International Journal of Mathematics and Mathematical Sciences

Volume 2014 (2014), Article ID 196391, 5 pages

http://dx.doi.org/10.1155/2014/196391

Research Article

Coproximinality in the Space of Bounded Functions

Eyad Abu-Sirhan and Zuhier Altawallbeh

Department of Mathematics, Tafila Technical University, Tafila 6610, Jordan

Received 16 January 2014; Accepted 25 March 2014; Published 7 April 2014

Academic Editor: Ram N. Mohapatra

Copyright © 2014 Eyad Abu-Sirhan and Zuhier Altawallbeh. 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.

As a counterpart to best approximation in Banach spaces, the best coapproximation was introduced by Franchetti and Furi (1972). In this paper, we shall consider the relation between coproximinality of a nonempty subset of a Banach space and of in .

1. Introduction

Let be a nonempty subset of a Banach space . We say that is proximinal in if each there corresponds at least to one point such that

Recently, another kind of approximation from has been introduced by Franchetti and Furi [1] who have considered those elements (if any) satisfying

Such an element is called best coapproximant of in . The set of all such elements, satisfying above inequality, is denoted by . The subset is called coproximinal in if is nonempty for any . If is singleton for any , then is called coChebyshev; see [2–4].

It is clear that is convex if is convex and closed. The kernal of is the set defined by

Many results in the theory of best coapproximation have appeared since Franchetti and Furri’s paper, 1972. These results are largely concerned with the question of existence and uniqueness of best coapproximation (see for example [5–8]). Let and denote the Banach spaces of all bounded (resp., continuous) functions from a topological space into a Banach space and let be a closed subset of . It should be remarked that if is a compact space, then is a subspace of . In this paper, we discuss the coproximinality of and in and , respectively. For uniqueness and existence of best coapproximation in , see [7].

Definition 1. Let be a coproximinal subset of a Banach space . A map which associates with each element in one of its best coapproximation in (i.e., for all ) is called a coproximity map.

Remark 2. Let be a coproximinal subset of a Banach space . We state some basic properties of a proximity map .

(1)If is coChebyshev, [5], then

(a), for any .(b), for any scalar and . (i.e., is homogeneous).(c), for any and .

(2)If is a subspace of , then

(a), for any and .(b), for all . (set and in (a) and take into account that ).(c) is continuous at .(d)If is linear, then is continuous.

Theorem 3 (see [3]). If is coproximinal hyperplane or -dimensional subspace of a Banach space , then has a continuous coproximity map.

Lemma 4 (see [3]). Let be a coproximinal subspace of . Then the following are equivalent.

(1) has linear coproximity map.(2) contains a closed subspace such that . Moreover, if (2) holds, then the linear coproximity map for can be defined by , , and .

Definition 5. Let be a Banach space. Consider the following.

(a)A subspace of is called one complemented in if there is a closed subspace such that and the projection is contractive.(b)A linear projection is called an -projection if for all . A closed subspace of is called an -summand if it is the range of an -projection, [9].(c)A linear projection is called an -projection if for all . A closed subspace of is called an -summand if it is the range of an -projection.

Clearly every -summand is one complemented.

Theorem 6. Let be a subspace of a Banach space . Then the following are equivalent.

(1) is one complemented in .(2) is coproximinal in and has linear coproximity map.

Proof. : Let , a contractive projection, and . Then ,

Hence is coproximinal in . Now let . Then for any ,   . Thus and has linear coproximity map by Lemma 4.

: If has a linear coproximity map, then contains a closed subspace such that . Moreover, , is a linear coproximity map for and so is the projection of along by Lemma 4. To show that is contractive, if , then for all , and so which means that is contractive.

Corollary 7. If is -summand, then is coproximinal and has continuous coproximity map.

For more information about coproximinal sets, optimal sets, and contractive sets, the reader is referred to [10–18].

2. Coproximinality in and

Let be a compact Hausdorff space and be a Banach space. We denote and to be the Banach space of all bounded ( resp., continuous) functions from into .

Lemma 8 (see [19]). Let be a continuous map of a Banach space into a closed subset of . If is compact Hausdorff space, then the map defined by is continuous from to .

Lemma 9. Let be a closed subset of a Banach space . If there is a continuous coproximity map of onto , then is coproximinal in and in fact it has a continuous coproximity map.

Proof. Let be a continuous coproximity map for . Define

by , then is continuous by Lemma 8. Let , then

for all ,

Hence

for all and is a continuous coproximity map from onto .

Lemma 10. Let be a closed subset of a Banach space . If is coproximinal in , then is coproximinal in . Moreover, if has a continuous coproximity map, then has a continuous coproximity map.

Proof. For , define by for all . The map . By assumption, there exists such that

for all . In particular,

for all . For any fixed ,

for all and so is a best coapproximation of . Now let be a continuous coproximity map. For any fixed , define by . Then is a proximity map. Suppose that in . Then . and

Hence is continuous.

Theorem 11. Let be a closed subset of a Banach space . Then the following are equivalent.

(1) is coproximinal in and has a continuous coproximity map.(2) is coproximinal in and has a continuous coproximity map.

Proof. The proof follows from Lemmas 9 and 10.

Theorem 12. Let be a closed subset of a Banach space and let be a topological space. Then the following are equivalent.

(1) is coproximinal in .(2) is coproximinal in .

Proof. . Let . Since is coproximinal in , then is nonempty for any . By axiom of choice, we may define a function such that for all . Since    for all and is a best coapproximation of in .

. For , define by for all . Since is coproximinal in , the map . By assumption, there exists such that

for all . In particular,

for all . For any fixed ,

for all and is a best coapproximation of .

We recall that.

Definition 13. Let be a set valued mapping, taking points of a topological space into the family of all subsets of a topological space . The mapping is said to be lower semicontinuous if, for each open set in , the set is open.

Theorem 14 (see [19, Michael Selection Theorem]). Let be a lower semicontinuous of a paracompact space into the family of nonvoid closed convex subsets of a Banach space . Then has a continuous selection; that is, there exists a continuous map such that for all .

Definition 15. A sequence of sets in a topological space is convergent and has the limit if and only if where is the set of those elements which are limits of points in for infinitely many and is the set of those elements which are limits of points in for cofinitely many (for all but finitely many) .

Lemma 16. Let be a compact Hausdorff space, a Banach space, and a compact subset of . For , we define by

Then is lower semicontinuous.

Proof. Let be an open set in . We need to prove that

is open. For this purpose, let be a sequence in and as , so for all which means that

Thus for all , and is a sequence of subsets of . We shall show that as . First, we show that

Let . Then there is a sequence of points and a subsequence of such that and as . By the definition of , we get

Taking and , and since is continuous, we have

so . This proves that . Now we shall show that . Let . In this case, if as then for infinitely many , and, so,

We may assume that, without loss of generality, as , for some , since is compact. Taking , we have that

so and we have as a conclusion the follwing result:

Thus, . Finally, we know that is a closed set and the members of the sequence of sets are all subsets of for all . Now, to complete the proof, we still need to show that

To do so, let . By the definition of , there is a sequence of points and a subsequence of the sequence such that

Since is closed, we get . This proves that . This means that , and so . Thus is open set and so is lower semicontinuous.

Theorem 17. Let be a compact convex subset of a Banach space and be a compact Hausdorff space. Then the following are equivalent.

(1) is coproximinal in .(2) is coproximinal in .

Proof. . The proof is similar to that given in Lemma 10.

. Let and as defined in Lemma 16. Then is lower semicontinuous by Lemma 16. Since is closed, nonvoid, and convex, then is closed, nonvoid, and convex. By Michael Selection Theorem, has a continuous selection . Then and

Hence is a best coapproximation of in .

Corollary 18. Let be a compact Hausdorff space and let be a reflexive Banach space. Then (the closed unit ball of ) is coproximinal in if and only if is coproximinal in .

Proof. Since is reflexive, then is compact by Alaoglu Theorem. As is closed and convex subset of , the result follows from Theorem 17.

Problem 19. If is a compact Hausdorff space and is a Banach space, when is coproximinal in .

Conflict of Interests

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

Acknowledgment

Finally the author would like to thank the referees for their valuable advice.

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