---

Dual Kadec-Klee norms and the relationships between

arXiv:math/9302212v1 [math.FA] 9 Feb 1993Dual Kadec-Klee norms and the relationships betweenWijsman, slice and Mosco convergenceJ.M. Borwein 1Department of Combinatorics and OptimizationUniversity of WaterlooWaterloo, Ontario, Canada N2L 3G1J.

Vanderwerff2Department of Pure MathematicsUniversity of WaterlooWaterloo, Ontario, Canada N2L 3G1Abstract. In this paper, we completely settle several of the open questions regarding therelationships between the three most fundamental forms of set convergence.

In particular,it is shown that Wijsman and slice convergence coincide precisely when the weak star andnorm topologies agree on the dual sphere. Consequently, a weakly compactly generatedBanach space admits a dense set of norms for which Wijsman and slice convergence coincideif and only if it is an Asplund space.

We also show that Wijsman convergence impliesMosco convergence precisely when the weak star and Mackey topologies coincide on thedual sphere. A corollary of these results is that given a fixed norm on an Asplund space,Wijsman and slice convergence coincide if and only if Wijsman convergence implies Moscoconvergence.1.

Research supported in part by an NSERC research grant.2. NSERC postdoctoral fellow.AMS Classification.

Primary: 46B20, 46B03, 46N10; Secondary: 46A55.Key Words: Set convergence, convexity, Kadec-Klee norms, Mosco convergence, slice con-vergence, Wijsman convergence, Asplund spaces.

0. Introduction.All Banach spaces considered here are assumed to be real.Let X be an infinitedimensional Banach space with a given norm ∥· ∥.

When considering a subspace Y ofX, we will always assume it is endowed with the relative norm unless stated otherwise.The ball and sphere of X are defined and denoted as follows: BX = {x : ∥x∥≤1} andSX = {x : ∥x∥= 1}. We also use the notation Br = {x : ∥x∥≤r}.

For x ∈X, A, B ⊂X,let d(x, A) = inf{∥x −a∥: a ∈A} and let d(A, B) = inf{∥a −b∥: a ∈A, b ∈B}. If A = ∅,the convention is that d(x, A) = ∞; similarly, d(A, B) = ∞if A or B is empty.

We aregoing to consider the following three notions of set convergence. Let Cα, C be closed convexsubsets of X.

If limα d(x, Cα) = d(x, C) for all x ∈X, then Cα is said to converge Wijsmanto C. More restrictively, Cα is said to converge slice to C, if limα d(W, Cα) = d(W, C) forall closed bounded convex sets W. We will say Cα converges Mosco to C, if the followingtwo conditions are satisfied.M(i) If x ∈C, then d(x, Cα) →0.M(ii) If xαβ ∈Cαβ for some subnet is such that {xαβ}β is relatively weakly compact andxαβw⇁x, then x ∈C.Notice that M(i) and M(ii) reduce to the usual definition for Mosco convergence in thecase of sequences (for M(ii) we use that a weakly convergent sequence is relatively weaklycompact). Moreover this definition is compatible with the Mosco topology as defined in[Be2].

As is the usual practice, we only consider these notions for closed convex sets. It isalso clear that Wijsman, Mosco and slice convergence coincide in finite dimensional spaces,so we will only consider infinite dimensional spaces.

As a matter of terminology, we willsay that given a fixed norm on X, Wijsman convergence implies Mosco (slice) convergenceif Cα converges Mosco (slice) to C whenever Cα converges Wijsman to C with respect tothe given norm on X (if Cn converges Mosco (slice) to C, whenever Cn converges Wijsmanto C, we will say sequential Wijsman convergence implies Mosco (slice) convergence).It is crucial that we stipulate which norm is being used on X when speaking of Wijsmanconvergence because it depends on the particular norm (see [Be4, BF1, BL]). However, itfollows from the definitions that Mosco and slice convergence do not depend on the normbeing used.

One can also easily check, using the definitions, that slice convergence impliesMosco convergence in every space and they coincide in reflexive spaces. Moreover, if anet of sets converges slice to some set in a Banach space X, then it is not hard to checkthat the convergence is Wijsman with respect to every equivalent norm; Beer ([Be4]) has1

recently shown the converse holds. See [Be3, Be4, BF1, BL, BB1, BB2] for further resultsand examples.Historically, the notion of Wijsman convergence was introduced by Wijsman in [W]where it is shown to be a useful tool in finite dimensional spaces.

Mosco’s fundamentalpaper [M] on set convergence introduced the concept of Mosco convergence which hasproved to be a very useful notion in reflexive spaces. Unfortunately, it has several defectsin nonreflexive spaces; see [BB1].

However, a recent paper of Beer ([Be3]) shows that manyof the nice properties of Mosco convergence in reflexive spaces are valid for slice convergencein nonreflexive spaces. Because of this and the fact that Wijsman convergence is simpler tocheck, it is desirable to know when Wijsman convergence implies slice convergence.

Recallthat [BF1, Theorem 3.1] shows that Wijsman and Mosco convergence coincide if and onlyif the space is reflexive and the weak and norm topologies coincide on the dual sphere (atopolgical version of this is proved in [Be2, Theorem 2.5]). This and the fact that Moscoand slice convergence coincide in reflexive spaces leads to the following natural questionposed in [BB2].

Do Wijsman and slice convergence coincide whenever the w∗and normtopologies agree on the dual unit sphere? This paper will provide an affirmative answer tothis question.In the first section, we present some basic facts about dual Kadec-Klee norms (whichfor brevity we call Kadec norms).

Let τ denote the Mackey topology on X∗, that is thetopology of uniform convergence on weakly compact sets.We will say a norm ∥· ∥isw∗-τ-Kadec if ∥x∗α∥→∥x∗∥and x∗αw∗⇁x∗imply x∗ατ⇁x∗; if this holds for sequences, then∥· ∥will be called sequentially w∗-τ-Kadec. These notions coincide when the dual ball isw∗-sequentially compact (see Corollary 1.2).

On the other hand, there are norms that aresequentially w∗-τ-Kadec, but not w∗-τ-Kadec; see Remark 1.5(a). Many of our results willdeal with the following property of a norm which is more restrictive than w∗-τ-Kadec; seeRemark 1.5(b).

A norm ∥·∥is said to be w∗-Kadec if ∥xα −x∥→0 whenever ∥xα∥→∥x∥and xαw∗⇁x; if this holds for sequences, ∥· ∥is said to be sequentially w∗-Kadec.Section 2 begins by showing that Wijsman convergence has the defect of not beingpreserved in superspaces while Mosco and slice convergence are. We also show that therelationship between Wijsman and slice convergence is separably and sequentially deter-mined (this allows us to restrict our attention to sequences of sets in separable subspacesin the third section).

Some relationships between set convergence, dual Kadec norms anddifferentiability are also presented in the second section.2

The third section contains the main results.It is shown that Wijsman and sliceconvergence coincide precisely when the dual norm is w∗-Kadec. Let us mention that wideclasses of Banach spaces can be renormed so that the dual norm is w∗-Kadec.

A Banachspace is said to be weakly compactly generated (WCG) if it contains a weakly compactset whose linear span is norm dense. It is clear (from the definition) that WCG spacesinclude all separable and all reflexive spaces.

It follows from [F1, Theorem 1] that everyWCG Asplund space can be renormed so that the dual norm is w∗-Kadec. If X∗is WCG,then there is a norm on X whose dual norm is w∗-Kadec; see [DGZ, F1, F2] for moreand stronger results on renorming.We also give some conditions on the limit set forwhich one can deduce slice convergence from Wijsman convergence in certain spaces withFr´echet differentiable norms (whose dual norms are not necessarily w∗-Kadec).

This willbe done by working with functionals in the subdifferentials of distance functions. Recallthat the subdifferential of a convex function f at x0 in the domain of f, is defined by∂f(x0) = {Λ ∈X∗: ⟨Λ, x −x0⟩≤f(x) −f(x0) for all x ∈X}.1.

Dual Kadec norms.The purpose of this section is to gather a few facts about dual Kadec norms which wewill need later. See also [DS, GM, JH] for some other properties of spaces related to dualKadec norms.Proposition 1.1.

For a Banach space X, the following are equivalent. (a) The dual norm on X∗is w∗-τ-Kadec (w∗-Kadec).

(b) For each Y ⊂X, the dual norm on Y ∗is w∗-τ-Kadec (w∗-Kadec). (c) For each separable Y ⊂X, the dual norm on Y ∗is sequentially w∗-τ-Kadec (sequen-tially w∗-Kadec).Proof.We prove this for the w∗-τ-Kadec case only.

(a) ⇒(b): Suppose Y is asubspace of X and its dual norm ∥· ∥is not w∗-τ-Kadec. Then there is a weakly compactset K, an ǫ > 0 and a net y∗αw∗⇁y∗such that ∥y∗α∥= ∥y∗∥= 1 andsupK|y∗α −y∗| > ǫfor all α.Let x∗α denote a norm preserving extension of y∗α.

By Alaoglu’s theorem, for some subnet,one has x∗αβw∗⇁x∗for some x∗∈BX∗. Observe that x∗|Y = y∗and so it follows that∥x∗∥= 1 andsupK|x∗αβ −x∗| > ǫfor all β.3

Hence the dual norm on X∗is not w∗-τ-Kadec.It is clear that (b) ⇒(c), so we prove (c) ⇒(a). Suppose the dual norm ∥· ∥on X∗is not w∗-τ-Kadec.

Thus we can choose a net {x∗α} and a weakly compact set K ⊂X suchthat x∗αw∗⇁x∗, ∥x∗α∥= ∥x∗∥= 1, andsupK|x∗α −x∗| > ǫfor all α and some ǫ > 0.Let {un} ⊂X be such that ∥un∥= 1 and ⟨x∗, un⟩> 1 −1n. Let α1 ≤α2 ≤.

. .

be chosenso that ⟨x∗α, un⟩≥1 −1n whenever α ≥αn. Now choose {x∗i } ⊂{x∗α} and {xi} ⊂Kas follows.

Let x∗1 = x∗α1 and x1 ∈K be chosen so that |⟨x∗1 −x∗, x1⟩| > ǫ. Suppose{x∗1, x∗2, .

. .x∗k−1} and {x1, x2, .

. .xk−1} have been chosen such that |⟨x∗j −x∗, xj⟩| > ǫ for1 ≤j ≤k −1.

Because x∗αw∗⇁x∗, we can choose x∗k = x∗α where α ≥αk and|⟨x∗k −x∗j, xj⟩| ≥ǫfor j = 1, 2, . .

., k −1.Now choose xk ∈K such that |⟨x∗k −x∗, xk⟩| > ǫ.Let Y = span{xi}∞i=1 ∪{ui}∞i=1. Then Y is separable and K1 = K ∩Y is a weaklycompact subset of Y .

Now let y∗i = x∗i |Y . Because Y is separable, BY ∗is w∗-sequentiallycompact and so y∗ijw∗⇁y∗for some subsequence and some y∗∈BY ∗.

Observe that⟨y∗, un⟩= limj ⟨x∗ij, un⟩≥1 −1n.Thus it follows that ∥y∗∥= 1.Moreover, for n > m, |⟨y∗n −y∗m, xm⟩| ≥ǫ; and sincexm ∈K1, this means y∗ij does not converge Mackey to y∗. Thus the dual norm on Y ∗isnot sequentially w∗-τ-Kadec.Corollary 1.2.

Suppose that BX∗is w∗-sequentially compact. If the dual norm on X∗issequentially w∗-τ-Kadec (sequentially w∗-Kadec), then it is w∗-τ-Kadec (w∗-Kadec).Proof.

We prove the w∗-τ-Kadec case only. Using w∗-sequential compactness, onecan show as in the proof of (a) ⇒(b) in Proposition 1.1 that the dual norm on Y ∗issequentially w∗-τ-Kadec for each subspace Y of X.

Therefore, by Proposition 1.1, thenorm on X∗is w∗-τ-Kadec.Recall that a space is said to have the Schur property if weakly convergent sequencesare norm convergent.Remark 1.3. (a) If X has the Schur property, then every dual norm on X∗is w∗-τ-Kadec.4

(b) There are spaces X such that X∗has a dual w∗-τ-Kadec norm, but BX∗is notw∗-sequentially compact.Proof. (a) Since weakly compact sets are norm compact, it is clear that w∗-convergenceis the same as τ-convergence.

(b) This follows from (a) because X = ℓ1(Γ) is Schur but BX∗is not w∗-sequentiallycompact whenever Γ is uncountable (see [Du, p. 48]).A Banach space X is called sequentially reflexive if every τ-convergent sequence in X∗is norm convergent (see [Bor, Ø]). It is shown in [Ø] that X is sequentially reflexive if andonly if X ̸⊃ℓ1 (by X ̸⊃ℓ1, we mean that X does not contain an isomorphic copy of ℓ1).This result provides a nice connection between w∗-Kadec and w∗-τ-Kadec norms.Theorem 1.4.

For a Banach space X, the following are equivalent. (a) The dual norm on X∗is w∗-Kadec.

(b) X is Asplund and the dual norm on X∗is sequentially w∗-Kadec. (c) BX∗is w∗-sequentially compact and the dual norm on X∗is sequentially w∗-Kadec.

(d) BX∗is w∗-sequentially compact, X ̸⊃ℓ1 and the dual norm on X∗is sequentiallyw∗-τ-Kadec. (e) X ̸⊃ℓ1 and the dual norm on X∗is w∗-τ-Kadec.Proof.

(a) ⇒(b): Let Y be a separable subspace of X and suppose ∥· ∥is a dualw∗-Kadec norm on X∗.Let {fn}∞n=1 be a fixed w∗-dense sequence in BY ∗.Now letf ∈SY ∗be arbitrary. For some {fj} we have fjw∗⇁f.

According to Proposition 1.1, ∥· ∥is sequentially w∗-Kadec on Y ∗, hence it follows that ∥fj −f∥→0. Thus Y ∗is separablesince its sphere has a countable norm dense subset.

Consequently X is an Asplund space(see [Ph, Theorem 2.34]). This shows (a) ⇒(b).

Now (b) ⇒(c) follows from the factthat Asplund spaces have w∗-sequentially compact dual balls ([Di, p. 230]). Accordingto Corollary 1.2, (c) ⇒(a) and hence (c) ⇒(b) which means X ̸⊃ℓ1, thus (c) ⇒(d).Moreover, Corollary 1.2 shows (d) ⇒(e).

Finally, we show (e) ⇒(a). By [Ø], if X ̸⊃ℓ1,then X is sequentially reflexive.

Combining this with Proposition 1.1, shows that for everysubspace Y of X, the dual norm on Y ∗is sequentially w∗-Kadec. Invoking Proposition 1.1again, shows that the dual norm on X∗is w∗-Kadec.As a note of comparison with weak Kadec properties, if X is separable and X ̸⊃ℓ1,then sequentially weak Kadec norms are weak Kadec while on ℓ1 there are sequentially weakKadec norms that are not weak Kadec; see [Tr].

Also, several spaces have w∗-sequentially5

compact dual balls. Indeed, Asplund, WCG and more generally Gateaux Differentiabilityspaces have w∗-sequentially dual balls; see [Di, Chapter XIII] and [LP].Remark 1.5.

(a) There is a norm on ℓ∞whose dual norm is sequentially w∗-τ-Kadec butnot w∗-τ-Kadec. (b) No dual norm on ℓ∗1 is sequentially w∗-Kadec, but every dual norm on ℓ∗1 is w∗-τ-Kadec.Proof.

(a) Since w∗-convergent sequences in ℓ∗∞are w-convergent ([Di, p. 103]) andsince ℓ∞has the Dunford-Pettis property ([Di, p. 113]), it follows that w∗-convergentsequences in ℓ∗∞are τ-convergent (see Proposition 2.5). Thus the dual of every norm onℓ∞is sequentially w∗-τ-Kadec.

Let ||| · ||| be an equivalent norm on c0 whose dual is notsequentially w∗-Kadec (see [BFa]). It follows from Theorem 1.4 that the dual norm of |||·|||is not sequentially w∗-τ-Kadec.

Now let ||| · ||| denote the second dual of this norm on ℓ∞.Then the dual of ||| · ||| is not w∗-τ-Kadec on ℓ∗∞by Proposition 1.1. (b) This is clear from Remark 1.3(a) and Theorem 1.4.Finally, let us mention that the following is not clear to us: if the dual norm on X∗issequentially w∗-Kadec, then must it be w∗-Kadec?

This, of course is true if the dual ballis w∗-sequentially compact (Theorem 1.4). So this question is equivalent to: if the dualnorm is sequentially w∗-Kadec, is the dual ball w∗-sequentially compact?2.

Basic properties of set convergence.One of the nice things about Wijsman convergence is the simplicity of its definition.However, this leads to the drawback that Wijsman convergence is not necessarily preservedby superspaces.Proposition 2.1. (a) Let Y be a Banach space and suppose Cα converges slice (Mosco)to C in Y .

If X is a superspace of Y , then Cα converges slice (Mosco) to C in X. (b) Wijsman convergence in X is not necessarily preserved in X × IR.Proof.

(a) It is clear from the definition that this holds for Mosco convergence. Wewill prove that slice convergence is preserved in superspaces.

Suppose that Cα, C ⊂Y ,where Y is a subspace of X and that Cα does not converge slice to C in X. We will showthat Cα does not converge slice to C in Y .

We may suppose Cα converges Wijsman to C inY , since otherwise we are done. Given any set B ⊂X and ǫ > 0, we can choose b ∈B andc ∈C such that ∥b −c∥≤d(B, C) + ǫ.

By Wijsman convergence, d(c, Cα) →d(c, C) = 0.Hence lim sup d(B, Cα) ≤lim sup(∥b −c∥+ d(c, Cα)) ≤d(B, C) + ǫ. Thus because Cα6

does not converge slice to C, by passing to a subnet if necessary, we find a bounded closedconvex W ⊂X such thatd(W, Cα) + 3δ ≤d(W, C)for all α and some δ > 0.Let r = d(W, C) −2δ. Then (W + Br) ∩Cα ̸= ∅for all α, while (W + Br+δ) ∩C = ∅.Using the separation theorem, we find a Λ ∈SX∗such thatsup{⟨Λ, x⟩: x ∈W + Br+δ} ≤inf{⟨Λ, x⟩: x ∈C}.Let a = sup{⟨Λ, x⟩: x ∈W + Br}, thena + δ = sup{⟨Λ, x⟩: x ∈W + Br+δ} ≤inf{⟨Λ, x⟩: x ∈C}(2.1)andW + Br ⊂{x : ⟨Λ, x⟩≤a}.Now let m > 0 be chosen such that W + Br ⊂Bm.

Because (W + Br) ∩Cα ̸= ∅, thereexists yα ∈Cα ⊂Y such thatyα ∈{y ∈Y : ⟨Λ, y⟩≤a} ∩Bm.We set W1 = {y ∈Y : ⟨Λ, y⟩≤a}∩Bm. Hence, W1 ∩Cα ̸= ∅for all α.

However, accordingto (2.1), d(W1, C) ≥δ and so Cα does not converge slice to C in Y . (b) Let X be c0 endowed with the norm ||| · ||| which is defined for x = (xn)∞n=0 asfollows:|||x||| = |x0| ∨|x1| ∨(supn≥2|xn + x1|).Let Y = {x ∈X : x0 = x1} and define ˆfn ∈X∗byˆfn(x) = x1 + xnandˆf∞(x) = x1.Then||| ˆfn||| = sup{x1 + xn : |x1 + xn| ≤1, |x1| ≤1} = 1,and||| ˆf∞||| = sup{x1 : |x1 + xn| ≤1, |x1| ≤1} = 1.Now ˆfn(x) →ˆf∞(x) for all x since xn →0.

It follows directly that ˆf −1n (1) convergesWijsman to ˆf −1∞(1); see [Be1, Theorem 4.3].7

Let fn = ˆfn|Y , then |||fn||| = |||f∞||| = 1, and so similarly it follows thatf −1n (1) converges Wijsman to f −1∞(1) in Y.However, f −1n (1) does not converge Wijsman to f −1∞(1) in X.Indeed, consider z0 =(0, 12, 0, 0, . .

.) and zn = 12e0 + 12e1 + 12en.

Then zn ∈f −1n (1) and12 = |||z0 −zn||| ≥dz0, f −1n (1).On the other hand, if x = (xi)∞i=0 ∈f −1∞(1), then x0 = x1 = 1 and consequently one has|||x −z0||| ≥|x0| = 1 which means dz0, f −1∞(1)≥1.A partial redress to Proposition 2.1(b) will be given in Corollary 3.3. We next showthat the relationship between Wijsman and slice (Mosco) convergence is separably andsequentially determined.Theorem 2.2.

Suppose Cα converges Wijsman but not slice (Mosco) to C in some sub-space E of X, then there is a separable subspace Y of E and a subsequence Cαn such thatCαn ∩Y converges Wijsman but not slice (Mosco) to C ∩Y as subsets of Y .Proof. Since Cα converges Wijsman to C in E, as in the proof of Proposition 2.1,lim supα d(B, Cα) ≤d(B, C) for any B ⊂E.

Thus, because Cα does not converge slice toC, by passing to a subnet if necessary, there is a bounded closed convex subset W of Eand a δ > 0 such thatd(W, Cα) + δ < d(W, C)for all α. (2.2)Let Z be an arbitrary separable subspace of E and set Z1 = Z.

Fix a dense subset {z1,i}∞i=1of Z1 and choose α1 such thatd(z1,1, C) −1 < d(z1,1, Cα1) < d(z1,1, C) + 1. (2.3)Using (2.2) and (2.3), one can choose w1 ∈W, c1 ∈Cα1, x11,1 ∈Cα1 and y11,1 ∈C suchthat:∥w1 −c1∥≤d(W, C) −δ;∥z1,1 −x11,1∥≤d(z1,1, C) + 1;∥z1,1 −y11,1∥≤d(z1,1, C) + 1.Suppose α1 ≤α2 ≤.

. .

≤αn−1, {zi,j}∞j=1, ci, wi, for i ≤n−1 and {yki,j}, {xki,j} for i, j ≤k,k ≤n −1 have been chosen. Then setZn = spanZn−1 ∪{yni,j} ∪{xni,j} ∪{cn} ∪{wn} : i ≤n, j ≤n.

(2.4)8

Fix a dense collection {zn,i}∞i=1 ⊂Zn and choose αn ≥αn−1 such thatd(zi,j, C) −1n < d(zi,j, Cαn) < d(zi,j, C) + 1nwhenever i ≤n, j ≤n. (2.5)Using this, for i ≤n, j ≤n we choose xni,j ∈Cαn and yni,j ∈C such that∥xni,j −zi,j∥≤d(zi,j, C) + 1n;(2.6)∥yni,j −zi,j∥≤d(zi,j, C) + 1n.

(2.7)According to (2.2), let wn ∈W and cn ∈Cαn be chosen so that∥wn −cn∥≤d(W, C) −δ. (2.8)Finally let Y be the norm closure of ∪∞n=1Zn.We now show that Cαn ∩Y converges Wijsman but not slice to C ∩Y as subsetsof Y .

Let ǫ > 0 and let y ∈Y . Since ∪nZn is norm dense in Y , for some zi,j we have∥zi,j −y∥< ǫ.

Observe thatd(y, C) ≤d(y, C ∩Y )≤∥y −zi,j∥+ d(zi,j, C ∩Y )≤ǫ + lim infn∥zi,j −yni,j∥≤ǫ + lim infnd(zi,j, C) + 1n[by (2.7)]≤ǫ + d(zi,j, C)≤2ǫ + d(y, C).Hence d(y, C ∩Y ) = d(y, C). Now,lim infnd(y, Cαn ∩Y ) ≥−∥y −zi,j∥+ lim infnd(zi,j, Cαn)≥−ǫ + lim infnd(zi,j, C) −1n[by (2.5)]= d(zi,j, C) −ǫ ≥d(y, C) −2ǫ.Thus, lim infn d(y, Cαn ∩Y ) ≥d(y, C).

On the other hand,lim supnd(y, Cαn ∩Y ) ≤∥y −zi,j∥+ lim supn∥zi,j −xni,j∥≤ǫ + lim supnd(zi,j, C) + 1n[by (2.6)]≤2ǫ + d(y, C).9

Consequently, limn d(y, Cαn ∩Y ) = d(y, C) = d(y, C ∩Y ). So Cαn ∩Y converges Wijsmanto C ∩Y in Y .However, for W1 = W ∩Y , we have wn ∈W1 and cn ∈Cαn ∩Y for all n (as chosenin (2.8) and notice that wn, cn ∈Y by (2.4)).

Hence,d(W1, C ∩Y ) ≥d(W, C)≥∥wn −cn∥+ δ[by (2.8)]≥d(W1, Cαn).So Cαn ∩Y does not converge slice to C ∩Y in Y .The Mosco case can be proved in a similar manner. It is clear that M(i) holds becauseof Wijsman convergence.

If M(ii) fails, then there is a net yβ ⊂Cαβ such that yβw⇁ywhere y ̸∈C and {yβ} is contained in a weakly compact set. Now one can strictly separatey from C, thus taking a further subnet, we may assume that no subsequence of {yβ} hasa limit point in C. Thus using the above construction and weak sequential compactness,one can obtain the result for Mosco convergence.The following proposition, which is based on ideas from [Be1], relates set convergenceto properties of the dual norm.

If Wijsman convergence of Lα to L implies Mosco conver-gence of Lα to L for any net (sequence) of sets Lα = {x : ⟨x∗α, x⟩= a}, L = {x : ⟨x∗, x⟩= a}where x∗α, x∗∈X∗and a ∈IR, then we will say Wijsman convergence implies Mosco con-vergence for (sequences of) level sets of functionals.Proposition 2.3 (a) If in X, Wijsman convergence implies Mosco convergence for (se-quences of) level sets of functionals, then the dual norm on X∗is (sequentially) w∗-τ-Kadec. (b) If in X, Wijsman convergence implies slice convergence for (sequences of) levelsets of functionals, then the dual norm on X∗is (sequentially) w∗-Kadec.Proof.

We only prove (a) for nets since the other part is similar and is essentially in[BF1, Theorem 3.1]. Moreover, (b) may be essentially found in [Be1, Be4].Suppose the dual norm is not w∗-τ-Kadec, then we can find x∗α, x∗∈SX∗such thatx∗αw∗⇁x∗but x∗α ̸τ⇁x∗.

Let Cα = {x ∈X : ⟨x∗α, x⟩= 1} and C = {x ∈X : ⟨x∗, x⟩= 1}.Since d(x, Cα) = |⟨x∗α, x⟩−1|, it follows that Cα converges Wijsman to C.We now proceed as in the proof of [BF1, Theorem 3.1]: by passing to a subnetif necessary, there is a weakly compact set K ⊂BX and {xα} ⊂K such that |⟨x∗α −x∗, xα⟩| ≥ǫ for some ǫ > 0. Let x0 ∈X be such that ∥x0∥≤3 and ⟨x∗, x0⟩> 2.

Now10

⟨x∗α, x0⟩→⟨x∗, x0⟩and so by considering only a tail of the net we may assume ⟨x∗α, x0⟩≥2for all α. Consider vα = x0 + xα.

Then ⟨x∗α, vα⟩≥1. Since ∥vα∥≤4, we can choose14 ≤λα ≤1 such that ⟨x∗α, λαvα⟩= 1.

By passing to a subnet we have λαβ →λ where14 ≤λ ≤1 and vαβw⇁v where ∥v∥≤4. Since ⟨x∗α, x0⟩→⟨x∗, x0⟩, it follows thatlim infβ|1 −⟨x∗, λαβvαβ⟩| = lim infβ|⟨x∗αβ −x∗, λαβvαβ⟩| ≥lim infβλαβǫ ≥ǫ4.Now, ⟨x∗, λαβvαβ⟩→⟨x∗, λv⟩and so we have |1 −⟨x∗, λv⟩| ≥ǫ4.

Consequently, λv ̸∈Cand so M(ii) fails. This completes the proof.Corollary 2.4.

Suppose in each separable subspace of X that sequential Wijsman con-vergence implies Mosco (slice) convergence, then the dual norm on X∗is w∗-τ-Kadec(w∗-Kadec).Proof. This follows from Proposition 2.3 and Theorem 2.2 (or Proposition 1.1).The next two results show a connection between set convergence and differentiability.Recall that a function is said to be weak Hadamard differentiable at a point if its Gateauxderivative exists at the point and is uniform on weakly compact sets.

The following propo-sition shows that this notion is related to Wijsman and Mosco convergence in non-Asplundspaces. Indeed, notice that property (∗) below ensures that X contains an isomorphic copyof ℓ1; see [Ø, BFa].Proposition 2.5.

Let X be a Banach space, then the following are equivalent. (a) For every equivalent norm on X, Wijsman convergence implies Mosco convergence forsequences of level sets of functionals.

(b) The following property is satisfied.⟨x∗n, xn⟩→⟨x∗, x⟩whenever x∗nw∗⇁x∗and xnw⇁x. (∗)(That is, w∗-convergent sequences in X∗are τ-convergent.

)(c) Weak Hadamard and Gateaux differentiability coincide for continuous convex func-tions on X.Proof. The equivalence of (b) and (c) follows from the results of [BFa]; see also [BFV].

(a) ⇒(b): Suppose (∗) fails, thus we can find x∗nw∗⇁x∗and xnw⇁x but |⟨x∗n, xn⟩−⟨x∗, x⟩| ≥ǫ for all n and some ǫ > 0. We now show that X admits an equivalent normwhose dual is not sequentially w∗-τ-Kadec.

Notice that we may assume ∥x∗n∥≤1 for alln. If ∥x∗∥= 1, then ∥· ∥is not sequentially w∗-τ-Kadec.

So suppose ∥x∗∥< 1. We may11

assume x∗= 0 and that ∥x∗n∥≤1 for all n. Now let y ∈X satisfy ∥y∥= 1. By replacing ywith −y if necessary, we have ⟨x∗j, y⟩≤0 for all j ∈J where J is an infinite subset of IN.Now choose y∗∈X∗satisfying ⟨y∗, y⟩= ∥y∗∥= 1.

Define a convex w∗-compact subset ofX∗byB = {Λ ∈X∗: |⟨Λ, y⟩| ≤1} ∩{Λ ∈X∗: ∥Λ∥≤2}.Let |||·||| denote the dual norm on X∗whose unit ball is B. Observe that |||y∗+x∗||| = 1 and|||y∗+x∗j||| ≤1 for all j ∈J.

Hence |||·||| is not sequentially w∗-τ-Kadec since y∗+x∗jw∗⇁y∗+x∗but ⟨y∗+ x∗j, xj⟩̸→⟨y∗+ x∗, x⟩. Thus (a) does not hold by Proposition 2.3(a).

(b) ⇒(a): Let ∥· ∥be any equivalent norm on X. If Cn converges Wijsman to Cwhere Cn = {x : ⟨x∗n, x⟩= α} and C = {x : ⟨x∗, x⟩= α}.

Then by [Be1, Theorem 4.3]x∗nw∗⇁x∗and ∥x∗n∥→∥x∗∥. Now suppose xj ∈Cj for j ∈J ⊂IN and xjw⇁x.

By property(∗), we have ⟨x∗j, xj⟩→⟨x∗, x⟩which means ⟨x∗, x⟩= α and x ∈C. Thus M(ii) holds.Since M(i) always holds in the presence of Wijsman convergence, we are done.Corollary 2.6.

Suppose that every separable subspace of X is contained in a comple-mented subspace whose dual ball is w∗-sequentially compact.Then the following areequivalent. (a) For every equivalent norm on X, Wijsman convergence implies Mosco convergence forsequences of level sets of functionals.

(b) X has the Schur property.Proof. (a) ⇒(b): This follows from Proposition 2.5 and [BFV, Corollary 3.5].

(b) ⇒(a): This is always true.The condition in the preceding corollary is, of course, satisfied in all spaces whose dualballs are w∗-sequentially compact (in particular WCG spaces) and in much more generalcases; see [BFV]. In addition, there are many Grothendieck C(K) spaces which satisfyproperty (∗) but are not Schur; see [BFV] and the references therein.3.

Dual Kadec norms and set convergence.The proof of our main result will use the following proposition which is essentially dueto Attouch and Beer (part (a)—for sequences—is contained in [AB, Theorem 3.1]). Wewill also need the following intermediate notion of set convergence.

For closed convex setsCα, C, we will say Cα converges weak compact gap to C, if d(W, Cα) →d(W, C) for allconvex weakly compact subsets W of X.12

Proposition 3.1. Suppose Cα and C are closed convex sets in a Banach space X. Considerthe following three conditions:(i) if x0 ∈C, then d(x0, Cα) →0;(ii) if x∗0 ∈SX∗attains its supremum on C, then there exist x∗α ∈BX∗such that ∥x∗α −x∗0∥→0 andlim supα{supCαx∗α} ≤supCx∗0;(iii) if x∗0 ∈SX∗attains its supremum on C, then there exist x∗α ∈BX∗such that x∗ατ⇁x∗0,andlim supα{supCαx∗α} ≤supCx∗0.

(a) If (i) and (ii) hold, then Cα converges slice to C.(b) If (i) and (iii) hold, then Cα converges weak compact gap to C.Proof. We prove only (b) since the proof of (a) is almost identical.

Let W be a weaklycompact convex set in X. According to (i), lim supα d(W, Cα) ≤d(W, C).

So we showthat lim infα d(W, Cα) ≥d(W, C). If d(W, C) = 0, there is nothing more to do, so supposed(W, C) > 0.

Let ǫ > 0 satisfy 2ǫ < d(W, C) and set r = d(W, C) −2ǫ. By the separationtheorem, there exists Λ ∈SX∗such thatsup{⟨Λ, x⟩: x ∈C} ≤inf{⟨Λ, x⟩: x ∈W + Br+ǫ}= inf{⟨Λ, x⟩: x ∈W + Br} −ǫ.By a general version of the Bishop-Phelps theorem ([BP, Theorem 2]), there is an x∗0 ∈SX∗which attains its supremum on C and strictly separates C and W +Br (one can also obtainthis from the Brøndsted-Rockafellar theorem [Ph, Theorem 3.18]).

ThusinfW x∗0 −supCx∗0 ≥r.Let x∗α be given by (iii) and let α0 be such thatsupCαx∗α ≤supCx∗0 + ǫ,andinfW x∗α ≥infW x∗0 −ǫfor α ≥α0.¿From this it follows thatd(Cα, W) ≥infW x∗α −supCαx∗α ≥r −2ǫ ≥d(W, C) −4ǫfor α ≥α0.Since ǫ > 0 was arbitrary, we are done.13

With Proposition 3.1 at our disposal, we are now ready for our main result.Theorem 3.2. For a Banach space X, the following are equivalent.

(a) The dual norm on X∗is w∗-Kadec. (b) Sequential Wijsman and slice convergence coincide in every separable subspace of X.

(c) Wijsman and slice convergence coincide in every subspace of X.Proof. (a) ⇒(b): Let Z be any separable subspace of X.

Suppose that Cn convergesWijsman to C as subsets of Z. We wish to show that (i) and (ii) in Proposition 3.1 hold.Clearly (i) follows from Wijsman convergence so we show (ii).

Let z∗0 ∈SX∗attain itssupremum on C, say supC z∗0 = ⟨z∗0, z0⟩where z0 ∈C.Let α0 = ⟨z∗0, z0⟩and let L = {z : ⟨z∗0, z⟩= α0 +1}. Since Z and hence L is separable,we can choose a sequence of compact convex sets {Kn} such that Kn ⊂L for each n,K1 ⊂K2 ⊂K3 ⊂.

. ., d(Kn, z0) < 1 + 1n andL is the norm closure of ∪∞n=1 Kn.

(3.1)Since compact sets have finite ǫ-nets and since Cn converges Wijsman to C, we deducethatlimj→∞d(Kn, Cj) = d(Kn, C)for each n(in other words, Wijsman convergence is precisely compact gap convergence). Thus wemay choose j1 < j2 < j3 < .

. .

such thatd(z0, Cj) < 1nand1 −1n < d(Kn, Cj) < 1 + 1nfor j ≥jn. (3.2)It follows that (1 −1n)BZ ∩(Kn −Cj) = ∅for j ≥jn.

Thus by the separation theorem,there exists Λn,j ∈SZ∗such thatsup{⟨Λn,j, z⟩: z ∈(1 −1n)BZ} ≤inf{⟨Λn,j, z⟩: z ∈Kn −Cj} for j ≥jn.This impliessupCjΛn,j + (1 −1n) ≤minKn Λn,jfor j ≥jn. (3.3)Now set z∗j = 0 for j < j1 andz∗j = Λn,jfor jn ≤j < jn+1.14

Claim. z∗jw∗⇁z∗0.Assume temporarily that the claim is true.

Because d(z0, Kn) < 1 + 1n, it follows thatminKn z∗j < ⟨z∗j , z0⟩+ (1 + 1n). (3.4)For j ∈IN, let nj denote the number n such that jn ≤j < jn+1.

Thus by (3.3) and (3.4)one hassupCjz∗j + (1 −1nj) ≤minKnjz∗j< ⟨z∗j , z0⟩+ (1 + 1nj).In other words,supCjz∗j < ⟨z∗j , z0⟩+ 2nj.Since ⟨z∗j , z0⟩→⟨z∗0, z0⟩= supC z∗0, this immediately yieldslim supj{supCjz∗j } ≤supCz∗0.Moreover, by Proposition 1.1, the dual norm on Z∗is w∗-Kadec, hence ∥z∗j −z∗0∥→0.Thus (ii) holds provided our claim is true.Let us now prove that the claim is true by showing every subsequence of {z∗j } has asubsequence which converges w∗to z∗0. By abuse of notation, let {z∗j } denote an arbitrarysubsequence of {z∗j }.

¿From the w∗-sequential compactness of BZ∗, by passing to anothersubsequence if necessary, we have z∗j′w∗⇁Λ for some Λ ∈BZ∗. We now show that ∥Λ∥= 1.Again, we use nj to denote the n such that jn ≤j < jn+1; because d(z0, Cj) ≤1n forj ≥jn, it follows thatsupCjz∗j ≥⟨z∗j , z0⟩−1nj.Let m ∈IN and z ∈Km be fixed.

Because z ∈Kn for n ≥m, the above inequality yields⟨Λ, z −z0⟩= limj′ ⟨z∗j′, z −z0⟩≥lim infjminKnjz∗j −(supCjz∗j + 1nj)≥lim infj1 −1nj−1nj[by (3.3)]= 1.15

Consequently, we haveminKm Λ ≥⟨Λ, z0⟩+ 1for all m ∈IN. (3.5)Since limn d(Kn, z0) →1, it also follows that ∥Λ∥= 1.It now suffices to show that Λ = z∗0.

Let H = {z : ⟨z∗0, z⟩≥0}. We claim that⟨Λ, z⟩≥0 for all z ∈H.

So suppose that ⟨Λ, h⟩≤−δ for some δ > 0 and some h ∈H with∥h∥≤1. Now consider z0 + h, then α0 ≤⟨z∗0, z0 + h⟩≤1 + α0 and so d(z0 + h, L) ≤1.Thus by (3.1), we can find ˜z ∈Km for some m such that ∥˜z −(z0 + h)∥≤1 + δ2.

Hence itfollows that⟨Λ, ˜z⟩≤∥˜z −(z0 + h)∥+ ⟨Λ, z0 + h⟩≤(1 + δ2) + ⟨Λ, z0 + h⟩≤⟨Λ, z0⟩+ 1 −δ2.This contradicts (3.5). Therefore ⟨Λ, h⟩≥0 for all h ∈H.

But since ∥Λ∥= ∥z∗0∥= 1, thismeans Λ = z∗0. This shows that the claim holds and thus (a) ⇒(b).Now, (b) ⇒(c) follows from Theorem 2.2 and (c) ⇒(a) is a consequence of Proposition2.3(b).¿From Theorem 3.2 and Proposition 2.1(a) we immediately obtainCorollary 3.3.

If the dual norm on X∗is w∗-Kadec and Cα converges Wijsman to C inX, then Cα converges slice to C in any superspace of X.We’ve also essentially proved the following variant of [AB, Theorem 3.1].Remark 3.4. Suppose X is a separable Banach space, then Cn converges Wijsman to Cif and only if the following two conditions hold.

(i) If x0 ∈C, then there exist xn ∈Cn such that ∥xn −x0∥→0. (ii) If x0 ∈SX∗attains its supremum on C, then there exist x∗n ∈BX∗such that x∗nw∗⇁x∗0and lim supn{supCn x∗n} ≤supC x∗0.Proof.

If (i) and (ii) hold, then the proof of Proposition 3.1 shows that Cn convergesWijsman to C (take W to be an arbitrary singleton).Conversely, (i) follows directlyfrom Wijsman convergence; moreover the z∗j ’s constructed in the proof of Theorem 3.2satisfy (ii) with z∗0 = x∗0 (the w∗-Kadec property was used only to tranform w∗into normconvergence in (ii) which by Proposition 3.1(a) then yields slice convergence).Another way of stating Theorem 3.2 is that Wijsman (compact gap) convergencecoincides with slice (bounded gap) convergence precisely when the dual norm is w∗-Kadec.The analog for Wijsman and weak compact gap convergence is also valid.16

Theorem 3.5. If X is a Banach space, the following are equivalent.

(a) The dual norm on X∗is w∗-τ-Kadec. (b) For each subspace Y of X, Wijsman and weak compact gap convergence coincide.

(c) Wijsman convergence implies Mosco convergence in X. (d) For each separable subspace Y in X, Wijsman convergence implies Mosco convergencefor sequences of level sets of functionals.Proof.

(a) ⇒(b): Let Y be a separable subspace of X. According to Proposition 1.1,the dual norm on Y ∗is w∗-τ-Kadec.

Using this with Remark 3.4 and Proposition 3.1(b)shows that Wijsman and weak compact gap convergence coincide in Y . Combining thiswith a Wijsman versus weak compact gap variant of Theorem 2.2 (the same proof works)shows that (b) holds.

To prove (b) ⇒(c), observe first that M(i) clearly holds. We nowshow M(ii): let xβ ∈Cβ for some subnet and suppose {xβ} is relatively weakly compact.If xβw⇁x and x ̸∈C, then there is an open halfspace containing x and a tail of {xβ}which is strictly separated from C. Let W be the closed convex hull of this tail.

Thend(W, C) > 0, but limβ d(W, Cβ) = 0. This contradicts (b).

Hence we have (b) ⇒(c);this also shows (b) ⇒(d). By Proposition 2.3(a), (c) ⇒(a) and (d) ⇒(a) follows fromProposition 2.3(a) and Proposition 1.1.Corollary 3.6.

If BX∗is w∗-sequentially compact, then each of (a)—(d) in Theorem 3.5is equivalent to the following condition.Sequential Wijsman convergence implies Mosco convergence in X.Proof. This follows from Theorem 3.5, Proposition 2.3(a) and Corollary 1.2.Corollary 3.7.

For a Banach space X, the following are equivalent. (a) BX∗is sequentially compact and sequential Wijsman and slice convergence coincide.

(b) In any subspace of X, Wijsman and slice convergence coincide. (c) X is Asplund and Wijsman and weak compact gap convergence coincide.

(d) X ̸⊃ℓ1 and Wijsman convergence implies Mosco convergence. (e) BX∗is w∗-sequentially compact, X does not contain an isomorphic copy of ℓ1 andsequential Wijsman convergence implies Mosco convergence.Proof.

Using Theorem 1.4, Proposition 2.3, Theorem 3.2 and Theorem 3.5, one cansee that each of (a)—(e) is equivalent to the norm on X∗being w∗-Kadec.On one hand, Corollary 3.7 shows that for a fixed norm on an Asplund space, Wijs-man convergence implies Mosco convergence if and only if Wijsman and slice convergence17

coincide. On the other hand, this does not mean that a sequence of sets converges sliceif and only if it converges Mosco and Wijsman (even in Asplund spaces).

Indeed, [BL,Theorem 6] shows that any separable Banach space can be renormed so that a decreasingsequence of subspaces converges Wijsman and Mosco but not slice. However, it is not clearto us whether Cn converges slice to C whenever X ̸⊃ℓ1 and Cn converges weak compactgap to C.In light of Corollary 3.7, let us mention that there are spaces that are neither Asplundnor Schur which can be renormed so that the dual norm is w∗-τ-Kadec.Example 3.8.

Let Ωbe a σ-finite measure space, then there is a dual w∗-τ-Kadec normon L1(Ω)∗.Proof. According to [BF2, Theorem 2.4], there is a norm on L1(Ω) whose dual normis w∗-τ-Kadec, in fact it is locally uniformly Mackey rotund.We need some more terminology before we can present further corollaries of Theorems3.2 and 3.5.

A norm ∥· ∥is said to be locally uniformly rotund (LUR) if ∥xn −x∥→0whenever 2∥xn∥2+2∥x∥2−∥xn +x∥2 →0. It follows immediately from the definitions thata dual LUR norm is w∗-Kadec.

On the other hand, the dual norm to the usual supremumnorm on c0 is w∗-Kadec but not LUR. For a Banach space X one can define a metric ρon the space P of all equivalent norms on X as follows.

Fix a norm on X with unit ballB1. For µ, ν ∈P, define ρ(µ, ν) = sup{|ν(x) −µ(x)| : x ∈B1}.

It is shown in [FZZ], that(P, ρ) is a Baire space.Corollary 3.9. If X admits a norm for which Wijsman convergence implies slice (Mosco)convergence, then the collection of norms on X for which Wijsman convergence impliesslice (weak compact gap) convergence is residual in (P, ρ).Proof.

The proof of [FZZ, Theorem 2] shows that if the set of norms on X whose dualsare w∗-Kadec (w∗-τ-Kadec) is nonempty, then it is residual in (P, ρ). This with Theorems3.2 and 3.5 proves the corollary.Corollary 3.10.

If X is a WCG Banach space, then the following are equivalent. (a) X is Asplund.

(b) There is a residual collection of norms in (P, ρ) for which Wijsman convergence impliesslice convergence (in any subspace of X). (c) X ̸⊃ℓ1 and there is an equivalent norm on X for which sequential Wijsman conver-gence implies Mosco convergence.18

Proof. (a) ⇒(b): It follows from [F1, Theorem 1], that there is a norm on X whosedual is LUR.

Hence [FZZ, Theorem 2] shows that the collection of norms with dual LURnorms is residual in (P, ρ). Invoking Theorem 3.2 shows that (b) holds.Clearly (b) ⇒(c).

Also, BX∗is w∗-sequentially compact, because X is WCG (see[Di, p. 228]). From this and Corollary 3.7 we conclude that (c) ⇒(a).A weakly countably determined (WCD) space is a more general type of space thanWCG spaces; see [DGZ, Chapter VI] and [F2] for further details and note that [F2] usesthe term Vaˇs´ak space instead of WCD space.Corollary 3.11.

Suppose X is a Banach space such that X∗is WCD. Then there is aresidual collection of norms in (P, ρ) for which Wijsman and slice convergence coincide.Proof.

According to [F2, Theorem 3] and [FZZ, Theorem 2], the collection of normson X with dual LUR norms is residual in (P, ρ).In Remark 3.14 it is observed that there are non-WCG Asplund spaces which cannotbe renormed so that Wijsman and slice convergence coincide. So the assumption that Xis WCG in Corollary 3.10 is not extraneous.

Moreover, there are non-WCG spaces X suchthat X∗is WCG ([JL]) so Corollary 3.11 covers some cases not included in Corollary 3.10.The following theorem shows that if we put some restrictions on the limit set C, wecan obtain slice convergence from Wijsman convergence in spaces whose dual spaces neednot admit any sequentially w∗-Kadec dual norm; see Remark 3.14. Recall that a Banachspace is said to have the Radon-Nikod´ym property (RNP) if every closed convex subsethas slices of arbitrarily small diameter.

See [Bou] for a comprehensive treatment of RNPspaces.Theorem 3.12. Suppose the norm ∥· ∥on X is Fr´echet differentiable and let f : X →IR ∪{+∞} be a convex ℓsc function such that lim∥x∥→∞f(x)∥x∥= ∞.

Suppose further thatfα are ℓsc convex functions such that epifα converges Wijsman to epif in Y = X × IRendowed with the ℓ2 product of the norms. (a) If X has the RNP, then epifα converges slice to epif.

(b) If f has weakly compact level sets, then epifα converges slice to epif.Proof. (a) If epif = ∅, then the result is clear.

Thus we may assume that f is proper.So let f(x0) < +∞. By [Ph, Proposition 3.15], f has an ǫ-subgradient at x0.

Using thiswith the fact that there is an n for which f(x) ≥0 whenever ∥x∥≥n, one can show easilythat f is bounded below.19

Since f is bounded below, we assume (by making appropriate translations) that f ≥0and f(0) < 1. Let us first deal with a sequence {fn} such that epifn converges Wijsmanto epif.

Let Cn = epifn and C = epif. As in the proof of Theorem 3.2, it suffices to prove(ii) in Proposition 3.1 in order to show that Cn converges slice to C.So let y∗0 ∈SY ∗attain its supremum on C. Now write y∗0 = (x∗0, t0) such that x∗0 ∈X∗and t0 ∈IR.

Since C is an epigraph of a function on X and y∗0 is bounded above on C,it follows that t0 ≤0. Let Λn = (x∗0, t0 −1n).

Then t0 −1n ≤−1n and since we assumedf ≥0, we havesupCΛn ≤supCy∗0for all n.(3.6)Moreover,∥Λn −y∗0∥= 1nand∥Λn∥≤1 + 1n. (3.7)Using the growth assumption on f, we choose an ≥2 such that f(x) ≥8n∥x∥whenever∥x∥≥an.

Define the sets Dn byDn = C ∩{(x, r) : x ∈X and r ≤8nan + 2}and let dn = sup{∥y∥: y ∈Dn}. Note that Dn is bounded and (0, 1) ∈Dn so that dn ≥1.Because Y = X × IR has the RNP, according to [Bou, Corollary 3.5.7], we can choosev∗n ∈Y ∗such that∥v∗n −Λn∥≤1ndn,(3.8)and moreover v∗n attains its supremum on both BY and Dn, say,supDnv∗n = ⟨v∗n, yn⟩where yn = (xn, rn) ∈Dn.

(3.9)¿From (3.8), one hassupDnv∗n ≤supDnΛn + 1n. (3.10)Writing v∗n = (x∗n, tn), we have −1 −1n −1ndn ≤tn ≤−1n +1ndn , and so for n ≥2, we havetn ≥−2.

From now on, for convenience, we will assume that n ≥2. Since (0, 1) ∈Dn,one has⟨v∗n, (0, 1)⟩= ⟨x∗n, 0⟩+ tn ≥−2 and thus supDnv∗n ≥−2.

(3.11)If rn ≥8nan, then because tn ≤−12n we have−2 ≤⟨v∗n, (xn, rn)⟩= ⟨x∗n, xn⟩+ rntn ≤2∥xn∥−8nan12n20

and so2an < −2 + 4an ≤2∥xn∥that is ∥xn∥> an.But, by the choice of an, if ∥xn∥≥an, then rn ≥f(xn) ≥8n∥xn∥and consequently⟨v∗n, (xn, rn)⟩≤⟨x∗n, xn⟩−12n8n∥xn∥≤2∥xn∥−4∥xn∥< −2.This contradicts (3.11). Thus we have rn ≤8nan.Recall that v∗n attains its norm on BY .

Now let vn = (˜xn, ˜rn) be such that ∥vn−yn∥=1 and⟨v∗n, vn −yn⟩= ∥v∗n∥. (3.12)Observe that ˜rn ≤rn + 1 ≤8nan + 1 and so if (c, t) ∈C and ∥(c, t) −(˜xn, ˜rn)∥≤1, thent ≤8nan + 2 which means (c, t) ∈Dn.

Since d(vn, Dn) ≤1, we have d(vn, Dn) = d(vn, C)and moreover,1 =1∥v∗n∥⟨v∗n, vn −yn⟩≤infC1∥v∗n∥⟨v∗n, vn −c⟩≤d(vn, C) ≤∥vn −yn∥= 1.Now let Λn,k ∈∂d(·, Ck)(vn). Thenlim supk⟨Λn,k, yn −vn⟩≤lim supkd(yn, Ck) −d(vn, Ck)= d(yn, C) −d(vn, C) = −1.Therefore limk⟨Λn,k, vn −yn⟩→1 = ⟨v∗n∥v∗n∥, yn −vn⟩.

As ∥Λn,k∥≤1 and ∥· ∥is Fr´echetdifferentiable, from ˇSmulyan’s criterion (see [DGZ, Theorem I.1.4]), one obtains(i) limkΛn,k −v∗n∥v∗n∥ = 0 and in particular (ii) limk ⟨Λn,k, vn⟩=1∥v∗n∥⟨v∗n, vn⟩. (3.13)Let zk ∈Ck be arbitrary, because Λn,k ∈∂d(·, Ck)(vn) we havelim supk⟨Λn,k, zk −vn⟩≤lim supkd(zk, Ck) −d(vn, Ck)= −d(vn, C) = −1.

(3.14)21

Consequently, we obtainlim supk⟨Λn,k, zk⟩≤lim supk⟨Λn,k, vn⟩−1[by (3.14)]=1∥v∗n∥⟨v∗n, vn⟩−1[by (3.13(ii))]=1∥v∗n∥⟨v∗n, vn⟩−∥v∗n∥=1∥v∗n∥⟨v∗n, yn⟩[by (3.12)]=1∥v∗n∥supDnv∗n[by (3.9)]≤1∥v∗n∥supDnΛn + 1n[by (3.10)]≤1∥v∗n∥supCy∗0 + 1n. [by (3.6)]This with (3.13(i)) shows that there exists kn ∈IN such that whenever k ≥kn one has:supCkΛn,k ≤supCy∗0 +1∥v∗n∥−1supCy∗0 +1n∥v∗n∥+ 1n,and∥Λn,k −y∗0∥≤Λn,k −v∗n∥v∗n∥ + v∗n∥v∗n∥−y∗0 ≤1n + v∗n∥v∗n∥−y∗0.By replacing kn with a larger number in necessary, we may assume kn > kn−1 for all n.For k < k2, let y∗k = 0 and for kn ≤k < kn+1 let y∗k = Λn,k.

According to (3.7) and(3.8), ∥v∗n∥→1 and ∥v∗n −y∗0∥→0. Thus it is clear from the above inequalities that∥y∗k −y∗0∥→0 andlim supksupCky∗k≤supCy∗0.Therefore epifn converges slice to epif.

The statement for nets follows from Theorem 2.2since the RNP and Fr´echet differentiable norms are inherited by subspaces. This completesthe proof of (a).

(b) Notice that the RNP was used only to obtain a dense set of functionals whichsimultaneously support Dn and BY . Since the level sets of f are assumed to be weaklycompact, it follows that every functional attains its supremum on Dn.Hence by theBishop-Phelps theorem there is a dense set of support functionals which simultaneouslysupport Dn and BY .

Therefore (b) follows from the proof of (a).By considering indicator functions in the above theorem, we immediately obtain thefollowing result.22

Corollary 3.13. (a) Suppose X has the RNP and its norm is Fr´echet differentiable.

If Cis a closed bounded convex set and Cα converges Wijsman to C, then Cα converges sliceto C.(b) Suppose the norm on X is Fr´echet differentiable and C is weakly compact. If Cαconverges Wijsman to C, then Cα converges slice to C.Remark 3.14.

There are spaces admitting Fr´echet differentiable norms for which thereis no equivalent norm such that Wijsman and slice convergence coincide. Indeed C[0, ω1],admits a Fr´echet differentiable norm ([Ta, Theorem 4]), while the techniques of [Ta, Theo-rem 3] can be used to show C[0, ω1]∗does not admit any dual sequentially w∗-Kadec norm.Hence it is necessary to have restrictions on the limit set in Theorem 3.12 and Corollary3.13.

Also, any space with a separable second dual is an example of a space with the RNPthat admits a Fr´echet differentiable norm.Finally, let us mention that one can use subdifferential techniques as in the proof ofTheorem 3.12 to show that Wijsman and slice convergence coincide if the dual norm isLUR. This proof is somewhat simpler than the w∗-Kadec case.

Also, similar techniquescan be used to provide a simple direct proof that Wijsman convergence implies Moscoconvergence when the dual norm is w∗-τ-Kadec; moreover, the proof of [Be2, Theorem 2.5],without modification, shows this result. However, these proofs do not appear to providethe additional information that Wijsman and weak compact gap convergence coincide inthis case.

As a follow up to the question stated at the end of Section 1, we should mentionthat we don’t know if sequential Wijsman convergence implies slice convergence providedthe dual norm is only assumed to be sequentially w∗-Kadec.References[AB] H. Attouch and G. Beer, On the convergence of subdifferentials of convex functions,Sem. d’Anal.

Convexe Montpellier 21 (1991) Expos´e No 8 (to appear in Arch. Mat.).

[Be1] G. Beer, Convergence of continuous linear functionals and their level sets, Arch. Math.52 (1989), 482–491.

[Be2] G. Beer, Mosco convergence and weak topologies for convex sets and functions, Math-ematika 38 (1991), 89–104. [Be3] G. Beer, The slice topology, a viable alternative to Mosco convergence in nonreflexivespaces, J. Nonlinear Analysis: Theory, Methods, Appl.

19 (1992), 271–290. [Be4] G. Beer, Wijsman convergence of convex sets under renorming, preprint.

[BB1] G. Beer and J.M. Borwein, Mosco convergence and reflexivity, Proc.

Amer. Math.Soc.

109 (1990), 427–436.23

[BB2] G. Beer and J. Borwein, Mosco convergence of level sets and graphs of linear func-tionals, J. Math.

Anal. Appl.

(in press). [BP] E. Bishop and R.R.

Phelps, The support functionals of a convex set, Proc. Sympos.Pure Math., Vol.

VII, pp. 27–35, Amer.

Math. Soc., Providence, Rhode Island, 1963.

[Bor] J.M. Borwein, Asplund spaces are “sequentially reflexive”, preprint.

[BFa] J.M. Borwein and M. Fabian, On convex functions having points of Gateaux differen-tiability which are not points of Fr´echet differentiability, preprint.

[BFV] J.M. Borwein, M. Fabian and J. Vanderwerff, Locally Lipschitz functions and bornolog-ical derivatives, preprint.

[BF1] J.M. Borwein and S. Fitzpatrick, Mosco convergence and the Kadec property, Proc.Amer.

Math. Soc.

106 (1989), 843–851. [BF2] J.M Borwein and S. Fitzpatrick, A weak Hadamard smooth renorming of L1(Ω, µ),Canad.

Math. Bull.

(in press). [BL] J.M.

Borwein and A.S. Lewis, Convergence of decreasing sequences of convex sets innonreflexive Banach spaces, preprint. [Bou] R.D.

Bourgin, Geometric Aspects of Convex Sets with the Radon-Nikod´ym Property,Lecture Notes in Mathematics 993, Springer-Verlag, 1983. [DGZ] R. Deville, G. Godefroy and V. Zizler, Smoothness and Renormings in Banach Spaces,Longman Monographs in Pure and Applied Mathematics (to appear).

[Di] J. Diestel, Sequences and Series in Banach Spaces, Graduate Texts in Mathematics93, Springer-Verlag, 1984. [Du] D. v.Dulst, Characterizations of Banach Spaces not Containing ℓ1, CWI Tract,Amsterdam, 1989.

[DS] D. v. Dulst and I. Singer, On Kadec-Klee norms on Banach spaces, Studia Math.

54(1976), 203–211. [F1] M. Fabian, Each weakly countably determined Asplund space admits a Fr´echet dif-ferentiable norm, Bull.

Austral. Math.

Soc. 36 (1987), 367–374.

[F2] M. Fabian, On a dual locally uniformly rotund norm on a dual Vaˇs´ak space, StudiaMath. 101 (1991), 69–81.

[FZZ] M. Fabian, L. Zaj´ıˇcek and V. Zizler, On residuality of the set of rotund norms on aBanach space, Math. Ann.

258 (1982), 349–351. [GM] N. Ghoussoub and B. Maurey, The asymptotic-norming property and the Radon-Nikodym properties are equivalent in separable Banach spaces, Proc.

Amer. Math.Soc.

94 (1985), 665–671. [JH] R.C.

James and A. Ho, The aysmptotic-norming and Radon-Nikodym properties forBanach spaces, Ark. Mat.

19 (1981), 53–70. [JL] W.B.

Johnson and J. Lindenstrauss, Some remarks on weakly compactly generatedBanach spaces, Israel J. Math.

17 (1974), 219–230. [LP] D.G.

Larman and R.R. Phelps, Gateaux differentiability of convex functions on Ba-nach spaces, J. London Math.

Soc. 20 (1979), 115–127.

[M] U. Mosco, Convergence of convex sets and of solutions of variational inequalities, Adv.Math. 3 (1969), 510–585.

[Ø] P. Ørno, On J. Borwein’s concept of sequentially reflexive Banach spaces, preprint.24

[Ph] R.R. Phelps, Convex functions, monotone operators and differentiability, LectureNotes in Mathematics 1364, Springer-Verlag, 1989.

[Ta] M. Talagrand, Renormages de quelques C(K), Israel J. Math.

54 (1986), 327–334. [Tr] S. Troyanski, On a property of the norm which is close to local uniform rotundity,Math.

Ann. 271 (1985), 305–313.

[W] R. Wijsman, Convergence of sequences of convex sets, cones and functions, II, Trans.Amer. Math.

Soc. 123 (1966), 32–45.25

---

출처: arXiv:9302.212 • 원문 보기