On the topology of pointwise convergence on the ... - CARMA

<strong>On</strong> <strong>the</strong> <strong>topology</strong> <strong>of</strong> <strong>pointwise</strong> <strong>convergence</strong>on <strong>the</strong> boundaries <strong>of</strong> L 1 -predualsWarren B. MoorsDepartment <strong>of</strong> Ma<strong>the</strong>maticsThe University <strong>of</strong> AucklandAucklandNew ZealandDedicated to J ohn R. Giles

IntroductionWe shall say that a Banach space (X, ‖·‖) is an L 1 -predual ifX ∗ is isometric to L 1 (µ) for some suitable measure µ. Someexamples <strong>of</strong> L 1 -preduals include (C(K), ‖ · ‖ ∞ ), and moregenerally, <strong>the</strong> space <strong>of</strong> continuous affine functions on a Choquetsimplex endowed with <strong>the</strong> supremum norm. The o<strong>the</strong>rnotion we shall consider in this talk is that <strong>of</strong> a boundary.Specifically, for a non-trivial Banach space X over R we saythat a subset B <strong>of</strong> B X ∗, <strong>the</strong> closed unit ball <strong>of</strong> X ∗ , is aboundary, if for each x ∈ X <strong>the</strong>re exists a b ∗∈ B suchthat b ∗ (x) = ‖x‖. The prototypical example <strong>of</strong> a boundaryis Ext(B X ∗) - <strong>the</strong> set <strong>of</strong> all extreme points <strong>of</strong> B X ∗, but <strong>the</strong>reare many o<strong>the</strong>r interesting examples. In a recent paper byMoors and Reznichenko <strong>the</strong> authors investigated <strong>the</strong> <strong>topology</strong>on a Banach space X that is generated by Ext(B X ∗) and,more generally, <strong>the</strong> <strong>topology</strong> on X generated by an arbitrary

oundary <strong>of</strong> X. In this talk we continue this study.To be more precise we must first introduce some notation. Fora nonempty subset Y <strong>of</strong> <strong>the</strong> dual <strong>of</strong> a Banach space X we shalldenote by σ(X, Y ) <strong>the</strong> weakest linear <strong>topology</strong> on X thatmakes all <strong>the</strong> functionals from Y continuous. In <strong>the</strong> paperby MR <strong>the</strong>y showed that “for any compact Hausdorff spaceK, any countable subset {x n : n ∈ N} <strong>of</strong> C(K) and anyboundary B <strong>of</strong> (C(K), ‖ · ‖ ∞ ), <strong>the</strong> closure <strong>of</strong> {x n : n ∈ N}with respect to <strong>the</strong> σ(C(K), B) <strong>topology</strong> is separable withrespect to <strong>the</strong> <strong>topology</strong> generated by <strong>the</strong> norm”.In this talk we extend this result by showing that if (X, ‖ · ‖)is an L 1 -predual, B is any boundary <strong>of</strong> X and {x n : n ∈ N}is any subset <strong>of</strong> X <strong>the</strong>n <strong>the</strong> closure <strong>of</strong> {x n : n ∈ N} in <strong>the</strong>σ(X, B) <strong>topology</strong> is separable with respect to <strong>the</strong> <strong>topology</strong>generated by <strong>the</strong> norm whenever Ext(B X ∗) is weak ∗ Lindelöf.

Preliminary ResultsLet X be a topological space and let F be a family <strong>of</strong> nonempty,closed and separable subsets <strong>of</strong> X. Then F is rich if <strong>the</strong> followingtwo conditions are fulfilled:(i) for every separable subspace Y <strong>of</strong> X, <strong>the</strong>re exists aZ ∈ F such that Y ⊆ Z;(ii) for every increasing sequence (Z n : n ∈ N) in F,⋃n∈N Z n ∈ F.For any topological space X, <strong>the</strong> collection <strong>of</strong> all rich families<strong>of</strong> subsets forms a partially ordered set, under <strong>the</strong> binary relation<strong>of</strong> set inclusion. This partially ordered set has a greatestelement, namely,G X := {S ∈ 2 X : S is a closed and separable subset <strong>of</strong> X}.<strong>On</strong> <strong>the</strong> o<strong>the</strong>r hand, if X is a separable space, <strong>the</strong>n <strong>the</strong> partiallyordered set has a least element, namely, G ∅ := {X}.

The raison d’être for rich families is revealed next.Proposition 1 Suppose that X is a topological space. If{F n : n ∈ N} are rich families <strong>of</strong> X <strong>the</strong>n so is ⋂ n∈N F n.Throughout this talk we will be primarily working with Banachspaces and so a natural class <strong>of</strong> rich families, given a Banachspace X, is <strong>the</strong> family <strong>of</strong> all closed separable linear subspaces<strong>of</strong> X, which we denote by S X . There are however many o<strong>the</strong>rinteresting examples <strong>of</strong> rich families.Theorem 1 Let X be an L 1 -predual. Then <strong>the</strong> set <strong>of</strong> allclosed separable linear subspaces <strong>of</strong> X that are <strong>the</strong>mselvesL 1 -preduals forms a rich family.

Lemma 1 Let Y be a closed separable linear subspace <strong>of</strong> aBanach space X and suppose that L ⊆ Ext(B X ∗) is weak ∗Lindelöf. Then <strong>the</strong>re exists a closed separable linear subspaceZ <strong>of</strong> X, containing Y , such that for any l ∗ ∈ L and any x ∗ ,y ∗ ∈ B Z ∗ if l ∗ | Z = 1 2 (x∗ + y ∗ ) <strong>the</strong>n x ∗ | Y = y ∗ | Y .Using this Lemma we can obtain <strong>the</strong> following <strong>the</strong>orem.Theorem 2 Let X be a Banach space and let L ⊆ Ext(B X ∗)be a weak ∗ Lindelöf subset. Then <strong>the</strong> set <strong>of</strong> all Z in S X suchthat {l ∗ | Z : l ∗ ∈ L} ⊆ Ext(B Z ∗) forms a rich family.Pro<strong>of</strong>: Let L denote <strong>the</strong> family <strong>of</strong> all closed separable linearsubspaces Z <strong>of</strong> X such that {l ∗ | Z : l ∗ ∈ L} ⊆ Ext(B Z ∗). Weshall verify that L is a rich family <strong>of</strong> closed separable linearsubspaces <strong>of</strong> X. So first let us consider an arbitrary closedseparable linear subspace Y <strong>of</strong> X, with <strong>the</strong> aim <strong>of</strong> showingthat <strong>the</strong>re exists a subspace Z ∈ L such that Y ⊆ Z. We

egin by inductively applying Lemma 1 to obtain an increasingsequence (Z n : n ∈ N) <strong>of</strong> closed separable linear subspaces<strong>of</strong> X such that: Y ⊆ Z 1 and for any l ∗ ∈ L and any x ∗ , y ∗ ∈B Z∗n+1if l ∗ | Zn+1 = 1 2 (x∗ + y ∗ ) <strong>the</strong>n x ∗ | Zn = y ∗ | Zn .We now claim that if Z := ⋃ n∈N Z n <strong>the</strong>n l ∗ | Z ∈ Ext(B Z ∗)for each l ∗ ∈ L. To this end, suppose that l ∗ ∈ L andl ∗ | Z = 1 2 (x∗ + y ∗ ) for some x ∗ , y ∗ ∈ B Z ∗.Then for each n ∈ N,l ∗ | Zn+1 = (l ∗ | Z )| Zn+1 = 1 2 (x∗ +y ∗ )| Zn+1 = 1 2 (x∗ | Zn+1 +y ∗ | Zn+1 )and x ∗ | Zn+1 , y ∗ | Zn+1 ∈ B Z∗n+1Therefore, by constructionx ∗ | Zn = y ∗ | Zn . Now since ⋃ n∈N Z n is dense in Z and both x ∗and y ∗ are continuous we may deduce that x ∗ = y ∗ ; which inturn implies that l ∗ | Z ∈ Ext(B Z ∗). This shows that Y ⊆ Zand Z ∈ L .To complete this pro<strong>of</strong> we must verify that for each increasingsequence <strong>of</strong> closed separable subspaces (Z n : n ∈ N) in

L , ⋃ n∈N Z n ∈ L . This however, follows easily from <strong>the</strong>definition <strong>of</strong> <strong>the</strong> family L .❦ ✂✁Let X be a normed linear space. Then we say that an elementx ∗ ∈ B X ∗ is weak ∗ exposed if <strong>the</strong>re exists an element x ∈ Xsuch that y ∗ (x) < x ∗ (x) for all y ∗∈ B X ∗ \ {x ∗ }. It isnot difficult to show that if Exp(B X ∗) denotes <strong>the</strong> set <strong>of</strong> allweak ∗ exposed points <strong>of</strong> B X ∗ <strong>the</strong>n Exp(B X ∗) ⊆ Ext(B X ∗).However, if X is a separable L 1 -predual <strong>the</strong>n <strong>the</strong> relationshipbetween Exp(B X ∗) and Ext(B X ∗) is much closer.Lemma 2 If X is a separable L 1 -predual, <strong>the</strong>n Exp(B X ∗) =Ext(B X ∗).Let us also pause for a moment to recall that if B is anyboundary <strong>of</strong> a Banach space X <strong>the</strong>nExp(B X ∗) ⊆ B ∩ Ext(B X ∗) ⊆ Ext(B X ∗) ⊆ B weak∗ .

The fact that Ext(B X ∗) ⊆ B weak∗ follows from Milman’s<strong>the</strong>orem and <strong>the</strong> fact that B X ∗ = co weak∗ (B); which in turnfollows from a separation argument.Let us also take this opportunity to observe that if B X denotes<strong>the</strong> closed unit ball in X <strong>the</strong>n B X is closed in <strong>the</strong> σ(X, B)<strong>topology</strong> for any boundary B <strong>of</strong> X.Finally, let us end this part <strong>of</strong> <strong>the</strong> talk with one more simpleobservation that will turn out to be useful in our laterendeavours.Proposition 2 Suppose that Y is a linear subspace <strong>of</strong> a Banachspace (X, ‖ · ‖) and B is any boundary for X. Then foreach e ∗ ∈ Exp(B Y ∗) <strong>the</strong>re exists b ∗ ∈ B such that e ∗ = b ∗ | Y .Pro<strong>of</strong>: Suppose that e ∗ ∈ Exp(B Y ∗) <strong>the</strong>n <strong>the</strong>re exists anx ∈ Y such that y ∗ (x) < e ∗ (x) for each y ∗ ∈ B Y ∗ \ {e ∗ }.By <strong>the</strong> fact that B is a boundary <strong>of</strong> (X, ‖ · ‖) <strong>the</strong>re exists a

∗ ∈ B such that b ∗ (x) = ‖x‖ ≠ 0. Then for any y ∗ ∈ B Y ∗we havey ∗ (x) ≤ |y ∗ (x)| ≤ ‖y ∗ ‖‖x‖ ≤ ‖x‖ = b ∗ (x) = (b ∗ | Y )(x).In particular, e ∗ (x) ≤ b ∗ | Y (x). Since b ∗ | Y ∈ B Y ∗ andy ∗ (x) < e ∗ (x) for all y ∗ ∈ B Y ∗ \ {e ∗ }, it must be <strong>the</strong> casethat e ∗ = b ∗ | Y .❦ ✂✁The Main ResultsTheorem 3 Let B be any boundary for a Banach space Xthat is an L 1 -predual and suppose that {x n : n ∈ N} ⊆ X,<strong>the</strong>n {x n : n ∈ N} σ(X,B) ⊆ {x n : n ∈ N} σ(X,Ext(B X ∗)) .Pro<strong>of</strong>: In order to obtain a contradiction let us suppose that{x n : n ∈ N} σ(X,B) ⊈ {x n : n ∈ N} σ(X,Ext(B X ∗)) .Choose x ∈ {x n : n ∈ N} σ(X,B) \ {x n : n ∈ N} σ(X,Ext(B X ∗)) .Then <strong>the</strong>re exists a finite set {e ∗ 1, e ∗ 2, . . .,e ∗ m} ⊆ Ext(B X ∗)

for all j ∈ N and all 1 ≤ k ≤ m. Thus,⋂{y ∈ X : |b ∗ k(x) − b ∗ k(y)| < ε} ∩ {x n : n ∈ N} = ∅.1≤k≤mThis contradicts <strong>the</strong> fact that x ∈ {x n : n ∈ N} σ(X,B) ; whichcompletes <strong>the</strong> pro<strong>of</strong>.❦ ✂✁Corollary 1 Let B be any boundary for a Banach spaceX that is an L 1 -predual. Then every relatively countablyσ(X, B)-compact subset is relatively countably σ(X, Ext(B X ∗))-compact. In particular, every norm bounded, relatively countablyσ(X, B)-compact subset is relatively weakly compact.Pro<strong>of</strong>: Suppose that a nonempty set C ⊆ X is relativelycountably σ(X, B)-compact. Let {c n : n ∈ N} be any sequencein C <strong>the</strong>n by Theorem 3∅ ≠ ⋂ {c k : k ≥ n} σ(X,B) ⊆ ⋂ {c k : k ≥ n} σ(X,Ext(B X ∗))n∈Nn∈N

Hence C is relatively countably σ(X, Ext(B X ∗))-compact. In<strong>the</strong> case when C is also norm bounded <strong>the</strong> result follows froman earlier result <strong>of</strong> Kharana.❦ ✂✁Recall that a network for a topological space X is a familyN <strong>of</strong> subsets <strong>of</strong> X such that for any point x ∈ X andany open neighbourhood U <strong>of</strong> x <strong>the</strong>re is an N ∈ N suchthat x ∈ N ⊆ U, and a topological space X is said tobe ℵ 0 -monolithic if <strong>the</strong> closure <strong>of</strong> every countable set has acountable network.Corollary 2 Let B be any boundary for a Banach space Xthat is an L 1 -predual and suppose that {x n : n ∈ N} ⊆ X.Then {x n : n ∈ N} σ(X,B) is norm separable whenever X isℵ 0 -monolithic in <strong>the</strong> σ(X, Ext(B X ∗)) <strong>topology</strong>. In particular,{x n : n ∈ N} σ(X,B) is norm separable whenever Ext(B X ∗) isweak ∗ Lindelöf.

Proposition 3 Let B be any boundary for a Banach space Xthat is an L 1 -predual and suppose that A is a separable Bairespace. If X is ℵ 0 -monolithic in <strong>the</strong> σ(X, Ext(B X ∗)) <strong>topology</strong><strong>the</strong>n for each continuous mapping f : A → (X, σ(X, B))<strong>the</strong>re exists a dense subset D <strong>of</strong> A such that f is continuouswith respect to <strong>the</strong> norm <strong>topology</strong> on X at each point <strong>of</strong> D.Pro<strong>of</strong>: Fix ε > 0 and consider <strong>the</strong> open set:O ε := ⋃ {U ⊆ A : U is open and ‖ · ‖ − diam[f(U)] ≤ 2ε}.We shall show that O ε is dense in A. To this end, let Wbe a nonempty open subset <strong>of</strong> A and let {a n : n ∈ N} be acountable dense subset <strong>of</strong> W. Then by continuityf(W) ⊆ {f(a n ) : n ∈ N} σ(X,B) ;which is norm separable by Corollary 2. Therefore <strong>the</strong>re existsa countable set {x n : n ∈ N} in X such that f(W) ⊆

⋃n∈N (x n+εB X ). For each n ∈ N, let C n := f −1 (x n +εB X ).Since each x n + εB X is closed in <strong>the</strong> σ(X, B) <strong>topology</strong> eachset C n is closed in A and moreover, W ⊆ ⋃ n∈N C n. Since Wis <strong>of</strong> <strong>the</strong> second Baire category in A <strong>the</strong>re exist a nonemptyopen set U ⊆ W and a k ∈ N such that U ⊆ C k . ThenU ⊆ O ε ∩ W and so O ε is indeed dense in A. Clearly, f is‖ · ‖-continuous at each point <strong>of</strong> ⋂ n∈N O 1/n.❦ ✂✁Theorem 4 Suppose that A is a topological space with countabletightness that possesses a rich family F <strong>of</strong> Baire subspacesand suppose that X is an L 1 -predual. Then for anyboundary B <strong>of</strong> X and any continuous function f : A →(X, σ(X, B)) <strong>the</strong>re exists a dense subset D <strong>of</strong> A such that fis continuous with respect to <strong>the</strong> norm <strong>topology</strong> on X at eachpoint <strong>of</strong> D provided X is ℵ 0 -monolithic in <strong>the</strong> σ(X, Ext(B X ∗))<strong>topology</strong>.

Pro<strong>of</strong>: In order to obtain a contradiction let us supposethat f does not have a dense set <strong>of</strong> points <strong>of</strong> continuity withrespect to <strong>the</strong> norm <strong>topology</strong> on X. Since A is a Baire spacethis implies that for some ε > 0 <strong>the</strong> open set:O ε := ⋃ {U ⊆ A : U is open and ‖ · ‖-diam[f(U)] ≤ 2ε}is not dense in A. That is, <strong>the</strong>re exists a nonempty opensubset W <strong>of</strong> A such that W ∩ O ε = ∅. For each x ∈ A, letF x := {y ∈ A : ‖f(y) − f(x)‖ > ε}.Then x ∈ F x for each x ∈ W. Moreover, since A has countabletightness, for each x ∈ W, <strong>the</strong>re exists a countablesubset C x <strong>of</strong> F x such that x ∈ C x .

Next, we inductively define an increasing sequence <strong>of</strong> separablesubspaces (F n : n ∈ N) <strong>of</strong> A and countable sets(D n : n ∈ N) in A such that:(i) W ∩ F 1 ≠ ∅;(ii) ⋃ {C x : x ∈ D n ∩ W } ∪ F n ⊆ F n+1 ∈ F for all n ∈ N,where D n is any countable dense subset <strong>of</strong> F n .Note that since <strong>the</strong> family F is rich this construction is possible.Let F := ⋃ n∈N F n and D := ⋃ n∈N D n. Then D = F ∈ Fand ‖ · ‖-diam[f(U)] ≥ ε for every nonempty open subsetU <strong>of</strong> F ∩ W. Therefore, f| F has no points <strong>of</strong> continuityin F ∩ W with respect to <strong>the</strong> ‖ · ‖-<strong>topology</strong>. This however,contradicts Proposition 3.❦ ✂✁

Corollary 3 Suppose that A is a topological space with countabletightness that possesses a rich family <strong>of</strong> Baire subspacesand suppose that K is a compact Hausdorff space. Then forany boundary <strong>of</strong> (C(K), ‖ · ‖ ∞ ) and any continuous functionf : A → (C(K), σ(C(K), B)) <strong>the</strong>re exists a dense subset D<strong>of</strong> A such that f is continuous with respect to <strong>the</strong> ‖ · ‖ ∞ -<strong>topology</strong> at each point <strong>of</strong> D.——————————– The End ——————————–