Functional Analysis by
Harro G. Heuser University �f Karlsruhe, Federal Republic of Germany
translated by John Horvath University of Marylan.t
A Wiley-Interscience Publication
JOHN WILEY & SONS Chichester . New York . Brisbane . Toronto . Singapore
This edition is published by permission of Vedag B G Teubner, Stuttgart. and is the sole authorized English translation of the original German edition. Copyright © 1982by John Wiley & Sons Ltd. All rights reserved. No part of this book may be reproduced by any means, nor transmitted. nor translated into a machine language without the written permission of the publisher. British Library Cataloguing in Publication Data: Heuser, Harro Functional analysis. I.
Functional analysis
I. Title 515.7 ISBN
0471
II. Funktionanalysis.
English
QA320 280526 (cloth)
0471100692 (paper) Filmset by Composition House Limited. Salisbury, Wiltshire. England. Printed in the United States of America.
To my mother
Preface A science as vitally important as functional analysis can fortunately not be
defined. But its great 'Leitmotiv' can nevertheless be indicated: It is the fusion of algebraic and topological structures. This seemingly abstract and anaemic subject has developed in such a rich and lively manner that nowadays functional analysis has a strong influence on a great number of completely different fields inside and outside of mathematics: systems engineers and atomic physicists cannot do without it, just like mathematicians working in numerical analysis, differential or integral equations, the theory of approximation or representation theory-just to name a few. The purpose of this book is to give a lively and thorough exposition of the basic concepts, the essential statements, the main methods and finally also of the way of thinking of functional analysis and to satisfy the needs of a large circle of users, in content as well as didactically. To achieve this purpose I chose an orientation toward problems. The present book starts out, whenever possible, from questions and facts of classical analysis and algebra, and tries to get to their core by leaving aside what is accidental in order to obtain general concepts and assertions. Conversely, it aims at making the newly acquired tools fruitful for the classical fields. Naturally in the course of this process so many questions of a 'purely functional analytic nature' accumulate that finally functional analysis is propelled by its own problems. But the main text and the exercises return constantly to the familiar world of analysis and algebra. I hope to enhance in this way the motivation and the intuitive understanding of the reader and to save him from that particular feeling of being lost, which occurs so easily and annoyingly when studying an abstract theory. I think that in this way I also acquit myself best of the duty imposed on an author by the word 'Leitfaden,l, i.e., leading thread. The first leading thread was that of Ariadne; Plutarch reports about it in his biography of Theseus the following2: ' I
•
•
•
after he (Theseus) arrived in Creta, he slew the
The original German edition was published in the 'Mathematische Leitfaden' series of
B.
G.
Teubner. Plutarch: The Lives of the Noble Grecians and Romans, translated by Thomas North, the None such Press, London, 1929.
2
vii
viii Minotaur ... by the means and help of Ariadne, who being fallen in fancy with him, did give him a clue of thread, by the help whereof she taught him how he might easily wind out of the turnings and cranks of the Labyrinth'. An organization directed towards problems does not try to represent a science as it evolved-this is the purpose of the genetic method-but how it could also have evolved. It is the coming together offortunate circumstancesmaybe even more-that one of the foundations of this book, the concept of a bilinear system, already is tacitly the basis of the pioneering works of Fredholm concerning integral equations, which started functional analysis. The problem of the solvability of Fredholm's integral equations will therefore playa central role in this book. It leads us through the Neumann series to the theory of Banach algebras and through the concept of a bilinear system and of normal solvabili ty to the extension principle of Hahn-Banach and to the duality theory which follows from it, a crown jewel of functional analysis. It should be observed emphatically that the investigation of Fredholm's integral equation originates from a very concrete situation: many boundary value problems of physics and of technology can be transformed into an equation of this kind. The table of contents gives detailed information about the subjects treated and their interdependence. I want to indicate here only a few major blocks. The first seven chapters are dominated by the problem of equations: under what conditions are equations in general spaces solvable, how can they be effectively solved, and how do the solutions depend on certain initial data? Mainly linear problems will be considered; the non-linear domain will be represented by the fixed-point theorems of Banach and Schauder; the latter occur, however, first only in §106. In Chapter VIII approximation problems will arise. The great subject of orthogonality will emerge here, and will be developed in the following two chapters (orthogonal decomposition and spectral theory in Hilbert spaces). Chapter XI to XIII reach the summit of the progress towards abstraction: they show how linear and topological structures are fusioned in the concept of a topological vector space. This fusion leads, if the structures are rich enough, back to a quite concrete situation: commutative, complex B*-algebras are nothing else but algebras ofcontinuous functions. With this theorem of Gelfand and Neumark the book concludes. The monumental work of Dunford and Schwartz 'Linear Operators' has 2592 pages; the theory of topological vector spaces is only touched in it. It is needless to say that my book had to sacrifice many important or only attractive subjects. I did not want to sacrifice, however, a copious motivation, an illustration from several angles of the central facts, and the details of the proofs. F or long stretches only the elements of analysis and linear algebra are required as prerequisites. The concept of a metric space and its continuous maps will be developed in the book. I listed in §81 and §lOl without proofs the few topological facts which will be needed from §82 on. The spectral theory in Chapter VII cannot be studied profitably without some knowledge of the theory of analytic functions of a complex variable: one needs Cauchy's integral theorem, Liouville's theorem, the Taylor and the Laurent expansions. At a few places theorems
IX
from real and complex analysis will be used which might not be familiar to every reader (e.g., the Stone-Weierstrass theorem); in these cases I have indicated references to the literature, where the proofs can easily be read. The Lebesgue integral and the theory of partial differential equations will not be used. I have to thank cordially the Universidad de los Andes in Bogota (Colombia) and the University of Toronto in Toronto (Canada): they gave me through visiting positions the possibility to work intensively on this book. Mrs. Mia Miinzel put her home, which lies quietly above Lake Garda, at my disposition for the last chapters; I am deeply indebted to her. My special thanks go to Dr. H. Kroh, Dr. U. Mertins and Mr. G. Schneider, further to Mrs. Y. Paasche and Mrs. K. Zeder. The three gentlemen were at my side during the preparation of this book with advice and help, they improved it and cleaned it up through valuable indications, and read all the proofs; the two ladies transformed with unbelievable patience a miserable manuscript into a clean typescript. I thank the Teubner-Verlag cordially for the pleasant collaboration and for its watchful help and assistance. Nastiitten in Taunus, August 1975
HARRO HEUSER
Table of contents vii
Preface List of symbols Introduction I.
II.
III.
xv
Banach's fixed-point theorem ~ 1 Metric spaces . ~2 Banach's fixed-point theorem §3 Some applications of Banach's fixed-point theorem
6 15 17
Normed spaces §4 Vector spaces §5 Linear maps §6 Normed spaces §7 Continuous linear maps §8 The Neumann series . §9 Normed algebras . §1O Finite-dimensional normed spaces *11 The Neumann series in non-complete normed spaces. §12 The completion of metric and normed spaces §13 Compact operators
24 29 33 41 46 51 55 60 65 69
Bilinear systems and conjugate operators §14 Bilinear systems §15 Dual systems . §16 Conjugate operators §17 The equation (I - K)x = y with finite-dimensional K §18 The equation (R - S)X = Y with a bijective Rand finite-dimensional S . §19 The Fredholm integral equation with continuous kernel §20 Quotient spaces §21 The quotient norm §22 Quotient algebras . xi
75 78 83 90 95 98 100 103 105
XII
IV.
V.
VI.
VII.
Fredholm operators §23 Operators with finite deficiency §24 Fredholm operators on normed spaces . §25 Fredholm operators in saturated operator algebras §26 Representation theorems for Fredholm operators. §27 The equation Ax = y with a Fredholm operator A
107
110 114 122 124
Four principles of functional analysis and some applications §28 The extension principle of Hahn-Banach . §29 Normal solvability §30 The normal solvability of the operators I - K with compact K . §31 The Baire category principle. §32 The open mapping principle and the closed graph theorem §33 The principle of uniform boundedness . §34 Some applications of the principles of functional analysis to analysis §35 Analytic representation of continuous linear forms §36 Operators with closed image space . §37 Fredholm operators on Banach spaces .
144 151 155 158
The §38 §39 §40 §41 §42 §43
160 164 166 169 173 176
Riesz-Schauder theory of compact operators Operators with finite chains . Chain-finite Fredholm operators The Riesz theory of compact operators. The bidual of a normed space. Reflexivity The dual transformation of a compact operator Singular values and eigenvalues of a compact operator
Spectral theory in Banach spaces and Banach algebras §44 The resolvent . §45 The spectrum . §46 Vector-valued holomorphic functions. Weak convergence §47 Power series in Banach algebras. §48 The functional calculus . *49 Spectral projectors §50 Isolated points of the spectrum §51 The Fredholm region. §52 Riesz operators §53 Essential spectra . §54 Normaloid operators .
128 133 136 137 138 142
181 183 186 194 200 204 207 209 217 221 223
XllI
VIII.
IX.
X.
XI.
Approximation problems in normed spaces §55 An approximation problem §56 Strictly convex spaces. §57 Inner product spaces . §58 Orthogonality §59 The Gauss approximation §60 The general approximation problem §61 Approximation in uniformly convex spaces §62 Approximation in reflexive spaces
230 233 235 239 244 247 249 252
Orthogonal decomposition in Hilbert spaces §63 Orthogonal complements §64 Orthogonal series . *65 Orthonormal bases §66 The dual of a Hilbert space §67 The adjoint transformation
254 255 258 260 263
Spectral theory in Hilbert spaces §68 Symmetric operators . §69 Orthogonal projectors §70 Normal operators and their spectra. §71 Normal meromorphic operators. §72 Symmetric compact operators §73 The Sturm-Liouville eigenvalue problem §74 Wielandt operators §75 Determination and estimation of eigenvalues §76 General eigenvalue problems for differential operators §77 Preliminary remarks concerning the spectral theorem for symmetric operators §78 Functional calculus for symmetric operators §79 The spectral theorem for symmetric operators on Hilbert spaces . Topological vector spaces §80 Metric vector spaces §81 Basic notions from topology. §82 The weak topology §83 The concept of a topological vector space. Examples . §84 The neighborhoods of zero in topological vector spaces §85 The generation of vector space topologies . §86 Subspaces, product spaces and quotient spaces §87 Continuous linear maps of topological vector spaces §88 Finite-dimensional topological vector spaces §89 Fredholm operators on topological vector spaces .
265 270 272 277 280 282 286 291 296 301 303 305
309 312 318 321 327 330 332 334 336 338
xiv XII.
XIII.
XIV.
Locally convex vector spaces ~90 Bases of neighborhoods of zero in locally convex vector spaces ~91 The generation of locally convex topologies by seminorms ~92 Subspaces, products and quotients of locally convex spaces ~93 Normable locally convex spaces. Bounded sets Duality and compactness §94 The Hahn-Banach theorem §95 The topological characterization of normal solvability §96 Separation theorems §97 Three applications to normed spaces §98 Admissible topologies §99 The bipolar theorem. §100 Locally convex topologies are 6-topologies §101 Compact sets. §102 The Alaoglu-Bourbaki theorem § I 03 The characterization of the admissible topologies §104 Bounded sets in admissible topologies. §105 Barrelled spaces. Reflexivity §106 Convex, compact sets: The theorems of Krein-Milman and Schauder The representation of commutative Banach algebras §107 Preliminary remarks on the representation problem. ~108 Multiplicative linear forms and maximal ideals §109 The Gelfand representation theorem . §11O The representation of commutative B*-algebras
340 342 345 346
349 350 351 353 355 356 358 360 363 364 365 367 372
381 384 387 389
Bibliography
393
Index .
401
List of symbols 86,336 85 84,263 31
A A+ A* A(E)
39 39
B(T) BV[a, b]
1 310 39 39,310 39 383 39 39 28 341
C qa, b) C[a, b] c
or
F
f: {E -->fF ) x f-+ (x
(x f-+ f(x) means that with the element x one associates the image f(x». The notation f: Xf-+ f(x) or even simpler Xf-+ f(x) is also used; then the set of definition and the target set must be given separately, if they are not evident. At times it will be handy-and harmless-to infringe on these notational conventions. Thus, for instance, we shall speak of the function sin x (instead of x f-+ sin x), of the polynomial x 2 (instead of x f-+ x 2), and of the kernel k(s, t) (instead of (s, t) f-+ k(s, t» of an integral equation. A self-map of E is a map from E into E. The identity map iE of E is the map x f-+ x (x E E). Two mapsf,: E, --> F"f2: E2 --> F2 are said to be equal if E, = E 2 , F, = F2 and f, (x) = f2(X) for all x E E ,. Let f: E --> F be given. For AcE, B c F the set f(A) := {f(x) E F: x E A} is the image of A, f-'(B):= {x E E: f(x) E B} is the preimage of B. f is sa,id to be surjective if f(E) = F, injective if f(x) = f(y) implies x = y, and bijective if it is both surjective and injective. The locution 'f maps E onto F' means that f is surjective. A family (a,: IE J) is only another name and writing for the map I f-+ a, from a set J of indices into a set A. When J = N one rather speaks of a sequence than of a family. A sequence ai' a2, ... of elements of A will be denoted briefly by (an) or (an) c A and occasionally also by (ai' a2, ... ). The wordfunction (and also functional) will in general be used only for maps from a set E into the field of scalars K (scalar-valued or K-valued maps). If the maps f: E --> F, g: F --> G are given (observe that the target set of f is the set of definition of g), then their composition (product) is the map 9 f from E into G which associates with each x E E the image (g f)(x) := g(.f(x» in G. For a self-map f of E, the iterates (powers) I" are defined by fO := iE, I" := f 1"- '(n = 1,2, ... ). Every injective map f: E --> F has an inverse map 0
0
0
f-
One has f-
1
0
f
1:
{f(E) --> E f(x) f-+ x.
= iEJ f- ' = i/(E)' 0
3
f: E --+ F is then and only then bijective, if there exist maps g: F --+ E, h: F --+ E withgof = iE,foh = iF. In thiscaseg = h = f-I. The following rules will be used frequently for the study of maps f: E --+ F (let A, A I be subsets of E, while B, B, are subsets of F ;f- I(B) is the above defined preimage, where f is not assumed to be injective):
U
f(U AI) = f(A,); 'EJ 'EJ
f-I(f(A» => A; f injective ¢> f- 1(f(A» = A for every AcE; f(f- t(B» If the map g: F
c
B; f surjective ¢> f(f-l(B»
--+
G is also given, then
= B for every Be
F.
for every C c G. The cartesian product of a family (E,; I E J) of non-empty sets E, is the set of all families (x, E E ,: IE J), i.e., the set of all maps I f--+ X, E E, defined on J. It will be denoted TI'EJ E" and in the case of a finite or countable index set also by EI x ... x En or EI x E2 X ... , respectively. EI x ... x En is the set of all n-tuples (XI' ... , x n), Xk E Ek for k = 1, ... , n, while E t x E2 x ... is the collection of all sequences (x I' X2, ... ) with Xk E Ek for kEN. There is no risk of confusing a couple (XI' X2) E EI x E2 with an open interval. For a given X := (x,: IE J) E TI'EJ E, the element x, is called thecomponentofx in E,; the maps 11: , : (x,: IE J) f--+ X, are called the projections or projectors onto the components.
Rules of complementation
M
Let (A,: I E J) be a family of subsets of E and M' := E\M the complement of c E in E. Then
4
Zorn's lemma For certain pairs x, y of elements of a set 9)1 '" 0 let a relation 'x defined, which satisfies the following axioms: 1.
2. 3.
« y' be
x « x for every x E 9)l. If x « y and y « x, then x = y. If x « y and y « z, then x « z.
Such a relation is called an order on 9)1 and 9)1 itself is said to be an ordered set. An order is called total, and m a totally ordered set, if two elements x, y ofm are always comparable, i.e., if either x « y or y « x holds. Every subset of 9)1 becomes through the order on 9)1 an ordered, possibly even a totally ordered set. y E 9)1 is called an upper bound for 91 c 9)1 if x « y for every x 6 ~n. z E 9)1 is a maximal element if z « x holds only for x = z. Zorn's lemma is as follows: If every totally ordered subset ofan ordered set m has an upper bound in 9)1, then there exists in m at least one maximal element.
Inequalities The quantities iX k, 13k, f(x), g(x) which occur in what follows are complex numbers. The sums are finite or infinite; in the latter case it will be supposed that every series, which is on the right hand side of an inequality, converges. In the integral inequalities it is enough to assume for our purposes that the integrands are continuous functions on a finite interval of the real line. Proofs can be found in [180]. HOlder's inequalities: If p > 1 and lip + 11q = 1, then
L IiXkf3k I ~ (I IlXkIP)IIP(I I13k Iq)llq, f1f(X)9(X)ldX
~ (f'f(X)IP dx r'P(f,g(X),q dx r
1q
;
for p = q = 2 one obtains the Cauchy-Schwarz inequalities: I IOC k f3kl ~ (I IlXkI2)1/2(I If3kI 2)112,
f1f(X)9(X)ldX
~ (flf(X W dx r/2(flg(X)12 dXr/2.
Minkowski's inequalities: For p ;?; lone has
(L IlXk + f3kI P)I/P ~ (I IOCkIP)IIP + (Ilf3kI P)IIP,
(b )l1P (iab (Ja1f(x)+g(x)IPdx If(x)IPdx )l/P + (ibalg(x)IPdx ) ~
lip
.
For p > 1 equality holds in the Minkowski inequalities if and only if one of the sequences (lXk), (13k) or one of the functionsf, 9 is a non-negative mUltiple of the other.
5 Labeling of results, references
To indicate the significance of the results, we use the hierarchy 'Lemma, Proposition, Theorem, Principle'. The .statements within each class will be numbered consecutively in each section, thus Lemma 27.1 is the first lemma in §27, Proposition 27.1 the first proposition in §27, and Theorem 27.1 the first theorem in §27. The examples are numbered correspondingly. The exercises are at the end of each section; cross-references like '§16 Exercise 6' or 'Exercise 6 in §16' need no explanation. The sections (paragraphs) are numbered consecutively (§1 to §11O). Square brackets refer to the bibliography. For the easier orientation of the reader it is divided into several sections, which are of course not completely independent of each other. We call the reader's attention especially to the section 'expository articles'; in them he will be made familiar easily and thoroughly with historical development, the essential problems, the fundamental notions and methods of certain fields offunctional analysis; most of these articles also contain a very detailed bibliography. Exercises
The exercises form an essential part of this book. They serve to get practice in the knowledge and methods of the main text (and give thereby also an opportunity for the reader to check whether he has understood it), they prepare for subsequent developments and communicate further interesting facts from functional analysis. Some exercises will be needed in the course of the main text, they are marked with a star in front of their number (e.g., *5). The reader is earnestly urged to work the exercises, the starred ones as well as the unstarred ones; the very hasty user of this book should at least glance through them. Those unstarred exercises which contain particularly interesting facts from functional analysis, which are not treated in the main text, are indicated by a plus sign in front of the number (e.g., + 2). Guide
Basic are Chapters I-III and sections 28, 31-33. The following division according to subjects does not contain them any more. The division is to be understood only as an orientation; it does not show the existing logical interdependence: (A) Geometry of spaces: 1. Normed spaces: Chapter VIII (§41 must be read before §61) 2. Inner product spaces: §§57-59, Chapter IX 3. Topological vector spaces: Chapters XI-XIII (B) Banach algebras: §§44-48, Chapter XIV (C) Operators on normed spaces: §§29, 36,41,44-50,53,54,97 (0) Operators on inner product spaces: §67, Chapter IX (E) Fredholm operators: Chapter IV, §§37, 39, 51, 89 (F) Compact and Riesz operators: §30, Chapter VI, §§52, 72, 75.
I
Banach's fixed point theorem
§ 1. Metric spaces One of the basic concepts of classical analysis is the concept of convergence of a sequence of numbers. This in turn is based on the concept of distance: in fact the convergence of the sequence (Xk) to x means that the distance IXk - x I of the k-th term xdrom the limit x will be arbitrarily small when k increases beyond all bounds. A corresponding definition is given for sequences of elements of Kn, where the distance d(x, y) between x = (~1' ... ' ~n) and y = (111, ... , I1n) is for instance defined by (1.1)
If one wants to build a theory of convergence, which is valid for sequences of
numbers as well as for sequences of vectors, and which eventually disregards completely the nature of the terms of the sequences, then one cannot use definitions of a distance like (1.1) (which make sense only for concretely given objects), but rather one will have to work with some properties, which intuitively every reasonable concept of distance must have. Such properties are for instance the following, where we call 'points' the objects between which the distances are defined: The distance of a point from itself and only from itself is 0, the distance of a point x from a point y is exactly as large as conversely the distance of the point y from the point x (symmetry property of distance), and finally a 'deviation property': If one does not go from the point x directly to the point y but first to the point z and then from there to x, then one has not made a shortcut, possibly one covered a larger distance. We need now only to express these properties precisely in the language of mathematical formulas to obtain the fundamental concepts of a metric and of a metric space: 6
7 A function d, which associates with any two elements x, y of a non-empty set E a real number d(x, y), is called a metric on E, if it possesses the following properties: (Ml) (M2) (M3)
d(x, y) ~ 0, where d(x, y) = 0 if and only if x = y, d(x, y) = d(y, x), d(x, y) ;;;; d(x, z) + d(z, y).
A metric space (E, d) is a (non-empty) set E on which a metric is defined. The elements of a metric space are usually called points, the number d(x, y) is the distance between the points x and y. (M3) is called the triangle inequality. As we shall soon see, many different metrics can be defined on a set E. The notation (E, d) for a metric space takes this fact into account: it indicates not only the underlying set E but also the metric d given on it. This careful notation is not always necessary, we shall take therefore frequently the liberty to speak simply of the metric space E instead of (E, d). The metric of the metric space E will then always be denoted by d. The best known metric space is the field K with the distance d(x, y) := Ix - y I. The examples which now follow are only the first samples, many others will be presented later. Basically we have to show for the distances defined below that axioms (Ml)-(M3) are satisfied. However, if an axiom can be trivially verified, then we will not mention it at all. Example 1.1. Let p ~ 1 be a fixed real number. We define a metric d on K" by associating with two points x = (~I' ... , ~"), y = (1] 1, ... , 1]") of K" the distance (1.2)
d(x, y)
:=
CtII~. -
1]. IP
riP.
The triangle inequality (M3) follows directly from Minkowski's inequality, since with z = (' 1, . . . , '") one has
; ct
I~. -
,. IP
riP +
ct
1'. -
riP = d(x, z) + d(z, y).
1]. IP
Our example shows that infinitely many metrics can be defined on K". For p = 2 we obtain the euclidean metric already mentioned in (1.1). If on K" we introduce the distance (1.2), then we denote the metric space so obtained by IP(n). Example 1.2. We obtain the metric space lOO(n) by defining on K", with the notations of Example 1.1, the so-called maximum-metric through (1.3)
d(x, y):= m~x I~. - 1].1. .=1
8 ~ p ~ 00.
Thus the spaces IP(n) have been introduced for 1
Example 1.3. Let T be a non-empty set and B(T) the set of all bounded functions x: T ~ K. B( T) becomes a metric space by means of the supremummetric, which we define by
(1.4)
d(x, y):= suplx(t) - y(t)l. lET
If T = N, then B(T) is the setl oo of all bounded sequences x = with the definition of distance
(~n)' Y =
(YIn), ...
00
d(x, y):= suPI~n - Ylnl. n=l
(1.5)
For T = {I, ... , n} the space B(T) is obviously the lOO(n) of Example 1.2. Example 1.4. Let C[a, b] be the set of all functions x with values in K which are continuous on the (finite closed) interval [a, b]. On C[a, b] we introduce the maximum-metric by
d(x, y):= max Ix(t) - y(t) I.
(1.6)
a~t~b
Example 1.5. by
On the set (s) of all sequences with terms in K we define a metric
d(
(1.7)
x, y
~ 1
) :=
~ 2n 1
n=l
I~n
-
+ I'-on J:
YIn I I' YIn
_
where x = (~1' ~2'·· .), Y = (Yl1' Yl2,.· .). The convergence of the series L 1/2n ensures that d(x, y) exists for all points x, y in (s). To prove (M3) we observe that the function t 1---+ t/(1 + t) has a positive derivative for t > -1, and is therefore increasing. From here follows the estimate (1.8)
liX + 131 < liXl + 1131 < liXl + 1131 1 + liX + 131 = 1 + liXl + 1131 = 1 + liXl 1 + 1131
for arbitrary scalars iX,
13, and so for any z =
(' 1,
'2, ... ) the triangle inequality
It is noteworthy that d(x, y) ;;; 1 for all x, y in (s). The metrics introduced in the above examples are canonical in the sense that e.g., B(T) will always be equipped with the supremum-metric-unless the contrary is
explicitly stated.
9 The reader will have noticed that it does not appear from the symbols lP(n), B(T), C[a, b] and (s) whether the field R or the field C was used to construct these spaces. Indeed, our results do not depend on whether e.g., the functions in B(T) have real or complex values, hence a distinction is unnecessary. The next example shows that every non-empty set can be made into a metric space.
Example 1.6.
On any set E 1=
0
the so-called discrete metric can be defined
by (1.9)
d(x, y):=
{~
if x 1= y, if x = y.
Metric spaces should serve to obtain the concept of convergence in its pure form. This can be achieved through the following definition. A sequence (Xk) in a metric space E converges to a point x E E-in symbols: Xk -> x or lim Xk = x -, if d(Xb x) converges to 0, i.e., if given t; > there exists an index ko(t;) such that d(Xk' x) ~ e for all k ~ ko(e). The point x is called the limit of the sequence (Xk)' The limit x is uniquely determined. Indeed, if (Xk) converges also to y, then it ~ d(x, y) ~ d(x, Xk) + d(Xk, y) = d(Xb x) + d(Xb y) -+ that follows from d(x, y) = 0, hence x = y. The spaces lP(n), B(T), C[a, b] and (s) arefunction spaces, i.e., their elements are (scalar-valued) functions with a common set of definition D. If D = N, then the function space is most often called a sequence space; thus 100 and (s) are sequence spaces. In a function space there exists a natural concept of convergence, that of pointwise convergence (in a sequence space or in K" it is preferable to speak of componentwise convergence): The sequence of functions (Xk) converges pointwise to the function x if xk(t) -+ x(t) for every t in the common set of definition D. It is a useful exercise, which we will perform at once, to compare the pointwise convergence in the above listed function spaces with the convergence in the sense of their canonical metric-the metric convergence. One sees without difficulty that in lP(n) metric convergence is equivalent to componentwise convergence, i.e., Xk = (~\k), . .. , ~~k» converges to x = (~1' ... , ~") in the sense of the metric of /P(n) if and only if ~~k) -+ ~v for v = 1, ... , n. This remark shows furthermore, that convergence in the sense of the metric of lP(n) is equivalent to convergence in the sense of the metric in lQ(n), more precisely: X k -+ x in [P(n) if and only if X k -+ x in /q(n). From the point of view of a pure theory of convergence, all the metrics of K" defined by (1.2) or (1.3) perform the same service; we shall see soon that it can, however, be useful to consider them separately. If Xk -+ x in B(T), then to every e > there exists a ko = ko(e) so that for k ~ ko one has d(xb x) = SUPtE T IXk(t) - x(t) I ~ t;, and so a fortiori
°
°
°
°
(1.10)
for all
k ~ ko
and all
t E T.
10 Thus the sequence of functions X k converges not only pointwise but even uniformly on T to the function x, since the index ko depends only on B but not on t. Conversely, if the sequence (Xk) does converge uniformly to x, i.e., to every B > 0 there exists ko such that (1.1 0) holds, then obviously d(x k , x) = sup IXk(t) - x(t)1 ~
B
lET
~ ko, i.e., Xk - x in the sense of the metric of B(T). Thus in B(T) metric convergence is equivalent to uniform convergence on T. The same is true in qa, b]; we only have to observe that C[a, b] is a subset of B([a, b]) and that the maximum-metric on C[a, b] is nothing but the supremum-metric of B([a, b]) restricted to C[a, b].-In the future we shall abbreviate B([a, b]) by B[a, b]. Next we show that metric convergence in (s) is equivalent to componentwise convergence, i.e. that
for k
Xk = (~lk), ~~k), ... ) -+ X = (~1' ~2' ... )
in the sense of the metric of (s) if and only if
n = 1,2, ....
for First it follows from
that a fortiori
I~~k)
1
-
+ I~~k)
~nl
~n I -+ 0
-
and thus also limk _ 00 I ~~k) - ~n I = O. Now assume conversely that (Xk) tends componentwise to x. After having chosen an arbitrary B > 0 determine no = no(B) . so that 1
00
B
L n~-· n=no+12 - 2 Then for k = 1,2, ... one has a fortiori
L -
I~~k)
00
n=no+1 2 n
-
~n I
B
0 and center Xo. The distinction between an open and a closed ball is often inessential; if we speak simply of a ball, then it may be open or closed. With the help of the triangle inequality one sees immediately that a subset of E which lies entirely in a ball around x o , is also contained entirely in an appropriate ball around any other center Xl' This fact makes it possible to call a subset (or a sequence) of E bounded if it lies in a ball. Every Cauchy sequence (x n ), in particular every convergent sequence, is bounded. Indeed, for e = 1 there exists an m such that d(xn' xm) ~ 1 for n ~ m. Then for any n we have obviously m-l
d(x n, xm) ~
I
d(xll' Xm)
+
1,
Il;l
so that (xn) lies in a ball around
•
Xm •
In a metric space the so-called quadrangle-inequality (1.12)
Id(x, y) - d(u, v)1
~
d(x, u)
+ dey, v)
holds. Indeed, applying twice the triangle inequality, one obtains d(u, v) d(u, x) + d(x, y) + dey, v), hence d(u, v) - d(x, y)
~
d(x, u)
+ dey, v).
~
d(x, u)
+ dey, v).
~
Similarly, one has also d(x, y) - d(u, v)
(1.12) follows immediately from the last two inequalities.
•
Distance is a continuous function, i.e., xn --+ x and Yn --+ Y imply that d(xn' Yn) --+ d(x, y). To see this one only has to set u = X n, V = Yn in (1.12). The concept of continuity will be defined in generality and investigated more closely in §7. The concepts introduced in this section are very few; they consist essentially of those of a metric space, convergence and completeness. Furthermore we studied some examples of complete metric spaces. In the next two sections we will show that this material is already sufficient to obtain interesting and nontrivial results.
Exercises
Every subsequence of a sequence converging to x converges again to x. *2. If a Cauchy sequence has a subsequence converging to x, then the sequence itself converges to x. 3. In a space E with the discrete metric, a sequence converges if and only if it is constant from a certain term on. A ball in E contains either only its center or it coincides with the whole space. *1.
13
4. Besides the canonical maximum-metric, one can introduce a further metric on qa, b] by dl(x, y):=
f'X(t) -
y(t)ldt
(justify carefully why (M 1) holds). With this metric qa, b] is not complete. The same holds if we define distance by dix, y):=
( Jfba Ix(t) -
y(t)1 2 dt
)1/2
(use Minkowski's inequality to prove (M3». 5. If on E two metrics d 1 and d 2 are defined, we say that d 1 is stronger than d 2 if dl(x n , x) --+ implies that dix n , x) --+ 0, i.e., if convergence with respect to d l entails the convergence with respect to d 2 to the same limit; in this case one also says that d 2 is weaker than d l . If d l is at the same time both stronger and weaker then d 2 (i.e., if convergence with respect to d I is equivalent to convergence with respect to d 2 ), then the two metrics are said to be equivalent. Show the following: (a) The discrete metric is the strongest metric on E (use Exercise 3). (b) d l is stronger than d2 if and only if for all x E E the following condition is fulfilled: Every open ball around x with respect to d 2 contains an open ball around x with respect to d I. (c) On every metric space (E, d) a metric equivalent to d can be defined by
°
d(x, y) dl(x, y):= 1 + d(x, y)
(cf. Exercise 14). *6. If (EI' d l ), ••• , (En' dn) are metric spaces, then the cartesian product E = E I X ... x En becomes a metric space if we define distance by n
d(x, y):=
L dk(X
b
Yk);
k=1
here x = (XI' ... , x n), y = (YI' ... ' Yn). Convergence in E is equivalent to componentwise convergence. E is complete if and only if each Ek is complete. Define other metrics on E which are equivalent to d (cf. the spaces [Pen»~. *7. If (E I , d l ), (E2' d 2), ... are metric spaces, then the cartesian product E = 1 Ek becomes a metric space if we define distance by
nk'=
14 here x = (XI' X2, •.• ), Y = (YI' Y2, .. .). Convergence in E is equivalent to componentwise convergence. E is complete if and only if every Ek is complete (cf. the space (s». *8. If the convergent sequence (xn) lies in the closed ball Kr[z], then its limit also lies there. 9. One has
~~~(t/~l' - t/v/pr * 10.
IP
=
~=;~/~l' - t/l'/'
Like in classical analysis, a subset M of a metric space E is said to be
open if around each point of M there exists a ball which lies entirely in M. We say that M is closed if E\M is open. Show the following;
(a) M is closed if and only if the limit of every convergent sequence in M lies again in M. (b) An open ball is an open, a closed ball is a closed set. *11. In a metric space E the empty set 0 and the whole space E are open; the intersection of finitely many open sets and the union of an arbitrary number of open sets is open. We have the complementary statements; 0 and E are closed; the union of finitely many closed sets and the intersection of an arbitrary number of closed sets is closed. 12. If E is equipped with the discrete metric, then any subset of E is open as well as closed. 13. In a metric space E an open neighborhood of Xo is an open set containing Xo' A neighborhood of Xo is a set which contains a ball with center Xo. It follows from Exercise 10 that a neighborhood of Xo can also be defined as a subset of E which contains an open neighborhood of Xo' Show the following; (a) A subset of E is open if and only if it is a neighborhood of each of its points. (b) (xn) converges to X if and only if for every neighborhood U of X there exists no = no( U) such that Xn E U for all n ~ no (briefly; if almost all terms Xn lie in any neighborhood of x). 14. The set U(x) of all neighborhoods ofx in a metric space E (see Exercise 13) is called thejilter of neighborhoods of x. Let two metrics d I and d 2 be introduced on E, let U I(x) and U2 (x) be the corresponding filters of neighborhoods, .0 1 and .0 2 the corresponding systems of open sets. Show with the help of Exercise 5 that the following assertions are equivalent; (a) d I is stronger than d 2 • (b) UI(x) => Uix)foreveryxEE. (c) .0 1 => .0 2 , It follows that d\ and d 2 are equivalent if and only if U\(x) = U2 (x) for all x E E or, equivalently, if.o l = .0 2 , *15. The closure or closed hull of a subset M of the metric space E is the intersection of all closed subsets of E containing M; it is denoted by M. One has M c: M; M is closed; and M is closed if and only if M = M. M is the set of all limits of convergent sequences from M.
15
§2. Banach's fixed-point theorem An important problem in analysis is the determination of all zeros of a real function f, i.e., the solution of the equation f(x) = O. This can often be achieved by putting the equation in the form
x = g(x)
(2.1)
and then using an iteration procedure: one chooses an Xo from the definition set of g, one sets Xl := g(xo), X2 := g(XI), generally x n+ I := g(x n) for n = 0, 1,2, ... and one hopes that the sequence (xn) of iterates converges to a solution of (2.1). This is indeed the case when we make certain hypotheses which will be indicated now. Suppose that the function 9 maps the interval [a, b] into itself and that it satisfies a Lipschitz-condition with a Lipschitz-constant q < 1, i.e., that there exists a number q with 0 ~ q < 1 such that (2.2)
Ig(x) - g(y) I ~ qlx - yl
for all
x, y in
[a, b].
Since 9 maps the interval [a, b] into itself, for any starting point Xo in [a, b] the above defined sequence (x n) of iterates exists. We show now that (xn) converges to a solution X of the equation (2.1) and that, furthermore, X is the only solution in [a, b] of this equation. It follows from (2.2) by repeated estimates that IX2 - XI I = Ig(xI) - g(x o) I ~ qlx l - xol, IX3 - x21 = Ig(X2) - g(xl)1 ~ qlX2 - xII ~ q21xI - xol, and in general
From here one obtains for k = 1,2, ... with the help of the triangle inequality that (2.3)
IXn+k - Xn I
~
.t
IXn+v - xn+v-II
l_ qk
= qn -1-- lxl - xol -q
~ ~
Ct
qn+v-I) IXI - Xo I
qn - - I x i - xol. l-q
Because 0 ~ q < 1, this estimate shows that (xn) is a Cauchy sequence in [a, b] and converges therefore to a point X E [a, b]. With this X one has Ig(x) - xl ~ Ig(x) - x n+ II ~ qlx - xnl
+
+ IXn+ I - xl = Ig(x) - g(x n) I + IXn+ I - xl IXn+1 - xl,
and since the last term of this inequality tends to 0, we have Ig(x) - x I = 0, i.e., indeed g(x) = x. If for a y in [a, b] we also have g(y) = y, then Ix - yl = Ig(x) - g(y) I ~ qlx - yl implies Ix - yl = O,hencex = y.Soalltheassertions made above are proven. •
16
If we replace the i.iJterval [a, bJ by an arbitrary complete metric space E and g by a map A from E into itself which is contracting, i.e., which for a fixed q, ~ q < 1, satisfies the Lipschitz-condition
°
(2.4)
d(Ax, Ay)
~
for all
qd(x, y)
x, y
in
E,
then we can repeat the above argument word for word; we only have to replace the distance Iu - v I in [a, bJ by the distance d(u, v) in E. Thus if Xo is an arbitrary point in E and we define the sequence of iterates (x n) by x n+ 1 := AXn (n = 0, 1,2, ... ), then we obtain the estimate analogous to (2.3) (2.5) from where it follows that (x n ) is a Cauchy sequence in E, which-because ofthe completeness of E-converges to a point x in E. One sees exactly as above that Ax = x and that x is the only fixed point of A in E, i.e., the only point which remains fixed under the map A. From (2.5) we obtain furthermore an error estimate if we let k -+ 00 and use the continuity of the metric (§l): (2.6)
qn d(x, x n) ~ -1-
-q
d(x 1, XO )·
We collect our results in the following theorem, known under the name of Banach's fixed-point theorem or the theorem on contracting maps. Theorem 2.1 If A is a contracting self-map of the complete metric space E, then the equation x = Ax has exactly one solution in E. This solution can be obtained by iteration: ifone chooses an arbitrary starting point Xo in E and sets x n+ 1 := Ax", n = 0, 1, 2, ... , then (x n) converges to x. Furthermore the error estimate (2.6) is valid. Ifone uses the powers or iterates An of A, then Xn = Anxo; under the hypotheses of the theorem the sequence (Anxo) converges to the unique fixed point of A in E. One says that the fixed point of A was found by iteration of A. In the next section we shall be able to convince ourselves of the great possibility of practical applications ofthe fixed-point theorem. Let us emphasize here only that the theorem not only guarantees the existence of a fixed point but also permits its construction. Exercises 1. The function g satisfies the Lipschitz condition (2.2) if it is differentiable and /g'(x)/ ~ q < 1 in [a, b]. + 2. Let L IXn be a convergent sequence with non-negative terms and A a map of the complete metric space E into itself, which for every natural number nand for all x, yin E satisfies the condition d(Anx, Any) ~ IXnd(X, y). Then A has exactly
17 one fixed point x in E, this fixed point is the limit of the sequence of iterates (A"xo) for an arbitrary Xo in E, and one has the error estimate d(x, A·xo)
~ (J.iXv)d(AXo, xo).
§3. Some applications of Banach's fixed-point theorem We consider first systems of n linear equations with n unknowns. Such a system can always be written in the form ~i
(3.1)
•
L iXik~k = Yi k=1
-
(i
= 1, ... , n)
and thus brought into the form of a fixed-point problem ~i =
(3.2)
L" iXik~k + Yi
(i
= 1, ... , n);
k=1
indeed, to solve the system (3.2) means to find a fixed point of the map A: K" defined by (3.3)
A(~I"'" ~"):=
Ct iXlk~k
+ YI,""
JI iX"k~k
-+
K"
+ Yo)'
To be able to apply the fixed-point theorem of §2 we have to make first a complete metric space out ofK·. For this purpose we have e.g., the metrics defined by (1.2) and (1.3) which transform K" into the metric spaces [pen). For the cases p = 1,2, oc; we want to write down the conditions which tell us that A is contracting. Let x = (~I, ... , ~") and y = (111, ... ,11.) be two arbitrary points in K". In the case p = lone has
=
.
L I~k k= 1
"
11k I L liXikl i= 1
and so (3.4)
In the case p
d(Ax, Ay) =
~ (~=;~ J11 iXik l)d(X, y).
2 one obtains with the help of the Cauchy-Schwarz inequality
d(Ax, Ay) =
and so (3.5)
d(Ax, Ay)
~
•
L
i,k= 1
IiXik1 2 d(x, y).
18
Finally, in the case p
d(Ax, Ay) =
=
00
we find the estimate
~~: IktIO(ik(~k -l1k)1 ~ ~~:CtIIO(ikl'~=~~I~k -l1kl).
and so (3.6)
d(Ax, Ay)
~ (~=~IX JIIO(ikl)d(X, y).
If we take into account that convergence in IP(n) is equivalent to component wise convergence (see §1), then we obtain from the fixed-point theorem of §2, with the help of the above estimates, the following statement: If one of the numbers n
(3.7)
ql:= max k= I
n
n
L 100ikl,
i= I
L
q2:=
n
qoo
IO(ik1 2 ,
:=
i.k= I
n
L
max 100iki i= I k= I
is less than 1, then the system ofequations (3.1) possesses exactly one solution. This solution x can be determined by iteration of A; the sequence of iterates converges componentwise to x. The reader can verify with simple examples that the three conditions are independent from each other, i.e., that none of them implies another one. We see that it is indeed useful to consider different metrics on the same set. So far in our considerations we made use only of the metric properties of Kn. But Kn has also an algebraic structure: its elements can be added together (componentwise) and multiplied by numbers from K, and these operations satisfy the known rules of vector algebra. We want to take advantage of this circumstance to describe the solution by iteration ofthe system (3.1) in a more striking fashion. To this end we define the map K: Kn -+ Kn by
K(~I"'" ~n):= (± O(ik~k"'" k=1
(3.8)
±O(nk~k)'
k=1
set x:= G I, ... , ~n)' g:= (YI' ... , Yn) and can now write (3.3) in the form
Ax = g
+ Kx.
As a matrix map K is linear, i.e., for all x, y in Kn and all
K(x
+ y) =
Kx
+ Ky,
0(
in K one has
K(O(x) = O(K(x).
If we choose Xo = g, then the iterated vectors x n+ I := AXn (n = 0, 1, ... ) can be
represented as follows:
XI = Axo = Ag = g + Kg, X2 =
AXI = g
+ KXI
=
g
+
K(g
+
Kg) = g
+ Kg +
K2g,
in general (3.9)
Xn
= AXn- 1 = 9 + Kg + K2g + ... + Kng
for
n = 1,2, ....
19
Thus (x n ) is the sequence of partial sums of the series (3.10)
g
+ Kg + K2g + ...
00
=
I
Kng.
n=O
If we write the system (3.1) in the form (3.11)
x - Kx
= g,
then we can summarize our result as follows: If the map K is defined by (3.8) and one of the numbers qt, q2, qoo of(3.7) is less than 1, then the equation (3.11) possesses exactly one solution x which one obtains as the sum of the componentwise convergent Neumann series (3.10). As the next application of the fixed-point theorem we consider a functional equation which is an exact analogue of (3.1), namely the Fredholm integral equation (of the second kind) (3.12)
x(s) -
f
k(s, t)x(t)dt
= g(s).
For the sake of simplicity we suppose that g is continuous on the interval [a, b] and the kernel k on the square [a, b] x [a, b], furthermore we want to seek only solutions x which are again continuous on [a, b]. Ifwe define a map A: era, b] --+ era, bJ by (3.13)
(Ax)(s)
:=
g(s)
+
f
k(s, t)x(t)dt,
then we see that the problem to solve the integral equation (3.12) by a continuous x is equivalent to finding a solution of the equation x = Ax, that is, a fixed point of A. We introduce in era, bJ the canonical maximum-metric (see Example 1.4), which turns era, bJ into a complete metric space and we obtain with it the estimate d(Ax, Ay)
f ~ a~::b f, =
a~::b I
k(s, t) [x(t) - y(t)Jdt
k(s, t)
la~~:bl x(t) -
I
y(t) Idt,
and so (3.14)
d(Ax, Ay)
~ C~::b f'k(S, t)ldt)d(X, y),
an inequality which is completely analogous to (3.6). From the fixed-point theorem of §2 now follows: If (3.15)
qoo
:=
max rb1k(s, t)ldt < 1
a~s;;§b
Ja
20 then the Fredholm integral equatioll (3.12) possesses exactly olle solution x which is continuous in [a, b]. This solution can be obtained by iteration of A; the sequence of iterates converges in the sense of the maximum-metric, i.e., uniformly on [a, b].
Similarly as in the preceding example, we want to make use also here of the fact that C[a, bJ has not only metric but also algebraic properties: functions from C[a, b J can be added (pointwise) and multiplied by numbers from K. Similarly as in (3.8) we define an integral transformation K: C[a, b J -+ C[a, b J by (3.16)
(Kx)(s):=
f
for
k(s, t)x(t)dt
a ~ s ~ b,
so that we have Ax = g + Kx. The map K is again linear, i.e., for all x, y in C[a, b J and for all oc in K we have K(x
+ y) =
Kx
+ Ky,
K(ocx) = ocKx.
As a consequence we can write the sequence of iterates (x n ) beginning with Xo = g like in (3.10) as a sequence of partial sums of an infinite series and obtain the following result: If the integral transformation K with a continuous kernel k is defined by (3.16), and if condition (3.15) is satisfied, then the equation
x - Kx = g,
g E C[a, bJ,
i.e., the Fredholm integral equation (3.12), possesses exactly one solution x in C[a, b J which can be obtained as the sum (?l the uniformly convergent Neumann
series (3.17)
g
+ Kg + K2g + ... =
00
L Kng. n=O
One is led to Fredholm integral equations among others through the study of boundary value problems; these occur frequently in physics and technology, e.g., in buckling and vibration problems, see [152J; there-as in any other book on integral equations-one can find justifications of the facts which we want to describe briefly now. By a linear differential operator of order n we understand a map L which according to the definition n
(3.18)
(Lx)(t):=
L f,,(t)x(V)(t)
(fv E C[a, bJ, f,,(t) f:. 0 on [a, bJ)
v=o
associates with every n times continuously differentiable function x on [a, bJ a function Lx E C[a, b]. Boundary conditions for such a function are equations of the form n-\
(3.19)
Rllx:=
I
v=o
[ocllvx(V)(a)
+
Pllvx(V)(b)]
=0
(/-I
= 1, ... , n)
21 with given numbers rt.1IV> Pllv' A boundary value problem for L is the task to find an n times continuously differentiable function x on [a, b] which for a given y E C[a, b] satisfies the eq uations (3.20)
Lx = y
and
Iffor a fundamental system Xl' Lx = 0 the determinant
Rllx = 0 .•. , Xn
for
/1
= 1, ... , n.
of the homogeneous differential equation
(3.21 ) then there exists one and only one Green's function G, which is continuous on the square a ~ s, t ~ b and with whose help the solution x-which is uniquely determined under the present hypotheses-of the boundary value problem (3.20) can be represented in the form (3.22)
x(s)
=
f
G(s, t)y(t)dt.
We consider now the boundary value problem with a parameter;. (3.23)
Lx -t.rx
= 0,
(/1
= 1, ... , n),
where r E C[a, b] and the product rx is defined, as is customary, pointwise: (rx)(t) := r(t)x(t). Every value ;., for which (3.23) possesses a non-trivial (i.e., not identically vanishing) solution x, is called an eigenvalue (or proper value) of the problem, and x is called an eigenso/ution corresponding to ;.. Eigenvalues and eigensolutions, or expressions which depend in a simple way on them, have often an important significance in physics and technology (buckling loads, frequencies of resonance, energy levels, vibration figures, etc.). This fact explains the basic role of the theory of eigenvalues in various branches of science. Clearly, A is an eigenvalue of (3.23) if and only if for a fundamental system X'(E, F) becomes a vector space (over the common field of scalars of E and F), if we define sums A + B and multiples aA pointwise: (A
+ B)x := Ax + Bx,
(aA)x:= a(Ax)
for all
x
E
E.
The zero element of this space is the zero map 0 which associates with every element in E the zero element of F. The composition BoA of the linear maps A: E --+ F, B: F --+ G is a linear map from E into G. We denote it more shortly by BA and call it usually the product of B by A.
30 In case the following products exist, they satisfy these rules: A(BC)
+ C) + B)C
A(B (A
= (AB)C, = AB + AC, = AC + BC,
oc(AB) = (ocA)B = A(ocB).
Products are in particular always defined if the factors lie in !Jl(E). It follows that /I'(E) is an 'algebra' over the scalar field K, since a set R is called an algebra over K if R is a vector space over K and iffor any two elements a, b in R a product ab E R is defined in such a way that the following rules hold: a(bc) = (ab)c, a(b + c) = ab + ac, (a + b)c = ac + bc, oc(bc)
= (oca)b = a(ocb).
The algebra R is said to be commutative if one always has ab = ba. An element e in R such that ae = ea = a for all a E R is called a unit element (identity) of R. . In an algebra there exists at most one unit element. The algebra !Jl(E) is in general not commutative but possesses a unit element, namely the identity map I defined by Ix:= x for all x E E. Sometimes we write IE instead of I to emphasize that I operates on E. We recall that the powers or iterates of an endomorphism A are defined successively by AO := I, An := AA n- 1 (n = 1,2, ... ). The importance of linear maps (and so also of linear spaces) is based among others on the fact that many problems concerning equations are linear problems in the following sense: Given a linear map A: E --+ F, determine for which 'righthand sides' y E F do.es the equation (5.1)
Ax
=Y
have a solution (i.e., describe the image of A) and check whether the solution, if it exists, is determined uniquely; if not, describe the set of solutions. The most satisfying situation occurs when our equation has exactly one solution for each y E F, i.e., when A is bijective. In §3 we wrote the system of linear equations (3.1) and the Fredholm integral equation (3.12) with the help of an endomorphism K of Kn or C[a, b], respectively, in the form (I - K)x
=
Y
and gave sufficient conditions for the bijectivity of the endomorphism I - K; moreover, we could also construct the solution. Further examples for linear problems are given by the boundary value problems of the form (3.20). Using the notations and hypotheses from there, if we set E:= {x E c(n)[a, b]: RJlx = 0 for J-L = 1, ... , n} and define the linear map A: E -+ C[a, b] by Ax := Lx, then
31
the boundary value problem (3.20) is precisely the task to solve the equation Ax = y. In this section we shall collect some simple concepts and facts which serve to investigate 'operator equations' of the form (5.1), and which in particular are concerned with the questions whether (5.1) has a unique solution or has a solution for every y E F, i.e., with the injectivity or surjectivity of A, respectively. We first remark that the image A(E):= {Ax: x E E} and the nullspace N(A)
:=
{x E E: Ax
=
O}
of A are linear spaces, in particular 0 E N(A), i.e., AO = o. If the equation (5.1) has the two solutions X o , XI' then A(xo - XI) = Axo - AXI = Y - Y = 0, i.e., XI - Xo E N(A) and so XI E Xo + N(A). Conversely, if XI is contained in Xo + N(A), then clearly AXI = Axo = y. With this we have a first assertion on the set of solutions of(5.1): Proposition 5.1. The set of all solutions of(5.1) can be represented with the help of one solution Xo in theform Xo + N(A). The map A is injective ifand only if N(A) = {O}, i.e., if Ax = 0 implies X = O.
If A is injective, the inverse map (see introduction) defined on A(E) is denoted by A-I and we call it shortly the inverse of A. It is easy to see that A - I is linear, hence lies in 9"(A(E), E). From here and from the discussion of inverse maps in the introduction we immediately obtain the following: Proposition 5.2. The linear map A: E --+ F is bijective if and only if there exist linear maps Band C from F into E such that
BA=IE In this case B
and
AC=I F •
= C = A - I.
A bijective linear map A: E" --+ F is also called an isomorphism (of the spaces E and F) and we say that E and F are isomorphic. This symmetric way of expression is justified because together with A also A-I is an isomorphism. The product BA of two bijective linear maps A: E --+ F and B: F --+ G is again bijective and has as inverse (BA)-I = A - I B- 1 • A particularly important class of endomorphisms consists of the so-called = F ~ G, i.e., if for every X E E one has a decomposition X = Y + z with uniquely determined components y E F, Z E G, then one can define a map P: E --+ F by Px := y. The map P is linear and is called projector because, taking over the language of geometry, P projects the space E along (or parallel to) G onto F. Then I - P projects E along F onto G. We have clearly projectors. If E
P(E) = F,
N(P)
= G and
p2
= P.
32 If we call an endomorphism A which satisfies A 2 = A idempotent, then the last equation asserts that projectors are idempotent. This property is characteristic for projectors. Indeed, if P is an idempotent endomorphism of E, then because of x = Px + (I - P)x, the space E is the sum of the two subspaces P(E) and (I - P)(E), and this sum is even direct since from Z E P(E) n (I - P)(E), i.e., Z = Px = (I - P)y it follows that z = Px = p 2 x = P(l _ P)y = Py _ p 2 y = Py - Py = o. It can be seen immediately that P is the projector of E onto P(E) along (I - P)(E). Because of Proposition 4.1, to every subspace F of E there exists a projector which projects E onto F. The image of a projector P of E is given by P(E) = {x E E: Px = x},
i.e., it is that subspace on which P acts as the identity. Indeed, if x E P(E) then x = Py and so Px = p 2 y = Py = x, and conversely from x = Px it follows trivially that x lies in P(E). With the help of projectors we can describe algebraically the discrepancy of a linear map A: E ---+ F from bijectivity. Let P be a projector of E onto N(A) and Qo a projector of F onto A(E); let the corresponding decompositions of E and F be E
= N(A) E£>
V,
F = A(E) E£> V.
We now define a linear map Ao: V ---+ A(E) by Aox:= Ax for x E U. The map Ao is clearly surjective, and since Aox = 0 implies that x E N(A) n V = {O}, it is also injective, hence AD I: A(E) ---+ V exists. Thus B:= AD IQo is a linear map from F into E. If we represent an arbitrary x E E in the form x = y + z with y E N(A), z E V, then we obtain BAx = AD IQoA(Y + z) = AD IQoAz = AD I Az = z = x - y = x - Px, hence BA = IE - P. Similarly one gets the equation AB = Qo. With the projector Q:= IF - Qo, complementary to Qo, which projects F along A(E) onto V, we have AB = IF - Q. Thus we proved the following proposition: Proposition 5.3. If A: E ---+ F is linear, P a projector of E onto N(A) and Q a projector of F along A(E), then there exists a linear map B: F ---+ E such that
the equations BA = IE - P,
AB
=
IF - Q
are valid.
Comparing this proposition with Proposition 5.2, one sees in what way one can describe with the help of projectors the deviation of a linear map from bijectivity. Ultimately this depends on the fact that the deviation from injectivity is given by the nullspace and the deviation from surjectivity by a complementary subspace of the image space. The linear subspace F of E is said to be invariant under A E 9'(E) if A(F) c F; in this case the restriction of A to F is a self-map of F. We say that the pair (F, G)
33 of subspaces reduces A if E = F EB G and both F and G are invariant under A. The reader will prove easily the following proposition:
Proposition 5.4. Let P be a projector of E onto F parallel to G. The subspace F is invariant under A if and only if AP = PAP holds. A is reduced by (F, G) if and only
if AP =
P A holds.
Exercises 1. Let A: E --+ F be a linear map. Show: (a) If {Ax,,: ,t E L} c A(E) is linearly independent, then also {x,,: ,t E L} c E is linearly independent. (b) If A is injective and {x,,:,t E L} c E is linearly independent, then {Ax,,: ,t E L} is also linearly independent. 2. Let E and F be two vector spaces having the same finite dimension and let A E [I'(E, F). With the help of exercise I show that the map A is bijective if it is either injective or surjective (cf. Exercise 3). 3. Construct an endomorphism of (s) which is injective but not surjective (cf. Exercise 2). *4. Let E and F be two vector spaces over K and let E be finite-dimensional. E and F are isomorphic if and only if dim E = dim F. *5. Let P be a projector of E and Me E. Then P(M) c M + N(P). *6. To A E //'(E, F) there exists aBE [I'(F, E) with ABA = A.
§6. Normed spaces When we discussed in §3 systems of linear equations and Fredholm integral equations, we encountered the linear structure of the spaces IP(n) and C[a, b]. This linear structure is closely related to the canonical metric structure of IP(n) and C[a, b]. We first define some useful concepts. The metric d on the vector space E is said to be translation-invariant if
= d(x + z, y + z) for all x, y, z in E. If in this case we set Ixl:= d(x, 0), then d(x, y) = d(x - y, y - y) = d(x - y,O) d(x, y)
= Ix - y I, i.e., the distance of two points of E can be described with the help of the 'absolute value' Ix I, just like the distance of two points or K with the help of the absolute value given there. From the axioms (M!) through (M3) in §l we obtain immediately the following properties of absolute value: (AI) (A2) (A3)
Ixl ~ 0, where Ixl = 0 holds exactly when x = 0, I-xl = lxi, Ix + yl ~ Ixl + Iyl (triangle inequality).
Conversely, if a vector space E is a valued space, i.e., if with every x in E a real number lxi, its absolute value, is associated in such a way that the axioms (AI) through (A3) hold, then (6.1)
d(x,y):=
Ix - yl
34 defines a translation-invariant metric on E, the canonical metric of a valued space. Thus the valued spaces are exactly the vector spaces with a translationinvariant metric. A sequence (x n ) in a valued space converges to x, if IXn - xl---+ 0; it is a Cauchy sequence iffor every l: > 0 there exists an no = no(l:) such that Ixn-xml ~ l: for n, m ~ no. A valued space is complete if it is complete as a metric space. The canonical metrics of the vector spaces lP(n), loo, B(T), C[a, b] and (s) are all translation-invariant (cf. Examples 1.1 through 1.5), so these spaces are valued spaces with the absolute values
1~ p
0 there exists ko = ko(e) so that for every partition Z of [a, b] and for all k, I ;:::; ko the estimate IXk(a) - x,(a) I +
•
L Ixk(tv) -
x,(t.) - [Xk(t v-
1) -
X,(tV-l)] I ~ IIxk - x,lI ~
e
v= 1
holds. If we let k -+ 00 and observe that the inequality one obtains is valid for every partition Z, then it follows that Ix(a) - x,(a) I + V(x - x,)
(6.11)
~
c
for
I;:::; ko .
In particular x - Xko and so also x = (x - Xko) + Xko are of bounded variation. (6.11) now states exactly that (x,) converges in the metric of BV[a, b] to x, which proves our assertion. • With the supremum-norm coming from B[a, b] the space BV[a, b] is not complete (see Exercise 3); for this reason we have not used it. We list the Banach spaces defined so far in Table 1 and emphasize again that these spaces will be equipped always with the norms indicated there-the canonical norms-unless the contrary is explicitly stated. Further examples of normed spaces will be found in the Exercises of this section. A linear subspace F of a normed space E becomes a normed space if we equip its elements with the norm that they already have as elements of E; this norm on F is called the norm induced by E (or by the norm of E) and F is then said to be a subspace of E. Thus, e.g., (co) is a subspace of (c) and (c) a subspace of 100 ; on the other hand, though BV[a, b] is a linear subspace of B[a, b], however, it is not a subspace because the norm of BV[a, b] is not induced by the norm of B[a, b]. The subspaces of valued spaces are defined correspondingly. We have seen at the beginning of this section that the translation-invariance of the metric given by the norm (which yields the triangle inequality (N3» and its homogeneity property (N2) establish a close connection between the linear and the metric structure of a normed space. This interplay of the two structures has among others the following consequences: Proposition 6.1. In a normed space addition, multiplication by a scalar and the norm are continuous, i.e.,from x. -+ x, y. -+ y and rx. -+ rx it follows that
x.
+ y. -+ x + y,
The proof of the first two assertions follows from the estimates lI(x.
+ Yn)
IIrxnxn - rxxll
- (x
+ y)1I
= IIrxn(xn -
=
x)
lI(xn - x)
+ (rxn
+ (Yn
- y)1I ~ IIxn - xii
- rx)xll ~ Irxnl· IIxn - xII
+
+ Irxn
llYn - yll, - rxl· IIxli.
39 Table I. Notation IP(n), 1 ~ p
~ %
Banach spaces Canonical norm
Definition Set Kn of all n-tuples
if
x = (~I'···' ~n)
Ilxll = maxl~vl
1~ p
0 there exists no = no(e, M) so that IIAnx - Axil ~ e for all n ~ no and all x in M. 8. Define on II for each natural number n a continuous endomorphism An by An(el, e2,"'):= (el, e2,"" en, 0, 0, ... ). Show that An converges pointwise but not uniformly to I. 9. Let A be a map from the metric space E into the metric space F. Show that the following assertions are equivalent (for the concepts used see Exercise 13 in §1): (a) A is continuous at the point Xo E E. (b) To every e > 0 there exists a b > 0 so that d(Ax, Axo) ~ e whenever d(x, x o) ~ b.
46 (c) To every ball V with center Axo there exists a neighborhood U of Xo so that A( U) c V. (d) To every neighborhood W ofAxo there exists a neighborhood U of Xo so that A(U) c W. (e) The pre image A -1(V) of every ball V with center Axo is a neighborhood of Xo. (f) The preimage A - I(W) of every neighborhood W ofAxo is a neighborhood of Xo. * 10. Under the hypotheses of Exercise 9 the map A: E --+ F is continuous (on E) ifand only ifthe preimage A -1(M) of every open set Me F is open in E, or equivalently if the preimage A - I(N) of every closed set N c F is closed in E. 11. If E is equipped with the discrete metric and F is an arbitrary metric space, then every map A: E --+ F is continuous.
§8. The Neumann series In §3, at the investigation of systems of linear equations and of Fredholm integral equations of the second kind, we met equations of the form x - Kx = Y with a linear map K. In certain cases we could solve the equations with the help of Neumann series. The concepts and theorems we have listed so far make it possible to obtain again the results from there in the abstract framework of normed spaces, and thereby to unify them. For this purpose we define an irifinite series L~ 0 Xv with elements Xv from a normed space E, exactly like in classical analysis, as the seq uence of its partial sums Sn := Xo + XI + ... + x n • The series is said to be convergent with sum s, in symbols L~= 0 Xv = s, if Sn -+ s. It is called a Cauchy series, if (sn) is a Cauchy sequence, i.e., if to every B > 0 there exists an no = no(e) so that for n > m ~ no one has
Iisn - smll
=
Ilxm+ 1 + ... + xnll
~
e.
A convergent series is a Cauchy series, hence the sequence of its terms tends to O. In a Banach space every Cauchy series converges.
Lemma 8.1. If L~=o Ilxvll converges, then the series L~=o Xv is a Cauchy series. Thus it converges if the space is complete, and in this case the generalized triangle inequality (8.1) holds.
Because of III~=m xvii ~ I~=m IIxvll it follows that I Xv is a Cauchy series. The generalized triangle inequality follows, because of the continuity of the norm, from the elementary triangle inequality IIL~=o xvii ~ L~=o Ilxvll as n -+ 00, provided that I Xv converges. •
47 Let now K be a continuous endomorphism of the Banach space E. We want to study the equation x - Kx
(8.2)
= y or (l - K)x = y.
Ifwe bring it to the form x = y + Kx and define the (nonlinear) self-map A of E by Ax := y + Kx, then (8.2) becomes the fixed-point equation Ax = x. Because of
= IIAx. - AX211 = IIKx. - KX211 = IIK(x. - x2)11 ~ IIKllllx. - x21 = IIKlld(x., x 2 ), according to Theorem 2.1 this equation possesses in the case q := II K II < 1 d(Ax.,Ax 2 )
exactly one solution x in E, i.e., equation (8.2) has for every y in E exactly one solution x in E, or in other words: The inverse (l - K)-I exists on the whole space E. If Xo is arbitrary in E and Xn := AXn_ 1 (n = 1,2, ... ), then Xn converges to x. If we choose Xo = y, then we obtain (cf. (3.9» Xn
=
Y
+ Ky + K2y + ... + Kny
~
x,
i.e., 00
L Kny
x = (l - K)-·y =
(8.3)
n=O
or 00
(8.4)
(l - K) -. =
L K n, n=O
if convergence of this series is understood in the sense of pointwise convergence of its partial sums. The series (8.3) and (8.4) are called Neumann series. However, the Neumann series in (8.4) converges not only pointwise but even uniformly. Indeed, from the error estimate (2.6) with Xo = y, x. = y + Ky it follows that
n ) II II ( (l - K)-· - v~o KV y
=
+.
IIKll n IIKll n IIx - xnll ~ 1 _ IIKII IIKYII ~ 1 _ IIKII Ilyll,
and this inequality shows that the linear map (l - K)-· - L~=o K V is bounded-hence (l - K)-I is also bounded-and that we have the estimate (8.5)
II
(l - K)-I -
± v=o
KV
II
~ IIKll n + 1-
1
IIKII
•
Because of II K II < 1 the partial sums L~ = 0 K V converge indeed uniformly to (I - K) - I , i.e., the expansion (8.4) holds, as asserted, also in the sense of uniform convergence. With the aid of Lemma 8.1 it further follows that
48 We state these important results:
Theorem 8.1. Let K be a continuous endomorphism of the Banach space E such that II K II < 1. Then the inverse (I - K) - 1 exists on E, is continuous with 1 II (I - K) - 1 II ~ 1 _ II K II '
(8.6)
and can be expanded into the uniformly convergent Neumann series 00
(I - K)-l
(8.7)
=
I
Kn;
n= 0
an error estimate is given by (8.5). The expansion (8.7) is obviously analogous to the geometric series 00
(l - q)-l
(8.8)
=
I
qn
for
Iql < 1.
n=O
We want to give now a new proof for (8.7) which goes similarly as the proof of (8.8). Under the hypothesis of Theorem 8.1 it follows from IIL~=m KVII ~ I~=m IIKvl1 ~ I~=m IIKllv that I:'=o K is a Cauchy series in 2(E). Since !£(E) is complete by Proposition 7.4, there exists an S in 2(E) with S = L:'=o KV. Because of the continuity of multiplication (see (7.8», it follows that SK = KS = I:'=o K n+ 1 = S - I, hence V
(I - K)S
= S - KS = I,
S(l - K) = S - SK = I.
Proposition 5.2 now implies that I - K is bijective and has S as its inverse.
•
This new proof shows in particular that (I - K) - 1 always exists on E and is continuous, if the Neumann series converges at all. Now one can prove exactly as in the scalar case the root test, i.e., the assertion that the series I Xn with terms xn in a Banach space converges or diverges according as oc := lim sup~ < 1 or > 1; in the case oc = 1 the test does not give any information concerning convergence or divergence. The convergence behavior of the Neumann series is thus determined by the quantity lim supllKnll lin . We first prove that here lim sup can be replaced by lim; Proposition 8.2 will then be obvious.
Proposition 8.1.
For every continuous endomorphism A of a normed space = 1,2, ....
limn~<x' IIA nll lin exists and is ~ IIAklll/k for k
Proof For oc n := IIAnll we have 0 ~ OCn+ m ~ OCn(J(m; hence the sequence converges to its greatest lower bound (see [183], Section I, Problem 98).
(oc~/n)
•
Proposition 8.2. If K is a continuous endomorphism of the Banach space E, then I - K has a continuous inverse on E whenever the Neumann series L:'= 0 K n
49 converges (uniformly); in this case the expansion (8.7) is valid. The Neumann series converges or diverges according as limllKnlll/n < 1 or > 1.
We show by an example that the convergence condition ofthe last proposition is much weaker than the earlier condition 11K II < 1. We define the Volterra integral transformation K on C[a, bJ by (Kx)(s):= fk(S, t)x(t)dt,
(8.9)
where the kernel k is continuous in the triangle a ~ t ~ s ~ b. This transformation is a linear and continuous self-map of C[a, bJ and the Volterra integral equation xes) - fk(S, t)x(t)dt = yes)
(8.10)
can be written with its help in the form (/ - K)x = y;
here y is supposed to be in C[a, bJ and we ask only for solutions x lying in C[a, b] equipped with the canonical maximum-norm. We examine now the iterates Kn. Setting fl.:= maxa~/~s~b Ik(s, t)1 we get I(Kx)(s) I = I(K 2 x)Cs)1 =
I fk(S, t)x(t)dt I ~ I
is a
jlllxll(s - a),
I
is
(s - a)2
k(s, t)(Kx)(t)dt ~ / . fl.llxllCt - a)dt = fl. 2 11xll ~-,
in general (s - a)n
I(Knx)(s) I ~ fl.nllxll -
_.n!
(n
= 1,2, ... ).
Thus (b - a)n IIKnxil = max ICKnx)(s) I ~ fl.n ---,- IIxll, a~s~b
hence IIKnil ~ fl.n 0~, a)n. n. Because of ~! (8.11)
---+
w we obtain
iimllKnlll/n = 0,
n.
50
hence (I - K)-I exists on C[a, b], i.e., the Volterra integral equation (8.10) has for any continuous right-hand side y exactly one continuous solution, however large II K II is. Exercises 1. Let the linear and continuous self-map of the Banach space I'X(n) be given by K(~I' ... ' ~n):= (2~1'.··' 2~n)· The corresponding Neumann series diverges, nevertheless the inverse (I - K)- I exists on l''''(n) and is continuous there. 2. The finite-dimensional analogue to a Volterra integral transformation of C[a, b] is a self-map K of /OO(n) by means of a triangular matrix (~ik) which has only zeros above the main diagonal, so that the components 1]i of the image vector (1] I' ... , 1]n) := K( ~ I' ... , ~n) are given by i
1]i:= I>ik ~k
(i
= \, ... , n).
k=1
The Neumann series does not converge always (see Exercise \), and the inverse (I - K) - I does not always exist either. It exists on lOO(n) if and only if all entries on the main diagonallXjj :f. I (observe that the determinant of a triangular matrix is the product of its entries on the main diagonal). 3. For the integral transformation K: qo, I] --+ qo, \] defined by (Kx)(s):= Ls. x(t)dt,
one has IIKII = 1 but
IimllK n ll 1 'n = 1-
(C[O, I] is equipped with the maximum-norm). Consequently, the Neumann series 0 Kny converges for every y E CEO, I] and yields a continuous solution of the integral equation x(s) sx(t)dt = y(s). This solution can be given explicitly. 4. If we introduce on CEO, 1] the analogue of an [I-norm by Ilxlll:= SA Ix(t) Idt, then the integral transformation K of Exercise 3 has norm II Kill = !. However, Theorem 8.1 cannot be applied, since CEO, 1] is not complete with this norm (see §I, Exercise 4). + 5. For A E 2(E) (E a normed space) the following assertions are equivalent:
I:,=
H
(a) An = 0. (b) There exists an mEN such that IIAml1 < 1. (c) L:'=o IIAnll converges. + 6. The series L:'= I x. in a Banach space E is absolutely convergent if L:'= 1 IIx.11 converges, it is unconditionally convergent if all its rearrangements converge to the same vector x. Absolute convergence implies convergence. In
51
finite-dimensional spaces a series is absolutely convergent if and only if it is unconditionally convergent. In [52] it is shown that this theorem is false in the case dim E = 00, in fact it is characteristic for finite-dimensional spaces.
§9. Normed algebras In the second proof for the expansion (S.7) in §S it was unimportant that K is a linear map. It was based only on formal properties of !e(E) which can also be found in other sets and which we make explicit in the following important definition: An algebra R is called a normed algebra if it is a normed vector space and the inequality (9.1)
Ilabll ~ Ilallllbll
holds for products. In a normed algebra all algebraic operations and the norm are continuous, i.e., from an --> a, bn --> b, ()(n --> IX it follows that
(9.2)
the continuity of the linear operations (addition, multiplication by a scalar) and of the norm are guaranteed by Proposition 6.1, the continuity of multi plication follows from (9.1) just like (7.S) followed from (7.6). For powers an we have, according to (9.1), Ilanll ~ Iiali n
for
n
=
1,2, ....
If a-not necessarily normed-algebra R has a unit element e and ab = e, then a is called a lefi-inverse of band b a right-inverse of a; we also say that b is left-invertible and a is right-invertible. If a is left- as well as right-invertible, i.e., if there exist elements band c with ba = ac = e, then a is said to be invertible or regular; in this case b = c and this uniquely determined element b is called the inverse of a and denoted by a - I. The unit clement is invertible, and the set of invertible elementsforms obviously a group with respect to multiplication. If a and b are invertible, then (ab)-I = b-1a- l • In an algebra R with unit element e we set aO := e. If R is normed and i= {O}, then Ilell = IIe 2 11 ~ II ell 2 immediately implies Ilell ~ 1. In the literature a normed algebra, which as a normed space is complete, is Usually called a Banach algebra. For the sake of simplicity we shall use this Word only for complete normed algebras which have a unit element e with Ilell = 1. Thus a Banach algebra consists of at least two elements.
52 /I'(E) is an algebra, 2(E) a normed algebra and by Proposition 7.4 even a Banach algebra, provided that E is a Banach space 0/= {O}. Observe that in the algebra Y(E) the word 'inverse' has two meanings, an operator-theoretical one and one from the theory of algebras: If A E //,(E) has an inverse in the sense of the theory of algebras, i.e., if there exists aBE .c1'(E) so that BA = AB = I, then by Proposition 5.2 the map A is in particular injective, thus has an inverse as a map, namely the inverse map A-I, and we have B = A-I: the inverse as a map coincides with the inverse in the sense of algebra-theory. If we only know that A has an inverse as a mapthe inverse map-then A does not have necessarily an inverse in the algebra Y(E); such an inverse exists according to Proposition 5.2 only if A is also surjective. Expressed differently: The inverse map A - I is the inverse of A in the
sense of the algebra /I'(E) if and only if A-I is defined on the whole space E. If we speak of the inverse A-I of a map A E Y'(E), we mean, according to the convention made in §5, the inverse map, even ifit is not defined on all of E. An analysis of the second proof in §8 for the Neumann expansion (8.7) shows immediately that it only uses properties of 2(E) which are present in every Banach algebra; the same holds for Propositions 8.1 and 8.2. Therefore we can state without a new proof the following two propositions:
Proposition 9.1. (9.3)
For every element x of a normed algebra the limit
lim Ilx"III/"
exists and is
~ Ilxklll/k
for
k = 1,2, ....
Proposition 9.2. The element e - x in a Banach algebra E has an inverse in E whenever the Neumann series L:'=o x" converges; in this case 00
(9.4)
(e - x)-
1
=
LX". "=0
The Neumann series converges or diverges according as lim..yjj~1 < 1 or > 1; in particular, it converges for all x with II x II < 1.
We want to discuss now the question whether the invertibility of an element Xo of a Banach algebra is perturbed if Xo undergoes a small change. If x lies sufficiently close to x o , more precisely: if
1 Ilx - xoll < Ilxolll'
(9.5)
then it follows from x
= Xo - (xo - x) = xo[e - Xo I(XO - x)]
that x is invertible; indeed, in the product on the right-hand side the first factor is invertible according to hypothesis, and the second factor has an inverse
53 according to Proposition 9.2 because Ilxo I(XO - x)11 ~ Ilxo 1IIIIxo - xii < 1. The inverse X-I is given by 00
X-I = [e - XOI(XO - X)]-IXOI = XOI
+ L [XOI(XO
- X)]nxOI,
n= I
from where we obtain the estimate (9.6)
It follows that the inverse x - I depends continuously on x: If Xn --+ Xo then x;; I converges to Xo I. We recall the concept of an open set (see §1, Exercise 10): A subset M of a metric space is said to be open if around each point of M there exists a ball which lies entirely in M. With the help of this concept we can summarize our results as follows: Proposition 9.3. The group of invertible elements of a Banach algebra is open, and the inverse X-I is a continuous function of x. Quantitatively: if x satisfies condition (9.5), then x possesses an inverse and this inverse satisfies the inequality (9.6). This proposition can be of importance for the solution of equations of the form Ax = y, where A is a continuous linear self-map of the Banach space E. In §7 we replaced a 'difficult' right-hand side y by 'simple' right-hand sides Yn which converged to y. One can, however, think of replacing a 'difficult' transformation A by a 'simple' transformation lying close to A and to obtain approximate solutions for the original problem by solving the modified problem. If we combine the two procedures, we obtain with the help of the preceding proposition easily the following result: If A has a continuous inverse on E, if the continuous linear maps An = A and the right-hand sides Yn --+ y, then the equations Anxn = Yn have unique solutions Xn at least for n greater than some no, and the sequence of solutions Xn converges to the solution of the equation Ax = y. As an example we want to consider the Fredholm integral equation (9.7)
x(s) -
f
k(s, t)x(t)dt
= y(s)
or
(l - K)x
=y
in C[a, b], where the kernel k should again be continuous and the integral transformation k is defined as usually. According to the Weierstrass approximation theorem, there exists a sequence of polynomials Pn(s, t) := Li.k OCl~)Sitk in two variables which converges uniformly on [a, b] x [a, b] to k(s, t), i.e., (9.8)
max Ipis, t) - k(s, t)1 a~s.t~b
-->
0
as
n
--> 00.
54 If we define the continuous endomorphisms P n of C[a, b] by
f
(Pnx)(s):=
then
II(Pn
K)xll = m:x
-
~ (b -
pis, t)x(t)dt,
f
I
[Pn(s, t) - k(s, t)]x(t)dt
a)max Ipis, t) - k(s,
I
t)lllxll,
.' K
I - Pn => I - K.
and thus also
If we already know that (I - K)-' exists and is continuous on C[a, b], then it follows from the remark after Proposition 9.3 that for sufficiently large n the Fredholm integral equation with polynomial kernel (9.10) is solvable by an Xn belonging to C[a, b], and that the sequence (x n ) converges in qa, b]-i.e., uniformly on [a, b]-to the solution of (9.7). A polynomial kernel p(s, t) can obviously be written in the form m
p(s, t)
=
L uj(s)Vj(t),
Uj
and
Vj
in
C[a, b];
j; ,
the corresponding integral transformation P is then given by (Px)(s) =
f jt,
Uj(s)vj(t)x(t)dt =
or (9.11 )
Px
=
jt,f
J, (f j.
vj(t-)x(t)dt·
vj(t)x(t)dt )Uj(S)
U
The image space of P is contained in the finite-dimensional linear hull [u" ... , urn] of the functions U" ••• , Urn and is therefore itself finite-dimensional. Linear operators with finite-dimensional image spaces are called finite-dimensional operators or operators offinite rank. For such operators, the equation (I - P)x = y can be solved, as we shall see in § 17, by purely algebraic methods. If we anticipate this result, we can say that the solution Xn of (9.10) can be determined in an elementary fashion and lim Xn is the solution of (9.7). We return again to Proposition 9.3. A subset M of the metric space E is said to be closed if its complement in E is open. M is closed if and only !f the limit oj every convergent sequence in M lies again in M (§I, Exercise to). From Proposition 9.3 we now obtain: Proposition 9.4. The set of non-invertible elements in a Banach algebra is closed; thus the limit ofa sequence of non-invertible elements is itselfnon-invertible.
55 A sequence of invertible elements can also have a non-invertible limit, as the sequence «(l/n)e) shows. Exercises
1. The Banach space C[a, b] becomes a commutative Banach algebra if we define the product X}" pointwise: (xy)(t):= x(t)y(t). 2. The Banach space II becomes a commutative Banach algebra if the product xy is defined by the convolution of the sequences x = (~o, ~I' ..• ), Y = (1]0' 1]1'···) thus:
xy:= (~o1]o, ~01]1
+
~11]0'···' ~o1]n
+
~11]n-1
+ ... + ~n1]o,·· .);
it is useful to start indexing of the sequences by 0 to make the connection with the Cauchy multiplication of power series easily. The properties of II which have to be proved can be obtained readily from the theory of power series. 3. With the aid of Exercise 2 and Proposition 9.2 prove the following wellknown Theorem: If the power series I:."'=o rxn(n has a convergence radius different from zero and (xo i= 0, then in a certain neighborhood of 0 the reciprocal (L:'=o :Xn(n)- I can be expanded into a power series I:'=o Pn(n. *4. The set L of left-invertible elements of a Banach algebra E is open and contains with any two elements their product (L is a multiplicative semi-group). The same holds for the set of right-invertible elements. Hint: If y is a left-inverse of x, then y(x + z) = e + yz is invertible for all z with sufficiently small norm.
§10. Finite-dimensional normed spaces
In the next sections we shall frequently need some properties of finite-dimensional normed spaces. We list therefore these facts briefly here and begin with the fundamental Lemma 10.1. Given finitely many linearly independent elements XI' ... ' xn in a normed vector space, there exists 11 > 0 so that (10.1)
for all numbers rx I, •.. , (Xn. For the proof, we first consider only linear combinations (XIXI with I(XII + ... + Irxnl = 1 and set }':=
II(Xlx l
inf 1: I ~v I =
+ ... + (XnXn
+ ... + (Xnxnll.
I
By the definition of the greatest lower bound, there exists a minimising sequence, i.e., a sequence of elements n
n
Yk:=
"L.
,,=
1
CX (k) v Xv
with
L \1=
1
Icx~k)1
= 1 and
56
Since ICt~~) I ~ I, there exists by the Bolzano-Weierstrass theorem an increasing sequence of integers k l , k 2 , ••• , such that Ct~kl) -> Pv (v = 1, ... , n). Obviously IPI I + ... + IPn I = I, hence by the linear independence of x I' ... ,Xnone has x:= {3IXI + ... + Pnxn f= 0 and so Ilxll > O. From the continuity of the operations and of the norm (see Proposition 6.1) it follows furthermore that
Since on the other hand Ilyd and by the definition of}'
-> }',
we have
Ilxll
= }'
> 0, thus also 11:= Ify > 0
(10.2) If now the Ct l , ... ,Ctn are arbitrary (but not all equal to zero, to exclude the trivial case), then we obtain (10.1) immediately if we replace in (10.2) the coefficients :Xv by
• The lemma just proved is the source from which we derive the next five propositions. Proposition 10.1. In afinite-dimensional normed space convergence is equivalent to componentwise convergence, i.e., (I" {x I, . . . , x n } is a basis ~I" the space then Yk:= L~= 1 Ct~k)Xv converges to y:= L~ = I Ct v Xv if and only if :x~k) -> Ct v for \' = 1, ... , n. Because of the continuity of the operations (Proposition 6.1) it is clear that component wise convergence implies convergence in the sense of the norm. Conversely, it follows from the above lemma that L~= I ICt~k) - Ct v I ~ IlllYk - YII·
•
This proposition immediately implies: Proposition 10.2. lent.
All norms on a finite-dimensional vector space are equiva-
This result goes far beyond the statement we could make at the end of §7 that all the IP(n)-norms on Kn are equivalent. Proposition 10.3. Every linear map from a finite-dimensional normed space into an arbitror j normed space is continuous.
57 Indeed, if {XI, ... , xn} is a basis of the first space, A the given map and if . Yk:= L.,.v= I (Xv(k) Xv converges t 0 y:= L.,v= I (XvXv, th en (Xv(k) converges t 0 (Xv (P ropOSltion 10.1), and because of the continuity of the operations it follows that
"n
"n
n
AJ'k = v~I(X~k)Axv
n
---+
(
n
)
v~,(XvAxv = A V~I(Xvxv = Ay.
•
On infinite-dimensional normed spaces there do exist discontinuous linear maps, see Exercise 1. Proposition 10.4. In a finite-dimensional normed space the BolzanoWeierstrass theorem holds, i.e., every bounded sequence contains a convergent subsequence.
Proof Let {x I' ... ,xn } be a basis of the space, Yk := L~ = I (X~k)xv and IIYk II ~ Y for k = 1,2, .... According to Lemma 10.1 for an appropriate 11 > 0 we have
.t11(X~k)1 ~ 1111 vtl (X~k)xv
II
= I1llYkl1
~ I1Y,
hence each numerical sequence «(X~k), v = 1, ... , n, is bounded. By the theorem of Bolzano-Weierstrass for number sequences, there exists an increasing sequence of integers k I, k 2 , ••• such that (X~~ll---+ f3v (v = 1, ... , n). Consequently Yk, = L~= I (X~k,)xv converges to L~= I f3vxv, thus (Yk,) is a convergent subsequence
•
~W
Proposition 10.5. Every finite-dimensional normed space is complete; hence every finite-dimensional subspace of a normed space is closed. Indeed, if (Yn) is a Cauchy-sequence in a finite-dimensional space, then (Yk) is bounded (see §1), hence by Proposition 10.4 contains a convergent subsequence. Its limit is also the limit of (Yk)' The closedness of finite-dimensional subspaces is now clear. • Let us return to Proposition 10.4. In infinite-dimensional normed spaces the Bolzano-Weierstrass theorem need not hold any more, see Exercise 2. It is reasonable to inquire whether among all normed spaces the finite-dimensional ones are distinguished by the validity of this important theorem. The answer to this question will be affirmative. The basis of our investigation is the following:
Lemma 10.2 (Lemma of F. Riesz). If F is a proper closed subspace of the normed space E, then to every '1 with 0 < '1 < I there exists a vector X/I in E With
IIx'Ill
= 1
and
Ilx - x'Ili
~ '1
for all
X E
F.
58 Proof There exists a vector y in E which does not lie in F. Let d := infxEF Ilx - yll and (xn) be a minimising sequence in F, i.e., Ilxn - yll --> d. It follows that d > 0 (since otherwise (d = 0) we would have Xn --> Y and because of the closedness of F the vector y would lie in F in contradiction to its choice). From here and from the hypothesis 0 < 1'/ < 1 it follows that dllJ > d, thus there exists z E F with 0 < liz - yll ::;;; d/I'/. If we set y:= l/llz - yll and x~:= y(y - z), then Ilx~ II = 1 and for all x E F we obtain because of y ~ I'//d and (l/y)x + z E F that Ilx -
x~11
= Ilx - y(y - z)1I = lI(x + yz) - yyll = ~
1'/
d' d =
1'/.
yll Gx + z) - y I
•
We are now in the position to characterize the finiteness of dimension . an algebraic property-of a normed space by a metric property-the validity of the Bolzano-Weierstrass theorem. The deep reason that this is possible is the fact that the algebraic and the metric structure of a normed space do not stand unrelated side by side but are intertwined through continuity. Theorem 10.1. The Bolzano- Weierstrass theorem is valid in a normed space, i.e., every bounded sequence contains a convergent subsequence, iland only if the space has finite dimension. In one direction the assertion is nothing but Proposition 10.4. Conversely, suppose that the normed space E is infinite-dimensional, so that no finitedimensional subspace of E coincides with E. Let XI E E, IlxI11 = 1 and F I := [XI] the linear hull of XI' As a finite-dimensional subspace, FI is distinct from E and by Proposition 10.5 it is closed. According to the lemma of F. Riesz, there existsanx 2 E Ewith IIx211 = 1 and Ilx - x 2 11 ~ Horallx E Fl' Let F2 := [XI' x 2 ] be the linear hull of XI' X2' Again F 2 i= E and F 2 is closed, hence there exists X3 E E such that IIx311 = 1 and Ilx - x 3 11 ~ t for all X E F 2 • One continues in this way, more precisely: if XI"'" Xn have been constructed, then Fn:= [x l' . . . ,Xn] i= E and F nis closed, hence there exists Xn + 1 E E such that II Xn + I II = 1 and IIx - Xn+ III ~ t for all X E Fn. The sequence (xn) is bounded since Ilxnll = 1 but contains no convergent subsequence, not even a Cauchy subsequence • because for n i= m one has Ilxn - xmll ~ t. Because of the fundamental importance of the Bolzano-Weierstrass theorem we distinguish in the theory of metric spaces those sets in which it is valid by a sp~cial name: A subset M of a metric space is relatively compact-or relatively sequentially compact-if every sequence in M contains a convergent subsequence. Its limit does not have to lie in M; if, however, it always belongs to M, we say that M is compact or sequentially compact.
59 A compact set M is always closed. Indeed, ifthe sequence (xn) from M converges to x, then because of compactness there exists a subsequence (xnJ which converges to an element y in M. Since y = x, the point x lies in M. •
It follows immediately that a set is compact if and only ifit is relatively compact and closed. A relatively compact, and a fortiori a compact set is bounded. If M were unbounded and y arbitrary in M, there would exist for every natural number n a point Xn in M such that d(xn' y) ~ n. On the other hand, because M is relatively compact, there would exist a subsequence (xnJ with x nk --+ x. Because of the continuity of distance (§1) it would follow that d(x nk , y) --+ d(x, y) which • contradicts d(x nk , y) ~ nk' A set U in a normed space is called a neighborhood of zero if it contains a ball around (cf. §1, Exercise 13). With this terminology, the above observations concerning compact and relatively compact sets, Exercise lOb in §1 and Theorem 10.1, the reader should have no difficulty in proving the following proposition.
°
Proposition 10.6.
For a normed space E the following assertions are equiva-
lent:
(a) (b) (c) (d) (e)
E isfinite-dimensional. Every closed and bounded subset of E is compact. The closed unit ball K 1 [0] is compact. Every bounded subset of E is relatively compact. There exists in E a relatively compact neighborhood of zero.
Exercises l. Define the linear map A: 12 --+ 12 by Ax := (L~= 1 (~v!v), 0, 0, ... ) and show that A is continuous (Hint: Cauchy-Schwarz inequality). If we introduce, however, on 12 c: 100 the lOO-norm Ilxlloo = supl~vl, then A is not continuous (consider the sequence of elements Xn := (1, 1, ... , 1,0,0, ... )-the first n terms = 1, all others =0). 2. From the bounded sequence of unit vectors en := (0, ... ,0, 1,0, ... ) in F-the nth term = 1, all others =O-no Cauchy subsequence can be extracted. 3. A distance problem: Let F be a proper subspace of the normed space E and 0. Does there exist ayE E so that Ilx - yll ~ Mor all x E F? Show on an appropriate subspace of 12 that the answer is negative if F is not closed. For a closed F it is, however, affirmative. *4. Let M be a non-empty subset of the metric space E; with the metric taken from E also M is a metric space. If M is complete as a metric space, then it is closed as a subset of E. If E itself is complete, then M is complete if and only if it is closed.
60 5. Let E, F be metric spaces, A: E ~ F a continuous map and M c E compact. Then A(M) too is compact, briefly: the continuous image of a compact set is compact. 6. A real-valued continuous function on a compact metric space is bounded and assumes its greatest lower bound and its least upper bound (use Exercise 5). 7. Let T be a compact metric space and C(T) the set of all continuous functions x: T --+ K. With the usual pointwise definition of addition and multiplication by scalars C(T) is a vector space over K. Defining the norm by Ilxll := maXtETlx(t)1 (cf. Exercise 6) C(T) becomes a Banach space. S. A compact metric space is complete. We make an observation before the remaining exercises. A relatively compact subset of a metric space is bounded and in finite-dimensional normed spaces, according to Proposition 10.6, exactly the bounded sets are relatively compact. In the following Exercises we shall state precise conditions which have to accompany boundedness in order to make subsets of certain Banach spaces relatively compact. To show that the conditions are sufficient, choose by the diagonal process from a given sequence a subsequence which at first converges componentwise (for the diagonal process see the proof of Proposition 13.1). 9. M c IP, 1 ~ p < 00, is relatively compact if and only if M is bounded and L~nl~vIP --+ 0 as n --+ 00, uniformly for all (~1' ~2"") in M. 10. M c (co) is relatively compact if and only if M is bounded and sUPv~" I~vl--+ 0 as n --+ 00, uniformly for all (~1' ~2"") in M. 11. Me (c) is relatively compact if and only if M is bounded and sUPv,I'~" I~v - ~I' I --+ 0 as n --+ 00, uniformly for all (~1, ~2'" .) in M. §11. The Neumann series in non-complete normed spaces The investigations in §S concerning the solution of the equation (11.1)
(l - K)x = Y
I:'=
with the aid of the Neumann series 0 K"y were always performed in complete normed spaces. In this section we want to investigate how far we can liberate ourselves from this sometimes obnoxious condition of completeness. The first question will be whether the Neumann series-if it converges at all for a certain y-yields a solution of equation (11.1). The following completely elementary but fundamental proposition gives an affirmative answer to this question. Proposition 11.1. If K is a continuous endomorphism of the normed, not necessarily complete space E and if the Neumann series 00
(11.2)
LK"y "=0
converges for a certain y in E to a vector x in E, then (J - K)x
= y.
61
Because of the continuity of K it follows from x = L:'=o Kny that Kx = Kny = x - y, i.e., indeed x - Kx = y. •
L:'=o K n+ ly = L:'= 1
Thus the Neumann series (11.2) can converge only for vectors y from the image space (/ - K)(E). Under certain hypotheses concerning K convergence will take place for all these vectors. The following proposition presents a condition of this kind. Proposition 11.2. For the map K from Proposition 11.1 the following assertions are equivalent: (a) K n --+ 0 as n -+ 00. (b) 1- K is injective, and the Neumann series L~o Kny converges for all vectors y from the image space of I - K.
If (a) holds, then from (/ - K)x = 0 it first follows that x = Kx, and then successively x = Knx for n = 1,2, ... , hence x = o. Thus I - K is injective. If y is from (/ - K)(E), then there exists a vector x-and, according to what we have just proved, exactly one x-such that x - Kx = y, i.e., x = y + Kx. It follows that x = y + K(y':.j- Kx) = y + Ky + K 2 x, and in general x = y + Ky + ... + K n- 1 y + Knx for all natural numbers n. Because of Knx --+ 0 we have x = L:'=o Kny. Now suppose that (b) is valid, and let x be an arbitrary vector in E. Then y = x - Kx lies in (/ - K)(E) and we see as above that x = y + Ky + ... + Kn-1y + Knx. Since the series L:'=o Kny converges by hypothesis, its sum solves equation 01.1) and since by the injectivity of I - K the only such solution is the x we started out with, it follows that y + Ky + ... + Kn-l y -+ x and so Knx --+ 0: thus (Kn) converges indeed pointwise to O. •
Towards the end of §7 we discoursed on the special importance which the continuity of the inverse has in applications. Therefore we do not want to omit the discussion of the continuity of (/ - K) - 1, and prove in this direction the following: Proposition 11.3. If for the map K of Proposition 11.1 we have K n => 0, then the inverse (/ - K)-l is continuous on its space of definition (/ - K)(E) (cf. §8, Exercise 5).
Proof (/ - K) - 1 exists by Proposition 11.2 since Kn => 0 implies a fortiori that K n --+ o. If m is a natural number such that II Km II < I-such an m exists according to hypothesis-then with 8:= 1 - IIKmll > 0 for every x E E we have (11.3)
for the first estimate we used (6.12). If (/ - K)- 1 were not continuous on (1 - K)(E), we would have by Proposition 7.5 a sequence (x n) with Oln)lIxnll > 11(1 - K)xnll. For Yn:= xn/llxnll we would thus have (11.4)
llYn II = 1 and
(I - K)Yn -+ O.
62 Because of I - K m = (I + K + '" + Km-1)(I - K) it would follow that (I - Km)Yn ---+ 0 as n ---+ 00, in contradiction to (11.3) according to which 11(1 - Km)Ynll ~ elly.11 = e > 0 for all n. Thus (I - K)-l is indeed continuous .
•
When applying Proposition ILl, one can often verify without any effort that for a certain Y the Neumann series (11.2) is a Cauchy series. It will be, in general, more difficult to show that this Cauchy series converges. This leads to the question whether one can determine classes of operators K for which the convergence of the Neumann series follows already from its being Cauchy. We shall see below that this is the case for operators with complete image space, for finite-dimensional operators, and for compact operators-a concept yet to be defined. Proposition 11.4. If the image space K(E) of the map Kfrom Proposition ILl is complete, then the Neumann series (11.2) converges whenever it is a Cauchy series. Its sum solves equation (ILl). We only have to modify slightly the proof of Proposition 11.1. If (11.2) is a Cauchy series, i.e., if the sequence of partial sums s.:= Y + Ky + ... Kny is a Cauchy sequence, then obviously also the elements KS n = Ky + K2y + ... + K· + 1Y form a Cauchy sequence in the complete subspace K(E). Consequently the sequence (Ksn) converges, i.e., the series 1 Kny, and with it also the Neumann series (11.2). The last assertion of the proposition follows from Proposition 11.1. •
I:'=
Proposition 11.4 can be invoked in particular if K is a continuous operator of finite rank, since then by Proposition 10.5 its finite-dimensional image space K(E) is complete. The next proposition shows that under weaker hypotheses we can prove even more. Proposition 11.5. Let K be a finite-dimensional endomorphism of the vector space E. If the Neumann series (11.2) is a Cauchy series with respect to some norm 11·110 on E, then it converges with respect to every norm 11·11 on E to a solution of the equation (11.1)-even ifit is not continuous with respect to this norm. If (11.2) is a Cauchy series with respect to the norm 11·110 for every y in E, then the inverse (I - K)- 1 exists on E and is continuousfor any norm on E with respect to which K itself is continuous. Proof We set F:= K(E) and denote by (E, 11·110) the space E equipped with the norm 11·110' We define (E, 11-11), (F, 11·110) and (F, HI) analogously. Th.:! restriction KF of K to F is obviously a linear self-map of F. Since F is finitedimensional, KF must be continuous for every norm of F according to Proposition 10.3. Assume now that the Neumann series (11.2) is a Cauchy series with respect to the norm II . II 0, i.e., that the sequence of partial sums Sn := y + K y + .. ,
63
+ K'y is a Cauchy sequence. Because of the completeness of(F, 11·110) it follows, as in the proof of the preceding proposition, that the sequence (Kso) converges to some z E F. Since by Proposition 10.2 all norms on F are equivalent, we have Ks. ~ z also in (F, 11·11). If we set x:= y + I KOy = y + z, then x is the sum of (11.2) in (E, 11·11) and from the continuity of KF on (F, 11·11) it follows that
I:';
Kx = Ky
+ KFC~I K'Y) =
Ky
+ '~I Kf·K'y =
Ky +
.~I K·+1y =
X -
y,
so that indeed x - Kx = y. Now let (11.2) be a Cauchy series for every yin (E, 11·110). Then by what we have just proved I - K is surjective, furthermore obviously K· ~ O. Similarly as at the beginning of the proof of Proposition 11.2 it follows that I - K is also injective, hence bijective (we cannot apply Proposition 11.2 directly because there K was supposed to be continuous). Now let K be also continuous on (E, 11·11) but assume that {I - K)-I is not continuous there. As it was shown in the proof of Proposition 11.3, there exists then a sequence (y.) in (E, 11·11) for which (11.4) holds. Because of
11 - IIKY.III = IIIY.II - IIKy.111
~ IIY. - Ky.11
it follows that IIKy.11 ~ 1. Thus (Ky.) is a bounded sequence in (F, Proposition 10.4 it contains a convergent subsequence (Ky.):
Ky.",
~
z,
Ilzll =
11·11).
By
1.
With the help of (11.4) it follows that
Y.", = {I - K)y.",
+
Kyo",
~
z.
Because we assumed that K is continuous with respect to the norm 11·11, we have Ky ... ~ Kz and so {I - K)yo .. ~ {I - K)z. Going back again to (11.4) we get {I - K)z = 0 and so z = 0 (since I - K is injective) in contradiction to Ilzll = 1. Thus {I - K)-l must be continuous on (E, 11·11). • Exercise 1 of §W shows that a finite-dimensional operator can very well be continuous with respect to one norm but discontinuous with respect to another one. The above proof of the continuity of (I - K)- 1 used, besides the continuity of K, only the following property: If (x.) is a bounded sequence, then (Kx.) has a convergent subsequence. Every linear map K of a normed space E into itself-or into another normed space F -with this property is said to be compact or completely continuous. A compact transformation is always continuous; otherwise there would exist a sequence (x o ) with Ilx.11 = 1 and IIKx.1I ~ 00, thus in spite of the boundedness of (x.), the sequence (Kx.) could not have a convergent subsequence. Thus we can state the following proposition: Proposition 11.6. For a compact self-map K of a normed space the inverse (I - K)-l is continuous provided it exists at all.
64 We shall see in the second next section examples of compact operators. For our present purposes it is decisive that compactness of an operator guarantees the convergence of certain sequences, even in non-complete spaces. Let us suppose for instance that the Neumann series (11.2) with a compact self-map K of the normed space E is a Cauchy series, i.e., that the sequence of the partial sums Sn:= I~=o KVy is a Cauchy sequence. Then the sequence of the KS n = I~: 1 KVy is also a Cauchy sequence. (sn) is bounded, hence (Ksn) contains a convergent subsequence; but then the Cauchy sequence (Ks n ), i.e., the series 1 Kny must also converge. This obviously implies that the series (11.2) converges, its sum is according to Proposition 11.1 a solution of the equation (11.1). Let us state this result:
I:'=
Proposition 11.7. If K is a compact self-map of a normed space, then the Neumann series (11.2) converges to a solution of equation (11.1) whenever it is a Cauchy series. We can follow another way to obtain a solution for equation (11.1)-or at least for an equation closely related to it-from a Cauchy series (11.2). We only have to think of the procedure with which one makes solvable the equation X2 - 2 = 0, which does not have a solution in the set Q of rational numbers: One enlarges the incomplete metric space Q to the complete metric space R, extends the function x ~ x 2 to R (i.e., defines the squares of real numbers x), and shows that the 'extended equation' x2 - 2 = 0 has solutions in R. The space R is a particularly economical extension of Q: Every x in R is the limit of a sequence (xn) in Q. We say therefore that Q is dense in R. In the next section we shall deal with the question whether one can in a similar way enlarge a noncomplete normed space E to a complete normed space £, extend equation (1Ll) to £, and solve it there with the aid of the Cauchy series (11.2) (see Proposition 12.4). Exercises 1. 2.
Consider Exercises 3,4 in §8 in the light of Proposition 11.5. If K is a finite-dimensional self-map of the vector space E over K and {x1, ... ,x,} is a basis of K(E), then there exist r linear mapsfp:E-+K with which Kx =
I
j~(x)xp
for all
x
III
E.
1)= 1
Thus we have Knx = I~= 1 flKn- 1 x)xp for n = 1,2, .... Use this observation to prove the convergence statement of Proposition 11.5 under the sole application of Proposition 10.1. *3. The zero transformation 0 is compact on all normed spaces, the identity transformation I only on finite-dimensional ones (thus there exist continuous transformations which are not compact).
65 + 4. The linear map K: E --+ F (E, F normed) is compact if and only if one of the following equivalent assertions is true: (a) {Kx: x E E, Ilxll ~ I}, i.e., the image of the closed unit ball in E, is relatively compact. (b) There exists a neighborhood U of zero in E whose image K(U) is relatively compact. *5. Scalar multiples and sums of compact operators are compact. The product of a compact operator with a continuous operator is compact, independently of the order of the factors (observe that the continuous image of a bounded sequence is again bounded). *6. A compact operator on a infinite-dimensional normed space does not have a continuous inverse. Hint: Exercises 3 and 5.
§12. The completion of metric and normed spaces
We investigate first the completion of general metric spaces and use the method ofthe Cantor fundamental sequences, with the help of which one extends the incomplete space Q of the rational numbers to the complete space R of real numbers, as is well known. To describe this method adequately, we need the concept of an isometry. A map A from the metric space (EI' d l ) into the metric space (E2' d2 ) is said to be isometric or an isometry if it preserves distances, i.e., if for two points x, y in EI one has dl(x, y) = diAx, Ay). An isometry is obviously injective and continuous; the inverse map A -I: A(E I) --+ E 1 is again isometric. Therefore we may give the following symmetric definition: Two metric spaces are said to be isometric if there exists an isometric map from one onto the other. From the point of view of a metric theory, isometric spaces differ only by the names of their elements and can therefore be identified. A subset M of the metric space E is dense in E if every point in E is the limit of a sequence in M, or equivalently, if for an arbitrary B > 0 and for every x E E there exists ayE M with d(x, y) ~ B. A non-empty subset M of a metric space E is itself a metric space if one assigns to any two elements of M the distance they already have as elements of E. One then calls M a subspace of E and says that the metric of M is induced by the metric of E, or simply by E (cf. the definition of a subspace of a normed space in §6 and Exercise 4 in §1O). Let now E be an incomplete metric space. We call two Cauchy sequences (x n), (Yn) from E equivalent, and write (x n) - (Yn), when d(xn' Yn) --+ O. This relation in the set F of all Cauchy sequences from E is reflexive, symmetric and transitive, i.e., an equivalence relation. It decomposes F into classes X, y, ... of equivalent Cauchy sequences. Let E be the set of all these equivalence classes. It follows from the quadrangle inequality that for two Cauchy sequences (x n), (Yn) the limit lim d(xn' Yn) always exists and that it does not change if we
66 pass to equivalent Cauchy sequences. Therefore if x, yare equivalence classes with representatives (x n), (Yn), the definition d(x, y) := lim d(xn' Yn) makes sense. It is easy to see that d is a metric on £. We now indicate a subspace of £ which is isometric with E. For this we observe that an equivalence class x can contain at most one constant sequence (x, x, x, ... ). Let Eo be the set of all equivalence classes which contain a constant sequence. The map A: E -+ Eo which associates with each x E E the equivalence class X E £ containing (x, x, x, ... ) is obviously an isometry of the spaces E and Eo. Eo is dense in E: To a representative (xn) of X E E and to a preassigned e> 0 there exists an mEN so that d(xn' xm) ~ I; for n ~ m. The equivalence class Y of the sequence (xm' X m, . .. ) lies in Eo and satisfiesd(J" x) = limn_co d(xm' xn) ~ t;. We show now that £ is complete. Let (xn) be a Cauchy sequence in £. Since Eo is dense in £, to each natural number n there corresponds a Yn E Eo so that d(xn' Yn) ~ lin. Because of d(Yn' Ym)
~ d(Yn, Xn) + d(x n, Xm) + d(x
m,
Ym)
~ ~n + d(x n, Xm) + !, m
the Yn form a Cauchy sequence. Let (Yn, Yn,"') be the constant sequence in Yn, i.e.,AYn = YnwiththeabovedefinedisometryA.Thend(Yn,Ym) = d(AYn' AYm) = d(Yn' Ym), and thus (Yl, Yz, ... ) is a Cauchy sequence in E. If we denote its equivalence class by y, then
hence lim d(xn' y) = O. The Cauchy sequence (xn) has therefore the limit y. Finally, we imbed E into £ with the aid of the isometry A. This means: If x is in Eo, and A - I X = x, i.e., (x, x, ...) is the constant sequence in x, then we remove x from £ and replace it by x. Thus we obtain a new set if = E u (£\Eo). On it we define a metric d I so that the situation on £ shall be conserved: for
x, Y in
d1(x, y):= d(x, y) = d(Ax, Ay)
for
x, Y III E.
d1(x, y):= d(Ax, y)
for
x in
d I (x, y)
:=
d(x, y)
£\Eo·
E
and
yin £\Eo.
Now (E, d I) is a complete metric space which contains (E, d) as a dense subspace. Assume that (E, d 2 ) is'a second complete metric space which contains (E, d) as a dense subspace. We show that (E, d 2 ) and (E, d 1 ) are isometric. Let x be an arbitrary element from E. Then there exists a sequence (xn) in E so that d 2 (x n , x) -+ O. Since (xn) is a Cauchy seguence in E, there exists an i in E so that d1(x n, i) -+ O. The point i is independent of the particular choice of the approximating sequence (xn). Indeed, if (Yn) is another sequence in E which converges to x in E, then it follows from d 1 (Yn, i) ~ d(Yn' xn)
+ d1(xn, i)
~ d 2 (Yn, x)
+ d 2 (x, xn) + d1(xn, i)
67 that also d, (Yn, x) -> O. We associate with each i in E the element x in E determined in this way. This map B: E --+ E is obviously surjective, and leaves the elements of E fixed. It follows further from the continuity of the metric that B is an isometry; indeed, if i, yare two points in E and (x n), (Yn) approximating sequences in E whose limits x, ji in E are given by x = Bi and ji = By, then
dii, y) = lim d 2 (x n, Yn) = lim d(xn' Yn) = lim d,(x n, Yn) = d,(x, Y)· We summarise now these results: Proposition 12.1. To every non-complete metric space E there exists a complete metric space E, determined uniquely up to an isometry, in which E is dense and which induces on E the original metric of E. We call E the completion or the complete hull of E. Let now E be a non-complete normed space. Let us denote the metric of its completion E by d, thus for x, y in E we have d(x, y) = Ilx - YII. We want to make out of E first a vector space and then a normed space. For x, Y in E there exist sequences (x n), (Yn) in E which converge to x and y, respectively. From II(xn + Yn) - (xm + Ym)11 ~ Ilx n - xmll + llYn - Ymll we deduce that (xn + Yn) is a Cauchy sequence, and has therefore a limit z in E. It is easy to see that z depends only on x and y and not on the choice of the approximating sequences (x n), (Yn). This observation justifies the definition x + Y := lim(x n + Yn). If x and y lie in E, then this sum coincides with the sum already defined in E: set Xn = x and Yn = Y for all n. In a similar way the multiplication by a scalar already defined on E can be extended to E by (Xx := lim«(Xx n). The reader should check that E is now a vector space. Next we introduce a norm on E by IIxll := d(x, 0), which is clearly an extension of the norm which existsalreadyonE.Itistrivialthat Ilxll ~ oand that Ilxll = Oifandonlyifx = O. The verification of the other properties of a norm will be based on the continuity of the metric (see §1) from which it follows in particular that Xn --+ x implies d(xn' 0) --+ d(x, 0), i.e., Ilxnll --+ IIxll. If the sequences (x n), (Yn) from E converge to x, Y from E, then Ilxn + Ynll ~ Ilxnll + IIYnl1 implies for n --+ 00 the triangle inequality Ilx + yll ~ Ilxll + Ilyll; the equality II(Xxll = 1(Xlilxll is proved completely analogously. Finally, we have d(x, y) = lim d(xn' Yn) = limllx n - Ynll = Ilx - ylI, i.e., the metric d on E derives from the norm introduced above. In summary, we say that E can be completed into a Banach space E: If E is another Banach space which is a completion of E, then the map B defined in the proof of Proposition 12.1 is not only an isometry from E onto E-so that in particular x and Bx always have the same norm- but it is also a linear transformation, i.e., an 'isomorphism' of normed spaces. The precise definition of this important concept is as follows: The linear map T from the normed space F onto the normed space G is an isomorphism of normed spaces or an isometric isomorphism, if
IITxll = IIxll
for all
x in
F.
68 An isomorphism of normed spaces T is because of
IITx -
Tyll = IIT(x -
y)11
=
Ilx - yll
an isometry, hence injective, thus an (algebraic) isomorphism of the vector spaces F, G. T - 1 : G ~ F is again an isomorphism of normed spaces; thus one is justified to say with a symmetric locution that two normed spaces are isomorphic as normed spaces (or isometrically isomorphic) if there exists a linear map which is an isomorphism of normed spaces from one space onto the other. From the point of view of the theory of normed spaces, isometrically isomorphic spaces differ from each other only in the names of their elements and can therefore be identified. The result of our investigations above can be described as follows: Proposition 12.2. To every non-complete normed space E there exists a Banach space E, determined uniquely up to an isomorphism ofnormed spaced, such that E is a subspace dense in E. We call E the completion or the complete hull of E. The next proposition asserts that continuous linear maps between normed spaces can be uniquely extended to their completions with preservation of the norm. Proposition 12.3. Let E, F be two normed spaces and E, F their completions. If A is a continuous linear map from E into F, then there exists exactly one continuous linear map Afrom E into F with Ax = Axfor all x E E. This extension preserves the norm, i.e., IIAII = IIA II. Proof If x is in Eand (x n) a sequence from E converging to x, then (Axn) is a Cauchy sequence because of IIAxn - Axmll = IIA(xn - xm)11 ~ IIA 1IIIxn - xmll, hence converges to an element y of F. It is easy to see that y depends only on x and not on the approximating sequence (x n). The expression Ax := lim AXn is thus well defined and yields a map A: E ~ F which is obviously linear and an extension of A. From IIAxn11 ~ IIAllllxnll the estimate IIAxll ~ IIAllllxll follows as n ~ ex); thus A is bounded and IIAII ~ IIA II. Since conversely also IIA I ~ IIAII (see §7, Exercise 1), we must have IIAII = IIA II. Finally, if A: E ~ F is a second continuous and linear extension of A to E, then Ax = lim Ax n, hence A = If.
•
It does not present any difficulties any more to answer the question which we posed at the end of §11 concerning the solution of the equation x - Kx = y. From the last two propositions and Proposition 11.1 we have immediately:
Proposition 12.4. normed space E and
Let K be a continuous endomorphism of the non-complete to the completion
K the continuous and linear extension of K
69
E of E. If the Neumann series I:'=o Kny is a Cauchy series for a y in E, then its sum X E E satisfies the equation x - Kx = y. Exercises 1. The isometric image of a complete metric space is again complete (completeness is preserved under isometries). 2. The image of a Cauchy sequence under a continuous map does not need to be a Cauchy sequence. 3. The map A from the metric space E into the metric space F is said to be uniformly continuous if for every E: > 0 there exists a J > 0 such that d(x, y) ~ J implies d(Ax, Ay) ~ E. If A is uniformly continuous, then the image (Ax n) of a Cauchy sequence (x n) in E is a Cauchy sequence in F. 4. Let A: E -+ F be a map of the metric spaces E, F. If A is uniformly continuous and bijective, and if the inverse map A-I is continuous, then, together with F, also E is complete (use Exercise 3). 5. A map with bounded dilation is uniformly continuous (see Exercise 3). 6. A continuous map from a compact metric space into a second metric space is uniformly continuous (see Exercise 3). +7. Every metric space (E, d) is isometric to a subset of the Banach space B(E). Hint: Let a be a fixed element of E. For every x E E define the function fx:E -+ R by
fx(t)
:=
d(x, t) - d(a, t)
(t
E
E)
and show successively with the help of the quadrangle inequality: (a) I fAt) I ~ d(x, a) for all tEE; thus fx E B(E). (b) For all x, y E E we have I fx - ,/;.11 = d(x, y); here 11-11 denotes of course the supremum-norm on B(E). It follows that the map x 1--+ fx from (E, d) onto the metric space {.f~: x E E} c B(E), equipped with the distance Ilfx - fyli induced by B(E), is an isometry. 8. With the help of Exercise 7 give a second proof of Proposition 12.1 concerning completion. Hint: Exercise 15 of §1. Observe that a closed subset of a complete metric space is itself complete. Remark. This second proof uses the completeness of B(E) and so ultimately the completeness of R. The proof given in the main text does not do this; it even given us the possibility to construct R as the completion ofQ (a procedure which goes back to Georg Cantor).
§13. Compact operators We recall at the outset the definition of compact operators in §11: A linear map K from a normed space E into a normed space F is said to be Compact or completely continuous if the image (Kx n) of every bounded sequence (x n ) from E contains a convergent subsequence.
70 We observed in §11 (and the exercises that followed it) that compact operators are continuous and that scalar multiples and sums of compact operators are compact. The product of a compact operator with a continuous one is compact, whatever the order of the factors. The identity transformation I is compact only on finite-dimensional spaces, and a compact operator can have a continuous inverse only if its space of definition is finite-dimensional. We denote by .~(E, F) the set of all compact operators K: E -+ F; the set of all compact self-maps of E is denoted by f(E) := f(E, E). By the above results f(E, F) is a linear subspace of the vector space 2(E, F), and f(E) is even a (two-sided) 'ideal' in the algebra 2(E)-as is well known, a subset M of an algebra R is a (two-sided) ideal ifit is a linear subspace of R and for every x in R and m in M the products xm and mx lie in M. The next proposition shows that f(E, F) is even a closed subspace of 2(E, F) if F is complete; thus if E #- to} is complete, then f(E) is a closed ideal in the Banach algebra 2(E). Proposition 13.1. If the sequence of compact maps Knfrom a normed space E into a Banach space F converges uniformly to K, then K is compact.
For the proof, let (Xi) be a bounded sequence in E: Ilx;!1 ~ y. Then there exists a subsequence (Xli) of (Xi) so that (K1Xli) converges, next a subsequence (X 2i ) of (Xli) so that (K 2x 2;) converges, etc. The diagonal elements Yi:= Xii form a sequence which from a certain index on is a subsequence of everyone of the sequences (Xk1' Xk2, ... ), and therefore the sequence (KnY;) converges for every operator Kn. Choose now an arbitrary I> > 0 and determine an no so that IIKno - K II ~ 1>. If we fix an io so that for i, k ~ io one has IIKnoYi - KnoYkl1 ~ r., then for these subscripts i, k IIKYi - KYkll ~ IIKYi - KnoYil1 + IIKnoYi - KnoYkll ~ I>lIy;l1 + f, + I>IIYkll ~ (2y + 1)1>.
+
II K noYk - KYkll
Thus (Ky;) is a Cauchy sequence in the Banach space F and so a convergent subsequence of (KxJ • A continuous operator K of finite rank is compact. Indeed, because of IIKxl1 ~ 11K IIllxll, the image of a bounded sequence is a bounded sequence in the finitedimensional image-space of K, and contains therefore by Theorem 10.1 a convergent subsequence. Ifwe also take into account Proposition 13.1, we may state: Proposition 13.2. A continuous operator of finite rank is compact, and a uniformly convergent sequence of such operators Kn: E -+ F has always a compact limit operator if F is complete.
The question whether conversely every compact operator K: E -+ F (F complete) is the limit of a uniformly convergent sequence of continuous, finite-
71 dimensional operators Kn: E --+ F has a negative answer (cf. [53], [48]). A totally different formulation of this approximation problem is given in [46]. Based on Proposition 13.2 we now give two important examples of compact operators. Example 13.1. The integral transformation K: C[a, b] continuous kernel k
(13.1)
(Kx)(s):=
--+
C[a, b] with
f
k(s, t)x(t)dt
is compact. Indeed, in §9 (see from (9.7) on) we observed that K is the uniform limit of a sequence of continuous operators of finite rank. Example 13.2.
Let
(lXik) be an infinite matrix with 00
I
(13.2)
IlXik 12
0 the estimate If(x) I ;;;; q Ilx II is valid for all x E E. The space E', equipped with the norm Ilfll
= sup I f(x) I Ilxll
= I
of a linear map, is a Banach space (Proposition 7.4). Exercises
1. A family F of continuous functions on [a, b] is said to be equicontinuous if to every B > 0 there exists a () > 0 such that for every x E F and for any two points t I' t2 from [a, b] with It I - t21 ;;;; () one has Ix(t I) - x(t 2)1 ;;;; e(observe that () should not depend on x). The theorem of Arzela-Ascoli (cf. [185], p. 144) asserts that a subset of C[a, b] is relatively compact if it is (norm-) bounded and equicontinuous. Use this theorem to give a new prooffor the compactness of the integral transformation K in Example 13.1. Hint: Show with the help of the continuity of the kernel k that the image K(M) of a bounded set M is an equiContinuous family of functions. 2. Apply Exercise 9 of §10 to supply a new proof for the compactness of the matrix transformation K in Example 13.2.
74
3. Show with the help of an appropriate diagonal matrix that hypothesis (13.2) is not necessary for the compactness of the matrix transformation K in Example 13.2. 4. Every linear map from a finite-dimensional normed space into a second normed space is compact. 5. Let E, F be non-complete normed spaces, E, F their completions. If K: E -+ F is compact, then also the continuous linear extension K: E -+ F of K to E is compact, and K(E) c F. If F = E and I the identity transformation on E, then N(l - K) = N(l - K) and (l - K)(E) = (I - K)(E) n E. 6. If E is a non-complete normed space and E its completion, then the duals E' and (E)' are isomorphic as normed spaces. + 7. Let E and F be Banach spaces. The linear map A: E -+ F is said to be nuclear if it can be represented in the form 00
Ax =
L Iv(x)yv
00
with
Iv E E', Yv
E
F,
L I1/v1111Yv11
Example 14.1. Every pair (E, E+) of vector spaces is a bilinear system with respect to the trivial bilinear form (x, x + >:= 0 for all (x, x +).
77
Example 14.2. If r is the smaller one of the natural numbers m, n and if is chosen arbitrarily, then (Km, K") is a bilinear system with respect to the bilinear form
(IX I , ••• , IXr)
r
<x, x+):= ~>v~v~:.
(14.5)
v=
For r = m = nand form in (14.3).
IXI
= ... =
IX"
I
= I we obtain precisely the second bilinear
Example 14.3. Let pea, b] be the vector space of all polynomials on [a, b] and WE qa, b]. Then (C[a, b], Pea, b]) and (C[a, b], C[a, b]) are bilinear systems with respect to the bilinear form (14.6)
<x, x+):=
f
w(t)x(t)x+ (t)dt;
for w(t) == I on [a, b] we obtain precisely the first bilinear form in (14.3). Example 14.4. bilinear form (14.7)
(C[a, b], BV[a, b]) is a bilinear system with respect to the
<x, x +):=
f
x(t)dx + (t).
Example 14.5. Let I < p, q < ro and lip bilinear system with respect to the bilinear form
+
l/q
= 1. Then (lP, lq) is a
00
(14.8)
<x,x+):= L~v~:.
v=
I
The convergence of the series follows from the Holder inequality. Example 14.6. Let E be a vector space and E+ an arbitrary subspace of E*, the space of all linear forms on E. Then (E, E +) becomes through the canonical bilinear form
(14.9)
a bilinear system. If x* is a linear form on E, then henceforth <x, x*) will mean as in (14.9) the value ofx* at the point x-even if we do not speak explicitly of a bilinear system.
The bounded ness of a linear form x* on a normed space E is then expressed by the inequality I<x, x*) I ~ IIxllllx*11 for all x in E. The concept of a bilinear system is symmetric: If (E, E+) is a bilinear system with respect to the bilinear form <x, x+), then (E+, E) is a bilinear system with respect to the bilinear form (14.10)
this is the canonical bilinear form for (E+, E); we shall always use it except when the contrary is explicitly stated.
78 Example 14.6 is particularly important for our purposes. In connection with (13.5) it shows that for every finite-dimensional operator K: E -+ F there always exists a bilinear system (E, E+), e.g., the bilinear system (E, E*), so that with appropriate vectors YI' ... ' Yn from F and xt, ... , x: from E+ the representation n
Kx
(14.11)
=
L <x, x;)Y.
.=
I
is valid for every x E E. Conversely, an operator defined by (14.11) has obviously finite rank. If K # 0, then the vectors xt, . .. , as well as the vectors YI' ... ,Yn' can be thought of as being linearly independent (see Exercise 3).
x:,
Exercises 1. If (E, E+) is a given bilinear system, then in general not every finitedimensional operator K: E -+ F will be representable in the form (14.11) by means of (E, E+). 2. The set of all self-maps of E which can be written in the form (14.11) by means of a given bilinear system (E, E+) is an algebra. *3. Show that if K # 0, then the vectors x x: and the vectors Y I' ... , Yn can be chosen linearly independent. Hint: If for instance Yn = OCIYI + ... + OCn-1Yn-l, then
t ,... ,
n-I
Kx
=
.=L <x, x; + oc.x:)Y.; I
continuing in this way reduce the number ofterms in the sum until the remaining y,. are linearly independent. One can reduce further in a similar fashion, if the are still linearly dependent. vectors from E+ occurring in +4. Assume that the normed spaces E, E+ form a bilinear system with the bilinear form <x, x+). The bilinear form is said to be continuous if <x n , x:)-> <x, x+) whenever Xn -+ x, x: -+ x+ ; it is called bounded if there exists a constant y > so that I<x, x+) I ~ yllxllllx+ II for all x E E, x+ E E+. Show that a bilinear form is continuous if and only if it is bounded.
= 0 for all x E E, implies that x+ = O. In this case E+ is isomorphic to the subspace A(E+) of E*, and can be identified with it. In a completely analogous way every x in E generates by means of
(15.2) a linear form F x on E +, and the correspondence x t-+ Fx is an isomorphic map from E into (E+)* if and only iffrom (x, x+ > = 0 for all x+ E E+ it follows that x vanishes. We call a bilinear system (E, E+): a left dual system iffrom (x, x+ > = 0 for all x in E it follows that x+ = 0, a right dual system if from (x, x + > = 0 for all x + in E + it follows that x = 0, a dual system if it is both a left and a right dual system. Using a self-explanatory language we can say that (E, E+) is a left or a right dual system if the 'left' space E or the 'right' space E+, respectively, contains 'many' elements, or is 'large', cf. also Proposition 15.1. If we make the above-described identifications, then in the case of a left dual system every x+ E E+ is a linear form on E with the values (15.3)
XE
In the case ofa right dual system every x
E
E.
E is a linear form on E+ ; it is defined by
(15.4) Observe that we identify E+ with a subspace of E* and E with a subspace of (E+)* only through the canonical imbeddings (15.5) and not with the help of any other isomorphism which might possibly exist. In other words: to consider x + as a linear form on E and x as a linear form on E + means always to use definitions (15.3) and (15.4), respectively. For the proof of the important Proposition 15.1, and for some other purposes, We need: Lemma 15.1.
Iffl, ... ,f" are linear forms on E and
if for afurther linear form
f on E we have
n .=1 n
(15.6) i.e., if fl(x) = ... tion of fl'" ., f".
= f,,(x) =
°
N(f.) c N(f),
implies that f(x)
= 0, then f is a linear combina-
80 Without restricting the generality, we may assume that the ll"",f" are linearly independent, and we use mathematical induction. Let first n = 1. Since J; "# 0, there exists an XI with II(X I ) "# O. Obviously, for every X E E the vector Yx'= X - (fl(x)/II(XI»X I lies in N(J;), hence x can be represented in the form (15.7)
Since by hypothesis l(yJ vanishes, we have I(x) = (f(x 1)/ II (xl»/I (x), i.e.,
1= IXII • Now let us assume that the Lemma is already proven for n - 1 linearly independent linear forms. Then n~= I. V*/1 N(f.) is not a subset of N(f/1)' since otherwise by the induction hypothesis 1/1 would be a linear combination of the other linear forms, in contradiction to the assumed linear independence of II' ... ,I,,· Thus for every J1 there exists an x/1 with J~(x/1) = 0 for v "# J1 and 1/1(x/1) "# 0; for reasons of homogeneity we may even assume that lix/1) = 1. For every x the vector Yx'= x - L~= I f,.(x)x v lies obviously in n~= I N(f,,), hence by (15.6) also in N(f), so thatf(x) = L~= tf(xv)J~(x), i.e.,
f
=
•
L" l(xv)/v'
v= I
Proposition 15.1. The bilinear system (E, E+) is a left dual system if and only iflor .finitely many linearly independent vectors xt, ... , in E+ there always exist vectors x I' ... , x" in E such that
x:
(15.8)
lor
i, k = 1, ... , n;
it is a right dual system if and only iflor.finitely many linearly independent vectors XI' ... , x" in E there always exist vectors xt . ... , x: in E+ so that (15.8) is valid. The elements Xl' ... , x" and xt, ... , x,,+, respectively, so determined are linearly independent.
We have to give the proof only for a left dual system since (E, E+) is a right dual system if and only if (E +, E) is a left dual system. Thus let (E, E +) be a left system. Then the linearly independent vectors xt, ... , x: from E+ are also linearly independent linear forms on E according to definition (15.3). The elements x I, ... , x" which satisfy (15.8) can be constructed, because of Lemma 15.1, just as it was already done in the proof of the Lemma. They are linearly independent, since from IXI x I + ... + IX"X" = 0 it follows because of (15.8) that IXk = (IXIX I + ... + IXkXk + ... + IX"X", = (0, =0
xt>
xt>
for k = 1, ... , n. Conversely, let (15.8) be valid, to put it briefly. Then in particular for every x+ "# 0 in E+ there exists an x in E such that (x, x+ >= 1, SO (E, E+) is a left dual system. •
81
(E, E+) is a left dual system if E+ is any subspace of E*, since a linear form is the zero form exactly ifit vanishes identically. But (E, E+) does not need to be a right dual system, as one can easily see by taking 'small' spaces E+. However, E* itself contains sufficiently many linear forms to make a right dual system out of (E, E*), as it will result from the following theorem. Theorem 15.1. If {x;.: A. E L} is a basis of the non-trivial vector space Ethere always exists one by Theorem 4.1-, then there exists exactly one linear form f on E which assumes at the points x;. arbitrarily prescribed values 0(;..
For the proof we represent every x in E in the form L;'d ~;.x;. with uniquely determined coefficients ~;., of which at most finitely many are #0, and define f by f(x):= L;'eL 0(;. ~;.. Obviously f is the only linear form which satisfies the requirement. • Proposition 15.2.
(E, E*) is a dual system.
It is sufficient to show that for every Xo # 0 in E there exists a linear form f on E so thatf(x o) # O. For this purpose one extends according to Theorem 4.1 the set {xo} to a basis of E and one knows by Theorem 15.1 that there is an f E E* with f(xo) = 1. •
On the basis of this theorem we can consider E as a subspace of the vector := (E*)* of all linear forms on E*: the element x from E becomes through space E**
for
xU):= f(x)
fEE*
a linear form on E* (see (15.4)). And finally, Proposition 15.2 in connection with (14.11) shows that to every finite-dimensional operator K: E --+ F there always exists a dual system (E, E+), e.g., the system (E, E*), with which one has n
(15.9) for all x
Kx =
L <x, x:)Y.
.=
1
t ,...
E E; in case K # 0 one can take the vectors x ,xn+ in E + as well as the vectors Yl' ... ,Yn in F linearly independent, as we already observed in connection with (14.11). What is interesting for the applications, to which we shall return, is just that frequently for the representation of K we do not have to take the not very handy system (E, E*) but we can use other systems (E, E+) where the elements of E are explicitly known, e.g., the system (C[a, b], C[a, b]) in the case of the integral transformation with degenerate continuous kernel discussed in §14. For certain normed spaces E, the elements of E', i.e., the continuous linear forms, can be precisely described, so that in these cases it might be favorable to use the system (E, E'). We shall see in §28 that (E, E') is a dual system; this important result will be much more laborious to prove than Proposition 15.2.
82 Exercises 1. Check which bilinear systems in the Examples 14.1 through 14.5 are left or right dual systems, respectively. 2. Let E = 100 or = (s). Construct a dual system (E, E+). 3. Given the linearly independent functions x I' ... ,Xn continuous on [a, b] there exist functions YI"'" Yn with the same properties so that
for i, k = 1, ... , n. 4. If fl' ... ,j~ are linearly independent linear forms on E, then the system A(x) = ~k (k = 1, ... , n) of equations has a solution x in E for any right-hand side. 5. If the finite-dimensional operator K: E -+ F is represented in the form (15.9) with linearly independent vectors xt, ... , x n+ with the help of a dual system, or only a left dual system (E, E+), then K(E) = [YI"'" Yn] (see Lemma 25.1). +6. Whenever (E, E+) is a dual system-or only a right dual system-and XI"'" Xn are linearly independent vectors in E, YI"'" Yn arbitrary vectors in F, there exists a finite-dimensional operator K: E -+ F of the form (15.9) so that Kxv = Yv for v = I, ... , n. *7. For every linear formf #- 0 on E the nullspace N(f) is a hyperplane through O. (Hint: see (15.7». Conversely, if H is a hyperplane through 0 in E, there exists a linear form f #- 0 on E so that H = N(f). +8. If the linear forms j~, ... ,In in E* are linearly independent, then codim n~= I N(fv) = n. (Hint: consider the vector Yx towards the end of the proof of Lemma 15.1.) Conversely, if F is a subspace of E of codimension n, there exist n linearly independent forms j~, ... ,fn on E with which n
F =
n N(fv)'
v= I
*9. Let {x A: A. E L} be a basis of the vector space E over K and let f run through E*. The correspondence f ~ (f(x): A. E L) is an isomorphism of E* onto the product space K A , where K A := K. 10. Kn is isomorphic to its algebraic dual (Kn)*. 11. We know that in the sense of the canonical imbedding x ~ F x (see (l5.5»we have E c E**. Show that E = E** ifand only if E is finite-dimensional. + 12. Iffo is a linear form on the proper subspace F of E, then there exists a linear formf on E withfo(x) = f(x) for all x in F (Extension theorem for linear forms). Hint: extend a basis of F to a basis of E. *13. (E, E+) is a left dual system if from (x, x+) = (x, y+) for all x E E it follows that x+ = y+; on the other hand, (E, E+) is a right dual system if <x, x+) = (y, x+) for all x+ E E implies that x = y.
DAEL
83 + 14. A linear formfon the algebra E is said to be multiplicative iff(xy) = f(x)f(y) for all x, y E E. Show that two multiplicative linear forms on E coincide if and only if they have the same nulispaces. Hint: Lemma 15.1.
§16. Conjugate operators
The importance of bilinear and dual systems is not based only on the fact that finite-dimensional operators can be represented with their help; more importantly they playa significant role in the study of operator equations Ax = y. As a preparation we recall a known theorem from the theory of systems of linear equations which anyway will be proved in a much more general form (see Proposition 16.2): The system of equations n
L rJ.ikek = '1i
(16.1)
(i = 1, ... , n)
k=1
is solvable if and only if
(16.2) i=1
for all solutions
(e i , ... , e:> of the transposed homogeneous system n
~>iket = 0
(16.3)
(k = 1, ... , n).
i= 1
The matrix A+
:=(rJ.~1 rJ.21 rJ. 1n
rJ.2n
of the system (16.3) is obtained from the matrix
. . :J
through reflexion in the main diagonal. We can, however, describe this 'transPosition' of A in a way which makes sense for arbitrary operators, not only lllatrix operators. If we consider namely the bilinear system (K n, Kn) with the bilinear form (x, x+ 1 then for all x and x+ we have
>:= Li'= eiet,
(16.4)
(Ax, x+>
=
JJkt1rJ.ik¢k)et = ktJit1rJ.iket )ek = (x, A+x+>;
it is also easy to see that A + is uniquely determined by this relation. The above SOlvability theorem can now be stated more shortly as follows:
84 The equation Ax = y is solvable x+ of the equation A + x+ = O.
if and
only
if (y,
x+) = 0 for all solutions
In general, if A is an endomorphism of an arbitrary vector space E, then we shall try to describe the solvability of the equation Ax = y in a similar way. Thus one will pick a bilinear system (E, E+), check whether to A there exists a linear self-map A+ of E+ so that we have (Ax, x+) = (x, A+x+) as in (16.4), and whether with this operator A + there exists a solvability theorem like the one above. This program is easy to carry out if we use the dual system (E, E*). For any fixed x* in E* the map x f--+ (Ax, x*) is again a linear form on E, which we denote by A*x* and for which by definition (Ax, x*) = (A*x*)(x) = (x, A*x*)
holds. A* is a linear self-map of E*; indeed, for all x in E and all x*, y* in E* we have (x, A*(x* + y*» = (Ax, x* + y*) = (Ax, x*) + (Ax, y*) = (x, A*y*) + (x, A*y*) = (x, A*x* + A*y*)
and since (E, E*) is a left dual system, it follows that A *(x* A*y*. One sees in an analogous way that A*(tXx*) = tXA*x*. A * is uniquely determined by the equation (16.5)
(Ax, x*) = (x, A*x*) for all x in E and all x* in
+ y*) =
A *x*
+
E*.
Indeed, if for some self-map B of E we have identically (Ax, x*) = (x, Bx*), then also (x, A *x*) = (x, Bx*), and, since (E, E*) is a left dual system, A*x* = Bx* for all x* in E*, i.e., B = A*. The linear self-map A* of E* uniquely determined by (16.5) is called the algebraic dual map (transformation, operator) of A: E -+ E. We can indeed state now a solvability theorem for the equation Ax = y with the help of the algebraic dual transformation, which coincides completely with the one formulated above. For its proof we need the following proposition on the richness of E*. Proposition 16.1. IfF is a proper subspace of E and ify does not lie in F, then there exists a linear formf on E which vanishes at all points of F and assumes the value 1 at y.
Indeed, if {X.l.: A. E L} is a basis of F, then {X.l.: A E L} u {y} is a linearly independent set and thus can be extended by Theorem 4.1 to a basis of E. According to Theorem 15.1 there exists a linear form f on E with f(x.l.) = 0 for all A E L • and f(y) = 1. f fulfills the requirements. Proposition 16.2. equation (16.6)
For every endomorphism A of the vector space E the Ax = y,
YEE
85 is solvable in E (16.7)
if and only if the following holds:
(y, x*) = 0
for all solutions x* ~rthe equation A*x*
= 0;
here (x, x*) is the canonical bilinear form of the dual system (E, E*). Proof If (16.6) has a solution, say x = xo, and if A*x* = 0, then x~ v= I
(see Example 16.3). Since (E, E+) is a dual system, for the linearly independent of E+ vectors XI' •.• , Xn there exist, by Proposition 15.1, elements with <xv, >= bvw Consequently we have
yt, ... , y:
y:
f.1.
= 1, ... ,no
But since K* maps E+ into itself(Proposition 16.4), we obtain from this equation that all the x/l* lie already in E+. Thus K is indeed representable in the form (16.11). • If a bilinear system (E, E+) and an endomorphism A of E are given, then one c~nnot
expect a solvability criterion of the kind of Proposition 16.2 even if A is conjugable. If e.g., A is continuous on the normed space E and A' the dual transformation from Example 16.2, then it is easy to see that such a criterion can be valid only if the image space A(E) is closed. We shall see in §29 that in this case the criterion does indeed hold. The problem will therefore consist in
89 finding, for a given bilinear system, classes of operators for which a solvability theorem of the kind we described can be proved. We shall consider such classes of operators in the next sections. For the sake of simplicity we considered in this section only self-maps of a vector space. It is clear how the concept of a conjugate map has to be formulated for linear transformations A: E -+ F: If (E, E+), (F, F+) are bilinear systems, and iffor the linear operator A: E -+ F there exists a linear operator A + : F + -+ E + such that for all x in E and all y+ in F+, then we say that A is conjugable and A + is called an operator conjugate to A or a conjugate of A. If (E, E+) is a left dual system, then there can exist at most one conjugate operator A + to A, whose linearity does not have to be postulated explicitly
because it is guaranteed by (16.17). It should be clear from Examples 16.l and 16.2 how the algebraic dual of A: E -+ F, and in case of a continuous A, the dual operator, are to be defined. The reader will easily provide the necessary modifications to Proposition 16.3, see Exercise 7.
Exercises 1. Let E be the vector space of all sequences x = (~1' ~2' ... ), where only finitely many components are ;60, let E+:=12 and <x,x+>:=L'%l~n~:. With this bilinear form (E, E+) is a dual system. Define A E !f1(E) by Ax:= (~1' 2~2' 3~3'··.) and show that A is not conjugable. 2. Under the hypotheses and with the notations of Exercise 1 let BE !f1(E) be defined by Bx:= (~1' t~2' t~3' ... ). B is conjugable and bijective. However, B- 1 = A is not conjugable (see Proposition 16.3, cf. also Exercise 3). 3. (K2,K3) with the bilinear form <x,x+>:=~I~i + ~2~i is a right but not a left dual system. Every endomorphism A of K2 is conjugable and has infinitely many conjugates. Among these one can find, if A is bijective, bijective ones as well as not bijective ones (see Proposition 16.3, cf. also Exercise 2). Hint: Use the matrix representation of A. + 4. If (E, E +) is a bilinear system, then for every conjugable endomorphism A of E the conjugate operator A + is uniquely determined if and only if (E, E+) is a left dual system. Hint: If(E, E+) is not a left dual system, then
for all x in E} is a non-trivial subspace of E +. The identity map 1+, as well as every projector of E+ parallel to F+, are conjugate to I. 5. If (E, E+) is a bilinear system, then not every endomorphism of E+ is necessarily conjugate to some endomorphism of E. Hint: Use the bilinear system from Exercise 3.
90 6. Let (E, E+) be a bilinear system, A an endomorphism of E and A + conjugate to A. If the equation Ax = y is solvable, then
>
<x, x+>:= fX(t)X+(t)dt
be given. A finite-dimensional endomorphism of C[a, b] is conjugable if and only if it is an integral transformation with degenerate kernel n
L xv(s)x:(t) v~
(xv, x:
E
C[a, b]).
1
§17. The equation (/ - K)x = Y with finite-dimensional K We observed in §9 that the Fredholm integral equation (17.1)
x(s) - fk(S, t)x(t)dt = y(s)
with continuous kernel
k(s, t)
can be solved in the following way if we already know that it is solvable for every y E C[a, b] and that the solution depends continuously on the right-hand side: One approximates k(s, t) uniformly by a sequence of degenerate kernels and solves the integral equations corresponding to these kernels for the given y; the sequence of these solutions converges then to the solution of(17.1). Our considerations following Example 13.2 show us that in the above assumptions we do not have to postulate explicitly the continuous dependence of the solution on the right-hand side: It is enough to suppose that (17.1) is uniquely solvablefor all yin C[a, b]. At the end of§19 we shall see that we can drop even this requirement of unique solvability.
91 According to these remarks, it will be useful for the investigation of the integral equation (17.1) to study equations of the form (/ - K)x = y with a finite-dimensional operator K. We assume that K :F 0 maps a vector space E into itself and that (E, E +) is a bilinear system with the help of which K can be represented in the form n
L (x, Xt)Xi'
Kx =
(17.2)
i=1
where the vectors XI' ... , Xn from E and the vectors xi, ... , x n+ from E + should be linearly independent. Such a bilinear system always exists, see (15.9). The equation (/ - K)x = y
(17.3) can now be written in the form
n
(17.4)
X-
L (x, xt )Xi =
y.
i= 1
Every solution x of this equation has the form n
X= Y
(17.5)
+ L ~iXi i= 1
and substituting this solution into (17.4) we get
JJ~i
(17.6)
- (y, xt) - ktl
~k(Xk' xt) JXi =
O.
With (17.7)
'1i := (y, xt ),
it follows from (17.6) because of the linear independence of XI' ... , Xn that the coefficients ~ I, ... , ~n satisfy the system of equations n
~i -
(17.8)
L rxik~k = '1i
(i = 1, ... , n).
k=1
Conversely, if (~I' ... ' ~n) is a solution of this system with the right-hand side "Ii := (y, xt) (i = 1, ... , n), and if we define x by (17.5), then x solves obviously the equation (17.4). We record this result: n
(17.9)
x = y
+
L(xisolves(l7.4)¢>(~I' ... '~n)solves(17.8)with i= 1
'1i
:=
(y, xt)·
The solution of equation (17.4) is thus indeed, as it was already indicated in §9, Possible by elementary algebraic methods.
92 If we consider the homogeneous problems n
(17.10)
(I - K)x = x -
L <x, xt )Xi = 0, i= 1
n
~i
(17.11)
LIXik~k = 0
-
(i = 1, ... , n),
k= 1
then we get from (17.9): n
(17.12)
x
=
L ~iXi solves (l7·1O)¢>(~1'"'' ~n) solves (17.10. i= 1
L?=
Since vectors of the form z" = 1 ~!")Xi (/1 = 1, ... , m) are linearly independent if and only if this is true for the coefficient vectors (~ F is linear and E = N(A) EEl U (cf. the proof of Proposition 5.3), then the vector spaces EIN(A) and U are isomorphic. *3. If E = FEEl G, then the vector spaces ElF and G are isomorphic. Therefore codim F = dim ElF. Hint: Apply Exercise 2 to the canonical homomorphism h: E -> Ej F and use Exercise 4 in §5. Observe that hereby Proposition 4.2 is proved again. *4. If f is a non-zero linear form on the vector space E over K, then dim EIN(f) = 1, and there exists an Xo E E so that E = [xoJ EEl N(f) (see Exercise 3).
§21. The quotient norm If A: E -> F is a continuous linear map of the normed spaces E and F, then one will ask whether one can introduce a norm on the quotient space EIN(A) so that also the canonical injection A: EIN(A) -> F is continuous. Since for all representatives x of the residue class x the estimate IIAxl1 = IIAxl1 ~ IIA 1IIIxil is valid, we also have the inequality IIAxl1 ~ IIAII . infxeX Ilxll. From it we see that A is certainly continuous if Ilxll := infxEX Ilxll does define a norm on EIN(A). The following proposition will show that this is indeed the case (observe that because of the continuity of A the nullspace N(A) is closed).
Proposition 21.1. If F is a closed subspace of the normed space E, and every equivalen(e class x in ElF one sets
Ilxll
:=
if for
inf Ilxll, xex
then a norm, the so-called quotient norm, is defined hereby on the quotient space ElF. The normed space ElF is complete if E itself is complete.
From the properties of a norm we only prove that Ilxll = 0 implies x = O. From Ilxll = 0 it follows that there exists a sequence of vectors Yn E X so that IIYnll -> 0, hence also Yn -> O. Because of the closed ness of F also the residue class x = x + F is closed as a subset of E, hence the limit 0 of the sequence (Yn) from lies in x, and so indeed x = O. Now let E be complete and (x n ) a Cauchy sequence in ElF. Then there exists a sequence n 1 < n z < n3 < ... of subscripts such that IIxn - xnkll < 1/2k for n ~ nk and in particular
x
IIxnk +
I
-
xnJ
the residue class A E ,c/,(E)jt&"(E) of A is left-invertible. (b) A E dfJ( E) ¢> there exist C E .Cf?( E) and K E t&"( E) such that A C = 1 - K ¢> the residue class A E .Cf?(E)jt&"(E) of A is right-invertible. (c) diE) and dfJ(E) are (multiplicative) semi-groups.
§24. Fredholm operators on normed spaces If A is a continuous endomorphism with finite deficiency of the normed space E, then it would be desirable to be able to choose the operators B, C, K 1 and K 2 in the characterizing equations (23.1) also continuous, so that we do not have to go outside the algebra 2(E). In order to study this situation more closely, let us return to the proof of Proposition 5.3 from which we obtained the equations (23.1). With the notation of this proof, we shall assume first that there exists a continuous projector P from E onto N(A), and an equally continuous projector
III
Qo from E onto A(E). The operator B:= AO" IQo is certainly continuous if AO" I is continuous, i.e., if Ao is open (see Exercise 5 in §21). But Ao is open if A itself is open; indeed, if Po = 1 - P is the (continuous) projector onto U along N(A) and if M c U is open in the su bspace U, then by Exercise lOin §7 also PO" I (M) = {x + y: x E M, y E N(A)} is open, hence, because of the assumed openness of A, the image A(PO" I(M» = {Ax: x E M} = Ao(M) is open in A(E) = Ao(U) and thus Ao is an open map. In summary, we can state: If A E ieeE) is open and if there exists a continuous projector P from E onto N(A) and a continuous projector Qo from E onto A(E)-hence also a continuous projector Q:-I - Qo of E along A(E)-then there exists a B Eie(E) such that
(24.1)
BA = I - P,
AB
= 1 - Q.
For an A with finite deficiency the projectors P, Q are of finite rank, so that in this case all the operators which figure in (23.1) can indeed be chosen to be continuous. If we multiply the first equation in (24.1) from the left by A, then we obtain (24.2)
ABA
= A,
independently of the dimension of the nullspace N(A). If we call a subspace F of E continuously projectable if there exists a continuous projector P with peE) = F, and furthermore if we call an operator A E ieeE) relatively regular if (24.2) holds with some B E ieeE), then we can now say the following: An open operator A E ieeE) with continuously projectable nullspace and image space is relatively regular. The converse of this assertion is also valid. Indeed, if A is relatively regular, i.e., if (24.2) holds, then (AB)2 = ABAB = AB
and
(BA)2 = BABA = BA,
hence as idempotent operators AB and BA are continuous projectors. From A(E) = (ABA)(E) c (AB)(E) c A(E) it follows that (AB)(E) = A(E), hence the image space of A is continuously projectable. From N(A) c N(BA) c N(ABA) = N(A) one obtains N(BA) = N(A), hence (/ - BA)(E) = N(A), and thus also the nullspace of A is continuously projectable. Finally, to show that A is open, we prove the identity (24.3)
A(G)
= B- I(G + N(A» n A(E)
for every
GeE.
If we take into consideration that the projector AB operates on A(E) as the
identity, then we obtain (24.4)
B-I(G
+ N(A» n
A(E) = (AB)[B-I(G c A(G
+ N(A»
+ N(A» =
n A(E)]
A(G).
On the other hand, since also BA is a projector, we can represent every g E G as a sum g = x + y with x E N(BA), y E (BA)(E); it follows that BAg = BAy = y = g - x, hence (BA)(G) c G + N(BA) c G + N(ABA) = G + N(A) and so A(G) c B- 1 (G + N(A» n A(E). From this inclusion and from (24.4) we obtain the asserted identity (24.3). If now G is an open subset of E, then for every x E E
112
+ x is open, hence also the sets G + N(A) = UXEN(A) (G + x) and N(A» are open in E (§I Exercise II, §7 Exercise 10). It follows, with the aid of(24.3), that A(G) is an open subset of the subspace A(E) (see Exercise 7
obviously G B-I(G
+
in §21). We summarize:
Proposition 24.1. A continuous endomorphism of a normed space is relatively regular ifand only if it is open and its nullspace and image space are continuously projectable. By virtue of this proposition we can formulate as follows the result found above concerning operators with finite deficiency: For a relatively regular operator A E :t'(E) with finite deficiency there exist continuous endomorphisms B, C and continuous, finite-dimensional endomorph isms K I, K 2 such that (24.5) one can even choose B = C. We call a continuous endomorphism of a normed space E a Fredholm operator ifit is relatively regular and has finite deficiency. We denote the set of all Fredholm operators on E by (E). For a Fredholm operator A on E the equations (24.5) hold, hence its residue class A in the quotient algebra 2(E)/~(E) is invertible (observe that 2(E)/~(E) is in general not normed, since the ideal ~(E) of continuous operators of finite rank is in general not closed). The question arises now whether conversely a continuous endomorphism A, for which the equations (24.5) hold (i.e., whose residue class A is invertible in 2(E)/~(E», is a Fredholm operator. We shall treat this question in the next section in a much more general form and answer it positively. Right now we want to turn again to the concept of continuous projectability. If E = F EB G and if the projector Ponto F along G is continuous, then we call G a topologically complementary space (or a topological complement) of F; of course F is then a topologically complementary space of G. It is precisely the continuously project able subspaces which have topological complements. Such subspaces are necessarily closed because they are nullspaces of continuous projectors. However, closed subspaces are not necessarily continuously projectable; cf. [62]. But Exercise 3 in §21 shows that a closed subspace F is continuously projectable if it has a finite codimension; every algebraic complement of F is then a topological complement. This assertion brings up the question whether also the simplest subspaces, the finite-dimensional ones, are continuously projectable. If XI"'" Xn is a basis of the finite-dimensional subspace F of E and P a continuous projector from E onto F, then P has according to Proposition 13.4 the representation n
(24.6)
Px =
L <x, X~>Xk k= I
with appropriate
x~ E E'.
113
Since P operates on P(E) = F as the identity, we have n
PXi
=
L <Xi' X~>Xk = Xi' k=1
hence (24.7)
for
i, k = 1, ... , n.
Conversely, if to X I' ... ,Xn there exist continuous linear forms X'I,' .. ,x~ so that (24.7) is valid, and one defines P by (24.6), then PX i = Xi for 1 ~ i ~ n, from where p 2 = P and P(E) = F follow: P is a continuous projector from E onto F. Because of Proposition 15.1 we can thus say: Every finite-dimensional subspace of E is continuously projectable if and only if (E, E') is a right dual system. In §28 we shall give, with the help of the fundamental Hahn-Banach theorem, a positive answer to the question whether (E, E') is a right dual system, and thus a dual system. Anticipating this theorem we can summarize our results as follows: Proposition 24.2. Finite-dimensional and closed finite-codimensional subspaces of a normed space are continuously projectable; every algebraic complement of a closed subspace offinite codimension is even a topological complement.
With the aid of Proposition 24.1 from here we obtain immediately: Proposition 24.3. A continuous endomorphism ofa normed space is a Fredholm operator if and only if it is open, has finite deficiency and a closed image space.
Exercises 1. A E !l'(E, F) is called relatively regular if there exists aBE !l'(F, E) with ABA = A. Show that A is relatively regular if and only if A is open and has a continuously projectable nullspace and image space. 2. A E !l'(E, F) is called a Fredholm operator ifit is relatively regular and has finite deficiency. To a Fredholm operator A there exists aBE !l'(F, E) and K\ E ff(E), K2 E .9'(F) so that BA = IE - K I , AB = IF - K 2 • 3. An element a of an arbitrary algebra R is called relatively regular if for some b E R the equation aba = a is valid. Show that the following elements of R are always relatively regular (in the last three examples it is assumed that R possesses a unit element e): The zero element 0; every a with a 2 = a (idempotent elements); every mUltiple of a relatively regular element; the unit element e; every a for which there exists an n ~ 1 with an = e (elements of finite order); every left- or rightinvertible element. 4. If the element a of the algebra R is relatively regular (see Exercise 3), then there exists abE R such that aba = a and bab = b (b is said to be relatively inverse to a).
114
5. The element a of the algebra R is relatively regular (see Exercise 3) if and only if aba - a is relatively regular for some b E R. + 6. Let the ideal J of the algebra R consist of only relatively regular elements. Show: a E R is relatively regular if and only if the residue class of a is relatively regular in RfJ. Hint: Exercise 5.
a
§25. Fredholm operators in saturated operator algebras Fredholm operators on normed spaces are essentially defined purely algebraically as relatively regular operators with finite deficiency. This suggests the idea to abandon completely metric hypotheses for the investigation of Fredholm operators. For this purpose we consider an algebra deE) of endomorphisms of a vector space E. We say that A E sl(E) is relatively regular (in dee»~, if there exists BE .91(E) with ABA = A (cf. Exercise 3 in §24); A is called a Fredholm operator (in d(E» if A is relatively regular in .91(E) and has finite deficiency. Because of Exercise 6 in §5, every element of the algebra .Yi'
i~
I
is a projector in .91(£) whose image space, according to Lemma 25.1, is equal to G; furthermore we have (25.5)
N(Q) = {xEE:(x,yt> =Ofori= l,oO.,n}.
Since {xt, ... , x n+} and {yt, ... , Y:} are bases of G+, it follows from (25.3) and (25.5) that
F = N(P) = {x
E
E: <x, x+> = 0 for all x+
E
G+} = N(Q);
therefore Q projects E parallel to F onto G, i.e., G is indeed an d(E)-complement of F.
117
Now we prove the second assertion of the lemma. Given the subspace H there exist finite-dimensional spaces M and N such that H
= F EB M
and
E
:=l
F,
= H EB N.
Let {YI"'" Ym} be a basis of M and {Ym+ I,··., Yn} a basis of N. Then {YI"'" Ym' Ym+ I"'" Yn} is a basis of the complement G := M EB N of Fin E. With the map Q defined by (25.4) we can project E parallel to F onto G; we also recall that for the vectors Yi' yt in (25.4) we have (Yi' Y:) = i5 ik · The operator defined by Qox:= L?=m+1 (x,yt)y;is a projector in .9I(E); according to Lemma 25.1 we have Qo(E) = [Ym+ I"'" Yn] = Nand N(Qo)
= {xEE:(x,yt) = Ofori = m + I, ... ,n}.
If we take into consideration that every vector x of E can be written in the form x = !XIYI
+ '" + !XmYm + !Xm + IYm+ I + ... + !XnYn + Z
with
Z
EF
and that x belongs to H if and only if !X m + I = ... = !Xn = 0, then we see immediately that N(Qo) = H. Thus Qo projects E parallel to H onto Nand consequently H is .9I(E)-projectable by I - Qo. • We now arrive at the main theorem of this section. Theorem 25.1. The operator A .from the E + -saturated algebra .9I(E) is a Fredholm operator ({and only i.fthere exist operators B, C in .9I(E) and K I' K z in ~(.9I(E))
(25.6)
so that the equations BA=I-K I ,
hold, i.e., ({and only if the residue class
AC=I-Kl
A o.fA is invertible in .9I(E)j.#'(.#(E)).
Proof We assume first that A is a Fredholm operator, i.e., that it is relatively regular and has finite deficiency. Then there exists aBE .#(E) with ABA = A, and one sees like in §24 (between (24.2) and (24.3)) that BA and AB are projectors in .9I(E) and that E is projected onto N(A) by K I := I - BA and onto a complement of A(E) along A(E) by K z := I - AB. Since dim N(A) and codim A(E) are finite, the operators K I' K 1 have finite rank. Thus the existence of equations (25.6) is proved. Let us now conversely assume (25.6), and let us suppose again without restricting the generality that (E, E+) is a left dual system. First it follows by Proposition 23.1 that A has finite deficiency. Furthermore we have A(E) :=l (AC)(E) = (I - K1)(E) :J N(K 1), from where we obtain with the aid of Lemmas 25.2 and 25.3 the .9I(E)-projectability of the image space A(E). Thus there exists also a projector P E .9I(E) which projects E along A(E) onto a Complement of A(E); in particular, P and also Kl - P are finite-dimensional. Since .9I(E) is saturated, we can represent K 1 - P in the form n
(25.7)
(K2 - P)x
=
L (x, xt)y; ;=1
118 with linearly independent vectors xi, ... ,x;; from E+. By Lemma 25.1 we have therefore (25.8) Because of the second equation in (25.6) we have (25.9)
AC
+ (K2
- P) = J - P,
and since J - P projects the space E onto A(E), we obtain with the help of(25.8) that [YI,"" Yn] = (Kl - P)(E) = (J - P - AC)(E) c A(E). Since thus every Yi lies in A(E), there exist Zi with Yi = Az i . The operator K defined by n
Kx:=
L <x, xt )Zi i= 1
lies in ,9/(E) because this algebra is saturated, and so, with the aid of (25.7), it follows that AKx =
n
n
i=1
i= 1
L <x, xt )AZi = L <x, xt )Yi =
(Kl - P)x
for all
x
E
E,
i.e., Kl - P = AK. From here we get, using (25.9), J - P = AC
+ (Kl
- P) = AC
+ AK
=
A(C
+ K)
and therefore A(C
+
K)A = (J - P)A = A - PA = A
(P A vanishes because of N(P) = A(E». Thus A is relatively regular. With this everything is proved. •
We want to state separately an observation made in the first part of the proof: Proposition 25.1. The nullspace and the image space of a relatively regular endomorphism A belonging to an operator algebra d(E) containing J are d(E)projectable. More precisely: There exists BE ,9/(E) so that E is projected by J - BA onto N(A) and by AB onto A(E). The next proposition specializes Theorem 25.1 to the E' -saturated algebra 2(E). Proposition 25.2. A continuous endomorphism of the normed space E is a Fredholm operator if and only if its residue class in 2(E)j.Q;;(E) is invertible. Since the invertible elements of an algebra form a multiplicative group, we obtain from Theorem 25.1 immediately the following analogue of Proposition 23.2:
119
Proposition 25.3. The following assertions are valid concerning the set cI> := cI>(sY'(E)) of Fredholm operators in the saturated algebra d(E): (a) (b) (c)
cI> is a multiplicative semi-group. If the product AB lies in cI>, then either bothfactors lie in cI> or none. With A also A + K lies in cI> for every K E ~(d(E)).
The finite-dimensional operators in an E+ -saturated algebra are constructed with the help of the space E+. Thus one will obtain 'many' operators of finite rank if E+ is 'large'. The following proposition makes this indication precise. Proposition 25.4. If d(E) is an E +-saturated algebra and (E, E +) a right dual system, then the following assertions are valid: (a) Given the linearly independent vectors x I' ... ,Xn and the arbitrary vectors YI"'" Yn' there exists a finite-dimensional operator K in sl(E) with KXi = Yifor i = I, ... , n. (b) Every finite-dimensional subspace of E is .r;1(E)-projectable. (c) Every finite-dimensional operator in d(E) is relatively regular.
In order to prove (a) we determine according to Proposition 15.1 vectors
xi, ... ,X: in E+ with (Xi' x k+) = bik . Then the operator K defined by n
(25.10)
Kx:=
L (x, Xk+)Yk k=1
clearly satisfies the requirement. To prove (b) let {XI"'" xn} be a basis of the finite-dimensional subspace F. If we replace in (25.10) every Yk by X k , then K projects the space E onto F (we have already made such a consideration immediately before Proposition 24.2). Now we prove (c). Let {XI"'" xn} be a basis of the image space of A E ~(d(E)). Then there exist vectors YI" .. ,Yn with AYi = Xi and by (a) an operator B E ~«r;1(E)) such that BXi = Yj. Then ABxi = AYi = Xi' and since Ax has for every X the form Ax = I ()(i(X)Xi, it follows that
L?=
ABAx =
and so ABA = A.
n
n
i= I
i= I
L ()(i(x)ABxi = L
()(i(X)Xi =
Ax,
•
Between the saturation of an algebra and the conjugability of the operators belonging to it there are close and important connections. First we prove: Proposition 25.5. If(E, E+) is a left dual system and d(E) an E+ -saturated algebra of operators, then every A E d(E) is E+ -conjugable. Proof Let A * be the operator algebraically dual to A and x + an arbitrary element of E+. Because of Proposition 16.4 we only need to show that A*x+ lies in E+. Since this is trivial in the case A*x+ = 0, we may assume that A*x+ oF O.
120 With a vector y¥-O from E we define the operator K by Kx:= <x, x+ )y. Since d(E) is E+ -saturated, K lies in .91(E). Then also the finite-dimensional operator KA belongs to d(E). We can represent KA in the form (25.11)
KAx = Xi
i= I
with linearly independent vectors
Xl"'"
x n ; thus we have
>= 0 for i = I, ... , n}. From here it follows on the one hand that (x, A + xt >= (Ax, xt >= A(E) = N(P) = {z E E: #- 0 for at least one i. This contradiction to (27.7) shows that Y must lie in A(E). Now we prove (27.3). For this we assume that in (27.4) also the vectors are chosen to be linearly independent; by Lemma 25.1 we have then F = P(E) = [Xl"'" xnJ, and so P(A) = n. With the aid of (27.6) it follows from here that P(A) :;;; Q(A +). In order to prove the opposite inequality, let yt, . .. , y~ be linearly independent vectors from N(A +). By Proposition 15.1 there exist linearly independent elements Yl,"" Ym in E with = J ik · If a linear combination y:= Q(IYI + ... + Q(mYm of these elements lies in A(E), i.e., if Y = Ax, then we have
xt, ... ,x:
Q(k
=
(Q(IYI
=0
+ ... + Q(mYm,Y:> = for
(Ax,y:> = <X,A+Yk>
k = 1, ... , m,
hence Y = O. Therefore [Yl,' .. , YmJ n A(£) also Q(A +) :;;; P(A).
=
{OJ and so m :;;; P(A) and thus •
126
Corollary. Ifunder the hypotheses of Proposition 27.1 the vectors XI"'" Xn and xt, . .. , x: in (27.4) are linearly independent, then xt, ... ,x: is a basis of N(A +). We want to examine now A + more closely, and for this we make some preparations. Since the algebraic dual operator of a finite-dimensional endomorphism is itself again finite-dimensional, as (15.9) and Example 16.3 show, we obtain from Proposition 16.4 immediately: Lemma 27.1. !f (E, E+) is a left dual system and K a finite-dimensional, E + -conjugable endomorphism of E, then K + is also finite-dimensional. Now we want to agree on one more notation. If d(E) is an algebra of E+conjugable operators-and (E, E+) again a left dual system-then denote by d+(E+) the set of all operators of E+ which are conjugate to endomorph isms from d(E): .91+(E+):= {A +: A
E
d(E)}.
Because of Proposition 16.3 d+(E+) is an algebra of operators it contains the identity transformation r of E+ whenever d(E) contains the identity transformation I of E. If E is a normed space with the (normed) dual E', then 2'(£') := {A': A E 2(E)} is a subalgebra of the algebra 2(£,) of all continuous endomorphisms of the Banach space E'; the continuity of the dual operator A' was asserted in Example 16.2. Proposition 27.2. If (E, E+) is a dual system and d(E) an E+ -saturated algebra, then d+(E+) is an E-saturated algebra and . $'(d+(E+»:= {K+: K
E
$'(d(E»}.
Proof Every finite-dimensional endomorphism K of E+ of the form Kx+ = I <x+, x;)xt lies in d+(E+) since it is conjugate to the operator K from d(E) defined by Kx := I <x, xt )x j (see Example 16.3). Now let A + be an arbitrary operator of finite rank from .91 +(E +) conjugate to A E d(E). Then A is E-conjugate to A +, hence by Lemma 27.1 finite-dimensional and so representable in the form Ax = I <x, xt )Xj; by Example 16.3 already mentioned, we have A + x+ = I <x+, Xj)xt, i.e., A + has the required representation . •
L?;
L?;
L?;
L?;
If now under the hypotheses of the last proposition A is a Fredholm operator in .91(E), then it follows from Theorem 25.1 by conjugation of the equations (25.6) that A + is a Fredholm operator in the E-saturated algebra .91+ (E+). If we apply Proposition 27.1 to the Fredholm operator A + and to its E-conjugate operator A -all this in the framework of the left dual system (E+, E) and the algebra .91+(E+)-then we obtain the following theorem on solvability:
Theorem 27.1. If(E, E+) isadual system and d(E) an £+ -saturated algebra, then for every Fredholm operator A in .91(£) the conjugate operator A + is a
127 Fredholm operator in.r;;1 + (E+)-and thus also in every operator algebra ;Jd(E+) .91+ (E+)-furthermore we have the equations (27.8)
a(A)
= fJ(A+),
and so
:::J
ind(A) = -ind(A +).
Finally: (27.9)
(27.10)
Ax A + x+
= Y is solvable
= y+ is solvable
>= Ofor all x+ E N(A +), (x, y+ >= ajar all x E N(A). (y, x+
If we want to apply this theorem to Fredholm operators in Y(E), then the question arises again, as in §24, whether a normed space E together with its dual E' forms a dual system (E, £'). We shall attack this question in the next section and answer it positively. If we anticipate this result for a moment, we can state the following proposition: Proposition 27.3. IfE is a normed space and A a Fredholm operator in Y(E), then the dual operator A' is a Fredholm operator in Y(E'), and the equations (27.8), as well as the solvability criteria (27.9) and (27.10) are valid-of course with A' in the place of A + . Fredholm operators on Banach spaces and on topological vector spaces will be studied in §37 and §89, respectively; see also §39 and §51. Exercises 1. The assertions of Theorem 27.1 are valid for every endomorphism A with finite deficiency of E if E+ = E* and A + = A*. 2. Under the hypotheses of Theorem 27.1 the equations fJ(A +) = fJ(A *) and IX(A +) = a(A *) are valid. 3. Under the hypotheses of Theorem 27.1 A is a Fredholm operator in d(E) if and only if A + is a Fredholm operator in .91+ (E +).
v Four principles of functional analysis and some applications §28. The extension principle of Hahn-Banach
In the course of our investigations we have encountered several times the question whether a normed space E, together with its dual E', forms a dual system (E, E') with respect to the canonical bilinear form defined by <x, x') := x'(x). Since the bilinear system (E, E') is a left dual system, we only need to examine the question whether it is a right dual system, i.e., whether to every Xo i= 0 in E there exists a continuous linear form f for which, after suitable normalization, we have f(x o) = 1. In the case of a positive answer the restriction fo off to [xoJ = {O(xo: 0( E K} is a continuous linear form with the valuesfo(x) = fo(O(xo) = O(fo(xo) = 0(. If we define conversely a linear form fo on [xoJ by fo(O(xo) = 0(, then fo(xo) = I and fo is continuous by Proposition 10.3. If we can always, i.e., for any initial vector Xo i= 0, extend this continuous linear form fo to a continuous linear form f on the whole space E, then f(xo) i= 0; hence (E, E') is a dual system. The extension theorem of Hahn- Banach ensures that continuous linear forms which are defined on arbitrary subspaces, not only on finite-dimensional ones, can always be extended to continuous linear forms on the entire space E and implies thus that (E, E') is a dual system. We prove it immediately in the form in which we shall need it later when studying locally convex vector spaces. Here the concept of a semi-norm will enter, which we have already mentioned in §6. We recall it once more: The map p from a vector space into its scalar field is called a semi-norm if it has the following properties: (HNI) (HN2)
p(x) ~ 0, p(O(x) = 100Ip(x),
p(x + y) ~ p(x) + p()} For a semi-norm p one has obviously p(O) vanishes only at the origin.
(HN3)
128
=
0, and p is a norm if and only if p
129 The following extension theorem of Hahn-Banach is one of the fundamental principles of functional analysis. Principle 28.1. Let E be a vector space over K, p a semi-norm on E and Fa linear subspace of E. If the linear form f defined on F satisfies the estimate
I f(x)1
(28.1)
~
p(x)
for all
x
F,
E
then there exists a linear form 9 on E with the following properties:
=
(28.2)
g(x)
(28.3)
Ig(x)1
~
f(x)
for all
x
p(x)
for all
x
F,
E
E.
E
Thus 9 extendsfto the entire space E conserving the semi-norm estimate (28.1).
The proof will be divided into two main sections. (I) Let E be real, i.e., K = R. In this case for a linear form h on the subspace H of E from an estimate of the form
Ih(x)1 ~ p(x)
(28.4)
for all
x
H
E
the estimate (28.5)
h(x)
~
p(x)
for all
x
E
H
follows trivially. Conversely, if(28.5) is valid, then -h(x) = h( -x) ~ p( -x) = p(x) for all x E H, so (28.4) holds. The two estimates (28.4) and (28.5) are thus equivalent. We may therefore replace the hypothesis (28.1) by the assumption (28.6)
f(x)
~
for all
p(x)
x
F
E
and need to prove instead of (28.3) only the estimate (28.7)
g(x)
~
for all
p(x)
x
E
E.
We first show in partial step (la) thatfcan be extended in the required way to a 'small' superspace of F. (la) Let Xo rt F and H be the linear hull of {x o } u F, i.e., H = {IXXo
+ Y:IXER,YEF};
here the element IXXo + Y determines its components IX E Rand Y E F uniquely. A linear extension h off to H must have, because of h(x) = h(IXXo + y) = IXh(xo) + f(y), the form (28.8)
h(x) =
lX~o
+
f(y)
with
~o E
R.
Conversely, every h which, with an arbitrary ~o E R, is defined by (28.8), extends flinearly to H. We will be able to extendfto H in the required way, i.e., conserving linearity and the estimate (28.6), if there exists a ~o E R with (28.9)
lX~o
+
f(y) ~ p(IXXo
+ y)
for all
IX E R
and all Y E F.
130
For such an
~o
we have necessarily ~o ~
(28.10)
p(xo
+ y)
for all
- f(y)
yEF
(set ex = 1), and also ~o ~
(28.11)
-p(xo
+ y)
for all
- f(y)
yE F
(replace ex by - 1 and y by - y). Conversely, if ~o satisfies the last two inequalities then it also satisfies (28.9). Indeed, if we replace in (28.10) first y by (l/ex)y with ex> 0, we get
~o ~ p(Xo + ~ y) - f(~ y) = ~ p(exxo + y) - ~ f(y), hence (28.9) holds for ex> o. If we now replace in (28.11) y by (l/ex)y with ex < 0, we see quite similarly that (28.9) holds also for ex < O. Finally, (28.9) holds for ex = 0 by the hypothesis (28.6). We can thus extendfin the required way to H exactly if there exists a ~o which satisfies the estimates (28.10) and (28.11). Such a ~o exists if and only if sup {-p(xo + y) - f(y)}
~
yeF
inf {p(xo + y) - f(y)} yeF
holds. But this inequality holds indeed, since for u, v in F we have f(u) - f(v) = f(u - v) ~ p(u - v) = p«xo ~ p(xo + u) + p(xo + v)
+ u) + (-xo
- v»
and so - p(xo
+ v)
- f(v)
~
p(xo
+ u)
- f(u).
Thusfcan be extended to H as required. The extendability to the whole space E will be proved in part (lb) with the help of Zorn's lemma. (Ib) Let!Ul be the set of all linear forms h with the following properties: h is defined on the space Dh , h(x) = f(x) h(x)
~
p(x)
for all for all
Fe
x x
E
E
Dh
c
E,
F, Dh ;
briefly, !Ul is the set of all extensions of the required kind off to superspaces of F. Since f lies in !Ul, we have !Ul i= 0. In !Ul we define an order relation '-«' by agreeing that
.=
For every totally ordered subset 5l of!Ul the union D Uhe.R Dh is obviously a linear subspace of E. On D we define a map ho by ho(x).= h(x) if x E Dh for some hER The total order of 5l ensures that ho is uniquely defined and linear. Trivially ho(x) ~ p(x) for all x E Dho = D, thus ho lies in !Ul. Furthermore h -« ho for all h E 5l, thus ho is even an upper bound of 5l in !Ul. By Zorn's lemma
131
m possesses a maximal element g, i.e., a
linear form which does not have a proper extension bounded from above by p. Because of (Ia)we must haveDg = E. Thus the proof for the case of a real vector space is concluded. (II) Let now E be complex, i.e., K = C. For the linear form h on E let h(x)
= hi (x) + ihz(x)
be the decomposition into real and imaginary part. Then h(ix)
= hl(ix) + ihz(ix) = ih(x) = ihl(x) - hz(x)
thus hl(ix) = -hz(x) and so (28.12) hi is a linear form on the real space Er belonging to the space E (see Exercise 2 in §4). Conversely, for every linear form hi on Er a linear form h on E is defined by (28.12). Let us now represent fin the formf(x) = fl(x) - ifl(ix). For the linear form fl on the real space F r belonging to F we have, because of (28.1), the estimate Ifl(x)1 ~ If(x)1 ~ p(x) for all x E Fr. By (I) there exists a linear form g I on Er such that (28.13)
gl(x) = fl(x)
for all
x
E
IgI(x)1
Fr ,
~
p(x)
for all
x
E
Er •
If we define according to (28.12) the linear form g on E by g(x):= gl(x) igl(ix), then g(x) = f(x) for x E F. Further we obtain with the help of the polar representation g(x) = pei"', p ~ 0, that g(e-i",x) = e-i"'g(x) = p is real and thus g(e-i",x) = gl(e-i'Px ). By (28.13) we have for all x E E Ig(x)1 = Ie-i'Pg(x) I = Ig(e-i'Px ) I = IgI(e-i'Px)1
~ p(e-i'Px )
= p(x).
Thus we have proved that g extends the linear formfin the required way. The proof of Principle 28.1 is now concluded. • The following theorem is the Hahn-Banach extension theorem for normed spaces. Theorem 28.1. Given a continuous linear form f on the subspace F of the normed space E, there exists a continuous linear form g on the whole space E with the following properties: g(x)
= f(x)
for all
x
E
F,
IIgll = II ill.
Proof We define on E a semi-norm p by p(x):= Ilfllllxll.Then we have I!(x) I ~ p(x) for x E F. By the above principle there exists an extension g off to E which satisfies the estimate Ig(x)1 ~ IIfllllxll for x E E. Thus g is continuous and IIgll ~ II ill· On the other hand, I f(x) I = Ig(x) I ~ IIgllllxll for x E F and
so lIiII ~ IIgll, thus altogether IliII = IIgll (because of this equality, g is also called a norm-preserving extension off). •
132 On the basis of our remark made at the beginning of this section, we immediately obtain from the above Theorem that a normed space E with its topological dual E' forms a dual system (E, E') with respect to the canonical bilinear form <x, x') := x'(x). In Theorem 28.3 we will be able to make this result more precise. We remind the reader that now Propositions 24.2, 24.3 and 27.3 are completely proved. Theorem 28.2. Let F be a subspace of the normed space E and Xo ¢ F. If F is closed or if at least
C):= inf Ilx - xoll > 0, XEF
then there exists a continuous linear form f on E with the following properties: f(x) = 0
for
x
E
F,
IIfll = 1.
Proof Let H be the linear hull of {x o } u F. The elements of H are the vectors of the form IXXo + x (IX E K, x E F); here IX and x are uniquely determined. We define the linear form h on H by
h(IXXo + x) := IXC). Clearly
h(x) = 0
for
h is continuous: For y := IXXo
x
E
and
F
h(xo) = C).
+ x with IX :j: 0, X E F, we
have
hence Ilh(Y)11 ~ Ilyll, and since this inequality is obviously valid also for IX = 0, we see that h is continuous and satisfies Ilhll ~ 1. We now show that also Ilhll ~ 1 and therefore IIhll = 1. To every e > 0 there exists an x E F with IIx - xoll ~ b + e. With this x we form the vector z:= (x - xo)!llx - xoll for which Ilzll = I, hence
Ilhll
~
1
Ih(z)1 = II x - Xo II C)
C)
~ -~- . u
+e
Letting e --+ 0 we get I h I ~ I. The linear formf of our theorem is now obtained by extending h to E according to Theorem 28.1. • If we choose in Theorem 28.2 the closed subspace F:= {O}, then we obtain: Theorem 28.3. To every vector Xo :j: 0 of the normed space E there exists a continuous linear formfon E such thatf(xo} = IIxoll and II!II = 1. In particular, (E, E') is a dual system.
133 Exercises 1. Compare Principle 28.1 and Theorem 28.1 with Exercise 12 in §15, furthermore Theorem 28.2 with Proposition 16.1. *2. A map p from the vector space E into the scalar field R is called a sublinear functional on E if p(a.x) = a.p(x) for a. ~ 0 and p(x + y) ~ p(x) + p(y) hold. Prove the following Extension theorem: Let E be a real vector space, p a sublinear functional on E, and F a linear subspace of E. If the linear form f defined on F satisfies the inequality f(x) ~ p(x) for all x E F, then there exists on E a linear form 9 with the following properties: g(x) = f(x) for all x E F, g(x) ~ p(x) for all x E E. *3. A vector Xo of the normed space E can be approximated arbitrarily closely by linear combinations of elements of the set M c E (i.e., Xo E [M]) if and only if every continuous linear form on E, which vanishes on M, also vanishes on Xo. + 4. Let E be a normed space over K and')' > O. There exists a continuous linear form f on E with I f I ~ ,)" which assumes at given points Xn prescribed values f(x n ) = a.n (n = 1, 2, ... ) if and only if for arbitrary n E Nand Pv E K the inequality IL~=1 Pva.vl ~ ')'IIL~=1 Pvxvll holds.
§29. Normal solvability The solvability criteria of Theorem 18.1 and Proposition 27.1 motivate the following definition. The endomorphism A of the vector space E is said to be normally solvable with respect to the left dual system (E, E+) or briefly E+ -normally solvable, if it is E+ -conjugable and the equation (29.1)
Ax
=Y
possesses a solution exactly if for all
(29.2)
x+
E
N(A +).
In this case we also say that the equation (29.1) is E+ -normally solvable. In this definition we assume (E, E+) to be a left dual system to make sure, . among others, that the conjugate operator A + is uniquely determined. If E is a normed space, then we call an endomorphism A of E shortly normally solvable if it is continuous and E' -normally solvable. According to Proposition 16.2 a linear self-map of the vector space E is always E*-normally solvable; a Fredholm operator on a normed space is always normally solvable (Proposition 27.3). The definition of normal solvability can be stated more concisely if we use the concept of orthogonal space. If (E, E +) is a bilinear system and M a non-empty subset of E, then M.l:= {x+
E
E+: <x, x+
>=
0 for all x
E
M}
134 is called the space orthogonal to M in E+; similarly for M c E+ the space orthogonal to M in E is defined by M1.:= {x E E: <x, x+)
= 0 for all x+ EM}.
M1. is a linear subspace of E+ or E, respectively. Under the above hypotheses on (E, E+) and on A the endomorphism A is E+ -normally solvable if and only if (29.3)
A(E) = N(A +)1..
We first make some remarks on orthogonal spaces. We fix a bilinear system (E, E+) and for (M1.)1. we write more briefly MH. The following is trivial: Lemma 29.1. (a) If MeN then N1. We have M c MH.
c
M1..
(b)
A set M such that M = MH is said to be orthogonally closed; such a set is always a linear subspace (of E or of E+). The spaces E and E+ are orthogonally closed. Lemma 29.2. A linear subspace M of E is orthogonally closed if and only if to every Xo ¢ M there exists an x+ E E+ so that <x, x+) = 0 for all x E M but <x o , x +) =F O. The orthogonally closed subspaces of E + can be characterized analogously. In the first place, if M = MH and Xo ¢ M, then also Xo ¢ M H , so not for all x + EM 1. is <x o , x +) = 0; this is precisely the assertion. Conversely, assume that the condition of the lemma is satisfied. If Xo ¢ M, then there exists an x+ E M1. with <xo, x+) =F 0, hence Xo ¢ MH. This means that MH c M; because of Lemma 29.lb we have M Lemma 29.3.
= MH.
•
Every orthogonal space M1. is orthogonally closed.
Because of Lemma 29.1 b we have M1. c (M1.)H =: M1.H. On the other hand, from Me M H , by assertion (a) of that lemma, the opposite inclusion M1.:::J MH1. follows. • Proposition 29.1. If (E, E+) is a left dual system and A an E+ -conjugable endomorphism of E, then A is E+ -normally solvable if and only ifits image space A(E) is orthogonally closed. If A is E+ -normally solvable, i.e., if (29.3) holds, then A(E) is an orthogonal space and thus, by Lemma 29.3, it is orthogonally closed. If, conversely, A(E) is orthogonally closed, then one obtains the E+ -normal solvability of A by reproducing the proof of Proposition 16.2 with some unessential modifications; Proposition 16.1 used there has to be replaced by Lemma 29.2. •
135 The fact whether a subspace is orthogonally closed depends on the choice of the bilinear system which has been fixed. For two particularly important bilinear systems the following proposition determines all orthogonally closed subspaces.
Proposition 29.2. Every subspace of the vector space E is orthogonally closed with respect to (E, E*). If E is a normed space, then a subspace of E is orthogonally closed with respect to (E, E') if and only if it is closed. The first assertion and one direction of the second assertion follows from Lemma 29.2, Proposition 16.1 and Theorem 28.2. The other direction is almost trivial: An orthogonally closed subspace of E is the orthogonal space of some • subset of E' and as such closed. Orthogonally closed subspaces of E' will be discussed in §41. . From the last two propositions it follows again that an endomorphism of E is always E*-normally solvable. Furthermore, we obtain the important
Proposition 29.3. A continuous endomorphism of a normed space is normally solvable if and only if its image is closed. We conclude this section with a proposition on the norm of the dual of an operator.
Proposition 29.4. A' E .P(F', E')
Let E, F be normed spaces with dual spaces E', F'. If
is the operator dual to A E .P(E, F), then IIA'II = IIA II.
Proof The inequality IIA'II ~ IIAII is proved as in example \6.2. We now show IIAII ~ IIA'II. To every x with Ax # there exists according to Theorem 28.3 a y' E F' such that (Ax, y') = IIAxl1 and 11y'11 = 1. With this y' we have
°
IIAxll = (Ax, y') = (x, A'y')
IIA'IIIIy'llllxll = IIA'llllxll, and since this inequality trivially also holds when Ax = 0, we have IIAII ~ IIA'II . ~
IIA'y'lllIxll
~
•
Exercises * 1. have
Under the hypotheses and with the notations of Proposition 29.4 we A(E).l A'(F').1
=
=
N(A'),
A(E) = N(A').l;
N(A),
A'(F') c: N(A).l;
in the last inclusion we have' =' whenever A'(F') is orthogonally closed (the orthogonals are taken with respect to the dual systems (E, E'), (F, F'».
136 + 2. Let E be a normed space and F a closed subspace of E. Then (ElF)' is isomorphic as a normed space to F1-. Hint: Associate with every f E (ElF)' the linear form x' E E' defined by x'(x) := f(x); here x is the residue class of x in ElF. 3. Let (E, E+) be a bilinear system. The intersection of arbitrarily many orthogonally closed subspaces is again orthogonally closed.
§30. The normal solvability of the operators /-K with compact K Operators of the kind named in the title arise in the study of the Fredholm integral equation (30.1)
(I - K)x = y.
or
x(s) - fk(S, t)x(t)dt = y(s)
If the kernel k is continuous, then the integral operator K is a compact self-map of the Banach space C[a, b] (see Example 13.1). In Theorem 19.1 we saw, among others, that equation (30.1) is normally solvable with respect to the dual system (C[a, b], C[a, b]) with the bilinear form <x, x+):= fX(t)X+(t)dt.
In this section we study the normal solvability of I - K with respect to the dual system (E, E'). Proposition 30.1. IfK is a compact self-map ofthe normed space E, then I - K is normally solvable or, equivalently, (Proposition 29.3): The image space of I - K is closed. Proof Let
We set A
:=
I - K and show that y.
:=
Ax.
-+
y implies y
E
A(E).
oc.:= inf Ilx. - ull; "EN(A)
for every n there exists a u.
E
N(A) such that
Ilx. - u.11
~
2oc•.
Setting v.:= x. - u. we have
y.
=
Av. and
IIv.11
~
2oc•.
The sequence (v.) is bounded: Otherwise it would contain a subsequencewhich we will denote again by (v.)-for which Ilv.1I -+ 00. If we set W.:= v./llv.ll, then (30.:) Because of Ilw.11 = 1, the sequence (Kw.) contains a convergent subsequence; let us say that KW. j -+ z. It follows from (30.2) that wllj = (l - K)w. j + KW. j =
137
AWnj + KW nj --+ z. From here it follows, if we apply (30.2) once more, that Az = lim AWnj = 0, hence z E N(A). Consequently we have
Ilw n - zil = II
Xn - Un II Ilvnll - z =
I fvJ Ilx n -
IXn
(Un
I
+ Ilvnllz)11 ~ Ilvnll ~ 2'
in contradiction to lim wnJ. = z. Thus the sequence (v n ) is indeed bounded. Therefore (Kv n) contains a convergent subsequence (Kv nJ), thus also the sequence of the vectors vnj = AVnj + KV nj = Ynj + KV nj converges to some VEE. Consequently Y = lim Yn = lim Yn.J = lim AvnJ. = Av E A(E). • Exercise Under the hypothesis of Proposition 30.1 the image spaces of all the powers (/ - K)n (n = 0, 1, 2, ... ) are closed. §31. The Baire category principle After we have posed and answered in the last section the question, suggested by Proposition 29.3, whether the image space of a special class of operators is closed, we will, in the next paragraphs, examine this problem in a general way. If E and Fare normed spaces and E is, furthermore, complete, then the image space of an operator A E 2(E, F) is certainly closed when A is injective and the inverse A - I is continuous on A(E); indeed, if the sequence Yn := AXn converges to Y E F, then (xn) is a Cauchy sequence because of II Xn - Xm II = II A - I(Yn - Ym) II ~ IIA -III llYn - Ymll, and converges thus to an x E E, so that Yn = AXn --+ Ax. Consequently, Y = Ax and lies therefore in A(E). What wejust proved, combined with Proposition 21.1, shows that A(E) is already closed if the inverse of the canonical injection A: E/ N(A) --+ F is continuous; indeed, its image space coincides with A(E). (A)-I is continuous according to Exercise 5 in §21 if and only if A is open; because of Proposition 21.3 this is the case exactly when A is open. We state: A E 2(E, F) has a closed image space whenever E is complete and A itself is open. This result leads us to the question: when is a continuous map open. An answer will be given by the open mapping principle in the following section. In order to prove it, we need the Baire category theorem, to which this paragraph is devoted. First we define the diameter (j(M) of a non-empty subset M of the metric space E by (j(M):= sup{d(x, y): x, Y E M} and prove the Cantor intersection theorem: FI
Proposition 31.1. Let a sequence of non-empty closed subsets Fn with ::> F2 ::> ... and (j(Fn) --+ 0 be given in the complete metric space E. Then 1 Fn contains exactly one point x E E.
n:;,
138 Proof. We choose an arbitrary. c > 0 and determine an no so that b(Fn) ~ c for n ;;:; no. If we choose for every n an Xn E Fn' then clearly d(x n, xm) ~ c for all m, n ;;:; no, thus (xn) is a Cauchy sequence. Because of the completeness of E, the sequence (xn) converges to an x E F. Since for every k = 1,2, ... the subsequence (Xk, Xk+ I,···) also converges to x and is contained in the closed set Fk , x, too, belongs to Fk and so x lies in F:= I Fk. Since for every y E F we obviously have d(x, y) ~ b(Fn) and since b(Fn) tends to zero, d(x, y) must vanish: F contains only the one point x. •
nf=
We now formulate one of the fundamental theorems offunctional analysis, the Baire category theorem: Principle 31.1. If the complete metric space E is represented as the union E = U:'=I Fn of countably many closed sets Fn, then at least one Fn contains a closed (and afortiori an open) ball. We make first a preliminary remark: If a closed set FeE contains no closed ball, then every closed ball Kr[xO] contains a closed ball disjoint from F. Indeed, the ball K r / 2 [xo] contains certainly a point XI ¢ F, and since F is closed, there exists a positive r l ~ r/2 so that Kr,[XI] n F = 0. Furthermore Kr,[xI] c Kr[xo] since for x E Kr,[x,] we have d(x, xo) ~ d(x, Xl) + d(x" Xo) ~ r l + r/2 ~ r. Now let us assume that Baire's theorem is false, i.e., that no Fn contains a ball. Then there exists, according to our preliminary remark, for an arbitrarily chosen closed ball K(O) a closed ball K(I) which is disjoint from F I; obviously we may assume b(K(t) ~ 1. To K(t) there exists-again according to the preliminary remark-a closed ball K(2) with K(2) c K(l), b(K(2» ~ ! and K(2) n F 2 = 0. Proceeding in this manner we obtain a sequence of closed balls K(n) such that K(1) ~ K(2) ~ ... , b(K(n» -+ 0 and (31.1 )
for
n = 1,2, ....
n:,=
By Proposition 31.1 the intersection I K(n) contains (exactly) one point Xo· Because of (31.1) the point Xo lies in none of the F n , in contradiction to the assumed representation E = U:'=I Fn. Thus Principle 31.1 is proved. • Exercises 1. Let J be the set of the irrational numbers in the interval [0, 1]. Show that a representation of the form J = U:;' I Fn with closed sets Fn is impossible. 2. There exists no real-valued function on the interval [0, 1] which is continuous at every rational point and discontinuous at every irrational point. Hint: Use Exercise 1.
§32. The open mapping principle and the closed graph theorem The following theorem, called the open mapping principle or the BanachSchauder theorem, shows that numerous continuous linear maps are open.
139
Principle 32.1. Every continuous linear map A from the Banach space E onto the Banach space F is open. We divide the proof into several steps. (a) First we make a preliminary remark which we will need in the next step and of which the reader can convince himself immediately. If M is a subset of a normed space and IX "# 0, then IXM = IXM; if, furthermore, M is open, then also the subsets IXM and x + M, for any element x of the space, are open. (b) Next we show: If U is an open ball around 0 in E, then A(U) contains an open ball around 0 in F. Let U have radius 2r and let K := Kr(O). Obviously E = U:'= 1 nK, hence also F = A(E) = U:'=l nA(K) and a fortiori F = U:'=l nA(K). According to the Baire category principle (Principle 31.1), one of the sets nA(K), say mA(K), contains an open ball S. Because of mA(K) = mA(K)-see (a)-we have (l/m)S c A(K). Since, furthermore, U => K - K, and thus also A(U) => A(K) - A(K), we obtain the inclusions
-
-
-
1 1=
A(U) => A(K) - A(K) => A(K) - A(K) => - S - - S m m
U
(1) x - - S .
xe(1/m)S
m
(l/m)S - (l/m)S is open as the union of the sets x - (l/m)S, which are open according to (a) (see Exercise 11 in §1), and contains the origin of F. Hence there exists an open ball around the origin of F contained in (l/m)S - (l/m)S and
thus also in A( U). (c) Now we prove a sharpening of (b): If U is an open ball around 0 in E, then already A(U) contains an open ball around 0 in F. In the proof let Er and Fr be open balls around 0 with radius r in E or F, respectively. We set U := E 2ro and choose positive numbers rn such that 00
L rn < roo
(32.1)
n=l
According to (b) there exist an > 0 which satisfy (32.2)
F"n
c
A(ErJ
we may obviously assume that an
-+
for
n = 0, 1, 2, ... ;
o. We will show that
(32.3) which will prove (c). By (32.2) each y E F"o lies in A(Ero), hence there exists an E ro such that Ily - Axo II < ai' i.e., y - Axo E F ".' Again by (32.2) we see that y - Axo E A(Er.), and so there exists Xl E Er. so that Ily - Axo - AXlll < a2' i.e., y - Axo - AXI EF"2· Continuing in this way we obtain a sequence of vectors Xn E Ern with Xo E
(32.4)
Ily - A(xo +
Xl
+ ." + xn)1I
Conversely, for arbitrary chosen scalars CXk a (continuous) linear formfis defined by (35.1) on E. With the aid of the correspondencef ~ (cx(, ... , cx") it is easy to see that E' is isomorphic to K". Example 35.2. We now determine the continuous linear forms on IP, 1 ~ p < 00. Let q be the number conjugate to p, i.e., let 1
I
p
q
- +- =
if p> 1
if p = I. Furthermore, let ek := (0, ... , 0, 1,0, ... ) be the sequence which has I at the kth Position and zeros everywhere else. ek lies in [P, and every x = (~k) E [P can be represented in the form q
=
I, 00,
152
For a continuous linear formfon IP we get therefore, if we set rtk:= f(e k), 00
(35.2)
f(x) =
L rtkek· k=1
a:= (rt l , we have
rt 2 , ... )
lies in lq according to Example 34.2 and Exercise I in §34, and for
p> I
for
p
(35.3)
=
l.
Conversely, if a:= (rtl' rt 2 , ... ) is an arbitrary vector in lq, then (35.2) defines a continuous linear formf on IP (in the case p > I use the Holder inequality); the norm off is given again by (35.3). With this we have proved the essential content of the following theorem (the reader will convince himself easily of the correctness of the assertions not yet proved): Every continuous linear form f on IP (I ~ P < (0) can be represented with the help of one and only one sequence a := (rtl, rt2, .. . )from lq in the form (35.2); the norm offis given by (35.3). The correspondencef ~ (rt l , rt2' ... ) is an isomorphism of the normed spaces (lP)' and lq; in the sense of this isomorphism ofnormed spaces we have therefore (lP)' = lq. Example 35.3. We now set ourselves the task to find an analytic representation of the continuous linear forms f on C[a, b]. The basic idea of our procedure is the following: We approximate the function x E C[a, b] by functions y for which f(y) can be represented in a simple fashion and obtain thenf(x) by the passage to the limit y ~ x. We will see that this simple idea requires a modification which will be made possible by the Hahn-Banach extension theorem.
For every partition Z: a = to < tl < ... < tn = b of the interval [a, b] we define the step function yz E B[a, b] by yz(t) := {
x(a)
k = 2, ... ,n.
X(t k - I )
Since x E C[a, b] is even uniformly continuous on [a, b], there exists for every e > 0 a J > 0 such that n
(35.4)
sup Ix(t) - yz(t)1
~
e
provided that p(Z) := max Itk
a~t~b
-
lk-
II
~
J.
k=1
We now define a family of functions ua(t) := 0 us{t):=
{~
Us E
B[a, b] by the following equalities:
for
a~t~b
for for
a~t~s
s = Ofor all x
E
N(A)}.
Proof N(A)-L is obviously closed; hence the closed ness of A'(E') will have been shown as soon as we have proved (36.3). The inclusion A'(E') c N(A)-L can be seen immediately. We now show, conversely, that every x' E N(A)-L lies in A'(E'). For this we define with the help of such an x' a linear formf on A(E) as follows: To y E A(E) we choose an x E E with Ax = y and set fey)
:=
<x, x'>.
fis uniquely determined; indeed, if we also have AXI
=
y, then
Xl -
X E
N(A),
157 hence <Xl' X') = <X, X'). It is trivial thatfis lineaLfis also continuous: For every u E N(A) we have fey) = <x - u, x'), hence I fey) I ~ II x'llllx - u II and thus also I f(y) I ~ Ilx'lld(x, N(A». Since by Proposition 36.1 the minimal modulus satisfies yeA) > 0, we get from here the estimate
I f(y) I ~
Ilx'III~:;~ = ~~~I~ lIyll.
We now extend f according to the Hahn-Banach theorem (Theorem 28.1) to a continuous linear form y' on E. Then for every x E E we have <x, x')
= f(Ax) = y'(Ax) = 0 so that for all S E 2(E) with IISII < p also A + S is a Fredholm operator and ind(A + S) = ind A. Thus the set (E) of the Fredholm operators on E is open in !.f(E). Proof
To A there exists by (24.5) aBE 2(E) and a K E ff(E) such that BA = I - K.
(37.1)
Because of ind(l - K) = 0 (Proposition 17.1), it follows with the aid of Lemma 23.1 that B has finite deficiency and ind(B) = - ind(A).
(37.2)
If we set p:= 1/IIBII and if IISII < p, then IIBSII ~ IIBIlIiSIl < 1; thus I + BS is bijective by Theorem 8.1, and has therefore a vanishing index. Because of Proposition 23.3 we obtain from B(A
+ S) =
BA
+ BS =
I - K
+
BS
=
(I
+
BS) - K
159 that also ind(B(A + S» vanishes. Since B has a finite deficiency, it follows with the aid of Lemma 23.1 that also A + S has finite deficiency and that the equation (37.3)
ind(B) = -ind(A
+ S)
must be valid. Thus A + S is indeed a Fredholm operator (Proposition 37.1) whose index coincides with ind(A) because of (37.2) and (37.3). •
Exercises 1. Let A be a Fredholm operator on the Banach space E and let B satisfy (37.1). Then A + S is a Fredholm operator for every S E ,;tJ(E) with infFej«E) liS - FII· infFd'-(E) liB - FII < 1, and ind(A + S) = ind(A). 2. Under the hypotheses of Exercise I we have infFej«E) liB - FII = infce.P(E) {IICII: to C there exists an FE .'F(E) with CA = I - F}. In the case of an infinite-dimensional E this greatest lower bound is positive. + 3. Let E be a normed space. An Atkinson operator in the (normal) algebra ,;tJ(E) (see Exercise 8 and 9 in §25) is also called an Atkinson operator on E. Show that to every Atkinson operator A on the Banach space E there exists a number p = p(A) > 0 so that for all S E ,;tJ(E) with IISII < P also A + S is an Atkinson operator and ind(A + S) = ind(A) (see also Exercise 4). + 4. Let E be a Banach space and A E 2(E) a semi-Fredholm operator, i.e., let at least one of the deficiencies oc(A), f3(A) be finite and let A(E) be closed. Then there exists a number p = p(A) > 0 so that for all S E 2(E) with IISII < p also (A + S)(E) is closed, furthermore oc(A + S) ~ oc(A), f3(A + S) ~ f3(A) and ind(A + S) = ind(A). For a proof see [54], p. 113.
VI
The Riesz-Schauder theory of compact operators §38. Operators with finite chains The nullspaces of the powers An of an endomorphism A on the vector space E form an increasing sequence N(AO) = {OJ c: N(A) c: N(A2) c: ... which we call the nullchain. If for a certain n ~ 0 we have N(A n) = N(A n+I), then also N(An+ I) = N(A n+2), and so N(An) = N(An+m) for m = 1,2, ... ; indeed from x E N(An+2) it follows that A n+lAx = 0, hence Ax E N(An+ 1) = N(An) and thus A n+IX = 0, i.e., x E N(An+ I). The smallest integer n ~ 0 for which this occurs will be called the length of the nullchain of A and will be denoted by p(A). If there isno such integer, i.e., if N(An) "# N(An+ 1) for all n, then we set p(A) = 00. The image chain of A is the decreasing sequence of image spaces AO(E) = E ~ A(E) ~ A2(E) ~ .... If for a certain n ~ 0 we have An(E) = A n+ I(E), then An(E) = An + m(E) for m = 1,2, ... ; the smallest integer n ~ 0 for which this occurs is called the length q(A) of the image chain~f A. Ifalways An(E) "# A n+I(E), then we set q(A) = 00. Clearly p(A) = 0 means that A is injective, and q(A) = 0 that A is surjective. For endomorph isms A := I - K with finite-dimensional K the lengths of both chains arefinite. Indeed, the nullspace of
lies in the image space Kn(E) and this space lies obviously in the finite-dimensional space K(E), so that the nullchain of A must eventually break off. Since, furthermore, Kn is finite-dimensional, and thus by Proposition 17.1 one has !X(I - Kn) = P(I - K n), it follows that p(An) = P(I - Kn) becomes a constant from the same exponent on as !X(An) = !X(I - Kn). Thus we have q(A) = p(A) < 00. • 160
161
The following two propositions give precise conditions for the breaking off of the nullchain or the image chain of an endomorphism A of E.
Proposition 38.1.
We have p(A)
~
m
AP(E) with p:= peA) = q(A).
AP is a Fredholm operator by Proposition 25.3, hence AP(E) has finite codimension and is d(E)-projectable (Proposition 25.1). From here and from (39.2) it follows, with the aid of Lemma 25.3, that there exists in deE) a projector P which projects E parallel to AP(E) onto N(AP); furthermore P is finite-dimensional. With the projector Q:= I - P, complementary to P, we set R := AQ - P and K:= AP + P.
R lies in deE), K even in ff(d(E», and obviously A = R
+ K.
Since AP(E) and N(AP) are mapped by A into themselves, P and Q commute with A (Proposition 5.4), hence Rand K also commute. It remains to show that R has an inverse in deE). If Rx = 0, then AQx = Px, hence also QAx = Px and thus (39.3)
QAx
=
° and
Px
= 0.
Because of the last one of these equations, x lies in AP(E), hence Qx = x; it follows with the help of the first equation (39.3) that Ax = AQx = QAx = 0, hence x lies also in N(A). A glance at (39.2) now tells us that x = and so IX(R) = 0. Since by Proposition 23.3 we have ind(R) = ind(A - K) = ind(A), and since ind(A) vanishes because of Proposition 38.6, we have also {3(R) = IX(R) - ind(R) = 0, thus R is bijective. Furthermore, R = A - K is by Proposition 25.3 a Fredholm operator, in particular it is relatively regular. Consequently we obtain from Proposition 26.1 that R has indeed an inverse in .s:i(E).
°
•
Proposition 39.2. If (E, E+) is a dual system and deE) an E+ -saturated algebra, thenfor every Fredholm operator A in deE) we have (39.4)
Iffurthermore A is chain-finite, then also A + is chain-finite and we have thefollowing equalities: (39.5)
IX(A)
= {3(A) = (l(A +) = peA +),
166 Since by Proposition 25.3 together with A also An is a Fredholm operator, (39.4) follows immediately from equations (27.8) if one replaces A by An. For the proof of the remaining assertions we only have to invoke Propositions 38.3 and 38.6. •
§40. The Riesz theory of compact operators In this section (with the exception of Proposition 40.2) let E be a normed space and K a compact self-map of E. We shall explain the essential traits of the theory which F. Riesz [133] has developed for operators of the form J - K, starting from the Fredholm integral equation ofthe second kind. To abbreviate, we set J-K=:A, N(An)=:Nn' An(E)=:Bn and observe that because the compact operators form an ideal, the operator Kn in the expansion (40.1)
An = ( / - K)n = J - [nK - (;)K2
+ ... + (_I)n+lKn]
= J - Kn
is compact. From here it already follows with the aid of Propositions 13.3 and 30.1 that for n = 1, 2, ... the nullspaces N n are finite-dimensional and the image spaces Bn closed. We now show that both chain lengths of A are finite. If we had p(A) = 00, then N n - 1 would be a proper closed subspace of N n for n = 1,2, .... According to the Lemma of F. Riesz (Lemma 10.2) there would then exist in each N n a vector Xn such that Ilxnll = 1 and Ilxn - xii ~ t for all x E N n- 1 • Since (40.2) and for m = 1, ... , n - 1 the element in parentheses lies in N n _ 1, we would have for these subscripts m that IIKxn - KXml1 ~ t, thus (Kxn) could not have a convergent subsequence, in contradiction to the compactness of K: therefore p(A) must be finite. If we had q(A) = 00, then because of the closedness of Bn for every n there would exist an Xn E Bn with II Xn II = 1 and I Xn - x II ;£ t for all x E Bn + 1. Since for m > n the element in parentheses in (40.2) belongs to Bn+ 1, we would have IIKxn - KXmll ~ t for all m > nand (Kxn) could not contain a convergent subsequence. This contradiction to the definition of compactness proves that q(A) is finite. With the help of Propositions 38.3 and 38.6 we now see that (40.3)
p(A) = q(A)
and
oc(A) = P(A),
and from Proposition 38.4 follows the decomposition (40.4)
E = Np EEl Bp
with
p'= p(A).
167 Since Bp has finite codimension and is closed, there exists a continuous (finitedimensional) projector P which projects E along B ponto N p (Proposition 24.2); then Q := I - P projects E along N ponto Bp. Setting S:= PK Eff(E),
V:= QK
we obtain, because of K = (P (40.5)
+ Q)K
=
and
E
'%(E)
R:= I - V
S
+ V, the decomposition
A = I - K = R - S.
From Rx = (l - V)x = 0 it follows on the one hand that x = Vx = QKx E Bp and on the other that Ax = (R - S)x = - Sx = - P Kx E N(A P), hence xEN(Ap+l) = N(AP) = N p, so that by (40.4), finally, x = 0; thus R = I - V is injective. Since V is compact, it follows by what has been proved abovesee the second equation in (40.3)-that R is even bijective. Furthermore, Proposition 11.6 shows that R - 1 lies in 2(E). And now it follows from (40.5) with the aid of Proposition 26.2 that I - K is a Fredholm operator (the E+saturated algebra .9I(E) of that proposition is 2(E) in the present case; E+ is E'). We observe that I - K is open by Proposition 24.3, and summarize our results:
Theorem 40.1. If K is a compact endomorphism of the normed space E, then I - K is a chain-finite Fredholm operator in 2(E) with vanishing index, in particular, I - K is open and (l - K)(E) closed. From this result we obtain immediately, if we invoke Theorem 27.1 and Proposition 39.2:
Proposition 40.1. If K is a compact endomorphism of the normed space E, .then J' - K' is a Fredholm operator in 2(E') and the equations (40.6)
rx(l - K)
= {J(l - K) = rx(J' - K') = {J(l' - K'),
(40.7)
p(l - K)
= q(l - K) = p(l' - K') = q(l' - K')
are valid; in particular, the operators I - K and J' - K' are bijective whenever one of them is either injective or surjective. Furthermore, we have: (l - K)x = y is solvable
¢>
(y, x') = 0 for all x' E N(l' - K'),
(J' - K')x' = y' is solvable ¢> (x, y') = 0 for all x
E
N(l - K).
The above solvability criteria and (40.6) can, because of (40.5), also be obtained from Proposition 18.1. Although the integral operator K in the Fredholm integral equation (19.1) is compact (Example 13.1), we cannot obtain Theorem 19.1, which describes the solvability behavior of the Fredholm integral equation, from Proposition 40.1, because this theorem is based on the dual system (C[a, b], C[a, b]), not on the topological dual system (C[a, b], (C[a, b])'). We shall see, however, that
168 we can obtain Theorem 19.1, in fact even a much stronger assertion, solely by applying the fundamental index relation ind(l - K) = O. The next proposition is a preparation for this. Proposition 40.2. If (E, E+) is a dual system and A an E+ -conjugable endomorphism of E with
(40.8)
ind(A) = -ind(A +),
then
(40.9)
oc(A)
=
fJ(A +),
fJ(A)
=
oc(A +),
and the operators A, A + are normally solvable, i.e., (40.10) (40.11) Proof
O. Let {x 1, .•. ,xn } be a basis of N(A). By Proposition 15.1 there exist in E+ linearly independent elements xi, ... , x;; with (40.13)
for
i, k = 1, ... , n.
We have (40.14)
+ ... + rxnx;; = A + y+ it follows, because of (40.13), that rx i = <Xi' rxlxi + ... + rxnX;;) = <Xi' A+y+) =
i:- O.
With the help of this y~ and the map A E .5t'(E', F'), which associates with each E E' its restriction to the subspace F, we define a linear form x~ E E" by
x'
(x',
x~>
:=
(Ax',
Since E is reflexive, written in the form
x~
(41.8)
(xo, x'> = (Ax',
y~>
coincides with an
Xo
y~>
for x' E E'. E E; thus the last equation can be for
x' E E'.
Xo even lies in F; otherwise there would exist by Theorem 28.2, since F is closed, an x' E E' with (xo, x'> i:- 0 and Ax' = 0; from (41.8) we get immediately a contradiction. If we extend y' E F' by Theorem 28.1 to an x' E E', then Ax' = y', hence by (41.8)
(xo, y') = (xo, x'> = (Ax', y~> = (y', y~>.
Because of (41.7) it follows that for every y' EM: (xo, y'> = (y',
y~>
= 0,
while
(xo, y~> = (y~, y~> i:- O.
Thus we have indeed (41.6).
•
We shall further pursue the topic of reflexivity in §62 and §105. Exercises *1. If E, Fare normed spaces, then the bidual operator A" to A E 2(E, F) is defined as the dual operator to A'; it lies in 2(E", F") and satisfies the equation (A'y', x"> = (y', A"x"> for all y' E F' and x" E E". The restriction of A" to E is A. 2. Let E be a normed space and A E 2(E). Show the following: (a) If the dual operator A' is E-normally solvable, then it is also E"-normally solvable. (b) If E is complete and A is E'-normally solvable, then A' is E-normally solvable (see Proposition 36.4). + 3. A reflexive Banach space has a reflexive dual. 4. Every finite-dimensional Banach space is reflexive. +5. The spaces IP are reflexive for 1 < p < 00. 6. (co), (c), P and 100 are not reflexive. Hint: Show first the non-reflexivity of (co) with the help of Exercise 1 of §35, then use Proposition 41.7 and Exercise 3. §42. The dual transformation of a compact operator
The following considerations prepare the answer to the question raised in §40 whether the dual transformation of a compact operator is again compact. A set (or a sequence) M in a metric space E is said to be totally bounded iffor
everye > 0 there exist finitely many open balls with radii e and centers in M, Whose union covers M.
174 A totally bounded set is bounded (in K the converse of this is also true). A Cauchy sequence (xn) is totally bounded since for sufficiently large m every Xn lies in one of the balls Kr.(x,J, Jl = 1, ... , m.
Lemma 42.1. and only
A set (or a sequence) M of a metric space is totally bounded has a Cauchy subsequence.
if
if every sequence (xn)from M
Proof Let M be totally bounded. If we cover M with finitely many balls with radius 1, then in at least one of them there are infinitely many terms of the sequence (x n); these form a subsequence (Xli' X12, ... )of(x n) withd(x ln , xlm)~2. If we now cover M with finitely many balls with radius 1-, then one of them contains a subsequence (X21' X22, ... ) of (Xln) with d(x2n, X2m) ~ 1. One continues in this way. The diagonal sequence (xnn) is a Cauchy subsequence of (xn). If M is not totally bounded, then there exists an Co > 0 so that M lies in no finite union of balls Kr. (x), X E M, Starting from an arbitrary XI EM we can therefore find successively points Xn E M such that d(x n, xm) ~ Co for n =f. m. Obviously (xn) contains no Cauchy subsequence. • [)
It follows from the Lemma that every relatively compact set is totally bounded. A metric space E is said to be separable if there exists an at most countable set, which is dense in E.
Lemma 42.2.
A totally bounded metric space E is separable.
Indeed, if we cover for k = 1, 2, ... the space E with finitely many balls with radius 11k, then the countable set of the centers of these balls is dense in E. •
In the following investigations let E and F be normed spaces. Proposition 42.1. If K: E -+ F is compact, then the image ofany bounded set is relatively compact and therefore totally bounded. The image space K(E) is separable. The first assertion is immediately clear. Let Bn:= {x E E: Ilxll ~ n}. Then K(Bn) is totally bounded, hence separable (Lemma 42.2). It follows that K(E) = I K(Bn) is also separable. •
U:,=
The following theorem ofSchauder answers the question raised at the beginning of this section (see [137J).
Proposition 42.2. is compact.
The dual transformation K' ofa compact operator K : E
-+
F
Proof Let (yD be a bounded sequence from F', i.e., let IIYkll ~ y for all k, and set B 1 := {x E E: II XII ~ I}. For a given Ii > 0 there exist, because of Propo-
175 sition 42.1, elements XI' ... ' Xn in BI with the following property: to every X E B I there exists an Xi such that
(42.1)
IIKx - KXil1
~ 8.
It follows from the estimate I(KXi' yi> I ~ IIKxdly that (yi) contains a subsequence (zi) such that limk .... oo (KXi' z;'> exists for i = 1, ... , n. Thus the sequence Uk := «KXI' z;'>, . .. , (Kxn' z;'» converges in [I(n) and is therefore totally bounded. Consequently there exist finitely many Uk' say Uk" ••. , Uk m , with the following property: to every Uk there exists an Ukl" such that n
Iluk -
Uk..l! =
L I(KXi' zi> -
(Kx;, zi) I ~
8.
i= I
It follows that for every z;. there exists a zil" with
(42.2)
I(KXi' z;'> - (Kx;, zil"> I ~
8
for
i
=
1, ... , n.
From (42.1) and (42.2) we obtain the assertion: There exist linear forms zi" ... , zi m with the following property: for every zi there exists a zi.. so that for all X E B J we have the following estimate: I(x, K'zi - K'zi) I = I(Kx, zi> - (Kx, zi .. > I ~ I(Kx, zi> - (KXi' zi> I + I(Kx;, z;,) - (KXi' zi .. > I + I(KXi' zi.. (Kx, z;',,> I ~ IIKx - Kxdlllzill + I(Kxi' zi> - (KXi' zi)1 + IIKxi - Kxllllzi.. 11 ~ 8y + 8 + 8y = (2y + 1)8;
>-
and this implies IIK'z;' - K'z;'J
=
sup I(x, K'z;' - K'zi .. > I ~ (2y
+ 1)8.
xeB,
Thus the sequence (K'zi) is totally bounded, hence it contains, by Lemma 42.1, a Cauchy subsequence which even converges because of the completeness of E'. Thus the compactness of K' is proved. • Proposition 42.2 has a converse if F is complete: Proposition 42.3. Let E be a normed space, F a Banach space and K If K' is compact, then so is K.
E
2'(E, F).
Proof By Proposition 42.2 the bidual operator K": E" --+ F" is compact. Its restriction to the subspace E of E" (see Proposition 41.2 and §41 Exercise 1) is then compact as a map from E into F". Since, however, F is complete and thus a closed subspace of F", we obtain the compactness of K as a map from E ~~
.
176
Exercises + 1. A linear map K: E -+ F (E, F normed spaces) is said to be precompact if the image of every bounded sequence contains a Cauchy subsequence. Show the following: (a) K is precompact if and only if the image of the unit ball (or of any bounded set) is totally bounded. (b) A precompact operator is continuous. (c) A compact operator is precompact; if K: E -+ F is precompact and F complete, then K is compact. (d) If the sequence of precompact operators Kn E ~(E, F) converges uniformly to K, then also K is precompact. (e) Sums and scalar multiples of precompact operators are precompact; the product of a precompact operator with a continuous operator is precompact. (f) The precompact endomorphisms of a normed space E form a closed ideal in ~(E). (g) A continuous operator K is precompact if and only if K' is compact. 2. The following assertions concerning a normed space E are equivalent: (a) E is finite-dimensional. (b) Every bounded sequence in E has a Cauchy subsequence. (c) Every bounded set is totally bounded. (d) The unit ball is totally bounded. 3. A set M in the normed space E is totally bounded if and only if for every open ball U around 0 (for every neighborhood U of 0) there exist finitely many points Xl' ... ' xn in M so that M c: I (Xi + U). 4. A E ~(E, F) maps every totally bounded set M c: E onto a totally bounded set. 5. Every subspace of a separable metric space is again separable. 6. The following spaces are separable: Finite-dimensional normed spaces, IP for 1 ~ p < 00, (c), (co), C[a, b]. 7. The following spaces are not separable: 100 , and more generally B(T) if T is an infinite set, BV[a, b]. *8. A normed space E is separable if its dual E' is separable. The converse is false. Hint: Let {x'I> X2, ••• } be dense on {x' E E': Ilx'll = 1}. Choose Xn E E with Ilxnll = 1 and I<xn, x~>1 ~ i. Show with the help of Theorem 28.2 that [Xl' X2,···J = E.
Ui=
§43. Singular values and eigenvalues of a compact operator When boundary value problems of mathematical physics lead to a Fredholm integral equation, then frequently a parameter it occurs in these in the following way (see (3.28»: X(s) - it
f
k(s, t)x(t)dt = y(s).
177
Therefore we consider operator equations of the form (/ - A.K)x = y,
(43.1)
where K is an endomorphism of a vector space E. The solvability behavior of equation (43.1) will be different for different values of A.. We say that A. is a regular value of K when (43.1) is uniquely solvable for all y; a non-regular value is said to be singular. For A. "# 0 (this is the only interesting case) and J1. := 1/A. we have (43.2)
I - A.K = A.(J1.I - K)
and
rt.(/ -
A.K) = rt.(J1.I - K).
Every number J1. (even J1. = 0) with rt.(J1.I - K) "# 0 is called an eigenvalue (or proper value) of K. A necessary and sufficient condition for J1. to be an eigenvalue of K is that there exist in E an x "# 0 with (J1.I - K)x = 0, i.e., Kx = J1.X; every such x is called an eigenvector (proper vector) or an eigensolution (proper solution) of K corresponding to the eigenvalue J1.. The space N(J1.I - K) is called the eigenspace of K corresponding to the eigenvalue J1., and dim N(J1.I - K) is the multiplicity of J1.. Clearly N(J1.I - K)\ {O} is the set of all eigensolutions of K corresponding to the eigenvalue J1.. Observe that 'eigenvalue of a boundary value problem' (§3) and the here defined 'eigenvalue of an operator' are two different concepts. For a compact endomorphism K every multiple A.K is compact, therefore the singular values of such a K are exactly the A. with rt.(/ - A.K) "# 0 (Theorem 40.1), and A. "# 0 is singular exactly if 1/A. is an eigenvalue. Thus instead of studying the singular values of a compact operator, we can just as well investigate the eigenvalues. We start with some algebraic considerations. For a polynomial f(A.):= rt.o + rt.tA. + ... + rt.nA.n in the variable A. E K with coefficients rt.v E K and for an endomorphism A of the vector space E over K we set f(A):= rt.oI
+ rt.IA + ... + rt.nAn.
The correspondence f ~ f(A) is a homomorphism of the algebra of all polynomialsf over K onto the (commutative) algebra of all operatorsf(A). Proposition 43.1.
For an endomorphism A of the vector space E the following
assertions are valid: (a) then
If fl' ... ,fn are pairwise relatively prime polynomials and iff:= fl ... fn' N{f(A» = N{fI(A» ••• N{fn(A».
(b) Eigensolutions of A corresponding to different eigenvalues are linearly independent. (c) For relatively prime polynomialsflJ2 we have
N(fI(A»
c:
f2(A)(E).
178
Proof (a) Let first n = 2, N:= N(f(A» and N;:= N(/;(A». Because of N; c N we have N 1 + N 2 C N. The converse inclusion is obtained as follows. Since fl' f2 are relatively prime, there exist polynomials gl> g2 so that gl(A)fl(A) + g2(A)f2(A) = 1, hence gl(A)fl(A) + g2(A)fz(A) = I. Every x E E admits thus a decomposition
(43.3) For x
E
N we havefl(A)f2(A)x = f2(A)fl(A)x = f(A)x = 0, hence
fz(A)x
E
Nl
and fl(A)x
E
N 2,
and since every polynomial in A maps the spaces N 1, N 2 into themselves, we obtain
With the aid of (43.3) it follows that x = Xl + X2, thus indeed N c Nl + N2 and therefore N = N 1 + N 2' Equation N 1 n N 2 = {O} follows immediately from (43.3). For n > 2 one proves (a) by mathematical induction; we only have to observe that iffl,"" J" are pairwise relatively prime, then the polynomials fl ... j~-l andJ" are relatively prime. (b) Let Al , ... , An be (pairwise) different eigenvalues of A and Xl'"'' X. the corresponding eigensolutions. Since the polynomials fl(A):= A - Al , ... , J.(A) := A - An are pairwise relatively prime, by (a) the sum N(A11 - A) + ... + N(A.I - A) is direct. From ClC1Xl + ... + ClC.Xn = 0 it follows therefore that ClC;X; = 0 and thus also ClC; = 0 for i = 1, ... , n. (c) Since for an arbitrary X E E we have the decomposition (43.3), we have in particular for every x E N(fl(A» the relation
x = g2(A)f2(A)x = f2(A)gz(A)x Thus indeed N(fl(A» c fz(A)(E).
E
fz(A)(E).
•
The next proposition shows that the eigenvalues (hence also the singular values) of a compact operator have a very simple distribution. Proposition 43.2. For every eigenvalue fl. of the continuous operator K on the normed space E we have
(43.4)
if K
is even compact, then its eigetlvalues form either a finite (possibly empty) set or a sequence which converges to zero.
Proof From Kx = fl.x, x¥: 0, it follows that 1fl.1 ~ IIKxll/llxll ~ IIKII which already proves (43.4). The second assertion of the proposition is equivalent to
179
saying that the eigenvalues of K have no limit point #0. If ~ were such a limit point, then there would exist a sequence (I1n) of eigenvalues with I1n # ~ and J.ln -+ ~, furthermore a sequence (xn) with KX n = I1nXn and Ilxnll = 1. Then we would have (43.5)
(0 - K)x n = (J.ln l - K)x n + (~
-
I1n)Xn = (~ - J.ln)xn -+ O.
If p is the chain length of ~(I - (1/~)K) = 0 - K, which according to Theorem 40.1 is finite, then by Proposition 43.1c we have Xn E F:= (0 - K)P(E). The subspace F is mapped into itself by K. If we denote by 1, K the restrictions of I, K to F, then it follows from the closedness of F (§40) that K is compact as an endomorphIsm of F and Proposition 38.4 implies the existence of the inverse K) - I on F. By Proposition 11.6 the inverse is continuous. From (43.5) it would follow that
(a -
•
in contradiction to Ilxnll = 1.
Exercises 1. If A is a chain-finite continuous endomorphism with finite deficiency of a Banach space, then rx.(AI - A) = f3(AI - A) = 0 for all ,1.#0 in a certain neighborhood of o. 2. For every eigenvalue 11 of K E 2(E) we have IJlI ~ limllKnlll/n (this improves (43.4». 3. Let A be an endomorphism of the finite-dimensional complex vector space E (thus A has at least one eigenvalue !). Then there exists exactly one polynomial m(A)-the minimal polynomial of A -with the following properties: (a) the highest coefficient ofm is 1, (b) m(A) = 0, (c) m(A) divides every polynomialf(A) such thatf(A) = o. If m(A) = (A - AI)p' ... (A - Ak)P" is the canonical decomposition into factors, then {AI, ... , Ad is the set of eigenvalues of A and Pi the chain length of ,1.;1 - A. + 4. Assume that some power K n (n ~ 1) of the continuous endomorphism K of the complex normed space E is compact. Then JlI - K is a chain-finite Fredholm operator with vanishing index for every 11 # 0, and the eigenvalues of K. fo~m; if t~e~e are inpnitnely many of them, a sequence converging to zero. Hmt. A - 11 - (A - ~ 0.=2 (A - A.), hence n
K n - I1 nl = (K - 111)
0 (K .=2
- ,1..1).
Use Exercise 1 in §40 to establish the Fredholm property of K - 111. The finiteness of the chain lengths is obtained with the help of Exercise 2 of §38 or of Proposition 43.1.
180
5. If .f~, ... ,f" are pairwise relatively prime polynomials, if f := fl'" fn and A E Y'(E), then f(A)(E) = n~= 1 f.(A)(E), fv(A)(E) + fiA)(E) = E for 1 ~ v < Jl ~ nand n
codim f(A)(E) =
L codim J:.(A)(E). v= 1
6. The eigenvalues of the boundary value problem (3.25) can accumulate only at infinity if 0 is not an eigenvalue.
VII
Spectral theory in Banach spaces and Banach algebras §44. The resolvent In §43 we investigated the invertibility of the operator AJ - K where A. varies; there K was compact. In this section we study the same question for an arbitrary continuous K on a Banach space. It will turn out to be not essential that K is a map; what is decisive is that K lies in a Banach algebra. We shall therefore situate our investigation very soon in a Banach algebra. First we define some basic concepts: The resolvent set p(K) of a continuous endomorphism K on the Banach space E consists of all scalars A. for which AJ - K has an inverse R). E 2(E). We call R). the resolvent operator of K; the map ,1,1---+ R). defined on p(K) is called the resolvent of K. The set u(K) := K\p(K) is the spectrum of K. Because of Theorem 32.2, A. lies in p(K) if and only if AJ - K is bijective. With the help of the deficiencies and chain lengths p(K) can be described as follows: (44.1)
p(K)
= {A. E K: oc(AJ = {A. E K: p(AJ -
K) = P(AJ - K) = O} K) = q(AJ - K) = OJ.
The completeness of E enters in an essential way into this characterization of the resolvent set. The eigenvalues of K all lie in u(K); they form the point spectrum a p(K) of K. Theorem 40.1 implies directly: Proposition 44.1. For the compact endomorphism K of a Banach space we have u(K)\ {OJ = a iK)\ {OJ. We get very easily also Proposition 44.2. For the continuous endomorphism A of the Banach space E we have a(A) = u(A'). 181
182
Proof From the conjugation rules of Proposition 16.3 we get immediately peA) c peA'). Now let A E peA'). Then AI - A has a continuous inverse according to Proposition 41.3, thus (AI - A)(E) is closed by Lemma 36.1, and from here it follows with the aid of Proposition 29.3 that AI - A is normally solvable, hence that (AI - A)(E) = N(AI' - A').l = {O}.l = E. Thus A lies also in peA), hence taken together we have peA) = peA') and thus O'(A) = O'(A'). •
Resolvent set, resolvent and spectrum are concepts which make sense also for elements of a Banach algebra E (the unit element of E will always be denoted bye):
For x E E the resolvent set is p(x):= {A E K: Ae - x is invertible in E}, the resolvent is the map A ~ r).:= (.1.(' - X)-l defined on p(x), the spectrum O'(x) is the complement of p(x) in K. From the propositions and formulas of §9 we obtain immediately assertions (a) and (b) of the following theorem; we only have to take into consideration the trivial transformations (Ae - x) - (.1.0 e - x) = (A - Ao)e,
(c) follows immediately from (a) and (b); (d) will be proved following the statement of the theorem. Theorem 44.1.
For an element x of a Banach algebra E the following asser-
tions are valid: (a)
If .1.0
E
p(x) and IA - .1.0 I < I/lholl, then A E p(x), 00
(44.2)
L (-1)"(.1. -
r). =
Ao}"rlo+ 1
"=0
and
(44.3)
I r). - r).o I
IA - .1.0 I 2. IA _ .1.0 I I r).o II I r).o I ,
limllx"111!" we have A E p(x) and
xn
00
(44.4)
r). =
L "=0
,"+ 1 . II.
For IAI > I x I we have therefore
(44.5)
IIr).1I
I (
~m
IIxII)-1 I-III
and, in particular, r). tends to 0 for IA.I .....
00.
183 (c)
u(x) is closed and because of
1..1.1 ~ limllxnl1 1/n
(44.6)
for all
A E u(x)
also bounded, thus compact. (d) If A and fJ. lie in p(x), then we have the resolvent equation
(44.7) furthermore
(44.8) We still need to prove (d). We have r;. = r;'(fJ.e - x)r/J = r;.[(fJ. - A)e + Ae - x]r/J = (fJ. - A)r;.r/J + r/J'
hence the resolvent equation (44.7) holds. From it follows the commutation relation (44.8). • If F is a non-trivial Banach space, then E Theorem 44.1 can be applied to !l'(F).
:=
!l'(F) is a Banach algebra; thus
Exercises 1. The spectrum of a compact operator on an infinite-dimensional Banach space contains at least O. +2. It is occasionally useful to denote the resolvent of T E !l'(E) more precisely by R;.(T). Show that on p(A) = p(A') we have R;.(A') = (R;.(A»'. 3. (44.8) also follows from the fact that e - Ax and e - fJ.X commute. +4. If ~(fJ.) := inf~Ea(x) IfJ. - I is the distance of the point fJ. E p(x) from u(x), then Ilr,J ~ 1/~(fJ.); in particular, IIr/J1l -+ r:f) when fJ. approaches u(x). Hint: Theorem 44.1a. + 5. Besides the point spectrum u p(A) of an operator A, we have as another interesting subset of u(A) the continuous spectrum uiA) := {A E u(A): AI - A is injective and (AI - A)(E) is dense in E but ,eE} and the residual spectrum ur(A) := {A E u(A): AI - A is injective and (AI - A)(E) is not dense in E}. oiA), uiA) and ur(A) are pairwise disjoint and their union is u(A). To any A E uiA) there exists a sequence (x n ) with Ilxnll = 1 such that (AI - A)xn -+ o. Hint: Proposition 36.1. +6. If A is normally solvable then 0 does not lie in uiA).
e
§45. The spectrum An endomorphism of afinite-dimensional space is already bijective when it is injective, thus its spectrum consists only of eigenvalues and may therefore be empty if the space is real. In the case of a complex space, however, the fundamental theorem of algebra (which is a theorem of complex function theory) ensures the existence of eigenvalues. If we want to prove that u(x) is not empty, we will
184 have to assume that the Banach algebra E containing x is complex; this assumption will open the possibility to put to work far-reaching theorems from complex function theory. If (l(x) is not empty, we define the spectral radius of x by r(x):= sup IAI. AEU(X)
Points of the spectrum are also called spectral points. Theorem 45.1.
For every element x of a complex Banach algebra E we have (l(x) #
(45.1)
0
and
rex) = limllxn I1 1/".
Proof With the continuous linear form x' on E we setf(A) := x'(rA)' From the resolvent equation (44.7) and the continuous dependence of r A on A (Theorem 44.1 a) it follows that f(A) - f(Ji)
as
A-Ji
A ~ Ji.
Thusfis a holomorphic function on p(x) which vanishes at infinity because of (44.5) and the inequality I f(A) I ~ Ilx'llllrAII. If (l(x) were empty, i.e., p(x) = C, then Liouville's theorem would imply thatf(A) = x'(rA) = 0 for all A-and this for all x' E E. Thus r A would vanish on C (Theorem 28.3). However, in an algebra with e # 0 every inverse is #0. This contradiction shows that (l(x) is not empty. Because of (44.6) we have rex) ~ r:= limllx"11 1/". Assume now that rex) < Ji < r. By Theorem 44.1 b, for every x' E E the series ,
(45.2)
00
f(A) = x (r A) =
x'(xn)
"~O A"+ 1
converges for
IAI > r;
it is the Laurent expansion of the functionf(A) which, according to the above, is holomorphic in p(x), hence certainly for IAI > r(x). The theorem concerning the domain of convergence of a Laurent expansion tells us that (45.2) must converge even for IAI > rex), hence for A = Ji. But then x'(xn/Ji" + 1) -+ 0 and so by Proposition 41.4 we have IIX"/Ji"+ 111 ~ Y < 00 for all n, hence r = limllx"111/" ~ lim(YJi"+ 1)1/" = Ji < r. The false assertion 'r < r' shows that we must have rex)
=
r.
•
The closed disc {.A. E C: IAI ~ r(x)} is called the spectral disk of x (even if rex) = 0). Since (l(x) is not empty and closed (Theorems 45.1, 44,1), on the houndar), of the spectral disk there lies at least one spectral point. Proposition 45.1.
For two commuting elements x, y of a complex Banach
algebra E we have
(45.3)
rex
+ y)
~ rex)
+
r(y) and
r(xy) ~ r(x)r(y).
185 Proof
From Theorem 45.1 it follows immediately that
r(xy) = limll(xy)"lli/n = limllxnynlli/n ;::;; Iim(llx nll i /nllynlli/n) = r(x)r(y). For the proofofthe first inequality in (45.3) let the numbers a > r(x)andp > r(y) be given arbitrarily and let mEN be such that for n ~ m we have Ilxnll ;::;;
an,
Ilynll ;::;;
pn.
Setting ~:= Ilxll, 11:= Ilyll we have Ilxkll ;::;; ~\
i
+
v=n-m+ 1
where }' :=
Ilill ;::;;
11k
for
k
=
0, I, ....
(n)a vpn _v(11)n-V f3
V
max -~)V O~v~m-l a
(
From here it follows that rex rex + y) ;::;; rex) + r(y).
+1+
+ y)
;::;; (a
max (11)V -. O~v~m-l f3
+ {3)limjY =
a
+ p;
thus also
•
For every natural m we have limn~oo II(xm)nlli/n = (limn~oo Ilxmnlli/mnr, hence (45.4) x is said to be nilpotent if xn = 0 for a natural number n, it is quasi-nilpotent if rex) = O. Nilpotent elements and Volterra integral operators (see (8.9» are quasi-nilpotent. The spectrum of an element x in a complex Banach algebra is reduced to 0 if and only ifx is quasi-nilpotent. Quasi-nilpotent endomorphisms of a Banach space are also called Volterra operators. The structure of those complex Banach algebras which are at the same time division algebras, i.e., in which every element x "# 0 is invertible, is surprisingly simple:
Proposition 45.2. A complex Banach algebra E, which is at the same time a division algebra, consists o/the mUltiples o/the unit element: E = {A.e: A. E C}.
186
Proof. If a certain x E E were not a multiple of e, then Ae - x would be different from zero for all A E C and thus invertible. In contradiction to Theorem 45.1, the element would have no spectral points. •
Exercises + 1. For commuting x, y we have Ir(x) - r(y) I ~ r(x - y). 2. The nilpotent elements of an arbitrary commutative algebra R form an ideal in R. 3. The quasi-nilpotent elements in a commutative Banach algebra E form a closed ideal in E. Use Exercise 1. 4. If x is a nilpotent element in an arbitrary algebra R with unit element e, then Ae - x has an inverse for all A =1= 0 which can be represented by a terminating Neumann series.
§46. Vector-valued holomorphic functions. Weak convergence For the proof of Theorem 45.1 we have put into action twice essential theorems of the theory of holomorphic functions. The passage from the vector-valued function Af-+ r;. to a complex-valued one was accomplished with the help of continuous linear forms. In this section we set ourselves the task to develop the fundamental concepts (derivative, integral) and essential assertions (Cauchy's theorem and integral formula, Taylor and Laurent expansions, Liouville's theorem) of complex function theory for the case of functions which depend on a complex variable and whose values lie in a complex Banach space. In principle it would be sufficient to assert that the above-mentioned concepts and theorems (together with their proofs) can be obtained from classical function theory by a purely formal transfer; we only need to replace absolute values by norms. But we can also use linear forms to obtain from the classical theoremswithout repeating their proofs-the corresponding assertions concerning vector-valued functions. We shall illustrate this method of transfer by some examples. For the needed facts from complex function theory we refer to [177]. Let E be a normed space over K, ~ a (non-empty) open subset of K. The vector-valued function f: ~ --+ E is said to be differentiable at the point ..1.0 E ~ if there exists anf'(Ao} E E such that (46.1)
II
f(A1
={o(Ao) -
f'(Ao)
11--+ 0
f'(Ao) is called the first derivative off at the point . 1.0 , It is evident how the higher derivatives are defined. It follows from (46.1) that for every x' E E' the limit
(46.2)
lim x' [f(A) - f(Ao)J = lim x'[f(A)] - x'[f(A·o)] A - ..1.0 ;.~;.o A - ..1.0
;.~;.o
187
exists. This fact is described by the locution that f is weakly differentiable at Ao. We shall show that conversely weak differentiability implies differentiability if E is complex and complete. First we must define and explain the concept of weak convergence. Let (E, E+) be a bilinear system. We say that a sequence (xn) from E converges E+ -weakly to x E E (in symbol: Xn ~ x) if (46.3)
(xn'x+)--+(x,x+)
forall
x+EE+.
If(E, E+) is a right dual system, then the E+ -weak limit is uniquely determined. In this case the E+ -weak convergence is nothing but the pointwise convergence of the sequence of linear forms (Xn) to the linear form x, because by §15 we can consider E as a subspace of(E+)*. We call (xn) an E+ -weak Cauchy sequence if«xn' x+» is a Cauchy sequence for every x+, or equivalently, iflim(x n, x+) exists for every x+. Observe that an E+ -weak Cauchy sequence does not need to have an E+ -weak limit. The Eweak convergence of a sequence (x;) from E+ to x+ E E+ is defined analogously
and we have analogous statements. If E is a normed space and we fix (E, E') as our basic dual system, then we call E'-weakly convergent sequences from E more shortly weakly convergent, and E' -weak Cauchy sequences more shortly weak Cauchy sequences. Then Xn ~ x means that the sequence of continuous linear forms Xn on E' tends pointwise to the linear form x (see Proposition 41.2). From here and from Proposition 41.4 and 33.1 we obtain:
Proposition 46.1. A weak Cauchy sequence (xn) in the normed space E is x it follows that IIx II :;;:;; lim inf Ilxnll. bounded. From Xn --'0.
If we want to prove a corresponding theorem about E-weak convergent sequences of continuous linear forms x~ E E', then we must assume the completeness of E; see Exercise 6. Convergence in the sense of the norm implies weak convergence; the converse, however, is not true in general. It is the more remarkable therefore that weakly convergent power series in Banach spaces are even convergent in the sense of the norm. We prepare this assertion (Proposition 46.3) by the following proposition (where we agree, in order to stay close to the classical notation, that x(x := (Xx for xEE,
(xE
K).
Proposition 46.2.
Let E be a Banach space, and assume that the power series
(46.4)
with coefficients
ak
E
E
is for A1 #= Ao a weak Cauchy series, i.e., that co
(46.5)
L x'(ak)(A -
k=O
A.O)k
exists for all
x'
E
E'.
188 Then L:;'o ak(A. - A.O)k converges in the sense of the norm for IA. - A.ol
0 with for all
Sl E Y(c").
Consequently, for every natural number m r(Sl) ~ IISlmlll/ m ~ IISlmll.~m ~ yl/mllSlmlll/m
(Theorem 45.1 and Proposition 8.1) and thus r(Sl)
= limIISlmll1£m.
Thus we have r(Sl) < 1, i.e., all eigenvalues of Sllie in the interior of the unit disk, if and only if one of the numbers IISlmll.., is < 1. Proposition 47.3 now turns into: Proposition 47.4. If Sl is an (n, n)-matrix over C and 11·119' is a matrix-norm on Y(c"), then the convergence assertions of Proposition 47.3 are valid if and only if one of the numbers II Slm 119' is < 1.
We return to our general considerations and define now functions which depend on elements of a Banach algebra. If the complex-valued functionfis holomorphic for IAI < r, i.e., iffor these A
(47.9) if, furthermore, x is an element of a complex Banach algebra E with r(x) < r, then we definef(x) by 00
(47.10)
L ~nxn;
f(x):=
n;O
the convergence of the series is ensured by Proposition 47.1. In §43 we had setf(A):= ~oI + ~IA + ... + ~nAn for polynomials f(A):= ~o + ~IA + ... + ~nAn and endomorphisms A of a vector space; for an A from the Banach algebra of all continuous endomorph isms of a Banach space this definition coincides with the one given above since a polynomial is a power series with an infinite radius of convergence. One of the most important functions of analysis is the exponential function exp(A):=
00
An
n;O
n.
L ,.
198 It is holomorphic in all of C, consequently for every x of a Banach algebra we have, by definition,
00
(47.11)
exp(x):=
xn
L -.
n! Obviously exp(O) = e. In the exercises we shall prove the functional equation of n=O
the exponential function :
(47.12)
exp(x
+ y)
= exp(x)exp(y)
for commuting elements x, y.
For an element A of the special Banach algebra 51'(E) (E =1= {O} a Banach space) one writes usually eA instead of exp(A). On the other hand it is not advisable to use the symbol e A in an unspecified Banach algebra because of the possibility of confusion with the unit element e. The functionfin (47.9), which is holomorphic in 1..1.1 < r, can be represented for all A inside a positively oriented circle r around 0 with radius < r also by the Cauchy integral formula (47.13)
1. = -2
f(A)
TCI
JIr f(O«(
- ..1.)- 1 d(.
We ask whether the equation 1. f(x) = -2
I f(O«(e
1. - X)-l d( = -2
I f(Or( d(,
TClJ r TClJ r obtained from (47.13) by substituting x for A, is correct; here we suppose of course that r(x) < r and the radius of r is not only < r but > r(x), so that r lies in the disk of holomorphy of f and contains the whole spectrum of x in its interior. The answer is positive: In the first place we have for n = 0, 1, 2, ... (47.14)
~ I (nr, d( = ~ I (n(~ + ~ + x 2 + ... )d( = 2m Jr 2TC1 J ( e (3
x n,
r as can be seen immediately by termwise integration (cf. Exercise 4 in §46) taking into consideration the integral formula
(47.15)
Ir ;k d( = {02ni
JI ..
for k = 1 for every other integer k;
thus we get, again by term wise integration,
11
1L
1
1 ( 0IXn(nr, 0 ) d( = 2------;1 00 ':X n (nr, d( f(Or, d( = 2------:2------:m r m r n=O TCI n=O r
L
00 IXn Xn = f(x). = L n=O
We state this result: Proposition 47.5. If the complex-valued function f is holomorphic for IAI < r and if x is an element of the complex Banach algebra E with r(x) < r, then the equation
(47.16)
r
1 . f(A)r;. dA f(x) = -2 m Jr
199
is valid; here r A is the resolvent of x and r is a positively oriented circle around 0 whose radius lies strictly between r(x) and r.
In the subsequent investigations we shall direct our attention to the dependence of the expressionf(x) onfwhen x is fixed. It will be useful to definef(x) not only for functionsfwhich are holomorphic in sufficiently large disks around O. This will be done in the next section based on equation (47.16).
Exercises In what follows E will always be a complex Banach algebra; x, y, an' bn etc., will be elements from E. 1.
If the series L:'=o Ilanll, L:'=o Ilbnll converge, then L:'=o Ilaob n + a1bn- 1
+ ... + anboll also converges, and we have Ctoan)Ctobn)
= nto(aobn + a1bn- 1 + ... + anbo)(Cauchymultiplication).
Hint: Because of Ilaob n + ... + anboll ~ Ilaollllbnli + ... + Ilanllllboll one can use the known theorem concerning multiplication of absolutely convergent series. 2. If L:'= 0 (J(nAnconvergesforlAI < randifr(x) < r,thenalsoL:'=ol(J(nlllxnll converges. 3. Use Exercises 1 and 2 to prove the functional equation (47.12) of the exponential function. 4. If f(A) := L:'= 0 (J(nAn and g(A) := L:'= 0 13nAn are holomorphic for IAI < r and if r(x) < r, then we have
«(J(f)(x) = (J(f(x),
(f + g)(x) = f(x) + g(x)
and
(fg)(x) = f(x)g(x),
in particular, f(x)g(x) = g(x)f(x) .. The correspondence ff-+f(x) is thus a homomorphism from the algebra H r of all functions holomorphic in {A.: 1.1..1 < r}
onto the commutative algebra {f(xJ: f E Hr; fixed x E E with r(x) < r}. This homomorphism is continuous in the following sense: iffn,f E Hr and if fn(A)
-+
f(A)
uniformly in a disk {A:IAI ~ ro} with r(x) < ro < r, thenfn(x) -+ f(x). Hint: Exercise 1 and 2 for the conservation of products, Proposition 47.5 for continuity. 5. If under the assumptions of Exercise 4 f(A) =1= 0 for 1.1..1 < r or only for A. E u(x), then the inversef(x)- I exists and is given by f(x)-l
=
2~ r Xl J r
f(\) r;. dA.::;:: A
f Yn xn,
n=O
200 where L:~ 0 Yn)..n is the expansion of II J()..). Briefly (but perhaps ambiguously): J(x)- I = J-I(x) (f-I is here the reciprocal not the inverse, function). Hint: Exercise 4 and a Heine-Borel covering argument; cf. also Theorem 48.1e. 6. The power series L:~o n!)..n is nowhere convergent. If K is the Volterra integral operator defined by (8.9), then L:~ 0 n! K n converges, however, if 11 < l/(b - a) (the meaning of 11, a, b is defined in connection with (8.9); there the reader will also find the further tools to solve the exercise). +7. Let F be a Banach space, U o E F and A E !.e(F). Then u(t):= elAuo is the unique solution of the initial value problem u'(t) = Au(t) (t ~ 0), u(O) = u o . Specialize this assertion to linear systems of differential equations with constant coefficients (F = en, A a matrix operator).
§48. The functional calculus Taking our cue from the last remark of the last section, we now describe a class offunctionsJfor whichJ(x) is to be defined. We denote by !J..(f) c C the set of definition off If x is an element of a complex Banach algebra, then let .Jr(x) be the set of all complex-valued functions J which are locally holomorphic on an (open) set !J..(!) :::) a(x). To .Jr(x) belong in particular all functions which are holomorphic on an open disk containing a(x). Observe that the set of definition !J..(f) may vary withf For functionsJfrom,}f'(x) we shall defineJ(x) by the integral formula (47.16). For this purpose we must, however, make some preliminary remarks concerning the technique of integration. The region Bee is said to be admissible (with respect to x) if the following hold: (a)
a(x) c B;
(b) B is open and bounded; (c) the boundary aB of B consists of finitely many closed, rectifiable Jordan curves C 1, ... , Cn which are pairwise disjoint; (d) the (positive) orientation of aB is given by the orientation of each C;: we will describe C i counterclockwise if the points of B which are adjacent to Ci lie in the interior of C i , otherwise C; will be oriented clockwise. It is easy to see that the points of C i cannot come arbitrarily close to the spectrum of x; thus C i lies completely in the resolvent set p(x). Figures 1 and 2 give examples of admissible domains B, their boundary curves C i and their orientation. B is crosshatched simply, a(x) doubly. For every J E.Jr(X) there exists an admissible region B with a(x) c Be. B c !J..(f). If B' is a second region of this kind, then according to the CauchY integral theorem (Proposition 46.4)
f
iJB
f()..)r;. d)" =
f
iJB'
f()..)r;. dX
201
Figure 1
Figure 2
These facts make the following definition possible: Iff E.ll(X) and B is an admissible region with a(x) (48.1)
f(x)
:=
1. -2 m
c
B c Be /1(f), then let
r f(A)r). dA.
JcB
Because of Proposition 47.5 this definition agrees with the definition of f(x) when f is holomorphic in a disk around 0 containing a(x). We will, however, not be satisfied with functions ofthis kind, because particularly valuable, namely idempotent, elements f(x) are obtained only from such functions f which are equal to 1 on certain parts of a(x) and equal to 0 on others; such anfcan be locally holomorphic but not holomorphic. If two functions f and 9 from Jft'(x) coincide at every point of an open set containing a(x)-in which case we say that they are equivalent (with respect to x)-then obviously f(x) = g(x). The usual algebraic operations are at first not defined in Jft'(x) because its elements do not have a common domain of definition. However, the observation we just made concerning equivalent functions suggests the following definitions of rt.f, f + 9 andfg (J, 9 E Jft'(x»: (rt.f)(A)
(f
+ g)(A) :=
f(A)
+ g().)
:=
for
rt.f(A)
and
A E /1(f)
(fg)().):= f(A)g(A)
for
A E /1(f)
11
/1(g).
After these definitions we can formulate the central theorem of the functional calculus:
Theorem 48.1.
The correspondencef 1---+ f(x) defined in (48.1) has the follow-
ing properties:
(a) (b) (c) (d) (e)
(rt.f)(x) = rt.f(x). + g)(x) = f(x) + g(x). (fg)(x) = f(x)g(x), hencef(x) commutes with g(x). For f(A) = An we havef(x) = xn (n = 0, 1,2, ... ). Iff().) =1= Ofor all A E a(x), thenf(x) has an inversef(x)-l = (11.f)(x).
(f
Proof (a) and (b) are trivial, (d) was shown in (47.14). For the proof of (c) we choose admissible regions Bf, Bg with (48.2)
Bf
C
Bg
C
Bg
C
/1(f)
11
/1(g).
202 Then
=
2~ nl
1
oBf
1. f(A)[-2 nl
1
oBg
g(J1)r;.r/l dJ1JdA.
From the resolvent equation r;. - r/l = (J1 - A)r;.r/l (Theorem 44.1) we obtain for the product r;.r/l in the second integral the representation
r;. r;.r/l = - - , J1-11.
r/l 1I.-J1
+ -,--
and thus the following continuation of our computation:
.1
1 f(x)g(x) = -2 m
1 1 [1
1. f(A)r;'[-2
oBf
+ (2m1.)2
m
oBg
g(J1), dJ1JdA J1 - II.
-,f(A) dA ] dJ1 J1 (why is the interchange of the order of integration in the second term permitted ?). Because of A E aB J and J1 E aBg we obtain from (48.2), with the aid of Cauchy's formulas,
~ 2m
1 oBg
oB g
g(J1)r /l
g(J1) dJ1 = g(A) J1 - A
and
oBf II. -
1
oBf
f(A) dA = O. A - J1
Thus we have, finally,
1. f(x)g(x) = -2 m
1
f(A)g(A)r;. dA,
oBf
which proves (c). For the proof of (e) we observe thatf(A) does not vanish on an open set ~ ::::J a(x), hence II f is locally holomorphic there and belongs thus to Jt"(x); (e) now follows with the help of (c) and (d) from the equation
f(A)(1/f)(A) = 1
for
•
A E~.
The first important consequence of Theorem 48.1 is the spectral mapping theorem:
Theorem 48.2.
For every f EJt"(X) we have a(f(x»
=
f(a(x».
Proof Let J1 E a(f(x». Ifwe had J1 ¢f(a(x», i.e., J1 - f(A) # 0 for all A E a(x), then J1e - f(x) would have an inverse (Theorem 48.1e) in contradiction to the hypothesis. Thus a(f(x» c: f(a(x». Now let, conversely, J1 E f(a(x», i.e., J1 = f(O with' E a(x). The function g defined on ~(f) by gel)
:=
f(A) - f(O J.-'
for J. # "
g(O := r(O
203 lies in.Yf (x); from Y(A)(( - A) = f«) - f(A) it follows therefore that y(x)«(e - x) = f(Oe - f(x) = Ile - f(x) (Theorem 48.1c). If Il were in p(.f(x», then we would have [(Ile - f(x»- 19(x)]«(e - x) = «(e - x)[(J1e - f(X»-lg(X)] = e,
thus (e - x would have an inverse, which is impossible because by hypothesis ( lies in u(x). Therefore we have alsof(u(x» c u(f(x». • Now we are in the position to investigate composite functions. Proposition 48.1.
rtf lies in .ff(x),
y il1 .Yf'(f(x» and ilwe define h hy
h(A) := g[f(A)] for all A withf(A)
E ~(g),
then h belongs to Yl'(x) and we have h(x)
=
g[f(x)].
Proof ~(h):= {A E ~(f): f(A) E ~(g)} is open (Exercise 10 in §7). From A. E u(x) it follows by Theorem 48.2 that f(A) E f(u(x» = u(.f(x», thus f(A) lies a fortiori in ~(g), hence a(x) c ~(h) and h belongs to Yl'(x). Let now Bx and By be regions which are admissible with respect to x and y:= f(x), respectively, and which satisfy the following conditions: u(y) c By c
By
c ~(g),
Then
= _1
r
r
= _1
r
1. = -2
r g(A)[Ae -
[_1 g(A) dA]«(e _ x)- 1 d( 2ni Ji'Bx 2ni J cBy A - f(0
r
g(A) [_1 «(e - X)-l d(]dA 2ni JCB y 2ni JCBx A - f(O
m JeBy
= g[f(x)].
f(x)r 1 dA •
A subset a of a(x) is called a spectral set of x if a and u(x)\a are closed. It is clearly equivalent to say that a has a positive distance from a(x)\a, or that there exist open sets ~ ::::J U and n ::::J u(x)\a which do not intersect. With a also a(x)\u is a spectral set of x.1t is called the spectral set complementary to a. If aI' a2 are complementary spectral sets of x and ~1' ~2 open sets which cover a 1> a 2, respectively, and do not intersect, then we define on ~ := ~ 1 V ~2 the functions f1 J2 by (48.3)
f10.):=
g
204 Both functions belong to £(x), consequently the elements (48.4)
PI
:=
fl(x)
and
P2:= fix)
are defined. By Theorem 48.1 we have pf = Pk (thus Pk is idempotent), PIP2 = P2PI = and PI + P2 = e. We state this result with a modification which is only formal:
°
Proposition 48.2. Let a I, a 2 be complementary spectral sets of the element x of a complex Banach algebra E, and let r b r 2 be simple, closed integration paths, oriented counterclockwise, which lie in p(x) and contain in their interior a I and a 2, respectively, but nofurther parts of a(x). Then the elements k
(48.5)
= 1,2
lie in E and satisfy the following equations: (48.6)
pf
= Pk'
PIP2
= P2PI = 0,
PI + P2 = e.
It is permitted that one of the spectral sets be empty.
Exercises + 1. Formulate and prove a continuity property of the correspondence ff-+ f(x) defined by (48.1) (cf. Exercise 4 in §47). 2. The equivalence of two functions from .Jt'(x) defined in §48 establishes an equivalence relation in £(x). If we define the algebraic operations in the corresponding set of equivalence classes £'(x) in the usual fashion (namely by means of representatives), then Jt'(x) becomes an algebra over C, which is homomorphic to a commutative subalgebra of E (cf. Theorem 48.1). 3. Iff E £(x) andf(x) = 0, thenf(A) vanishes for all A E a(x). 4. ForfE£(x) we have r(f(x» = maxAEO"(x) I f(A) I ;::;;; maxIAI~r(x)lf(A)I. 5. Let x be an element ofthe complex Banach algebra C[a, b] (see Exercise 1 in §9). Then a(x) = {x(t): t E [a, b]}, r(x) = Ilxll, and for f E £(x) we have [f(x)](t) = f[x(t)]. For instance [eX](t) = exIt). 6. Define the continuous endomorphism A of the complex Banach algebra C[a, b] by (Ax)(t):= g(t)x(t), where g is a fixed element of C[a, b]. Then a(A) = {g(t): t E [a, b]},r(A) = Ilgll,and forf E £(A) we have [f(A)x](t) = f(g(t»x(t) (cr. Exercise 5).
§49. Spectral projectors Let E be a non-trivial complex Banach space. We now apply the functional calculus to the Banach algebra 2(E). If a is a (possibly empty) spectral set, rer a curve enclosing a of the kind descrihed in Proposition 48.2 and RA is the resolvent of A E 2(E), then (49.1)
r
1. Per:= -2 RA dA m Jr ..
20S
is a continuous projector of E, the spectral projector or Riesz projector associated with u; occasionally we shall denote it more precisely by PiA). The most important properties of projectors can be found in §S. The projector complementary to P" is / - P"
=
Pt
=
2~
( R" dA.
Jrt
7rl
with
!:=
u(A)\u.
For an empty u we have P" = 0 and P = /. Let M,,:= P,,(E) = {y E E: p"y = y} and N,,:= N(P,,); M t and Nt are defined correspondingly. These spaces are all closed and we have t
M t = N" and Nt = M". E can be represented as the direct sum E = M" EB N,,; if x = y + z is the corresponding decomposition of x, then-since P" and Pt commute with A, just like any f(A)and correspondingly Thus the spaces M" and N" are invariant under A. The restrictions A", At of A to M(1 and N(1 = M" are consequently continuous endomorphisms of M" and N '" respectively. The same proof shows, by the way, that M" and N" are invariant even under everyf(A), a fact we shall soon need. To a given Ji ¢: u there exist open, mutually disjoint sets Ll(1' Ll t covering a and !, respectively, and such that Ji ¢: Ll". We now define j~ and f by
fo(l)
~
t :::::~
f(l)
~ ~~ {
1
for
A. E Ll t
fa andf lie in £(A) and we have (Ji - A.)f(A.) = f(1(A.) for all A. E Lla u Ll" and consequently by Theorem 48.1 (Ji/ - A)f(A)
=
f(A)(Ji/ - A)
=
f,,(A)
=
Pa ·
Since M a is invariant under all the operators of this equation and since P" is the identity on M a , we obtain by restriction to M" that Jilies in p(A,,). Thus We have a(A,,) c: a and analogously a(At) c: !. Since, however, as it is very easy to see, p(A) = p(A,,) n p(A t ), hence a(A) = u(A,,) u a(A t ), We must have u(A,,) = u and a(At) = !. Thus we have proved the following decomposition theorem: (49.2)
Theorem 49.1. Let E be a complex Banach space and u a (possibly empty) Spectral set of the continuous endomorphism A. Then the projector P" in (49.1) generates a decomposition
(49.3)
E = M" EB Na
where
Ma:= Pa(E)
and
Na:= N(P,,);
the spaces M" and N a are invariant under A, and even under every f(A) with f E £(A); the spectrum of the restriction of A to M" is a and the spectrum of the restriction of A to N a is u(A)\u.
206 If the spectral set (J lies in a particularly simple way in (J(A), then M" can be characterized analytically. We say that a subset (J of (J(A) is circularly isolated if there exists a circle
r:= {k IA-
(49.4)
- IXI
= r}
with the following properties: (a) r lies in p(A); (b) the interior of r contains (J but no further points of (J(A). Obviously such a (J is a spectral set of A and (49.5)
for
n
= 0, 1, 2, ...
(cf. Exercise 6), hence also (49.6)
(IXI - AtP"x
(see Exercise 1). For x
E
1. = -2 m
Jrr (IX -
A-)"R;.x dA-
M" we have P"x = x and thus by (46.13) 1 Atxll ~ 2- 2n:r· r" max IIR,dl
II(cd -
n:
AEr
IIxll·
Since this estimate also holds for some r' < r (r lies in p(A», we have lim supll(IXI - A)"xll II" < r.
(49.7)
Conversely, if this inequality is valid, then the series I:,=o [(IXI - A)/(IX - A-)]"x converges for all A- E r and for its sum y(A-) we have, according to Proposition 11.1, - A) ( I - IXI IX _ A- Y(A-)
so that, as a simple calculation shows, Y(A-) 00
R"x = -
n~o
=
(IXI - A)"x (IX _ A-)"+ I
= x, (A- - IX)R"x and therefore
for all
A-
E
r.
A licit termwise integration now yields P x "
=
r
_1_ R;.xdA2n:i r
J
= __1_
r
_1_ xd A2n:i Jr IX - A-
=x
(use an integration formula corresponding to (47.15». Thus x lies in M". We summarize: Proposition 49.1. If under the hypotheses of Theorem 49.1 the set (J C (J(A) is circularly isolated through IA- - a I = r, then the vector x lies in M" if and only if (49.7) is valid. In particular, iflX is an isolated point of the spectrum of A, then
(49.8)
M!~l =
{x: limll(al - A)nxlli/n = O}.
207 Exercises *1. Formulate and prove a theorem which justifies the passage from (49.5) to (49.6); the theorem has the form A{it)dit)x = A(it)x dit, A(it) E 2(E). *2. Let the Banach space E be the direct sum of the closed subspaces F 1, F 2 invariant under A E 2(E); let Ak be the restriction of A to F k . Then p(A) = P(Al) n p{A 2 ), hence u(A) = u(A l ) U u(A 2 ) (generalization of (49.2». 3. Under the hypotheses of Theorem 49.1 the eigenvalues of Aa are precisely the eigenvalues of A lying in u (Aa is the restriction of A to Ma)' 4. A spectral set u of A is also a spectral set of A', and we have Pa(A') = (Pa(A»,. Hint: Proposition 44.2, Exercise 2 in §44 and Exercise 10 in §46. *5. If U1>"" Un are pairwise disjoint spectral sets of an operator T, if PI"", Pn are the corresponding spectral projectors and if
(Ir
Ir
u(T) = Ul U ... U Un' then we have: (a) P'P Il = b. Il P v and I = L~= 1 P., (b) E = Pl(E) EB··· EB Pn(E), (c) P.(E) is invariant under T (v = 1, ... , n). *6. We assume the hypotheses necessary for the definition of P a in (49.1) and we use the notation employed there. Show that for every f E Jt"(A) we have
r
1. -2 f(it)R;. dit m Jra
=
f(A)P a .
§50. Isolated points of the spectrum We begin this section with a representation theorem. In it we speak of 'functions of A'. By this we mean operators of the form f(A) with f E Jt"(A). For the notation employed (e.g., P a , Ma) we refer to Theorem 49.1. In the whole section let E be a complex Banach space and A E 2(E). Proposition 50.1. Assume that A is not quasi-nilpotent and that it has the spectral set u:= {ito} consisting of one point. Then there is a representation (50.1)
A=U+S
of A with the following properties: (a)
U and S arefunctions of A, hence they commute with each other and with A; (b) ito lies in p( U) and in u(S), so that, in particular, R := ito 1 - U has an inverse in 2(E) and Aol - A can be represented in theform Aol - A = R - S with R- l E2(E); (c) S maps E, and already the subspace Ma, onto Ma; (d) we have limll(AoI - A)nSxlll/n = 0 • (50.3) for all x E E. (50.2)
208 For quasi-nilpotent operators a representation of the described kind is not possible. Proof Let .::\0" .::\t be open, mutually disjoint sets which cover u and u(A)\u, respectively, and let the functions fa'!t from ~(A) be defined as follows: T :=
A E.::\a A E.::\;
for for
/.(A):=
{~
finally, let p:= PO' and Q:= Pt = I - P. For the proof of the representation (50.1) and of assertion (a) we now set (50.4) (50.5)
U := AQ
and
S := AP
if Ao =F 0;
+P
and
S:= AP - P
if AO
U := AQ
= O.
With the operators so defined (50.1) holds, furthermore U and S are functions of A (e.g., in the case Ao = 0 obviously U = f(A) withf(A) := Aft(A) + fa(A». (b) For Ao =F 0 we have g(A):= Ao - A/.(A) =F 0 for A E u(A), thus g(A) = AO I - AQ = AO I - U has an inverse in 2(E) by Theorem 48.1e, i.e., Ao lies in p(U); and since S = h(A) with h(A) := .ifa(A), the set u(S)
= {Afa(A): AE u(A)}
contains, because of Theorem 48.2, the point AO = AO fa(AO)' In the case AO = 0 let g(A):= A/.(A) + fa(A) and h(A):= AfiA) - fa(A). Just like above, one sees that g(A) = AQ + P = U has an inverse in 2(E), and u(S) = u(h(A» = h(u(A» contains the point 0 = Ao (observe that T =F 0 because A is not quasinilpotent !). (c) The inclusion S(E) c Moo = P(E) is trivial; the equation S(Ma) = M" becomes obvious if we take into account that u(A,,) = {AO} (Theorem 49.1) and that S coincides on M" with A" or A" - I, respectively. (d) now follows from Proposition 49.1. Let us suppose, finally, that for a quasi-nilpotent A we have the representation (50.1) with the described properties, where of course {AO} = u(A) = {O}. Then in particular S is a function of A, say S = f(A), consequently u(S) is the set {f(0)} consisting of one point. Since 0 lies in u(S) by hypothesis, we have u(S) = {O} and thus S is quasi-nilpotent. From Proposition 45.1 it follows nOw that r(U) = r(A - S) ~ r(A) + r(S) = 0, hence also U is quasi-nilpotent in contradiction to the assumption that 0 should lie in p(U). Thus Proposition 50.1 is completely proved. • An isolated point Ao of the spectrum of A is a non-removable singularity of the resolvent RA (end of §46). According to Proposition 46.7, there exists a
209 Laurent expansion of R.. in a neighborhood of Ao:
P
ao
(50.6) R ..
= n~l
(A _
ao
~o)" + n~o Qn(A - Ao)"
for
0 < IA - Ao I < r;
the coefficients are calculated according to the formulas (50.7) (50.8)
Qn
=
r
1 R .. 2ni r (A _ Ao)"+ 1 dA,
J
where r is a sufficiently small, positively oriented circle around Ao. We turn our attention to the principal part of the Laurent expansion. For (J := {Ao} it follows, with the help of the functional calculus, from (50.7) immediately that (n = 1,2, ...).
(50.9)
These equations show that either all Pn =F 0, or that there exists a natural number p such that
P n =F 0
for
n = 1, ... , p,
but
Pn = 0
for
n = p
+ 1, p + 2, ....
In the first case Ao is an essential singularity of R .. , in the second case a pole of order p. The next proposition shows that in the discussion of the poles of R .. the chain conditions playa decisive role. We recall again that an endomorphism T is injective or surjective if the length p(T) of its nullchain or the length q(T) of its image chain vanish, respectively (§38). Proposition 50.2. Ao is a pole of the resolvent of A if and only if A - Ao I has positive finite chain lengths; the common length p of the chains is the order of the pole. In this case Ao is an eigenvalue of A;for the spectral projector P corresponding to {Ao} we have
(50.10) Proof
P(E)
=
N[(A - AoI)P],
N(P)
=
(A - AoI)P(E).
Let Ao be a pole of order p of R .. ; setting T:= A - AoI we have thus p-1p =F 0,
(50.11)
PP = O.
It follows that P(E) c N(P), P(E) =F N(P- 1). Since on the other hand N(P) lies in P(E) by Proposition 49.1, we have altogether N(TP- 1) =F N(P) = P(E), thus the nullchain of T has length p > 0 (consequently Ao is an eigenvalue of A), and the first one of the equations (50.10) holds. If A is not quasi-nilpotent, we can, according to Proposition 50.1, represent T in the form (50.12)
T
=
R+ S
with bijective
Rand
S
=
PB,
210 where BE .P(E) is determined by (50.4) or (50.5); but such a representation also holds trivially for a quasi-nilpotent A: In this case we have, namely, P = I, hence A = (A - I) + S with a bijective A - I and S:= P = I. From (50.11) and (50.12) it follows that P+ 1 = PR + PPB = PR; because of R(E) = E we obtain p+ '(E) = peE), hence the image chain of T has also finite length. The two chain lengths of T coincide by Proposition 38.3. Because of Theorem 49.1 and Proposition 38.4 the decompositions (50.13) (50.14)
E E
= peE) E8 N(P)
= N(P) E8 peE)
hold, where T is bijective on N(P) and therefore N(P) = P[N(P)] c peE). The decomposition x = y + z of an element x from peE) according to (50.13) is because of peE) = N(P) and N(P) c peE) at the same time a decomposition according to (50.14); but then y = 0 and x = Z E N(P), hence peE) c N(P), so that also the second equation in (50.10) is valid. Let us now suppose conversely that T has the finite chain length p > o. Then ,10 is an eigenvalue of A and we have the decomposition (50.14) with the closed nullspace N(P). With the aid of Proposition 36.2 it follows that also peE) is closed. Let T" T2 be the restrictions of T to the Banach spaces N(P), peE), respectively. Since T, is nilpotent and T2 bijective (Proposition 38.4), peT,) contains C\ {o} and p(T2) contains a disk around 0, and it follows because of peT) = peT,) II P(T2) (Exercise 2 in §49) that 0 is an isolated point of aCT), hence ,10 is an isolated spectral point of A. As above, let P be the spectral projector corresponding to {Ao}. Because of (50.9), we only have to prove that peE) c N(Tm) for some natural number m. If x E peE) and x = u + v is the decomposition of x according to (50.14) with U E N(P) and v E peE), then T"x = T"v for n ~ p; with the aid of Proposition 49.1 it follows that v lies in peE), thus taken together we have v E TP(E) II P(E).lfwe can prove that this intersection is equal to {O}, then x = U E N(P) and our proof is ended. We set D := peE) II peE) and prove the equation D = {O} by showing that the restriction Ao of A to the Banach space D, invariant under A, has an empty spectrum (see Theorem 45.1). To every XED there exists exactly one y E P(E) with x = Ty (Proposition 38.4); with the aid of Proposition 49.1 it follows that y lies in D, i.e., that T is bijective on D and so ,10 is in p(Ao). Now let A¥- ,10. Then Theorem 49.1 implies that A lies in the resolvent set of the restriction of A to peE), consequently to every XED there exists exactly one y E peE) such that x = (A - AI)y· If y = U + v with U E N(P), v E peE) is the decomposition of y according to (50.14), then x - (A - AJ)(u + v) = 0, hence x - (A - AI)v = (A - AI)u, and since the element on the left-hand side belongs to TP(E), the one on the right-hand side to N(P), we have (A - AI)u = 0, hence U E N[(A - A.oI)"J II N(A - A./). With the aid of Proposition 43.1a it follows that u = 0, thuS y =V E TP(E), and so y lies in D. This means that also Ao - A./ is bijective, i.e., that A. lies in p(Ao). Therefore we have indeed p(Ao) = C and thus a(Ao) = eJ. Our proposition is now completely proved. •
211 Proposition 50.3. A - Aol is a Fredholm operator with positive finite chain length if and only if Ao is an isolated spectral point of A and the corresponding spectral projector P is finite-dimensional. In this case Ao is a pole of R l .
Proof
If Ao is isolated and P finite-dimensional, then because of N[(A - Ao l)n] c P(E)
-see Proposition 49.l-IX(A - Ao l) and p(A - Ao l) are finite. Let A 1 be the restriction of A to N(P). According to Theorem 49.1 we have AO E p(A 1 ), hence for n = 1,2, ... (50.15)
N(P)
=
(Al - Aol)n(N(p» c (A - Ao/)n(E).
If we take into account that in the decomposition E = P(E) Ef) N(P) the first summand has a finite dimension, i.e., that codim N(P) is finite, then it follows from (50.15) that also {3(A - Aol) and q(A - Aol) must be finite. With the aid of Proposition 37.1 we obtain the assertion in one direction. The other direction • follows immediately from Proposition 50.2.
From the Riesz theory of compact operators we obtain very easily: Proposition 50.4. Every spectral point Ao different from zero of a compact operator K on E is a pole of the resolvent, and the corresponding spectral projector is finite-dimensional.
For the proof we observe that Aol - K is a chain-finite Fredholm operator (Theorem 40.1) whose chain length must be positive because Ao is a spectral • point. The assertion now follows from Proposition 50.3. Exercises + 1. Ao is a pole of the resolvent of A if and only if Ao is a pole of the resolvent of A'; in this case the orders of the poles are the same. Hint: Proposition 44.2, Exercise 2 in §44 and Exercise 10 in §46. 2. Investigate the case when 0 is a pole ofthe resolvent of a compact operator. In §52, Exercise 2 a similar problem will be discussed.
§51. The Fredholm region
In this section let E be a Banach space over K and let A befrom 2(E) (unless something else is explicitly said). We shall investigate the so-called Fredholm region of A, i.e., the set A := {A E K: AI - A E (E)}.
Because of Proposition 37.1 we have A = {A E K: AI - A has finite deficiency}. Obviously p(A) c: A- For a compact A we have A ::> K\ {OJ. The points of
212 the Fredholm region A are called the Fredholm points of A. Before we formulate the next proposition, we recall that an open subset M =f. 0 of K can be decomposed into maximal open, connected, pairwise disjoint non-empty sets, the components of M. Proposition 51.1. The set A is open, and on every component of A the function ind(A.J - A) is constant.
The openness of A follows immediately from Proposition 37.2. The assertion concerning the index follows from the same proposition with the help of the method of chains of disks, familiar from the theory of analytic continuation: Join a fixed point ..1.0 of the component C with an arbitrary point ..1.1 E C by a polygonal line Pee and associate with each J-t EPa disk in which ind(A.J - A) is equal to ind(J-tI - A). Since by the Heine-Borel theorem already finitely many of these disks cover P, we have ind(A.I1 - A) = ind(A.oI - A). • For further investigation and classification of the components of A we draw on the length p(A.J - A) of the nullchain of A.J - A. We shall make ample use of the propositions of §38. The reader should remember that A. is an eigenvalue of A if one--and thus each -of the assertions
'rx.(A.J - A: =f. 0',
'p(A.I - A) =f. 0'
holds. Proposition 51.2.
For ..1.0
E
A we have the following alternative:
Either p(A.oI - A) is finite-then ..1.0 is not a limit point of eigenvalues of Aor p(A.oI - A) is infinite-then there exists a neighborhood V on ..1.0 whose points are all eigenvalues, and in such afashion that for A. E U\ {A. o} one has rx.(A.J - A) = const
~
rx.(A.oI - A).
Proof. We set T := ..1.0 I-A. Obviously A. is an eigenvalue of A if and only if J-t:= ..1.0 - A. is an eigenvalue of T. In the case p(T) < 00 we must therefore prove that the eigenvalues of T cannot accumulate at the point 0, in the case p(T) = 00, however, that there exists a neighborhood V of 0 which consists entirely of eigenvalues, and that
n:=
rx.(J-tI - T)
=
const
~
rx.( T)
forall
J-tEV\{O}.
Let F:= 1 Tn(E) and T the restriction of T to F. F is a Banach space because the powers T" lie all in (E) and so their image spaces are closed (Propositions 25.3 and 24.3). In the first case of the alternative (p(T) < 00) we have by Proposition 38.7 and Lemma 38.1 that rx.(T) = {3(T) = 0, so that 0 and with it also a neighborhood {J-t E K: IJ-tl < r} of 0 lie in p(T). None of these values J-t =f. 0 is an eigenvalue of T; otherwise we would have Tx = J-tX for some x =f. 0, hence Tnx = J-tnx (n = 1, 2, ...). Thus x would lie in F and we would have Tx = J-tX which is impossible. Thus 0 is indeed not a limit point of eigenvalues
213 of T. In the second case of the alternative (p(T) = (0) we have, by Proposition 38.7, that 0 < a(T) and trivially a(T) ~ !Y.(T), furthermore by Lemma 38.1 that {J(T) = O. Thus T E (F) and
0 < ind(T)
(51.1)
= a(T) - {J(T) = a(T) ~ aCT).
Because of Proposition 37.2 there exists an r! > 0 with ind(p/ - T) = ind( T) > 0
(51.2)
Since Proposition 26.1 ensures the existence of a right inverse R there exists an r2 > 0 with (51.3)
E
S£l(F) of
T,
{J(p/ - T) = 0
(Exercise 4 in (51.3)
~9).
For I!ll < min(r!.
o
-ideals. In (c) at first r(A) is not defined when E has finite dimension (Banach algebras should have unit elements different from zero !). Of course, A exists in the algebra !£l(E)/" = {O}. If we set r(A) := 0, then the proposition trivially also holds in the finite-dimensional case. We can now invoke Proposition 45.1 in order to make assertions concerning sums and products of Riesz operators. Here the following generalization of the concept of commutativity will be useful: If ,I is an ideltl in 2(E), then we say that A ,I-commutes with B if AB - BA E,I, i.e., if the residue classes of A and B commute in 2(E)/,I. We can leave the proof of the following proposition to the reader; observe only that operators which,l-commute also J -commute, and that the closure of a cI>-ideal is again a cI>-ideal (Proposition 51. 7). Proposition 52.3. Let E be a complex Banach space and ,I a cI>-ideal in 2(E). T hen the following invariance assertions hold: (a) Scalar multiples, sums and products of ,I-commuting Riesz operators are again Riesz operators; in the case of products it is even sufficient that only one of the factors lies in P(E). (b) The sum of a Riesz operator and of an arbitrary operator from ,I is a Riesz operator. The Riesz property is to a large extent stable with respect to the formation of functions. More precisely we have Proposition 52.4. Let E be a complex Banach space,j E £(A) and f(O) = O. Then together with A alsof(A) is a Riesz operator. The converse is true whenever f vanishes only at O. Proof. Let A be a Riesz operator. Because off(O) = 0 there exists agE £(A) such thatf(il) = ilg(il), i.e.,f(A) = Ag(A). Since A commutes with g(A), it followS from here with the aid of Proposition 52.3a thatf(A) is a Riesz operator.
219 Now letf(A) be a Riesz operator and assume thatfvanishes only at O. Then there exists an h E ~(A) and a natural number m such that for all A. of the set of definition offwe have: f(A.) = A.mh(A.)
and
h(A.) # O.
Consequently f(A) = Amh(A), and h(A) has an inverse in 2(E). Sincef(A) and h(A)- I commute, we obtain from here with the help of Proposition 52.3a that Am = f(A)h(A)-1 is a Riesz operator. It now follows from Proposition 52.2c in combination with (45.4) that A itself is a Riesz operator. • If K is from a (J}-ideal, then I - A.K is a Fredholm operator for all A.. With the aid of Proposition 52.1 we obtain therefore immediately:
Proposition 52.5.
If E is complete, then every operator from a (J}-ideal is a
Riesz operator.
Because of this proposition (J}-ideals are also often called Riesz ideals or ideals of Riesz operators. In general, P(E) is not a Riesz ideal.
The requirement of ,I-commutation is the weaker-and thus the efficiency of Proposition 52.3 the greater-the larger ,I is. It is therefore reasonable to ask whether there exists a Riesz ideal containing all (J}-ideals. The answer is positive and we are led to it as follows. In the first place we may disregard finite-dimensional spaces since for them P(E) = 2(E). Thus let dim E = 00 so that 2' := 2(E)/ ff(E) is an algebra with a unit element # the zero element. If ,I is a (J}-ideal and K E ,I, then I - AK is a Fredholm operator for every A E 2(E), consequently 1 - AK is invertible for every A E 2' (Proposition 25.2) (we denote by f the residue class of T E 2(E) in 2'). The set r:= {i{ E jj: 1 - Ai{ is invertible for every A E jj} is the 'radical' of jj; namely, if d is an algebra with unit element e # 0, then we call {r
E
.Si1: e - ar is invertible for every a E d}
the radical of d. We are interested here in the fact that the radical is a proper (two-sided) ideal of d (see e.g. [14]). If h is the canonical homomorphism from .ct'(E) onto !i, then the image h(f) of every (J}-ideal f lies in the radical r of !i; conversely h-I(r) is obviously a (proper) (J}-ideal. Thus h-l(r) is a maximal ell-ideal, containing all (J}-ideals, and is so uniquely determined and closed (otherwise the closure h-I(r) would be, according to Proposition 51.7, a -ideal which contains h-I(r) properly). We summarize:
r(A) the resolvent RA of A possesses the representation RA = I:'~o Anjiln+ 1 (Theorem 44.1) from where R;,.(F) c F immediately follows. For every x E F and x' E Flo = {y' E E': (y, y') = 0 for all y E F} the function il f--+ (RAX, x'), which is holomorphic on p(A), vanishes outside the spectral disk of A. Since p(A) is connected, it follows from here, by the identity theorem for holomorphic functions, that (RAX, x') = 0 for all il E p(A). Therefore RAx lies in Fl.l. = F (Proposition 29.2). We conclude from here that every element x E F is the image of an element of F under ill - A, namely of RAx. Since on the other hand F is invariant under ill - A, we obtain the equation (ill - A)(F) = F for all il E p(A). It follows immediately that p(A) c p(A) and therefore u(A) c a(A), and furthermore we obtain that for il E p(A) the resolvent RA of A is the restriction of RA to F. Now let ilo # 0 be a spectral point of A. Because of u(A) c a(A), the point ilo is isolated in u(A) and in u(A). The spectral projector P E 2(F) corresponding to {ilo} and to A is given by
P-
1 = -2. 1tI
fr
RA dil,
where r is a circle around ilo which isolates ilo from a(A)\ {iloJ. For every x E F we thus have, according to Exercise 1 in §49, Px
1 = -2. 1tI
fr
RAx dil
1 = -2. 1tI
f r
RAx dil
= Px,
where P is the spectral projector corresponding to {il o} and A. Thus P is the restriction of P to F. It now follows immediately from Proposition 52.2 that 1 is a Riesz operator. • We can now clarify without effort the behavior of Riesz operators with respect to duality: Proposition 52.S. A continuous endomorphism A of the complex Banach space is a Riesz operator if and only if the dual transformation A' is a Riesz operator.
Proof Let A be a Riesz operator. Then ill - A is for every il # 0 a Fredholm operator in 5t'(E) (Proposition 37.1), consequently ill' - A' is for all these }. a Fredholm operator in 2(E') (Proposition 27.3). Since E' is complete, A' must be a Riesz operator by Proposition 52.1. Now let conversely A' be a Riesz operator. By what we have just proved, the bidual transformation A" is then also a Riesz operator. Since its restriction to the closed subspace E of E" is exactly A because of (41.5), it follows from • Proposition 52.7 that A itself must be a Riesz operator.
221 Exercises + 1. Prove with the help of only Proposition 52.2 the following assertion, which follows of course also from Proposition 52.4: If a power An is compact, then A is a Riesz operator (we are in a complex Banach space). 2. Cf. §50 Exercise 2. The spectrum of a Riesz operator A on a complex, infinite-dimensional Banach space contains at least the point O. The point 0 is a pole of the resolvent of A if and only if a power An (n ~ 1) is finite-dimensional. + 3. The limit of a uniformly convergent sequence of commuting Riesz operators on a complex Banach space is a Riesz operator. Can the assumption of commutation be weakened? Hint: Exercise 1 in §45. 4. In a saturated operator algebra deE) there exists exactly one -ideal which contains all -ideals. + 5. A continuous endomorphism A of a Banach space is a Riesz operator if and only if A! - A is an Atkinson operator for every A #- O. Hint: Exercise 3 in §37. If one invokes Exercise 4 in §37 then one can even say: A is a Riesz operator if and only if A! - A is a semi-Fredholm operator for all A #- O. As a consequence, because of Proposition 36.3, the following proposition is valid: A is a Riesz operator if and only if peAl - A) is finite for all A #- O.
§53. Essential spectra In this section let E be a Banach space over C and A E 2(E). The points of the resolvent set of A are in a certain sense the 'good' points of A: For A E peA), and only such a A, the equation (A! - A)x = y is uniquely solvable for an arbitrary y (furthermore the solution depends continuously on the right-hand side). Now one could think to consider, besides the elements of peA), also such points A as 'good' for which A! - A (or the equation (A! - A)x = y) has a behavior which can be easily described; the points which are not 'good' would then form a subset of the spectrum, an 'essential spectrum'. Examples of sets of 'good' points are for instance the following: PI(A):= {A: ind(A! - A)
= 0 and peAl -
A)
= q(A! -
A) < oo}
= P A,
P2(A) := {A: ind(AI - A) = O}, piA):= {A: ind(A!- A) exists}
= A'
piA):= {A: A! - A is normally solvable}.
Setting CTk(A) := C\Pk(A) (k We have hence
=
1, ... ,4), we then obtain four 'essential spectra'.
222 Each of these 'essential spectra' is obtained by removing from a(A) points which according to a certain criterion are still considered to be 'good'. Obviously
A the residue class of A
in 2(E)/%(E).
If E is finite-dimensional, then all 'essential spectra' are empty; in the case of infinite dimension (k = 1,2,3) for every Riesz operator A.
In this section we want to consider a2(A) more closely. We call aiA) the essential spectrum of A and denote it henceforth by a..(A). Correspondingly, let Pe(A) := piA), i.e., p,,(A):= {A: ind(U - A)
= O},
a,,(A) is also called the Weyl spectrum of A.
Proposition 53.1. ae(A) is closed and lies entirely in the closed disk around 0 with radius r~(A) (see (51.7». If E is infinite-dimensional, then at least one point of ae(A) lies on the boundary of this disk. Proof Pe(A) consists of all the components of the Fredholm region A on which ind(U - A) vanishes (Theorem 51.1), it is therefore open; consequently ae(A) is closed and lies in the disk indicated. Now let dim E = 00. Then, with the notation ofthe proof of Proposition 51.8, we have r~(A) = r(A) (see (51.9», on the circle IAI = r(A) there lies a point Ao from a(A) (see §45), AO does not belong to A (Proposition 51.6), and a fortiori not to p,,(A). •
The next proposition will show that the points of the essential spectrum of A are firmly rooted in a(A): They cannot be removed from a(A) by adding to A operators from a -ideal (e.g., compact operators). For the proof we need a simple proposition concerning the preservation of the index: Proposition 53.2.
For every T E (E) andfor every Kfrom a -ideal in 2(E)
we have
ind(T
+ K) = ind T.
This proposition is proved exactly like Proposition 23.3; observe only that all operators which figure there can in the present case be chosen to be continuous and that ind(I - K) = 0 (Proposition 52.5). • Proposition 53.3.
For any -ideal J in 2(E) we have ae(A) =
n
Ke?
a(A
+
K).
223 We prove the proposition in the following formulation: A. E Pe(A) there exists a K E " with A. E p(A + K). If A. E Pe(A), then by Proposition 26.2 there exists a bijective R E !l'(E) and a K E ffi"(E) with U - A = R + K. Consequently U - (A + K) = R is bijective, hence A. E p(A + K). Now let conversely A E p(A + K) for some K E,I. Then ind(U - A - K) = 0, hence by Proposition 53.2 we also have ind(U - A) = 0 and thus A lies in Pe(A). •
c"
Exercises + 1. A -ideal " in the operator algebra d(E) (E an arbitrary vector space) is said to be a o-ideal if ind(I - K) vanishes for all K E " . If d(E) is saturated, then for every Fredholm operator A in .4(E) and for every K from a -ideal " also A + K E (A(E»; if " is even a o-ideal, then we have ind(A + K) = ind(A). Hint: Proposition 51.6 and proof of Proposition 23.3. Examples of o-ideals are ffi"(E) and %(E); cf. also Proposition 52.5. 2. Let d(E) be a saturated operator algebra on E and
0 := {A E (d(E»: ind(A)
=
O}.
Show the following: (a) An ideal" in d(E) is a o-ideal 0 + " c 0' (b) There exists exactly one maximal o-ideal in d(E); every o-ideal is contained in " m' Hint for (b): Apply Zorn's lemma to the non-empty set of 0ideals ordered by inclusion and use (a) for the proof of uniqueness. 3. The representation theorems 26.2 and 39.1 for Fredholm operators in saturated operator algebras are valid without change if we replace the o-ideal ffi"(d(E» by an arbitrary o-ideal. Use Exercise 1. + 4. Let d(E) be a saturated operator algebra on the vector space E over K and A E d(E). Let the spectrum o{A) of A be the complement of
"m
p(A)
:=
{A: (U -
A)-l E d(E)}
in K, the essential spectrum oAA) the complement of Pe(A) := {A: U - A
E
(d(E», ind(U - A) = O}.
Show that for every -ideal " in d(E) one has O'e(A) Hint: Exercises 1 and 3.
=
nKe.f O'(A
+ K).
§54. Normaloid operators Let A be throughout a continuous endomorphism of the complex Banach space E. The inequality r(A) ~
(54.1)
IIA"lll!"
~ IIA II
Suggests to consider operators A for which (54.2)
r(A)
= IIAII
or equivalently
IIA"II = IIAII" for n = 1,2, ... ;
such A are called normaloid. The explanation of the name is that so-called normal operators on complex Hilbert spaces satisfy equation (54.2), as we shall see
224 later. At present we will content ourselves with proving that hermitian matrices (or operators) A on E = [2(n) are norma[oid (we identify again (n, n)-matrices with the operators they generate; see the remarks following the proof of Proposition 47.2). As known, A := (lXjk) is said to be hermitian if one of the following equivalent conditions is satisfied: for j, k = 1, ... , n (54.3) or
x, y
for all
(AxIY) = (xIAy)
E
[2(n);
here (x Iy) := L~= 1 ~YIfY is the canonical inner product of the vectors
The inner product is related to the norm on [2(n) by the equation IIxl12 = L~= 11 ~y 12 = L~= 1 ~y~y = (x Ix) and by the Cauchy-Schwarz inequality
For every continuous endomorphism A we have IIA211 ~ IIAI12; if A is furthermore hermitian on [2(n) then IIAxl12 = (AxIAx) = (A2Xlx) ~ IIA2Xllllxli ~ IIA21111xl1 2 and therefore IIAI12 ~ IIA211, thus IIA211 = IIAI12. Since together with A obviously every power An is hermitian, it follows that IIA 2n li = IIAI1 2., hence r(A) = limIIA2"lll/2" = IIAII. Thus a hermitian A is indeed normaloid. Hermitian operators on [2(n) have the valuable property that an expansion theorem holds for them, more precisely: There exist n eigensolutions y 1, ... , y. to the eigenvalues 1X1"'" IX. so that (yyIY/l) = bY/l and that for every x E [2(n)
•
(54.4)
X= L(xlyy)yy y=1
hence
•
Ax= LlXy(xIYy)Yv;
y=1
the second equation is called the expansion theorem. In the finite sequence lXI' ... , IX. each eigenvalue of A occurs as often as its multiplicity indicates. If ..1.1' ... ,Am are the different ones among the 1X 1, ... , IX. and if in (54.4) the order is fixed in such a way that..1. 1 = IXI = ... = IXk"..1.2 = IXk,+1 = ... = IXk,,··· then we define by k,
P 1 x:=
L (xIYy)Y" y=1
k2
P 2 x:=
L
y=k,+1
(xlyy)y" .. ·
projectors PI, . .. , Pm; P/l projects [2(n) onto the eigenspace of ..1./l' Thus we have AP/lx = ..1./lP/lx and Ax = L~=I ..1./lP/lx or shorter: m
(54.5)
L ..1./lP/l
for It = 1, ... , m. and AP/l = ..1./lP/l /l=1 The question arises whether for normaloid operators on an arbitrary complex Banach space E there also exist projectors P /l so that equations of the forIll A =
225 (54.5) are valid (where, of course, the finite sums are replaced by infinite series). We shall show that under certain conditions this is indeed the case. First we present a new characterization of normaloid operators. Proposition 54.1. A¥-O is normaloid if and only IIAnxll/rn(A) (n = 0, 1, ... ) is monotone decreasing. Namely, if A is normaloid, then
IIAn+ lxll < IIAIIIIA"xll = IIAnxll r"+ leA) = r(A)r"(A)
for
rn(A)
if the sequence of numbers
n = 0, 1, ....
Conversely, assume that the mono tonicity condition is satisfied. Then in particular IIAxll/r(A) ~ IIxll for all x, hence IIA I ~ rCA) and thus IIA I = rCA) .
•
The set O',,(A) = O'(A) n {AEC: IAI = rCA)} is called the peripheral spectrum of A. Interesting assertions can be made concerning the peripheral spectrum of a normaloid operator. Proposition 54.2.
If A¥-O is normaloid, then for A E O',,(A) we have
p(M - A)
~
1
and
°
fJ(M - A) > 0.
Proof If (M - A)2X = but (M - A)x ¥y := (11 A)(M - A)x is different from zero and
°
for a certain x E E, then
Anx = [M - (M - A)]nx = AnX - n(M - A)An-IX =
nAn(~ -
y).
hence .
I _y I ~- n 1M - I III nY,
IIAnxll = IIAnxll rn(A) lAin = n ~n
from where it follows because of y¥-O that IIAnx II/rn(A) -+ 00. Since this contradicts Proposition 54.1, there cannot exist an x of the above kind, we have rather N[(M - A)2] c N(M - A) and therefore p(M - A) ~ 1. If we had fJ(M - A) = 0, then it would follow with the help of Proposition 38.5 that also a(M - A) = 0, and therefore the spectral point A would lie in peA). This contradiction shows • that we must have fJ(M - A) > 0. From Proposition 54.2 and 50.2 we get immediately: Proposition 54.3. If A¥-O is normaloid, then A. E O',,(A) is a pole of the resolvent R;. if and only if q(M - A) isfinite; in this case the pole A. has order 1.
The operator A is said to be meromorphic if its non-zero spectral points are poles of the resolvent. Compact and, more generally, Riesz operators are meromorphic. A meromorphic operator possesses at most countably many spectral
226 points AI' A2,'" which we always consider arranged according to decreasing absolute values: (54.6) With this ordering we have obviously a,,(A) = {A'I"'" An} for an appropriate index n. Each Av is an eigenvalue of A (Proposition 50.2). Let P v be the spectral projector associated with {Av} in what follows. In the case Av E a,,(A) we can make an important assertion concerning its norm: Proposition 54.4. !f A :f. 0 is normaloid and meromorphic, then the spectral projectors P v corresponding to the poles Av E a,,(A) have norm 1,furthermore (54.7)
Ilxll;£ Ilx
+ yll
for
x
E
N(AvI - A),
Y E (Av I - A)(E).
if y is a linear combination of eigenvectors
This inequality is valid in particular belonging to eigenvalues :f. Av.
Proof Let a,,(A) = {AI"'" An} and C v a circle around Av with such a small radius r that C vlies entirely in peA) and Av is the only spectral point in its interior. Since the pole Av is simple according to Proposition 54.3, it follows from (50.10) for ..1.0 = Av and p = 1 that APvx = AvPvX for all x E E; consequently, AP v = Av P v and more generally (54.8)
Akp v =
A~Pv
hence
1. -2 1!1
i
AkR;. dA =
Cv
A~Pv
(k = 0, 1, ... )
(see Exercise 6 in §49). We can represent Ak in the form
i
Ak =
v=
I
2~ 1!1
i
Cv
AkR;. dA
+ 2~ 1!1
i
C
AkR;. dA =
i A~Pv + 2~ i AkR;. dA;
v=
I
1!1
C
here C is an appropriate circle around 0 with radius p < IAl I = rCA). If as an abbreviation we set A k := (l/21!i) AkR;. dA, then
Sc
n
Ak =
L: A~Pv + Ak ,
v= I
(54.9)
IIAkll ;£ ppk max IIR;.II = ypk ;'eC
and therefore
II
L:m k=1
Ak k II :s; y Al
-
p )k :s; --'--Y L:m ( -k=1
lAd
for all natural
m.
-1--.!!...-
1..1.1 I Furthermore, setting ex:= min~=211 - (Av/A I ) I and J1.:= max~=2 for m = 1,2, ...
IlPvll, we have 2nJ1. < -. = ex
227 From (54.9) it now follows with the aid of the last two estimates that (54.10) as m
f
-.!.. (~)k mk=1 Al
--+ 00.
=
PI
+ -.!..
f
mk=1
±
v=2
(Av)kpv Al
f
A: ~ PI mk=1 At
+ -.!..
Since
f (0)k IAll
(~)k II -mk=1 ~ -.!.. I -.!.. mk=1 Al
f
=
-.!.. m m
= 1,
we obtain from (54.10) that also liP 111 ~ 1. But then we have even lIP 111 = 1 since because of 0 < lIP 111 = IIP~ II ~ lIP 1112 we have lIP 111 ~ 1. Thus the first assertion of the proposition is proved. From it follows (54.7) with the aid of Propositions 50.2 and 54.3, and, if we further invoke Proposition 43.1c, also • the last assertion of the proposition. In order to be able to make assertions also concerning spectral points which do not lie on the boundary of the spectral disk, we must restrict the class of normaloid operators. We say that A is spectrally normaloid if for every spectral set (1 C (1(A) the restriction of A to the invariant subspace M" := P,,(E) (see Theorem 49.1) is again normaloid. Hermitian operators on 12(n) are spectrally normaloid: indeed, because of the second equation in (54.3) they are hermitian even on every invariant subspace. A second example will be given in Proposition 54.6.
Proposition 54.5. If A is spectrally normaloid and meromorphic, then every pole of the resolvent R;. is simple. Furthermore
nN(P v),
n- 1
(54.11)
IlPnxll
~
Ilxll for x E
•
n>l, and
v=1
(54.12)
IIPnl1
~
2n- 1 for
n
=
1,2, ....
Proof Obviously we may assume that A # O. By Proposition 54.3 the pole Al has order 1. We first prove the analogous fact for the poles An (n > 1) which are #0. Let (1:= (1(A)\{A 1, ... ,An-d. According to Theorem 49.1, An is an isolated, peripheral spectral point of the restriction A" of A to M". By restricting the Laurent expansion of R;. around An to M" we see that An is a pole of the resolvent of A" and that An as a pole of the resolvent of A has the same order as it has as a pole of the resolvent of A". But the latter order is 1 since A # 0 is normaloid (Proposition 54.3). Now we consider the case that 0 is a pole of R;.. Let p be its order and set (1:= {O}. Then by Proposition 50.2 we have M" = N(AP), the normaloid restriction A" of A to M" is nilpotent and thus IIA"II = r(A,,) = O. It follows that N(AP) = N(A); hence p = 1. Bya similar splittingoff procedure, whose details can be left to the reader, we obtain with the aid of Proposition 54.4 and Exercise 5 in §49 the estimate (54.11). We now pass to the proof of (54.12). For n = 1 we obtain the assertion from Proposition
228 54.4. Now let n > 1. Because of PIlPV = JIlVP Il the vector (I - L~= ~ Pv)x lies in ~ N(P Il ); from §49 Exercise 5 and (54.11) it follows therefore that
n:=
for all
x
E
E,
and thus
IIPnl1
~ II I -
:t: P v II
~ 1 + :t>Pvll.
With the help of this estimate we obtain (54.12) by mathematical induction on n.
•
We now state the announced expansion theorem which represents the analogue to (54.4):
Theorem54.1. If A =I- 0 is spectrally normaloid and meromorphic, and if p. is the spectral projector corresponding to the eigenvalue A. n =1-.0 onto the eigenspace N(A..I - A), then the uniformly convergent expansion (54.13)
.=1
is valid whenever one of the following conditions is satisfied:
(a) (b) (c)
The sequence of norms IIL~= 1 P.II (n = 1,2, ... ) is bounded; 2·.1.. -+ 0 as n -+ 00; IIP.II = 1 for n = 1,2, ... , and nA.. -+ O.
The last two conditions are meaningful only if A has infinitely many eigenvalues. If the set of eigenvalues is finite, then the series (54.13) is to be interpreted as a finite sum. Proof
We assume that A has infinitely many eigenvalues. If a:= a(A)\ {A. 1 , ••. , .1..- d
and A~ is the normaloid restriction of A to M u , then IIAul1 = r(Au) = 1.1..1 (see Theorem 49.1). It follows that
consequently
II Ax - :t>vPv x I
~ /.1.. (1 + :t: IIPvll) IIxll· 1
From here we obtain directly or with the help of the estimate (54.12) the uniform convergence of the expansion (54.13) in case (a), (c) or (b) are verified, respectively. The reader will now easily settle the case of a finite set of eigenvalues. •
229 A is said to be paranormal iffor every x
E
E the inequality
(54.14) holds. At the beginning of this section we have seen that hermitian operators on 12(n) are paranormal. We shall now show that paranormal operators are normaloid on every closed invariant subspace; a fortiori they are spectrally normaloid. We first prove a lemma.
Lemma 54.1. Proof
If A is paranormal, then so is every power An (n = 1,2, ... ).
From (54.14) we get for k = 0, 1, ... the estimate
IIAk+ lxll IIAk+ 2xll II A k x I ~ -: -11A-;-k--;-+ x-CCr '---1
From here follows that
IIAnx11 Ilxll
IIAxl1 IIA 2xII Ilxll IIAxl1
- - = - - - - - ...
and with it the asserted inequality
IIA nx11 IIAn-lxll
IIA nxl1 2 ~
IIA 2nxii IIAnxl1 ' II(An)2xllllxll.
•
With this lemma we obtain exactly like in the discussion of hermitian operators at the beginning of this section: Proposition 54.6. A paranormal operator A is normaloid, and since A is obviously paranormal on every closed invariant subspace, A must be normaloid also on all these subspaces. In particular, A is spectrally norma/oid. In this section we presented essentially unpublished investigations of F. V. Atkinson and H. Heuser. Exercises
1. A spectrally normaloid Riesz operator, which satisfies one of the conditions (a) through (c) of Theorem 54.1, is compact. 2. The inverse of a paranormal operator is also paranormal if it exists on all of E.
VIII
Approximation problems in normed spaces §55. An approximation problem Our functional analytic studies have been motivated so far mainly by the problem of solving equations. We want to approach now a completely different group of questions and begin by formulating two important problems of analysis and by extracting their functional analytic core. The Weierstrass approximation theorem states that a continuous function x on the interval [a, b] can be approximated arbitrarily well by polynomials, uniformly on [a, b]. For computational reasons, in applied mathematics it is not so much the arbitrarily close approximability by all polynomials on what the interest is centered, but rather on the problem for a given natural number n to approximate x uniformly as well as possible by polynomials p(t):= (Xo + (Xlt + ... + (Xntn of degree ~n; we ask therefore whether in the set Pn of all polynomials of degree ~ n there exists a q such that (55.1)
max Ix(t) - q(t) I ~ max Ix(t) - p(t)1 a~t~b
If we use the canonical maximum-norm on C[a, b], then the question can be restated: Does there exist in the finite-dimensional subspace Pn of C[a, b] a q such that (55.2)
Ilx - qll
~
IIx - pll
for all
p E Pn ,
or, expressed differently, can the variational problem (55.3)
IIx - pil = min,
be :>olved by an element of Pn? The next problem is one of the starting points for the theory of Fourier series. Let C := C[ - n, n] be the real vector space of the real-valued continuous 230
231
functions on [ -n, n]. The question is: Given an x set T,. of trigonometric polynomials
(55.4)
~n
f"
C, does there exist in the
n
p(t):= ~ of degree
E
+ '~I (IX, cos vt + p, sin vt)
a q, such that
[x(t) - q(t)]2 dt
~
for all
f,,[X(t) - p(t)]2 dt
PET,. ?
If, contrary to our convention, we introduce on C the norm Ilxll
:=
if"
[X(t)]2 dt
(Exercise 4 in § I), then we can formulate this problem of approximation in the square mean as follows: Does there exist a q in the finite-dimensional subspace T,. of C such that (55.5)
Ilx - qll
~
Ilx - p II
for all
PET,.,
or, expressed differently, can the variational problem (55.6)
Ilx - pil
= min,
PET,.
be solved by an element of T,.? It is very simple to deal with this problem computationally. If we set
a, (55.7)
J" J"
x(t)cos vt dt
(v = 0, 1, ... )
b, := -1 x(t)sin vt dt n _"
(v = 1,2, ... )
:= -1
n _"
(these numbers are called the Fourier coefficients of x) and if we take into account the so-called orthogonality relations
f~"cos vt sin JLt dt = 0 (55.8)
for
v, JL = 0, 1, ... ,
f_""cos vt cos JLt dt = f"_"sin vt sin JLt dt = {On
for for
v =F JL v = JL
~
1,
then we obtain (omitting the variable t) f,.<x - p)2 dt
= {"x 2 dt - 2 f"x P dt + f"p 2 dt = f"x 2 dt
+ n[!
0'.
235
Proof Assume that E is strictly convex and let Ilx + yll = Ilxll + IIYIi. Without restricting the generality we may assume that Ilyll - Ilxll :;::; 0 and x # O. Then
2 :;::;
1111:11 +
II~IIII :;::; 1111:11 + II~IIII-IIII~II - "~,,II
Ilxll + Ilyll _II (Ilyll - Ilxll)y II = 2 Ilxll IlxllllYIl ' hence II(x/llxll) + (y/llyll)1I = 2. Because of (56.1) we have x/llxli = yllyll and =
therefore (56.2) holds. Now assume conversely that (56.2) is valid and let Ilxll = Ilyll = II(x + y)/211 = 1. Then Ilx + yll = 2 = Ilxll + Ilyll from where we get, by hypothesis, that x = lXy with IX :;::; O. It follows that IX = Ilxll/llyll = 1, hence x = y. Thus E is strictly convex. • If we take into account the remark concerning the validity of the equal sign in the Minkowski inequality, then we obtain from Proposition 56.2 immediately:
Proposition 56.3.
The spaces IP are strictly convex for 1 < P
0) we obtain from it that IIx + yl12 < 4, which proves strict convexity. If we equip C[a, b] with the norm IIxll := (S~ Ix(tW dt)I/2, and compute as above Ilx + yl12 and Ilx - y112, then we see that also in this case (57.1) holds, so that the space C[a, b] with the euclidean norm is again strictly convex. In both examples the decisive parallelogram law follows from the formal properties of the expression
!i 00
I
n
(xly):=
=:
~niin
for
x, y E [2
for
x, yE C[a, b]
1
b
x(t)y(t)dt
and its relation to the norm: For all vectors x, y, z and all scalars oc we have namely: (JPI) (x + ylz) = (xlz) + (ylz) (IP2) (ocxly) = oc(xly) (lP3) (xly) = (ylx) (lP4) (xix) ~ 0, where (xix) = if and only if x = 0,
°
furthermore IIxll = j(xlx).
(57.2)
From (IPl) and (lP2) it follows with the help of (lP3) that (57.3)
(xly
+ z) = (xly) + (xlz)
and
(xlocy)
= li(xIY).
From these rules we obtain Ilx
+ yl12 =
(x
+ ylx + y) =
(xix)
Ilx - yl12 = (x - ylx - y) = (xix)
+ (xly) + (ylx) + (yly) - (xly) - (ylx) + (yly)
from where indeed the parallelogram law follows. A vector space E over K is called an inner product space or a pre hilbert space if with every pair (x, y) of elements of E a number (xly) from K is associated so that (IP I) through (IP4) are valid. (Y; Iy) is called the inner product of the vectors x, y. The space E equipped with the inner product (,1,) is sometimes denoted more carefully by (E, (,1,», Our considerations can now be summarized as follows: If an inner product space E is at the same time a normed space, and if the inner product and the norm are related by (57.2), then in E the parallelogram law (57.1) is valid and E is strictly convex.
237 It is now a fundamental fact that in every inner product space E a norm can be defined by (57.2); equipped with this norm E is thus strictly convex. That by (57.2) a real number is associated with every x E E and that this correspondence satisfies the norm axioms (Nl), (N2) in §6, follows immediately from (lP4), (lP2) and (57.3). Before proving the triangle inequality (N3), we prove the capital Schwarz inequality: For all vectors x, y of an inner product space we have
(57.4) This assertion is trivialfor y = o(observe that (xIO) = (xIO·O) = O·(xIO) = 0); therefore we assume y =I- O. For all a we have 0 ~ (x + aYlx + ay) = (xix) + a(ylx) + ~(xly) + a~(yly); from here we obtain (57.4) by setting a = - (x Iy)f(y Iy). • From the definition (57.2) we now obtain, with the help of the Schwarz inequality, I!.\
+ yl12
= (x
+ ylx + y)
= IIxll2
= IIxl12
+ (xly) + (ylx) + IIyl12
+ 2Re(xly) + IIyl12
~ IIxl12
+ 211xllilyll + lIyl12
= (11xll + IIYII)2, hence also (N3) holds. We can therefore summarize our results in the following. Proposition 57.1. In every inner product space E, a norm, the canonical norm of E, is defined by IIxll := J(xlx). For this norm the parallelogram law (57.1) holds; thus E is not only a normed space but even strictly convex.
One can show that a normed space E over K, in which (57.1) holds, is an inner product space, i.e., that one can introduce an inner product (·1·) in E which generates the existing norm 11·11 according to Ilxll = J(x Ix). For this purpose one writes if K = R if K = C. We cannot go into the proof which employs elementary but long winded Computations (see [80]). Summarized briefly, we can say that exactly those normed spaces are inner product spaces in which the parallelogram law is valid. For euclidean R2 the parallelogram law expresses the theorem from elelllentary geometry that in a parallelogram the sum of the squares of the sides is equal to the sum of the squares of the diagonals. An inner product space will be equipped from now on always with its canonical norm.
238 The Schwarz inequality (57.4) can be written more briefly in the form l(xly)1 ~ IIxlillyll·
(57.5)
From it we obtain the continuity of the scalar product: Proposition 57.2. In an inner product space Xn -+ X, y. -+ y implies (xnIY.) (xly)·Inparticular,wehave(xly) = L:'=I (x.IY) whenever x = Ln"'=1 x n •
-+
The proof follows from the estimate
+ (xnIY. - y)1 xlillyll + IIx.lllly. -
l(x.IYn) - (xly)1 = I(x. - xly) ~
Ilx. -
YII·
•
An inner product space, which as a normed space is complete, is called a Hilbert space. 12(n), 12 and L 2(a, b) with the inner products (x Iy):= L~= I ~.ii., (x Iy) := L:'= I ~. ii. and (x Iy) := x(t)y(t)dt are Hilbert spaces; C[a, b] with (xly):= x(t)y(t)dt is a noncomplete inner product space. A linear subspace F of the inner product space E becomes in a natural wayby the restriction of the inner product to F -an inner product space, whose canonical norm coincides with the one induced by E. The product EI x ... x E. of finitely many inner product spaces (E., (·1·).) becomes with the inner product
S:
S:
• (xly):= L (x.IY.).,
.=
where
x = (XI' ... , x.),
y = (YI,···,Y.),
I
an inner product space. Its norm is Ilxll = (L~= I Ilx.II;)1/2; it is a Hilbert space if and only if each E. is complete. Exercises
If OCI, ••• , OCn are given positive numbers, then an inner product will be defined on K· by (xly):= L~= I oc.~.ii.; in this connection (oc l , ... , oc.) is called a weight vector. Define a new inner product on [2 by means of a 'weight sequence'. 2. If p(t) > 0 is continuous on [a, b] ('weight function'), then (x Iy) :== p(t)x(t)y(t)dt is an inner product on C[a, b]. + 3. Let T be an arbitrary non-empty set and 12(T) the set of all functions x: T -+ K with the following properties: (a) x(t) i= 0 for at most countably many t E T; (b) LtET Ix(t)12 converges (the sum is taken with respect to only those t for which x(t) i= 0). Then [2(T) is a linear function space over K and becomes a Hilbert space through the introduction of an inner product (xly):= LtE1' x(t)y(t). For every SET let Ys E [2(T) be defined by ys(t) = (jst. We have (Yr IYs) = (jrs, and (x IYs) = 0 for all SET implies x = O. 4. In the Schwarz inequality the equality sign holds if and only if x, yare linearly dependent. 1.
S:
239
5. A normed space, which is isometrically isomorphic to an inner product space, is itself an inner product space. 6. An algebraic isomorphism A between two inner product spaces E, F is an isomorphism of normed spaces if and only if (Ax lAy) = (x Iy) for all x, Y E E. 7. The vectors x I' ... , Xn of the inner product space E are linearly dependent if and only if their Gram determinant I(Xi Ix k ) I vanishes.
§58. Orthogonality As is well known, two vectors x = (~I' ~2' ~3)' Y = ('11' '12' '13) of R3 are perpendicular if their inner product (xly) = L~=I ~v'1v vanishes. This suggests the following definition of a new structural element which distinguishes the inner product spaces among the normed spaces: Two vectors x and y of an inner product space are said to be orthogonal or perpendicular to each other if(x Iy) = 0. The symmetry in the expression is justified by (lP3). Because of the importance of the matter, we will not refrain from documenting with a few examples that the concept of orthogonality penetrates many mathematical considerations. Example 58.1. If we want to represent the vector x from Kn as a linear combination of n basis vectors YI' ... , Yn' then we must calculate in general the coefficients (Xv in x = (XIYI + ... + (XnYn from a system of linear equations. If, however, {y I, . . . ,Yn} is a so-called orthonormal basis, i.e., if (yv IY,.) = bVIl for all V,Il, then (xly ll ) = (L~=I (XvYvly ll ) = L~=I (Xv(Yvly ll ) = (XII and x has the representation which can be written down immediately: n
(58.1)
X
=
L (xIY.)Yv. v= I
Example 58.2. D. Bernoulli and L. Euler were led through their investigations of the vibrating string to the problem of harmonic analysis: Given a realvalued function x E C[ - n, n] with x( -n) = x(n), determine real numbers a., b v so that for all t in [ -n, n] the representation (58.2)
x(t) =
~o + vtl (av cos vt + bv sin vt)
is valid. If we assume for a moment that the expansion (58.2) is possible and even that it converges uniformly in [ -n, n], then we can determine immediately the coefficients avo b v multiplying the series by cos vt or sin vt, respectively, integrating termwise and taking into account the orthogonality relations (55.8) of the trigonometric functions; we obtain
fit = - fit
a v = -1 n bv
x(t)cos vt dt
(v
= 0, 1, ... ),
x(t)sin vt dt
(v
= 1,2, ... ),
_It
1
1t
-"
240
i.e., the Fourier coefficients we defined already in (55.7). Decisive for this procedure was that the functions Yl(t):=
I ;;C' y' 2n
Y2(t):=
cos t
;::' yit):=
y'
sin t
n
;::' Y4(t):=
y'
n
cos 2t
;::' Ys(t):=
y'
n
sin 2t
;::' '"
y'
n
form an orthonormal sequence in the (real) function space C[ -n, n], equipped with the inner product (xIY):= S~1t x(t)y(t)dt, i.e., that (Yvly It ) = (JVlt for all v, J1.. The result of our considerations can now be formulated as follows: If G is the linear subspace of all functions x from C[ -n, n] which can be expanded on [ -n, n] into a uniformly convergent series 1 IXvyv, then we have
L:'=
00
(58.3)
x
=
L (xlyv)Yv'
v= 1
This expansion of x is so similar to (58.1) that we may call the sequence (Yl' Y2,···) an 'orthonormal basis' of the space G. By the way, also the numbers (xly It ) are called Fourier coefficients of x.
In the theory of Fourier series one inverts these considerations. One starts with a function x, integrable on [ -n, n], forms wi~ the Fourier coefficients (x Iy.) the so-called Fourier series 1 (x IYv)Yv and asks whether it converges in the interval [ -n, n] to x. As long as one has pointwise convergence in mind, this question is extraordinarily difficult. It is, however, reasonable to replace pointwise convergence by convergence in the sense of the norm IIxll:= (S~1t x 2(t)dt)1/2, which is generated by the inner product (convergence in quadratic mean), i.e., to ask whether for n -+ 00 one has
L:'=
We shall discuss this problem later in a very general form. Compare with these considerations also the approximation of continuous functions by trigonometric polynomials, which we treated in §55, and where the orthogonality relations (55.8) of the trigonometric functions ensured the explicit solvability of the approximation problem and the uniqueness of the solution. Example 58.3. We now investigate the distribution of temperature u(x, t) of a finite thin rod going from x = 0 to x = L, whose left end-point is kept at the constant temperature 0, while the right end-point is subject to an exchange of heat with a surrounding medium of temperature #0; at time t = 0 let the rod have the given temperature f(x) at the point x (observe that in this example x is a space variable, not a function). The distribution of temperature is then that solution of the heat equation
(58.4)
which satisfies the boundary conditions (58.5)
u(O, t)
=
0,
ux(L, t)
+ O'u(L, t) = 0
for all
t ~
°
241
and the initial condition (58.6)
U(x,O) = f(x)
for
0
~
x
~ L;
a and a are positive constants depending on the material of the rod. If we substitute into (58.4) a solution which we assume tentatively to be of the form u(x, t) = v(x)w(t), then we obtain the equation v"(x)
1
wet)
vex)
a2
wet)
which can hold only if each of its sides is equal to the same constant - A. Taking into consideration the boundary conditions (58.5), we obtain that v is a solution of the boundary value problem (58.7)
v" + AV = 0,
v(O) = 0,
v'(L)
+ av(L) =
0,
while w has only to satisfy the differential equation (58.8)
w+ a 2 Aw =
0
without further conditions. Conversely, we can check immediately that a solution v of (58.7) and a solution w of (58.8) always yields a solution U := vw of (58.4) which satisfies the boundary conditions (58.5), and which is continuous together with its partial derivatives U t and UXX ' The boundary value problem (58.7) does, however, not have a non-trivial solution for all A.; if, namely, v is a non-trivial solution, thus because of v(O) = 0 also a non-constant one, then it follows from AV = - v" that
hence that A must be > O. For positive A the general solution of the differential equation (58.7) is given by v(x):= A sin(fix) + B cos(fix) with arbitrary constants A, B. Because of v(O) = 0 we have B = 0 and so vex) = A sin(fix). With the second boundary condition v'(L) + av(L) = 0 we obtain tan(fiL) :::: -fila. This equation for A has countably many solutions An > 0 which increase strictly to + 00. Only for these eigenvalues of the boundary value problem (58.7) do there exist non-trivial solutions; the eigenfunction Vn corresponding to the eigenvalue An is given up to a multiplicative constant by vlI(x):= sin.,jI,;x. These eigensolutions have, like the trigonometric functions, the basic orthogonality property (58.9)
if n =F m if n=m.
242
From
+ Anvn) = 0, vn(v;;' + Amvm) = 0 we obtain first (An - Am)vnvm - vnv;;') = 0 and hence, using the boundary conditions,
vm(v~
+ (vmv~
0= (An - Am) LLVnV m dx
+
L\VmV~ -
= (An - Am) L\nVm dx
+
LL
= (An - Am) L\nVm dx
+ vm(L)v~(L)
Vnv;;')dx
~ (VmV~ -
vnv;")dx
- vn(L)v;"(L) -
Vm(O)v~(O)
+ vn(O)v;"(O) = (An - Am) LLVn Vm dx,
• The general solution corresponding to An of the differential equation (58.8) is given by C n exp( -Ana2t). For every choice of the constant Cn the function Un defined by (58.10) is a solution of the heat equation which satisfies the boundary conditions (58.5). It will, however, in general not verify the initial condition (58.6). In order to
satisfy (58.6) we will reason as follows: Equation (58.4) and the boundary conditions (58.5) are linear and homogeneous, thus every (finite) sum L Un of solutions of the form (58.10) satisfies again (58.4) and (58.5) for every choice of the Cn. The same is true for (58.11)
u(x, t):=
co
co
n= 1
n= 1
L un(x, t) = L Cn sin(Ax). exp( -An a2t )
provided that the coefficients Cn are chosen in such a way that (58.11) converges and the derivatives U,' U xx can be obtained by termwise differentiation. And the decisive question, whether the solution U can be fitted to the initial condition u(x, 0) = f(x), amounts now to the problem of determimng the C n so that co
f(x) =
LC
n
sin(Ax).
n=1
Thus we face again the problem to expand a givenfunction according to pairwise orthogonal functions, where orthogonality in CEO, L] is defined by means of the inner product (g Ih) := J&g(x )h(x )dx. After these examples we return to general considerations. Let E be an inner product space throughout. We express orthogonality by writing x .1 y. The
243 symbol N .1 M means that every vector of N is orthogonal to every vector in M. If N consists of only one element x, then we write more briefly x .1 M. The set M1-:={XEE:x.lM}
(58.12)
is a closed linear subspace of E, the space orthogonal to M; there is no danger of confusing it with the orthogonal space defined in §29: In prehilbert spaces we shall only use the orthogonal space (58.12). If M is itself a linear subspace of E, then we have obviously M n M1- = {O}; we shall often make use of this simple fact. The zero vector is the only vector orthogonal to all elements of E. If x is perpendicular to the vectors XI' ... ' X n, then x.l [XI' ... ' xnl From the continuity of the inner product it follows that a vector which is orthogonal to all terms of a convergent sequence is also orthogonal to the limit. From these remarks it follows that X .1 M implies X .1 [M]. We conclude this section with the Theorem of Pythagoras. First the simplest version: From uj .1 Uk for j =f. k it follows that (58.13) We have indeed
A non-empty subset S of E is called an orthogonal system if two different elements of S are always orthogonal to each other. If furthermore the vectors of S are normalized, i.e., if lIull = 1 for every U E S, then S is called an orthonormal system. A countable orthogonal system is also called an orthogonal sequence; it is now clear what is meant by an orthonormal sequence. An orthogonal system S does not have to be linearly independent, since it can contain the zero vector; if 0 does not lie in S (if e.g., S is an orthonormal system), then the linear independence is ensured: From IXIU I + ... + IXnUn = 0 (Uk E S) it then follows that 0= CtlXkUklUm) =ktllXk(UkIUm)
= IXm(umlum),
hence IXm = 0 for m = 1, ... , n. We can now generalize considerably the Theorem of Pythagoras: Proposition 58.1. Converges, then also
If (u l , U2'· •• ) is an orthogonal sequence and I II Uk 112 converges, and we have
Lk"=
if Lk"= I
Uk
244 For the proof let Sn:= Ul + ... + Un and U:= Lt'= 1 Uk = lim Sn. Because of the continuity of the inner product (Proposition 57.2), we have (sn Isn) = Lk=l I ukl1 2 ~ (ulu)lIuI1 2 , which proves the assertions. • Exercises 1. The unit vectors ek := (c5 jk ) form an orthonormal sequence in [2. From x 1- ek for all k it follows that x = o. 2. The functions Ys (s E T) of Exercise 3 of §57 form an orthonormal system in [2(T). From x 1- Ys for all S it follows that x = o. 3. The trigonometric functions Yl' Yz, ... in Example 58.2 form, as we already observed there, an orthonormal sequence in the real (and also in the complex) space C[ -n, n]. The functions (l/~)einr (n = 0, ± I, ± 2, ... ) form an orthonormal sequence in the complex space C[ - n, n]. Every x, which is orthonormal to this sequence, vanishes; the reader can find a proof for this theorem, which is valid also in L 2( - n, n), and which is basic in the theory of Fourier series, e.g., in [l81]. +4. For (ML)L we write more briefly MH. Show: (a) Ml c: M2 = Mt c: Mi. (b) M c: M H , even [M] c: MH. (c) ML = MLH. 5. We consider the Fredholm integral equation (19.1) under the hypotheses stated there; let K be the corresponding integral operator. If k(s, t) = k(t, s) or k(s, t) = k(t, s) for all s, t, according as C[a, b] is real or complex, then (Kx Iy) = (x IKy) for all x, y. It follows from here that every eigenvalue of K is real (trivial if C[a, b] is real) and that eigensolutions corresponding to different eigenvalues are perpendicular to each other (see Proposition 68.1). + 6. Orthogonality in an inner product space E over K can be characterized alone with the help of the norm: x 1- Y¢>
Ilxll
~
Ilx - IJ(YIl
for all
IJ( E K
(geometric meaning?). By the inequality on the right we can define the orthogonality of x to y, in symbol: x 1- y, in an arbitrary normed space. Introduce in R2 the maximum norm, determine to which x E R2 with Ilxll = I the vector (0, I) is orthogonal to, and show that x 1- Y does not imply y 1- x and that x 1- Yl' Y2 does not imply x 1- (Yl + Y2)· §59. The Gauss approximation We saw in §55 that the problem to approximate in quadratic mean as closely as possible a continuous function by trigonometric polynomials of degree ~ n could be solved easily and explicitly because the trigonometric polynomials form an orthogonal system. Also the uniqueness of the best approximation was obtained from the orthogonality relations (by now we know that every Cebisov
245 approximation problem (55.10) in an inner product space E has only one solution because E is strictly convex). It is therefore reasonable to consider orthonormal vectors in connection with the Cebisov approximation in an inner product space. In this case we speak of a Gauss approximation problem. The following theorem is essentially nothing but an abstract repetition of the trigonometric approximation in §55. Proposition 59.1. Let S:= {u l , ... , un} be an orthonormal system in the inner product space E. Then the Gauss approximation problem to determine,for a given xEE, numbers IXI, ... ,lXn so that Ilx - L~=llXvUvll is minimal, has IXv:=(xluv) as its unique solution. The vector x - L~ = I (x IuJu v is perpendicular to Sand thus also to [u l , ••• , un]. Furthermore we have the Bessel identity
(59.1)
Ilx - vt(XIUv)vr = IIxl12 -'- vtll(XIUvW
and the Bessel inequality n
L l(xluJl 2 ~
(59.2)
Ilx112.
v= I
Proof.
For arbitrary IXv we have
o~
Ilx -
t
v-I
n
= (xix) = IIxl12 -
= IIxl12 -
v-I
IXvUvl x -
t
/l-I
n
v= 1 n
n
v=1
v=1
IX/lU/l)
n
L IXv(uvlx) - L 1l/l(XI u/l) + L
IXva/l(uvlu/l)
n
L IXv(xlu v) - L av(xluJ + L IXvav n
= IIxl12 -
t
IXvUvl12 = (x -
v=1
n
L l(xluvW + v=L [(xlu v) -
v= 1
IXv][(xlu v) - a.]
1
n
n
v= 1
v= 1
L l(xluvW + L I(xlu v) -
IXv12.
The distance IIx - L~= 1 IXvuvll is therefore minimal if we choose IXv = (xlu v) for every v. The Bessel identity follows immediately from our computation, the Bessel inequality follows from (59.1) because the left-hand side is non-negative. y .= x - L~ = 1 (x I uJu v is perpendicular to S because (y I u/l) = (x I u/l) L~=1 (xluJ(uvlu/l) = (xlu/l) - (x I u/l) = 0 for Ii = 1, ... , n. • The converse of the last assertion also holds: If x - L~= 1 IXvUv is perpendicular to S, then IXv = (x I u v ); indeed, 0=
(x - /l-1t IX/lU/lIUv)
= (xlu v )
-
i
/l=1
IX/l(u/llu v) = (xlu v)
-
IXv·
246 The best approximation y E [UI, ... , un] of the point x is thus characterized by thefact that x - y is orthogonal to [UI, ... , un] (y is the 'foot of the perpendicular' dropped from x onto [u l , ••• , un]). In applications the case occurs frequently that though the vectors XI'···' Xn of problem (55.10) are linearly independent, they do not form an orthonormal system. This situation can be remedied by determining an orthonormal basis {u l , ••• , un} in [XI' ... , x n], obtaining the best approximation y of the point X in the form y = I~=I(xluv)uv and-if desired-writing y as a linear combination of the original vectors XI' ... ' Xn (for which it is necessary to know how the vectors U v can be expressed as linear combinations of the X I' ... , xn). That, starting from the elements XI' ... ' Xn, one can easily determine such an orthonormal basis, and expand each U v simply in terms of the basis {XI' ... , x n}, is ensured by the Gram-Schmidt orthogonalization process, which is described in the following proposition. Proposition 59.2. From an at most countable and linearly independent subset {XI' X2, ... } of the inner product space E we can obtain an orthonormal system {u l , u 2, ... } whose construction will be given in the proof, and which has the following properties: For every n = 1, 2, ... we have
(59.3) Thus, in particular, {UI' ... ' un} or {UI' U2' ... } generates the same subspace as {XI'· .. ' xn} or {XI' x 2, .. . }, respectively. Proof.
We define the sequence {u l , u2, ... } inductively:
(a) Set U I :=xtlllxlli. This is possible because of x l "# O. Now (59.3) is trivially true for n = 1. (b) If an orthogonal system {UI' ... , Ur-I} is already defined so that (59.3) holds for n = 1, 2, ... , r - 1, then r-I (59.4) Zr:= Xr - I (xrlup)up.l {u l , · · · , Ur- d· p=1 zr does not vanish since otherwise we would have by our construction
r-I r-I p r-I Xr = I (xrlup)u p = I (xrlu p) I (XpaXa = I YpXp' p=1 p=1 a=1 p=1 which contradicts the linear independence of the set {XI' x 2 , •• • }. If we set Ur := zr/ll Zr II , then {UI, ... , ur} is an orthonormal system according to (59.4), • and (59.3) is valid for n = 1 ... , r. In [178] Proposition 59.2 is applied to obtain from simple functions the polynomials of Legendre, Cebisov, Jacobi, Hermite and Laguerre (see also Exercise 2).
247
Exercises 1. Let XI"'" Xn be linearly independent vectors in the inner product space E. We have Ilx - I~= I (Xvxvll = min for a fixed given X E E if and only if all (Xv
I:
satisfy the system of the so-called normal equations = I (Xix" IXv) = (x Ix.) (v = 1, ... , n). The solvability of the system is ensured by Exercise 7 of §S7. Discuss the case of linearly dependent vectors XI, ... , Xn applying the normal equations. 2. Define for real-valued continuous functions on [ -1, 1] an inner product by (x IY):= J~ I x(t)y(t)dt and obtain by orthogonalization of the elementary polynomials 1, t, (2, ... the polynomials (()n(t):= y'(2n + l)/2Pn(t) (n = 0, 1, ... ,4), where Po(t):= 1, PI(t):= t, P 2(t):= tt 2 - t, P 3(t):= ~t3 - tt, P4 (t):= 3b4 - 1,1t2 + l The polynomials Pn(t) are called Legendre polynomials.
§60. The general approximation problem The Cebisov approximation problem seeks to a given point x E E a point Y lying closest in a finite-dimensional subspace F c: E. In the general approximation problem we may take F to be an arbitrary non-empty subset of E. In this generality the problem is, however, not solvable; we must subject F to certain restrictions. In prehilbert space it is enough if e.g., F is convex and complete (completeness is to be understood in the sense of the theory of metric spaces: every Cauchy sequence in F converges to an element of F): Theorem 60.1. If K =1= 0 is a convex and complete subset of the inner product space E (e.g., a complete subspace of E), thenfor every x E E the problem (60.1)
IIx - yll
=
min,
YEK
is uniquely solvable in K, i.e., there exists exactly one Yo (60.2)
IIx - Yoll
~
IIx - yll
forall
E
K such that
YEK.
Proof Let y := infYEK IIx - yll and let (Yn) be a minimizing sequence in K, i.e., lim II x - Ynll = y. In the parallelogram law lIu + Vll2 + lIu - vl12 = 211ull 2 + 211vll2 we set u:= X - Ym and v:= X - Yn' Since u
+v=
(x - Ym)
+ (x
- Yn) = 2[X - Ym ; YnJ.
u - V = (x - Ym) - (x - Yn) = Yn - Ym'
and since (Ym
+ Yn)/2 lies in K because of the convexity of K, we have
llYn - Ymll 2 = 211x - Ymll 2 + 211x - Ynl1 2 ~ 211x - Ymll 2
+ 211x -
-
411x - Ym ; Ynr
Ynll 2 - 4y2,
248 hence if n, m ~ 00 then llYn - Ymll ~ O. Thus (Yn) is a Cauchy sequence in K and has therefore a limit Yo E K. Since IIx - Ynll converges to y and to Ilx - YolI, we must have IIx - Yoll = Y and Yo is a solution of (60.1). We have obtained in particular that every minimizing sequence in K (with respect to x) is a Cauchy sequence. If now u, v E K are two solutions of (60. 1), i.e., if Ilx - ull = IIx - vii = y, then obviously the sequence u, v, u, v, ... is a minimizing sequence, thus a Cauchy sequence, and therefore u = v. • The hypotheses of the proposition are satisfied in particular when E is complete and K a convex, closed, non-empty subset of E. In the case of a Gauss approximation (K finite-dimensional subspace) Yo is the 'foot of the perpendicular' from x to K (x - Yo 1. K). This holds also for arbitrary subspaces, more precisely: Proposition 60.1. If F is an arbitrary linear subspace of the inner product space E and iffor an x E E the problem
IIx - yll = min,
(60.3)
has a solution Yo dicular to F.
E
F, then Yo is the unique solution in F and x - Yo is perpen-
Proof. We first show that x - Yo 1. Y for all Y =f. 0 in F. For all we have
Ilx - Yol12
IX E K, Y E F
Ilx - (Yo + lXy)11 2 = «x - Yo) - lXyl(x - Yo) - IXY) = Ilx - Yoll2 - cX(x - YoIY) - IX(Ylx - Yo) + IXcXllyl1 2. ~
If Y =f. 0, we set
IX'=
(x - YoIY) IIyII 2
_
thus
IX =
(Ylx - Yo) IIyII 2 '
and obtain from the above estimate that
I _ 112 _ 2 1(x - Yo1Y)12 + I(x - YoIyW I x _ Yo 112 ~ _ x Yo lIyII2 IIyII 2 hence
I(x - YoIyW < 0 IIyI1 2 = and therefore (x - Yo Iy) = 0, as asserted. For any further solution Yl E F of (60.3) we now have Yo - Yl = (x - Yl) - (x - Yo) E F (\ Flo = {O}, which proves the uniqueness. •
249 Exercise Proposition 60.1 has a converse: If Yo E F and x - Yo 1. F, then Yo is the (unique) solution of (60.3). Taken together, and expressed briefly: the feet of the perpendiculars (provided they exist) are the best approximations. §61. Approximation in uniformly convex spaces The proof of the Approximation Theorem 60.1 was based on the fact that every minimizing sequence is a Cauchy sequence: this fact in turn followed from a property of the inner product norm, namely the parallelogram law. One would therefore think of looking for normed spaces whose norms have the property to make a Cauchy sequence out of every minimizing sequence of the problem
(60.1 ). From the parallelogram law in the form Ilx; yr
+
IIx
~ yr = ~ IIxl12 + ~ IIyl12
we obtain 'from Ilx.11 (61.1)
~ 1, IIY.II ~ 1
it follows that
IIx. - y.11
and
II x. ; y. 11-+ 1
-+ 0'.
This property is, similarly to strict convexity, a rotundity property of the closed unit ball: If the midpoints of segments in K 1 [0] crowd to the surface, the endpoints will be pushed together arbitrarily. A normed space is said to be uniformly convex if (61.1) is valid in it. In uni-· formly convex spaces the problem (60.1) will turn out to be solvable. The proof of t~e following proposition can be left to the reader: Proposition 61.1. Inner product spaces are uniformly convex. Uniformly convex spaces are strictly convex. Figure 6 illustrates the hierarchy of the types of spaces studied so far. We arrive now at the decisive approximation assertion: Proposition 61.2. If K "# 0 is a convex and complete subset of the uniformly convex space E, thenfor every x E E the problem
(61.2)
Ilx - yll = min,
YEK
is uniquely solvable in K. Proof. Let y:= infYEK IIx - YII. Ify = 0, then x E K and x itself is the unique Solution of the problem. Let now y > 0 and let (Y.) be a minimizing sequence from K, i.e., limllx - y.11 = y. If we set Xn:=
X -
Yn
and
(1n:= IIxnll - y,
250
Normed spaces Strictly convex spaces Uniformly convex spaces product I I Innerspaces
Figure 6
then
Xn
=f. 0, an
~
0, furthermore
Ilxnll
-+ y
an
and
Because of the convexity of K the point (Yn
y ~ I x - Yn + 2 Ym I
-+ 0.
+ Ym)/2lies in K, and consequently
I Xn +2 Xm I ~ 21 IIxnll + 21 Ilxmll, from where II(xn + xm)/211 -+ y follows, as n, m -+ 00. For xn:= xn/llxnli we have =
thus
Ilxnll = Ilxmll = 1 and
I xn ~ xm 11-+ 1
Since E is uniformly convex, we obtain Ilxn - xmll Because of
for
n, m
-+ 00.
-+ 0.
llYn - Ymll = Ilxn - xmll = 1IIIxnllxn - Ilxmllxmll = :I(y + an)xn - (y + am)xmll ~ yllxn - xmll + anllxnll + amllxmll it follows that the minimizing sequence (Yn) is indeed a Cauchy sequence. The remaining demonstration proceeds now exactly as in the proof of Theorem 60.1.
•
For the proof of the next proposition we need the inequality (61.3)
Ilx + yliP + IIx - yIIP
which we will verify now. For s, t > 0, IX := (S2 +
~ 2P-l(lIxIi
t2)1/2 and p
P+ IlyIIP) for x, Y E [P, 2 ~ p
2 follows from the Holder inequality (in which p is to be replaced by p/2 and q by p/(p - 2». From (61.5) and (61.6) we obtain
(Ie
+ 11IP +
Ie - 11I P)l/P ~ 2(P-l)/p(leI P +
l11n~/p
and thus also the inequality (61.7)
for
p
~
2.
en,
If we replace in it ~, 11 by the components 11n of the vectors x, YElP, then we • obtain by addition of the inequalities the estimates (61.3). Proposition 61.3.
The spaces IP are uniformly convex for 1 < p
0 such that Ilxll ~ I, lIyll ~ I
and
IIx - yll
~
e
imply
II x ;
y
1/
~ 1-
(j(e)
(~(e) is called a convexity modulus of E).
+3. If E is uniformly convex, then every subspace, every quotient space and the completion of E is uniformly convex. 4. If we introduce on C[a, b] the norm Ilxll := (J: Ix(tW dt)l/ p , 1 ~ P < 00, (see Exercise 8 in §6), then inequality (61.3) also holds in this case; thus C[a, b] is uniformly convex with respect to the above 'LP-norm' at least for p ~ 2.
252 5. Oneprovestheinequality(sP + tP)I/P ~ (sq + tq)l/qfors, t ~ 0, 1 ~ q < P exactly like (61.4). With its help one can see that lq is a proper linear subspace of IP for 1 ~ q < p < 00, and that Ilxll p ~ IIxll q for all x E lq. 6. If K '# 0 is a convex subset of the strictly convex space E, then problem (61.2) has at most one solution in K. §62. Approximation in reflexive spaces The solution of the approximation problem (55.10) was based on the fact that every minimizing sequence is bounded and thus-since it lies in a finitedimensional subspace-has a convergent subsequence. If we look for best approximations in an infinite-dimensional subspace, then this selection procedure cannot be applied any more because of Theorem 10.1. (Problems (60.1) and (61.2) were treated consequently by a different method: We used the fact that, under the hypotheses given there, a minimizing sequence is always a Cauchy sequence.) The spaces lP (1 < p < 00) suggest, however, to use a selection procedure with respect to a notion of convergence which is different from convergence with respect to the norm. Namely, if (Yn) is a minimal sequence in a set M c IP, hence a bounded sequence, then with the help ofthe diagonal process we see immediately that (Yn) contains a subsequence which at any rate converges componentwise to an element Yo E IP; and about this Yo we can prove that it is a best approximation provided it lies in M at all. By Exercise 9b of§46 a bounded and componentwise convergent sequence in IP-and only such a sequence-is weakly convergent; we should therefore try to obtain the solution of the general approximation problem with the help of a selection procedure with respect to weak convergence. Indeed, this way leads to success in reflexive spaces. We first make some preparations. Proposition 62.1. Every bounded sequence (x~) in the dual E' of a separable normed space E has an E-weakly convergent subsequence. Proof. Let {Xl' x 2 , ••• } be a dense subset of E. The sequence of numbers is bounded and contains therefore a convergent subsequence (X~I(XI»· For the same reason (X~l) contains a subsequence (X~2) which converges at X2· One continues in this way. The diagonal sequence (y~) := (x~n) converges then at every X k • Since {x 1, X 2, ... } is also dense in the completion E of E, and since y~ can be extended to E with conservation of the norm (Proposition 12.3), it follows with the help of Theorem 33.1 that (y~) converges pointwise on E to • a certain y' E E'. X~(XI)
We can now prove without great effort a 'weak' selection theorem for reflexive spaces. Proposition 62.2. Every bounded sequence (xn) in a reftex.ive Banach space E has a weakly convergent subsequence.
253
Proof. The closed subspace F:= [Xl, x 2 , ••. ] of E is obviously separable and by Proposition 41.1 it is reflexive, it can therefore be identified through the canonical imbedding with its bidual F". Consequently F" and thus also F is separable (Exercise 8 in §42). Since (x n ) is a bounded sequence in F", it follows from Proposition 62.1 that there exists a subsequence (xnJ and an X E F" = F such that So in E. If we order 9Jl by inclusion, then we obtain the assertion from Zorn's lemma. The existence of Illaximal orthonormal systems follows now if one extends an orthonormal system • of the form So = {x/llxll}, x # 0, to a maximal one. A maximal orthonormal system S can be useful even when it is not an orthonormal basis, because the Fourier coefficients (x Iu) of an element x with respect
260 to S determine this element uniquely. Indeed, from (xlu) = (ylu), i.e., (x - ylu) = 0 for all u E S, it follows namely that x - y = o. In [71] examples of non-complete inner product spaces are given which have no orthonormal basis.
Exercises + 1. The following assertions concerning the orthonormal system S in the inner product space E are equivalent: (a) S is an orthonormal basis. (b) S is a basic set (i.e., [S] = E). (c) For all x, yin E one has (xly) = LUEs (xlu)(Ylu) (also this equality is called the Parseval identity). 2. Let S:= {u l , U2' ••• } be a maximal orthonormal system in the inner product space E. Show that E is complete if and only if I O£. u. converges for every sequence of numbers (0£.) for which I 10£.1 2 < 00. Hint for the proof of the sufficiency: Show that every x E E can be expanded with respect to Sand construct an isomorphism of normed spaces from E onto 12. 3. Let {u t : t E T} be a maximal orthonormal system in the inner product space E. Show that E is complete if and only if LtET O£,U t converges for every family of numbers (O£t: t E T) which has the following properties: (a) At most countably many terms of the family (O£t) are #-0 (and only these terms shall occur effectively in the above series.) (b) LtET IO£tl 2 < 00. Hint: Proceed as in Exercise 2, replace, however, F by the space F(T) (see §57 Exercise 3). + 4. A Hilbert space E with the orthonormal basis S is isomorphic as a normed space to 12(S) (see Exercise 3 in §57). Special cases: (a) If dim E = n < 00, then E is isomorphic as a normed space to [2(n). (b) If E is infinitedimensional but separable, then E is isomorphic as a normed space to 12. +5. Two maximal orthonormal systems SI, S2 in the inner product space E have the same cardinality (which is also called the Hilbert dimension of E). Hint: With every UES I associate the (at most countable) set Sz(u):= {VES2: (ulv) #- o} and show that S2 = UUES, Sz(u).
L:'=
L:'=
§66. The dual of a Hilbert space The dual of a Hilbert space is described by the following representation theorem of Frechet and F. Riesz:
Proposition 66.1. For each fixed element z of the Hilbert space E a continuous linear form is defined on E by f(x):= (x Iz). Conversely,for each continuous linear form f on E there exists exactly one vector z E E such that f(x) = (x Iz) holds for all x E E. Furthermore Ilfll = IIzll· Proof.
Trivially, for a fixed z E E a linear form is defined on E by f(x):===
(xlz) which is continuous because of I f(x) I = l(xlz)1 ~ IIxllllzll, and satisfies IIfll ~ IIzll· Because of If(z)1 = l(zlz)1 = IIzllllzll, we even have IIIII = IIzll (this part of the proof is valid for arbitrary inner product spaces). Now let,
261 conversely, f be an element of E'. If f = 0, then Z = 0 satisfies the requirements. If f "# 0, then because of Theorem 63.1 there exists a vector Xo "# 0 which is orthogonal to the closed nullspace N(f}; in particular f(xo} "# O. By (15.7) every x E E can be written in the form x = 1XXo + Y with uniquely determined 1X E K and y E N(f}. Thus we have f(x} = 1Xf(xo}·
(66.1)
Setting Z := [f(xo}/(xo Ixo}Jxo E N(f}J., we have (xlz) = (1XX o
+ ylz) =
1X(xolz}
= 1X(xolf(xo} (X~:o}) = 1Xf(xo} = f(x};
the last equality in this chain is nothing but (66.1). obviously the only vector in E which does so.
Z
thus generates f, and is •
For greater clarity we denote the vector generating fEE' by Z f' By Proposition 66.1 the correspondence f ....... Z r defines a norm-preserving, bijective map A from E' onto E. Obviously zf+g = zf + Zg' hence A is also additive. But because of (x Iz"f} = (1Xf)(X) = 1Xf(x) = (x Iaz f)' we have z,f = ,jz f' i.e., A(1Xf) = fiAf, hence the map A is only in the case of a real Hilbert space linear, and thus an isomorphism of normed spaces between E' and E. If E is complex, then we define by means of an orthonormal basis S of E (Theorem 65.1) a self-map K of E (a conjugation) by Kx:= (xlu)u. Obviously K has the following properties:
LueS
(a) (b) (c) (d)
(e)
K(x + y) = Kx K(1XX} = aKx,
+ Ky,
K2 = J, IIKxll = IIxll, , K is bijective and K- 1 = K.
It now follows that f ....... Kz f is an isomorphism of normed spaces between E' and E. We state this result: Proposition 66.2.
Every Hilbert space is isometrically isomorphic to its dual.
On the dual E' of a Hilbert space E one defines by (flg):= (Zglzf) an inner product which generates the norm of E' by Proposition 66.1: thus E' is a Hilbert space. Consequently every continuous linear form F on E' is generated by a well-determined gEE (Proposition 66.1): F(f) = (fIg) for all fEE'. By the definition of the inner product (fIg) and of the imbedding isomorphism J: E -+ E" (see (41.1», we have thus F(f) = (Zglzf) = f(zg) = (Jzg)(f); thus E" = J(E} and we obtain Proposition 66.3.
Every Hilbert space is reflexive.
Because of Proposition 66.1, weak convergence means that (x n Iz) -+ (x Iz) for every Z E E.
Xn ----'"-
x in a Hilbert space E
262
From Propositions 62.2 and 66.3 we obtain immediately: Proposition 66.4. Every bounded sequence (xn) in a Hilbert space E contains a weakly convergent subsequence, i.e., there exists a subsequence (xnJ and an x E E such that (x nk Iz) converges to (x Iz) for every z E E. Some important propositions of the present and the preceding section could be proved only under completeness hypotheses. In the case of a non-complete space E we will often be able to remedy the situation by constructing the completion of E; it is a Hilbert space, more precisely: Proposition 66.5. For the non-complete inner product space E there exists a Hilbert space E. unique up to an isomorphism of normed spaces, so that E is a dense subspace ofE (in particular, the inner product ofE is induced by the one in E). Proof. Let E be the completion of the normed space E according to Proposition 12.2. If x, yare vectors in E and (x n), (Yn) are approximating sequences in E, then from
+ l(xmlYn xmllllYnl1 + IIxmllllYn
l(xnIYn) - (xmIYm)1 ~ I(xn - xmlYn)1 ~ Ilxn -
Ym)1
- Ymll
it follows that lim(x nIYn) exists. A similar estimate shows that this limit is independent of the particular choice of the approximating sequences. Consequently, we have the right to define (x Iy) := lim(xn IYn). It is easy to see that hereby an inner product is introduced on E. This inner product obviously extends the inner product given on E. and the corresponding norm Ilxll = J(xlx) = lim J(xnlx n) = limllxnll coincides with the norm on E. Thus E is a • Hilbert space. Exercises 1. The second assertion of Proposition 66.1 is false in non-complete inner, product spaces. On such spaces there exist more continuous linear forms than are generated by the elements of the space. 2. An orthonormal basis of the inner product space E is also an orthonormal basis of its completion E. Hint: Use Exercise 1 in §65. 3. Every orthonormal sequence in a Hilbert space converges weakly to O. Hint: Exercise 1 in §64. + 4. Let E be an inner product space over K. A map (x, y) ~ sex, y) froIll E x E into K is called a sesquilinear form if the following hold: sex I + X2' Y)
= S(XI, y) + S(X2, y), sex, YI + Y2) = sex, Yt) + sex, Y2), S(IXX, y) = IXS(X, Y), sex, IXY) = as(x, y). For every A E !/'(E), e.g., sex, y):= (AxIY) is sesquilinear. Continuity and boundedness of a sesquilinear form is defined as for bilinear forms
263 (Exercise 4 in §14); the bounded sesquilinear forms are precisely the continuous ones. Show the following: (a) Let A EY(E); then s(x,y):=(Axly) is continuous if and only if A is continuous. In this case IIAII c= SUPx,y"O Is(x, y)l/llxIIIIYIi. (b) Let E be complete and sex, y) a continuous sesquilinear form. Then there exists an A E 2(E) such that sex, y) = (Ax Iy) for all x, y E E. §67. The adjoint transformation Let E and F be two Hilbert spaces over K. The transformation A' dual to 2(E, F) maps F' into E'. Proposition 66.1 suggests, however, not to consider the map g ~ f:= A'g of linear forms, but the corresponding map yg ~ x J of the generating vectors (one does not have to leave then the spaces E and F). The map A*: F -+ E so defined is characterized by the equation A
E
(67.1)
(Axly)
= (xIA*y)
for all
x E E, Y E F,
and is called the transformation (map, operator) adjoint to A or shortly the adjoint of A (a confusion with the algebraically dual transformation, which we also denote by A *, is not to be feared: In the con text of Hilbert spaces A * will always denote the adjoint of A). One will expect that A * has properties similar to A'; indeed the reader can easily convince himself of the truth of the following proposition in which A and B denote continuous linear maps of Hilbert spaces. Proposition 67.1.
A*
is
a continuous linear map with
IIA*II = IIAII.
The
following rules hold:
= A* + B*, (ocA)* 1* = I, 0* = 0;' (A +B)*
= CiA*,
(AB)*
= B*A*,
together with A also A * is bijective, and we have (A *) - 1 A**
= (A - 1)* ;
= A.
In the assertion before the last one, observe that A-I is continuous by Proposition 32.2. In the last assertion, A ** is a short way of writing (A*)*. Also the simple proof of the following proposition is left to the reader (cf. Exercise 1 in §29). Proposition 67.2. If A is a continuous linear map from the Hilbert space E into the Hilbert space F, then the following identities hold: (67.2)
A(E).l
(67.3)
A *(F).l
=
N(A *),
= N(A),
= N(A*).l, A*(F) = N(A).l. A(E)
From Proposition 29.3 and the second equation in (67.2) we obtain imlllediately
264
Proposition 67.3. Under the hypotheses of Proposition 67.2 the map A is normally solvable if and only if A(E) = N(A*).1 holds, i.e., if the equation Ax = y has a solution exactly in the case '(ylz) = 0
for all
ZE
N(A*)'.
Thus in the theory of normal solvability A* takes over the role of A', the inner product the role of the canonical bilinear form. We close this section with a useful identity concerning norms.
Proposition 67.4. We have IIA*AII = IIAA*II = IIAII2. Proof. With the aid of Proposition 67.1 we get IIA*Axll ~ IIA*IIIIAllllxll = IIA11211xlI, hence IIA*AII ~ IIAI12. On the other hand, IIAxl12 = (AxIAx) = (xIA*Ax) ~ IIA*AllllxI1 2, hence IIAII2 ~ IIA*AII. Altogether we have IIA*AII = IIAII2.1t follows further that IIAA*II = IIA**A*II = IIA*112 = IIA112. •
Exercises +
2.
1.
For A E 2(E) we have u(A*) = {I: A. E u(A)} (cf. Proposition 44.2). N(A*) = N(AA*), A(E) = AA*(F).
3. A compact self-map of a Hilbert space has a compact adjoint (cf. Proposition 42.2). 4. If the endomorphism A of the space 12(n) or [2 is generated by the matrix (lXjk) (i.e., if ('71, '72",,):= A(~l' ~2'···) with '7j:= L IXjk~k' j = 1,2, ... ), then A* is generated by the matrix (tXk). *5. A closed subspace F of a Hilbert space E is invariant under A E 2(E) if and only if F.l is invariant under A*; (F, F.l) reduces A if and only if F is invariant under A and A*.
x Spectral theory in Hilbert spaces §68. Symmetric operators We remind the reader of the definition of a symmetric operator in §63: An endomorphism A of the inner product space E is said to be symmetric if (68.1)
for all
(Axly) = (xIAy)
x, y in E.
We already saw an example in §58 Exercise 5. Real multiples, sums and (pointwise) limits of symmetric operators are again symmetric. The product of the symmetric endomorphisms A, B is symmetric if and only if A and B commute.
Proposition 68.1. For a symmetric operator A every eigenvalue and (Ax Ix) is real. Eigenvectors corresponding to distinct eigenvalues are orthogonal to each other. Proof (Ax Ix) is real because of (Axlx) = (xIAx) = (Axlx). For an eigenvalue A we have Ax = AX with an appropriate x =I 0 and thus A = (Axlx)/(xlx) is real. If, furthermore, Ay = p.y, y =I 0 and p. =I A, then A(xly) = (Axly) = (Axly) = (xIAy) = (xlp.y) = p.(xly), hence (A - p.)(xly) = 0 and therefore (x Iy) = O. •
We say that the symmetric operator A is positive, and write A ~ 0, if (Ax Ix) 0 for every x E E. In this case [x Iy] := (Ax Iy) defines on E a so-called semiinner product, i.e., an expression which has all the properties of an inner product with one possible exception: From [xix] = 0 it does not have to follow that x = O. However, the Schwarz inequality (57.4) is validfor such semi-inner products, as we can see by modifying slightly the proof. We can therefore state the following proposition: ~
Proposition 68.2. (68.2)
If A ~ 0, then the generalized Schwarz inequality
I(AxIYW ~ (Axlx)(Ayly)
holds.
265
for all
x, yin
E
266
From the generalized Schwarz inequality it follows that a symmetric endomorphism is uniquely determined by its quadratic form (Axlx); more precisely, we have: Proposition 68.3. Iffor the symmetric endomorphisms A and B one always has (Axlx) = (Bxlx), then they are equal.
Indeed, in this case T:= A - B is symmetric and satisfies (Tx Ix) = 0 for all x. From (68.2) it follows that (Tx Iy) = 0 for all x, y, hence also I Tx 112 = (Tx I Tx) = 0 for all x and thus T = O. • For arbitrary endomorphisms this proposition is correct only if the space is complex. For the proof we use the representation of the hermitian form (Ax Iy) of A, which is valid for any endomorphism A of a complex inner product space:
+ y)lx + y) - (A(x - y)lx - y) + i(A(x + iy) Ix + iy) - i(A(x - iy) Ix - iy).
4(Axly) = (A(x
(68.3)
This formula expresses, stated briefly, the hermitian form of A by the corresponding quadratic form. Thus if the quadratic forms of A and B coincide, then (Axly) = (Bxly) for all x, y from where A - B = 0 follows as above. We consider along with (68.3) the analogous representation
=
4(xIAy)
(68.4)
+ ylA(x + y» - (x - YIA(x - y» + i(x + iYIA(x + iy» - i(x - iYIA(x
(x
- iy».
If (Azlz) is always real, then it follows that (Azlz) = (zIAz); from (68.3) and (68.4) we obtain then that (Axly) = (xIAy) for all x, y, hence A is symmetric. If we take into account Proposition 68.1, then we can state the following result: Proposition 68.4. Let A and B be endomorph isms of the complex inner product space E. Then the following assertions are valid:
(a) From (Ax Ix) (b)
= (Bx Ix) for all x it .follows that A = B.
A is symmetric
if and only if (Ax Ix) is realfor all x E E.
The next proposition shows that the norm of a continuous symmetric operator can be determined with the help of its quadratic form. Proposition 68.5.
(68.5)
For the continuous and symmetric operator A we have
IIAII
I(Axlx)l·
= sup Ilxll = 1
Proof.
(68.6)
For every normalized x we have v(A):= sup Ilxll
=1
I(Axlx)1
I(Axlx)1
~
~
IIAllllxll 2 = IIAII, hence
IIAII.
267 For an arbitrary A > 0 we have 4//Ax// 2 = (A(
Ax + ~ AX) lAx + ~ AX) -
(A(
Ax - ~ AX) lAx - ~ AX)
~ V(A{IIAx + ~ AXr + IIAX - ~ Ax112] = 2v(A) [A2/1X/l2
+
;2 /l AX /l 2]
(the last equality follows from the parallelogram law). If /lAx/i =F 0, and if one sets A2 = /lAx/il/ix/l, then it follows that /lAx/i ~ v(A)/ix/i. This inequality is trivially true also in the case /lAx/i = 0 and yields the estimate /lA/I ~ yeA) for • the norm of A, from where the assertion follows with the aid of (68.6). If J is a symmetric operator, if xn
-+
x and AXn -+ y, then for every z we have
(Axn/z)
-+
(y/z)
and (Axn/z) = (xn/Az)
-+
(x/Az) = (Ax/z),
so that Ax = y and thus A is closed. With the aid of the closed graph theorem (Theorem 32.3) we obtain from here the important Hellinger- Toeplitz Theorem (see Exercise 6 in §63): Proposition 68.6. Every symmetric operator on a Hilbert space is continuous. For a symmetric operator A we have (x/A*y) = (Ax/y) = (x /Ay) for all x, y, hence A = A *. Conversely, from A = A * the symmetry of A follows. Continuous endomorphisms A of a Hilbert space, which coincide with their adjoint A *, are called selfadjoint. With this terminology an endomorphism of a Hilbert space is symmetric if and only ifit is selfadjoint. Because their quadratic form is real, symmetric operators have some properties which remind us of the situation in the real number field. Thus e.g., an order can be introduced: For two symmetric operators A, B the symbol A ~ B (or B ~ A) means that (Ax/x) ~ (Bx/x) for all x (for the verification of the axioms of order one needs Proposition 68.3). Obviously A ~ B is equivalent to B - A ~ O. A sequence An of symmetric operators is said to be monotone increasing or decreasing if At ~ A2 ~ ... or At ~ A2 ~ ... , respectively; it is said to be bounded from above (below) if there exists a symmetric operator B with An ~ B (An ~ B) for all n. A sequence is said to be bounded if it is bounded from above and from below. Analogously to the well-known convergence theorem for monotone sequences of numbers, we have: Proposition 68.7. Every monotone and bounded sequence of symmetric operators on a Hilbert space converges pointwise to a symmetric operator.
268 In the proof we consider a monotone increasing sequence: A1 ~ A2 ~ ... B. For n > m we have An - Am ~ 0; with the aid of Proposition 68.5 we obtain first ~
IIAn - Amll
sup (Anx - Amxlx)
=
Ilxll=1
~
sup [(Bxlx) - (A 1xlx)] =: oc, Ilxll=1
and then with the aid of Proposition 68.2 the estimate IIAnx - Amxl14
=
«An - Am)xl(An - Am)x)2
~ «An - Am)xlx)(An - Am)2xl(An - Am)x) ~ [(Anxlx) - (Amxlx)]oc 3 1IxI12.
The sequence of numbers (Anxlx) is monotone increasing and bounded, hence convergent. Our estimate shows therefore that (Anx) is a Cauchy sequence. Since the space is complete, this sequence converges, i.e., (An) tends pointwise to a (symmetric) operator A. • We return to the order relation between symmetric operators. If A, B, Care symmetric, then A ~ B implies that A + C ~ B + C and ocA ~ ocB if oc ~ o. The problem, whether the inequality A ~ B 'may' be multiplied by a positive operator, is more difficult. For its investigation we need Reid's inequality (see [132]): Proposition 68.8.
If A, B are continuous operators, A
~
0 and AB symmetric,
then for all x we have
(68.7)
I(ABxlx)1 ~ IIBII (Axlx).
Proof. From (68.2) (generalized Schwarz inequality) and the inequality between the arithmetic and geometric mean we obtain
I(Axly)1 ~ t[(Axlx) + (Ayly)]'
(68.8)
Because of (ABnxly) = (B n- 1xIABy) = (AB n- 1xIBy) = (Bn-2xIAB2y) = ... = (x IABny), the operator ABn is symmetric; from (68.8) we obtain for n = 1, 2, ... the estimate I(ABnxlx)1 = l(xIABnx)1 = I(AxIBnx)1 ~ t[(Axlx)
+ (AB 2nxlx)],
By mathematical induction we get now the inequality (68.9)
I(ABxlx)1
~
G+ ~ + ... +
;n)(AXIX)
+ ; .. (AB 2"xlx).
If IIBII = 1, then because of I(AB 2"xlx)1 ~ IIAII·llxIl 2 the last term in (68.9) tends to 0; thus for n -+ 00 we obtain I(ABxlx)1 ~ (Axlx). From this special case of Reid's inequality we obtain (68.7) if we replace B by B/IIBII. •
269 We can now show that the order relation between continuous symmetric operators is compatible with multiplication: Proposition 68.9. If A, B, C are continuous symmetric operators, then A ;£ B and C ~ 0 imply AC ;£ BC whenever C commutes with A and B. In particular, the product of two continuous, positive and commuting operators is always positive. We prove first the last assertion. Let A, B be continuous, positive and commuting. We may assume 0 ;£ I - B ;£ I; ifthis does not already hold, we replace B by /3B with an appropriate positive factor /3. By Proposition 68.5 we have then I I - B I ;£ 1; since A commutes with I - B, and since therefore A(/ - B) is symmetric, it follows from Proposition 68.8 that (A[I - B]xlx) ;£ (Ax Ix),
hence
A - AB ;£ A
and thus 0 ;£ AB. The first assertion now follows easily from what we have just proved; one only has to multiply the positive operator B - A by C. • Exercises
*1. For every continuous positive operator we have IIAxll2 ;£ IIAII (Axlx). + 2. If [x Iy] is a continuous semi-inner product on the Hilbert space E with the inner product (xly) (i.e., if Xn -+ x, Yn -+ Y implies that [xnIYn]-+ [xly]), then there exists a continuous, positive operator S with [xly] = (Sxly) for all x, y E E (see §66 Exercise 4). *3. On the vector space E let a semi-inner product [xly], and with it a semi-norm Ix I := J[x Ix], be given. Furthermore let A E [I>(E) be symmetric, i.e., let [Ax Iy] = [x IAy] for all x, y E E. Show the following: (a) [Ax Ix] E R for all x E E. _ (b) If A is even fully symmetric, i.e., if Iu I =1= 0 for all eigensolutions u of A corresponding to non-zero eigenvalues, then all the eigenvalues are real, and the eigensolutions corresponding to distinct eigenvalues are orthogonal to each other. (c) From v(A) := sup Ixl = 1 I [Ax Ix] I < CIJ it follows that A is bounded and that its norm IA I is equal to v(A), provided that there exist at all elements x with Ixl =1= 0; otherwise we have trivially IAI = v(A) = 0 (boundedness and norm of A are defined like in normed spaces). +4. The endomorphism A of the Hilbert space E is said to be symmetrizable if there exists a continuous semi-inner product [x Iy] on E, with respect to which A is symmetric; A is said to be fully symmetrizable if we have furthermore Ixl:= J[xlx] =1= 0 for all eigensolutions x corresponding to non-zero eigenvalues. Show the following: (a) A is symmetrizable if and only if there exists an H ~ 0 such that HA is . SYmmetric with respect to the inner product of E. (b) The eigenvalues of a fully symmetrizable operator are -all real; for eigensolutions u, v corresponding to distinct eigenvalues we have [u Iv] = o.
270
(c) A symmetrizable and bounded operator A is also bounded with respect to the semi-norm Ix I, and we have IA I ~ II A II. Hint: Exercises 2,3; Proposition 68.8. + 5. The definition of symmetrizability and of full symmetrizability from Exercise 4 can be carried over word for word to the case when E is a Banach space (cf. [60]). Show the following: (a) A semi-inner product [x IyJ on E is continuous if and only if there exists a y ~ 0 such that Ixl:= J[xlx] ~ yllxll holds for all x. (b) For a fully symmetrizable A assertion (b) in Exercise 4 holds. (c) For a symmetrizable and bounded A assertion (c) of Exercise 4 holds. Hintfor(c):FromIA nxI 2 ~ IxllA 2nxiitfollowsthatlAxl/lxl ~ (IA2kxl/lxI)I/2~ if Ix I =1= 0. Now use (a). See also the beginning of §74. §69. Orthogonal projectors In this section let E be a Hilbert space.
We have defined orthogonal projectors in §63; in Proposition 63.1 we asserted that precisely the symmetric projectors are the ones which are orthogonal. The terminology introduced in §5 concerning reducing subspaces will be somewhat simplified in the context of Hilbert spaces where only orthogonal decompositions and orthogonal projectors playa role: We say that the closed subspace F of E reduces the operator A E f/(E) if F and F.l are invariant under A (observe that A = FEB F.l by Theorem 63.1). Proposition 5.4 takes now the following form: Proposition 69.1. Let P be the orthogonal projector from E onto the closed subspace F. Then the following assertions hold: (a) (b)
F is i.lvariant under the endomorphism A if and only if AP = PAP. F reduces A if and only if AP = P A.
The following theorems deal with commuting projectors and order relations between them. Proposition 69.2. Let PI, P 2 be the orthogonal projectors onto the closed subspaces F 1, F 2 of E. Then the following assertions hold: . (a) If PI and P 2 commute, then P IP2 is the orthogonal projector onto F 1 ( l F 2' (b) If PIP 2 = 0, then also P 2 PI = 0, the subspaces F 1, F 2 are orthogonal to
each other, and PI + P 2 is the orthogonal projector onto the subspace F 1 (which is closed by Exercise 4 in §63).
EB F 2
Proof·
(a) From the assumption that PI and Pz commute, it follows that P 1Pz is idempotent and symmetric, hence an orthogonal projector. The image space of PIP z is {x E E: PIP z X = x} = {x E E: P z PIX = x} = F 1 ( l F 2 •
271
(b) We have P 2 P I = P!Pi = (P I P 2)* = 0* = O. For xEF I ,YEF2 we have furthermore (xly) = (P l xIP 2y) = (XIP I P 2y) = 0, i.e., F I , F2 are orthogonal to each other. PI + P 2 is symmetric and because of (PI + P 2)2 = pi + 2P I P 2 + P~ = PI + P 2 idempotent, hence an orthogonal projector, and it is easy to see that its image space coincides with F I El3 F 2. • For every orthogonal projector P we have (Pxlx) = (P 2xlx) = (PxIPx), hence (69.1)
because of 0 ~ IIPxl1 2 ~ 1IPI1211xl12 ~ (xix) (we have IIPII ~ 1 by Proposition 63.1), it follows furthermore that
o ~ P ~ I.
(69.2)
A more general assertion concerning order is made by: Proposition 69.3. Under the hypotheses of Proposition 69.2 the following assertions are equivalent: (a) (b)
PI ~ P 2 • IIPIXII ~ 11P2xll for all x P I P 2 = P 2P I = Pl·
E E.
(c) (d) FI C F 2 . (e) P 2 - PI is an orthogonal projector. Proof.
Because of (69.1) we have (a) _ (b). From (a) it follows that I -
P2 ~ I - PI' thus because of (69.1) we have 11(1 - P 2)PIX11 2 = «(I P 2)P IX IPlx) ~ «(I - PI)PIX I PIX) = 0 for all X E E and therefore (I - P 2)P I = 0, hence P 2P I = Pl· But then also P I P 2 = PiP! = (P 2 P I )* = Pi = Pl·
Thus we have proved (a) => (c). (c) => (d) since, because of Proposition 69.2a, we have F I = PI(E) = (P I P 2)(E) = FI n F 2 • We now show (d) => (b). With the aid of the orthogonal complement G of F I in F 2 we obtain in E = F I + G + F"i a decomposition of E into pairwise orthogonal subspaces. If x = Xl + y + X2 is the corresponding decomposition of an arbitrary vector x, then IIP l xl1 2 = IIxll12 ~ IIxll12 + IIyl12 = Ilxl + yl12 = 11P2X112, hence (b) holds. So far we have proved the equivalence of the assertions (a) through (d). (a) => (e): P2 - PI is symmetric, and since together with (a) also (c) holds, it follows that (P 2 - P I )2 = P~ - 2P 2P I + pi = P 2 - 2P I + PI = P 2 - PI, hence P 2 - PI is an idempotent, and thus (e) is true. With the help of (69.2) from (e) we get (a) immediately. • Exercises *1.
Under the hypotheses of Proposition 69.2 we have: F I .1. F 2 implies
PI P 2 = P 2 P I = o. *2. Let PI' ... ' Pn be the orthogonal projectors onto the closed subspaces . F h ... , Fn of E. Then the following holds: p.= PI + ... + P n is an orthogonal
272
*
projector if and only if PjPk = 0 for j k. In this case P(E) = F I + ... + F •. 3. If under the hypotheses of Proposition 69.2 PI and P 2 commute, then p:= PI + P 2 - P I P 2 is an orthogonal projector and P(E) = F I + F 2' §70. Normal operators and their spectra We consider first endomorphisms A of an n-dimensional Hilbert space E. On an eigenvector U the operator A acts as a multiplication by a number: Au = AU. Consequently it will always be easy to describe A when E has a basis UI' ... , U. consisting of eigenvectors Uk' Indeed, if AI' ... , A. are the corresponding (not necessarily distinct) eigenvalues and if one represents x in the form x = = I A* and with the aid of (70.3) it follows immediately that A is normal. If the closed subspace F reduces the normal operator A, then F is also invariant under A * (Exercise 5 in §67), and obviously the restriction of A* to F is the adjoint of the restriction of A to F. It follows immediately that this restriction is again normal: Normalcy is preserved under restriction to reducing subspaces. In §54 we called an operator A on a complex Banach space normaloid if its spectral radius r(A) is equal to IIA II. The following proposition motivates this terminology.
Proposition 70.2. normaloid.
Proof. (70.4)
Every normal operator is paranormal and thus also spectrally
For a normal A we have IIAxl12
= (Ax lAx) = (A*Axlx)
~ IIA*Ax~ll!xi:
= IIA 2xllllxll,
.
hence A is paranormal. For the remainder of the assertion invoke Proposition ~~
Since together with A also A.J - A is normal, from (70.3) and Proposition 67.2 we obtain immediately
For a normal A and an arbitrary A we have N(A.J - A) = N(A - A)l., and we have the orthogonal decomposition E = (A.J - A)(E) E!3 N(A.J - A). Proposition 70.3.
= N(A.J - A*); consequently (AI - A)(E)
From this proposition, together with Proposition 67.3, we obtain that a normal operator A is normally solvable if and only if the equation Ax = y has a solution exactly when y 1- N(A). Furthermore we get from it (see the proof of the corresponding assertion in Proposition 68.1): Proposition 70.4. Eigenvectors of a normal operator corresponding to distinct eigenvalues are orthogonal to each other. We now turn to the investigation of isolated points AD in the spectrum of a normal operator A. The subspace M a invariant under A, belonging to the .spectral set (J := p.o} (see Theorem 49.1), coincides by Proposition 49.1 with the set {x: limll(Aol - A)"xIl 1 / n = O}. Since A* commutes with A and therefore also with (,101 - A)" (Proposition 70.1), it follows from this representation that Ma is also invariant under A*, and from here we obtain, as above, that the restriction Aa of A to Ma is normal. The spectrum of Aa is {AD} (Theorem 49.1), ,the spectrum of the normal operator Aola - Aa is therefore {O}, hence we have ;"'A -, 0 I - A ) = 0 from where with the aid of Proposition 70.2 it follows that lJlAola - Aall = 0, i.e., that Aa = Aola· Consequently A.o is an eigenvalue of A ~nd Ma = N(A.ol - A). From (50.9) and the remark following it we obtain now (f
(f
274 that Ao is a pole of first order of the resolvent R;.. With the aid of Propositions 50.2 and 70.3 we obtain furthermore that (Ao I - A)(E) is closed and that the spectral projector P tT is an orthogonal projector. With this we have proved the following: Proposition 70.5. An isolated spectral point AO of the normal operator A is an eigenvalue of A, even a pole of first order of the resolvent. Ao! - A is normally solvable, and the spectral projector belonging to {A O} is an orthogonal projector; it projects E onto N(AoI - A) along (Ao! - A)(E).
This proposition and the following lemma, which is an immediate consequence of (70.4), will enable us to give a simple characterisation of the essential spectrum of a normal operator.
Lemma 70.1.
The length of the nullchain of a normal operator A is at most 1.
Proposition 70.6. The essential spectrum ae(A) of a normal operator A consists of the limit points of a(A) and of the isolated spectral points having irifjnite multiplicity.
By the multiplicity of an isolated spectral point we mean, of course, its multiplicity as an eigenvalue. For the proof we assume first that Ao is an isolated spectral point with finite multiplicity. Then we have E = (Ao I - A)(E) EB N(AoI - A) by Proposition 70.5, hence ind(AoI - A) = 0: thus Ao does not lie in ae(A). Now let conversely AO be in a(A)\ae(A). Then we have 0 < a(AoI - A) = f3(AoI - A) < ooandbyLemma70.1thelengthofthenullchainisp(Ao!- A) = 1. With the aid of Proposition 38.6 it follows from these two assertions that also the length q(Ao I - A) of the image chain is equal to 1. Proposition 50.2 now yields that Ao is an isolated spectral point (its multiplicity is finite by assumption) .
•
We call A an approximate eigenvalue of A if there exists a sequence of normalized vectors Xn with AX n - AXn --+ 0; the set of the approximate eigenvalues is the approximative point spectrum. It contains a p(A) and lies in a(A). For normal operators we have furthermore Proposition 70.7. coincides with a(A).
The approximate point spectrum of a normal operator A
We only have to show that every spectral point A of A, which is not an eigenvalue, belongs to the approximate point spectrum. For such a A the subspace (A./ - A)(E) is not closed (otherwise not only a(A.J - A), but because of Proposition 70.3 also f3(A.J - A), would be zero, hence A would lie in p(A». By Proposition 36.1 the minimal modulus is Y(AI - A) = inf II(A.J - A)xll = 0 xiO
IIxll
from where the assertion immediately follows. • If the normal operator A has no eigenvector, then its spectrum has no isolated points (Proposition 70.5). Since it is also closed, it is a perfect set and has therefore
275 the power ofthe continuum (see [181], p. 72. We assume, of course, that E =F {O}, hence that a(A) is not empty). In this case one says that A possesses a pure segment spectrum. Consider now the other extreme case: let the set of all eigenvectors be so large that its closed linear hull is all of E. We want to show that then the spectrum of A consists of all eigenvalues of A and their limit points; we say that A has a pure point spectrum. For every eigenvalue il we determine an orthonormal basis of the eigenspace N(ill - A)-see Theorem 65.1-we combine these bases and obtain so an orthonormal system S (Proposition 70.4), which is obviously maximal, hence an orthonormal basis of E (Proposition 65.3). We now show-and with this our assertion concerning a(A) will be proved -that a point il, which is neither an eigenvalue of A nor a limit point of eigenvalues, lies in p(A). For this it is sufficient to prove that the equation
y has for every y E E a solution x E E. We assume first that (70.5) has a solution x. Then there exists a sequence (Uk) in S such that (ill - A)x
(70.5)
=
00
x =
L (x IUk)Uk k=1
00
y =
and
L (yIUk)Uk' k=1
If ilk is the eigenvalue belonging to Uk' then 00
(ill - A)x =
L (il -
00
ilk)(xluk)U k =
k=1 hence (x! Uk) = (y Iuk)/(il - ilk) for all k, and so
(70.6)
_ x -
~ L-
(ylu k)
, _
k=11L
,
ILk
L (yIUk)Uk'
k=1
Uk'
In order to prove the solvability of (70.5) for a given y E E, we write the tentative solution x in the form (70.6). According to our hypothesis concerning il there exists f. > 0 such that I il - ilk I ~ e for all k, consequently the series
f
1 (ylu k) 12 k=1 il-ilk
converges (Proposition 64.2), and by Proposition 64.la the series
I
(y Iu k )_ Uk k= I lil- ilk! has a sum x in E. Obviously x is a solution of equation (70.5).
•
If A does have eigenvalues but not a pure point spectrum, then let F be the closed linear hull of the set of all eigenvectors of A. The subspaces F and F.l are invariant under A and A*, the restrictions A p , As of A to F, F\ respectively, are normal and a(A) = a(Ap) u a(As) (Exercise 2 in §49). According to construction, Ap has a pure point spectrum; As has a pure segment spectrum (an eigenvector x of As would lie in F.l, on the other hand-as eigenvector of A-also in F, thus We would have (x! x) = 0 in contradiction to x =F 0). Thus a(A) is obtained as the union of a pure point spectrum and of a pure segment spectrum. a(As) is also called the segment spectrum of A.
276 The term 'segment spectrum' can be better understood if we consider symmetric operators, because the spectra of these lie in intervals of the real axis which are easily described. We now turn to the examination of these facts. If A is a symmetric operator on E, then we call the numbers (70.7)
m(A):= inf (Ax Ix)
and
M(A):= sup (Ax Ix)
Ilxll = 1
IIxll
= 1
the lower and the upper bound of A, respectively. It follows from Proposition 68.5 that (70.8)
IIAII = max{lm(A)I, IM(A)I}
(observe in these considerations that (Ax Ix) is real according to Proposition 68.4 and that A is continuous according to Proposition 68.6). The bounds of A make it possible to describe more precisely the position of the spectrum a(A): Proposition 70.S. The spectrum of the symmetric operator A lies in the closed interval [m(A), M(A)] ~fthe real axis; the bounds m(A), M(A) belong to a(A). If A is not real, then for the resolvent operator R;. = (AI - A)- 1 the estimate
1
(70.9)
IIRJ
~ 11m AI
is valid. Pro~l We first show that a(A) is real. For a non-real A and for every vector x#-o we have
0< 1,1 - Alllxl1 2 = I([AI - A]xlx) - ([AI - A]xlx)1
= I([AI -
A]xlx) - (xl [AI - A]x)1 ~ 211(AI - A)xllllxll,
thus A does not lie in the approximate point spectrum of A, hence, because of Proposition 70.7, not in a(A). Also (70.9) follows immediately from the above estimate. Let us now assume temporarily that m(A) > O. Then 0 lies in p(A); otherwise there would exist because of Proposition 70.7 a sequence Xn of normalized vectors such that AX n -4 0 in contradiction to the inequality < I1l(A) ~ (Axnlx n) ~ IIAxnll-lfwe apply this result in thecaseA. < m(A) orA. > M(A)to the operators A - AI or AI - A, respectively, we see that these values A belong to p(A). Thus a(A) lies in [m(A), M(A)]. Let now A = m(A). Then A - AI ~ 0, and from the generalized Schwarz inequality (68.2) we get
°
II(A - AI)xl1 4 = ([A - AI]xl[A - AI]X)2 ~ ([A - AI]x Ix)([A - AI]2X I [A - AI]x)
~ ([A - A/]xlx)IIA - AI1I311x1l2.
Since by the definition of m(A) there exists a sequence Xn of normalized vectors such that (Ax nIx n) -4 m(A) = 11., this estimate shows that A belongs to the approximate point spectrum, hence, by Proposition 70.7, also to the spectrum of A. One sees analogously that M(A) lies in a(A). •
277
Exercises I. Show with the help of the propositions of this section that for a normal operator A on the complex space /2(n) the expansion (70.1), with pairwise orthogonal eigenvectors Uk corresponding to the eigenvalues Ak , is valid. This expansion, which motivated our definition of normal operators, is thus characteristic for normalcy in the finite-dimensional case. 2. A normal operator is normally solvable if and only if the length of its image chain is finite. +3. Every complex number A can be written in the form A = rx + if3 with rx, 13 E R; the real part (J( and the imaginary part 13 are uniquely determined: rx = !(A + X), 13 = Oj2i)(A - A} A-I exists if and only if (rx 2 + 13 2 )-1 exists; in this case A-I = X(rx 2 + 13 2 ) - I. Show that analogous assertions hold for (in some cases normal) operators: (a) L E 2(E) can be written in the form L = A + iB with A = A*, B = B*; the (selfadjoint) operators A, B are uniquely determined: A = !(L + L*), B = (lj2i)(L - L *). (b) L is normal if and only if the 'real part' A and the 'imaginary part' B commute. (c) If L is normal, then L - 1 exists on E if and only if (A 2 + B2) - 1 exists on E; in this case L - I = L*(A 2 + B2)-I. + 4. A is an eigenvalue of the normal operator A if and only if (A./ - A)(E) is not dense in E. + 5. The definition of the approximate point spectrum applies to any continuous endomorphism A of a Banach space. Show that the approximate point spectrum of A is contained in a(A) and that it is closed. 6. The linear map of L 2(a, b) defined by (Ax)(t) := tx(t) is symmetric but has no eigenvalue. +7. The continuous endomorphism U of the complex Hilbert space E is said to be unitary if U U* = U* U = I. In what follows, let U be a unitary operator. Show the following: (a) U is normal. (b) The unitary operators form a multiplicative group. (c) U preserves the inner product: (Uxl Uy) = (xly). (d) U is isometric: I Uxll = IIxli. (e) a( U) lies on the boundary of the unit disk. (f) For every symmetric operator A the operator eiA is unitary.
§71. Normal meromorphic operators In this section let A be a continuous endomorphism of the complex Hilbert space E. If A is meromorphic and normal, hence also spectrally normaloid (Proposition 70.2), then the eigenvalues A" =I: 0 of A are simple poles of the resolvent RA (Proposition 54.5); consequently the spectral projector P" belonging to A"
278
projects E along the (closed) image space (). "I - A)(E) onto N(A"I - A) (Proposition 50.2), and is an orthogonal projector because of Proposition 70.3. Since P" Pm vanishes for n =1= m, it follows from Exercise 2 in §69 that also P 1 + ... + P" is an orthogonal projector, and has therefore a norm ~ 1 (Proposition 63.1). Thus A satisfies all the hypotheses and condition (a) of Theorem 54.1, so that we can state the following proposition: Proposition 71.1. If A =1= 0 is normal and meromorphic, if {A1' A2, ... } is the set of =1= 0 eigenvalues of A ordered according to decreasing absolute values, and if P" is the orthogonal projector of E onto the eigenspace N(A"I - A), then for A the uniformly convergent expansion
(71.1) is valid, hence because of Proposition 70.4 we have
(71.2)
for every
x
E
E.
If one knows the eigenvalues and the eigensolutions (hence also the spectral projectors) of A, then Proposition 71.1 is very helpful in solving operator equations. We have namely: Proposition 71.2.
Under the hypotheses of Proposition 71.1 the equation
(71.3)
(AI - A)x = Y
with
A =1= 0
is solvable if and only if y is orthogonal to N(AA - I); in this case a solution is given hy
(71.4) Proof The solvability criterion follows immediately from Proposition 70.3 since (AI - A)(E) is closed. In the case of solvability one first observes by means of (71.2) that
converges, hence by Proposition 64.1 the series in (71.4) has a sum x. We noW obtain with the help of the expansion (71.1)
=
1 y - - Ay A.
1
00"
+ 1" ,,=L1 A"P"y = I\.
y.
•
279 Because of Proposition 71.1, a normal Riesz operator is the uniform limit of a sequence of compact operators, since its projectors Pn are finite-dimensional (Proposition 52.2), thus it is compact itself (Proposition 13.1). If for every eigenspace NO'nl - A) = 0 we determine an orthonormal basis {un,"', Un }, then , "" we have Pnx = (xlun,)u n, + ... + (x Iunk)u nk" , hence Ax = I:'=I An[(xlun,)un\ + ... + (x Iunk )unk ]. In order to simplify this cumbersome notation, we agree on the following c;nvention: Let {u I, U2, ... } be the union of the above orthonormal bases of the eigenspaces corresponding to eigenvalues =f. 0, and let Iln be the eigenvalue belonging to Un' In the sequence (Ill' 1l2' ... ) every eigenvalue =f. 0 will occur thus as many times as is indicated by its multiplicity, while in the set {A'l' A2, ... } the eigenvalues An were pairwise distinct. If we take into account that Ilixlu n) = (xl,unun) = (xIA*u n) = (Ax Iun), then from Proposition 71.1 we obtain immediately: Proposition 71.3. The eigenvalues different from zero of a normal compact operator A =f. Oform a nonempty sequence Ill' 1l2' ... ,ifone lists every eigenvalue as many times as is given by its multiplicity; this sequence tends to zero if it is infinite. To this sequence there corresponds an orthonormal sequence UI, U2' ... of eigensolutions (so that AUn = IlnUn for all n), with the aid of which the expansion (71.5)
Ax =
I
Iln(X IUn)U n =
I
(Ax IUn)U n
for all x
E
E
is valid. If (Iln) is a finite sequence, or a sequence tending to zero, and {u l , U2' ... } is an orthonormal system, then (71.5) defines a continuous endomorphism A on E whose adjoint is given by A*x = I:'=I ,un(xlun)un. It follows that IIAxl1 = IIA *x I for all x, thus A is normal. If we define the continuous, finite-dimenI Ilk(xluk)Uk, then sional operator An by Anx:=
Ik=
n>k n>k consequently An ~ A; by Proposition 13.1 we obtain that A is compact. The representation (71.5)-with the stated properties of (Iln) and (un)-is therefore characteristic for normal compact operators. Under the hypotheses of Proposition 71.3, formula (71.4) for the solution of the equation (AI - A)x = y has the form (71.6) here we assume that y..l N(AI - A). Exercises + 1.
u2 ,
••• }
ofA.
Under the hypotheses of Proposition 71.3 the eigensolutions {u 1 , of A form an orthonormal basis of E if and only if 0 is not an eigenvalue
280
*2. Assume that for an endomorphism A of the inner product space E over K one has Ax = L J1.n(x Iun)u n (x E E) with a sequence of numbers J1.n =I 0 and an orthonormal sequence (un). Show the following: (a) AUn = J1.nUn. (b) A is bounded if and only if (J1.n) is. (c) If A is bounded and A is the extension of A to the Hilbert space completion E of E, then Ax = J1.n(x Iun)u n for all x E E; A is normal. (d) A is precompact if and only if (J1.n) is finite or converges to zero. (e) A is symmetric if and only if all J1.n are real.
L
§72. Symmetric compact operators In applications it is of considerable importance that for a symmetric compact operator A =I 0 the expansion (71.5) is valid even if the space E is not complete. One can see this e.g., by extending A to the completion of E, where one will invoke Exercise 5 in §13. We want to expound here, however, another method which uses directly the compactness of A. E is a real or complex inner product space. By Proposition 68.5 we have sUPllxll=l I(Axlx)1 = IIAII, consequently there exists a sequence (xn) and a number J1. with I J1.1 = II A II > 0 such that Ilxnll = 1 and
(Axnlxn)
-+
J1..
From 0 ~ IIAxn - J1.xnll2 = IIAxnll2 - 2J1.(Ax nlx n) + J1. 211xnll 2 ~ IIAII2 - 2J1.(Ax nlxn) + IIAII2 it follows now that (72.1) Because of the compactness of A the sequence (Axn) has a convergent subsequence (AxnJ; it follows from (72.1) that also (xnJ converges to a (normalized) element U and that Au - J1.U = 0, hence U is an eigensolution corresponding to the eigenvalue J1. = ± IIA II. Obviously (72.2)
I(Aulu)1 = sup I(Axlx)l,
Ilxll = 1
and conversely, every vector u which satisfies (72.2) is an eigensolution of A corresponding to the eigenvalue ± IIA II (choose Xn = u). Let now J1.1 := J1., U1 := U and E1 := [u1F. The restriction A1 of A to E1 is a symmetric compact endomorphism of E 1 ; if it is =10, then by what we have just proved it has an eigenvalue J1.2 such that 0 < 1J1.21 = IIA1 II ~ IIAII = 1J1.11· Let U2 be the corresponding normalized eigensolution. If A2 is the restriction of A to E2 = [Ub U2J.l, then the same arguments yield, if A2 "# 0, an eigenvalue J1.3 of A2 such that 0 < 1J1.31 = IIA211 ~ IIA111 = 1J1.21, and a corresponding eigensolution U3; trivially U2, U3 are also eigensolutions of A corresponding to the eigenvalues J1.2, J1.3. It is clear now how the process continues. One obtains a possibly terminating sequence (J1.n) of eigenvalues such that I J1.1 I ~ IJ1.21 ~ ...
281
> 0, and an orthonormal sequence (un) of corresponding eigensolutions. (J1.n) terminates with J1.rn precisely if A vanishes on Em := [u \' ... , urn].l; in this case E = [u\, ... , um] EEl Em (§63), hence x = (x IUk)Uk + Y with y E Ern and so
I::,= \
m
Ax =
I
J1.k(X IUk)Uk·
k= \
°
If (J1.n) does not terminate, then J1.n ~ (Proposition 43.2). For an arbitrary xEE the vector Yn:=X-li:=l(xluk)Uk lies in En, consequently IIAYnl1 ~ IIAnllllYnl1 = lJ1.n+ tlllYnl1 and IIYnl1 2 = IIxl12 - Ii: = 1 l(xlukW ~ Ilx112. It follows that AYn ~ 0, hence 00
Ax =
I
k=\
J1.k(X IUk)Uk·
In the sequence (J1.k) every eigenvalue #0 of A occurs as many times as indicated by its multiplicity. Otherwise there would exist an eigensolution U with Au # and u Uk (k = 1,2, ... ), with which we would then have Au = J1.k(U IUk)Uk = 0, which is a contradiction. We summarize:
I
°
°
Proposition 72.1. If A # is a symmetric compact endomorphism of the (real or complex, complete or non-complete) inner product space E, then one obtains an orthonormal sequence of eigenvectors Un by determining first a solution Ul of the variational problem
under the side condition Ilxll I(Axlx)1 = max and then successively for n = 2, 3, ... a solution of the problem
I(Axlx)1 = max
under the side conditions for
k
= 1,
IIxll =
=
I,
1, ... , n - 1,
as long as this maximum is positive; the absolute value of the eigenvalue J1.n corresponding to Un is equal to this maximum. The described procedure yields each eigenvalue #0 of A as often as its multiplicity indicates, and one has the expansion (72.3)
for all
x
E
E.
Formula (71.6) is valid also under the more general hypotheses of Proposition 72.1, and we can make an even more general statement:
Proposition 72.2 Assume that for the symmetric operator A of the inner when A. # 0, and let the expansion product space E one has ind(A./ - A) = (72.3) be valid with an orthonormal sequence of eigensolutions Un and a sequence of corresponding eigenvalues J1.n' which can accumulate only at 0. Under these hypotheses the equation (Al - A)x = Y is solvable if and only if y 1. N(Al - A); and in this case a solution x is given by
°
(72.4)
1 1 x :=I y + I
L
Iln# 0 and
r(t) >
(OC I , !(2)
and
q,
r E qa, b],
°
on
[a, b],
(fJ" fJ2) are real vectors 0/= (0,0)
(see (3.23) and the considerations made there). u E C 0 we have by (74.2) the estimate i5{3 = Iv(a,
{3)1 =
IV(A)I ~ If(A)1 ~
IR,dlxl 2 ~
Ixl2
73'
we must have i5 = 0 and thus v(a, {3) == 0 in the upper half-plane. It follows now from (74.3) that Iyl = 0 (y = RAX) and so
Ixl2
= =
[xlxJ = [(AI - A)ylxJ A[ylxJ - [ylAxJ ~ IAllYllxl + lyllAxl
=
o.
Thus Proposition 74.1 is proved. From it we obtain now very easily an existence theorem for eigenvalues. Let it be preceded by a definition: An endomorphism A of a complex vector space with a semi-inner product is called a Wielandt operator if it is fully symmetric and if ind(AI - A) vanishes for all A =I o. Proposition 74.2. The Wielandt operator A on E has a non-zero eigenvalue if and only !f IAy I does not vanish jar every y E E. Proof. The condition is trivially necessary. Assume that it is satisfied. If A did not have a non-zero eigenvalue, then AI - A would be bijective for every A =I 0, consequently x:= A2y (y arbitrary in E) would lie for every real ~ in (0 - A)2(E), hence we would have Ixl = 0 (Proposition 74.1) and thus IAyl2 = [AyIAyJ = [A 2 ylyJ = [xlyJ ~ Ixllyl = O. Since y was arbitrary, we obtain a contradiction to our assumption. •
For a vector space E with semi-inner product [x Iy J the notion of an orthonormal system can be defined as usual. Since Bessel's identity and the results which follow from it do in reality not depend on the strict definiteness of the inner product, we have for an orthonormal system S in E the following assertions: (74.6)
UvES; n
(74.7)
L
l[xluvJl 2 ~ IxlZ,
UvES;
v= 1
(74.8)
for a fixed x at most countably many [x Iu] are =10 (u E S).
If S is an uncountable orthonormal system, then we list, as earlier, in a series in which Fourier coefficients [x Iu] occur, only those terms for which [x Iu] =I O. It
follows then from Bessel's inequality (74.7) that
LueS l[xlu]1
2
converges and
289 is ~ I X 12 , and with the aid of the Cauchy-Schwarz inequality one obtains that the expression (74.9)
<xly>:= [xly] -
L
[xlu] [uly]
ueS
exists for all X,.r E E. It obviously is a semi-inner product on E. If til' tl 2 , ••. are the clements II from S with [x Ill] "# 0, then it follows from the Bessel identity (74.6) that
<xix>
vanishes if and only if
In this case we write briefly
L
x=
(74.10)
[xlu]u,
ueS
but we keep in mind that the sum of an infinite series is not uniquely determined because of the lack of definiteness of our semi-norm. We state: (74.11)
(74.10) holds ifand only if<xlx> vanishes.
The full symmetry of an operator A has as a consequence that on every eigenspace N(O - A) corresponding to a (real) eigenvalue "# 0 the semiinner product [,1-] is an inner product and that N(O - A) n (eI - A)(E) = {O}. If A is a Wielandt operator, then one can therefore construct an orthonormal basis S~ for N(O - A); the union S:= U~*O S~ is an orthonormal eigensystem of A; furthermore we have the decomposition
e
(74.12) so that 0 position
E
-
= N(O -
A) Ef)
(0 -
A)(E)
A is chain-finite by Proposition 38.4 and that even the decom-
(74.13) is valid; the assertions from (74.12) on are valid trivially also for non-eigenvalues "# o. With the aid of the above eigensystem S we introduce on E the semiinner product (74.9) and the corresponding nullspace N:= {w E E: <wlw> = O}. As an endomorphism of the space (E, the operator A obviously satisfies the hypotheses of Proposition 74.1, and because of N(O - A) c N, we have by (74.13) the decomposition E = N + (0 - A)2(E) for every "# o. Thus every vector y:= A 2 x belongs for every real to N + (0 - A)2(E), hence by Proposition 74.1 it lies in N, and therefore = <xIA 2x> ~ <XIX>1/21/2 = O. It follows now from (74.11) that the expansion
e
m+r
n=m
where the first sum does not vanish and /In < /l for n > m for k --> 00 we have
+ r.
Consequently
hence
IIAk+1xll IIAkXl1 One sees similarly that AkX
AkX /lk
/lk
--=-.---->
IIAkXl1
IIAkXl1
I;:o:,:; (xl un)u n . III;:o:':; (xl un)unll '
the limit is a normalized eigensolution corresponding to the eigenvalue Ji.. By similar considerations we see that the sequence of the Schwarz quotients converges: (A k + lxlx)
(AkXlx) -->/l.
This sequence has the advantage of being monotone increasing, since by Proposition 68.2 we have (Ak+1Xlx)2 = (A k(Ax)lx)2 ~ (Ak(Ax) IAx)(AkX Ix) = (A k+2xlx)(A kxlx). If A is not positive any more, then we can apply our results to the positive operator A 2 and obtain thus all the assertions of the following proposition, except the last one. Proposition 75.1.
If Ax # 0, then
IIA 2k +2xll IIA2kXll
--> IX2
> 0,
At least one of the vectors v:= U
1 + - Au or IX
w:=
1 U -
-
Au
IX
is an eigensoiution of A corresponding to the eigenvalue
IX
or -
IX,
respectively.
293
The last assertion of the proposition follows from the equations Av = lXV, = -am:, which are easy to verify, and from the fact that because of v + w = 2u =I 0, at least one of the vectors V, w does not vanish. •
Aw
We now turn to estimating and comparing eigenvalues. We concentrate on positive eigenvalues. One obtains assertions concerning negative eigenvalues by applying Propositions 75.2 through 75.6 to the operator - A. Proposition 75.2.
least r eigenvalues such that
If IX is a positive and r a natural number, then A has at if and only if there exists an r-dimensional subspace F of E
~ IX
(Axlx)
~
oc(xlx)
x E F;
for all
the eigenvalues are counted according to their multiplicity. Proof. We first assume that A possesses r eigenvalues ~IX; by changing subscripts we may suppose that these are the eigenvalues )11' •.. , )1,. For every element x = I~= I ~pup of the r-dimensional subspace F:= [UI, ... , u,] we have then (Axlx) = L~=I Jlpl~pl2 ~ IX I~=I l~pl2 = IX(xlx). Now let conversely an r-dimensional subspace F = [YI' ... , y,] exist on which (Ax Ix) ~ IX(xlx). We assume that A has only q < r eigenvalues ~IX; by changing subscripts we can obtain that )1n ~ IX for n = 1, ... , q and )1n < IX for n > q. We now determine a non-trivial solution (rl!, ... , '1,) of the system of equations
,
I
~p(Yplu,,) = 0,
(J
= 1, ... , q
p=1
(this is possible because of q < r); for z:= I~= I '1pYP E F we have then for
(J
= I, ... , q,
hence n>q
n>q
(the last estimate follows from Bessel's inequality). Since the two extreme terms of this chain of inequalities agree, we obtain (75.5)
n>q
n>q
hence
L (IX -
n>q
)1n)l(z 1un) 12 = 0,
°
°
from where because of IX - Jln > (n > q) we obtain (z 1 un) = for n > q. . From (75.5) we obtain therefore that (z 1z) = 0, in contradiction to z =I 0. Thus A must indeed have at least r eigenvalues ~ IX. •
294 We now decompose (Jin) into a monotone decreasing sequence of positive and a monotone increasing sequence of negative eigenvalues: Ji1
~
Ji2
~
... < 0
u:,
(each of the two sequences can be finite and one of them even empty); let u;; be the corresponding eigensolutions. From Proposition 75.2 we obtain immediately:
Proposition 75.3. The eigenvalue Jir+ exists and is ~ oc > 0 if and only if there exists an r-dimensional subspace of E on which we have (Ax Ix) ~ oc(x Ix).
If for every non-zero vector x of an r-dimensional subspace F we have (Axlx) > 0, then also oc(F):= min{(Axlx): xEF, Ilxll = I} > 0 (the minimum exists because the function x f--+ (Ax Ix) is continuous on the set {x E F: Ilxll = I}, which according to Proposition 10.6 is compact); then Jir+ exists and is ~oc(F) by Proposition 75.3. For F0:= [ui, ... , ur+] we have J!.r+ = ?-~), since on the one hand for every x = L~=I ~fJu: with Ilxll = ylL~=1 l~fJl2 = lone has (Axlx) = L~=I Ji: l~fJl2 ~ Ji:, and on the other (Au: lun = Ji;. Through these considerations we have proved the Courant maximum-minimum principle: Proposition 75.4.
We have +
Jir
=
. (Ax Ix) max mm -(-1-) ,
F
O*xeF
X X
where F runs through aI/ r-dimellsiollal suhspllces (}f" E 011 which (Ax Ix) > O.fiJr x "# O. The maximum is attainedfor F = [ui, ... , u/]. In particular, Jii exists and is given by +
(Ax Ix) x x
Jil = max -(-1-) x*O
provided that (Ax Ix) is positive for at least one x E E. (Axl x) .
-(--)- IS
xix
..
called the Rayleigh quotient.
The following proposition, which is complementary to Proposition 75.4 in so far as finite-codimensional subspaces occur in it instead of finite-dimensional ones, is called the Courant minimum-supremum principle. Its proof can be left to the reader. Proposition 75.5. then we have
If the right-hand side of the follOWing equation is positive,
Ji+ r
(Ax Ix)
.
= mm sup - - - , F
O*xeFL
(xix)
295 where F runs through all (r - I)-dimensional subspaces of E. The minimum is attained for F = [u u,+_ 1].
t, ... ,
From the minimum-supremum principle we obtain in a simple fashion the Weyl comparison theorem: Proposition 75.6. If the operators A, Band C satisfy the hypotheses (75.1) and (75.2), if a;;, p;; and are their positive eigenvalues arranged in a monotone decreasing order, and if
y;;
A
=
B
+ C,
then the estimates + < a,+.-1 =
p+r + Y.+
are valid.
For the proof let
y;; and
v;;, w;; be the eigensolutions of B, C corresponding to p;;, H:= [wt, ... , w:- 1 ].
Then we obtain with the help of Proposition 75.5 the following chain of inequalities, in which x should satisfy the condition Ilxll = 1: a,++.-1 ~ sup (Axlx) ~ sup (Bxlx)
+
~ sup (Bxlx)
sup (Cxlx) xeFl.
xeFl.
+
xeGl
sup (Cxlx) = xeHl.
P: + y:.
•
To conclude we present an inclusion theorem: Proposition 75.7. If for an x EE with IIxl12 = L~1 l(xlun)1 2 > 0 and for a real polynomial p(t):= a o + a 1t + a2t2 we have (p(A)xlx) = ao(xlx)
+ a 1(Axlx) + a2(A 2xlx)
~ 0,
then the set {t E R: p(t) ~ O} contains at least one non-zero eigenvalue of A. Proof. 00
With the help of (75.3) we obtain from our hypotheses the estimate 00
L P(Jln)l(x I Un) 12 = L [aol(xlun)1 2 + alJlnl(xlun)1 2 + a2Jl;I(xlu n)1 2] n=1 n=1 = (P(A)xlx)
~
0,
and since by assumption at least one I(x Iun) 12 does not vanish, not all P(Jln) can be negative. •
296
Exercises Let the operator A satisf)' the assumptions (75.1), (75.2).
1. State the propositions corresponding to Propositions 75.2 through 75.6 for the negative eigenvalues Jl.;; of A. 2. The number of eigenvalues ~IX > 0 of A is equal to sup{dim F: FeE, (Ax Ix) ~ IX(X Ix) for all x E F}. 3. A is precompact.
§76. General eigenvalue problems for differential operators We study in this section a generalization of the boundary value problem (3.23). Let two differential operators L, M be defined by I
(Lx)(t)
(76.1)
:=
L fv(t)x(V)(t)
for
x E E l := CO)[a, b],
for
x E Em:= c<m)[a, b].
v:O m
(76.2)
(Mx)(t)
:=
L g,,(t)x(")(t) ,,:0
We assume that all functions have values in K, and that iv, gIl are continuous on [a, b]. Under these circumstances the image spaces of Land M lie in E:= C[a, b].
Besides the two differential operators, let two linear maps (76.3)
P: El -+ Kl,
Q: Em -+ Kl
be given; e.g., P could be a boundary value operator defined by [: 1
(76.4)
with Rllx:=
Px:= (RIX, ... , RIX)
L [1X1l\"x(V)(a) + PIl\.x(\")(b)]
v: 0
(cf. (3.19) and (73.2». We now consider the problem to find non-trivial solutions x of the system of equations (76.5)
Lx
=
A.Mx,
Px = A.Qx.
Every A. E K for which such a solution x exists is called an eigenvalue of the problem (76.5), x itself is called an eigensolution corresponding to the eigenvalue A.. Again one has to distinguish between the eigenvalue of.a problem and that of an operator. In order to have something definite before our eyes, we assume I > m and f,(t)
(76.6)
'* 0
for all
t E
[a, b].
Under these assumptions we have (a)
El
C
Em
C
E,
(b) (L - )"M)(EI ) = E for every)" E K, (c) N),:= {x EEl: (L - A.M)x = O} has for every A. E K the same dimension I.
297 The last two assertions result from the theory of linear differential equations. The following considerations are based uniquely on the properties (a) through (c). We will therefore assume, more generally than up to now, that E" Em, E are arbitrary vector spaces over K and that M: Em
L: E,-+ E,
-+
E,
are linear maps; furthermore let assertions (a) through (c) be valid. Under these hypotheses we study the eigenvalue problem (76.5) which with the help of the linear operators (76.7)
L: E,
-+
E
(76.8)
M:Em
-+
Ex K',
X
K'
Lx:= (Lx, Px) Mx := (Mx, Qx)
can be transformed into the equivalent eigenvalue problem Lx = AMx.
(76.9)
We start our investigations with two lemmas. Let
Ni . := {x E E,: (L - AM)x = O}. Lemma 76.1. dim N;. = codim(P - AQ)(N ;.). Proof.
Let {X;'I' ... , x;.,} be a basis of N;. and
(76.10)
for
v = 1, ... , l.
x lies in N;. if and only if
,
x =
L OCvXi."
v=1
,
and
(P - AQ)X =
L ocva;.v =
O.
v= 1
The maximal number of linearly independent elements in N ... is thus equal to the maximal number of linearly independent vectors (OCI' ... , oc,) which satisfy the system of equations L~= 1 ocva;.v = 0.1t is well-known that the latter number is 1- dim[a;'b"" a)./] = 1- dim(P - AQ)(N;.) = codim(P - AQ)(N;.). •
Lemma 76.2. If A = 0 is not an eigenvalue of the problem (76.9), then L exists on E x K'.
1
Proof. Obviously we only have to show that for an arbitrary element (y, 1) from E x K' there exists an x E E, with Lx = (y, 1). Because of (b) there exists in the first place an XI E E, such that Lx 1 = y, and since codim P(N 0) = 0 (Lemma ,76.1), there exists an Xo E No = N(L) such that PXo = 1) - PXI' Consequently 4xI + xo) = y, P(XI + xo) = 1), hence L(xi + xo) = (Y,1). •
If A = 0 is not an eigenvalue of the problem (76.5), then by Lemma 76.2 the linear map
298
exists; we call it the Green operator of problem (76.5). The eigenvalues of G are precisely the reciprocals of the eigenvalues of the problem (76.5); the corresponding eigenspaces are the same. The eigenvalue problem (76.5) amounts
thus to determining the eigenvalues and the eigensolutions of the Green operator. The system of equations (L - AM)x
=
(P - AQ)X
y,
=
1)
(y E E, 1) E K')
is equivalent to the operator equation (I - AG)X = z,
(76.11)
L - l(y, 1).
Z :=
The most important result of this section is: Proposition 76.1.
For all A we have ind(l - AG) = O.
Proof. Because of G(Em) c E, c Em, it is sufficient by Proposition 23.4 to prove the asserted equation for the restriction of G to E,. The subspace N). contains a basis {Ul,"" Ud} of N)., where d:= codim(P - AQ)(N;.) (Lemma 76.1); if d = 0, let {Ul' ..• ,Ud} = 0. We complete this basis of N;. by r:= I - d elements Z1"'" Zr to a basis U1"'" Ud' Z1"'" Zr of N;... We then have (P - AQ)U. = 0
for
3. := (P - AQ)Z. =F 0
v = 1, ... , d,
for
v = 1, ... , r.
The vectors 31' •.. , 3r are linearly independent: From tX131 + ... + tX r 3r = 0 it follows namely that (P - AQ)(tX1Z1 + ... + tXrzr ) = 0, hence tX 1Z 1 + ... + tX r Zr EN;.; according to the construction of the Z. all tX. must vanish. We complete the set {31,"" 3r} by d = 1- r vectors 1)1' ... , l)d to a basis {l)1"'" l)d' 31' ... , 3r} ofK'. By Lemma 76.2 for 1). there exists exactly one y. E E, such that Ly. = (0,1).), v = 1, ... , d. Furthermore by Proposition 16.1 there exist d linear forms xt, ... , x1 on E, such that (Ui' xt) = (jik and (zp, xt) = 0 for i, k = 1, ... , d, p = 1, ... ,r. With these y. and xt we define the finite-dimensional operator S on E, by d
Sx:=
L (x, x:)y• •= 1
(if d = 0 we set S = 0; in this case the remainder of the proof becomes simpler). Finally, let G be the restriction of G to E, and
R
:=
I - AG - S,
hence I - AG = R + S.
We first show that R is injective. From Rx hence (76.12) (L - AM)x
= LSx
=.t1
=
0 it follows that (I - AG)X = Sx,
(x, x:) (0, 1).) = (0,
.~ (x, x:)1).);
299 consequently x lies in N). and has therefore the form
x=
OC 1 Z 1
+ ... + OCrZr + PIUl + ... + {JdUd
so that (P - AQ)X =
(J(131
+ ... + OCr 3r
<x, x:> = P•.
and
It thus follows from (76.12) that
oc 1 31
hence
OC 1
= ... =
(J(r
+ ... + OCr 3r = Pl1)1 + ... + Pd1)d,
= PI = ... = Pd = 0, and so x = O. Next we show that
R(E I ) = E" i.e., that for an arbitrary y E E[ the equation
Rx = y
(76.13)
= y + Sx
(I - AG)X
i.e.,
can be solved by an x EEl' (76.13) is equivalent to the equation (L - AM)x = L(y
+ Sx)
= Ly
(0, .tl <x, x:>1).),
+
which decomposes into the two equations d
(76.14)
(L - AM)x = z,
(P - AQ)X = a
+
L <x, x:>1).,
.=
1
if we set Ly = (z, a). Because of (b) there exists a solution Xo to the first equation; then all its solutions can be represented in the form (76.15)
X = Xo +
r
d
p=1
.=1
L ocpzp + L P.u•.
If we write :to := (P - AQ)X o , then for such an
x
we have
r
(P - AQ)X = :to
+ L (J(p3 p,
<x, x:> = <xo, x:> + P., p=1 so that the second equation in (76.14) takes the form d
r
:to
+ L ocp3p = a + L [<xo, x:> + P.]1). p=1
.=1
or r
d
L OC p3p - .=1 L P.1). = p=1
d
a -
:to
+
L <xo, x:>1) •. .=1
Since {31' .. ·,3" 1)1' ... , l)d} is a basis of Kl, this equation can be satisfied by . appropriate numbers (J(p. P•. The element x formed with these oc p• P. according . to (76.15) then solves (76.13). Taking all together R is thus bijective. We finish the proof with the observation that by Proposition 23.3 we have • ind(I - AG) = ind(R + S) = ind(R) = O.
300
The applicability of the Green operator G is improved by the following fact which is obtained immediately from the results up to now and from Proposition 23.4: Proposition 76.2. Let the hypotheses (a), (b), (c) befuljilledfor the eigenvalue problem (76.S). Assume that A = 0 is not an eigenvalue, and let G be the Green operator of problem (76.5). Furthermore let E be an arbitrary vector space between G(Em) and Em' and let G be the restriction of G to E. Then for all A the relatiun ind(/ - AG) = is valid, and x is an eigensolution of problem (76.5) corresponding to the eigenvalue A if and only if x is an eigensolution of G corresponding to the eigenvalue 1/A.
°
Owing to this propositIOn, the eigenvalue problem (76.S) amounts toexpressed briefly-finding the eigenvalues #0 of any G and the corresponding eigensolutions. The greater flexibility we have gained thereby is felt particularly pleasantly when we want to make the connection with the eigenvalue theory of symmetric operators: In order to make this theory fruitful for dealing with problem (76.S), it is sufficient to introduce an inner or semi-inner product on any vector space E between G(Em) and Em in such a way that G becomes symmetric or fully symmetric, respectively. In particular we have the following: If the hypotheses of Proposition 76.2 are satisfied and if [xly] on E, then G is a Wielandt operator. Consequently (Proposition 74.3) every vector x = Gy with y E E can be expanded into a series of the form
Proposition 76.3.
G isfully symmetric with respect to a semi-inner product
x =
L -,1- [ylu]u = L [xlu]u
ueS A
ueS
with respect to an orthonormal system S of eigensolutions of the problem (76.5) corresponding to its (real) eigenvalues A. The expansion is to be understood in the sense of the limit assertion (74.1S).
We want to investigate more precisely the distribution of eigenvalues in the case that L, M are the differential operators (76.1), (76.2) which satisfy hypothesis (76.6); let K be the complex number field, furthermore let P and Q be boundary value operators, i.e., of the form (76.4). It is well known that the functions x;.!(t), ... , x;./(t) which satisfy the equations (L - AM)x;..
= 0,
x~-I)(a)
= b".
(v, J1.
= 1, ... , I)
form a basis of the nullspace N;. of L - AM, and for every fixed t depend together with their derivatives differentiably on the complex parameter ;.. Consequently the determinant D(it), whose rows are the vectors 0)..:::::: (P - itQ)x;.., is a holomorphic function of A on C. Since because of Lemma 76.1 exactly the zeros of D(A) are eigenvalues of problem (76.S), we have the following situation: Either every complex number is an eigenvalue, or the
301 eigenvalues form an at most countable-possibly empty-set without finite limit points. If A is not an eigenvalue-this is the situation considered from Lemma 76.2 on, in which the Green operator exists-then the second case of this alternative holds and we can apply Proposition 76.3, Proposition 74.3 and the propositions of §75, provided that an inner product can be found on the space E which makes G symmetric. The investigations of this section present in an abstract and generalized form considerations which figured in the course on eigenvalue theory at Ttibingen by H. Wielandt in 1952 we already mentioned (§74). For further developments of these ideas we refer to [157]. Exercise Let L, M be the differential operators (76.1), (76.2) which satisfy the hypotheses (76.6). Let P be the boundary value operator (76.4) and Q = O. Furthermore assume that 0 is not an eigenvalue of the problem (76.5), and let G(s, t) be the Green function of the boundary value problem Lx = y, Px = 0 (see the considerations from (3.20) on). Then the Green operator G is given by (Gy)(s) = J~ G(s, t) [My] (t)dt, and (76.5) is equivalent to finding a non-trivial solution of the equation xes) - J~ G(s, t)[Mx](t) dt = 0 (cf. (3.27), (3.28) for (Mx)(t) := r(t)x(t)). In the case m ~ 1 this equation is a so-called integro-differential equation. §77. Preliminary remarks concerning the spectral theorem for symmetric operators A symmetric operator A of can be represented in the form
en (in other words: a hermitian
(n,
n)-matrix)
(77.1)
here ,11, ... , Ar are the (real and pairwise different) eigenvalues of A, and P p is the orthogonal projector of en onto the eigenspace N(Ap! - A) (cf. §54). Since these eigenspaces are pairwise perpendicular to each other, we have for
(77.2)
p
=1=
(J
(Exercise 1 in §69), furthermore
P 1 +···+Pr =/· . For normal meromorphic, in particular symmetric compact operators on , Ifilbert spaces we have found expansions which are analogous to (77.1). In : the last analysis this was possible because the said operators are richly endowed , With eigenvectors. There exist, however, symmetric operators which have no
302
eigenvalues at all. Naturally the question arises, whether also in such cases an analog of (77.1) exists, and in case of a positive answer, what it looks like. To approach the answer to this question, we recall that the natural generalization of a (finite or infinite) sum is the Stieltjes integral. We will write (77.1) therefore in the form of such an integral and examine then whether a corresponding representation is possible for arbitrary symmetric operators. For this purpose we consider the AI' arranged in monotone increasing order: A, < ..1.2 < ... < Ar. By Proposition 70.8 we have A, = meA), Ar = M(A) (m(A) and M(A) are the bounds of A (see (70.7»; observe that (T(A) = {A" ... , Ar}). We set m := meA), M := M(A), and define a family of endomorphisms E;. for - 00 < A < 00 by
(77.3)
for
A < A, = m
[0P,
for
A,
E;':=I~,+P2
for
..1.2 ~ A < ..1.3
for
Ar-, ~ A < Ar
for
A ~ Ar = M.
IP'+ ... +P.-, P, + ... + Pr = I
~
A < ..1.2
Now let Z.:= {110' ... ' 11.} be a (generalized) partition of the interval i.e., let (77.4)
Em, M],
110 < m < 11, < ... < 11.-, < 11. = M.
Then Eilk - Ellk - = L PI" where the sum is taken with respect to those subscripts p for which I1k- I < AI' ~ 11k. With the aid of (77.1) we obtain immediately, that for any choice of intermediate points AI. E [l1k-" 11k] and for any sequence of partitions Zn with max (11k - I1k-') -+ 0 we have 1
L• Ak(Ellk -
k='
Ellk _,)
=>
A.
Postponing a rigorous definition of the Stieltjes integral with an operator-valued integrating function, we write for this limit relation more briefly (77.5)
A
=
f:-/
dE;..
We also state some properties of the 'spectral family' (E;.): E;. is an orthogonal projector,
(77.6)
(E;.) is monotone increasing: E;. ;;;i Ell for A ;;;i 11, (E A) is continuous to the right: E H EA = 0 for A < m, EA = I for A.
~
• -+
EA for e -+ +0,
M.
The first property follows because of (77.2) from Exercise 2 in §69, the second follows from Proposition 69.3, the third is trivial and the fourth is already listed
303 in (77.3). We shall see that for every symmetric operator A of a complex Hilbert space a spectral family (E;) can be found, which has all the properties (77.6), and which with the representation (77.5) is valid. In the case of the operator A in (77.1), one can represent E .. very easily as a function of A. Since for every polynomial cp(A):= 0(0 + 0(1.1 + ... + O(nAn one has obviously cp(A) = I~= 1 cp(Ap)P p' for a polynomial cpiA) such that for Ap ~ J1. l" 1 lor Ap > J1. we get immediately E/1 = cpiA). In the general case we cannot expect to obtain E .. as a polynomial in A. The considerations so far suggest, however, the following procedure: Define e/1(A) for functions I cpiAp) = { 0
for A ~ J1. for A> J1.'
(77.7)
-00 < J1. < +00,
set E/1 := e/1(A) and check whether (E/1) has the properties (77.6) and whether A can be represented in the form (77.5). This method of pro offor equation (77.5) can be found in [1OJ; its seems to yield the most elementary approach to the 'spectral theorem' (77.5) and therefore we shall make use of it. §78. Functional calculus for symmetric operators
In this section let A be a symmetric operator of the complex Hilbert space H. A is continuous by Proposition 68.6. Let m:= m(A) and M:= M(A). In what follows, all polynomials p(A) will have real coefficients; then p(A) is always symmetric. Basic is the following: Lemma 78.1. If two polynomials p, q satisfy the inequality p(A) all A E [m, MJ, then p(A) ~ q(A).
~
q(A) for
We obviously only have to show thatp(A) ~ oforA. E [m, MJ impliesp(A) ~ O. For this we represent p(A) in the form p(A) = 0( (A - A.), where every zero A. Occurs as often as its multiplicity indicates; to avoid the trivial case, we assume IX :f= O. We introduce:
n
IX j
the zeros
fh the zeros
~
m,
~M,
Yl the zeros in the interval (m, M), (in
= 'n + i'1n the non-real zeros.
Every Yl has even multiplicity, and together with (in also bn is a zero. With appropriate indexing we have (.A. - YI)(.A. - y,+ 1) = (.A. - YI)2
(.A. - (in)(.A. - (in + 1)
=
(.A. - (in)(.A. - ()n)
=
for
(.A. - 'n)2
1 = 1, 3, ... ,
+
'1;
for
n
= 1, 3, ...
304
and so peA) = :1.'
n(A - n(13k - A) n(A rx)
hence also p(A)= rx'
,'/)2
n [(A -
+ ry;] with
(n)2
'Yo'
> 0
k i n
j
n(A -rxjl) n(13k I - A) n(A - i'J)2 n[(A - (n1)2 + ry; 1],
rx' > O.
k i n
We have thus represented peA) as a product of positive, commuting operators; by Proposition 68.9 we have peA) ~ o. • Let now K I be the class of all functions f: [m, M] -+ R for which the following holds: There exists a sequence of continuous functions fn with fn(A) ~ J,.+ 1(,1) ~ 0 and fn(A) -+ f(A)-all this for A E Em, M]. If one approximates the functions gn(A):= fn(A) + lin sufficiently closely by polynomials according to the Weierstrass approximation theorem, then one sees that in this definition 'sequence of continuous functions' can be replaced by 'sequence of polynomials', while everything else remains unchanged. Every function in K I is non-negative, and all non-negative continuous functions on Em, M], but also the step functions ell defined in (77.7), lie in K I' We leave to the reader the simple proof of the following lemma by means of the Heine- Borel covering theorem. Lemma 7S.2. If the functions f, g E K I are approximated according to definition by continuous functions fn, gn, and if f(A) ~ g(A) for all AE Em, M], then for every natural number k there exists an index n such that
for
A E Em, M].
If f is from K I, and (J,.) is a monotone decreasing sequence of polynomials converging to j; then we have by Lemma 78.1 fl(A)
~
f2(A)
~
...
~
0,
consequently (J,.(A» converges pointwise to a symmetric operator B (Proposition 68.7). With the aid of Lemma 78.2 one sees that B does not depend on the choice of the approximating sequence of polynomials. We have thus the right to denote the limit operator B by f(A). The correspondence fr-+ f(A) has the following properties, which follow from the two lemmas or immediately from the definition off(A), respectively: . From f(A)
~
g(A) for all A E Em, M] itfollows thatf(A)
~
g(A);
rxfr-+ rxf(A) for rx ~ 0, f + g r-+ f(A) + g(A), fg r-+f(A)g(A); leA) commutes with g(A) and even with every continuous endomorphism which commutes with A; if f is approximated according to its definition by continuous functions fn' then fn(A) -+ f(A).
305
Let the class K 2 consist of all differences h = f - g of functions f, g E K 1. We set h(A) := f(A) - g(A) and ask the reader to convince himself that this definition is unique, and that K2 and the correspondence h 1--+ h(A) have the following properties (for the proof use the assertions listed above concerning Kl):
K2 is a real algebra, containing C[m, MJ; from g(A.)
~
h(A.) for all A. E em, MJ it follows that g(A)
~
h(A);
IXh 1--+ IXh(A) for all IX E R, g + h 1--+ g(A) + h(A), gh 1--+ g(A)h(A); h(A) commutes with g(A) and even with every continuous endomorphism which commutes with A. We already observed that the functions for for
A. ~ fl A. > fl
lie in K 1. Consequently
E Il := ell(A) exists for every real fl. The operator Ell is symmetric, and because of e; = ell it is idempotent, thus it is an orthogonal projector (Proposition 63.1). It is just as easy to see that the family (Ell) possesses all the other properties listed in (77.6); one only has to use the corresponding properties of the basic functions ell" We call (Ell) the spectral family of the operator A. Exercise +For a positive operator A there exists exactly one positive operator B such that B2 = A. B is called the square root of A and is denoted by A 1/2.
§79. The spectral theorem for symmetric operators on Hilbert spaces
In this section let again A be a symmetric endomorphism of the complex Hilbert space H, and let m := m(A), M := M(A). We consider a (generalized) partition Zn of the interval em, M]: flo<mfor all x E E, y+ E F+ and so it is conjugate to A. Now let A be conjugable and Axo
+
V:= Axo
+ {YEF:
~~~ II ~ e}
an arbitrary (weak) basic neighborhood ofAxo . Then Xo
+
U:= Xo
+ {x E E: ~~~ I<x, A +y:- >I ~
is a (weak) neighborhood of Xo and A(xo + U) = Axo Thus A is weakly continuous at the arbitrary point Xo.
e}
+ A(U) c
Axo
+ V. •
Since A is conjugate to A + with respect to the bilinear systems (F +, F), (E +, E), from Proposition 82.1 we obtain immediately: Proposition 82.2. If (E, E+) is a left dual system and (F, F+) a right dual system, then the weak continuity of A: E -+ F implies the weak continuity of A+: F+ -+ E+. By weak continuity of A + we mean continuity with respect to the topologies a(F+, F) and a(E+, E).
For every subspace G+ of E+ the restriction of the bilinear form of (E, E+-) establishes (E, G +) as a bilinear system, and by the Hausdorff criterion a(E, G +) -< a(E, £+). If (E, E+)-and with it also (E, G+)-is a left dual system, then the set of all a(E, G+)-continuous linear forms coincides with G+, and the set of all a(E, E+)-continuous linear forms coincides with E+. This observation yields: Proposition 82.3. If (E, E+) is a left dual system and G+ a proper subspace of E+, then the topology a(E, E+) is strictly finer than a(E, G+).
321 Exercises 1. If (E, E+) is a dual system, then the finite-dimensional endomorphism K of E is weakly continuous if and only if it can be represented in the form n
Kx
=
L <x, x:)x. v= 1
Hint: Proposition 16.5. +2. If (E, E+) is a left dual system and 4(E) an E+ -saturated algebra of operators, then every A E 0 and a neighborhood U of Xo so that IXX E W
for
IIX - lXo I ~ (j
and
x E U.
Example 83.1. Normed spaces and metric vector spaces are topological vector spaces when they are equipped with their metric topology (consider Exercise 9 in §81). Example 83.2.
The weak topology a{E, E+) is a vector space topology on
E. We first observe that
for a fixed
x+
E
E+
322 defines a semi-norm p on E. The typical basic neighborhood ofthe origin is then given by (83.1)
UPI"",pn;£:={XEE:~~~Pv(X)~t:}
where
p.(x):=I<x,x:>1
(Theorem 82.1). We shall handle it similarly as a ball K£[O] in a normed space. Let now W:=zo + UP1, ... ,Pn;£ be an arbitrary basic neighborhood of Zo:= Xo + Yo (Theorem 82.1). The neighborhoods U:= Xo + U P1 , ... ,Pn;£/2 of Xo and V:=Yo + U P1 , ... ,Pn;£/2 of Yo obviously satisfy U + VC Wand therefore addition is continuous. Now let W:= 1X0Xo + U
with
U:= Up1, ... ,Pn;£
be an arbitrary basic neighborhood of 1X0Xo. We will have proved the continuity of multiplication by scalars if we can exhibit a (j > 0 and a neighborhood of zero V such that (83.2) IXX -
1X0 Xo E
U
whenever
IIX -
1X0 I ~ (j
and
x - Xo E V.
For this we write (83.3)
IXX - 1X0Xo = 1X0(x - xo)
+ (IX -
1(0)xo
+ (IX -
1(0)(x - xo)
and try to make each of the three summands 'sufficiently small'. Setting VI U P1 , ... ,Pn;£/3 we have (83.4)
:=
and setting if if
1X0 = 1X0 =#=
0 0
we have (83.5) Clearly, either Vl is contained in V2 or V2 is contained in Vl, consequently (83.6) is the smaller one of the neighborhoods VI' V2 ; because of (83.5) we have thus (83.7) Finally, there exists a (j > 0 so that IIX - 1X0 I ~ IIX - 1X0 IPv(xo) ~ t:/3 for v = 1, ... , n, hence (83.8) We may obviously choose (j (83.9)
~
1, but then
(j
implies
Pv«1X -
1(0)xo) ==
323
From (83.3) we now obtain with the aid of (83.7) through (83.9) and of (83.4) that ax - ao Xo E VI + VI + VI C V whenever Ia - aO I ~ 0 and x - Xo E V; thus we have proved the continuity of multiplication by scalars. • Example 83.3. The proofs of continuity in the last example were based solely on the semi-norm properties of the P•. This suggests the following generalization: Let a family P of semi-norms p be given on a vector space E. For every finite subset {PI, ... , Pn} of P and every e > 0 let V Pl, ... ,Pn;e:=
{x
E
E:
m~x p.(x) ~ e} v=1
(cf. (83.1». The collection U(y), which consists of all sets of the form y
+
UP1, ... ,Pn;e
= {XEE:
m~xpv(x -
y)
v= I
~ e}
and their supersets, satisfies the neighborhood axioms (Ul) through (U4). The topology defined on E by the neighborhood filters U(y) is called the topology generated by the family P. It is a vector space topology for E, and since for seminorms p we always have Ip(x) - p(y) I ~ p(x - y),
(83,10)
all PEP are continuous with respect to it. One sees, as in the case of weak topologies, that it is Hausdorff if and only iffrom the fact that p(x) = 0 for all PEP, itfollows that x = 0; in this case we say that P is total. The convergence Xn -+ Y is equivalent to p(xn - y) -+ 0 for all pEP. Topologies of normed spaces are included in our construction: We obtain them when P consists only of the given norm. A total sequence P = (PI' P2' ... ) of semi-norms on E generates, besides the vector space topology t I just described, also a metric vector space topology 't 2 by means of the absolute value 1 Pv(x) (83.11) Ixi :=I 2' 1 + Pv{x)
(see Example 80.4, and the remarks made there). 't I coincides with t 2: Indeed, if for the ball Kr[y] we first choose a natural n such that I:':n+1 (1/2') ~ r/2, and then an e > 0 so that t/(1 + t) ~ r/2n for 0 < t ~ e, then for all x E V:= Y + UP1, ... ,Pn;e we have the estimate
Ix - yl
n
=
1
I - - .v=1 21
Pv(x-y) +p.(x-y)
+
ex>
1
p{x-y)
I .' v=n+121+pv(x-y)
r ~ n2n
r
+-
2
= r,
thus V lies in Kr[y] and so 't 2 -< 't l' If now a basic neighborhood V:= y + Up1 .... 'Pn;. of the point y with respect to the topology '1 is given, then it first follows from Ix - yl ~ (1/2n)(e/(I + -1, that P.(x - y) ~ e for v = 1, ... , n, hence
324 X E V. The ball Kr[yJ with r:= (1/2")(8/(1
+ 8»
thus lies in V, and therefore also
'1- 0 and PI' ... , Pn in P so that q(x) ~ y max~~ I Pv(x) for all x E E. Hint: Proof of (82.5). (c) Under the hypotheses of (b), the family of all continuous semi-norms also generates the topology of E. 5. If P = {PI"", Pn} is a finite total set of semi-norms on E, then Ilxll := I~; I Pv(x) defines a norm on E which generates the same topology as P. 6. No system IS of subsets SeN generates the normed space topology of IP, 1 ~ P < 00. +7. Let E be a normed space. On 2(E) we define semi-norms by (a) Ps(A):= SUPxeS IIAxl1 for every bounded subset S of E, (b) qx(A):= IIAxl1 for every x E E, (c) rx.AA):= I1 ~ Il} and observe that it is exactly the
332 F -bounded sets SeT which yield absorbing sets Us .•. Since the collection of all sets Us; .. which belong to a system 6 of F -bounded subsets of T, does in general not form a filter basis, we consider the set 91 of all finite intersections USt ..... Sn ;.:=
nUSv;.={XEF:SuPIl~eforv=
v= 1
1, ... ,n}.
tESv
91 satisfies all conditions of Theorem 85.1, defines therefore a vector space topology on F, which is, of course, nothing but theZ-topology. Exercises 1. Let the collection 9Jl of subsets of the vector space E have the following properties: (a) Every U E 9Jl is absorbing. (b) Every U E 9Jl is balanced. (c) For every U E 9Jl there exists a V E 9Jl such that V + V c U.
n7.=
Then the finite intersections I U,. (U" E 9Jl) form a basis of neighborhoods of zero for a uniquely determined vector space topology on E (cr. Examples 85.1, 85.2). 2. To Example 85.2: If for every S 1, S2 E 6 there exists an S3 E 6 with Sl uS: c S3, then already the sets Us;. (S E 6, l: > 0) form a basis of neighborhoods of zero for the topology generated by 6. 3. Again to Example 85.2: Let 6 1 be the collection of all subsets of the S E 6, and let 6 2 be the collection of all finite unions of sets from 6. Then 6, 6 1 and 6 2 generate the same topology. If T:= E, F:= E+ and (E, E+) is a bilinear system, then let 6 3 be the collection of all sets IXS (IX E K, S E Z), and let 6 4 be the collection of all balanced hulls of the S E 6. Show that Z, 3 3 , 3 4 generate the same topologies on E+. 4. Let (E, E+) be a bilinear system. Among all the 6-topologies on E+ which satisfy USE!! S= E, the topology (1(E+, E) is the coarsest. Hint: §84 Exercise 5b and the above Exercise 3. §86. Subspaces, product spaces and quotient spaces In this section we deal with the question whether certain algebraic operations in and with topological vector spaces (the formation of linear subspaces, product spaces and quotient spaces) again yield topological vector spaces. The proofs of the first two propositions are left to the reader as simple exercises. Proposition 86.1. For a linear subspace F of a topological vector space E the following assertions hold: (a) F is a topological vector space with respect to the induced topology. A basis of neighborhoods of zero 91 in E yields the basis of neighborhoods of zero {UnF: UE91} inFo (b) The closure F is a linear subspace of E.
333 Proposition 86.2. The product of an arbitrary family of topological vector spaces, equipped with the product topology, is a topological vector space. Proposition 86.3. If the quotient space ElF of a topological vector space E with respect to a linear subspace F is equipped with the quotient topology, then the following assertions are valid: (a) (b) (c) (d) h(x). A of zero
ElF is a topological vector space. ElF is Hausdorff if and only ifF is closed. The canonical homomorphism h: E --+ ElF is continuous and open. A neighborhood V of x E E is mapped by h onto a neighborhood h(V) of basis of neighborhoods of zero 91 in E generates the basis of neighborhoods {h(U): U E 9l} in ElF.
We first prove (c). By Example 81.4 the map h is continuous. 1£ M c E is open then h- 1(h(M» = M + F = UXEF (M + x) is also open, from where it follows that h(M) is open; thus h is an open map. (d) follows from (c). (a) Let Wbe a neighborhood of h(x) + h(y) = h(x + y). Then h- 1(W) is a neighborhood of x + y, hence there exist neighborhoods U of x, V of y so that U + V C h-l(W). It follows that h(U) + h(V) = h(U + V) c h(h- 1(W» c W. Because of (d) we obtain from here the continuity of addition. The continuity of multiplication by scalars can be seen just as easily. (b) If ElF is Hausdorff, then {h(O)} and thus also F = h- 1(h(O» is closed. Let, conversely, F be closed and h(x) #- O. Then x does not lie in F, hence there exists a neighborhood V of x such that V n F = 0. For every y E V we have therefore y ¢ F and so h(y) #- O. Thus h(V) is a neighborhood of h(x) (see (d» which does not contain O. Then h(x) - h( V) is a neigh borhood of zero which does not contain h(x), and it follows now from Proposition 84.3 that Ej F is Hausdorff.
•
Unless something else is said explicitly, we always equip subspaces, products and quotients of topological vector spaces with the topologies we gave them in this section: the relative -topology, the product topology and the quotient topology.
Exercises + 1. The product of at most countably many metric vector spaces is again a metric vector space. +2. The product of infinitely many normed spaces is not normable (i.e., the product topology does not derive from a norm). Hint: Otherwise there would exist a neighborhood of zero V:= VI c {x: Ilxll ~ I}. Choose now an x#-O With IXX E U for all IX. + 3. Let F be a closed subspace of the normed space E. Then the quotient norm on ElF generates the quotient topology.
n
334
§87. Continuous linear maps of topological vector spaces If E, F are topological vector spaces over K, then we define, as in the case of normed spaces: 2(E, F) the set of all continuous linear maps A: E
-+
F,
2(E) := 2(E, F), E'
:=
2(E, K).
E', the set of all continuous linear forms on E, is also called the (topological) dual of E.
Proposition 87.1. (a) A linear map is everywhere continuous if it is continuous at the origin. (b) Sums, scalar multiples and products of continuous linear maps are again continuous. In particular, 2(E, F) is a vector space and 2(E) is an algebra. (c) The nullspace of A E 2(E, F) is closed whenever F is a Hausdorff space. Thus the nullspaces of continuous linear forms are always closed. Proof (a) Let W = Ax + V(VaneighborhoodofzeroinF)beanarbitrary neighborhood of Ax. Since A is continuous at 0, with V we can associate a neighborhood of zero U c E so that A(U) c V. For the neighborhood x + U of x we have then A(x + U) c Ax + V = W. (b) Let A, BE 2(E, F) and let W be a neighborhood of zero in F. For W there exists a neighborhood V of zero in F such that V + V c W, and for V there exist neighborhoods U 1, U 2 of zero in E so that A(U 1) C V, B(U 2) C V. Then U:= U 1 n U 2 is a neighborhood of zero in E for which (A + B) (U) c A(U) + B(U) c V + V c W. It follows that A + B is continuous at 0, and because of (a), everywhere. We leave the proof of the remaining assertions in (b) to the reader. (c) {O} c F is closed if F is Hausdorff, the same is then valid for A -1({0}) =NW •
Continuity and openness of a linear map A can be recognized by means of its canonical injection A (cf. Proposition 21.3): Proposition 87.2. A E geE, F) is continuous or open if and only if the corresponding canonical injection A: E/N(A) -+ F is continuous or open, respectively. Proof we have (87.1)
With the aid of the canonical homomorphism h from E onto E/N(A) A =
Aoh.
The statement about continuity follows now from the proposition following Example 81.4, and the assertion concerning openness by means of Proposition ~k
•
335 Continuously projectable subspaces and topological complementary spaces (or complements) are defined as in §24 and just like there, it is true that exactly the continuously projectable subspaces possess topological complements. The proof of Proposition 24.1 can be taken over word for word. When doing this, one sees that it also covers the generalization in Exercise 1 of §24; thus we have:
Proposition 87.3. The continuous linear map A: E -> F is relatively regular, i.e., there exists BE st'(F, E) such that ABA = A, if and only if'it is open and its nullspace and image space are continuously projectable. Through an evident generalization of Proposition 26.1, we obtain from here immediately: Proposition 87.4.
Given A
E
2(E, F), there exists
(a) BE st'(F, E) such that BA = IE if and only if A is injective, open and A(E) is continuously projectable, (b) C E st'(F, E) such that AC = IF if and only if A is surjective, open and N(A) is continuously projectable, (c) BE st'(F, E) such that BA = IE and AB = IF if and only if A is bijective and open. We now turn to linear forms on a topological vector space. The basis of our investigations is the simple
Lemma 87.1. Every non-trivial linear form on a one-dimensional topological vector space E is open. If the topology is Hausdorff, then it is also continuous. We can represent f in the form f(rl.Xo) = rx where Xo #- o. The map rx ~ rxxo is trivially continuous. Therefore f must be open. Now let E be Hausdorff. Then for every e > 0 there exists a balanced neighborhood V of zero in E which does not contain eXo (Proposition 84.3 and Theorem 84.1). If rxxo lies in V, then If(rxx) I = Irxl ~ e (which proves the continuity of f according to Proposition 87.1a); otherwise we would have le/rxl < 1, hence (e/rx)U c V and so eXo = (e/rx)rxxo E V in contradiction to the choice of U. • Proof.
f is bijective, and the inverse map
Proposition 87.5. A non-trivial linear formf on the topological vector space E is open. It is continuous if and only ifit has a closed nullspace. Proof. By §15 Exercise 7 the space E/N(A) is one-dimensional. The assertion Concerning the openness of f now follows immediately from Lemma 87.1 in combination with Proposition 87.2. To see the correctness of the statement about continuity, invoke also Propositions 87.1c and 86.3b. •
336 With the help of Proposition 87.l b the reader immediately sees that for every A E 2(E, F) there exists a map A': F' -+ E' such that (87.2)
(Ax, y') = (x, A'y')
forall
XEE
and
y'EF'.
A' is linear and it is uniquely determined by (87.2) because (E, E') is a left dual system (see end of §16). Thus A is conjugable. By the conjugate of A we understand always, unless something else is explicitly said, the map A' just defined.
Exercises + 1. If E is a complete metric vector space and if the subspaces E 1 , E2 are closed and algebraically complementary, then they are also topologically complementary. Hint: Closed graph theorem (last assertion in Theorem 80.l). 2. A continuous projector of a topological vector space is always open. + 3. Every finite-dimensional subspace of a Hausdorff topological vector space E has a topological complement if and only if (E, E') is a dual system. Hint: Proof of Proposition 25.4b.
§88. Finite-dimensional topological vector spaces
Basic for this section is: Theorem 88.1. On a finite-dimensional linear space there exists only one separated vector space topology; it is generated by a norm. xn} is a basis of the linear space E, then 11~IXl + .. , + defines a norm on E; let 'N be the topology it generates. A basis ofrN-neighborhoods of zero is formed by the balls Kr:= {x E E: Ilxll ~ r}. Now let, be a second Hausdorff vector space topology on E, and let 91 be a basis of r-neighborhoods of zero with the properties (a) and (b) listed in Theorem 84.1. By Lemma 85.1 for every U E 91 there exists a V E 91 such that Proof.
~nxnll
(88.1)
:=
If
{Xl, ••. ,
max~= 1 I~vl
----V+ ...
+ VC
U,
n terms
and for this (absorbing and balanced) V there exists r > 0 so that rxv E V for v = 1, ... , n. It follows that Kr lies in U; indeed, if ~ lXl + ... + ~nxn E K r , then (IU/r) ~ I, hence ~vxv = (~v!r)rxvE V for v= l, ... ,n and thus ~IXl + ... + ~nxn E U (see (88.1». With the aid of Proposition 84.4 it follows that, -< rN' We now show that conversely 'N -< " i.e., that an arbitrary Kr is always a ,neighborhood of zero. For this it is sufficient to prove that there exists a Vo E which is bounded with respect to the norm, i.e., lies in a K ro , since in this case Kr = (r/ro)Kro contains the ,-neighborhood of zero (r/ro)Vo , and is thus itself a ,-neighborhood of zero. Let us assume that WE 91 lies in no ball around O.
m
337 Then for every natural number k there exists a Wk E W such that Ilwkll ~ k. Because W is balanced, also Vk := (k/llwkll)w k lies in Wand thus Uk := (1/k)Vk lies in O/k)W. Obviously Ilukll = 1; by Proposition 10.4 there exists a subsequence (ukJ which converges with respect to 'N' and a fortiori with respect to the coarser topology" to an element Uo i: O. Because W is balanced, we have W :) t W :) t W :) ... ; it follows therefore from Uk m E (1/k m ) W that Uk m lies in (1/k)Wfor all sufficiently large m. Since O/k)Wis closed, also Uo lies in (1/k)W, hence kuo E W for k = 1, 2, .... From here it follows, if we use again that W is balanced, that W contains the one-dimensional subspace [uo]. The existence of a neighborhood Vo E 91 which is bounded with respect to the norm will therefore certainly be proved, if we can find a ,-neighborhood of zero in which no onedimensional subspaces lie. We now show that such a neighborhood does indeed exist. Since, is Hausdorff, there exists aVo E 91 which does not coincide with E. Let m be the largest dimension of the subspaces contained in V 0; obviously o ~ m ~ n - 1. If m = 0, then we are ready. If, however, m ~ 1, then we choose V E 91 so that V + V c V o. Either V contains only subspaces of dimension ~ m - 1, or it contains exactly one m-dimensional subspace Fm. In the second case there exists an x i: 0 in F m' and aWE 91 which does not contain x (Proposition 84.3). V n W contains a VIE 91, and this V 1 can contain only subspaces of dimension ~ In - 1; indeed, an m-dimensional subspace lying in VI and thus also in V would have to coincide with F m' so that V 1 would contain the vector x in contradiction to the construction of V 1. Thus starting with V 0 we have found in 91 a neighborhood of zero, namely V or VI' which contains only subspaces of dimension ~ In - I. We only have to continue this procedure in order to obtain finally a Vo E 91 in which no on.e-dimensional subspaces lie .
•
Proposition 88.1. Every finite-dimensional subspace F of a Hausdorff topological vector space E is closed. Proof. Let Xo E F and let F 0 be the linear hull of F u {xo}. Then F 0 is a finite-dimensional Hausdorff topological vector space for the relative topology '0; by Theorem 88.1 this topology '0 is generated by a norm (on Fo). Since obviously Xo is also a 'o-closure point of F, the assertion of Proposition 88.1 follows immediately from Proposition 10.5. •
Proposition 88.2. If F is a closed and G a finite-dimensional subspace of the topological vector space E, then F + G is closed. Proof. Let h be the canonical homomorphism from E onto the separated topological vector space E/F (Proposition 86.3b). h(G) is finite-dimensional, hence, because of Proposition 88.1, closed in E/F. Thus F + G = h- 1 [h(G)] is also closed (Proposition 86.3c). •
338 Proposition 88.3. Every linear map A from a finite-dimensional Hausdorff topological vector space E into an arbitrary topological vector space is continuous. In particular E' = E*. Proof. Let {XI"'" xn} be a basis of E. The topology of E is generated by the norm 11~lxI + ... + ~nxnll:= max~=1 I~vl (Theorem 88.1). If Wis an arbitrary neighborhood of zero in F and Va neighborhood of zero such that V + ... + V c: W(sum with n terms; see Lemma 85.1), then there exists a b > 0 such that ~v Axv E V whenever I~v I ~ b, hence
Thus A is continuous at 0 and therefore on E (Proposition 87.1a).
•
Proposition 88.4. If F is a closed, finite-codimensional subspace of the topological vector space E, then every algebraic complement G of F in E is also a topological complement. Proof. Let P project E onto G along F and let P: ElF -+ G be the injection associated with P. Then ElF is a finite-dimensional separated topological vector space (Proposition 86.3b), because of Proposition 88.3 the map P is thus continuous, from where the continuity of P follows with the aid of Proposition 8~2 •
Exercises
1. If E is a finite-dimensional Hausdorff topological vector space with the basis {XI"'" x n }, then the map A defined by A(L~= I ~vxv) := (~I' ... , ~n) is an isomorphism from E onto IP(n), 1 ~ p ~ 00, which is continuous in both directions. 2. Let E be a topological vector space with a 'large' dual space, i.e., let (E, E') be a dual system. Show the following: (a) E is Hausdorff. (b) A continuous endomorphism of E with finite deficiency is relatively regular if and only if it is open and its image space is closed. Hint: Exercise 3 in §87.
§89. Fredholm operators on topological vector spaces In this section let E be a topological vector space. A Fredholm operator in !i'(E) is also called a Fredholm operator on E. Because of Proposition 87.3, the endomorphism A of E is a Fredholm operator on E if and only if it has finite deficiency, it is continuous and open, and has a continuously projectable nullspace and image space. Basic for the theory of Fredholm operators on E is
339 Proposition 89.1.
If E is Hausdorff, then the algebra 2(E) is E'-saturated.
Proof. Let the finite-dimensional endomorphism A of E be represented in the form Ax = L~= 1 (x, x~>yv with the help of linearly independent vectors Yl"'" Yn from E and certain linear forms xt, ... , x:. If all x~ are continuous, then every map x ~ (x, x~> Yv and therefore also their sum A is continuous. Now let conversely A be continuous. The linear form fv on F := [Yl, ... , Yn] defined by f(I~= 1 el'YI') := ev is continuous by Proposition 88.3, thus x~ = fv A is also continuous. • 0
Proposition 89.1 permits to make use of the results of §25, 26, 27 and 39 for Fredholm operators on Hausdorff topological spaces: Theorem89.1. Let E be a separated topological vector space. Then we obtain assertions concerning Fredholm operators on E if in the Propositions and Theorems of §25, 26,27 and 39 we replace d(E) by 2(E), .'F(d(E» by the ideal .1'(E) of all continuous, finite-dimensional endomorphisms, E+ by E', and A + by A'. Where (E, E+) is supposed to be a dual system, we must require that (E, E') is a right dual system. In the latter case-(E, E') a right dual system-A E 2(E) is a Fredholm operator if and only if it has finite deficiency, it is open and has a closed image space (see §88 Exercise 2). Exercise Formulate the theorems, propositions and exercises of §25, 26, 27 and 39 for the case that E is a separated topological vector space, d(E) = 2(E), .~(d(E» = .1'(E) and E+ = E'.
XII
Locally convex vector spaces § 90. Bases of neighborhoods of zero in locally convex vector spaces In the most important examples of topological vector spaces we have investigated so far, the topology was generated in the sense of Example 83.3 by a family P of semi-norms on the vector space E. The canonical basis of neighborhoods of zero in such a topology consists of the sets (90.1)
U Pl • .... Pn;' := {x
E
E:
m~x p.(x) ~ e}, v= I
where {PI' ... , Pn} runs through all finite subsets of P, and e through all positive numbers. The basic neighborhoods (90.1) have a property whose fundamental importance will become increasingly clear in the following sections: they are convex. Topological vector spaces, which have a basis of neighborhoods of zero consisting of convex sets, are called locally convex. They are the most important spaces of functional analysis. The topology of a locally convex space is also said to be locally convex. Topologies which are generated by semi-norms, in particular normed topologies, weak topologies and 6-topologies, are locally convex. For the following investigations we need three lemmas. Lemma 90.1. assertions hold: (a) (b) (c)
(d)
For convex sets K in a vector space E over K the following
x + K and rxK are convex for every x E E and Q:E K. The intersection of convex sets is convex. If K contains the point zero, then the balanced core of K is convex. For rx, {3 ;?; 0 we have (rx + {3)K = rxK + 13K.
Proof (a) and (b) are trivial. (c) follows from (a) and (b) with the aid of Lemma 84.1d. To prove (d) we may assume that rx, {3 > O.Obviously(rx + {3)K c rxK + [3K. But because of the convexity of K we also have (rx/(rx + {3»K + ([3I(rx + [3»K c K, hence rxK + 13K c (rx + {3)K. •
340
341 Because of the above assertion (b), there exists for every M c E a smallest convex set containing M, namely the intersection of all convex sets K :::> M. It is called the convex hull of M, and is denoted by co(M). A sum of the form 1X1Xl + ... + IXnXn with IX. ~ 0 (v = 1, ... , n), 1X1 + ... + IXn = 1, is called a convex combination of the vectors Xl' ... , X n . We leave the simple proof of the next lemma to the reader. Lemma 90.2. The convex hull of a set M consists of all convex combinations of the elements of M. The convex hull of an open set in a topological vector space is open.
A balanced and convex set is called absolutely convex (see Exercise 1). By Lemma 90.1c, the balanced core ofa convex set containing 0 is absolutely convex. Lemma 90.3.
The closure of a convex set K in a topological vector space is
convex. Proof Let x, y E K, IX, f3 > 0 and IX + f3 = 1. We must show that every neighborhood W of IXX + f3y intersects the set K. For this we choose neighborhoods U, V of x, y, respectively, so that IXU + f3V c W. Since x, yare closure points of K, there exist in U n K, V n K points u, v, respectively. Obviously IXU + f3v lies in W n K, hence this intersection is not empty. •
Proposition 90.1. In a locally convex space the absolutely convex, closed, and the absolutely convex, open neighborhoods of zero form bases of neighborhoods of zero. Proof According to Proposition 84.2d, every neighborhood W of zero contains a closed neighborhood V of zero, and by hypothesis V contains a convex neighborhood U of zero. The closure Dlies in V, and is convex according to Lemma 90.3. The balanced core of D is closed by Lemma 84.lf, convex by Lemma 90.1c, and by Proposition 84.2c it is a neighborhood of zero which lies in D, hence a fortiori in W. According to what we have just proved, every neighborhood W of zero contains an absolutely convex neighborhood V of zero, and V contains by Theorem 84.1a' an open neighborhood U of zero. The balanced hull D of U is open by Lemma 84.lfand lies in V. By Lemma 90.2 the set co(D) is also open and is a subset of V, hence a fortiori of W. Since co(D) is obviously balanced and a neighborhood of zero, we have proved also the • second assertion.
With the aid of Lemma 90.1 we obtain from Proposition 90.1 immediately Proposition 90.2. Every point of a locally convex space has a basis of neighborhoods which consists of convex closed sets, and one which consists of convex open sets.
342 Exercises
1. A subset K of a vector space E is absolutely convex if and only if x, y E K ~~ and IIXI + IPI :::£ 1 imply that IXX + py E K. *2. Let A be a linear map from E into F, let M be a convex set in E and N a convex set in F. Then the sets A(M) and A - I(N) are also convex. The corresponding statements are true for absolutely convex sets M, N. *3. If K I, ... , Kn are convex subsets of the vector space E, then co(
UKv)
v=1
= {IXIXI
+ ... + Q(nXn: Xv E Kv, IXv
~ 0 (v =
1, ... , ±IXv I}. n),
=
v=1
4. Let the filter basis ~ on the vector space E consist of absorbing, absolutely convex sets. Then there exists exactly one locally convex topology on E for which 91:= {t:V: t: > 0, V E ~} isa basis of neighborhoods of zero. Example: P a family of semi-norms,
(see (90.1)). §91. The generation of locally com'ex topologies by semi-norms We know that every family P of semi-norms on the vector space E generates a locally convex topology for which all the PEP are continuous (Example 83.3). We will show in this section that conversely every locally convex topology on E is generated by a family of semi-norms. With every absorbing subset V of the vector space E we associate its M inkowski functional p, which is defined by (91.1)
p(X):= inf{Q( > 0: X E IXV}.
p is also called the distance function or gauge of V and will be denoted occasionally more precisely by Pu. Obviously 0 :::£ p(x) < 00. The (open or closed) unit ball V of a normed space generates the Minkowski functional p(x) = Ilxll. Every neighborhood of zero ofa topological vector space has a Minkowski functional.
Proposition 91.1. The Minkowskifunctional p of an absorbing and absolutely convex set V c: E is a semi-norm on E, and we have
(91.2)
{x: p(x) < I} c: V c: {x: p(x) :::£
n
Proof Given two vectors x, y and an arbitrary r. > 0, there exist positive numbers IX, {J such that • p(x) :::£ Q( :::£ p(x)
P :::£
+ t: +e
and
x
E
lXV,
pV. With the aid of Lemma 90.1d it follows that x + y E (IX + p)V, hence p(x + y) ;2; Q( + P ~ p(x) + p(y) + 2e. Since t: was arbitrary, we obtain p(x + y) p(y) :::£
p(y)
and
y
E
:::£ :::£
343
p(x) + p(y). Next we prove the homogeneity property p(AX) = IAlp(x). Because of 0 E V we have p(O· x) = p(O) = inf{IX > 0: 0 E IXV} = 0 =
o· p(x).
For A > 0 we obtain
p(h) = inf{IX > 0: AX
E
IXV: = inf{IX > 0: x
E
i
V};
setting {3:= IX/A we have thus p(AX) = A inf{{3 > 0: x E {3V} = Ap(X). Now let IAI = 1. Since V is balanced, we have in this case AX E IXV X E lXV, from where p(h) = p(x) follows. For an arbitrary A#-O we have, according to what has been shown so far,
(91.3)
p(AX) =
P(IAII~I x) = IAIP(I~I x) =
IAlp(x).
Now we prove (91.2). From p(x) < 1 it follows that there exists a positive IX with IX < 1 such that x E rxV. Since V is balanced, x lies also in V. The second inclusion in (91.2) is trivial because of V = I . V. •
Proposition 91.2.
In a topological vector space E the following assertions
are valid: (a) An absorbing, absolutely convex set VeE is a neighborhood of zero if and only if the semi-norm Pu is continuous. (b) For any absolutely convex neighborhood of zero V we have V = {x: Pu(x) ~ I} if V is closed, and V = {x: Pu(x) < I} if V is open.
Proof (a) Let V be a neighborhood of zero, e > 0 and p:= Pu. The eV is also a neighborhood of zero, and x E eV implies (l/e)x E V, hence p«(l/e)x) ~ 1, (Proposition 91.1) and so p(x) ~ e. Hence p is continuous at 0 and therefore on E (Exercise 4 in §83). Now let p be continuous. Then {x: p(x) < 1} is an open neighborhood of zero, which lies in V because of Proposition 91.1. Consequently V must also be a neighborhood of zero. (b) Let V be an absolutely convex, closed neighborhood of zero and p:= Pu. By Proposition 91.1 we have V c {x: p(x) ~ I}. Now let p(x) ~ 1. Then for 0< rx < 1 we have p(IXX) = rxp(x) < I, hence (91.4)
IXX
E
V
for all
IX
E
(0, 1)
(Proposition 91.1). If x were not in V, then there would exist, since V is closed, a neighborhood V of x such that V (] V = 0. Since the vectors rxx lie in V for all numbers rx sufficiently close to 1, these vectors could not lie in V, in contradiction to (91.4). Thus p(x) ~ 1 implies indeed x E V. Now let the absolutely convex neighborhood V of zero be open. By Proposition 91.1 we have {x: p(x) < I} c V. If x lies in U, then so does a neighborhood x + V of x, therefore there exists a {3 > 0 such that (1 + {3)x E V, from where p(x) ~ 1/(1 + {3) < 1 follows. •
344 From the propositions of this section we obtain easily the following theorem which makes the locally convex topologies in a certain sense analyticaJly ~. accessible. Theorem 91.1. The topology of a locally convex space E can always be generated by a family P of continuous semi-norms on E (e.g., by the family of all continuous semi-norms). E is Hausdorff if and only if P is total.
Proof Let P be the family of all continuous semi-norms on E, let 91 be the canonical basis of neighborhoods of zero for the topology generated by P (i.e., the collection of sets U PI ..... Pn;e:= {x: max~= 1 P.(x) ~ e}, where P. E P and e > 0), and 9Jl the set of all absolutely convex, closed neighborhoods of zero in E. By Proposition 90.1 the collection 9Jl is a basis of neighborhoods of zero in E. Obviously 91 c 9Jl. But the opposite inclusion is also valid: indeed, for every U E 9Jl we have Pu E P by Proposition 91.2, and U = {x: Pu(x) ~ I} E ~1. It follows now with the aid of Proposition 84.4 that P generates the topology of E. The condition for E to be Hausdorffhas already been proved in Example 83.3 .
•
It is very well possible that not every continuous semi-norm is needed to generate the topology of E. In normed spaces, for instance, it is sufficient to consider the single semi-norm p(x):= Ilxll. The canonical basis of neighborhoods of zero 91 can be described particularly easily if P is saturated, i.e., if for any Pl" .. , Pn in P also the semi-norm x 1--+ max~= 1 p.(x) is in P. Then 91 consists of all sets U p;e = {x: p(x) ~ e}, where PEP, e > O. The family of all continuous semi-norms on E is saturated; thus every locally convex topology can be generated by a saturated family of semi-norms. As the first application of Theorem 91.1, we show that the continuity of linear maps on locally convex spaces can be described by means of semi-norms. Proposition 91.3. Let the topologies of the locally convex spaces E, F be generated by the families of semi-norms P, Q. In this case the linear map A: E - F is continuous if and only iffor every q E Q there exists a y > 0 and finitely many Pl"'" Pn from P such that n
(91.5)
q(Ax)
~
y max P.(x)
for all x
E
E.
.=1
Proof Let (91.5) be satisfied, and let v:= {y E F: milX; = 1 qJx) ~ e} be a basic neighborhood of zero in F. For every qp. there exists a yp. > 0 and a continuous semi-norm p(p.) on E such that qp.(Ax) ~ yp.p(p.)(x) for all x E E. It follows that the neighborhood of zero U := n;= 1 {x E E: p(p.)(x) ~ ejyp.} is mapped by A into V, so that A is continuous at 0 and therefore on E. Now let A be continuous and q E Q. Then there exist Pl' ... ,Pn E P and e > 0 so that q(Ax) ~ 1 whenever p(x):= max~= 1 p.(x) ~ e. From here one obtains (91.5) with y:= lie by • the same argument which led from (82.4) to (82.5).
345
From Proposition 91.3 we obtain immediately: Proposition 91.4. If the topology on E is generated by the family of seminorms P, then a linear form f on E is continuous if and only if there exist y > 0 and finitely many PI' ... , Pn from P such that n
(91.6)
If(x)l;,£ ymaxp.(x)
for all
x E E.
• =1
The estimates (91.5), (91.6) become simpler when P is saturated: in this case the semi-norm x ~ max~= I Pv(x) can be replaced by apE P. Exercises + 1. The set of all semi-norms on the vector space E generates a locally convex topology T on E for which the collection of all absorbing, absolutely convex subsets of E is a basis of neighborhoods of zero. T is the finest locally convex topology on E. It is Hausdorff. +2. Let the families of semi-norms P, Q generate the topologies Tp, TQ on E. We have TQ -< Tp if and only if for every q E Q there exists a y > 0 and finitely many PI' ... ,Pn from P so that q(x) ;'£ y max~= I P.(x) on E .. 3. Let P be the set of all continuous semi-norms on tlte locally convex space E. Then the collection of all neighborhoods U p := {x: p(x) ;'£ I} (p E P) is a basis of neighborhoods of zero of E. + 4. A sequence (x n) in a topological vector space E is called a Cauchy sequence, if for every neighborhood U of zero-or for every neighborhood from a basis of neighborhoods of zero-there exists an index no(U) such that Xn - Xm E U for n, m ~ no(U). Every convergent sequence is a Cauchy sequence. E is said to be sequentially complete if every Cauchy sequence in E converges to an element of E. If the topology of E is generated by the family of semi-norms P, then the following hold: (a) (x n) is a Cauchy sequence if and only if p(x n - x m ) --+ 0 as n, m --+ 00 for every pEP. (b) Xn --+ x holds if and only if p(xn - x) --+ 0 for every pEP. (c) Eissequentiallycompleteifandonlyiffromp(xn - xm) --+ Oasn,m --+ 00 for every PEP, it follows that there exists an x E E such that p(xn - x) --+ 0 for every pEP.
§92. Subspaces, products and quotients of locally convex spaces From Proposition 86.1a one obtains immediately: Proposition 92.1. Let the topology of the locally convex space E be generated by the family P of semi-norms. Let F be a linear subspace of E and let P be the family of restrictions p ofp E P to F (i.e., p(x) = p(x)for x E F). Then the relative topology on F is generated by P, it is thus locally convex.
346
n'EJ
.
If we consider that a product K, of convex sets K, c E, is convex in the product space E1 E" then we obtain: it
n,
Proposition 92.2.
c,
The product of locally convex spaces is locally convex.
If p is a semi-norm on the vector space E and if F is a linear subspace of E, then p(x) := inf p(y)
for
x E ElF
}'EX
defines a semi-norm p on ElF, called the quotient semi-norm. The image of V:= {x: p(x) < f.} under the canonical homomorphism h: E -> ElF is then h( V) = {x: p(x) < f.}. From here follows, with the aid of Proposition 86.3d: Proposition 92.3. Let E be a locally convex space whose topology is generated by a saturated family P of semi-norms, and let F be a linear subspace. Then the quotient topology on ElF is locally convex, and is generated by the family of quotient semi-norms p (p E P). Exercise +The product of countably many normed spaces is not a normed space but it is a metric locally convex vector space (see Exercise 1 and 2 in §86). §93. Normable locally convex spaces. Bounded sets A topological vector space is said to be normable if its topology arises from a norm. A normable space is locally convex and possesses a neighborhood of zero which is bounded (for the norm), namely the unit ball V; the collection {f. V: B > O} is a basis of neighborhoods of zero. In this section we shall define the concept of a bounded set in topological vector spaces and show that a separated locally convex space, in which there exists a bounded neighborhood of zero, is always normable. A subset M of the topological vector space E is said to be bounded iffor every neighborhood V of zero-or for every neighborhood from a basis of neighborhoods of zero--there exists an rx > 0 such that M c rxV. In a metric vector space this topological notion of boundedness does not have to coincide with the metric notion of boundedness (being contained in a ball) we define earlier. Thus e.g., (s) is bounded metrically but not topologically. In a normed space, however, the two concepts of boundedness are trivially identical. The following proposition is usually called the Ko/mogorov l1ormability theorem.
Proposition 93.1. A locally convex space E is normable Hausdorff and has a bounded neighborhood of zero.
if and
only
if it
is
347
Proof Since the necessity ofthe conditions is clear after the above considerations, we turn to the proof oftheir sufficiency. Let U be a bounded neighborhood of zero and Van absolutely convex, closed neighborhood of zero contained in U (Proposition 90.1); by Proposition 91.2 we have V = {x: Pv(x) ~ I}. Since V is obviously bounded, for every neighborhood of zero W there exists an I: > 0 so that I:V c W. Thus the collection of sets I:V = {x: Pv(x) ~ I:} is a basis of neighborhoods of zero, and the topology of E is therefore generated by the seminorm Pv. Because E is Hausdorff, Pv is even a norm (Theorem 91.1), and the proof is complete. •
The simple proofs of the next two propositions are left to the reader. Proposition 93.2. bounded:
I n a topological vector space the following sets are always
(a) finite sets and subsets of bounded sets; (b) scalar multiples, unions and sums offinitely many bounded sets; (c) balanced and closed hulls of bounded sets; (d) continuous linear images of bounded sets.
Proposition 93.3. Let the topology of the locally convex space E be generated by the family of semi-norms P. The subset M c E is bounded if and only if every PEP is bounded on M. From here it follows that if (E, E +) is a bilinear system, then M c E is (T(E, E+)-bounded if and only if every linear form x 1--+ (x, x+), x+ E E+, is bounded on M (i.e., if Mis E+ -bounded in the sense of Example 83.6).
Exercises 1. We keep the notation and the hypotheses of Example 83.5. Let the collection 6 of F-bounded subsets of T generate the locally convex topology T on F. The set M c F is T-bounded if and only if the family of functions t 1--+ (tlx) (x E M) is uniformly bounded on every S E 6. . +2. Every Cauchy sequence, in particular every convergent sequence, in a topological vector space is bounded. 3. A subset M of the topological vector space E is said to be B-bounded if for every sequence (x.) in M and every sequence (IX.) converging to zero, one has IX. x. --+ o. Show that a set is bounded if and only if it is B-bounded. + 4. Let E, F be topological vector spaces. A E Y'(E, F) is said to be hounded if in E there exists a neighborhood U of zero with bounded image A( U). Show the following: (a) Bounded maps are continuous. (b) The identity transformation I of a Hausdorff locally convex space E is bounded if and only if E is normable. (c) A continuous finite-dimensional operator A : E --+ F is bounded whenever F is Hausdorff.
348
(d) Sums and scalar multiples of bounded operators are bounded. (e) The product of a bounded operator by a continuous operator is bounde~, + 5. Let the topology of the locally convex space E be generated by tile family P = (p,: I E J) of semi-norms. Show that A E 9'(E) is bounded (see Exercise 4) if and only if there exists apE P and numbers y, > 0 so that p,(Ax) ~ y,p(x) for alII E J and all x E E. In this case A is also said to be p-bounded. + 6. Let the hypotheses of Exercise 5 be given and let A be p-bounded. Set
IAI,:= inf{y, > 0: p,(Ax)
~ y,p(x)}
and
IA 1:= inf{y > 0: P(Ax)
~ yp(x)}.
of course, IA I, and IA I depend also on p; we fix, however, p in this discussion. Show that the set of p-bounded endomorphisms of E forms an algebra .tJ4 p(E), and for A, B E .94 p(E) we have IA + B I, ~ IA I, + IB I" locA I, = Ioc II A I" IAB I, ~ IA 1,1 BI, in particular IAni, ~ IA 1,1 A In-l for n = 1,2, .... If E is even sequentially complete (Exercise 4 in §91) and IA I < 1, then I - A has a continuous inverse defined on E. Hint: For every y E E the sequence Xn:= Y + Ay + ... + Any converges; it appears thatl - A is bijective and that (I - A) - 1 - I is p-bounded. +7. Let A be a bounded endomorphism of the sequentially complete locally convex space E. Then there exists an r > 0 such that U - A has for every IA.I > r a continuous inverse defined on E. Hint: Exercise 6.
XIII
Duality and compactness §94. The Hahn-Banach theorem The following extension theorem of Hahn-Banach is a generalization of Theorem 28.1. It is of basic importance for our further work. Theorem 94.1. For every continuous linear formf on the subspace F of the locally convex space E there exists a continuous linear form g on E such that g(x) = f(x)for all x E F. The proof consists in generating the topology of E and F by semi-norms (Theorem 91.1 and Proposition 92.l) and then applying Proposition 91.4 and Principle 28.1. • A counterpart to Theorem 28.2 is: Xo
Theorem 94.2. Let F be a closed subspace of the locally convex space E and ¢ F. Then there exists a continuous linear formf on E such that f(x) = 0 Proof
for
x
E
F
and
f(xo) = 1.
On the linear hull H of {xo} u F we define a linear form h by h(rxxo
+ x) := rx
(rx
E
K, x
E
F)
(cf. the beginning of the proof of Theorem 28.2). Obviously h(x) = 0
for
x
E
F
and
h(xo) = 1.
We now prove that h is continuous. Xo + F is closed and does not contain o. Consequently there exists an absolutely convex, closed neighborhood of zero U which does not intersect Xo + F, and since U = {x: p(x) ~ I} (Proposition 91.2 with p:= Pu), we must have p(xo + x) > 1 for x E F. In the case rx "# 0, for all x E F we obtain Ih(rxxo
+
x)1
= Irxl
~ IrxlP(xo +~) = p(rxxo +x), 349
350
and this inequality holds obviously also for rx = O. By Proposition 91.4 the continuity of h follows since p, and with it also its restriction to H, is continuo;.~ (Proposition 91.2). We now only need to extend h to E according to Theorem 94.1. • In a separated space E the subspace F := {O} is closed. Therefore we immediately obtain from Theorem 94.2: Theorem 94.3. A separated locally convex space E forms, together with its dual E, a dual system (E, E'). There exist separated, non-locally convex spaces E which have a trivial dual E' = {O}; see [8], vol. I, p. 158. Exercise + A continuous endomorphism of a separated locally convex space E is a Fredholm operator on E if and only if it has finite deficiency, it is open and has a closed image space.
§95. The topological characterization of normal solvability The continuous endomorphism A of the topological vector space E is said to be normally solvable ifit is E-normally solvable in the sense of§29 (observe that the E'-conjugate A' always exists). An (E, E')-orthogonally closed subspace of E is obviously always closed. In the case of a locally convex E also the converse holds (Lemma 29.2 and Theorem 94.2). Because of Proposition 29.1 we can state: Proposition 95.1. The continuous endomorphism A of the locally convex space E is normally solvable if and only if its image space is closed. With the aid of Proposition 82.1 we obtain from Proposition 95.1 immediately Proposition 95.2. Let (E, E+) be a left dual system. The E+ -conjugable endomorphism A of E is E+ -normally solvable if and only if its image space is weakly (i.e.Jor the topology a{E, E+» closed. Propositions 82.1 and 95.2 describe the conjugability and normal solvability, i.e., the central concepts of the theory of solving equations, with the help of the weak topology. Exercises 1. Let E be a separated locally convex space and A a continuous, open endomorphism of E with finite deficiency, whose image space is closed. Then rx(A) = f3(A'), f3(A) = rx(A'), A is E'-normally solvable, A' is E-normally solvable. Hint: Theorem 27.1, Exercise in §94.
351 2. Let E be a topological vector space. A E 2(E) is normally solvable whenever one of the following conditions is satisfied: (a) f3(A) is finite and A(E) closed; (b) (E, E') is a right dual system and A(E) is continuously projectable. + 3. If E is a complete metric vector space, A E 2(E) and f3(A) finite, then A(E) is closed and A is normally solvable. Hint: E = A(E) EEl C. Then E x C is a complete metric vector space, P(x, c) := x, Q(x, c) := C define continuous linear maps P, Q from E x C onto E, C. The map B := AP + Q is continuous. Show with the aid of Theorem 80.1 that E\A(E) = B«E x C)\(E x {O}» is open.
§96. Separation theorems In this section let E be a topological vector space. We recall that every neighborhood of zero U generates a Minkowski functional Pu. The concept of sub linear Junctional occurring in the next lemma has been defined in §28 Exercise 2.
Lemma 96.1. The Minkowski Junctional p oj a convex neighborhood oj zero U in E is sublinear. IJU isJurthermore open, then we have (96.1)
eU = {x: p(x) < e} Jor all e > 0, in particular
lJ
= {x: p(x)
< I}.
Proof The first assertion is obtained by repeating word for word the beginning of the proof of Proposition 91.1. We need to prove the second assertion obviously only in the indicated special case (e = I). If p(x) < I, then there exists an oc such that p(x) ~ oc < 1 and x = ocy for some y E U, consequently also x = ocy + (I - oc)· 0 lies in U; thus we have {x: p(x) < I} c U. The converse inclusion is obtained as in the proof of Proposition 91.2. •
A set M := Xo + F, where F is a linear subspace of E, is called a linear manifold. It is called a hyperplane (through x o) if codim F =·1; in this terminology F itself is a hyperplane through O. It follows immediately from Exercise 7 in §15 that ifJis a linear Jorm # 0 on E andJ(xo) = oc, then Xo
+ N(f) =
{x: J(x) = oc}
represents a hyperplane through x o , and conversely every hyperplane through Xo can be described in thisJashion.f(x) = oc is called the equation of H. The hyper-
plane H is closed if and only ifJis continuous (Proposition 87.5). Theorem 96.1. For every convex, open set K # 0, and every linear manifold M in E which does not intersect K, there exists a closed hyperplane H which contains M and does not meet K. In the first part of the proof let E be real. We may assume without restricting the generality that 0 lies in K, hence that K is a neighborhood of zero. Let p
352 be its Minkowski functional. Furthermore let M = Xo subspace of E. It follows from K n M = 0 that (96.2)
p(xo
+ y)
~
for all
1
+ F, where F is a linear ".-
yEF
(Lemma 96.1) and Xo rt F. Because of the second assertion the scalar 0( in the representation x = O(x o + y of a vector x E F 0 := {O(xo + y: 0( E R, y E F} is uniquely determined, hence we can define a linear form [ on F 0 by [(rxxo
+ y) :=
0(.
We have (96.3)
[(O(x o + y)
this is trivial for :x
~
~
p(O(xo
0, while for
[(O(x o + y) =
0(
0(
+ y)
(0(
E
R; y
E
F);
> 0 it follows from (96.2) that
~ O(p(Xo + ~) =
p(rxxo
+ y).
By Exercise 2 in §28 there exists a linear form g on E such that (96.4) g(x) = [(x)
for
x E F0
and
g(x) ~ p(x)
for
x
E
E.
We now show that g is continuous. For this, let K. be the balanced core of eK (e > 0). The set K. is a neighborhood of zero (Proposition 84.2) such that K. c: eK and K. = - K •. From x E K. it follows because of Lemma 96.1 that Ig(x) I = {g(X) ~ p(x) < e -g(x) = g( -x)
~
p( -x) < e
if g(x) ~ 0 if g(x) < 0,
thus g is indeed continuous, so that H:= {x: g(x) = I} is a closed hyperplane. From (96.1) and (96.4) it follows that K c: {x: g(x) < I}, and since g(x) = 1 whenever x E M, we have M c: H. The hyperplane H satisfies therefore all the requirements of the theort;m. Now let E be complex and let E, be the real space corresponding to E (see Exercise 2 in §4). E, with the topology of E is a topological vector space, M is a linear manifold and K a convex, open set in E,. We assume without restricting the generality that 0 lies in M, i.e., that M is a linear subspace. By what we have proved already, there exists a continuous linear form g on E, such that {x: g(x) = O} contains M and docs not intersect K. If we now define a continuous linear form [on E by[(x):= g(x) - ig(ix) (cf. Part II of the proof of Principle 28.1), then H:= {x: [(x) = O} is a closed • hyperplane which contains M = M n (iM) and does not meet K. Theorem 96.2. Let E be locally convex and K c: E not empty, closed and convex. Thenfor every y rt K there exists a continuous linear form [ on E and an 0( E R such that (96.5)
Re fey)
1. From the hypothesis ~ c: A(K I) it follows that VP'" c: A(K,") for every e E (0, 1) and n = 0, 1, .... One now sees as in the proof of Principle 32.1 (from (32.2) on) that for every y E Vp there exists an x E K 1/(1 _ ,) such that Ax = y, which proves the assertion. • Proposition 97.1. The continuous linear map Afrom the Banach space E into the Banach space F is surjective if and only ({ A' possesses a continuous inverse.
354
Proof We assume that A' has a continuous inverse and show first that (with the notation of the above lemma) • ~.' 1
(97.1 )
Vp c A(K 1 )
for
p:= II(A')-lil'
To do this, we consider a y which lies in Vp but not in A(K 1). The set A(K 1) is convex, being the linear image of a convex set (Exercise 2 in §90), hence by Lemma 90.3 also A(K 1) is convex. Consequently there exists a y' E F' such that Re(y, y') > Re(z, y') for all ZE A(K 1) (Theorem 96.2), in particular Re(y, i) > Re(Ax, y') for all x E K 1 • Let (Ax, y') = rei"" r ~ O. Since with x also e-i",x lies in K 1 , it follows that
Re(y,l) > Re(Ae-i"'x,y') = Ree-i"'(Ax,y') = r = I(Ax,i)l, from where we obtain plly'11 = pll(A')-1 A'y'll ~ pll(A')-IIIIIA'y'11 = IIA'y'1i = sup I(x, A'y') I = sup I(Ax, y') I ~ Re(y, i)
~
I(y, y') I ~ lIylllly'll.
xeKl
Thus we have p ~ Ilyll in contradiction to our assumption. With the aid of Lemma 97.1 it follows from (97.1) that Vp c A(K I), from where we obtain immediately that A(E) = F. Now let A be surjective. If A' would not have a continuous inverse on A'(F'), then by Proposition 7.5 there would exist a sequence (y~) c F' such that Ily~11 = I and IIA'y~11 -> O. If we set (Xn:= max{JIIA'y~ll, I/Jn} and z~:= y~/(Xn' then (97.2)
IIz~11 -> 00,
IIA'z~11 ->
0,
consequently (Ax, z~) = (x, A'z~) -> 0 for every x E E. Because of the surjectivity of A, we obtain from here, with the aid of Proposition 33.1, that (1Iznll) is bounded, in contradiction to (97.2). •
The counterpart to the proposition just proved is Proposition 41.3. Compare also with the proof of Proposition 36.5 and with Exercise 3 in §36. Let A be a continuous endomorphism of the Banach space E. If A(E) is closed, then it follows from Proposition 36.4 that also A'(£') is closed. We now show that the converse is also valid. Let A'(E') be closed. We set F := A(E), define BE !feE, F) by Bx:= Ax for x E E, and show that B(E) = F, which will prove that A(E) is closed. Because of Proposition 97.1, we only have to verify that B': F' -> £' has a continuous inverse. From B(E) = F the injectivity of B' follows in a very simple fashion (see Exercise 1 in §29). Furthermore we obtain from the Hahn-Banach extension theorem that B'(F') = A'(E'), hence B' has a closed image space. With the aid of Lemma 36.1 we now see that (B,)-I is continuous. •
355 We can summarize this proposition and Propositions 29.3, 36.4 in: Theorem 97.1. Let A be a continuous endomorphism of the Banach space E. Then the following assertions are equivalent: (a) (b) (c) (d)
A(E) is closed. A'(E') is closed. A is E'-normally solvable: A(E) = {y: (y, x') = Ofor all x' E N(A')}. A' is E-normally solvable: A'(E') = {y': (x, y') = Ofor all x E N(A)}.
As the last application, we supply the proof of Proposition 62.3 for the case the closed subset F 'f 0 of the reflexive Banach space E, which occurs there, is not linear but only convex. Obviously, the following hint is sufficient: The minimizing sequence (Yn) c: F, which was introduced in the first part ofthe proof of Proposition 62.3, has a subsequence (YnJ which converges weakly to a vector Yo of E; the assumption that Yo does not lie in F leads immediately to a contradiction because of Theorem 96.2. •
Exercises
1. Let E, F be Banach spaces. The image space of the compact operator K: E -+ F is closed if and only if K has finite rank. Hint: Consider the compact operator K 1: E -+ K(E) defined by K 1 X := Kx for x E E, and use the Propositions 97.1 and 42.2. 2. A compact endomorphism of a Banach space is relatively regular if and only if it has finite rank. Hint: Exercise 1.
§98. Admissible topologies A locally convex topology T on the vector space E IS said to be admissible with respect to the left dual system (E, E+) if E+ is the space of all r-continuous linear forms of E. Obviously u(E, E+) is the coarsest admissible topology. The topology of a locally convex space E is, by definition, admissible with respect to (E, E'). The importance of admissible topologies is based on the fact that sets M in a locally convex space E may have topological properties which can be described with the help of continuous linear forms alone. If M has such a property in one admissible topology, for instance in the weak one, then it has the same property in all of them. We now give an example.
K
Proposition 98.1. Let (E, E+) be a left dual system. Then a convex set c: E is closed either for all admissible topologies or for none of them.
356 Proof We may assume that E and E+ are real, because convexity and closedness are not influenced when we pass to the real space E,. Let K be non-empty and closed with respect to an admissible topology T. A set of the form
{x E E: f(x)
~ ~}
or {x
E
E: f(x)
~ ~},
wheref =F 0 is a T-continuous linear form, is called a closed half-space determined by f and ~. It follows from Theorem 96.2 that K is the intersection of all closed half-spaces H => K. From this representation of K it follows immediately that K is closed for all admissible topologies. • The aim of the following sections is to characterize the admissible topologies and to construct them as S-topologies. Our most important tool in these investigations will be the bipolar theorem to which we now turn. Exercise A closed linear manifold M in a locally convex space is the intersection of all closed hyperplanes H => M. §99. The bipolar theorem Let (E, E+) be a linear system, 6 a family of a{E, E+)-bounded (i.e., E+bounded) subsets of E (see Proposition 93.3). The typical neighborhood of zero U for the S-topology on E+ (Example 83.6) can be written with the aid of certain sets S1"'" Sn from 6 in the following form: U = {x+ E E+: sup I<x, x+) I =
e
n
xeS v
~ e, v =
{x+ EE+: supl<x,x+)1
v= 1
xeSv
1, ... ,
n}
~ 1}.
With every subset M of E we associate its polar MO:={X+EE+:SUP1<x,x+)1 xeM
~ 1}
in E+ (and correspondingly with every subset N of E+ its polar N°:= {x E E: sup I<x, x+)1 x+ eN
in E). With this, U assumes the form n
U = e nS~. v= 1
~ 1}
357 IfEisanormedspace,E+ = E' and S = K1[O], then SO = {X/EE' : Ilx'll ~ 1}: The polar of the closed unit ball ofE is the closed unit ball of E'. We now turn to the study of polar sets. We emphasize that the concept of polarity makes sense only in connection with a bilinear system; this must in principle always be given in advance. MOO := (MO)O is called the bipolar of M. For M Ooo := (MOO)O we do not introduce a special name. Proposition 99.1. Let (E, E+) be a bilinear system and M, M, subsets of E. Then the following assertions hold: (a) (b) (c) (d) (e) (f) (g) (h)
(i) (j)
From Ml C M2 itfollows that M? ::::l M~. Me MOO. MO = MOoo. (r:x.M)O = (l/r:x.)MO for r:x. =I O. (U,eJ M,)O = n,eJ M? MO is absolutely convex. For the balanced hull D of M we have DO = MO. MO is a(E + , E)-closed. MO is absorbing if and only if Mis aCE, E+)-bounded. For linear subspaces M we have MO = M 1-.
Proof (a), (b), (d), (e), (f) and (j) are easy to see; (c) is proved similarly as the equation M1- = M.LH (see Lemma 29.3). (g) -Because of M c D, we have DO c MO by (a). The converse inclusion follows from Lemma 84.1e with the aid of (d) and (e). (h) For every x E E the linear form x+ f--+ <x, x+) is weakly continuous on E+, hence N x := {x+ E E+: I<x, x+) I ~ I} and thus also -Mo = nxeM N x is weakly closed. (i) Let M be weakly bounded. Then by Proposition 93.3 for every x + E E + there exists an r:x. > 0 such that SUPXE M I<x, x + ) I ~ r:x., from where it follows immediately that x + E r:x.Mo: the set MO is absorbing. The implications are reversible. • The following principle, the so-called bipolar theorem, is the central theorem ofthe theory of polarity. We first introduce a definition: If M is a subset of the vector space E, then the intersection of all absolutely convex sets NeE which contain M is itself absolutely convex and is called the absolutely convex hull of M. If E is a topological vector space, then the intersection of all absolutely convex, closed sets NeE which contain M is called the absolutely convex, closed hull of M; it is absolutely convex and closed. Principle 99.1. Let (£, £+) be a bilinear system. Then the bipolar MOO of a non-empty subset M of E is the absolutely convex, a{E, E+)-closed hull of M.
Proof
Let H be this hull. From Proposition 99.1b, f and h, it follows that
H c MOO. Thus we only have to prove that y ¢ H implies y ¢ MOO. By Theorem
358
96.2 there exists a a(E, E+)-continuous linear form f and a real r:t. such that Re f(x) < oe < Re f(y) for all x E H. Because of 0 E H we have 0 = f(O) < oe; for the a(E, E+)-continuous linear form g := (l/oeJf we have therefore Re g(x) < 1 < Re g(y),
(99.1 )
xEH.
If we set g(x) = rei
1 ~ Re(y,x+> = Reg(y) > 1 (see (99.1), we obtain from here that y ¢ MOo. • The following frequently used result is only a special case of the bipolar theorem: Proposition 99.2. Let (E, E+) be a bilinear system and M a linear subspace or only a non-empty, absolutely convex subset of E. Then MOO is the a(E, E+)closure of M. Exercises + 1. Let (E, E+) be a bilinear system and F a linear subspace of E. Then F1is a(E+, E)-closed, and FH is the a(E, E+)-closure of F. 2. Let (E, E+) be a bilinear system and (M,: I E J) a family of non-empty, absolutely convex and a(E, E+)-closed subsets of E. Then (n,EJ M,)O is the absolutely convex, a(E+, E)-closed hull of M? 3. The absolutely convex hull of a set M is the convex hull of the balanced hull of M. 4. Let (E, E+) be a dual system and A an E+ -conjugable endomorphism of E. Then for M c E, N C E+ we have
U,eJ
and §100. Locally convex topologies areS-topologies Let (E, E+) be a bilinear system and 6 a collection of a(E+, E)-bounded subsets of E+. Then the sets U of the form
n n
(100.1)
U
:=
B
S~,
where
B
> 0, Sv E 6
for
v = 1, ... , n
v= 1
form a basis of neighborhoods of zero for the 6-topology on E (see beginning of §99). We now raise the following question: Can a given locally convex
-i,"
359
topology , on E be represented as an 6-topology, where 6 is a system of (1(E', E)-bounded subsets of the ,-dual E' of E? We assume that such a system 6 exists. Then for f. > 0 and S E 6 the set Ue'=f.S o = {XEE:SUP1(X,X')1 x'eS
is a ,-neighborhood of zero in E. For an arbitrary neighborhood of zero V (namely V ,= U e) such that (100.2)
Ix'(x)1
= l(x,x')1
~ G
for all
x
E
~ f.} G
> 0 there exists thus a and all x'
V
E
S.
This situation will remind the reader of the concept of equicontinuity. As is well known, a family J offunctions f: [a, b] -+ K is equicontinuous at the point Xo E [a, b] if for every G > 0 there exists a b > 0 such that If(x) - f(xo) I
~
f.
for
Ix -
Xo
I ~ b and all f
E
J.
If E, F are topological vector spaces, then a family :F of maps A: E -+ F is said to be equicontinuous at the point Xo E E if for every neighborhood of zero (or only for every neighborhood of zero belonging to a basis of neighborhoods of zero) V in F there exists a neighborhood of zero V in E such that Ax - Axo
E
V
for
x -
Xo E
V
and (11/
A
E
.Y.
The family will be called shortly equicontinuous if it is equicontinuous at every point of E. Obviously a family of linear maps is equicontinuous if and only if it is equicontinuous at the point zero. Thus we oblain from (100.2) the assertion: The system 0'), which generates the topology" consists of equicontinuous subsets of E'. We will now prove the following theorem: Theorem 100.1. The topology, of the locally convex space E is generated by (f) of all equicontinuous subsets of E'.
the system
The proof will be preceded by two lemmas. The first one shows essentially that (f) generates a topology, because the sets in Gj are all (1(E', E)-bounded. Lemma 100.1. Let E be locally convex. SeE' is equicontinuous if and only if there exists a continuous semi-norm p on E such that (100.3)
I f(x) I ~ p(x)
for all
x
E
E
and all
f
E
S.
In this case S is also (1(E', E)-bounded.
The first assertion is proved similarly as Proposition 91.4. The second assertion follows from (100.3) in combination with Proposition 93.3, since (100.3) shows that S is E-bounded. • Lemma 100.2. Let E be a topological vector space. The set SeE' is equicontinuous ifand only ifthere exists a neighborhood ofzero V in E such that S c Vo.
360
Proof If S is equicontinuous, then there exists a neighborhood of zero U in E such that I (x, x') I ~ 1 for all x E U and all x' E S, thus we have S c Uo. Now let S c UO for a certain neighborhood of zero U in E. For an arbitrary e > 0 we have then I(eX, x') I ~ e for all x E U and x' E S, or expressed differently: We have I(y, x')1 ~ e for all y in the neighborhood of zero eU and all x' E S. Consequently S is equicontinuous. •
We now turn to the proof of Theorem 100.1. By Proposition 90.1 the set 9l of all absolutely convex, closed neighborhoods of zero of E is a basis of r-neighborhoods of zero. Because of Proposition 98.1, every U E 9l is also a(E, E)-closed, hence we have UOO = U (Proposition 99.2). Because of the equicontinuity of UO (Lemma 100.2) it follows that U is a neighborhood of zero in the (fj-topology y and so r -< y. Now let V:= e n~= 1 S~ (e > 0, Sv c E equicontinuous) be a y-neighborhood of zero belonging to the canonical basis. By Lemma 100.2 and Proposition 90.1 there exists r-neighborhoods of zero w., and absolutely convex, closed r-neighborhoods of zero U v such that Sv c W~ and U v c w.,. Because of U~o = U v (see the first part of the proof), it follows first, with the aid of Proposition 99.1a, that Sv c U~, then S~ :::l U v' and finally that V :::l e n~= 1 U v ' Consequently y -< r. • Exercises 1. Let (J.: I E J) be an equicontinuous family oflinear forms on the subspace F of the locally convex space E. Then there exists an equicontinuous family (g,: I E J) oflinear forms on E such that g,(x) = J.(x) for all x E F and alii E J. *2. Let E, F be normed spaces, AlE 2(E, F) for IE J. The family (A,: IE J) is equicontinuous if and only if the family (II A,II : I E J) is bounded.
§101. Compact sets
We first recall some concepts and facts from general topology. Let M be a subset of the topological space E. Every system (fj of open sets GeE such that M C UGe(f; G is called an open cover of M; every subsystem (fj' c (fj, which also covers M, is called a subcover of (fj. The set M is said to be compact if every open cover of M contains a finite subcover. Every subset of a compact set is said to be relatively compact. The subset M of a metric space is compact or relatively compact, respectively, in the sense just defined if and only ifit is (sequentially) compact or (sequentially) relatively compact, respectively, according to the definition given earlier in §1O. Finite sets and finite unions of compact sets are compact. Every closed subset of a compact set is compact. Compact subsets of separated topological spaces are closed, in such spaces a subset M is relatively compact jf and only if M is compact. Compactness is an 'intrinsic property', more precisely: The set M c E is compact if and only if it is compact for its relative topology.
,,'•
361
Let mbe a system of non-empty, closed subsets of a compact set, which is totally ordered with respect to set-theoretical inclusion. Then nMe!IJ M =1= 0. Of basic importance is the theorem of Tikhonov: The topological product of compact topological spaces is compact. Finally, we shall use the fact that continuous images of compact sets are compact, that consequently compactness is preserved when the topology becomes coarser, and a real-valued, continuousfunction on a compact topological space has a maximum and a minimum. We now turn to compact and relatively compact
sets in topological vector spaces. Proposition 101.1. Every compact subset, hence a fortiori every relatively compact subset of a topological vector space E is bounded.
Proof Let M c E be compact, U a neighborhood of zero, and V a balanced, open neighborhood of zero such that V + V c U (Theorem 84.1). Since x + V is open and M c UxeM (x + V), there exist finitely many vectors Xl"'" xn in M such that M c U~= I (xv + V). Furthermore, since V is absorbing, there exists an IX ~ 1 such that Xv E IX V for v = 1, ... , n. Because V is balanced, we have (l/IX)V c V, and so we obtain finally
M c
v~/Xv + V) c
IXV
+ V = IX{VI + ~ : VI' V2 E
V} c IX(V
+ V)
c IXU .
• Proposition 101.1 cannot be reversed. We have, however: Proposition 101.2. If we equip the algebraic dual E* of the vector space E with the weak topology o{E*, E), then a subset of E* is relatively compact if and only if it is bounded.
Because of Proposition 101.1, we only have to show that a bounded set Me E* is relatively compact. If {x,: I E J} is an algebraic basis of E, then the spaces E* and K J:= n,eJ K, (K,:= K for alii E J) are isomorphic to each other by means of the map x* 1--+ «x" x*): IE J) (Exercise 9 in §15), hence they can be identified. Thus E* carries in a natural way two locally convex topologies, namely the weak topology a:= a(E*, E) and the product topology r of KJ.
We now show that these two topologies coincide. For this purpose, we first prove the relation a -< • by showing that every X E E is a .-continuous linear form on E*. Let 8 > 0 be given arbitrarily. We represent x in the form x = L,eJo e,x" where J 0 is a subset of J having n elements and =1= ofor I E J o . Then
e,
U:= {(IX: I E J)'I'" I
•
""I
I< = _8_ nle,l
for
IE
J o}
362 is a ,-neighborhood of zero. For every linear form x* = (ex,:
Ix(x*)1 = l<x,x*)1 =
I E
I~>i~,1 ~ L lex,II~,1 ~ leJa
J) in U we have
..
~
e,
lEJa
which proves the ,-continuity of x. Now, in order to show that , -< a, let U:= {(ex,): lex,,.! ~ e for v = 1, ... , n} be a ,-neighborhood of zero in E*. Because of U X't",X,,,;< = {x* E E*: I<Xlv' x*) I ~ e for v = 1, ... , n} = U, we see that U is also a a-neighborhood of zero, thus, is indeed coarser than a. Because of the a-boundedness of M, and by Proposition 93.3, the sets M,:= {x.(x*): x* E M} c K are bounded, hence their closures M, are compact. Tikhonov's theorem now shows that O'EJ M, is ,-compact, hence by what has been proved above, also a-compact. Thus M as a subset ofO'EJ M, is relatively • compact. Proposition 101.3. If K I , .•• , Kn are compact and convex (absolutely convex) subsets of a topological vector space, then also the convex (absolutely convex) hull of U~= I Kv is compact. Let the sets Kv be convex. The collection A of all vectors E Kn with ex v ~ 0 (v = 1, ... , n), ex l + ... + exn = 1 is compact; thus by Tikhonov's theorem the product A x K 1 X .•. x Kn is compact, hence also its image under the continuous map (a, Xl, ••• , X n ) f-+ L~= 1 exvXv' But this image is co(U~= 1 Kv) (Exercise 3 in §90). Now let the Kv be even absolutely convex. Then co(U~= 1 Kv) is balanced, hence it is already the absolutely convex hull of U~ = 1 Kv and so it is compact by what has been just proved . Proof
a:= (ex l , ... , exn )
•
Exercises 1. The balanced hull of a compact set in a topological vector space is compact. 2. Let K be a compact, A a closed subset of a topological vector space and let K n A = 0. Then there exists a neighborhood of zero V such that (K + V) n (A
+
V) =
0.
3. If K is a compact and A a closed subset of a topological vector space, then K + A is closed. Hint: Exercise 2. + 4. The subset M of the topological vector space E is said to be totally bounded or precompact if for every neighborhood of zero U there exist finitely many points Xl"'" xn in M so that M c Ui= 1 (Xi + U) (cf. Exercise 3 in §42). Show the following: (a) Every compact subset of E is precompact. (b) Precompact subsets of E are bounded. (c) Subsets and finite unions of precompact sets are precompact. (d) The continuous linear image of a precompact set is precompact.
363 5. A subset M := {( ~V), ~~), ... ): I E J} of (s) is relatively compact if and only if for every index k the set {~~): I E J} is bounded. Hint: Diagonal process. The map (~1' ~2' ... ) f--+ ~k is continuous. *6. Let E be a topological space. y E E is called a limit point of the sequence (x n ) c E if in every neighborhood U of y there lie infinitely many terms of the sequence (i.e., if for every neighborhood U of y there exist infinitely many indices n 1 < n 2 < ... such that x nk E U for k = 1,2, ... ). Show that every sequence from a compact set K c E has at least one limit point in K. Hint: Proof by contradiction.
§102. The Alaoglu-8ourbaki theorem It reads as follows: Theorem 102.1. Every equicontinuous subset of the dual E' of a topological vector space E is relatively (J(E', E)-compact. Because of Lemma 100.2, this theorem is proved when we show that the following assertion is valid: Theorem 102.2. Let E be a topological vector space. Then the polar UO formed in E' of a neighborhood of zero U c E is (J(E', E)-compact. Proof Together with the bilinear system (E, E'), we consider also the dual system (E, E*), and denote by UP the polar of U formed in E*. Since U is absorbing and U c UPP (Proposition 99.1b), also UPP is absorbing. Thus because of Proposition 99.1 i the set UP is certainly (J(E*, E)-bounded and thus even relatively (J(E*, E)-compact (Proposition 101.2). But since UP is also (J(E*, E)-closed by Proposition 99.lh, and since (J(E*, E) is furthermore separated, we now obtain the (J(E*, E)-compactness of Up. Obviously UO c UP; but we also have UP c UO. Indeed, for x* E UP it follows from x E IOU that I<x, x*) I ~ 10, whatever 10 > 0 is; consequently x* is continuous and lies therefore in UO. Thus the polar UO, since it coincides with UP, is (J(E*, E)-compact. Now (J(E', E) is the topology induced by (J(E*, E) on E', so that UO must be also (J(E', E)compact. •
The polar in E' of the closed unit ball in the normed space E is the closed unit ball in E' (§99). Thus we obtain from Theorem 102.2: Proposition 102.1. The closed unit ball in the normed dual E' of the normed space E is (J(E', E)-compact. This proposition is the basis of the following representation theorem: Proposition 102.2. Let E be a normed space over K. We equip the closed unit ball K' of E' with the topology induced by (J(E', E); by Proposition 102.1
364
the set K' is compact. Let C(K') be the vector space of all continuous functions f: K' -+ K equipped with the norm Ilfll := sUPx'eK' If(x')I. Then E is isomorphic .;. as a normed space to a subspace of C(K'). Proof For every x E E we define the O'(E', E)-continuous linear form Fx on E' by Fix'):= <x, x'). The restriction fx of Fx to K' belongs then to C(K'). The map x ~ fx is obviously linear and preserves the norm: Ilfxll = supl fix') I = sup IFx(x') I = IlFxll = Ilxll (see (41.3». Ilx' II
x'eK'
•
~1
In an abbreviated form, Proposition 102.2 states that every normed space can be considered as a subspace of a space of continuous functions, which are defined on a compact set and equipped with the supremum-norm.
§103. The characterization of the admissible topologies Let r be an admissible topology on E with respect to the left dual system (E, E +). By Theorem 100.1 the sets of the form e n~= S~ (e > 0, S. a r-equicontinuous subset of E+) form a basis of neighborhoods ofzero for r. With the aid of Lemma 100.2 and Proposition 99.1a, b, one sees immediately that already the neighborhoods e n~=1 u~o (e> 0, U. a r-neighborhood of zero) form a basis of neighborhoods of zero, hence that r is generated by the system 6)0 := {Uo: U is a r-neighborhood of zero}. UO is absolutely convex by Proposition 99.lf and by Theorem 102.2 it is O'(E+, E)-compact. It follows that the system IDl of all absolutely convex and O'(E+, E)-compact subsets of E+ generates a topology on E which is finer than any topology admissible with respect to (E, E+). It is called the Mackey topology and will be denoted by r(E, -E +). We now show that r(E, E+) is admissible with respect to (E, E+), whereby it will be proved that the Mackey topology is the finest admissible-topology (while the weak topology O'(E, E+) is the coarsest admissible topology). For this purpose we prove first
1
Lemma 103.1. The polars of the absolutely convex, O'(E+, E)-compact subsets of E + form a basis of neighborhoods of zero for r(E, E +).
Proof The typical neighborhood of zero, belonging to the basis by which the Mackey topology has been defined, is e n~= K~ (e > 0, Kv c: E+ absolutely convex and O'(E+, E)-compact). The absolutely convex hull H of U~= 1 K. is by Proposition 101.3 again O'(E+, E)-compact, hence the same is true also for the absolutely convex set K := {l/e)H. The assertion follows now from the inclusions obtained from Proposition 99.1:
1
e
nn
K~
.=1
=
e
(n)o U K. .=1
::::>
eHo
(1e )0 =
= -
H
KO.
•
365 For the proof of the admissibility of T(E, E+) we remind the reader that E+ is a linear subspace of E*, hence every subset of E+ is also a subset of E*. Po lars will be formed in the dual system (E, E*). Let now Xo be a T(E, E+)-continuous linear form on E. By Lemma 103.1 there exists an absolutely convex, I1(E+, E)compact subset K of E+ so that I(x, xo) I ~ 1 for x E KO. Consequently Xo lies in KOo. Since I1(E+, E) is the topology induced by I1(E*, E), we can assert the I1(E*, E)-compactness, hence also the I1(E*, E)-closedness of the absolutely convex subset K of E*. From Proposition 99.2 it follows that KOO = K, consequently Xo lies in K, hence also in E+, i.e., the T(E, E+)-dual of E is contained in E +. The converse inclusion follows very simply from the fact that an x + E E + is always I1(E, E+)-continuous, hence a fortiori T(E, E+)-continuous. • We summarize our essential results concerning admissible topologies in the Mackey-Arens theorem: Theorem 103.1. A locally convex topology on E is admissible with respect to the left dual system (E, E+) if and only if it is finer than I1(E, E+) and coarser than T(E, E+). The strong topology peE, E+) as the finest 6-topology on E is finer than T(E, E+) (see Example 83.6 in combination with Proposition 93.3). It is in general not admissible. If for instance F is a normed space, then P(F', F) is the topology v defined by the norm on F' (see end of Example 83.6); the dual of (F', v)-i.e., the bidual F"-does in general not coincide with F.
Exercise +Let (E, E+) be a dual system. Then every E+ -conjugable endomorphism of E is T(E, E+)-continuous. Hint: Let V be an absolutely convex, closed neighborhood of zero in E, where E is equipped with the topology T(E, E+). Then VO is I1(E+, E)-compact. Apply now Propositions 82.1, 82.2, the definition of the Mackey topology, and Exercise 4 in §99. §104. Bounded sets in admissible topologies It follows from Propositions 41.4 and 93.3 that the weakly bounded subsets of a normed space are even bounded. We shall now generalize this assertion to separated locally convex spaces. We begin with a preliminary consideration. Let V be a non-empty, absolutely convex subset of a vector space E. It follows from Lemmas 84.1b and 90.1d that [V] = U:'=l nV. Consequently V is absorbing in [VJ. By Proposition 91.1 the Minkowski functional p of V defined on [V] is a semi-norm. Let Ey be the vector space [V] equipped with the locally convex topology -r:(p) generated by p. By Example 83.3 the semi-norm p is continuous with respect to -r:(p), consequently V is a -r:(p)-neighborhood of zero (Proposition 91.2) and with the aid of Proposition 91.1 it follows now that {eV: e > O} is a basis of the -r:(p)-neighborhoods of zero in Ey.
366 Now let E be equipped with an additional separated vector space topology • and let V be .-bounded. Then for every .-neighborhood of zero W in E there exists an t: > 0 such that t; V c W. It follows that .(p) is finer than the topology induced by • on [V]. Since the latter is Hausdorff (Exercise 5 in 81), a fortiori .(p) must be Hausdorff, hence p is a norm (Theorem 91.1). Now let us assume that V is even .-compact. Then the normed space Ey is complete. To see this, let us consider a Cauchy sequence (x n) in Ey. For every t: > 0 there exists an index no = no(t:) such that (104.1)
for all
n, m ~ no.
Without restricting the generality we assume that (x n) c V. Because of the .-compactness of V the sequence (xn) has a .-limit point Xo E V (Exercise 6 in §101). Since Xm + eV is .-closed, it follows from (104.1) that Xo E Xm + eV for all m ~ no, and from here we obtain that Xm -+ Xo for the topology defined by the norm on Ey. We state this result: Lemma 104.1. Every non-empty, absolutely convex and compact subset V of a separated topological vector space E generates a Banach space E y , whose construction is described above. The topology given by the norm on Ey is finer than the topology induced by E.
We can now prove the theorem announced at the outset: Theorem 104.1. (Theorem of Mackey). (a) If (E, E+) is a dual system, then all admissible topologies on E yield the same bounded sets. (b) A subset of a separated locally convex space E is bounded if and only ifit is aCE, E')-bounded. Proof Since the assertions (a) and (b) are equivalent, it is sufficient to prove (b). For this purpose we only have to show that every aCE, E')-bounded set is also bounded for the topology. of E. Let V be a .-closed, absolutely convex neighborhood of zero in E. Then V is also aCE, E')-closed (Proposition 98.1). Thus we obtain from Proposition 99.2 that V = V Oo • Furthermore V O is absolutely convex and aCE', E)-compact (Proposition 99.1 and Theorem 102.2), so that (E')uo is a Banach space (Lemma 104.1). Now let M c E be aCE, E')-bounded. By Proposition 93.3 every x' E E' is bounded on M, in particular we have
(104.2)
supl<x, x')1
0 such that M c lXV, from where it follows that VO C IXMo. Since MO is the typical neighborhood of zero for the strong topology (J(E', E), we see that VO is bounded with respect to (J(E', E), hence also with respect to (T(E', E"). Consequently the polar (VO)P of VO, formed with respect to (E', E"), is a (J(E", E')-neighborhood of zero. Since V is also (T(E, E')-cIosed (Proposition 98.1), it follows now with the aid of Proposition 99.2 that
(Voy n E = {x E E: I<x, x') I ~ 1 for all x' E VO} = Voo = V. We conclude that V is a y-neighborhood of zero and thus y is finer than r. We now show that conversely r is finer than y (whereby the lemma will be proven completely). The typical (J(E", E')-neighborhood of zero is the polar MP, formed with respect to (E', E"), of a (T(E', E")-bounded subset M of E'. The set M is also (T(E', E)-bounded; hence V:= MO is, because of Lemma 105.1, a barrel in E. But E is barrelled, hence V must be a r-neighborhood of zero, and because of
MP n E = {xEE: I<x, x')1 ~ 1 for all x' it follows that r
E
M}
>- y.
= MO =
V,
•
The main theorem of the theory of reflexive spaces is:
Theorem 105.3. A Hausdorfj'locally convex space is reflexive if and only ifit is barrelled and every bounded subset of E is even relatively (T(E, E')-compact. Proof (a) Let E be reflexive. Then we have r = (J(E, E') by definition, hence E is barrelled by Proposition 105.1. Because of E = E" = (E'p)' the topology (J(E', E) is admissible with respect to (E', E), hence coincides with the Mackey topology r(E', E) (Theorem 103.1 and the remark following it). Now let M be a bounded subset of E. By definition MO is a (J(E', E)-neighborhood of zero, hence also a r(E', E)-neighborhood of zero in E'. By Lemma 103.1 there exists an absolutely convex, (T(E, E')-compact subset K of E such that KO C MO.
371 It follows that Me MOO c K OO = K, hence M is indeed relatively a{E, E')compact. (b) Now let E be barrelled and every bounded subset M of E even relatively aCE, E')-compact. We have then {3(E', E) = T(E', E), hence the strong topology {3(E', E) is admissible on E' with respect to the dual system (E', E). It follows that E" = E, i.e., E is semi-reflexive. Since {3(E", E') = {3(E, E') coincides with the • topology of E because of Proposition 105.1, E is even reflexive.
The reader can convince himself without effort that the following theorem holds: Theorem 105.4. A Banach space E is reflexive if and only if its closed unit ball is aCE, E')-compact.
Proposition 10.6 and the last theorem say that compactness properties of the closed unit ball K of a Banach space E determine to a certain extent its destiny: K is compact for the norm ¢> E is finite-dimensional, K is weakly compact
¢>
E is reflexive.
The reader should recall that the closed unit ball in the dual E' is always weakly compact, i.e., aCE', E)-compact (Proposition 102.1). We obtain from Proposition 62.2 that the closed unit ball K of a reflexive Banach space is weakly sequentially compact, i.e., that from every sequence (x n ) c K one can select a subsequence which converges weakly to an x E K. We want to mention without proof that the converse also holds. Thus reflexive Banach spaces can also be characterized by the weak sequential compactness of their closed unit ball; see [8], vol. I, p. 315. Exercises + 1. (s) is a Frechet space. On (s) all 6-topologies formed with respect to the dual system «s), (s)') coincide. 2. Find examples ofFrechet spaces. Hint: Proposition 83.1 and the Examples and Exercises of §83. +3. Every E'-conjugable endomorphism of a separated barrelled space E (e.g., a Frechet or Banach space) is continuous (cr. Proposition 41.5). Hint: Exercise in §103. 4. Let T be a barrel and K an absolutely convex, compact subset of a Hausdorff locally convex space E. Then there exists an IX > 0 such that K c IXT. Hint: TO is T(E', E)-bounded, KO is a T(E', E)-neighborhood of zero. + 5. A continuous endomorphism A of a Fn!chet space is a Fredholm operator if and only if the deficiencies IX(A), {3(A) are finite. Hint: Exercise in §94, Exercise 3 in §95.
372
6. By A(el> e2,"'):= (e2, e3,"') we define a continuous endomorphism oftheFrechetspace(s).Forall..1. =F Owehaveex(A.I - A) = landp(,u - A) = o. It follows from Exercise 5 that,u - A is for all A. =F 0 a Fredholm operator with ind(A.I - A) = 1; cf. Proposition 52.1 and Exercise 8. +7. Let A be a bounded Fredholm operator on a Fn!chet space E. Then for all sufficiently small A. also A - ,u is a Fredholm operator on E with ind(A - ..1.1) = ind(A). Hint: Exercise 6 in §93, proof of Proposition 37.2. + 8. Let A be a bounded endomorphism of a Fn!chet space, and let ex(,u - A), P(,u - A) be finite for all ..1. =F O. Then AI - A is for all A. =F 0 a Fredholm operator with vanishing index. Investigate how far the theory of the Fredholm domain (§51) can be carried over to the present case. Hint: Exercise 7 in §93 and the above Exercise 7. 9. A non-complete metric and locally convex vector space does not have to be barrelled. Hint: Exercise 1 in §33. 10. We consider the complete metric vector space H(/1) of all holomorphic functions on the simply connected domain /1 c C (see Exercise 4 in §80). Show the following: (a) H(11) is a Fn!chet space, hence also barrelled. (b) M c H(/1) is (topologically) bounded if and only if the family offunctions M is uniformly bounded on every compact subset of /1. (c) Every bounded and closed subset of H(/1) is compact (this is precisely the assertion of the famous theorem of MonteZ concerning normal families of holomorphic functions; observe that in a metric space exactly the sequentially compact sets are compact). + 11. Exercise 10 is the motivation for the following definition: A separated locally convex space E is called a MonteZ space if it is barrelled and if every closed, bounded subset is compact. Show the following: (a) H(11) (see Exercise 10) is a Montel space. (b) Normed Montel spaces are finite-dimensional. (c) A Montel space is reflexive. (d) Together with E also E'p is a Montel space.
§l06. Convex, compact sets: The theorems of Krein··Milman and Schauder In this section we shall be concerned first with so-called extremal points and at the end we shall return to the problem of fixed points. If x, yare elements of the vector space E, then S(x, y):= {exx
+ (1 - ex)y: 0 < ex < I}
is called the open and S[x, y]
:=
{exx
+ (1 - ex)y: 0
~
ex
~
I}
373
the closed segment with end-points x, y. The end-points are allowed to coincide; if this is, however, not the case, we speak of a proper segment. Let now M be a subset of E. A point Xo E M is called an extremal point of M, ifit lies on no proper open segment whose end-points belong to M, Le., ifthe assumptions Xo E Sex, y) and x, y E M imply that x = y = Xo. This formulation suggests the following generalization: N c: M is called an extremal subset of M if N is not empty and if the assumptions N n Sex, y) i= 0 and x, y EM imply that x, YEN. Obviously Xo is an extremal point of M if and only if {xo} is an extremal subset of M. The relation 'N is an extremal subset of M' creates an order relation' one can find finitely many points Xl" .. , XII in C so that the following is valid:
°
(106.10)
For every x E C there exists an Xv so that IIx - xvii < e.
We now define functionsfl, ... ,fn on C by
{o,
fv(x):= e - Ilx -
Xvii
Obviously fv is continuous, fv(x) ~ Thus
f(x)
:=
°
if if
Ilx - xvii Ilx - xvii
and L~= 1 fv(X) >
~
e
~ e.
°
(observe (106.10)).
L~= 1 fv(x)xv L~= 1 !v(x)
defines a continuous mapffrom C into Ko:= CO(Xl,"" xn); we have
Ilf(x) - xii ~ e
(106.11)
for all
x
E
C.
B:= f A is a continuous map from K into Ko c K, hence the restriction B of B to Ko is a continuous self-map of the convex, compact subset Ko of the finite-dimensional normed space [Xl"'" XII]. By Lemma 106.2 there exists now a z E Ko such that Bz = /jz = z, hencef(Az) = z. Because of (106.11) we have IIAz - zll = IIAz - f(Az)11 ~ e. We summarize this intermediary result: For every e > there exists a z = z(e) E K such that IIAz - zll ~ e. We now determine according to this assertion for every mEN a point Zm E K such that 0
°
(106.12)
Since AZm lies in the compact set C, there exists a subsequence (zmk) and an x E C so that AZmk -+ x. With the aid of (106.12) it follows that zmk -+ x, and since A
378 is continuous, (AzmJ converges to Ax. The limits x and Ax of (Azm) must • coincide: x is a fixed point of A. From Theorem 106.3 we obtain immediately: Proposition 106.2. Let A be a continuous self-map of the non-empty convex subset K of the normed space E. The map A has afixed point whenever one of the following conditions is satisfied: (a) (b)
K is compact. K is closed and A(K) is relatively compact.
In contrast to Banach's fixed-point theorem, the fixed-point theorems of this section do not require that A shall be a contracting map. They do, however, not yield uniqueness of the fixed point, nor do they give a constructive procedure for obtaining it. Tikhonov has shown that Proposition 106.2 (with condition a) is valid also for locally convex spaces. An easily accessible proof can be found in [4]. It is shown in Exercise 1 that one can obtain the Peano existence theorem from Schauder's Proposition 106.2. We shall prepare a completely different application of the fixed-point theorem by the following considerations. Every eigenspace of an endomorphism A of E is obviously invariant under A. If A has no eigenvalue, then one can still ask whether there exists at least one subspace i=E, {O}, invariant under A, more briefly: whether A has a non-trivial invariant subspace. For a compact A this question got a positive answer in 1954 in [121]. An eigenspace of A has even the property to be invariant under any endomorphism which commutes with A. The question immediately arises whether every compact A has a non-trivial subspace, which is even irivariant under all continuous endomorphisms which commute with A. The answer was given by Lomonosov in 1973 in [131J: Proposition 106.3. For every compact endomorphism A i= 0 of a complex normed space E there exists a non-trivial, closed subspace of E which is invariant under all continuous endomorph isms of E which commute with A. Proof 9l:= {B E 2(E): AB = BA} is obviously an algebra over C. We shall prove by contradiction, and assume therefore that the assertion is false. Then A has certainly no eigenvalue, in particular Ay i= 0 for all y i= o. It follows that Fy:= {By: BE 9l} is for every y i= 0 a closed subspace i= {O}, which is invariant under all BE 9l. For the points x = Xo + y, lIyll ~ 1 of the closed unit ball K around Xo we have IIAxl1 ~ IIAxoll - IIAYII ~ IIAxoll - IIAII; because of A i= 0 we can choose Xo so that the zero vector does not lie in A(K) and thus Fy i= {O} for every y E A(K). We fix Xo in such a way. For every y E A(K) there exists aBE 9l such that IIBy - xoll < 1: If we had, namely, IIBy - xoll ~ 1 for all BE 9l, then we would also have liz - xoll ~ 1 for all Z E F y , thus Xo would not lie in
379
Fy and so Fy would be a non-trivial, closed subspace, invariant under all BE!7l, in contradiction to our assumption that such subspaces do not exist. If we also invoke the continuity of B, we can say: For every Y E A(K) there exists an open ball K(y) around y and a By E!7l so that IIByz - xoll < 1 for Z E K(y). Because of the compactness of A the set A(K) is sequentially compact, hence also compact; consequently A(K) is covered already by finitely many among the balls K(y). It follows that there exist in !7l finitely many operators B l , .. ·, Bn with the following property:
For every y
(106.13)
E
A(K) there exists a Bi such that IIBiy -
xoll
0 whenever y E A(K), because of (106.13). Consequently we can define a map f: A(K) --+ E by f(y):=
(106.14)
Ii=l cp(IIBiy Ii= 1 cp(IIBjy
xoll)Bjy - xoll)
f is continuous, hence f(A(K» is compact. Furthermore sincef(y) is a convex combination of the Bly, ... , Bny, and since Bjy lies in K because of (106.13), we have f(A(K» c f(A(K» c K, so that f "A maps the convex set K continuously into a compact subset of K. By Theorem 106.3 there exists an x E K such thatf(Ax) = x. Thus because of (106.14) we have Ii=l (XjBjAx = x with certain numbers (Xj, and since x, as an element of K, is distinct from the vector zero, this equation means that B := Ii= 1 (Xi B jA has the eigenvalue 1. If we take into account that B is compact, we can assert that dim N(I - B) is positive and finite. Since N(I - B) is obviously invariant under A, the restriction of A to N(l - B), and thus also A itself, has an eigenvalue, in contradiction to our hypothesis. • Exercises 1. Let T:= [to - (x, to + oc], X:= [~o - {3, ~o + {3] and f : T x X --+ R continuous. Then the initial t:alue problem dx/dt = f(t, x), x(t o) = ~o has at least one solution defined in a neighborhood of to (existence theorem of Peano). Hint: The problem is equivalent to solving the (non-linear) integral equation
x(t) =
~o +
I'
fer, x(r»dr.
10
Set J1. := max Tx x I f(t, K
:=
~) 1,15
{x E E: Ix(t) -
:=
min(/3/J1., !X), E
~o I ~ {3},
:=
(Ax)(t)
C[t o - 15, to :=
+ 15],
~o + I'I(r, x(r»dr. 10
K is convex and closed, A a continuous self-map of K. With the help of the
Arzela-Ascoli theorem we see that A(K) is relatively compact.
380 2. We use the notations and hypotheses of Exercise 1 (with the exception of !5). Let J satisfy additionally a Lipschitz condition with respect to the second variable, i.e., let there exist a constant A > 0 so that IJ(t, ~) - J(t, '1) I ~ AI ~ - '11. Let 0 < !5 < min(/3//l, a, 1/..1). Show that A is a contracting map of the complete metric space K. Banach's fixed-point theorem shows now that the initial value problem of Exercise 1 can be solved uniquely and constructively (existence and uniqueness theorem of Picard-LindeI6f). 3. Let F be a closed subset of the metric space E and A a continuous map from F into E with a relatively compact image A(F). Furthermore for every G > 0 let there exist an x = x(c) E F such that d(Ax, x) ~ c. Then A has a fixedpoint.
XIV
The representation of commutative Banach algebras § 107. Preliminary remarks on the representation problem We have seen in Proposition 102.2 that a normed space E is nothing but a subspace of the normed space C(T) of all continuous functions on a certain compact set T. The question arises, when does E coincide with the whole space C(T)? But since C(T) is not only a normed space, but even a commutative Banach algebra (Exercise 1), E can be made in this case in a natural way into a commutative Banach algebra, which differs from C(T) only in the notation of its elements. We shall therefore pose the problem when is E = C(T) only for commutative Banach algebras E. First we make some general comments on the representation problem, which will motivate our procedure and also throw new light on Proposition 102.2. The ge~eral representation problem consists in mapping a set E which is equipped with some structure L (e.g., a group, a normed space, a Banach algebra) onto a set C of concrete, familiar objects (which must of course also carry the structure L, i.e., which must also be e.g., a group, a normed space, a Banach algebra) in such a way that the structure be preserved, i.e., so that the map is a homomorphism. If such a map is possible, then one says that E is represented by C, or that one has found a representation for E. The representation is faithful if the representing homomorphism from E onto C is injective, i.e., if it is even an isomorphism. Only in this case will one have a complete grasp of E by means of C and consider the representation problem as solved. Which set C one chooses is, within certain bounds, arbitrary. In the representation theory of grou~s, C is frequently a set of matrices. To the analyst (real or complex-valued) functions are the most familiar 'concrete' objects, and therefore he will try to represent his structures E by sets offunctions C, whenever possible. The decisive question will then be how one can obtain a function from an element x E E, or somewhat more pointedly, how one can convert x into a function. This conversion is always possible if on E there is defined an-at first entirely arbitrary-non-empty set of K-valued functions. The conversion procedure, 381
382
considered canonical hereafter, consists in associating with every x E E the function Fx: -. K whose 'table of values' is given by the family (f(x): f E