A First Course in Functional Analysis

AFirst Course in Functional Analysis MARTIN DAVIS PROFESSOR EMERITUS COURANT INSTITUTE OF MATHEMATICAL SCIENCES NEW YORK UNIVERSITY

Copyright Copyright© 1966, 1994 by Martin Davis All rights reserved.

Bibliographical Note This Dover edition, first published in 2013, is an unabridged republication of the work originally published by Gordon and Breach, New York, in 1966.

Library of Congress Cataloging-in-Publication Data Davis, Martin, 1928A first course in functional analysis I Martin Davis, professor emeritus, Courant Institute ·of Mathematical Sciences, New York University.

p.cm. Originally published: New York: Gordon and Breach, 1966, in the series Notes on mathematics and its applications. Includes index. ISBN-13: 978-0-486-49983-3 ISBN-I 0: 0-486-49983-9 1. Functional analysis. I. Title. QA320.D32 2013 515'.7-dc23 2012041596 Manufactured in the United States by Courier Corporation 49983901 www.doverpublications.com

To Harold and Nathan

Preface

The undergraduate mathematics major nowadays encounters modem mathematics as a collection of more or less unconnected subjects. The aim of this little book is to demonstrate the essential unity of twentiethcentury mathematics without assuming more mathematical knowledge or maturity than can reasonably be presumed of seniors or beginning graduate students. The contents may be described as an exposition, starting from scratch, of Gelfand's proof via maximal ideals of Wiener's famous result that when an absolutely convergent trigonometric series has a non-vanishing sum, the reciprocal of the sum can likewise be expanded into an absolutely convergent trigonometric series. En route one is led to prove Zorn's lemma and the Hahn-Banach extension theorem and to find the geometric series representation of inverces in an AbelianBanach algebra. And each of these leads to brief detours: to a discussion of the functional equationf(x+y) =f(x)+f(y), of the existence of Green's functions, and of existence and uniqueness theorems for certain linear integral and 4ifferential equations. The book is almost entirely self-contained. The reader should have had some experience with e- ~. A prerequisite (or corequisite) would be complex analysis through Cauchy's integral formula, unless the instructor is willing to take the extra time needed to develop this material. Also it would be advisable for the student to have seen the quotient group or quotient ring construction before encountering it here. I taught a three-credit-hour course covering the present material during the Spring 1959 semester at the Hartford Graduate Center of Rensselaer Polytechnic Institute to beginning graduate students with quite modest preparation. No suitable text being available, dittoed lecture notes were prepared from the students' own notes, the students, in tum, each being responsible for a specific portion of the course. The present book consists of these very notes, lightly edited. The problems ix

X

PREFACE

were taken from those assigned as homework as well as from the final examinations. I am grateful to Jack Schwartz for suggesting that these notes be included in the series under his editorship, and I will feel most pleased if this book leads others to enjoy Gelfand's beautiful proof. MARTIN DAVIS

New York City Apri/1966

Contents

1. SET THEORETIC PREUMINARIES 1. Sets and Members 2. Relations 3. Equivalence Relations 4. The Principle of Choice 5. Zorn's Lemma 6. The Functional Equation Problems 2. NORMED LINEAR SPACES AND ALGEBRAS 1. Definitions . 2. Topology irt a Normed Linear Space 3. Rasa Normed Linear Space 4. The Cartesian Product of Normed Linear Spaces Problems 3. FUNCTIONS ON BANACH SPACES 1. Continuous Functions 2. The Spaces CF(a, b) 3. Continuous Operators 4. Spaces of Bounded Linear Operators 5. Existence Theorems for Integral Equations and Differential Equations . 6. The Hahn-Banach Extension Theorem 7. The Existence of Green's Function Problems 4. HOMOMORPHISMS ON NoRMED LINEAR SPACES 1. Homomorphisms on Linear Spaces 2. Norms in a Quotient Space 3. Homomorphisms on Normed Linear Algebras 4. Inverse~ of Elements in Normed Linear Algebras Problem XI

1 1 4 5 7 8 15 19 23 23

28 32 34 36 39

39 42 44 45 51 60 67 74

77 77 79

82 84

89

xii 5.

CONTENTS

ANALYTIC FUNCTIONS INTO A BANACH SPACE

91 91 92

1. Derivatives . 2. Integrals of Banach-Space Valued Functions 3. Line Integrals and Cauchy's Theorem 4. Banach Algebras which are Fields . 5. The Convolution Algebra P Problems

100 101 107

INDEX

109

96

CHAPTER I

Set Theoretic Preliminaries

I. Sets and Members In this section the terms class, set, collection, totality and family will be used synonomously. The symbol, e (epsilon), will be used to denote membership in a class. The symbol,¢, will denote non-membership in a class, i.e., x e C means x is a member of set C x ¢ C means x is not a member of set C

Example: If Cis the class of all even numbers, then 2 belongs to C, (2 e C), 4 belongs to C, (4 e C), and 3 does not belong to C, (3 ¢ C). If a set is finite, then it can be described by listing its members. For example: Let C be the set {1, 3, 5} then 1 e C, 3 e C, 5 e C and all other elements are not members of the set C. The description of an infinite set, however, is not so simple. An infinite set can not be described by simply listing its elements. Hence, we must have recourse to defining the set by a characteristic property. The set can then be described as the set of all elements which possess the property in question. If the property is, say, P(x), then the set will be written {xI P(x)}, thus to . write C = {xI P(x)}, is to say that Cis the set of all elements x, such that, P(x) is true. For example, {a, b} = {xI x =a or x = b} A set may have only one element: {a} = {xI x =a}.

Definition 1.1. The union of set A with set B, A u B is the set of all those elements which belong either to set A or to set B or to both. Symbolically1 AuB = {xlxeA or xeB} 1

2

FUNCTIONAL ANALYSIS

Definition 1.1. The intersection of two sets, An B, is defined to be the set of all those elements which are common to both set A and set B. Symbolically, AnB = {x!xeA and xeB} Definition 1.3. The difference between two sets, denoted by A- B is defined to be the set A-B = {x!xeA and x¢B}

~

.u AnB

AuB

A-B

We shall use the symbols: :::;. to mean "implies"

to mean "if and only if" and

3

to mean "there is a"

Actually a certain amount of care is necessary in using the operation: E.g., let P(x) be the property: x ¢ x. (T~us, we could plausibly claim as an x for which P(x) is true, say, the class of even numbers, whereas the class of, say, non-automobiles, or the class of entities definable in less than 100 English words, could be claimed as x's making P(x) false.) Suppose we can form:

{xI·.. . .. }.

A={xlx¢x} i.e. Applying this to x

= A, yields AeAA¢A

a contradiction. This is Russell's paradox. In an.axiomatic treatment of set theory (cf. Kelley, General Topology, Appendix) suitable restrictions have to be placed on the operation:

SET THEORETIC PREUMINARIES

3

I.. . ... }

{x in order to prevent the appearance of the Russell paradox or similar paradoxes. In these lectures we shall use this operation whenever necessary. However, all of our uses could be justified in axiomatic set theory. The symbol = of equality always will mean absolute identity. In particular, for sets A, B, the assertion A = B means that the sets A and B have the same members. Definition 1.4. The set A is a subset of a set B, written A c: B, if each element of A is also an element of B, that is if x e A => x e B. From this definition it follows that A c: A. Definition 1.5. The empty set, represented by the symbol set which contains no elements. For example:

0, is the

0={xlx=Fx}

0 ={xI x = 0 and

x

= 1}

The empty set is a subset of any set, i.e. 0 c: A. Definition 1.6. 2A is the class of all sets B, such that B is a subset of A, i.e. For example, consider the set A, where

A={L,M,N} Then the elements of 2A are the empty set 0, the sets containing only one element, {L}, {M}, {N}, the sets containing two elements, {L, N}, {L, M}, {M, N}, and finally the single set containing all three elements {L, M, N}. Note that A contains 3 elements and that 2A contains 8 = 23 elements. Definition 1.7. The ordered pair of two elements is defined as (a, b)= {{a}, {a, b} }. It can easily be shown that (a, b) =(a', b') =>a = a' and b = b'. (Cf. Problem 1.) It is this which is the crucial property of the ordered pair. Any other construct with this property could be used instead. Note that (a,_b) is quite different from {a, b}. For, {a, b} is always equal to {b, a}.

4

FUNCTIONAL ANALYSIS

Definition 1.8. The Cartesian product of sets A and B, written, A x B, is the set of all ordered pairs (a, b), such that, a belongs to the set A, and b belongs to the set B, i.e.,

I

Ax B ={(a, b) aeA and beB}

For example, if: A= {L,M,N} B= {P, Q}

and

then the Cartesian product is the set (L, P), (L, Q) } (M, P), (M, Q) =Ax B

{

(N, P), (N, Q)

Note that there are 6 ( = 3 x 2) elements in the set.

2. Relations Definition 1.1. A relation between sets A and B is a set C such that

cc

A X B. Examples of such a Care {(L, P), (L, Q)}, {(M, P), (M, Q)}, and {(N, P), (N, Q)} as taken from the above example.

Definition 1.2. A relation on A is a relation between A and A. Definition 1.3. A mapping, transformation or function rx.from A into B is a relation between A and B, such that, for each x e A, there is exactly one y e B such that (x, y) e rx.. We write: rx.(x) = y to mean (x, y) e rx.. For example, let A be a set of positive integers and C be the set of ordered pairs (a, b) such that a-b is even, i.e. C = {(a, b) a-b is even}. Then, (1, 3) e C

I

(2, 4) e C (1, 6) ¢

c

Specifically, (a, b) e C.- a= b mod 2. Cis a relation on A.

SET THEORETIC PRELIMIN,ARIES

5

Example: Let A be the set of human beings, and let IX be the set of pairs (x, y) where x e A, yeA and y is the father of x. Then, ex is a relation on A and is also a function from A into A. However (cf. Def. 2.4 below), it is not a function from A onto A. Definition 2.4. A function IX is called a mapping from A onto B, if for each y e B there is at least one x e A such that (x, y) e IX. Definition 2.5. A function IX is called a one-one mapping from A onto B, if for each y e B, there is exactly one x e A such that (x, y) e IX

Example: Let R be the set of real numbers, and let P be the set of non-negative real numbers, let . IX= {(x,y)lxeR andy= x 2 }

P={(x,y)lxeP and y=x 2} Then, IX is a function from R into R. It is also a function from R into P and in fact is a function from R onto P. However IX is not one-one. Pis a one-one mapping from P onto P. We should remark that, if IX and p are functions, then IX c p turns out to mean simply that 1X(x) = y => P(x) = Y

In this case p is called an extension of IX.

3. Equivalence Relations As already noted, a relation on a set A is a subset of A x A, so that a relation consists of ordered pairs of elements of A. The fact that an element a e A bears the relation R to b e A may be expressed in the form (a, b) e R, or, as is more usually written, aRb.

Example: Let A be the set of all integers. Consider "< ".


yRx transitive if xRy and yRz => xRz A relation R on a set A is called an equivalence relation on A if it is reflexive, symmetric, and transitive. Such a relation can always be replaced by the equality relation between suitable sets. · If R is an equivalence relation on A, we define

I

[x] ={yeA xRy}

Theorem 3.1. Let R be an equivalence relation on a set A. Then: (1) xRy[x] = [y]. (2) xe[x]. (3) [x] n [y] #: 0=> [x]

= [y].

(That is, the equivalence classes [ x] divide the set A in a manner such that: (1) two elements bear the relation to each other if and only if they are in the same equivalence class; (2) each element is in one equivalence class; (3) the equivalence classes do not overlap.)

I

Proof: (2) [x] = {we A xRw} xRx (R is reflexive) :. xe[x]. (3) Let z0 e [ x] n [y] and t e [ x] xRt, xRzo, YRzo z 0 Rx (R is symmetric) z 0 Rt (R is transitive) yRt (R is transitive) te [y]

[x]

c

[y].

SET THEORETIC PRELIMINARIES

7

The same argument, with x and y interchanged, may be used to show that [y] c [x].

[x]

=

[y]

(1) =>

Let xRy

Let [x]

ye[x]

ye[y]

ye[y]

= [y] by (2)

·ye[x] ([x] = [y])

by (2)

ye[x] n [y]

xRy

:. [x] = [y] by (3) 4. The Principle of Choice This topic is controversial: the Principle of Choice will be proved using the Multiplicative Principle, which some people say is obvious and which others say is false. We shall use these principles freely throughout these lectures.

Convention: By a family we shall mean a collection of sets. Many sets are collections of sets, to be sure, but we shall use the term family when we wish to emphasize that its members are sets. Definition 4.1. The family :F of sets is called a family of disjoint sets if: A e :F and Be :F and A #= B => A n Br 0. Example: {1, 2, 3} n {2, 4, 6} = {2} so these two sets are not disjoint. Example: :F = {{0, 3, 6, 9, ...}

{1, 4, 7, 10, ... }

{2, 5, 8, 11, ... } } These three sets are disjoint. They are, indeed, equivalence classes (the relation being congruence modulo 3), and equivalence classes are disjoint (cf. Theorem 3.1 (3) ).

8

FUNCTIONAL ANALYSIS

The Multiplicative Principle: Let :F be a family of non-empty disjoint sets. Then there is a set M which has exactly one element in common with each set of :F. People object to the Multiplicative Principle: sometimes no principle of selection from each set can be stated, because the elements in the set can not be characterized in a usable fashion. Some people feel that this inability to specify the individual elements of M invalidates the principle; others feel that it does not. Example: Take A

= {x e R I0 ~ x

~ 1} and xRy means that

x- y is rational. R is clearly an equivalence relation. Here we have sets whose elements can be unnamable or even unimagined. Yet the multiplicative principle applies for the family :F of all equivalence classes. Theorem 4.1. (Principle of Choice).

Let :F be a family of non-empty sets. Then there is a function f such that for each A e :F, we havef(A) eA.

I

Proof· For each A e :F, define TA = {(A, x) x e A}. Define

I

'D = {TA A e:F} f'§ is a family of disjoint sets, so the multiplicative principle applies. There is a set f which has exactly one element in common with each TA. This f satisfies the requirements of the theorem, since y = f(A) means (A, y) ef.

5. Zorn's Lemma

We define

U A= {xI for some A,xeA and Ae:F} Ae!F

n A= {xl for all A,AeP=>xeA}

Ae!F

Definition 5.1. A chain, rti, is a non-empty family of sets such that A e rti and B e rti => A c B or B c A. Example: { {0}, {0, 1}, {0, 1, 2}, {0, 1, 2, 3} } is a chain but { {0}, {0, 1}, {0, 2} } is not a chain.

SET THEORETIC PRELIMINARIES

9

Note that a family which is not a chain may have subsets which are chains. In the second example above, { {0}, {0, 1}} and { {0}, {0, 2} } are chains yet subsets of a family which is not a chain.

Definition 5.2. !F is called an inductive family of sets if whenever rti is a subset of !F and rti is a chain then U A is in !F. Ae'it

Examples of non-inductive sets: (1) { {0}, {0, 2}, {0, 2, 4}, ... ' {0, 2, 4, ... }. {0, 2, 4, ...• 1}. {0, 2, 4, ... ' 1, 3, ... } } ; (2) { {0}, {0, ~}. {0, 1, 2}, {0, 1, 2, 3} ... } Note that in the second example each subset is finite yet the union of all subsets contains all positive integers and hence is infinite, therefore cannot be a member of the Himily of sets.

Theorem 5.1. Let !F be an inductive family of sets. Let f be a function on !F such that if A e !F, then f(A) e !F and /(A) c A. Then there is a set A 0 e !F such thatf(A 0 ) = A 0 • . · As a preliminary example, consider the inductive family

!F={{a,b}, {a,b,c}} and define

/({a, b}) ={a, b, c} /({a, b, c}) = {a, b, c}

Therefore, the theorem is satisfied trivially. It can be observed that this will always be the case when !F is a family consisting of a finite nwnber of sets. Proof· Let K e !F. Then a family f'§ is a K-support if:

(1) Kel§

(2) Aef'§=>f(A)ef'§ (3) rti c f'§ and rti is a chain => (

U A) e l§, Ae'it

i.e. l§ is inductive. Note that !F is a K-support.

10

FUNCTIONAL ANALYSIS

Now let

r

= {'D If'§ is a K-support}, and define :1l' =

n

f'§

This

!fer

means that :1l' is a sub-family of every K support. An alternative way of defining :1l' are: .;K =

{A I'D is a K support=> Aet!i}

I

or

:1l' ={A A belongs to every K-support}

At this point, the completion of the proof rests upon two assertions about :1l'. These are: (a) :1l' is a K-support, (b) :1l' is a chain.

The procedure now will be to assume these two assertions and complete the proof on this basis. Then, a series of 6lemmas will be proved, the 1st of which proves :1l' is a K-support and the 6th that :1l' is a chain. Therefore, assuming Jlf is a K-support and a chain, let A0 = U A. Now A 0 e :F since :1l' is a chain and :1l' c :F. Ael Thereforef(A 0 ) => A 0 by hypothesis. But since :1l' is a chain and a e :1l'. In other words 0 e :1l'. K-support and :1l' c :1l', then ( U ·

Ael

A)

Since :1l' is a K-support, f(A 0 ) e :K. :. f(A 0 ) c

A

U A, i.e. Ael

/(A 0 ) c A 0

But /(A 0 ) => A 0 and /(A 0 ) c A 0 • Hence, /(A 0 ) = A 0 , which is the conclusion of the theorem. Now the remaining task is to prove the assertions (a), (b). Lemma I. :1l' is a K-support.

Proof· In order to show that :1l' is a K-support, it must be shown that :1l' has the three definipg properties of a K-support. (1) K belongs to every K-support, so Ke :1l'

(2) If A e :K, then A belongs to every K-support. Therefore /(A) belongs to every K-support, by the definition of a K-support.

f(A) e :1l'; i.e. A e :1l' => f(A) e :1l'.

11

SET THEORETIC PRELIMINARIES

(3) If~ c 31' and ~ is a chain, then ~· c Then [ U AJ ef"§ for every K-support f"G.

f"§

for every K-support

f"§.

Ae~

:. [ U AJ e#; i.e. 31' is inductive. Ae~

This completes the proof of Lemma 1 since 31' satisfies all of the conditions required for a family to be a K-support. Lemma 2. (The principle of induction for

.*'.) Let &'

c 31'

such

that (1) Ke&', (2) Ae&'=>/(A)e&', (3) ~is a chain and~ c &'=> [

U A] e&'.

Ae~

Then&'=#. Proof: By the definition of a K-support, &' is a K-support. Then 31' c &',since by definition 31' is a sub-family of all K-supports. But &' c 31' by hypothesis. Therefore&'= 31'. Note. The usual principle of mathematical induction is quite analogous to this lemma. It can be stated as follows: Let N be the set of all positive integers and let S c N such that:

(1) 1eS,

(2) xeS=>x+leS. Then, S = N. Lemma 3. Ae# =>A:::. K

I

Proof: LetJ! = {Ae# A:::. K}. : . .4!;:) 3{'

(1) Ke# and K:::. K. :. KeJ!

(2) If A e .4! then A :::. K. But f(A) :::. A since A e 31' c §. :::. K. Also /(A) e 31' since 31' is a K-support. Thereforej(A) e .4!, i.e. A e .4! => f(A) e .4!.

Therefore/(~)

12

FUNCTIONAL ANALYSIS

(3) Let ~ be a chain where ~ c Jt, then A e ~ => A :::. K.

:. -[ U

A.e~

But [

A]:::. K.

U AJ e :# since :# is a K-support.

A.e~

:. [ U

Ae~

A] e.A

Then .A=:# by Lemma 2, or:#= {Ae:# IA:::. K}. Therefore A e :# => A :::. K which is the conclusion of this lemma. Definition: Let

.!t'={Ae:#I[BeJ"t' and B#=A and BcA]=>/(B)cA} Thus, !l' c :#. Lemma 4. A e !l' and Be .Yt' => B c A orB:::. /(A). Proof. Let A 0 e !l'. Now define

.IV= {Be:#jB cA 0 orB :::.f(A 0 )} : . .IV c .Yt' (I) Ke :#. Since A 0 e :#, K c A 0 by Lemma 3. :. Ke.!V

(2) If Be .AI", does /(B) e .AI"? First of all /(B) e :# since :#is a K-support. There are now 3 cases to consider. B c A 0 gives (a) B = A 0 or, (b) B c A 0 and B #: A 0 ,

and the other case is (a) If B = A 0

(c) B :::.f(A 0 ) (b) If B c A 0

(c) B =>f(A 0 ) f(B) => B

f(B) =f(Ao)

and B

:. f(B) :::. /(Ao)

then f(B) c A 0

since Be§

:. f(B)e.!V

since A 0 e !l'

:. f(B) :::.f(Ao)

:. f(B)e.!V

:. f(B)e.l(

Therefore Be%=>/(B)e%.

:;6 Ao

SET THEORETIC PRELIMINARIES

(3) If

f(j

is a chain and

f(j :::.

13

U BJ e.A'? Note that [ Be f(B) c: f(A)? The situation now is Be .1t' and A e 2. Hence, from Lemma 4, B c A or B:::. f(A). Note that f(A) e ;/I' since .1t' is a K-support. Again we have the 3 cases to consider:

(a) B =A f(B) =f(A) :. f(B) c: /(A) .". f(A) E !l'

(b) B c: A and B =I= A, then f(B) c: A since A e 2 and B satisfies the conditions in the definition of 2. Also, A c: f (A) since

Ae§. :. f(B) c f(A)

or f(A) e !l'. (c) B :::./(A). But B c: f(A) by hypothesis. Therefore B = f(A) but this is impossible since B =1= f(A) by hypothesis. Therefore, this case does not satisfy the hypothesis and the assertion is tr1,1e vacuously. Thereforef(A) e 2. Then in all cases A e !l' => f(A) e !l'.

14

FUNCTIONAL ANALYSIS

(3) If ~ is a chain and ~ c !l', does T = [

U

AJ e !l'? Note that

Ae~

T e ;If since ;If is a K-support. Suppose there exist B e ;If such that B #:- T and B c T. Since A e ~ c !l', B c A or B => f(A) :::. A, since A e:F. (a) Can it be that for all A e ~. B => A? If so, B:::. U A, i.e. Ae~

B :::. T. But B #:- T and B c T by hypothesis. Therefore this situation is impossible andf(B) c Tis true vacuously. Therefore Te !l'. (b) There is some A 0 e ~ such that B c A 0 and B #:- A 0 • Now since A 0 e I£ and Be ;If, thenj(B) c A 0 since B satisfies the conditions in the definition of !l'. But A 0 c U A, i.e. A 0 c T. Thereforef(B) c T and T e !l'. Ae~ Then in all cases,~ is a chain and~ c I£=> L~~ e!l'. The con-

A]

clusion now follows at once from Lemma 2.

Lemma 6. ;If is a chain. · Proof· Let A, Be ;/f. Then A e !l' since !l' = ;If by Lemma 5. Therefore by Lemma 4, either B c A or B:::. f(A) ~ A. Hence, ;If is a chain.

Definition 5.3. M is called a maximal element of a family :F if: A e§ and

A :::. M =>A = M

Example: Let§= { {1}, {1, 2}, {4}, {4, 5} }. :F has the maximal elements {1, 2} and {4, 5}.

Theorem 5.2. (Zorn's Lemma.) Let Then, § has a maximal element.

§

be an inductive family.

Proof (by contradiction): Suppose § has no maximal element. Then for each A e § there is a B e § such that: A c B and A #:- B. For each A e §,let TA = {Be§ B:::. A and B #:-A}. Then for each A E §, TA #:- 0. Let r = {T.A A E §}. By Theorem 4.1 (Principle of choice), there is a function g defined on r such that· for each TA, we have g(T.A) e T.A. Define/so thatj(A) = g(TA); soj(A) eTA. I.e., f(A) :::. A,.f(A) #:- A. But, this is a contradiction, by Theorem 5.1.

I

I

SET THEORETIC PRELIMINARIES

15

6. The Functional Equation f(x+y) =f(x)+f(y)

In this section all functions are real-valued functions of a real variable.

Example: Letf(x) = kx. f(x+y) = k(x+y) = kx+ky =f(x)+f(y)

We will find all functions/that satisfy this equation.

Theorem 6.1. Ifjis defined for all real x, and if f(x+y)

= f(x)+f(y)

thenf(x) = /(1) · x for all rational numbers x. Proof:

Thus if n is a positive integer f(n) =/(1+1+1+ ... +1) (nl's) =/(1)+/(1)+ ... +/(1) (nf(1)'s) =/(1)· n

Finally, for -n where n is a positive integer 0 =f(O) =f(n+( -n)) =f(n)+/{-n)

f(-n) = -f(n) = -/(1)· n =/(1)· (-n)

Next,

f(O) = /(1 +( -1)) = /(1)+/( -1) = /(1)-/(1)

=0 = 0·/(1)

16

FUNCTIONAL ANALYSIS

Thus the result has been demonstrated for all integers x. For x rational, we have x = mfn where n is positive. Hence,

nr(:) =/(i)+f(:)+ ... +/(~) (n !(:)'s) m m) (n =f -+-+ ... +m (n

n

n

=f(m) =/(1)·m; !(:)=/(1)·:

so

Theorem 6.2. If f(x+y) =f(x)+f(y) and f(x) = f(1) • x for all x.

.f is continuous, then

Proof: Let x 0 be any real number. Let x: 0 is rational then,

= lim rn

f(x 0 )

= lim f(x:) = x--+.xo

where each rn

n-+oo

limf(rn) na+cc

= lim (!(1) · rJ = f(l) lim r" = /(1). X:o

In order to prove the existence of discontinuous solutions of the functional equationf(x+y) =f(x)+f(y), we require the notion of Hamel basis.

Definition 6.1. A set r of real numbers is called a Hamel basis if 1 e r and for each real number x there are uniquely determined numbers x: 1 , x 2 , ••• , xn e r and non-zero rational numbers r1o r2 , ••• , r" such that:

Theorem 6.3. There is a Hamel basis. Proof: A set of real numbers

(1) 1 er,

r

will be called nice if:

17

SET THEORETIC PRELIMINARIES

(2) r 1 , r 2 , ferent and

••• ,

r" rational and xh x 2, ••• , x" e r and x 1's all difn

L, r,x, = 0=> r., r2 , ••• , rn =

0

1=1

Let fF be the family: , = Let~

Lemma I.

{rj r

is nice}

be a chain. Let A., A 2 ,

••• ,

An e ~- Then,

A1 uA 2 u ... u4e~ Or, the union of a finite number of elements of a chain belongs to the chain. Proof: The proof will be by mathematical induction on n. For n = 1 it is obvious-the union is just the one set. Suppose the result is true for n = k. Then for n = k+ 1, Ak> Ak+ 1 e~.

let

A., A 2 ,

let

R = A 1 u A 2 u ... u A,.,

• ••

and letS =RuA,.+ 1 To prove: S e ~.

R e f(i by induction hypothesis. Ak+ 1 e~.

R c Ak+ 1 or Ak+ 1 c R since ~ is a chain. If R c Ak+ 1 then S = Ak+ 1 and Se~. If Ak+ 1 c R then S =Rand Se ~. Lemma 2. § is inductive.

Proof· Let f(f c 9', where f(f is a chain. Let T = U A To prove Ae~ • that T e § or that Tis a nice set: (I) 1 e Tis obvious because: 1 e A for each A e ~.

(2) Let r., r2 , n

••• ,

r" be rational. Let x., x 2 ,

••• ,

different. Let L, r1 x 1 = 0. To prove that the ri's = 0: 1=1

Eachx 1 eT~

Therefore x, e A 1 e ~. i

= 1, 2, ... , n

x, e T,

x,'s all

18

FUNcnONAL ANALYSIS

Let M = A1 u A 2 u ... u A,.. Then x 1 e M, i = 1, 2, ... , n. Me~ (By Lemma 1). ~ c:: §. Me§. Thus, M is nice.

Therefore r1 , r2 , •• • , r,. = 0, so that Tis nice. This completes the proof of the lemma. Since §is inductive, Zorn's lemma now tells us that §has a maximal element. Call this H. (We want to prove His a Hamel basis.) Suppose there is some such that:

e

n

e#: L. r,x;, X;EH, e¢H 1=1

Let H' = Hu {e}. Then H c:: H'. Claim: H' is nice. (I) 1 e H. Therefore 1 e H' .

..

(2) Let L, r1 x 1 = 0, x;'s different, x 1 e H'. Suppose 1=1

eis any one

of

the x's say x 1 • Then

e

and this is impossible. So is not any one of the x's. Therefore X to x 2 , ••• , x,. e H. But, H is nice. Therefore r., r2 , ••• , r,. = 0. Therefore H' is nice. Therefore H' e §. But H is a maximal element, and H c:: H'. H'e§.

ThenH= H'. , But ¢ H and e H'. Therefore for each = L, r 1x 1, x 1's all different and x 1 e H. 1 It remains only to prove that this representation is unique. Suppose. for x., ... , x,. e H, x = r1 x 1 =r2 x 2 + .. .+r,.x,.

e

e

e, e

X=

s1 X 1 +s2 x 2 + ... +s,.X,.

•=

19

SET THEORETIC PRELIMINARIES

Here some of the r1's and s1's may be 0. Then, subtracting:

0 = (r1-s 1)x1 +(r2.-s 2)x2.+· .. +(r11 -S11)X11 Since His nice, these coefficients are all = 0. I.e., r 1 r,. = s,..

= sl> r 2

=

s2 , ••• ,

Theorem 6.4. Let H be any Hamel basis. Let f(x) be defined arbitrarily for x e H and let II

/(x)

= 1=1 L: rd(xi)

Then,

II

for

x=

L: r1 x~o

1=1

x 1eH

f(x+y) =f(x)+f(y)

Proof: Let X= r 1 X 1 +T2 X2 + ..• +TnXn

y = s1 x 1 +s2 x 2 + ... +s11 x.

= L: r;/(xl)

Then,

f(x)

and

/(y) = L;sd(x1)

Now,

x+y = (r1 +s1)X1 +(r2.+Sz)X2.+· .. +(r11 +S11)X11

and therefore f(x+y)

Thus,

= L;(r1+s1)j(x1)

f(x+y) =f(x)+f(y)

Problems 1. Which of the following statements are true and which are false? (a) {1, 2, 3} c {1, 2, 3, 4}.

(b) {1} E (1, 2).

(c) {1, 2, 3} n {3, 4, 5} = 3. (d) {1, 2} u {2, 3} = {1, 2, 3}. (e) {1} e 2{~, 2, 3}. (/) (1, 2) E {1, 2, 3} X {1, 2, 3}.

20

FUNCTIONAL ANALYSIS

2. Let: j be a certain definite dog named Jimmy,

D be the class of all dogs, S be the set of all species of animals, and A be the set of all animals.

(a) Write all true statements of the forms x e y and x c y where x andy can be j, D, S, or A. (b) Find a solution in the set {j, D, S, A} for the equation: 2x = y. 3. Prove using the definition of ordered pair that (x, y) = (x', y') implies x = x' and y = y'. 4. (a) For each of the following relations tell whether it is (i) reflexive, (ii) symmetric, and (iii) transitive:

1.

(1) I xI = I y x, y complex numbers. (2) x < y, x, y real numbers. (3) x' = yx, X > 0, y > 0. (4) x c y, x, y sets of real numbers. (b) Which are equivalence relations? (c) For those which are, describe the equivalence classes.

S. (a) Which of the following families are inductive? (1) The family of all finite sets of positive integers. (2) The family of all sets of positive integers. (3) The family of all sets which are residue classes of positive integers modulo some integer. (E.g. the set of even positive integers, and the set of multiples of 3, but not the set of perfect squares, belongs to this family.) (b) Choosing one example in (a) which is inductive, show how the proof of Theorem 5.1 would work out. (You may ignore the Lemmas.) 6. Let Q be some definite set of real-valued functions defined for -l;;!x~l. Let, §a= {

{!, -!} lfeQ}

2

(E.g. ifj(x) = x is in Q, then the set{/, g} e §a where g(x) = -x2 .)

SET THEORETIC PRELIMINARIES

21

By the multiplicative principle, there is a set R such that for each

I e Q, either I e R or -I e R but not both. Can you define a set R with this property, but not using the multiplicative principle, if: (a) Q is the set of linear functions? (b) Q is the set of polynomials? (c) Q is the set of functions expressible as convergent power series 00

L anx" n=O

for

lxl < 1

(d) Q is the set of functions continuous for - 1 ~ x ~ 1. (e) Q is the set of all functions defined for -1 ~ x ~ 1.

(f) Q is the set of solutions off(x+y) =l(x)+l(y).

(This problem was communicated by Dr. Marvin Minsky, who attributed it to J. von Neumann.)

CHAPTER 2

Normed Linear Spaces and Algebras

I. Definitions We shall let R denote the real number field, C denote the complex number field, and F denote either R or C. Definition 1.1. A linear space over F is a set, L, taken together with a function, +, from L x L into L, and a function · from F x L into L such that (1) X, YeL =>X+ Y = Y+X, (2) X, Y, Z eL => X+(Y+Z) =(X+ Y)+Z,

(3) there is an element 0 in L such that X+ 0 = X, (4) XeL => 3(-X)eL such that X+(-X) = 0.

( (1)- (4) means that L, with the operation +, forms an Abelian Group.) (5) a, b e F and X e L => a(bX) = (ab)X, (6) aeFandX, YeL=>a(X+Y)=aX+aY, (7) a, b eFand XeL => (a+b)X = aX+bX, (8) XeL => 1 ·X= X, (9) X e L => 0 · X = 0. The dot · will usually be omitted. Definition 1.1. L is a normed linear space over F if L is a linear space over F and for each X e L there is a number X such that:

I I

I I I

(1) X II~ 0, (2) X+ y ~ X + y (3) aX =: a X (4) ~XII =0-X=O.

I I I I I' I I Ill I • 23

24

FUNCTIONAL ANALYSIS

Definition 1.3. A sequence of elements of a normed linear space is a function from the positive integers into L. If X is a sequence we write X,= X(n) and X= {X,}

!I

Definition 1.4. X,-+ X, or lim X, =X means lim X,- X II = 0. n .... oo

n .... oo

Corollary 1.1. If X,-+ X and X,-+ X', then X= X'.

Proof: ~X-X' II= IIX,-X'-X,+XII = IIX,-X'-l·X,+l·XII

= IIX,-X'+(-l)X,-(-l)XII = II<X,-X')+(-lXX,-X)~ ~ IIX,-X'~ + 11(-l)(X,-X)II

IIX-X'II ~ IIX,-X'II + IIX,-XII But,

~X,-X'II-+0;

Thus,

II X -X' II= 0, i.e. X= X'

IIX,-XII-+0

Corollary 1.2. X,-+X andY,-+ Y=>aX,+bY,-+aX+bY.

Proof: II (aX,+bY,)-(aX +bY) II= II a(X,-X)+_b(Y,- Y) I ~II a(X,-X) II+ II b(Y,- Y) ~

=I a 1·11 X,-X II+ I b 1·11 Y,- Y 11-+0 Thus,

llIi -+0

Therefore,

aX,+bY,-+ aX +bY

Definition 1.5. {X,} is a Cauchy sequence if lim II Xm-Xn II = 0; m-+oo n-+ex>

that is, if for every e > 0, there is an N such that: m,n >N=> IIXm-Xnll <e Definition 1.6. {X,} converges ifthere exists an X such that X, -+ X. Corollary 1.3. If {X,} converges, then {X,} is a Cauchy sequence.

Proof: Let X, -+ X.

25

NORMED LINEAR SPACES AND ALGEBRAS

Consider

IIX,-Xmll

IIX,-X-Xm+XII = II(X,-X)-(Xm-X)II ~ IIX,-XII + IIXm-XII-+0

=

as m, n-+ oo. Therefore, lim m-+oo n-+oo

I X,-Xm II= 0, i.e. {X,} is a Cauchy sequence.

Definition 1.7. A normed linear space Lis called complete if every Cauchy sequence in it converges. A complete normed linear space is also called a Banach Space. Several examples of Banach spaces will be studied later. In fact, functional analysis, is largely concerned with the theory of Banach spaces. 00

Definition 1.8.

L X,= X

n=l

.

Y,=

means Y,-+ X, where

n

L Xk=Xt+X2+· . .+X, k=l

00

00

n=l

n=l

LX, converges means that LX,= X for some X. 00

Corollary 1.4. In a Banach space, if 00

L II X, I n=l

converges, then

L X, converges.

n=l

n

Proof· Let Y, =

n

L xk and let t, =k=l L II xk II· k=l

Now we show that

the terms of the sequence I Y,, ·_ Y, I can be made small. Without loss of generality, let m > n. Then, m

n

II Ym-¥..11 =II k=l L Xk- k=l L Xkll =II Xn+l +Xn+2+Xn+3+• • .+Xm II ~ II X,+ 1 I + I Xn+211 + • • · + II Xm I = I tm- t, 1-+ 0 as m, n -+ oo

{k=l± k} X

00

is a Cauchy sequence, and since this is a Banach space,

L X, converges.

n=l

26

FUNCTIONAL ANALYSIS

Definition 1.9. A linear space L over F is called a linear algebra over F if there is defined a function o from L x L into L such that: (I) X, Y,ZeL=> Xo(YoZ) = Xo Y)oZ

(2) There is an element e e L such that eoX=Xoe=X (3) X o (Y+Z) = Xo Y+X oZand (X+ Y) oZ =X oZ+ YoZfor X, Y,ZeL. (4) a(X o Y)

= (aX) o

Y.

Definition 1.1 0. A linear algebra is called Abelian if X o Y = Yo X. Definition 1.11. A normed linear algebra is a normed linear space which is also a linear algebra and such that:

I e I = 1, I Xo Yll ~II Xll·ll Yll

and

Note. The symbol o will usually be written · and will often be omitted altogether. Corollary 1.5. Proof·

Thus

X., _. X ~

I X,, I _. I X I

I X,. II= I x .. -X+XII ~ I x.. - x I + I x I I x.. I - I X I ~ I x.. - X I

Applying the same process to

I! X I

I XII= I X-X,,+Xn I I X- x.. I + I x.. I = I x.. - X I + I x.. I I X I - I x.. I ~ l x.. - X I Ill x.. I - I X Ill ~ l x.. - X l _. 0 ~

Thus That is to say Therefore

0~

I !IX.. I -

~ X

Ill .-. 0,

or ~

x .. I

_. I X l

27

NORMED UNEAR SPACES AND ALGEBRAS

Corollary 1.6. In a normed linear space:

X"-+ X and Y"-+ Y=> X" Yn-+ XY

I Xn Yn-XYII =II Xn Yn-Xn Y+Xn Y-XYII =II xn (Yn- Y)+(Xn-X)Y I ~II Xn 11·11 Yn-YII +II Xn-XII·II Yll

Proof·

-+0

Definition 1.11. A normed linear space which is complete is called a Banach algebra. Definition 1.13. In a linear algebra, Y is: (1) an inverse of X if XY = YX = e. (2) a right-inverse of X if XY =e. (3) a left-inverse of X if XY =e. If Y and Z are both inverses of X, we have Y= YXZ=eZ=Z.

Convention. We write Y = x-t to mean that X has an inverse, and that its value is Y. Theorem 1.7. In a Banach algebra:

I e-X I

< I

=> X has an inverse

The proof of this fact may be motivated by the following identity valid for real or complex numbers a for which -a < 1 :

II

1 a

a- 1 =-=

1 1-(1-a)

I

=1+(1-a)+(1-a)2 + ... 00

Proof Consider, then, the series

L

(e-X)", with the purpose

n=O 00

first of showing that it converges. The series

L I e-X I " converges, n=O ·

because it is simply a geometric series of real numbers and because e-X < 1 by hypothesis. Now

I

I

I (e-X)" I = I (e-x)· <e-X)· ... · (e-x) I ~ I e-X I · I e-X I ' · · · · I e-X I = I e-X I "

28

FUNCTIONAL ANALYSIS 00

Thus the series

L II (e-X)" II converges, by the ordinary comparison n=O 00

test. Therefore, by Corollary 1.4,

L (e-X)"

converges.

Let

n=O 00

Y=

L (e-X)", and let us show that it is an inverse of X.

n=O

YX = Y[e-(e·-X)] = Y+ Y[ -(e-X)] XY = [e-(e-X)]Y = eY +[-(e-X)]Y = Y +[-(e-X)]Y n

Remembering that Y =lim

L (e-Xt, let us compute

n-+oo k=O

Y[-(e-X)] and [ -(e-X)]Y n

L (e-X)k

Y[-(e-X)] = lim

(e-X)

n-+oo k=O

n

n+l

L -(e-X)k+t =lim L -(e-X)k

=lim

n-+oo k=>O

n-+oo k=l

00

=-

L (e-X)" =e-Y

n=l n

[ -(e-X)]Y =lim

L -(e-Xt

n .... a:>k=O n

=lim

n-+oo k=O

Therefore YX

oo

L -(e-X)k+t =lim L -(e-X)k = e- Y n-+oo k=l

= Y +(e- Y) = e

XY=Y+(e-Y)=e

2. Topology in a Normed Linear Space

In this section, L is some fixed definite normed linear space. Definition 2.1. A c L is called closed if X,, e A for all n implies that and Xn -+ X=> X e A Definition 2.2. A c L is called open if X e A => {y E L X- y < E } c A.

Ill

I

Note. A set may be neither open nor closed.

3 e > 0 such that

29

NORMED LINEAR SPACES AND ALGEBRAS

Theorem 2.1. A is open -L-A is closed. Proof·=> Let {Xn} be some sequence such that X"eL-A where X" X. To prove X e L-A. Suppose that X e A. Since A is open then 3 e > 0 such that { Y e L X- Y < e} c: A. But,

-+

Ill

I

I Xn-X II-+ 0 Therefore, 3 N such that n > N => II X"- X all n, X" e L-A which is a contradiction. Proof· X" e A. But for

3 e > 0 such that

{YeLIIIX-YII <e}c:A.

Suppose this is not true. Then, for every e > 0, 3 Y e L such that II X- Y II < e, but Y ¢ A. In particular we may pick e = 1/n Call the Y, Y". Therefore, II X- Y" II < 1/n, but Y" ¢ A, i.e. Y" e L-A. Now, Y"-+ X. Hence, since L-A is closed, XeL--A, which is a contradiction. Definition 2.3. Let A c: L. Let !F = {B c: L I B => A and B is closed}. A= B. A is called the closure of A.

n

Bell

Corollary 2.2. A c: .A. Proof· A =

n B, but every B e fF includes A.

Bell

Corollary 2.3. A is closed. Proof·· Let X" e A and X" -+ X. To prove, X e .A. Since Xn e .A, for each closed B => A, X" e B, which implies that for each closed B => A, X e B. Thus, by Definition 2.3_ X eA.

Corollary 2.4. If B => A and B is closed then B => A. Proof· Let X e A; then, by Definition 2.3, A

XeB. Definition 2.4. Interior (A)

= L·-(L...:_A).

=

n B.

Be.l'

Hence,

30

FUNCI'IONAL ANALYSIS

Corollary 2.5. that Xn-+ X.

If X e .A, then there exists a sequence X" e A such

Proof. Let B be the set defined by: B = {X e L X"-+ X}.

I there exists a

sequence {X"}, such that X" e A, and

First note that A c B. (For, if X e A, then X is the limit of the sequence X, X, X, •.. so that X e B.) Next, we note that B is closed. For, let Y" e B, Y"-+ Y. Then we must show that Y B. Since Y" B, there is a sequence, Xc;? A, such that Xc;?-+ Y" as m-+ oo. Hence, for each n, there is an integer mn such that

e

e

I X!,:'~

- Y, I
N ~ I X,- X I < 1

Then

II(X,-X)+XII ;:;i IIX,-XII +II XII X

~

M.

Theorem 3.1. If A has an upper bound, then there is exactly one number S such that: (1) Sis an upper bound of A. (2) If X < S, then X is not an upper bound of A. This theorem expresses the so-called "completeness" property of the real numbers, and we accept it here without proof.

Definition 3.1. We writeS= sup A if Sis the number of Theorem 3.1. Sis called the supremum or least upper bound of A~ If A has no upper bound, we write sup A= oo. Similarly we may define S = inf A where S is the greatest lower bound of A. Details are left to the student. Definition 3.3. {Xn} is monotone increasing if Xn

~

Xn+t·

NORMED LINEAR SPACES AND ALGEBRAS

33

Theorem 3.2. If {Xn} is monotone increasing and bounded, then it converges. Proof: Let A be the set of all terms of {Xn}· Let X= sup A. Then Xn ~ X. Let e > 0 be any number. Then X- e is not an upper bound for A. Therefore, there exists an N such that XN > X -e. Now, n > N => Xn ~ XN > X-e. n > N=> X-e < Xn ~X.

n > N => - e < Xn- X

I

~

0.

I

n > N => Xn- X < e. Thus, Xn ~ X. The definition of monotone decreasing sequence and the statement and proof of an analogue of Theorem 3.2 are left to the student.

Theorem 3.3. Every sequence has either a monotone increasing or a monotone decreasing subsequence. For the purpose of proving this theorem we define: Xk is a peak of the sequence {Xn} if n > k => Xn ~ Xk. Lemma. If {Xn} has no monotone increasing subsequences, then for each ro. there is a k > ro such that xk is a peak. Proof: Suppose otherwise. Then there is some r0 such that for no k > ro is xk a peak. XNr = x,D+I is not a peak. There is N2 > N~o XN 2 > XNc Continuing this process we see that {Xn} has the monotone increasing subsequence XNr• XN 2 , • • • • This is a contradiction. Proof of Theorem 3.3: Let {Xn} -be a sequence. Suppose {Xn} doesn't have a monotone increasing subsequence. Then, (take r = I in lemma) 3 k 1 such that Xk, is a peak. Next taker= k 1 , 3 k 2 > k 1 such that xk2 is a peak. There exist kl < k2 < k3 < k4 ... such that each xk, is a peak. .Hence Xk 1 ~ X;c 2 ~ Xr. 3 ~ • • • is a monotone decreasing subsequence.

Theorem 3.4.

(Bolzano-Weierstrass theorem) R

is piecewise

compact. Proof: Let A c::: R be bounded and closed. Let Xn e A; {Xn} has a monotone subsequence {XnJ (Theorem 3.3). {XnJ is bounded because A is bounded. Hence {X""} converges. (Bounded monotone sequences converge.) Say X"" ~ X. But X e A since A is a closed set. Hence A is compact.

34

FUNCTIONAL ANALYSIS

Corollary 3.5. (Cauchy convergence criterion.) R is a Banach space. Proof: By Theorem 3.4, R is piecewise compact, and hence, by · Theorem 2.11 it is a Banach space.

4. The Cartesian Product of Normed Linear Spaces In this section L and Q are two normed linear spaces over the same field F. The Cartesian product L x Q can be made into a normed linear space by defining for X, X' e L, Y, Y' E Q, and a E F: (X, Y)+(X', Y') = (X+X',y

=

Y')

a· (X, Y) =(aX, aY)

I (X, Y) I = .j II X I 2 + I y I 2 It is left to the reader to show that L x Q is a normed linear space and that I XII~ I (X, Y) I • I Yll ~II (X, Y) II· Theorem 4.1. If Land Q are Banach spaces, then so is L x

Q.

Proof: Let {(X,,, YJ} be a Cauchy sequence in L x Q.

I X,-Xm I

~

I (X,,

Y,)-(Xm, Ym)

I

Thus, {X,} is a Cauchy sequence. Similarly, { Y,} is a Cauchy sequ~nce. Let X,

-+

X, Y,

-+

Y. Then,

I (X,, Y,)-(X,

Y)

II= I (X,- X, Y,- Y) I = .j I X,- X p + I Y,- y I

. Co~ollary 4.2. C is a Banach space. Proof: C = R x R.

2

-+

0

35

NORMED LINEAR SPACES AND ALGEBRAS

Theorem 4.3. If L and Q are piecewise compact, then so is L x Q. Proof: Let A c L x Q where A is bounded and closed. Let M be such that (X, Y) e A=> I (X, Y) I ~ M. Let {(Xn, Y,,)} be a sequence such that (Xn, Y,) eA. Consider {Xn}· Then II Xn I ~ II (X"' Yn) II ~ M. Let B be the set of terms of {Xn}· Then B is closed and bounded. Therefore B is compact. Therefore there exists a subsequence.

X""__,. XeB Now let E be the set of terms of { Y""}, the corresponding subsequence of { Yn}· E is closed and bounded. Therefore E is compact. Therefore there exists a subsequence

Yn

"•

-t-EE

Therefore (Xn , Yn ) __,.(X, Y) e A since A is closed.

"•

"•

Corollary 4.4. Cis piecewise compact. Proof· C = R x R and R is piecewise compact. Notation (1) Then dimensional Cartesian product

L1

X

L2

X £3 X • • • X

Ln = L1

X (£2 X (£3 X • • • X (Ln-1 X

Ln)

...)

(2) If each of the L 1 above are the same, then we write

L" = L

X

L

X ••• X

L,

to n factors.

Corollary 4.5. If L is piecewise compact then. L" is piecewise compact. If Lis a Banach Space, then L" is a Banach Space. Proof: By induction: (1) L 1 is piecewise compact by hypothesis. (2) If L" is piecewise compact, then L"+ 1 = L x L" is piecewise compact by Theorem 4.3. The same argument works for being a Banach space. The space R" is called n-dimensional Euclidean Space. The space C" is called n-dimensional Complex Euclidean Space.

36

FUNCTIONAL ANALYSIS

Problems 1. Prove that in an Abelian Banach algebra: n

{X+ Y)" =

L "C"X" yn-k k=O

where n is a positive integer and nck

n! = k!(n-k)!

2 (a) Prove that in any Banach algebra the following series converge for all X:

(b) Calling the sum in (a), exp (X), prove that exp (X+ Y) = exp (X) exp ( Y) in any Abelian Banach algebra. (You may assume with-

out proof that multiplication of absolutely convergent series is legitimate in a Banach algebra.) Definition for Problem 3

An inner product space I over R is defined as follows: (1) I is a linear space over R. (2) There is an operation (called "inner product") from I x I into R whose value for given (X, Y) e I x I will be written [X, Y], with the following properties: (a)

[X 1 +X2 , Y] =

(b) [X, Y]

[X~o

Y]+[X2 , Y]

= [Y, X]

[aX, Y] = a[ X, Y] [X, X] ;;; 0 (e) [X, X] = 0 => X = 0.

(c)

(d)

3. (a) In an inner product space prove Schwarz's inequality:

I [X, J:'] I ;a! .J<x. X) ../lY, Y)

37

NORMED LINEAR SPACES AND ALGEBRAS

II X II = J(X, X) an inner product space becomes a normed linear space. (c) Show that using the "dot product" of elementary vector analysis, R 3 may be regarded as an inner product space and that the definition in (b) gives the usual norm. (d) What about R"? (b) Show that with the definition

4. Prove that: (a) A= A (b) A is closed A = A (c) A uB =.Au B (d) Interior (A) is open (e) Prove that if A and Bare closed, then so are Au Band An B. 5. (a) Define inf A, where A c R

(b) Prove using Theorem 3.1, an analogue of Theorem 3.1 for infremums. 6. Define monotone decreasing sequence of real numbers and prove an analogue of Theorem 3.2 concerning such sequences. 7. (a) Show that L x Q is a normed linear space under the definitions given. (b) Show that

II

X

II

~

II (X,

Y)

II ; II

Y

II

~

II (X,

Y)

II ·

CHAPTER 3

Functions on Banach Spaces

I. Continuous Functions

In this section L and Q are two normed linear spaces over the same field F. Definition 1.1. Letfbe a function from A c L into Q. Then.fis continuous on A if:

XneA and Xn-+X=>j(Xn)-+f(X) Definition 1.2. Let.fbe a function defined on A, then the image of A under.f.f[A], is: f[A] = {YI Y =/(X) and XeA}

Theorem 1.1. Iffis continuous on a compact set A thenf[A] is compact. Proqf: Let Y" ef [A] be a sequence. We must show that { Yn} has a convergent subsequence. For each n, then exists X" e A such that

Y, =f(XJ hence for a suitable subsequence Xnk-+XeA Therefore Corollary 1.2. Let.fbe continuous on a compact set A. Then there is a number M > 0 such that

XeA=> llf(X)II ~M Proof: f [A] is compact => .f [A] is bounded. 39

40

FUNCTIONAL ANALYSIS

Corollary 1.3. Let Q be R, and let f be continuous on a compact set A. Thenftakes on a maximum and a minimum value on A. Proq.f: .f[A] is compact=> f[A] is bounded and closed. Let M = supf[A]. We must show that Mef[A]. For each n, there is Yn ef[A] such that

M

~

Y, > M-1/n

I Y,-M I< 1/n

i.e. Therefore,

Y, __,. M; and Y,, ef[A]

Therefore Mef[A] sincef[A] is closed, i.e. M=f(X) for some X e A. A similar proof holds for the minimum. Definition 1.3. f is uniformly continuous on A means that for each e > 0 there is a > 0 such that X',X"eA and IIX'-X"II llf(X'}-f(X")II <e

(Note that the value of l> is independent of the points X', X".) Theorem 1.4. on A.

Iffis uniformly continuous on A, then.fis continuous

Proof: Let Xn e A, X.,__,. Xe A. We must show thatf(Xn) __,. f(X).

Choose e > 0. Then there is a > 0 as in Definition 1.3. Now, there exists on N such that n >N-Il X,-XII < {J n > N -llf(X,,)-f(X) II< e

So, or

II

f(Xn) __,. f(X)

Corollary 1.5. If .f is uniformly continuous on A, X,, X,- X~ then l!f(X,)-f(X~

11-0,

11-0.

X~

e A, and

Proof: Choose e > 0. Then there exists a > 0 such that Definition 1.3 holds. But, there exists an N such that

Hence,

n>N=>I!Xn-X~II N => l!f(X,)-f(X~) II < e

41

FUNCTIONS ON BANACH SPACES

Faulty Proof of Corollary 1.5 using only continuity: Let

X~-+X.

Then, f(X11 ) -+ f(X) f(X~) -+ f(X)

Thus,

Hence, f (X11) - f (X:,) -+ 0; the fault lies in the fallacious assumption that X exists such that X~ -+ X. Example. Let A= {X I 0 < X~ 1} and let f(X) = 1/X. Then A. However it is not uniformly continuous since Corollary 1.5 is not satisfied.

f is continuous on For, let Then but

/(X11)-/(X~)

= n2 -n = n(n-l)-+ oo

Note. A is not a closed set and hence is not compact.

Theorem 1.6. Let f be continuous on a compact set A. Then f is uniformly continuous on A. Proof: (by contradiction). Suppose f is continuous on A, but not uniformly continuous on A. Then there is an e > 0, such that for every ~ > 0, there are X, X' e A such that II X- X' II < ~. but llf(X)-f(X')

II

~ E

Lete0 be such an e. Setting~= 1/n,for everypositiveintegern, there are numbers X11 , X~ eA such that II X"- X~ II < 1/n and llf(X.,) -/(X~) II ~ e 0 • From this we have X11 - X~-+ 0. But, since A is compact, {X~} has a subsequence {X~,.} such that X~,.-+ X eA. Then, X 11,. = (X11,. -X~,.)+ X~,.-+ 0 +X = X. Therefore,by the definition of continuity,/(X~,.)-+ f(X) andf(X11,.)-+ f(X). Thus, /(X11,.)-/(X~,.) -+/(X)-/(X) = 0. Hence, there surely is a number N ( = nk, for large enough k) such that IIJ(XN) -f(X~) II < e 0 • This is a contradiction, since llf(X.,)-f(X~) II ~ e 0 for all n.

42

FUNCTIONAL ANALYSIS

l. The Spaces

cF (a. b)

If a, b are real numbers such that a < b, then we write:

[a,b] =

The set [a,

b]

{XeRja ~X~ b}

is called a closed interval.

Definition 1.1. CF(a, b)= {!If is a continuous function from [a, b] into F}. (Note that in the case F = C, the members of CF(a, b) may be regarded as parametric representation of continuous curves in the complex plane.) In CF(a, b), we define: {1) (f+g) f(X)+g(X) (2) (af) (X) = af(X), (where a e F)

(X)=

(3)

III I =

sup lf(X)

a;1;X;1;b

I· I I

Note that, by Corollary 1.3, f

= lf(X0 ) I for some X 0 e [a, b].

With the above definitions, it is easy to see that the spaces CF(a, b) are normed linear spaces, e.g.:

ilf+g II= sup j(f+g)XI a;1;X;1;b

=

sup If( X)+ g(X) a;1;X;1;1>

~ sup [IJ<X)I a~X;1;b

~ sup al1!X;1;b

I

+ lg(X)j]

lf(X) I +

sup al1!X~I>

Ig(X) I

II! I + I g I il afll = sup I(af)(X) I = sup Iaf(X) I = sup lai·IJ(X)I al1!Xl1!1> =

Also:

a;1;Xl1!1>

al1!X;1;b

= jaj sup lf(X)I a;1;X;1;b

=

lal·ll!ll

The rest of the verification is left to the student.

43

FUNCTIONS ON BANACH SPACES

Theorem 1.1. ll!n-fll

-+

0 ¢> J, (X)-+ f(X) uniformly in (a, b].

Proof~: ll!,-!11 = sup lf,(X)-f(X)I. Choose

e > 0.

Then

a;!!X~b

there is N such that n > N~ ll!,-!11 <e

Then, for each X e [a, b], n > N ~ lfn(X)-f(X) I~ sup lf,(X)-f(X) :

a;!!X;!!I>

I= ll!,-!11 < e

Proof 0, then, by the definition of uniform convergence there is an N such that X e [a, b] and

n> N~

lin (X)-f(X) I < e

For each n, there exists a number, x e [a, b] such that sup lf,(X)-f(X) I= lf,(X")-f(X") I

Xe[a,l>)

Hence, n > N ~ llf,-fll <e.

Theorem l.l. CF(a, b) is a Banach space. Proof: Let {J,} be a Cauchy sequence. Then lim llfn-fm n-+oo m-+oo

II= 0

Now, for each Xe[a,b], lfn(X)-fm(X)I ~ llfn-fmll-+0. Therefore, for each X e [a, b] the sequence {J, (X)} of elements of F is a Cauchy sequence. For each X e [a, b], there is a number f(X) e F such thatf,(X)-+ f(X). But, for each e > 0, there is anN such that m, n > N ~ lfn(X)-fm(X)

I< e

Letting m-+ oo: n > N~ lfn(X)-f(X) I~ e. Therefore {J,} converges uniformly to f. Therefore, by an elementary theorem, f is continuous. (Cf. e.g. Taylor's Advanced Calculus.) In other words, feCF(a,b). F.inally, by Theorem 2.1 ll.f..-fll-+0. This proves that CF(a, b) is a :Qanach Space.

44

FUNCTIONAL ANALYSIS

3. Continuous Operators Let L and Q be given normed linear spaces over F.

Definition 3.1. An operator is a function from L into Q. Note that an operator must be defined on the whole space. Definition 3.2 Afunctional is an operator from L into F. Definition 3.3. A operator T is called linear if: (1) T(X+ Y) = TX+TY. (2) T(aX) = aTX.

Definition 3.4. An operator is bounded if there is an M > 0 such that ~Mil For linear operators. boundedness and continuity are equivalent as is shown in the following theorem:

I TXll

XII·

Theorem 3.1. The linear operator Tis continuous Tis bounded. Proof=:

I TX-TXn I = I T(Xn-X) I ~ M I (Xn-X) I

Therefore, X,

--+

X

~

TX,

-+

TX.

Proof~.- Suppose Tis continuous, but not bounded. Then, for each positive integer n, there is a point X" e L such that

Let

I TXn I > n I X, I 1

Y,=

nllX,Ilx"

1 1 n I Xn I I TY, I =II~ I X, I T(X,) I = n I X,~ I TXn I > n I X, I = 1 I.e. I TY,Il > 1 1 1 But. I Y, I = I n X, I · I X,. I = ;; so that Yn -+ 0. By continuity, TY, -+ 0 so that I TY" I ultimately

Then.

must be < 1. This is a contradiction.

45

FUNCTIONS ON BANACH SPACES

Corollary 3.2. If T is linear and Yn continuous operator.

-+

0

~

TY"

-+

0, then T is a

Proof: The proof just above that continuity of a linear operator implies boundedness uses only the fact that Yn -+ 0 ~ TY" -+ 0. Hence the proof applies to any T satisfying the hypothesis of the present Corollary. Hence, such a T must be bounded, and therefore, by Theorem 3.1, continuous.

4. Spaces of Bounded Linear Operators The linear operators considered in this section will be bounded and hence continuous. (By Theorem 3.1 continuity and boundedness are equivalent for linear operators.)

Definition 4.1. [L, Q] = {T I Tis a bounded linear operator from L into Q}. For the sake of brevity, we will denote [L, L] by [L ]. In this section it will be shown that, with suitable definitions of +, ·, and norm, [L, Q] forms a normed linear space if Land Q are normed linear spaces, and if in addition Q is a Banach space, then [L, Q] forms a Banach space, whether or not L is one. It will further be shown that L a normed linear space implies [L] a normed linear algebra, and L a Banach space implies [L] a Banach algebra. In what follows Land Q are normed linear spaces unless otherwise stated. We must first define a function + from ([L, Q] x [L, Q]) into [L, Q] and a function· from (F x [L, Q]) into [L, Q].

Definition 4.2. Let T, T' be operators. Then: T + T' is defined by (T + T')X = T X+ T' X, and for a e F, aT is defined by (aT)X = a(TX). Theorem 4.1. If T, T' are linear, so are T+T' and aT. Proof: We must show that the two requirements of Definition 3.3 are satisfied. First for T + T' : (I) (T+T')(X+ Y) = T(X+ Y)+T'(X+ Y)

(from Def. 4.2)

= (TX+TY)+(T'X+T'Y) (since each operator

is linear)

= (TX+T'X)+(TY+T'Y) (Q is a linear space) (from Def. 4.2) = (T+T')X+(T+T')Y

46

FUNCTIONAL ANALYSIS

= T(bX)+T'(bX)

(2) (T+T')(bX)

= b(TX+T'X) = b(T+T')X

(Def. 4.2) ~each operator linear) (Q a linear space) (Def. 4.2)

= a(T(X+ Y)) = a(TX+TY) = a(TX)+a(TY) = (aT)X+(aT)Y

tDef. 4.2) (Tlinear) (Q a linear space) (Def. 4.2)

= a(T(bX) = ab(TX)

(Def. 4.2) (Tlinear) (F abelian) (Def. 4.2)

= bTX+bT'X

Next for aT: (1) (aT)(X + Y)

(2) (aT)(bX)

= ba(TX)

= b(aT)X)

Theorem 4.2. T, T' e (L, Q] => T+T' e (L, Q] and aTe (L, Q].

Proof: To be in [L, Q] means to be linear and bounded. Theorem 4.1 shows that these elements are linear; we must now show them to

be bol,lllded. ll 0. Then there is anN such that: m,n >

N-=:.11 Tm-Tn II< E

m, n > N

=:.II TmX -T.,X I = I (Tm-T.,)X I ~ I Tm-Tn 11·11 X I < E I X I

50

FUNCTIONAL ANALYSIS

Letting, n

-+

oo, m >N=> II TmX-TXII ~e~XII m > N=> IICTm-T)XII ~ell XII

m > N=> IICTm-T)XII ~e IIXII m > N =>II Tm- T II ~ e (by Corollary 4.6) Therefore, Tm-+ T, and [L,

Q] is a Banach space.

Definition 4.4. If T, T' e [L] then (T · T')X = T(T' X). Corollary 4.9. T, T' e (L] => T · T' e [L]. Proof: (1) (T· T')(X+ Y) = T(T'(X+ Y)) = T(T'X+T'Y) = T(T' X)+ T(T' Y) = (T· T')X+(T· T')l' (2) (T · T')(aX)

= T(T'(aX))

= T(aT'X) = a(T· T')(X)

II ~

II T 11·11 T'X II ~ IITII·IIT'II·IIXII

(3) II (T· T')(X) II = II T(T'X)

which proves that T · T' is bounded. But (1 ), (2), (3) imply that

T· T'e [L]. The proof of (3) just given proves:

Corollary 4.10. II TT'

II

~

II T 11·11 T' II·

Definition 4.5. The operator I o([L] is defined by IX

= X.

Corollary 4.11. lis a bounded linear operator. Corollary 4.12. IT= TI the student.)

= Tfor Te [L]. (The proofs are left for

Corollary 4.13. II I II = 1 Proof: III II =

sup II IX II =

IIXII=l

sup II X II = 1

IIXII=l

Cor:ollary 4.14. [L] is a normed linear algebra, (the proof is left for the student).

51

FUNCTIONS ON BANACH SPACES

Corollary 4.15. If Lis a Banach space, [L] is a Banach algebra. Proof: [L] is a normed linear algebra from Corollary 4.14 and a Banach space from Theorem 4.8. Cor~llary 4.16. If Tn e [L] and Te [L] and Tn-+ T and XeL then TnX-+ TX.

I

Proof· TnX-TXII But Tn -+ T => Tn-T

I

=II

I·

(Tn-T)XII ~II Tn-T II·~ X -+ 0. Therefore

I

II TnX-TXII-+ 0

XII·

implying TnX-+ TX.

5. Existence Theorems for Integral Equations and Differential Equations In this section we write C(a, b) to mean CR(a, b).

Definition 5.1. An acceptable Fredholm kernel function is a continuous mapping, k, from [a, b] x [a, b] into R such that M(b-a) < 1

where M is the maximum value taken on by k(x, y) in [a, b] x [a, b ]. (Note that [a, b] x [a, b] is a square in the Cartesian plane. Since k(x, y) is a continuous function on a compact set, it assumes a maximum value.) We now define K, the Fredholm operator associated with k. Given Xe C(a, b), KX = Ymeans: Y(s)

=

S:

k(s,t)X(t)dt

The question arises: Does Y e C(a, b)? Let sn Y(sn)- Y(s) =

S:

-+

s then

[k(sn, t)-k(s,t)] X(t)dt

Since k is continuous on a compact set, k is uniformly continuous on [a,. b] x [a, b]. Choose e > 0. Then there is a ~ > 0 such that if II (sno t)-(s, t) < ~then

I

E

Ik(sn, t)-k(s, t) I< 211 X I (b-a)

52

FUNCTIONAL ANALYSIS

(Note that if II X II

= 0, Y = 0 and proof ~s trivial.)

II (s11, t)-(s, t) II = ..j(s11 -s)2 = IS11 -S I Therefore

ls~~-sl < D=> II k(s ,t)-k(s,t)ll < lll X ~~b-a) 11

But Sn-+ s. Therefore there is anN such that n > N =>I Therefore

n > N =>I Y(s11) - Y(s)l =I

J:

Sn-S

I 1, where a;:;;; t;:;;; s;:;;; b

kn(s,t)X(t)dt

Proof' By mathematical induction: (a) for n = 1, the result is obvious from the definition of K. (b) assume the result is true for n = q then Y

= Kq+tx =

Y(s) =

J: ff

K(KqX)

k(s,u{J: kq(u,t)X(t)dt]du

from the definition of K and the induction hypothesis. Hence, Y(s) =

k(s, u) kq(u, t}X(t)dtdu

Now the order of integration can be changed if the limits are selected properly. If this is done, (see figure below) Y{s) =

a.

Ef

k(s,u)kq(u,t)X(t)dudt

s

57

FUNCTIONS ON BANACH SPACES

Hence,

Y(s) =

or

Y(s) =

J:

X(t{J: k(s,u)kq(u,t)du]dt

J: kq+ (s,t)X(t)dt 1

by the definition of kn(s, t). Theorem 5.11. Let M = sup I k(s, t) I where the supremum is taken over all (s, t) on which k is defined, i.e. all (s, t) such that a ~ t ~ s ~ b. Then M"(s-t)"- 1 lkn(s,t)l~ (n-1)! Proof: By mathematical induction:

(a) For n = 1, the result is obvious from the definition of M and the definition of supremum. (b) Assume the result is true for n = q. Then jkq+t(s,t)l

~ 1J: k(s,u)kq(u,t~dul

by the definition of kn. Therefore

f

Ikq+t (s, t) I~ Ik(s, u) 1·1 kq(u, t) Idu ~ (:-~;!r (u-t)q-t du by the induction hypothesis. Hence Mq+t (u-t)J 5 Mq+l lkq+ 1 (s,t)l ~( _ ) 1- = -1-(s-t)q q 1. q ' q.

Corollary 5.12. Proof· Theorem 5.11 and the fact that (s-t)

Theorem 5.13. Proof: Kn

=

sup

IIXII=l

II Knx I

~

(b-a).

58

FUNCTIONAL ANALYSIS

Now let Y

= K"Xfor II

Then

X

II =

f • I

Y(s) =

k.,(s, t)X(t)dt

IY(s)l~"

Hence,

l.

by Theorem 5.10.

M"(s t)"- 1 (n-=-l)! (l)dt

by Theorem 5.11 and since II X II = l. M" I -(n-1)!

I Y{s) :S - -

[-(s-t)"]" n

,.

=

M"(s-a)" n!

M"(b-a)"

:S ---'--n!

II K"X il ~ M"(b~a)"

Hence for all X such that

II X II

Therefore,

n.

= 1.

~ M"(b,-a)"

II K" 11

from the definition of II K"

n.

II . 00

L K" converges to an operator H such

Theorem 5.14. The series '

n=O

= (1-K)H =I.

that H(I-K)

Proof" By Theorem 5.13 and the comparison test 00

(

since

) L M"(b-a)" converges to eM , 1 11.

n=O

00

00

L

II

L

K" II converges. Hence,

n=O

K" converges, since K e [ C(a, b)]

n=O

which is a Banach space and from Corollary 2-1.4, in a Banach space, if

00

00

n=O

n=O

L II X, II converges, then L X, converges.

The rest of the proof follows as in Theorem 2-1.7: 00

00

n=O

n=1

H=

L K" = 1 + L

m

Hence, -HK =-lim

m

r K. K"

m~oon=O

m

m

-KH =-lim

oo

L K"·K =-lim L K"+ 1 = - n=l L K"

m~~n=O

Also

K"

m~oon=O

= -lim

oo

L K"- 1 =- n=l L K"

m~oon=O

59

FUNCTIONS ON BANACH SPACES 00

Thus

-HK= -KH

and H(I-K)=(l-K)H=H-

L Kn n=l

Therefore, H(I-K) = (1-K)H =I. Theorem 5.15. (Existence and uniqueness theorem for the Volterra equation.) For each Y e C(a, b) there is exactly one X e C(a, b) such that Y = (I- K)X. In fact, X= HY =

(~ 0K")Y = n~o Knl'

Or Proof: Exactly as for the Fredholm equation. See Corollary 5.4, Theorem 5.5, Corollary 5.7 and Corollary 5.9.

Theorem 5.16. (Existence and uniqueness theorem for second order linear differential equations with initial conditions.) Let X, A1o A 2 e C (a, b) and let m and n be any real numbers. Then there exists exactly one function Y e C(a, b) such that: (1) Y"(s)+A 1 (s)Y'(s)+A 2 (s)Y(s) = X(s), {2) Y(a) = m,

(3) Y'(a)

= n.

Proof: Assume that we have a Y satisfying these conditions and define Z(s) = Y"(s).

f

Then Also,

Z(t)dt

= Y'(s)-l''(a) = Y'(s)-n

ffz(t)dtdu= f[Y'(u)-n]du= Y(s)-Y(a)-n(s-a)

fI

Hence

Z(t)dtdu = Y(s)-m-n(s-a)

However, by changing the order of integration

ff

Therefore,

Z(t)dtdu

=

ff

Z(t)dudt

Y(s)-m-n(s-a) =

f

=

f

Z(t)(s-t)dt

Z(t)(s-t)dt

60

FUNCTIONAL ANALYSIS

Substituting in the differential equation,

Z(s)+A 1 (s)[n+ Ez(t)dt] +A 2 {s{m+n(s-a)+

I

Z(t)(s-t)dt] = X(s)

This can be written,

Z{s)-

I [-A 1 (s)-A~(s)(s-t)]Z(t)dt = X(s)-nA 1 (s)-A 2 (s)[m+n(s-a)]

But this is a form of the Volterra integral equation

~(s)-

I

k(s,t)Z(t)dt

= V(s)

which has a unique solution as given in Theorem 5.15. Moreover, as is easily seen, if Z(s) satisfies this equation, then

Y(s) =

J:[f

Z(u)du+n]dv+m

satisfies (1), {2), and {3).

6. The Hahn-Banach Extension Theorem In this section L is a fixed normed linear space over F, where F can be either R or C. Suppose that we have V1 c: V2 c: L where V1 and V2 are linear subspaces of L. Let/1 be a function defined on V1 and/2 a function defined on V2 • Then/1 c: / 2 means (X, Y)eft =>(X, Y)efz or

!1 (X) = Y => fz (X) = Y

That is, / 2 is an extension of / 1 if for each X for which / 1 (X) is defined,f2 (X) is also defined andf2 (X) = / 1 (X).

61

FUNCTIONS ON BANACH SPACES

Theorem 6.1. Let Vt c: V 2 c: L, where Vt and V 2 are linear subspaces of L. Let ft and / 2 be bounded linear functionals (cf. Definition 3.2) on Vt and V 2 respectively, and letf1 c:/2· Then lift II~ IIJ2ll· Proof:

lift II

= sup lit (X) I (by definition) XeVt

IIXII=t

j/2 (X) I

sup

=

XeVt

(because / 1 = / 2 on Vt)

IIXII=t

~

sup lf2(X)I

=

IIJ2II

XeVz

IIXII=t

Theorem 6.2. (Hahn-Banach extension theorem.) Let V be a linear subspace of L and let g be a bounded linear functiona1 on V. Then, there is a bounded linear functional G => g, on L, such that II II g II·

Gil=

Proof: (Case 1, F = R.) Let fF be the family of all.fsuch that: (1) /is a bounded linear functional on W which is a subspace of L.

(2) g c: f.

II 1 II = II

o II .

Lemma. fF is inductive. Proof ofLemma: Let~ be a chain,~ c: !F. Let k =

U f.

We want

le'G

to prove k e !F. k is a function (that is, (X, a) e k and (X, b) e k => a = b)

because; (X, a) eft e ~ and (X, b) e/2 e ~. and since~ is a chain one of these functions must be an extension of the other. Therefore, a = b. Now, k(X) =a there is anfc: ~such thatf(X) =a. For each fe~. let W1 be the space on which/is defined. Let 1f" = {W1 lfe ~}. Let Wk = U W1 = U W. Wk is the set on which k is defined. W" fe'G

We11'

is a linear subspace of L. For, let X, y E wk. Then X E Wit• y E w/2. If, say, W 1 , c: W 12 , then X, Ye W 11 • Therefore, aX+bYe W 1 , and aX +byE wk so wk is a linear subspace of L.

62

FUNCTIONAL ANALYSIS

To complete the proof of the lemma we must show (1) k is a linear functional on W", (2) g c: k, and (3) k is bounded and I k II = II g II· X, YeW"=> X, Ye W1 z

=>.f2(aX+bY) = af2(X)+b/2(Y) => k(aX +bY)

= ak(X)+bk(Y)

Thus k is a linear functional on W,.. Let g(X) = a for some X e V. For each .fe ~. f(X) = a, that is (X, a) ej. Thus, (X, a) e k. Therefore, k(X) =a, or in other words k ::::>g.

For each X e W", there is anfe ~such that

I k(X) I =

lf(X) _I

;;;; II f I . II X II

=

I g II . I

X

II

Thus, k is bounded and I k II ;;;; II g II· But by Theorem 6.1, 11 ~ 11 g 11 so that 11 k 11 = 11 g 11 . So k e fF and this completes the proof of the lemma. Returning to the proof of the theorem, we see by Zorn's lemma that fF has some maximal element G. Since G e !IF, I G I = II g !!.and 11

k

G::::>g.

Let Wa be the space on which G is defined. We want to prove Wo =L.

Suppose there is a Z 0 e L and Z 0 ¢ W 0 • Let Q = {X+aZ 0 IaeR and Xe W 0 }. Then, Q is a linear subspace of L because (a(X1 +atZ0 )+b(X2+a2Z0 ) ) = (aX1 +hX2)+(aa1 +ba2)Z0 e Q Anytime X 1 +a 1 Z 0 = X 2+a2Z 0 then X 1 = X 2 and a 1 = a 2 because

Then, But, we are supposing that Z 0 ¢ W0 • Hence, a 1 = a 2 and X 2-X1 = 0 or X 2 = X1 • We defineR: H(X +aZ 0 )

= H(X)+aH(Z 0 ) =

G(X)+at

63

llUNCTIONS ON BANACH SPACES

where t is some real number which will be determined. Obviously with any choice oft, His a linear functional on Q. H will also be bounded if we can arrange matters so ~hat

IH(X + aZ0 ) I ~ I g 11·11 X+ aZ0 li Consider the case when a > 0. Multiplying the inequality H(X+aZ0 ) ~

1 g 11·1] X+aZ0 I

by 1/a we get

!H(X +aZ0 ) ~~II g 11·11 X +aZ0 I a a

~H(~+Zo) ~ llo 11·11~+Zo II !loll·llu+Zoll forallue W 0 ~G(u)+t ~ llo 11·11 u+Z 0 I for all ueW0 t ~ -G(u)+ llo 11·11 u+Z 0 I for all ue W0 = bg(aX0 )

FUNCTIONS ON BANACH SPACES

67

show that g is a linear functional on V. The equalities

Ig(aXo) I = Ia I Xo Ill = Ia Ill Xo I = I aXo I show that g is bounded. By the Hahn-Banach Extension Theorem then, there exists G a bounded linear functional on L.

~ g,

Corollary 6.4. Let X 0 e L, X 0 =F 0. Then, there is a bounded linear functional G on L such that G(X0 ) =F 0. Proof: Use the G given by Corollary 6.3.

Corollary 6.5. If X 0 eLand G(X0 ) = 0 for every bounded linear functional G on L then, X 0 = 0. Corollary 6.6. Let X1 , X 2 e L and X1 =F X 2 • Then, there is a bounded linear functional G on L such that G(X1) =F G(X2 ). Proof. Let X = X 2 - X to X =F 0. Use Corollary 6.4 to obtain a bounded linear functional G such that G(X) =F 0. Then G(X2-X1) =F 0. G(X2)-G(X1) =F 0. G(X2) =F G(X1 ).

7. The Existence of Green's Function

We now show how the Hahn-Banach extension theorem can be applied to a classical problem: the existence of Green's function. A bit more demand on the reader's knowledge of analysis will be made in this section than in the remainder of the book. This section can be entirely omitted without disturbing continuity. As usual·we write for the Laplacian operator

fP

fP

v2 = ox2+ oy2 Deflnltion·7.1. u is harmonic in the open set D c R 2 means = 0 on D, and o2ufox2, o2ufoy 2 are continuous in D.

V2u

68

FUNCTIONAL ANALYSIS

Theorem 7.1. u is harmonic in D (analytic) in D.

~

3 v such that u+iv is regular

Proof: See any book on complex variable theory. Domain means open connected set. (Cf. Problem 14 below.) By the boundary of an open set D, we mean the set 15-D.

Theorem 7.1. (maximum modulus theorem). Letf(z) be analytic

in the bounded domain D and its boundary. Then,lf(z) I takes on its maximum value on the boundary of D. Proof: Cf. proof of Theorem 7.1.

Theorem 7.3. If u is harmonic on 15 where D is a domain, (i.e. in some open set E:=J15) then u takes on its maximum and minimum values on the boundary of D. Proof: It suffices to consider the maximum, since u is harmonic if and only if -u is. Choose v such that u+iv is analytic in D. Then, if+lv is analytic in D. llf+fv = ff' = eu takes .on its maximum on the boundary; hence so does u. Let D be a domain and let its boundary be the curve M; let QeD. Then, GQ is a Green's Function on D if: (1) GQ(P) = -In I P- Q I +k(P, Q) in D. (2) GQ(P) = 0 on M. (3) GQ is continuous on 15- {Q} and harmonic on D- {Q}. Let f be continuous on M, and let Gp be a Green's function on D. Furthermore, let

I

I"., I

u(P) = _!._ f(Q) oG~(Q) ds 2nJM on

f

Then, u = f on M, and u is harmonic in D: Cf. Nehari, Conformal Mapping. Thus, if the existence of a Green's function on D can be demonstrated, it will follow that Laplace's equation V 2 u = 0 is solvable subject to arbitrary continuous boundary conditions. It is easy to see that such boundary conditions determine a unique solutien. For otherwise, iful> u2 are two solutions, then V 2(u 2 -u 1) = 0 in D and u2 -u1 = 0 on M. Hence, by Theorem 7.3, u2 -u 1 = 0 in D.

FUNCTIONS ON BANACH SPACES

69

We shall sketch a proof, due to Peter Lax (Proc. Am. Math. Soc., 3 (1952), pp. 526-31) of the existence of a Green's function on any Cauchy domain D. (For the meaning of the term Cauchy domain, cf. Definition 5-3.2.)

Theorem 7.4. Let D be a Cauchy domain with boundary M. Then for each Q e D, there is a Green's function GQ on D. Proof· Let B be the set of all continuous functions from M into R. Then, exactly as for CR [a, b], it is easy to see that under the definitions:

{f+g)(X) =J(X)+g(X) (af)X = a·j(X)

III I = Xsup !f(X) I eM B becomes a normed linear space and, in fact, a Banach space over R. Let B 0 be the set of allje B for which 3u such that u = f on M and u is harmonic in D. (E.g. Ifjis a constant! e B0 .) By the above remarks on the uniqueness of solutions, for each f e B 0 , there is exactly one harmonic function u1 which extends f to D. Let us note that (since V 2 (au+bv) = a V 2 u+b V2 v), B 0 is a linear subspace of B. Let Q be some fixed point in D. Let rQ be defined forfe B 0 , by

Since, u J+g = u 1 + u,, u" 1 = au 1 , by uniqueness and by linearity of the Laplacian, rQ is a linear functional on B 0 • Moreover, by Theorem 7.3,

~ sup lf(z)l ::eM

=

11!11

Hence rQ is a bounded linear functional; moreover, II rQ II ~ l. But, I rQ(1) I = l. Hence, II rQ II = 1. By the Hahn-Banach extension theorem (Theorem 6.2), there is a bounded linear functional RQ ~ rQ, defined on B, such that II RQ II = I.

70

FUNCTIONAL ANALYSIS

Next, for each P in the plane, let g P be defined by gp(z)

for z eM. If P ¢ to D, u9 P where

=In Iz-PI

15, then gp e B 0 , since it has the harmonic extension u9P(Q) =In IQ-PI

Thus, Whether or not P e that

rQ(gp) =In

IQ-PI

15, so long asP¢ M,

gp is continuous on M, so

P¢M=>gpeB

Hence, for P ¢ M, we may define kQ by kQ(P) = RQ(gp)

For P e M, we define

I

kQ(P) =In Q-PI

Now, the Green's function GQ may be defined by GQ(P) =

-In!P-QI

+kQ(P)

Then, GQ clearly satisfies conditions (I) and (2) in the defining characterization of the Green's function. In order to verify condition (3), we shall require the lemmas:

Lemma I. The operators RQ and V2 commute: V 2 RQ = RQV 2 • Lemma 2. Let z eM, and P 0 eM. For each P e D, let P' be the mirror image of P in the tangent line to M at the point of M nearest to P. Then, z-P'! lim l = 1

P-+Polz-P

I

uniformly in z. Assume for the moment, the validity of these lemmas; we may proceed as follows: In D, V 2 kQ(P) = V 2 RQ(gp) = RQ \i 2 gp =RQO

=0 Hence, GQ is harmonic in D- { Q}.

FUNCTIONS ON BANACH SPACES

71

We shall show that GQ is continuous on 15- { Q} by showing that it is continuous across M. By Lemma 2, if P and P' are related as in the hypotheses of Lemma 2,

-~~yJm\:=~,11 j-.o That is Hence,

as P-+P 0

I gp-gp·ll -+ 0 asP-+ P IkQ(P)-kQ(P') I= IRQ(gp)-RQ(gr) I =I RQ(gp-gr) I ~II RQII·II gp-gp·ll =II gp-gr 11--. o 0•

MP-+~.

.

It remains only to prove Lemmas I and 2.

Lemma I is an immediate consequence of: Lemma 3. Let/ be a continuous function on M x C. Let L be a bounded linear functional on B. For z e C let z = x+iy. For each z e C, let of fox be continuous on M. Then ofox(Lf) exists and equals L(of/ox). Proof of Lemma 3. Let g(x,y)

= Lf(P, x+iy). Then

g(x,yo)-g(xo.Yo) =_I_ [Lf(P,x+iy )-Lf(P,x0 +iy0 )] 0 x-x0 x-x0 1 X-X 0

= - - L[f(P,x+iy0 )-f(P,x0 +iy0 )] = L[.f(P,x+ iy 0 )-f(P,x 0 + iy 0)]

x-x 0 g(x,yo)-g(xo.Yo) x-x 0

Hence,

Lf (P x

,Xo

+'tYo ·)

-f (P . )] ,Xo+ryo L[f(P,x+iyo)-f(P,xo+iyo) X-X .

x

0

-+ L (0) = 0 because L is continuous.

72

FUNCTIONAL ANALYSIS

Proof of Lemma 2: Given e > 0. We must determine a number {) > 0 such that P-P 0 < {) ~

I

I

11:=~'1-ll < Since

.J; is continuous at x lx-llO~I.J;-ll<e

. 4sint 1nn-2-=0 t-+0 cos t there is a number q 2 such that 4lsintl ltl T(X+ Y) = 0 =>X+ YeKT.

(3) Suppose X e KT. XeKT => TX = 0 => aTX = 0 => T(aX)

= 0 => aXeKr.

Definition 1.3. L and Q are isomorphic if there is a one-one homomorphism from L on to Q. Problem: To find all (up to isomorphism) homomorphic images of L. 77

78

FUNCTIONAL ANALYSIS

Definition 1.4. Let V be a subspace of L. Then, X= Y mod V means X- Y e V. This is a relation on L

{(X, Y) I X= Ymod V} Corollary 1.1. X

c:

L xL

= Y mod V is an equivalence relation on L.

Proof: (1) X= X mod V since X- X

= 0e

V. Hence the relation is reflexive.

(2) X= Ymod V=> X- Ye V => Y-Xe V=> Y Hence it is symmetric. (3) X

= Xmod

V.

= Y mod V and Y =Z mod V => X- Y e V and

Y-Ze V=> X-Z =(X- Y)+{Y-Z)e V=> X= Zmod V Hence it is transitive. Recalling the notation of section 3, Chapter 1 :

[X]= {YeLl X= Ymod V} we have at once by Theorem 1-3.1:

Corollary 1.3. (1) X= Y mod V ~[X] = [ (2) Xe [X]. (3) (X] n (Y] -# 0 =>[X]= (Y]. Definition 1.5. L/V = {[X]

I X e L},

i.e.

Y].

Lf V is the set of all

equivalence classes.

Corollary 1.4. (l) X= X' mod Vand Y Y'mod V=> X+ Y =X'+ Y'mod V.

=

(2) X= X'mod V=> aX= aX' mod V.

Proof: (1) (X+ Y)-(X'+ Y') = (X-X')+(Y- Y')e V. (2) aX -aX' = a( X- X') e V.

Definition 1.6. [X]+[Y] =[X+ Y] and a [X]= [aX]. Corollary 1.5. L{Vis a linear space with the definitions in Definition 1.6.

· Proof: Left to the reader.

HOMOMORPHISMS ON NORMED LINEAR SPACES

79

Theorem 1.6. The operator Tdefined by TX = [X] is a homomor· phism from L on to L/V with kernel V. Proof: We must show that Tis a linear operator.

= [X+ Y] = [X]+[Y] = TX+TY T(aX) = (aX] = a[ X] = aTX

T(X+ Y)

I

I

Finally, KT = {XeL TX = 0} = {XeL [X]= 0} = {X e L X 0 mod V} = .{X e L X -0 e V} = {X E L X E V} = v.

=

I I

I

Theorem 1.7. LetS be a homomorphism from Lon to Q. Then

Q is isomorphic to L/ K8 • Lemma I. (X]= (X']=> SX

=

SX'.

=

Proof of Lemma: [X] = [X'] => X X' mod Ks => X- X' e Ks => S(X-X') = 0 => SX -SX' = 0 => SX = SX'.

Lemma 1 enables us to define: R[ X] = SX. Lemma 1. R[ X] = R[ X'] => [X] = [X'], i.e. R is one-one. Proof of Lemma 2: R[ X] = R[ X'] => SX = SX' => S(X- X') => X- X' e Ks => (X] = (X'].

=

0

To complete the proof of the theorem, we must finally show that sums and scalar products are preserved. R([X]+[Y]) = R([X+ Y]) = S(X+ Y) = SX+SY = R[X]+R[Y] Also

R(a[ X])

=

R([aX])

=

S(aX)

=

aS X == aR[ X].

Therefore Q is isomorphic to L/ K8 •

1. Norms in a Quotient Space In this section L is a normed linear space and Vis a subspace.

Definition 1.1.

ll[xJ II= inf I z I ZeX

80

FUNCTIONAL ANALYSIS

Theorem 2.1. Let V be a closed subspace of L. Then Definition 2.1 makes L/V a normed linear space. Proof: L/V is a linear space by Corollary 1.5. We must show that LfV is normed, i.e. must show:

I [X] I = inf I Z I ~ 0. Obvious, since I Z I ~ 0. (2) I [X]+[Y] II~ I [X] II+ I [Y] I By definition, I [.A1 + [Y] II= I [X+ Y] II= inf I Z I (1)

Ze[X]

Ze[X+f]

But,

X1 e [X] and Y1 e ( Y] => X1 + Y1 e (X+ Y]. Hence, we have

II[X]+[Y] II~ r,e[f] inf II X +Ytll ~ inf [II Xdl +II YdiJ r,e[YJ 1

X 1 e[X]

IIXdl + inf II Ydl = !I[XJII + II[YJ II !l[aXJII=Iaiii[XJII. Ze[aX] =>Z =aX mod V=>~ Z =X mod V=>Z =a [~z] = inf

Xt e[X]

(3)

X 1 e[X] ft e[f]

1

where So

-ZeX

a II [aX] II

= Z e[aX] inf I Z II = iuf II aX 1 11 X e[X] 1

==I a Ix,inf I X til= Ia Ill [XJII I [X] II= o~[x] = 0. II

I

I

I

IIYn+p-Ynll

=II (Yn+p- Yn+p-t)+(Yn+p-1- Yn+p-2)+. • .+(Yn+l- Yn) I

I

82

FUNCTIONAL ANALYSIS

and remembering that we had p


Q =>II. xp-XN I < e/2. Therefore p > Q => I XP- X I < e.

3. Homomorphisms on Normed Linear Algebras Definition 3.1. Let L, Q be normed linear algebras; Tbe a homomorphism from L as a linear space onto Q as a linear space. Then T will be called a homomorphism from the algebra L onto the algebra Q if T(XY) = (TX)(TY)

Definition 3.2. A proper subspace V of a normed linear algebra is called an ideal in L if XeV

and

YeL=>XYeV

and

YXeV.

Note that L itself is not considered an ideal in L.

Theorem 3.1. Let T be a homomorphism from the normed linear algebra L onto the normed linear algebra Q. Then KT is an ideal in L. Proof: Letting X e Kn Y e L, we must show that XY e KT and YXeKT. Since X e Kn TX = 0, so that T(XY) = TX· TY = 0 · TY = 0. Therefore XY e KT. Similarly T(YX) = TY · TX = TY · 0 = 0 and YXeKT.

HOMOMORPHISMS ON NORMED LINEAR SPACES

83

It remains to be shown that KT "I= L. Suppose that KT = L. Then, e e KT. Hence Te = 0 in Q. ButTe must be the multiplicative identity in Q. (For TX · Te T(Xe) = TX.) Hence Te = 1, whereas

=

I 0I =

I

I

0 which is a contradiction. In what folJows let L be a normed linear algebra and let V be an ideal in L. Theorem 3.2. If X XY= X'Y'mod V.

= X' mod

V and Y

= Y' mod

V then

Proof: XY-X'Y' = (XY-XY')+(XY'-X'Y')

= X(Y- Y')+(X-X')Y'e V The notation [X] will be understood as above. Definition 3.3. [X] · ( Y] = [ XY). Theorem 3.3. If Vis· nn ideal in L, then L/ V forms a linear algebra.

Proof· Left to the reader. Theorem 3.4. The mapping TX = [X] is a homomorphism from L on to L/ V with kernel V.

Proof· Left to the reader. Theorem 3.5. LetS be a homomorphism from Lon to Q. Then Q is isomorphic to LfK 5 •

Proof· Left to the reader. Theorem 3.6. Let L be a normed linear algebra and V a closed ideal in L. Then, under the previous definition Lf Vis a normed linear algebra. Moreover, [X] ~ X

I

I

I II·

Proof" We have only to prove the following statements:

I [e] II= 1,

and

I [X] II~ I XII I [X]. [Y] II~ I [X] 11·11 [Y] II·

84

FUNCTIONAL ANALYSIS

I [X]. [Y] I = I [XY] I = Ze[Xf] inf IIZII

But,

~ inf

z,e[X]

I Z1 Z2 I

Z2 e[Y]

~ inf

Zte [X]

I Z 1 I Z2inf e

[f]

II Z2ll

~ I [X] 11·11 [Y] I Also,

II[XJII =

inf

Ze[X]

liZ II~ !lXII

In particular,

I [eJ II~ I e II= 1 Moreover

I [e] I

"I= 0 since

!l[e]!I=O=>[e]=[O] whereas [e] has an inverse and

[0] doesn't.

II [e] · [e] I ~II [e] 11·11 [e] II II [eJ II ~ I [eJ 1 2

And,

so that

1 ~ !l[e] II·

i.e.

I [eJ II ~ t

Since

and

I [eJ I ~ t; II [eJ I =

t.

Corollary 3.7. If in Theorem 3:6, L is a Banach algebra, so is L/ V.

4.

lnve~es

of Elements in Normed Linear Algebras

In this section L is an Abelian normed linear algebra.

Theorem 4.1. If X e Vand Vis an ideal then X has no inverse. Proof: Let X have an inverse x- 1. Then, X E v => e = xx-l E v. If e e Y, eYe V for all Y e L and V = L. This contradicts the definition of ideal.

85

HOMOMORPHISMS ON NORMED LINEAR SPACES

Theorem 4.2. If for every ideal V, X tf: V, then X has an inverse.

I

Proof: Let Q = { XY Y e L} and the following are true: (l) Xe Q, (2) XY1 + XY2 = X( Y 1 + Y2), (3) a(XY) = a(YX) = (aY)X = X(aY), (4) (XY) Y'

= X( YY').

(2) and (3) imply that Q is a linear subspace of L. (2), (3) and (4) imply that Q is an ideal or Q = L. By (1) and hypothesis, Q is not an ideal. Hence Q = L. In particular, there is a YeL such that XY =e. Therefore X has an inverse. Corollary 4.3. X has an inverse X belongs to no ideals. Proof: Cf. Theorem 4.1 and 4.2.

Definition 4.1. M is a maximal ideal in L if: (1) M is an ideal in L and, (2) there are no ideals l => M, l "I= M. Theorem 4.4. For every ideal I, there is a maximal ideal M => I. Proof· Let Jt be the family defined by:

I

Jt = {J J =>I and Jis an ideal} Claim. Jt is inductive! Let rl be a chain, rl c: Jt. Let J 0 =

U J. Je'tt

To show that J 0 e Jt we must first prove that J 0 is a superset of/. This can be easily seen from the fact that J 0 => J => I, since J 0 is the union of all sets, J e rl. Next, we must prove that J 0 is an ideal. Let X, YeJ0 • Since J 0 is the union of all sets, Jerl, then XeJxe

XeJx

=> aXeJx

=>aXeJ0

86

FUNCTIONAL ANALYSIS

Hence J 0 is a linear subspace of L. To show that J 0 is an ideal, we must show that J 0 is closed under multiplication from the outside. Let XeJ0 and YeL. XeJxeCC. SinceJxisanideal,

XY = YXeJxeCC Therefore, XY = YX e J 0 , since J 0 is the union of all J e CC. Finally we must show that J0 "I= L. We will do this by contradiction. Suppose J 0 = L. Then, in particular, eeJ0 • Then eeJeCC. But e has an inverse, namely, e. Hence, e t/: J, because J is an ideal. Therefore this leads to a contradiction. So we have proved the claim that Jt is inductive. By Zorn's lemma, Jt has a maximal element M. To complete the proof of the theorem, we must prove that M is a maximal ideal. Since Me Jt, M => I and M is an ideal. Now, wet have only to show that there are no ideals K => M, K "I= M. We will do this by contraction. Suppose K is an ideal and K => M. Since M => I, then K => I. By the definition of Jt, K e Jt. Therefore, K = M. Thus we have shown by contradiction that there are no ideals, K => M, K -:1= M.

Corollary 4.5. X has an inverse X belongs to no maximal ideal. Proof==>: Suppose X has an inverse. This implies that X belongs to no ideal, which in turn implies that X belongs to no maximal ideal. Proof aXn e I and XnYei. Letting n-+ oo, aXel and XYel. Finally, we must prove that i "I= L. Again, we will prove this by contradiction. Suppose 1 = L. Then eel. There exist {Xn} with the properties Xn e I, Xn -+ e. There exists an N such that II XN- e II < I. Therefore, by Theorem 2-17, XN has an inverse. From Corollary 4.3, X N ¢: I. Contradiction.

HOMOMORPHISMS ON NORMED LINEAR SPACES

87

Corollary 4.7. A maximal ideal is closed. Proof: M c: M. Since M is a maximal ideal, M = M. Therefore,

Mis closed. Definition 4.2. An Abelain linear algebra is afield, if X #= 0 implies X has an inverse.

Theorem 4.8. If I is a maximal ideal, then L/1 is a field. Proof(by contradiction): To prove that L/1 is a field, we will show that L/Ihas no ideals except {[0]}. This will imply that each non-zero element of L/Ihas an inverse and therefore that L/1 is a field. Given I is a maximal ideal. Suppose k is an ideal in L/1, and k #= {[0]}. We define: K = .{XeL [X] ek}

I

First, we must show that K is an ideal. To accomplish this, we will show that K satisfies all the closure properties of an ideal. Suppose X 1 , x2 E K. Then [x.], [X2] ek. Sincekisanideal, then [Xt]+ [X2]ek. I.e. [ x. + X2] E k. Thus, x. + x2 E K. Next, suppose that then

XeK

[x] ek a[X]ek [aX] ek;

therefore,

aXeK.

Finally, suppose

XeK, YeL

then

[X]ek [X][Y] ek [XY] ek;

therefore,

XYeK.

Thus, we have shown that all the closure properties of an ideal are valid forK. Next, we ask, is K = L? If it is, then, e e K which implies [e] e k. Then, X E L => [X] = TX] [ e] E k, i.e., k = L/I. This is a contradiction, because k·is an ideal. Hence; K "I= L. We, therefore, conclude that K is an ideal of L. ·

88

FUNCTIONAL ANALYSIS

Now, Z 0 e I - [Z0 ) = [0] e k - Z 0 e K. I.e, K => I. But, this is impossible since I is a maximal ideal. This contradiction proves the theorem. Theorem 4.9. If Lfi is a field, then I is maximal. Proof: Suppose I is not maximal. Then I is an ideal, K #: L.

c:

K and I #: K, where K

Let k ={[X] I XeK}

[X], [ Y] e k - X, Y e K- X+ Y e K => [X+ Y] e k => [X]+ [ Y]e k [X] ek- X eK- aX,XYeK- [aX], [XY]ek -a[X], [X]· [Y]ek Thus either k is an ideal or k = LfI. But since LfI is a field, any ideal is {[0]}. So either k = {[0]} or k = L/1. But k #: {[0]} by the hypothesis I#: K, and k #: L/I by the hypothesis K #: L. We have arrived at a contradiction, proving that it is impossible that I not be maximal. Definition 4.3. X is called regular if it has an inverse; otherwise it is called singular. Theorem 4.10. In a Banach algebra, the set of all regular elements is an open set. Proof: Let X be regular. Let Y satisfy the relation .

1

IIX-YII < Consider II

e-x-• Yll

llx-111

=II

x- 1 x-x- 1 Yii

~:n

x- 1 11·11 X- Yii

=II

x- 1(X- Y) II

< 1

x- Yhas an inverse, say Z x-tyz = e = x- ZY

Therefore, by Theorem 2-1.7,

1

1

But then

x-tz is an inverse of Y, and

Yis regular.

Corollary 4.11. The set of singular elements of a Banach algebra is closed. (This set is the union of all maximal ideals.)

HOMOMORPHISMS ON NORMED UNEAR SPACES

89

Theorem 4.12. Let L be a Banach algebra. Let Vbe the set of regular elements of L. Letf (X) = x-• for X e V. Then, fis continuous on V. Lemma. If X,e V and X,-+ e then X;; 1 -+ e. Proof of lemma. There exists an N such that n > N=> II

x,-ell
n>N=>X;; 1 = _L(e-X,)q q=O

n > N =>II x;;•11 Let

q

~ q~) e-X, II ~ q~a(~)q = 2 1

1

1

M=max 0 there is a fJ > 0 such that

z~zo [ex(z)-ex(z I < 0)]

E

Definition 1.2. ex is differentiable on the set D if ex'(z) exists for all ZED.

Theorem 1.1. Let ex be differentiable on D. Let f be a bounded linear functional on L. Let q be defined on D by q(z) = f(ex(z) ). Then q is differentiable on D and q'(z) = f(ex'(z)) in D. (Note that q is an ordinary complex-valued function of a complex variable.) Proof. Let r0 = f(ex'(z 0 ) ) for some z 0 ED; consider

1-

I

1 q(z)-q(zo)- r 0 1 = - [f(ex(z) )-f(ex(z0 ) )] -f(ex'(z o)) z-z 0 z-z 0

as z

-+ z

=

k(z~zo [ex(z)-ex(z

=

k(z~zo [ex(z)-ex(z )]-ex'(z ))1

0 )] )-f(ex'(z 0 ))\

0

~·111 ll·llz~zo [ex(z)-ex(z 0,

by hypothesis. 91

I

0

0 )] -ex'(z 0 )

II-+ 0

92

FUNCTIONAL ANALYSIS

Theorem 1.2. If oc is differentiable in D_. then oc is continuous in D. Proof: Let Xm ED;

Xm-+ XED.

1 P(z) = -[oc(z)-oc(x)]-oc'(x)

Let

z-x

(where zED; z :f. x; x and z are complex numbers), and let P(x) = 0. Then, oc(z) = oc'(x) · (z-x)+oc(x)+P(z) · (z-x). (Note that this equation remains valid for z = x.)

= oc'(x)(xm-x)+oc(x)+P(xm}(xm-x) • oc(xm) -+ oc'(x) · 0 +oc(x)+O·O = oc(x), i.e. oc(xm)-+ oc(x).

oc(xm)

2. Integrals of Banach-Space Valued Functions

Let oc be a function into L defined on define

J:

[a, b]. We will show how to

oc(t)dt

Definition 2.1. A partition, of [a, b] is a finite set of points, P= {tht2•••tm-d

where we take t 1 < t 2 < t 3 < ... < tm_ 1 ; a< write t0 = a, tm = b. We write

III' Let oc be a function from

11

ti

< b. We usually

= max (ti-t 1_ 1) 1;jii;jim

[a, b] into L.

Definition 2.2. A Riemann sum for oc over Pis a sum: m

L oc(q;)(t1-t1_ 1)

i=1

where

t 1_ 1 ~

q1 ~

t1

Definition 2.3. oc is Riemann integrable over [a, b] with integral 1 if the conditions: (l)

I pm II-+ 0,

93

ANALYTIC FUNCTIONS INTO A BANACH SPACE

(2) for each m, Sm is a Riemann sum for oc over Pmimply: Sm

In this case, we write,

I=

s:

-+

1.

oc(t) dt

We wish to show that if oc is continuous on integrable over [a, b].

[a, b]. then it is Riemann

Lemma I. Let oc be continuous on a, b. Then for every E > 0 there is a [) > 0 such that if S, S' are Riemann sums for oc over partitions P, P' respectively, and if II P II < b, I P' II < b, then II S-S' II <E.

Proof: Let P" = P u P'.

Let

P' = {tJ.,

t2,

t;, ... ,t~,-1}

P" "= {t"1• t''2• t"3•···• t"r-1 } n

S=

L oc(q;)(t;- t;- 1); i=

t;-1 ~ q; ~ t,.

1

m

Similarly

S' =

L ct(qi)(ti-t;_t).

i= 1

r

s = 2:

Now

k=1

oc(pk)(r;;- t;:_ 1 >

where each Pk is one of the qi. r

and

S'=

L ct(p/,)(ti:-ti:-1)

k= 1

where each

p~

is one of the qi.

~

r

2:

k=l

11

oc-oc II (ti:- t;-1>

94

FUNCfiONAL ANALYSIS

Since [a, b] is a compact set, oc is uniformly continuous on [a, b]. (Theorem 3-1.6.) Let E > 0 be given. Then there is a number rt > 0 such that, if t, t' E [a, b] and It-t' < rt then

I

E

Let {J = ti-l

I oc(t)-oc(t'> II< b-a rt/4, let I P I < (), I P' I < b. Let Pk = q;.

~ Pk ~ t; and t}-1 ~Pic~

Let pj. = qj where

tj. [t;- 1 , t;] and [t}- 1, tj] overlap,

since if they were disjoint, Pk would have some Pi =I= pi, corresponding to Pk· Then: Pk- Pk ~ 2() < 4() = '1

I

I

II oc(pk)-oc(pD II < E/(b-a> r

E

E

-a

-ak=1

r

I S-S' II< L -b-(t/;-ti:-1) = -b- L (tk-tk-1) k=1

E

Therefore

= --(b-a) (b-a)

I S-S' I <E.

=E.

Lemma 2. Let oc be continuous on [a, b] and let {Sm} be as in Definition 2.2. Then {Sm} is a Cauchy sequence. Proof. Choose

E

> 0; obtain () as in Lemma I. Then there exists

I P I < fJ, since I P I -+ 0. m, n > N => I Sm- sn I < E

N such that m > N =>

m

m

Lemma 3. Let oc be continuous on Definition 2.3. Then S"- S~ -+ 0. Proof: Choose N 1 , N 2 such that

E

[a, b] let {S,.},

{S~} be as in

> 0. Obtain () as in Lemma I. Then there exist

I I max(N 1 ,N 2 )=> I S"-S~II <E.

n > N 1 => P n

Then:

Then

n > N 2 => P~

Theorem 2.1. If ex is continuous on on [a, b].

[a, b ], oc is Riemann integrable

Proof: Immediate from Lemmas 2 and 3.

95

ANALYTIC FUNCTIONS INTO A BANACH SPACE

Theorem 2.2. Let f be a bounded linear functional on L, and let oc be a continuous function from [a, b] into L. Then,

1[ I:

I:

ct( t) dtJ =

Example: Let L be CR(c, d) For XeL.let f(X)

=I:

f( ct( t)) dt

X(t)dt

Let oc be a continuous function, oc: [a, b] -+ CR(c, d). Setting oc(t) = and letting Y(t, u) = X,(u)

X,,

it is easily seen that Y is simply a continuous function from

[a, b] x [c, d] into C. Theorem 2.2 asserts about such a Y, that

ff

Y(t,u)dtdu

=

ff

Y(t,u)dudt

i.e. that the order of integrations in an iterated integral may be interchanged. Proof of Theorem 2.2. Lett" -+ t. Then oc(tn) -+ oc(t), andf(oc(tn)) f(oc(t) ).

-+

Thereforef(oc(t)) is continuous. Let Sn be a sequence of Riemann sums such that sn -+

f

oc(t) dt

m

s" =

2: oc(q;)(t;-ti-1> i=l

where m, q;, ti all depend on n. Then m

J(Sn)

= i=l L f(oc(q;) )(t;- t;-1).

S:

Therefore

f(S")-+

But,

f(Sn)-+ f(f ct(t)

f(oc(t)) dt.

dt).

96

FUNCTIONAL ANALYSIS

3. Line Integrals and Cauchy's Theorem Definition 1.1. A smooth Jordan arc is a function z(t) from [a, b] into C such that

(l) z'(t) is continuous on [a, b] (2) z(t 2 ) = z(t 1) => t 2

= t 1•

Example: z(t) =

fl', 0 ~ t ~ !n, is a quadrant of a circle.

Definition 3.2. A set D c: C is called a Cauchy domain if: (l) Dis an open set (2) z 1, z 2 E D => there exists a smooth Jordan arc which contains z 1 and z 2 and lies entirely in D.

(3) The boundary of D consists of a finite number of smooth Jordan arcs.

a

Definition 3.3. Let M be a smooth Jordan arc given by z(t), t ~ b. Let oc be a continuous function from the points on M into

~

L. Then,

IM oc(z) dz = S: a(z(t) )z'(t) dt Definition 3.4. Let oc be defined on a set P c: C. Then oc is regular or analytic inP if there is an open set D => P on which oc is differentiable. Definition 3.5. Let a curve M be made up of the finite number of smooth Jordan arcs M 10 M 2 , • •• , M". Then,

r ct(z)dz JM, r ct(z)dt+ JMz r ct(z)dz+ ... + JMn r a(z)dz

JM

=

ANALYTIC FUNCTIONS INTO A BANACH SPACE

97

Theorem 3.1. Let/be a bounded linear functional on L, and let ex be continuous on each of the smooth Jordan arcs which make up the curve M. Then

Proof: For M a smooth Jordan arc, the result follows at once from Theorem 2.2 and Definition 3.3. For the general case, we need only employ Definition 3.5 and the linearity off.

Theorem 3.2. (Classical Cauchy Theorem.) Let ex be a complexvalued function which is analytic in the Cauchy domain D plus its boundary M. Then, fM ex(z)dz = 0 Proof: See any book on complex variable theory. Note. A boundary curve is always oriented so that the region is on the left as one advances along the curve.

Theorem 3.3. (Cauchy's Theorem for Banach Spaces.) Let ex be a Banach-space valued function which is analytic in the Cauchy domain D plus its boundary M. Then, JM cx(z)dz = 0 Proof: Let X 0 = fM cx(z)dz

Let f be any bounded linear functional on L. Then

/(X0 ) =f(JM cx(z)dz) = JMf(ex(z)dz = 0

by Theorems. 3.2 and 1.1. Hence, by Corollary 3-6.5 (a consequence of the Hahn~ Banach Theorem), X 0 = 0.

98

FUNCTIONAL ANALYSIS

Theorem 3.4. (Cauchy's integral formula). Let ct be analytic in the Cauchy domain D plus its boundary M, and let z E D. Then, oc(z) = -1.

i

oc(w) - dw

2m Mw-z

i

Proof: Let

1 oc(w) X 0 =oc(z)--. --dw 2m Mw-z

Let/be any bounded linear functional on L. Then,

=f(oc(z))-~

f(X 0 )

f /(oc(w)~dw

2mJM w-z

=0 by the classical Cauchy integral formula. Therefore by Corollary 3-6.5, X 0 = 0 and 1 oc(z) = -2. oc(w) dw.

r

mJMw-z

Lemma. If II oc(z) II ~ M on the curve Nand the length of N is I then

II

L

oc(z)dz I

~ Ml

z = z(t) = x(t) + iy(t :

Proof:

I=

L f oc(z)dz =

oc[z(t)]z'(t)dt.

Now S"-+ /where m

L oc[z(q1)] z'(q1)(t1-t1_ 1).

S" =

1=1

Also,

l

=

f

"[x'(t)] 2 + [y'(t)] 2 dt =

m

Hence

IISnll ~

And letting n

-+

III II

fI

L lloc[z(q;)JII·Iz'(q,)l·(t;-t;-1) i=1 m

~M

L !z'(qJI(t;-t;-1). 1=1

oo,

f

~ M Iz'(t) l_dt = Ml.

I

z'(t) dt.

99

ANALYTIC FUNCTIONS INTO A BANACH SPACE

Theorem 3.5. (Liouville's Theorem.) If oc is analytic in the entire complex plane and II cx(z) II ~ M, then oc is constant. Proof l (Using the classical Liouville's Theorem): Suppose oc is not constant; then there exist z 1 and z 2 such that oc(z 1) -:1: cx(z2 ). Hence there must be a bounded linear functional/ on L such that f(oc(z 1) ) -:1: f(oc(z 2 ) ) by Corollary 3-6.6. Butf(oc(z)) is analytic in the complex plane, by Theorem 1.1. Now l/(oc(z)) I ~ IIJ II · II oc(z) II ~ M IIJ II Hence by the classical Liouville's Theorem,f(oc(z)) is constant. This is a contradiction; hence oc must be constant.

Proof2 (Using Cauchy's integra/formula): Suppose oc is not constant; then there exist z 1 and z2 such that oc(zt) -:1: oc(z2 ). Let N be a circle with center at z 1 and radius r so large that I z2 -z 1 I < r. Then by Theorem 3.4

Hence,

II oc(z 2 )-oc(z 1) II=

1 1 1 II - -]oc(w)dw II 2 n JN w-z 2 w-z 1

f[f

_ 1 II (z 2 -z 1)oc(w) d II - 2n IIJN(w-z 2 )(w-z 1) w II oc(w) II~ M, -1 w-zd

Now, lw-z 2

= r,

and

1= l<w-z 1)-(z2 -z 1)1 !;;llw-zd -lz2 -zdl = r-lz 2 -zd

Hence by the Lemma lloc(zz)-oc(zt)ll

~21n ~Zz~ztiM·f;r --.o r- z -z r 2

Hence cx(z2 ) constant.

==;

as

r--.oo

1

cx(z 1) which is a contradiction and oc must therefore be

100

FUNCTIONAL ANALYSIS

4. Banach Algebras which are Fields To say that a Banach algebra L is a field is to say that Lis Abelian and that for X e L, X -:1: 0, X has an inverse, or what comes to the same thing: If X, UeL, X -:1: 0, 3VeL, V= U/X, i.e. VX=XV= U.

Theorem 4.1. A Banach algebra over C, which is also a field, is isomorphic to the field of complex numbers C. Proof· Common hypotheses for all lemmas: L is a Banach algebra over C, which is also a field.

Lemma I. For each X e L, there exists an a e C such that X= ae. Proof· Suppose that there exists an X e L such that X -:1: ae for any ae C. Let oc(z) = (X-ze)- 1 • Note that X-ze is non-zero by hypothesis and the inverse exists because L is a field.

Now,aslzl-+ oo,

Xz-

1

-+0, i.e., Xz- 1 -e-+ -e.

By Theorem 4-4.12 (continuity of the inverse function)

Hence Therefore, there is a number p > 0 such that

lzl>p=>II<X-ze)-111
I (X -zer 1 II ;£ K. But this means that cx(z) is analytic everywhere and II cx(z) II ;£ max (K, 1). Therefore ex is constant by Liouville's Theorem. Note that cx(z)

-4

0 as z

-4

oo.

Hence cx(z) = 0. But this means that (X -ze)- 1 = 0 and this implies that e = (X-ze)(X-ze)- 1 = 0. This is impossible; hence the lemma is proved.

Lemma 2. The mapping a+-+ae is an isomorphism between C and

L. Proof· Let a+-+ ae, b +-+be, then a+b +-+ (a+b)e = ae+be, a· b +-+(a· b)e = (ae) · (be) = a(be)

I a I = I ae II

since

II e \1

= 1;

finally suppose ae = be and a =F b. This implies (a-b)e = 0, (a-b) =F 0, hence e = 0 which is impos~ sible. Therefore ae = be => a = b and a +-+ ae is an isomorphism be~ tween C and L. Theorem 4.1 follows at once from Lemma 2.

Corollary 4.2. Let L be a Banach algebra over C and let M be a maximal ideal in L. Then LfM is isomorphic to C. Proof· LfM is a Banach algebra which is a field, (Theorem Hence LfM is isomorphic to C.

~4.6).

5. The Convolution Algebra 11 Definition 5.1. l

1

={{an}"\n=O,

±1, ±2, ... and

aneC

and

n=~jan\oo

IIX -Xp\l = 0 11

p->-ac:? I = 0

n-t-co m p->-a~>

n->oo

I= 0

p-> = bm uniformly in m.

\IXn-XII =I:Ia~>-bml II

{bm} such that

104

FUNCTIONAL ANALYSIS

Therefore,

q

I

lim X,.-X

n-.co

II= lim q_,.co lim L Ia~>-bml m= -q n-t-oo

q

L-q Ia~>- bm I

= lim lim q~co

n-t-co m=

=lim0=0 q->oo

where the interchange of limit operations is justified by uniformity.

I I I I

lim a~> = bm uniformly in m.

Also,

11->00

I:m Ibm I = I:m n-.oo lim Ia~> I = limi;Ia~>l n-t-co m

= lim n->oo

I {a,.} I < oo

which proves the completeness of P and completes the proof of the Theorem. Let us introduce G = {c5~}el 1 • We assert the following identity: 00

{a,.}=

L a,G" n=-co

Let us verify this identity. Certainly

0 -1 = {o;l} {c5~} · {c5; 1 } =

for

{c,.}

where except whenp = 1 and n-p = -1, and thus when n = 0. Therefore c, = c5~, and {c,.} = {c5~} = e. Also,

G2 = {c,.}, c,. = I;c5~c5!-p = c5~ and Gk = {c5~} p

Now let us compute

= { ... ,0, 0, a_r, a-r+l• ... 'ao, au az, ... 'ar-1• ar,O, 0, ...}

Thus, letting r-+ oo, we obtain the desired result.

105

ANALYTIC FUNCTIONS INTO A BANACH SPACE

Let Tbe any homomorphism from 11 on to C. Let TG = t, a complex number. Now G = 1, and remembering that

I I

li[X] I = inf

IzI ~ IX I

ze[X)

we have I t I ~ 1.

TG- 1 = 1ft, and since ~ G- 1 II = 1, it also follows that

Thus

ll/t I ~

1.

It I= 1.

We may represent TG

=t=

e; 8• Then

n=-co r

L a,G" r-t-oo n= -r

= Tlim

r

L a,G" n= -r

= lim T r~co

r

L a,(TG)" r-.co n= -r

= lim 00

L

=

a,eniB

r=-w

Which elements {a,} e 11 have inverses? Those {a,} which belong to no maximal ideals (cf. Corollary 4-4.5). In other words, those {a,} which belong to the kernel of no homomorphisms from 11 onto C, i.e., those {a,} which are mapped into 0 by no homomorphism.

Conclusion: {a,} has an inverse there is no 0, 0 that This leads to:

Theorem 5.2. (Wiener.) Let 00

f(t)

= L

a,e"i', 0 ~ t < 2n

n=-co 00

where

:L Ia, I< oo n=-oo

~

0 < 2n, such

106

FUNCTIONAL ANALYSIS

Let/(t) vanish for no t, 0 ;;?; t < 2n. Then there are numbers bn such that 1 J(t) =

00

•

00

n=~ oo bn tf"', n=~ oo I bn I