A Reverse Theorem on the - Continuity of the Dual Map

Abstract

Given a Banach space , , and , we define the set of all for which there exist two sequences and such that converges to , has a subnet -convergent to , and for all . We prove that if is separable and reflexive and enjoys the Radon-Riesz property, then is contained in the boundary of relative to . We also show that if is infinite dimensional and separable, then there exists an equivalent norm on such that the interior of relative to is contained in .

1. Preliminaries and Background

Recall that a point in the unit sphere of a real or complex normed space is said to be a smooth point of , provided that there is only one functional in attaining its norm at . This unique functional is usually denoted by . The set of smooth points of the (closed) unit ball of is usually denoted as . is said to be smooth provided that . If is smooth, then the dual map of is defined as the map such that and for all . It is well known that the dual map is - continuous and that for all and all . We refer the reader to [1, 2] for a better perspective on these concepts.

On the other hand, recall that a normed space is said to be rotund (or strictly convex) provided that its unit sphere is free of nontrivial segments. It is well known among Banach Space Geometers that smoothness and rotundity are dual concepts in the following sense: if a dual space is rotund (smooth), then the predual is smooth (rotund). The converse does not hold though. Next, we will gather some of the most relevant results in terms of rotund and smooth renormings into the following (see [3, Theorem 1 (VII.4)] and [1, Corollary 4.3]).

Theorem 1 (see [1, 3]). Let be a real or complex normed space. Then one has the following. (i)If is separable, then admits an equivalent norm so that both and are rotund.(ii)If is reflexive, then admits an equivalent norm so that is rotund and smooth.

In case is not a smooth point then is defined as , that is, the set . We will now continue with a brief introduction on faces and the impact of surjective linear isometries on them. The following definition is very well known amid Banach Space Geometers.

Definition 2. Let be a real or complex normed space and consider a nonempty convex subset of . Then one has the following. (i) is said to be a face of provided that verifies the extremal condition with respect to ; that is, if and with , then .(ii) is said to be an exposed face of provided that there exists such that , where .

It is immediate that every exposed face is a proper face, and every proper face must be contained in the unit sphere. Also notice that for every ; that is, is an exposed face of . We will refer to these exposed faces as -exposed faces.

A face, an exposed face, or a -exposed face, which is a singleton, is called an extreme point, an exposed point, or a -exposed point, respectively.

Remark 3. Observe the following, since we will use it in upcoming sections (see [1, 4]).(1)A point if and only if is a singleton, and, in this situation, . So with this notation if is a smooth point, then is a -exposed point.(2)Also note that for every proper face .(3)Assume now that is a surjective linear isometry between the real or complex normed spaces and . It is not difficult to check that if , then

As mentioned before, in a smooth space if converges to , then the sequence is -convergent to (see [1]). The main result in this paper is the converse to the previous statement (see Theorem 12).

Theorem 4. Let be a separable real Banach space with . Consider . Then one has the following. (1).(2)If is reflexive and has the Radon-Riesz property, then .(3)If is infinite dimensional, then there exists an equivalent norm on such that .

More precisely, for a given real or complex Banach space and we will define the set of all for which there exist two sequences and such that converges to , has a subnet -convergent to , and for all . As expected, and denote the boundary and the interior, respectively, of relative to . The set denotes the localized boundary of (see Definition 9), that is, the set of all for which there exists a dense subset of with and satisfying that for all , where (recall that stands for the density character of ).

To conclude, we draw the reader’s attention to the fact that if is real and , then for all .

2. The Geometric Tools

Note that no complex Banach space admits proper faces with nonempty interior relative to the unit sphere in its unit ball (see [5, Theorem 2.1]). This is the reason for considering real Banach spaces only. On our journey to proving the main result, we will utilize several technical lemmas and remarks. The first lemma involves the intersections of -exposed faces and the disjointness of their interiors.

Lemma 5. Let be a real Banach space and consider . Then one has the following. (1).(2) if and only if .(3)If , then .

Proof. Assume that there exists . As mentioned in the introduction, we have that ; therefore which contradicts the hypothesis of the lemma.
If , then and for all . Conversely, if , then the Hahn-Banach Separation Theorem assures the existence of such that which immediately implies that .
The proof uses a similar argument as in the previous item.

Remark 6. No extreme point can lie in the interior of a proper face relative to the unit sphere. Therefore, Every -exposed face is -closed and thus -compact; therefore the Krein-Milman Theorem allows the existence of extreme points in every -exposed face (even more, every -exposed face is the -closed convex hull of its extreme points).
The second technical lemma is a slight generalization of the well-known fact that the dual map is - continuous in smooth spaces.

Lemma 7. Let be a real Banach space with and fix an arbitrary element . Then . As a consequence, if , then .

Proof. Let . We can find two sequences and such that converges to , has a subnet -convergent to , and for all . Next observe that for all we have which implies that and thus . In order to show that , we simply start off with two sequences and such that converges to and for all . The -compactness of assures that has a subnet -convergent to . By definition .

In general, no convexity properties are verified by even in finite dimensions. Indeed, if and , then .

We will finish this section with a characterization of a smooth point in terms of localizing (linearly and topologically).

Proposition 8. Let be a real Banach space. For a given point , the following are equivalent. (1).(2)For every , is a smooth point of the unit ball of .(3)For every dense subset of , is a smooth point of the unit ball of for all .(4)There exists a dense subset of with such that is a smooth point of the unit ball of for all .

Proof. (1)⇒(2) Let . By the Hahn-Banach Extension Theorem we may assume that , which implies that .
(2)⇒(3) Immediate.
(3)⇒(4) Immediate.
(4)⇒(1) Let . For every we have that thus by hypothesis for all . As a consequence, and so due to the density of in .

In [6, Remark ] it is shown that if is a Hausdorff locally compact topological space with more than one point, then the constant function 1 is a uniformly nonsmooth point of (recall that is said to be a uniformly nonsmooth point provided that is not a smooth point of the unit ball of for all ).

We refer the reader to [7, page 168] where a precise description of the smooth points of is given, for is a Hausdorff locally compact topological space.

3. The Main Result

We first define a new boundary for the -exposed faces other than the regular topological boundary.

Definition 9. Let be a real Banach space and consider . We define the localized boundary of as the set of all for which there exists a dense subset of with and satisfying that for all , where .

Note that the localized boundary of a -exposed face is always contained in the topological boundary of that face relative to the dual unit sphere.

Proposition 10. Let be a -dimensional real Banach space. For every we have that .

Proof. The -dimensionality of assures that for some . It is clear that we assume that is not a smooth point in virtue of Lemma 7. In accordance to Mazur’s Theorem (see [8]), the separability of allows for the density of the smooth points in the unit sphere; therefore We can then find two sequences both converging to . Because of the compactness of we can assume without loss of generality that the two sequences and are, respectively, convergent to some (recall the proof of Lemma 7).

Remark 11. In Proposition 10 we have proved something more. Indeed, under the hypotheses of Proposition 10, if , then a sequence exists such that -converges to .

Prior to stating and proving the main result, we recall the following fact: a real or complex Banach space is said to have the Radon-Riesz property provided that the following condition holds: if is -convergent to and is convergent to , then converges to .

Theorem 12. Let be a separable real Banach space with . Consider . Then one has the following. (1).(2)If is reflexive and has the Radon-Riesz property, then .(3)If is infinite dimensional, then there exists an equivalent norm on such that .

Proof. Let be an element in . By hypothesis there exists a dense sequence such that for all , where . By Proposition 10 for every , we have that , so we can find and such that , and . By the Hahn-Banach Extension Theorem we may assume that for all . It only remains to show that has a subnet -convergent to . In fact, we will show more; we will prove that the whole sequence is -convergent to . Fix an arbitray . Notice that for every we have that At this stage, it is easy to understand that converges to .
Assume to the contrary that there exists which is also in the interior of relative to . Notice that in this case since the interior of relative to is not empty (which implies that cannot be a singleton). Now consider two sequences and such that converges to , has a subnet -convergent to , and for all . Observe that , so since the topology is metrizable in (due to the fact that is separable), there exists a subsequence of which is -convergent to . Since is reflexive, we have that the and the topologies coincide on ; therefore is -convergent to . Since has the Radon-Riesz property, , and , we deduce that is convergent to . This means that there exists such that . This is impossible by Lemma 5.
Fix an arbitrary such that . Observe that can be equivalently renormed so that and , so we will assume from now on that is already endowed in such a way. Since is infinite dimensional we have that is -dense in and so is since is complete, where denotes the set of norm-attaining functionals on . We refer the reader to [9, Theorem 3.11] in order to take into consideration the fact that Therefore, fix which is also in the open unit ball of and denote . We can find a sequence which is -convergent to (remember that the -topology is metrizable in due to the separability of ). For every there exists such that . Finally, consider for every the elements and . By construction we have that is convergent to and is -convergent to . As a consequence, .

Note that if a (nonnecessarily convex) subset of verifies the extremal condition (see Definition 2) and for some proper face of , then .

Corollary 13. Let be an infinite dimensional separable real Banach space. For every there exists an equivalent norm on satisfying that does not verify the extremal condition.

By combining Proposition 10 and (2) of Theorem 12 we obtain the following corollary.

Corollary 14. Let be a -dimensional real Banach space. For every , .

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.