Grundlehren der mathematischen Wissenschaften
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A Series of Comprehensive Studies in Mathematics
Errett Bishop Doug...
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Grundlehren der mathematischen Wissenschaften
279
A Series of Comprehensive Studies in Mathematics
Errett Bishop Douglas Bridges Constructive Analysis
SpringerVerlag Berlin Heidelberg NewYork Tokyo
Errett Bishop
Douglas Bridges
Constructive Analysis
SpringerVerlag Berlin Heidelberg NewYork Tokyo
Errett Bishop t Department of Mathematics University of California, San Diego, USA Douglas Bridges Department of Mathematics University of Buckingham Buckingham, MK18 1EG, England
Mathematics Subject Classification (1980)' 03F65, 46R05, 28C05, 28C10
ISBN 3540150668 SpringerVerlag Berlin Heidelberg New York Tokyo ISBN 0387150668 SpringerVerlag New York Heidelberg Berlin Tokyo
Library of Congress Cataloging in Publication Data Bishop, E rret t 19281983 Con· ,
structive analysis (Grundlehren der mathematischen Wissensch a ften
,
279) An outgrowth
of Foundations of constructive analysis I Errett Bisho p [1967] Bibliography p Inclu des index 1 Mathematical analysis  Foundations I 1945
II
Bridges, D S (Douglas S ),
Bishop, Errett, 19281983 Foundations of constructive analysis
III Title IV Series QA299 8 B57
1985
515
85·2828
ISBN 0·387·15066·8 (U S)
This work is subject to copyright All rights are reserved, whether the whole or part of the material is concerned, specifically those of translation, reprinting, re·use of illustrations, broadcasting, reproduction by photocopying machine or similar means, and storage in data banks Under § 54 of the German Copyright Law where copies are made for other than private use a fee is payable to "Verwertungsge,ellschaft Wort", Munich © Springer· Verlag Berlin Heidelberg 1985
Printed in Germany Typesetting Universitatsdruckerei H Sturtz AG, Wiirzburg Printing GmbH Berlin Bookbinding Liideritz & Bauer·GmbH, Berlin ,
2141/3020543210
MercedesDruck
This book is dedicated to the memory of
Errett Bishop in the hope that it will promote the achievement of a man remarkable as a person and as a mathematician. vivida vis animi pervicit, et extra processit longe Flammantia moenia mundi atque omne immensum Peragravit, mente animoque
.. the vital strength of his spirit won through, and he made his way far outside the flaming walls of the world and ranged over the measureless whole, in both mind and spirit
Preface
This work grew out of Errett Bishop's fundamental treatise' Foundations of Constructive Analysis' (FCA), which appeared in 1967 and which contained the bountiful harvest of a remarkably short period of research by its author. Truly, FCA was an exceptional book, not only because of the quantity of original material it contained, but also as a demonstration of the practicability of a program which most mathematicians believed impossible to carry out. Errett's book went out of print shortly after its publication, and no second edition was produced by its publishers. Some years later, 'by a set of curious chances', it was agreed that a new edition of FCA would be published by Springer Verlag, the revision being carried out by me under Errett's supervision; at the same time, Errett generously insisted that I become a joint author. The revision turned out to be much more substantial than we had anticipated, and took longer than we would have wished. Indeed, tragically, Errett died before the work was completed. The present book is the result of our efforts. Although substantially based on FCA, it contains so much new material, and such full revision and expansion of the old, that it is essentially a new book. For this reason, and also to preserve the integrity of the original, I decided to give our joint work a title of its own. Most of the new material outside Chapter 5 originated with Errett. In particular, there is a full and much improved account, in Chapter 6, of the BishopCheng theory of integration; and there is a completely new approach to the theory of Banach algebras, in Chapter 9 Of special interest also is the last section of Chapter 5, in which are found necessary and sufficient conditions for the existence of a Riemann mapping function. One important part of FCA has been omitted from our new book: the work on martingales and ergodic theory. Errett had some ideas for improving that section, but unfortunately he had no time to put them on paper before he died. Following his advice, and not wishing to delay the publication of this book any longer, I decided to leave
VIII
Prefdce
out that material altogether Jane Bishop has suggested that I revise Errett's ergodic theory for publication separately in an expository paper, I intend to do this in the near future I have retained (with minimal changes) as a Prolog Errett's testamental preface to FCA; I have also retained his first chapter, 'A Constructivist Manifesto'. Errett saw, contributed to, and approved most of the main text of the book outside Chapter 7. In Chapter 7 he made substantial improvements to my original draft of the material on approximation theory (in Section 2) and the RadonNikodym theorem (in Section 3); he never saw the rest of the chapter. The other parts of the book which he did not see are the first section of Chapter 2; the discussion of the Jordan curve theorem preceding Lemma (7.9) of Chapter 5, and the proof of that lemma (which was based on a suggestion of his); Sections 810 of Chapter 6; and the footnotes at the end of each chapter. I do not know if Errett would have included a proof of the Jordan curve theorem. There is a strong case for its inclusion, in order to make the book selfcontained. However, bearing in mind that the intuition underlying the Jordan curve theorem is clear, and that there is an excellent presentation of its proof in a paper by Julian et al. [5], I decided to follow the usual practice in texts on complex analysis, and omit its proof. The perceptive reader will notice that the literary style of this book is different from that of FCA. This is hardly surprising: most of the drafts, and the entire final manuscript, were prepared by me, and it seemed both natural and sensible to use my style, rather than adapt it to Errett's. In doing this, I was not implying any criticism of Errett's style; nor, incidentally, did he ever criticize mine. I take full responsibility for all the material in this book which Errett did not see, and for any errors and omissions in the final version of the text. Although in theory the prerequisites for understanding this book are few (some familiarity with elementary calculus, linear algebra, and the basic notions of abstract algebra), in practice it requires a level of mathematical maturity achieved by few undergraduates. Indeed, a better appreciation of the similarities and contrasts between the classical and constructive developments will be gained by the reader with some experience of classical functional analysis and measure theory. A word about internal references: within the text, a citation of the form (m n) refers to (Theorem, Proposition, or Lemma) n in Section m of the current chapter; a citation from another chapter will have the form' by (m.n) of Chapter ... '
Preface
IX
During the writing of this book, support was provided by the University of Buckingham, New Mexico State University, and Massey University. In addition, I enjoyed the hospitality of the Bishop family, of Bill and Nancy Julian, and of Edna and Bruce Mawson. Bill Julian, Ray Mines, Fred Richman, and Garth Dales read parts of the manuscript, and have given me good advice and much encouragement over the years. Several improvements to the book were brought about by the good offices of Michael Dummett. Julie Cakebread took out of my hands a large share of the burden of typing, which she carried with great patience and skill. Special thanks are due to my wife and children, who have borne with great fortitude my limited contribution to our family life over the past two years. Buckingham, May 1985
Douglas S. Bridges
Contents
Prolog
1
Chapter 1 A Constructivist Manifesto
4
1. The Descriptive Basis of Mathematics 2. The Idealistic Component of Mathematics 3. The Constructivization of Mathematics Notes
13
Chapter 2 Calculus and the Real Numbers
14
1. 2. 3. 4. 5. 6. 7.
14 18
Sets and Functions . . . . . . . . The Real Number System . . . . . Sequences and Series of Real Numbers Continuous Functions Differentiation . . . . . . Integration ...... . Certain Important Functions Problems Notes
Chapter 3
Set Theory
4
6 9
28 35 44'
5c1
55 62 64
67
1. Some Basic Notions of the Theory of Sets 2. Complemented Sets . . . . . . . . . . 3 Neighborhood Spaces and Function Spaces Problems Notes
67 72
Chapter 4
81
Metric Spaces
1. Fundamental Definitions and Constructions 2. Associated Structures . . . . . . . 3. Completeness ......... . 4. Total Boundedness and Compactness
75 78 79
81
87 89 94
Contents
Xl
5. Spaces of Functions 6. Locally Compact Spaces Problems Notes
100 109 121 125
Chapter 5 Complex Analysis
128
1. 2. 3. 4. 5. 6. 7.
128 130 134 142 152 163 180 210 213
The Complex Plane Derivatives Integration The Winding Number Estimates of Size, and the Location of Zeros Singularities and Picard's Theorem The Riemann Mapping Theorem Problems Notes
Chapter 6 Integration
215
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
216 222 232 236 245 252 259 265 277 282 292 297
Integration Spaces . . . . . . Complete Extension of an Integral Integrable Sets . . . . Profiles . . . . . . . . . . . Positive Measures on IR . . . . Approximation by Compact Sets Measurable Functions . . . . . Convergence of Functions and Integrals Product Integrals Measure Spaces Problems Notes
Chapter 7 Normed Linear Spaces
299
1. 2. 3. 4. 5. 6. 7. 8.
299 306 313 334 342 350 357 363 390 396
Definitions and Examples FiniteDimensional Spaces . . The Lp Spaces and the RadonNikodym Theorem The Extension of Linear Functionals Quasinormed Linear Spaces; the Space Loo Dual Spaces . . . . . . . . . . . . Extreme Points . . . . . . . . . . . Hilbert Space and the Spectral Theorem Problems Notes . . . . . . . . . . . . . . .
XII
Contents
Chapter 8
Locally Compact Abelian Groups
399
1. Haar Measure Convolution Operators 3 The Character Group . 4. Duality and the Fourier Transform Problems Notes
419 424 434 447 449
Chapter 9
450
2.
Commutative Banach Algebras
399
1. Definitions and Examples . . . . . 2. Linear Equations in a Banach Algebra Problems Notes .
450
References
463
Symbols
467
Index
453 461 462
. 470
Prolog
Most mathematicians would find it hard to believe that there could be any serious controversy about the foundations of mathematics, any controversy whose outcome could significantly affect their own mathematical activity. Their attitude well represents the actual state of affairs: during a halfcentury of splendid mathematical progress there has been no deviation from the norm The voices of dissent, never much heeded, have long been silent Perhaps the times are not conducive to introspection Mathematics flourishes as never before, its scope is immense, its quality high. Mathematicians flourish as never before, their profession is respectable, their salaries good. Mathematical methods are more fashionable than ever before: witness the surge of interest in mathematical logic, mathematical biology, mathematical economics, mathematical psychology  in mathematical investigations of every sort. The extent to which many of these investigations are premature or unrealistic indicates the deep attraction mathematical exactitude holds for the contemporary mind. And yet there is dissatisfaction in the mathematical community. The pure mathematician is isolated from the world, which has little need of his brilliant creations. He suffers from an alienation which is seemingly inevitable. he has followed the gleam and it has led him out of this world. If every mathematician occasionally, perhaps only for an instant, feels an urge to move closer to reality, it is not because he believes that mathematics is lacking in meaning. He does not believe that mathematics consists in drawing brilliant conclusions from arbitrary axioms, of juggling concepts devoid of pragmatic content, of playing a meaningless game. On the other hand, many mathematical statements have a rather peculiar pragmatic content Consider the theorem that either every even integer greater than 2 is the sum of two primes, or else there exists an even integer greater than 2 that is not the sum of two primes. The pragmatic content of this theorem is not that if we go to the integers and observe, we shall see certain things happening
XII
Contents
Chapter 8 1. 2.
3 4.
Locally Compact Abelian Groups
Haar Measure Convolution Operators The Character Group . Duality and the Fourier Transform Problems Notes
Chapter 9
Commutative Banach Algebras
399 399
419 424 434 447 449 450
1. Definitions and Examples . . . . . 2. Linear Equations in a Banach Algebra Problems Notes .
450 453 461 462
References
463
Symbols
467
Index
. 470
Prolog
Most mathematicians would find it hard to believe that there could be any serious controversy about the foundations of mathematics, any controversy whose outcome could significantly affect their own mathematical activity. Their attitude well represents the actual state of affairs: during a halfcentury of splendid mathematical progress there has been no deviation from the norm The voices of dissent, never much heeded, have long been silent Perhaps the times are not conducive to introspection Mathematics flourishes as never before, its scope is immense, its quality high. Mathematicians flourish as never before, their profession is respectable, their salaries good. Mathematical methods are more fashionable than ever before: witness the surge of interest in mathematical logic, mathematical biology, mathematical economics, mathematical psychology  in mathematical investigations of every sort. The extent to which many of these investigations are premature or unrealistic indicates the deep attraction mathematical exactitude holds for the contemporary mind. And yet there is dissatisfaction in the mathematical community. The pure mathematician is isolated from the world, which has little need of his brilliant creations. He suffers from an alienation which is seemingly inevitable. he has followed the gleam and it has led him out of this world. If every mathematician occasionally, perhaps only for an instant, feels an urge to move closer to reality, it is not because he believes that mathematics is lacking in meaning. He does not believe that mathematics consists in drawing brilliant conclusions from arbitrary axioms, of juggling concepts devoid of pragmatic content, of playing a meaningless game. On the other hand, many mathematical statements have a rather peculiar pragmatic content Consider the theorem that either every even integer greater than 2 is the sum of two primes, or else there exists an even integer greater than 2 that is not the sum of two primes. The pragmatic content of this theorem is not that if we go to the integers and observe, we shall see certain things happening
2
Prolog
Rather, the pragmatic content of such a theorem, if it exists, lies in the circumstance that we are going to use it to help derive other theorems, themselves of peculiar pragmatic content, which in turn will be the basis for further developments. It appears then that there are certain mathematical statements that are merely evocative, that make assertions without empirical validity. There are also mathematical statements of immediate empirical validity, which say that certain performable operations will produce certain observable results: for instance, the theorem that every positive integer is the sum of four squares. Mathematics is a mixture of the real and the ideal, sometimes one, sometimes the other, often so presented that it is hard to tell which is which The realistic component of mathematics  the desire for pragmatic interpretation supplies the control which determines the course of development and keeps mathematics from lapsing into meaningless formalism. The idealistic component permits simplifications, and opens possibilities which would otherwise be closed. The methods of proof and the objects of investigation have been idealized to form a game, but the actual conduct of the game is ultimately motivated by pragmatic considerations. For 50 years now there have been no significant changes in the rules of this game. Mathematicians unanimously agree on how mathematics should be played. Accepted standards of performance suffice to regulate the course of mathematical activity, and there is no prospect that these standards will be changed in any significant respect by a revision of the idealistic code. In fact, no efforts are being made to impose such a revision. There have been, however, attempts to constructivize mathematics, to purge it completely of its idealistic content. The most sustained attempt was made by L.E.1. Brouwer, beginning in 1907. The movement he founded has long been dead, killed partly by compromises of Brouwer's disciples with the viewpoint of idealism, partly by extraneous peculiarities of Brouwer's system which made it vague and even ridiculous to practising mathematicians, but chiefly by the failure of Brouwer and his followers to convince the mathematical public that abandonment of the idealistic viewpoint would not sterilize or cripple the development of mathematics. Brouwer and other constructivists were much more successful in their criticisms of classical mathematics than in their efforts to replace it with something better. Many mathematicians familiar with Brouwer's objections to classical mathematics concede their validity but remain unconvinced that there is any satisfactory alternative. This book is a piece of constructivist propaganda, designed to
Prolog
3
show that there does exist a satisfactory alternative. To this end, we develop a large portion of abstract analysis within a constructive framework. This development is carried through with an absolute minimum of philosophical prejudice concerning the nature of constructive mathematics. There are no dogmas to which we must conform. Our program is simple' to give numerical meaning to as much as possible of classical abstract analysis. Our motivation is the wellknown scandal, exposed by Brouwer (and others) in great detail, that classical mathematics is deficient in numerical meaning. Some familiarity with Brouwer's critique is essential. Following Brouwer, Chapter 1 is primarily devoted to an examination of the defects of classical mathematics, and a presentation of the thesis that all mathematics should have numerical meaning. Chapter 3 presents constructive versions of the fundamental concepts of sets and functions, and examines some of the obstacles to the constructivization of general topology The remaining chapters are primarily technical, and constitute a course in abstract analysis from the constructive point of view. Very little formal preparation is required of the reader, although a certain level of mathematical sophistication is probably indispensible. Every effort has been made to follow the classical development as closely as possible; digressions have been relegated to notes at the ends of the various chapters. The task of making analysis constructive is guided by three basic principles. First, to make every concept affirmative. (Even the concept of inequality is affirmative.) Second, to avoid definitions that are not relevant. (The concept of a pointwise continuous function is not relevant; a continuous function is one that is uniformly continuous on compact intervals.) Third, to avoid pseudogenerality. (Separability hypotheses are freely employed.) The book has a threefold purpose: to present the constructive point of view, to show that the constructive program can succeed, and to lay a foundation for further work. These immediate ends tend to an ultimate goal  to hasten the inevitable day when constructive mathematics will be the accepted norm Weare not contending that idealistic mathematics is worthless from the constructive point of view. This would be as silly as contending that unrigorous mathematics is worthless from the classical point of view. Every theorem proved with idealistic methods presents a challenge: to find a constructive version, and to give it a constructive proof. Errett Bishop
Chapter 1. A Constructivist Manifesto
1. The Descriptive BaSIS of Mathematics Mathematics is that portion of our intellectual actIvIty which transcends our biology and our environment The principles of biology as we know them may apply to life forms on other worlds, yet there is no necessity for this to be so. The principles of physics should be more universal, yet it is easy to imagine another universe governed by different physical laws. Mathematics, a creation of mind, is less arbitrary than biology or physics, creations of nature, the creatures we imagine inhabiting another world in another universe, with another biology and another physics, will develop a mathematics which in essence is the same as ours In believing this we may be falling into a trap. Mathematics being a creation of our mind, it is, of course, difficult to imagine how mathematics could be other than it is without our actually making it so, but perhaps we should not presume to predict the course of the mathematical activities of all possible types of intelligence. On the other hand, the pragmatic content of our belief in the transcendence of mathematics has nothing to do with alien forms of life. Rather, it serves to give a direction to mathematical investigation, resulting from the insistence that mathematics be born of an inner necessity The primary concern of mathematics IS number, and this means the positive integers. We feel about number the way Kant felt about space. The positive integers and their arithmetic are presupposed by the very nature of our intelligence and, we are tempted to believe, by the very nature of intelligence in general. The development of the theory of the positive integers from the primitive concept of the unit, the concept of adjoining a unit, and the process of mathematical induction carries complete conviction. In the words of Kronecker, the positive integers were created by God. Kronecker would have expressed it even better if he had said that the positive integers were created by God for the benefit of man (and other finite beings). Mathematics belongs to man, not to God. We are not interested in properties of the
1 The Descriptive Basis of Mathematics
5
posItIve integers that have no descriptive meaning for finite man. When a man proves a positive integer to exist, he should show how to find it. If God has mathematics of his own that needs to be done, let him do it himself. Almost equal in importance to number are the constructions by which we ascend from number to the higher levels of mathematical existence. These constructions involve the discovery of relationships among mathematical entities already constructed, in the process of which new mathematical entities are created. The relations which form the point of departure are the order and arithmetical relations of the positive integers From these we construct various rules for pairing integers with one another, for separating out certain integers from the rest, and for associating one integer with another Rules of this sort give rise to the notions of set and function. A set is not an entity which has an ideal existence. a set exists only when it has been defined To define a set we prescribe, at least implicitly, what we (the constructing intelligence) must do in order to construct an element of the set, and what we must do to show that two elements of the set are equal. A similar remark applies to the definition of a function: in order to define a function from a set A to a set B, we prescribe a finite routine which leads from an element of A to an element of B, and show that equal elements of A give rise to equal elements of B Building on the positive integers, weaving a web of ever more sets and more functions, we get the basic structures of mathematics: the rational number system, the real number system, the euclidean spaces, the complex number system, the algebraic number fields, Hilbert space, the classical groups, and so forth. Within the framework of these structures most mathematics is done. Everything attaches itself to number, and every mathematical statement ultimately expresses the fact that if we perform certain computations within the set of positive integers, we shall get certain results. Mathematics takes another leap, from the entity which is constructed in fact to the entity whose construction is hypothetical To some extent hypothetical entities are present from the start whenever we assert that every positive integer has a certain property, in essence we are considering a positive integer whose construction is hypothetical. But now we become bolder and consider a hypothetical set, endowed with hypothetical operations subject to certain axioms In this way we introduce such structures as topological spaces, groups, and manifolds. The motivation for doing this comes from the study of concretely constructed examples, and the justification comes from the possibility of applying the theory of the hypothetical structure to the
6
Chapter 1 A Constructivist Manifesto
study of more than one specific example Recently it has become fashionable to take another leap and study, as it were, a hypothetical hypothetical structure ~ a hypothetical structure qua hypothetical structure. Again the motivations and justifications attach themselves to particular examples, and the examples attach themselves to numbers in the ultimate analysis Thus even the most abstract mathematical statement has a computational basis The transcendence of mathematics demands that it should not be confined to computations that I can perform, or you can perform, or 100 men working 100 years with 100 digital computers can perform. Any computation that can be performed by a finite intelligence ~ any computation that has a finite number of steps ~ is permissible. This does not mean that no value is to be placed on the efficiency of a computation An applied mathematician will prize a computation for its efficiency above all else, whereas in formal mathematics much attention is paid to elegance and little to efficiency. Mathematics should and must concern itself with efficiency, perhaps to the detriment of elegance, but these matters will come to the fore only when realism has begun to prevail. Until then our first concern will be to put as much mathematics as possible on a realistic basis without close attention to questions of efficiency
2. The Idealistic Component of Mathematics Geometry was highly idealistic from the time of Euclid and the ancients until the time of Descartes, unfolding from axioms taken either to be selfevident or to reflect properties of the real world. Descartes reduced geometry to the theory of the real numbers, and in the nineteenth century Dedekind, Weierstrass, and others, by the arithmetization of the real number system, brought space into the concrete realm of objects constructed by pure thought Unfortunately, the promise held out to mathematics by the arithmetization of space was not fulfilled, largely due to the intervention, around the turn of the century, of the formalist program. The successful formalization of mathematics helped keep mathematics on a wrong course. The fact that space has been arithmetized loses much of its significance if space, number, and everything else are fitted into a matrix of idealism where even the positive integers have an ambiguous computational existence. Mathematics becomes the game of sets, which is a fine game as far as it goes, with rules that are admirably precise. The game becomes its own justification, and the
2 The Idealistic Component of Mathematics
7
fact that it represents a highly idealized version of mathematical existence is universally ignored. Of course, idealistic tendencies have been present, if not dominant, in mathematics since the Greeks, but it took the full flowering of formalism to kill the insight into the nature of mathematics which its arithmetization could have given To see how some of the most basic results of classical analysis lack computational meaning, take the assertion that every bounded nonvoid set A of real numbers has a least upper bound. (The real number b is the least upper bound of A if a ~ b for all a in A, and if there exist elements of A that are arbitrarily close to b.) To avoid unnecessary complications, we actually consider the somewhat less general assertion that every bounded sequence (x k ) of rational numbers has a least upper bound b (in the set of real numbers). If this assertion were constructively valid, we could compute b, in the sense of computing a rational number approximating b to within any desired accuracy; in fact, we could program a digital computer to compute the approximations for us. For instance, the computer could be programmed to produce, one by one, a sequence ((b k , m k )) of ordered pairs, where each b k is a rational number and each m k is a positive integer, such that Xj ~ bk + k  1 for all positive integers j and k, and x mk ~ bk  k  1 for all positive integers k. Unless there exists a general method M that produces such a computer program corresponding to each bounded, constructively given sequence (x k ) of rational numbers, we are not justified, by constructive standards, in asserting that each of the sequences (x k ) has a least upper bound. To see the scope such a method M would have, consider a constructively given sequence (n k ) of integers, each of which is either 0 or 1. Using the method M, we compute a rational number b 3 and a positive integer N == m3 such that (i) nj~b3 +! for all positive integers j, and (ii) nN~b3 j. Either nN=O or nN = 1. If nN =0, then (i) and (ii) imply that
for all j. Since each nj is either 0 or 1, it follows that nj = 0 for all j Thus for each of the sequences (n k ) being considered, the method M either produces a proof that the n k are all equal to 0, or produces a positive integer N such that nN = 1. Of course, such a method M does not exist, and nobody expects that one will ever be found Such a method would solve most of the famous unsolved problems of mathematics  in particular, Fermat's last theorem, the Goldbach conjecture, and the Riemann hypothesis, since each of these problems can be reduced to finding, for a certain sequence (n k ) of the type being
8
Chapter 1 A Constructivist Manifesto
considered, either a proof that nk = 0 for all k or a proof that nk = 1 for some k. For another instance, consider the intuitively appealing theorem that every continuous realvalued function f on the closed interval [O,IJ, with f(O) i In the first case, a ~ 0, and therefore the first nonzero term of the sequence (n k ), if one exists, equals 1. Similarly, in the second case, the first nonzero term, if one exists, equals 1. Thus our theorem gives a method, which, applied to each of the sequences (n k ) being considered, either (i) proves that any term that equals 1 is preceded by a term that equals 1, or (ii) proves that any term that equals 1 is preceded by a term that equals 1. Nobody believes that such a method will ever be found. Brouwer fought the advance of formalism and undertook the disengagement of mathematics from logic He wanted to strengthen mathematics by associating with every theorem and every proof a pragmatically meaningful interpretation. His program failed to gain support He was an indifferent expositor and an inflexible advocate, contending against the great prestige of Hilbert and the undeniable fact that idealistic mathematics produced the most general results with the least effort More important, Brouwer's system itself had traces of idealism and, worse, of metaphysical speculation. There was a preoccupation with the philosophical aspects of constructivism at the expense of concrete mathematical activity A calculus of negation was developed which became a crutch to avoid the necessity of getting precise constructive results. It is not surprising that some of Brouwer's precepts were then formalized, giving rise to socalled intuitionistic
3 The Constructivization of Mathematics
9
number theory, and that the formal system so obtained turned out not to be of any constructive value. In fairness to Brouwer it should be said that he did not associate himself with these efforts to formalize reality, it is the fault of the logicians that many mathematicians who think they know something of the constructive point of view have in mind a dinky formal system or, just as bad, confuse constructivism with recursive function theory Brouwer became involved in metaphysical speculation by his desire to improve the theory of the continuum. A bugaboo of both Brouwer and the logicians has been compulsive speculation about the nature of the continuum In the case of the logicians this leads to contortions in which various formal systems, all detached from reality, are interpreted within one another in the hope that the nature of the continuum will somehow emerge. In Brouwer's case there seems to have been a nagging suspicion that unless he personally intervened to prevent it, the continuum would turn out to be discrete. He therefore introduced the method of freechoice sequences for constructing the continuum, as a consequence of which the continuum cannot be discrete because it is not well enough defined This makes mathematics so bizarre it becomes unpalatable to mathematicians, and foredooms the whole of Brouwer's program. This is a pity, because Brouwer had a remarkable insight into the defects of classical mathematics, and he made a heroic attempt to set things right.
3. The Constructivization of Mathematics A set is defined by describing exactly what must be done in order to construct an element of the set, and what must be done in order to show that two elements are equal. There is no guarantee that the description will be understood; it may be that an author thinks he has described a set with sufficient clarity but a reader does not understand. For an illustration, consider the set of all sequences (n k ) of integers. To construct such a sequence we must give a rule which associates an integer nk with each positive integer k in such a way that for each value of k the associated integer nk can be determined in a finite number of steps by an entirely routine process Now this definition could perhaps be interpreted to admit sequences (n k ) in which n k is constructed by a search, the proof that the search actually produces a value of nk after a finite number of steps being given in some formal system Of course, we do not have this interpretation in mind, but it is impossible to consider every possible interpretation of our definition
10
Chapter 1 A Constructivist Manifesto
and say whether that is what we have in mind. There is always ambiguity, but it becomes less and less as the reader continues to read and discovers more and more of the author's intent, modifying his interpretations if necessary to fit the intentions of the author as they continue to unfold. At any stage of the exposition the reader should be content if he can give a reasonable interpretation to account for everything the author has said. The expositor himself can never fully know all the possible ramifications of his definitions, and he is subject to the same necessity of modifying his interpretations, and sometimes his definitions as well, to conform to the dictates of experience The constructive interpretations of the mathematical connectives and quantifiers have been established by Brouwer. To prove the statement (P and Q) we must prove the statement P and prove the statement Q, just as in classical mathematics. To prove the statement (P or Q) we must either prove the statement P or prove the statement Q, whereas in classical mathematics it is possible to prove (P or Q) without proving either the statement P or the statement Q. The connective "implies" is more complicated. To prove (P implies Q) we must show that P necessarily entails Q, or that Q is true whenever P is true. The validity of the computational facts implicit in the statement P must ensure the validity of the computational facts implicit in the statement Q, but the way this actually happens can only be seen by looking at the proof of the statement (P implies Q). Statements formed with this connective  for example, statements of the type ((P implies Q) implies R)  have a less immediate meaning than the statements from which they are formed, although in actual practice this does not seem to lead to difficulties in interpretation. The negation (not P) of a statement P is the statement (P implies (0 = 1)) Classical mathematics makes no distinction between the content of the statements P and (not (not P)), whereas constructively the latter is a weaker statement. Brouwer's system makes essential use of negation in defining, for instance, inequality and set complementation. Thus two elements of a set A are unequal according to Brouwer if the assumption of their equality somehow allows us to compute that 0 = 1. It is natural to want to replace this negativistic definition by something more affirmative, phrased as much as possible in terms of specific computations leading to specific results. Brouwer himself does just this for the real number system, introducing an affirmative and stronger relation of inequality in addition to the negativistic relation already defined. Experience shows that it is not necessary to define inequality in terms of negation. For those cases in which an inequality relation in needed,
3 The Constructivization of Mathematics
11
it is better to introduce it affirmatively; the same remarks apply to set complementation. Van Dantzig and others have gone so far as to propose that negation could be entirely avoided in constructive mathematics Experience bears this out: in many cases where we seem to be using negation  for instance, in the assertion that either a given integer is even or it is not  we are really asserting that one of two finitely distinguishable alternatives obtains. Without intending to establish a dogma, we may continue to employ the language of negation but reserve it for situations of this sort (at least until experience changes our minds) and for counterexamples and purposes of motivation This will have the advantage of making mathematics more immediate, and in certain situations forcing us to sharpen our results Proofs by contradiction are constructively justified in finite situations. When we have proved that one of finitely many alternatives holds at a certain stage in the proof of a theorem, to finish the proof of the theorem it is enough to show that the theorem is a consequence of each of the alternatives. Should one of the alternatives lead to a contradiction  that is, imply (0 = 1)  either we may say that the alternative in question is ruled out and pass on to the consideration of the other alternatives, or we may be more meticulous and prove that the theorem is a consequence of the equality 0 = l. A universal statement, to the effect that every element of a certain set A has a certain property P, has the same meaning in constructive as in classical mathematics. To prove such a statement we must show by some general argument that if x is any element of A, then x has property P Constructive existence is much more restrictive than the ideal existence of classical mathematics. The only way to show that an object exists is to give a finite routine for finding it, whereas in classical mathematics other methods can be used. In fact, the following principle is valid in classical mathematics: Either all elements of A have property P or there exists an element of A with property (not P) This principle, which we shall call the principle of omniscience, lies at the root of most nonconstructivity in classical mathematics. This is already true of the principle of omniscience in its simplest form. if (n k ) is a sequence of integers, then either nk = 0 for some k or nk =1= 0 for all k. We shall call this the limited principle of omniscience. Theorem after theorem of classical mathematics depends in an essential way on the limited principle of omniscience, and is therefore not constructively valid. Some instances of this are: the theorem that a continuous realvalued function on a closed, bounded interval attains its maximum; the fixedpoint theorem for a continuous map of a closed cell into
12
Chapter 1 A Constructivist Manifesto
itself, the ergodic theorem, and the HahnBanach theorem Nevertheless these theorems are not lost to constructive mathematics: each of these theorems P has a constructive substitute Q which is a constructively valid theorem Q implying P in the classical system by a more or less simple argument based on the limited principle of omniscience. For example, the statement "every continuous function from a closed cell in euclidean space into itself admits a fixed point" finds a constructive substitute in the theorem that such a function admits a point which is arbitrarily near to its image. The extent to which good constructive substitutes exist for the theorems of classical mathematics can be regarded as a demonstration that classical mathematics has a substantial underpinning of constructive truth When a classical mathematician claims he is constructivist, he probably means he avoids the axiom of choice. This axiom is unique in its ability to trouble the conscience of the classical mathematician, but in fact it is not a real source of nonconstructivity in classical mathematics. A choice function exists in constructive mathematics, because a choice is implied by the very meaning of existence. Applications of the axiom of choice in classical mathematics either are irrelevant or are combined with a sweeping use of the principle of omniscience. The axiom of choice is used to extract elements from equivalence classes where they should never have been put in the first place. For example, a real number should not be defined as an equivalence class of Cauchy sequences of rational numbers; there is no need to drag in the equivalence classes. The proof that the real numbers can be well ordered is an instance of a proof in which a sweeping use of the principle of omniscience is combined with an appeal to the axiom of choice. Such proofs offer little hope of constructivization: it is not likely that the theorem "the real numbers can be well ordered" will be given a constructive version consonant with the intuitive interpretation of the classical result. Almost every conceivable type of resistance has been offered to a straightforward realistic treatment of mathematics, even by constructivists. Brouwer, who has done more for constructive mathematics than anyone else, though it necessary to introduce a revolutionary, semimystical theory of the continuum Weyl, a great mathematician who in practice suppressed his constructivist convictions, expressed the opinion that idealistic mathematics finds its justification in its applications to physics. Hilbert, who insisted on constructivity in metamathematics but believed that the price of a constructive mathematics was too great, was willing to settle for consistency. Brouwer's disciples joined forces with the logicians in attempts to formalize
Notes
13
constructive mathematics Others seek constructive truth in the framework of recursive function theory Still others look for a short cut to reality, a point of vantage which will suddenly reveal classical mathematics in a constructive light. None of these substitutes for a straightforward realistIc approach has worked It is no exaggeration to say that a straightforward realistic approach to mathematics has yet to be tried It is time to make the attempt.
Notes Errett Bishop was never happy with the standard constructive interpretation of implication (the one given in Section 3). Among the alternatives he felt worthy of serious investigation is "Godel implication", as discussed in [9]. Bishop also worked on a deeper study of implication, but unfortunately he left only fragmentary notes on his ideas At first sight, Bishop' remark, "A choice function exists in constructive mathematics, because a choice is implied by the very meaning of existence", appears to be contradicted by counterexamples of the sort discussed in connection with the leastupperbound principle. In fact, there is no contradiction here. To see this, consider a paraphrase of Bishop's remark: if to each x in a set A there corresponds an element y of a set B such that a given property P(x, y) holds, then it is implied by the very meaning of existence in constructive mathematics that there is a finite routine for computing an appropriate y in B from a given x in A; although this routine may not be a function relative to the given equality relation on A, it is a function relative to the equality relation of identity (intensional equality) on A, in which two elements are equal if and only if they are given as identically the same object
Chapter 2. Calculus and the Real Numbers
Section 1 establishes some conventions about sets and functions. The next three sections are devoted to constructing the real numbers as certain Cauchy sequences of rational numbers, and investigating their order and arithmetic. The rest of the chapter deals with the basic ideas of the calculus of one variable. Topics covered include continuity, the convergence of sequences and series of continuous functions, differentiation, integration, Taylor's theorem, and the basic properties of the exponential and trigonometric functions and their inverses Most of the material is a routine constructivization of the corresponding part of classical mathematics; for this reason it affords a good introduction to the constructive approach.
We assume that the reader is familiar with the order and arithmetic of the integers and the rational numbers. For us, a rational number will be an expression of the form plq, where p and q are integers with q*O. Two rational numbers plq and p'lq' are equal if pq'=p'q. The integer n is identified with the rational number nil. There are geometric magnitudes which are not represented by rational numbers, and which can only be described by a sequence of rational approximations. Certain such approximating sequences are called real numbers. In this chapter we construct the real numbers and study their basic properties. Then we develop the fundamental ideas of the calculus.
1. Sets and Functions Before constructing the real numbers, we introduce some notions which are basic to much of mathematics. The totality of all mathematical objects constructed in accordance with certain requirements is called a set. The requirements of the
1 Sets and Functions
15
construction, which vary with the set under consideration, determine the set. Thus the integers form a set, the rational numbers form a set, and (we anticipate here the formal definition of 'sequence') the collection of all sequences of integers is a set. Each set will be endowed with a binary relation = of equality. This relation is a matter of convention, except that it must be an equivalence relation, in other words, the following conditions must hold for all objects x, y, and z in the set: (11)
(i)
X=X
(ii) If X= y, then y=x (iii) If x = y and y = z, then x = z. The relation of equality given above for rational numbers is an equivalence relation. In this example there is a finite, mechanical procedure for deciding whether or not two given objects in the set are equal Such a procedure will not exist in general: there are instances in which we are unable to decide whether or not two given elements of a set are equal; such an instance, in the theory of real numbers, will be given later. We use the standard notation aEA to denote that a is an element, or member, of the set A, or that the construction defining a satisfies the requirements a construction must satisfy in order to define an object of A. We also use the notation {aI' a 2 , ... } for a set whose elements can be written in a (possibly finite) list. The dependence of one quantity on another is expressed by the basic notion of an operation An operation from a set A into a set B is a finite routine f which assigns an element f(a) of B to each given element a of A. This routine must afford an explicit, finite, mechanical reduction of the procedure for constructing f(a) to the procedure for constructing a. If it is clear from the context what the sets A and B are, we sometimes denote f by a 1+ f(a), in order to bring out the form of f(a) for a given element a of A. The set A is called the domain of the operation, and is denoted by dmn! In the most important case, we have f(a}=f(a') whenever a,a'EA and a=a'; the operation f is then called a function, or a mapping of A into B, or a map of A into B For two functions f, g from A into B, f = g means that f(a) = g(a) for each element a of A. Taken with this equality relation, the collection of all functions from A into B becomes a set. The notation f: A + B indicates that f is a function from the set A to the set B. A function x whose domain is the set 7l+ of positive integers is called a sequence. The object x.=x(n) is called the nth term of the
16
Chapter 2
Calculus and the Real Numbers
sequence. The finite routine x can be given explicitly, or it can be left to inference: for example, by writing the terms of the sequence in order (X),X 2 ,· ) until the rule of their formation becomes clear. Different notations for the sequence whose nth term is Xn are: nt>x n, (x),x 2 , ..• ), (xn):;O=), and (xJ Thus the sequence whose nth term is n 2 can be written nt>n z, or (1,4,9, ... ), or (nZ):;o= I' or simply (nZ). A subsequence of a sequence (xn) consists of the sequence (xn) and a sequence (nk)r= I of positive integers such that n 1 0, p =1= 0, and p is relatively prime to q, we obtain a sequence (1 4)
(0/1, 1/1,1/1,
which has the property that for any given rational number r there exists exactly one term equal to r. F or each positive integer n, let lln be the set {O, 1, ... , n l} If there is a mapping of lln onto the set A, then we say that A has at most n elements. A set with at most n elements for some n is said to be subfinite, or finitely enumerable. Note that every subfinite or countable set has at least one element. Before we introduce stronger notions than countability and sub· finiteness, we must discuss the composition of functions The composition of two functions f: A + Band g: B + C is the function g of: A + C defined by (gof)(a)==g(f(a))
(aEA).
Composition is associative: h a (g oj) = (h a g) of
whenever the compositions are defined If f. A + B, g: B + A, and g(f(a)) = a for all a in A, then the function g is called a left inverse of f, and the function f is called a right inverse of g (Note that f has a left inverse if and only if it is oneone, in the sense that a = a' for all elements a, a' of A with f(a) = f(a')) When g is both a left and a right inverse of f, then it is simply called an inverse of j; f is then called a oneone correspondence, or a bijection, and the sets A and B are said to be in oneone correspondence with each other A set which is in oneone correspondence with the set ll+ of positive integers is said to be countably infinite. For example, let f be the sequence (1.4), and define a function g from the set n 1 for some n in '1.+. Choose N in '1.+ with
Then Xm~Xn Ix m xnl ~xn _m 1 _n 1 ~xnnlNl
>N 1
whenever m~ N. Therefore (2.8.1) is valid. Conversely, if (2.8.1) is valid, then (2.71) holds with n=N+l Therefore XE1R +.
22
Chapter 2 Calculus and the Real Numbers
Assume next that xElR o +. Then for each positive integer n, (m~n).
Therefore (2.8.2) is valid with Nn == n. Assume finally that (2.8.2) holds. Then if k, m, and n are positive integers with m ~ Nn , we have xk~xmIxmxkl ~
_n 1 _k 1 _m 1•
Since m and n are arbitrary, this gives x k ~ xElR o +. 0

k 1. Therefore
As a corollary of Lemma (2.8), we see that if x and yare equal real numbers, then x is positive if and only if Y is positive, and x is nonnegative if and only if y is nonnegative. It is not strictly correct to say that a real number (xn) is an element of IR +. An element of IR + consists of a real number (xn) and a positive integer n such that xn > n 1, because an element of IR + is not presented until both (xn) and n are given. One and the same real number (xn) can be associated with two distinct (but equal) elements of IR + Nevertheless we shall continue to refer loosely to a positive real number (xJ On those occasions when we need to refer to an n for which Xn > n 1, we shall take the position that it was there implicitly all along. The proof of the following proposition is now easy, and will be left to the reader. For convenience, IR * represents either IR + or IR 0 +. (2.9) Proposition. Let x and y be real numbers. Then
(a) x + YEIR * and XYEIR * whenever xEIR * and YEIR * (b) X+YEIR+ whenever xEIR+ and YElR o+ (c) IxiElR o + (d) max {X,Y}EIR* whenever xEIR* (e) min {X,Y}EIR* whenever XEIR* and YEIR*. We now define the order relations on IR. (2.10) Definition. Let x and Y be real numbers. We define x>y(ory<x)
ifxYEIR+
X~y
if XYElR o +.
and (or y;£x)
A real number x is negative if x < 0*  that is, if  x is positive.
2 The Real Number System
23
Consider real numbers x, x', y, and y' such that (i) x=x', y=y', and x>y We have X'y'=XYEJR+ and therefore (ii) x' > y'. We express the fact that (ii) holds whenever (i) is valid by saying that> is a relation on JR. More formally, a relation on a set X is a subset S of X x X such that if x, x', y, y' are elements of X with x=x', y=y', and (X,Y)ES, then (X',Y')ES. We express the fact that x> y if and only if y < x by saying that > and < are transposed relations. Similarly, ~ and ~ are transposed relations. If x < y or x = y, then x ~ y. The converse is not valid: as we shall see later, it is possible that we have x ~ y without being able to prove that x < y or x = y. For this reason it was necessary to define the relations < and ~ independently of each other. The following rules for manipulating inequalities are easily proved from Proposition (2.9). We omit the proofs. (2.11) Proposition. For all real numbers x, y, z, and t, (a) xnl. Write
1X=!(X 2n + Y2n)·
Then Also, YIX~Y2n1X IY2n yl ~!(Y2nX2n)(2n)1 >0.
Therefore x < IX < y.
D
As a corollary, for each x in 1R and r in 1R + there exists with Ix IXI < r. Here is another corollary. (2.16) Proposition. If
XI' ••. ' xn are real numbers with then x;>O for some i (1 ~i~n).
XI
IX
in 0,
Proof. By (2.15), there exists a rational number IX with 0!IX. ;=1
;=1
;=1
Therefore a; > (2n)11X for some i. For this i it follows that x;~a;lx;a;I>O.
D
(2.17) Corollary. If x, y, and z are real numbers with Y < z, then either x y. Proof: Since zx+xy=zy>O, either zx>O or xy>O, by
(2.16).
D
The next lemma gives an extremely useful method for proving inequalities of the form x ~ y. (2.18) Lemma. Let x and y be real numbers such that the assumption x> y implies that 0 = 1. Then x ~ y. Proof· Without loss of generality, we take y=O. For each n in Z+, either Xn ~ n lor Xn > n I. The case Xn > n  1 is ruled out, since it implies that x> o. Therefore  Xn ~  n  I for all n, and so  x ~ o. Thus x~O. D
2 The Real Number System
27
(2.19) Theorem. Let (an) be a sequence of real numbers, and let Xo and Yo be real numbers with Xo < Yo Then there exists a real number x such that xo~x~Yo and X =1= an for all n in 7l+.
Proof: We construct by induction sequences (xn) and (Yn) of rational numbers such that (i) xO~xn~xma n or Yn Y = x and so an =1= x. Thus x has the required properties. 0 Theorem (2.19) is the famous theorem of Cantor, that the real numbers are uncountable. The proof is essentially Cantor's" diagonal" proof. Both Cantor's theorem and his method of proof are of great importance. The time has come to consider some counterexamples. Let (n k ) be a sequence of integers, each of which is either 0 or 1, for which we are unable to prove either that nk = 1 for some k or that nk = 0 for all k. This corresponds to what Brouwer calls "a fugitive property of the natural numbers". For example, such a sequence can be defined as follows. Let nk be 0 if ul + Vi =1= Wi for all integers u, v, w, t with 0 < u, v, w ~ k and 3 ~ t ~ 2 + k. Otherwise let nk be 1. Then we are unable to prove nk = 1 for some k, because this would disprove Fermat's last theorem. We are unable to prove n k = 0 for all k, because this would prove Fermat's last theorem. Now define xk==O if nj=O for all j~k, and x k==2 m otherwise, where m is the least positive integer such that nm = 1. Then x == (x k ) is a
28
Chapter 2 Calculus and the Real Numbers
nonnegative real number, but we are unable to prove that x> 0 or x = O. Since nothing is true unless and until it has been proved, it is untrue that x> 0 or x = O. Of course, if Fermat's last theorem is proved tomorrow, we shall probably still be able to define a fugitive sequence (n k ) of integers. Thus it is unlikely that there will ever exist a constructive proof that for every real number x~O either x>O or x=O. We express this fact by saying that there exists a real number x ~ 0 such that it is not true that x>O or x=O. In much the same way we can construct a real number x such that it is not true that x~O or x~O.
3. Sequences and Series of Real Numbers We develop methods for defining a real number in terms of approximations by other real numbers. (3.1) Definition. A sequence (x.) of real numbers converges to a real number Xo if for each k in 7l+ there exists Nk in 7l+ with (3.1.1)
The real number Xo is then called a limit of the sequence (xJ To express the fact that (x.) converges to Xo we write lim X.=Xo .00
or
as n+oo or simply x. + xo' A sequence (x.) of real numbers is said to converge, or be convergent, if there exists a limit Xo of (xJ It is easily seen that if (x.) converges to both
Xo
and
x~,
then
Xo
=x~.
A convergent sequence is bounded: there exists r in IR. + such that n A convergent sequence of real numbers is not determined until the limit Xo and the sequence (Nk ) are given, as well as the sequence (x n ) itself. Even when they are not mentioned explicitly, these quantities are implicitly present. Similar comments apply to many subsequent definitions, including the following.
Ix.1 ~r for all
3 Sequences and Series of Real Numbers
29
(3.2) Definition. A sequence (x.) of real numbers is a Cauchy sequence if for each k in 7l+ there exists Mk in 7l+ such that (3.2.1)
IXmx.l~kl
(m,n~Mk)'
(3.3) Theorem. A sequence (x.) of real numbers converges
if and
only
if
it is a Cauchy sequence. Proof' Assume that (x.) converges to a real number xo' Let the sequence (Nk ) satisfy (3.1.1). Write M k ==N2k • Then
Ixmx.1 ~ IX mxol +Ix.xol ~(2k)1 +(2k)1 =k 1 for m, n~Mk' Therefore (x.) is a Cauchy sequence. Assume conversely that (x.) is a Cauchy sequence. Let the sequence (M k ) satisfy (3.2.1). Write Nk == max {3k,M 2k }. Then IXmx.I~(2k)1
(m,n~Nk)'
Let Yk be the (2kyh rational approximation to XNk ' For m~n, IYmy.I~IYmXNJ+lxNm xNJ+lxN• Y.I ~(2m)1
+(2m)1 +(2n)1 +(2n)1 =m 1 +n 1.
Therefore Y==(Y.) is a real number. To see that (xn) converges to y, we consider n ~ Nk and compute lyx.1 ~ Iy Y.I + Iy. xN.1 + Ix N• xnl ~nl +(2n)1 +(2k)1 ~(3k)1 +(6k)1 +(2k)1 =k 1.
D
A subsequence of a convergent sequence converges to the same limit If a sequence converges, then any sequence obtained from it by modifications (including, perhaps, insertions or deletions) which involve only a finite number of terms converges to the same limit If x == (x.) is a regular sequence of rational numbers, then (x:) converges to x, by (2.14). A sequence (x.) is increasing (respectively, strictly increasing) if x.+ 1 ~ x. (respectively, x.+ 1> x.) for each n. Decreasing and strictly decreasing sequences are defined analogously, in the obvious way. A theorem of classical mathematics states that every bounded increasing sequence of real numbers converges. A counterexample to this statement is given by any increasing sequence (x.) such that xn = 0 or x. = 1 for each n, but it is not known whether x. = 0 for all n It is useful to supplement Definition (3.1) by writing limx.= 00 .~oo
or
x. + 00
as
n + 00
30
Chapter 2 Calculus and the Real Numbers
to express the fact that for each k in 7l+ there exists Nk in 7l+ with xn>k for all n~Nk. We also define limxn= 
00
n~oo
or Xn+ 
00
as
n+ 00
to mean that lim Xn= 00. n~oo
The next proposition shows that we may work with real numbers constructed as limits by working with their approximations. (34) Proposition. Assume that Xn+XO as n+oo, and Yn+Yo as n+oo, where Xo and Yo are real numbers. Then (a) xn+Yn+xO+Yo as n+oo (b) XnYn+XoYo as n+oo (c) max{x n, Yn} +max{xo, Yo} as n+oo (d) Xo = c whenever Xn = c for all n (e) if Xo =!= 0 and Xn =!= 0 for all n, then x; 1+ xi) 1 as n+ 00 (f)
if Xn~Yn
for all n, then xo~yo·
Proof: (a) For each k in 7l+ there exists Nk in 7l+ such that
Then Therefore xn+Yn+xO+Yo as n+oo. (b) Choose m in 7l+ such that IYol ~m and IXnl ~m for all n. For each k in 7l+ choose Nk in 7l+ with IXnxol~(2mk)1,
IYnYol~(2mk)1
(n~Nk)·
Then for n ~ Nk , IX nYnXo Yol ~ IXn(Yn  Yo)1 + IYo(xnxo)1 ~m(lYn  Yol + IXnxol)~k1. Therefore xnyn+xoYo as n+oo. (c) Since Imax{x n, Yn} max{x o, Yo}1 ~max{lxn xol, IYn  Yol}, it follows that max {xn,Yn}+max{x o, Yo}
as n+oo.
3 Sequences and Series of Real Numbers
31
(d) If Xn = c for all n, then (xn) converges to c. Therefore Xo = c. (e) Since IXol >0, IXnl ~ IXollxnxol >1lxol whenever n is large enough, say for n ~ no' Let k and n be positive integers such that n~no and IXnxol «2k)llxoI2. Then Ix; I_XO II = IXnlIlxolllxn xol ~ 2lxol 2(2k)llxoI2 =k I. Therefore x; 1+ Xo I as n + 00. (f) We compute
Yo Xo = lim Yn lim xn= lim (Ynxn)= lim IYn xnl n~oo
.~oo
= lim max{Ynxn, x n Yn} =max{yo X o, Xo  Yo} ~O, n~oo
by (a), (b), (c), and (d).
0
For each sequence (xn) of real numbers the number
is called the nth partial sum of (x.), and (sn) is called the sequence of partial sums of the sequence (xn)' A sum So of (xn) is a limit of the sequence (sn) of partial sums. We write
to indicate that So is a sum of (xn). A sequence which is meant to be summed is called a series. A series is said to converge to its sum. Thus the sequence (2 n):."'= I converges to 0 as a sequence, but as a series it 00
converges to
L 2  n= 1.
n= I A convergent series remains convergent, but not necessarily to the same sum, after modification of finitely many of its terms. 00
The series (x.) is often loosely referred to as the series 00
If the series 00
L
L xn converges, then x.+O as n+oo. 11=1
L x n. 11=
1
00
L
A series Xn is said to converge absolutely when the series IXnl converges. .= I n= I In classical analysis a series of nonnegative terms converges if the partial sums are bounded. This is not true in constructive analysis However, we have the following result.
32
Chapter 2
Calculus and the Real Numbers 00
L Yn
(3.5) Proposition. If
is a convergent series of nonnegative terms,
n= 1
and
if IXnl :;:;; Yn for
00
L Xn converges.
each n, then
n= 1 00
Proof Since
L Yn
is convergent, the sequence of partial sums is a
n= 1
Cauchy sequence. Therefore for each kin '1.+ there exists an Nk in '1.+ with '" Yj:;:;;k 1
L
j=n+ 1
Then
00
Therefore the sequence of partial sums of the series sequence. By (3.3), the series converges. D
L Xn is
a Cauchy
n= 1
The criterion of Proposition (3.5) is known as the comparison test It follows from the comparison test that every absolutely convergent 00 series is convergent. The terms of an absolutely convergent series L Xn may be ren= 1
ordered without affecting the sum
So
of the series. More precisely, if A:
00
'1.+ +'1.+ is a bijection, then
L x;'(n)
exists and equals so. This may
n= 1
00
L Xn is merely convergent.
not be true if the series
11=
1
A sequence (xn) is said to diverge if there exists E in 1R. + such that for each k in '1.+ there exist m and n in '1.+ with m, n~k and Ix", xnl ~ E. The motivation for this definition is, of course, that a sequence cannot be both convergent and divergent. A series is said to diverge if the sequence of its partial sums diverges. 00
The series
L n
1
diverges, because
n= 1
00
The series
L xn
diverges whenever there exists r in 1R. + such that
n= 1
IXnl ~r
for infinitely many values of n. 00
Let
L Xn "=1
00
and
L Yn
be series of nonnegative terms. The compari
"=1
son test for divergence is that
00
L Xn 11=
00
diverges whenever
1
and there is a positive integer N with
L Yn 11=
Xn ~ Yn
for all n ~ N.
1
diverges
3 Sequences and Series of Real Numbers
33
The following very useful test for convergence and divergence is called the ratio test. 00
(3.6) Proposition. Let
L Xn be a series,
c a positive number, and N a
n= 1 00
positive integer. Then
L Xn converges if c < 1 and n= 1
(3.6.1)
and diverges
if c> 1 and
(3.6.2)
Proof. Assume that c< 1 and that (3.6.1) is valid. Then Ixnl ~cnNlxNI 00
for n ~ N. By the comparison test,
L xn converges. n= 1
Next, assume that c> 1 and that (3.6.2) holds. Then IXnl~cnN1IxN+11~lxN+11
(n~N+l)
and 00
Therefore
L Xn diverges.
0
n= 1
A corollary of the ratio test is that if the limit L= limlxn+1x;11 n_oo 00
exists, then L Xn converges whenever L < 1 and diverges whenever L>1. n=l The ratio test says nothing in case L = 1. To handle this case, we introduce stronger tests based on Kummer's criterion. (3.7) Lemma. Let (an) and (xn) be sequences of positive numbers, c a 00
positive number, and N a positive integer. Then an Xn + 0 as n > 00 and (3.7.1) 00
while (3.7.2)
00
L xn diverges if L a;
1
diverges and
L Xn
converges
if
34
Chapter 2 Calculus and the Real Numbers
Proof: Assume that anxn+O and that (3.7.1) is valid. Let e be an arbitrary positive number, and choose an integer v ~ N so that a k X k ajxj~ce whenever j>k~v. For suchj and k we have j
L
j
L
xn~c1
Xn(an_1Xn_1X,;1an) n=k+1 =c 1(a k x k ajxj)~ e.
n=k+1
Thus ttl Xn): 1 is a C:uch y sequence, and so Next assume that
L a,;l ft=
diverges and that (3.7.2) holds. Then for
1
00
each n~N, Xn~aNxNa,;l. Thus
L Xn
diverges, by comparison with
n= 1
00
D
n= 1
n~l xn converges.
(3.8) Lemma. Let (Yn) be a sequence of positive numbers, c a positive number, and N a positive integer such that n(YnY';_,\ 1)~c
(n~N).
Then limYn=O. n~oo
Proof: For each n>N, YN y; 1 =(YN YN~ l)(YN+ 1YN~ 2)'" (Yn1 Y'; 1) ~(1 +cN 1) ... (1 +c(n1)1) n1 ~1+c L k 1. k=N
n1
Given e>O, choose an integer v>N so that
L k 1 >c 1(e 1YN1)
k=N
for all n~v. Then for such n we have Yn<e. Hence Yn+O as n+oo.
D
The next convergence test is known as Raabe's test. 00
(3.9) Proposition. Let
L Xn ft=
1
be a series of positive numbers such that 00
n(Xnx';;11) converges to a limit L. Then L xn converges n= 1 diverges if L< 1.
if
Proof: First note that n(n xn/(n + 1)xn+ 1 1) =n(n+ 1)1 (n(XnX';; 1 1) 1) +L1 as n+ 00.
L>1, and
4. Continuous Functions nXn~O
If L>1, it follows from (3.8) that
35
as n+oo. We then obtain the
00
convergence of
L Xn
an == n
by taking
(nEZ+) in Kummer's criterion.
n= 1
00
The same choice of an yields divergence of
L Xn in case L < 1.
0
n= 1
Important real numbers represented by series are 00
e=1+ L(n!)l n= 1
and 00
n=4
L (1)"(2n+ 1)1.
n=O
The series for e converges by the ratio test. The convergence of the series for n is a consequence of the general result that a series 00
L (1)" Xn converges whenever (i) Xn ~ 0 for all nand (ii) the sequence n= 1
is decreasing and converges to O. To see this, consider positive integers m and n with m ~ n. Then
(Xn)
O~(xnxn+ 1)+(X n + 2 Xn+3)+'"
+( _l)m+n Xm
m
=( 1)"
L (_1)kXk
=Xn(X n + 1 X n + 2 )  .. ·
+( _1)m+n Xm~Xn'
It follows that the sequence of partial sums of the series is a Cauchy sequence. Therefore the series converges.
4. Continuous Functions A property P which is applicable to the elements of a set S is defined by a statement of the requirements that an element of S must satisfy in order to have property P. To construct an element of S with property P we must construct an element of S, perform certain additional constructions which depend on the property P, and prove that the entities constructed satisfy certain requirements that are characteristic of the property P. Each property P applicable to elements of a set S determines a subset of S which is denoted by {x: XES,
X
has property P}
{XES:
X
has property Pl.
or
36
Chapter 2
Calculus and the Real Numbers
When the context makes it clear which set S is under discussion, we also write simply {x: x has property P}. Properties applicable to elements of a set S, and subsets of S, are essentially the same things regarded from different points of view. Among the most important subsets of lR are the intervals. (4.1) Definition. For all real numbers a and b we define (a,b)=={x: xElR, a<x 0 there exists N. in 7l + such that (4.9.1)
I/n{x)  f(x)1 ~ e
(xEI, n"?, N
E).
A sequence Un) of continuous functions on an arbitrary interval J converges on J to a continuous function I if it converges to I on every compact subinterval I of J; in that case, I is called the limit of the sequence Un). Definition (4.9) can be recast to bear a closer resemblance to Definition (3.1). To this end, we define the norm 11/111 of a continuous function I on a compact interval I to be the supremum of Ilion I. Then (In) converges to I on I if and only if for each k in 7l+ there exists N" in 7l+ with II!.  IIII~kl (n"?,Nk) (4.1O) Definition. A sequence (In) of continuous functions on a compact interval I is a Cauchy sequence on I if for each e > 0 there exists ME in 7l+ such that (4.10.1)
1/",(x)/n(x)l~e
(xEI; m,n"?,ME).
A sequence of continuous functions on an arbitrary interval J is a Cauchy sequence on J if it is a Cauchy sequence on every compact subinterval of J. The sequence (!.) is a Cauchy sequence on the compact interval I if and only if for every k in 7l+ there exists M" in 7l+ such that 11/",/nIII~kl
(n,n"?,M,,).
Notice that a sequence (c n) of real numbers converges if and only if the corresponding sequence of constant functions, which we also denote by (c n), converges on a given nonvoid interval I, and that a sequence of real numbers is a Cauchy sequence if and only if the corresponding sequence of constant functions is a Cauchy sequence on I. Because of these remarks, the following theorem is a generalization of Theorem (3.3).
42
Chapter 2 Calculus and the Real Numbers
(4.11) Theorem. A sequence (fn) of continuous functions on an interval 1 converges on 1 if and only if it is a Cauchy sequence on 1.
Proof. Assume that (fn) converges to f on 1. Let I be any compact subinterval of 1. For each 6>0 choose N. in Z+ satisfying (4.9.1), and write M.=N./ 2 . Then whenever m, n~M. and xEI, we have If". (x)  fn(x)1 ~ Ifm(x)  f(x)1
+ Ifn(x) 
f(x)1
~6/2+6/2 =6.
Therefore (fn) is a Cauchy sequence on I. It follows that (f,.) is a Cauchy sequence on 1. Assume conversely that (fn) is a Cauchy sequence on 1. Then for each x in 1, (fn(x)) is a Cauchy sequence of real numbers, whose limit we denote by f(x). We shall show that f: 1 >JR. is a continuous function and that (fn) converges to f on 1. It is enough to show that f is continuous on each compact subinterval I of 1, and that (fn) converges to f on I. To this end, choose the positive integers M. such that (4.10.1) is valid, and for each n in Z+ let Wn be a modulus of continuity for fn on I. For each 6>0 write
where M=M./ 3 • Then whenever x,YEI and
IxYI~w(6),
we have
If(x)  f(Y)1 ~ If(x)  fM(X) I+ IfM(X)  fM(Y) I+ IfM(Y)  f(Y)1
= lim Ifn(x)  fM(X) I + IfM(X)  fM(Y) I + lim IfM(Y)  fn(Y) I 666 :=;;++=6. 3 3 3 Therefore f is continuous, with modulus of continuity w. Finally, if xEI, 6>0, and n~M .. then If,. (x)  f(x) I = lim Ifn(x)  fm(x)1 ~ 6. 11100
0
Hence (fn) converges to f on I.
Notations to express the fact that (fn) converges to fare limfn= f and f,,+f
We also write simply f,.+f
as n>oo.
4. Continuous Functions
43
To each sequence (J.) of continuous functions on an interval 1 corresponds a sequence (gn) of partial sums, defined by n
gn==
Lk k~l
If (gn) converges to a continuous function g on 1, then g is the sum of co
L fn'
the series
co
n~l
co
and the series is said to converge to g on 1. If
n=1
00
then
L J. n~
L Ifni converges on 1,
is said to converge absolutely on 1. An absolutely con
1
vergent series of functions converges. The comparison test and the ratio test for convergence carryover to 00
series of functions. The comparison test states that if
L gn
is a con
n~l
vergent series of nonnegative continuous functions on an interval 1, 00
then the series
L fn
of continuous functions on 1 converges on 1
n~l
whenever Ifn(x)1 ~gn(x) for all n in Z+ and all x in 1. 00
The ratio test states that if
L fn is a series of continuous functions n~l
on an interval J such that for each compact subinterval 1 of J there exist a constant c[, O 0, x, YEI, and Iy xl ~ {)(e). Then f is said to be differentiable on I, g is called a derivative of f on I, and {) is called a modulus of differentiability for f on I. If f and g are continuous functions on a proper interval J, then g is a derivative of f on J if it is a derivative of f on every proper compact subinterval of J; f is then said to be differentiable on J. To express that g is a derivative of f we write g=J,'
or
g=DJ,
or
df(x)
g(x)=~.
One way to interpret Definition (5.1) is that the difference quotient (f(y)  f(x)) (y _X)1
approaches g(x) as y approaches x. In other words, g is the rate of change of f If f has two derivatives on I, then they are equal functions. (5.2) Theorem. Let f1 and f2 be differentiable functions on an interval I. Then It + f2 and f1 f2 are differentiable on I. In case f1 is bounded away from 0 on every compact subinterval of I, then 11 1 is differentiable on I. The function Xl+X is differentiable on JR.. For each c in JR. the function Xl+C is differentiable on JR.. The derivatives in question are given by the following relations: (a)
D(f1 + f2)=Df1 +Df2
(b)
(c)
D(f1 f2)= f1 Df2 + f2 Df1 1 =  It 2 Df1
(d)
=1
(e)
dc =0 dx .
dx dx
DIt
5. Differentiation
45
Proof" It is enough to consider the case in which I is compact. Let b1 and b2 be moduli of differentiability for ft and f2' respectively, on I, and W1 a modulus of continuity for ft on I. (a) Whenever x, YEI and Iy xl ~b(B)=min{b1(B/2), b 2 (B/2n, we have If1(Y) + f2(Y) (f1(X) + f2(X)) (fleX) + f2(X))(Y x)1 ~ Ifdy)  f1(X)  f{(x)(y x)1 + If2(Y)  f2(X)  f2(X) (yx)1
Thus f1 + f2 is differentable on I, with derivative fi + f2 and modulus of differentiability b. (b) Let M be a common bound for If11, If21, and If21 on the interval I. (For instance, define M=max{llftIII, Ilf211I' Ilf21II}.) Then whenever x, YEI and Iy  xl ~ b(B) = min {b 1«3 M)l B), b 2 «3M)1 B), W1 «3 M)l Bn, we have Ifl(Y) f2(Y)  ft(x) f2(X) (fleX) f2(X) + f2(X) flex)) (y x)1 ~lfl(Y)llf2(Y)f2(X)f2(X)(yx)1
+ Ifl(Y)  f1(X)llf2(X)lly xl + If2(X)llft(y)  fleX)  f{(x)(y x)1 ~3M(3M)1 Blyxl =Blyxl. Therefore fl f2 is differentiable on I, with derivative ft f2 + f2 modulus of differentiability b. (c) For each B>O write
fl and
b(B) =min {b 1(! M 2 B), W1(! M 4 Bn
where M=max{llft 11II,llfiIII}. Then whenever x, YEI and Iyxl ~ b(B), we have If1 1(y)  ftl(X)+ ft2(X) fleX) (y x)1 = Iftl (x) fl l (y)llf1(Y)  fleX)  fl(Y) fll(X) fleX) (y x)1 ~ M21fl (y)  ft (x)  fleX) (y  x)1 +M 2 If{(x)fl(X)11Ifl(y)f1(X)llyxl ~M2(!M2 B) Iyxl +M 4 (!M 4 B) Iyxl =Blyxl. Therefore ft 1 is differentiable on I, with derivative  ft 2 f{ and modulus of differentiability b.
46
Chapter 2 Calculus and the Real Numbers
(d) This is obvious. (e) This is obvious too.
D
(5.3) Corollary. For all positive integers n, dx" Tx=nx"l.
(5.3.1)
Proof: The proof is by induction on n. When n = 1, (5.3.1) is just (d) of Theorem (5.2). If (5.3.1) is true for a given value of n, then dx"+l
d(x·x")
dx
dx
by (b) of Theorem (5.2). Therefore (5.3.1) is true for all n.
0
Theorem (5.2) and its corollary imply the formula D(f1 f2 1) = f2 2(f2 D f1
it D f2)
for the derivative of a quotient, and the formula D
(± an_kxk) f. ka _k xk  1 =
k~
0
n
k~
1
for the derivative of a polynomial. The next theorem is the socalled chain rule for the derivative of a composite function. Its intuitive meaning is that the rate of change of quantity C with respect to quantity A is the product of the rate of change of C with respect to some third quantity B by the rate of change of B with respect to A. (5.4) Theorem. Let f: I +lR and g: J +lR be differentiable functions such that f maps each compact subinterval of I into a compact subinterval of J. Then go f is differentiable, and
(5.4.1)
(go f), =(g' 0 f) J'.
Proof. It is no loss of generality to assume that I and J are compact. Let bf be a modulus of differentiability and wf a modulus of continuity for f on I. Let bg be a modulus of differentiability for g on J.
For each B>O write where
5. Differentiation
Then for x, y in 1 and Iy  xl ~ c5(e) we have If(y)  f(x)1
~ c5 g (IX),
~ IX If(y)
 f(x)l.
Ig(f(y»  g(f(x»  g'(f(x» (f(y)  f(x»1
47
so that
Also If(y)  f(x)1 ~ 111'111 Iy xl + If(y)  f(x)  f'(x) (y xli
and
If(y)  f(x)  f'(x) (y  x)1 ~ PIy  xl·
Using these inequalities and noting that IX 111'111 <e/2, we compute Ig(f(y»  g(f(x»  g'(f(x» f'(x) (y xli ~
Ig(f(y))  g(f(x»  g'(f(x» (f(y)  f(x»
I
+ Ig'(f(x»llf(y)  f(x)  f'(x) (y  x)1 ~ IX If(y)
 f(x)1 + Ilg'IIJ If(y)  f(x)  f'(x) (y xli
~ IX 111'111 Iy xl + (IX +
e
Ilg'IIJ) If(y)  f(x)  f'(x) (y xli
e
O. We may assume that f'(a)~m. For each x in [a,b] we have f'(x)~m. For if f'(x)<m, then f'(x) ~  m, so that, by the intermediate value theorem (4.8), there exists ~ in [a,b] with If'WI<m; this contradicts the definition of m. Now choose points a=xo~xl~ ... ~xn=b so that Xk+lxk~c5(!m) (O~k~n 1). Then 0= feb)  f(a) nl
=
L (f(Xk+l) f(Xk»
k=O n1
=
k= 0 nl
~
n1
L f'(Xk)(Xk+l xd+ L (f(Xk+1) f(Xk) f'(Xk)(Xk+l Xk»
L
k= 0
nl
L
m(xk+l Xk)!m(Xk+l Xk) k=O k=O =!m(ba»O.
48
Chapter 2 Calculus and the Real Numbers
This contradiction ensures that m=O. The desired conclusion follows immediately. 0 Rolle's theorem implies the mean value theorem, which gives a basic estimate for the difference of two values of a differentiable function. (5.6) Theorem. Let f be differentiable on the interval [a, b]. Then for each E > 0 there exists x in [a, b] with If(b)  f(a)  f'(x)(b a)1 ~E. Proof Define the function h on [a, b] by h(x) == (x  aHf(b)  f(a))  f(x)(b  a)
(XE [a,
b ]).
Then h(b)=h(a)= f(a)(ba). By (5.5), there exists x in [a,b] with E~
Ih'(x)1 = If(b)  f(a) f'(x)(b a)l.
0
A function f on strictly increasing) if x, YEI and x> y We decreasing) if  f is
a proper interval I is increasing (respectively, f(x) ~ f(y) (respectively, f(x) > f(y)) whenever say that f is decreasing (respectively, stricdy increasing (respectively, strictly increasing). It follows from Theorem (5.6) that if f: I > 1R is differentiable on I and f'(x) ~ 0 (respectively, f'(x) ~ 0) for all x in I, then f is increasing (respectively, decreasing) on I.
(5.7) DefinitiolL Let f, Pll,f(2), ... , P"l) be differentiable functions on a proper interval I such that Df=P 1 ), DP 1 )=P 2), . ,Df(n2)=f(nl),
and set pn)==Df("l). Then PO) is called the nth derivative of f on I, and is also written D"f; f is then said to be n times differentiable on I. The function f itself may be written PO) or DO! A natural way to simplify a continuous function and set it up for computation is to replace it by a polynomial approximation. The basic result on polynomial approximation of differentiable functions is Taylor's theorem «5.10) below). To see the motivation for Taylor's theorem, consider an n times differentiable function f on an interval I, and a point a in I. It is natural to approximate f by a polynomial of degree n whose derivatives of orders 0,1, .. ,n at a have the same values as the corresponding derivatives of f at a. The unique such
5. Differentiation
49
polynomial is obviously (5.8)
This approximation is useful for a given value b of x only when there exists a good estimate for the remainder n jO and each pair of points x, y in 1 with Ix  yl ~ w(e), we have Ig(y)  g(x)  f(x)(y  x)1 =
II II
f(t) dt 
I
f(x) dtl
= (f(t)  f(x)) dtl ~ely xl· Therefore g is differentiable on 1, with derivative f and modulus of differentiability w. If also go = f on 1, then (g  go)' = 0 on 1. By the meanvalue theorem, (ggoHa)=(ggo)(b) whenever a, bEl and a 0 we want dx d d d 1 = dx = dx (exp (In (x)) = exp (In (x)) dx In(x)=x dx In (x).
This gives (7.6)
d
dx In(x)=x 1
(x >0).
We therefore define In (x) to be the integral of XI; specifically x
(7.7)
In (x) = St 1 dt
(x >0).
1
By (6.8), In is differentiable on (0, (0) and (7.6) is valid. Let y be any positive real number. By (5.4.1), the derivative of the function (7.8)
x ~ In (x y) In (y)
is
XI. Also, (7.8) vanishes at 1. By the last statement of (6.8), In(x) =In(xy)ln(y); in other words,
(7.9)
In(xy)=ln(x)+ln(y)
(x,y>O).
This is the functional equation for the logarithmic function, corresponding to the functional equation (7.4.1) for the exponential function.
58
Chapter 2 Calculus and the Real Numbers
By (7.5) and (5.4), the composite function In 0 exp exists and differentiable everywhere on IR, and d dx(ln(exp (x))=1
IS
(XEIR).
Since also In(exp(O»=ln(I)=O, we have In (exp(x»=x
(7.10)
(XEIR),
by the last part of (6.8). Consider x> 0 and write y == exp (In (x». Then In (y) = In (exp (In (x») = In (x), by (7.10). Thus y
0= In (y) In (x) = St 1 dt. x
If y>x, this gives O;;:;yl(yX), a contradiction. By (2.18), we therefore have y~x. Similarly, x~y; whence x=y. Thus exp(ln(x»=x for all x> o. It follows that the functions exp: IR + IR + and In: IR + + IR are inverse to each other. For any a> 0 we now define
aX == exp (x In (a»
(XEIR).
We leave the reader to confirm that the map X1+ aX has the familiar properties. Note that for a fixed x> 0 the map t 1+ t of IR + into IR + extends to a continuous map of IR 0 + into IR 0 + with OX = 0, this follows from the fact that t is arbitrarily small for all positive numbers t sufficiently close to 0 The trigonometric functions sin and cos also can be approached via an intuitive analysis of their rates of change. From this analysis we are led to believe that X
X
(7.11)
d
.
COSX= Slnx, dx
d . dx smx=cosx
(XEIR).
We therefore define these functions by power series constructed in such a way that (7.11) will hold. Remembering that cos (0) = 1 and sin (0) = 0, we are forced to define 00
(7.12)
cosx==
n=O 00
sinx==
x2n
L (_l)n _ _
(2n)! ' X2n+1
L (I)n(2 n+. 1)1
n=O
Theorem (6.10) implies that (7.11) is valid.
(XEIR).
7 Certain Important Functions
59
(7.13) Proposition. The functions sin and cos satisfy the functional equations
(7.13.1)
sin (x + y) = sin x cos y + cos x sin y
(7.13.2)
cos (x + y) =cos x cos y sinx sin y
for all x, y in 1R. Proof Consider both sides of (7.13.1) as functions of x, say f(x) and g(x). The functions f and g have Taylor series about x = 0 which
converge on 1R. The coefficients of these series are expressed in terms of the various derivatives of f and g at x = O. Therefore to show that f = g on 1R it is enough to show that f(n)(o) = g(n)(o) for all nonnegative integers n. Since JO for all x with
0~xX n. Consider a value of t with xn cos Xn 
(Xn+l  x.) =0.
Since cos is a continuous function and cos t > 0 whenever 0 ~ t ~ Xn or cost>O whenever O~t~Xn+l. By induction, (7.15.1) holds for all n. Since cos Xn > 0 for all n, the sequence (xn) is increasing. From the meanvalue theorem and (7.15.1) we obtain
xn 0, by (7.15.1). Taking limits on both sides of (ii) as n + 00 we see that cos (n/2) =0. 0
By (7.15.2), sin(n/2)~0. Since cos (n/2) =0, it follows from (7.14) that sin (n/2) = 1, and hence from (7.13.1) that sin(x+n/2)=cosx for each x in IR.. Similarly, sin (n/2 x) = cos x and cos (x + n/2) = sin x. Hence sin (x + n) = cos (x + n/2) =  sin x, and thus sin (x + 2n) = sin x. Similarly, cos(x+n)= cosx and cos(x+2n)=cosx. On the other hand, if a> 0 and sin (x + tI) = sin x for all x, then cos tI = sin (n/2 + a) = sin (n/2) = 1. Since (as the reader may prove) cos is a strictly decreasing function on [0, n] and a strictly increasing function on [n,2n], and since cos 0 = cos 2n = 1, it follows that a ~ 2n. Thus the function sin, and similarly the function cos, is periodic, with period 2n.
7 Certain Important Functions
61
Next we study the function arc sin, which is inverse to the function sin. For motivation we take the derivative of the equation sin (arc sin x) = x and get d
cos (arcsin x) (arc sin x) = 1
dx
Since 1 = cos 2 (arc sin x) + sin 2(arc sin x) = cos 2(arc sin x) + x 2 , it follows that
With this motivation, we define
" 2 )tdt arcsinx=J(1t
(1<x. ,X n are real numbers such that Xl .. Xn < 0, then Xi < 0 for some i. 6. Call a pair (S, T) of nonvoid subsets of the set Q of rational numbers a Dedekind cut if s < t for all s in Sand t in T, and if for arbitrary rational numbers x,y with xO there exists c in [a, b] such that
II
f(x) g(x) dx  f(c) I g(x) dxl < E.
22. Prove that under the hypotheses of Taylor's theorem, b f(n+l)(t) R= J , (bttdt. a
n.
Hence prove that for each E> 0 there exists c with min {a, b} ~ c ~max{a,b}, such that
IR
I
f(n+1)(c) (bat+ 1 <E. (n+ I)!
23. Prove that if f is infinitely differentiable on the open interval I=(at,a+t), and if rnpn l +0
n!
then the Taylor series for
as n+oo (O o.
2 Complemented Sets
73
Note that this inequality has the additional property: if x =1= Y entails 0 = 1, then x = y. An inequality with this property is said to be tight. (2.2) Definition. Let X be a nonvoid set with an inequality =1=. A complemented set in X is an ordered pair A == (A 1, A 0) of subsets of X such that x =1= y whenever xEA 1 and YEA 0. The characteristic function of a complemented set A is the function XA: A 1 U A + {a, I} defined by if xEAt,
°
if xEAo The union (respectively, intersection) of a sequence
(An)~ 1
of com
plemented sets in X is the complemented set n?l An (respectivelY, n01 An) with characteristic function nY/An (respectively, n0/An). The
union and intersection of a finite sequence (A1' ,AN) of complemented sets are defined similarly. We often write, for example, N
A1 v ...
V
AN instead of
V An n=l
If A and B are complemented sets in X, then the complement of B in A is the complemented set A  B which has characteristic function XA (1  XB). The complement of B is the complemented set  B with characteristic function 1  XB (Thus if B == (B\ BO), then  B = (BO, B1 ).) A complemented set A is a subset of a complemented set B if Ale B1 and BO c A 0; we then write A < B. The complemented sets A and B are equal if A1=B1 and AO=Bo.
Unless we state otherwise, we shall assume that if A is a complemented set, then A 1 is the first component of A, and A ° is the second. We shall also write xEA, and say that x belongs to A, when we mean that xEA1. The following algebraic laws are valid for complemented sets: (A)=A, <Xl
 V An= "=1
<Xl
/\ (An), n=1
and <Xl
 /\ An= n=1
<Xl
V (An). "=1
74
Chapter 3 Set Theory
In addition, all the usual finite algebraic laws which do not involve the operation of set complementation are satisfied. For instance, the distributive law v v (V A W) = (V v V) A (V V W) is valid for complemented sets V, V, and W. On the other hand, the law V A (V V  V) = V is not valid. The infinite distributive law 00
0()
(2.3)
Vv
1\
V.=
n=1
1\ (Vv V.) PI=1
also fails. To see this, take X == JR and let (nk)k..l be a sequence of integers, each of which is either 0 or 1, for which it is not known if nk = 0 for any k. Write Vk =(JR, qJ) if nk = 1, =(qJ,JR)
if nk=O
and
Then V v Vk = (JR, qJ) for each k, and thus the right side of (2.3) equals (JR, qJ); but we have no way of constructing any element of JR which 00
belongs to V v
1\
Vk. Hence (2.3) is not valid.
k~1
The invalid law (2.3) has the constructive substitute (2.4)
(V v  V)
A
(V
V
ZI
V.) =(V V

V)
A
.01
(V v v.),
whose proof is left to the reader. The dual law (2.5)
(V
A 
V)
V
(V
A
.Yl V.) =(V
A 
V)
V
.Yl (V
A
V.)
is also valid. If to each complemented set A we assign the set A I , then we obtain a map j from the set of complemented sets in X to the set of subsets of X; this map preserves all the above operations except complementation. Since (S, qJ) is a complemented set for each subset S of X, j is onto the set of subsets of X. Consider the special case X == JR., with the standard operations. Then (JR 0 +, JR ) is a complemented set, where JR.is the set of negative numbers. Thus we can consider JR.to be the complement of JR 0 +. A different result would be obtained by basing the notion of complementation on negation  that is, by defining x to be in the complement of JR 0 + precisely when the assumption xEJR 0 + leads to a
3 Neighborhood Spaces and Function Spaces
75
contradiction  since we have no way of showing that XElR  whenever xElR and the assumption XElR 0 + is contradictory. On the other hand, it is true that xElR 0 + if the assumption XElR  leads to a contradiction; this follows from Lemma (2.18) of Chapter 2. Thus if we were to base complementation on negation, the complement of lR would be lR 0 +, but the complement of lR 0 + would not be lR . Hence a complemented set would not necessarily equal the complement of its complement.
3. Neighborhood Spaces and Function Spaces Sometimes a set X comes with a family of subsets that can be used to define a notion of proximity, for instance, the real numbers come with the family of all open intervals. Classically, this observation leads to the idea of a topological space, but constructively it seems more natural to introduce the concept of a neighborhood space, as follows. (3.1) Definition. A neighborhood space is a pair consisting of a set X and a family of subsets of X, called neighborhoods, such that if U and U' are neighborhoods and XE U n U', then there exists a neighborhood V with XEVC U n U'. When it is clear what the neighborhoods are, we usually refer to X itself as a neighborhood space. If x is an element of a neighborhood U, then U is called a neighborhood of x. A subset of X which is equal to the union of some family of neighborhoods is called an open set. Thus a set V is open if and only if each x in V has a neighborhood U with XE U c V. The intersection of finitely many open sets is open, and the union of an arbitrary family of open sets is open. The union of all neighborhoods which are included in a given subset Y of X is an open set called the interior of Y; it is the largest open subset of Y, and is written yo. A subset Y of X is closed if XE Y whenever x is a point of X such that each neighborhood of x contains a point of Y. The intersection of any family of closed sets is a closed set. The closure of a subset Y of X consists of all x in X each of whose neighborhoods contains a point of Y; it is the smallest closed set containing Y, and is written Y. A subset Y of X is dense in X if its closure equals X. Associated with each neighborhood space X is a subset Cont(X, lR) of F(X, lR) whose elements are called the weakly continuous functions on X. A function f: X > lR belongs to Cont(X, lR) if
76
Chapter 3
Set Theory
and only if for each x in X and each e > 0 there exists a neighborhood U of x such that If(x)  f(x')1 ~ e (x' E U) In practice, the set Cont(X, IR) turns out to be of little interest, because it is not possible to get a good hold on its structure. To see this by an example, consider the set IR of real numbers and let F denote the set of continuous functions from IR to IR as defined in Chapter 2 It is easy to see that every continuous function is weakly continuous. Therefore Fe Cont(X, IR) c F(IR, IR).
As we remarked previously, Brouwer asserts that all functions in F(IR, IR) are continuous, so that these inclusions are actually equalities; he also contends that this assertion is a theorem We support the first claim and reject the second; while reflection makes it extremely plausible that F = F(IR, IR), to accept Brouwer's arguments as a proof would destroy the character of mathematics. It seems likely that we are in the tantalizing situation in which the equalities F = Cont(lR, IR) and Cont(lR, IR) = F(IR, IR) will never be proved and never be counterexampled. This situation is typical. For most sets X we shall be interested in a certain wellstructured subset of F(X, IR) (rather than in F(X, IR) itself) which it is not possible to realize as the set of weakly continuous functions relative to any neighborhood structure on X. Therefore it makes sense to focus attention on subsets of F(X, IR) instead of on neighborhood structures For convenience these subsets are required to satisfy certain minimal restrictions. (3.2) DefinitiOlL A function space is a pair consisting of a set X and a subset F of F(X, IR), called the topology, such that the following conditions are satisfied. (a) F contains the constant functions. (b) The pointwise sum and product of two elements of F belong to F. (c) The composition hof of an element f of F and a continuous function h: IR + IR is in F. (d) Uniform limits of elements of F are in F; in other words, if cP maps X into IR and for each e > 0 there exists f in F such that IcP(x)f(x)I~E for all x in X, then cPEF. The elements of F are then called the continuous functions of the topology.
3 Neighborhood Spaces and Function Spaces
When there is no confusion over the topology in usually refer to X itself as a function space. The set F(X, lR) is a topology on X. In particular, topology on lR. A more interesting topology on lR continuous functions as defined in Chapter 2. Let F be a topology on a set X. For all f and g in F,
77
question, we F(lR, lR) is a is the set of
since
max{1, g} = f +max{g  1, O} = f +t(g  f + Ig fl), max{1,g} belongs to F; similarly, min{1,g} belongs to F. Since fg=i((f + g)2 _ P _g2),
the requirement that products of elements of F belong to F is actually superfluous A topology F for a set X introduces a notion of proximity into X. As we remarked above, a notion of proximity is introduced into X classically not by giving a family of functions, but by giving a family of subsets, either open sets or neighborhoods. Classically, this is equivalent to giving a family of functions from X to {O, I}. Constructively, there is a vast difference: since functions are sharply defined, whereas most sets are fuzzy around the edges, only the alltoorare detachable subsets of X correspond to functions from X to {O, I}. The fuzziness of sets is another reason to focus attention on function spaces instead of on neighborhood spaces. The truth of the matter is that neither function spaces nor neighborhood spaces are as important as certain related structures  metric spaces and normed linear spaces  which will be studied 10 later chapters. One can pass from a neighborhood structure on a set X to a topology, or go the other way. (3.3) Proposition. The set Cont(X, lR) of weakly continuous functions on a neighborhood space X is a topology for x.
Proof The proof is routine, and is left to the reader.
D
Conversely, a topology F on a set X induces a neighborhood structure on X, obtained by taking neighborhoods to be sets of the form Vf=={XEX: f(x»O} where f is an element of F. To see this, consider elements 1, g of F and define h == min {1, g}. Then Vh = Vf n Vg , so that the intersection of neighborhoods is a neighborhood. Thus X is a neighborhood space.
78
Chapter 3 Set Theory
Any set X can be topologized: take F == F(X, IR). This is the finest topology, and the most natural; its lack of structure keeps it from being important. At the other extreme, X can be topologized by letting F consist of the constant functions and nothing else; this topology is too small to be useful. Any subset Fo of F(X, IR) can, of course, be used to generate a topology F for X; the topology in question is the smallest subset of F(X, IR) containing Fo and satisfying conditions (a){d) of Definition (3.2). In fact, F is just the subset of F(X, IR) obtained inductively from the following principles of construction. (i) (ii) (iii) (iv)
The elements of Fo and the constant functions are in F. The pointwise sum and product of two elements of F are in F. hof is in F whenever fEF and h: IR+IR is continuous. A uniform limit of elements of F is in F.
Problems 1. Prove the algebraic laws stated after Definition (1.3).
2. A family of sets consists of an index set T, a finite routine A. that assigns to each element t of T a set A.(t), and a finite routine ¢J that assigns to each ordered pair (t', t) of elements of T with t = t' a function ¢J(t', t) from A.(t) to A.(t') such that (i) ¢J(t, t) is the identity map on A.(t) and (ii) ¢J(t", t')o¢J(t', t)=¢J(t", t) whenever t=t'=t". (This definition is due to Richman. The sets A.(t) are not necessarily subsets of some fixed set.) Define the exterior union (or disjoint union) of such a family. In the case where the sets A.(t) are all subsets of a fixed set S, relate the exterior union to the interior union defined at (1.5). 3. Show that the family of detachable subsets of a set X need not be closed with respect to countable unions. 4. Let S be a discrete set, and f an element of F(S x s, {O, I}) such that elements sand s' of S are equal if and only if f(s, 8') = 1. Show that the relation =1=, defined by setting s =1= 8' if and only if f(8, 8') = 0, is an inequality relation on S.
Notes
79
5. Let S be a discrete set with inequality defined as in Problem 4, and let (T, A) be the family of detachable subsets of S. Show that if a, bES and a*b, then there exists t in Twith aEA(t) and bE A(t). 6. Call two elements a and b of a set A weakly unequal if there exists a function f. {a, b} .lR with f(a) f(b). Show that a and b are weakly unequal if and only if a = b entails 0 = 1.
*
*
7. Let A and B be arbitrary sets, and let be an inequality relation on B. Show how to define an inequality relation on F(A, B). 8. Let A be a function from a set A into F(A, {O, I}). Show that there exists f in F(A, {O, I}) such that f *A(a) for all a in A, where is the inequality on F(A, {O, I}) defined in Problem 7. (This is a constructive version of Cantor's result that a set is always smaller than the set of its subsets.)
*
9. Prove (2.4) and (2.5). 10. Show that the union of two closed subsets of lR need not be closed. 11. A subset S of lR is convex if tx+(lt)y belongs to S whenever x, YES and 0 ~ t ~ 1. A function space (X, F) is connected if for each f in F the closure of f(X) is a convex subset of lR. Show that a product X =Xl x .. X X N of connected spaces (Xl, Fd, ... , (X N, FN) is connected, where the topology for X is the topology generated by the set of all functions (Xl' ... , xN)t+(fkEFk' 1'5;, k '5;, N). 12. Prove that a function space (X, F) is connected if and only if for arbitrary x, y in X, arbitrary bounded elements fl, ... , fn of F, and an arbitrary e > 0, there exist Xl> ... , XN in X such that Xl = X, XN = y, and
Notes Contrary to classical usage, the notion of equality is a convention. We define what it means for elements of a given set to be equal, but the notion of equality of elements of different sets is not defined. The only way in which elements of different sets A and B can be regarded as
80
Chapter 3 Set Theory
equal is by realizing A and B as subsets of a third set S. For this reason we define the operations of union and intersection only for sets which are given as subsets of a given set. (See, however, Problem 2.) An alternative approach to Definition (1.4) would be to define a family of subsets of a set S as a subset of the cartesian product S x T, where T is the index set for the family. The concept of a complemented set suffers from what appears to be a glaring deficiency· if X is a set with many elements, it seems unnatural to call (qJ, qJ) a complemented set in X, especially since (qJ,X) is also a complemented set. On the other hand, there is nothing to force us to consider complemented sets like (qJ, qJ) In the applications there will always be some condition that keeps the complemented sets A with which we are concerned from having the property that both Aland A 0 are ridiculously small. Brouwer's contention that all elements of F(1R,1R) are continuous seems to contradict claims of certain recursive function theorists, who give examples of elements of F(1R, 1R) that are not continuous In both instances, the claims are based on extramathematical considerations. Brouwer analyzes all possible techniques for constructing elements of F(1R,1R), and comes to the conclusion that all such elements are continuous. The recursive function theorists analyze the possibilities for constructing real numbers, and come to the conclusion that they all possess a certain property (they are recursive); in addition, they show how to construct a discontinuous function on the set of recursive real numbers. These two positions are, in fact, compatible; they do not contradict each other, because it is possible to believe both that all constructive real numbers are recursive, and that without making use of some unprovable hypothesis (such as the hypothesis that all constructive real numbers are recursive), the only elements of F(1R, 1R) that can be constructed are continuous. Extramathematical considerations of both types (especially the first) are useful in indicating that we should not try to do certain things constructively; but they have no place in the actual development of constructive mathematics. Definition (3.2) should not be taken too seriously. The purpose is merely to list a minimal number of properties that the set of all continuous functions in a topology should be expected to have. Other properties could be added, to find a complete list seems to be a nontrivial and interesting problem.
Chapter 4. Metric Spaces
The concept of a metric is defined, some examples are studied, and various techniques for constructing metrics are developed The neighborhood structure of a metric space is defined, and the notions of weakly continuous and uniformly continuous functions are introduced. Completeness is defined in Section 3, and the construction of the completion is carried through. Following Brouwer, we define a compact space to be a metric space that is complete and totally bounded. Compact and locally compaU spaces are studied in Sections 46. Constructivizations of various classical results, such as Ascoli's theorem, the Stone Weierstrass theorem, and the Tietze extension theorem, are given. The concept of a located set, due to Brouwer, plays an important role. Crucial for later developments is Theorem (49), a partial substitute for the classical result that a closed subset of a compact space is compact.
The sets that occur in analysis are constructed from the set of real numbers by certain standard methods, for instance by the formation of subsets or cartesian products of sets already defined. They often carry a structure, such as a function space structure or a neighborhood structure, built up at the same time the sets themselves are constructed. Of special importance is the structure given by a metric, where a metric on a set X is a function on X x X satisfying certain properties that we intuitively associate with distance. A metric is yet another way of introducing a notion of proximity into X.
1. Fundamental Definitions and Constructions Guided by our knowledge of the real numbers and our intuition, we introduce the notion of distance as follows. (1.1) Definition. A metric on a set X is a function p: X x X such that for all x, y, and z in X
+ 1R 0 +
82
Chapter 4
(1 1.1)
(1.1.2) ( 1.1.3)
Metnc Spaces
p(x,y)=O
if and only if x=y
p(x, y) = p(y, x) p(x, y) ~ p(x, z) + p(z, y)
(the triangle inequality).
A pair (X, p) where p is a metric on X is called a metric space. When there is no confusion over the metric, we often refer to X itself as a metric space. The standard inequality relation on a metric space (X, p) is defined by x =1= y if and only if p (x, y) > O. It often happens that in place of (1.1.1) the weaker condition p(x,y)=O
if X=y
is satisfied. In this case p is called a pseudometric, and (X, p) is called a pseudometric space. Associated with each pseudo metric space (X, p) is a metric space (X 0, p) obtained by taking X 0 to be the set X with the equality relation x = y modified to mean that p(x, y) =0. The theory of pseudometric spaces therefore is essentially no different from the theory of metric spaces. The most important example of a metric space is the set of real numbers, metrized by defining p(x, y) == Ix  yl
(x, YElR)
In order to define metrics for other sets, it is best to establish some general methods of metrization. (1.2) Definition. Let f: Y+X be a function from a set Y to a pseudo metric space X The induced pseudometric p* on Y is defined by (1.2.1)
p* (y, i) == p(f(y), f(i»
(y, i
E
Y).
For p* to be a metric it is sufficient that p be a metric and that y = y' whenever fey) = fey')· As an example of the use of Definition (1.2), the inclusion map i: Y+ X from a subset Y of a metric space X into X induces a metric on Y. In particular, the above metric on 1R induces metrics on CQ, 71., and the various intervals. As another example, if f: X +1R is any realvalued function on a set X, the pseudometric p* is given by p* (x, y) = If(x)  f(y)l.
1 Fundamental Definitions and Constructions
83
(1.3) Definition. Let (Xl> pd, ... , (Xn' Pn) be metric spaces. The product metric P on the cartesian product space X == Xl X . • x X n is given by
•
p(X, y) == L Pi(Xi, Yi)
(1 3.1)
i=l
There are other ways of metrizing a finite product of metric spaces. In fact, JR" is usually given the metric d defined by d(x,y)== (tl(XiYi)2r,
to correspond to the geometry of lR· as we know it from experience. This metric is so important that we digress at this point to give the proof that it actually is a metric. To this end, we establish two basic inequalities. The first is known as the CauchySchwarz inequality. (1.4) Proposition. Let
Xl> ...
,x.,
Yl. ...
,Y. be real numbers. Then
(1.4.1) Proof: We compute
PI
PI
PI
PI
PI
PI
=iLxt LyJ+iLxJ LYt LXiYi LXjYj i=l
j=l
j=l
i=l
i=l
j=l
= L i(xt yJ +xJ yt  2XiYjXjYi) i.i=l n
= Li(XiYjXjYi)2~O. i,j=l
This is equivalent to (1.4.1).
D
The next inequality is called Minkowski's inequality. (1.5) Proposition. Let (1.5.1)
Xl.
.,
X.,
Yl.
, Y.
be real numbers. Then
84
Chapter 4
Metnc Spaces
Proof Using (1.4.1), we compute n
PI
PI
PI
I(XiYi)2= LX?+ Ly?2L x iYi
i=l
i=1
~itIX?+it/?+2 (~x?)t (t/?r =(
(tl x?
This is equivalent to (1 5 1).
r (t/?rr· +
0
(1.6) Corollary. The function d is a metric on JR". Proof Clearly, d satisfies Minkowski's inequality, d(x,
conditions
and
(1.1.1)
z) =(tl «Xi  Yi) (Zi  Yi))2)
(1.1.2).
By
t
~ (tl (Xi  Yi)2 r +(tl (Yi  zif r = d(x, y) + d(y, z)
X==(XIo
for all ,x"), Y==(YIo d also satisfies (1.1.3). 0
,Yo), and
Z==(ZIo ... ,z") in JR". Therefore
Related to the metric d on JR" is a certain metric on the set F of continuous, realvalued functions on a compact interval [a, b]. For arbitrary elements f. g of F we define d(f.
g) == (! If(x)  g(xW dx f·
Minkowski's inequality for integrals, d(f. g) ~ d(f. h) + d(h, g)
(f. g, hE F),
follows from (1.5.1) by approximation. There is also a CauchySchwarz inequality for integrals,
IJ f(x) g(x) dxl ~ (J f(X)2 dx)t (J g(X)2 dx)"t
(f. gEF),
which follows from (1 4 1) by approximation. The CauchySchwarz and Minkowski inequalities admit many other generalizations, as will be seen later. In the meantime, we end our digression, and return to the consideration of metric spaces in general
1. Fundamental Definitions and Constructions
85
A metric space (X, p) is called bounded if there exists a real number C > 0, called a bound for (X, p), such that p(x, y) ~ C for all x and Y in X. A subset Y of a non void metric space X is bounded if, for all (equivalently, some) x in X, the set Yu {x} with the induced metric p* is a bounded metric space. A subset Y of IR can be bounded as a metric space but not bounded as a subset of IR. A countable product of metric spaces can be metrized when each of the spaces is bounded. (1.7) Definition. Let
«Xn,Pn))~l
be a sequence of metric spaces, each
wi th bound 1 The product metric p on X ==
TI X
n
is defined by
n=1
L2 npn(x n,Yn)
P«Xn),(Yn))==
«xn), (Yn)EX).
11=1
Definition (1 7) is useful even when the .spaces (X n, Pn) are not necessarily bounded, because there is a standard method for transforming an arbitrary metric into a metric bounded by 1. (1.8) Proposition. Let (X,p) be a metric space. Let h IR o + +IR o + satisfy the conditions
if and only if u=O h(u + v) ~ h(u) + h(v) (u, VE IR 0+).
(1.8.1)
h(u)=O
(1.8.2)
Then d == hap is a metric on X Proof Condition (1.1.1) is satisfied because of (1.81). Condition (1.1.2) is obviously satisfied To check (1.1.3) note that d(x, z) = h(p{x, z)) ~ h(p(x, y) + p(y, z)) ~ h(p(x,
for all x, y, z in X.
y)) + h(p(y, z)) = d(x, y) + dey, z)
0
(1.9) Corollary. If p is any metric on a set X, then the function p' defined by p'(x,y)==min {p(x,y), I} is a bounded metric on X.
Proof. The proof is obvious.
0
(X,YEX)
86
Chapter 4
Metnc Spaces
This corollary leads us to define the product of an arbitrary sequence ((X n, Pn)) of metric spaces to be the product of the associated sequence ((X n, p~)). The definition of a continuous function on a compact interval generalizes in the context of a metric space, and leads to the important idea of a uniformly continuous function. (1.10) Definition. A function f: X ~ X' from a metric space (X, p) to a metric space (X', p') is uniformly continuous if there exists an operation w from IR+ to IR+ such that p'(f(x),f(Y))~E whenever X,YEX, E>O, and p(x, y) ~ wee). The operation w is called a modulus of continuity for f on X For example, if Xo is a point in an arbitrary metric space X, then the function XI+p(x,xo) is uniformly continuous on X, with modulus of continuity EI+E. To see this, consider points x and y in X We have p(x, xo) ~ p(x, y) + p(y, xo) and therefore p(x, xo)  p(y, xo) ~ p(x, y). Similarly, p(y, xo)  p(x, xo) ~ p(x, y) Therefore Ip(x, xo)  p(y, xo)1 ~ p(x, y)
It follows that Ip(x, xo)  p(y, xo)1 ~ E whenever p(x, y) ~ E. For each x==(x n ) in the product (X,p) of a sequence ((Xn,Pn)) of metric spaces (each with bound 1), and each k in Z+, write nk(x) == Xk. Then the projection nk: X ~Xk of X onto X k is uniformly continuous,
with modulus of continuity
EI+2 k E.
(1.11) Definition. A sequence (1.) of functions from a set S to a metric space (X, p) converges uniformly to a function f: S ~ X if for each E > 0 there exists N. in Z + such that p(J.(X),f(X))~E
(n~N.,xES).
+ g and product fg of uniformly continuous functions f" X ~IR and g: X ~IR are uniformly continuous, and f 1 is uniformly continuous if If(x)l~c for all x in X and some c>O. The composition f2 0 h of uniformly continuous functions f1: X 1 ~ X 2 and f2: X 2 ~ X 3 is uniformly continuous. The limit f: X 1 ~ X 2 of a uniformly convergent sequence (fn) of uniformly continuous functions from a metric space Xl to a metric space X 2 is uniformly continuous. (1.12) Proposition. The sum f
Proof. The proofs of the various statements are similar to certain proofs in Chapter 2, and are left to the reader. D
2 Associated Structures
87
A uniformly continuous function f from a metric space X I onto a metric space X 2 which has a uniformly continuous inverse is called a metric equivalence, and X I and X z are said to be equivalent metric spaces. Two metrics Pl and pz on the same set X for which the identity function from X to itself is a metric equivalence of (X, PI) with (X, pz) are called equivalent metrics on X. The product metric P on JR' is equivalent to the metric d. An arbitrary metric P is equivalent to the bounded metric P' of Corollary (1.9). If (X.,P.) and (Y",,o.) are equivalent metric spaces for each n in 7L+, then the products of the sequences «Xn,Pn» and «Y",,on» are equivalent.
2. Associated Structures Every metric space has a neighborhood structure and a topology. (2.1) Definition. The open sphere of radius r>O about a point x in a metric space X is the subset S(x, r)=:o {YEX: p(x,y) IR which are uniformly continuous on each member of a given family of subsets of X. For example, this was done when we defined the continuous functions on an arbitrary interval in IR
3. Completeness The construction which led from
p(x,A)
p(x, A) > (2n)1
+
{O, I} such
and With z any point of A, construct a sequence (an) in A such that: if ),(1) = 1, then an = z for all n; if ),(1) = 0 = },(n), then p(x, an) < n 1 , while if },(1)=0 and },(n) = 1, then an=anl> where m is the unique positive integer with l(m) =0 and l(m + 1) = 1. Then (an) is a Cauchy sequence of elements of A: in fact, p(am,an)~2nl whenever m~n. It therefore converges to a point a in A with p(a,an)~2nl for each n. Suppose that p(x,a»O, and compute N in 7l.+ so that p(x,a»3N 1 • Then
p(x, aN) ~ p(x, a)  p(a, aN) > N 1 •
3 Completeness
93
Therefore A(N) cannot be 0, so that A(N)=l Hence p(x,A»(2N)1 >0. 0 An immediate corollary of Lemma(3.S) is that p(x, A) >0 whenever A is a complete, located subset of a metric space X, and x is a point of X such that x =1= y for all y in A We now arrive at the Baire category theorem. (3.9) Theorem. Let (Un) be a sequence of dense open sets in a complete
n Un is also dense in X. 00
metric space X. Then the intersection U ==
n~l
Proof. Let S(xo, ro) be any open sphere in X. Since U1 is dense, there exists a closed sphere S c (x1,rd with 0 < r1 ~ 1 such that Sc(x1,rd c S(xo,ro)n U1·
Continuing inductively, we construct a sequence (S c(xn, rn)) of closed spheres such that 0 < rn ~ n 1 and Sc(xn' rn)cS(xn_ 1, rn_ 1)n Un
for all n in Z+. Whenever m~n, Xm belongs to S(xn,rn), and therefore p(x nlO xn) < rn ~ n 1. Hence (xn) is a Cauchy sequence. Since all except finitely many terms of this sequence lie in any given closed sphere S c(xn, rn), the limit x lies in each of these spheres Therefore cc
XES(Xo, ro)n nUn. n=l
It follows that U is dense in X.
0
Theorem (3.9) is one of the most useful versions of Cantor's diagonal technique, which was used in the proof of Theorem(2.19) of Chapter 2. In fact, the latter theorem is a special case of Corollary (3.10) below. A located subset Y of a metric space X is nowhere dense in X if the metric complement  Y of Y is dense. With this definition we reformulate Theorem (3.9) (3.10) Corollary. Let (y") be a sequence of nowhere dense subsets of a complete metric space X. Then every open sphere in X contains a point y whose distance to each Y" is positive. Proof" Let Sex, r) be an open sphere in X. By (3 9), there exists a point
n  Y". The distance from y to Y" is positive because yE 00
y in Sex, r) n
 Y".
0
n~l
94
Chapter 4 Metnc Spaces
4. Total Boundedness and Compactness A more significant concept for metric spaces than boundedness is the concept of total boundedness, defined as follows. (4.1) Definition. Let (X, p) be a metric space. An e approximation to X is a subset Y of X such that for each x in X there exists y in Y with p(x, y) < e. We say that X is totally bounded if for each e > 0 there exists a finite e approximation to X. The reader can show as an exercise that X is totally bounded if for each e > 0 there exists a subfinite e approximation {XI. ... , x n } to X. A subset of a metric space is totally bounded if and only if its closure is totally bounded The property of total boundedness is preserved under passage to an equivalent metric. The product of a sequence of totally bounded metric spaces is totally bounded. We call a metric space X separable if it contains a countable dense subset. The product of a sequence of separable metric spaces is separable. A totally bounded metric space X is separable. for if Sn is a <Xl
finite n 1 approximation to X (nEZ+), then USn is a countable dense n=1 set in X. (4.2) Proposition. The image f(X) of a totally bounded metric space X under a uniformly continuous mapping f: X ~ Y is totally bounded.
Proof Let w be a modulus of continuity for f For each e>O let {XI. ... , x n } be an w(ie) approximation to X. Then if x is any point of X, we have p(x, Xi) <w(te) for some i with 1 ;;£i;;£n. For this i we have p(f(x), f(Xi»;;£te < e. It follows that {f(xd, ... , f(x n )} is a subfinite e approximation to f(X). 0 (4.3) Corollary. Let f: X ~ 1R. be a uniformly continuous function on a totally bounded metric space X. Then the supremum and infimum of f exist.
Proof. By (4.2), we see that f(X) is totally bounded. The result now follows from (4.4) of Chapter 2. 0 It follows that if X is totally bounded, then the diameter
diamX=sup{p(x,y): X,YEX} of X exists.
4. Total Boundedness and Compactness
95
(4.4) Proposition. A totally bounded subset Y of a metric space X is located. Proof. If x is any point of X, then yf> p(x, y) is a uniformly continuous function on Y, as we have seen. By (4.3), its infimum p(x, Y) exists. Thus Y is located. 0 (4.5) Proposition. A located subset Y of a totally bounded metric space X is totally bounded.
Proof: Consider 8> 0, and let {XI, ... , xn} be an 8/3 approximation to X. For each i choose Yi in Y with P(Xi,y;) y) < 8/3. This gives p(y, Yj) ~ p(y, Xj) + p(Xj, Yj) < 8/3 + 8/3 + 8/3 = 8.
Thus the subfinite set {Yl> . . , Yn} is an 8 approximation to Y. Since arbitrary, it follows that Y is totally bounded. 0 (4.6) Proposition. A subset X of JR" is totally bounded
8
is
if and only if it
is located and bounded. Proof Proposition (4.4), together with the fact that a totally bounded metric space is bounded, implies that the conditions are necessary for X to be totally bounded. Conversely, assume that X is located and bounded. Because X is bounded, there exists c>O such that Xc [  c, C ]n. The product space [  c, cJ" is totally bounded. As a located subset of a totally bounded metric space, X is totally bounded, by (4.5).
0
One of the most important concepts of classical analysis is that of compactness, effective application of which cuts infinite sets down to finite, and therefore manageable, size. Regrettably, most such applications use either the 'open cover' or the 'sequential' forms of compactness, neither of which is appropriate in a constructive setting. For example, there is no known case of a nontrivial metric space for which there exists a routine method for extracting finite subcovers from arbitrary open covers. We are therefore relieved to find that the other common form of compactness of metric spaces is acceptable, fruitful, and hence adopted as definitive in constructive analysis. (4.7) Definition. A compact metric space, or simply a compact space, is a metric space which is complete and totally bounded.
96
Chapter 4
Metric Spaces
The compact intervals of real numbers are compact in the sense of the above definition. A subset of 1R.. is compact if and only if it is closed, located, and bounded. The product of a sequence of compact spaces is compact. Total boundedness is the more important of the two properties which occur in the definition of compactness, since a metric space which is totally bounded but not necessarily complete can always be compactified by embedding it in its completion and taking its closure. Let X be an arbitrary metric space. For each located subset B of X the function x I> p(x, B) is uniformly continuous, by (2.4). Therefore the supremum m(A,B)=sup.{p(x,B): xEA}
exists for each compact set A c X. The function p defined by p(A,B)=max {m(A,B), m(B,A)}
is well defined on Cx C, where Cis the set of all compact subsets of X. We shall show that p is a metric on C. Clearly, peA, B) = pCB, A), and p(A,A)=O. Assume that p(A,B)=O. Then m(A,B)=O, so that p(x,B) = 0 for all x in A. Since B is closed, it follows that XE B whenever xEA, whence AcB. Similarly, BcA. Thus A=B whenever p(A,B)=O. It remains to check that p satisfies the triangle inequality To this end, consider arbitrary compact subsets A, B, and C of X, and an arbitrary x in X. By (2.4), p(x, C)~inf{p(x,y)+p(y, C):YEB} ~inf{p(x,y)
YEB} +m(B, C)~p(x,B)+p(B, C).
Therefore meA, C)~m(A,B)+p(B, C)~p(A,B)+p(B, C).
Similarly, m( C, A) ~ peA, B) + pCB, C).
Therefore peA, C)~p(A,B)+p(B, C),
and so p satisfies the triangle inequality. Turning now to the study of sufficient conditions for a subset of a compact space to be compact, we first prove that every compact space is a union of finitely many compact sets of arbitrarily small diameter. (48) Theorem. Let X be a compact space, and e a positive number. Then there exist finitely many compact subsets X b ... , X. of X, each with diameter at most E, whose union is X.
Proof. By the total boundedness of X, there exist subsets XL ... , X! of X, each consisting of one point, such that for each x in X at least one
4 Total Boundedness and Compactness
97
of the numbers p(x, XJ) (1 ~j~n) is less than 3 2 e. We define inductively sequences (XD~ b ... , (X~)~ 1 of subfinite subsets of X such that for 1 ~j ~ nand i"?; 1, (i)
X} cXJc
(ii) (iii)
if XEX~+l, then p(x,X))O, represent X as a finite union X
=
U Xj of compact j=l
sets, each of diameter less than e/3. If sup {p(x, Xj). XEX} < e/3 for some j (1 ~j ~ N), then p(g(x), g(y»
~ diam
X <e
(x, yEf(X».
In proving g uniformly continuous, we may therefore assume that sup {p(x, Xj): XEX} >0 for each j; whence, by (4.9), there exists r such that 0 < r < e/3 and each of the sets Sj=={XEX·
p(x,X)~r}
={XEX:
p(x,Xj)~
r}
(l~j~N)
100
Chapter 4
Metric Spaces
is compact. Define and A.(O)=O whenever i*j. By (5.5), the set B={U=(Ul, ... ,U.)EIRn. IUil:;:::;K, IUiujl:;:::;rij (1:;:::;i,j:;:::;n)}
is totally bounded. Therefore it is enough to show that A = B, or (since obviously AcB) that BcA. Since X is totally bounded, (Xl, ... ,X.) can be extended to a sequence (Xk)k= 1 of elements of X which is dense in X. Consider an element U of B. We continue the finite sequence (Ul, . . ,u.) to an infinite sequence (Uk) such that for all i,j in 7L+, (5.6.1) This is done inductively. Certainly, (5.6.1) holds for 1 :;:::;i,j:;:::;n. Assume that Ulo ., Urn have already been constructed to satisfy (5.6.1). Then, by (5.4), Urn+l can be constructed to satisfy (5.6.1) because the hypotheses on 2 ensure that
lUi ujl:;:::; 2(P(Xi' Xj)):;:::; 2[P(Xi' xm+d + P(Xm+lo Xj)] :;:::;A.(P(Xi' xrn+d)+2(p(xjo Xm+l))
(1 :;:::;i,j:;:::;m).
Thus the sequence (Uk) is constructed inductively. Let Y be the dense subset {Xl, X2, ... } of X. Define h: Y +JR. by h(xi)=ui (iEZ+). By (5.6.1), we see that h is uniformly continuous. It follows from (3.7) that h has a uniformly continuous extension f to X. By (5.6.1), the fact that Y is dense, and the continuity of the various functions involved, we see thatfES. Since u=(f(xd, .,f(x.)), U belongs to A. Thus BcA, as was to be proved. 0 (5.7) Corollary. Let X be compact, and let K, c, and (X be positive numbers with (X:;:::; 1. Let S be the set of all f in C( X) with II f II :;: :; K which satisfy the Lipschitz condition
If(x)  f(y)l:;:::; c p(x, y)~
(x, YEX)
Then S is compact. Proof· The result follows from (5.6) by taking 2(t)=ct~ for each t~O. (It is a simple exercise to show that (s + t)~ ~ s~ + t a whenever s;;::: 0, t;;::: 0, and O so small that II cf II ~ 1. Let Ii> 0, and let p be the polynomial function of (5.9) Then
°
Ilcf(x)l p(cf(x))1 ~ Ii
Since po(cf) belongs to H, and Therefore IfIEH, and so
Ii
(XE X).
is arbitrary, it follows that IcflEH
max {f,g} =t(f + g+ If  gI)EH. Similarly, min {f,g}EH.
0
106
Chapter 4 Metnc Spaces
(5.12) Lemma. Let G be a set of continuous functions on a compact space X, and let H be the closure of d(G) in C(X). Let h be a function in H with O 0, choose N in 7l+ so 00
L
2 n p(x, y) 
e/2 ~ e/2.
n=1
Therefore P.(x n,Yn»e/2 for some n with n~N. Write g=::fnonn, where the continuous function fn: X n + IR is defined by fn(zn)=::(Pn(x n,Yn))1 Pn(xn,z.) If Z=::(Zk)EX and
p(x,z)~c5(e),
(znEXn)
then
Ig(z)1 = If.(zn)1 ~ 2e 1P.(x n, zn) ~2e12np(x,z)~2cI2N
Similarly, if
p(y,z)~c5(e),
Ig(z) 11
c5(e)=e.
then
= Ifn(zn)  fn(Y.) 1 ~ 2e 1IPn(x n, z.)  Pn(x n, Yn)1 ~ 2e 1Pn(Yn,
Therefore G is separating
zn) ~2c 12N c5(e) = e.
0
(5.16) Corollary. Let X be a compact space with positive diameter, and let G consist of all functions of the form Xf+p(x,x o) with XoEX. Then d(G) is dense in C(X). Proof' Let d=::diamX and c=d/7. Let F be a finite c approximation to X Consider any x in X, and suppose that p(x,x') c for some x' in F Let e>O, and write b(e)=min{e 2 ,ce}. Consider any x and y in X with p(x,y)~e. Define g in d(G) by g(z)=p(X,y)l p(x,z)
(ZEX)
If p(x, z) ~ b(e), then Ig(z)1 ~ e 1 p(x, z) ~ e,
and if p(y, z)~b(e), then Ig(z) 11 = p(x, y)llp(x, z)  p(x, y)1 ~C1
p(y, z)~e.
On the other hand, if YEX is arbitrary and we choose y' in F so that p(y,y'»c, then (5.13.2) holds with g the function Zf+p(y,y')l p(Z,y') in d(G). Hence d(G) is separating, and so, by (5.14), d(G)=d(d(G)) is dense in C(X). D The case n= 1 and X = [ 1,1] of our third corollary is the famous Weierstrass approximation theorem. (5.17) Corollary. Every continuous function on a compact set Xc lR" can be arbitrarily closely approximated on X by polynomial functions. Proof: First consider the case in which n = 1. By (5 9), the function xf+lxxol (with XoEX) can be arbitrarily closely approximated on X
by polynomials. The theorem then follows from (5 16). Next consider the case where X = [a, b]" for some compact interval [a, b]. The result then follows from (5.15) and the case just proved. Finally, consider the general case. Since X is bounded, there exists a compact interval [a, b] with X c [a, b]". By the case just considered, each of the functions Xf+p(x,x o) (with xoE[a,b]") can be arbitrarily closely approximated by polynomials on [a, b ]". The result then follows from (5.16). D
6. Locally Compact Spaces Many important metric spaces, such as the euclidean spaces lR", that are not compact have a property almost as good.
110
Chapter 4
Metnc Spaces
(6.1) Definition. A nonvoid metric space X is locally compact if every bounded subset of X is contained in a compact subset. In that case, a function f: X . Y from X to a metric space Y is continuous if it is uniformly continuous on every compact subset of X, or, equivalently, on every bounded subset of X. A continuous function g: X .IR such that the set X,,={XEX:
g(x)~ex}
is bounded for all cx in IR is called a compactifier for X. If Xo is any point in the locally compact space X, then the function XI+ p (x, xo) is a compactifier. The term 'compactifier' comes from the fact that if g is any compactifier, then Xu. is either void or compact for all except countably many real numbers cx. This follows from Theorem (4.9). Since a locally compact space is a countable union of compact spheres (by Theorem (4.9)), it is separable. Since any Cauchy sequence is bounded, a Cauchy sequence in a locally compact space is contained in a compact set, and therefore converges. Hence a locally compact space is complete. (6.2) Proposition. A locally compact subset Y of a metric space X is located. Proof: Let x be any point of X. Let YoEY, and choose c>2p(x,Yo) so
that
Y,,= {YE Y: p(y, yo)~c} is compact Since p(x, YcJ~p(x,yo) 0 so that B c: Y n Xc and X 4c is compact. Note that if XEX and p(x, Yo) < 2c, then
6 Locally Compact Spaces
III
p(X, YnX 4c ) exists and equals p(x, Y): for in that case, if YEY and p(x, Y) < p(x, y) + 2c  p(x, Yo), then p(y,yo)~p(x,y)+p(x,Yo) Y  {w}; and let f be an element of Coo (X). Then there exists a unique element f# of C(Y) such that
(611.1)
for all x in X,
f#(i(x»=f(x)
and
f#(w)=O.
Proof. In view of (3 7) and the fact that i(X) u {w} is dense in Y, it will suffice to prove that the map f#. i (X) u {w} > lR defined by condition (6111) is uniformly continuous To this end, let j: i(X)>X be the inverse of the mapping i Given c > 0, choose a compact set K c X such that If(x)lO. Choose c with 00 there exists N in 71+ such that whenever n1 0, we write
Kr== {ZECC: p(z, K)~r}. A totally bounded set K c CC is well contained in an open set U c CC if Krc U for some r>O. We then write Kcr::, U. If K c CC is totally bounded, then Kr is compact for each r > O. To see this, we note that Kr is then closed and bounded, and we refer to (63) of Chapter 4 and the following lemma. (2.3) Lemma. Let K be a located subset of CC, and r a positive number. Then Kr is located, and (2.3.1)
p(z, Kr)=max{O, p(z, K)r}
for each z in CC. Proof Consider an arbitrary complex number z If Z1 EK and IZ21 ~ r, then
2. Derivatives
131
Therefore IZ(Zl +z2)1 ~tx(z), where tx(z) is the righthand side of (2.3.1). On the other hand, let I: be any positive number, and choose , in K with Iz'I O. In the latter case, let " be that unique nonnegative multiple of z  , for which lei =min{r,lzW. Then ,+" belongs to K., and Iz(' +01 =max{lz'Ir, O} <max{p(z, K)r, O} +1:. Since I: > 0 tx(z). D
IS
arbitrary, it follows that p(z, Kr) exists and equals
(2.4) Definition. A function f: U > (c defined on an open subset U of (c is continuous if it is continuous on every compact set K which is well contained in U. If f and g are continuous on U and I' = g on every compact set K ~ U, then g is called a derivative of f on U, and f is said to be differentiable on U.
The derivative of f on a compact or an open subset of (C is unique, if it exists. Many of the results obtained in Chapter 2 for differentiation in the real domain carryover to the complex numbers: for instance, the rules for differentiating sums, products, quotients, and polynomials. The chain rule, (2.5)
(go fY
= (g' of) 1',
is valid whenever f is a differentiable function from a compact subset K of (C into a compact subset L of (C and g is a differentiable function on L In case f is a differentiable function from an open subset U of (C into an open subset V of (C, and g is differentiable on V, equality (2.5) is valid under the extra hypothesis that f(K)~ V for each compact K~ U. (See Theorem (5.17) below for conditions under which this extra hypothesis holds.) It is easily seen that if f and g are complexvalued functions on an open set U c (C such that for each compact set K ~ U there exists an operation /j from lR+ to lR + with respect to which (2.1.1) holds, and such that either f is continuous or g is bounded, then f and g are continuous, and therefore g = 1'. There is another kind of derivative in the complex domain (2.6) Definition. Let f and g be continuous complexvalued functions on a compact set K c (C, and /j an operation from lR + to lR + such
132
Chapter 5 Complex Analysis
that li(xz+iy) f(x l +iy)g(x l +iY)(XZXI)I~Elx2Xll
whenever Xl +iYEK, Xz+iYEK, E>O, and IXZXII~c5(E) Then g is called the partial derivative of f with respect to X on K  written g =Dx!' or g=fx, or g(z)=("1f(z)/(ix  and c5 is called a modulus of xdifferentiability for f on K We say that g = fx on an open set U if g = fy on every compact set K rt:, U. The partial derivative fy of f with respect to y is defined similarly. It is clear that a differentiable function f on a compact set K (respectively, on an open set U) has partial derivatives fx and fy on K (respectively, on U), and that
fx= 1',
1,.= if'.
The following converse to this simple remark is a basic criterion for differentiability. (2.7) Theorem. Let f be a continuous function on an open set U c O with K,~ V, and let J be a modulus of continuity for f on K,. Then if e > 0, Zl ' Z2 E K, and IZ2 zll ~min {r, J(e)}, we have
Ig(Z2)  g(Zl)  f( Zl)(Z2  zl)1
II
= (f(z)  f(zl» dzi ~sup
{If(z)  f(zl)l: ZEK" Iz  zll ~ J(e)} IZ2  zll
~elzz zll·
Therefore g' = f on K. Since K is an arbitrary compact set well contained in V, we have g'= f on U. 0 (3.8) Lemma. Let V be a convex open set in tb(e). Then 1(2 zl>tb(e), both (1 and (2 belong to K, and so Ih«(dh«(2)I~e. Since e>O is arbitrary, it follows that h is uniformly continuous on B {z}. By (3.7) of Chapter 4, we see that h has a unique continuous extension to B. Hence, clearly, h has a continuous extension to U. To show that the extended function h is analytic on U, consider points Zl,Z2,Z3 in U with K=span(zl,z2,z3)~ U. Choose r>O so that K.~U. Either p(z,K»r/2 or p(z,K) 0 and using the continuity of h on K, we can find ex> 0 so that (4.6.2)
II h«() d(l < e
whenever m=min{lz2zl,lz3zl}<ex. Either m<ex or m>O. In the latter case, choose a point w of lin (z, Z2) so that
p(z, span (w, Z2, Z3)) > 0 and
II h«() d( 
I h«() d(l < e, l
where A=poly(w,z2,z3' w). By (3.6), I h(Od(=O, whence (4.6.2) holds. l
Thus (4.6.2) holds in all cases. As e>O is arbitrary, (4.6.1) obtains, and so h is analytic on U. 0 The following result, embodying Cauchy's integral Jormula, uses the winding number to establish a representation of a differentiable function by means of an integral.
4 The Winding Number
147
(4.7) Theorem. Let f be a differentiable function on an open set U c ce, let Z be a point of U, and let y be a closed path in U  {z} which is nullhomotopic in U. Then
j(y,z)f(z)=(2ni)1 Jf(C)(C _Z)l de. Proof: The function CI+(f(C) f(z))(C _Z)l extends to an analytic function h on U, by (46). Since Jh(C)dC=O, 1
(2nO 1 J f(C)(C _Z)l dC=(2ni)1 J f(z)(C _Z)l dC=j(y,z)f(z).
D
Cauchy's integral formula is often used in conjunction with the following proposition (4.8) Proposition. Let h be a continuous function on the carrier of the
path y in by
ce, and
let n be a positive integer. Then the function f defined
is differentiable on the metric complement  car y of car y, and where g(z)=(2ni)lnJ h(C)(C _z)n1 dC.
l' = g,
Proof: Let K be any compact set well contained in  car y. Write C=sUP{lh(C)lkt1ICZIk: CEcary, ZEK}.
Let w be a common modulus of continuity for the continuous functions (C,z)I+(C _Z)k (1 ~k~n) on the product space (cary)xK. Consider e>O and Zl,Z2 in K with IZ1  z21 ~ w(e). Then If(z2)  f(zl)  g(Zl)(Z2  zl)1
= (2 n)lIJ h(O((C  z2)n  (C  zl)n  n(C  Zl)n1(Z2  Zl)) dCI =(2n)1Iz2 zl'ls h(O 1
(i (C Zl)k(C Z2)n1+k k~l
 n(C  Zl)n1 )
dcl
148
Chapter 5 Complex Analysis
= (2 n)llz 2  Zlllf h(()ktl ((  ZI)k(((  Z2)n1 +k _(( _ ZI)nl +k) dcl ~(2n)llz2ZIII')I1 CEo
lt follows that
l' = g on
K, and therefore on  cary.
0
(4.9) Lemma. Let U c O with Kr~ U, and let {C l , ... ,(n} be an r/2 approximation to K. Let b be a modulus of differentiability for f, and w a common modulus of continuity for f and g, on the union of the closed spheres Sc(Ci,r) (1~j~n) Consider points ZI and Z2 in K, and a positive number e, such that
Iz 2  zil ~ min {r/2, beE), w(e)}.
We have IZ1() r with S c(zo, t)~ U. Let y be the circular path of radius t about Zoo By (4.7) and (4.8), we see that f'(z)=(2ni)1 Sf(O(Cz)2dC
whenever Iz  zol < t. Hence, again by (4.8), f' is differentiable on Sc(zo,r). It follows from (4.9) that f' is differentiable on U. The formula (4.10.1) now follows from (4.7) and (48) 0 (4.11) Corollary. A function f analytic on an open set U c R is arbitrary, we now see that function. 0
ZI+
L bnzn is
an entire
n~ 0
5. Estimates of Size, and the Location of Zeros It is a famous principle of classical analysis, called the maximum principle, that an analytic function attains its maximum on the boundary. The next proposition is the constructive version of this result.
(5.1) Definition. If K is a compact subset of 0, the proposition is proved.
K
D
For each compact set K c 3£5, and let n be a positive integer. Let f be differentiable on a closed sphere Let Z1""'Z. be points of S with IZjzkl~d (1 ~j 315, and let f be a nonzero differentiable mapping on a closed sphere S=Sc(w, r)c CC. Then there exists s such that 15  I: < S < 15 and m(f, r(w, s)) > O. Proof' To begin with, assume that IlfIIT>O, where T=Sc(w,r3t5). Choose an integer N;:;: 2 so that
lI!IIs(r  2t5t (r  t5)N If((')1 ~ m(f, SC(Zk' t k))
and so m(f, SC(Zk' tk))=O, by (5.3). Thus we can take W=Zk and to=tk' D
(5.8) Lemma. Under the hypotheses of Lemma (5.7), there exists with f(z)=O.
Z
in K
Proof Using (5.7), construct recursively a sequence (Sn);~)~l of closed spheres with K:;)SI:;)Sz:;)'" such that m(f,~»m(f,Sk)=O, where ~ is the boundary of Sk' and such that the radius of Sk is at most k 1 . Then these spheres have a common point Z with f(z)=O. D (5.9) Definition. A polynomial z~ao Z· + a l znI least k if an_/oF 0 for some j ~ k.
+ ... + an
has degree at
We now prove the fundamental theorem of algebra. (5.10) Theorem. If the polynomial p(z)=aoz n+ ... +a. has degree at least k, then there exist complex numbers Zl' ... , Zk and a polynomial q such that p(Z)=(ZZI) '" (ZZk) q(z) (ZE 0 with
jI
lan_)r j > Ilan_mlrm+lp(O)I. m~O
5 Estimates of Size, and the Location of Zeros
157
Then either an _ m =1= 0 for some m > j or else iI n (5.10.1) lan_ilri > Ilan_mlrm + I lan_mlrm+lp(O)I. m~O
m~i+1
In the former case, we replace j with m and repeat the above construction. Eventually, this process leads to values of j and r for which n
(5.101) holds, where
follows that
I
lan_ml rm is taken as 0 if j=n. It then
m~i+1
inf{lp(z)l: Izl=r}>lp(O)I; whence, by (5.8), there exists a complex number ZI with P(Zl) = O. Using the process of polynomial division, we find a polynomial ql(z)==boz n  1+ ... +b n _ 1 and a constant c in O. If m(gn' K) > 0, then the process stops and we are through. Otherwise, m(gn' K) = 0, and we continue to the (n + l)th stage. Either the process proceeds for N stages or it terminates at some stage n < N. Consider the former case. By the choice of N, there exist v of the points Zl' ... ' ZN whose mutual distances are less than fJ. Call these points w1 , ... , w Then V •
(5.11.1)
f(Z)=(ZW 1 )
•••
(zw.) h(z)
(zEK)
for some differentiable function h on K. Since !(w1)=0 and 1!(z)1 >~m(f, B) for all Z in K nB36' we have ro=p(wl,B»3fJ.
Hence, by (2.3), there exists, in B36 with (5.112)
"w113fJ, we have (5.11.4)
(r  3fJ/2) (r  fJ)l  (ro  3fJ/2) (ro _ fJ)l =~fJ(r 
As Iwiw11 Ilfllr(r)'
D
6. Singulanties and Picard's Theorem
163
6. Singularities and Picard's Theorem In this section we prove two constructive versions of a famous theorem of Picard concerning the range of a differentiable function on the punctured disk {ZEt1 and NE7L+, then N
«( _Z)l =«( _Z)l«( ZO)N(ZZO)N 
L (zzo)n«( ZO)"1
whenever I(  zol = t 1 ; so that N
I S f(0«( _Z)1 d( + L ( S f(O«( zo)"l d()(zzo)nl yet,)
n= 1
y(tj)
=I(ZZo)N S f(0«(Z)1«(ZO)Nd(l
yet,) ~ 2n IIf IIr(zo.t,)t 1(lz  zol t1)1(tdlz  ZOI)N.
nth
6 Singulanties and Picard's Theorem
165
As N > CXJ this last expression converges to 0 uniformly on each compact set K ~ T. It follows from (3.12) that the series (6.2.5), with coefficients bn given by (6.2.2), converges to the expression (62.6) uniformly on K Conversely, if co
co
converges uniformly to f on each compact set well contained in A, then for each integer m we have y(r) 00
00
=
L ak I ('zot+m1d,+ L bk I ('zo)k+m1d,. k~O
k~1
y(r)
y(r)
For each nonzero integer j. the derivative of the map 'Hr1(,Zo}' on car'}'(r) is 'H('zo)i1.1t follows from (3.4) that
I f(0('zot
1
y(r)
d,={2ni)a_ m =(2ni)bm
if m~O, if m>O.
Hence the coefficients an' b nare given by (62.1) and (622) respectively
0
In Theorem (6.2) the series 00
ex,
n~O
n~1
L an(zzot+ L bn(zzo)n, with coefficients given by (6.2.1) and (6.2.2), is called the Laurent series, or Laurent expansion, of f in A. It is convenient to write a_n=b n for each n in lL+, so that the Laurent series of f is 00
n= 
00
with coefficients an given by an={2ni)1
I f{O(' zo)nl d,
(nElL).
y(r)
Theorem (6.2) and these remarks apply with the obvious modifications when f is a differentiable function on an unbounded open annulus A={z: r1 O. In the latter case, as f is nonvanishing in A, we must have Z1 = ... =zn=zO; whence f has a pole of determinate order n at zoo 0 (6.7) Lemma. Let f be a nonvanishing differentiable function on an open annulus A={ZE r 1 so that
anz n be the 00
In~l anznl < el2
whenever Izl (41X 3 )1. In the latter, 1(1IWI>(21X)1 and so either "I >(21X+ 1)/2 IX or "1< (21Xl)/21X. Hence, by (6.13.2), Ig(O)1
~ Iln(lml~min {In e~: 1), In (2:: 1 )}>(41X )1. 3
Thus in both cases, (41X 3 )1 ~ Ig(O)1 ~ lIn (IWI + 11m g(O)1 ~ln(lX) + ~41X3,
n+ 1
(since 1X 1 ~ "I ~ IX)
172
Chapter 5 Complex Analysis
by (6.13.3). Hence
lIn (lg(O)1)1 ~ In (40(3) = 3In (0() + 2In (2)
0
(6.14) Lemma. Let cjJ be a continuous map of [0,1) into IR 0+, and let K
be a positive constant, such that Then (6.14.1)
for each r in [0, 1). Proof. First consider the case where cjJ is bounded by some constant M>O. Let O~r=ro (oc+3)1 or min {If(O)I, 11f(0)1} 2. To begin with, take 1f(0)1 If(O)1 or m(f, r(r)) b(l);;;; b(m(f, r(s))).
We now see from (6.20) that f takes at least one of the values 0, 1 in the annulus {ZE 1 we have F(z)=zexp =Z
(t (azz
2
+ Jzbnz Zn ))
(l t (azz 2 + n~2bnZZn) + terms in Z4),
where the final expression converges to F(z) uniformly on each compact set well contained in A. Hence the Laurent expansion of F in A is of the form F(z) = z taz Zl + terms in z 3. By (7.10), we have Ha21~1 and therefore II"(O)I=2Ia21~4. We now prove the Koebe covering theorem.
0
190
Chapter 5 Complex Analysis
(7.14) Theorem. If f is a normalized equivalence of S(O, 1) with an open set, then S(O, i) c f(S(O, 1)). Proof Consider an arbitrary complex number w with Iwl O. Then elementary calculations show that g(z) == f(r z)/r(1  w 1 f(r z))
defines a normalized equivalence g on S(0,1) with g"(O) = r(f"(O) +2w 1 ). By (7.13), we have rlf"(0)+2wll~4 and If"(0)1~4. Thus Iwl 1 ~H4r 1 + 11"(0)1) ~ 2(r 1 + 1), and so Iwl ~ r/2(1 + r), a contradiction. Hence m(f  w, S c(O, r)) < m(f  w, r(r)), so that there exists z in S (0, r) with f (z) = w 0 The Riemann mapping theorem will provide conditions under which there exists an equivalence of a simply connected open set U c O Then there exists a path y: [0, 1] + 8c 1: in fact, if {Cl' ... 'C} is a 2 g e approximation to S(D, 9 c 1), then we may take N == 1 + v. Suppose that M j > 2 5 e(lrj)l
(7.22.1)
~j ~ N).
(1
For eachj with l~j~N choose Zj in Aj with 1f'(zj)I>2 5 e(lrj)1, and define a normalized equivalence hj on S(D, 1) by h(z)_f(zj+(rj + 1 rJz)!(Zj) J f'(zj)(rj+ 1 rj)
(zES(D,I».
We show that (7.22.2) Suppose, on the contrary, that p(f(Zj).!(r(rj+ 1») 2 8 e.
It follows that m(f w,Sc(D,rj+d) we have Ig'(f(z))I 1= If'(z) I < 2 4 e(IR)1 =2 4 eIR 1 z _ZI1 and therefore 14(R 1 z  z)/eg'(f(z)) I
By (7.14), there exists w in S(O, 1) such that ¢(w) = 4(R 1 z  z)/eg'(f(z)).
Hence g(!Bw+f(z))=R1z; so that IR 1zl ... , WN} be an E/4 approximation to J(r(rd) in the metric d. By the foregoing, for each k (1 ;£ k;£ N) there exists 'k in E E/2 E/4=E/4.
7 The Riemann Mapping Theorem
199
Hence d*(y,f(r(rd»'i;;6/4. By (725), y lies in U. Thus {ell .. ,eN} is an 6border of U relative to O. 0 The converse of Theorem (7.26) is the Riemann mapping theorem. In order to prove the latter, we first investigate a new property of open sets. (7.27) Definition. Let U be a nonvoid subset of f,l(z). In that case, U is open, f,l is a function on U, and the positive number f,l(z) is called the maximal extent of U about z. The next two results show that mappability and the maximal extent property are equivalent properties of a simply connected open set. (7.28) Proposition. Let U be a mappable set. Then U has the maximal extent property. Moreover, if ZoEU, 6>0, Be is an 6border of U relative to zo, and f,l is the maximal extent of U about zo, then (7.28.1)
Proof: Consider any point Zo in U, and recall that Zo will serve as the distinguished point of U. For each 6> 0 let Be be any 6border of U relative to zoo Then for all positive band 6 we have (7.28.2)
e
To see this, suppose that d(zo,Be»d(zo,B~)+6, and choose in B~ with d(zo,B e»d(zo,e)+6. Then eEU, by (7.24). But this is absurd, as eEB~c U. Thus (7.28.2) holds. Hence
Id(zo, Be) d(zo, B~)I ~ max {b, 6}
(b, 6 > 0).
Since lR. is complete, it follows that there exists a nonnegative number f,l such that (7.28.1) holds for each 6>0. Consider any r>f,l. By (7.28.1), we have d(zo,Be)0. Hence SOO(zo,r)nB., and therefore SOO(zo, r) U, is nonvoid. It follows that if t > 0 is chosen with SOO(zo, t) c U, then we must have f,l'i;; t>O. Finally, consider any z in SOO(zo,f,l). Choosing b>O so that d(z,zo)e(1 + lal)(lrlal)l. Choose also R>1 so that if Iwl~R, then Ihb(w)l~r, where b==oca. Then if ZES(O, 1) and Iz'l ~r, we have (7.31.2)
Ih.(z)  h.{z')1 = 1(1 a* a)(z  z')(1  a* Z)l (1 a* z')ll ~ (1laI 2 )(1 lal)l(1  r lal)llz  z'l
= (1 + lal)(lrlal)llz z'l. With Zo and B as above, let K==och.(B). Then KcS{O, 1), by (7.7). Let y:[O,I]+([; be a path with left end point och.(zo), such that p*(y,K)~e'. Consider any t in [0,1] such that ly(t)I~R. With Z'==OC*hb(y(t)), we see from (7.31.2) that for each z in B, loc* hb(y(t))  zl = Iz'  zl
+ lal)l(1rlal) loch.(z) och.{z')1 = (1 + lal) 1(1 rial) loch.(z)  y{t)l.
~ (1
Since OCh.(Z)EK, it follows that loc* hb(y(t))  zl ~ (1 + lal) 1(1 r lal)e' > e
(zEB).
As B is an eborder of (V,p), it follows that if t>O and y([O,t]) c S c{O, R), then the restriction of the path oc* hb 0 I' to [0, t] lies in V. Now, either 111'11 0. Hence the restriction of oc* hb 0 I' to [0,.] lies in V; so that 1'(.) = och.(oc* hb(y(.))) E V c S(O, 1),
which is absurd. Thus the case Ilyll > 1 is ruled out, and so 111'11 ~~~ ~~~ =2 (1 + ly(tW)t(1 + Iwl2)t (1 + ly(tW)t(1 + Iwl2)t ~ td*(y, BO)2 ~ te 2
since both wand  w belong to Bo. Thus d* (so ° y, B) ~ te 2, so that So ° Y lies in V. Hence y=so(sooy) lies in V o , by (517). Thus Bo is an eborder of Vo relative to s(Q. Since e is arbitrary, Vo is mappable. D It is left to the reader to prove.
(7.33) Lemma. With V, s, and Vo as in Lemma (732), suppose also that V is sequestered. Then Vo is sequestered. Moreover, if'EV, e>O, and B is an E2border of (V,p) relative to " then Bo=':{W:W 2 EB} is an Eborder of (Vo,p) relative to sm; and if BcS(O, 1), then BocS(O, 1).
(7.34) Definition. Let V be a sequestered set containing 0, and let J1 be the maximal extent of (V, p) about (which exists, by (7.28)) Note that J1 e the set
205
° By (7.31) and
Bt == {ct*hb(w)' w 2EcthaCB)}
is an e' border of (vt, p) relative to 0, and Bt c S(O, 1). Since B = t/I (Bt), we see from (7.35.1) that p(O, B);;::; p(O, Bt). It follows from (7.28), with reference to the remarks preceding (7 30), that fl;;::; fl t + E. Since e is arbitrary, we therefore have fl;;::;/1t. Now let O r/4 for all w in B.
°
7 The Riemann Mapping Theorem
209
Proof Consider any e>O. We may assume that e is so small that d(w,w'»e whenever IwClr/4 for all W in S, and IwCl 1 or 1. Theorems (6.21) and (6.22) are equivalent classically, but provide totally different constructive information. Problem 18 shows that a further classically equivalent formulation of Picard's theorem does not hold. (The material in Section 6 is based on joint work of Bridges, Calder, Julian, Mines, and Richman. An earlier constructive presentation of the Picard theorems was given by Belinfante [4].) Note that the Picard theorem on entire functions (Problem 17) does not follow constructively from Theorem (6.22). It can be proved by an argument analogous to that used for Theorem (6.22). Details of the generalization of the notion of winding number, and a particularly elegant proof of the Jordan curve theorem, are found in K
,~
[5]. The notion of mappability (Definition (7.23)) replaces Bishop's original notion, which is not a necessary condition for the existence of an equivalence with S(O, 1). See Problem 22, where the original notion is called strong mappability. The work in Section 7 amplifies and corrects the material in [29]. Not every simply connected, bounded, open set U is equivalent to S(O, 1). To see this, consider any sequence (nJ:'= 1 in {a, I} such that we 00
do not know if there exists k with nk = 0, and take U == U Uk' where Uk==S(O, 1) if nk=l=O, and Uk==S(I, 1) if nk=O k=l The reader is invited to give a routine for computing the value of the mapping function ¢ of a given mappable set U at a given point z to within a given accuracy E. He may also give a routine for computing a modulus of continuity for ¢ on a given compact set K ~ U The mapping function ¢ on a mappable set U containing is obtained classically as an analytic function from U into S(O, 1) which vanishes at and has the maximum possible value of I¢'(O)I. Constructively, this approach does not work, because it is not a priori evident that such a function exists. Once the mapping function is known to exist, it can easily be shown to be such a function (Problem 23).
°
°
Chapter 6. Integration
An integration space consists of a set X with an inequality relation, a set L of partial functions from X to JR, and a function I: L + JR, called an integral, which has certain properties classically equivalent to those of a Daniell integral. Integration spaces are introduced in Section 1, and several examples are given. The most important example occurs when X is a locally compact space, L is the set of test functions on X, and the integral is a positive measure on X  that is, a linear map Ji.: L + JR such that Ji.(f) ~ 0 whenever f is a nonnegative test function. The goal is to define the class Ll of integrable functions, construct the integral, and investigate its properties. The construction of the integral is carried out in Section 2. Integrable sets are discussed in Sections 3 and 4; an integrable set is a complemented set whose characteristic function is integrable The classical theorem that every compact set in a locally compact space X is integrable with respect to a positive measure on X fails in the constructive setting. However, Theorem (4.11) says that there are sufficiently many integrable sets for our needs, this theorem is proved using the properties of profiles, which are introduced in Section 4 In Section 5 we show that every positive measure on JR is induced by a monotone function. In Section 6 we consider a positive measure on a locally compact space, and show that integrable sets can be approximated arbitrarily closely from within by compact integrable sets Measurable functions and simple functions are discussed in Section 7. In the next section we prove various results about the convergence of sequences of integrable functions .. in particular, we prove the monotone and dominated convergence theorems associated with the name of Lebesgue. In Section 9 we construct product integrals and prove Fubini's theorem. The final section of the chapter deals with measure spaces, in which the attention is focussed on certain complemented sets with properties abstracted from those of the integrable sets in an integration space, every integration space gives rise to a measure space, and vice versa. Any constructive approach to mathematics will find a crucial test in its ability to assimilate the intricate body of mathematical thought
216
Chapter 6 Integration
called measure theory, or the theory of integration. This subject was initiated in 1904 with a book by Lebesgue, and now pervades abstract analysis in a way so essential that there is little point in trying to give meaning to that branch of mathematics without laying a constructive foundation for measure theory. It was recognised by Lebesgue, Borel, and other pioneers in abstract function theory that the mathematics they were creating relied, in a way almost unique at that time, on settheoretic methods, and led to results whose constructive content was problematical. Today it is true more than ever that much of abstract analysis has no ready constructive interpretation. It is to be expected that such work will be seen in a proper constructive light only as the result of an investigation which recreates analysis from the very beginning in accord with constructive principles. We are at the stage in such an investigation in which we can properly take up the study of integration, basing our work on the theory of complemented sets from Chapter 3 and the metric space theory of Chapter 4. In two respects this study is crucial. First, as already indicated, measure theory and its techniques underlie many parts of modern analysis. Second, measure theory provides the proper framework for a discussion of discontinuous functions, which in most cases seem to be realized best in a measuretheoretic framework, as functions defined on a full set with respect to the appropriate integral.
1. Integration Spaces In this section we introduce the notion of an integration space, and give some examples of this fundamental concept. Throughout our discussion of integration theory, when we refer to a function f: X + Y between sets with inequality relations, we shall assume that f has the property of strong extensionality for all x and x' in X, if f(xHf(x'), then x =1= x'. We define .9'(X, Y) to be the set of all strongly extensional partial functions from X to Y, elements f and g of .9'(X, Y) being equal if they are equal as partial functions. In the special case where Y is IR., we write .9'(X) == .9'(X, IR.) If L1 is any of the symbols 1:, U, V, or /\, then fj, will denote 00
n,
n
fj, n= 1
Another convention that we shall adopt in the context of integration theory is that if (fn) is a sequence of realvalued partial functions on a set X, and A is any nonvoid subset of dmnfn' then Lfn is the
n n
n
1 Integration Spaces
217
function defined on A by
provided that the series on the right converges for each x in A This usage of the notation I fn differs from that introduced in Section 4 of n
Chapter 2; however, the intended interpretation should be clear from the context (1 1) Definition. A triple (X, L, 1) is an integration space if X is a nonvoid set with an inequality =1=, L is a subset of ~(X), and I is a mapping of L into IR such that the following properties hold. (111) If f,gEL and rx,/3EIR, then rxf +/3g, If I, and f /\ 1 belong to L, and [(rxf +/3g)=rx/(!)+/3I(g). (1.1.2) If f ELand Un) is a sequence of nonnegative functions in L such that I I (fn) converges and I I (fn) < I (f), then there exists x in X
•
such that If.(x) converges and If.(x) 0, and K a compact support of f Then f(x»O for some x in K. Proof. It is enough to prove that II f II > O. Define a test function g by g(x)=:(lp(x,KW
(XEX).
Then Ilfllg f~O, so that
J
J
O~J0, and therefore Ilfll >0.
0
220
Chapter 6 Integration
(110) Theorem. If fJ. is a positive measure on a locally compact space X, then (X, C(X), fJ.) is an integration space. Proof It is clear that (1.1.1) and (1 1 3) hold, and that lim S U n~
1\
n)dfJ.
00
= S f dfJ. for each f in C(X). If f E C(X) and K is a compact support of f, define an element g of C(X) by (1.1 0.1)
g(X)=(1p(x,K))+
(XEX).
Then for each n in 7l+ we have 0;:::;lfll\n1;:::;nlg. Since fJ. is positive, it follows that O;:::;S (If I 1\ n1)dfJ.;:::;n 1 gdfJ.;
J
whence lim J(lfll\n1)dfJ.=0. Thus (1.1.4) holds To complete the proof, it remains to verify (1 1.2). Accordingly, consider f in C(X) and a sequence Un) of nonnegative test functions such that Jfn dfJ. converges to a sum less than Jf dfJ.. Let K be a
L n
compact support of f Applying (1 8) recursively, construct a sequence (gk) of nonnegative test functions such that gk has a compact support of diameter less than k 1 , and n
By (1.9), for each k in 7l+ there exists x k in X with k
0;:::;
L Ungl"
gk)(Xk)«fgl···gk)(X k)·
For 1 ;:::;j;:::;k we have giXk»O; thus Xj and Xk belong to each compact support of gj' and therefore p(xj,Xk);:::;jl. Hence (x k) is a Cauchy sequence in X; it therefore converges to a point x of X. Since the j
functions fn and gi are nonnegative, we have 1 ;:::;j;:::;k, so that
L fn(x k) < f(x k)
for
n= 1
j
L In(x);:::; f(x) n= 1
Also, f(xk»O and therefore XkEK (kE71+); so that xEK Define g as in (1101) Choose oc>O so that
L JfndfJ.+ oc + oc JgdfJ.< Jf
dfl·
Choose also a strictly increasing sequence (N(k));;'~ 1 of positive tegers such that L in d flO, it follows that Lln(x) < f(x). This completes the proof. D Here are some important examples of positive measures. Every point x in a locally compact space X gives rise to a positive measure bx , the point mass at x, defined by Sldbx=/(x)
(fEC(X)).
The set 71. of integers is locally compact, and C(71.) consists of all functions I: 71. .1R which vanish outside a finite subset of 71.. Every nonnegative function ot:: 71. ~ 1R determines a positive measure Jla on 71., defined by 00
SldJl7=
L
ot:(n)f(n)
(fEC(71.)),
11=  0 0
where, of course, the sum is actually finite. Conversely, if positive measure on 71., then Jl = Jl. for a unique ot:, in fact, ot:(n)
=S andJl
)J.
is a
(nE7I.),
where an is the element of C(71.) whose value at n is 1, and whose value elsewhere is O. In case ot:(n) = 1 for all n in 71., the measure Jla. is called counting measure on 71.. b Lebesgue measure Jl on 1R is defined by S I d)J.= S I(x)dx, where a
[a, b] is any compact interval supporting the test function f Test functions and positive measures have some useful properties relative to certain maps. To see this, let X and Y be locally compact spaces, and ).: X . Y a continuous map. Then for each I in C(Y) the mapping 10). is continuous on X. Suppose that). is proper, in the
222
Chapter 6 Integration
sense that
A:: {XEX: A(X)EB}
is a bounded subset of X for each bounded subset B of Y. Then f A belongs to C(X): for if B is a compact support of f, then any compact subset of X containing A is a support of f 0 A. Thus A induces a map A*: C(Y)~ C(X) given by 0
A*(f)::foA
In turn, A* induces a map A*.
(fEC(Y)).
M+(X)~M+(Y),
Jf dA*(Ji.):: p*(f) dJi.
given by
(Ji.EM+(X),fEC(Y)).
A positive measure Ji. on X can be multiplied by a nonzero continuous function h: X ~ IR 0 +, to give a positive measure h Ji. defined by
Jf
d(hJi.):: Jfh dJi.
(f E C(X)).
A positive measure Ji. on X is said to be supported by a locally compact subset Y of X if f dJi. = 0 whenever f E C(X) and f(y) = 0 for all y in Y. (For example, the measure hJi. is supported by Y if h vanishes on  Y.) If this is the case, then consider any function g in C(Y). By the Tietze extension theorem, g has a continuous extension g* to X with compact support. Moreover, if g is nonnegative, then g* can be chosen nonnegative. Since Ji. is supported by Y, the integral Jg * d Ji. does not depend on the choice of the extension g *. Therefore the equation Jgdv::Jg*dJi.
J
defines a positive measure v on Y, called the restriction of Ji. to Y.
2. Complete Extension of an Integral Let (X, L, I) be an integration space. In order to develop the theory of integration, we need to extend the domain of the integral I. (2.1) Definition. An element f of jii"(X) is an integrable function if there exists a sequence (fn) of functions in L such that I(lfnD con
L n
verges, and f(x) = Ifn(x) whenever
Ilfn(x)1 converges. The sequence n
(fn) is then called a representation of f
by elements of L, or a representation of f in L We write Ll for the class of integrable functions.
2 Complete Extension of an Integral
223
Under these conditions, since 11(1.)1 ~ 1(11.1) for each n, the series L I(f.) converges absolutely to a real number. Moreover, if (g.) is n
another representation of I by elements of L, then LI(g.)= LI(f.). To see this, suppose that
L I(g.) oF L 1(1.). 00
•
Then
•
.~N+
<Xl
L .~
•
L
•
•
Choose in turn
tx,
N so that
00
1(11.1) < tx,
and
L .~N+
1
1(11.  g./) < tx. I
<Xl
1(11.1)+
N+ I
L .~
I(II.g./)I(lfl), then for some n in 7l+,
which contradicts Proposition (1.5). To complete the description of LI as a set, we must introduce an appropriate notion of equality of its elements. This involves a discussion of sets containing the domain of an integrable function. (2.4) Definition. A subset F of X is called a full set if there exists an integrable function f with dmn f c F. (2.5) Proposition. In order that F c X be a full set, it is necessary and sufficient that there exist a sequence (fn) of integrable functions with dmnfn cF.
n n
Proof: The necessity of the condition is clear. To prove the sufficiency, suppose that such a sequence (fn) exists. For each n in 7l+ choose a
2 Complete Extension of an Integral (fn,k)';'~ 1
representation
225
of fn in L, and set 00
IXn == 1 + L 1(lfn,kl), k~
1
Arrange the terms
2 nIX; 1 fn,k
(n, kEZ+)
into a single sequence (¢n)':~ l ' Then the series that the function ¢ == L ¢n with domain {XEX.
L 1(I¢nl) converges,
so
L l¢n(x)1 converges} 00
is integrable. For each x in dmn ¢ and each n, the series L Ifn,k(X)1 converges Hence k~ 1 dmn¢cn dmnf.cF n
and so F is a full set.
0
(2.6) Corollary. A countable intersection of full sets is a full set. Proof: This follows immediately from (2.4) and (2.5).
0
In constructive mathematics, full sets playa role analogous to that of the complements of sets of measure zero in classical measure theory; roughly, what happens outside a full set may be ignored. To make the last remark more precise, we need a lemma. (2.7) Lemma. Let (f.) be a sequence of elements of L such that L 1(lfnl) n
converges, and L fn(x) ~ 0 whenever L Ifn(x)1 converges. Then L 1(fn) ~ O. n
n
Proof: Since L 1(lf.l) converges, the series L 1(f/) and L 1(1.) both
•
n
converge, and L1(f.)= L1(f/) L1(fn) n n n
Also, Llfn(x)1 converges if and only if both Lf/(x) and Ifn(x) n
converge, in which case Lfn+(x) I f..(x) = Ifn(x)~O. n
n
226
Chapter 6 Integration
Now suppose that
I
l(fn) O so that
n
00
Choose also N in 7L+ with
I
l(fn) N we
~I (Ifni):;:; 1(\ktl gk\) + E/2 = I (Iktl gkl) + 1(Iktl gklIktl gkl) + 1:/2 :;:;1
(Iktl gkl) + 1(lk~ t+ gkl) + e/2
:;:;I(lktl
I
gkl) + k~t+
I
1(lgkl)+e/2 00, by (2.15). 0
I.
fn EL for each N; and
n~ I
Ilf  I. fnll n~ I
>0 as I
(2.17) Corollary. LI is complete with respect to the metric arising from the norm II III' Proof: Let Un) be a Cauchy sequence in L I , and construct a strictly increasing sequence (n(k»~~ I of positive integers such that II fn(j)  fn(k) I I 00. Since Un) is a Cauchy sequence, it follows that II f  fn III > 0 as n>oo. Hence LI is complete. 0 (2.18) Theorem. (X,LI,I) is an integration space. Proof" We have already seen that (X,LI,I) satisfies (1.1.1). To verify (1.1.2), let Un) be a sequence of nonnegative elements of L I , and f an arbitrary element of L I , such that L IUn) converges to a sum less than n
IU)· To begin with, assume that f ~ O. Choose
0(
> 0 so that
L IUn) n
+30«IU). By (2.14), for each n in 7l+ there exists a representation Un,k)~= I of fn in L such that
L 1(lfn)
< IUn} + 2  n 0(.
k
On the other hand, if (cPk) is a representation of f in L, then we can find
N
so that
IU)(An)· 11
11
11
PI
Proof: For arbitrary positive integers j, k with k > j, it follows from (3.5)(3.7) that
Hence
(/llY1An)):1 is a Cauchy sequence, and so converges to a V An is integrable, and
limit in 1R.. By (3.8),
n
(3.11) Proposition. If A is an integrable set and f is an integrable function, then the functions XAf and xAf are integrable. Proof: Since f = f+  f, we may assume that f ~ O. Choose a strictly increasing sequence (nk)f~ 1 of positive integers such that O~I(f  fA n k )O, we see that h(xo,xl,.)o! is integrable whenever 0< Xo < XI Hence A(h(xo, Xl' .» == J(h(xo, Xl' .) of)
(0 <XO < Xl)
defines an increasing map A. C+lR, and the pair (C,A) is a profile the (I,f)profile for lR °+.
4 Profiles
237
(4.2) Definition. Let (t&', A) be a profile for the interval J, let [a, b] be a compact subinterval of J, and let E>O We say that [a,b] has profile lower than E relative to (t&', A) if there exist points a', b' of J with a'O, we say that [a,b] has arbitrarily low profile relative to (t&', A). There now follows a succession of lemmas which will enable us to prove the fundamental theorem in the theory of profiles: relative to a given profile on an interval J, {x} has arbitrarily low profile for all but countably many points x of J. (4.3) Lemma. Let (t&', A) be a profile for an interval J, E a positive number, and a, b, c points of J such that a;£bi£c, [a,b]~E, and [b,C]~E.
Then
[a,c]~2E.
Proof: Since [a, b] ~ e, there exist points 0(, p of J with 0( < a;£ b < p, and functions ¢I' t/ll in t&', such that ¢1(x)=l whenever XEJ and x~O(, t/ll(X)=O whenever XEJ and x;£p, and A(¢I)A(t/ll)<E Likewise, there exist points ,}" fJ of J with'}' < b i£ c < fJ, and functions ¢2' t/l2 in t&', such that ¢2(X) = 1 whenever XEJ and x~'}', t/lz{x)=O whenever XEJ and x;£fJ, and A(¢2)A(t/l2)<E. It will suffice to prove that A(¢l) A(t/l2)
We begin the induction by setting XO,1=X O,2=",=X O,M==X() Now assume that Xn k has been defined and satisfies the appropriate parts of (I)(III) for O~' n ~ Nand 1 ~ k ~ M. Then for each e with lei = N there are exactly M. values of k for which XN,k=X •. For M •• o of these values (it doesn't matter which) we define XN+ 1,k==X.*O; for the remaining M"l values we define XN+ 1,k==X.*1' Taken with our induction hypothesis, this construction ensures that the appropriate parts of (1)(III) hold for 0 ~ n ~ N + 1 and 1 ~ k ~ M. This completes the definition of the sequences Sk' Since Xn,k and Xn+1,k both belong to the same I. with lel=n, we have IXn,k  xn+ 1,kl ~ 11.1 < (3/4)n l(b a). Hence for all m> n, 00
IXn,k  xm,kl
5(3/4)n1(ba) for 1~k~M. Then for such k we have Ixx n,kl>(3/4)"1(ba). Since the points Xn,k (1~k~M) are the midpoints of those intervals I. with lel=n and M. > 0, it follows from (iv) above that x has a positive distance from all such intervals. Therefore, by (vii), x has a distance 0 from the union of those intervals I. with lel=n and M.=O. Since I.~(M.+1)e=e for each such e, it follows that {x} ~e. 0 Here, at last, is the result for which we have been preparing (4.8) Theorem. If (tC, A) is a profile for an interval J, then for all but
countably many points x of J, the set {x} has arbitrarily low profile.
4 Profiles
Proof' Write 1" ==
r.
"" i., U
241
where each i. is a proper compact interval
"=1
contained in For each pair (n, k) of positive integers there are finitely many points x~~L x~~L ... ,x~~Z of i. such that {X}~kI whenever xEi. and x =!= X~)k (1 ~j ~ N). Let (x.) be a rearrangement of all the points x~L together with those end points (if any) of i that belong to i, in a single sequence. Then {x} has arbitrarily low profile whenever XEi and x =!= x. for all n D Note that if {x} has arbitrarily low profile, then x is an interior point of i; this follows from Definition (42). (49) Definition. Let (C, A) be a profile for an interval i. A point x of i is smooth for (C, A) if there exists a real number c with the property: for each e>O there exists 15>0 such that 1..1(4)) CI<e whenever 4>EC, 4>(t)=O for all t in i with t~xc5, and 4>(t)=1 for all t in i with t~x+c5.
Taken with Theorem (48), the following theorem shows that all but countably many points of i are smooth. (4.10) Theorem. If (C, A) is a profile for an interval i, then any point of i with arbitrarily low profile is smooth for (C, A) Proof: Let x be a point of i with arbitrarily low profile. Chose r > 0 so that x  r and x + r are interior points of i Construct a strictly decreasing sequence (IX.) of numbers in (0, r) converging to 0, and sequences (4).), (1/1.) of elements of C, such that for each n, 4>.(t) = 1 whenever tEi and t~XIX., 1/I.(t)=O whenever tEi and t~X+IX., and A(4).)A(I/I.)O and t=\=xn for all n, then {t} is smooth relative to the (I,f)profile for IR 0 +. Call such t admissible relative to the excluded sequence (xn). Consider an arbitrary admissible t > O. Choose a positive integer N with t>2 N • For each integer n~N and each x in X, define ¢n(x) =h(t 2 n, t 2 n
1,
f(x))
4 Profiles
243
and t/ln(x)=h(t+2 n
1,
t+2 n,f(x)).
Then ¢n?;, ¢n+ 1 ?;, t/I n+ 1?;, t/I n on dmn f Also, since t is smooth, the sequences (I(¢n))';:N and (l(t/ln)':~N converge to a common limit c in 1R Thus ()()
()()
Hence ()()
X=¢N
I
(¢n¢n+l)
defines an integrable function on the full set F= {XEX:
n~}¢n(X)¢n+ l(X)) converges},
and I(X)=c. For each x in F, either X(x»O or X(x)< 1. In the former case we have ¢n(x»O, and therefore f(x)?;,t2 n, for all sufficiently large n; so that f(x)?;,t, ¢n(X) = 1 for all n?;,N, and therefore X(x)=1. On the other hand, if X(x) < 1, we can choose m with ¢m(x) < 1. We then have f(x)~t2ml; so that f(x)t) is an
n~N
integrable set, also with measure c. Now let 1'>0 be arbitrary. Choose k?;,N so that I(¢k)I(t/lk)<E, and write c5=2 k  1 . Consider any admissible point t' with Itt' I 0 In case J.l(B) <J.l(A), choose x with XA(X) = 1 and XB(X)=O. Then f(x»O, and so xEA1 for some N. Were A(N)=O we would have A1 =B1cBI, and therefore XB(X) = 1, a contradiction Thus J.(N)=l and J.l(AN»O. This completes the proof. 0
In order to prove the converse of Proposition (4.13) we need two useful lemmas. (4.14) Lemma. Let f be a nonnegative integrable junction, and (tn) a decreasing sequence of positive numbers converging to 0 such that the
5 Positive Measures on IR
245
complemented set (f ~ t n ) is integrable for each n. Then
lim I(XC! 0 such that II(XAf)l 0 such that if f is integrable, o~ f ~ 1, and f(t) = 0 for all t in dmn f with Ix  tl ~ c'5 x (e), then Sf dJl<e. Proof. First note that every compact interval J in IR is contained in an integrable proper compact interval K Indeed, if g(x)o:=max{O,1p(x,J)}
(XEIR),
then gE C(IR) and we can take K to be any integrable set of the form (g ~ r) with 0 < r < 1. It follows that IR is the union of an increasing sequence of integrable proper compact intervals. Hence it will suffice to prove that for each integrable compact interval J 0:= [a, b] with a < b, there exists a sequence (x.) of real numbers such that (i)(iii) hold whenever x and y belong to (a, b) and are distinct from each x •. To this end, construct h in C(IR) so that h(x)=x for all x in J. Applying (4.11) to the integrable functions (bh)xJ and (ha)xJ, we can find a sequence (x.) in IR such that if a < x < b and x =t= x. for each n, then the sets [a,x] and [a,x) are integrable and have the same measure; the sets [x,b] and (x,b] are integrable and have the same measure; and for each e > 0 there exists wx(e) > 0 such that
IJl([a, x]) Jl([a, x'])1 < e and IJl([x, b])  Jl([x', b])1 < e
whenever a<x'b, and, by (5.3) and (54),
IS fj d,u 
2~j~n
and x 0 such that for each E > 0, there exists c5 > 0 such ISfdJlcl<E whenever fEC(X), O~f~l, f(x)=l for all x in K, f(x)=O for all x with p(x,K)~c5.
said X if that and
Under the conditions of Definition (6.1), the constant c is uniquely determined by the strongly integrable set K. This follows from our next result. (6.2) Proposition. Let K be a strongly integrable set with respect to a
positive measure Jl on a locally compact space X. Then K is integrable and, in the notation of Definition (6.1), Jl(K)=c Proof" There exists a decreasing sequence (c5 n) of positive numbers converging to 0 such that ISfdJlcl 0, X(¢,r)= {xEK: ¢(x)~r}.
By (4.9) of Chapter 4 and (4.11) above, for all but countably many real numbers r with O0, and e a positive number. Then there exist a closed estrongly integrable set C, a finite set HeX, and a positive number r < e, such that C 1 cAt, J.L(A)  J.L( C) < e, S c(y, r) is integrable and J.L( C " S c (y, r)) > for each y in H, and p(x, H) < e for each x in C 1 •
°
Proof: We may assume that e<min H, J1.(A)}. Let (fn) be a representation of XA by elements of C(X). The proof hinges on the idea that for a sufficiently large k and for a suitable c in (0,1), the complemented set
(tl fn ~ c) will approximate A as closely as we wish. By (2.15), we
can choose integers 1 0
and A(k) = 1 => Il(B "S C(Xk' r)) < e/2N.
Let Since ktt(B "S C(Xk' r))
~ IllY1 (B "S C(Xk' r))) ~ Il(B) >0,
we may assume that H is nonvoid. By (6.3), there exists 0( in (r,e) such that S= {xEB 1: p(x, H)~O(} = {xEB 1. (1 p(x, H))+ ~ 1 O(} is strongly integrable. Then the complemented set
C=B"S is integrable, C 1 is closed, C 1 cB1 cAl, and p(x,H)~O«e for each in C 1 . Also, if YEH, then
X
Il( C "S c(y, r)) = Il(B "S c(y, r)) > o. N
Now, if XB_dx)=1, then xEB 1c U Sc(xk,r) and p(x,H)~O(>r, so that XE U S C(Xk' r). Thus k= 1 }'(k) = 1
(6.5.3)
Il(B)Il(C)~Il(B
,,(
V }'(k)
~
L }'(k)
=
SC(Xk,r))) 1
Il(B" SC(Xk' r)) < Ne/2N= e/2
=1
and so Il(A)  Il( C) = Il(A)  11 (B) + 11 (B)  Il( C) < 1'..
Finally, C 1 c S, S is strongly integrable, and, by (6.5.2) and (6.5.3), Il(S)  Il( C) ~ Il(B 1)  11 (B) + 11 (B)  Il( C) 0 so that 0 ~f ~ c on a full set F. Compute real numbers ao==O, a 1 , •.. ,an with an>c, such that for each k (l~k~n), 0< a k  ak _ I < e and Bk == (f~ aJ is integrable. Then the complemented
260
Chapter 6 Integration
sets Ao==ABI and Ak==BkBk+l(l~k~nl) are integrable. Write Xk == XA. (0 ~ k ~ n  1), and define the simple function X by nI X== L akXk· k~O
Let x belong to the full set S==Fn(AI uAO)ndmnxo ndmnXI n ... ndmnXn_l.
nI
V
If
Xk(x)=l, then there is a unique i (O~i~nl) with Xi(x)=l; so
k~O
that
ai~f(x) 00. Proof. Given E > 0, choose A and N as described above. There exists no ~ N such that to each n ~ no there corresponds an integrable set K with K 1 cA1, Jl(AK)oo.
0
We now arrive at Lebesgue's dominated convergence theorem. (8.8) Theorem. Let (fn) be a sequence of integrable functions converging in measure to an integrable function f, and let g be an integrable function such that Ifn I ~ g for all n. Then lim I (I f  fn I) = o. Proof" By (8.7), it is enough to show that for each E > 0 there exist an integrable set A and a positive integer N such that I(XBlffnl)<E whenever n~N and B is a measurable set with Jl(AI\B) 0, choose an integrable set A with I(LAg)<Ej4. By (4.l5), there exists N in 7l+ such that I(XKg)<Ej4 for each integrable set K with Jl(K) < N  1. Consider any measurable set B with Jl(A 1\ B) < N 1. Since XB= XAAB+ XBA~XAAB+ XA on a full set, for any integer n ~ 1 we have I(XB Iffnl)~I(XA"B Iffnl)+I(LA Iffnl} ~2I(XA" Bg) + 2I(LAg) < E Since
E
is arbitrary, the proof is complete.
0
Before going any further, we introduce two important types of integration space.
g Convergence of Functions and Integrals
269
(8.9) Definition. The integration space (X, L 1 , I), or just the integral I itself, is finite if the constant function 1 is integrable. The integration space, or just I, is afinite if there exists a sequence (Kn) of integrable sets such that K~cKic ... , UK~ is a full set, and (XK) converges in
•
measure to 1; in that case, the sequence (Kn) is called an I basis of X. Clearly, if I is finite, then it is afinite. The reader is invited to verify that the complete extension of a positive measure on a locally compact space is afinite. (8.10) Proposition. Suppose that I is afinite, let (Kn) be an Ibasis of X, and let f be a nonnegative measurable function. Then f is integrable if and only if fxKn is integrable for each nand t == lim I (j XKJ exists .. in n~a: which case I(f)=t. Proof: It readily follows from the appropriate definitions that the sequence (fxK) converges to f in measure, and pointwise on the full set
The result now follows from the monotone convergence theorem, (3 11), and (2.9) 0 A special case of Proposition (8.10) occurs when f is the characteristic function of a measurable set A. In that case, A is an integrable set if and only if each of the complemented sets A A K. is integrable and lim J1(A A Kn) exists. n~oo
(8.11) Lemma. Let I be afinite, and let (A.) be a sequence of measurable sets such that J1(E) = lim J1(E A An) for each integrable set E. Then UA~ is a full set. n~oo n
Proof: First observe that E A An and E  A. are integrable for any n and any integrable set E, by (7.12). Let (Kn) be an Ibasis of X. Write S== V A •. For all j, kin '1.+ we have
•
J1l~1 (K j 
A.»)
~J1(Kj A
k
)= J1(K)  J1(K j A Ak)·
Since J1(K j AA k ) ..... J1(K j) as k ..... oo, it follows that
!~~ J1C~1(KjA.»)
=0. By (3.8), KjS= /\ (KjA.) is integrable and has measure O. It n
270
Chapter 6
Integration
follows from (8.10) that  S is integrable, with J1( S)= lim J1(K j S)=O. joo
Hence (_S)O is a full set. Since (S)o=SleUA~, the result follows. 0 (8.12) Lemma. Let I be (ifinite, and let (in) be a sequence of measurable functions converging in measure to each of two measurable functions f and g. Then f = g on a full set. Proof. Let (Kn) be an Ibasis of X. For each pair (n,j) of positive integers choose an integrable set A(n,j) such that A(n,N e K~, J1(K n  A(n,j» 0 there corresponds an integrable set B with BI e A I and J1(A  B) < e, such that for each (; > 0 there exists N in 'I. + with Ifmfnl0. (10.1.4) If (A.) lim Jl ( koc
is
a
sequence
of elements
of M
AA.) exists and is positive, then nA; is nonvoid.
n= 1
n
such
that
10. Measure Spaces
283
We then call Il the measure, and the elements of M the integrable sets, of the measure space (X, M, Il). For each A in M the nonnegative number Il(A) is called the measure of A. An integration space (X, L, I) gives rise to a natural measure space (X, M, Il), in which M is the set of integrable sets relative to I, and Il is
the map AI+I(XA) on M. This measure space is said to be induced by (X,L,I). Consider an arbitrary measure space (X, M, Il). The following sequence of lemmas will enable us to show that (X, M, Il) gives rise to an integration space. (10.2) Lemma. If A is an element of M such that A 1 is void, then Il(A) =0.
Proof· Suppose that Il(A) > O. Write A. == A for each positive integer n. Then, by (10.1.4), A 1 = A~ is nonvoid. This contradiction ensures that Il(A)=O. D •
n
n•
(10.3) Lemma. If K 1 , ... ,K. are elements of M, and F== (Kt uK?), then «(/), F) also belongs to M. ;~ 1 Proof: By (10.1.2), for each i we have
«(/),Kf uK?)=K;K;EM. Hence (g),F) =
V(0,K;1 uK?) belongs to M, by (10.1.1).
D
i= 1
(10.4) Lemma.
•
Let A, K l '
n (K;l uK?). Then (A
1
... ,
K. be elements of M, and F ==
nF, AO nF) belongs to M and has measure equal
;= 1
to Il(A). Proof. By (10.3), B == (O, F) belongs to M. It follows from (10.1) that A 1\ B and A  B belong to M. Moreover, Il(A)=Il(A I\B)+Il(AB) =1l«(/),(A 1 uAO)nF)+Il(A 1 nF,Ao nF) =1l(A 1 nF,Ao nFl,
by (10.2).
D
(10.5) Lemma. If A, B, and K1,.··,K n belong to M, and
•
n(Kt uK?), then Il(A)~Il{B).
i= 1
if
XA~XB
on
284
Chapter 6 Integration n
Proof. Write F== n(KfuK?). By (10.4), A'==(AlnF,AOnF) and i~1
B'==(B I nF,BOnF) belong to M, J.l(A')=J.l(A), and J.l(B')=J.l(B). Since XA ~ XB on F, we have A' A B' = A' and therefore (by (10 1.2)) J.l(A) = J.l(A') = J.l(B')  J.l(B'  A') ~ J.l(B') = J.l(B).
0
The notion of a simple function extends in a natural way from the context of an integration space to the present one: a simple function is n
L ai XAW
a function of the form X ==
where ai' .. , a. are real numbers,
i= I
are
A(l), ... ,A(n)
n •
elements
of M,
and
the
domain
of X is
(A (i)1 u A (i)0).
i= 1
Lemma (7.8) remains valid for measure spaces when interpreted appropriately Thus a simple function X can be written in the form
•
L ai XA(i)
where the complemented sets A (i) belong to M and are
i= I
disjoint, in the sense that XA(i)XA(j)=O on (A(i)1 uA(i)O)n(AU)1 uAU)°) whenever i j. (Note that if A and B are disjoint elements of M, then J.l(AAB)=O: this follows from (104) and (10.2).) Using an argument like that in the proof of Lemma (9.4), the reader can prove the following lemma.
*'
(106) Lemma. Let a l , ... , a. and b l , ... , bm be real numbers, and let A(l), ... ,A(n) and B(l), .. ,B(m) be elements of M. Suppose that n
m
i= I
j= I
L ai XA(i) ~ L b
j
F==
XB(j) on the set
(C\ (A (i) UA(i)0)) n CC\ (B(j)1 UBU)°)) I
n
Then
m
L aiJ.l(A(i))~ L bjJ.l(B(j)). j~
i= I
I
It follows from Lemma (10.6) that
J(tl
(10 7)
ai XA(i») ==
itl
ai J.l(A(i))
defines a mapping J from the set of simple functions into lR. For each simple function X we call J (X) the integral of X. In order to justify this use of the word 'integral', we need two estimates of J(X). (10.8) Lemma. Let f be a nonnegative simple function, and Ii a positive number Then there exists B in M such that BI u BO c dmnf, f < Ii on
BO, and
J.l(B)~2c
I
J(f).
10 Measure Spaces
285
n
Proof" We may assume that f
=
L a i XA(i)' where a p i~
""
an are non
1
negative numbers and A (1), ... , A (n) are disjoint elements of M. Write {I, .. , n} as a union of two sets Sand T such that ai < E for all i in S, and ai > E/2 for all i in T. If G i < E for all i, then the desired conclusion is obtained by taking B=(0,dmn!). So we may assume that T is nonvoid. Let E= V A(i) and B=(E 1 ndmnf, EO ndmn!). Then iET
Bl uBo cdmnf Also, by (10.1.1) and (10.4), BEM and /l(B)=/l(E)= L/l(A(i))
Consider any x in BO Since xEdmnfn(nA(i)O), either f(x)=O or ieT
there exists a unique i in S with XA(i)(X) = 1. In the latter case, f(x) =ai<E. Thus f<E on BO. D (10.9) Lemma. Let f be a simple function, let c be a positive number such that f ~ c on dmnf, and let A be an element of M such that f ~ 0
on A °n dmn! Then for each E > 0 there exists B in M such that Bl uBo cdmnf, f >E on Bl, and /l(B)?;c1(1(j)2E/l(A)). n
Proof We may assume that f=LaiXA(i)' where A(l), ... ,A(n) are i= 1
disjoint elements of M. Let E > 0 Write {I, ... , n} as a union of two disjoint sets Sand T, where ai < 2 E for each i in S, and ai > E for each i in T. To begin with, assume that T is nonvoid Let E= V ACi) and iET
B=(E1ndmnf, EOndmn!). Then B1uBocdmnf, f>E on Bl, BEM, and /lCB) = /l(E). Thus I(!)~I(XA!)
(by (10.6))
= Lai/l(A 1\ A(i))+ Lai/l(A 1\ A (i)) ieS
~2E L/l(A
1\
A(i)) +C L /l(A
ieS
1\
A (i))
ieT
~2E/l(A)+C/l(B).
Hence /l(B)?;c1(1(f)2E/l(A)). In the case where T is void we need only take B
=(0, dmn!).
D
(10.10) Theorem. Let (X, M, /l) be a measure space, L the corresponding set of simple functions, and I the mapping defined on L by (10.7). Then (X, L, 1) is an integration space.
286
Chapter 6 Integration
Proof. To verify properties (1.1.1) and (114), we need only consider n
L CiXqi)
simple functions of the form
with C(1), .. :, C(n) disjoint ele
i~1
ments of M, the appropriate arguments are then straightforward. To verify (1.1.3), choose A in M with Jl(A»O; then P==Jl(A)1 XA belongs to L, and I(P) = 1. It remains to verify (1.1.2). Accordingly, let (fn) be a N
sequence of nonnegative elements of L, and let
f ==
L i~
element of L such that
aiXA(i)
be an
I
L I(fJ converges to a sum less than I(!). Write n
n
and N
A==
V A(i).
Then AEM, and f=O on AOndmnj Choose c>O so that.f~c on dmn f, and write and ex == c l(r  2 e Jl(A)).
Choose a strictly increasing sequence with n(l)=l, such that I(
n(k+ I )  I
L /;
)
(n(k));;"~
~22klexe
I
of positive integers,
(k~2).
i~n(k)
Define n(k+
1)1
L /;
gk==
(k~ 1).
By (10.8), for each k ~ 2 there exists Bk in M such that B~ u B~ cdmngk , gk 1/4, k~
n
and therefore xEA 1. Thus B1 c A 1. N~w suppose that lL{A) > lL{A 1) A
2 00
+ 1L
lY2 A
k ).
V A k • Thus k~
1 00
1 =XA{y)~g1(Y)+ L gk{Y) k~2
~IX+ 1/4JR a continuous mapping such that a == inf f < b == sup f Let 8 be the set of all continuous functions from [a, b] into [0,1], let x o , Xl be any two points of X, and for each h in It define A(h) ==  inf {max {p(x, x o), P(Xl' x o)(l h(f(x)))}: XEX}
Then (8, A) is a profile for [a,b]. Show that p(xo,X r) exists for each smooth point t for this profile, where Xr=={XEX: f(x)~t}. Show also that for each smooth point t and each 8> 0 there exists c5 > 0 such that Ip(x o, X r )  p(x o, Xr")1 Jl.(X) Prove that there exists a subfinite set SeX such that for each x in  S there exists f~O in qX) with f(x) = 1 and SfdJl. 0 there exists 6> 0 such that whenever P == (a o , ... , am) and Q== (b o , ... , bn) are partitions of [a, b] with mesh less than 6, and Xl' ... ,Xm' YI' ... 'Y n are points of F with aO~xl ~al ~ ... ~am_l ~xm~am and bO~YI ~bl ~ ... ~bn_l ~Yn~bn' then IS(f, P)  S(f, Q)I ~ 8, where m
b
and S(f, Q) is defined similarly. Define the Riemann integral
Jf(x)dx a
of f, and show that f is Riemann integrable if for each 8> 0 there exist 6> 0 and a compact integrable set KeF with fl(  K) ~ 8, such that If(x) f(Y)1 0, and a sequence Un) of test functions with a common compact support, such that IIfnll ~c for all n, and Un) converges almost everywhere to f 26. An integral 1 is weakly ufinite if there is a sequence (K(n));;"~ 1 of integrable sets such that K(l)l c K(2)1 c.. and U K(n)l is a full set. n
Show that 1 is weakly ufinite if and only if there exists an integrable function h such that h > 0 on a full set 27. Let 1 be a weakly ufinite integral, and let Un) be a sequence of measurable functions converging in measure to a measurable function f Suppose that there is an integrable function g such that Ifni ~ g for each n. Show that f is integrable (whence, by the dominated convergence theorem, 1(lf  fnl) > 0 as n > (0). (Hint You may assume that g>O on a full set. Look at the proof of Theorem (7.11).) 28. Let 1 be a weakly ufinite integral, and f a nonnegative measurable function. Prove that f is integrable if and only if sup {I(n
1\
XAf): nEZ+, A is an integrable set}
exists. 29. Let 1 be a finite integral, f an integrable function, and S a locally compact subset of IR with nonvoid metric complement, such that J1.(A)I/(XAf)ES for each integrable set A with positive measure. Prove that f(X)ES for all x in a full set. (Hint. Consider any r > 0 such that Sr={tEIR: p(t,S);:;:r} is locally compact, and show that p U(x), Sr);:;: r for all x in a full set) 30. Let (fn) and (gn) be sequences of nonnegative measurable functions. We say that (fn) dominates (gn) in measure if to each integrable set A and each G > 0 there corresponds N in 7l+ such that for each integer n;:;: N, there exists an integrable set B with Bl c A I, J1.(A  B) < G, and fn;:;: gn on Bl Prove that if (i) fn and gn are integrable for each n, (ii) Un) dominates (gn) in measure, and (iii) (gn) converges in measure to a nonnegative integrable function g, then for each G> 0 there exists N in 7l+ such that IUn) > I(g)  G for all n;:;: N.
Notes
297
31. Let X and Y be locally compact spaces, Jl a positive measure on X, and v a positive measure on Y. For each test function I on the locally compact product space X x y, yl+ SI(x, y) dJl is a test function on Y, and so u(f) == S(S I(x, y) dJl) dv
is a well defined real number Show that u is a positive measure on X x Y, that u(f)= S(S f(x,y)dv)dJl
(IEC(X x Y)),
and that the complete extension of u is the product integral Jl x v as defined in Section 9. 32. Let I be a nonnegative measurable function with respect to a product integral I x J, and let the iterated integral c == J(l(f)) exist. Show that (l x J)(f) exists and equals c.
Notes The work of this chapter is based on the treatment of the integral given by Bishop and Cheng [l3]. The proof of Theorem(1.10) is due to YK Chan. Related to Theorem (1.10) is Problem 1, which originated with Newcomb Greenleaf. Strictly speaking, an integrable function is a pair (1, (fn)) consisting of a function I and a representation (In) of I by elements of L. It is simpler to concentrate on the function 1, and emphasize the representation (fn) when we need it. The existence of
;~~
Jll01 An)
in the hypotheses of Proposition
(3.8) is superfluous from a classical viewpoint. As Problem 15 shows,
it is necessary in the constructive setting. The obvious example of a profile is obtained by taking tC to be the set of all continuous functions from a proper compact interval [a, b] b
to [0, 1], and defining A(f) == Sf(x) dx (the Riemann integral) for each I in tC. The proof of Lemma (4.7) is a modification of one given by YK Chan.
298
Chapter 6
Integration
It can be shown that if J1. is a general measure (not necessarily positive) on JR, then there exist a set S consisting of all but countably many real numbers, and a function IX: S > JR of bounded variation, such that J1. = J1.~ (where J1.~ is constructed as in Problem 12). Further information on general measures can be found in [23]. Note that whereas a function of bounded variation is classically defined at every point of the interval under consideration, constructively this is usually not possible. It is easy to give an example of a compact set K c JR and a positive measure J1. on JR with respect to which K is not integrable: take K == {OJ and let J1. be the point mass at a, where a = 0 is impossible but we do not know that a =l= O. A more interesting example is required in Problem 4. Note how the constructive definition (Definition (8.1)) of convergence almost everywhere differs from the classical definition (pointwise convergence on a full set). The classical definition would be of no use constructively. Lemma (8.12) extends Proposition (8.3) in the case where I is (Jfinite. Definition (10.13) differs slightly from the definition of a complete measure space given by Bishop and Cheng. Certain parts of measure theory in JR 2 have been developed by Brouwer, with a different approach from the one used here.
Chapter 7. Normed Linear Spaces
In Section 1 we introduce normed linear spaces and bounded linear mappings. Section 2 is concerned with finitedimensional spaces and with the problem of best approximation by elements of a finitedimensional subspace. In Section 3 we discuss Lp spaces. we prove the completeness of L p , and determine the form of the normable linear functionals on Lp in case p> 1. (In contrast to the classical theory, a bounded linear functional need not have a norm.) We then apply these results to the proof of the RadonNikodym theorem. In Section 4 we prove the separation theorem and its corollary, the HahnBanach theorem, under the assumption that a certain convex set is located. Section 5 introduces the notion of a quasinorm, and discusses the quasinormed space Loo and !ts linear functionals. Next we prove the compactness of the unit ball of the dual, and a result (6.8) which makes it plausible that every constructively defined linear functional on the dual X* is determined by an element of X a curious form of reflexivity! Section 7 deals with extreme points, and contains a proof of the KreinMilman theorem; in contrast to the classical development, detailed estimates are needed. In the final section of the chapter we define Hilbert space and prove the spectral theorem for hermitian operators; we also prove versions of the Gelfand representation theorem for algebras of normable operators. Most classes of functions that arise in analysis are endowed with both a linear structure and a norm. This makes it natural to introduce the concept of a normed linear space.
1. Definitions and Examples To be able to discuss real and complex linear spaces simultaneously, we let IF denote either the real or the complex number field. The elements of IF are called scalars.
300
Chapter 7
Nonned Linear Spaces
(1.1) Definition. A linear space, or vector space, over IF is an abelian group X with a multiplication operation (a,x)t+ax from IF x X to X such that for all a, aI' a 2 in IF and x, XI' X2 in X,
(a l +a 2 )x=a l x+a 2 x, a(xI +x 2 )=ax I +ax 2 ,
a l (a 2 x) = (a l a 2 )x, and 1x=x.
The elements of X are called vectors. We assume the reader to be familiar with the elementary theory of linear spaces (1.2) Definition. A seminorm I lion a linear space X over IF is a function xt+llxll from X to JR o + such that for all a in IF and X,XI' X2 in X, Ilaxll = lalllxli and IlxI +x 2 11 ~ IlxI11 + Ilx 211· The pair (X, II II) is called a seminormed linear space When there is no confusion over the seminorm, we often refer to X itself as a seminormed linear space. If x=O whenever Ilxll =0, then the seminorm II II is called a norm, and we refer to X as a normed linear space, or simply a normed space There is no loss of generality in confining our attention to norms, since a seminorm II lion a linear space X becomes a norm if the equality of XI and X2 in X is changed to mean that IlxI x211 =0. A seminorm II lion a linear space X determines a pseudo metric p on X, given by
Clearly, p is a metric when II II is a norm A normed space X is thus a metric space. The standard inequality =1= on X is the inequality defined relative to the metric p. A vector X is nonzero if X=1= o. It is easily seen that when X is a normed space, addition and subtraction are continuous functions from X x X to X, and that multiplication of a vector by a scalar is a continuous function from IF x X to X. The notion of an infinite series with terms in a normed space X is defined in the natural way Tests for the convergence of such series, similar to tests for the convergence of series in JR, can be derived. We omit the details.
1. Definitions and Examples
301
The norm on a normed linear space X can be described geometrically in terms of the unit sphere Sc(O, 1) of X (Problem 1). (1.3) Definition. A mapping u of a vector space X into a vector space Y is said to be linear if
u(ax)=au(x) and
U(XI +XZ)=u(xI)+u(X Z) whenever aEIF and x, Xl' XzEX. (If Y=IF, then u is called a linear functional on X.) In case X and Y have norms and therefore metrics, we say that a linear mapping u: X + Y is continuous if it is uniformly continuous on each bounded subset of X (1.4) Lemma. For each element
X
of a normed linear space X,
Ilxll = inf{lal l : aEIF, a =1= 0, Ilaxll ~ 1}.
Proof. For each nonzero scalar a, Ilaxll ~ 1 ¢> lallixil ~ 1 ¢>
The result now follows.
lal l ~ Ilxll·
D
(1.5) Proposition. The following are equivalent conditions on a linear map u of a normed linear space X into a normed linear space Y.
(i) (ii) (iii) (iv) (v)
u is continuous. u is uniformly continuous. u is bounded on the unit sphere of x. u is bounded on each bounded subset of x. There exists a positive number c, called a bound for u, such that Ilu(x)11 ~c Ilxll (XEX).
Proof. Suppose that u is continuous. Then u is uniformly continuous on the unit sphere S of X, and so there exists r > 0 such that Ilu(x)11 = Ilu(x)u(O)11 ~ 1
(11xll ~r)
Consider an arbitrary x in X. For each nonzero scalar a with Ilaxll ~r we have Ilu(x)11 =lalIllu(ax)11 ~Iall. Hence Ilu(x)1I ~rIllxll, by (1.4). Thus (i) implies (v). It is clear that (v) implies (iv), and that (iv) implies (iii). Now assume (iii), and choose c>O so that Ilu(x)11 ~c for each x in S.
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Chapter 7 Normed Linear Spaces
Consider arbitrary Xl' x 2 in X. For each nonzero scalar a with a(xi X 2)ES, we have Ilu(x l )u(x 2 )11 = Ilu(x l x 2)11 =IalIllu(a(xi x 2))11 ~c lal l
It follows from (1.4) that Ilu(x l )u(x 2)11 ~c Ilxl x211.
Hence u is uniformly continuous, and (iii) implies (ii). Finally, it is obvious that (ii) entails (i). 0 In view of property (v) of Proposition (1.5), we often refer to a continuous linear map u: X > Y as a bounded linear mapping on the normed linear space X. (1.6) Definition. The kernel, or null space, of a linear mapping u: X + Y between vector spaces X and Y is the set ker u == {XEX: u(x)=O}. In case Y has an inequality relation we say that u is nonzero.
=1=
and u(x) =1= 0 for some x in X,
Of particular interest are the bounded linear functionals on a normed linear space. These are attached to special linear subsets known as hyperplanes. (1.7) Definition. A hyperplane in a normed linear space X is a linear subset H of X with the property: there exist Xo in X and c > 0 such that Ilxxoll ~c
(xEH)
and such that each element x of X can be written in the form (1.7.1)
x=axo+Y
with aEIF and YEH. The vector Xo is then said to be associated with H. If H is a hyperplane in X, then for each x in X the expression (1.7.1), with aEIF and YEH, is unique. (1 8) Proposition. If u is a nonzero bounded linear functional on a normed linear space X, then ker u is a hyperplane. Conversely, if H is a hyperplane with associated vector x o, theY,! there exists a unique bounded linear functional u on X with ker u = Hand u(x o) = 1.
1 Definitions and Examples
303
Proof" Let u be a nonzero bounded linear functional, and Xo any vector with u(x o)=1. By (1.5), there exists c>O such that clu(x)I;;;;llxll for all x in X. If x is any vector in ker u, this gives
Ilx xoll
~c lu(x
xo)1 =c lu(x)u(xo)1 =c.
On the other hand, for each vector x in X we have x=ax o + y, where a=u(x) and y=xu(x)x o ; clearly, YEkeru. Thus keru is a hyperplane with associated vector Xo' Conversely, consider a hyperplane H with associated vector xo' If the required linear functional u exists, then it must be given by u(x)=a, where x=ax o + y, a ElF, and YEH. Define u in this way. Clearly, u(x o)=l and H=keru. Choose c>O so that Ilxxoll~c for all x in H. With x, a, and y as above, and 8> 0 arbitrary, we prove that (1.8.1 )
lu(x)1 ;;;;c 1 Ilxll +8
Either (1.8.1) holds or lu(x)I>O. In the latter case, a = u(x)=t= 0 and so Ilxll = lallix o ( _a 1 y)11 ~ lal c= lu(x)1 c. Hence lu(x)l;;;;c 1 1Ixll, and (1.8.1) holds in this case also. Since 8 is arbitrary, it follows that lu(x)l;;;;c 1 1Ixll. Thus u is bounded. D (1.9) Definition. Let u be a bounded linear map of a normed linear space X into a normed linear space Y. We say that u is normable if
Ilull =sup{llu(x)ll: XEX, Ilxll;;;; I} exists as a real number. In that case, Ilull is called the norm of u, and is the smallest bound for u. If X is nontrivial, in the sense that it contains a nonzero vector, and if u is normable, then Ilull =sup{llu(x)ll: XEX, Ilxll = I}. Although no general criterion of normability has been found, there is a very important criterion for normability of bounded linear functionals. (1.10) Proposition. A nonzero bounded linear functional u on a normed linear space X is normable if and only if ker u is located.
Proof: Assume that u is normable. Since u is nonzero, II u I > N 1 for some N in Z +. For each integer k ~ N let X k be a vector with I X k I = 1 and u(x k ) > Ilullk l . For each x in X and y in keru we have IlxYII~llullllu(x)l. On the other hand, for each k~N the vector Zk=XU(x)U(Xk)l Xk
304
Chapter 7 Normed Linear Spaces
belongs to ker u, and Ilx  zkll = lu(x)1 U(X k)1 < lu(x)l(11 ull  k 1) 1. Thusp(x,keru) exists and equals Ilull 1 Iu(x)l. Conversely, assume that ker u is located. Since u is nonzero, there exists a vector Xo with u(x o) = 1. As u is uniformly continuous (1 5), p(xo,keru»O. Thus
inf{llyll : u(y) = 1} = inf{ Ilxo zll : zeker u} = p(xo, ker u) exists Also lu(z)1 >0 and Ilzll = 1, where z= Ilx o ll 1xo. Hence sup{lu(x)l: Ilxll = 1} = sup {llyll1: y = U(X)1 x for some x with Ilxll = 1 and u(x)=l= O} =sup{ Ilyll1: u(y)= 1} exists and equals p(xo, keru)1. It now follows that u is normable, with Ilull =p(x o, keru)1. 0 The next definition introduces the most useful type of normed linear space. (1.11) Definition. A normed linear space which is both separable and complete is called a Banach space.
Recall that a metric space is separable if it has a countable dense set There are no important constructively defined normed linear spaces that are not separable As far as existing mathematics is concerned, there is no loss of generality in postulating separability, and there is a great gain in power and convenience. The fact that every metric space can be completed leads us to hope that the same is true of every normed linear space X. This turns out to be the case, and the completion of X as a normed linear space can be identified with its completion as a metric space. Here is an outline of the construction. We take X to consist of all Cauchy sequences in X, with termwise addition and termwise scalar multiplication. Two elements (x n) and (Yn) of X are equal if lim Ilx n Ynll =0 Clearly X is a linear space. Under the norm n 00
it is a complete normed linear space, called the completion of X. The inclusion map t·: X ~X defined by i(x)=(x n), where Xn=X for each n, preserves norms and realizes X as a dense linear subset of X. We thus have the following result.
1 Definitions and Examples
305
(1.12) Proposition. If X is a normed linear space, then there exist a complete normed linear space g and a normpreserving linear inclusion map t from X onto a dense subset of g
In case X is separable, g is also separable and therefore a Banach space. The bounded linear functionals on X are in oneone correspondence with the bounded linear functionals on g. if u is any bounded linear functional on X, and (xn) is any point of g, then (u(xn)):;O~ 1 is a Cauchy sequence in IF whose limit u((x.)) defines an extension of u to X. An important example of a Banach space is the space C,u(X, IF) of continuous functions vanishing at infinity on a locally compact space (X, p). In this case, the norm is given by Ilfll =sup{lf(x)l: XEX}
(fE Coo(X, IF)).
Recall from (611) of Chapter 4 that if (Y, d) is a onepoint compactification of X with point at infinity wand inclusion map i. X + Y, then there is a normpreserving algebra isomorphism ff+f" which identifies Coo(X,IF) with the metric space G of all elements f in c(Y,IF) with f(w)=O. As with every normed linear space, we are interested in the bounded linear functionals on CDC (X, IF). It will suffice to consider the case IF=lR, since a bounded linear functional u on Coo(X,d. D In the proof of our next theorem we use the notion of a quotient space, which will appear again later in the chapter.
(2.9) Definition. Let Y be a located linear subset of a normed linear space X. Then Ilxllx/y=p(x, Y) (XEX) defines a seminorm on X. Taken with this seminorm and the corresponding equality relation = XIY, X is a normed linear space  called the quotient space of X by Y, and written XIY. (2.10) Lemma. Let X be a normed linear space, n a positive integer, and
F an (n + 1)dimensional subspace of X. Let {e 1 , ... , en + d be a basis of F, and Y= IFe n + 1 . Then F is an ndimensional subspace of the quotient space XIY, with basis {e1, ... ,en }.
Proof" Let the coordinate functionals for the basis {e1, ... ,en+tl of F be U l ' ... , Un + l ' and let c > 0 be a common bound for these functionals n
For each x in F we have x=x/Y i
*' j). If also x'
L ui(x)e i, where ui(e)=O i~
EF
(1 ~i, j~n,
1
and x = X/Y x', then as Y is closed, n
L (ui(x) 
ui(x')) ei E Y,
i= 1
and so for 1 ~k~n we have uk(X)Uk(x')
=
Uk
(tl (ui(x) Ui(X'))ei) =0.
Hence Uk is a function on F relative to the equality relation on XIY. Since Uk is linear relative to = X/Y, it remains to prove that Uk is bounded on F relative to I Ilx/y' To this end, consider an arbitrary
312
Chapter 7 N armed Linear Spaces
element x of F with II x II X/y ~ 1. Since p (x, Y) < 2, there exists a in IF such that Ilxaen + 1 11 p(a,F) => Ilazkllx/y=p(az k, IRe n+ 1 »p(a,F).
Since
II(zl +'l e n+l)(z2+'2 e n+l)11 = II(ZI z2)+('I'2)en+ 1 11 ~ Ilzl z21Ix/y>O,
we have
max {ila Zl  , I e n+ III, Iia Z2  '2en+ III} > pea, F) Hence, by (2.12.1) and (2.12.2), max{lla zlllx/y, Ilaz 21Ix/y} > pea, F)=inf{llazllx/y: zEF}. Since ZI' Z2 are arbitrary, it follows that a has at most one closest point in the ndimensional subspace F of XIY. By our induction hypothesis, there exists Zo in F with p(az o, Y)= Ilazollx/y=inf{llazllx/y: zEF} =p(a, F).
Applying (2.8) with x=az o, .e=en+ l , and d=p(a,F), we now construct a real number to such that Ilazotoen+111 =p(a,F).
=
Then b Zo + to e n + I is a closest point to a in F relative to the original norm on X. Finally, if b' is any closest point to a in F, then assuming that b =l= b' , we have max{lla bll, Ilab'll} > pea, F), a contradiction. Hence b = b'.
D
3. The Lp Spaces and the RadonNikodym Theorem Throughout this section, unless we state otherwise, I will be a completely extended, afinite integral on x.
314
Chapter 7 Normed Linear Spaces
Our next goal is to introduce certain basic normed spaces of measurable functions. This requires some preliminary inequalities. (3.1) Lemma. Let a, b be nonnegative numbers, and cx, p positive numbers with cx + p = 1. Then (3.1.1 )
Proof: For fixed values of b, cx, and p, consider f(a)= cxa + pb a~bfJ.
We have f'(a) = cx  cxa~l bfJ = cx(l (ball)
if a =1= O. Thus f'(a)~O if a>b, and f'(a)~O if O O. By continuity, (3.1.1) is valid for a~O. D Successive applications of Lemma (3.1) lead to (3.2) Corollary. If ai' ... , a. are nonnegative numbers, and cx l ' positive numbers with ex 1 + ... + cx. = 1, then
... '
cx. are
The proof of our next lemma is another simple application of differentiation, and is left to the reader. (3.3) Lemma. Let x, y be nonnegative numbers, and let 0 < t < 1. Then (x + y)' ~x' + y'.
.t.
(3.4) Theorem. Let fl' ... be nonnegative integrable functions, and cx l ' ... ' cx. positive numbers whose sum is 1. Then ff' . .. f.~n is integrable, and
(3.4.1) Proof: The integrability of ff' .. f.~n follows from (3.2), and (7.11) of Chapter 6. Let Cl> ... , Cn be positive numbers with Ci ~ I(f;) (1;£ i;£ n). By (3.2), I((cllflt' ... (c;;lf.)~n);£1(CXl Cllfl + ... +cx.c;;lfn)
= cx l c l l 1(fl) + ... + CXnc;; 1 I(fn) ~ CX l + ... + CX n = 1.
3 The Lp Spaces and the RadonNikodym Theorem
315
Thus
l(ff' ... g.n)~c~' ... c~n. Letting ci +l(J;) (1 ~ i ~ n), we obtain (3.4.1).
D
Inequality (3.4.1) is called HOlder's inequality. It is a basic tool in the theory of certain spaces  the Lp spaces  which we now define. (3.5) Definition. For each p ~ 1 the set Lp consists of all measurable functions f such that Ifl P is integrable. The norm of a function f in Lp is (3.5.1 ) Elements f and g of Lp are equal if f(x) = g(x) on a full set. We now derive a remarkable inequality relating certain Lp spaces, which is actually a reformulation of Holder's inequality. (3.6) Proposition. Let p> 1, q> 1, and
r
1 +q1
= 1. Let fELp and
gELq. Then fgEL 1 and l(fg)~ Ilfllp Ilgll q.
Proof: Applying (3.4) with f1=lfI P, f2=lglq, IX1=rl, and we see that If gl, and therefore f g, is integrable, and that
IX 2 =q1,
l(fg)~I(1fgl)=I(lfIP«1Iglq«2)
~I(lfIP)«1 1(lglqt 2
= Ilfllp Ilgll q.
D
HOlder's inequality leads to another remarkable inequality (3.7.1), known as Minkowski's inequality. (3.7) Theorem. Let p ~ 1, and let f, gELp. Then f (3.7.1)
+ g is
in L p' and
Ilf +gllp~ Ilfllp+ Ilgll p·
Proof Since f + g is measurable and If +gIP~(2max{lfl,
Igl})P~2P(lfIP +
IgjP),
f
+ g belongs to L p ' by (7.11) of Chapter 6. Let 0 < IX < r 1. Then by (3.3), If+gIP«~lfIP«+lgIP«. By (3.4), the functions IfIP«lf+gIP(l«) and IgIP«lf+gI P(1«) are integrable, and
1(lf + gIP)=l(lf +gIP« If + gIP(1«) ~ 1(lfl P« If + gIP(1«) +1(lgIP« If + gIP(1«) ~ (l(lfI P)« + l(1gIP}«) l(1f + gIP}1«.
316
Chapter 7
Nonned Linear Spaces
Thus (3.7.2) Now for each e>O, either (3.7.3)
Ilf +gllp~e+ Ilfllp+ Ilgll p
Ilf+gllp>O In the latter case, dividing both sides of (3.7.2) by Ilf + gll~(Ia) and then letting (X+pl gives Ilf + gllp ~ I flip + Ilgllp. There
or
fore (3.7.3) holds in all cases. Since e>O is arbitrary, inequality (3.7.1) now follows. D It is now clear that Lp is a normed linear space relative to the Lpnorm I lip given by (3.5.1). Note that the equality relation corresponding to I lip is the same as the equality on Lp defined in (3.5). So far, we have not made use of the assumption that I is afinite.
This property of I comes into play when we prove that Lp is complete. (3.8) Lemma. If (f.) is a Cauchy sequence in the normed linear space Lp (where p ~ 1), then (f.) is Cauchy, and hence convergent, in measure. Proof: For each e>O there exists N in Z+ such that I(lfmf.n~el+P
(m,n~N).
Consider integers m, n with m ~ n ~ N. By (8.20) of Chapter 6, there exist (X in (e,2e) and a measurable set A such that Ifmf.IO on a full set. Write q=pj(p1). By (3.15), I((f + g)p)=I(f(f +g)P1)+I(g(f +g)p1) ~ I(fP)l/ p I((f + g)p)l/q + I(gP)l/ p I((f + g)p)l/q.
Since I((f + g)p) > 0 by (4.13) of Chapter 6, we can divide through by I((f + g)p)l/q to obtain (3.17.1). 0 (3.18) Lemma. If OO, 00
f(t)=t 1(exp( IXt)l)=
L (_1)"IX n t n  1jn! n= 1
so that 00
f'(t)=
L (1)"(n1)IX n t n  2 jn! n= 2
Since IX < 0, all the terms of the last series are positive, and so f'(t) > O. The result now follows. 0 (3.19) Lemma. Let x, y, p, and q be real numbers, with p 1 Then
(3.19.1) for
p~2,
and
(3.19.2) {or 1 Iyl >0. Dividing by Ixl q and setting c=:lyx 1 1 (so that 0n~N.,
and choose r in (IX, 2 IX) so that
is both 1 and Iintegrable Then m
I(xd~rI
L
I(lfkl) 0 and any integrable set A. By (7.2) of Chapter 6, there exist c > 0 and an integrable set K such that K1c:Al, I(XA_K)<e/2, and O~fo~c on Kl. Since I(gAn)>I(g) O~(gAn)fo~gfo.
as n > 00, the sequence (g A n) converges to g in measure Thus for all sufficiently large n there exists an integrable set B such that Bl c: Kl, I(XK_B) <e/2, and IggAnl 0, the vectors x + y and t \' belong to S. An important type of cone is associated with convex subsets of X. Recall that K c X is convex if (1  t)x + t Y belongs to K whenever x, y E K and ~ t ~ 1. In that case, the set c(K)== {tx: t>O,xEK}
°
is a cone, the cone generated by K. To see this, it is enough to show that if Xl' X2EC(K), then Xl +X2EC(K). Writing Xl == tl Yl and x 2 == t 2Y2 with t l , t2 >0 and Yl' Y2EK, we have Xl +X2 =(t l + t 2)z, where Z==tl(t l +t 2)1 Yl +t 2(t l +t 2)1 YlEK. Hence
Xl
+x 2 belongs to c(K).
(4.2) Lemma. Let K be a bounded, located, convex subset of a normed
linear space X, whose distance from located.
°is positive. Then the cone c(K) is
Proof: For each t>O write tK=={tX:XEK}.
336
Chapter 7
Normed Linear Spaces
Then for each Y in X we have p(y, tK) = tp(t 1 y, K) ~ t p(O, K) IIYII,
Since p(O,K»O,we can choose or: > 1 so that p(y,tK»p(y,K) whenever t > or:  1. It now follows that p(y, c(K»
which exists by (4 1).
= inf{p(y, t K): 0 < t < or:},
D
The fundamental geometric fact about normed linear spaces is that under certain conditions two convex sets can be separated by a normable linear functional. This result, which we now prove, is called the separation theorem. (43) Theorem. Let F and G be bounded convex subsets of a separable normed linear space X, whose algebraic difference {yx: xEF, YEG} is located, and whose mutual distance
d==:inf{ Ily xii: xEF,YEG} is positive. Then for each E > 0 there exists a normable linear functional u on X of norm 1 such that (4.3.1)
Reu(y»
Re u(x)+d E
(xEF,YEG)
Proof We may assume that E < d. Consider first the case IF = JR.. By a generalization of (2.3) of Chapter 5, the bounded, open, convex set K==:{yXZ:XEF,YEG, Ilzll 'r, it follows that p(O, K,;) ~min {tp(O, K n_ 1), (2n)1 'r} > 0.
Hence K,;, and similarly K;;, satisfies (4.3 3). Now suppose that max {p(xo, c(K,;)), p(x o, c(K;;))} < IX =. (1  2 n 1) p(x o, c(Kn_ 1)). Choose Zl in c(K,;) and Z2 in c(K;;) so that
338
Chapter 7
Normed Linear Spaces
Then ZI=tIXn+YI and Z2=t 2 Xn+Yz, YI Ec(K n _ I ), and Y2EC(Kn_I)· We compute
where
tl>O,
t 2 >0,
(1  2 nI)(t 1 + t 2) p(Xo, c(Kn_ 1))
> tl Ilxo ( t 2 Xn + Yz)[f + t211 xo(tl Xn + YI)II ~ II(t l +t 2)X O (t 2YI +t l Y2)[f =(tl +t 2) Ilx o(t 2(t1 +t 2)1 YI +tl(t l +t 2)1 Yz)11 ~(tl
+t 2 )p(x O,c(Kn _
I
»·
This contradiction ensures that max {p(xo, c(K;;», p(xo, c(K;;»} > (1  2 n) p(xo, c(K n_ I»' Therefore either K;; or K;; satisfies (4.3.2). In the former case define Kn == K;;, and in the latter define Kn == K;; This completes the construction of the sequence (Kn)~ o. The set 00 K", == U c(Kn) n= 1
is an open convex cone By (4.3.2), (p(xo, c(Kn»)~ 1 is a Cauchy sequence. As c(Kn_l)cc(K n) for each n, the limit of this sequence is p (xo, K ",). Moreover, again by (4 3.2), p(xo, c(Kn)) >
and so Let
COy
2 k ») p(xo, c(Ko)) >8/8
(nEZ+)
p(xo,K",)~8/8.
K_oo=={XEX: xEKoo}'
Since XoEKoo, it follows that xoEK_ oo ' By (4.3.4), KoouK_ oo is dense in X. Also, Koo and K_", are disjoint, in the sense that if xEKoo and YEK_ oo , then IlxYII >0: for, choosing n so that xEc(K n) and YEc(K n), we see that xYEc(Kn); so that xy=tz for some t>O and z in K n, and therefore IlxYII ~tp(O,Kn»O, by (4.3.3). Let N be the intersection of the closure of K 00 and the closure of K_ oo • Then N is closed. Since both K", and K_ oo are cones, N is a cone; since also XEN if and only if xEN, N is a linear subset of X. Consider an arbitrary vector x in Koo' We show that there is a unique vector ¢(x) in NnL, where L== {tx+(1t)xo:
O~t~
1}
is the line segment joining Xo to x. As K 00 and K _ '" are open, there exists b in (0,1) such that x+zEK oo and xo+zEK_oo whenever Ilzll 0 about a point x in X is the set Sex, r) consisting of all y in X such that there exists r' < r with II x  y II i < r' for all i. The neighborhood structure of X is defined by taking the neighborhoods to be the open spheres. When applied to subsets of X, the notions of open set, closed set, dense set, interior, and closure refer to this neighborhood structure. A subset Y of X is closed if and only if it contains all points that are limits of sequences of points of Y, in the sense of (5.3) Definition. A sequence (xn) of elements of a quasinormed linear space (X, (II II ;)iEI) converges to an element x of X if for each B > 0
344
Chapter 7
N armed Linear Spaces
there exists N. in '1.+ such that Ilx x.lli~8
(n~N., iEI).
The point x is then called the limit of (x.) in X, and we write either lim x. =X .~oo
or x. +x
as n+ 00.
The closed sphere of radius r ~ 0 about a point x in a q uasinormed linear space (X, (II II i)iEf) is the closed set Sc(x,r)={YEX: Ilxylli~r for each i in I}.
Of special interest will be the unit sphere Sc(O, 1) (5.4) Definition. A Cauchy sequence of elements of a quasinormed linear space (X,(II IUiEl) is a sequence (x.) such that for each 8>0 there exists a positive integer N. with Ilxmx.lli~8
(m,n~N., iEI).
A subset A of X is said to be complete if every Cauchy sequence of elements of A converges to a limit in A. Naturally, we are interested in bounded linear maps between quasinormed linear spaces. (5.5) Definition. Let u be a linear map of a quasinormed linear space (X, (II II;liEf) into a quasinormed linear space (Y,(II Ilj)jEJ). We say that u is bounded if there exists a positive number c, called a bound for u, such that if XEX, 8>0, and jEJ, then c Ilxll i > Ilu(x)ll j 8 for some i in I. We say that u is uniformly continuous if for each 8>0 there exists 8 for some j in J, then Ilxx/lli>b for some i in I. The following result generalizes Proposition (1.5). (56) Proposition. A linear map between quasinormed linear spaces is bounded if and only if it is uniformly continuous. Proof Let (X,(II IIJiEf) and (Y,(II II)jEl) be quasinormed linear spaces, and u: X + Y a linear mapping. Suppose that u is bounded, with bound c>O. Let x, X'EX, 8>0, jEJ, and Ilu(x)u(x')ll j >8. Choose i in I so that c Ilx x' Iii> Ilu(x x')ll j 8/2 >8/2.
Then Ilx x' Iii >(2C)1 8. Hence u is uniformly continuous.
5 Quasinonned Linear Spaces, the Space La>
345
°
Conversely, assume that u is uniformly continuous. Choose r> so that for each j in J, if Ilu(x)ll j > 1, then there exists i in 1 with Ilxll;>r. Consider x in X, E>O, and j in J. Either Ilu(x)ll j >E/2 or Ilu(x)llj<E. In the former case, setting a =(llu(x)ll j E/2) \
we have Ilu(ax)llj=lalllu(x)llj> 1; whence IlaxII;>r for some i. Then Ilxll i = lal 1 IlaxII; > r lal 1 > r(llu(x)ll j E) and therefore (56.1)
In case Ilu(x)llj<e, inequality (56.1) clearly holds for all i in 1. Hence r 1 is a bound for u. 0 For all quasinormed linear spaces X and y, we shall use the notation Hom(X, Y) to denote the set of all bounded linear mappings of X into Y. We shall consider Hom(X, Y) as a quasinormed linear space with the quasinorm (II 11.}"es given by Ilullx= Ilu(x)11
(uEHom(X, Y), XES),
where S is the unit sphere of X. The neighborhood structure corresponding to this quasinorm on Hom(X, Y) will be called the uniform neighborhood structure on Hom(X, Y). Of special interest is the case Y=IF. (5.7) Definition. The dual space of a quasinormed linear space X is the quasinormed linear space Hom(X,IF) of all bounded linear functionals on X. The dual space is also written X*. From a classical point of view, two quasinorms on a vector space X are essentially the same if they give rise to the same norm. We capture this notion in a more general form by means of (5.8) Definition. Let (X, (II II ;);El) and (Y, (II UjEJ) be quasinormed linear spaces, and u: X + Y a linear map. We say that u is a linear isometry if (i) for all e > 0, x in X, and i in 1, there exists j in J such that Ilu(x)ll j > IlxlliE, and (ii) for all E>O, x in X, and j in J, there exists i in 1 such that Ilxll;> Ilu(x)lljE If u: X + Y is a linear isometry, then it is injective, in the sense that u(x)=I=u(x') whenever x, X'EX and x=l=x'. Also, u is normpreserving, in
346
Chapter 7 N ormed Linear Spaces
that if XEX is norm able, then u(x) is normable in Y and Ilu(x)11 = Ilxll Finally, 1 is a bound for u, and so U is uniformly continuous. For example, if (X, II II) is a nontrivial separable normed space, then the identity map is a linear isometry of (X, II II) onto (X, (II Ilu)uEs')' where S* is the unit sphere of X* and Ilxllu=lu(x)1
(UES*).
This follows from Corollary (4.5) and the definition of S*. For another example, consider a ufinite integral I on a set X, and positive numbers p, q with p> 1 and p 1 + ql = 1. By Theorem (3.25), the linear map gl+U g defined by u g (f)=I(fg)
(fELp, gELq)
is a linear isometry of Lq onto the normed linear space of normable elements of L~. In case p = 1, the normable elements of L~ are similarly related to the normable elements of a certain quasinormed linear space. In order to discuss this space, for the rest of this section we take I as a u1inite integral on X. (5.9) Definition. A realvalued function f defined on a full subset of X is essentially bounded relative to I if there exists a nonnegative number c, called a bound for f, such that If I ~ c on a full set. Two essentially bounded functions f and g are equal if f = g on a full set. Taken with this notion of equality, the collection of all essentially bounded, measurable functions relative to I is a set, which we denote by Loo. Let d denote the set of all integrable sets with positive measure, where two elements of d are equal if their characteristic functions are equal elements of L l . If AEd, fELoo, and c~O is a bound for f, then since If XA I ~ C XA on a full set, IlfII A =Jl.(A)ll(lfl XA) exists and is at most c. Thus (II IIA)A""" is a quasinorm on Loo. From now on we shall consider Loo as a quasinormed linear space relative to (II IIA)A""". Note that the equality induced on Loo by this quasinorm coincides with the notion of equality introduced in Definition (5.9). We shall write II f II 00 for the norm of a norm able element f of Loo. Thus Ilfll oo =sup{Jl.(A)l 1(lfl XA): XAEL1' Jl.(A) > O} if the right side exists. The next two lemmas will enable us to prove the completeness of Loo·
5 Quasinormed Linear Spaces, the Space Loo
347
(5.10) Lemma. If fELoo, and IlfllA~c for each A in .91, then Ifl~c on a full set.
Proof: By (8.20) of Chapter 6, for each n in 7l+ there exist cn III (c,c+n 1) and a measurable set An' such that Ifl0, then IlfIIA~,u(A)1 I(cnXA)=cn>c,
which is absurd. Hence ,u(KiAn)=O for each i; so that, by (8.10) of Chapter 6,  An is integrable, and ,u(  An) = O. Thus A~ is a full set for each n, and so F ==: A~ is a full set. Clearly, If I ~ C on F. D
n n
(5.11) Lemma. If (fn) is a Cauchy sequence in L oo ' then (fn) converges in measure to a measurable function.
Proof. Consider any integrable set A and any E > O. Either ,u(A) > 0 or ,u(A) < E. In the latter case, setting B ==: A  A, we have B integrable, BlcAl, ,u(AB)<E, and Ifmfnl<E on Bl for all m, n in 7l+. In case ,u(A) >0, write 0 and A in d. We construct 1 in LI such that 11/111~1 and Ilf II )1> IlgllAe Either IlgIIA<e, when we can take 1=.0; or, as we a~~lIJllc. IlgII A>eI2. By (820) of Chapter 6, there exist a measurable set B and a number b in (0,1'/2) such that Igl > b on Bl and Igl ~ bon BO. Thus
el2 < Ilgll A = tL(A) I (I (Igl XA B) + I (Igl XA AB)) ~ b tL(A)1 tL(A  B) + c tL(A)  I tL(A 1\ B)
< 1'/2 + c tL(A)1 tL(A 1\ B). Hence tL(A
1\
B) > O. Now define
1 on
the full set
F=.dmngndmnXAAB
by [(x) = tL(A
1\
B)I
if xEAlnBlndmng and g(x»b,
= tL(AI\B)l
if xEAlnBlndmng and g(x)< b,
=0
if xE(A I\B)Ondmng
Then
1 = tL(A 1\ B)l (gXB + XB)l
Igl XAA B
on F; whence 1 is measurable, by (7.7) and (7.10) of Chapter 6. Since also 1/1=tL(AI\B)I XAAB on F, we see that IELI and that 11/111=1. Also, Ig=tL(A 1\B) l lgIXAAB on F, and so lug(f)I=I(fg) =tL(A
1\
B)I I(lgIXAAB)
~ tL(A) I (I (Igl XA)  I (Igl XAB)) ~ Ilgii AtL(A)1 btL(A B)
> Ilgii Ae. To complete the proof that 2 is a linear isometry, we must show that for each I' > 0 and each 1 in LI with III III ~ 1, there exists A in d with (5.13.2)
Consider such e and f There exist a measurable set B and a number a such that lug(f)Ie 0, and choose in turn positive
00
L
integers N, v so that
k~N+
2~k<E/4
Ilu(x k)u n(x k)II<E/2
Then for
n~
and
1
(n~v,l~k~N).
v we have
Illuunlll ~
N
00
k~l
k~N+l
L 2~k Ilu(xk)un(xk)11 +2 L
2~k
N
oo, and so B is doublenorm complete.
0
6 Dual Spaces
353
We now focus our attention on the unit sphere S*={UEX*: lu(x)l~ Ilxll for all x in X}
of the dual X* of a separable normed space X. The following lemma extracts some of the sting from the proofs of the succeeding theorems. (6.6) Lemma. Let (xn) be a dense sequence in a separable normed space X, and IIIIII the corresponding double norm on X* Then for each e>O there exists a finitedimensional subspace X 0 of X such that Illu  vIII < e whenever u, v belong to the unit sphere of X* and lu(x)  v(x)1 < e/3 for all x in the unit sphere of X o. 00
Proof" Choose a positive integer N with
L
2 n < e/6. By (2.5), there
n=N+ 1
exists a finitedimensional subspace X 0 of X such that p(x n , X 0) < min {1, e/6}
For each n (1 ~n~N), choose Then
x~
(1 ~ n ~ N).
in Xo with
Ilxnx~11
<min{1,e/6}.
Let u, v be elements of the unit sphere of X* such that lu(x)v(x)l<e/3
(xEXo,llxll~l)
Then N
00
Illuvlll~ L2 n(1+llx nll)1Iu(x n)v(x n)I+2 11=
L
2 n
1
N
~
L 2n(l(u v)«1 + Ilxnll)l x~)1 H=
1
+(1 + Ilxnll)11(uv)(XnX~)I)+E/3 N
M(v.,K)n 1 <ex.<M(v.,K)
(7.4.2)
S(v.,exn,K) is compact
and such that for each n ~ 2, (7.4.3)
S(v., ex., K)c S(vn_ l' ~ex._1 +1 M(v n_ l' K), K)
(7.4.4)
the variation of u. on S(v n , ex n , K) is at most n 
(7.4.5)
Iv.(x)v n_ 1 (x)1 vn(xo)~(M(vn' K)lXn) ~~lXn +~M(vn' K)~(M(vn' K)lX n)
=~lXn +~M(vn' K).
Choose IX with ~lXn+~M(vn' K) v(x)~(M(vn' K)lX n) >~lXn +~M(vn' K) ~(M(vn' K)lXn)
=lXn· Hence S(v,IX,K)cS(vn' IXn' K). Since n is arbitrary, the desired conclusion follows from (7.4.7). 0
8. Hilbert Space and the Spectral Theorem
363
The next result is called the KreinMilman theorem. (7.5) Theorem. If x is a point in a compact convex subset K of a separable normed space X over JR, then there exist extreme points Xl' ... , Xn of K, and nonnegative numbers IX l' ... , IXn with sum 1, such that 1X1 Xl + ... +lXnX n is arbitrarily close to x.
Proof' Let e be any positive number. With Xl any extreme point of K, we define inductively a finite sequence (Xl'"'' Xn) of extreme points of K as follows. Once Xj has been constructed, let K j be the convex subset of K spanned by Xl'"'' X/
K j ={1X 1X1 + ... +lXjXj:
lXi~O (1 ~i~j),.t lXi= I}. ,~
1
Since K j is totally bounded, the distance d j from X to K j exists. Either dj<e or d j >e/2. In the first case, stop the construction and take n=j. In the second case, by the separation theorem there exists u in the unit sphere of X* such that u(x»sup{u(y): YEK) +e/2.
By (7.4), there exists an extreme point Xj + 1 of K with u(xj+ d>sup{u(y): YEK) +e/2.
Then p(X j + l' Kj)~ inf{lu(x j + 1) u(Y)I: YEK j } = u(Xj+ 1) sup {u(y): YEK j } > e/2.
It follows that II Xi  Xj I > e/2 for all i, j with i =t= j, as long as the construction proceeds. Since K is totally bounded, the construction must stop at some stage. Therefore dj<e for some j. Taking n equal to this value of j, we then have
Ilx(1X1x1 + ... +lXnxn)ll <e for some choice OfIX1, ... ,lXn~O with LlXi=l. i~
D
1
8. Hilbert Space and the Spectral Theorem One of the central problems of functional analysis is to analyze the structure of a bounded linear map of a Banach space into itself. (8.1) Definition. A bounded linear map u of a normed space X into itself is called an operator on X. If u and v are operators, then we'
364
Chapter 7 Normed Linear Spaces
write UV==UOV, and call uv the product ofu and v. We say that u and v commute, or that u commutes with v, if u v = v u. We also define powers of u by setting u 1 = U, Uz = uu, and so on. The operator X 1+ X is called the identity operator on X The structure of an operator is best understood when X is a Hilbert space, which is a special case of an inner product space. (8.2) Definition. An inner product on a linear space X is a mapping (x,y)+(x,y) of X x X into IF such that for all x, y, z in X and a, b in IF, (8.2.1)
(x,y)=O, this gives
y) + (y,x»)=ab(2ab  (x,y)  (y,x»).
(x,y)+(y,x)~2ab.
Letting e+O, we then have
(x,y)+(y,x)~21Ixllllyll·
In this inequality replace x by ax, where a is any element of IF with lal = 1; then a(x, y) + a*(x, y)* ~21Ixllllyll. Dividing by 2 and taking the supremum with respect to a gives (8.4.1). Using (8.4.1), we compute (8.4.5)
Ilx+ yl12 = IIxl12 + Ilyll2+ (x,y)+(y,x) ~(llxll + IlyW.
This is equivalent to (8.4.2). Equality holds in (8.4.2), or equivalently, in (8.4.5), if and only if Re (x,y) =t(x, y) + (y, x») = Ilxllllyll. In view of (8.4.1) this condition is equivalent to (x, y) = Ilxllllyll. Taking e = 0 in (8.4.4) gives Ilxllllyll (21Ixllllyll (x,y) (y,x»)~O,
366
Chapter 7 Normed Linear Spaces
with equality if and only if Ilyllxllxlly=O. Therefore (x,y)= Ilxllllyll. Conversely, if Ilyllx= Ilxlly, then
Ilyllx= Ilxlly if
Ilyll (x,y)= Ilxll (y,y)= IIxlillyl12 and so
Ilyll (1Ixllllyll (x, y»)=O. Therefore
1(llxllllyll (x,y)W ~(llxllllyll + l(x,y)1) 1(llxllllyll (x,y»)1 ~21Ixllllylll(llxllllyll (x,y»)1
and thus (x,y)= Ilxllllyli. The fact that I I is a norm follows from
(8.4.2).
=0
D
It follows from (8.4.1) that the inner product is a uniformly continuous map on each bounded subset of X x X. (8.5) Definition. A Hilbert space is a complete separable inner pro
duct space. Our next theorem, describing a fundamental geometric property of a Hilbert space, makes use of the identity
(8.6)
IIx+ y112+ Ilx y112=211x11 2+21IYI12,
whose proof is left to the reader. (8.7) Theorem. If M is a subspace (that is, a closed located linear subset) of a Hilbert space H, then to each x in H there corresponds a unique closest point P x in M. The vector P x can also be characterized as the unique vector y in M such that (x  y, z) = 0 for all z in M. The function P is an operator with range M, and satisfies p 2 = P. If M contains a nonzero vector, then P is normable, and IIPII = 1.
Proof Consider an arbitrary vector x in H. Let d=p(x,M), and let (y.) be a sequence of elements of M such that
(8.7.1)
IIYnxll+d
as n+oo.
Using (8.6) and the fact that IlzxI12~d2 for all z in M, for each m and n we compute
IIYm Y.112 = IIYm x (y._x)11 2 =211Ym _x11 2+21Iy.xI1 2411!(ym+ Yn)xI1 2 ~ 2(IIYm  xl12 _d 2 )+ 2(lly.xI1 2_d 2 ).
8. Hilbert Space and the Spectral Theorem
367
It follows that (y.) is a Cauchy sequence. Since H is complete and M is closed, (y.) converges to a limit y in M. By (8.7.1), Ilx yll =d. If y' is any element of M with Ilx y'11 =d, then by (8.6), Ily y'I12 ~21IyxI12 +211y' xI12_411t(y+ y')_xI12 = 4(d 2 IIt(y+ y')_xI1 2) ~O. Hence y' = y. Thus P x == y is the unique closest point to x in M. For all z in M and a in IF, we have
(xPx+az, xPx+az)?,d 2 =(xPx, xPx) and therefore lal 2 11z112 +2 Re (a* (x  P x, z»)?, O. Were Re (x  P x, z) =1= 0 we could contradict the last inequality by taking a sufficiently small real number a with aRe(xPx,z)0; whence P is normable, and IIPII = 1. 0 The operator P: H + M in Theorem (8.7) is called the projection of H on the subspace M. For each x, Px is called the projection of the vector x on M. Vectors x and y in an inner product space X are orthogonal if (x, y) = O. We then write x..ly. For orthogonal vectors x and y we
368
Chapter 7 Nonned Linear Spaces
have a generalization of Pythagoras's theorem:
Ilx+ yl12 = IIxl12 + Ily112. The orthogonal complement KJ. of a subset K of X consists of all x in X such that x.1.y for all y in K. Clearly, KJ. is a closed linear subset of X Consider a subspace M of a Hilbert space H. By Theorem (8.7), every vector x in H has a unique representation X=X l +x 2 as the sum of an element Xl of M and an element X 2 of MJ.; in fact, Xl =Px and x 2 =xPx, where P is the projection of H on M. For each vector y in MJ. the vector z=(xPx)y belongs to MJ., and so z.1.Px. Therefore
Ilx yl12 = IIPx+zI1 2= IIPxl1 2+ IIzl12 ~ IIPxl1 2= IIx(xPx)11 2. Thus X  P X is a closest point to x in M J.. Since x is arbitrary, it follows that MJ. is located. Thus MJ. is a subspace of H, and the projection of H on M J. is 1 P, where I is the identity operator on H. (8.8) Definition. A sequence (en) of vectors in a Hilbert space H is
orthonormal if e m .1.en whenever m4=n, and if for each neither Ilenll = 1 or Ilenll =0. Such a sequence is called an orthonormal basis of H if each vector x can be written uniquely in the form
where (an) is a sequence of scalars such that an = 0 whenever en = O. The scalars an are then called the coordinates of x with respect to the orthonormal basis (en). An orthonormal basis can always be obtained, as in the proof of the next theorem, by the GramSchmidt orthogonalization process (8.9) Theorem. Every Hilbert space H has an orthonormal basis. If (en)
is such a basis, if a l ,a 2 , ••• are the coordinates of x, and are the coordinates of y, then (i)
if
b l , b2 ,
••.
an = (x, en> and b n= (y, en> for each n 00
(ii) (x, y> =
L anb: 00
(iii)
Ilx112=
L lanl
2•
n= 1
Proof Let (Yn) be a dense sequence in H. With eo=O, we define inductively an orthonormal sequence (e n ):'= 0 such that for each n ~ 1
8. Hilbert Space and the Spectral Theorem
there exists a linear combination
y~
of e 1 ,
369
en with
••• ,
(8.9.1) Assume that eo, .. ,en have been defined. Let M n be the subset of H consisting of all linear combinations of eo, ... , en. Then M n is a finitedimensional normed space, and so is a subspace of H. Let P" be the projection of H on Mn. Either IIYn+1P"Yn+111«n+1)1 or IIYn+1P" Yn+ 1II > O. In the first case, set en+ 1== 0 and y~ + 1== P" Yn+ 1· In the second, set
Then Ilen+ 111 = 1, and the required inequality is satisfied with y~+ 1== P"Yn+ 1+ IIYn+ 1 P"Yn+ 111 en+1= Yn+ 1·
The vector en + 1 is orthogonal to eo,···, en because Yn+ 1 P"Yn+ 1 belongs to by (8.7). This completes the inductive construction of the orthonormal sequence (e O ,e 1 , ••• ). To finish the proof, consider arbitrary vectors x and Y in H For each n ~ 1 we have unique representations
M;;,
n
p"x =
L aie i
n
L biei
P"y =
and
i= 1
i= 1
with ai=bi=O whenever ei=O. If 1 ~k~n, then taking inner products with e k , we obtain n
ak = Lai(ei,e k ) = (p"x, ek ) = (x, ek )

(x  p"x, ek ) = (x, ek ).
Similarly, b k = (y, e k ). Thus ak and b k are independent of n. We have n
(p"x, P"y) =
L aibr i= 1
Since (Yn) is dense in H, it follows from (8.9.1) that p(x,Mn)+O and p(y,Mn)+O as n+oo; that is, p"x+x and P"Y+ Y as n+oo. Therefore ()()
(x, y) = lim (p"x, P"y) =
L aibr
Setting Y = x gives ()()
IlxI1 2 =(x,x)=
L la;12. i= 1
0
370
Chapter 7 Normed Linear Spaces
The following lemma enables us to characterize the dimensionality of a Hilbert space in terms of an orthonormal basis. (8.10) Lemma. Let V be a finitedimensional subspace of a Hilbert space H, and (en) an orthonormal basis of H. Then there exists N in Z+ such that p(e., V»O whenever n~N and Ilenll =1.
Proof: Using (2.3), construct a t approximation {XI' ... ,xm } to the unit sphere of V. Choose N in Z+ so that I(x k , en>1 < 3/8 whenever n~ N and 1 ~ k ~ m. Consider n ~ N with I en II = 1. Since V is a subspace of H, the projection P of H on V is defined, and IIPenl1 ~ 1. Choose k so that IIPenxkll Then
+ IIxkl12 ~121(xk,en>I>·L
so that peen, V)= IlenPenil ~
IlenXkllIIPenXkll
>tt=O. D (8.11) Proposition. Let (en) be an orthonormal basis of a Hilbert space H. Then H is finitedimensional if and only if en = 0 for all sufficiently large n. Proof: The necessity of the stated condition follows by taking V = H in (8.10). Conversely, if en=O for all n>N, where NEZ+, then H is the finitedimensional Hilbert space consisting of all linear combinations ofel, .. ·,e N · D
(8.12) Definition. A normed space X is infinitedimensional if for each finitedimensional subspace V of X there exists a vector x with p(x, V»O.
(8.13) Proposition. Let (en) be an orthonormal basis of a Hilbert space H. Then H is infinitedimensional if and only if Ilenll = 1 for infinitely many values of n. Proof" Suppose that H is infinitedimensional. Consider an arbitrary positive integer N. Let V be the finitedimensional subspace of H consisting of all linear combinations of e 1 , ... , eN' and let P be the projection of H on V. Choose X in H with p(x, V»O. Then 00
L i~N+
l(x,e)1 2 = IIxPxI1 2 =p(x, V)2>O, I
8 Hilbert Space and the Spectral Theorem
371
so that l(x,ek)I>O for some k>N. It follows from (8.4.1) that Ilekll >0, whence Ilekll = 1. Since N is arbitrary, it follows that Ilenll = 1 for infinitely many n. If, conversely, this last condition holds, then the infinitedimensionality of H follows from (8.10) 0 Two operators A and B on a Hilbert space H are called adjoint if (Ax,y)=(x,By)
(x,YEH).
In that case, the operator B is uniquely determined in terms of A by the above property, and is written A *; B is called the adjoint of A; and we have (A*)* =B* =A. If A and A are adjoint  that is, if (Ax,y) = (x, Ay) for all vectors x and y  then A is said to be hermitian or selfadjoint. If A is hermitian, then for each x in H, (Ax,x)
= (x, Ax) = (Ax,x)*,
so that (Ax,x) is a real number. It is clear that a sum of hermitian operators is hermitian, a real multiple of a hermitian operator is hermitian, and a product of commuting hermitian operators is hermitian. The identity operator I is hermitian. More generally, let M be a subspace of H, with projection PM. For all vectors x and y we have (PMx, y) = (PMx, PM y) + (PMx, y  PM y) = (PMx, PMy) = (PMx, PMy) + (xPMx, PMy) =(x,PMy)·
Thus PM is hermitian. Conversely, consider any hermitian operator P on H with p 2 = P. Write M= {xEH: Px=x}. Clearly, M is a closed linear subset of H. For each y in Hand z in M we have (yPy, z) = (y, z)  (Py, z) = (y,z)  (y,Pz) =0,
since Pz=z. Hence yPyl.M. Since P(Py)=Py, we also have PYEM. Thus for each x in M,
IlyxI1 2= IlyPy+PyxI1 2 = lIyPyIl2+ IIPy_xIl2~
IlyPyI12,
since PyxEM and yPyl.M. Hence Py is a closest point to y in M. It follows that M is located, and that P is the projection of H on M. An important role in the classical analysis of the structure of a hermitian operator is played by its eigenvalues and eigenvectors. In a
372
Chapter 7
Normed Linear Spaces
constructive setting, where we may be unable to lay hands on exact eigenvectors, it is natural to look for a good approximation instead. (8.14) Definition. Let A be an operator on a Hilbert space H, and e a positive number. A vector x in H is called an e eigenvector of A if Ilxll =1 and IIAx(Ax,x)xll<e. The construction of the functional calculus given later will depend on the existence of e eigenvectors common to finitely many commuting hermitian operators on H. (8.15) Lemma. Let A be a hermitian operator with bound 1 on a Hilbert space H. Let x be any unit vector  that is, a vector of norm 1  in H, and write y == Ax  (Ax, x)x. Then y.lx, and y is the projection of the vector Ax on the orthogonal complement of {x} lltite also XA
== Ilx+hll1(x+h)·
Then
(AXA' x A)  (Ax,x);?;i IIAx  (Ax,x)xI12;
if B is any hermitian operator which commutes with A, then IIBxAl1 ~21IBxll· Proof: It is trivial to verify that y.lx and that Ax  y is orthogonal to
and
each vector in {x}l.. Thus, by (8.7), y is the projection of the vector Ax on {x}l. As (Ax,y) = (Ax(Ax,x)x,y)= IIyl12 and IIAyl1
Ilyll, we have (A(x+h), x+h)= (Ax,x) +~(Ax,y) +~(Ay, x) +i(Ay,y) ;?; (AI x, x) e.
Proof There is no loss of generality in assuming that 1 is a bound for each of the operators AI' ... ' An. Consider an arbitrary e > o. If n = 1, then by (8.16), there exists y in H with IIYII=l, IIA 1 y(A l y,y)yll<e, and (Aly,y)~(AIX,X). Assume therefore that n>l, and that the lemma is true for all smaller values of n. Choose a positive integer N> 32e 2. By the induction hypothesis, there exists a vector u of norm 1 such that (Al U, u) > (Alx,x) e12 and
374
Chapter 7 Normed Linear Spaces
Define sequences (Uk)' (v k) as in Lemma (8.16), with A==An and UI==U Choose k with 1 ~ k ~ Nand (8.17.1)
and set y == Uk. We may assume that k> 1, since if k = 1 there is nothing to prove. If 1 ~ i ~ n 1, then the hermitian operator Ai O. Since f is continuous, there
If(x)f(X l ,X 2 , ""XN'O,O, .. ·))I<E for all x == (x n):'= 1 in !!l'. By the Weierstrass approximation theorem «5.17) of Chapter 4), there exists a polynomial function p: [1, l]N + lR such that
If«x l , .. ·, x N, 0, 0, ... )) p(x l , ... , xN)1 <E (Xl' ""XNE[ 1,1]).
Hence Ilf  qll
~ 2E,
where q is the element of f!J' defined by
q(x)==p(xl, .. ·,X N)
(x ==(Xn)E!!l').
0
If A == (An):'= 1 is a sequence of commuting hermitian operators on a Hilbert space H, let f!J'(A) be the real subalgebra of Hom (H, H) generated by the operators An and the identity operator I. Denote by fHf(A) the unique algebra homomorphism ¢: f!J'+f!J'(A) with ¢(1) =1 and ¢(7l: n)=An for each n in Z+. The mapfHf(A) is called the canonical homomorphism of f!J' into f!J'(A). (8.20) Lemma. Let A == (An) be a sequence of commuting hermitian operators on a Hilbert space H, with common bound 1. Then for each j
in f!J', f(A) is a hermitian operator with bound II fII.
376
Chapter 7 Normed Linear Spaces
Proof. Given f in &, write N
L
f= ii,
a(i 1, ·.. ,U1Oi'
.. · 10;
,im=O
where N ~ 1. Since each coefficient a(i 1 , ... , im) is a real number, and since sums and products of commuting hermitian operators are hermitian, N
L
f(A) = il.
a(i 1 , ... , im)A\'.
A;
,im=O
is hermitian Clearly, f(A) commutes with each operator An Assume to begin with that H is nontrivial. Consider any unit vector x and any e > O. By (8.17), there exists a unit vector U such that 0 and IIfgll' ~ IIf11' IIgll' for all 1 and g in C(X,IF). (8.24) Proposition. Let X be a compact space, and I II' an algebra seminorm on C(X,IF) such that IIf11'~ 1If11 for all f in C(X,IF). Then there exists a compact set K c X such that I f II' = I f I K for all f in C(X,IF)
Proof' Let K consist of all x in X such that If(x)1 ~ IIf11' for all f in C(X, IF). We first show that if fo is any element of C(X, IF) with 111011'>0, and Ko is any compact support for fo, then KonK is nonvoid. To this end, we define inductively elements fo, fl' f2"" of C(X,IF) and compact sets Ko~Kl~'" so that, for all n~l, IIfnll'>O, Kn supports fn' and diamKn~nl. The function fo and the set Ko have already been defined Assume therefore that n~ 1, and that fO""'/nl' K o ,···,Kn_ 1 have already been defined. Using (6.15) of Chapter 4, choose elements g I' ... , gN of C(X, IF), each of which is supported by some compact subset of K n _ 1 of diameter at most n I , such that gl + ... + gN= fnl' Since
IIg111'+ +lIgNII'~lIfnIII'>O, Set I n =gi' and take Kn to be a compact support
IIgill'>O for some i for gi with KncK n_ 1 and diamKn~n1 This completes the induction. The sets Kn have a unique point Xo in common Trivially, XoEKo. To show that xoEK, consider an arbitrary f in C(X,IF). Given 8>0, construct g in C(X,IF) so that IIfgll0, K is nonvoid. To see that K is totally bounded, fix 8>0, and choose functions fl""'/n in C(X,IF) with 11+ ... +fn=l, IIfll'~
"182
Chapter 7
Normed Linear Spaces
such that /; is supported by some compact set Ki of diameter less than E. The set of positive integers i with 1 ~ i ~ n is the union of finite subsets Sand T such that 11/;II'«2n)1 for all i in S, and 11/;11'>0 for all i in T By the above, for each i in T there exists a point Xi in KJl K. We shall show that these points Xi form a subfinite E approximation to K To this end, consider any X in K. Since fl (x) + ... + f.(x) = 1, we have /;(x) > (2n) 1 for some i. Therefore lin' > (2n)I, by the definition of K, and so iET. Also, xEK i . Since XiEKi and diamKi<E, we have p(X,X;l<E. Thus {Xi: iET} is a subfinite E approximation to K. Hence K, which is clearly complete, is compact. Now consider any f in C(X,IF). For each E>O either Ilfll' < IlfilK +E or Ilfll'> IlfiIK. In the latter case, Ilfll> IlfilK and so K is nonvoid. Construct g in C(X,IF) such that IIgll ~ IlfIIK+E and f  g is supported by a compact set Lc  K. If Ilf  gil' >0, then by the above, L n K is non void ; this contradiction shows that II f  gil' = O. Hence Ilfll'~ Ilgll' + Ilf gll'= Ilgll'~ Ilgll ~ IlfIIK+ E
in this case also. Since E is arbitrary, Ilfll'~ IlfilK As it follows from the definition of K that IlfIIK~ IIfll', we see that Ilfll'= IlflIK· 0 Let K be compact space. A linear functional u on C(K, IF) is said to be multiplicative if u(x y) = u(x) u(y) for all X and y in K A nonzero bounded multiplicative linear functional u on C(K,IF) is normable, with Ilull = 1. For if fE C(K,IF), Ilfll < 1, and lu(f)I> 1, then
lu(f")1 = lu(j)I' + 00
as n + 00
but II f" II + 0 as n + 00, which contradicts the boundedness of u. Thus lu(f)1 ~ Ilf II for all f in C(K,IF). On the other hand, since u(f) =u(1)u(f) for all f, and u is nonzero, we have u(1)=u(1)2=F0, and thus u(1)= 1. Hence lIull exists and equals 1. Our first application of Proposition (8.24) describes the form of a nonzero bounded multiplicative linear functional on C(K, IF). (8.25) Proposition. Let K be a compact space, and r the set of all nonzero bounded multiplicative linear functionals on C(K,IF). For each x in K define the element U x of r by
ux(f)=f(x)
(fEC(K,IF)).
Then r is compact in the metric induced by the double norm on C(K,IF), and the map xr+u x is a metric equivalence of K with r. Proof· Consider an arbitrary element u of r. Applying (8.24) to the algebra seminorm fr+ lu(f)I, construct a compact set L c K such that
8 Hilbert Space and the Spectral Theorem
383
lu(f)I= IlfilL for each f in C(K,IF). Suppose that diamL>O. Then there exist g, h in C(K, IF), each with support contained in L, such that Ilgll = 1, Ilhll = 1, and gh=O; so that lu(gh)1 =0< IlgllL IlhilL = lu(g)llu(h)l.
This contradiction shows that diam L = 0, so that L consists of a single point x. Thus lu(f)1 = If(x)l, and u(f) = 0 whenever f(x) = O. Hence u(f)= u(f  f(x))+ f(x) u(1) = f(x) = ux(f)
(fE C(K,IF)
and so u = U x Thus every element of r is of the form U x for some x in K. Let ). denote the mapping x H U x on K. Let (fn) be a dense sequence in C(K,IF), and 111111 the corresponding double norm on C(K,IF)* Then 00
IlIuxlll =
L 2 (1 + Ilfnll)llfn(x)1 n
(xEK).
n=l
Since each fn is uniformly continuous, it follows that ). is uniformly continuous. Clearly, ). is injective and thus a bijection of K onto r To show that ).  1 is uniformly continuous, consider E> 0, and let {x 1 , ••• ,X N } be an E approximation to K. By (6.3), there exists c5>0 such that Ip(x,x,)p(y,x')I<E (1~i~N) whenever x,y are points of K with 111).(x)  ).(y)11I ~ c5. For such x and y, if we choose i with p(x, x,) < E, then p(y, xJ < 2E and so p(x, y) < 3E Thus). 1 is uniformly continuous on r. Hence ). is a metric equivalence, and therefore r is compact.
D Another application of Proposition (8.24) leads to the following supplement to the spectral theorem. (8.26) Proposition. Let H be a nontrivial Hilbert space; let A == (An) be a sequence of commuting hermitian operators with bound 1, such that each operator in [ljJ(A) is normable; and let (/1, cpHcp(A)) be a functional calculus for A. Then f(A) is normable for each f in C(8l"), and there exists a compact full set K c 8l" such that (8.26.1 )
IIf(A)11 = IlfilK
(fEC(8l")).
Proof: If (fn) is a sequence of elements of C(!f"), then since (by (8.20))
[ljJ
converging to a limit f in
Illfm(A)IIllfn(A)111 ~ Ilfm(A) fn(A) II ~ Ilfm  f.11 (1If.(A)II)~
(m,n~ 1),
1 is a Cauchy sequence in 1R Also, (I.) is a bounded sequence converging to I in measure relative to /1, so that (I.(A)) converges strongly to I(A), by (8.22). It follows that f(A) is normable,
lb4
Chapter 7
Normed Linear Spaces
with
Ilf(A)11 = lim II jn(A) II ~ lim IIf"11 = IIfll. noc
nQ(.
Applying (8.24) to the algebra seminorm j 1+ II f (A) lion C(f!l'), construct a compact set K c f!l' satisfying (8.26.1). The continuous function f defined by j(x)==p(x,K)
(XEq[)
vanishes on K, so that f(A)=O, by (8.26.1). Hence J1.(f)=0, by (8.22.1). Since f~O, we have p(x, K)=O on a full set. Thus K contains a full set, and is therefore full. 0 We now show that certain algebras of operators on a Hilbert space H can be represented as spaces of continuous functions. (This is essentially a reformulation of Proposition (8.26)) To do this, we first recall from Section 5 that the neighborhood structure of Hom (H, H) relative to the standard quasinorm is called the uniform neighborhood structure. For each A in Hom (H, H) and each e > 0, there is a uniform neighborhood of A consisting of all B in Hom (H, H) such that A  B has a bound less than e. Such concepts as 'uniformly closed' and 'uniformly separable' pertain to this neighborhood structure. The set of normable operators is uniformly closed, and on any linear subset of the set of normable operators the uniform neighborhood structure is induced by the metric (A, B) 1+ II A  B II. The spectrum of an algebra 9f of operators on H is the set of all nonzero bounded multiplicative linear functionals on 9f. Recall that the identity operator on H is denoted by I. (8.27) Theorem. Let 9f be a commutative algebra oj normable hermitian operators on a nontrivial Hilbert space H, with I E9f, which is uniformly separable and uniformly closed. Then each u in the spectrum E of 9f i~ normable, with Ilull = 1, and E is compact in the metric induced by the double norm. Moreover, the map y defined by y(U)(u) == u(U)
(uEE, U E9f)
is a normpreserving isomorphism of the algehra f!Il onto the algebra C(E).
Proof Note that 9f is only a real subalgebra of Hom (H, H), since iA is not hermitian for any nonzero A in 9f. Let A == (An);,"'~ 1 be a sequence of elements of 9f, with common bound 1, whose linear combinations are uniformly dense in 9f. Let (/l,qJl+qJ(A)) be a functional calculus for A. Since the map qJl+qJ(A) is a boundpreserving
8 Hilbert Space and the Spectral Theorem
385
homomorphism and ?fJ is dense in C(,q[), for each f in C(8l') the operator f(A) is a uniform limit of polynomials in the An' and so f(A)ErJt. Let K be a full compact subset of El' such that (8.26.1) holds. By Theorem (6.16) of Chapter 4, each element of C(K) extends to an element of C(,q[). If f E C(K) and g is any such extension of f, then since K is full, f(A) == g(A) defines an element f(A) of rJt which does not depend on g. From (826.1) and our choice of A, we see that the map ft+ f(A) from C(K) to Hom (H, H) is a normpreserving isomorphism F of C(K) onto a dense subset of PIt, since C(K) is complete, it follows that F maps C(K) onto PIt. Let (fn) be a dense sequence in C(K); let C(K)* be given the corresponding double norm; and let rJt* be given the double norm defined by the dense sequence (fn(A» in rJt. Let r c C(K)* be as in Proposition (8.25). Then the map which carries an element u of r to the mapping f(A)t+u(f) is a bijection of r onto L which preserves the double norm. Hence L is compact in the double norm, by (8.25), and each u in L is normable, with Ilull = 1. Now let A be the metric equivalence xt+u x of K with r. Then the map ft+ f 0 AI is a normpreserving isomorphism Q of C(K) onto C(T). On the other hand,
cP(g)(u)==g(uoF)
(gEC(l), UEL)
defines a normpreserving isomorphism cP of C(T) onto C(L) Thus it is enough to prove that y = cP 0 Q 0 F I. For all f in C(K) and u in L we have
(cP Q F 1 )(f(A»(u) = (cP Q) (f)(u) = cP(f AI )(u) =(fo AI)(u o F) 0
0
0
0
= (u a F)(!) = u(f(A» = y(f(A»(u) Since every element of PIt is of the form f(A) for some f in C(K), the result follows. D An algebra rJt of operators on H is selfadjoint if each element of rJt has an adjoint belonging to rJt The first corollary to Theorem (8.27) is the Gelfand representation theorem. (8.28) Corollary. Let rJt be a complex commutative algebra of normable operators on a nontrivial complex Hilbert space H, with I ErJt, which is
uniformly separable, uniformly closed, and selfadjoint. Then each u in the spectrum L of rJt is normable, with Ilull = 1; L is compact in the metric induced by the double norm. and the map y defined by y(U)(u)==u(U)
(UEL, UErJt)
386
Chapter 7
N ormed Linear Spaces
is a normpreserving algebra isomorphism of
~
onto C(L, u is a metric equivalence of 17' with 17 0 , Therefore 170 is compact. By (69) of Chapter 4, 17 is locally compact under the metric (830.1), and 170 is a onepoint compactification of 17 with point at infinity 0 and the natural inclusion map. The restriction y~ of y' to q{ is a normpreserving isomorphism of q{ into the set G of elements f in CC17', CC) with f(w) = 0, where w is that element of 17' which vanishes on q[. In fact, y~ is an isomorphism of i!lt onto G' for if f EG, then f = y'(V + 21) for some V in q{ and), in CC; so that 0= f(w)=w(V +),1)=w(V)+),=)., and therefore f = y'( V)E')"(q{). Now, by (6.9) of Chapter 4, 17'  {w} is locally compact under the metric p', where p'(u', v'):= Illu' v'lll + 1(lIlu' wlll1lllv' _wlll1)1
(u', v'E17'),
and 17' is a onepoint compactification of 17'  {w} with point at infinity wand the natural inclusion map. Identifying G with C",,(17'  {w}, CC) as on page 118, we see that y~ is a normpreserving isomorphism of q{ onto Coo(17'  {w}, CC) Since u'r>u is a metric equivalence of 17' with 170 which maps w to 0, its restriction is a homeomorphism of 17'  {w} onto 17, and induces a normpreserving isomorphism C/J of C oo (17'{w},CC) onto C oo (17,CC). Since y is the composition of y~ with C/J, it follows that y is a normpreserving isomorphism of !!It onto C",,(17, CC). 0
390
Chapter 7 Nonned Linear Spaces
Problems 1. Let X be a linear space over IF. A subset S of X is circular if for each x in X the set Sx =:0 {aEIF: aXES} is locally compact, contains a nonzero scalar, and contains all scalars a with lal ~ Ibl for some b in Sx' Show that if S is convex and circular, then
Ilxll'=:oinf{lal l : aES x, a=t=O}
(XEX)
defines the unique seminorm on X relative to which S is the closed unit sphere. Under what conditions is II II' a norm on X? 2. Two norms II II and II II' on a linear space X are equivalent if the identity map from (X, II II) onto (X, II II') is a metric equivalence. Construct a nontrivial norm on IR 2 which is not equivalent to the euclidean norm 3. Give an example of a closed linear subset V of a finitedimensional Banach space X such that V is not finitedimensional. Construct a bounded linear functional on V that is not normable 4. Let X be a Banach space. Discuss the relation between the following statements. (i) X is finitedimensional. (ii) All elements of X* are normable. 5. Vectors Xl' ... , xn in a normed linear space X are said to be metrically independent if for each e>O there exists 15>0 such that IAll+ ... +IAnl~e whenever Al, ... ,A n are scalars with IIAlx l + .+Anxnll~!5 Prove that the following conditions are equivalent. (i) Xl' ... , xn are metrically independent. (ii) The linear combinations of Xl"'" xn form a finitedimensional subspace of X. (iii) There exists c>O such that IIAlx l + ... +Anxnll;:;;c whenever AI'"'' An are scalars such that IAll + . + IAnl = 1.
6. Let XI" "xn+1 be vectors in a normed linear space X such that XI'''''Xn are metrically independent. Prove that xI,,,,,xn+1 are metrically independent if and only if p(xn+ I' V) > 0, where V is the finitedimensional subspace of X generated by Xl'" , X n · 7. Prove that vectors XI'" , Xn in a normed linear space X are metrically independent if and only if they are linearly independent, in the
Problems
391
sense that )"IXI + ... +Anx.=l=O whenever AI, ... ,An are scalars such that 1,111 + ... + IAnl >0. 8. Construct vectors XI' ••. , Xn in a normed linear space such that (i) if are scalars with A1 X 1 + ... +Anxn=O, then Al = =An=O, and (ii) XI' ••• , xn are not linearly independent.
AI' ... ,An
9. Construct a onedimensional subspace V of a normed linear space X and a point x of X such that x does not have a closest point in V. 10. Let V be a nonvoid linear subset of a normed linear space X such that each x in X has at most one closest point in V. Need V be located? 11. Let K be a closed located convex subset of a uniformly convex Banach space X Show that each x in X has a unique closest point in K. 12. An integration space is said to be separable if there exists a sequence (An) of integrable sets such that for each integrable set A and each e > 0, there exists n in lL + with .u(A  An) + .u(A n A) < e.
Prove that if .u is a positive measure on a locally compact space X, then the integration space (X, L 1 ,.u) is separable. 13. Prove that if (X, L I , 1) is a separable integration space, then Lp is a separable normed space for each p ~ 1 14. The sequential Lp space Lp(lL) consists of all sequences (an) of elements of IF such that II(an)11 =(L: lanlP)llp exists. Show that for p= 1 n
the dual of Lp(lL) can be identified with the set of all bounded sequences of elements of IF. Show that for p> 1 the dual of Lp(lL) can be identified with the set of all sequences (b n ) of elements of IF such n
that
L: lakl q is k~
a bounded function of n, where pl + ql = 1. What
I
about the general case p ~ 1? 15. Show that the complex Lp spaces are uniformly convex for p> 1, and exhibit the form of the normable linear functionals. 16. Let I and J be completely extended integrals with a common initial set L, such that J is ofinite, and absolutely continuous relative
192
Chapter 7
Normed Linear Spaces
to I Let f be a nonnegative function in L] (I) n L] (J) with an 1representation (fn) by nonnegative elements of L Show that (1n) is a Jrepresentation of f 17. Let I and J be completely extended, finite integrals with a common initial set. Suppose that there exists a nonnegative Imeasurable function fo such that for each f in L 1(I)nL 1(J), ffoEL1(1) and I(ffo) = J(f). Show that J is absolutely continuous and normable relative to I.
18 Construct finite integrals I and J such that J is absolutely continuous, but not norm able, relative to I 19 In the notation of Theorem (3.34), show that foEL1(1) if and only if the sequence (XsJ:,"'~ 1 converges to 1 in measure relative to J. Use this to give an alternative proof of Corollary (3.36)
20. Prove the uniform boundedness theorem If (An) is a sequence of bounded linear mappings from a Banach space X to a normed linear space Y, and if (x.) is a sequence of unit vectors in X such that I A.xnll + 00 as n + 00, then there exists a vector x in X such that the sequence (1IAnxll) is unbounded. (Hint· Write Uk=={XEX:
IIAnxl1 >k for some
n}.
Show that the sets Uk are open and dense in X, and apply the Baire category theorem.) 21. Prove that there exists a continuous function Fourier series tx.
f
on [n,n] whose
1t
L S f(x)e
ikx
dx
k=OO7[
at the point 0 diverges. (Hint. Define the linear functional u. on C([ n, n]) by 1t
u.(f) ==
L S f(x)e
ikx
dx
k=n7[
Show that there exists a sequence (fn) of unit vectors m C([  n, n]) such that lun(fn)I+ 00 as n + 00.) 22 Let X and Y be normed linear spaces, and u a map of X onto Y We say that u is open if u(S(O, 1)) contains an open sphere S(O, r) in Y. Prove that if Y is finitedimensional, then every bounded linear map from X onto Y is open.
Problems
393
23 A map u: X + Y between normed linear spaces X and Y is unopen if there exists a sequence (Yn) of unit vectors in the range of u such that Yn=l=u(x) whenever nE7L+, XEX, and Ilxll O such that if x is a unit vector with Ilx  yll < r for some Y in ker u.
Ilu(x)11 <s, then
Show also that if (*) holds for some r in (0, 1), then u is open If (*) holds with r= 1, is u open? 26. With V and X as in Problem 3, construct a bounded linear functional on V that can not be extended to X 27. A linear map u of a quasinormed space (X, (II I i)iEl) into a quasinormed space (Y,(II II)jEJ) is weakly bounded if there exists c>O such that Ilu(x)t;;=;c whenever jEJ, XEX, and Ilxll i ;;=; 1 for all i in I. Construct a weakly bounded linear map between quasinormed spaces that is not bounded 28. Let Un) be a bounded sequence in Loo. Prove that Un) converges in measure to a measurable function I if and only if 11/ IniiA +0 as n + 00 for each A in d. 29. Construct an integral I and a norm able linear functional u on L1 (I) such that u is not of the form ug for some g in Loo. 30. Show that the linear subset of Lex generated by the normable elements is dense in Loo' 31. Call a Banach space B reflexive if (i) the set V c B* of normable linear functionals on B is closed with respect to addition, and (ii) every normable linear functional on V (relative to the norm I vii == sup {lv(x)l: x E B, Ilxll;;=; 1}) is of the form vl>v(x) for some x in B. Is every uniformly convex Banach space reflexive? (This is an unsolved problem) 32. Let K be a closed convex subset of a Banach space B such that aXEK whenever aEIF, lal;;=; 1, and XEK. Show that K is located if and
394
Chapter 7
Nonned Linear Spaces
only if K' == {uEB*: lu(x)1 ~ 1 for all x in K} is a located subset of B* relative to the double norm. 33. A Banach dual is a pair consisting of a normed linear space V and a compact convex subset K of V such that (i) aUEK for all a in IF with lal ~ 1 and all U in K, and (ii) for each nonzero U in V there exist positive constants C1 and C2 with C1UEK and p(c 2u,K»0. Let B be the set of all linear maps from V to IF that are continuous on K, and for each f in B write Ilfll==sup{lf(u)l: uEK}. Show that B is a Banach space Identify V with B*, and K with S*. 34. Give an example of a compact convex subset K of IR. 2 with at most three extreme points, such that (i) the set of extreme points of K is not located, and (ii) not every point of K is a linear combination of extreme points. 35. Find the extreme points of the unit sphere of C(X)* for a compact space X. 36. Show that a bounded linear functional U on a Hilbert space H is normable if and only if there exists a vector a in H such that u(x) = (x, a) for all x. 37. Let M and N be orthogonal linear subsets of a Hilbert space H (so that Me Ni), with M+N=={x+y: xEM, YEN} dense in H. Prove that M and N are both located. 38. Construct a hermitian operator A on a Hilbert space H such that ker A is located but the range of A is not located. 39. Construct a bounded linear operator A on a Hilbert space such that the adjoint of A does not exist. 40. Let A be a hermitian operator on a Hilbert space H, and let C> 0 be a bound for A. Show that there exist a dense set S e [  C, c] and a map A.I+p;' from S to the set of projection operators, such that
(i) (ii)
p;. ~ = 1';. (A., Il E S,
A. ~ Il) CES, CES, P_c=O, and p"=I c
(iii) A =
S A. dp;., in the sense that for all x and
y in H the function A.
I+(iicx,y) is of bounded variation, and (Ax,y)=
JA. d(1';.x, y) c
(where this integral is defined as in Problem 12 of Chapter 6). (Hint. Use the spectral theorem.)
Problems
395
41. A linear map u between normed spaces X and Y is compact if it maps the unit sphere of X onto a totally bounded subset of Y. Give an example of a normable linear map u on a Hilbert space such that the range of u is finitedimensional, but u is not compact. 42. Construct a compact linear map u of a Hilbert space into a finite dimensional linear space such that ker u = {O} but the range of u is not finitedimensional. 43. Show that if U is a compact operator on a Hilbert space H, with located range, then U has an adjoint and U* is also compact. (Hint. Use Problem 36 and Ascoli's theorem.) 44. A bounded linear operator U on a Hilbert space H is unitary if U* exists and U U* = U* U = I Show that a unitary operator U preserves norms. 45. Let U be a unitary operator on a Hilbert space H, and let x be any vector in H. For each positive integer n write xn=nl(x+ Ux+ ... + unIx).
Show that the sequence that for each n and each for all m~N.
(1Ixnll) is essentially decreasing, in the sense > 0, there exists N such that I xm I ~ I Xn I + I>
I>
46. Continuing Problem 45, prove the mean ergodic theorem of von Neumann: (x n ) converges if and only if (1Ixnll) converges. 47. A locally convex space is a pair consisting of a linear space X over IF and a set N of seminorms, called admissible seminorms, on X such that if I 111 and I 112 are admissible and if c> 0, then every seminorm such that Ilxll ~c(llxlll + Ilx11 2) (XEX) is admissible. A defining set of admissible seminorms for X is a set S eN such that for each I I in N there exist I 111' .. ·' I I n in Sand c> such that Ilxll ~ c(llxlll + ... + Ilxll n} (XEX).
°
Extend the notions of bounded linear functional, dual space, and bounded subset of the dual space to the context of a locally convex space X. (The dual space X* should be defined so that it is also a locally convex space.) Extend Theorems (6.7) and (6.8) to the case where X is locally convex.
196
Chapter 7 Norrned Linear Spaces
48 A locally convex space is said to be anormed if it has a countable defining set of admissible seminorms Extend the KreinMilman theorem to the case where X is a anormed locally convex space.
Notes Some of the Banach spaces of classical analysis (for example, Lev) are not Banach spaces in our sense, because the norm is not defined
constructively. Proposition (1.10) has the following generalization: A bounded linear map of a normed linear space onto a finitedimensional space is compact if and only if its kernel is located. See [18] and Problems 4142.
Note the care with which a finite basis is defined (Definition (2.1)) In fact, as Problem 7 shows, a normed linear space X is finitedimensional if and only if it is generated by vectors Xl' .. , xn that are linearly independent, in the sense that IAII + ... + IAnl > 0 whenever AI' ... ,An are scalars with Al Xl + .. +}'nxn=!=O Note that a subspace of a normed linear space is located, by definition. Theorem (2.6) is a special case of the more general result that a Banach space which is generated algebraically by a compact set is finitedimensional [76] Theorem (2.12) could be proved much more easily, without the use of Lemma (2.8), if we could prove that every continuous map from a compact space into 1R + has a positive tnfimum For further developments in constructive approximation theory see [17]. One might wish to complete Theorem (3 25) by determining the form of all bounded linear functionals on Lp  not just the normable ones Perhaps there is no better answer than that given by Theorem (6.7). For a generalization of Theorem (325) in the context of Orlicz spaces see [46]. The classical notion of absolute continuity (namely, that every set with Imeasure 0 also has Jmeasure 0) is classically equivalent to our notion (Definition (326)) when the integrals are afinite, but it is of no use constructively. The normability condition in Theorem (3.34) cannot be removed (see Problems 17 and 18) For an extension of the RadonNikodym theorem to the case where I and J are both afinite, see [15].
Notes
397
It is not always pos!>ible to choose the normable linear functional u of Theorem (4.6) so that II u II = II vii. The notion of a quasinorm (which was introduced under a different name by D.L Johns in his Liverpool University Ph.D. thesis) is classically equivalent to that of a norm, and enables us to handle spaces like L"" in a natural way. The proof of Proposition (5.6) actually shows that a linear map between quasinormed linear spaces is bounded if and only if it is continuous at 0 (in the obvious sense). It would be interesting to extend the theory in Sections 4, 6, and 7 to the context of quasinormed spaces. Problem 29 seems to be nontrivial; a solution is found in D.L. Johns's Ph.D. thesis Note that the constructive definition of an extreme point (Definition (7.1)) constitutes a strengthening of the classical version. Since a Hilbert space may be neither finitedimensional nor infinitedimensional, an orthonormal basis must be allowed to have some of its vectors equal to O. Lemma (8.10) provides a particularly simple proof that a locally compact Hilbert space is finitedimensional. The definition of strong convergence (preceding Lemma (8.18)) could have been phrased in terms of the double norm on Hom (H, H), and Lemma (8.18) could have been derived from Proposition (6 2) It is simpler to work directly. The map xt+u x ' where U x is the linear functional given by uAy) == (y, x), from a Hilbert space to its dual need not be onto, as Problem 14 shows. This raises interesting questions about the constructive interpretation of various classical results. (See also Problem 36.)
It would be interesting to find a good constructive substitute for the closed graph theorem, which says that a linear map u from a Banach space X into a Banach space Y is continuous if and only if {(x, u(x)). XEX} is a closed subset of X x Y. Another important principle of classical analysis whose constructive status is uncertain is the open mapping theorem, which states that a bounded linear map from a normed space onto a Banach space maps open sets to open sets. Problems 22 and 23 indicate some partial substitutes for this theorem A detailed, but inconclusive, constructive investigation of the open mapping theorem has been carried out by Stolzenberg [85] A Banach space X is classically called reflexive if X = X**. Since X* generally is not a Banach space, this definition does not make constructive sense. Problem 31 is an attempt to find a constructive substitute for reflexivity.
398
Chapter 7
Normed Linear Spaces
Even classically, many linear spaces have a topology which requires more than one seminorm for its full description. To deal with such examples we are led to the concept of a locally convex space (Problems 47, 48). The simplest example of a locally convex space is a normed space: in this example the given norm is the sole element of the defining set of seminorms. A more interesting example arises in the following way. Let!?) denote the set of all infinitely differentiable functions ¢: IR + IR with compact support. For each sequence N == (Nk)':~ 1 of positive integers and each ¢ in !?), write
where, of course, the outer sum is actually finite. The seminorms II liN define a locally convex structure on !?); with this structure, !?) is called the space of test functions, and its dual !?)* is called the space of distributions on IR. At first glance, the concept of a locally convex space would appear to be important for constructive mathematics, since examples exist in profusion. However, in most cases of interest it seems to be unnecessary to make use of any deep facts from the general theory of locally convex spaces. For example, the dual X* of a separable normed space X is most conveniently studied in terms of the double norm, rather than the natural locally convex structure. Another example is given by the theory of distributions Classically, the space!?) of test functions is complete (in the sense of Problem 24 of Chapter 4), but the constructive completion of !?) consists of all infinitely differentiable functions ¢: IR + IR such that the norm II ¢ II N exists for each sequence N of positive integers. This difficulty could be overcome by redefining !?), although such a procedure seems slightly artificial. The same difficulty arises, in a more acute form, with the dual space !?)* (the space of distributions): the constructive completion of !?)* cannot be identified with !?)*, and in this case it definitely seems artificial to enlarge !?)* to make it complete. Although detailed studies are needed, tentatively we conclude that the constructive theory of distributions should rely heavily on the concept of sequential convergence, defined ad hoc for sequences in fi) and for sequences in fi)*, rather than on general theorems about locally convex space~
Chapter 8. Locally Compact Abelian Groups
Section 1 constructs Haar measure on a locally compact group G, by a method of H. Cartan. Certain least upper bounds must be proved to exist in order to make the classical proof constructive; this adds length to the classical treatment. In Section 2 convolution is defined and the group algebra is studied. Specializing to a commutative group G, we prove the fundamental fact that the convolution operator of an integrable function is normable. The chapter closes with a study of the dual group G*. The spectral theorem is used to establish basic properties of the Fourier transform, in particular the inversion theorem. By standard methods, we use the inversion theorem to get the Pontryagin duality theorem.
Many groups of special mathematical interest  for instance, various important groups of geometrical transformations  have a locally compact metric structure with respect to which the group operations are continuous. Such groups are called locally compact groups. Every locally compact group has a leftinvariant measure We construct this measure, and use it for a detailed study of a locally compact commutative group. In particular, we construct the dual group, prove the duality theorem, and study the Fourier transform.
1. Haar Measure In this section we define the concept of a locally compact group, prove the remarkable fact that it admits a leftinvariant measure, and study convolution on the group. (1.1) Definition. A locally compact group is a set G which is both a
group and a locally compact metric space, such that the operation (X,y)HX 1 y
from G x G to G is continuous.
400
Chapter 8
Locally Compact Abelian Groups
Setting y equal to the identity element e, we see that the operation xt+x I is continuous. Since the multiplication operation (x, y)t+xy is the composition of the operations (x, y)t+(x 1 , y) and (x, y)t+x 1 y, this operation is also continuous. In particular, for each x in G the maps yt+xy and yt+yx are continuous Since the inverse maps are continuous too, each of these maps is a homeomorphism of G with itself. Since a continuous function from one locally compact space into another takes bounded sets into bounded sets, the sets KL== {xy: xEK,YEL},
K 1 L== {x 1 y: xEK,YEL},
and KL 1 == {xyI: xEK, YEL}
are bounded subsets of G whenever K and L are bounded subsets of G By the same token, KL, K  I L, and KL I are totally bounded whenever K and L are totally bounded. For brevity we let p(x) be the distance p(x, e) from an arbitrary element x of G to the identity element e. (1 2) Proposition. Let K be a bounded subset of a locally compact group G. For each e>O there exists 6>0 such that p(x,y)~e whenever x and yare elements of K with p(x 1 y)~6; and for each 6>0 there exists e>O such that p(x I y)~6 whenever x and yare elements of K with p(x,y)~e. Proof. Consider e > O. Since K x K  1 K is bounded, multiplication is uniformly continuous on K x K 1 K, and we can choose 6 so small that p(x,y)=p(x(x 1 x),x(x 1 y))~e
whenever xEK, YEK, and p(x 1 y)~6. Consider conversely any 6>0. Since K x K is bounded, the map (x, y)t+ XI Y is uniformly continuous on K x K. It follows that p(x 1 y)=p(x 1 y)_p(x 1 X)~6
whenever xEK, YEK, and p(x, y) is sufficiently small.
0
(1 3) Corollary. A subset K of a locally compact group G is totally bounded if and only if for each 6>0 there exist x1, ... ,x n in K such that for each x in K we have p(X i 1 x) ~ 6 for some i (1 ~ i ~ n).
Proof If K is totally bounded, then K is bounded. By (1.2), for each 6>0 there exists e>O such that p(x 1 y)~6 whenever xEK, YEK, and
1 Haar Measure
401
p(X,y)~e
Therefore any subfinite e approximation {Xl' ... ,xn } to K has the stated property. Assume conversely that for each b > 0 the points XI' ... , Xn exist Choose ~ I' . , ~N in K such that for each X in K there exists i with P(~il x)(x takes bounded sets into bounded sets, each of the sets {XEG:P(~i1 x)O such that p(X,y)<e whenever x, YEK and p(x I y) < b. Choosing XI' ... ' Xn in K so that for each X in K there exists i with p(XiIX) SjEY, J..(i)=l
p(sj,LK1»O,
=>
and J..(i)=O for some i. If XEK and J..(i) = 1, then p(sj,Lx1»0. By the continuity of the map Yl+yx 1, p(s;x, L) > 0; whence e/J(Sjx) =0. Thus f~
L
c; T(sJe/J
"(;)~O
So
L "(j)~O
c;T(SJe/JEA. Clearly,
L
C;~LC;. Thus we may take C;==c j
"(i)~O
and s;==Sj for each i with J..(i)=O.
0
It follows from Lemma (1.6) that
(f: e/J)= inf{L cj: L Cj T(s;le/JEA, provided that the righthand side exists.
SjEY
for each i}
1 Haar Measure
403
(1.7) Lemma. For each j in C+ and each ¢ in f!J, the quantity (f: ¢) exists. Proof Let K be a compact support of f, and L a compact support of
¢. Choose r> 0 so that Y= {XEG: p(x, LKl)~r}
is compact. Then we need only consider elements L C; T(sJ¢ of A with each S; in Y. Since ¢Ef!J, we have ¢(t»O for some t in G. Choose 1'>0 so that ¢(tx»y whenever p(x), and 13>0 is chosen so that Ilfll >eI14111, then (f:41)~e If fEC+, g E flJ>, and 41 E flJ>, then (f: 41) ~ (f: g)(g: 41)· Hence if fEflJ, then (1.8)
(g. f) 1
As a function of f, (f:
41)
~ (f:
41 )(g: 41) 1 ~ (f: g).
is homogeneous,
(Af: 41)=A(f: 41)
(AElR o+),
and subadditive, (fl
+ f2: 41) ~ (fl : 41) + (f2 : 41)·
It is also leftinvariant, in the sense that (T(s)f: 41)=(f: 41)
(SEG).
Now fix a particular function fo in flJ for the rest of this section, for purposes of normalization. For each f in C+ and each 41 in flJ> we write I ",(f) == (f: 41)(fo: 41)1 In case fEflJ>, (1.9) (fo :f)1 ~I",(f) ~ (f· fo), by (1.8). A function 41EflJ is small of order c (where c is a positive constant) if 41(x)=0 whenever p(x)~c. The following lemma shows that I", is approximately linear if 41 is sufficiently small. (1.10) Lemma. Let 13 and M be positive numbers, and fl' ... In elements of C +. Then there exists c > 0 such that, whenever AI' ... ' An are real numbers with 0 ~ Ai ~ M (1 ~j ~ n) and 41 is small of order c, we have
]; A/ ",(ij) ~ I ",(]; AJ) + e. Proof" Consider such AI' ... , An. Since I",(];AJ)~I",(];Mf),
the quantities I",(];AJ) are bounded independently of the values of the Ai. To prove the lemma it is therefore enough to show that for each e>O we can choose c>O independent of the Ai' such that ]; A/",(f) ~ (l ",(]; Aiij) + e)(1 + e)
whenever
41
is small of order c.
406
Chapter 8 Locally Compact Abelian Groups
Let K be a compact set which supports each of the functions fj. Let g be any element of ?J! such that g(x) = 1 for each x with p(x,K)~l. Write c5 == c(g:fO)I. By (1.9), c5 1",(g) ~ c5(g :fo) = c. Define test functions h j by (1 ~j ~ 11, xEG)
h/x) ==f/x) (2:' A;./;(X) + c5 g(x) +min {1, p(x, K)})1
Then the functions hj have a common modulus of continuity independent of the values of AI' ... , An. By (1.4), we may therefore choose c>O independent of the Aj , such that
Ih/s 1)h/x)1 <M 1n 1c
(1.10.1)
(1 ~j~n)
whenever p(sx)c or p(s;x)O so small that Ig(x)g(y)I~'1 whenever p(xyl)~C. Let s l ' ... , s. be elements of K 1 such that for each x in K we have P(SiX). In view of the equality g(Sl x) =g(x 1s) = T(xl)g(s)
408
Chapter 8
Locally Compact Abelian Groups
and the leftinvariance of I, we obtain (1.11.5)
(f(x)  E)l",(g) YJ 1",(f):£ l",(E hJg(SiX)) :£ (f(x) + E) l",(g) + YJ 1",(f).
Since, by (1.8), 1",(f) 1",(g)1 :£ (f: g) = 1X2~G ,
dividing (1.11.5) by l",(g) gives (1.11.6)
f(x)  /3:£ l(l",(g) 1 E g(SiX) hJ):£ f(x) + /3,
where /3==E+~(il(E)
and t I ",(g) ~ I ",(f) + (1 + I ",(h))E
whenever ¢ is small of order c For such ¢ it follows from (1.9) that It  I",(f) I ",(g)II ~ I",(g)I (1 + I ",(h))E ~(fo: g)(l+(h:fo))E
D
(1.19) Theorem. There exists a positive leftinvariant measure )1 on G such that Sf d)1 > 0 for all f in ~. If v is any positive leftinvariant measure on G, then v = C)1 for some c in IR 0 +.
Proof: Fix an element g of ~. Consider f in C+, and choose h in fljJ so that h(f +g)= f +g. For each kin '1.+ choose tk>O so that f~tkg+kIh
(1.19.1)
Consider any h' in that
~
and
tkg«f+kIh.
and any sequence
f«t~g+kI h'
and
(t~)
of positive numbers such
t"g«f +k I h'
(kEZ+).
By (1.18), for each pair of positive integers j, k we have (1.19.2)
and It~  I",(f)I",(g)11 ~ (fo:
whenever ¢
E fljJ
g)(l + (h': fonk I
is of sufficiently small order. Hence
Itjt~1 ~(fo:
g)(l +max {(h:fo), (h':fo)})U 1 +k 1).
In particular, Itjtkl ~(fo: g)(l +(h· fo))U1 +k 1).
It follows that (tn) is a Cauchy sequence whose limit does not depend on the choice of hand (tJ Denote this limit by )1 (f). It is clear that )1(rxf)=rx)1(f) for all rx~O Consider elements fl and f2 of C+. By the foregoing and (1.10), we can choose ¢ in ~ so that the quantities )1(fI)' )1 (f2)' )1(f1 + f2)' and I",(fl + f2) are arbitrarily close to I",(fl)I",(g)t, I",(f2)I",(g)1,
1 Haar Measure
413
I",(fl + f 2)I",(g)1, and I",(fl) + I",(f2)' respectively. Therefore Jl(fl
+ f2) =
Jl(fl)+ Jl(f2)'
For each test function f choose functions flJ2 in C+ with f = fl  j 2' and define Jl(f) == Jl(fl)  Jl(f2) If also f = f;  f; with f;, f; in C+, then fl + f; = f; + f2' and so Jl(fl) + Jl(f;) = Jl(f;) + Jl(f2)' It follows that Jl(f) does not depend on the choice of fl and f2' and that Jl is a well defined linear functional from C(G) to JR. By construction, Jl(f)~O when fE C+. Therefore Jl is a positive measure on G Given f in C+ and s in G, choose h in fjJ so that h(f + T(s)f + g) =f+T(s)f+g. For each k in 7l+ choose tk>O so that (1.191) holds Then as T(s) f < f < T(s)f, (1.19.1) holds with f replaced by T(s)f. Hence ST(s)fdJl=SfdJl and Jl is leftinvariant. Taking f = g in (1.192), we see that JgdJl= 1. For each f in fjJ we can find C1,. 'Cn in IRo + and 05 1, ""sn in G such that g~L:ciT(sJf. Since Jl is positive and leftinvariant, it follows that
1=SgdJl~L:cJfdJl
and so S fdJl>O. It remains to show that any positive leftinvariant measure v on G is a multiple of Jl For all fl and f2 in C+ with fl~f2 we have Sfl d v ~ Sf2 d v, by the left invariance and positivity of v. In particular, with f in C+ and tk as in (1.19.1) we have Sfdv~tkS gdv+k 1 Shdv
and tkJgdv~Sfdv+kl Shdv
for all k in 7l+. Letting kHJJ, we obtain Sfdv=SfdJlSgdv. Thus v = (S gdv)Jl.
D
The measure of Theorem (1.19) is called Haar measure. When we work with a locally compact group G, we shall assume that Jl is Haar measure on G. For clarity, we sometimes write S f(x)dJl(x) instead of S fdJl; thus, for example, S f(sx)dJl(x) stands for S T(s)fdJl. We also denote by Lp(G) the Lp space associated with the Haar measure Jl. If f is integrable with respect to Haar measure, and sEG, then T(s)f is integrable, and S T(s) f dJl = Sf dJl. This follows from Theorem (1.19) applied to a representation of f in C(G). Another important consequence of Theorem (1.19) is that Jl(A) > 0 whenever A is an integrable set such that A 1 has nonvoid interior To
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Chapter 8
Locally Compact Abelian Groups
prove this, it suffices to observe that we can construct an element j of f1jJ with O~I~xA. Group multiplication induces an important operation, called convolution, on test functions. In defining this operation we work with complexvalued test functions; unless we state otherwise, all functions on G from this point on will be permitted to have complex values. (For the meaning of I I dJ.l. when I is complex, the reader should refer to the discussion at (8.3) of Chapter 7.) For each test function I define the test function J by J(x)=I(x 1 )*. In case G is commutative, the linear functional IH I IdJl (when restricted to real test functions) is a positive leftinvariant measure on G. By Theorem (1.19), there exists c > 0 such that II dJ.l. = c I I dJ.l. for all real test functions f. Since = I, we have I(I +J)dJ.l.=c I (f + j)dJ.l. (fE C(G)).
J
Therefore c = 1, and thus
for all real test functions
f. For complex test functions I we have
(1.20)
In general, a group G with Haar measure J.l. satisfying (1.20) for all test functions I is called unimodular Thus every commutative locally compact group is unimodular. For each I in Ll(G) the function J defined by j(x)=I(x 1 )* belongs to LdG); also, if G is unimodular, then I satisfies (1.20). These facts readily follow from the observation that if I is a realvalued integrable function on G and (fn) is a representation of I by elements of C(G), then Un) is a representation of ! by elements of C(G) If G is unimodular and IE C(G), then
Jj(xs)dJ.l.(x) = I f(x 1 s)dJ.l.(x)
(by (1.20)) = Jj«Sl x)l)dJ.l.(x)
= Jj(x 1 )dJ.l.(x)
(by leftinvariance)
= II(x)dJl(x). Thus Haar measure on a unimodular group is rightinvariant. By the StoneWeierstrass theorem, any test function f(x,y) on the product G x G of a locally compact group G with itself can be uniformly approximated by finite sums l:'Ii(x)gi(Y)' where the J; and gi are test functions on G whose support lies in some compact set K c: G
1 Haar Measure
depending only on (1.21)
f
It follows that
S(S f(x, y)dll(X»dll(Y) = S(S f(x, y) dll(y))dll (x)
The convolution f (1.22)
415
*g
(f * g)(x)
of test functions f and g is defined by
=Sf(y) g(yI x) dll(Y)
(XEG)
If K is a compact support of f, and L is a compact support of g, then the closure of KL is a compact support for the continuous function f * g. Thus f * g is a test function. If f, g, and h are test functions, then
«(f * g) * h)(x) = S(f * g) (y) h(y 1 x)dll(Y) = Sf f(z) g(z 1 y) h(y 1 x) dll(Z) dll(Y) = SS f(z) g(z 1 y) h«Z1 y)1 Z1 x) dll(Y)dll(Z) = SS fez) g(y) h(y 1 z 1 x)dll(y)dll(z) (by leftinvariance of 11) = S f(z)(g * h) (z 1 x)dll(Z) = (f * (g * h»(x). Thus convolution is associative. In case G is commutative (and hence unimodular),
(f * g) (x)= S fey) g(y 1 x)dll(Y) = S f(y 1) g(y x) dll(Y) = Sf(x 1 y)I)g(Xl yx)dll(Y) = S f(yl x) g(y)dll(Y) = (g * f)(x) and thus convolution is commutative. (1.23) Proposition. Let f and g be test functions on a unimodular group G Let p, q, and r be constants with p~1, q~1, r~l, and
pl+ q l=1+r 1. Then (1.23.1)
Proof" We may take p> 1 and q> 1, as the general case is then handled by an approximation argument using the fact that for a test function f, Ilf Iisn + I f I s whenever (sn) is a strictly decreasing sequence converging to s ~ 1. Clearly we may also take f~ 0 and g ~ O. The numbers O(=r I , (J=pl_ r \ and y=qI_rI are positive and sum to 1. Holder's inequality «3.4) of Chapter 7) gives
(f * g)(x) = S(f(y)P g(y 1 X)q)a f(y)pfJ g(y 1 X)qy dll(Y) ~(S f(y)P g(y 1X)qdll(Y))" IlfII~fJ Ilgll:Y.
416
Chapter 8
Locally Compact Abelian Groups
(Note that Jg(yl x)qdJl(Y) = Ilgll:, by (1.20) and the rightinvariance of Jl.) Therefore, using (1.21), we have
fII:
P Ilgll:Y(Jjf(y)P g(y 1 x)q dJl(Y) dJl (xW/' I f * gil, ~ I = IlfII: P Ilgll:Y(11 fII: Ilgll:)l/, = Ilfllpllgll q. D
In order to extend Proposition (1.23) to more general functions, we need the following result. (1.24) Lemma. Let f be either a test function or a simple function on the locally compact group G. Then there exists c>O such that If¢dJl exists and If¢ dJl ~ cI ¢ I q whenever q ~ 1 and ¢ E Lq (G). Proof Choose an integrable set K of positive measure such that XKf = 0 on a full set. Let M > 0 be a bound for f, and write c=(1+Jl(K»M.
Consider any q~l. For each n in Z+ define q(n)=q+n 1 and p(n)=(1_q(n)1)1 Then
I f I pen) = (I IfIP(n) dJl)l/p(n) ~ M Jl(K)l/p(n) < c, K
since p(n»
1 If ¢ is a simple function, we now have I f¢dJl ~ I f
IIp(n) I ¢llq(n) ~cll¢llq(")'
by (3.6) of Chapter 7. Letting n+oo, we obtain If¢dJl~cll¢llq. Now let ¢ be an arbitrary element of Lq(G) We may assume that f and ¢ are nonnegative. By (3.10) of Chapter 7, we can find an increasing sequence (¢n) of nonnegative simple functions converging to ¢ in Lq(G). Since f¢ is measurable and (f¢)q~Mq¢q, we see that f ¢ELq(G). Also,
Ilf¢ f¢nllq~M II¢ ¢nllq
(nEZ+),
so that the sequence (f ¢n) converges to f ¢ in Lq(G). It follows from (3.8) of Chapter 7 that (f ¢n) converges to f ¢ in measure. On the other hand, by the case already considered,
II f¢mdJl I f¢ndJlI ~cll¢m ¢nllq
(m, n~ 1),
and so (I f¢ndJl) converges in lR. Hence, by the monotone convergence theorem and the ufiniteness of Jl, f¢ is integrable, and If¢dJl= lim Jf¢ndJl~c lim nno
II¢nllq=cll¢ll q. D
1 Haar Mea,ure
417
(1 25) Proposition. Let p, q, and r be constants with p ~ 1, q ~ 1, r ~ 1, and pl +ql = 1 +r 1 Let f and g belong to Lp(G) and Lq(G), respectively, where G is a unimodular group Then the integral (1.22) exists for all x in a full set, the function f * g so defined belongs to Lr(G), and (1 23.1) is valid.
Proof It is enough to consider the case where f~O and g~O In case f and g are hoth test functions. the result is already established Cons ide I the case in which f IS a test function and g is arbitrary By (3 13) of Chapter 7, there exists a sequence (gn) of test functions converging to g in Lq(G). By (1.23 1), we see that (f * gn) is a Cauchy sequence in Lr(G), whose limit we call h In view of (3.8) of Chapter 7 and (8.16) of Chapter 6, we may assume that (f * gn) converges to h almost everywhere, and pointwise on a full set S. Now, by (1.24), the unimodularity of G, and the rightin variance of J1., for each x in G the map FI+ Sf(y) F(y  1 x) d J1.(y)
is a bounded linear functional on Lq(G). Hence (f * g)(x) = lim (f * gn)(x) = h(x)
(XES)
n~oo
Thus j
* gELr(G) and
~
IIfllp lim Ilgnllq (by (1.23)) n~ao
=
IIfllp Ilgll q.
Next consider the case in which f is a simple function vanishing on the metric complement of a compact set B, and g is an arbitrary element of Lq(G). Since C(G,0 as n> 00 We may assume that (fn) converges to f almost uniformly. Choose c >0 so that c/2 is a bound for f Then replacing fn by (fn A c) V  c, if necessary, we may also assume that each fn has bound c. (Note that «I. A c) v c);: 1 converges to f in Lp(G), by the dominated convergence theorem.) Given e>O and x in G, now choose an integrable set A such that Ale K, S Ig(y 1 x)1 dJ1.(Y) < (2c) 1 e, KA
418
Chapter 8 Locally Compact Abelian Groups
and Un) converges to f uniformly on A 1; this is possible by the definition of almostuniform convergence, and (4.15) of Chapter 6. Then for all sufficiently large n we have
I(f * g)(x)  Un * g)(x)1 ~ 2c J Ig(yl x)1 dJl(Y) + S If(y)  fn(y)llg(yl x)1 dJl(Y) KA
A
< G(l + JIg(y 1 x)1 dJl(Y)). Hence Un * g)(x)'U * g)(x) as n+ 00. On the other hand, Ilfm*gfn*gll,~llfmfnllpllgllq+O
as m,n+oo,
so that (fn * g) is a Cauchy sequence in L,(G). Hence Un * g) converges to a limit in L,(G). As some subsequence of (fn * g) converges to this limit pointwise on a full set, f * gEL,(G) and lim I f
* g fn * gil, = o.
Letting n' 00 in the inequality
II f * gil, ~ I f * g  fn * gil, + I fn I p I gI q' we obtain (1.23.1). Next consider the general case. Let Un) be an increasing sequence of nonnegative simple functions converging to f in the Lpnorm. Let (Kn) be a Jlbasis consisting of compact subsets of G. Replacing fn by XKnfn' we may assume that fn vanishes on the metric complement of Kn. By the case just considered, (fn * g) is a Cauchy sequence in L,(G), whose limit we denote by h. We may assume that Un * g) converges to h almost everywhere, and pointwise on a full set S. For each x in S the monotone convergence theorem shows that the sequence (y I+fn(Y) g(yl x)):;O~ 1 converges almost everywhere, and pointwise on a full set, to an integrable function (p with ScjJ dJl = h(x). Clearly, cjJ (y) = f(y) g(yl x) on a full set. Thus
h(x) = Sf(y) g(yl x) dJl(Y) = (f * g)(x). So
f * g is defined on a full set, f * gEL,(G), and Ilf * gll,= Ilhll, = lim 11f.. * gil, ~ lim Ilfnllp Ilgll q = Ilfllp Ilgll q • 0 "00
n+oo
The reader may check that in case fELl (G) and gELq(G), Proposition (1.25) holds for an arbitrary locally compact group (not necessarily unimodular). He may also confirm the associativity (and, in case G is commutative, the commutativity) of the convolution operation * when the functions convoluted are not necessarily test functions.
2. Convolution Operators
419
2. Convolution Operators In the sequel G will be a locally compact abelian group with Haar measure p.. We see from (1.25) that if fELl (G) and gEL 2(G), then f*gEL 2(G) and lIf*g[[2~llflltllgI12' Thusg~f*g is a bounded linear operator T(f) on L2 (G), and
Since convolution is commutative and associative, the operators T(f) and T(f') commute for all f and f' in Ll (G) Our first task is to show that the operators T(f) are normable. For this we need some preliminaries. Let W be a compact integrable subset of G For each g in C(W) define p.(g)=p.(g), where g is any element of Ll(G) such that X~wg=O and the restriction of g to W equals g. Then p. is a positive measure on W When we consider integrable functions on W, we will have in mind the complete extension of this measure on W We note the following elementary facts, whose proofs are left to the reader. If hEL1(G) and X~wh=O, then the restriction h of h to Wndmnh is integrable, and p.(h) = p.(h). If FeW is full relative to the measure p. on W, then F u  W is full relative to the measure p. on G. For each p ~ 1 let Lp(W) denote the Lp space associated with the measure p. on W; then for each g in Lp(W) the function g defined on dmngu  W by g(x) = g(x)
(2.2)
=0
if xEdmng, if XE W
belongs to Lp(G), and Ilgll p= Ilgll p. For each f in C(G) and each g volution f * g by
III
L2 (W), we define the con
(f * g)(x)=(f * g)(x) = (g *f)(x) = Sg(y)f(y~ I x)dp.(y)
(XEG),
w
where g is given by (2 2) Next we prove an elementary, but fundamental, result in Hilbert space theory. (2.3) Proposition. Let u be a bounded linear funaional on a Hilbert space H. Then u is normable if and only if (2.3.1 )
u(x) = (x, b)
(xEH)
for some vector b in H. in which case b is unique, and II u I = II b II.
420
Chapter 8
Locally Compact Abelian Groups
Proof Suppose first that (2.3.1) holds. It is clear that b is unique. By the CauchySchwarz inequality, lu(x)1 ~ IIbll for all x in the unit sphere S of H On the other hand, if e > 0, then either II b II > 0 or II b II < e In the first case, Ilbll1bES and u(llbll1b)=lIbll>llblle; in the second, u(O)=O> IlbllE. Since e is arbitrary, it follows that Ilull exists and equals Ilbll COl1\el~el~ ~lIpp(l~C th.lt /I i~ l10rmahle COI1~idel rir~t the ca~c where II u II > 0 Then ker u is located, by (1.10) of Chapter 7. Using the projection of H on (ker u)~, we can choose a vector Xo in (ker u)~ with u(x o) = 1. Since p (xo' ker u) > 0, and since each x in H has a representation x = u(x)x o +(x  u(x)x o)
with (xu(x)xo)Ekeru, we see that keru is a hyperplane with associated vector xo' Write b==llx o ll 2 x o ' Then v(x)==(x,b) defines a bounded linear functional v on H with ker v = ker u and v(x o) = 1. It follows from (1 8) of Chapter 7 that u = v. No'A let II bc .Ill arbitrary normable linear funcllol1dl on H Define an increasing sequence A: Z+ +{O, 1} such that A(n)=O => Ilull Ilull >0.
We may assume that .1.(1)=0. If A(n)=O, set b.==O. If ),(n) = 1, then by the foregoing, there exists a unique vector b such that (2.3.l) holds; write b. == b in this case Then (b.) is a Cauchy sequence in H: in fact, Ilbmb.11 0 or II u II = 0, then (2.3 1) holds with b == boo. Thus for each x in H the possibility u(x) =l= (x, boo> is ruled out. The result now follows. 0 We now prove a succession of technical lemmas which will enable us to establish the normability of the convolution operators T(f). (2.4) Lemma. Let f be a test function on the locally compact abelian group G, and let W be a compact integrable subset of G. Then sup{llf*hI1 2 :hES} exists, where S is the unit sphere of L 2 (W). Proof Let S* be the unit sphere of the dual of the Hilbert space L2(W), By (2.3), the normable elements of S* are precisely those of the
form
Uh'
where hES and
2 Convolution Operators
421
From (6 7) of Chapter 7 we see that in the metric induced by the double norm, S* is compact and r={uh:hES}
is dense in S*. Therefore r is totally bounded. Hence it will suffice to prove that the function Uhf+ II f * h 112 is uniformly continuous on r. Let E be any positive number. By (1.4), there exists 15>0 such that If(x)f(y)I~E whenever x, yEG and p(yI x)~15. Thus (241)
l(f * h)(x)  (f * h)(y)1 = I Jh(Z)(f(Z1 x) f(Z1 y)) d/i(:::)1 w
J
~ E Ihl djl~ E IlhI12/i(W)"~ ~ E/i(W)t
whenever hES, xEG, YEG, and p(yI x)~15. Let U be a compact support of f, and choose a compact integrable set K containing WU. Then the functions f * h (h E S) vanish on  K. Let x I' ... , xn be elements of K such that for each x in K there exists k with p(x;; I x) ~ 15. For each x in G let fx be the restriction to W of the test function y f+ j (y I x). Since the maps Uf+u(f,,) are uniformly continuous on S*, there exists w > 0 such that
I(f * hI )(x k) 
(f * h 2)(X k)I = Iuh, (fxJ  u h , (fxJI ~ E
whenever 1 ~k~n, hI ES, h 2ES, and IlIu h, uh,111 ~w. For such hI and h2' it follows from (2.4 1) and the choice of points X k that l(f * hl)(x) (f * h 2 )(x)1 ~ E(1 + 2/i(W)t)
(xEK)
Hence
Ilf * hI  f * h2112 = (J If * hI f * h212 d/i)t ~ E(1 + 2jl(W)t) jl(K)t, K
and so Uhf+ Ilf * hl12 is uniformly continuous on
r.
0
(2.5) Lemma. Let U be a compact neighborhood of the identity e in G,
and n a positive integer. Then there exists 15 > 0 such that whenever p(x, un) < 15.
XE U n + I
Proof" Choose r>O so that S(e,r)c U. There exists 15 in (0,1) such that p(yIx,e) 0 so that Kr=={XEG: p(x,K)~r} is compact and integrable. Given e>O, compute 0>0 so that for all x, y in Kr with p(x,y)
(gEL2(G))
and so I T(f.*) I = I TUn) II. As rx  1 (fn) = rx (f.*)* for each rx follows that p*(rx 1, /3 1) = pi(rx, /3) (rx, /3 E G*).
In
G*, it
Since p* and pi are equivalent metrics on M*, we now see that rx I+rx 1 is uniformly continuous on M*. It now suffices to show that for each e > 0 there exists e' > 0 such that p*(rx/3, rxo/3o)<e whenever rx, /3, rx o' and /30 belong to M*, p*(rx,rxo)<e', and p*(/3,/3o)<e'. Note first that if M*cN*(K,t) for some neighborhood K of e, then (M*)2 c N*(K, t); hence (M*)2 is a bounded subset of G*, by (3.15). In view of (3.16), to prove the existence of e' it is enough to show that for each compact set KeG and each 15 > 0 there exist a compact set K' c G and a constant 15' > 0 such that Ilrx/3rxo/3oIIK0 Hence II
443
J f 112 = 0, and so J= f
on a full set
as n>oo. D
Our final task is to prove the Pontryagin duality theorem, which states that the natural map XI+X of G into G** is both a homeomorphism and an isomorphism onto G**. To do this we must first prove some lemmas .. The most important of these are analogs of Lemmas (3.15) and (3.16), and characterize the metric p on G in terms of the behavior of the elements x of G** on certain subsets of G*; the first lemma prepares the way for these. (4.21) Lemma. Let X be an integrable compact set, and L, Y compact subsets of G such that XX1yy1cL; let f be an element of Ll(G)nL2(G) such that Lxf=O; and let g be a test function on G with support Y such that II g 112 = 1 and II f * g 112 > 2  t II T(f) II. Then for each x in  L there exists 0( in G* such that 11  0( (x) I> 1 and (4.21.1)
110((y)1 ~211 T(f)1I 1 1If  T(y)flll
(YEG).
Proof Let x belong to  L. Since Lx f = 0 and g has support Y, for each y with (f*g)(y)=I=O we have YEXY; so that if also (f*g)(xy)=!=O, then XYEXY and therefore XEXX 1 yy 1 cL. This contradiction shows that if (f * g)(y)=!= 0, then (f * g)(x y) = O. It follows that f * g and T(x)f * g are orthogonal element~ of L2(G). Thus
II f/'(f  T(x)f) II ~ = II T(f  T(x)f) 112 ~ II f * g  T(x)f * g II ~ = Ilf*gll~+ IIT(x)f*gll~=21If*glli >2(2 t IIT(f)llf= IIT(f)112. SO there exists 0( in G* with (4.21.2)
II T(f)11 < 10((f  T(x)f)1 = 10((f) O((x)* 0((f)1
and therefore, by (3.7), 110((x)1 =IIO((x)*1 >10((f)111IT(f)11 ~ 1. From (4.21.2) and the inequality 110((x)*1::::;2 we have 10((f)1 ~211IT(f)II. Thus for all y in G, Ilf  T(y)flll ~ 100(f  T(y)f) I =IO((f)IIIO((y)1 ~2111 T(f)IIIIO((y)l. This immediately leads to (4.21.1).
D
444
Chapter 8 Locally Compact Abelian Groups
For each 6> 0 and each K* c G* define
N(K*,6)=={XEG: 1(J(x)11;;;;;6 for all (J( in K*}. (4.22) Lemma. For each 6 in (0,1) and each neighborhood K* of the identity 1 in G*, the set N(K*,6) is bounded Conversely, if M is any
bounded subset of G and 6 any positive number, then there exists a neighborhood K* of 1 in G* such that McN(K*,6) Proof. Consider a real number 6 with 0 < 6 < 1, and a neighborhood K* of 1 in G*. Let KI be any compact neighborhood of e in G. By (3.15), N*(K I,!) is a bounded subset of G*. By (3.16), there exist a compact set K c G and a positive number b such that (J(EK* whenever aEN*(K I,!) and aEN*(K,b). We may assume that b;;;;;! and KIcK, so that N*(K,b)cK*. We may also assume that K=K I and that K is integrable. By (2.7), there exist compact integrable neighborhoods V and X of e in G such that KV c X and .u(X  V);;;;; (bj4) .u(X). Write f == Xx, and notc~ that as 11111 00 ;;;;; Ilflll and 1(1)= S fd.u= Ilflll' we have II TU)II =llfll:x=llflll=.u(X). For each y in K we have T(y)f=Xy1x, VcyI X, and VcyX Hence (4.22.1)
IlfT(y)ft=SIXxxy lxld.u =.u(X  yI X)+.u(yI X X) = .u(X  y I X) + .u(X  y X) (by leftinvariance) ;;;;;2.u(X  V) ;;;;; (bj2) .u(X) = (bj2) II TU) II
Take gin C(G,CC) with IIg112=1 and Ilf*gI12>2!IITU)II. Let Y be a compact support for g, and L a compact set containing X X  I yy  I. Then by (4.21) and (4221), for each x in L there exists (J( in G* with 11(J(x)I>1 and 111aIIK;;;;;b Thus 11a(x)I>6 and (J(EN*(K,b)cK*, so that x cannot belong to N(K*,6). Hence N(K*,6)cL, and so N(K*,6} is bounded Conversely, consider any bounded subset M of G and any 6> 0 Let K be any compact subset of G with M c K, and M* any compact neighborhood of 1 in G* By (3 16), there exists b > 0 such that if aEM* and p*«(J(,I);;;;;b, then Ila11I K;;;;;6. Thus K*=={aEM*: p*(a, 1);;;;; b} is a neighborhood of 1 in G* such that M c N(K*, 6). 0 (4.23) Lemma. Let X be a neighborhood of e in G, and b a positive number. Then there exist compact integrable neighborhoods U, V, and W of e in G, and a positive number r, such that U c X, U I V C W, .u(V).u(W)1 > Ib, and XEX {or all x with p(x, W);;;;;r.
4 Duality and the Fourier Transform
445
Proof By (4.9) of Chapter 4 and (411) of Chapter 6, there exist numbers r, s with 0 < s < r such that See, 3r) e X, the sets V S c(e, s) and W=Sc(e,r) are compact and integrable, and Jl(W)Jl(V) Ib, and XEX whenever p(x, W)~r. Compute t in (O,r) so that U=Sc(e,t) is compact and integrable, and p(yI x) 0 and each compact set K* e G* there exists I; > 0 such IlxyIIK*~b whenever x,YEM and p(x,y)~I;.
1;>0 that each that
Proof. Consider any I; > O. By (1.2), there exists a compact neighborhood L of e in G such that p(x,y)~1; whenever x,YEM and xyI EL. Let X be an integrable compact neighborhood of e in G with XZ X z e L. By (4.23), there exist integrable compact neighborhoods U, V, and W of e, and a positive number r, such that U e X, U I Ve W, Jl(V)Jl(W)1 >i, and XEX for all x with p(x, W)~r Let f be a nonzero test function with support U, and h a test function with Ilf * hllzllhl121 > (2j3)t II T(f)II·
By (2.8), there exists y in G such that Ilf * Xw T(y)hllzIIXw T(y)hI121 >(i)t Ilf * hl1211hl121 >2 t IIT(f)II.
Since Xw T(y)h vanishes on  W, and XEX whenever p(x, W)~r, there exists a test function g on G, supported by X, such that Ilgllz = 1 and Ilf*gI12>2 t IIT(f)II. Using (32), choose a neighborhood K of e so that IlfT(y)flll 1 and, for each y in K,
=
11 ex(y)1 ~ 211 T(f) III II f  T(y)f III ~211 T(f)IIli II T(f)11 =t·
So ex belongs to N*(K,t), and therefore to K* Since 11ex(x)l> 1, x cannot belong to N(K*, t). It follows that N(K*, t) e L. Now consider elements x and y of M with IlxYIlK*~t. We have IlxyIIIIK*~t so that xyIEN(K*,t); hence xyIEL. By the choice of L, we therefore have p(x,y)~s. Thus we may take b=t
446
Chapter 8
Locally Compact Abelian Groups
Conversely, consider ,,>0 and a compact set K*cG*. By (3.15), there exists a neighborhood K of e in G such that K* c N*(K, "). Choose 8>0 so small that xy1EK whenever x, YEM and p(X,y)~8 Then for all x and y in M with p(X,y)~8, and all oc in K*, we have loc(x) oc(y)1 = IOC(xyl) 11 ~ ". Hence IlxYIIK.~"
D
We can now prove the Pontryagin duality theorem (425) Theorem. The mapping Xl+X is both a homeomorphism and an isomorphism of G onto G**. Proof Denote the mapping in question by i. We have already observed (4.6) that x E G** for each x in G By the first part of (4.24), X= Y whenever x, YEG and x= y. On the other hand, the identity xy=xy shows that e is a homomorphism Hence i is an isomorphism of G onto £(G). By (4.22), a subset M of G is bounded if and only if there exists a neighborhood K* of 1 in G* such that
Ix(oc)11 = loc(x)11 ~~
(xEM, ocEK*).
By (3.15), this holds if and only if £(M) is bounded in G**. It follows from (3 16) and (4.24) that on each bounded set MeG, the metric p of G is equivalent to the metric (x,y)l+p**(x,y) induced on M by the metric p** of G**. Hence e (G) is a locally compact subset of G**, and i is a homeomorphism of G onto t(G). It remains to show that t(G) =G**; since e·(G) is closed, it is enough to prove that p**(u,e(G))=O for all u in G** To this end, consider an arbitrary u in G**, and assume that p**(u,e"(G))>2r>0. Construct h~O in C(G**), with support contained in Sc(u,r)cG*, such that h(u»O. Choose a compact neighborhood K** of the identity e in G** so that p**(wv, v)~r whenever p**(v, u)~r and wEK**. Then if vEG**, wEK**, and h(w1v»0, we have p**(v,t (G)) ~ p**(u,e (G))  p**(u, W 1 v)  p**(v, w 1 v) >2rrr=0
and thus vEe·(G). Hence h(w1v)=0 whenever VEt(G) and wEK**. Let g be any nonnegative test function on G** with g(e»O and with support K**. Then g*h vanishes on t(G); also, by (119), (g*h)(u»O, and therefore IIg*hIl 2 >0 Let qJ be the inverse Fourier transform of g*h; then qJEL 2 (G*). Since qJ=gh is a product of functions in L 2 (G*),
Problems
447
we also have cpELl(G*). For each x in G, ~*(cp)(x)=cP(x*)
=(g*h)(x*) =0,
(by (4.11» (by (416»
since X*Et{G). Hence
llg * hl12 > O. This contradiction ensures that p**(u,t{G» = 0, as we required. 0=
11~*(cp)112 =
llcp 112 =
D
Since the map XHX* is continuous on G**, we can identify G with G** by means of the map XHX*. Under this identification, the inverse Fourier transform ~*: Ll(G*)+ Coo(G, 0 there exists a compact subset K of G such that for each x in G there exists y in K with If(xz) f(yz)I::£8
(ZEG).
Notes
449
Show that an almost periodic function f is uniformly continuous on G, in the sense that for each E>O there exists b>O such that If(x)  f(y)1 ~ E whenever x, YEG and p(y 1 x) ~ b. 16. Show that a continuous realvalued function f on a locally compact group G is almost periodic if and only if for each E > 0 there exist points Xl' ... , xn in G such that for each X in G there exists j (1 ~j ~ n) with If(xz) f(xjz)1 ~E (ZEG).
Notes The existence of (f: cp) is trivial classically The proof of Lemma (2.7) is due to Bernard Kripke. Theorem (2.9) is a key fact, but has no classical content. It would be possible to use Lemmas (3.15) and (3.16) to construct directly a metric on G* homeomorphic to the metric p* induced by the identification of G* with L. Further developments in the constructive theory of almost periodic functions can be found in [20], [38], and [56].
Chapter 9. Commutative Banach Algebras
In this chapter we introduce commutative Banach algebras and their spectra. A substitute is found for the classical result that every ideal is contained in a maximal ideal. We obtain as corollaries substitutes for other classical results, such as the compactness of the spectrum and the standard expression for the spectral norm. It is indicated in the exercises that, as far as the theory is carried, it has the same applications to analysis as its classical counterpart.
1. Definitions and Examples Many function spaces form an algebra: the elements can be multiplied as well as added. To regard such a function space only as a Banach space is to ignore some of its structure. To recover this structure in an abstract setting, we introduce the notion of a Banach algebra. (1.1) Definition. A (complex) Banach algebra fJll is an algebra over the
field of complex numbers, with a multiplicative identity e, which endowed with a norm I I satisfying the following conditions: (i) Ilell = 1 (ii) fJll is a Banach space relative to the norm I I (iii) IlxY11 ~ Ilxllllyll for all x and y in fJll.
IS
Examples come readily to mind. The simplest examples, of course, are the function algebras C(X, CC) (for an arbitrary compact space X), endowed with the usual norm. More generally, any closed subalgebra of C(X, CC) that contains the constant functions is a Banach algebra. Various important algebras of analytic functions arise in this way. Another important example of a Banach algebra comes from the group algebra Ll (G) of a locally compact abelian group G, endowed with the L1norm I 111' Now, L 1 (G) itself may not be a Banach algebra, because it may not have an identity element for con
1 Definitions and Examples
451
volution. However (by a simple construction which we shall not present), an identity element can be adjoined to Ll (G), and the resulting algebra is a Banach algebra. The above examples are all commutative Banach algebras. An example that may not be commutative is given by an arbitrary algebra of normable linear operators from a complex Banach space B into itself that contains the identity operator and is complete and separable with respect to the operator norm. For group algebras of locally compact abelian groups and for selfadjoint commutative operator algebras, the spectrum E has already played a vital role. Here is the general definition of this central concept. (1.2) Definition. The spectrum E of a commutative Banach algebra !fft consists of all u in the dual space qt* of !fft such that u(e) = 1 and u(xy) = u(x) u(y) for all x and y in qt. In other words, E consists of all bounded homomorphisms of !fft onto 0 be a bound for u. For each x in qt and each n in 7l+, we have
lu(x)1 = lu(x")l l/" ~ (c Ilx"ll)l/n ~ C lln Ilxll. Since cl/n + 1 as n + 00, we see that 1 is a bound for u. Since lu(e)1 = 1 = Ilell, it follows that u is in fact normable, and that Hull = 1. Our study of locally compact abelian groups and selfadjoint commutative operator algebras makes it plausible that E is compact. Unfortunately this is not true in general. The following proposition is the most we can say. Proposition. Let !fft be a commutative Banach algebra with spectrum E. Then there exist subsets E 1 ::::J E 2 ::::J ••• of the unit sphere S* of !fft*, each of which is either compact or void, such that E= En. (1.3)
n n
Proof. Since Ilull = 1 for all u in E, we have E c S*. Let (x~):,~ 1 be a dense sequence of elements of !fft. Choose positive numbers c 1> C 2 > ... converging to 0 so that each of the sets En= {UES*: lu(x? xJ)u(x?) u(xJ)1 ~ cn (1 ~ i,j ~ n), 11u(e)1 ~ cn}
is compact or void; we can do this because for each x in !fft the map ul+u(x) is continuous on S*. Then each En is a subset of S*,
452
L 1:=J
Chapter 9
L
2 :=J ••. ,
Commutative Banach Algebras
and
Len Ln· In fact, L = nLn· For if nL n, then u(e) UE
n
"
n
=1, and u(x?xf)=u(x?)u(xf) for all i andj; since (x~) is dense in~, it follows that u is multiplicative, and hence that UEL. D For the rest of this chapter, ~ will be a commutative complex Banach algebra, (x~) will be a fixed dense sequence of elements of ~, and the sets Ln will be as in the proof of Proposition (1.3). We shall see later that each Ln is actually compact (Proposition (2.7)). Here is an example of a Banach algebra d for which the spectrum L is not compact. Fix integers ro ~ r] ~ .. so that for each n either rn=O or rn=l, and so that we are unable to rule out either the possibility that r. = 1 for all n or the possibility that rn = 0 for some n. Let d consist of all sequences x == (x"):."'~ 0 of complex numbers for which 00 (1.4)
Ilxll ==
L
r.lxnl
"~ 0
exists Define the elements x and y == (Yn):~ 0 of d to be equal if Ilx  yll = o. Then d is a Banach space with norm given by (1.4). Moreover, if we define the product on d by xY== (to
XkYnkr~o'
then d is a Banach algebra. In case rn = 1 for all n, every complex number z with Izl ~ 1 defines an element U z of the spectrum L of d by ()()
uz(x) ==
L xnzn n~
(xEd).
0
On the other hand, if there exists n with rn = 0, then L consists of a single element u o, given by uo(x) == Xo (xEd). Since neither of these possibilities is ruled out, L is not totally bounded, and so is not compact We shall need to consider differentiable functions with values in a normed linear space B. Let K be a compact subset of 211yll the inverse of yze exists and equals
= L ZkIl, 00
bO(z)
k=O
and we have IIbO(z)II ~(2I1yll)1(1 +t+i+ ... )= IIYIlI.
Choose an integer K with 4C11yll+l zo' Let (J be the circular path tt>re it with domain [0,2 n]. If bO were an entire function, then by the Cauchy integral formula, the quantity u==(2ni)1 S(zzo)lbO(z)dz
would equal bO(zo). This is almost the case: by (2.1.4), if we patch together all of the bi , we obtain a function on the whole of 1. Hence n
I
u=(2ni)1 S(zzo)lbO(z)dz(]
j
(2ni)1 S(zzo)lbi(z)dz 1
=
IJ
because each of the integrals under the summation sign vanishes, by the Cauchy integral theorem. By Cauchy's integral formula, we have
Thus
I
u=bO(zo)+(2ni)1 S(zzo)lbO(z)dzi
Yo
~
(2ni)1 S(zzo)lbi(z)dz. 1
Yj
If bO were entire and each bi were equal to bO, all integrals in this last formula would cancel, giving u = bO(zo) As it is, for 1 ~j ~ N we can split each of the integrals overy i into four parts  the integrals over the sides of the square S;. For j=O we can split the integral over Yo into many parts, each an integral over a side of one of the squares Si for 1 ~ i ~ N. Let T be the set of all paths t formed by splitting the paths "Ii in this way for 0 ~j ~ N. The last expression for u becomes u=bO(zo)+(2ni)1
I J(zzo)l(b i'(z)b i'(z))dz, lET
t
where for each t, it and j, are certain integers in the set {O, . ,N}. From (2.1.4) it follows that n
u=bO(zo)+
I
dix i ,
i= 1
where the numbers IldJ are bounded by a positive number depending only on C, n, and Ilyll On the other hand, standard estimates give
Thus we see from the definition of u that
n
and therefore u=O. Hence bO(zo)= by yzoe, we get e= 
I
I
dix i . Multiplying this equality
i~ 1
(yzOe)dix i ·
a:=
Thus we have derived (2.1.2) with (yzOe)d i . Since the numbers Iidill are bounded by some number depending only on C, n, and IIYII.
2 Linear Equations in a Banach Algebra
457
and since Ilyzoell ~ Ilyll + IZol = Ilyll + C I K ~ Ilyll +411yli + 3 C I , the proof is complete
D
We now give an obvious extension of Theorem (2.1) (2.2) Corollary. Let
XI' "Xn and Yl' ""YN be elements of [1Jl such that for all AI' ... ,AN in (C there exist ai' ... ,a n and b l , ... ,b N in fR with Ilaill ~ C (1 ~i~n), Ilbjll ~ C (1 ~j~N), and n
N
i~1
j~1
L aix i + L biYjAje)=e.
(2.2.1 )
Then there exist a positive number C' (depending only on n, N, C, and IIYIII, ... ,IIYNII), and elements a~, ... ,a~ oj [1Jl with Ila;II~C' for each i, n
such that
L a;xi = e. i= 1
Proof The case N = 1 is just (2.1). Also, (2.1) implies that if (22) is true for a given value of N, then it is true for the next value. This proves (2 2) by induction. 0
Now consider a positive integer N, and elements x I' ... , Xn of Let
[1Jl.
M= 1 +max {llx?ll, ... , Ilx~II}.
For each Ntuple A=(AI, ... ,A N) of complex numbers, let H .. be the totally bounded set of all linear combinations of the form (23)
h=
n
N
N
i~1
j~1
i.j~1
L ZiXi+ L Wj(xJAje)+ L (;jX?(xJAje)
with IZil~M (1~i~n), and with Iw)~M, l'ijl~M (1~i,j~N)
(2.4) Lemma. Let 0 1 2e. By (4.4) of Chapter 7, there exists u in the unit sphere S* of [Jl* such that u(e»12e+lu(h)1 for all h in H;.. Since u(e)~l,
this gives (2.5.1)
lu(e)11 1 2e, we must have cr < 1 e. The desired conclusion now follows from (2.4). D Applications of Theorem (2.6) will be found in the problems at the end of the chapter.
460
Chapter 9
Commutative Banach Algebras
(2.7) Proposition. For each positive integer N the set L N is compact. Proof Since L N is either compact or void, it is enough to show that LN contains at least one element. To do so, take n= 1 and Xl =0 in
the definition of the sets H).., and define a by (2.62). If a 0, there exists N in 7l+ such that bn~am+e whenever b n is a term of the other sequence with n ~ N Proof: Let e be any positive number, and Ilxmll l /m any term of the first sequence. We may assume that m~2. Choose 6>0 so that
(1lxmll
+6)I/m~
Ilxmlll/m+e
By choosing N in 7l+ sufficiently large, we can ensure that the elements x,x 2 , ••• ,xm of!Jfl are arbitrarily close to the set {x?, ... ,x~}, and that u(x? xJ) is arbitrarily close to u(x?) u(xJ) for 1 ~ i, j ~ Nand all u in LN It follows that for all n~N, all u in L n, and 2~J~m, u(x') is arbitrarily close to u(x) U(X,l) Therefore, since lu(xr)u(xYI ~ lu(x')u(x)U(X'I)1 + Ilxlllu(xrI)u(xyll,
we can make u(xm) arbitrarily close to u(xr. In particular, we can ensure that lu(x m)  u(x)ml < b. We then have lu(x)lm~lu(xm)I+6~
Ilxmll +6
Problems
461
and therefore lu(x)1 ~(llxmll + b)l/m~ Ilx m I1 1/m +e. Since u is an arbitrary element of L n , it follows that Ilxllrn~ IIx ml1 1/m
+e. Conversely, consider a term Ilxllr~ of the second sequence Choose v;;; m so that e)llxllr~ +e/2)<e/4,
where (en) is the sequence of positive numbers used to define (L n) in the proof of (1.3). By (2.6), there exists R > 0 such that y 1 exists, and Ilylll~R,whenever YE~ and lu(y)l;;;e/4 for all u in Lv, Consider a complex number C with ICI> Ilxllr~ +e/2. For each u in Lv we have
lu(x  Ce)1 ;;; ICllu(e)Ilu(x)l;;; (11xll r~ + e/2)(1 eJ  Ilx Ilr~ = e/2  ev(llxllr~ + e/2) > e/4. Hence (xCe)1 exists, and II(xCe)111 ~R. In view of (1.8), it readily follows that zf>(ezx)1 is differentiable on the open set U =:0 {ZE(llxllr~ +e/2)\ we have Ilx"11 ~ c(llxllr~ +e/2)", where
C=:oR(lixll rm +e/2).
Since
lim C I / n=l,
it
follows
that
n~()()
Ilxnlll/"~ Ilxllr~ +e for all sufficiently large
n.
0
Problems 1. A subset S of a Banach algebra ~ generates ~ if ~ is the smallest closed subalgebra of ~ containing Su {e}. Show that if ~ is a commutative Banach algebra with a single generator x o , then the map u
462
Chapter 9
I>U(X o)
Commutative Banach Algebras
is a metric equivalence of the spectrum 1: with {u(xo):
uE1:} cO as n>oo (where p is the usual metric on the set of compact subsets of 1:1.) Show that if (jf has a single generator and 1: is compact, then 1: is firm. 3. Let (jf be a commutative Banach algebra whose spectrum 1: is firm. Let XI' ... , x. be elements of (jf, and {) a positive number, such that lu(x1)1 +.
+ lu(x.)1 ~ {)
for all U in 1:. Show that there exist +x.y.=e.
Yl' ... ,Y.
in
(jf
such that
X1YI
+ ...
4. Let (jf be the Banach algebra consisting of all continuous functions on {ZEt)
11. AxB M
IF keru c(K)
X* Hom(X, Y) .91
II 11" III III S* M(u,K) S(u,IX,K)
A* p(x) T(s)
C+ P/
(f cp) i .. (f)
