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 AeA<=>A¢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 "< ".
<
= {(a, b) Ia less than
b}
(2, 6) e < , more usually written 2 < 6
6
FUNCTIONAL ANALYSIS
There are many kinds of relations, having many different properties. Three possible properties of relations are of special importance. A relation R on a set A is called:
reflexive if xRx for all x e A symmetric if xRy => 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<&
.AI, does
U BJ e .1t' since .1t' is a K-support. [ Be<&
There are two cases to consider: (a) For every Be f(j, it is the case that B c: A 0 • In this case
U B] [ Be<&
c:
A
0•
.".
[
U
Be~
B] e .AI
(b) There is some Be f(j such that B c: A 0 is false. But Be .AI. Therefore B:::. f(A 0 ). However, [ U BJ :::.B. ·
Be~
:. [ Be<& U BJ :::.f(A 0 ) Therefore,
f(j
or
is a chain and ., c: .IV=> [
[
U BJ e%
Be<&
U BJ e .AI". From Lemma 2,
Be<&
the conclusion is that .IV = #. In other words, .1t' = {Be#IB c: A 0
or B :::.f(Ap)}
when A 0 e !l'. Therefore A e 2 and Be ;/I'=> B c A or B :::. f(A 0 ) which is the conclusion of this lemma. Lemma 5. !l' = #. Proof: (1) K e !l' is true vacuously because by Lemma 3, if Be .1t' then B :::. K. Hence the hypothesis is never satisfied. (2) Does A e !l' => f(A) e !l'? That is, does Be ;If and B =I= f(A) and B c: f(A) => 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,
ll
Ii -+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·<= Let Xe A. To prove
II
< e => 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 <
e
~
Let Z" = X~!- Then Z" e A, and
II Zn- Y II~ I Zn- Yn II + I Yn- Yn I
~~+IIY"-¥11-o n Therefore Y e B. Thus, B is closed. However by Corollary 2.4, A c B and B closed implies A c B. Hence A c B.
Definition 2.4. such that
A c L is bounded means there exists an M > 0
XeA=>IIXII ~M
This concept can be thought of in terms of a geometric picture. If the set can be enclosed in a sphere, it is bounded.
Definition 2.5. A c L is called compact if whenever X" e A, there is a convergent subsequence of {Xn} whose limit is in A. Corollary 2.5. A compact set is closed. Proof: Let A be compact. Let X" e A. X" -+ X. Then it must be shown that X e A . . ·· This is obvious, since every subsequence of X" approaches X. So X
eA.
OefJnition 2.6. The sequence {Xn} is bounded means that there exists an M > 0, such that II X" II ~ M.
NORMED LINEAR SPACES AND ALGEBRAS
31
Corollary 2.7. A convergent sequence is bounded. Proof: Let X,
~
Th~
X.
n
there is an integer N such that
> N ~ I X,- X I < 1
Then
II(X,-X)+XII ;:;i IIX,-XII +II XII ·
n>N=>IIX~II = Now let M = Then,
I X, I
~ M.
Corollary 2.8. A compact set is bounded. Proof· Let A be compact. Suppose A is unbounded. Then, for each positive integer n, there is an X, e A such that I X, I > n (using principle of choice)
Since A is compact, some subsequence {X,k} converges
I x,k I ~ M (by Corollary 2.7). However I X,k I > nk > M (for some value of k).
Therefore
And so, we arrive
at a contradiction. Definition 2.7. The space L is called piecewise compact if every closed, bounded set in L is compact. Corollary 2.9. A Cauchy sequence is bounded. Proof· Let {X,} be a Cauchy sequence. Then, there is an integer N
such that:
> N => I X,- Xm I < I n > N~ I X,-XN+t II< 1 m, n
Hence,
The rest of the proof is identical to that of Corollary 2. 7 with the role, played there by X, being played here by X N + 1 • Corollary 2.10. If A is bounded, then A is bounded. Proof· Let Y e A ~ I Y I ;?. M. Let X e A. Then, by Corollary 2.5 there exists X, e A, Xn ~ X. Now, I X, ~ ;:;i M. Thus,
Jim
n->oo
I X, I
which completes the proof.
~ M, and so
I XI
~ M,
32
FUNCflONAL ANALYSIS
Theorem 2.11. If Lis piecewise compact, then Lis a Banach space. Proof: Let {Xn} be a Cauchy sequence. Then {Xn} is bounded (Corollary 2.9). Let A be the set of all terms of {Xn}· Then, A is bounded. Hence A is bounded. (Corollary 2.10) .. Hence the closure of A is compact because A is closed (Corollary 2.3) and bounded. Therefore {Xn} has a subsequence {Xnk} which converges to a point X e A (Definition of compactness). Let e > 0 be given. Choose N such that m,n > N=> Xn-Xm e (definition of Cauchy sequence). There is an integer k 0 such that n110 > N and X""o- X ej2 since X"k ~ X. Thus, 11 > N implies
I
I Xn-X I Hence Xn space.
~
II<
~
I
I Xn-Xn,. I 0
I<
+ I X:'ko -X II< !e+!e = e.
X. This implies that L is complete, i.e. that it is a Banach
3. R as a Normed Linear Space In this section, all elements are in R, normed linear space over R.
I X I = I X I, so that R is a
Definition 3.1. Miscalled an upper bound ofthe set of real numbers A if X e A => 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 > 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<=: Choose e > 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
~Mil XII+
I (aT)(X) II
M'll XII= (M+M'>ll XII·
= II aTX I = I a I ·
~
II TX II
lalMIIXII.
Since there is closure under + and ·, [ L, Q] is itself a linear space. We now define norm in L, Q. Definition 4.3~
sup
IIXII=l
I TX II
for Te [L, Q].
I T II < oo. II TX II ~ M II X II (Tis bounded). I X II = 1 => II TX I ~ M. II Til~ M < 00.
Corollary 4.3.
Proof:
II T II =
FUNCTIONS ON BANACH SPACES
47
I T I = sup I TX II· Proof: Let K = sup I TX II· Corollary 4.4.
IIXII~l
IIXII~l
II T I
~ K
I I
( T is defined as the supremum of a subset of the set of which K is the supremum.)
IIXoll ~ 1.
Let
1
Let
X=
IJXo[IXo
I XII= 1 Xo= JIXoiJX I TXo II= IJ T(jl Xo I X) I = JJXo Jil TXIJ ~l·llrll·
IITjj. K=Jirjj. K~
Thus, Hence,
Theorem 4.5.
I TXjj ~II Tll·ll XII·
Proof: For X= 0, the result is clear, since
T(O)
= T(O+O) = T{O)+T(O)
T(O) = 0
Otherwise, let
1
=II XIlX JJXoll = 1 I TXoll ~II Til Xo
I
1 rx o 11 = rC1 ~ 11)x 1
_= 1JXlf I TX I IITXII ~ IIXII·II Til
I
=
11,1 ~ , ~~II
48
FUNCTIONAL ANALYSIS
Corollary 4.6. II Til=~~~ Proof: Let K
= sup II TX II
.
X#O
Taking
II
Letting X 0
IITXII lXf
IIXII
X
II =
¢
0, we have
1, II T
II
~ K
II TX o II ~ II T
11·11 X 0 II (Theorem 4.5)
I!TXoll<\jTjj I!Xoll = K~IITII Therefore, II T II = K. Theorem 4.7. [L, Q] is a normed linear space.
Proof: [L, Q] has already been shown to be a linear space. Now we must show that the four conditions of Definition 2-1.2 are satisfied. (1) II T II =
sup II TX II ~ 0
IIXII=1
(2) II T+T'II =
~
~
sup II
IIXII=l
sup
IIX (1=1
sup
II X' 11=1
II TX'Il +
sup
II X" 11=1
II T'X"II
=IITII+IIT'II (3) llaTII=
sup ll
IIXII=1
= jaj sup
II X II= I
sup jal·IITXII
IIXII=1
II TXjj
=Ia!· II TII (4)
li Til= 0¢>SUpll TXII = o¢>11 TX II= 0 for all XeL X#O !lXII !lXII ¢>II TX II = 0 for all X eL, ¢>TX = 0 for all XeL,
¢>T=O
49
FUNCTIONS ON BANACH SPACES
Theorem 4.8. lf Q is a Banach space, so is [L, Q]. Proof· Let {T.,} be a Cauchy sequence lim m-+ex>
I Tm-Tn II= 0
Consider {T.,X}, where XeL and T.,Xe Q.
I
II= I {Tm-T,.)XII
Tm-T..II· 11 X
TmX -T.,X ~II {T., X} is therefore a Cauchy sequence. Let,
11
-+0 as m,n-+ co
T.,X-+ TX (Q is a Banach space).
Next we inquire: is Tin [L, Q]? To be in [L, Q] it must be linear and bounded. (1) T(X + Y) = lim
T,.(X + Y) = lim (T.,X + T, Y)
n-+ex>
n-+ex>
=TX+TY. (2)
T(aX)
= lim T.,(aX) = lim aT., X =a lim
T,.X
n-+ex>
=aTX. Therefore, T is linear. {T.,} is bounded {because it is a Cauchy sequence).
liT.. II~ M
I T.,X I ~ I T., II· I X I ~ M I X I I TX I ~ M I X II, so that Tis bounded.' There remains now only to prove that Tis the limit of {T.,} as n-+ co. Choose e > 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
[k(s11,t)-k(s,t)]X(t)dtl
e
~ 211 Xll
Theorem 5.1. Ke [C(a, b)]. (That is, K is a bounded linear operator on C(a, b).) The proof is left for the student. The Fredholm integral equation is Y= (1-K)X
where I is the identity operator and K is the Fredholm operator. Problem. Given Ye C (a, b) to find X e C (a, b) such that Y = (1- K)X. In integral form, the Fredholm equation is: Y(s)
= X(s)-
Corollary 5.2.
II K II < 1.
Proof: II K II =
sup II KX II IIXII=l
J:
k(s,t)X(t)dt
Let Y= KXwhere II XII= 1. Then,
Y(s)
=
J:
k(s, t) X(t) dt
I Y(s) I ~ M(b-a) · 1 II Y II ~ M(b-a), II KX II ~ M(b-a) Hence
II K II ~ M(b-a) < I
FUNCTIONS ON BANACH SPACES
53
Corollary 5.3. The operator I- K has an inverse given by the formula: (I-K)- 1 = l+K+K 2 + ...
I 1-(1-K) I =II K I < 1
Proof:
Now [C(a, b)] (the algebra of all bounded linear operators on C(a, b)) is a Banach algebra, (by Corollary 4.15) so this result is an immediate consequence of Theorem 2-1.7. Corollary 5.4. For each Y e C(a, b), there is a unique X e C(a, b) such that Y = (1-K)X, namely X=
C~o Kn)·Y
0
(Here K is understood to mean/.) Proof: Let 00
H=
By Corollary 5.3, H(I-K) (1) Existence. (1-K)(HY)
LK" n=O
= (1-K)H = 1.
= [(1-K)· H]Y =I· Y= Y.
(2) Uniqueness. Let X satisfy: (1-K)X
Then:
= Y.
H[(I-K)X] = HY [H(I-K)]X = HY IX=HY X=HY
Theorem 5.5. In Corollary 5.4, we may write 00
X=
L (KnY) n=O
Proof: Let
54
FUNCTIONAL ANALYSIS
Then, using Corollary 4.16
( f Kn)y =(lim Hn)Y n-~ooo
n=O
=lim (Hn Y) n-+co
== lim
(f. Kk)y
=lim
L (KkY)
n-+co
k=O n
n-+cok=O co
=
L (KnY)
n=O
Definition 5.2 k 1 (s, t) = k(s, t) kn(s,t) = [ k(s,u)kn_ 1 (u,t)du,
n
>1
Theorem 5.6. Y=Knx ._ Y(s) = J: kn (s, t) X (t) dt
Proof (by mathematical induction): Obvious for n = l. Assume the result known for n = q. We must verify it for n = q+ I. Y = Kq+l X.- Y = K(K'lX) Hence using the induction hypothesis,
Y = Kq+tx .- Y(s) = =
S: k(s,u>[[ kq(u,t)X(t)dt]du
J: f
k(s, u)kq(u, t)X(t)dtdu
= J: J: k(s,u)kq(u,t)X(t)dudt
=
S: X(t)[[ k(s,u)kq(u,t)du]dt
= J:x(t)kq+ 1 (s,t)dt
FUNCTIONS ON BANACH SPACES
55
Corollary 5.7. For the X of Corollary 5.4, X(s)
Corollary 5.8.
= n~of kn(s,t)Y(t)dt
Ikn{s, t) I~ M"(b-at- 1
Proof(by induction): Obvious for n = 1. Assume known for q.
Ikq+ 1 (s, t) I =
IJ:
k(s, u) kq(u, t)dtl
~
(b-a)M · Mq(b-a)q- 1 = Mq+ 1(b-a)q
Corollary 5.9. For the X of Corollary 5.4, X(s) Proof:
S:
= n~o kn(s, t) Y(t)dt
Ikn(s, t) I~ Mn(b-a)"- 1
00
L kn(s, t) converges uniformly in n=O
[a, b] x [a, b] by the Weierstrass
00
M-test. Hence
L
kn (s, t) Y( t) converges uniformly in [a, b] x [a, b],
n=O
and term-by-term integration is legitimate. We shall define the Volterra operator K so that for X e C(a, b) Y
= KX
¢>
Y(s)
=
s:
k(s, t)X(t)dt
The conditions on k will now be weaker than in the case of the Fredholm operator. Definition 5.3. k is an acceptable Volterra kernel if it is a continuous real valued function on the triangular region: {(s, t) a ~ t ~ s ~ b}
I
t
=t ~s
56
FUNCTIONAL ANALYSIS
Definition 5.4. k 1 (s, t) kn(s,t) =
J:
= k(s, t).
k(s,u)kn_ 1 (u,t)du for n
Theorem 5.10. Y=K"X<=>Y(s)=
f
> 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.
<3> 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 <;=t~ inf [-G(u)+lloll·!lu+Zo]IJ=q
<;=H(u+Z0 );£
ueWG
Next consider the case when a < 0. Multiplying this time by -1/a we get
_!H(X +aZo) ~ _!a l 9 11· il X +aZo I a
~ H ( - ~- Zo) ~ llo 11·11- ~- Zo II <;= H( -u-Z0 ) ~II g
ll·ll-u-Z 0 I
for all ueW0
<=> -G(u)-t ~II g 11·11 u+Z0 I for all ue W0 <=>t ~ -G(u)-llo 11·11 u+Z 0 I for all ueW0 <;=t:G; sup [-G(u)-lloll·llu+ZoiiJ=p ueWG
Note thstt q must be greater than p if our statements about t are to be satisfied. Thus it is necessary to show that for all u 1 , u2 in Wa
- GCut)-11 g 11·11 Ut + Zo I ~ - G(u2) +II g li ·II t/2 +Zo I <=>G(u2)-G(u 1 ) ~II g I [il u1 +Z 0 II+ I U2 +Zo IIJ
64
FUNCTIONAL ANALYSIS
The last statement is true because G(u2)-G(u1) = G(u2-u 1 )
I ;;;; I g 11·11 u2-utll = llo II· I (u2 +Zo)-(ul +Zo) I ~II u II· [II u2+Zo II+ I ul +Zo IIJ
;;i G(u2-ut)l
Following the steps backward we see that for all a (including the case a = 0, for which it is obvious)
H(X+aZo) ~II 9
11·11 X+aZo I
I H(X+aZ I= ±H(X+aZ
Now
0)
0 );
-H(X+aZ0 ) = H(-X-aZ0 )
and,
I 9 11·11 -X-aZo I =II 9 11·11 X+aZo I Thus IH(X+aZo) I~ I 9 11·11 X+aZo I for all X +aZ e Q, i.e., His bounded, and I H I ~ I g I . Hence, by Theorem 6.1, I H I = I g II· Moreover H;::) G thus He !IF. ~
0
Since G is maximal, H = G, and since (Z0 , t) e H, it must also be true that (Z0 , t) e G which contradicts the assumption that Z 0 ¢ WG· Therefore W0 = L. This completes the proof for the case F = R. Case 2. F = C (Bohnenblust-Sobczyk) Note that for each X e V, g(X) is a complex number. For the complex number a+bi, where a, be R, we write a = Bl(a+bi), b = J(a+bi).
Let
h(X) = !H(g(X) ),
k(X) = J(g(X)) Vand L can be regarded as normed linear spaces over R. Now, g(X) = h(X)+ik(X), so that g(iX) =-= ig(X) = ih(X)-k(X) g(iX) = h(iX) + ik(iX)
65
FUNCTIONS ON BANACH SPACES
I hus h(-Y)
= k(iX) and k(X) =
-h(iX), so that we may write
g(X) = h(X)-ih(iX) h(X+ Y)
= Bl{g(X+ Y)) = Bl(g(X)+g(Y)) = Bl(g(X))+Bl(g(Y)) = h(X)+h(Y)
Also, for real a, h(aX) = Bl(g(aX)) = Bl(ag(X)) = a9l(g(X))
h is therefore a linear functional on V taken over 91.
Ih(X) I = IB~(g(X)) I ~ I g(X) I ~ I 9 I · I X I Thus his bounded; I h I ~ I 9 II· Also
By case 1, there is a bounded linear functional H on L over Bl such that H c: h and H = h
I I I I·
Let
G(X) = H(X)-iH(iX) G(X+ Y) = H(X+ Y)-iH(i(X+ Y)) = H(X)+H(Y)-i[H(iX+iY)] = H(X)+H(Y)-i[H(iX)+H(iY)]
= H(X)-iH(iX)+H(Y)-iH(iY) = G(X)+G(Y)
G( (a+bi)X) = H( (a+bi)X)-iH(i(a+bi)X) = H(aX +biX)-iH(iaX-bX) = H(aX)+H(biX)-iH(iaX)-iH(bX)
= aH(X)+bH(iX)-iaH(iX)+biH(X) = (a+bi)H(X)-iH(iXXa+bi) = (a+bi)(H(X)-iH(iX)) = (a+bi)G(X)
Therefore G is a linear functional on L over C, and G
~ g.
66
FUNCTIONAL ANALYSIS
Choose a fixed X e L. Then, for suitable t, G(X) =
G(X1)
=
H(Xl) =
Thus,
1 '
X 1 = e- 1rx
Let, Then,
I G(X) I e
I G(X) I Bf!(G(Xl)) = I G(X) I
e- 1'G(X)
=
I G(X) I = H(Xt) = I H(X1) I ~ II H I · I X1 I
=II h 11·11 Xtll
Thus,
I G II
~
II g II . II Xt II
=
lloll·llxll
~
I g II , so that by
I g I .l . I X I
=
Thoerem 6.1,
II Gil= II oil· Corollary 6.3. Let X0 eL. Then, there is a bounded linear functional G on L such that G(X0 ) = II Xo II .
I
Proof. Let V = {aX0 a e F}, V c: L. The equations a 1 X 0 +a2 X 0
= (a1 +a2 )X0
b(aX0 ) = (ba)X0
show that Vis a linear subspace of L. Let g(aX0 ) = a II X 0 I . The equations g(a 1 X 0 +a2 X 0 ) = g((at +az)X0 ) = (al +az)
= al II Xo II =
and
g(b(aX0 ))
I Xo I
+azll Xo g(a1 X0 )+g(azXo)
I
= g( (ba)Xo) = ba I Xo I = b(a I X0 II> = 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-ll
E
= 1, there is a number q 1 such that
and x>O~I.J;-ll<e
. 4sint 1nn-2-=0 t-+0 cos t there is a number q 2 such that 4lsintl ltl
Since
Let Mbe given by y = f(x) near P 0 • Then Arctanf'(x) is continuous, hence uniformly continuous in a neighbourhood of P 0 • Thus, there is an q 3 such that in this neighbourhood I Xt-x 2 1 < q 3 ~ 1Arctanf'(xt)-Arctanf'(x2)1 < qz We distinguish two cases: P'
p
Case 1:
lz-P0 1< q 3 /2
Letting x., be the abscissa of z, etc., we have: lx.,-Xp0 1 < q3f2 Then, if P is sufficiently close to P0 that I xN- Xp0 I < q3 /2, then xN-.x., < q 3 • Now, the inclination of line NT to the x-axis is Arctanf'(xN). By the mean value theorem, that of NZ is Arctanf'(x 1)
I
I
73
FUNCTIONS ON BANACH SPACES
I
where I x 1-xN I < I xN-Xz I < q3 • Hence, I ex < q2 • Using the law of cosines and the fact that I N- P' I = I N-P , lz-P'\ 2 -Iz-PI 2 = 4\N-PI·I N-z\sinex
l[
lz-P'I]z lz-P I
-11
= 41 N -Pl. IN -zll sinexl I z-PI I z-PI
sinm cos(ex-m)l . I smex cos ex cos ex
= 4- · ~
II
z- P' I jz-P I
Hence,
jsinexl 4 -2-
-11 <
E
Thus, in this case, it suffices to take b = q3 /2. Case 2: ·
Then,
Iz-P0 I~ q 3 /2
II z- P' I -11 = llz-P I
I'
z- P' I -I z- p lz-PI
'I
< l
IP-P'I I (z-P0 )-(P-P0 ) I
:-:;;; IP-P'I -llz-P0 I-IP-P0 II
< jP-Pol =
+ IP'-P0 j
j(q 3 /2)-l P'-Po
II
But, as P-+ P 0 , P'-+ P 0 • Hence, this last expression (which is inde-
pendent of z)-+ 0. Hence, 3 q4 such that IP-Pol <
q4=>11:=~'1-11 < e
Thus, to satisfy both cases take b = min (q 3 /2, q4 ).
74
FUNCTIONAL ANALYSIS
Problems 1. Complete the proof that CF(a, b) is a normed linear space. 2. Prove Corollaries 4.ll, 4.12, and 4.14. 3. Prove Theorem 5.1. 4. Using Corollary 5.4 solve the integral equation -5s = X(s)-
6
Answer: X(s)
1 1
0
1 X(t)dt -st·
2
= s.
5. Solve the integral equation, s = X(s)-
J:
(t-s)X(t)dt
Answer: X(s) = sins.
6. For X, YeR", X= (X 1 , X 2 , define the "dot product":
••• ,
X,),
Y
= (Y1 , Y 2 , ••• , Y,)
[X, Y] = X 1 Y 1 +X2 Y 2 + ... +X,Y,
(Cf. Problem 3, Chapter 2.) (a) For fixed a, letf(X) = [a, X]. Show that/is a bounded linear functional on R". (b) Prove that for each bounded linear functional/ on R", there is an a e R" such that: f(X) = [a, X].
Hint forb: What doesfdo to the elements
(I, 0, ... , 0), (0, l, ... , 0), ... , (0, 0, ... , I)?
7. (a) For fixed Me CR(a, b), let f(X) =
J:
X(t) M(t) dt
Show thatfis a bounded linear functional on CR(a, b) (b) Are there any others?
FUNCTIONS ON BANACH SPACES
75
8. (a) Let B(a, b) consist of all real-valued functions f defined on [a, b] for which sup lf(t) < oo o~r~b
I
Show how to define the appropriate notions so as to make B(a, b) a Banach space with subspace CR(a, b). (b) Let f(X)' =
f
X(t) dt for
X e CR(a, b)
Show thatfis a bounded linear functional on CR(a, b). Evaluate
IIIII· (c) Using (a), (b) and the Hahn-Banach theorem, what can we conclude? 9. Below are sets A in normed linear spaces as listed. In each case find A. A Space (a)
R
{xl 0 < x < 1}
(b)
c
{zl1 < lzl <2}
(c) CR(O, 1}
{xiJ:
(d) CR(O, 1)
{XjX(O) = 2}
X(t)dt <
1!}
10. Show that each of the following are bounded linear functionals on CR(O, 1) and find their norms: (a) f(X) = X(O),
(b) f(X) =
J:
X(t)e-rdt,
(c) f(X) = 2X(0)+3X(1).
11. Show that each of the following are bounded linear operators from R 2 into R 2 and find their norms: (a) T(a, b) = (b, a), (b) T(a, b}
= ( (a+b)/2, (a-b)/2),
(c) T(a, b) = (6a, 6b).
76
FUNCTIONAL ANALYSIS
12. Solve the integral equation: u(s) =
e -t(e-18
s:
u(t)dt)
13. (a) For which values of q do our results guarantee the existence of a solution for the integral equation: X(s)-q
s:
e•+r X(t)dt = e8
(b) Solve the equation in (a) for the indicated range of values of q.
(c) For which values of q is the solution obtained in (b) actually valid? 14. Let L, Q be normed linear spaces. Definition l. A
c:
L is called separated into B and C if B
c:
L,
Cc:LB~~C~~A=Bu~BnC=a
Definition 2. A c: L is called connected if it can not be separated into sets B and C. Theorem. Let A c: L be connected, and let .f be a continuous function from A into Q. Thenf[A] is connected. Problem. Prove the theorem.
CHAPTER 4
Homomorphisms on Normed Linear Spaces
I. Homomorphisms on Linear Spaces In this section, L and Q are linear spaces over F. Definition 1.1. A homomorphism from L on to Q is a linear operator T on L such that for each Y e Q, there is X e L such that TX = Y. In this case Q is called a homomorphic image of L. Definition 1.2. KT, the kernel of a homomorphism T, is defined by KT= {XeL!TX=O}
Theorem 1.1. KT is a linear subspace of L. Proof: We must show that those properties in the definition of a linear space which may be in doubt do hold. (I) 0 E KT. (2) Suppose X, Y e KT. X, YeKT => TX = 0 and TY = 0 =>TX+TY= 0 => 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. <=is obvious. e[X]
(4)
To verify the implication in the converse direction, we first note that [X] is closed for each X. For, suppose that Z, e [X], Z, --+ Z, and let Y, = Z,- X. Then, Y, e V, Y, --+ Z- X. Since Vis closed, Z- X e V. I.e. Z e [X]. Thus, X is closed. Now, suppose that [X] = 0. Then, inf Z = 0. Hence,
II
I
ZeX
I I
I
I
for each positive integer n, there is a Z, e [X] such that Z, < lfn. Thus, Z, --+ 0. Since X is closed, 0 e X, i.e. [X] = [0 ]. We now consider the problem of determining under what conditions a homomorphism Tis bounded (i.e. continuous). For when it is, the operations of analysis will be preserved under it.
HOMOMORPHISMS ON NORMED LINEAR SPACES
81
Theorem 2.2. Let V be a closed subspace of L, and let T be defined by TX = [X]. Then, Tis bounded (i.e. continuous).
I TX I ~ M I X I for some M. I TX II= I [XJII = inf I z II~ I X II·
Proof: We have to show that
Ze(X]
(In fact, T has norm 1.)
Theorem 2.3. Let L, Q be normed linear spaces, and let T be a continuous homomorphism from L on to Q. Then KT is closed. Proof: Let Xn e Kn Xn -+ X. Then TXn -+ TX since Tis continuous. But TX" = 0. Therefore TX = 0, so that X e Kp Hence, KT is closed.
Theorem 2.4. Let V be a closed subspace of the Banach Space L. Then, L/Vis a Banach Space. Proof. For this proof, we will adopt the notation
x,y, zeL;
X, Y,Z eLfV
The proof consists of showing that a Cauchy sequence in L/ V converges. Let {Xn} be a Cauchy sequence. Then for each n, 3 Nn such that Nn < Nn+l and r,p ~ Nn Xp-Xr < 1/2". In particular, XNn+,-XN" < 1/2". Now, let Yn = XNn· Choose any y 1 E Yl' Choose some y 2 e Y2 such that I y 2 - y 1 I < 2 Y2 - Y1 II • Choose some y 3 e Y 3 such that II y 3 -y2 11 < 211 Y 3 - Y2 11 and in general, choose Yn+t e Yn+l such that I Yn+t-Yn < 211 Yn+l- Yn II· Note that such a choice is always possible since otherwise, e.g., 2 I Y 2 - Y1 would be a lower bound for I z I , z e Y2 - Y1 which is greater than I Y2 - Y1 I , the infremum, or greatest lower bound, of this class. The sequence of y's just described is a: Cauchy sequence in L since
I
I
=>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
<
2
1
IX 1
Nn+ p
L 2n+k-1 =2n-1 L
1<=1
1-
k=l
X Nn 1
2k-1
I < 1/2", this becomes
1 <2n-1·2-+0
Since {Yn} is a Cauchy sequence in a Banach space, it converges; so let the limit bey. Let Tbe given by Tx = [x]; then Yn = Ty"-+ Ty = X, say. Now we need to show that Y" -+ X. Choose e > 0. Then there is a Q such that p, r > Q =>II Xp-Xr I < e/2 by definition of Cauchy sequence. Then there is anN> Q such that for some·n, N = N" such. that I XN-XII < E/2. Further 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<=: Suppose X belongs to no maximal ideal, then X belongs to no ideal (Theorem 4.4). Then X has an inverse.
Theorem 4.6. If lis an ideal, so is i. Proof· To prove the theorem we must first prove that i is a linear subspace. Let X, Y e i. Then, Xn __.. X and Yn -+ Y where Xn, Yn e I; then Xn + Yn e I. Letting n -+ oo, X+ Y e i. Similarly, if X e i: Xn -+ X, where Xn e I. Hence, a e F and Y e L ==> 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
Then
II
IIX;;
1
x;; 1 11
-ell = IIX;;
1
~ M for all n
-X;; 1 X,II = IIX;; 1(e-X,)II
~ IIX;; 11·11 e-X, II~ 1
M·ll e-X,II-+0
Proof of Theorem 4.12: Let X, -+ Z, X, e V, Z e V.
Then By the lemma
X,Z- 1 -+ZZ- 1 =e.
(x,z- 1)- 1 -+ e x;;•z-+ e x;;•zz- 1 - z-• x;; 1 -z-•.
Problem LetL = R 3 , Q = R. Let Tbe defined by: T(a, b, c)= a. (a) Show that Tis a homomorphism from L on to Q. (b) Find KT.
CHAPTER 5
Analytic Functions into a Banach Space I. Derivatives In this chapter L is some fixed Banach space over C, and we shall consider functions ex from D c: C into L.
Definition 1.1. ex'(z 0 )
1
= lim - - [ex(z)- ex(z0 )] z-+z0 Z-Zo
if this limit exists, i.e., if for every
Iz-z I< c5 =>II ex'(z 0
E
0)-
> 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 <(J;
I I < b. n > 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 <
1
ANALYTIC FUNCTIONS INTO A BANACH SPACE
101
But {z I I z I ;£ p} is compact. Hence by Corollary 3-1.2 there is a number K such that I z I ;£ p => 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\
102
FUNCTIONAL ANALYSIS
An example of such an {a,.} is the sequence where 1
a,= n2+1: 1 1
1
1
1 1
1
· · ·• to' 5' 2' ' 2' 5' to···· In this space we introduce the definitions: (1) {a,} +{b,} = {a,+b,} (2)
c {a,}
= {ca,}
CJO
(4)
c, = L apbn-p
{a,}{b,.} = {c,}, where
p=-oo
Theorem 5.1. Under the operations just defined, / 1 is a Banach algebra. Proof: The proof that / 1 is a linear algebra is trivial. Notice that
0 ={a,} where a,= 0 for all n e = { c5°} where n
1 for n =m c5m = { n 0 for n :/:: m
For, <Xl
where
c, =
L a,c5,_P = p=-oo
a,.·1 =a,.
I {a,}+{b,.} II= I {a,+b,} I =
I;la,+b,l
~ I: [I a, I + I b, IJ = I:la,l +I:I
b,l
= I {a,.} I + I {b,} I
ANALYTIC FUNCTIONS INTO A BANACH SPACE
103
where we have omitted the index and limits of summation.
I {an}· {bn} II= II {en} I = LILapbn-pl p
II
~ 1:1: Iapll bn-pl II
p
=I:I;!apllbn-pl p
II
= ~ [laPI~I bn-pl]
~ [lapl~l bn\]
=
I:laPI· L\ bnl =II {an} 1\·1\ {bn} \1 II e I = I: I c;~ I = t =
p
II
II
1
Thus / is a normed linear algebra. Now we show it is also complete. Let {X11 } be a sequence, X 11 e / 1 , and suppose lim
n->oo
IIX -Xp\l = 0 11
p->
Now let each X 11 = { aC;?}. Then lim
L Iac;.>-ac:? I = 0
n-t-co m p->
For fixed m,
Thus
I
lim a~>-a~>
n->oo
I= 0
p->
uniformly in m. It then follows that there must exist a sequence X= lim a~> = 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 <
00,
0 ;;?; t < 27t.
Proof: Let X= {an}, Xezt. By the conclusion above, X has an inverse in / 1 ' 00
x- 1 = {bn}.
I: 1bnl < oo n=-co
Let n 00
g(t)f(t)
= L
00
an tnt·
n=:-co
L bn tnt n=-co
n=-co
Let us define a sub-algebra of / 1 : Definition 5.2.
A= {{an} I{an} el 1 and an= 0 for n < 0} We have easily Corollary 5.3. A is a closed sub-algebra of / 1 and hence is also a Banach algebra.
In the algebra A,
What does a homomorphism T from A on to C look like? Let
TG=zeC 00
T{an} =
L anzn, n=O
lzl;;?; 1
Thus as in the proof of Theorem 5.2, we have:
ANALYTIC FUNCTIONS INTO A BANACH SPACE
107
Theorem 5.4. Let <X)
f(z) =
L a, z"
for
n=O
Iz I ~ 1
where
I I
and letf(z) have no zeros for z ~ 1. Then 1 f(z) = ..~o b, z", Ib, I < <X)
Io <X)
oo
Example. ""z"
f(z)=2+I:-z n=l n
Problems
1. Complete the proof of Theorem 5.4, 2. (a) Show that under the definition: (X· Y)(t) = X(t) · Y(t)
the space CC(O, I) (which we already know is a Banach space) becomes a Banach algebra. (b) Show that for each maximal ideal Min cc(O, 1), there is a point t 0 in [0, 1], such that M
= {Xj X(t0 ) = 0}
(Hint: Use the Weierstrass approximation theorem which states: Let X be continuous on [a, b ]. Then, there is a sequence {P,} of polynomials such that P,(t) -4 X(t) uniformly on [a, b]. )
Index
Abelian linear algebras 26 analytic functions 96 axiom of choice 7 Banach algebras 27 Banach spaces 25 Bolzano-Weierstrass theorem 33 bounded operators 44 bounded sequences 30 bounded sets 30 Cartesian product spaces 34 Cartesian products 4 Cauchy convergence criterion 34 Cauchy domains 69, 96 Cauchy sequences 24 Cauchy's integral formula 98 Cauchy's theorem 97 chains 8, 9 closed intervals 42 closed sets 28 closure, of a set 29 compact sets 30 completeness, of a normed linear space 25 continuity of a function 39 of an operator 44 convergent sequences 24 convolution algebra 101, 102
fields 87, 88, 100 Fredholm integral equation 52 Fredholm kernel, acceptable 51 Fredholm operator 51 functionals 44 functions 4 greatest lower bounds 32 Green's functions 68 Hahn-Banach extension theorem 6067, 69, 75, 97 Hamel bases 16 harmonic functions 67, 68 homomorphisms 77 ideals 82, 83, 85, 86 image, under a function 39 inductive family 9 inner product space 36 intersection of sets 2 inverse elements 27, 84-86 isomorphisms n kernel, of a homomorphism 77 Laplace's equation 68 Lax, Peter 69 least upper bound 32 left-inverses 27 linear algebras 26 linear differential equations 59 linear operators 44 linear spaces 23 Liouville's theorem 99, 101
derivatives 91, 92 difference of sets 2 domains 67 empty set 3 equivalence relations 5, 6 existence theorems 51-60 109
110
INDEX
mappings 4 maximal ideals 85-88 maximum modulus theorem 68 Minsky, Marvin 21 monotone increasing sequences 32, 33 multiplicative principle 8
Riemann integrability 92, 93 Riemann sums 92 right-inverses 27 Russell's paradox 2
n-dimensional Euclidean space 35 normed linear algebras 26 normed linear spaces 23
Schwarz's inequality 36 singular elements 88 smooth Jordan arcs 96 subsets 3 supremum 32 symmetric relations 6
one-one mappings 5 onto mappings 5 open sets 28 operators 44 orderedpairs 3
transformations 4 transitive relations 6
partitions 92 peak, of a sequence 33 piecewise compactness 31, 33, 35 principle of choice 7, 8 quotient spaces 78 reflexive relations 6 regular elements 88, 89 regular functions 96 relations 4
union of sets 1 upper bounds 32 Volterra integral equation 59, 60 Volterra kernel, acceptable 55 Volterra operator 55 von Neumann, J. 21 Wiener, Norbert lOS Zorn's lemma 14, 62, 86
A CATALOG OF SELECTED
DOVER BOOKS IN SCIENCE AND MATHEMATICS
co
CATALOG OF DOVER BOOKS
Mathematics-Bestsellers HANDBOOK OF MATHEMATICAL FUNCTIONS: with Formulas, Graphs, and Mathematical Tables, Edited by Milton Abramowitz and Irene A. Stegun. A classic resource for working with special functions, standard trig, and exponential logarithmic definitions and extensions, it features 29 sets of tables, some to as high as 20 places. 1046pp. 8 X 10 112, 0-486-61272-4 ABSTRAGr AND CONCRETE CATEGORIES: The joy of Cats,Jiri Adamek, Horst Herrlich, and George E. Strecker. This up-to-date introductory treatment employs category theory to explore the theory of structures. Its unique approach stresses concrete categories and presents a systematic view of factorization structures. Numerous examples. 1990 edition, updated 2004. 528pp. 6 1/8 x 9 1/4. 0-486-46934-4 MATHEMATICS: Its Content, Methods and Meaning, A. D. Aleksandrov, A. N. Kolmogorov, and M.A. Lavrent'ev. Major survey offers comprehensive, coherent discussions of analytic geometry, algebra, differential equations, calculus of variations, functions of a complex variable, prime numbers, linear and non-Euclidean geometry, topology, functional analysis, more. 1963 edition. 1120pp. 5 3/8 x 8 112. 0-486-40916-3 INTRODUCTION TO VECTORS AND TENSORS: Second Edition-Two Volumes Bound as One, Ray M. Bowen and C.-C. Wang. Convenient single-volume compilation of two teXts offers both introduction and in-depth survey. Geared toward engineering and science students rather than mathematicians, it focuses on physics and engineering applications. 1976 edition. 560pp. 6 112 x 9 1/4. 0-486-46914-X AN INTRODUCTION TO ORTHOGONAL POLYNOMIALS, Theodore S. Chihara. Concise introduction covers general elementary theory, including the representation theorem and distribution functions, continued fractions and chain sequences, the recurrence formula, special functions, and some specific systems. 1978 edition. 272pp. 5 3/8 x 8 112. 0-486-47929-3 ADVANCED MATHEMATICS FOR ENGINEERS AND SCIENTISTS, Paul DuChateau. This primary text and supplemental reference focuses on linear algebra, calculus, and ordinary differential equations. Additional topics include partial differential equations and approximation methods. Includes solved problems. 1992 edition. 400pp. 7 112 X 9 1/4. 0-486-47930-7 PARTIAL DIFFERENTIAL EQUATIONS FOR SCIENTISTS AND ENGINEERS, StanleyJ. Farlow. Practical text shows how to formulate and solve partial differential equations. Coverage of diffusion-type problems, hyperbolic-type problems, elliptic-type problems, numerical and approximate method~. Solution guide available upon request. 1982 edition. 414pp. 6 1/8 x 9 1/4. 0-486-67620-X VARIATIONAL PRINCIPLES AND FREE-BOUNDARY PROBLEMS, Avner Friedman. Advanced graduate-level text examines variational methods in partial differential equations and illustrates their applications to free-boundary problems. Features detailed statements of standard theory of elliptic and parabolic operators. 1982 edition. 720pp. 6 1/8 X 9 1/4. 0-486-47853-X LINEAR ANALYSIS AND REPRESENTATION THEORY, Steven A. Gaal. Unified treatment covers topics from the theory of operators and operator algebras on Hilbert spaces; integration and representation theory for topological groups; and the theory of Lie algebras, Lie groups, and transform groups. 1973 edition. 704pp. 6 118 x 9 1/4. 0-486-47851-3
CAD!LOG OF DOVER BOOKS
A SURVEY OF INDUSTRIAL MATHEMATICS, Charles R. MacCluer. Students learn how to solve problems they'll encounter in their professional lives with this concise single-volume treatment. It employs MATLAB and other strategies to explore typical industrial problems. 2000 edition. 384pp. 5 3/8 x 8 1/2. 0-486-47702-9 NUMBER SYSTEMS AND THE FOUNDATIONS OF ANALYSIS, Elliott Mendelson. Geared toward undergraduate and beginning graduate students, this study explores natural numbers, integers, rational numbers, real numbers, and complex numbers. Numerous exercises and appendixes supplement the text. 1973 edition. 0-486-45792-3 368pp. 5 3/8 X 8 1/2. A FIRST LOOK AT NUMERICAL FUNCTIONAL ANALYSIS, W. W. Sawyer. Text by renowned educator shows how problems in numerical analysis lead to concepts of functional analysis. Topics include Banach and Hilbert spaces, contraction mappings, convergence, differentiation and integration, and Euclidean space. 1978 edition. 208pp. 5 3/8 x 8 1/2. 0-486-47882-3 FRACTALS, CHAOS, POWER LAWS: Minutes from an Infinite Paradise, Manfred Schroeder. A fascinating exploration of the connections between chaos theory, physics, biology, and mathematics, this book abounds in award-winning computer graphics, optical illusions, and games that clarify memorable insights into self-similarity. 1992 edition. 448pp. 6 l/8 x 9 1/4. 0-486-47204-3 SET THEORY AND THE CONTINUUM PROBLEM, Raymond M. Smullyan and Melvin Fitting. A lucid, elegant, and complete survey of set theory, this three-part treatment explores axiomatic set theory, the consistency of the continuum hypothesis, and forcing and independence results. 1996 edition. 336pp. 6 x 9. 0-486-47484-4 DYNAMICAL SYSTEMS, Shlomo Sternberg. A pioneer in the field of dynamical systems discusses one-dimensional dynamics, differential equations, random walks, iterated function systems, symbolic dynamics, and Markov chains. Supplementary materials include PowerPoint slides and MATLAB exercises. 2010 edition. 272pp. 6 1/8 X 9 1/4. 0-486-47705-3 ORDINARY DIFFERENTIAL EQUATIONS, Morris Tenenbaum and Harry Pollard. Skillfully organized introductory text examines origin of differential equations, then defines basic terms and outlines general solution of a differential equation. Explores integrating factors; dilution and accretion problems; Laplace Transforms; Newton's Interpolation Formulas, more. 818pp. 5 3/8 x 8 112. 0-486-64940-7 MATROID THEORY, D. J. A Welsh. Thxt by a noted expert describes standard examples and investigation results, using elementary proofs to develop basic matroid properties before advancing to a more sophisticated treatment. Includes numerous 0-486-47439-9 exercises. 1976 edition. 448pp. 5 3/8 x 8 1/2. THE CONCEPT OF A RIEMANN SURFACE, Hermann Weyl. This classic on the general history of functions combines function theory and geometry, forming the basis of the modem approach to analysis, geometry, and topology. 1955 edition. 208pp. 5 3/8 X 8 1/2. 0-486-47004-0 THE LAPLACE TRANSFORM, David Vernon Widder. This volume focuses on the Laplace and Stieltjes transforms, offering a highly theoretical treatment. Topics include fundamental formulas, the moment problem, monotonic functions, and Tauberian 0-486-47755-X theorems. 1941 edition. 416pp. 5 318 x 8 1/2.
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Mathematics-Logic and Problem Solving PERPLEXING PUZZLES AND TANTALIZING TEASERS, Martin Gardner. Ninety-three riddles, mazes, illusions, tricky questions, word and picture puzzles, and other challenges offer hours of entertainment for youngsters. Filled with rib-tickling drawings. Solutions. 224pp. 5 3/8 x 8 1/2. 0-486-25637-5 MY BEST MATHEMATICAL AND LOGIC PUZZLES, Martin Gardner. The noted expert selects 70 of his favorite "short" puzzles. Includes The Returning Explorer, The Mutilated Chessboard, Scrambled Box Tops, and dozens more. Complete solutions included. 96pp. 5 3/8 x 8 1/2. 0-486-28152-3 THE LADY OR THE TIGER?: and Other Logic Puzzles, Raymond M. Smullyan. Created by a renowned puzzle master, these whimsically themed challenges involve paradoxes about probability, time, and change; metapuzzles; and self-referentiality. Nineteen chapters advance in difficulty from relatively simple to highly complex. 1982 edition. 240pp. 5 3/8 x 8 1/2. 0-486-47027-X SATAN, CANTOR AND INFINITY: Mind-Boggling Puzzles, RaymondM. Smuliyan. A renowned mathematician tells stories of knights and knaves in an entertaining look at the logical precepts behind infinity, probability, time, and change. Requires a strong background in mathematics. Complete solutions. 288pp. 5 3/8 x 8 112. 0-486-47036-9 THE RED BOOK OF MATHEMATICAL PROBLEMS, Kenneth S. Williams and Kenneth Hardy. Handy compilation of 100 practice problems, hints and solutions indispensable for students preparing for the William Lowell Putnam and other mathematical competitions. Preface to the Frrst Edition. Sources. 1988 edition. 192pp. 0-486-69415-1 5 3/8 X 8 1/2. KING ARTHUR IN SEARCH OF HIS DOG AND OTHER CURIOUS PUZZLES, Raymond M. Smullyan. This fanciful, original collection for readers of all ages features arithmetic puzzles, logic problems related to crime detection, and logic and arithmetic puzzles involving King Arthur and his Dogs of the Round Table. 160pp. 5 '3/8 x 8 1/2. 0-486-47435-6 UNDECIDABLE THEORIES: Studies in Logic and the Foundation of Mathematics, Alfred Tarski in collaboration with Andrzej Mostowski and Raphael M. Robinson. This well-known book by the famed logician consists of three treatises: "A General Method in Proofs of Undecidability," "Undecidability and Essential Undecidability in Mathematics," and "Undecidability of the Elementary Theory of Groups." 1953 edition. 112pp. 5 3/8 x 8 112. 0-486-47703-7 LOGIC FOR MATHEMATICIANS,]. Barkley Rosser. Examination of essential topics and theorems assumes no background in logic. "Undoubtedly a major addition to the literature of mathematical logic." -Bulletin ofthe American Mathematical Society. 1978 edition. 592pp. 6 1/8 x 9 114. 0-486-46898-4 INTRODUCTION TO PROOF IN ABSTRACT MATHEMATICS, Andrew Wohlgemuth. This undergraduate text teaches students what constitutes an acceptable proof, and it develops their ability to do proofs of routine problems as well as those requiring creative insights. 1990 edition. 384pp. 6 1/2 x 9 1/4. 0-486-47854-8 FIRST COURSE IN MATHEMATICAL LOGIC, Patrick Suppes and Shirley Hill. Rigorous introduction is simple enough in presentation and context for wide range of students. Symbolizing sentences; logical inference; truth and validity; truth tables; terms, predicates, universal quantifiers; universal specification and laws of identity; more. 288pp. 5 3/8 x 8 1/2. 0-486-42259-3
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Mathematics-Algebra and Calculus VECTOR CALCULUS, Peter Ba.xandall and Hans Liebeck. This introductory text offers a rigorous, comprehensive treatment. Classical theorems of vector calculus are amply illustrated with figures, worked examples, physical applications, and exercises Q-486-46620-5 with hints and answers. 1986 edition. 560pp. 5 3/8 x 8 1/2. ADVANCED CALCULUS: An Introduction to Classical Analysis, Louis Brand. A course in analysis that focuses on the functions of a real variable, this text introduces the basic concepts in their simplest setting and illustrates its teachings with numerous examples, theorems, and proofs. 1955 edition. 592pp. 5 3/8 x 8 112. 0-486-44548-8 ADVANCED CALCULUS, Avner Friedman. Intended for students who have already completed a one-year course in elementary calculus, this two-part treatment advances from functions of one variable to those of several variables. Solutions. 1971 edition. 0-486-45795-8 432pp. 5 3/8 X 8 1/2. METHODS OF MATHEMATICS APPLIED TO CALCULUS, PROBABiliTY, AND STATISTICS, Richard W. Hamming. This 4-part treatment begins with algebra and analytic geometry and proceeds to an exploration of the calculus of algebraic functions and transcendental functions and applications. 1985 edition. Includes 310 figures Q-486-43945-3 and 18 tables. 880pp. 6 1/2 x 9 1/4. BASIC ALGEBRA I: Second Edition, Nathan jacobson. A classic text and standard reference for a generation, this volume covers all undergraduate algebra topics, including groups, rings, modules, Galois theory, polynomials, linear algebra, and associative algebra. 1985 edition. 528pp. 6 1/8 x 9 114. 0-486-47189-6 BASIC ALGEBRA II: Second Edition, Nathan jacobson. This classic text and standard reference comprises all subjects of a first-year graduate-level course, including in-depth coverage of groups and polynomials and extensive use of categories and functors. 1989 edition. 704pp. 6 118 x 9 1/4. 0-486-47187-X CALCULUS: An Intuitive and Physical Approach (Second Edition), Morris Kline. Application-oriented introduction relates the subject as closely as possible to science with explorations of the derivative; differentiation and integration of the powers of x; theorems on differentiation, antidifferentiation; the chain rule; trigonometric func0-486-40453-6 tions; more. Examples. 1967 edition. 960pp. 6 1/2 x 9 1/4. ABSTRACT ALGEBRA AND SOLUTION BY RADICALS, John E. Maxfield and Margaret W Maxfield. Accessible advanced undergraduate-level text starts with groups, rings, fields, and polynomials and advances to Galois theory, radicals and roots of unity; and solution by radicals. Numerous examples, illustrations, exercises, appen0-486-47723-1 dixes. 1971 edition. 224pp. 6 1/8 x 9 1/4. AN INTRODUCTION TO THE THEORY OF LINEAR SPACES, Georgi E. Shilov. Translated by Richard A. Silverman. Introductory treatment offers a clear exposition of algebra, geometry, and analysis as parts of an integrated whole rather than separate subjects. Numerous examples illustrate many different fields, and problems include hints or answers. 1961 edition. 320pp. 5 3/8 x 8 1/2. 0-486-63070-6 LINEAR ALGEBRA, Georgi E. Shilov. Covers determinants, linear spaces, systems of linear equations, linear functions of a vector argument, coordinate transformations, the canonical form of the matrix of a linear operator, bilinear and quadratic forms, and more. 387pp. 5 3/8 x 8 1/2. 0-486-63518-X
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Mathematics-History THE WORKS OF ARCHIMEDES, Archimedes. Translated by Sir Thomas Heath. Complete works of ancient geometer feature such topics as the famous problems of the ratio of the areas of a cylinder and an inscribed sphere; the properties of conoids, Q-486-42084-1 spheroids, and spirals; more. 326pp. 5 3/8 x 8 1/2. THE HISTORICAL ROOTS OF ELEMENTARY MATHEMATICS, Lucas N. H. Bunt, Phillip S. Jones, and Jack D. Bedient. Exciting, hands-on approach to understanding fundamental underpinnings of modem arithmetic, algebra, geometry and number systems examines their origins in early Egyptian, Babylonian, and Greek Q-486-25563-8 sources. 336pp. 5 3/8 x 8 1/2. THE THIRTEEN BOOKS OF EUCLID'S ELEMENTS, Euclid. Contains complete English text of all 13 books of the Elements plus critical apparatus analyzing each definition, postulate, and proposition in great detail. Covers textual and linguistic matters; mathematical analyses of Euclid's ideas; classical, medieval, Renaissance and modem commentators; refutations, supports, extrapolations, reinterpretations and historical notes. 995 figures. Total of 1,425pp. All books 5 3/8 x 8 1/2. Vol. I: 443pp. 0486-60088-2 Vol. II: 464pp. Q-486-60089-0 Vol. III: 546pp. D-486-60090-4 A HISTORY OF GREEK MATHEMATICS, Sir Thomas Heath. This authoritative two-volume set that covers the essentials of mathematics and features every landmruk innovation and every important figure, including Euclid, Apollonius, and others. 5 3/8 x 8 112. Vol. I: 461pp. Q-486-24073-8 Vol. II: 597pp. 0-486-24074-6 A MANUAL OF GREEK MATHEMATICS, Sir Thomas L Heath. This concise but thorough history encompasses the enduring contributions of the ancient Greek mathematicians whose works form the basis of most modem mathematics. Discusses Pythagorean arithmetic, Plato, Euclid, more. 1931 edition. 576pp. 5 3/8 x 8 l/2. 0-486-43231-9 CHINESE MATHEMATICS IN THE THIRTEENTH CENTURY, Ulrich Ubbrecht. An exploration of the 13th-century mathematician Ch'in, this fascinating book combines what is known of the mathematician's life with a history of his only extant work, the Shu-shu chiu-chang. 1973 edition. 592pp. 5 3/8 x 8 1/2. 0-486-44619-0 PHILOSOPHY OF MATHEMATICS AND DEDUCTIVE STRUCTURE IN EUCLID'S ELEMENTS, Ian Mueller. This text provides an understanding of the classical Greek conception of mathematics as expressed in Euclid's Elements. It focuses on philosophical, foundational, and logical questions and features helpful appendixes. 400pp. 6 1/2 X 9 1/4. 0486-45300-6 BEYOND GEOMETRY: Classic Papers from Riemann to Einstein, Edited with an Introduction and Notes by Peter Pesic. This is the only English-language collection of these 8 accessible essays. They trace seminal ideas about the foundations of geometry that led to Einstein's general theory of relativity. 224pp. 6 1/8 x 9 1/4. 0-486-45350-2 HISTORY OF MATHEMATICS, David E. Smith. Two-volume history - from Egyptian papyri and medieval maps to modem graphs and diagrams. Non-technical chronological survey with thousands of biographical notes, critical evaluations, and · contemporary opinions on over 1,100 mathematicians. 5 3/8 x 8 1/2. Vol. I: 618pp. 0-486-20429-4 Vol. II: 736pp. 0-486-20430-8
