A characterization of subspaces of weakly compactly ... - CiteSeerX

A characterization of subspaces of weakly compactly generated Banach spaces M. Fabian∗, V. Montesinos,†and V. Zizler

‡

Abstract We prove that a Banach space X is a subspace of a weakly compactly generated Banach space if and only if, for every ε > 0, X can be covered by a countable collection of bounded closed convex symmetric sets the weak∗ closure in X ∗∗ of each of them lies within the distance ε from X. As a corollary, we give a new, short functional-analytic proof to the known result that a continuous image of an Eberlein compact is Eberlein. ∗ †

ˇ 201/01/1198 (Czech Republic). Supported by grants AV 1019003, 1019301, and GACR Supported in part by Project pb96-0758 (Spain), Project BFM2002-01423 and by the

Universidad Polit´ecnica de Valencia, ‡ Supported by NSERC 7926 (Canada). Key words: weak compact generating, Eberlein compacts, projectional resolutions. AMS Subject Classification: 46B20, 46B03.

1

1

Introduction

A Banach space X is called weakly compactly generated (WCG, in short) if it contains a weakly compact set whose linear span is dense in X. A Banach space C(K) of continuous functions on a compact space K is WCG if and only if K is an Eberlein compact, i.e., K is homeomorphic to a weakly compact set in some c0 (Γ) [1]. Any continuous image of an Eberlein compact is an Eberlein compact [2]. Thus a Banach space is a subspace of a WCG space if and only if its dual unit ball BX ∗ in its weak∗ topology is an Eberlein compact. However, not all subspaces of WCG Banach spaces are WCG [15], [4, Section 1.6]. The purpose of this note is to characterize subspaces of WCG spaces via a covering property involving ε−weakly compact sets. Given ε ≥ 0, we say that a subset M of a Banach space X is ε−weakly compact if it is bounded ∗

and its weak∗ closure M in the second dual X ∗∗ lies in X + εBX ∗∗ . Note that 0−weak compactness coincides with the relative weak compactness. Theorem 1 A Banach space X is a subspace of a WCG Banach space if and only if for every ε > 0 the space X can be covered by a countable family Bε of bounded, convex, symmetric, and ε−weakly compact sets.

2

Remarks

(i) After this note had been submitted for publication, it has been proved in [7] that the convex hull of an ε−weakly compact set is 2ε−weakly compact. Thus the assumption of the convexity and symmetry of the sets in Theorem 1 can be dropped. 2

(ii) Let X be a WCG space with a weakly compact convex symmetric set L ⊂ X witnessing for this. Then the families Bε = {nL + εBX ; n ∈ IN}, ε > 0, clearly satisfy the condition in Theorem 1. On the other hand, assume that L ⊂ X is a bounded set such that X =

S∞

n=1

(nL + εBX ), with some fixed

ε > 0, and assume that the sets nL + εBX are 1−weakly compact for every n ∈ IN. Then, necessarily, L is relatively weakly compact. Indeed, take any ∗

∗

∗

x∗∗ ∈ L . For every n ∈ IN we have nx∗∗ ∈ nL ⊂ nL + εBX ⊂ X + BX ∗∗ , and hence x∗∗ ∈ X + n1 BX ∗∗ . Therefore x∗∗ ∈ X and so L must be relatively weakly compact. (iii) Let M ⊂ X be an ε−weakly compact set. It is natural to ask if there is a constant k > 0 such that M is a subset of C + kεBX , with a suitable weakly compact set C ⊂ X. The answer is negative. One possible way to see that is the following: Let X be a subspace of a WCG Banach space which is not WCG. We claim that for every k ∈ IN and every ∆ > 0 there exist a set M ⊂ BX and ε ∈ (0, ∆) such that M

w∗

⊂ X + εBX ∗∗ and yet

M \ (C + kεBX ) 6= ∅ for every weakly compact set C ⊂ X. Indeed, let Bε , ε > 0 be the covering from Theorem 1. For every p ∈ IN let {Mn,p ; n ∈ IN} be an enumeration of the family {B ∩ BX ; B ∈ B1/p }. Assume that our claim is false. Then there exist k ∈ IN and ∆ > 0 such that whenever 0 < ε < ∆ and M ⊂ BX is an ε−weakly compact set, then M ⊂ C + kεBX for a suitable weakly compact set C ⊂ X. Thus, in particular, for every n, p ∈ IN, with p >

1 , ∆

we can find a weakly compact set Cn,p ⊂ X such that

Mn,p ⊂ Cn,p + kp BX . Clearly, we may assume that every Cn,p lies in (1+k)BX . 1 Put C = { np Cn,p ; n, p ∈ IN, p >

1 } ∆

∪ {0}; this is a weakly compact set. It

remains to verify that C is linearly dense in X. So fix any x ∈ BX and any 3

ε ∈ (0, ∆). Take p ∈ IN so large that p > kε ; then p > that x ∈ Mn,p . Then dist (x, span C) ≤ dist (x, Cn,p )
0 and for every x ∈ X, there is B ∈ B such that x ∈ B and B ⊂ X + εBX ∗∗ . Indeed, if Bε , ε > 0, are the families from Theorem 1, then the (countable) ∗

family B = {B ; B ∈ B1/p , p ∈ IN} clearly satisfies the condition above. Conversely, let B be the family from our condition. Given ε > 0, let Bε0 ⊂ B be a subfamily such that X ⊂

S

Bε0 ⊂ X + εBX ∗∗ . Then the families Bε :=

{B ∩ X; B ∈ Bε0 }, ε > 0, do the job. (v) A Banach space X is called a Vaˇs´ ak space (or weakly countably determined space) if there is a countable family B of weak∗ compact sets of X ∗∗ so that for each pair of points x ∈ X and x∗∗ ∈ X ∗∗ \ X there is B ∈ B such that x ∈ B and x∗∗ 6∈ B [18], cf., e.g. [3], or [4]. The sets in the definition of Vaˇsa´k spaces can be taken convex and symmetric [18]). It follows from the condition in Remark (iv) that every subspace of a WCG space is a Vaˇsa´k space. Indeed, given a pair x ∈ X and x∗∗ ∈ X ∗∗ \ X, it suffices to consider ε > 0 such that dist (x∗∗ , X) > ε. An example of a Vaˇs´ak space that is not a subspace of a WCG space is in [16], cf. [4, Section 4.3].

4

3

Preliminary facts

Before starting on the proof of Theorem 1 we need to introduce some notation and state a few standard facts. We will provide proofs to some of them for the reader’s convenience. Given a dual pair hX, Y i of vector spaces, we denote by w(X, Y ) the topology on X of the pointwise convergence on the elements of Y . Let M be a w(Y, X)−bounded subset of Y . We define a convex function on X for x ∈ X by kxkM = sup{hx, yi; y ∈ M }. We will keep the notation standard in Banach space theory (cf., e.g. [6]). If Y ⊂ X ∗ is a subspace of (X ∗ , k · k), we will write k · kY instead of k · kBY ; note that this is a continuous seminorm on X. The space Y is called norming if k · kY is an equivalent norm on X. The space Y is said to be 1-norming in the case when k · kY = k · k. It follows that Y is norming if and only if BY if and only if BX if and only if BY

w(X,Y ) w∗

w∗

contains λBX ∗ for some λ > 0,

(= B(X,k·kY ) ) is bounded. Likewise, Y is 1−norming

= BX ∗ , if and only if B(X,k·k) is w(X, Y )−closed, i.e.,

k · k is w(X, Y )-lower semicontinuous. Obviously, any norming subspace Y of X ∗ is 1−norming for k · kY , and this norm is w(X, Y )-lower semicontinuous. Observe also that kx∗ kY = kx∗ k for every x∗ ∈ Y . Proposition 2 (i) The kernel of every x∗∗ ∈ X ∗∗ \ X is a norming hyperplane of X ∗ (see, e.g., [6, Exercise 3.43]. (ii) If X is a subspace of a Banach space Z and Y ⊂ X ∗ a norming subspace, then q −1 (Y ) is a norming subspace of Z ∗ , where q : Z ∗ → X ∗ is the canonical quotient mapping (see, e.g. [8]).

5

Proposition 3 Let X be a Banach space. Consider x∗∗ ∈ X ∗∗ and denote by H ⊂ X ∗ its kernel. Then 1 ∗∗ kx kB w∗ ≤ dist(x∗∗ , X) ≤ 2kx∗∗ kB w∗ H H 2 Proof Fix any x ∈ X and any x∗ ∈ BH

w∗

(1)

. Then for every h∗ ∈ BH we have

hx∗∗ , x∗ i = hx∗∗ , x∗ − h∗ i = hx∗∗ − x, x∗ − h∗ i + hx, x∗ − h∗ i ≤ 2kx∗∗ − xk + hx, x∗ − h∗ i. ∗

Hence hx∗∗ , x∗ i ≤ 2kx∗∗ − xk for every x ∈ X and every x∗ ∈ BH . Thus the left hand side inequality is proved. It remains to prove the right hand side inequality in (1). We may assume that kx∗∗ k = 1. Put H ⊥ = {u∗∗ ∈ X ∗∗ ; hu∗∗ , h∗ i = 0 ∀h∗ ∈ H}. Using the canonical isometry between H ∗ and X ∗∗ /H ⊥ , we get kxkH = dist (x, H ⊥ ) for every x ∈ X. Then dist (SX , H ⊥ )kxk ≤ kxkH ≤ kxk for every x ∈ X.

(2)

The parallel hyperplane lemma (see, e.g., [6, Exercise 3.1]) yields min kx∗∗ ± xk ≤ 2kxkH

for every x ∈ SX .

Thus dist(x∗∗ , SX ) ≤ 2 inf {kxkH ; x ∈ SX } = 2 inf {dist(x, H ⊥ ); x ∈ SX } = 2 dist(SX , H ⊥ ).

(3)

Due to the assumption kx∗∗ k = 1, the bidual form of (2) gives that dist(SX , H ⊥ ) ≤ kx∗∗ kB 6

w∗ H

.

(4)

This estimate, inequality (3) and the fact that dist (x∗∗ , SX ) ≥ dist (x∗∗ , X), gives the right hand side inequality in (1). Lemma 4 Let Z be a Banach space, C ⊂ Z a non-empty convex set and z ∗∗ ∈ C

w∗

. Then dist(z ∗∗ , C) ≤ 2 dist(z ∗∗ , Z).

Proof Take any δ such that dist(z ∗∗ , Z) < δ and find z ∈ Z such that kz ∗∗ − zk < δ. Then z ∈ C C + δBZ

k·k

w∗

+ δBZ ∗∗ ⊂ C + δBZ

w∗

. It follows that z ∈

. Therefore, given ε > 0 there exists c ∈ C and b ∈ BZ such

that kz − c − δbk < ε; so kz − ck ≤ ε + δ. Finally we get kz ∗∗ − ck = kz ∗∗ − z + z − ck ≤ 2δ + ε, and then dist(z ∗∗ , C) ≤ 2δ + ε. As ε > 0 was arbitrary, dist(z ∗∗ , C) ≤ 2δ. Therefore dist(z ∗∗ , C) ≤ 2dist(z ∗∗ , Z).

Proposition 5 Let (Z, k · k) be a Banach space, X a subspace of Z, ε > 0, and let M be an ε−weakly compact subset of Z. Then M ∩ X is a 2ε−weakly compact subset of X. Proof Inject canonically X ∗∗ into Z ∗∗ . Let z ∗∗ be any element of the weak∗ closure of the set M ∩ X in the space X ∗∗ . Then z ∗∗ belongs to the weak∗ closure of the set M in the space Z ∗∗ . Then it is enough to use Lemma 4 with C := X to get dist(z ∗∗ , X) ≤ 2 dist(z ∗∗ , Z) ≤ 2ε. Lemma 6 Let (X, k·k) be a Banach space, ε ≥ 0, and let |·| be an equivalent norm X whose unit ball is ε−weakly compact (with respect to k · k). Let Y be a subspace of X ∗ and consider x∗ ∈ BY x∗ to Y is at most 2ε. 7

w∗

. Then the | · |−distance from

Proof Let us denote the distance in question by ∆. If ∆ = 0, we are done. Assume now that ∆ > 0. By [6, Proposition 2.8] find x∗∗ ∈ X ∗∗ such that |x∗∗ | = 1, x∗∗ ∈ Y ⊥ , and hx∗∗ , x∗ i = ∆. Recalling that x∗∗ ∈ B(X,|·|)

w∗

we

get, from the assumptions, that the k · k−distance from x∗∗ to X is at most ε. Then, by Proposition 3, kx∗∗ kB

w∗

H

x∗∗ , so kx∗∗ kB

4

w∗ Y

≤ 2ε, where H ⊂ X ∗ is the kernel of

≤ 2ε, too. Therefore ∆ = hx∗∗ , x∗ i ≤ 2ε.

Proof of Theorem 1

Assume that X is a subspace of a WCG Banach space (Z, k · k). Let K be a linearly dense and weakly compact subset of Z. By Krein’s theorem (see, e.g., [6, Theorem 3.58]), we may and do assume that K is convex and symmetric. For ε > 0 put Bε = {(nK + 2ε BZ ) ∩ X; n ∈ IN}. Since each set nK + 2ε BZ is 2ε −weakly compact in Z, Proposition 5 guarantees that the members of the family Bε are ε−weakly compact in X. The remaining properties of the families Bε stated in our theorem are obvious. Conversely, assume that the covering condition of Theorem 1 holds. We will show that BX ∗ in its weak∗ topology is an Eberlein compact. This will imply that C(BX ∗ ) is a WCG space [1]) which contains X, and the proof of the sufficiency will be finished. First of all, we will prove the following Claim: The covering condition in our theorem implies the existence of a symmetric and linearly dense set Γ ⊂ BX such that for every ε > 0 there are 8

sets Γεm ⊂ Γ, m ∈ IN, with

S∞

m=1

Γεm = Γ, such that for every x∗ ∈ BX ∗ and

for every m ∈ IN, #{γ ∈ Γεm ; |hγ, x∗ i| > ε} < ℵ0 . Once having this proved, we can define a mapping T : X ∗ → IRΓ by T x∗ = (hγ, x∗ i; γ ∈ Γ), x∗ ∈ X ∗ , which is injective and weak∗ to pointwise continuous (IRΓ is endowed with its pointwise topology). We observe that each T x∗ is countably supported in IRΓ . Thus the compact set T (BX ∗ ) satisfies the assumptions of (the easier part of) [10, Theorem 2.9]. Therefore, this compact, and hence also (BX ∗ , w∗ ), is an Eberlein compact. We will prove the Claim by induction on the density of X. If X is separable, we can take for Γ any countable symmetric dense subset of BX . Further, let ℵ be any uncountable cardinal and assume that we have verified the Claim for all Banach spaces whose density is less than ℵ. Now, let X be a Banach space of density ℵ and assume that we have the families Bε , ε > 0 as in Theorem 1. A simple gymnastics with these families produces sets Mn,p ⊂ BX , n, p ∈ IN, which are convex, symmetric, p1 −weakly compact, have nonempty interior, and

S∞

n=1

Mn,p = BX for every p ∈ IN.

As mentioned in Remark (v) following Theorem 1, the space X is Vaˇs´ak. By, e.g., [4, Proposition 7.2.1], X has a projectional generator (W, Φ), with W = X ∗ , see [4, Definition 6.1.6 and Remark 6.1.8]. For n, p ∈ IN, let k · kn,p be the Minkowski functional of the set Mn,p . Clearly, each k · kn,p is an equivalent norm on X. Thus, using the proof of [4, Proposition 6.1.10], we can find a “long sequence” (Pα ; α ≤ µ) of projections on X which form simultaneously a projectional resolution of the identity on X with respect to the original norm k · k as well as with respect to each norm k · kn,p , n, p ∈ IN, 9

see, e.g., [4, Definition 6.1.5]; here µ means the first ordinal whose cardinality is equal to the density of X. Then for every α < µ and for every n, p ∈ IN we have kPα kn,p = 1, and so Pα (Mn,p ) ⊂ Mn,p . Fix any α < µ and put Qα = Pα+1 − Pα . Define Bεα = {B ∩ Qα (X); B ∈ Bε/2 },

ε > 0.

These families, in the subspace Qα (X), satisfy the condition in Theorem 1 (see Proposition 5 for the last one). Now, since the subspace Qα (X) has density less than ℵ, there exists, by the induction assumption, a set Γα ⊂ BQα (X) which is symmetric, linearly dense in Qα (X), and has the property that for every ε > 0 there are sets Γα,ε m ⊂ Γα , m ∈ IN, with

S∞

m=1

Γα,ε m = Γα ,

such that ∗ ∀y ∗ ∈ B(Qα X)∗ ∀m ∈ IN #{γ ∈ Γα,ε m ; |hγ, y i| > ε} < ℵ0 .

Put Γ =

S α 0. Find p ∈ IN so large that p > 6ε . Put then Γεm,n =

[ α ε} is finite for every x∗ ∈ BX ∗ . So fix an arbitrary x∗ ∈ BX ∗ . We 10

Subclaim: For every α ≤ µ #{γ ∈ Γεm,n ; |hγ, Pα∗ x∗ i| > ε} < ℵ0 .

(6)

Due to the symmetry, it is enough to show that for every α ≤ µ #{γ ∈ Γεm,n ; hγ, Pα∗ x∗ i > ε} < ℵ0 .

(7)

Having this proved and recalling that Pµ∗ x∗ = x∗ , the Claim will also be verified. The subclaim will be proved by induction over the ordinal α. (7) is true for α = 0 since P0∗ = 0. Consider any ordinal β ≤ µ and assume that (7) was verified for every α < β. Denote S = {α < µ; hγ, Pβ∗ x∗ i > ε for some γ ∈ Γα,ε m ∩ Mn,p }. Assume that S is infinite. Then there exist ordinals α1 < α2 < · · · < µ and γi ∈ Γαmi ,ε ∩ Mn,p such that hγi , Pβ∗ x∗ i > ε for every i ∈ IN. We observe that β > αi for every i ∈ IN. Indeed, β ≤ αi would imply that hγi , Pβ∗ x∗ i = hPβ γi , x∗ i = 0 as γi ∈ Qαi (X) and Pβ ◦ Qαi = 0. Put λ = limi→∞ αi ; thus λ ≤ β. If λ < β, then we would have (ε < ) hγi , Pβ∗ x∗ i = hPβ γi , x∗ i = hPλ γi , x∗ i = hγi , Pλ∗ x∗ i for every i ∈ IN, which contradicts (7), valid for α := λ (indeed, i 6= j implies αi 6= αj and hence γi 6= γj ; thus the set {γ1 , γ2 , . . .} is infinite). Therefore λ = β. Now, we will show that there exists j ∈ IN such that 6 sup hMn,p , Pβ∗ x∗ − Pα∗j x∗ i < . p 11

(8)

Once having this proved, then for all i ∈ IN, with i > j, we have (ε < ) hγi , Pβ∗ x∗ i = hγi , Pβ∗ x∗ − Pα∗j x∗ i ≤ ≤ sup hMn,p , Pβ∗ x∗ − Pα∗j x∗ i