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Connectedness and Compactness of Weak Efficient Solutions for SetValued Vector Equilibrium Problems
Journal of Inequalities and Applications volumeÂ 2008, ArticleÂ number:Â 581849 (2008)
Abstract
We study the setvalued vector equilibrium problems and the setvalued vector HartmanStampacchia 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 singlevalued vector variational inequalities in finite dimensional Euclidean space. Gong [3â€“5] has studied the connectedness of the various solutions set for singlevalued vector equilibrium problem in infinite dimension space. The setvalued 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 FloresBazÃ¡n [13], Lin et al. [14] and Long et al. [15] have studied the existence of solutions for setvalued vector equilibrium and setvalued vector variational inequalities problems. However, the connectedness and the compactness of the set of solutions to the setvalued vector equilibrium problem remained unstudied. In this paper, we study the existence, connectedness, and the compactness of the weak efficient solutions set for setvalued vector equilibrium problems and the setvalued vector HartmanStampacchia 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 setvalued map, and let be a closed convex pointed cone in .
We consider the following setvalued vector equilibrium problem (SVEP): find , such that
Definition 2.1.
Let A vector satisfying
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
be the dual cone of .
Definition 2.2.
Let . A vector is called an efficient solution to the SVEP if
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 setvalued map is called to be convex in its second variable if, for each fixed , for every , , the following property holds:
Definition 2.4.
Let be a nonempty convex subset in . A setvalued map is called to be concave in its first variable if, for each fixed , for every , , the following property holds:
Definition 2.5.
Let be a nonempty subset of . Let be a setvalued 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 setvalued map
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
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 setvalued 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 setvalued map.
Theorem 3.1.
Suppose that , and that is a convex set for each . Then
Proof.
It is clear that
Now we prove that
Let . By definition, , for all . Thus
As is a convex pointed cone, we have
By assumption, is a convex set. By the separation theorem of convex sets, there exist some , such that
By (3.6), we obtain that and
Therefore, . Hence . Thus we have
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 setvalued 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 setvalued map by
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
As is convex in its second invariable, we can get that
By (4.3), we know that there exist , , such that
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
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
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 setvalued 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 setvalued maps , by
respectively. As for each , we have , then . The proof of the theorem is divided into four steps.

(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 existâ€‰ such that
Since is convex in its second variable, we have
Let , for each . By (4.9), we know there exists , such that
As , we have
While by (4.8), we have . This is a contraction. Hence is a KKM map on .

(II)
for all and is a KKM map.
By the pseudomonotonicity of , for each , we have . Since is a KKM map, so is .

(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
As , and the continuity of , then for each , we have
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, .

(IV)
.
Because , we have . Now let us show that . Let . For each , and each , we have
For any and for each fixed , define the setvalued mapping by
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
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 setvalued vector equilibrium problems and the setvalued 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 setvalued 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 setvalued map by
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
Because is concave in its first variable, for each fixed , and for above , and , we have since is convex, and
Hence for each , , there exist , , and , such that . As and by (5.2), we have
Thus
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
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
By the continuity of and , we have
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 whenâ€‰ , we have . Therefore, there exists , and whenâ€‰ , we have
Hence
Consequently, by (5.8), (5.10), we have
By (5.7), we have . So for any and for each , we have . Hence
This means that
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
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
Thus there exists such that
Hence is a neighborhood of . Since is lower semicontinuous at , there exists some neighborhood of such that
Since , there exist some , and whenâ€‰ , we have . By (5.17),
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 setvalued 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 setvalued map by
By Theorem 4.2, for each , we have and . Hence and . Clearly, is a convex set, hence it is a connected set. Define the setvalued maps , by
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
Then, for each , , and , we have since is convex and
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
Then, for each , we have that
By assumption, for each , is pseudomonotone, and by (5.24), for each , for the above , and for each , we have
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 whenâ€‰ , we have . Therefore, there exists , and whenâ€‰ , we have
Hence
Then
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
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
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 setvalued mapping
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
of , there exists a neighborhood of such that
For the above , since , when , there exists , and whenâ€‰ , we have By (5.33), we have
Thus there exist such that , for all . By (5.32), we have
that is
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 setvalued 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
that is
Thus there exists such that
Hence there exists such that
where is the unit ball of . Set
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
Since there exists such that when , we have Thus by (5.42) we have
Thus there exist for all . We have , and hence
Therefore,
Since and , we have . Hence there exists , and whenâ€‰ we have . This combining (5.45) implies that
Hence
By (5.40) and (5.47), we have
On the other hand, since , we have
This contradicts (5.48), because Thus . This means that is a closed subset of .
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Acknowledgments
This research was partially supported by the National Natural Science Foundation of China and the Natural Science Foundation of Jiangxi Province, China.
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Chen, B., Gong, XH. & Yuan, SM. Connectedness and Compactness of Weak Efficient Solutions for SetValued Vector Equilibrium Problems. J Inequal Appl 2008, 581849 (2008). https://doi.org/10.1155/2008/581849
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DOI: https://doi.org/10.1155/2008/581849