Treatise on Analysis: 002

TREATISE ON ANALYSIS Volume I1

Enlarged and Corrected Printing

This is Volume 10-11 in PURE AND APPLIED MATHEMATICS A Series of Monographs and Textbooks

EILENBERG AND HYMAN BASS Editors: SAMUEL A list of recent titles in this series appears at the end of this volume.

Volume 10 TREATISE ON ANALYSIS 10-1. Chapters I-XI, Foundations of Modern Analysis, enlarged and corrected printing, 1969 10-11. Chapters XII-XV, enlarged and corrected printing, 1976 10-111. Chapters XVI-XVII, 1972 1 0-IV. Chapters XVIII-XX, 1974

TREATISE ON

ANALYSIS J. DI'EUDONN~ Membre de l'lnstitut

Volume I /

Enlarged and Corrected Printing

Translated by

1. G. Macdonald

University of Manchester Manchester, England

ACADEMIC PRESS

New York San Francisco London

A Subsidiary of Harcourt Brace lovanovich, Publishers

1976

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“Treatise on Analysis,” Volume I1 Enlarged and Corrected Printing First published in the French language under the title “Eldments d’Analyse,” tome 2, 2“ edition, revue et augment& and copyrighted in 1974 by Gauthier-Villars, Gditeur, Paris, France.

SCHEMATIC PLAN OF THE WORK

U I. E l e m e n t s of the theory of sets

X I Elementary spectral

C XXII.

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CONTENTS

Notation

............................

Chapter XI1

TOPOLOGY AND TOPOLOGICAL ALGEBRA

. . . . . . . . . .

1 . Topological spaces. 2. Topological concepts. 3. Hausdorff spaces. 4. Uniformizable spaces. 5 . Products of uniformizable spaces. 6. Locally finite coverings and partitions of unity. 7. Semicontinuous functions. 8. Topological groups. 9. Metrizable groups. 10. Spaces with operators. Orbit spaces. 1 1 . Homogeneous spaces. 12. Quotient groups. 13. Topological vector spaces. 14. Locally convex spaces. 15. Weak topologies. 16. Bake's theorem and its consequences.

Chapter Xlll

INTEGRATION

.......................

1. Definition of a measure. 2. Real measures. 3. Positive measures. The absolute value of a measure. 4. The vague topology. 5. Upper and lower integrals with respect to a positive measure. 6. Negligible functions and sets. 7. Integrable functions and sets. 8. Lebesgue's convergence theorems. 9. Measurable functions. 10. Integrals of vector-valued functions. 1 1 . The spaces L' and Lz. 12. The space L". 13. Measures with base p. 14. Integration with respect to a positive measure with basep. 15. TheLebesgue-Nikodym theorem and the order relation on MR(X). 16. Applications: I. Integration with respect to a complex measure. 17. Applications: 11. Dual of L'. 18. Canonical decompositions of a measure. 19. Support of a measure. Measureswith compact support. 20. Bounded measures. 21. Product of measures.

Chapter XIV

INTEGRATION IN LOCALLY COMPACT GROUPS

ix

1

98

. . . . . . . . 242

1. Existence and uniqueness of Haar measure. 2. Particular cases and examples. 3. The modulus function on a group. The modulus of an automorphism. 4. Haar measure on a quotient group. 5. Convolution of measures on a locally compact

vii

CONTENTS

viii

group. 6. Examples and partkular cases of convolution of measures. 7. Algebraic properties of convolution. 8. Convolution of a measure and a function. 9. Examples of convolutions of measures and functions. 10. Convolution of two functions. 11. Regularization. Chapter X V

NORMED ALGEBRAS A N D SPECTRAL THEORY

. . . . . . . . 304

1. Normed algebras. 2. Spectrum of an element of a normed algebra. 3. Characters and spectrum of a commutative Banach algebra. The Gelfand transformation. 4. Banach algebras with involution. Star algebras. 5. Representations of algebras with involution. 6. Positive linear forms, positive Hilbert forms, and representations. 7. Traces, bitraces, and Hilbert algebras. 8. Complete Hilbert algebras. 9. The Plancherel-Godement theorem. 10. Representations of algebras of continuous functions. I I. The spectral theory of Hilbert. 12. Unbounded normal operators. 13. Extensions of hermitian operators.

. . . . . . . . . . . . . . . . . . . . . . . . . . . 444 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447

References Index

NOTATION

In the following definitions, the first number is the number of the chapter and the second number is the number of the section within that chapter.

A-'

AB

support of a function: 12.6 characteristic function of a subset A of a set: 12.7 upper and lower envelopes of a family of real-valued functions: 12.7 upper and lower limit of a sequence of real-valued functions: 12.7 opposite of a group G: 12.8 group of automorphisms of a topological vector space E: 12.8 set of elements x-' where x E A (where A is a subset of a group in which the group law is written multiplicatively): 12.8 set of elements xy with x E A and y E B, where A and B are subsets of a group in which the group law is written multiplicatively: 12.8 normalizer and centralizer of a subset H of a group: 12.8 set of finite products xlxz . . . x, (with n arbitrary) of elements of a subset V of a group: 12.8 ix

x

NOTATION

ZL.

G*A

set of p-adic integers: 12.9, Problem 4 p-adic solenoid: 12.9, Problem 4 homogeneous space of left (right) cosets of a subgroup H in a group G : 12.10 and 12.11 union of the orbits of the points of a subset A of E, with respect to an action of G on E: 12.10 space of orbits with respect to G: 12.10 space of continuous complex-valued functions on X: 12.14 quotient normed space of a normed space E by a vector subspace F: 12.14 x'(x) for a vector x E E and a continuous linear form x' E E' (the dual of E): 12.15 transpose of a continuous linear mapping: 12.15 unitary group of a Hilbert space E: 12.15, Problem 8 space of continuous complex-valued functions with support contained in the compact subset K of X : 13.1 space of continuous complex-valued functions on X with compact support: 13.1 Dirac measure at the point x: 13.1 measure with density g with respect to p: 13.1 and 13.13 image of a measure p under a proper continuous mapping n: 13.1 and 13.1, Problem 8 measure induced on an open set U by a measure p (or restriction of p to U): 13.1 space of complex measures on X : 13.1 integral of f € X ( X ) with respect to p : 13.1 space of continuous real-valued functions with support contained in the compact subset K of X: 13.2

NOTATION

IPI

$59"

PY

xi

space of continuous real-valued functions on X with compact support: 13.2 conjugate of the measure D: 13.2 space of real measures on X: 13.2 real and imaginary parts of a complex measure: 13.2 set of positive measures on X: 13.3 positive and negative parts of a realvalued function J': 13.3 order relation between real measures : 13.3 absolute value of a complex measure: 13.3 set of lower semicontinuous functions on X which are bounded below by a function belonging to XR(X):13.5 upper integral off: 13.5 sum of a sequence (t,) of elements 20 of 8: 13.5 set of upper semicontinuous functions on X which are bounded above by a function belonging to X.(X): 13.5 lower integral off: 13.5 outer and inner measures of A c X: 13.5 equivalence class off (with respect to p): 13.6 order relation between equivalence classes: 13.6 sum and product of equivalence classes : 13.6 integral of a p-integrable function: 13.7 space of (finite) real-valued p-integrable functions: 13.7 measure of a p-integrable set: 13.7 integral of an equivalence class: 13.7 integral off over A: 13.9 upper integral off over A: 13.9 measure induced by p on a closed subspace Y: 13.9

xii

NOTATION

entropy of a finite partition a : 13.9, Problem 27 entropy of a finite partition a relative to a finite partition P : 13.9, Problem 27 least upper bound of a finite sequence of finite partitions: 13.9, Problem 27 entropy of a mapping u relative to a finite partition a : 13.9, Problem 28 entropy of a mapping u: 13.9, Problem 28 integral of a vector-valued function : 13.10 space of complex-valued p-integrable functions : 13.10 s* If I dp: 13.11 (Y If l2 dp)"': 13.1 1 space of (finite) real-valued squareintegrable functions : 13.11 space of p-negligible functions : 13.1 1 space of classes of pth power integrable functions: 13.11 and 13.11, Problem 12 N,U) for anyfin the classf: 13.11 space of complex-valued square-integrable functions: 13.11 space of classes of complex-valued square-integrable functions: 13.11 maximum and minimum in measure off: 13.12 space of real-valued p-measurable functions bounded in measure: 13.12 space of complex-valued p-measurable functions bounded in measure : 13.12 space of classes of real-valued p-measurable functions bounded in measure : 13.12 space of classes of complex-valued pmeasurable functions bounded in measure : 13.12 space of (finite) real-valued p-measurable functions: 13.12, Problem 2 space of classes of (finite) real-valued pmeasurable functions: 13.12, Problem 2

NOTATION

xiii

space of real-valued locally p-integrable functions : 13.13 space of complex-valued locally pintegrable functions : 13.13 space of classes of real (complex) valued locally p-integrable functions : 13.13 integral with respect to a complex measure : 13.16 support of a measure: 13.19 norm of a measure: 13-20 space of bounded real (complex) measures: 13.20 space of real (complex) valued continuous functions which ' tend to 0 at infinity: 13.20 integral with respect to the product measure A Q p : 13.21 product measure: 13.21 upper integral with respect to a product measure: 13.21 lower integral with respect to a product measure: 13.21 the function ( x , y ) H f ( x ) g ( y ) : 13.21 product of n measures: 13.21

m

0 n= 1

integral with respect to the product of the n measures: 13.21 product of an infinite sequence of measures: 13.21, Problem 9 translates of a function: 14.1 translates of a measure: 14.1 transforms of a function or measure under SHS-': 14.1

xiv

NOTATION

T AG(S),

mod,(u), mod(u)

one-dimensional torus: 14.2 modulus function on a group: 14.3 modulus of an automorphism of G: 14.3

field of p-adic numbers: 14.3, Problem 6 average off over a fibre: 14.4, Problem 2 convolution of n measures: 14.5 space of measures with compact support: 14.7

centralization of a measure: 14.7, Problem 6 convolution of a measure and a function: 14.8

convolution of two functions: 14.10 spectrum of an element x of an algebra A: 15.2 spectral radius of x : 15.2 spectrum of an algebra A: 15.3 Gelfand transformation: 15.3 Hardy spaces: 15.3, Problem 15 adjoint (in an algebra with involution): 15.4

Hilbert-Schmidt norm : 15.4 involution in MS(G): 15.4 algebra of Hilbert-Schmidt operators : 15.4

Problem 18 regular representation of a complete Hilbert algebra: 15.8 space of hermitian characters: 15.9 multiplication by the class of u : 15.10 space of bounded universally measurable complex-valued functions on K: 15.10 ( H e x J x )2 0 for all x : 15.11 function of a normal operator: 15.1 1 square root of a positive hermitian operator: 15.11 absolute value of an operator: 15.11, Problem 6 trace of a nuclear operator: 15.11, Problem 7 ( ~ ( X * X ) ) ” ~ : 15.4,

HZO

f (N) abs(T)

NOTATION

dom(T)

T*

xv

nuclear norm: 15.11, Problem 7 space of nuclear operators: 15.11, Problem 7 approximative point-spectrum of an operator: 15.1 1, Problem 9 domain of an unbounded operator: 15.12 adjoint of an unbounded operator with dense domain: 15.12 spectrum of an unbounded operator: 15.12 function of an unbounded normal operator: 15.12

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CHAPTER XI1

TOPOLOGY AND TOPOLOGICAL ALGEBRA

As we said in the Introduction, we have sought to limit the material in this chapter to the minimum that the reader will require. We shall not stay to explore the refinements of general topology (filters, uniformities, separation axioms); instead, we shall pass as soon as possible to the category of uniformizable spaces, which are the only ones we shall meet in later chapters. Usually these spaces occur merely as “ ambient spaces ” in which it is convenient to operate, and we shall be mainly concerned-in conformity with the spirit of this book-with their separable metrizable subspaces. $or this reason we have included many criteria of metrizability and separability (12.3.6,12.4.6,12.4.7,12.5.8,12.9.1,12.10.10,12.11.3, 12.14.6.2,12.15.7,12.15.9, 12.15.10). These, together with Bake’s theorem and its consequences, are the only results in the chapter whose proofs are not straightforward. We have also included an account of various purely topological techniques which were not needed in the first volume (partitions of unity (12.6), semi-continuous functions (12,7)), and more than half the chapter is devoted to elementary concepts of topological algebra (topological groups, spaces with operators, topological vector spaces). All of this will be used constantly in the succeeding chapters.

1. TOPOLOGICAL SPACES

A topology on a set E is a set 0 of subsets of E (in other words, a subset of

p(E))satisfying the following two conditions :

(0,) The union of any family (AJA of sets belonging to D belongs to 0; (OJ The set E belongs to D,and the intersection of any two sets belonging to D belongs to D. 1

2

XI1 TOPOLOGY AND TOPOLOGICAL ALGEBRA

A topologicalspace is a set E endowed with a topology 0.The subsets of E which belong to 0 are called the open sets of the topological space E. Taking L = @ in (O,), it follows that the empty subset @ of E is always an open set, and from (0,Jwe see that E itself is an open set. Examples (12.1.1) The set D of open sets in a metric space E satisfies (0,)and (Oil) (3.5.2 and 3.5.3). This topology D is called the topology of the metric space E (or is said to be dejined by the distance given on E). Two topologically eguiualent distances (3.12) define the same topology. A topological space is said to be metrizable if its topology can be defined by a distance (and then this topology is also said to be metrizable). On any set E, the set 0 = E} is a topology, called the chaotic topology. It is not metrizable if E has at least two elements, because otherwise there would exist an open ball containing one of these elements and not the other, and this is impossible because E is the only nonempty open set. On a set E = {a, b } consisting of two elements, the set 0 = (0, { a } , E} is a nonmetrizable topology. If D,,0, are two topologies on the same set E, we say that 0, isfiner than O1 (or that 0, is coarser than D2)if 0, c D2. Two topologies on E are said to be comparable if one is finer than the other. The chaotic topology is coarser than all others. The discrete topology (i.e., the topology defined by the metric (3.2.51, for which D = V(E)) is finer than all others. Two topologies on E are not necessarily comparable; for example, let E = { a , b } be a set with two elements, and consider the two topologies 0, = (0, { a } , E} and 0, = {@, (61, El.

{a,

2. TOPOLOGICAL CONCEPTS

We have already remarked (3.12), in the context of metric spaces, that the notions of closed set, cluster point of a set, interior point of a set, frontier point of a set, neighborhood of apoinr (or of a set), dense set, continuous.function and homeomorphism are defined entirely in terms of the notion of an open sel in a metric space. Their definitions can therefore be transferred, without any change, to the context of arbitrary topological spaces. Moreover, all the properties involving these notions which were proved for metric spaces, and whose statements do not involve the distance (cf. (3.5) to (3.12)) remain valid for arbitrary topological spaces (and we shall therefore make use of them in the general case) with the following exceptions:

2 TOPOLOGICAL CONCEPTS

3

(1) The properties (3.8.11) and (3.8.12) are not true for arbitrary topological spaces, as is shown by the example of the space E = {a, b } with the topology (0, { a } ,E}. The existence of a distance is essential to the proof of these two propositions, although it does not feature in the enunciation. (2) We may define as in (3.9) the notion of a basis of open sets (or of the topology) of a topological space E. The criterion (3.9.3) is valid in general. We can also prove as in (3.9.4) that if there exists a denumerable basis for the topology of a topological space E, then there exists an at most denumerable set which is dense in E; but the converse is not valid for arbitrary topological spaces (Section 12.4, Problem 6).

By definition, a basis following property:

B of the topology of a topological space E has the

The intersection of any two sets of B is a union of sets of 8. Conversely, let B be a set of subsets of a set E. If B has the above property and if E E 8,then the set SJ of (arbitrary) unions of sets of B is a topology for which B is a basis. For it is immediately seen that 0 satisfies (0,)and (O,,). If E is a topological space and if F is any subset of E, then the set of intersections U n F, where U runs through the set of open subsets of E, satisfies the axioms (0,) and (0") and is therefore a topology on F. This topology on F is said to be induced by the topology on E. The set F, endowed with this topology, is called a subspace of E. With these definitions, all the propositions of (3.10) are valid for arbitrary topological spaces, with the exception of (3.10.9), which has to be stated in the following form: if the topology of E has a denumerable basis, then so does the topology induced on any subset of E. We have already remarked that the properties of compactness and local compactness for a metric space depend only on the topology, and not on the distance which defines the topology. We may therefore speak unambiguously of compact and locally compact rnetrizable spaces. We observe also that all the definitions and results of (3.19) relating to connectedness are valid for arbitrary topological spaces. (12.2.1) Let SJ1, SJ2 be two topologies on a set E. Then the followingproperties are equivalent:

(a) D2 is$ner than SJl; (b) if Ei denotes the topological space obtained by giving E the topology Di (i = 1, 2), then the identity mapping of E, onto E, is continuous; (c) for all x E E, every neighborhood of x in the topology Dlis a neighborhood of x in SJ2.

4

XI1 TOPOLOGY A N D TOPOLOGICAL ALGEBRA

The equivalence of (a) and (b) follows from the criterion of continuity (3.11.4(b)), and the equivalence of (b) and (c) from the definition of a continuous function.

Remark (12.2.2) Let (UJaE I be an open covering (3.16) of a topological space E. For a set G c E to be open in E it is necessary and sufficient that each of the sets G n U, should be open in the subspace U, . This follows immediately from the and (O1Jand the relation G = (G n U,). Taking complements, axioms (0,)

u

as1

it follows that a set F c E is closed in E if and only if each of the sets F n U, is closed in the subspace U, . (12.2.3) Let L be a subset of a topological space E. Then the followingproperties are equivalent: (a) L nV (b) (c)

For each x E L there exists a neighborhood V of x in E such that is closed in V ; L is an open subset of the subspace L (the closure of L in E); L is the intersection of an open subset cind a closed subset of E.

It is clear that (b) implies (c), L being the intersection of L with an open subset of E. Equally clearly, (c) implies (a). Let us show that (a) implies (b). For each x E L, we have V n L = V n L, because V n L is closed in V; this shows that in the subspace L the point x is an interior point of L, and therefore L is open in L. When L satisfies the equivalent conditions of (12.2.3), it is said to be a locally closed subset of E. (12.2.4) A procedure for constructing topological spaces which is frequently used is that of “patching together” a family of topological spaces in such a way that, in the topological space E so obtained, the Ed are identified with open sets of E. Since pairs of these sets may very well intersect, this identification requires that, for each pair of indices (A, p), we are given a homeomorphism of an open subset of Ed onto an open subset of E, . To be precise, suppose that for each pair (A, p) E L x L we are given:

(I) an open subset A,, of E,; (2) a homeomorphism AMa: Alp-+ A,, , satisfying the following conditions :

2 TOPOLOGICAL CONCEPTS

5

(I) A,, = E,, and h,, is the identity mapping l E A ; (11) for each triple of indices (A, p, v) and each x E A,, n A,, , we have h,,W E A,, and (12.2.4.1 )

hv,(x) = hv,(h,,(X))

(patching condition). Let F be the sum of the E,, which are therefore pairwise disjoint subsets of F (1.8). In F, consider the relation

R(x, y ) : " there exist A, p such that x y = h,,(x)."

E

A,, , y E A,, , and

This is an equivalence relation. It is reflexive by virtue of condition (I); it is symmetric because h,, and h,, are inverses of each other (by applying (11) with v = A, and then (I)); finally, it is transitive, because if x

E A,,

Y

y

= h,,(x) E A,, n A,,

9

z = h,,Ot),

then we have x = h,,(y) and therefore x E A,, n A,, by (11), and it follows now from (12.2.4.1) that z = h,,(x). We remark also that, by condition (I), the intersection of an E, and an equivalence class of R consists of at most one point. If E = F/R is the set of equivalence classes of R and if

n : F -+F/R = E is the canonical mapping, then each of the restrictions n, = n I E, : E, -,E is injective. Moreover, the sets n,(E,) form a covering of E. Now consider the set 0 of subsets X of E with the following property: for each A E L, the set n;'(X n n,(E,)) is open in E, . It is clear that 0 satisfies axioms (0,) and (01,), and is therefore a topology on E. We shall show that in this topology the sets n,(E,) are open in E and that n, is a homeomorphism of E, onto the subspace n,(E,) of E, for each A E L. In view of the definition of D and of the fact that n, is a bijection of E, onto n,(E,), it is enough to establish the equivalence of the following two properties of a subset X , of E,: (a) X, is an open set in the topological space El; (b) for each p E L, the set n;'(n,(X,) n n,(E,)) is open in E,. Taking p = A, we see that (b) implies (a). Conversely, if (a) is satisfied, then n n,(E,)) = A,, by definition, the condition (b) since we have n;'(n,(E,) signifies that, for each p E L, the set h,,(X, n A,,) is open in E,. Now X, n A,, is open in A,, , and h,, is a homeomorphism of A,, onto an open subset of E, . Hence (a) implies (b). The topological space E so defined is said to be obtained by patching together the E, along the A,, by means of the h,, .

6

XI1 TOPOLOGY AND TOPOLOGICAL ALGEBRA

3. HAUSDORFF SPACES

The notion of a limit is defined as in (3.13) for arbitrary topological spaces, and the criterion (3.13.1)is merely another formulation of the definition. However, the conclusion of (3.13.3)is no longer necessarily true: for example, if E is a set endowed with the chaotic topology (12.1.1),every sequence o f points of E has every point of E as a limit. A topological space E is said to be Hausdorff, and its topology is said to be a Hausdorf topology, if it satisfies the following " Hausdorff axiom":

Given any two distinct points a, b in E, there exists a neighborhood U of a and a neighborhood V of b which do not intersect. Every metrizable space is Hausdorff

(12.3.1) Let E be a topological space, A a subset of E, and let a be a cluster point of A. Then a mapping f of A into a Hausdorff space E ' has at most one limit at the point a with respect to A. For if a', b' were two distinct limits off at the point a, there would exist neighborhoods U', V' of a', b', respectively, having no point in common. But by hypothesis there would be a neighborhood W of a in E such that f ( W n A) c U' and f (W n A) c V', and this is absurd because W nA # 0. Hence we may continue to use the notation

lim x E A ,x

f ( x ) to denote the +

a

unique limit o f f a t the point a. The propositions of (3.13) which do not refer to distances will remain valid for mappings (in particular for sequences) from an arbitrary topological space to a Hausdorf space, with the exception of (3.13.13)and (3.13.14).

(12.3.2) Every subspace of a Hausdorfspace is Hausdorf. (12.3.3) Every topology which isfiner than a Hausdor# topol5gy is Hausdorff. These are immediate consequences of the definitions.

(12.3.4) In a Hausdorff space E, every finite subset is closed. For if ( a i ) l s i s nis a finite sequence of points in E, a point b distinct from all the a, cannot be a cluster point of the set of the a i , because for each i there

3 HAUSDORFF SPACES

exists a neighborhood V i of b which does not contain a i , and V = then a neighborhood of b which does not contain any of the ai .

7

n Vi is n

i= 1

(12.3.5) LetS, g be two continuous mappings of a topological space E into a Hausdorffspace E’. Then the set A ofpoints x E E such that f ( x )= g(x) is closed in E.

The proof is similar to that of (3.15.1) (which is a special case of (12.3.5)). If a # A, then f ( a ) # g(a), and so there are disjoint neighborhoods U’ of f ( a ) and V’ of g(a). Since f -‘(U’) and f -‘(V’) are neighborhoods of a in E, the same is true of their intersection W , and it is clear that f ( x ) # g(x) for all x E W . Hence E - A is open. The principle of extension of identities (3.1 5.2) therefore remains true for continuous mappings of any topological space into a Hausdorfs space. Proposition (3.1 5.3) and its corollary (3.15.4) (the “principle of extension of inequalities ”) are also valid-the proofs are the same-for arbitrary topological spaces. (12.3.6) Let E be a compact metrizable space, F a Hausdorff space, and f a continuous mapping of E into F . Then f ( E ) is closed in F. I f f is injective, it is a homeomorphism of E onto the subspace f ( E ) of F.

Let y be a point of the complement off (E). For each z € f ( E ) ,there exist disjoint open neighborhoods V ( z ) of z and W ( z ) of y . The inverse images f - ’ ( V ( z ) ) form an open covering of E as z runs through f ( E ) (3.11.4). Hence by compactness there exists a finite number of points zi Ef ( E ) such that the V ( z i )form an open covering of f ( E ) in F. But then U = W ( z i )is an open i

neighborhood of y which does not intersect f ( E ) . This shows that f ( E ) is closed in F. It follows that if A is any closed subset of E, thenf(A) is closed in F (and therefore also in f ( E ) ) , because A is compact (3.17.3). Iff is injective, this establishes the continuity of the inverse mapping f (E) .+ E (3.11 -4). (12.3.7) I f a Hausdorff topology is coarser than the topology of a compact metrizable space, then the two topologies coincide. (12.3.8) Let E be a compact metrizable space, F a HausdorfSspace, f a continuous mapping of E into F . If b is any point o f f ( E ) and U is any open set in E containing f -‘(b), then there exists a neighborhood V of b in F such that f-yv) c

u.

8

XI1 TOPOLOGY AND TOPOLOGICAL ALGEBRA

For E - U is closed in E, hencef(E - U) is closed in F by (12.3.6). We have b #f(E - U) by definition, hence the complement V off(E - U) in F is an open neighborhood of b, and clearly U xf-'(V).

PROBLEMS

1. Find all possible topologies on a set of 2 or 3 elements. 2.

Let 5 be a set of subsets of a set E. Show that 5 is the set of closed sets for a topology on E if and only if 5 satisfies the following two conditions: (1) every intersection of a family of sets belonging to 8 belongs to 8 ; (2) the empty set belongs to 5, and the union of two sets belonging to 8 belongs to 8.

3. Let E be a set and for each x E E let B(x) be a set of subsets of E. Then there exists a topology F on E such that, for all x E E, B(x) is the set of neighborhoods of x with respect to Y, if and only if the sets B(x) satisfy the following conditions: (V,) Every subset of E which contains a set belonging to B(x) belongs to B(x). (V,,) The intersection of two sets belonging to %(x) belongs to B(x). (VIII) For all x E E and all V E B(x), we have x E V. (VIv) For all x B E and all V E B(x), there exists a set W E B(x) such that V E B(y) for all y E W. The topology Y is then unique. 4.

Let A be a commutative ring with an identity element 1 # 0. An ideal p of A is prime if A/@is an integral domain (and therefore #O). The set of prime ideals of A is called the spectrum of A and is denoted by Spec(A) (it can be shown to be nonempty). If a is an ideal in A, the set of prime ideals p 2 a is denoted by V(a). Show that V(a n b) = V(a) u V(b) for any two ideals a, b in A. Deduce that the subsets V(a) of Spec(A) are the closed sets in a topology, called the spectral topology, on Spec(A). If x , y are two distinct points of Spec(A), show that either there is a neighborhood of x which does not contain y, or else there is a neighborhood of y which does not contain x. Under what conditions is a set ( x } consisting of a single point closed in Spec(A)? Consider the case where A = Z, and the case where A is a discrete valuation ring. Let A' be another ring and let h: A+A' be a ring homomorphism such that h(1) = 1. Show that the mapping p'wh-'(@') of Spec(A') into Spec(A) is continuous with respect to the spectral topologies.

5. (a) The following conditions on a nonempty topological space E are equivalent: (1) the intersection of any two nonempty open sets in E is nonempty; (2) every nonempty open set is dense in E; (3) every open set in E is connected. The space E is then said to be irreducible. A nonempty subset F of a topological space E is said to be an irreducible set if the subspace F is irreducible. (b) Show that in a Hausdorff space every irreducible set consists of a single point.

3 HAUSDORFF SPACES

9

(c) In a topological space E, a subset F is irreducible if and only if its closure F is irreducible. In particular, for each x E E, the set is irreducible. If an irreducible then x is said to be a generic point of F. set F in E is of the form (d) Let A be an integral domain. Show that the topological space Spec(A) (Problem 4) is irreducible and that {0}is its unique generic point. In Spec(Z), show that the complement of the generic point is an irreducible set which has no generic point. (e) If E is an irreducible space, every nonempty open set in E is irreducible. (f) Let (Urn),A be an open covering of a topological space E, such that U. n UB# 0 for all pairs of indices (a,6). Show that if the sets U, are irreducible, then E is irreducible. (g) Let E, F be two topological spaces and letfbe a continuous mapping of E into F. If A is an irreducible subset of E, show thatf(A) is an irreducible subset of F. (h) Let E, F be two irreducible spaces, each of which has at least one generic point. Suppose also that F has a unique generic point b. Let f be a continuous mapping of E into F. Show thatf(E) is dense in F if and only iff(x) = b for each generic point x of E.

E,

a

6. A topological space E is said to be quasi-compact if it satisfies the Borel-Lebesgue axiom (3.16) and compact if it is quasi-compact and Hausdorff. A subset F of a topo-

logical space E is said to be a quasi-compact set (resp. a compact set) if the subspace F of E is quasi-compact (resp. compact). Every finite set is quasi-compact. (a) In a quasi-compact space, every closed set is quasicompact. (b) Let E be a Hausdorff space and let A, B be two disjoint compact sets in E. Show that there exist two disjoint open sets U, V, such that A c U and B c V. (Consider first the case where A consists of a single point.) Deduce that a compact set in a HauSdorff space is closed. (c) In a topological space E, any finite union of quasi-compact sets is quasi-compact. (d) If E is a quasi-compact space andf: E + F is a continuous mapping, then the set f(E) is quasi-compact. (e) A filter base on a set E is a set 8 c b(E) of nonempty subsets of E such that, whenever X and Y belong to 8, there exists 2 E 8 such that Z c X n Y. Show that if E is a quasi-compact space and all the sets of 8 are closed, then the intersection of the sets of B is nonempty (consider the complements of the sets of 8, and argue by contradiction). (f) If A is a commutative ring with an identity element 1 # 0, show that the topological space Spec(A) (Problem 4) is quasi-compact. (g) Let E be the union of the interval 10, I ] and two elements a, 6. Let 8 be the set of finite intersections of sets of the forms ]a, I], {a}u 10, a [ , @ } u 10, a [ , where 0 < a < 1 . Show that 8 is a basis of a non-Hausdorff topology on E, for which E is quasi-compact and every set consisting of a single point is closed in E. Every point of E has a compact metrizable neighborhood, but there are such neighborhoods of CL (or 8) which are not closed in E. Show that the intersection of a compact metrizable neighborhood of a and a compact metrizable neighborhood of is not quasi-compact. The topology of E has a denumerable basis consisting of separable metrizable subspaces. (h) Let E be the sum (1.8) of N and an infinite set A. For each x E N, write Q(x) = ( x } . For each x E A, let Sx+ denote the union of { X I and the set of integers z n , and let G(x) denote the set of sets Sx, as n runs through N. Show that there exists a non-Hausdorff topology on E such that G(x) is a fundamental system of neighborhoods of x , for each x E E. In this topology, every set consisting of a single point is closed, and E has a dense compact subspace, although E itself is not quasi-compact.

10

XI1 TOPOLOGY AND TOPOLOGICAL ALGEBRA

4. UNIFORMIZABLE SPACES

A pseudo-distance on a set E is a mapping d of E x E into R which is such that d(x, x ) = 0 for all x E E, and satisfies axioms (I), (HI), and (IV) of (3.1), but not necessarily axiom (11). It is clear that a pseudo-distance satisfies the inequalities 4 x 1 , xn) 5 4 x 1 ,

~

2

+) 4 x 2

3

~

14x9 2 ) - d(y, 41

3

+ )

* * *

+

s 4x9 Y )

4 X n - 1 , Xn),

for all x i and x , y , z in E. Examples (12.4.1)

Iffis any mapping of a set E into R,the mapping ( x , r)b IfW - f(Y>I

is a pseudo-distance. Let cp be a mapping of the interval [0, co[ into itself which satisfies the following three conditions : (1) cp(0) = 0 ; (2) cp is increasing; (3) cp(u v ) 5 cp(u) cp(u) for all u >= 0 and v 2 0. Then if d is a pseudo-distance on a set E, the same is true of the composite mapping cp d : ( x , y ) ~ c p ( d ( xy ,) ) . We have only to check the triangle inequality, and by conditions (2) and (3) we have

+

+

+

0

cp(d(x, 4) 5

cp(d(X,

Y ) + d(Y, 4) 5 cp(&

Y>)+ cp(d(x, 2)).

If (d,) is a sequence of pseudo-distances on a set E, such that the series E E x E, then its sum d(x, y ) is a pseudo-

1dn(x,y ) converges for all ( x , y ) n

distance on E (3.15.4). Now consider a family (da)aE , of pseudo-distances on a set E. For each a E E, each finite family ( a j ) l j j m of elements of I and each finite family (rj)l s j s m of real numbers >O, put

B(a; (aj), (r,)) = { x E E I daj(a,x ) < r j for 1 5 j 5 m}, B’(a; ( a j ) ,( r j ) ) = { x E E I da,(a, x) S r j for 1 S j 5 m}. Let r) denote the set of subsets U of E such that, for each x E U, there exists a finite family ( c t j ) , S j s m of elements of I and a finite family ( r i ) l s j S m of strictly positive real numbers such that B(x; (mj),( r j ) ) c U. It is immediately verified that D is a fopology on E. This topology is said to be defined by the

4 UNIFORMIZABLE SPACES

11

family of pseudo-distances (d,), I . A topology which can be defined by a family of pseudo-distances is said to be uniformizable, and a space equipped with such a topology is said to be uniformizable. It is obvious that a metrizable space is uniformizable, but the converse is false (for example, the family consisting only of the pseudo-distance d = 0 defines the chaotic topology). There exist Hausdorff topologies which are not uniformizable (Problems 3 and 4); but throughout this book (and in the majority of situations in analysis) uniformizable spaces are the only ones which we shall have to consider. (12.4.2) Let E be a uniJormizable space whose topology is defined by a family of pseudo-distances (d,), E,. Then the sets

B(a; ( a j ) , (rj))

(resp. B‘(a;( a j ) , (rj)))

are open (resp. closed) in E.

Let x E B(a; (aj), (rj)),and put sj = d,,(a, x ) . Then sj c r j for 1 s j 5 m, hence B(a; ( a j ) , ( r j ) ) contains the set B(x; ( a j ) , ( r j - s j ) ) by virtue of the triangle inequality for each of the d., . This shows that B(a; (aj), ( r j ) ) is an open set. If x 4 B‘(a;(aj), ( T i ) ) , then there exists an index k such that 1 k g m and d,,(a, x ) = s, > r, , and hence the triangle inequality for d,, shows that the set B(x; ak , s, - r,) does not meet B’(a; (aj), (rj)). This shows that B’(a;(aj),(rj)) is closed.

s

(1 2.4.3) In a uniformizable space, the closed neighborhoods of a point form a fundamental system of neighborhoods of the point.

This follows from (12.4.2) and the fact that

B’(a;

( a j ) , (rj/2))c

B(a; ( a j ) , (rj))*

(12.4.4) Let E be a uniformizable space and let (d,), I be a family of pseudodistances defining the topology of E. Then E is Hausdorfifand only $for each pair of distinct points x, y in E , there exists p E I such that d,(x, y ) # 0.

If E is Hausdorff, then by definition there exists a set B ( x ; (aj), ( r j ) ) not containing y , and therefore for at least one indexj we have d,,(x, y ) 2 rj > 0. Conversely, if d,(x, y ) = t > 0, then the open neighborhoods B(x; p, t / 2 ) and B(y; p, t/2) of x and y , respectively, are disjoint, by virtue of the triangle inequality for d, .

I f (d,) is a family of pseudo-distances defining a Hausdorff topology on a set E, then to say that a sequence (x,,)of points of E has a point a as limit with respect to this topology means that lim d,(a, x,,) = 0 for all LY. n-rm

12

XI1 TOPOLOGY AND TOPOLOGICAL ALGEBRA

Two families of pseudo-distances (dJ, (di)A on the same set E are said to be topologically equivalent if they define the same topology on E. (12.4.5) If ( d J U E I is any family of pseudo-distances on a set E , there exists a topologically equivalent family (d;), E , such that 0 5 dl 5 1 for all 01 E I .

+

Let cp(u) = inf(u, 1) for 0 5 u < co.Then cp(0) = 0, and cp is an increasing function on [0, + co[ and satisfies the inequality

+ cp(4.

cp(u + 0) 5

=+

+

(This last assertion is clear if u 1 or u > 1 or if u v 5 1 ; and if u 5 1 and v 5 1 and u v > 1 we have cp(u v ) = 1 < cp(u) + cp(v).) It follows rlow from (12.4.1) that di = cp d, is a pseudo-distance on E, for each 01 E 1. That the topologies defined by the families (d,) and (d;) are the same follows from (12.2.1) and the fact that, when 0 < r < 1, the relations d,(x, y ) < r and di(x, y ) < r are equivalent.

+

0

(1 2.4.6) A Hausdorfluniformizable space E , whose topology can be deJined by a sequence (d,,) of pseudo-distances, is metrizable.

Without loss of generality we may assume that the sequence (d,) is infinite, and by (12.4.5) that 0 5 d,, 5 1 for all n. Then the series (12.4.6.1)

1 d(x, y ) = 2 di(x, y )

1

1

+ 27 d,(x, y ) + + 2. d,,(x, y ) + * *

* * *

converges for all pairs (x, y ) of elements of E, and d is a pseudo-distance on E (12.4.1). Moreover, d is a distance, because d(x, y ) = 0 implies that d,,(x, y ) = 0 for all n, and therefore that x = y because E is Hausdorff (12.4.4). If Bo(x;r ) is the open ball with center x and radius r with respect to the distance d, then it follows immediately from (12.4.6.1) that B,(x; r ) c B(x;n,2"r) for all n. Conversely, if n is so large that 2"-' 2 l/r, then we have

C 2-"-k k= 1 W

d,,+k(x,y ) 5 2-" 5 +r

for all x , y in E , and therefore

B(x; (192, . .

9

4,( t r , . . ,h))= B o b , r). *

By (12.2.1), the proof is complete. (12.4.7) Let E be a topological space and (U,) aJinite or denumerable open covering of E such that the subspaces 0, are separable and metrizable. Then E is separable and metrizable.

4 UNIFORMIZABLE SPACES

13

We begin by showing that the topology of E is Hausdorf. Let x , y be two distinct points of E. If there exists an index n such that x E U, and y 4 On, then V = U, and W = E - 0, are open neighborhoods of x and y respectively which do not intersect. If on the other hand x E U, and y E 0, for some n, then in the metrizable subspace 0, there exists an open neighborhood V, of x and an open neighborhood W, of y which do not intersect. The set V = V, n U, is an open neighborhood of x in E, and there exists an open set W in E such that W n 0, = W, . Hence W is an open neighborhood of y in E, and VnW=@. For each n, let d, be a distance defining the topology of O n ,and let (Vrn,JmLbe a basis for the topology of U, , where V,, is an open ball with center a,, and radius r,, (3.9.4). Let A,,,, be the function which is equal to rmn- d,(a,,, x ) in V,, and is 0 in the complement of V,,,, in E. Then.f,, is continuous on E, since it vanishes at the frontier points of V,, . Now let &n(X,

Y ) = Ifmn(x) -fmn(Y)l

for all x, y in E. By virtue of (12.4.6), the proof will be complete if we show that the pseudo-distances d,,, define the topology of E. For each x , E E , every set defined by an inequality of the form d,,(xo, x ) < CI is open in E (3.11.4). Conversely, there exists an integer n such that x , E U, , and for each neighborhood W of x , in E there exists an integer m such that V,, is an open neighborhood of xo contained in W (3.9.3). Hence fmn(xo)= B > 0, and the set of all x E E such that d,,(xo, x ) < +B is therefore contained in W. By virtue of (12.2.1), this completes the proof. We remark that the conclusion of (12.4.7) is not valid without the hypothesis of denumerability on the open covering (U,) (Section 12.16, Problem 22); again, the hypotheses cannot be weakened by assuming merely that the open sets U, are separable and metrizable (section 12.3, Problem 6(f)).

PROBLEMS

1. Let E be a topological space such that, for each x E E, the neighborhoods of x which are both open and closed in E form a fundamental system of neighborhoods of x. Show that E is uniformizable.(Observe that the characteristicfunction of a set which is both open and closed in E is continuous on E.)

2. Let E be a denumerable Hausdorff uniformizable space. Show that, for each x E E, the open-and-closed neighborhoods of x form. a fundamental system of neighborhoods of x.

14

XI1 TOPOLOGY AND TOPOLOGICAL ALGEBRA

3. Let 6 be a positive irrational number. For each point (x, y ) E Q x Q+ (where Q+ = Q n R,) and each integer n > 0, let B.(x, y ) denote the set consisting of (x, y ) and the points (z, 0) E Q x Q + such that Iz - (x 6y)l < I/n or Iz - (x - 6y)l < I/n. As (x, y ) runs through Q x Q ,and n runs through the set of strictly positive integers, show that the sets B.(x, y ) form a (denumerable) basis of a topology .T on Q x Q + Show that .T is Hausdorff but that, if a and b are any two points of Q x Q + , every closed neighborhood of a meets every closed neighborhood of 6. Deduce that Q x Q ,, endowed with the topology .T, is connected and that every continuous mapping of Q x Q, into R is constant. This shows that the topology .T is not uniformizable.

+

.

4.

Let Q be the set of rational numbers endowed with the topology I .induced by that of R. Let !Dl be the set of subsets A of Q such that the closure A of A (in the topology .To)has only finitely many nonisolated points (3.10.10). Let B be the set consisting of all open intervals in Q, all complements of sets belonging to Fm and all intersections of these complements with open intervals in Q. Show that B is a basis for a topology F on Q, and that .T is finer than .To and therefore Hausdorff. Show that, with respect to this topology, a convergent sequence in Q has only finitely many distinct terms, although the topology I is not discrete. In the topology I , no point of Q has a denumerable fundamental system of neighborhoods, although Q is denumerable and every point of Q is the intersection of a denumerable family of neighborhoods of the point. By using (1 2.4.3), show that I is not uniformizable. although it is finer than a metrizable topology.

5. Let E be a topological space satisfying the following condition: for each x ‘EE and each neighborhood V of x in E, there exists a continuous mapping f: E -+ [O,11 such that f(x) = I and f(y) = 0 for all y E E - V. Show that E is uniformizable. Deduce that if E is a topological space such that every point of E has a closed neighborhood in E which is a uniformizable subspace, then E is uniformizable. Can the word “closed” be omitted from this proposition? (Cf. Section 12.3, Problem 6(f).) 6.

For each x E R and each integer n > 0, let U.(x) denote the union of the intervals [x, x l/n[ and ] - x - I/n, -XI.Show that there exists a topology 9on R such that the U.(x) form a fundamental system of neighborhoods of x, and that .T is not comparable with the usual topology of R. Show that I is uniformizable (use Problem 5). In this topology, every point has a denumerable fundamental system of neighborhoods, and there exists a denumerable dense subset. But there exists no denumerable basis of open sets for I and , consequently I is not metrizable. Show also that the topology induced by .T on the interval E = [ - I , 11 of R is not metrizable, although E is quasicompact and Hausdorff in this topology.

+

5. PRODUCTS O F U N I F O R M I Z A B L E SPACES

Let El, E, be two topological spaces. In the product set E = El x E, let 0 be the set of (arbitrary) unions of sets of the form Al x A , , where A, is open in El and A, is open in E, . The set 0 is a topology on E, because it clearly satisfies axiom (Of), and it satisfies (0,Jby reason of the relation

5

PRODUCTS OF UNIFORMIZABLE SPACES

15

where Ai and B, are open sets in E i ( i = 1,2). This topology 0 is called the product of the topologies on El and E,, and the set E endowed with this topology is called the product of the topological spaces El, E, . If x = ( x , , x,) is any point of E, the sets V, x V, (where V i runs through a fundamental system of neighborhoods of xi, f0r.i = 1, 2) form a fundamental system of neighborhoods of x in E (3.9.3). From this it follows that the continuity criterion (3.20.4)is valid for the product of two arbitrary topological spaces. The relation A, x A, = A, x A, (where A, c El and A, c E,) also remains true; for if (a,, a,) E E and if Vi is a neighborhood of a, in Ei (i = 1,2), the set (V, x V,) n (A, x A,) = (V, n A,) x (V, n A,) is empty if and only if one of the sets V in Ai is empty. For each a, E El, the set ((a,} x E2) n (A, x A2) is either empty or equal to {a,} x A , , and therefore the mapping x 2 ++(a,, x 2 ) is a homeomorphism of E, onto the subspace (a,} x E, of E. Likewise, if a2 E E,, the mapping x,H(x,, a2) is a homeomorphism of El onto the subspace El x { a 2 }of E. From this it follows immediately that the propositions (3.20.12)and (3.20.13), and the continuity criteria (3.20.14)and (3.20.15),are valid in general. If El and E, are Hausdorf then so is El x E, . For if x = (x,, x 2 ) and y = (yl, y 2 )are distinct, then either x, # y , or x, # y , . If the latter, there is a neighborhood U of x2 and a neighborhood V of y , in E, which do not intersect, and then El x U and El x V are disjoint neighborhoods of x and y , respectively, in E. IfE, = El, the canonicalsymmetry (x,, x , ) ~ (x, , xl) is a homeomorphism of El x El onto itself, and is equal to its inverse. The product of any finite number of topological spaces is defined in the same way. It follows immediately from the definition that the canonical “associativity” mappings such as (El x E,) x E, 4 El x (E, x E,) are homeomorphisms. We shall be concerned especially with the case in which El and E, are uniformizable. In this case the product space E is also uniformizable. To be precise, let (d$l))LL, ( d r ) ) r be families of pseudo-distances which define the topologies of’ El, E, respectively, and let

Then the el1) and )e: are pseudo-distances on E, and the definition of neighborhoods given above shows that the family of pseudo-distances ey) and ff), where I E L and p E M, defines the product topology on E. This leads to a generalization of the notion of a product to arbitrary (not necessarily finite) families of topological spaces. We shall restrict ourselves to uniformizable spaces.

16

XI1 TOPOLOGY AND TOPOLOGICAL ALGEBRA

Let (E,),

I

be any family of uniformizable spaces, and let E =

For each a E I, let (d,, topology of E, . If we put

fl E, .

,El

be a family of pseudo-distances defining the

eu, Ax, Y ) = d,, n(P'uX7 PlhY)

(12.5.1)

for each pair (x, y ) of elements of E =

nE,, then it is immediately checked

. E l

that the e,, are pseudo-distances on E. The product of the topologies of the E, is then the topology on E defined by the e,, as a runs through I and, for each a E I, Iz runs through L, . The set E endowed with this topology is called the product of the spaces E, . When I = { 1,2} this definition agrees with that given above for two spaces El, E, . of For every finite family ( a i , Ai), and every finite family (ri)l strictly positive real numbers, the set B(x; ( ( a i ,Ai)), (ri)) can be written in B, where, if a = ai, we have B, = B(pr,, x ; (Aj), (rj)),where j the form

n

UEI

runs through the indices for which aj = t l i , and B, = E, if a is not equal to any ai. From this and (3.6.4) we deduce immediately: (12.5.2) For each a E I, theprojection pr, is a continuous mappingof E onto E, , and the image under pr, of any open set in E is an open set in E, .

(,VH )

(12.5.3) For each finite subset H of I, and each family (UJUEH,where U, is U, x ( , B H E . ) is open in E. open in E, for each a E H, the set

For it is the intersection of the sets pr;'(U,), are open (12.5.2) and finite in number.

where a E H, and these sets

The sets described in (12.5.3) are called elementary sets in E. The description given above of the neighborhoods shows that (having regard to (3.9.3)) these sets form a basis for the topology of E. Incidentally, this shows also that the topology of E depends only on the topologies of the E, and not on the families of pseudo-distances (d,, Moreover, if b, is a basis for the topology of E,, the elementary sets such that U, E b, for all a form a basis for the U, with topology of E. It should be noted that, if I is infinite, a product

n

u61

U, open in E,, nonempty and # E, for a f fc1 E I, is not open in E. (12.5.4) Let A, be a subset of E,, for each a, and let A

A=

=

n A,.

Then

n A,. In particular, $each A, is closed in E, , then A is closed in E.

,El

QCI

5

PRODUCTS OF UNIFORMIZABLE SPACES

17

We have pr,(A) c pr,(A) = A, for all u E I ((12.5.2) and (3.11.4)), hence A,. Conversely, if a = (a,) E A,, and if we consider an ele-

Ac

n

n

as1

mentary set U =

n U, n E, containing a, its intersection with A is n n A,. as1

x

aeH

asl-H

aeH

(Aan U a ) x

asl-H

Since none of the A, is empty, no A, is empty and therefore U n A # which shows that a E A.

a,

(12.5.5) Let z~ f ( z ) = (f,(z)), I be a mapping of a topological space F into E. For f to be continuous at a point z, E F , it is necessary and sufficient that each f, should be continuous at zo .

For any elementary set U =

n f,-1(uU).

n U, n E,,

asH

x

asl-H

we have f-'(U) =

aeH

(12.5.6) Let (x("))be a sequence of points of E. For a = (a,) to be a limit of the sequence (x'")), it is necessary and suficient that a, should be a limit of the sequence (prax(n)),,tl, - for each CI E I .

This is an immediate consequence of (12.5.5) and the definition of a limit. In particular, if all the E, are equal to the same space F, then to say that a mapping u E F' of I into F is a limit of a sequence of mappings ( u , , ) , , of ~ ~I into F, with respect to the product topology, means that for each c1 E I the sequence (u,,(M))has u(a) as a limit in F. In this situation we say that the mapping u is a simple limit of the sequence (u,,), or that the latter converges simply to u. (12.5.7) v e a c h space E, is Hausdorfl, then so is the product space

E.

For if x # y , there exists an index u E I such that prax # p r a y ; hence, using (12.4.4), there exists A E La such that d,, I(pr,x, p r a y ) # 0, and therefore e,,,(x, y ) # 0. The result now follows from (12.4.4). (12.5.8) The product E of a denumerable family of metrizable (resp. separable metrizable) spaces is metrizable (resp. separable and metrizable).

First of all, E is Hausdorff by (12.5.7). To show that E is metrizable, apply (12.4.6) and the definition of pseudo-distances on a product. Now suppose that E =

fr En,where the Enare separable and metrizable. Then each

n= 1

18

XI1 TOPOLOGY A N D TOPOLOGICAL ALGEBRA

Enhas a denumerable basis (Um,JmLO of open sets. For each n

n uf(j), n n

the set of elementary sets in E of the form

x

a

k=n+ 1

j= 1

1, let 23, be Ek, where f is

any mapping of { 1, 2, . . . , n} into N. Since N" is denumerable (1.9.3), the set 23, is at most denumerable (1.9.2). We have seen that the union 23 of the sets 23, is a basis for the topology of E (12.5.3); since $'3 is denumerable (1.9.4), it follows that E is separable (3.9.4). The product of a denumerablefamily of compact metrizable spaces is compact and metrizable. (1 2.5.9)

Let (En) be a sequence of compact metric spaces. Since we know already (12.5.8) that E =

m

fl E,

is metrizable, it is enough to show that every

,= 1

sequence (x,) of points of E has a cluster value (3.16.1). We define by induction on m 2 0 a family of sequences ( x ? ) ) , , ~ of ~ points of E, as follows: x:)' = x,; for m 2 1, the sequence ( x : ~ ) ) , ,is-a ~ ~ subsequence of the sequence (xp- I)),, (in other words, there exists a strictly increasing mapping q m : N + N such that xkm)= x$",;nf)) such that the sequence ( p r m x ~ ) ) n z l converges in Em to a point a,,,. This construction is possible because Em is compact. Now consider the sequence (y,) in E, where y, = xf) (Cantor's q l , we have y , = x,Jn); "diagonal trick"). If we put $,, = (P, q n - l 0 since $,(n) > $n-l(n - 1) by definition, it follows that (y,) is a subsequence of (x,). Furthermore, for each m, the sequence (y,Jn2, is a subsequence of the sequence ( x : ~ ) ) , , ~and ~ , therefore the sequence (prmy,JnLm converges to a,". Hence so does the sequence (prmy,),;L,, since it differs from the former by only a finite number of terms. Hence the sequence (y,) converges to the point a = (a,,,) (12.5.6). 0

0

PROBLEMS

1. Show that the product of two quasi-compact topological spaces (Section 12.3, Problem 6) is quasi-compact. 2. Show that a topological space E is Hausdorff if and only if the diagonal (1.4.2) of E X E is closed in E x E.

3. Let (Ea)as, be a family of arbitrary (not necessarily uniformizable) topological spaces, and let E = E.. Show that the elementary sets, defined as in (12.5.3), form a basis

n

OlEl

of a topology on E. This topology is called the producr of the topologies of the E, Generalize (12.5.2) to (12.5.7) to this situation.

.

5 PRODUCTS OF UNIFORMIZABLE SPACES 4.

Let E =

n E.

19

be a product of Hausdorff topological spaces, such that each E.

= E l

contains at least two distinct points a , , 6 , . For each (1: E I, let c. be the point of E such that pr.c. = b. and prac. = aa for all /3 # a. Show that every point of the set { c ~ } is~ an ~ , isolated point. Deduce that the topology of E has a denumerable basis if and only I is denumerable and each of the E. has a denumerable basis. Show that if I is not denumerable, the point a = (a,) has no denumerable fundamental system of neighborhoods. If E. = {arn, ba}for each a, and if F c E is the set of all x t E such that pr,x = b. for all but denumerably many indices a, show that F is dense in E and that a = (an)is not a limit of any sequence of elements of F.* 5. Let K be the discrete space {0, l}, let A be an infinite set, and let E be the product space KA. Let V be a nonempty elementary set in E. If h is the (finite) number of indices o! such that pr,V # K, let p(V) = 2 - h (cf. (13.21)). (a) Show that, if U,, . . . , U. are nonempty mutually disjoint elementary sets, then

"

&=1

p(Uk)5 1. (Write Urin the form Wx x KB,where B is the same for all k and is the

complement of a finite subset of A.) (b) Deduce from (a) that if (U,A6 is a family of nonempty pairwise disjoint open sets in E, then L is at most denumerable, although there exist sets C of arbitrary cardinal in E (for a suitable choice of A) all of whose points are isolated. Show also that if the cardinal of A is strictly greater than that of B(N), then E contains no denumerable dense subset.

6. With the notation of Problem 5, let 9 be the set of all subsets of KA of the form M a , where M. = K except for ar most denumerably many indices or. Show that 8

n

.EA

is a basis for a Hausdorff topology on KA which is finer than the product topology, and which is not discrete provided that A is infinite and non-denumerable.In this topology, every denumerable intersection of open sets is open; no point has a denumerable fundamental system of neighborhoods; and every quasi-compact subset is finite. The projections pr. are continuous with respect to this topology. Deduce that KA is uniformizable with respect to this topology (Section 12.4, Problem 5).

7. Let 1 be the interval [0, 11 in R,with the induced topology. Show that every separable metrizable space E is homeomorphic to a subspace of the product IN. (Reduce to the case where the distance d on E is 5 1, and consider a sequence (0.) which is dense in E, and the functions x H d ( a . , x).) 8. With the notation of Problem 7, show that in the uniformizable product space I', the subspace of continuous mappings of I into I is dense. Deduce that I' has a denumerable dense subset, although there is no denumerable basis of open sets (Problem 4). 9.

Show that if I is a nonempty open interval in R,there exists no nonconstant mapping f o f I into the product space NN with the following property: for each x E I and each

*This example shows that in general topological spaces (and even in Hausdorff uniformizable spaces) the convergent sequences do not determine the topology, as they do in metrizable spaces (cf. (3.1 3.3) and (3.1 3.4)). To get corresponding results in general, it is necessary to replace the notion of sequence by the more general notion of a filter (cf. [5]).

20

XI1 TOPOLOGY AND TOPOLOGICAL ALGEBRA

integer n > 0, there exists a neighborhood V of x in I such that, for all y E V, the first n terms of the sequence f(y) are the same as the first n terms of the sequence f ( x ) . (Show that this condition implies that f i s continuous, and use (3.19.7).) 10. (a) Let E be a topological space and A a nonempty closed subset of E. Let E’ be the sum of E - A and a set { w } consisting of a single element, and let 0’be the set of subsets of E’ which are either of the form U where U is an open set in E - A, or of the form (V - A) u { w } , where V is an open set in E which contains A. Show that 0’is a topology on E’, and denote by E/A the set E’ endowed with this topology. Let p(x) = x if x E E - A, and ~ ( x = ) w if x E A. Show that y is a continuous mapping of E onto E/A. Every continuous mapping f of E into a topological space F, which is constant on A, can be written uniquely in the form f = g 0 p, where g: E/A -+ F is continuous. (b) Suppose that E is metrizable and compact; let d be a distance defining the topology of E, and let (Wn)nblbe a basis for the topology of E - A. Let f o ( x )= d(x, A), f.(x) = d(x, E - W,) for n 2 1, and f ( x ) = ( f , ( ~ ) ) e. R~ N~ . Show that f(E) is a compact subspact: of RN, homeomorphic to E/A. This shows that E/A is compact and metrizable. (c) Show that if we take E = R and A = Z, then the space E/A defined in (a) is not metrizable (cf. Section 3.6, Problem).

6. LOCALLY FINITE COVERINGS A N D PARTITIONS O F U N I T Y

In a topological space E, a family (A,),., of subsets of E is said to be locallyfinite if for each point x E E there exists a neighborhood U of x in E such that U n A, = $3 for all but a finite number of indices c1 E I . If E is metrizable and K is a compact subset of E, it follows that there is a covering of K by afinite number of neighborhoods of points of K in E, each of which meets only afinite number of the sets A,. In particular, K meets only afinite number of the A,. If ( A J L E L ,(B,),EM are two coverings of a topological space E, the covering (B,) is said to be finer than (A,) if, for each p E M, there exists A E L such that B, c A,. (12.6.1) Let E be a separable, locally compact, metrizable space and let 23 be.a basis of open sets in E. If (A,),. is any open covering of E, there exists a denumerable locallyfinite open covering (B,) of E which isfiner than (A,), and such that the sets B, are relatively compact and belonq to 23. Consequently each B, meets only finitely many of the sets B, .

From (3.18.3) we know that there exists an increasing sequence (U,JnLO of relatively compact open sets such that 0, c U,,, for all n, and E = U,.

u n

6

LOCALLY FINITE COVERINGS AND PARTITIONS OF UNITY

21

Put K, = 0, - Un-l, which is a compact set for all n 2 0 (we put U,, = for all n < 0). For each n 2 0, the open set U,+, IS a neighborhood of K, . Hence, for each x E K,, there exists an open neighborhood W v ) E 23 of x, contained in Un+l- On-2and contained in one of the sets A,. There exists a finite number of points x i E K,, (I 5 i 5 p,) such that the sets WE) cover K, . Arrange the sets WLy) (m 2 0, 1 5 i 5 p m for each m) in a sequence in any way, and let (B,) denote this sequence. It is clear that (B,) is an open covering of E which is finer than (A,), L , and that the sets B, are relatively compact. Thus all that has to be checked is that the covering (B,) is locally finite. Let z be any point of E, and let n be the least integer ~ U hX that z E U, . Since z 4 U,-l, there exists a neighborhood T of z contained in U, and not meeting Consequently T can intersect only the sets W:) for which n - 2 m 5 n + 1, and the number of these sets is finite. (12.6.2) Let (A,,) be a denumerable locally$nite open covering of a metrizable space E. Then there exists an open covering (B,) of E such that B, c A, for all n.

B,

We shall define the family (B,) by induction on n, in such a way that

c A,, for each n, and such that for each n the family consisting of the B, with k 5 n and the A j with j > n is an open covering of E. Suppose that the

sets B, have been defined for n < m. Then the B, with n < m and the A j with

j 2 m cover E. Let C be the open set which is the union of the B, with n < m

and the Aj withj > m + I , so that we have E - A, c C. We shall show that there exists an open set V such that E - A,, c V c 0 c C. If E = A,, we may take V = 0. If C = E, we may take V = E. If neither E - A,, nor E - C is empty, there exists a continuous mapping f of E into the interval 10, 11of R,which is equal to 0 on E - A,, and to 1 on E - C (4.5.2).We then take V to be the open set of points y such that f ( y ) < f , and then 0 is contained in the closed set of points y such that f(y) 5 +, and hence ? c C. Put B, = E - 0. Then we have B,, c E - V c A,,, and B,, u C = E, so that the sets B,with n 5 m satisfy the required conditions, and the induction can proceed. For each x E E, there exists an integer n such that x 4 A,, for all m > n, and therefore x E B, for some k 5 n. Hence the sets B, cover E.

It is clear that the same argument will apply if the covering (A,) isfinite.

I f E is a topological space andf'is a mapping of E into a real vector space F or into R, the support off (denoted by Supp(f)) is defined to be the smallest closed set S in E such thatfvanishes on E - S. In other words, Supp(f) is the closure in E of the set of points x E E such that f(x) # 0 ; or again it is the set of points x E E with the property that every neighborhood of x in E contains a point y such that f ( y ) # 0.

XI1 TOPOLOGY A N D TOPOLOGICAL ALGEBRA

22

Let (f,),. I be a family of mappings of E into F (resp. R) whose supports form a locally finite family. Then the sum C f a ( x ) is defined for all x E E, aa I

because it contains at most finitely many non-zero terms. We denote by f , the function x + + CfR(x).If F is B or a normed space (or more generally

c

as1

aE1

a topological vector space (12.13)), and if each f , is continuous on E, then so is f = f a :for given x E E there exists a neighborhood V of x which meets only

c

as1

finitely many of the Supp(f,), and hence there is a finite subset H of I such that f ( y ) = f a ( y ) for all y E V.

c

asH

A continuous partition of unity on E is by definition a family ( f a ) , ,I of continuous mappings of E into [0, I], such that the supports of the f a form a locally finite family, and such that f a ( x )= 1 for all x E E. If (Aa)aE I is an

1

aal

open covering of E indexed by the same set I, then the partition of unity (fa), I is said to be subordinate to the covering (AJa I if we have Supp(f,) c A, for all u E I. (12.6.3) Let (A,) be an at most denumerable 1ocally.finite open covering of a metrizable space E. Then there exists a continuous partition of unity (f,)on E subordinate to (A,,).

Let (B,) be an open covering of E, such that B, c A, for all ii (12.6.2). It is clear that the covering (B,) is locally finite. By (4.5.2), for each iz there exists a continuous mapping h,: E -+ [0, I] such that h, is equal to 1 on B, and equal to 0 on E - A,. If we put g , = (h, - +)+, then Supp(g,) is contained in the set of points x such that h,(x) 2 t,and hence is contained in A,. Let g = g, .

1 n

Since the sets B, cover E, we have g ( x ) > 0 for all x E E, and therefore the functionsf, = g,/g are defined and continuous on E, and form a partition of unity with the required properties. (12.6.4) Let E be a metrizable space, K a compact subset of E and afinite covering of K by open sets of E. Then there exist m continuous mappings f,: E --* [O, I] such that supp(fk) c A, for 1 5 k 5 m, and such that m

1

fk(x)

k= 1

5 1 for all x E E and

m

fk(x) = 1 for all x E K.

k= 1

Take a continuous partition of unity (fk)Odksn, subordinate to the open covering of E consisting of A,, = E - K and the A, (1 5 k 5 m).

7 SEMICONTINUOUS FUNCTIONS

23

Remark (12.6.5) Let F be a subset of the set of continuous mappings of E into R,with the following three properties :

(1) for each pair of disjoint nonempty subsets M, N of E, with M compact and N closed, there exists a function f 2 0 in F which is equal to 0 on N and 21 on M ; ( 2 ) if (f U ) E, , is a family of functions in 9 whose supports form a locally finite family, then f, belongs to 9; U E l

(3) i f f € 9 is such thatf(x) > 0 for all x E E, then g/fe 9for all g E 9.

Then the conclusions of (12.6.3) and (12.6.4) are still valid, when the A, are relatively compact, if we impose the extra condition that the functionsh should belong to 9. The proofs are unaltered.

7. SEMICONTI NUOUS FUNCTIONS

In this section, and above all in Chapter XIII, we shall need to consider mappings of a set A into the extended real line 8. Such mappings we shall call (by abuse of language) real-valuedfunctions on A. Such a function f is said to be finite if its value at each point a E A is finite, i.e., iff(A) c R.The functionf is said to be majorized or bounded above (resp. minorized or bounded below) if there exists a finite majorant (resp. a finite minorant) of f ( A ) . Iff is both majorized and minorized, it is said to be bounded (which clearly implies that f i s finite). We recall ((4.1.8), (4.1.9), and (3.15.5)) that in R x R the function (x, y ) x +~y (resp. (x,y ) H x y ) is defined and continuous except at the points (+ co, - co) and (- co, 00) (resp. (+ co, 0), (- co,O), (0, co), (0, - co)). The function X H I / X on R is defined and continuous except a t the point x = 0. In the interval [0, + co] of R, the function X H l/x (defined on 10, +a])can be extended by continuity by giving it the value f c o at the point 0. Let E be a topological space and letfbe a mapping of E into the extended real line R. The function f is said to be lower (resp. upper) semicontinuous at a point x o E E if, for each a E R such that a c f ( x o ) (resp. a > f ( x , ) ) , there exists a neighborhood V of xo in E such that, for all x E V, we have a < f ( x ) (resp. c( > f ( x ) ) . The functionfis said to be lower (resp. upper) semicontinuous on E if it is so at every point xo E E.

+

+

24

XI1 TOPOLOGY A N D TOPOLOGICAL ALGEBRA

Clearly, iff is lower semicontinuous at xo , then -f is upper semicontinuous at this point. Hence we need consider only lower semicontinuous functions. If (P is a mapping of a topological space F into E which is continuous at a point yo E F, and iff is lower semicontinuous at the point xo = (~(y,,),then f (P is lower semicontinuous at y o . In particular, if F is a subspace of E, and iff is lower semicontinuous at a point yo E F, then so is the function f I F. 0

Examples

(12.7.1) A mapping f : E .+ a is said to have a relative minimum at a point xo E E if there exists a neighborhood V of xo such that f ( x ) 2 f ( x o ) for all x E V. If so, thenfis lower semicontinuous at the point xo . This is clearly the case whenever f ( x o ) = - 03. For each x E R, put f ( x ) = 0 if x is irrational, and f ( x ) = l/q if x is rational and equal to p / q (where p , q are coprime integers and q > 0). For each integer n 2 0, the subspace of rational numbers p / q with q S n is closed in R and discrete. Hence for each irrational x there exists a neighborhood V of x such that f ( y ) l / n for all y E V, and therefore f is continuous at the point x . On the other hand, f has a relative maximum at each rational point, and is therefore upper semicontinuous on R. (12.7.2) A mapping f : E + is lower semicontinuous on E if and only $ for each a E R, the set f - ' ( ] a , 031) of points x at which f ( x ) > a is open in E (or, equivalently, the set f -I([ - 03, a ] ) of points x such that f ( x ) 5 a is closed in E).

+

For to say that f is lower semicontinuous on E signifies that, for each a E W, the set f - ' ( ] a , +a])is a neighborhood of each of its points (3.6.4).

If A is any subset of a set E, the characteristic function of A (usually denoted by qA)is the mapping of E into R such that ( P ~ ( x= ) 1 for all x E A and pA(x) = 0 for all x E E - A. So we have q E = 1, ( P = ~ 0, and the formulas (PE-A=

I1 2.7.3)

(PA

-(PA,

+ (PB = (PA v B + (PA

nB

(PA(PB=(PAnB,

,

(PA n CB

= (PA

- (PA (PB,

where A, B are any two subsets of E, and (A,) is any family of subsets of E.

7 SEMICONTINUOUS FUNCTIONS

25

(12.7.4) A subset A of a topological space E is open (resp. closed) in E ifand only if qAis lower (resp. upper) semicontinuous on E.

This follows immediately from (12.7.2). (12.7.5) Let f, g be two mappings of a topological space E into R, each of which is lower semicontinuous at a point xo E E . Then the functions sup(f, g ) and inf(f, g ) are lower semicontinuous at xo . The same is true ,forf + g if the sum f ( x ) + g(x) is defined for all x E E (4.1.8), and for f g i f f and g are both 20 and the product f (x)g(x)is defined for all x E E (4.1.9).

We shall give the proof for f + g ; the other cases are analogous. The result is obvious if f ( x o ) or g(xo) is equal to -03. If not, then we have f ( x o ) + g ( x o ) >- co. Every number u E R such that u m such that c < inf x , + ~< c', hence there exists P20

p 2 0 such that c < x,+, < c'. This shows that a is a cluster value of the sequence (x,) (3.13.11). 'If a = co (resp. a = -a),the argument is the same if we replace c' (resp. c) by a . Conversely, if b is a cluster value of (x,), then it is also a cluster value of every sequence ( x , + p ) p z o ,and therefore b 2 inf x , , + ~(3.13.7). Hence b 2 a.

+

P20

This result shows that the existence of a limit of the sequence (x,) is equivalent to the relation lirn inf x, = lim sup x, , and that the common value n+m

n+ m

of the two sides is then the limit of (x,) (3.16.4). It follows from the definition that if (x,,) is any subsequence of a sequence R, then we have

(x,) in

(12.7.12)

lirn inf x,, 2 lirn inf x, k+ m

n-rm

(12.7.13) Let E be a Hausdorf topological space and letf'be a mapping of E into R which is lower semicontinuous at a point a E E. Then for every sequence (x,) of points of E such that lirn x, = a, we have ,+ a,

(12.7.13.1)

lirn inf f ( x , ) 2 f(a). n+m

For every u < f ( a ) there exists a neighborhood V of a in E such that f ( y ) 2 u for all y E V ; but there exists no such that x, E V for all n 2 n o . Hence f (x,) >= u for all n 2 n o , which proves the result.

PROBLEMS

1. Let E be a topological space. For each closed subset A of the product space E x R, show that the mapping xHinf(A(x)) of prlA into R is lower semicontinuous. Conversely, iff: E -+R is lower semicontinuous, then the subset B of E x R consisting of all pairs ( x , y ) such thatf(x) 5 y is closed in E x R.

7 SEMICONTINUOUS FUNCTIONS 2.

29

(a) Let E, F be two metrizable locally compact spaces and let T :E -+ F be a continuous map. The mapping n is said to be proper if, for each compact subset K of F, the inverse image n--I(K) is compact. Show that, if T is proper and y E r(E), every neighborhood in E of the set r - ' ( y ) contains a neighborhood of the form T-'(U), where U is a neighborhood of y in F. (b) Let g be a lower semicontinuous real-valued function on E. For each y E F, let f ( y ) be the greatest lower bound of g on the set ~ - ' ( y (so ) that f ( y ) = 03 if n--'(y) = 0).Show thatfis lower semicontinuous on F. (c) Let g be the continuous function ( x ~ , x ~ ) H [ x ~ x11z on R x 10, +a[.For each x E R,letf(xl) = inf g(xl, xz). Show thatfis not lower semicontinuous on R.

+

XZ>0

+ +

+ +

3. For each point t = (tl, . . . , I,) E C", put P,(X) = X" tlX"-' . .. t.. Show that there exists a continuous mapping r + + x ( t )of C" into R and a lower semicontinuous mapping t t-+y(t) of C" into R, such that P,(x(r) iy(t)) = 0 for all r E C". (Take x ( t ) to be the largest of the real parts of the roots of P, , and use (9.17.4)). 4.

Let f be any mapping of a metric space E into a metric space F. For each x E E, let n ( x ) be the oscillation offat x with respect to E (3.14), so that n(x) is a positive real number or co. Show that the mapping XH n(x) of E into R is upper semicontinuous.

+

5.

Let E be a metric space and letf: E + R be lower semicontinuous. Let a E E be a point at which the oscillation n(a)o f f is finite. Show that for each E > 0, there exists a neighborhood V of a in E such that inf n ( x ) 5 E . (Show that otherwise there would X E V

exist points x arbitrarily close to a at which f ( x ) took arbitrarily large values.)

6. Let (an)n3lbe an infinite sequence of distinct points in the interval [0, I[. For each integer N > 1, let b l , . . . , bN denote the sequence obtained by arranging the set {u,, . . ., aN} in increasing order. The intervals [0, b,[, [ b ~b ,~ [... . , [ b -1, ~ b ~ [ , [ b1[~ , are said to form the Nrh subdivision of [0, I[ corresponding to the sequence (a"). Let uN (resp. uN)denote the minimum (resp. maximum) of the lengths of the intervals of the Nth subdivision. Let h = lim inf NUN, p = lim sup NUN. N- m

N-m

+

(a) For each E > 0, let no be such that uN 2 (A - E)/N and uN 2 ( p E)/N for all N 2 n o . Show that, for each N 2 no and each integer r such that 0 5 r 5 N, there exist 2r intervals of the (N r)th subdivision whose lengths are greater than or equal to the 2r numbers

+

h-E h-E h-E -h.- .E. N+l'

N+l'

N+2'

N+2'

h-E 'N+r'

h - E -

N+r

+

and such that the other intervals of the (N r)th subdivision are intervals of the Nth subdivision (use induction o n r). (b) Show that, if N 2 n o , the intervals of the Nth subdivision can be arranged in order in such a way that their lengths are less than or equal to the N 1 numbers

+

p +_ E _ pL-E -

N

' N+l'""

2N

'

(For each interval I of the Nth subdivision, let s(1) be the largest integer 5 2 N such that 1 is an interval of the s(1)th subdivision, and arrange the intervals I in order of

increasing s(I).)

30

XI1 TOPOLOGY AND TOPOLOGICAL ALGEBRA

(c) Deduce from (a) and (b) that

h 5 l/log 4, (use the fact that

N

1/(N

,= 1

p

2 I/log 2

+ r ) tends to log 2 as N -+ + co).

(d) Take a. = log2(2n - 1) - [log2(2n - I)]. Show that we have h = l/log 4 and p = l/log 2 for this sequence. (Observe that, if N = 2p, the set {al, , , . , a N }is the same as the set {CN , . . . , c ~ -N1) where c. = log, n - [log, n].) (Cf. Section 13.21, Problem 15.)

7. Let E be a metrizable space. Show that a nonempty closed subset S of E is the support of a lower semicontinuous (resp. continuous) real-valued function 2 0 on E if and only if S is the closure of its interior. 8.

Let E be a metrizable space and f a mapping of E into R. The lower semicontinuous regularization off is defined to be the upper envelope of the continuous functions g on E such that g f. (a) For each x E E we denote by lim inff(y) the greatest lower bound of the numbers Y-x

lirn inff(y.), for all sequences (y.) tending to x. Show that the function xwlim inff(y) Y-x

"-I m

is the lower semicontinuous regularization off. (b) Suppose that E is an open set in R. For each x greatest upper bound of the numbers lirn inff(y.) y , + x and y .

E

E, let lim inf f(y) denote the Y'X.

Y>X

for all sequences (y.) such that

n-m

2 x for all n. Show that the set of all x E E such that lirn inf f ( y ) > lirn inff(y) Y-x,

Y-IX

Y>X

is at most denumerable. (For each pair ( p , q) of rational numbers with p > q. show that the set of points x E E such that

lirn inf f(y) > p > q > lirn inff(y) Y-x.

Y>X

Y+X

is at most denumerable, by following the method of Section 3.9, Problem 3.) 9. A Dirichlet series is a series whose general term is of the form a.e-'"S, where (h.) is an increasing sequence of real numbers, tending to co, and (a")is any sequence of complex numbers, and s is complex. (a) Show that if the series is convergent for s = s o , then it is uniformly convergent in the angular sector consisting of the points s = so pele with p 2 0 and -a 2 6 a, where a is such that 0 < a < n/2. (Reduce to the case so = 0; show first that if 9 s = u, then

+

+

le-os-e-b*l

5

I~Io-'(e-O"--e-b")

whenever a and b are real and a < b, by considering the integral

+

s."

cXs dx. For each

integer m and each n l m , put S,,,= ~ , , + a , + ~* . . + a , and use the identity (Abel's partial summation formula)

7 SEMICONTINUOUS FUNCTIONS

31

(b) Deduce from (a) that there are the following three alternatives: either the series ( ~ " e - ~ does " ' ) not converge for any value of s, or it converges for all values of s, or else there exists a real number uo such that the series converges for 9 s > uo and does not converge when 9 s < uo. In the first (resp. second) case we put uo = m (resp. - 03). In all cases, the (extended real) number uo is called the abscissa of convergence of the series. The sum of the series is an analytic function of s in the region 92s > uo. (c) Show that if uo 2 0, then uo = lirn sup (log ISo. .I)/&. (Show first that if the

+

n-m

series with general term a.e-"' = b, is convergent (with s real and >O), then we have ISo, 5 KeAnswhere K is a constant, by writing a, = b.eAnS.Then show that if y = lim sup (log ISo, "\)/An,the series (ane-Afls) converges when s = y 6, where 6 is

+

n-m

real and >0, by arguing as in (a).) (d) Let u1 be the abscissa of convergence of the Dirichlet series with general term la.le-Ans.Then u1 2 uo. If uo < m , show that

+

log n u1 - uo 5 lim sup n-m

A.

I

(Remark that, given E > 0, we have lan\5 e'"o+e'Anfor all sufficientlylargen.)Consider the case where A,, = n (cf. Section 9.1, Problem 1). Show that, for the series with general term

wehaveuo=-mandal=+co. 10. Letfbe a real-valued function defined on the interval [0,

+

m[, satisfying the following conditions: (1) f ( s + t ) Z f ( s ) + f ( t ) ; (2) there exists a number M > 0 such that If(t)l 5 Mt for all t. Show that the limits

exist and are finite, and that at sf(r)5 /3r for all t 2 0. (To establish the existence of a it is enough to show that, if we take a = limsupf(r)/t (Problem 8), then 1-0

sf(r)/rfor

all t > 0. Take a decreasing sequence (1.) tending to 0, such that limf(rn)/tn = a, and let k, be the integral part of r/t.. Show that a

The proof of the existence of lim

f(r)/t

is similar; put

t-+m

/3 = lim inff(r)/t.) r-+m

11. Let f b e a continuous mapping of an open set U in a Banach space E into a Banach space F. For each x E U, let Ilf(Y) -f(x)ll

D+f(x) = lim sup y-x.y+x

D - f ( x ) = lim inf Y-x.Y#x

We have 0 2 D-f(x)

5 D+f(x)

+ m.

IlY-XI1

'

/I f(Y)- f ( x )II IlY-XI1

.

32

XI1 TOPOLOGY AND TOPOLOGICAL ALGEBRA

(a) If the function f is differentiable at a point x E U, then D+f(x) = Ilf’(x)ll. If f:(x) is not a linear homeomorphism of E onto a subspace of F,we have D-f(x) = 0 . In the contrary case, we have D-f(x)= ~ ~ ( J ’ ( x ) ) - ~where ~ ~ - ~(,f ’ ( ~ ) ) - ~denotes the inverse homeomorphism. (b) Suppose that the segment with endpoints a, b is contained in U, and that D+f(x) M for all points x in this segment. Prove that llf(b) -f(a)ll 5 Mllb - all (see the proof of 8.5.1).) (c) Take U = E =R3 and F =R3,and letfbe defined as follows:

if (fl, &) # (0,O); and f(0,O) = 0. Show that the greatest lower bound of D-fon U is strictly positive, but that there is no neighborhood of 0 in U on whichfis injective. (d) For the remainder of this problem, suppose that there exist two numbers m, M with 0 < m < M < co, such that m 5 D-f(x) 5 M for all x E U. Suppose also that for each x E U there exists an open neighborhood V of x*in U such that f l V is a homeomorphism of V onto an open set in F ; thenf(U) is open in F. Let a be a point of U, and for each line D c F containing f ( a ) let ID be the connected component of the point f(a) in the open set D n f ( U ) in D. The union S, of the sets I D is the largest star-shaped open set with respect to f ( a ) contained in f(U). For each line D in F passing through f ( u ) , tnere exists a unique continuous mapping go : ID--f U such that g o ( f ( a ) )= a and f ( g D ( y ) )= y for all y E ID (the proof is the same as in Section 10.2, Problem 6(c).) If y, y’ are points of ID, we have I ~ ~ D ( Y ‘ ) - S D ( Y ) ~S~ m--’Ily’ - Y I I . Deduce that as y tends to an endpoint of ID (when there is one), g&) has a limit belonging to Fr(U). (e) Let y : J + U be a path in U with origin a and extremity b. Show that, if f(y(J)) C S., then we have b =gD(j(b)), where D is the line throughf(a) and f(b), and II f ( b )-f(a)ll 2 mll b - all. (f) Deduce from (d) and (e) that if U = E then S, = F, and f is a homeomorphism of E onto F. (9) Let k = M/m. Let a, b be two points of U such that the set Ek7a,b of points z E E such that I/z- all llz - bl/ 5 k/l b - all is contained in U. If L is the closed segment with endpoints a, b, show that f(L) c S,, and hence that

+

+

IlfW -f(a)ll 2 mll b - all.

+

(Proof by contradiction: consider the least I E [0, I ] such that y = f ( u t(b - a)) $ S., If D is the line throughf(u) and y, there is an endpoint u of ID belonging to the open segment with endpoints !(a) and y . When f ‘ < t tends to t, there is a point u‘ of the open segment with endpoints f(a) and y‘ = f ( a r’(b - a)) which tends to u. Let D’ be the line through f ( a ) and y’, and let z’ = go&’). Using (e), show that z’ E Ek,(I,b , and obtain a contradiction by making r’ tend to I and using (d).) (h) Suppose that E and F are Hilbert spaces and that U is the ball llxll < 1. Deduce (kZ- 1)’l2), the restriction off to B is a from (g) that if B is the ball llxll < 1/(1 homeomorphism onto f(B),and that Ilf(x’) -f(x)il 2 m ljx’ - XI( for any two points x, x’ in B. ( 9 With the hypotheses of (h), suppose also th at 4 < ((1 +2/5)/2)1/2. Show that f is then injective on U and, more precisely, that for any two points x, x’ in U we have Ilf(x’) -f(x)II 2 p Ilx’- x/I, where - M(M2 - m2)1/2 p = m + (Mz - m2)1/2 ‘

+

+

8 TOPOLOGICAL GROUPS

33

8. TOPOLOGICAL GROUPS

If G is a group, in which the law of composition is written (for example) multiplicatively, a topology on G is said to be compatible with the group structure if the two mappings ( x , y ) H x y of G x G (endowed with the product topology) into G, and X H X - ' of G into G, are continuous. A group endowed with a topology compatible with its group structure is called a topological group. We leave it to the reader to transcribe this definition (and all the results which follow) into additive notation. An isomorphism of a topological group G onto a topological group G' is by definition an isomorphism of the group G onto the group G' which is bicontinuous. If G' = G, we say automorphism instead of isomorphism. If G is any group, the law of composition ( x , y ) ~ y defines x a group structure on the set G (and this structure is different from the given one unless G is commutative). The group so defined is denoted by Go and is called the opposite of G. Any topology which is compatible with the group structure of G is also compatible with that of Go, and hence makes G o into a topological group. The mapping XHX-' is then an isomorphism of G onto Go.

Examples (12.8.1) The discrete topology and the chaotic topology (12.1.1) are compatible with the structure of any group. The topology of a normed space (in particular R or C) is compatible with its additive group structure. On the additive group Q of rational numbers, the topology defined by the p-adic distance d (3.2.6) is compatible with the group structure, because by virtue of the definition of this distance and (3.2.6.4) we have

4x0

+ yo x + Y ) s MaxMxo 4, dcvo ,v)) 5

Y

and

d(-xo, -x)

= d ( x o ,x ) .

If E is a (real or complex) Banach space and GL(E) the set of linear homeomorphisms of E onto E, then the topology induced on GL(E) by that of Y(E; E) is compatible with the group structure ((5.7.5) and (8.3.2)).

34

XI1 TOPOLOGY AND TOPOLOGICAL ALGEBRA

In particular, the topology induced on the multiplicative group R* (resp. C*)of real (resp. complex) numbers # O by the topology of R (resp. C) makes R* (resp. C*)into a topological group. If G is a topological group (in multiplicative notation), the mappings (x, y ) H xy-' and ( x , y ) H x - ' y are continuous mappings of G x G into G (3.11.5). For each a E G , the left and right translations x H ax and x H xa

are homeomorphisms of G onto G, because they are bijective and continuous ((3.20.14) and (12.5))and so are the inverse mappings x H a - ' x a n d x H x a - ' .

For any a, b in G, the mapping x H a x b (and in particular the inner automorphism x ~ a x a - ' is ) therefore a homeomorphism of G onto G (3.11.5). Since the mapping X H X - ' is bijective and equal to its inverse, it also is a homeomorphism of G onto G. (12.8.2) Let G be a topological group.

(i) For each open (resp. cloJed) subset A of' G, and each x E G, the sets x A , Ax and A - ' (the set of ally-', where y E A) are open (resp. closed) in G. (ii) For each open set A in G and each subset B of G , the sets AB (the set of all products yz, where y E A and z E B) and BA are open in G. The assertions in (i) follow immediately from the preceding remarks, and (ii) follows from (i), since AB = A z is a union of open sets.

u

ZEB

(12.8.2.1) On the other hand, if A and B are closed in G, it does not necessarily follow that AB is closed (cf. (12.10.5)). For example, if 9 is an irrational number, consider the subgroups Z and OZ of R . Then the subgroup Z 8Z is not closed in R. To see why, observe that this subgroup is denumerable, and therefore is not the whole of R (2.2.17). Hence it is enough to prove the following result:

+

The only closed subgroups of R are R itseyand the subgroups of the form crZ, where cr E R. (12.8.2.2)

Granted this, we cannot have both 1 = nz and 9 = ma with m and n integers, because 8 is irrational, and the assertion of (12.8.2.1) then follows. To prove (12.8.2.2), we shall first show that a subgroup H of R is either discrete or dense in R. If H is not discrete, then for each E > 0 there exists x # 0 in the set H n [ - E , S E ] .Since the integer multiples n x ( n E Z) belong to H, every interval of length greater than E in R contains one of these points and therefore H is dense in R.

8 TOPOLOGICAL GROUPS

35

To complete the proof of (12.8.2.2), it remains to be shown that if H is discrete it is of the form a Z . We may assume that H # ( 0 ) . Since H = -H, the intersection H n 10, +a[is not empty. If b > 0 belongs to H, the intersection H n [O, b ] is compact and discrete and therefore$nire (3.16.3). Let a be the smallest element > O in this set, and for any X E H let m = [ x / a ] be the integral part of x / a . Then x - ma E H, and 0 5 x - ma < a. This implies that x - ma = 0 and so we have H = a Z . (12.8.3) Let G be a topological group. (i) Let a E G . r f V runs through a fundamental system of neighborhoods of the neutral element e of G , the sets a V (resp. V a )form a fundamental system of neighborhoods of a. (ii) For each neighborhood U of e, there exists a neighborhood V of e such that VV-' c U. (iii) For each neighborhood U of e and each a E G , there exists a neighborhood W of e such that aWa-' c U. (iv) G is Hausdorflifand only if the set { e } is closed in G . The assertion (i) follows from the fact that translations are homeomorphisms; (ii) expresses that the mapping (x, y ) ~ x y - 'is continuous at ( e , e ) , having regard to the definition of open sets in G x G (12.5); (iii) expresses the continuity of the mapping X H axa- at the point e . As to (iv), it is clear that if G is Hausdorff the set { e } is closed (12.3.4). Conversely, if { e } is closed and if x , y are distinct points of G, then there exists a neighborhood V of e such that e # x-'yV, i.e., such that x # y V . If W is a neighborhood of e such that WW-' c V (which is possible by (ii)), then we have XW n yW = 0, because the relation xw' = yw" with w' and W" in W would imply x = yW"W'-1

Eyww-'

cyv

Hence G is Hausdorff. A neighborhood V of e is said to be symmetric if V-' = V. The symmetric neighborhoods of e form a fundamental system of neighborhoods of e in G, because if U is any neighborhood of e, then so is U-' (12.8.2) and therefore U n U-' is a symmetric neighborhood of e contained in U. For each integer n > 0 and each subset V of G we define V" inductively by the rule V" = V"-' V = V V"-'. (The set V" is not the set of X" with x E V.) If V is a neighborhood of e , then so is V" 3 V for all n 2 I , and it follows from (12.8.3(ii)) that if U is any neighborhood of e and n is an integer > 1, there exists a symmetric neighborhood V of e such that V" c U.

36

XI1 TOPOLOGY AND TOPOLOGICAL ALGEBRA

(12.8.4) For a homomorphism f of a topological group G into a topological group G' to be continuous, it is necessary and suficient that f should be continuous at one point.

For iff is continuous at a E G and if V' is a neighborhood of f ( a ) , then f -'(V') = V is a neighborhood of a. For each x E G, we have

f(xa-'v) = f ( x > ( f ( a ) > - ' f ( Vc)f ( x > ( f ( a ) > - ' v ' , which establishes the continuity off at the point x , by virtue of (12.8.3(i)). If H is a subgroup of a topological group G, the induced topology on H is clearly compatible with the group structure of H. Whenever we consider H as a topological group, it is to be understood in this sense unless the contrary is explicitly stated. (12.8.5) The closure R of a subgroup (resp. normal subgroup) H qfa topological group G is a subgroup (resp. normal subgroup) of G. If G is Hausdorff and H is commutative, then R is commututive.

'

The image of R x R = H x H under the continuous mapping (x,y ) Hxy of G x G into G is contained in R, because the image of H x H under this mapping is contained in H (3.11.4). Hence R is a subgroup. Likewise, if H is normal and a E G, the image of H under the mapping x H a x a - ' is contained in H, hence the image of R is contained in R, and therefore R is normal. Finally, if G is Hausdorff and H is commutative, the continuous functions x y and y x are equal on H x H and therefore also on R x R by virtue of the principle of extension of identities ((3.15.2), 12.3, and 12.5). (12.8.6) ( i ) In a topological group G, the normalizer N ( H ) of a closed subgroup H (i.e., the set of all x E G such that xHx-' c H) is a closed subgroup. (ii) In a Hausdorff topological group G , the centralizer d ( M ) of any subset M ofG (i.e., the set of all x E G which commute with every element of M ) is a closed subgroup of G. In particular, the center of G i s closed.

For each z E H, the set of elements x E G such that xzx-' E H is the inverse image of H under the continuous mapping X H X Z X - I , hence is a closed set (3.11.4). Since N ( H ) is the intersection of these sets as z runs through H , it follows that N(H)is closed (3.8.2). Again, if G is Hausdorff then for each z E M the set of x E G such that zx = xz is closed (12.3.5) and hence so is S ( M ) since it is the intersection of these sets. (12.8.7) (i) In a topological group G, every Iocally closed subgroup is closed. Every subgroup which has an interior point is both open and closed. (ii) In a HausdorfSgroup, every discrete subgroup is closed.

8

TOPOLOGICAL GROUPS

37

(i) Let H be a locally closed subgroup of G. Then R is a subgroup of G, and H is an open subgroup of R (12.2.3). Hence it is enough to prove the second assertion. Now if H has an interior point, then by translation it follows that every point of H is interior, and therefore H is open. Hence the left cosets x H are also open sets, and therefore CH is open in G (because it is a union of cosets xH). Consequently H is closed in G. (ii) If G is Hausdorff and H is a discrete subgroup of G, then there exists a symmetric open neighborhood V of e such that V n H = { e } . If X E R ,we have xV n H # fa. Now if y ~ x V n H, then x ~ y V and the set { y } is closed in the open set yV, because G is Hausdorff. Since yV n H = { y } (because y E H), we must have x = y , and therefore R = H. (12.8.8) If G is a connected group and V is a symmetric neighborhood of e, then G is equal to the union V" of the sets V"for n > 0.

The set V" is clearly symmetric, and since V"V" = Vm+"it follows that V"V" c V". Hence V" is a subgroup of G. Since e is an interior point of V", this subgroup is both open and closed (12.8.7) and therefore is the whole of G . (12.8.9) In a topological group G , the connected component K of the neutral element (3.19) is a closed normal subgroup (called the neutral component or identity component of G). For each x E G , the connected component of x in G is XK = Kx.

If a E K, the set a - ' K is connected and contains e, hence K-'K c K. This shows that K is a subgroup of G. It is invariant under all automorphisms of the topological group G, in particular under all inner automorphisms, hence K is normal in G. Also K is closed in G (3.19). Finally, the last assertion follows from the fact that left and right translations are homeomorphisms of G onto G. (12.8.10) If G,, G, are two topological groups, then the product topology on the product group G = G, x G, is compatible with the group structure of G. For, identifying canonically the product spaces G x G and (GI x G,) x (G, x G,) (12.5), the mapping

((x19 x2)9 ( Y l , Y 2 ) ) +-+(XlY13 x2 Y2) is continuous ((3.20.15) and (12.5)), and the mapping (xl, x,)H(x;~, x;') is continuous for the same reason. The group G, x G , , endowed with the product topology, is called the product of the topological groups G , and G , . If G is a commutative topological group, the mapping (x, y ) ~ x is y a continuous homomorphism of G x G into G.

38

XI1 TOPOLOGY AND TOPOLOGICAL ALGEBRA

PROBLEMS

1. Let G be a group and let 'x1 be a set of subsets of G , satisfying conditions (V,) and (V,,) of Section 12.3, Problem 3, together with the following:

(GV,) For all U E % there exists V E 3 such that V . V c U. (GV,,) For all U E ti we have U - ' E %, (GV,,,) For all U E % and a E G, we have aUa-I E ti. Show that there exists a unique topology on G , compatible with the group structure of G, for which % is the set of neighborhoods of the neutral element e . 2.

Every topology compatible with the group structure of a finite group G is obtained by taking as neighborhoods of the neutral element the sets containing a normal subgroup H of G. Give an example of a non-Hausdorff topological group in which the center is not closed and has as its closure a noncommutative subgroup.

3.

Let G be a connected topological group. Show that every totally disconnected normal subgroup D of G is contained in the center of G. (If a E D, consider the mapping XHXUX-' of G into D.)

4.

Show that the commutator subgroup of a connected topological group is connected (use (3.19.3)and (3.19.7)).

5.

Let G be a topological group and let H , K be subgroups of G such that H 3 K 2 H', where H' is the commutator subgroup of H. Show that R contains the commutator subgroup of R . Deduce that if G is Hausdorff, the closure in G of a solvable subgroup is solvable (induction on the length of the derived series).

6. Let G be a topological group and let H be a closed normal subgroup of G , containing the commutator subgroup of G. If the identity component K of H is solvable, show that the identity component L of G is solvable. (Show by using Problem 4 that K contains the commutator subgroup of L.)

T be the canonical homomorphism and let 6 be an irrational number. O n the topological space G = RZ x T2,a group law is defined by

I. Let cp : R

--f

In this way G becomes a locally compact group (even a Lie group). Show that the commutator subgroup of G is not closed in G. 8.

Let (G&cl be any family of topological groups. Show that the product topology on G = G , is compatible with the product group structure (Section 12.5, Problem 4).

n

.El

The topological group so defined is called the product of the topological groups G, . Let H be the normal subgroup of G cocsisting of all x = (x,) such that for all but a finite number of indices x. is the identity element of C , . Show that H is dense in G .

9

METRIZABLE GROUPS

39

9. METRIZABLE G R O U P S

If G is a group, a function f on G x G is said to be left (resp. right) invariant if f ( x y , x z ) =f ( y , z ) (resp. f ( y x , zx) =f ( y , z ) ) for all x, y , z in G. When G is commutative, these two conditions coincide, andfis then said to be translation-invariant. A distance d o n G is left (resp. right) invariant if and only if the left (resp. right) translations are isometries with respect to d. Iff is a left-invariant function on G x G, then the function (x,y ) ~ f ( x - ly,- ' ) is right-invariant, and vice versa. For example, if E is a normed space, the distance / / x- y/1 on E is translation-invariant. (12.9.1 ) In order that the topology of a topological group G should be metrizable (in which case G is said to be a metrizable group) it is necessary and suficient that there should exist a denumerable fundamental system of neighborhoods of the neutral element e, whose intersection consists of e alone. When this condition is satisfied, the topology of G can be defined by a leftinvariant distance, or by a right-invariant distance.

It is enough to show that if there exists a denumerable fundamental system of neighborhoods (U,) of e in G such that U, = { e } ,then the topology of G n

can be defined by a left-invariant distance. We define inductively a sequence (V,) of symmetric neighborhoods of e in G such that V, c U, and

VL, v,

n

u,

for all n 2 1 (12.8.3), so that (V,) is also a fundamental system of neighborhoods of e. Now define a real-valued function g on G x G as follows: g(x, x ) = 0 ; if x # y , then either x - ' y 4 V,, in which case we takeg(x, y ) = 1; or else there exists a greatest integer k such that x - ' y E Vk (because x - ' y # e cannot belong to all the V,), in which case we define g(x, y> = 2-k. It is clear from this definition that g ( x , y ) = g(y, x), that g(x, y ) 2 0, and that g(zx, z y ) = g(x, y ) for all x , y , z in G. Now let (12.9.1 . I )

where the infimum is taken over the set of all finite sequences (zo , z l , . . . , z,,) (with p arbitrary) such that z,, = x and z p = y . We shall show that d is a leftinvariant distance on G and satisfies the inequalities (1 2.9.1.2)

M X ?Y)

5 4x9 Y ) I g(x, Y).

40

XI1 TOPOLOGY A N D TOPOLOGICAL ALGEBRA

From the definition of d, it follows immediately that d is left-invariant (because g is), satisfies the triangle inequaiity, is symmetric and positive. Moreover, the right-hand inequality in (12.9.1.2) is obvious, and shows that d(x, x ) = 0 for all x E G. Hence d is a pseudo-distance on G. T o prove the lefthand inequality of (12.9.1.2) we shall show by induction on p that, for each finite sequence ( z i ) o s i s po f p 1 points in G such that zo = x and z p = y , we have

+

P- 1

(12.9.1.3)

1 g(zi

i=O

7

This inequality is obvious if p

zi+1) 2 +g(x, Y ) .

1g(zi, ziti). The

P- 1

1. Let us write a =

=

i=O

inequality (12.9.1.3) is true if a 2 +, because g(x, y ) 1, If a = 0, then zi = zi+l for 0 5 i S p - 1, hence x = y and so (12.9.1.3) is trivially satisfied. So suppose that 0 < a < and let h be the greatest index such that g ( z i , zi+l>5 +a. Then we have g ( z i , z ~ +>~+a, ) so that

1

+,

i O such that 2 - k 5 a. Then k 2 2, and the elements X-'zh, Z ; ' Z ~ + ~ ,Z;>~Y are all in Vk, by virtue of the definition ofg. Hence x - ' y E V l t Vk-l, which implies that g(x, y ) 5 2l - k 5 2a, and proves (12.9.1.3). Hence (12.9.1.2) is established, and shows first of all that d is a distance on G. Also, if r > 0, the ball B'(e; r ) (with respect to d ) contains vk for all indices such that 2 - k < r, and conversely each v k contains the ball B'(e; 2-k-1). Since d is left-invariant, it now follows from (12.8.3(i)) and (12.2.1) that d defines the topology of G. In general it is not possible to define the topology of G by means of a distance which is both left- and right-invariant (Problem 1 and Section 14.3, Problem 11). (12.9.2) Let G be a metrizable group. A necessary and su8cient condition for a sequence (x,) in G to be a Cauchy sequence with respect to a left-invariant distance defining the topology of G is that, for each neighborhood V of e, there exists an integer no such that x i ' x , E V for all m 2 no and n 2 n o .

For if d is a left-invariant distance, we have d(x,, x,) = d(e, X ,

'Xrn),

and the proposition follows from the fact that d defines the topology of G.

9 METRIZABLE GROUPS

41

Hence the property of being a Cauchy sequence with respect to some leftinvariant distance is independent of the choice of distance, and depends only on the topology of G. We shall say simply that such a sequence is a left Cauchy sequence in G. A right Cauchy sequence in G is defined analogously by replacing x; ' x , in (12.9.2) by x, x i It can happen that a left Cauchy sequence is not a right Cauchy sequence (Problem 8); but if (x,) is a left Cauchy sequence, then (x,-') is a right Cauchy sequence. Since the mapping X I - + X - ~ is continuous it follows that, if every left Cauchy sequence in G converges, then so does every right Cauchy sequence. In this case G is said to be a complete metrizable group: it is complete with respect to every leftinvariant distance and every right-invariant distance defining the topology of G. (12.9.3) Let G, G' be two metrizable groups. Then every continuous homom0rphism.f of G into G' is unformly continuous with respect to left- (resp. right-) invariant distances on G and G'. Let d, d' be two such distances on G, G , respectively. By hypothesis, for each E > 0 there exists 6 > 0 such that the relation d(e, z ) 5 6 implies d'(e', f ( z ) ) S E . Hence if d(x, y ) = d(e, x - ' y ) 5 6 , then d'(f(x), f ( y ) ) S E , because d ' ( f ( X ) , f ( Y ) ) = d'(e'9 ( f ( x ) ) - ' f ( y ) )= d'(e',f(x-'y))

since f is a homomorphism. (12.9.4) Let G,, G, be two metrizable groups such that G, is complete, and let H, (resp. H,) be a dense subgroup of G , (resp. G,). Then every continuous homomorphism u : H1 -+ H, can be extended uniquely to a continuous homomorphism ii : G , -+ G, . If G, is also complete and if u is an isomorphism (of topological groups) of H, onto H, , then ii is an isomorphism (of topological groups) q f G , onto G, . There exist left-invariant distances on G, and G, , and the existence of the continuous extension ii then follows from (12.9.3) and (3.15.6).The fact that ii is a group homomorphism follows from the principle of extension of identities applied to the two functions ( x , y ) ii(xy) ~ and (x,y)wE(x)U(y)on G, x G, (having regard to (3.20.3)). Finally, if G, is complete and if v : H, -+ H, is the inverse of the isomorphism u, then v can be extended to a continuous homomorphism fi of G, into G,. Since V 0 U and E fi agree with the identity mappings on HI and H, respectively, they are the identity mappings on G , and G, respectively, by virtue of the principle of extension of identitiesand (3.11.5). This completes the proof. 0

42

XI1 TOPOLOGY AND TOPOLOGICAL ALGEBRA

It should be carefully noted that, if we suppose only that u is injective (resp. surjective), then 6 is not necessarily injective (resp. surjective) (Problem 9). (12.9.5) Let G be a metrizable group in which there exists a neighborhood V of e which is complete (with respect to some left- or right-invariant distance). Then G is complete. In particular, a locally compact metrizable group is complete.

Let d be a left-invariant distance on G and let (x,,) be a left Cauchy sequence in G. Let E > 0 be such that theclosed ball B’(e; E ) is contained inV. By hypothesis there exists an integer no such that d(x,,, x,) 5 E for all m 2 no and n 2 n o , hence the sequence (x,,),~,,,, is contained in the closed ball B’(x,,,; E ) . But this closed ball is a complete subspace, because it is obtained by left translation from B’(e; E ) which is closed in V (3.14.5). Hence the seconverges in G. The last assertion follows from (3.16.1). quence (x,,),,~~ (12.9.6) In a Hausdorff topological group G , every locally compact metrizable subgroup H is closed.

Let x E R, let V be a neighborhood of e in G such that V n H is compact, and let W be a symmetric neighborhood of e in G such that W2 c V. Then XW n H is nonempty and relatively compact in H, because if yo E XW n H then for each y E xW n H we have y o ‘ y E Wz n H c V n H and therefore y E yo(V n H), which is a compact set. It follows (12.3.6) that the closure of XW n H in G is contained in H. and therefore x E H. Let G be a Hausdorff commutative topological group, written additively. All the material on series in Section 5.2 which involves only the topology of C remains valid without any change. The same is true of Cauchy’s criterion (5.2.1) if C is metrizable, by replacing the norm llxll by d(0, x), where d is an invariant distance on G. (12.9.7)

PROBLEMS

1. Let G = GL(2, R) be the multiplicative group of all real 2 x 2 nonsingular square

matrices. For each integer

n

> 0, let V. be the set of matrices X =

(z

t )

E

G such

that jx - I I 6 l/n, ly] 5 l / n , IzI 5 l/n, and jt - 11 5 I/n. Show that the family of sets V, is a fundamental system of neighborhoods of the neutral element I of G for a topology .Y compatible with the group structure of G (cf. Section 12.8, Problem I). The group G is locally compact in the topology .T, but 9-cannot be defined by a

9

METRIZABLE GROUPS

43

distance which is both left and right-invariant (cf. Section 14.3, Problem ll).) (Use the fact that if the topology of a metrizable group G can be defined by a distance which is both left and right-invariant, then for every neighborhood V of the neutral element e of G there exists a neighborhood W c V of e such that xWx-' c V for all X E G.) 2.

Let S be a compact subset of a metrizable group G, such that xy E S whenever x E S and y E S . Show that for each x E S we have xS = S. (Consider a cluster value y of the sequence (x")., in S, and show that yS is the intersection of the sets Y S; deduce that yxS = y S . ) Deduce that S is a subgroup of G.

3

Let G be a locally compact and totally disconnected metrizable group. (a) Show that every neighborhood of e in G contains an open compact subgroup of G. (Every neighborhood V of e contains a neighborhood U of e which is both open and closed (Section 3.19, Problem 9). If B = CU, show that there exists a symmetric open neighborhood W of e in G such that W c U and UW n BW = 125, and deduce that the subgroup generated by W is contained in U.) (b) Suppose that the topology of G can be defined by a left- and right-invariant distance. Show that every neighborhood of e in G contains a compact open normal subgroup of G (remark that every neighborhood V of e contains a symmetric open neighborhood W such that XWX-'c V for all x E G).

4.

Let p be a prime number. Consider the family of finite groups Z/p"Z (n 2 I), each with the discrete topology, and their product G (Section 12.8, Problem 8) which is compact and totally disconnected. For each n, let y.: Z/p"Z+Z/p"-'Z be the canonical homomorphism. = z. for all n is a closed (a) Show that the set of all z = (z.) E G such that q&.) (hence compact) subgroup Z, of G. If a+bnis the restriction to Z, of the projection pr, : G + Z/p"Z, then is a surjective homomorphism. (b) Foreachz=(z.)EZ,,put lz(,=Oifz=O,and ~ ~ ~ , = p ' - ~ i f m i s t h e s m a l l e s t integer such that z, # 0. Show that 1z - 2'1, is a translation-invariant distance which defines the topology of Z, . (c) For each n 2 1, let f n : Z + Z/pnZ be the canonical homomorphism. Show that the homomorphismf: ZH( ~ , ( Z) ) , , ~of Z into Z, is injective and that its imagef(Z) is a dense subgroup of Z,. If we put d(z, z') = I f ( z ) -f(z')l,, show that d is the p-adic distance on Z (3.2.6).The elements of Z, are called p-adic integers.

5.

Let p be a prime number, and let G be the compact group which is the product of an infinite sequence (G,),,, of groups G. all equal to T =R/Z. For each n, let vnbe the homomorphism XHPX of G . into G.-'. The compact subgroup of G consisting of all z = (2.) such that y,(z.) = zn-' for all n is called the p-adic solenoid and is denoted by Tp. (a) For each n , let fn:T, -+ G , = T be the restriction of pr. to T, . Show that f. is a surjective homomorphism with kernel isomorphic to the group Z, (Problem 4). (b) Let y : R -+T be the canonical homomorphism. For each x E R, show that the point 8(x) = (cp(x/pn)).,, belongs to T,. Prove that 6 is an injective continuous homomorphism of R into T,, and that 8(R) is a dense subgroup of T, . Deduce that T, is connected. (c) Let I be an open interval in R, with centre 0 and length < 1. Show that the subspace f 0, and hence get a contradiction.) 7.

Let G be a locally compact metrizable group with no small subgroups, and let V be a compact symmetric neighborhood of e containing no subgroup of G other than {el, and such that for all x , y E V the relation x 2 = y2 implies x = y (Problem 6(a)). (a) Show that if (an)is any sequence of points of V with e as limit, there exists a subsequence (b.) of (a,) and a sequence (k.) of integers > O such that the sequence (b:")converges to a point #e. (Consider the smallest of the integers k such that .:+I 6 V.) (b) Show that if r, s are two real numbers such that the sequences ( b p ' ) , (b:'"') converge to x, y, respectively, in G, then the sequence (b!,(""""') converges to xy. (Here [t] denotes the integral part of the real number t . ) (c) Using (a), (b), and the uniqueness of the square root, show that for every dyadic number r E [0, I ] the sequence (by'"') converges in G. (d) Let W c V be a neighborhood of e in G. Show that, for every real number r E [0, I], there exists a dyadic number s such that b[,'+s)knlE W for all n (use Problem 6(b)). Deduce that, for each r E [0, 11, the sequence (by'"') has a limit X(r). (e) If -1 2 r $0,put X ( r ) = ( X ( - r ) ) - I . Show that, if r, s and r s are all in the interval [- 1, 11, we have X ( r ) X ( s ) = X(r s) and that the mapping r t+X(r) of [-1, 11 into G is continuous. Deduce that this mapping can be extended to a nonconstant continuous homomorphism of R into G.

+

8.

+

Let I be the compact interval [0,1] in R,and let G be thegroup of all homeomorphisms of I onto I, which is contained in the Banach space '#,(I) (7.2.1). Show that the

10 SPACES WITH OPERATORS. ORBIT SPACES

45

(metrizable) topology induced on G by that of VR(I) is compatible with the group structure of G, and that right Cauchy sequences in G are the same as Cauchy sequences (7.1). &Give an example of a right Cauchy (in G) with respect to the norm on ‘if sequence in G which is not a left Cauchy sequence. 9.

(a) Let G’ be a complete metrizable group and let G o # G‘ be a dense subgroup of G’. Let G be the topplogical group obtained by giving Go the discrete topology. The identity mapping G + G o is a continuous bijective homomorphism, but its extension by continuity to a homomorphism G --t G’ (12.9.4) is not surjective. (b) Let G be the dense subgroup QZ of RZ,let 0 be an irrational number, and let u be the continuous homomorphism ( x , ~ ) H X 0y of G into R. Let G’ = u(G).The mapping u, considered as a homomorphism of G into G’, is bijective, but its continuous extension to a homomorphism of RZ into R is not injective. (c) Use (a) and (b) to construct an example of two complete groups G I , G2 and a continuous bijective homomorphism u of a dense subgroup HI of G I onto a dense subgroup Hz of Gz , such that the continuous extension of u to a homomorphism of G I into GZis neither injective nor surjective.

+

10. Let f be a continuous homomorphism of a subgroup H # {O} of R into a locally compact metrizable group G. Show that i f f is not an isomorphism of H onto the

subgroup f(H) of G, then f(H) is relatively compact in G. (Reduce to the case where H is closed in R andf(H) is dense in G . Begin by showing that, for each neighborhood W of the neutral element e in G, the set f-’(W) is unbounded, and deduce that f(H n R,) is dense in G. If V is a compact symmetric neighborhood of e in G, show that there exist a finite number of elements ti > 0 in H such that the neighborhoods f ( f l ) Vcover V. For each x E G, let A, be the set of all t E H such that f ( t ) E xV. Show that, if t E A,, there exists a ti such that t - ti E A,, and deduce that, if I is the largest compact interval with origin 0 in R containing all the t l , then I n A, is not empty. Deduce that G Cf(1) . V is compact.) 11. Let G be a commutative metrizable topological group, and let d be an invariant

distance defining the topology of G. Let h be the Hausdorff distance on %(G)corresponding to d (Section 3.16, Problem 3). Show that if M, N, P, Q are four bounded closed subsets of G, then h(MP, NQ) h ( M , N) h(P, Q).

+

10. SPACES WITH OPERATORS. ORBIT SPACES

Let G be a group and E a set. An action (or left action) of G on E is a mapping (s, x ) H S . x of G x E into E satisfying the following conditions:

-

(1) If e is the neutral element of G, then e x = x for all x E E. (2) For all s, t in G we have s ( t * x ) = ( s t ) x for all x E E.

-

- -

These conditions imply that s-’ (s x ) = x for all s E G and all x E E; hence for each s E G the mapping x ~ x sis a bijection of E onto E, and the inverse bijection is XHS-’ . x. For each x E E, the set G x of elements s . x where s E G is called the orbit of x (for the given action of G on E). The set S , of elements s E G such

XI1 TOPOLOGY AND TOPOLOGICAL ALGEBRA

46

that s * x = x is a subgroup of G called the stabilizer of x. The relation s * x = t * x is equivalent to t - ' s E S,. The mapping SHS x of G onto G x factorizes as follows: (12.10.1)

G A G/S,

3G

x

where G/S, is the set of left cosets sS, of S, in G ; p is the canonical mapping SHSS,; and cp is the bijection s S , w s x. The group G is said to act faithfitlly on E if the intersection of the stabilizers S,, as x runs through E, consists only of e. The group G is said to act freely on E if the stabilizer of every x e E consists only of e, or equivalently if for every x E E the relation s * x = t ' x implies s = t . The relation " y belongs to the orbit of x" is an equivalence relation on E, for which the equivalence classes are the orbits of the elements of E. The set of orbits is denoted by E/G (it is a subset of Cp(E)). If this set consists of a single element (in other words if, given any two elements x , y of E, there exists s E G such that y = s . x), then G is said to act transitively on E. The union G * A of the orbits of the elements of a subset A of E is called the saturation of A with respect to G ; the restriction to G x (G. A) of the mapping (s, X)HS * x is an action of G on G A. If : E -+ E/G is the canonical mapping (so that n(x) = G * x for all x E E), then G * A is equal to x-'(x(A)). The relation G A = A is equivalent to G ' A c A.

-

Now suppose that G is a topological group and E a topological space. Then G is said to act continuously on E if the mapping (s, X)HS * x of the product space G x E into E is continuous.

Examples (12.10.2) Let H be a subgroup of a topological group G . Then each of the actions (s, x)++sx and (s, X)HSXS-' of H on G is continuous. For the former action, the stabilizer of each x E G is the identity subgroup { e } , and the orbit of x is the right coset Hx; for the latter action, the stabilizer of x is the intersection H A b(x), where b(x) is the centralizer of x in G (12.8.6), and the orbit of x is the set of its conjugates hxh-' by elements h E H. If H, K are subgroups of G, the product group H x K acts continuously on G by ((s, t ) , x)Hsx~-'. The orbit of x is the double coset H x K of x with respect to H and K . If E is a real (resp. complex) topological vector space (12.13), the group R* (resp. C*) acts continuously on E by (A, x ) ~ A . xThe . orbits are {0} and the sets D - {0}, where D is a line (i.e. a one-dimensional subspace) in E.

10 SPACES WITH OPERATORS. ORBIT SPACES

47

If E is a Banach space, the group GL(E) of linear homeomorphisms of E onto E (12.8.1) acts continuously on E by (u, X ) H u(x) (5.7.4). Let G be a topological group and E a topological space. The mapping (s, X)HX of G x E into E is a continuous action of G on E, called the triuial action. Let G be a topological group acting continuously on a topological space E. If p is a continuous homomorphism of a topological group G' into G, then G' acts continuously on E by (s', x ) ~ p ( s '* )x. (12.10.3) r f G is a topological group acting continuously on a topological space E, then for each s E G the mapping x H s * x is a homeomorphism of E onto E.

For it is a continuous bijection, and the inverse bijection XHS-' also continuous.

x is

(12.10.4) If G is a topological group acting continuously on a Hausdorf topological space E, the stabilizer of each point of E is a closed subgroup of G . This follows immediately from (12.3.5). (12.10.5) Let G be a metrizable group acting continuously on a metrizable space E. Let A be a compact subset of G, and B a closed (resp. compact) subset of E. Then A * B is closed (resp. compact) in E.

The second assertion follows from the fact that A B is the image of the compact set A x B (3.20.16(v)) by the continuous mapping (8, X)HS * x (3.17.9). As to the first assertion, consider a sequence (s, * x,) of points of A * B (where s, E A and x, E B) with a limit z E E. By hypothesis, the sequence (s,,) has a subsequence (s,,,) converging to some point a E A . Since x,, = s";' (s,, * x,,,), the sequence (x,,,)converges to a-' * z. But B is closed in E and therefore a-' * z E B, hence z = a * (a-' * z ) E A * B. By (3.13.13), the proof is complete. Let G be a topological group acting continuously on a topological space E. Let E/G be the set of orbits and let TC : E + E/G be the canonical mapping, so that n(x) is the orbit G * x for each x E E. Let D be the set of subsets U of E/G such that n-'(U) is open in E. It follows immediately from the formulas (1.8.5) and (1.8.6) that r) is a topology on E/G. The set E/G, endowed with

48

XI1 TOPOLOGY AND TOPOLOGICAL ALGEBRA

this topology, is called the orbit space of the action of G on E. The mapping UH x-'(U) is a bijection of the set of open sets of E/G onto the set of saturated open sets of E. A subset F of E/G is closed in E/G if and only if n-'(F) is closed in E, because n-'((E/G) - F) = E - n-'(F).

(12.10.6) (i) The canonical mapping x : E -+ E/G is continuous. (ii) The image under n of every open set in E is open in E/G. (iii) A mapping f of E/G inzo a topological space E' is continuous if and only i f f n : E -,E' is continuous. 0

The first assertion follows from the definition of the topology of E/G and from (3.11.4(b)).To prove the second, it is enough to show that if V is open in E then its saturation G * V = n-'(n(V)) is open in E; and this is clear because G V = s * V, and each s V is open in E by virtue of (12.10.3). Finally, if

u

SEG

f:E/G + E' is continuous, then so is f

0 x by (i); conversely, i f f 0 x is continuous, then for every open set U' in E' the set n-'(f -'(U')) is open in E, hence f -'(U') is open in E/G. This proves (iii).

If x is any point of E and if V runs through a fundamental system of neighborhoods of x in E, the sets x(V) form a fundamental system of neighborhoods of the point x(x) in E/G.

(12.10.7) Let A be a subset of E and let A' = G A = n-'(n(A)) be its saturation with respect to G. Then the canonical bijecfion cp of the subspace n(A) of E/G onto the orbit space A'/G is a homeomorphism. The mapping cp is defined as follows: if x E A, the image under cp of the orbit G x is the same orbit considered as an element of A'/G. As U runs through the set of open subsets of E/G, the mapping cp takes U n n(A) to the canonical image of n-'(U) n A' in A'/G. Now U n n(A) runs through the set of open subsets of the subspace x(A) of E/G, and cp(n-'(U) n A') runs through the set of open subsets of A'/G. For if V' is a saturated open subset of .the subspace A' of E, then V' is of the form V n A', where V is open in E; but also V' = (G V) n A' = x-'(n(V)) n A', and x(V) is open in E/G by (12.10.6).

(12.10.8) Let G be a topological group acting continuously on a topological space E. In order that E/G should be Hausdorff it is necessary and suflcient that, in the product space E x E , the set R of pairs (x, y ) belonging to the same orbit be closed. When this is the case, every orbit is closed in E.

10 SPACES WITH OPERATORS. ORBIT SPACES

49

Let n(x) and n(y) be distinct points of E/G. If E/G is Hausdorff, there exist saturated open sets V, W in E such that V n W = fa and x E V and y E W. It is clear that the open set V x W in E x E contains ( x , y ) and does not meet R. Hence R is closed in E x E. Conversely, if R is closed in E x E, there exists an open neighborhood S of x and an open neighborhood T of y in E such that (S x T) n R = @. By (12.10.6), n(S) and n(T) are neighborhoods of n(x) and n(y), respectively. I f they intersected, there would exist s E S and t E T belonging to the same orbit, which means that (s, t ) E R ; and this is absurd. Hence E/G is Hausdorff. The last assertion follows from (12.3.4) and the continuity of n. (12.10.9) Let E be a metrizable space, G a topological group acting continuously on E, and n : E + E/G the canonical mapping. Suppose that E/G is metrizable. Then :

(i) ifE is separable, E/G is separable. (ii) ifE is locally compact (resp.' compact), E/G is locally compact (resp. compact); (iii) ifE is locally compact and K is any compact subset of E/G, there exists a compact subset L of E such that K = n(L).

If D is a denumerable dense subset of E, then n(D) is dense in E/G (3.11.4(d)). This proves (i). If V is a compact neighborhood of x E E, then n(V) is a compact neighborhood of n(x) in E/G ((3.17.9) and (12.10.6)); hence (ii). Finally, for each z E K , let V(z) be a compact neighborhood of a point of n-'(z) in E, so that n(V(z))is a compact neighborhood of z. There exist a finite number of points zi E K such that the n(V(z,)) cover K. Let L, be the compact set V(zi) in E. We have K c n(L,), hence the set L = L, n n-'(K)

u i

is compact (because it is closed in L, : (3.11.4) and (3.17.3)) and we have x(L) = K. (12.10.10) Let E be a locally compact, separable metrizable space and G a topological group acting continuously on E. Let n : E --+ E/G be the canonical mapping. Suppose that (1) E/G is Hausdorff, (2)for each x E E there exists a compact subset K ( x ) of E , containing x and such that the restriction of n to K ( x ) is injective and n ( K ( x ) )is a neighborhood o f n ( x ) . Then E/G is metrizable (and therefore, by (12.10.9), locally compact and separable).

We shall first show that there exists a sequence (K,) of compact subsets of E such that the restriction of n to K, is injective and the interiors of the sets n(K,,) cover E/G. By (3.18.3)there exists an increasing sequence (A,,) of compact subsets of E, whose union is E. For each z E n(A,), let x be a point of

50

XI1 TOPOLOGY AND TOPOLOGICAL ALGEBRA

n-’(z), and let K(x) be a set with the properties enunciated above. Then V(z) = (n(K(x)))O is an open neighborhood of z, and therefore n-’(V(z)) is an open neighborhood of n-‘(z) n A, . Since A,, is compact, there are a finite number of points zi E n(A,,) such that the sets n-’(V(z,)) (1 g i 6 p,,, say) cover A,. Let Ki,, be the set K(x) corresponding to V(zi), so that the interiors of the sets n(K,,,) form an open covering of n(A,,)in E/G. Then it is clear that the K i n( n 2 1 , 1 5 i 5 p , for each n ) satisfy the required conditions. This being so, the restriction of n to K, is a homeomorphism of K, onto the subspace n(K,,) of E/G, because E/G is Hausdorff (12.3.6), and this subspace is therefore compact and metrizable. The result therefore follows from (12.4.7). (12.10.11) Let G (resp. G‘) be a topological group acting continuously on a topological space E (resp. E’). Then G x G’ acts continuously on E x E‘ by (s,s‘) (x, x’) = (s x, s’. K‘), and the mapping w defined by

-

-

(G x G’) (x, x’) H ((G . x), (G’ . x’)) is a homeomorphism of(E x E’)/(G x G’) onto (E/G) x (E’/G’).

It is immediately checked that o is bijective, and it is continuous by virtue of (12.10.5). Moreover, every open set in (E x E’)/(G x G’) is the image, under the canonical mapping p : E x E’ + (E x E’)/(G x G’),of an open set U in E x E‘, and we have o ( p ( U ) ) = n(prl U) x n’(pr, U ) , where n : E -+E/G and 71‘ : E’ -+ E’/G’ are the canonical mappings. The set w ( p ( U ) ) is therefore open in (E/G) x (E’/G’), and the proof is complete. (12.10.12) Let G be a connected topologicalgroup acting continuously on a topological space E. If the orbit space E/G is connected, then E is connected.

Since the mapping s w s x of G onto G x is continuous, it follows

(3.19.7) that every orbit is connected. Suppose that there exist two non-empty open sets U, V in E such that U n V = @ and U u V = E. For each x E E, the sets U n (G * x) and V n (G * x) are open in G * x; their union is G * x

and their intersection is empty. Hence one of them is empty; in other words, U and V are saturated. But this implies that n(U) and n(V) are nonempty open sets in E/G whose intersection is empty and whose union is E/G, and this contradicts the hypothesis that E/G is connected. Remark (12.10.13) A right action of a group G on a set E is a mapping (s, x) Hx * s of G x Einto E such that x . e = xfor all x E E, and(x * t ) s = x . (ts)for all

10 SPACES WITH OPERATORS. ORBIT SPACES

51

x E E and all, s, t in G. Everything we have said can be immediately transposed to this situation; the set of orbits is sometimes denoted by G\E. For example, a subgroup H of a topological group G acts continuously on the right on G by the action (s,x) Hxs.

PROBLEMS

1. Let G be a locally compact metrizable group and E a locally compact metrizable space on which G acts continuously. For each pair of subsets K, L in E let P(K, L) denote the set of all s E G such that (s . K) n L # 0. (a) Show that if K is compact and L is closed in G, then the set P(K, L) is closed in G. The group G is said to act properly on E if P(K, L) IS a compact subset of G whenever K and L are compact subsets of E. (This will always be the case if G is compact.) (b) Show that, if G acts properly on E, then F . K is closed in E whenever F is closed in G and K is a compact subset of E. In particular, for each x E E, the orbit G * x is closed in E. (c) Under the same hypotheses, for each x E E the stabilizer S, of x is a compact subgroup of G, and the canonical map G/S, + G . x is a homeomorphism. (d) Under the same hypotheses, the orbit space E/G is Hausdorff. 2.

For each pair (a,t) of real numbers such that a 2 1, the pointf.(r) follows : if

t


0, where Db is the set of points (t, -1.0) with t E R , and D:(is the set of points ( t , 1, z) with t E R . The additive group R acts continuously on E as follows: (1) s . ( t , - 1,O) = (s f , - 1,O); (2) s . (f.(t), z)= (f.(s -tt ) , z ) ; (3) s . ( t , 1, z ) = ( t - s, 1, z). Show that the orbits of this action have the same properties as in Problems 2 and 3, and that the orbit space E/R is Hausdorff but not metrizable.

+

5.

Let E be a locally compact metrizable space and let G be a topological group acting continuously on E. Let w : E + E / G be the canonical map. Suppose that E/G is Hausdorff. (a) Let K be a compact subset of E and let U be an open neighborhood of K. Show that there exists a continuous mapping of E into [0,1] which takes the value 1 on w - ’ ( r ( K ) ) and the value 0 on the complement of a-‘(a(U)). (b) Deduce from (a) that there exists a continuous mapping of E/G into [0, 11, taking the value 1 on T ( K ) and the value 0 on the complement of w(U). (Show first that there exists a relatively compact open neighborhood U, of K in E such that 0, c U. Deduce that there exist two continuous mappings fi,fi of E into [0, t ] such that f i takes the value 4 on T-’(T(K)) and the value 0 on E - w-’(T(U,)),and fi takes the value 4 on T - ~ ( T ( ~ ~and ) ) the value 0 on E - T-~(T(U)). Consider the functionf, f2 Iterate this “interpolation” indefinitely and pass to the limit.) (c) If E is separable, show that there exists a sequence (U,) of relatively compact open sets in E such that the n(UJ form a basis for the topology of E/G. (d) Suppose that E is separable. Show that E/G is metrizable. (For each pair of indices m, n such that 0, c U., consider a continuous mappingf,. of E/G into [0, 11 which takes the value 1 on T(U,) and the value 0 on the complement of n(U,). Consider the continuous mapping x ~ ( f , . ( x ) ) of E/G into the product space RN N).

+ .

6. If M is a monoid with neutral element e , we define an action of M on a set E as at the beginning of (12.10). Suppose that E is a compact metric space and that, for each s E M, the mapping XHS . x is continuous. The closed orbit of x with respect to M is defined to be the set M * x ; it is stable under the action of M. Show that for each x E E there exists in M . x a minimal closed orbit Z (i.e., such that M . z = Z for all z E Z). (For each y E M . x let h(y) be the least upper bound, as t runs through M . y, of the distances from t to a closed orbit contained in M . t . Show that the greatest lower bound of the numbers h(y) with y E M . xis 0, by showing that otherwise E would not be precompact. -Deduce that there exists a sequence (y.) of points of M * x such that y , , , E M * y . and the

-

sequence (h(y.)) tends to 0. Now use the compactness of E to complete the proof.)

11. HOMOGENEOUS SPACES

Let G be a group, H a subgroup of G.Recall that the set of left cosets xH (resp. right cosets Hx)of H in G is denoted by G/H(resp. H\G). For each

11 HOMOGENEOUS SPACES

53

coset J = xH (resp. J = Hx) in G/H (resp. H\G) and each s E G, write s 3 = (sx)H (resp. 1 s = H(xs)). Then it is clear that G acts on the left (resp. on the right) on G/H (resp. H\G), and that this action is transitive. The set G/H (resp. H\G) together with this action of G is called the homogeneous space of left (resp. right) cosets of H in G. We shall work throughout with G/H. Everything we shall do can be transposed in an obvious way to apply to H\G. Let n denote the canonical mapping XHXH of G onto G/H. Clearly G/H is the set of orbits of elements of G for the right action (h, x ) w x h (12.10.13) of H on G. If G is a topological group, we can therefore topologize G/H with the topology defined in (12.10), which is called the quotient by H of the topology of G. If x, E G and if 3, = x,H is its image in G/H, we obtain a fundamental system of neighborhoods of 3 , in G/H by considering the canonical images in G/H of the neighborhoods V of x, in G (i.e., for each V, the set of cosets x H of the elements x E V, or equivalently the image of V (or of VH) under the canonical mapping n : G + G/H). Whenever we consider G/H as a topological space, it will always be this topology that is meant, unless the contrary is expressly stated. (12.11.1)

The group G acts continuously on G/H.

Let (so, a,) be a point of G x (G/H) and let xo be an element of the coset 3,, . Every neighborhood of so 3 , is of the form n(V), where V is a neighborhood of soxo in G . There exists a neighborhood U of so and a neighborhood W of xo such that the relations s E U and x E W imply sx E V; hence the relations s E U and 3 E n(W) imply s J E n(V).

-

-

(12.11.2) Let G be a topological group, H a subgroup of G. (i) G/H is Hausdorfl if and only if H is closed. (ii) G/H is discrete ifand only i f H is open. (iii) If H is discrete, then every x E G has a neighborhood V such that the restriction of n to V is a homeomorphism of V onto the neighborhood n(V) of n(x) = 3 in G/H.

(i) If G/H is Hausdorff, then { n ( e ) }is closed in G/H (12.3.4) and hence H = n-'(n(e)) is closed in G. Conversely, if H is closed in G, the set of pairs (x, y ) E G x G belonging to the same orbit under the action of H on the right is the set of (x, y ) such that x - ' y E H, and is therefore closed, being the inverse image of H under the continuous mapping (x, y ) w x - ' y . Hence G/H is Hausdorff (12.10.8). (ii) If G/H is discrete, the set (n(e)) is open in G/H and therefore H = n-'(n(e)) is open in G. Conversely, if H is open in G, then so is each

54

XI1 TOPOLOGY AND TOPOLOGICAL ALGEBRA

coset x H , and therefore { n ( x ) }= 7c(xH)is open in G/H (12.10.6). Hence G/H is discrete. (iii) Let U, be a neighborhood of e in G such that U, n H = { e } , and let V , be a symmetric open neighborhood of e such that V$ c.Uo (12.8.3). Then, for each x E G, the restriction of 7c to V = x V , is injective, because if h, h’ in H are such that xzh = xz’h’ with z , z’ in V,, we have h’h-’ = z ’ - l z E Vg c U, , so that h’ = h and z’ = z. Since the image under 71 of any open set in V is open in n(V) (12.10.6), and 7c is continuous, it follows that the restriction of I[ to V is a homeomorphism onto n(V). (12.11.3) Let G be a metrizable group, H a closed subgroup of G, and let d be a right-invariant distance dejning the topology of G (12.9.1). For any two points 1, j in G/H, put do(i,j ) = d(xH, y H ) (3.4). Then do is a distance

defining the topology of G/H. If G is complete (12.9.2), then G/H is complete with respect ro d o .

We remark first that if x E f and y

Ej

, we have do(f,j ) = d(x, yH). For

d(x,y H ) = inf d(x, yh) and therefore, for every h’ E H, we have d(xh’, yH)

=

hEH

d(x,y H ) by virtue of the right-invariance of d. Hence, for each z E G, we have ldo(f, i) - dO(3,ill

=

14%zH) - dcv, zH)I 6 4x9 Y )

by (3.4.2). Since this is true for all x , y in the respective cosets f, j , it follows that Id,(& 2 ) - do(j, i)l6 do(& j ) , which shows that do is a pseudo-distance on G/H. The relation do(f, 3 ) < r is equivalent to the existence of a point y E 3 such that d ( x , y ) c r . Since H is closed in G, this proves that do is a distance, and that if B(x; r ) is the open ball with center x and radius r (with respect to d ) , then its image under the canonical mapping n : G + G/H is the open ball with center n(x) and radius r with respect to do. Hence do defines the topology of G/H. Now suppose that G is complete. Let (1,)be a Cauchy sequence with respect to do. We shall show that, by passing to a subsequence of (1,)if necessary, we can reduce to the case where there exists a right Cauchy sequence (x,) in G such that n(x,) = 1,. Since by hypothesis the sequence (x,) converges, this will show that ( i nhas ) a cluster value and therefore converges (3.14.2), and hence that G/H is complete. There exists in G a denumerable fundamental system (V,) of neighborhoods of e such that V;,, c V, for all n (12.9.1 and 12.8.3). Let (&), be a sequence of real numbers > O such that the relation d(e, z ) < E, implies that z E V, . The hypothesis implies that we can define inductively a subsequence (Ank) of (a,) such that d o ( i n pin4) , < E~ for all p 2 k and q 2 k. Replacing the original sequence (1,)by (ink), we can therefore assume that d O ( i pkq) , < E, for

12 QUOTIENT GROUPS

55

all p 2 n and q >= n. Since d is right-invariant, this means that for all y E Rp the intersection of R, and the neighborhood V ,y is not empty. We can then define, by induction on n, a sequence (x,) in G such that x , E R , and x,, E V ,x, for all n. Hence it follows, by induction on p , that for all p > 0 we n ~ V n - V,2+1 1 ~ n c V , for have x,+*EV,+ p - 1 V n + p - 2 . . . V n t l V n ~ (because all n). Hence the sequence (x,) is a right Cauchy sequence in G. Q.E.D. Let G be a topological group, and let E be a topological space on which G acts continuously and transitively. Let x E E and let S, be the stabilizer of x in G; then (12.10.1) the continuous mapping h, : SHS * x of G into E, which by hypothesis is surjective, factorizes canonically as (12.1 1.4)

h, : G 2 G/S, 2 E,

where n, is the canonical mapping of G onto the homogeneous space G/S,, and f, is the bijection SS,H s x . It follows from (12.10.6) that f, is a continuous bijection, but it is not necessarily a homeomorphism (Section 12.12, Problem 2). (12.1 1.5) With the notation above, the bijection f, : G/S, -,E is a homeomorphism for each x E E if and only if there exists a point xo E E such that the mapping h,, : SHS * xo transforms each neighborhood of e in G into a neighborhood of xo in E.

In view of (12.10.6) we need only prove that the condition is sufficient. Notice first that each x E E can by hypothesis be written in the form x = t xo for some t E G. I f V is a neighborhood of e, it follows from the hypothesis that V . x = ( V t ) xo is a neighborhood of x, because we can write ( V t ) x o = t ( ( t - l v t ) . xo), and t -'Vt is a neighborhood of e in G (12.8.3), and ~ H Sy is a homeomorphism of E onto E, for each s E G (12.10.3). Since each open set in G/S, is the image under n, of some open set in G, it is enough to show that for each open set U in G, and each x E E, the set h,(U) =f,(n,(U)) = U * x is open in E. Now, for each t E U, the set t -'U is a neighborhood of e, and therefore ( t -'U) x is a neighborhood of x in E, from the first part of the proof. Hence U * x = t . ( ( t- ' U ) x ) is a neighborhood of t * x ; by (3.6.4), the proof is complete.

-

-

12. QUOTIENT GROUPS

If H is a normal subgroup of a group G, then xH = H x for each x E G and therefore the sets G/H and H\G are identical; and G/H has a group structure

56

XI1 TOPOLOGY AND TOPOLOGICAL ALGEBRA

(the quotient of G by H) for which the canonical mapping n : G -+ G/H is a homomorphism. (12.12.1) Let G be a topologicalgroup and H a normalsubgroup of G. Then the quotient by H of the topology of G is compatible with the group structure of G/H.

Let (,to,j o ) be a point of (G/H) x (G/H), and let xo (resp. yo) be an element of the coset f, (resp. jo).Every neighborhood of i o j 0is of the form n(U), where U is a neighborhood of xo y o . There exists a neighborhood V of xo and a neighborhood W of yo in G such that the relations X E V and y e W imply x y U. ~ Hence the relations , t ~ n ( V )and j ~ n ( W imply ) i j E n(U), and therefore the mapping ( i j,) ~ + , t jis continuous at the point (a,, yo). The proof of continuity of the map 2 w i - l is similar. Whenever we speak of the quotient G/H o f a topological group G by a subgroup H, it is to be understood that G/H carries the quotient topology. (12.12.2) Let G be a topologicalgroup and let H , K be two normal subgroups of G such that H 3 K. Then the canonical bijection cp : G/K -+ (G/H)/(K/H) is an isomorphism of topological groups.

We know that cp is the unique mapping which makes the following diagram commutative : G

A G/K

where n, d ,and n" are the canonical homomorphisms. The definition of a quotient topology (12.10) shows that a subset U of (G/H)/(K/H) is open if and only if n'-'(n''--'(U>)is open in G, that is, if and only if n-'(cp-'(U)) is open in G ; but this last condition signifies that cp-'(U) is open in G/K. Since cp is bijective, the result follows. (12.12.3) Let G be a topologicalgroup, H a normal subgroup of G , 71: G + G / H the canonical mapping, and A a subgroup of G. Then the canonical bijection of n(A) onto the quotient group AH/H is an isomorphism of topological groups.

This is a particular case of (12.10.7), applied to the right action (h, x ) ~ x of h H on G.

12 QUOTIENT GROUPS

57

There is also a canonical bijection II/ of the quotient group A/(A n H) onto the subgroup n(A) of G/H ;this bijection II/ is the unique mapping which makes the following diagram commutative: A

(1 2.12.4)

/n

b n o j

A/(A n H)+

n(A)

where n' is the canonical mapping and j : A + G the canonical injection. It follows from (12.10.6) that is continuous, but it is not necessarily bicontinuous (Problem 2). However:

(12.12.5) Let G be a metrizable group and H a compact subgroup of G . For each closed subgroup A of G , the canonical bijection II/ : A/(A n H) -+ n(A) is an isomorphism of topological groups. It is enough to show that the image under II/ of an arbitrary closed subset F of A/(A n H) is closed in n(A). Now n'-'(F) = B is a closed subset of A, hence closed in G , and $(F) = n(j(B)) = n(BH). But since H is compact, the set BH is closed in G (12.10.5) and saturated, hence n(BH) is closed in G/H (12.10) and therefore also closed in n(A).

(12.12.6) Let G,, G2 be two topological groups and HI (resp. H,) a normal subgroup of GI (resp. G 2 ) .Then the canonical bijection 0 : (GI

x G2)/(H, x H2)

-+

(G,/H,) x (G2/H2)

is an isomorphism of topological groups. This is an immediate consequence of (12.10.11 ) .

(12.12.7) Let G, G' be two topological groups and u : G -+ G' a continuous homorphism. Then u factorizes canonically as G 5 G/N 1;u ( ~2) G! where N is the kernel of u, p is the canonical homomorphism, and j the canonical injection. The mapping u is a continuous bijection (12.10.6(iii)), but is not necessarily bicontinuous. When u is bicontinuous, the homomorphism u is said to be a strict morphism of G into G ' ; for this to be the case, it is necessary and sufficient that for each neighborhood V of e in G, the set u(V) should be a neighborhood of e' = u(e) in u(G) (12.11.5).

58

XI1 TOPOLOGY AND TOPOLOGICAL ALGEBRA

PROBLEMS

1. (a) Let G be a topological group, K its identity component, H a subgroup of G contained i n K. Show that the connected components of the space G/H are the images of the connected components of G under the canonical mapping T : G + G / H . Show that K is the smallest of the subgroups L of G such that G/L is totally disconnected. (b) Let G be the additive subgroup of the Banach space d , ( N ) (7.1.3) consisting of the mappings nt-+f(n) such that f(n) E Q for all n E N and lim f ( n ) exists in R. The

"- uo

distance induced on G by that on d R ( N ) defines a topology compatible with the group structure of G , and with respect to this topology G is totally disconnected. Let H be the subgroup of G consisting of all f~G such that lim f(n) = 0. Show that H is closed n- m

in G and that G / H is isomorphic to R,and therefore connected. (c) Let G be a locally compact metrizable group, H a closed subgroup of G, and rr : G + G/H the canonical mapping. Show that the connected components of G/H are the closures of the images under T of the connected components of G. (Reduce to the case where G is totally disconnected, and use Problem 3 of Section 12.9.) 2.

In the additive group R,let H be the subgroup Z and let A be the subgroup 0Z, where 0 is an irrational number. Show that the canonical bijection A/(A n H) +(A H)/H is not an isomorphism of topological groups.

+

3. Let p be a prime number and let (GJ be an infinite sequence of topological groups all equal to the discrete group Z / p 2 Z .Let H. be the subgroup p Z / p 2 Z of G, . Let G be the subgroup of the product g r o u p n G. consisting of all x = (x.) such that xn E H. for all n

but a finite set of values of n. Let 23 be the set of neighborhoods of 0 in the product group H, = H c G. Show that W is a fundamental system of neighborhoods of 0

n

for a topology on G compatible with the group structure, and that in this topology G is metrizable and locally compact and G/H is discrete. Show that the homomorphism u : X H ~ Xof G into G is not a strict morphism of G into G, and that u(G)is not closed in G. 4.

Let G be a metrizable group, K a closed normal subgroup of G. If K and G/K are complete, show that G is complete.

5. (a) Let G be a topological group, K a normal subgroup of G. Show that if K and G/K have no small subgroups (Section 12.9, Problem 6), then G has no small subgroups. (b) Deduce from (a) that if H I , H 1are two normal subgrouos of a topological group G, such that G/H1 and G/H, have no small subgroups, then G/(H, n H,) has no small subgroups. 6. Let G be a connected, locally compact, metrizable group, K a compact normal subgroup of G , and N a closed normal subgroup of K, such that K/N has no small subgroups (Section 12.9, Problem 6). Show that N is a normal subgroup of G. (Observe that the hypothesis on K / N implies the existence of a neighborhood U of e in G such that xNx-' c N for all x E U.)

7. Let G be a connected metrizable group and H a compact normal commutative subgroup of G, with no small subgroups. Show that H is contained in the center of G .

13 TOPOLOGICAL VECTOR SPACES

59

(Observe that, if s is close to e in G, then the image of H under the mapping is close to e.)

xt-+sxs-'x-'

8. Let G be a locally compact metrizable group and H a closed normal subgroup of G , such that G/H is not discrete and has no small subgroups. Show that there exists a f continuous homomorphism f:R + G such that the composition R + G +G/H is nontrivial (see Section 12.9, Problem 7).

13. T O P O L O G I C A L V E C T O R SPACES

We shall adhere to the conventions of (5.1), so that all vector spaces under consideration have as field of scalars either the real field R or the complex field C. If E is a vector space over R (resp. C),a topology on E is said to be compatible with the vector space structure if the mappings ( x ,y ) - x + y of E x E into E, and (A, X ) H A X of R x E (resp. C x E) into E are continuous. A vector space endowed with a topology compatible with its vector space structure is called a topological vector space. Since -x = (- l)x, the conditions above imply that the topology is afortiori compatible with the additive group structure of E, and all the notions and properties established for topological groups in the preceding sections will apply in particular to topological vector spaces (provided of course that these properties are transcribed into additive notation). An isomorphism of a topological vector space E onto a topological vector space F is a linear bijection of E onto F which is a homeomorphism. If llxll is a norm on a vector space E, the topology on E defined by the distance IIx - yll is compatible with the vector space structure of E (5.1.5) and therefore makes E a topological vector space. Two equivalent norms (5.6) define the same topology. In a topological vector space E, the translations XHX a and the homotheties XH Ax ( A # 0 ) are homeomorphisms of E onto E, the inverse mappings being, respectively, the translation X H x - a and the homothety xwa-lx. In a vector space E over R (resp. C) a set M is said to be balanced if, for each x E M and each scalar A such that 1A1 =< I , we have Ax E M (in other words, if AM c M whenever 111 5 1). The set M is said to be absorbing if, for each X E E , there exists a real number a > 0 such that A X E M whenever (A16 a. Every absorbing subset of E generates the vector space E. For a balanced subset M of E to be absorbing, it is enough that for each x E E there should exist a scalar A # 0 such that Ax E M.

+

60

XI1 TOPOLOGY A N D TOPOLOGICAL ALGEBRA

(1 2.13.1 ) In a topological vector space E, the absorbing balanced neighborhoods of 0 form a fundamental system of neighborhoods of 0.

For each xo E E, the mapping AH AxO of the scalar field into E is continuous at the point 0; hence, for each neighborhood V of 0 in E, there exists ct > 0 such that the relation 1 11 S CI implies Ax, E V. This shows that every neighborhood of 0 is absorbing. On the other hand, the continuity of the mapping (A, x) I--, Ax at the point (0,O) implies that, for each neighborhood V of 0 in E, there exists ct > 0 and a neighborhood W of 0 in E such that the relations 111 6 ct and x E W imply Ax E V. The union U of the sets 1W, where 1A1 6 CI, is clearly a balanced neighborhood of 0 in E, and is contained in V. Hence the result. Let F be a vector subspace of a topological vector space E. Then the topology induced on F by the topology on E is compatible with the vector space structure of F. Whenever we consider a vector subspace F as a topological vector space, it is always the induced topology that is meant, unless the contrary is expressly stated. The proof of (5.4.1) applies without any change and shows that F is a vector subspace of E. The definition of a total subset of E is the same as in (5.4). On the quotient vector space E/F, the quotient topology (12.11) is compatible with the vector space structure. For if n : E + E/F is the canonical mapping, let ( A o , 2,) be any point of R x (E/F) (resp. C x (E/F)) and let xo be any point of 2,. Then every neighborhood of 1, i o contains a neighborhood of the form n(V),where V is a neighborhood of 1, *yo.There exists a neighborhood U of A, in R (resp. C ) and a neighborhood W of x, in E such that the relations 1 E U and x E W imply Ax E V; consequently the relations 1 E U and R E n(W) imply 1i E n(V), and the assertion is proved. Whenever we consider E/F as a topological vector space, it is always with the quotient topology, unless the contrary is expressly stated. The criteria for continuity in a product ((3.20.15) and (12.5)) show immediately that if El, E, are two topological vector spaces, the product topology on El x E2 is compatible with the vector space structure of El x E, . The vector space El x E 2 , endowed with this topology, is called the product of the topological vector spaces El and E 2 . Proposition (5.4.2) and the definition of the topological direct sum of two subspaces are valid without modification for arbitrary topological vector spaces. If E is the direct sum of two subspaces F,, F, , and if p 1 : E + F, is the corresponding projection, then the kernel of p 1 is F, , and p1 therefore has a

13 TOPOLOGICAL VECTOR SPACES

61

canonical factorization E + E/F2% F,, where jl is a bijection. To say that E is the topological direct sum of F1 and F2 signifies that .jl is continuous (12.10.6), or equivalently that the inverse bijection, which is the restriction to F, of the canonical mapping n2 : E -+ E/F2 (and which is always continuous) is bicontinuous. (12.1 3.2) (i) Let E be a topological vector space and let H be a hyperp1ane.h E , with equation f ( x ) = 0. Then H is closed in E ifand only if the linear form f is continuous on E. (ii) Let E be a HausdorJf topological vector space over R (resp. C ) of finite dimension n. I f ( a i ) 1 6 i s nis a basis of E, the mapping

of R" (resp. C") onto E is an isomorphism of topological vector spaces. (iii) Let E be a topological vector space, M a closed vector subspace of E , and F a finite-dimensional vector subspace of E. Then M F is closed in E. In particular, if E is Hausdorf, every finite-dimensional vector subspace of E is closed in E. (iv) Let E be a topological vector space and let M be a closed vector subspace of finite codimension in E. Then every algebraic supplement N of M in E is a topological supplement.

+

We shall prove (ii) first. Since u is continuous and bijective, we have to show that u is bicontinuous, or equivalently (considering the inverse images under u of open sets in E) that if D is a Hausdorf topology compatible with the vector space structure of R" (resp. C"), and coarser than the product topology Do, then 0 = Do. Put JIxIJ= ~ u p l 0 such that p'(x) _I c . sup pi(x) i for aN x E E.

Apply (12.14.11) to the identity map u

=

l E (12.2.1).

(12.14.13) Finally, we remark that the notions of series and convergent series are defined in a topological vector space E just as in a normed space (5.2); propositions (5.2.2), (5.2.3), (5.2.4) and the fact that the general term of a convergent series tends to 0 remain valid without change. If E is locally convex and its topology is defined by a family ( p J of seminorms, then for a series with general term x, to converge in E it is necessary that, for each E > 0 and each scalar I , there should exist an integer no such that

for all n 2 no and q > 0. If E is a FrCchet space, this condition is also sufficient, as follows from Cauchy's criterion (12.9).

PROBLEMS

1. Let E be a vector space over R (resp. C) and let 8 be a set of subsets of E satisfying conditions (V,) and (V,,) of Section 12.3, Problem 3, together with the following: (EV,) Every set V E 8 is balanced and absorbing.

14 LOCALLY CONVEX SPACES

69

(EV,,) For all V E 23 and all h # 0 in R (resp. C), we have XV E B. (EVIII) For each V E B, there exists W E B such that W W c V. Show that there exists a unique topology on E, compatible with the vector space structure, for which 8 is a fundamental system of neighborhoods of 0.

+

2.

Let E be a vector space over R with a denumerable infinite basis (en), and let 5 be the set of aN balanced absorbing sets in E. Show that 5 does not satisfy axiom (EVIII) of Problem 1. (For each n > 0, let A. be the set of points

n 1=1

he, such that [tIl6 l/n for

1 5 i 5 n, and let A be the union of the sets A,, . Show that there exists no set M € 5such that M M C A.)

+

3. Let E be the vector space VR(R) of continuous mappings of R into R. For each continuous function m such that m(f) > 0 for all r E R,let V, denote the set offunctions YE E such that If(r)l 5 m(f) for all t E R. (a) Show that the V, form a fundamental system of neighborhoods of 0 in a topology .Toon E which is compatible with the additive group structure of E, but not with the vector space structure. (b) Let D be the vector subspace of E consisting of functions with compact support (12.6). Show that the topology Y induced by T oon D is compatible with the vector space structure of D, and is not metrizable. 4.

Let (EJaG1be any family of topological vector spaces, and E =

n E. the product a.1

vector space. Show that the product of the topologies of the Em(12.5) is compatible with the vector space structure of E. If each E. is locally convex and its topology is defined by a family of seminorms (p& Lm, then the topology of E is locally convex and is defined by the family of seminorms par pr. (for all a E I and all A E Le).In particular, every (finite or) denumerable product of Frkhet spaces is a Frkchet space. 0

5. Let p be a seminorm on a (real or complex) vector space E. Show that the set N of x E E such that p(x) = 0 is a vector subspace of E. The seminorm defined by (12.14.8.1) on E/N is a norm. Let E be a Hausdorff locally convex space whose topology is defined by a family (pa) of seminorms, and for each a let N. be the subspace of E consisting of those x E E for which p.(x) = 0. Show that E is isomorphic to a subspace of the product of the normed spaces E/N, (the norm on E/N. being i.). In particular. every Frkchet space is isomorphic to a closed subspace of a denumerable product of normed spaces. 6. A subset A of a vector space E over R is said to absorb a subset B of E if there exists a > 0 such that hB c A whenever 5 a. If E is a topological vector space, a subset

1x1

B of E is said to be bounded(with respect to the topological vector space structure of E) if every neighborhood of 0 absorbs B. (a) The closure of a bounded set is bounded. Every finite set is bounded. If A, B are bounded, then so are A u B and A B. If E is metrizable, every precompact set in E is bounded. (b) Suppose that E is Hausdorff. Then a subset B of E is bounded if and only if, for every sequence (x,,) in B and every sequence (A,) of real numbers tending to zero, the sequence (h.~.) tends to 0 in E. (c) Let (E.) be a family of topological vector spaces and E E. their product

+

=n 8

(Problem 4). Show that a subset B of E is bounded if and only if pra(B) is bounded in E. for each a.

70

XI1 TOPOLOGY AND TOPOLOGICAL ALGEBRA (d) Let E, F be two topological vector spaces and f:E + F a continuous linear mapping. Then for each bounded subset B of E, the image set f(B) is bounded in F.

7. (a) Let E be a (real or complex) Hausdorff topological vector space. Show that, if there exists a bounded neighborhood V of 0 in E (Problem 6), the sets n-'V (n an integer > 0 ) form a fundamental system of neighborhoods of 0 in E, and consequently E is metrizable. If furthermore E is locally convex, then its topology can be defined by a single norm. (b) Let E be a metrizable locally convex space whose topology is defined by an increasing sequence ( p , ) of seminorms. Show that the topology of E can be defined by a single norm if and only if there exists an integer N such that, for all n 2 N, there exists a real number k. 2 0 for which pn 2 k n p N . (c) Let E be the real vector space of indefinitely differentiable real-valued functions on the interval I = [0, I ] in R.For each n 2 0 and eachfe E, put

(where Dof=f). Show that each p . is a norm on E and that the topology defined by the sequence of norms (p,) cannot be defined by a single norm (cf. (17.1)). (d) Let E be a Hausdorff locally convex space whose topology is defined by a sequence of seminorms but cannot be defined by a single norm. If d is a translationinvariant distance defining the topology of E (1 2.9.2) show that every set in E which is bounded in the sense of the topological vector space structure of E is bounded with respect to the distance d , but there exist subsets of E which are bounded with respect to d but are unbounded with respect to the topological vector space structure of E. 8. Let E be a metrizable topological vector space. (a) Show that every balanced subset of E which absorbs all sequences in E which converge to 0 is a neighborhood of 0 in E. Deduce that, if u is a linear mappingof E into a topological vector space F, and u transforms every sequence converging to 0 in E into a bounded sequence in F, then u is continuous. (b) Let (B,) be any sequence of bounded subsets of E. Show that there exists a sequence (A,) of nonzero scalars such that the union of the sets A,B. is bounded. 9. Let E be a topological vector space. A set b of bounded subsets of E is called a fundamental system of bounded subsets if every bounded set in E is contained in a set belonging to b. Show that, if E is a metrizable locally convex space whose topology

cannot be defined by a single norm, no fundamental system of bounded subsets of E is denumerable (use Problem 8).

10. Let (En)nslbe a sequence of normed spaces such that Elis not separable. Let F be a E,, such that pr.(F) = E, for each n. Show that there exists vector subspace of E

=n n

a real number 6 > 0 and a bounded sequence (x,,Jmal in F such that llprlxl- pr1xAl

28

whenever i # j . (Remark that, for each index n and each nondenumerable subset A of F, there exists a nondenumerable subset B of A such that pr.(B) is bounded in En.) Deduce that a Frkchet space is separable if every bounded subset is relariuely compact. 11. A subset A of a real or complex vector space E is convex if, whenever x and y belong to A and 0 5 A 6 1, we have kr + (1 - A)y E A (Section 8.5, Problem 9). It follows

71

14 LOCALLY CONVEX SPACES

that, if A is convex and ( x , ) ~ , < ,is, 0, show that there exists a continuous mapping s : L -+ E such that p ( z - f ( s ( z ) ) ) E for all z E L. (Consider a locally finite denumerable open covering (U,) of L such that p(x' - x) 2 &e whenever x and x' lie in the same U,, a continuous partition of unity (h,) subordinate to (U"),and for each n a point a,,E U, nf(E). If b, ~ f - ' ( a . )show , that the function s defined by s(z)

=Cb,,h,(z) satisfies the required conditions.) I

15 WEAK TOPOLOGIES

73

15. WEAK TOPOLOGIES

(12.15.1) Let (E,),e I be a family of locally convex spaces, and f o r each a let Lm be a family of seminorms dejining the topology of E, . Let E = E,

n

be the product vector space, and for each a E I and each I (12.1 5.1 . l )

&(x)

= pal(prorx)

E L,

, put

UEE

for all x E E.

Then the pad are seminorms on E and dejine the product topology (12.5) on E. It is straightforward to check that the pLl are seminorms, and then the second assertion follows from the definition (12.5) of a product of uniformizable spaces. In particular, when all the E, are equal to the same locally convex space F, then the product topology on the vector space F' of all mappings of I into F is defined by the seminorms

f

HP ( f

(4)

where a E I and p belongs to a set r of seminorms defining the topology of F. This topology is called the topology of simple (or pointwise) convergence on the vector space F', or on any of its vector subspaces (cf. (12.5)). With respect to this topoiogy, each mapping f w f (a) (a E I) is a continuous linear mapping on F', and maps each open set in F' to an open set in F (12.5.2). The topology of simple convergence is Hausdorf if F is Hausdorff (12.5.7). A mapping t wft of a topological space T into F' is continuous with respect to the topology of simple convergence if and only if the mapping t w f , ( a ) of T into F is continuous, for each a E I (12.5.5). When F is a Banach space over K (= R or C) and T is an open subset of K", a mapping t wft of T into F' is said to be p times diferentiable (resp. indefinitely diferentiable, resp. analytic) with respect to the topology of simple convergence if, for each index a E I, the mapping t w f t ( a ) of T into F has the corresponding property. We shall be particularly concerned with the case in which F is the field of scalars K, and I is a vector space E over K. Let V be a subspace of KEwhose elements are linear forms on E. The topology of simple convergence on such a subspace V is called the weak topology; it is defined by seminorms (12.1 5.2)

f-

If(X>l,

where x runs through E. For each x E E, the mapping f ~ f l x is) a continuous linearform on V with respect to the weak topology. A sequence (f.)of elements

74

XI1 TOPOLOGY AND TOPOLOGICAL ALGEBRA

of V which converges to f E V with respect to the weak topology (i.e., is such that lim f,(x) =f ( x ) for all x E E) is also said to be weakly conuergent. On n-) m

the same principle, we shall speak of subsets of V being weakly closed (in V), weakly compact, etc. A subset B of KE is said to be weakly bounded if all the seminorms (12.15.2) are bounded on B, that is if sup I f ( x ) l < 00 for

+

JEB

all x E E. We shall also use the phrases weakly continuous (resp. weakly differentiable, weakly analytic) or scalarly continuous (resp. scalarly differentiable, scalarly analytic) in place of “continuous (resp. differentiable, analytic) with respect to the weak topology.”

If E is a locally conuex space (over K = R or C), the vector space of continuous linear forms on E is called the dual of E, and is often denoted by E’. If x’ E E is a continuous linear form on E, its value at a point x E E is often denoted by the expression (x, x’) or (x’, x) instead of x’(x). Since E‘ is a vector subspace of KE, it is endowed with the weak topology defined by the seminorms X ’ H I(x, x’)l. For each x E E, the function X ’ H ( x ,x‘) is a weakly continuous linear form on E’. If E is a FrCchet space and E, a dense vector subspace of E, then it follows from (12.9.4) and the principle of extension of identities (3.15.2) that the mapping X ’ H X ’ I E, is an (algebraic) isomorphism of the dual E‘ of E onto the dual Eb of E, . (12.15.3) Let E, F be two locally convex spaces, E‘ and F’ their respective duals, and u : E -+ F a continuous linear mapping. Then Y ‘ H Y ‘ u is a linear mapping of F’ into E’ which is continuous with respect to the weak topologies. 0

If y’ is a continuous linear form on F, then y’ u is a continuous linear form on E. Given x, E E, there exists yo E F such that (x,, , y’ u ) = ( y o , y’) for all y’ E F’, namely yo = u(x,); the continuity of the mapping y ’ y’ ~ u is therefore an immediate consequence of (12.14.11) and the definition (12.15.2) of the seminorms which define the weak topology. 0

0

0

, the linear mapping ‘u : F’ -+ E’ is called the We write y’ u = ‘ ~ ( y ’ )and transpose of the continuous linear mapping u. Thus we have 0

(u(x), Y‘> = (x, ‘4Y’))

(12.15.4)

for all x E E and y’ (12.15.5)

E

F’. Clearly

yu1

+ u2) = +

tU2,

‘(Au)= A . ‘u

for any scalar 1,and for any continuous linear mappings u l , u2 of E into F. If

15 WEAK TOPOLOGIES

75

v is a continuous linear mapping of F into a locally convex space G, then (12.15.6)

‘(0

0

u ) = 5.4 0 %.

The weak topology on the dual of a locally convex space is not metrizable in general (Problem 2). However, there is the following result: (12.15.7) Let E be a separable metrizable locally convex space. If H is any equicontinuous (7.5) subset of the dual E’ of E, then the weak closure of H in E‘ is a compact metrizable space in the weak topology.

We shall first prove the following lemma: (12.15.7.1 ) Let E be a metrizable vector space and F a normed space. In order that a set H of linear mappings of E into F should be eguicontinuous, it is necessary and suficient that there should exist a neighborhood V of 0 in E and a real number c > 0 such that IIu(x)I/I c for all x E V and all u E H.

(If E is normed, this condition is also equivalent to sup llull < +a UEH

by (5.7.1).)

The condition expresses that H is equicontinuous at the point 0. If this is the case, then for each xo E E and each x E xo EV,we have

+

IIU(4

for all u

E

- u(xo>ll= l l 4 X

- X0)Il

I EC

H, and therefore H is equicontinuous at xo .

In proving (12.15.7), we may first of all restrict ourselves to the case where H is weakly closed in E’. For if R is the weak closure of H in E’, the continuity of the mapping X’H ( x , x’) on E’ implies that if I(x, x ’ ) ] 5 c for all x E V and all x’ E H, then also I(x, x’)l 5 c for all x E V and all x’E R (3.15.4), whence the result by (12.15.7.1). Now let (a,,) be a sequence which is dense in E. We shall show that the weak topology on H is defined by the pseudo-distances

d,,(x’, y’) = I & , ,

x’ - Y Y .

Clearly the topology defined by these pseudo-distances is coarser than the weak topology; hence it is enough to show that it is alsofiner. By hypothesis, every neighborhood of a point xb E H in the weak topology contains a subset W of H consisting of elements x’ E H satisfying ajinite number of inequalities (12.15.7.2)

[ ( x i ,x’- xb)l < r

(1

5 i 5 m)

76

XI1 TOPOLOGY A N D TOPOLOGICAL ALGEBRA

where the x i are points of E, and r > 0. Suppose that H satisfies the condition (12.15.7.1). Choose E > 0 such that ~ E < C +r. For each index i, there exists an a,,(i)such that x i - a,,(i)E EV. By virtue of (12.15.7.1), it follows that for each xf E H we have [ ( x i - a,,(i),xf - xb)l

5~

< +r,

E C

and the relations x’ - xb)l < f r therefore imply the inequalities (12.15.7.2). Since the weak topology is Hausdorff, we have shown that H is metrizable (12.4.6). More precisely, the above argument shows that the map~ H into the product space KN (where K = R or C) ping x’t-+((a,, x ’ ) ) , , ~of is a homeomorphism of H onto a subspace L of KN.So it remains to be shown that L is compact. In the first place, the projections of L are boundedsubsets of K (and therefore relatively compact ((3.17.6) and (3.20.16))): for each index n there exists A,, > 0 such that AnanE V and therefore [(a,,, x’)l 5 c/A,, for all x’ E H. By virtue of (12.5.9), (12.5.4), and (3.17.3), it remains to be shown that L is closed in KN.Now if (xh) is a sequence of points in H such that for each n the sequence ((a,,, ~ k ) ) , , , > converges ~ in K, then it follows from (7.5.5) that the sequence (xk) converges simply in V. Since for each Z E E there exists a scalar y such that yz E V (so that (z, x i ) = y-l(yz, xh)), the sequence (xh) converges simply in E: its limit y’ is a linear mapping, by the principle of extension of identities, and is continuous by (7.5.5), and is therefore a point Q.E.D. of H. We recall that if E is a normed space, its dual E’ = 2 ( E ; K) is a Banach space with respect to the norm IIx’II = sup I(x, x’)l (5.7.3); since IIXII 5

1

I ( x , x’)I 5 ((x(( I(xf((, the topology defined by this norm (sometimes called the strong topology on E‘) isjner than the weak topology on E’.

(12.15.8) Let E be a normed space. The mapping X’H I(x’I( of E’ into R is lower semicontinuous with respect to the weak topology.

For it is the upper envelope of the weakly continuous mappings

x’H((x, x’)], where x runs through the ball (1x115 1 in E. Hence the result follows from (12.7.7).

If (x;) is a sequence in E’ which converges weakly to a’, then by (12.7.13) we have (12.15.8.1)

lim inf (lxLll 2 Ila’ll. n-tm

15 WEAK TOPOLOGIES

77

(12.15.9) Let E be a separable normed space. Then every closed ball B': IIx'II

5 r in E' is metrizable and compact with respect to the weak topology.

By (12.15.7) and (12.15.7.1) it is enough to show that B' is weakly closed, and this follows from the fact that the function x ' w llx'II is lower semicontinuous on E' (12.7.2). Now let E be a Hilbert space (6.2). For each x E E, let j ( x ) denote the continuous linear form y ~ ( I x) y on E. It follows from (6.3.2) that X H ~ ( X ) is a semilinear isometry (i.e., we have Axl

+ x2) = j(xl> +Ax2),

j(Ax) = JAx)

for all scalars A) of E onto its dual E'. We can therefore " transport" to E the weak topology on E': the weak topology on E is therefore the topology defined by the seminorms x ~ l ( a l x ) as l a runs through E. This topology is coarser than the topology defined by the norm on E (which is called the strong topology of the Hilbert space E). From (12.15.9) we have: (12.15.10) In a separable Hilbert space, every closed ball is metrizable and compact with respect to the weak topology.

If u is a continuous endomorphism (with respect to the strong topology) of a Hilbert space E, then by virtue of (12.15.4) and (11.5.1) MX)

I Y ) = (u(x), iW> = ( x , ' 4 A Y ) ) ) = (x l~-1(r4AY)N) = (x I U*(Y))

for all x, y in E , and therefore (6.3.2) (12.15.11)

u*

=j-1

o'uoj.

This relation shows immediately that 'u is strongly continuous and that = IIu*II = (lull.Replacing u by u* in (12.15.11), we see that every strongly continuous endomorphism of E is also weakly continuous (cf. (12.16.7)).

II'uII

(12.1 5.12) In a Hilbert space E , a sequence (x,) converges strongly to apoint a

if and o n b if it converges weakly to a and lim llxnll = Ilall. n+ w

The conditions are clearly necessary. To see that they are sufficient, consider the formula IIX,

- allZ = IIXnII' - 29W, I a) + IIaI12.

Since by hypothesis the sequence ((x, 1 a))converges to (a I a ) = 11a1I2,it follows that IIx, - all tends to 0.

78

XI1 TOPOLOGY AND TOPOLOGICAL ALGEBRA

PROBLEMS 1. Let E be a vector space over K (= R or C) and let F be a vector subspace of KEwhose elements are linear forms on E. We write ( x , x ’ ) in place of x’(x) for x E E and x’ E F. Given a finite sequence ( x i ) l 6 l a n of elements of F, show that for this sequence to be free (A. 4.l)it is necessary and sufficient that there should exist n vectors x1 (1 i 5 n) in E such that ( x , , x ; ) = (Kronecker delta). The vector space E is then the direct sum of the subspace V of dimension n generated by the X I , and the subspace W of all x E E such that ( x , x i ) = 0 for 1 5 n. A linear form x’ E F is such that ( x , x’> = 0 for all x E W if and only if x’ is a linear combination of the x { . (Prove these assertions simultaneously by induction on n.)

s

sr,

is

2.

The notation and hypotheses are the same as in Problem 1. Suppose also that there exists no vector x # 0 in E such that ( x , x ‘ ) = 0 for all x’ E F. (a) Show that the weak topology on F is metrizable if and only if E has an at most and ( z , ) ~ ~are , ~ two ~ finite denumerable basis (A. 4.1). (Observe that if (yf)161sm sequences of vectors in E such that the relation sup I ( y l , x’)l 5 1 implies the relation

sup I (z,,

x’)l

51

ISl6rn

in F, then the z, are linear combinations of the y , .

16J6n

Show that if we have ( y , , x’) = 0 for 1 5 i 5 m,then we must also have (z,, x ’ ) = 0 for 1 6j 5 n, and apply the result of Problem 1 by identifying E with the set of linear forms X‘H ( x , x’) on F.) (b) Deduce from (a) that, in an infinite-dimensional separable Hilbert space, the weak topology is not metrizable (cf. Section 5.9, Problem 2). (c) Show that every linear form on F which is continuous with respect to the weak topology can be written x ’ w ( x , x ’ ) for a uniquely determined x E E (argue as in (a)).

3. (a) In a Hilbert space E, let A be a closed convex set (Section 12.14, Problem 11) and a a point of E. Show that there exists a unique point b E A such that d(a, A) = d(a, b) (the projection of a on A). For each z E A show that (z - b I b - a) 20 (argue as in (6.3.1)). (b) Deduce from (a) that every closed convex set A in E is the intersection of closed half-spaces whose frontiers are hyperplanes of support of A (Section 5.8, Problem 3). Deduce that A is weakly closed, and that if C is any convex subset of E, the strong and weak closures of C are the same. Every bounded closed convex subset of E is weakly compact. (c) Show that the weak closure in E of the sphere S : ((xi(= 1 is the ball ( / X I / 0 for all n.) (e) Suppose that E is separable, and let (en)nblbe basis (6.5.2) of E. Let A be the closed convex hull (Section 12.14, Problem 13) of the set of points e./n. Show that A is compact and that the interior of A is empty, and that A has no closed hyperplane of support containing the (frontier) point 0, although there exist lines D passing through 0 and such that D n A = {O}.

15 WEAK TOPOLOGIES

4.

79

Let E be a separable real Frechet space a n d p a continuous seminorm on E. (a) Let (an)n2O be a total (12.13) free sequence in E, and let E. be the subspace of dimension n 1 generated by a o , a ] , . . . , a,. Show by induction on n that for each n there exists a linear formf, on E, such that f. is the restriction of fn+] to En, fo(ao)= p ( a o ) and fn(x) s p ( x ) for all x E E.. (Consider the hyperplane H in given by the equation f n - l ( x )=fn -l(ao), and use Problem 3(d) in the plane which is the quotient of E, by the hyperplane Ho : j m - I ( x= ) 0 in En-l.) (b) Deduce from (a) that there exists a continuous linear form f o n E such that f ( a 0 ) =p(ao) and If(x)i s . p ( x ) for all x E E (Hahn-Banach theorem for a separable Frechet space). In particular, if E is a separable Banach space, then for each x E E we have

+

IIX II = S U P I <x, Y’> I IlY’ll s I (the norm on the dual E being that defined in (5.7.3)). Deduce that if E and F are separable Banach spaces, E and F’ their duals, and u : E -+ F a continuous linear mapping, then llfu/1= IIu/l. (c) Deduce from (b) the geometrical form of the Hahn-Banach theorem: given any nonempty convex open set A in E, and a point x o $ A, there exists a closed hyperplane with equation g ( x ) - a = 0 such that g(x) > a for all x E A, and g ( x o ) 5 a. (Reduce to the case where 0 E A and x o is a frontier point of A. We can then assume that A is symmetrical about 0 and therefore is the set of all x such that p ( x ) < 1 , where p is a continuous seminorm on E.) (d) Let A, B be two disjoint closed convex sets in E, one of which (say, A) is compact. Show that there exists a closed hyperplane H, with equation g(x) - a = 0, which strictly separates A and B: that is to say, such that g ( x ) > a for all x f A and g ( x ) < a for all x E B. (e) Deduce from (d) that every closed convex set in E is an intersection of closed half-spaces, and that every closed linear variety in E is an intersection of closed hyperplanes. (f) Deduce that for a vector subspace F not to be dense in E it is necessary and sufficient that there should exist a continuous linear formff 0 on E such that f ( x ) = 0 for all x E F.

5. Let A be a convex set in a real vector space. A point a E A is said to be an extremal point of A if there exists no line-segment containing more than one point, which is contained in A and has a as an interior point: in other words, if the relations b E A, ~ E Aa ,= X b $ ( l - & c , O < h < l imply b = c = a .

(a) Let E be a Hilbert space, A a bounded closed convex set in E, 6 > 0 the diameter of A, and a, b two points of A such that Ilb - all 2 &dE. 6 . Show that there exists a supporting hyperplane H of A orthogonal to the vector b - a and such that the set H n A has diameter 548. (b) Deduce from (a) that A has at least one extremal point (proof by induction). (c) Show that A is the closed convex hull of the set of its extremal points (KreinMilman fheorem for Hilbert space). (First deduce from (b) that every closed hyperplane of support of A contains at least one extremal point. Then argue by contradiction, assuming that the closed convex hull B of the set of extremal points of A is distinct from A; by using Problem 4(e), show that there would then exist a hyperplane of support of A which did not meet B.) (Cf. Section 13.10, Problem 8.)

80

XI1 TOPOLOGY AND TOPOLOGICAL ALGEBRA

(d) If C is a convex set in E, the intersection of C with a closed (resp. open) halfspace is called a closed (resp. open) cap in C . If a E A is an extremal point, show that the open caps in A which contain a form a fundamental system of neighborhoods of a in A. (Remark that the convex hull of the union of two closed caps in A is weakly compact and does not contain a if neither of the two closed caps contains a ; then use Problem 4(e), and Section 12.3, Problem 6(e).) (e) Suppose that there exists a weakly compact set K c A such that the closed convex hull of K is A. Show that every extremal point of A belongs to K (use (d)). (f) In the Banach space (co) (Section 5.3, Problem 5) show that the closed ball llxll 1 has no extremal points. 6. Let E, F be two locally convex spaces. Let (p.) be a family of seminorms defining the topology of F, and let 6 be a set of bounded subsets of E (Section 12.14, Problem 6). For each continuous linear mapping u of E into F, and each B E (5, put p., .(u) = supp.(u(x)). Show that p., is a seminorm on the vector space P ( E ; F) of conZ€B

tinuous linear mappings of E into F. If the set of homothetic images of the sets of 6 covers E, then the topology defined by the seminorms p a , on 9 ( E ; F)is Hausdorff. If E and F are normed spaces and if 6 consists only of the closed unit ball /\XI\ 5 1 in E, then the corresponding topology on 9 ( E ; F) is that defined by the norm (5.7.1).

7. Let E be an infinite-dimensional separable Hilbert space. Show that, with respect to the topology defined by the norm (5.7.1) on P ( E ; E) (this topology is called the norm topology), the unit ball V consisting of the operators U with norm l\Ul\ 1 (the "contractions" of E) is not separable. (Let (en),>,, be an orthonormal basis of E. For each point z = ({.).20 of the Banach space 1" (Section 5.7, Problem l), let U,E 9 ( E ; E) be the mapping defined by U,. en= [.en for all n 2 0. Show that zt-+ Uzis an isometric linear mapping of I" onto a subspace of 9 ( E ; E), and use the Problem of Section 5.10.) 8. Let E be a separable Hilbert space of infinite dimension. The strong topology on P(E; E) is that obtained by the procedure of Problem 6 by taking (5 to be the set of all finite subsets of E, so that this topology is defined by the seminorms p,(U> = llU. x(I for all x E E. The strong topology is coarser than the norm topology.

(a) Show that the strong topology is not metrizable. (Observe that 9 ( E ; E) contains vector subspaces which, when given the topology induced by the strong topology, are isomorphic to E endowed with the weak topology.) (b) The unit ball V of 9 ( E ; E), endowed with the strong topology, is separable and metrizable. (Proceed as in (12.5.7), by embedding V in EN.) (c) Show that the mapping (U,V ) w UV is continuous in the strong topology on Q x 9;p(E;E) and not continuous on 9(E: F,)X Q. (d) Show that the mapping UHU' of V into V is not continuous with respect to the strong topology. (Let (en)nzObe an orthonormal basis of E, and consider the sequence of operators uk such that uk * e, = 0 if n < k and U k. e . = en-k if n 2 k . ) (e) Let U(E) c V be the unirury group of E, consisting of all linear isometries of E onto E. Show that the topology induced on U(E) by the strong topology is compatible with the group structure. Give an example in this group of a right Cauchy sequence which is not a left Cauchy sequence. Show that U(E) is closed in V. 9.

Let E be a separable Hilbert space of infinite dimension. The weak ropology on Y(E; E) is that obtained by the procedure of Problem 6 by takingG to be the set of

16 BAIRE'S THEOREM A N D ITS CONSEQUENCES

81

finite subsets of E, but taking E with the weak topology, so that this topology is defined by the seminorms p x ,,(U)= I (U . x I y ) 1 for all x , y in E. This topology is coarser than the strong topology. (a) Show that the weak topology is not metrizable (same methodas in Problem 8fa)). Ib) The unit ball Q in Y(E; E) is metrizable and compact in the weak topology (method of (12.5.7)). (c) The bounded sets in 9 ( E ; E) are the same for the weak topology as for the strong topology. (d) The mapping U H U * of Y(E; E) into itself is continuous with respect to the weak topology. (e) Show that the mapping (U, V ) 4 U V of V x V into V is not continuous for the weak topology. (IF is a Hilbert basis of E indexed by Z, consider the "shift operator" U defined by U .en = en+l for all n E Z, and its powers U" and U+.) The topology induced on the unitary group U(E) by the weak topology is identical with that induced by the strong topology, and hence is compatible with the group structure. Also U(E) is not closed in G9 for the weak topology. (f) Show that, for each Uo f Y(E; E), the linear mappings VH Uo V and V++ VUO are continuous on Y(E; E) for the weak topology. (9) Let uly be a submonoid of V (i.e., uly is stable under multiplication). Show that the closure of uly with respect to the weak topology on 9 ( E ; E) is a compact monoid, which is commutative if uly is commutative. 10. The notation is the same as in Problem 9.

(a) Let U E Q. Show that the relations U * x = x , (x [ (I.x ) = (x [ x), and U* . x = x are equivalent. (b) Deduce that the idempotent operators in are the orthogonal projections (Sections 6.3 and 11.5.)

11. Let E be a separable Hilbert space, A a submonoid containing the identity l E which is weakly closed (and therefore weakly compact) in V (notation of Problem 8). For each x E E, the orbit of x under A is the set .I. n of vectors U * x where U E 4; it is weakly compact in E. We say that x is a flight vector with respect to uly if 0 E uly * x, and that x is a reversibfe vector with respect to &if for each U EX there exists V E X such that VU . x = x . Let F(uly) (resp. R(uly)) denote the set of flight (resp. reversible) vectors with respect to uly. If x is reversible, then IIU .xII = llxll

for all U E A. (a) Every orbit uly * x contains a minimal orbit N (Section 12.10, Problem 6), and every y E N is reversible with respect to uly. Let U E uly be such that U . x = y E N, and let d be the weakly closed submonoid generated by l E and U : this submonoid d is Commutative. Let P C d . y c N be a minimal orbit with respect to d , and let V E d be such that V .y = z E P. Show that there exists W E d such that VU W . z = I;deduce that the vector x - W . z is a flight vector, and hence that every x E E is the sum of a flight vector and a reversible vector belonging to A * x . (b) Show that if x E E is reversible with respect to X then x is also reversible with respect to every weakly closed submonoid M . (Split up x into the sum of a vector y E R( N)belonging to N . x and a vector z E F(N);use the fact that if a , b are two b)ll lim infl/x.-all.

”-.

0)

16. BAIRE’S THEOREM A N D ITS CONSEQUENCES

(Baire’s theorem) Let E be a topological space in which eoery (12.16.1) point has a neighborhood homeomorphic to a complete metric space. If(U,) is a sequence of dense open sets in E, then the intersection of the U, is dense in E. It is enough to prove that, for each x E E and each neighborhood V of x, the intersection of V and the U, is not empty. Hence we may assume that E itself is a complete metric space and (bearing in mind (3.14.5)) prove that the intersection G of the U,, is nonempty. Let d be a distance defining the topology of E, with respect to which E is complete. We shall define by induction on n a sequence (x,) of points of E and a sequence (r,) of real numbers > O , as follows: x1 E U1; I , < l / n for each n >= 1 ; the closed ball B’(x,,; r,) is contained in U, n B’(xnWl;r,,-l). This is possible because U, n B(xn-l; r,,-l) is a nonempty open set in E, since U, is dense in E. Clearly we have d(x,, x,+,,) r, < l / n for each n 2 1 and each p > 0, and therefore (x,) is a Cauchy sequence in E. By hypothesis it converges to a point a E E, and since x,+,, E B’(x,; r,) for each p > 0, we have a E B’(x,; I,,) c U, for each n since B‘(x,; r,) is closed in E. The point a therefore belongs to G . Q.E.D. We shall apply Baire’s theorem mainly when E is an open subspace of a complete metric space or a locally compact rnetrizable space (by virtue of (3.16.1)). In a topological space E, a subset A is said to be nowhere dense if the open set E - i% is dense (or, equivalently, if A contains no nonempty open set, or if has empty interior). For example, in a Hausdorff space E, a set { a } consisting

84

XI1 TOPOLOGY AND TOPOLOGICAL ALGEBRA

of a single point is nowhere dense unless it is open, i.e., unless a is isolaied in E (3.10.10). In a topological group G, every closed subgroup H which is not open is nowhere dense (12.8.7). In a topological space E, a subset B is said to be meager if it is the union of at most denumerably many nowhere-dense sets. Since the closure of a nowheredense set is nowhere dense by definition, Baire's theorem can be restated in the form that, in a space E satisfying the conditions of (12.16.1), the complement of a meager set is dense in E. For example, in a Frtchet space E, a denumerable union of closed linear varieties, each distinct from E, i.e., sets of the form a, + V,, where V, # E is a closed vector subspace of E, has a dense complement in E. For it follows from (12.13.1) that a vector subspace V of a topological vector space E cannot be open in E unless V = E. In particular, a denumerable set in a Frtchet space is meager, but it should be noted that such a set may also be dense (if the space is separable). (12.16.2) Let E be a topological space in which every point has a neighborhood homeomorphic to a complete metric space, and let u be a lower semicontinuous function on E. Ifu(x) < + 00 for each x E E , then given any nonempty open set U in E there exists a nonempty open set V c U such that sup u(x) < 00. xov

+

It is enough to prove the result when U = E. For each integer n > 0, let F, be the set of all x E E such that u(x) 5 n. By hypothesis, F, is a closed set (12.7.2), and E is the union of the F,; hence at least one of the F, is not nowhere dense (12.16.1), and therefore contains a nonempty open set. Q.E.D. Notice that under the hypotheses of (12.16.2) it can happen that sup u(x) = xsE

+ 00.

Consider for example the real-valued function on R which is equal to 0 at x = 0, and equal to 1/x2when x # 0. In particular: (12.1 6.3) In a Fre'chet space E , every lower semicontinuous seminorm is continuous.

By definition, a seminorm p on E is finite at all points of E. If p is lower semicontinuous, it follows from (12.16.2) that there exists a point x,, E E, a

16 BAIRE’S THEOREM AND ITS CONSEQUENCES

85

neighborhood V of 0 in E, and a real number c > 0 such that p(x) 5 c for all

x E xo + V. Hence

P(Z)

for all z

E V,

5P b O )

+ P(X0 + z ) 2 c + P ( X 0 )

which proves the result (12.14.2).

(Banach-Steinhaus theorem) Let E be a Frkchet space, F a (12.16.4) normed space. Let H be a set of continuous linear mappings of E into F. Suppose that sup IIu(x)II < + co for all x E E . Then H is equicontinuous. ueH

The function p(x) = sup IIu(x)II, being finite for all x E E, is a seminorm usH

on E (12.14.1). Since each of the functions X H IIu(x)I[ is continuous, p is lower semicontinuous (12.7.7). The result therefore follows from (12.1 6.3) and (12.15.7.1).

As a consequence, we have: (12.16.5) (i) Under the hypotheses of (12.16.4), let (u,) be a sequence of continuous linear mappings of E into F which converges simply in E to a mapping u of E into F. Then u is a continuous linear mapping of E into F. (ii) More generally, let Z be a metric space, A a subset of Z , and z w u, a mappie3 of A into the space of continuous linear mappings of E into F. Let zo E Z be a point in the closure of A, and suppose that the limit

lim u,(x) = v(x)

z-+zo,zeA

exists in F, for each x E E. Then v is a continuous linear mapping of E into F .

(i) From the hypotheses we have sup IIu,(x)II < +co for all x

E

E, hence

n

the Banach-Steinhaus theorem shows that the sequence (u,) is equicontinuous, and the continuity of u follows from (7.5.5). The fact that u is linear is an immediate consequence of the principle of extension of identities. (ii) The point zo is the limit of a sequence of points (z,) in A (3.13.13), hence u(x) = lim U J X ) for all x E E. Now apply (i). n-tm

(12.16.6) Let E be a Frkchet space, F a Banach space, I an open interval in R. Let Y ( E ; F ) denote the space of continuous linear mappings of E into F , endowed with the topology of simple convergence (12.15). I f a mapping t t - f , of I into Y(E; F ) is diferentiable with respect to the topology of simple convergence (12.15), then there exists a mapping t H f j of I into 2 ( E ; F ) such that f ; ( x ) = D(f,(x))for all x E E.

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XI1 TOPOLOGY AND TOPOLOGICAL ALGEBRA

For D(f,(x)) is the limit, as h # 0 tends to 0, of the mapping

h + + ( f r + h -fr)lh, and the result therefore follows from (12.16.5). The mapping f iois said to be the derivative at the point to of the mapping t w f , (with respect to the topology of simple convergence), or the weak derivative when F is the field of scalars (and therefore Y(E; F) = E’). Remark

(12.16.6.1) Let A be an open set in C and let Z H ~ , be a weakly analytic mapping (12.15) of A into the dual E’ of a complex FrCchet space F. Then the same argument as in (12.16.6) shows that there exists a weak derivative 2-f: of Z H ~ ,, with fL(x) = D(f,(x)) for all x E E, and this weak derivative is weakly analytic. Moreover, for each a E A and each circuit y contained in A - { a } , we have the Cauchy formula (12.16.6.2) Conversely, let y be a road in C defined in an interval I = [b, c ] in R,and let z H g , be a weakly continuous mapping of y(1) into E’. Then, for z # y(I), the mapping

(12.16.6.3) is a linear formf, on E, belonging to E’. Indeed, for each x E E the right-hand side of (12.16.6.3) is the limit of a sequence

where

((3.16.5) and (8.7.8)),hence our assertion is a consequence of the BanachSteinhaus theorem. By virtue of (9.9.2),f, is weakly analytic in C - y(I), and for a 4 y(1) we can write (12.16.6.4)

m

f Z ( 4

=

1 cn(x)(z - a)”,

n=O

the series being convergent in any disk with center a not meeting y(1). The

16 BAIRE’S THEOREM AND ITS CONSEQUENCES

87

coefficients c,, E E’ are given by

Applying this result to the situation where y is a circuit t H a + reit (0 5 t 5 2n) in A - { a } we obtain, for any weakly analytic function ZI+L on A, the Taylor expansion (12.16.6.5)

convergent in the disk ) z - a1 < r (independent of z E E), and the derivatives are given by (12.16.6.6)

More generally, suppose that in A - { a } we have (12.16.6.7)

where the x: are linear forms on E (a priori, not necessarily continuous), and zHg, is a weakly analytic function on A - { a } which is weakly bounded in a neighborhood of a. Then, by (9.15), for each x E E, the function zt-tg,(x) can be extended by continuity to the point a, and the extended function is analytic in A. Hence there exists a weakly analytic function z H h, in A whose restriction to A - ( a } is zI+g,. Furthermore, the x: are continuous linear forms on E (i.e., they are elements of E’). Indeed, for each x E E we have ( Z - u ) ~ - ’ ~ = ( dz x)

where y is a circuit tt+a + rei‘ (0 5 t 2n) with r sufficiently small and independent of x (9.14); our assertion therefore follows from the considerations above (12.16.6.3). Finally, we remark that the principle of analytic continuation (9.4.2) remains valid for weakly analytic functions. This is an immediate consequence of the definitions. (12.16.7) Let E be a Hilbert space and u an endomorphism of the (nontopological) vector space E. Then the following three conditions are equivalent: (a) u is continuous; (b) u is weakly continuous; (c) u has an adjoint (11S).

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XI1 TOPOLOGY A N D TOPOLOGICAL ALGEBRA

We have already seen (12.15.11) that (a) implies (b). If u is weakly continuous, then for all y E E the linear form X H (u(x) I y ) is weakly continuous, and therefore a fortiori continuous with respect to the strong topology, which is finer than the weak topology. Consequently (6.3.2), for each y E E there exists a unique point u*(y) such that (u(x) I y) = (x I u*(y)) for all x E E: in other words (11.5), u has an adjoint, and therefore (b) implies (c). Finally, if u has an adjoint, then we have IIU*(Y)ll = SUP

llxll 5 1

IbIU*(Y))l

= SUP

llxll 6 1

I(u(x)lv)l,

and each of the linear forms y ~ ( I u(x)) y = (u(x) l y ) is continuous, hence

YH IIu*(y)(l is a lower semicontinuous seminorm on E, and is therefore continuous in the strong topology (12.1 6.3). But this is equivalent to saying that

u* is (strongly) continuous, and therefore so is u = u** (11.5.2).

We remark that it is essential for the truth of (1 2.16.4), (12.16.5), (12.1 6.6), and (12.16.7) that the space E should be complete (cf. Problems 21 and 22). (12.1 6.8) (Banach's theorem) Let E, F be two Fre'chet spaces and let u be a continuous linear mapping of E into F. Then either u(E) is meager in F, or else u(E) = F. In the latter case, if N is the kernel of u and E --+ E/N & F the canonical factorization of u, the mapping v is an isomorphism of the Frtchet space E/N (12.14.9) onto F (in other words (12.12.7), u is a strict morphism of E onto F).

Suppose that u(E) is not meager in F. By virtue of (12.13.1) and (12.12.7), the theorem will be proved if we can show that, for each neighborhood V of 0 in E, u(V) is a neighborhood of 0 in F. There are two steps to the proof. (12.16.8.1) Let E, F be two topological vector spaces, and u a linear mapping of E into F such that u(E) is not meager in F. Thenfor each neighborhood V of 0 in E, the closure u(V) of the image of V is a neighborhood of 0 in F.

Let W be a balanced neighborhood of 0 in E such that W + W c V ((12.13.1) and (12.8.3)). Then for each x E E there exists an integer n 2 1 such that x E nW. Consequently u(E) is the union of the sets u(nW) = n u(W). Since u(E) is not meager, at least one of the sets n * u(W) has an interior point, and therefore so does u(W). But since __ -u(W) = u(W), we have -~ - u(W) = u(W). So if yo is an interior point of u(W), so is - y o , and -then therefore 0 = yo ( y o ) is an interior point of u(W) u(W) (12.8.2). But _ _ u(W) u(W) is contained in theclosure of u(W) + u(W) = u(W W ) c u(V), and so (12.16.8.1) is proved.

+

+

+

+

16

BAIRE'S THEOREM A N D ITS CONSEQUENCES

89

Now suppose that E, F are equipped with translation-invariant distances d, d , respectively, compatible with their topologies (12.9.1). By (12.16.8.1), the hypothesis that u(E) is not meager in F implies that, for each r > 0, there exists p = p(r) > 0 such that B'(0; p) c u(B'(0; r)). By translation it follows that for each x E E we have B'(u(x); p ) c u(B'(x; r)). Hence it is enough to prove the following lemma: (12.16.8.2) Let E be a complete metric space, F a metric space, and u a continuous mapping of E into F with the following property: for each r > 0 there exists p = p(r) > 0 such that B'(u(x);p ) c u(B'(x;r ) ) for all x E E. Then B'(u(x);p ) c u(B'(x; 2r))for all x E E.

For each integer n 2 1, there exists by hypothesis a number p, > 0 such that B'(u(x); p,) c u(B'(x; 2-"+'r)) for all x E E. We may take p1 = p and (replacing p, if necessary by inf(p, , 2 - 7 ) assume that lirn p, = 0. Let xo be n+m

any point of E, and let y E B'(u(x,); p). We shall prove that y E u(B'(x,; 2r)). For this we define inductively a sequence (x,),,, of points of E such that, for each n 2 1, we have x, E B'(x,,-,; 2 - " + l r ) and u(x,) E B'(y; pn+,). Suppose that xl, . . . , x,-, have been chosen to satisfy these conditions. Then we have y E B'(u(x,,-,);p,), and since B'(u(x,,-,); p,) c u(B'(x,-,; 2-"+'r)), there exists a point x, E B ' ( X , , - ~2; - " + ' r ) such that u(x,,) E B'(y; p,+J. So the inductive construction can continue. Now the sequence (x,) is a Cauchy sequence in E, because for each n 2 0 we have + 2-"-p+' r 5- 2-"+'r d(x,, x,,~) 2-"r 2 - " - l r +

s

+

-

for all p > 0. Since E is complete, the sequence converges in E to a point

x such that d(x,, x ) 5 2r. Since u is continuous, we have u(x) = lirn u(x,,);

and since d'(u(x,), y )

s pnil,

it follows that lim u(x,) = y . n+m

n-+m

Q.E.D.

This theorem has the following corollaries : (12.1 6.9) Let E, F be two Frkchet spaces. Then every continuous bijective linear mapping of E onto F is an isomorphism.

In this proposition it is essential to assume that both E and F are complete. For example, if we take E = I; (6.5) and F to be the canonical image of E under the identity mapping E-tBR(N) (7.1), endowed with the topology induced by that of gR(N), then this mapping is continuous because

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XI1 TOPOLOGY AND TOPOLOGICAL ALGEBRA

C 4; 2 sup (,2 n

n

(5.5.1),but the inverse bijection of F onto E is not continuous:

otherwise F would be complete, and therefore closed in g,(N) (3.14.4),which is absurd, because its closure in gIR(N)is the set of all sequences (5,) such that lim 4, = 0. n-, m

(1 2.16.10) Let E, F be two Banach spaces, and u a continuous surjective linear mapping of E to F. Then there exists a number m > 0 such that for each x E E there exists x' E E,for which u(x) = u(x') and IIu(x)II 2 m * IIx'II. Bearing in mind (12.14.8)and (5.5.1),this expresses that the bijection F .+ E/u-'(O), the inverse of the bijection E/u-'(O) F induced by u, is continuous on F. -+

(12.16.11) (Closed graph theorem) Let E, F be two Frtchet spaces. For a linear mapping u of E into F to be continuous, it is necessary and sufficient that the graph (1.4)of u should be closed in the product space E x F. In general, iff is a continuous mapping of a topological space X into a Hausdorff topological space Y, the graph offis closed in X x Y, because it is the set of all z E X x Y satisfying the relation pr, z =f(prlz), and theassertion follows from (12.3.5).To show that the condition stated in (12.16.11)is sufficient, remark that it implies that the graph G of u, being a closed vector subspace of the FrCchet space E x F (3.20.16(iv)),is a FrCchet space (3.14.5). The projection z w p r l z of G onto E is therefore a continuous bijective linear mapping, hence an isomorphism (12.16.9). Since the inverse mapping is u : X H ( X , u(x)), it follows that xt-,u(x) = pr,(u(x)) is continuous on E. The condition of (12.16.11)may also be expressed by saying that if a sequence (x,, u(x,)) in E x F tends to a point (x, y ) , then y = u(x). Replacing x, by x, - x, and using the linearity of u, an equivalent formulation is that i f a sequence (x,) in E tends to 0 and is such that the sequence ( ~ ( x , )tends ) to a limit y , then y = 0. It is this criterion that we shall apply in practice to verify the continuity of u. Finally, the following consequence of Baire's theorem allows us to use the criterion (12.11.5):

(12.16.12) Let G be a separable, metrizable, locally compact group, acting continuously and transitively on a Hausdorff topological space E in which every point has a neighborhood homeomorphic to a complete metric space. For each x E E, let S, be the stabilizer of x. Then the canonical bijection f, : G/S, -+ E is a homeomorphism.

16 BAIRE’S THEOREM A N D ITS CONSEQUENCES

91

Let xo E E. We have to show that, for each neighborhood V of e in G, the set V xo is a neighborhood of x, in E (12.11.5). Let W be a compact symmetric neighborhood of e in G such that W2 c V, and let (s,) be a sequence which is dense in G, such that (s,,W) is a covering of G. Each of the sets s,,W . x, is closed in E, because st+s x, is continuous (12.3.61, and E is the union of the denumerable sequence of closed sets s, W xo . By Baire’s theorem (12.16.1), there exists an index n such that s,,W * xo has an interior point s,s xo, with s E W. It follows (12.10.3) that x, is an interior point of the set a

s-’s,’(s,W.

xo) = s-’w* xo

in other words, V * x, is a neighborhood of xo .

c

v . x,; Q.E.D.

(12.16.13) Let G be a separable, metrizable, locally compact group, let G’ be a metrizable group, and let f : G -+ G’ be a continuous surjective homomorphism. Then f is a sfrict morphism (12.12.7) (in other words, if H is the kernel off; then the canonical bijection g : G/H G’ is an isomorphism of topological groups). --f

For we may consider G’ as a space on which G acts continuously and transitively by the rule (s, t ’ )f (~s ) f ’ ,and the stabilizer of the neutral element e‘ of G’ is H. The result therefore follows from (12.16.12).

PROBLEMS 1. Let E, F be two metric spaces, A a dense subspace of E, and f a continuous mapping of A into F. If F is complete, show that the set of points in E at whichfhas no limit relative to A (3.1 3 ) is meager in E. (For each n , consider the set of points x E E at which the oscillation offwith respect to A (3.14) is > l/n.) 2.

Let E, F be two complete metric spaces, and f a homeomorphism of a dense subspace A of E onto a dense subspace B of F. Show that there exists a subspace C 3 A in E (resp. D 3 B in F) which is a denumerable intersection of open sets, and an extension offto a homeomorphism g of C onto D. (Apply Problem 1 to f and its inverse.)

3. Let E be a complete metric space, F a metric space, and (fn) a sequence of continuous mappings of E into F which converges simply (pointwise) in E to a mapping f. Show that the set of points x E E at which f is not continuous is meager in E. (Let G p ,,, be the set of x E E for which the distance between f,(x) and &(x) is 5 1/2n for all q z p . Show that the union of the interiors of the sets G,, for p 2 1 is a dense open set, by using Baire’s theorem. Deduce that the set of points at which the oscillation of f’is 0, let A. be the set of points x E [a, b[ such that thereexist pointsy, z E [a, h[ satisfyingx < y < z < x l / n and

+

Show that A. is a dense open subset of [a, b[.)

16 BAIRE'S THEOREM AND ITS CONSEQUENCES

97

28. Let f b e an indefinitely differentiable real-valued function defined on an open interval ]u,b[ of R.Suppose that for each point x E Ju,b[ there exists an integer N(x) such that DN("'f(x)= 0. Show that f i s a polynomial. (One may proceed as follows: (a) Let G be the open set of points x E ]a, b[ such that in some neighborhood of x the function fcoincides with a polynomial, and let F be the complement of G in ]u,b[. Show that F has no isolated points (3.10.10). (b) For each integer n, let En be the closed subset of F consisting of the points x E F such that D"f(x)= 0. If F is not empty, show that there exists a nonempty open interval 1 C ]a, b[ and an integer N such that F n I is nonempty and is contained in EN (use Baire's theorem). Then deduce from (a) that F n I c E, for all n > N. (c) Deduce from (b) that F n I is nowhere dense in I, and then that DNf(x)= 0 on every component interval of G n I. Hence derive a contradiction of the hypothesis F f 121.) 29. Let E, F be two separable Banach spaces, F' the dual of F. Suppose that F is contained

in a Hausdorff locally convex space G , and that the topology on F induced by that of G is coarser than the weak topology of F . Let u : E -+ G be a continuous linear mapping. (a) Show that, for every ball B in the Banach space F', the inverse image u-'(B) is a closed subset of E. (Use (12.15.9) and (12.15.8.1).) (b) Suppose that there exists a nonmeager subset A of E such that u(x) E F' for all x E A. Show that u(E) c F' and that u is continuous for the topology defined by the norm on F'.

30. Let E be a Frechet space, and let (u.) be a family of continuous linear mappings of E into a normed space F. Suppose that there exists a nonmeager subset A of E such that for each x E A the set of points tl.(x) is bounded in F. Show that the family (u,) is equicontinuous. (For each integer n 1, consider the set of points x E E such that l/u,(x)lI5 n for each index a,)

CHAPTER XI11

INTEGRATION

The theory of integration which we shall develop in this chapter is restricted to separable metrizable locally compact spaces, this being sufficient for our purposes in later chapters. We have followed fairly closely the exposition of N. Bourbaki [22], with the simplifications afforded by our more restricted hypotheses. The key results of the theory of integration are Lebesgue’s convergence theorems (1 3.8), the Fischer-Riesz theorem (13.11.4), the Lebesgue-Nikodym theorem (13.1 5.5) and the Lebesgue-Fubini theorem (1 3.21.7). Unfortunately it is necessary to include rather a lot of material on upper integrals, measurable functions and negligible functions, which are indispensable technical tools. The important properties of certain particular measures on locally compact groups or on differential manifolds will be examined in Chapters XIV and XVI. We have also included, amongst the problems, applications of integration which are not dealt with in the text, especially to ergodic theory and orthogonal systems. The reader who wishes to go further in these directions should consult [21], [26a], [28], [30], and [31]. Nowadays the purposes of a theory of integration are very different from what they were at the beginning of this century. If the aim was only to be able to integrate “ very discontinuous ” functions, integration would hardly have gone beyond the rather narrow confines of the “ fine ” theory of functions of one or more real variables. The reasons for the importance that Lebesgue’s concept of integral has acquired in modern analysis are of quite a different nature. One is that it leads naturally to the consideration of various new complete function spaces, which can be conveniently handled precisely because they are spaces offunctions (or of classes of “equivalent ” functions) and not just abstract objects, as is usually the case when one constructs the completion of a space. Another is that the theorem of Lebesgue-Nikodym 98

1 DEFINITION OF A MEASURE

99

and the properties of measures defined by densities (13.15) give us a method for dealing with denumerable families of measures on a locally compact space, by fixing a basic measure and working with the densities relative to this basic measure (hence again withfunctions); this again proves to be extremely convenient. Here the modern point of view emerges: given a p-integrable function5 what is important is not the values taken byfso much as the way in whichf operates on the space of bounded continuous functions by means of the linear mapping g w fg dp (this mapping depends only on the equivalence class off and therefore does not change when we modifyfat the points of a set of measure zero). The development of this point of view will lead in Chapter XVII to the theory of distributions, which is a natural generalization of the notion of measure on differential manifolds. Throughout this chapter, the phrase “ locally compact space ” will always mean ‘‘ separable metrizable locally compact space.” 1. DEFINITION OF A MEASURE

To begin with, let X be a compact (metrizable) space. A measure (or complex measure) on X is by definition an element of the dual of the Banach space gc(X) of complex-vaiued continuous functions on X (7.2),that is to say ((12.1 5) and (5.5.1 )) it is a linear form f tt p(f) on VC(X)which satisfies an inequality of the form Icl(f)l 2 a I l f

(13.1.1)

for a l l y €%,(X) (recall that

llfll

= sup

II

1f ( x ) l ) .

XEX

Now let X be a locally compact space (metrizable and separable, in accordance with our conventions). For every compact subset K of X, let X ( X ; K) (or X,(X; K)) denote the vector subspace of g,-(X) consisting of the functions whose support (12.6) is contained in K (and is thereforecompact). We shall denote by X,(X) (or X ( X ) )the union of the X c ( X ;K ) as K runs through all compact subsets of X. In other words, X,(X) is the vector space of (complex-valued) continuous functions with compact support. Clearly .X,(X) = V:cm(X). A measure (or complex measure) on X is by definition a linear form p on X,(X) with the following property: for each compact subset K of X, there exists a real number aK >= 0 (in general depending on K) such that (13.1.2)

for allfE X ( X ;K).

Ip(f)I

5 OK l l f l l

XI11 INTEGRATION

100

This definition agrees with the preceding one when X is compact. It expresses that the restriction of p to X ( X ;K) is continuous with respect to the topology induced by that of %?,?(X).We remark that X ( X ; K) is closed in %?F(X),and therefore a Banach space (3.14.5). I n general, a measure is not necessarily continuous on X,(X) with respect to the topology induced by that of %?F(X)(i.e., the topology defined by the norm 11 f 11). We shall examine this question later (13.20). Examples of Measures (13.1.3) Let X be a locally compact space, and let X E X .The mapping f t +f ( x ) of X ( X )into C is a measure, for it is linear and we have If(x)l 5 i l f l l for each compact subset K of X such that f E X ( X ; K). This measure is called the Dirac measure at the point x , or the measure defined by the unit mass at thepoint x , and is denoted by E,. More generally, let (a,) be a sequence of distinct points in X , and (t,,) a sequence of complex numbers, such that for each compact K c X the subsequence formed by the t,, f o r which a, E K is absolutely summable (5.3). Let cK = ItJ. Then, for each function f E X ( X ;K), the series t,,f(a,) is a,

E

1

K

n

absolutely convergent, because the only nonzero terms are those for which a, E K, and we have

C ItnSta,JI S IfII C = CK . IISIIThis also shows that f H C t,,f (a,) is a measure on X . This measure is said to *

U"EK

a,eK

tn

n

be defined by the masses t, at the points a,, for all n (cf. (13.18.8)). (13.1.4) Let f~ X c ( R ) . For each interval [a, b ] containing the support off, the value of the integral Jabf(t)dt (8.7) is the same, and we denote it by

s'^

f ( t ) dt. The mappingfwJ+OOf(t) -00 dt is a linear form on XX,(R),and it is a measure because, for each compact interval K = [a, b ] in R and each functionfE X ( R ; K) we have -m

by the mean value theorem (8.7.7). This measure is called Lebesgue measure on the real line R. (1 3.1.5) Let p be a measure on X and let g E %',(X). Then for each function f E X ( X ) it is clear that g f E , X ( X ) , and the mapping f w p ( g f ) is therefore

1 DEFINITION OF A MEASURE

101

a h e a r form on X(X). It is a measure, for if K is any compact subset of X and i f f E X ( X ;K), then IIgfII g 11 f 11 sup Ig(x)l, and consequently

-

XEK

Ip(gf)l 5 bK (1 f 11, where bK = aK sup Ig(x)l. This measure is denoted by g * p, X€K

and it is called the measure with density g relative to p (cf. (1 3.13)). (13.1.6) Let n : X + X' be a homeomorphism of X onto a locally compact space X'. For each functionfE X(X'),the function f 0 n belongs to X(X), and we have Supp(f0 n) = n-'(Supp f). It follows immediately that, if p is any measure on X, then f Hp( f 0 n) is a measure on X', called the image of p under n and denoted by n(p). (13.1.7) Let Y be a closed subset of X (and therefore a locally compact subspace of X (3.18.4)) and v a measure on Y. For each f E X ( X ; K), the restriction f I Y belongs to X ( Y ; K n Y), and hence there exists a constant CK such that If(y)l 5 c K l l f l l Iv(flY>I5 cK ' YE

Y

for all f E X ( X ;K). The mapping f Hv( f I Y) is therefore a measure on X, called the image of v under the canonical injection Y + X, or the canonical extension of v to X. (1 3.1.8) Let U be an open subset of X (and therefore again a locally compact subspace (3.18.4)). For each compact subset K of U, it is clear that the mappingft+ f I U is an isometry of X ( X ;K) onto X ( U ; K). The image under the inverse isometry of a function g E X ( U ; K) is the function g' which agrees with g on U and is zero on X - U. (By abuse of notation, we shall often write f in place off/ U when Supp(f) c U, and g in place of g'). The mapping g w g U of Xc(U) into X,(X) is therefore injective. If p is any measure on X , the mapping g H p(g') is a measure on U, said to be induced by p on U, or the restriction of p to U, and denoted by pu or p I U. It should be noted that a measure v on U is not necessarily the restriction of a measure on X (Section (13.4), Problem l), and that an " extension " of v to X, if it exists, is not in general unique. However, there is the following result: (13.1.9) Let (U,),€, be an open covering of X. For each a E I let pa be a measure on U, such that, for each pair of indices a, p, the restrictions of pa and p, to U, n U, (1 3.1.8) are equal. 'Then there exists a unique measure p on X whose restriction to U, is p a , for each a E 1.

We shall first show that each function , f X,(X) ~ can be written in the n

form f = cfiwhere, for each index i, there exists ai E I such that f i E Xc(X) i= 1

102

Xlll INTEGRATION

and Supp(fi)c U,, . For this purpose we observe that if K is the support of f, then there exist finitely many indices cli E I (1 5 i 5 n ) such that the U,, cover K. Hence ((3.18.2) and (12.6.4)) there exist n continuous mappings hi : X -+ [0, I] such that Supp(hi) is compact and is contained in U,, for 1 5 i 5 n, and such that

n

h,(x) = 1 for all x E K. Then the functions f i =f hi

i=1

satisfy the required conditions. This already proves the uniqueness of p: for by definition we must have

~ ( f=)i= C1pU(fhi)= i2= 1pa,Uhi). n

To prove the existence of a linear form p on X,(X) whose restriction to X,.(U,) is p, for each a E I, it is enough to establish the following assertion: given two finite sequences (gJl s i j m and j g n of functions belonging to X,(X), such that Supp(g,) c U,, for 1 5 i 5 m and Supp(hj) c U,, for 1 S j S n, and such that

for all x E Supp(f),then we have

Now, we have

and therefore

Similarly,

But since Supp(fgihj) is contained in U,, n U,,, it follows from the hypotheses that pu,(f g i hi) = ppj(fgihj),and our assertion follows. It remains to be shown that the linear form p so defined is a measure. Let K be a compact subset of X, and define the U,, and the hi as at the beginning of the proof. If H i = Supp(hi), then by hypothesis there exists a

2 REAL MEASURES

number a, 2 0 such that lpai(g)l 5 a i llgll for all g Hence, for each function f E X ( X ;K), we have Ipai(.fhi>l5 aiiifhiil

so that

I ~ W5I

E

.X(X; Hi) (13.1.2).

5 aiilfil,

(,faijli/ii* I=

103

Q.E.D.

1

(13.1.10) If I and p are two measures on X, then so are A + p and a l for any scalar a E C. The set of all measures on X is therefore a uecior subspace of CXC('), which we denote by M,(X) or M(X). By analogy with the example (13.1.4), if p is a measure on a locally com-

s

s

pact space X, we write f dp or f ( x ) dp(x) (or also (f, p ) or ( p , f ) ) in place of p[f), for any f E X ( X ) , and we call this number the integral o f f with respect to p. 2. REAL MEASURES

Let X be a locally compact space. Let X , ( X ) denote the set of all realualued continuous functions on X with compact support, and X,(X; K) the set of those whose support is contained in K. Clearly XR(X)is a real vector subspace of X,(X), and we can write YC(x)=

x,(x) @ iXR(x)

(direct sum). For every (complex) measure p , the restriction of p to XR(X) is an R-linear mapping p o of X R ( X ) into C; moreover p,, determines p uniquely, for i f f = f l ifz with f i t f z in .X,(X),then p ( f ) = p o ( f i ) ipo(fz). Conversely, if an R-linear mapping p o : X,(X) + C is such that, for each compact subset K of X, there exists aK > 0 with the property that Ipo(f)l 5 uK llfll for allfE X , ( X ; K), then it is immediately obvious that the mapping

+

+

fi + $2

HP O ( f 1 )

+ iPO(f2)

is a (complex) measure on X. Hence we may identify each measure on X with its restriction to .X,(X).

Let p be a (complex) measure on X . It follows immediately from (13.1.2) that the m a p p i n g f t + X ) is also a measure on X, called the conjugate of p and denoted by ji. We have p = / i , and if I , .iiare measures o n X and a, b are any two complex numbers, then aI = bp = ZX hii. More generally, if g is any function belonging to %,(X) and p is any measure on X, then we have F p = S ji (13.1.5).

+

-

104

Xlll INTEGRATION

A measure p on X is said to be real if ji = p, or equivalently if p ( f ) is real for everyfe X,(X). We may therefore identify the set of real measures on X with a vector space of linear forms on the real vector space X,(X). This vector space is denoted by MR(X). Lebesgue measure and all Dirac measures are real. If p is any complex measure, then the measures p1 = ( p + ji)/2 and p 2 = ( p - F)/2i are real. They are called respectively the real and imaginary parts of p, and are denoted by 92p and Yp respectively. For each function f E X,(X), we have

and by definition (13.2.2)

p

= 92p

+ iYp,

ji

= 9?p - iYp.

3. POSITIVE MEASURES: T H E ABSOLUTE V A L U E O F A MEASURE

A measure p on a locally compact space X is said to be positive if, for each function f~ ,X,(X) such that f 2 0, we have p(f) 2 0. Consequently, i f f and g are two functions belonging to X,(X) such that f 5 g, we have p ( f ) 5 p(g). Since each f E X,(X) can be written in the form f = f + -f (where f + ( x ) = ( f (x))' and f - ( x ) = ( f ( x ) ) - (2.2)), it follows that a positive measure is a real measure. We denote by M+(X) the set of all positive measures on X. Surprisingly, the property of positivity alone implies the defining property (13.1.2) of a measure: (13.3.1) p(f)

Let p be a linear form on the real vector space X,(X) such that is a (positioe) measlire on X.

2 0 whenever f 2 0 . Then p

We have to show that (13.1.2) is satisfied. There exists a function g E X,(X) with values in [0, I ] and equal to I throughout K ((3.18.2) and (4.5.2)). Hence for all f E X,(X; K), we have

0 S . f + 5 ll.fll* g ,

0 5 . r 5 llfll . g

and therefore 0

s A f + )5 llf'll . A g > ,

so that finally

Icc(f'>I S 2 llfll . A g ) .

05 K

- )s I l f II .

3 POSITIVE MEASURES: THE ABSOLUTE VALUE OF A MEASURE

105

The notion of a positive measure enables us to define an order relation on the vector space MR(X) of real measures on X. We write p S v if the measure v - p is positive. Since the relations p 2 0 and p S 0 imply that p ( f ) = p ( f + ) - p ( f - ) = 0 for allfc X,(X), and therefore that p = 0, it follows that p v is indeed an order relation on MR(X)(not, in general, a total ordering). It is clear that p 5 v implies that I + p 5 I + v for all real measures I , and ap =< av for all real scalars a 2 0. (For a study of this order relation, see (13.15).) (13.3.2) Let p be a (complex) measure on X . Then there exists a smallest positive measure p on X such that Ip(f)l 5 p(1fl)for a l l f c X,(X).

For every positive measure v such that Ip(f)l S v(lfl) for all f~ .X,(X), the relations g 2 0 and Ihl 2 g (9,h E ,X,(X)) imply Ip(h)l Iv(lhl) 5 v(g). We shall show that there exists a positive measure p on X such that, for each functionfz 0 in .X,(X), we have (13.3.2.1)

This p will then clearly satisfy the conditions of (13.3.2). To begin with, we remark that the right-hand side of (13.3.2.1) isJinite; for if K = Supp(f), then Supp(g) c K, and Il*(g)l 5 a K ' llgll

5 aK llfll

whenever 1g(s A by virtue of (13.1.2). Also it is clear that, for any real scalar a 2 0, we have p(af) = ap(f). We shall show next that, if f , and f 2 are two functions 2 0 belonging to ,XR(X),then

For each E > 0, there exists g i E X,(X) such that lgil 5 fi and

IAgi>lZ p ( f i ) - E (i = 192).

Multiplying gi by a complex number with absolute value 1, we may assume that p ( g i ) = Ip(gi)l, and then we have P(91

+ g2) = IP(Sl>l + 1AS;)l 2 P(f1) + p(f2)

- 2%

+

and since lgl + g215 f l + f 2 we have p(f, + f 2 ) 2 p(f,) p(f2) - 2 ~ Since . is arbitrary, it follows that p(f,) + p ( f 2 ) 2 p ( f i + f 2 ) . On the other hand, let h E .X,(X) be such that Ihl 5fl + f 2 . Let hi be the function which is equal to hf,/(fi + f 2 ) at the points x where f i ( x ) + f 2 ( x ) # 0 and is zero E

106

Xlll INTEGRATION

elsewhere (i = 1, 2). Then hi is continuous on X, because fi/(fl + f 2 ) is continuous at the points x wheref,(x) +fz(x) > 0, and also Ihi(x)l 2 Ih(x)l for all x E X, which proves that hi is continuous at the points x where f i ( x ) +fz(x)= 0, because h(x) also vanishes at these points. It is clear that / h i /Sfi (i = 1, 2) and that h = h, + h,; hence lP(h>l I lP(M

+ lP(h2)l I P ( f , ) + P(.fz)*

Since Ip(h)l may be taken to be arbitrarily close to p ( f , +f2),it follows that p(fl +fz) I p ( , f l ) p(f,). Hence (13.3.2.2)is proved. We shall now extend the definition of p ( f ) to all functionsf€ X,(X). To do this we write p ( f ) = p(f’) - p ( f ” ) , wheref=f’ -f ” is any decomposition off as the difference of two functions f ’ , f “ 2 0 belonging to X,(X). The value of p ( f ) so obtained is independent of the decomposition, because if f = f ; - f ; = f ; - f i , then f ; +f;’ =f;’ + f ; and therefore p(f;) p(f;) = p(f;’) + p(f;) by virtue of (1 3.3.2.2). With this definition, the formula (13.3.2.2) is valid for allfl,fz in X,(X). For we can writef, =f ; -f ; , f 2 =f ; - f i wherefi’,fi” (i = 1,2) are 20 and belong to X,(X); sincef, + f z = ( f ; + f ; ) - ( f : +f;’), our assertion follows from the definition above and from (13.3.2.2)for functions 2 0 . Finally, the above definition shows that, for each scalar a 2 0, we have p(uf) = up(f ) ; and if a < 0, then we have

+

+

P ( d > = p(af’

- a ! ” )= d - a f ” )

+ p(af’>

- p(-af‘) = ( - M f ” )- (- 4 P C f ’ ) = (-a)p(f”)

= ap(f).

Hence the relation p(uf) = ap(f) is valid for all real scalars a, and therefore we have proved that p is a (real) linear form on X,(X). Hence by (13.3.1)it is a positive measure. Q.E.D. The measure p defined by (13.3.2)is called the absolute value of the complex measure p, and is denoted by IpI. Hence by definition we have

It is immediately seen that if a E C and p is a measure on X, we have

(13.3.4)

laPl = I4 - IPl.

If p is a positive measure on X, then

(1 3.3.5)

IPI = P.

3

POSITIVE MEASURES: THE ABSOLUTE VALUE OF A MEASURE

107

By virtue of (13.3.2.1) it is enough to show that Iy(f)l S y(lf1) for all = 1 and Iy(f)l = [y(f) = p ( ( f ) ; since p is real, we have y(Tf) = y(9(cf)); and since y 2 0 and 9(Cf)5 Ifl, it follows that y(B([f)) 5 y(lfl), which establishes the assertion.

f~ X,-(X).Givenf, there exists [ E C such that l[l

If y is any real measure on X, it follows from (13.3.3) that y 5 lyl, and therefore :

(13.3.6) Every real measure on X is the diference of two positive measures (for a more precise result, see (13.15)). If y is any (complex) measure on X, it follows from (13.3.3) and (13.3.2) that we have

Also if y, v are any two measures on X we have

(13.3.8)

IP +

4 5 IPI + I4

by virtue of (1 3.3.2). Finally, it follows immediately from the definitions that, if n : X -+ X’ is a homeomorphism and y is any measure on X, then

PROBLEMS

1. Let X be a locally compact space, E a vector subspace of eR(X)and P a conuex cone in U,(X) (i.e., a subset of this space such that the relations f~ P and g E P imply f+ g E P and a f P~ for all scalars a > 0). Suppose that for each function h E S R ( X ) there exists f~ E such that f- h E P. Let u be a real linear form on E such that the relation f~ E n P implies u ( f ) 2 0. Let h E XR(X)and let PA be the set of all f~ E such that h - f P, ~ and Pi the set of a l l f s E such thatf- h E P. Show that these two sets PA, Pi are nonempty, and that if a’ is the least upper bound of the u ( f ) with f~ P A and a“ the greatest lower bound of the u ( f ) with f s Pl , then a’ and a” are finite and a’ 5 a”. Deduce that there exists a linear form u1 on the subspace El = E + Rh of VR(X) which extends u, such that the relationf, E El n P implies ul(fi)2 0. Show that a’ 2 u,(h) 6 a# for any such extension u1 of u, and that the extension is unique if and only if a’ = a”.

108 2.

XI11 INTEGRATION

Let X be a compact space and let p be a real-valued function on VR(X) satisfying the following two conditions: (i) p ( f + g ) g p ( f ) + p ( g ) ; (ii) p(uf) = u p ( f ) for all u > 0. Then the set P of functions f~ VR(X) such that p ( f ) 5 0 is a convex cone. Suppose furthermore that (iii) inff(x) s p ( f ) 5 supf(x) for all f~ V,(X), so that we have X E X

X E X

p(l) = 1. Show that there exists a positive measure p of mass 1 on X such that p ( f ) sp ( f ) for allfc VR(X).If moreover we are given a linear form u on a vector subspace E of VR(X) such that u ( f ) ( p ( f ) for all f~ E, then there exists a measure (L of the above type which extends the form u. (Consider a denumerable total set ( g n ) n ) Oin WR(X), with go = 1 ; use the result of Problem 1 inductively to obtain, on the subspace G of 'e,(X) generated by E and thegn, a linear form u such that -p( -f) 2 u (f)5 p c f )

for allfg G, and deduce that u extends by continuity to a measure on X.) Show that the measure p i s unique if and only ifp(f) p ( - f ) = 0 for allfe E.

+

4. T H E V A G U E T O P O L O G Y

Since the space M,(X) is a subspace of CXC('), we can define the weak topology, i.e., the topology of simple convergence in X c ( X ) (12.15). This topology on M,(X) is called the vague topology. To say that a sequence (p,) of measures on X converges cyaguely to a measure p therefore means that, for each function f E X,(X), the sequence ( p , ( f ) ) converges to p ( f ) in C. (13.4.1 ) Let (p,,)be a sequence of measures on X such that,for each f E X,(X), the sequence (pn(f )) tends to a limit p( f ) in C . Thenf Hp ( f ) is a measure on X and is the vague limit of the sequence (p,). If the pn are all positive, then so is p.

We have already remarked (13.1) that, for each compact subset K of X, X ( X ;K) is a Banach space and the restrictions of the p, to X ( X ;K) are continuous linear forms. Hence it follows from the Banach-Steinhaus theorem (12.16.5) that the restriction of p to X ( X ;K) is continuous, and therefore that p is a measure on X (and clearly a positive measure if the ,u, are positive). We recall (12.15) that a subset H of M,(X) is said to be vaguely bounded (or just bounded, if there is no risk of ambiguity) if, for each f E X,(X), we have sup I p ( f ) l < +a.Every vaguely convergent sequence is vaguely bounded.

PEH

(13.4.2) Let H be a bounded subset of M,(X). (i) For each conipact subset K of X, there exists a real number cK > 0 such that, for each p E H and each,f E .X,(X), we have Ip(.f>I 5

cK

(and therefore (13.3.2.1), 1p1(1 f I) 5 cK 11 f 11).

llfil

4 THE VAGUE TOPOLOGY

109

(ii) The vague closure of H in M,(X) is a compact metrizable space with respect to the vague topology. (i) This is an immediate consequence of the Banach-Steinhaus theorem (12.1 6.4). (ii) Let (U,) be a sequence of relatively compact open sets in X which cover X and are such that 0, c U,,+l (3.18.3). Since every compact subset K of X is contained in some U, (3.16), each space X ( X ;K) can be identified hence with a closed with a closed subspace of one of the Banach spaces W(u,,), subspace of X ( X ;0,).Now we know (7.4.4) that %(On)is separable, hence the same is true of X ( X ;0,) (3.10.9). Let (fmn)mL1 be a dense sequence in X ( X ;On).To show that H is metrizable it is enough to show that the vague topology on H can be defined by the pseudo-distances I(f, p - v ) l (12.4.6). This means that if g i (1 5 i 5 p ) is a finite sequence of functions belonging to X,(X), if p o is an element of H and r a real number >O, there exists a finite number of functions f,,,, (1 =< k =< q ) such that the relations p E H and I(fmknk P - ~ o > 5 l ! ~ r(1 5 k _I 4 ) imply I(gi P - PO)I 5 r (1 5 i 5 P ) . But the g i all belong to X ( X ;0,)for some fixed n, and the set of restrictions is equicontinuous, by (i) above and (12.15.7.1). of the p E H to X ( X ;D,,) Hence the assertion follows from (12.15.7). It remains to show (by the same reasoning as in (12.15.7)) that, when we identify H with its image L in the product space C N x Nby means of the mapping PI--+(( f,, , p)), L is closed in CNx .' If ( p k ) is a sequence of points of H such that each of the sequences ((fmn, p k ) ) k L 1 is convergent, then it follows from (12.15.7) that, in each X ( X ;On),the restrictions of the pk converge to a continuous linear form. Hence the sequence (pk) converges vaguely to a measure on X. Q.E.D. 3

9

We shall see later (13.20) that condition (i) in (13.4.2) can also be written in the form IpI(K) 5 cK for each measure p E H. Notice also that (13.4.2(ii)) implies (13.4.1) as a particular case. In particular: (1 3.4.3) Let v be a positive measure on X . Then the set of complex measures p such that lpl 5 v is metrizable and compact with respect to the vague topology.

For this set is evidently bounded and closed in M,.(X) in the vague topology. It should be noted that a bounded set of measures does not necessarily satisfy the hypothesis of (13.4.3).

110

Xlll INTEGRATION

(13.4.4) Let (pn) be an increasing sequence of real measures on X such that, for each function f 2 0 belonging to XR(X), the sequence ( p n ( f ) )is bounded above in R. Then the sequence ( p n ) has a vague limit in MR(X) which is also its least upper bound f o r the order-relation on MR(X).

For each function f 2 0 in XR(X), the sequence (pn(f)) is increasing and bounded above in R, hence (4.2.1) has a limit p ( f ) in R which is equal to sup p n ( f ) .Since every function belonging to X,(X) is a linear combination n

of four functions 2 0 belonging to X,(X), it follows that the sequence (p") is vaguely convergent (13.4.1), and it is clear that p is its least upper bound, by the definition of the order-relation on MR(X). (13.4.5) I f a series of positive measures on X , with general term p n , is such that, for each f 2 0 in X,(X), the series with general term p,,(f) 2 0 is convergent in R, then the series with general term p n is vaguely convergent in MR(X), and its sum p = p,,is such that p( f)= 2 p n ( f ) f o rall f E X,(X).

1 n

n

Apply (13.4.4) to the partial sums of the series.

PROBLEMS

+

1. Let be Lebesgue measure on R, let ,u be its restriction to R t =lo, m [, and let g be the function X H l/x on Rf Show that the measure g . ,u cannot be extended to a measure on R (cf. (17.9)).

.

2. On the real line R,show that the sequence of Dirac measures E. (unit mass at the point +n) converges vaguely to 0. Give an example of a sequence (f.) in Xc(R) which converges to 0 in the Frkchet space Qc(R) (12.14.6) but is such that the sequence ( where ak, ,, = (1 - (k - l ) / n ) ~ ~ -and ~ , uo, = 1, and Px. is a polynomial of degree s k - 1 whose coefficients are less than Ck/n in absolute value, where ck is a constant independent of n. (Differentiate the equation ( * k ) with respect to t , then multiply through by t . ) (c) Deduce from (b) and Weierstrass' theorem that, for every continuous function f c %'(I), the sequence (B".,) converges uniformly t o f o n 1. 10.

Let X be a metrizable compact space, let (f.).,,, be a sequence of complex-valued continuous functions on X, and let ( c . ) . , ~ be a sequence of complex numbers. (a) In order that there should exist a complex measure p on X such that p(f.) = c, for all n , it is necessary and sufficient that there should exist a number A > 0 with the following property: for every finite sequence of complex numbers, we have

(Use the Hahn-Banach theorem.) (b) Suppose that the fk are real-valued, the numbers ck real, and fo = I . Then there exists a positiue measure p on X such that p(fn)= c. for all n if and only if, for each sequence (A,),,,

2

I=O

11.

/\k

ck

of real numbers such that

"

1X,fk(x) 2 0 for all x E X, we have

k=O

2 o (cf. Section I3.3, Problem 2).

In Problem 10, take X = [0, I ] and For each sequence (c,) of scalars, put

f.(r)

= 1"

("Hausdorff's moment problem").

(a) Show that there exists a complex measure p on X such that p(fn)= c, for all n if and only if there exists a number A > 0 such that

(Observe that Akc. must be the value of p for the polynomial t"(1 - t ) k , and use Problem 9(b) by remarking that, for each real polynomial P, there exists a constant Cp such that I 1 Bn,P - - CPS P 5 B",P - CP

+

for all n.) (b) There exists a posiriue measure p on X such that p(f,)= c. for all n if and only if Akc. 2 0 for all k 2 0 and all n >= 0 (same method).

5 UPPER A N D LOWER INTEGRALS

113

12. Let X be a compact space. Show that in M(X) the set of positive measures with finite support and total mass 1 is dense (with respect to the vague topology) in the set P of all positive measures with total mass 1. (Let U be a neighborhood of p E P with respect to the vague topology, consisting of measures v E P such that Ip(fi)- v(fi)l 5 6, with fi E %‘,(X).Consider a continuous partition (g,) of unity and points U, E X such that Ifi(x) -Cfi(u,)g,(x)l 5 6 for all i . ) J

13. Let X be the unit interval [0, 11 in R.Show that the set K of Dirac measures E , (x E X) is vaguely compact, and that in M(X) the Lebesgue measure lies in the vague closure of the convex hull of K, but does not lie in the convex hull of K.

5. UPPER A N D LOWER INTEGRALS WITH RESPECT TO A POSITIVE MEASURE

In Sections (13.5) to (13.14)(including the problems), p denotes a positive measure on a locally compact space X. We shall show (13.7.3)that p can be extended from .XR(X) to a vector subspace YA(X, p) of RX,depending on p and containing (and in general distinct from) .XR(X),in such a way that this extension (also denoted by p) is apositive linear form on 2’A(X, p) (i.e., p takes values 2 0 on functions 20 belonging to Y ; ( X , p)) and possesses the fundamental property of passage to the limit for increasing sequences: that is to say, if (f,) is an increasing sequence of functions in 9A(X, p) whose upper envelopef (12.7.5)also belongs to Y A ( X , p), then we have lim p(fn) = &-). n+m

Let 9 (or 4(X)) be the set of all functions f : X + which are lower semicontinuous on X and bounded below by a function belonging to X,(X) (this implies that f ( x ) > - cx) for all x E X ; but we can have f ( x ) = + 00 at some points x E X : indeed, the constant function equal to + 00 belongs to 9). Every function f 2 0 which is lower semicontinuous on X belongs to 4 . For every f E 9,we put

this is a real number, or +a. Clearly iff E .X,(X) we have p * ( f ) = p ( f ) ; iff, g E 9 a n d f s g , we have p * ( f ) 5 p*(g); and for any real number a > 0, we have p*(af) = a p * ( f ) for all f E 4 .

(13.5.2) Let (f,) be an increasing sequence of functions belonging to 4 , and = supf, (so that f E 4 (12.7.6)).Then

let f

n

Xlll INTEGRATION

114

Suppose first of all that f E .X,(X) and that f,E X,(X) for all n. It is clear that the supports of all thef, are contained in the same compact set K = Supp(f) u Supp(f,). By Dini's theorem (7.2.2), the sequence (f.)converges uniformly t o f o n X. The formula (13.5.2.1) then follows from the fact that the restriction of p to X ( X ;K) is a continuous linear form on this Banach space. Now pass to the general case. Clearly we have p*(f,) 5 p * ( f ) for all n. Hence it is enough to show that, for each function u E X,(X) such that u 55 we have p*(u) S sup p*(f,). Now by (12.7.8) we know that, for each function,f, , there exists an increasir.5 sequence (gm,JnlZlof functions belonging to X,(X) such that f, = sup gmn. We have f = sup gmn, hence also m

m, n

f = sup h, , where h, = sup g p q. The functions h, clearly belong to X,(X) n

p l n , qsn

and form an increasing sequence. Since u SJ, we have u = sup(inf(u, h,,)); the n

sequence of functions inf(u, h,) is increasing and belongs to X,(X);since also u E X,(X), the first part of the proof shows that p*(u) = sup p*(inf(u, h,,)); n

but since h, S.d, we have p*(inf(rr, h,)) 5 p*(f,), and therefore finally P*@) 5 SUP P*(f,). Q.E.D. n

We remark that because the functions in 9 never take the value -03, the sumfl +fi of two such functions is defined at every point of X, and belongs to 9 (12.7.5).

We may write . f l

=

lim g, and f2 = lim h,, where (g,) and (h,) are two

n-. w

n+ a,

increasing sequences of functions belonging to X,(X) (12.7.8). Hence we have fl +f2= lim (g, h,) (4.1.8). Since p ( g , h,) = p(g.) + p(h,), the ti+

m

+

+

result follows from (13.5.2) and (4.1.8). Let (t,) be any sequence whose terms are real numbers 20, or +a. Since the partial sums s, = t , + -.-+ r, are defined (4.1.8) and form an increasing sequence, this sequence has a limit in R (4.2.1), denoted by

m

C t,

n= 1

and called the sum of the series with general term t,. For every sequence

(f,)of positiue functions belonging to I, the function

m

XH

c f , ( x ) is therefore n= 1

5 UPPER A N D LOWER INTEGRALS m

defined, and is denoted by

115

f n ; also it belongs to 9 (12.7.6). If we apply

n= 1

(13.5.2) to the sequence of partial sums

cfn,and (13.5.3) to each of the N

n= 1

terms in this partial sum, we obtain the following corollary: (13.5.4)

Zf(fn)ntl is a sequence of functions 20 belonging to 9,then

Now consider an arbitrary mapping f of X into R. There always exist functions h E 9 such that h 2 J for example the constant function equal to +co. Put p*(f) =

(13.5.5)

inf

htj,he/

p*(h);

this number p * ( f ) is called the upper integral o f f with respect to the measure p. Iff E 9 it is clear that this definition agrees with the preceding one. Here the value of p * ( f ) can be any element of R. The relation f 6 g implies p * ( f ) 5 p*(g). For any scalar a > 0 we have p*(af) = ap*(f).

+

(13.5.6) Ifthe sum f l f , of two mappings of X into R is definedat allpoints of X , and if p*( f l ) > - 00 and p*( f,) > - 03, then

+ f J 5 P*(fl) + P*(f,).

P*(fl

This is obvious if one of the numbers p * ( f i ) , p * ( f 2 ) is +a.If not, given any a > p * ( f i ) and b > p*(f2), there exist h,, h, in 9 such that h,, f , 5 h, and p*(hl) 5 a, p*(h,) b. It follows that h, + h, 2 f l +f , fi and p*(h, + h,) 5 a + b by (13.5.3). Hence the result. (I3.5.7) Zf (f.) is any increasing sequence of mappings of X into p*( f n ) > - 03 for all suficiently large n, then

1

P* SUPfn (

n

The inequality p* supf ( n

W

such that

= SUP P*(fn) = lim p*(fn)* n+

n

m

2 sup p*(fn) is clear. Let us prove the reverse n)

n

inequality. We may assume that sup p*(fn) < n

+

03,

otherwise there is

nothing to prove. By hypothesis, we may therefore assume that sup p * ( f . ) n

and all the p * ( f . ) arefinite. For each E > 0, we shall show that there exists an

116

Xlll INTEGRATION

increasing sequence (g,) of functions belonging to 9 such that for each n we have f , 5 gn and p*(gn) 5 p*(f,) + E . Then, if we put f = supf, and g = sup g, , we shall have g E 9,f 5 g, p*(g) = sup p*(gn) 5 sup p*(f,)

+E

by virtue of (13.5.2),and finally p * ( f ) 5 p * ( g ) 5 sup p * ( f , ) + E . Since

was

n

n

n

E

n

arbitrary, this will complete the proof. By definition, for each n there exists h, E Y such that f n 5 h, and p*( f , ) 5 p*(hn)5 p*(f,) 2 - k Put g, = sup(h,, . . . , I?,,), so that gnE 9 (12.7.5).Clearly the sequence (g,) is increasing, and f,5 gn. We shall show by induction on n that

+

This is evident from the definitions when n = 1. Suppose that (13.5.7.1)is true, and remark that gn+l= sup(hn+lrg,) and f , 5 inf(h,+,, gn). Since the functions h,+l and g, do not take the value -co,we have SUp(hn+ 1,

gn>

+ inf(hn+

13

Sn) = h n +

1

+ gn

and therefore, by virtue of (13.5.3)and the fact that p*(g,,) and p*(h,+l) are finite,

by the inductive hypothesis (1 3.5.7.1).

Q.E.D.

It should be noted that for a decreasing sequence (fn), it is no.t necessarily true that p* infS, = inf p*(f,), even if all the numbers p*(fn) are $finite

L1

(Section 13.8,Problem 13). (13.5.8) Z f ( f n ) is any sequence of functions 20,then

5

UPPER AND LOWER INTEGRALS

117

For each integer N 2 1 it follows from (13.5.7) that

now apply (12.5.7) to the increasing sequence of partial sums

N

Cfn.

n= 1

R, the mapping

If (f n )is any sequence of mappings of X into

x H l i m inff,(x)

(resp. x H l i m supfn(x))

n+w

n-t m

is defined for all x E X. It is denoted by lirn inff, (resp. lirn supf,). n+m

(13.5.9)

n- m

(Fatou’s lemma) r f ( f n )is any sequence ofjiunctions 2 0 , then p* lirn inff,,

(

n-tm

1 s lirn inf p * ( f , ) . n+m

For each n 2 1 , let gn = inf (f,+,). Clearly we have p*(gn) 5 inf p * ( f , + & , P20

and gn 2 0; since the sequence (gn) is increasing and lirn inf f ,

p* lirn inff, n+m

= sup g n , it

n+ m

foIlows from (13.5.7)that

(

P20

= sup p*(g,) 1

5 sup (inf p*(fn+,))

n

pZ0

n

= lirn inf p*(f,). n+ w

Iff: X + R is any mapping, we define p * ( f ) = - p * ( - f ) ; this number is called the lower integral off with respect to the measure p. All the properties above which were proved for the upper integral can be immediately translated into properties of the lower integral. In particular, if we put -9 = 9 (or Y ( X ) ) , then Y is the set of all upper semicontinuous functions on X which are bounded above by a function belonging to X,(X) (which implies that they do not take the value + 00). For all functions f E 9, we have

CL*U)=

inf

9 b f 9 9 E XR(X)

P(S>

and for all f:X + R we have P*(f) =

(13.5.10)

SUP P*(h).

hsf,heY

I f f is any mapping of X into R, then p * ( f ) 5 p * ( f ) .

118

Xlll INTEGRATION

By virtue of the definitions, it is enough to show that if u E 9, u E 9 and u 6 u, then p*(u) p*(u). Since - u E 9 it follows that u - u = u ( - u ) is defined throughout X,belongs to Y and is 20;hence

+

0 5 p*(u - u) = p*(u

+ (- u)) = p*(u) + p*( =p*(d

- u)

- P*b)

by virtue of (13.5.3). We shall oftenwrite j * f d p or s * f ( x ) dp(x) (resp. S*fdpor f t f ( x ) dp(x)) instead of p * ( f ) (resp. p*(f)). For any subset A of X, we put p*(A) = p*(qA) and p,(A) = p*(qA), where qA is the characteristic function of A (12.7). These numbers p*(A) and p,(A) are 20 (possibly +a);they are called, respectively, the outer measure and inner measure of A. Example (13.5.11) Let I be Lebesgue measure on R and let I = ] a , b[ be an open interval in R. We shall show that I*(I) = b - a (which is equal to + 03 if b = + 03 or a = - a).If a < a' < b' < b, then there exists a continuous mappingfof R into the unit interval [0, 11, with support contained in [a, b ] , and equal to 1 throughout [a', b'] (4.5.2). We have J -00 + m f ( t )dt 2 b' - a'. Conversely, for each function g E XX,(R) such that 0 5 g 5 q r ,we have / +- mm g ( t )dt 5 b - a. Hence it follows that I*(I) = b - a. Now let U be any open set in R. Then its connected components (3.19) are open intervals ((3.19.1) and (3.19.5)), and the set of connected components is at most denumerable, because each component contains a point of the denumerable set Q, and the components are pairwise disjoint. Hence if they are r k = ] & , bk[, then q u = C qlk,and therefore ((12.7.4) and (13.5.4)) k

n*(u)= 1 ( b k - ak). k

(13.5.11.1)

6. NEGLIGIBLE F U N C T I O N S A N D SETS

A mapping f : X --t R is said to be negligible (with respect to the measure

p), or p-negligible, if p * ( l f l ) = 0. Then ufis also negligible for all a # 0 in

and if 191 5

If[,

then g is negligible.

R;

6 NEGLIGIBLE FUNCTIONS AND SETS

(13.6.1) I f ( f . ) is a sequence of negligible functions 20, then

m

119

fn is negligible.

n=l

This follows immediately from (13.5.8).

A subset N of X is said to be negligible (with respect to p) or p-negligible if its characteristic function cp, (12.7) is p-negligible. Clearly any subset of a negligible set is negligible. (1 3.6.2)

A denumerable union of negligible sets is negligible.

For if ( N k )is a sequence of negligible sets and N (P,

= sup cpNk k

=

5 1 q N kand , the result follows from (13.6.1).

u

N k , we have

k

k

For example, with respect to Lebesgue measure A, every set { t } consisting of a single real number is negligible. For if E > 0 there exists a function f E XX,(R)with values in [0, I] which is equal to 1 at the point t and vanishes on the complement of the interval [t - E , t + E ] , and therefore L*((P{f,)

I 4f1 I 28.

Hence, by (13.6.2), it follows that every denumerable subset of R (for example, the set Q of rational numbers) is negligible with respect to Lebesgue measure. One can also give examples of nondenumerable sets in R which are negligible with respect to Lebesgue measure (Section 13.8, Problem 4).

A property P(x) of the points of X is said to be true almost everywhere (with respect to p) if the complement of the set of points for which P(x) is true is pnegligibie. (13.6.3) A mapping f :X -+ everywhere.

(P,

is negligible if and only

if it is zero almost

Let N be the set of all x E X such that If(x)l > 0. Then we have 5 sup nl f 1, and If1 5 sup ncp,, hence the result follows from (13.5.7). n

n

(1 3.6.4) I f a mapping f : X -+ R issuch that p*( f)< + cx) (resp. p * ( f ) > - a), then f ( x ) < co (resp. f ( x ) > - a)almost everywhere.

+

It is enough to prove the assertion relating to the upper integral. By hypothesis, there exists a function h E 9 such that f 5 h and p*(h) < + 00. We may therefore limit ourselves to the case where f E 9,and since there is

120

Xlll

INTEGRATION

then a function u E .X,(X) such thatf - u >= 0, we may also assume t h a t f z 0. Now let N be the set of points x E X such thatf(x) = + 03 ; we have ncp, 5 f for all integers n > 0, hence np*(cp,) 5 p*(f), and the hypothesis therefore implies that p*((pN) = 0. The relation “f(x) = g(x) almost everywhere in X ” is an equivalence relation between mappings of X into a set E, because the union of two negligible sets is negligible. We say then that f and g are equivalent (with respect to p) or p-equivalent, and we shall denote by f the equivalence class of a mapping f :X --+ E. A mapping f :A -,E, where A is a subset of X, is said (by abuse of language) to be defined almost everywhere in X if X - A is negligible. The equivalence class of such a mapping f is then defined to be the equivalence class of any mapping of X into E which extendsf; as before, we denote the equivalence class by f. Two functions f, g defined almost everywhere are said to be equivalent i f f = 8; this means that the set of points x E X at which f and g are both defined andf(x) = g(x) has a negligible set for its complement. (1 3.6.5)

Let f,g be two equivalent mappings of X into 8. Then p*( f) = p*(g).

Let N be the negligible set of points x E X such that f (x) # g(x). Since the functions f, g and sup(f, g ) are equal on X - N, we may assume that f 5 g. Let h be the negligible function which is equal to co at the points of N, and to 0 elsewhere. If v E 9 is such t h a t f s v, the function v h is defined at all points of X, and we have g 5 v + h, so that

+

+

by virtue of (13.5.6), because p*(v) > - co. From the definition of p * ( f ) , it follows that p*(g) 5 p * ( f ) and hence that p*(g) = p*(f). Iffis a mapping into 8 which is defined and finite almost everywhere in X, then f is equivalent to afinite function defined on the whole of X. Iff, g are functions defined almost everywhere in X, with values in R, and finite almost everywhere, then the same is true o f f + g andfg, and the equivalence classes of these functions depend only on f and 8. They are denoted by f + and fg, respectively. Iff(x) 5 g(x) almost everywhere, thenfl(x) 5 gl(x) almost everywhere for any functions fl, g1 equivalent to f, g, respectively. In this case we write f ij , and this defines an order relation on the set of equivalence classes (with respect to p ) of mappings of X into W.

7 INTEGRABLE FUNCTIONS A N D SETS

121

7. INTEGRABLE F U N C T I O N S A N D SETS

Wehaveseen (13.5.10) that every function f:X + R satisfies p * ( f ) S p * ( f ) . The function f is said to be integrable (with respect to p) or p-integrable if p * ( f ) and p * ( f ) are finite and equal. Their common value is then called the integral off with respect to p, and is written p ( f ) , or (S, p), or j f dp, or

1f ( x ) dp(x). Clearly every function f

.X,(X) is integrable and its integral is the value of p atf, so that our notation is consistent. An integrable function is therefore finite almost everywhere (13.6.4), but a bounded function is not necessarily integrable. For example, a constant nonIxl), is not integrable zero function on R, or the continuous function 1/(1 (with respect to Lebesgue measure) (cf. (13.20)). In view of the definitions in (13.5), we have the following criterion for integrability : E

+

(13.7.1) (i) For a function f :X -+ R to be integrable, it is necessary and sufficient that, given any E > 0, there should exist g E 9’and h E J such that g S f h and p*(h) - p*(g) 5 E (or, equivalently (13.5.3), p*(h - g ) 6 e). (ii) I f f is integrable, there exists a decreasing sequence (h,) of functions belonging to 9 and an increasing sequence (g,) of functions belonging to Y such that g, 5 f 5 h, for all n and lim p*(hn) = lim p*(gn) = Af).

n+ w

n+

m

(i) The condition implies that p*(h) and p J g ) are necessarily finite (and therefore so are p*( f ) and p*( f ) ) , and that p*( f ) = p*( f ) . (ii) For each n there exists h; E J and g; E 9’ such that g; S f S hi and p ( f ) - n-l S p*(gA) 5 p ( f ) 5 p*(h:) 5 p ( f ) + n-’. The required conditions are then satisfied by taking h, = inf(h;, . . . , h@ and g, = sup(g;, .. ., g;). Iff is integrable, then clearly so is any function f l equivalent to f (13.6), and we have f i dp = f dp (which we denote also by p ( f ) ) . It is now clear how integrability should be defined for functions defined almost everywhere in X: such a function f is integrable if the functions equivalent to f and defined on the whole of X are integrable. The number p(f) is also denoted by f dp,

s

s

f

or j f W dP(X), or A f ) ,or (S, P>. (13.7.2) For a mapping f : X + to be integrable it is necessary and suficient that, given any E > 0, there exists a function u E .X,(X) such that cc*(lf - 4) S E.

Xlll INTEGRATION

122

Necessity. Suppose that f is integrable, then there exist functions g E 9’and h E 9 such that g 5 f 5 h and p*(h - g ) 5 +E. Also ((13.5,l) and (13.5.3)) there exists u E X,(X) such that u 5 h and p*(h - u) +&. Since If - uI 5 Ih - uI + Ih - 91, it follows by (13.5.3) that

s

s

p*(lf - 4) p*(lh - 4)+ p*(lh - g l ) S

6.

Suficiency. If u E .X,(X) is such that p * ( l f - ul) 5 E , then by definition (13.5.5) there exists a function v E 9 such that If - uj v and p*(v) 5 2 ~ . But the relation - u 5 f - u 5 v can be written in the form

s

+ u5f5 v + u have - u + u E 9’and v + u E .f (12.7.5), -v

(u being finite); also we p*(v u - (- u u ) ) = 2p*(v)

+

+

5 4 ~ hence ; f is integrable.

and

s

The set L?i(X, p) (also denoted by L?A(p) or 2’:) offinite p-integrable functions on X is a vector space over R,and the mapping f H f dp is a positive ( I 3.7.3)

s

linear form on 9; (i.e.) the relation f 2 0 implies f dp 2 0).

Iff is finite and integrable, then clearly so is af for every scalar a E R, and we have af dp = a f dp. Iff and g are finite and integrable, it follows from (13.5.6) applied to f and g and to -f and -9 that

I

s

s s s* f&+ B 4 - I

(f+B)dPS

which completes the proof.

s*

(f+B)dPS

s s f4+

94.4

It follows that iff and g are integrable functions (finite or not) on X with values in R, then f + g (which is defined almost everywhere) is integrable, and that

J ( f -k 9 ) dP = / f

+ 1 9 4.

(13.7.4) I f f is integrable, then so are If I, f

+

and f -, and we haue

(13.7.4.1)

I f f and g are integrable, then so are sup(f, g ) and inf(A 9).

If

uE

X,(X) is such that p * ( l f - ul) 5 E , then since

I If1 - I4 I S I f - 4,

7 INTEGRABLE FUNCTIONS AND SETS

)I

it follows that p*(I If1 - ( u J 5 E , which by (13.7.2) shows that rable. Moreover, we have -If[ 5 f 5 If1 and hence

If1

123

is integ-

Ifl) = -PL(ISI) 5 P*(f) = C lm 5 P*(lfl> = A l f I)? which establishes (13.7.4.1). Since f = f ( f + If[) and f - = +(If1 -f), it /A*(-

+

follows thatf' and f - are integrable (13.7.3). Iff and g are integrable, then f - g is defined almost everywhere and integrable, and the functions (everywhere defined) sup(f, g ) and inf(f, g ) are equivalent respectively to the functions (defined almost everywhere) + ( f g Ifgl) and +(f g - I f - 91). Hence they are integrable.

+ +

+

(13.7.5) For a function h E 9 (resp. g E 9) to be integrable, it is necessary and suflcient that p*(h) < + 00 (resp. p*(g) > - 00).

For if p*(h)
0 there exists u E XR(X)such that

by (13.5.1) and (13.5.3). The result now follows from the definition of integrable functions. E,

A subset A of X is said to be integrable if its characteristic function cpA is integrable, or equivalently if p*(A) and p*(A) are jinite and equal. Their common value cp d is then denoted by p(A) and is called the measure of A. We have p ( 0 ) = 0. The negligible sets are the same as the integrable sets with measure zero. If A is integrable, then so is every set B for which A n eB and B n EA are negligible, and p(B) = p(A).

1.p

(1 3.7.6) I f A and B are integrable sets, then A v B, A n B and A n CB are integrable.

This follows from the formulas (12.7.3). (13.7.7) Every compact set is integrable. For an open set U to be integrable it is necessary and suficient that p*(U) < + 00. In particular, every relatively compact open set is integrable.

This follows immediately from (1 3.7.5) and (12.7.4).

Example (13.7.8) With respect to Lebesgue measure on R, every bounded interval with endpoints a, b is integrable and its measure is Ib - a / , by virtue of

124

Xlll

INTEGRATION

(13.5.11) and the fact that all finite sets are negligible (for Lebesgue measure) (13.6). If I = [a, b] is a bounded interval, the “Dirichlet function” (3.11) (P, - ( P , ( P ~ (which is 0 at the rational points of I and on the complement of I, and 1 at the irrational points of I) is integrable, and its integral is b - a, since the set Q is negligible (13.6). (13.7.9) A subset A of X is integrable i f and only i f , given any E > 0, there exists a compact set K and an open set G such that K c A c G and p*(G - K) E.

The condition is sufficient by virtue of the definition of integrable functions and (12.7.4). Suppose conversely that A is integrable, and let E be such that 0 < E < 1. By hypothesis, there exists h E 9 such that ( P 5 ~ h and

s

- (PA) dp 5

Let G be the set of points x E X such that h(x) > 1 - E ; then G is open (12.7.2) and contains A. Clearly h 2 (1 - E ) ( P ~ ,hence

I

Also there exists g E Y such that g 5 ( P and ~ ( q A- g ) dp 5 E , and by the definition of Y the set of points x E X such that g ( x ) > 0 is relativelycompact. Choose 6 > 0 so that Sp(A) 5 E , and let K be the set of all x E X such that g ( x ) 2 S; then K is closed in X (12.7.2) and therefcre compact (3.17.3), since it is contained in a relatively compact set. Clearly we have K c A and g 5 ( P ~ Sq,, where B = A - K ; hence

+

j g dp

S p(K) + M B ) S p(K) + M A ) S

and finally p(A) 5 j g dp

p(K)

+ E,

+ I p(K) + 2 ~ .

Q.E.D.

E

There exist bounded subsets of R which are not integrable with respect to Lebesgue measure (Section 13.21, Problem 6). (13.7.10) Let n : X -+X‘ be a homeomorphism. It follows directly from the definitions that iff’ is any function on X with values in R, we have

j*Yd(nW) = J>ft

.

O

dP.

125

8 LEBESGUE’S CONVERGENCE THEOREMS

The function f ’ is n(p)-integrable if and only if ,f’ that case we have J f ‘ d(n(p))= f ’ o 71) dp.

I(

0

71

is p-integrable, and in

8. LEBESGUE’S C O N V E R G E N C E T H E O R E M S

(13.8.1) Let (f,) be an increasing sequence of integrable functions. For supf, to be integrable, it is necessary and sufficient that n

and if this condition is satisfied we have

(13.8.1.1) Since J*f,dp > - co for all n, we have

j*(sup f.) n

dp

= sup

/fndp

n

(13.5.7).This already shows that the condition is necessary. Conversely, if the condition is satisfied, let f = supf , . Then, given any E > 0, there exists n n

such that the functi0n.f -f, (which is defined almost everywhere because f and f n are finite almost everywhere (13.6.4))satisfies

(13.5.6). But there exists a function (13.7.2),hence by (13.5.6)we have

uE

X,(X) such that J Ifn - uI dp 5 E

and the result follows by (13.7.2). There is of course a corresponding theorem for decreasing sequences of integrable functions, obtained by replacing f , by -A in (13.8.1). (13.8.2) Let (f,) be any sequence of integrablefunctions. In order thatf = sup fn

should be integrable, it is necessary and suficient that there should exist a g dp < + 00 and f n 5 g almost everywhere. function g 2 0 such that

I*

The condition is obviously necessary (take g =f+).Conversely, if it is satisfied, let gn = sup fk . Then g, is integrable (13.7.4) and f = sup gn . 1S k S n

n

Xlll INTEGRATION

126

<s*

Since the sequence (gn) is increasing and s g n dp = follows from (1 3.8.1).

gn dp
0. Show that the set B of points of X which n

belong to infinitely many of the sets A. is integrable, and that p(B) 2 rn. 8.

Let A be Lebesgue measure on the interval 1 = [0, 11 and let A. be a sequence of integrable sets in I such that inf h(A.) = rn > 0. For each integer k , let %k denote the

"

+

set of 2k intervals of the form [ j . 2-,, ( j 1) .2-,] in I(0 5 j < 2'9, and CSk the set of all the unions of sets of a,. Show that there exists a decreasing sequence ( I k ) k r O of subsets of I such that 1, = I , I, E Ek for all k , and a subsequence (Ank) of the sequence (A,) with the following properties: (i) &A,, n (I - I k ) ) 5 &rn;\(l - I t ) for all r >= k ) ; (ii) &Anvn J) 2 trnA(J) for all r 2 k and all J E Dk contained in I,. (Proceed by induction on k , using the diagonal trick.) Deduce that there exists a subsequence (B.) of (A,) such that, for all k and all J E ' D k contained in I, , the intersection of J with the sequence (B,) is not empty. (Use Problem 7 and the diagonal trick.) Deduce that B, contains a non-empty compact set with no isolated points (and therefore n

nondenumerable (Section 4.2, Problem 3(c))).

9. Letfbe a real-valued function 2 0 on X. Show that the mapping p ~ p * ( f 'of) M+(X) into R is continuous with respect to the vague topology (13.4) if and only iff is continuous and compactly supported. The mapping p ~ p * ( fis) lower semicontinuous with respect to the vague topology if and only iffis lower semicontinuous. 10. Let (pn)be an increasing sequence of positive measures on X. Suppose that the sequence is bounded above in M+(X), and let t~ be its least upper bound (13.4.4). Show that, for every functionfz 0 on X, we have p * ( f ) = lim pcL:(/). = "-1

Let (pn)be a decreasing sequence of positive measures on X, and let p be the greatest lower bound of the sequence in M+(X) (13.4.4). Iff is a function 2 0 on X such that pz(f)< +to for all sufficiently large n, show that p*(f)= limp,*(f).

11. (a)

+

"+ m

(If g 2 0 is lower semicontinuous and p*(g) < a,observe that there exists a sequence (h,) of continuous functions 2 0 with compact supports, such that h,, s g and p.*(g)

=?p.*(h,)

x m

for all n.)

(b) On the discrete space N, let p n be the measure such that p n ( { m }= ) 0 for rn < n and 1 for rn 2 n. Show that the greatest lower bound of the decreasing sequence (p.) in M+(N) is 0, but that p,*(N)= to for all n.

+

8 LEBESGUE'S CONVERGENCE THEOREMS

133

12. Let X, Y be two locally compact spaces and T : X -+Y a proper continuous mapping (Section 12.7, Problem 2). Let p be a positive measure on X, and let v = ~ ( p(Section ) 13.4, Problem 8). Show that v * ( g ) = p*(g T)for all functions g 2 0 on Y. (Consider first the case in which g has compact support, and then use Section 12.7, Problem 0

2(b).) Show that N c Y is v-negligible if and only if n-l(N) is p-negligible.

13. Assume that there exists in I = [0, I ] an increasing sequence (H.) of nonmeasurable H. = I and h,(H,) = 0 for all n. sets (with respect to Lebesgue measure h) such that

u"

Show that h* inf (1 - yHn)) = 0, but that h*(l - vHn) = 1 for all n.

("

Let F be a convex function defined on a convex open set A c R" (Section 8.5, Problem 8) and let fk (1 6 k 6 n) be n integrable functions on X, such that the mapping x ~ ( f * ( x ) takes ) ~ ~its~ values ~ ~ in A, and such that the composite function x ~ F ( f i ( x .).,. , f . ( x ) )is integrable. Show that if u , is bounded (13.9.16) and p(X) = 1, then

14. (a)

(Use the Hahn-Banach theorem applied to the convex set of points (ti, ..., ~ , , , u ) E A X R

such that u 2 F(r,, . . . , t,,).) Consider in particular the case n = 1, F ( t ) = e', and deduce the inequality of the means uyu;2

. . . u:m

s

U1Ul

where the uk are >O, the uIiare 2 0 and

1-u2 u2

+ ..+ '

umum,

m k= t

uk= 1.

(b) Suppose that p is bounded and p ( X ) = l . Show that if two functions .f2_0, g 2 0 are integrable and such thatfg 2 1, then ( l f d p ) ( s g d p ) 2 1 (use (a)). 15. For each finite sequence (A,)

,5 , d m of p-integrable sets, put

if none of the A, is pnegligible. (Remark that a point of D(AI, . . . , An) belongs to a t least n - 1 of the sets D(A,, A,) (i <j).)

h is Lebesgue measure on the unit interval I = [0,1] and if S(1) is the set of nonempty closed subsets of I (Section 3.16, Problem 3), show that the function A H ~ ( A ) is not continuous with respect to the Hausdorff distance on S(1).

16. If

134

Xlll

INTEGRATION

9. MEASURABLE FUNCTIONS

(1 3.9.1) Let A be an integrable subset of X . Then there exists a partition of A consisting of a sequence (K,) of compact sets and a negligible set N . We define the sets K, inductively as follows, using (13.7.9), (13.7.7), and

u

n-1

(13.7.6):K, c A and p(A - K,) 5 1 ; K, c A -

i= 1

K i and

if n > 1. Then the intersection N of the sets A -

(13.8.7).

u Ki n

i= 1

is negligible, by

A subset of A is said to be measurable (with respect to p) or p-measurable if there exists a partition of A consisting of a sequence of compact sets and a negligible set. Jf K' and K are compact sets such that K' c K, then K - K' is integrable ((13.7.6) and (13.7.7)), and therefore, by virtue of (13.9.1), an equivalent statement is that A is the union of a sequence of compact sets and a negligible set. The same reasoning shows also that another equivalent statement is that A is the union of a sequence of integrable sets. A set A c X is said to be universally measurable if it is measurable with respect to every positive measure on X.

(13.9.2) For a subset A of X to be measurable it is necessary and suficient that A n K should be integrable,for each compact set K in X . The condition is necessary, by virtue of (13.8.7(ii)). It is sufficient, because X is the union of an increasing sequence (K,) of compact sets (3.18.3), and therefore A I S the union of the sequence (A n K,) of integrable sets. In particular, we see that the space X itself is measurable.

(13.9.3) (i) The complement of a measurable set is measurable. (ii) Denumerable unions and denumerable intersections of measurable sets are measurable. (iii) All open set,r and all closed sets are universally measurable. Assertion (i) follows from (1 3.9.2) and (1 3.7.6); assertion (ii) from (13.9.2) and (13.8.7); assertion (iii) from (13.9.2) and (13.7.7) for closed sets, and then by (i) for open sets.

9

MEASURABLE FUNCTIONS

135

The sets obtained from the open (or closed) sets of X by iterated application of the operations (i) and (ii) above are therefore also universally measurable. The definition of a p-measurable set A also shows that there exists a universally measurable set B contained in A such that A - B is p-negligible. By applying this result to X - A, we see that there also exists a universully measurable set C containing A such that C - A is p-negligible. A mapping u of X into a topological space Y is said to be measurable (with respect to p) or p-measurable if there exists a partition of X into a sequence of compact sets K, and a p-negligible set N such that each of the restrictions u I Kn is continuous. The mapping u is said to be universally measurable if it is measurable for all measures on X. A continuous function is universally measurable. For a subset A of X to be measurable (resp. universally measurable) it is necessary and sufficient that its characteristic function q A should be measurable (resp. universally measurable). For if we have a partition of X consisting of a sequence of compact sets K, and a negligible set N, such that qAI K, is continuous for all n, then K, is the union of two disjoint compact subsets, namely A n K, and K, - (A n K,)((3.11.4) and (3.17.3)), and A is the union of the sets A n K, and A n N, and is therefore measurable. Conversely, if A is measurable, then so is X - A (13.9.3), hence there exists a partition of X consisting of a sequence of compact sets K, and a negligible set N, such that either K, c A or K, c X - A for all n. Consequently qAI K, is continuous for all n. (13.9.4) Let u be a mapping of X into a topological space Y . Then the following three properties are equivalent: (a) u is p-measurable. (b) For each compact subset K of X and each E > 0, there exists u compact subset K' of K such that p(K - K') 5 E and such that the restriction of u to K' is continuous. (c) For each compact subset K of X, there exists apartition of K consisting of a sequence (L,) of compact sets and a p-negligible set M, such that each of the restrictions u I L, is continuous. To show that (a) implies (c), we remark that the hypothesis (a) implies the existence of a partition of X consisting of a sequence (K,) of compact sets and a negligible set N, such that u I K, is continuous for all n ; hence (c) is satisfied by taking L, = K n K, and M = K n N. To show that (c) implies (b), we remark that it follows from (c) that p(K)

a,

=

1 p(L,) fl=

1

(13.8.7), and hence that there is an integer n such that

p(K - K') 5 E , where K' =

u Li .

icn

Clearly K' satisfies the conditions of (b).

136

Xlll

INTEGRATION

Finally, we have to show that (b) implies (a). The space X is the union of an increasing sequence (H,) of compact sets (3.18.3). For each n we use the hypothesis (b) together with (13.7.9)to define a sequence (Km,),21 of compact sets such that K,, c H, - H,-,, K,, c (H, K i n , and

u

i<m

and such that the restrictions u I K,, are continuous. The complement of the union of the sets K,,, for n 2 1 and m 2 I is then negligible, and consequently u is measurable. It follows that, for a mapping f : X -+ B to be measurable, it is necessary and sufficient that, for each compact subset K of X, the function f q K (which by convention is taken to be zero at all points of X - K, even iff is infinite at such points (cf. (13.11)) should be measurable. For this condition is sufficient by virtue of (13.9.4),and it is necessary by (13.9.4)and the fact that X - K is measurable (13.9.3). The same reasoning shows that iff is a mapping of X into R and K is a compact subset of X such that f I K is continuous, then f q K is measurable.

(13.9.5) Let (Y,)be a sequence of topological spaces, and f o r each n let u,, be a measurable mapping of X into Y, . Then, given any compact subset K of X andany E > 0, there exists a compact subset K' of K such that p(K - K') 5 E and such that u, I K' is continuous for all n. Using (13.9.4)we define inductively a decreasing sequence (K,) of compact subsets of K such that p(K - K,) 5 f~ and p(K, - K,+l) 5 e/2"+', and such that u, 1 K, is continuous. Then by (13.8.7)it is clear that K' = K, satisfies

0 n

the required conditions.

(1 3.9.6) Let (Yi)l be afinite sequence of topological spaces, Z a topological space, and v a continuous mapping of Y i into Z. Let :ii : X -+ Y i (1 2 i 5 n)

n

be measurable. Then the mapping

i

XF-+ v(u,(x),

... , u,,(x)) is measurable.

This follows immediately from (1 3.9.5) and (1 3.9.4).

(13.9.7) V J g are two measurable functions with values in 8, then sup(f, g ) and inf(f, g ) are measurable. r f u, v are measurable mappings of X into a (real or complex) vector space, then u + v and au (where a is any scalar) are measurable.

9 MEASURABLE FUNCTIONS

137

(13.9.8) A function equivalent to a measurable function is measurable.

This follows from (13.9.4)and the fact that if K is compact and N c K is negligible, then there exists a compact set K' c K - N with measure arbitrarily close to the measure of K (1 3.7.9). A mapping f defined almost everywhere in X , with values in a topological space Y, is said to be measurable if every mapping of X into Y which is equivalent to f (13.6)is measurable. Clearly (13.9.8)it is enough for one such mapping to be measurable.

(13.9.8.1) In particular, it follows from (13.9.6)and (13.9.8)that iff and g (with values in R) are defined andjinite almost everywherein X and measurable, then f g and f g (which are defined almost everywhere) are measurable.

+

(13.9.9) Let u be a measurable mapping of X into a topologicalspace Y . I f M is any open or closed subset of Y , then u-'(M) is measurable. Consider a partition of X consisting of a sequence of compact sets Kn and a negligible set N, such that the restrictions u I K, are continuous. If M c Y is closed, then u-'(M) n K, is compact (3.11.4),hence u-'(M) is measurable. If M is open, then X - u-'(M) = u-'(Y - M) is measurable, and therefore so is u-'(M) (13.9.3).

(13.9.10) (Egoroffs theorem) Let Y be a metric space and let (f,)be a sequence of measurable mappings of X into Y such that, for almost all x E X , the sequence (f,(x)) converges to a limit f (x). Then ( 1 ) f is measurable; ( 2 ) f o r each compact subset K of X and each E > 0, there exists a compact subset K' of K such that p(K - K ' ) S E and such that the restrictions f,I K' are continuous and converge uniformly in K' to f I K'. Clearly it is sufficient to prove the second assertion ((1 3.9.4)and (7.2.1)). Let KObe a compact subset of K such that p(K - KO)5 +E and such that the restrictionsf, I KO are continuous for all n (13.9.5). If d denotes the distance fmction on Y, let B,,, be the set of all x E KO such that d(fp(x),fq(x))2 l / r for ar least one pair of integers ( p , q ) such that p 2 n and q 2 n. If A p,q,r is the set of points x E KOsuch that d(f,(x),f,(x)) 2 l/r, then we have Bn,r

=

U

phn. s t n

Ap.q,r*

138

Xlll INTEGRATION

Now the sets A p , q ,are r closed (3.11.4),hence B,,,r is integrable (13.8.7). Moreover, by hypothesis, for each integer r 2 l we have p( B,,, = 0, and the

() .)

sequence (B,,,,),,? is decreasing, so that by (13.8.7) we have lim p(BnPr)= 0. n-m

Hence there exists an integer n, such that P ( B , , , ~ 2 ) 142"'~.Let B the set B is integrable, and we have p(B) 5

m

C142'''

=&

=

u r

B,,,.,,;

by (13.8.7). The

I=

set C = KO- B is also integrable, and by definition the sequence ( f , I C ) converges uniformly to f l C . Hence we may take K' to be any compact subset of C such that p(C - K') 5 4 6 (13.7.9). (1 3.9.1 1) Let (f,)be any sequence of measurable functions on X with values in R. Then the functions inff, , sup f,, , lim sup f, and lim inff, are measurable. n

n

n-+ w

n-m

For supf, is the limit of the increasing sequence of functions g,

=

sup f;.,

l 0, there exists a measurable set A, an integrable function g 2 0, and an integer no such that, for all n 2 n o , we have If.(x)l

s

5 Ig(x)l for all x E A. Show that f i s integrable and that lim If-f.1 n-m

dp= 0.

Consider the converse. (c) Show by examples that the conditions of (a) are not sufficient and that the con-

I

I

ditions of (b) are not necessary for f to be integrable and f d p = lim f. dp. n-m

If A is measurable and B is any subset of X, show that

p*(B) = p*(B n A)

+

+ p*(B n (X -A)).

(If p*(B)< m, consider an integrable set B1 3 B such that p*(B) = p(BI).) Conversely, show that if A satisfies this condition (for all B c X), then A is measurable (cf. Section 13.8, Problem 3). Suppose that X is compact. A bounded real-valued function f on X is said to be continuous almost everywhere (with respect to p) on X if the set of points of discontinuity off is negligible. (a) Give an example of a functionfwhich is continuous almost everywhere and such that there exists no continuous function g which is equal to falmost everywhere. (b) Suppose that the support of p is equal to X. Show that a bounded real-valued function fdefined on X is equal almost everywhere to an almost everywhere continuous function on X if and only if there exists a subset A of X such that X - A is negligible and f l A is continuous. (To show that the condition is sufficient, observe that A is dense in X and that the lower semicontinuous extension of f JA to X is continuous at every point of A.) Deduce that f is measurable. Show that there exists a sequence (A) of continuous functions on X which converges at every point of X and whose limit is almost everywhere equal tof(cf. Section 13.11, Problem 3). (c) Show that a real-valued function f on R which is continuous on the right (i.e., such thatf(x+) = f ( x ) for all x E R) is continuous except at the points of an at most denumerable. set, and is therefore continuous almost everywhere with respect to Lebesgue measure. (Apply Section 3.9, Problem 3 to the set A, of points x E R at which the oscillation offis > l/n.) A partition w = (A.) of a set E is said to be finer than a partition m’ = (A;) of E if, for each index a , there exists B such that A, c A;. The partitions of E form an ordered set with respect to this relation. (a) Let X be a compact metric space. For each finite partition zu = (Ak)of X consisting of integrable sets and each bounded real-valued function f on X, put

(“Riemann sums” relative tofand the partition

and that if

w

w ) . Show that

is finer than m’, then s , # ( f ) s , ( f ) and S , ( f )

6 S,,(f).

9

MEASURABLE FUNCTIONS

145

(b) A sequence (m,)of finite partitions of X is said to be fundamental if ZU,,+~ is finer than w, for all n and if the maximum of the diameters of the sets of w, tends to 0 as n tends to m. Show that iffis a bounded integrable function on X, then there exists at least one fundamental sequence (m,) of finite partitions of X, consisting of

+

integrable sets, and such that the sequences ( s W n ( f and ) ) [S,"(f)) both tend to

s

f dp.

(c) Letfbe bounded and continuous almost everywhere on X (Problem 6). Show that, for all fundamental sequences (m,) of finite partitions of X consisting of integrable

f dp. (Observe that if A. is the J X at which the oscillation off (Section is Ll/n, then

sets, the sequences (s,Jf)) and ( S m n ( f ) ) both tend to

closed set of points x E 3.14) p(A,,) tends to 0 as n tends to m ; each A,, has an open neighborhood V. such that the points of V,,have a distance < l/n from A,, and such that p(V,) tends to 0. For a partition m k whose sets all have diameter < l/n, consider separately the sets of the partition which meet V. and those which do not.) Iffis bounded and lower semi-

+

s

continuous on X, show that s,,(f) tends to f d p for every fundamental sequence (m,) of finite partitions of X consisting of integrable sets. (d) A subset A of X is said to be quadrable (with respect to p) if its characteristic function va is continuous almost everywhere, or equivalently if the frontier of A is p-negligible. Show that every point xo in X has a fundamental system of quadrable open neighborhoods. (For each neighborhood V of xo , let f :X + [0, I ] be a continuous mapping such that f ( x o ) = 1 and f ( x ) = 0 for all x E X - V. For each a E 30, 1 [, consider the set of points x E X such that f ( x ) > E . ) Deduce that there exists a fundamental sequence of finite partitions of X consisting of sets which are either open or negligible. Give an example of a nonquadrable closed set (Section 13.8, Problem 4(a); cf. Section 13.21, Problem 2). (e) Let (w,) be a fundamental sequence of finite partitions of X consisting of sets which are either open or negligible. Iffis a bounded function on X, let g be the largest lower semicontinuous function sfi show that s&) = s,(g). Deduce that the sequences ( s m n ( f ) and ) (S,"(f)) tend to the same limit if and only iff is continuous almost everywhere. (f) Use (e) to give an example of a negligible functionfand a fundamental sequence (w") of finite partitions of X consisting of integrable sets, such that the sequences (s,Jf)) and ( S m n ( f ) )do not tend to the same limit. 8.

Let f be a measurable mapping of X into a complete metric space E, and let K be a compact subset of X. Show that f l K can be approximated uniformly by measurable step-functions if and only iff(K) is relatively compact in E.

9. Suppose that X is compact. Let (f.) be a sequence of p-measurable finite real-valued functions on X. Show that the following properties are equivalent: (1) There exists a subsequence of (A) which tends to 0 almost everywhere in X. (2) There exists a sequence (I,) of finite real numbers, such that lim sup 1t.l > 0 n-m

and such that the series with general term f,f.(x) converges almost everywhere in X. (3) There exists a sequence (t.) of finite real numbers such that C 1r.l = m n

+

and such that the series with general term t.f.(x) converges absolutely almost everywhere in X.

146

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INTEGRATION

(To show that (I) implies (2) and (3), use Egoroff's theorem. To show that (3) implies (1). show that (3) implies that there exists a n increasing sequence of integrable subsets A, of X and a subsequence ( f n k ) of the sequence (fn), such that p(Ak)tends to

p ( X ) and

s*, Ifn,.[

d p tends to 0, as k tends to

+

03.)

10. Let X, Y be two locally compact spaces and 7r : X + Y a proper continuous mapping. ) 13.4, Problem 8). Let p be a positive measure on X, and let v = ~ ( p(Section (a) A mapping g of Y into a topological space is v-measurable if and only if g 0 T

is p-measurable. (b) A mapping g : Y -+ R is v-integrable if and only if g we have s g dv = / ( g

0

7r)

0

7r

is p-integrable, and then

dp. (Use Problem 12 of Section 13.8.)

(c) Show that the support of v is the closure of

7r

(Supp(p)).

11. Let X be a locally compact space, u : X + X a continuous mapping, and p a positive

measure on X. Suppose that for each p-negligible set N, the inverse image u - ' ( N ) is p-negligible (cf. Problem 17). We shall write u-"(A)for (u")-'(A),for each integer n 1 and each subset A of X, and uo = lx. If A is any p-measurable subset of X, let A,,,

u u-"(A) be the set of points m

=

"=O

which "enter A at least once," and let A,,, = A n u-'(A,,,) be the set of points of A which "return at least once to A". Also put A,,,

inf =

An

n u-"(A,,,), the set of m

"=O

points of A which "return to A infinitely often;' The set A is said to be wandering (relative to u) if the sets rc-"(A)( n 2 0) are pairwise disjoint. The mapping u is said to be incompressible if, for each p-measurable subset A of X such that A c u.-'(A),the set u-'(A) - A is negligible; or equivalently if, for each p-measurable subset B of X such that u-'(B) c B, the set B - u-'(B) is negligible. If u is not incompressible it is said to be compressible. (a) Show that the following four properties are equivalent: ( a ) u is incompressible. (p) The wandering sets (relative to u) are negligible. ( y ) For each measurable set A, the set A - A,,, is negligible. is negligible. (6) For each measurable set A, the set A - A,,, (Observe that u-'(Ae0,) c A,,, and that A - Are, is wandering.) (b) Show that u is compressible if and only if there exists a measurable real-valued function f on X such that f ( u ( x ) )>f(x) for all x E X and such that the set of points x E X at whichf(u(x)) > f ( x ) has measure >O. (To show that the condition is sufficient, argue by contradiction to show that it implies that there exists a rational number r for which the set of points x E X such thatf(x) < r < f ( u ( x ) ) is not negligible.) Deduce that if u" is incompressible for some integer n > 1, then u is incompressible. (c) Suppose that X is compact and that the measure p is inoariant with respect t o u (cf. Section 13.4, Problem 8). Show that u is incompressible (" Poincarb's recurrence theorem "). (d) Show that if u is incompressible, then there exists a negligible set N in X such that, for every x $ N and every neighborhood V of x in X, there are infinitely many integers n such that u"(x) E V. (Consider the sets U - U,,, i n f , as U runs through a denumerable basis of open sets of X.)

9

MEASURABLE FUNCTIONS

147

Let (t,)oarsnbe a sequence of real numbers, and let m be an integer ( n . Let L, be the set of indices i with the following property: there exists an integer p such t12_0.(Observethat t h a t O _ I p s m a n d t f + t i + l + ~ ~ ~ + t i + 2p 0- .- Show l that

12. (a)

lELm

if i E L,, and if p is the smallest of the integers with the above property, then i + 1, ..., i + p - 1 belong to L,,,.) (b) Let X be a locally compact space and p a positive measure on X. Let u : X + X be a proper continuous mapping such that u ( p ) = p. Iff’is any p-integrable function on X, putf, = f a n d f x = f o up ( k 2 1). For each integer m , let A, be the measurable set of points x E X such that one of the sumsfo(x) +fi(x) ...+f,(x), wherep 5 m,

+

is

2 0. Show that

JAM

f(x) dp(x) 2 0. (For each integer n > 0 and each x E X, consider

the sequence ( f r ( ~ ) ), and ~ ~apply ~ ~ (a) ~ to+ this ~ sequence, denoting the corresponding set of indices by L,(x). For each k 5 n m, let Bkbe the set of x E X such that

+

k

E

Lm(x), and deduce from (a) that

n+m

C

k=O

B!,

fx(x) dp(x) 2 0. Now use the fact that

Bk= Kk(Am)for 0 5 k 5 n to deduce that

for all n.) If A is the union of the Am, conclude that

f ( x ) dp(x) 2 0 (“maximal /A

ergodic theorem ”). (c) Let a be a real number and C a n integrable set such that

x

1”-1 a < Iim sup fx(x) n-+m

for all x E C . Show that ap(C) 6

nk=o

1

If(x)l dp(x). (Apply (b) to the function f- avc .)

(d) Let a , b be two real numbers such that a < b. If E is the set of points x E X such that 1”-1

Iim infn-m

1”-1

1fk(x) < a < b < Iimn + msup -n kC= o /i(x),

k=O

show that E is p-negligible. (Deduce first from (c) that E is integrable, and then apply (b) to each of the functions (f- b)vEand (a -f)yE, using the fact that u(E) C E.) Hence show that,for almost aNx E X, the sequence

converges to a limit f*(x), that f* is integrable and that f*(u(x)) = f * ( x ) almost everywhere (G. D. Birkhoff’s ergodic theorem). (Take a , b to be all pairs of rational numbers such that a < b.)

s

s

f* dp = f dp. (Reduce to the case f 2_ 0, and consider first the case wherefis bounded; then pass to the general case by observing

(e) If X is compact, show that that

/ If’l

dp 5

If1 dp.1

148

Xlll INTEGRATION (f) Suppose that X i s compact and p(X) = 1. Consider a sequence (gn)of pintegrable functions such that lim g,(x) = f ( x ) almost everywhere. Suppose moreover that there nr'"

exists a p-integrable function G 2 0 such that /gnI 5 G almost everywhere, for each integer n. Show that under these conditions the sequence

converges almost everywhere to f * ( x ) . (Reduce to the case f = 0. For each e E 10, 1 [, there exists 6 > 0 such that, for each p-integrable subset B of X satisfying p(B) 5 6, we have of points x

G d p e2 (13.15.5). Next, there exists an integer m such that the set A, X satisfying

E

SUP Ig,(x)l

nbm

Ie2

has measure p(A,) 2 1 - 6. Let G , be the function which is equal to G on X - A,; then for n 2 m and all x E X, we may write

Deduce from (c) that the set of points x

1:

lim sup "'rn

has measure 13.

E

6'

on A, and to

X such that g,,(uk(x))

2e

5 26.)

Let X be a locally compact space, p a positive measure on X, and u : X +X a continuous mapping such that, for each p-negligible set N, the set u-'(N) is p-negligible. A measurable real-valued function f on X is said to be pinvariant with respect to u if f a u and J' are equal almost everywhere (relative to p). A p-measurable subset A of X is said to be p-inuarianr with respect to u if its characteristic function is pinvariant with respect to u. (a) For a measurable function f to be pinvariant with respect t o u, it is necessary and sufficient that, for each rational number r , the set of points X E X such that f ( x ) 2 r should bc p-invariant with respect to u. (To show that the condition is sufficient, consider the set of points x E X such thatf(u(x)) # f ( x ) . ) (b) From now on, suppose that u is proper and the measure p-invariant with respect to u (i.e., u ( p ) = p). The mapping u is said to be ergodic with respect to p (or p is ergodic with respect to u) if the only p-measurable functions which are pinvariant with respect to u are constant almost everywhere. Then, for every p-integrable function f on X, the function

(sf(.)

is constant almost everywhere. If X is compact, the value of the constant is

dp(x))/p(X). Conversely, if X is compact and i f f * is constant almost every-

where, for every p-integrable function A then u is ergodic.

9

MEASURABLE FUNCTIONS

149

(c) Take X to be the circle U : IzJ= 1 in C , and p to be the image of Lebesgue measure under the mapping t w e Z n i rof [0, 11 onto U. If 6 is irrational, show that the mapping u : z w e Z n i e zleaves p invariant and is ergodic with respect to p. (Use Bohl's theorem, Section 13.4, Problem 7.) (d) Suppose that X is compact and u ergodic with respect to p. If A and B are p-measurable subsets of X, show that

Conversely, if this formula is satisfied for all pairs of measurable subsets A, B in X, then u is ergodic with respect to p. (e) Suppose that X is compact and u ergodic with respect to p. If a measurable 0 is such that function

fz

exists almost everywhere, thenfis integrable. (Consider f a s the limit of an increasing sequence of bounded functions.) (f) Suppose that X is compact, that u is ergodic with respect to p, and that Supp(p) = X. Show that, for almost all x E X, the set of points u"(x) (n 2 0) is dense in X. 14.

Let X be a compact space, u : X + X a ccntinuous mapping, and p a positive measure on X, invariant with respect to u and of total mass 1. Let A be a p-measurable subset of X. For each x E A, let n(x) denote the smallest integer n 2 1 such that u"(x) E A, If we put A, = u - " ( A ) and B. = u-"(X - A), then the set of points x E A for which n(x) = rn is A,, n B, n Bz n ... n B,-, n A,,,, and hence the function x ~ n ( x is) measurable. (a) Put a0 = 1 and a,

= p(Bo

n B1 n . . . n B",-,) (rn 2 1). Show that

p(Ao n B1 n B, n ... n B,,-l n B, n A,,,,)

=

+

a,,, - 2a,,,+, a m f 2

for all m 2 0. (b) Show that the series whose general term is a,,, - a,+1 is convergent and that its terms form a decreasing sequence (interpret a,,, - urn+,as the measure of a set). Deduce that lim rn(tL, = 0.

n B..

m-m

(c) Let B,

=

Show that IAnLL(x)dp(x) = 1 - pCL(Bm)

("Kac's theorem": use (a) and (b)). Consider the case where u is ergodic with respect to p, and p(A) > 0. (d) Suppose that u is bijective and ergodic with respect to p, and that p(A) > 0. Let En, be the set of points x t A such that n(x) = rn. Show that the complement in

A of

u Emis negligible (cf. Problem Il(a)). Show that the sets up(E,) for rn 2 1 m

m=l

and 0 ' p < rn are mutually disjoint, and that the complement of their union in X is negligible (" Kakutani's skyscraper "1.

150

Xlll

INTEGRATION

15.

Let f be a lower (or upper) semicontinuous real-valued function on R". Let X be a locallycompact space and ul, . . . , universally measurablefinite real-valued functions. Show thatf(ul, . . . , u,) is universally measurable (cf. Problem 17).

16.

With the notation of Section 12.7, Problem 3, show that there exist n universally measurable mappings f w z k ( t )of C" into C (1 5 k 5 n) such that P,(X)=X"+ r,X"-'+

...+ t . - n ( x - z & ) ) . k=O

(By induction on the degree n , using Problem 15.) 17. Let f be the continuous function on I = [0, I ] which was defined in Section 4.2, Problem 2(d), such that the restriction o f f t o the Cantor set K is a bijection of K onto 1. Let g be the function defined by g (x ) = x -tf(x); then g is a homeomorphism of 1 onto 21= [0,2], such that g(K) is a compact set of measure 1 (with respect to Lebesgue measure), although K is negligible. If h(x) = 3-'(2x), then h is a homeomorphism of I onto I such that h-'(K) is not negligible with respect to Lebesgue measure. Moreover, if A is a nonmeasurable set (with respect to Lebesgue measure) contained in g(K), then the set B = h(&A)is negligible and therefore measurable. The function F~ h is not measurable, although h is continuous and pB is measurable (cf. (16.23)). 0

R. Show that there exists a real-valued function g on R which is universally measurable and such thatf(g(x)) = x for all x E ~ ( R(cf. ) Section 12.7, Problem 1). (b) Assume that there exists a partition of I = [0, 11 into a family (HJnSz of nonmeasurable sets (with respect to Lebesgue measure) such that Card (H.) = Card(R) for all n E 2. Let K be the Cantor set (Section 4.2, Problem 2) and F = (K t n),

18. (a) Let f be a finite continuous real-valued function on

u

n s z

so that F is a negligible closed set. Show that there exists a bijection f:R .+R such thatffF n In, n 1[) = H, for all n, and such that f is linear in each of the component intervals of IF. Then f is measurable with respect to Lebesgue measure, but f - I is not.

+

19.

Let X be a locally compact space and p a positive measure on X. Let D be a denumerable dense subset of R.Show that a mappingf: X R is p-measurable if and only if, for each r E D, the set of points x E X such thatf(x) 2 r is p-measurable.

20.

Suppose p bounded and p ( X ) = I . If f

20

s

is an integrable function, show that

(1 -tf2)''2is integrable and that, if A = fdp, then

(Use Problem 14(a) of Section 13.8.) 21.

-

Let X, Y, Z be three locally compact spaces. iff: X + Y and g :Y Z are universally measurable, show that g of is universally measurable. (To show thatg o f is p-measurable where p is any positive measure on X, reduce to the case where X is compact and consider the measure f(p) on Y (Section 13.4, Problem 8).) In particular, if B c Y is universally measurable, then f -'(B) is universally measurable.

9 MEASURABLE FUNCTIONS 22.

151

Let X be a locally compact space, Y a compact space, p a positive measure on X, and (fn)">, a sequence of p-measurable mappings from X to Y . Show that there exists a mapping ( p , X)HU~(X) of N x X into N with the following properties: (1) for each integer p 2 0 and each integer n 2 0, the set upl(n) is p-measurable; (2) for each x E X, the sequence (f~,cx,(x)),soconverges in Y . (One method is as follows. Taking a distance function on Y , define universally measurable subsets A, of Y satisfying conditions (i) and (ii) of Section 4.2, Problem 3(a) and such that (in the same notation) A,, n A , - = a . Then define subsets B, of X inductively as follows: if B, has been defined in such a way that, for each x E B,, infinitely many terms of the sequence ( f , ( x ) ) belong to A,, we define B,. as the set of all points x E B, such that infinitely many terms of the sequence (fn(x)) belong to A,, , and we define B,. to be the complement of B,, in B,. Now let x E X and let (&,JP>, be the sequence of 0's and 1's such that, if s, = then x E B,, for all p . Define up(x) by induction as follows: uo(x) is the smallest integer n such that fn(x) E A,, , and up(x)is the smallest integer n > O,-~(X) such that f.(x) E A,, . Use Problem 21 to verify that this definition of u,(x) satisfies the conditions of the problem.)

23. Let p be a bounded positive measure on X and let f > 0 be a pintegrable function.

Show that there exists a lower semicontinuous function g such that g 2 l/f (with the convention that l / O = to) and such that gf is integrable (with the convention that the product of 0 and cc is 0). (Reduce to the case wherefis bounded; consider the sets A. of points x E X such that f ( x ) 5 Ijn, and apply (13.7.9) to these sets to construct 9 . )

+ +

24. Let X, Y be locally compact spaces, p a positive bounded measure on X, and 7r a p-measurable mapping from X to Y . For each f c Xw,(u>, show that f o n is p-integ-

rable and that the mapping f~

s

( f 0 n)dp is a posiriue measure v on Y (called the

image of p under T , and denoted by ~ ( p ) )Generalize . the results of Section 13.8, Problem 12, and Section 13.9, Problems 10 to 14.

25.

Let X be a compact space and let (m,)be a sequence of finite partitions of X consisting of integral sets, such that m ,is finer then w. whenever m > n (Problem 7). An elementary martingale relative to the sequence (m,)is by definition a sequence (f.) of integrable functions 2 0 such that: (i) fn is constant on each set of the partition m,; (ii) If m > n, for each set A em. we have

(in other words, on each set A E wnthe value offn is equal to the average offm over A). (a) Let a , b be real numbers such that 0 5 a < b, and let Eabbe the integrable set of points x E X such that

lim inffn(x) < a < b < lim supfn(x). n-m

n-m

Show that E,, is negligible. (Suppose if possible that p(E.J > 0. For each integer p > 0, let F, be the union of the sets A E m. (n > p ) such that A n Eubi 0; if F = F,, then p(F) 2 p(Enb)> 0. Show that for each integer p and each A E m, (n 2 p )

n P

such that A n Eabf

a,we have b . p(A n F,)

a . p(A) for all q

2 n, by using the

152

XI11 INTEGRATION

+

+

definition of a martingale; now let q+ co, then p - + co, and so obtain a contradiction.) (b) Deduce from (a) that the sequence (f.) converges almost everywhere to a n integrable functionf: (Give a , b rational values, and use Fatou’s lemma.)

J^

(c) Give an example where f dp is not equal to the (constant) value of the integrals

J^hdp. (Take X

~7

[0, 11 and take p to be Lebesgue measure; take wnto consist of the

intervals [0, 2-”[, [2-”, 2-”+’[, ..., [B, 11.1 (d) Let F(x) = supf.(x). Show that there exists a constant C > 0 such that, for all n

a > 0, if B“ is the set of points x such that F(x) > a, we have p(B.) that B, is the union of the sets Ba, = { x E X : f.(x) > a}.)

5 Cia. (Remark

26.

With the hypotheses of Problem 25, let (T be another finite partition of X into pintegrable sets. For each x E X and each integer n, let B E u and A E w. be the sets of the partitions u and m, which contain x ; put f.(x) 0 if p(A) = 0, and f&) = p(B n A)/p(A) if p(A) > 0. Show that the sequence ( f . ( x ) ) converges almost everywhere to a limit f ( x ) 5 I , and that the sequence (f.)converges in mean tof: (For each set B t u, observe that the functions ysfA form an elementary martingale for the measure yB. p, and use Problem 25.)

27.

Let X be a compact space with measure p(X) = 1 , and let a = (Ai)1E16nbe a finite partition of X into integrable subsets. For each x t X, if Ai E a is the set containing x, the number i(a;x ) =

- log

p(Ad

is called the information at the point x corresponding to the partition a . The number

(with the convention that f log t = 0 when t = 0) is called the entropy of the partition a (relative to the measure p). It is the “average” of the information corresponding to a. I f / = (BJ, Eks,,, is another finite partition of X into integrable subsets, the number H(a//)

=

- C p(AI n Bk) log(p(A, n Bk)/p(BxN2 0 i, k

(in which the terms for which p(BJ = 0 are replaced by 0) is called the entropy ofa relative ro p. If (0is the partition consisting of the single set X, then H ( a / w )= H(a). (a) If ( C L , ) ~ ~is, 0, then h(u) = 0.

10. INTEGRALS O F VECTOR-VALUED F U N C T I O N S

Consider first a mapping f of X into ajnite-dimensional real vector space

E. Let (ei)l5is,n be a basis of E, and put f(x) =

cfi(x)ei for each x m

i= 1

E X.

Then f is said to be integrable (with respect to p ) or p-integrable if each of the real-valued functions fi is integrable, and we define (1 3.10.1) It is immediately checked that this definition is independent of the basis chosen. If llzll is a norm defining the topology of E, then we have the following criterion: (13.10.2) A mapping f : X -+E is integrable ifand only iff is measurable and XH Ilf(x)JIis then integrable.

j*llf(x)ll dp(x) < + co. Thefunction

For f is measurable if and only if the f;: (1 S i 5 n) are measurable, by (13.9.6); also there exist two constants a, b such that a

Ci Ifi(x) I IIIf(x>IIS b 1 I fi(x>I i *

by (5.9.1). The result now follows from (13.9.13) and (13.9.6). In particular, a complex function f on X is integrable if and only if Bf

10 INTEGRALS OF VECTOR-VALUED FUNCTIONS

155

and 9 f are integrable, and we have

from which it follows immediately that S f d p = J Z

The set Y,!.(X, p) of p-integrable functions on X with values in C is a vector space over C , and f~ I f d p is a linear form on this vector space. Furthermore, iff is any complex-valued integrable function, then I f [ is integrable and

(13.10.3)

+

For since 1.f I = ( ( W f ) 2 ( Y f ) 2 ) ” 2it, follows that If I is measurable (13.9.6),and we have p * ( l f l ) 5 p*(lWfl) p*(lXfl). Hence the first assertion, by (13.9.13).Also there exists a complex number c of absolute value 1 such that Mf)= Mf)L and hence IP(f’)l = W ( P ( c f ) ) = P ( W ( i f ) ) 5 Alfl), which establishes (13.1 0.3).

+

If E‘ is the dual of the finite-dimensional vector space E, then to say that thef, are integrable is equivalent to saying that, for each z’ E E’, the mapping x w ( f ( x ) , z ’ ) is integrable (because it is a linear combination of the A). We generalize this as follows: if I is any set, then a mapping X H ~ ,of X into the vector space K’ (where K = R or C) is scalarly p-integrable if for each c1 E I the function x ~ f , ( a )is integrable.

(1 3.1 0.4) Let E be a Frkchet space, E‘ its dual (1 2.15). Let X H f , be a scalarly integrable mapping of X into E’ such that, for each convergent sequence (a,,) in E , there exists a ,function g 2 0 defined on X such that p*(g) < + 03 and Ifx(a,)l 5 g(x)for all n, almost everywhere in X . Then there exists a continuous linear form z’ E E’ on E such that (Z , z’) = / f x ( z )dp(x)f o r all z E E.

sf,(.)

By (3.13.4)it is enough to show that if (a,,) is a sequence in E with limit z then dp(x)= lim sfx(a,) dp(x);and this follows from (13.8.4). n+m

The linear form z’ so defined is called the integral (or weak integral) of the function x-f, with respect to p , and is denoted by dp(x). Hence, for

sf,

Xlll INTEGRATION

1%

all x E E, we have (13.10.5)

If in particular E is a Hilbert space, then since there is a semilinear isometry z w j ( z ) of E onto its dual E’ (12.15), we can define the notion of a scalarly integrable mapping xl-*f(x) of X into E: this means that, for each z E E, the complex-valued function x.-,(f(x) Iz) is integrable. If the function XH(I f(x)ll is integrable, then by (13.10.4) there exists a unique element of E, denoted by jf(x) &(x) and called the integral (or weak integral) of f, with the property that (1 3.1 0.6)

for all z E E. PROBLEMS 1. If a mapping x H f , of X into R’is scalarly p-integrable, then the element

j

H fx(a) dp(x)

of R’ is called the integral of the mapping and is denoted by f,

J

f,

dp(x). If, for all x

E

X,

belongs to a weakly closed convex set A in R’ and if X is integrable and p(X) = 1,

then the integral jf,dp(x) also belongs to A. (Use Problem 13 of Section 12.15.) 2.

Let E be a separable real Frkhet space and E’ its dual (12.15). It follows from the Hahn-Banach theorem (Section 12.15, Problem 4) that the linear mapping cE which maps each z E E to the linear form Z’H (I,z’) on E’, is an injective mapping of E into RE’and is continuous with respect to the product topology on RE’.If K is any compact subset of E, the restriction CE I K is a homeomorphism of K onto the subspace ce(K) of Re’(12.3.6). (a) Suppose that K is a compact conuex subset of E. A mapping f of X into K is said to be scalarly p-integrable if, for each z’ E E’, the real-valued function XH (f(x), z’) is p-integrable. Using the remarks above and Problem 1, show that there then exists a unique vector z E K such that (z, z is denoted by

s s

2’) =

I

0 for all k . If nc > 1 and E > c(1 - c)/(n - l), show that there exist two distinct indices i, j such that p(Afn A,) cz - E . (Argue by contradiction, using the CauchySchwarz inequality.)

5.

Let (fn)be a sequence of functions belonging to -Yk(X, p) ( p = 1 or 2) which converges almost everywhere to a functionf. (a) Show that, if N,(f.) 5 a for all n, then f~ U k ( X , p) and N,(f) 5 a. (Use Fatou's lemma.) Give an example (with p = 1) where N,(f,) does not tend to N,(f). (b) Now suppose in addition that N,(f.)+N,(f) as n-, S m . Show that N,(f-f,) + 0. (Show that for each E > 0 there exists a compact subset K of X and an

If. 1

dp 5

E

for all n 2 no .)

6 . In the space -Yi(I,A), where I = 10, 11 and h is Lebesgue measure, consider the functions t' ( a real), which belong to this space provided that a > - 3. Show that a sequence of distinct exponents a, > -3 is such that the functions tanform a total sequence in Yi(1, A) if and only if the sequence (an) satisfies one of the

following three conditions: there exists a subsequence of the sequence (an) tending to a finite limit > - 4; 1 (2) lima,=- - - and n-m 2 lim a. = m a n d z a;' = + co. (3) (1)

n-r m

+

"

(Using Weierstrass' theorem (7.4.1), calculate for each integer m > 0 the minimum of Nz(t"'- f ( r ) ) asfruns through the set of all linear combinations of the first n functions tuk. For this purpose, use Problem 3(b) of Section 6.6, and the formula (Cauchy's determinant) (or - aj)(bi - 6,) det - = (at 6,) ' 0,

(c) Deduce that, for each measurable subset A of X and each E > 0, there exists an integer n > 0 with the property that for each integer m > 0, there exists an integer k such that m 5 k =< m t n - 1 and p(A n u - ~ ( A )>=) (p(A))’ - E (“Khintchine’s statistical recurrence theorem ”). (d) For each f E and each integer n , put

Show that the limit F A X ) = lim

m-rm

1

m-1

- hC ( ( U h.f,>(x) - (P . f ) ( x ) Y , =o

m

s

(where P .fdenotes a function in the class P . f ) exists almost everywhere and that lim F, dp = 0. (Use Birkhoff’s ergodic theorem.)

n-r m

(e) LetfeY’(X, p) be such thatf(x)z 0 almost everywhere. Show that, for almost all x E X , either f ( x ) = 0 or ( P . f ) ( x ) > 0 . (Consider the set N of points x E X at which ( P . f ) ( x )is either undefined or equal to zero.) 11. Let X be a metrizable compact space and let u : X + X be a homeomorphism. The set I of measures 2 0 on X of total mass 1 and invariant under u is then a nonempty vaguely compact subset of M(X) (Section 13.4, Problem 8). A point x E X is said to be quasi-regular (relative to u) if the sequence of measures

+

converges vaguely as n + 03 (necessarily to a measure px E I). Let Q denote the set of quasi-regular points of X. A point x E Q is said to be ergodic (relative to u ) if the measure pxis ergodic (Section 13.9, Problem 13). Let E be the set of ergodic points x E Q. A point x E Q is said to be dense if x belongs to the support of px. Let D be the set of dense points x E Q. The points belonging to R = E n D are said to be regular (relative to p). (a) Show that the complement of R (and hence also the complement of Q, E and D) is negligible with respect to any invariant measure v E I. (To show that Q has measure 1 for any measure v E I, apply Birkhoff’s ergodic theorem to the functions belonging to a dense sequence in V?(X). To show that D has measure 1, consider a denumerable basis (U,) for the topology of X, and for each pair (m.n ) such that 0, c U,, a continuous

168

Xlll INTEGRATION

mapping gmn: X + [0, 11 such that gmn(x)= 1 for all x E U, and gmn(x)= 0 for all x E X - U.; apply Problem 10(e) to each of the functions gmn. Finally, to show that E has measure 1, show (with the notation of Problem 10) that, for any measure v E I,

j((P . f ) ( Y ) - (P. f ) ( x V W Y ) = 0 for almost all x E X and allfE U ( X ) ,by applying Problem 10(d); now letfrun through a dense sequence of functions in U( X) .) (b) With respect to the vague topology on M(X) show that, for every measure v E I,

in the sense of (13.10), and that the external points of I are the measures which are ergodic with respect to u (cf. Section 13.10, Problem 9). (c) For every ergodic measure v E I, let QV (resp. R,) be the set of points x E Q (resp. x E R) such that px = v . The measure v is concentrated on R, (13.18). The sets QYand R, are called the quasi-ergodic set and the ergodic set corresponding to v . The Qv(resp. R,) form a partition of Q (resp. R), and we have u(Q,)= Q. and u(RJ = R,. Show that, for each closed set F such that u(F) = F, we have either F n R,= R, or F n Ri= 0. (d) For every nonempty closed set F such that u(F) = F, show that R n F # @ (consider the points which are regular with respect to u [ F). (e) Let A be the monoid generated by l x and u. Show that if 2 is any minimal closed orbit with respect to A (Section 12.10, Problem 6) we have u(2) = 2 and hence R n 2 # @. Deduce that, for each x E X, the closed orbit O(x) of x with respect to A intersects R, and that O ( x ) n R is a union of ergodic sets R, . (f) For each x E X,let p be a measure which is a cluster point of the sequence (pn,x) in M(X). Show that p(O(x))= 1. (g) Show that for a measure p E I to be ergodic it is necessary and sufficient that there should exist an ergodic set R, such that p(RJ = 1, and that then we have p = v . (Use (b) to prove that the condition is sufficient.) (h) If x E X is such that O(x) contains only one ergodic set, show that x is an ergodic point (use (f) and (g)). (i) Suppose that I consists of a single measure y o . Then Q = X, and R = D is the support of vo (use (h)) and is the only minimal closed orbit with respect to M. More1”-l

over, for each function f~ %‘(X),the sequence of numbers f(uk(x)) converges to n k=O f / d v , uniformZy in x

E

X. (Observe that, for each n , the mapping x ~ p . ,of X into

M(X) is continuous with respect to the vague topology, and use (7.5.6))

(s’

12. For every real number p such that 0 < p
1, put q = p / ( p - I). I f f , g are any two mappings of X into R,show that N l ( f g ) 5 N ,(f)N,( g) (“Holder’s inequality”). (Show that the set of points (rl, tz) E R2 such that t 1 2 0 and t 2 2 0 and t : l P t Y 2 1 is convex, and argue as in Section 13.8, Problem 14(c), making use of (13.5.6). Notice that the proof of (13.11.2.2) is a particular case.) (b) I f f 2 0,g 2 0 and p > 1, then N,( f ’+ g ) I N , ( f ) N,(g) (“ Minkowski’s inequality”). (Same method as (a): consider the set of points (II, t z ) such that t l L 0 and t 2 2 0 and ti’’ I:/’ 2 I . )

+

+

11 THE SPACES L’ AND

L2

169

(c) Extend the results of Section 13.11 to the case where p is any number such that 1 < p < +a. (d) Suppose that p > 1. Let f~ U i , g E -3’2 be two functions 2 0. Show that we have Nl(fg) = N,(f)N,(g) if and only if there exist two constants a > 0, p > 0 such that a(f(x))” = p(g(x))‘I almost everywhere. (e) Suppose that the measure p is bounded and that p(X) = 1. For each r > 0 and each measurable function 0 such that f‘ is integrable, show that the mapping p - N p ( f ) is an increasing function on the interval 10, r ] (use HGlder’s inequality).

fz

As p - 0 , its limit isexp(J* log

If1

d p ) , or 0 if

I’

log

If1

dp = - co. If the limit is

# 0, it follows that f ( x ) # 0 almost everywhere. If exp(l* log

the function (f) Let

If1

if1

dp)

=

/If1

dp,

is constant almost everywhere.

fz0 be a p-measurable function, and for each a > 0 let A, be the set of all

2 a. I f f e &;, then app(A,) 6/fp dp. Conversely, if the measure p is bounded and if there exist constants C > 0, E > 0 such that p(A.) C . a - p - c for all a > 0, then f~ 9: (cf. Section 13.9, Problem 3).

x E X such thatf(x)

+

f be a real valued function 2 0 defined on R*, = ] 0, to [, which is Lebesguemeasurable and such that I f l p is Lebesgue-integrable (1 < p < co). (a) The functionfis integrable on every compact subset of LO, co [,and in particular

13. Let

c +

F(x) =

f ( f ) dr

+ +

is defined for all x > 0 (use Holder’s inequality). As x tends to 0

or t o co, the quotient F ( X ) / X ( ~ tends -~)’~ to 0 (same method). (b) Show that the function F(x)/x is pth-power-integrable on 10,

+ co [ and that

(“Hardy’s inequality”). (Consider first the case where f~ X(R+,).For each compact interval [a, b] c R*, , majorize the integral using Holder’s inequality.)

Jab

(F(r)/fy df, by integrating by parts and

f be a complex p-measurable function such that ( I ) I f l p is p-integrable and ( 2 ) e r f is p-integrable for 0 < f < f o . Show that N p ( r - ’ ( e r f - 1) -f) -0 as r +O. (Let f = u iv where u and v are real; observe that if we put wI= erY- 1 - ru, we have w,/s 2 wI/f for 0 6 s 5 t. If 0 c p < 1, use the elementary inequality (a b)’ 5 u p bP when n , b are 2 0 . Finally, apply (13.8.1) and (1 3.8.4).)

14. Let p be a real number > O and let

+

+

15.

+

Let (An)”s1be a strictly increasing sequence of integers >0, and let h be a rational integer. For each integer N 2 I , put

Show that the series

170

Xlll

INTEGRATION

converges. Note that if mz 6 N < (rn

+ l)',

we have

~ to 0 for almost all x E [0, I ] (with and deduce that the sequence ( & ( x ) ) ~ >tends respect to Lebesgue measure). Hence show that, for almost all x E [0, I], the sequence (x.x - [x,x]) is equirepartitioned with respect to Lebesgue measure (Section 13.4, Problem 7).

16. Let U be a continuous endomorphism of the space LA(X, p) with norm 5 1 (in other words, such that NI(U . f ) 5 N,(f)) and such that the relationfz 0 implies U .f& 0 (13.6). l f f i s any function in the class we denote by U . f a n y function in the class V .J: (a) Let (f,) be a sequence of functions in 2 L p :which converges almost everywhere to a function f and is such that there exists a function h E Y i satisfying 1f n l 5 h for all n. Show that under these conditions the sequence ( U .f,) converges almost everywhere to U -f (Reduce to the case f = 0, and consider the sequence of functions

r:

9. = SUP

lf"+Pl.)

P

(b) For each finite sequence (rk)l

k 0, then for all (11

tl

r,,,

E

R we have

+ s,(tz, . . . , t n + d1s.(tl, ..., td.

(c) For each functionfE U:, show that

...

S"(U.f,

)

un+1.f)z U.S"(f,

. . ' ) U..f)

almost everywhere. (Use (I) and the fact that for any n functions f , , we have sup

IqkQn

u'fksu'

sup

...,f , E 14:

fk

I&k 0 and each integer n 2 0, let E , , , t ( f )denote the set of points x E X such that

t ...+ (u"-' .f)(x)>nt,

f(x)+(U.f)(x)

and let E,(f) denote the union of the En,,(f). Show that each of the sets E,,,(f) is integrable and that

(Observe that E.,,(f) is contained in the set of points x E X satisfying one or other of then relations ( U k . f ) ( x ) > t, where 0 5 k n - 1; then continue as in Problem 16, by remarking that


= n o , we have I f ( x ) -f,(x)I 5 l / m for all X E CH,. The union H of the sets H, is negligible, and f , ( x ) tends to f ( x ) uniformly in GH. (13.12.4) The normed space L,"(X, p) (resp. L,"(X, p)) is complete (Lea, a Banach space).

Let (,f,) be a sequence such that (f,)is a Cauchy sequence in L; (resp. L,"). For each integer n >= 1, there exists an integer k , such that N,(L

-f)5 1/n

12

THE SPACE L"

175

for r 2 k, and s 2 k , . For each pair (r, s) such that r 2 k , and s 2 k,, let A,,, be the negligible set of points x E X such that If,(x) -f,(x)I > I/n, and let A be the union of the sets A,,, (n 2 1, r 2 k , , s 2 k,), so that A is a negligible set. It is clear that in X - A the sequence ( f , ( x ) ) converges uniformly to a limitf(x). The functionf, which is defined almost everywhere, is measurable (13.9.10) and bounded in X - A. Hence f~ Lg (resp. f~ L,"). Clearly N , ( f - f , ) 5 I/n for all r 2 k , , and therefore 7 is the limit of the sequence (f,).

(13.12.5) r f f E S[(X, p ) (where p = 1 or 2) and g E =.YF(X,p), then PI, and N,(fg) 5 N p ( f ) N m ( 9 ) .

fgE

ym

For fg is measurable (13.9.8.1), and If(x)g(x)l everywhere; hence the result.

If(x)IN,(g)

almost

Remark

(13.12.6) Evidently we have %;(X) c S g ( X , p); but in general the canonical image of %?;(X) in L,"(X, p ) is not dense in the latter space, and L,"(X, p ) is not separable in general (Problem 1). PROBLEMS

1. If h is Lebesgue measure, show that the space Lg(R, h) is not separable. (If (A,) is an infinite sequence of nonnegligible measurable sets, no two of which intersect, consider the functions which are constant on each A, and take only the values 1.) 2.

(a) For every p-integrable subset A of X ahd every 6 > 0, let V(A, 6) denote the set of p-measurable real-valued functions f such that the set M of points x E A for which If(x)l > 6 has measure p(M) 2 6. Let Y ( X , p) denote the vector space of (finite) real-valued p-measurable functions on X. Show that the sets V(A, 6) form a fundamental system of neighborhoods of 0 for a topology on Y(X, p) compatible with its vector space structure (Section 12.14, Problem 1). This topology is called the topology of convergence in measure (with respect to p). A sequence (fn)which tends to a limitf in this topology is said to converge in measure to f. (b) Show that the intersection of all the neighborhoods V(A, 6) is the subspace .N of negligible functions, and that the quotient space S(X, p) = 9 ( X , p ) / N is metrizable. (c) If (fn) is any sequence in Y ( X , p) (12.9) such that (h)is a Cauchy sequence in S(X, p), show that there exists a subsequence (jJ of (f,)such that the sequence (fnk(x)) converges for almost all x E X. Deduce that the metrizable vector space S(X, p) is complete. (d) Every sequence (1;)of measurable real-valued functions which converges almost everywhere to a functionf, converges in measure tof. (e) For every finitep 2 1, show that the space U g ( X , p) is dense in Y ( X , p),and that the topology induced on 3g(X,p) by the topology of convergence in measure is coarser than that defined by the seminorm N, .

Xlll INTEGRATION

176

(f) Suppose that X is compact and that the measure p is diffuse (13.18). Show that, for every neighborhood V of 0 in .V(X, p) and every function f E Y(X, p), there exists an integer n such that all the functions af(where a is any real number) belong to V ... V ( n summands). Deduce that every continuous linear form on .V(X, p) is identically zero, and hence that every vector subspace of finite codimension in .V(X,p) is dense in Y ( X , p).

+ +

3. Suppose that X is compact and the measure p diffuse (13.18). (a) Let be a Hilbert basis of Lk(X, p). Show that, for each real number 6 > 0, there exists a compact subset Y of X such that p(X - Y) 5 6 and the sequence is total in L;(Y, py).(By using Problems 2(e) and 2(f), show that there exists a sequence of linear combinations of the f. ( n >= 2) which converges in measure to fl ; then use Problem 2(c) and Egoroffs theorem.) (b) Show that there exists a bounded measurable function h such that the sequence (&),,2 is total in L&(X,p). (Choose h > 0 such that f J h $ U&(X,p),and then show that no nonnegligible function can be orthogonal to hf. for all n >= 2.) 4.

Let p be a finite real number if, for each

E

>= 1. A subset H of

.PJLpg(X, p) is said to be equi-integrable

sA1fIp

> 0, there exists a compact subset K of X such that

for all f~ H, and a real number 6 > 0 such that

6.

integrable sets A of measure p(A)

sx-,lf1P

d p =( E

dp 5 E for all f~ H and all

(a) On an equi-integrable set H, show that the topology of convergence in measure is the same as that defined by the seminorm N, . Is the conclusion true if H is merely bounded in Y p ? (b) A sequence (f.) in .P&(X,p) is convergent if and only if it is equi-integrable and convergent in measure. (c) Suppose that the measure p is bounded and that p(X) = 1. Let (f,)be a sequence of functions belonging to Y & ,and suppose that lim lim Nl(I -If.[)

= 0.

"+a

s s

Show that lim

n-r m

n-r m

I Yfnldp = 0. (Use (b).)

(d) With p as in (c), let (f.) be a sequence of functions belonging to 9 ';such that

If.1

d p = lim n-r m

s

lfnll'z

dp = 1.

Show that lirn N1(l -f,)= 0. (Reduce to the situation of (c) above by using the n-r m

Cauchy-Schwarz inequality; write 11 - L l S 11 - ILII

and

+ 1If.I -f,l

11 - I L I ~= 11 - I f n I 1 / z I 5.

11

+ I.Lnt1'21.)

Let F be a real-valued function with period 1 on R which is integrable on the unit interval I = [0, 11 (with respect to Lebesgue measure). (a) Iff is any bounded measurable function on [0,1], then

12 THE SPACE L"

177

If also F is bounded on [O, 11, then this equality is valid for every integrable functionfon [0,1]. (Start with the case where F is bounded a n d f i s continuous on [0, 11, then (forfE 2")approximatefby continuous functions. When F is unbounded, reduce to the case F 2 0 and approximate F by an increasing sequence of bounded functions.) (b) Deduce from (a) that, for every function f which is integrable on an interval [a, b ] in R , we have lim j a b f ( t )sin nt dt = lim

I-m

j a b f ( t )cos nt

dt = 0.

"-4"

(c) Let q~ be the canonical mapping of R onto T = R/Z, and let p be the measure on T which is the image under q~ I I of Lebesgue measure on I; then p is invariant under translations in the compact group T (cf. (1 4.4)). Let k be an integer > 1, and let u : T + T be the continuous mapping such that u(v(t))= u ( k t ) for t E 1. The mapping u is not injective, but p is invariant with respect to u. Deduce from (a) that u is ergodic with respect to p (use Problem 13(b) of Section 13.9). Deduce that lim sup F(k"t) n-tm

is almost everywhere equal to the constant ess sup F(t), and that the same is true of

sup F(nr).

n21

6. Let

(cI

f~ 9 ; ( p ) . Show that

Isf I =!if] dp

161

d p if and only if there exists c E C with

= 1 such thatf(x) = clf(x)[almost everywhere.

S

7. Suppose that p is bounded and that p ( X ) = 1. LetfE 9;.If we have (1 for every complex number [, show that

s

+ [fl

dp >= 1

f d p = O . (For a fixed [ E C , consider

8. (a) If l s p < r < + m , then L:n L, "cLg n L: and L ; n L , " c L L n L , " . On the space L;: n L," , the function N,,, = N, + N, is a norm with respect to which the space is complete. Give an example in which the norm induced on L&n L," by n Lg we always have N,,,(f) 5 2Np,,(f).) NP., is not equivalent to N,,, . (In The space L;: n L," is a Banach algebra relative to the product 38 defined in Section 13.6; it posesses an identity element only if p is bounded (in which case it is identical with L,"). (b) The set I(X, p) (or I(p)) of idempotents in the algebra L> n L2 is independent of r , and consists of the classes @ A , where A runs through the set of integrable subsets of X. If A, B are integrable subsets of X, the relation @A = qB signifies that the set D(A, B) = A u B -A n B (Section 13.8, Problem 15) is negligible; the relation p(D(A, B)) = 0 is an equivalence relation on the set of integrable subsets of X, and the quotient set may be identified with I(X, p). We shall write A in place of qA. All the norms N,," induce the same topology on I(X, p), and this topology is therefore defined by the distance d(& 8) = NI(vA - yB)= p(D(A, B)), with respect to which I(X, p) is a complete space. The mappings (A, 8)Hsup(& 8) = (A u B)" and (A, 8)Hinf(A, 8 )= (A n B)" are continuous for this topology. (c) Let Y be another locally compact space, v a positive measure on Y. If there exists an isometry U of I(X, p) onto I(Y, v) such that U ( 4 ) = the measures p and v are said to be isometric. For each number p E [l, m [, there then exists a unique

+

4,

178

Xlll INTEGRATION linear isometry U,,of LE(X, p) onto L((Y, v) which extends U. (Show that U has a unique extension to a linear bijection of the space E(X, p) of classes of p-integrable step functions on X, onto the space E(Y, v) of_classes of v-integrable step functions on Y . Begin by proving that if inf(A, 8 ) = $, then also inf(U(A), U(8))= C$ and U(sup(A, 8))= sup(U(A), U ( B ) ) ; then deduce that U ( A ) 5 U(B)whenever A 5 8 ; finally, show that for arbitrary A,B we have U(inf(A, B)) = inf(U(A), U(@) and U(sup(A,B)) = sup(U(A), U(8)).Observe then that N,(U(f)) = N,(f) forfE E(X, p).) Show that the image of L[(X, p) n LF(X, p) under U p is L:(Y, v) n L,"(Y, v), and that the restriction of U p to Lg(X, p) n L,"(X, p) is an isomorphism of Banach algebras.

+

(d) Conversely, for a p such that 1 s p < m , let V be a linear isometry of Lg(X, p) onto Lg(Y, v) such that the restriction of V to Lg(X, p ) n L,"(X, p) is an algebra isomorphism onto L;(Y, v) n L,"(Y, v). Then the restriction of V to 1(X, p) is an isometry onto 1(Y, v) which maps I$ to

4.

9. Let X be a compact space such that p(X) = I . Let 2 be a set of integrable subsets of X, such that the relations A E 5 and B E 5 imply that X - A E 2 and A n B E 5 . Suppose also that the set {A:A E Z} is dense in the metric space I(X, p) (Problem 8). (a) Let (Cj)lsjCn be a finite partition of X into p-integrable subsets. Show that for each E > 0 there exists a partition (Aj)1Qj4,,consisting of sets in 5 such that p(D(Cj, A,)) : E for 1 5j 5 n. (Observe that if p(D(Cj, Bj)) 5 6 for 1 5 j 5 n - I , then p ( B j n B,) 5 26 for I 51j < k 5 n - 1 ; if N is the union of the Bj n B,, consider

the sets Aj =: B,

-N

(1

5 j 5 n - 1) and A,, = X

u Aj).

"- 1

-

j=1

of X into integrable (b) Deduce from (a) that for each finite partition y = (C,), subsets, and each E 1 0 , there exists a finite partition a = (Aj)lajsn consisting of sets in Z and such that (in the notation of Section 13.9, Problem 27)

H(a/y) -i-H(y/a) 5 E . (Reduce to the case where none of the C j is negligible, and use the fact that the function t ~ +t log t is zero and continuous at f = I . ) 10. Let X be a compact space such that p ( X ) = 1, and let u : X --z X be a p-measurable mapping such that u ( p ) = p . (a) Let a be a finite partition of X into integrable subsets, and let Z be the set of all finite unions of subsets of X belonging to one or other of the partitions

n-

1

V [)-'(a).

j = O

Suppose that the set of classes {A: A E 2) is dense in 1(X, p) (Problem 8). Show that we then have h(u) : h(u, a). (It is enough to show that h(u, /3) 5 h(u, a ) for every finite partition /3 of X into integrable subsets; observe that we have

and use Problem 28(d) of Section 13.9, and Problem 9 above.) is (b) Under the hypotheses of (a), show that if in addition u is bijective and p-measurable, then h(u)= 0. (Observe that the set of classes { ( u - * ( A ) ) - :A E Z} is again densc in I(X, p), and use Problem 28(c) of Section 13.9.) (c) S~tpposethat u is bijective and N - I is p-measurable. Let 5' be the set of finite

" V uj(a),and suppose J = -n

unions of subsets of X belonging to one or other of the partitions

13 MEASURES WITH BASE p

179

that the set of classes A, as A runs through 5', is dense in 1(X, p). Show that h(u) = h(u, a ) (Kolmogoroff-Sinai theorem). (Same method as in (a).) (d) Take X to be the unitcircle U : 1 z 1 = 1, let p be the image of Lebesgue measure under the mapping t H e Z Z iof t [0, I ] onto U, and let N be the mapping z H e Z n L a z . Show that h(u) = 0. (Distinguish two cases according as 8 is rational or irrational. In the latter case use (b), by taking a to be a partition into two half-open semicircles.) 11. Let X, Y be two compact spaces and let p (resp.v) be a positive measure on X (resp. Y ) such that p(X) = v(Y) = 1. A p-measurable (resp. v-measurable) mapping u : X HX (resp. u : Y t,Y) such that u ( p )= p(resp. u(v) = v) defines an endomorphism U : ~ H(f0 u) - of L&(X,p) (resp. an endomorphism V :B H (g 0 v)- of Li(Y, v)) (Section 13.1 1, Problem 10). The mappings u, u are said to be conjugate if there exists an isometry T of I(X, p) onto I(Y, v) (Problem 8) such that V o T = T o U. Show that if this is the case then h(u) = h ( v ) .

12. Suppose that the measure p is bounded, and let p E [ l , +a[. Let (Un) be a sequence of continuous linear mappings of L&(X)into the space S(X, p) (Problem 2). For any function of f e -Yk we denote by U, .f any function belonging to the class U, .f and pu$

For each a > 0 let E.,N(f) denote the set of all x E X such that (U,*. f ) ( x ) > a, and let E,(f) denote the set of all x f X such that (U* . f ) ( x )> a. (a) Show that the set of f~ Lg such that p(E.,N(f)) 5 E is closed, for each E > 0. (b) Assume that for aN f E 2 k the function U* . f i s finite almost everywhere. Show that under these conditions the number

+

tends t o 0 as a+ co (Banach's principle). (Use Bake's theorem in the complete space Lg .) (c) Under the hypotheses of (b), show that the set H of classes f such that the sequence ((U, . f ) ( x ) )converges everywhere in X is closed in Lk. (Write

RVXX) = lim .-a

remark that R(f)(x)

=

(

SUP

m,n3,

I (u,,,. f ) ( x ) - (u,. f ) ( x ) lj,

R(f- g ) ( x ) almost everywhere for @ E H, and use (b).)

13. MEASURES WITH BASE p

n.

Then the following con(13.13.1) Let g be a mapping of X into C or into ditions are equivalent: (a) For each x E X,there exists a neighborhood V of x in X such that gqv is integrable.

180

Xlll INTEGRATION

(b) The function g is measurable, and f o r each compact set K c X we have J*lc?l(P.

dP < + a *

(c) For each function h E X,(X), the function gh is integrable.

+

Since we may write g = g1 ig, , where g 1 and 9 , are 8-valued, and then g1 = g : - g ; , g 2 = g t - g ; , we reduce straightaway ((13.9.6) and (13.10)) to the case where g is a mapping 2 0 of X into To show that (a) implies (b), cover K by a finite number of open sets V j such that g q is integrable for VJ each j . Then sup(gcp,,) is integrable (13.7.4) and hence so I S

w.

j

g(PK = (PK * sup(g'PVj) i

(13.9.14). Hence (b) follows from (13.9.13). To show that (b) implies (c), observe first that g is almost everywhere finite (X being a denumerable union of compact sets), hence gh is measurable (13.9.8.1). Moreover, if L = Supp(h), we have 1gh( 5 lgpLl * Ilhll, hence it follows from (b) and (13.9.1 3) that gh is integrable. Finally, to show that (c) implies (a), consider a compact neighborhood V of x e X and a continuous mapping h : X -+ [O, 1) which is equal to 1 on V and has compact support ((3.18.2) and (4.5.2)). By hypothesis gh is integrable, hence so is g p , = (gh)pv (13.9.14).

When the equivalent conditions of (13.13.1) are satisfied, the function g is said to be locally integrable (with respect to p) or locally p-integrable. Clearly every integrable function is locally integrable. Every measurable function whose restriction to every compact subset of X is bounded almost everywhere (in particular, every function belonging to 9 2 or 2'2) is locally integrable. Every function belonging to 9;or 9 6 is locally integrable, by (13.11.7). Every function equivalent to a locally integrable function is locally integrable. We have remarked in the course of the proof of (13.13.1) that every locally integrable function is finite almost everywhere. The function on R which is equal to l/lxl when x # 0 and is 0 when x = 0 is lower semicontinuous but not locally integrable with respect to Lebesgue measure. For a complex-valued function g to be locally integrable,,it is necessary and sufficient that Seg and 9 g should be locally integrable. For a real-valued function g to be locally integrable, it is necessary and sufficient that g+ and g- should be locally integrable.

i

Let g be a locally p-integrable function. Sincef H fg dp is defined on the whole of .X,(X), it is a linear form on this complex vector space. Moreover

13 MEASURES WITH BASE p

181

this linear form is a (complex) measure on X, because if K is any compact subset of X,we have

for allfE X,(X; K), by virtue of (13.10.3). The measure so defined is called the measure with density g relative to p, and is denoted by g * p ; when g is continuous, this agrees with the definition (13.1.5). Measures of the form g p are also called measures with base p. It follows immediately from this definition that if g takes values in R (resp. 20 almost everywhere), then the measure g * p is real (resp. positive). Furthermore, g p does not change if we replace g by an equivalent function (with respect to p), and therefore we may restrict ourselves to the case where g is everywhere finite and universally measurable (1 3.9.12). If g 1 and g2 are two locally integrable functions, then g 1 + g2 and ag, are locally integrable (a being any complex scalar), and we have

-

-

For every complex-valued locally integrable function g, we have

(13.13.4) The set 9;oc, R(X,p) (resp. 9’;oc, c(X, p)) of real-valued (resp. complex-valued) locally p-integrable functions is a real (resp. complex) vector space, often denoted by 910c(X,p ) or 2’loe(p) or U,,,(X) or Y p l O c . For every compact subset K of x, the mapping p K :g w )g(pKI dp is a semi-norm on this space. We shall always suppose YlOc to be endowed with the topology defined by these seminorms. If (K,) is an increasing sequence of K, = X and K, c K,,, (3.18.3), then it is compact subsets of X, such that

i

u n

immediately seen that the topology of 9,0c is defined by the seminorms pK,. In particular, if, X is compact, then 9’;oc, R(X,p) (resp. $Plot, c(X, p)) is identical with 9 k ( X , p) (resp. 9’A(X, p)). The set of locally integrable realvalued functions g such that p&) = 0 for all compact subsets K of X is the space Jfr of p-negligible functions. We define Lioc,

p) = 9’lloc,

p)/ Jfr

and L;oc,c(X, p ) analogously. The seminorms pK(g) depend only on the class of g: if we put pK(g”)= p K ( g ) ,then we obtzin seminorms defining the topologies of these spaces, which are therefore metrizable and locally convex.

Xlll INTEGRATION

182

(13.13.5) Zfg is locallyp-integrable, then so is 191, and wehave 1g pi

=

-

19) p.

The first assertion follows immediately from (1 3.1 3.1) and (1 3.7.4).To prove the second, note first that i f f 2 0 is a function belonging to X,(X) and if u E X,(X) is such that ( u (SJ then we have

(13.10.3), and therefore (13.3.2.1)1g . pl S 19) . p. To prove the reverse inequality, let L denote the (compact) support 0f.f. Let A be the set of all x E L such that g ( x ) # 0; the set A is integrable ((13.9.9)and (13.9.2))and therefore (1 3.9.1) there exists an increasing sequence of compact subsets K, of A such that A - K, is p-negligible. Hence, by (13.8.4),for any E > 0

u n

there exists an integer n such that /A-Kn f1g1 dp S E . The same reasoning, using the measurability of g and (13.8.4),shows that there exists a compact subset KA of K, such that g I KA is continuous and JA-K,, f l g l d p 5 2 ~Now . consider on KA the continuous function x t t Ig(x)l/g(x);by applying (3.18.2) and the Tietze-Urysohn theorem (4.5.1)to the real and imaginary parts of this function, we see that there exists a function w E X,(X) such that w(x) = ( g ( x ) j / g ( x for ) all x E KA. The function u = w . inf(1, l / J w J )(with the convention 1/0 = + 03) is then continuous on X, hence belongs to X,(X), and is such that v(x) =: Ig(x)l/g(x) in KA and [u(x)l 5 I throughout X. Hence If4 S f a n d

but

and

so that finally

Since E > 0 was arbitrary, this completes the proof.

13

MEASURES WITH BASE p

183

PROBLEMS

1. Show that the locally convex metrizable space L;oc,R(X, p) (resp. Lloc,&X, p)) is complete (in other words, is a Frkhet space). 2.

Let A“ be a vector subspace of .2’:oc,Rcontaining the subspace Jy of pnegligible Jy is endowed with the structure functions. Suppose that the quotient space H = P/ of a real Hilbert space, in which the scalar product is denoted by (318) and the norm 131.Suppose moreover that, for every compact subset K ofX, there exists a constant uK 2 0 such that

for all II E 2 (cf. Section 15.11, Problem 26). (We shall also use the notation (u I u) andlulfor (ci I d) andltilwhen N and u belong to 2.)Let 2; denotethe set of bounded measurable functions on X with compact support. (a) Given any function f~ 9 2 , show that there exists a function U’ E 2 such that

s

(Uf / u) = u f d p for all u E 2 ;and that the class of U’ in H is uniquely determined by the class off. The function Uf is called the potential of f. Show that the set of classes of potentials Uf is dense in the Hilbert space H (use (6.3.2)). If J g are two elements of

s

9 2 ,then (U’ I Us) = guydp =

i

f U g dp.

(b) As f runs through the functions 2 0 belonging to YE, the set of potentials U’ is a convex cone in 2. We denote by B the closure of this cone (with respect to the topology defined by the seminorm 161 on X ) . The elements of 9’ are called pure potentials. Let P denote the image of 9’ in H. For each element ri E H let d be the projection of ti on P (Section 12.15, Problem 3(a)). Show that we have lal’ = (61li), that u(x) 2 u ( x ) almost everywhere, and that d is the only element of P satisfying these conditions. Also we have 161 5 151. Deduce that, for each L? E H , the projection of 0 on the closed convex set of points 4 E H such that u(x) 2 u ( x ) almost everywhere belongs to P. (c) Deduce from (b) that an element u E P belongs to B if and only if ( u l w ) 2 0 for all w E 2 such that w ( x ) 2 0 almost everywhere (consider the difference between d and its projection on 9). Equivalently, Id i-3 I 2 151 for all w E X such that w ( x ) 2 0 almost everywhere. (d) Suppose that for all u E 2 we have IuI E 2 and I(lu1)- I5 Iril. Show that every pure potential u is 2 0 almost everywhere (use (c)). If u, u are two pure potentials, show that inf(u, u ) is a pure potential. (Among the elements of A“ which majorize inf(u, u), consider an element w such that 1 6 1 is a minimum; then w is a pure potential, by (b), and we have ( u w 1 u - w) 5 ( u w 1 Iu - w l ) . By calculating I(inf(u, w))” 12, deduce that I(inf(u, w ) ) - I $ 131; likewise that I(inf(u, w))- 15 161;and hence that w = inf(u, w) = inf(u, w ) almost everywhere.) (e) With the same hypotheses as in (d), show that i f f € 9 2 is 20 almost everywhere, and if u E B is such that Uf(x) 5 u ( x ) almost everywhere in the set of points x E X such rhat f ( x ) > 0 , then Uf(x) 5 u ( x ) almost everywhere in X (“principle of domination”). (Observe that u = inf(Uf, u ) is a pure potential, that (Uf 1 Uf - u ) = 0 and (u I U’ - u) 2 0, and deduce that u = Uf almost everywhere.) (f) Suppose that the hypotheses in (d) are satisfied and also that for all u E 2‘we have inf(u, 1) E 2 and I(inf(u, 1))- 15 I t i l . Show then that if u is a pure potential,

+

+

184

Xlll INTEGRATION

then so is inf ( u , 1) (same method as in (d)). Deduce that, if u and u are pure potentials, then inf(u, ti 1) is a pure potential, by remarking that u inf(u, I ) is a pure potential 2 2 is 20 almost everywhere and if u E B is and using (d). Finally show that, if such that U f ( x )5 u(x) 1 almost everywhere in the set of points x E X such that f ( x ) > 0, then U f ( x )5 u(x) 1 almost everywhere in X ("complete maximum principle": same method as in (e)).

+

+

+

+

14. I N T E G R A T I O N WITH RESPECT TO A POSITIVE MEASURE WITH BASE p

(13.14.1) Let g be a locally p-integrable function which is 2 0 on X, and let v = g p. Then iff 2 0 is any 8-valued function on X, we have (13.14.1 . I )

s'f

dv

= s"fg

4

where, on the right-hand side, the value of f g is by dejinition taken to be 0 at every point x E X at which one of f ( x ) , g ( x ) vanishes (even if the other factor is +a (13.11)).

The proof consists of several steps. (13.14.1.2) Suppose first of all that f E 3. Then (12.7.8) there exists an increasing sequence ( f , ) of functions belonging to X,(X) such that f = sup f , . n

In view of the convention about products, this implies that the sequence (f,g) is increasing and that f g is equivalent to sup(f,g), because g is almost n

everywhere finite (with respect to p ) . Moreover, the functions f . g are p integrable (13.13.1), and 1f.g dp = . j f n dv. Hence it follows from (13.5.7) that

(1 3.1 4.1.3)

Every p-negligible set N is also v-negligible.

Suppose first of all that N is relatively compact. Then ((3.18.2) and (1 3.7.9)) there exists a decreasing sequence of relatively compact open sets U, containing N, such that inf p(U,) = 0. Since g q u , is p-integrable n

by virtue of (13.1 3.1), we have Sgq", dp = v ( ~ , ) by (13.14.1.2)

and

14 INTEGRATION WITH RESPECT TO A POSITIVE MEASURE

185

(13.7.7). But, by (13.8.4), if N ' x N is the intersection of the U,, then inf lg(pun dp = /g(pN.dp = 0; and since v(N) S v(U,) for all n, we have n

v(N) = 0. Now let N be any p-negligible set, and (K,) a denumerahle covering of X by compact sets. Then the sets N n K, are v-negligible, by what has just been proved, and hence so is their union N. (13.14.1.4) Suppose now that K = Supp(f) is compact and that f l K is continuous, with values in R (and therefore bounded (3.17.10)). Then there exists a decreasing sequence (U,) of relatively compact neighborhoods of K such that K = U, (3.18.2); also, by virtue of the Tietze-Urysohn theorem

n n

(4.5.1), there exists for each n a function f, E .X,(X), with support contained in U,, which extendsf and is such that [If,[] = Ilfll. Hence we have If,dv = / f , g dp for all n. Bearing in mind (1 3.13 4 , it follows from (1 3.8.4)

that f is v-integrable and fg is p-integrable, and that Ifdv = /fg dp. (13.14.1.5) The set A of points x E X such that g(x) = 6 is v-negligible.

Since g is p-measurable (13.13.1), A is p-measurable (13.9.9), and hence is the union of a sequence (K,) of compact sets and a p-negligible set N. By virtue of (13.14.1.4) applied t o f = qK,, we have v(K,) = sg(pKn dp = 0, and v(N) = 0 by (1 3.14.1.3) ; hence v(A) = 0. (13.14.1.6) Consider now the case where Supp(f) = K is compact andfl K is lower semicontinuous on K. Then it follows from (12.7.8) that there exists an increasing sequence of finite real-valued functions u, which are continuous and >, 0 on K and such that f l K = sup u, . Let f, be the function which is n

equal to u, on K and zero on X - K, so thatf= supf,. By virtue of (13.5.7) n

and (13.14.1.4), we have

j*

f dv = s u p n

j*

f, dv = sup n

j* *! f,g d p =

fg dp

sincefg is equivalent (with respect to p) to supf,g, by virtue of the convention n

about products and the fact that g is finite almost everywhere with respect to p.

186

Xlll

INTEGRATION

End of the Proof (13.14.1.7)

For every function u E 9 such that f 5 u, we have f g 5 vg, and

J’* v dv = J’* vg dp

J’*f g dp 5 J‘*vg dp = J’* v dv.

by (13.14.1.2), so that

Hence, by definition of the upper integral, we have it remains to establish the opposite inequality

I*

f g dp S

s*

f dv. Hence

(13.14.1.8)

Let h E 9 be such that h 2 f g . Then it is enough to show that

s* s* f dv

(13.14.1.9)

5

h dp.

The set X - A is the union of a denumerable increasing sequence of compact sets H, and a p-negligible set N, such that g I H, is continuous, finite and > O for all n. We define a mapping u of X into R as follows: u = h/g in the union of the sets H,, and u(x) = 00 in N and in A. In each of the sets H, we have ug = h, and by virtue of (13.14.1.6)

+

;:j

dv = 1:;g

dp =

f/I d p .

But v(N) = 0 by virtue of (13.14.1.3), and v(A) hence, as h 2 0 , (13.5.7) j * u dv = sp :

:1;

dv = s:p

/:/I

=0

by virtue of (13.14.1.5),

dp 5 j * h dp.

Since f 5 u, we obtain the required inequaliLy (13.14.1.9).

Q.E.D.

(13.14.2) Let g be a locally p-integrable function which is 2 0 on X, let S be the p-measurable set of points x E X such that g(x) > 0, and let v = g p . Let f be a mapping of X into E.Then the following conditions (with the conventions of (1 3.11) for products) are equivalent : (a) f is v-measurable ; (b) f q s is pi-measurable : (c) f g is p-measurable.

We may suppose g to be finite. With the conventions we have made, we have f g = ( f cps)(gcps),and since the two factors on the right-hand side never

14 INTEGRATION WITH RESPECT TO A POSITIVE MEASURE

187

take the values 0 and kc0 simulta~ieously,it follows from (13.9.8.1) tnat (b) implies (c). Conversely, it is immediately seen that the function g’ which is equal to g-l on S and vanishes on X - S is p-measurable: for there exists a partition of S (resp. X - S) into compact sets L, (resp. M,) and a p-negligible set P (resp. Q) such that the restriction o f g to each L, is continuous, and it follows that the restrictions of g’ t o L, and M, are continuous. Furthermore, we have (fg)g’ =f q s , hence by the same argument it follows that (c) implies (b). Hence it is enough to prove that (a) and (b) are equivalent. Suppose first that f q s is p-measurable. The set GS is v-negligible, by virtue of (13.14.1.5); also the hypothesis implies the existence of a partition of S into a sequence of compact sets L, and a p-negligible set N, such that f l L , is continuous for each n. Also N is v-negligible, by (13.14.1.3); hence f is v-measurable. Conversely, suppose thatfis v-measurable. Then there exists a partition of X consisting of a sequence of compact sets H, and a v-negligible set N, such thatfl H, is continuous for all n. Also there exists a partition of S consisting of a p-negligible set L and a sequence of compact sets K, such that g I K, is continuous (and >O) for all n. For all n, we have inf g(x) = a, > 0. Finally, XEK,

there exists also a partition of X - S into a p-negligible set L’ and a sequence of compact sets KI,. It is clear that the restriction of f i s to each of the sets H, n K, and KL is continuous, and it is therefore enough to prove that N n S is p-negligible. Now if p*(N n K,) > 0, then we should have g dp 2 a,p*(N n K,) > 0, which is absurd because

S”K,

0 = v(N n S)

=lNnSg dp

by virtue of (13.14.1). Hence each set N n K, is p-negligible, therefore so is N n S, and hence finally f qs is p-measurable. (13.14.3) Let g be a locally p-integrable function uhich is 2 0 on X, and let v = g . p. Then a mapping f : X -+ R is v-integrable f a n d only f ( w i t h the concentions of (13.11)) f g is p-integrable; and in that case we hare

J

P

(1 3.14.3.1)

fdv=

J

P

fgdp.

For f to be v-iniegrable it is necessary and sufficient thatf’ and f - should be (1 3.7.4) ; and since we have (fg)’ =f ‘ g , ( f g ) - =f -9, with the conventions of 13.11 about products, it is enough to prove (13.14.3) when f 2 0. The first assertion is now a consequence of (1 3.1 4.1 ), (13.1 4.2), and (1 3.9.13); and the relation (13.14.3.1) is just (13.14.1 .I).

188

Xlll INTEGRATION

(13.14.4) With the hypotheses of (1 3.14.3), the measure v is bounded ifand only i f g i s p-integrable; and v = 0 i f and only i f g is p-negligible. (13.14.5) Let g,, g 2 be two mappings of X into R, such that g1 is 2 0 and locally p-integrable. Then g2 is locally ( g l * p)-integrable Sf and only f ( w i t h the product convention of (13.11)) g 2 g 1 is locally p-integrable, and in that case we have

By considering g: and g;, we reduce immediately to the case where g 2 2 0. To say that g 2 is locally ( g l * p)-integrable signifies (13.13.1) that, for every f E XR(X), the function g 2f is ( g , p)-integrable, or equivalently (13.14.3), that g 2 g l f is p-integrable; which in turn means that g 2 g 1 is locally p-integrable. Also, if we put v = g1 . p and 1 = g 2 - ( g l * p ) , we have f dA = s f g 2 dv f g 2 g 1dp by (13.14.3); hence the relation (13.14.5.1). a

s

=s

PROBLEMS

1. Let X be a locally compact space, p a positive measure on X, and (/") a sequence of

= f. I

functions belonging to U k ( X ,p). Write p n =fn . p and pn(A)

dp for any

p-measurable set A. (a) Show that if X is not compact it can happen that the sequence (A) is unbounded in L', but that the sequence (p,) is vaguely bounded. (b) Suppose that, for every subset A of X consisting of a single point, and for every open subset A of X such that the measure induced by p on the frontier of A has finite support, the sequence &(A)) is bounded. Show that the sequence (A) is bounded in L'. (Show first that every point xo E X has an open neighborhood U such that the sequence of numbers Ip,l(U) is bounded. To do this, argue by contradiction, by showing that otherwise it would be possible to define a strictly increasing sequence of integers (nk), a decreasing sequence (uk) of open neighborhoods of xo , and a sequence (wk) of p-quadrable (Section 13.9, Problem 7) open sets, with the following properties: o k c u k - i , ~cL(ukC{Ixo})~1/k, l p n ~ l ( u k - { X o } ) ~ 1for i < k , \kkcUk--k+,, and finally

Consider the union W of the Wk, and obtain a contradiction. Then show by a similar argument that there exists a compact subset K of X such that the sequence (Ip.l(X - K)) is bounded.) (c) Deduce from (b) that if the sequence (p.(A)) is bounded for every open set A in X, then the sequence (A) is bounded in L'. (d) On the interval [0,1], take p. to be the measure defined by a mass n at the point

189

14 INTEGRATION WITH RESPECT TO A POSITIVE MEASURE

0 and a mass -n at the point I/n. Then the sequence &(A)) is bounded for every open set A which is 1p.I-quadrable for all n , but the sequence of norms Ilp.jl is unbounded. Again, if we take pnto be the measure defined by a mass n at the point l / n and a mass --n at the point l / ( n l), then the sequence (p.(A)) is bounded for every interval A c [0,1] (and therefore also for every finite subset A of [0, l]), but the sequence of norms IIpL.llis not bounded.

+

2.

For each integer n 2 1, let E. be an at most denumerable closed subset of I = 10, 1 [, and let I, be the family of component intervals of I - En. Suppose that the maximum length d. of the intervals J E 9"tends to 0 as! n + m. Let h be Lebesgue measure on I and let A be a A-measurable subset of I. Suppose that there exists a number k E 10, 1[ such that, for all n and all J E I,,we have h(A n J) < kh(J). Show that h ( A ) = 0. (By using (13.7.9), show that h(A) 5 E k ( h ( A ) E ) for all E > 0.)

+

+

+

3. For each integer n > 0, the Farey series of order n is the set F. of all rational numbers which, when expressed in their lowest terms p/q, are such that 0 $ p 5 q 5 n, and arranged in increasing order. The distance between two consecutive terms of F. is therefore 5 l/n. (a) Show that if two rational numbers r = p / q and r'=p'/q' are such that qp' - pq' = & 1, then for every pair of integers (p", q") there exist integers x, y such that p" = px p'y and q" = qx 4'y. The fraction p"/q" belongs to the closed interval with endpoints r, r' if and only if x, y are of the same sign. (b) Deduce from (a) that if r = p/q and r' = p'/q' are two rational numbers belonging to the interval [0, 11, such that q > 0, q' > 0 and qp' - pq' = f1, then r and r' are consecutive elements of the Farey series FsUpc,, ,,). Moreover the smallest of theintegers m such that the open interval with endpoints r, r' contains a point of F, is the integer q q', and this open interval contains only one point of F,,,., namely(p +p')/(q 4'). (c) Conversely, if r and r' are consecutive terms of F., show that qp' - p4' = f1 (induction on n).

+

+

4.

+

+

For each x E I = 10, I ] , put p(x) = (I/x) - [I/x], where [t] denotes the integral part of the real number t. Ifp"(x) = p(p"-'(x)) is defined and nonzero, putqn+I(x)= [l/p.(x)] (with 41(x) = [l/xl). (a) For every n such that p"(x) is defined, show that

where A.(x) and B.(x) are integers k n - 1. The fractions An-l(x)/Bn-l(x) and A,(x)/B.(x) are two consecutive terms of a Farey series (Problem 3) and x belongs to the closed interval with these as endpoints. Deduce that p"(x) is defined for all n 2 1 if and only if x is irrational. The (finite or infinite) sequence 01 numbers q.(x) is called the continued fraction expansion of x, and the fractions A.(x)/B.(x) are the convergents to x . The A.(x) and B.(x) are constant on the complement of the denumerable closed set En of points x such that P.+~(X) is not defined, and on each of the component intervals of I - E, the function p. is monotone and varies from 0 to 1 . (b) Let h be Lebesgue measure on I and let A be a h-measurable subset of I consisting entirely of irrational numbers. Suppose that is h-invariant with respect to p (Section

190

Xlll INTEGRATION 13.9, Problems 13 and 24). Show that &A) is either 0 or 1 . (Suppose that d = &A) t 10, 1 [. Show that, for each component interval J of I - En,we have &A n

J) 5 (2d/(l + d))h(J).

For this purpose use (a) above and Problem l(c) of Section 13.21 to show that

+

for suitably chosen integers p , q, r , s with p s - qr = 1. Use Problem 2 to complete the proof.) x), is invariant with respecf (c) Show that the measure p = g A, where g(x) = l/(l fo p. Deduce from (b) and the ergodic theorem (Section 13.9, Problems 13 and 24) that, for every &integrable function f on I and almost all irrational x E I, we have

+

(Gauss-Kuzmin formula). (d) Deduce from (c) that, for each integer p k n such that qx(x)= p (x irrational), then

2 1,

if v,(x, p ) is the number cf indices

for almost all irrational x E I. (e) Deduce from ( c ) that

for almost all irrational x .

15. T H E L E B E S G U E - N I K O D Y M T H E O R E M A N D T H E ORDER R E L A T I O N O N MR(X)

(13.15.1) Let p and v he two positire measures on a focally compact space X, such that v 5 p. Then there exists a focally pintegrable function g such tl~at v=g-/&.

We distinguish two cases. (1) Suppose first that v is bounded. Since the function 1 is then v-integrable, it follows from (13.11.2.2) that for everyfE X,(X)

(13.15.1.1)

Iv(.f)12

5 V(l)V(f2)

s V(l)Pc(f*).

191

15 THE LEBESGUE-NIKODYM THEOREM

This shows first of all that iff is p-negligible then v ( f ) = 0. Passing to the quotient by the subspace Jlr of p-negligible functions, the linear form v on X,(X) therefore defines a linear form ?++v(?) on the subspace g R ( X ) of Li(X, p ) which is the canonical image of X,(X) in Li(X, p). The inequality (13.5.1.1) shows that this linear form is conrinuous (5.5.1) on the subspace g R ( X ) ;since the latter is dense in Li(X, p) (13.11.6), the linear f o r m f w v ( f ) extends by continuity to a linear form on the Hilbert space L:(X, p ) (5.5.4). Hence there exists a function g E Yi(X, p) such that v ( f ) = p(gf) for all f E X,(X) (6.3.2). Sincegislocallyp-integrable (13.13),it follows that v = g . p . (11) General case. There exists a partition of X into a sequence of compact sets K, and a p-negligible set N (13.9.2). If we put M = X - N = K,,

u n

then q M is locally p-integrable and 1 - qM is p-negligible, hence (13.14.4) q M . p = p . Put p, = qK, p and v, = qK,. v. Since qKnis the lower envelope of a decreasing sequence of functions 2 0 belonging to X,(X), the relation v 5 p implies v, p, for all n ; furthermore, v, is bounded (13.14.4). Hence, by (I) above, there exists a locally p,-integrable function g, such that v,, =g, - p , = (g,qK,) . p = )gnqK,I p ((13.13.5) and (13.14.5)), and thefunction g,qK, is locally p-integrable. Let g denote the function which is equal to Jg,qK,I on each Kn and is zero on N. Then g is the sum of the series

-

m

C

lg,(PK,(.I f f is any function 2 0 in X,(X), and m is any integer, then

and therefore ((13.8.4) and (13.13.1)) the function g is locally p-integrable.

__

m

Also, since f = zfqKn ilmost everywhere (with respect to

sfs

,= 1

/I),

we have

(13.8.4) d p = j f q n dv. To show that v = g * p , it remains to be proved that N is also v-negligible. But by definition (13.5.5), for each E > 0 there exists a function h E 9 such that q N 5 h and p(h) 5 E . Since h is the upper envelope of an increasing sequence of functions 2 0 belonging to X,(X), the inequality v 5 p implies that v*(h) S p(h) 5 E , and the proof is complete.

The following lemma will be generalized later (1 3.1 5.8) : (13.15.2) r f ( p j ) ,j j j r is any$nite sequence of complex measures on X , there exists a positiile meamre A such that each p j is a measure with base 2.

192

Xlll INTEGRATION

By writing each p j as a linear combination of four positive measures (13.3.6), we may assume that the measures p j are all positive. Then we take r

1 = 1p j , and apply the result of (13.15.1) to each p j

A.

j= 1

(13.15.3) (i) With respect to the order relation on MR(X), m y two real measures p, v have a least upper bound sup(p, v) and a greatest lower bound inf(p, v). For each real measure p put p' = sup@, 0) and p- = sup( - p , 0); then we have (13.15.3.1)

inf(p+,p - ) = 0,

lpl = p + + p - = sup(p,-p)

p = p+-p-,

and, for any two real m e a w e s p, v,

(13.15.3.2)

(ii) Let 1 be a positive measure and let g l , g2 be real-valued locally I-integrable functions on X.r f p , = g1 1 and p2 = g2 * A, then (1 3.15.3.3)

and in particular, for any real-valued locally A-integrable function g , (1 3.1 5.3.4)

W e have g1 * 1 respect to 1.

( g . A ) += g +

*A,

g 2 1 if and only

( g ' A ) - = g - -1.

if gl(x) 5 g2(x) almost everywhere with

(iii) Let (gn) be an increasing sequence of locally A-integrable real-valued functions. Then the increasing sequence of measures gn ' A is bounded above in MR(X)$and only if the function sup gn is locally A-integrable, and in that case n we have (13.15.3.5)

SUP n

( g n * 1)=

(SUP n

*

A.

To prove (i), we remark that the measures p, v can be written in the form p = p1 - 1 2 , v = v1 - v 2 , where pl, p 2 , v1 and v 2 are positive (13.3.6). Applying (13.15.2) to the four measures p l , p 2 , vl, v 2 we see (by using (13.13.2)) that p =: u , Iand v = u . A, where A is a positive measure and u, u 1

15 THE LEBESGUE-NIKODYM THEOREM

193

are locally A-integrable real-valued functions. Hence the assertions of (i) are consequences of those of (ii). To prove (13.15.3.3), we reduce to the case where pz = 0; for if sup(pl - p 2 ,0) exists and is equal to (gl - gz)+ 1, it follows immediately (13.3) that we shall have

This reduces us to proving (13.15.3.4). For this, we shall begin by proving the last assertion of (ii), or equivalently that the relation g A 2 0 implies g(x) 2 0 almost everywhere with respect to A. Let N be the set of points x E X such that g(x) < 0. If we put v = 9-1,it follows from (1 3.14.1.5) that v(X - N) = 0; so if we can show that v(N) = 0, it will follow that v = 0, and this will establish our assertion (13.14.4). Clearly it is enough to prove that v(N n K) = 0 for each compact subset K of X. If U is any relatively compact open set g - dA g + dA. For by hypothesis containing N n K, then we have /(g' - g - ) f dA 2 0 for every function f 2 0 belonging to X,(X), and it is therefore sufficient to remark that g - dA = sup g - f dA, g f dA =

1"

ju

i

ju

s sf5

lu

sup g'lf dA, the supremum being taken over all functions f E X,(X) such that 0 qrl(13.5.1). Since g'(x) = 0 for x E N n K, we have g + dA = inf U

consequently, J N (13.14.3).

g - dA = inf U

I

ju

g f dA = 0;

g - dA = 0,which shows that v(N n K) = 0

+

Now let p be a measure such that p 2 0 and p 2 g A. Putting A p = 0 , we have (13.15.1) p = u * a and A = u 8 , where u and u are locally a-integrable and u 2 0 and u 2 0 almost everywhere with respect to 8. I t follows that u 2 gu almost everywhere with respect to a, hence that u 2 (gv)' = g'v almost everywhere with respect to a ; but this implies that p =u .a

>= (g+v) - a = g + - A

by virtue of (13.14.5). Finally, we have to prove (iii). If the sequence (9,. A) is bounded above in MR(X), then for each function f 2 0 belonging to .X,(X) we have

.s

sup fgn dA
0, there exists 6 > 0 such that the relations 0 2 h 5 f and l * h dp 5 6 imply / * h dv 5 E . (c’) For every compact subset K of X and every real number E > 0, there exists 6 > 0 such that the relations A c K and p*(A) 5 6 imply v*(A) 5 E (“absolute continuity” of v with respect to p ) . (d) v = sup (inf(v, np)). n

The fact that (a) implies (b) follows immediately from (13.14.1) and

(13.6.3). Clearly (b) implies (b’). Conversely, suppose that (b’) is satisfied, and let N be a p-negligible set. Since X is a denumerable union of compact sets K,, it is enough to show that each of the sets N n K, is v-negligible, and we may therefore assume that N is relatioely conzpact. Since N then has a compact neighborhood (3.18.2),it follows from (13.7.9)and (13.8.7,(i)) that we may restrict ourselves to the case where N is a denumerable intersection of relatively compact open sets; but in this case N is v-integrable ((13.7.7)and (13.8.7,(i)), hence is the union of a sequence (H,) of compact sets and a v-negligible set P. Since by hypothesis v(H,) = 0 for all n, it follows that N is v-negligi ble. To prove that (b) implies (d) and that (d) implies (a), we remark that we may express p and v in the form p = u * A and v = v 1, where 1 is a positive measure and u, v are finite, 2 0 and locally A-integrable ((13.15.2) and (13.15.3)).If we put v’ = sup(inf(v, nu)) and o” = u - ZI’, then we have n

21’

2 0, v“ 2 0 and sup(inf(v, np)) = u‘ . A (13.15.3).Hence to show that (b) n

implies (d) it is enough to show that (b) implies that v” . A = 0. Now the set A of points x E X at which v”(x) > 0 is contained in the set of points x at which

196

XI11 INTEGRATION

u(x) = 0,and therefore is p-negligible (13.14.1), hence v-negligible by hypothesis. On the other hand, U” = u q , , hence ((13.14.1) and (13.6.3)) the relation v(A) = 0 is equivalent to u“ being hegligible, and therefore implies that u” A = 0 (13.14.4). To show that (d) implies (a), it is enough to remark that the condition (d) signifies that u” * 1 = 0, and hence (13.14.4) the set A is &negligible. Hence if we put g(x) = u(x)/u(x) at points where u(x) > 0, and g(x) = 0 elsewhere, we shall have u(x) = g(x)u(x) at all x 4 A, hence almost everywhere with respect to 1.Consequently (1 3.14.5), g is locally p-integrable, and we have v = g * p. It remains to establish the equivalence of (b), (c), and (c’). Clearly (c’) is a consequence of (c), applied to f = q K .Also (c‘) implies that v*(A) = 0 for every relatively compact p-negligible set A; since every p-negligible set is the union of relatively compact p-negligible sets, this proves that (c’) implies (b). To prove that (b) implies (c) we shall argue by contradiction. Suppose therefore that there exists a function fo 2 0 which is p-integrable and v-integrable, and a real number a > 0 such that, for each integer n > 0, there exists a function 9, for which 0 5 g, S fo, j * g , dp S 2-” and / * g n dv > a . By virtue of (13.5.5), we may replace g , by inf(fo, 9:) for a suitably chosen function g: E 9,without disturbing the above properties, hence (13.9.13) we may assume that 9, is p-integrable and v-integrable. Now let

-

h = lim sup g, = inf h , , n-tm

n

where m

Since g, sfofor all n, it follows that h, is p-integrable and v-integrable for all n (13.8.2), apd we have

(13.5.8). Hence J h d p

=0

(13.8.1), and the hypothesis (b) therefore implies

that h is v-negligible (13.6.3). But we have

s

h, dv

>= a

(13.8.1),

which gives the required contradiction. Hence (b) implies (c), and the proof is complete. (13.15.6) Let p and v be two positive measures on X. Then the,followii?g conditions are equivalent: (a) The negligible sets are the same f o r p and f o r v .

15 THE LEBESGUE-NIKODYM THEOREM

197

(b) v = g . p7 where g is locally p-integrable and g(x) > 0 almost everywhere with respect to p. If (a) is satisfied, it follows from (13.15.5) that v = g * p and p = h * v, where g (resp. h) is positive and locally integrable with respect to p (resp v), Hence (13.4.5) the function hg is locally p-integrable, and we have p = (hg) * p, which implies (13.15.3) that hg is equivalent (relative to p) to the function 1, so that g(x) > 0 and h(x) = l/g(x) almost everywhere with respect to p. Conversely, suppose that v = g . p, with g(x) 0 almost everywhere with respect to p. Since (l/g(x))g(x)is equal to 1 almost everywhere with respect to p7 the function I/g(x) is locally v-integrable, and we have p = ( l / g ) v (13.14.5), so that the condition (a) is satisfied. If p and v satisfy the equivalent conditions of (13.15.6) they are said to be equivalent positive measures on X. Clearly we have here an equivalence relation on the set of positive measures on X. The notion of a measurable function is the same for two equivalent measures (1 3.9.4). (1 3.15.7) If p is any positive measure on X , there exists a continuousfunction h such that h(x) > 0 for all x E X and such that the measure v = h * p (which is equivalent to p by virtue of (13.15.6)) is bounded.

Let (U,,) be an increasing sequence of relatively compact open sets in X with the properties of (3.18.3), and for each n let f,, be a continuous function on X with values in [0, 11, such that f,,(x) = 1 for all x E U,, andf,(x) = 0 for all x E X - Un+l(4.5.2). Let (a,,) be a sequence of real numbers > O such that m

1a,, < +

03.

Then the series h =

n= 1

m

1a,f,

is normally convergent in X (7.1),

n= 1

hence h is continuous on X (7.2.1), and h(x) > 0 for all x E X . If v = h * p, then v*(l) = l * h dp 5 Ca,,J'/;. dp ((13.14.1) and (13.5.8)). If for example we take n

a,, =

2-" a,,
0 is at most denumerable.

Since X is the union of a sequence of compact sets K,, it is enough to show that each of the sets A n K,, is at most denumerable. For this, it is enough to show that, for each integer m 2 I , the set A,, of points x E A n K, such that I p I ( { x } ) 2 I / m i s j n i t e ; now this is immediate, for if B c A,,, consists of p points, then we have p / m 5 IpI(B) 5 lpl(K,). A measure p on X is said to be atomic if it is concentrated on an at most denumerable set. From this definition and from (13.1 8.1) it follows immediately that an atomic measure and a diffuse measure are always disjoint. The sum of two atomic measures is atomic. The least upper bound of a set of atomic positive measures which is bounded above is atomic ((13.15.4) and (13.15.9)).

(13.18.6) Every measure p can be expressed uniquely in the form p, where p1 is a diffuse measure and p2 an atomic measure.

+p2,

The uniqueness of the decomposition follows from the fact that an atomic measure and a diffuse measure are disjoint. To establish the existence of the decomposition, it is enough to consider the at most denumerable set

208

Xlll INTEGRATION

D of points x E X such that ( p ( ( { x } > ) 0 (13.18.5). This set is clearly p-measurable, and if we write p = h . ) p ( )(13.16.3), the measure p z = ( q D h ) lpl is concentrated on D (13.14.1), and pl = p - p2 is diffuse. For we have (p21 = qD * / p ( ((13.13.4) and (13.16.3)), hence lpll= /pi- J p z ( ;for each x E D, we have I p l ( { x } ) = Jpz(({x})by construction, and for each x 4 D we have ( p l ( { x ) )= ( p z ( ( { x )=) 0 by the definition of D; hence, by (13.16.1), I p t l l ( { x ) )= 0 for all x E X. Example (13.18.7) Let A be Lebesgue measure on R. From (13.18.4) and (13.18.6) it follows that every measure p on R is uniquely of the form pi + p z + p 3 , where pl = g . A is a measure with base A, p 2 is an atomic measure and p 3 is a diffuse measure disjoint ,from 1, and therefore concentrated on a A-negligible set (necessarily nondenumerable).

Remarks (13.18.8)

(i) If p is an atomic measure, the denumerable set of points

x E X such that Ipl({x>)> 0 is the smallest set on which p is concentrated. On the other hand, for a diffuse measure v # 0, there exists no smallest set on which v is concentrated: in other words, there exists no largest v-negligible set,

because every set consisting of a single point is v-negligible. (ii) Let p be an atomic measure concentrated on a denumerable set D, and let (a,,)be the sequence consisting of the distinct points of D arranged in some order. Put Ip(({u,)) = y,, > 0. If D,, = {ul, . . . , a,,],then 'pD = sup qD,; fl

since ( p (= qD.Ipl, then by (13.14.1) and (13.5.7) we have

for any function f 2 0 (the sum on the right is either a finite real number or +a).This implies in particular that, for any compact set K, we have C yn < -too.Furthermore, since every set consisting of a single point is aneK

p-measurable and since X

- D is p-negligible, it follows that

every mapping

of X into a topological space is p-measurable. The p-integrable functions

are therefore those for which

m n= 1

ynlf(an)l
O such that C yn < + 00 for all a,EK

compact subsets K of X. We have already seen (13.1.3) that m

f

H

v(f) =

C rnf(an>

n= 1

is a positive measure on X. With the same notation as above, it is clear that v = sup(rpDn.v), and since q D , . v is concentrated on D, and hence on D, it n

} ) yn. For if K is a follows that v is concentrated on D. Also we have ~ ( { a , , = compact neighborhood of a,, then for each E > 0 there exists an integer m such y p 5 E . I f f : X + [0, 11 is a continuous mapping which takes that ap E K , p z m

the value 1 at a,,, and the value 0 on X - K and at points a, such that k # n and k 5 m (4.5.2), then we have v({a,}) 5 v ( f ) 5 7, + E ; on the other hand we may choose the neighborhood K such that v(K) 5 v( {a,}) E (13.7.9), and a fortiori yn 5 v ( f ) 5 v({a,}) + E . Since E was arbitrary, our assertion is proved.

+

PROBLEMS 1. Let p, u be two atomic measures on a locally compact space X, and let M, N be the smallest sets carrying lpl, 1 ~ 1 ,respectively. Show that p, u are disjoint if and only if M n N = 0 . Hence give an example of an atomic measure v on the interval I = [0, I ]

of R such that I is the support (13.19) of v + and of v-.

Let p be a positive measure on a locally compact space X. If A c X is universally measurable and not p-negligible, show that there exists a positive measure v carried by A such that v # 0 and v 5 p. (Observe that A contains a compact set K which is not p-negligible.) (b) Let M be a universally measurable subset of X. For a positive measure p to be carried by M it is necessary and sufficient that v should be disjoint from every positive measure carried by X - M (use (a)). (c) If M is closed in X, show that the vector subspace of M(X) consisting of the measures carried by M is vaguely closed in M(X). (d) Show that Lebesgue measure on I = [0, I ] is the vague limit of a sequence of atomic measures carried by a fixed denumerable subset D of I. The subspace of M(I) consisting of measures carried by D is therefore not vaguely closed.

2. (a)

Xlll INTEGRATION

210

3. (a) Let p be a positive measure on a locally compact space X, and let A be a p-integrable set such that p(A) > 0. Suppose that, for each p-integrable subset B of A, we have either p(B) = 0 or p(B) = p(A). Show that there exists a point a E A such that p({a})= p(A). (Consider the intersection of the compact sets K c A such that p(K) = p(A): show that it is not empty, has measure equal to p(A) and consists of a single point.) (b) Suppose that p is a diffuse measure. For each p-integrable set A such that p(A) > 0 and each E > 0, show that there exists a p-integrable subset B of A such that 0 < p(B) 5 E (use (a) to show that there exists an integrable subset C of A such that 0 < p ( C ) 5 fp(A)). Deduce that, as B runs through the set of p-integrable subsets of A, the set of values of p(B) is the closed interval [0, p(A)]. (For each real number b such that 0 < 6 < p(A), let c be the least upper bound of the measures of measurable subsets C of A such that p(C) 5 6. Show that c = 6 by using the preceding result, and then show that there exists an increasing sequence (C.) of measurable subsets of A such that limp(C.)=b.) n+ m

4.

(a) Let v be a positive atomic measure on a locally compact space X,and let A be a v-integrable subset of X. Show that the set of values of v(B), as B runs through the set of v-integrable subsets of A, is closed in R. (Let P be the smallest set carrying v. Assuming that A n P is infinite, and arranging the points of A A P in a sequence (xJ, consider the mapping p of the product space {0, into R defined by p(~,)= E.Y({x.)), and show that p i s continuous.) (b) Deduce from (a) and Problem 3(b) that if p is any positive measure on X and if A is a p-integrable subset of X, then the set of values of p(B), as B runs through the set of p-integrable subsets of A, is closed in R. Extend this result to the situation where p is any real measure on X. (c) Deduce from Problem 3(b) that if p is a diffuse real measure on X, then the set of values of p(A) =- p+(A) - p-(A), where A runs through the set of 1pl-integrable subsets of X, is a closed (possibly unbounded) interval of R. (d) Give an example of an atomic positive measure v on a locally compact, noncompact space X, such that the set of values of v(A), where A runs through the set of v-integrable subsets of X, is not closed in R.(Take v so that inf v ( { x } )> 0.) I E X

5. (a) Let X be a compact space and p # 0 a diffuse positive measure on X. Let (f,) be a total orthonormal sequence in Y&(X,p). Show t h a t x [f&)lz = i- a, for almost n

all x E X. (Argue by contradiction, using Problem 3(b), Bessel's inequality, and the Cauchy-Schwarz inequality for series: show that there would exist a measurable set B with measure :,O and arbitrarily small, contained in the set of points x such that C Ifn(x)lz < i-la, such that, if c, = (rpBIfn), the series c n f n ( x )converges almost

"

sh;

n

everywhere in B to a function then observe that by virtue of the hypothesis that the sequence ( f n ) is total, together with Parseval's identity, we have tiin n-r

m

S, 1 I -

s,(x)lz ~ ( x =) 0 ,

where

y. =

2

ckh.

k= 1

Hence obtain a contradiction.) (b) Under the same hypotheses, show that there exists a sequence (b.) of scalars such that lim 6, = 0 and /b,,lzlfa(x)/z = fa, almost everywhere. (Use Egoroff's theorem "-1

m

x n

18 CANONICAL DECOMPOSITIONS OF A MEASURE

211

to obtain an increasing sequence of sets A,,, C X such that p(X - A,”) + O as m -+ and in which the partial sums s,(x)

=xlf~(x)12tend unifornily to k=l

$-co. Then

a,

remark

that the series with general term If.(x)12/s.(x) is divergent almost everywhere (Section 5.3, Problem 6). (c) Under the same hypotheses show that for almost all x o E X there exists a function g E Y&(X,p) (depending on xo) such that the series with general term (.y Ij;,)fn(xo) has sum equal to co (use Problem 23 of Section 12.16).

+

6. (a) Let 6 : R R be increasing and continuous on the right. Show that there exists a unique positive measure v on R such that v(]a,b ] ) -&b) - 6(u) for every half-open interval ]a,b ] . (This measure is called the “Stieltjes measure defined by 8”.and we --f

s

s

write f d 6 in place of f d v . ) Conversely, every positive measure on R may be obtained in this way, and two functions 6 , , Q z on R (both of them increasing and continuous on the right) define the same measure if and only if 8, 8,is constant. Under what conditions is the measure Y diffuse? What is then the image of v under 8? (b) Let K be the Cantor set (Section 4.2, Problem 2). Show that there exists a diffuse positive measure v on R with support K and total mass equal to 1 (Section 4.2, Problem 2(d)). Deduce that there exists on R a di’irse positive measure, disjoint from Lebesgue measure, with support equal to I = [0, 11. (In each component interval J of I - K choose a measure proportional to the image of v under an affine linear mapping of I onto J; then proceed by induction.) (c) Deduce from (a) that if p is a positive measure on R , a n d f a function belonging to Y:oc, c(R,, p), and (1.) a dense sequence in R such that f dp = 0 for all n , ~

!

LO,

then f i s p-negligible.

r.1

7. Let X be a compact space and p a diffuse positive measure on X with total mass equal

to 1. (a) Construct a family (U(t))oe,s of p-quadrable open sets, such that U(0) = 0, U(1) = X, u(t)C U(t‘) whenever t < r ’ , and p(U(1))= t for all 1. (First define the U(t) for r of the form k/2”, by induction on n. Use Problem 7(d) of Section 13.9 to prove that if V, Ware two quadrable open sets in X such that V c W a n d p(V) < p(W), then there exists a quadrable open set U such that c U c 0 c W and

v

:p(w - V)< p ( U - V) < $p(W - V). Also use the result of Problem 3(b).)

(b) Deduce from (a) that there exists a continuous mapping T of X onto I = [0, 11 such that the image of p under T is Lebesgue measure A on 1. (c) Show that there exists a p-negligible subset N of X, a A-negligible subset M of I and a homeomorphism r0 of I - M onto X - N such that, for every A-measurable subset A of 1 - M , the set rro(A) is p-measurable and of measure p ( r 0 ( A ) )= X(A). (Use Problem 7(d) of Section 13.9 to show that for each integer n , 0 there exists a finite partition of X consisting of a p-negligible set and quadrable open sets of diameter < l / n (with respect t o a distance defining the topology of X) and of measure s l / n . Proceeding by induction on n and passing to the limit, obtain a homeomorphism f of I - D onto a subset of X of measure I , where D is a denumerable subset of I.) 8. Let X, Y be two compact spaces, U a continuous linear mapping of the Banach space ‘dR(X) into the Banach space ‘CR(Y),and T : X + Y a continuous mapping.

212

Xlll

INTEGRATION

for all y E Y and allfE ‘C,(X). Show that under these conditions there exists a vaguely continuous mapping u: y w u Y of Y into M+(X) such that uy is carried by n - l ( y ) for all y E Y, and such that ( U . f ) ( ~ ( x )=) 0 and each integer n > 0, there exists a p-measurable subset A of X such that (i) the sets uJ(A) (0 5 j 5 n - 1 ) are pairwise --f

disjoint, and (ii) the complement of

u uJ(A) has measure 5

n- 1

E

(Rokhlin’s theorem).

j=o

(Choose a p-measurable set B such that 0 < p(B) < ~ / nand , construct the corresponding “Kakutani skyscraper” consisting of the sets uP(Em)for m 2 1 and 0 z p < rn (Section 13.9, Problem 14(d)). Show that we may take A to be the union of the sets d”(E,,) for all integers j > 0 and integers m such that m 2 ( j -t 1)n.)

19. SUPPORT OF A MEASURE. MEASURES WITH COMPACT SUPPORT

(13.1 9.1 ) If p is any measure on X, the union of all p-negligible open sets is p-negligible (and hence is the largest p-negligible open set). For if U is this union, it follows from (13.1.9) that the measure induced by p on U is zero. The complement in X of the largest p-negligible open set is called the support of p, and is denoted by Supp(p). To say that x E Supp(p) signifies that Ipl*(V) > 0 for eoery neighborhood V of x, or equivalently (13.5.1) that IpI ( Ifl) > 0 for every f~ Xx,(X)such that f ( x ) # 0; or equivalently again, that for each neighborhood V of x there exists a function , f ~ .X,(X) with support contained in V, such that p ( f ) # 0. If Supp(p) = X, the only p-negligible continuous function is the constant 0. From the definition we have Supp(p) = Supp( ]PI), and it is clear that Supp(ap) = Supp(p) for all scalars a # 0. More generally, if g is any locally Ipl-integrable function, then we have Supp(g p) c Supp(g) n Supp(p); for if we put v = g lpl, and if an open set U does not intersect Supp(g) or does not intersect Supp(p), then Ivl*(U) = 0 (13.14.1).

-

(1 3.1 9.2)

(i)

If p and v are positive measures, then SUPP(P + 4

= SUPP(P) u

SUPP(V).

19 SUPPORT OF A MEASURE. MEASURES WITH COMPACT SUPPORT

213

(ii) Zf H is any majorized family of positive measureh and v = sup H (13.15.4), then Supp(v) is the closure of the union of the supports of the measures pEH. This follows immediately from (13.16.1) and (13.15.9), applied to the characteristic 'function of a relatively compact open set (having regard to (13.1 5.4)). The complement of Supp(p) can also be defined as the largest of the interiors of p-negligible sets. Hence Supp(p) is the intersection of the closures of all sets on which the measure p is concentrated (13.18). But it should be realized that two disjoint measures can have the same support: for example, an atomic measure on R,concentrated on a denumerable dense set, and such that each point of this set has nonzero measure (13.18.8), has the same support as Lebesgue measure. (13.19.3) For a measure p on X to be such that every continuous complexualued,function on X is p-integrable, it is necessary and suflcient that Supp(p) hould be compact. The mapping fH p( f ) is ther. a continuouh linear form on the Frkchet space cS,(X) (12.14.6), and conversely every continuous linear form on %,.(X) is of this type.

If p has compact support S , then X - S is p-negligible, hencefq, is equal to f almost everywhere. Since fq, is measurable (13.9.6) and 1fqsl is bounded above by a multiple of qs (3.17.10), it follows that fqs,and hence also f , is integrable. Let us show conversely that if S = Supp(p) is not compact, then there exist continuous real-valued functions f 2 0 such that IpI*(f) = + co.By hypothesis (3.18.3) there exists an increasing sequence of relatively compact open sets U, in X such that 0, c Untl and the U, cover X. Since X is not compact, we have U, # X for all n. We now define inductively a sequence (an) of points of X and a sequence (V,) of relatively compact open sets, as follows: a, E S ; V, is a neighborhood of a, ; a,,, belongs to the intersection of S and the complement of the union of 0, and the v k with k 5 n (by hypothesis, this intersection is not empty); Vntlis a relatively compact neighborhood of a,,, such that does not intersect O,, nor any v k with k 5 n. For each n, let g, be a continuous mapping of X into [0, 13 which takes the value 1 at a, and the value 0 throughout X - V, (4.5.2). The hypothesis implies that lpI(g,) > 0. Put f, = g,/ lpl(gn), so that Ipl(f,) = 1. Then the function

v,,,

m

f=

f, is everywhere finite and continuous (because every point

n= 1

XE

X

214

Xlll INTEGRATION

belongs to some U,, which intersects only finitely many of the V k ) ,and we have Ipl*(f)

=

+ co,because Ipl*(f) 2..

n

-

k= I

Ipl(fk) for all n.

If S = Supp(p) is compact, the first part of the proof shows that E bP,(X), and therefore p is a

lpl ( f ) 5 Ipl(S) sup If(x)l for all functions f xeS

continuous function on the Frechet space U,(X). Conversely, let I be a continuous linear form on this space, so that there exists a compact set K c X and a constant c > 0 such that II(f)i 5 c * sup If(x)l for allJE bPC(X) (12.14.6). If XEK

L is any compact subset of X, we then have Il(f)l5 c * llfll for all , f ,X,(X; ~ L), and hence the restriction p of 1 to Xc(X)is a meamre on X. Moreover, if h is a continuous mapping of X into [O, 13, with compact support and such that h(x) = 1 for all x E K ((3.18.2) and (4.5.2)), then l(fh) = l ( f ) for all functionsf€ bPg,(X),becausef-fh is zero on K. From this it follows immediately that the support of p is contained in Supp(h); hence everyfe U J X ) is p-integrable, and p(f) = p ( f h ) = I ( f h ) = I ( f ) . Q.E.D. (13.19.4) If n : X -+ X’ is a homeomorphism and ,u is a measure on X, we have Supp(n(p)) = x(Supp(p)). This follows directly from the definitions. PROBLEM

Let p be a positive measure on X, and A a p-measurable set. Let i(A) be the set of points x E X such that there exists a compact neighborhood V of x in X for which p(V n (X - A)) = 0. Show that i(A) is open and that &(A) n (X - A)) = 0. (Consider S u p p ( ~ .~p).) -~

20. BOUNDED MEASURES

For every (complex) measure p on X, we define

which is a finite real number, or

+ co.

(13.20.2) We have 11p11 = lpl*(l). Hence llpll isjnite ifandonly measure lpl is bounded (13.9), and in that case 11PIl = lPI(1) = IPKW

is the total mass of X with respect to p.

if the positive

20

BOUNDED MEASURES

215

For every functionfE X,(X) we have (13.3.3) IP(f)I

s 1pK If11 s I l f

II lPI*(l) *

with the usual convention about products in [0, +a] (13.11), hence I(p(( a (13.5.1), and hence a function f~ .X,(X) such that If I 5 g 5 1 and Ip[f)l > a (13.3.2.1). This completes the proof. If 11p11 is finite, the measure p is said to be bounded. If p is a positive measure, this definition coincides with that given in (13.9). For a measure p to be bounded it is necessary and sufficient that 1p1 should be bounded, and we have II lpl II = 11p11. The set Mk(X) (also denoted by M’(X)) of bounded complex measures on X is a vector subspace of the space M,(X). We write MA(X) = M;(X) n MR[X) for the space of bounded real measures on X. The definition (13.20.1) shows that it comes to the same thing to say that the bounded measures on X are the continuous linear forms on the space .X,-(X), endowed with the norm (5.5.1), and that llpll is the usual norm (5.7.1) on the dud ML(X) of this normed space (12.15). Moreover, ML(X) is complete with respect to this norm (5.7.3). If p is any measure on X and g a locally p-integrable function, then the measure g p (13.16) is bounded if and only if g is p-integrable, and we have (1 3.20.3)

119 * pll = N,(g) =

s

191 4 P l .

- lpl (13.16.6).

This follows immediately from the relation ( g * pi = (91

It is clear that every measure p with compact support is bounded, because if S = Supp(p) then lpI*(l) = Ipl(S). If p is a bounded positive measure, then we have

because by (1 3.9.17) every bounded measurable function is integrable, and the Cauchy-Schwarz inequality (13.11.2.2) applied to g = 1, together with (13.9.13), shows that everyfe Uz(X, p) is integrable and that

N , ( f ) 5 NAf) ’ P(x>1’2. This shows also that the canonical injection Ug(X, p) + Y;(X, p) is continuous with respect to the seminorms on these two spaces; the same is true

216

Xlll INTEGRATION

for the injection 9 F ( X , p) + 9 i ( X , p) (with respect to the corresponding seminorms) because it follows from (13.12.5) that N2(f) 5 N,(f) p(X)”* for allfE 9 F ( X , p). (13.20.5) In particular (13.9.17), if p is a bounded (complex) measure, every function f~ %;(X) is p-integrable, and we have / f d p l 5 11p11 * llfll by

I

I

(13.16.5). In other words, f H f dp is a continuous linear form on the Banach space %;(X). But it should be noted that, in general, there exist continuous linear forms on this space which are not of this type. (13.20.6) The space Mk(X) is also the dual of the closure %,!(X) of X,(X) in the Banach space %$‘(X) (12.15). A function f belonging to %g(X) may be characterized by the following property: for each E > 0, there exists a compact subset K ofX such that If(x)l E for all x E X - K. For i f f € %g(X), then for each E > 0 there exists by definition a function g c X,(X) such that [ I f - 911 I E ; if K is the support ofg, then If(x)l 5 E for all x q! K. Conversely, suppose that f has the above property, and let h be a continuous mapping of X into [0, I], with compact support and equal to 1 on K ((3.18.2) and (4.5.2)); it is clear that Ilf-fhll 5 E and that fh E X J X ) , hence f E %g(X). The functions belonging to %g(X) are called (complex-valued) continuous functions which tend to 0 at infinity. (When X = R, they are indeed the continuous functions f such that lim f ( x ) = lim f ( x ) = 0.) We put X‘+W

x-+-w

%?i(X)=%g(X) n %,(X) for the corresponding space of real-valued functions. When X is compact, we have

%F(X)= %g(x)= X,(X) = %,(X). PROBLEMS

1. The space M:(X) of bounded real measures on a locally compact space X can be considered as a space of linear forms on each of the following vector spaces: (1) the space El = X d X ) of continuous functions with compact support; (2) the space E2= Wg(X) of continuous functions which tend to 0 at infinity; (3) the space E3 = Y g ( X ) of bounded continuous functions; (4) the space E, of linear combinations of (upper or lower) semicontinuous bounded functions ; ( 5 ) the space Es = @p(X)of universally measurable bounded functions.

Moreover, if v is any positive measure on X, the space M:(X,v) of bounded measures with base Y (which may be identified with the space L:(X,v) by virtue of 13.14.4)) can be considered as a space of linear forms on the vector space E4.,of bounded functions which are continuous almost everywhere with respect to v.

20

BOUNDED MEASURES

217

Let F Ldenote the weak topology on M:(X) corresponding to the vector space T4, the weak topology on M:(X, v) (or on L:(X, v)) corresponding to E4.* (Cf. (12.15)). The topology Y t on M:(X) is coarser than TIif i < j . The topology induced by F3on M:(X, v) is coarser than T4,which in turn is coarser than the topology (Consider the lower semicontinuous regularization of a function which induced by F5. is continuous almost everywhere with respect to v (Section 12.7, Problem 8).) (a) Let (p.) be a sequence of bounded real measures on a locally compact but not compact space X. Give an example in which (p.) tends to 0 vaguely (Le., for the topology Y1) but does not converge for the topology F2. A sequence (p.) which vaguely converges to 0 also converges to 0 with respect to Fzif and only if the sequence of norms (!lp.!l)is bounded (use the Banach-Steinhaus theorem). (b) Give an example of a sequence (p.) which converges to 0 for the topology T 2 but not for F 3 . A sequence (p.) of bounded real measures which converges vaguely to a measure p also converges to p for the topology T3if and only if, for each E > 0, there exists a compact subset K of X such that lp&X - K ) Z E for all n. (To show that the condition is necessary, argue by contradiction, by using the method of Section 13.14, Problem 1.) (c) Give an example of a sequence (p.) in M:(X, v) which converges to 0 for the topology F3but does not converge to 0 for the topology Y4, (take X = [0,1I). A sequence (p.) of measures belonging to M:(X, v) which converges to 0 for the topolalso converges to 0 with respect to F4,if and only if it satisfies the following ogy F3 condition: (C4.") For each v-negligible compact subset K of X and each E > 0, there exists an open neighborhood U of K such that /p,J(U)=< E for all n. (To prove that the condition is sufficient, reduce to the case where X is compact and apply (C,, ") to the set K, of points x E X at which the oscillation of an almost everywhere continuous function is Z E .To prove that the condition is necessary, argue by contradiction as in (b) above.) (d) Show that a sequence (p.) in M:(X) which is a Cauchy sequence for one of the topologies TI,F2, F3is convergent for this topology. (For F3, use (b) and argue by contradiction: form a sequence of measures which tends to 0 with respect to .T3 without satisfying the condition given in (b).) (e) A sequence (pJ of measures belonging to M:(X, v) is a Cauchy sequence with respect to the topology F4,if and only if it satisfies the condition ((24, ") (argue by contradiction as in (d)); the sequence (p,)then converges with respect to T4.? to a measure belonging to M&X, v). (f) Let (p,,) be a sequence of measures belonging to M:(X, v). For each subset A of X which is either finite or else open and v-quadrable (Section 13.9, Problem 7), suppose that the sequence (p.(A)) has a finite limit; show that (p")is then a Cauchy sequence for the topology T4. ". (Use Problem 1 of Section 13.14, and argue by contradiction.) Show that the hypothesis relative to finite sets A cannot be omitted. (g) Let (pn)be a sequence of measures belonging to M:(X, v) which converges in , ~ ~then the topology Y 3to a measure p belonging to M:(X, v). If also lim ~ ~ p=, IlpII,

EL( i # 4), and

n+ m

the sequence ( p c )converges to p in the topology F4,(use (e)). Give an example of a sequence (pn)ofpositiue measures belonging to M:(X, v), where the space X is compact, such that (pn)converges vaguely to a measure which does nof belong to Mi(X,v). 2.

The notation is the same as in Problem 1. Let (p")be a sequence of measures belonging to Mi(X). Then there exists a positive measure v on X such that p..E M:(X, v) for all n.

Xlll INTEGRATION

218

(a) Show that the following properties are equivalent: (a) The sequence (p.) is convergent for the topology r 6 . (8) For each closed subset A of X, the sequence @.(A)) has a finite limit. ( y ) The sequence (pn)converges for the topology Y3 and satisfies the following condition: (C,) For each compact subset K of X and each E > O , there exists an open neighborhood U of K such that Ip,J(U - K) 5 E for all n. (6) The sequence (pn)converges for the topology F 3and satisfies the following condition : (C,) For each E > 0 there exists 6 > 0 such that, for each universally measurable set A satisfying v(A) 5 6, we have Ipnl(A) 2 E for all n. (To show that (p) implies (y), use Problem 1 of Section 13.14 to show that the sequence l(p.11 is bounded. To establish (C5),argue by contradiction: first consider the case in which the sequence (p.) tends to 0 for the topology Ys, and then pass to the general case as in Problem l(d). Next show that ( y ) implies that the sequence (p.) is convergent for the topology F , , and in particular implies (p). To show that (6) implies (a), use the definition of measurable functions. To show that ( a ) implies (S), argue by contradiction. Finally, to show that ( y ) implies (a),consider first the case where lim lIp.ll = jlpI/,where p is the limit of (p,) for the topology Y 5 and , argue by n-m

contradiction to prove that ( y ) implies (6). To pass to the general case, reduce to the situation where p = 0, and argue by contradiction: we may assume that there exists a functionfe F-6 such that the sequence ( p n ( f ) )has a limit #O. On the other hand, we can pick a subsequence (p,,))of the sequence (p,) such that the sequence (&), and hence also (p;)), converges vaguely. Hence arrive at a contradiction, by observing that these two sequences also satisfy ( C 5 ) . ) (b) Give an example of a sequence (,urn) of measures belonging to M:(X,v) which convergesfor the topology f 4 , but not for the topology Y, (cf. Section 13.18, Problem 2(d)). on MA(X) are distinct (c) Take X = [O, 11. Show that the topologies Y5and (cf. Section 12.15, Problem 2(c), and Section 13.11, Problem 3). (d) Extend the results of Problems 1 and 2 to bounded complex measures. 3. Let X be a locally compact space, p a positive measure on X. Let (9.) be a sequence of p-integrable functions such that (1) the sequence (9")converges in measure (Section 13.12, Problem 2) to a function g; (2) the sequence of measures (gn . p) converges for the topology F6(Problem 2). Show that under these conditions the function g is p-integrable and that the sequence ( g n ) tends to g in 9 A ( X , p). 4.

Let X be a compact space, p a positive measure on X, and (fn)an orthonormal sequence in .!Z&(X, p) consisting of functions which are uniformly bounded on X. (a) Show that, for each function g E &(X, p), the sequence of numbers ( g If.) = /g(x)f(x)dp(x) tends to 0. (Reduce to the case where g is bounded and hence belongs to 9 $ ( X , p).) (b) Let xo E X . For the series C ( g I f . ) f n ( x o ) to converge for euery function n

3 g ( X , p), it is necessary and sufficient that (in the notation of Section 13.17, Problem 2) the sequence of bounded measures K.(xo , .) . p should converge for the topology 9 - 6 (Problem 2) to a bounded measure h,, . p ; hence in particular we have

gE

=I

UXO)hxO(x)fn(x)dp(x) for all n.

21 PRODUCT OF MEASURES

219

(c) Show that there exists a measurable subset A of X with measure >0, such that for each xo E A there exists a function g E -Yg(X, p) (depending on xo) for which the series with general term (g If.)f(xo) does not converge. (Take A to be the set of points x E X for which the sequence (f,(x)) does not converge to 0, and use Problem 7(a) of Section 13.11. Then argue by contradiction, using (a) and (b).) Let S be a closed subset of R. Given a sequence ( c , ) . ~ ~of real numbers, show that there exists a positive measure p on R with support contained in S and such that

j x ” +(x) = c. for all n 2 0 if and only if, for every polynomial P(X) =

2 &Xk such

k=O

that P(x) >= 0 for all x E S, we have

n

fkck k=O

2 0. (As in Problem 2 of Section

13.3,

show that there exists a positive measure p with support S which extends to a linear form u defined on .f,,(R) and on the space of polynomials on R, and such that U(P) =

C &ck

for each polynomial P(X) =

k=O

is p-integrable and that

s

x” +(x)

= c,

n

C k=O

fk

Xk.Then prove that each power xn

. For this purpose, remark that for each in-

teger n >= 0 and each E > 0 there exists a number R > 0 such that If.,R(x)l 5 EX^"+^, where f,. is the function equal to 0 for 1x1 < R and to x” for [XI 2 R.) Particular cases: (1) S = R (“Hamburger’s moment problem”): the condition is that the quadratic forms

1. k = O

c,+,&[x should be positive for all n (observe that every

+

polynomial P(x) which is 20 on R is the sum of two squares (Pl(x))’ (PZ(x))’). (2) S = [0, +a[(“Stieltjes’ moment problem”): the conditiori is that the quadratic forms

5

J. k = O

c,+ktjtk

and

5 cj+k+L[]& should be positive for all n 2 o (remark

1. k = O

that every polynomial P(x) which is 2 0 on [0, +m[ can be written in the form (Pdx))’ (PZ(X))’ x((p3(x))z (P&))*).

+

+

+

Let v be a positive measure on X and (fa)a sequence of functions belonging to .Y;(X, v) such that the sequence of measures Cf. . v) converges to f. v (where f E 9;(X, v)) for the topology Y6. Show that f(x) 5 lim supfn(x) almost everywhere n- m

with respect to v. (Use (13.8.3).)

Let p, v be two bounded complex measures on X. Show that the following conditions are equivalent: (1) p and v are disjoint; (2) JJp =tvI/= IlpIl llvll; (3) IIP YII + llp - YII = 2(llpll llvll).

+

+

+

21. PRODUCT O F MEASURES

(13.21.1) Let X, Y be two locally compact spaces, 1 a measure on X and p a measure on Y. Then there exists a unique measure v on the product space X x Y such that, for each pair of functions f E Xx,(X)and g E .X,(Y), we have

220

Xlll

INTEGRATION

(1) Uniqueness. Every compact subset of X x Y is contained in one of the form L x M, where L and M are relatively compact open subsets of X and Y respectively ((3.20.17) and (3.18.2)). We shall show that, for each function 12 E Xc(X x Y) with support contained in L x M, the value of v(h) is well-determined. This will follow from (13.21 .I.2) /f L c X and M c Y are relatiueIy compact open sets, then every function h E X,(X x Y ) with support contained in L x M is a cluster point, in the Banach space Xc(X x Y ; L x M ) , of the set of functions of the form (x, y)”cfi(x)gk(y), where (A) is a finite sequence of functions in .X,(X) i, k

with support contained in L, and (gk) is afinite sequence of functions in X C ( Y ) with support contained in M .

Assuming this for the moment, by hypothesis there exists a number c > 0 such that I v(u) I 5 c . llull for all u E X,(X x Y) with support contained in L x M. On the other hand, for each E > 0 there exist two finite sequences of functions f i E X c ( X ) gk , E X , ( Y ) such that

I h(x,r) - cfi(x)gk(Y)I 5 i, k

for all (x, y ) E L x M , the support of the left-hand side being contained in L x M. Using (13.21 .I.I), it follows that i, k

and since E is arbitrary, this establishes the uniqueness of v. To prove (13.21 .I .2), let E be a positive real number. Then for each z = ( x , y ) E L x M there exists a compact neighborhood U c L (resp. V c M) of x in X (resp. of y in Y) such that the oscillation of h in U x V is S E (3.20.1). The projections S c L and T c M of Supp(h) are compact, hence for each x E S there exists a finite number of points y j ( x )( 1 5.j 5 n(x)) of T , and for e a c h j a compact neighborhood U j ( x )c L of x in X and a compact neighborhood Vj(yj(x))c M of y j (x ) in’Y, such that the oscillation of h in U j ( x ) x Vj(yj(x)) is S e , and such that the interiors of the sets V j ( y j ( x ) ) cover T. The set U’(x) = U j ( x ) is a compact neighborhood of x in X. j

Hence there exists a finite number of points x i (I 5 i 5 m ) in S such that the p the interiors of the sets U’(xi) cover S. Put A i = U ’ ( x i ) ,and let ( B k ) l s k S be family of sets obtained as follows: for each point y E T, let W ( y ) be the intersection of the (finitely many) interiors of sets V j ( y j ( x i )which ) contain y ; the sets W(y) are open and nonempty, and form a finite open covering of T which we denote by (Bk)ls k s p . Notice also that, by construction, the oscillation of h in each set A j x B, is 2 E. Now let (A),s8s,n(resp. (gk),‘ k c p ) be continuous mappings of X (resp. Y) into [0, I], such that Supp(,fi) c Ai and Supp(gk) c Bk

21 PRODUCT OF MEASURES

for each pair of indices i, k , and such that c f i ( x ) 5 1 for all x for all y E Y, and Then we have

LA(.)

i

=1

for all x E S,

i

y) =

for all x E X and all y fore imply that

E Y.

5

E X,

221

Ck gk(y)S

1

1k g k ( y )= 1 for all y E T (12.6.4).

y)fi(xbk(y)

i. k

If yk is any point of B,, the hypotheses there-

xfi(x>gk(Y>

i,k

for all (x,y ) E A x B, where A

=

u i

=
c i ~ ( x is ) continuous on Y, and supp(g) c M.

MY’) - d Y ) l

=

IS

v>>W x )

( N x , Y’) - 0,

I ULE

We return to the proof of (13.21 .I). For each function h E X,(X x Y), the number v(h) = p(g) (which by abuse of notation is also written in the form p ( j h ( x , y ) dA(x))) is defined. Furthermore, with the notation used above, there exist by hypothesis two numbers aL and b, such that, for each

222

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INTEGRATION

function u E X ( X ; L) (resp. v E X ( Y ; M)), we have IA(u)l 5 aL llull (resp. Ip(u)l 5 b , Ilvll). Hence, for each function h ISX(X x Y; L x M), we have

for all y

E Y,

and by virtue of (13.21.1.3) and the definition of v(h),

Iv(h)l 5 aLb, llhll. Since every compact subset of X x Y is contained in the product of its projections on X and on Y, the proof of (13.21.1) is complete. The number v(h) = p ( / h ( x , y ) dA(x)) is also denoted by SdP(Y) Sh(x, Y ) M x ) .

s

Since we can clearly interchange the roles of X and Y, it follows that (13.21.2)

h(x, Y )

4

x

9

s s

h(x, Y ) M

Y)

= j d P ( Y )p x , Y ) W

x)

Y)=

d4x)

for all h E Xc(X x Y). By reason of this formula we write ssh dL d p or s / h ( x , y ) dA(x) d p ( y )

s

instead of h(x, y ) dv(x, y ) , and we may also interchange A and p in these notations. The measure v is called the product of 1- and p, and is denoted by A 0 p. It is clear that the mapping (2, p) H A 0 p of M,(X) x M,(Y) into Mc(X x Y) is bilinear: in other words, we have

(11 + A21 0 (PI

+ p 2 ) = A1 0 p1 + 1, 0 p 2 + A2 0 p1 + 2 2 0 p 2

and ( a 4 0 p = A 0 (up) = 4 10 p)

for any scalar a. Furthermore, if A and p are real (resp. positive) measures, then so is A@p. (13.21.3) Let A, p he two positive measures on X, Y respectively, and v = A @ p their product. For each function h E J ( X x Y), the function

21

PRODUCT OF MEASURES

223

belongs to Y(X), and we have

{*h dv = .i*(s”h(~, Y> d p W )

(1 3.21.3.1)

(with an abuse of notation analogous to those above). By (12.7.8) there exists an increasing sequence (h,) of functions belonging to X,(X x Y ) such that h = sup h, . Foieach n, it follows from (13.21 .I.3) n

that the function f n ( x )= jh,(x, y ) dp(y) belongs to .X,(X); hence f = supf, n

belongs to Y(X), and moreover we havef(x) = j*h(x, y ) d p ( y ) for all x E X, by virtue of (13.5.2). Since v(h,) = A(fn)by definition, another application of (13.5.2) completes the proof. From now on, up to and including (13.21.16), we shall assume that the measures i? E M(X) and p E M(Y) are positive, and we shall write v = i? @I p. (1 3.21.4)

If h is any mapping of X x

Y into R, then

{ * h dv 2 i?’(j*h(X,

(13.21.4.1)

Y ) dP(Y)).

Let ~ E Y ( X x Y) be such that h 5 u. Then for each j * h ( x , y ) dp(y) 5 u(x, y ) dp(y), and consequently

/*

XE

X we have

by (13.21.3). Hence the inequality (13.21.4.1) follows from the definition of v*(h) (13.15.5). We shall write /[*A di? dp or //*h(x, y ) dA(x) dp(y) instead of v*(h), and /*di?(x)/*h(x, y ) d p ( y ) in place of A*(/*h(x, y ) dp(y)). Similarly, we shall use the notations J/*h dA dp and //*h(x,y ) di?(x) dp(y) for lower integrals. Thus the inequality (13.21.4.1) takes the form (1 3.21.4.2)

j{*h(X,

Y ) W x ) d A y ) 2 j * d A ( X ) J*h(X, Y> M Y )

with equality if h E 9 ( X x Y). There is of course an analogous inequality (resp. equality) obtained by interchanging the roles of X and Y, and an

224

Xlll INTEGRATION

inequality in the reverse direction (with equality when h E 9’(X x Y)) for lower integrals :

(1 3.21.5) I f N is a v-negligible subset of X x Y, then the set of points x E X such that the section N(x) c Y of N is not p-negligible is I-negligible (in other words, we have p(N(x)) = 0 almost everywhere with respect to I).

This follows immediately from (13.21.4.2) applied to h = q N . (13.21.6) I f h is a v-measurable mapping of X x Y into a topologicul space E, then the set of points x E X for which the partial mapping y Hh(x, y ) is not p-measurable is A-negligible (in other words, the mapping y H h ( x , y ) is pmeasurable almost everywhere with respect to A).

By hypothesis, there exists a partition of X x Y consisting of a v-negligible set N and a sequence (K,),,, of compact sets, such that each of the restrictions h I K, is continuous (13.9). Let M be the I-negligible set of points x E X at which the section N(x) of N is not p-negligible (13.21.5). For each x $ M, Y admits a partition consisting of compact sets K,(x) (n 2 1) and the p-negligible set N(x), such that the restriction of y w h ( x , y ) to each of the sets K,(x) is continuous. Hence the result. It should be noted that it can happen that for each x E X the function y w f(x,y ) is p-measurable, and for each y E Y the function X H f (x, y ) is A-measurable, but that .f is not v-measurable. (13.21.7) (Lebesgue-Fubini Theorem) Let A, p be positive measures on X and Y, respectively, and v = I Q p their product. For each +-integrable mapping h of X x Y into the set of points x E X such that the partial mapping

w,

y w h(x,y ) is not p-integrable is I-negligible; the,,function

s

h(x, y ) dp(y), which is defined almost everywhere with respect to 1, is A-integrable; and X H

(12.21.7.1 )

It follows from (13.21.4.1) that the function x w J * J h ( x y)l , dp(y) is finite on the complement of a A-negligible set N, (13.6.4). On the other hand, the set N, of points x E X such that y w h(x, y ) is not p-measurable is I-negligible

21 PRODUCT OF MEASURES

225

(13.21.6). Hence, from (13.9.13), it follows that for each x $ N = N, u N, the mapping y Hh(x, y ) is p-integrable, and therefore the function X Hj

h k Y ) dP(Y)

is defined almost everywhere with respect to A. The fact that it is I-integrable, and the relation (13.21.7.1), then follow from (13.21.4.2), and (13.21.4.3). Interchanging the roles of X and Y we have also, under the hypotheses of (1 3.21.7),

But here again it needs to be said that the right-hand sides of (1 3.21.7.1) and (13.21.7.2) can be defined and equal without h being v-integrable (even if h is v-measurable) (Problem 3). (13.21.8) Let h 2 0 be a v-measurable function. Then the mapping

is A-measurable, and

Let (K,) be an increasing sequence of compact subsets of X x Y which cover X x Y (3.18.3). Then we have h = sup h , , where h, = inf(h, nqPK,), n

and h, is v-integrable ((13.9.7) and (13.9.13)); on the other hand (13.5.7), v*(h) = sup v(h,). By (1 3.6.2) and (13.21.7), there exists a A-negligible set N n

s

such that, for all x 4 N, all the functions y~ h,(x, y ) are p-integrable; also (13.21.7) the functions XH h,(x, y ) d p ( y ) are A-integrable and we have

.s

v(h,) = j d A ( x ) j h , ( x , y ) d p ( y ) ‘for all n. Hence it follows from (13.5.7) that j * h ( X , Y ) 4 0 )= SUP

h,(% Y ) d A Y )

for all x $ N. Consequently the function XH h(x, y ) dp(y) is A-measurable, s* by (13.9.11), and the relation (13.21.8.1) follows by another application of (1 3.5.7).

226

Xlll

(1 3.21.9)

INTEGRATION

A v-rneasurable,function h is v-integrable if and only

/* d W / * Ih(x, Y)l dP(Y) < + a.

if

This follows from (1 3.21.8) and (1 3.9.13). (13.21.10) Let A be a v-measurable set in X x Y . (i) The set M of all x E X such that the section A(x)is not p-measurable is I-negligible; the function x H p*(A(x)) is I-measurable; and

v*(A) = /*p*(A(x)) dI(x). In particular, i f A(x) is p-negligible except on a A-negligible set of values of x , then A is v-negligible. (ii) If A is v-integrable, then the set of all x E X such that A(x) is not p-integrable is &negligible; the ,function XI--,p ( A ( x ) ) (which is dejned almost everywhere Kith respect to A ) i3 A-integrable; and v(A) = S p ( A ( x ) )dA(x). These assertions are particular cases of (13.21.6), (13.21.8) and (13.21.7). (13.21.11). Let f (resp. g ) be a mapping of X (resp. Y) into [0, +a].With the convention of (13.11) for products, we have

By virtue of (13.21.4), we have

On the other hand, for each x E X we have

and

21 PRODUCT OF MEASURES

227

with the product convention referred to above. Hence it is enough to prove the inequality

Now this inequality is clearly valid when the right-hand side is equal to + co. So consider the case in which each of the factors on the right is finite. Then there exist two decreasing sequences (fn), (gn) such that f;,E Y(X), g n E 9 ( Y ) ,f S f , , g S g . f o r a l l n , a n d j*fd2

=

:!z j*fnd2,

[*g dp = lim j * g n dp. n-

03

By reason of our conventions and of (12.7.5), the function (x, y)wf,(x)g,(y) belongs to 9 ( X x Y), and we have f(x)g(y) 5J8(x)g,(y) for all n and all (x, y ) E X x Y. But, by virtue of (13.21.3),

L [I*f(X)S(Y) W x ) dcc(Y), from which the desired inequality follows by letting n tend to +a. Finally, to deal with the case where (for example) 1’ is A-negligible, it is enough (by virtue of our conventions) to prove that the function (x, y ) Hf(x)g(y) is v-negligible. This is a consequence of the following proposition: (13.21.12) If N is any 2-negligible subset of X, then the set N x Y is vnegligible.

For since Y is the union of a denumerable sequence (L,) of compact sets, it is enough to show that each of the sets N x L, is v-negligible. But since we have proved that (13.21.11.1) is valid whenever both of the factors on the right-hand side arejnite, we may take f = q N and g = qL, in this formula. (13.21.13) Let E, F, G be three topological spaces and u : E x F + G a continuous mapping. Let f (resp. g) be a I-measurable mapping of X into E (resp. a p-measurable mapping of Y into F). Then (x, y ) w u ( f ( x ) ,g(y)) is a v-measurable mapping of X x Y into G.

228

Xlll INTEGRATION

By (13.9.6) it is enough to show that (x, y ) ~ f ( x is) a v-measurable mapping of X x Y into E. There exists a partition of X consisting of a sequence (K,) of compact sets and a %negligible set N such that the restriction off to each K, is continuous. For each compact subset L of Y, the restriction of the mapping ( x , y )f ( x~) to each of the compact sets K, x L is continuous, and the set N x L is v-negligible (1 3.21.12), whence the result follows (13.9.4). Given two mappings f : X --f R and g :Y -+R, we denote by f Q g the mapping ( x ,y )f (x)g(y) ~ of X x Y into R (with the convention of (13.1 1) for products in R). Similarly for mappings into C. (13.21.14)

(resp. Y ) into

f (resp. g ) is a A-integrable (resp. p-integrable) mapping of X

a or C , then the function f Q g is v-integrable, and we have

By linearity we reduce to the case where f and g are mappings into R. By virtue of (13.21.12), the set of points ( x ,y ) E X x Y at which f(x) or g(y) is infinite is v-negligible, hence it follows from (1 3.21.I3) and (1 3.9.6) that f @ g is v-measurable. The fact that f Q g is v-integrable then follows from (13.21 .II) and (13.9.1 3). Finally, the formula (1 3.21.14.1) is a consequence of the Lebesgue-Fubini theorem. (13.21.15) Let A be a subset of X , and B a Jubset of Y . Then

(i) v*(A x B) = A*(A)p*(B) (with the product Convention of (13.11)). (ii) If A is I-measurable and B is p-measurable, then A x B is vmeasurable. (iii) If A is A-integrable and €3 is p-integrable, then A x B i~ v-integrable, and v(A x B) = A(A)p(B). These assertions are particular cases of (1 3.21 .II ) , (13.21.I3) and (13.21.I 4). (13.21.I 6) If ,f (resp. g ) is a locally A-integrable mapping of X (resp. locally p-integrable mapping of Y ) into or C , then f Q g is locally v-integrable, and

a

Since the set of points (x, y ) at which f (x) or g(y) is infinite is v-negligible (13.21.12), it follows from (13.21 .I3) and (13.9.8.1) thatfQ g is v-measurable.

Moreover, for each compact subset K (resp. L) of X (resp. Y) the function

21 PRODUCT OF MEASURES

229

(f@ g)'pK = (fqK)0 (g(pL)is integrable, by virtue of (13.21.14), hence f @ g i s locally v-integrable (13.13.1). Also, for each u E .X,(X) and each u E X c ( Y ) ,we have

by virtue of (1 3.21.I4) ;hence the formula (1 3.21.I6.1) follows from (1 3.21.I). (13.21.17) Let A, p be complex measures on X , Y, respectively. Then In 0 PI = I4 @ IPI.

-

We may write A =f. \A1 and p = g lpl, where If1 = / g )= 1 (13.16.3). Hence If 0 g l = 1, and the result therefore follows from (13.21.16) and (1 3.13.4). (1 3.21.18) Let A be a complex measure on X and p a complex measure on Y. (i) If A is concentrated on A c X and p is concentrated on B c Y (13.18), then A 0 p is concentrated on A x B.

( 4 Sup.p(A0 P) = S U P P ( x~ SUPP(P). (jii) With the product conuention qf (13.11) we have (13.21.18.1)

Ilf 0 PI1 = 1141- 11FIl.

In particular, $ A and p are hounded, then so is A@ p.

By virtue of (1 3.21. I 7) we may restrict ourselves to the case in which A and p are positive. Then (i) follows from the fact that X x Y - A x B is the union of the (A 0 p)-negligible sets (X - A) x Y and X x (Y - B) (13.21.12). It follows from (i) that Supp(A 0 p) c Supp(A) x Supp(p). On the other hand, if x E Supp(A) and y E Supp(p), then for each compact neighborhood V of x in X and each compact neighborhood W of y in Y, we have A(V) > 0 and p(W) > 0, whence by (13.21.15) (A 0 p)(V x W) = A(V)pL(W)> 0. This establishes (ii), by virtue of the definition of neighborhoods in X x Y . Finally, (iii) follows from (13.21.11) applied to f = 4px and g = 'py, except when one of the factors on the right-hand side is 0 and the other i s +a,; but in this case we have A 0 p = 0 by (13.21.12), and therefore the formula (13.21.I8.1) remains valid in this case. (13.21.19) There are analogous definitions and results for the product of any finite number of measures: if (Xi)1 i s any finite sequence of locally compact

230

Xlll

INTEGRATION

spaces, and if pi is a (complex) measure on X ifor each i, then the unique measure v on X

nXi which satisfies the relation n

=

i=1

is called the product of the measures p i (1 5 i 5 n), and is denoted by n

pl @ p2 @

*

@ p,, or @ p i , The

existence and uniqueness of v are proved

i=1

@ P , , - ~ )@ p,,, and observing that if by induction on IZ, putting v = (pl@ v' also satisfies the required conditions, then we must have

v'(h Of,) = ( ( P I O * ' . @ P n - 1Xh))pn(fn) = v(h Of,)

for all h E X

n Xi),by the inductive hypothesis. This characterization of the

n- 1

(i=1

product of measures shows immediately that the product is associative; in particular, we have also pi 0 ~2 0

*.

+

0 P n = pi 0 (PZ0

*

*

0 pn>.

We write

s

instead of f d v , and analogous notations for upper and lower integrals. We shall leave to the reader the task of formulating and proving the results corresponding to those established above for the case n = 2 ; the proofs require nothing but simple arguments by induction on n. The product of the Lebesgue measures on the n factors of R" is called Lebesgue measure on R".

PROBLEMS

1. Let X be a locally compact space, p a positive measure on X, and f a real-valued function 2 0 defined on X. In the product space X x R,let D, denote the set of points (x, t ) such that 0 5 t 5f(x). Also let denote Lebesgue measure on R.

(a) F o r f t o be p-measurable it is necessary and sufficient that D, should be measurable with respect to the product measure v = p Oh. (To show that the condition is necessary, prove that iffis p-measurable then D, is the union of a v-negligible set and a denumerable family of sets of the form A x I, where A is p-measurable and I is an interval in R + To show that the condition is sufficient, prove that if it is satisfied there exists a dense subset H of R such that f - ' ( [ a . i-031) is p-measurable for all tl E H, by using (13.21.10).)

.

21

PRODUCT

OF MEASURES

231

(b) Show that for f to be p-integrable it is necessary and sufficient that

D, should be v-integrable, and that we have then v(D,)

=s+a

s

=

fdp. Moreover,

if g is the decreasing real-valued function on R + defined by g ( r ) = p * ( f - ' [ t , then

if

dp

+a]),

g ( r ) dr.

(c) Suppose that X is an interval [0, a ] (where a > 0) and take p to be Lebesgue measure on X. Iffis a decreasing p-integrable function, show that

for every p-measurable subset A of X. (Use (b).) 2.

(a) With the notation of Problem 1, let I?, be the graph off, i.e., the set of all points ( x , f ( x ) ) in X x R.Iff is p-measurable, show that rf is v-negligible (use (13.21.13) and (13.21.10)). (b) Let 0 < a < 1 and let Q(a) be the subset of R2 which is the complement of the union of 1- I , 1 [ X I -a, a[and I-a, a [ x 1- 1 , I [in the squareQ = [- 1 , 1 ] x [- 1,1]. The set Q(a) is the union of its four connected components Ql(a)= [ - I ,

-a1 x [ - I ,

Qda)= 1-1,

-a1 x [a, 11, Q4(a)= [a, 11 x [ - I , -a].

-a],

Qda) = [ a , 11 x [ a , 1 I,

For i = 1, 2 , 3 , 4 let hi, denote the similitude mapping Q to Q1(a)for which

h1,=(-1, - 1 ) = ( - 1 , - I ) , h2..(-1, - I > = (-1, 4, h P I ( - l , --I)=(a,a), h4.d-1. - - 1 ) = ( 1 , -4,

h l . d , -])=(--I, -a), h Z . . ( l , - 1 ) = ( - a , a>. ha,.(l, - 1 ) = ( l , a), h4,d1, - l ) = ( l , -1).

Also, in the interval [0,71 in R, put I k = [ k, k i 11 for 0 6 k 5 6 and let uk be the increasing similitude which maps [0,7] to 1,. Let f. be the continuous mapping of [0,7] into Q which is affine-linear on each of the intervals I k and is such that the images of 0, 1, 2, ... , 7 are respectively the points(--l, - 1 ) , ( - - 1 , - - 4 , ( - 1 , a ) , ( - = , a ) , ( a ,a ) , ( l , a ) , ( l , --a),(l, -1). We shall now define a sequence (gn)of continuous mappings of [0, 71 into Q as follows. Let be a decreasing sequence of numbers belonging to the interval 10, 1 [. Define go =fa, . For n 2 1, suppose that 9.- has been defined, and consider all sequences s = (il, . . ., in)in which each ix is one of the integers 0, 1, . . , 6. Put us = uI10 . . . UI.. Then it is sufficient to define gn(us(f))for 0 5 t 5 7 and for each =g.-l(us(r)). of the 7" sequences s. If at least one of the it is odd, we put gn(us(f)) If on the other hand il = 2j, for 1 S I5 n (with 0 5 j r 2 3), we putgn(us(t)) = ws(hn(r)), where w ~ = ~ J ~ o + ~ I J, ~~ +~0 I , ~h ,~n + l , a n .Show that thesequence(g,)converges uniformly to an injecfiue confinuous mapping g : [O, 71 + R2,and that the simple arc g([O, 71) (Section 4, Appendix to Chapter IX) is nonnegligible with respect to Lebesgue measure on R2, if the sequence (an)is suitably chosen.

.

0

3. (a) Give an example of two compact spaces. X and Y , positive measures h and p on X and Y, respectively, and a h 0p-measurable function f such that the two

integrals

1

dp(y) f ( x , y ) dh(x) and

unequal (cf. Section 5.2. Problem 5).

i s

dh(x) f ( x , y ) d p ( y ) are both defined and are

Xlll INTEGRATION

232

(b) For each integer n > 0 , let A;= [2-", 3 . 2-"-'[ and A = [ 3 . 2-"-', 2-"+l[. Let BA=AAxAA, B:=A:xA;, CA=ALxA:, C:=A;xAA in R2. Define f:RZ+ R as follows: f ( x , y )= 4"+' for ( x , y ) E BA LJ B:; f(x, y ) = -P+' for ( x , y ) E CA u C:, for each integer n > 0; f ( x , y ) = 0 otherwise. Show that f is measur-

ss

ss

able and that the two integrals dy f ( x , y ) dx and dx f ( x , y ) dy are defined and equal, but thatfis not integrable with respect to Lebesgue measure on R2. 4.

Let X, Y be two locally compact spaces, A a positive measure on X and p a positive measure on Y . Let f be a mapping of X x Y into a metrizable space G, such that: ( I ) for each x E X, the mappingf(x, .) is p-measurable; (2) for each y E Y, the mapping f(.,y ) is continuous. Show that under these hypotheses f is (A 0p)-measurable. (Reduce to the case where X and Y are compact; by using Egoroff's theorem and the fact that X is metrizable, show that f is almost everywhere (with respect to A 0p) the limit of a sequence of ( A @3 measurable functions.)

5. Let X, Y be two locally compact spaces, a positive measure on X and p a positive measure on Y. Letfbe a real-valued function 2 0 on X x Y which is bounded on every compact subset of X x Y and such that: ( I ) for almost all x E X the function f ( x , .)

is p-measurable; (2) for each function h E X ( Y ) ,the function

XH

s

f ( x , y)h(y) d p ( y ) ,

which is defined almost everywhere, is A-measurable. Show that under these conditions there exists a (I\ 0p)-measurable function g such that for each x E X we have f ( x , y ) = g ( x , y ) except at the points of a p-negligible set A, (depending on x ) . (Show that, for each function U E X(X x Y), the function f ( x , . ) u ( x , .) is p-integrable for almost all x

E

X, and that the function

XH

s

f ( x , y ) u ( x , y ) d p ( y ) , defined

s s

almost everywhere, is A-integrable; for this purpose approximate u by functions of the form u(x)w(y). Then remark that the linear form

I( H

d/\(x) f ( x , y)u(x, y ) d p ( y )

is a positive measure on X x Y with base 0p,and apply the Lebesgue-Nikodym theorem. Finally use the fact that Y is a denumerable union of relatively compact open sets U, and that there exists in X ( Y )a denumerable set of functions D such that every function belonging to X ( Y ) is the uniform limit of functions belonging to D and having their supports contained in some Un.) (b) Show that the conditions of (a) are satisfied if: (1) for almost all y E Y,the function f( . ,y ) is hneasurable; (2) for almost all x E X, the function f ( x , . is continuous almost everywhere with respect to p. (Use Problem 7(c) of Section 13.9.) (c) Take X = Y = [0, 1 ] and take A and p to be Lebesgue measure. Assuming the continuum hypothesis, let x < y be a well-ordering on X for which there exists no greatest element and such that, for each x E X,the set of all z E X such that z < x is denumerable. Show that the characteristic functionfof the set of pairs ( x , y ) for which x

< y satisfies the conditions of (a), but that

jf(x, y ) +(y)

=

1 for all x

E

y ) dh(x) = 0 for all y

E

Y and

X.

6 . Let u, u be two increasing real-valued functions on R, each continuous on the right, and such that u(x) = u(x) = 0 for x < 0. Let w be the increasing function on R, continuous on the right, defined by w ( f )= u(t)u(t) for t 2 0 and w ( t ) = 0 for f < 0. Let A, p,v be the Stieltjes measures associated with u, u , w , respectively (Section 13.18, Problem 6).

21 PRODUCT OF MEASURES

233

To each real-valued function f o n R we associate the function fo on R2defined by fo(x, y ) = f ( x ) if y < x , f o ( x ,y ) =f(y) if y 2 x . Show that f i s v-integrable if and only

I SIf

if fo is ( h 0p)-integrable, and that in this case f d v =

dh d p . (Prove the result

first for characteristic functions of intervals.) Hence deduce the formula

If(4M x ) =I f ( X ) v ( x - 1du(x) + S f ( x ) u ( x + 14 x 1 . In particular, if u and u are contintrous on by purrs :

R,we have the formula of integration

IObu(x)du(x) = u(b)u(b)- u(a)u(u) - s b u ( x ) du(x). II

Consider the case where u and v are constant on each interval [n,n integer n 2 0 ("Abel's partial summation formula").

+ 1 [ for each

7. Let p be a finite real number 2 I , let X and Y be locally compact spaces, h a positive measure on X and p a positive measure on Y . Let 0 be a function on X x Y such that f and f p are ( h 0p)-integrable. Prove that

fz

( [ * ( J f d p ) ' dh)'"

5

J*(Ifp

dh)'" d p .

(For each X E X ,apply Holder's inequality (13.11, Problem 12(a)) to the function

y ~ f ( xy), , in the forrnf(x, y ) = g ( x , y ) conjugate t o p . ) 8.

l/Prl

f p ( x ,y ) dh(x))

, where 9 is the exponent

Let X I(1 2 i 5 n ) be locally compact spaces and pi a positive measure on X i for 1 5 i 5 n. For each i , let E, denote the product XI. Let fi be a function 2 0 which

"

is measurable with respect to p = @ p I on X i=I

n n X i , and does not depend on

J+i n = I=I

f;-' is integrable with respect to the measure @ p,, for 1 5 k function

5 n, show

xi. If

that the

J+k

nfiis p-integrable and that n

I= 1

where J x =

S. . . I

f;- dpl

'

. .d p k -

+&+

.. . d p . for 1 5 k 5 n. (By induction on

n , using the Lebesgue-Fubini theorem and Holder's inequality.) Deduce that if A is a p-measurable subset of X, and Ai its projection on E l , and if Al is integrable and of measure m,with respect to the measure @ pJ on El, then A J#i

is p-integrable and p(A) 2 ( m l. . . mn)'/("-'). Generalize to the case in which, instead of considering the products of the XI n - 1 at a time, we consider the (g) products of the X Ip a t a time, and integrateoverx a product of (;) functions 20, each of which depends on only p of the variables X I , . . . , X" .

Xlll

234

INTEGRATION

9. Let (X,),,, be an infinite sequence of compact spaces, and let pa be a positive measure on X, with total mass 1 , for each n.

n X, there exists a unique positive measure m

(a) Show that on X =

of functions f;E %',,(XI),we have

each integer n and each finite sequence ( f i ) 0 6

p(f) =

p such that, for

"=O

n p,(fi), where

I=O

f ( x ) = n f , ( p r i x ) . (Observe that the continuous functions 1=0

of the form n f i ( p r i x ) , for all choices of n and off, E ~ R ( X , )form , a total set in the 1=0

Banach space 'ZR(X).)The measure p is called the product of the family (p.) and is

"

written p = @ p.. *=O

(b) For each n, let A. be a p,-measurable subset of X,. Show that the product

n m

A=

"=O

A. is p-measurable, and that p(A) =

m

"=O

p.(An).

n

(c) Letf? 0 be a p-integrable function on X . For each subset L of N, put L' = N - L, and identify X with the product XL x X,, , where XL= X. and X,. = X,. If x

"€L

E

n

E

L'

X, put xL =: prLx and x L p= prL,x, and identify x with (xL,xL.). Finally, let pL,

J

0 , p, on XL,, and putfL(x) = ~ ( x LXL,) , dpLt (x,~).

be the product measure

"EL

Now let (L.) be an increasing sequence of subsets of N, and put g = supf,, h = supf,,. . For each c > 0, let A, be the set of points x

E

X at which g(x) > c,

and B, the set of points x E X at which h(x) > c. Show that c . p(A,) c

and

5

i

f dp and

. p(B,) 5 s f d p . (Remark that A, is the set of points x E X for which at least one

$IG"

of thefln(x) is >c, and express A, as a denumerable union of mutually disjoint sets G . such that c p(G.)

fdp.)

(d) Suppose that (L.) is an increasing sequence of finite subsets of N whose union is N. Show that the fLn tends to f almost everywhere and that f ~ tends " to the constant

[fdp almost everywhere. (For each E > 0, consider a continuous function g, depending on only finitely many variables, and such that to the function

If-

gl.)

s

If-gI

dp

s . ~and, apply (c)

On each factor X. ( :D) 10. Let D be the discrete space (0, 1) and X the product space DN. of X let p,, be the measure for which pn(0)= p.(l) = 1, and let p = @ pn be the n e N

product measure on X (Problem 9). (a) For each point x = (x,),,,

in X with x,

=0

or x.

=

I , put T(X) =:

Lo

C ~.2-"-~.

"=O

Show that T is a continuous mapping of X onto the unit interval I [C, I ] in R, and that the image n(p) of p is Lebesgue measure on I . For each t E I, the set n - ' ( t ) consists of a single point, unless I is of the form k . 2-" with k integral and 0 < k < 2". Hence the mapping f w / o T induces, on passing to the quotients, an isometric isomorphism of LE(I,x) onto LE(X, p) for 1 2 p 5 I a. (b) For each t E 1 and each n 2 I , put r,(t) = I - 2pr,- , ( r - ' ( t )if) f is not of the form k . 2-" with k integral and 0 k 5 2". and r.(t) =: 0 otherwise. The function rnis called the nth Radeniacherfitncfion.These functions form a nontotal orthonormal 7

21

PRODUCT OF MEASURES

235

system in 9&(1, A) (show that they are all orthogonal to the functions cos 2kn-r for k integral and 2 0 ) . Show that r,(t) = sgn(sin 2 ” d ) . Show that the measure p is (c) For each x = (xJns0in X, put u ( x ) = (x.+ invariant with respect to u and that u is ergodic with respect to p (use (a) and Problem 5(c) of Section 13.12). Deduce that lim

n-tm

I n

- ( r l ( t )i . .

+ r.(t)) = 0

almost everywhere with respect to Lebesgue measure (“ Borel-Cantelli theorem ”). (Observe that r.(l - t ) = -u,(t) for 0 5 t 5 4.) (d) For each finite strictly increasing sequence (6), s f s k of integers >= 1, show that

jolr n l ( t ) r n z ( f.). . r m k ( fdt) (e) For each finite sequence > 0, show that

bxbn

=

0.

of complex numbers, and each real number

p

(“Khintchine’s inequality”). (First consider the case where p =: 2h, with h an integer 2 1, by using (d). For 2h - 2 < p 5 2h, use problem 12(e) of Section 13.11.) (f) With the notation of (e), show that

(use (e) for p

=

4,and Holder’s inequality for p

=

3 and q = 3).

11. (a) The hypotheses are those of Section 13.17, Problem 2, and in addition the functions f. are assumed to be real. Let XH~(X)be a p-measurable mapping of X into the set I = { I , 2, . . . , n}, and let k~ w ( k ) be an increasing function on I with values >O. Then the mapping (s, u)t+KJ(.)(s, u ) / w ( j ( s ) ) of X x X into R is p 0p-

measurable. For each p-measurable subset A of X, show that

where h(s, t ) = inf(j(s),j(r)), (Express the square of the integral on the left as a double integral, so as to reduce the problem to the evaluation of a triple integral over A x A x X, and use the fact that the j,are orthogonal.) Show next that

(split up A x A into two measurable sets, defined respectively by j ( s ) <j ( r ) and for (s, t ) E A x A and use the fact that K.(s, t ) = K.(t, s)). (b) Suppose that there exists an increasing sequence n Hw(n) of numbers > O and a p-measurable subset A of X such that IH.(s)/ 5 c * w(n) for all s E A and all n 2 1, where c > 0 is a constant. Show that, for each function g E 9 i ( X , p), the sequence (s.(g)(r))/w(n) is bounded for almost all t E A. (Consider the increasing sequence of p-integrable functions u.(t) = sup (sk(g)(t))/w(k), and show that the sequence of

j(s) >j(t)

lbkBn

236

Xlll

INTEGRATION

=jA u.(t) + ( t )

integrals J.

is bounded above. For this purpose, note that we may

write u,(t) = ( s j c , , ( g ) ( t ) ) / w ( j ( r ) ) for a suitably chosen p-measurable mapping j of X into I, and majorize J,' by using the Cauchy-Schwarz inequality and (a) above.) (c) Suppose that the hypotheses of (b) are satisfied and also that lim w h ) = f a . If the a. are real and if the series

c m

n-m

a.' w2(n)converges, show that the series

n=l

c a.f.(t) m

"21

converges almost everywhere in A. (Start by using Problem 8(b) of Section 13.11. Then, in order to majorize Is.(t) - s.,(t)l for nk I n < n k + l ,determine an increasing sequence (c,) of numbers >O such that lim c.= n-m

+ a and

m

10.2 w'(n)c,' < +co

"= 1

).fI

(Section 5.3, Problem 6), so that the numbers b. = a. w(n)c. are of the form (9 for some g E Yi(X, p). Then use the fact that, by virtue of (b), the partial sums

Ik1:, I b&t)

are bounded for almost all t E A, and apply Abel's partial summation

formula.)

(d) Deduce from (c) that if IH.(s)l $ c for all s E A and all n, and if the series converges, then the series

m

m

a.'

"= 1

a.f.(t) converges almost everywhere in A. (Use Problem

I= 1

6 of Section 5.3 again.) 12. (a) Let (c.) be a sequence of real numbers such that

series

m

;I:e,' < fa.Show that

the

I =1

m

c. r.(t)

(the notation is that of Problem 10) converges almost everywhere in

I=1

I with respect to Lebesgue measure h. (Express the Rademacher functions as linear combinations of the functions of the Haar orthonrmal system (Section 8.7, Problem 7), and note that Problem 1 Id) is applicable to the latter orthonormal system (cf. Section 13.17, Problem 2)) (b) Given any sequence (a") of real numbers, a real E > 0 and a measurable set A c I, show that there exists an integer no such that

whenever rn > n 2 n o . (Use the Cauchy-Schwarz inequality, the fact that the functions r l ( t ) r j ( t ) with ic j form an orthonormal system (Problem 10(d)) and Bessel's inequality applied to this system and to the function (c) Let (amn)be a double sequence of nonnegative numbers such that for each m the set of integers n 2 1 for which anla# 0 is finite, and such that lim am"= 1 for each m-m

n. Also let ( c ) be a sequence of complex numbers such that, if we put

the sequence (S,"(t)) is convergent in an integrable set A c I of positive measure.

c m

Show that

n=

lc.I2

< + m . (By virtue of Egoroff's theorem, there exists an integer

1

no and a set B C A of positive measure, such that for all rn

2 no and q > p 2 n o , we

21

have

1

a m k c k r k ( t5 ) 1 for

all t

E

OF MEASURES

PRODUCT

237

B, and therefore

Minorize this integral with the help of (b).) (Rademacher-Kolmogoroff theorem.)

X be a locally compact space, p a positive measure on X, and (uJnao a sequence of p-integrable complex functions such that, as H runs through the set of finite subsets of

13. Let

N, the set of numbers m

I

JIZ"u.(x)

dp(x) is bounded. Show that in these conditions the

x lun(x)lz is convergent almost everywhere in X. (Observe that, for all

series

Klc

n=D

I

ak(x)rk(t) dp(x) is

t E I = [0, 11, the set of numbers

bounded above by a number

independent of n and t ; then use Problem 10(f) and the Lebesgue-Fubini theorem.)

X be a compact space, p a positive measure on X, and (f,) an orthonormal sequence in (t&(X,p) which is uniformly bounded in X.

14. Let

(a) Let (6.) be a sequence of real numbers such that not possible that

x Ib,(g If,)l

m

2 b,2 = + 03.

"=l

Show that it is

< + 03 for all continuous functions g on X.(Argue by

m

"=I

contradiction: the functions u.(x) = b,fn(x) would satisfy the conditions of Problem 13, by virtue of the Ranach-Steinhaus theorem applied to the linear forms g

Hz b.(g If,) "EH

on V(X). Hence

zb;lfn(x)Iz < + m

*=I

co almost everywhere in X.

Using Egoroff's theorem, the fact that the functions f. are uniformly bounded and

If,(x)1' dp(x) = 1, obtain a contradiction of the hypothesis

5 b.2

"= 1

=

+

03.)

(b) Let 9 be a real number such that 1 < 9 < 2. Show that it is not possible that m

I(g

"=l

1f J q

< +a, for

all

continuous functions g on X.(If p

= 949 -

I), show that

if the assertion were false there would exist a sequence (b.) of real numbers > 0 such that b.2 = co, 6,"< a and C I b,(g Ifn)l < m for all continuous

x n

+

c

+

+

n

I

functi0ns.g.) (c) Deduce from (b) that there exists a continuous function g on X such that, for all 9 satisfying I < q < 2, we h a v e x I(g ljJq = +a.(Use the principle of condensation

"

of singularities (Section 12.16, Problem 14).) 15. On the space R",let h be Lebesgue measure and let ljxll be a norm for which the unit ball llxll 6 1 has measure 1. Let ( x k k a l be an infinite sequence of distinct points in an integrable bounded set B such that h(B) = 1. For each integer m > 1, let d,,, denote the smallest of the numbers jlx, - xjll with 1 6 i < j 5 m. Show that lim inf m d,,,5 c; where m+ m

l(1-

t)"

dt

238 XI11 INTEGRATION

(cf. Section 12.7, Problem 6). (Argue by contradiction, and assume that for some E > 0 there exists mo such that md," > h: for all m 2 m o , where h: = c;' 4E . For 1 i 2 m,let B, be the ball with center x f and radius )h. m-"",and for m 2 i 6 2"m let B, be the ball with center x i and radius )h.(2i-"" - m-""). Show that the 2"m balls B, are pairwise disjoint, and calculate the measure of their union by the EulerMacLaurin summation formula.) 16. Let X, Y be two locally compact spaces and A a universally measurable subset of x x Y. (a) For each x E X, show that the section A(x) of A at x is a universally measurable subset of Y. Also, for each measure p 2 0 on Y , the function x ~ p * ( A ( x )is) universally measurable (use the Lebesgue-Fubini theorem). (b) If Y is compact and if A is closed in X x Y, then the function x-p*(A(x>> is upper semicontinuous. (c) Let p be a positive measure on Y such that the section A-'(y) is denumerable for almost all y E Y (with respect to p ) . Show that the set N of points x E X for which p*(A(x)) > 0 contains no nondenumerable compact subset. (By using Section 3.9, Problem 4, reduce to the case where this compact set contains no isolated point, and show that such a set is the support of a diffuse measure #0, by using Section 13.18, Problem 6(b) and Section 4.2, Problem 3(b).) 17. Let p be a bounded positive measure on X, with total mass 1 . I f f € 9; is such that

/log11

+ Cfl

d p 5 0 for every complex number (, then f is p-negligible. (Use the

formula

(which is valid for all

5 E C) together

with the Lebesgue-Fubini theorem, to evaluate

the integral jlog+lRfl d p , where R > 0, and deduce that this integral must be zero.)

..., m - 1}. Let X denote the compact product space Iz, and p a positive measure on I with total mass 1 (so that p is defined by the

18, Let I denote the discrete space {O, 1,

finite sequence of masses p ,

=

p ( { j } )such that p , 2 0 and

X, of X let p" denote the measure p , and let

Y

m-1

I=o

p,

=

1). On each factor

be the product measure @ p,, on X n € Z

(Problem 9). (a) For each x = ( x ) " ~in~X , put u(x) = ( x . + ~ ) Show ~ ~ ~ that . u is a homeomorphism of X onto X and that the measure Y is invariant under u. The triple (X,Y , u) is called the Bernoulli scheme B ( p o , .., p a (b) Consider in particular the Bernoulli scheme B(),i). For each x = ( x J n G ZE X , where x. = 0 or 1 for all n E Z, letf(x) denote the canonical image in TZof the point

.

( y , z ) E R1,where y =

m

"-0

x - n 2 - " - 1 and z =

Q

n=l

xn2-".Let

fl be the normalized

Haar

measure (14.3) on the compact group T2.Show thatfis continuous and that f(v) = !?t; also that the set of points t E T2such that f - l ( r ) does not consist of a single point is fi -negligible.

21

PRODUCT OF MEASURES

239

Let v be the restriction of the canonical mapping T : R2+ T 2to the set K of points ( x , y ) such that 0 2 x < 1 and 0 y < 1, so that p is a continuous bijection of K onto TZ.For t E T2and ( x , y ) = v-I(t),put

Show that the mapping u : T2 + T2(“baker’s transformation”) is continuous almost everywhere and that u o f = f o u (where u is the mapping defined in (a)). We shall see later that u and 2) are ergodic (relative to v and p, respectively) (Section 15.1 1, Problem 16). (c) Show that for the Bernoulli scheme B ( p , , . . . ,p b - , ) , the entropy h(u) (Section 13.9, Problem 28) is given by h(u) ==

+

IogpO

-(PO

“’

f p k - 1 logpk-1).

(Use the Kolmogoroff-Sinai theorem, by starting with the partition cc = ( A j ) o s j q k - l , where A, is the set of (x.),.~ such that xo j . Use Problem 9(a) and the definition of u, and remark that the set :

Ajon u - ’ ( A J 1 n ) . . . n U P +(AJII-1) ’

is the set of ( x n ) n s zsuch that x,, =j o , x , calculate the sum

=j , ,

.. . , x,- I

=

The problem is then to

For this purpose, observe that

Deduce from this calculation that the bijections u of the Bernoulli schemes B(1, 8) and B(i, 1, a) are not conjugate (Section 13.12, Problem 11). 19. Let X be a locally compact space, let p be a positive measure on X, and let A g be two nonnegative p-measurable functions on X. For each cc > 0, let A, be the set of x E X such that f ( x ) > u . Suppose that: (i) p(A,) < 03 for all u > 0; (ii) g E =Y;(X, p) for some p E ]I, m[;

+

(iii) for each cc > 0, p(A.) 5

U

1’

+

g dp.

A.

Show that these conditions imply that f~ =Yg(X, p) and that

(Wiener’s inequality). (Consider first the case where f is bounded and of compact support, and integrate from 0 to co the inequality

t P - ’ j V + t tdp 5

tp-’!gP)At

dp

240

Xlll INTEGRATION

(cf. Problem 1). For the general case, consider the functions ~ u p ( n , f g ) ~ ~where ), (K.) is an increasing sequence of compact sets whose union is X.)

20. Let U be a continuous endomorphism of the space L:(X, p) satisfying the hypotheses of Section 13.1 1, Problem 17, so that U extends to an endomorphism of each of the spaces Le(X, p), 1 < p 5 CD (loc. cit.). For each functionfe 9;,put

+

(Use Problem 19 above, and Section 13.11, Problem 17(c).) (b) For each functionf E =Y;, let P be the limit in L: of the sequence ((R.(f))”) (Section 12.15, Problem 12(c)). We have U P = P U = P . Show that the sequence (R.(f)) converges almost everywhere to P .f (Dunford-Schwartz ergodic theorem). (Put

S(f) = lim SUP RAf), n-m I(f)

= lim

inf R.(f).

n- m

Observe that I(f) 5 S(f) 5 R*(f), and that S(R,(f) - P .f)= S(f) - P .f and I(R,(f) - P -f)= I(f) - P .ffor all rn 2 1, and use (a) with R,(f) - P * f i n place of f.) (c) Show that N I ( P .f)5 N l ( f ) for all f~ 9; n Y;, and hence that P extends to a contraction on the space LA. Deduce that the sequence (R.(f)) converges almost everywhere to f, for each function f~ 2:.(Put L(f) observe that L(f) R*(f- g) Section 13.11, Problem 7(d).)

I R A f ) - P .fl,

= lim SUP n-m

+ P . If-

g

1

for all g

E 9: n 9

g , and make use of

21. The notation is that of Section 13.17, Problem 7. For each t~ > 0 and each function g E 9P:o,,c(Y, v), let A&) denote the set of points y E Y such that ig(y)/ > a. If p E [I, +a[,the endomorphism U is said to be of weak type (p,p) if there exists a constant C > 0 such that, for each p-integrable step function f and each CL > 0, we have

21

PRODUCT OF MEASURES

+

241

Let p, q be such that 1 z p < q < w . Show that, if U is of weak type ( p , p ) and also of weak type (q, q), then U is of type (r, r) (Section 13.17, Problem 7) for all r such that p < r < q (Marcinkiewicz's interpolation theorem). (For each f~ 9A(X, p) and a > 0, let f: be the function which is equal to f ( x ) if If(x) I > a, and zero otherwise, and put f = f -f:. Show that, for each t > 0, t'-'p(Azt(U .f))5 C't'-"' Integrate from 0 to

00

1 If:lp

and use Problem 1.)

dp

+ C"t'-@-'

CHAPTER X I V

INTEGRATION IN LOCALLY COMPACT GROUPS

Haar measure and convolution on arbitrary locally compact groups have become indispensable tools for the modern analyst, as they were in classical analysis on the real line and in finite-dimensional Euclidean spaces. Together with convolution of distributions, which generalizes convolution of measures and which we shall introduce in Chapter XVII, they are the fundamental notions in harmonic analysis (Chapter XXII) and the theory of linear representations of compact groups (Chapter XXI). We have again followed, though in less detail, the exposition of N. Bourbaki [22]. I n fact, since in this treatise practically the only locally compact groups we shall consider will be Liegroups (Chapters XVI, XIX, and XXl), for which there is a much simpler proof of the existence of a Haar measure, we could have restricted ourselves entirely to Lie groups. Nevertheless, it seemed worthwhile to bring out the fact that the theory of integration on a locally compact group is entirely independent of any differential structure; and the totally discontinuous locally compact groups have ceased to be mere curiosities since the advent of p-adic and “adelic” groups in the theory of numbers [36]. Throughout this chapter, the phrase L L locally compact group separable metrizable locally compact group.”

”

will mean

1. E X I S T E N C E A N D U N I Q U E N E S S O F H A A R M E A S U R E

Let G be a (separable, metrizable) locally compact group. Iff is any mapping of G into a set E and if s is any element of G, we define the left and right translations y(s)fand 6(s)foffby s, which are mappings of G into E, by 242

1 EXISTENCE AND UNIQUENESS OF HAAR MEASURE

243

the formulas

It follows immediately from this definition that we have

for all s, t E G . If p is a (complex) measure on G, we denote by -y(s)pand S(s)p the measures on G which are the images of p under the homeomorphisms X H S X and x w x s - ' , respectively (13.1.6), so that (14.1.2)

(f,Y ( ~ ) P = ) ( ~ ( s - ' ) fp), ,

(f,WS)P)

= (W-')f, r ~ )

for all functions f E .fX,(G).From this definition it follows that

for all s, t E G. The measure p is said to be /eft (resp. right) inuuriant if Y(4P = P (resp. W

(14.1.2.2)

P = PI

for all s E G. If a measure p # 0 on G is left invariant, then Supp(p) = G, because Supp(y(s)p) = s Supp(p) for all s E G by virtue of (1 3.19.4), and Supp(p) # 0. Similarly for right-invariant measures. Let p be a left-invariant measure on G and letfbe a p-integrable mapping of G into R or C. Then for each ~ E the G function X H ~ ( S - ' X ) is also p-integrable, and we have

-

(14.1.2.3)

J W X )

44x) =

s

f(x)44x)

by (13.7.10). In particular, if A is a p-integrable set, then so is sA, and

Iffis any mapping of G into a set E, we write (14.1.3)

f(x) =f(x-')

for all x E G.

244

XIV

INTEGRATION

IN LOCALLY COMPACT GROUPS

If p is any measure on G, we denote by ji the image of p under the mapping X H X - ' of G onto G, so that we have

F> ( J P>

(14.1.4)

(f, =

for all f E Xc(G).It follows immediately from the definitions that we have (y(s) f ) - = 6(s)f, each side being the function X H f ( s - ' x - ' ) ; and therefore, for any measure p on G, we have (y(s)p)" = S(s)fi. Hence if p is a leftinvariant measure, ,ilis a right-invariant measure, and vice versa. (14.1.5) Let G be a locally compact group. Then there exists a nonzero leftinvariant positive measure p on G , and every other left-invariant measure on G is of the form ap, where a E C. (1) Existence. Let Xx*,denote the set of functions g E XR(G) which are 2 0 and not the zero function. For each f E XR(G) and each g E .X: , there exist positive real numbers cl, . . . , c, and elements sl, . . . ,s, in G such that

(14.1.5.1)

(Le., such that f ( x ) S

r

1 c,g(s;'x)

for all x

E G).

For there exists a non-

i=1

empty open subset U in G such that a = infg(x) > 0; since Supp(f) is comXOU

pact, there exist a finite number of points si E G (1 5 i 5 r ) such that the s,U cover S u p p ( f ) , and then (14.1.5.1) is satisfied by taking ci = llfll / a for all i. We shall denote by ( f :g) the greatest lower bound of the numbers i=1

c i , for all systems (cl,

. .. , c,,

sl,

. . . ,s,) satisfying (14.1.5.1).

The symbol

(f:g) has the following properties:

(0 ( W f g:) = (f:9)

for all f E XR(G), g E S ' f , S E G ; (ii) ( a f :g) = a(f:g) for all f ' XR(G), ~ g E X*, , a 2 0 ; (iii) (fl+ f 2 :9) 5 (fl :g) + ( f 2 : g ) for h,f2 S'R(G), g X*,; (iv) (f:g) 2 supf(x)/sup g(x) for all f~ XR(G), g E X : ; xsG

x s 0

.

for all S , f o , g in .X*,

1 EXISTENCE AND UNIQUENESS OF HAAR MEASURE

245

Properties (i), (ii), and (iii) are immediate consequences of the definitions. Given (14.1.5.1), there exists (3.17.10) s E G such that

and (iv) follows. For (v), observe that if we have g5

c bjy(rj)h,then f 5 1a i b j i

i, i

fr Caiy(si)g

and

y(si tj)h, and therefore

(f:h)sCaibj= i, i

(T

a,

Chi.

l ( j

1

Since we may take c a i (resp. c b j ) arbitrarily close to ( f : g) (resp. (g :A)), i

i

we have (v). Finally, (vi) follows from (v) applied to fo,f, g and to f, j b , g. By hypothesis, there exists a denumerabli fundamental system (V,) of neighborhoods of the neutral element e in G. For each n, let g, be a function belonging to X*, such that Supp(gn)cV, (4.5.2). Let fo be a function in X * ,, fixed once and for all, and put (14.1.5.2)

In(f)

= (f: gn>/(fo: gn)

for ali f~ X*,; let I,(O) = 0. From (ii) and (iii) above it follows immediately that the mappings f-I,( If/) are seminorms on XX,(G),and from (i) that I,(y(s)f) = I,(f) for all s E G. Next, there exists a sequence of relatively compact open sets Up which cover G and are such that 0, c (3.18.3). The space %(Up)is separable (7.4.4) and therefore so is X ( G ; 0,) n X r (3.10.9). Hence there exists a dense sequence (fmp),,,21 in the latter space. Since the sequence of values In(fmp) (n 2 1) lies in the compact interval with endpoints l/(fo : f m p ) and (fmp :fo)by (vi), it follows from (12.5.9) that if we replace the sequence (9,) by a suitably chosen subsequence, we may assume that the sequence (In(fmp)),,2 tends to a limit > O for all m,p . Also it is clear that iff, f ’ in X,* are such that then I,(f) 0 be a real number, and let h be a function 5 0 belonging to XR(G) such that h(x) 2 1 on the union of the (compact) supports offandf’ ((3.18.2) and (4.5.2)). Then it is enough to prove that there exists a compact neighborhood V of e in G, such that, for all g E %*, with Supp(g) c V, we have

(f:g ) + (f‘:g) 6 (f+f’: 9) + 4 h : g).

(1 4.1.5.4)

To prove this, put u = f + f ’ + + E / I , and let u (resp. u’) be the function which coincides withflu (resp. f ’ / u ) on Supp(f+f’) and is zero on the complement of this set. Since at every frontier point x of Supp(f+f’) we have f ( x ) + f ’ ( x ) = 0, and therefore f ( x ) = f ’ ( x ) = 0, it follows that u and u’ belong to XR(G) and are 2 0 , The functions u and 11’ are therefore uniformly continuous with respect to a left-invariant distance defining the topology of G ((12.9.1) and (3.16.5)), and therefore, for each q > 0, there exists a compact neighborhood V of e such that Iu(s) - u(t)l 6 q and Iu’(s) - u’(r)l 6 q for all pairs (s, t ) such that s - ’ t E V. Now let g EX: be such that Supp(g) c V. For each s E G we have u y(s)g 6 (u(s) q) y(s)g. (This is obviously true at points where y(s)g vanishes, hence at points outside sV;and if r E sV we have u ( t ) u(s) q.) Similarly, we have D‘. y(s)g 6 (u’(s) q)y(s)g. This being so, let ci (1 5 i 511)be real numbers 2 0 , and si (1 5 i 5 n) elements of G such

-

+

1

+

that u 6

+

c ciy(si)g.Then we have n

i= 1

n

n

and a similar set of inequalities withfreplaced byf’. Consequently

c n

(f : 9) + (f’: 9) 5 i = 1C

i ( W

+

+ 2rl) 6 (1 + 277) c c i , n

“’(Si)

i= 1

because u + u’ 5 I . From the definition of u, and properties (ii), (iii), and (v), we obtain

(f:

g)

+ (f’:8) 5 (1 + 2q)(u: g) 5 (1 + 2 q ) ( ( f + f ’ : g) + +E(h :9)) 6 ( f + f ’ : g) + t 4 h : 9)+ 2 t l ( ( f + f ’ ) : h)(h : g) + El](h : g)

so that, by taking q such that

q(2((f +f ’ ) : h) + E ) 5 38, we get (14.1.5.4)

1 EXISTENCE AND UNIQUENESS OF HAAR MEASURE

247

Now extend I to the whole of X X , ( Gby ) putting I(0) = 0 and, for each -f2 with fI, f i in if:, I(f) = I(fi) - I(fi). From (14.1.5.3) it follows immediately that I(f) depends only o n f a n d not on the choice of expression f = f i - f2. Clearly (14.1.5.3) remains valid for all J; f' in X R ( G ) Further. more, the original definition of I shows immediately that I(If) = I I ( f ) for all f E XT and all real numbers I > 0; and this relation obviously extends to the case where f~ XR(G) and I E R. We may therefore conclude from (13.3.1) that I is a positive measure on G ; also I is not zero and, by construction, I(y(s)f) = I(f) for all f e X R ( G ) . In other words, we have constructed a nonzero left-invariant positive measure on G.

f=fi

(2) Uniqueness. Let p (resp. v) be a nonzero left (resp. right) invariant measure on G. Then 5 is a left-invariant measure. We shall show that p and 5 are proportional, and this will complete the proof of (14.1.5). Let f~ X,(G) be such that p ( f ) # 0, and consider the function D, on G defined by

D,(s> = pc(f>- J f ( t - 1 r ) We shall show that D, is continuous on G. In fact this will be a consequence of the following more general result: (1 4.1.5.5) Ler G be a locally compact group, H a closed subgroup, a a measure on H, and f : G -+C a continuous mapping. Suppose that either Supp(f) is compact or Supp(a) is compact. Then the mappings SH

s

f(st)da(t)

and

SH

are continuous on G .

s

f(ts)da(t)

Consider, for example, the first of these integrals. Let so E G and let V, be a compact neighborhood of so. Given E > 0, we have to find a neighborhood V c V o of so such that for all s E V we have f ( s t ) - f ( s o 1 ) ) d a ( t ) )S E . If K = Supp(f) is compact and L = VG'K, then

I/(

n

, l

since f is uniformly continuous (with respect to a right-invariant distance on G) (3.16.5), there exists a neighborhood W of e in G such that the relation s E Ws, implies I f ( s t ) -f ( s o t)l &/IaI(L)for all t E G, and we may take V = V, n Ws, . If on the other hand S = Supp(a) is compact, we have

248

XIV

INTEGRATION IN LOCALLY COMPACT GROUPS

j ( f ( s t ) - f(so 0 )Wt)= (f(st) - f ( s 0 0) d 4 0 , 1s

and if r E S and s E V,, then st E V, S , which is compact. The restriction f l Vo S is uniformly continuous and therefore we may take W so that the relation s E Ws, implies If(sf)- f ( s o f ) l S E / Ial(S) for all t E S ; as before, we take V = V, n Ws,. Now let g be any function belonging to X,(G). Then the function (s, t ) H f ( s ) g ( t s ) is continuous on G x G and has compact support. By (13.21.7) and the left-invariance of p, we have

and since p ( f ) # 0 by hypothesis, it follows that vtg) = AD,

- s).

This shows first of all that D, does not depend onA for iff’ is another function in X,-(G) such that p ( f ’ ) # 0, then it follows from above that D, p = D,. p, and hence (13.15.3) that D, and D,. are equal almost everywhere with respect to p. But because p # 0 and is invariant, its support is the whole of G; also we have seen that D, and D,, are continuous on G; hence the set of s E G at which D,(s) # Df@) is open and negligible, therefore empty. In other words, D, = Df, = D, say. We have therefore, by the definition of the function D,

-

PL(f)W) = $ ( f ) for every f~ X,(G) such that p ( f ) # 0. This formula, being true in the complement of a hyperplane in Xc(G),is true in the whole of Xc(G),since both sides are linear forms in J Since v # 0 we have D(e) # 0; therefore p and t are proportional, and the proof of (14.1.5) is complete. Q.E.D. Any left (resp. right) Invariant positive nonzero measure on G is called a feft (resp. right) Haar measure on G . From (14.1.5), any two left (resp. right) Haar measures on G are proportional.

2 PARTICULAR CASES AND EXAMPLES

249

PROBLEMS

Let G be a locally compact group, A a dense subset of G, p a left Haar measure on G, and H a p-measurable subset of G with the following property: for each s E A, the sets s H n CH and H n CsH are p-negligible. Show that either H or its complement is negligible (prove that the measure P)" . p is left-invariant). Let G be a locally compact group, p a left Haar measure on G, and A, B subsets of G. (a) Suppose that one of the following two conditions is satisfied: (a)A is p-integrable; (p) p*(A) < co,and B is p-measurable. Show that in either case the function f ( s ) = p*(sA n B) is uniformly continuous on G with respect to a right-invariant distance on G. (For any two subsets M, N of G, put

+

p(M, N) = p*((M n CN) u (N n CM)).

Consider first the case where A is compact, and show that for each E > 0 there exists a neighborhood U of e in G such that p(sA n B, stA n B) 5 E for all s E G and all t E U. Then apply Problem 5 of Section 13.9. If B is p-measurable and p*(A) < co, observe that there exists a decreasing sequence (A,) of p-integrable subsets of G containing A and such that inf(p(A,)) = p(A), and show that

+

p*(sA n B) = inf(p*(sA. n B)) (Section 13.9, Problem 2(a)). Use the fact that p(A. - Am+')tends to 0 with I/a) (b) If A is p-integrable and p*(B) < co, then the function f is also uniformly continuous with respect to a left-invariant distance on G. If also A-' is p-integrable, then

+

lo

f(s) ds = p(A-')p*(B). (Reduce to the case where B is pintegrable, and

observe that in that case p(sA n B) = p(A n s-'B), and that V,A B = vSApnand VsA(t) = P E A - 1(s).) (c) Deduce from (a) that in the two cases considered there, the interiors of AB and BA are nonempty if A and B are not p-negligible. (d) In the group G = SL2(R), give an example of a compact set A and a p-measurable set B such that the functionf(s) = p(sA n B) is not uniformly continuous with respect to a left-invariant measure on G. (Observe that there exists a sequence (t.) of elements of G tending to e and a sequence (s.) of elements of G such that the sequence s; It. s. tends to the point at infinity.) Let G be a locally compact group, p a left Haar measure on G, and A an integrable subset of G such that p(A) > 0. Show that the set H(A) of elements s E G such that p(A) = p(A n sA) is a compact group. (Use Problem 2 to show that H(A) is closed in G. To show that H(A) is compact, consider a compact subset B of A such that p(B) > &(A), and prove that H(A) c BB-I.) Let G be a commutative locally compact group, written additively. Let p be a Haar measure on G and let A, B be two integrable subsets of G. (a) For each s E G let B = T,(A, B) = (A - s)nB. A' = u,(A, B) = A u (B s),

+ +

Show that p(A) + p(B') = p(A) p(B) and that A A + 4 = 4 for all subsets A of G.)

+ B'

C

A

+ B.

(Note that

250

XIV

INTEGRATION IN LOCALLY COMPACT GROUPS

(b) Suppose that 0 E A n B. A pair (A', B') of integrable subsets of G is said to be derived from (A, B) if there exists a sequence ( s ~ ) , $ x Qof~ elements of G and two Seand ( B k ) O Q k of Q nsubsets of G such that quences (Ak)oQkan Ao=A,

Bo-B,

Ax- Osk(Ak-1,%-1),

Bk-Tsk(Ak-1,Bk-l)

for 1 5 k 5 17, and sk E A x - ] for 1 5 k 5 n , and A' =~A,, B' = B,. Show that there exists a sequence (En, F.) of pairs of subsets of G such that (i) Eo = A and Fo B, (ii) ( E n + , ,F n + l )is derived from ( E n ,F,J, ( i i i ) p((E, - s) n F,) 2 p(F"+])- 2-" for En,F, = F, . Show that for each s E E, we have all n and all s E En. Let E, 7

u

n

p((E* - s) n Fa,)= p(F&). (c) Suppose that p(F,) =. 0. Show that the function

f ( s ) = p((E,.

- .P)

nF d

takes only the values 0 and p(Fn,). Let C be the set of elements s E G such that f ( s ) = p(F,). Show that C is open and closed, that p ( C ) == p(Em,)and that C is the closure of E m (use Problems 2(a) and 2(b)). Let D be the set of all s E F, such that the intersection of F, with every neighborhood of s has measure > O . Show that p(D) = p(Fm)and that E, + D c C. Deduce that D is contained in the subgroup H(C) defined in Problem 3 , and that H(C) is a compact open subgroup of G . Show also that C H(C) = C, that p(C) > p ( A ) p(B) p(H(C)) and that C C A 4 B (consider the measure of E, n (c - F,) for each c E C). (d) Let A, B be two integrable subsets of G . Deduce from ( c ) that either p+{A B)? p(A) p(B), or there exists a compact open subgroup H in G such that A -t- B contains a coset of H, and that in this case p*(A ~1 B) 2 p(A) -t p(B) - p(H). Consider the case when G is connected.

+

+~

5.

+

~

+

(a) Let A (resp. B) be the set of real numbers x - xo

+ ,xx , 2 - ' ou

,=I

where xo is an

integer, each xi ( i 2 1) is 0 or I , and x i = 0 for all even i > 0 (resp. x i = 0 for all odd i > 0). Show that each of A, B has zero Lebesgue measure but that A B = R. (b) Deduce from (a) that there exists a Hamel basis H of R (over Q) contained i n A u B and theiefore of measure zero. The set PI of numbers of the form rh, where r E Q and h E H, is also of measure 0. (c) Let P, denote the set of real numbers which have at niost 17 nonzero coordinate5 relative to the basis H . Show that if P. is negligible and P.,, Lebesgue-measurable, in then P,. I is negligible. (Let h, E H , and show that the set S of numbers x E P., which the coefficient of h , is nonzero, is negligible. Using Problem 2(c), show that if Pn+,were not negligible there would exist two points x', x'' in P.+l n I S such that (x' - - x")//iOwere rational. and hence obtain a contradiction.) (d) Deduce from (b) and (c) that there exist two negligible sets C, D in R such that C .I- D is not measurable (with respect to Lebesgue nieasure).

+

6.

Let G be a group acting (on the left) on a set X. A subset P (resp. C ) of X is said to be a G-parking (resp. a G-rouering) if for each s # e in G we have s . P n P = 0 (resp. if X - s . C). A subset P which is both a G-packing and a G-covering is called a

u

S C G

G -tessellation. (a) Suppose that X is separable, metrizable and locally compact, that G is at niost denumerable and acts continuously on X (with respect to the discrete topology on G)

PARTICULAR CASES AND EXAMPLES

2

251

and that there exists a positive nonzero G-invariant measure p on X. Let P (resp. C) be a G-packing (resp. a G-covering) such that P and C are p-integrable. Show that p(C) 2 p(P). (Remark that p(C) 2 1 p(C n s . P) = p(s-l . C n P).)

1

S C G

S E G

(b) Suppose that there exists a G-invariant distance d on X defining the topology of X. Let A(G) denote the greatest lower bound of the numbers p(C), where C runs through all integrable G-coverings of X. Let r be a real number > O such that there exists a point a E X for which p(B(a; Y)) > A(G). Show that there exists s f e in G such that d(a, s . a ) < 2r. (c) Suppose that X is a locally compact group, p a left Haar measure on X, and G a denumerable subgroup acting on X by left translations. Show that if A is an integrable subset of X such that p(A) > A(G), then there exists s E G n AA-' such that s # e. (d) With the same hypotheses as in (a), let F be a p-integrable G-tessellation and let Go be a subgroup of finite index h in G . If sl,. . . , s h are a system of representatives of sL . F is a Go-tessellation. the right cosets of Go in G, show that Fo =

u

1S14h

(e) Same hypotheses as in (a). L e t f > 0 be a p-integrable function on X. Show that there exist two points a, b in X such that p(C)

Z f ( s ' 42 Jxf(x)dp(x)

and

(Observe that if g

2 0 is an integrable

there exists c E E such that

SE

p(P) C f ( s . b) 5 I x I ( x ) W x ) . S E G

S S G

function and E an integrable subset of X, then

g(x) +(x) 5 g(c)p(E), and

c' E

E such that

IEg(x)dp(d 2 dC')p(E).)

2. PARTICULAR CASES A N D EXAMPLES

(14.2.1) On the additive group R , Lebesgue measure (13.1.4) is a Haar measure (left and right, since R is commutative). This follows from the formula for change of variable (8.7.4) applied to the function ( ~ ( 5 = ) 5 + E, which gives J -' m " f ( f + ct) df dt for allfEXX,(R) and all ct E R.

=StI,f(t)

(14.2.2) Now consider the multiplicative group RT of real numbers >O. This is a locally compact commutative group ((13.18.4), (4.1.2), and (4.1.4)) For each function f~ XX,(R*,),there exists a compact interval [a, b] with 0 < a < b, containing the support o f f . Hence for each interval [c, d] in RT containing Supp(f), the integral S c d ( f ( f ) d t ) /isf defined and its value,

which we denote by / o + m ( , f ( f ) df)/f, is independent of the choice of interval

[c, d] containing Supp(f). We assert that f ~ / ~ + ~ ( dt)/t f ( l )is a Haar measure on R,*. From the formula for change of variable (8.7.4), it follows that, for each s > 0,

252

XIV

INTEGRATION IN LOCALLY COMPACT GROUPS

which proves our assertion. (14.2.3) Let G be a locally compact group with neutral element e, and let p be a (left or right) Haar measure on G. Then G is discrete if and only i f p( { e } )> 0 ; and G is compact ifand only ifp*(G) < + co (i.e., if and only if p is a bounded measure (13.20)).

It is clear that if G is compact then p is bounded. If G is discrete, { e } is an open neighborhood of L-,and hence p ( { e } )> 0 because the support of p is the whole of G. Conversely, let V be a compact neighborhood of e. If p ( { e } )> 0, we have p ( { s } ) = p ( { e } ) for all s E G, because p is invariant; hence the number of points of V i s j n i t e and s p ( V ) / p ( { e } ) Since . G is Hausdorff it follows that G is discrete. Suppcse now that p is bounded and (say) left-invariant. Consider the set (5 of finite subsets {sl,s 2 , . . . , s,} in G such that s i V n sjV = Q whenever j i # j . We have np(V) = p(s,V u A,V u - * * u s,V)

p(G),

hence I I 5 p(G)/p(V). Hence (5 contains a subset {sl,. . . , s,} having the largest possible number of elements. For each s E G it then follows that sV must meet at least one of the siV, that is to say, s E siVV-'. Hence G is the union of the sets siVV-', which are compact (12.10.4) and so G is compact. (14.2.4) Let G be a locally compact group, V an open subset of G, and p a nonzero positive measure on V, having the following property: if U is an open subset of V and if s E G i s such that sU c V , then the image of the measure pu, inducedbyp on U (13.d.8) under the homeomorphism XHSX (13.1.6). isthe measure psuinduced by p on sU.Then there exists a unique left Haar measure a on G which induces p on V.

For each S E G , let ps be the image of p under the homeomorphism x Hsx of V onto sV. The restriction of ps to V n sV is the image of ps- lv A under the restriction of XHSX to s-'V n V. By hypothesis, this image is pv s v . By translation it follows that for all s, t in G the measures ,usand p t have the same restriction to sV n tV. By virtue of (1 3.1.9), there exists a positive measure r on G which induces ps on sV for all s E G. Clearly a is leftinvariant, and is therefore the unique left Haar measure on G which induces p on V.

2 PARTICULAR CASES AND EXAMPLES

253

We shall use this local definition of a Haar measure in Chapter XIX to construct a left Haar measure on a Lie group. Here we note the following consequence of (14.2.4) : (14.2.5) Let G be a locally compact group, H a discrete normal subgroup of G , and n : G -+ G/H the canonical homomorphism. Also let V be an open neighborhood of the neutral element of G such that the restriction of n to V is a homeomorphism of V onto the neighborhood n(V) of the neutral element of G / H (12.11.2). Ldt A be u left Haar measure on G. I f p is the image under n I V of the restriction Av of A to V, then p is the restriction to n(V) of a left Haar measure on G/H.

For every open set in n(V) is of the form n(U), where U c V is open, and the relation n(s)n(U)c n(V) is equivalent to sU c V ; hence it follows immediately from the definitions that p satisfies the condition of (14.2.4).

Example (14.2.6) The mapping cp : t w e Z n i ris a strict morphism (12.12.7) of R onto the compact group U of complex numbers of absolute value 1, by virtue of (9.5.2) and (9.5.7). The kernel of cp is the discrete subgroup Z consisting of the integers, and U may therefore be canonically identified with the quotient group R/Z = T (also called the 1-dimensional torus or the additive group of real numbers modulo 1). Apply (14.2.5) to the case where V = 3-4, +[; bearing in mind that a Haar measure p on U must be diffuse (14.2.3) and that the complement of cp(V) in U consists of a single point, we see that a function f' on U is p-integrable if and only if the function t H f(e2"") is Lebesgue-

integrable on

1-4,

,::I

$[, and that we then have I f dp =

f(e*"") dt.

(14.2.7) Let G,, G, be two locally compact groups, and p1 (resp. p,) a left Haar measure on G, (resp. G,). Then pl @ p 2 is a left Haar measure on G, x G 2 .

For each function.fe ,X,(G, x G2) and each (s,, s,) jJ/(S1Xl9

=p

E

G, x G , we have

S Z X J 4 4 x 1 ) &Z(XZ)

P ' ( X J j/O'Xlj

S Z X J dPZ(X.2)

254

XIV

INTEGRATION I N LOCALLY COMPACT GROUPS

by virtue of (13.21.2). Hence the result.

I n particular, Lehesgue measure on R" (13.21.19) is a Haar measure on the additive group R".

PROBLEMS 1.

Let G be a locally compact group, p a left Haar measure on G, A a subset of G , and B a relatively compact p-integrable subset of G such that p(B) > 0. Show that, if p*(AB) 1, and let A be a symmetric convex set in R" with nonempty interior. Show that, for each r > 0 satisfying h ( A ) r " 2 Pp"', there exists a point x # 0 in r A with integer coefficients, such that ui(x) = 0 (mod p ) for 1 5 i 5 nt. (Apply Minkowski's theorem to the subgroup Go of Z" consisting of all z E 2 such that u i ( z ) 0 (mod p ) for I 5 i 5 m , and use Problem 6(d) of Section 14.1.) In particular, show that if cl, c2 are any two integers, there exist integers x , , x 2 , not and c,xl -tc 2 x z = O (mod p ) (Thue's both zero, such that lxllg d i , lx21

sdi

theorem),

(c) Let a, h be two integers. Use (b) to show that there exist integers x , , x2,x3,x4, not all zero, such that fixI -1- hx2 = x 3 (mod p ) ,

bxl

- ax2 =- x4 (mod p

)

and y=x:

I x: I-x: t - x :

IdZp.

+ +

Show that if p is prime one can find two integers N , 6 such that a* b2 1 -0 (mod p ) (assuming p is odd, observe that when z takes the h(p I I ) integer values

3 THE MODULUS FUNCTION ON A GROUP

255

0 , 1, 2, . . . , - l ) , the residues mod p of the numbers zz are all distinct). Deduce that in this case y = p . Hence, using the multiplicative property of norms of quaternions, deduce that every positive integer is the sum of at most four squares (Lagrange's theorem).

(a) Let h be Lebesgue measure on R. Letfbe a real-valued function 2 0 on R which is h-integrable, bounded and of compact support. Put y= supf(t). For each w E R, tER

let U,(w) denote the set of t E R such that f ( t ) 2 w , and put v f ( w )= h*(U,(w)). Show that, for all (Y > 1,

-+ m (b) Let g be another function satisfying the same conditions asf, and put 6 = sup g(t). fER

Let h be the function on RZ defined by h(u, u) = f ( r t ) i -g(u) if f ( u ) g ( u ) # 0, and h(u, Y) = 0 otherwise; also let k ( t ) = sup h(u, u), so that k is positive, h-integrable, "+"=I

bounded and compactly supported. Show that, for all

j-

+m m

k"(t) dt

2 ( y i-6)61

(+/-+I

CL

f " ( t ) df -I-

> 1,

-/ 6a I

+m

g"(t) d t ) .

-m

+

(Observe that if 0 < w 5 1 we have Uk(yw-t 6 w ) 3 U,(yw) U,(6w), and use (a) above and Problem 4(d) of Section 14.1). (c) Let h, be Lebesgue measure on R",and let A. €3 be two A,,-integrable subsets of R".Show that ((&(A

i- B))""

2 (hn(A))l'n-1- (AJB))''"

(Brunn-Minkowski irrequalify).(Reduce to the case where A and B are compact. Then use induction on ti, Problem 4(d) of Section 14.1, the theorem of Lebesgue-Fubini, the inequality established i n (b) above, and Holder's inequality.)

Let p be a prime number. The normalized Haar measure p on the compact group Z, of p-adic integers (Section 12.9, Problem 4) is such that the measure of any closed ball of radiusp-' is equal top-i. (Show that Z, is the union ofpXclosed balls of radius p - ' , no two of which intersect.)

3. THE MODULUS FUNCTION O N A GROUP; T H E MODULUS OF A N AUTOMORPHISM

Let G be a locally compact group, p a left Haar measure on G. For all s, t E G we have Y(f)(WP) = W(Y(t)Pc) = W P directly from the definitions (14.1.2), and therefore 6(s)p is also a left-invariant positive nieasure on G. Hence there is a unique real number AG(s) > 0 (also written A(s)) such that S(s)p = Ati(s)p (14.1.5), a n d this number is clearly, by

256

XIV

INTEGRATION I N LOCALLY COMPACT GROUPS

virtue of (14.1.5) again, independent of the choice of left Haar measure p. The mapping S H AG(s) is called the modulus function on G. Iff is any p-integrable function, the function XH f ( x s ) is therefore also p-integrable, and we have r

P

(14.3.1 . I )

In particular, if A is any p-integrable subset of G, then As is also p-integrable, and AS)

(1 4.3.1.2)

(14.3.2) The mapping s-AG(s) multiplicative group R*,.

= A(s)p(A).

is a continuous homomorphism of G into the

This. follows immediately from the formula (14.3.1.1) and the lemma (1 4.1.5.5).

The group G is said to be unimodular if AG(s) = 1 for all s E G. In this case there is no distinction to be made between left and right Haar measures, and we call them simply Haar measures on G. Ifthere exists in G a compact neighborhood V of'the neutral element e which is invariant under all inner automorphisms of G, then G is unimodular. This is the case, in particular, when G is compact, or discrete, or commutative. (1 4.3.3)

For each s E G we have p(V) = p(s-'Vs) = AG(s)p(V) by (14.3.1.2) and (14.1.2.4); hence the result, since p(V) # 0 (14.1.2). (14.3.4) I f p is any left Haar measure on G , then fi = A-' * p. Iff is any p-integrable function on G, then the function X H f (x-')A(x)- is p-integrable, and we have

I

~ f ( X - ' ) A ( x ) - ' d P ( x= ) f ( x >4-0).

(14.3.4.1)

For each s E G we have ti(s)(A-'

*

p) = (ti(~)A-') . ( & ( ~ ) p=) (A(s)-'A-')

*

(A(s)p)

= A-'

p,

which shows that A - ' * p is a right Haar measure on G. Since jl is also a right Haar measure, there exists a constant a > 0 such that fi = aA-' * p. It follows that p = a(A-' . p)" = aA * fi = a2p (13.14.5), whence a2 = 1 and therefore

3

THE MODULUS FUNCTION ON A GROUP

257

a = 1 (because a > 0). This proves the first assertion. The second then follows (13.14.3). We deduce that iff is a locally p-integrable function, then

(f.P)"

(1 4.3.4.2)

=

*

P.

In particular, if G is unimodular, then for each p-integrable function the functions y(s)f, 6(s)f and f are p-integrable, and we have

(14.3.5)

(14.3.5.1)

s

f ( s x ) dp(x) =

s

I

s

f ( x s > dp(X) = f ( x - ' ) d p ( x ) = f ( x ) d p ( ~ ) .

In particular, if A is any p-integrable set in G, then (14.3.5.2)

~(sA= ) AS)

= p(A-') = p(A)

for all s E G. When G is injnite and compact (resp. injinite and discrete) and therefore unimodular (14.3.3), the normalized Haar measure on G is the unique Haar measure p on G for which p(G) = 1 (resp. p ( ( e ) )= 1). (14.3.6) Now let u be an automorphism of the topological group G . It is clear that the image u-'(p) of a left Haar measure p on G (1 3.1.6) is another left Haar measure; hence (14.1.5) there exists a number a > 0, independent of the choice of p, such that u-'(p) = up. This number a is called the modulus of the automorphism u, and is denoted by mod,(#) or mod@). For every p-integrable function f we have therefore

and in particular, for any p-integrable set A, (14.3.6.2)

L44-9= (mod W A ) .

In particular, for each s E G, let isbe the inner automorphism X H S - ~ X S . Then we have i; = G(s)y(s), and therefore i; '(p) = w

p =W p ,

which proves that (14.3.7)

mod(&) = A($).

258

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INTEGRATION I N LOCALLY COMPACT GROUPS

If G is either compact or discrete, then we have mod(u) = 1 for euery automorphism u of G. For it is clear that u(G) = G and u({e}) = {e}, and we may apply (14.3.6.2) with A = G and A = {e}, respectively. We deduce also from (14.3.6.2) that if u and ZI are two automorphisms of G, then

(14.3.9)

-

mod(u u) = mod(u) rnod(z1)

(1 4.3.8)

0

If u is any automorphism of the uector space R", then mod(u)

=

ldet uI.

Let U = ( a i j ) be the matrix of the automorphism u with respect to the canonical basis of R". Let E i j denote the n x n matrix which has the element in the ( i , j ) place equal to 1 and all other elements equal to 0. If i # j and A E R, Put Bij(A)= I,, + AEij.

Then we have the following lemma: (14.3.9.1) Every invertible n x n matrix U is a product of matrices of the form Bij(A)and a matrix of the form I,, + (a - l)En,,.

Consider invertible matrices of the form 1 0 0 1

X=

"'

* * '

0 0

t1,n-h r2,n-h

* * .

52"

...................................... " '

tn-h-1,"-h"'

....................................... 0 0

"'

0

0, in the notation of Section (3.6). (Observe that V,(A)

8.

=A

+ B(0; r).)

Let H c R" be a hyperplane passing through the origin 0, and let T be a rotation transforming H into R"-'. For each universally measurable bounded subset A of R", put uH(A) = T - ' . I J " - ~T( A) (the "Steiner symmetrization" of A with respect to H). (a) Suppose that A is closed and contained in a closed ball B with center 0. Show that, if W is an open subset of the sphere S which is the boundary of B and if W n A = then uH(A) does not intersect W nor the image of W under the symmetry with respect to H. (b) Deduce from (a) that, under the same hypotheses, there exist finitely many hyperplanes H I , ..., H, passing through 0, such that the set uHruHr- . . . uH~(A) is contained in B and does not meet S unless A = B (use the compactness of S). (c) Suppose that A is compact. If Bo is the closed ball with center 0 such that &(A) = X.(Bo), show that there exists a sequence (HnJ of hyperplanes passing through 0, such that the sequence of compact sets A,,, = uHmI J ~ , , , - ~ ... uHI(A) tends to Bo in the topology defined in Section 3.16, Problem 3. (Use the result of this problem, by showing first that the sequence (A,,,) can be assumed to have a limit A' such that A' C B(0; R), where R is the greatest lower bound of the radii of closed balls with center 0 containing a transform of A under the composition of a finite number of Steiner symmetrizations uH with respect to hyperplanes H passing through 0. Then argue by contradiction and use (b) above to show that B'(0; R) = Bo.)

a,

9.

(a) Let A be a nonempty compact subset of the space R" endowed with the Euclidean scalar product. For each unit vector u E R",put h(A; u) = sup+ I u), and b(A; u) =

+

YEA

h(A; u) h(A; - u ) (the width of A in the direction u). The least upper bound of the numbers b(A; u ) as u varies on the unit sphere S n - , is the diameter &A) of A (Section 6.3, problem 2). For any pair of compact sets A, B and any real number tc > 0 we have h(A B ; u ) = h(A; u ) h(B; u ) and h(aA; u ) = ah(A; u) , from which it follows that &A B) 5 &A) 6(B). (b) Let s be a finite sequence ((a,, U,))l j d , , 2 of pairs in which the aJare real numbers 2 0 such t h a t x a, = I , and the U j are rotations about 0 (i.e., elements of SO(n, R)).

+

+ +

+

I

The rotnfionol memi of A corresponding to the finite sequence s is defined to be the compact set p,(A) tlJ U,(A). We have h,(p,(A)) 2 AJA) (Section 14.1, Problem

=c J

4(d)). For each u E Snv1, show that h(pdA);

11)

=x

CLJ

h(A; U y '(u)).

(c) Let A be a nonempty compact set contained in a closed ball B with center 0, and let S be the frontier of B (a sphere). Show that if there exists a nonempty open subset W of S which does not meet A, then there exists a rotational mean p,(A) such that p,(A) c B and S n p,(A) = 0 (same method as Problem 7; use the compactness of S). (d) Let R be the greatest lower bound of radii of closed balls with center Ocontaining a transform of A under the composition of a finite number of rotational means. If Bo = B(0; R), show that there exists a sequence of sets A, = p.,p.,. . . p,,(A) tending to Bo , relative to the distance defined in Section 3.16, Problem 3. (Argue by

3 THE MODULUS FUNCTION ON A GROUP

265

contradiction, by first showing that the sequence (A,) may be assumed to have a limit A' c Bo , and then using (c) above to prove that A' contains the frontier So of Bo; finally remark that $3, &So = Bo .) (e) If A is any compact subset of R",show that

+

where the diameter &A) is relative to the Euclidean distance, and V. is the Lebesgue measure of the ball B, ("Bieberbach's inequality"). (Use (d) and the inequality &(A)) 5 10. With the notation of Problem 7, for any compact subset A of R" the upper (resp. lower) Minkowski area of A is defined to be the number

.+(A)

= lim

sup (X.(V,(A)) - h.(A))/r

,-0

(resp.

m-(A)

= lim

inf (X.(V,(A))

- h.(A))/r)

,-0

where r tends to 0 through positive values. (a) Give an example of a compact set A such that a+(A)= 1 and a-(A) = 0. (Take n = 2, and consider a union of finite sets of the form ck A*, where the sequence (ck) tends to 0,and for each k the set Ak is a product 1, x J k . where I k , Jk are finite (qbk)O 0 and that the measures of the sets 's . r ( U ) are all equal.) 5.

Let G be a locally compact group and p a right Haar measure on G. (a) In order that a positive measure a # 0 on G, with support S , should be such that S is a closed subgroup of G and the measure induced on S by a a right Haar measure on S, it is necessary and sufficient that &)a = a for all s E S. The set of elements t E G such that S(t)a = a is then equal to S. (b) Let I? denote the set of positive measures a on G satisfying the condition of (a) above. Show that I' is vaguely closed in M+(G) - {O}. I f f € X + ( G )is such that f ( e ) > 0, show that the set I?, of measures a

compact (prove that sup a ( K ) < =Err

E

I

I' such that f(s) da(s) = 1 is vaguely

+ m for every compact subset K of G). Deduce that

the mapping a + + ( a ( f ) ,a / a ( f ) )is a homeomorphism of I? onto the product space

RI, X F f .

(c) For each a E r, let H. = Supp(a): it is a closed subgroup of G. Let d be a rightinvariant distance on G such that d(x, y ) 2 1 for all x, y in G (12.9.1). Let (K,) be an increasing sequence of compact sets, covering G and such that K, is contained in the interior of K,+ I for each n (3.18.3). If h is the Hausdorff distance on G corresponding to d (Section 3.16, Problem 3), then for any two nonempty closed sets M, N in G we define h.(M, N ) = 1 if either M n K. or N n K. is empty, and h,(M, N ) = h(M n K,, N n K.) otherwise. Endow the set S(G) of nonempty closed subsets of G with the topology defined by the pseudo-distances h,. Show that the mapping a w H . is a homeomorphism of I71 (endowed with the vague topology) onto the set C of closed subgroups of G , endowed with the topology induced by that of %(G). 6. With the notation of Problem 5 , consider the subset I'O of I? consisting of measures a E I' such that H. is unimodulur. This is also the set of measures a E I' such that a ( f ) = a ( f ) for all f~ X ( G ) ,and is therefore vaguely closed in I?. For each a E Po, put Q, = G/H.; then there exists a relatively G-invariant measure p. on Q. such that

IGf(~)

dp&)

=

1

H .

a

f ( x s )d a b )

for everyfE X ( G ) (where R is the coset xH.) (Problem 2). (a) If a E ro and f~ -f(G), put f.(x)

Ilf,ll

=

5

f ( x s ) da(s). Show that the mapping

H.

is vaguely continuous, and deduce that the mapping a++ ljp.jl is lower semicontinuous with respect to the vague topology. (b) Let g 2 0 be a real-valued p-integrable function, and let ro(g) be the set of a-

measures a E rosuch that a+-+IIp.ll of

I*

g(xs) da(s) 2 1 for all x

ro(g)into R is vaguely continuous.

E

G . Show that the mapping

(It is enough to show that the map-

ping is upper semicontinuous. Let h E Y + ( G )be such that

I

1g(x) - h(x)l dp(x) 5 E ,

and let K = Supp(h). If r D: G + Q a is the canonical mapping, show that pu(Qu-- r u ( K ) )5 E for all fl E I'O(g). On the other hand, consider a functionfEX+(G) such that j G f ( x s )da(s) 5 1 for all s E G and lem 2(a)). If U, is the set of measures all x

E

SG

f ( x s ) da(s) = 1 for all s E K (use Prob-

fl E I ' O ( g )

K, show that (1 - E ) ~ B ( T # ( K 5)l\pall.) )

such that

f ( x s d/?(s)> 1 - E for

5 CONVOLUTION OF MEASURES ON A LOCALLY COMPACT GROUP

271

(c) Let r: be the set of measures O L EI'O such that G/H. is compact. Show that r"(g)c for all g E Z + ( G ) . Let a E,:'I and let g E Z + ( G ) be such that / g ( x s ) da(s) = 2 for

all x

E

G (Problem 2(a)). For each compact L c G , the set W

of measures @ E r0 such that

ro

s

g(xs) d@(s)2 1

for all x

E

L is a neighborhood of a

in with respect to the vague topology. Show that if G is generated by a compact neighborhood U of e , and if we take L = U K above, then W c ro(g). Deduce that the restriction to roof the mapping a- lIpmllis vaguely continuous in this case. (d) Let rdC robe the set of measures a such that H, is discrete, and let N be the subset of rd consisting of measures a such that a ( { e } )= 1. For each relatively compact open neighborhood U of e in G, let N u be the set of all a E N such that H, n U = { e } . Show that NU is compact (observe that the relation a E N u is equivalent to a({e})2 1 and a(U) 5 1). As U runs through the set of relatively compact open neighborhoods of e in G, the interiors of the sets N u cover N. A subset M of N is relatively compact in N if and only if there exists a relatively compact open neighborhood U of e in G such that M c N u .

7. With the notation of Problems 5 and 6 , suppose that there exists a neighborhood of e in G which contains no finite subgroup of G other than { e } . Show that the mapping Q H a({e})of r d into R*,is vaguely continuous. (Show that there exists a neighborhood V of e in G and a neighborhood W of a in rd such that the relation E W implies (Vz- V) n Ho= 0 :argue by contradiction.)

8. With the notation of Problems 5 and 6 , suppose that G is cornrnirtatiue and generated by a compact neighborhood of e . Let N, denote the subset of N consisting of measures a such that Q. = G/H, is compact, so that N, = N n I':; the set N, is open in N (Problem 6(c)). A subset A of N, is relatively compact in N, if and only if it satisfies the following two conditions: (1) there exists an open neighborhood U of e in G such that H, n U = { e } for all a E A ; (2) there exists a constant k such that pL,(G/H,)5 k for all a E A. (Use Problem 6 and the fact that p a is a Haar measure on G/H. .)

9. Translate the results of Problems 6 to 8 into statements about subspaces of the space of closed subgroups of G (Problem 5(c)): in particular, the subspace of unimodular of subgroups H E cosuch that G / H is compact, closed subgroups, the subspace In particular, the subspace D of discrete subgroups, and the subspace D, = D n obtain Mahler's criterion: if for each discrete subgroup H E D, we denote by u(H) the total mass of G/H relative to the measure p a corresponding to the Haar measure a on H such that @({el)= 1 , then a subset A of D, is relatively compact in D, if and only if (1) there exists a neighborhood U of e in G such that H n U = { e } for all H E A; and (2) there exists a constant k such that u(H) 5 k for all H E A. Consider the case G = R".

c

xo

x,"

x:.

5. CONVOLUTION O F MEASURES O N A LOCALLY COMPACT GROUP

Let pl,. . . , p,,be a finite number of (complex) measures on a locally compact group G. The sequence (pl,. . . , p,,) is said to be conuolvabfe if, for each functionfE X,-(G), the function (XI,

..

' 9

X,,) H f ( X I X 2

. . . x,>

272

XIV

INTEGRATION IN LOCALLY COMPACT GROUPS

--

is integrable with respect to the product measure p1 0 p z 0 . 0 p, on G". It follows immediately from this definition and from (13.21.1 7) that the sequence (pi)l5 is convolvable if and only if the sequence ( Ipil)lsis,, of absolute values of these measures is convolvable. Moreover, it is clear ?hat

J*

fH

Sf(x1x2

* * *

xn) d 1p1I

.

* *

d I P ~(xn> I

is then a positive linear form on XR(G) and therefore a positive measure on G (13.3.1); and for all f E XR(G) we have

by virtue of (13.16.5) and (13.21.17). It follows directly (13.1.1) that

f~

j--

--

/ f ( x l x 2 x,) dpl(xl) * dp,,(x,) is also a (complex) measure on G. This measure is denoted by pl * p 2 * * * * p, and is called the convolution product or convolution of the sequence (pJljiSn. The formula (14.5.1) also shows that

-

For each functionfe X,-(G),the function ( x l , . . . , x,) w f ( x 1 x 2* . x,,) is continuous and therefore measurable with respect to every measure on G"; hence, by virtue of (13.21.lo), the sequence ( p l , . . . , p") is convolvable if and only if 4

for some permutation 0 of (1, 2, . . . , n } . An equivalent condition is that, for each compact subset K of G, the (closed) set A c G"of points (xl,x2, . . . , x,,) such that x l x 2 * * x,, E K is (pl€3 p 2 €3 * . * €3 p,,)-integrable. If a sequence (p,v ) of two measures is convolvable, we say that p and v (in this order) are convolvable, or that p is convolvable on the left with v , or that v is convolvable on the right with p. If p and v are convolvable and p', v' are two measures such that lp'l 5 lpl and lv'l 5 IvI, then it is clear from (14.5.3) that p' and v' are convolvable.

PARTICULAR CASES OF CONVOLUTION OF MEASURES

6

273

Zfthe sequence ( p i ) l 6 i 6 nis convolvable, and i f S i = Supp(pi) (13.19), S, , and the two sides are equal if all then Supp(pl * p 2 * * * p,) t S,S2 the measures p j are positive. (14.5.4)

- -

Let x be a point not in the'closure of S1S2 *.. S , , and let V be an open neighborhood of x not meeting SlS2 * S, . Iff is any continuous function whose support is contained in V, we can write (1 3.21. I 8)

-

But if xi E Sjfor 1 5 i 5 n, we have f(xlx2 x,) = 0 by hypothesis, and the first assertion of (14.5.4) is proved. If all the p i are positive, then so is p =p,

* p2 *

*-.

* p,.

If U is a p-negligible open set, K a compact subset of U, andfE XX,(G)a function with values in [0, 11 which is equal to 1 on K and to 0 on 6U (4.5.2), then we have

J. *

*

Jf(xlx2

* * *

xn)

djLI(x1) . . * dpn(xn) = 0;

therefore the open set of points (x,, . , . , x,) such that f(xlx2 * * x,) > $ is (p, 0 p 2 0 * * 0 p,J-negligible, and consequently does not intersect the support S, x S2 x ... x S, of this measure (13.21.18), Since the mapping (xl, .. . , X , ) H X ~ X ~ .. 'x, is continuous, we conclude that K does not intersect SIS, * * S, , and therefore that S,S2 . . S, is contained in Supp(p). This completes the proof.

-

-

6. EXAMPLES A N D PARTICULAR CASES OF C O N V O L U T I O N O F MEASURES

(14.6.1) A Dirac measure E, (13.1.3) on G is convolvable (on either side) with every measure p on G, and we have

(1 4.6.1.2)

E,

* E , = E,, .

274

XIV

INTEGRATION IN LOCALLY COMPACT GROUPS

For example, let us verify the first of the formulas (14.6.1.1). We have to show that, for every functionfe Xc(G),the continuous function ( x , y )f~( x y ) is (6, 6p)-integrable. Now the function x ~ f ( x y )is &,-integrable, and

1f ( x y )dc,(x)

(sy). Since the function y~ f (sy) is continuous and has compact support, it is p-integrable. Our assertion therefore follows from (13.21 .lo), and then the theorem of Lebesgue-Fubini (13.21.7) gives f ( x y ) d.cS(x)dp(y) = f (sy) dp(y). The formula to be proved follows now from (14.1.1) and (14.1.2). =f

//

(14.6.2) Every jinite family (pl, . . , , p,,) of bounded measures on G is convolvable. The measure pl * p2 * * . . * p,, is bounded, and

To prove the first assertion, observe that the measure pl 0 * * * 0 p,, on G" is bounded (13.21.18). For every function f E X c ( G ) ,the function ( x l , . . . , x,,)w f ( x I x 2* * x,,)is continuous and bounded on G", and therefore 0 p,)-integrable (13.20.4). To prove (14.6.2.1), it is enough to (pl 0 remark that iff E X c ( G )and IIf II 5 1, then

by virtue of (13.21.18). (14.6.3) A left Haar measure ,I on G is c o n ~ ~ o l on ~ ~the ~ ~right l e with any bounded measure p on G , and p * 1 = p( 1)i.

We may restrict ourselves to the case where p 2 0 (cf. (14.7.1.2)). For each function f E X + ( G )we have

and therefore, by virtue of (13.21.9), the function ( x ,y ) Hf ( x y ) is ( p 0 1)integrable and its integral is equal to 1(f) ( I p ( ( . The same calculation shows that I is not convolvable with itselfif G is not compact, for then the function 1 is not A-integrable (14.2.3). (14.6.4) Let ( p l , . . . , p,,) be a jinite sequence of measures on G , all of which except possibly for one have compact support. Then the sequence (pl, . . . , p,,) is convolvable.

7 ALGEBRAIC PROPERTIES OF CONVOLUTION

275

Let S i = Supp(pi), and suppose that S i is compact except possibly for one index j . Letff Xx,(G)and let K be the compact support off. It is enough to show that the set of points ( x i , . . .,x,,) E G" which belong to the support

n n

S i of p 1 0 * *

6 pn (1 3.21.18) and are such that x l x z * * . x,

E

K, is relatively

i= 1

compact in G". Now, the conditions x i E S i for all i, and x l x l imply that xi E S,7-', * * * S;'KS,' . . . ST' J+1

x,, E K,

7

and this set is compact (12.10.5). Since by hypothesis S iis compact whenever i # j , our assertion is proved (3.20.16). The result of (14.6.2) or (14.6.3) shows that two measures can be convolvable without either of them having a compact support. Later (14.10.7) we shall see examples of unbounded measures which are convolvable.

7. ALGEBRAIC PROPERTIES OF C O N V O L U T I O N

(14.7.1) Let A, p, v be three measures on G, and suppose that the pairs (A, p) and (A, v) are convolvable. Then so is the pair (A, p + v) and we have

A * (p + v) = A * p

+ A * v. For by virtue of the relation Ip + vJ S Ipl + IvI we may restrict ourselves

(14.7.1.1)

to the case where A, p and v are positive, and in this case the result follows immediately from (13.1 6.1). Similarly, if (A, v) and (p,v) are convolvable, then so is (A p, v) and we have

+

(14.7.1.2)

(A + p) * v = A * v + p * v.

Also it is clear that if the pair (A, p) is convolvable, then so is (an, bp) for all scalars a and b, and

(14.7.1.3)

(an) * ( b p ) = (ab)A* p.

(14.7.2) Let A, p, v be three measures #O on G. (i) If the sequence (A, p, v) is convolvable, then so are the sequences (A,p), ( 14 * Id, v), ( P , 4,(A 1p1 * Ivl), and we have (14.7.2.1)

A * p * v = (A * p) * v = A * ( p * v).

276

XIV

INTEGRATION IN LOCALLY COMPACT GROUPS

(ii) Zf the sequences (A,p) and (111 * lpl, v) are convoluabfe, then so is Likewise $the sequences (p, v) and (A, lpl * Ivl) are conuolvable.

(A,p, v).

We may restrict ourselves to the case in which A, p, and v are positive. Suppose that the sequence (A, p, v) is convolvable; then for every compact subset K of G the set of triples (x,y, z) such that xyz E K is (A 0 p 0 v)integrable. Let A be the set of pairs (x,y) such that xy E K. For each compact subset K‘ of G, the set A x K’ c G 3 is contained in the set of triples ( x , y, z ) such that xyz E KK’, and since KK’ is compact, it follows that A x K‘ is ((A 0 p) 0 v)-integrable. Since v # 0, this implies that A is (A 0 p)-integrable (13.21.11) and hence that (A, p) is convolvable. Consequently, for any f E X+(G),it follows from the hypothesis and the Lebesgue-Fubini theorem that

[hw p c t 44

A*

=

s’

Mz)

ss

= JJJf(XYZ)

f(xyz)

W) MY)

W) M Y )Wz)

since the function I ~ f ( t z is) in X + ( G ) for , each x E G. This shows that A * p and v are convolvable. One proves in the same way that (p,v) and (A, p * v) are convolvable. The formula (14.7.2.1) is then a consequence of the LebesgueFubini theorem. Conversely, suppose that (A, p) and (A * p, v) are convolvable, and let f be a function belonging to X + ( G ) .For each z E G, the function t H f ( t z ) belongs to X+(G),hence is (A * p)-integrable, and we have

J l ( t 4 41 * A(f)=

ss

f(xyz)

W) MY).

Hence, by Lebesgue-Fubini, it follows that

which proves that (A, p, v) is convolvable. One can give examples of measures A, p, v on G such that the pairs (A, p), ( A * p, v), (p,v) and (A, p * v) are convolvable but (A * p) * v # A* ( p * v) (Problem 1).

7 ALGEBRAIC PROPERTIES OF CONVOLUTION

277

(14.7.3) I f the sequence (pl, . . . , p,) is convohable, then so is the sequence

(fi,, . . . , PI), and we have (14.7.3.1)

(pl*pLZ*-..*pn)"=fin*ii,-l *...*fil.

For (14.1.4 and 13.7.10) we have

if and only if

and these two integrals are equal. On the other hand, if the sequence ( A , p ) is convolvable, it does not necessarily follow that ( p , A) is convolvable (Problem 2). But if G is commutative this will be the case, and we shall have A * p = p * A. In particular, it follows from the preceding results that (14.7.4) On the set MS(G) of bounded measures on G, the law of composition ( A , p ) H1 * p (together with the vector space structure) defines a C-algebra structure; the unit element is the Dirac measure E, at the neutral element e ojG. The set Mk(G) of compactly supported measures on G is a subalgebra of M,!.(G). The algebra Mh(G) is commutative ifand only ifG is comniutative.

The fact that G is commutative if Mh(G) is commutative follows from the formula (14.6.1.2). If G is discrete, the algebra M$(G) consists of all linear combinations

1U S E S , where a, = 0 for all but a finite number of points s E G (3.16.3), and

scG

the formula (14.6.1.2) shows that

This is what is called in algebra the group algebra of the group G over the field C .

278

XIV

INTEGRATION I N LOCALLY COMPACT GROUPS

PROBLEMS 1. On the additive group R, let h be Lebesgue measure, let p = pI . h, where 1 is the interval

+

and let n < b be two distinct points of R. Show that the convolution product * p) * h is defined, but that p and are not convolvable. Show that the convolutions p * ( ( F . - & b ) * h ) and ( p * (E. - E ~ ) )* h a r e both defined and are unequal. [0,

001,

((E. - Eb)

2. Let G be a locally compact group which is not unimodular. (a) Show that there exists a bounded positive measure p on G such that AG . p is not bounded (take p to be discrete). (b) Let h be a left Haar measure on G. By (14.6.3), p and h are convolvable. Show that h and p are not convolvable.

3. Let G be a locally compact group. (a) Let p, v be two positive measures on G . If p * v = E , , show that p = a & , and v = u-'E,- 1 for some x E G and a # 0 (cf. 14.5.4). (b) Give an example of a positive measure on the group 2 / 2 2 whose support is the whole group and which has an inverse (with respect to convolution product). 4.

(a) In the set M:(R), consider the two sequences of boundzd measures p n = F , and 2 c - " , both of which tend to 0 in the topology F2defined in Section 13.20, Problem I . Show that the sequence of measures pn * v. does not tend to 0 with respect to the topology 9,. (b) Let G be a locally compact group and let (p.), (v,) be two sequences of real bounded measures on G . Suppose that pn+ p in the topology Y2, and that v,, + v in the topology Y3 (notation of Section 13.20, Problem 1). Show that the sequence p n * v, tends to p * v i n the topology . T 2(Observe . that iff, g E .X(G), the function (x, y ) ~ y ( ~ , ) f ( x has . v ) compact support and can be uniformly approximated by a linear combination of functions u i@ u , , where uLand u, are i n Ty(G).) Give an example (with G = R) where p == v = 0 and pn * v. does not tend to 0 with respect t o the topology Y 3 .Show that if p n 0 with respect to Y 3and if the sequence of norms (]lv.ll) is bounded, then pn * v. + O with respect to F 2 .If pn+ p with respect to T3, and v. + v with respect to Y 3 show , that p n * v, + p * v with respect to F3(similar method). ,n + p and v,+v, (c) With the same notation, show that if, in the topology Y b p then p n * v, p * v (use Problem 2 of Section 13.20, and Egoroff's theorem). (d) Take G t o be the group RZ.Let a, 6 be the vectors of the canonical basis of G over R; let p. be the measure on I =- [O, 74 c R whose density with respect to Lebesgue measure is the function sin(2"x); let pn be the measure p. 0E~ on G, and let v n be the ~ nE~ on C . Show that pn+ 0 with respect to 9 6 and that v,,+ 0 with measure ~ b , respect to Y 3 ,but that the sequence (p. * v.) does not tend to 0 with respect to 9 6 .

"

-j

--f

5.

Let G be a compact group and p a positive measure on G such that Supp(p) = G , and p * p = p. Show that p(G) = I and then that p is a Haar measure on G. (Let

I

j'e K + ( G ) ,and put g(x) = f(yx) d p ( y ) , which is a continuous function on G. Show

J

that y(x) = g(yx) dp(y), and deduce that g is constant, by considering the set of points at which it attains its upper bound.)

8 CONVOLUTION OF A MEASURE AND A FUNCTION

279

(a) Let p be a measure on a locally compact group G. Show that p * v = v * p for all measures v such that both p * v and v * p are defined, if and only if p * E, = E, * p for all x E G . (b) Suppose that G is compact. If p is any measure on G, show that the measure pb defined by

(where

p is a Haar measure on G) satisfies pb * v = v * pb for every measure

v on G .

Let G be a compact group and p the Haar measure on G for which p(G) = 1 . Let p be a positive measure on G such that p(G) = 1 and p 2 cp, where 0 < c < 1. Show that lip*" - PI1 5 2"(1 - c)", where p*" denotes the convolution product of n measures equal to p (use (14.6.3)). Let r be a real number such that 0 < r 5 4. For each integer n > 1 , let P,,,~ denote the measure HE,. on R,and let pn.,denote p l , , * pz., * . . . * p.. ,. (a) Show that the sequence (pn.,)"? converges vaguely to a measure p on R with support contained in I = [ - I , I ] (prove that for every interval U c R the sequence (pn.W ) )converges). (b) Show that, if r < &, the measure p is disjoint from Lebesgue measure on R, but that p"lz is the measure induced on 1 by Lebesgue measure. t R. Show that (c) Let v l l J be the image of pI14under the homothety f ~ 2 on p1/4 * v , , = ~ p l I z ,although pI14and v1,4 are each disjoint from Lebesgue measure (use Problem 4(b)).

+

8. CONVOLUTION OF A MEASURE AND A FUNCTION

In the rest of this chapter we shall fix once and for all a left Haar measure P on the group G. I f f is any mapping of G into R or C, the norm N , ( f ) ( p = 1, 2 or + co)is taken with respect to the measure P. (14.8.1) Let p be a measure on G and let f be a (complex) locally P-integrable function on G (13.13.1). For the measures p and f . to be conuolrable, it is necessary and suficient that there should exist a P-negligible set N such that the function S H f ( s - ' x ) is p-integrable f o r all x $ N,and the function

(dejined almost everywhere with respect to 8) is locally P-integrable. When this condition is satisfied, the function g(x) = f (s- 'x)dp(s), defined almost ecerywhere with respect to p, is locally /I-integrable, and p We shall first prove the following lemma:

* ( f 8) is equal to g P. *

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INTEGRATION IN LOCALLY COMPACT GROUPS

(14.8.1.1) The image of the measure p @ /3 under the homeomorphism (s, x) H (s, s- ' x ) of G2 onto G2 is the measure p 0 p.

We may assume that p 2 0. Then, for every F E X + ( G 2 ) ,the function (s,x ) w F ( s , s - ' x ) belongs to X + ( G 2 ) and , we have

s s

F(s, s- 'x) dp(s) dp(x) = dp(s)

F(s, s- 'x) d&x)

by virtue of the left-invariance of fi and the Lebesgue-Fubini theorem. This proves the lemma. Now suppose that p and f * p are convolvable. Since If- PI = If I * p (13.13.4) we may limit ourselves to the case where p 2 0 and f 2 0. For each function h E X c ( G ) ,the function (s, x ) w h ( s x ) f ( x ) is ( p 0 P)-integrable by hypothesis (13.21.16 and 3.14.3); by virtue of (14.8.1.1), the same is true of the function (s,x,t+h(x)f(s-'x) (13.7.10). If A, is the set of points x E G such that h(x) # 0, it follows from the theorem of Lebesgue-Fubini that there exists a P-negligible set Nh in A, such that, for each x E A, n IN,, the function s ~ f ( s - ' x is ) p-integrable. Taking a sequence of functions h in X + ( G )such that the corresponding sets A, cover G (4.5.2), we see that the function s ~ f ( s - ' x is) p-integrable except at the points x of a a-negligible set N. Furthermore, it follows from the Lebesgue-Fubini theorem that the function x~ h(x) / f ( s - ' x ) dp(s), defined almost everywhere with respect to p, is fl-integrable, and that

s s

j J W I ( X ) d A s ) dP(x) = h ( x ) d N x ) f(s- 'x) 4 4 s ) . This proves that the condition of (14.8.1) is necessary, and that p * (f.p) = g * fi (13.13.1). Conversely, suppose that the condition is satisfied. We have to show that, for each function h E X+(G), the function (s, x ) h(sx) ~ f ( x ) is ( p 0 p)integrable. Now this function is ( p 0 @-measurable because h is continuous (13.21.13), and therefore it is enough to show that

8

CONVOLUTION OF A MEASURE AND A FUNCTION

281

But the lemma (14.8.1.1) shows that this is equivalent to the relation

ss'

h(x)f(s-'x)

dp(s) d p ( x )
0, there exists a compact subset K of G such that p([K) S E . Let V, be a compact neighborhood of x, . The function f is uniformly continuous on the compact set K-'Vo, hence there exists a neighborhood V c V, of x, in G such that the relation x E V implies

s

If(s-'x) -f(s-'xo>I

5 &/I*(K)

284

XIV

INTEGRATION IN LOCALLY COMPACT GROUPS

for all s E K (3.16.5). Hence, for all x

E

V, we have

s 4 1 + 2 Ilfll),

from which (ii) follows.

(iii) Again we may suppose that p 5 0. By hypothesis, for each E > 0, there exists a compact subset H of G such that If ( x ) l 6 E for all x 4 H. Take K as in the proof of (ii) above, and suppose that x 4 KH. Then if s E K we have s-'x 4 H and therefore

(14.9.3) Every measure p on G is convolvable with every function f E Xc(G); the integral on the right-hand side of (14.8.2) is defined for all x E G, and the ,function x H f (s-'x) dp(s) is continuous on G.

s

Since the measure f p has compact support, p and f * p are convolvable (14.6.4), and it is clear that the integral f ( s - ' x ) dp(s) is defined for all x E G. The continuity of the function

XH

s

s

f ( s - ' x ) dp(s) follows from (14.1.5.5).

We shall leave to the reader the task of stating the corresponding propositions for the convolution f * p. It should be noticed in particular that (14.9.2) and its analog f o r f * p prove that if G is unimodular (14.3), then 9 S ( G )is a left and right module over the algebra MKG), and the external laws of composition of these two module structures are compatible by virtue of (14.7.2). (14.9.4) Let p , v be two measures on G, and let f E Y c ( G ) .Suppose that

fi and v are concolvable. Then the function p * f is v-integrable and (14.9.4.1)

( f , fi

* v> = ( p *"A v>.

Likewise, if p and v are bounded and f E %'g(G), the function p * f is continuous and bounded (hence v-integrable) and the formula (14.9.4.1 ) is valid. For the hypothesis implies that the function (s, x) +I f (s-lx) is (p 0 v)integrable, and the result therefore follows from the theorem of LebesgueFubini.

9 EXAMPLES OF CONVOLUTIONS OF MEASURES AND FUNCTIONS

285

PROBLEMS

1. Let G be a locally compact group and p a positive nonzero bounded measure on G , such that p * p = p. (a) Show that S = Supp(p) is compact. ( I f f € X + ( G )is not identically zero, remark that p *fmust be constant on S, and use (14.9.2(iii)).) (b) Show that S is a compact subgroup of G and that p is the Haar measure on S for which p(S) = 1. (Use fa), Problem 2 of Section 12.9 and Problem 5 of Section 14.7.) 2.

+

(a) Generalize (14.9.2(i)) to the case where 1 < p < 03, by using Holder's inequality (Sectipn 13.11, Problem 12). (b) Let p be a bounded measure on G . Show that the norm of the continuous endo~ * fp(14.9.2) is equal to IIpII. (Let (f,)be a morphism of L1(G, p) induced by f sequence satisfying the conditions of (14.11.1). If the norm in question were 11p/1- a with a > 0, we should have NI(p * f * g) 5 (Ilpll - a)N~(f.* d 5 (Ilpil - a)Nl(g). Deduce that N m ( p*f.) I/pI/- a , and obtain a contradiction by letting n tend to fa.) (c) Under the hypotheses of (b), show that the norm of the continuous endomorphism of Lm(G,p) induced by f ~ * j 'pis equal to IlpI1. (Reduce to the case where p has compact support and has a continuous density with respect to IpI.) (d) Suppose that G is compact and p is positive. Show that, for 1 < p < 03, the norm of the continuous endomorphism of Lp(G, p) induced by f ~ * fp is equal to llpll. (e) Let G be a cyclic group of order 3. Give an example of a measure p on G such that the norm of the endomorphism of LZ(G,p) induced by f ~ * fpis strictly less than llpll.

+

3. (a) Let p be a bounded measure on G . Show that we have NI(p *f)= N,(f) for all f E 9 ( G , p) if and only if p is of the form c . E~ with IcI = 1. (Using Problem 2(b), show that for each f~ X(G) we must have

1[ I

f dp = [ l f l d l p l . Deduce first that

p = clpl, where c is a constant such that IcI = 1, and then that p is a point-measure.) (b) Take G to be a cyclic group of order 3. Give an example of a measure p on G which is not a point-measure and is such that N2(p * f )= N2(f) for all real-valued functionsfon G . 4.

Let G be a,locally compact group and p a left Haar measure on G . Let f be a bounded real-valued function on G, uniformly continuous with respect to a left-invariant distance on G . If p is any bounded measure on G , show that p *f(relative to 8) is uniformly continuous with respect to a left-invariant distance on G .

5.

Let G be a locally compact group and p a left Haar measure on G . (a) With the notation of Section 13.20, Problem 1, let ( p n )be a sequence of bounded measures on G which converges to t~ with respect to the topology F3, and let (f.) be a sequence of functions in Y ( G , p) such that the sequence of bounded measures (f. . p) converges to f?.! , with respect to the topology 9 6 . Show that the sequence of bounded measures (p. * (fn . p)) converges to (p,,* (f.p)) with respect to 9 6 (cf. Problem 4(d) of Section 14.7). (Use Problems 1 and 2 of Section 13.20.) (b) Let E,/z denote the space of bounded real-valued functions on G which are uniformly continuous with respect to a left-invariant distance on G , and let

286

XIV

INTEGRATION IN LOCALLY COMPACT GROUPS

denote the weak topology on M:(G) corresponding to the vector space E5,Z; the topology F5,* is finer than Fzand coarser than F 3Give . an example of a sequence (p,) which tends to 0 with respect to F5,*, and a sequence (f,)of functions in P ( G , B) such that the sequence (f,. p) converges to 0 with respect to F6, but such that the sequence (p, * (f,. p)) does not tend to 0 with respect to F A (Take . G -= R andf,(t) to be the function which is equal to sin nt in the interval [O, n] and is zero elsewhere.) (c) Let (pn)be a sequence of bounded measures on G which converges to p with let, f E 9 ( G , 6) and let (fa)be a sequence of functions in P ( G , p) respect to F S i z such that N,(f-f,)+O. Show that NI(p.*fn-p*f)+O. (Reduce to the case where p = 0 and f. = f for all n , and then to the case where f~ X ( G ) . Show that the sequence (p, * (f.8)) tends to 0 with respect to F3, by remarking that if g is a bounded continuous function on G , the functionf* g is uniformly continuous with respect to a left-invariant distance on G.) Show that the result is no longer valid if 9-5,zis replaced by F z . (d) Let (p,) be a sequence of bounded measures on G which tends to 0 with respect 2 , (v.) a sequence of bounded measures on G which tends to Y with respect to 9 5 /and Show that the sequence (p. * v,) tends to 0 with respect t o F5/2. (By using to Fz. (14.11.1), reduce to proving that -to for any f~ X ( G ) and g uniformly continuous with respect to a left-invariant distance on G . Then use (c).) 6.

The notation is that of Problem 5. (a) Let (p.) be a sequence of bounded measures on G. Suppose that, for each function f~ P ( G , p), the sequence ( p m* (f.p)) tends to 0 with respect to F2. Show that the sequence of norms (IIpnll)is bounded. (Apply the Banach-Steinhaus theorem to the sequence of mappings f ~ ( p*f). of L1(G, 8) into L'(G, p), and use Problem 2(b).) Deduce that the sequence (p.) tends to 0 with respect to F2. (b) Suppose that, for eachfe Y ' ( G ,p), the sequence (p. * (f.p)) converges vaguely to 0. Show that the sequence (p.) tends vaguely to 0. (If K is any compact subset of G , show as in (a) that the sequence (Ip,I(K)) is bounded.) Give an example in which the sequence (IIpnll)is not bounded (take G = Z). (c) Suppose that, for each f~ 9 ( G ,p), the sequence (p. * (f.p)) converges to 0 with respect to 9 5 / 2 . Show that pn-0 with respect to 9 5 / 2 (use (a)). Give an example where N,(pn*f)+ O for each f e 9 ( G ,p) but the sequence (p.) does not tend to 0 with respect to F 3 .

10. CONVOLUTION OF TWO FUNCTIONS

(14.10.1) Let .f and g be two (conzplex) locally 8-integrable functions on G. Then the measures f 1 and g . B are convolvable if and only i f there exists a P-negligible set N such that the function s H g ( s - ' x )f ( s ) is P-integrable f o r all x $ N and the function

10 CONVOLUTION OF TWO FUNCTIONS

(defined almost everywhere, relative to condition is satisfied, the function

s

p) is locally p-integrable. When this

h(x) = s ( s - ' x ) f ( s >M defined almost everywhere relative to

s),

p, is locally /I-integrable, and

( f PI * ( 9 rc3) *

287

*

=h

*

8.

This is a particular case of (14.8.1). When the conditions of (14.10.1) are satisfied, the functions f and g are said to be convolvable (with respect to p). Any function which is equal almost everywhere (with respect to p) to the function h above is called a convolution o f f a n d g (with respect to p) and is writteiIf* g (orf * pg, where it is necessary to bring j3 into the notation). Thus for almost all x we have (14.10.2)

and likewise, using (14.8.4), (14.10.3)

s s

(f* s ) ( x ) = d s - ' x > f ( s >4 T s )

(f * g ) ( x ) =

f ( x s - ')s(s)&(s-

'1 a

s )

almost everywhere with respect to p. When one of the functions equal to h almost everywhere is continuous on G, we adopt the same convention as in (14.8) and call this function the convolution off and g. In particular, when f * g is continuous, we have n

(14.10.4)

It should be remarked that the property off and g of being convolvable does not depend on the choice of left Haar measure p, but their convolution f* pg does. If p is replaced by ap, where a > 0, then we have f * 'pg = a . f * pg. When G is discrete, to say that f and g are convolvable signifies that the family (g(s-'x)f(s)),,G is absolutely summable (5.3.3) for all x E G, and we have (f * s)(x) = s ( s - ' x ) f ( s )

1

S E G

if the Haar measure

p is such that /?({e})= 1.

288

XIV

INTEGRATION IN LOCALLY COMPACT GROUPS

From the results of (14.9) we have, in particular: (14.10.5)

Suppose that f andg are locally P-integrable. Ifone of the functions

f,g is continuous, and if one has compact support, then f and g are convoluable, the right-hand sides of (14.10.2) and (14.10.3) are dejned for all x E G , and the function f * g is continuous. If both f and g belong to X,-(G), then so does f*s. This follows from (14.9.1) and (14.9.3). (14.10.6) Let f be a P-integrable function. (i) For p = 1, 2 or co,the function f is convolvable with every function g E 49g(G,P); the function f * g belongs to Yg(G,P); and

+

(14.1 0.6.1)

(ii) Zfp

=

+

00,

and g E Yg(G, P), the integral

s

J g ( s - ' x ) f ( s >d P 6 ) = f ( x s - l)g(s)A(s-l)dP(s)

is dejned for all x E G, and the function XH lf(xs-')g(s)A(s-') dP(s) is uniformly continuous with respect to every right-invariant distance on G. (iii) If p = 1, we have

(iv) Zfg

E

Wg(G), then also f

* g E @(G).

Parts (i) and (iv) follow from (14.9.2) and the relation 11 f * PI1 = N , ( f ) (13.20.3). T o prove (14.10.6.2), we remark that by virtue of (14.8.1.1) and the fact that the function (s, x ) ~ g ( x ) f ( sis) (P 0 P)-integrable (13.21.14),

the function (s, x ) w g ( s - ' x ) f ( s ) is also (P 0 /?)-integrable. The formula (14.10.6.2) then comes immediately from the Lebesgue-Fubini theorem and the left-invariance of P. As to (ii), for each x E G the function S w g ( s - l x ) belongs to 49Z(G),and therefore the integral on the right-hand side of (14.10.2) is defined for all x E G. If we put v = A-' P, then v is a right Haar measure (14.3.4), and we may write (14.10.3) i n the form

10 CONVOLUTION OF TWO FUNCTIONS

Consequently,

s s

tcf * g>(x>- ( f * g)(x')i 5 N d g ) If (xs- '1 - f (x's-')l = N,(g)

289

dv(s)

If(s-') - f ( X ' x - ' S - ' ) l

~v(s).

The result will therefore be a consequence of the following more general lemma: (14.10.6.3) For p = 1 or p = 2, every right Haar measure v on G and every function h E YE(G,vj, the mapping s ++S(s)h of G into Yg(G,v) is continuous and satisfies N,(S(s)h) = N,(h).

The second assertion is an immediate consequence of the right-invariance of v . To prove the first, suppose first of all that h E X c ( G ) ;then the continuity of S H 6(s)h follows from (14.1.5.5). In the general case, if (h,) is a sequence of functions in Xc(G )which converges to h in Y $ ( G ,p) (13.11.6), the relation N,(S(s)h - 6($)h,) = N,(h - h,) shows that the sequence of functions S H ~ ( S ) ~ ,converges uniJormIy on G to the function s ~ 6 ( s ) h Hence . the result (7.2.1). (14.10.6.4) In the same way, one shows that if h E 2 g ( G ,p) wherep = 1 or 2, the mapping s H y(s)h of G into Yg(G,p) is continuous, and that N,(y(s)h) = N,(h). (14.10.7) Let f e B,?(G, p) and let g E YZ(G,8). Then the integral

f g ( s - ' x ) ( f (s) dp(s) is dejined for all x E G, and the function f %g(G)and satisfies the inequality

*g

belongs to

For each x E G, the function st+g(s-'x) belongs to YZ(G,j?), and therefore the first assertion follows from (1 3.11.7). Moreover, again from (1 3.11.7),

290

XIV

INTEGRATION I N LOCALLY COMPACT GROUPS

which is (14.10.7.1). If both f and g are in Xc(G),then so is f * g ; since the formula (14.10.7.1) shows that the bilinear mapping (f,g ) H f * g of 9 i ( G ,P) x Y:(G, into 9?,(G) is continuous (5.5.1), and since X c ( G )is (13.11.6), the values of the mapping dense in Y;(G, P) and in Y;(G, (f,g ) t + f * g belong to the closure of X c ( G )in Bc(G) (3.11.4): that is, they belong to @(G) (1 3.20.5).

B)

B)

Proposition (14.10.7) implies the following corollary: (14.10.8) Let A and B be two 8-integrable sets in G. Then the function xt+fl(A n xB) is continuous on G and tends to 0 at injnity (13.20.6). If B-' is also P-integrable, the function x H P(A n xB) is P-integrable, and we have

"A

n xB)

= P(A)P(B-').

If moreover neither A nor B is @-negligible, the set AB-' has a nonempty interior. Finally, for every subset A which is P-measurable and not P-negligible, the set AA- is a neighborhood of the neutral element e of G.

'

We have qAE 9 ' ( G , P) and q B E 9'(G, P), hence we may apply (14.10.7) with f = qA and g = 3jB. Since 3jB(s-'x) = qB(x-'s) = (pXB(s), we have

I

( q *~~ B ) ( x = ) (PA x&) dP(s) = P(A n xB). This proves the first assertion. If B-' is also /?-integrable, then both qAand ( p B - l belong to 9 ' ( G , P), and

we can apply (14.10.6.2), which gives us the formula P(A n xB) dP(x) = P(A)P(B-'). If the right-hand side of this formula is nonzero, it follows that P(A n xB) is not identically zero, and hence there exists a nonempty open set U in which the continuous function P(A n xB) is >O. This implies that U c AB-'. Finally, to prove the last assertion, we observe that there exists a compact set K c A which is not P-negligible, so that we may assume that A is compact; but then /?(A)= (qA* 3jA)(e)is >O, and it follows as above that there exists a neighborhood of e contained in AA-'. (14.10.9) Let A g be two locally /I-integrable functions, p n measure on G. The two following conditions are equivalent: (a) .f andg' are convolvable and the function If I * Ig'I is P-integrable: (b) ,Z u n d f are conuoluable and the function g(l,ilI * I j l ) is /I-intqgrable. If these conditions are satisjied, we have (14.10.9.1 )

( g ?ii

* ( f . P)>

=

( f * g'? P ) .

From (13.21.9) and (14.8.1), it follows that condition (a) means that the

10 CONVOLUTION OF TWO FUNCTIONS

291

function (s, x) + + g ( x - ' s ) f ( x ) = g(s-'.x)f(x) is (p 0 /?)-integrable, and the right-hand side of (14.10.9.1) is then equal to (14.10.9.2)

Similarly, condition (b) means that the function (s, x ) H g ( x ) f ( s x ) is (p @ /?)integrable, and the left-hand side of (14.10.9.1) is then equal to nn

(14.1 0.9.3) But from the theorem of Lebesgue-Fubini and the left-invariance of /?,it follows that one of the integrals (14.10.9.2), (14.10.9.3) exists if and only if the other one exists, and then both are equal.

PROBLEMS

1. (a) Let g be a /%measurable function of compact support on G. Show that f a n d g are convolvable and that f * g is rontinltort.7 on G in the following two cases: (i) g is essentially bounded a n d f i s locally /%integrable; (ii) g E -4Pf(G,/If) i s,/&measurable a n d f 2 is locally ,%integrable. (b) Assume that G is unimodular. Show that if f e -Yg(G,8) and g E Yg(G,P), 1 1 where p 2 1 , 9 2 1 and - - 2 1 , then f and g are convolvable, f * g E -Y;(G, P) P 9 1 1 1 where - = - - - 1 , and N,(f* g) 5 N,(f)N,(g) (W. Young's inequality). (Consider r P 9

+

+

1

+1

first the case where - - = 1, then use (14.10.6.1) and the Riesz-Thorin theorem P 9 (Section 13.17, Problem 7).) 2.

Let G be a compact group, p the Haar measure on G for which P(G) = 1 , and let A, B be two p-integrable sets. Show that for each 8 > 0 there exists x E G such that P(A n xB) 2 (1 s)P(A)p(B). Deduce that, for each integer n such that B(A) 2 l/n, there exist n points x,,. . . , x, in G such that @(xlAu x2 A u ... u x , A ) 2 6. (Consider the sets x, . CA.)

+

3.

Let G be a compact commutative group, P the Haar measure on G for which P(G) = 1, and g a measurable function on G. Let (r,) be the orthonormal system of Rademacher functions (Section 13.21, Problem 10). Let CL > 0 and let A be the set of x E G such that Ig(x)l > a . Let n be an integer 2 1 such that n . P(A) 2 I . Show that there exists a P-measurable set B such that @(B) 2 i,and n points s,, . . . , s, in G such that, if we put F(x, t ) =

2

k=l

rk(f)g(.skx),the following is true: for each x E

B, there e x i m a finite union l(x) of

intervals in [O, 11 such that (i) x(l(x)) 2 4 (A being Lebesgue measure); (ii) for all

292

XIV

INTEGRATION IN LOCALLY COMPACT GROUPS

t E I(x), F(x, t ) 'i. a . (Use Problem 2, and observe that, by virtue of the relation r.(l - t ) : - r n ( f ) , for a given integer h E [ I , n] the set of t E [0, 11 such that the x ) the same sign, has measure 2 4.) numbers rh(f)g(shx)and ~ r k ( ? ) g ( s khave k#h

4, Let G be a compact commutativegroup, @theHaar measure on G for which p(G) = 1 , Let (UJ be a sequence of continuous endomorphisnis of L:(G, p), each of which commutes with all translationsf~(y(s)f)" for all s E G . For each f~ Y:(G, p) we denote by U. .fany function belonging to the class U , .f, and we put

u* .f = sup U" .f: n For each a > 0, let E,(f) be the set of points x E G such that ( U * .f ) ( x ) > a. (a) Let f~ Y i ( G ,6) be such that N2(f) 5 1, and let n be an integer such that n.p(E,(f))> ] . L e t s l , ..., s,bepointsofGsuchthat theunionBofthesetss;'E,(f) has measure p(B)

2 4 (Problem 2), and let F(x, 1 ) =

"

k=l

r k ( t ) f ( s k x )Show . that for each

a finite union of intervals I(x) c [0, 11 such that )\(I(x)) 2 1 and such that x E E,(F( . , t ) ) for all t E I(x). (Observe that if x E B there exists an integer m and an integer j E [ I , n] such that ( U , .f ) ( s j x )= 0, and apply Problem 3 to g =

x E B there exists

u,. .f:)

(b) Let S c [0, 1 1 be a A-integrable set such that h(S) 2 i. Show that there exists t E S such that P(E.(F(. , t ) ) ) > &. (If H is the set of (x, t ) E G x [0, 11 such that (U*. F( . ,t ) ) ( x ) > a , remark that for each x E B we have A(H(x)) 2 2, and deduce that (/3 OAHH) > t.1 (c) For each M > 0, let S, be the set of t E [0, I] such that N,(F(. , t ) ) 5 M . Show that h(S,) 2 1 - n/M2. Deduce from (b) that if M2 4n there exists r E [0, 11 such that both N,(F(. , t ) )5 M and p(E.(F(. t ) ) )> &. (d) Suppose that for eachfE gi(C, p) the function U* .fis finite almost everywhere. Show that there exists a constant C > 0 such that, for each f E -LP:(G, p) and each a > 0, we have (P(E,(f)))"* 5 Ca-'N,( f ) (E. Stein's theorem). (Deduce from Section 13.12, Problem 12 that there exists a constant c > 0 such that for each function h E Y i ( G ,p) satisfying N2(h) 5 M, we have /3(EcM(h))< &. Then make use of (c) above, taking M = LY/C, h = F( . , t ) and n = [iM'].) 5. Let A be Lebesgue measure on R and let f be a compactly supported h-integrable function. Put

for each F

> 0, and 1

rh

these are lower semicontinuous functions on R (Problem 1) with compact support. @( f ) is the Hardy-Littlewood maximal function relative to f . (a) For each LY > 0, let Ee,=(f) be the set of x E R such that @ , ( f ) ( x )> a . Every compact set K c E,,.(f) is contained in the union of a finite number of compact intervals

Ik

(1 5 k

5 n) such that

a

'

5

j Ik

If(t)l dt.

11

293

REGULARIZATION

(b) Show that there exists a sequence of indices ( k , ) l s , s , such that the intervals I, are mutually disjoint and such that

(We may assume that no 11,is contained in the union of the I, such that h # k . Show br] and arrange the Ix so that a, 5 a k + l , then necessarily that if we put Ix = [ak, ax< a k + l , b2,-, < a z r + l and b 2 k < a Z X + 2by , considering three intervals with consecutive indices. Deduce that the intervals are mutually disjoint, the Izr are mutually disjoint and that the k, may therefore be taken to be either the even indices or the odd indices.) (c) Deduce from (a) and (b) that if E.(f) is the set of x E R such that 8 ( f ) ( x )> a, then

s

(d) Let g be a nonnegative function in -YA(x) such that (i) g ( - t ) decreasing on [0,

+ a,[;

(iii)

g ( t ) dt = 1 . Prove that

=g(t);

(ii) g is

I (g * f ) ( x )I 5 8 ( f ) ( x ) for

all

R . (For each CL > 0, let ]-h(a), h(cr)[be the largest open interval in which&) > a . Show that x

E

=lo+s””’

(9 * f ) ( x )

and observe that

5

da

f ( x - t ) dl

-h(a)

+m

h(a)da = 4 .)

0

(e) State and prove analogous results for functions which are integrable with respect to Haar measure on the torus T.

11. RECU LARl ZATlON

(14.11.1) Let (f,) be a sequence of P-integrable functions which satisfy the following conditions:

(a) the sequence of integrals If,(x)I dp(x) is bounded;

s

1 V

(b) the sequence of integrals f,(x) dp(x) tends to 1; (c) for each neighborhood V of e, the sequence of If,(x)l dp(x) tends to 0.

integrals

Then (i) For each bounded continuous function g on G , the sequence (f,* g ) converges uniformly to g on every compact subset of G. g g is uniformly continuous with respect to a right-invariant distance on G , then the sequence (f.* g ) converges uniformly to g in G.

294

XIV

INTEGRATION IN LOCALLY COMPACT GROUPS

(ii) Zfp = 1 or 2 and g E Z P P ( Gthe ) , sequence of norms N,((f, * g ) - g) tends to 0 as n tends to +a.For each function g E L,"(C), the sequence ( ( f ,. g ) " ) converges weakly to y" in L,"(G), considered as tlze dual of Lk(G) (12.15 and 13.17). (iii) Suppose in addition that the supports of the f , are contained in a fixed compact subset of G . Then for each measure p on G , the sequence of measures

cc * (f,*PI = (cc *f,). P converges vaguely to p (1 3.4).

(i) For each x from the definitions

E

g(x) - ( f n

G and each compact neighborhood V of e we have

* g)(x)

= g(x)

+

(1 -

I

I

fn(s)(g(x)

- lci/.(S)g(s-

1

f n ( s > dP(s)

- g(s-

'x)) dP(s)

1x1 d ~ ( s ) .

Let Vo be a compact neighborhood of e and let L be a compact subset of G . Since V i ' L is compact (12.10.5), the restriction o f g to V;'L is uniformly continuous with respect to a right-invariant distance on G (3.16.5). Hence, for each E > 0, there exists a compact neighborhood V c V, of e such that Ig(x) -g(s-'x)( =< E for all X E L and all ~ E V Now . choose no so that If,(s)I dP(s) 5 E and ( 1 - Sf. dP(s)I 0 5 E for all n 2 n o . We have then

scv

and therefore, for all x E L,

I 1 REGULARIZATION

295

Ifg is uniformly continuous with respect t o a right-invariant distance on G, the same argument applies, taking L = G. (ii) There exists a function h E X,(G) such that N,(g - h) 5 E (13.1 1A). By (14.10.6.1), it follows that N,((f, * g ) - ( f , * h)) 5 aN,(g - h) a&.Thus we are reduced to proving (ii) when g E X c ( G ) .Let S be the support of g . If V is a compact neighborhood of e, we have seen in (i) that f, * g - g converges uniformly to 0 in the compact set K = S u VS. Next, if x # K, we have

(f" * g ) ( x ) - g ( x ) =

f,(s)g(s-

XI

dP@)

Since, on the other hand, the integral I ( f , * g ) ( x ) - g ( x ) I @ ( x ) tends t o 0 SK with l/n, we see that

and as N,(f,

*g

-g)

5 (a + l ) ~ ~the g ~same ~ , is true of

Suppose now that g E 2'g(G), and let 11 E 2'A(G). If V is any compact neighborhood of e in G, we have

296

INTEGRATION IN LOCALLY COMPACT GROUPS

XIV

But we have n

and therefore,

For each E > 0, it follows from (14.10.6.4) that there exists a compact neighborhood V of e in G such that N,(h - y(s-')h) 5 E for all s E V, and hence,

Examples (14.11.2) Let (V,) be a fundamental system of neighborhoods of the neutral element e in G . Since the support of p is the whole of G, there exists for each n a function f n e X + ( G ) with support contained in V, and such that /fn(x) @(x) # 0 (13.19.1). Multiplying each f , by a suitable constant, we

s

may assume that fn(x)@ ( x ) = 1, and the sequence (f,) now satisfies the conditions of (14.11.1). We see therefore that every measure p on G may be approximated (in the sense of the vague topology) by a sequence of "regularizations " which are measures having a continuous density with respect to /I(14.9.3).

I1

REGULARIZATION

297

(14.11.3) Take G = R, and /3 to be Lebesgue measure. Put (14.11.3.1)

gn(x) =

(I:-

if X E [-I, if 1x1 > 1.

x2)"

11,

=I1

Let a, g,(x) dx, and f, = ai'g,. The sequence (f,)satisfies the conditions of (14.11.1). For 1 - x2 2 1 - 1x1 for 1 S x 5 1, hence a,

2 2 Jol(l- x)"

2 n+1'

dx = -

+

and therefore f,(x) 5 ( n 1)(1 - x')" for all x E [ - 1, 11, which proves that f,(x) + 0 uniformly on every compact interval not containing 0. Let p be a measure on R with support contained in [-f, f]. Then we have

and if x E [ -$,+I

this gives (p * jn)(x) = a,

5

112

1

- 112

(1 - (x - Y)')" ~ A Y ) ,

showing that the function p * f, is equal to a polynomial function on [ -3, f]. In particular, if p = h p, where h is a continuous function with support contained in [-$, $], we obtain from (14.11.l(i)) the theorem of Weierstrass on uniform approximation of continuous functions by polynomials on a compact interval (7.4.1).

-

PROBLEMS

1.

If a locally compact group G is such that the algebra S ( G ) is commutative with respect to convolution, show that G is commutative. (Show by regularization that the algebra of measures with compact support is commutative.)

2.

Let G be a locally compact group, p a left Haar measure on G. Show that the algebra L'(G, p) has a unit element if and only if G is discrete. (Suppose that G is not discrete, and let fo E -LP'(G,p); then there exists a compact neighborhood V of e such that

< 1. Show that, if U is a compact symmetric neighborhood of e such that Uz t V, then I(rpu * f o ) ( x ) I < I for almost all x E U, and hence that f0 cannot be Jv Ifo(x)l # ( x )

the unit element of L'(G,

6))

298

3.

XIV

INTEGRATION IN LOCALLY COMPACT GROUPS

Let G be a locally compact group, /? a left Haar measure on G. If G I { e } , the algebra L'(G, /?) has zero-divisors # 0. To construct two nonnegligible functions f,g in Y ' ( G ,p) such that f * c/ is negligible, we may proceed as follows: ( I ) The case where G has a compact subgroup H #- { e } . Take for f a characteristic function cpa and for y a function of the form cpSR - cpR, where A, B, and s are suitably chosen, and remark that A,(x) = I for all x E H . (2) The case G = Z. Show that we may take

for all 17 t Z, and y =f. (3) The general case. Prove first of all that there exists (I # e in G such that A(a) = 1. The closure H in G of the subgroup generated by a is then either compact or isomorphic to Z (Section 12.9, Problem 10). In the former case, use the result of case ( I ) ; in the latter, take

where the set U and the sequences case (2). 4.

5.

(4 (p.) ,are

suitably chosen with the help of

Let G be a locally compact group, /3 a left Haar measure on G. In order that a subset H of P ( G , 8) (I s p < 0 0 ) should have a relatively compact image R in LP(G,p), it is necessary and sufficient that the following conditions should be satisfied: (1) R is bounded in Lp(G,/?); (2) for each E > 0, there exists a compact subset K of G such - ~E )for all f~ H; (3) for each E > 0, there exists a neighborhood V that N , ( ~ $ J ~ ;g of e in G such that N,((y(s)f) -f) 5 E for aHff H and all s E V. (To prove that these conditions are sufficient, observe that if g E X ( G ) and if L is a compact subset of G, then the image, under the mappingfwg *f,of the set of restrictions to L of functions belonging to H is an equicontinuous subset of X(G).)

+

Let G be a locally compact group, /? a left Haar measure on G , and p a positive measure on G. Show that if A is a p-integrable set and B is a universally measurable set in G, the function u : s ~ p ( n A sB) is /3-nieasurable on G. If moreover B-I is /?-integrable, then so is u and we have j p ( A n sB) d/?(s) = p(A)p(B-'). (Use the Lebesgue-Fubini theorem and Problem 21 of Section 13.9.) Give an example of a measure p such that the function s w p ( A n sB) is not continuous. If p is a measure with basis /? and if A is p-integrable and B is universally measurable, then the function s ~ p ( n A sU) is continuous on G.

6.

Let G be a locally compact group, G' a topological group, f a homomorphism of G into G' which is /?-measurable, where /? is a left Haar measure on G. Show thatfis continuous. (Observe that if we put y(x) = f ( x - ' ) , there exists a compact non-pnegligible subset K of G such that the restrictions o f f and y to K are continuous. Deduce that the restriction o f f t o K K - ' is continuous, by using (12.3.8); then apply (14.10.8).)

11

REGULARIZATION

299

7. Let G be a locally compact group. For each t E R*,, let y, .f 0 be a positive bounded

measure on G. Suppose that the mapping t H p L ,is continuous with respect to the topology .F2(in the notation of Section 13.20, Problem I), and that p-,,, = ps * pt for all s, t in Rf . (a) Show that there exists a real number e such that llp,ll = ecr. (Observe that the mapping I H IIpJ is lower semicontinuous, and apply Problem 6.) (b) Show that, as t + O , pr converges (with respect to Y2) to a Haar measure of total mass 1 on a compaci subgroup of G. (Use (12.15.9) to show that there exists a sequence (s.) tending to 0 such that the sequence (ySn)tends to a limit ~1 in the topology Y2, and that y, * y = yr = y * yr for all t > 0. Deduce that ps+y as s + 0, and that y * u , = y. Complete the proof by using Problem 5 of Section 14.7.)

8.

Let h denote Lebesgue measure on R“.Let A be a bounded convex open set in R” (Section 8.5, Problem 8), and let D(A) = A - A be the set of all x - y where x , y E A. The set D(A) is convex, open, symmetric and bounded. For each x f 0 in D(A), let p(x) be the unique real number, belonging to 10, I[, such that p ( x ) - * x lies in the frontier of D(A). If we put p(0) = 0, then p is a continuous function on D(A) (Section 12.14, Problem 12). (a) For each x e D(A), show that h(A n (A -t x ) ) >. ( I - p(x))”h(A). (Observe that if x # 0 and p ( x ) - I x (1 - p(x))A

=

b - a where a, b

t p(x)b = (1

- p(x))A

E

A, then

+ p(x)a + x ) .

(b) Show that

(c) Deduce from (a) and (b) that

Prove that this integral is equal to A(D(A)) J 1 n t n - l ( l 0

-

t)” dt

= -h(D(A)). (nYZ

(2n)!

(Split up the interval [0, 11 into rn parts by means of an increasing sequence (tk)o 0. Put

clly(z)f-/ll, Show that for each z E G we have l(y(z)g - g(/I

I

where c = f ( s ) dP(s),

and that g(z) < g(e) if z # e . (If not, we should have y(z-')f'=f, which would imply that zx E V for all k.) (b) Let (f.) be a uniformly bounded equicontinuous family of positive functions with support contained in V, and suppose that for each n there exists a subset A. c V such that Ily(z)fI -fJl 5 M for all n and all z E An. Show that the set of functions

is equicontinuous, and that the same is true of the set of functions y(z)g. 2 E An, for each n.

- gn where

10. Let G be a locally compact group and let u, u be two positive bounded functions on G. For each x E G let

w(x) = sup u(y)u(y-'x). Y E C

(a) Show that w(x) = sup u(xy-')u(y),and that for each s E G we have YEG

Ily(s)w - wll

s IIull . Ily(s)u- uII.

If A, B are the supports of u, u respectively, show that the support of w is contained in the closure of AB. (b) Suppose that u ( x - ' ) = v(x). Show that, for all x , x' in G, we have

I w ( 4 - w(x'>l SIly(x)v -Y(X')4 (Observe that for each y

EG

. ll4I.

we have

llull ' l l Y ( X b - Y(X')Ull

2 u(Y)(u(Y-'x) - d y - ' x ' ) ) 2 u(y)u(y-'x) - w(x').)

In particular, if u is continuous and compactly supported, then so is w. 11.

Let G be a locally compact group, and V a symmetric compact neighborhood of e in G such that V 2 contains no subgroup # { e } . For each integer i, let UI be the set of points x such that x x E V for 1 5 k 5 i. As i+ -t- co, the sets U, form a fundamental system of neighborhoods of e in G (Section 12.9, Problem 6b)). Let n, be the largest integer n such that U: c V. Define a function u , on G as follows: ui(e)=- I ,

k u I ( x )= 1 - ni

if x E U:

-

and uI(x)= 0 if x $ U;'. (a) Show that for all x E U, we have Ily(x)ui - uil/ 5 1 h .

U:-'

(1

5 k 5 n,),

11

REGULARIZATION

301

(b) With the help of Problem 10, construct a sequence (w,) of continuous positive functions, with supports contained in V2,and such that IIw,ll 2 1 for all i, the sequence (w,) being uniformly bounded and equicontinuous and such that the set of functions n,(y(z,)wl- w,) is uniformly bounded as zl runs through UI for each i 2 1. (c) With the help of Problem 9, construct a sequence (g,) of positive functions with supports contained in V4, which is uniformly bounded and equicontinuous and such that the set of functions n,(y(zi)gr- g l ) is uniformly bounded and equicontinuous as z, runs through U, for each i 2 1 ; and such that for each function f which is the uniform limit of a subsequence of the sequence (g,), we have f ( x ) < f ( e ) whenever x # e. 12. Let G be a locally compact group with no small subgroups, and let V be a symmetric compact neighborhood ofe, containing no subgroup of G other than {e), and such that for x, y E V the relation x 2 = y 2 implies that x = y (Section 12.9, Problem 6(a)). A

continuous homomorphism r w X ( r ) of R into G is called a one-parameter subgroup of G (Section 12.9, Problem 7). The constant one-parameter subgroup r H e is denoted by 0. For each t E R,the one-parameter subgroup r HX ( t r ) is denoted by t X . If X is a one-parameter subgroup of G, a continuous real-valued function f on G is said to be X-di’erentiable if the function r - ’ ( y ( X ( r ) ) f - f ) converges uniformly on G to a limit Dxf as r+O in R. If so, then f is also tX-differentiable for all real t , and we have D,xf= tDxJ If f is X-differentiable, then for each x E G the function r H f ( X ( - r ) x ) is differentiable on R, and its derivative is the function rHDxf(X(-r)x). A continuous bounded real-valued functionfon G is said to be adequate if there exists a neighborhood V of e in G such that Ily(z)f-fll # 0 for all z E V. The functions fconstructed in Problem ll(c) as limits of subsequences of the sequence ( g l ) are adequate. Show that, if f is adequate and if DJexists for some one-parameter subgroup X # 0, then D x f # 0. If on the contraryfis not adequate, then there exists a oneparameter subgroup X # 0 such that Dxf= 0. (Consider a sequence (a,) of elements # e in G, tending to e, and such that ]ly(a,)f-fli = 0. Show that it can be assumed that there exists a sequence of integers mj with the property that for all r E R the sequence a?’ converges to X(r), where X # 0 (Section 12.9, Problem 7).)

13. The notation and assumptions on G and V remain the same as in Problem 12. Let (f,) be a sequence of bounded continuous functions on G which converges uniformly on G to a function J Let (a,) be a sequence of elements of G tending to e, and (mi) a sequence of integers >O.

(a) Suppose thatfis adequate and that the functions m,(y(a,)& -f,) are uniformly bounded. Then, for each neighborhood U of e, there exists a number E > 0 such that the relation Irl < E implies a?” E U. (Suppose that U2 c V. Let a = SUP m,II Y(aJlf,- A It ; J

by taking j sufficiently large, we may suppose that a, E U and ily(z)fi -fill 2 > 0 for all z E V - U. Show that we cannot have a ) E V - U for k < a m j / P , by using the inequality

(b) Suppose that the functions m,(y(a,)f, - f i ) are uniformly bounded and uniformly tends to X(r), equicontinuous, and also that for each r E R the sequence (a?’) where Xis a one-parameter subgroup. Show that Dxfexists and is the limit of the

302

XIV

INTEGRATION IN LOCALLY COMPACT GROUPS

sequence (m,(y(a,)f, -f,)). (It is enough t o prove that if the sequence of functions (m,(y(a,)h -f,)) converges to a function F, then Dxf exists and is equal to F. Fix a neighborhood W of e such that / l y ( z ) F- FII 5 BE for all z E W, then fix a real number 6 > 0 such that at E W for 0 k < 6 m,. Now show that for large j , and k < a m , , we have

and deduce that I l r - ’ ( y ( X ( r ) ) f - f ) - FII 2 e for 0 5 r < 6.) In particular, if the sequence (a”””) tends to e for all r E R, then the sequence (m,(y(a,M; -A)) tends uniformly to 0. 14.

(a) Consider the sequence ( g , ) constructed in Problem 1 l(c). Show that the sequence of numbers i/nl is bounded above. (Argue by contradiction, by supposing that there exists a subsequence ( i ( k ) ) k e Nsuch that lim nI(*,/i(k)= 0. Reduce to the case where k+ m

the sequence (gr(k))converges uniformly to a limit g in G , and there exists a sequence of points q ( k ) E Uy:;!) converging to an element c # e . Then there exists an element O ~ ( LE) ul(k)such that

IIY(cl(k))gl(k,- gI(k,//5 nl(k)dY(‘l(k))gl(k)- gl(k)II. Now use the hypothesis

hi

k-r m

nf(k)/i(k)= 0, Problem 7 of Section 12.9, and the definition

( ,r;f(”’])

tends to e. Using Problem of the U1 to deduce that for each r E R the sequence 13(b), show that this implies that IIy(c)g -glj = 0 , contrary to the fact that g is adequate.) (b) Show that if g is the uniform limit of a subsequence of the sequence (gi),then X-di’erentiirble for every one-parameter subgroup X . (Using (a) and Problem I I(c), show that the functions i ( y ( X (l / i ) ) g l- g l ) form a uniformly bounded and uniformly equicontinuous set, and apply Problem 13(b).)

y is

15.

Let G be a unimodular locally compact group and /3 a Haar measure on G . Show that a functionfe -Epz(G,8) is equal almost everywhere to a function which is uniformly continuous with respect to a right-invariant distance on G if and only if the function s++y(s)A with values in .YZ(G, P), is continuous at the point e . (To show that the condition is sufficient, consider a sequence (u.) of functions belonging to X + ( G ) such that Supp(u,) c V, (notation of (14.11.2)) and

s

u.(x) d P ( x ) = 1. Prove that

N,,,(u, *f-f) tends to 0 with l/n, by using (13.17.1) and the Lebesgue-Fubini theorem.) 16.

If p is any bounded measure on R,the integral

is defined for every z E C such that 9 z # 0, and is an analytic function of z in each of the half-planes 9 z =. 0, .P’Z < 0. It is analytic also at all x E R which do not lie in the support of p. The function F, is called the Stieltjes trnnsjorm of p. (a) Let x o E R. Suppose that the restriction of p to an open neighborhood V of 0 has a continuoits density g with respect to Lebesgue measure on V. Show that as y tends to 0 through positive values, Y(F,(xc -1 iv)) tends to a limit equal to xg(x0).

REGULARIZATION

11

303

Show that if the analytic function F, is identically 0 on the half-plane Y z > 0, then p = 0. (Replacing p by a regularization p * (f.p), show that (b)

and use (a) to conclude that p

* (f.p) is zero.)

17. Let G be a holomorphic function in the disk B : 1z( < I . Show that .%G(z)2 0 for all z E B if and only if there exists a positive measure v on the interval [ 0 , 2 ~ such ] that

where c have

E

R. (To show that the condition is necessary, note that for Izl < r < 1 we G(z) = i . SG(0)

+

PSV) dp,

where p r ( q ) = WG(re"). Observe that the measure p , . h (where A is Lebesgue measure on [0, 2 ~ 1is) positive and of total mass MG(O), and use (13.4.3).) Deduce that for a function F holomorphic in the upper half-plane 9 2 > 0 to be such that d F ( z ) 2 0 for all z in this half-plane, it is necessary and sufficient that there should exist a bounded positive measure p on R and two constants a 2 0, b E R such that

for 9 z > 0. (Map B onto the upper half-plane by means of z-i(l

+ z)/(l

- z)).

18. (a) Let h be Lebesgue measure on R, and let f be a A-integrable function. For each h > 0, put

j

I * f h ( x )= f(x - t ) dr. 2h - h

Show that, for almost all x E R, &(x) +f(x) as h -+ 0 i - (Lebesgue's theorem). (If we put (Section 12.7, Problem 8)

R(f)(x)

= lim SUP h-0

lfh(x)-f(x)

I

o(f) +

then R(f-g) = R(f) for all g E .X(R); also R(f) 5 Ifl, with the notation of Section 14.10, Problem 5. Use Section 14.10, Problem 5(c) to deduce that the set of x E R such that R(f)(x) > CL has measure zero, for all a > 0.) (b) Let (gn)be a sequence of nonnegative functions in X ( R ) , satisfying conditions (a), (b), and (c) of (14.11.1); assume also that g n ( - t ) == gn(f)and that gn is decreasing in 10, a[.Show that as n + co,(9. *f)(x) +f(x) almost everywhere in R (Lebesgue's theorem). (Same method, using Section 14.10, Problem 5(d).) (c) State and prove analogous results for Haar measure on the torus T.

+

CHAPTER X V

NORMED ALGEBRAS A N D SPECTRAL THEORY

The spectral theory of operators, which we have already encountered in an elementary aspect in Chapter XI, is one of the masterpieces of modern analysis. Its main object is to obtain, for linear operators on a Hilbert or prehilbert space satisfying suitable continuity conditions, an analog of the classical theorem of algebra which assigns a canonical form (by means of “Jordan matrices”) to an n x n matrix over C (or, equivalently, to an endomorphism of aJnite-dimensional complex vector space). In Chapter XI we have seen how this result, suitably modified, can be extended to compact operators. But there is another much less obvious generalization, due to Hilbert and his successors, which applies to a wider class of operators: in particular, to continuous self-adjoint operators (11.5), and more generally to normal operators (15.11). Just as in the classical case a self-adjoint (or normal) matrix over C has a canonical form which is a diagonal matrix, the continuous normal operators can all be described in terms of a single model: namely multiplication M,(u) : . f ~ ( u f )in- a space Li(p) by the class of an essentially boundedfunction u (15.10). The Lebesgue theory intervenes here in an essential way (even if we begin with a self-adjoint operator which comes from a differential equation as regular as we please (Chapter XXIII)), and it is no exaggeration to say that the occurrence of Lebesgue theory in spectral theory and related fields such as harmonic analysis and the theory of representations of locally compact groups is the principal reason of its importance in Analysis. A modern account of spectral theory does not follow the path mapped out by Hilbert, but uses a much more elegant and powerful method based on the theory of normed algebras inaugurated by Gelfand and his school. In this chapter we shall concentrate mainly on normed algebras with involution (15.4), because these are the ones which arise in spectral theory. But the general theory of normed algebras, and especially the fundamental concepts of 304

1 NORMED ALGEBRAS

305

spectrum and Gelfand transformation (15.3), have found many other applications in modern analysis, notably in the theory of analytic functions. Some of these applications are touched on in the problems, and the reader interested in this aspect is referred to [35] and [29]. The central part of this chapter is the study of representations of algebras with involution, which enables such an algebra, given “ abstractly,” to be “ realized ” as an algebra of operators on a Hilbert space. The essential notion in the modern development of this theory is that of a Hilbertform, which is closely related to that of a Hilbert algebra (15.7). We shall study in detail only two particular aspects of the theory of Hilbert algebras: the first (1 5.8) prepares the way for the theory of representations of compact groups (Chapter XXI), and the second (15.9) for spectral theory and harmonic analysis (Chapter XXII). The reader who wishes to go further (notably in view of the deep and difficult theory of representations of locally compact groups) is warmly recommended to read the two beautiful volumes by J. Dixmier ([24] and [25]) which dominate the subject. The Hilbert spectral theory (15.10 and 15.11) appears in our treatment as an immediate particular case of the general theorem of Bochner-Godement (15.9). It can be reached more directly and rapidly (Section 15.10, Problem 2) from the Gelfand-Neumark theorem (15.4), but it seemed to us to be more instructive to deduce it from a much more powerful theorem which is the cornerstone of harmonic analysis, even if this requires a small additional effort. The applications of spectral theory are not limited to those referred to above. Among the most celebrated we should mention at least the following: (1) one of the most elegant theories in Analysis, namely the “moment problem ” inaugurated by Stieltjes, with its many ramifications (analytic functions, orthogonal polynomials, Jacobi matrices, continued fractions, etc), which fits admirably into the theory of unbounded Hermitian operators; (2) the interesting relations between ergodic theory and spectral theory; (3) perturbation theory. Some of the important results of these theories are mentioned in the problems, and the reader is referred for more ample information to the works [20], [281, [301, and [321 in the References. I. N O R M E D ALGEBRAS

When we speak of algebras in this chapter, we shall always mean algebras over the field of complex numbers C . A normed algebra is defined to be an algebra A endowed with a norm X H llxll (5.1) satisfying the inequality (15.1.1) for all x, y in A.

llwll 5 llxll * llyll

306

XV

NORMED ALGEBRAS AND SPECTRAL THEORY

If also A has a unit element e and is not the zero algebra (in other words, if e # 0) we shall always assume that the norm satisfies the additional condition (15.1.2)

llell = 1.

The inequality (15.1.1) shows that the bilinear mapping (x, y ) ~ x of y A x A into A is continuous (5.5.1).

A complete normed algebra (i.e., a normed algebra in which the underlying normed vector space is a Banach space) is called a Banach algebra. It is clear that every subalgebra B of a normed algebra A (if A has a unit element e, we require that B contains e), endowed with the restriction to B of the norm (IxII, is a normed algebra. If in is a closed two-sided ideal of A, then the quotient algebra A/in, endowed with the norm (12.14.10.1) induced by IIxII, is again a normed algebra. For if i, j are two elements of A/m, then for each E > 0 there exist x E i and y E j such that

+

+

from which it follows that llxyll 5 ( 1 1 1 li e)(IIj(l E ) ; since xy E i j and since E was arbitrary, this shows that Ilijll 5 llfll . 11j11. If moreover A has a unit element e, and if in # A, then i is the unit element of A/in and is ZO,hence /(C((S (le((= I by definition, and on the other hand I(t(( = (lt2(I S j(tj(2, which implies that \[ti] >= 1. Hence llell = 1, and so A/in is a normed algebra. (15.1.3) Let A be a normed algebra. Then the closure in A of a subalgebra of A (resp. of a commutative subalgebra, a left ideal, a right ideal) is a subalgebra (resp. a commutative subalgebra, a left ideal, a right ideal).

By virtue of (5.4.1) and the principle of extension of identities, this follows directly from the continuity of multiplication in A, by the same proof as in (5.4.1). If A, B are two normed algebras, an algebra isomorphism u : A + B is said to be a topologicalisomorphism if it is bicontinuous, that is if (5.5.1) there exist two real numbers a > 0 and b > 0 such that a llxll iIIu(x)II 5 b IIxI/ for all x E A (5.5.1). The isomorphism u is said to be isometric if in addition we have IIu(x)ll = (Jx(1 for all x E A.

1 NORMED ALGEBRAS

307

Examples of’ Normed Algebras (15.1.4) For each nonempty set X, the set ,%,(X) of bounded complex-valued functions on X is a commutative Banach algebra with respect to the norm (7.1 .I) and ordinary muItiplication of functions: the inequality (15.1 .I) is clearly satisfied, the constant function 1 is the unit element of &?,-(X) and satisfies (15.1.2). If X is a topological space the subspace %‘F(X)of bounded continuous functions on X is a closed subalgebra of W,(X). If X is metrizable, separable and locally compact, the space @(X) (13.20.5) of continuous functions which tend to 0 at infinity, and the space .X,(X) of continuous functions with compact support, are ideals in the algebra g;(X), the former being the closure of the latter. (15.1.5) If X is the closed disk IzI 5 1 in C, then the set &(X) c U,(X) of continuous functions on X which are analytic in the interior IzI < 1 of X is a closed subalgebra of g,(X), by virtue of the theorem of convergence of analytic functions (9.12.1 ). (15.1.6) We have already seen (11.1) that the algebra 9 ( E ) of continuous endomorphisms of a complex normed space E is a normed algebra (in general noncommutative) with the identity mapping 1, of E as unit element (the inequality (15.1.1) in this case is just (5.7.5), and the relation (15.1.2) is obvious). Furthermore, if E is a Banach space, then 2 ( E ) is a Banach algebra (5.7.3). The set of compact operators in 2 ( E ) is a closed two-sided ideal by virtue of (11.2.6) and (11.2.10). (15.1.7) Let G be a separable metrizable locally compact group. The set Mh(G) of bounded measures on G, endowed with the convolution product (A,p ) A *~ p and the norm (13.20.1) is a Banach algebra. For in this case the inequality (15.1.1) is just (14.6.2.1) with n = 2; the unit element E, (the Dirac measure at the neutral element e of G) is such that lleell = 1 ;and M;(G), being the dual of the normed space Wg(G) (13.20.6), is complete (5.7.3). The subspace of MA(G) consisting of measures with base a left Haar measure on G can be identified (together with its norm) with the Banach space LA(G, p), because N,(f) = llf. fill (13.20.3), and it follows from (14.9.2) that if G is unimodular, LA(G, 8) is a closed two-sided ideal in Mr(G). We shall always identify LA(G, p) with this ideal (the identification of course depends on the choice of p). Under this identification, X,(G) is identified with a subalgebra (14.10.5) of MA(G), which is not, in general, an ideal (unless G is compact (14.9.1)) and, in general, is not closed, because its closure is LA(G, p) (13.11.6).

a

308

XV

NORMED ALGEBRAS AND SPECTRAL THEORY

Remark (15.1.8) An algebra A over C , endowed with a topology, is said to be normable if its topology can be defined by a norm with respect to which A becomes a normed algebra in the sense defined above. For this it is necessary and sufficient that the topology of A should be definable by a norm, and should be compatible with the vector space structure of A, and that the mapping ( x , y ) x y ~of A x A into A should be continuous for this topology. Clearly these conditions are necessary. To see they are sufficient, observe that if llxll is a norm defining the topology of A, then there exists a constant a > 0 such that llxyll 5 a * [I.v[l . (lyl( for all x , y in A (5.5.1). If we replace the norm IIx(I by the equivalent norm (5.6.1) allxll = IIxII1, then clearly llxylll 5 llxlll * llylll. This establishes our assertion in the case where A has no unit element. If A has a unit element e # 0, we consider for each x E A the left translation L, : YH xy; this is a continuous endomorphism of the normed space A, and we have L,+,, = L, + L,. and L,, = I L , for all I E C , and L,,, = L, L,,. In other words, the mapping X H L, is a homomorphism of the algebra A into the algebra Y(A) of continuous endomorphisms of the normed space A. Moreover, this homomorphism is injective, because the relation L , = 0 implies x = xe = L,(e) = 0. Now we have 11L,11 g a [Ixll (5.7) and on the other hand, llxll = llxell 5 llLxll * Jlel(.It follows that if we put IIxI12 = llL,ll, then llxl12 is a norm on A which is equivalent to the given norm llxll (5.6.1), and for which A is a normed algebra. 0

PROBLEMS 1. Let A be the C-algebra of complex-valued functions defined and k times continuously differentiable on [0, 11. Show that the function

on A is a norm for which A is a Banach algebra. 2. Let (a,,) be a sequence of real numbers > O such that a,, = 1, a,+. lim a:'"

n-. m

= 0.

Let A denote the set of all formal power series x

complex coefficients & such that

2 m

"=O

m

=

5 a.,a.

and

&Tn with the

"=O

< + a.Show that A is a subalgebra of the

algebra C[[T]] of formal power series in T, and that llxll = for which A is a Banach algebra with unit element.

m "PO

~~15.1is a norm o n A

1 NORMED ALGEBRAS

309

3. Let I ( X ) be the set of continuous functions on the closed unit disk X : IzI 5 1 in C which are analytic in the interior Irl < 1 of X. For each pair of elements x , y E I ( X ) we define (x * Y ) ( 5 ) = 5

I'4 5

- t51Y(t5) dt

for all 5 E X. Show that this product makes d ( X ) a commutative algebra without a unit element, and that with respect to the norm induced by the norm on VC(X) the algebra d ( X ) is a Banach algebra. 4. (a) Let !2 be the set of finite real-valued functions w on R belonging to L'(R,h) (where h is Lebesgue measure) such that w ( - t ) = w ( r ) , w ( t ) 2 0 for all t E R, and such that w is decreasing on [0, a)[. Every function w E !2 is the uniform limit of an increasing sequence (w,) of step functions belonging to 52. Put

+

+1

+m

V(w) = w(0)

w ( t ) dt.

-m

If wl, w z E !2, show that w I * w 2 E and that V(wl * w z ) V ( w l ) . V(w2) (consider first the case where one of wl, w 2 is a step function). (b) Let d denote the set of &measurable complex-valued functionsf on R such that, for at least one function w E R, the function If12w-' is A-integrable (with the convention that 0 . (+ 00) = 0). We have R c d.For each functionf6 d,put

Show that also

Deduce that d is a complex vector space and that N, is a seminorm on A. (Observe that if a, b, a , p a r e real and >0, then

( a + b)2 c(+p

az

bZ

I-+-.) - a j3

Show that I c YA(R, h) n 9 $ ( R , A) and that Nl(f) 2 NB(f) and N2(f) 5 N B ( ~ ) . (c) L e t f s Y 1n Y 2be the function defined as follows:f(x) = l/n2 if 2"- 1 5 x 2" (n an integer L l ) , f ( x ) = 0 otherwise. Show t h a t f e d. (d) For each function f e d,show that there exists wf E 0 such that

(Consider a sequence (w.) of functions belonging to R such that V(w.) such that the sequence of integrals

s

= N,(f)

and

Iflzw;' dh tends to NB(f). Then consider the

function lim sup w , , and use Fatou's lemma.) n-r m

(e) The subspace -4'" of hegligible functions is contained in d.By passing to the quotient, N, induces a norm on the space A = d / N . Show that, with respect to this norm (denoted by N,(f) or I l f l l ) , the space A is a Banach space. (If (f.) is a sequence in

310

XV NORMED ALGEBRAS AND SPECTRAL THEORY

I such that (A) is a Cauchy sequence in A, show that the sequence (wrn)converges in 9, and that the sequence of classes of the functions If.12wi,,’ converges in L’.) (f) Let I. be the interval [-n, n ] in R and d 1 I. the set of functions belonging to d whose support is contained in I,. Show that, for eachfE I ,the sequence of functions f~,,, tends to f i n d .The topologies defined on d I I. by Ne and by the seminorrn N, are equivalent. Deduce that the space S ( R ) is dense in d ,and that the Banach space A is separable. (g) Iffand g are two functions belonging to d ,show that

- 0, then y" = 1 x , and we write

+

+

y = (e

+

+

x)lIrn,

12. With the same hypotheses as in Problem 1 1 , let (Kj)l*j 0 such that llyxll 5 kllxyll for all x , y in A, then A is commutative. (Apply the inequality with e-{' in place of x and ecxy in place of y, and use (b).) (e) Let a E A be such that

+

+

ll(a 5 . 1)xIl 5 llx(a 5 . 1)Il for all 5 E C and all x E A. Prove that a belongs to the center of A. (Prove that, for each 5 E C and all integers n 2 no (where no depends on C),

for a l l y E A ; then let n tends to

+

00

and use (b).)

15. Let A be a Banach algebra with unit element e, and let C be a commutative Banach subalgebra containing e. Show that there exists a commutative Banach subalgebra B of A containing C such that SpR(x) -.Sp,(x) for each x t C.

3. CHARACTERS A N D SPECTRUM O F A COMMUTATIVE B A N A C H ALGEBRA. T H E GELFAND TRANSFORMATION

Let A be a commutative algebra (over C). A character of A is defined to be any homomorphism x of the algebra A into C which is not identically zero. Because x(1x) = Ax(x) for all scalars 1, this condition is equivalent to x(A) = C. If A has a unit element e # 0, then x(e) # 0 (otherwise ~ ( x=x(ex) ) = x(e)x(x)= 0 for all x E A); it follows that x(e) = 1, because x(e)2 =x(e2) = x(e). An ideal in # A in A is said to be maximal if there exists no ideal n such that in # n, it # A and in c n. If A has a unit element, in is maximal if and only if A/in is a (commutative)jeld. (15.3.1) (i) (ii) (iii) onto the

Let A be a commutative Banach algebra with unit element e # 0. For each x E A and each character x of A, we have ~ ( x E) SpA(x). Every character x of A is a continuous linear form with norm 1. The mapping x++x-'(O) is a bijection of the set of characters of A set of maximal ideals of A (which are therefore closed).

Assertion (i) is a particular case of (15.2.8(i)). It follows (15.2.4(iii)) that [x(x)l 5 ((XI(, which shows (5.5.1) that x is a continuous linear form with norm 5 1 ; and since x(e) = 1 and //el/= 1, we have lIxil = 1. Finally, since

3 SPECTRUM OF A COMMUTATIVE BANACH ALGEBRA

319

x(A) = C, the quotient algebra A/m is a field isomorphic to C, hence in is maximal. Conversely, let it be a maximal ideal of A. The fact that n is closed is a consequence of the following lemma: (15.3.1 . I )

If a # A is an ideal in A, then its closure ti is an ideal # A .

For the complement C a contains the set G of invertible elements of A , which is open in A (15.2.4(i)), and hence G is contained in the complement of ii. Applied to the maximal ideal n, this lemma shows that ii= it, because n and ii# A. The quotient normed algebra (15.1) A/n is therefore a Banach algebra (12.14.9) which is a field because n is maximal, hence bv the Gelfand-Mazur theorem (15.2.5) is isomorphic to C. In other words, there exists a unique homomorphism x : A -+ C such that x(e) = 1 and x - ' ( O ) = 11, and the proof of (iii) is complete. -

11

=)

(15.3.2) Let A be a Banacli algebra with unit element e # 0. The set X(A) of characters o j A i s a subset of the unit ball J/x'// 5 1 in the dual A' of the Banach space A, and is closed in A' with respect to the weak topology (12.1 5 ) . A is separable then X(A) is metrizable and compact for the weak topology.

In view of (12.15.9) we need only prove that X(A) is a closed subset of A', or equivalently that if u lies in the closure of X(A) in A' for the weak topology, then ( x y , u ) = (x, u ) ( y , u ) for all x,y E A, and ( e , u ) = 1. But this follows from the continuity of the mapping U H (x, u ) of A' into C (for each x E A) and the principle of extension of identities. The set X(A), endowed with the topology induced by the weak topology on A', is called the spectrum of A. For each x E A, the mapping x H ~ ( x of ) X(A) into C is denoted by 9 x or 9* x, and is called the Gelfand transform of x; the mapping X H ~ Xof A into CX(A)is called the Gelfand transformation. Hence we have by definition, for all x E A and all x E X(A), (15.3.3)

(Bx)(x) =

xw

(15.3.4) Let A be a separable commufatii3e Banach algebra with unit element e # 0. (i) The Gelfand transformation XH 9 x is a continuous homomorphism of the Banach algebra A into the Banach algebra g,-(X(A)) such that ii9xlJ= p(x) 6 llxi/and such that %e is the constantfunction I . (ii) The set of values of the continuous function 9 x on X(A) is equal to SpA(x).In particular, for x to be incertible it is necessary and s u ~ c i e n tthat 9 x should not vanish on X(A).

320

XV

NORMED ALGEBRAS AND SPECTRAL THEORY

For each x E A, the mapping xw (.Yx)(x) = ~ ( x is) continuous on X(A), by the definition of the weak topology. Moreover, we have

( % v J ) ) ( x )= x(xA = x(x)x(y>= ((Yx)(x))((Yy)(x)) and therefore 9(xy) = (Yx)(Yy).The relation Y e = 1 is obvious. The equality )19x))= p ( x ) will follow from (ii) and the definition of p(x) (15.2.7). Consider therefore the first assertion in (ii). We know already that ~ ( xE) Sp(x) for all x E A and x E X(A) (15.3.1).Let us show conversely that, for each A E Sp(x), there exists a character x such that ~ ( x= ) 1, or equivalently that x(x - Ae) = 0. Since x - 1e is not invertible, the ideal A(x - 1e) generated by this element is distinct from A. In view of (15.3.1(iii)) we are therefore reduced to proving the following lemma:

(15.3.4.1) In a separable commutative Banach algebra A with unit element e # 0, every ideal a # A is contained in a maximal ideal. Let (x,) be a dense sequence in A. We define inductively an increasing sequence (a,,) of closed ideals # A (n 2 0) as follows. Take a, = ii (which is distinct from A , by (15.3.1.1)). In the quotient Banach algebra A / a , - , consider the image 2, of x, and an element A, of its spectrum (which by (15.2.4) is not empty), so that X, - A, C? (where 2 is the canonical image of e in A/ a,is not invertible in A / a , - , . It follows that the ideal a: in A generated by a, - and x, - A, e is distinct from A, and therefore a,, = i: is distinct from A. The union in' of the ideals a,, is therefore also an ideal #A, and hence the closure in of m' is also an ideal #A. We shall show that m is maximal, which will complete the proof. If e" is the unit element of A / m , and x; the image of x, in A/m, then by construction we have x: = A,e"; in other words, all the x i belong to the closed subalgebra (5.9.2) Ce" of A" = A/in. Since they form a dense subset of A" (3.11.4), it follows that A" = Ce", and therefore in is maximal.

Remarks (15.3.5) It should be noted that the Gelfand transformation x w Y x is not necessarily injective. For example, if x is nilpotent but not zero, say xk = 0, then the image of Sp(x) under cw is (0) ((15.2.3.1) and (15.2.1)) and therefore Sp(x) = { O } , hence ~ ( x = ) 0 for all characters x, although x # 0. This example also shows that IIYxll need not be equal to ((x((. But even if 9 is isometric (in which case the image %(A)is closedin %',-(X(A)) (3.14.4)), %(A)may not be equal to %,-(X(A)) (15.3.8).

c

3 SPECTRUM OF A COMMUTATIVE BANACH ALGEBRA

321

(15.3.6) Suppose that there exists an element xo E A which, together with e, generates a dense subalgebra of A (in other words, as P runs through the set C[X] of polynomials in X with complex coefficients, the subalgebra consisting of the P(x,) is dense in A). Then the mapping x H x(xo)is a homeomorphism of X(A) onto Sp,(x0).

Notice first that the elements P(xo) for which the coefficients of the polynomial P are of the form LY + pi with LY and p rational, form a dense subset of A, and therefore A is separable. The mapping X H X ( X ~ )is continuous and surjective (15.3.4), and X(A) is compact and metrizable (15.3.2) ; hence it is enough to show that this mapping is also injective (3.17.12). But if xl(xo)= x2(x0) for two characters xi, x 2 , then also xl(P(xo)) = x2(P(xo)) for all polynomials P; since x1 and x2 are continuous on A, we deduce that x1 = x2 (3.15.2), and the proof is complete.

Examples of Spectra and Gelfand Transformations (15.3.7) Let X be a metrizable compact space and A = qC(X) the Banach algebra of continuous complex-valued functions on X. Then the characters of A are the Dirac measures E, (x E X) (13.1.3), and the mapping XH E, is a homeomorphism of X onto the spectrum X(A). For in this situation we have A' = M,(X), since every measure on X is bounded (13.20), and the weak topology on A' is by definition the vague topology (13.4). A character is therefore a measure p # 0 (because p(1) = 1). We shall show that Supp(p) consists of a single point. If Supp(p) contained two points a # b, then a and b would have disjoint neighborhoods U and V, respectively, hence we should have functionsf and g belonging to A, with supports contained in U and V, respectively, and such that p ( f ) # 0 and p(g) # 0. But then fg = 0 so that p ( f ) p ( g )= p(fg) = 0, giving a contradiction. If Supp(p) = {x}, then the relation ~ , ( f )= 0 implies p ( f ) = 0, and therefore (A.4.15) there exists a scalar c1 such that p = a&,; since p(1) = 1, it follows that a = 1. The continuity of the mapping X H E , is immediately seen: for each f e A, we have (f, E, - E , ~ ) =f ( x ) -f ( x o ) , and since f is continuous, for each S > 0 there exists a neighborhood V of xo in X such that If(x) -f(xo)l 5 S for all x E V. In order to prove that XHE, is a homeomorphism, it is enough (12.3.6) to show that this mapping is injective; and this is clear, because E,, and E , ~have distinct supports if x1 # x 2 . We have ( % f ) ( ~ , = ) ~ , ( f )=f(x) for all x E X and all , f ~A, and therefore when we identify X and X(A) by means of the mapping X H E, , the Gelfand transformation becomes the identity mapping. (15.3.8) Let D be the closed unit disk 5 1 in C, and let A = d ( D ) be the closed subalgebra of gC(D)consisting of the continuous functions on D which

322

XV

NORMED ALGEBRAS AND SPECTRAL THEORY

are analytic in the interior of D (1 5.1 S).We shall show that the characters of A are again in this case the restrictions to A of the Dirac measures E,(X E D). From this it will follow (15.3.7) that X W E , is again a homeomorphism of D onto the spectrum X(A). Letf,, be the canonical injection of D in C;then it is enough to show thatf, and the unit element I of A generate a dense subalgebra of A (for it is clear that SpA(f,,)= D, and the result will then follow by (1 5.3.6)). Now, let A, denote the subalgebra of A consisting of the functions which are restrictions to D of functions analytic on some neighborhood of D in C.This subalgebra is dense in A, because for each integer n > 0 and each function f E A, the functi0n.h which is the restriction to D of the mapping

r ++f((n/(n+ l)>O

of ((n + I)/n)D into C clearly belongs to A,,, and .f,-+ f uniformly on D as n + + co (3.16.5). On the other hand, every function f~ A, is the uniform limit (in D) of a sequence of polynomials P, , namely the partial sums of the Taylor series for f at the point 0 (9.9.1 and 9.9.2). Hence the set of polynomials in [ is dense in A, and our assertion is proved. We remark that here the Gelfand transformation is isometric, but %(A) is distinct from gC(X(A)). (15.3.9) Let B be the Banach algebra consisting of the restrictions to the unit circle U : = 1 of the functions belonging to the algebra A of (15.3.8). Then the restriction mappingfwfl U is an isometric isomorphism of A onto B: this is an immediate consequence of the maximum modulus principle (9.5.9), which implies that llfll = l l f l UII. Here the mapping XH E, of U into X(B) is again continuous and injective; but it is no longer a homeomorphism, because all the characters of A can be identified with characters of B, and therefore X(B) I= X(A) = D.

PROBLEMS 1.

Let X be a metrizable compact space and B a subalgebra of U,(X) containing the unit element. Assume that B is equipped with a norm llxlie which makes it a Banach algebra. (a) Show that llxlls2 /IxIl (the norm on U,(X)), by remarking that V ( t ): x HX(?) is a character of B for all t E X. Deduce that 0 is the only quasi-nilpotent element of B. (b) Show that 9) is a continuous mapping of X into X(B). If B separates the points of X (7.3) then is a homeomorphism of X onto a closed subset of X(B). (c) Suppose that, for each function x E B, the conjugate 2 belongs to B, and that if x E B is such that x ( t ) # 0 for all t E X, then the inverse x - I of x in VC(X) belongs to B. Show that in these conditions the mapping 9) is surjective. (Let tn be a maximal ideal of B, and let Z be the set of all E X such that x ( t ) = 0 for all x E m.Show that

3 SPECTRUM OF A COMMUTATIVE BANACH ALGEBRA

323

Z is not empty: if Z were empty there would exist a finite open covering (V,) of X , and for each j a function xl E rn such that xi(?) # 0 for all t E V l ; now consider the function x, Z 1 , which belongs to m.) I

(d) Deduce from (b) and (c) that the spectrum X(A) of the algebra of Section 15.1, Problem 1 can be canonically identified with the interval [0,1]. 2. Show that the spectrum of the algebra A of Section 15.1, Problem 2 consists of a single point, and that the unique maximal ideal of A is the radical of A (Section 15.2, Problem 7).

3. In the algebra d ( X ) of Section 15.1, Problem 3, show that llx"II 5 Ilxll"/(n- l)!, and deduce that d ( X ) is equal to its radical. Deduce that d ( X ) has no characters (use Problem 5 of Section 15.1). 4.

Let B be the Banach subalgebra of Vc(D) (notation of (15.3.8)) generated by d ( D ) and the function Ifo[ : [ H 151. Show that X(B) is homeomorphic to the set of points (x,, xz, xJ)E R3 such that x: x i 5 x: and 0 5 x3 1. (Remark that B contains all functions of the form [+ig(l[I), where g is a continuous function on (0, 11, and that B also contains the subalgebra A. of d ( D ) introduced in (15.3.8). To show that Ix(fo)l 5 x(/fol)for all characters of B, consider the function [([ - (If1 which belongs to B for all E > 0.

+

+

x

5. The space 1; (6.5) becomes a Banach algebra without unit element under the multiplication (fn)(vn)= (&vn). Let A be the Banach algebra obtained by adjoining a unit element to 1: (Section 15.1, Problem 5). Show that X(A) may be identified with the compact subset of R consisting of co and the integers 21 (use the fact that 16 can be canonically identified with its dual). The Gelfand transformation 9, then becomes the identity mapping, and 9 J A ) is a nonclosed dense subalgebra of V,(X(A)).

+

Show that the dual of the underlying Banach space of the Beurling algebra A (Section 15.1, Problem 4) can be identified with the space of classes of A-measurable complex-valued functions g such that

6. (a)

I

r+m

where the supremum is taken over the set of functions w E i2 such that V ( w )> 0. Then IlQll is the norm on the dual A' of A . Show that

The canonical bilinear form 0. (f) Let M be a proper closed B-submodule of =.Y'(p).Show that M is of the form q W ( p ) , where 141 = 1 . (Let N = M n Z 2 ( p ) ,which is a closed B-submodule of P2(p).I f f € M is not in the closure of BoM in .ILp'(p),deduce from (e) that f = g h , with g E N and h E X 2 ( p ) ;moreover g is not in the closure of BoN in .ILp2(p).Use (c) to show that N _=q3Ep2(p)with 191 = 1, and deduce that f~ qX'(p). Applying this , thatf, E 9-@'(p) result tof+f,, wheref' is in the closure of BoM in Y 1 ( p ) deduce and hence that M c 9 2 ' ( p ) . ) 16.

We shall apply the results of Problems 10 to 15 to the Dirichlet algebra B of (15.3.9), where p is the normalized Haar measure dp(6) = ( 2 ~ ) - 'd6 on the unit circle U, and is the characterft+f(O), so that p is the unique representative measure for (Problem 9(d)).

x

x

(a) B is the closure in gc(U) of the algebra of trigonometric polynomials

"

k=O

ckek"

3

SPECTRUM OF A COMMUTATIVE BANACH ALGEBRA

with exponents k 2 0. To each function function in the disk 1 z 1 < 1 defined by

f t Xp(p)

331

there corresponds the analytic

For 1 < p = < co (resp. p = i-co) f is the limit (resp. weak limit) in W ( p ) of the functions B+-tF(re") as r - I , so that HP(p) may be identified with the space of analytic functions F corresponding to the functions f E -P(p). (Use the fact that x p ( p )is the closure (resp. weak closure) of B in P ( p ) . ) In particular, H2(p) is identified with the space of functions m

f(z) == E C" z" "=a

which are analytic for IzI < 1 and such that

m "=O

'/,cI

< 4-m (cf. the problem in

Section 9.13). (b) Let u be a measure disjoint from p which is quasi-representative for Show that e - ' @. u is also a quasi-representative measure for and hence by induction that e - n i e . v is a quasi-representative measure for for all integers n > 0. Hence show that u = 0 and consequently (Problem 14) that every quasi-representative measure for has base p. (c) Let f~ , P 1 ( p )be nonnegligible (with respect to p). Then the function loglfl is p-integrable, or equivalently, J,(lfl) > 0, and consequentlyfis the product of an exterior function and an interior function. (This follows from Problem ll(b) if

x,

s / d p # 0. If

x.

x,

x

.c

f dp = 0 then f,identified with an analytic function on the unit disk,

is of the form f(z): z"g(z) where ir > 0 and g(0) f 0. We have y E Y l ( p ) and log If1 = log 191 on U.)In particular, a function belonging to Z ' ( p ) which vanishes on a non-p-negligible subset of U is p-negligible. Every function in 2 ' ( p ) is the product of two functions in . F 2 ( p )and conversely (Probleni 15(e)). A function f> 0 belonging to .4p1(p)is equivalent to a function of the form 1.4 I ', wherey t Xi"(p), if and only if J,(f) > 0 (or equivalently if and only if log If1 is p-integrable). (See Section 22.19, Problem 19.) 17. Let B be the Banach algebra considered in Problem 16, and BI the Banach subalgebra of B consisting of functionsfcontinuous on Iz/ S 1 and analytic on 1.7 < 1, and such thatf'(0) = 0. For each complex number c such that j1 , let p c denote the measure F, . p, where F,(z) = 1 - W(Ez). Show that the measures pc are representative measures for the character :f ~ f ( 0 ) the ; representative measures which are extremal points of are the pCsuch that / c / = 1 ; the nieasure po = p is the unique Jensen measure for and hence there exists no hyperextremal representative measure for

IcI

%(x) x;

18.

x

x.

Let A be a commutative Banach algebra with unit element. (a) Let u E A be such that p ( u ) < 1. Then for -each character of A the element 1 - x(u)u is invertible. If v = ( ~ ( u-) u)(l - x(u)u)-', then /x(u)I < 1. (b) If xI, x 2 are two characters of A, let o(x,,x.) denote the least upper bound of the number Ix2(u)l as u runs through the set of elements of A such that ~ ( u5) 1 and xl(u) = 0. Show that if p(u) 5 1 we have

x

IXI(4 -

X Z W 5 O(y.1,

Xdll

-

xI(u)xz(u)l

332

XV

NORMED ALGEBRAS AND SPECTRAL THEORY

(use (a)). Deduce that

dx1, XZ) = d x 2 ,xI) and that

Deduce from this inequality that, if

xl, xz, x3 are three characters of A, we have

and hence that the relation o(x', x") < 1 between characters x', x'' of A is an equiuulence relurion on X(A). The equivalence classes for this relation are called Gleuson parts of X(A). (c) Let u E A be such that p(u) 5 1, and let xl, x2 be two characters of A, such that xl(u)= 0. Show that there exists a complex number such that Ihl < 1 and such that, if we put v = (A - u)(1 - xu)-' E A, then xl(u) = A and x2(v)= -A. Deduce that

(d) I f

xI, xz are two characters of A, put

Show that

(use (c)). The equivalence relation defined in (b) is therefore equivalent to T(x', x") < 2. (e) Let u E A be such that Wx(u) 2 0 for all x E X(A). Show that, for any two characters xl, x 2 ,we have

(Apply (*) to the element e-'", where f > 0, and let f tend to 0.) 19. With the hypotheses of Problem 8, let xl, x2 be two characters of B. (a) Let pI,p 2 be representative measures for xl, x2 respectively (Problem 9). If a(xI, x2)= 1 (Problem 18), show that p1 and p2 are disjoint (13.18). (Write p2 = h . p1 Y where v is disjoint from pl.and by majorizing Ix2(u)- x~(u)Ifor u E B

+

1

2 I , show that the hypothesis implies that (h + 1) dpl $ j l h - 11 dp1.) (b) If o(x1,x2) < 1, show that there exist representative measures pI for x1 and and IIull

p 2 for

x2 such that

+ u(xl, ,y2))-'. Use Problem

(Put c = (1 - cr(xI,x2)(1

18(e) and the Hahn-Banach

3 SPECTRUM OF A COMMUTATIVE BANACH ALGEBRA

333

theorem to show that there exist two positive measures a, /3 on X such that

xJu> - c x I ( u )= / u dol and xl(u)- cx&)

=/ u

dp.) Deduce that if

x

x is a character

of B, then the Gleason part (Problem 18(b)) containing is the set of characters for which a representative measure has base p, for at least one representative measure p for x. (c) Suppose x is a character for which the representative measure is unique. If belongs to the Gleason part containing x , show that the representative measure p' for is also unique, and that

x'

x'

x

(d) Suppose that there exists a hyperextremal representative measure p for (Problem 11). Show that if a character x'admits a representative measure p' with base p, then p' is the only representative measure for x' with base p. (If p ' = f . j and = g . p are two representative measures for show that f- g is zero almost everywhere with respect to p, by using Problem 13(j).)

x',

PI'

20. Let A and B be two commutative Banach algebras having unit elements and h : A

+B

a homomorphism which sends unit element to unit element. show that the (a) Let I' be the graph of h in A x B. For each character x of mapping (x,y ) Hx(h(x)) - x(y) is continuous on A x B. Deduce that if (a. b) is in the closure of I?, then ,-&(a)) = ~ ( b ) . (b) Deduce from (a) that if B is without radical (Section 15.2, Problem 7) then h is necessarily continuous (use the closed-graph theorem). In particular, for a commutative C-algebra having a h i t element, in which the intersection of the maximal ideals

n,

is {O},two norms which define Banach algebra structures are necessarily equivalent.

21.

Let 9 be the algebra of indefinitely differentiable complex-valued functions on [0, I]. (a) Let A be a subalgebra of 9, containing the unit element and endowed with a norm which makes A a Banach algebra. Show that there exists a sequence (mn)n30 of finite real numbers 20 such that, for all functions x E A, we have sup lx(")(t)l = O(mA

O O on X(A); use (15.3.4). If a character x of A is not hermitian, then there exists a self-adjoint y E A such that x(y) = i, whence x(1 yy*) = 0, and 1 + y y * is not invertible.) Hence give an example of a commutative Banach algebra with involution, having a unit element, for which there exist nonhermitian characters.

+

+

+

4.

(a) If A is a star algebra with unit element, then llx112 = p ( x * x ) for all x E A (prove this first when x is self-adjoint). Deduce that A is without radical. (b) Let A be the algebra of Problem 2 of Section 15.1. If x m

=z

&T", define x* t o be

"=O

&T".This defines an involution on A for which A is a commutative Banach algebra

"20

with involution, having a unit element, and such that the only character of A (Section 15.3, Problem 2) is hermitian. 5.

-

With respect to the involution f*(r) = f ( - r ) on the Eeurling algebra (Section 15.1, Problem 4 and Section 15.2, Problem 6), show that all the characters are hermitian.

6. On the algebra A of Problem 7 of Section 15.3, consider the involution induced by

f ~ f and , extend this involution to A by putting e * with involution. Find its hermitian characters.

= e.

Then A is a Banach algebra

4 BANACH ALGEBRAS WITH INVOLUTION. STAR ALGEBRAS 7.

341

(a) In a star algebra, every hermitian quasi-nilpotent element (Section 15.2, Problem

5) is zero.

(b) Let A be a star algebra with unit element, and let x be an element of A which is not left-invertible. Then x*x is not invertible in the star subalgebra B of A generated by 1 and x*x, and therefore there exists a sequence (y.) of elements of B such that lIynll = 1 and lim y.x*x = 0 (use the Gelfand-Neumark theorem). Deduce that a n+m

noninvertible element of A is necessarily a (left or right) topological zero-divisor.

8. Let A be a star algebra with unit element, x a normal element of A, and S = SpA(x). Then there exists a unique homomorphism of the star algebra V,(S) into A, transforming unit element into unit element, and such that ~ ( 1 , ) = x. This homomorphism of V,(S) onto the star subalgebra of A genis an isometry, which sends f to erated by 1, x and x*; all the elements of this subalgebra are therefore normal. Put ~ ( f=f(x), ) and show that if f i s analytic in a neighborhood of S, then the element f(x) is equal to the element so denoted in Section 15.2, Problem 11.

v(f)*,

9.

Let A be a star algebra with unit element. Let P denote the set of self-adjoint elements of A such that SpA(x) C [0, cc [. (a) Show that if x E A is self-adjoint and jle - X I / 5 1, then x E P. If x E P and llxll 5 1, then lle -xIl 5 1. (Consider the subalgebra generated by e and x . ) Show that a self-adjoint element x belongs to P if and only if IIx - IIxlleIl 5 IIxII. (b) Deduce from (a) that P is a closed convex cone in A, such that P n (-P) = (0). (c) Show that the relation x*x E - P implies that x = 0. (Observe first that also xx* E -P, and by writing x = u iu, where u and u are self-adjoint, deduce that x*x E P, whence x*x = 0.) (d) Deduce from (c) that x*x E P for all x E A. (Write x*x = u - v, where u, u are hermitian and belong to P, and uu = uu = 0 (use Problem 8). If z = xu, show that z*z E -P, and deduce that u = 0.) Deduce that e x*x is invertible in A (use Problem 8).

+

+

+

Let A be a commutative Banach algebra with unit element,and let XHX* be any involution on A. For each character x of A, put x*(x) = x(x*). Show that XI--+X* is an involutory homeomorphism of the space X(A). (b) Let X be a metrizable compact space, and v an involutory homeomorphism of X onto X. For eachfE FC(X),putf*(x) =f(v(x)). Show that this defines an involution on Vc(X), and that every involution on V,(X) may be so obtained. (c) Let X be the compact subspace of RZ which is the union of the segment y = 0, - 1 5 x 5 2, the segment x = 0, 1 5 y 5 2, the circle x 2 y z = 1 and the open halfdisk D : y > 0 , xz y z < 1. Assume that the only involutory homeomorphism of X onto X is the identity (Chapter XXIV). Let A be the Banach subalgebra of %‘,(X) consisting of functions which are analytic in D. Show that the spectrum of A can be canonically identified with X, and that the algebra A has no involution other than the identity.

10. (a)

+

+

11. Let A be a noncommutative star algebra with unit element.

(a) Show that there exists a hermitian element y E P (notation of Problem 9) such that y is invertible and y’ is not in the center Z of A. (Using Problem 9, show that if the result were false, the intersection of Z with the real vector subspace H of A consisting of the hermitian elements would contain a neighborhood of e in H, which would imply that A was commutative.)

342

XV

NORMED ALGEBRAS AND SPECTRAL THEORY

(b) Deduce from (a) that, if we put x'=y-'x*y for all X E A, then continuous involution on A such that (x')* # (x*)' for some x E A. 12.

is a

Let E be a Hilbert space. Show that a self-adjoint compact operator U is a HilbertSchmidt operator if and only if, (h.) being the sequence of eigenvalues of U each counted according to its multiplicity, the s u m z Ih.1* is finite. The sum is then equal to

13.

XHX'

I1

w:.

n

(a) The norm llullz is defined as in (15.4.8) when E is a finite-dimensional Hilbert space. In a Hilbert space of dimension n, give an example of an operator u such that

(b) Let E be a Hilbert space of infinite dimension. Give an example of a sequence (u,) of Hilbert-Schmidt operators on E such that the sequence of numbers ~ ~ u ~ u ~ tends ~ ~ z to/ 0 ~ as ~ nu tends , J ~ to ~ m. 14. Let X be a (metrizable, separable) locally compact space, and let p be a positive measure on X. Let K(x, y) be a function on X x X, belonging to 2&(X x X, p 0p).

For each function f~ 2 6 ( X , p), the function

XH

s

K(x, y ) f ( y ) dp(y) is defined al-

most everywhere, and its class belongs to L&(X,p) (use the Lebesgue-Fubini theorem and the Cauchy-Schwarz inequality). If this class is denoted by U . A show that U is a Hilbert-Schmidt operator; U is said to be associated to the kernel K. Show that IILIIIz = N,(K) and that I/* is associated to the kernel ( x , y ) ~ K ( x). y , If U1, Uzare associated to kernels K I , K2 respectively, belonging to Y & ( X x X, p @ p), then U,Uzis associated to the kernel

~ ( xy ,) = s ~ xt)K2(r, , y ) dp(t). 15.

16.

Let E be an infinite-dimensional Hilbert space and U a Hilbert-Schmidt operator on E. If h is a regular value for U,show that the operator (U- h . lE)-l is of the form V - h-' . l E , where V is a Hilbert-Schmidt operator. Deduce that the spectrum of U in the algebra Y(E) is the same as in the algebra obtained by adjoining a unit element to Y,(E). For every function f which is analytic in an open set containing Sp(U), the operatorf(U) (Section 15.2, Problem 11) is a Hilbert-Schmidt operator. Let HI,. ..,H. be self-adjoint operators, each pair of which commute, on a separable Hilbert space E, and suppose that liHJll,5 1 and that IIHJHkll5 dJ-lrl for all j , k, where E is a real number such that 0 5 E < 1. Then lIH1

+ . . . + H"ll5 (1 + +)A1

-E)

(" Cotlar's lemma").

(Using the Gelfand-Neumark theorem, reduce to proving the same property for real numbers ul,. . . , u.. For each t 2 0, let v(r) be the number of indices j such that Ju,I 2 t . Remark that

2 lul

J= I

+

=Jol v ( t ) dt

and prove that v ( t ) 5 1 [2 log tllog E ] , by observing that if lull 2 t and then dJ-*l 2 IuJukl 2 t 2 . )

IUkl

2 I,

4

BANACH ALGEBRAS WITH INVOLUTION. STAR ALGEBRAS

343

17. Let A be a Banach algebra with unit element e, and let x-x*

be an involution on A which is not necessarily continuous. (a) Show that the radical ‘31of A (Section 15.2, Problem 7) is a self-adjoint subset of A . Passing to the quotient, it follows that the involution on A defines an involution (also written zt+z*) on A/%. (b) Let a = a* be a self-adjoint element of A, such that SpA(a)is contained in the open half-plane 9[ > 0. Then there exists an element b E A, which is the limit of a sequence (u,) of self-adjoint elements which are polynomials in a, and which satisfies b Z = a. (Write a = o r ( e - ( e - .-‘a)) for a suitable a > O , and use Problem Il(f) and Section 15.2, Problem 15.) Show that 6 and b* commute, and that we may therefore write b = u iu, where u and v are self-adjoint and commute with each other and with a. Deduce that uu = 0 and u2 - v 2 = a. (c) Let C be a commutative Banach subalgebra of A which contains e, a, b, b* and is such that Sp,(b) = SpA(b) (Section 15.2, Problem 15). Let C be a commutative Banach subalgebra of A containing the image of C under the homomorphism : x-x* of C into A. If W’ is the radical of C and T : C C / W ’ the canonical homomorphism, then the homomorphism T 0 9 : C + C’/W’ is continuous (Section 15.3, Problem 20).

+

--f

1

Deduce that u = - (6* - b) E and thence that SpA(a)= SpA(uz). (Observe that 2i Sp,(a) = S p c , , a ( . ( ~ ( u-zu z ) ) = SpC./st.(7i(u2)).)Consequently u is invertible, u = u - ’ u u = 0, and finally b = u is self-adjoint. Furthermore, if Sp(a) is contained in 10, a[,then the same is true of Sp(b). %I,

+

18. With the hypotheses of Problem 17, put pA(x)= (p(x*x))’l2 (also denoted by p(x)). We havep(x*) = p(x); also if W is the radical of A and if T :A + A p t is the canonical homomorphism, then P ~ / ~ ( T ( X=)pA(x). ) The involution x-x* on A is said to be hermitian, and the involutive algebra A is said to be hermitian, if Sp(a) c R for all self-adjoint elements a E A. If A is commutative, an equivalent condition is that every character of A is hermitian (Problem 2). Assume for the rest of this Problem that A is hermitian. (a) Let x E A be such that p(x) < 1 . Show that (e i-x*)(e - x) is invertible. (Remark that e - x*x = w2, where w is self-adjoint and invertible, by virtue of Problem 17; hence show that the spectrum of w-l(x* - x)w-’ is contained in iR.) Deduce that

for all x E A (Ptak’s inequality). Consequently, if x is normal, we have p(x) = p ( x ) . (Remark that if x, y are two ) by (15.3.4).) elements of a Banach algebra which commute, then ~ ( x y 0 such that Iln(x*)Il 5 clla(x)ll, and that pA(x) = ~ A / w ( T ( x ) ) . ) (h) Show that, for each x E A, Sp(x*x) is contained in [0, m[. (Argue by contradiction: suppose that there exists x E A such that - 1 E Sp(x*x). Write z = x * x , and for each positive integer n let w. be a self-adjoint element of A which commutes with z,

+

+

+

and is such that w:

= z’

+ -n1 e and Sp(w,) c 10, +

03

[ (Problem 17). Let b, = w, - z.

By considering a commutative Banach subalgebra B of A containing w, and z, and such that the spectra of z, w., and b. are the same in B and in A, show that

+

Show also that p(b,) 5 1 2p(z) = a, independent of n (use (c)). Put y. = xb,, so that y : y n : - biw. - 1 ~( l / n ) b n ;deduce that Sp(y,fy,) is contained in the interval 1- a),a/n],and then that Sp(y,y,*) is contained in the interval [- a/n, m[ (use (c)); hence that p(y,*y.) -< a/n (Section 15.2, Problem 2(b)), and consequently that 1x(y,*y.)) 5 a/n for all characters x of B. Finally, obtain a contradiction by noting that y:yn = biz, that ~(w,,) 2 0 and that there exists a character x such that x ( z ) = - 1 .)

+

19.

Let A be a Banach algebra with unit element, and let x - x * be a (not necessarily continuous) involution on A. Show that the following properties of A are equivalent: ( a ) A is hermitian. (fl) p(x) ( p ( x ) for all normal x E A. ( y ) p(x) = p ( x ) for all normal x E A. (6) p(x* x ) 5 2p(x) for all x E A. ( E ) P(X v) 5 p ( x ) P(P) for all x , y E A. (5) The set of numbers p(x), where x runs through the unitary elements of A, is bounded. (7) For each x E A the element e t-x*x is invertible in A. (To show that (7) implies (a), argue by contradiction by showing that the spectrum of a hermitian element cannot contain the number i. To show that ( y ) implies (a), argue as in (15.4.12). To show that (6) implies (fl), remark that (8) implies that p(x)” = p(x”) 5 2p(x)” for x normal and n 2 I . Finally, to show that ([) implies (a), notice first that (5) implies that Sp(x) c U for all unitary elements x , by considering the powers x” ( n E Z). Next observe that if a is self-adjoint and p(a) < 1, there exists a self-adjoint element h, commuting with a, such that b2 = e - a a 2 (Problem 17), hence a ib is unitary. Then consider a commutative Banach subalgebra containing a and band such that the spectra of a, b and a ib are the same in A and B).

+ +

+

+

+

5 REPRESENTATIONS OF ALGEBRAS WITH INVOLUTION

345

20. Let A be a Banach algebra with unit element e, and let x - x * be a (not necessarily continuous) involution on A. Show that if IIx/l= < p ( x ) ,then A is a star algebra.

5. REPRESENTATIONS O F ALGEBRAS WITH I N V O L U T I O N

Let A be an algebra with involution (not necessarily endowed with a norm, and not necessarily having a unit element), and let H be a Hilbert space. A representation? of A in H is a homomorphism SH U(s) of A into the algebra 9 ( H ) of continuous endomorphisms of H, such that (15.5.1)

U(s*) = (U(s))*

This implies in particular that if s is self-adjoint, then so is U(s). If A has a unit element e, we require in addition that (15.5.2)

U(e) = 1.,

The representation U is said to be faithfulif the homomorphism SH U(s)is injective, that is if the relation U(s) * x = 0 for all x E H implies that s = 0. Let H, H’ be two Hilbert spaces. A representation SH U(s)of A in H and a representation SHU’(S) of A in H’ are said to be equivalent if there exists a Hilbert space isomorphism T : H --+ H’ such that U‘(s)= TU(s)T-’ for all s E A. When H = H’, the Hilbert space automorphisms T of H are precisely the unitary elements of the star algebra 9 ( H ) . For T must be invertible and satisfy(T.xlT.y)= (xIy)forallx,yinH,so that(T*T.xly)=(xly)and therefore T*T = l,, hence T* = T-’; the converse is immediate. Let H,, H, be two Hilbert spaces, S H Ul(s) a representation of A in H,, and SH U,(s) a representation of A in H,. Let H be the Hilbert sum of H, and H, , so that H, and H, are identified with supplementary subspaces of H. If x = x, + x2 and y = y , + y , are two elements of H (where x i , y i in Hi for i = 1, 2), then (6.4) (x I Y>= (x1 I Y1)

+ (x2 I Y 2 ) -

If we put U(s) . (xl + x,) = U , ( s ) . x , + U,(s) . x, , it is immediately verified that U(s)E 9 ( H ) for each s E A, and that SH U(s) is a representation of A, called the Hilbert sum of the given representations.

t Strictly speaking, a unitary representation. Since we shall not consider other types of representations, we shall suppress the word unitary,” by abuse of language. ‘I

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Let SHU(S) be a representation of A in a Hilbert space H. A vector subspace E of H is said to be stable with respect to this representation if U(s)E c E for aN S E A . If E is stable with respect to U , then so is the closure E of E (3.11.4). If E is closed and if E’ is the orthogonal supplement of E in H (6.3), then E’ is also stable with respect to U. For if x E E and x’ E E‘, we have (x 1 U(s) . x’) = ((U(s))*xI x’) = (U(s*)x I x ’ ) = 0 by hypothesis, hence U ( s ) .x’ is orthogonal to all x E E and therefore belongs to E‘. If U,(s)and U,(s) are the restrictions of U(s) to E and E‘, respectively, then the representation U is the Hilbert sum of U , and U , .

(15.5.3) For a closed subspace E of H to be stable with respect to U , it is necessary and sujicient that PEU(s) = U(s)PEf o r all s E A, where PE is the orthogonal projection on E (6.3). The condition is necessary, for if X E E we have P,. x = x and P,. (U(s).x) = U ( s ) . x, because E is stable with respect to U ; and if x E E’ we have P , x = 0 and P, * (U(s) * x) = 0, because E’ is also stable with respect to U. Conversely, if the given condition is satisfied, then U ( s ) x = P, U(s)* x E E for all x E E and all s E A.

On a Hilbert space H, an orthogonal projector is by definition any continuous operator on H which is an orthogonal projection onto a closed subspace of H. The importance of such projectors is due to (1 5.5.3) and to their characterization in terms of the structure of algebra with involution of Y(H) : (15.5.3.1) A continuous operator P on a Hilbert space H is an orthogonal projector if and only if it is idempotent and hermitian. The necessity of these conditions has already been proved (1 1.5). Conversely, if P z = P = P*, then ( P . X I y - P . y ) = (XI Pa y - P 2 y ) = 0 for all x , y E H. Since P(H) is also the kernel of 1, - P , it is closed, and H is the Hilbert sum of P(H) and P -‘(O). Hence the result.

-

Suppose that H is the Hilbert sum of an infinite sequence (H,) of subspaces which are stable with respect to the representation U. Let U,(s)denote the restriction of U(s) to H,, so that for each n the mapping SH U,(s)is a representation of A in H, By abuse of language, the representation U is said to be the Hilbert sum of the representations U, . For each s E A and each x = x, E H, where x, E H, for each n, we have U(s) * x = U,(s) x, , and

.

1 n

n

1 IIUn(s).xnII2 n

=

IIu(s).xIIZ.

5

REPRESENTATIONS OF ALGEBRAS WITH INVOLUTION

347

A representation SH U ( s )of A in H is said to be topologically irreducible if there exists no closed vector subspace E of H, other than (0) and H, which is stable with respect to U. From (15.5.3) we obtain the following irreducibility criterion: In order that U should be topologically irreducible, it is necessary and suficient that the only orthogonal projections P such that PU(s) = U(s)P for all s E A should be 0 and 1,. (15.5.4)

For this condition expresses that ( 0 ) and H are the only closed subspaces of H which are stable under U .

(15.5.5) Let SH U(s) be a representation of A in H , let E be the closure in H of the vector subspace generated by the elements U(s) . x for s E A and x E H , and let E‘ be the set of all x E H such that U(s) . x = 0 .for all s E A. Then E and E’ are stable with respect to U and are orthogonal supplements of each other in H . Since U(st) = U(s)U(t),it is clear that E and E’ are stable subspaces of H. Let E” be the orthogonal supplement of E in H. We have seen earlier that E” is stable with respect to U. But if x E E”, we have U(s) x E E by definition, hence U(s). x E E n E” = (0) for all s E A, so that E” c E’. Conversely, if x E E’, s E A and y E H, we have (x I U ( s ) .y ) = (U(s*) . x 1 y ) = 0 by definition, so that x is orthogonal to E, and therefore E’ c E”. The subspace E is called the essential subspace for U . If E’ = { 0 } , the representation U is said to be nondegenerate. An equivalent condition is that the elements U(s) . x should form a total subspace of H ; by (15.5.2), this is always the case if A has a unit element.

A vector xo E H is a totalizer or totalizing vector for a representation U of A in H if the vector subspace of H generated by the transforms U ( s ) .xo of xo , as s runs through A, is dense in H . (In any case, this subspace is stable with respect to U . ) A representation which admits a totalizer is said to be topologically cyclic. If U is topologically irreducible, every nonzero vector xo E H is a totalizer, and conversely.

(15.5.6) Suppose that A has a unit element. Let SH U(s) be a nondegenerate representation of A in a separable Hilbert space H . Then H is the Hilbert sum of a sequence (H,) ( f n i t e or infinite) of closed subspaces, stable with respect to U and such that the restriction of U to each H , is topologically cyclic.

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Let ( x , ) , ~be~ a dense sequence in H. We define the H, inductively, as follows. HI is the closure of the vector subspace of H generated by the U(s) . xl, as s runs through A; we have x1 E H I , because A has a unit element. If H,, .. . , H,-l have been constructed, it may be the case that H is equal to the (direct) sum L of the H i , and the construction stops. If not, let L' be the orthogonal supplement of L in H, and let p(n) be the smallest integer such that, if xi is the orthogonal projection of x,(,) on L', the vector subspace HL of L' generated by the U(s) xi is not equal to (0): there exists such an index because the representation U is nondegenerate, and we take H, to be the closure of H i . Since x; E H, , the subspaces H, satisfy the required conditions, by virtue of (6.4.2). (15.5.7) If A is a Banach algebra with involution, having a unit element, then every representation SH U(s) of A in a Hilbert space satisjies IIU(s)ll 5 llsll (and in particular U is a continuous mapping of A into Y(H)).

We have IlU(s)l12 = IIU(s)*U(s)ll = p(U(s)*U(s))in the star algebra 9 ( H ) (15.4.14.1). Since U(s)*U(s)= U(s*s),it follows from (1 5.2.8(i)) that P(W*S))

5 P(S*4 5 IIs*sII 2 IIs*II . llsll = llS1l2.

In particular, we recover in this way (15.3.1(ii)),

6. POSITIVE LINEAR FORMS, POSITIVE HILBERT FORMS AND REPRESEN T A T l O NS

Let A be an algebra with involution (not necessarily normed, and not necessarily having a unit element). A linear form f : A + C on A is said to be positive if (15.6.1 )

f(x*x) 20

for all x E A. (15.6.2) Let f be a positive linear form on an algebra A with involution. (i) The mapping ( x , y ) H g ( x , y ) =f ( y * x ) is a positive hermitian form on A x A: in other words, (1 5.6.2.1)

for all x , y in A.

f(x*r) =f(v*x)

6 POSITIVE LINEAR FORMS, POSITIVE HILBERT FORMS

349

(ii) For all x, y in A we have

(iii) If A has a unit element e , then (1 5.6.2.3) (1 5.6.2.4)

To prove (i), express that g ( x + Y , x + Y ) = g(x9 4

+ g(x9 Y ) + 9(Y, 4 + 9(Y, Y )

is real: by virtue of the hypothesis (15.6.1), we obtain

m ( Y , 4)= -JW,39); changing x into ix, this becomes %?(g(x,y ) ) = &(g(y, x)), hence g ( y , x) = g(x, y ) . Assertion (ii) is the Cauchy-Schwarz inequality (6.2.1) applied to g . The relations (iii) are obtained by replacing y by e in (15.6.2.1) and (15.6.2.2), and using (15.4.2). The hermitian form g so obtained from f is not arbitrary, because it clearly satisfies the relation

for all x, y , z in A. We therefore make the following definition: A positive Hilbert form on the algebra with involution A is a positive hermitian form on A x A which satisfies the relation (15.6.3). If A has a unit element, every positive Hilbert form g comes as above from a positive linear f o r m 8 we have only to definef(x) = g(x, e). But we shall see later that this is no longer the case when A has no unit element (15.7.4). We shall now show that there are remarkable relationships between Hilbert forms on A x A and representations of A. In the first place, every representation SH U(s) of A in a Hilbert space H gives rise to positive linear forms (and hence to positive Hilbert forms) in the following way: for each xo E H, if we put

350

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NORMED ALGEBRAS AND SPECTRAL THEORY

then it is clear thatf,, is a linear form on A, and we have .fx,(s*s) = (W*)W) * xo I xo) = ( U s ) xo I U(S>

XO) =

II W ) xo112 > 0 *

by virtue of (15.5.1). The corresponding Hilbert form is

For example, if A = Y(H) where H is a Hilbert space offinite dimension n, and 1, : TH T is the identity representation of A in H, then the form f,, is calculated explicitly as follows: if ( e J l s i 5 n is an orthonormal basis of H, and

(cij) the matrix of T with respect to this basis, then for x , = 1A,ei we have i= 1

(15.6.6) When studying positive linear forms of the type (15.6.4) we may always assume that xo is a totalizer for U (15.5), because f , , is unchanged when we replace U by its restriction to the stable subspace of H which is the closure of the stable subspace generated by x o and the U ( s ) .xo (s E A). Under this additional assumption, the form f,, determines the representation U up to equivalence : (15.6.7) Let U, U p be two representations of A in Hilbert spaces H, H’, respectively, and suppose that U (resp. U ’) has a totalizer xo (resp. xb). Then if (U(s). xo I x o ) = (U’(s) . xb I xb) for all s E A, the representations U and U ‘are equivalent . For all s. t in A we have (Lys) . xo I U ( t ) . xo) = (U(t*s) * xo I xo)

(15.6.7.1)

= ( U ’ (t*s) ’ x; I xb)

= ( U ’ ( s ) . xb I U ’ ( t ) * xb).

Since the vectors U ( t ) . x o (resp. U ’ ( t ) . x b ) form a dense subspace H, (resp. Hb) of H (resp. H’), this proves already that U(s). xo = 0 if and only if U’(s) . xb = 0 . It follows that, for each z E H, and each s E A such that U ( s ) . x o= z , the vector U ’ ( s ) * x b is constant and equal to say z‘ = T z E Hb . By (15.6.7.1) the mapping T is an isomorphism of the prehilbert space H, onto the prehilbert space Hb , which extends uniquely to an isomorphism (also denoted by T ) of the Hilbert space H onto the Hilbert space H’ (by virtue of (5.5.4) and the principle of extension of identities). It

POSITIVE LINEAR FORMS, POSITIVE HILBERT FORMS

6

351

remains to be shown that T . (U(s) . z ) = U ’ ( s ) . ( T . z), and by the principle of extension of identities we may assume that z is of the form U ( t ) * x, . But then U(s). ( U ( t ) . xo) = U(st ) x, , and therefore 1

T . ( U ( s t ) .x,)

=

U ‘ ( s t ) .X,

=

U ’ ( S ) (. U ’ ( t ) .x,)

=

U ’ ( S )* ( T . z).

This completes the proof. Furthermore, the prehilbert space H, can be deJned directly in terms of the form f x o, and even in terms of the Hilbert form gxo. We have the following general proposition : (15.6.8) Let g be a positive Hilbert form on A. Then the set n of elements s E A such that g(s, s ) = 0 is a left ideal of A, and is equal to the set of s E A such that g(s, t ) = 0 for all t E A. If II : A + A/n is the canonical linear mapping, then there exists a unique structure of prehilbert space on A/n such that ( I I ( S ) I rc(t)) = g(s, t)for all s, t E A.

The Cauchy-Schwarz inequality Ig(s, t)I2 5 g(s, s)g(t, t ) (6.2.1) shows that n is the set of all s E A such that g(s, t ) = 0 for all t E A. The relation (1 5.6.3) then proves that n is a left ideal in A. By virtue of the relation g(t, s) = g(s, t), we have g(t, s) = 0 for s E n and t E A ; it follows that if s - s ‘E n and t - f’ E n (that is, if z(s) = n(s’) and n(t) = n(t’)) then g(s, t ) = g(s’, t’), because g(s, t ) - g(s , t’) = g(s, t - t’) + g(s - s’,t’). Since g(s, t ) depends only on I I ( S ) and n(t), we may write g(s, t ) = (n(s)In(t)),and it is straightforward to check that the function (x I y ) so defined on (A/n) x (A/n) is a non-degenerate positive hermitian form. In the case where g = gxo, we have gxo(s,s) = 11 U(s) . xO(l2,hence n is the kernel of the linear mapping u : SH U ( s ) .x, of A onto H,. The formula (15.6.5) shows that the bijective mapping S : A/n + H, induced by u is a prehilbert space isomorphism. This shows, as stated earlier, that the Hilbert space H is determined by gxo up to isomorphism. Further, the representation S H U(s) can be reconstructed from the algebra structure of A and the form gxo, because

-

-

U(s) (U(t) x,)

=

U(S) (S . n(t))= S . n(st);

and once U(s) is known in H,, it extends uniquely by continuity to H (5.5.4).

Nevertheless, positive Hilbert forms of the type gxo are not the most general ones, for they satisfy an additional condition which expresses that each

352

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NORMED ALGEBRAS AND SPECTRAL THEORY

U(s)is a continuous operator on Ho . In view of (15.6.5) and (5.5.1), this continuity is equivalent to the following condition on the form g = gxo : (U) For each s E A there exists a real number M, 2 0 such that

for all

tE

A.

Conversely : (15.6.10) Let g be a positioe Hilbert form on A, satisfying the condition (U), and suppose that (with the notation of (15.6.8)) the prehilbert space A/n is separable, so that (6.6.2) it can be identijied with a dense vector subspace of a Hilbert space H. Then for each S E A the endomorphism n ( t ) H n ( s t )of A/n extends to a continuous endomorphism x H V(s) . x of H, and s H V(s)is a representation of A in H. If n(t) = n(t’), then n(st) = n(st)’) because n is a left ideal. Hence the endomorphism of H, under consideration is well-defined, and the definition of the scalar product on A/n, together with (15.6.9), shows that this endomorphism is continuous (5.5.1). Hence the existence of the continuous operator V(s) (5.5.4). Since n((ss’)t) = z(s(s’t)), we have V(ss’) = V(s)V(s’). Also, by virtue of (1 5.6.3)

(V(s*)* n(t)I n(t’)) = g(s*t, t’) = g(t, s t ’ ) = g(st’,

t ) = (V(s). n(t‘)I n(t)) = (n(t)I V ( S ). n(r’))

which shows that V(s*)= (V(s))*. Finally, if A has a unit element e, then evidently V(e)= 1, , and therefore S H V(s) is a representation of A in H. It is useful to know when the representation SH V(s)so defined is nondegenerate (15.5.5). This is equivalent to the following condition on g : (N) The elements n(st)form a total set in the prehilbert space A/n. This condition is trivially satisfied when A has a unit element. Under the conditions of (15.6.10),there does not in general exist an element

xo E H such that g(s, t ) = (V(s) . xo I V(t) . xo) (cf. Section 15.9, Problem 3).

However, such a vector always exists when A has a unit element e: we may take xo = n(e).

6

POSITIVE LINEAR FORMS, POSITIVE HILBERT FORMS

353

When A has a unit element and is a Banach algebra with involution, not only does the theory of positive Hilbert forms reduce to that of positive linear forms, and the condition (N) is automatically satisfied, but also the condition (U) is satisfied. Precisely, (15.6.11) Let A be a Banach algebra with involution, having a unit element e # 0. Let f be a positive linear form on A. Then (i) f i s continuous and 11 f II = f ( e ) . ( 4 If(Y*XY)I IIlXllf(Y*Y).

We shall use the following lemma: (15.6.11.1)

adjoint y

E

If X E A is self-adjoint and llxll < 1, then there exists a selfA such that y 2 = e + x.

If t is real and It1 < 1, Taylor’s formula (8.14.3) gives

where

(-1)” 1 . 3 - . . . ( 2 n - 1) J: (%)” r,(t) = 2 2 . 4 . ~2n . 1+ s

ds

(1

+ s)1’2’

This leads immediately to the estimate

Irn(t)l 6

and therefore we have (15.6.11.2)

(1

+ t)”’

=1

Ill”

+ + t + ( 4 1 ’ + ... + (;)P +

. . a ,

the series on the right being absolutely convergent for I tl < 1 (9.1.2). It follows that in the Banach algebra A the series

converges absolutely for llxll < 1, and its sum y is self-adjoint when x is selfadjoint. Furthermore, since the square of the power series on the right-hand side of (1 5.6.1 1.2) is 1 + t, it follows that y 2 = e + x, by virtue of (5.5.3). This lemma shows that, if x E A is self-adjoint and llxll < 1, there exists such that y*y = e - x, so thatf(e - x) 2 0, or equivalentlyf(x) S f ( e ) . If now z E A is such that llzll < 1, then also 11z*z11 < 1 and therefore, using

y

EA

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NORMED ALGEBRAS A N D SPECTRAL THEORY

(15.6.2.4), we have If(z)l’ S f ( e ) f ( z * z )sf(e)’. This proves that f is continuous and that llfll s f ( e ) (5.7.1). Since also I(el(= I , we havef(e) 5 Ilfll, and assertion (i) is proved.

To prove (ii), observe that for y E A the linear form x ~ f ( y * x yon ) A is positive, because f ( y * x * x y ) = f ( ( x y ) * ( x y ) )2 0; by virtue of (i), the norm of this linear form isf(y*y). This completes the proof.

PROBLEMS 1. Let A be an algebra with involution and let U be a representation of A on a Hilbert

space H. In order that U should be irreducible, it is necessary and sufficient that the subalgebra B of Y(H) consisting of the endomorphisms V such that VU(s)= U(s)V for all s E A should be equal to C . I , ( . (To show that the condition is necessary, observe that B is an’involutive closed subalgebra of Y(H), hence is a star algebra. If S is a hermitian operator belonging to B, consider the closed commutative subalgebra C of B generated by S, which is separable and therefore isomorphic to ‘c,(X) for some compact metrizable space X. Show that C has no divisors of zero, other than 0, and conclude that X consists of a single point.) 2.

Let A be a Banach algebra with unit element e, and let x ~ x be * a (not necessarily continuous) involution on A . (a) Let f be a linear form on A such that f ( a z ) 2 0 for all self-adjoint elements a E A. Show that if also p ( a ) < 1, then f ( a ) is real and I < f ( e ) (use Section 15.4, Problem 17). Show that, for each self-adjoint element a E A , f ( a ) is real, If(a)i f(e)p(a) and f(a)’ 5J(e)f(a’). (Consider f ( ( a -i-&)’) where 6 E R.) Deduce that If(x)l 0 such that /lull 5 p . p(a) for all hermitian elements a E A. (To show that (6) implies (a),show that (6) implies that /IxlI 5 2pp(x), by writing x =a ib with a, b hermitian, and using Section 15.4, Problem 18(d) and 18(g). To show that (p) implies (S), show first that if a is hermitian, then Allall < ( a ) , by using (15.2.7); deduce that if x is normal, then h2llxn1l~ !l(x*)”II 1, and use Section 15.4, Problem 19 to show that A is hermitian. Finally, to show that ( y ) implies (6), observe that for each hermitian element a such that p(a) i1 there exists in A a hermitian element b, commuting with a, such that bZ = e - a’ (Section 15.4, Problem 17); consequently u = a ib is unitary, and a = t ( u f u*).)

+

+

5.

Let A be a Banach algebra with unit element e, endowed with a hermitian involution XHX*.

(a) Let x E A be such that p(x) < 1 . Show that x(e - x*x)’/’ = ( e - x x * ) ’ / ’ x (cf. Section 15.2, Problem ll(f)). (b) Under the same hypotheses, show that the function F( 0, and that F( 0 such t h a t c h : converges but hn does not. Use (a) by considering

c

the restriction of a hypothetical tracefto the algebra of endomorphisms of the vector subspace of H generated by the e, with k 5 n.)

8. COMPLETE HILBERT ALGEBRAS

Throughout this section we shall assume that the Hilbert algebra A is complete (and thus a Hilbert space) with respect to the norm llxll = (x I x)'". We shall also assume that the bilinear mapping (x, y ) ~ x of y A x A into A is continuous with respect to this norm (it can be shown (Section 12.16, Problem 8c)) that this is in fact a consequence of the other hypotheses). Every closed self-adjoint subalgebra B of A is a complete Hilbert algebra.

360

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NORMED ALGEBRAS AND SPECTRAL THEORY

(1 5.8.1) Let b be a closed left ideal in A, and put ub(x)* y = x y for all y E b and all x E A. Then X H U ,(x) is a representation of the algebra A on the Hilbert space b. First of all, U b ( x )is a continuous operator on b, by (15.7.5.3).Clearly U,(xx') = U,(x)U,(x'). For y , z in 6,we have

(ub(x)* * y I z,

=(y

I Ub(x> z>= ( y I xz) = (x*y I z> = (ub(x*) y I '

*

by virtue of (15.7.5.2), hence U,(x)* = ub(x*). Also if A has a unit element e, then U,(e) is the identity transformation. When b = A, we write U ( x ) in place of U,(x); the representation U is called the regular representation of A. It is.faithful(15.5)by virtue of (15.7.5.7). Moreover, X H U(x) is a continuous linear mapping of A into LZ(A), by the continuity of (x,y ) H xy. The study of complete Hilbert algebras is founded on the consideration of their minimal left ideals and the idempotents which generate them. For each left ideal I of A, we denote by I* the image of I under the involution S H S * ;clearly I* is a right ideal.

(1 5.8.2) For each left ideal I of A, the orthogonal supplement I' of the closure I of I is a left ideal. Since i is a left ideal (15.1.3),we may as well assume that 1 is closed. If

y E 1' and z E A, then for each x E 1 we have ( z y I x ) = ( y I z*x) = 0 because I is

a left ideal, whence it follows that zy E 1'.

(15.8.3) Let e # 0 be an idempotent in A. Then (9 llell 2 1; (ii) e* is idempotent; (iii) the left ideal Ae is the set of all X E Asuch that x = xe, and is closed in A. The first assertion follows from the inequality llell = lle211 5 Ilel12. The second is trivial. As to (iii), it is clear that if x = xe then x E A e , and conversely if x E A e , then x = ye for some y E A, hence xe = ye2 = ye = x. The fact that A e is closed then follows from the continuity of the mapping X H X - xe, and (3.15.1). Consider in particular the self-adjoint idempotents (cf. (1 5.5.3.1)):

(1 5.8.4) If el, e2 are self-adjoint idempotents in A, then the following properties are equioalent: (a) (el I e,) = 0 ; (b) e1e2 = 0; (c) e2 el = 0.

8 COMPLETE HlLBERT ALGEBRAS

361

Since (e1e2)* = efe: = e, el, it is clear that (b) and (c) are equivalent. If (el le,) = 0 then by (15.7.5.2) and (15.7.5.5) we have 0 = (e: le:) = (e1e2 I e1e2)= lle1e211', so that (a) implies (b). Conversely, if e1e2 = 0, then (el I e,) = (e: I e,) = (el I e l e z ) = 0, and therefore (b) implies (a). (15.8.5)

Every left ideal I # (0) in A contains a self-adjointidernpotent #O.

Let x be a nonzero element of I. Then z = x*x # 0 (15.7.5.7), and z is a self-adjoint element of I. Multiplying z by a suitable nonzero real scalar, we may assume that IIU(z)((= 1, where U is the regular representation of A (15.8.1). Hence, by (15.8.1) and (11.5.3), IIU(z')l( = 1, and therefore IIU(z2")11= 1 by induction on n. On the other hand, we have

II U(Zk+l)II= II U(z)U(zk)ll5 II U(z)II II U(z"Il

= II ~(z"II(

for all k , hence the sequence (IIU(zk))ll)is decreasing. Since it has infinitely many terms equal to 1, it follows that 11 U(zk))I= 1 for all integers k , and hence that I(zk(l4 1/ 11 UII for all k. We shall show that the sequence ( z Z kis) a Cauchy sequence in the Hilbert space A. Let n, p be two integers > 0 , and put m = n f p . Then (ZZm

I z2n)

=(

z 2 P p

I p )= ( z P + Z n I Z P + 2 n )

= 11 U ( Z P )

- (ZZn I and

*

z 2 y 2

IZ Z n )

1- II ~ Z( ~ ) . Z ~ + ~ n l l Z j ( Z ~ + Z n ~ Z =~( +Z ZZmn~)Z Z n )

( Z Z m l Z2 m

so that, for all m > n, l / U112 5 ( Z Z m I Z Z m ) 5 ( z 2 m 122") 5 (zZn I z'"). This shows that the sequence moreover, we have IlzZm

- ZZnllZ

( 1 1 ~ ~ ~ 1 is 1 ~ decreasing ) and has a limit a > 0;

= (z2m

5

I z 2 m ) - 2(22" IZZn)

llzZn112

+ (z2n 1 z2")

-a

which proves that ( z z k )is a Cauchy sequence, as asserted. Hence the sequence a limit e. By continuity, ez = l i m ( ~ = ~ e, ~ )and e* = lim(z*)Zk= e

( z z k ) has

k-+ do

k+ m

because z is self-adjoint, and finally ez2 = limzZk+2 =e, so that e E I. Finally, since llzk(lL I /

k- m

11 (Ill for all k 2 1, we have llell > 0, and the proof is complete.

A self-adjoint idempotent e # 0 is said to be reducible if there exist two orthogonal self-adjoint idempotents e l , e , , each nonzero, and such that e = el e , . In this case, by (15.8.4), we have eel = e,e = el and ee, = e, e = e2 . If e # 0 is not reducible, it is said to be irreducible.

+

362

XV

NORMED ALGEBRAS AND SPECTRAL THEORY

(15.8.6) (i) Every self-adjoint idempotent e # 0 is the sum ofafinite number of irreducible self-adjoint idempotents belonging to Ae. (ii) Every left ideal 1 # (0) contains an irreducible self-adjoint idempotent. Clearly (ii) is a consequence of (i) and (15.8.5). Hence it is enough to prove (i). If l\e1I2 < 2, then e must be irreducible, for otherwise we should have llellZ = lle11I2 lle2\12,where el and e, are self-adjoint idempotents # O , hence lle11* 2 2 by (15.8.3). We shall proceed by induction on the smallest integer n such that l\ellZ< n. If e = el + e, is reducible, where el and e2 are orthogonal self-adjoint idempotents #O, then el = ele and e, = e2 e, so that el and e, lie in A e ; moreover, we have

+

lle,ll

2

=

2

llell - Ile,lIZ 5 Ilel12 - 1 < n - 1

and similarly Ile,IIZ < n - 1, so that the inductive hypothesis can be applied to el and e, . This completes the proof. A left ideal I in A is said to be minimal if I # ( 0 ) and if there exists no nonzero left ideal I' # I contained in 1. Similarly for minimal right ideals.

(1 5.8.7) A left ideal I in A is minimal if and only if it is where e is an irreducible self-adjoint idempotent # 0.

of

the form I

= Ae,

If 1 is minimal, it contains an irreducible self-adjoint idempotent e # 0 (1 5.8.6). We have A e c I, and e = ez E A e , hence A e = I. Conversely, let e be an irreducible self-adjoint idempotent # O , and let I = A e . Suppose that I contains a left ideal I' distinct from (0) and I. Let e' be a self-adjoint idempotent # O contained in I' (1 5.8.5). Then we can write e = el e2 , where e2 = ee' and el = e - ee'. We shall show that e, and e2 are orthogonal selfadjoint idempotents. Since e' E A e , we have e' = e'e (15.8.3), hence

+

e2Z = eeleel

= eel2 = ee' = e2 Y

ee, = e2,

e, e

= ee'e = ee' = e,

,

and therefore elez = ( e - eJe2 = 0, e: = ( e - e2)'

e, el = e2(e - e,) = e - e, = el ;

= 0,

finally eT = (ee'e)* = ee'e = e, , and hence e, = e - e2 is also self-adjoint. We shall obtain a contradiction if we can show that e, and e, are both nonzero. If el = 0 , it follows that e = ee' E 1', hence I = 1', contrary to hypothesis. On the other hand, e'e2 = e'ee' = er2= e' # 0, and therefore e , # 0. This completes the proof.

8 COMPLETE HILBERT ALGEBRAS

363

In view of (15.8.6) and (15.8.3),we obtain the following corollary: (1 5.8.8) Every left ideal in A contains a minimal left ideal. Every minimal left ideal is closed.

(15.8.9) (i) If e, e‘ are two orthogonal self-adjoint idempotents, then the left ideals A e and Ae‘ are orthogonal. be a finite family of pairwise orthogonal, self-adjoint (ii) Let (ei)l

idempotents. Then for each x E A, the element x A e j (1 5 j 5 n).

-

n

i= 1

xe, is orthogonal to each

(i) We have (xe I ye’) = (xe I ye”) = (xee’ I ye’) = 0 , by (15.7.5.5) and (15.8.4). (ii) Likewise,

for all j . (1 5.8.10) For each x E A , there exists a finite or infinite sequence (en)of pairwise orthogonal, irreducible self-adoint idempotents belonging to the closure I of the ideal A x , such that x = xe,, (this series being convergent in A), and

llx112 =

1 Ilxen1I2.

n

n

We may assume that x # 0. Then there exists at least one self-adjoint idempotent e E I such that xe # 0, for the construction in (1 5.8.5) produces a self-adjoint idempotent e # 0 in I such that e(x*x)’ = e, and therefore ex* # 0, so that xe = (ex*)* # 0. Since e is the sum of a finite number of irreducible self-adjoint idempotents belonging to A e (I5.8.6), there exists at least one of these idempotents whose product with x is #O. We next remark that if (ei)l is a finite sequence of pairwise orthogonal, self-adjoint idempotents such that IIxeil12 2 u > 0 for all i, then by virtue of (15.8.9) we have

and therefore n S llxl12/a. Now define inductively an increasing sequence (cp(n))nz, of integers 2 0 ; a (finite or infinite) sequence (ek) of pairwise orthogonal, irreducible self-adjoint idempotents belonging to I; and a

364

XV

NORMED ALGEBRAS AND SPECTRAL THEORY

sequence ( x , ) , ~ of ~elements o f A, as follows: xo = x , q(0) = 0 ; suppose that cp(n) has been defined and that the ek have been defined for 1 5 k 5 cp(n), and let x,

=x -

v(n)

1 xek . If x, =0, the sequence (ek)is finite and has cp(n)elements;

k=

1

we take x,, = x, = 0 for m 2 n, and cp(m) = q(n) for m 2 n. If x , # 0, take a finite (possibly empty) sequence i s r of irreducible self-adjoint idempotents belonging to 1, orthogonal in-pairs and orthogonal to each ek for 1 k 5 cp(n), such that Ilxe:I12 2 I I x ~ ( ~ / ~ “ + ~ , and that the number r is as large as possible among all finite sequences having the above properties. (We have seen above that this number is s2“+’.) Then put cp(n + 1) = q(n) + r ;

s

ek = ei for k

= cp(n)

r

+ i, 1 2 i 2 r ; and x , + ~= x, - 1xe: . If no x, vanishes, i= 1

the sequence (cp(n)) tends to +a;for if it were bounded, we should have q ( m ) = q ( n ) for some n and all m 2 n, and by definition this would mean that for each irreducible self-adjoint idempotent e’ in the orthogonal supplement

F of

1 Ae,,

v(n)

we should have xe’

= x,e’ = 0 ;

but x,

E

F, x, # 0, and F is a

k= 1

closed left ideal (15.8.2), so that this would contradict what was established at the beginning of the proof. This being so, it is clear from the construction that Ilxe,(IZI llx1I2 by

1

virtue of (15.8.9); hence if the series

n

1 xe, is not a finite sum, it converges in n

any case to an element y E I which is the orthogonal projection of x on the closure of the left ideal a which is the sum of the Ae, (for x - y , being the limit of x -

m

1 xek, is orthogonal

k= 1

to all the en) (6.5.2). If x - y # 0,

there would exist in the left ideal a’ an irreducible self-adjoint idempotent e” E 1 such that xe” = ( x - y)e” # 0. If n is the smallest integer such that Ilxe”l12 2 ll~11~/2”+’, the existence of e” would contradict the maximality of Q.E.D. the family of e, such that cp(n) < i 5 cp(n + 1). (1 5.8.1 1) Suppose that the algebra A is separable. Then every closed lefi ideal

b is the Hilbert sum of a (jinite or injinite) sequence of minimal left ideals 1, = Ae, , where e, is an irreducible self-adjoint idempotent. For each x E b we have x = 1 xe, , and for all x , y in b we have ( x I y ) = 1 (xe, I ye,). n

n

The second and third assertions are consequences of the first and of the definition of a Hilbert sum (6.4), since xe, is the orthogonal projection of x on Ae, (15.8.9). To prove the first assertion, let ( x , ) , , ~be~ a dense sequence in b (3.10.9). We define inductively, for each n, a (finite or infinite) sequence (en,i ) i s , , of irreducible self-adjoint idempotents, as follows. For (el,J i e , ,

8 COMPLETE HILBERT ALGEBRAS

365

we take any sequence of pairwise orthogonal, irreducible self-adjoint idempotents belonging to b and such that x1 = q e , , (15.8.10). Now suppose that

1 i

the em, have been defined for m 5 n in such a way that they are pairwise orthogonal and belong to 6,and are such that the x, with m 5 n belong to the closure a, c b of the left ideal which is the sum of the ideals Ae,, for all m S n and i E I, for each m. Let x;+ be the orthogonal projection of x,+, on a' n b; then take for (e,+,,i)iE,n+la sequence of pairwise orthogonal irreducible self-adjoint idempotents which belong to a' n b and are such that xA+, = X A + ~ ~ , + ~ , (15.8.10). It is clear that the double family (en,i),

1 I

arranged in a single sequence, has the required properties, by virtue of (15.8.9) and (6.4.2). This theorem applies in particular when b = A . In this case we get a decomposition of A as a Hilbert sum of minimal left ideals. It shouId be remarked that in general there will exist infinitely many such decompositions (see later (15.8.14)). More precisely: (15.8.11 .I ) Suppose that A is separable, and let 1 be a minimal left ideal of A. Then there exists a decomposition of A as a Hilbert sum of minimal left ideals I,, such that I , = 1.

Apply (15.8.11) to b = 1'. (15.8.12) Let e, e' be two irreducible self-adjoint idempotents, and let 1 = Ae, 1' = Ae' be the corresponding minimal left ideals. (i) Every homomorphism of the A-module I into the A-module I' is of the form fa : x ~ x a where , a E eAe' = e A n Ae'; it is either zero or bijective, and the mapping a w f a is an isomorphism of the complex vector space eAe' onto Hom,(l, 1') such that fa6 =f b o f , . (ii) The C-algebra eAe, isomorphic to End,(l), is afield, equal to Ce (and therefore isomorphic to C). (iii) If I and I' are not isomorphic as A-modules, then e and e' (and consequently I and 1') are orthogonal, and 11' = 1'1 = (0). If I and 1' are isomorphic as A-modules, then eAe' is a complex vector space of dimenston I , and 11' = 1'. (iv) Zf x E A , then Ix is a leji ideal which is either zero or isomorphic (as A-module) to 1.

(i) If g : I -+ I' is an A-module homomorphism and a = g(e), then for all x E I we have g(x) = g(xe) = xg(e) = xu (15.8.3(iii)). Since a E I' we

366

XV

NORMED ALGEBRAS AND SPECTRAL THEORY

have a = ae’ (15.8.3(iii)), and on the other hand a = g(e2)= eg(e) = ea, so that a E eAe’, and g = f a . It is clear that eAe’ c e A n Ae’; conversely, if y E e A n Ae.‘, then y =ye’ and y = ey (15.8.3(iii)), so that y = eye’ E eAe‘. The image g(1) is a left ideal contained in I’, so it is either ( 0 ) or 1’; likewise, the kernel g-’(O) is a left ideal contained in I, hence is either ( 0 ) or I. If g-l(O) = I, then g(I) = ( 0 ) ; if g-’(O) = {G}, then we must have g(1) # {0}, hence g(1) = I’ and g is bijective. Finally, if f a = 0, we have f,(e) = ea = 0 ; but a E eAe’, so that ea = a, and consequently a = 0. (ii) The C-algebra eAe is a closed subalgebra of A (15.8.3(iii)). Since we have seen above that every element of End,([) is either zero or invertible, it follows that End,(I) is a (possibly noncommutative) field, and hence the same is true for eAe. Clearly e is the unit element of eAe, and because A is a normable algebra (1 5.1.8) it follows from the Gelfand-Mazur theorem that eAe = Ce. (iii) If I and I’ are not isomorphic, we have eAe’ = {0} by (i) above, and in particular ee’ = 0; hence e and e’ are orthogonal (15.8.4). The same is true of I and I’ (15.8.9), and II’ = ( 0 ) . If 1 and I’ are isomorphic, and if g is an isomorphism of I onto 1’, then every homomorphism of I into I’ is of the form g u, where u is an endomorphism of 1. Hence eAe‘ is a complex vector space of dimension 1, by virtue of (i) and (ii) above. Clearly 11’ is a left ideal contained in 1‘; since it contains eAe‘ # {0}, it must be equal to 1’. (iv) Since Ix is the image of 1 under the homomorphism y w y x of I into A, it is a left ideal isomorphic to I/[’, where I’ is the kernel of the above homomorphism. But I’ must be equal to either {0} or 1, and so y w y x is either zero or injective. 0

(15.8.1 3) Suppose that A is separable. Then : (i) There exists afinite or infinite sequence ( I k ) k e J of minimal left ideals, no pair of which are isomorphic, such that every minimal left ideal of A is isomorphic (as an A-module) to some Ik . (ii) For each index k E J, the closure of the sum of all the minimal left ideals of A which are isomorphic to I , is a self-adjoint two-sided ideal ak . Every minimal left ideal of the Hilbert algebra ak is a minimal left ideal of A, isomorphic to I,, and the algebra ah contains no closed two-sided ideals other than ( 0 ) and a k . (iii) Each of the algebras a k is the Hilbert sum of a (finite or infinite) sequence of minimal left ideals isomorphic to I,. The algebra A is the Hilbert sum of the sequence (akIkE,, and ah ah = ( 0 ) whenever h # k. Start with a decomposition of A as a Hilbert sum of minimal left ideals 1; (15.8.11). We take I, = I; and define inductively & + I to be equal to I;,

8 COMPLETE HILBERT ALGEBRAS

367

where m is the smallest integer such that ;1 is not isomorphic to any of the ideals I,, . . . , I,. (If all the 1; are isomorphic to one or other of I,, . . . , Ik, the induction stops at Ik .) Let J be the sequence of indices k so obtained, and for each k E J let Ij be the sequence of integers n such that I: is isomorphic to 1., We define ak to be the Hilbert sum of the 1; for n E 1,. Clearly A is the Hilbert sum of the left ideals ak (6.4.2). Let I be any minimal left ideal in A . Then I must be isomorphic to one of the Ik, for otherwise it would be orthogonal to all the 1; (15.8.12(iii)) and hence orthogonal to A itself, which is absurd. The same argument shows that I is orthogonal to all the ah with h # k . Hence, as ak is the orthogonal supplement of the Hilbert sum of the a,, such that h # k, we must have I c ak. From this it follows already that a k is the closure of the sum of all the minimal left ideals of A which are isomorphic to Ik , and therefore ak is independent of the decomposition of A as the Hilbert sum of the I;, from which we started. Moreover, for each x E A and each n E I,, 1;x is a left ideal which is either (0) or isomorphic to 1; (15.8.12(iv)), hence is contained in ak. This proves that ak is a two-sided ideal. If 1; = Ae;, where e; is an irreducible self-adjoint idempotent, then I;* = e; A , hence a: = ak. Let I” be a minimal left ideal of the Hilbert algebra ak . We have I” = ake”, where e“ is a self-adjoint idempotent (15.8.7), and e; e” cannot vanish for all n E I,, otherwise I” would be orthogonal to all the 1; (n E Ik) and therefore to the closure of their sum, namely to ak : which is absurd because I” # ( 0 ) . Hence there exists at least one index n E 1, such that 1; I” # ( 0 ); since 1; I” is a left ideal in a k , we must have 1; I” = I”, which shows that I” is a minimal left ideal of A, necessarily isomorphic to 1; and therefore to I, (15.8.12(iii)). If now b is a nonzero two-sided ideal of the algebra ak, it contains at least one minimal left ideal I” of this algebra (15.8.8), hence also contains all the I”1; ( n E lk). But 1’7; = 1; (15.8.12(iii)), and therefore b contains the sum of all the 1; (n E Ik). If b is closed, it follows that b = a k . Finally, a h a k c ah n ak = ( 0 ) if h # k , because ah and ak are two-sided ideals.

A complete Hilbert algebra A is said to be topologically simple if it contains no closed two-sided ideal other than A and ( 0 ) .It follows from (15.8.13) that the study of the structure of a separable complete Hilbert algebra A reduces completely to the study of the a k , that is to the case where A is topologically simple. (15.8.1 4) Let A be a topologically simple, complete, separable Hilbert algebra. Thenfor each minimal left ideal 1 of A, the representation X H U,(x)of A in the Hilbert space I is faithfur. If A is injnite-dimensional, then so is I. The image of A under U , is the algebra Y2(I)of Hilbert-Schmidt operators on I (15.4.8), and there exists a

368

XV

NORMED ALGEBRAS AND SPECTRAL THEORY

constant y > 0 such that

(for the scalar product defined in (1 5.7.4)). If A is finite-dimensional, then the image of A under U , is the algebra End,-(I) of all endomorphisms of the vector space I , and the relation (15.8.14.1) remains valid, for the scalar product defined in (15.7.4.1) (from the scalar product on I which is the restriction of that on A). We can assume that A is the Hilbert sum of a (finite or infinite) sequence of minimal left ideals I,, = Ae,, where I = I , (15.8.11.1) and all the I, are isomorphic (15.8.13). If x # 0 and Li,(x) = 0, that is, if X I = {0}, we should have (Ax)I = {0}, and since the ideal A x is nonzero, it contains a minimal left ideal I’ (15.8.8), which must be isomorphic to I (15.8.1 3); hence 1’1 = (O), contrary to (15.8.12(iii)). Hence the representation U, is faithful. Put P,,= U,(e,,), which is the orthogonal projection of I = A e , onto the one-dimensional subspace enA e , (15.8.12), because (xel - enxe, I eny e , ) = (enxe, - ei xel lye,) = 0. Since erne,= 0 if m # n, we have PmP, = 0 and therefore the subspaces enA e , are orthogonal in pairs. Moreover, I i s the Hilbert sum of the subspaces enAe,, because, if xe, is orthogonal t o all these subspaces, we have P,,(xe,) = 0 for all n, so that e,,xe, = 0 for all n, and therefore xe, belongs to the right annihilator of A, which is zero (15.7.5.7). This shows that the sequence (I,,) is finite if and only if 1 (and therefore each I,) is finite-dimensional over C, or equivalently if and only if A is finite-dimensional. Let (a,,) be an orthonormal basis of I such that a,, E e,Ae, for each n. Then a,,a,* E e,Ae,,, hence ana: = A, en for some A,, E C* (15.8.12). Likewise a,* a,, = A; el. We have A; = A,, , because on the one hand and on the other hand

a,, aza, a: =,.:A

a,, a,*a, a: = A; a, e,a,* = A; a,, a,* = A; A, e n ,

8 COMPLETE HILBERT ALGEBRAS

for all n, and all the 2, have the same value y A, we have

369

I e l ) - ' . Hence, for all x , y in

= (el

(xa, I van) = ( Y * X I a, a,*) = (Y*X I ye,)

=ybe,

I ye,);

since the series with general term (xe, I ye,) is absolutely convergent, with sum (x I y ) (1 5.8.1 I ) , it follows that if A is infinite-dimensional then U,(x) is a Hilbert-Schmidt operator, and the relation (15.8.14.1) is valid. Since A is a Hilbert space, so is its image under U , , and to show that this image is the whole of the Hilbert space Y z ( l(15.4.8), ) it is enough to show that U,(A) is dense in Yz(l). Now, for each pair, m, n with m # n, we have

emAel, * e, Ae, = e,(Ae,)(Ae,)

= emA e ,

(15.8.12(iii)), and since e,Ae, = C a n , it follows that there exists em, E elnAe, such that emnun= a, (which implies that em,,= y-'a,a,*), and clearly emnap= 0 if p # n. We conclude from this that Em, = U,(e,,) is the continuous endomorphism of the Hilbert space 1 such that Em,, * a,, = a, and Em,,ap = 0 if p # n. Our assertion now follows from the fact that the finite linear combinations of the El,, are dense in Y2(I) (15.4.8). The proof is analogous but simpler when A is finite-dimensional.

-

We remark that, in all cases, the U#,) consist of endomorphisms of the form U(x) o P,,and therefore consist of endomorphisms of rank 1 of I.

Under the hypotheses of (1 5.8.1 4), if there exists an element # 0 in the center of A, then A isjinite-dimensioiial. In that case the center of A is Cu, where u is the unit element of A. (1 5.8.1 5)

If c E A belongs to the center of A , then U,(c) is an endomorphism of the A-module I, hence is a homothety X H ~ X - ,where 2 E C (15.8.12). But clearly a homothety cannot be a Hilbert-Schmidt operator on an infinite-dimensional Hilbert space, unless it is zero. (15.8.16) Let A be a separable complete Hilbert algebra and let ak ( k E J) be the topologically simple Hilbert algebras which are the Hilbert summands of A (1 5.8.1 3 ) . For each k E J, let 1, be a minimal left ideal of ak . Let V be a nondegenerate representation (15.5.5) of A in a separable Hilbert space H, such that V : A H 2 ( H ) is continuous. (i) H is the Hilbert sum of subspaces H, ( k E J) stable with respect to V , such that if V , is the restriction of V to H, we have v&) = 0 for all s E ah and all h # k : so that V , may be considered as a representation of a k on H, . (ii) If ah isjinite-dimensiond over c, the representation v k is the Hilbert sum of a (finite or injinite) sequence of irreducible representations, each equivalent to the representation U,, of ak (15.8.14).

370

NORMED ALGEBRAS AND SPECTRAL THEORY

XV

Let H, be the closure of the vector subspace of H generated by the vectors v(sk) . x, where sk E a, and x E H. Since each s E A can be written in the form s = sk with s, E a,, and since V is continuous, we have

1 k

k

(5.5.2), and therefore H is the closure of the sum of the H k . Also, if I? # k a n d E ah, sk E a k , we have

sh

(V(sh)

’

I v(sk> Y ) = (V(s,*sh> I Y ) = ’

’

because a,, is self-adjoint and ak ah = to). This proves (i). Now assume that A is topologically simple and finite-dimensional over C, and therefore has a unit element; we can then restrict ourselves to the case where there exists a totalizer xo for V (1 5.5.6). The vector subspace of H generated by the V ( s ) xo is finite-dimensional, hence closed (5.9.2) and so equal to H. We may therefore argue by induction on the dimension of H. Since A is the sum of a finite number of minimal left ideals, there is at least one minimal left ideal, say I, such that the subspace E = V(1) xo is nonzero. The surjection S H V(s)* xo of I onto E is then an A-module homomorphism, and since its kernel is a left ideal I‘ contained in I and distinct from I, we have 1’ = (0). Hence E is a stable subspace of H with respect to V , such that the restriction of V to E is equivalent to U , . Since the orthogonal supplement H’ of E in H is stable with respect to V and of dimension strictly less than the dimension of H, we have only to apply the inductive hypothesis to complete the proof. The result of (15.8.16(ii)) can be shown to be valid without assuming that ak is finite-dimensional (Problem 1). PROBLEMS 1. Let A be a topologically simple, separable, complete Hilbert algebra. Let V be a nondegenerate representation of A in a Hilbert space H. With the notation of the proof of (15.8.14), put En= V(e.) and A, = V(a,). The Enare orthogonal

projectors on subspaces H, of H, such that H is the Hilbert sum of the H,, and we have A.(H,) = H, and A:(H.) = H I . If (bkl)kEI is a (finite or infinite) orthonormal basis of H,, then for each n the vectors bx. = y - 1 ’ 2 A , ( b r l form ) an orthonormal basis for H, as k runs through the index set 1. Deduce that, if H; is the subspace of H generated by the bk, (n 2 l), then Hi is stable with respect to V , and V is the Hilbert sum of the restrictions Vkof V to H; . Each of those representations is topologically cyclic, bkl being a totalizer because El . b k l = b k l . Show that this representation is equivalent to the representation Ul . 2. Let H be an infinite-dimensional separable Hilbert space, A = 9 A H ) the Banach

algebra of Hilbert-Schmidt operators on H, and B a closed self-adjoint subalgebra of

9 THE PLANCHEREL-GODEMENT THEOREM

371

A . Show that there exists a decomposition of H as a Hilbert sum of a subspace Ho and a (finite or infinite) sequence of closed subspaces Hkstable with respect to B, with the following properties: (1) the restrictions to Ho of the operators belonging to B are all zero; (2) each HI,is the Hilbert sum of a Jinife sequence (HLp),d p d r k of subspaces of the same dimension (finite or not) which are stable with respect to B; the restrictions to H x l of the operators belonging to B form the algebra of Hilbert-Schmidt operators on H k l ;moreover, for 2 5 p 5 r k , there exists an isometric isomorphism T, of Hkt onto H,, such that, if U, is the restriction to Hkl of an operator U E B, then the restriction of U to HkDis T,UIT;'.

9. T H E PLANCHEREL-GODEMENT T H E O R E M

In this section we shall study commutative (but not necessarily complete) Hilbert algebras. More generally it will be convenient, in view of applications, to consider a commutative algebra A with involution, endowed with a bitrace g (15.7) satisfying conditions (U) and (N) of (15.6). We shall denote by it, the two-sided ideal of elements s E A such that g(s, s) = 0, and by ng : A + A/n, the canonical mapping. We recall that A/n, is canonically endowed with a structure of Hilbert algebra (15.7). We shall assume throughout that the prehilbert space A/n, is separable, and hence is a dense subspace of a separable Hilbert space, which we denote by H,. Thus, starting from g, we obtain canonically a representation of A in H, (15.6.10), which we denote by U , . The image of A under U , is a commutative subalgebra with involution (15.6.1) of Y(H,). Let d,denote its closure in Y(H,), so that d, is a commutative star algebra (and therefore consists of normal operators (15.4.11)). We shall assume that this algebra d,is separable (this is not a consequence of the separability of A/n,, cf. Problem 1). A particular example of a trace on A (which therefore gives rise to a bitrace, by the canonical procedure (15.6.2)) is provided by the hermitian characters of A, i.e., the characters x of A which satisfy

Xb*)

(15.9.1 )

=

xo

for all x E A. It follows that x(x*x) = I x(x)lz 2 0, and if we put g(x3

v> = X ( Y * X )

= XOX(X),

the condition (U) of (15.6) is trivially verified, because

On the other hand, the ideal n,, which is here the kernel of X, is a hyperplane in A, so that the algebra A/n, can be identified with C, and the condition (N)

372

XV

NORMED ALGEBRAS AND SPECTRAL THEORY

follows immediately from the fact that ~ ( x#)0 implies ~ ( x ’ )# 0. The corresponding representation U, is evidently irreducible. We denote by H(A) the set of hermitian characters of A ; it is a subset of the product space CA and is closed in the product topology (3.15.1). We give H(A) the topology induced by the product topology (i.e., the weak topology (12,lS)). When A is a separable commutative Banach algebra with involution, having a unit element e # 0, the space H(A) is a compact metrizable subspace of X(A) (15.3.2). In this situation it can happen that H(A) # X(A) (Section 15.4, Problem 3), but H(A) is equal to X(A) if A is a separable star algebra (1 5.4.14).

We shall show that the bitraces g satisfying the conditions at the beginning of this section can all be obtained, by a canonical process of “integration ”, from hermitian characters: (Plancherel-Godement theorem) Let g be a bitrace on a coin(15.9.2) mutative ulgebra A with inriolution, satisfying (U) and (N),and such that the prehilbert space A/n, and the star algebra d,c U(H,) are separable. (I) We can dejine canonically: (1) a subspace S, of H(A), whose closure in CA is either S, or S, v (0) and is metrizable and compact (so that S, is locally

compact, metrizable and separable); (2) a positive measure my on S, , with the follo MYng properties: ) to $P:(S,, m,) (i) For each x E A, the function X H ~ ( X ) = ~ ( xbelongs and we hal;e

for all x , y in A. (ii) As x runs through A, the set of functions i is contained in %?;(S,) (1 3.20.6) and is dense in this Banach space.

(iii) The support of the measure m, is the whole of S , . (iv) The mapping X H i factorizes into x H x,(x) 2 2,and the mapping To of A/n, into %$(S,) extends to an isomorphism T o f the Hilbert space H, onto L@,, m,) such that for all x E A we have U,(x) = T-’M(i?)T, where M ( 2 ) is multiplication by the class of i? in L$(S,, m,) (cf. (13.21.5)). Also 1 1 1 I”, a

for all 1 E C arid x

E

+ U,(X>II = I l l + 41

A.

(11) Conversely, let S be a subspace of H(A) such that S u ( 0 ) is compact and metrizable, and let m be a positive measure on S, with support equal to S,

9 THE PLANCHEREL-GODEMENT THEOREM

such that for all x E A the function

373

x i(x) = ~(x)belongs to H

Y ; ( S , m) n %?g(S).

Then g'(x, y ) = fs R(x)y(x)dm(x) is a bitraee on A saridying (U) and (N), such that A/n,, and d,, are separable; and we have S,, = S and m,. = m . The proof of this theorem is in several steps. (1 5.9.2.2)

Construction of S, and proof of (ii).

The subalgebra C * IH9 + d,= ,Pei is closed in 9(H,) (5.9.2) and hence is a commutative star algebra with unit element. For each character 5' E X ( d & it follows from (15.4.14) that 5' Or, is identically zero on A or else is a hermitian character of A : in other words, w : ('-5' U, is a mapping of X ( d i ) into H(A) u (0). This mapping o is injectiw, because we have t'(lHg) = 1 for all 5' E X ( d i ) , and since t;' is continuous on di, the values of x)/(f(e) -ffz*z))

belong to P,, and that i f f is an extremal point then f=f, = f 2 . Hence we have f ( z * r x ) - z f ( x ) f ( z * z ) ; replacing z by r(e y ) where y is small, deduce that f is a character.) (c) Show that for each f~ P, there exists a unique positive measure pf on P I , with niass 1, such that for each x E A we have

+

f ( x ) = Jpnx(*:(x)

&Ax).

(Use Section 13.10, Problem 2(b). For the uniqueness of pr, use the Stone-Weierstrass theorem.)

10. R E P R E S E N T A T I O N S O F ALGEBRAS O F C O N T I N U O U S F U N C T I O N S

Let K be a coinpact nzetrizable space. The application of the results of Scction 15.9 to the case where A = %‘,-(K) allows us to describe very simply all the representations of this algebra. We consider first the topologically cjdic. representations u H T(u). (15.10.1) Let K be a compact metrizable space and A = %,(K). Then every topologically cyclic representation of the commutative algebra with involution A, in a sc>jiarabie Hilbert space E, is equivalent to one of the representations UH M,,(u) (lefined as,follows: let p be a positive measure on K, and for each u E A, let M J u ) dcnote the continuous operator on Lz(K, p) induced by multiplication by u : that is, .for each f E 2’t(K, p), M,(u) * f is the class of uf in

G ( K >10.

Let a be a totalizing vector for a representation U H T(u) of A in a separable Hilbert space E. The representation Tis determined up to equivalence by the positive linear form f,(u) = ( T ( u ) a I a) on A (15.6). Since A is separable (7.4.4),the Bochner-Godement theorem applies ; all the characters of A are hermitian, and the spectrum X(A) can be canonically identified with K (15.3.7).Hence the proposition is an immediate consequence of (1 5.9.qiv)). It follows from the definition (15.10.1) of M , that

10 REPRESENTATIONS OF ALGEBRAS OF CONTINUOUS FUNCTIONS

387

(with respect to the measure p). For it follows from (13.12.2) that //M,(u)/l2 N,(u); also, for every positive real number a < Nm(u), there exists a nonnegligible integrable subset P of K such that lu(t) I 2 c( for all t E P; hence N,(ucp,) 2 ci * N2(qP),from which (15.10.2) follows. The measure p is not uniquely determined. We shall come back to this point a little later (15.10.7). The definition of M,(u) given in (15.10.1), and the formula (15.10.2), still make sense if u is not necessarily continuous on K, but simply p-nieasurable and bounded in measure (by virtue of (13.12.5)). The mapping u w M J u ) extended in this manner is a representation of the algebra with involution Y,"(K, p) on the Hilbert space Lf(K, p). Clearly, if u1 and u2 are equal almost etlerywhere with respect to p, we have M,(u,) = M,(u,). Usually we shall restrict the representation M , to a self-adjoint subalgebra of Y","K, p) which does not depend on the choice of p, namely the algebra diC(K) of universally measurabfe, bounded complex-cafued ,functions on K (13.9). By (13.9.8.1) this is indeed a C-algebra with involution, and it is a Banach subalgebra of BC(K) by virtue of Egoroff's theorem (13.9.1 0). (15.10.3) Again let u- T(u) be a representation of %',-(K) in a separable Hilbert space E, admitting a totalizing vector a. (These hypotheses will be in force up to and including (15.10.7).) The preceding remarks show that the representation T may be extended to the algebra with involution @c(K). The extended representation (also denoted by T ) does not depend on the choice of totalizing vector a. To prove this assertion, let x and y be two vectors belonging to E. For each u E Vc(K), we have I/ T(u)I/ 5 /Iull (15.5.7) and therefore

consequently the linear form U H ( T ( u ). x I y ) is a measure p x , yon K such that llPx,yll 5 llxll . llull. From this definition it follows immediately that (15.10.3.2)

-

Also, if x = T(v) a and y = T(w) a, where 2) and each function u E q C ( K ) we have

-

14'

are in q C ( K ) , then for

( T(u) x I y ) = (T(uo). a I T(w) * a) = ( T(M'ur)* a I a) =

388

XV

NORMED ALGEBRAS AND SPECTRAL THEORY

and therefore

The independence asserted above will be a consequence of the following more precise result: (15.10.4) (i) For all x, y E E and all u E %,-(K) we have

s

. x IY ) = 4 0 d P x , y ( o .

(15.10.4.1)

(ii) Zf(u,) is a uniformly bounded sequence offunctions in %,(K) which conoerges to u, then

for all x , y in E. It is enough to prove (15.10.4.2) for x, y lying in a rota1 subspace of E: the sesquilinear functions (x,y ) H(T(u,) x I y ) form an equicontinuous set, because

(7.5.5). Take x by definition,

=

-

T(s,) a and y = T(s2) a, with s1 and s2 in %,-(K); then,

(T(u,) . x I Y ) = (T(3,unsI) * a I a ) =

s

~2

u,si d p ?

and it is enough to apply the dominated convergence theorem (13.8.4) to the measure p. As to (15.10.4.1), it is valid by definition when u E %,(K), and in general it follows by applying the dominated convergence theorem twice to the measure p L x , yand , using (13.7.1). (15.10.5) For the operator T(u) to be hermitian (resp. positive hermitian (11.5), resp. zero, resp. unitary) it is neces’sary and suficient that u(x) should be real (resp. u(x) 2 0, resp. u(x) = 0, resp. lu(x)l = 1) almost euerywhere with

respect to the measure p .

389

10 REPRESENTATIONS OF ALGEBRAS OF CONTINUOUS FUNCTIONS

It is immediately verified that if U is a hermitian (resp. positive hermitian, resp. unitary) operator on a Hilbert space E, and if S is an isomorphism of E onto a Hilbert space E', then the operator SUS-' on E' has the same property (since these properties involve only the Hilbert structure). Hence we may take T(u) = M,(u), and in this case the sufficiency of the conditions stated is clear. On the other hand, if for example there exists a measurable subset X of K such that p ( X ) = a > 0 and 9 ( u ( x ) )2 fl > 0 for all x E X, then we have 9(( T(u) . qXI qX)) = u dp) 2 ctfl and therefore T(u) is not hermitian. The other cases are dealt with similarly.

9(Ix

(15.10.6) (i) The orthogonal projectors belonging to the algebra T ( @ J K ) ) are the operators of the form T(cpx),where X is a universally measurable subset of K . (ii) T(@,(K)) is a maximal commutative subalgebra of,ri4(E). (iii) A closed vector subspace F of E is stable under T ifand only if it is of the form T ( q x ) ( E ) ,where X is a universally measurable subset of K .

(i) By virtue of the characterization of orthogonal projectors on E (15.5.3.1) and by (15.10.5),it follows that T(u)is an orthogonal projector if and only if u is almost everywhere equal to a p-measurable bounded real-valued function u such that u2 = u. Hence v = cpx, where X is a p-measurable subset of K, and the result follows. (ii) It is clearly enough to prove that a continuous operator V E 9 ( E ) which commutes with T(u) for all u E @&) is of the form T(u). If a is a totalizing vector for T, then for each universally measurable subset X of K we have

5 IIVII . llT('px). all2 = IIVII

*

s

' p x dP.,a = 1 I VII.

.s

dP

because T('px)is an orthogonal projector which commutes with V. The Lebesgue-Nikodym theorem (13.15.5(c')) therefore shows that p v . o , r r= h * p for some p-integrable function h. Since also it follows from the inequality above that

we conclude that the linear form for every step function WE&&), WH h w d p defined on the vector space of these step functions extends by

s

390

XV

NORMED ALGEBRAS A N D SPECTRAL THEORY

continuity to L?A(K, p) ((13.9.12) and (13.9.13)). Hence, by (13.17.1), h is bounded in measure with respect to p, and we may assume that 12 E %,-(K). For all s, t E @c(K) we have (b‘T(s). u I T(t). a )

= (T(s)V * u

=

pl

1 T ( t ) .U ) =

s

si dp“..,.

dp = ( T ( h ) T ( s )u. I T ( t ) .a ) ;

since the vectors T ( s ). a form a total set in E, it follows that V = T ( h ) . (iii) To say that F i s stable with respect to T means that the orthogonal projector f’, commutes with T(u) for all u E %,(K) (15.5.3); hence P, E T(J2/c(K))by virtue of (ii), and is of the form T(cp,) by virtue of (i). We can also characterize all the totalizing vectors of the given representation: (15.10.7) For a cector Lj E Li(K, p) to be totalizing f o r the representation M,, of ‘e,(K), it is necessary and sufficient that g ( t ) # 0 alyost everywhere with respect to p . In order that the classes (fg)-,where f runs through V,(K), should generate a subcpace which I S not dense i n the Hilbert space LE(K, p), it is necessary and sufiicienl that there should exist a nonnegligibIe h E 9’$(K, p) such

i

that fgh dp = 0 for all f E %‘,(K) (6.3.1). Also we have gh E P,!.(K, p) (1 3.11.7) ; hence the measure ( g h ) * p is zero, which implies that g ( t ) h ( t )= 0 almost everywhere with respect to p (13.14.4). But since by hypothesis the integrable set A of points t E K at which h ( t ) # 0 is not negligible, we must have g ( t ) = 0 almost everywhere in A. This proves the proposition. To such a totalizing vector Lj for M,, corresponds by definition (15.10.1) the positive measure 1gI2 . p on K. Since 1gI2 is p-integrable and # O almost everywhere, we obtain in this way all the positive measures on K which are equivalent to 11 (13.15.6). In other words, the measure p of the statement of (1 5.10.1) is determined on/^ up to equicalence by the representation T.

(1 5.10.8) Now let us consider an arbitrary nondegenerate representation Tof ‘ec(K) in a separable Hilbert space E. From (15.5.6) we know that E is the Hilbert sum of a sequence (En)of closed subspaces stable with respect to T , and such that the restriction T,, of T to En admits a totalizing vector a,,, for each n. The definition of the measures p x , ygiven in (15.10.3) applies without

10 REPRESENTATIONS OF ALGEBRAS OF CONTINUOUS FUNCTIONS

391

change. Also, for each function u E %&), we define T,(u) on En (15.10.1), and by virtue of (15.10.2) we have llT,,(u)~lI llull for each n. Now we have the following lemma: (15.10.8.1) Let E be a Hilbert space which is the Hilbert sum of a sequcnce (En) of closed subspaces. For each n let U , be a continuous operator on En, and suppose that the sequence of norms (11 U J ) is bounded. Then there exists a unique continuous operator U on E whose restriction to En is U,, for euch Also, the restriction of U * to En is U,* . Suppose that /I U,ll all n and llxll2 =

5 a for all n. For each x

c llx,l12 (6.4), we have

=

n.

1x, E E, where x, E En for n

n

C IIUn

xn/125

n

a2 C IIxnI12 =a211xI12, n

which shows that the series 1U , * x, converges in E. If U . x denotes its sum, n

it is clear that U is linear and that, from above, 11 U . xi1 5 allxll, so that U is a continuous operator (5.5.1). The uniqueness of U follows from the fact that the union of the subspaces En is a total set in E (6.4). Finally, we have 11 U,*l/ = /I U J 5 a for all n, and therefore there exists a continuous operator Y on E whose restriction to Enis U,* for all n . If y = y,, with y , E En for all n

and

c

1

llynllZ= lly(I2,then (6.4) we have

n

(u . x I Y > = Cn ( u n

*

xn I yn) =

C (xn I UT . yn> = (X I I/

Y),

n

which proves that V = U *. Applying (15.10.8.1), we see that there exists a unique normal continuous operator T(u)on E whose restriction to Enis T,(u), for each n. It is immediate that UI+ T(u) is a representation of @&) in E which extends the representation T of qC(K). Next, the proposition (15.10.4) generalizes without any change in the proof: we have only to observe that we can take as a total subset of E the set of all T,(s) a,, where s E VC(K) and n is arbitrary. In general there exist infinitely many decompositions of E as a Hilbert sum of subspaces with the properties of (15.10.8). However, there is the following result : (15.10.9) There exists a decomposition of E as a Hilbert sum of a (finite or infinite) sequence (En) of closed subspaces, stable with respect to T , such that the

392

XV

NORMED ALGEBRAS A N D SPECTRAL THEORY

restriction of T to Enadmits a totalizing vector a, and such that, if p, is the positive measure on K corresponding to a, (15.10.1), then p, + is a measure with base p,, (13.1 3) f o r each n. We begin with a decomposition of E as the Hilbert sum of any sequence (F,) as described above. Let b, be a totalizing vector for the restriction of T to F, , and let v, be the corresponding positive measure on K. Since the measure v, is determined only up to equivalence (15.10.7), we may multiply it by a strictly positive constant so as to ensure that the series of norms ~ ~ v ,con,~~ verges. By induction on n 1 2 we define two sequences (v;), (v:) of positive measures on K, as follows. The measures v; and vl; are those which appear in the Lebesgue canonical decomposition of v 2 relative to v,, v; being a measure with base vl, and vl; disjoint from v, (13.8.4). For k > 2, the measures v; and v; are likewise such that vk = v; + v;, where v; is a measure with base (v, vl; v;- 1) and v; is disjoint from v, vl; + * * . + v;To each of these decompositions there corresponds a partition of K consisting of two universally measurable sets B; and B; such that v; is concentrated on B; and v: is concentrated on B; . Let Fk= F; 0 FL be the corresponding decomposition of F, as a Hilbert sum of mutually orthogonal subspaces. If F, is identified with Lg(K, vk) (15.10.1) then F; and F; admitb; = @,.,and 6; = @s..,astotalizing vectors, respectively (15.10.6). The sequence of measures v1 + v'; + . . + v; is increasing, and the norms of these measures are bounded above by

+ + +

,.

+

-

m

I(v,ll. Hence (13.4.4) this sequence has a least upper bound p1 in M,(K),

n= 1

and p, is also the vague limit of the sequence. The preceding construction allows us to assume that, if v1 is concentrated on B,, then the sets B,, Bl;, . . . , B;, . . . are pairwise disjoint, and vg is identified with q,.., * p,. If El is the Hilbert sum of F, and the F; for k 12, then E is the Hilbert sum of El and the F; ( k 1 2). The subspace El is identified with Lg(K, p J ; since + 6; 11b~1I2= vk(B;) I IIVkll, the series 6, + bl; + converges in El, and its sum a, is identified with @*,,where A, is the union of B, and the B; (k 2 2). Clearly a, is a totalizing vector for the restriction of T to El. Thus, starting with the given decomposition (F,) of E, we have constructed a decomposition of E into El and the F;, where for each k 2 2 the measure v; corresponding to F; is a measure with base pl.Repeating the construction, we define for each n a decomposition of E as the Hilbert sum of subspaces El, . . . , En,F!,'J , . . . , F$!,, . . . , all stable under T, such that the restriction of T to each subspace admits a totalizing vector, and such that if the corresponding measure is pk for E, (1 5 k j n) and $1,' for F?jk, then pk+l is a measure with base pk for 1 5 k 5 n - 1, and v!,'jk is a measure with base p, for all k 2 1. To achieve this we have only to apply the previous reasoning to

+

,

10

REPRESENTATIONS OF ALGEBRAS OF CONTINUOUS FUNCTIONS

the Hilbert sum E,, @ FP! @ . . * @ FPJ, @ . that

F, 0 F,

0

s

.

.

. + .

393

It is clear from the construction

0 F, c El 0 E, @ ... @ E n

for each n ; since E is the Hilbert sum of the Fk,it follows that E is also the Hilbert sum of the E, (6.4.2), and the subspaces E, and measures C(k therefore satisfy the required conditions. Put pk = gk p,, and let Sk be the set of points t E K such that gk(t) > 0. Evidently we may assume that the sequence ( s k ) is decreasing and that each s k is universally measurable (13.9.3). Put M, = K - S, and M, = s k - s k + , fGr k 2 2; also put M, = S,. For 1 5 i S k, let Hi, = T(qMk)(Ei). Then

n

kgl

it follows from (15.10.6) that the restrictions of T to the k subspaces Hi, (1 5 i 5 k ) are equivalent, and we put Gk= H,, 0 - . . @ H,, . Clearly Gk is the Hilbert sum of the subspaces H i , (1 2 i 5 k ) . Similarly, put Hi, = T(qM,)(Ei)for each i 2 1. The restrictions of T to all the subspaces Him( i 2 1) are equivalent, and we denote by G , the Hilbert sum of the H i m for all i 2 1. Then it is clear that E is the Hilbert sum of the G, ( k 2 1) and G, . It can be shown (Problem 5) that the measures are determined up to equivalence, and hence that the S, (and consequently the M,) are determined up to a p1-negligible set. The subspaces G, = T(q,,)(E) are uniquely determined by T. The restriction of T to Gk(resp. to G,) is said to have multiplicity k (resp. injinite multiplicity). A representation of multiplicity 1 is therefore topologically cyclic.

(15.10.10) Since the measures pk are determined only up to equivalence (15.10.7), we may suppose that pk = 'psk p 1 . It follows that, up to equivalence, the most general representation u ~ T ( u of ) %,-(K) on a separable Hilbert space E may be described as follows: Consider a positive measure v on K, and a decreasing sequence (Sj)15j). Now, if M is the intersection of Sp(H,,) with the complement 1- co,O [ of R, in R, then the relation M # @ would imply p,,(M) > 0 (15.1.14). Since ]-co:O[ is the union of the intervals I-m, - l / n ] , there would exist m > 0 such that

/

n 1- 00, - ( l / m ) I )= ~1

> 0,

and consequently [qH([) dp,,(() 5 -u/m < 0, contrary to hypothesis. Finally, the relation (15.11.7.2) follows from (1 5.11.7.1) and (1 5.4.14.1), because the spectral radius of H is equal to the larger of linf(Sp(H))(, Isup(Sp(H))I. (15.11.8) (i) For each ,function f E @,-(Sp(N)), the spectrum of f ( N ) is contained inf(Sp(N)) (closure in C), and (1 5.11.8.1 )

400

XV

NORMED ALGEBRAS AND SPECTRAL THEORY

For every eigenvalue a of N, f (a) is an eigenvalue o f f ( N ) , and the eigenspace E ( a ; N ) is contained in E(f(a);f(N)). Iff is continuous, then Sp(f(N)) =f(Sp(N)). (ii) More precisely, with the notation of (15.11.3), i f f E 4Yc(Sp(N,,)), then the spectrum o f f (N,,) consists of the complex numbers fi such that

ess inf 1p -f([)l = 0 c E Sp",) (relative to the measure p,,). (iii) I f f € 4Yc(Sp(N)) and i f g is a continuous mapping off(Sp(N)) into C , then g(f(" = (9 o f ) ( N ) . (iv) If a sequence (fk) offunctions belonging to @,-(Sp(N)) is uniformly bounded and converges simply t o f , then for every x E E the sequence (f k ( N ) * x) converges in E to f ( N ) x.

-

We shall start by proving (ii). We have seen that f(NJ may be identified with multiplication by the class of the function f in L;(Sp(N,), p,,). Hence P # Sp(f(N,,)) if and only if there exists a real number a > 0 such that N2((fi- f ) u ) 2 a N,(u) for all functions u E Y;(Sp(N,,), p,,) (551). If ess inf Ip -f([)I > 0, then we may take a to be equal to this number, by

< E SP(Nn)

virtue of (13.12.2), and therefore

p # Sp(f(N,)).

ess inf

< E Sp(N,)

Ifi - f ( ( ) l

Conversely, if

= 0,

then for each E > O the set M of complex numbers [ E S ~ ( N , , )such that Ifi -f(T)l 5 E is not p,,-negligible, and we have N,((P - f ) q M ) 6 &N,(q,), hence P E SP(f(Nn))* It follows from (ii) that Sp(f(N,,)) cfISp(N,,)). If, moreover, .f is continuous, then f(Sp(N,,)) is compact (3.17.9), and for each fi = f ( a ) , where CL E Sp(N,,), every compact neighborhood of c1 has pn-measure > O (15.11.4), so thatf(Sp(N,,)) = Sp(f(N,,)). To show that Sp(f(N)) cf(Sp(N)) in general, and that Sp(f(N)) =f(Sp(N)) when f is continuous, we have only to use (15.11.5) and the fact that f(Sp(N)) is compact i f f is continuous. Finally, the assertions about eigenvalues are evident, again by reducing to the case of simple operators and using (15.11.6). To prove (iii), note first that g o f is a universally measurable mapping of Sp(N) into C (13.9.6). The relation 9(f"

= (9 o f ) ( N )

cPc4

3

follows from (15.11.1) in the case where g ( [ ) = ( p , q being in gers 2 0 ) . Now, for each E > 0, there exists a polynomial h in [ and 4 such that Ig(0 - &)I 5 E for all C € f ( S p ( W (7.3.2).Since Ig(f(T)>- h(f(l))I S E for

11 THE SPECTRAL THEORY OF HILBERT

401

iE Sp(N), it follows from (15.11.8.1) that IIg(f(N)) - h(f(N))I) 5 E and that 11(g f)(N) - (h f)(N)II 5 E . Since E was arbitrary, this gives the required 0

0

result. Finally, to prove (iv), note that because the sequence of norms 11 fk(N)II is bounded, by virtue of (1 5.11.8.1),it is enough to prove the convergence of the sequence (fk(N) - x ) for x belonging to a total subset of E ((12.15.7.1) and (7.5.5)). With the notation of (15.11.3), we may therefore restrict ourselves to proving the assertion for each N,,. But since fk(N,,) may be identified with multiplication by the class of the restriction of fk to Sp(N,,) in the space Li(Sp(N,,), p,,), the result follows from (13.1 1.4(iii)). (15.11.9) For each universally measurable subset M of Sp(N), let E(M) be the closed subspace of E, stable with respect to N and N*, which is the image of E under the orthogonalprojector PE(M) = qM(N) (15.10.6). Then the spectrum of the restriction of N to E(M) is contained in (closure in C ) .

For each n, let E,,(M) be the image of En under the orthogonal projector cp,(N,,). It is immediately seen that E(M) is the Hilbert sum of the E,,(M), hence (1 5.11.5) it is enough to prove the proposition for each N,, . If CI $ there exists a continuous function g on Sp(N,,) such that g([)(cc - [)(pM([) = qM([)for all E Sp(N,) (4.5.1). It follows that, if Ni is the restriction of N,, to E,,(M), then C L I ~ , (-~ N; ) has an inverse equal to the restriction ofg(N,,) to En(M).

a,

Remark (15.11.lo) Note that the argument which proves (ii) in (15.11.8) shows that, for each positive measure p on C with compact support IC, multiplication by the class of 1, in Li(K, p ) is a normal continuous operator N such that Sp(N) = K (converse of (15.11.3)). 1) Let f be a homeomorphism of a closed subset M of C onto a closed (15.11.I subset N of C containing Sp(N). Then there exists a unique normal continuous operator N' on E whose spectrum is contained in M and which is such that f(N') = N.

If h is the homeomorphism of N onto M which is the inverse off, then by virtue of (15.11.8(iii)) we must have N' = h(f(N')) = h ( N ) ; and since conversely the spectrum of h ( N ) is h ( S p ( N ) ) c M, it follows thatf(h(N)) = N , and the result follows.

402

XV

NORMED ALGEBRAS A N D SPECTRAL THEORY

In particular: (15.11.12) If H is any positive, self-adjoint operator then there exists a unique-positive self-adjoint operator H ’ such that H = H. l2

Apply (15.11.11) with M = N

= R,

andf(()

=

t2.

The unique positive self-adjoint operator H ’ defined in (15.11.12) is denoted by HI/’. Example (1 5.11 .I 3) Let E be the Hilbert space Lg(R; A), where 1is Lebesgue measure. Since the function f ( t ) = e-l‘l is I-integrable, the convolution g H f * g defines, on passing to the equivalence classes, a continuous operator H on E (14.10.6) with norm N , ( f ) = 2. As in (11.6.1), it is immediately seen that H is self-adjoint. It can be shown directly (Problem 5) that the interval [0, 21 in R is equal to Sp(H); this also follows from the general theorems of harmonic analysis (Chapter XXIl). Note that, for each a E R, the function g,(t) = eintis such that the convolution f * ga is defined and equal to 29,/(1 + a2). However, it is not the case that the g, are “eigenfunctions” of H , because they do not belong to Yg(R, A). In Chapter XXIII we shall obtain a generalization and an interpretation of this phenomenon. (15.11.14) The case of normal operators whose spectrum contains no nonisolated point # 0. In this case (which is that of compact normal operators (11.4.1)) let (A,,) be the (finite or infinite) sequence of points of Sp(N), other than 0. These are the eigenvalues of N (15.11.6). The eigenspace E(An;N ) corresponding to A,, is just the space E({A,,})defined in (15.11.9), and these closed subspaces are therefore pairwise orthogonal. Moreover, we have E({O}) = Ker(N), the spectrum of the restriction of N to this subspace being reduced to 0 (15.1 1.9). Finally, E is the Hilbert sum of E( (0)) and the E( {A,,}). For it is enough to apply (15.11.8(iv)) to the sequence of functions (f,), where f,(() = 1 for ( = 1, with k 5 n , f,(() = 0 for ( = lk and k > n, and f,(O) = 1 ; this shows in particular that every x E E is the limit of the sum of its projections on E({O}) and the E({A,}) with k 5 n. Hence the result (6.4). In particular, if E isfinite-dimensional, a normal operator on E may be defined to be an operator whose matrix is diagonal, with respect to a suitably chosen orthonormal basis of E.

11 THE SPECTRAL THEORY OF HILBERT

403

PROBLEMS

1.

Show that a continuous operator N on a Hilbert space E is normal if and only if /IN. X I / = /IN* . xi1 for all x t: E.

2. Let N be a continuous normal operator on a Hilbert space E. (a) For each x E E, consider the open sets W c C such that there exists a continuous

mapping f w of W into E satisfying the relation

(A 1 - N) fw(h) =x for all h E W. Show that there exists a largest open set n ( x ) with this property, that all the functions f w are restrictions of a unique mapping f of n ( x ) into E, and that f i s analytic in n ( x ) . (Use (15.5.6) to reduce to the case where the normal operator N is simple. In this case, E being identified with L&(Sp(U),p) and x with the class of a function g, the set n(x) is the interior of the set of all h E C such that the function - C)F1g([) belongs to -%SP(N), p).) (b) Put @(x) = C - n ( x ) . Show that, for each closed subset M of Sp(U), the space E(M) = vM(U)(E) is the set of all x E E such that @(x) c M. (Again reduce to the case of a simple normal operator.) (c) Show that every continuous operator V EL(E) which commutes with N also commutes with all the operators g(N), where g E *&p(N)), and in particular commutes with N* (Fuglede's theorem). (First show that Q ( V .x) 3 Q(x) for all x E E, by considering the function f defined in (a) and the function h H V . f ( h ) . Using (b), deduce that, for each closed subset M of E, the operator Vcommutes with = &N), and conclude that V commutes with g(N) for all conthe projector PE(M) tinuous functions g on Sp(N).) (d) Deduce from (c) that, if NI and N2 are commuting normal operators, then N,Nz is normal. 3. Let p be normalized Haar measure on the group U : IzI = 1 (so that d p ( @ = (27r)-l do) and let E = L&(p). Then the operator M,,(lc) = U is a unitary operator on E. Every closed subspace of E which is stable under N is either of the form vM(N)(E),where M c U is of measure < 1 ; or else is of the form @ .H2(p), where 141 = 1 (Section 15.3, Problem 15) (Beurling's theorem). Deduce that there exist closed subspaces F of E which are stable under U but not under U* = U-',and are such that the orthogonal projector PF does not commute with N (compare with (15.15.3)). 4.

Without using Riesz theory (11.4), prove that the spectrum of a compact normal operator has only isolated points, except for the point 0 (reduce to the case of a simple normal operator).

5.

Show that the spectrum of the self-adjoint operator H considered in (15.11.13) is the interval [0,2] in R. (Observe that llHI1 5 2; to prove that H 2 0, that is to say

( H . u I u) 2 0 for all u E E, consider first the functions

r+

u(t) =

elrXdx

and their

linear combinations. To show that every number of the form 2/(1 a') belongs to the spectrum of H , approximate g. by functions u,g,, where u. E Y 2is 2 0 and the sequence (u,) is increasing and tends t o 1.)

404 6.

XV

NORMED ALGEBRAS AND SPECTRAL THEORY

Let E be a separable Hilbert space and T a continuous operator on E. Let R and L be the positive hermitian operators which are the square roots (1511.12) of T*T and TT*,respectively. We write R = abs(T), and call it the "absolute value" of T. Then L = abs(T*). (a) . . Show that Ker(T) = Ker(R) and that L(E) = T ? ) . There exists a unique isometry Vof R(E) onto T(E) such that T = YR. If we extend V by continuity to R(E),and then to an operator U E P ( E ) by putting U(x) = 0 on the orthogonal supplement of R(E), then we have also T = UR (polar decomposition of T ) . Show that R = U*T= U*UR = RU'U,

L

=

URU*,

T = LU*.

(b) For T to be invertible it is necessary and sufficient that R = abs(T) and L = abs(T*) are invertible. (To prove necessity, consider the spectra of R and L. To prove sufficiency, use the closed graph theorem.) (c) Nis normal if and only if abs(N) = abs(N*), and if this condition is satisfied there exists a unitary operator Wsuch that N = W . abs(N).

7. A compact operator Ton a separable Hilbert space E is said to be nuclear if, denoting by (A,) the full sequence of eigenvalues of abs(T) (Section 11.5, Problem 8), we have ( a j Use polar decomposition (Problem 6) to show that the product SISz of two Hilbert-Schmidt operators is nuclear. Conversely, if T is nuclear, then abs(T)"* is a self-adjoint Hilbert-Schmidt operator, and T is the product of two HilbertSchmidt operators. Consequently T* is also nuclear. If A is any continuous operator on E, then A T and TA are nuclear. (b) If A and B are two Hilbert-Schmidt operators and if (em) is a Hilbert basis of E, then the seriesx (AB . en I en) a n d C (BA . en I en) are absolutely convergent, and

"

"

their sums are equal. (Write B . en =C ( B . en I e,)e, .) Consequently, for every unitary m

operator U and every nuclear operator T, we have

Deduce that, for a nuclear operator T, the sum C ( T . enI en) is independent of the n

Hilbert basis (en) chosen. This sum is called the trace of T and is denoted by Tr(T). If A , B are two Hilbert-Schmidt operators, then Tr(AB) = Tr(BA) = (A I B*). (c) If T is nuclear, show that

where the supremum is taken over all pairs of Hilbert bases (a"), (b.) of E (use the polar decomposition of E). If we put IITlll = Tr(abs(T)), then the set 6L01(E) of nucIear operators on E is a vector space on which ljTlll is a norm, such that

ll~ll5 z IITllx.

(d) If ( T J is a sequence of nuclear (resp. Hilbert-Schmidt) operators on E which converges weakly (Section 12.15, Problem 9) to an operator T, and which is such that the sequence of norms (llTvlll)(resp. (liTvilz))is bounded, then T is a nuclear (resp. Hilbert-Schmidt) operator. (e) Show that 6L01(E) is a Banach space with respect to the norm llTlji.

11 THE SPECTRAL THEORY OF HILBERT

405

(f) Let (h.) be the sequence of eigenvalues of a nuclear operator T, each counted IA,l 2 IITlll. (For each according to its algebraic multiplicity (11.4.1). Show that integer p , consider the sum V of the p spaces N ( p X ;T ) (1 5 k < p ) , where p,, . . . , pr are the first p distinct eigenvalues in the sequence (A,,). Take a Hilbert basis of V with respect to which the matrix of TI V is triangular, and use (c).) Deduce that if T is a Hilbert-Schmidt operator and if (A,) is the sequence of its eigenvalues, each counted with its algebraic multiplicity, t h e n x lh,I2 5 11 Tll:. n

(9) If a continuous operator T E Y ( E ) is such that, for each pair of Hilbert bases (a"), (b.) of E, the series ( T . a, I b,) is convergent, then T is nuclear. (Write

"

T = LU* (Problem 6(a)), and by choosing (an)and (b,) suitably show that L' / z is a

Hilbert-Schmidt operator.) (h) For a continuous operator T E Y(E) to be nuclear, it is necessary and sufficient IIT. enll should converge. that, for a t least one Hilbert basis (en) of E, the series (Write T = UR (Problem 6(a)) and note that (R . e, I en)2 11 T . enll. This proves that the condition is sufficient. Conversely, take for (en) a basis consisting of eigenvectors of R.) (i) In the space E = /&, let (en) be the canonical Hilbert basis, and let a (l/n)e.. If F is the subspace C . a of dimension 1, show that the projector PF

=c n

is nuclear, but that the seriesx lipF.enll does not converge. 8.

Show that, for every normal operator N on a separable Hilbert space E, there exists = N.Give examples where there exist infinitely many such operators.

a normal operator N' such that N''

9. Let T be a continuous operator on a separable Hilbert space E. (a) For T to be a topological left zero-divisor (Section 15.2, Problem 3) in the algebra Y ( E ) , it is necessary and sufficient that there should exist a sequence (xn)of vectors

in E such that 1 ~ ~ = ~ 1 1 1for all n and such that ( T .x,) tends to 0 . The complex numbers 5 such that T - 5 . 1 is a topological left zero-divisor in 9 ( E ) form what is called the approximative point-spectrum Spa(T). Thus 5 $ Spa(T) means that T - 5 . 1 is injective and a homeomorphism of E onto a closed subspace of E. Show that Spa(?") is closed in C,and contains the frontier of Sp(T). If P is any polynomial, show that Spa(P(T))

= P(Spa(T)).

1) (b) Let T E Y(E). For each E C , let m(T, A) denote the dimension of Ker(T* (so that m(T, A) is either a nonnegative integer or w),equalalso to thecodimension of ( T - A . 1)(E). Let To E Y(E) and let A, be a complex number not belonging to Spa(To). Show that there exists a number E > 0 such that m(T, A) = m(To,A,) whenever I1T- Toll 5 E and Ih - hol 5 E . (c) Deduce from (b) that if K is a compact subset of C which does not intersect Spa(To), then there exists E > 0 such that m(T, h) = rn(To,h) for all E K and all T E Y(E) such that IlT- Toll 5.s. (d) Let T E Y(E) and let K be a compact subset of Sp(T) - Spa(T) with the following properties: (1) 0 $ K ; (2) the inverse image K' of K under the mapping 5 ~ 5 is convex; (3) m(T, h) = 1 for all A E K. Then there exists no operator T'E Y(E) such that T'' = T . (Supposing the contrary, let L = K n Sp(T'); then we have L C Sp(T') - Spa(T'), and L u (-L) = K'. Show that h E L implies that -A $ L

+

'

406

XV

NORMED ALGEBRAS AND SPECTRAL THEORY

xz

a,

so that L n (-L) = which is a contradiction. Observe that if h E L, then is an eigenvalue of T* and m(T, A') = 1.) Show also that, if T is invertible, there exists E > 0 such that every operator T I E P ( E ) satisfying liTl - TI1 5 E is invertible and such that there exists no operator T'E Y(E) for which T" = TI (use (c).) 10. Let R be a bounded open subset of C, and let H be the Hilbert space of analytic

functions f o n R such that

(cf. Problem in Section 9.13). Let T be the operator which maps each f~ H to the function [++[f([). Show that, for each h E R and each function g E H such that g(h) = 0, there exists a unique f~ H such that ( T - h1) .f= 9. Also, if the disk I[ - hI < 6 is contained in R, then /1g1/2 2 & 8 z ~ ~ Deduce f ~ ~ z that . Sp(T) is the closure of R in C,and that i2 is contained in Sp(T) - Spa(T). Deduce from these results that if R is taken to be the open annulus rl < IzI < r2 (where r l > 0), then the operator T is invertible and has no square root in Y(H) (cf. Problem 9). 11. Let E be a separable Hilbert space, K a compact subset of C, and (x,y)~m,,,

a continuous sesquilinear mapping of E x E into the space M,(K) of complex measures on K. Suppose that

and that the measure mx, is positiue for all x on E such that

E

E. Let T be the continuous operator

for all x, y in E. For each functionfE 9,(K), letf(T) denote the operator defined by

(f(T). x I Y ) = If([)dm,,A) for all x , y in E (Section 15.10, Problem 1). The mapping T + f ( T ) of %,-(K) into Y(E) is linear and such that T*= c(T), where c([) = but this mapping is not in general an algebra homomorphism. Prove that there exists a separable Hilbert space H, the Hilbert sum of E and another Hilbert space F, and a representation f++ V(f) of Q,(K) on P ( H ) such that, if P is the orthogonal projection of H on E, we have f(T)= P V(f)I E. (Apply Problem 6 of Section 15.9, by taking r to be the set of all finite products of characteristic functions va,(n E N) of universally measurable sets in K, chosen in such a way that these functions form a total set in each of the spaces Yc(K, mx,.,,), where (x,) is a dense sequence in E.) ("Neumark's theorem.")

I,

12.

Let E be a separable Hilbert space. Let H be a self-adjoint operator on E such that 0 5 H 5 l E. Show that there exists a separable Hilbert space G which is the Hilbert sum of E and a Hilbert space F, and an orthogonal projector Q on G such that H = PQ I E, where P is the orthogonal projection of G on E. (For x , y in E, define mx, t o be the measure carried by the set of two points (0, 1) such that mX.,({O)) = ((IE - H ) . x I y ) and mx,,({l)) = (H* x I y), and apply Problem 11.)

11 THE SPECTRAL THEORY OF HILBERT

407

13. Let E be a separable Hilbert space, and (H.) a sequence of self-adjoint operators

with the following property: there exists an interval [-M, MI in R such that, if P(X) = a. a l X . . . a.X" is any polynomial with real coefficients and P ( 0 2 0 for - M 5 6 5 M, then also aoI a l H l . . . a. H. 2 0 (which implies inter alia that -M . I =< H, =< M . I for all n). Show that there exists a separable Hilbert space G which is the Hilbert sum of E and a Hilbert space F, and a self-adjoint operator H on G such that H. = PH" I E for all n 2 1, where P is the orthogonal projection of G on E. (Using Problem 5 of Section 13.20, prove that for each pair (x, y ) of elements of E there exists a real measure inx, on [-M, MI such that

+

+ +

+

(H. . x I Y ) =

for all n

+ +

j5"dm,, y(t),

( x I Y ) = / d m x , y(6)

2 1.) Deduce that Ht

5 Hz,

H;n+i 5 IIH2nIIHzn+z.

14. Let E be a separable Hilbert space and T a continuous operator on E such that /)TI15 1. Then there exists a separable Hilbert space H , the Hilbert sum of E and a

Hilbert space F, and a unitary operator U on H such that, if P is the orthogonal projection of H on E, then T" = PU" 1 E for all n 2 1. (Apply Problem 6 of Section 15.9, taking r = Z, the involution on I' being n w -n, and the representation of r in H such that U(n)= T" for all n 2 0. Observe that, for all y E E and all 5 E C such that 151 < 1, we have

W(1

(*)

+

. Y I (1 - @) . Y ) 2 0,

and note that every x e E can be written in the form (I- [ T ) . y for some ~ E E . +m

In particular, this is so for every linear combination x =

x,e-"''+', where the x. E E

-m

are zero except for finitely many indices n, and where express C (rI"-'"'U(n- m) x. 1 x,)

5 = re'@ with

r > 0. Then

m. n

in terms of the left-hand side of (*), and let r tend to 1.) Deduce that, if

m

Iz( = 1, and if u(T)

cn z" is a power series which converges absolutely on the circle

=cc,T", then the relation

"=O

m

lu(z)l

51

(resp. Wu(z) LO) for

Irl 5 1 implies that ilu(T)II 2 1 (resp. u(T) + u(T*) 2 0).(Note that u(T) = Pu(U) I E.) "=O

15. If N is a normal continuous operator on a separable Hilbert space E, and K a compact

subset of C containing Sp(N), then we have a (nonfaithful) representation f ~ + f ( N ) of V,(K) in E; the measures m x , ycan be considered as measures on K satisfying (15.11.2) and (15.11.2.1). In particular, if U is a unitary operator on E, then f.-.f(U) is a representation of U,(U) in E. Let (en) be a Hilbert basis of E indexed by Z, and let U be the unitary operator on E such that U .en= en+ Show that the representationf++f(U) of U,-(U) on E is topologically cyclic and that eo is a totalizing vector; and that p, = me,,, eo is the normalized Haar measure on U (cf. (7.4.2)). Deduce that Sp(U) is the whole of the circle U, and also give a direct proof of this fact. Give examples of closed vector subspaces of E which are stable under U but not under U*= U-'.

408

XV

NORMED ALGEBRAS AND SPECTRAL THEORY

16. Let X be a locally compact space, p a bounded positive measure on X with total mass 1, and u a p-measurable mapping of X into X such that p is inuariunt with respect to u (Section 13.9, Problem 24). Let U be the unitary operator on L;(X, p) such that U . f = (fou)" (Section 13.11, Problem 10). (a) The mapping u is said to be mixing (resp. weakly mixing) with respect to if,

for each pair of p-measurable subsets A, B of X we have

lim (cL(u-"(A)n B)) = p(A)p(B)

n-. m

(resp. 1 "-1

Every mixing mapping is weakly mixing. Every weakly mixing mapping is ergodic (Section 13.9, Problem 13(d)). (b) Show that u is mixing (resp. weakly mixing) if and only if, for each pair of functions f,g in Y&(X, p), we have I im( u n . f lO) = ( f l

I-. m

i)(ip)

(resp. lim n-m

n

5' ~(un.fI8) - (fI i)(i1 8) 1 k=O

= 0).

An equivalent condition is that, for each f E U;(X, p) such that ( f l y ) (i.e.,/fdp

= 0),

=0

we have Iim (un

n+m

=0

(resp.

(Replace f by f+ g, and remark that if a sequence (an) satisfies the condition -1 n - 1 1 n-1 lim lak[* = 0,then also Iim lakl = 0, by the Cauchy-Schwarz inequality.)

n-.m Tl k = o

m+m

k=O

(c) For u to be ergodic with respect to p, it is necessary and sufficient that 1 should be an eigenvalue of U with multiplicity 1. If this is so, then all the eigenvalues of U have multiplicity 1 and form a subgroup of the group U of complex numbers of absolute value 1. For each eigenvector / E L&(X,p) of U,the function If\ is constant almost everywhere. (Remark that if U ./= hf and U .8 = hfi, then u .(Slf)" = (Sf)".) (d) Show that the following properties are equivalent: (a) u is weakly mixing with respect to f ~ . (B) u x u is an ergodic mapping of X x X into X x X with respect to the measure p, @ p. ( y ) The only eigenvalue of U is I. (To show that (a) implies (p), consider subsets M x N of X x X, where M and N are p-measurable subsets of X. To prove that (p) implies (y), observe that i f f i s an

11 THE SPECTRAL THEORY OF HILBERT

409

eigenvector of U,then (f@f)- is an eigenvector of the unitary operator corresponding to u x u, for the eigenvalue 1. To prove that ( y ) implies (a), use the last criterion of (b); if (f If) = 0, introduce the measure v = m7.7 (15.11.1), and observe that this measure on U is diffuse (15.11.6). Then we are reduced to proving that lim n-rm

1"-1

-

C

nk=o

1."

tkdv([)

1

= 0.

Write this relation in the form

and remark that the diagonal of U x U is (v @ +negligible.) (e) With the notation of Section 13.9, Problem 13(c), show that if 0 is irrational, the mapping z w e Z n f Sof z U onto U is not weakly mixing. (Calculate the spectrum of the corresponding unitary operator U.) (f) Suppose that the space Li(X, p) is the Hilbert sum of the subspace C . 1 (the classes of the constant functions) and an at most denumerable family (HI),., of Hilbert spaces, where each H, has a Hilbert basis (en,)nEZ such that U .en, = en+ for all n E Z (cf. Problem 15). Then the mapping u is mixing. In particular, if. (X, p, p) is the Bernoulli scheme B(), 4) (Section 13.21, Problem 18) then u is mixing. (If, for each n E Z, f. is the function on Iz such that h(x) = -1 if pr.x = 0, and f.(x) = 1 if pr.x = 1, then the classes of the finite productsf.,f,, . . .Lk,in which all the indices are distinct, form together with the class of 1 an orthonormal basis of L&(X,p).) Likewise show that, if X is the torus T2,7r : R + T the canonical homomorphism and p the normalized Haar measure on X, then u defined by

444, T(YN = ( 4 x + Y ) , d x + 2Y)) is mixing. (9) Suppose that u is ergodic with respect to p, so that (by virtue of (c)) the eigenvalues of U form an at most denumerable subgroup G of U,the eigenspace corresponding to an eigenvalue a E G being a line D(a) in L:(X, p). Show that there exists a family of eigenvectorsf, E D(a) of U such that . f I I = 1 almost everywhere in X and such thatf., = f a f a almost everywhere,for all pairs (a, @)ofpoints of G. (Let ha E D(a) be such that (h.1 = 1 almost everywhere, for all a E G ; for each pair of points a, /3 E G we may write h., = r(a, @)h.h, almost everywhere, where r(a, p) E U is a constant. Denote by A the subgroup of Ux generated by the ha for a E G and the group of constant functions from X to U (which may be identified with U). Show that there exists a homomorphism 0 : A -+ U such that e([) = 6 for all 5 E U. For this purpose, arrange the ha in a sequence (h.) and proceed by induction: if 0 is already defined on the subgroup A. generated by U and the h, such that j < n, distinguish two cases according as h; is not constant almost everywhere for any nonzero k E Z, or on the contrary that there exists a smallest positive integer k such that hi is constant almost everywhere; use the fact that for all [ E U and all integers k > 0, there exists 7) E U such that 7' = 5. Then take f . = 8(h.)ha.) (h) Let v : X + X be another mapping which is ergodic with respect to p, and let V be the corresponding unitary operator. Show that if u satisfies the hypotheses of (g) and if U, V have the same eigenvalues, then u and v are conjugate (Section 13.12, Problem 11).

410

XV

NORMED ALGEBRAS AND SPECTRAL THEORY

17. Let E be an infinite-dimensional separable Hilbert space and N a normal continuous

operator on E. (a) Show that E is the Hilbert sum El E2 of two infinite-dimensional subspaces, each of which is stable under N and N*. (Reduce to the case of a simple normal operator M,,(lc); with the notation of (15.11.9), observe that if M is a closed subset of Sp(N), then E(M) can be finite-dimensional only if M is a finite set for which each point has p-measure f-0. Then distinguish two cases, according as there exist infinitely many points of measure f O or not; in the second case, use Problem 3(b) of Section

+

13.18.)

(b) Deduce from (a) that there exists a decomposition of E as the Hilbert sum of a n infinite sequence (E,) of infinite-dimensional subspaces, each of which is stable under N a n d N*. Let E be a separable Hilbert space which is the Hilbert sum of an infinite family of infinite-dimensional subspaces. There exists a unitary operator S on E such that S(E,) = E n + ,for all n E Z. Show that if P (resp. Q ) is the operator which is equal to S1-2"(resp. S-2") on Enfor each n, then P 2 = Q 2 = I E , and S = P Q . (b) Deduce from (a) and from Problem 17 that every unitary operator on an infinite-dimensional separable Hilbert space is the product of four involutoryunitary operators. (c) Jxt w be a complex cube root of unity and let U be the homothety with ratio w on E, which is a unitary operator. Show that U is not the product of three involutory unitary operators. (In general, in a group G, if t is in the center of G , andifthereexist x , y , z in G such that t = x y z and x2 = y 2 = z2 = 1, then also t = y z x , t = xyxy and t 3 = xzy = rl.)

18. (a)

Let E be a n infinite-dimensional separable Hilbert space and let (e,),*o be a Hilbert basis of E. The continuous operator V such that V . en= enfl for all n 2 0 is called the one-sided shqt operator; it is an isometry of E onto the hyperplane orthogonal to eo; its spectrum is the disk 151 6 1 and contains no eigenvalue of V (Section 11.1, Problem 4); its approximative point-spectrum (Problem 9) is the circle U : 161 = 1. The spectrum of the adjoint operator V* is also the disk151 5 1, and every 1such that 151 < 1 is an eigenvalue of V*. (b) Let T be a continuous operator on E which is an isometry of E onto a (necessarily closed) subspace T(E). Show that there exists a decomposition of E as the Hilbert sum of subspace L and an at most denumerable family (F,), I of subspaces, where L and each F, is stable under Tand are such that (1) TI L is unitary and (2) each F, is infinitedimensional and TI FLis the one-sided shift operator, for a suitably chosen orthonormal basis. (Consider the orthogonal supplement N of T(E), and show that E is the Hilbert sum of the T"(N) (n >= 0) and L = T"(E).)

19. (a)

n

(c) Deduce from (a) and (b) that if T is any nonunitary isometry of E onto a subspace of E, then Sp(T) is the unit disk 151 5 1, and that IIT- UII = 2 for all unitary operators U.(Observe that IlT- Uil = / / U * T - lE/I and that U*T is not unitary, so that the point 5 = - 1 belongs to its spectrum.) 20.

(a) Let E be a separable Hilbert space, T a continuous operator, and C a compact operator on E. Show that the points of Sp(T+ C ) which do not belong to Sp(T) are eigenvalues of T C. (Reduce to the case where 5 = 0 is such a point and observe that, if T i s bijective, we may write T + C = T(IE T - l C ) , where T-lCis compact; if - 1 E Sp(T-'C), it follows that - 1 is an eigenvalue of T-IC.)

+

+

11 T H E SPECTRAL T H E O R Y OF HILBERT

411

(b) With the notation of Problem 15, let C be the operator of rank 1 defined by C . x = - ( x 1 e-&o. Show that Sp(U+ C ) is the disk 151 5 1 while Sp(U) is the circle 151 = 1. (Consider separately the restrictions of U C to the subspace generated by thee. with n 2 0 and to its orthogonal supplement.) (c) Let N be a normal operator on E and C a compact operator on E. If Sp(N) is nondenumerable, show that the same is trueof Sp((N C)*(N C))(use(a)aboveand (15.11.8(i)). Deduce that the one-sided shift operator V (Problem 19(a)) cannot be of the form N C (observe that V* V = lE).

+

+

+

+

21.

Let E be an infinite-dimensionalseparable Hilbert space. (a) Show that every nonzero two-sided ideal 3 of the ring P ( E ) contains the ideal 6 of operators of finite rank. (If T # 0 belongs to 3, show that every operator of rank 1 can be written in the form BTCfor suitably chosen operators Band C.) (b) Show that the only closed two-sided ideal of the Banach algebra P(E), other than P ( E ) and {O},is the ideal B of compact operators. (First observe that Q is the closure of 6,and then that if a two-sided ideal contains a noncompact operator, then it also contains a noncompact positive hermitian operator H (Problem 6). For such an operator, show that there exists an interval M = [a, 03 [ with a > 0 such that thespace E(M) (in the notation of (15.11.9)) is infinite-dimensional. If Vis an isometry of E onto E(M), show that V*HV is invertible.)

+

22.

Let E be an infinite-dimensional separable Hilbert space. An operator with index on E is a continuous operator T such that (1) T(E) is closed and of finite codimension; (2) T-I(O) is finite-dimensional. (a) If T is an index operator, show that there exists a continuous operator A such that l E- ATand l a - TA are of finite rank. (Show that Tis a homeomorphism of the orthogonal supplement F of T-'(O) onto T(E), and take A to be the inverse homeomorphism on F and zero on the orthogonal supplement of F.) (b) Conversely, suppose that Tis a continuous operator on E, for which there exists a continuous operator A such that l E- AT and l E - TA are compact. Show that T is an index operator. (Using (11.3.2), show first that the kernels of Tand T* are finitedimensional, and hence that T(E) has finite codimension. Then use the fact that the restriction of AT to the orthogonal supplement F of the kernel of AT is a homeomorphism onto its image, and finally use (12.13.2(iii)).)

23. Let E be an infinite-dimensional separable Hilbert space. (a) Let T be a continuous operator on E such that T-'(O)is of infinite dimension. Then E is the Hilbert sum of an infinite sequence (En)"%,of infinite-dimensional subspaces, such that the En with n 2 1 are contained in T-'(O). For each n 2 1, let S. denote an isometry of E, onto En. Let A be the continuous operator which on Eo is equal to S1, and on En is equal to S. + S;', for all n 2 1. Also let V be the operator ; I on En for all which is zero on Eo , is equal to Si ' on El,and is equal to Sn- S n 2 2. Let Todenote the restriction of P,,T to Eo , and let B be the continuous operator which is equal to VT on Eo , to -To 5'1' on E l , and to -Sn- ,TOX'on E. for all n 2 2. Prove that T = AB - BA. (b) Deduce from (a) that for any continuous operator T on E there exist four operators A, B, C, B such that

T = (AB - BA)

+ (CD -DC).

(Write T as the sum of two continuous operators, each of which has an infinitedimensional kernel.)

412

NORMED ALGEBRAS AND SPECTRAL THEORY

XV

24.

Let E be a separable Hilbert space and H a positive self-adjoint operator; then Z+ AH is invertible for all > 0. For x E E and > 0 let F,(x) = (h(1-t A H ) - ’ . x I x). Show that FA(x)is increasing as a function of A, and that it is bounded if and only if x E H”’(E). (Reduce to the case where H i s a simple operator (15.11.3))

25.

Let Eo be a real Hilbert space and E the Hilbert space obtained by extending the field of scalars of Eo to C, so that every element of E is uniquely of the form x iy with x, y E Eo, and

+

(x‘ -tiy‘ I x”

+ i f ’ ) = (x’ I x ” ) + (y’ I y ” ) + i(y‘I x”) - i(x’ I y”).

Show that every self-adjoint operator HO on Eo extends uniquely to a self-adjoint operator H on E, having the same spectrum. 26. With the notation of Problem 2 of Section 13.13, suppose that for each compact subset K of X there exists a constant bL2 0 such that

for all u E 2.This condition implies condition (A) of Problem 2, Section 13.13, but is not equivalent to (A). (a) In Problem 5(b) of Section 6.6, suppose that X is compact and that the functions f. are real-valued, bounded and measurable with respect to a positive measure p on X and that they satisfy the condition

where

lI.f1

=

sup lfn(x)l. Show that the space .%? of functions which are p-equivalent X E X

to the functions belonging to the space denoted by E in the problem referred to, is such that H = .%?/N is a Hilbert space isomorphic to E, and satisfying condition (B) above. (b) For every functionfe 9 ; ( X , p) with compact support K, there exists a function

U’

E

.@ such that

s

(U’ I u) = uf dp for all u E

.@. The class of

U’ is uniquely deter-

mined by the class off, and we have lUfl I b:l’N2(f). Then the set 9 defined in Section 13.13, Problem 2(b) is also the closure of .%? in the set of the Uf for which f is 20, compactly supported and belongs to 9:. Generalize the result of part (e) of this problem to the case where f E 9: is compactly supported and 2 0 almost everywhere. Likewise, generalize part (f) of the same problem. (c) Suppose that X is compact. Then Uf is defined for all functions f E -Ep:(X, p), and we have NZ(U’) b , N , ( f ) . If G . f i s the class of U’, then G is a continuous positive selfadjoint operator on L:(X, p). If F is the closure in L i of G’”(L:) (which is the orthogonal supplement of Ker(G’”) = Ker(G)), then the restriction of G’/’ to F is an isometry of the subspace F of L : onto the Hilbert space H (equipped with the norm lfil). Hence H = G’’*(Li). (d) Suppose that X is compact and that the “domination principle” is satisfied, in the form of (b): that is to say, iff f 9: is 2 0 almost everywhere, and if u E 9’ is such that Uf(x) 5 u(x) almost everywhere in the set of points x where f ( x ) > 0, then Uf(x) 5 u(x) almost everywhere in X. For each A > 0, put R1= G(I hG)-’. I f f € 9: is 20 almost everywhere, and if g is a function whose class is equal to

+

12 UNBOUNDED NORMAL OPERATORS

413

RA.f : then g(x) >= 0 almost everywhere. (Observe that

AUP+(X)2 U’(X) + hus-(x)

almost everywhere in the set of x such that g + ( x ) > 0.) Deduce that, for all U E H , we have lul E H and I(lul)”l 5 I&/. (If F,(V^) = (h(1 -t hG)-’ . B 16) for tr E 9’:, show that FA(lul-) 5 I i i I *, and use (c) above and Problem 24.) (e) Generalize the results of (d) to the case where X is locally compact. (Let (K.) be a sequence of compact subsets which cover X, and such that each is contained in the interior of the next. For each n, consider the space X nof restrictions to K, of functions of the form U’, where f E Y : ( X , p) and Supp(f) C K.; apply (d) to each of these spaces.)

12. U N B O U N D E D N O R M A L OPERATORS

(15.12.1) Let E be a separable Hilbert space, and let Z denote the identity mapping of E. A (not necessarily continuous) linear mapping T of a subspace dom(T) of E (the “domain” of T, which need not be closed) into E will be called, by abuse of language, a not necessarily bounded operator on E, or simply an unbounded operator on E. The graph I-( T ) (1.4) is a vector subspace of E x E, and T is said to be a closed operator if I-( T ) is closed in the product space E x E. The kernel Ker( T ) of a closed operator is closed in E, because it may be identified with the intersection of T(T) and E x (0) in E x E. Throughout, it is to be understood that an equality T I = T , between two unbounded operators on E implies the equality

dom(T,) = dom(T,). (15.12.2) Let T be an unbounded operator on a Hilbert space E. Then, of the

three properties ; (i) dom(T) is closed in E; (ii) T is closed; (iii) T is continuous; any two imply the third.

If T is continuous, then I-(T)is closed in dom(T) x E and therefore also in E x E if dom(T) is closed in E. Next, if T is continuous on dom(T), then it extends by continuity to a linear operator T‘ which is continuous on dom( T’) = dom( T ) (5.5.4), and I-( T‘)is the closure of I-( T ) in E x E. Hence if Tis closed we have I-(T’) = I-(T),so that dom(T) is closed in E. Finally, if dom( T ) is closed in E and if I-( T )is closed in E x E, then T is continuous by the closed graph theorem (12.16.11). (15.12.3) In what follows we shall be concerned with unbounded operators T such that dom( T ) is dense in E. Let F be the set of all y E E such that the

414

XV

NORMED ALGEBRAS AND SPECTRAL THEORY

linear form X H (T * x I y ) is continuous on dom( T ) ;when this is so, this linear form extends by continuity t o the whole of E (5.5.4) and therefore can be written as xt-i(xI T * ' y ) for a uniquely determined vector T* y , because dom( T ) is dense in E (6.3.2). This uniqueness shows that T* is a linear mapping of F into E, and therefore an unbounded operator, called the adjoint of T . (When dom(T) = E and T is continuous, this definition clearly agrees with that of (11.5).) Hence we have

-

(1 5.1 2.3.1)

(T-xly) = ( X I

T* ' y )

for all x E dom( T ) and y E dom(T*). If T, is an unbounded operator such that dom(Tl) 3 dom(T), then dom( TT) c dom( T*). In what follows we shall endow E x E with the structure of a Hilbert space such that

so that E x E is the Hilbert sum of its two subspaces E x (0) and (0) x E, each of which is isomorphic to E. Also we denote by J the continuous operator ( x , y ) ~ ( y- x, ) ; clearly J is a unitary operator on E x E, and 5' = - I . (1 5.12.4) Let T be an unbounded operator on E, such that dom( T ) is dense in E. (i) The adjoint -operator T* is closed, and its graph I-( T*) is the orthogonal supplement of J ( r ( T ) )in the Hilbert space E x E. (ii) The following properties are equivalent: (a) T can be extended to a closed operator. (b) dom(T*) is dense in E. If these conditions are satisfied, the graph of any closed operator extending T contains the graph of T**, and r(T**) = r(T). (Thus T** is the smallest closed operator which extends T, and in particular, T** = T if T is closed.) Moreover, (T**)*= T*.

(i) If a sequence (y,) of points in dom(T*) converges to y E E and is such that the sequence (T* y,) converges to z E E, then the sequence of continuous linear forms xt-i(xI T* * y,) converges for all x E E to the continuous linear form X H ( X ~ Z ) .But if x ~ d o m ( T ) , we have ( x I z ) = l i m ( T . x ( y , ) = ( T . x ( y ) ,hence yEdom(T*) and z = T * - y by n-, m

definition, which shows that T* is closed. On the other hand, to say that ( y , z ) E E x E is orthogonal to all the vectors (T * x, -x) with x E dom( T ) signifies that ( T * x I y ) = (x I z), i.e., that x I+ (T x I y ) is continuous, hence y E dom( T*)and z = T* * y .

12 UNBOUNDED NORMAL OPERATORS

415

(ii) A closed vector subspace G of E x E is not the graph of a closed operator if and only if, for some x E prl(G), there exist at least two distinct points ( x , yl) and (x, y 2 ) belonging to G, or equivalently (since G is a vector subspace) that (0,y1 - y 2 ) E G. But to say that dom(T*) is not dense in E means that there exists z # 0 in E orthogonal to dom(T*) (6.3.1), or equivalently - that (z, 0) is orthogonal to r(T*),or again that (0, - z ) belongs to T(T). It follows that (0, - z ) cannot be contained in the graph of any closed operator which extends T. Conversely, if dom(T*) is dense in E, then T** is defined, and r(T**)is the orthogonal supplement of J(T(T*));but this orthogonal supplement is also equal to

(6.3.1).

We say then that T** is the closure of T.

(15.12.5) Given two not necessarily bounded operators U, V on E, the vector U * x V * x is defined for all x E dom( U ) n dom( V ) , and we denote by U V the linear mapping X H U * x V x of dom(U) n dom( V ) into E. In particular, if V is everywhere defined, then dom(U V ) = dom(U), and the graph T(U + V ) is the image of r(U ) under the linear mapping ( x , y ) H(x, y + V x ) of E x E into E x E. If V is continuous and U is closed, it follows therefore that U + V is closed, since the mapping

+

+

+ -

+

-

(X,Y)H(X,Y

+ Y.x)

and its inverse ( x , y ) H ( x , y - V * x) are continuous. Again, the vector U ( V x ) is defined for the set of all x E E such that x ~ d o m ( V )and V e x ~ d o m ( U ) This . set is a vector subspace which we denote by dom(UV), and UV denotes the linear mapping XH U ( V . x ) of dom(UV) into E. If T is a not necessarily bounded operator which is an injective mapping of dom( T ) into E, we denote by T-' the inverse mapping of T(dom(T)) = dom(T-') into E. The graph T ( T - ' ) is the image of T ( T ) under the mapping (x, y ) ~ ( yx)., Hence T-' is closed if T is closed (and injective).

-

(15.12.6) (von Neumann) Let T be a closed operator on E such that dom(T) is dense in E. Then dom(T*T) is dense in E; the operator T*T is closed; and the operator Z + T*T (defined on dom(T*T)) is a bijection of dom(T*T) onto E. The operator B = ( I + T*T)-' is defined on E, continuous, self-adjoint and injective, and its spectrum is contained in the interval [0, 11 of R.Furthermore, the hermitian form ( x , y ) +-+ (B . x I y ) is positive and nondegenerate, and C = TB is a continuous operator defined on E, such that C(E) c dom(T*). Finally, (T*T)* = T*T.

416

XV

NORMED ALGEBRAS AND SPECTRAL THEORY

We have seen in (15.12.4) that T(T) and J(T(T*)) are orthogonal supplements of each other in E x E. Hence for each X E E there exists a unique y E dom( T) and a unique z E dom( T*) such that (x, 0) = ( y , T y )

(15.12.6.1)

+ (T*

*

Z,

- z).

-

Put y = B x and z = C x. Clearly Band Care linear operators defined on the whole of E, and we have B(E) c dom(T) and C(E) c dom(T*). Also, by (15.12.6.1),

-

11xIl2= /lYl12 + l1T.YIl2 + lIz1l2 + IlT* - zl12,

- s

so that IIB xII llxll and IIC xII IIxII. Hence Band Care continuous. The relation (15.12.6.1) is equivalent to x=B*x+T*C*x

and

O=-C.x+TB.x,

so that C = TB and T(B(E))c dom(T*), hence B(E) c dom(T*T). Consequently T*TB is defined on all of E, and we have Z = B + T*TB=(Z+ T*T)B, which shows that B is injective and I + T*T surjective. For each w E dom(T*T) we have (15.12.6.2)

(W

+ T*T. w 1 W ) = ((w((’+ ( T * T *w 1 W ) =

llW1l2 3. lIT* WII’

because T = T**; this shows that the relation w + T*T. w = 0 implies that w = 0, and hence that Z + T*T is a bijective mapping of dom( T*T) onto E. Also, since T ( B ) is closed in E x E (1 5.1 2.2), the same is true of T(I + T* T) (15.12.5), and it follows immediately (15.12.5) that T*T is closed. We next remark that, for all, u, u in E, we have

(B* U I

U)

= (B*

UI

B u

+ T*TB - U )

( B * u I T* TB * u) = ( B u l B * u) ( T B . uI TB. u) = (B u I B V ) + ( T*TB * u I B * U ) = ((I T* T)B * u I B * U) = (U I B * u). = ( B * u IB

-

+

*

-

V) f

+

-

Hence B is self-adjoint. Also, replacing w by B x in (15.12.6.2) we obtain, for each x E E, (XI B X) = IIB * ~ 1 1 ’ llTB * xllz 2 0;

-

+

12 UNBOUNDED NORMAL OPERATORS

-

417

since I1B xII 5 IIxII, it follows from the above and from (15.11.7) that Sp(B) is contained in [0, 11. Moreover, the relation ( x I B x) = 0 implies B * x = 0, hence x = 0, and therefore the hermitian form ( x , y ) I+ ( B x Iy ) is nondegenerate. We shall prove next that dom(T*T) is dense in E. If T‘is the restriction of T to dom(T*T), it will be enough to show that r(T’)is dense in r ( T ) ,since dom(T*T) is the first projection of r(T’),and dom(T) is dense in E. To see that the subspace r(T’) of the Hilbert space r(T )is dense in r(T ) , it is enough to show that if a vector (u, T * u) E r(T )is orthogonal to r(T’), then it is zero. Now this condition is ((u, T * U ) I (v, T v)) = 0

-

-

-

for all u E dom(T*T), or equivalently ( u [ u) + ( T . u 1 T * u) = 0; or, since T * u E dom( T*),(u I u) ( u I T* T . v ) = 0, that is to say,

+

(u I ( I

But I

+ T*T )

*

u) = 0.

+ T*T maps dom( T* T ) onto E. Hence u = 0, as required.

Finally, since B is self-adjoint, T(B) is the orthogonal supplement of J(T(B)).Since T ( B ) is the image of T ( I + T* T )under the symmetry operator S : ( x , y ) ~ + ( yx), , and since JS = -SJ, it follows that T(Z+ T * T ) is the orthogonal supplement of J(T(Z + T*T)); in other words (15.12.4) ( I + T* T)* = I + T* T, or equivalently (T*T)* = T* T. (15.12.7) A not necessarily bounded operator T is said to be normal if it is closed, if dom(T) is dense in E and and if T*T= TT* (which, we recall, implies by definition that dom(T*T) = dom(TT*)). We say that T is seyadjoint if dom(T) is dense in E and if T* = T (which implies that T is closed (15.12.4)). Clearly a self-adjoint operator is normal. It follows from (15.12.6) that, if T is any closed operator such that dom( T )is dense in E, then T* T and TT*are self-adjoint. The following theorem reduces the problem of the structure of (unbounded) normal operators to that of continuous normal operators:

Let E be a separable Hilbert space. I f N is a not necessarily bounded normal operator, then dom(N) = (i) dom(N*) and IIN.xll = IIN* ex11 for all X E dom(N). The space E is the Hilbert sum of a family (En) of closed subspaces, such that En c dom(N) and En ir;stable under N and N * for all n, so that the restriction N, of N to En is a continuous normal operator. (ii) Conversely, let Enbe a sequence ojclosed subspaces of E , such that E is the Hilbert sum of the En.For each n, let N, be a continuous normal operator on (15.12.8)

418

XV

NORMED ALGEBRAS AND SPECTRAL THEORY

En. Then there exists a unique normal operator N on E such that En c dom(N) and N , = N I En, ,for all n. The set F = dom(N) is the set of all x = x ,

1 n

(where x , E Enfor all n) mch that

c IIN,, . n

x,Il2 < +a,

and we have

N .x

N , .x,

=

N* .x =

and

n

N , * .x,. n

We shall begin by proving the first assertion of (i). We have seen in the course of the proof of (15.12.6) that the graph of the restriction of N to dom(N*N) is dense in T(N); hence, for each x ~ d o m ( N ) there , exists a sequence (y,) in dom(N*N) such that lim y , = x and lim N * y , = N . x . fl-

But for each z

E dom(N*N)

we have

n-. m

30

( I N .zllZ = ( z I N * N * z ) = ( z I N N *

*

Z)

=

I/N*

ZI~',

because N*N = NN*. Applying this result with z = y , - y m , it follows that the sequence (N* * y,) is a Cauchy sequence and therefore converges in E. Since N* is closed (15.12.4), we conclude that x E dom(N*) and N* x = lim N* * y, , whence IIN XI\ = [IN* * xII. We have therefore proved

-

n- m

that dom(N) c dom(N*). Since N** = N , the operator N * is normal and dom(N*) c dom(N). Now consider (ii). We shall first show that, if N satisfies the conditions of (ii), then N*(E,) c En. For all m # n, all y, E E, and all x , E Em, we have (x, I N* . y,) = ( N x, I y,) = 0, because N(Em)c Em.Hence N * * y , is orthogonal to all the Em( m # n), hence must lie in En.Next we shall show that, if x E dom(N), then P," N x = N P E , * x = N , ' ( P E , ' X). For all y,

E

(PE,

En, we have *

and since N * * y , (PE,N

'

I yn> = ( N = PEn N*

I yn> = ( P E , .

*

'

x I pEn

'

Yn)

=(N

'

I yn) = (x I N * ' Yn)?

y , from above, N*

'

Yn) =(NPE,

*

I Yn)

=(Nn

'

(PE,

*

x)l Y n ) ;

-

since this holds for all y , E En, it follows that the two elements P E n N x and N , (PE, x) of E are equal.

-

We shall now prove the following lemma, which generalizes (15.10.8.1):

12 UNBOUNDED NORMAL OPERATORS

419

(15.12.8.1) Let (En) be a sequence of closed subspaces of E, such that E is the Hilbert sum of the En.For each n, let T , be a continuous operator on En. Then there exists a unique closed operator T o n E such that, for all n,

En c dom(T), TI En = T, , PE,T x = TPE, x (x E dom( T ) ) . Moreover, dom(T) i$ the set F of all x = x, (with x, E E,,for all n) such that

c IlT,

< +coy and we have T * x

*

1 T,

n

=

x,.

n

It

Since we must have

1

*

(7’. x)ll’ = 11 T . x1I2
0. (Express y ( T ) in terms of Ti', where T I is the operator considered in (e).) Show that y(T*) 2 y(T).

1 2 UNBOUNDED NORMAL OPERATORS

425

( 9 ) Suppose that Im(T) is closed in F. If M 3 Ker(T) is a closed subspace of E, show that T(M n dom(T)) is closed in F. (Consider the restriction of T to M n dom(T) as an unbounded operator from M to F, and use (f).) (h) Let N be a closed subspace of F such that N n Im(T) = {O}. If N Im(T) is closed in F, then Im(T) is closed in F. (Consider the operator T2from E x N to F defined on dom(T) x N by T2* (x, y ) = T . x y , and remark that y ( T )2 y(T,).) In particular, if Im(T) has finite codimension in F, then Im(T) is closed in F.

+

+

2.

Let E, F, H be three Hilbert spaces, Tan unbounded operator from E to F, and U an unbounded operator from H to E. Suppose that T is closed, dom(T) is dense in E, Ker(T) is finite-dimensional and Im(T) is closed in F. (a) Show that if U is closed then so is TU (defined on dom(U) n U-' (dom(T)).) (Suppose that z. E H tends to z and that TU . 2. tends to y E F. Using the relation y(T) > 0 (Problem l), show that there exists a sequence (x.) in Ker(T) such that CJ * z. x, has a limit x in E. Use the fact that Ker(T) is locally compact to show that the sequence (x.) must be bounded (argue by contradiction) and then pick a convergent subsequence of (x").)

+

(b) If U is closed and Im(CJ) is closed in E, then Im(TU) is closed in F. (Use Problem l(g) and (5.9.2)) (c) Suppose that Ker(U) is finite-dimensional. Show that Ker(TU) is finite-dimensional and that dim(Ker(TU)) = dim(Ker(U))

+ dim(Im(U) nKer(T)).

(d) Suppose that Uis closed, doni(U) dense in H and Im(U) offinire codimension in E (which implies, by virtue of Problem l(h), that Im(U) is closed in E). Show that dom(TU) is dense in H. (Consider the orthogonal supplement HI of Ker(U) in H , and the restriction U, of U to the dense subspace dom(CJ) n H I of H I ; observe that U-' is continuous on Im(U,) = Im(U) and that dom(T) n Im(U)is dense in Im(U).) (e) Suppose that Im(T) and Im(U) have finite codimension in F and E, respectively, and let v be the codimension of Im(U) n Ker(T) in Ker(T). Show that Im(TU) has finite codimension in F and that codim(Im(TU)) = codim(Im(T))

+ codim(Im(U)) - v.

(Remark that E is the direct sum of Im(U), a supplement N1 of Im(U) n Ker(T) in Ker(T), and a subspace N, of finite dimension contained in dom(T), and that the restriction of T to N, is injective.) 3. Let E, F be two Hilbert spaces. An unbounded operator Tfrom E to F is said to be an operator with index (Section 15.11, Problem 22) if T is closed, dom(T) dense in E, Ker(T) finite-dimensional and Im(T) of finite codimension in F (in which case T(E) = Im(T)is closed in F by Problem l(h)). The index ofTis defined as the number

i(T) = dim(Ker(T)) - codim(Irn(T)). (a) If Tis an index operator from E to F, then T* is an index operator from F to E, and i(T*)= -i(T) (use Problem l(e)). (b) Deduce from Problem 2 that if U : H -+ E and T :E -+F are index operators, then TU : H + F is an index operator and i(TCJ)= i(T)+ i ( U ) . (c) Let TI be an unbounded operator from E to F, which extends T and is such that dom(T,)= dom(T)@M, where M is a finite-dimensional subspace of E. If T is

426

XV

NORMED ALGEBRAS AND SPECTRAL THEORY

closed, show that TI is closed; if Im(T) is closed in F, then so is Im(Tl); if Tis an index operator, then so is T , , and i(Tl) = i(T) dim(M). (By induction on dim(M).)

+

4.

Let E, F, be two Hilbert spaces and let T be an unbounded operator from E to F. Suppose that T is closed and dom(T) dense in E. Let B be a continuous operator from E to F. (a) Show that, for each x E Ker(T+ B),

Deduce that, if Ker(T) has finite dimension and y ( T )> 0 and IlBll < y(T), then dim(Ker(T+ B)) 5 dim(Ker(T)) (use Problem 9 of Section 6.3). Furthermore, Im(T+ B ) is closed in F (consider the restriction of T + B to the orthogonal supplemeiit of Ker(T) Ker(T+ B) in E). rr. (b) Suppose that T is an index operator and that (lBI1< y(T). Show that codim(Im(T+ B)) =< codim(Im(T)), and that i( T+ B ) = i(T). (To prove the first inequality, consider (T+ B)*. For the equality of the indices, reduce to the case where Tis injective, by considering the orthogonal supplement of Ker(T) and using Problem 3(c). Then observe that y(T+ hB) is a continuous function of h for 0 5 h 5 1.)

+

5.

With the same general hypotheses as in Problem 4, suppose that B is a compact operator from E to F. (a) Show that, if Ker(T) is finite-dimensional and Im(T) is closed, then Ker(T+ B) is finite-dimensional and Im(T B) is closed. (Consider the orthogonal supplement M of Ker(T), and show first that M n Ker(T+ B) c M n dom(T) is finite-dimensional, by noting that there exists m > O such that \IT. xII 1mllxlI for all x E M n dom(T), and using (59.4). Then take the orthogonal supplement N of M n Ker(T+ B ) in M, and prove that there exists no sequence (x,) in N n dom(T) such that lIx.IJ = 1 for all n and T(xJ B(x,) tends to 0.) (b) Deduce from (a) that if T is an index operator then so is T + B , and that i(T+ B ) = i(T).(To prove that Im(T+ B) has finite codimension, consider T* B*; then observe that h ~ i ( T hB) + is a finite continuous function of h on [0, I ] by virtue of Problem 4.) (c) Suppose that E = F and that there exists a regular value for T such that (Tis a compact operator. Show that, for all 5 E C , the operator T- [I has index 0; the spectrum of T is a denumerable discrete subset of C, all of whose points are eigenvalues of T, and (T- 5I)-' is a compact operator for all regular values 5 of T . (Note that

+

+

+

to

51-

T= (I+

(5 - 50)(50~ - T ) - ~ ) ( ~T~I I -

and use the theory of Riesz (11.4.1).) Hence give an example of an operator of index 0 whose spectrum is empty (use Problem 9 of Section 11.6). 6. With the notation of the proof of (15.12.8), prove that the spectrum of the operator N,, is contained in the annulus S, : ( n - I)'/' 5 151 5 n"'. For each pair of points (x, , y.) in F, , there corresponds to N , a measure mg! Yn with support contained in S , . If x x, and y y , are two points of E, with x. E F. and y , E F., then the

=c "

=c n

12 UNBOUNDED NORMAL OPERATORS

sum m x , y= C mjmn).y,is a bounded complex measure on C, with norm

427

l l 1: express this difference in terms of the measure m,, (Problem 6), and use the Lebesgue-Fubini theorem and the dominated convergence theorem.) 10. Show that, for each closed operator T on E such that dom(T) is dense in E, there

exists a positive self-adjoint operator R such that dom(R) = dom(T) and an isometry V of R(dom(R)) onto T(dom(T)), such that T = VR. (As in the proof of (15.12.8), define the spaces F,, c dom(T*T) c dom(T) so that the restriction of T*T to F. is a continuous positive self-adjoint operator H,, for which we may take the square root R,. We have 1) T . x,(J= llRn . x.11 for all x, E F,,, hence there exists an isometry V. of R.(F.) onto T(FJ such that TI F, = V. R.. Let R be the positive self-adjoint operator whose restriction to each F, is R.. Show that the V , are the restrictions of an isometry V of R(dom(R)) onto T(dom(T)), by considering the finite sums of the F. and using the facts that Tis closed and dom(R) is dense in dom(T). Finally, show that V-lTis a closed operator which coincides with R, on F, for each n,and hence deduce that dom(R) = dom(T) and V-'T= R.) 11. Let T be a closed operator on E such that dom(T) is dense in E and such that

T(dom(T)) c dom(T*). Show that dom(T)= E and that E is continuous. (If

428

XV

NORMED ALGEBRAS AND SPECTRAL THEORY

+

B = ( I TT*)-'and if for each y E E we put x = B . y , show that the hypotheses imply that y E dom(T*), by using the fact that x E dom(T*).) 12.

Let N be a normal operator on E. If dom(NZ)= dom(N), show that N is continuous. (If not, there would exist a closed subspace F of E, such that F is the Hilbert sum of an infinite sequence (F,,) of nonzero subspaces stable under N a n d N * , such that the restriction N. of N to F, is a continuous operator and such that

for x.

E

F., where the sequence (an)is increasing and tends to

+

tc, .)

13. E X T E N S I O N S OF H E R M I T I A N O P E R A T O R S

(15.13.1) Let E be a separable Hilbert space. A not necessarily bounded operator H on E is said to be hermitian if: (1) dom(H) is dense in E; (2) dom(H) c dom(H*), and the restriction of H* to dom(H) is equal to H ; in other words, for all x , y in dom(H),

In particular, ( H . x I x ) is real for all x E dom(H). It should be noted that in general H* is not hermitian. When H is conrinuous, this definition agrees with that of (11.5), and a continuous hermitian operator is the same thing as a continuous self-adjoint operator. An unbounded self-adjoint operator is always hermitian, but as we shall see below there exist closed hermitian operators which are not self-adjoint. Example (15.13.2) Let 9 be the vector space of indefinitely differentiable complexvalued functions on R with compact support. If p denotes Lebesgue measure on R, then 9 may be identified with a subspace of L$(R, p) (13.19), which is dense in Li(R, p) (17.1.2). It follows immediately from the formula for integration by parts (8.7.5) and the fact that the functions belonging to 9 vanish outside a compact set, that for all x, y in 9

s

( D x I J J ) = Dx(t)y(t) dt

= -

s-

~(t)Dy(t)d t = -(x I Dy).

This may be expressed by saying that iD = H is a hermitian operator. The conclusion is the same if we replace Lg(R, p) by Li(I, p), where I is a (bounded

13 EXTENSIONS OF HERMITIAN OPERATORS

429

or unbounded) open interval in R,and 9 by the subspace 9(I) of indefinitely differentiable complex-valued functions on 1 with compact support. It is clear that the graph of D is not closed: for example, it contains the pairs (x, D x ) where x is of class C' but not of class C 2 ,and Supp(x) is compact (17.1.2). (15.13.3) Let H be a hermitian operator. Since dom(H*) is dense in E, the closure H** of H exists, and since its graph is the closure of the graph of H (15.12.4), it follows from (15.13.1 .I)by continuity that H** is hermitian. We may therefore restrict our considerations to closed hermitian operators. (15.13.4) (i) Let H be a closed hermitian operator on E. For each real number c1 # 0, the closed operator H + a i l : x HH x + mix is an injective mapping of dom(H) into E, whose image Fa is a closed subspace of E , and the operator ( H aiZ)-' (15.12.5) is continuous on Fa. (ii) The linear mapping

+

(15.13.4.1)

V:xH(H-iZ)(H+iZ)-'.x

of F1 into E is an isometry of F, onto the closed subspace V(F,) = F - l . The mapping Z - V : F, = dom( V )+ E is a bijection of dom( V ) onto dom(H), and we have (1 5.13.4.2)

H . y = i ( i + V ) ( I - V)-' * y

for all y E dom(H-). (iii) Conversely, let F be a closed subspace of E , and U an isometry of F onto a subspace of E , such that the image G of F = dom(U) under Z - U is dense in E. Then the operator I - U is a bijection of F onto G ; i f , f o r each y E G, we put

(15.1 3.4.3)

-

-

H y = i(Z + U ) ( I - U ) - ' y ,

then H is a closed hermitian operator, with dorn(H) = G ; moreover, the operator V defined in (1 5.1 3.4.1 ) is equal to U . By virtue of (15.13.1.1) we have

430

XV

NORMED ALGEBRAS AND SPECTRAL THEORY

for all y E dom(H), which shows that H + ail is injective and that the inverse mapping of F, onto dom(H) is continuous (5.5.1). Also, since the operator ( H ail)-' is continuous and its graph is closed (15.12.5), F, is a closedsubspace of E (15.12.2). If in (15.13.4.4) we put a = - 1 and y = ( H + iZ)-' * x, where x E F,, then we get ll(H - i l ) :yl12 = ll(H iZ) y(I2,hence 11 V . xII = llxll for all x E F,. Since the subspace V(F,) is therefore isometric to F,, it is complete and therefore closed (3.14.4). On the other hand, the relations H a y iy = x and H - y - iy = V - x give

+

+

+

i y = t ( x - Vsx),

H-y=f(x+ V*x)

-

for all x E F,. This shows that, if x - V x = 0 for x E F,, then we must have x V * x = 0, hence x = 0. Hence (i) and (ii) are proved. We shall now prove (iii). The hypothesis that U is isometric implies that

+

(15.13.4.5)

(xly- U*y)+(x- U.xlU*y)=O

for all x , y in F; hence, if x - U * x = 0, x is orthogonal to G, and since G is dense in E we have x = 0. Next, let us show that H defined by (15.13.4.3) is hermitian. If x , y E G, then we may write x = u - U u and y = v - U v, with u, v in F; hence, as ( U u I U * v) = (u I v), we obtain

-

-

( H x I y ) = (i(u + U u) I v - U v) = i(( U * u I v) - (u I U * v)) = (u -

U *uI i(v

+U

*

v)) = (XI H . y )

which proves our assertion. Next we show that H i s closed. If (y,) is a sequence of points in G which tends to y E E and is such that the sequence ( H * y,) converges to z E E, then the sequence of points x,=H.y,+

iy,=2i(I-

U)-'.y,,

which belong to F, converges to x = z + iy E F; hence, as U is continuous on F, the sequence of points ( I - U ) * x, = 2iy, converges to a point in G. This shows that y E G and that ( I - U ) x = 2iy. On the other hand, the sequence of points ( I + U ) * x, = 2 H y , converges to ( I + U ) * x = 2z for the same reason, and therefore H * y = z. Hence H is closed. Finally, if x = u - U * u with u E F, then we have H x = i(u + U * u), and it is immediately verified that V * u = U * u.

-

The isometric linear mapping Vdefined in (15.13.4.1) is called the Cayley transform of H. Proposition (15.13.4) reduces the study of hermitian operators to that of their Cayley transforms.

13 EXTENSIONS OF HERMITIAN OPERATORS

431

(15.13.5) With the notation of (15.13.4), let EA be the orthogonal supplement (dom(V))' of the closed subspace F, = dom(V), and EH the orthogonal supplement (V(dom(V)))' of the closed subspace F-, = V(F,). (15.1 3.6) (i) The subspace EA (resp. EJ is the set of ail x E dom(H*) such that H * x = ix (resp. H * * x = -ix), and dom ( H * ) is the direct sum ofdom(H), E i , and EH .

(ii) Let G + (resp. G - ) be the subspace oj' T(H*) whose first projection in E is EL (resp. EG). Then G+ and G- are closed in E x E and T(H*) is the Hilbert sum of'T(H), G + ,and G - . We have x

E

E i if and only if

-

for all y E dom(H), or equivalently if and only if ( H y I x) = ( y I i x ) ; by definition (15.12.3), this signifies that x E dom(H*) and H* x = ix. Similarly for x E EH , if we observe that V(F,) is the set of all H * y- iy with y E dom(H). Let x be an arbitrary point in dom(H*); as E is the Hilbert sum of dom( V ) and E; , we may write H* x + ix = H * y + iy + z with y E dom(H) and Z E E;. As we have H y = H* * y and H* z = iz, we may write z = H* * z1 + iz,, with z1 = z/2i, hence H* x + ix = H* y + iy + H* - z1 + iz,; this shows that z2 = x - y - z1 satisfies H* z 2 = - i z 2 , in other words z2 E EH , and therefore dom(H*) is the sum of dom(H), EG, and EH . This implies that T(H*) is the sum of T(H), G+ and G - . That G+ (resp. G-) is closed follows from the fact that this subspace is the intersection of the closed subspace T(H*) of E x E and of the subspace consisting of the points (x,v) such that y = ix (resp. y = -ix), which is obviously closed. It remains to be shown that G + , G - , and T(H) are mutually orthogonal. The fact that G + and G- are orthogonal follows from the relation

-

-

Furthermore, if x E dom(HI and y E E,, we have

and as H* y = -iy, we see that G- is orthogonal to T ( H ) ; the fact that G + is orthogonal to r ( H ) is proved in the same way, and this ends the proof of (i) and (ii).

432

XV

NORMED ALGEBRAS AND SPECTRAL THEORY

(15.13.7) The defect of a closed Hermitian operator H is defined to be the pair (m, n) where m (resp. n) is equal to the dimension of E i (resp. E;) if this dimension is finite, and is l-co otherwise; m is called the positive defect and n the negative defect of H. (1 5.1 3.8) For a closed hermitian operator H to be self-adjoint, it is necessary and suficient that its defect should be (0, 0), or in other words that its Cayley transform V should be unitary. For H to be capable of extension to a selfadjoint operator, it is necessary and suficient that its defect should be of the form (m, m). The first assertion follows trivially from (15.13.6). Next, if H is the restriction of a self-adjoint operator A , then the Cayley transform V of H is the restriction of the Cayley transform U of A , which is unitary. Hence by definition we have EH = U(Ei) because U is an automorphism of the Hilbert space E, and the defect of H is therefore of the form (m, m). Conversely, if this is so, then there exists an automorphism U of the Hilbert space E which coincides with V on dom(V) and is such that U(EA) = EH (because two Hilbert spaces of the same finite dimension, or two separable infinitedimensional Hilbert spaces, are isomorphic (6.6.2)). Since U is a unitary operator on E, and the image of E under Z - U (which contains dom(H)) is dense in E, it follows that U is the Cayley transform of a self-adjoint operator which extends H .

A hermitian operator H whose closure H** is self-adjoint is said to be essent ialIy self adjoint .

Examples (15.13.9) Let (en),,>,,be a Hilbert basis of an infinite-dimensional separable Hilbert space E. Let F be the closed hyperplane in E generated by the en with n 2 1 (the orthogonal supplement of the line Ce,) and let U‘ be the isometry of F onto E such that U’ * en = en-l for all n 2 1. Then the image G of I - U’ is dense in E. For G’ contains the vectors en - en+, for all n 2 0, Q,

a n d i f a v e c t o r x = x 0, and use Problem 16 of Section 14.11.)

3. (a) With the hypotheses and notation of Problem 2, consider two sequences (y"), (2.)

of complex numbers satisfying the recurrence relations

(1)

A h = b"-lY"-l

(2)

pzn = b,-

1

z.-

1

+any"

+ a,

2"

+ bd.,

1,

-k b. Z"+

1,

where A, p are any complex numbers. Show that (3)

( p - A)

n- 1

C

k=m

y,

zk =

bn- I ( Y -~ I zn - yn zn-

1)

- bm

- 1( Y m -

(b) Deduce the following formulas from (a): (4)

Pn- i(h)Q.(A)

(show that the Qn(A)satisfy (I));

1

- Pdh)Q.- i(A) = -

bn-

1

1 Zm

- Y m z m - 1).

438

XV

NORMED ALGEBRAS AND SPECTRAL THEORY

Show that &(A) the point

is a closed disk contained in the half-plane Y w > 0, with center a t

Qn-

~(h)p.(h)

- Qn(h)P,-

i(h)

P"(h)P"-l(h)- Pn-l(h)p.(h) and radius 1

the frontier of K,(h) being the closure of the set of points

where t E R. We have K, + I(h)c Kn(h),and the frontiers of these two disks have a point in common. The intersection K,(h) of these disks consists of a single point if

and otherwise is a disk of radius 1

(d) Show that the following three properties are equivalent: ( a ) The operator H deduced from the Jacobi matrix J is self-adjoint. (p) There exists only one positive measure v which extends the linear form aF (in other words, the "moment problem" for the sequence (c,) has only one solution). ( y ) The set K,(h) consists of a single point, for some h E C with Y h > 0 (and consequently for all h E C with Y h > 0). (To show that (8) and ( y ) are equivalent, consider the matrices J. with n rows and n columns obtained by deleting the rows and columns with indices > n in J ; consider the corresponding problem in C", and then pass t o the limit.) (e) If the equivalent conditions i n (d) are satisfied, then the P, form a total system in L&(v).

13

4.

EXTENSIONS OF HERMITIAN OPERATORS

439

c"

With the notation of Problems 2 and 3, show that if the moment problem for the sequence (c.) has two distinct solutions, then both the series \Pn(A)l*and

1 IQ,,(&l'

converge for =FOX# 0 (use formula (4)of Problem 2).

n

Deduce that the moment problem has only one solution in each of the following cases: 1 (a) (Use formula (4) of Problem 3); n b.

c-=+co

(use the relation

(c) there exists a finite number r such that b.-i +a,

+ b, < r

for all n. (Note that equation (1) of Problem 3, with

M Y n + 1 - Y.)

- b,-l(y.

- yn-d = ( r

= r , takes

- b,-l -a.

the form - bnlynr

and deduce that the sequence (Q.(r)) is an increasing sequence of numbers > O . ) 5. (a) If (u.) is a convergent series of real numbers >0, prove Carleman's inequaliry

+

by writing u l u z ... u. = (uIaluza2. . . u.a.)/(n I)" for suitably chosen a n , to be determined, and using the inequality of the means (Section 13.8, Problem 14). (b) With the notation of Problem 2, show that if

then the corresponding moment problem has only one solution (" Carleman's criterion"). (Observe that bo bl

. . . bawl/(P&))'

and deduce that bo bl . . . bn-

i

dv(t)= t"P,(t) dv(t), then use (a).)

6 . Let H be a simple unbounded self-adjoint operator (Problem 1). Show that every closed subspace of dom(H) which is stable under H i s of the form dom(H) n E(A), where E(A) is the closed subspace of E which is the image of E under the orthogonal projector qa(H),and A is a universally measurable subset of E. (By using 3(d) and 3(e), remark that we may assume that H is of the form M , , where the bounded measure v on R is such that polynomials are dense in ..Y,$(v).) 7.

By using the existence of closed hermitian operators with defects (1,O) or (0, 1) (15.13.9), give examples of closed hermitian operators with defect (m, n) where m, n are arbitrary integers 20,or co.

+

440

XV

NORMED ALGEBRAS A N D SPECTRAL THEORY

8. A conjugation in a complex Hilbert space E is a semilinear bijection C of E onto E (i.e., such that C . (ax fl y) = G C . x PC. y ) satisfying (C . x 1C. y ) = ( y 1 x ) and C-I = C. Show that if a closed operator T o n E commutes with C (which implies that C(dom(T)) c dom(T)), and if dom(T) is dense in E, then C also commutes with T*. Deduce that, if a hermitian operator H commutes with C , then C(E4) = E,, and consequently the defects of H are equal. Apply this result in particular to the case where E is obtained by extension of scalars from a real Hilbert space Eo, and H is the extension to E of an unbounded operator Ho defined on a dense subspace dom(Ho) of Eo, such that (Ho . x I y ) = ( x 1 Ho . y ) for all x , y in dom(Ho).

+

9.

+

Show that a surjective hermitian operator H is self-adjoint. (For each y E dom(H*), remark that there exists z E dom(H) such that H * . y = H . z, and show that z must equal y.)

10. Let E be an infinite-dimensional separable Hilbert space. (a) If (a,) is any infinite sequence in E, then the vector subspace generated by the a. (i.e. the set of all (finite) linear combinations of the a,) is not equal to E (cf. (12.16.1)). (b) Show that there exist two Hilbert bases (a"), (b,) of E such that, if F and G are the vector subspaces of E generated by the a, and the bn respectively, we have F n G = {O}. (Starting with an arbitrary Hilbert basis (a"), construct inductively a total sequence (c,) such that the subspace G it generates is such that F n G = {0}, by using (a). Then orthonormalize the sequence (c").) (c) Suppose that the Hilbert bases (a"), (b,) have the property stated in (b). Consider two compact self-adjoint operators A , B on E, defined by A . a, = hnan and B b, =A. b., where (h,) is a sequence of real numbers > O and tending to 0. Show that the sequence (h,) can be chosen so that A(E) n B(E) = {O}. (Proceed by induction: let S. be the set of points of F of the form

"

k=

fkhkak 1

with

"

Ifll/

k= 1

=

1,

and let d, be the distance (strictly positive) of S, from the subspace G, generated by bl, . . . , b,; choose the h k with k > n such that h, < dJn. Then prove that, if U is k>n

the closed ball with center 0 and radius 1 in E, we have A(U) n B(U) = { O } . ) (d) The operators A and Bare injective; A-' and B-' are therefore self-adjoint closed operators (Problem 8) such that dom(A-') n dom(B-') = YO}. (e) Deduce from (d) an example of a self-adjoint operator H a n d a unitary operator U on E such that U z= l E and dom(H) n dom(ll-'HU) = (0).(Take E = F @ F and U .( x , y ) = ( y , x), and take H to be equal to A - ' on one of the summands F, and to B-' on the other, where A and Bare defined as above.) Hence give an example of a closed operator Tsuch that dom(T)is dense in E, but dom(T2) = {O}. 11. Let T be a closed operator on a separable Hilbert space E. The essential spectrum of T is the set o f f E C such that Im(T- 61) is not closed; it is a subset of the spectrum

of T. (a) Show that if N is an unbounded normal operator on E, then the isolated points of Sp(N) do not belong to the essential spectrum of N . (If h is isolated in Sp(N) and if M = Sp(N) - (A}, show that Irn(N- h l ) is the image of E under the projector P = v M ( N )(Section 15.12, Problem 7), by observing that there exists a continuous bounded function f on C such that f ( N ) y E dom(N) and ( N - hZ) . ( f ( N ). y ) = y for all y E P(E). (b) Conversely, show that for each unbounded self-adjoint operator A on E, a point h E Sp(A) which does not belong to the essential spectrum of A is isolated in Sp(A)

13

EXTENSIONS OF HERMITIAN OPERATORS

441

and is an eigenvalue of A . The essential spectrum of A is therefore the set of nonisolated points of Sp(A). (Reduce to the case h = 0, and by using Problem 1 of Section 15.12 reduce further to the case where Ker(A) = {O}; then show that 0 is a regular value for A , and use (15.12.11)) (c) If H is a closed hermitian operator of defect (m,n), then Sp(H) contains the halfplane 9 z 2 0 (resp. 9 z 5 0) if m > 0 (resp. 17 > 0). The essential spectrum of H is contained in R.If m and n are finite, and if H, is a closed hermitian operator which extends H, then the essential spectra of H and HI are the same (Section 15.12, Problem l(h)). 12.

Let W be an unbounded hermitian operator on a separable Hilbert space E. (a) If HI is the restriction of H * t o the subspace dom(H,) = dom(H) Ker(H*), show that HI is hermitian. (b) Show that if Im(H) is closed in E, then H I is self-adjoint. (If xcdom(H:), show that HT . x is orthogonal to Ker(H*), and therefore H: . x E Im(H) (Section 15.12, Problem l(c)); deduce that x E dom(H,).) If H i s closed and if there exists a real number which does not belong to the essential spectrum of H, deduce that the defects of H a r e equal. (c) Show that E&, (resp. E i l ) (notation of (15.13.6) is the intersection of E&(resp. EL) with the orthogonal supplement of Ker(H*). Deduce that if there exists h E R not belonging to the essential spectrum of H , and if the defects of H (which are necessarily equal) have a finite value m,then dim(Ker(H* - AZ)) 2 m. (d) Suppose that H i s closed, and let x be an eigenvector of H * corresponding to a real eigenvalue h. Put x = x,, y z, where xo E dom(H), y E EJ and z E E,. Show that llyli = I/zIl (reduce to the case h 0). (e) Suppose that H i s closed, and let h be a real number which does not belong to the essential spectrum of H ; suppose that the defects of H are equal and finite, say m, and that dim(Ker(H - hf)) is finite, say k . Then dim(Ker(H* - hr)) = m k . (We may assume that h = 0. By considering the restriction of H to the orthogonal supplement of Ker(H), reduce to the case where k = 0. Then deduce from (d) that Ker(H*) cannot have dimension > m , by using the hypothesis Ker(H*) n dom(H) = {O}, and complete the proof by using (c).)

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:

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13.

Let H be a closed hermitian operator whose defects are equal and finite, say m. (a) If Vis defined as in (15.13.4), then the self-adjoint extensions A of H are of the form A = i ( f + U)(Z- U ) - , , where U is a unitary operator extending V, such that U ( E & )= E i ; dom(A) is therefore the direct sum of dom(H) and the subspace ( I - U)(EH+)of dimension m,contained in dom(H*). (b) For a real number h to be an eigenvalue of a self-adjoint extension A of H, it is necessary and sufficient that h should be an eigenvalue of H * (use Problem 12(d)). Show that if h E R is not an eigenvalue of H, then there exists a self-adjoint extension A of H for which h is not an eigenvalue. (Use Problem 12(d) and (e), and choose suitably the unitary operator U of (a).) (c) Suppose that m > 0. For each h E R,show that there exists a self-adjoint extension A of H such that h E Sp(A). (Remark that, if A,, is a self-adjoint extension of H and h 4 Sp(Ao), then it follows from Problems 12(e) and ll(c) that h is an eigenvalue of H*.) (d) Let A , , A z be two self-adjoint extensions of H. If P+ (resp. P-)is the orthogonal projection onto Ef (resp. E;), show that the continuous operator

D

=(AZ

+ iZ)-I

-(A,

+ iz)-I

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NORMED ALGEBRAS AND SPECTRAL THEORY

is such that D = P-D = DP+, and is therefore of rank m (note that, if y X I = (A1 iZ)-' * y , then x1 E dom(H*) and y = (H* + iZ) . XI.)

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E

E and

14. A closed unbounded hermitian operator H on E is said to be majorized (resp. minorized) if there exists a real number c such that ( H ' x l x )