded derivative ' which is
continuous everywhere on [a, b] except (possibly) at a fiqite number of points,
At these exceptional points it is required that both right- and lea-hand derivatives
exist. Every piccewise smooth path is rectifiable and its are length is given by the
integral  l,'(t)l dt,
A piecewise smooth closed path will be called a circuit.

Def. 16.:l
lmre 16,2
DeluSion 16.2. lf a  C and r > O, the path ? defined by the equation
(0)  a + re , 0  0  2,
i called a positively ,rited circle with center at a d ri r
. ¢ omec mning of (0) is shown in ig. 16.2. As 0 varies from 0
to 2n, the int 0) moves lkwisv around t I¢.
16.3 CONTOUR INTEGRALS
Contour imegrals will be defined in terms of complex Riemann-Stieltjes integrals,
discussed in Section 7.27.
Definition 16.:L Let }, be a path in the complex plane with domain [a, hi, and let f
be a complex-valued function defined on the graph of . The contour integral off
along , denoted by Sr f, is defined by the equation
whenever the Riemann-Stiettjes integrat on the right exists.
NOTATION. We also write
f(z) dz or f(z)
for the integral. The dummy symbol z can be replaced by any other convenient
symbol. For example, Jrf(z) dz =-- Jr f(w) dw.
If  is rectifiable, then a sufficient condition for the existence of fis thatfbe
continuous on the graph of , (Theorem 7.27).
The effect of replacing " by an equivalent path (as defined in Section 6.12) is,
at worst, a change in sign. In fact, we have:
Theorem I6.4. Let  and 3 be equivalent paths describing the Same curve F. If
Jf exists, then Sof aiso exists. Moreover, we have
"I'k 16.6 Colltollr 
if  and 5 trace out F in the same direction, whereas
if  and 5 trace out F in oppesite directions.
Proof. Sulose 5(t) -- ,[u(t)] where u is strictly monotonic on [c, d]. From
the change-of-variable formula for Riemann-Stieltjes integrals (Theorem 7.7) we
have
fly(t)] d?(t) -- f[(t)] dfi(t) - f. (1)
J c)
Ifu is increasing then u(c) = a, u(d) = b and (l) becomes ]f = f.
If u is decreasing then u(c) = b, u(d) - a and (I) becomes - f = , f.
The reader can easily verify the following additive properties of contour
integrals.
Theorem 16.5. Let ? be a path with domain [a, hi.
i) If the integrals ,f and , y exist, then the integral  (f + fig) exists for every
pair of complex numbers a, fl, and we hae
ii) t ? d ?z denote the restrictio of ? to [a, c] d [c, hi, resctivefy,
where a < c < b. If two of the three integrals in (2) exBt, then the third also exists
 we e
In practice, most paths of integtion a tifiable. For such ths the
folloffing theom is often ud to tima the aolu value of a contour inteal.
Theorem 16.6. Let t be a rectifiable path of length A(). If the integral , f exists,
and f If(z)l < M for all z on the graph of , then we have the inequality
Proof. We simply observe that all Riemann-Stieltjes sums which occur ira the
definition of J f['/(t)] dy(t) t/ave absolute value not exceeding
Contour integrals taken over piecewise smooth curves can be expressed as
Riemann integrals. The following theorem is an easy consequence of Theorem 7.8.

have
16.4 T INTRAL ALONG A C/CULAR PATH AS A FUNCTION OF
THE RADIUS
Consider a crcuIar path 7 of radius r   ad cemer a given by
(0) = a + re , 0  0  2
In this sti.on we study e inteal yfas a thncfion, of the radius r.
Let (r) = j'f,. Snce. 7'(0) = fre:, Theorem t6.7 gves us
O(r) = f(a + re)ire e dO. (3)
As r varies over an nterval It,, r, where 0 :g r < ra the points y(O} trace out
an annulus which we denote by A(a; G, re). (See Fig. [6.3,) Thus,
lf.r = 0 the annulus is a closed disk of radius r> I(fis continuous on tte
then  is continuous o the inte'al Ira, ta]. Iffis analytic on the annulus, then
is differentiaNe on [r, ra]. The next theorem shows thai .p is censtam on [r,
iff is a.naly6c everywhere on the annulus excep possibly on a finite subset,
vided thatfis continuous oa this subset.
Figure 163
Theorem l&8o Assurne f is analytic on the annulus d{a; r, rz) , excep possib& at a
fin#e number of points, dg these exceptionM points assume hat f is contieuous
Tken the .[hncion  d@ed 0:" {3) is constant on the intereal [r, r2.] Atoreover
ff'r:. = 0 the const.ar gs O
groo¢ Let z ..... , z. deoe the exptionat, points wh.eref fai.!s to be analytic.
Lal h.ese points according to increasing distances from the center say
and let R = : - a.. A1so(!et R = e, R.+ = r>
The union of the irtervals [&, &. ] \>r k = O, , 2 ....... e, is {he interva
', gel. We wi]{ shove tha  is constant on each imerva [R, &...]. We write
(3) in the lbrm
(r) = { {r,.  ,0). d0.. where g{r,. ,0),. = ./'(a..a,
An easy application of the chan ru¢ shows iha we have
=,- = ir .......... , (4)
(I.e reader should verify this fi)rraua.) .Commuity of J" implies conthmty of the
partial derivatives @/& and <ic., Therefore. on each ope imerval (N, R+
we can calculate ('() by difl?retiation nder the ntegrai sign {Theorem
and then use (4) and the second fundamental georem of caicu}us (Teorem 7.34}
to obtain
Applying Theorem 2I0, we see ,hat p is constam on each open subinterval
(R. R.+ ). By continuity, p is constant on each c]od subinterval. [R, R + ] and
hence on heir unio [r, r]. From (3) we see t.at e(r).-* 0 as r .-.--. 0 so the
consam vatue of p is 0 f r = 0
16.5 CAUCHYS INT;EGRAL THEOREM FOR A CIRCLE
The fbilowing sfcial case of Theorem t68 is of particIar importance.
Theorem 16.9 (Cauchy', inteyral teorem Jar a circle), If f is" analygic on a disk
B(a; R) #xcepg po;dbdy fbr a finite number of points at whick f¢ is continuous, gken
fbr every c#cu[ar pah 7 wit center at a and radius" r < R.
Pro@ Choose r e so that r < r a < R and apply Theorem 6,8 with r = 0.
NOWe. There {s a moe genera1%rm of Cauc.hy's imegrat theorem in which the
circular pa{h  s replaced by a more general c!e.d path. These more general paths
wi] be atrodced through the concept of komot@y.
16.6 f-|OM()TOPIC CURVILS
Figure 6,4 shows three arcs having the same edpomts d and B and lying in an
open regior Do Arc I cae be comirmously deformed into arc 2 through a collection
ofitermediate arcs, each of which }ies m D. Two arcs with this property are said

Theorem ar,A e Resle C|e|s
,o be hom:,)wpic in D, Are : cannot be so deformed imo arc 3 (beca:se. of he
.hoe separa{ig hem) so gey are o bomotopc ie D
I {is section we give a h:-rma defiio oF homotopy, Then we stow hat, if
.fis analy:c  D., the cog:our mteal of,fi{om A {o B has the same value along
any two homotopic patios in D. n oher words, the va:e of a conVaur n,egal
f is gai:tered under a continuous deformation .of e path, provided be
termed{ate contours remai wkhin, tbe region of analyt/city off This prorly
of contour imegrals is of utmost .impor:.ace ig the applcations of complex
megration,
Oefialtioa 16.10. Let 7c, and y be lwo pahs with a common domain [.a, b]. A.sume
that eitker
a) % and-;, have the same en@oints: ?da) = y(a.} and 7o(b)  *&(b), or
b) 70 ad 7: are hoh ctosedpatks: 7(a) = y(b) and -'(a.) =
Let D be o subse f C congaimq ke graphs  gfYc, and y. Then ;: and y. are said
to he kometopic in D :" :here exis:s a fimcAon k, continuous on the rec,faqle
In add#ion g'e requ#e that fi>r each s in [0, ] we hove
3a) /:(s, a) = 77(a) am: k(s. b) = y<)=(b), in case (a);
3b) k(s', a) = k(s; b), in case
The concept of homoQpy has a smp{e geometric interpretation, For each
fixed s in [0, 1], le 7,,(:) = k(a, z). The y, can be regarded as an imermediate
moving path which sar:s t}om Ye whe s = 0 aed ends a.: . whet s = .
Em,le L t[:mot(:py o apeA¢, tf 7: iS a consan fuaclkm, so ha: is graph s a
Example 2. Lbwar kemoto??, [fi for each t m [o> hi, the Hne segmet jo:ing y0(r} and
?(e) its fn D, hen 70 and }q are homotopz n D cause the
pa.rtcuar, any v,o vaths w.Rh domain [a, b] am .inearly h.omotopic i:: C (he comple>
paee.) or, more genera/y, n any convex st: cor:ainng lhe:r graphs,
The new theorem shows that ke{ween any .wo homo{opic pa:hs we can inter-
palate a n.ie nmber of a{ermedae palygoeal paths, each of which is linearly
homo,opec to is eighbor.,
Theorem 16.11 (PoIyWonM interpMatm theorem). Let Yo and 7 he
paths in an open st: D Then. there e.:ri37 a fim?e number (f paghs > : : ...... % sck
that:
b) : is a potygona: park for : g j   --
Proqf Since y, agd ? are homoopic i D, tere is a homotopy h sai¢}'ing te
conditions in Definition I6.10, Cosider partkios
{so, s: ..... &.} of [0, t] and {..  .... , Q} of [<
into n equa parts, choosing n so large hat the {mage of each rectangle Is.e, s+ a]
[a, r . ] nder h is contained {n an open disk. D: contained in D. (The reader
shoutd verify, thal vhis is possible cause of uniform continuity of
On the ietermediae pa{h y, given by
y.,it) = hi: O, t) for 0 < j <
we iasc:ribe a polygonal path %. with vet{ices at the pole,s h(%, @. That
::.() = k(s, t) for k = 0, I .....
a.d j is liear o.n each submterva! [e,  +  for 0 N k N  - *, We also
<,: = % aed  = y.. (An exampte is shown in Fig.
The four re,ices :i(¢), %(:a,, :}, + :(*) and :+ ,{+ ) al :it in .:he disk
Since D.: is convex, the line segments joining them aso ie ix D.t  and hence the
pois
+ (I - (5)
a,,,..3= . 7 0

442 Cauchy'  anti  Residue C_Azlcn Th. 16.12
lie in DTt for each (s, t) in [0, t] × [tt, tt+'[. Therefore the points (5) lie in D
for all (s, t) in [0, 1] x in, hi, so j+ l is linearly homotopic to tj in D.
16.7 INVARIANCE OF CONTOUR INTEGRALS UNDER HOMOTOPY
]'heorem 16.12..4ssumefi$ analytic on an open set D, except possibly for a flnite
number of points where it .is continuous. If y o and ?1 are piecewise smooth paths
which are homotopic in D we have
Proof First we consider the case in which , and yt are linearly homotopic. For
each s in [0, 1] let
r(t) = s3,,(t) + (1 - S)7o(t) if re [a,
Then 3', is piecewisc smooth and its graph lies in D. Write
r,(t) = to(t) + where
and define
9(s)=,f=[f[?(t)]dyo(t)+s;f['(t)]drt(t),
for 0 < s _< 1. We wish to prove that p(O) = q(1). We will in fact prove that ¢p
is constant on [0, I].
We use Theorem 7.40 to calculate q'(s) by differentiation under the integral
sign. Sinee
=
this gives us
= ot(b)f[?,(b)] -- (a)f[y(a)],
by the formula for integration by parts (Theorem 7.6). But, as the reader can easily
verify, the last expression vanishes because yo and 't are homotopie, so ¢p'(s) = 0
for all s in [0, 1]. Therefore q is constant on [0, 1]. This proves the theorem when
Yo and 71 are linearly homotopic in D.
If they are homotopic in D under a general homotopy h, we interpolate poly-
gonal paths , as described in Theorem 16.1 !. Since each polygonal path is pieee-
wis smooth, we can repeatedly apply the result just proved to obtain
16.8 GENERAL FORM OF CAUCHY'S INTF_,GRAL THEOREM
The general form of Cauchy's theorem referred to earlier can now be easily deduced
from Theorems 16.9 and 16.12. We remind the reader that a circuit is a piecewise
smooth closed path.
Tlorem 16.13 ( Cauchy' s integral theorem for circuits hometopic to a lnt}. Assume
f is anelytic on an open set D, except possibly for a finite number of ponts at which
we assume f is continuous. Then for every circuit y which is homotopic to a point in
D we have
Proof. Since y is homotopic to a point in D, y is also homotopic to a circular
path fi in D with arbitrarily small radius. Therefore  f =  f, and ,f = 0 by
Theorem 16.9.
Definition 16.14. An open connected set Dia catled simply connected f every closed
path in D ix homotopic to a point in D.
Geometrically, a simply connected region is one without holcs Cauchy's
theorem shows that, in a simply connected region D the integral of an analytic
function is zero around any circuit n D.
16.9 CAUCHY"S INTEGRAL FORMULA
The next theorem reveals a remarkable property of analytic functions. It relates
the value of an analytic function at a point with the values on a closed curve not
containing the point.
Theerem 16.I (Caueh¥'s iegrl formtak). Assume f is analytic on an open set D,
and let  be any circuit which is homotopic to a point in D. Then for any point z m
D which is not on the graph of 7 we have
; f(w) dw= f(z) ;, I dw" w- z w- z (6)

Proof Define a new function g on D as follows:
a(w) --- - z
[f (z) if w ---- z.
Theng is analytic at eachpoint w  zin D and, at the point z itself, g is conlinuous.
Applying Cauehy's integ theorem to g w have J'r g = 0 for ever circuit
homotopie to a point in D. But if z is not on the graph of ? we caa write
f,a = f,l  w)-f(z) aw-- f,w- z w- zf(W) dw- f(z) f, , _1 z dw,
which prov (6).
NO. The same proof hows that (6) is also valid if there is a finite subet T of
D on whlch fis not analytic, provided thatfis continuous on T and z is not in T.
The integral j' (w - z)-l dw which appears in (6) plays an important role in
complex integration theory and is discussed further in the next 8tion. We can
easily calculate its value for a circular path.
Fxl. If ? is a lsitively oricnl! circular path with center at z and radius r,  can
write 7(0) = z + re m, 0 < 0 <_ 2. Th y'() = ire m = i{7(0) - z}, and we find
1O1. In this ease Cauchy's integral formula (6) takes the form
2lf(z) = f (--W)-z dW.
Aain writing (0) - z + re , we can put this in the form
f(z) =  f(z + re ) dO. (7)
This can be interpreted as a Mean-Value Theorem expressing the value off at the
center of a disk as an average of its value at the boundary of the disk. The function
fis assumed to be analytic on the closure of the disk, except possibly for a finite
subset on which it is continuous.
I6.10 TIIE WINDING NUMBER OF A CIRCUIT WIT8 RgPECT I"O A POINT
lftteem 16.16. Let ? be a circuit and let z be a point not on the graph of . Then
there i an integer n (dependin on y and on z) such that
; dw = 2tin.
(8)
I)I£ l&17  Wtmlillg Nml¢ [ It Ortalt 
Proof. Suppos r has domain [a, hi.
in (8) as a Rimann integral,
By Theorem 16.7 we can express the integral
J. (0 - 
Define a complex-valued function on the interval [a, b] by the equation
-- ira _< x _< b.
To prove the theorem we must show that F(b) -- 2nin for some integer n. Now P
is continuous on [a, b] and has a derivative
F'(x)= f(x) ,
(x) - z
at each point of continuity of ?' Therefore the function G defined by
G(t) = e-¢°{y(t) - z} if t  [a, b],
is also continuous on [a, b]. Moreover, at each point of continuity of?' we have
G'(t) = e-r¢oy'(t) - F'(t)e-eO{?(t) - z} = O.
Therefore G'(t) ---- 0 for each t in
points. By continuity, G is constant throughout [a, b]. Hence, Crib) -- G(a). In
other words, we have
Since y(b).-- ?(a)  z we find
e
which implies F(b) = 2nin, where n is an integer. This completes the proof.
 16d7. If
defined by (8)
denoted by
2hi J  - z
Nor Canhy's integral formula (6) can now be restated in the form
n(r, z)f(z) = . - - d.
The term "winding nmber" is used bceanse n(?, z) gives a mathematically
pveiw way of counting the number of times the point t) "'winds around" the
point z as t varies over the interval a, hi. For example, if , is a positively oriented

TI 15.18
circle given by ?(0) = z + re i°, where 0 < 0  2g, we have already seen that the
winding number is I. This is in accord with the physical interpretation of the
point y(0) moving once around a circle in the positive direction as 0 varies from
0 to 27t. If 0 varies over the interval [0, 21n], the point y(0) moves n times around
the circle in the positive direction and an easy calculation shows that the winding
number is n. On the other hand, if 6(0) = z + re- * for 0 < 0 g 2nth then 6(0)
moves n times around the circle in the opposite direction and the winding number
is -n. Such a path 6 is said to be negatively oriented.
16AI THE UNDEDNESS OF THE SET OF POllY£S Y(ITH
NUME ZIRO
Let F denote the graph of a circuit 7- Since F is a compact set, its complement
C - F is an open set which, by Theorem 4.44, is a countable union of disjoint
open regions (the components of C L 1-). If we consider the components as
subsets of the extended plane C*, exactly one of these contains the ideal poirtt co.
In other words, one and only one of the components of C - F is unbounded.
The-next theorem shows that the winding number n(y, z) is 0 for each z in the
unbounded component.
Theorem 16,18. Let ? be ¢ circuit with graph F. Divide the set C - F into two
subsets:
E = {z : n(r, z) = 0} and I = {z : n(, z) # 0},
Then both E and I are open. Moreover, E is unbounded and I is bounded.
Proof Define a function 9 on C L F by the formula
g(z)= n('e,z)=ml " dw
2hi ,], w -- z
By Theorem 7.38, g is continuous on C - F and, since g(z) is atways an integer,
it follows that g is constant on each component of C - F. Therefore both E and
I are open since each is a union of components of C - F.
Let U denote the unbounded component of C - F. If we prove that E con-
tains U this will show that E is unbounded and that I is bounded. Let K be a
constant such that I?(t)l < K for all t in the domairt of ?, and let c be a point in
U such that Icl > K + A(?) where A(y) is the length of 3,. Then we have
I 1 < 1 1
Icl - K
Estimating the integral for n(?, c) by Theorem 16.6 we find
o  lo(c)l <- A(r) < 1,
Since g(c) is an integer we must have g(c) = O, so g has the constant value 0 on U.
Hence E contains tl point c, so E contains all of U.
There is a general theorem, called the Jordan curve theorem, which states that
if F is a Jordan curve (simple dosed curve) described by 7, then each of the sets
E and/in Theorem 16.18 is connected. In other words, a Jordan curve F divides
C - 1" into exactly two components E and I having F as their common boundary.
The set I is ealIed the inner (or interior) region of F, and its points are said to be
/rtside F. The set E is called the outer (or exterior) region of F, and its points are
said to be outside F.
Although the Jordan curve theorem is intuitively evident and easy to prove for
certain familiar Jordan curves such as circles, triangles, and rectangles, the proof
for an arbitrary Jordan curve is by no means simple. (Proofs can be found in
References 16.3 and 16.5.)
We shall not need the Jordan curve theorem to prove any of the theorems in
this chapter. However, the reader should realize that the Jordan curves occurring
in the ordinary applications of complex integration theory are usually made up of
a finite number of line segments and circular arcs, and for such examples it is
usually quite obvious that C - F consists of exactly two components. For points
z inside such curves the winding number n(3,, z) is + 1 or - 1 because  is homo-
topic in/to some cireular path 6 with center z, so n(7, z) - n(6, z), and n(6, z) is
+ 1 or -I depending on whether the circular path  is positively or negatively
oriented. For this reason we say that a Jordan circuit 3, is positively oriented if,
for some z inside F we have n(y, z) = + 1, and negatively oriented if n(y, z) = - 1.
16.12 ANALYTIC FUNCTIONS DEFINED BY CONTOUR INTEGRALS
Cauchy's integral formula, which states that
n(r,z)f(z) = l_ f f(w) dw,
2hi J w - z
has many important consequences. Some of these follow from the next theorem
which treats integrals of a slightly more general type in which the integrand
f(w)/(w - z) is replaced by ¢(w)/(w -- z), where p is merely continuous and not
necessarily analytic, and 3' is any rectifiable path, not necessarily a circuit.
Theorem 16.19. Let 3' be a rectifiable path with graph F. Let ¢p be a complex-tyalued
]'unction which is continuous on F, and let f be defined on C - 1  by the equation
f(z) = f ¢(w) dw if z ¢ r.
Then f has the following properties:
a) For each point a in C - F, f has a power-series representation
f(z) = _, c,(z - a)',

T 16.]9
where
f  ')
c, = (w -- -)?+1 dw for n -- O, 1, 2,... (10)
b) The series in (a) has a positive radius of convergence >_ R, where
R =inf{I w- al:wF}. (11)
c) The functitm f hus a derivative of every order n on 12 - V given by
(w- z) "÷i
(12)
Proof First we note that the number R defined by (i l) is positive because the
flmetion 9(w) = Iw - at has a minimum on the compact set F, and this minimum
is not zero since a ¢ F. Thus, R is the distance from a to Ih¢ nearest point of F.
(See Fig, 16.6.)
16.6
Then l/(l - t) = (w- a)/(w - z).
integrating along 7, we lind
f(z) = f (w)- z dw
.--0 O' - a) "÷
To prove (a) we begin with.the identity
l  t *+z
..........  t" + --, 03)
l-t ,=o
valid for alI t  1. We take t = (z - a)/(w - a) where [z - al < R and we r.
Multiplying (13) by q(w)/(w- a) and
where c, is given by (I0) and E is given by
dw + f, () (z - a +
w - z \--a/ dw
E, = f (w) (z - a) +'
w - z \w - a/
(14)
Now we show that E - 0 as k - co by estimating the integrand in (14). We have
- R Iw-zl Iw--a4-a- zl-R- la-zl
Let M = max {l(w)l : w  F}, and let A0) denote the lengah of 7. Then (14)
gives us
MA() (.,z-al) .+l
IEt ---  _ id-- z l
Since Iz - al < - we fred that E, -- 0 as & --, . Thiz proves (a) and (b).
Applying Theorem 9.23 to (9) we lind that fha drivative of eve order on
the dik B(a; R) and thatf*)(a) = n!c.. Sin a in an arbitrary point of (7 - F
this proves (¢).
Noa-. The series in (9) may have a radius of convergence greater than R, in which
case it may or may not represent fat more distant poims.
16.13 POWER-SERIES EXPANSIONS FOR ANALYTIC FUNCTIONS
A combination of Cauhy's integral formula with Theorem 16.I9 gives us:
Tkeorem 16.2. Assume f is anal)tic on an open set S in C, and let a be any
of S. Then all derivatives f'(a) exist, and f can be represented by the conrgent
power series
f(z) =   f)(a) (z - a)% (15)
in every disk B(a; R) whose closure lies in S. Moreover, for every n >_ 0 we have
n" f, f(') dw, (16)
/(*)(a) --- 2ti (w -')'÷
where  is any positively oriented eireutar path with center at a and radius r < R.
OTE. The series in (15) is known as the Taylor expansion off about a. Equation
(16) is called Cauchy"s integral formula forf*)(a).
Proof. Let 7 be a circuit homotopic to a point ia S, and let F be the graph of
7. Define .q on C - F by the equation
J,
W  Z
If z • B(a; R), Cauchy's integral formula tells us that 8(z) = 2n/n(?, z)f(z).

How let (0) ffi a + re , where [z- a[ < r < R and 0< 0 < 2x. Then
n(', z) ffi I, so by applying Theorem 16.19 to q(w) = f(w)[(2i) we find a series
representation
convergent for Iz - al < R, where c, = f('(a)/n L Also, part (¢) of Theorem
16.19 gives (16).
Theorems 16.20 and 9.23 cx tell us that a necessary and sufficient con-
dition for a complex-valued function f to be analytic at a point a is that f be
representable by a po,at series in ome neighborhood of a. When such a power
series exists, its radius of convergence is at least as large as the radius of any
disk B(a) which lies in the region of analyticity off Since the circle of convergence
eanaot contain any points m its interior whcrcffaJls to be analytic, it follows that
the radius of convergence is exactly equal to the distance from a to the nearest
point at which flails to be analytic.
This observation gives us a deeper insight concerning power-series expansions
for real-valued functions of a real variable. For example, letf(x) -- 1/(1 + x ) if
x is tea[. This function is defined .everywhere in R 1 and has derivatives of every
order at each point in R 1. Also, it has a power-series expansion about the origin,
namely,
1 1 x ± + x 't x  +
However, this rprcntation is valid only in the open interval (- 1, I). From the
standpoint of ral-variable thcory, there is nothing in the behavior off which
explains this. But when we examine the situation in the complex plane, we see at
once that the function f(z) = !/(1 + z 2) is analytic everywhere in C except at
the.points z ffi :1: i. Therefore the radius of convergence of the power-series
expansion about 0 must equal 1, the distance from 0 to :" and to -L
Examples. The following power series expansions are valid for all z in C:
l ).z z-
c) cos = )7
,=0"-¢ (2n)!
- I Y'z z'÷ i
b) sinz =
• o (2n + 1)! "
16.14 CAUCHY'S INF_UALITIES. LIOUVILI'S THEOREM
If f is analytic on a closed disk B(a; R), Cauchy's integral formula (16) shows that
2i (w +  dw,
where  is any positively oriented circular path with center a and radius r < R.
We can write (0) = a + re , 0 < 0 < 2n, and put this in the form
2r  f(a + re ¢) e -t* dO. (17)
This formula expresses the nth derivative at a as a weighted average of the valen
off on a circle with center at a. The special case n = 0 was obtained earlier in
Section i6.9.
Now, let M(r) denote the maximum value of Ill on the graph of y. Estimating
the integral in (17), we immediately obtain Cauchf inequalities:
I < M(r)n__[ (n ffi 0, I, 2,... ). (18)
r n
Te next theorem is an easy tonsequence of the case n = I.
Tkcen,cm 16,21 (Liemeille's t&eorem). If f is analytic eerywhere on C and bnded
on C, then f i conaant.
Proof. Suppose [f(z)l - M for all z in C. Then Cauchy's inequality with n = 
gives us [f'(a)l _< M]r for every r > 0. Letting r - +o, we findf'(a) = 0 for
every a in C and hence, by Theorem 5.23,fis constant.
tOT. A function analytic everywhere on C is called an entire function. Examples
are polynomials, the sine and cosine, and the exponential. Liou#ille's theo[em
states that every bounded entire function is constant.
Liouville's theorem leads to a simple proof of the Fundamental Theorem of
Al  br
Tkeorem 16.22 (Fmedametal Tteerem of ,'Jlebm), Ery polynomial of deree
n 3 1 hasazero.
Proof. Let P(z) = ao + az + ... + a,z , where n _> 1 and a  0. We assume
that P has no zero and prove that P is constant. Let f(z) ffi liP(z). Then f is
analytic everywhere on C since P is never zero. Also, since
'
we see that ]P(z)[ as Izl --, sof(z) - 0 as [z[ -- +. Therefore
fis bounded on C so, by LiouviIle's theorem, fand hence P is constant.
16.15 ISOLATION OF THE ZEROS OF AN ANALYTIC FUNCTION
If f is analytic at a and iff(a) = 0, the Taylor expansion of f about a has constant
term zero and hence assumes the following form:

16.23
This is valid for each z in some disk B(a). Iffis identically ro on this disk [that
is, iff(z) = 0 for every z in B(a)]. then each c. = 0, since c. -- f<"(a)ln!. If f is
not identically zero on this neighborhood, there will be a first nonzero coefficient
c in the expansion, in which case the point a is said to be a zero of order k. We
will prov© next that there is a neighborhood of a which contains no further zeros
off This property is described by saying that the zeros of an analytic function are
isolated.
l'i,orem 16.23, Assume that f is analytic on an open set S in C. Suppose f(a) = 0
for some point a in S and assume that f is not identically zero on any neighborhood
of a. Then there exists a disk B(a) in which f has no further zeros.
Proof. The Taylor expansion about a becomesf(z) = (z - a).q(z), where k 7 I,
g(z) = c + c÷l(z - a) +. -.., and g(a)  c # O.
Since g is continuous at a, there is a disk B(a) _ S on which g does not vanish.
Therefore, f (z) # 0 for all z # a in B(a).
This theorem has several important consequences÷ For example, we can use
it to show that a function which is analytic on an open region S cannot be zero
on any nonempty open subset of S without being identically zero throughout S.
We recall that an open region is an open connected set. (See Definitions 4.34
and 4.45.)
Ttem, em 16,24, Assume that f is analytic on an open region S in C. Let A denote the
set of shose points z in S for which there exists a disk B(z) on which f is identically
zero, and let B = S - A. Then one of the two sets .4 or B is empty and the other
one is S itself.
Proof. We have S  A w B, where A and B ate disjoint sets. The set A is open
by its very definition, If we prove that B is al open, it will follow from the
cnnectedness of S that at least one of the two sets .4 or B is empty.
To prove B is open, let a be a point of B and consider the two possibilities:
f(a) # O, f(a) =- O. Iff(a) # 0, there is a disk B(a) _ S on which f does not
vanish. Each point of this disk must therefore belong to B. Hence, a is an interior
point of B iff(a) ¢ 0. But, if f (a) = 0, Theorem 16.23 provides us with a disk
B(a) containing no further zeros off This means that B(a) _ 11. Hence, in either
case, a is an interior point of B. Therefore, B is open and one of the two sets A or
/i' must be empty.
16.16 THE IDENTITY THEOREM FOR ANALYTIC FUNCTIONS
Theorem 16.25. Assume that f is analytic on an open region S in C. Let T be a
subset of S haviny an accumulation point a in S. If f (z) = 0 for ever. z in T, then
f(z) = O for every z in S,
Proof There exists an infinite sequence {z.}, whose terms are points of T, such
that lim_. z. -- a. By continuity, f(a) = [im._f(z) = 0. We will prove
Maaimum  MiaEmum Mlnh
next that there is a neighborhood of a on which f is identically zero. Suppose
there is no such neighborhood. Then Theorem 16.23 tells us that there must be a
disk B(a) on whichf(z) # 0 if z # a. But this is impossible, since every disk B(a)
contains points of T other than a. Therefore there must be a neighborhood of a
on which f vanishe identically. Hence the set A of Theorem 16.24 cannot 
empty. Therefore, A = S, and this meansf(z) = 0 for every z in S.
As a corollary we have the following impoant result, sometimes referred to
as the identity theorem for analytic functions:
Tleorem 16.26. Let f and O be analptic on an open region S in C, If T is a subset
of S having an accumulation point a in S, and if f (z) : g(z) for every z in T, then
f(z) = 9(z) for every z in S,
Proof. Apply Theorem 16.25 tof- g.
t6.17 THE MAXIMUM AND MINIMUM MODULUS OF AN ANALYTIC
FUNCTION
The ab,olute value or modulus Ill of an analytic function f is a real-valued non.
negative function. The theorems of this section refer to maxima and minima of
]rl-'orem 1,27 (Local tnaxlnmm  peincile). Assume f is analytic and not
constant on an open region S. Then Ill has no local maxima in S. That is, every
disk B(a; R) in S contains points z such that [f(z)l > If(a)l.
Proof We assume there is a disk B(a; R) in S in which If(z)l  If(a)l and prove
thatfis constant on S. Consider the concentric disk B(a; r) with 0 < r < R.
From Cauchy's integral formu|a, as expressed in (7), we have
If(a)l < r If(a + re')l dO. (19)
Now If(a + rea)l  If(a)l for all 0. We show next that we cannot have strict
inequality If(a + re)l < If(a)l for any 0. Otherwise, by continuity we would
have If(a + re)l <_ If(a)l  e for some  > 0 and all 0 in some subinterval I of
[0, 2x] of positive length h, say. Let J = [0, 2hi  L Then J has measure
2n - h, and (19) gives us
2nlf(a)l<_ Jr If(a+ rdn)[ dO+ L If(a+ re)l dO
<- hlf(a)l - e} + (2 - h)If(a)l = 2r If(a)l - he < 2t If(a)l.
Thus we get the contradiction If(al < If(a)l- This shows that if r  R, we
cannot have strict inequality If(a + re)l < lf(a)l forany 0. Hence If(z)] =lf(a)l
for every z in B(a; R). Therefore Ifl is constant on this disk so, by Theorem 5.23,
fitself is constant on this disk. By the identity theorem,f is constant on S.

Tlorem 16.28 (,4bselut¢ maximum moda pei). Let T be  compact subset
of the mpte ple C. e f  continuo on T d alytic on t imerior of
Z Th the solute  of [f on T is attained on T, the ury of T.
Prf Sin T is com, If atlns its alute mmum somew on T,
y ata. Ira  Te is nong to pro. Ifa int T, let S e mpont
oft Tcontaining a. Since Ill h a lal mimum at a, m 16.27 imp
atfis nsnt on S. By continuity, f is constant on 0S  T,  te mimum
value, If(a)l, i attn on 0S. But S  gT(Why) so the mmum is atoned
on Z
Tm 1629 (M 1 pp). Asu f is ltic ad not conxtt
on  on re, ion S. If Ifl h a locM mnimum in S  a, 1hen f(a) = O.
Proof. Ifa) # 0 we apply eorem 16.27 to 9  Ill en g is analytic in me
o sk B(a; R) and g s a lal mimum a a. fo  and hencefis
constant on this disk d thefo on S, contradicting e hyesis.
16.18 THE OPEN MAPPING THEOREM
Nonconstant analytic functions are open mappings; that is, they map open sets
onto open sets. We prove this as an application of the minimum modulus
principle.
Theorem 16.:10 (Open mpitt 0 theorem). If f is analytic and not constant on an
open region S, then f is open.
Proof. Let A be any open subset of S. We are to prove thatf(A) is open. Take
any b in f(A) and write b = f(a), where a e A. First we note that a is an isolated
point of the inverse-imagef-({b}). (If not, by the identity theorem f would be
constant on S.) Hence there is some disk B = B(a; r) whose closure B lies in A
and contains no point off- l({b)) excep! a. Since f(/) _ fb4) the proof .will be
complete ff we show thatf(B-) contains a disk with center at b.
Let 0B denote the boundary of B, OB = {z : [z - al - r}. Then f(B) is a
compact set which does not contain b. Hence the number m defined by
m =inf{If(z)-bl:z
is positive. We will show thatf(B) contains the disk B(b; m/2). To do lhis, we
take any w in B(b; m/2) and show that w = f(zo) for some z 0 in .
Let g(z) ffi f(z) - w ifz e J. We will prove that g(z0) = 0 for some zo in 8.
Now [gl is continuous on  and, since  is compact, there is a point z o in  at
which Ig[ attains its minimum. Since a 6 B, this implies
Ig(zo)l _ Ig(a)l = If(a) - wl = Ib - wl < m.
2
But if z e OB wc have
Ig(z)I --If(z) - b + b - wl  If(z) - bl - Iw - bl > m - 
m m
Hence, Zo ¢OB so zo is an interior point of B. In other words, [gl has a local
minimum at z0 Since g is analytic and not constant on B, the minimum modulus
principle shows that g(z0)  0 and the proof is complete.
16.19 LAURF_,NT EXPANSIONS FOR FUNCTIONS ANALYTIC IN AN
ANNULUS
Consider two functionsft and gt, both analytic at a point a, with g(a) = 0. Then
we have power-series expansions
and
=1
for Iz - al < r,
A(z) = , c.O - ), ¢or [z - aI < r,.
Let ]' denote th composite function given by
A(© = g - + a .
(20)
Then f2 is defined and analytic in the region Iz - al > r and is represented there
by the convergent series
A(z) =  bo(z -
for Iz - al > rt- (21)
Now if r < r2, the series in (20) and (21) will have a region of convergence in
common, namely the set of z for which
r < Iz - af < "2-
In this region, the interior of the annulus A(o; r a, r2) , bothft andfz are analytic
and their sumf + fz is given by
MO
The sum on the right is written more briefly as
. c,(z - a)",
where c_, --- bo for n = 1, 2,... A series of this type, consisting of both positive
and negative powers of z - a, is called a Laurent series. We say it converges if
both parts converge separately.
Every convergent Laurent series epresents an analytic function in the interior
of the annulus/t(a: rt, r2). Now we will prove that, conversely, every function
fwhich is analytic on an annulus can be represented in the interior of the annulus
by a convergent Laurent series.

Tlteerem 16J1. Assume that f is analy:ic on an annulus A(a;
every interior point z of this annulus we hae
f(z) = flCz) +
where
Th£n for
(22)
Z(z) = 22, c,(z - a)" and
The coeicients are gioen by the formul
c -- dw
2 ( : r +'
(n=0, +1, +2,...),
(23)
where  is any positively oriented circular path with center at a and radius r, with
rl < r < r2. The function fl (called the realar part off at a) is analytic on the
disk B(a; r2). The function f2 (called the principal part of fat a) is analytic outside
the closure of the disk B(a;
Proof Choose an interior point z of the annulus, keep z fixed, and define a function
g on A(a; rt, r2) as follows:
f. )w - f(z) if w # z
o(w) = z
[f (z) if w = z.
Then g i, analytic at w if w ¢ z and g is continuous at z. Let
(r)
where y, is a positively oriented circular path with center a and radius r, with
r t  r _< r z. By Theorem I6.8, p(r,) = q(r2)so
fr a(w) dw = £ g(w) aw,, , (24)
where 7 = , and z -- ),,. Since z is noton the graph oft or of, in each of
these integrals we can write
W -- Z
Substituting this in (24) and tranlsing terms, we find
(25)
But , (w - z)-  dw -- 0 since the integrand is analytic on the disk B(a;
and r (w - z) -t dw = 2ri since n(7, z) = I. Therefore, (25) gives us the
equation
f(z) = it(z) + fz(Z),
where
if, f(w) d w and fz(z) if, f(W) dw"
f(z) = 2xi , w - z 2hi . w- z
By Theorem [6.lg, ft is analytic on the disk B(a; rz) and hence we have a Taylor
expansion
A(z) = , c,(z - a)" for Iz - al < r,
where
1  f(,) dw. (26)
M0t'cover, by Theorem 16.8, the path ,2 can be replaced by , for any r in the
intervat r _< r < r z.
To find a series expansion iotA(z), we argue as n the proof of Theorem 16.19,
using the identity (13) with t = (w - a)/(z - a). This gives us
! ' (w- a" (w-a''+'(z-a) (27)
 - (, - a)/(, - a) = ,=o \ - a: + \z - a,: - "
Ifwis on the graph ore,, we have Iw - al = r, < lz - al, so Itl < 1. Nowwe
multiply (27) by -f(w)l(z - a), integrate along ,, and let k - m to obtain
A(z) = _, b.(z- a)-" rot I - al > r,
where
1  f(w) dw. (28)
bo = . ,(w-a)'-°
By Theoi'em 16÷8. the path y can be replaced by  for any r in [r, r]. Lt'we take
the same path y, in both (28) and (26) and if we write c_. for b,, both formulas
can be combined into one as indicatl in (23). Since z was an arbitrary interior
point of the annulus, this completes the proof.
NOTE. Formula (23) shows that a function can have at most one Laurenl ex-
pansion in a given annulus.
16.20 ISOLATED SINGUIARITIES
A disk B(a; r) minus its center, that is, the set B(a; r) -- {a}, is called a deleted
neighborhood of a and is denoted by B'(a; r) or B'(a).

D¢ 16.32. A point a is called an isolated singularity off if
a) f is analytic on a deleted neighborhood of a,
and
b) f is not analytic at a.
NOTE. /need not le defined at a.
If a is an isolated singularity off, there is an annulus A(a; rz, rz) on which/is
analytic. Hence/has a uniquely determined Laurent expansion, say
f(r) =  %(z - o)" +  c_.¢z - a)-'. (29)
Sin the inner radius r c  arbitrarily small, (29) is valid in c deleted
neighrhood B'(a; rz), The singularity a is classified into one of thr tys
(de,haling on the fo ofe principal par 0 as follows:
If no negative powe apar in (29), that is, ifc_, = 0 for eve n = I, 2,...,
the int a is called-a rale singular#y. In this case,z)  co as z  a and
the sinology can  removed by defining f at a to have the value f(a) = co.
(S Enample I low.)
If only a finite humor of netive po ap, that is, if c_, 0 for me
n but c_, = 0 for eve m > n, the ini a is said to  a pole of order n. In this
ca, the pfinpal part is simply a fite sum, namely,
c_, + _.__e-z
z-a (z-a) +''"
A pole of order 1 is usually called a simple pole. If there is a pole at a, then
If(z)l --,  as z --, a.
Finally, if e_,  0 for infinitely many values of n, the point a is called an
es.sentialingularity. In this case, f (z) does not tend to a limit as z -} a+
ExamOie 1. Removable singularity. Lt/(z) = (sin z)/z ifz = O,f(O) = O. This func-
tion is analytic everywhere except at O. (It is discontinuous at O, sinc (sin z)/z - 1 as
z --, 0.) The Laurem expansion about 0 has the form
sinz_ 1 ---4 +--.
z 3!
Since no negative powers of z appear, the point 0 is a removable singularity. If we re-
de/me./to have the value 1 at 0, the modified fuaction becomes analytic at 0.
Example 2. Pole. Letf(z) = (sin z)/z  if z  0. The Laurent expansion about 0 is
sin z _, 1 _ l 1
-z ---z + zZ+..-
z n 3! 5[
In this case, the point 0 is a pole of order 4. Note that nothing has been said about the
value of fat O.
In. 16.33 Rcsio at an Isolated Simcalar Point 459
Example 3. Essential singularity. Let f(z) = e 't" if z ¢: 0. The point 0 is an ¢,scntial
singularity, since
tUz I + z -t + ! z z + + !
= ..... Z--n __
2! n!
Theorem 16.33. Assume that/is analytic on an open region S in C and define g by
the equation g(z) = ill(z) if f(z)  O. Then/has a zero of order k at a point a in
S if, and only if, g has a pole of order k at a.
Proof. If/has a zero of order.k at a, there is a deleted neighborhood B'(a) in
which/does not vanish. In the neighborhood B(a) we hayer(z) = (z - a)h(z),
where h(z)  0 if z  B(a). Hence, 1/h is analytic in B(a) and has an expansion
1 1
h(z) be + bl(Z - a) + , where b o = h(a) # O.
Therefore, if z e B'(a), we have
I _ b o b t. __
() = (z -- agh(z) (z - a)  + (--- a)P -t +
and hence a is a pole of order k for g. The converse is similarly proved.
16.1 THE RESIDUE OF A FUNCTION AT AN ISOLATED SINGULAR POINT
If a is an isolated singular point off, there is a deleted neighborhood B'(a) on
which f has a Laurent expansion, say
= - + c_o(z - a)-'. (3o)
The coefficient c_  which multiplies (z - a)-  is called the resMue off at a and
is denoted by the symbol
c_ -- Resf(z).
Formula (23) tells us that
f(z) dz = 2hi Res/(z), (31)
if ? is any positively oriented cireular path with center at a whose graph lies in the
disk
In many cases it is relatively easy to evaluate the residue at a point without
the use of integration. For example, if a is a simple pole, we can use formula (30)
to obtain
gesf(z) = lim (z - a)f(z), (32)

Similarly, if a is a pole of order 2, it is easy to show that
Res/(z) = tl'(a), where (z)
In cases like this, where the residue can be computed very easily, (31) gives us
simple method for evaluating contour integrals around circuits.
C.auchy was the first to exploit this idea and he developed it into a powerful
method known as the residue calculus. It is based on the Cauchy residue theorem
which is a generalization of (31).
I,22 THE CAUCHY RESIDUE THEOREM
Tleoeern 16,34, Let f be analytic on an open region S except for a fin#e number of
isolated singularities z 1 ..... z, in S. Let 7 be a circuit which is homotopic to
point in S, and assume that none of the singularitiea ties on the graph of y. Then we
f(z)dz = 2i ,=t n(,. z)=,Res f(z), (33)
where n(7, zt) is the winding number of  with respect to
Proof. The proof is based on Ihe following formula, where m denotes an integer
(positive, negative, or zero):
if m ¢: - I. (34)
The formula for m = - I is just the definition of the winding number n(y, z).
Let [a, b] denote the domain of:y. Ifm # ,- 1, letg(t) = {y(t) - zt} "÷ for t in
[a, b]. Then we have
1
- {a(b) - a(a)} = 0,
m+l
since g(b) = g(a). This proves (34).
To prove the residue theorem,'letf denote the principal part of fat the point
z. By Theorem 16.31 ,f is analyticeverywhere in C except at z. Thereforef - f
is analytic in S except at z .... , z,. Similarly, f - fl  f is analytic in S except
at za .... , z and, by induction, we find that f - _=  fk is analytic everywhere
in S. Therefore, by Cauchy's integral theorem,  (f - ;  f) - 0, or
Now we epress f as a Laurent series about z and integrate this series term by
term, using (34) and the definition of residue to obtain (33).
ro. If y is a positively oriented Jordan curve with graph F, then n(7, z 0 = 1
for each z inside F, and n(, zD = 0 for each z outside F. In this case, the
integral off along . is 2hi times the sum of the residues at those singularities lying
inside F.
Some of the applications of the Cauchy residue theorem are given in the next
few sections.
16,23 COUNTING ZFAtOS AND POLES IN A REGION
Iffis analytic or has a pole at a, and iffis not identically 0, the Laurent expansion
about a has the form
s(=) = E  Xz-
where c,, # 0. If m > 0 there is a zero at a of order m; if m < 0 there is a pole
ata of order -m, and ifm - 0 there is neither a zero nor a pole at a.
uo. We a/so write re(f; a) for m to emphasize that m dependi on bothfand a.
Theorem 16.35. Let f be a function, not identically zero, which is analytic on an
open region S, except possibly for a finite number of poles. Let y be a circuit which is
homotopic to a point in S and whose graph contains no zero or pole off. Then we
1 [" = a  raf/; a), (35)
where the sum on the right contains only a finite number of nonzero terms.
rcrr: If  is a positively oriented Jordan curve with graph F, then n(y, a) -- !
for each a inside F and (35) is usually written in the form
I
 f'(z)dz = N - P, (36)
where N denotes the number of zeros and P the number of poles off inside F,
each counted as often as its order indicates.
Proof. Suppose that in a deleted neighborhood of a point a we have f(z) --
(z - a)'g(z), where g is atmlytic at a and g(a) : 0, m being an integer (positive
or negative). Then there is a deleted neighborhood of a on which we can write
f'(z) m
f(z) z - a #(z)
the quotient g'/g being analytic at a. This equation tells us that a zero off of
order m is a simple pole off'ffwith residue m. Similarly, a pole off of order m
is a simple pole off'if with residue -m. This fact, used in conjunction with
Cauchy's residue theorem, yields (35).

16.24 EVALUATION OF iREAL-VALUF. llNrrEGRAL$ BY MEANS OF
RltmUF
Cauehy's residue theorem can sometimes be used to evaluate real.valued Riemann
integrals. There are several techniques available, depending on the form of the
integral. We shall describe briefly two of these methods.
The first method deals with integrals of the form o z* R(sin 0, cos 0) dO, where
R is a rational function* of two variables.
Tleorem 16.J6. Let R be a rational function of two variables and let
whenever the expression on the right is finite. Let ? denote the posilively oriented
unit circle with center at O. Then
f2. R(sin O, s O) dO  f f(z) dz,
(37)
d iz
proofed that f  no poles on the yraph
Proof S (0) = e  with 0  0  2n, we have
r'(O) iO), r(0) - I sin 0, (0): + l
................
2;r(o) 2r(o)
and (3 follo at on from eom t6.7.
 To evuate e intel on
idu of t ted at those poles wMeh lie inside the unit circle.
 Evaluate I = [ t(a + cos 0), wh a i , la] > 1. Applying (37), we
find
zz + 2az + 1
e intd h mple o1 at  roots of the ua6on z z + z + 1 = 0. e a
• e ts
z = -a+ a - 1,
zz = -a- 4a  - 1.
function P defined on C × C by an equation of the form
is ca/led a polynomial in two rr/ab/es. The ¢o¢lienl a. I may be reaI or complex The
quotient of two such polynomials is called a rationalfimetion of two variables.
The corrponding residues R and R2 are given by
then
lira f[,f(x) dx = 2hi  Resf(z). (39)
where z, ..., z e the l of f which lie  
Proo t   e sitively ofient th foed by  a on of the 1
i8 from - R to R and a miIe in T ha@ng [- R, R]  its diamer, whe R
i  l enou to enclo @! the pol zx,..., z,. 
2hi  Res/(z) = f(z) dz = f(x) dx + i f(Rei R  dg
 R  + , e last Jnteal t to ro by (38) d  obn (39).
m. Equation (38) is automafilly satisfi  f is e querier of two poly-
nomials, my f = P[Q, ded t e  of Q exs the  of P by
at tst 2. ( eise 16.36,)
I,,tle. To evaluate I_ dx/(t + x*), It f(z) = l/(z* + 1).  affz) = I,
Q(z) -- 1 + z ", and henc (38) holds. The poles of f are the roots of the equation
1 + ' = 0. These ar zx, z, z, z, where
z, = e (z-;t (k = 1, 2, 3,4).
Of these, only z and z lie in the upper half-plane. The resJdtm at z is
R¢$ f(z) = Iiffl (z -- zi)f(z) =
If a > I, z is inside the unit circle, zz is outside, and 1 = 4/(z - zz) = 2;z,tx/a -T - I.
Many improper integrals can be dealt with by means of the following theorem:
Tlurorem 16.37. Let T = {x + iy : y > 0} denote the upper half-plane. Let S be
an open region in C which contains T and suppose f is analytic on S, except, possibly,
for a finite number of potes. Suppose further that none of these poles is on the real
lira f'f(Re ) Re  dO = 0,

SillilarD',we find gC___g.f(z) = (l/4i)t g|P. Therefore..
4i 4
16.2.5 EVALUATION OF GAUSS'S SUM BY RESIDUE CALCULUS
The residue theorem is often uf, ed to evaluate sums by integration. We illustrate
with a famom example called Gams's um G(n), defined by the formula
(40)
where n > I. This sum occurs in various ps of the Theory of Numbers. For
small values of n it can easily be computed from its definition. For example, we
have
G(I) : 1, G(2) = O, G(3) = ix/-, G(4) = 2(1 + i).
Although each term of the sum has absolute value I, the sum itself has absofute
value 0, xn, or "4. In fact, Gauss proved the remarkable formula
for every n  i. A number of different proofs of(41) are known. We wilt deduce
(41) by considering a more eaeral sum S(a, n) introduced by Oiriehlet,
s(o, ,,) = e
If the product na is even, we have
rO
where n and a ate positive integers, ff a - 2, then S(2, n) = G(n). Dirichlct
proved41) as a corollary of a reciprocity law for S(a, n) which can be stated as
follows:
T/g-orem 16.3&
(42)
where the bar denotes the complex conjugate.
NO'. To deduce Gauss's formula (41}, we take a = 2 in (42), and observe that
$(n,  -- 1 + e -'uz.
Proof The proof given here is particularly instructive because it illustrates several
techniques used in complex anMysis. Some minor computational details are left
as exercises for the reader.
Let  be the function defined by the equation
O(z) =  • `+'pt". (43)
Th. 16.3g E,alhl#lim d £llllia  4[
Then g is analytic everywhere, and g(0) -- S(a, n). Since na is even we find
( + l) - () = e''( ' - 1) = '(' - l) ..
m=O
(Exei 6.41). ow finef by the uxfion
f(z) = g(z)/("" -- 1).
Then f is analytic eye, here except for a trder pole at each intent, and f
fies the eqtion
f(z + !) = Az)+ (z), ()
where
= e e (45)
e function  is anMc evhe.
At z = 0 the midue of f is (0)/(2i) (Exei 16.41), and hen
S(a, n)  o(O) = 2hi Resf(z) =  f(z) dz, ()
whe  is any positively oriented fimpl¢ cl  wh ph nns only e
pole z = 0 in i infior re#on. We 11 choo   that it deri a pl-
lom with vec ,  + 1, B + I, B, wh
  -{- R u* d B=  + Re =q4,
B+I
16.7
as shown in Fig. 16.7. Integrating f along ? we have
f= f+ f+ f+ f.
JA+I
In the integral j+[S÷ 11 fwe make the change of variable w = z + ! and then use (44)
to
f(w) dw = z + 1) dz = (z) dz + ¢p(z) dz.

Therefore (46) becomes
s(a, n) -- (=) az + f(=) dz - f(=) dz. (47)
Now we show that the integrals along the horizontal segments from ,4 to ,4 ÷ 1
and from B to B + 1 tend to 0 as R  + o, To do this we estimate the integrand
on these segments. We write
and estimate the numerator and denominator separately.
On the segment joining B
7(t) = t + Re-q4. where -½ _< t < ½.
From (43) we find
IO[NO]I-< exp , (49)
r=O
where exp z -- eL The expression in braces has real part (Exercise 16.41)
Since te+" t = e  and exp {-ax/-2R/n) < I, each term in (49) has absolute
value not exceeding xp {-aRZ/n) ¢xp (-fatR/n]. But -½ < t < ½, so
we obtain the estimate
For the denorfiinator in (48) we use the triangle inequality in the form
Since Ixp (2#(t))l -- exp [-2R sin (t#4)) = exp (-x/R), we find
Therefore on the line segment joining B to B + I we have the estimate
lle r4¢R/(2s) e
lf(z)l -< !- e_¢i, a - o(1) as R -. +0,
Here o(1) dos a fon of R w¢h tends to 0  R  + .
A similar ment o that e tegand nfls to 0 on-¢ gment joining
A to A + 1 as R  + . Sin ¢ Ih of e  of ingafion is 1 in ch
, this ows at the nd and  ins on the fit of (4 tend t0 0
A
as R - + . Therefore we can write (47) in the form
S(a, n) -- ¢(z) dz + o(1) a R --,. +o. (0)
To deal with the integral  q we appiy Cauchys theorem, integrating ¢p around
the parallelogram with vertices A, B, , -c t, where • = B + ½ = Re 114. (See
Fig. 16.8.) Since  is analytic everywhere, its integral around this parallelogram
is O, so
¢ +  +  + ¢ = O, (51)
Ilecause of the exponential factor e "mt" in (45), an argument similar to that given
above shows that the integral of ¢ along each horizontal segment -0 as R  + oo
Therefore (51) gives us
and (50) becomes
q(z)dz+o(l)
whe e - Re . Using (45) we find
asR +o
as R - + , (52)
e "*/° e = dz = _, e-'i"'/* l(a, m, n,
where
Applying Caucby's theorem again to the parallelogram with vertices -,,, a,
o - nm/a, - - nm[a, we find as before that the integrals along the horizontal

segments -0 as R -, + o% so
l(a, m, n, R) = exp z + d + o(1) as
 ¢han of variable w = (z + nm/a) puts ts imo the form
l(a, m, n, R)  e " d + 1) as R
fing R  +  in (52), we find
S(a, n) =  e-'""t" lira __ e "" dw. (53)
By ting T  /nR, we  that the la limit is equ to
Tl
lim e " dw = L
y, where I is a num inddt of a and n. Therefo (53) Ores us
Te evNte 1  ke = 1   in (54). Then N(1,2)
(2, !)  I, e (54) ipl I =  + i)/, and (N)  to (42).
16.26 APPLICATION OF THE RESIDUE THEOREM TO THE INVERSION
FORMULA FOR LAPLACE TRANSFORMS
The following theorem is, in many eases, the easiest m6thod for evaluating the
limit which appears in the inversion formula for Laplace transfor,..Is. (See Exercise
1.38.)
Tkeorem 16.39. Let F be a function analytic everywhere in C except, possibly, for
a finite number of poles. SEopoae there exist three positive constants M, b, c such that
M
IF(z)l <-
wheneer [z[ .. b.
Let a be a positive number such that the vertical line x = a contains no pole of F
and let z j,..., z, denote the poles of F which lie to the left of this line. Then, for
each real t > O, we have
lim | e t'+''' F(a + h,)dv = 2n  Res {e'F(z)}.
T+e(r JIT gi'z=zk
(55)
16.9
Proof We apply Cauehy's residue theorem to the positively oriented path F
shown in Fig. 16.9, where the radius T of the ci.,dar part is taken large enough
to endose all the poles of F wWr.h lie to the left of the Ume x ffi a, and aLo T> b.
Now write
where A, B, C, D, E axe t/xe points iBdicatod in Fi 16.9, and denote  integrals
by It, I2, la, I,, I. We will prove that It --, 0 m T --, + o when k > 1.
F'trst, we have
Since T aresin (a/T)  a as T --, + o, it follows that I --, 0 as T --, +
the same way we prove I, --, 0 as T  +
Next, consider I;. We have
In
But sin ¢  2/ if 0 <_ ¢p < nf2, and hence
M f./2e -z'rot'dg= M(I_ e ='r)O
llsl <  Jo 2t---
Similarly, we find I, - 0 as T  + o. But as T  + o the righthand side of

(56) remains unchanged, Hence fimr_, +  I, exists and we have
lira I, = lim e t'+'")' F(a + iv) i dv -- 2ti Rcs
Example. Let F(z) = zl(z" + az), wher  is real. Them F has simple poles at + i.
Since zl(z 2 + a ) = ½l/(z + i) + l](z - ia)]. we find
R (d'V(z)} =/r e a,
Thcxct'or th limit in (55) has the value 2:ti cs ¢tt. From Excrci 11.38 we s that the
function f continuous on (0, +o), whose Laplav¢ tansform is F, is given by f0) 
16.27 CONFORMAL MAPPINGS
An analytic function f wi[i map two line ¢gments, intersecting at a point c, into
two curves intersecting at f(c). In this section we show that the tangn lines
these curves intersect at the same angle as the given line segments iff'(c)!  0.
This. property is geometrically obvious for linear functions. For example,
suppose f(z) =  + b. Thig represents a translation which moves every line
parallel to itseIf, and it is clear that angles are prcsrvvd. Another example is
f(z) = az, where a # 0. If [a[ = , then a = e  and this rpresvnts a rotation
about the origin through an angle a. If lal # l, then tt = Re " and f represents
a rotation composed with a stretching (if R > I) or a contraction (if R < 1).
Again, angles are presumed. A general linear functionf(z) = az + b with a  0
is a composition of these types and hence also preserves angles.
In the senctal case, differentiability at c means that we have a linear approx-
imation near c, say f(z) = f(c) + f'(cXz - c) + o(z - c), and if f'(c)  0 we
can epect angles to be preserved near c,
To formalize these ideas, let 7 and 'z be two piecewise smooth paths with
respective graphs F t and Fz, intersecting at c. Suppose that y is one-to-one on
an interval containing t, and that ?e is one-to-one on an interval containing
where ?(q) = T(Q) ffi c. Aume also that ?'(q) # 0 and "][(tz) # 0. The
difference
is called the arle from F t to F at c.
Now ass,me thatf'(c) # O. Then (by Theorem 13.4) there is a disk B(c) on
which f is o.e-to-one. Hence the composite functions
,(t) = f[y(t)] and w(t) =
will be locally one-to-one near t, and tz, respectively, and will describe arcs C,
and C intersecting atf(c). (See Fig. 16.10.) By the chain rule we have
v'](q) = f'(c)?;(t;)  0 and w'z(t) = f'(c)?(t) ¢. 0.
rl
r C
tJo
Therefore, by Theorem 1.48 there exist integers n and n such that
ar s [wi(q)] ffi arg [f'(c)] + arg [?(t,)] +
arg [w(tz)] = arg [if(c)] + arg [(t)] + 2nn,
so the angle from C to Cz atf(c) is equal to the angle from F, to Fz at c plus
an integer multiple of 2. For this reason we ey thatfpreservea anles at c. Such
a function is also said to be conformal at c.
Angles are not preserved at points where the derivative is zero. For example,
ill(z) = z 2, a straight fine through the origin making an angle u with the real axis
is mapped byfonto a straight line making an angle 2 with the real axis. In general,
whenf'(c) : 0, the Taylor expansion off assumes the form
= - cy[a + - c) + ---],
where k > 2. Using this equation, it is easy to see that angles between curves
intersecting at c are multiplied by a factor k under the mapping f
Among the important examples of conformal mappings are the Mtbius
transformations. These are functions f defined as follows: If a, b, c, d are four
complex numbers such that ad - bc  O, we define
az+ b
/(z) - , (57)
cz + cl
whenever ' cz + d ¢ 0. It is convenient to define f everywhcr on the extended
plane C* by setting f(-d/c) = oo and f(co) = a/c. (If c = O, these last two
equations are to be replaced by the single equationf(co) = co.) Now (57) can be
solved for z in trms off(z) to get
+
Z 
This means that the inverse function f-  exists and is given by
-clz + b
f-'(z) = ,
CZ  G

with the understanding that f-I(a]c) = o andf-t(o) --- -die. Thus we see
that M6bius transformations are one-to-one mappings of C* onto itself. They are
also conformal at each finite z # --d/c, tfince
bc -- ad
o.
(ez + d) z
One of the most important properties of these mappings is that they map circles
onto circles (including straight lines as spoAal cases of circles). The proof of this
is sketched ia Exercise 16.46. Further properties of M6bius transformations are
also described in the exercises near the end of the chapter.
I6.1 Let r be a tficwise smoot path with domain In, b] and graph r. Assunm .flaat the
integral J7/exists. Let S be an open region containing F and 1¢t g be a function s0ch that
a'(z) exists and equals/(z) for each z on F. Prove Ihat
In particular, if y is a gh'cuit, then A = B arKt the integral is 0. Hint. Apply Theorem 7.34
to each intm, val of cmfinuity of
16.2 Let ,he a tmsitively oriemedeitxtiarpathwith ¢amtr0 and radius 2. Verify eada
of the following by using one of C.auchy's imegral formulas.
a) f,  dz = 2,i. h) f dz  ni.
z(----l) dz--2ni(e- t). f) z2( - t)
16.3 l:t f ---- u + iv b¢ analytic on a disk B(a; R). If 0 < ¢ < R, pro that
16.4 a) Prove the following stronger version of Liouvill,'s heorem: lff
function such tt lirn= ff(z)/z l = o, the fin a constant,
b) What tan you conclude about an entire furction which satisfies an inequality of
the form [/(z)[ _< M[z[  for every compIex
473
16_ Assume that/is analytic on B(0; R). Let 7 deno the oositively oriented circle
with center at 0 and radius r, where 0 < r < R. If a is inside , show that
If a = Ae , show that this reduces to the formula
By iing t ml  of h uation we obn  ion own 
integral formula.
l&6 Au atfis ic  ¢ clu oft diB(0; 1). Iflal < !, sh at
(1  lal=)f(a)--, f(z) ----- a d,
where 7 is the positively oriented unit circle with omter at 0. Deduce the inequality
(1 - lal If(a)f __- _n If(e'[ d.
16"/ Let fCz) = ,_=o 2az'/3- if lzl < 3/2, and It #(z.) =
7 b¢ the positively oriented c/ocular path of radius [ and center 0, and ddine h(a) for
[ol # as follows:
Prove that
= a 2 2 - uz/ dz"
3 if lal < 1,
3-2a
h(a) = 2a z
-- if ]aJ > 1.
1 - 2a
Taylor eximnsi0m
16.8 Define fort the disk B(O; 1) by the equation f(z) = -%o z=- Find the Taylor
epansion of lahore the point a = ½ and also about the point a = -½. Determine the
radius of convergmce in each case.
16.9 Assume that fhas the Taylor cxpansionf(z) = ,%o (n)z", valid in B(0; a). Let
.p--I
]
e( z) -- - .f(ze'O.
Prove that the Taylor eXlansion of  consists of every pth term in that off. That is, if'
z e B(O; R) we have
e(z) =  a(en)z ='.

474
16.10 Assume that/has the Taylor expansionf(z) = -,o a.a", valid in B(0; R). Let
s(z) = =o at t- IfO < r < R and if Izl < r, show that -
1 ff(w) w "+a --
s.(z)-- . w + w
where ? is the positively oriented circle with center at 0 and radius r.
16.11 Given the Taylor ¢pansions f(z)  ,V_., 0 a.z" and g(z) -- 0 b , valid for
[z] < Rt and ]z] < R2, respectively. Prove that if [z[ < RIR  we have
whca-¢ y is the positively oriented circle of radius R l with center at 0.
16.12 Assume that fhas the Taylor expansionf(z) = _=o adz - oy', valid in 8(a; R).
a) If 0 : r < R, deduce Parseat' identity:
b) Use (a) to deduce the ineclity .,o io.I
maximum of Iff on the cirde Iz - al = r.
c) Use (b) to give another proof of the local maximum modulus principle rem
16.13 Prove Schwarz'a lemma; Let f be analytic on the disk B(O; 1). Suppose tt f(O) = 0
tf [f'(0)l = 1 r f If(o)[ = [zol for at east one o in '(0; ),
f(z) = ez, where  is real.
Hint. Apply the maximum-modulus theorem to g, where g(0) - f'(0) and g(z) = f(z)/z
ifz O.
16.14 Let fand g be analytic on an open region S. Let 7 be a Jordan circuit with graph F
such that both F and its inner region lie within S. Suppose that [9(z)l < If(z)l for every
zon F,
a) Show that
I f f'(z) + g'(a) dz = I .if(z)
zi.2, fO) + o) z, ,  d.
Hint. Let m = inf {[f(z)l - [9(z)l :   V}. Then m > 0 and hence
[/(z) + ta()l -> m > o
for each t in [0, I ] and each z on V. Now let
I
f f'(z) + tg'(z)dz, if O < t -g I.
Then  is continuous, and hcnoe constant, on [0, 1 ]. Thus, 0) = (1).
b) Use (a) to prove that f and f + g haVe the sal'tl¢ number of zeros inside F
(RoueA's theorem).
16.15 Let p be a polynomial of degree n, say p(z)  a o + az + .-. + aaz , where
a  O. Take f(z) = aw', g(z)= p(z) -- f(z) in Rouch.'s theorem, and prove that p
has exactly n zeros in C.
16.16 Let f be armlytic on the closure of the d/sk B(0; 1) and suppose If(z)l < 1 if
[z[ -- 1. Show that there is one, and only one, point eo in B(0; 1) such that f(zo)
Hint. Use Rouch6N theorem.
16.17 Le! p(z) denote the nth partial sum of the Taylor expansion e 
Using Rouch6's theorem (or otherwise), prove Ihat for every r > 0 there exists an N
(deiwding on r)such that n > Nimpliesp.(z)  0 for every z in B(0;
16.18 Ira > e, find the number of zca'os of the functionf(z) = e   az  which lie inside
the circle ]zl = 1,
16.19 Give an example of a function which has all the following properties, or else explain
why there is no such function: f is analylic everywhere in C except for a pole of order
2 at 0 and simple poles at i and -i; f(z) = f(-z) for all z; f(l) = 1; the function
9(z) = f(1/z) has a zero of oeder 2 at z = 0; and Resf(z) =
16.20 Show that each oftl following Laurent expansions is valid in the region indicated:
a)(z_ 1X2- z)  + -- ill < I z] < 2.
1
,, 1 -- -.2"- if Izl > 2.
16.1 For   t in C, dne Y(t)   c t of z  in I t on
Sh that for   0 wv ve
=- c(tO- )
1  's tom: gz o   ted aity off  gl/I  d
• e intels for e  a, in the ut exon offd ow tt   0 f
n <0.
f  t c   biary tex nr, n, f ery e > 0  e k
there ex:saint z in B(zo)sh t: If0) - cl < . Hint, tt t  
hl and ve at a ion by applg  16, to g, w g0) =

416
163,4 Thepgintatiafmit. A functioafis said to be analyth; at  iflhe functiongdelined
by the equation g(z) = f(l/z) is analytic at the o6gin. Similarly, wesay that fhas a zero,
a pole, a removable singulariw, or a essential sinsularity at 0o ifgJas a , a pole,
at 0. Liouville's theorem states that a function which is analytic everywhere in C* must
be a constant. Prove that
a)/is a polynomial if, and only if, the only singularity of fin C* is a pole at
in which case the order of the pole is equal to the degree of the polyaomial.
b) f is a rational function if, and only if, f has no singularities in C* other than
16.2q Derive the following "short cuts*' for computing residues:
a) If a is a first order pole for f, then
Ref(z) = lim(z - a)f(z).
b) If a is a pole of order 2 for £ then
Re f(z) =
where a( z ) --
Suplxfando are both analytic at a, withf(a)  0 and a a first-order zero for
thai
RCS f(z) _ f(a) Res f(z) _ f'(a)g/(a) - f(a)g"(a)
,-. g(z) a'(a)' --, [a(z)P
d) If land 9 are as in (e), except that a is a second-order zero for g, then
Rs f(z) _ 6f'(o)g"(a) --
• --, ¢(z) 3 [g°(a)] 2
126 Compute the re, dues at the_poles of f if
26  e z
a) f(z) - z2 _ 1 " b) f(z) z(z - 1) '
e) f(z) - sin z 4) f(z) - 1
z cos z 1 - 
1
e) f(z) --
1 - Z"
(where n is a iositiv¢ integcx)
16.,T/ If y(a; r) denotes the positively oriented circle with center at a and tad/us r, show
that
f - f( 2z dz = 4hi,
a) 3z 1 dz = (mi, b) z
0. (z + l)fz - 3) 0:z) + !
c) .-- dz = 2i, d) e' dz = 2ie z.
f:* dt
(a + boos t)  (a 2  bz) 
Evaluate the integrals in Exercises 16.28 through 16.35 by mtns of rlducs.
2ha
- if0< b< a.
= cos 2t dt 2ga 2
16.29 = -- if a  < 1.
1 - 2a COS t + a z I -- a 
 (1 + cos 3t)dt n(a - a + 1)
16.30 = if0 < a < I.
1 - 2a.cost + a  1 - a
161 -2, sinZtdt _ 2n(a- [a z- b 2) if0 < b <
a+ boost b 
f:. 1 dx-
16,32 x ' + x + I 3
1633 16
16.34
16,35
. X 2
(x  + 4)(x  + 9) dx '
Hint. Integrate z/(1 + z s) around the boundary of the circular sector
S : (re °:0 <_ r <_ R, 0 < 0  2/5),andletR--, .
b) -- -- dx = -- sin - n , m,. integer% 0 < m <
i + x " t
16.36 Prove that formula (38) holds if f is the quotient of two polynomials, say/" =
where the degr¢ of Q exeeods that of P by 2 or more.
16,37 Prove that formula (38) holds if/"(z) = e'P(z)lQ(z), where ra > 0 and P and Q
are polynomials such that the degree of Q exccods that of P by 1 or mor. This mak it
possible to evaluate integrals of the form
f: e ' P(Xl dx
x)
by the method described in Theorem 16.37.
16.38 Use the method suggested in Ecreis¢ 16.37 to evaluate the following integrala:
f: sinmx dx = n(1 - e -m) if am > O,a > 0.
a) x( a2 + x ) 2a 2 _
b) x + a   + ifm > 0, a>0.

16.39 Let w = e 2"g:3 and let y be a positively oricated circle whose graph does not pass
through 1, w, orw z. (The numbers 1, w, w2 are the cu be roots of I.) Prove that the integral
is qual to 2i(m + nw)/3, whcrc m and n arc integers. Determine the possible values of
m and n and describe how they delXald on y.
16.40 Let 7 bc a positively oriented circle with coates 0 and radius < 2n. If a is complex
and n is an inter, let
lov¢ that
l(0, a)=-a,
l(n, a) = 1  :-- -e-
2ni.[ 1 - e z
l([,a) = -1, and /(n,a) = 0 ifn > 1.
Calculate l(-n, a) in terms of Be.xnoulli polynomials wh¢ n _> 1 (s¢¢ Exercise 9.38).
16.41 This exci'cis¢ ,qucsts some of the details of the proof of Theorem 16.38.:..._t
() =  :'+", /(z) = (z)/(e '" - 1),
wh©r¢ a and n arc psitive integers with n even. Prove. that:
a) g(z + l)- g(z) -e='[(eZgz- l)_e zx..
b) Res::o.f(z) =
c) The real part ofi(t + Re""* + r) z is -(',/2tR + R z + /rR).
One-to-one amdytlc functions
16.42 Let S be an open subset of C and assume that fis analytic and one-to-one on S.
Pxove that:
a) f'(z) # 0 for each  in S. (Henccfis conformal a each point of
b) If g is the inverse off, then g is analytic on f(S) and '(w) = I/f'((w)) if
16.4 Let f; C -, C be anyic and one*to-one on C. Prove thatf()  z ÷ b where
  O. What can you conclude if f is one-to-one on C* and analytic on C* except possibly
for a finite numl of poles?
1,44 If f and g arc Mbius transformations, show lhal lh comsition fo is alv a
M6bius transformation.
16.4 Describe geometrically what happens to a point z when il is carried into f(z) by the
following special MSbius transformations:
a) f(z) = z + b (Translation).
b) f(z) = az, where a > 0 (Stretching or contradion)
c) f(z) = e¢z, where a is rcal (Rotation),
d) f(z) = 1/z (Inversion).
16.46 Ifc- 0, wchave
az + b a be - ad
cz + d c c(cz + d)"
Henc every MiSbius transformation can be expressod as a composition of the special cases
dcscrihed in Exercise 16.45. Use this fact to show that M6bius transformations carry
circles into circles (where straight lines am considca'ed as special cases of clinics).
16.47 a) Show that all MObius transfomaations which map th¢ uptcr half-gRant T =
{x + iy : y  0} onto the closure of the disk B(0; 1) can be cxprcssod in the
form f(z) = e(z - a)/(z - ), wher ,, is real and a  T.
b) Show that a and  can always b chosen to map any three given points of the
real axis onto any three ghren points on the unit circle.
16.48 Find all M6bins transformations which map the fight half-plane
S= (x + iy:x  O)
onto the closure of B(0; 1).
16.49 Find all MObius transformations which map the closure of B(0; 1) onto itself.
1620 The fixed points of a M6bius transformation
az+b
]'(z) - (ad - bc ; 0)
cz+ d
are those points z for which f(z) = z. Let D = (d - a)  + 4be.
a) Dctcmain¢ air fxcd points wh¢ c = 0.
b) If e # 0 and D # 0, prove that f has exactly 2 fix¢l points z and z2 (both
finite and that they satisfy the equation
.f(z) - z _ Re z  z where R > 0 and 0 is real.
f(z) - z z- z'
¢) If c  0 and D - 0, prove that f has exactly one fixed point z, and that it
satisfies the equation
1 1
- + C for some C  0.
d) Given any M/Shins transformation, investigate the successive images of a given
point w. That is, let
w = ffw), w2 =.f(w,), ..., w. - f(w,) .....
and sltldy tbe behavior of the sequence {w.}. Consider tl'a: special  a, b, c. d
real, ad-bc- I.
16.$1 Determine all complex z such that

16_ ]ff() -- _m__o , iS art entire on  at If(re'l   t U r > O,
ests (bly t¢) if, and oy if,  ex an inf n d a fulion g, ayfic
on 0; a), 0)  0, s tt f(z) = g(z) in(0; a).
  z) ffi -o a t  a polynol of  n th 1 ts lfyg
ove tz) = 0 imp [z[ > 1. Hinl. nsif (1 - z)z).
16. A funiA d on a dk a; r), is d to  a  of ini¢ ord at a if,
on B(a: r). Iff a  of infini o at a, prove lf  0  a;r).
1 e M's theo: lf f  ti on an on rion S  C  f  f  0
for ery toI circuit 7 in S, then f  aiytic on S .
SUGGESTED REFERFCES FOR FURTHER STUDY
16.1 Ahlfor L. V., Complex Analysis, 2nd ed. McGraw-Hill, New York, 1966.
16.2 Carathodory, C., 77teory of Functions of a Complex Variable,2 vols F. Steinhardt,
translator, Cbc].ca, New York, 1954.
16.3 Estermann, T., Complex Numbers and Funeton÷ Athlone Pr¢, London, 1962.
16.4 HeMs, M., Corfptex Function Theory. Academic Pre New York, 1968.
16.5 Heins, M., Selected Topica in the Classical Theory of Functions of a Complex Vari-
able. Holt, Rinehart, and Winston, New York, 1962.
I6.6 Knopp, K., Tkeory of Functions, 2 vols. F. Bagemihl, travator. Dover, New
York, 1945.
16.7 Sa, S., and ZygrnurKI, A., Analytic Furwlions, 2rid ed. E. J. Scott, translator.
. Monografie Matemmyczne 28, Warsaw, 1965.
16.8 Sansone, G., alld Gerrelu:n, J, Lecture on tlw Theory of Functions of a Complex
Variable, 2 vols. P. Noordhoff, Grningen, 1960.
16.9 Titchmarsh, E. C., Theory of Functlonx, 2rid ed. Oxford University Press, 1939.
INDEX OF SPECIAL SYMBOLS
e, ¢, belongs to (does not belong to), I, 32
0 is a subset of, I, 33
R, set of real numbers, 1
It +, It-, set of positive (negative) numbers, 2
{x: x satisfies P}, the set of x which satisfy property P. 3, 32
(a, b), [a, b], open (closed) interval with endpoints a and b, 4
Ia, b), (a, b], haW-open intervals, 4
(a, + co), [a, + ), (-co, a), (-o, a], infinite intcrval, 4
Z + , set of positive integers, 4
Z, set of all integers (positive, negative, and zero), 4
Q, set of rational numbers, 6
max S, ndn S, largest (smallest) element of S, $
sup, inf, supcemum, (infimum), 9
Ix], greatest integer _< x, 1 [
R*, extended veal-number systcn, 14
C, the set of complex numbcTs, the complex plane, 16
C*, extended complex-number sysm, 24
A x B, cartesian product of I and B, 33
F(S), image of S under F, 35
F: S  T, function from S to T, 35
{Fj}, sequence whose nth term is F,, 37
U, , union, 40, 41
[, , intersection, 41
B - A, the set of points in B but not in I, 41
f-(Y), inverse image of Y under f, 44 (Ex. 2.7), 8
It", n-dimensional Euclidean space, 47
(x,. ., x), point in R , 47
]txl[, norm or length of a vector, 48
ut, kth-unit coordinate vector, 49
B(a), B(#; r), open n-baLl with center #, (radius r), 49
int S, interior of S, 49, 61
(a, b), I'm, b, n-dimensional open (closed) interval, 50, 52
, closure of S, 53, 62
S', set of accumulation points of S, 54, 62
(M, d), metric space M with metric d, 60
48t

d(x,y), distance from x to y in metric space, 60
BM(a; r), ball in metric space M, 61
aS, boundary of a set S, 64
tim, lira, right- (left-)hand limit, 93
f(c + ), f(c -), right- (left-)hand limit off at c, 93
fly(T), osdIlation off on a set T, 98 (Ex. 4.24), 170
w,(x), oscillation of fat a point X, 98 (Ex. 4.24),
f'(c), derivative of fat c, 104, 114, 117
Df, partia! derivative off with respect to the kth coordinate,
Dr.if, second-order partial derivative, 116
[a, hi, set of all partitions of [a, b], 128, 141
Vf, total variation off, 129
At., length of a rectifiable path f, 134
S(P,f, tQ, Riemann-Stieltjes sum, 141
fe R() on [a, b],fis Riemann-integrable with respect to • on In, b], I41
fe R on In, b],fis Riemann-integrable on ['a, b], 142 
0t ., on In, b], • is increasing on In, hi, 150
Uff',f, t), L(P,f, ), upper (lower) Stieltjes sums, 151
lim sup, limit superior (upper Iimit), 184
lim inf, limit inferior (lower limit), 184
ao = O(b,), a, = o(b),-big oh (little oh) notation, t92
l.i.m, f = f, {fo} converges in the mean to f, 232
fe C,fhas derivatives of every order, 241
a.e., almost everywhere, 172
f /" fa.e. on S, sequence {f) increases on S and converges tofa.e, on S, 254
S(1), set of step functions on an interval/, 256
U(1), set of upper functions on an interval I, 25.6
L(I), set of Lebesgue-integrable functions on an interval I, 260
f+, f-, positive (negative) part of a function f, 261
M(1), set of measurable functions on an interval/, 279
(s, characteristic function of S, 289
/(S), I_¢besgue measure of S, 290
(f, g), inner product of functions fond g, in L(i), 294, 295
Ill[I, L2-norm off, 294, 295
L(I), set of square-integrable functions on 1, 294
f* ,, convolution off and ,, 328
f'(¢; u), directional derivative of f at e in the direction u, 344
To, f'(c), total derivative, 347
V f, gradient vector off, 348
re(T), matrix of a linear function T, 350
I)f(e), Jacobian matrix of f at c, 351
L(x, y), line segment joining x and y, 355
det [a], determinant of matri [a], 367
J, Jacobian determinant of f, 368
f  C', the components off have continuous first-order partials, 371
tf(x) dx, multiple integral, 389, 407
KS), (S), inner (outer) Jordan content of S, 396
c(5"), Jordan content of S, 396
,f, contour integral offatong 1', 436
A(a; rx, r:), annulus with center a, 438
n(1', z), winding number o1' a circuit 1' with respect to z, 445
B'(a), B'(a; r), deleted neighborhood of a, 457
Res $(z), residue of fat a, 459

Abel, Nei{s
Abel, l tl,
A i 13,
 438
3
  of a JO
t of pl , 2I
o for  n I, Z 9
BeJI, tn a metric spece, 6!
in R', 49
478
i,  t (. 9
i , 3  ll.I
i 251 (. 9. 4 .
INDEX
Cantor, GeOl (1845-1918), 8, 32, 56, 67,
I, 312
Cantor inlet'so theorem, 56
t- , 67 .. 3.25)
   (. 7.32)
 , 312
 , 475 .

Cauchy condilion,
for producls, 207
for SeXlUOnCeS, 73, 183
for sris, 186
for uform  , 223
uchy, inili, 451
instal futa,
al t,
pfipa[ v, 277
product,
sMue thegn,
uen, 73
Cauy-Ri uatis, 118
Cauchyh
for impels,
for ss,.
o, s,
summability of Fo ,
Chain l¢, compl funis
maix fo of, 353
t-vfl fi, 114
Ch  b, in a  1,
in a mtip
in a Ri int, I
cfisfic fion, 289
Cuit, 435
, ll, 67 x. 3.31)
, 435
inll, 4,
mapping,  (Ex. 4.32)
, 3, 62
os of a , 53
Cumti law,
Compt t, 59, 63
mp tt,
ol¢mmt, 4i
mpl¢ c s 74
p on  $36 ( !1.6)
Compls , 9
Compl num,
Complex pl 17
Comnt, at, 51
of a ic s. 87
of a  47
Comfite fion, 37
Condensation point, 67 (E. 3.23)
Conditional coneq=nt series, 189
rearrat,crncnt of, 197
Conformal mapping, 471
Conjugate complex number, 28 (Ex. 1.29)
Connected, metric spao, 86
set, 86
Content, 396
Continuity, "/8
uniform, 90
Continuously differcatiable function, 371
Contour intogral. 436
Contraction, constant, 92
fixed-point theorem,
mapping, 92
Convergence. absolute., 189
bounded, 227
conditional, 189
in a metric space, 70
maJ 232
of a product, 207 c:." .... "
of a sequence, 183
of a series,
pointwis¢, 21
uniform, 22I
Converse of a relation, 36
Convex set, 66 (Fx. 3.14}
Convolution integral, 328
Convolution theorem, for Fouri trans-
forms, 329
for Laplac¢ transforms, 342 (F_.x. 11.36)
Coordinate mmsformation, 417
Countabl additivity, 291
Countable set, 39
Covering of a
Covering theorem, Hein-Borl, 58
Lindel6f, 57
Cramcr's rule, 367
Ckaw, closed, 435
Jordan, 435
piewissmooth, 435
roctifiablc, 134
Danieli, P.J. (1889-1946), 252
Darboux, Gaton (1842-1917), 152
DccinIs, I1, 12, 27 (Ex. 1L22)
Dodekind, Richard (1831-1916), 8
Deleted neighborhood, 457
Dc Moivre Ham (I667-1754), 29
pc Moivr's thoorom, 29 (F, 1.44)
Dense f-t, 68 (Ex. 3.32)
Demtmerable set, 39
Derivative(s), of complex functions, 117
directional, 344
partial, 115
of real-valuo functions, 104
total, 347
of vctor-valucd functions, 114
Dezivod t, 54, 62
Dotetminant. 367
Difference of two sets, 4I
Differentiatlo, of integis, 16-, 167
of seqtnces, 229
of series, 230
Dini, Ulise (18451915), 248, 312, 319
Dini's theorem, on Fourier  319
on uniform convergem 248 (Ex. 9.9)
Dirichlet, Peter Gttav  (I05-
1859), 194, 205, 215, 230, 317, 464
Dirichlet, inlegrals, 314
kernel, 317
product, 205
series, 215 (Ex. 8,34)
Didchlet's tesl, for convergence of series,
194.
for uniform convergence of series, 230
Disconnected set, 86
Discontinuity, 93
Discrete metric spaee 6I
Disjoint sets, 41
collection of, 42
Dik, 49
of convergence, 234
Dislance function (metric), 60
Distributive law, 2, 16
Divergent, product, 207
sequence, 183
series, 185
Divisor, 4
greatest common, 5
Domain (open region), 90
Domain of a function, 34
Dominated convergence theorem, 270
Dot product, 48
Double, integral, 390, 40"/
Double sequence,, 199
Double series, 200
Du BoiReymond, Paul (1831-1889), 312
Doplication formula for the Gamma
fimction, 34I (Ex. 11,31)
e, ixrationality of, 7
Element of a set, 32
Empty  33
Equ of  i 36
relation, 43 (E. 2.2)
F_sential ingtflarity, 458
Euclidean, metr, 48, 61
space R', 47
Euclid's lemma, 5
Euler, Leonard (1707-1783), 149, 192,
Eut-r's, constant, 192
product for ¢:(a), 209
summation folmula, 149
theorem on homogeneous functions, 365
Exponential form, of Fourie integral
theorem, 325
of Fourier 'ie, 323
Exponential function, 7, 19
Extended complex plane, 25
Extended real-numbe system, 14
Extemion of a function, 35
Exterior (or outer rion) ofa Jorda curve,
447
Extremum problems 375
Fatou, Pien-e (I878-1929L 299
Fatou's lenuna, 299 (Ex. 10.8)
Fej6r, Leopold (I880-1959), 179, 312, 320
Fejr's theorem, 179 (Ex. 7.23), 320
Yekete, Michel, 178
Field, of complex number 116
of real numa, s, 2
Finite set, 38
Fischer, Ernst (1875-1954), 297, 311
Fixed point, of a function, 92
Fixed-point themrern, 92
Fourier, Joseph (1758-I80), 306, 309,
312, 324, 326
Fourier coefficient, 309
Fourier integral theorem, 324
Fourier , 309
Fourier tramform, 326
Fubini, Guido (1879-I943), 405, 410, 413
Fulfini's theorem, 410, 413
Function, ¢knition of, 34
Fundamental theorem, of a/lbra, 15, 451,
475 (F.x. 16.15)
of intral calculus, 162

Gamma run,ion, continui of, 282
defmithn of, 277
derivative of, 284 03 (Ea. 109)
duplication formula for, 341 (E 11.3I)
functional equation for, 2"/8
series for, 30 (Ex. 10.31)
Gauss, Karl  0777-1855), 17,
Geometric  190, 195
Gibbs" phenomenon, 338 (Ex. 11.19)
Global prope, 79
Gunrsat, lkkuard (1858-1930, 434
Gradient, 348
Gram, Jrgen Pedersen (18.0-1916), 335
klt proce 335 (I it 1
Greatest lower boml, 9
Hadamard, Jacques (186.%1963), 386
Hadamard determinant theocem, 386
13.16)
txty, Godfrey throld (1877-947), 30,
206, 217, 251, 312
Heine, Eduard (1821-1881), 58, 91, 312
Heine-Betel  theorem, 58
Heine's theorem, 91
Hobson, Ernest William (1856-1933),
312, 415
Homoemous fuaction, 364 (F. I2.18)
Hyperplane, 394
Identity theorem for analytic fulx:tiom, 452
 35
Irnaginary parg 15
Imaginary unit, 18
Implicit-function theorem, 374
Improper Riemann integral, 276
Increasing seqnce, of functions. 254
of mbcs, .71 185
Independt t of functions, 335 (Ex. 11.2)
hducion Winciple, 4
Induce set, 4
CatrJay-Schwarz, 14, 177 (Ex. 7.163, 294
Minkowski, 27 (Ex. 1
triangle, 13, 294
Infinite, derivative, 108
prod 2
serk 185
set, 38
Iafmity, u C , 2
 R*, 14
 Jo m 3
 4
t, 191
, 42
 (or  )
7
. of a  49, 6t
In g . 61 .. ......
f , 112
 of  41
In,   4
h , 36
Ini  372
 ,  (. 2.,
Ini foa, for Fo
327
for 1  2 ( 11.38),
h t, 53
Ilat silW, 458
Ilat  452
I Iit, 1
Jacobi, Carl Gustav Jacob (1804-1851),
351,368
jacobian, determhumg 368
ntrix, 351
Jordan, Camille (I838-1922), 312,
396, 435, 447
Jordan, arc, 435
content, 396
curve 435
curve theonn, 417
theorem on Fourier series, 319
Jordan-measurable set, 396
Jump, discontinuity, 93
of a function, 93
Kestetman, Hyman, 165, 182
Kronecker delta, o, 385 (Ex. 13.6)
La-norm, 293, 295
Lagra.ug, Joseph Iuis 0736-1813), 27,
30, 30
Lagrange, identity, 27 (Ex. 1.23), 30
L48), 38O
multipliers, 380
Landau, F.dmnnd (1877-1938), 31
Laplac Pierre Simon (1749-1827), 326,
342, 468
Laplace transform, 326, 342, 468
Laurent, Pierre Alphonse (1813-1854),
455
Laurent expansion, 455
I_east upper bou,.d. 9
Lcbesgue Henri (1875-1941), 141, 171,
260, 270 273, 290, 292;312, 391,405
bounded convergence theorem, 273
criterion for Riemann intcgrabiIity, 171,
391
dominated-convergence theorem, 270
integral of complex functions, 292
integral of real functions, 260, 407
Legendre, Adrien-Marie (1752-1833), 336
l_egendre po|ynomials, 336 (Ex. 11.7)
Lcibniz, Gottfried WilheJm (t6461716),
121
l_¢ibniz' formuta, 121 (Ex. 5.6)
Length of a path, ! 34
Levi, IkV (1S75-1961), 265, 267, 268,
407
Levi monotone convergence theorem, for
sequences, 267
for series, 268
for step functions, 265
Limil, inferior, 184
in a metric space, 71
superior, 184
Limit function, 218
Limit theorem of Abel 245
LindelOf, Ernst (1870-1946), 56
489
Linddff covering theorcca, 57
Linear function, 345
Linear space, 48
of functions, 137 (Ex. 6.4)
Line sgment in R 88
Linearly dependent s of ftmcfions, 122
(Ex. 5.9)
Liouville, Joseph (1809-1882), 451
Liouville's theorem, 451
Lipschitz, Rudolph (1831-1904), 121,137,
312, 316
Lipschitz condition, 121 (Ex. 5.1), 137 (F_.x.
6.2), 3116
Littlewood, John Edensor (1885-
312
Local extremum, 98 (E. 4.25)
 property, 79
Localization theorem, 318
Logthm, 23
Lower bound, 8
Lower integral, 152
Lower limit, 184
Mapping, 35
Matrix, 350
product, 351
Maximum and minimum, 83, 375
Maximum-modulus principle, 453, 454
Mean convergence, 232
M-V Th for
of l-ucd funcfio, 110
of t.valu fo, 355
M-Value eo for in
multiple imps, 1
Rien ials, 1, .165
Rianttj inls,
Mumble fion, 2, 
Muble t, , 
Mu of a , , 
, 169, , 39t, 5
Me, F 01927),
Mt* theom, 
M¢, 
Mmum-mul iple,
Minko, Hn (181), 27
Mi's iuaiity, 27 (Ex. 1.)
Mbius, Austus Fdina (1
 8), 471
Mbius oti 471
Modut of a mpl n, 18
Moaot fi 

Monolonic sequence, 185
Multiple intcglXI, 389, 407
Multiplicative function, 216 (F.x. 8.45)
Neighborhood, 49
of infinity, ! 5, 25
Niven, Ivan M. (1915- ), 180 (Ex.
7.33)
Nonemy set, 1
Nonmeasurable function, 304 (Ex, 10.37)
Nonmeasurable set, 304 (Ex. 10.36)
Non_negative, 3
Norm, of a function, ]O2 ff.x. 4.66)
of  partition, 141
of a vector, 48
O, o, oh notation, 192
One-to-one function, 36
Onto, 35
Operator, 327
Open, covering, 56, 63
interval in R, 4
mapping, 370, 454
mappi tlmmn, 37 I, 454
set in a metric spaoe, 62
set ia R', 49
Order, of pole, 458
of zero, 452
Ordered n-tupl 47
Ohod pair, 33
Ordr-p fmtion, 38
Orat , 403 . 14AI)
Orthoormal ! of uom, 306
Oillafio of a fm0fi 98 (Ex. ,I.24, 170
Outer Jordan contenl, 396
Parallelogram law, 17
Paneval, Mark-Antoine (circa 1776-
1836), 309, 474
ParscvaPs formula, 309, 474 (El. 26.12)
Partial derivative, !15
of higher order, 116
Partial sum, 185
Partial summation formula, 194
Partition of an interval, 128, 141
Path, 88, 133, 435
Peano, Ginseppo (1558-1932), 224
Perfect set, 67 (Ex. 3.25)
Periodic function, 224, 317
Pi, x, irrationality of, 180 (EX. 7.33)
Piecewise-smooth path, 435
Point, in a metric stm.ce, 60
in R', 47
Poinlwise oonvergem:e, 218
Poisson, Sim6on Dm3is (I781-1840), 332,
473
Poisson, integral formula, 473" (Ex. 16.5)
summation formula, 332
Polar rdinates, 20, 418
Polygonal cnrve 89
Polygonally coma:clod set, 89
Polynomial, 80
in two variables, 462
zm-os of, 451, 475 (Ex. 16.15)
Powers of complex numbers, 21 23
Prime number, 5
Prime-numlx:r theorem, 175 (Ex. 710)
Principal pa 456 :...
Projection, 394
Quadratic form, 378
Quadric surface, 383
Quotient, of complex numbers, 16
of real numbers, 2
Radius of eonvergen 234
Ran of a function, 34
Ratio test, 193
Rational function, 81, 462
Rational number, 6
Real number, 1
ReaJ part, 15
Rt of series, 196
Reciprocity law for Gauss sums, 464
Rectifiable path, 134
Reflexive relation, 43 (Ex. 2.2)
Region, 
Relation, 
Removable discontinuity, 93
Removable s/ngularity, 458
Residue, 459
Residue thooan, 460
Restriction of a fanction. 35
(1826-1866), 17, I42, 153, 192, 209,
312, 313, 318, 389, 475
condition, 153
integral, 142, 389
localization theorem, 318
sphere, 17
theorem on angularities, 475 (Ex. 16.22)
zeta function, ! 92, 209
Riemarm-Lehesgue lemma, 3] 3
Riesz, Frigyes (1880-1956), 252, 297, 305,
311
Riesz-Fischer theorem, 297, 311
Righthand derivative, 108
Righthand limit, 93
Rolle, Michel (1652-1719), 110
Rolle's' theorem, 110
Root test, 193
Roots of complex numbcxs, 22
Rouch6, Eug/me (I832-1910), 475
Rouchb-'s theorem, 475 (Ex. 16.14)
Saddle point, 377
Scalar, 48
Schmidt, Erhard (1876-1959), 335
Schoenberg, Isaac J., (i903- ), 224
Schwarz, Hermann Amandus (1843-1921),
I4, 27, 30, 122, 177, 294
Schwarzian derivative, 122 (Ex. 5.7)
Schwarz's lemrn 474 (E. 16.13)
Second-derivative test for cgtrean0h 378
S¢ond Mean-Value Theorem for Ricmann
integrals, 165
Semimic spao, 295
Se4hatabl¢ metrie spa 68 (Ex. 3.33)
Scqutmce, de.ilion of, 3"/.
Similar (¢quinumerous) sets, 38
Simple curve, 435
Simply c rogion, 443
Singularity, 458
essential, 459
poI¢, 451]
removable. 458
$1obbovian integral, 249 (Ex. 9.17)
Space-ling cur. 224
Spherical coordinates, 419
Stationary point, 377
Sz.p function, ]48, 406
$r¢og'aChi¢ projection, 17
Sfi¢ltjes, Thomas Jan (1956-1894), 140
Stieltjcs integral, :140
Smae, Marshal [-I. (1903- ), 2.52
Strictly increasing function, 94
Subsequenee, 38 .... 
Subset, 1, 32 ...... .._
Substitution theorem for power serka, Z
Sup norm, 102 (Ex. 4.66)
Supremum, 9 . :. ...
Symmetric quadratic form, 378
Symmetric relation, 43 (Ex. 2.2)
Tannery, Jules (1848-]910), 299 •
Tannery's theorem, 299 (F.x. 10.7) " -:
Taubsr, Alfred (1866-c/rca 1947),.
Tauberian theorem, 246, 251 (Ex,
Taylor, Brook (I685-1731), 113, 241,
361,449
Taylor's formula with remainder, 1
for functions of several variables, 361
Taylor's sexies, 241,449
Telescoping series, 186
Them function, 334
Tonelli, Leonida (1885-1940, 415
Tonelli-Hobson test, 415
Topological, n'tapping. 84
property, 84
Topology, point set, 47
Total variation, I29, 178 (Ex. %20)
Transformation, 35, 417
Transitive relation, 43 (F..x. 2.2)
Triangle inequality, 13, 19, 48, 60, 294
Trigonometric series, 312
Two-valtaxl function, 86
Uncountable set, 39
Uniform bound, 221
Uniform continuity, 90
Uniform convergence, of sequenh 221
of series, 223
Uniformly bounded sequence, 201
Union of sets, 41
Unique factorization theoron, 6
Unit crdinate vectors, 49
Upper bound, 8
Upper half-plane, 463
Upper function, 256, 406
Upper integral, 152
Upper limit, 1,
Vall6e-Poussin, C. J. de ia
312
Value of a ftmction, 34
(18e6-196),

Variation. boundl. 12
total, 129
Vecto¢, ,$7
Vor-valued funon, 77
Volute, 3, 397
Wromki,.J.M.H. (177g-lg53), 122
Wromkian, 122 (Ft. 5.9)
Yours, William Hamry (1963-1942), 252,
312
Zero mea 169, 391, 40_
Zero of an analytic function, 42
Zero --'tor, 48
Zeta. function, Euler product fo, 209
stries rtpretentafion, 192