Rudin - Analisys

This book was set in Times New Roman.The editors were A. Anthony Arthur and Shelly Levine Langman;the production supervisor was Leroy A. Young.R. R. Donnelley & Sons Company was prinler and binder.

This book is printed on acid-free paper.

Library of Congress Cataloging in Publication Data

Rudin, Walter, datePrinciples of mathematical analysis.

(International series in pure and applied mathematics)Bibliography: p.Includes index.1. Mathematical analysis. I. Title.

QA300.R8 1976 515 75-17903ISBN 0-07-054235-X

PRINCIPLES OF MATHEMATICAL ANALYSIS

Copyright © 1964, 1976 by McGraw-Hili, Inc. All rights reserved.Copyright 1953 by McGraw-Hill, Inc. All rights reserved.Printed in the United States of America. No part of this publicationmay be reprodu(''ed, stored in a retrieval system, or transmitted, in anyform or by any means, electronic, mechanical. photocopying. recording. orotherwise, without the prior wriHen permission of the publisher.

Preface

ChapiN t The Real and Complex Number Systems

IntroductionOrdered SetsFieldsThe Reai FieldThe EXlended Real Number SyslemThe Complex FieldEuclidean SpacesAppendixExercises

Chapter 2 Basic Topology

Finite, Countable, and Uncountable SetsMelric SpacesCompact SelsPerfect Sels

CONTENr

IJ

JI2

vi CONTENTS viiCONTENTS

Connected Sets 42 Chapter 6 The Riemann-Stieltjes Integral 120Exercises 43

Definition and Existence of the Integral 120Properties of the Integral 128

Chapter 3 Numerical Sequences and Series 47 Integration and Differentiation· 133

Convergent Sequences 47 Integration of Vector-valued Functions 135

Subsequences 51 Rectifiable Curves 136

Cauchy Sequences 52 Exercises 138

Upper and Lower Limits 55 Chapter. 7 Sequences and Series of Functions. 143Some Special Sequences 57Series 58 Discussion of Main Problem 143Series of Nonnegative Terms 61 Uniform Convergence 14~

The Number e 63 Uniform Convergence and Continuity 149The Root and Ratio Tests 65 Uniform Convergence and Integration 151Power Series 69 Uniform Convergence and Differentiation 15~

Summation by Parts 70 Equicontinuous Families of Functions IS·'Absolute Convergence 71 The Stone-Weierstrass Theorem IS'Addition and Multiplication of Series n Exercises WRearrangements 75

Chapter 8 Some Special Functions 17:Exercises 78Power Series 17:The Exponential and Logarithmic Functions 17:

Chapter 4 Continuity 83 The Trigonometric Functions 18_Limits of Functions 83 The Algebraic Completeness of the Complex Field 18-Continuous Functions 85 FDurier Series 18Continuity and Compactness 89 The Gamma Function 19Continuity and Connectedness 93 Exercises 19'Discontinuities 94Monotonic Functions 95 Chapter 9 Functions of Several Variahles 20Infinite Limits and Limits at Infinity 97 Linear Transformations 20Exercises 98 Differentiation 21

The Contraction Principle 22-

Chapter 5 Differentiation 103The Inverse Function Theorem 22The Implicit Function Theorem 22

The Derivative of a Real Function 103 The Rank Theorem 22Mean Value Theorems 107 Determinants 23The Continuity of Derivatives 108 Derivatives of Higher Order 23L'Hospital's Rule 109 Differentiation of Integrals 23Derivatives of Higher Order 110 Exercises 23

.Taylor's Theorem 110Differentiation of Vector-valued Functions III Chapter 10 Integration of Differential Forms 24Exercises 114 Integration 2,1

Primitive MappingsPartitions of UnityChange of VariablesDifferential FormsSimplexes and ChainsStokes' TheoremClosed Forms and Exact FormsVector AnalysisExercises

Chapter 11 The Lebesgue Theory

Set FunctionsConstruction of the Lebesgue MeasureMeasure SpacesMeasurable FunctionsSimple FunctionsIntegrationComparison with the Riemann IntegralIntegration of Complex FunctionsFunctions of Class !i'2Exercises

Blbliograpby

List of Special Symb<!ls

Index

248251252253266273275280288

300

300302310310313314322325325332

335

337

339

PREFACI

This book is intended to serve as a text for the course in analysis that is usual]taken by advanced undergraduates or by first-year students who study math,matics.

The present edition covers essentially the same topics as the second on'with some additions, a few minor omissions, and considerable rearrangement.hope that these changes will make the material more accessible amd more attra,tive to the students who take such a course.

Experience has convinced me that it is pedagogically unsound (thou~

logically correct) to' start off with the construction of the real numbers from t1,rational ones. At the beginning, most students simply fail to appreciate the necfor doing this. Accordingly, the real number system is introduced as an order<field with the least-upper-bound property, and a few interesting applications.this property are quickly made. However, Dedekind's construction is not omited. It is now in an Appendix to Chapter I, where it may be studied and enjoy.whenever the time seems ripe.

The material on functions of several variables is almost completely r,written, with many details filled in, and with more examples and more motivlion. The proof of the inverse function theorem-the key item in Chapter 9-

x PREFACE

,i'."plified by means of the fixed point theorem about contraction mappings.DIfferentIal forms are dIscussed in much greater detai\. Several applications ofStokes' theorem are included.

As regards othe~ changes, the chapter on the Riemann-Stieltjes integralhas been trimmed a bIt, a short do-it-yourself section on the gamma functionhas been a?ded to Chapter 8, and there is a large number of new exercises mostof them wIth fairly detailed hints. '

J ha~e also included se~eral references to articles appearing in the AmericanM.athemalt~alMonthl~ and In Mathematics Magazine, in the hope that studentsWIll develop the habIt of looking into the journal literature: Most of thesereferences were kindly supplied by R. B. Burcke\. :, ~ver the. ~e~rs, many people, students as well as teachers, have sent mecorre~tlOns, CritiCisms, and other comments concerning the previous editionsof thIs book. I have appreciated these, and [ take this opportunity to expressmy SIncere thanks to all who have written me.

WALTER RUDIN

1THE REAL AND COMPLEX NUMBER SYSTEMS

INTRODUCTION

A satisfactory discussion of the main concepts of analysis (such as convergence,continuity, differentiation, and integration) must be based on an accuratelydefined number concept. We shall not, however, enter into any discussion ofthe axioms that govern the arithmetic of the integers, but assume familiaritywith the rational numbers (i.e., the numbers of the form min, where m and nare integers and n i' 0).

The rational number system is inadequate for many purposes, both as afield and as an ordered set. (These terms will be defined in Sees. 1.6 and 1.12.)For instance, there is no rational p such that p2 = 2. (We shall prove thispresently.) This leads to the introduction of so-called "irrational numbers"which are often written as infinite decimal expansions and are considered to be"approximated" by the corresponding finite decimals. Thus the sequence

I, lA, 1.41, 1.414, 1.4142, ...

"tends to }2." But unless the irrational number }2 has been clearly defined,the question must arise: Just what is it that this sequence "tends to"?

I 2 PRINCIPLES OF MATHEMATICAL ANALYSI~THE REAL AND COMPLEX NUMBER SYSTEMS 3

(I)

1.1 Example We now show that the equation

This sort of question can be answered as soon as the so-carred "realnumber system" is constructed.

is not satisfied by any rationalp. If there were such ap, we c~uld writep ,= mInwhere m and n are integers that are not both even. Let us aSSUID.e "this "is done.Then (I) implies

1.4 Definition Throughout Chap. I, the set of all rational numbers will bedenoted by Q.

ORDERED SETS

1.5 Definition I.:et S be a set. An order on S is a relation, de.,oted by <, withthe following two properties:

(i) If XES and yES then one and only one of the st~tements

In order to elucidate its structure, as well as that of the complex numbers,we start with a brief discussion of the general concepts of ordered set and field.

Here is some of the standard set-theoretic terminology that will be usedthroughout this book.

1,3 DefiDitions If A is any set (whose elements may be numbers or any otherobjects), we write x E A to indicate that x is a member (or an element) of A.

If x is not a member of A, we write: x ¢ A.The set which contains no element will be called the empty set. If a set has

at least one element, it is called nonempry.If A and B are sets, and if every element of A is an element of B, we say

that A is a subset of B, and write A c B, or B:::> A. If, in addition, there is anelement of B which is not in A, then A is said to be a proper subset of B. Notethat A c A for every set A.

If A c Band B c A, we write A = B. Otherwise A oF B.

m 2 = 2n2,(2)

This shows that m' is even. Hence m is even (ifm were odd, m' would be odd),and so m' is divisible by 4. It follows that the right side of (2) is divisible by 4,so that n2 is even, which implies that n is even.

The assumption that (I) holds thus leads to the conclusion that both mand n are even, contrary to our choice of m and n. Hence (I) is impossible forrational p.

We now examine this situation a little more closely. Let A be the set ofall positive rationals p such that p' < 2 and let B consist of all positive rationalsp such that p' > 2. We shall show that A contains no largest number and B contains no smallest.

More explicitly, for every p in A we can find a rational q in A such thatp < q, and for every p in B we can find a rational q in B such that q < p.

To do this, we associate with each rational p > 0 the number

1.2 Remark The purpose of the above discussion has been to show that therational number system has certain gaps, in spite of the fact that between anytwo rationals there is another: If r < s then r < (r + s)/2 < s. The real numbersystem fills these gaps. This is the principal reason for the fundamental rolewhich it plays in analysis. '

y<xx=y,x<y,is true.

(ii) If x, y, Z E S, if x < y and y < Z, then x < z.

The statement "x < y" may be read as "x is less than y" or "x is smallerthan i' or H x precedes y".

It is often convenient to write y > x in place of x < y.The notation x :'E: y indicates that x < y or x = y, without specifying which

of these two is to hold. In other words, X:'E: Y is the negation of x> y.

1,6 Definition An ordered set is a set S in which an order is defined.For example, Q is an ordered set if r < s is defined to mean that s - r is a

positive rational number.

1.7 Definition Suppose S is an ordered set, and E c S. If there exists afJ E S such that x :'E: fJ for every x E E, we say that E is bounded above, and callfJ an upper bound of E.

Lower bounds are defined in the same way (with ~ in place of :'E:).

p'-22p+2q=p---=-_.

p+2 p+2

Then

(3)

(4) , _ 2 _ 2(p' - 2)q - (p + 2)' .

If p is in A then p' - 2 < 0, (3) shows that q > p, and (4) shows thatq' < 2. Thus q is in A. ,,'

If p is in B then p' - 2 > 0, (3) shows that 6< q < p, and (4) shows thatq' > 2. Thus q is in B. '

4 PRINCIPLES OF MATHEMATICAL ANALYSIS

,1.8 Definition Suppose S is an ordered set, E c S, and E is bouuded above.Suppose there exists an a E S with the following properties:

(i) a is an upper bound of E.(ii) If 1 < ex then 1 is not an upper bound of E.

Then a is called the least upper bound ofE [that therds at most one sucha is clear from (ii)l or the supremum ofE, and we write

ex = supE.

The greatest lower bound, or infimum, of a set E which is bounded belowis defined in the same manner: The statement

ex = infE

means that ex is a lower bound of E and that no p with p > ex is a lower boundof E.

1.9 Examples

(a) Consider the sets A and B of Example 1.1 as subsets of the orderedset Q. The set A is bounded ahove. In fact, the upper bounds of A areexactly the members of B. Since iJ contains no smallest member, A hasno least upper bound in Q.

Similarly, B is hounded below: The set of all lower bounds of Bconsists of A and of all r E Q with r ,,; O. Since A has no lasgest member,B has no greatest lower bound in Q. .(b) If ex = sup E exists, then ex mayor may not be a member of E. Forinstance, let E1 be the set of all r E Q with r < O. Let E, he the set of allrE Q with r ~ O. Then

sup E1 = sup E, = 0,

and 0 ¢ E1, 0 E E,.(c) Let E consist of all numbers lin, where n = I, 2, 3, .... Thensup E = I, which is in E, and inf E = 0, which is not in E.

1.10 Definition An ordered set S is said to have the [east-upper-bound propertyif the following is true:

If E c S, E is not empty, and E is bounded above, then sup E exists in S.Example 1.9(0) shows that Qdoes not have the least-upper-bound property.We shall now show that there is a close relation between greatest lower

bounds and least upper bounds, and that every ordered set with the least-upperbound property also has the greatest-lower-bound property.

THE REAL AND COMPLEX NUMBER SYSTEMS 5

1.11 Theorem Suppose S is an ordered set with the least-upper-bound property,Be S, B is nat empty, and B is bounded below. Let L be the set of all lower

bounds of B. Thenex = supL

exists in $, and IX = inf B.In particular, inf B exists in S.

Proo( Since B is bounded below, L is not empty. Since L consists of. exactly those yES which satisfy the inequality y ~ x for every x E B, wesee that every x E B is an upper bound of L. Thus L is bounded above.Our hypothesis ahout S implies therefore that L has a supremum in S;call it ex.

If 1 < ex then (see Definition 1.8) y is not an upper bound of L,hence 1 ¢ B. It follows that ex ~ x for every x E B. Thus a E L.

If a ex. In other words, a

is a lower bound of B, but Pis not if P> a. This means that ex = inf B.

FIELDS

1.12 Definition A field is a set F with two operations, called addition andmultiplication, which satisfy the following so-called "field axioms" (A), (M),and (D):

(A) Axioms (or addition

(AI) If x E F and y E F, then their sum x + y is in F..(A2) Addition is commutative: x + y = Y + x for all x, y E F.(A3) Addition is associative: (x + y) + z = x + (y + z) for all x, y, Z E F.(A4) F contains an element 0 such that 0 + x = x for every x E F.(AS) To every x E F corresponds an element -x E F such that

x +( -x) = O.

(M) Axioms for multiplication

(M I) If x E F and y E F, then their product xy is in F.(M2) Multiplication is commutative: xy = yx for all x, y E F.(M3) Multiplication is associative: (xy)z = x(yz) for all x, y, z E F.(M4) F contains an element I '" 0 such that Ix = x for every x E F.(MS) If x E F and x'" 0 then there exists an element I{x E F such that

x . (1Ix) = I.


6 PRINCIPLES OF MATHEMATICAL ....N....LYSIS

(-x)(-y) = -[x(-y)] = -[-(xy)]=xy

by (c) and 1.14(d).

a contradiction. Thus (b) holds.The first equality in (c) comes from

(-x)y + xy = (-x + x)y = Oy = 0,

combined with J.l4(c); the other half of (c) is proved in the same way.

Finally,

(a) Ox = O.(b) If x # 0 and y # 0 then xy # O.(c) (-x)y = -(xy) = x( -y).(d) (-x)( -y) = xy.

Proof Ox + Ox = (0 + O)x = Ox. Hence J.l4(b) implies that Ox = 0, and

(a) holds.Next, assume x # 0, y # O,hut xy = O. Then (a) gives

(i) x + y < x + z if x, y, Z E F and y < z,(ii) xy > 0 if x E F, y E F, x > 0, and y > O.

If x > 0, we call x positive; if x < 0, x is negative.For example, Q is an ordered field. . . .All the familiar rules for working with inequalitIes apply m every ordered

field: Multiplication by positive [negative] quantities prese~ves [~eversesl mequalities, no square is negative, etc. The followmg propOSItIOn lIsts some of

these.

,. I = m(~)Xy=m(~)o~o,

Definition An ordered field is afield F which is also an ordered set, such1.17that

1.15 Proposition The axioms for multiplication imply tire fol/owing statements.

(a) If x # 0 and xy = xz tlren y = z.(b) Ifx # 0 and xy = x then y = 1.(c) If x # 0 and xy = I then y = I/x.(d) If x # 0 tlren 1/(1/x) = x.

The proof is so similar to that of Proposition 1.14 that we omit it.

1.16 Proposition Thefield axioms imply tire fol/owing statementsJor any x, y,

zEF.X 2 3x-y,-,x+y+z,xyz,x,x ,2x,3x, ...y

Proof If x +Y = x + z, the axioms (A) give

y=O+y=(-x+x)+Y= -x+(x+Y)= -x+(x+z)=(-x+x)+z=O+z=z.

This proves (a). Take z = 0 in (a) to obtain (b). Take z = -x in (a) toobtain (c).Since -x + x = 0, (c) (with -x in place of x) gives (d).

x(y + Z) = xy + XZ

(a) Ifx +Y = x + z then y = z.(b) If x +Y = x then y = O.(c) I!x+y=Otheny= -x.(d) -(-x) = x.

(a) One usually writes (in any field)

in place of

x + (-y), x . G), (x + y) + z, (xy)z, xx, xxx, x + x, x + x + x, ....

(b) The field axioms clearly hold in Q, the set of all rational numbers, ifaddition and multiplication have their customary meaning. Thus Q is afield.(c) Although it is not our purpose to study fields (or any other algebraicstructures) in detail, it is worthwhile to prove that some familiar propertiesof Q are consequences of the field axioms; once we do this, we will notneed to do it again for the real numbers and for the complex numbers.

1.14 Proposition The axioms for addition imply the fol/owing statements.

Statement (a) is a cancellation law. Note that (b) asserts the uniquenessof the element whose existence is assumed in (A4), and that (c) does the samefor (AS).

holds for all x, y, Z E F.

(D) The distributive law

1.13 Remarks


1.18 Proposition The following statements are true in everyorderedfield.

(a) If x> 0 then -x < 0, and vice versa.(b) Ifx>Oandy<zthenxy<xz.(c) [fx < 0 and y < z then xy>Jxz'(d) If x # 0 then x' > O. In par icular, I > O.(e) If0 < x <y then 0 < Ify < fx.

Proof(a) If x> 0 then 0 = -x + x > -x + 0, so that -'x < O. If x < 0 then'0= -x + x < -x + 0, so that -x> O. This proves (a).(b) Since z > y, we have z - y > y - y = 0, hence x(z - y) > 0, andtherefore

xz = x(z - y) + xy > 0 + xy = xy.

(c) By (a), (b), and Proposition 1.16(c),

- [x(z - y)] = (-x)(z - y) > 0,

so that x(z - y) < 0, hence xz < xy.(d) If x > 0, part (ii) of Definition 1.17 gives x' > O. If x < 0, then-x> 0, hence (-x)' > O. But x' = (-x)', by Proposition 1.l6(d).Since I = 1',1 >0.(e) Ify > 0 and v :S 0, then yv :S O. But Y . (Ify) = I > O. Hence Ify > O.Likewise, Ifx > O. If we multiply both sides of the inequality x < y bythe positive quantity (Ifx)(lfy), we obtain Ify < Ifx.

THE REAL FIELD

We now state the existence theorelll which is the core of this chapter.

1.19 Theorem There exists all orderedfield R which has the least-upper-boulldproperty.

Moreover, R contaills Q as a subfield.

The second statement means that Q c R and that the operations ofaddition and multiplication in R, when applied to members of Q, coincide withthe usual operations on rational numbers; also, the positive rational numbersare positive elements of R.

The members of R are called real numbers.The proof ofTheorem 1.19 is rather long and a bit tedious and is therefore

presented in an Appendix to Chap. 1. The proof actually constructs R from Q.


The next theorem could be extracted from this construction with verylittle extra effort. However, we prefer to derive it from Theorem 1.19 since thisprovides a good illustration of what one can do with the least-upper-bound

property.

1.20 Theorem

(a) If x E R. Y E R. and x > 0, tlren there is a positive integer n such that

nx > y.

(b) Ifx E R, Y E R, and x < y, then there exists apE Qsuch that x < p < y.

Part (a) is usually referred to as the archimedean property of R. Part (b)may be stated by saying that Q is dense in R: Between any two real numbers

there is a rational one.

Proof(a) Let A be the set of all nx, where II runs through the positive integers.If(a) were false, then y would bean upper bound of A. But then A has aleast upper bound in R. Put ~ = sup A. Since x> 0, ~ - x <~, .a~da _ x is not an upper bound of A. Hence a - x < lUX for some pOSitIveinteger m. But then ~ < (III + l)x E A, which is impossible, since ~ is an

upper bound of A.(b) Since x < y, we have y - x > 0, and (a) furnishes a positive integer

n such thatII(Y - x) > I.

Apply (a) again, to obtain positive integers III, and III, such that III, > IIX:

m2 > -nx. Then-m2 <nx<m l •

Hence there is an integer m (with -1112 ~ m ~ 1111) such that

m - 1 :::; fiX < m.

If we combine these inequalities, we obtain

nx < m :S I + nx < ny.

Since II > 0, it follows that

mx < - <yo

n

This proves (b), with p = mfn.

10 PRINCIPLES OF MATHEMATICAL ANALYSIS THE REAL AND COMPLEX NUMBER SYSTEMS 11

We shall now prove the existence of nth roots of positive reals. Thisproof will show how the difficulty pointed out in the Introduction (irration-ality of J'i) can be handled in R. '

But y - k < y, which contradicts the fact that y is the least upper boundof E.

Hence]l' = x, and the proof is complete.

1.21 Theorem For every real x > 0 and every integer; n > 0 there is oneand only one po.itive real y such Ihal ]I' = x.

Corollary If a and b are positive real numbers and n is a positive integer, then

(ab)l/. = al/"b'/".

-oo<x<+oo

THE EXTENDED REAL NUMBER SYSTEM

Then x = sup E. The decimal expansion of x is

(k ~ 0, 1,2, ...).

1.23 Definition The extended real number system consists of the real field Rand two symbols, +00 and - 00. We preserve the original order in R, anddefine

Conversely, for any infinite decimal (6) the set E of numbers (5) is boundedabove, and (6) is the decimal expansion of sup E.

Since we shall never use decimals, we do not enter into a detaileddiscussion.

for every x E R.

ab = a"p" = (ap)",

since multiplication is commutative. [Axiom (M2) in Definition 1.12.]The uniqueness assertion of Theorem 1.21 shows therefore that

(ab)I/" = ap = al /"b 1/".

Proof Put a = al/", P= b1/". Then

(6)

nl nitno + 10 + ... + 10' ;<;; x.

Let E be the set of these numbers

(5)

1.22 Decimals We conclude this section by pointing out the relation betweenreal numbers and docimals.

Let x > 0 be real. Let no be tbe largest integer such that 110 ;<;; x. (Note thatthe existence of 110 depends on the archilDedean property of R.) Having chosenno, n, .... , nt-I' let nt be the largest integer such that

b" - 0" < (b - a)nb"-1

This number y is written':;; or x l / ft,

Proof That there is at most one such y is clear, since 0 < Yl < y, implies'>":<>'2.

Let E be the set consisting of all positive real numbers I such thatt ft < x.

If I = x/(l + x) then 0;<;; I < I. Hence I" ;<;; I < x. Thus lEE, andE is not empty.

If I > 1 + x then I" :2: I > x, so that I ¢ E. Thus 1 + x is an upperbound of E.

Hence Theorem 1.19 implies the existence of

y = sup E.

To prove that ]I' = x we will show that each of the inequalities ]I' < xand yft > x leads to a contradiction.

The identity b" - a" = (b - a)(b"-1 + b"-'a + ... + 0"-1) yieldsthe inequality

when 0 < a < b.Assume ]I' < x. Choose h so that 0 < h < 1 and

h< x-]I'n(y + Ir1

Put a = y, b = y + h. Then

(y+iJ)"-]I'<hn(y+hr l <hn(y+ Ir1 <x-]I',

Thus (y + h)" < x, and y + h E E. Since y + h > y, this contradicts thefact that y is an IIpper bound of E. '

Assume ]I' > x. Put

]I' -xk=--·n]l'-1

Then 0 < k < y. If 1:2: Y - k, we conclude that

]I' - 1";<;;]1' - (y - k)" < kn]l'-1 =]1' - x.

Thus I" > x, and I ¢ E. It follows that y - k is an upper bound of E.

12 PRINCIPLES OF MATHEMATICAL ANALYSISTHE REAL AND COMPLEX NUMBER SYSTEMS 13

The extended real number system does not form a fieId, but it is customaryto make the following conventions:

(a) If x is real then

It is then clear that + 00 is an upper bound of every subset of the extendedreal number system, and that every nonempty subset has a least upper bound.If, for example, E is a nonempty set of real numbers which is not boundedabove in R, then sup E = + 00 in the extended real number system.

Exactly the same remarks apply to lower bounds.

(b) Ifx>Othenx'(+oo)= +oo,x'(-oo)= -00.(c) Ifx<Othenx'(+oo)= -oo,x'(-oo)= +00.

When it is desired to make the distinction between real numbers on theone hand and the symbols + 00 and - 00 on the other quite explicit, the formerare called finite. Then

(A3) (x + y) + z = (a + c, b + d) + (e,f)= (a + c + e, b + d + f)= (a, b) + (c + e, d +f) = x + (y + z).

(A4) x + 0 = (a, b) + (0, 0) = (a, b) = x.(A5) Put -x = (-a, -b). Then x + (-x) = (0, 0) = O.(MI) is clear.(M2) xy = (ac - bd, ad + be) = (ca - db, da + cb) = yx.(M3) (xy)z = (ac - bd, ad + bc)(e,f)

= (ace - bde - adf - be/, acf - bdf+ ade + bee)= (a, b)(ce - df, cf+ de) = x(yz).

(M4) Ix = (I, O)(a, b) = (a, b) = x.(M5) If x # 0 then (a, b) # (0, 0), which means that at least one of thereal numbers a, b is different from O. Hence a' + b' > 0, by Proposition1.18(d), and we can define

;=C':b" a,~bb')'

x- 00 = -00,x + 00 = +00,

I (ax . - = (a, b) ""--b"x a +

-b )-,--, = (I, 0) = I.a + b ,

1.26 Theorem For any real numbers a and b we have

(D) x(y + z) = (a, b)(c + e, d + f)

= (ac + ae - bd - bf, ad + af + be + be)

= (ac - bd, ad + be) + (ae - bf, af+ be)

= xy + xz.

THE COMPLEX FIELD

1.24 Definition A complex number is an ordered pair (a, b) of real numbers."Ordered" means that (a, b) and (b, a) are regarded as distinct if a # b.

Let x = (a, b), y = (c, d) be two complex numbers. We write x = y if andonly if a = c and b = d. (Note that this definition is not entirely superfluous;think of equality of rational numbers, represented as quotients of integers.) Wedefine

x + Y = (a + c, b + d),

xy = (ac - bd, ad + be).

(a, 0) + (b, 0) = (a + b, 0),

The proof is trivial.

(a, O)(b, 0) = (ab, 0).

1.25 Theorem These definitions of addition and multiplication tum the set ofall complex numbers into afield, with (0, 0) and (1,0) in the role of0 and I.

Proof We simply verify the field axioms, as listed in Definition 1.12.(Of course, we use the field structure of R.)

Let x = (a, b), y = (c, d), z = (e,f).(AI) is clear.(A2) x + y = (a + c, b + d) = (c + a, d + b) = y + x.

Theorem 1.26 shows that the complex numbers of the form (a, 0) have thesame arithmetic properties as the corresponding real numbe,rs a. We can therefore identify (a, 0) with a. This identification gives us the real field as a subfieldof the complex field.

The reader may have noticed that we have defined the complex numberswithout any reference to the mysterious square root of - I. We now show thatthe notation (a, b) is equivalent to the more customary a + bi.

1.27 Definition i = (0, 1).

14 PRINCIPLES OF MATHEMATICAL ANALYSISTHE REAL AND COMPLEX NUMBER SYSTEMS 15

1.28 Theorem i' = -I.

1.31 Theorem If z and ware complex, then

Proof i' = (0,1)(0, I) = (-1,0) = - I.

1.29 Theorem If a and b are real, then (a, b) = a + bi.

Proof

-

Proof (a) and (b) are trivial. Put z = a + bi, w = c + di, with a, b, c, dreal. Then

Izwl' = (ac - bd)' + (ad + be)' = (a' + b')(c' + d') = Izj'lwl'

or Izwl' ~ (Izllw\)'. Now (c) follows from the uniqueness assertion ofTheorem 1.2I. .

To prove (d), note that a' " a' + b', hence

lal = J~'" J~j+bj.

To prove (e), note that zw is the conjugate of ziV, so that ziV + zw =2 Re (ziV). Hence

lz + wi' = (z + w)(z + iV) = zz + ziV + ZW + wiV

~ jz I' + 2 Re (ziV) + IwI'

s; Izl' +2lzwl + Iwl'

= Izl' + 21zll wl + Iwl' = (izi + Iw\)'.

Now (e) follows by taking square roots.

•Xl + X2 + ... + xn = L xj •

j= 1

We conclude this section with an important inequality, usually known asthe Schwarz inequality.

1.34 Notation If Xl' ... , x" are complex numbers, we write

b = Im(z).a = Re(z),

(a) z + w = z+ iV,(b) zw = z· w,(c) z + z = 2 Re(z), z - z = 2i Im(z),(d) zz is real and positive (except when z = 0).

Proof (a), (b), and (c) are quite trivial. To prove (d), write z = a + bi,and note that zz = a' + b'.

a + bi = (a, 0) + (b, 0)(0, I)

= (a, 0) + (0, b) = (a, b).

1.30 Definition if a, b are real and z = a + bi, then the complex numberz = a - bi is called the conjugate of z. Tbe numbers a and b are the real partand the imaginary part of z, respectively.

We shall occasionally write

1.32 Definition If z is a complex number, its absolute value Izl is the nonnegative square root of zz; that is, Izi = (ZZ)I/'.

The existence (and uniqueness) of Izi follows from Theorem 1.21 andpart (d) of Theorem 1.31.

Note that when x is real, then x = x, hence Ixl =p. Thus Ixl = xifx;e,O, Ixl = -xifx<O.

1.33 Theorem Let z and w be complex numbers. Then

1.35 Theorem If a I' ... , an and bl , ... , bn are complex numbers, then

I. I" .j~lajDJ "J~llajl'j~llbjl'.

Proof Put A = LI ajl', B = L Ibjl', C = LajDj (in all sums in this proof,j runs over the values I, ... , n). 1f B = 0, then b. = ... = b. = 0, and theconclusion is trivial. Assume therefore that B > 0. By Theorem 1.31 we

have

(a) Izi > °unless z = 0, 101 = 0,(b) Izi = Izl,(c) Izwl=lzllwl,(d) IReils;lz!.(e) Iz + wi" Izi + Iwl.

L IBaj - Cbjl' = L (Baj - CbJ)(Biij - CbJ)

= B' L lajl' - Be L ajDj - BCL iijbj + ICj' L Ibjl'=B'A-BICI'

= B(AB -ICI').


Since each term in the first sum is nonnegative, we see that

B(AB - IGI') ;;, O.

Since B > 0, it follows that AB - IGI' ;;, O. This is the desired inequality.

EUCLIDEAN SPACES

1.36 Definitions For each positive integer k, let R' be the set of all orderedk-tuples

where Xl' ... , X, are real numbers, called the coordinales of x. The elements ofR' are called points, or vectors, especially when k > I. We shall denote vectorsby boldfaced letters. Ify = (Y" •.. , y,) and ifoc is a real number, put

x + Y = (Xl + Yl' ... , X, + y,),

ax = (ocx1 , .•. , axl )

so that x + y E R' and ocx E R'. This defines addition of vectors, as well asmultiplication of a vector by a real number (a scalar). These two operationssatisfy the commutative, associative, and distributive laws (the proof is trivial,in view of the analogous laws for the real numbers) and make R' into a veclorspace over Ihe realfield. The zero element of R' (sometimes called the origin orthe null veclor) is the point 0, all of whose coordinates are O.

We also define the so-called "inner product" (or scalar product) of x andy by

and the norm of x by

Ixl =(X'X)I/'=(tx;)"'.The structure now defined (the vector space R' with the above inner

product and norm) is called euclidean k-space.

1.37 Theorem Suppose x, y, z E R" and oc is real. Then


Proof (a), (b), and (c) are obvious, and (d) is an immediate consequenceof the Schwarz inequality. By (d) we have

Ix + yj' = (x + y). (x + y)

=x'x+2x'y+y'y

~ lxi' + 21xllyj + jyj'

= <Ixl + Iyll',so that (e) is proved. Finally, (I) follows from (e) if we replace x byx - yand y by y -z.

1.38 Remarks Theorem 1.37 (a), (b), and (f) will allow us (see Chap. 2) toregard R' as a metric space.

R' (the set of all real numbers) is usually called the line, or the real line.Likewise, R' is called the plane, or the complex plane (compare Definitions 1.24and 1.36). In these two cases the norm is just the absolute value of the corresponding real or complex number.

APPENDIX

Theorem 1.19 will be proved in this appendix by constructing R from Q. Weshall divide the construction into several steps.

Step 1 The members of R will be certain subsets of Q, called culs. A cut is,by definition, any set oc c Q with the following three properties.

(I) oc is not empty, and oc ~ Q.(ll) If P E OC, q E Q, and q < p, then q E oc.

(Ill) IfP E oc, then p < r for some r E oc.

The letters p, q, r, ... will always denote rational numbers, and oc, p, y, ...will denote cuts.

Note that (III) simply says that oc has no largest member; (II) implies twofacts which will be used freely:

If p E oc and q ~ oc then p < q.Ifr ~oc and r < s then s~oc.

Step 2 Define "oc < P" to mean: oc is a proper subset of p.Let us check that this meets the requirements of Definition 1.5.If oc < Pand P< Yit is clear that oc < y. (A proper subset ofa proper sub

set is a proper subset.) It is also clear that at most one of the three relations

(a) Ixl;;'O;(b) Ixl =0 if and only ifx =0;(c) locxl = locllxl;(d) IX'yl ~ Ixllyl;(e) jx+yl~lxl+lyl;

(I) Ix-zl ~ Ix-yl + ly-zl· oc < P, oc = P, P<oc


can hold for any pair a, p. To show that at least one holds, assume that thefirst two fail. Then a is not a subset of p. Hence there i~ apE a with p I{; p.If q E p, it follows that q < p (since p I{; P), hence q E a, by (II). Thus pea.Since p # a, we conclude: p < a.

Thus R is now an ordered set.

Step 3 The ordered sel Rhos Ihe leasl-upper-bound properly.To prove this, let A be a nonempty subset of R, and assume that pER

is an upper bound of A. Define y to be the union of all a E A. In other words,p E Yif and only if pEa for some a E A. We shall prove that y E R and that

y = sup A.Since A is not empty, there exists an ao E A. This ao is not empty. Since

ao c y, y is not empty. Next, yep (since a c p for every a E A), and thereforey # Q. Thus y satisfies property (I). To prove (II) and (III), pick p E y. ThenpEa, for some a, E A. If q < p, then q E a" hence q E y; this proves (II). IfrEa, is so chosen that r > p, we see that r E y (since a, c y), and therefore y

satisfies (III).Thus y E R.It is clear that a :5: y for every a E A.Suppose 0 < y. Then there is an s E y and that s I{; 0. Since s E y, sEa

for some a E A. Hence 0 < a, and 0 is not an upper bound of A.This gives the desired result: y = sup A.

Step 4 If a E Rand pER we define a + p to be the set of all sums r + s, where

rEa and s E p.We define O· to be the set of all negative rational numbers. It is clear that

O' is a cut. We verify Ihal Ihe axioms for addition (see Definition 1.12) hold in

R, with O· playing the role ofO.(AI) We have to show that a + p is a cut. It is clear that a + p is anonempty subset of Q. Take r' I{; a, s' I{; p. Then r' + s' > ~ + s for allchoices of rEa, s E p. Thus r' + s' I{; a + p. It follows that a + p hasproperty (I).

PickpEa+p. Thenp=r+s, with ,rEa, SEP. Ifq<p, thenq _ s < r, so q - SEa, and q = (q - s) + sEa + p. Thus (II) holds.Choose I Ea so that t > r. Then p < I + sand t +.s E a + p. Thus (III)

holds.(A2) a + p is the set of all r + s, with rEa, s E p. By the same definition,p + a is the set of all s + r. Since r + s = s + r for all r E Q, SEQ, wehavea+p=p+a.(A3) As above, this follows from the associative law in Q.(A4) Ifr Ea andsE 0', thenr + s < r, hencer + SE a. Thus a + O' ca.To obtain the opposite inclusion, pick pEa, and pick rEa, r > p. Then


p - rEO', and p = r +(p - r) E a + 0'. Thus lc c a + 0'. We concludethat a + O' = a.(AS) Fix a E R. Let Pbe the set of all p with the following property:

There exisls r > 0 such Ihal - p - r I{; a.

In other words, some rational number smaller than - p fails tobe in a.

We show Ihal pER and Ihal a + p = 0'.Ifs I{; a and p = -s - I, then - p - I I{; a, hence pEP. So P is not

empty. If q E a, then -q I{; p. So P# Q. Hence p satisfies (I).Pick PEP, and pick r > 0, so that -p - r I{; a. If q < p, then

-q - r > -p - r, hence -q - r I{; a. Thus q E p, and (II) holds. Putl=p+(r/2). Thenl>p, and -1-(r/2)= -p-rl{;a, so that IEP.Hence psatisfies (III).

We have.proved that pER.IfrEaand SEP, then -sl{;a, hence r<-s, r+s<O. Thus

a + pcO'.To prove the opposite inclusion, pick v EO', put W ~ -v/2. Then

w > 0, and there is an integer n such that nw E a but (n + l)w I{; a. (Notethat this depends on the fact that Q has the archimedean property!) Putp= -(n+2)w. ThenpEp, since -p-wl{;a,and

v = nw + pEa + p.Thus O· c a + p.

We conclude that a + p= 0'.This Pwill of course be denoted by - a.

SteP 5 Having proved that the addition defined in Step 4 satisfies Axioms (A)of Definition 1.12, it follows that Proposition 1.14 is valid in R, and we canprove one of the requirements of Definition 1.17:

If a, p, y E Rand p < y, Ihen a + p < a + y.

Indeed, it is obvious from the definition of + in R that a + pea + y; ifwe had a + p= a + y, the cancellation law (Proposition 1.14) would implyp= y.

It also follows that a > O· if and only if - a < 0'.

Step 6 Multiplication is a little more bothersome than addition in the presentcontext, since products of negative rationals are positive. For this reason weconfine ourselves first to R+, the set of all a E R with a > 0'.

If a E R+ and p E R+, we define ap to be the set of all p such that p ,; rsfor some choice of rEa, s Ep, r > 0, S > O.

We define I' to be the set of all q < I.

20 PRINCIPLES OF MATIlEMAllCAL ANALYSIS THE REAL AND COMPLEX NUMBER SYSTEMS 21

Then the axioms (M) and (D) of Dejinition 1.12 hold, with R+ in place ofF,and with 1* in the role of I.

The proofs are so similar to the ones given in detail in Step 4 that we omitthem.

Note, in particular, that the second requirement of Definition 1.l7 holds:If a > 0* and fJ > 0* then afJ > 0*.

Step 7 We complete the definition of multiplication by settingetO* = O*a = 0*,and by setting

Conversely, suppose p E (r + s)*. Then p < r + s. Choose t so that2t = r + s - p, put

r' = r - t, s' = s - t.

Then r' Er*, s' ES*, and p = r' + s', so thatp Er* + s*.This proves (a). The proof of (b) is similar.If r < s then rEs*, but r ¢ r*; hence r* < so.If r* < so, then there is apE s* such that p ¢ r*. Hence r:S; p < S, so

that r < s.This proves (c).

ap + ay = a(p + y).

ay = a(p + y) + a . (- fJ).

oul a' (- P) = -(afJ). Thus

The orner cases art: hand.led iii ttc ;;~;:::.e V.'2)'.·

We have now completed the proof that R is an ordered jield with the leastupper-bound property.

St~F S We ass0ciate with each r E Q the set r* which COiislsts of all p € Qsuch that p < r. It is clear that each r* is a cut; that is, r* E R. TheSf; CUt5 satistythe folloo;.ing relations:

Step 9 We saw in Step 8 that the repiacement of the rational lluwti\:;;;· ~y ta'lecorresponding "rational cuts" r* E R preserves sums, products, and order. Thisfact may be expressed by saying that the ordered field Q is isomorphic to theordered field Q* wh()se elements are the rational cuts. Of course, r* is by nomeans the same as r, but the properties we are concerned with (arithmetic andorder) are the same in the two fields.

It is this identificatioll of Q with Q* which allows us to regard Q as asubjield of R.

The second part of Theorem 1.19 is to be undersiood in terms of thisidentification. Note that the same phenomenon occurs when the real iiiimber;;are regarded as a subfield of the complex field, and it also occurs at a muchmore eiementary level, when the integers are identified with a certain subset of Q.

It is a fact, which we will not prove here, that any two ordered jields withthe least·upper-bound property are isomorphic. The first part of Theorem 1.19therefore characterizes the reai field R completely.

The books by Landau and Thurston cited in the Bibliography ai'C entirdydevoted to number systems. Chapter 1 of Knopp's book contains <i merele:surely description of ho~' R can be obtained from Q. Another COj1structi.::m,in which each real number is defined to be an equivalence ciass ur Cl:tud:,Ysequences of rational numbers (see Chap. 3), is carried out in Sec. 5 of the bookby Hewitt and Stromberg.

The cuts in Q which we used here were invented by Dedekind. The:::cI".l:tr'..lcticr:. of R from Q hy means of Cauchy sequences is dUf'tu Cdiiwr.Both Cantor and Dedekind published their constructions in j 872.

if a. <0*, f3 <0*,

ifa<O*,fJ>O*,

if a > O*, P< 0*.

[(-aX- f.)

af! = - [( -'a)fJ]

t -[Ct'( -fJn

The Foduots On the right were defined in Step 6.Having proved (in Step 6) that tne axioms (M) hold in R+, it is now

perfectly simple to prove them in R, by repeated application of the identityy = -( - y) which is part of Proposition 1.14. (See Step 5.)

The proof of the distributive law

a(p + y) = ap + ay

breaks into cases. For instance, suppose a > 0*, fJ < 0*, p+ Y > 0*. Theny = (p + y) -;- (- Pi, and (since we already know that the distributive law holdsin R+)

(a) r* + s* = (r + s)',(b) r*s* = (rs)*,(c) r* < s* ifand only if r < s.

To prove (a), choose p Er* + so. Then p = u + v, where u < r, v < s.Hence p < r + s, which says that p E (r + s)*.

EXERCISES

Unless the contrary is explicitly stated, all numbers that are mentioned in these exercises are understood to be real.

1. If r is rational (r =F 0) and x is irrational, prove that r + x and rx are irrational.


2. Prove that there is no rational number whose square is 12.3. Prove Proposition 1.15.

4. Let E be a nonempty subset of an ordered set; suppose 0: is a lower bound of Eand ft is an upper bound of E. Prove that" ,;; ft.

5. Let A be a nonempty set of real numbers which is bounded below. Let -A bethe set of all numbers -x, where x E A. Prove that

inf A = -sup(-A).

6. Fix b > I.(a) If m, n, p, q are integers, n > 0, q > 0, and r = mIn = plq, prove that

(b"')lfll = (b p)l/'!.

Hence it makes sense to define br = (blll)II•.

(b) Prove that b'",,"1 =b'b· if rand s are rational.

(r) Ifx is real; define B(x) to be the set of all numbers bt, where t is rational andt ~ x. Prove t.~at

b' ~ sup B(r)

when r h: ratio!lal. Hence it makes sense to define

b' ~ SUp B(x)

for every real x.(d) Prove that bH

, = bW for all real x and y.7. Fix b > I, Y > 0, and prove that L,.ere is a unique real x such that t'" = y, by

compl~ting the following outline. (This x is called the logarithm oly to the base b.)(a) For any positive iilteger n, Oil - 1 ~ n(b - 1).

(b) Hence b - I ~ n(b'" - I).

(e) If I > I and n > (b - 1)/(1- I), then b'" < I.(d) If w !s such-that b';;' < Y. then b.... +(lm < y for sufficiently large n; to see this,appjy part (e) with t =Y' ,- .....

(c; If t: ... •";;. y, ~j:r. b.:::~I~J > y for sufficiently large t!.

<I) Let A be the set of ail w such that b.... < y, and show that x = sup A satisfiesb"'=y.(q) Prove that this x is unique.

e. PrG'v'c tha~ n('. ::-.;d~.; Cf.n be def.ned in the complex field that turns it !!1to 2.rl erdered

field . ..I-J:r:t: -1 is 3. square.9. SUPP05t" Z = a ....;.. hi, W = c +di. Define z < W if a < c, and also i~ a = c but

h < d. Prove that this turns the set of all complex numbers into an ordered set.(Til:;; type :::f crde~ rel2.tic~ !s called a dictionary order, or lexicographic order, forobvious reasons.) Does this ordered set have the least-upper-bound property?

lO. Suppose z = a + hi, W = U + iv, and


Prove that Z2 = Wif v~ 0 and that (Z)2 = w if v:::;: O. Conclude that every complex

number (with one exception!) has two complex square rootS.11. If z is a complex number, prove that there exists an r~ 0 a!1d a complex number

w with Iwi = 1 such that z = TW. Are wand r always uniquely determined by z?

12. If Z I, ••• , ZII are complex, prove that

Iz,+z,+···+z.1 ';;Izd + Iz,1 + ... + \z.l·

13. If x, y are complex, prove that

Ilxl-lyll';; Ix-yl·14. If z is a complex number such that IzI = I, that is, such that zi= I, compute

11 +z!z-7:" !l-Z!2

15. Under what conditions does equality hold in the Schwarz inequality?

16. Suppose k ~3, X. yE R', Ix- yl =d>O, and r >0. Prove:(a) If 2r > d, there are infinitely many Z E Ri such that

iz-xl =iz-yl =r.

(b) If 2r = d, there is exactly one such z.

(c) If 2r < d. there is no such z.How must these statements be modified if k is 2 or 1?

17. Prove that

Ix + yl' + Ix - yl' = 21xl' + 21yl'

if XE R i and y E Rt • Interpret this geometrically, as a statement about paraHel

ograms..18. If k :2: 2 and x E R i , prove that there exists y E Rt such that Y=:'=-O but :x . y = O.

Is this also true if k = I ?19. Suppose a E R i , bERt. Find CE Rt and r > v'such thal

Ix-al =2jx-bl

It and oniy if jx-ci =(.

(Solulion: 3c =4b -., 3r = 21b - al·)

20. With reference to the Appendix, suppose that property (III) were omitted from theacfmition of a cut. Keep the same definitions of order and addition. Show thatt:-:'e res~Iti!"!g crdered ",et has the least-upocr-bound property, that addition satisfiesaxioms (AI) to (A4) (with a slightly different zero-element I) but that (AS) iaiis.

2BASIC TOPOLOGY

FINITE. COUNTABLE, AND UNCOUNTABLE SETS

We begin this section with a definitIOn of the function concept.

l..l Definiiioli Consider two ;;ets A aild B. whose elements may be any ohjectswhatsoever. and suppose that with each element X of A there is associated, insome manner, an element of B, which we denote by [(x). Then[is said to be a[unction from A to B (or a mapping of A into B). The set A is called the domain-off (we also s~y fj~defined on A), and the elementsf(x) are called the valuesoff Tht Sd of ;,.U Values G.ff is cilled the range off

2.2 Definition Let A and B be two sets and let [be a mapping of A into B.If E -= A,f(E) is defined to be the set of all elements [(x), for X E E. We call[(E) the image of E under f In this notation, [(A) is the range off It is clearthat[(A) c B. 1f[(A) = B. we say that[maps A onto B. (Note that, accordingto this usage, onto is more specific than into.)

IfEc B.r'(E) denotes the set of all XEA such that[(x)EE. We call[-I (E) the inverse image of E under f If y E B,f-l(y) is the set of all X E A

BASIC TOPOLOGY Z5

such that [(x) = y. If, for each y E B,f-l(y) consists of at most one elementof A. then [ is said to be a I-I (one-to-one) mapping of A into B. This mayalso be expressed as follows: [is a I-I mapping of A into B provided that[(Xl) # [(X2) whenever Xl # X2 , Xl E A, x 2 EA.

(The notation Xl # X, means that Xl and X2 are distinct elements; otherwise we write Xl = X2')

2.3 Definition If there exists a I-I mapping of A onto B, we say that A and Bcan be put in 1-1 correspondence. or that A and B l\ave the same cardinal number.or, briefly, that A and B are equivalent, and we write A ...... B. This relationcleariy has the foliowing properti~s:

It is reflexive: A-A.It is symmetric: If A ,"-' B, t.'len B", A.It is transitive: If A - Band B - C, then A - C.

Any relation with these three properties is called an equivalqnce relation.

2.4 Definition For any positive integer n, let J. be the set whose eiements arethe integers 1, 2, ... , n; letJ be the set consisting of all positive integers. For anyset A, we say:

(0) A is finite if A - J. for some n (the empty set is also considered to befinite).

(b) A is infinite if A is not finite.(c) A is countable if A -j(d) A is uncoulltoble if A is neither finite nor countable.(e) A is-a! most count[ible if A is finite or countable.

Countable- sets are sometimes called _enumerable, or denumerable.For two finite sets A and B, we evidently have A - B if and only ifA and

B.contain thesam·e number of elements. For infinite sets,.however, the idea of"having the Su.iilC number of elements" becorr.es quite vague, wh~!.·ea~ the noHonof I-I correspondence retains its clarity.

2.5 Example Let A be the set of all integers. Then A is countable. For,consider the following arrangement of the sets A and J:

A: 0, I, -1,2, -2,3, -3, ...J: 1,2,3.4.5,6,7....

26 PRINCIPLES OF MATHEMATICAL ANALYSIS BASIC lOPOLOGY 2.7

We can, in this example, even give an explicit formula for a function ffrom J to A which sets up a I-I correspondence:

The set whose elements are the sets E, will be denoted by {E,}, Insteadof speaking of sets of sets, we shall sometimes speak of a collection of sets, ora family of sets.

The union of the sets E, is defined to be the set S slIch that XES if and onlyif x E E, for at least one a EA. We use the notation

r~fen) =)

! n - I(--2

(n even),

(n odd). (I) S= UE,.,,,'

If A is the set of all positive integers. the usual notation i'

If A consists of the integers I, 2, ... , n, one usually writes

The symbol 00 in (4) merely indicates that the union of a countable collection of sets is taken, and should not be confused with the symbols + 00, - 00,

introduced in Definition 1.23.The intersection of the sets E. is defined to be the set P such that x E P if

and only if x E E, for every a E A. We use the notation

•S = U Em

m""!

S .::' E1 u E2 U ... u E•.

or

(2)

(3)

(5)

(4)

2.7 Definition By a sequence. we mean a function f defined on the set J of allpositive integers. If f(n) = x., for n E J, it is customary to denote the sequencejby the symbol (x.), or sometimes by Xl> X2, x", ... The values off, that is,the elements x., are called the terms of the sequence. If A is a set and if x. E Afor aL! n ~ J, then {x,) is said to be a sequence in A, or a sequence ofelements ofA.

Note that the terms x" x2 ' x, , ... of a sequence need not be distinct.Since every countable set is the range of a I-I function defined on J, we

;nay regard every countable set as the range of a sequence of distinct terms.Speaking more loosely, we may say that the elements of any countable set canbe Harranged in a sequence."

Sometimes it is convenient to replace J in this definition by the set of alluDi1iicgative integers, i.e., to start with 0 rather than with 1.

2.6 Remark A finite set cannot be equivalent to one, of its proper subsets.That this is, however, possible -for -infinite sets, is shown by Example 2.5, inwhich J is a proper sUDset of A.

In fact, we could replace Definition 2.4(b) by the statement: A is infinite ifA is equivalent to one of its proper subsets.

as for unions. If An B is not empty, we say that A and B intersect; otherwisethey are disjoint.

2.10 Examples

(a) Suppose E, consists of I, 2, 3 and E2 consists of 2, 3, 4. ThenE, u E2 consists of I, 2, 3, 4, whereas E1 n E2 consists of 2, 3.

2.8 Theorem Every infinite subset ofa countable set'A is countable.

Proof Suppose E c A, and E is inhmte. A~range rhe-elements x ofA ina sequence (x.) of distinct elements. Construct a sequence (n,) as follows:

Let n, be the smallest positive integer-such that x., E E. Havingchosean,,:c., n'_1 (k = 2,3,4, ...), let n, be the s'!!allest integer greater·h H C1lf'h that '}' E Ew. "'_1 •• - "" • . _

Puttingf(k) = x., (k = I, 2, 3, ...), we obtain a I-I correspondencebetween E and J.

The theorem shows that, roughly speaking, countable sets representthe "smallest" infinity: No uncountable set can be a subset of a countableset.

2.9 Definition Let A and n be sets, and suppose that with each element a ofA there is associated a subset of n which we denote by E•.

or

(6)

or

f7'\ ..:

•p = nEm = E, (', £, n ... n E",

m=1

~

p= nEm,m"""1

18 PRINCIPLES OF MATHEMATICAL ANALYSISBASIC TOPOLOGy 29

Thus the omission of parentheses in (3) and (6) is justified.The distributive law also holds:

(b) Let A be the set of real numbers x such that 0 < x ~ I. For everyx E A, let Ex be the set of real numbers y such that 0 < Y < x. Then

'3 14

23 x2 •

x" X 34

X.3 x~~

.,S = UE•.

n= 1

in which the elements of E. form the nth row. The array contains allelements of S. As indicated by the arrows, these elements can bearranged in a sequence

If any tWO of the sets E. have elements in common, these wi!! appear morethan once in (17). Hence there is a subset T of the set of all positiveintegers such that S - T, which shows that S is at most countable(Theorem 2.8). Since E, c: S, and Et is infinite, S is infinite, and thuscountable.

(15)

Then S is countable.

Proof Let every set E. be arranged in a sequence {x.,}, k = I, 2, 3, ... ,and consider the infinite array

2.12 Theorem Let {E.}, n = 1,],3, ... , be a sequence ofcountable sets, andput

(16)

A nB=BnA.

(A n B) n C = A n (B n C).

Ex c: E, if and only if 0 < x ~ z ~ I;UEx=E,;

zoAnEx is empty;uA

A n (B v C) = (A n B) v (A n C).

AvB=BvA;

(A u B) v C = A v (B v C);

(i)(ii)

(iii)

(i) and (ii) are clear. To prove (iii), we note that for every y > 0, Y ¢ Exif x < y. Hence Y;n"A Ex·

(8)

(9)

(10)

2.11 Remarks Many properties of unions and intersections are quite sirriilar. to those of sums and products; in fact, the words sum and product were some

times used in this connection, and the symbols ~ and n were written in placeofU and n.

The commutative and associati,e laws are trivial:

To prove this, let the left and right members of (10) be denoted by E and F,respectively.

Suppose x E E. Thcn x E A and x E B u C, that is, x E B or x E C (possibly both). Hence x E A n B or x E A n C, so that x E F. Thus E c: F.

NeAt, :iuppcse x ~ F. Then x E A n B or x EA (~c. That is, x € A, andx E B v C. Hence x E A n (B v C), so that F c: E.

It follows that E = F.We list a few more relations which are easily verified:

(il) A c:A v B,

Corollary Suppose A is at most countable, and, for er;.:r.... .x E A, B:z. is at mostcountable. Put

Then T is at most countable.For Tis equivalent to a subset of(15).

2.13 Theorem Let A be a countable set, and let B. be the set of all n-tuples(a" ... , a.), where a, E A (k = I, ... ,11), and the elements a" ... , a. need not bedistinct. Then B" is countable.

(I~ AnBc:A.

If 0 denotes the empty set, then

(13)

If A c: B, then

(14)

AvO =A,

AvB=B,

A nO=O.

AnB=A.

(18)

Proof That B, is countable is evident, since Bt = A. Suppose B._, iscountable (n = 2, 3, 4, ...). The elements of B. are of the form

(h, a) (b E B.- t , a E A).

For every fixed h, the set of pairs (b, a) is equivalent to A, and hencecountable. Thus B. is the union of a countable set of countable sets. ByTheorem 2.12, B. is countable.

The theorem follows by induction.

30 PRINCIPLES OF MATHEMATICAL ANALYSIS BASIC TOPOLOGY 31

2.16 Examples The most important examples of metric spaces, from ourstandpoint, are the euclidean spaces R', especially R' (the real line) and R2 (thecomplex plane); the distance in R' is defined by

Corollary The set ofall rational numbers is countable.

Proof We apply Theorem 2.13, with n = 2, noting that every rational ris of the form bfa, where a and b are integers. The set of pairs (a, b), andtherefore the set of fractions bfa, is countable. (19) d(x, y) = Ix - yl (x, Y E R').

In fact, even the set of all algebraic numbers is countable (see Exercise 2).

That not all infinite sets are, however, countable, is shown by the nexttheorem.

2.i4 Theorem Let A be the set nf all sequences whose elements are the digits 0and I. This set A is uncountable.

The elements of A are sequences like 1,0,0, 1,0, I, 1, I, ....

Proof Let E be a countable subset of A, and let E consist of the sequences 51' S2' 33 , .... We construct a sequence_os as follows. If the nthdigit in s. is I, we let the nth digit of s be 0, and vice versa. Then thesequence s' differs from every member of E in at least one place; hences ~ E. But clearly sEA, so that E is a proper subset of A.

We have shewn that every'countable subset of A is a proper subsetof A. It follows that A is uncountable (for otherwise A would be a propersubset of A, which is absurd).

The idea of the above proof was first used by Cantor, and is called Cantor'sdirr~enal process' for, if the sequences s" S2' s" ... are placed in an array like(16), it is the ele~ents on the diagonal which are involved in the construction ofthe new sequence.

Readers who are familiar with the binary representatIOn of the realnumbers (base 2 instead of 10) will notice that Theorem 2.14 implies thatthe5d of ,,11 ;-eal numbe!s i~ uncountable. V'e._sJ!al~ give a second proof of thisfact in Theorem 2.43.

METRIC SPACES

2.15 DefiniHon A set X. whose elements we shall cali poims, is said to be a",etric space if with any two points p and g of X there is associated a realnumber d(p, g), called the distance from p to g, such that

(a) d(p,g»Oifp#g;d(p,p) =0;(b) d(p, g) = d(g,p);(c) d(p, g) :$ d(p, r) + d(r, g), for any rEX.

Any function with these three properties is called a distance function, or

a metric.

By Theorem 1.37, the conditions of Definition 2.15 are satisfied by (19).It is important to observe that every subset Y ofa metric space X is a metric

space in its own right, with the same distance function. For it is clear that ifconditions (a) to (c) of Definition 2.15 hold for p, g, rEX, they also hold if werestrict p, g, r to lie in Y.

Tb us every. subset of a euclidean space is -a metric space. Other cxa~ple:;

are the spaces '"6(K) and fe2(p), which are discussed in Chaps. 7 and 11, respectively.

2..17 Definiiion By the -segment (a, b) we mean the set of all- real TItimbcn: :'-i

such that a < x < b. .By the interval [a, bJ we mean the set of all real numbers x such that

a:$ X:$ b.Occasionally we shall abo encounter "halfcopen intervals" [a, b) and (a, b];

the first consists of all x such that a :$ x < b, the second of all x such thata < x:$ b.

If a, < b, for i = I, ... , k, the set of all points x = (x" ... , x,) in R' whosecoordinates satisfy the inequalities iI,:$ x,:$ b, (I :$ i:$ k) is called a k-cell.Thus a I-cell is an interval, a 2-cell is a rectangle, etc.

If x E R' and r > 0, the open (or closed) ball B with center at x and radius ris defined to bethe set of ali y E R' such that rY - xl < r (or Iy - "i:$ r).

We call a set E c: R' convex if

AX + (i - A)Y E;;

whenever x E E, y E E, and 0 < }. <: I.For example, balls are convex. For if Iy - xl < r, Iz - xl < r, and

o< }, < \, we have

!Ay + (J - A)Z - xl = j),(y - x) + (\ - }.)(z - x)1

:$Aly-xl +(I-}.)!z-x! <},r+(\ -}.)r

=r.

The same proof applies to closed balls. It is also easy to see that k-cells areconvex.

32 pRINCIPLES OF MATHEMATICAL ANALYSIS

2.18 Definition Let X be a metric space. All points and sets mentioned beloware understood to be elements and subsets of X.

(a) A neighborhood of p is a set N,(p) consisting of all q such thatd(p, q) < r. for some r > O. The number r is called the radius of N,(p).

(b) A point p is a limit point of the set E if every neighborhood of pcontains a point q 0# p such that q E E.

(c) If pEE and p is not a limit point of E, then p is called an isolatedpoint of E.

(d) E is closed if every limit point of E is a point of E.(e) A point p is :m in!!!rior point of E if there-is a neighborhood N ofp

such that N c: E.(f) E is open if every point of E is an interior point of E.(g) The complement of E (denoted by E') is the set of all points p E X

such that p ¢ E.(il) E is perfect if E is closed and if every point of E is a limit point

of E.(i) E is bounded if there is a real number M and-a point q E X such that

d(p, q) < M for all pEE.Ul E is der,:;. in X if every point of X is a limit point of E, or a point of

E (or both).

Let us note that in R' neighborhoods are segments, whereas in R2 neighborhoods are interiors of circles.

2.19 Theorem Every neighborhood is an open set. '

Proof Consider a neighborhood E =N,(P), and let q be any point of E.Then there is a positive real number h such that

d(p,q) = r - h._

For all points s such that d(q, s) < h, we have then

d(p, s) ;!; d(p, q) + d(q, s) < r - h + h = r,

so that s : E. Tnus q is an inttrior point of E.

2.20 Tneorem If p is a limit point of a set E, then every neighborhood of pcontains infinitely many points ofE.

Proof Suppose there is a neighborhood N of p which contains only afinite number of points of E. Let q" ... , q. be those points of N " E,which are distinct from p, and put

r = min d(p, q,..)lsm$n

BASIC TOPOLOGY 33

[we use this notation to denote the smallest of the numbers d(p, q,), ... ,d(p, qJl. The minimum of a finite set of positive numbers is clearly positive, so that r > O.

The neighborhood N,(p) contains no point q of E such that q 0# p,so that P is not a limit point of E. This contradiction establishes thetheorem.

Corollary A finite point set has no limit points.

2.21 Examllles Let us consider the following subsets of R2:

(a) The set orall complex z such that Izl< I.(b) The set of all complex z such that Izl ;!;l.

(c) A nonempty finite set.(d) The set of all integers.(e) rne set consisting of the numbers lin (n = 1,2,3, ...). Let us notethat this set E has a limit point (namely, z = 0) but that no point of E isa limit point of E; we wish to stress the difference between having a limitpoint and containing one.(f) Tne set of ali complex numbers (that is, R 2 ).

(g) The segment (a, b).

Let us note that (d),(e), (g) can be regarded also as subsets of RI •

Some properties of these sets are tabulated below:

Ckised Open Perfect Bounded(a) No Yes No Yes(bl Yes No Yes Yes(c) Yes No No YesCd) Yes No No No(e) No No No Yes(f) Yes Yes Yes No(g) No No Yes

In (g), we left the second entri blank. The reason is that the segmeut(a, b) is not open if we regard it as a subset of .R,2, but·it is an open subset of 1(1.

2.22 Theorem Let {E.} be a (finite or infinite) collection ofsets E.. Then

(20) (V E.)' = ~ (E~).

Proof Let A and B be the left and right members of (20). If x E A, thenx; U. E., hence x; E. for any «, hence x E E~ for every «, so that x E nE~ .Thus A c: B.


Conversely, if x E B, then x E E~ for every 0:, hence x f E. for any 0:,

hence x f U. E., so that x E ( U. E.)'. Thus B c:; A.It follows that A = B.

2.23 Theorem A set E is open ifand only if its complement is closed.

Proof First, suppose E' is closed. Choose x E E. Then x fE', and x isnot a limit point of E'. Hence there exists a neighborhood N of x suchthat E' ,.... N is empty, that is, N c:; E. Thus x is an interior point of E,

and E isopen... _ .Next, suppose E is open. Let x be a iimitpoi~t of E'.Then every

neighborhood of x contains a point of E', so that x is not an interior pointof E. Since E is open, this means that x E E'. It follows that E' is closed.

Corollary A. set F is closed if and olJly if its complemeizt is open.

2.24 Theorem

(a) For any coUeclioli (G.] ofopen sets, U. G. is opeli.(b) For any col/eclion {F.} of closed sels, n. F. is closed.(c) For anyfmite collection G1! •••• Gn of open sets, n~= I Gi is open.(d) . For any finite col/eclion F" ... , F. of closed sels, Uf;, F, is closed.

Proof Put G = U. G•. If x E G, theil x E G. for some 0:. Since x is aninterior !Joint of Goo x is also an interior point of G, and G is open. This

proves (a).By Theorem 2.22,

(2 '!'j' 11\ ;;-\' -I : (F'"\\Il"~) -- '-: v.h·

and F: is open, by Theorem 2.23. Hence (a) implies that (21) is open so

that (\ F, is closed. .Next, put H':;:; n~=: 1 Gj • For any x E H,·.there exist neighborhoods

N. of x with radii Til such that ~h,;icGi~i=l, ... ,n). Put. ,r = min (r1 , ... , r~),

and let N be the neighborhood of x of radius r. Then N ;:: G, for i = I,... , n. so that N c:; H, and H is open.

By taking complements, (d) follows from (c):

(. )' .,~, F, = [),(F[).

BASIC TOPOLOGY 3S

2.25 Examples In parts (c) and (d) of the preceding theorem, the finiteness of

the collections is essential. For let G. be the segment ( - ~, ~) (n = 1,2, 3, ...).

Then G. is an open subset of R'. Put G = n;:,,=, G•. Then G consists of a singlepoint (namely, x = 0) and is therefore not an open subset of R'.

Thus the intersection of an infinite collection ofopen sets need not be open.Similarly, the union of an infinite collection of closed sets need not be closed.

2.26 Definition If X is a metric space, if E c:; X, and if E' denotes the set ofall limit points of E in X, then the closure of E is the set E = E u E'.

2.27 Theorem IfX is a metric space and E c:; X, Ihen

(a) E is closed,(b) E = E ifand only (f E is closed,(c) E c:; Ffor every closed sei F c:; X such thai E c:; F.

By (a) and (c), E IS the smal/est closed subset of X that contains E.

Proof(a) Ifp E X and Pf E then p is neither a point of E nor a limit point of E.Hence p has a neighborhood which does not intersect E. The complementof E is therefore open. Hellce E is closed.(b) If E = E, (a) implies that E is closed. If E is closed, then E' c:; E[by Definitions 2.18(d) and 2.26], hence E =' E.(c) If F is closed and F 00 E, then F;:, F', hence F 00 E'. Thus F 00 E.

- 2.28 Theorem Let E be a nonempty sci ofreal numbers which is bounded above.Let y = sup E. Then Y E 1':. Hence y E E ifE is closed.

COillpare this v!ith the exampies in Sec. 1.9.

Proof If y E E then y E E. Assume y f E. For every h > 0 there existsthen a point x E E such thai )'~- h .( x< y, for otherwise y - h would bean upper bound of E. Thus y is a-limit point of E. Hence Y E E.

2.29 Remark Suppose E c:; Y c:; X, where X is a metric space. To say that Eis an open subset of X means that to each point pEE there is associated apositive number r such that the conditions dip, q) < r, q E X imply that q E E.But we have already observed (Sec. 2.16) that Y is also a metric space, so thatour definitions may equally well be made within Y. To be quite explicit, let ussay that E is open relative 10 Y if to each pEE there is associated an r > 0 suchthat q E Ewhenever dIP, q) < rand q E Y. Example 2.21(g) showed that a set


may be open relative to Y without being an open subset of X. However, thereis a simple relation between these concepts, which we now state.

2.30 Theorem Suppose Y c X. A subset E of Y is open relative to Y if andonly ifE = Y n G for some open subset G of X.

Proof Suppose E is open relative to Y. To each pEE there is a positivenumber r p such that the conditions (/(p, q) < r p' q E Y imply that q E E.Let Vp be the set of all q E X such that d(p, q) < rp' and define

G= U Vp .poE

Then G is an open subset of X, by Theorems 2.19 and 2.24.Since p E Vp for all pEE, it is clear that E c G n Y.By our choice of Vp , we have Vp (1 Y c: E for every pEE, so that

G nYc E. Thus E = G n Y, and one half of the theorem is proved.Conversely, if G is open in X and E = G n Y, every pEE has a

neighborhood Vp c G. Then Vp nYc E, so that E is open relative to Y.

BASIC TOPOLOGY 37

2.33 Th eorem Suppose KeY c X. Then K is compact relative to X if andonly ifK is compact relative to Y.

By virtue of this theorem we are able, in many situations, to regard compact set,s as metric spaces in their own right, without paying any attention toany embedding space. In particular, although it makes little sense to talk ofopen spaces, or of closed spaces (every metric space X is an open subset of itself,and is a closed subset of itself), it does mak.e sense to talk of compact metricspaces.

Proof Suppose K is compact relative to X, and let {Vol be a collectionof sets, open relative to Y, such that K c Uo Vo ' By theorem 2.30, thereare sets Go, open relative to X, such that Vo = Y n Go, for all~; and sinceK is compact relative to X, we have

(22) K eGo, U ... U Go.

for some choice of finitely many indices ~l' ... , ~o' Since KeY, (22)implies

(23)

COMPACT SETS

2.31 Definition By an open cover of a set E in a metric space X we mean acol1ection {GJ of open subsets of X such that E c Uo Go.

2.32 Definition A subset K of a metric space X is said to be compact if everyopen cover of K contains afinite subcover.

More explicitly, the requirement is that if {Gol is an open cover of K, thenthere are finitely many indices ~l' ... , ~. such that

K c Gal U ... U Gan ·

The notion of compactness is of great importance in analysis, especiallyin connection with continuity (Chap. 4).

It is clear that every finite set is compact. The existence of a large class ofinfinite compact sets in R' will follow from Theorem 2.41.

We observed earlier (in Sec. 2.29) that if E eYe X, then E may be openrelative to Y without being open relative to X. The property of being open thusdepends on the space in which E is embedded. The same is true of the propertyof being closed.

Compactness, however, behaves better, as we shall now see. To formulate the next theorem, let us say, temporarily, that K is compact relative to X ifthe requirements of Definition 2.32 are met.

This proves that K is compact relative to Y.Conversely, suppose K is compact relative to Y, let {Gol be a col

lection of open subsets of X which covers K, and put Vo = Y n Go. Then(23) will hold for some choice of ~I' ... , ~.; and since Vo c Go, (23)implies (22).

This completes the proof.

2.34 Theorem Compact subsets ofmetric spaces are closed.

Proof Let K be a compact subset of a metric space X. We shall provethat the complement of K is an open subset of X.

Suppose p E X, P ¢ K. Ifq E K, let V, and W, be neighborhoods ofpand q, respectively, of radius less than td(p, q) [see Definition 2.18(0)].Since K is compact, there are finitely many points qt •... , q" in K such that

Kc W" V··· U W,. = W.

If V = V" n ." n V,.' then V is a neighborhood of p which does notintersect W. Hence V c K<. so that p is an interior point of K<. Thetheorem follows.

2.35 Theorem Closed subsets ofcompact sets are compact.

Proof Suppose Fe K c X, F is closed (relative to X), and K is compact.Let {Vol be an open cover of F. If P is adjoined to (vo), we obtain an


open cover n of K. Since K is compact, there is a finite subcollection <1>of n which covers K, and hence F. IfF C is a member of <1>, we may removeit from <1> and still retain an open cover of F. We have thus shown that afinite subcollection of {V,} covers F.

Corollary If F is closed and K is compact, then F n K is compact.

Proof Theorems 2.24(b) and 2.34 show that F n K is closed; sinceF n K c K, Theorem 2.35 shows that F n K is compact.

2.36 Theorem /f{K,} is a collection ofcompact subsets ofa metric space X suchthat the intersection of every finite subcollection of {K,} is nonempty, then n K,is nonempty.

Proof Fix a member K, of {K,} and put G, = K;. Assume that no pointof K 1 belongs to every K,. Then the sets G, form an open cover of K,;and since K1 is compact, there are finitely many indices tXt, ... , tXII suchthat K 1 c G" U ... u G,•. But this means that

K1 n Klilj

n ... r. KtJ..

is empty, in contradiction to our hypothesis.

Corollary If {K.} is a sequellce of nonempty compact sets such tl1at K. :0 K.+,(n = I, 2, 3....), then nl" K. is not empty.

2.37 ,Theorem If E is an i/ifinite subset of a compact sel K, thell E has a limitpoint in K.

Proof If no point of K were a limit point of E, then each q E K wouldhave a neighborhood V, which contains at most one point of E (namely,q, if q E E). It is clear that no finite subcollection of {V,} can cover E;and the same is true of K, since E c K. This contradicts the compactness~K .

2.38 Theorem If {I.} is a sequence of intervals in R', such that I. :0 1.+,(n = 1,2,3, ., .), then nl" I. is not empty.

Proof If I. = [a", b.l, let E be the set of all a.. Then E is nonempty andbounded above (by b,). Let x be the sup of E. If m and n are positiveintegers, then

so that x ,,; bm for each m. Since it is obvious that am ,,; x, we see thatx E 1m for m = I, 2, 3, ....

BASIC TOPOLOGY 39

2.39 Theorem Let k be a positive integer. If {I.} is a sequence of k-cells suchthat I. :0 I.+l(n = I, 2, 3, ...), then nl" I. is not empty.

Proof Let I. consist of all points x = (x" ... , x,) such that

(I ,,;j,,; k; n ~ I, 2, 3, ...),

and put I•. j = [a••j • b•. j ]. For each j, the sequence {I.) satisfies thehypotheses of Theorem 2.38. Hence there are real numbers x;(1 ";j,,; k)such that

(1 ,,;j,,; k; n = 1,2,3, ...).

Setting x' = (xi, ... , xi), we see that x· E I. for Il = 1,2,3, .... Thetheorem follows.

2.40 Theorem Every k-cell is compact.

Proof Let I be a k-cell, consisting of all points x = (x, . ... , x,) suchthat OJ ";Xj ,,; bj (1 ,,;j,,; k). Put

~ = (~(bj - a)')"'.

Then Ix - y I ,,;~, if x E I, y E I.Suppose, to get a contradiction, that there exists an open cover {G,}

of I which contains no finite subcover of I. Put cj = (OJ + b;l12. Theintervals [OJ' cjl and [cj , bj ] then determine 2' k·cells Qi whose union is I.At least one of these sets Q" call it I,. cannot be covered by any finitesubeolleetion of {G,} (otherwise I could be so covered). We next subdivideI, and continue the process. We obtain a sequence {I.} with the followingproperties:

(a) 1:0 I, :0 I, :0 I, :0 •.. ;

(b) I. is not covered by any finite subcollection of {G,};(c) if xEI. and y E I., then Ix - yl ,,; Z-.~.

By (a) and Theorem 2.39, there is a point x· which lies in every I •.For some tx, X*E Ga. Since Ga is open, there exists r> 0 such thatIy - x'l < r implies that y E G,. If Il is so large that 2-·~ < r (there issuch an n, for otherwise 2""; 8fr for all positive integers II, which isabsurd since R is arehimedean), then (c) implies that I. c G" which contradicts (b).


The equivalence of (a) and (b) in the next theorem is known as the HeineBorel theorem.


2.41 Theorem If a set E in R' has one of the following three properties, then ithas the other two:

BASIC TOPOLOGY 41

PERFECT SETS

(a) E is closed and bounded.{oJ E is compact.

(e) Every infinite subset ofE has a limit point' in E.

Proof If (a) holds, then E c: I for some k-cell I, and (b) follows fromTheorems 2.40 and 2.35. Theorem 2.37 shows that (b) implies (e). ItremaIns to be shown that (e) implies (a).

If E is not bounded, then E contains points x. with

The set S consisting of these points x" is infinite and clearly has no limitpomt 10 R', hence has none in E. Thus (c) implies that E is bounded.

If E is not closed, then there is a point X o E R' which is a limit pointof E but ~ot a point of E. For n = 1,2,3, ... , there are points x. E Esuchthat IX. - xol < lin. Let S he the set of these points x•. Then Sis,.nfiOlte (otherwise Ix. - XoI would have a constant positive value, formfiOltely many n), S has Xo as a limit point, and S has no other limitpoint in R'. For if y E R', y # xo, then

Ix.1 > n (n = I, 2, 3, ...).

2.43 Theorem Let P be a nonempty perfect set in R'. Then P is uncountable.

Proof Since P has limit points. P must be infinite. Suppose P is countable, and denote the points of P by x" X2, x" .... We shall construct asequence {V.l of neighborhoods, as follows.

Let V, be any neighborhood of x,. If V, consists of all y E R' suchthat Iy - x,l < r, the closure V, of V, is the set of all y E R' such thatly-x,l5,r.

Suppose V. has been constructed, so that V. n P is not empty. Sinceevery point of P is a limit point of P, there is a neighborhood V.+, suchthat (i) v,,+, c: V.. (li) x. ¢ v,,+!, (iii) V.+, n P is not empty. By (iii),V.+, satisfies our induction hypothesis, and the construction can proceed.

Put K. = v" n P. Since r" is closed and bounded, V. is compact.Since x. ¢ K.+" no point of P lies in n;o K•. Since K. c: P, this impliesthat n;o K. is empty. But each K. is nonempty, by (iii), and K.::> K.+"by (i); this contradicts the Corollary to Theorem 2.36.

Corollary Every interval [a, b) (a < b) is uncountable. In particular, the set ofall real numbers is uncountable.

Ix.-YI;;o: IXo-yl-lx.-xol1 I

;;0: IXo-YI-~;;O:2Ixo-YI

for all but finitely many n; this shows that y is not a limit point of S(Theorem 2.20).

Thus S has no limit point in E; hence E must be closed if (e) holds.

. We should remark, at this point, that (b) and (e) are equivalent in anymetnc space (ExerCISe 26) but that (a) does not, in general, imply (b) and (e).Examples are furnished by Exercise 16 and by the space 2'2, which is discussed in Chap. 11.

2.42 Theorem (Weierstrass) Every bounded infinite subset of R' has a limitpoint in R'.

Proof Being bounded, the set E in question is a subset of a k-cell I c: R'.By Theorem :,4(; J is compact. ~l1d S0 £ h:h 2. limit Dcint in i byTheorem 2.37. . .

2.44 The Cantor set The set which we are now going to construct showsthat there exist perfect sets in R' which contain no segment.

Let Eo he the interval [0, 1]. Remove the segment (t, !), and let E, bethe union of the intervals

[0, tJ [!, 1].

Remove the middle thirds of these intervals, and let E2 be the union of theintervals

[O,·n [i, n [~, H [%, I].

Continuing in this way, we obtain a sequence of compact sets E., such that

(a) E,::> E2 ::> E, ::> ••• ;

(b) E. is the union of 2" intervals, each of length r".The set

~

P_lIr:... -; 1 ...... 11

11= 1

is caned the Camor seE. P is cleariy compact. and Theorenl :.36 shows that F15 not empty.

42 PRINCIPLES OF MATHEMATICAL ANALYSIS BASIC TOPOLOGY 43

CONNECTED SETS

No segment of the form

where k and m are positive inregers, has a poini in comttlon with P. Since everysegment (IX, P) contains a segment of the form (24), if

3_. P- IX

<--,6

Since x E A, and y E B" A and B are nonempty. Since A, c (- 00, z) andB, c (z, 00), they are separated. Hence E is not connected.

To prove the converse, suppose E is not connected. Then there arenonempty separated sets A and B such that A u B = E. Pick x E A, y E B,and assume (without loss of generality) that x < y. Define

z = sup (A" [x, yl).

By Theorem 2.28, z E.4; hence z ¢ B. In particular, x ;<; z < y.If z ¢ A, it follows that x < z < y and z ¢ E.If Z E A, then z ¢ E, hence there exists z, such that z < z, < y and

ZI ¢ B. Then x < Zt < Y and z, ¢ E.

EXERCISES

1. Prove that the empty set is a subset of every set.2. A complex number z is said to be algebraic if there are integers 00 , ••. , an, not all

zero, such thatOor' + atz.. - t + ... + a._Iz + an =0.

Prove that the set of all algebraic numbers is countable. Hint: For every positive

integer N there are only finitely many equations with

n+ la,[ + la.[ +".+ [a.1 =N.

#3. Prove that there exist real numbers which are not algebraic.4. Is the set of all irrational real numbers countable?5. Construct a bounded set of real numbers with exactly three limit points.

, 6. Let E' be the set of all limit points of a set E. Prove that E' is closed. Prove thatE and E have the same limit poirits. (Recall that £ = E u E'.) Do E and E' alwayshave the same limit points?

I'. 7. Let At. A z , AJ, ... be subsets of a metric space.(a) If B. ~ ur•• A" prove that B. ~ ur=, .4" for n = I, 2, 3, " ..(b) IfB~U?,A"provethat1i::> U?,.4,.Show, by an example, that this inclusion can be proper.

8. Is every point of every open set E c RZ a limit point of £? Answer the same

question for closed sets in R Z•

9. Let E' denote the set of all interior points of a set E. [See Definition 2.18(e);

E C is called the interior dE.](a) Prove that EC is always open.(b) Prove that E is open if and only if E' = £.(c) If G:::: E and G is open, prove that GeE".Id) Prove that the complement of E'"- is the closure of the complement of E.{e) Do E and E ai;;'aY5 na-v"e the same interiors"fen Do E and E~ always have the same closures?B, = E n (=., oc.)..4, = E n (- oc.. Z),

(24)

P contains no segment.To show that P is perfect, it is enough to show that P contains no isolated

point. Let x EP, and let S be any segment containing x. Let I. be that intervalof E. which contains x. Choose n large enough, so that I. c S. Let x. be anendpoint of I" such that x. '" x.

It follows from the construction of P that x. EP. Hence x is a limit pointof P, and P is perfect.

One of the most interesting properties of the Cantor set is that it providesus with an example of an uncountable set of measure zero (the concept ofmeasure will be discussed in Chap. \I).

2.45 Definition Two subsets A and B of a metric space X are said to beseparated if both A " E and A " B are empty, i.e., if no point of A lies in theclosure of B and no point of B lies in the <:losure of A.

A set E c X is said to be connected if E is not a union of two nonemptyseparated sets.

2.46 Remark Separated sets are of course disjoint, but disjoint sets need notbe separated. For example, the interval [0, 1] and the segment (I, 2) are natseparated, since I is a limit point of (I, 2). However, the segments (0, 1) and(1,2) are separated.

The connected subsets of the line have a particularly simple structure:

2.47 Theorem A subset E of the real line R' is cannected ifand only if it has thefol/owing propert)': If x E E, )' E E, and x < z <)', then ~ E E.

Proof If there exist x E E. y E E. and some Z E (x, y) such that z Ii E. thenE = A_ u B, where

44 PRINCIPLES OF MATHEMATICAL ANALYSISBASIC TOPOLOOY 45

ure separated.

10. Let X be an infinite set. For p E X and q E X, define

p(t) = (l - (la, '7" to

Prove thai ihb i:s i1 illt:ui..:. \Vl-tich subsets of the resulting metric space are open?

Which are closed? Which are compact?11. ForxER' andYER', define

(a) Prove that Ao and Bo are separated subsets of RI.

(b) Prove that there exists t, E (0, I) such that pit,) $ A v B.

(c) Prove that every convex subset of R" is connected.

22. A metric space is called separable if it contains a countable dense subset. Showthat Ri< is separable. Hint: Consider the set of points which have only rational

coordinates.

23. A collection {V.l of open subsets of X is said to be a base for X if the followingis true: For every x E X and every open set G c X such that x E G, we havex E V. c G for some IX. In other words, every open set in X is the union of a

subcollection of {V",}.

Prove that every separable metric space has a cOllntable base. Hint: Takeall neighborhoods with rational radius and center in some countable dense subset

of X.

24. Let X be a metric space in which every infinite subset has a limit point. Prove thatX is separable. Hint: Fix () > 0, and pick Xl E X. Having chosen Xl, ... , XJ E X,

choose Xj+l E X, if possible, so that d(xi, xj+d~ S for i = 1, ... ,j. Show thatthis process must stop after a finite number of steps, and that X can therefore becovered by finitely many neighborhoods of radius S. Take 8 = 1/n (n = 1,2.3, ...'l,and consider the centet:s of the corresponding neighborhoods.

25. Prove that every compact metric space K has a countable base. and thdt K istherefore separable. Hint: For every positive integer n, there are finitely man~

neighborhoods of radius lin whose union covers K.

26. Let X be a metric space in which every infinite subset has a limit point. Provethat X is compact. Hint: By Exercises 23 and 24, X has a countable base. It

follows that every ope~ cover of X has a COUll table subcover {Gn :, 11 = 1, 2, 3, ....

If no finite subcolJection of fGn : covers X, ther: the complement Fo of GJ U ... v Gn

is nonempty for each n. but nFn is empty. If E is a set which contains a pointfrom each Fn , consider a limit point of E. and obtain a contradiction.

27. Define a point p in a metric space X to be a condensaTion poinT of a set E -= X ifevery neighborhood of p contains uncoumablymany points of E.

Suppose E c R\ E is uncountable, and let P be the set of all condensationpoints of E. Prove that P is perfect and that at most countably many points of E

are not in P. In other words. show that pI: n E is at most countable. Hint: Let

~_Vn~- be a countable base of Rk, let i'V be the union of those Vn for which E n '-'"i; at tr.05~ ~ou:l::.b!e. and shO'.." that P = H/I:.

28. Prove that every dosed set in a separable metric space is the u!1ion of a (possiblyempty\ perfect set ;;.nd a set which is at most countable. (Cw(lit'ary: Every countabie ciosed set in R'" has isolated poims.) Hiiil: Use Exercise :'7.

(ifp;<q)

(ifp~q).d(p, q) = (~

d,(x, y) = (x - y)',

d,(x, y) ~ v'lx- yl,d,(x, y) ~ Ix' - y'l,

d4(X, y) = Ix- 2yl,, Ix- yl

d,(x, YJ = I ' 1 1•T x-y

(b) Prove the same for diSjoint open sets.

(c) Fix p E X, S > 0, define A to be the set of all q E X for which d(p. q) < S, defineB simiiariy, with> in place of <. Prove that A and B are separated.(d) Prove that every connecled metric space with at least two points is uncountable. firnt: Use (c).

20. Are closures and interiors of connected sets aiways connected? (Look at subsetsof R~.)

";'1. Let A and B be separated subsets of some RI'. suppose a -5 A. b -5 E. and der.ne

Determine. for each of these, whether it is a metric or not.12. Let K c: R' consist of 0 and the numbers I/n, for n = I, 2, 3, .... Prove that Kis

compact directly from the definition (without using the Heine-Borel theorem).• 13. Construct a compact set of real numbers whose limit points form a countable set.

14. Give an example of an open c~ver of the segment (0, I) which has no finite subcover.

15. Show that Theorem 2.36 and its Corollary become false (in R', for example) if theword "compact" is replaced by "closed" or by "bounded."

16. Regard Q. the set of all rational numbers, as a metric space, with dip, q) = ip - ql.

Let E be the set of all p E Q such that 2 <p' < 3. Show that E is closed andbounded in Q, but that E is not compact. Is E open in Q?

... 17. Let E be the set of all x E [0.1] wb.ose decimal expansion contains only the digits4 and 7. Is E countable? Is E dense in [0, I]? Is E compact? Is E perfect?

• 18. Is there a nonempty perfect set in R 1 which contains no rational number?19. (a) If A and B are disjoint closed sets -in some metric space X, prove that they

6 PRINCIPLES Of MATHEMATICAL ANALYSIS

'0. Imitate the proof of Theorem 2.43 to ohtain the following result;

If Rt = UfF., where each F. is a closed subset of Rt• then at least one F.

has 3. nonempty interior.

Equivalent statement: If G. is a dense open subset of Rl, for n = 1, 2, 3, ... ,then niG. i~ not empty (in fact, it is dense in R l

).

(This is a special case of Baile's theorem; see Exercise 22, Chap. 3, for the generalcase.)

3NUMERICAL SEQUENCES AND SERIES

As the title indicates, this chapter will deal primarily with sequences and seriesof complex numbers. The basic facts about convergence. however, are just aseasily explained in a more general setting. The first three sections will thereforebe concerned with sequences in euclidean spaces, or even in metric spaces.

CONVERGENT SEQUENCES

3.1 Definition A sequence {p,} in a metric space X is said to converge if thereis a po;ntp E ;r with the following property: For every e > 0 there is an integerN such that II 2: jV implies that d(Pn' p) < z. (Here d der..ctes th~ dista!!ce in X.)

In this case we also say that {p,} converges to p, or that p is the limit ofL-"/l} (see Theorem 3.2(b)]. and we \\Tite Pn -po or

lim p, = p.

If{;/I} does not converge, it is said to dh·erpe.

48 PRINCIPLES OF MATHEMATICAL ANALYSISNUMERJCAL SEQUENCES AND SERIES 'tY

(b) Let. > 0 be given. There exist integers N, N' such that

For sequences in R' we can study the relation between convergence, onthe one hand, and the algebraic operations on the other. We first considersequences of complex numbers.

Hence if n ~ max (N, N'), we have

d(p,p') $: d(P,p.) + d(p"p') < •.

Since. was arbitrary, we conclude that d(p, p') = O.(c) Suppose p. -> p. There is an integer N such that n > N

implies d(p.. p) < 1. Put

r = max (I, d(p" p), ... , d(PN' pl}.

Then d(p.. p) $: r for n = I, 2, 3, ....(d) For each positive integer n, there is a point P. E E such that

d(p" p) < lin. Given. > 0, choose N so that N. > 1. If n > N, itfollows that d(P.. p) < •. Hence P. -> p.


It might be well to point out that our definition of "convergent sequence"depends not only on {p.l but also on X; for instance, the sequence "{Ilnl converges in R' (to 0), but fails to converge in the set of all positive real numbers[with d(x,y) = Ix - yll. In cases of possible ambiguity, we can be moreprecise and specify ";convergent in X" rather than "convergent."

We recall that the set of all points p. (n = 1,2, 3, ...) is the range of (p,l.The range of a sequence may be a finite set, or it may be infinite. The sequence{p,l is said to be bounded if its range is bounded.

As examples, consider the following sequences of complex numbers(that is, X = R'):

(a) If s. = lin, then lim,_oo s. = 0; the range is infinite, and the sequenceis bounded.

(b) If s. = n', the sequence {s.l is unbounded, is divergent, and hasinfinite range.

(c) If s. = I + [( - 1l"lnl, the sequence {s,l converges to I, is bounded,and has infinite range.

(d) If s, = i', the sequence (s.l is divergent, is bounded, and has finiterange.

(e) If s, = 1 (n = I, 2, 3, ...), then (s.l converges to I, is bounded, andhas finite range.

\Vc now summarize some important properties of convergent sequencesin metric spaces.

n ~ N implies

n ~ N' implies

•d(P.. p) < 2'

d(p.. p') <~.

3.2 Theorem Let (p.l be a sequence in a metric space X.3.3 Theorem Suppose {s.l, (t.l are complex sequences, and lim._ oo s. = s,limn_ ce tn = 1. Then

Proof

(c) lim s,t, = st;

lim ~ = ~, provided s, # 0 (n = 1, 2, 3, ...), and s # O."-<:e SrI 5

'-00lim CSn = CS~ lim (c + 5/1) = c + Sl for any number c;

(d)

(a) lim (s, + t.) = s + t;

(b)

I "n ~ J.V1 implies SrI - 51 < 2"1

(a) Given. > 0, there e,i~! integers N" N, such that

Proof (a) Suppose p, -> p and let V be a neighborhood of p. Forscme e > 0, the conditions dCq, p) < '. q E X impiy q E V. COrlespond·ing to this " there exists N such that n ~ N implies d(p,. p) <.. Thus11 ~ N implies p" E V.

Conversely. suppose every neighborhood of p comains all butfinitely many of the Pn' Fix B > O. and let V be the set of all q E X suchtr.m d(p. '-J'j <:.. Iiy ;;;;timptic::-~. tl~~re exists N {corresp-andi'P.g to this V)

such that Pr. E V if 11 2: N. Thus d(Pn' p) < t: if n ~ N: hence Pn - p.

(a) {p.l converges to p E X if and only if every neighborhood ofp comainsp, for all but finitely many n.

(b) Ifp E X, p' E X, and if{p,l converges to p and to p', then p' = p.(c) If{p.l converges, then (p,) is bounded.(d) If £ c X and ifp is a limit poim ofE, then there is a sequence (p.l in £

such that p = lim P•.

50 PRINCIPLES OF MATHEMATICAL ANALYSISNUMERICAL SEQUENCES AND SERIES 51

(b) Suppose {x,,}, {y,} are sequences in 1/', {fJ,} is a sequence ofreal numbers,and x" -+ x, y, -+ y, p, -+ p. Then

(a) If x" -+ x, the inequalities

jlXj ,,-IX,1 ~ Ix,-xJ,

which follow immediately from the definition ofthe norm in R', show that(2) holds.

Conversely, if (2) holds, then to each • > 0 there corresponds aninteger N such that n ~ N implies

(1 )

If N = max (N" N,), then n ~ N implies

1(s, + I,) - (s + I)I ~ Is, - s1+ II, - II < •.

This proves (a). The proof of (b) is trivial.

(c) We use the identity

S,I, - sl = (s, - S)(I, - t) + s(l, - I) + I(s, - s).

Given. > 0, there are integers N" N, such that

n~ N, implies Is, - sl < J;,n ~ N, implies It, - 1\ <,fl.

If we take N = max (N" N,), n ~ N implies

I(s, - S)(I, - t)1 <e,

lim (x" + Y,) = x + y,

Proof

lim x" . y, = x . y,,~~

lim p,x, = px.

so that

lim (s, -:- S)(I, ...,. I) = O.•IIX)." - IX). 1 < -r.-

..;k(I ~j~k).

We now apply (a) and (b) to (I), and conclude that

lim (S~I, - Sl) = O.

Hence n ~ N implies

(

, \ '/2IX,-xj =2: jlXj,,-ojl'/ <.,

}=1

(d) Choosing m such that Is, - sl < !\sl if n~ m, we see that

Is,l > telsj (n ~ m).

so that x, -+ x. This proves (a).Part (b) follows from (a) and Theorem 3.3.

Given. > 0, there is an integer N > m such that n ~ N implies

IS,-sl <!Isj'•.Hence. for n ~ N,

1

'1 II Is,-sl 2,- - - = -- <.,..--,"Is, - sl < •.Sn S SnS iSI

3.4 Theorem

(a) Suppose x, E R' (n = 1,2,3, ...) and

SUBSEQUENCES

3.5 Definition Given a sequence {p,l, consider a sequence {n,} of positiveintegers, such that n, < n, < n, < .... Then the sequence {p"l is called asubsequence of {p,}. If {p,,} converges, its limit is called a subsequenlial limilof {p,l.

It is clear that {p,l converges to p if and only if every subsequence of{p,l converges to p. We leave the details of the proof to the reader.

3.6 Theorem

Then Ix,) com'erges 10 x = I." .. ". ,,) tf and only if

Ern '1.J,r: = :J.}I~- :r:

il,sj,sk).(a) If {PIl} is a se(?uence in a compact metric space X. then some sub

sequence of{Pn} cont'erges TO a point of X.(0) Every bounded sequence in Rk contains a converflenr subsequence.


Proof

(a) Let E be the range of {p.l. If E is finite then there is ape E and asequence {nil with n, < n2 < n3 < "', such that

P"l = p"" = ... = p.

The subsequence {P..l so obtained converges evidently to p.If E is infinite, Theorem 2.37 shows that E has a limit point P e X.

Choose n, so that d(p, P.,) < 1. Having chosen n" ... , nj_" we see fromTheorem 2.20 that there is an integer nj > n,_, such that d(p, P..) < I/i.Then {p..l converges to p.

(b) This follows from (a), since Theorem 2.41 implies that every boundedsubset of R' lies in a compact subset of R".

3.7 Theorem The subsequential limits of a sequence {P.l in a metric space Xform a closed subset of X.

Proof Let E' be the set of all subsequential limits of {p.l and let q be alimit point of E'. We have to show that q e E'.

Choose n, so that P., #- q. (If no such"n, exists, then E' has onlyone point, and there is nothing to prove.) Put b = d(q, P.,). Supposent • ... , nj_l are chosen. Since q is a limit point of E*, there is an x E E*with d(x, q) < 2-'b. Since x e E', there is an n, > nj_, such thatd(x, P..) < 2 -i b. Thus

d(q, p~,):s 21-'b

for i = I, 2, 3, .... This says that {P.,} converges to q. Hence q e E'.

CAUCHY SEQUENCES

3.8 Definition A sequence {p.l in a metric space X is said to be a Cauchysequence if fOT every e > 0 there is an integer N such that d(p". Pm} < e if n ~ NaBel rn ~ .1.\'.

In our discussion of Cauchy sequences, as well as in other situationswhich wiii arise later. the foHowing geometric concept will be useful.

3.1) Do?-finition let E he a nonemotv subset of a metric space X, and let S bethe set of all real numbers ot the form dtp. q). with pEE and q E E. The: ~llP

0-: 5. is .:alied the diameter of E.

NUMERICAL SEQUENCES AND SERIES S3

If{p.l is a sequence in X and if EN consists of the points PN, PN+"PN+2' "',It IS clear from the two preceding definitions that {p.l is a Cauchy sequenceifand oniy if

lim diam EN = O..... -0.>

3.10 Theorem

(a) If E is the closure ofa set E in a metric space X, then

diam E = diam E.

(b) If K. is a sequence of compact sets in X such that K.::> K.+ 1

(n = 1,2,3, ...) and iflim diam K. = 0,

then n~K,. consists of exactly one point.

Proof

(a) Since E c E, it is clear that

diam E:s diam E.

Fix', '> O. and choose pEE. q E E. By the definition of E. there arepoints~suchthat dip, p') < ,. d(q, q') < 8. Hence

dip, q):s dip. p') + dip' q') + d(q', q)

< 28 + dip', q') :S 28 + diam E.

It follows thatdiam E:s 2, + diam E,

and since' was arbitrary. (a) is proved.(b) Put K = n~K,. By Theorem'1.36, K is not empty. If K containsmore than one point. then diam K > o. But for each 11. K" =' K. so thatdiam K, ~ diam K. This contradicts the assumption that diam K, ~ O.

3.11 Theorem

(a) Iii any mNric space X, every conrerg!!n! sequence is (J COllrh.l' sequence.(bl U X is a compacT metric space and iffp,,} is a Caucll)· st:quence iii X.

ihen {Pr.; conrerges to some poim of X.(c) In Ric.. aery Cauch.1· sequence conrerges.

NOTe: The difference between the definition of conYergence andthe definition c.~ a. Calleny :;~qllem:~ is t.h~t. t.he limil i~ ·.::·:::-li:::t!y in\"·")j\"{'~

in the former. but not in the bner. Tbus Theorem 3.llil' may enabie us

54 PRINCIPLES OF MATHEMATICAL ANALYSIS NUMERICAL SEQUENCES AND SEIUES 55

to decide whether or not a given sequence converges without knowledgeof the limit to which it may converge.

The fact (contained in Theorem 3.11) that a sequence converges inRic if aDd onJy if it is a Cauchy sequence is usually called the Cauchycriterion for convergence.

Theorem 3.2(c) and example (If) of Definition 3.1 show that convergentsequences are bounded, but that bounded sequences in R" need not converge.However, there is one important case in which convergence is equivalent toboundedness; this happens for monotonic sequences in R'.

Proof

(a) If p, -+ p and if 6 > 0, there is an integer N such that d(p, p.J < £

for all n ;0: N. Hence

d(p, , Pm) 5: d(p, , p) + d(p, p",) < 26

as soon as n ;0: Nand m ;0: N. Thus {p,} is a Cauchy sequence.

3.13 Definition A sequence is,} of real numbers is said to be

(a) monotonicaily increasing if s, 5: s,., (n = 1,2, 3, );(b) monotonically decreasing if s, ;0: S,+1 (n = 1, 2, 3, ).

The class of monotonic sequences consists of the increasing and thedecreasing sequences.

3.14 Theorem Suppose is,} is monotonic. Then is,} converges if and only if itis bounded.

Proof Suppose s, 5: s,., (the proof is analogous in the other case).Let E be the range of is,}. If is,} is bounded, let s be the least upperbound of E. Then

s-e<SII:S s,

Since {s,,} increases,

(n = 1, 2, 3, ...).$" SS

for otherwise S - 6 would be an upper bound of E.n ;0: N therefore implies

For every £ > 0, there is an integer N such that

by Definition 3.9 and Theorem 3.10(a). Being a closed subset of thecompact space X. each EN is compact (Theorem 2.35). Also E.::> E•• "so that EN =:l £1'1+1'

Theorem 3.10(b) shows now that there is a unique P E X which liesin every EN'

Let '> 0 be given. By (3) there is an integer No such thatdiam EN <, if N;o: No. Since pEEN' it follows that d(p, q) < £ for

. every q E EN. hence for every q E EN' In other words, d(p, p,) < 6 ifn ;0: No. This says precisely that p, -+ p.

(b) Let {PRJ be a Cauchy sequence in the compact space X. ForN = I, 2,3, ... , let E. be the set consisting of P., P•• " PN.2' .. , .Then

(3) lim diam EN = 0,!'i_a:, ---

(c) Let {x,} be a Cauchy sequence in R'. Define EN as in (b), with x,in place of P" For some N, diam EN < 1. The range of {x,,} is the unionof EN and the finite set {Xl' ... ' XN_'}' Hence {x,,} is bounded. Sinceevery bounded subset of R' has compact closure in R' (Theorem 2.41),(c) follows from (b).

which shows that is,} converges (to s).The converse follows from Theorem 3.2(c).

UPPER AND LOWER LIMITS

3.12 i>efinition A melfic space in which every Cauchy sequence converges issaid to be complete.

""hus Theorem 3.11 says that all compact metric spaces and all Euclideanspaces arc complete. Theorem 3.11 implies also that everJ' closed subset E of awmpleie metric spac.: .X is CC!7Zp!e!!!. !1:very Cauchy sequence in E is a Cauchy:_:;;;-.;'j;::r::.: i;: X. h('nc:" i~ cO!1\'e:;~~ to 50me J' t: X. and actually pEE since E iscjosed.) An exampie of a metric space which is not complete is the space of aii

3.15 Definition Let is,} be a sequence of real numbers with the fol1owingproperty: For every real M there is an integer N such that n 2: N impliesSll ~ M. W'e then write

s,.- +':IJ.

Similarly, if for every real Af there is an integer jV such that 11 ~.N implies.':,~.5 M. we w!'i!:e

:-:.:ion3.: !1umbe!"s. with d~.\", .... ~ = x - y ,

56 PRINCIPLES OF MATHEMATICAL ANALYSISNUMERICAL SEQUENCES AND SSlES 57

It should be noted that we now use the symbol -+ (introduced in Definition 3.1) for certain types of divergent sequences, as well as for convergentsequences. but that the definitions of convergence and of limit, given in Definition 3.1, are in n0 w~y changed.

3.16 Definition Let is,} he a sequence of real numbers. Let E be the set ofuumbers x (in the extended rea! number system) such that s" -+ x for somesubsequence {s,,}, This set E contains all subsequential limits as defined inDefinition 3.5, plus possibly the numbers +00, -00.

We now recall Definitions 1.8 and 1.23 and put

s* = sup El

$* = inf E.

The numbers s·, s. are called the upper and lower limits of {s:};notation

lim sup s, = I, lim inf S, = -1.

(b) Let s, = (-I')![I + (lin)]. Then

lim sup s, = lim inf S, = s."-ClO ,,_cc

(c) For a real-valued sequence {so}, lim S, = s if and only if,-'"

II"'CC II_ClO

3.18 . Examples

(a) Let {so} be a sequence containing all rationals. Then every realnumber is a suhsequentiallimit, and

lim sup s" = + co, lim inf s" = - co.

We close this section with a theorem which is useful, and whose proof isquite trivial:

we use the

lim inf $n = $*.

,-'"lim sup $" = s*,,-'"

3.17 Theorem Let {so} be a sequence of real numbers. Let E and s' have thesame meaning as in Definition 3.16. Then s· has the follOWing two properties:

(a) s· E E.(D) If x> s·. there is an integer N such that n ~ N implies s, < x.

Moreover, s· is the only number with the properties (a) and (b).

3.19 Theorem If s, :S t, for n ~ N, where N is fixed, then

lim inf S, :S lim inf t.,n"'o- II-ClO

lim sup s, :S lim sup t,.

3.20 Theorem

te) ~l' lx, < 1. then lim)(' = O.

SOME SPECIAL SEQUENCES

(b) If p > 0, then lim,j'p = 1.

n%If P > 0 and !J. is real, ilien lim -(-.-n = O.

,-x 1 T p)(d)

(a) If P > 0, then lim ~ = O.n-cr; n

(c) lim .:;/;; = 1.,-x

We shall now compute the limits of some sequences which occur frequently.The proofs will all be based on the following remark: If 0 :S x, :S s, for n ~ N,where N is some fixed number, and if s, -+0, then x, -+0.

Of course, an analogous result is true for s•.

Proof

(a) Ifs· = + 00, then E is not bounded above; hence {so} is not boundedabove, and there is a subsequence {s••} such that s,• .:. + 00.

If s* is real! then E is bounded above, and at least one subsequentiallimit exists. so that (a) follows from Theorems 3.7 and 2.28.

If s· = - 00, then E contains only one element, namely - 00, andthere is no subsequential limit. Hence, for any real M, S, > M for atmost a finite number of values of n, so that Sn -+ - 00.

1lns establisnes (a) in ail cases.(D) Suppose there is a number x > s· such that s, ~ x for infinitelymany values of n. In that case. there is a number y E E such thaty :;;:::: x > s*. contradicting the definition of s*.

Thlis s* satisfies (a) and (b).Teo .;1;..')".:<..' ~:'_'" l:nic!'.:eness. suppose the!'e are two numbers, p and q.

which satisfy (0:1 and (b). and suppose p < q. Choose x such that p < x < q.Since p satisfies (Dl, we nave s~ < x for Jl ,;;:: X. BUllhen q cannot satisfy tp).

58 PRINCIPLES OF MATHEMATICAL ANALYSIS NUMERJCAL SEQUENCES AND SERIES 59

Proof

(a) Take n > (I/e)'!P. (Note that the archimedean property of the realnumber system is used here.)

(b) If p> 1, put x, =:jp - 1. Then x, > 0, and, by the binomialtheorem,

1 + nx. 5 (1 + x.)" = p,

so that

p-Io<x <--.,- n

Hence x, .... 0. Ifp = 1, (b) is trivial, and if 0 < p < 1, the ~esult is obtainedby taking reciprocals.

(c) Put x, = :j-;' - 1. Then x. :2: 0, and, by the binomial theorem,

n(n - 1)n= (1 + xJ' :2: 2 x;.

G;Y Definition Given a sequence {a,l, we use the notation

•L a, (p ~ q)'.p

to denote the sum ap + ap+1 + '" + a.. With {a,l we associate a sequence{s.l, where

•s. = La,.l-'" 1

For {s.l we also use the symbolic expression

a, + az + a, + ...or, more concisely,

(4)

The symbol (4) we call an infinite series, or just a series. The numberss, are called the partial sums of the series. If {s,l converges to s, we say thatthe series converges, and, write

(n :2: 2).

Hence

05X.5J 2n -1

(d) Let k be an integer such that k > a, k > O. For n > 2k,

, " n(n - 1) ... (n - k + 1), n'p'(1 +p) >G)p = k! p >2'k!'

~

La. =s.n'" 1

The number s is called the sum of the series; but it should be clearly understood that s is the limit of a sequence of sums, and is not obtained simply byaddition.

If {s,l diverges, the series is said to diverge.Sometimes, for convenience of notation, we shall consider series of the

form

in Exercise 15.

In the remainder of this chapter, all sequences and series under consideration'.\~:: be comolex-valued. l,;:.flJ~~S th: C0?:.:r2.';· h explicitly stated. Exte~sicns cfsome of the~theorems which foliow~ to series with terms in RI<.. are mentioned

~

La•.n"'O

:Ean converges if and only (f for every E > 0 there is an integer

(5)

And frequently, when there is no possible ambiguity, or when the distinctionis immaterial, we shall simply write ~a, in place of (4) or (5).

It is clear that every theorem about sequences can be stated in terms ofseries (putting 01 = Sl, and an = Sn - SII-l for n > 1), and vice versa. But it isnevertheless useful to consider both concepts.

The Cauchy criterion (Theorem 3.11) can be restated in the followingform:

@ TheoremS such that

(n > 2k).

Hence

nIX. Zkk!_0<---<-. n'Z. k

(1 +p)" p'

Since a - k < 0, n'-' .... 0, by (a).(e) Take a = °in (d).

SERIES

60 PRINCIPLES OF MATHEMATICAL ANALYSIS NUMERICAL SEQUENCES AND SERIES 61

In particular. by taking m = n. (6) becomes

la,l S;e (n"?N).

In other words:

The comparison test is a very useful one; to use it efficiently, we have tobecome familiar with a number of series of nonnegative terms whose convergence or divergence is known.

o Theorem Ij'La, converges. then lim,~ro a, = O.

The condition a, -+0 is not, however. sufficient to ensure convergenceof ka,. For instance. the series

SERIES OF NONNEGATIVE TERMS

The simplest of all is perhaps the geometric series.

ro II -

n=1 n3.26 Theorem If 0 S; x < I. then

diverges; for the proof we refer to Theorem 3.28.ro IIx"=-·

n=O 1 - x

Theorem 3.14. concerning monotonic sequences. also has an immediatecounterpart for series.

3.24 Theorem A series of nonnegative' terms converges if and only if itspartial sums form a bounded sequence.

If x "? I. the series diverges.

Proof Ifx", I,

n l-x"+1s=Ix'= .

II 1:=0 I-x

We now tum to a convergence test of a different nature, the so-called"comparison test."

The result follows if we let n -+ 00. For x = I, we get

1+1+1+ .. ·•

which evidently diverges.

Proof By Theorem 3.24. it suffices to consider boundedness of thepartial sums. Let

3.27 Theorem Suppose 01 ~ al ~ 03 ~ ..• ~ O. Then the series I~ 1 an converges if and only if the series

5n= a~ ~ G: ~ .. , + en'

:.1; = a1 ...:... 2a2 -!- ... -7- 11.:01 1<.

00

I 2'a,. = a, + 2a, + 4a4 + 8a, + ...1:=0

In many cases which occur in applications. the terms of the series decreasemonotonically. The following theorem of Cauchy is therefore of particularinterest. The striking feature of the theorem is that a rather "thin" subsequenceof {a,l determines the convergence or divergence of ka,.

converges.

(7)by the Cauchy criterion. Hence

l.t,akIS;J, !ak I S;J,ck S; e,

m

I Ck S; e.k=n

Note that (b) applies only to series of nonnegative terms a,.

Proof Given e > O. there exists N "? No such that m "? n "? N implies

and (0) follows.Next, (b) follows from (a), for if ka, converges. so must kd, [note

,;';at (b) :lisa fo110\",'5 from Th;;arem 3.24).

1 The expression ;. nonnegative" a.lways reiers to reai numbers.

3.25 Theorem

@ If Ia,l S; c, for n "? No. where No is some fixed integer, and if kC,converges, then ~on converges.(b) If a, "? d, "? 0 for n "? No, and ifkd, diverges, then ko, diverges.


Forn<2",

s,. ~ Q1 + (Q2 + Q3) + ... + (Q2 k + ... + °2.. + 1-1)

~ 0t + 202 + ... + 210 2k

NUMERICAL SEQUENCES AND SERIES 63

Proof The monotoruclty of the logarithmic function (which will bediscussed in more detail in Chap. 8) implies that {log n} increases. Hence{lin log n} decreases, and we can apply Theorem 3.27 to (10); thisleads us to the series

= Ii:,

so that(11) ~2" I _ ~~_I~

'~1 2'(log 2')' - '~1 (k log 2)'

I ~ I

(log 2)'J. k P '

(8) anll Theorem 3.29 follows from Theorem 3.28.

converges.

We may now observe that the terms of the series (12) differ very littlefrom those of (13). Still, one diverges, the other converges. If we continue theprocess which led us from Theorem 3.28 to Theorem 3.29, and then to (12) and(13), we get pairs of convergent and divergent series whose terms differ evenless than those of (12) and (13). One might thus be led to the conjecture thatthere is a limiting situation of some sort, a "boundary" with all convergentseries on one side, all divergent series on the other side-at least as far as serieswith monotonic coefficients are concerned. This notion of "boundary" is ofcourse quite vague. The point we wish to make is this: No matter how we makethis notion precise, the conjecture is false. Exercises Il(b) and 12(b) may serveas illustrations.

We do not wish to go any deeper into this aspect of convergence theory,and refer the reader to Knopp's "Theory and Application of Infinite Series,"Chap. IX, particularly Sec. 41.

diverges, whereas

On the other hand, if n > 2"S.;;" 01 + 0, + (0, + 0.) + ... + (0"-'+1 + ... + 0,.)

;;" tOI + 0, + 20. + ... + 2'-10,.

= ttl'

so that

(9) 2s. ;;" I,.By (8) and (9), the sequences :{S:l and {t,} are either both bounded

or both unbounded. This completes the proof.

3.28 Theorem I ~ converges if p > I ond diverges ifp :5: I.nP

Proof If p:5: 0, divergence follows from Theorem 3.23. If p > 0,Theorem 3.27 is applicable. and we are led to the series

~ 1 oc.

I2"~ = I 2(I- P)'.

k=O 2kP. 1;;=0

Now, 21 - P I,

(12)

(13)

This procedure may evidently be continued. For instance,

~ I

.~, n log n log log n

~ I

.~, n log n(log log n)'

(10)

cont'erges; ifP :::; 1~ the series direrges.

RE'mark "1m! n" denotes the logarithm of 11 to the base e (compare Exercise 7.ChaD. 1); the-number e wiii be defined in a moment (see Definition 3.30). \Veiet the s~ries sra'n with n = 2. since log 1 = O.

THE NUMBER e

cr· 13.30 Definition e = )' _.

117 01l!

Here 11! = 1 . :; . 3 ... 11 if n ~ 1. and O! = 1.

64 PRINCIPLES OF MATHEMATICAL ANALYSIS NUMElUCAL SEQUENCES AND SERIES 65

the series converges, and the definition makes sense. In fact, the series convergesvery rapidly and aliows us to compute e with great accuracy.

If is of interest to note that e can also be defined by means of anotherlimit process; the prc0f pr0vides a good illustration of operations with limits:

3.31 Theorem lim (1 + ~).= e.Il~OX: n

Hence til ;S; SII I so that

Since

lim sup t. :S e,.~oo

I0<::' e - 811 < -,-'

n.n

10< q!(e - Sq) < -.

q

. By our assumption, q ie is an integer. Since

q's =q'(1 .'-\ ... 2- + 00' + 2-). , . , '2! q!1

3.32 Theorem e is irrational.

Proof Suppose e is rational. Then e = plq, where p and q are positiveintegers. By (16),

so that

(17)

(16)

Thus s,o, for instance, approximates e with an error less than 10-'. Theinequality (16) is of theoretical interest as well, since it enables us to prove theirrationality of e very easily.

The rapidity with which the series L ~ converges can be estimated asn!

follows: If s. has the same meaning as above, we have

I I I- = + + +00.

e s. (n + I) i (n + 2)! (n + 3)!

< I (1+_1_+ 1+ 00 .)=_1_(n + I)! n+ I (n + 1)2 n!n

( -+).t. = I + ~ .

• Is. = L kl '

k=O •

By the binomial theorem,

I I Is =1+1+-+--+'00+;-;;--• 1'2 1'2·3 1'2 00 'n

I I I< I + I + 2+ 22 + 00. + 2.-' < 3,

Proof Let

+~ (t- ~\(I - ~) 00 . (I -~) .n. --nl- n , n

1-(" I) . ( I' ( 2)t. = I + I + - I - - +..: I - -) I - - +. 002! n 3! n n

(14)

by Theorem 3.19. Next, if n ~ m,

.." 1 r---m (' , 1\t. ~ I + I +..: It - -'-) + 00. + "":"'/1 - - -:-:. j:... .... - ).21 \ n m! \ n n

Let n'" 00, keeping m fixed. We get

is an integer, we see that q!(e - Sq} is an integer.Sinceq ~ I, (17) implies the existence of an integer between 0 and 1.

We have 11ms reached a contradiction.

Actually, e is not even an algebraic number. For a simple proof of this,see page 25 of Niven's book, or page 176 of Herstein's, cited in the Bibliography.

. '1

liminft.,;~1 + 1 +2\+"" +~~II~~ • m.

so that THE ROOT AND RATIO TESTS

Letting m ... 00, we finally get

(15) e :Slim inf t•.

The theorem follows from (14) and (15).

. ., .. /1"':"1Theorem (Root Test) GIVen La., put a = 11m sup 'oj 1u. I'.-00

Then

(a) if a < I, 1:a. converges;(b) if a> I, 1:a. diverges;(c) if a = I, the test gives no information.

" PRINCIPLES OF MATHEMA1JCAL ANALYSJS

Proof If a < I, we can choose p so that a < p< I, -and an integer Nsuch that

-1l a.I<Pfor n ~ N [by Theorem 3.l7(b)]. That is, n ~ N implies

la.1 < po.

Since 0 < P< I, '£po converges. Convergence of I:a,. follows now fromthe comparison test.

If a> I, then, again by Theorem 3.17, there is a sequence {nJ suchthat

Hence Ia.1 > I for infinitely many values of n, so that the conditiona• .... 0, necessary for convergence of '£a., does not hold (Theorem 3.23).

To prove (c), we consider the series

I IL~' Ln2'

For each of these series a = I, but the first diverges, the second converges.

~ Theorem (Ratio Test) The series '£a,

(a) converges ifli~~p \a::'1 < I,

(b) diverges ifla::'1 ~ I for all n~ no, where no is some fixed integer,

Proof If condition (a) holds, we can find P< I, and an integer N, suchthat

for n ~ ]v. In particular,

NUMERICAL SEQUENCES AND SEIlIES 67

That is,

la.1 < laNIp-N .p'for n ~ N, and (a) follows from the comparison test, since '£p' converges.

If Ia.+1 j ~ Ia,.1for n ~ "", it is easily seen that the condition a" .... 0does not hold, and (b) follows.

Note: The knowledge that lim a.+,Ia,. = I implies nothing about theconvergence of '£a• . The series '£lln and '£lln2 demonstrate this.

3.35 Examples

(a) Consider the series

1111111I2+ :3 + 22 + 32 + 2' + 3' + 24 + 34 +

for which

. ·ofa.+1 I' (2)' 0hm. -- = .m - =,n~lI) an n~lI) 3

I, 'of·r:: I' 2~1 I1M1 van= lID. -;;= j_tn~lI) n~1X) 3 V 3

lim sup~ = lim 2. {T = ~,n~1X) n--IX) ~r "\/2

. a.+, ' I(3)'hmsup--= hm - - = +00.n--IX) an n--«l 2 2

The root test indicates convergence; the ratio test does not apply.(b) The same is true for the series

11111 I 12+ I + 8+ 4+ 32 + 16 + 128 + 64 + ...

where

I, , f a,TI I1mln --=-,

n-IX) an 8

but•• 1:' •

llm';'iQr:,=--;.


3.36 Remarks The ratio test is frequently easier to apply \!Ian the root test,since it is usually easier to compute ratios than nth roots. However, the roottest has wider scope. More precisely: Whenever lhe ratio test shows convergence, the roct test does too; whenever the root test is inconclusive, the ratiolest is loo. Tills 1> a cOJl..quen~ of Theorem 3.37, and is illustrated by theabove examples.

Neither of the two tests is subtle with regard to divergence. Both deducedivergence from the fact that a, does not tend to zero as n --+ ro.

3.37 Theorem For any sequence {c,} ofpositive numbers,

I · . f CII +1 ne1m m --::;; lim inf.v CII'II-GO Cn " ..... GO

1· ,r:: I' CII +l1m sup y CII ~ 1m sup--'II~OO II"'" co CII

Proof We shall prave the second inequality; the fraof of the first isquite similar. Put

1, CII+lIX = 1m sup-_·

11 ..... 00 CII

If IX = + 00, there is nothing to prove. If IX is finite, choose fJ > IX. Thereis an integer N such that

C,,+1 ::;; Pc,

for n ;:0: N. In particular, for any p > 0,

(k = 0, 1, ... ,p - 1).

Multiplying these inequalities, we obtain

or


by Theorem 3.20(b). Since (18) is true for every fJ > IX, we have

limsup~~ a.'_00

POWER SERIES

3.38 Definition Given a sequence {c,} of complex numbers, the series

(19)

is called a power series. The numbers c, are called the coefficients of the series;z is a complex number.

In general, the series will converge or diverge, depending on the choiceof z. More specifically, with every power series there is associated a circle, thecircle of convergence, such that (19) converges if z is in the interior of the circleand diverges if z is in the exterior (to cover all cases, we have to consider theplane as the interior of a circle of infinite radius, and a point as a circle of radiuszero). The behavior on the circle of convergence is much more varied and cannot be described so simply.

3.39 Theorem Gillen the power series I.cllzH, put

IX = lim sup~I,

(If IX = 0, R = + 00; if IX = + 00, R = 0.) Then Lc, z' converges if [zl < R, anddiverges if Iz1 > R.

Proof Put a, = c,z', and apply the root test:

lim sup~, ~ = != [lim sup ~/ ic, III"'" co II-oc;

(n;:O: N). Note: R is called the radius of convergence of Lc,z".

Hence

'c- < "C fJ- N • fJv ,,_"\.' .1... ,

"",llCI

so L~at

iiIll sup ~: (" 5, {1,r:-,,;

3.40 Examples

(a) The series Ln' z" has R = 0....(b) The series'>:'" has R = .,- x. (In lhis case the ratio lest is easier to

-!1 ~

appiy than the root test.)


(c) The series U" has R = I. If Izi = I, the series diveFges, since {1"}does not tend to 0 as n -+ 00.

(d) The series L:: has R = I. It diverges if z = I. It converges for alln

other z with IzI = L (The last assertion will he proved in Theorem 3.44.)

(e) The series L:' has R = I. It converges for all zwith Izi = I, by

the comparison test, since 11"In21 = I/n2 •

SUMMATION BY PARTS

3.41 Theorem Given two sequences {a,}, ibn}, put

ifn ~ 0; put A_I = O. Then, if 0 ,,;p,,; q, we have

NUMER,ICAL SEQUENCES AND SERIES 71

Proof Choose M such that IA,I ,,; M for all n. Given < > 0, there is aninteger N such that bH ,,; «12M). For N ,,; p ,,; q, we have

It a,b,1 = l'fA'(b, - b'+l) + A,b, - Ap_lbpi"::::p "ep

,,; M I'f(b, -b'+1) + b, + bpi'-p

= 2Mbp ,,; 2MbH ,,; <.

Convergence now follows from the Cauchy criterion. We note that thefirst inequality in the above chain depends of course on the fact that

b" - b,,+l ~ O.

3.43 Theorem Suppose

(~ I~I~I~I~I~I~"';(b) C2m-l ~ 0, C2m"; 0 (m = 1,2,3, ., .);(c) lim,_", c, = O.

Then 1:c, converges.

(20)q q-l

L aRb, = L A,(b, - b'+l) + A,b, - Ap_lbp.lI;::p n;::p

Proof

and the last expression on the right is clearly equal to the right side of(20).

Series for which (b) holds are called "alternating series"; the theorem wasknown to Leibnitz.

Proof Apply Theorem 3.42, with a, = (_1)'+1, b, = Ic,l.

3.44 Theorem Suppose the radius oj convergence oj 1:c,1" is I, and supposeCo ~ Ct ~ C2 ~"', lirn,. ... oo clI = O. Th_cn Icnzll com,'erges at erer)' point on the

circle IzI = I, except possibly at z = I.

Proof Put a, = 1", b, = c,. The hypotheses of Theorem 3.42 are thensatisfied, since

Formula (20), the so-called "partial summation formula," is useful in theinvestigation of series of the form l:a,b.. particularly when {b,} is monotonic.We shall now give applications.

!i;S Theorem Suppose

(a) the partial sums A, oJ1:a,Jorm a bounded sequence;(b) bo ~ bl ~ b2 ~ ••• ;

\c) lim b, = O.

Then Y.ar:br: converges.

I, I II-z'+ll 2IA,I = L zm = < ,moO I-z -II-z[

iflzl =1,z#1.

ABSOLUTE CONVERGENCE

The series Ian is said to converge absolutely if the series ~ ~ an: converges.

§ Theorem {rr.ar: conrerges absolUTely. then ;::ar: converges.

72 PRINCIPLES OF MATHEMATICAL ANALYSIS NUMElUCAL SEQUENCES AND SERIES 73

Proof The assertion follows from the inequality

IJ.a'l $ .t.la,l,plus the Cauchy criterion.

Thus two convergent series may be added term by term, and the resulting series converges to the sum of the two series. The situation becomes morecomplicated when we consider multiplication of two series. To begin with, wehave to define the product. This can be done in several ways; we shall comiderthe so-called "Cauchy product."

3.48 Definition Given:Ea. and rb" we put3.46 Remarks For series of positive terms, absolute convergence is the sameas convergence.

If :Ea. converges, but r Ia.1 diverges, we say that ra. converges non-absolutely. For instance, the series

•Cn = I akbn - k

k=O(n=0,1,2,Oo')

and call rc. the product of the two given series.This definition may be motivated as follows. If we take two power

series ra.=" and rb.=", multiply them term by term, and collect terms containing the same power of z, we get

:>: ce

L a.z·· L b.="=(ao+a,z+a,z' +Oo·)(bo +b,z+b2 z2 + 'Oo)11=0 11=0

= aobo + (oob, + a,bo)z + (oob2 + o,b, + a,bo)z' + ...= Co + clz + c2Z2 + ....

Setting z = I, we arrive at the above definition.

( -10)"L-n

converges nonabsolutely (Theorem 3.43).The comparison test, as well as the root and ratio tests, "is really a test for

absolute convergence, and therefore cannot give any information about nonabsolutely convergent series. Summation by parts can sometimes be used tohandle the latter. In particular, power series converge absolutely in the interiorof the circle of convergence.

We shall see that we may operate with absolutely convergent series verymuch as with finite sums. We may multiply them term by term and we maychange the order in which the additions are carried out, without affecting thesum of the series. But for nonabsolutely convergent series this is no longer true,and more care has to be taken when dealing with them.

3.49 Example If

•An = I ilk'

11.=0

,Bn = ~ bk ,

11.=0

,en = L Ck ,

k=O

ADDITION AND MULTIPLICATIO!\' OF SERIESa Theorem If ro, = A, and rb,.= B, then r(o, + b.) = A + B,rca. = cA,for any fixed c.

Proof Let

and

and A, ~ A, B, ~ B, then it is not at all clear that {C,l will converge to AB,since we do not have en = An Bn. The dependence of {en} on {An} and {Bn} isquite a complicated one (see the proof of Theorem 3.50). We shall now showthat the product of two convergent.~eries may actually diverge.

The series

lim (A, .,. B,) = A .;- B.

The proof of the second 2.5senion is even simDier.

•A, + B, = )" (a, .;- b,).

k~O

Since lim".... T: An = A and limn .... T: En = B. we see that

I ' \-t~------------- --,

'. ~ "'l., ,.::,- ,-,-~"\

f (-I)' =I_~+~_~+'"n";"o,,'n+l ,;2 v'3 ,/4

converges (Theorem 3.43). We form the product of this series with itself andobtain

•B, = )" b,.

k";"0

,A, = L a"

k=O

Then

,.. .t'Kl'''\'U'J...t:.:> vi' MATHE.MATICAL ANALYSIS


we have

Since

so that


Put

'"ac = L: /a,l·n;::::O

lim y, = O.,~'"

since a. -0 as k -+ ro. Since" is arbitrary, (21) follows.

Y, = aop, + alP,_l + ... + a,po·

We wish to show that C, -+AB. Since A,B -+AB, it suffices toshow that

[It is here that we use (a).] Let" > 0 be given. By (c), P, -+ O. Hence wecan choose N such that /P, I 5;" for n ~ N, in which case

IY,I 5; IPoa, + + ffNa,-NI + IPN+la,-N-l + ... + p,aol

5; IPoa, + + PNa,-NI + "ac.

Keeping N fixed, and letting n -+ ro, we get

lim sup IY,I 5; "ac,

Another question which may be asked is whether the series Le.. if convergent, must have the sum AB. Abel showed that the answer.is in the affirmative.

Put

3.51 Theorem If the series La" Lb.. Le, converge to A, B, C, ande, = aob, + ... + a,bo, then C = AB.

Here no assumption is made concerning absolute convergence. We shallgive a simple proof (which depends on the continuity of power series) afterTheorem 8.2.

(21)

(n = 0, 1,2, ...).

'"I c,=AB.n;::::O

'"(a) L: a, converges absolutely,n;::::O

'"(b) L: a, = A,n;::::O

'"(c) I b, = B,n;::::O ,

(d) c, = I a.b,_.k;::::O

In view of the next theorem, due to Mertens, we note Jhat we have hereconsidered the product of two nonabs61utely convergent series.

1e,1 ~ ±_2_ = 2(n + I),.=on+2 n+2

so that the condition c, -+ 0, which is necessary for the convergence of Le" isnot satisfied.

, Ic, = ( -I)' L: J~=;=="~~

'=0 (n k+I)(k+l)

(n - k + I)(k + I) = (~+r-(~-k)' 5; (~ + Ir·

That is, the product of two convergent series converges, and to the rightvalue. if at least one of the two series converges absolutely.

Proof Put

Then

C, = aobo + (aob, + a,bo) + ... -I- (aob, + a,b,_, + ... + a,bo)

= aoBn+ 01 Bn- 1 + ... + anBo

= ao(B + fl,) + al(B + fi'-I) + ... + a,(B + Po)

Then

,c, = I c.,

k;::::OP, = B, -B.

REARRANGEMENTS

3.52 Definition Let {k,L n = I. 2. 3. .. .• be a sequence in which everypositive integer appears once and only once (that is, (k,l is a I-I function fromJ onto J, in the notation of Definition 2.2). Putting

a~ = at" (n = !, 2, 3, .. ,),

we say that !a~ is a rearrangement of ran.


and one of its rearrangements

Then p. _ q. = a. , p. + q. = Ia. I,P. ~ 0, q. ~ 0, The series r.P., r.q.must both diverge.

For if both were convergent, then

r.{p. H.) = r.1 a. Iwould converge, contrary to hypothesis. Since

N N N N

L a. = L (P, - q.) = L p, - L q.."=1 "=1 "=1 11=1

. T'··· -bl' e ,p 'od' 0 rlivpro-pand continue in thIS 'Nay. AnlS IS pOSSI e SInC - n -~ -_" - ·_·t-·

If XII' J'II denote the partial sums of (25) whose last terms are Pmn '

- Q", then

Since p .. ----l-O ~Jid roo ----l-O as n --+ X', we see that X;; ----l- [3, .1'.; --+ ':1...

Finall\'. it is dear that no number less than :t. or greater th:::.n /1 canbe a subseq~entiallimit of the panial sums of (15).

divergence of r.p, and convergence of r.q, (or vice versa) implies diver-gence of r.a" again contrary to hypothesis. .

Now let PI' P" P" .. , denote the nonnegative terms of r.a., 10 theorder in which they occur, and let Q" Q" Q" .'. be the absolute valuesof the negative terms of r.a., also in their original order.

The series r.P.. r.Q, differ from r.P.. r.q, only by zero terms, and

are therefore divergent.We shall construct sequences {"'.}, {k,}, such that the series

(25) PI + .. , +P", - Ql - , .. - Qt, +P",+l + .,.+Pm2 - Qkl+l - ... - Qk2+ "',

which clearly is a rearrangement of La" satisfies (24).Choose real-valued sequences {:1n}, {fin} such that :1n --+:L, fJn --+ P,

~, < P.. PI > 0.Let "'" k1 be the smallest integers such that

P1 +"'+Pm1 >{31'

P, +". + Pm, - Q, -'" - Qt, < ~,;

let "'" k, be the smallest integers such that

PI + +Pm1

- Ql -'" - Qk1+Pltlt + 1 + +Pm1 > {32'

PI + +Pm1 - Ql - ... - Qk 1+Pml+l + +Pm2 - Qkl+l- ..• - Qk2 < 0:2;

If s is the

(n = 1. 1. 3....).

II ..... oc

lim sup s; = p,

I-±+t-t+t-i+'"

lim inf s~ = :x,

I + t - ±+ t ++- i + t + 1\ - i + ...

T ••~~,Proof

so that (23) certainly does not converge to s [we leave it to the reader to verifythat (23) does, however, converge].

This example illustrates the following theorem, due to Riemann,

- 00 :0; a :0; P:0; 00.

lim sup s~ > s; = i,'-00

Then there exists a rearrangement ra~ with partial sums s~ such that

(22)

(24)

If {s.}, is;} are the sequences of partial sums of U .. u;-;-it is easily seenthat, in general, these two sequences consist of entirely different numbers.We are thus led to the problem of determining under what conditions allrearrangements of a convergent series will converge and whether the sums arenecessarily the same.

3.54 Theorem Let Lall be a series of real numbers which converges, but notabsolutely. Suppose

for k ~ I, we see that s; < S6 < s, < ... , where s; is nth partial sum of (23).Hence

in which two positive terms are always followed by one negative.sum of (22), then

Since

s<I-±+t=i-

3.53 Example Consider the convergent series

(23)

78 PRINCIPLES OF MATHEMATICAL ANALYSIS NUMElUCAL sEQUENCES AND SERIES 79

3.55'J) Theorem [JUl. is a series ojcomplex numbers which.E0nverges absolutely,then every rearrangement of!.a" converges, and they all converge to the same sum.

Proof Let La~ be a rearrangement. with partial sums s~. Given e > 0,there "xists an integer N such that m ;<: n ;<: N implies

(a) L n'Z",

(26)m

L: lad,; e.i=n

Now choose p such that the integers 1,2, ... , N are all contained in theset k" k" "', k p (we use the notation of Definition 3.52). Then if n > p,the numbers a1•.•• , aN will cancel in the difference srI - s~, so thatIs. - s;1 ,; e, by (26). Hence Is;} converges to the same sum as {s.}.

8. If :Ea" converges. and if {bIll is monotonic and bounded, prove that :Eallb" con

verges.9. Find the radius of convergence of each of the following power series:

2"(b) l: ,Z",n.

2" n3

(c) Ln,Z", (d) L 3'Z'"

10. Suppose that the coefficients of the power series L.a. z" are integers. infinitely manyof which are distinct from zero. Prove that the radius of convergence is at most 1.

11. Suppose a.. > 0, s" = a l + ... + a., and ~. diverges.

EXERCISES

1. Prove that convergence of {SII} implies convergence of fls"l}. Is the converse true?

2. Calculate lim (Vn' + n - n).

3. If s, = y'Z, and

"" a. d'(a) Prove that £...-- Iverges.1 + all

(b) Prove that

~+ ... +aN+i:~l_~SN+l SN+i: 5N+i:

prove that {sIll converges, and that 5" < 2 for n = 1, 2, 3, ...4. Find the upper and lower limits of the sequence is,} defined by

5,,+1 = V2+ vs.; (n ~ 1, 2, 3, ...), ""a'd'and deduce that L- - Iverges.s,

(e) Prove that

5. For any two real sequences {a.), {b.}, prove that

lim sup (a. + b,) $ lim sup a, + lim sup b.,,~.

provided the sum on the right is not of the form co - co.6. Investigate the behavior (convergence or divergence) of :Ea. if

(a) all =,/n ~ 1 - v';;;

(b) a. ~ V;=J - V~ ;n

(c) all=(-V/;;-l)n;

(d) all = 1 : Zll' for complex values of z.

7. Prove that the convergence of ~all implies the convergence of

if a. :2: 0,

"" a.and deduce that L- 1: converges.s.

(d) What can be said about

12. Suppose an > 0 and 2:a"converges. Put

•r,,= Lalli.

'" "'"

(a) Prove that

a", aliT._-;-"'-,-- >1--r", rll Till

if m < n. and deduce that I:~ diverges.r.


(b) Prove that

a.and deduce that L . j- converges.

v r.

(n = I, 2, 3, ...).

so that, setting fJ = 2-/;;,

XII.I = ~ (X" + ~).(a) Prove that (x.) decreases monotonically and that lim x. = -/;;.

(b) Put '. = x. - ,I;;, and show that

15. Definition 3.21 can be extended to the case in which the all lie in some fixed R".Absolute convergence is defined as convergence of :E IaliI. Show that Theorems3.22, 3.23, 3.25(a), 3.33, 3.34, 3.42, 3.45, 3.47, and 3.55 are true in this moregeneral setting. (Only slight modifications are required in any of the prom-s.)

16. Fix a positive number ct. Choose Xl> V;:;:, and define X2. X3, X4• ••• , by the

recursion formula

(n = 0, I, 2, ...).11""7"1

So -T- SI -7- '" + s"0,,=

(a) If lim SIt = s, prove that lim 0" = s.

(b) Construct a sequence {SII} which does not converge, although lim 0" = O.(e) Canithappeo that s. > ofor all nand that lim sup s. = oc,although lim •• = O?(d) Put a. ~ s. - s._" for n"21. Show that

13. Prove that the Cauchy product of two absolutely convergent series convergesabsolutely.

14. If {sIll is a complex sequence, define its arithmetic means Oil by

1 •5,,-0,,=-,- :Lkai..

lI,lt=1

Assume that lim (na,,) = 0 and that ·:o,,} converges. Prove that {sIll converges.[This gives a converse of (a), but under the additional assumption that na,,""'" 0.](e) Derive the last conclusion from a weaker hypothesis: Assume M < ::c,

inan \.::::;: M for aU n, and lim a" = o. Prove that lim Sn = a, by completing thefollowing outline: .

If m <n, then

(e) This is a good algorithm for computing square roots. since the recursionformula is simple and the convergence is extremely rapid. For example, if oX = 3and Xl = 2, show that cl/fi < -to and that therefore

FS <4 '10- 16• £6 <4.10- 32

•

17. Fix 0: > 1. Take XI> v';, and define,

"x-x" . :r-x"X,,4-1 =-1--= x.- -1--' .

- x" ,.\"

m- 1 I's,,-o.=--(O"-Olll) +-- L (5,,-S;).

JI-m n-m ;=11I+1

For these i,

. (n-ilM (n-m-I)M15,,-Si < < ...", i-I m~~

Fix £ > 0 and associate with each 11 the integer m that satisfies

(a) Prove that Xl > X3 > X5 > .(b) Prove that Xl < X4 < X 6 < .

(c) Prove that lim x" ="\ IX.

(d) Compart: the rapidity of convergence of this process \,-,ith the one described

in Exercise 16.18. Replace the recursion formula of Exercise 16 by

p-I 7X".,.,---X ........ _\",,-1' ... 1- P ." p.

1/-£m<-- <m-':-l.

--.' - f

Tbenim-:-llili-m;:::;;;l cand 'Sr.-5, <AlE. Hence

where p is a fixed positive integer. and describe the behavior uf the re:;ul.ing

sequences ·:x.:.19. Associate to each sequence a = .::1:":. in which ::1:" is 0 or 2. the reai number

;il:'-: s~p 5, - v: ::;'_\1:"r._,--.

p,O\'" tic", tn'• sO' 0;" "'ll XI,7l j;: orecisei\" rhe Camor set ciesc:-i"oec in Se:. :.4-+.~ . ... ..."', '''' ... ' '.... - . .


20. Suppose {PII} is a Cauchy sequence in a metric space X, and some subsequence(P.,} converges to a point p E X Prove that the full sequence {P.} converges to p.

21. Prove the following analogue of Theorem 3.1O(h): If {Eo} is a sequence of closed!.'l.cnempt~' and bcundee. sets in a-complciE metric space X, if E. =- ElI+ll and if

lim diam Eo = 0,

then il fEll consists oi exactly one point.22. Suppose X is a nonempty complete metric space, and {G.} is a sequence of

dense open subsets of X. Prove Haire's theorem, namely, that nfG. is notempty. (In fact, it IS dense in X) Hint: Find'a shrinking sequence of neighborhoods E. such that E. c: G.. and apply Exercise 21.

23. Suppose (P.} and {q.} are Cauchy sequences in a metric space X. Show that thesequence (d(P., q.)} converges. Hint: For any In, n,

d(p.. q.) CS;;d(p.. p.) +d(p.,q.) +d(q., q.);

it follows that

Id(p.. q.) - d(p., q.)1

is small if m and n are large.24. Let X be a metric space.

(a) Call two Cauchy sequences {P.}, {q.} in X equivalent if

lim d(p., q.) = o....Prove that this is an equivalence relation.(b) Let X· be the set of all equivalence classes so obtained. If P EX', Q EX',{p.l E P, {q.} E Q, define

!!.(P, Q) = lim d(p.. q.);

by Exercise 23, this limit exists. Show that the number !!.(P, Q) is unchanged if{p.} and (q.l are replaced by equivalent sequences, and hence that!!. is a distancefunction in X·.(c) Prove that the resulting metric space X· is complete.(d) For each pE X, there is a Cauchy sequence all of whose terms are p; let Pp

be the element of X· which contains this sequence. Prove that

for all p, q E X In other words, the mapping <p defined by <pep) = P, is an isometry(i.e., a distance-preserving mapping) of X into X·.(e) Prove that l:p(X) is dense in X·. and that cptX') = X* if X is complete. By (d),

n'C may identify X and ep(X) c:.lld thus regard X as embedded in the complete

25. Let X be the metric space whose points are the rational numbers, with the metricd(x.. )':1 = x - y' \v'hat is the completion of this space? (Compare Exercise 24.)

4CONTINUITY

The function concept and some of the related terminology were introduced inDefinitions 2.1 and 2.2. Although we shall (in later chapters) be mainly interestedin real and complex functions (i.e., in functions whose ~a!ues.are rea! or complexnumbers) we shall also discuss vector~valued f~nctlOns (t.e., functIOns WIthvalues in Rk) and functions with values In an arbttrary metrIc space. The theorems we shall discuss in this general setting would not become any easIer If werestricted ourselves to real functions, for instance, and it actua!ly simplifies andclarifies the picture to discard unnecessary hypotheses and to state and provetheorems in an appropriately general context.

The domalDS of delmition of our functions wiil also be metric spaces,

suitably specialized in various instances.

LI~lITS OF FUNCTIONS

;:!,1 Definition L~t X and Y be metric 5;::2.~~3; 3~ppose E;:: X~f!i1ap3 E l!!~':

Y. and p is a limit point of E. We write lex) - q as x ~ p, or

iimf(x-j =q;:;-p

... PRINCIPLES OF MATHEMATICAL ANALYSISCONTINUITY 85

if there is a point q E Y with the following property: For-every e > 0 thereexists a a> 0 such that

The symbols dx and d, refer to the distances in X and Y, respectively.. If X and/or.Yare replaced by the real line, the complex plane, or by some

euclidean space R , the distances dx , d, are of course replaced by absolute valuesor by norms of differences (see Sec. 2.16). '

It should be noted that p E X, but that p need not be a point of Ein the above definition. Moreover, even if pEE, we may very well haveI(p) of-lim._,/(x).

We can recast this definition in terms of limits of sequences:

(2)

for all points x E E for which

(3)

d,(f(x), q) < e

0< dx(x,p) < O.

4.3 Definition Suppose we have two complex functions,f and g, both definedon E. By 1+ 9 we mean the function which assigns to each point x of E thenumber I(x) + g(x). Similarly we define the difference 1- g, the product Ig,and the quotient/lg of the two functions, with the understanding that the quotient is defined only at those points x of E at which g(x) of- O. IfI assigns to eachpoint x of E the same number c, then I is said to be a constant function, orsimply a constant, and we write 1= c. If I and 9 are real functions, and ifI(x) ~ g(x) for every x E E, we shall sometimes write I~ g, for brevity.

Similarly, if f and g map E into R" we define f + g and f· g by

(f + g)(x) = f(x) + g(x), (f· g)(x) = f(x) • g(x);

and if }. is a real number, (}.f)(x) = }.f(x).

. 4.4 Theorem Suppose E c X, a metric space, p is a limit point 01 E, I and 9

are complex lunctions on E, and

lim I(x) = A, lim g(x) = B.

Tnis follows from Theorems 3.2(b) and 4.2.

4.2 Theorem Let X, Y, E, f, and p be as in Definition 4.1. Then

Coronary {ff has a limit at p. thi.1i limit is unique.

Proof Suppose (4) holds. Choose {p.l in E satisfying (6). Let 8> 0be given. Then there exists a> 0 such that d,(f(x), q) < e if x E Eand 0 < dx(x, p) < O. Also, there exists N such that n > N implies0< dx(p. ,p) < O. Thus, for n > N, we have dy{f(P.), q) < e, whichshows that (5) holds.

Converselv, suppose (4) is false. Then there exists some 8 > 0 suchLoa! for every 0 > 0 there exists a point x EE (depending on a), for 'whichdy(f(x). q) ~ e but 0 < dx(x, p) < b. Taking O. = lin (n = I. 2, 3, ...), wethus find a sequence in E satisfying (6) for which (5) is false.

Then (a) lim (f + g)(x) = A + B;.~,

(b) lim (fg)(x) = AB;.~,

(c) lim (£)(x) =:i, if B of- O.:x-p 9 B

Proof In view of Theorem 4.2, these assertions follow immediately fromthe analogous properties of sequences (Theorem 3.3).

Remark Iff and g map E into R" then (a) remains true, and (b) becomes

(b') lim (f· g)(x) = A' B.

CONTINUOUS FUNCTIONS

d,(f(x)J(p)) < e

for all poims x E E for which dx(x. p) < b.If (is continuous at every point of E~ thenfis said to be cominuous 011 E.It··should be noted that/has to at: Jefined at thE p0int pin crder 1::) ~e

cominuous at p. (Compare this with the remark follo\ving Definition 4.1:i

4.5 Definition Suppose X and Yare metric spaces, E eX, pEE. and/mapsE into Y. Then f is said to be continuous at p if for every t > 0 there exists a

b > 0 such that

(Compare Theorem 3.4.)

lim P. =p..~oo

lim/(x) =q.~,

lim 1(P.) =q

P. of- p,

if and only if

(4)

(5)

(6)

lor every sequence {p.l in E such that


Ifp is an isolated point of E, then our definition implies tIiat every functionI which has E as its domain of definition is continuous at p. For, no matterwhich e > 0 we choose, we can pick fJ > 0 so that the only point x e E for whichdx(x,p) < fJ is x = p; then

dy(f(x),[(P)) = 0 < e.

4.6 Theorem In the situation given in Definition 4.5, assume also that p is alimit point 01 E. Then I is continuous at p if and only iflimx_pl(x) = 1(P).

Proof This is clear if we compare Definitions 4.1 and 4.5.

We now turn to compositions of functions. A brief statement of thefollowing theorem is that a continuous function of a continuous function iscontinuous.

4.7 Theorem Suppose X, Y, Z are metric spaces, E c X, I maps E into Y, gmaps the range 01f, I(E), into Z, and h is the mapping 01 E into Z defined by

hex) = g(f(x)) (x e E).

III is continuous at a point peE and ifg is continuous at the point I(p), then h iscontinuous at p.

This function h is called the composition or the composite of I and g. Thenotation

h =g 01is frequently used in this context.

Proof Let e > 0 be given. Since g is continuous at I(p), there exists11 > 0 such that

dz(g(y), g(f(p))) 0 such that

dy(f(x)J(P)) < ~ if dx(x, p) < fJ and x e E.

It follows that

CONTlNU1TY 87

Proof Supposelis continuous on X and Vis an open set in Y. We haveto show that every point of I-'(V) is an interior point of I-I(V). SO,suppose p e X and/(P) e V. Since V is open, there exists B > 0 such thatye V if dy(f(P), y) < e; and since I is continuous at p, there exists 11 > 0such that dy(f(x),[(p)) 0, let V be the set of all y e Y such that dy(yJ(P)) < e.Then V is open; hence/-I(V) is open; hence there exists fJ > 0 such thatx e/-'(V)as soon as dxCP, x) < fJ. But if x e I-'(V), then I(x) E V, sothat dy(f(x),[(P)) < B.


Cornllary A mappingI 01a metric space X into a metric space Y is continuous ifand only if1-'(C) is closed in X lor every closed set C in Y.

This follows from the theorem, since a set is closed if and only if its complement is open, and since/-l(E') = [r'(E)]' for every E c Y.

We now turn to complex-valued and vector-valued functions, and tofunctions defined on subsets of R'.

4.9 Theorem Let I and g be complex continuouslunctions on a metric space X.Then I + g,lg, andlfg are continuous on X.

In the last case, we must of course assume that g(x) '" 0, for all x E X.

Proof At isolated points of X there is nothing to prove. At limit points,the statement follows from Theorems 4.4 and 4.6.

4.10 Theorem

(a) Let I" ... , fk be real I unctions on a metric space X, and let f be themapping 01 X into R' defined by

The functl(!!l~Ji' .... h. are caned the c(m-!pon~.'1!:i· ·.)f f. Nut~ Lhatf -"- 0 is a maoDing- into Ri<. \vhereas f· g is a real function on X.e ~~ ~

then f is continuoLls ifand only ifeach oltheILlnctions/" ... ,[, is continuous.(b) If fond gore cominuous mappings 01 X into R', then f + g and f· gare continuous on X.

dz(l;(x), h(p» = dz(g(f(x», g(J(p») < f;

if dx(x. p) < ii and x e E. Thus h is continuous at p.

4.8 Theorem A mapping f of a metric space X imo a meiric space Y is conTinuous 011 X [f and only ijf~ i (V) is open in X for every open set l~ in Y.

(I:1Ye~se images are defined in Definilion 2.2.) This is a very useful charac·i:,:rizalion of continuity.

(7) f(x) = (f,(x), ... ,fk(x)) (x e J;');

90 PRINCIPLES Of MATHEMATICAL ANALYSIS CONTINUITY 91

The conclusion may also be stated as follows: There exist points p and qin X such that f(q) ":;'f(x) ,,:;,f(P) for all x e X; that is, f attains its maximum(atp) and its minimum (atq).

Proof By Theorem 4.15,/(X) is a closed and bounded set of real numbers; hencef(X) contains

4.19 Theorem Let f be a continuous mapping of a compact metric space Xinto a metric space Y. Then f is uniformly continuous on X.

Proof Let e > 0 be given. Since f is continuous, we can associate toeach point p e X a positive number <p(p) such that

M=supf(X) and m =inff(X),(16) q EX, dx(P, q) < <p(p) implies dr(f(p), f(q)) <~.

by Theorem 2.28.

4,17 Theorem Suppose f is a continuous I-I mapping of a compact metricspace X onto a metric space Y. Then the inverse mapping f- 1 defined on Y by

rV(x)) =x' -(x eX)

Let J(p) be the set of all q e X for which

(17) dx(P, q) < t<P(p)·

Since p e J(p), the collection of all sets J(p) is an open cover of X; andsince X is compact, there is a finite set of points PI' ... , P. in X, such that

(18)

(20) dx(p, pml < t<P(Pm),

is a continuous mapping of Yonto X.

Proof Applying Theorem 4.8 to f- 1 in place off, we see that it sufficesto prove that f( V) is an open set in Y for every open set V in X. Fix sucha set V.

The complement V' of V is closed in X, hence compact (Theorem2.35); hence f( V ') is a compact subset of Y (Theorem 4.14) and so isclosed in Y (Theorem 2.34). Since f is one-ta-one and onto, f( V) is thecomplement off(V'). Hencef(V) is open.

(19)We put

b = t min [<P(Pl)' ... , <p(P.)].

Then b > O. (This is one point where the finiteness of the covering, inherent in the definition of compactness, is essential. The minimum of afinite set of positive numbers is positive, whereas the inf of an infinite setof positive numbers may very well be 0.)

Now let q and p be points of X, such that dx(p, q) < b. By (18), thereis an integer m, I,,:;, m":;' n, such thatpeJ(Pm); hence

4.18 Definition Letfbe a mapping of a metric space X into a metric space Y.We say thatfis uniformly continuous on X if for every e > 0 there exists b > 0such that

for all p and q in X for which dx(p, q) < b.Let us consider the differences between the concepts of continuity and of

uniform continuity. First, uniform continuity is a property of a function on aset, whereas continuity can be defined at a single point. To ask whether a givenfunction is uniformly continuous at a certain point is meaningless. Second, iff is continuous on X, then it is possible to find, for each e > 0 and for eachpointp of X. a number b > 0 having the property specified in Definition 4.5. Thisb denends on c and onp. Iffis. however. uniformly continuous on X then it ispossible, for each e > O. to find one number 6 > 0 which will do for all points

;.' '_'i ·~E\'identiY. every uniformly continuous function is continuous. That the

tWO .:oncepts are equivalent on compact sets follows from the next theorem.

(15) drCf(p),f(q)) < e

and we also have

dx(q, Pm) ,,:;, dx(p, q) + dx(p, Pm) < b + t<P(Pm) ,,:;, <P(Pm)'

Finally, (16) shows that therefore

dr(f(P),f(q)),,:;, dr(f(P),f(Pm)) + dy(f(q),f(Pm)) < e.


An alternative proof is sketched in Exercise 10."Vt now proceed to show that compactness is essential in the hypotheses

of Theorems 4.14, 4.15, 4.16, and 4.19.

4,20 Theorem Let E be a Ilollcompact set ill R 1 • Theil

(a) there exists a cominuous funCTion on E which is not bounded:~-b) there exists c; cominuous and bounded funCTion 011 E which has nomaximum.

1< ill addition. E is bounded. rile!!

CONTINUITY 9392 PRINCIPLES OF MATHEMATICAL ANALYSIS

(c) there exists a continuous function on E which is not uniformlycontinuous.

Proof Suppose lirst that E is bounded, so that there exists a limit pointXo of E whicb is not a point of E. Consider

This is continuous on E (Theorem 4.9), but evidently unbounded. To seethat (21) is not uniformly continuous, let E > 0 and J > 0 be arbitrary, andchoose a point x e E such that Ix - Xo I < J. Taking t close enough toxo, we can then make the difference if(t) - f(x) I greater than E, althoughIt - x I < J. Since this is true for every J > 0, f is not uniformly continuous on E.

(21)1

f(x)=--x-xo

(xe E).

4.21 Example Let X be the half-open interval [0,2,,) on the real line, andlet f be the mapping of X onto the circle Y consisting of all points whose distancefrom the origin is I, given by

(24) f(t) = (cos t, sin t) (O:S t < 2,,).

The continuity of the trigonometric functions cosine and sine, as well as theirperiodicity properties, will be established in Chap. 8. These results show thatf is a continuous 1-1 mapping of X onto Y.

However, the inverse mapping (which exists, since f is one-to-one andonto) fails to be continuous at the point (I, 0) = flO). Of course, X is not compact in this example. (It may be of interest to observe that f -1 fails to becontinuous in spite of the fact that Y is compact!)

CONTINUITY AND CONNECTEDNESS

¥le conclude this section by showing that compactness is also essential inTheorem 4.17.

is continuous on E, and is bounded, since 0 < g(x) < 1. It is clear that

sup g(x) = I,HE

whereas g(x) < 1 for all x e E. Thus g has no maximum on E.Having proved the theorem for bounded sets E, let us now suppose

that E is unbounded. Thenf(x) = x establishes (a), whereas

Proof By Theorem·2.47, [a, b] is connected; hence Theorem 4.22 showsthat/era, b]) is a connected subset of [II, and the assertion follows if weappeal once more to Theorem 2.47.

4.24 Remark At first glance. it might seem that Theorem 4.23 has a converse.That is~ one might think that if for any two points Xl < X 2 and for any number cbetweenf(x~) andf(x~) there is a point x in (Xl' x:) such thM ((x) = c. thenfmust be conunuous.

That this is not so may be concluded from Example 4.27(dt

4.22 Theorem Iff is a continuous mapping of a metric space X into a metricspace Y, and if E is a connected subset of X, thenf(E) is connected.

Proof Assume, on the contrary, thatf(E) = A v B, where A and Barenonempty separated subsets of Y. Put G = E n f-'(A), H = E n f-'(B).

Then E = G v H, and neither G nor H is empty.Since A c:.4 (the closure of A), we have G c:f-l(.4); the latter set is

closed, sincefis continuous; hence G c:f-'(.4). It follows thatf(G) c:.4.Since f(H) = Band .4 n B is empty, we conclude that G n H is empty.

The same argument shows that G n H is empty. Thus G and Hareseparated. This is impossible if E is connected.

4.23 Theorem Let f be a continuous real function on the interval [a, b]. Iff(a) <feb) and if c is a number such that f(a) < c <feb), then there exists apoint x e (a, b) such that f(x) = c.

A similar result holds, of course, if f(a) > feb). Roughly speaking, thetheorem says that a continuous real function assumes all intermediate values onan interval.

(xeE)

(xeE)

1

sup hex) = 1HE

1 + (x - XO)2

x 2

h(x)=-1+ x2

g(x)

The function g given by

establishes (b), since

and hex) < 1 for all x e E.Assertion (c) would be false if boundedness were omitted from the

hypotheses. For, let E be the set of all integers. Then every functiondefined on E is uniformly continuous on E. To see this, we need merelytake J < 1 in Definition 4.18.

(23)

(22)

94 PRINCIPLES OF MATHEMATICAL ANALYSIS CONTINUITY 95

Then f is continuous at x = 0 and has a discontinuity of the secondkind at every other point.(c) Define

DISCONTINUITIES

If x is a point in the domain of definition of the function f at which f is notcor..tinuous, we say that f is discontinuous at x, or that f has a disconrinuity at x.Iij is defined on an interval or on a segment, it is customary to divide discontinuities into two types. Before giving this classification, we have to define theright-hand and the left-hand limits offatx, which we denote byf(x+) andf(x-),respectively.

(

x+ 2

f(x)= -x-2

x+2

(-3<x< -2),

(- 2:;; x < 0),

(0 :;; x < I).

4.25 Definition Let f be defined on (a, b). Consider any point x such that

a :S x < b. We write

Then f has a simple discontinuity at x = 0 and is continuous atevery other point of ( - 3, I).(d) Define

Since neither frO +) nor fro -) exists, f has a discontinuity of thesecond kind at x = O. We have not yet shown that sin x is a continuousfunction. If we assume this result for the moment, Theorem 4.7 impliesthat f is continuous at every point x # O.

f(x+) =q

iff(t.) -+ q as n -+ 00, for all sequences {t.} In (x, b) such that t. -+ X. To obtainthe definition of f(x-), for a < x :S b, we restrict ourselves to sequences {t.} In

~4 .. .It is clear that any point x of (a, b), limf(t) eXIsts If and only If,-x

f(x+) =f(x-) = limf(t).,-x

{

. Ism-

f(x) = 0 x(x 0# 0),

(x = 0).

4.26 Definition Letfbe defined on (a, b). Iffis discontinuous at a point x,and iff(x+) andf(x-) exist, thenfis said to have a discontinuity of the firstkind, or a simple discontinuity, at x. Otherwise the discontinuity is said to be of

the second kind.There are two wavs in which a function can have a simple discontinuity:

either f(x+) # f(x-) [in which case the valuef(x) is immaterial], or f(x+) =

J(x-) # f(x).

4.27 Examples(a) Define

MONOTONIC FUNCTIONS

We shall now study those functions which never decrease (or never increase) ona given segment.

4.28 Definition Let J be real on (0. b). Then J is said to be monotonicallyincreasing on (a, b) if a < x < y < b implies f(x) :Sf(y). If the last inequalityis reversed, we obtain the definition of a monotonically decreasing function. Theclass of monotonic functions consists of both the increasing and the decreasingfunctions.

J(x) = (~,.(x rational),(x irrational).

4.29 Theorem Let J be monotonically increasing on (a, b). Then f(x+) andf(x- ) exist at every point of x oj (0. b). More precisely,

Then f has a discontinuity of the second kind at every point x. sinceneither J(x+) nor f(x-) exists.(b) Define

(25) sup J(t) = f(x-) :sJ(x) :sJ(x+) = inf fit).D<t<X x<t<b

Fllrthermore. if a < x < y < b, then

f(x) = ;~"

(x rational!'(x irrational). Analogous results evidently hold for monotoni~aliy decreasing functions.

CONTINUITY 9796 PRINClPLfS OF MATHEMATICAL ANALYSIS

Corollary Monotonic functions have no discontinuities of the second kind.

This corollary implies that every monotonic function is discontinuous ata countable set of points at most. Instead of appealing to the general theoremwhose proof is sketched in Exercise 17, we give here a simple proof which isapplicable to monotonic functions.

Comparison of (29) dnd (30) gives (26).

f(y-) = sup f(t) = sup f(t).o<t<y x<t<y

(a < x < b).f(x) = LC,xn<x

(31)

The summation is to be understood as follows: Sum over those indices nfor which x, < x. If there are no points x, to the left of x, the sum is empty;following the usual convention, we define it to be zero. Since (31) convergesabsolutely, the order in which the terms are arranged is immaterial.

We leave the verification of the following properties off to the reader:

(a) fis monotonically increasing on (a. b);(b) f is discontinuous at every point of E; in fact.

f(x,+) - f(x,-) = C,.

(c) f is continuous at every other point of (a, b).

Moreover. it is not hard to see thatf(x-) = f(.') at all points of (a, b). Ifa function satisfies this condition, we say that f is continuous from the leji. Ifthe summation in (31) were taken over all indices 11 for which x,:S x, we wouldhavef(x-t) = f(x) at every point of (a. b); that is. fwould be continuous fromthe right.

Functions of this sort can also be defined by another method; for anexample we refer to Theorem 6.16.

4.31 Remark It should be noted that the discontinuities of a monotonicfunction need not be isolated. In fact, given any countable subset E of (a, b),which may even be dense, we can construct a function f, monotonic on (a, b),discontinuous at every point of E, and at no other point of (a, b).

To show this, let the points of E be arranged in a sequence {x,l,n = I, 2, 3,.... Let {c,l be a sequence of positive numbers such that LC,converges. Define

Since x, <x, implies f(x,+):sf(x,-), we see that r(x,)#r(x,) ifXI:F X2'

We have thus established a I-I correspondence between the set E anda subset of the set of rational numbers. The latter, as we know, is countable.

(x - fJ < t < x).f(x - fJ) :Sf(t) :S A

Since f is monotonic, we have

Hencef(x-) = A.The second half of (25) is proved in precisely the same way.Next, if a < x < y < b, we see·from (25) that

f(x+) = inf f(t) = inf f(l)·x<t;<b x<r<y

The last equality is obtained by applying (25) to (a, y) in place of (a, b).

Similarly,

Combining (27) and (28), we see that

If(t) - A I < e (x - fJ < t < x).

Proof By hypothesis, the set ofnumbersf(t), where a"< t < x, is boundedabove by the number f(x), and therefore bas a least upper bound whichwe shall denote by A. Evidently A ~f(x). We have to show thatA =f(x-).

Let e > 0 be given. It follows from the definition of A as a leastupper bound that there exists fJ > 0 such that a < x - fJ < x and

A - e <f(x - fJ) :S A.

(30)

(29)

(28)

(27)

4.30 Toeorem Let f be monotonic Oil (a. b). Then the set 0/ points of (a, b) at

which / is discontinuous is at most countable.

Proof Suppose. for the sake of definiteness. that f is increasing. andlet E be the set of points at which / is discontinuous.

\Vith every point x of E we associate a rational number rlx) such

I:-\FINITE LIMITS AND LIMITS AT I:-\FINITY

To enable us to operate in the extended real number system. \"le shall nowenlarge the scope of Definition 4.1. by reformulating it in terms of neighborhoods.

ror any real number x. ;;"'E ha ....-e (i,trcaciy ricTlTi.:,i " n,,:ighbnrh,)().-i of ,.....De any segment (x - d. x ....... b)'

98 PlUNCIPLES Of MATHEMATICAL ANALYSISCON'l'lNU1TY 99

EXERCISES

1. Suppose f is a real function defined on R I which satisfies

4.34 Theorem Let f and g be defined on E c: R. Suppose

a"d let E be a dense subset of X. Prove that [(E) is dense in [(X). If g(P) = [(P)for all pEE, prove that g(P) = [(P) for all p E X. (In other words, a continuousmapping is determined by its values on a dense subset of its domain.)

5. Iff is a real continuou~ function defined on a dosed set E c R 1, prove that there

exist continuous real functions 9 on R' such that g(x) = [(x) for all x E E. (Suchfunctions 9 are called continuous extensions of i from E to RI.) Show that theresult becomes false if the word "closed" is omitted. Extend the result to vectorvalued functions. Hint: Let the graph of g be a straight line on each of the segments which constitute the complement of E (compare Exercise 29, Chap. 2).The result remains true if Rl is replaced by any metric space, but the proof is notso simple.

;1pIf[is defined on E, the graph of[is the set of points (x,[(x)), for x E E. In partie,~ ular, if E is a set of real numbers, and [is real-valued, the graph of[is a subset of

the plane.Suppose E is compact, arid prove that! is continuous on E if and only if

its graph is compact. .tr.)If E c X and if[is a function defined on X, the restriction of[to E is the function'~ g whose domain of definition is E, such that g(p) ~ [(P) for pEE. Define [ and 9

on R' by: [(0, 0) ~ g(O, 0) ~ 0, [(x, y) = xy'!(x' ..,.. y'), g(x, y) = xy'!(x' + y')if (x, y) *(0, 0). Prove that [is bounded on R', that g is unbounded in everyneighborhood of (0, 0), and that [is not continuous at (0, 0); nevertheless, therestrictions of both!and g to every straight line in R 2 are continuous!

8. Let! be a real uniformly continuous function on the bounded set E in R I. Prove

that [is bounded on E.

Show that the conclusion is false if boundedness of E is omitted from thehypothesis.

9. Show that the requirement in the definition of uniform continuity can be rephrasedas follows, in terms of diameters of sets: To every E >0 there exists a lJ >0 such'that diam[(E) < E for all E c: X with diam E < 8.

10: Complete the details of the following alternative proof of Theorem 4.19: If[is not-......-./ uniformly continuous, then for some e >0 there are sequences {p,,}, {q,,} in X such

that dx(p., q.) ~ 0 but d,(f(P.),[(q.)) > E. Use Theorem 2.37 to obtain a Contradiction.

11. Suppose! is a uniformly continuous mapping of a metric space X into a metricspace Yand prove that {!(x,,)} is a Cauchy sequence in Y for every Cauchy se·quence {x,,} in X. Use this result to give an alternative proof of the theorem statedin Exercise 13.

t2:A uniformly continuous function of a uniformly continuous function is uniformlycontinuous.

,~ State this more preciseiy and prove it.®Le~ r be ::.. dense subs.e! of..;;. !netfIe ~p3.ce X. and let /' be :i l..:.nif~rm!·.' ;:~~~:.fc.~c~s

reai function defined. on E. Prove that f has a continuous extension from E to X

as t-+x.g(t)-+Bf(t)-+A,Then

(a) f(t) -+ A' implies A' = A.(b) (/+ g)(t) -+A + B,(c) (/g)(I) -+ AB,(d) (/Ig)(t) -+ AlB,

provided the right members of (b), (c), and (d) are defined.Note that 00 - 00,0' 00, 00/00, AIO are not defined (see Definition 1.23).

4.33 Definition Letfbe a real function defined on E c: R. We say that

f(t)-+A as t-+x,

where A and x are in the extended real number system, if for every neighborhoodU of A there is a neighborhood V of x such that V () E is not empty, and suchthatf(t) E U for all t E V () E, t >F x.

A moment's consideration will show that this coincides with Definition4.1 when A and x are real.

The analogue of Theorem 4.4 is still true, and the proof offers nothingnew. We state it, for the sake of completeness.

for every set E c X. (E denotes the closure of E.) Show, by an example, that

f(£) can be a proper subset of f(E).

3. l.etfbe a continuous real function on a metric sDace X. Let Z(n (the zero set off)be the set of all p e X at WhlChf(pl = o. Prove that Zef) is dosed.

4. Let f and p De continuous mappings of a metric space X into a metric soace

lim [f(x -'- h) - [(x - h)] = 0••0

for every x e R1• Does this imply that f is continuous '1

2. Iflis a continuous mapping of a metric space X into a metric space Y, prove that

4.32 DefinItion For any real e, the set of real numbers .r- such that x > e iscalled a neighborhood of + 00 and is written (e, + 00). Sinnlarly, the set (- 00, c)is a neighborhood of - 00.

100 PRINCIPLES OF MATH£MAnCAL ANALYSIS

(see Exercise 5 for terminology). (Uniqueness follows fromExereise 4.) Hinl: Foreach p E X and each positive integer n, let V.(P) be the set of all q E E withd(p, q) < lin. Use Exercise 9 to show that the intersection of the closures of thesets f(V,lp», f( V,(P», '" , consists of a single point, say g(P), of R'. Prove thatthe function 9 so defined on X is the desired extension off.

Could the range space R' be replaced by R'? By any compact metric space?Bv any comolete metric space? By any metric space?

14. ~t I ~ [0, I'] be the closed unit interVal. Suppose /is a continuous mapping of Iinto 1. Prove that f(x) = x for at least one x E 1.

~, Call a mapping of X into Yopen if f( V) is an open set in Y whenever V is an openset in X.

Prove that every continuous open mapping of R1 into Rl is monotonic.16: Let [xl denote the largest integer contained in x, that is, [xl is the integer such

that x-I < [x] ";x; and let (x) = x ---, [xl denote the fractional part of x. Whatdiscontinuities do the functions [xl and (x) have?

17) Let f be a real function defined on (0, b). Prove that the set of points at which f_/ has a simple discontinuity is at most countable. Hint: Let E be the set on which

f(x-) <f(x+). With each point x of E, associate a triple (p, q, r) of rationalnumbers such that(0) f(x-) p.The set of all such triples is countable. Show that each triple is associated with atmost one point of E. Deal similarly with the other possible types of simple discontinuities.

18. Every rational x can be written in the form x = min, where n > 0, and m and n areintegers without any common divisors. When x = 0, we take n = 1. Consider the

function f defined on R' by

(x irrational),

Prove thatfis continuous at every irrational point, and thatfhas a simple discont!ml1ty at every rational point.

l~C:Suppose f i; a real fur.ction \....it.~ domain R l which has the intermediate valueproperty: If /(0) < c <f(b), then [(x) ~ c for some x between a and b.

Suppose also~ for every rational r, that the set of all x withf(x) = r is closed.Prove that f is continuous.Hint: If x, ~ x, but f(x,) > r > f(x,) for some r and all n, then [(t.) = r

lor :SUllll;; i" utiv.,cc[J ':':0 .iii"::: ;,;~: thU$ In -+ Xu. F:~i 2. conu-.:idictiou. iN. J. Fin.:-,.,.mer. ]vfath. Alonrhly, vol. i3. 1966. p. 782.)

CONTINUITY 101

20. IfE is a nonempty subset of a metric space X, define the distance from x E X to Eby

p.{x) = inf d(x, z)....(a) Prove thatp.{x) = 0 if and only if x E E.(b) Prove that PE is a uniformly continuous function on X, by showing that

Ip.(x) - ph) I ,,; d(x, y)

for all XE X, yE X.Hint: p.{x) ,,; d(x, z) ,,; d(x, y) + d(Y, z), so that

p.(x) ,,; d(x, y) + ph).

21. Suppose K and F are disjoint sets in a metric space X, K is compact, F is closed.Prove that there exists 8> 0 such that d(p, q) > 8 if p E K, q E F. Hint: pr is acontinuous positive functio~ on K.

Show that the conclusion may fail for two disjoint closed sets if neither iscompact.

22. Let A and B be disjoint nonempty closed sets in a metric space X, and define

f(P) = PA(P) (P EX).PA(P)+ P.(p)

Show thatfls a continuous function on X whose range lies in [0,1], thatf(p) =0precisely on A andf(p) = 1 precisely on B. This establishes a converse of Exercise3: Every closed set A c X is Z(f) for some continuous real f on X. Setting

show that Vand Ware open and disjoint, and that A c V, Be W. (Thus pairs ofdisjoint closed sets in a metric space can be covered by pairs of disjoint open sets.This property of metric spaces is called normality.)

23. A real-valued function f defined in (a, b) is said to be convex if

f(!Ix + (I - A)Y) ,,; AI(x) + (I - A)f(y)

whenever a < x < b, a < y < b. 0 < A< 1. Prove that every convex function iscontinuous. Prove that every increasing convex function of a convex function isconvex. (For example, if fis convex, so is ef.)

Iffis convex in (a, b) and if a <s < t < u <b, show that

((11- ((s) < flu) - f(s) < nu) - f(t)t-s u-s - u-t .

24. Assume thatjis a continuous real function defined in (a. b) such that

[(x -::- y) ,,; f(x) -; f(v)

" .. .

for all x, y E (a. oj. Prove thatfis convex.


25. If A c R' and B c R', define A + B to be the set of aU sums x + y with x E A,YEB.(a) If K is compact and C is closed in R'. prove that K + C is closed.

Hi"r: Take z'; K + C. put F~z - C, the set of all z - y with y E C. ThenK and F are disjoint. Choose ~ as in Exercise 21. Show that t:he open ball withcenter z and radius 0 does not intersect K + C.

(b) Let 0:: be an irrational real number. Let C t be the ~t I)f ".il integers. iet t;: ~the S~I of all na. with n E Ct. Show that Ct and C2 are dosed subsets of R1 whosesum CJ + C 2 is not closed, by showing that C t + C2 is a countable dense subsetof R'.

26. Suppose X. Y. Z are metric s!,aces. and Y is compact. Let f map X into Y, letg be a continuous one-to--one mapping of Y into Z, and put h(x) = g(f(x» forx 6 X.

Prove that f is unifor!!l!Y cond!!!.!ot!s if h is uniforr.1!Y continuous.Hint: g-' has compact domain g(Y), and f{x) =g-'(h(x)).Prove also that f is continuous if h is contirruous.Show (by modifying Example 4.21, or by finding a different example) that

the compactness of Y cannot be omitted from the h)''Potheses, even when X andZ are compact.

5DIFFERENTIATION

In this chapter we s~all (except in the final section) confine our arrel1ti0n to realfunctions defined on interva+s or segments. This is not just a matter of convenience, since genuine differences appear when we pass from real functions tovector-valued ones. Differentiation of functions defined on Rio. will be discussedin Chap. 9.

THE DERIVATIVE OF A REAL FUNCTION

5.1 Definition Let/be defined (and real-valued) on [a, b]. For any x E [a, b]form the quotient

(I)

and define

</>(1) =/(1) - /(x)t - x

(a < t < b, t "" x),

(2) f'(x) = lim </>(1),I-x


provided this limit exists in accordance with Definition 4.1.We thus associate with the function f a function f' whose domain

is the set of points x at which the limit (2) exists; f' is called the derivativeoff '

If f' is defined at a point x, we say that f is differentiable at x. If f' isdefined at every point of a set E c [a, b], we say that f is differentiable on E.

It is possible to consider right-hand and left-hand limits in (2); this leadsto the definition of right-hand and left-hand derivatives. In particular, at theendpoiIits a and b, the derivative, if it exists; is a right-hand or left-hand derivative, respectively. We shall not, however, discuss one·sided derivatives in anydetail.

If f is defined on a segment (a, b) and if a < x < b, thellf'(x) is definedby (1) and (2), as above. Butf'(a) andf'(b) are not defined in this case.

5.2 Theorem Let f be defined on [a, bf,7Jfis differentiable at a point x E [a, b],then f is continuous at x.

Proof As t -+ x, we have, by Theorem 4.4,

DIFFERENTIATION lOS

If we divide this by t - x and note thatf(t) -+ f(x) as t -+ x (Theorem 5.2),(b) follows. Next, let h = fig. Then

h(t) - h(x) = I [g(x/(t) - f(x) _ f(xl(t) - g(X)].t - x g(t )g(x) . t - X . t - x

Letting t -+ x, and applying Theorems 4.4 and 5.2, we obtain (c).

5.4 Examples' The derivative of any constant is clearly zero. Iff is definedby f(x) = x, thenf'(x) = 1. Repeated application of (b) and (c) then shows thatx" is differentiable, and that its derivative is nx"-t, for any integer n (if n < O.we have to restrict OUfs~:YeS to x #" 0). Thus every polynomi3.! i~ differentiable.and so is every rational function, except at the-points whei"c the dCiioffiinator iszero.

The following theorem is known as the "chain rule" for differentiation.It deais with differentiation of composite functions and is prcbably the mostimportant theoieJTI C\hout derivatives. We shalf ineet more general versions of itin Chap. 9.

5.5 Theorem Suppose f is continuous on [a, b],f'(x) exists at some pomtx E [a, b], 9 is defined on an interval I which contains the range vf f, and g isdifferentiable at the point f(x). If

h(t) = g(f(t) (a 5, t 5, b),

then h is differentiable atx, and

f(t) - f(x)f(l) - f(x) = . (t - x) -+ f'(x) . 0 = O.

t - x

The converse of this theorem is not true. It is~easy to construct continuousfuncuons which fail to re differentiable at isolated points. In Chap. 7 we shalleven become acquainted with a function which is continuous on the whoie iinewithout being differentiable at any point!

5.3 Thenrem Suppose f and 9 are defined on [a, b] and are differentiable at apoint x E {a, bj. Then! + g,jg, and fig are differentiable at x, arul

(a) (f + g)'(x) =f'(x) + g'(x);

(h) (fg)'(x) = f'(x)g(x) + f(x)g'(x);

(c) (£\' (x) = g(x)f'(x)_- g'(x)f(x).g" " g"'{x) __

In (c), we assume of course that g(x) oF O.

Proof (a) is clear, by Theorem 4.4. Let h = fg. Then

h(:) - h(x) = f(t)[g(t) - g(x)] + g(x)[f(t) - f(x)].

(3)

(4)

(5)

(6)

h'(x) = g'(f(x»f'(x).

Proof Let y = fix). By the definition of the derivative, we have

f(t) - f(x) = (t - x)[f'(x) + u(t I],

g(s) - g(y) = (5 - y)[g'(y) + v(s)],

where t E [a, b], s E I, and u(t) -+ 0 as t -+ x, v(s) -+ 0 as s -+ y. Let s = f(t).Using first (5) and then (4), we obtain

h(t) - h(x) = g(flt)) - g(f(x»

= i flli - fIx)] . [g'(y) + vIs)]= (t - x)· [f'(x) + u(t)] . [g'(y) + VIS)],

Of, if t ¢ x,

hi!) - hex) = [g'(y) + v(s)] . [f'(x) + u(t Ii.t - x

Letting t -+ x, we see that s -+ y, by the continuity' off, so that the rightside of (6) tends to g'(y)f'(x), which giveS' (3).


5.6 Examples(a) Let/be defined by

DIFFERENTIATION 107

MEAN VALUE THEOREMS

At x = O. these theorems do not apply any longer, since I/x is not definedthere, and we appeal directly to the defioition: for I # 0,

Taking for granted that the derivative of sin x is cos x (we shalldiscuss the trigonometric functions in Chap. 8), we can apply Theorems5.3 and 5.5 whenever x # 0, and obtain

I(t) -/(0) . I'-'--'--::-~ = SID -.

t - 0 t

A5 ; ~ 0 this does not tend to any limit, so that 1'(0) does not exist.,(b) Let/be defined by I

(7)

(8)

(9)

{

. Ixsm-

I(x) = 0 x

I I II'(x) = sin - - -cos-

x x x

{

. Ix2 sm-

I(x) = 0 x

As above, we obtain

(x #0),

(x = 0).

(x # 0).

(x # 0),

(x = 0),

5.7 Definition Let I be a real function defined on a metric space X. We saythat/has a local maximum at a point p E X if there exists 0 > 0 such that/(q) :s;1(P) for all q E X with d(p, q) < 0.

Local minima'are defined likewise.Our next theorem is the basis of many applications of differentiation.

5.8 Theorem Let I be defined on [a, b]; ifI has a local maximum at a pointx E (a, b), and iff'(x) exists, thenf'(x) = O.

The analogous statement for local minima is of course also true.

Proof Choose 0 in accordance with Definition 5.7, so that

a < x - 0 < x < x + 0 < b.

If x - 0 < t < x, then

I(t) - I(x) 0.-'-''---':...:....: ~ .1- X

Letting I ~ x, we see thatf'(x) ~ O.If x < t < x + 0, then

1(1) - I(x) :s; 0,1- x

which shows thatf'{x):S; O. Hencef'(x) = O.

At x = 0, we appeal to the definition, and obtain

5.9 Theorem II I and 9 are continuous real funclions on [a, oj which aredifferentiable in (a, b), then there is a poinl x E (a, b) at which

[f(b) - I(a)jg'(x) = [g(b) - g(a)jf'(x).

Note that differentiability is not required at the endpoints.

Proof Put

(10)I I

1'(:;;) = 2x sin - - cos-x x

i(I) - f(O)1 'I . II "I- I = Ism - :s; IIII 1-0' tI , ,

(x # 0).

(I # 0);

h(t) = [f(b) - I(a)]g(t) - [g(b) - g(a)]/(t) (a :s; t :s; b).

(II)

ieiting t -+ 0, we see that

1'(0) = o.

Thus I is differentiable at all points x, but I' is not a continuousfunction, since cos (l/x) in (10) does not tend to a limit as x ~ O.

(12)

Then h is continuous on [a, bj, h is differentiabie in (a, 0), and

h(a) =/(b)o(a) - I(a)g(b) =h(b).

To prove the theorem, we have to show that h'(x) = 0 for some x E (a, b).If h is constant, this holds for every xE(a,b). If h(t) > h(a) for

some I E (a, b), let x be a point on [a, bj at which h attains its maximum

108 PRINCIPLES OF MATIfEMATlCAL ANALYSISDIFFERENTIATION 109

(Theorem 4.16). By (12), x E (a, b), and Theorem 5.8 shows that h'(x) = O.If h(t) < heal for some t E (a, b), the same argument applies if we choosefor x a point on [a, b] where h attains its minimum.

Corollary II I is differentiable on [a, bj, then [' cannot have any simple discontinuities on [a, bj.

But[' may very well have discontinuities of the second kind.

This theorem is often called ageneralized mean value theorem; the followingspecial case is usually referred to as Hthe" mean value theorem:

5.10 Theorem IIlis a real continuouslunction on [a, b] which is differentiablein (a, b), then there is a point x E (a, b) at which

L'HOSPITAL'S RULE

The following theorem is frequently useful in the evaluation of limits.

I(b) - I(a) = (b - a)/1.ll..

Proof Take g(x) = x in Theorem 5.9.

5.13 Theorem Supposeland 9 arereal and differentiable in (a, b), and g'(x) '" 0lor all x E (a, b), wher<! - 00 :'> a + 00. Suppose

The analogous statement is of course also true if x ..... b, or if g(x) ..... - 00

in (IS). Let us note that we now use the limit concept in the extended sense ofDefinition 4.33.

g(x) .... +00 as x ..... a,

l(x) .... Oand g(x) .... Oas x .... a,

I(x)--+A asx-+a.g(x)

(a <y < c).

['(x)-- -+ A as x -+ a.g'(x)

['(x)--<r.g'(x)

1(Y)g(y):'> r <q

I(x) - 1(Y) ['(t)=--<r.

g(x) - g(y) g'(t)

Suppose (14) holds. Letting x .... a in (18), we see that

Proof We first consider the case in which - 00 :'> A < + 00. Choose areal number q such that A < q, and then choose r such that A < r < q.By (13) there is a point c E (a, b) such that a < x < c implies

Ifa < x-< y < c,then Theorem 5.9 shows that there is a point t E (x, y)such that

(19)

(18)

(17)

(16)

(13)

If

(14)

or if(IS)

then

which is valid, for each pair of numbers x" Xl in (a, b), for some x betweenXl and X,2.

5.12 Theorem Suppose I is a real differentiable lunction on [a, b] and suppose1'(0) < A</,(b). Then rhere is a point x E (a, b) such rhat ['(x) = i ..

A simiiar result holds of course if['(a) > ['(b).

Proof Put get) = I(t) - At. Then g'(a) < 0, so that get,) < g(a) for somet, E (a, b), and g'(b) > 0, so that g(tl ) < g(b) for some t l E (a, b). Helice9 attains its minimum on [a, b] (Theorem 4.16) at some point x such thata < x < b. By Theorem 5.8, g'(x) = O. Hence['(x) = A.

5.11 Theorem Suppose I is differentiable in (a, b).

(a) IIj'(x) ;:: '() for all x E (a, b), then I is monotonically increasing.

(b) 11[,(x) = olor all x E (a, b), then I is constant.

(c) If[,(x):'> 0 for all x E (a, b), then I is monotonically decreasing.

Proof All condusions can be read off from the equation

I(x l ) - I(x,) = (Xl - x,)['(x),

THE CONTINUITY OF DERIy.ATIVES

We have already seen [Example 5.6(b)] that a function I may have a derivativej' which exists at every point, but is discontinuous at sQme point. However, notevery function is a derivative. In particular, derivatives which exist at everypoint of an interval have one important property in common with functionswhich are continuous on an interval: Intermediate values are assumed (compare

.Theorem 4.23). The precise statement follows.

i 10 PRINCIPLES OF MATHEMATICAL ANALYSISDIFFERENTIATION 111

DERIVATIVES OF HIGHER ORDER

5.14 Definition If/has a derivativef' on an interval, and iff' is itselfdifferentiable. we denote the derivative off' by f" and call!" the second derivative offCOTItie.uing in this manner, we obtain functions ,

f,f',f",f(31, ... ,flo),

~ach of which is the derivative of the preceding one. j(n) is called the nth derivative, or the derivative of order n, off

in order for/(":I (x) to exis.t ai a point X,/{n-I) (t) must exist in a neighborhood of x (or in a one-sided neighborhood, if x is an endpoint of the intervalon wbichlis defined), and/,"-I) must be differe'!tiable at x. Since/("-I) mustexist in a neighborhood of X,!'"-2) must be differentiable in that neighborhood.

(a';' t,;, b).

1(f3) = PcP) + M(f3 - x)"

get) =/(t) -pet) - M(t - a)"

and put

DIFFERENTIATION OF VECTOR-VALUED FUNCTIONS

(26)

We have to show that n!M = /,")(x) for some x between a and B. By

(23) and (26),

(27) 9'")(t) = /'")(1) - n!M (a < t < b).

Hence the proof will be complete if we can show that 9'")(x) = 0 for somex between a and p.

Since p(l)(a) =l(l)(a) for k = 0, ... , n - I, we have

(28) g(a) = g'(a) = ... = g("-I)(a) = O.

Our choice of M shows that g(P) = 0, so that g'(xl ) = 0 for some XI

between a and p, by the mean value theorem. Since g'(a) = 0, we concludesimilarly that g'(X2) = 0 for some X2 between a and XI' After n steps wearrive at the conclusion that r/n)(x::) = 0 for some X n between a and Xn - Il

that is, between x and p.

(25)

For n = I, this is just the mean value theorem. In general, the theoremshows that I can be approximated by a polynomial of degree n - I, and that(24) allows us to estimate the error, if we know bounds on Itt")(x) I·

Proof Let M be the number defined by

(24)

Then there exists a point x between a and fJ such that

I (")(x)1(f3) = P(f3) + -- (f3 - a)".

n!(a < x < CI)'

(a < x < C2)'I(x)-<gg(x)

Next, suppose (15) holds. Keeping y fixed in (18), we can choosea point CI E (a, y) such that g(x) > g(y) and g(x) > 0 if a < x < CI' Multiplying (18) by [g(x) - g(y))/g(x), we obtain

I(x) < r _ , g(y) + 1(Y)g(x) g(x) g(x)

If we let x --> a in (20), (15) shows that there is a point C2 E (a, cI)such that

Summing up, (19) and (21) show that for any g, subject only to thecondition A < g, there is a point C2 such that/(x)!g(x) < g if a < x < c2 •

In the same manner, if - cr.. < A :::; + 00 1 and p is chosen so thatp < A, we can find a point c, such that

I(x)p < - (a < x < C,),

g(x)

and (16) follows from these two statements.

20)

(22)

e21)

TAYLOR'S THEOREM

5.15 Theorem Suppose f is a real function on [a, b], n is a positiL'e integer,(("-I) is continuous on [a, b),/,")(1) existslor every t E (a, b). Let a, pbe distinct,oints 01 [a, b], and define

pet) = "f/(')(a) (t - all.1-0 k!

5.16 Remarks Definition 5.1 applies without any change to complex fu~ct;ons

f defined on [a, b], and Theorems 5.2 and 5.3, as well as their proofs, re!!lairrvalid. IfII and J~ are the real and imaginary parts off, that is, if

I(t) = fi(t) + if2(t)

for a"; t,;, b, where/l(t) and/2(t) are real, then we clearly have

(29) f'(x) =I{(x) + ifi(x);

also,fis differentiable at x if and only if both II and 12 are differentiable at x.

112 PRINCIPLES OF MATHEMATICAL ANALYSISDIFFERENTIATION 113

and f' is again a function with values in R".If Ji, ... , /;. are the components of f, as defined in Theorem 4.10, then

Passing to vector-valued functions in general, i.e., to functions f whichmap [a, b] into some R", we may still apply Definition 5.1 to define f'(x). Theterm cf>(I) in (l) is now, for each I, a point in R" and the limit in (2) is taken withrespect to the norm of R'. In other words, f'(x) is that point of R' (if there isone) for which

If'(X) I 1 xg'(x) = Ig'(x)1 ~ 2 - x

I2'1 2Ig'(x) I ~ 2x - ~ - 1 ~ -; - I.

(0 < x < I),

. f'(x)hm--=O.x-o g'(x)

, /2x 2i) 'Ix'9 (x) = 1 + \ - -; e

Hence

(40)

and so

(39)

(38)

so that

Next,

(37)

f' = (f;, .,. ,ft),

lim If(l) - f(x) - f'(x) I= 0,''''x t-x

(31)

(30)

By (36) and (40), L'Hospital's rule fails in this case. Note also that g'(x) -# 0

on (0, 1), by (38). .,'" fHowever. there is a consequence of the mean value theorem WhlCh.•or

purposes of applications, is almost as useful as Theorem 5.l0, and which remains true for vector-valued functions: From Theorem 5.10 It follows that

and f is differentiable at a point x if and only if each of the functions f" .•. , /;.is differentiable at x.

Theorem 5.2 is true in this context as well, and so is Tneorem 5.3(a) and(b), iffg is replaced by the inner product f· g (see Definition 4.3).

When we turn to' the mean value theorem, however, and to one of itsconsequences, namely, L'Hospital's rule, the situation changes. The next tWOexamples will show that each of these results fails to be true for complex-valueufunctions.

(41) If(b) - f(a) 1 ~ (b - a) sup 1f'(x)l·s<x<b

5.18 Example On the segment (0, 1), definef(x) = x and

5.17 Example Define, for real x,

Since Ieit I = I for all real I, we see that

so that lfl(x) i = i for all real x.Tnus Tneorem 5.10 fails to hold in this case.

5.19 Theorem Suppose f is a continuous mapping of [a, h] into R' and f isdifferentiable in (a, b). Then there exists x E (a, b) such that

If(b) - f(a) I ~ (b - a)lf'(x)!.

Proof' Put z = feb) - f(a), and define.

cp(t)=z'f(t) (a~ t~b).

Then (j) is a real-valued continuous function on [a. b] which is differentiable in (a, b). The mean value theorem shows therefore that

<pCb) - <pea) = (b - a)<p'(x) = (b - a)z . f'(x)

for some x E (a, b). On the other hand,

<pCb) - <pea) = z . feb) ~Z' f(a) =z' z = Izi '.

The Schwarz inequality now gives

Izl'=(b-a)lz'f'(x)! ~(b-a)izilf'(x)l·

Hence Izl ~ (b - a) If(x) I, which is the desired conclusion.

f'(x) = ie'x,

g(x) = x + x'e'ix'.

f(x) = eix = cos X + i sin x.

f(27.) - f(O) = 1 - 1 =0, --

(35)

(34)

(32)

(33)

but

(The last expression may be taken as the definition of the com plex exponentiale'x; see Chap. 8 for a full discussion of these functions.) Then

(36). f(x)hm-()=l.x ...09 x

1 v, P. Havin translated the second edition of this book into Russian and added this

proof to the original one.

114 PRINCIPLES OF MATHEMATICAL ANALYSISDIFFERENTIATION 115

EXERCISES

and prove that g is monotonically increasing.

.. Suppose f'(;.:), g'{x) exist, g '(x) ¥= 0, and f(x) = g{x) = O. Prove that

lift) - [(x) I'(X)I < e

t-x

(if x ""0),

(iCY = 0).(x' sin (ixl-')

[(x) ~ °

Compare with Example 5.18. Hint:

r J(x + h)+ [(x - h) - 2[(x) - r( ),L~ h'J. - x.

J(x) ~ (J(x) _ A) .~ + A.~.g(x) \ x g(x) g(x)

Apply Theorem 5.13 to the real and imaginary parts ofJ(x)ix and g(x)!x.11. Suppose/is defined in a neighborhood of x, and suppose ["{x) eXIsts. Show that

Show by an example that the iimit may exist even if["(x) does not.

Hint: Use Theorem 5.13.12. If[(x) = Ix I', compute I'(x), ['(x) for all real x, and show that [(3l(O) does not

exist.ii Suppose a and e are real numbers, e > 0, andJis defined on [-1, IJ by

whenever 0< It-xl <0, as.xs.b, as.ts.b. (This could be expressed bysaying thatJis uniformly differentiable on [a, bl ifI' is continuous on [a, b].) Doesthis hold for vector~valued functions too?

9. ':, Let / be a continuous real function on R1, of which it is known that /'(x) exists

for all x ""°and thatl'(x) ... 3 as x ... 0. Does it follow that 1'(0) exists?10. SupposeJand g are complex differentiablefunction,s on (0, I),[(x) ... O,g(x) ... 0,

I'(x) ... A, g'(x) ... B as x ... 0, where A and B are complex numbers, B "" O. Provethai

Prove the following statements:(a) f is continuous if and only if a > O.(b) 1'(0) exists if and only if a > I.(e) I' is hounded if and only if a ;;" I + e.

(d) I' is continuous if and only if a > I + e.(e) 1"(0) exists if and only if a> 2+ c.

ef) f" is bounded if and only if a ~ 2 + 2c.(g) r is continuous if and only if a > 2 + 2e.

14. Let f be a d:fferentiable real function defined in (0, b). Prove that ris convex ifand only if f' is monotonically increasing. Assume next that fH(X) exists for

" every x E (a, b), and prove that [is convex if and only ifr(x);;"°for all x E (a, b).lS.;Supposea E Rl,[is a twice-differentiable real function on (a, 00), and M o, M1 , Mz

are the least upper bounds of IJ(x)l, II'(x)!, Ir(x)!. respectively, on (a, 00).

Prove that

(a <x < b).

(x >0)g(x) =J(x)x

g'(f(x)) = I'~x)

C C1 . C"_1 C~O+-+···+--+--~O

2 n n+l '

where Co, ... , C~ are real constants, prove that the equation

Co + C1x -+- ... + CII_1X,,-1 + CII;x'! = 0

has at least one real root between 0 and 1.5. SupposeJis defined and differentiable for every x> 0, andl'(x) ... 0 as x ... + 00.

Put g(x) = J(x + I) - J(x). Prove that g(x) ... 0 as x ... + 00.

~ Suppose(a) f IS continuous for x ;;;::: 0,(b) I'(x) exists for x > 0,(e) j(O) = 0,

(d) J' is monotonicaiiy llcro...asing.Put

1. Let J be defined for all real x, and suppose that

IJ(x) - J(y) I s. (x - y)'

for all real x and y. Prove that/is constant.

2. Supposel'(x) >°in (a, b). Prove thatJis strictly increasing in (a, b), and let g beits inverse function. Prove that g is differentiable, and that

3:; Suppose g is a real function on R\ with bounded derivative (say Ig' i, -::; M ,. Fix

t > 0, and defineJ(x) = x + eg(x). Prove thatJis one-to-one if e is small enough.(A set of admissible values of e can be determined which depends only on M.)

if; If

lim Jet) =I'(x).... ,. g(t) gf(X)

(This holds aiso jor complex functions.)

8. Suppose I' is continuous on [a, bJ and e > O. Prove that there exists 0 > 0 suchthat

116 PRINCIPLES OF MATHEMATICAL ANALYSIS DIffERENTIATION 117

Hint: If h > 0, Taylor's theorem shows that

for some gE (x, X + 2h). Hence

1['(x)1 5,hM, + ~o.

has no fixed point, although 0 <['(I) < 1 for all real r.(e) However, if there is a constant A < 1 such that i['(I) I $A for all rcall, provethat a fixed point x off exists, and that x = lim X'II where Xl is an arbitrary realnumber and

1(1) ~ t + (1 + e')-'

for n = 1, 2, 3, '" .(d) Show that the process described in (e) can be visualized by the zig-zag path

23. The function j defined by

Prove the following statements:(a) If ". < 0 < f3., then lim D. = ['(0).(b) If 0 < ". < f3. and {f3.f(f3. - ",)} is bounded, then lim D. = ['(0).(e) If[' is continuous in (-I, I), then lim D. =['(0).

Give an example in which/is differentiable in (-1,1) (but!, is not continuous at 0) and in which IX ft , fJft tend to 0 in such a way that lim D ft exists but is different from ['(0).

20. Formulate and prove an inequality which follows from Taylor's theorem andwhich remains valid for vector-valued functions.

:U. Let E be a closed subset of R'. We saw in Exercise 22, Chap. 4, that there is are~l continuous function! on Rl whose zero set is E. Is it possible, for each closedset £, to find such an f which is differentiable or.. R 1

, or C!1e which is f1 timesdifferentiable, or even one which has derivatives of all orders on R 1 ?

22. Suppose [is a real function on (-00, <0). Can x a,fixedpoint of[if,f(x) ~x.(aliffis differentiable and ['(I) "" 1 for every real t, prove that [has at most onefixed point.(b) Show that the function [defined DY

lUi - 1(fJ)t - f3

Q(I)

['(x) ~ ;h [f(x + 2h) - [(x)]- hr(f)

To show that 1\1 f = 4M0 M 2 can actually happen, take a = - I, define

f2x' - 1 (-1 < x < 0),

"j(xj = !x' - I. \X2 +1 (0:5:£<00),

and show that M o = 1, M 1 =4, M 2 =4.

Does J.,f~ :5:4Mo Jl.12 hold for vector-valued fun~tions too?16. Supposejis twice-dilferentiable On (0, <0), r is bounded on (0, <0), and[(x) ~ 0

as X-+ 0). Prove thatf~(x) -+0 as x -+ 00.

Hint: Let a ...".. 00 in Exercise 15.

17. SupposeJ is a real, three times differentiable function on [-I, 1], such that

[(-1)=0, I(O)~O, [(1)=1, ['(O)~O,

Prove that J<:!)(x) :::: 3 for some x € (-1. 1).Note that equality holds for t(x' + x').

Hiu!: Use Theorem 5.15, with lX .= 0 and p= ± 1, to show that there exists ErG, i) and I E (-1, 0) such that

["J(s) +["'(I) ~ 6.

18. Suppose f is :l real function on ta, b1, n is a positive integer, and [1ft -1) exists forevery t E [a, bl. Le!:x, 13, and P be as in Taylor's theorem (5.15). Define

for t E [a, bl. t "" f3, differentiate

/(1) - i(PJ = (t - fJ)Q(t)

n - 1 times at t = lX, and derive the following version of Taylor's theorem:

[(8) ~ P(ft) + ~~"..:.';i~) (f3 - ,,)'.

19, Suppose I is defined in (-1,1) and ['(0) exists, Suppose -1 < ". <f3. < I,lXn-+ 0, and 13n -+ 0 as n -+ co. Define the difference quotients

, ' 1[(x) = ::....:::.

3

For arbitrarily chosen Xli define {Xn} by setting X/H 1 = f(x N).

(a) If Xl < ~ prove that XII -+ - co as n -+ 00.

(b) If" < x, < Y. prove that x. -+ f3 as n -+ 00.

(e) If y < Xh prove that x" .... + co as n -+ 00.

Thus f3 can be located by this method, but" and y cannot.

1 <Y < 2.0< f3 < 1,-2<1X<-1,

has three fixed POL.'1.!S, say lX, ,8, y, where

[(f3.) - [(",)fJn ctft

D.

118 PRINCIPLES OF MATHEMATICAL ANALYSIS DIFFERENTIATION 119

is, by definition, ~ d!ffere!1tiable{unction/on [a, b) such that/Cal = c, ex ~f(x) ~ (J,and

u= 1, ... , k).

(a~x:sb).

Y/'.o) = CJ

rex) = q,(x,[(x))

yj = ~I.x, YIt '" ,Y.l),

Note that this can be rewritten in the form

Prove that such a problem has at most one solution if there is a constant A suchthat

M, = sup Ifix) I, M, ~ sup Ir(x) Ifor a ::;: x ::;: -'to. For any such x,

If(x)! :S M,(x, - a) :S A(x, - a)M, .

Hence M, = 0 if A(x, - a) < 1. That is,f= 0 on [a, x,]. Proceed.27. Let ep be a .real function defined on a rectangle R in the plane, given by a::;: x ::;: b,

0:: ::;: Y ::;: f3. A solution of the initial-value problem _

y'~q,(x,y), y(a)=c (a~c~f3)

26. Suppo~efis differentiable on [a,bJ,J(a) ~O, and there is a real number A suchthat Ir(x) I :S A If(x) I on [a, b]. Prove that fix) = 0 for all x E [a, b]. Hint: Fixx, E [a, b], let

I.p(x, y,) - .p(x, Y,)i :S A Iy,- Y,:

whenever (x, y,) E R and (x, y,) E R.Hint: Apply Exercise 26 to the difference of two solutions. Note that this

uniqueness theorem does not hold for the initiai-value problem

which has two solutions: fix) = 0 and fix) ~ x'/4. Find all other solutions.28. Formulate and prove an analogous uniqueness theorem for systems of ditTerentiai

equations of the form

a+xg(x)=--.l+x

(c) Use Taylor's theorem to show that

AII+1 = X _ /(x ll) •

" j'(x")

Interpret this geometrically, in terms of a tangent to the graph off,(b) ,Prove that X n+l < X n and that

Both f and 9 have v; as their only fixed point in (0, CYJ). Try to explain, on thebasis of properties off and g, why the convergence in Exercise 16, Chap. 3, is somuch more rapid than it is in Exercise 17. (Compare f' and g', draw the zig-zagssuggested in Exerc!se 22.)

Do the same when 0 < 0:: < 1.

Suppose f is twice differentiable on [a, bJ, f(a) < 0, feb) > oJ'(x) ~ S > 0, ando ~r(x) ~ M for all x" [a. b]. Let g be the unique point in (a, b) at whichl(gj ~O.

Cc:nplcte the details in !he fcHo\·.'~ng ou!!ine of Newlon's method for C0m

puting r(a) Choose Xl E a, b), and define {XII} by

24. The process described in part (c) of Exercise 22 can of course also be applied tofunctions that map (0, CYJ) to (0, CYJ).

Fix some a > 1, and put·

(j~l, ... , k-l),

where y ~ (y" ... ,Y.) ranges over a k-cell, cj> is the mapping of a (k + I)-cellinto the Euclidean k-space whose components are the iunctions 4>1, ... , ,pi, and cis the vector (CIt ... ,Ci)' Use Exercise 26, for vector-valued functions.

29. Specialize Exercise 28 by considering the system

yea) ~ cy' ~ cj>(x, y),("(I),- - f = _._"_ ( _ C)'• II-tl .. 2/,(x

n) x~ ~

for some '" E (g, x").(d) If A ~ MI2S, deduce that

y; ~ f(x) - :t gj,x)y"J"

wheref. 91, ... , Uk are continuous real functions on [a, bl. and derive a uniquenesstheorem for solutions of the equation

y''> +g.(x)y"-" +... + g,(x)y' + g,(x)y =f(x),

subject to initial conditions

(Ccmpare with Exercises 16 and 18, Chap. 3.)(e) Show that Newton's method amoUnts to finding a fixed point of the function

g dcfincG by

fix)g(x)=x- --.rex)

How doesg'(x) behave for x near g?(f) Putf(x) = x'I> on (- ro, CYJ) and try Newton's method. What happens? Yea) ~ c" y'(a) = C" ... , yCi-ll(a) =Ck.

6THE RIEMANN-STIELTJES INTEGRAL

THE RlEMANN~STJELTJESINTEGRAL 121

Now suppo~e/is a bounded real function defined on [a, bj. Corresponding toeach partition P of [a, bj we put

M, = sup/(x)

m, = inf/(x)

"U(P,f) = L M,!lx"

. i= 1

"L(P,f) = Lm, !lx"i=l

(Xi-I::;; X ~ Xi),

(X'_l "X " x,),

and finally

(I)

(2)

J:I dx = inf U(P,f),

bJI dx= sup L(P,f),-'

The present chapter is based on a definition of the Riemann integral whichdepends very explicitly on the order structure of the real line. Accordingly,we begin by discussing integration of real-valued functions on intervals. Extensions to complex- and vector~valued functions on intervais follow in later'Eeciions. Inl~g:i·iition over sets other than intervals is discussed in Chaps. 10and 11.

where the inf and the sup are taken over all partitions P of [a, bj. The leftmembers of (l) and (2) are called the upper and lower Riemann integrals of Iover [a, bj, respectively.

If the upper and lower integrals are equal~ we say that f is Riemannintegrable on [a, b], we write IE ffI (that is, {iI denotes the set of Riemannintegrable functions), and we denote the common value of (I) and (2) by

(3) (f dX,

or by

b

(4) J. f(x) dx.

This is Ihe Riemunn integral of f over [a, bj. S!!!!:e f!~ bounded, thereexist two numbers, m and M, such that

DEFINITION AND EXISTENCE OF THE INTEGRALHence, for every P:

m "/(x)" M (a" x " b).

We write

6.1 Definition Let [a, b] be a given interval. By a partition P of [a, b] wemean 3 finite set of points Xv, Xl' ... , Xr. ~ where

(i = I, ... , n).

m(b - a) " L(P,f) " V(P,f) " M(b - a),

so thRt the numbers L(P,f) and V(P,f) form a bounded set. This shows thatthe upper and lower integrals are defined for every bounded function f. Thequestion of their equality, and hence the question of the integrability off, is amore delicate one. Instead ofinvestigating it separately for the Riemann integral,we shall immediately consider a more general situation.

122 PRINCIPLES OF MATHEMATICAL ANALYSIS THE RIEMANN~STlELTJESINTEGRAL 123

where Mil IIlj have the same meaning as in Definition 6.1, and we define

"U(P,J, a) = L M, /).(1."i= 1

"L(P,J, a) = L m, Aa"i= I

the inf and sup again being taken over all partitions.If the left members of (5) and (6) are equal, we denote their common

value by

6.3 Definition We say that the partition p* is a refinement of P if P * ::> P(that is, if every point of P is a point of P *). Given two partitions, P, and P2,we say that p* is their common refinement if p* = PI U P2 .

" "L c" L Cki= 1 J:=1

6.4 Theorem If p* is a refinement ofP, Ihen

mean the same thing~ since each means CI + C2 + ... + CII •

Of course, I}o harm is done by inserting the variable of integration, andin many cases it is actually convenient to do so.

We shall now investigate the existence of the integral (7). Without sayingso every time, f will be assumed real and bounded, and a monotonically increas-

ing on [a, b]; and, when there can be no misunderstanding, we shall write f in

place of r..

The integral depends on J, a, a and b, but not on the variable of integration,which may as well be omitted.

The role played by the variable of integration is quite analogous to thatof the index of summation: The two symbols

J: f da = inf U(P,J, a),

bLfda = sup L(P,J, a),(6)

(5)

6.2 Definition Let a be a monotonically increasing function on [a, b] (sinceala) and alb) are finite, it follows that a is bounded on [a, b]). Corresponding toeach partition P of [a, b], we write

Aa, ~ a(x,) - a(x,_,).

It is clear that Aai ~ O. For any real function f which is bounded on [a, b]we put

(7)

r. flY) dalY)·

This is the Riemanll-Stielljes ill/egral (or simply the Slieltjes integral) offwith respect to a, over [a, b].

If (7) exists, i.e., if (5) and (6) are equal, we say that f is integrable withrespect to a, in the Riemann sense, and write f E 9!(a).

By taking a(x) = x, the Riemann integral is seen to be a special case ofthe Riemann-Stieltjes integral. Let us mention explicitly, however, that in thegeneral case ex need not even be continuous.

A few words should be said about the notation. We prefer (7) to (8), sincethe leller x which appears in (8) adds nothing to the content of (7). It is immaterial which leller we use to represent the so-called "variable of integration."For instance, (8) is the same as

L(P,J, a) ,;; L(P*,f, a)

(x,_, ,;; x ,;; x*),

(x* ,;; x ,;; x.J.

(x,_,';; x,;; x,).

w, = infj(x)

W2 = inff(x)

m, = infj(x)Hence

Clearly WI ~ mi and w2 ~ mi' where, as before,

L(P*,J, a) - L(P,J, a)

= w,[a(x*) - a(xl-l)] + w2[a(x,) - a(x*)]- m,[a(x,) - a(x,_,)]

= (w, - m,)[a(x*) - a(xi_,ll + (w 2 - m,)[a(x,) - .(x*)] ~ O.

If P' contains k points more than P, we repeat this reasoning ktimes, and arrive at (9). The proof of (10) is analogous.

U(P*,J, a)';; U(P,J, a).

Proof To prove (9), suppose first that p* contains just one point morethan P. Let this extra point be x*, and suppose x i _, < x* < x" wherex j _ 1 and Xi are two consecutive points of P. Put

(9)and(10)

s: f(x) da(x).

or sometimes by

(8)


. -.6.5 Theorem J. fda" f. fda.

Proof Let p* be the common refinement of two partitions P, and P z .By Theorem 6.4,

L(P"f, a) "L(P*,f a) " U(P*,!, a) " U(Pz,f, a).

THE RIEMANN-STIELTJES INTEGRAL 125

We choose P to be the common refinement of P, and Pz . Then Theorem6.4, together with (14) and (15), shows that

U(P,f, a)" U(Pz ,f, a) < f fda + i < L(P"f, a) + 8" L(P,f, a) + 8,

so that (13) holds for this partition P.

The theorem follows by taking the inf over all Pz in (12).

(11)

(12)

Hence

L(P,,!, a) " U(Pz ,f, a).

If Pz is fixed and the sup is taken over all P" (11) gives

Ifda" U(Pz ,f, a).

Theorem 6.6 furnishes a convenient criterion for integrability. Before weapply it, we state some closely related facts.

6.7 Theorem(a) If(13) holds for some P and some 8, then (13) holds (with the same 8)

for every refinement of P.(b) If (13) holds for P = (x., ... , x,l and if s" t, are arbitrary points ill

[XI_I' xil, then

6.6 Theorem fE 9I'(a) all [a, b] if and only if for every 8> 0 there exists apartitioll P such that

(13) U(P,f, a) - L(P,f, a) < 8.

Proof For every P we have

L(P,f, a)" Ifda " Jf da" U(P'[' a).

Thus (13) implies

Hence, if (13) can be satisfied for every 8 > 0, we have

Jfda = jfda,

,I I/(s,) - f(t,) I l'J.a, < 8.

1= 1

(c) IffE 9I'(a) and the hypotheses of (b) hold, then

Ii f(t,) 1'J.a., - f f da.j < 8.1= 1 a

Proof Theorem 6.4 implies (a). Under the assumptions made in (b),bothf(s,) andf(t,) lie in [m" M,], so that If(s,) - f(tl) I "M, - mi' Thus

,I If(s,) - f(t,lI l'J.a, " U(P,f, a) - L(P,f, a),

I"" 1

which proves (b). The obvious inequalities

L(P,f, a)" If(t,) 1'J.a.," U(P,f,a)and

L(P,f, a.) " Jf da " U(P,f, a)prove (c).

(14)

(15)

that is,fE 9I'(a).Conversely, suppose f E 9I'(a), and let 8 > 0 be given. Then there

exist partitions P, and Pz such that

U(Pz,f, a) - f f da <~,

f f da - L(P,J, a) < i·

6.8 Theorem' Iff is continuous all [a, b] then f E 9I'(a.) on [a, b].

Proof Let 8> 0 be given. Choose ~ > 0 so that

[a(b) - a(a)]~ < 8.

Since f is uniformly continuous on [a, b] (Theorem 4.19), there exists a8 > 0 such that

(16) If(x) - f(t)1 <1/

126 PRINCIPLES Of MATHEMATICAL ANALYSIS THE RIEMANN-STIELTJES INTEGRAL 127

6.9 Theorem 1[[ is monotonic on [a. b], and if" is continuous on [a. bl. then[E 9l(,,). (We still assume. o[ course. that" is monotonic.)

Proof Let. > 0 be given. For any positive integer n. choose a partitionsuch that

(17)

ifxe[a.b]. te[a.b]. and Ix-tl <{).

If P is any partition of [a. b] such that /ix, < {) for all i. then (16)implies that

(i-I..... n)

and therefore•

U(P.f, IX) - L(P,f, IX) = I (M, - m,) /).'"1= I

•,;; q I !!.<x, = q[,,(b) - ,,(a)] < •.

i=1

By Theorem 6.6,[E 9l(,,).

Remove the segments (uj • Vj) from [a. b]. The remaining set K iscompact. Hence [is uniformly continuous on K. and there exists {) > 0such that I[(s) - [(t)1 <. if se K. t E K. Is - II < {).

Now form a partition P = {xo. x" .... x.} of [a. b], as follows:Each uj Occurs in P. Each vj occurs in P. No point of any segment (uj ' v)occurs in P. If Xj-l is not one of the uj ' then .6.x j < D.

Note that M j - m j ::; 2M for every i, and that M, - nJj :$ E unlessXi_l is one of the uJ . Hence, as in the proof of Theorem 6.8,

. U(P.f, IX) - L(P.f, IX) ,:; [,,(b) - ,,(a)]. + 2M•.

Since. is arbitrary. Theorem 6.6 shows that[E 9l(,,).Note: If[and" have a common point of discontinuity. then[need not

be in 9l(,,). Exercise 3 shows this.

6.n Theorem Suppose [E 9l(,,) on [a, b]. 111 ,;;[:5, M, </> is continuous on[m. M], and h(x) = </>(f(x» on [a, b]. Then hE 9l(,,) on [a. b].

This is possible since" is continuous (Theorem 4.23).We suppose that[is monotonically increasing (the praofis analogous

in the other case). Then

Proof Choose. > O. Since</> is uniformly continuous on [111, M]. thereexists {) > 0 such that {) <. and 1</>(s) - </>(1)1 <. if Is - tl ,;{) ands. t E [m. M].

Since[E 9l(,,). there is a partition P = {xo. XI • ..•• x.} of [a. b] such

U(P,f, IX) - L(P.f, IX) < {)2.

that

(18)(i = I ..... n).

(i=I, .... n)./).". = "....:(....:b)_-_"....:(....:.a)• n

Since. was arbitrary. Theorem 6.6 implies that h E 9l(,,).Remark: This theorem suggests the question: Just what functions are

Riemann-integrable? The answer is given by Theorem 11.33(b).

so that,,(b) - ,,(a) •

U(P.f, IX) - L(P.f, IX) = I [f(x,) - [(x,_.)]n 1= 1

= ,,(b) - ,,(a) . [f(b) _ [(a)] < •n

if n is taken large enough. By Theorem 6.6. [ E 9l(,,).

6.10 Theorem Suppose [is bounded on [a. b]. [has only finitely many pointso[ discontinuity on [a. bl. and" is continuous at every point at which [is discontinuous. Then [E 9l(,,).

Proof Let. > 0 be given. Put M = sup 1[(x)l. let E be the set of pointsat which [is discontinuous. Since E is finite and" is continuous at everypoint of E. we can cover E by finitely many disjoint intervals [uj • vj ] C

[a. b] such that the sum of the corresponding differences "(Vj) - ,,(uj ) is.less than.. Furthermore, we can place these intervals in such a way thatevery point of E " (a. b) lies in the interior of some [uj ' Vj].

(19)

Let l\Jj, lUi have the same meaning as in Definition 6.1, and let Mt. m~be the analogous numbers for h. Divide the numbers 1, ...• It into twoclasses: iEA if Mj-mj <0, iEBif M j -ml:2:o.

For i E A, our choice of 0 shows that Mi - mi ::; s.For i E B. M,- - m~ ,; 2K. where K = sup I</>(t) I. m ,; t,;; M. By

(18). we have

{) I l'J."i';; I (M, -mil I'J.", < {)2i€B isB

so that Ii,. /).'" < {). It follows that

U(P. h. IX) - L(P. Ii, IX) ~ I (Mi - mil l'J."i + I (M~ - m~)!'J.<Xii€A iEB

,; .[,,(b) - ,,(a)] + 2K{) < .[,,(b) - ,,(a) + 2K].

128 PRINCIPLES OF MATHEMATICAL ANALYSIS TIlE RIEMANN-S'IIELTJES INTEGRAL 129

6.14 Definition The unit step lunction I is defined by

U=I,2);

(x,;. 0).

(x> 0).I(x) ={~

eJld.;"O.

I Jid.1 = c Jid. = Jcld. ,;.·Jlfl d••

U(P.Jj,.) < JJj d. +.hence (20) implies

Jid.,;. U(P,J..) < Jfi d. + Ji, d. +28.

Then

since cl,;. III.

4fg = (f+ g)' - (f_g)'

completes the proof of (a).Ifwe take 4>(t) = Itl, Theorem 6.11 shows similarly that III e 91(.).

Choose c = ± I. so that

The-se inequalities persist if Pi and P2 are· replaced by their commonrefinement P. Then (20) implies

U(P,J. .) - L(P,J..) < 2••

which proves that Ie 91(.).With this same P we have

Since e was arbitrary, we conclude that

(21) Jid.,;. Jfid.+ Ji,d•.

If we replace 11 andl, in (21) by -fi and -I" the inequaiity isreversed, and the equality is proved.

The proofs of the other assertions of Theorem 6.12 are so similarthat we omit the details. In part (e) the point is that (by passing to refinements) we may restrict ourselves to partitions which contain the point c,in approximating JI d•.

6.13 Theorem IIIe 91(.) and g e 91(.) on [a. bl, then.(a) Ig e iJl(.);

(b) III e 91(.) and II.'I d.1 ,;. fill d•.

Proof If we take 4>(1) = t', Theorem 6.11 shows thatl' e iJl(.) iffe iJl(.).The identity

J:ld«+ (Id.= fl d•.

(d) Ille iJl(.) on [a. b] and if I/(x) I ,;. M on [a. b]. then

If>d·1 ,;. M[.(b) - .(0)].

(e) lfle iJl(.I) and/e iJl(.,). then Ie iJl(.1 + .,) and

f.ld(.I+.,)= f.ld«l+ f. ld.,;

ifle iJl(.) and c is a positive constant, then Ie iJl(c.) and

f Id(c.) = cf Id•.

Proof Ifl= fi +1, and P is any partition of [a. b]. we have

(20) L(P.!I'.) +L(P,!, • • ) ,;. L(P,!, .)

,;. U(P,J..) ,;. U(P,fi••) + U(P,f, • • ).

If 11 e 91(.) and I, e 91(.). let. > 0 he given. There are partitions PJU= I. 2) snch that

6.12 Theorem(a) lffi E iJl(.) andl, E iJl(.) on [a, b]. then

fi +I, E iJl(.).

cfE 9l(rx) for every constant c, and

f.(fi +I,)d.= f./,d.+ f.I,d••

f. cld. = c f>d«·

(b) lffi(x) ,;.I,(x) on [a. b]. then

f fi d. ,;. f.I, d•.

(c) IfIe iJl(.) on [a. b] and if 0< C < b, then Ie iJl(.) on [a, c] and on[c, b], and

PROPERTIES OF THE INTEGRAL

130 PRlNCJPLI.$ OF MATHEMATICAL ANALYSIS THE RIEMANN-STIELTJIS INTEGRAL 131

Let f be continuous on [a, b]. Then

6.16 Theorem Suppose CII ~ 0 for 1, 2, 3, ... , rc" converges, {sIll is a sequenceof distinct points in (a, b), and

6.15 Theorem 1/ a < s < b, / is bounded on [a, bl, f is continuous at s, and.(x) = l(x - s), then

•I I.'(s,) - .'(t,) I dX, < e,1=1

• •I f(s,) d., = I f(s,).'(t,) /;.x,i= 1 1=1

by (28) and Theorem 6.7(b). Put M = suplf(xll. Since

where M =suplf(x)l. Since a =', + a" it follows from (24) and (25)that '

(26) Iffda -,~, c.f(s.)Is. Me.

If we let N - 00, we obtain (23).

(29)

6.17 Theorem Assume IX increases monotonically and cl E!:it on (a, b]. Let fbe a bounded realfunction on [a, b].

Theil f E Bi(.) ifand only iff.' E Bi. In that case

• •(27) f. f d. = f. f(x).'(x) dx.

Proof Let E; > 0 be given and apply Theorem 6.6 to (x': There is a partition P = {x" ... , x.1 of [a, b] such that

(28) U(P, .') - L(P, .') < e.

The mean value theorem furnishes points t i E [Xi-I, Xi] such that

d., = .'(t,) /;.x,for i = 1, ... , n. If Sl E [Xi-l> XI]. then

it follows from (29) that

(30) I.t/(S,) da, - ,t/(s,)a'(s,) /;.x,1 s. Me.

L(P,J..) = m,.

~

a(x) = I c.l(x - s.).11= 1

U(P,J. a) = M"

Proof The comparison test shows that the series (22) converges forevery x. Its sum .(x) is evidently monotonic, and a(a) = 0, .(b) = lc•.(This is the type of function that occurred in Remark 4.31.)

Let E; > 0 be given, and choose N so that

Proof Consider partitions P = {xo, Xl' X2 , x3}, where Xo = a, andXl = S < X2 < X3 = b. Then

Since f is continuous at s, we see that M 2 and m2 converge to f(s) asXl -+s.

~

Ie. < e.N+l

f: f da = f(s).

(22)

(23)

PutN

.,(x) = I c.l(x - s.),I!= 1

a,(x) = Ie. l(x - s.).N+'

In particular,

•I f(s,) d., S. U(P,J.') + Me,I=!

for all choices of Si E [XI_I. Xl], so that

U(P,/..) S. U(P,Ja') + Me.

The same argument leads from (30) to

U(P,Ja') S. U(P,/. .) + Me.

By Theorems 6.12 and 6.15,• N

(24) f. f d., = '~,c.f(s;).

Since .,(b) - a,(a) < e,

(25)(31)

ThusIU(P,/..) - U(P,J.') I S. Me.

131 PRINCIPLES OF MATH~MA.T1CA1" ANALYSIS THE RIEMANN-sTIELTJES INTEGRAL 133

Then 9 e iJ/(P) and

Since/e Bf(a), P can be chosen so that both U(P,f, a) and L(P,f, a)are close to Jfda. Hence (38), combined with Theorem 6.6, shows thatg e Bf(P) and that (37) holds. This completes the proof.

Proof To each partition P = {xo, ... , XII} of [a, b] corresponds a partitionQ = {Yo, ... , Y,} of [A, B], so that x, = q>(yi). All partitions of [A, B]are obtained in this way. Since the values taken by f on [Xl_I' xil areexactly the same as those taken by g on [Yl-l,yd, we see that

L(Q, g, P) = L(P,f, a).U(Q, g, P) = U(P,f, a),

Let us note the following special case:Take a(x) = x. Then P= q>. Assume q>' e !Jl on [A, B]. If Theorem

6.17 is applied to the left side of (37), we obtain

s: f(x) dx = S; I(q>(y»q>'(y) dy.

(37)

(39)

(38)

(32) J>da = J: I(x)a'(x) dx,

for any bounded f The equality of the lower integrals follows from (30)in exactly the same way. The theorem follows.

Now note that (28) remains true ifP is replaced by any refinement.Hence (31) also remains true. We conclude that

II:I da - I: f(x)a'(x) dxI~ Me.

But E is arbitrary. Hence

6.18 Remark The two preceding theorems illustrate the generality andflexibility which are inherent in the Stieltjes process of integration. If IX is a purestep function [this is the name often given to functions ofthe form (22)], theintegral reduces to a finite or infinite series. If a: has an integrable derivative,the integral reduces to an ordinary Riemann integral. This makes it possiblein many cases to study series and integrals simultaneously, rather than separately.

To illustrate this point, consider a physical example. The moment ofinertia of a straight wire of unit length, about an axis through an endpoint, atright angles to the wire, is

(33)

where m(x) is the mass contained in the interval [0, x]. !fthe wire is regardedas having a continuous density p, that is, if m'(x) = p(x), then (33) turns into

INTEGRATION AND DIFFERENTIATION

We still confine ourselves to real functions in this section. We shall show thatintegration and differentiation are, in a certain sense, inverse operations.

Thus (33) contains (34) and (35) as special cases, but it contains muchmore; for instance, the case in which m is continuous but not everywheredifferentiable.

6.19 lheorem (chaDge of variable) Suppose cp is a strictly increasing continuousfunction that maps an interval [A, B] onto [a, b]. Suppose a: is monotonicallyincreasing on [a,b] andle!Jl(a) on [a,b]. Define p andg on [A,B] by

On the other hand, if the wire is composed of masses m, concentrated atpoints x,, (33) becomes

(35) ~:XI mi'I

f>' p(x)dx.

F(x) = rI(t) dt.

6.20 Theorem LetfeBfon [a,b]. Fora ~x~b,put

Then F is continuous on [a, b]; furthermore, iff is continuous at a point Xo of[a, b], then F is differentiable at xo, and

F'(xo) = I(xo).

Proof Since Ie Bf, I is bounded. Suppose If(t) I ~ M for a ~ t ~ b.Ifa~x<y~b, then

IF(y) - F(x)1 = If: f(t) dtl ~ M(y - x),

by Theorem 6.12(c) and (d). Given. > 0, we see that

IF(y) - F(x)1 «,g(y) = I(q>(y»·p(y) = a(q>(y»,

(34)

(36)

134 PRINCIPLES OF MATHEMATICAL ANALYSIS THE RlEMANN-STIELTJES INTEGRAL US

that

provided that Iy - xl < <1M. This proves cOlltinuity (and, in fact,uniform continuity) of F.

Now supposef is continuous at xo. Given i > 0. choose fJ > Osuch

we have, by Theorem 6.12(d),

IF(t) - F(s) I I I f' It _ s - f(xo) = t _ s • [I(u) - f(xo)] du < <.

It follows that F'(xo) = f(xo)'

If(t) - f(xo)I<e

if It - Xo I< ~, and a;S; t ,;, b. Hence, if

INTEqRATION OF VECTOll.-VALUED FUNCTIONS

ff da = U:It da, ... , f f, do).In other words, Sf da is the point in R' whose jth coordinate is Jjj da.

It is clear that parts (a), (c), and (e) of Theorem 6.12 are valid for thesevector-valued integrals; we simply apply the earlier results to each coordinate.The same is true of Theorems 6.17, 6.20, and 6.21. To illustrate, we state theanalogue of Theorem 6.21.

6.23 Definition Letlt, ... ,f, hareal functions on [a, b], and let f = (ft, ... ,I.)be the corresponding mapping of [a. b] into R". If a increases monotonicallyon [a, b], to say thalf E !Il(a) means thatjj E !Il(a) for j = I, ... , k. If this is theease, we define

a;S; s < t ;S; b,andXo - {) <s:s;; xo;:s; t <xo + {)

6.21 The fundameotal tb""rem of ealculus If f E !Il on [a, b] and if there isa differentiable function F on [a, b] such that F' = f, then

f f(x) fix = F(b) - F(a).

Proof Let < > 0 be given. Choose a partition P = (xo, ... , x.J of [a, b]so that U(P,fJ - L(P,fJ < <. The mean value theorem furnishes points'i E [X i _ ll XI] such that

6.24 Theorem Iff andF map [a, b] into R', iff E!Il on [a, b], and ifF' = f, then

s:f(t) dt = F(b) - F(a).

The analogue of Theorem 6.13(b) offers some new features, however, atleast in its proof.

6.25 Theorem Iff maps [a, b] into R' and iff E !Il(a)forsome monotonicallyincreasing function. on [a, b], then Ifl E !Il(.), and

for i= 1, ...• n. Thus

•I fit;) lui = F(b) - F(a).'~1

It now follows from Theorem 6.7(c) that

IF(b) - F(a) - s:fix) dxl «.

Since this holds for every < > 0, the proof is complete.

6.22 Theorem (integration by parts) Suppose F and G are differentiable functions on [a, b], F' =fE!Il, and G' =g E!Il. Then

s:F(x)g(x) fix = F(b)G(b) - F(a)G(a) - ff(x)G(x) fix.

Proof Put H(x) = F(x)G(x) and apply Theorem 6.21 to H and its derivative. Note that H' E!Il, by Theorem 6.13.

(40)

(41)

(42)

Proof Ifflo ... ,f, are the components of f, then

IfI = (f; + ... +ft)">.By Theorem 6.11, each ofthefunctions!i belongs to !Il(a); hence so doestheir sum~ Since x2 is a continuous function of x. Theorem 4.17 showsthat the square-root function is continuous on [0, M]. for every real M.If we apply Theorem 6.11 once more, (41) shows that IfIE !Il(.).

To prove (40), put y = (Ylo ... ,Y,), where YJ = Sjj d•. Then we havey = Sf d., and

iYI' = Ii = IYj ffjd. = f eLY,jj) d•.

By the Schwarz inequality.

(a';' t ,;,b);

~ PRINCIPLES OF MATHQoIATICAL ~ALYSJSTHE JUEMANN-STIELTJES II\TEGRAL 137

(43)

hence Theorem 6.l2(b) implies

Iyl':s Iyl Ilflda.

If y = 0, (40) is trivial. If Y to 0, division of(43) by Iyl gives (40).

6.27 ,Theorem Ifi is continuous on [a, b), then 1 is rectifiable, and

A(y) = I: 11'(1) I dt.

Proof If a:S;; Xi_I < xl:S;; b, then

Iy(x,) - y(Xl-l) 1 = 1(/(1) dtl:s C, ly'(t)ldt.

RECTIFIABLE CURVES

We conclude this chapter with a topic of geometric interest which provides anapplication of some of tbe preceding theory. The case k = 2 (i.e., the case ofplane curves) is of considerable importance in the study of analytic functionsof a complex variable.

6.26 DefinitioD A continuous mapping y of an interval [a, b] into R' is calleda curve in K. To emphasize the parameter interval [a, b), we may also say thaty is a curve on [a, b].

If Y is one-to-one, y is called an arc.lf yea) = y(b), y is said to be a closed curve.

It should be noted that we define a curve to be a mapping, not a point set.Of course, with each curve 1 in J<l there is associated a subset of R·, namelythe range of 1, but different curves may have the same rac.ge.

We associate to each partition P = {xo, ... , x.1 of [a, b] and to eachcurve y on [a, b] the number

•A(P, y) = I Iy(xJ - y(xi-l)l·,-,

The ith term in this sum is the distance (in R') between the points y(Xl-l) andy(xJ. Hence A(P, y) is the length of a polygonal path with vertices at y(xo),y(x,), ... , y(x.), in this order. As our partition becomes finer. aqd finer, thispolygon approaches the range of 1 more and more closely. This makes it seemreasonable to define the length of y as

A(y) = sup A(P, y),

where the supremum is taken over all partitions of [a, b].If A(y) < 00, we say that y is rectifiable.In certain cases, A(y) is given by a Riemann integral. We shall prove this

for continuously differentiable curves, Le., for curves 1 whose derivative l' iscontinuous.

Hence•

A(P, y) :S I. I1'(t) I dt

for every partition P of [a, b]. Consequently,

A(y):s fi 1'(t)1 dt.

To prove the opposite inequality, let £ > 0 be given. Since l' isuniformly continuous on [a, b)~ there exists ~ > 0 such that

ly'(s)-y'(t)l« ifls-tl,<5.

Let P = {xo, ... , x.1 be a partition of [a, b], with lu, < 5 for all i. IfXl-I :s;; t ~ X" it follows that

11'(t)l:s I1'(x,) I+e.Hence

r I1'(t)1dt:S Iy'(x,) I lu, + e lu,Xl_ I

= IC}1'(I) + 1'(x,) - y'(t)] dtl +e lu,

:S 1(/(1) dtJ + Ic,[y'(X,) - y'(t)] dtl + e Ax,

,;; Iy(x,) - y(Xl-l) I +2e lu,.

If we add these inequalities, we obtain

I: Iy'(t)1 dt:S A(P, y) + 2e(b - a)

:S A(y) + 2e(b - a).Since e was arbitrary,

I:I1'(I)1 dt:sA(y).



EXERCISES

1. Suppose tx increases on [a, b], a:S;;; Xo :s;;; b, tx is con.inuous at Xo, /(xo) = I, andf(x) -0 if x "" x •. Prove thatfe 9f(a) and that f/da ~ O.

2. Suppose /;;::::0, lis continuous on [a, b], and J: lex) dx = O. Prove th,at/(x) = 0

for all x E [0, b]. (Compare this with Exercise 1.)3. Define three functions ,8",8",8, as follows: ,8Ax)~ 0 if x <0, ,8,(x) ~ I if x> 0

forj~ I, 2, 3; and ,8,(0) ~ 0, ,8,(0) ~I, ,8,(0) - i. Letfbe a boundedfunetion on[-I,IJ.(a) Prove thatfe 91(,8,) if and only ifl(O+) ~ f(O) and that then

Ifd,8, ~ f(O).

(b) State and prove a similar resuJt for fJ2.(c) Prove that/e bf(P3) if and only if/is continuous at O.(d) If/is continuous at 0 prove that

I fd,8, ~Ild,8, ~Ifd,8, ~l(O).

4. Iflex) ~ 0 for aU irrational x,/(x) ~ I for all rational x, prove thatfl' 91 on [a, bJfor any o<b.

S. Suppose f is a bounded real function on [a, bJ, and f' e 91 on [a, bJ. Does itfonow that / e [JI? Does the answer change if we assume that /3 E fJl?

6. Let P be the Cantor set constructed in Sec. 2.44. Let/be a bounded real functionon [0, 1] which is continuous at every point outside P. Prove that/ E fJl on [0, 1].Hint: P can be covered by finitely many segments whose total length can be madeas small as desired. Proceed as in Theorem 6.10.

7. Suppose/is a real function on (0,1] and/e fit on [c, 1] for every c > O. Define

f., f(x) dx = lim f.' f(x) dx

o c... o "

if this limit exists (and is finite).(a) If/e dI on [0,1], show that this definition of the integral agrees with the oldone.(b) Construct a function! such that the above limit exists, although IT fails to existwith Ifl in place off.

8. Supposef e 91 on [a, b] for every b > a where a is fixed. Define

r f(x) dx = ~ f f(x) tbr

if this limit exists (and is finite). In that case, we say that the integral on the leftconverges. If it also converges after / has been replaced by Ill, it is said to converge absolutely.

THE RWdANN-STIELTIES INTEQRAL 139

Assume that lex) ;;';?: 0 and that I decreases monotonically <In [1, co). Prove

that

ff(X)dx

converges if and only if

tf(n)..,converges. (This is the so-called uintegral test" for convergence of series.)

9. Show that integration by parts can sometimes be applied to the U improper"integrals defined in Exercises 7 and 8. (State appropriate hypotheses, formulate a

theorem, and prove it.) For jnstance show that

f.=cosx dx=f.= sinx fix.o 1 + x 0 (1 +X)2

Show that one of these integrals converges absolutely, but that the other does not.10. Let p and q be positive real numbers such that

~+!.~l.p q

PrOve the following statements.(a) If u;" 0 and v ;" 0, then

u· v'UV";p+q'

Equality holds if and only if uP = v".

(b) If fe 9f(a), 9 E 9f(a).!;" 0, 9 ;" 0, and

f' /'da ~ I ~rg'de<,. .then

rfgda";l.•

(c) If/and 9 are complex functions in 9l(tx), then

Iff9lkHf Ifl'dar{£: Igl·dar·,

This is Holder's inequality. When p=q=2 it is usually called the Schwarzinequality. (Note that Theorem 1.35 is a very special case of this.)(d) Show that Holder's inequality is also true for the "improper" integrals de

scribed in Exercises 7 and 8.


11. Let a be a fixed incroasing function nn la, b]. For u e iJl(a), define

lIull,~ ({IUI'd<·f'·Suppose/, g. h e !1t(<<.). and prove the triangle inequality

IIf-hll,';; IIf-gll, + IIg-hll,as a consequence of the Schwarz inequality. as in the proof of Theorem 1.37.

12. With the notations of Exercise 11, suppose Ie £1(<<.) and e > O. Prove thatthere exists a continuous function 9 00 [a, b] such that Ilf- gll2 < e,

Hint: Let P={xo,., '. XII} be a suitable partition of [a. b], define

x,-t t-XI 1g(t) = A::- f(XH) +~ fix,).~, ~I

ifX,_1S;t::5:X,.

13. Define

S'··fix) =, sin (t') dt.

(a) Prove that If(x)1 < I/x if x >0.

Hint: Put t 2 = u and integrate by parts, to show that/ex) is equal to

cos (x') cos [(x + I)']~- 2(x+l)

Replace cos u by -I.(b) Prove that

W(X) = cos (x') - cos [(x +I)') + rex)

where Ir(x)] < c/x and c is a constant.(c) Find the upper and lower limits of x/ex). as x _ co.

(d) Doesf: sin (I') dl converge?

14. Deal similarly witb

S'··fix) -, sin (e') dl.

Sbow that

<"I f(x) 1<2

and that

<"f(x) ~ cos (e') - e-' cos (e"') + r(x),

where !r(x)1 < Ce-", for some constant C.

THE RIEMANN-sTIELTJES INTEGRAL 141

15. Supposejis a real, continuously differentiable function on (a. b].f(a) c=j(b) = 0,and'

{f'(X) dx - I.

Prove that

(v(X)!'(X) dx ~ - i

and that

{(f'(X)]' dx' {X'f'(X) dx > i.

16. For 1 < s < co. define

" I'(3)= L-'

11 .. 1 ".

(This is Riemann's zeta function, of great importance in the study of the distri·butioo of prime numbers.) Prove that

and tbat

3 f"X-[X](b) '(3) = 3 _ I - 3 •~ dx,

where [x] denotes the greatest integer ~ x.Prove that the integral in (b) converges for all s > O.Hint: To prove (a), compute the difference between the integral over [1. N]

and-the Nth partial sum of the series that defines '(3).17. Suppose «. increases monotonically on [a, b], 9 is continuous. and g(x) = G'(x)

for a s: x ::5: b. Prove that

{a(x)g(X) dx = G(b)a(b) - G(a)a(a) - {G da.

Hint: Take 9 real, without loss of generality. Given P = {xo, Xl> ••• , XII}'

choose t, E (x'_1> x,) so thatg(t,) Ax, = G(X,) - G(X'_l). Show that

:t a(x,)g(t,) tJ.x, - G(b)a(b) - G(a)a(a) - f G(x. _0) <la,.'''1 l~

18•.Let ;'1' ;'2, ;'3 be curves in the complex plane, defined on [0.2.".] by

;'1(t) = e", Y2(t)=e%lr. ",(t) = e2KII 11K (l/r).

Show that these three curves have the same range, that ;'1 and "2 are rectifiable,that the length of"l is 21T, that the length of;'2 is 41T, and that;'3 is not rectifiable,

142 PRINCIPLES OF MA:rHEMAnCAL ANALYSIS

19. Let ')'1 be a curve iII Ri• defined on [a, b); let <P be a continuous 1-1 mapping of

(c. d] onto [a, bl. such that "'(c) ~ a: and define y,(s) ~'YM(s». Prove that y, isan arc. a closed curve. or a rectifiable curve if and only if the same is true of ')'1'

Prove that ')'2 and ')'1 have the same length.

7SEQUENCES AND SERIES OF FUNCTIONS

In the present chapter we confine our attention to complex-valued functions(including the real-valued ones, of course), although many of the theorems andproofs which follow extend without difficulty to vector-valued functions, andeven to mappings into general metric spaces. We choose to stay within thissimple framework in order to focus attention on the most important aspects ofthe problems that arise when limit processes are interchanged.

DISCUSSION OF MAIN PROBLEM

7.1 Definition Suppose (/J. n = 1,2,3, ...• is a sequence of functionsdefined on a set E, and suppose that the sequence of numbers (j,(x)} convergesfor every x E E. We can then define a function I by

(1) I(x) = lim[.(x)"~~

(x E E).

144 PRlHClPu.s OF MATHEMATICAL ANALYSIS SEQUENCES AND SERIES OF FUNCTIONS 145

Under these circumstances we say that tf.} collverges on E and that f isthe limit, or the limit function, of{,(,.}. Sometimes we shall use a more descriptiveterminolngy and shall say that "{fJ converges tnfppintwise on E" if (I) holds.Similarly, if IJ,.(x) converges for every x E E, and if we define

On the other hand, for every fixed m,

Urns.,,, = 0,,-~

(2)~

f(x) = LJ.(x),-I

(x E£),

so that

(5) lim lim s.,,, = O.lII __ ca "... co

the function f is called the sum of the series T.J. .The main problem which arises is to determine whether important

properties of functions are preserved under the limit operations (I) and (2).For instance, if the functions f. are continuous, or differentiable, or integrable,is the same true of the limit function? What are the relations betweenf~ and f',say, or hetween the integrals ofJ. and that off?

To say that/is continuous at a limit point x means

limf(t) = f(x).,~.

7.3 Example Let

x'J.(x) = (I + x')'

and consider

(6)

(x real; n = 0, I, 2, " .),

Hence, to ask whether the limit of a sequence of continuous functions is continuous is the same as to ask whether

Sincef,(O) = 0, we havef(O) = O. For x '" 0, the last series in (6) is a convergentgeometric series with sum I + Xl (Theorem 3.26). Hence

Le., whether the order in which limit processes are carried out is immaterial.On the left side of (3), we first let n .... co, then t .... x; on the right side, t .... xfirst, then n- 00.

We shan now show by means of several examples that limit processescannot in general he interchanged without affecting the result. Afterward, weshall prove ~t under certain conditions the order in which limit operationsare carried out is immaterial.

Our first example, and the simplest one, concerns a "double sequence."

so that a convergent series of continuous functions may have a discontinuoussum.

(x = 0),

(x '" 0),f(x) = (~+ x'(7)

7.4 Example For m = I, 2, 3, ... , put

f.(x) = lim (cos m!nx)".,~~

lim limJ.(t) = lim limJ.(t),(3)

f(x) = lim f.(x).

When m!x is an integerJ.(x) = I. For aU other values of xJ.(x) = O. Now let

For irrational x, f.(x) = 0 for every m; hence f(x) = O. For rational x, sayx =pjq, where p and q are integers, we see that m!x is an integer if m ~ q, sothatf(x) = I. Hence

7.2 Example For m = 1,2,3, ... , n = 1,2,3, .. " let

ms =-_.-," m+n

Then, for every fixed n,

lim s.,. = 1,.~~

so that(8) lim lim(COSm!nx)"=(~

III'" co ,, __ co

(x irrational),(x rational).

(4) lim lim s.,,, = 1..... ca ."'00

We have thus obtained an everywhere discontinuous limit function, whichis not Riemann-integrable (Exercise 4, Chap. 6)•

146 PRINCIPLEs OF MATHI!MA~CAL ANItLYSIS SEQUENCES AND SERIES OF PlJ"CTIONS 147

[(x) = liml.(x) = O.

7.5 Example Let

(9)

and

[.( )_sinnx.x _

.jn (xreal, n = 1,2,3, ...),

Thus !lie limit of the integral need not be equal to the integral of the limit,even if both are finite.

After these examples, which show what can go wrong if limit processesare interchanged carelessly, we now define a new mode ofconvergence, strongerthan pointwise convergence as defined in Definition 7.1. which will enable us toarrive at positive results.

Then('(x) = 0, and

whereas

A simple calculation shows that

lim!.(x) = 0,

Thus, in spite of (11),

•I fi(x) = s.(x)1=1

so that

•I!.(x) - [(x) I :s; 2'

UNIFORM CONVERGENCE

converges uniformly on E.The Cauchy criterion for uniform convergence is as follows.

7.8 Theorem The sequence offunctions U;,}. defined on E, converges uniformlyon E if and only if for every s > 0 there exiSts an integer N such that m ~ N,n ~ N, x E E implies(13) I!.(x) - [mix) I :s; •.

Proof Suppose {!.} converges uniformly on E, and let [ be the limitfunction. Then there is an integer N such that "~ N. x e E implies

7.7 Definition We say that a sequence of functions (f.), n = 1,2,3, ... ,converges uniformly on E to a functionfif for every s > 0 there is an integer Nsuch that n ~ N implies(12) II.(x) - fix) I :S;.for all x E E.

It is clear that every uniformly convergent sequence is pointwise convergent. Quite explicitly, the difference between the two concepts is this: If(!.)converges pointwise on E, then there exists a function / such that, for everys > 0, and for every x E E. there is an integer N, depending on s and on x, suchthat (12) holds if n :? N; if(!.} converges uniformly on E, it is possihle, for each• > 0, to find one integer N which will do for all x E E.

We say that the series rl.(x) converges uniformly on E if the sequence{s.} of partial sums defined by

(O:s; x:s; I, n = 1,2,3, ...).

7.6 Example Let

(10) I.(x) = n'x(1 - xl)·

For 0 < x .s; 1, we have

(1 n2J. !.(x) dx = -- ~ + ex>o 2n +2

[:(x) = J;; cos nx,

so that [f~} does not converge tof'. For instance.

f:(O) = .jn~ + ex>

as n ~ ex>, whereas ('(0) = O.

f 1 IJ, x(1 - x')· dx = -_.o 2n+2

by Theorem 3.20(d). Since!.(O) = 0, we see that

(II) liml.(x) = 0 (O:s; x :s; I)..~~

as n_ 00.

If, in (10), we replace n' by n, (11) still holds, but we now haveI

lim i !.(x) dx = lim _n_ = ~,""'co 0 ,._«) 2n + 2 2

{ [liml.(X)] dx = O.o ,,-to«)

I!.(x) - [mix) I :s; I!.(x) - [(x) I + If(x) - [mix) I ,;.ifn ~ N,m ~ N. x eE.

148 PRINCIPLES Of MATHEMATICAL ANALYSIS SEQUENCES AND SERIES Of j'UNCTIONS 149

Conversely. suppose the Cauchy condition holds. By Theorem 3.11.the sequence (f.(x)) converges. for every x, to ~ limit which we may callJ(x). Thus the sequence (f.) converges on E, tof, We have to prove thatthe convergence is uniform.

Let' > 0 be given, and choose N such that (13) holds. Fix n, andlet m _ 00 in (13). Since J.(x) _ J(x) as m - 00. this gives

UNIFORM CONVERGENCE AND CONTINUITY

7.11 Theorem Suppose I,. --+ juni/ormly on a set E in a metric space. Let x bea limit point of E, and suppose "that(15) limf.(l) = A. (n = 1.2,3, ...)..-.Then {All} converges, and

(16)(14) If.(x) - J(x) I ".

for every n :2: N and every x e E, which completes the proof.

The following criterion is sometimes useful. (17)

limJ(I) = lim A•.

In other words, the conclusion is that

lim limf.(l) = lim limf.(I).f ... .II: .... ao ,,"ao '-+.11:

Then J. - J uniformly on E ifand only if M. - 0 as n - 00.

For series, there is a very convenient test for uniform convergence, due toWeierstrass.

Since this is an immediate consequence of Definition 7.7, we omit thedetails of the proof.

M. = sup If.(x) - f(x) I.u.

•If(l) - 1.(1) I "3(20)

Proof Let' > 0 be given. By the uniform convergence of (f.). thereexists N such that n ;,: N. m ;,: N, leE imply

(18) 11.(1) - J.(I) I " •.Letting I _ x in (18). we obtain

IA. - A.I'"for n;': N, m;" N. so that (A.) is a Cauchy sequence and therefore

converges, say to A.Next,

(19) If(l) - A I " If(l) - J.(t) 1+ If.(I) - A.I + IA. - A I·We first choose n such that

(xeE).limf.(x) = J(x).-~7.9 Theorem Suppose

Put

7.10 Theorem Suppose{J,.} is a sequence offunctions defined on E, and suppose

provided m and n are large enough. Uniform convergence now followsfrom Theorem 7.8.

Then T.IlI converges uniformly on E lfIMJI converges.

Note that the converse is not asserted (and is, in fact, not true).

Proof If IMlI converges, then, for arbitrary e > 0,

(xeE.n=I.2.3....).

Then, for this n, we choose a neighborhood V of x such that

for all teE (this is possible by the uniform convergence), and such that

•lAo - A I" 3'(21)

if t e V" E. ti'x.Substituting the inequalities (20) to (22) into (19). we see that

If(l) - A I "'.provided I E V" E. ti'x. This is equivalent to (16).

•(22) 11.(1) - A.I "3

(x eE).l.tpX)I",~M,,, •

If.(x) I "M.

150 PRINCIPLES OF MATHI!.MATICAL ANALYSIS

7.12 Theorem If{!.l is a sequence of continuous functions on E, and iff• .... funiformly on E, then f is continuous on E.

This very important result is an immediate corollary of Theorem 7.11.The converse is not true; that is, a sequence of continuous functions may

converge to a continuous function, although the convergence is not uniform.Example 7.6 is of this kind (to see this, apply Theorem 7.9). But there is a casein which we can assert the converse.

7.13 Theorem Suppose X is compact, and

(a) Lf.J is a sequence of continuous functions on X,(b) (!.l converges pointwise to a continuousfunctionfon K,(c) f.(x) ;;,;!.+,(x)for all x eK, n = 1,2,3, ....

Then!. .... f uniformly on K.

Proof Put gn =III - f. Then gil is continuous, gil - 0 pointwise, andgn ~ g,,+I' We have to prove that gil _0 uniformly on X.

Let • > 0 be given. Let K. be the set of all x e K with g.(x) ;;,; •.Since gil is continuous, X" is closed (Theorem 4.8), hence compact (Theorem2.35). Since UII ~ U,,+h we have XII ~ X,,+I' Fix x E X. Since gll(x) -0,we see that x ¢ K. ifn is sufficiently large. Thus x ¢ nK•. In other words,nK. is empty. Hence KN is empty for some N (Theorem 2.36). It followsthat 0 ,; g.(x) < • for all x e K and for all n ;;,; N. This proves the theorem.

Let us note that compactness is really needed here. For instance, ifI

!.(x) =.nx + I (0 < x < I; n = I, 2, 3, ...)

thenf.(x) .... 0 monotonically in (0, I), but the convergence is not uniform.

7.14 Definition If X is.a metric space, 'l(X) will denote the set of all complexvalued, continuous, bounded functions with domain X.

[Note that boundedness is redundant if X is compact (Theor.cm 4.15).Thus 'l(X) consists of all complex continuous functions on X if X is compact.]

We associate with eachfe 'l(X) its supremum norm

IIfll = sup If(x) I·uX

Since f is assumed to be bounded, Ilfll < 00. It is obvious that Ilfll = 0 only iff(x) = 0 for every x e X, that is, only iff= O. Ifh = f + g, then

Ih(x) I ,; If(x) I+ Ig(x) I ,; Ilfll + Ilgll

for aU x E X; hence

Ilf+ gil ,; Ilfll + Ilgll·

SEQUENCES AND SElUES OF FUNCTIONS 151

If wedefine the distance between fe 'l(X) and.g e'l(X) to be III-gil,it follows that Axioms 2.15 for a tnetric are satisfied.

We have thus made fI(X) int9 a metric space.Theorem 7.9 can be rephras~d as follows:

A sequence Lf,,} converges to I with respect to the metric 01 ~(X) if andonly if!. .... I uniformly on X.

Accordingly, closed subsets of 'l(X) are sometimes called uniformlyclosed, the closure of a set d c ~(X) is called its uniform closure, and so on.

7.15 Theorem The above metric makes 'C(X) into a complete metric space.

Proof Let (!.l be a Cauchy sequence in 'l(X). This means that to each• > 0 corresponds an N such that II!. - 1.11 <. if n;;" Nand m ;;,; N.It follows (by Theorem 7.8) that there is a functionfwith domain X towhich (!.l converges uniformly. By Theorem 7.12, I is continuous.Moreover, f is bounded, since there is an n such that If(x) - I.(x) I < Ifor all x e X, and/. is bounded.

Thus Ie 'C(X), and since !. ....I uniformly on X, we have111-!.II .... Oas n .... 00.

UNIFORM CONVERGENCE AND INTEGRATION

7.16 Theorem Let a be monotonicolly increasing on [a, b]. Suppose!. e BI(a)on [a, b),jor n = I, 2, 3, ... , andsuppose!. ....funiformly on [a, b]. Thenf e BI(a)on [a, b), and

• •(23) f Ida = lim f !. da.

II .... ao /I

(The existence of the limit is part of the conclusion.)

Proof It suffices to prove this for real!.. Put

(24) '. = sup I!.(x) - f(x) I,

the supremum being taken over a :S x .s b. Then

f" - t" s.f:sf" + til'

so that the upper and lower integrals of/(see Definition 6.2) satisfy. -.(25) f (f. - •.) da ,;fI da " fIda"f (f. +e.) da.. - .

Hence

o,;Jfda - fIda" 2••[a(b) - a(a)].

152 PIUNCIPLES OF MA~nCAL ANALYSISSEQUENCES AND SERIES OF FUNCTIONS 1S3

Since ..... 0 as ~ ... 00 (Theorem 7.9), the upper ¥nd lower integrals offare equal.

Thusfe !R(a). Another application of (25) now yields

(26) If> cia - s: f.daI,;; •.[a(b) - a(a)].

This implies (23).

CoroUary Iff, e !R(a) on [a, b] and if~

f(x) = L f,(x)._1 (a';;x,;;b),

(30)

Ifwe apply the mean value theorem 5.19 to the function!. - fm, (29)shows that

Ix - II. •I!.(x) - fm(x) - !.(I) + fm(t) I ,;; 2(b _ a) ,;; 2.

for any x and I on [a, b], if n~ N, m ~ N. The i~equality

I!.(x) - fm(x) I ,;; I!.(x) - fm(x) - !.(x.) + fm(x.) I + I!.(x.) - fm(x.) Iimplies, by (28) and (30), that

I!.(x) - fm(x) I <. (a ,;; x ,;; b, n ~ N, m ~ N),

so that {f,} converges uniformly on [a, b]. Let

Let us now fix a point x on [a, b] and define

the series converging uniformly on [aJ b], then

In other words, the series may be integrated term by term.

UNIFORM CONVERGENCE AND DIFFERENTIATION

(31)

(32)

f(x) =limf,(x)

4>,(t) = !.(t) - !.(x) ,I-x

for a ,;; I ,;; b, I '" x. Then

lim 4>,(1) = f:(x)I~x

(a,;;x';;b).

4>(1) = f(l) - f(x)I-x

(n = 1,2,3, ...).

We have already seen, in Example 7.5, that uniform convergence of{f,} impliesnothing about the sequence (f;}. Thus stronger hypotheses are required for theassertion thatf: ... f'iff, ...f.

The first inequality in (30) shows that

•14>,(1) - 4>m(t) I ,;; 2(b _ a) (n~N,m~N),

7.17 Theorem Suppose (f,} is a sequence offunctions, differentiable on [a, b]and such Ihal (!.(x.)} converges for some poinl x. on [a, b]. If (f;} convergesuniformly on [a, b], then {f,,} converges uniformly on [aJ b], to afunction/, and

so that (4),} converges uniformly, for t '" x. Since (!.} converges to f, weconclude from (31) that

(33) lim 4>,(1) = 4>(1)

•If,(x.) - fm(x.) I < 2.

Proof Let. > 0 be given. Choose N such that n ~ N, m ~ N, implies

(27)

(28)

f'(x) = limf:(x) (a';;x,;;b).uniformly for a ,;; I ,;; b, I '" x.

If we now apply Theorem 7.11 to (4),}, (32) and (33) show that

lim 4>(1) = limf;(x);

and this is (27), by the definition of 4>(t).

(29)

and

eIf;(I) - f~(t) I< 2(b _ a) (a';;I';;b).

Remark: If the continuity of the functionsf: is assumed in addition tothe above hypotheses, then a much shorter proof of (27) can be based onTheorem 7.16 and the fundamental theorem of calculus.

154 PRINCIPLES OF MATIiEMATlCAL ANALYSIS SEQUENCES AND SEJUES Of f~NcnONS ISS

(39)

By Lebesgue's theorem concerning integration of boundedly convergentsequences (Theorem 11.32), (40) implies

IJ.(x) I < q,(x) (x E E, n = 1,2,3, ...).

We say that (I.} is uniformly bounded on E if there exists a number Msuch that

7.19 Definition Let (I.} be a sequence of functions ~efined on a set E.'We say that(I.} ispoin/wise bounded on Eifthe sequence (J.(xl) is bounded

for every x E £, that is, jf there; exists a finite-valued function r/J defined on Esuch that

(0 :s; x :s; 2n).

(O:s; x :s; 2n);

2,lim J (sin nllx - sin nk:+1x)1 dx = O.

.I; ... ao 0

lim (sin n"x - sin nk:+1x)1 = 0.-~

lim (sin nkX - sin n.t+1x) = 0.~~

If,(x) I < M (x E E, n = 1,2,3, ...).

Now if {f,,} is pointwise bounded on E and £1 is a countable subset of E,it is always possible to find a subsequence (f.,} such that (J.,(x)} converges forevery x E E,. This can be done by the diagonal process which is used in theproof of Theorem 7.23.

However, even if {f,,} is a uniformly bounded sequence of continuousfunctions on a compact set E, there need not exist a subseG:Jence which converges pointwise on E. In the following example, this would be quite troublesome to prove with the equipment which we have at hand so far, but the proofis quite simple if we appeal to a theorem from Chap. 11.

hence

7.20 Example Let

J.(x) = sin nx (0 :s; x :s; 2n, n = 1,2,3, ...).

Suppose there exists a sequence {naJ such that {sin n.tx} converges, for everyx E [0, 2n]. In that case we must have

(40)

(41)

(34) q>(x) = Ix I (-I :s; x ~ I)

and extend the definition of q>(x) to all real x by requiring that

(35) q>(x + 2) = q>(x).

Then, for all sand t,

(36) Iq>(s) - q>(t) I :s; Is - / I.In particular, cp is continuous on R1. Define

7.18 Theorem There exists a real continuous function on the rea/line which isnowhere differentiable.

Prool Define

~

(37) f(x) = L W'q>(4'x).,_0

If(X + b.) - f(x)I= If (~)'y,1t5", ,,=0 4

.-1

;>:3·-L3',-0

= t(3· + I).

As m -+ 00, b. -+ O. It follows thatf is not differentiable at x.

Since O:s; q> :s; I, Theorem 7.10 shows that the series (37) convergesuniformly on R1

• By Theorem 7.I2,fis continuous on R1•

Now fix a real number x and a positive integer m. Put

(38) b. = ± t· 4-·

where the sign is so chosen that no integer lies between 4"'x and 4"'(x + ()",).This can be done, since 4· lb. I = t. Define

q>(4'(x + b.)) - q>(4'x)

"I" ()'"

When n > m, then 4"()". is an even integer, so that 'Y. = O. When 0::::;; n ::::;; m,(36) implies that ly,1 :s; 4'.

Since l'YIn I = 4"', we conclude that

EQUICONTINUOUS FAMILIES OF FUNCTIONS

In Theorem 3.6 we saw that every bounded sequence of complex numberscontains a convergent subsequence, and the question arises whether somethingsimilar is true for sequences of functions. To make the question more precise,we shall define two kinds of boundedness.

But a simple calculation shows that

2,J (sin n.l;x - sin nlt.+lx)1 dx = 211',o

which contradicts (41).

156 PR,lNCIPLES OF MATHEMAnCAL ~ALYSIS

Another question is whether every conyergent sequence contains auniformly convergent subsequence. Our next example will show that thisneed not be so, even if the sequence is uniformly bounded on a compact set.(Example 7.6 shows that a sequence of bounded functions may convergewithout being uniformly bounded; but it is trivial to see that uniform convergence of a sequence of bounded functions implies uniform boundedness.)

7.21 Example Let

but

(n = 1,2,3, ...),

so that no subsequence can converge uniformly on [0,1].

The concept which is needed in this connection is that of equicontinuity;it is given in the following definition.

7.22 Definition A family ~ of complex functions f defined on a set E in ametric space X is said to be equicontinuous on E if for every e > 0 there exists a6 > 0 such that

If(x) - f(y) I «

whenever d(x,y) < 6, x e E, y eE, andfe~. Here d denotes the metric of X.It is clear that every member of an equicootinuous family is uniformly

continuous.The sequence of Example 7.21 is not equicontinuous.Theorems 7.24 and 7.25 will show that there is a very close relation

between equicontinuity. on the one hand, and uniform convergence of sequencesof continuous functions. on the other. But first we describe a selection processwhich has nothing to do with continuity.

7.23 Theorem q{fJ is a pointwise bounded sequence of complex functions on. a countable set E, then {f.} has a subsequence {f.,} such that {f.,(x)} convergesforevery X eE.

SEQUENCES AND SERIiOS OF FUNCTIONS 157

Prool Let {x,}, i = l, 2, 3, ... , be the points of E, arranged in a sequence.Since (I.(x,)} is bounded,. there exists a subsequenc&, which we shalldenote by U; .•}, such that U; .,(xl )} converges as k _ <Xl:

Let us now consider sequences SI> S2' SJ, ... , which we representby the array .

S,: I, .' 1,., 1", I, "S2: f,.1 f", f", f".S,: f"l f", f", f".

..and which have the following properties:

(a) S" is a subsequence of S"-1' for n = 2, 3, 4, ....(b) (I. .•(x.)} converges, as k - <Xl (the boundedness of (I.(x.)}makes it possible to choose S" in this way);(c) The order in which the functions appear is the same in each sequence; Le.• ifone function precedes another in SIt t"ey are in the samerelation in every SII' until one or the other is deleted. Hence. whengoing from one row in the above array to the next below. functionsmay move to the left but never to the right.

We now go down the diagonal of the array; i.e.• we consider thesequence

S: 1", f", f,., f •.• ···,

By (c), the sequence S (except possibly its first n - I terms) is a subsequence of S., for n = 1,2,3, .... Hence (b) implies that (I. .•(x,)}converges. as n -+ co, for every X, e E.

7.24 Theorem IfK is a compact metric space, iff,. e fl(K)jor n = 1,2,3•....and if{fJ converges uniformly on K, then{fJ is equicontinuous on K.

Prool Let e > 0 be given. Since {.£.} converges uniformly, there is aninteger N such that

(42) III. - fN11 « (n > N).

(See Definition 7.14.) Since continuous functions are uniformly con~

tinuous on compact sets. there is a b > 0 such that

(43) If'(x) - jj(y) I «

if 15, i,;,N and d(x,y) < O.If n > Nand d(x, y) < 0, it follows that

II.(x) - I.(y) I ,;, II.(x) - fNCx) I + IMx) - fN(y) I + If.(Y) - I.(Y) I < 3•.

In conjunction with (43), this proves the theprem.

158 PaINCIPLES OF MATHEMATICAL ANALYSISSEQUENCES AND SERJES OF FUNCTIONS 19

7.'1.5 Theorem If K is compact, IfJ. e 'e(K) for n ~ I, 2, 3, ••.• and if{J.} i'pointwise bounded and equicontinuous on K, then

(a) {J.} is uniformly bounded on K,(b) {!.} containa a uniformly convergent subsequence.

Proor

(a) I.et. > 0 be given and choose 6 > 0, in accordance with Definition7.22, so that

(44) I!.(x) - !.(Y) I <£

for all n. provided that d(x,y) < 6.Since K is compact, there are finitely many points PI • ... , p, in K

such that to every x e K corresponds at least one P, with d(x.pJ < 6.Since{!.} is pointwise bounded, there exist M, < ClO such that I!.(PJ I < M,for all n. If M ~ max (M" ... , M,), then 1!.(x)1 < M +. for everyx e K. This proves (a).(b) Let E be a countabie dense subset of K. (For the existence of such aset E. see Exercise 25, Chap. 2.) Theorem 7.23 shows that {!.} has asubsequence {I.,} such that{J.,(x)} converges ror every x e E.

Put f., ~g .. to simplify the notation. We shall prove that (gilconverges uniformly on K.

I.et • > 0, and pick 6 > 0 as in the beginning of this proof. LetVex, 6) be the set of all y e K with d(x, y) < 6. Since E is dense in K, andK is compact, there are finitely many points XI' ... , x. in E such that

THE STONE·WEIERSTRASS THEOREM

uniformly on [a, b]. Iff is real. the P, may be taken real.

This is the form in which the theorem was originally discovered byWeierstrass.

7.26 Theorem If/is a cotttinuous complex/unction on [a,b], there exists asequence 0/polynomials PII su~h that

iim P,(x) ~ f(x)

(O:>:x:>: i).

(n ~ i; 2. 3, ...),

(n ~ i. 2. 3, ...).

g(x) ~ f(x) - f(O) - x[f(l) - flO)]

Q,(x) ~ c.(i - x')'

where CII is chosen so that,f Q,(x) dx ~ i-1

Here g(O) ~ g(I) ~ 0, and if 9 can be obtained as the limit of a uniformlyconvergent sequence of polynomials, it is clear that the same is true for t.since/ - 9 is a polynomial.

Furthermore. we definef(x) to be zero for x outside [0, i]. Thenfis uniformly continuous on the whole line.

We put

Proof We may assume, without ioss of generality, that [a. b] ~ [0, I].We may also assume thatf(O) ~f(l) ~ O. For if the theorem is provedfor this case, consider

(48)

(47)

K c V(x" 6) u ... u V(x., 6).(45)

for every i. If i <? N andj <? N, it follows from (46) that

(46) Ig,(x.) - gjx.) I <.

Ig,(x) - g.(x;) I <.

whenever i ~ N,j ~ N, 1 :::;; s :::;;m.If x e K. (45) shows that x e V(x.. 6) for some a, so that

We need some information about the order of magnitude of CII' Since

f (I - x')' dx ~ 2( (I - x')' dx <? 2I'/JO (I - x')' dx-I 0 0

II/.fft

<? 2 (I - nx') dxo

4~3';;'

1>-..;;.

it follows from (48) that

Since {g ,(x)} converges for every x e E. there is an integer N suchthat

Ig,(x) - gjx) I :>: Ig.(x) - g.(x.) I+ Ig.(x.) - gjx;) I+ Igjx;) - gjx) I<3£.

This compietes the proof.(49)

160 PJUNCIPLES OF MATHEMATICAL ANALYSIS SEQUENCES AND SERIES OF FV!"CTlONS 161

Our assumptions about/show, by a simple change of variable, that

The inequality (I - X2)1I ~ I - nx2 which ~e used above is easily

shown to be true by considering the function

(I-x'l"-l +nx'

which is zero at x = 0 and whose derivative is positive in (0, I).For any fJ > 0, (49) implies

(50)

(51)

Q.(x) ",; ,;n (1 - fJ'l"

so that Q. ~ 0 uniformly in fJ ",; Ix I ",; l.Now set

1

P.(x) = I f(x + I)Q'(I) dl-1

(a",; Ixl ",; I),

(O",;x"';I).

In the proof of Theorem 7.32 we shall not need the full strength ofTheorem 7.26, but only the following special case, which we state as a corollary.

7.27 Corollary For every interval [- a, aJ there is a sequence of real poly-nomials P II such that PiO) = °and such that .

lim P'(x) = Ix Iuniformly on [ - 0, a].

Proor By Theorem 7.26, there exists a sequence {P:} of real polynomialswhich converges to Ix I uniformly on [- a, oj. In particular, P:(O) -+ °as n -+ 00. The polynomials

p.(x) = P:(x) - P:(O) (n = I, 2, 3, ...)

have desired properties.

We shall now isolate those properties of the polynomials which makethe Weierstrass theorem possible.

where c is any constant, the convergence being uniform in each case.Hence f + g e ilI,fg e ill, and cfe ill, so that ill is an algebra.By Theorem 2.27, ill is (uniformly) closed.

7.29 Theorem LeI ill be the uniform closure of an algebra d of boundedfunctions. Then ill is a uniformly closed algebra.

Prool If f E til and g E U1, there exist uniformly convergent sequences{/.},{g.} such that/. ....f,g.~g and/. ed,g. Ed. Since we are dealingwith bounded functions, it is easy to show that

c/II -+ cj;/. +g.....f +g,

7.28 Definition A family ,r;/ of complex functions defined on a set E is saidto be an algebra if (i)f+ g ed, (ii)fg ed, and (iii) cfed for allfed, g e dand for all complex constants c, that is, if d is closed under addition, multiplication, and scalar multiplication. We shall also have to consider algebras ofreal functions; in this case, (iii) is of course only required to hold for all real c.

If d has the property that fed whenever /. e d (n = I, 2, 3, ...) andf • ....f uniformly on E, then d is said to be uniformly closed.

Let til be the set of all functions which are limits of uniformly convergentsequences of members of d. Then til is caned the uniform closure of d. (SeeDefinition 7.14.)

For example, the set of all polynomials is an algebra, and the Weierstrasstheorem may be stated by saying that the set of continuous functions on [a, hJis the uniform closure of the set of polynomials on [a, h].

If(Y) - f(x) I <~.

Let M = sup If(x) I. Using (48), (SO), and the fact that Q.(x) ~ 0, wesee that for 0 ",; x ",; I,

lP'(x) - f(x) 1= 1([f(X + I) - f(X)]Q.(I)dll

1,;; t If(x + I) - f(x) IQ'(I) dl

I-' 8I' Jl,;; 2M -1 Q.(I) dl + 2 _, Q.(I) dl + 2M. ,.Q-!:) dl

8",; 4M,;n (I - a')' +2

<8

1-% 1

p.(x) =L. f(x + I)Q.(I) dl = I/(t)Q.(t - x) dl,

and the last integral is clearly a polynomial in x. Thus {p.} is a sequenceof polynomials, which are real iff is real.

Given 8 > 0, we choose a> 0 such that Iy - x I < a implies

for all large enough n, which proves the theorem.

It is instructive to sketch the graphs of QII for a few values of n; also,note that we needed uniform continuity of / to deduce uniform convergenceof{P.}.

161 PRINCIPLES OF MATHEMATICAL ANALYSIS SEQUENCES AND SER.IES OF FUN<-1I0NS 163

(-a:O;y:O;a).

7.30 DeliDitiOD Let '" be a family of fUDCtiOns OD a set E. Then'" is saidto separate points on E if to every pair of distinct points xJ' x2 E E there corresponds a functionfE'" such thatf(x,) "'f(x,).

If to each x E E there corresponds a function 9 E'" such that g(x) ,., 0,We say that sI vanishes at no point ofE.

The algebra of all polynomials in ODe variable clearly has these propertieson R'. An example of an algebra which does not separate points is the set ofall even polynomials, say on [-I, I), since f( - x) = f(x) for every even function f

The following theorem will illustrate these concepts further.

(53)

and let e > 0 be given. By Corollary 7.27 there exist real numberscJ, ••• , CII such that

l,tctY'- IYII <e

Since 11 is an algebra, the function

g= tc,f',-,

u =gk - g(x,)k, v =gh - g(x,)h.

Then u Ed, v Ed, u(x,) = v(x,) = 0, u(x,),., 0, and vex,) ,., O. Therefore

7.31 Theorem Suppose'" is an algebra offunctions on a set E, '" separatespoints on E, and sI vanishes at no point of E. Suppose XJ' X2 are distinct pointsof E, and c1, C2 are constants (real if d is a real algebra). Then d contains afunction f such that

·!(x,) = c" f(x,) = c,.

Proof The assumptions show that '" contaiDs functions g, h,such that

hex) = ff(x)Ig(x)

By max (f, g) we mean the function h defined by

iff(x) ;,: g(x),if/ex) < g(x),

is a member of 111. By (52) and (53), we have

!g(x) - If(x) II <e (x E K).

Since iJI is uniformly closed, this shows that IfI E 111.

Proof Step 2 follows from step 1 and the identities

and min (f, g) is defined likewise.

STEP 2 IffE iJI and 9 E iJI, then max(f,g) E iJI and min(f, g) E 111.

and k

k(x,)'" O.g(x,) ,., g(x,),

Put

STEP 3 Given a real function J, continuous on K, a point x E K, and 6 > 0, thereexists a function g. E iJI such that g.(x) = f(x) and

Proof Since d c iJI and d satisfies the hypotheses of Theorem 7.31 sodoes 11. Hence, for every y E K, we can find a function hy E!JI such that

By iteration, the result can of course be extended to any finite setof functions: Iff" ... J. E iJI, then max (f" ... ,[.) E 111, and

, min (Itt ... ,In) E lB.

(t E J().g.(t) > f(t) - e

f+g If-glmax (f, g) = -2- + -2-'

. f+g If-glmm(f,g)=-2-- -2-'

(54)

STEP I Iff E 111, then IfI E 111.

Proof Let

We now have all the material needed for Stone's generalization of theWeierstrass theorem.

has the desired properties.

7.32 Theorem Let d be an algebra of real continuous functions on acompactset K. If'" separates points on K and if d vanishes at no point ofK, then theuniform closure 1M ofd consists ofall real continuous functions on K.

We shall divide the proof into four steps.

(52) a = sup If(x) I (xEK) (55) h,(x) = f(x), h,(Y) = f(Y).

SEQUENCES AND SERIES OF fUNcnONS 165

Since fJI is uniformly closed, this statement is equivalent to the conclusionof the theorem.

STEP 4 Given a realfunction/, continuous on K, and e > 0, there exists afunctionh e fJI such thaI

Put

By the continuity of h, there exists an open set J" containing y,such that

Proof Let d R be the set of all real functions on K which belong to d.Iffe d andf = u + iv, with u, v real, then 2u =f+I, and since d

is self-adjoint, we see that u E .91R' If Xl ;j:: X2, there exists f E.sJ suchthatf(x,) = l,f(x,) = 0; hence 0 = u(x,) '" u(x,) = I, which shows thatd R separates points on K. If x e K, then 9(X) '" 0 for some 9 ed, andthere is a complex number A. such that 19(x) > 0; iff= ).g,/= u + iv, itfollows that u(x) > 0; hence d R vanishes at no point of K.

Thus d R satisfies the hypotheses of Theorem 7.32. It follows thatevery real continuous function on K lies in the uniform closure of .91R,

hence lies in fJ. If f is a complex continuous function on K, f = u + iv,then u e fJI, v e fJI, hence f e fJI. This completes the proof.

Theorem 7.32 does not hold for complex algebras. A couIJterexample isgiven in Exercise 21. However, the conclusion of the theorem dqes hold, evenfor complex algebras, if an extra condition is imposed on sI, namely, that ,r;Ibe self-adjoint. This means that for every fed its complex conjugate I mustalso belong to d; I is defined by lex) = f(x). .

7.33· Theorem Suppose.rtf is a self-adjoint algebra of complex continuousfunctions on a compact set K, .91 separates points on K, and d vanishes at nopoint of K. Then the uniform closure 1M of.rtl consists of all complex r.ontinuousfunctions on K. In other words, .rtf is dense <t(K).

(I eJ,).

(x e K).

h,(t) > f(l) - •

Ih(x) - f(x) I <.

Since K is compact, there is a finite set of points Yh ... , y" such that

Kc:J'lU···uJ,...

g~ = max (h'l ' .. . I h,,).

By step 2, 9, e fJI, and'tha relations (55) to (57) show that 9, has the otherrequired properties.

Proof Let us consider the functions g¥, for each x E K, constructed instep 3. By the continuity of 9,0 there exist open sets V.x containing x,such that

(56)

(57)

(58)

Finally, (58) follows from (61) and (62).

h = min (gZI" .• ,9;1&...)'

(60) K<= V" u'" u V,.'

Put

For what values of x does the series converge absolutely? On what intervals doesit converge uniformly? On what intervals does it fail to converge uniformly? Is Icontinuous wherever the series converges 1 Is I bounded '1

• I!(x)= :E-1 + ' .

• _1 n x

t. Prove that every unifonnly convergent sequence of bounded functions is uniformly bounded.

2. If tr.} and {g.} converge uniformly on a set E, prove that {III + gil} convergesuniformly on E. If, in addition, {!..} and {g.} are sequences of bounded functions,prove that {/.g.} converges uniformly on E.

3. Construct sequences {/.,}, {g.} which converge uniformly on some set E, but suchthat {/"U.} does not converge uniformly on E (of course. {/"U.} must converge onE).

4. Consider

EXERCISES

(t e K),

(I e V,).9,(1) <f(t) +.Since K is compact, there exists a finite set of points Xl • ... , x.

such that

(59)

(61)

By step 2, h e fJI, and (54) implies

h(l) > f(t) -.

whereas (59) and (60) imply

(62) h(l) <itt) +s (t e K).

SEQUENCES AND SERIES OF FUl"cnONS 167166 PRINCIPLES OF MATflEMATICAL ANALYSIS

5. Let

-f(~) ~ :E c.I(~ - ~.)"-.

(neal).f(x) ~ f (~)R.I n

Prove that

(See Exercises 7 and 8 of Chap. 6 for the relevant definitions.)This is a rather weak form of Lebesgue's dominated convergence theorem

(Theorem 11.32). Even in the context of the Riemann integral, uniform convergence can be replaced by pointwise convergence if it is assumed that IE fit. (Seethe articles by F. Cunningham in Math. Mag.• vol. 40, 1961. pp. 179-186, andby H. Kestelman in Amer. Math. Monthly, vol. 77, 1970, pp. 182-187.)

13. Assume that {J,.} is a sequence of monotonically increasing functions on R1 witho';;f.,(x)';; I for all x and all n.(0) Prove that there is a function/and &; sequence (ni} such that

f(x) ~ lim /.,(x)..-

J- g(x) <lx < <Xl.,

lim J- /.(x) <lx = J-f(x) <lx.R"CO 0 0

for every x E R 1• (The existence of such a pointwise convergent subsequence is

usually called Helly's selection theorem.)(b) If, moreover, / is continuous. prove that J,.t - /uniformly on compact sets.

Hin!: (i) Some subsequence {f,.,} converges at all rational points r, say, tof(r). (ii) Define f(~), for any x E R', to be supf(r), the sup being taken nver ailr ::s;; x. (iii) Show that J,.,(x) - lex) at every x at which / is continuous. (This iswhere monotonicity is strongly used.) (iv) A subsequence of {f,.,} converges atevery point of discontinuity of / since there are at most countably many suchpoints. This proves (0). To prove (b), modify your proof of (iii) appropriately.

Find aU discontinuities of f. and show that they form a countabll: dense set.Show that / is nevertheless Riemann-integrable on every bounded interval.

11. Suppose (/'), {g.) are defined on E, and(a) 1:.[,. has uniformly bounded partial sums;(b) gR _0 uniformly on E;(c) gl(x) ;;:::g](x) ;;:::gJ(x);;:::'" for every x E E.

Prove that ~ f,.gR converges uniformly on E. Hint: Compare with Theorem

3.42.U. Supposeg and/.(n ~ I, 2, 3, ...) are defined on (0, <Xl), are Riemann-integrable on

[t, T] whenever 0 < t< T< co, IJ,.I :5:g,/.-/uniformly on every compact sub-set of (0, <Xl), and

10. Letting (x) denote the fractional part of the real number x (see Exercise 16. Chap. 4,for the definition). consider the function

(~ ,;; 0),(~ >0),

(~ <~).

(~< n~I)'

( I ~ I)n+l$X::::'~.

if {XR} is a sequence of distinct points of (a. b). and if :ElcRI converges. prove thatthe series

co x1+n:E( -1)"-,-R_l n

~

/.(~)-I +~,.

Show that {f,,} converges uniformly to a function f. and that the equation

f'(~) - limf;(~)"--

converges uniformly in every bounded interval, but does not converge absolutely

for any value of x.7. For n = 1,2.3•..• , x real. put

converges uniformly. and that / is continuous for every x i= X R •

9. Let {f,.} be a sequence of continuous functions which converges uniformly to a

function / on a set E. Prove that

lim /'(x.) = f(x)..-for every sequence of points XR E E such that X. _ x, and x E E. Is the converse of

this true?

Show that {f,.} converges to a continuous function, but not uniformly. Use theseries ~ /R to show that absolute convergence, even for all x, does not imply uni

form convergence.6. Prove that the series

is correct if x =1= 0, but false if x = O.

8. If

168 PRINCIPLES OF MATHEMATICAL ,l.NALYSlSSEQUENCES AND SERIES OF FUNCTIONS 169

14. Let f be a continuous l'e41 function on R 1 with ttJ,e following properties:

05:. I(t) 5:.1.I(t + 2) - I(t) for every t, and

Prove that C!> is contin~Us and that C!> maps 1= [0, 1] onto the unit square [2 c R2•

If fact, show that C!> maps the Cantor set onto [2.

Hint: Each (xo, )'0) e J2 has the form(0 real).

(n = 0, I, 2, ...),

20. Iff is continuous on [0, 1l and if

{ I(x)x" dx ~ 0•

prove that I(x) - 0 on [0, I]. Hint: The integral of the product of I with any

polynomial is zero. Use the Weierstrass theorem to show thatf: f2(X) dx = O.

21. Let Kbe the unit circle in the complex plane (i.e.• the set of all z with Izi = I), andlet .RI be the algebra of all functions of the form

•y(t) ~ L 2-·/(3'·t)...,.

x(t) ~ L 2-~(3"-'t),..,

I(t) ~ (~

Put <I)(t) ~ (x(t), y(t)), where

.xO=LZ- II0211-lJ..,

where each 0, is 0 or 1. If .t. ~ L 3 -'-'(20,),.,

Then sI separates points on K and JII vanishes at no point of K, but neverthelessthere are continuous functions on K which are not in the uniform closure of d .Hint: For every fe sI

,.f I(e'')e'' dO ~ 0,•

(a 5:. x 5:. b).

show that/(3"to) = 0", and hence that x(to) = Xo, y(to) =)'0.

(This simple example of a so-called U space.filling curve" is due to I. J.Schoenberg, Bull. AM.s., vol. 44, 1938, pp. 519.)

15. Supposelis a real continuous function on R1 ,f,,(t) = I(nt) for n =1, 2, 3, ... , and{f,,} is equicontinuous on [0, 1]. What conclusion can you draw about I?

16. Suppose {f.} is an equlcontinuous sequence of functions on a compact set K, and{f,,} converges pointwise on K. Prove that {f.} converges uniformly on K.

17. Define the notions of uniform convergence and equicontinuity for mappings intoany metric space. Show that Theorems 7.9 and 7.12 arc valid for mappings intoany metric space, that Theorems 7.8 and 7.11 are valid for mappings into anycomplete metric space, and that Theorems 7.10, 7.16, 7.17, 7.24, and 7.25 hold forvector-valued functions, that is, for mappings into any R".

18. Let {f,.} be a uniformly bounded sequence of functions which are Riemann·inte~

grable on [a, bl, and put

F.(x) ~ (I.(t) dt

Prove that there exists a subsequence {F".} which converges uniformly on [0, b].19. Let Kbe a compact metric space, let Sbeasubset of~(K). Prove that Sis compact

(with respect to the metric defined in Section 7.14) if and only if S is uniformlyclosed, pointwise bounded, and equicontinuous. (If S is not equicontinuous,then S contains a sequence which has no equicontinuous subsequence, hence hasno subsequence that converges uniformly on K.)

and this is also true for every f in the closure of sI.n. Assume I e !R(et:) on [0, bl, and prove that there are polynomials P. such that

•lim f I/-P.I'd«=O."... "" .

(Compare with Exercise 12, Chap. 6.)23. Put Po = 0, and define, for n = 0, 1,2, ... ,

X 2 -P;(x)P...(x) ~ P.(x) + 2 .

Prove that

!~~ P.(x)~ lxi,

uniformly on [-I, 1].(This makes it possible to prove the Stone-Weierstrass theorem without first

proving Theorem 7.26.)Hint: Use the identity

IXI_P.,,(X)~[IXI_P,(X)][I_IXI~P.(X)]

to prove that 0 5:.P.(x) 5:. p...(x) 5:. Ixl if Ixl 5:.1, and that

Ix l -P.(x)5:. lx l (l- ~)' <_2_2 .+1

if Ixl 5:.1.


24. Let X be a metric space. with metric d. Fix a point a ~: X. Assign to each p e Xthe function I. defined by

f.(x) ~ d(x,p) - d(x, a) (x e X).

Prove that Ilix) I:::;; d(a, p) for all x e X. and that therefore I. e <C(X).Prove that

IIf.-t.II-d(p,q)for all P. q e X.

If l!>(p) = I. it follows that iI> is an isometry (a distance-preserving mapping)of X onto ~(X) co 'If(X).

Let Y be the closure of lI>(X) in tj'(X). Show that Y is complete.Conclusion: X is isometric to Q dense subset 01 a complete metric space Y.

(Exercise 24, Chap. 3 contains a different proof of this.)

25. Suppose '" is a continuous bounded real function in the strip defined byo:::;; x :::;; 1, - 00 < y < 00. Prove that the initial-value problem

SEQUENCES AND SER.IES OF '¥NeIlONS 171

(f) Hence

f(x) ~ c + {.p{I,f(I» dl.o

This1 is a solution of the given problem.26. Prove an analogous existence theorem for the initial-value problem

y' ~ 4>(x, y), yeO) - c,

where now C E R t, Y E Rt, and 4' is a continuous bounded mapping of the part of

RHI defined by 0:::;; x:::;; 1. y E R t into R t . (Compare Exercise 28, Chap. 5.) Hint:Use the vector-valued version of Theorem 7.25.

y(0) - c

has a solution. (Note that the hypotheses of this existence theorem are less stringentthan those of the corresponding uniqueness theoremj see Exercise 27, Chap. 5.)

Hint: Fix n. For i= 0, .... n, put X, = ;In. Let/.. be a continuous functionon [0, 1] such that [.(0) ~ c,

f~(t)=¢>(x,,/.(x.» ifx,<t<x,+h

and put

8.(1) - f:(1) - .p(I,f.(I»,

except at the points XI, where .6..(1) = O. Then

[.(x) = c+ {[.p(I,[.(/»+ 8.(/)] dl.o

Choose M < 00 so that ItP I :::;; M. Verify the following assertions.

(a) It:1 $M,18.1 $2M,8.e91,and If.1 $Icl +M~M.. say,on [0,1], forall n.

(b) ([.) is equicontinuous on [0,1], since It:1 $M.(e) Some {/.t} converges to some/, uniformly on [0, 1].(d) Since tP is uniformly continuous on the rectangle 0:::;; x:::;; 1. Iyj :::;; M I ,

.p(/,[.,(/))~ .p{1,f(/»

uniformly on [0, 1].(e) 8.(t)~0 uniformly on [0,1], since

8.(/) - .p(x, ,[.(x,» - .p(/,[.(/»

8SOME SPECIAl. FUNCTIONS 173

8.1 Theorem Suppose the series

SOME SPECIAL FUNCTIONS

converges/or Ixl < R, and define

(4)~

J(x) = I c, x",.0

(Ixl < R).

Then (3) converges uniformly on [-R + e. R - el, no matter which e > 0is chosen. The function / is continuous and differentiable in ( - R, R), and

(5)~

['(x) = I nc,x"-111=1

(Ixl < R).

Proof Let e > 0 be given. For Ixl';; R - e, we have

Ic"x" I,;; Ic,(R - e)' I;and since

POWER SERIES

In this section we shall derive some properties of functions which are representedby power series. Le., functions of the form

These are called analytic functions.We shall restrict ourselves to real values of x. Instead of circles of con

vergence (see Theorem 3.39) we shall therefore encounter intervals of convergence.

If (1) converges for all x in (-R, R), for some R > 0 (R may be + co),we say that/is expanded in a power series about the point x = O. Similarly, if(2) converges for Ix - al < R, Jis said to be expanded in a power series aboutthe point x = a. As a matter of convenience, we shalt often take a = 0 withoutany loss of generality.

In particular,

(7) J"'(O) = k!c, (k = 0, I, 2'r" .).

(HereJ'O) meansf, andJ'" is the kth derivative off, for k = 1,2,3, ...).

~

[",(x) = I n(n - 1) ... (n - k + I)c,x"-'.,.,

!.c,(R -e)'

converges absolutely (every power series converges absolutely in theinterior of its interval of convergence, by the root test), Theorem 7.10shows the uniform convergence of (3) on [- R + e, R - eJ.

Since .!jn -+ 1 as n -+ co, we have

lim sup 1n Ic, I= lim sup ifC,J,II"" co 11"'00

so that the series (4) and (5) have the same interval of convergence.Since (5) is a power series, it converges uniformly in [- R + 8,

R - e], for every e > 0, and we can apply Theorem 7.17 (for series instead of sequences). It follows that (5) holds if Ixl,;; R - e.

But, given any x such that IxI< R, we can find an 8 > 0 such thatIxl < R - e. This shows that (5) holds for Ixl < R.

Continuity ofJ follows from the existence of[' (Theorem 5.2).

Corollary Under the hypotheses oj Theorem 8.1, J has derivatives oj allorders in ( - R. R), which are given by

(6)

~

J(x) = I C,x",-0

~

J(x) = I C,(x - a)'.,-0

or, more generally,

(2)

(I)

174 PlUNCIPLES or MATJlO(Anc.u. ANALYSlI

SOME SPECIAL FIDlcnONS 175

I~ INS

If(x) -.1 = (1- x) L (', -.)X" ,;; (1- x) L I., - '11xl' + -2';; S8=0 8~0

for 0 :s; x:s; I. For x < 1, these series converge absolutely and hence may bemultiplied according to Definition 3.48; when the multiplication is carried out,we see that

if x > I - J, far some suitably choseo J > O. This implies (8),

As an application, let us prove Theorem 3.51, which asserts: /frail, '£b".'£c", converge to A, B, C, andifc" = aob" + ... +a"bo• then C=AB. We let

~

f(x) ~ L a,x",8=0

~

hex) = L c,x",,.0

(Ixl < I),

~

g(x) = L b.x",8=0

~

(l-x)Lx"~ I._0we obtaiu from (9)

Then, since

8.2 Theorem Suppose Ic" converges. Put

Proof Equation (6) follows if we apply Theorem 8.1 successively to f,f',J', .... Putting x = 0 in (6), we obtain (7). '

Formula (7) is very interesting. It shows, on the one hand, that thecoefficients of the power series development of/are dotermined by the valuesoffand of its derivatives at a single point. On the other hand, if the coefficientsare given, the values of the derivatives of/at the center of the interval of con·vergence can be read off immediately from the power series.

Note, however, that although a function f may have derivatives of allorders. the series Ic"X-, where c. is computed by (7), need not converge to/ex)for any x ;I:: O. In this case,/cannot be expanded in a power series about x = O.For if we hadf(x) = 14.x", we should have

n!a. = f(')(O);

hence all = c". An example of this situation is given in Exercise I.If the series (3) converges at an endpoint, say at x = R, then/is continuous

not only in ( - R, R), but also at x = R. This follows from Abel's theorem (forsimplicity of notation, we take R = I):

Proof Let s" = Co + ... + cll , S-1 = o. Then

~

(8) limf(x) = L C";1;~1 " ... 0

as x _ I. Equations (10) and (11) imply AB ~ C.We now require a theorem concerning an inversion in the order of sum

mation. (See Exercises 2 and 3.)

hex) -c

(0';; x < I).

g(x) _B,

f(x) , g(x) = hex)

By Theorem 8,2,

(10)

(11)

(-I <x< I).~

f(x) = L c.x",_0

Then

'" III .-1

L c.x" = L ('. - '._1)x" = (I - x) L ',x" + ••x'".8"'0 ._0 8 ... 0

8.3 Theorem Given a double sequence {al}}' i ~ I, 2, 3, ... , j = I, 2, 3, ... ,suppose that

For Ixl < I, we let m - 00 and obtain

~

(9) f(x) = (I - x) L .,x".,-0

Suppose. = lim ',. Let s > 0 be given. Choose N so that n > N.~~

implies

(12)

and r.b, converges. Then

(13)

(i = 1,2,3, ...)

Proof We could establish (13) by a direct procedure similar to (althoughmore involved than) the one used in Theorem 3.5S. However. the followingmethod seems more interesting.

176 PRl1'lCIPLES OF MATQEMATICAJ. ANALYSpJSOME SPECIAL FUNCIIONS 177

Let E be a countable set, consisting of the points Xo, Xl' X2 , •.. , andsuppose x.. -+ Xo as n -+ 00. Define

This is the desired expansion about the point X = a. To prove its validity,we have to justify the change which was made in the order of summation.Theorem 8.3 shows that this is permissible if

~

(14) J.(xo) = I a'J (i = 1,2,3, , ..), (18) I tic, (n) r"(x - 0)"/J-l, .. =Om"'O m

(15) J.(xJ = I au (I, n = 1,2,3, ...), converges. But (18) is the same as)=1

~~

(16) g(x) = I f,{x) (XE E). (19) I Ic,I'(lx-al +Ia[)','_I ,,=0

Now, (14) and (15), together with (12), show that each f, is continuous at xo' Since lJ.(x) I ,;; b, for x E E, (16) converges uniformly, sothat g is continuous at Xo (Theorem 7.11). It follows that

and (19) converges if Ix - 01 + 101 < R.Finally, the form of the coefficients in (17) follows from (7).

8.5 Theorem Suppose the series I:allx" and I.b"x" converge in the segmentS ~ (-R, R). Let E be the set of all XES at which

It should be noted that (17) may actually converge in a larger interval thanthe one given by Ix - 01 < R - 101.

If two power series converge to the same function in ( - R, R), (7) showsthat the two series must be identical, Le., they must have the same coefficients.It is interesting that the same conclusion can be deduced from much weakerhypotheses:

~ ~ ~

I I a;J = I f;(xo) = g(xo) = lim g(x,)i"l)=l 1"'1 "-00

~ ~ ,= lim I f.{x,) = lim I I a;J

II-CO 1=1 If-OO i"'l )"'1

II co coco

= lim I Iau= I Ia'J'11-(0)-11-1 )-11-1

8.4 Theorem Suppose

~

f(x) ~ I C,x",,-0

(20)~ ~

I a,x" = I b,x".,,"'0 ,,-0

the series converging in IxI< R. If - R < a < R, then f can be expanded in apower series about the point x = a which converges in Ix - aI< R - Ia I, and

IfE has a limit point in S, then a, = b,for n = 0, I, 2, .... Hence (20) holds forall XES.

This is an extension of Theorem 5.15 and is also kn~~ as Taylor'stheorem.

Proof We have

~

f(x) = I C,[(x - a) + oj',-0

~ '(n)= I c, I "-"(x - a)"1l-0 11I=0 m

= f rf (n) c, ..-"] (x - a)".m"O l,."'1JI m

(x E S).~

f(x) ~ I C,x"1f=0

Proof Put e" = a" - b" and

Then f(x) ~ 0 on E.Let A be the set of all limit points of E in S, and let B consist of all

other points of S. It is clear from the definition of "limit point" that Bis open. Suppose we can prove that A is open. Then A and B are disjointopen sets. Hence they are separated (Definition 2.45). Since S ~ A u B,and S is connected, one of A and B must be empty. By hypothesis, A isnot empty. Hence B is empty, and A = S. Since f is continuous in S,AcE. Thus E = S, and (7) shows that c, = 0 for n = 0, I, 2, ... , whichis the desired conclusion.

(21)

(lx-al<R-lal)·~ r"(a)

f(x) = I --(x - a)'.. -0 n!

(17)

178 PRINCIPLES OF MATHEMATICAL ANALYSASOME SPECIAL FUNCllONS 179

Thus we have to prove that A is open. If X o F A, Theorem 8.4 shows

We claim that d" = 0 for all n. Otherwise, let k be the smallest nonnegative integer such that d" ¢ O. Then

This shows that E(z) '" 0 for all z. By (25), E(x) > 0 if x> 0; hence (27) showsthat E(x) > 0 for all real x. By (25), E(x) _ +00 as x _ +00; hence (27) showsthat E(x) _ 0 as x _ - 00 along the real axis. By (25), 0 < x < y Implies thatE(x) < E(y); by (27), it follows that E( - y) < E(-x); hence E is strictly increasing on the whole real axis.

The addition formula also shows that

(22)

(23)

that~

J(x) = I d,(x - Xo)','0

J(x) ~ (x - xo)'g(x)

(lx-xol<R-lxol)·

<Ix - xol < R - Ixoi), (28) lim E_(,-z_+_h-:-)_-_E",,(z-,)"DO h

E(z) li~ E(h~- I E(z);

Since 9 is continuous at Xo and

g(xo) = d. '" 0,

there exists a ~ > 0 such that g(x) '" 0 if Ix - Xo I<~. It follows from(23) that f(x) '" 0 if 0 < Ix - Xo I<~. But this contradicts the fact thatXo is a limit point of E.

Thus d, = 0 for all n, so thatf(x) = 0 for all x for which (22) holds,i.e., in a neighborhood of xo. This shows that A is open, and completesthe proof.

the last equality follows directly from (25).Iteration of (26) gives

Ifp = nlm, where n, m are positive integers, then

Let us take Zt = ... = Z" = 1. Since E(l) = e, where e is the number definedin Definition 3.30, we obtain

(n = 1,2,3, ...).

[E(p)]· = E(mp) = E(n) = e",

E(n) = e"

E(z, + '" + z,) = E(z,) ..• E(z,).(29)

so that

(30)

(31)

~

g(x) = I d...(x - Xor..'0

where

(24)

(32) E(P) = eP (p > 0, P rational).

THE EXPONENTIAL AND LOGARITHMIC FUNCTIONS

We define

It follows from (27) that E( -p) = e- P if p is positive and rational. Thus (32)holds for all rational p.

In Exercise 6, Chap. I, we suggested the definition

which gives us the important addition formula

The ratio test shows that this series converges for every complex z. ApplyingTheorem 3.50 on multiplication of absolutely convergent series, we obtain

eo z" eo w'" 00 " zkw" -"E(z)E(w) ~ I - I - = I I ~---'77

,,=0 n! .-0 m! ,,"0 "=0 k!(n - k)!

= I .!.- t (n) z'Mr'= f ~,,,"0 n! 11:=0 k ,,=0 n!

(z, w complex).

the continuity and monotonicity properties of E, together with (32), show that

(p < x, p rational),

x' = sup x',

c=supe"

where the sup is taken over all rational p such that p < y, for any real y, andx > I. If we thus define, for any real x,

(33)

(34)

(35) E(x) = eX

for ~ll real x. Equation (35) explains why E is called the exponential function.The notation exp (x) is often used in place of e;J;;, expecially when x is a

complicated expression.Actually one may very well use (35) instead of (34) as the definition of eX;

(35) is a much more convenient starting point for the investigation of theproperties of eX. We shall see presently that (33) may also be replaced by amore convenient definition [see (43)].(z complex).

~ z'E(z) = I -

,,_0 n!

E(z)E(-z) = E(z - z) = E(O) = I

E(z + w) = E(z)E(w)

One consequence is that

(25)

(26)

(27)

180 PRINCIPUS OF MA~T1CAt. ANALYSIJ

SOME SPECIAL fUNCTIONS 181

if x >0 and n is an integer. Similarly, if m is a positive integer, we have

Quite frequently, (39) is taken as the starting.point of the theory of the logarithmand the exponential function. Writing u = E(x), v ~ E(y), (26) giv~

L(uv) =L(E(x) . E(y)) =L(E(x + y)) =x + y,

so that

(40) L(uv) = L(u) + L(v) (u> 0, v > 0).

This shows that L has the familiar property which makes logarithms usefultools for computation. The customary notation for L(x) is of course Jog x.

As to the behavior of log x as x -+ + 00 and as x -+ 0, Theorem 8.6(e)shows that

x" = E(nL(x))

as x -+ +00,

asx-+O.

logx-+ +00log x -+ -00

It is easily seen that

(41)

We now revert to the customary notation, e", in place of E(x), and summarize what we have proved so far.

for x > 0, so that

~ _. (n + I)!"e <---,

x

. 8.6 Theorem Let e" be defined on R' by (35) and (25). Then(a) e" is continuous and differentiable for all x;(b) (e")' = e";(c) e" is a strictly increasing function of x, and e" > 0;(d) e"+' = e"e';(e) c -+ + co as x -+ + 00, C -+ 0 as x -+ - co;(f) limz.... +oox"e-z = O,lor every n.

Proof We have already proved (a) to (e); (25) shows that

x"+'e">---

(n + 1)1

(42)and (J) follows. Part (f) shows that e" tends to +00 "faster" than anypower of x, as x -+ + 00.

Since E is strictly increasing and differentiable on R1, it has an inverse

function L which is also strictly increasing and differentiable and whose domainis E(R'), that is, the set of all positive numbers. L is defined by

X'/.=E(~L(X)),

since each term of (42), when raised to the mth power, yields the correspondingterm of (36). Combining (41) and (42), we obtain

(43) XO = E(aL(x)) = eO'o"

Differentiating (37), we get (compare Theorem 5.5)

(36)

or, equivalently, by

(37)

E(L(y)) =y

L(E(x)) = x

(y> 0),

(x real).

for any rational ex.We now define XII, for any real ex and any x > 0, by (43). The continuity

and monotonicity of E and L show that this definition leads to the same resultas the previously suggested one. The facts stated in EXercise 6 of Chap. 1, aretrivial consequences of (43).

If we differentiate (43), we obtain, by Theorem 5.5,

Writing y ~ E(x), this gives us

L'(E(x)) . E(x) ~ I. (XOY ~ E(aL(x)) . ~ = ax"-'.x

(44)

N'?te that we have previously used (44) only for integral values of ex, in whichcase (44) follows easily from Theorem 5.3(b). To prove (44) directly from thedefinition of the derivative, if x" is defined by (33) and a is irrational, is quitetroublesome.

The well-known integration formula for ~ follows from (44) if ex i:- -1,

~and from (38) if a = -I. We wish to demonstrate one more property of Jog x,

amely,(~. lim x'O!ogx=O

JC: .... + 110

(y > 0).

f'dXL(y)= -., x

(38) L'(y) = ~y

Taking x ~ 0 in (37), we see that L(l) = O. Hence (38) implies

(39)

182 PRINCIPLES OF MATHEMATICAL ANALYSIS SOME SPECIAL FUNCTIONS -183

Then C(nI2) = 0, and (48) shows that S(nI2) = ± 1. Since C(x) > 0 in(0, n12), S is increasing in (0, nI2); hence S(nI2) ~ 1. Thus

Let Xo be the smallest p~sitive number such that C(xo) = o. This exists,since the set of zeros of a co:ntinuous function is closed, and C(O) =F O. Wedefine the number 1t by

for every ex > O. That is, log x -+ + 00 "slower" than any positive power of x,as x-+ +00.

For if 0 I, then

x-a log X = x-a r" t- 1 dt < x-It f' ('-1 dt, I

X' - 1 xI-«=x-«·_-<-,

• •and (45) follows. We could also have used Theorem 8.6(f) to derive (45).

(51)

and the addition formula gives

1t= 2xo·

THE TRIGONOMETRIC FUNCTIONS

Let us define

(46)I

C(x) = '2 [E(ix) + E(- Ix)],I

Sex) = 2i [E(ix) - E( - Ix)].

(52)

hence

(53)

8.7 Theorem

E(ni) = -I,

E(z + 2ni) = E(z)

E(2ni) = I;

(z complex).

We shall show that C(x) and Sex) coincide with the functions cos x and sin x,whose definition is usually hased on geometric considerations. By (25), E(z) ~

E(z). Hence (46) shows that C(x) and Sex) are real for real x. Also,

(47) E(ix) = C(x) + is(x).

Thus C(x) and Sex) are the real and imaginary parts, respectively, of E(ix), ifx is real. By (27),

IE(ix) I, = E(ix)E(ix) = E(ix)E( -Ix) = I,

so that

(48) IE(ix) I = I (x real).

From (46) we can read off that C(O) = I, S(O) = 0, a?d P8) shows that

(49) C'(x) = -S(x), Sex) = C(x).

We assert that there exist positive numhers x such that C(x) = O. Forsuppose this is not so. Since C(O) = I, it then follows that C(x) > 0 for allx> 0, hence Sex) > 0, by (49), hence S is strictly increasing; and since S(O) = 0,we have Sex) > 0 if x> O. Hence if 0 < x < y, we have

(SO) S(x)(y - x) <rS(I) dl = C(x) - C(y) S; 2.x ,

The last inequality follows from (48) and (47). Since Sex) > 0, (SO) cannot betrue for large y, and we have a contradiction.

(a) The junction E is periodic, with period 2ni.(b) TheJunclions C and S are periodic, wilh period 2n.(c) If0 < I < 2n, Ihen E(il) '" 1.(d) If z is a complex number wilh Izi = I, Ihere is a unique I in [O,2n)

such Ihal E(il) = z. .

Proof By (53), (a) holds; and (b) follows from (a) and (46).Suppose 0 < I < nl2 and E(il) = x + iy, with x, y real. Our preceding

work shows that 0 < x < I, 0 < y < 1. Note that

E(4il) = (x + iy)4 = x4 _ 6x'y' + y4 + 4ixy(x' _ y').

If E(4il) i~ real; it follows thatx' - y' = 0;. since x' + y' = I, by (48),we have x = y = t, hence E(411) = -1. ThIS proves (c).

IfO S; I, < I, < 2n, then

E(il,)[E(il,Jr I = E(il, - itl ) '" I,

by (c). This establishes the uniqueness assertion in (d).To prove the existence assertion in (d),:fix z so that Iz) = 1. Write

z = x + iy, with x and y real. Suppose first that x ;;, 0 and y ;;, O. On[0, n12], C decreases from I to O. Hence C(I) = x for some IE [0 nI2].S· C' S' 'mce + = I and S;;' 0 on [0, nI2], it follows that z = E(il).

If x < 0 and y ~ 0, the preceding conditions are satisfied by - iz.Hence -iz = E(il) for some IE [0, n12], and since i = E(ni/2), we obtainz = E(i(1 + nI2)). Finally, if y < 0, the preceding two cases show that


- z ~ E(il) for some IE (0, n). Hence z = - E(il) ~ E(i(1 + n».This proves (<i), and hence the theorem.

It follows from (d) and (48) that the curve y defined by

(54) y(l) = E(il) (0';; I ';;2n)

is a simple closed curve whose range is the unit circle in the plane. Sinceitt) = iE(il), the length of y is

"fo

Iy'(I) I dt = 2n,

by Theorem 6.27. This is of course the expected result for the circumference ofa circle of radius J. It shows that ", defined by (51), has the usual geometricsignificance.

In the same way we see that the point y(l) describes a circular arc of length10 as I increases from 0 to 10 , Consideration of the triangle whose vertices are

ZI =0, z, = y(to), z, =C(to)shows that C(t) and Set) are indeed identical with cos t and sin t, if the latterare defined in the usual way as ratios of the sides of a right triangle.

It should -be stressed that we derived the basic properties of the trigonometric functions from (46) and (25), without any appeal to the geometric notionof angle. There are other nongeometric approaches to these functions. Thepapers by W. F. Eberlein (Amer. Math. Monthly, vol. 74, 1967, pp. 1223-1225)and by G. B. Robison (Math. Mag., vol. 41, 1968, pp. 66-70) deal with thesetopics.

THE ALGEBRAIC COMPLETENESS OF THE COMPLEX FIELD

SOME SPECIAL FUNCTIONS 185

The right side of (56) tends to 00 as R -. 00. Hen(e there exists Ro suchthat [P(z) I> I' if Izl > R o• Since IPI is continuous on the closed discwith center at 0 and radius Ro , Theorem 4.16 shows that IP(zo) I=1' forsome zo.

We claim that I' = O.If not, put Q(z) = P(z + zo)/P(zo). Then Q IS a nonconstant poly

nomial, Q(O) = I, and IQ(z) I~ I for all z. There.is a smallest integer k,1 :5: k :5: n, such that

(57) Q(z) ~ I + b,z' + ... + b,z', b, .. O.

By Theorem 8.7(d) there is a real 8 such that

(58) e"'b,=-lb,l.

If, > 0 and ,'lb,l < I, (58) implies

I1+ b,,'e;" 1= I - ,'lb,l,so that

[Q(,e;')I,;; I - r'{lb,l- rlbk+ll- '" - ,'-'lb,I}.For sufficiently small r, the expression in braces is positive; henceIQ(re") I< I, a contradiction.

Thus I' = 0, that is, P(zo) = O.

Exercise 27 contains a more general result.

FOURIER SERIES

8.9 Definition A trigonometric polynomial is a finite sum of the form

where ao, ...• aN, bI , ••• , bN are complex numbers. On account of the identities(46), (59) can also be written in the form

We are now in a position to give a simple proof of the fact that the complexfield is algebraically complete, that is to say, that every nonconstant polynomialwith complex coefficients has a complex root.

8.8 Theorem Suppose ao • ... , alt are complex numbers, n~ t. alt 7'=0,

(59)N

f(x) = ao .,. L (a, cos nx + b, sin nx)n=l

(x real),

which is more convenient for most purposes. It is clear that every trigonometricpolynomial is periodic, with period 2".

If n is a nonzero integer, e ilt): is the derivative of eilt):jin, which also hasperiod 2". Hence

N

f(x) = L Cit eilt):-N

,P(z) = L a,z'.

oThen P(z) = 0 far some complex number z.

Proof Without loss of generality, assurne a, = J. Put

(55) I' = inf IP(z)1 (z complex)

If Izi = R, then

(56) IP(z) I~ R"[I -lao-lIR-1 - ... - laoIR-'j.

(60)

(61) I f' (I- e1udx=21t _. 0

(x real),

(if n = 0),(if. = ± I, ±2, ...).

186 PRINCIPLES OF MATHEMATICAL ANA~YSISSOME SPECIAL FUNCTIONS 187

Let us multiply (60) by e- ifllX, where m is an integer; if we integrate the

product. (61) shows that

for Iml" N. If Iml > N. the integral in (62) is O.The following observation can be read off from (60) and (62): The

trigonometric polynomial f, given by (60). is real if and only if c_, = Co forn=O, ... ,N.

In agreement with (60), we define a trigonometric· series to be a series ofthe form

Then {q,,} is said to be an orthogonal system offunctions on [a. b]. If, in addition.

(n = I. 2. 3....).

,f Iq,,(x) I' dx = I•

(65)

for all n. {q,,} is said to be orthonormal.For example, the functions (2n) -Ye inx form an orthonormal system on

[- x. x]. So do the real functions

1 cos x sin x cos 2x sin 2x

j2;;' -f1r . J~-' J~ . -f1r .If {q,,} is orthonormal on [a. b] and if

c. = (f(t )q,,(I) dl•

(66)(x real);

I •Cm= - f f(x)e- J

•• d.x2n _~

(63)

(62)

we call c, the nth Fourier coefficient off relative to {q,,}. We write

be the nth partial sum of the Fourier series off, and suppose

8.11 Theorem Let {q,,} be orthonormal o~ [a. bl. Let

, ,(70) f If - s.I' dx" f if- t,I' dx.

• •and equality holds ifand anly if

~

fix) - I c, q,,(x)I

•s,(x) = I Cmq,m(X)

m-I

,I,(x) = I Ym q,m(X)'

m_1

(67)

and call this series the Fourier series off (relative to {q,,}).Note that the symbol - used in (67) implies nothing about the conver

gence of the series; it merely says that the coefficients are given by (66).The following theorems show that the partial sums of the Fourier sories

of/have a certain minimum property. We shall assume here and in the rest ofthis chapter thatfE 91. although this hypothesis can be weakened.

(68)

(69)

Then

the Nth partial sum of (63) is defined to be the right side of (60).Iffis an integrable function on [-x. xl. the numbers cm defined by (62)

for all integers m are called the Fourier coefficients off, and the series (63) formedwith these coefficients is called the Faurier series off

The natural question which now arises is whether the Fourier series of/converges tot, or, more generally, whether/is determined by its Fourier series.That is to say, if we know the Fourier coefficients of a function, can we findthe function, and if so, how?

The study of such series, and, in particular, the problem of representing agiven function by a trigonometric series, originated in physical problems suchas the theory of oscillations and the theory of heat conduction (Fourier's"TMorie analytique de la chaleur" was published in 1822). The many difficultand delicate problems which arose during this study caused a thorough revisionand reformulation of the whole theory of functions of a real variable. Amongmany prominent names, those of Rieinann, Cantor, and Lebesgue are intimatelyconnected with this field, which nowadays. with all its generalizations and ramifications, may well be said to occupy a central position in the whole of analysis.

We shall be content to derive some basic theorems which are easilyaccessible by the methods developed in the preceding chapters. For morethorough investigations, the Lebesgue integral is a natural and indispensabletool.

We shall first study more general systems of functions which share aproperty analogous to (61).

8.10 Definition Let {q,,} (n = I. 2. 3, ...) be a sequence of complex functionson [a. bl. such that

That is to say, among all functions t,., s,. gives the best possible meansqu~.re approximation to f.(64) (q,,(X)q,m(X) dx = 0

•(n #m).

(71 ) (m=I ..... n).

SOMI. SPECIAL FUNCTIONS 189

(76)

Proof Let I denote the integral over [a, b], 1: the sum from I to n. Then

ffl, = ffI Ym'Pm = I CmYm

by the definition of {cm},

since {<Pm} is orthonormal, and so

is the Nth partial sum of the Fourier series ofJ. The inequality (72) now takesthe form

If' N If'2n _.lsN(xW dx = ~)c,I';5; 2n _,If(xJ!' dx.

In order to obtain an expression for SN that is more manageable than (75)we introduce the Dirichlet kernel

The first of these equalities is the definition of DN(x). The second follows ifboth sides of the identity

(e lx _ l)DN(x) = ef(N+l)x _ e- 1Nx

are multiplied by e- 1x !'2.

By (62) and (75), we have

f If- t, I' = f IfI, - f fl, - f Jt, + fit, I'

= f IfI, - I CmYm - I em Y. + I Y. Ym

= fW- I Icml'+ I IYm-c.I',

which is evidently minimized if and only if I'll' = cm •

Putting "I. = c. in this calculation, we obtain

, . f'f Is,(xW dx = I Ic.I';5; If(xW dx,

• 1 •

since Ilf- t,l' ~ o.

8.12 Theorem If {<p,} is orthonormal on [a, bj, and if

(77) D () _ ~ 'ox_sin(N+t)xNX-t... e - .

,. -N sin (xI2)

~

f(x) ~ I c. <p,(x),.-1then

(73)

In particular,

(74)

~ ,I Ic,l' ;5; f.lf(x)I' dx.

lim Cn = O.

so that

(78)

The periodicity of all functions involved shows that it is immaterial over whichinterval we integrate, as long as its length is 21T. This shows that the two integralsin (78) are equal.

We shall prove just one theorem about the pointwise convergence ofFaurier series.

Proof Letting n - <Xl in (72), we obtain (73), the so-called "Besselinequality. ,.

g(t) f(x - t) - f(x)sin (tI2)

lim s.{f; x) = f(x).N~~

If(x + r) - f(x)! ;5; MI tl

8.14 Theorem If, for some x, there are constants b > 0 and M < 00 such that

(79)

for all t E ( - 6, 6), then

(80)

(81)

Proof D.fineN

SN(X) = SN(f; x) = L Cn e11lX

-N(75)

8.13 Trigonometric series From now on we shall deal only with tlie trigonometric system. We shall consider functions f that have period 2n and that areRiemann-integrable on [-n, n] (and hence on every bounded interval). TheFourier series off is then the series (63) whose coefficients en are given by theintegrals (62), and

190 PRINCIPLES OF MATHEMA.TICAL ANALYSIS SOME SPECIAL FUNcnONS 191

8.16 Parseval's th~orem Suppose f and 9 are Rit.mallll-j11legrable junctiollswith period 2n, and

Proof Let us use the notation

I ';~":, 2n t IJ(x) - SN(J; x)I' dx = 0,

J J' - Q

2n _/(x)g(x) dx = 1c, Y"

I J' i2n _ IJ(x) I' dx = i 1e,I'·, -00

(If' )'/'Ilhll, = 2n _,lh(xJ!' dx .

Let. > 0 be given. Since JE iJt andJ(n) = J( -n), the construction~escnbe~ In ExercIse 12 of Chap. 6 yields a continuous 2tr-periodic functIOn h wIth

00

g(x) - I Y,e'''.-00

00

I(x) -- I ·c" e iflX,

-00

Then

(82)

(83)

(84)

(85)

(86)

CoroUary IJ J(x) = 0 Jar all x in some segment J, then lim SN(J; x) = 0 Jarevery x eJ.

Hence (78) shows that

I· , (I)s.(J; x) - J(x) = - J g(t) sin N + - t tlt2n -11 2 .

I '[ t] I J' [ t]= - J g(t) cos - sin Nt dt + - g(t) sin -2 cos Nt dt.2n -11: 2 :an -11:

Here is another formulation of this corollary;

IfJ(t) ~ g(t) Jar all t in some neighborhood oj x, then

SN(J; x) - SN(g; x) = SN(J - g; x) -+008 N -+ 00.

for 0 < ItiS n, and put g(O) ~ O. By the definition (77),I ' .- J DN(x) dx= I.2n _11 .

By (79) and (81), g(t) cos (t/2) and g(t) sin (t/2) are bounded. The lasttwo integrals thus tend to 0 as N -+ 00, by (74). This proves (80).

This is usually called the localization theorem. It shows that the behaviorof the sequence {SN(f; x)}, as far as convergence is concerned, depends only onthe values ofJ in some (arbitrarily small) neighborhood of x. Two Fourierseries may thus have the same behavior in one interval, but may behave inentirely different ways in some other interval. We have here a very strikingcontrast between Fourier series and power series (Theorem 8.5).

We conclude with two other approximation theorems.

8.15 Theorem IfJ is continuous (with period 2n) and if. > 0, then there is atrigonometric polynomial P such that

IP(x) - J(x) I< •Jar all real x.

Proof If we identify x and x + 2", we may regard the 2n-periodic functions on R1 as functions on the unit circle T, by means of the mappingx ~ e i

%. The trigonometric polynomials, i.e., the functions of the form(60), form a self-adjoint algebra d, which separates points on T, andwhich vanishes at no point of T. Since T is compact, Theorem 7.33 tellsus that d is dense in 'C(T). This is exaclly what the theorem asserts.

A more precise form of this theorem appears in Exercise 15.

(87)

(88)

(89)

(90)

(91)

(92)

IIJ - hll, < •.

By Theorem 8.15, there is a trigonometric polynomial P such thatIh(x) - Pix) I <. for all x. Hence IIh - PII, <.. If P has degree NTheorem 8.11 shows that 0,

Ilh - sN(h)II, Sllh - PII, <.

for all N';? No· By (72), with h - J in place ofJ,

IIsN(h) - sNUlII, = IIsN(h - fllb S IIh - JII, < •.

Now the triangle inequality (Exercise II, Chap. 6), combined with(87), (88), and (89), shows that

IIJ - sN(J)II, < 3.

This proves (83). Next,

I f' - N I' _ N

2n _!N(J)g dx ~ ~ c, 2n t, e'ox g(x) dx = ~ c, Y..

and the Schwarz inequality shows that

/JJg- fSN(J)u/sflJ-SN(J)1191 s{JIJ-SNI' Jlgl't',

(94)


which tends to 0, as N -+ ce, by (83). Comparison of (91) and (92) gives(84). FinaUy, (85) is the special case 9 =f of (84).

A more general version of Theorem 8.16 appears in Chap. II.

THE GAMMA FUNCTION

This function is closely related to factorials and crops up in many unexpectedplaces in analysis. Its origin, history, and development are very well described

'in an interesting article by P. J. Davis (Amer. Math. Monthly, voJ. 66, 1959,pp. 849-869). Artin's book (cited in the Bibliography) is another good elementary introduction.

Our presentation will be very condensed, with only a few comments aftereach theorem. This section may thus be regarded as a large exercise, and as anopportunity to apply some of tbe material that has been presented so far.

8.17 Definition For 0 < x < ce,

(93) rex) = r t'-le-' dt.

The integral converges for these x. (When x < I, both 0 and 00 have tobe looked at.)


8.19 Theorem Iff is a positive function On (0, co'l such that(a) f(x + I) = xf(x), •(b) f(l) ~ I,(c) logf is convex,

then f(x) ~ rex).

Proof Since r satisfies (a), (b), and (c), it is enough to prove thatf(x) isuDlquely determmed by (a), (b), (c), for all x > O. By ( ) 't' hdo thiS for x E (0, 1). a,' IS enoug to

Put 'P ~ logi Then

'P(x + 1) = 'P(x) + log x (0 < x < 00),

~(l)9= 0, and <p is convex. Suppo~e °< x < 1, and n is a positive integer., Y( 4), 'P(n + I) ~ log(n!). ConSIder the difference quotients of'P on themtervals [n, n + I], [n + I, n + I + x], [n + I, n + 2]. Since <p is convex

log n < 'P(n + I + x) - <pen + I)- x ::;; log (n + I).

Repeated application of (94) gives

'P(n + 1 + x) ~ <p(x) + log [x(x + I)· .. (x + n)].

Thus

As a by-product we obtain the relation

8.20 Theorem If x> 0 and y > 0, then

(95) rex) = lim n!n'"~~ x(x + I) .. · (x + n)

at least when 0 < x < I; from this one can deduce that (95) holds for all x > 0smce rex + I) = xr(x). '

8.18 Theorem(a) The functional equation

rex + I) = xr(x)

holds if0 < x < 00.

(b) r(n + I) = n!for n = I, 2, 3, ....(c) log r is convex on (0, ce).

Proof An integration by parts proves (a). Since r(1) = I, (a) implies(b), by induction. If I < p < 00 and (lIp) + (1Iq) = I, apply Holder'sinequality (Exercise 10, Chap. 6) to (93), and obtain

r(; +~) ::;; r(x)'/'r(Y)'/'.

This is equivalent to (c).

It is a rather surprising fact, discovered by Bohr and MoUerup, thatthese three properties characterize r completely.

(96)

[ n' ' ]o< <p(x) _ log .n ( I)- x(x + I) '" (x + n) "x log I +;; .

The last expression tends to °as n .... ceo Hence (). d . d<p X IS etermme, andthe proof is complete.

,f t'-'(I - t),-1 dt = r(x)r(Y).o rex + y)

This integral is the so-called beta function H(x, y).

194 PRINCIPLES OF MATHEMATICAL ANJ\LYSISSOME SPECIAL FUNCTIONS 195

Determine h(u) so that h(O) = I and

if -1 < u < 00, u # O. Then

It follows that h is continuous, and that h(u) decreases monotonically from 00

to 0 as u increases from - I to 00.

The substitution u = s J2iX turns (104) into

Proof Note that BO, y) = Ily, that log B(x, y) is a convex function ofx, for each fixed y, by Holder's inequality, as in Theorem 8.18, and that

x(97) B(x + I y) = -- B(x, y)., x+y

To prove (97), perform an integration by parts on

B(x + I, y) = f' (_I_)X (I - 1)"+'-' dt.o I - 1

These three properties of B(x, y) show, for each y, that Theorem 8.19

applies to the functionJ defined by

J(x) = [(x + y) B(x, y).'r(y)Hence J(x) = [(x).

8.21 Some consequences The substitution 1 = sin' B turns (96) into

xl' [(x)r(y)(98) 2 f

o(sin O)'x-' (cos B),,-1 dB = [(x + y)

The special case x = y = t gives

(104)

(105)

(106)

(107)

where

Here is a proof. Put 1 = x(1 + u) in (93). This gives

r(x+ 1)=xx +'e-' f~ [(I +u)e-'J'du.-,

[u' -,

(I + u)e-' = exp - - h(lI) I2 I

~

2h(u) = --, [u - log (I + u)].

u

rex + I) = x'e-x j'ber ljI,(s) ds-~

(99)

(100)

rm =-.r;..The substitution t = s' turns (93) into

[(x) = 2 s: S'x-1 e-" ds (O<x< (0).

(-jXj2< S < (0),

(s ,; - JXi2).

Note the following facts about IjIx(s):

(103)

The special case x = t gives

(101) r e-" ds =.fie.-~

By (99), the identity

(102) [(x) = 2; rmr(X; I)follows directly from Theorem 8.19.

8.22 Stirling's formula This provides a simple approximate expre~sion for[(x + I) when x is large (hence for nl when n is large). The formula is

. f(x+I)_1lIm - .x~~(xle)"~

(a) For every s, I/JAs) -+ e- s2 as x -+ 00.

(h) The convergence in (a) is uniform on [- A, A], for every A < 00.

(c) When s < 0, then 0 < ljI,(s) < e-".(d) When s > 0 and x > I, then 0 < IjIx(s) < 1jI,(s).(e) SO' I/I,(s) ds < 00.

The convergence theorem stated in Exercise 12 of Chap. 7 can thereforebe applied to the integral (107), and shows that this integral converges to .fieas x -+ 00, by (101). This proves (103).

A more detailed version of this proof may be found in R. C. Buck's"Advanced Calculus," pp. 216-218. For two other, entirely different, proofs,see W. Feller's article in Amer. Malh. Monthly, vol. 74, 1967, pp. 1223-1225(with a correction in vol. 75,1968, p. 518) and pp. 20-24 of Artin's book.

Exercise 20 gives a simpler proof of a less precise result.

196 PRINCIPLES Of MATHEMATICAL ANALYSISSOME SPECIAL FUNCTIONS 197

EXERCISES

1. Define

(e- 11"'2

[(x) = 0(x'" 0),(x= 0).

5. Find the following limits

. e - (1 + X)l /'"(a) hm .

", ... 0 x

(b) lim _n_ [11 1/11 - 1]..... <Xl logn

Prove that f has derivatives of all orders at x = 0, and that P")(O) = 0 for

n= 1,2,3, ....

1. Let all be the number in the ith row and jth column of the array

-I 0 0 0

t -I 0 0

i t -I 0

t i t -I..........................

1. tanx-x

(c) 1m •", ... 0 x(1 - cos x)

(d) lim x - sin x .,,,,"'0 tan x-x

6. Suppose [(x)[(y) = [(x + y) for all real x and y.(a) Assuming that / is differentiable and not zero, prove that

[(x) = eO>

so that

Prove that

3. Prove that

a,,={~121 - 1

LL a" = -2,, J

(i <j),(i-j),(i > j).

where c is a constant.(b) Prove the same thing, assuming only that/is continuous.

"7. IfO<x<Z' prove that

2 sin x I-<--< ." x

8. For n = 0, I, 2, ... , and x real, prove that

if a'J ;;0, 0 for all i andj (the case +'" = +'" may occur).4. Prove the foHawing limit relations:

b"'-l(a) Ibn---log b

JI"O X

(b) Ibn log (I + x) I.JI"O X

(c) Ibn (I + x)'" = e.,..(d) Ibn (I + ~)' = e',

... <0 n

(b >0),

Isinnxl ";nlsinxj.

Note that this inequality may be false for other values of n. For instance,

Isin ",I >tlsin"j,

9. (a) Put SN = 1 +m+ ... + (lIN). Prove that

lim (SN -log N)N••

exists. (The limit, often denoted by y, is called Euler's constant. Its numericalvalue is 0.5772 .... It is not known whether y is rational or not.)(b) Roughly how large must m be so that N = 10"' satisfies SN > lOO?

10. Prove that:E IIp diverges; the sum extends over all primes.(This shows that the primes form a fairly substantial subset of the positive

integers.)

198 PRINCIPLES OF MATHEMATICAL ANALYSISSOME SPECIAL FUNCTI01'OS 199

Hint: Given N, let Ph , .. , Pt be those primes that divide at least one integer ~N. Then

'1 '( 1 1 )2:-,;;IT 1+-+,+'""=1 n 1~1 P1 P1

14. If/(x) ~ (rr - lxi)' on [-rr, rrj, prove that

7TZ

'" 4I(x) ~ - + I --; cos nx

3 11= I n-

and deduce that

'( 1)-'- IT 1--1 .. 1 P1

, ,::S;:exPL -=-.

J=1 P1

The last inequality holds because

"" 1 1l'4-I -~-n"'ln

4 90'

(A recent article by E. L Stark contains many references to series of the formL n-', where-s is a positive integer. See Afath. ;Haq., vol. 47, 1974, pp. 197-202.)

15. With D. as defined in (77). put

if 0 :5: x ,;; t.(There are many proofs of this result. See, for instance, the article by

L Niven in Amer, Math, ,~fonrhiy. voL 78, 1971, pp. 272-273, and the one by

R. Bellman in Amer. Math. Monthly, vol. 50, 1943, pp. 318-319.)

11. Suppose / E fX on [0, A] for all A < 00, and/(x) --I as x -- + 00. Prove that

1 •K.(x) ~ -N' 1 I Dix).

-:- ~*O

Prove that

1 l-cos(N+ l)xK",(x) = N + 1 1 - cos x

(c) Deduce from Parseval's theorem that

12. Suppose 0 < 8 < rr,j(x) ~ 1 if Ixl ,;; o,j(x) ~ 0 if 0 < Ix I :5: rr, and I(x + 2rr) ~I(x) for all x.(a) Compute the Fourier coefficients of f.(b) Conclude that

N+l

if 0<8.,;:;; ixl :::;;1l'.

SO---"-SI--;-"'+S-"CT,V =

and that(a) K,;:, 0,

1 f'(b) -2 K,,(x) dx ~ 1,rr _.

1 2(e) K,,(x) ,; N -'- 1 . 1 - cos 0

If SN = sllf; x) is the Nth partial sum of the Fourier series of J. consider

the arithmetic means

(I> 0)•

(0 < 0 < rr).

lim I J" e-"/(x) dx ~ 1• _0 0

(d) Let (5 ---+ 0 and prove that

Prove that

1 •a.,(f; x) ~ 2;; t. !(x - I)K.,(I) dt,

(~) Put SC-='Ir/2 in (c-;. Wh.:p '.i'J- you get ~ -

13. Put fex) = x if 0 ::s;: x < 211', and apply Parseval's theorem to conclude that

'" 1 1T1

I -~-... 1 n1 6 '

and hence prove Fejer's theorem:If/is continuous, with period 27T, then 17.-.;(/; x) __/(x) uniformly on [-1l', 7T].

Hint: Use properties (a), (b), (c) to proceed as in Theorem 7.26.Ii), Prove a pcir.twise ver~ion [If Feir'<; theorem:

Iff E 3l and fex +), f(x -) exist for some x, then

lim a.(f; .<) ~ t[!(x -'-I +!(x-)].." ... 00

200 PRINCIPLES OF MATHEMATICAL A.....ALYSIS SOME SPECIAL FU-SCTIONS 201

Prove that there exists a constant C > 0 such that

17. Assume I is bounded and monotonic on (-71', 71'), with Fourier coefficients c~, asgiven by (62),

(a) Use Exercise 17 of Chap. 6 to prove that {nc~} is a bounded sequence.(6) Combine (a) with Exercise 16 and with Exercise 14(e) of Chap. 3, to concludethat

21. LeI

1 ['L. ~ - ID,(t) I dl211'. _!<

(n ~ I, 2, J, .. ,),

lim SH(I; x) ~ Hf(x+) + f(x-)JH•• L" > Clogn (n~ I, 2, J, .. ,),

for every x.

(c) Assume only that IE 2i on [-77, 71'J and that I is monotonic in some segment(x, f3)c [-7T, 7T]. Prove that the concl~sion of (b) holds for every x E (ct, f3).

(This is an application of the localization theorem.)18, Define

I(x) = xJ- sin 2 x tan x

g(x) = 2x 2- sin2 x - x tan x.

Find out, for each of these two functions, whether it is positive or negative for allx E (0, 7T,12), or whether it changes sign. Prove your answer.

19. Suppose I is a continuous function on R 1, I(x + 271') = I(x), and Cl/1T is irrationaL

Prove that

I ., 1 [';~~ NJ, f(x';- na) ~ 2rr, _. f(l) dl

for every x. Hint: Do it first for I(x) = eil:<.

20. The following simple computation yields a good approximation to Stirling'sformula.

For m = 1,2. 3, ... , define

fix) ~ (m-I- x) log m.,. (x- m) log (m+ I)

if m:::;; x::;;: m + I, and define

g(x) ~ =. - 1 + log mm

if m- -t::;;:x <m+ t. Draw the graphs of/andg. Note that/(x) s;;: log x S;;:g(x)if x ~ 1 and that

f· ,f(x)dx~log(n,)-,logn> -1+ r g(x)dx,, , ,

Integrate log x over [1, n]. Conclude that

t < log (n!) - (n + t) log n + n < 1

for n = 2, 3,4, .... (Note: log v'L"'; '" 0.918 ....) Thus

n'e7

/8 < --'-,_ <e.

(nie)~v'n

or, more precisely, that the sequence

is bounded.22. If ~ is real and -1 < x < I, prove Newton's binomial theorem

Hint: Denote the right side by I(x). Prove that the series converges. Prove that

(I T x)f'(x) ~ ,fIx)

and solve this differential equation.Show also that

• r(n.c,)(l-x)-'= L -,-'-x'

"=0 n. fect)

if -1 < x < 1 and ct > O.23. Let y be a continuously differentiable closed curve in the complex plane, with

parameter interval [a, bl, and assume that yet) '7"':0 for every t E [a, b]. Define theindex of y to be

1 I~ y'(t)Ind (y) ~ 2- ' -(I·) dl.

"I "a y

Prove that Ind (1) is always an integer.Hint: There exists l' on [a, b] with rp' = y'/y, gJ{a) = O. Hence y exp( ~<p)

is constant. Since yea) = reb) it follows that exp <pCb) = exp 1'(0) = 1. Note that<P(b) ~ 2"i Ind (y),

Compute Ind (y) when y(l) = el"', a = 0, b = 21T.

Explain why Ind (y) is often called the winding number of y around O.

24. Let i' be as i:l Exo;T'clse 23, ar'.d assume in addition that lhe range of y does not

intersect the negative real axis. Prove that Iod (y) = O. Hint: For 0 s;;: c < :lJ,

Ind (y + c) is a continuous integer-valued function of c. Also, Ind (y --:- c) -+ aasc-+'XJ.

for every real constant e.31. In the proof of Theorem 7.26 it was shown that

f' 4(1 - XZ)~ dx :2:-;=

_1 3"\' n

for n = 1, 2, 3, .... Use Theorem 8.20 and Exercise 30 to show the more precise

result

202 PRINCIPLES OF MATHEMATICAL A:-IALYSIS

25. Suppose Yl and Y2 are CUI\les as in Exercise 23, and

ly,(t)~y,(')I< !y,(t)1 (as;tS;b).

Prove that Ind (y,) = Ind (yz).

Hint: Put Y= YZ/Yl. Then 11 - yl < 1, hence Iod (y) = 0, by Exercise 24.Also,

L=ri_Y2Y yz Yl

26. Let y be a closed curve in the complex plane (not necessarily differentiable) withparameter interval [0,2,,], such that y{t)? 0 for every t E [0, 211].

Choose 8 > 0 so that Iy(t)i > 8 for all t E [0, 2':1"]. If Pi and P z are trigonometric polynomials such that IFlt) - y(r)1 <014 for all t E [0, 211] (their existence is assured by Theorem 8.15), prove that

Ind (P,) ~ Ind (P,)

by applying Exercise 25.Define this common value to be Iod (y).

Prove that the statements of Exercises 24 and 25 hold without any differentiability assumption.

27. Let f be a continuous complex function defined in the complex plane. Suppose

there is a positive integer n and a complex number e i= 0 such that

lim r"/(z) ~ c.1%1-'"

Prove thatf(z) = 0 for at least one complex number z.

Note that this is a generalization of Theorem 8.8.Hint: Assume/(z) #0 for all z. define

y,(l) = f(re/l)

for 0:::; r < x, 0:::;;; t::;:: 27., and prove the following statements about the curvesy.:

o(a) Ind (Yo) ~ O.

(b) Ind (Yr) = n for ail sufficiently large-r.

(c) Ind (Yr) is a continuous function of r, on [0, :0),

[In (b) and (e), use the last part of Exercise 26.]

Show that (a), (b), and (c) are contradictory, since n > O.28. Let Jj bot the closed unit disc in the complex plane. (Thus z E [j if and oniy if

!z I :::;;; 1.) Let 9 be a conrinuous";':i1apping of [) into the unit circle T. (Thus,]g(z)1 = 1 for every =ED.)

Prove that g(z) = -z for at least one z E T.

Hint: For O::;;:r::;;: 1,0:::;;;t::;::2r., put

y.(t) ~ g(rc"l.

and put !fi(t) = e-i'y,(t). If g(=);;:f=. -z lor every =E T, then !f(t) ? -1 for every

t E [0, 211]. Hence Ind (,p) = 0, by Exercises 24 and 26. It follows that Ind (Yl) = 1.But Ind (yo) = O. Derive a contradiction, as in Exercise 27.

29.

30.


Prove that every continuous mapping/of D into D has a fixed point in D.(This is the 2-dimensional case of Brouwer's fixed-point theorem.) .Hint: Assume fez) oF z for every zED. Associate to each zED the pomt

g(z) E T which lies on the ray that starts at fez) and passes through z. Then 9

maps fJ into T, g(z) = z if z E T, and 9 is continuous, because

g(z) ~ z - ,(z)[f(z) - zl.where s(z) is the unique nonnegative root of a certain quadr~tic equation whose

coefficients are continuous functions of / and z. Apply ExerCIse 28.

Use Stirling's formula to prove that

. rex ~ c) 1hm---~..._~ xcr(x)

r' -lim v'~ (I - x')" dx ~ v'~.II-~ ~-1

9FUNCTIONS OF SEVERAL VARIABLES

LINEAR TRANSFORMATIONS

We begin th~s chapter with a discussion of se-ts of vectors in euclidean n-space R".The algebraIC facts presented here extend without change to finite-dimensionalv~ct0.r spaces ove~ a~y field o~ .scalars. However, for our purposes it is quiteSuffiCIent to stay wIthm the famIlIar framework provided by the euclidean spaces.

9.1 Definitions

(a) A nonempty set Xc: R" is a vector space if x + YE X and ex E Xfor all x E X, YE X, and for all scalars c.(b) If Xl .. , .. " "*" E R" and c1 , ••• , c" are scalars, the vector

i::. ...:altooa litnar cJmhDwtion of Xl' ... , xk • If 5 c R" and iF E is the s.::(of all linear combinations of elements of S, we say that S spans E or thatE is the span af S. '

Observe that every span is a vector space.

FU:SCTIO:SS OF SEVERAL YARl.-\BLES 205

(c) A set consisting of vectors Xl' ... , Xk (we shall use the notation{Xl> ... , x/;J for such a set) is said to be independent if the relationCtX t + '" + CkXk = 0 implies that c1 ='" = Ck = O. Otherwise {Xl> .•. , xJis said to be dependent.

Observe that no independent set contains the null vector.(d) If a vector space X contains an independent set of r vectors but can·tains no independent set of r + 1 vectors, we say that X has dimension r,and write: dim X = r.

The set consisting of 0 alone is a vector space; its dimension is O..(e) An independent subset of a vector space X which spans X is calleda basis of X.

Observe that if B = (Xl"'" Xr } is a basis of X. then every x EXhas a unique representation of the form x == Icjx j . Such a representationexists since B spans X. and it is unique since B is independent. Thenumbers C1, •.. , c~ are called the coordinates of x with respect to thebasis B.

The most familiar example of a basis is the set {e 1•... , ell}' whereej is the vector in R" whosejth coordinate is 1 and whose other coordinatesare all 0. IfxE R", X =(x 1, ••• , x,,), then x ==Ixjej . We shall call

{e, .... ,e,}

the standard basis of R".

9.2 Theorem Let r be a positire integer. If a rector space X is spanned by aset of r vectors, then dim x::;: r.

Proof If this is false. there is a ,'eclor space X which contains an independent set Q= {Yl .... ' Y~+l}and which is spanned by a set So consistingof r vectors.

Suppose °::; i < r. and suppose a set 5, has been constructed whichspans X and which consists of all Yj with 1 ::;j ::; i plus a certain coHectionof r - i members of So' say Xl' ... , Xr _,. (In other words. Si is obtainedfrom So by replacing i of its elements by members of Q. without alteringthe span.) Since Si spans X. }'i+ 1 is in the span of 5,; hence there arescalars G1, , •. , ai +1> b1• •••• br - i , with ai+l == 1. such that

i..,. 1 r-i

IajYj+ Ib,x,=O.j= 1 k= 1

If all bk's were O. the independence of Q would force all a/s to be 0, ac0~t!"adicti0n. Jt f,,!lows thM some X;. E Si is fl linear combination of theother members of Ti == Si U {Yi+1}' Remove this x" from T i and call theremaining set Sj+ 1- Then Si+ 1 spans the same set as Ti • namely X, sothat Si+ 1 has the properties postulated for Si with i + 1 in place of i.

206 PRINCIPLES OF MATHE.\tATICAL ANALYSIS

Starting with So, we thus construct sets 5 1"",Sr' The last ofthese consists of y10"" Yr' and our construction shows that it spans X.But Q is independent; hence Yr+ t is not in the span of Sr' This contradiction establishes the theorem.

Corollary dim RII = n.

Proof Since {e 1, ••• , elf} spans RII, the theorem shows that dim RII ~ n.Since {e lo "" elf} is independent, dim RII ~ n.

9.3 Theorem Suppose X is a vector space, and dim X = n.

(a) A set E ofn vectors in X spans X if and only if E is independent.(b) X has a basis, and every basis consists ofn vectors.(c) If I ::;: r ::::;: nand {Yt> "" Yr} is an independent set in X, then X has a

basis containing {Yt, .. ·, Yr}.

Proof Suppose E ~ (x" "', x,). Since dim X = n, the set (x" ... , x" y)is dependent, for every YE X. If E is independent, it follows that Y is inthe span of E; hence E spans X. Conversely, if E is dependent, one of itsmembers can be removed without changing the span of E. Hence Ecannot span X, by Theorem 9.2. This proves (a).

Since dim X = n, X contains an independent set of n vectors, and(a) shows that every such set is a basis of X; (b) now follows from 9.I(d)and 9.2.

To prove (c), let (x" .... x,) be a basis of X. The set

spans X and is dependent, since it contains more than n vectors. Theargument used in the proof of Theorem 9.2 shows that one of the x/s isa linear combination of the other members of S. If we remove this Xi fromS, the remaining set still spans X. This process can be repeated r timesand leads to a basis of X which contains {y1, .•. , Yr}, by (a).

9.4 Definitions A mapping A of a vector space X into a vector space Y is saidto be a linear transformation if

A(cx) = cAx

for .?!1 x, .X:, Xl e X .a.:1d ell scalar:s c. . N-ote that one often write~ -AA insteadof A(x) if A is linear.

Observe that AO = 0 if A is linear. Observe also that a linear transformation A of X into Y is completely determined by its action on any basis: If

FUNCnO:-iS OF SEVERAL VARJABLES 207

{X t , ... , XII} is a basis of X, then every X E X has a unique representation of theform

,X=LCjXi'

i: t

and the linearity of A allows us to compute Ax from the vectors Ax!> .,., Ax"and the coordinates cb ••. , C" by the formula

,Ax=LciAxi •

j: I

Linear transformations of X into X are often called linear operators on X.If A is a linear operator on X which (i) is one~to-one and (ii) maps X ontoX, we say that A is invertible. In this case we can define an operator A-Ion Xby requiring that A- 1(AX) = X for all x E X. It is trivial to verify that we thenalso have A(A -IX) = x. for all x E X. and that A -1 is linear.

An important fact about linear operators on finite-dimensional vectorspaces is that each of the above conditions (i) and (ii) implies the other:

9.5 Theorem A linear operator A on a finite-dimensional vector space X isone-Io-one if and only if the range of A is ali of X.

Proof let {Xl' .. " X,,} be a basis of X. The linearity of A shows thatits range 2l(A) is the span of the set Q = (Ax" ... , Ax,). We thereforeinfer from Theorem 9.3(0) that 2l(A) = X if and only if Q is independent.We have to prove that this happens if and only if A is one-to-one.

Suppose A is one-to-one and ICiAxi = O. Then A(Ic;xJ = 0, henceIcjxi = 0, hence Cl ='" = C" = 0, and we conclude that Q IS independent.

Con .....ersely, suppose Q is independent and A(IcixJ = O. ThenIC i AX j = 0, hence Cl = ... = C" = 0, and we conclude: Ax = 0 only ifx ~ O. If now Ax = Ay. then A(x - y) = Ax - Ay = 0, so that x - y ~ 0,and this says that A is one-to-one.

9.6 Definitions

(a) Let L(X, Y) be the set of all linear transformations of the vector spaceX into the vector space Y. Instead of L( X, X), we shall simply write L( X).If At, A z EL(X, Y) and if C I , Cz are scalars. define ctA 1 + Cz Az by

(c,A, + c, A,)x = c,A,x + c, A,x (x EX).

It is then dear that cIA l + Cz Az E L(X, Y).(b) If X, Y, Z are veCi.cr spaces. and if A E L(x' Y) aild Be L(Y, Zj, ~J,ie

define their product BA to be the composition of A and B:

(BA)x = B(Ax) (x EX).

Then BA E L( X, Z).

208 PR1:-';CIPLfS OF MATHEMATICAL ANALYSIS

Note that BA need not be the same as AB. even if X = Y = Z(c) For A EL(R', Rm), define the norm IIAII of A to be the sup of allnumbers lAx], where x ranges over all vectors in R" with Ix]:::::; 1.

Observe that the inequality

lAx!,; IIA'llxl

holds for all x E R', Also, if i is such that IAx I ,; ),1 x! for all x E R',then I! A!! :$ I..

9,7 Theorem

(a) If A E L(R', Rm), then !IAII < 00 and A is a umformly continuousmapping of R' into Rm,

(b) If A, BEL(R', Rm) and c is a scalar, then

IIA + BII ,; !IAII + IIBI!, IlcA11 = Icl IIAii.

With the distance between A and B defined as i!A - Bli. L(R", Rm ) is ametric space.

(c) If A E L(R', Rm) and B E L(Rm, R'), then

rlBA11 ,; IIB!I ilAil·

Proof

(a) Let {e l ....• ell} be the standard basis in R" and suppose x = :Ecie i ,

Ixl ,; 1, so that Ic,l ,; I for i = 1, ... , n. Then

IAxi =iL: c,Ae,j,; L: Ic,! IAe,i ,; L: IAe,[

FUNCTIONS OF SEVERAL VARIABLES 209

and it is easily verified that IIA - Eil has the other properties of a metric(Definition 2.15).(c) Finally, (c) follows from

[(BA)xl ~ IB(Ax) [ ,; IIBIIIAxl,; 11B'111Alllxl.

Since we now have metrics in the spaces L(R", Rm), the concepts of openset, continuity, etc., make sense for these spaces. Our next theorem utilizesthese concepts.

9.8 Theorem Let Q be the set of all invertihle linear operators 011 R".

(a) If A En, BE L(R'), and

liB - A II '1IA- 1 11 < 1,

then BEn.(h) .n is an open subset ofL(R"), and the mapping A ........ A-I is continuous

on n.(This mapping is also obviously a 1 - 1 mapping of .Q onto n,

which is its own inverse.)

Proof

(a) Put li['I! = 1/" put liB - All = p. Then p<,. For every x E R',

,ix! = ,!['Axl ,; ,I[['I!' IAxl

= lAx!,; I(A - B)xl + IBxl,; Pixl + IBx[,

so that

Since, - P> 0, (1) shows that Bx # 0 if x # O. Hence B is 1 - 1.By Theorem 9.5, BEn. This holds for all B with liB - A II <,. Thuswe have (a) and the fact that .n is open.(b) Next, replace x by B-'y in (I). The resulting inequality

(2) ('-P)IB-'Y!';IBB-1YI~IYI (YER')

shows that iIB-'II,; (, - P)-l. The identity

B-' - A-I = B- 1(A - B)A -I,

combined with Theorem 9.7(c), implies therefore that

IIS-' - ['II ,; IIB-1

1111A - BII IIA- 1Ii ,; * ~ P)'

so that

,IIAII,;2: IAe,:< 00.

i= 1

Since lAx - AYI'; :IAII Ix - YI if x, YE R', we see that A is uniformlycontinuous.(b) The inequality in (b) follows from

i(A + B)xj = lAx + Bxl:$ jAxi -+- IExi ~ (!!Ail + !IBil) ixl.

The second part of (b) is proved in the same manner. If

A, 13, C E L(R", Rm),

we have the triangle inequality

(I) (, - P)lx!,; IBxl (x E R').

!IA - q = II(A - B) + (B - C)II ,; ilA - Bil + liB - q, This establishes the continuity assertion made in (b), since f3 ........ 0 as B - A.

210 PRI~CIPLES OF MATHEMATICAL A1"ALYSIS

FUNCTIO:-;S OF SEVERAL VARIABLES 211

It is convenient to visualize these numbers in a rectangular array of m rowsand n columns, called an m by n matrix:

(l SkSp, I SJSn).(5)

'IAII «> a' I'!21 - ti Il} .

If we apply (6) to B - A in place of A, where A, BE L(R", Rm ), we seethat if the matrix elements aij are continuous functions of a parameter, then thesame is true of A. More precisely:

(6)

the independence of {Zt, "', zp} implies that

If S is a metric space, if au, ... , am" are real continuous functions on Sand If, for each PES. A p is the linear transformation of R" into Rm whose matri.~has entries aij(p), then the mapping p - Ap is a continuous mapping of S intoL(R", Rm

).

Thus

This shows how to compute the p by n matrix [BA] from [B] and [A]. If wedefine the product [B](A] to be [BA], then (5) describes the usual rule of matrixmultiplication.

Finally, suppose {Xl' ... , X,,} and {y 1> ••. , )'m} are standard bases of R" andRm, and A is given by (4). The Schwarz inequality shows that

jAxl' = I (2: aljcj)' 2 S I II a;J· I cJ)' ~ I a;~ixi'.I 1 I \ 1 j I, j

DIFFERENTIATION

9.10 Preliminaries In order to arrive at a definition of the derivative of afunction whose domain is R" (or an open subset of R"), let us take another lookat the familiar case n = 1, and let us see how to interpret the derivative in thatcase in a way which will naturally extend to n > 1.. Iff is a real function with domain (a, b) c: R' and if x E (a, b), thenj'(x)15 usually defined to be the real number

(7) 1. f(x + h) - f(x)1m ,

h-O h

provided, of course, that this limit exists. Thus

(lsjSn).

(BA)x j =I ekj Zk'k

m

AXj = L aijYii: 1

(3)

9.9 Matrices Suppose {Xl' ... , X,,} and {Yl' .. " Ym} are bases of vector spacesX and Y, respectively. Then every A E L(X, Y) determines a set of numbersau such that

Observe that the coordinates au of the vector Ax; (with respect to the basis(Y""" Ymll appear in the jth column of [A]. The vectors AXj are thereforesometimes called the column vectors of [A]. \Vith this terminology, the rangeof A is spanned by the column "ectors of[A].

If x = LC j x j , the linearity of A, combined with (3), shows that

(4) Ax =J, Ct,a;jCj) Y,·

Thus the coordinates of Ax are Ljaijc;. Note that in (3) the summationranges over the first subscript of aij , but that we sum over the second subscriptwhen computing coordinates.

- Suppose next that an m by n matrix is given, \vith reaI entries a ij' If A isthen defined by (4), it is clear that A E L(X, Y) and that [A] is the given matrix.Thus there is a natural 1-1 correspondence between L(X. Y) and the set of all

:J real m by n matrices. We emphasize, though, that [A] depends not only On Abut also on the choice of bases in X and Y. The same A may give rise to manydifferent matrices if we change bases, and vice versa. \Ve shall not pursue thisobservation any further, since we shall usually work with fixed bases. (Someremarks on this may be found in Sec. 9.37.)

If Z is a third vector space, with basis {Zl' "" zpj, if A is given by (3),and if

where the "remainder" r(h) is small, in the sense that

th.::n A E L(X, Y), BE L(Y. Z), BA E L(X, Z). and since

B(Ax) = B I a'jY' = I aljBy,; .

(8) f(x + h) - f(x) = j'(x)h + r(h)

(9) . r(hlhm-' =0.h-O h

212 PRI:-lCIPLES OF :MATHEMATICAL A.....ALYSLS FUNCTIONS OF SEVERAL VARIABLES 213

as t -0.

9.12 Theorem Suppose E and f are as in Definition 9.11, x E E. and (14) holdswith A =A 1 and with A =A 2 . Then Al =A 2 •

Proof If B = A 1 - A" the inequality

IBhl S If(x + h) - f(x) - A,hl + If(x + h) - f(x) - A, hi

shows that IBh I/Ih I ~ 0 as h~ O. For fixed h '" O. it follows that

IB(th) ~O

Ithl

The linearity of B shows that the left side of (16) is independent of t.Thus Bh = 0 for every hER'. Hence B = O.

It is of course understood in (14) that hER'. If Ihl is small enough, thenx + h E E. since E is open. Thus f(x + h) is defined, f(x + h) E RO, and sinceA EL(R', RO), Ah E RO. Thus

f(x + h) - f(x) - Ah E RO.

The norm in the numerator of (14) is that of R"'. In the denominator we havethe R"-norm of h.

There is an obvious uniqueness problem which has to be settled beforewe go any further.

(16)

9.13 Remarks

f(x + h) - f(x) ~ hy + r(h),

. (f(X + h) - f(x) .. )hm - y = o." .... 0 h

We can again rewrite this in the form

Note that (8) expresses the difference I(x + h) - I(x) as the sum of thelinear function that takes h to f'(x)h, plus a small remainder.

We can therefore regard the derivative of f at x, not as a real number,but as the linear operator on R' that takes h to j'(x)h.

[Observe that every real number 0: gives rise to a linear operator on R 1;

the operator in question is simply multiplication by!J.. Conversely, every linearfunction that carries R 1 to R 1 is multiplication by some real number. It is thisnatural 1-1 correspondence between R 1 and L(R1

) which motivates the preceding statements.]

Let us next consider a function f that maps (a, b) c R1 into R"'. In thatcase, f'(x) was defined to be that vector y E R'" (if there is one) for which

(10)

(11)

where r(h)/h~O as h~O. The main term on the right side of (11) is again alinear function of h. Every y E R'" induces a linear transformation of R1 intoR"', by associating to each hE R1 the vector hy E.R'". This identification of R'"with L(R' , R") allows us to tegard f'(x) as a member of L(R' . RO).

Thus, iff is a differentiable mapping of (a, b) c: R' into RO, and if x E (a, b),then f'(x) is the linear transformation of R1 into R'" that satisfies

If f is differentiable at every x E E, we say that f is differentiable in E.

9.11 Definition Suppose E is an open set in R", f maps E into R"', and x E E.If there exists a linear transformation A of R" into R'" such that

where the remainder r(h) satisfies

lim Ir(h)i ~O.b-O ihl

We may interpret (17), as in Sec. 9.10, by saying that for fixed x and smallh, the left side of (17) is approximately equal to f·(x)h. that is. to the valueof a linear transformation applied to h.(b) Suppose f and E are as in Definition 9.11, and f is differentiable in E.For every x E E, f'(x) is then a function, namely. a linear transformationof R" into Rm

. But f' is also a function: f' maps E into L(R", R"').(c) A gianct Cil (I 7) shows that f i~ continuous at any point at which f isdifferentiable.(d) The derivative defined by (14) or (17) is often called the differentialof f at x, or the total derivative of f at x, to distinguish it from the panialderivatives that will occur later.

(a) The rdation (i4) can be rewritten in the form

(18)

(17) f( x + h) - f(x) = f'(x)h + r(h)

f'(x) ~ A.

. f(x + h) - f(x) - f'(x)hhm h = 0,'-0

. If(x + h) - f(x) - f'(x)hlhm I I =0.h ..... O h

We are now ready for the case n > 1.

. If(x + h) - f(x) - Ahl;'~ [hi =0,

then we say that r is differentiable at x, and we write

or, equivalently,

(15)

(13)

(14)

(12)

214 PRINCIPLES OF MATHEMATICAL ANALYSIS FUNCTIONS OF SEVERAL VARIABLES 215

9.16 Partial derh'atives We again consider a function f that maps an openset E c R" into Rm

. Let {e l , . .-., en} and {u 1, ... , Um} be the standard bases ofR" and R m

. The components of f are the real functions 11, .. .• 1111 defined by

Hence (22) and (23) imply, for h " 0, that

IF(xo + h) -;~(xo) - BAhI ,; IIBII e(h) + [IIA II +e(h) ]~(k),

Let b~ O. Then e(h) ~ 0, Also, k ~ 0, by (23), so that ~(k) ~ O.It follows that F'(xo) = BA, which is what (21) asserts,

9.14 Example We have defined derivatives of functions carrying R" to Rm tobe linear transformations of R" into Rm

. What is the derivative of such a lineartransformation? The answer is very simple.

If A EL(R", Rm) and ifx E R", then

(19) A'(x) ~ A.

Note that x appears on the left side of (19), but not on the right. Bothsides of (19) are members of L(R", Rm), whereas Ax E Rm.

.- The p.roof of (19) is a triviality;. since

(20) A(x + h) - Ax = Ah,

by the linearity of A. With f(x) ~ Ax, the numerator in (14) is thus 0 for everyhER". In (17), r(h) = O. (24)

•f(x) ~ I j,(x)u,

i= I(x E E),

We now extend the chain rule (Theorem 5.5) to the present situation.

9.15 Theorem Suppose E is an open set in R", f maps E into Rm, f is differentiable

at Xo E E, g maps an open set containing f(E) into Rk, and g is differentiable at

f(xo)· Then the mapping F of E into R' defined by

or, equivalently, by f,(x) ~ f(x) . Uj , 1 ,; i,; m.for x E E, I ~ i ~ m, 1 ~ j ~ n, we define

(25)

F(x) = g(f(x»

is differentiable at xo, and

provided the limit exists. Writing J;(Xt, ... , xn) in place ofh(X), we see tha~

Djli is the derivative ofJ; with respect to xj , keeping the other variables fixed.The notation

On the right side of (21), we have the product of two linear transformations, as defined in Sec. 9.6.

Proof Put Yo ~ f(xo), A = f'(xo), B = g'(yo), and define

u(h) ~ f(xo + h) :... f(xo) - Ah,

v(k) = g(yo + k) - g(yo) - Bk,

for all hER" and k E Rm for which f(xo + h) and g(yo + k) are defined,Then

F'(xo) ~ g'(f(xo»f'(xo)'

and'

where e(h) ~ 0 as h ~ 0 and ~(k) ~ 0 as k ~ O.Given h. put k = f(xo + h) - f(xo). Then

Ik! = !Ah + u(h)I,; [IIAII + e(h)] IhI.

(1 ,;j,; n)..

f'(x),) = I (D)j,)(x)u,i= 1

(26)of,OXj

is therefore often used in place of D j/i' and Djft is called a partial derivative.In many cases where the existence of a derivative is sufficient when dealing

with functions of one variable, continuity or at least boundedness of the partialderivatives is needed for functions of several variables. For example, thefunctions I and 9 described in Exercise 7, Chap. 4, are not continuous, althoughtheir partial derivatives exist at every point of R2

. Even for continuous functions.the existence of aU partial derivatives does not imply differentiability in the senseof Definition 9,11; see Exercises 6 and 14, and Theorem 9.21.

However, if f is known to beaifferentiable at a point x, then its partialderivatives exist at x, and they determine the linear transformation f'(x)completely:

9~17 Theor~m Suppose f maps '1.-: open set E c: R" lnto R.'II, andf is differentiableat a point x E E. Then the partial derivatiues (Djft)(x) exist, and

(27)

Iv(k)1 ~ ~(k)! kl,lu(h)1 =e(h)lhl,

F(xo + h) - F(xo) - BAh ~ g(yo + k) - g(yo) - BAh

= B(k - Ah) + v(k)

~ Bu(b) + v(k).

(21)

(22)

(23)

216 PRlNCIPLES OF MATHEMATICAL ANALYSIS FUNCTIOSS OF SEVERAL VARIABLES 217

This is a frequently encountered special case of the chain rule. It can berephrased in the following manner.

Associate with each x E E a vector, the so·called "grariient" of f at x,defined by

Since i(t) E L(R' , R") and f'(y(l)) E L(R", R' ), (32) defines g'(t) as a linearoperator on RI

. This agrees with the fact that 9 maps (a, b) into R I• However,

g'(I) can also be regarded as a real number. (This was discussed in Sec. 9.10.)This number can be computed in terms of the partial derivatives off and thederivatives of the components of y, as we shall now see.

With respect to the standard basis {e"

... , e"} of R", [y'(t)] is the n by Imatrix (a "column matrix") which has "I; (t) in the ith row, where 1'1' "" "III arethe components ofy. For every x E E, [f'(x)] is the I by n matrix (a "row matrix")which has (Djf)(x) in thejth column. Hence [g'(t)J is the 1 by 1 matrix whoseonly entry is the real number

Here, as in Sec. 9.16, {e"

... , e"} and {u"

... , um } are the standard basesof R/I and Rm•

Proof Fixj. Since f is differentiable at x,

f(x + te;) - f(x) = f'(x)(le;) + r(le j )

where Irete;) I/t ~ 0 as t ~ O. The linearity of f'(x) shows therefore that

(28) I' f(x + te;) - f(x) ,1m = f (x)ej .

1-0 t

If we now represent f in terms of its components, as in (24), then (28)becomes

(29) I' ~ f,(x + Ie;) - J:(x) f'( )1m L Uj = x ej .t ..... O i; I t

It follows that each quotient in this sum has a limit, as t _ 0 (see Theorem4.10), so that each (DJ,)(x) exists, and then (27) follows from (29).

(33) "g'(t) ~ 1: (DJ)(tCt))y; (I).i: I

Here are some consequences of Theorem 9.1 7:Let (f'(x)] be the matrix that represents f'(x) with respect to our standard

bases, as in Sec. 9.9.

(34)

Since

"(Vf)(x) ~ 1: (DJ)(x)e,.,: I

Then f'(x)e j is thejth column vector of [f'(x)], and (27) shows thereforethat the number (DJ,)(x) occupies the spot in the ith row and jth column of[f'(x)]. Thus

9,18 Example Let y be a differentiable mapping of the segment (a, b) c R'into an open set E c R", in other words, "I is a differentiable curve in E. Let!be a real-valued differentiable function with domain E. Thus f is a differentiablemapping of E into R I . Define

[f'(x)] ~ [(~lf~)(X) (D:J.I)(X)].

(Ddm)(x) (DJm)(x)

If h ~ 'Lhj ej is any vector in R", then (27) implies that iul = 1), and

(- 00 < I < cc).

g'(I) = (Vf)(y(t)) . i(l),

y(t) ~ x + tu

lim f(x + tu) - f(x) ~ (Vf)(x) . U.

1-0 t

Hence (38) gives

"(35) let) = 1: y; (t)e"1=1

(33) can be written in the form

the scalar product of the vectors (Vf)(y(I)) and let).Let us now fix an x E E, let U E RII be a unit vector (that is,

specialize 'I so that

(36)

(37)

(39)

Then let) = u for every t. Hence (36) shows that

(38) g'(O) ~ (Vf)(x) . u.

On the other hand, (37) shows that

get) - g(O) = f(x + tu) - f(x).

(a < t < b).

get) = fCt'(t» (a < i < b).

The chain rule asserts then that

(30)

(3! )

(32) g'(I) ~ f'(y(t))i(l)


The limit in (39) is usually called the directional derivative of/at x, in thedirection of the unit vector D, and may be denoted by (Duf)(x),

Iff and x are fixed, but u varies, then (39) shows that (D.f)(x) attains itsmaximum when u is a positive scalar multiple of (Vf)(x). [The case (Vf)(x) ~ 0should be excluded here.]

If u ~ LU, e" then (39) shows that (D.f)(x) can be expressed in terms ofthe partial derivatives offat x by the formula

FUNCTIONS OF SEVERAL VARJABLES 219

9.20 Definition A differentiable mapping f of an open set E c R" into Rm issaid to be continuously differentiable in E if f I is a continuous mapping of Einto L(R", Rm

).

More explicitly, it is required that to every x E E and to every e > 0corresponds a c:5 > 0 such that

Ilf'(y) - f'(x)11 < ,

(40) "(D.J)(x) ~ I (DJ)(x)u,.i=l

ifYEEand Ix-yl<b.If this is so, we also say that f is a ~'-rnapping, or that f E 11'(E),

Some of these ideas will playa role in the following theorem,

9.19 Theorem Suppose f maps a convex open set E c: R" into R"', f is differentiable in E, and there is a real number .?vI such that

lIf'(x) II ,; M

for every x E E. Then

If(b) - f(a) I ,; Mlb - al

foral/aEE, bEE.

Proof Fix a E E, bEE. Define

((t) = (I - l)a + Ib

for all IE R' such that ((I) E E. Since E is convex, ((I) E E if 0 ,; I ,; 1.Put

g(t) ~ f(((I)).

Then

9,21 Theorem Suppose f maps an open sel E c K' into Rm• Then f E 'C'(E) if

and only if the partial derivatives Djfi exist and are continuous on E for 1 ::;; j :s;: m,1 ::;;j ::;; n,

Proof Assume first that f E 'C'(E). By (27),

(DJ,)(x) = (f'(x)e) . u,

for all i, j, and for all x E E. Hence

(Djf,)(y) - (Djf,)(x) = ([f'(y) - f'(x)Jej}· u,

and since lu,1 ~ lejl ~ I, it follows that

I(DJ,)(y) - (Djf,)(x) I ,; 1[f'(y) - f'(x)Jeji

,; Ilf'(y) - f'(x)il·

Hence D jf: is continuous,For the convene, it suffices to consider the c.:ase m = 1. (Why'?)

Fix x E E and e > 0, Since E is open, there is an open ball SeE, withcenter at x and radius r, and the continuity of the functions Dj / showsthat r can be chosen so that

g'(t) = f'(((t))y'(I) = f'(((l))(b - a),

so that (41),

I(Djf)(y) - (DJ)(x) I < ~ (YES, I,;j';n).

ig'(I)I,; Ilf'(((t))lllb - al ,; Mlb - al

for all IE [0, I]. By Theorem 5.19,

Ig(I) - g(O) I ,; Mlb - ai.

But g(O) = f(a) and g(I) ~ feb). This completes the proof.

Corollary - If,-in addilion, f'(x) = 0 for ail x EE, Ihen f is conslanl.

Proof To prove this, note that the hypotheses of the theorem hold nowwith lvl = 0,

Suppose h = I.hjej, Ihi < r, put Vo = 0, and v,I: = h1e 1 +". + hke.\:>for 1 ::;; k ::;; n, Then

"(42) f(x + h) - f(x) = I [f(x + v) - f(x + Vj-I)]'

j~l

Since Ivlol < r for 1 :s:; k::; n and since S is convex, the segments with endpoints X-t-Vj_1 and x+vj lie in S. Since "j=vj _ 1 +;lj~j' the meanvalue theorem (5.10) shows that thejth summand in (42) is equal to

hj(DJ)(x + vj _ 1 + 8j hj e)

220 PRINCIPLES OF MATHEMATICAL Ai'iALYSISFU:-':C1l0~S OF SEVERAL VARIABlES 221

for some $jE(O, 1), and this differs from hj(DJ)(x) by less than Ihj!e!n,using (41). By (42), it follows that

I 'I l'I(x + h) - I(x) -J,hj(Djf)(x) ;; ~ J, Ihjle;; Ih[e

for all h such that [h I< r.This says that I is differentiable at x and that j'(x) is the linear

function which assigns the number l:h/Djf)(x) to the vector h ~ ,[hje}.The matrix [I'(x)] consists of the row (D1f)(x), ... , (D,f)(x); and sinceD11. ..-., DJ are continuous functions on·E,' the concluding iemarksoiSec. 9.9 show that/E~"(E).

If n < m, it follows that

m

d(x" xm);; L d(x i , Xi-I)i="+l

s:; (e" + e,,+l + ... + Cm- 1) d(x 1 , X o)

;; [(1 - C)-l d(x 1, XO)]C'.

Thus {x~} is a Cauchy sequence. Since X is complete, lim X~ = x for someXE X.

Since cP is a contraction, \fJ is continuous (in fact. unifC'fmly con~

tinuous) on X. Hence

rp(x) = lim <pix,) = lim X'd = X.

THE CONTRACTION PRINCIPLE

We now interrupt our discussion of differentiation to insert a fixed pointtheorem that is valid in arbitrary complete metric spaces. It will be used in theproof of the inverse function theorem.

9.22 Definition Let X be a metiic space, with metric d. If rp maps X into Xand if there is a number c < 1 such that

THE INVERSE FUNCTION THEOREM

The inverse function theorem states, roughly speaking, that a continuouslydifferentiable mapping f is invertible in a neighborhood of any point x at whichthe linear transformation f'(x) is invertible:

for all x, y E X, then <p is said to be a contraction of X into X.

9.23 Theorem If X is a complete metric space, and If r.p is a contraction of Xinto X, then there exists one and only one x E X such that <p(x) = X.

: In other words, <p has a unique fixed point. The uniqueness is a triviality,for if rp(x) = x and <p(y) = y, then (43) gives d(x, y) ;; c d(x, y), which can onlyhappen when d(x, y) ~ O.

The existence of a fixed point of cp is the essential part of the theorem.The proof actually furnishes a constructive method for locating the fixed point.

Proof Pick Xo E X arbitrarily, and define {x~} recursively, by setting

Writing the equation y = f(x) in component form. we arrive at the follow~

ing interpretation of the conclusion of the theorem: The system of n equations

(1;;i;;n)

(x E U).g(f(x)) = x

then g E 'O"(V).

9.24 Theorem Suppose f is a rt'-mapping of an open set E c R" into R'\ f'(a)is invertible for some a E E, and b = f(a). Then

(a) there exist open sets U and V in R" sueh that a E U, bE V, f is one-toone on U, andf(U) = V;

(b) if g is the inverse off [which exists. by (a)]. defined in V by

d(rp(x), rp(y)) ;; c d(x, y)(43)

Choose c < 1 so that (43) holds. For n ;, 1 we then have

d(x,,+ 1. x,,) = d(cp(xll ), CP()',,-l)) s: c d(x" , xn - 1).

Hence induction gives

(44)

(45)

(n = 0, 1,2, ...).

(n = 0, I, 2, ...).

can be solved for Xl' ... , x~ in terms of )'1' ... , y~ , if we restrict x and y to smallenough neighborhoods of a and b; the solutions are unique and continuouslydifferentiable.

Proof

(a) Put ('(0) = A, and choose ;. so that

(46)

222 PRDiCIPLES OF MATHEMATICAL A.."lALYSIS FUNCTIO",S OF SEVERAL VARIABLES 223

Since f' is continuous at a, there is an open ball U c E, with center at a,such that

Ilf'(x) - Ail <,l (x E U).

We associate to each YE Rn a function cp, defined by

rp(x) = x + [ley - f(x) (x E E).

Note that f(x) ~ y if and only ifx is a fixed point 01 '1'.

Since rp'(x) = [- A-If'(x) = ['(A - f'(x», (46) and (47) implythat

(47)

(48)

(49) IIrp'(x)jj < t (x E U).

By (46), (47), and Theorem 9.8, f'(x) has an inverse, say T. Since

g(y + k) ~ g(y) ~ Tk = h - Tk = - T(f(x + h) - f(x) - f'(x)h],

(51) implies

Ig(y + k) - g(y) - Tkl lin If(x + h) - f(x) - f'(x)hl=CO---'---clk'-;I~-----' ,,; -i.- . Ihi'

As k ~ 0, (51) shows that h ~ O. The right side of the last inequalitythu~ tends to O. Hence the same is true of the left. Vle have thus provedthat g'(y) = T. But Twas chosen to be the inverse off'(x) = f'(g(y). Thus

THE IMPLICIT FUNCTION THEOREM

Iflis a continuously differentiable real function in the plane, then the equationf(x, y) =°can be solved for y in terms of x in a neighborhood of any point

The hypotheses made in this theorem ensure that each point x E E has aneighborhood in which f is 1-1. This may be expressed by saying that f islocally one-to-one in E. But f need not be 1-1 in E under these circumstances.For an example, see Exercise 17.

9.25 Theorem IfC is a Cfl'-mapping of an open set E c R" into R" and iff'(x)is invertible for every x E E. then f( W) is an open subset oj Rn for every open setWeE.

In other words, f is an open mapping of E into Rn•

(y E V).g'(y) = {f'(g(y»)}-l

Finally, note that g is a continuous mapping of V onto U (since g

is differentiable), that f' is a continuous mapping of U into the set n ofall invertible elements ofL(Rn), and that inversion is a continuous mappingof n onto n, by Theorem 9.8. If we combine these facts with (52), we seethat g E 'f!"(V).


(52)

Remark. The full force of the assumption that f E 'f!"(E) was only usedin the last paragraph of the preceding proof. Everything else, down to Eg. (52),was derived from the existence of f'(x) for x E E. the invertibility of f'(a).andthe continuity off' at just the point a. In this connection, we refer to the articleby A. Nijenhuis in Amer. Math. Monthly, vol. 81,1974, pp. 969-980.

The foilov.-'ing is an immediate consequence of part (a) of the inversefunction theorem.

Hence

(50) Irp(x,) - rp(x,) I ,,; tix i - x,i (x" x, E U),

by Theorem 9.19. It follows that r.p has at most one fixed point in U, sothat f(x) = y for at most one x E U.

Thus Cis 1 - 1 in U.

Next, put V = feU), and pick Yo E V. Then Yo = f(xo) for someX o E U. Let B be an open ball with center at Xo and radius r > 0, so smalltha.t its closure 13 lies in U. V{e will show that y E V whenever Iy - Yo I< ).r.ThIs proves, of course, that V is open.

Fix y, Iy - yol < i.r. With 'I' as in (48),

Irp(xo) - xol = IA-I(y - Yo)1 < riA-'il!.r =~._ 2

If x E B, it therefore follows from (50) that

Irp(x) - Xo I ,,; irp(x) - rp(xo) i + Irp(xo) - Xo I

1 r<2Ix-xol +2,,;r;

hence rp(x) E B. Note that (50) holds if x, E B, X, E 8._ Thus rp is a contraction of 13- into B. Being a closed subset of R",B is_complete. Theorem 9.23 implies therefore that <p has a fixed pointx E B. For this x, I(x) = y. Thus y E feB) e f( U) = V.

This proves part (a) of the theorem.

(b) Pick y E V, Y+ k E V. Then there exist x E U, X + h E U. so thaty = f(x), y + k = f(x + h). With rp as in (48),

rp(x + h) - rp(x) = h + [l[f(x) - f(x + h») = h _ [lk.

By (50), Ih - [lkl,,; flh!. Hence IA-Ikl;, tlhl. and

(5l) Ihi,,; 2jjA- l illk! = i.-Ilki.

224 PIUNCIPLES OF MATHEMATICAL ANALYSIS

(a, b) at whichf(a, b) = 0 and efley # O. Likewise, one can solve for x in termsof y near (a, b) if eflex # 0 at (a, b). For a simple example which illustratesthe need for assuming efley # 0, consider f(x, y) = x 2 + yl - 1.

The preceding very informal statement is the simplest case (the casem = n = 1 of Theorem 9.28) of the so-called "implicit function theorem." Itsproof makes strong use of the fact that continuously differentiable transformationsbehave locally very much like their derivatives. Accordingly, we first proveTheorem 9.27, the linear version of Theorem 9.28.

FU1'iCTlO:-;S OF SEVERAL vARIABLES 225

Then there e~ist open sets U c R,,+m and We Rm, with (a, b) E U andb E W, having the following property :

To every yEW corresponds a unique x such that

(56) (x, y) E U and f(x, y) = o.

If this x is defined to be g(y), then g is a \f'-mapping of W into R", g(b) =-,

(5i) f(g(y), y) = 0 (y E W),

The function g is "implicitly" defioed by (57). Hence the name of the

theorem.The equation f(x. y) = 0 can be written as a system of n equations in

n + m variables:

[~'f~ D:!.,]Dlf" D"f"

evaluated at (a, b) defines an invertible linear operator in R"; in other w~rds,

its column vectors should be independent, or, equivalently, its determmantshould be *O. (See Theorem 9.36.) If, furthermore. (59) holds when x ~ a andy = b, then the conclusion of the theorem is that (59) can be sol~ed for Xl •.. '.' x"in terms of Yl"'" Ym. for every y near b, and that these solutIOns are contmu

ously differentiable functions of y.

Proof Define F by

(60) F(x, y) = (f{x, y), y) ((x, y) E E).

Then F is a 't§' -mapping of E into R,,+m. We claim that F'(a, b) is an

invertible element of L(K'+m):Since f(a, b) = 0, we have

f(a + b, h + k) ~ A(b, k) + r(h, k),

where r is the remainder that occurs in the definition of f'(a, b). Since

F(a + h, b + k) - F(a, b) ~ (f(a + h, b + k), k)= (A(h, k), k) + (r(h, k), 0)

9.26 Notation If x = (Xi> ... , x,,) E R" and y = (Yr . ... , YIr!) E Rm• let us write

(x, y) for the point (or vector)

(Xl' .. " X", Yi> ... , Ym) E R,,+m.

In what follows. the first entry in (x. y) or in a similar symbol will always be avector in R", the second will be a vector in RIr!.

Every A E L(R,,+m. R") can be split into two linear transformations Ax andA" defined by

(53) A.h = A(h, 0), A,k = A(O, k)

for any hER", k E Rm. Then Ax E L(R"), A, E L(Rm, R"), and

(54) A(h. k) ~ A.h + A,k.

The linear version of the implicit function theorem is now almost obvious.

9.27 Theorem ifA E L(K'+"'. R") and if Ax is invertible, then there-correspondsto every k E R m a unique hER" such that A(h, k) = O.

This h can be computed frcm k by the formulao

(55) h ~ -(A.r'A,k.

Proof By (54), A(h, k) ~ 0 if and only if

A.h+A,k~O,

which is the same as (55) when Ax is invertible.

The conclusion of Theorem 9.27 is. in other words. that the equationA(h, k) = 0 can be solved (uniquely) for h if k is given, and that the solution his a linear function of k. Those who have some acquaintance with linear algebrawill recognize this a3 a very fumiiiar statement abollt systems oflinear equations.

9.28 Theorem Let f be a f(j'·mapping of an open set E c R,,+m into R", suchthat f(a, b) = ofor some point (a, b) E E.

Put A = f'(3, b) and assume that Ax is invertible.

and

(58)

(59)

g'(b) = -(AX'A,.

f,,(X t , ...• X", Y1' ... , Ym) = O.

The assumption that Ax is invertible means that the n by n matrix

226 PRlNCIPLES OF MATHEMATICAL A..'1AJ.,YSlS FUNCTI01"S OF SEVERAL VARIABLES 227

it follows that F'(a, b) is the linear operator on R'·· that maps (b, k) to(A(h, k), k). If this image vector is 0, then A(b, k) = 0 and k = 0, henceA(h,O) = 0, and Theorem 9.27 implies that b = O. It follows that F'(a, b)IS I-I; hence It IS mvertible (Theorem 9.5).

The inverse function theorem can therefore be applied to F. It showsthat there exist open sets U and V in R"m, with (a, b) E U, (0, b) E V, suchthat F is a I-I mapping of U onto V.

We let W be the set of all y ERm such that (0, y) E V. Note thatbE W

It is clear that W is open since V is open.Ify E W, then (0, y) = F(x, y) for some (x, y) E U. By (60), f(x, y) =0

for thIS x.

Suppose, with the same y, that (x', y) E U and f(x', y) = O. Then

F(x', y) = (f(x', y), y) = (f(x, y), y) = F(x, y).

Since F is I-I in U, it follows that x' = x.This proves the first part of the theorem.

For the second part, define g(y), for YEW, so that (g(y), y) E U and(57) holds. Then

(61) F(g(y), y) = (0, y) (y E W).

If G is the mapping of Vonto U that inverts F, then G E C(j', bv the inversefunction theorem, and (61) gives "'

This is equivalent to (58), and completes the proof.

Note. In terms of the components of f and g, (65) becomes,I (DJ.J(a. b)(D,9j)(b) = -(D,.J,)(a, b)

j=: 1

or

i: (~fi) (~9j) = _ (~fi)j"l ,CXj 0Yk \Oh

where 1~ i .::;: n, 1 S; k S; rn.For each k. this is a system of n linear equations in which the derivatives

cgji"cyk. (l ~j:S;; n) are the unknowns.

9.29 Example Take n = 2, m = 3, and consider the mapping f = (/,.!,) ofR5 into Rl given by

f,(x" x" y" y" 13) = 2...' + X, Yl - 412 + 3

fz(xj, Xl' Yl' Y2, Y3) = Xl cos Xl - 6x 1 + 2YI - Y3'

If a = (0, I) and b = (3,2,7), then f(a, b) = O.With respect to the standard bases, the matrix of the transformation

A = f'(a, b) is

Since G E 'C', (62) shows that g E 'C'.Finally, to compute g'(b), put (g(y), y) = <illY). Then

(63) <il'(y)k = (g'(y)k, k) (y E W, k E Rm).

By (57), f(<il(y)) = 0 in W. The chain rule shows therefore that

f'(<il(y))<il'(y) = O.

When y = b, then <!ley) = (a, b), and f'(<il(y)) = A. Thus

(64) A<!l'(b) = O.

It now follows from (64), (63), and (54), that

Axg'(b)k + A,k = A(g'(b)k, k) = A<!l'(b)k = 0

for every k E RWI• Thus

Axg'(b) + A, = O.

(58) gives

'A] r 2 3 I -4-~ll = l I 2 0-6

Hence

[Ax] = [_~ ~l [A,] = [;-4

- ~]0

I [1[g'(3, 2,7)] = - 20 6

. _I] []_' I [1 -3][(Ax) = Ax = 20 6 2

We see that the column vectors of [AAl are independent. Hence A;c is invertibleand the implicit function theorem asserts the existence of act'-mapping g, definedin a neighborhood of (3, 2, 7), such that g(3, 2, 7) = (0, 1) and f(g(y), y) = o.

W; can use (58) to compute g'(3, 2, 7): Since

(y E W).(g()), y) = G(O, y)(62)

(65)

FUNCTIONS OF SEVERAL VARIABLES 229

228 P~CIPLES OF MATHEMATICAL ANALYSIS

In terms of partial derivatives, the conclusion is that

D,U, = tD,U, = -t

at the point (3, 2, 7).

THE RANK THEOREM

D,U, = tD 2 g2 =t

D 3 gl = - /0D3 g2 =fo

If Xl contains only 0, this is trivial: put Px = 0 for all x EX.Assume dim Xl = k > O. By Theorem 9.3, X has then a basis

{Ub ... , uII

} such that {u b .. " Uk} is a basis of Xl' Define

P(ClUl + + c"un) = Cl U 1 + ... + CkUk

for arbitrary scalars c" , cll •

Then Px ~ x for every x E X" and X, = &(P).Note that {Uk+ l' ...• u,,} is a basis of%(P). .Note al~o that the:e are

infinitely many projections in X, with range Xl, If 0 < dim Xl < dim X.

Define a mapping G of E into R' by setting

(69) G(x) = x + SP[F(x) - AX] (x E E).

Since F'(a) ~ A, differentiation of (69) shows that G'(a) = I. the identityoperator on Rn.. By the inverse function theorem. there are upen sets Uand V in R", with a E U. such that G is a 1-1 ma~ping of U o~to V whoseinverse H is also of class ri'. Moreover, by shrink~ng U~nd V.lfnecessary,we can arrange it so that V is convex and H'(x) is mveruble for every x E V.

(y E Y,).ASy ~y

for all scalars cl' ...• c,.Then ASYi ~ AZ i = Yi for I :s; i:; r. Thus

(68)

(67)

9.32 Theorem Suppose m, n, rare nonnegatipe integers. m ;?: r, n :.;::: r, F is a'ti'-mapping of an open set E c R" into Rm

, and F'(x) has rank r fo; every ~ E~.Fix a E E. put A = F'(a), let Yl be the range of A, and let P oe a projectIOn

in Rm whose range is Y1• Let Y2 be the null space of P. .Then there are open sets U and V in R", with a E U, U c E, and there IS a

1-1 rt'-mapping H of V onto U (whose inverse is also of class 16') such that

(66) F(H(x)) = Ax + <p(Ax) (x E V)

where qJ is a C4'-mapping of the open set A(V) c Yl into Y2 •

After the proof we shall give a more geometric description of the informa

tion that (66) contains.

Proof If r = O. Theorem 9.19 shows that F(x) is constant in a neighborhood U of a, and (66) holds trivially, with V = U, H(x) ~ x, <prO) ~ F(a).

From now on we assume r> O. Since dim Yl = r. Yl has a basIs

{ Y' Choose z· E R" so that AZ j = Yi (I .:s; i .:s; r), and define a iinearYI"'" ,J' ,

mapping S of Y1 into R" by setting

S(C1Yl + .,. + c,Y,) = ClZl + .. , + G,Ir

9.31 Projections Let X be a vector space. An operator P E L(X) is said to bea projection in X if p 2 = P.

More explicitly, the requirement is that P(Px) = Px for every x E X. Inother words, P fixes every vector in its range 2i(P).

Here are some elementary properties of projections:

(a) If P is a projection in X, then e~ery x E X has a unique representationof the form

Although this theorem is not as important as the inverse function theorem orthe implicit function theorem, we include it as another interesting illustrationof the general principle that the local behavior of a continuously differentiablemapping F near a point x is similar to that of the linear transformation F'(x).

Before stating it, we need a few more facts about linear transformations.

X=X 1 +X2

where x, E &(P), x, E %(P).To obtain the representation, put Xl = Px, X2 = X - Xl' Then

PX2 = Px - Px. 1 = Px. - p 2x = O. As regards the uniqueness, apply P t.othe equation x == x 1'+ x2 • Since Xl E 3f(P), PX t = Xl; since PX2 = 0, itfollows that Xl = Px.(b) If X is a finite-dimensional vector space and if Xl is a vector space inX, then there is a projection P in X with 2i(P) = Xl'

9.30 Definitions Suppose X and Yare vector spaces, and A E L(X, Y), as inDefinition 9.6. The null space of A, %(A), is the set of all x E X at which Ax = O.It is clear that %(A) is a vector space in X.

Likewise, the range of A, 2i(A), is a vector space in Y.The rank of A is defined to be the dimension of &(A).For example, the invertible elements of L(R") are precisely those whose

rank is n. This fellows from Theorem 9.5.IfA EL(X, Y)andA has rankO, then Ax =OforallxEA,hence%(A) = X.

In this connection, see Exercise 25.

230 PRINCIPLES OF MATHEMATICAL ANALYSIS FUl'OCTIONS OF SEVERAL VARIABLES 231

Define

Here is what the theorem tells us about the geometry of the mapping F.If YE F(U) then y = F(H(x» forsome x E V, and (66) shows that Py ~ Ax.

Therefore

This formula shows that qJ E qj' in W, hence in A(V), since Yo was chosenarbitrarily in A(V).

The proof is now complete.

so that g(l) = g(O). But g(l) = t/J(x,) and g(O) = t/J(x l ). This proves (74).By (74), t/J(x) depends only on Ax, for x E V. Hence (73) defines <p

unambiguously in A(V). It only remains to be proved that ({J E «i'.Fix Yo E A(V), fix X o E V so that Axo = Yo' Since V is open, Yo has

a neighborhood W in Yi such that the vector

X ~ Xo + SlY - Yo)

lies in V for all yEW. By (68),

Ax = Axo + Y - Yo = y.

(y E W).

Thus (73) and (79) give

<p(y) ~ t/J(xo - SYo + Sy)(80)

(79)

AG(x) ~ PF(x) (x E E).

In particular, (70) holds for x E U. If we replace x by H(x), we obtain

PF(H(x» = Ax (x E V).

t/J(x) = F(H(x» - Ax (x E V).

Since PA = A, (71) implies that Pt/J(x) ~ 0 for all x E V Thus t/J is a\f'-mapping of V into Y,. .

Since V is open, it is clear that A(V) is an open subset of its range9l(A) = Yl .

To comple~e the proof, i.e., to go from (72) to (66). we have to showthat there IS a Cf} -mapping ({J of A(V) into Y2 which satisfies

<p(Ax) = t/J(x) (x E V).

As a step toward (73). we will first prove that

t/J(x,) = t/J(x,)

. Note that ASPA ~ A, since PA = A and (68) holds. Therefore (69)gIves

(70)

(71)

(72)

(73)

(74)

DETERMINANTS

Determinants are numbers associated to square matrices. and hence to theoperators represented by such matrices. They art:: D if and aniy if tht corresponding operator fails to be invertible. They can therefore be used to decidewhether the hypotheses of some of the preceding theorems are satisfied. Theywill play an even more important role in Chap. 10.

This shows that y is determined by its projection Py, and that P, restrictedto F(U), is a I-I mapping ofF(U) onto A(V). Thus F(U) is an "r-dimensionalsurface" with precisely one point "over" each point of A(V). We may alsoregard F(U) as the graph of <p.

If (x) = F(H(x»), as in the proof. then (66) shows that the level sets of (these are the sets on which $ attains a given value) are precisely the level sets ofA in V. These are "flat" since they are intersections with V of translates of thevector space ,1V(A). Note that dim ....Y(A) = n - r (Exercise 25).

The level sets of F in U are the images under H of the flat level sets of <Din V. They are thus "(n - r)-dimensional surfaces" in U.

(75)

(76)

(78)

if Xl E V; X2 E V; AX1 =AX2'

Put \x) ~ F(H(x)), for x E V Since H'(x) has rank n for everyx E V, and F (x) has rank r for every x E U, it follows that

rank '(x) = rank F'(H(x»H'(x) ~ r (x E V).

Fix x E V. Let M be the range of '(x). Then Me RO, dim M = r.By (71),

P'(x) = A.

Thus P maps Manto iJ/(A) = Yl . Since M and Y, have the same dimension, it follows that P (restricted to M) is 1-1.

, Suppose now that Ah = O. 'Then P'(x)h = 0, by (76). But (x)h E M, and P 15 1·1 on M. Hence '(x)h = O. A look at (72) showsnow that we have proved the following:

ffx E V and Ah ~ 0, then t/J'(x)h = O.We can now prove (74). Suppose x E V X E V 4 - A

hi, 2 ,. X 1 ~ X 2 Put

= X 2 - Xl and define .

gir) = t/J(x, + th) (0',; t ,; I).

The convexity of V shows that Xl + th E V for these t. Hence

g'(t) = t/J'(x, + th)h = 0 (0 S t ,; I),

(81) y = Py + <p(Py) (y E F(U».

232 PRC-<CrpLES OF MATHEMATICAL A.....ALYS1S FUNCTIONS OF SEVERAL VARlABLES 233

9,34 Theorem

(a) Ifl is the identity operator on R", then

9.33 Definition If Cit> ... ,j,,) is an ordered n-tuple of integers, define

It will be convenient to think of det [A) as a function of the column vectorsof [A]. If we write

(90)

9.36 Theorlm A linear operator A on R' is invertible 'fand only ifdet [A] '" O.

Proof If A is invertible, Theorem 9.35 shows that

det [A] del [A-'] ~ det [A[l] ~ det [I] = I,

so lhat det [A] '" O.

If A IS not invertIble, the columns Xl' . . , Xt! of [A] are dependent(Theorem 9.5); hence there is one. say. XI:' such that

XI:+LCjXj=OUk

9,35 Theorem If(A] and [B] are n by n malrices, then

det ([BJ[A]) = det [B] det [A].

Proof If X" •.• , x, are the columns of [A], define

(85) "'.(x"

... , x,) = "'.[A] = det ([B][A]).

The columns of [B][AJ are the veclOrs Bx" , Bx,. Thus

(86) "'.(x" ... , x,) = det (Bx" , Bx,).

By (86) aod Theorem 9.34, "'. also has properties 9.34 (b) 10 (d). By (b)

and (84),

"'.[A] ="'. (~ aU, I)e" ~, , ... , x,) ~ ~ aU, I) "'.ce, , x, , ... , x,).

Repeating this process with X2, ... , Xn , we obtain

(87) "'.[A] = I aU" l)aU, , 2)'" aU" n) "'.ce", ... , e,J,

the sum being extended over all ordered n-tuples (iI' ...• it!) withI $ i, $ n. By (c) aod (d),

(88) "'.(e", ... , e,J = IU" .,., i,) "'.ce"

... , e,),

where I ~ I, O. or -I, and since [B][I] = [B], (85) shows lhat

(89) "'.(e"

... , e,) ~ det [B].

Substituting (89) aod (88) iota (87), we obtaio

det ([B][A]) ~ (I aU" 1) ... aU" n)tU" ... , i,)) det [B],

for all n by n matrices [A] and [BI. Taking B = I, we see lhat the abovesum in braces is det [A]. This proves the theorem.

for certain scalars cj . By 9.34 (b) and (d), Xk can ~ replaced by xI: + CjXj

without altering the determinant, if j i= k. Repeatmg, we see that x.\:: can

(lo5,jo5,n).

sUI> '" ,j,) ~ T1 sgn Uq - j,),,<q

,xj ~ I aU,j)e,

j: 1

det [A] = I SU"", ,j,)a(l,j,)a(2,j,)'" a(n,j,).

(82)

det (IJ = det (e" ... , e,) ~ l.

(b) det is a linear function ofeach of/he column vectors Xj' if the others areheld fixed.

(c) If [A], Is obtained fram [A] by interchanging Iwo columns, thendet [A], = -det [A].

(d) If [A] has Iwo equal columns, then det [AJ = O.

Proof If A ~ I, then aU, i) ~ 1 and a(i,j) ~ 0 for i "'j. Hence

det [I] = s(l , 2, .... n) = I,

which proves (a). By (82), SU" ... ,j,) ~ 0 if any two of the}'s are equal.Each of the remaining n~ products in (83) contains exactly one factorfrom each column. This proves (b). Part (c) is an immediate consequenceof the fact that s(jt . ... . j,,) changes sign if any two of the j's are inter~

changed, aod (d) is a corollary of (c).

where sgn x = 1 if x > 0, sgn x = - 1 if x < 0, sgn x = 0 if x = o. ThensUI> ... ,j,,) = 1, -1, orO, and it changes sign if any two ofthej's are interchanged.

Let [A) be the matrix ofa linear operator A on R", relative to the standardbasis {e l , .... e,,}, with entries aU,j) in the ith row andjth column. The determinant of [A] is defined to be the number

The sum in (83) extends over all ordered n-tuples of integers Ut, ... ,j,,) with1 5;.jr::::; n.

The column vectors x j of [A) are

(83)

(84)

det (x" "', x,) ~ det [A],

det is now a real function on the set of all ordered n-tuples of vectors in R".

234 PRISCIPLES OF MATHE.'dATICAL A.c~ALYSISFUNCTIONS OF SEVERAL VARIA1lLES 235

and also to

be replaced by the left side of (90), i.e., by 0, without altering the determinant. But a matrix which has 0 for one column has determinant O.Hence det [A] = O.

9.37 Remark Suppose {e l , ... , ell} and {u t , ... , ulI} are bases in R".Every linear operator A on RII determines matrices [AJ and [AJu, with entriesaij and :tij' given by

(i,j = 1, ... , n).

If all these functions Dijf are continuous in E, we say that! is of class r:c w in E,or that f E '?i"(E).

A mapping f of E into R'" is said to be of class '?i" if each component of fis of class ~".

It can happen that Dij!¥=- DjJat some point, although both derivativesexist (see Exercise 27). However, we shall see below that Duf = DjJwheneverthese derivatives are continuous.

For simplicity (and without loss of generality) we state our next twotheorems for real functions of two variables. The first one is a mean valuetheorem.

9.39 Definition Suppose f is a real function defined in an open set E c R",with partial derivatives D t /, ... , Dill If the functions Djf are themselvesdifferentiable, then the second~order partial derivatives off are defined by

DERIVATIVES OF HIGHER ORDER

9.40 Theorem Suppose f is defined in an open set E c R', and DJ and D,Jexist at every point of £. Suppose Qc E is a closed rectangle with sides parallelto the coordinate axes, having (a, b) and (a +h, b + k) as opposite vertices

(h '" 0, k '" 0). Put

!>(1, Q) ~ f(a + h, b + k) - f(a + h, b) - f(a, b + k) + f(a, b).

AUj = L:tijlli',

det [AJu ~ det [A].

Aej = Laije i ,,

(92)

If llj = Be j = 'Lbije io then Au} is equal to

I: "jBe, ~ I: "j I: bike, = I: II: b""j) e"k Ie i i \ Ie

The determinant of the matrix of a linear operator does therefore notdepend on the basis which is used to construct the matrix. It is thus meaningfulto speak of the determinant of a linear operator, without having any basis in mind.

ABej = A I: b'je, = I: (I: aikb'j) e,.Ie I Ie I

Thus 'Lb,1e Cf,tj = 'Laile blej , or

(91) [B][A]u = [A][B].

Since B is invertible, det [B] '" O. Hence (91), combined with Theorem 9.35,shows that

9.38 Jacobians If f maps an open set E c R" into R", and iff is differentiable at a point x E E, the determinant of the linear operator f'(x) is calledthe Jacabian of f at x. In symbols,

!>(f, Q) = u(a + h) - ural

~ hu'(x)

= h[(D,fJ(x, b + k) - (Dd)(x, b)]

=hk(D,d)(x, y).

Then there is a point (x, y) in the interior of Q such that

!>(f, Q) ~ hk(D2I f)(x, Y)·

Note the analogy between (95) and Theorem 5.10; the area of Q is hk.

"Proof Put u(t) = f(t, b + k) - f(t, b). Two applications of Theorem 5.10show that there is an x between a and a + h, and that there is a y betweenband b +k, such that

(95)

9.41 Theorem Suppose f is defined in an open set E c R2, suppose that Dit.

D2 d, and Dz.f exist at every point of E, and D21 f is continuous at some point(a, b) E E.

J,(x) = det f'(x).

We shall also use the notation

(93)

(94) C(YI' . ~., Y")c(x t , ... , x,,)

for Jr(x), if (Ylo ... , YII) = f(x l , ... , XII)'In terms of Jacobians, the crucial hypothesis in the inverse function

theorem is that J,(a) '" 0 (compare Theorem 9.36). If the implicit functiontheorem is st:Hcrl irr temlS of the fvticthilS (59), the assumption mad~ th..::r= onA amounts to

236 PRl:-;CIPLES OF MATHEMATICAL A~ALYSIS FU~CT[OSS OF SEVERAL VARIABLES 237

(c) cp' E !i(,) for every IE [e, d);(d) c < s < d, and to every e > 0 corresponds a (j > a such that

I(D, cp)(x, I) - (D, cp)(x, s)1 <'

for all x E [a, bJ and far all t E (s - b, s + b).

Then D 12 f exists at (a, b) and

(96) (DI,j)(a, b) = (D,d)(a, b).

Corollary D,d = DI,j iffE 'fj"(E).

Proof Put A = (D21 fJ(a, b). Choose, > O. If Q is a rectangle as In

Theorem 9.40,and if hand k are sufficiently small, we have

fA - (D,d)(x. Y)I <'

for all (x, y) E Q. Thus(100)

Define

,b

f(t) ~ I cp(x, t) d,(x)"

(e,;; I ,;; d).

DIFFERENTIATION OF INTEGRALS

Note that (c) simply asserts the existence of the integrals (laO) for allt E [c. d]. Note also that (d) certainly holds whenever D:cp is continuous on therectangle on which <p is defined.

Proof Consider the difference quotients

cp(x, t) - cp(x, s)</I(x, I) =--'-t---s-'--

for 0 < it - sl < b. By Theorem 5.10 there corresponds to each (x, t) anumber u between sand t such that

(97)

I~~~Q) - AI <e,

by (95). Fix h, and let k ~ O. Since D,j exists in E, the last inequalityimplies that

'I (D,j)(a + h, b) - (D,j)(a. b) Ih -AI,;;"

Since e was arbitrary, and since (97) holds for all sufficiently smallh ,. 0, it follows that (D 12 f)(a, b) = A. This gives (96).

(l01)

Then (D, cp)' E !i(,).j'(s) exists, and

.b

j'(s) = J (D, cp)(x, s) d2(X).,

is true? (A counter example is furnished by Exercise 28.)

It will be convenient to use the notation

(a,;;x';;b. 0< II-si <b).

.b

I 0(x. I) d,(x)."

Note that

I</I(x, t) - (D, cp)(x. s)1 <,

f(l) - f(s)

I - s

By (102), </I'~(D,cp)', uniformly on [a,b]. as 1~5. Since each</I' E !if'), the desired conclusion follows from (103) and Theorem 7.16.

Herree (d) implies that

(103)

</I(X. I) = (D, cp)(x. iI).

(102)

cp'(x) = cp(x, t).

(98)

(99)

Suppose cp is a function of two variables- which can be integrated with respectto one and which can be differentiated with respect to the other. Under whatconditions will the result be the same if these two limit processes are carried out

'in the opposite order? To state the question more precisely: Under whatconditions on cp can one prove that the equation

d rb .b ccp-:- cp(x, I) dx =J -:0- (x, I) dxat-a " ct

9.43 Example One can of course prove analogues of Theorem 9.42 with(_ co, 00) in piace of [a, b]. Instead of doing ths. lei. us simply bok at an

example. Define

Thus cpt is, for each t, a function of one variable.


(a) cp(x, I) is definedfor a :<; x :<; b. e,;; t,;; d;(b) :.t: is an increasing function on [a, b]; (104)

.~

f(t) = I e-,' cos (xt) dx._~

238 PRI:-.lCIPLES OF ~fATHEMATICAL A:"ALYSISFm-oCTIO:-;S OF SEVERAL VARIABLES 239

The integral (l04) is thus explicitly determined.

if (x, Y) "' (0, 0),

1. If S is a nonempty subset of a vector space X, prove (as asserted in Sec. 9.1) that

the span of S is a vector space.

2. Prove (as asserted in Sec. 9.6) that BA is linear if A and B are linear transformations.Prove also that A -I is linear and invertible.

3. Assume A E L(X, Y) and Ax = 0 only when x = O. Prove that A is then 1-1.

4. Prove (as asserted in Sec. 9.30) that null spaces and ranges of linear transformations are vector spaces.

5. Prove that to every A E L(R", R 1) corresponds a unique y E R" such that Ax = x' y.

Prove also thatiA = Iy! .Hint: Under certain conditions, equality holds in the Schwarz inequality.

6. If1(0, 0) ~ 0 and

EXERCISES

1\(5, t) = (b -;- a cos 5) cos t

11(5, t) = (b -;- a cos 5) sin t

J~(5, t) = a sin 5.

xyI(x, y) ~ -,--,

X 7 y.

prove that (DJ)(x, y) and (Dd)(x, y) exist at every point of R\ although f isnoc continuous at (0, 0).

7. Suppose that f is a real-valued function defined in an open set E c R", and thatthe partial derivatives Dtf, ... , DJ are bounded in E. Prove that f is continuous

in E.

Hint: Proceed as in the proof of Theorem 9.21.

8. Suppose that f is a differentiable real function in an open set E c R", and that fhas a local maximum at a point x E E. Prove thatj'(x) = O.

9. If f is a differentiable mapping of :l connected open set E c: R~ into R"'. and if

f'(x) = 0 for every x E E, prove that f is constant in E.

10. lf/is a real function defined in a convex open set E c R", such that (DJ)(x) = 0

for every x E E, prove that I(x) depends only on Xl, ... , x".

Show that the convexity of E can be replaced by a weaker condition. but

that some condition is required. For example, if n = 2 and E is shaped like a

horseshoe, the statement may be false.

11. If f and g are differentiable real functions in R", prove that

and that v(l'f) = - f-lvfwherever f:;i: O.

12. Fix two real numbers a and b, 0 < a < b. Define a mapping f = (fIJI.!3) of R 2

into R 3 by

(- ao < I < co).

g(l) ~ - f~ xe-" sin (XI) dx,-~

J'(t) =g(t)

and

(105)

(106)

for - ''lJ < t < 00. Both integrals exist (they converge absolutely) since theabsolute values of the integrands are at most exp (_x 2 ) and Ixl exp (_x 2 ),

respectively.

Note that 9 is obtained fromfby differentiating the integrand with respectto t. We claim thatfis differentiable and that

To prove this. let us first examine the difference quotients of the cosine:if f3 > 0, then

cos (x + p) - cos x 1 .%+6(107) {J + sin, = /J J, (sin, - sin I) dl.

Since Isin :x - sin t I :::; It - '1 i, the right side of (l07) is at most f3/2 in absolutevalue; the case f3 < 0 is handled similarly. Thus

(108) ICOS('+{J)-COS' . I

f3 +sm, ,,; lf3i

for all f3 (if the left side is interpreted to be 0 when f3 = 0).Now fix t. and fix h "0. Apply (108) with, = xl, f3 = xh; it follows from

(104) and (105) that

'1/(1 + h) - 1(1) I ,~" h - g(t) ,,; Ihl J_.x'e-".dx.

When h ~ 0, we thus obtain (106).

Le[ us go a step further: An integration by parts, applieft to (104), showsthat

(109) I() 2J" _,sin(xI)t = xe ;r ---dx.

- 00 I

Thus tl(l) ~ - 2g(l), and (106) implies now that I satisfies the differentialequation

(110) 21'(1) + tl(l) ~ O.

If we solve this differential equation and use the fact that f(O) = ,/~ (see Sec.8.21), we find that

, '2)I(t) = ,j~ exp l- ~ ,(i 11)

240 PRiSCIPLES OF ~tATHEMATlCAL ANALYSIS FU:-lCTIONS OF SEVERAL VARIABLES 241

(vf,)(f~'(p))~ O.

t:=2xy.

fl(x, y) = eX cos y,

f(x, y) = 2x 3 - 3x2 -+- 2y 3 + 3y 2.

(a) Find the four points in R 1 at which the gradient offis zero. Show thatfhas

exactly one local maximum and one local minimum in R2.

Show that the system of equations

3x -""-- y - z -;- u 2 = 0

x-y+2z+u=0

2.-c -.:..... 2y - 3z -7- 2u = °can be solved for x, y, u in terms of =; for x, z, Ii in terms of .v; for y, z. u in terms

of x; but not for x, y, z in terms of u.Take n = m = I in the implicit function theorem, and interpret the theorem (as

well as its proof) graphically.

Dell.m, I in 1\.2 by

(b) For 0::::;:: 8::::;::2r., -:::tJ < t< GO, define

__ g.(t)~f(tcos e, ISin e).Show that gi(O) = 0, g~(O) = 0, g;(O) = 2. Each gi has therefore a strict local

minimum at t = O.In other words, the restriction of f to each line through (0, 0) has a strict

local minimum at (0, 0).(c) Show that (0, 0) is nevertheless not a local minimum for f, since/(x, Xl) = _x4.,

Show that the continuity of f' at the point a is needed in the inverse function

theorem, even in the case n = 1: If

fit) ~ t ~ 2t' sin G)for t '" 0, and 1(0) ~ 0, then !'CO) ~ 1, f' is bounded in (-1, 1), but f is not

one·to-one in any neighborhood of O.Let f = (/"/1) be the mapping of R 2 into R1 given by

(a) What is the range of I?(b) Show that the Jacobian of lis not zero at any point of R 1

• Thus every pointof R 2 has a neighborhood in which I is one-to-one. Nevertheless, f is not one-to

one on R 1,

(e) Put a = (0, 17/3), b = f(a), let g be the continuous inverse of f, defined in a

neighborhood of b, such that g(b) = a. Find an explicit formula for g, compute

f'(a) and g'{b), and verify the formula (52).(d) What are the images under f of lines parallei to tne coordinale axes?

Answer analogous questions for the mapping defined by

20.

21.

18.

19.

17,

16.

if (x, y) "" (0,0)."fix, y) ~ X;-i- Y'

if {x, y) '" (0,0).(a) Prove, for aU (x, y) E R 2

, that

Conclude that I is continuous.

Describe the range K of f. (It is a certain compact subset of R 3 .)

(a) Show that there are exactly 4 points p E K such that

(vf,)(f~ '(q)) ~ O.

Find these points.

(b) Determine the set of all q E K such that

(e) Show that one of ~he points p found in part (a) corresponds to a local maximum of II, one corresponds to a local minimum, and that the other two areneither (they are so-called "saddle points").

Which of the points q found in part (h) correspond to maxima or minima?

(d) Let A be an irrational real number, and define g(t) = f(t, At). Prove that g is a1~1 mapping of R L onto a dense subset of K. Prove that

Ig'(t)1 1 = a2T A1(b -:- a cos t)2.

13. Suppose f is a differentiable mapping of R L into R3 such that If(l)1 = 1 for every t.Prove that f'(t)· fit) ~ O.

Interpret this result geometrically.

14. Define frO, 0) ~ 0 and

(a) Prove that Ddand Dzjare bounded functions in R 2• (Hence/is continuous.)

(b) Let u be any unit vector in R 1• Show that the directional derivative (D,J)(O, 0)

exists, and that its absolute value is at most l.

(e) Let y be a differentiable mapping of R L mto R 2 (in other words, y is a differentiable curve in R'), with )'(0) ~ 10, 0) and Iy'(O) I> o. Put git) ~ I(y(t)) andprove that 9 is differentiable for every t E R I

•

If y E "6", prove that g E Ci'.(d) In spite of this, prove that I is not differentiable at (0, 0).

Hint: Formula (40) fails.

15. Define frO, 0) ~ 0, and put

242 PRn-iCIPLES OF MATHEMATICAL ANALYSIS

(b) Let S be the set of all (x, y) E R 1 at which f(x, y) = O. Find those points ofS that have no neighborhoods in which the equation f(x, y) = 0 can be solved for

y in terms of x (or for x in terms of y). Describe S as precisely as you can.22. Give a similar discussion for

f(x, y) = 2x3 + 6xy2f_ 3x2+ 3y 2.

23. Define f in R 3 by

Show that /(0, 1, -1) = 0, (DIf)(D, I, -1) r 0, and that there exists therefore a

differentiable function 9 in some neighborhood of (1, -1) in R l, such that

g(1, -I) ~°and

/(g(y" y,), y,; y,) ~ 0.

Find (D,g)(1, -I) and (D,g)(1, -1).

24. For (x, y) '" (0, 0), define f ~ (f,,j,) by

xyfleX, y) = -,-.-, .

X ,y

Compute the rank of f'(x, y), and find the range of f.25. Suppose A E L(R~, R"'), let r be the rank of A.

(a) Define S as in the proof of Theorem 9.32. Show that SA is a projection in R~

whose null space is ",V(A} and whose range is 9t(S). Hint: By (68), SASA = SA.

(b) Use (a) to show that

dim %(A) + dim ':31(.1) ~ n.

26. Show that the existence (and even the conlinuity) of DIll does not imply the

existence of Dd. For example, let/ex, y) = g(x), whereg is nowhere differentiable.27. Put/Co, 0) ~ 0, and

FUNCTlOSS OF SEVERAL VARIABLES 243

Show that f.fJ is continuous on Rl, and

(D, <p)(x, 0) ~°for all x. Define

let) ~ r' <p(x. r) dx.. ~,

Show that l(t) ~ I ii IIi < t. Hence

/'(0) '" f,<D,<P)(X, 0) dx.

29. Let E be an open set in R". The classes CC'(E) and rtf"(E) are defined in the text.• By induction, ~(k)(E) can be defined as follows, for all positive integers k: To say

that/ E tCtkl(E) means that the partial derivatives Djf, ... , Dn/belong to ~( .. -1)(£).

Assume Ie C§Ctl(E), and show (by repeated application of Theorem 9.41)

that the kth-order derivative

is unchanged if the subscripts i lo ••• , h are permuted.

For instance, if n ;;::: 3, then

D ll13l= D31l1/

for e..'ery f E 0'(4).

30. Let f6: "8("'\(£), where E is an open subset of R~. Fix a E E, and suppose x E R~

is so close to 0 that the points

pe,) ~ a.,- tx

lie in E whenever 0 ::;; t ::; 1. Define

h(f) ~/(p(t)

xy(xl _ yl)f(x.y)~ , ,

X 7y

ii (x, Y) '" (0,0). Prove that(a) f. D.J: Dl/are continuous in Rl~

(b) Dll/and Dldexist at every point of Rl, and are continuous except at (0, 0);

(e) (D,,/)(O, 0) ~ 1, and (D"f)(O, 0) ~ -1.28. For t ~ 0, put

. (x ~<p(x, I) ~ l~x + ZVI

and put <p(x, I) ~ -<p(x. IIII if 1< O.

(0';; x ';;VI)(VI';;X';ZVI)(otherwise),

for all t E R l for which p(t) E E.(a) For 1 :::;; k s: m, shOW (by repeated application of the chain rule) that

h(~)(t) = L. (Dll ••• 1.J)(P(t» X'I ••• XI ...

The sum extends over all ordered k-tuples (ilt ... , ill in which each I) is one of the

integers 1, ... , n.

(b) By Taylor's theorem (5.15),

.. -1 htll(O) , h1m1(t)h(I) ~ I -,-.,- -,-,,,,,1} k. m.

for some t E lO, 1). Use this to prove Taylor's theorem in II variables by showing

that the formula

244 PRI~CIPLES OF MATHEMATICAL ANALYSIS

m-l 1I(a + x) = L k1 L (D'l ... ltf)(a)xll ... Xl t -:- rex)

t-O •

represents I(a i- x) as the sum of its so~called "Taylor polynomial of degreem - 1," plus a remainder that satisfies

. rex)lIm -I-,-, ~ o.>("0 x Im-

Each of the inner sums extends over all ordered k-tuples (it, ... , i k ), as inpart (a); as usual, the zero~order derivative of I is simply f, so that the constantterm of the Taylor polynomial of fat a isf(a).(c) Exercise 29 shows that repetition occurs in the Taylor polynomial as written inpart (b). For instance, DllJ occurs three times, as D llJ , D1Jl , D J l1,. The sum ofthe corresponding three terms can be written in the form

3(Df Dd)(a)xf XJ.

Prove (by calculating how often each derivative occurs) that the Taylor polynomialin (b) can be written in the form

Here the summation extends over all ordered n·tuples (Sit ... , s~) such that eachSl is a nonnegative integer, and Sl + ... + s~:::;; m - 1 .

31. Suppose IE Wm in some neighborhood of a point a E RZ, the gradient of f is 0at a, but not all second-order derivatives off are a at a. Show how one can thendetermine from the Taylor polynomial of fat a (of degree 2) ',l...he~her fhas a localmaximum, or a local minimum, or neither, at the point a.

Extend this to R~ in place of R2.

10INTEGRATION OF DIFFERENTIAL FORMS

Iategration can be studied on many levels. In Chap. 6, the theory 'NeS d~ve!oped

for ;easonably well~behaved functions on subintervals of the reat line. InChap. II we shall encounter a very highly developed theory of integration thatcan be applied to much larger classes of Junctions, whose domains are moreor less arbitrarv sets not necessarily subsets of RR. The present chapter isdevoted to tho;e asp~cts of integration theory that are closely related to thegeometry of euclidean spaces, such as the change of variables formula, lineinte!!rals, and the machinery of differential forms that is used in the statementand-proof of the n~dimensional analogue of the fundamental theorem of ca1cu!us,namely Stokes' theorem.

INTEGRAno",10.1 Definition Suppose]k is a k~cell in R\ consisting of all

x=(x", .. ,x,)such that(I) (i=I, ... ,k),

246 PRI:SCIPLES OF MATHEMATICAL ANALYSISINTEGRATION OF DIFFERENTIAL FOR.\1S 247

Ii is the j-cell in Rj defined by the first j inequalities (1), and I is a real continuous function on [Ic.

Putf=f" and definef,_, on 1'-1 by

A priori, this definition of the integral depends on the order in which thek integrations are carried out. However, this dependence is only apparent. Toprove this, let us introduce the temporary notation L(/) for the integral (2)and L'(f) for the result obtained by carrying out the k integrations in some

other order.

The uniform continuity of Ik on lk shows that h-t is continuous on lk-l.Hence we can repeat this process and obtain functions Jj, continuous on Ii, such

,thaJ-0"-t is the integral offj, with respect to xj ' over [aj , bJ After k steps wearrive at a number fo, which we call the integral oflover [Ie; we wrIte it in the

form

J. f= f fQk 11<

f f= r fRk ·Ik

(4)

10.4 Example Let Q' be the k·simplex which consists of all points x =

(x X)"n R' for which x + ... + x, ,n and Xi:2: 0 fOi 1= 1, ... , k. If

l' •.. , Ie I 1k = 3, for example, Q' is a tetrahedron, with vertices at 0, 0., e" 0,. IffE '6'(Q ),extendfto a function on I' by settingf(x) = 0 off Q" and define

The integral so defined is evidently independent of the choice of ]1, provided

only that ]1 contains the support off. . . k

It is now tempting to extend the defimtIOn of the Integral oyer R tofunctions which are limits (in some sense) of continuous functions with compactsupport. We do not want to discuss the conditions under .which this can bedone; the proper setting for this question is the Lebesgue Integral. We shallmerely describe one very simple example which will be used in the proof of

Stokes' theorem.

function with compact support, let]k be any k-cell which contains the support

of f, and define

(3)

r f. l'

orr f(x) dx• I'

(2)

Here [Ie is the "unit cube" defined by

05. Xi 5. 1 (1 5. i 5. k).

Since f may be discontinuous on ]1, the existence of the i~t~gral on theright of (4) needs proof. We also wish to show that th~s integral IS mdependentof the order in which the k single integrations are carned out.

To do this, suppose 0 < 6 < 1, put

(I - 6 < , 5. I)

10.2 Theorem For every f E '6'(1'), L(f) = L'(f)·

Proof If h(x) = hl(x l ) .•• h,(x,), where hi E '6'([ai , biD, then

k -.• !'(

L(h) = ,I] t hi(x;) dX i = L'(h).

If.;1 is the set of all finite sums of such functions h, it follows that L(g) =L'(g) for all 9 E d. Also, s1 is an algebra of functions on lie to which theStone-\Veierstrass theorem applies.,

Put V = TI (b i - oJ Iff E '6'(1') and, > 0, there exists 9 E s1 such1 '

that llf - gil < 'IV, where ill!! is defined as max If(x) I (x E I'). Then!L(f - g)1 < " IL'(f - g)1 < " and since

L(f) - L'(f) = L(f - g) + L'(g - f),

(5)

and define

(6)

('5.1-6)

(l < '),

F(x) = <p(x1 + .,. + x,)f(x) (x E I').

Then FE '1/(1'). , 1Put Y=(x"""'~'-I)' x=(y,x,). For each yEI -, the set of all x,

such that F(y, x,) ~ f(y; x,) is either empty or is a segment whose length does

not exceed~. Since 0 :s;; lp :s;; 1, it follows that

we conclude that! LU) - L'(f) 1 < 1"In this connection, [;.;trc;-sc 2 i" id~vant.

10.3 Definition The support of a (real or complex) function f on Rk is theclosure of the set of all points x E R le at which f(x) i: o. If f is a continuous (7)

248 PR1:-.'CIPLES OF MATHEMATICAL A:'iALYSIS I~TEGRArrON OF DIFFERE:-ITIAL FORMS 249

where il!11 has the same meaning as in the proof of Theorem 10.2, and Ft

_I

,

h.-I are as in Definition 10.1.

As 0-+0, (7) exhibitsh._t as a uniform limit of a sequence of continuousfunctions. Thusfk_l E o/(Ik-l), and the further integrations present no problem.

This proves the existence of the integral (4). Moreover, (7) shows that

Note that (8) is true, regardless of the order in which the k single integrationsare'carriea out. Since FE !ff(lk), JF is unaffe(,;ted by any change in this order.Hence (8) shows that the same is true of fJ

This completes the proof.Our next goal is the change of variables formula stated in Theorem 10.9.

To facilitate its proof, we first discuss so-called primitive mappings, and partitions of unity. Primitive mappings will enable us to get a clearer picture of thelocal action of a (If'-mapping with invertible derivative, and partitions of unityare a very useful device that makes it possible to use local information in aglobal setting.

(8) ItF(X)dX- L/(x)dxl ';;011/11.

10.6 Definition A linear operator B on R" that interchanges some pair ofmembers of the standard basis and leaves the others fixed will be called a jiip.

For example, the f1ip_B on R4 that interchanges e2 and e4 has the form

(13) B(xlel +X2el +x3 e3 +x4 e4)=x1 eI +X2e4 +xJeJ +X4e2

or, equivalently,

(14) B(xt el + X 2 e2 + xJ eJ + X 4 e4) = Xl el + X 4 e1 + XJ e3 + X2 e4 ·

Hence B can also be thought of as interchanging two of the coordinates, ratherthan two basis vectors.

In the proof that foUaws, we shall use the piojections Po, "', P" in RII,

defined by Po x =°and

(15) P",x=xl e1 +"'+xm em

for 1 :::; rn :::; n. Thus Pm is the projection whose range and null space arespanned by {el, "" em} and {em ... l , ••• , en}' respectively.

10.7 Theorem Suppose F is a rei-mapping ofan open set E c: R" into R II, 0 E E,

F(O) = 0, and F'(O) is invertible.Then there is a neighborhood of0 in R II in which a representation

PRIMITIVE :YIAPPINGS (16) F(x) = B, ... B'_lG,'"'' G,(x)

.nd we see (by Theorem 9.36) that G'(.) is invertible ifand only if(D. g)(.) ,,0.

If 9 is differentiable at some point a E E, so is G. The matrix [:Xij} of theoperator G'{a) has

10.5 Definition If G maps an open set E c: RII into R", and if there is aninteger_In and....a real function g wjth domain Esuch that

as its mth row. For j ¥= m, we have :J.jj = 1 and (Xij = 0 if i ::I: j. The Jacobianof G at a ~s thus given by

(x E V.).

,F~(O)e. = I (D. ,,)(O)e,.

i=m

,Fm(x) = Pm_IX + L ::xlx)e i ,

l"'m

P._,F.(x) = p._ 1 x

By (17), we have

where (Xm, ••• , (X" are real «j'-functions in V",. Hence

and

(19)

(18)

(17)

is valid.In (16), each Gi is a primiti!;e «j'·mapping in some neighborhood oj 0;

GlO) =-= 0, G~(IJ) is inrertible, and each B; is either ajiip or the identity operator.

Briefly, (16) represents F locally as a composition of primitive mappingsand flips.

Proof Put F = Fl' Assume 1::::; m ::::; n - I, and make the followinginduction hypothesis (which evidently holds for m = I):

V. is a neighborhood 0/0, F. E 't'W.) ,F.(O) ~ 0, F~(O) is invertible,

(x E E),

G(x) = x + [g(x) - x.le•.

Jot.) = det[G'(a)l = (D. g)(.),

G(x) = LX, e, + g(x)e.i*m

(DIg)(.), ... , (D. g)(a), ... , (D, g)(.)

(12)

(11 )

then we call G primitive. A primitive mapping is thus one that changes at mostone coordinate. :'\ote that (9) can also be written in the form

(10)

(9)

250 PRI~CIPLES OF MATHEMATICAL ANALYSIS 1~n:GRATIO~ .. OF .DlfFERf:"iTIAL I'OK~S 251

Since F~(O) is invertible, the left side of (19) is not 0, and therefore thereis a k such that m " k" nand (Dm >,)(0) " O.

Let Bm be the flip that interchanges m and this k (if k = m, Bm. is theidentity) and define

Then Gm E C6"(Vm), Gm is primitive, and G~(O) is invertible, since(Dm >,)(0) " O.

The inverse function theorem shows therefore that there is an openset Urn_ with OE Um.= Vm, such that Gm is a 1-1 mapping of Um onto aneighborhood Vm + l of 0, in which G,: I is continuously differentiable.Define Fm +, by ,

f= 'UJ.j= I

PARTlTlOi'iS OF U"ITY

Corollary IffE '6'(R") and the support off lies in K, then

(25)

10.8 Theorem Suppose K is a compact subset of R'\ and {V,,} is an open coverof K. Then there exist functions ifi" ... , ifi,EIf(R") such that

(a) 0" ifi,,, 1 for I " i" s;(b) each ljJj has its support in some V" and(c) ifil (x) + ... + tjI,(x) = 1 for every x EK.

Because of (c), {tPJ is called a partition of unity, and (b) is sometimesexpressed by saying that {ljJJ is subordinate to the cover {V,,}.

(x E Vm).Gm(x) ~ x + [>,(x) - xmlem(20)

(21)

(22) Pm Fm+t(Gm(x)) = Pm BmFm(x)

~ Pm[Pm-,x + >,(x)em+ '''j

=Pm-1x + 2k(x)em

so that

(x E R")., ,I ifi ,(x) = 1 - f1 [I - 'I',(x)];·1 i""l

ifi'+l = (I - '1',) ... (I - 'I',)'I'i+'

K c: B(x,) v ... v B(x,).

By (26), there are functions '1'" ~ ... , '1', E If(R"), such that 'I',(x) = I onB(x,), 'I',(x) = 0 outside W(x,), and 0 " 'I',(x) " I on R". Define ifi, = '1',and

B(x) c: W(x) c: W(x) c: V,(,).

Since K is compact, there are points XI' ... , Xs in K such that

for i = I, ... , s - I.Properties (a) and (b) are clear. The relation

ifi, + ... +ifi, ~ I - (I - '1',) ... (I - 'l'i)

is trivial for i ~ I. If (29) holds for some i < s, addition of (28) and (29)yields (29) with i + I in place of i. It follows that

(30)

(27)

(29)

Each tP J has its support in some V".

The point of (25) is that it furnishes a representation of f as a sum ofcontinuous functions ljJ,Iwith "small" supports.

Proof Associate with each x E K an index o-:(x) so that x E Vl:(x)' The;),there are open balls B(x) and W(x), centered at x, with

(28)

(26)

(x E Um ).

F = F1 = BIF1 i} G1

= RIEl F 3 ° Gz ° G1 = ",

=.81 '·'B,,-IFll oG.. _1 o'··oG1

If we apply this with m = 1, ... , n - 1, we successively obtain

Then Fm., E 1f'(Vm+,), Fm+,(O) ~ 0, and F~+,(O) is invertible (bythe chain rule). Also, for x E Um ,

Our induction hypothesis holds therefore with m + 1 in place of m,[In (22), we first used (21), then (18) and the definition of Bm , then

the definition of Pm, and finally (20).]

Since BmBm= I, (21), with y = Gm(x), is equivalent to

(23)

(24)

in some neighborhood of O. By (17), F" is primitive. This completes theproof.

If x E K, then x E B(x,) for some i, hence 'I',(x) = I, and the product in(30) is O. This proves (c).

252 PRINCIPLES OF MATHEMATICAL A."lALYSIS

CHANGE OF VARIABLES

We can now describe the effect of a change of variables on a multiple integral.For simplicity, we confine ourselves here to continuous functions with compactsupport, although this is too restrictive for many applications. This is illustratedby Exercises 9 to 13.

10.9 Theorem Suppose T is a I-I fC' -mapping of an open set E c Rk into Rk

such that JT(x) i= O/or all x E E. If/is a continuous/unction on Rk whose supportis compact and lies in T (E), then

(31) r J(y)dy=f i(T(x» IJT(x) I dx.~ RlJ. RlJ.

We recall that JT is the Jacobian of T. The assumption JT(x) # 0 implies.by the inverse function theorem, that T- L is continuous on T(E), and thisensures that the integrand on the right of (31) has compact support in E(Tbeorem 4.14).

The appearance of the absolute value of JT(x) in (31) may call for a comment. Take the case k = 1, and suppose T is a 1-1 'If'-m2pping of R 1 onto R I

•

Then JT(x) = T'(x); and if T is Increasing, we have

(32) r J(y) dy = f J(T(x»T'(x) dx,~ Rl RI

by Theorems 6.19 and 6.17, for all continuousJwith compact support. But ifT decreases, then T'(x) < 0; and if / is positive in the interior of its support,the left side of (32) is positive and the right side is negative. A correct equationis obtained if T' is replaced by IT' I in (32).

The point is that the integrals we are now considering are integrals offunctions over subsets of Rk

, and we associate no direction or orientation withthese subsets. We shall adopt a different point of view when we come to inte

-gration of differential forms over surfaces.

Proof It follows from the remarks just made that (31) is true if T is aprimitive ~'-mapping (see Definition 10.5), and Theorem 10.2 showsthat (31) is true if T is a linear mapping which merely interchanges twocoordinates.

If the theorem is true for transformationsP, Q, and if S(x) = P(Q(x»,then

IJ(Z) dz = I J(P(y»!JP(y)1 dy

=I J(P(Q(x»)) IJp(Q(x)) I iJ"(x)! dx

= I J(S(x» IJs(x) I dx,

INTEGRATIQS OF DIFFERESTIAL FORMS 253

since

Jp(Q(x»JQ(x) = det P'(Q(x» det Q'(x)

= detP'(Q(x»Q'(x) = det S'(x) ~ Js(x),

by the multiplication theorem for determinants and the chain rule. Thusthe theorem is also true for S.

Each point a E E has a neighborhood U c: E in which

where G i and Bi are as in Theorem 10.7. Setting V=..; T(U), it followsthat (31) holds if the support o[Jlies in V. Thus:

Each point y E T(E) lies in an open set V, c T(E) such thar (31) holdsfor all continuous functions whose support lies in Vy •

Now let/be a continuous function with compact support K c T(E).Since {Vy } covers K, the Corollary to Theorem 10.8 shc\l,IS that/= I.iftJ,where each l/!i is continuous, and each t./Ji has its support in some VY'Thus (31) holds for each IjJJ, and hence also for their sum!

DIFFERENTIAL FORI\lS

We shall now develop some of the machinery that is needed for the n-dimensional version of the fundamental theorem of calculus which is usually calledSlakes' theorem. The original form of Stokes' theorem arose in ~pplicationsofvector analysis to electromagnetism and was stated in terms of the curl of avector field. Green's theorem and the divergence theorem are other specialcases. These topics are briefly discussed at the end of the chapter.

It is a curious feature of Stokes' theorem that (he only thing that is difficultabout it is the elaborate structure of definitions that are needed for its statement.These definitions concern differential forms, their derivatives, boundaries, andorientation. Once these concepts are understood, the statement of the theoremis very brief and succinct. and its proof presents little difficulty.

Up to now we have considered derivatives of functions of several variablesonly for functions defined in open sets. This was done to avoid difficulties thatcan occur at boundary points. It will now be convenient, however, to discussdifferentiable functions on compact sets. We therefore adopt the followingconvention:

To say that f is a !(?'-rnapping (or a rt'''.mapping) of a compact setDc Rk into R" means that tht:re is a !&'.mapping (or a !(?"-mapping) g ofan open set We R' into 11:' such that DeW and such that g(x) = f(x) for

all XED.

254 PRINCIPLES OF MATHEMATICAL ANALYSISrNTEGRAnoN OF D1FFERE!'ITIAL FOR.o.tS 255

10.10 Definition Suppose E is an open set in R". A k-surface in E is a f.(j'

mapping 1> from a compact set D c R k into E.D is called the parameter domain of 11>. Points of D will be denoted by

U~(UI' ... ,uk).We shall confine ourselves to the simple situation in which D is either a

k-cell or the k·sirnplex Qk described in Example lOA. The reason for this isthat we shall have to integrate over D, and we have not yet discussed integrationover more complicated subsets of R k

• It will be seen that this restriction on D(which will be tacitly made from now on) entails no significant loss of generalityin the resulting theory of differential forms.

We stress that k-surfaces in E are defined to be mappings into E. notsubsets of E. This agrees with our earEer definition of curves (Definition 6.26).In fact, I-surfaces are precisely the same as continuously differentiable curves.

Then

I

Jw ~-f [7.(1)1,(1) +1,(1)7',(1)] dl = 11(1)Y,(l) - y,(O)Y,(O).y 0

Nate that in this example 5" (J) depends only on the initial point 1'(0)and on the end point y(l) of y. fn particular, L W = 0 for every closedcurve y. (As we shall see later, this is true for every I-form 00 which is

exact .)Integrals of I-forms are often called line integrals.

(b) Fix a > 0, b > 0, and define

y(l) = (a cos I,b sin I)

so that y is a closed curve in R1.. (Its range is an ellipse.) Then

10.11 Definition Suppose E is an open set in R". A differential form of orderk ~ 1 in E (briefly, a k:!ornz in E) is a function w, symbolically represented bythe sum

"Jxdy = J ab C052 tdt = Trab,

y 0

whereas

(35)

Then

OS; 8 S; n,

Hence

Dehne (r, e, rp) ~ (x, y, z), where

x = r sin ecos qJ

y = r sin () sin qJ

z=rcos(].

2,ry dx = - f ab sin 2 t dt = -nab.• 7 0

c(X, y, z)J.(r, 8, rp) ~ '( 8 )

c r, ,CP

Note that Lx dy is the area of the region bounded by y. This is aspecial case of Green's theorem.(c) Let D be the 3-cell defined byJw = JLai, '" i,((U)) c:x", , Xi) duo

III D o(u l , , Uk)

where D is the parameter domain of cD.The functions ail'" i" are assumed to be reai and continuous in E. if

cPI' ... , ¢" are the components of cD, the Jacobian in (35) is the one determinedby the mapping

(U I, ... , Uk) ~ (;\;,(u), ... , ;\,.(u)).

Note that the right side of (35) is an integral over D, as defined in Definition 10.1 (or Example 10.4) and that (35) is the definilion of the symbol f. w.

A k-form co is said to be of class <fj" or 't" if the functions all'" lie in (34)are all of class ct' or qj".

A Q-form in E is defined to be a continuous function in E.

(34) W = L ai, .,. i,(X) dXi, A .,. A dx i,

(the indices i l , ,." ik range independently from 1 to n), which assigns to eachk-surface 1> in E a number 00($) = f~ w, according to the rule

10,12 Examples(a) Let l' be a I-surface (a curve of class W') in RJ, with parameterdumain to, I).

Write (x. y, z) in place of (Xl) Xl, x), and put

co = x dy + y dx.

(36)

Note that maps D onto the closed unit ball of R J, that the mapping

is I-I in the interior of D (but certain boundary points are identified by<1», and that the integral (36) is equal to the volume of (D).

256 PRINCIPi.£S OF MATHEMATICAL ANALYSISiNTEGRATlONOF .DlHERENTIAL FORMS 257

for every k-surface <l> in E. As a special case of (37), note that - w is defined sothat

~O.13 EJem.entary properties Let w, WI' W 2 be k-forms in E. We write WI = W2

If and only If wl(<I» ~ w,(<I» for every k-surface in E. In particular, w ~ 0me~ns that w(¢l) = 0 for every k-surface ¢l in E. If c is a real number, thencw IS the k-form defined by

These forms dx[ are the so-called basic k-forms in R".It is not hard to verify that there are precisely n!jk!(n - k)! basic k·forms

in R Il; we shall make no use of this, however.Much more important is the fact that every k~form can be represented in

terms of basic k-forms. To see this, note that every k-tuple {JI' ... ,jd of distinctintegers can be converted to an increasing k-index j- by 2. fi'::Lite number 0f interchanges of pairs; each of these amounts to a multiplication by -1. as we sawin Sec. 10.13; hence

(44)

10.14 Basic k-forms If ii' ... , ill. are integers such that 1 :::;; i1 < iz < ...< ill. S n, and if I is the ordered k-tuple {il' ... , It}, then we call I an increasingk-index, and we use the brief notation

J cw ~ c r w,$ '$

(37)

and w = (£)1 + (£)2 means that

(38)

Consider a k-form

(40) w = a(x) dx" 1\ ... 1\ dx"

and let wbe the k-form obtained by interchanging some pair of subscripts in(40). If (35) and (39) are combined with the fact that a determinant chanaessign if two of its rows are interchanged, we see that .:>

and

where eUl, ... ,jk) is 1 or -I, depending on the number of interchanges thatare needed. In fact, it is easy to see that

'u" ... ,j,) = s(j" ... ,j,)

where s is as in Definition 9.33.For example,

dX I 1\ dxs /\ dx) /\ dX2 = -dx l /\ dx"}, /\ dX J /\ dx s

(45)

(46)

J(-w)~ -J' dw.$ $

jjj = -w.

(39)

(41)

holds for all i andj. In particular,

dX4 1\ dX2 /\ dX J = dX2 /\ dX J /\ dx4 .

If every k-tuple in (34) is converted to an increasing k~index, then weobtain the so~called standard presentation of w:

(42)

(43)

As a special case of this, note that the anticommutarive relation

dXi /\ dXj = -dxJ /\ dXi

(i= I, ... ,n).(47) w ~ 2: br(x) dXI'

I

More generally, let us return to (40), and assume that i, = is for somer =I:- s. If these two subscripts are interchanged, then w= whence w = 0 by(41). ",

In other words, if w is given by (40), rhen w = 0 unless the subscriptsii' "" ik are all distinct.

I.f co is as in (34), the summands with repeated subscripts can thereforebe omItted without changing w.

It follows thal a is the only k~form in any open subset of R'" if k > n.The anticommutativlty expressed by (42) is the reason for the inordinate

amOunt of attention that has to be paid to minus signs when studying differentialforms.

The summation in (47) extends over all increasing k-indices 1. [Of course, everyincreasing k~index arises from many (from k!, to be precise) k-tuples. Eachb [ in (47) may thus be a sum of several of the coefficients that occur in (34).]

For example,

Xl d"z 1\ dX l - Xl dXJ /\ dx z + xJ dX 2 /\ dXJ + dX l /\ dxz

is a 2-form in RJ whose standard presentation is

(l - x~) dX1 1\ dxz + (x: + -'(3) dx: /\ dx J •

The following uniqueness theorem is one of the main reasons for theintroduction of the standard presentation of a k-form.

258 PRDo:CIPLES OF MATHB1ATICAL ANALYSIS- n;TEGRAnON OF DIFFERENTIAL FOR,MS 259


(48) w ~ L. b)(x) dx)I

(54)

Move i p to the right, step by step, until its right neighbor is larger than ip •

The number of steps is the number of subscripts t such that ip <jt. (Note thatosteps are a distinct possibility.) Then do the same for ip-\ . ... , i1 • The totalnumber of steps taken is Ct. The final arrangement reached is [I, J]. Each step,when applied to the right side of (52), multiplies dx) A dXI by -1. Hence (53)holds.

Note that the right side of (53) is the standard presentation of dx) A dXJ.Next, let K ~ (k" ... , k,) be an increasing ,-index in {I, .. " n}. We shall

use (53) to prove that

(55) (dXI A dxJ) A dXK = dx) A (dXJ A dXK)'

If any two of the sets I, J, K have an element in common. then each sideof (55) is 0, hence they are equal.

So let us assume that I, J, K are pairwise disjoint. Let [I, J, K] denotethe increasing (p + q + r )-index obtained from their union. Associate j3 withthe ordered pair (J, K) and y with the ordered pair (I, K) in the way thatl wasassociated with (I, J) in (53). The lefr side of (55) is then

(-I)" dX[I,J] AdxK= (-lY( -l)P+' dX[I.J.K]

by two applications of (53), and the right side of (55) is

( - l)P dx f A dX[J. K] ~ ( - 1)P( - Iy+, dX[I. J, K]'

Hence (55) is correct.

If I and J have an element in common, then the discussion in Sec. 10.13shows that dx[ 1\ dX J = O.

If I and J have no element in common, let us write [I,1] for the increasing(p + q)-index which is obtained by arranging the members of I v J in increasingorder. Then dX,1,J) is a basic (p + g)-form. We claim that

(53) dx[ A dXJ = (-1)" dX[I,J]

where ~ is the number of differences jt - is that are negative. (The number ofpositive differences is thus pq - IX.)

To prove (53). perform the following operations on the numbers

(u ED).

J' w = r bI((u)) duo• 'D

dX 1 1\ dX2 + dX2 1\ dX 1 = O.

ProQf Assume, to reach a contradiction, that bAy) >0 for some vEEand for some increasing k-index J = Ul' ... ,jk}' Since bJ is continuous.there exists h > 0 such that bAx) > 0 for ali x E R" whose coordinatessatisfy IXi - vii s h. Let D be the k-ceil in Rk such that U ED if andonly if Iur Ish for r = 1, ... , k. Define

,(u) = v + L. ",ej ,

r= 1

Then is a k-surface in E, with parameter domain D, and bI((u)) > 0for every U E D.

We claim that

Since the right side of (50) is positive, it follows that w(<I» '" O. Hence(50) gives Ollr contradiction.

To prove (50), apply (35) to the presentation (48). More specifically,compute the Jacobians that occur in (35). By (49),

For any other increasing k-index I=/; J, the Jacobian is O. since it is thedeterminant of a matrix with at leas~ one row of zeros.

is the standard presentation of a k-form ill in an open set E c: R". II ill = 0 in E,then bix) = 0 for every increasing k-index I andfor every x E E.

Note that the analogous statement would be false for sums such as (34),since, for example,

(49)

(50)

where I and J range over all increasing p-indices and over all increasing q·indicestaken from the set {I, ... , n},

10.17 Multiplication Suppose wand; are p- and q-forms, respectiveiy, insome open set E c R", with standard presentations

10.16 Products of basic k-forms Suppose

(51) 1= {ii' ... , i,}, J = U" ... ,J,}

where 1 Sil < '" <il'snand 1 sit < ... <jqsn. The product of the corlespording basic forms dXr and dXJ in R" is a (p + q)-fonn in R\ cenured bythe symbol dx) A dxJ, and defined by

(52) dXr 1\ dX J = dX i1 1\ ..• I\. dx jp /\ dx j ( 1\ .•. 1\ dxjq

•

(56)

260 PRINCIPLES OF MATHEMATICAL ANALYSISINTEGRATION OF DrrFERENTlAL- FORMS 261-

By the chain rule, the last integrand is (J' I)'(t)· Hence

and we see that Ldfis the same for all}' with the same initial point and the sameend point, as in (a) of Example 10.12.

Comparison with Example 1O.l2(b) shows therefore that the I-form x dyis not the derivative of any D-formf This could also be deduced from part (b)

of the following theorem, since

d(x dy) = dx 1\ dy i= (J,

Their product, denoted by the symbol w /\ l, is defined to be

(57) w /\ A = 2: b[(x)cJ(x)dx[ /\ dxJ .[.J

In this sum, I and J range independently over their possible values, and dX r 1\ dXJ

is as in Sec. 10.16. Thus w /\ i. is a (p + g)-form in E.It is quite easy to see (we leave the details as an exercise) that the distribu

tive laws

(w, + w,) /\ A~ (w, /\ i.) + (w, /\ A)

and

(62) rdJ~ f(y(I)) - J(y(O)),"

W /\ (AI + ).,) = (w /\ i.,) + (w /\ A,)

hold, with respect to the addition defined in Sec. 10.13. If these distributivelaws are combined with (55), we obtain the associative law

10.20 Theorem

(a) If wand i. are k- and m:!orms, respectively, of class CCf

in E, then

It is customary to write/ro, rather than/ 1\ W, when/is a Q-form.

for arbitrary forms ro, i., a in E.In this discussion it was tacitly assumed that p .;::: 1 and q ;;:: 1. The product

of a D-form J with the p-form w given by (56) is simply defined to be the p-form

Jw = wJ = 2:J(x)b/.x) dx[.[

If w = r.b [(x) dx [ is the standard presentation of a k-form w, and b[E 'C'(E)for each increasing k-index I, then we define

10.19 Example Suppose E is open in R', J E '(f'(E), and y is a continuouslydiffe~cn~iable (,u;'1e ir-. E, ';.,ith domain [0, 1). By (59) and (35),

10.18 Differentiation We shall now define a differentiation operator d whichassociates a (k + I)-form dw to each k~form (j) of class CC' in some open setEcRn.

A Q-form of class CC' in E is just a real function / E "6'(£), and we define

(58)

(59)

(60)

(61)

(w /\ !.) /\ a ~ w /\ (i. /\ a)

,dJ = 2: (DJ)(x) dx,.

i= 1

'dw = 2: (db[) /\ dXJ.[

J' ,

dJ= r I (DJ)(y(t))y;(t) dt.y ·0 ;::: 1

(63)

(64)

dew /\ i.) ~ (dw) /\ ). + (-I)' w /\ dA.

(b) IJ w is oj ciass '(f" in E. then d'w ~ O.

Here d 2w means, of course, d(dw).

Proof Because of (57) and (60), (a) follows if (63) is proved for the

special case

i.=gdx1

where f, 9 E "4'(£), dx [ is a basic k·form, and dXJ is a basic m-form. (Ifk or ,;or both are 0, simply omit dx [or dXJ in (64); the proof that follows

is unaffected by this.] Then

w/\A~Jgdx[/\dxJ'

Let us assume that I and J have no element in common. [In the othercase each of the three terms in (63) is O.J Then, using (53),

dew /\ !.) ~ d(Jg dXJ /\ dxJ) ~(-I)' d(Jg dX[I. 1])'

By (59), d(Jg) ~ J dg -i- g df Hence (60) gives

dew /\ A) = (-1)' (Jdg + gdJ) /\ dX[I.J]

= (gdf +Jdg) /\ dx, /\ dXJ'

Since dg is a I-form and dx[ is a k-form, we have

dg /\ dx [ ~ (- I)' dx[ /\ dg,

INTEGRATION OF OIFFEP.E~TIAL FORMS 263

262 PRL""CIPLES OF MATHEMATICAL ANALYSIS

10.22 Theorem Wilh E and T as In Sec. 10.21. leI wand ;. be k- and m-forms

in V, respectively. Then

= (dfJr.

By the chain rule, it follows that

dUT) = I (DJT)(X) dXjj

= I I (D,f)(T(x»(Dj I,)(X) dXjJ i

= I (D,f)(T(x» dl,,

df= I (D,f)(y)dy,.fT(X) = f(T(x»),

(a) (w + !.lr = WT + ;'T if k = m;(b) (w /\ !·lr = WT /\ AT ;(c) d(WT) = (dwlr if w Is of class '11' and T Is of class 'G".

Proof Part (a) follows immediately from the definitions. Part (b) isalmost as obvious, once we realize that

(dy,! 1\ ... 1\ dY,JT = dt,! 1\ ... 1\ dti~

regardless of 'Nhether {i l , . • , i,} is increasing or not; (68) holds becausethe same number of minus signs are needed on each side of(68) to produceincreasing rearrangements.

We tum to the proof of (c). Iff is a G-form of class '11' in V, then

(69)

(68)

d'w = (d'f) /\ dx1= O.

10.21. Change of variabJes Suppose E is an open set in R", T is a CC'-mappingof E mto an open set V c: R"', and w is a k-form in V, whose standard presentation is

by (42). Hence

dew /\ A) ~ (df /\ dx I) /\ (gdxJ) + (-I)'(jdx l ) /\ (dg /\ dxJ)

= (dw) /\ A+ (-I)'w /\ dA,

which proves (a).Note that the associative law (58) was used freely.Let us prove (b) first for a G-formfE '11":

.d'f~ dCt/Djf)(X) dXj)

,~ I (D,J)(x) dx, /\ dxj •

i.J= 1

Since Dijf= DJJ (Theorem 9.41) and dX i 1\ dx) = -dxJ 1\ dx" we seethat d'f = O.

If w = f dx I' as in (64), then dw = (d]) /\ dx I' By (60), d(dx I) = O.Hence (63) shows that

(65) w = I bJ,y) dYf,I

(We use y for points of V, x for points of E.)Let I" ... , 1m be the components of T: If

Jf dYr = dY'1 1\ •.• 1\ dYi", then (d.vr)r = dt'l 1\ , .. 1\ dti~' and Theorem10.20 shows that

(70) d((dy Ilr) = o.

then Y, = I,(X). As in (59),

y = (y" ... , Ym) = T(x)

Thus each dt i is a I-form in E.The mapping T transforms ill into a k-form W T in E, whose definition is

(67) WT ~ I b J,T(x» dl" /\ ... /\ dl, .I •

In each summand of (67), 1= {fl , ... ~ it} is an increasing k-index.Our next theorem shows that addition, multiplication, and differentiation

of forms are defined in such a way that they commute with changes of variables.

(66) dl, = i: (Djl,)(X) dxj ' (1:;; I:;; m).j=l

(This is where the assumption T E CC" is used.)Assume now that w =f dy r' Then

WT = fT(X) (dy I)r

and the preceding calculations lead to

d(wr) = d(jT) /\ (dYIlr = (dflr /\ (d,VIlr

~ ((df) /\ dy Ilr ~ (dwlr·

The first equality holds by (63) and (70), the second by (69). the third bypart (b), and till::: last by the definition of dO).

The general case of (c) follows from the special case just proved, ifwe apply (a). This completes the proof.

264 Pim';CI?LES OP MATHEMATICAL ANALYSISI~l"EGRATlQN OF DIFFERE~TIAL FORMS 265

OUf next objective is Theorem 10.25. This will follow directly from twoother important transformation properties of differential forms, which we statefirst.

10.24 Theorem Suppose co is a k-form in an open set E c RIl, is a k-surface

in E, with parameter domain D c Rk., and 6. is the k-surface in Rk, with parameterdomain D, defined by 6.(u) ~ u(u E D). Then

10.23 Theorem Suppose T is a Cf!'-mapping of an open set E c RII into an openset V c Rm

, S is a ct'-mapping of V into an open set We RP, and co is a k-formin W, so that ills is a k-form in V and both (cosh and W ST are k-forms in E, whereST is defined by (ST)(x) = S(T(x)). Then

1\ •.. 1\ duk. = f W<p.,f W ~ f a(<1>(u))J(u) du

o D

= r a(<1>(u))J(u) du,-,

d1>" = L ,(p, g) dU q,

8(Xi , ... , XiJJ(u) ~ , ,

o(u" ... , Uk)

let [A] be the k by k matrix with entries

,(p,g) = (D q 1>,,)(u) (p,g=I, ... ,k).

so that

dA.. " ... " d.!.. ~",(I, g,)'" ,(k, g.) du" " ... " du".'t"11 '+'ll< L.-

In this last sum, q" ... , qk range independently over 1, ... , k. The anti

commutative relation (42) implies that

du " ... " du =s(g" ... ,gk)du,,,"·" dUk'ql ql<

where s is as in Definition 9.33; applying this definition. we see that

d.!.. " ... " J1>. = det [A] du, " ... " dUk;'+' II 'l<

and since J(u) = det [AJ, (72) is proved.

The final result of this section combines the two preceding theorems.

Then

sinct (72) implies

where

If 1>" ... , 1>. are the components of <1>, then

W o = a(<1>(u)) d1>" " .. , " d1>,•.

The theorem will follow if we can show that

d.!.. " ... " dd>· = J(u)du, " ... " dUk,'+'11 ' Ik

10.25 Theorem Suppose T is a Cf!'-mapping of an o~en set E c RII

into an open

set V c Rm , 4'1 is a k-surface in E. and OJ is a k-form in V.

Then

(72)

and

so that the chain rule implies

(wslr ~ L (Djs,)(T(x)) dtjj

= L (Djsq)(T(x)) L (D, t;l(x) dx,j i

= L (D,r,)(x) dx, = dr, = WST',

(w 1\ lhT = COST 1\ )·ST.

Thus if (71) holds for wand for )., ir follows that (71) also holds for W " )•.

Since every form can be built up from Q-forms and I-forms by additionand multiplication, and since (71) is trivial for O-forrns, it is enough toprove (71) in the case W = dz" g = I, "" p. (We denote the points ofE, V, W by x, y, z, respectively.)

Let t l , ••. , tm be the components of T, let Sl' ••• , sp be the components of S, and let r l , ••• , rP be the components of ST. If co = dzq , then

ws=dsq = L(Djs,)(y)dYj'j

(WS)T = WST'

.Proof If.ev,and), are forms in W, Theorem 10.22 shows that

((w " A)s)T = (ws " Aslr = (wslr " C!-slr

(71)

Proof We need only consider the case

co = a(x) dX z1 1\ ••. 1\ dXil<. f W = f WT'TO 0

266 PRf:SCIPLES OF :\lATHE:MATICAL ASALYSISINTIGRATION OF DIFFERE:"o'TIAL FOP_\1S 267

Proor Let D be the parameter domain of III (hence also of Tlll) anddefine L1 as in Theorem 10.24.

Then

We call 0' oriented to emphasize that the ordering of the vertices Po, ... , PI.:is taken into account. If

SIMPLEXES AND CHAINS

f W= f wT.= f (wT).=J wT ·~ 4 4 ~

such that (,(i ;::: 0 for i = 1, ... , k and L;(; :::;; l.

Assume now that Po, Pl' 0'" Pie are points of R"o The oriented affinek~simplex

r J = 'J(po)·",

where {io, it, "', ik } is a permutation of the ordered set fO, 1, ... , k}, we adoptthe notation

(80)

where s is the function defined in Definition 9,33. Thus a = fa, depending onwhether s = I or s = - I. Strictly speaking, having adopted (75) and (76) asthe definition of 0', \\o"e should not write a= (J unless io = 0, ... , ik = k, e:oen·fs(,· ,.) = I' what we have here isan equivalence relation, not an equahty.1 0' ... , Ie ,

However, for our purposes the notation is justified by Theorem 10.27. _If if = Ea (using the above convention) and if £ = I, we say that 0' and (J

have the same orient~tion; if E = -1, if and a are said to have opposite,orienta~tions. Note that we have not defined what we mean by the "orientatIon o.f asimplex." What we have defined is a relation between. pairs of simple,xes h~vIn~

the same set of vertices, the relation being that of "havmg the same onentatIon.There is, however, one situation where the orientation of a simplex can

be defined in a natural way. This happens when n = k and when the vectorsPi - Po (1 :::;; i::::;; k) are independent. In that cas,e, the Iin~ar ,transformation Athat appears in (78) is invertible, and its determmant (whIch IS .the sam,e as th,eJacobian of 0') is not 0. Then (J is said to be positively (or negatIVely) onented Ifdet A is positive (or negative). In particular, the simplex [0, e l , .,', ed in Rk

,

given by the identity mapping, has positive orientation.So far we have assumed that k;::: t An oriented O-simplex is defined to

be a point with a sign attached. We write (J = +Po or a = - Po. If (J = gpo(, = ± I) and ifJis a O-form (i.e., a real function), we define

(79)

Qk which is given

f(x) = f(O) + Ax

a = [Po, P" ... , p,]

(73)

for some A E L( X, Y).

An affine mapping of R" into Rn is thus determined if we know reO) andf(eJ for 1 :::;; i::::; k; as usual, {e l , ..• , e.\J is the standard basis of R".

We define the standard simplex Q" to be the set of all U E R k of the form,

U = LCt:je j

i=1

The first of these equalities is Theorem 10.24, applied to T<1> in place of <1>.The second folJows from Theorem 10.23. The third is Theorem 10.24,with Q)T in place of w.

(74)

is defined to be the k-surface in R" with parameter domainby the affine mapping

10.26 Affine simplexes A mapping r that carries a vector space X into avector space Y is said to be affine if f - f(O) is linear. In other words, the require~ment is that

. (75)

Note that 0' is characterized by

(76),

a(x1e1 + ... +Xkele)-= Po + L Xlpi - Po).1= I 10.27 Theorem 1/ (J is an oriented rectilinear k-simplex in an open set E c R"

and if if = Ea then

and that

(73) o(u) = Po + Au (u E Qk)

where A E L(R', R") and Aei = Pi - Po for I S; j S; k.

(77) 0(0) ~ Po, aCe,) ~Pi (for I S; j S; k), (81)

for every k-jorm w in E.

Proof For k = 0, (81) follows from the preceding definition. So weassume k ;::: I and assume that (J is given by (75).

268 PR1:"OCIPLES OF MArH~MATICAL ANALYSIS INTEGRATIO:-< Of DlFFERE:-iTlAL- FOR.I\.tS 269

Suppose 1:::;; j:::;; k, and suppose a is obtained from 0" by inter.changing Po and Pj' Then e = -1, and

where B is tb.e ~ine~r mapping. of Ric into R~ defined by Bej

= Po _ Pj'Bei = Pi - Pj If I '# J. If we wnte Aei = X j (l $ i:::;; k), where A is givenby (78). the column vectors of B (thar is. the vectors Be;) are

We may view a k-surface <1l in E as a function whose domain is the collection of all k-forms in E and which assigns the number J.:p w to w. Since real.valued functions can be added (as in Definition 4.3), this suggests the use of thenotation

,oa = I (-I)i[Po,···. Pi-" Pi"'.···. p,].

j=O

For ex.ample, if 0' = [Po, PI' P2], then

Be,~p,-po (if I ';'i';'j-I).

Be, = Pi"' - Po (if j,;, i,;, k - I).

aa = [Pl' p,] - [Po. p,] + [Po. Ptl = [Po. p,] + [Pl' p,] + [p,. Pol,

which coincides with the usual notion of the oriented boundary of a triangle.For 1 $):::;: k, observe that the simplex O'j = [Po, ... , Pj_1 . Pj+ l' ... , Pic]

which occurs in (85) has QIc-1 as its parameter domain and that it is defined by

(86) a/u) = Po + Bu (u E Q'-l).

where B is the linear mapping from Rk-

1 to R~ determined by

The simplex

(85)

10.29 Boundaries For k ~ 1, the boundary of the oriented affine k-simplex

a ~ [Po. P, •.... p.Jis defined to be the affine (k - I)-chain

aD ~ [Pl' P, •.... p.J.which also occurs in (85), is given by the mapping

ao(u) = P, + Bu,

of addition are possible. Originally, a was defined as an R~-valued functionwith domain Qk; accordingly, a l + a2 could be interpreted to be the function0" that assigns the vector O"l(U) + 0"2(U) to every U E Qk; note that 0" is then againan oriented affine k~simplex in R~! This is not what is meant by (83).

For example, if (12 = -0'1 as in (80) (that is to say, if 0'1 and 0"2 have thesame set of vertices but are oppositely oriented) and if r = 0'1 + 0'2' thenSr OJ = 0 for all w, and we may express this by writing r = 0 or 0"1 + 0'2 = o.This does not mean that O'I(U) + 0"2(U) is the null vector of R~.

(u E Q').

r = a 1 + ... + Or

O'(u) = Pi + Bu

10.28 Affine chains An affine k-chain r in an open set E c R~ is a collectionof finitely many oriented affine k-sirnplexes 0'1' ... ,0', in E. These need not bedistinct; a simplex may thus occur in r with a certain multiplicity.

If r is as above, and if w is a k-form in E, we define

(82)

Xl - Xj' ... , Xj _ I - Xj , - Xj , Xj + 1 - Xj , ... , XIc - xj .

If we subtract the jth column from each of the others, none of the determinants in (35) are affected, and we obtain columns Xl' ... , x

j_

1• -x"

xj + 1•..• , X k · These differ from those of A only in the sign of the jthcolumn. Hence (81) holds for this case.

Suppose next that 0 < i <) $ k and that a is obtained from (j byinterchanging Pi and Pj' Then a(u) = Po + Cu, where C has the samecolumns as A, except that the ith and )th columns have been interchanged. This again implies that (81 )holds, since' = _ I.

The general case follows, since every permutation of {O, 1, ... , k} isa composition of the special cases we have just dealt with.

or, more compactly,

(83)

(84) where Be j = Pi-d - PI for 1 $ i:$ k - 1.

to state the fact that (82) holds for every k-forffi w in E.To avoid misunderstanding, we point out explicitly that the notations

introduced by (83) and (80) have to bt: nandled with care. The point is thatevery oriented affine k-simplex a in R" is a function in two ways, with differentdomains and different ranges, and that therefore two entirely different operations

10.30 Differentiable simplexes and chains Let T be a et"·mapping of an openset E c R~ into an open set V c Rm

; Tneed not be one· to-one. If 0' is an orientedaffine k-simplex in E, then the composite mapping <D = I ~ (J (which we shallsometimes write in the simpler form TO') is a k-surface in V, with parameterdomain QIc. We call <1> an oriented k·simplex of class <&".

270 PRl~CIPLF.S OF MATHE\1.ATICAL A:-;ALYSIS

INTEGRATION OF DIFFER£:-<T1AL FORMS 271

or

A finite collection 't' of oriented k·simplexes $1' "', <1l. of class ffj" in Vis called a k-chain of class 71" in V. If w is a k-form in V, we define

Both 0"1 and 0"2 have Jacobian I > O. Since

I w-I WilT l ilT l

hold for every (n - I)-form w? The answer is yes, but we shall omit the proof.(To see an example, compare the end of this section with Exercise 17.)

One can go further. Let

n= Et U ... u E.,

where E j = T/Q"), each T, has the properties that Thad above, and the interiorsof the sets Ei are pairwise disjoint. Then the (n - I)-chain

oTt + ... + aT. = anis called the positively oriented boundary of n.

For example, the unit square 12 in R2 is the union of O"t(Q2) and 0"2(Q2),

where

An obvious question occurs here: If E = TI(Q") = T,(Q"), and if bothT 1 and T2 have positive Jacobians, is it true that aT! = iJT2 ? That is to say,does the equality

The boundary 0<1> of the oriented k-simplex = T, a is defined to be the(k - 1) chain .

(89) c<l> = T(oa).

and use the corresponding notation \f' = I<Di

.

If r = Ia i is an affine chain and if <1>, = To a j , we also write 't' = Tor,

In justification of (89), observe that if T is affine, then $ = To a is anoriented affine k-simplex, in which case (89) is not a matter of definition, but isseen to be a consequence of (85). Thus (89) generalizes this special case.

It is immediate that aq, is of class "e" if this is true of $.

Finally, we define the boundary c'i' of the k-chain 'i' = :1:<1>, to be the(k - I) chain

(90) 8'1' = L c<l>,.

(87)

(88)

10.31 Positively oriented boundaries So far we have associated boundaries tochains, not to subsets of R". This notion of boundary is exactly the one that ismost suitable for the statement and proof of Stokes' theorem. However, inapplications, especially in R l or R J

, it is customary and convenient to talkabout "oriented boundaries" of certain sets as well. We shall now descnbethis briefly.

Let Q" be the standard simplex in R", Jet 0"0 be the identity mapping withdomain Q". As we saw in Sec. lO.26, ao may be regarded as a positively orientedn-simplex in R". Its boundary cao is an affine (n - I)-chain. This chain iscalled the positively oriented boundary of the set Q".

For example, the positively oriented boundary of QJ is

[el' e" eJ ] - [0, e" eJ ] + [0, e[, eJ ] - [0, e[, e,].

Now let T be a I-I mapping of Q" into R", of class CC", whose Jacobian ispositive (at least in the interior of Q"). Let E = T(Q"). By the inverse functiontheorem, E is the closure of an open subset of R". We define the positivelyoriented boundary of the set E to be the (n - I)-chain

oT = T(oa,),

and we may denote this (n - I)-chain by oE.

we have

Ca[ = ret, e,] - [0, e,] + [0, ell,

00"2 = (e2, el ] - [eI + e2, el] + reI + ez, e2 ];

The sum of these two boundaries is

8[2 = [0, etl + ret, e[ + e,l + [e[ + e2 , e, ] + [e" 0],

the positively oriented boundary of 12. Note that [e~, ell canceled (e2' ed.If e:t> is a 2-surface in Rm

, with parameter domain [2, then q> (regarded asa function on 2·forms) is the same as the 2-chain

$00"{ +<1>oa2 .

Thus

0<1> = 0(<1>' at) + 0(<1> ' "2)

= <1>(001) + <1>(00,) = <1>(0[').

In other words, if the parameter domain of is the square 12, we need

not refer back to the simplex Q2, but can obtain a directly from Cl2.Other examples may be found in Exercises 17 to 19.

book describes some of the historical background.) These special cases will bediscussed further at the end of the present chapter.

Proof It is enough to prove that

272 PRl-SCIPLES OF MATHB-fATICAL A.....ALYSIS

10.32 Example For 0 :::; u.::; n, 0 :::; r: :s; 2n, define

l(u, v) = (sin u cos v, sin u sin v, cos u).

Then I is a 2-surface in R 3, whose parameter domain is a rectangle D c R 2,

and whose range is the unit sphere in R 3• Its boundary is

oL = L(cD) ~ {I + {, + {, + (4

where

(l(U) = L(u, 0) = (sin u, 0, cos u),

(,(v) = len, v) = (0,0, - 1),

(,(u) = l(n - u, 2rr) = (sin u, 0, -cos u),

(4(V) = L(O, 2" - v) = (0, 0, 1),

with [0, rr} and [0, 2n} as parameter intervals for u and D, respectively.Since Y2 and f-l- are constant, their derivatives are 0, hence the integral of

any I-form over {, or (4 is 0. [See Example !.I2(a).]Since {,(u) = (l(rr - u), direct application of (35) shows that

I w- I W13 - - '/1

for every I-form w. Thus fer W = 0, and we conclude that CL = O.(In geographic terminology, OL starts at the north pole N, runs to the

south pole 5 along a meridian. pauses at S, returnS to ~V along the same meridian,and finally pauses at lV. The two passages along the meridian are in oppositedirections. The corresponding two line ini'egrab therefore ~ancd each other.In Exercise 32 there is also one curve which occurs twice in the boundary, butwithout cancellation.)

STOKES' THEOREM

(92)

(93)

(94)

INTEGRATlO:-': OF DIFFERE:--ITIAL FOR."lS 273

f dw = r w~ ~o<Il

for every oriented k-simplex <1> of class CC" in V. For if (92) is proved andif 'I' ~ l<p" then (87) and (89) imply (91).

Fix such a and put

v = [0, e" ... , ek]'

Thus u is the oriented affine k-sirnplex with parameter domain Qk whichis defined bv the identity mapping. Since is also defined on Qk (seeDefinition 10.30) and 4> E '{j", there is an open set E c Rk which containsQk, and there is a et"·mapping T of E into V such that = T ~ u. ByTheorems 10.25 and 1O.22(c), the left side of (92) is equal to

Ir• dw ~ I. (dwJr = I. d(wr)·

Another application of Theorem 10.25 shows, by (89), that the right side

of (92) is

J- w = I w = r wr ·e(T~) r{e~l ·i~

Since (JJr is a (k - I)-form in E, we see that in order to prove, (92)

we mere(v have to show thN

for the special simplex (93) and for every (k - I)-form;. of class '(f' in E.

If k = 1, the definition of an oriented O-simplex shows that (94)

merely asserts that

10.33 Theorem If \fI is a k-chain of class CC" in an open set V c Rm and If (J)

is a (k - I )-form of class '(j' in V. then

The case k = m = ! is n.0tl;.!vg but- the ft!n':lame-ntal tbeorem of -calculus(with an additional differentiability assumption). The case k = m = 2 is Green'stheorem, and k = m = 3 gives the so-called "divergence theorem" of Gauss.The case k = 2. m = 3 is the one originally discovered by Stokes. (Spivak's

(96) J. ~ f(x) dx, " ... "dx'_1 " dx.. , " ... "dXk

since every (k - I)-form is a sum of these special ones. for r = 1, ... , k.

(91) r dw = r w.• 't' • i't'

(95)Ir I'(u) du = f(I) - f(O)

'0

for every continuously differentiable function f on [0, I], which is trueby the fundamental theorem of calculus.

From now on we assume that k> 1, fix an integer r (I :s; r ::; k),and choosejErc'(E). It is then enough to prcve (94) f0f the C:lse


By (85), the boundary of the simplex (93) is,

oa ~ [e" ... , e,] + L (-l)'r,I~ 1

where

INTEGRATION OF DIFFERE:-iTIAL FOR..\iS 215

On the other hand,

dJ. = (Drf)(x)dx, 1\ dX l 1\ .•. 1\ dX~_1 1\ dX~+l 1\ ... 1\ dXk

= (-I)'-'(DJ)(x) dx, A ••. A dx,

(97)

for i = 1, ... , k. Put

0fote that to ,is ob_t.ained from [el' ... , ekJ by r - I successive interchangesof e~ and its left neighbors. Thus

,oa=(-l)'-'ro + L(-I)'r,.

i= I

Each r i has Qk -1 as parameter domain.Ifx = ro(u) and UE Q'-', then

(103)

so that

r dl = (-Iy-' f (DJ)(x) dx..la Q"

We evaluate (lO3) by first integrating with respect to X~, over the interval

[0,1..-:- (Xl + ... + X,_l + X~+l + .,. + Xk)],

put (Xl' .... X~_I' .\r+1' •.. , Xk) = (u I , ... , Uk_I)' and see with the aid of(98) that the integral over QK. in (103) is equal to the integral over Qk~l

in (102). Thus (94) holds, and the proof is complete.

(98) fU.

x·= I'-(u + ... +u )J I I k-I

"uj _ 1

(I 75,J<r),(j= r),(r<J75,k).

CLOSED FORMS AND EXACT FORMS

If 175, i 75, k, U E Q'-', and x = r,(u), then

For 0 75, i 75, k, IetI, be the Jacobian of the mapping

(99)

(100)

(I 75,J < i),(j = i),

U<J75,k).

10.34 Definition Let w be a k-form in an open set E c R". If there is a (k - 1)form;' in £ such that w = dJ., then w is said to be exact in E.

If w is of class 'fl' and dw = 0, then (;J is said to be closed.Theorem 1O.20(b) shows that every exact form of class 't' is closed.In certain sets E. for example in convex ones, the converse is true; this

IS the content of Theorem 10.39 (usually known as Poincare's lemma) andTheorem 10040. However, Examples 10.36 and 10.37 will exhibit closed formsthat are not exact.

by (35) and (96). Consequently, (97) gives

induced by r,. When i = 0 and when i = r, (98) and (99) show that (100)15 the identity mapping. Thus Jo = 1, J~ = 1. For other i, the fact thatXI = a in (99) shows that J; has a row of zeros, hence J1 = O. Thus

10.35 Remarks

(101 ) [;=0"

U'" 0, i '" r),

(104)

(a) Whether a given k-form w is or is not closed can be verified bysimply differentiating the coefficients in the standard presentation of (;J,

For example. a I-form .w = If'(x) dx"

i= 1

(l02) [ A~(-IY-'f A+(-I)'J' Aell !o !~

~ (-1)'-' [ [f( ro(u)) - f( r,(u))] du.

with /; E 't'(E) for some open set E c R", is closed if and only if theequations

(105) (DJ,)(x) = (DJ)(x)

hold for all i, J in (I, ... , n) and for all x E E.

276 PRINCIPLES OF MATIlEMATICAL A."lALYSIS INTIGRATION OF DIFFERENTIAL FORMS 277

10.36 Example Let E = R' - {O}, the plane with the origin removed. The

I-form

xdy-ydx(I 10) ~ = x' + y2

is closed in R 2 - {O}. This is easily verified by differentiation. Fix r> 0, and

define

(111) y(l) = (rcos I, rsin I) (O;S; I;S; 2n).

Then y is a curve (an "oriented I-simplex") in R' - to}. Since y(O) ~ ,(2rr),

we have .. --

Note that (l05) is a "pointwise" condition,; it does not involve anyglobal properties that depend on the shape of E.

On the other hand, to show that (0 is exact in E, one has to provethe existence of a form A, defined in E, such that dA = w. This amountsto solving a system of partial differential equations, not just locally, butin all of E. For example, to show that (104) is exact in a set E, one hasto find a function (or O-form) 9 E '(f'(E) such that

(106) (D,g)(x) = J,(x) (x E E, 1 ;S; i;S; n).

Of course, (105) is a necessary condition for the solvability of (106),

(b) Let w be an exacl k-form in.E. Then there is a (k - I)-form i. in Ewith dJ.. = w, and Stokes' theorem asserts that

(107) fw=jdi.=f.i~'f If "a'l'

(112) a, ~ 0,

Direct computation shows that

for every k-chain 'II of class CC" in E.If'¥l and '¥2 are such chains, and if they have the same boundaries,

it follows that

j w= j w.'1'1 'i'1

In particular, the integral of an exact k-form in E is °over everyk-chain in E whose boundary is 0.

As an important special case of this, note that integrals of exactI-forms in E are 0 over closed (differentiable) curves in E.

(c) Let w be a ciosed k-form in E. Then dw = 0, and Stokes' theoremasserts that

(108) f w:'jdw=O,.. ..for every (k + I)-chain '¥ of class '{f" in E.

In other words, integrals of closed k.jorms in E are 0 over k-chainsthat are boundaries of(k + I)-chains in E.

(d) Let'¥ be a (k + I)-chain in E and let .i be a (k - I)-form in E, bothof class f.(j", Since d 2J.. = 0, two applications of Stokes' theorem show that

(109) f i. = f d.i = j d 2.i = O.,:Ia'P "'i' '¥

We conclude that a2'P = 6. In other words, the boundary of aboundary is O.

See Exercise 16 for a more direct proof of this.

(113) •

The discussion in Remarks IO.35(b) and (c) shows that we can draw two

conclusions from (113):

First, 11 is not exact in R 2 - {O}, for otherwise (112) would force the integral

(I I3) to be O. 2'"Secondly, , is not Ihe boundary oj any 2-chain in R - {O} (of class 6 ),

for otherwise the fact that 11 is closed would force the integral (II3) to be o.

10.37 Example Let E ~ R3 - {O}, 3-space with the origin removed. Define

x dy A dz + y dz /'oi·dx + z dx A dy(114) (= (x' +y2 +Z2)3/'

where we have written (x. y, z) in place of (Xl' X2' x 3). Differentiation showsthat d( = O. so that ( is a closed 2-form in R

3- {O}" _

Let L be the 2-chain in RJ - {O} that was constructed III Example 10...:>2;recall that L is a parametrization of the unit sphere in R

3. Using the rectangle

D of Example 10.32 as parameter domain, it is easy to compute that

(l15) f, (~ t sin u du dv = 4rr '" O.

As in the preceding example, we c~~ now conclude that ( i:s nor exad inR3 _ {O} (since cI: ~ 0, as was shown in l'xample 10.32) and that the s~here I:is not the boundary of any 3-chain in RJ

- to} (of class (t"), although 01: = 0,The following result will be used in the proof of Theorem 1O.39.

278 PRI:s"CIPlES OF MATHEMATICAL ANALYSIS INTEGRATION OF DIFFERENTIAL FORMS 279

10.38 Theorem Suppose E is a convex open set in Rn,f E CC'(E), p is an integer,1 ~p:::;; n, and

(116) (DJ)(x) = 0 (p <j:$ n, x E E).

Then there exists an FE CC'(E) such that

for all x E E.If p = I, V is a segment in R I (possibly unbounded). Pick C E V

and define

If p > I, let U be the set of all x' E RP-I such that (x', xp

) E V forsome x p' Then U is a convex open set in RP-l, and there is a function, E'1I'(U) such that (x', ,(x')) E V for every x· E U; in other words, thegraph of, lies in V (Exercise 29). Define

(p<j:$n).

"I I (Djfl)(x) dXj /\ dx [ ~ dw = O.I j:l

We shall proceed by induction on p.Assume first that WE Y I • Then W ~ f(x) dx,. Since dw = 0,

(DJ)(x) ~ 0 for 1 < j :$ n, x E E. By Theorem 10.38 there is an F E 'f!'(E)such that DIF = f and DjF~ 0 for I <j:$ n. Thus

dF ~ (DJ)(x) dX1 = f(x) dX1 = W.

Now we take p > I and make the following induction hypothesis:Every closed kform that belongs to Y p - l is exact in E.

Choose W E Yp so that dw = O. By (118),

w =, + If[(x) dx[o /\ dxp,10

where Cl. E Yp-l' each 10 is an increasing (k - I)-index in {I, ... , p - I},and I ~ (10' p). By (120), Theorem 10.38 furnishes functions F[ E 'f!'(E)such that

Consider a fixed j, with p <j:$ n. Each I that occurs in (118) lies in{I, ... , pl. If II' I, are two of these k-indices, and if II 1'1" then the(k + I)-indices (II,j), (I, ,j) are distinct. Thus there IS no cancellatlOn,and we conclude from (119) that every coefficient in (li8) satisfies

(Djf[)(x) = 0 (x E E,p <j 5. n).

We now gather those terms in (lI8) that contain dxp and rewrite win the form

(121)

(120)

(119)

(122)(x E E).

(x E E).

(p < j :$ n, x E E).

f(x) = 'I'(x', x p )

(DJ)(x) ~ 0

(X,

F(x) = '1'(1) dl"

jX,

F(x) ~ 'I'(x', I) dl.li{x')

(DpF)(x) = f(x),

Proof Write x = (x', xp' x"), where

x' = (Xl' ... , x p _ l ), X" = (x p + l , ••. , X,,).

(When p = 1, x' is absent; when p = n, Xl; is absent.) Let V be theset of all (x', x p ) E RP such that (x'. x

p, x") E E for some x~. Being a

projection of E, V is a convex open set in RP. Since E is convex and (116)holds, f(x) does not depend on x". Hence there is a function cp, withdomain V, such that

(117)

In either case, F satisfies (117). Put

(Note: Recall the usual convention that f~ means - Jb if b < a.)(123)

10.39 Theorem If E c: RR is convex and open, if k;;:: 1, if w is a k-jorm ofclass ct' in E, and if dw = 0, then There is a (k - I)-form ;. in E such that w = JL

Briefly, closed forms are exact in convex sets.

Proof For p = J, ... , n, let Yp denote the set of all k·forms w, of class'15' in E. whose standard presentation

(118)

does not involve dxp• l , ... , dx". In other words. I c {I .... , p} ifjAX) l' 0for some x E E.

and define y = w - (- I)'-I dp. Since p is a (k - I)-form, it follows that

p

y ~ w - I I (DJ[)(x) dx[o /\ dXj10 j= 1

p-I=. -I I (DjF[)(x) dx[o /\ dXj'10 j = 1

which is clearly in Yp - t ' Since dw = 0 and J2p = 0, we have dy = O.Our induction hypothesis shows therefore that y = dp for some(k - I)-form fl in E. If 1 ~ fl + (_l)'-Ip, we conclude that w ~ d)..

By induction, this completes the proof.

2.80 PRISCIPLES OF MATHEMATICAL ANALYSIS INn:GRATION OF DIFFERENTIAL FOR..M:S 281

Furthermore, if E is 'C"-equivalent to a convex set, then (a) and (b) haveconverses, in which we assume that F is a vector field in E, of class~':

10.42 Vector fields Let F = F1 e1 + F2 e2 + FJ e3 be a continuous mapping ofan open set E c R J into RJ • Since F associates a vector to each point of E, Fis sometimes called a vector field, especially in physics. With every such F isassociated a l·form

V' F = D,F, + D, F, + D, F,.

These quantities have various physical interpretations. \Ve refer to thebook by O. D. Kellogg for more detaiis.

Here are some relations between gradients, curls, and divergences.

10.43 Theorem Suppose E is an open set in R J, U E CC"(E), and G is a vector

field in E, of class C.

(a) ifF = Vu, then V x F = O.(b) IfF = V x G, then V . F = O.

Ar = F, dx + F, dy + F, dz

COy = F1 dy /\ dz + F2 dz /\ dx + F 3 dx 1\ dy.

Here, and in the rest of this chapter, we use the customary notation (x, y, z)

in place of (Xl' X2' xJ).

It is clear, conversely, that every I-form ), in E is J. y for some vector fieldFin E, and that every 2-form ill is w, for some F. In RJ

, the study of I-formsand 2-forms is thus coextensive with the study of vectOr fields.

If U E CC'(E) is a real function, then its gradient

Vu = (D,u)e, + (D,u)e, + (D, u)e,

is an example of a vector field in E.Suppose now that F is a vector field in E, of class '{f'. Its curl V x F is the

vector field defined in E by

V x F = (D, F, - D, F,)e, + (D, F, - D,F,)e, + (D,F, - D, F,)e,

and its divergence is the real function V . F defined in E by•

(125)

and a 2-form

(124)

w = (WT)S = (di·)s = d(i.s).

Since AS is a (k - I)-form in U, w is exact in U.

10.41 Remark In applications, cells (see Definition 2.17) are often more convenient parameter domains than simplexes. If our whole development hadbeen based on cells rather than simplexes, the computation that occurs in theproof of Stokes' theorem would be even simpler. (It is done that way in Spivak'sbook.) The reason for preferring simplexes is that the definition of the boundaryof an oriented simplex seems easier and more natural than is the case for a cell.(See Exercise 19.) Also, the partitioning of sets into simplexes (called "triangulation") plays an important role in topology, and there are strong connectionsbetween certain aspects of topology, on the one hand, and differential forms,on the other. These are hinted at in Sec. 10.35. The book by Singer and Thorpecontains a good introduction to this topic.., Since every cell can be triangulated, we may regard it as a chain. Fordimension 2, this was done in Example 10.32; for dimension 3, see Exercise 18.

Poincare's lemma (Theorem 10.39) can be proved in several ways. See,for example, page 94 in Spivak's book, orpage 280 in Fleming's. Two simpleproofs for certain special cases are indicated in Exercises 24 and 27.

10.40 Theorem Fix k, 1 :;; k :;; n. Let E eRn be an open set in which everyclosed k-form is exact. Let T be a 1-1 C(r-mapping of E onto an open set U eRnwhose inverse S is also of class CC".

Then every closed k-form in U is exact in U.

Note that every convex open set E satisfies the present hypothesis, byTheorem 10.39. The relation between E and U may be expressed by sayingthat they are '{f"-equivalent.

Thus every closed/orm is exact in any set which is '(f"-equivalent to a convexopen set.

Proof Let w be a k-form in U. with dw = O. By Theorem 1O.22(c),W T is a k-form in E for which d(wT) = O. Hence wT = d). for some(k - I)-form i. in E. By Theorem 10.23, and another application ofTheorem 1O.22(c),

VECTOR ANALYSIS

We conclude this chapter with a few applications of the preceding material totheorems concerning vectOr analysis in R3

• These are specIal cas"es of theoremsabout differential forms, but are usually stated in different terminology. Weare thus faced with the job of translating from one language to another.

(a') If V x F = 0, then F = Vufor some u E '6'"(E).(b') IfV . F = 0, then F = V x G for some vector field G in E, ofclass '6'"

Prooi If we compare the definitions of Vu, V x F, dnd Y·:F with thedifferential forms !.r and W r given by (124) aod (125), we obtain thefollowing four statements:

-- 282 PRINCIPLES Of MATHEMATICAL ANALYSIS INTEGRATION OF DIFFERENTIAL FORMS .. 283

F = Vu

VxF=O

F=VxG

V·F=O

if and only if A. = duo

if and only if dA. = O.

if and only if w. = dAG •

if and only if dw. = O.

Proof Put;. = , dx + Pdy. Then

dA = (D 2a) dy A dx + (D,P) dx A dy

= (D,P - D2a) dA,

and (127) is the same as

Now ifF = Vu, then ;.• = du, hence dA. = d'u = 0 (Theorem 10.20),which means that V x F = O. Thus (a) is proved.

As regards (a'), the hypothesis amounts to saying that d).• = 0 in E.By Theorem 10040, AF = du for some IJ..form U. Hence F = Vu.

The proofs of (b).nd (b') follow exactly the same pattern.

f A= JdA,ce 0

which is true by Theorem 10.33.

With a(x, y) = - y and P(x, y) = x, (127) becomes

10.44 Volume elements The k-form (128) t r (x dy - y dx) = A(Q),, ce

is called the volume element in R'. It is often denoted by dV (or by dV, if itseems desirable to indicate the dimension explicitly), and the notation

is used when c1> is a positively oriented k-surface in Rk and 1 is a continuousfunction on the range of ell.

The reason for using this terminology is very simple: If D is a parameterdomain in R'<, and if<!J is a 1-1 '\f'-mapping of D into Rk, with positive JacobianJ., then the left side of (126) is

f f(<!l(u))J~(u) du =f f(x) dx,D <!ltD)

A(<!l) = f IN(u, v)1 du dv.D

J' fdA = f f(<!l(u, v))IN(u, v)1 dudv.• D

In particular, whenf= I we obtain the area of<1J, namely,

(132)

The Jacobians in (129) correspond to the equation

(130) (x, y, z) = <!leu, v).

If f is a continuous function on <!J(D), the area integral of f over <!J is

defined to be

(131) .

the area of Q.With 0 = 0, P= x, a similar formula is obtained. Example 1O.l2(b) con

tains a special case of this.

10.46 Area elements in RJ Let c1> be a 2-surface in RJ, of class ee f

, with parameter domain Dc R 2

• Associate with each point (u, v) E D the vector

. e(y, z) e(z, x) e(x, y)(129) N(u,v) =,--()e,+.,--()e,+.,--()e,.u u, v c u, v 0 u, v

We already saw

.Jf(x) dx, A ... A dx, = f f dV~ ~

(126)

by (35) and Theorem 10.9.

In particular, whenf= I, (126) defines the volume of<!l.a special case of this in (36).

The usual notation for dV2 is dA.

10.45 Green's theorem Suppose E is an open set in R', 0 E 't'(E), PE \!i'(E),and Q is a closed subset 01 E. with positively oriented boundary en, as describedin Sec. 10.31. Then

(127) f ,(ap ao)(0 dx + Pdy) = J - - - dA.m 0: ox iJy

The following discussion will show that (131) and its special case (132)are reasonable definitions. It will also describe the geometric features of the

vector N.Write 11> = (l)le1 + (P2 e2 + cp~ e3' fix a point Po = (ut), vo) E D, put

N = N(po), put

(133) " = (D,'P,)(Po), P, = (D, 'P,)(Po) (i = 1,2,3)

284 PR1I'<CIPLES Of MATHEMATICAL Al'iALYSIS

A straightforward computation now leads to

and let T E L(R2, R 3

) be the linear tran..sformation given by

Note that T = <t>'(po), in accordance with Definition 9.11.Let us now assume that the rank of T is 2. (If it is I or 0, then N = 0, and

the tangent plane mentioned below degenerates to a line or to a point.) Therange of the affine mapping

0'::::;v.::::;2JT.O:$. U.::::; 271;,o~ t :$. a,

J~ = o(x, y, z) = t(b + , sin u)0(1, a, v)

vol ('I'(K)) = 2n'Q'b

which is positive on K, except on the face t = O. If we integrate J'i' over K, we

obtain

(l41)

x=(COSU

Y = (b + t sin u) cos u

z = (b + I sin a) sin v

describe a mapping \.fI of RJ into RJ which is l~l in the interior of K, such that

"¥(K) is a solid torus. Its Jacobian is

The equations

\Ve conclude that (132) is correct when is affine. To justify the definition(132) in the general case, divide D into small rectangles, pi~k a point (uo , vo)in each, and replace <1>. in each rectangle by the correspondmg, tangent .plane.The sum ofthe areas of the resulting parallelograms, obtained VIa (140), IS thenan approximation to A(<t». Finally, one can justify (131) from (132) by approxi

mating/by step functions.

10.47 Example Let 0 < a (po) + T(a, v)

is then a plane II, called the tangent plane to <t> at Po. [One would like to callIT the tangent plane at (po), rather than at Po ; if is not one-to-one, this runsinto difficulties.]

If we use (133) in (129), we obtain

(135) N = (a,p, - "p,)e, + ("P, - "p,)e, + ("P, - a,p,)e"

and (134) shows that

Hence N is perpendicular to IT. It is therefore called the normal to at Po.A second propeny of N, also verified by a direct computation based on

(l35) and (l36), is tbat the determinant of the linear transformation of R' thattakes (e" e" e,) to (Te" Te" N) is IN I' > 0 (Exercise 30). The 3-simplex

(134)

is therefore IN I.This parallelogram is the image under T of the unit square in R 2

• If Eis any rectangle in R2

, it follows (by the linearity of T) that the area of theparallelogram T(E) is

is thus positively oriented.The third property of N that we shall use is a consequence of the first two:

The above-mentioned determinant, whose value is !Ni 2, is the volume of the

parallelepiped with edges [0. Ted, [0, Te,], [0, N]. By (137), [0, N] is perpen- •dicular to the other two edges. The area of the parallelogram with vertices

(138)

(139)

(l40)

[0, Te" Te, , N]

A(T(E)) = INIA(E) = J IN(ao, vo)1 da dv.E

as the volume of our solid torus.Now.consider the 2-chain <t> = elf'. (See Exercise 19.) qJ maps the faces

u = 0 and u = 2IT of K onto the same cylindrical strip, but with opposite orien~ations. '¥ maps the faces v = 0 and v = 2n onto the same circular disc, ~ut WIthopposite orientations. If' maps the face ( = 0 onto a circ!~, ~hich contrIbutes 0to the 2-chain olf. (The relevant Jacobians are 0.) Thus <D IS sImply the 2-surfaceobtained by setting t = a in (141), with parameter domain D the square defined

by 0 ,; a ,; 2n, 0 ,; l' ,; 2n. .According to (129) and (141), the normal to <t> at (u, v) ED IS thus the

vector

N(u, v) ~ alb + a sin a)n(a, v)

where

n(u, v) = (cos u)e1+ (sin u cos v)e 2 + (sin u sin v)e 3 •

286 PRI:-iCrPLES OF ~ATHEMATICAL ASAtYSIS

INTEGRATIO:-O OF DrFFERESTIAL FOR~S 287

Since In(u, v)1 = I, we have IN(u, e)1 ~ a(b + a sin u), and if we integrate thisover D, (131) gives

A(<I» ~ 4n'ab

as the surface area of our torus.

H we think of N = N(u, v) as a directed line segment, pointing from<!:leu, v) to .<!:l~u, v) + l':(u, v), then N points outward, that is to say. away from'!'(K). ThIS IS so because J~ > 0 when I ~ a, ,

For example, take u = v = rr/2, t = a. This gives the largest value of z on'¥(K), and N = a(b + a)e 3 points "upward" for this choice of Cu, v).

I~.48 Integrals. of I-forms in R 3 Let y be a (g'-curve in an open set E c R3,wIth p~rameter interval [0, 1], let F be a vector field in E, as in Sec. 10.42, andde~ne )'r by (124). The integral of J. r over y can be rewritten in a certain waywhIch we now describe.

For any u E [0, 1],

leu) = yi(u)e I + y;(u)e, + y;(u)e,

is cal~ed t?e tangent vector to y at u. We define t = t(u) to be the unit vector inthe dlfectlOn of y'(u). Thus

y'(u) = Iy'(u) It(u).

[If y'(u) = 0 for some u, put t(u) = eI ; any other choice would do just as well.]By (35),

The right side of (143) should be regarded as juSt an abbreviation for thelast integral in (142). The point is that F is defined on the range of y, but t isdefined on [0, I]; thus F,' t has to be properly interpreted. Of course, when yis one-to-one, then t(u) can be replaced by t(y(u», and this difficulty disappears.

10.49 Integrals of 2-forms in R J Let <D be a 2-surface in an open set E c R 3,

of class C(j', with parameter domain D c: R 2• Let F be a vector field in E, anddefine WI' by (125). As in the preceding section, we shall obtain a differentrepresentation of the integral of W F over cD.

By (35) and (129),

f W, = r (F, dy A dz + F, dz A dx + F, dx A dy). '.= r I(F, 0 <1» ~(y, z) + (F, 0 <1» ~(z, x) + (F, ,<I» ~(x, y)) du dv

, D \ c(u, v) c(u, 0) c(u, v)

= r F((u, v)) , N(u, v) du dv.'D

Now let n = n(u, v) be the unit vector in the direction of N(u, v). [IfN(u, v) = 0 for some (u, 0) E D, take n(u, v) ~ e,.] Then N = IN!n, and there'fore the last integral becomes

f F((u, v))· n(u, v)IN(u, v)ldudv.D

By (131), we can finally write this in the form

With regard to the meaning of F . n, the remark made at the end of Sec. 10.48applies here as well.

We can now state the original form of Stokes' theorem.

(142)

f',I

i., ~ I J F,(y(u»)y,(u) duy '''' 1 ()

I

= f F(y(u»· leu) duo,I

~ I F(I(u», l(u)1 /(u) 1 duo• 0

(144) f w,=f (F'n)dA,. '.Theorem 6.27 makes it reasonable to call Ii(u) I du the element of arc

length along y. A Customary notation for it is ds, and (142) is rewritten i~ theform

10.50 Stokes' formula IfF is a vector field of ciass Cfj' in an open set E c R 3 ,

and If 11 is a 2·surface oj class rc" in E, then

1143) I 'r = J' (F' t)j1's., , (145) r (If x F) , n dA = r (F' I) ds.'41 'i"t>

Since t is a unit tangent vector to }', F . t is called the tangential component Proof Put H = V x F. Then. as in the proof of Theorem 10.43. we haveofF along y.

(146) wH=dJ."

288 PRl~CIPLES OF ~ATHEMATiCAL A~ALYSIS

Hence

J(V x F) . n dA ~ J(H' n) dA = J"'H• • •

= r dA,=J' ;.,= r (F·t)ds."$ ';$ 'c$

Here we used the definition of H, then (144) with H in place of F,then (146), then-the main step-Theorem 10.33, and finally (143),extended in the obvious way from curves to I-chains.

10.51 The divergence theorem If F is a vector field of class 0/' in an open setE c R J

, and if n is a closed subset of E with positively oriented boundary en(as described in Sec. 10.31) then

(147) J(V· F) dV = J' (F' n) dA.o en

Proof By (125),

dw, ~ (V . F) dx A dy A dz = (V . F) dV.

Hence

f (V· F) dV = J d"', = I "" = I (F' n) dA,o 0 "'0 ill

by Theorem 10.33, applied to the 2-form "'" and (144).

EXERCISES

1. Let H be a compact convex set in Rt, with nonempty interior. LetjEC(j(H), put

j(x) = 0 in the complement of H, and define Iii las in Definition 10.3.Prove that fHi is independent of the order in which the k integrations are

carried out.Hint: Approximate i by functions that are continuous on Rk and whose

supports are in H, as was done in Example lOA.2. Fori= 1,2,3, ... ,let tpIE'tf(R 1

) have support in (2- 1,2 1-'), such that j<pr= 1.

Put ,f(x, Y) ~ L [,/,,(x) - '/" H(X)],/,,(y)

/"'1

Then jhas compact support in R 2,lis continuous except at (0,0), and

INTEGRATION OF DiffERENTIAL FORMS 289

3. (a) If F is as in Theorem 10.7, put A ~ F'(O), F,(x) ~ A"F(x). Then F;cO) ~ [.

Show that

in some neighborhood of 0, for certain primitive mappings Gi> ... , Gft • Thisgives another version of Theorem 10.7:

F(x) ~ F'(O)G. 0 G•. , 0 ... " G,(x).

(b) Prove that the mapping (x,y)-+(y,x) of R 2 onto R 2 is not the compositionof any two primitive mappings, in any neighborhood of the origin. (This showsthat the flips B I cannot be omitted from the statement of Theorem to.7.)

4. For (x, y) E R2, define

F(x, y) = (e-" cosy - 1, e:r sin y).

Prove that F = G 2 :> GI> where

G,(x, Y) ~ (e' cos Y - 1, Y)

G,(a, v) ~ (a, (1 + u) tan v)

are primitive in some neighborhood of (0, 0).

Compute the Jacobians of G I , G2 , F at (0, 0). Define

and findR,(u, v) ~ (h(a, v), v)

so that F = H t 0 H 2 is some neighborhood of (0, 0).5. Formulate and prove an anaiogue of Theorem to.8, in which K is a compact

subset of an arbitrary metric space. (Replace the functions Pi that occur in theproof of Theorem 10.8 by functions of the type constructed in Exercise 22 ofChap. 4.)

6. Strengthen the conclusion of Theorem 10.8 by showing that the functions V;! canbe made differentiable, and even infinitely differentiable. (Use Exercise 1 ofChap. 8 in the construction of the auxiliary functions !Pl.)

7. (a) Show that the simplex Qt is the smallest convex subset of Rt that contains

0, el, ... , e•.(b) Show that affine mappings take convex sets to convex sets.

8. Let H be the parallelogram in R 2 whose vertices are (1,1), (3, 2), (4, 5), (2, 4).Find the affine map T which sends (0,0) to (I, 1), (1,0) to (J, 2), (0, 1) to (2,4).

Show that J T = 5. Use T to convert the integral

I dy If(x, y) dx~ 0 but I dx Jf(X, y) dy ~ 1.

Observe that f is unbounded in every neighborhood of (0, 0). to an integral over P and thus compute CI:.

290 PRINCIPLES OF MATHEMATICAL ANALYSIS INTeGRATION OF DIFFERENTIAL FOR.\1S 291

Show that

Show that T maps Ik onto Q\ that Tis 1-1 in the interior of I k, and that its

inverse 5 is defined in the interior of Qk by UI = Xl and

• •L: x, ~ 1 - [1 (1 - u,).1= 1 1= 1

1 - Xl - ••• - XI_Iu,

O-';;;'r~a,

9. Define (x, y) ~ T(r, B) on the rectangle

by the equations

x=rcosB, y=fsin8.

Show that T maps this rectangle onto the closed disc D with center at (0, 0) andradius a, that Tis one·to-one in the interior of the rectangle, and that J r(f, 8) = f.

If/e <'C(D), prove the formula for integration in polar coordinates:

f I(x, y) dx dy ~ r"f" f(T(r, O»r dr dB.D ·0 0

for i = 2, ... , k. Show that

Hint: Let Do be the interior of D, minus the interval from (0,0) to (0, a).

As it stands, Theorem 10.9 applies to continuous functions/whose support lies inDo. To remove this restriction, proceed as in Example 1004.

10. Let a -+ co in Exercise 9 and prove that

and

13. Let fit ... , fk be nonnegative integers, and prove that

If k:2:: 2 and (] = [Po, PI> ... , Pk] is an oriented affine k-simplex, prove that a21] = 0,

directly from the definition of the boundary operator 8. Deduce from this thatoZ\}" = 0 for every chain 'Y.

Hint: For orientation, do it first for k = 2, k'= 3. In general, if i <j, let l]i1

be the (k - 2).simplex obtained by deleting Pi and P1 from 1], Show that each I]IJ

occurs twice in a2 a, with opposite sign.Put JZ = r! + 7Z, where

Explain why it is reasonable to call j2 the positively oriented unit square in R 2•

Show that 8Jl is the sum of 4 oriented affine I-simplexes. Find these. What is8(T! - Tz)1

Consider the oriented affine 3-simplex18.

17.

16.

Hint: Use Exercise 12, Theorems 10.9 and 8.20.Note that the special case fl = ... = fk = 0 shows that the volume of Qk

is 11k!.14. Prove formula (46).15. If wand ,.\ are k- and m·forms, respectively, prove that

0<1 < 1O<S< :::0,

f I(x, y) dx dy ~ f" f'" fCT(r, B»r dr dB,AZ 0 0

for continuous functions / that decrease sufficiently rapidly asixi ....:..- Iyl ---+ 00.

(Find a more precise formulation.) Apply this to

by setting u = s - st, v = st. Show that T is a 1-1 mapping of the strip onto thepositive quadrant Q in R Z

• Show that Ir(s, t) = s.For X > 0, Y > 0, integrate

ICx,y) ~ exp(-x' - y')

to derive formula (101) of Chap. 8.11. Define (u, v) = T(s, t) on the strip

over Q, use Theorem 10.9 to convert the integral to one over the strip, and deriveformula (96) of Chap. 8 in this way.

(For this application, Theorem 10.9 has to be extended so as to cover certainimproper integrals. Provide this extension.)

12. Let r be the set of all u = (UI, "" Uk) E RIf. with 0 ~ Ul ~ I for all i; le[ Qk be theset of aU x = (XI, ... , Xk) E Rk with Xl :::::: 0, I:XI ~ 1. (Ik is the unit cube; Qk isthe standard simplex in R k

.) Define x = T(u) by

in R3. Show that 0"\ (regarded as a linear transformation) has determinant 1.Thus I]l is positively oriented.

292 PRlNCIPLES OF MATHEMATICAL ANALYSIS INTEGRATION OF DIFfERE:-lTlAL FORM:S 293

These are affine, and map R2 into R 3 •

Put fid = B,t(P), for r = 0, I, i = I, 2, 3. Each {3d is an affine-oriented2-chain. (See Sec. 10.30.) Verify that

Let a2, ... , (16 be five other oriented 3~simplexes, obtained as follows:There are five permutations (it, i:, i3 ) of (1, 2, 3), distinct from (I, 2, 3). Associatewith each (iI, i2 , i3 ) the simplex

Use Stokes' theorem to deduce that

Then is a 2-surface in R 2- {O} whose parameter domain is the indicated rect

angle. Because of cancellations (as in Example 10.32),

<1>(1, u) ~ (I - u) ret) -,- uy(t).

Hint: For 0:::;;: t:::;;: 27T, O:s;; u:S;; I, define

with parameter interval [0, 27T], with f(0) = r(2rr), such that the intervals [yet),

r(t)] do not contain 0 for any t E [0, 2..]. Prove that

f ~ ~ 2".r

because dry = 0.

(c) Take f(t) = (a cos t, b sin t) where a> 0, b > 0 are fixed. Use part (b) toshow that

(d) Show that

Bu(u, v) = (1, u, v),

BI2 (1l, v) = (u, 1, v).

B I3(u, v) = (u, v, 1).

Bot(u, v) = (0, u, v),

Bo2 (u, v) = (u, 0, v),

Bo3(u, v) = (u, v, 0),

sUit i2 , i3 )[0, ell' ell + e l1 , elJ + ei1 + ell]

where s is the sign that occurs in the definition of the determinant. (This is how 7"2

was obtained from 7"1 in Exercise 17.)

Show that (12, ••• , (16 are positively oriented.

Put J3 = a1 + ... --!- a6. Then j3 may be called the positively oriented unitcube in R 3

•.

Show that 8P is the sum of 12 oriented affine 2~simpiexes. (These 12 triangles cover the surface of the unit cube P.)

Show that x"= (Xl, X2, xJ) is in the range of (11 if and only if 0 ~ XJ ~ X 2~xI:S:L

Show that the ranges of ai, ...• (16 have disjoint interiors, and that theirunion covers P. (Compare with Exercise 13; note that 3! = 6.)

19. Let J2 and J3 be as in Exercise 17 and 18. Define

,oJ' ~ L (-l)'(~" - ~,,),,. , Tf = d(arc tan~)

in agreement with Exercise 18.'20. State conditions under which the formula

in any convex open set in which x oF 0, and that

Tf = d(- arc tan~)

21.

is valid, .and show that it generalizes the formula for integration by parts.Hint: d(fw) ~ (df).~ w.;.. fdw.

As in Example 10.36, consider the I-form

in any convex open set ir. which y i= O.Explain why this justifies the notation TJ = dB, in spite of the fact that TJ is

not exact in R 2- {OJ.

xdy- ydxTf = x2 + y2

(e) Show that (b) can be derived from (d).

(/) If r is any closed ~"·curve in R l- {O}, prove that

(a) Carry OUt the computation that leads to formula (113), and prove that en, = O.

(b) Let yet) = (r cos t, r sin t), for some r > 0, and let r be a ~"~curve in R 2 - {O}, (See Exercise 23 of Chap. 8 for the definition of the index of a curve.)

294 PRINCIPLES OF MATHEMATICAL ANALYSIS ~n:GRAnO~ OF DIFFERE:-inAL FORMS' 295

22. As in Example 10.37, define { in R 3 - {O} by

,~xdY A dz+ ydz A dx+zdx Adyr'

where Y = (Xl +yl + Zl)lll, let D be the rectangle given by 0:5: u:5: Tf, 0:::;;; v:::;;; 2n,and let;'; be the 2-surface in R3, with parameter domain D, given by

face $ to which (c) can be applied to show that J~~ = O. The same thing holdswhen u is fixed. By (a) and Stokes' theorem,

f '~f d, ~ O... .(e) Put ,\ = - (z!r)7). where

(a) Prove that d, ~ 0 in R' - (0).

(hj Let S denote the restriction of Z to a parameter domain E c D. Prove that

x = sin u cos v, y = sin u sin v, Z= cos u. xdv-vdx7} = "1 -.z '

x --:- y

as in Exercise 21. Then,\ is a I-form in the open set V c: R 3 in which Xl -:- y1 > O.Show that { is exact in V by showing that

where (u, v) E E, 0::;; t::::;; 1. For fixed v, the mapping (t, u) -+ 'F(t, u, v) is a 2-sur-

'FCt, u, v) ~ [1 - t + tf(u, v)J L (u, v),

(-I <t < 1).

and

g~(t) = J' (I - 5 1 ),t-3)/2 ds.-,Him: h. satisfies the differential equations

x . (~f.)(x) = (l

•Uh = (YI<)-I< L (_l)I-lx ; dXl ;\ .. ,\ d.~;_l 1\ dXi+l 1\'" ;\ dXI<.

j"'l

Note that W2 = 7), W3 = ;;, in the terminology of Exercises 21 and 22. Notealso that

(f) Derive (d) from (e), without using (c).

Hint: To begin with, assume 0 < Ii < Tf on E. By (e),

E, c E, c ... c E, ~ R' ~ {O}.

(a) Prove that dWI< = °in E•.(b) For k = 2.... , n, prove that Wt is exact in E._I. by showing that

w~ = d(f,.wt_d = (df..) i\ WI<_l,

where f.(x) = (-1)"' gl«x.!r.) and

and

Show that the two integrals of ,\ are equal, by using part (d) of Exercise 21, and bynoting that zir is the same at ~(u, v) as at O(u. r).

(g) Is { exact in the complement or"every line through the origin?23. Fix n. Define Yt = (xi +... ...;... xi)l/l for 1~ k::::;; n, let E. be the set of all x E W

at which '1< > 0, and let WI< be the (k - I)-form defined in EI< by

z ~ g(r)h,(s).

f ,~O,•

y ~ g(t)h,(s),x ~ g(t)h,(s),

Prove that

r'~J sin u du dv ~ A(S)," '

(c) Suppose g, hI> h l , h3 , are W"·functions on CO, 11. g > O. Let (x, y, z) = <fl(s, I)

define a 2-surface <fl, with parameter domain P, by

(Since S is the "radial projection" of 0 into the unit sphere, this result makes itreasonable to call fn{ the "solid angle" subtended by the range of n at the origin.)

}-li.'7t: COi:.side!' the J-surface q.' giYer. by

where A denotes area, as in Sec. 10.43. Note that this contains (115) as a special•case.

f'~f'~A(S)n ,

directly from (35),

Note the shape of the range of <fl: For fixed s, <fl(s, t) runs over an intervalon a line through O. The range of ep thus lies in a "cone" with vertex at the origin.

(d) Let E be a closed rectangle in D, with edges parallel to those of D. SupposefE 'C"(D),f> O. Let n be the 2-surface with parameter domain E, defined by

il(u, v) ~ feu, v) L (u, v).

Define S as in (b) and prove that

296 PRI:-.'CIPLES OF ~fATHnfATICAL A:-<ALYSISINriGRATIO~ OF DIFFER.E~TIAL FORMS -297

24.

(c) Is w" exact in E"?

(d) Note that (b) is a generalization of part (e) of Exercise 22. Try to extend someof the other assertions of Exercises 21 and 22 to w", for arbitrary n.

Let w =~ai(x) dXi be a I-form of class tg" in a convex open set E c:: R". Assumedw = 0 and prove that w is exact in E, by completing the following outline:

Fix pEE. Define

28. Fix b > a > 0, define

<1>(" 0) ~ (r cos 0, r sin 0)

for a S r ~ b, 0 S;; 8 :::;: 2IT. (The range of tf> is an annulus in R2 .) Put w = x 3 dy,and compute both

"Apply Stokes' theorem to affine-qriente,d 2-simplexes [p, x, yJ in E. DeduC'e that

. .,fey) - fix) ~L (y, - x,) J a,«l - t)x -+- ty) dt

,,. 1 0

f(x) ~ J W

[II,X]

(x E E).

and

to verify that they are equal.29. Prove t'he existence of a function iX with the properties needed in the proof of

Theorem 10.38, and prove that the resubng function F is of class rt'. (Bothassertions become trivial if E is an open cell or an open baH, since :x can then betaken to be a constant. Refer to Theorem 9.42.)

30. If N is the vector given by (135), prove that

for x E E, Y E E. Hence (D,f)(x) ~ a,(x).

25. Assume that w is a I-form in an open se[ E c:: R" such that

for every closed curve y in E, of class ~'. Prove that w is exact in E, by imitatingpart of the argument sketched in Exercise 24.

26. Assume w is a I-form in R 3- {OJ, of class f.(j' and dw =0. Prove that w is exact in

R 3- {OJ.

Hint: Every closed continuously differentiable curve in R J - (O} is the.boundary of a 2-surface in R3

- {OJ. Apply Stokes' theorem and Exercise 25.

27. Let E be an open 3-ceH in R3, with edges parallel to the coordinate axes. Suppose(a, b, c) E E,/, E 'If'(E) for j ~ 1,2,3,

W =/1 dy 1\ dz...:... Il dz i\ dx...;.. 13 dx i\ dy,

and assume that dw = 0 in E. Define

where

g,(x, y, z) ~ rf,(x, y, 5) ds - ( f,(x, t. c) dt• c • ~

g,(x, y, =) ~ - rf,(x, y, 5) d5,'<

for (x, y, Z) E E. Prove that dA = w in E.

Evaluate these integrals when w = ~ and thus find the form A that occurs inpan (e) of Exercise 22.

Also, verify Eq. (13i).

31. Let E c:: RJ be open, suppose g E CC"(E), II E 93"(E), and consider the vector field

F ~ 9 V h.

(a) Prove that

where 'Ph = \" . ('\17) = ::Jj 2hjcxf is the so-called "Laplacian" of h.

(b) If 0 is a closed subset of E with positively oriented boundary cD (as inTneorem 10.51), prove that

, I' 2hJ [gY'h+(Yg)'(Yh)]dV~ ga;;dA

{) 'Ul

where (as is customary) we have written chien in place of (vh) . n. (Thus Eh/enis the directional derivative of h in the direction of the outward normal to eo, the

so-called normal derimth"(! of h.) Interchange g and h, subtract the resultingformula from the first one, to obtain

r r ( 2h dB)(gY'h-hY'g)dV~ g-o- -h-o- dA.

"0 "~n en en

These two formulas are usually called Green·s identiries.

(c) Assume that h is harmonic in E; this means thaI Ph = O. Take g = 1 and con

clude that

298 PR.ll"CIPLES OF MATHEMATICAL A....ALYSLS

Take 9 = h, and conclude that h = a in n if h = a on EO.(d) Show that Green's identities are also valid in R1

•

32. Fix 8, 0 < 8 < 1. Let Dbethesetofall(B, t) eR1suchthatO 5.B 5.11'", -0 -:;:;'t 5.0.Let tfl be the 2-surface in RJ, with parameter domain D, given by

x~ (1- ISin e) cos zey~ (l - ISin e) sin zeZ= tcos 8

where (x, y, z) = ¢l(8, t). Note that ~(11'", t) = t!>(0, -0, and that <D is one-to-one

on the rest of D.The range lvf = 4l(D) of <fl is known as a lv!6bius band. It is the simplest

example of a nonorientabie surface.Prove the various assertions made in the following description: Put

p, ~ (0, -0), p, ~ (rr, -0), p, ~ (rr, 0), p. ~ (0, '), p, ~ p,. Put y, ~ [p" p,. d,i= 1, ... ,4, and put f/= t!> 0 Yl. Then

Put a ~ (1, 0, -0), b ~ (1, 0, 0). Then

(p,) ~ (p,) ~ a,

and e<1> can be described as follows.r 1 spirals up from a to b; its projection into the (x, y)-ptane has winding

number +-1 around the origin. (See Exercise 23, Chap. 8.)

f, ~ [b, aJ.r J spirals up from a to b; its projection into the (x,y) plane has winding

number -1 around. the origin.f. ~ [b, aJ.Thus c<1> = r 1 -i- r J --;-. 2f 2 •

If we go from a to b along r L and continue along the "edge" of ,\t[ until we

return to a, the curve traced out is

which may also be represemed on the parameter interval [0, 2;r] by the equations

x ~ (1 -,- 0 sin e) cos zey = (l -:- 0 sin 8) sin 28

z= -ocos8.

It should be emphasized that r 7"= c<P: Let TJ be the I-form discussed inExercises 21 and 22. Since dry = 0, Stokes' theorem shows [hat

But although r is the "geometric" boundary of ;\1, we have

In order to avoid this possible source of confusion, Stokes' formula (Theorem

10.50) is frequently stated only for orientable surfaces tfl.

11THE· LEBESGUE THEORY-JOt -

11.1 Definition A family Jf of sets is called a ring if A E ~ and BE Jf impliesTHE LEBESGUE THEORY

(1) Au B E r%, A - B E r%.

(2)

Since A n B = A - (A - B), we also have A n B E Jf if Jf is a ring.A ring 9i is called a (J·ring if

whenever All E Jf (n = I, 2, 3, ...). Since

~ ~nA, = A1 - U (A 1 - A,),II'" 1 11= 1

we also have

if;?) is a a-ring.

11.2 Definition We say that ¢ is a set function defined on Jii if 1> assigns toevery A E Jf a number ¢(A) of the extended real number system. 1> is additire

if A n B = 0 implies

and 4> is countably additive if A j n A j = 0 (i # j) implies

We shall always assume that the range of 4> does not contain both + 'J:)

and - 00; for if it did, the right side of (3) could become meaningless. Also,we exclude set functions whose only value is +::0 or - oc.

It is interesting to note that the left side of (4) is independent of the orderin which the An's are arranged. Hence the rearrangement theorem shows thatthe right side of (4) converges absolutely if it converges at all; if it does notconverge, the partial sums tend to + 00, or to -:0.

If 4> is additive, the following properties are easily verified:

if A j n A j = 0 whenever i #; j.

It is the purpose of this chapter to present the fundamental concepts of theLebesgue theory of measure and integration and to prove some of the crucialtheorems in a rather general setting, without obscuring the main lines of thedevelopment by a mass of comparatively trivial detail. Therefore proofs areonly sketched in some cases, and some of the easier propositions are statedwit~out proof. However, the reader who has become familiar with the techniques used in the preceding chapters will certainly find no difficulty in supplying the missing steps.

The theory of the Lebesgue integral can be developed in several distinctways. Only one of these methods will be discussed here. For alternativeprocedures we refer to the more specialized treatises on integration listed inthe Bibliography.

SET FUNCTIONS

If A and B are any two sets, we write A - B for the set of ail elements x suchthat x E A, x ¢ B. The notation A - B does not imply that Be A. We denotethe empty set by 0, and say that A and B are disjoint if An B = O.

(3)

(4)

(5)

(6)

¢(A u B) ~ ¢(A) + ¢(B),

¢(O) ~ O.

¢(A , u ... uA,) = ¢(A\) + ... + ¢(A,)

302 PRJNCIPLES Of MATHEMATICAL A:"ALYSIS

THE LEBESGUE THEORY 303

(7) <P(A, u A,) + <P(A, n A,) ~ <P(A,) + <P(A,).

If <P(A);;' 0 for all A, and A, c A" then

(8) <P(A,) ,; <P(A,).

If A is the union of a finite number of intervals, A is said to be an elemen·tary set.

If I is an interval, we define

Because of (8), nonnegative additive set functions are often calledmonotonic.

p

m(I) = T1 (b i - a),,"'1

if B c A, and I(<PB)I < + 00.

no matter whether equality is included or excluded in any of the inequalities (10).If A = II V ... u I". and if these intervals are pairwise disjoint, we set

(9) <P(A - B) ~ <P(A) - <P(B)

(11 ) meA) = m(I,) + ... + m(I,).

Then, as n -t co,

<P(A,) ~ <P(A).

00

A = UA,.,,"'1

11.3 Theorem Suppose ¢ is cot/mably additive on a ring lA. Suppose A" E &l(n ~ 1,2,3, ...), A, c A, c A, c "', A E~, and

We let g denote the family of all elementary subsets of RP.At this point, the following properties should be verified:

(12) t is a ring, but not a a-ring.(13) If A E,c, then A is the union of a finite number of disjoint intervals.(14) If A E t, meA) is well defined by (11); that is, if two different decompo-

sitions of A into disjoint intervals are used, each gives rise to the samevalue of meA).

(15) m is additive on t.

Note that if p = 1,2,3, then m is length, area, and volume, respectively.(n = 2, 3, ...).B,,=A

Il-A

II_ 1

Proof Put B, = A" and

11.5 Definition A nonnegative additive set function ¢ defined on rff is said tobe regular if the following is true: To every A E I! and to every e > 0 thereexist sets FE tff, GEtS such that F is closed, G is open, Fe A c G, and

Then B, n Bj = 0 for i i' j, A, = B, u ... u B" and A = UB,. Hence

,<P(A,) ~ I f(B,)

i= I

and(16) <P(G) - £'; <P(A) ,; <P(F) + £.

00

<P(A) ~ I <P(B,).j"'l

CONSTRUCTIO" OF THE LEBESGUE MEASURE

11.6 Examples

(a) The set function m is regular.If A is an interval, it is trivia! that the requirements of Definition

!l.S are satisfied. The general case follows from (13).(b) Take RP = R1

, and let 0: be a monotonically increasing function. defined for all real x. Put

11.4 Definition Let RP denote p-dimensional euclidean space. By an intervalin RP we mean the set of points x = (XI' ... , xp) such that

or the set of points which is characterized by (10) with any or all of the :;:;:signs replaced by <. The possibility that a j = b i for any value of i is not ruledout; in particular, the empty set is included among the intervals.

(10) U=I, ... ,p),

~([a, b» = ,(b-) - ,(a-),

~([a, b]) = ,(b+) - ,(a-),

~(a, b]) = ,(b +) - ,(a+),

~«a, b)) ~ ,(b-) - a(a+).

Here [a, b) is the set a .:s; x < b, etc. Because of the possible discon·tinuities of '::t, these cases have to be distinguished. If J1 is defined for

304 PRINCIPLES OF MATHEMATICAL ANALYSISTHE LEBESGUE THEORY 305

elementary sets as in (II), f1 is regular on 8. The proof is just like thatof Cal.

Our next objective is to show that every regular set function on 6' can beextended to a countably additive set function on a a-ring which contains t.

The regularity of fl shows that A contains a closed elementary set F suchthat pCF);' pCA) - ,; and since Fis compact, we have

Fc.A1v···uA.v

for some N. Hence

11.7 Definition Let f1 be additive, regular, nonnegative, and finite on C.Consider countable coverings of any set E c. RP by open elementary sets A,,:

N

pCA) :s; pC£) +, :s; pCA1 u ... u AN) + , :s; I pCA,) +, :s; p*CA) + 2,.1

the inf being taken over all countable coverings of E by open elementary sets.p*(E) is called the outer measure of E, corresponding to fl.

It is clear that p*CE) ;, 0 for all E and that

11.8 Theorem

Ca) For every A E e, p*CA) = pCA).00

Ch) If E = UE" Ihen!

In conjunction with (20), this proves Cal.Next, suppose E = UEn , and assume that f1*(E,,) < + 00 for all n.

Given e > 0, there are coverings {A"k}' k = 1,2,3, ... , of En by openelementary sets such that

~

I pCA,,):S; p*CE,) + 2-',..l:"'I

Then

p*CE):S; I f pCA,,):S; f p*CE,)+ e,n= 1 k= I "'" I

and (19) follows. In the excluded case, i.e., if p*CE,) = + ex: for some n,(19) is of course trivial.

(21)

11.9 Definition For any A c RP, B c. RP, we define

'Ne write An - A if

(22) SCA, B) = CA - B) u CB - A),

(23) dCA, B) = p*CSCA, B)).

p*CE1) :s; p*CE,)

00

p*CE) :s; I p*CE,).n=1

~

11*CE) ~ inf I pCA,),n'= I

Define

C17)

(18)

(19)

Note that (a) asserts that fl* is an extension of)1 from 0 to the family ofall subsets of R'. The property (19) is called slihaddilivily.

Proof Choose A E ,ff and, > O.

The regularity of jJ. shows that A is contained in an open elementaryset G such that pCG):s; pCA) + e. Since p*CA):s; pCG) and since' wasarbitrary, we have

(20)

The definition of jJ.* shows that there is a sequence {All} of opene!ementa!'"y ::cts whose unicn contains A, such t}l:it

lim dCA, A,) = O.

If there is a sequence {All} of elementary sets such that An - A, we saythat A is finitely jJ.-measlirable and write A E 9Jl f Cu).

If A is the union of a countable collection of finitely ,u-measurable sets,we say that A is f1.measurable and write A E 9Jl(fl).

SeA, B) is the so-called "symmetric difference" of A and B. We shall seethat d{A, B) is essentially a distance function.. . .

The following theorem will enable us to obram the deSired extensIon of fl.

U.10 Theorem 9Jl(u) is a (J-ring. and tl* is countablv additfL'e on 9J"l(tt).

00

I pCA,):S; p*CA) + ,."=1

Before we turn to the proof of this theorem, we develop some of theproperties of SCA. B) and dCA, B). We have

306 PRI:"<CIPLES OF MATHE~T1CAL ANALYSISTHE LEBESGljE THEORY 307

dCA, B) = 0,

By (23), (19), and (18), these properties of SeA, B) imply

The first formula of (26) is obtained from

(A, u A,) - (B, u B,) c (A, - B,) u (A, - B,).

~

A = UA,/I'" 1

~

II'(A) ,; )' II'(A,)./1::;"'1

We need one more property of dCA, B), namely,

I!"(A) - !"(B) I ,; dCA, B),

Then

is the required representation. By (19)

(30)

Letting n --jo co, we obtain, by (34) and Theciem 11.8(a),

I"(A) + I"(B) = I'*(A u B) + I"(A n B).

If A n B = 0, then I"(A n B) = O.It follows that 1" is additive on WIF(I').

Now let A E 9Jl(u). Then A can be represented as the union of acountable collection of disjoint sets of 9RFCu). For if A = UA~ withA; E W1,(!'), write A, = A{, and

A, = (A { u ... u A;) - (A; u ... u A;_,) (n = 2, 3, 4, ...).

d(A,O)';d(A, B)+ d(B, 0),

I"(A) ,; dCA, B) + !"(B).

I'(A,) + I'(B,) = !,(A, u B,) + I'(A, n B,).

that is,

Since j)*(B) is finite, it follows that

if at least one of !"(A), !"(B) is finite. For suppose 0,; I"(B) ,; I"(A).Then (28) shows that

!,'(A) - !"(B) ,; dCA, B).

Proor of Theorem 11.10 Suppose A E WI,(I'), BE 9)1,(!'). Choose {A,},{B,} such that A, E g. B, E g, A, ~ A, B, ~ B. Then (29) and (30) showthat

(31) A, u B, ~ A u B,

(32) A, n B, ~ A n B,

(33) A, - B, ~ A - B,

(34) I"(A,) ~ I"(A),

and !,'(A) < + co since d(A" A) ~O. By (31) and (33), 9)1,(!') is a ring.By (7),

(35)

(36)

(B - A) c (C - A) u (B - C).(A - B) c (A - C) u (C - B),

(26)

(24)

(25)S(A, B) = SIB, A), S(A, A) = 0,

S(A, B) c S(A, C) u SIC, B).

SIAl u A" B, u B')1SCAt n A" B, n B,)} c SIAl' B,) u S(A" B,).S(A,-A"B,-B,)

(24) is clear, and (25) follows from

(27)

(28)

(29)

Next, writing E C for the complement of E, we have

SeAl n A" B, n B,) = S(A~ u A" Bf u B2J

c S(A~, BD u S(Aj, B2J = SeAl' B,) u S(A" B,);

and the last formula of (26) is obtained if we note that

dCA, B) = d(B, A), dCA, A) = 0,

dCA, B),; dCA, C) + d(C, B),

dCA, u A" B, u B,))dCA, n A" B, n B,) ,; deAl> B,) + d(A" B,).dCA, - A" B, - B,)

The relations (27) and (28) show that dCA, B) satisfies the requirementsof Definition 2.15, except that dCA, B) = 0 does not imply A = B. For instance.jf J.l = m, A is countable, and B is empty, we have

dCA, B) = m'(A) = 0;

to see this, cover the nth point of A by an interval 1/1 such that

m(1,) < 2- ',.

But if we define two sets A and B to be equivalent, provided

we divide the sub<;ets of RP into eq'.:ivalence classes, a:id dCA, E) ;nahs the Sttof these equivalence classes into a metric space. 9J1

F(u) is then obtained as the

closure of cff. This interpretation is not essential for the proof, but it explainsthe underlying idea.

308 PRI:"iCIPLES OF MATHEMATICAL A~ALYSISTHE LEBESGUE THEORY 309

(37)

(38)

On the other hand, A::> A, U ... u A"; and by the additivity of~* on !JJlF(j<) we obtain

~*(A) ;>: ~*(A, u ... u A") ~ ~*(A,) + ... + ~*(A,).

Equations (36) and (37) imply

00

~*(A) = L jl*(A").,,"'I

Suppose ~*(A) is finite. Put B" ~ A, u ... u A,. Then (38) showsthat

00 00

d(A, BJ = ~*( U A,) = L ~*(A,)-->Oi"'l1+ 1 i"'l1+ 1

as n - co. Hence BII_A; and since BII E~£mf(jl), it is easily seen thatA E !JJlF(j<).

We have thus shown that A E !JJlF(j<) if A E !JJl(j<) and ~*(A) < + 00,It is now clear that fl* is countably additive on illlC,u). For if

A = UA",

where {A,} is a sequence of disjoint sets of 9Jl(j<), we have shown that (38)holds if ~*(A,) < + 00 for every n, and in the other case (38) is trivial.

Finally, we have to show that !JJl(j<) is a a-ring. If A" E 9Jl(j<), n = I,2,3, ' .. , it is clear that UA, E 9Jl(j<) (Theorem 2.12). Suppose A E !lJl(j<),BE !JJl(j<), and

11.11 Remarks

(0) IrA is open, then A E9Jl(j<). For every open set in R' is the unionof a countable collection of open intervals. To see this, it is sufficient toconstruct a countable base whose members are open inter.....als.

By taking complements, it follows that every closed set is in 9Jl(j<).(b) If A E 9Jl(~) and, > 0, there exist sets F and G such that

FcAcG,

F is closed, G is open, and

~(A - F) < e.

The first inequality holds since p* was defined by means of coveringsby open elementary sets. The second inequality then follows by takingcomplements.(c) We say that E is a Borel set if E can be obtained by a countablenumber of operations, starting from open sets, each operation consistingin taking unions, intersections, or complements. The collection :!J of allBorel sets in RP is a O"~ring; in fact, it is the smallest a-ring which containsall open sets. By Remark (a), E E 9Jl(~) if E E:2.

(d) If A E 9Jl(~), there exist Borel sets F and G such that F c: A c: G,and

00

A = UA"/1= 1

00

B ~ UB"11= 1

40)

and since P = nEll' PeE" for every n. so that me?) = O.

This follows from (b) if we take e = lin and let n ~ 00.Since A = F u (A - F), we see that every A E 9JI(~) is the union of a

Bore! set and a set of measure zero.The Borel sets are wmeasurable for every Jl. But the sets of measure

zero [that is. the sets E for which ~*(E) = 0] may be different for differentJ1"s.(e) For every Jl, the sets of measure zero form a a-ring.(/) In case of the Lebesgue measure, every countable set has measurezero. But there are uncountable (in fact, perfect) sets of measure zero.The Cantor set may be taken as an example: Using the notation of Sec.2.44, it is easily seen that

where A" B, E !JJlp(j<). Then the identity

00

A, n B ~ U (A, n B,)i'" 1

shows that A" n B E 9Jl(j<); and since

~*(A, n B) :> ~*(A,) < + 00,

A, n B E !JJlF(j<). Hence A, - B E !JJlF(~)' and A - B E 9Jl(j<) slllceA - B = U,;;. dA, - B).

\Ve now rep!a.ce Jl*(A) by Jl(A) if A ~ 9R(jt). Tr.as il, originaily only defined on G, is extended to a countably additive set function on the a~ring

~l(Jl). This extended set function is called a measure. The special case Jl = ~is called the Lebesgue measure on RP.

m(E,) = (t)" (n ~ 1,2,3, ...);

310 PRI:-<CIPLES OF ~ATHE:'fATICAL A~ALYSIS

THE LEBESGlJE THEORY 311

MEASURE SPACES

for the set of all elements x which have the property P.

11.12 Definition Suppose X is a set, not necessarily a subset of a euclideanspace, or indeed of any metric space. X is said to be a measure space if thereexists a Q"·ring 9)1 of subsets of X (which are called measurable sets) and a nonnegative countably additive set function J.l (which is called a measure), definedon 9Jt

If, in addition, X E 9J1, then X is said to be a measurable space.For instance, we can take X = RP, 9R the collection of all Lebesgue·

measurable subsets of RP, and !1 Lebesgue measure.

Or, let X be the set of all positive integers, 9J1 the collection of all subsetsof X, and /1(E) the number of elements of E.

Another example is provided by probability theory, where events may beconsidered as sets. and the probability of the occurrence of events is an additive(or countably additive) set function.

In the following sections we shall always deal with measurable spaces.It should be emphasized that the integration theory which we shall soon discusswould not become simpler in any respect if we sacrificed the generality we havenow attained and restricted ourselves to Lebesgue measure, say, on an intervalof the real line. In fact, the essential features of the theory are brought outwith much greater clarity in the more general situation, where it is seen thateverything depends only on the countable additivity of J.l on a a-ring.

It will be convenient to introduce the notation

(41) {xlPj

11.15 Theorem Each of the following four conditions implies the other three:

(43) (xlf(x) > aj is measurable for every real a.

(44) {x1f(x):?: aj is measurable for every real a.

(45) (xlf(x) < aJ is measurable for every real a.

(46) (xlf(x),; aj is measurable for every real a.

Proof The relations

{xlf(x):?: aj =,C\ (xlf{X) > a - ~l,

(xlf(x) < aj = X - (xlf(x):?: aj,

(xlf(x) ,; aj = IJI (xlf(x) < a +~),

(xlf(x) > aJ = X - (xlf(x),; aj

show successively that (43) implies (44), (44) implies (45), (45) implies(46), and (46) implies (43).

Hence any of these conditions may be used instead of (42) to define

measurability.

11.16 Theorem Iff is measurable, then ifi is measurable.

Proof

(xl If(x) I < aJ = (xlf(x) < aJ n (xlf(x) > - aj.

MEASURABLE FUNCTIONS

11.13 Definition Letfbe a function defined on the measurable space 'X, withvalues in the extended real number system. The functionfis said to be measur·able if the set

(42) (xlf(x) > oj

is measurable for every real a.

11.14 Example If X = R' and ill! = 9Jl (/1) as defined in Definition 11.9,every continuous/is measurable, since then (42) is an open set.

11.17 Theorem Let {j,,} be a sequence of measurable functions. For x E X, put

g(x) = supnx) (n = 1,2,3, ., .),

hex) = lim sUPf"(x).

Then g and h are measurable.

The same is of course true of the inf and lim info

Proof~

(xlg(x) > aJ = U (xlf"(x) > aj,"=1

hex) = inf gm(x),

wheregm(x) = sup/,,(x)(n:?: mI·

312 PRINCIPLES OF ~rHEIJATICAL ANALYSIS THE LEBESGUE THEORY 313

CoroJlaries

11.18 Theorem Let f and 9 be measurable real-valued functions defined on X,let F be real and continuous on R 2

, and put

(a) Iffand 9 are measurable, then max (f, g) and min (f, g) are measurable.If

it follows, in particular, that f+ andf- are measurable.(b) The limit ofa convergent sequence afmeasurable/unctions is measurable.

(xlf(x) > a}

is always a Borel set, without reference to any particular measure.

f is measurable and 9 is continuous, then h is not necessarily measurable.(For the details, we refer to McShane, page 241.)

The reader may have noticed that measure has not been mentioned inour discussion of measurable functions. In fact, the class of measurable functions on X depends only on the ,,-ring!JJl (using the notation ofDefinition 11.12).For instance, we may speak of Borel-measurable functions on RP, that is, offunction f for which

f- = - rnin (f, 0),f+ = max (f, 0),(47)

hex) = F(f(x), g(x)) (x EX).

Then h is measurable. SIMPLE FUNCTIONSIn particular, f + 9 and fg are measurable.

11.19 Definition Let s be a real-valued function defined on X. If the rangeof s is finite, we say that s is a simple function.

Let E c X, and put

that is, every simple function is a finite linear combination of characteristicfunctions. It is clear that s is measurable if and only if the sets £1' '." Ell aremeasurable.

It is of interest that every function can be approximated by simplefunctions:

Let

(x E E),(x ¢ E).

(i = I, ... , n).

,s = L cjKE(,

,,= I

E i = {x Is(x) = c i }

Then

(49)

KE is called the characteristic/unction of E.Suppose the range of s consists of the distinct numbers C1, .•• , cn •

(48)

where {In} is a sequence of open intervals:

I, = leu, v)la, < u< b" c, < v < d,}.

(xl a, <f(x) < b,} = (xlf(x) > a,} (") (xlf(x) < b,}

is measurable, it follows that the set

is measurable. Hence the same is true of

(xih(x) > a} = (xl,(f(x), g(x)) E G,}

G, = leu, v) IF(u, v) > a}.

Then Ga is an open subset of R 2, and we can write

Proof Let

Since

11.20 Theorem Let / be a real function on X. There exists a sequence {Sll} ofsimple functions such that s,lx) ~ f(x) as n ~ co, for every x E X. Iff is measurable, {s,,} may be chosen to be a sequence of measurable functions. If/"?. 0, {sIllmay be chosen to be a monotonically increasing sequence.

Proof If/~ 0, define

00

= U (x[(f(x),g(x)) E/,}.n=1

Summing up, we may say that all ordinary operations of analysis, including limit operations, when applied to measurable functions, lead to measurablefunctions; in other words, all functions that are ordinarily met with are measur~

able.· ..

That this is, however, only a rough statement is shown by the followingexample (based on Lebesgue measure, on the real line): If hex) = f(g(x»), where

I!i-I iiE" =\"" 2' 5, f(x) < Z;J ' F, = {xlf(x);" n}

314 PRINCIPLES OF MATHE....,{ATICAL ANALYSIS THE LEBESGUE THEORY 315

INTEGRATION

is measurable, and suppose E E Wl. We define

(I) Iff E 2'(~) on E, A E illl, and AcE, thenf E 2'CJl) on A.

(c) Iff and g E 2'CJl) on E, and if f(x) 5, g(x) for x E E, then

f fd~ =0.E

(e) If I'(E) = 0, andfis measurable, then

f fd~ 5, f gd~.E E

(d) If! E 2'CJl) on E, then ef E 2'(~) on E, for every finite constant e, and

(56)

11.23 Remarks The following properties are evident:

(a) If f is measurable and bounded on E, and if ~(E) < + co, thenf E 2'CJl) on E.

(b) If a 5,f(x) 5, b for x E E, and ~(E) < +co, then

a~(E) 5, f f d~ 5, b~(E).E

If at least one of the integrals (55) is finite, we define

I ef d~ = e I f d~.E E

If both integrals in (55) are finite, then (56) is finite, and we say that f isintegrable (or summable) on E in the Lebesgue sense, with respect to fJ.; we writef E 2'CJl) on E. If ~ = m, the usual notation is: f E 2' on E.

This terminology may be a little confusing: If (56) is +GO or - 00, thenthe integral of f over E is defined, although f is not integrable in the abovesense of the word; f is integrable on E only if its integral over E is finite.

We shall be mainly interested in integrable functions, although in somecases it is desirable to deal with the more general situation.

(x E x, e, > 0)•

sex) = I e,KE,(x)i=l

for n = 1,2,3, ... , i = 1,2, ... , n2n• Put

n2" j _ 1sn = i~l -z;- KEn; + nKF,,·

In the general case, letf = r - f-, and apply the preceding constructiontof+ and tof-.

It may be noted that the sequence {s.} given by (50) converg~

uniformly to f iff is bounded.

(54)

•(52) I,{s) = I e,~(E n E,).

i=l

Iff is measurable and nonnegative, \\Ie define

(53) Jf d~ = sup IE(s),E

where the sup is taken over all measurable simple functions s such that 0 :::; s s;fThe left member of (53) is called the Lebesgue integral of f, with respect

to the measure fJ., over the set E. It should be noted that the integral may havethe value + 'J:).

It is easily verified that

We shall define integration on a measurable space X, in which 9.Jl is the a-ringof measurable sets, and fJ. is the measure. The reader who wishes to visualizea more concrete situation may think of X as the real line, or an interval, and offJ. as the Lebesgue measure m.

(51)

11.21 Definition Suppose

(50)

for every nonnegative simple measurable function s.

11.22 Definition Letfbe measurable, and consider the two integrals

(55) fr d~,"E

wheref+ andf- are defined as in (47).

11.24 Theorem

(a) Suppose f is measurable and nonnegative on X. For A E 9Jl, define

(57) ",(A) = I f d~.A

316 PRINCIPLES Of MATHEMATICAL ANALYSIS

Then ¢ is countably additire on m.(b) The same conclusion holds ifJ E :£(;1) on X.

Proof It is clear that (b) follows from (0) if we write J = r -J- andapply (a) toJ+ and toJ-.

To prove (0), we have to show that

THE LEBESGUE THEORY· 317

It follows that we have, for every n,

(61) <P(A , u ... u A,) " <p(A , ) + ... + <p(AJ.

Since A ::> A, u ... u A" (61) implies

00

(62) <p(A) "I <p(A,),n'" I

(58)

(59)

(60)

00

<p(A) = I <p(A,)""'1

if A, E 9Jl (n = I, 2, 3, ...), A, r"1 Aj = 0 for i # j, and A = U~ A,.Iff is a characteristic function, then the countable additivity of 4> is

precisely the same as the countable additivity of IJ., since

f KE dl' = I'(A r"1 E).A

IfJis simple, thenJis of the form (51), and the conclusion againholds.

In the general case, we have, for every measurable simple function ssuch that 0 ,;; s ,;;J,

Therefore, by (53),

00

<p(A) s. I <p(A,).II'" 1

Now if <p(A,) = + ro for some n, (58) is trivial, since <p(A) " <p(A,).Suppose <p(A,) < + ro for every n.

Given e > 0, we can choose a measurable function s such thato ,;; s ,;;J, and such that

f s dl' "f J dl' - e,AI Al

Hence

<p(A, u A,) " f s dl' = f s dll +f s dl' " <p(A,) + <p(A,) - 2e,AI vA) AI A)

so that

and (58) follows from (59) and (62).

Corollary If A E WI, BE 9Jl, Be A, and I'(A - B) = 0, then

f J dl' =f J dl'.A B

Since A ~ B u (A - B), this follows from Remark 11.23(e).

11.25 Remarks The preceding corollary shows that sets of ;.1easure zero arenegligible in integration.

Let us write1- 9 on E if the set

{xIJ(x) # g(x)} r"1 E

has measure zero.

Then J - J; J - g implies g - J; and J - g, g - h implies J - h. That is,the relation", is an equivalence relation.

IfJ - g on E, we clearly have

f J dl' ~ f g dl',A A

provided the integrals exist, for every measurable subset A of E.If a property P holds for every x E E - A, and if I'(A) ~ 0, it is customary

to say that P holds for almost ail x E E, or that P holds almost everywhere onE. (This concept of "atmost everywhere" depends of course on the particularmeasure under consideration. In the literature, unless something is said to thecontrary, it usually refers to Lebesgue measure.)

IfJ E :£(;1) on E, it is clear thatJ(x) must be finite almost everywhere on E.In most cases we therefore do not lose any generality if we assume the givenfunctions to be finite-valued from the outset.

11.26 Theorem IJJ E :£(p) on E, then iJI E :£(1') on E, and

(63)



Proof Write E = A u B, where f(x) 20 on A and lex) < 0 on B.By Theorem 11.24,

f IfI dl' = f If! dl' +f If! dl' = f r dll +fr dll < + 00,E A. B A. B

so that IfI E :t'CI'). Sincef';; If! and -f';; Iff, we see that

f f dll ,;; f IfI dll,E E

(69)

Choose c such that 0 < c < 1, and let s be a simple measurablefunction such that 0 ,;; s,;;1 Put

E, = (xlf,(x) 2 cs(x)} (n = 1, 2, 3, ...).

By (64), E, c E, c £, c .. ' ; and by (65),

~

E= UE,.11=1

and (63) foHows. For every n,

Since the integrability of f implies that of If I ' the Lebesgue integral isoften called an absolutely convergent integral. It is of course possible to definenonabsolutely convergent integrals. and in the treatment of some problems it isessential to do so. But these integrals lack some of the most useful propertiesof the Lebesgue integral and playa somewhat less important role in analysis.

We let n -+ 00 in (70). Since the integral is a countably additive set function(Theorem 11.24), (69) shows that we may apply Theorem 11.3 to the lastintegral in (70), and we obtain

11.27 Theorem Suppose f is measurable on E, Ifi ,;; g, and g E :t'C!') on E.Thenf E :t'C!') on E.

Proof Wehavef+ ,;;gandf- ,;;g.

(71) oX;::: cJ s dp.E

Letting c -+ 1, we see that

11.28 Lebesgue's monotone cODYergence theorem Suppose £ E9Jt Let UII} bea sequence of measurable functions such thot

(64) o';;f,(x) ';;f,(x) ,;; ...

Let f be defined by

(x E E).

and (53) implies

U~ a2ff~E

The theorem follows from (67), (68), aod (72).

.:IS n - 00. Then

(65) f.(x) - lex) (x E E) 11.29 Theorem Suppose f ~ f, + f" where J, E :t'C!') on E (i ~ 1, 2). Thenf E :t'(Il) on E, and

Proof By (64) it is clear that, as n - 00,

(68) a ,;; f f dll·E

r s, dl' = f s; dll + f s; dll,~E E E

and (73) follows if we let n ~ 00 and appeal to Theorem 11.28.

Proof First, supposef,20,J, 20. Iff, andf, are simple, (73) followstrivially from (52) and (54). Otherwise, choose monotonically increasingsequences {s~}, {s~} of nonnegative measurable simple functions whichconverge to 11,/2' Theorem 11.20 shows that this is possible. Puts., = S; + s;. Then

(73)

(67) f f, dll ~aE

fur some et; and since Jill:;:; S.r. we have

(66)

320 PRINCIPLES OF MATHEMATICAL ANALYSIS THE LEBESGUE THEORY 321

Next, suppose f, :2: OJ, ;s; o. Put

A ~ (xlf{x) :2: OJ, B = {xlf(x) < OJ.

Strict inequality may hold in (77). An example is given in Exercise 5.

Proof For n = 1,2,3, ... and x E E, put

Then.f.fl' and -f2 are nonnegative on A. Hence 9.(X) = infj,(x) (i:2: n).

or

Similarly, - f,h, and - f2 are nonnegative on B, so that

(x EE)

r9. d~ -f fd~,• E E

j,(x) - f(x)

Then 9/1 is measurable on E, and

so that (77) follows from (79) and (81).

(81 )

(78) O;s; 9,(X);S; 9,(x) ;S; "',

(79) 9.(X) ;S;j,(x),

(80) 9.(X) - f(x) (n - (0).

By (78), (80), and Theorem 11.28,

11.32 Lebesgue's dominated convergence theorem Suppose E E 9Jl. Let {fll} bea sequence oj measurable functions such that

(82)(i = 1, 2, 3, 4),

(75) fBIt d~ = f/ d~ - f/, d~,

and (73) follows if we add (74) and (75).

In the general case, E can be decomposed into four sets E; on eachof whichfl(x) andf2(x) are of constant sign. The two cases we have provedso far imply

and (73) follows by adding these four equations. as n -+ co. If there exists a Junction 9 E !l'(p.) on E, such that

We are now in a position to reformulate Theorem 11.28 for series. (83) If.(x)! ;S; 9(X) (n = 1,2,3, ... , X E E),

11.30 Theorem Suppose E E 9)1. If{!,,} is a sequence of nonnegative measurablefunctions and

11.31 Fatou's theorem Suppose E E 9.R. If U:'} is a sequence of nonnegariremeasurable functions and

rf df.l ;S; lim inf f f. d~.¥E /1 ..... 00 £

lim r f. d~ = f f d~./1-4":0 .. E E

r f d~ ;S; lim inf r f. d~.• E /I ..... on .. £

or

rU + g) d~ ;S; lim inff U. + g) d~,~E ,,-471 £

then

(84)

Because of (83), U.j is said to be dominated by g, and we talk aboutdominated convergence. By Remark 11.25, the conclusion is the same if (82)holds almost everywhere on E.

Proof First, (83) and Theorem 11.27 imply that f. E fI!(ji) and f E fl!liL)on E.

SinceJ:. + g;:: 0, Fatou's theorem shows that

(85)

(x E E).

(x E E),oc

f(x) = L f.(x)11= 1

f(x) = lim inf.f.(x)

Proof The partial sums of (76) form a monotonically increasing sequence.

(76)

then

then

(77)

322 PRINCIPLES OF MATHEMATICAL ANALYSIS THE LEBESGUE THEORY 323

Since g - I". 2'.: 0, we see similarly that

J(g - I) dl' 5, lim infJ(g - /,) dl',E "'-:0 E

so that

- JI dl' 5, lim inf [-J /, dl']'E "'-:0 E

which is the same as

(86) J I dl' ;" lim sup J I dl'.E "'-00 E

The existence of the limit in (84) and the equality asserted by (84)now follow from (85) and (86).

Corollary If 1'(£) < + co, {/,} is uniformly bounded on E, andl,(x) - I(x) on E,then (84) holds.

A uniformly bounded convergent sequence is often said to be boundedlyconvergent.

for the Lebesgue integral of f over [a, bJ. To distinguish Riemann integralsfrom Lebesgue integrals, we shall now denote the former by

,!it J Idx.

•

11.33 Theorem

(a) III E!it on [a, b], then I E If on [a, b], and

, ,(87) J I dx = !it J I dx.

• •(b) Suppose I is bounded on [a, b]. Then/E!it on [a, b] if and only iflis

continuous almost everywhere on [a, b].

Proof Suppose I is bounded. By Definition 6.1 and Theorem 6.4 thereis a sequence {Pit} of partitions of [a, bJ, such that PHI is a refinementof Pk, such that the distance between adjacent points of Pit is less than11k, and such that

COMPARISON WITH THE RIEMANN INTEGRAL(88) ~~n;,L(p,J)=211Idx, lim U(P, J) = !itJI dx.

'-ro

for all x E [a, b], since Phi refines P,. By (90), there exist

U,(a) = L,(a) =/(a);

(In this proof, all integrals are taken over [a, b].)If Pk = {xo , Xt, ... , A".}, with Xo = a, x". = b, define

put Uk(x) = lvt j and Llx) = mi for x j - 1 < x $ Xi' 1 $ i $ n, using thenotation introduced in Definition 6.1. Then

U(x) = lim U,(x).'-ro

U(P,,f) = Ju, dx,

L(x) = lim L,(x),'_ro

L(P,,f) = JL, dx,

L,(x) 5, L,(x) 5,'" 5,/(x) 5, .,. ,; U,(x) 5, U,(x)

and

(89)

(90)

(91 )J Idmx

Our next theorem will show that every function which is Riemann-integrableon an interval is also Lebesgue-integrable, and that Riemann-integrable functions are subject to rather stringent continuity conditions. Quite apart from thefact that the Lebesg'J.e theory therefore enables us to integrate a much largerclass of functions, its greatest advantage lies perhaps in the ease with whichmany limit operations can be handled; from this point of view, Lebesgue'sconvergence theorems may well be regarded as the core of the Lebesgue theory.

One of the difficulties which is encountered in the Riemann theory isthat limits of Riemann-integrable functions (or even continuous functions)may fail to be Riemann-integrable. This difficulty is now almost eliminated,since limits of measurable functions are always measurable.

Let the measure space X be the interval [a, b] of the real line, with J.L = m(the Lebesgue measure), and 9Jl the family of Lebesgue-measurable subsetsof [a, b]. Instead of

Observe that Land U are bounded measurable functions on [a, bJ,it is customary to use the familiar notation,

JIdx• (92)

that

L(x) 5,/(x) 5, U(x) (a,; x 5, b),

324 PRINCIPLES OF MATHEMATICAL ANALYSISTHE LEBESGUE THEORY 325

and thatINTEGRATION OF COMPLEX FUNCTIONS

(95) L(x) = f(x) = U(x)

almost everywhere on [a, b], so that f is measurable, and (87) followsfrom (93) and (95).

Furthermore, if x belongs to no Pin it is quite easy to see that U(x) =

L(x) ifand onlyiffis continuous atx. Since the union of the sets Pk

is countable, its measure is 0, and we conclude thatfis continuous almost everywhere on [a, b] if and only if L(x) = U(x) almost everywhere, hence(as we saw above) if and only iff o:J/


IL dx = I U dx;

since L :5 U, (94) happens if and only if L(x) = U(x) for almost allx E [a, b] (Exercise I).

In that case, (92) implies thatIE IfI d,u < + 00,

J f dp = Ju dp + j Jv dpE E E

(97)

and we define

Suppose / is a complex-valued function defined on a measure space X, and/ = u + iv, where u and v are real. We say thatf is measurable if and only ifboth u and v are measurable.

It is easy to verify that sums and products of complex measurable functionsare again measurable. Since

IfI = (u' + ,,-')' /2,

Theorem 11.18 shows that IfI is measurable for every complex measurable fSuppose 11 is a measure on X, E is a measurable subset of X, and f is a

complex function on X. We say thatf E .5!'(p) on Eprovided thatfis measurableand

if (97) holds. Since IuI :5 IfI, IvI :5 IfI, and IfI :5 Illi + Iv I, it is clear that(97) holds if and only if II E .5!'(p) and v E.5!' (P) on E.

Theorems 11.23(a), (d), (e), U), II.24(b), 11.26, 11.27, 11.29, and 11.32can now be extended to Lebesgue integrals of complex functions. The proofsare quite straightforward. That of Theorem 11.26 is the only one that offersanything of interest:

Iff E .5!'(p) on E, there is a complex number v, Ivi = 1, such that

IUdx=:JI JfdX,IL dx =:JI Ifdx,

by (88), (90), and the monotone convergence theorem.So far, nothing has been assumed about/except thatfis a bounded

real function on [a, b].To complete the proof, note thatf E:JI if and only if its upper and

low~r Riemann integrals are equal, hence if and only if

(93)

(94)

The familiar connection between integration and differentiation is to alarge degree carried over into the Lebesgue theory. If/E::t on [a, b], and

then F(x) = f(x) almost everywhere on [a, b].Conversely, if F is differentiable at every point of [a, bJ ("almost every

where" is not good enough here!) and if F' E ff' on [a, bJ, then

(96) F(x) = rfdt, (a :5 x :5 b),

v r f dp ? O.'E

Put 9 = c/ = u + iv, u and v real. Then

Ir f dP! ~ v J' f dp = r9 dp = r u dp :5 r Ifi d,u.~E E °E "E "E

The third of the above equalities holds since the preceding ones show thatSg dp is real.

F(,) - F(a) = rF(I)"

(a"; x :5 b). FUNCTIONS OF CLASS .5!"

For the proofs of these two theorems, we refer the reader to any of theworks on integration cited in the Bibliography.

As an application of the Lebesgue theory, we shall now extend the Parsevaltheorem (which we proved only for Riemann-integrable functions in Chap. 8)and prove the Riesz-Fischer theorem for orthonormal sets of functions.

326 PRINCIPLES OF MATHEMATICAL ANALYSlSTHE. -UBesGUE- ~THWRY--327

More explicitly, this means that for any f E 2 2 on [a, b], and any E > 0,there is a function g, continuous on [0, b], such that

(

b 11/'Ilf-gil = J,lf-gl'dX/ <e.

Proof We shall say that f is approximated in g2 by a sequence {gIl} ifIlf - g,11 ~O as n ~ 00.

Let A be a closed subset of [0, b], and K.4 its characteristic function.

11.34 Definition Let X be a measurable space. We say that a complexfunctionf E .5!"(jl) on X iffis measurable and if

Jill' dJ1. < + 00.x

If J1. is Lebesgue measure, we say f E .5!". For fE .5!"(jl) (we shall omit thephrase "on X" from now on) we define

I )1/'11I11 = dx IfI' dJ1.

and call I1II1 the .5!"(jl) norm off

Put

and

t(x) = inf Ix - yl (y E A)

(98)

(n ~ I, 2, 3, .. Y11.35 Theorem Suppose f E .5!"(jl) and g E .5!"(jl). Then fg E .5!'(j1), and

J Ifgl dJ1. ,; I1I11 Ilgil.x

This is the Schwarz inequality, which we have already encountered forseries and for Riemann integrals. It follows from the inequality

0,; J (III +Algi)' dJ1. = lilli' + 2). J Ifgl dJ1. + )·'llgll',x x

which holds for every real 1.

11.36 Theorem Iff E .5!"(jl) and g E .5!"(J1.), then f + g E .5!"(j1), and

II!+ gil ,; II!II + Ilgll.Proof The Schwarz inequality shows that

1g,(x) = I + nt(x)

Then g, is continuous on [a, b], g,(x) = 1 on A, and g,(x) ~O on B,where B ~ [a, b]- A. Hence

by Theorem 11.32. Thus characteristic functions of closed sets can beapproximated in 2 2 by continuous functions.

By (39) the same is true for the characteristic function of anymeasurable set, and hence also for simple measurable functions.

Iff~ 0 and f E 2 2, let {s,,} be a monotonically increasing sequence

of simple nonnegative measurable functions such that s,,(x) - f(x).

Since If - s,,1 2 ~/l, Theorem 11.32 shows that ilf - s,,:! ...... O.The general case follows.

11.39 Definition We say that a sequence of complex functions {¢,,} is anorthonormal set of functions on a measurable space X if

(n '" m),(n = m).

IfIE .5!"(I') and if

(n ~ 1,2,3,. ..),

we write

In particular, we must have cP" E 2 2(p.).

III +gil' = Jill' + J19+ JJg + J Igl'

,; Illil' + 211/1111gl1 + Ilgll'= <11/11 + Ilgll)'.

11.37 Remark If we define the distance between two functions I and 9 in.5!"(j1) to be IIf - gil, we see that the conditions of blefinition 2.15 are satisfied,except for the fact that Ilf - gil = 0 does not imply that f(x) = g(x) for all x,but only for almost all x. Thus, if we identify functions which differ only on aset of measure zero, 2 2(p.) is a metric space.

We: now consider :c 2 on an interval of the rt:ai line, with respect [Q

Lebesgue measure.

11.38 Theorem The continuous functions form a dense subset of 2 2 on [a, b].as in Definition 8.lO.

328 PRINCIPLES OF MATHEMATICAL A~ALYSIS

The definition of a trigonometric Fourier series is extended in the sameway to !i'2 (or even to !i') on [- n, n]. Theorems 8.11 and 8.12 (the Besselinequality) hold for any f E !i'2(p). The proofs are the same, word for word.

We can now prove the Parseval theorem.


11.41 Definition Let f and f. E !i"(p) (n = I, 2,3, ...). We say that (f,)converges tofin !i'2(p) if 11f, - fll ~O. We say that {f,} is a Cauchy sequencein 2'2(p) if for every e> 0 there is an integer N such that n ~ N, m ~ N implies

Ilf. - f.11 ,;; e.

where f e 2'2 on [-Jr, It]. Let Sn be the nth partial sum of (99). Then

lim Ii! - 5,11 ~ 0,

11.40 Thenrem Suppose

(99)

(100)

~

fix) - L c,e''',-~

11.42 Theorem If (f,) is a Cauchy sequence in !i'2(p), then there exists afunction f E !i"(P) such that {f.} converges to f in !i"(p).

This says, in other words, that 2'2(p) is a complete metric space.

Proof Since {In} is a Cauchy sequence, we r.an find a sequence {n k},

k = 1,2, 3'; ... , such that

(k = 1,2,3, ...).(101) ~ I J'LIc,1 2 =- If['dx.

-:0 21t -It

Proof Let e > 0 be given. By Theorem 11.38,function 9 such that

,Ilf-gil <-.

2

there IS a continuous

I11f., -.f... ,II < 2'

Choose a function g E !i"(~). By the Schwarz inequality,

J' I (f, r ) I d < ilgilx 9 n",-JnJ<+l P-y'

Hence

"(103) Ig(x) I L IJ",(x) - f,,, ,(xli < + 00k~ 1

By Theorem 1IJO, we may interchange the summation and integration in(102). It follows that

.Moreover, it is easy to see that we can arrange it so that g(Jr) = g( - rr).Then 9 can be extended to a periodic continuous function. By Theorem8.16, there is a trigonometric polynomial T, of degree N, say, such that

, eIlg - Til <2'

Hence, by Theorem 8.11 (extended to !i'2), n;o, N implies

!Is, - fll ,;; 'liT - fil < "

and (100) follows. Equation (101) is deduced from (100) as in the proof ofTheorem 8,16.

Cnrollary Iff E!i" on [- n, n], and if

(/(X)e-'''dx=O (n =0, ii, i2, .. .),

then Ilfll = o.

Thus if two functions in 2'2 have the same Fourier series, they differ atmost on a set of measure zero.

(104)

almost everywhere on X. Therefore

x

L If".,(x) - f,,(x) [ < + 00k'" 1

almost everywhere on X. For if the series in (104) were divergent on aset E of positive measure, we could take g(x) to be nonzero on a subset ofE of positive measure, thus obtaining a contradiction to (103).

Since the kth partial sum of the series

"L (f,,, .ex) - f,,(X)),.1:=1

which converges almost everywhere on X, is

k ..(x) - J",(x),

330 PR1:-;CIPLES OF MATHEMATICAL· ANALYSIS


....'e see that the equationso that

Letting n -+ 00, we see that

and the proof is complete.

fix) = lim J..,(x),-~

defines f(x) for almost all x E X, and it does not matter how we definefix) at the remaining points of X.

We shall now show that this function f has the desired properties.Let e > a be given, and choose N as indicated in Definition llAI. Ifnk > N, Fatou's theorem shows that

(k = 1, 2, 3, ...),

Thus f - J.., E 2'<I'tand since f ~ if - J...J +J.." we see that f E 2'<1').Also, since e is arbitrary,

11.44 Definition An orthonormal set {¢/l} is said to be complete if, forf E 2'(1'), the equations -

lim Ilf - J...II = o.,-~

Finally, the inequality

(105) Ilf - J..II s; II! - J..,II + 111., - J..il

shows that U.l converges to f in 2'<1'); for if we take nand n, largeenough, each of the two terms on the right of (105) can be made arbitrarily small.

imply that II!II = O.In the Corollary to Theorem 11.40 we deduced the completeness of the

trigonometric system from the Parseval equation (101). Conversely, the Parsevalequation holds for every complete orthonormal set:

11.45 Theorem Let {<p.l be a complete orthonormal set. Iff E 2'<1') and 'f

then

Proof By the Bessel inequality, L Iclll 1 converges. Putting

S/l = C1¢1 + ... + C/lrP",

the Riesz~Fischer theorem shows that there is a function g E .f£I2(p.) suchthat

(106)

(107)

(108)

Proof For n > m,

Ils.-smll'= Icm";I'+"·+ Ic.I',

so that {S/l} is a Cauchy sequence in .f£I2(p.). By Theorem 11.42, there isa functionf E 2'(1') such that

11.43 The Riesz-Fischer theorem Let {<p,l be orthonormal on X. SupposeL! C/l i2 converges, and put S/l = C1¢1 + ... + cA)/l' Then there exists a functionf E 2'<1') such that {s.} conL'erges to f in 2'<1'), and such that

lim Ilf - s.11 = o.

Now, for n > k,

and such that Ilg - s,ll ~O. Hence Ils.11 ~ Ilgll. Since

Ils,lI' = [c,I' + ". + Ic.I',we have

(109)

- 332 PRINCIPLES OF MATHEMATICAL ANALYSIS THE-LEBESGUE THEORY 333

6. Let

this is· false. For instance, if

(Ix[ ';;n),

(Ixl >n).

(n ~ 1,2,3, ...).

Then fn(x) -+0 uniformly on RI, but

rJdx~2

1f(x) ~ 1--'-I ',IX

(We write f~21 in place of fRI') Thus uniform convergence does not imply dominated convergence in the sense of Theorem 11.32. However, on sets of finitemeasure, uniformly convergent seq.uences of bounded functions do satisfy Theo

rem 11.32.7. Find a necessary and sufficient condition thatfE .aNa.) on [a, b]. Hint: Consider

Example 11.6(b) and Theorem 11.33.8. If fE:Jt on [a. b] and if F(x) = 1:1(1) dr, prove that F'(.<) ~ f(x) almost every·

where on [a, b].9. Prove that the function F given by (96) is continuous on [a, b].

10. If ~(X) < +x and f E .!l"(~) on X, prove that f E .!l'(~) on X. If

~(X)~ ,x,

theofE2'l on R I, butf$.Pon R i

•

11. Iff, 9 E 2'(p.) 00 X, define the distance between [and 9 by

Now (106), (108), and the completeness of{,p,} show that 111- g[1 = 0,so that (109) implies (107).

~

f - 1:: c,,p,,/1=1

t.If/:20andfEfd/-L=O,provethat/(x)=OalmosteverywhereonE. Hint: LetE.be the subset of E on whichf(x) > l/n. Write A = UE•. Then ~(A) = 0 ifand onlyif f1-(E.) = 0 for every n.

2. If fA / df1- = 0 for every measurable subset A of a measurable set E, then/ex) = 0almost everywhere on E.

3. If {f,.} is a sequence of measurable functions, prove that the set of points x atwhich {fii(X)} converges i$ mea,>urable.

4. If / E If'(/-L) on E and 9 is bounded and measurable on E, then fg E 2'(fJ-) on E.

5. Put

Combining Theorems 11.43 and 11,45, we arrive at the very interestingconclusion that every complete orthonormal set induces a 1-1 correspondencebetween "the functions IE .!l"(P) (identifying those which are equal almosteverywhere) on the one hand and the sequences Ie,} for which L Ic, I' converges,on the other. The representation

together with the Parseval equation, shoWS that 2'2(jJ.) may be regarded as aninfinite-dimensional euclidean space (the so-called "Hilbert space"), in whichthe point/has coordinates C/I, and the functions 4>/1 are the coordinate vectors.

EXERCISES

Show that

but

g(x) ~ (~f,,(x) = g(x)

flk+i(X)=g(l-x) .

lim inf f.(x) ~ 0:

(O:S;:x:S;:!),

(!<x,;;I),(0 ,;; x ,;; I),

(O';;x,;; 1).

(0';; x ';;1),

Prove that .5!'(p.) is a complete metric space.

12. Suppose(a) 1 f(x,Y)[ ,;; 1 if 0 ,;;X,;; 1, O';;y';; I,(b) for fixed x,/(x, y) is a continu6Us function of y,

(c) for fixed y,/(x, y) is a continuolTh function of x.

Put

g(x) ~J: f(x;£) dy

Is .Q continuous?

13. Consider the functions

[Compare with (77).] f.(x) = sin nx (n£I.2,3, ...• -1r:s;:X~7T)

ask_co,


as points of !£2. Prove that the set of these points is closed and bounded, butnot compact.

14. Prove that a complex function f is measurable if and only if f-l( V) is measurablefor every open set V in the plane.

15. Let:Jt be the ring of all elementary subsets of (0, 1]. If 0 < a ::;;: b :::;; 1, define

<}I([a, bJ) ~ <}I([a, b» ~ <}I ( (a, bJ) ~ <}I«a, b» ~ b - a,

but define

<}I«O, b» ~ <}I((O, b]) ~ I .;.- b

if 0 < b ::;;: 1. Show that this gives an additive set function ,p on 9f, which is notregular and which cannot be extended to a countably additive set function on aa-ring.

16. Suppose {nt} is an increasing sequence of positive integers and E is the set of allx E (-7T, 7T) at which {sin nkx} converges. Prove that m(E) = O. Hint: For everyA eE,

t sin n.x dx.....,:..O,

and

2 fA (sin nlx)2 dx = t (1 - cos 2nkx) dx -meA)

17. Suppose EC(-7T,7i),m(E»O.S>O. Use the Bessel inequality to prove thatthere are at most finitely many integers n such that sin nx ~ :5 for all x E E.

18. Suppose f E i.t 2(JL), g E !£2(p.). Prove that

~f and only if there is a constant c such that g(x) = cf(x) almost everywhere.(Compare Theorem 11.35.)

BIBLIOGRAPHY

ARTIN, E.: "The Gamma Function," Holt, Rinehart and Winston, Inc., New York,1964.

BOAS, R. P.: "A Primer of Real Functions," Carus Mathematical Monograph No. 13,John Wiley & Sons, Inc., New York, 1960.

BUCK, R. C. (ed.): "Studies in Modern Analysis," Prentice·Hall, Inc., EnglewoodCliffs, N.J., 1962.

---: "Advanced Calculus," 2d ed., McGraw-Hill B"ook Company, New York,1965.

BURKILL, J. c.: ''The Lebesgue Integral," Cambridge University Press, New York, 1951.

D1EUDONNE, J.: "Foundations of Modem Analysis," Academic Press, Inc" New York,1960.

FLEMING, W. H.: "Functions of Several Variables," Addison·Wesley Publishing Com.pany, Inc., Reading, Mass., 1965.

GRAVE·S, L. M.: "Toe Theory c.·f Functions 0; Real Vatiables," 2d ed., ,\.fcGraw-H:;.i

Book Company, New York, 1956.

HALMOS, P. R,: "Measure Theory," D. Van Nostrand Company, Inc., Princeton, N.J.,1950.

LIST OF SPECIAL SYMBOLS

The symbols listed below are followed by a brief statement of their meaning and bythe number of the page on which they are defined.

336 PR1:{CIPU~S OF MATHEMATICAL A:-i.-\LYSTS

---: "Finite-dimensional Vector Spaces," 2d ed., D. Van Nostrand Company, Jnc.,Princeton, N.J., 1958.

HARDY, G. H.: "Pure Mathematics," 9th ed., Cambridge University Press, New York,1947.

--- and ROGOSI:-.iSKI, W.: "Fourier Series," 2d ed., Cambridge University Press,New York, 1950.

HERSTEIZ'I, I. N.: "Topics in Algebra," Blaisdell Publishing Company, New York, 1964.HEWITT, E., and STROMBERG, K.: "Real and Abstract Analysis," Springer Publishing

Co., Inc., New York, 1965.KELLOGG, o. D.: "Foundations of Potential Theory," Frederick Ungar Publishing Co.,

NEW York, 1940.KNOPP, K.: "Theory and Application of Infinite Series," Blackie & Son, Ltd.,

Glasgow, 1928.LANDAU, E. G. H.: "Foundations of Analysis," Chelsea Publishing Company, New York,

1951.MCSHANE, E. J.: "Integration," Princeton University Press, Princeton, N.J., 1944.NIVEN, 1. M.: "Irrational Numbers," Carus Mathematical Monograph No. 11, John

Wiley & Sons, Inc., New York, 1956.ROYDEN, H. L.: "Real Analysis," The Macmillan Company, New York, 1963.RUDIN, W.: "Real and Complex Analysis," 2d ed., McGraw·Hill Book Company,

New York, 1974.SIMMONS, G. F.: "Topology and Modern Analysis," McGraw-Hill Book Company,

New York, 1963.SINGER, I. M., and THORPE, J. A.: "Lecture Notes on Elementary Topology and Geom

etry," Scott, Foresman and Company, Glenview, IlL, 1967.SMITH, K. T.: "Pri:r.er of Modern Analysis," Bogden and Quigley, Tarrytown-cn

Hudson, N.Y., 1971.SPIVAK, M.; "Calculus on Manifolds," W. A. Benjamin, Inc., New York, 1965.THU~STON, H. A.: "The Number System," Blackie & Son, Ltd., London·Glasgow, 1956.

E belongs to .. . . . . . . . . . . 3~ does not belong to . . . . . 3c ::: inclusion signs 3Q rational field 3<, ~, >,;;::: inequality signs. 3sup least upper bound .. '" ., .... , 4inf greatest lower bound 4R real field 8+cc, -co,:cinfinities .11,27i complex conjugate 14Re (z) real part.......... .. .. 141m (z) imaginary part '" .14Iz I absolute value 14z:: summation sign 15, 59Rk t:uclidean k-space 16o null vector 16x .-y inner product 16Ixl norm of vector x 16

{x.} sel1uence... . .26U, u union 27

n, rl intersection 27(a, b) segment. 31[0, b] interval 31Ec complement of E 32E' limit points of E 35E closure of E 35lim limit. .47- converges to " , . .47, 98lim sup upper limit. . . . . . . .56lim inf lower limit. . . . . . . .56go f composition _.. 86f(x+ ) right-hand limit 94f(x-) left-hand limit 94/', f' (x) derivatives 103,112U(P,f), U(P,f, x), L(P,f), L(P,f, x)

Riemann sums ..... 121, 122

Jacobian .. , 234

338 PRI~CIPLES OF MATHEMATICAL ANALYSIS

;]i, ;]i(e<:) classes of Riemann (Stieltjes)integrable functions 121,122

'tf(X) space of continuousfunctions , , 150

norm 140, 150,326exp exponential function 179D... Dirichlet kernel 189rex) gamma function 192{e1,.,., en} standard basis 205L(X), L(X, Y) spaces of linear

transformations , , , 207[A] matrix " 210D1f partial derivative , 215v f gradient., , 217!'&'. t".(j" classes of differentiable

functions 219,235det [A] determinant..... . 232Jf(x) Jacobian , 2348Cy" , Yn)

c(x" , x n)

P k-cell , 245Q' k-simplex 247dXI basic k-form .2571\ multiplication symbol .254d differentiation operator 260WT transform of W ••••••••••.•• 262a boundary operator 269v x F curl , 281v .F divergence , 281g ring of elementary sets 303m Lebesgue measure ., 303, 308i-' measure , 303, 3089J1 F , m families of measurable sets 305{x!P} set with property P" 310f+,/- positive (negative) part

off 312K e characteristic function 3132',2'(1'),2",2"(1') classes of

Lebesgue-integrablefunctions ,.315, 326

Abel. N. H .. 75. 174Absolute convergence. 71. of integral. 138

Absolule value. 14Addition (see Sum)Addition formula. 178Additivity, 301Affine chain. 268Affine mapping, 266Affine simple'l. 266Algebra. 161

self-adjoint. 165uniformly closed. 161

Algebraic numbers. 43Almost everywhere. 3 17Alternating series, 7 [Analytic function. InAnticommutative law. 256Arc, 136 'Area element. 283Arithmetic means. 80. 199Artin. E., 192. 195Associative law, 5. 28. 259Axioms. 5

Baire's theorem. 46. 82Ball. 31Bl!.se. 45Basic fonn. 2.57Basis, 205Bellman. R.. 198Bessel inequality. 188. 328Beta function. 193Binomial series. 20 IBohr-MolieruD theorem. 193Borel-measu~ble function. 313

Burel sec 309Boundary. 269Bounded convelgence. 322Bounded function. 89Bounded sequence. 48Bounded· set. 3 '2Brouwer's theorem. 203Buck. R. C. 195

Cantor, G., 21. 30. 186Cantor set. 41. 81. 138. 168.309'Cardinal number, 25Cauchy criterion. 54. 59. 147Cauchy sequence. 21. 52. 82. 329Cauchy's condensation test. 61Cell. 31;g".equivalence. 280Chain. 268

affine. 168differentiable. 270

Chain rule. 105.214Change of variables. 132. 252. 262Characteristic function. 313Circle of convergence. 69Closed curve. 136Closed form. 275Closed set. 32Closure. ~5

unifonn. 151. 161Collection. 27Column matrix, 217Column vector. 210Common refinement. 123Commutative law. ~ 28Compact metric space. 36Compact set. 36

INDEX

Comparison test. 60Complement. 32Complete metric space. 54. 82.

151, 329Complete onhonormaJ set, 331Completion. 82Complex field. 12. 184Comple.'I number. 12Complex plane. 17COlllponent of a function. 81. 215Composition. 86. 105. 127.207Condensation point. 45Conjugate. 14Connected set. 42Constant function. 85Continuity. 85

uniform. 90Continuous functions. space of.,,,Continuous mapping. 85Conlinuollsly differentiable curve.

136Continuously differentiable map

ping. 219Contraction. 220Convergence. 47

absolute. 71bounded. 322do:nir,atd. n 1of integral. 138pointwise. 144radius of. 69. 79of sequences. 47of series. 59uniform. 147

Convex function. 10 ICOnVe1l. set. 31

340 INDEX

Coordinate function. 88Coordinates. 16.205Countable additivity. 301Counlable base. 45'Countable set. 25Cover. 36Cunningham. F .. 167Curl. 28lCurve. 136

closed. 136Continuously differentiable. 136rectifiable. 136space-filling. 168

Cut. 17

Davis. P.l.. 192Decimals. 11Dedekind. R .. 21Dense subset. 9. 32Dependent set. 205Derivative. 104

directional. 218of a form. 260of higher order. 110of an integral. 133.236.324integration of. 134. 324partial. 215of power series. 173tota!. 213of a transformation. 214ofa vector-valued function. 112

Determinant. 232of an operator. 234product of. 233

Diagonal process. 30. 157Diameter. 52DifferlC.ttidble function. 104. 212Differential. 213Differential equation. 119. 170Differential form (Set' Form)Diff~entiation lsee Derivative)Dimension. 205Directional derivative. 218Dirichlet's kernel. 189Discontinuities. 94Disjoint sets. 27Distance. 30Distributive law. 6. 20. 28Divergence. 281Divergence theorem. 253. 272.

~88

Divergent sequence. 47Divergent series. 59Domain. 24Dominated convergence theorem.

15i. 157. 32iDouble sequence. 144

e. 63Eberlein. W. F .. 184Elementary set. 303Empty set. 3Equicontinuity.IS6

Equivalence relation. 25Euclidean space. 16. 30Euler's constant. J97Exact form. 275Existence theorem. 170Exponential function. 178Extended real number system. IIExtension. 99

Family, 27Fatou's theorem. 320Feler's kernel. 199Fejer's theorem, 199Field axioms, 5Fine.N.l.. 100Finite set. 25Fixed point. 117

theorems. 117,203,220Fleming. W. H.. 280Flip. 149Form. 254

basic. 257of class W'.F", 254closed. 275derivative of. 260exact. 275product of. 258. 260sum of, 256

Fourier. 1. B.. 186Fourier coefficients. 186. 187Fourier series. 186. 187.328Function. 24

absolute value. 88analytic. 172Borel.-measurable. 313bounded. 89characteristic. .313component of. 87constant. 85continuous. 85

from left. 97from right. 97

continuously differentiable. 219convex. 101decreasing, 95differen!iable. 104.212exponential. 178harmonic. 297increasing. 95inverse. 90lebesgue-integrable. 315limit, 144linear, 206logarithmic, 180measurable. 310mono(Gr.ic. ~.:;

nowhere differentiable continu_ous. 154

one-to-one. 25orthogonal. 187periodic. 183produl;t of. 85rational. 88Riemann-integrable. 121

Function:simple. 313sum of. 85summable. 3 I5trigonomelric. 182uniformly continuous. 90uniformly differentiable, 115vector-valued. 85

Fundamental theorem of calculus,134. 324

Gamma function. 192Geometric series. 6 IGradient. 217. 281Graph. 99Greatest lower bound. 4Green's identities. 297Green's theorem, 253. 255. 272.

182

Half-open interval. 31Harmonic function. 297Havin. Y. P., 113Heine·Borel theorem. 39Hel1y's selection theorem. 167Herstein. I. N .. 65Hewitt, E.• 21Higher-order derivative. 110Hilben space. 332HOlder's inequality, 139

i. 13Identity operator. 232Image. 24Imaginary pan. 14Implicit function theorem. 224Improper integral. !39Increasing index. 257Increasing sequence. 55Independent set, 205Index of a curve. 20 IInfimum. 4Infinire series. 59Infinite set. 25Infinity. IIInitial-value problem, 119. ]70Inner product. 16Imegrable funcdons. spaces oi.

315.326Integral:

countable additivity of. 316differentialion of. lB. 236. 324lebesgue. 314JQWel. 121. 112Riemann. 121Stieltjes. 122upper. 121. 122

Integral test. 139Integration:

of derivative. 134. 324by parts, 134. 139. 141

Interior. 43

Interior poin!. 32Intermediate value. 93. 100. 108Intersection. 27Interval. 3 J. 302Into. 24Inverse function. 90Inverse function theorem. 221Inverse image. 24Inverse of linear operator, 207Inverse mapping. 90Invertible transfonnation. 207IITationaJ number, l. 10, 65Isolated point. 32Isometry. 82. 170Isomorphism. 21

Jacobian. 234

Kellogg, O. D .. 281Kestelman. H.. 167Knopp. K .. 11. 63

landau. E.G. H., 2!Laplacian. 297least upper bound. 4

propeny.4."18Lebesgue. H,L. 186Lebesgue·integrable function. 315Lebesgue integral. 314Lebesgue measure. 308lebesgue's theorem. 155. 167.

318. 321Left-hand limit. 94Leibnitz. G. W.. 71L:~llgl~'. 136L'Hospitars rule: 109. 113Limit. 47. 83. 144

left-hand. 94lower. 56poinlWise. 144right-hand. 94subsequential. 51upper. 56

Limit function. 144Limit point. 32Line. 17Line integral. 255Linear combimllion, 204Linear function. 206Linear mapping. 206Linear operator. 207Linear transformation. 206Local maximum. 107Localization !heor~m. 190LocaflY one-tQ-one mapping, !23Logarithm. 22. 180Logarithmic function. 180Lower bound. 3Lower integrai. 121. 122Lower limit. 56

McShane. E. L 313

Mapping. 24affine. 266contmuous. 85continuously differentiable. 219linear. 206open. 100. 223primitive. 248uniformly continuous. 90(See also Function)

Matrix. 210produc!. 2! I

Maximum. 90Mean square approximation, 187

"'- Mean value lheorem. 108. 235Measurable function. 310Measurable set, 305. 3 It)

Measurable space. 310Measure. 308

outer. 304Measure space, 310Measure zero. set of. 309. 317Mertens. F .. 74.'vIetric space. 30Minimum. 90Mobius band. 298Monotone convergence theorem.

318Monotonic function. 95, 302Monotonic sequence. 55Multiplication lsee Productl

Negative number. 7Negative orientation. 267Neighborhood. nNewtOn's method: 118Nijenhuis. A .. 223Niven. I.. 65. 198Nonnegative number. 60Nonn. 16. 140. 150.326

of operator. 208Normal derivative. 297Nonnal space. 101Normal vectOr. :!-84Nowhere differentiable function.

15'Null space. 228Null vector. 16Number:

algebraic. 43cardinal, 25complex. 12decimal. I Ifinite. 12irrational. J. 10. 65negative. 7nonnegatIve. 60poslllve. 7. 8rational.real. 8

One· to-one correspondence. 25Onto. 24Open cover. 36

INDEX 341

Open mapping. 100.223Open set. J2Order. 3. 17

lexicographic. 22Ordered field. 7. 20

k-tuple. 16pair. 12set. J. 18.22

Orienred simplex. 266Origin. 16Onhogonal set of functions. 187Orthonormal set. 187.327.331Outer measure. 304

Parameter domain, 254Parameter interval. ! 36Parseval's theorem, 191. 198. }28.

331Partial derivative. 215Panial sum. 59. 186Partition. 120

of unity. 251Perfect set. 32Periodic function. 18}. 190r.. 183Plane. 17Poincare's lemma. 275. 280Pointwise bounded sequence. 155Pointwise convergence. 144Polynomial. 88

trigonomelric. 185Positive orienration. 267Power series. 69. 172Primes. 197Primitive mapping. 248

Cauchy. 73of compiex numbers. Pof determinants. 233of field elements. 5of forms. 258. 260of functions. 85inner. 16of matrices. 211of real numhers. 19.20scalar. 16of series. 73of transformations. 207

Projection. 228Proper >unset. 3

Radius. 31. 32of convergence. 69. 79

Ral'ge. 24. 207Rank. L28Rank theorem. 229Ratio leSt. 66Rational funclion. 88Rational number. IReal field. 8Real line. 17Real number. 8Reallart. 14

342 INDEX

Rearrangement. 75Rectifiable curve. 136Refinemenl. 123Reflexive propeny. 25Regular set function. 303Relatively open set. 35Remainder. 211. 244Restriction. 99Riemann. B.. 76. 186Riemann integral, 121Riemann-Stieltjes integral. 122Riesz-Fischer theorem. 3)0Right-hand limit. 94Ring. 30 IRobison. G. 8.. 184Root. 10Root test. 65Row malri". 217

Saddle point. 240Scalar product. 16Schoenberg. L L /68Schwarz inequality. 15. 1J9. 326Segment. 3 ISelf.adjoint algebra. 165Separable space. 45Separated sets. 42Separation of points. 162Sequence. 26

bounded. 48Cauchy. 52. 82. 329convergent. 47divenzent. 47double. 144of functions. 143increasing. 55monotonic. 55pointwise bounded. 155poir;twise t:onVergent. 144uniformly bounded. 155uniformly Cunvergent. 157

Series. 59absolute!).' convergent. 71

~ alternating. 71conllergent. 59dillergent. 59geometric. 6 Inonabsolutely Conllergent. 72power. 69. 172product of. 73trigonometric. 186uniformly COnllergent, 1,57

Set. 3at most countable. 25Borel. 309bOunded. J2bOunded abolle. 3Cantor 41. Jj 1. 138. !I;~. ::09closed. 32compact. 36complete onhonormal. 3J Iconnected. 42convex. 3 ICountable. 25

Set,dense. 9. 32elementary. 3D3empty. 3finite. 25independent. 205infinite. 25measurable. 305. 310nOnemply. 3open. 32ordered. 3perfect. 32. 4 Irelatively open. 35unCountable. 25. 30. 41

Set function. 30 I(T-nng, ..01Simple diSContinuitv. 94Simple function. 313Simplex. 247

affine. 266differentiable. 269oriented. 266

Singer. I. .\L 280Solid angle. 294Space:

Compact metnc. 36compJete metric. 54connected. 42of cominuous functions. 150euclidean. 16Hilbert. 332

of integrable functions. 315. 326measurable. 310measure. 310metric. 30normal. 10 !separable. 45

Span. 204Sphere. 2n, 277.294Spivak. M.. 272. 280Square root. 2, 81. 118Standard basis, 205Standard presentation. 257Standard simplex. 266Stark. E. L.. 199Slep function. 129Slieltjes integral. 122Stirling's formula. /94. 200Stokes' theorem. 253. n2. 287Slone. Weierstrass theorem. 162.

J90.246Stromberg. K.. 2 ISubadditivily. 304Stibcbver. J6Subfield. 8. 13Subsequence. 5 ISUbsequential limit. 51Subs!:!. 3

dense. 9. 32proper. 3

Sum•. 5of Comple)( numbers. 12of field elements. j

of forms. 256of functions. 1i5 •

Sum,of linear transformations. 207of oriented simple)(es. 268of real numbers. 18of series, 59of vectors. 16

Summation by pans. 70Suppon. 246Supremum. 4Supremum norm. ISOSurface. "254Symmetric difference. 305

Tangent plane. 284Tangent '..ector. 286Tangential component. 286Taylor polynomial. 244Taylor's theorem. 110. 116.176.24Thorpe. 1. A .. 280Thurston. H. A.. 21Torus. 239-240. 285Total derivative. 213Transformation (see Function;

\-iapping)TransitivilY. 25Triangle inequality. 14. 16.30.140Trigonometric functions. 182Trigonometric polynomial. 185Trigonometric series. 186

Uncountable set. 25. 30. 41Uniform boundedness. 155Uniform closure. 151Uniform cominuity. 90Uniform CO'lVergence. 147Uniformly clcsed algebra. 161,..rniformly continuous mapping, 90Union. 27

Uniqueness theorem. 119. 258Unit cube. 247Unit Vector. 217Upper bound. 3Upper integral. 121. 122Upper limit. 56

Value. 24Variable of integralion, 122Vector. 16VeCtor field. 28 IVector space. 16.204Vector-valued function. 85

derivative of. 112Volume. 255. 282

Weierstrass tes!, 148Weierslrass theorem. 40. 159Winding number. 201

Zero set. 98. 117Zeta funclion. 14/

Documents

Rudin - Analisys