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International Journal of Mathematics and Mathematical Sciences
VolumeΒ 2008, Article IDΒ 196326, 6 pages
http://dx.doi.org/10.1155/2008/196326
Research Article

Farthest Points and Subdifferential in -Normed Spaces

Department of Mathematics and Center of Excellence in Analysis on Algebraic structures, Ferdowsi University of Mashhad, P.O. Box 1159, Mashhad 91775, Iran

Received 25 December 2007; Revised 9 March 2008; Accepted 13 March 2008

Academic Editor: Narendra KumarΒ Govil

Copyright Β© 2008 S. Hejazian et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We study the farthest point mapping in a -normed space in virtue of subdifferential of , where is a weakly sequentially compact subset of . We show that the set of all points in which have farthest point in contains a dense subset of .

1. Introduction

Let be a real linear space. A quasinorm is a real-valued function on satisfying the following conditions. (i) for all and if and only if .(ii) for all and all .(iii)There is a constant such that for all . The pair is called a quasinormed space if is a quasinorm on . The smallest possible is called the modules of concavity of By a quasi-Banach space we mean a complete quasinormed space, that is, a quasinormed space in which every Cauchy sequence converges in .

This class includes Banach spaces. The most significant class of quasi-Banach spaces, which are not Banach spaces, is -spaces for equipped with the -norms A quasinorm is called a -norm if for all . In this case, a quasinormed (quasi-Banach) space is called a -normed (-Banach) space. By the Aoki-Rolewicz theorem [1], each quasinorm is equivalent to some -norm. Since it is much easier to work with -norms than with quasinorms, henceforth we restrict our attention mainly to -norms. See [2–4] for more information.

If x* is in X*, the dual of and we write as . We also consider quasinorms with . The case where turns out to be the classical normed spaces, so we will not discuss it and refer the interested reader to [5–7] for analogue results concerning normed spaces.

In this paper, using some strategies from [5–7], we study the farthest point mapping in a -normed space in virtue of subdifferential of , where is a weakly sequentially compact subset of . We show that the set of all points in which have farthest point in contains a dense subset of .

Let be a -normed space and let be a nonempty bounded subset of . The mapping defined by is called the farthest point map of . We call a remotal (uniquely remotal, resp.) set if for each the set is not empty (is singleton, resp.) [8–10].

2. Main Results

Let be a -normed space and let be a bounded subset of . For each , we define the subdifferential of a function at by This set may be empty even if we consider to be a Banach space [7, Example 3.8]. In a -normed space, it may happen that , although we should note that a -normed space may have a trivial dual as well as it might have a nontrivial dual, see [11, Chapter 3], for some examples. To see the nonemptiness, suppose that is a -normed space, , and . Thus, and obviously . Also for each with , we haveIt follows that , and so . Throughout the rest, we assume when we deal with this set. For an arbitrary nonempty bounded subset of , finding the set of all for which remains an open question.

Lemma 2.1. Let be a -Banach space and let be a bounded subset in . Then for each , each element of has norm less than or equal to 1 and hence is -compact.

Proof. Let and . We have By definition of we have for all [10, 12].
Hence and therefore

Note that , thus

Now we have the following proposition which is interesting on its own right.

Proposition 2.2. Let be a -Banach space and let be a bounded subset of . Then the set for some is of the first category in .

Proof. Let Then We will show that for each , (i) is a norm closed subset of ;(ii) has empty interior. To see (i), let be a sequence in which converges to an element in . For each , choose such thatBy Lemma 2.1 for all . Without loss of generality, we assume that converges to For every , we haveThis shows that converges to Since ,or equivalentlyIt follows thatand henceThis shows that It follows from (2.4) thatWe use the fact that once more to obtain the inequalityTherefore . So is a closed subset of .
To see (ii), suppose that some has nonempty interior. Then, there exists an open ball in of radius for some and center at such that . Let , and and choose such thatLet thenChoose such thatThen there exists such thatWe will show thatThis will contradict the fact that is a subdifferential of at and the proof would be completed. To achieve a contradiction, we will consider four cases as follows. (i) and .(ii) and .(iii) and .(iv) and . We investigate case (i) in detail. The other cases can be studied similarly. First of all note thatNow, we have

We recall that a set is said to be weakly sequentially compact if each sequence of elements of contains a subsequence converging weakly to some element .

Theorem 2.3. Let be a weakly sequentially compact subset in a -Banach space . Then the set for some contains a dense -set in . In particular, the set of farthest points of is nonempty.

Proof. Let and be defined as in Proposition 2.2 and let . Thenwhere each is an open dense subset of . Hence is a dense -set in . For each and , we haveBy the weak compactness of , there exists a point with HenceThis shows that for some .

Acknowledgment

The authors would like to thank the referee for valuable comments.

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