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Connectedness and Compactness of Weak Efficient Solutions for Set-Valued Vector Equilibrium Problems

Journal of Inequalities and Applications20082008:581849

https://doi.org/10.1155/2008/581849

Received: 1 November 2007

Accepted: 5 September 2008

Published: 21 September 2008

Abstract

We study the set-valued vector equilibrium problems and the set-valued vector Hartman-Stampacchia variational inequalities. We prove the existence of solutions of the two problems. In addition, we prove the connectedness and the compactness of solutions of the two problems in normed linear space.

1. Introduction

We know that one of the important problems of vector variational inequalities and vector equilibrium problems is to study the topological properties of the set of solutions. Among its topological properties, the connectedness and the compactness are of interest. Recently, Lee et al. [1] and Cheng [2] have studied the connectedness of weak efficient solutions set for single-valued vector variational inequalities in finite dimensional Euclidean space. Gong [35] has studied the connectedness of the various solutions set for single-valued vector equilibrium problem in infinite dimension space. The set-valued vector equilibrium problem was introduced by Ansari et al. [6]. Since then, Ansari and Yao [7], Konnov and Yao [8], Fu [9], Hou et al. [10], Tan [11], Peng et al. [12], Ansari and Flores-Bazán [13], Lin et al. [14] and Long et al. [15] have studied the existence of solutions for set-valued vector equilibrium and set-valued vector variational inequalities problems. However, the connectedness and the compactness of the set of solutions to the set-valued vector equilibrium problem remained unstudied. In this paper, we study the existence, connectedness, and the compactness of the weak efficient solutions set for set-valued vector equilibrium problems and the set-valued vector Hartman-Stampacchia variational inequalities in normed linear space.

2. Preliminaries

Throughout this paper, let , be two normed linear spaces, let be a nonempty subset of , let be a set-valued map, and let be a closed convex pointed cone in .

We consider the following set-valued vector equilibrium problem (SVEP): find , such that
(2.1)

Definition 2.1.

Let A vector satisfying
(2.2)

is called a weak efficient solution to the SVEP. Denote by the set of all weak efficient solutions to the SVEP.

Let be the topological dual space of . Let
(2.3)

be the dual cone of .

Definition 2.2.

Let . A vector is called an -efficient solution to the SVEP if
(2.4)

where means that , for all . Denote by the set of all -efficient solutions to the SVEP.

Definition 2.3.

Let be a nonempty convex subset in . A set-valued map is called to be -convex in its second variable if, for each fixed , for every , , the following property holds:
(2.5)

Definition 2.4.

Let be a nonempty convex subset in . A set-valued map is called to be -concave in its first variable if, for each fixed , for every , , the following property holds:
(2.6)

Definition 2.5.

Let be a nonempty subset of . Let be a set-valued map, where is the space of all bounded linear operators from into (let be equipped with operator norm topology). Set

(i)Let be a convex subset of . is said to be -hemicontinuous if, for every pair of points , the set-valued map
(2.7)

is lower semicontinuous at 0.

(ii)Let . is said to be -pseudomonotone on if, for every pair of points , , for all , then , for all .

The definition of -hemicontinuity was introduced by Lin et al. [14].

Definition 2.6.

Let be a Hausdorff topological vector space and let be a nonempty set. is called to be a KKM map if for any finite set the relation
(2.8)

holds, where denoted the convex hull of .

For the definition of the upper semicontinuity and lower semicontinuity, see [16].

The following FKKM theorem plays a crucial role in this paper.

Lemma 2.7.

Let be a Hausdorff topological vector space. Let be a nonempty convex subset of , and let be a KKM map. If for each , is closed in , and if there exists a point such that is compact, then .

By definition, we can get the following lemma.

Lemma 2.8.

Let be a nonempty convex subset of . Let be a set-valued map, and let be a closed convex pointed cone. Moreover, suppose that is -convex in its second variable. Then, for each , is convex.

3. Scalarization

In this section, we extend a result in [3] to set-valued map.

Theorem 3.1.

Suppose that , and that is a convex set for each . Then
(3.1)

Proof.

It is clear that
(3.2)
Now we prove that
(3.3)
Let . By definition, , for all . Thus
(3.4)
As is a convex pointed cone, we have
(3.5)
By assumption, is a convex set. By the separation theorem of convex sets, there exist some , such that
(3.6)
By (3.6), we obtain that and
(3.7)
Therefore, . Hence . Thus we have
(3.8)

4. Existence of The Weak Efficient Solutions

Theorem 4.1.

Let be a nonempty closed convex subset of and let be a closed convex pointed cone with . Let be a set-valued map with for all . Suppose that for each , is lower semicontinuous on , and that is -convex in its second variable. If there exists a nonempty compact subset of , and , such that , for all , then, for any , , , , and .

Proof.

Let . Define the set-valued map by
(4.1)
By assumption, , for all , so . We claim that is a KKM map. Suppose to the contrary that there exists a finite subset of , and there exists such that . Then for some , , with , and , for all . Then there exist , such that
(4.2)
As is -convex in its second invariable, we can get that
(4.3)
By (4.3), we know that there exist , , such that
(4.4)
Hence . By assumption, we have . By (4.2), however, we have . This is a contradiction. Thus is a KKM map. Now we show that for each , is closed. For any sequence, and . Because is a closed set, we have . By assumption, for each , is lower semicontinuous on , then by [16], for each fixed , and for each , there exist , such that . Because , we have
(4.5)
Thus . By the continuity of and , we have . By the arbitrariness of , we have , that is, . Hence is closed. By the assumption, we have , and is closed. Since is compact, is compact. By Lemma 2.7, we have . Thus there exists . This means that
(4.6)

Therefore, . Next we show that . If , then . It follows from that , and by Theorem 3.1, we have .

Theorem 4.2.

Let be a nonempty closed convex subset of and let be a closed convex pointed cone with . Let . Assume that is a -hemicontinuous, -pseudomonotone mapping. Moreover, assume that the set-valued map defined by is -convex in its second variable. If there exists a nonempty compact subset of , and , such that , for all , then and .

Proof.

Let . Define the set-valued maps , by
(4.7)
respectively. As for each , we have , then . The proof of the theorem is divided into four steps.
  1. (I)
    is a KKM map on .
     
Suppose to the contrary that there exists a finite subset of , and there exists such that . Then for some , , with , and , for all . Then there exist  such that
(4.8)
Since is -convex in its second variable, we have
(4.9)
Let , for each . By (4.9), we know there exists , such that
(4.10)
As , we have
(4.11)
While by (4.8), we have . This is a contraction. Hence is a KKM map on .
  1. (II)
    for all and is a KKM map.
     
By the -pseudomonotonicity of , for each , we have . Since is a KKM map, so is .
  1. (III)
    .
     
Now we show that for each , is closed. Let be a sequence in such that converges to . By the closedness of , we have . Since , then for each , we have
(4.12)
As , and the continuity of , then for each , we have
(4.13)
Consequently, . Hence is closed. By the assumption, we have . Then is compact since is compact. By step (II), we know is a KKM map. By Lemma 2.7, .
  1. (IV)
    .
     
Because , we have . Now let us show that . Let . For each , and each , we have
(4.14)
For any and for each fixed , define the set-valued mapping by
(4.15)
We pick a sequence such that and set . Since is a convex set, for each . It is clear that . Let . We have Since is -hemicontinuous, is lower semicontinuous at 0. By [16], there exist , such that . As , there exist such that . By , we have . By (4.14), we have
(4.16)

Since , . Hence since is continuous and . Therefore, for any and for each , we have . Hence . Thus . This means that there exists , for each , we have , for all . It follows that , thus . By the proof of Theorem 4.1, we know . Since , we have . The proof of the theorem is completed.

5. Connectedness and Compactness of The Solutions Set

In this section, we discuss the connectedness and the compactness of the weak efficient solutions set for set-valued vector equilibrium problems and the set-valued vector Hartman- Stampacchia variational inequalities in normed linear space.

Theorem 5.1.

Let be a nonempty closed convex subset of , let be a closed convex pointed cone with , and let be a set-valued map. Assume that the following conditions are satisfied:

(i)for each , is lower semicontinuous on ;

(ii) is -concave in its first variable and -convex in its second variable;

(iii) , for all ;

(iv) is a bounded subset in ;

(v)there exists a nonempty compact convex subset of , and , such that , for all .

Then is a nonempty connected compact set.

Proof.

We define the set-valued map by
(5.1)
By Theorem 4.1, for each , we have , hence and . It is clear that is convex, so it is a connected set. Now we prove that, for each , is a connected set. Let , we have and
(5.2)
Because is -concave in its first variable, for each fixed , and for above , and , we have since is convex, and
(5.3)
Hence for each , , there exist , , and , such that . As and by (5.2), we have
(5.4)
Thus
(5.5)

that is . So is convex, therefore it is a connected set.

Now we show that is upper semicontinuous on . Since is a nonempty compact set, by [16], we only need to prove that is closed. Let the sequence and , where converge to with respect to the norm topology. As , we have
(5.6)
that is, , for all . As and is compact, we have . Since for each , is lower semicontinuous on , for each fixed , and each , there exist , such that . From , we have
(5.7)
By the continuity of and , we have
(5.8)
Let . By assumption, is a bounded set in , then there exist some , such that for each , we have . For any , because with respect to norm topology, there exists , and when  , we have . Therefore, there exists , and when  , we have
(5.9)
Hence
(5.10)
Consequently, by (5.8), (5.10), we have
(5.11)
By (5.7), we have . So for any and for each , we have . Hence
(5.12)
This means that
(5.13)
Hence the graph of is closed. Therefore, is a closed map. By [16], is upper semicontinuous on . Because is -convex in its second variable, by Lemma 2.8, for each , is convex. It follows from Theorem 3.1 that
(5.14)

Thus by [17, Theorem 3.1] is a connected set.

Now, we show that is a compact set. We first show that is a closed set. Let with . Since is compact, We claim that Suppose to the contrary that , then there exist some such that
(5.15)
Thus there exists such that
(5.16)
Hence is a neighborhood of . Since is lower semicontinuous at , there exists some neighborhood of such that
(5.17)
Since , there exist some , and when  , we have . By (5.17),
(5.18)

This contradicts . Thus This means that is a closed set. Since is compact and , is compact.

Theorem 5.2.

Let be a nonempty closed convex subset of , and let be a closed convex pointed cone with . Assume that for each , is a -hemicontinuous, -pseudomonotone mapping. Moreover, assume that the set-valued map defined by is -convex in its second variable, and the set is a bounded set in . If there exists a nonempty compact convex subset of , and , such that , for all , then is a nonempty connected set.

Proof.

We define the set-valued map by
(5.19)
By Theorem 4.2, for each , we have and . Hence and . Clearly, is a convex set, hence it is a connected set. Define the set-valued maps , by
(5.20)
respectively. Now we prove that for each , is a connected set. Let , then . By the proof of Theorem 4.2, we have , so . Hence for and for each , , we have
(5.21)
Then, for each , , and , we have since is convex and
(5.22)
Hence . Thus . Consequently, for each , is a convex set. Therefore, it is a connected set. The following is to prove that is upper semicontinuous on . Since is a nonempty compact set, by [16] we only need to show that is a closed map. Let sequence and , where converges to with respect to the norm topology of . As , we have
(5.23)
Then, for each , we have that
(5.24)
By assumption, for each , is -pseudomonotone, and by (5.24), for each , for the above , and for each , we have
(5.25)
As , we have , and . As , and is compact, we have . Let . By assumption, is a bounded set in . Then, there exists , such that for each , we have . For any , because with respect to the norm topology, there exists , and when  , we have . Therefore, there exists , and when  , we have
(5.26)
Hence
(5.27)
Then
(5.28)
Then, by (5.25), (5.28), we have . Hence for each , and for each , we have . Since is -pseudomonotone, for each , and for each , we have . Hence . Therefore, the graph of is closed, and is a closed map. By [16], we know that is upper semicontinuous on . Because is -convex in its second variable, for each , is convex. It follows from Theorem 3.1 that
(5.29)

Then, by [17, Theorem 3.1], we know that is a connected set. The proof of the theorem is completed.

Let denote the base of neighborhoods of 0 of . By [18], for each bounded subset and for each neighborhood of 0 in , we have
(5.30)

Lemma 5.3.

Let be a nonempty convex subset of , and let If is lower semicontinuous on , then is -hemicontinuous on .

Proof.

For any fixed , we need to show that the set-valued mapping
(5.31)
is lower semicontinuous at 0. For any and for any neighborhood of , there exists such that and there exists a neighborhood of in , such that . Since is lower semicontinuous at , for the neighborhood
(5.32)
of , there exists a neighborhood of such that
(5.33)
For the above , since , when , there exists , and when  , we have By (5.33), we have
(5.34)
Thus there exist such that , for all . By (5.32), we have
(5.35)
that is
(5.36)

This means that is lower semicontinuous at 0. By definition, is -hemicontinuous on .

Theorem 5.4.

Let be a nonempty closed bounded convex subset of , and let be a closed convex pointed cone with . Assume that for each , is a -pseudomonotone, lower semicontinuous mapping. Moreover, assume that the set-valued map defined by is -convex in its second variable, and the set is a bounded set in . If there exists a nonempty compact convex subset of , and , such that , for all , then is a nonempty connected compact set.

Proof.

By Lemma 5.3 and Theorem 5.2, is a nonempty connected set. Since is a compact set, we need only to show that is closed. Let It is clear that and . We claim that Suppose to the contrary that , then there exists such that
(5.37)
that is
(5.38)
Thus there exists such that
(5.39)
Hence there exists such that
(5.40)
where is the unit ball of . Set
(5.41)
By definition, is a neighborhood of zero of , since is bounded. Since is lower semicontinuous at , for the above and the above , there exists a neighborhood of such that
(5.42)
Since there exists such that when , we have Thus by (5.42) we have
(5.43)
Thus there exist for all . We have , and hence
(5.44)
Therefore,
(5.45)
Since and , we have . Hence there exists , and when  we have . This combining (5.45) implies that
(5.46)
Hence
(5.47)
By (5.40) and (5.47), we have
(5.48)
On the other hand, since , we have
(5.49)

This contradicts (5.48), because Thus . This means that is a closed subset of .

Declarations

Acknowledgments

This research was partially supported by the National Natural Science Foundation of China and the Natural Science Foundation of Jiangxi Province, China.

Authors’ Affiliations

(1)
Department of Mathematics, Nanchang University

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© Bin Chen et al. 2008

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