Journal of Convex Analysis Volume 2 (1995), No.1/2, 277–285

Norms that Generate the Same Wijsman Topology on Convex Sets

Pietro Poggi-Corradini

Math. Dept., GN-50, University of Washington, Seattle, WA 98195, USA. e-mail: poggi@math.washington.edu

Dedicated to R. T. Rockafellar on his 60th Birthday

In a Banach space X, we introduce a criterion for comparing the Wijsman topologies that are induced by two equivalent norms of X on the hyperspace of closed convex sets C(X). Thereafter, we study the duality map associated with the unit ball of a given norm of X in relation to its composition with the polarity map. This more geometrical description of the norm allows us to give a direct proof of a known theorem (see [3] and [1]): If X is reflexive and the duality map is n-to-n-usco, then the Wijsman topology coincides with the Mosco topology on C(X).

1. Introduction

Let X be a Banach space and let C(X) be the family of closed and convex subsets of X. The Wijsman topology on C(X) is defined to be the weakest topology for which the distance functionals d(x, .) : C(X) ⟶ IR are continous for all x ∈ X. This topology can also be viewed as the union of two other topologies: the lower Vietoris which is generated by all subsets of C(X) of the form V^{−} = {A ∈ C(X) | A ∩ V ≠ ∅}, where V is open in X; and the upper Wijsman which is generated by all sets of the form (B^{c})^{++} = {A ∈ C(X) | inf_{a∈A b∈B} ∥a − b∥ = D(A, B) > 0}, where B is a homothetic translation of the unit ball of X. Clearly the lower Vietoris depends only on the open sets of X, hence only on the topology of X, while the upper Wijsman depends also on the norm chosen on X. Our purpose is to give a geometric criterion for deciding when two equivalent norms of X generate the same upper Wijsman topology on C(X) and, in order to illustrate a possible application of this criterion, we will give another proof of a theorem of Borwein and Fitzpatrick (see [3]). This theorem says that, if X is reflexive and the norm p on X is sufficiently “smooth” (i.e. the duality map is n-to-n-usco), then the Wijsman topology associated to p is the maximum of all the Wijsman topologies induced by equivalent norms, i.e. is the Mosco topology (see [2] for this). In the following, U and S will always denote, respectively, the closed unit ball and the unit sphere. We will add an upper-star to objects lying in the dual space. In the next section we illustrate a geometrical criterion involving unit balls of X which allows us to compare the corresponding Wijsman topologies on C(X). In the third section we study the relationship between the unit ball and the polar of its associated duality map under the hypothesis that the duality map is norm-to-norm-upper-semicontinous and

ISSN 0944-6532 / $ 2.50 ◯^{c} Heldermann Verlag

278 P. Poggi-Corradini / Norms that generate the Wijsman topology on convex sets

norm-compact valued. Finally, in section 4 we give a different proof of the theorem of Borwein and Fitzpatrick.

2. A criterion for comparing Wijsman topologies

The criterion we propose in this section has been inspired by the one given in [4] in the case of the hyperspace of closed subsets of a metric space. Suppose that p_{0} and p_{1} are two equivalent norms on X. For any x^{∗} ∈ X^{∗} with ∥x^{∗}∥_{1} = 1 and any ϵ > 0, consider the closed half-space H(x^{∗}, ϵ) = {x ∈ X | ⟨x^{∗}, x⟩ ≤ 1 + ϵ}. Note that H(x^{∗}, ϵ) contains U_{1}, because sup_{x∈U_{1}}⟨x^{∗}, x⟩ = ∥x^{∗}∥_{1} = 1 < 1 + ϵ.

Definition 2.1. We say that p_{0} covers p_{1} if, for every x^{∗} ∈ S_{1}^{∗} and for every ϵ > 0, there are x_{1}, …, x_{n} ∈ X and λ_{1}, …, λ_{n} > 0 such that:

U_{1} ⊂ ⋃_{i=1}^{n}(x_{i} + λ_{i}U_{0}) ⊂ H(x^{∗}; ϵ)

In other words, p_{0} covers p_{1}, if, whenever we choose a hyperplane with positive distance from the ball U_{1}, it is always possible to find a finite number of p_{0}–balls whose union contains U_{1} but does not intersect that hyperplane. The next theorem will justify the choice of such a criterion, but first we need to state some preliminary well-known facts, which we will prove for the sake of completeness.

Lemma 2.2.

1. If D ⊂ X, then for all x ∈ X and λ > 0: x + λ(D^{c})^{++} = [(x + λD)^{c}]^{++}.

2. If D_{i} ⊂ X for i = 1, …, n, then: [(∪_{i=1}^{n}D_{i})^{c}]^{++} = ∩_{i=1}^{n}(D_{i}^{c})^{++}.

3. If E is closed and D ⊂ X: D ⊂ E ⟺ (E^{c})^{++} ⊂ (D^{c})^{++}.

Proof.

1. Note that D(x + λA, x + λD) > 0 ⇔ inf_{a∈A b∈D} ∥(x + λa) − (x + λb)∥ > 0 ⇔ inf_{a∈A b∈D} λ∥a − b∥ > 0 ⇔ D(A, D) > 0. So x + λA ∈ [(x + λD)^{c}]^{++} ⇔ A ∈ (D^{c})^{++} ⇔ x + λA ∈ x + λ(D^{c})^{++}.

2. If A ∈ [(∪_{i=1}^{n}D_{i})^{c}]^{++}, then D(A, ∪_{i=1}^{n}D_{i}) > 0. So D(A, D_{i}) > 0 for all i = 1, …, n. Hence A ∈ (D_{i}^{c})^{++} for all i and A ∈ ∩_{i=1}^{n}(D_{i}^{c})^{++}. Conversely, suppose D(A, D_{i}) > 0 for all i = 1, …, n. Then D(A, ∪_{i=1}^{n}D_{i}) = min_{i=1...n} D(A, D_{i}) > 0, so A ∈ [(∪_{i=1}^{n}D_{i})^{c}]^{++}.

3. Suppose D ⊂ E, then if A ∈ (E^{c})^{++}, D(A, D) ≥ D(A, E) > 0. So A ∈ (D^{c})^{++}. Conversely, suppose (E^{c})^{++} ⊂ (D^{c})^{++}. If D ⊄ E, pick x ∈ D \ E. Then, since E is closed, d(x, E) > 0. So {x} ∈ (E^{c})^{++} ⊂ (D^{c})^{++}, hence d(x, D) > 0, but x ∈ D, and this is a contradiction.

Theorem 2.3. For i = 0, 1, let τ_{W_{p_{i}}}^{+} be the upper Wijsman topology induced by the

norm p_{i} on C(X), the space of closed convex subsets of X. Then:

p_{0} covers p_{1} ⟺ τ_{W_{p_{0}}}^{+} ≥ τ_{W_{p_{1}}}^{+}

Proof. First we assume that p_{0} covers p_{1}. It is enough to show that the subbase of τ_{W_{p_{1}}}^{+},

as defined in Section 1, is contained in τ_{W_{p_{0}}}^{+}. Consider a homothetic translation of the

unit ball for the norm p_{1}: B = x + λU_{1}, with x ∈ X and λ > 0. By Lemma 2.2 (1), we have x + λ(U_{1}^{c})^{++} = [(x + λU_{1})^{c}]^{++}. Therefore, we only need to prove that (U_{1}^{c})^{++} is open in τ_{W_{p_{0}}}^{+}, because the topology τ_{W_{p_{0}}}^{+} is invariant under translations and positive

dilations (this is easily checked on its subbase). So, let Â be a fixed closed convex set in (U_{1}^{c})^{++}. Since D(Â, U_{1}) > 0, there is δ > 0 such that (1 + δ)U_{1} ∩Â = ∅. Then, the Hahn–Banach theorem gives us x^{∗} ∈ S_{1}^{∗} such that:

sup_{z∈U_{1}}⟨x^{∗}, z⟩ = 1 < 1 + δ < inf_{a∈Â}⟨x^{∗}, a⟩  (2.1)

We know that p_{0} covers p_{1}, therefore we can cover U_{1} with homothetic translations of U_{0} without exceeding H(x^{∗}; δ). That is to say, there are x_{1}, …, x_{n} ∈ X and λ_{1}, …, λ_{n} > 0 such that:

U_{1} ⊂ ⋃_{i=1}^{n}(x_{i} + λ_{i}U_{0}) ⊂ H(x^{∗}; δ)  (2.2)

Define B_{i} = x_{i} + λ_{i}U_{0} and B = ∪_{i=1}^{n}B_{i}. We claim that (B^{c})^{++} is an open set in τ_{W_{p_{0}}}^{+}

which contains Â and is contained in (U_{1}^{c})^{++}. From the second inclusion in (2.2) we have that B ⊂ H(x^{∗}; δ). Moreover, since H(x^{∗}; δ) is a half-space, for any a ∉ H(x^{∗}; δ), d(a, H(x^{∗}; δ)) = ⟨x^{∗}, a⟩ − (1 + δ). Hence, by the second inequality in (2.1), we have that: D(Â, B) ≥ D(Â, H(x^{∗}, δ)) = inf_{a∈Â}⟨x^{∗}, a⟩ − (1 + δ) > 0. Therefore Â ∈ (B^{c})^{++}. Moreover, by the first inclusion in (2.2), U_{1} ⊂ B, so by Lemma 2.2 (3), (B^{c})^{++} ⊂ (U_{1}^{c})^{++}. Thus, we proved that Â ⊂ (B^{c})^{++} ⊂ (U_{1}^{c})^{++}. Finally, by Lemma 2.2 (2), we have (B^{c})^{++} = [(∪_{i=1}^{n}B_{i})^{c}]^{++} = ∩_{i=1}^{n}(B_{i}^{c})^{++}, which ensures that (B^{c})^{++} is an open set in τ_{W_{p_{0}}}^{+}. Since Â was arbitrary, (U_{1}^{c})^{++} is also open in τ_{W_{p_{0}}}^{+}. So τ_{W_{p_{0}}}^{+} ≥ τ_{W_{p_{1}}}^{+}.

Conversely, suppose that τ_{W_{p_{0}}}^{+} ≥ τ_{W_{p_{1}}}^{+}. Fix x^{∗} ∈ S_{1}^{∗} and ϵ > 0. Then the closed half-

space A = {x ∈ X | ⟨x^{∗}, x⟩ ≥ 1 + ϵ} belongs to (U_{1}^{c})^{++} ∩ C(X), because D(A, U_{1}) = inf_{a∈A z∈U_{1}} ∥a−z∥ ≥ inf_{a∈A z∈U_{1}}⟨x^{∗}, a−z⟩ = inf_{a∈A}⟨x^{∗}, a⟩−sup_{z∈U_{1}}⟨x^{∗}, z⟩ = (1+ϵ)−1 = ϵ > 0. So there is an open set B in τ_{W_{p_{0}}}^{+} that contains A and is itself contained in (U_{1}^{c})^{++}.

Now, B can be chosen to be in the base of τ_{W_{p_{0}}}^{+}, i.e. B = ∩_{i=1}^{n}(B_{i}^{c})^{++}, where B_{i} are

homothetic translations of U_{0}. By Lemma 2.2, ∩_{i=1}^{n}(B_{i}^{c})^{++} = [(∪_{i=1}^{n}B_{i})^{c}]^{++} ⊂ (U_{1}^{c})^{++} implies U_{1} ⊂ ∪_{i=1}^{n}B_{i}; while, A ∈ [(∪_{i=1}^{n}B_{i})^{c}]^{++} implies ∪_{i=1}^{n}B_{i} ⊂ A^{c} ⊂ cl(A^{c}) = H(x^{∗}; ϵ). So U_{1} ⊂ ∪_{i=1}^{n}B_{i} ⊂ H(x^{∗}; ϵ).

Accordingly, if a norm p covers all other equivalent norms, then the Wijsman topology induced by p will be the maximum of all the Wijsman topologies induced by equivalent norms. This observation will be our main tool for proving the theorem of Borwein and Fitzpatrick. The intuition is that the unit ball U of a norm that covers any equivalent norm cannot have “sharp corners”, i.e. must satisfy some rotundity property, in order to finitely cover any other equivalent ball without exceeding any given hyperplane. In this context the right object to consider for studying the corners of U is the polar of the duality map.

3. The unit ball and the polar of its duality map

3.1. The smallest cones containing the unit ball

Let p be a given norm on X. The duality map associated with p is a multifunction J : S ⟶ S^{∗} defined as follows:

For all x ∈ S, J(x) = {x^{∗} ∈ S^{∗} | ⟨x^{∗}, x⟩ = 1}.

Note that J has always non-empty, weak^{∗}-compact and convex values (see [6]). Recall that for an arbitrary set A ⊂ X^{∗} we can define its polar set in X to be:

A^{∘} = {x ∈ X | sup_{a∈A}⟨a, x⟩ ≤ 1} = ⋂_{a∈A}{x ∈ X | ⟨a, x⟩ ≤ 1}

Clearly, A^{∘} is always closed and convex. In what follows our attention will focus on the polar set of the set J(x̄) for a fixed x̄ ∈ S. We will show that J(x̄)^{∘} is the smallest closed convex cone with vertex at x̄ containing the unit ball U. But first we need to better describe the latter. So, fix x̄ ∈ S. Define B_{λ} = x̄ + λ(U − x̄), for all λ > 0, and let B = ∪_{λ>0}B_{λ}.

Lemma 3.1. The B_{λ} form an increasing family of balls; x̄ is in every ∂B_{λ} = x̄+λ(S−x̄), for all λ > 0; also B = ∪_{λ>0}∂B_{λ} = x̄ + ∪_{λ>0}λ(S − x̄); and finally, cl(B) is the smallest closed convex cone with vertex at x̄ containing U.

Proof. Suppose x ∈ B_{λ}; then x = x̄ + λ(y − x̄) for some y ∈ U. Let μ > λ, then we also have x = μ[\frac{λ}{μ}y + (1 − \frac{λ}{μ})x̄] + (1 − μ)x̄. But 0 < \frac{λ}{μ} < 1, so by convexity \frac{λ}{μ}y + (1 − \frac{λ}{μ})x̄ ∈ U, hence x ∈ B_{μ}. Therefore B_{λ} is increasing in λ. Next, since x̄ ∈ S, x̄ = x̄ + λ(x̄ − x̄) ∈ ∂B_{λ} for all λ > 0. To see that B = x̄ + ∪_{λ>0}λ(U − x̄) = x̄ + ∪_{λ>0}λ(S − x̄), first note that S ⊂ U. So x̄+∪_{λ>0}λ(S−x̄) ⊂ B. Hence, it is enough to show that B ⊂ x̄+∪_{λ>0}λ(S−x̄). Let x ∈ B, then x ∈ B_{λ} for some λ > 0. Let λ_{0} = inf\{λ > 0 | x ∈ B_{λ}\} = inf\{λ > 0 | \frac{x−x̄}{λ} ∈ U − x̄\}.

We can suppose that x ≠ x̄, so, for λ < \frac{∥x−x̄∥}{2}, we have ∥\frac{x−x̄}{λ}∥ > 2, i.e. \frac{x−x̄}{λ} ∉ U − x̄, thus λ_{0} > 0. For all λ > λ_{0}, \frac{x−x̄}{λ} ∈ U − x̄, but U − x̄ is closed, so \frac{x−x̄}{λ_{0}} ∈ U − x̄. Moreover, for λ < λ_{0}, \frac{x−x̄}{λ} ∉ U − x̄. Therefore, \frac{x−x̄}{λ_{0}} ∈ (U − x̄) ∩ cl((U − x̄)^{c}) = S − x̄. Hence x ∈ x̄ + λ_{0}(S − x̄). Thus B = x̄ + ∪_{λ>0}λ(S − x̄). Now let us prove that B is a convex cone with vertex in x̄. For this we show that C = ∪_{λ>0}λ(U − x̄) is a convex cone with vertex at the origin. Consider x_{1}, x_{2} ∈ C, then there are λ_{1}, λ_{2} > 0 and y_{1}, y_{2} ∈ U such that: x_{i} = λ_{i}(y_{i} − x̄) for i = 1, 2. Therefore:

\begin{aligned}x_{1} + x_{2} & = λ_{1}y_{1} + λ_{2}y_{2} − (λ_{1} + λ_{2})x̄ \\ & = (λ_{1} + λ_{2})\frac{λ_{1}y_{1} + λ_{2}y_{2}}{λ_{1} + λ_{2}} − (λ_{1} + λ_{2})x̄ \end{aligned}

Since U is convex x_{1} + x_{2} ∈ (λ_{1} + λ_{2})(U − x̄) ⊂ C. Hence, C + C ⊂ C. Thus, C is a convex cone. Therefore, cl(B) = cl(x̄ + C) is a closed convex cone with vertex at x̄. Moreover, all convex cones containing U − x̄ must contain λ(U − x̄) for all λ > 0, hence must contain C. Therefore cl(B) is the smallest convex cone containing U with vertex at x̄.

Proposition 3.2. Let x̄ ∈ S and B be defined as above. Then J(x̄)^{∘} = cl(B).

Proof. Consider x ∈ B_{λ}; then x = x̄ + λ(y − x̄) with y ∈ U. Thus, for all x^{∗} ∈ J(x̄), ⟨x^{∗}, x⟩ = 1+λ(⟨x^{∗}, y⟩−1) ≤ 1. So x ∈ J(x̄)^{∘}. Therefore, B ⊂ J(x̄)^{∘}, hence cl(B) ⊂ J(x̄)^{∘}, since, by its definition, J(x̄)^{∘} is closed. Conversely, suppose that x ∈ J(x̄)^{∘} \ cl(B). Since cl(B) is closed and convex, by the Hahn–Banach theorem we can find d > 0 and x^{∗} ∈ S^{∗} such that: ⟨x^{∗}, x⟩ > d > sup_{b∈cl(B)}⟨x^{∗}, b⟩. Thus sup_{b∈cl(B)}⟨x^{∗}, b⟩ < ∞ and since cl(B) is a cone with vertex at x̄, this implies that sup_{b∈cl(B)}⟨x^{∗}, b⟩ = ⟨x^{∗}, x̄⟩. Moreover, U ⊂ cl(B), therefore we have:

1 ≥ ⟨x^{∗}, x̄⟩ = sup_{b∈cl(B)}⟨x^{∗}, b⟩ ≥ sup_{b∈U}⟨x^{∗}, b⟩ = ∥x^{∗}∥ = 1

So ⟨x^{∗}, x̄⟩ = 1, hence x^{∗} ∈ J(x̄). But then ⟨x^{∗}, x⟩ > d > 1 contradicts the fact that x ∈ J(x̄)^{∘}.

This proposition says that J(x̄)^{∘} is generated by positive linear combinations of U and x̄, so J(x̄)^{∘} is the smallest closed convex cone with vertex at x̄ containing U. In our “downto-earth” discussion we will say that J(x̄)^{∘} is the corner of U at x̄. Then, Proposition 3.2 can be restated by saying that the dilations of U spreading from x̄ form an exhaustion for the corner of U at x̄. Since in the sequel we will assume that J is norm-compact valued we need to study more carefully the polar of a compact subset of S^{∗}. The next section is devoted to this purpose.

3.2. The polar of a compact subset of S^{∗}

We defined the polar of a set A to be A^{∘} = ⋂_{x^{∗}∈A}{x ∈ X | ⟨x^{∗}, x⟩ ≤ 1}. In the next proposition we will look at the consequences of A being a compact subset of S^{∗}. It is easy to show that int(A^{∘}) is weakly open in X if and only if A is bounded and finite dimensional in S^{∗}. Here we will show that, if A is norm compact in S^{∗}, then the norm interior of A^{∘} is bounded-weakly open in X, that is to say, its intersection with any bounded set E in X is relatively weakly open in E (see [5] p.48).

Proposition 3.3. If A is a norm-compact subset of S^{∗}, then:

1. int(A^{∘}) = ⋂_{x^{∗}∈A}{x ∈ X | ⟨x^{∗}, x⟩ < 1}.

2. int(A^{∘}) is bounded-weakly open in X.

Proof. First suppose x ∈ int(A^{∘}), then there is r > 0 such that x + rU ⊂ int(A^{∘}) ⊂ A^{∘}. So for all x^{∗} ∈ A, sup_{y∈U}⟨x^{∗}, x+ry⟩ ≤ 1, hence ⟨x^{∗}, x⟩+r∥x^{∗}∥ ≤ 1, i.e. ⟨x^{∗}, x⟩ ≤ 1−r < 1. Conversely, suppose that for all x^{∗} ∈ A, ⟨x^{∗}, x⟩ < 1. Then, since A is norm-compact and f(x^{∗}) = ⟨x^{∗}, x⟩ is norm-continous on X^{∗}, we have max_{x^{∗}∈A}⟨x^{∗}, x⟩ = δ < 1. Let r = 1 − δ and consider z ∈ x + rU, then z = x + ry with y ∈ U, so, for all x^{∗} ∈ A, ⟨x^{∗}, z⟩ = ⟨x^{∗}, x⟩ + r⟨x^{∗}, y⟩ ≤ (1 − r) + r = 1. Therefore x + rU ⊂ A^{∘}, i.e. x ∈ int(A^{∘}).

To show that int(A^{∘}) ∩ E is relatively weakly open in E for any bounded set E, we will show that the complement is relatively weakly closed. Suppose x_{α} is a net in E weakly convergent to x_{0} ∈ E such that, for all α, x_{α} ∉ int(A^{∘}). By part (1), this means that for all α we can find x_{α}^{∗} ∈ A such that ⟨x_{α}^{∗}, x_{α}⟩ ≥ 1. Since x_{α} is bounded, and converges weakly to x_{0}, this implies that ⟨x_{0}^{∗}, x_{0}⟩ = lim⟨x_{α}^{∗}, x_{α}⟩ ≥ 1, for some x_{0}^{∗} ∈ A so x_{0} ∉ int(A^{∘}). Hence E \ int(A^{∘}) is weakly closed in E, so int(A^{∘}) ∩ E is weakly open in E.

This Proposition applied to our study of the duality map tells us that, when J is compactvalued, int(J(x̄)^{∘}) is bounded-weakly open in X for any x̄ ∈ S. That is to say, the corner of U at x̄ is sufficiently “flat”, since weakly open sets are in a sense “bigger” than normopen sets.

3.3. A stronger continuity property for the duality map

A multivalued map F between two topological spaces T and S is said to be uppersemicontinous if for each x ∈ T and any neighborhood V of F (x) in S there is a neighborhood W of x in T such that F (W) ⊂ V. Note that if S is a metric space, e.g. a normed linear space, and F (x) is compact, then the neighborhoods V of F (x) can be assumed to be ϵ-expansions of F (x): S_{ϵ}[F (x)] = ∪_{y∈F (x)}{z ∈ S | d(y, z) < ϵ}). Given a norm on a Banach space X, the duality map J associated with it is a multivalued map from the sphere S to the dual sphere S^{∗} which is uppersemicontinous if we endow S with the norm topology and S^{∗} with the weak^{∗}-topology (see [6]). In the following we require a stronger condition, i.e. we want J to be norm to norm uppersemicontinous (S^{∗} endowed with the dual norm topology). Moreover, the values of J are always weak^{∗}compact by definition, but here we ask that J be norm-compact valued. In short we say that J is n-to-n-usco. The next lemma is well-known.

Lemma 3.4. Suppose that J is n-to-n-usco. If {x_{k}}, x̄ ∈ S, z_{k}^{∗} ∈ J(x_{k}), and x_{k} ⟶ x̄ in norm, then there is x_{0}^{∗} ∈ J(x̄) such that, “up to subsequences”, z_{k}^{∗} ⟶ x_{0}^{∗} in norm.

Proof. Since J(x̄) is convex and weak^{∗}-compact, there is x_{k}^{∗} ∈ J(x̄) such that ∥z_{k}^{∗}−x_{k}^{∗}∥ = d(z_{k}^{∗}, J(x̄)). By assumption, J(x̄) is norm-compact, so, up to subsequences, x_{k}^{∗} converges in norm to x_{0}^{∗} ∈ J(x̄), i.e. given ϵ > 0 there is k_{0} such that ∥x_{k}^{∗} − x_{0}^{∗}∥ ≤ ϵ for k ≥ k_{0}. Moreover, J is norm-to-norm uppersemicontinous in x̄, hence there is k_{1} such that for all k ≥ k_{1}: J(x_{k}) ⊂ S_{ϵ}[J(x̄)] = {x^{∗} ∈ X^{∗} | d(x^{∗}, J(x̄)) < ϵ}. Therefore, for all k ≥ k_{1}, z_{k}^{∗} ∈ S_{ϵ}[J(x̄)], i.e. ∥z_{k}^{∗} − x_{k}^{∗}∥ = d(z_{k}^{∗}, J(x̄)) < ϵ, so ∥z_{k}^{∗} − x_{0}^{∗}∥ ≤ ∥z_{k}^{∗} − x_{k}^{∗}∥ + ∥x_{k}^{∗} − x_{0}^{∗}∥ ≤ ϵ + ϵ = 2ϵ, for k ≥ max{k_{0}, k_{1}}.

The map J being n-to-n-usco has an interesting consequence for the polar of J(x̄), for a fixed x̄ ∈ S. In Section 3.1 we showed that B_{λ} = x̄ + λ(U − x̄) is an increasing family of balls such that int(J(x̄)^{∘}) ⊂ ∪_{λ>0}B_{λ} ⊂ J(x̄)^{∘}. The same is true if we take the balls B_{λ} to be open, i.e. int(B_{λ}) instead of B_{λ}. So, given any norm-compact subset K of int(J(x̄)^{∘}), by compactness, there is λ_{0} such that K is contained in int(B_{λ}) for all λ ≥ λ_{0} > 0. In a similar way, when J is n-to-n-usco, given any weakly compact subset K of int(J(x̄)^{∘}), we can find λ_{0} so that K is contained in all the B_{λ}’s for λ ≥ λ_{0} > 0.

Proposition 3.5. Suppose J is n-to-n-usco at x̄. If K ⊂ int(J(x̄)^{∘}) is a weakly compact set, then there is λ_{0} > 0 such that K ⊂ B_{λ_{0}} = x̄ + λ_{0}(U − x̄).

Proof. Since X is a Banach space, by the Eberlein-Smulian Theorem (see [5] Theorem 1. p. 58), K is weakly sequentially compact. Suppose that for every natural number n > 0 there is x_{n} ∈ K \ B_{n}. Then we can extract a subsequence y_{k} which converges weakly to x_{0} ∈ K, where y_{k} = x_{n_{k}} ∈ K \ B_{n_{k}}. By Lemma 3.1, we have int(J(x̄)^{∘}) = int(cl(B)) = int(B) ⊂ x̄ + ∪_{λ>0}λ(S − x̄), since B = ∪_{λ>0}B_{λ} is convex and has nonempty interior. For each k, y_{k} ∈ K ⊂ int(J(x̄)^{∘}), so we can find λ_{k} > 0 such that y_{k} ∈ x̄ + λ_{k}(S − x̄) ⊂ B_{λ_{k}}. Note that y_{k} ∉ B_{n_{k}}, and since the sets B_{λ} are increasing with λ, we must have λ_{k} > n_{k}, so λ_{k} converges to infinity. Moreover, we can write y_{k} = x̄ + λ_{k}(z_{k} − x̄) for some z_{k} ∈ S. Now recall that K is weakly compact in X and that X is a Banach space so K is norm-bounded by a constant M > 0, so that z_{k} converges in norm to x̄, because ∥z_{k} − x̄∥ = \frac{∥y_{k}−x̄∥}{λ_{k}} ≤ \frac{M+∥x̄∥}{λ_{k}} ⟶ 0, as k ⟶ ∞. For each k pick z_{k}^{∗} ∈ J(z_{k}) ≠ ∅. Then by Lemma 3.4, there is x_{0}^{∗} ∈ J(x̄) so that, up to a subsequence, z_{k}^{∗} ⟶ x_{0}^{∗} in norm. Now, we compute ⟨x_{0}^{∗}, x_{0}⟩. Given ϵ > 0, eventually

\begin{matrix}⟨x_{0}^{∗}, x_{0}⟩ ≥ & ⟨x_{0}^{∗}, y_{k}⟩ − ϵ (y_{k} → x_{0} weakly) \\ = & ⟨z_{k}^{∗}, y_{k}⟩ + ⟨x_{0}^{∗} − z_{k}^{∗}, y_{k}⟩ − ϵ \\ ≥ & ⟨z_{k}^{∗}, x̄ + λ_{k}(z_{k} − x̄)⟩ − ∥x_{0}^{∗} − z_{k}^{∗}∥∥y_{k}∥ − ϵ \\ ≥ & λ_{k}⟨z_{k}^{∗}, z_{k}⟩ + (1 − λ_{k})⟨z_{k}^{∗}, x̄⟩ − ϵM − ϵ \\ ≥ & λ_{k} + (1 − λ_{k}) − ϵ(M + 1) = 1 − ϵ(M + 1) \end{matrix}

The last line follows from the fact that ⟨z_{k}^{∗}, x̄⟩ ≤ 1 and 1 − λ_{k} < 0. Now, since ϵ is arbitrary, we get ⟨x_{0}^{∗}, x_{0}⟩ ≥ 1. But x_{0} ∈ K ⊂ int(J(x̄)^{∘}), so, by Proposition 3.3, this is a contradiction. Therefore there is n_{0} ∈ IN such that K ⊂ B_{n_{0}}, hence K ⊂ B_{λ} for all λ ≥ n_{0}.

The Proposition we just proved tells us that, when J is n-to-n-usco, the corners of U are “nicely” exhausted by the dilations of U itself. With this more thorough description of the duality map at hand we are now able to prove the theorem of Borwein and Fitzpatrick in one direction.

4. Another proof of the B-F theorem

The idea of the proof is based on the fact that, when J is n-to-n-usco, the corners of the unit ball U are sufficiently flat (i.e. bounded-weakly open) and are nicely exhausted by the dilations of U (i.e. any weakly compact subset of int(J(x̄)^{∘}) is in reality contained in one dilated ball B_{λ}), together with the fact that in a reflexive space any equivalent ball is weakly compact. But first recall the following fact.

Lemma 4.1. If Y is a compact Hausdorff space and V_{1}, …, V_{n} are open sets that cover Y, then there is a finite refinement {W_{j}} of {V_{i}} with W_{1}, …, W_{j} closed in Y.

Proof. For each y ∈ Y, we have y ∈ V_{i} for some i. Since Y is compact, Y \ V_{i} is compact, and since Y is regular, there is an open neighborhood G_{y} of y such that y ∈ G_{y} ⊂ cl(G_{y}) ⊂ V_{i}. By compactness, Y = ∪_{j=1}^{m}G_{y_{j}} for some y_{1}, …, y_{m} in Y. Set W_{j} = cl(G_{y_{j}}). Then {W_{j}} is a closed finite refinement of {V_{i}}.

Theorem 4.2. If X is reflexive, with norm p and J its duality map, let π be the set of norms that are equivalent to p and C(X) be the set of closed and convex subsets of X; then:

J n-to-n-usco ⟹ τ_{W_{p}} = sup_{p′∈π} τ_{W_{p′}} on C(X)

Proof. We must show that for every norm p′ equivalent to p we have τ_{W_{p}}^{+} ≥ τ_{W_{p′}}^{+}.

Therefore, by Theorem 2.3, it’s enough to show that p covers every equivalent norm p′. So, let U′ be the unit ball of the norm p′. Fix x^{∗} ∈ S^{′∗} and ϵ > 0. Then U′ ⊂ H(x^{∗}, ϵ) = {x ∈ X | ⟨x^{∗}, x⟩ ≤ 1 + ϵ}. We must show that U′ can be covered by a finite number of pballs which remain contained in H(x^{∗}, ϵ). As a matter of notation, let N be the hyperplane {x ∈ X | ⟨x^{∗}, x⟩ = 1 + ϵ}, that is the boundary of H(x^{∗}, ϵ). Since the Wijsman topology is invariant for homothetic norms we can suppose, by dilating U if necessary, that the unit ball of the norm p is such that sup_{x∈U}⟨x^{∗}, x⟩ = 1 + ϵ, so that U is also contained in H(x^{∗}, ϵ). Also, since X is reflexive, there is x̄ ∈ S such that ⟨x^{∗}, x̄⟩ = 1 + ϵ, so that x̄ ∈ N. By Proposition 3.2, the corner of U at x̄, J(x̄)^{∘}, is the smallest closed convex cone with vertex at x̄ that contains U, therefore J(x̄)^{∘} ⊂ H(x^{∗}, ϵ), because H(x^{∗}, ϵ) is a closed convex cone containing U and has vertex at x̄, since x̄ ∈ N. Then, for each x ∈ N, consider the open cone V_{x} = int(J(x̄)^{∘}) + (x − x̄), i.e. the corner of U at x̄ translated so that its vertex is still in N. Since J(x̄)^{∘} ⊂ H(x^{∗}, ϵ), every V_{x} is also contained in H(x^{∗}, ϵ). The fact that, when J is n-to-n-usco, the corner of U at x̄ is sufficiently flat comes to use in the next Claim.

Claim: U′ ⊂ ∪_{i=1}^{n}V_{x_{i}}, for some x_{1}, …, x_{n} ∈ N.

Consider y ∈ U′. Then ⟨x^{∗}, y⟩ = k ≤ 1, because x^{∗} ∈ S^{′∗}. Let x = y + (1 − \frac{k}{1+ϵ})x̄. Then

⟨x^{∗}, x⟩ = ⟨x^{∗}, y⟩ + (1 − \frac{k}{1+ϵ})⟨x^{∗}, x̄⟩ = k + (1 − \frac{k}{1+ϵ})(1 + ϵ) = 1 + ϵ. So x ∈ N. Note that

we can write y = \frac{k}{1+ϵ} x̄ + (x − x̄). Moreover, for all y^{∗} ∈ J(x̄) ⊂ S^{∗}, ⟨y^{∗}, \frac{k}{1+ϵ} x̄⟩ = \frac{k}{1+ϵ} < 1, and since J(x̄) is norm-compact in X^{∗}, Proposition 3.3 (1) applied to J(x̄)^{∘}, tells us that \frac{k}{1+ϵ} x̄ ∈ int(J(x̄)^{∘}). Therefore y ∈ V_{x}. So U′ ⊂ ∪_{x∈N}V_{x}. Now, since U′ is bounded (in every equivalent norm), Proposition 3.3 (2) says that, for each x ∈ N, V_{x} ∩ U′ is weakly open in U′. So ⋃_{x∈N}(V_{x} ∩ U′) is a relatively weakly open cover of U′. But X is reflexive, so U′ is weakly-compact. Therefore, we can find x_{1}, …, x_{n} ∈ N such that U′ ⊂ ∪_{i=1}^{n}V_{x_{i}}. Now, by Lemma 4.1, we can find a finite refinement {W_{j}}_{j=1}^{m} of the cover {V_{x_{i}}} where the W_{j}’s are relatively weakly closed in U′. Hence, for each j = 1, …, m, W_{j} ⊂ V_{x_{i_{j}}}, for some 1 ≤ i_{j} ≤ n, and W_{j} is weakly compact. Proposition 3.5 says that each W_{j} is in fact contained in a p-ball, i.e. there is λ_{j} > 0 such that W_{j} ⊂ B_{λ_{j}} + (x_{i_{j}} − x̄) = λ_{j}U +(1−λ_{j})x̄+(x_{i_{j}} −x̄) = (x_{i_{j}} −λ_{j}x̄)+λ_{j}U. Call these p-balls D_{j} = (x_{i_{j}} −λ_{j}x̄)+λ_{j}U. Then, U′ ⊂ ∪_{j=1}^{m}W_{j} ⊂ ∪_{j=1}^{m}D_{j} ⊂ ∪_{j=1}^{m}V_{x_{i_{j}}} ⊂ H(x^{∗}, ϵ). Hence U′ ⊂ ∪_{j=1}^{m}D_{j} ⊂ H(x^{∗}; ϵ).

Since x^{∗} ∈ S^{′∗} and ϵ > 0 were arbitrarily fixed, the hypothesis of Theorem 2.3 are satisfied, thus τ_{W_{p}}^{+} ≥ τ_{W_{p′}}^{+}. So τ_{W_{p}} = sup_{p′∈π} τ_{W_{p′}}.

P. Poggi-Corradini / Norms that generate the Wijsman topology on convex sets 285

Acknowledgment. The author acknowledges support from INdAM (Istituto Nazionale di Alta Matematica) while studying at the University of Washington and wishes to thank Professors S. Levi and R. Phelps for their help and advice.

References

[1] G. Beer : Mosco convergence and weak topologies for convex sets and functions, Matematika 38, 1991, 89–104.

[2] G. Beer : Wijsman convergence of convex sets under renorming, to appear, Nonlinear Anal.

[3] J. Borwein, S. Fitzpatrick : Mosco convergence and the Kadec property, Proc. Amer. Math. Soc. 106, 1989, 843–849.

[4] C. Costantini, S. Levi, J. Zieminska : Metrics that generate the same hyperspace convergence, Set Valued Analysis, Vol. 1 n.2, 1993, 141–157.

[5] M. M. Day : Normed Linear Spaces, Ergebnisse der Mathematik. Band 21, third ed., Springer-Verlag, Berlin 1973.

[6] R. Phelps : Convex functions, monotone operators and differentiability, Lecture Notes in Mathematics 1364, second ed., Springer-Verlag, Berlin 1993.

286 P. Poggi-Corradini / Norms that generate the Wijsman topology on convex sets HIER :

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