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SUPPORT FUNCTIONALS AND THEIR RELATION
TO THE RADON-NIKODYM PROPERTY
I. SADEQI
Received 26 May 2002
In this paper, we examine the Radon-Nikodym property and its relation to the Bishop-Phelps theorem for complex Banach spaces. We also show that the Radon-Nikodym property implies the Bishop-Phelps property in the complex case.
2000 Mathematics Subject Classification: 46E99, 46J10.
1. Introduction. Let X be a complex Banach space and let C be a closed convex subset ofX. The set ofsupport points ofC, written as suppC, is the collection of all pointsz∈Cfor which there exists nontrivialf∈X∗such that supx∈C|f (x)| = |f (z)|.
Such anfis called support functional. The pointz∈suppCis called astrongly exposed point ofC if for all sequences{zn} ⊂C, limn→∞(Ref (zn))=supCRe(f )implies that
zn→z, where Re denotes the real part.
In this paper, we will show that the unit ball of an infinite-dimensional function algebra has no strongly exposed points. Lomonosov [4] constructed a closed, bounded, and convex subsetC of a complex Banach space such that the set of support points ofCis empty. This means that the Bishop-Phelps theorem fails to hold in the complex case. We show below that for Hardy spaces, the Bishop-Phelps theorem does hold.
Bourgain [1] proved that ifXis a real Banach space, then theRadon-Nikodym property (RNP) and theBishop-Phelps property (BPP) are equivalent. The precise definitions are given below.
It is natural to ask whether this equivalence remains true in the complex case. In this paper, we show that, appropriately defined, it does indeed hold for complex Banach spaces. Recall that a Banach spaceX is said to have the RNP, provided that for every measure space(Ω,Σ, µ)withµ(Ω) <∞, and everyµ-continuous measureT :Σ→Xof finite variation, there exists a Bochner integrable functionf:Ω→Xsuch thatT (E)=
Ef dµfor everyE∈Σ.
LetXand Y be Banach spaces and letL(X, Y )be the Banach space of all bounded linear operators fromX intoY. Suppose thatT ∈L(X, Y )and C is a nonempty and bounded subset of X, then we define T(C):=sup{T x:x∈C}. A Banach space X is said to have the BPP if for any nonempty, bounded, and closed subsetC ofX, any Banach spaceY, and anyT∈L(X, Y ), there is an approximating sequence(Tn)in
L(X, Y )for which each(Tn)achieves its max norm(Tn)(C)onC.
open-neighborhood ofx. In the final part of this paper, we discuss uncountability of the set of normalized support functionals, a question posed by L. Zajicek in 1999.
2. The RNP for complex Banach spaces. Bourgain proved in [1] that if the real Ba-nach spaceXhas the RNP and ifCis a closed, convex, and bounded subset ofX, then the set of support functionals that strongly expose some point ofC is dense in X∗. Suppose that the complex Banach spaceXhas the RNP and thatCis a bounded, closed, and convex subset ofX. Put
H:=eiθx: 0≤θ <2π , x∈C. (2.1)
LetB denote the closed convex hull ofH. Then, as cited above, the set of real parts of linear functionals which strongly expose some point ofBforms a dense subset of Xr∗. HereXr is the underlying real Banach space. By the standard isometryf→Ref
betweenX∗ and Xr∗, these are the real parts of a dense subset of X∗. The strongly
exposed points ofBare contained inHsinceHis closed, so by the theorem of Phelps,
sup|f|(C)=sup Ref (H) (2.2)
and the support functionals are dense in the complex case. Therefore, using the two theorems of Phelps and Bourgain, the RNP implies the Bishop-Phelps theorem in the complex case. If we show that the RNP implies the BPP in the complex case, the Bishop-Phelps theorem clearly holds for complex Banach spaces with the RNP without recourse to the theorems of Bourgain [1] and Phelps [6].
Definition2.1. LetB be a nonempty, bounded, closed, and convex subset of the
complex Banach spaceX. Let Y be a Banach space andT ∈L(X, Y ). Say thatT is an F-strongly exposing operator for the setBif there exists some pointx∈Bdepending onT such that every sequence(xn)⊂Bsatisfying
supT(B)=lim
n→∞T
xn (2.3)
has a subsequence converging toαxfor some complex numberαwith|α| =1.
IfX is a real Banach space, then the definition of anF-strongly exposing operator becomes Bourgain’s definition of anR-strongly exposing operator, as follows.
LetB be a nonempty, bounded, closed, and convex subset of the complex Banach spaceX. LetY be a Banach space andT ∈L(X, Y ).T is called anR-strongly exposing operator for the setBif there exists some pointx∈Bdepending onTsuch that every sequence(xn)⊂Bsatisfying (2.3) has a subsequence converging toxor−x[1].
Theorem2.2(Bourgain). LetBbe a nonempty, bounded, closed, and convex subset
of the real Banach spaceX. Assume that every nonempty subset ofBis dentable. Then for any Banach spaceY, the set of all R-strongly exposing operatorsT∈L(X, Y )for the setBis a dense subset ofL(X, Y ).
Banach space and let C be a circled, bounded, closed, and convex subset of X (i.e., αC⊆C if|α| =1). Suppose thatT achieves its max normT(C)onC. Then there existsx0inCsuch that
supT(C)=Tx0. (2.4)
Chooseαn=i; it is clear that
lim
n→∞T
αnx0=supT(C), (2.5)
and(αnx0)has no subsequence converging tox0or−x0. So the circled subsets ofX
have noR-strongly exposing operator. Also letXr andYr be the underlying real part
of the complex Banach spacesXandY, respectively. SinceL(Xr, Yr)is not isomorphic
toL(X, Y )in general, we cannot directly use Bourgain’s definition in the complex case.
Theorem2.3. LetXbe a complex Banach space with the RNP. ThenXpossesses the
BPP.
Proof. LetCbe a nonempty, closed, convex, and bounded subset ofX. LetY be a
complex Banach space. We must show that forT∈L(X, Y ), there is an approximating sequence(Tn)for which each(Tn)achieves its max norm onC. Forn∈N, define
Kn:=
T∈L(X, Y ):∃ξ≥0,∃t∈X, S(T , ξ)⊂ |α|=1
Nαt, n−1 , (2.6)
whereS(T , ξ):= {x∈C:T x ≥ T(C)−ξ}. SinceX has the RNP, every nonempty and bounded subset ofXis dentable, soKnis dense inL(x, y)(see [1]). IfT∈Knfor
eachn∈N, then there exists a sequence(xn)inCsuch that limn→∞T xn = T(C).
There exists alsot∈X such that(xn)is contained inF := {αt:|α| =1}. It follows
thatxn=αnt, and there is a subsequence of(xn)converging toα0tfor someα0with
|α0| =1. Therefore,T is anF-strongly exposing operator and achieves its max norm onC atα0t. It follows thatT is anF-strongly exposing operator if and only ifT∈Kn
for everyn∈N. Put K:=n∞=1Kn. Since anyKn is dense inL(X, Y )if we show that
anyKn is open, we conclude thatKis dense inL(X, Y ). LetT∈Kn and suppose that
F∈L(X, Y )withF−T ≤ξ/3. It is easy to see thatS(F , ξ/3)⊆S(T , ξ/3), soF ∈Kn
andKnis open inL(X, Y ). The fact that eachT∈Kis anF-strongly exposing operator
and achieves its max norm onCcompletes the proof.
Corollary2.4. The RNP implies the Bishop-Phelps theorem in the complex case.
Proof. Since the RNP implies the BPP in the complex case, the proof is clear.
That the BPP implies the RNP can be easily checked through Bourgain’s proof in [1], so we have the following result.
Theorem2.5. The RNP and the BPP are equivalent for complex Banach spaces.
We are ready to examine the RNP for some complex Banach spaces. Letτ:= {eiθ:
theorem is true for them. The complex Bishop-Phelps theorem is still open forL1; how-ever, we can verify this theorem for some subspaces ofL1. LetC(τ)be the set of all continuous functions onτ and letC0denote the functions inC(τ)which are analytic with mean value zero. It is well known that
C(τ)/C0∗≈H1(τ). (2.7)
SinceH1(τ)is a separable space with a predual, it has the RNP and, hence, the Bishop-Phelps theorem is true forH1(τ)⊂L1(τ). Also, allHp(1≤p <∞)are separable and
have a predual. The following theorem is an immediate consequence of this discussion.
Theorem2.6. The Bishop-Phelps theorem is satisfied for the Hardy spacesHp(1≤ p <∞)in the complex case.
As mentioned above, a separable dual Banach space has the RNP. SinceH∞is not separable, it is still unknown whether the Bishop-Phelps theorem is true forH∞. Hens-gen (see [3]) proved that the unit ball ofH∞has no strongly exposed points. Therefore, H∞ is guaranteed to lack both the RNP and the BPP. In the following, we will show that the unit ball of an infinite-dimensional uniform algebra has no strongly exposed point. As a result, such spaces do not have the RNP. So it is natural to ask whether the Bishop-Phelps theorem holds for infinite-dimensional uniform algebras.
LetXbe a nonempty set and letKbe a normed linear algebra. We denote by∞(X, K) the normed linear algebra of all bounded mappings ofXintoKwith pointwise addition and scalar multiplication and with the uniform norm. For a bounded mappingf, the uniform norm is defined by
f∞=supf (x):x∈X. (2.8)
A uniform algebra of functions onX is a subalgebra of the Banach algebra∞(X, F ), whereF is either the real or complex numbers. Given a nonempty topological space X, C(X, K)denotes the linear space of all continuous mappings ofXintoKwith point-wise addition and scalar multiplication. WhenXis compact,C(X, K)is a closed linear subspace of∞(X, K), and in particular,C(X, F )is a uniform algebra of functions. The notationC(X, F )is abbreviated toC(X).
Lemma2.7. LetXbe a compact Hausdorff space such that C(X)is
infinite-dimen-sional. Then the unit ball ofC(X)has no strongly exposed point.
Proof. IfXadmits no Baire diffuse measure (a nonnegative measureµonXis said
to be diffuse ifµ(V ) >0 for every nonempty open subsetVofX), then the unit ball of C(X)contains no exposed point (see [5]). So the proof is clear in this case. LetXadmit a diffuse measure and suppose thatλ∈C(X)∗exposesf∈U, whereUis the unit ball of C(X). Then|f (t)| =1(t∈X). SinceXadmits a diffuse measure andf =1, a result of Eberlein’s guarantees that accumulation points can be approximated by sequences (see [7]). We can assume that there exists a sequence(tn)⊂X such thatf (tn)→1.
Choose pairwise disjoint open setsUninXsuch thattn∈Un. Also choosehn∈C(X)
andgn(t)→1 for eacht∈X. Hencegnf (t)→f (t), t∈X. Also by theHahn-Banach
theorem, it is easy to see thatλmust be of the form
λ(g)=
X
gf dµ.¯ (2.9)
It is clear thatλ(gnf )→λ(f ), butgnf−f = hnf ≥1. Sofcannot be a strongly
exposed point. Also since any uniform algebra is a commutativeB∗-algebra, by the Gelfand-Naimark theorem,Ais an isometric isomorphism ofC(∆A), where∆A is the
maximal ideal space ofA. Therefore, as cited above,Ahas no strongly exposed point, and we have the following result.
Theorem2.8. The unit ball of an infinite-dimensionalC∗-algebra has no strongly
exposed point.
It is well known that ifX is a complex Banach space with the RNP, then the set of support functionals that expose some points of the unit ball ofX is norm dense in X∗[1].
Corollary2.9. An infinite-dimensonal separableC∗-algebra possesses neither the
RNP nor the BPP; hence it has no predual.
3. The set of normalized support functionals
Problem3.1. Suppose thatXis a real Banach space with dimX >1 andC⊆Xis a
bounded, closed, and convex subset ofX. Is the set of normalized support functionals Σ(C)an uncountable subset ofSX∗? (Due to L. Zajicek).
Phelps has made the following observations. Suppose thatΣ(C)is countable; then it has no interior point, soSX∗\Σ(C)is a denseGδset inSX∗. SinceΣ(C)is dense inSX∗,
from the Bishop-Phelps theorem, we conclude thatX∗ is separable. IfX is reflexive, then any linear functional ofSX∗ supports C, which is a contradiction. SinceΣ(C)is countable,Xmust be a nonreflexive space with a separable dual space. AlsoC must have an empty interior because otherwise we may assume that 0∈intC, and since dimX >1, it is possible to have a two-dimensional subspaceMofX. So iff∈SM∗, then
it supportsM∩C. By the Hahn-Banach theorem, there is an extentionfoff inΣ(C). That is, there is one-to-one map from the uncountable setΣ(M∩C)intoΣ(C), which is impossible.
Theorem3.2. LetXbe a weakly sequentially complete (w.s.c.) Banach space and let
Cbe a closed, convex, and bounded subset ofX. ThenΣ(C)is uncountable.
Proof. If Σ(C) is countable, since Σ(C) is dense in SX∗, SX∗ is separable. By
Dounford-Pettis theorem (see [2]),X∗ possesses the RNP. Let(xn)be a bounded
se-quence inX. ThenY, the closed linear span of(xn), is a separable subspace ofX. Since
X∗has the RNP,Y∗is separable. By a classical result of Banach,(xn)has a weak Cauchy
subsequence inY (again denoted by(xn)), which is also a weak Cauchy sequence inX.
SinceX is w.s.c., then(xn)is a weakly convergent sequence, therefore, any bounded
sequence(xn)is a weakly convergent sequence inX, and soXis a reflexive space. Thus
Remark 3.3. If any nonempty subset of a closed, convex, and bounded set C is
dentable, then by a result of Bourgain (see [6]), for such a set C, the set of support functionals isGσdense inX∗. And by a classical theorem that the Banach spaceXhas
no countableGσdense subset, we conclude thatΣ(C)is uncountable.
Acknowledgment. My sincerest thanks go to Professor R. Phelps for his
assis-tance.
References
[1] J. Bourgain,On the dentability and the Bishop-Phelps property, Israel J. Math.28(1977), 263–271.
[2] J. Diestel,Geometry of Banach Spaces, Springer-Verlag, Berlin, 1975.
[3] W. Hensgen,Exposed points in Lebesgue-Bochner and Hardy-Bochner, J. Math. Anal. Appl. 198(1996), 780–796.
[4] V. I. Lomonosov,A counterexample to the Bishop-Phelps theorem in complex spaces, Israel J. Math.115(2000), 25–28.
[5] R. R. Phelps,Extreme points in function algebras, Duke Math. J.32(1977), 267–278. [6] ,The Bishop-Phelps theorem in complex spaces: an open problem, Pure Appl. Math.
131(1991), 337–340.
[7] T. S. S. R. K. Rao,There are no denting points in the unit ball ofW C(K, X), Proc. Amer. Math. Soc.127(1999), no. 10, 2969–2973.
I. Sadeqi: Departement of Mathematics, Sahand University of Technology, P.O. Box 51335-1996, Tabriz, Iran
