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Journal of Function Spaces
Volume 2016, Article ID 2734947, 5 pages
http://dx.doi.org/10.1155/2016/2734947
Research Article

A Remark on the Stability of Approximative Compactness

1School of Mathematical Sciences, Huaqiao University, Quanzhou 362021, China
2School of Mathematical Sciences, Xiamen University, Xiamen 361005, China

Received 26 November 2015; Accepted 12 January 2016

Academic Editor: Giuseppe Marino

Copyright © 2016 Zhenghua Luo 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 stability of approximative -compactness, where is the norm or the weak topology. Let be an index set and for every let be a subspace of a Banach space . For , let and . We prove that (resp., ) is approximatively -compact in if and only if, for every , (resp., ) is approximatively -compact in .

1. Introduction

Let be a real Banach space and let be a subset of . We denote by either the norm or the weak topology on . The metric projection of onto is the set valued map defined by for , where denotes the distance from to . If, for every , we say that is a proximinal subset of . A sequence is called minimizing for , if .

The notion of approximative compactness was introduced by Efimov and Stechkin [1] in connection with the study of Chebyshev sets in Banach spaces and plays an important role in approximation theory (see, e.g., [2, 3]). Deutsch [4] extended this notion to define approximative -compactness.

Definition 1. Let be a -closed subset of and . We say that is approximatively -compact for if every minimizing sequence for has a -convergent subsequence. If is approximatively -compact for every , we say that is approximatively -compact in .

It is easy to verify that approximative -compactness implies proximinality. Clearly, compact sets or finite-dimensional subspaces of a Banach space are approximatively compact; weakly compact sets or reflexive subspaces of Banach spaces are approximatively weakly compact. Approximative -compactness has been studied in detail in [1, 37].

When it comes to the stability of approximative -compactness, we suppose that is an index set and for every is a subspace of a Banach space . And let , where . Bandyopadhyay et al. [5] proved that if is approximatively -compact in , then is approximatively -compact in for every . In this paper, we prove that the converse is also true. On the other hand, the proximinality of the unit ball of subspaces has been the subject in many recent papers (see, e.g., [811]). In this paper, under the above assumption, we also prove that the unit ball of is approximatively -compact in if and only if, for every , the unit ball of is approximatively -compact in .

For a real Banach space , we denote by the unit ball of and denote by the dual space of . Before we prove the main conclusions we first show a simple property on approximative -compactness of the unit ball of subspaces.

Proposition 2. Let be a subspace of a Banach space . If is approximatively -compact in , then so is . But the converse is not true.

Proof. Suppose that and is a minimizing sequence of in ; that is, . Then for sufficiently large . This means that and is also a minimizing sequence of in . By approximative -compactness of (which is equivalent to the one of ), has a -convergent subsequence.
To illustrate that the converse is not true, first, we show that is not approximatively weakly compact in . Take and for every , , where 1 appears times. Then and is a minimizing sequence of in . But has no weakly convergent subsequence. Hence is not approximatively weakly compact in .
Next, let and . For any , it is easy to see that and . Now, suppose is a minimizing sequence of in ; that is, This implies that . Hence . Therefore is approximatively compact in . But, by the above discussion, is not approximatively weakly compact in , and not in either.

In order to prove our conclusions, we need the following lemmas.

Lemma 3. Let be a sequence of Banach spaces and let be a subspace of , respectively, for . Consider and , where . Let and be a minimizing sequence of in . Then, for every , there exists some such that, for all , .

Proof. If the conclusion does not hold, then, for every , there exists infinitely many such that . We can choose some such that and infinite subset of such that for every . Therefore for every ,But ; then for sufficiently large . This is a contradiction.

Remark 4. In Lemma 3, if we replace by , that is, is a minimizing sequence of in , then the conclusion still holds.

Lemma 5. Under the assumption in Lemma 3, if, moreover, and for every , , then(1);(2).

Proof. (1) For every , by Lemma 3, there exists such that, for all , . For every fixed , we can choose such that and . Then By the arbitrariness of , we have .
(2) Note that ; hence . This implies that . To prove that , for every , we define , where for , and for . Then .
For arbitrary , by Lemma 3, there exists such that , and for all , . Further, we can choose some such that, for all , . Then for all , we haveBy the arbitrariness of , we have . This implies that Therefore . So we have .
For the second equality, first, it is obvious that On the other hand, let be given. For every , we can choose such that . Let ; then By the arbitrariness of , we have . Therefore the second equality holds.

The following is our main result.

Theorem 6. Let be an index set. For every , let be a subspace of a Banach space . For , let and . Then(1) is approximatively -compact in if and only if, for every , is approximatively -compact in ;(2) is approximatively -compact in if and only if, for every , is approximatively -compact in .

Proof. (1) Necessity has been proven in [5].
Sufficiency: let and be a minimizing sequence for . We will show that has a -convergent subsequence. Without loss of generality, we can assume and .
First, notice that if for every , then Hence . And for every , So This implies that, for every , is a minimizing sequence for in . Then has a -convergent subsequence by the approximative -compactness of . By employing the diagonal process, we can choose a subsequence of such that, for every , has a -convergent to some . Obviously, , and .
We still denote the subsequence as . Next, to complete the proof, we will prove that has a -convergent to .
Case  1. is the norm topology. For every , by Remark 4, there exists some such that and for all , . Then we can choose such that, for , . Hence for all , Therefore, by the arbitrariness of , we have that converges to .
Case  2. is the weak topology. Suppose with , where when and when . For every , again by Remark 4, we can choose some such that , and for all , . Note that, for every , weakly converges to ; hence there exists such that, for , . Then for all ,Again by the arbitrariness of , we have that weakly converges to .
(2) Necessity: fix . Suppose that , and is a minimizing sequence of in . Let , , where and for . Then , and Notice that Hence which implies that is a minimizing sequence of in . By approximative -compactness of in , has a -convergent subsequence . Therefore is -convergent.
Sufficiency: suppose that and is a minimizing sequence of in . Like the proof in (1), we will prove that has a -convergent subsequence and we can assume , , and By employing the diagonal process, we can choose a subsequence of (we still denote the subsequence as ) such that , and for every , . Then by Lemma 5, we have , andNext, for every , we will show that has a -convergent subsequence. We can assume that, for all and , . Otherwise, we can replace with which we define in the proof of Lemma 5(2).
Note that, for every , Then Hence, when , This implies that is a minimizing sequence of in . By approximative -compactness of in , has a -convergent subsequence.
Employing the diagonal process again, we can choose a subsequence of such that, for evrey , has a -convergents to some . Let ; then . We still denote as . Finally, just like the proof in (1), we can prove that has a -convergent to .

Remark 7. The above theorem does not hold for . Indeed, suppose that is an infinite-dimensional proper closed subspace of . By Theorem in [12], is approximatively compact in . Next, we show that is not approximatively compact in . Choose and such that . Furthermore, we take a sequence with satisfying that has no convergent subsequence. Note that and for any , This means that and is a minimizing sequence of in . But has no convergent subsequence.

Conflict of Interests

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

Acknowledgments

The authors thank colleagues and graduate students in the Functional Analysis group of Xiamen University for their very helpful conversations and suggestions. Zhenghua Luo was supported partially by the Natural Science Foundation of China, Grant no. 11201160, and the Natural Science Foundation of Fujian Province, Grant no. 2012J05006. Wen Zhang was supported in part by the Natural Science Foundation of China, Grant no. 11471270, and by the Natural Science Foundation of Fujian Province, Grant no. 2015J01022.

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