Journal of the Mathematical Society of Japan Vol. 7, No. 3, June, 1955
Representations of a compact group on a Banach space.
By Koji SHIGA
(Received Jan. 14, 1955)
Introduction. Let G be a compact group and IYIa complex normed linear space. We mean by a representation {, T(a)} a homomorphism of G into the group of bounded linear operators on such that a --~ T(a). The following types of representations are important : i) algebraic repre- sentation, ii) bounded algebraic representation,l' iii) weakly measurable representation,2' iv) strongly measurable representation,3' v) weakly continuous representation,4' vi) strongly continuous representation.5' Clearly each condition above becomes gradually stronger when it runs from i) to vi), except iv). When the representation space 1 of {IJI,T(a)} is a Banach space, it is called a B-representation. In considering a B-representation {8, T(a)}, what we shall call the conjugate representation {8*, T *(a)} arouses our special interest. For example, we can prove, by considering {**, T **(a)} of a weakly continuous B-representation {8, T(a)}, the equivalence of all four condi- tions (Theorem 2) : 1) {l3, T(a)} is completely decomposable, 2) {8, T(a)} is strongly continuous, 3) {3, T(a)} is strongly measurable, 4) for any given x e 3, the closed invariant subspace generated by T(a)x,
1) Algebraicreps esentation such that IIT(a) II< M for a certain M(> 0). 2) Boundedalgebraic representation such that f(T(a)x) is a measurable function on G for any xe3 and f€3,*where 9* meansthe conjugatespace of . 3) Boundedalgebraic representation such that T(a)x is strongly measurable (in the sense of Bochner)for any fixedX€ l1. 4) Boundedalgebraic representation such that f(T(a)x) is a continuousfunction on G for any x€31and f€31*. 5) Algebraicrepresentation such that IIT(a)x- T(ao)x II -'0 as as -- a0, (and then necessarilybounded). Representations of a compact group on a Banach space. 225
a e G, is separable. This explains naturally the classical result of Peter and Weyl [1O].6i The content of the present paper is briefly as follows. We first give a survey of unitary representations of a compact group (~ 1). In § 2 the finite dimensionality of an irreducible algebraic representation is proved under a weak assumption. Our discussions which follow are really based on two fundamental Propositions A and B (~ 3) ; in fact, Proposition A yields the essential part of the structure Theorem 2 ( 4). On the other hand, it is possible to deduce from Proposition B a formal structure of weakly continuous representations which might be regarded as a formulation of reduction of the weak continuity to the strong one (~ 5). The relations between the weak and the strong continuity are discussed in 6. The role played by the weak and the strong measurable representations becomes clear in the last section ( 7). The principal idea used here is to reduce an arbitrary repre- sentation to a unitary one. Since we make full use of the boundedness of Haar measure, it seems difficult to generalize further this method beyond the compact case. Throughout this paper we concern ourselves with the representa- tion of a compact group G. s , 3 and denote a Hilbert space, a Banach space and a normed linear space, respectively; Hilbert space is not assumed to be separable. Linear operators are assumed to be bounded and are denoted by S(a), T(a), U(a) etc., whose adjoint opera- tors are denoted by S*(a), T *(a) etc, respectively. We denote elements of a compact group G by a, b, c, and elements of , s~ etc, by x, y, z, f, g, The integral is always taken with respect to the Haar measure on G. LA(G) (1pC C co) and C(G) (the family of con- tinuous functions on G) denote the usual Banach spaces. For a given 9, we shall denote by [x JP(x)] the closed subspace of IYI,generated by all elements x having the property P(x). The author wishes to express his deepest thanks to Professors Y. Mimura and Y. Kawada for their constant encouragements and advises during the preparation of this paper. The author is also grateful to Mr. N. Iwahori for his valuable remarks.
6) Numbers in brackets refer to the bibliography at the end of the paper, a
226 K. SHIGA
1. Unitary representations. In this section we consider a unitary representation {, U(a)}. The following three propositions are well-known.7' Since Proposition 1 plays a fundamental role for our later discussions, we give here a direct proof of Proposition 1. PROPOSITION1. {5~,U(a) } is completely reducible, i, e. S3 is a dis- crete direct sum of mutually orthogonal finite-dimensional subspaces spa's, each of which is invariant and irreducible under U(a) (a e G) : ~~.= o ~, PROOF. Since the invariance of a closed subspace of implies that of its orthogonal complement, it suffices to prove this proposition only in the case where ~ is generated by U(a) (a e G) for some fixed ., I.H=1. For all x, y in 3, we define
[x, y]= (x, U(a))(y, U(a)) da. (1)
Normalizing Haar measure so that da=1, we have [x, y] < IIx I.jy. ~ Hence [x, y] defines a bounded bilinear form on ~. Let H be the Hermitian operator associated with [x, y]8':
[x, y]=(Hx, y) for all x, y. Then H is a positive definite operator in the strict sense. Now, for any ao e G and for every x, y in , we have (U(ao)Hx, y) = (Hx, U(ao)*y) = (x, U(a))(y, U(aoa)) da
_ (U(ao)x, U(a) -)(y, U(a)) da
= (HU(a0)x, y) , whence each U(a) commutes with H. Moreover, H is completely
7) See Godement[1], Hurevitsch[5], Ito [6], Murakami[8] and Wigner L14]. 8) This Hermitianoperator H was first introducedexplicitly by S, lto [6]. Representations of a compact group on a Banach space. 227
continuous. In fact, if xn -+ xo (weakly), Hxn --~Hxo (weakly) ; on the other hand, from (1) and Lebesgue's convergence theorem follows that tHx2n j -> ~~Hxo l 2, and thus Hxn -b Hxo (strongly). Consequently, if we perform the spectral resolution of H, we have H= Jx1P~~1, where Pjj1 is the projection whose range is the finite- dimensional eigenspace 91 corresponding to the eigenvalue x1. This, together with the positive definiteness of H in the strict sense, yields
=> wi r• Since each finite•dimensional subspace W 1 reduces every U(a), because of the commutativity of H and U(a), the above decomposition of ~ gives a decomposition into invariant subspaces. If {JJj UjJ(z(a)}9' is still reducible, it is possible further to decompose 91 into irreducible subspaces by the use of the well-known procedures. This completes the proof. We call such a decomposition of as stated in Proposition 1 a complete decomposition of {s~,U(a)}. Now we shall establish the rela- tionship between two complete decompositions of {, U(a)}. This is easily formulated in terms of the following notion : For a complete decomposition {3, U(a)} = +0 {, Us (a)} and for a fixed equivalent a class of irreducible representations of G, we denote by S~() the closed subspace generated by all those la's for which the irreducible representation U(a) belongs to . Thus, ~ = +0a and = Gs denoting two complete decompositions of {, U(a)}, they yield another kind of decompositions such that S3= >j 0+ () and = > respectively, with respect to the class of irreducible representations. Then from the orthogonality relations we get the following result. PROPOSITION2. ~()='().`~~ PROPOSITION3. Any irreducible unitary representation of a com- pact group is equivalent to one which is obtained in a complete decom- position of the regular representation. PROOF. Let {°, U(a)} be irreducible. Then making use of nota- tions in the proof of Proposition 1, we have H=xI (x> 0), that is 9) Un1(a)denotes the representationof G obtainedby restrictingU(a) in1. Such notationis usedin § 1; in the remainingsections, if it happenssuch a case, we write U(a) for Us1(a) only for simplicity.
Ifo(T(a)xo) J2da=0,
and thus fo(T(a)xo)=T*(a)fo(xo)=0 for all ae G; accordingly, putting J1= {x; T*(a)fo(x)=0 for all a e G}, we have xo e 1Y1o.Buto is clearly an invariant closed subspace, whence from the irreducibility of {9, T(a)} and fo = 0 it results o= {0}, and so xo=0 as desired. (x, y) defined in (2) therefore gives an inner product of 11. ,5 denotes the completion of 1 with respect to this inner product and III III denotes 10) Irreducibilityof a representationmeans the non-existenceof a proper invariant closedsubspace.
228 K. SHIGA
X!Lx 2= I(x, U(a)) I2da.
Hence if we assign to every x of an element f(a)=x (x, U(a) )/j/x of L2(G), we obtain an isomorphism between 3 and a closed subspace L of L2(G), through which U(a) can be identified with the regular repre- sentation in L. REMARK. This result can be generalized to a theorem for locally compact groups which asserts that any irreducible unitary representation corresponding to an integrable positive definite function (in Godement's sense [1]) is realizable in the regular representation.
2. Irreducible representations.lo> In his paper [7] I. Kaplansky remarked the fact without giving its proof that any irreducible strongly continuous B-representation is finite- dimensional. It is a purpose of this section to formulate this theorem in a more generalized form and to prove it. THEOREM1. Let {Y1,T(a)} be an irreducible bounded algebraic representation of G. If for some non-zero element fo e J* f0(T(a)x) is continuous on G for any x e fit, then 't is finite-dimensional. PROOF. Set
(x, y)= fo(T(a)x)fo(T(a)y) da (2) for x, y e J1. From the assumption the integrand being continuous, this definition is meaningful. We shall prove that (x0,xo)=0 implies x0= 0. In fact, (x0,x0)= 0 means
1II T(ao)x 1112= Ifo(T (aao)x)12dx=II xJ 112. Thus, being isometric, T(a) can be uniquely extended to a unitary operator U(a) on 3, which clearly induces an algebraic representation of G. We assert furthermore that {s~,U(a)} is a unitary representation of G. This follows immediately from the fact that the function f o(T(ba)x)f o(T(b)y) is continuous on G x G. Applying Proposition 1 to {, U(a)}, we obtain a complete decom- position of : c= EEP,
Iixn-xo 1112= I fo(T(a)xn)-fo(T(a)xo)12da --->o ; hence 91 is continuosly imbedded into in the sense of the strong topology. We denote the image of 91 into ~~ by Now consider T(a) as an operator on (). We have by the right invariance of Haar measure
Representations of a compact group on a Banach space. 229
the norm in . If x-> x0 (strongly), we have T*(a)fo(xn)-~T*(a)fo(xo) for each a, and so
a a where Pa means the projection operator associated with . Suppose that there is a non-zero x in 1) whose projection on some a, is zero, i. e. x = 0 and P, x=0. Putting this assumption is equivalent to = {0}. Let J i be the inverse image of (1) (\ s~l through the natural mapping of 9 onto (1Y1).Since this mapping is continuous,~1 is a closed subspace of different from {0}; mereover, the invariance of under U(a) implies that of under T(a). Thus the irreducibility of {IYI,T(a)} leads to J =J. But this means s~(Y1)=(J1) f\ S, and so ~( J) 1, which is contradictory to the fact that ~~(iJ) is dense in. Hence, from x e (91) and x == 0, we can conclude Pax =1=0 for all a. Taking any ao, let x correspond to Paox. Then this induces an isomorphic mapping of into Sao regarded as linear spaces. Now aa being finite dimensional, we find that J1 is also finite-dimensional. This completes the proof. After accomplishing the proof, observe that the finite dimensionality of 1 implies ~~= (R). This shows that the complete decomposition of 230 K. SHIGA
{, U(a)} given above is trivial. REMARK. In Theorem 1, if is furthermore assumed to be separable, the assumption on the continuity of f0(T(a)x) can be replaced by the measurability. This is shown by the use of the proof of Theorem 9.
§ 3. Main propositions. Suggested by the result of Peter and weyl, which clarifies the structure of the regular B-representation induced in C(G), we introduce DEFINITION 1. For a given algebraic representation {ifl, T(a)}, let {1Y1,,T(a)}aeA be the family of all the finitedimensional irreducible representations such that 'i's are invariant subspaces of IQ {i1, T(a)} (or J) is called completely-decomposable if the subspace generated by all {ta}aeA is dense in with respect to the strong topology. From Theorem 1 follows that, if {1Y,T(a)} is weakly continuous, the assumption on finite dimensionality of Ya in the above definition becomes superfluous. Now we shall prove two fundamental Propositions A and B, which seem to be, in a sense, in a dual relation. These propositions are very similar to each other not only in the statement but also in the proof. However, as is seen later, they possess in an essential feature the different significance for the theory of representations and will be used rather independently. PROPosITIoN A. Let {i, T(a) } be a bounded algebraic representa- tion of G, and let {C, T(a)} be a weakly continuous representation such that is an invariant subspace of JLI'' Then, regarding {J**, T **(a)} as a bounded algebraic representation of G, we can find an invariant closed subspace of J** with the following properties. i) {C~,T**(a)} is strongly continuous and completely decomposable. ii) Let an element f of~* have the property : F(f)=0 for all F e ~. Then f belongs to iL, where =[f If(x)=0 for all x e PROOF. Taking any fixed e C~, we define an inner product in 'J* by
(f, g)= f(~(a))g(T(a)) da. (3)
11) In such a case,we say that {C~,T(a) } is containedin {I, T(a) }.
I1,,,(f) I < Jill • IIIh lie, (5) hence Fe,h(f) definesa abounded linear functional on If h~= h2, we have (f, hl) =1=(f, h2) for at least one f, i.e. Fe,hl= Fe,h2.From (5) follows II F ,hII < IIIh II l. Put J* *=[Fe,hjhc]; ¶K* is a closed subspace of Denoteby the mappingof ~~into * such that h --~Fe ,h. Then we have shown that !v induces a univalentanti-homomorphism of ~~ into which is continuouswith respect to the strong topologyin each space. Now observe that the operator U(a) on S is transferred by ' to such an operator on SJ * as the one obtained by restricting T **(a) to J(*; in fact, FE, U(a~h(f) _ (f, U(a)h)e= (U*(a)f,h) =Fe,h(T*(a)f)=T**(a)Fe,h(f)
Representationsof a compactgroup on a Banachspace. 231 Let 'J1 be the closedsubspace of formed by all elements f with (f, f)=O; we obtain a Hilbert space by completionof the quotent space J */i . Denote the canonical mapping of * into , by f-~f, and the imbedded image of into ~~, by s~( J *)• Since the repre- sentation {I, T(a)} is bounded,we may assume by (3) lllf III < Ilf II, (4) where III III means the norm of . Now, from T *(a)i~ SJ, it follows that T *(a) induces an operator in (1Y1*);this operator being isometricby (3), we can extend it to a unitary operator U*(a) on. Rememberingthe proof of Theorem 1, we know that {S, U(a)} thus obtained is a unitary representationof G, and that admits a com- plete decompositionwith respect to this representationsuch that
Considerthe linear functional F(f) on JV*associated with each element h e , defined by F(f)=(f, h). We have by (4) 232 K. SHIGA
From this it follows thatis an irreducible invariant finite- dimensional subspace in {J**, T **(a)}, and that T **(a)} is strongly continuous. V being continuous, the subspace generated by all ) is dense in whence {*,~~ T **(a)} is strongly con- tinuous and completely decomposable, as is readily verified. Finally, put
We show that meets the requirements i) and ii) stated in the pro- position. i) follows from the fact that is generated by the closed subspaces * in which T **(a) is strongly continuous . ii) is proved as follows : Choose an f in ?* such that F(f) = 0 for all F e ~. If f 5~, we have f(x0) 0 for an xo e 5. Consider F 0f ; then we have F0, 1(f) = 0 and so
f (T(a)xo) J2da=0 , which implies f(x0)=O by the continuity of f(T(a)x0). This contradic- tion shows the validity of ii). Thus the proof of Proposition A is completed. Before we enter into Proposition B, for a given bounded algebraic representation T(a)} it is convenient to introduce the subspace iP of i defined by = {f; f(T(a)x) is continuous on G for any x} . Clearly '~ is a closed subspace which is invariant under T *(a). PROPOSITIONB. Let {s)1,T(a)} be a bounded algebraic representa- tion. Regarding {*,~~T *(a)} as a bounded algebraic anti-representation, we can find an invariant closed subspace ££ of J1 with the following properties. i) {, T *(a)} is strongly continuous and completely decomposable. ii) Let an element x of ~~ have the property : f(x)=0 for all fe £. Then x belongs to J, where J r= [x f(x)=0 for all feJlfl. PROOF. Although almost all parts of this proof go on simply by a slight modification, or rather simplification of the proof of Proposition A, we shall follow it briefly for the sake of later use. Representations of a compact group on a Banach space. 233
For any fixed f e we define an inner product in 'J1 by
(x, y)f = f(T (a)x)f(T (a)y) da . Denoting by 'f the closed subspace of 9 consisting of all elements (x, x), = 0, we obtain a Hilbert space f by completion of Consider a unitary representation U(a) } induced naturally from {c, T(a)}. Thenf is decomposed such that ~~f
For any z e f, the linear functional f, x on defined by = (x, z) belongs to *, and the anti-linear mapping 'f of into ~J1*such that z - 3 i f,x is univalent and continuous. Denote by i" the image of mapped by Pf. Then Y1f, is a finite-dimensional irreducible invariant subspace of with respcet to T *(a) ; furthermore the representation {i1,f , aT *(a) } is strongly continuous. Thus, as is readily seen, putting
-[~f ,x!ze~3f,fei ], we have the desired closed subspace, and this completes the proof. It is clear that £ possesses the subspace generated by all elements /if, y of * as a dense subspace, where
~f,~(x)= f(T(a)x)f(T(a)y)da (y e , f e i~e) . This fact becomes essential in 5.
§ 4. Structure of strongly continuous B-representations. In order to explain our aim in this section, let us consider Pro- position A. This suggests ; if a weakly continuous representation {, T(a) } is given, we have a strongly continuous representation which is contained in {9**, T **(a)}. This representation space is rather thickly contained in but not always in 5 if we regard ~~ as a subspace of While this situation will be pursued in § 6, we first examine 234 K. SHIGA
a case where such a strongly continuous representation is already realized in { J1,T(a)}. Conditions to guarantee us such a normal case lead naturally to a strongly continuous B-representation, and there we meet the theorem of Peter-Weyl. THEOREM2. Let {, T(a)} be a weakly continuous B-representa- tion. Then the four properties listed below ar<; equivalent to each other : 1) {8, T(a)} is completely decomposable. 2) {3, T(a)} is strongly continuous. 3) {B, T(a)} is strongly measurable. 4) For every x e 3, the invariant closed subspace [T(a)x I a E G] is separable.' ' PROOF. We shall prove this theorem by showing a sequence of implications : 1) -- 2) -k 4) -*3) - 1). PROOF OF 1) -~ 2). Assume that {, T(a)} fulfills i). Then for any x e 8 and for any positive number e, there exists an element y belonging to an invariant finite-dimensional subspacex such that IIx-y II
x, f = x, f(a)da , we have F~,f(g) = g(T(a))f(T(a)) da
=g f(T(a)) T(a)& da
=g(x , f) for all g e* . This means F ,f=x, f, as we wish to show. Thus all parts of Theorem 2 are completely proved. As easy applications of Theorem 2 to the regular representation in a function space : f(x) --~f(xa), we have the following well-known results. COROLLARY1 (Peter-Weyl). Denote by R(G) the linear set gene- rated by the coefficients of all finite-dimensional irreducible representa- tions of G. Then R(G) forms a dense linear set in LA(G) (1 p < oo) and in C(G) with respect to the proper topology in each space. We mean by Cr(Tn) the family of all functions of the class Cr on
ID~k>f(x1, ..., x,,) _ D'k)P(xl, ..., x,,) I C e for 0 < k < N, where D'k'f is the k-th derivative of f.
236 K. SHIGA
the n-dimensional torus Tn. COROLLARY2. Let f(x1, •••, x,) e (Y(Tn) (rC co). For a given e>0 and for an integer N such that NC r, there exists a trigonometrical polynomial P(x) with the properties :
5. Structure of weakly continuous representations. We shall prove that any weakly continuous representation is ex- pressed as a kind of a continuous direct sum of strongly continuous representations, each of which is realized as the regular representation in a function space. This aspect is very analogous to the situation that any W*-algebra (ring, according to a terminology of von Neumann) on a separable Hilbert space can be reduced to factors and is written as a direct integral of them, as is shown by von Neumann [9], Segal [12] and Godement [2]. THEOREM3. Let {9, T(a)} be a weakly continuous representation of G. Then we can find a locally compact space 9, and a non-trivial normed linear space I" corresponding to each point of 9 ; I' is a subfamily of continuous functions on G with a suitable norm, and is also anti-isomorphic to an invariant subspace of {SJ.*,T*(a)}. Further- more we have a linear mapping of ~~ onto 1' such that x -->x(a). We denote by 4 the element in corresponding to x(a). The totality of x (a) ( e 9, x eJ) can be considered as a linear space d, formed by such functions on 9 as taking values in J', at = o. Those have the following properties : 1) The linear mapping x-~ {x(a)} of onto d is univalent. This correspondence is denoted by x~x(a). 2) itx(a) III < I{x II, where II III means the norm in I'. 3) T(b)x~x(ab). 4) I I x(aba)--x(abo) II ---~0 as ba --~bo, i.e. the right regular repre- sentation in I' is strongly continuous. 5) For fixed a and x, x (a) as a function of belongs to L~(9),
IIxf(a)IIf=JJ0J,I1x - As is seen from (6) and (7), the nonmed linear space I' f whose con- stituents are I'f and II II, is anti-isomorphic to by means of the correspondence x f(a) Now we shall prove that 2, I' f and x --px (a) have the properties stated in the theorem.
Representationsof a compactgroup on a Banach space. 237 where L..CQ)means the function space formed by all continuous func- tions vanishing at infinity. 6) IIT(a)x I = max (x(a) 7) Let x~x(a), y~y(a). Then
y (x) = x(a) ye(a) da . 8) Let x~x(a). If a sequence {~} of 2 converges to , then x n converges weakly to 4,. PROOF. Let 12' be the unit circle of cJ~*. Then, as is well-known, 12' is compact with respect to the weak topology. Thus, putting 2 for S2'- {0}, we find that 2 is the locally compact space (the eliminated point 0 is considered to be at infinity). For any x e J1, we assign to each f e 2 the continuous function on G such that xf(a)=f(l(a)x). (6) For a fixed f e 2, let F'1 be the linear space consisting of all functions xf(a), where x runs over J ; I'f is clearly nontrivial. The mapping x -~ xf(a) gives rise to a linear mapping of 'iJ onto F'. In order to define a norm in I''f, we make use of Proposition B. Applying it to {iY,T(a)} in question, we see that 2 is an invariant closed subspaces of {9*, T~`(a)}which is generated by invariant sub- spaces f (feJ1*); J1,,,is formed by all elements such that = f( T(a)y)f(T(a)x)da (xe 1), (7) as is remarked at the end of Proposition B. If f(T(a)x) -0, then of course Of, x = 0 ; conversely cP1, x = 0 yields 01,x(x) = 0, that is, f ( T(a)x) IZda = 0 and so f(T(a)x) = 0. Therefore it is possible to introduce a norm in P''1 by Ilfli~l feo
238 K. SHIGA
1) x f(a) = f (T(a)x) = 0 for all fcQ, is equivalent to x=0. 2) By (7) and II f II< 1, we may assume II q f, x II< I I x JI, and so Ilxf(a)Ilf < lIxlI. 3) From x'-'-'x(a) and xf(a)= f(T(a)x), it results T(b)x~xf(ab). 4) Since by (7) ~pf, ~(b)x=Z T ~`(b-1)~f,x, we have II xf(ab) II f = II T(b_1)~ f, xl I, and so IIxf(aba)-xf(abo) 1/=11 T*(b;l)pj,_ II This tends to zero as ba-~ b0, because {91, T *(a) } is strongly con- tinuous. 5) This is easily seen by the definition of weak topology. 6) IIT(a)x 11=max If(l(a)x) I= max Ixf(a) I . 7) By (6) and (7) this is clear. 8) Suppose that f5-f0~(weakly); then f(T(a)x)-3fo(T(a)x)nfor each a, whence by (7) cjif,, (weakly). This completes the proof. REMARK. 12 does not always satisfy the first axiom of countability. But it is known that, in the case when 91 is a reflexive Banach space, 12 is locally sequentially-compact. Now it is desirable to take an available simple space as the base space 12 in the above theorem. Because we know by Theorem 2 the structure of strongly continuous representations, we arrive at DEFINITION2. Let {92,T(a)} be a bounded algebraic representa- tion. Assume that it is possible to find a strongly continuous repre- sentation {C~,S(a)} such that 1) There exists a continuous univalent linear mapping ?F of 91 onto C~. 2) r( T(a)x) = S(a) '(x). Then {91,T(a)} is called almost strongly-continuous. THEOREM4. For a given weakly continuous representation {9t, T(a) }, assume that we can select suitable countable elements f 1,f2, in 9* such that {T*(a)f; ; ae G, i=1, 2, ••. } forms a total subset in 91*.13) Then {91, T(a)} is almost strongly-continuous. PROOF. Denote by '' the linear set consisting of all elements
13) A subsetC~ in the conjugatespace of a normedlinear space is called total, when,if f(x) =0 for all f€,C~then x=0. Representations of a compact group on a Banach space. 239 such that cx= {x11(a),x 12(a), •••}, where x,.(a) is defined by (6). This is a subset of a sequential space, whose coordinates are continuous functions on G. We introduce a norm in C~' by
II I= ~ ~~xfZ(a)Ilfi . i=1 2 here we assume II fz I1=1 (i=1, 2, ..•) without losing generality. Denoting by the normed linear space {';C~II li}, we define a mapping V of onto by x--~ 1. Then ' is obviously linear and continuous by 2) i n Theorem 3. Now Ox= 0 means II x fi(a) II fi= 0 for all i and so x fi(a)=0, whence we have f=(T(a)x)=T*(a)f1(x)=0 for i=1, 2, ••. and for all a e G. This, together with the assumption on {T*(a)f;}, yields x=0. Hence ! is univalent. Now the representation {C~,S(a)} with S(b)c- {xf1(ab),x f2(ab), ... } is strongly continuous, onto which {,T(a)}.is imbedded by !v. Hence {, T(a)} is almost strongly-continuous. In § 7, we shall treat again almost strongly-continuous representa- tions from a somewhat different viewpoint.
6. Relations between two notions of continuity. In this section we shall treat the relations between the weakly- and the strongly-continuous representations.14) From Proposition .A and Theorem 2 we have THEOREM5. In the case where the representation space is separable or reflexive, the weak continuity of {, T(a)} necessarily implies the strong continuity. Let {, T(a)} be a bounded algebraic representation. Form an invariant closed subspace Jw of consisting of all those elements x for which f(T(a)x) is continuous on G for every fe J* ; { W,T(a)} is the greatest weakly-continuous representation contained in {, T(a)}, and is called the weakly continuous part of {, T(a)}. The strongly 14) We couldnot succeedin obtainingan exampleof a weakly,but not strongly, continuousrepresentation. 240 K. SHIGA
continuous part {Js, T(a)} is defined similarly : J s is the invariant closed subspace of, consisting of all those elements x such that x e ¶W and [T(a)x i a e G] is separable. It is clear that J c . We are now interested in studying a weakly continuous representa- tion {, T(a)} with = . LEMMA 1. Using the same notations as in Proposition A, if fit, then {C~,T(a)} is strongly continuous. PROOF. The assumption, combined with ii) of Proposition A, yields C~c ; {~, T* *(a)} being strongly continuous, {C~,T(a)} is of course strongly continuous. Consequentjy, if a weakly continuous representation {1, T(a)} with =1=9 s is given, applying Proposition A to {fl, T(a)} itself, we have p\ (J-1fl)~* = {0} and C~('\ (-J5)={O}.~Furthermore, by ii) of Proposition A, from F(f) = 0 for all F e it results f = 0. Thus we have shown THEOREM 6. For any weakly continuous representation {, T(a)}, the strongly continuous parts* of {* *, T**(a)} forms a total sub- space. IfS, thens s . COROLLARY.Let T(a)} be a bounded algebraic representation. Consider a closed subspace ~?c* of 9V, formed by all elements F such that F(T*(a)f) is a continuous function on G for every f eJ *. If then {, T(a)} is strongly continuous. The above discussions may be used to obtain a remarkable, but somewhat pathological property of a weakly continuous B-representation {8, T(a)} with 3 = To see this some preliminary considerations are necessary. Let ,3 be a non-reflexive Banach space. A theorem due to Milman asserts that a closed subspace of a reflexive Banach space is also reflexive, whence, repeating the process to construct's from 8, we get S 3(0) C S (1) c ... c 3(n)C ... ~ where 3(O)_ 3 and (n+1)=3(*S n). This process is further carried on ; by transfinite induction we shall define a Banach space(a' for any ordinal number a. Suppose that, for an ordinal number a0, we have already defined Vi(a) for a we define 3(a°' as 8 * *. If ao is a limit ordinal number, first put Q(a°'= Ua~a°a ; then £a°' becomes 1n a natural manner a normed linear space. Now we define 8`a°' as a Banach space obtained by completion ofThus we have obtained for any a, such that for ct<$. Assume that a weakly continuous B-representation {8, T(a)} is given. Then we can define an operator T" (a) on capin an obvious manner ; T (a) } is a bounded algebraic B•representation of G. We denote by { w', T (a) } and {s ', T (a) } the weakly and the strongly continuous part of T (a) }, respectively. Then from _ Proposition A and Lemma 1 we have THEOREM7. Let { 3, T(a)} be a weakly continuous B-representa- tion such that 3 = 3s. Then and 8 ~= W' for any So far we have mainly concerned with the relationship between T(a)} and {**, T **(a)}. From now on we shall consider the properties of {*, T *(a)} induced from a given representation {, T(a)}. If the representation space is reflexive, the situation is very clear, that is, the strong continuity of {3, T(a) } implies the strong continuity of {*, T*(a)} and vice versa. DEFINITION3. A bounded algebraic representation T(a)} is. called almost weakly-continuous whenc is a total set in 1*.15) From Proposition B and this definition we obtain : THEOREM$. Let {` , T(a)} be an almost weakly-continuous repre sentation. Then the strongly continuous part Js of {Yi*,T'~(a)} is total in . We shall give an example of {y, T(a)} such that Js is total in j1*. Let {, T(a)} be the regular representation induced in L'(G) ; then {gds, T*(a)} is equivalent to the regular anti-representation induced. in C(G), and C(G), regarded as a subspace of L°°(G), is total in L°°(G). As a result of the above theorem we have COROLLARY.I f a weakly continuous, representation {, T(a)} is. regarded as the conjugate representation of a certain representation,16' then {, T(a)} contains the strongly continuous part as a total subset.. 15) As to the definitionof 3, see p. 232. 16) Namely,we can findan algebraicanti-representation {C~, S(a)} such that T(a)=Sp(a). {R,T(a)} is calledthe conjugaterepresentation of {C~,S(a)}. 242 K. SHIGA From this corollary a problem arises : Does there exist a weakly continuous representation such that its strongly continuous part is {0} ? The author has no idea to approach this problem. Now let {*, T *(a)} be the conjugate representation of a given weakly continuous representation {~~,T(a)}. Concerning the formal structure of {J1*, T *(a)}, we can obtain the corresponding result to Theorem 3 by the use of Proposition A. The essential distinction arising here lies in the fact that in this case it is assured that the base space Q is only completely-regular instead of locally-compactness, contrary to Theorem 3. Since this assertion can be easily formulated, we omit the exact formulation. § 7. Mesurable B-representations. Concerning unitary representations, we know that, whenever the representation space is separable, the weak measurability of a repre- sentation necessarily leads to the strong continuity. We proceed as before to reduce our discussions on measurable B-representations to this result of the unitary representation. For that purpose the following lemma is useful, the devices of whose proof are due to E. Hopf [4]. LEMMA 2. Let { 8, T(a)} be strongly measurable. Then f0(T(a)x0) ~2da=0 implies f0(x0)= 0. PROOF. Owing to a lemma of Pettis, for a null set Ms,, `~xo= [T(a)x0 I a Mxo] is separable, where we may assume e Mxo. Let x1, x2, •• be a countable dense subset ofxo. Now consider the closed subspace of 8 which consists of all elements x such that f0(T(a)x)= 0 holds for a Nx, where Nx means a suitable null set. Clearly this subspace contains all T(a)x0 for a e G and thus a f ortiorixo. Putting N=UNxn, we see that is a null set and that f0(T(a)x)=0 for a N, n=1 irrespective of x e,. Hence, putting xo for x in this equality and then replacing xo by T(b)x0 for any b Mxo, we have f0(T(ab)x0)= 0 for a N. On the other hand, it follows "immediately from the boundedness of Haar measure that there exists an element c such that c N, If(x) I sup If If =IIxJI for all xe 3. fe~,fo f Representations of a compact group on a Banach space. 243 c 1 M 0. Hence we have f o(x0)= fo(T(cc 1)x0)= 0, as we wished to prove. THEOREM9. If is separable, the weakly measurable B-repre- sentation {3, T(a) } is strongly continuous. PROOF. This theorem holds for a finite-dimensional case. Applying the consideration similar to the one used in the proof of Theorem 2,, we see that it suffices to show the complete decomposability of {8, T(a)}. We first prove : a) Let {, U(a)} be the algebraic unitary representation whichh is obtained from {8, T(a)} by the procedures used in Proposition A. (as is easily seen, this is possible). Then {, U(a)} is a (continuous) unitary representation. b) If (f, f) = If (T(a)i) Izda=0, then f(s)=0. PROOF OF a). It is clear that the separability of 3 * implies thatt of ; besides (U(a)f,`) =J f(T(a_lb))g( T(b)&)db is measurable. Hence {, U(a)} is a continuous representation, ac~ cording to a known result for the unitary representation. PROOF OF b), Since is separable because of the separability of 3*, {8, T(a)} is strongly measurable and so Lemma 2 can be applied. Now, as is easily verified, the above two assertions enable us to carry over the proof of Proposition A and the proof of iii) -- i) of Theorem 2 to {3, T(a) } in question. Thus, if we follow this, we get the complete decomposability of {3, T(a)}, which completes the proof. In the above theorem the separability of is assumed. The author cannot determine whether this assumption may be replaced by a more weakened one such as the separability of 3. However we. shall show later that this is the case when {3, T(a)} is the conjugate representation of some representation. DEFINITION4. Let be a Banach space. If a subset of satisfies the following condition, is called strictly total: 244 K. SHIGA Obviously a strictly total set is total. We have examples of such sets : 1) 3, as a subspace of 2) C(G), as a subspace of L°°(G), which is conjugate to Ll(G), 3) The subspace in the conjugate space of C(G), consistng of all Radon measures {dµ} such that dµ=f(x)dx (f(x) e C(G)). DEFINITION 5. A bounded algebraic B-represeptation {8, T(a)} is called s-almost weakly-continuous in the case where is strictly total. In order to make the position of the following theorem clear, we first give a general property of almost strongly-continuous representa- tions. PROPOSITION4. An almost strongly-continuous representation {T(a)} as necessarily almost weakly-continuous. PROOF. By Definition 2 T(a)} is continuously imbedded onto a strongly continuous representation {C, S(a)}. Denote by ?v this linear univalent mapping of n2 onto ~, and by its adjoint mapping of C~k into 5Th. Observe that, for every F e ~* rt*(F)(T(a)x) is con- tinuous for each x e J2, i. e. t'*(C~.") ~ SJZ~. This is. immediately seen from the strong continuity of {C~,S(a)} and ~'~'(F)(T(a)x)=F(~y(T(a)x)) (from the definition of !E*) =F(S(a)!p-(x)) (from Definition 2) . Now if !V*(F)(x)=0 for all Fe C~*, then F(!v(x))=0, and so v(x)=0, x=0. This means that '*(C~*) is total, and thus SJ2~ is total. This completes the proof. THEOREM 10. Assume that {3, T(a)} is a strongly measurable B-representation. If {23, T(a)} is almost weakly-continuous, it contains a dense invariynt subspace C~such that {~, T(a)} is an almost strongly- continuous representation. I}' {~, T(a)} is further assumed to be s- almost weakly-continuous, {8, T(a)} becomes strongly continuous. PROOF OF THE FIRST ASSERTION. Consider the total closed sub- space 8j stated in Definition 3. For f, g e we define an inner product by (f, g) = f( T(a)) g( T(a)-) da. Then, by the repetition of the same arguments as in Proposition A, Representations of a compact group on a Banach space. 245 we can conclude that there exists a closed subspace C of (3)* such that is invariant under the operator induced by T *(a) for every a e G (this operator is denoted by the same notation) and {C~,T **(a)} is strongly continuous. The elements F~,g of C~corresponding to e, g e that is, defined by F~,g(f )= f(T(a)) g(T(a)) da (f Ec) (8) generate a dense invariant subspace in ~. Put x ,g= g(T(a)f) T(a)da. (9) From the assumption this Bochner integral exists. (8) and (9) yield f(x,g)=F~,g(f) for all f e~ . Since is total, x, g is uniquely determined by the above relation. From this, putting for the subspace generated by x g ( e 8, ge e), we know that l is an invariant subspace of {8, T(a)}, and the map- ping ' defined by F(x,g)=F~,g gives rise to a univalent linear mapping of onto ~. Now (x, g) It= Sup If(x,g) I -_ g(T(a)) f(T(a))da=0. This is written as g f (T(a)i) T(a) da =0 for all e , g e . 246 K. Sxrcn But ~~ being total, it results f j?T(a)) ~T(a)~da=0 for all ~ e ~ . Here applying f, we have f If(T(a)~) Jzda=0 for all ~ e ~ , whence by Lemma 2 we obtain f=0. This completes the proof. PROOF OF THE SECOND ASSERTION. Suppose that {~, T(a)} 1S strongly measurable and s•almost weakly-continuous. When we apply the above proof to {8, T(a)}, by Definition 4 we have the equality sign in (10), and so C~ may be identified with L, from which the strong continuity of {3, T(a)} follows. REMARK. The second part of this theorem is a strengthened form of Theorem 2, 3)-~2). COROLLARY. Suppose that {8, T(a)} is a weakly-continuous B- yepresentation and {3*, T'`(a)} is strongly measurable. Then T *(a)} is strongly continuous. This follows from the fact that {t7;'', T*(a)} is s•almost weakly- continuous because ~ is strictly total in 3*`~. Now, in order to derive some consequences from this corollary, we need a following result which is a slight generalization of Pettis' lemma [11]. LEMMA 3. Let 8 be a Banaeh space and C~ be a strictly total subspace in Let x(a) be a mapping of measure space (2, B, yn) into 3 such that x(a) is almost sepayably-valued. If f(x(a)) is a measurable function on .sz for every f e Cam,then x(a) is strongly measurable. PROOF17~ We shall first prove the measurability of IIx(a) II• Without losing generality, we may assume x(.Q)={x(a); ae,s2} is separable. Let {x}n_ {x(a)}nbe a countable dense set in x(.iz). Cam, being strictly total, contains elements {fnm},such that 17) See Hille [3], p. 36. II x(a) I(-- 1- -2e, Representations of a compact group on a Banach space. 247 Ilfn,mIHl, 11xn n na(xn) i 11xn - j_. m Put q'r(a) = SUP, Ifn, m(x(a))1. Then by the measurability of fn,m(X())a we see that q(c) is also measurable. For any x(a) and for any given E > 0 there exists an xn such that IIx(a)-xn II ~m(X) ~ Ifn,m(x(a)) I fn,rn(Xn)I - I fn,m(x(a)-xn) I m and so m(a) ~' 1~IIx(a)II~m(a). '~ m Hence we have supm q m(a)=11x(a) II, which means the measurability of 1I x(a) !I. Now the strong measurability of x(a) follows immediately from this result by the use of the familiar argument. We can now show that there exist some remarkable relations between {8, T(a)} and T'~(a)} under the assumption on the separ- ability of 8*. THEOREM 11. Let {8, T(a)} be a weakly measurable B-representa- tion. If is separable, then {, T (a)} is strongly continuous. PROOF. From Lemma 3 it results that {, T*(a)} is strongly measurable since is strictly total in 3K; on the other hand, Theorem 9 shows the strong continuity of {93, T(a)} under our assumption. Thus Corollary of Theorem 10 can be applied, which yields the desired result. COROLLARY1. If `` is separable, the weak measurability of {8, T(a)} implies the strong continuities of {' , T(a)} and {3*, Te(a)}. The following corollary should be compared with Theorem 9. COROLLARY2. Assume that a weakly measurable B-representation {3, T(a)} is the conjugate representation of a certain representation. Then {, T(a)} is strongly continuous whenever 3 is separable. THEOREM12. If is separable, the strong continuity of {33, 248 K. SnIGA T(a)} means the same property of {, T(a)} and vice versa . PROOF. If {, T (a) } is strongly continuous, it is of course weakly measurable, so that by Theorem 11 {, T *(a)} is strongly continuous . The converse part follows from Theorem 5, since 8 is separable. 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