Abstract
We study extended mixed vector equilibrium problems, namely, extended weak mixed vector equilibrium problem and extended strong mixed vector equilibrium problem in Hausdorff topological vector spaces. Using generalized KKM-Fan theorem (Ben-El-Mechaiekh et al.; 2005), some existence results for both problems are proved in noncompact domain.
1. Introduction
Giannessi [1] first introduced and studied vector variational inequality problem in a finite-dimensional vector space. Since then, the theory with applications for vector variational inequalities, vector equilibrium problems, vector complementarity problems, and many other problems has been extensively studied in a general setting by many authors; see for example [2–7] and references therein.
In 1989, Parida et al. [8] developed a theory for the existence of a solution of variational-like inequality problem and showed the relationship between variational-like inequality problem and a mathematical programming problem. The problem of vector variational-like inequalities is also one of the generalizations of vector variational inequalities studied by many authors; see [9–11] and references therein.
On the other hand, equilibrium problem was first introduced and studied by Blum and Oettli [12]. Many authors [13–15] have proved the existence of equilibrium problems by using different generalization of monotonicity condition and generalized convexity assumption. The main objective of our work is to study an extended weak mixed vector equilibrium problem and an extended strong mixed vector equilibrium problem and we prove existence results for both problems by using a generalized coercivity type condition, namely, coercing family. Both problems are combination of a vector equilibrium problem and a vector variational-like inequality problem. Our results presented in this paper improve and generalize some known results obtained by [12, 16–18].
2. Preliminaries
Throughout this paper, let and be the Hausdorff topological vector spaces. Let be a nonempty convex closed subset of and a pointed closed convex cone with . The partial order “” on induced by is defined by if and only if . Let , and be the mappings, where is the space of all continuous linear mappings from to . We denote the value of at by . In this paper, we consider the following problems.
Find such that We call problem (1) extended weak mixed vector equilibrium problem and problem (2) extended strong mixed vector equilibrium problem.
Let us recall some definitions and results that are needed to prove the main results of this paper.
Definition 1. A mapping is said to be(i)lower semicontinuous with respect to at a point , if for any neighborhood of in , there exists a neighborhood of in such that (ii)upper semicontinuous with respect to at a point , if (iii)continuous with respect to at a point , if it is lower semicontinuous and upper semicontinuous with respect to at that point.
Remark 2. If is lower semicontinuous, upper semicontinuous, and continuous with respect to at any arbitrary point of , then is lower semicontinuous, upper semicontinuous, and continuous with respect to on , respectively.
Definition 3 (see [19]). Let and be the mappings. Then(i) is said to be --pseudomonotone, if for any , (ii) is said to be strongly --pseudomonotone, if for any , (iii) is -hemicontinuous, if for any given and , the mapping is continuous at ;(iv) is said to be affine in the first argument, if for any and , with and any , we have
Definition 4 (see [20]). Consider a subset of a topological vector space and a topological space . A family of pair of sets is said to be coercing for a mapping if and only if(i)for each , is contained in a compact convex subset of and is a compact subset of ;(ii)for each , there exists such that ;(iii)for each , there exists with .
Remark 5. In case where the coercing family reduced to single element, condition (iii) of Definition 4 appeared first in this generality (with two sets and Z) in [21] and generalizes the condition of Karamardian [22] and Allen [23]. Condition (iii) is also an extension of coercivity condition given by Fan [24].
Definition 6. Let be a nonempty convex subset of a topological vector space . A multivalued mapping is said to be KKM mapping, if, for every finite subset of , where denotes the convex hull of and is a finite index set.
Theorem 7 (see [20]). Let be a Hausdorff topological vector space, a convex subset of , a nonempty subset of , and a KKM mapping with compactly closed values in (i.e., for all is closed for every compact set of ). If admits a coercing family, then
Lemma 8 (see [25]). Let be a Hausdorff topological space and nonempty compact convex subsets of . Then is compact.
3. Existence Results
In this section, we first present an existence result for extended weak mixed vector equilibrium problem (1).
Theorem 9. Let be a nonempty closed convex subset of a Hausdorff topological vector space , a Hausdorff topological vector space, and a closed convex pointed cone with . Let , and be the mappings satisfying the following conditions:(i) is affine in the second argument and continuous in the first argument;(ii), for all ;(iii) and , for all ;(iv) is affine in both arguments and continuous in the second argument;(v) is -hemicontinuous, --pseudomonotone, and continuous;(vi)the mapping , defined by , is upper semicontinuous on ;(vii)there exists a family satisfying conditions (i) and (ii) of Definition 4 and the following condition: for each , there exists such that Then, there exists a point such that
For the proof of Theorem 9, we need the following proposition, for which the assumptions remain the same as in Theorem 9.
Proposition 10. The following two problems are equivalent:(I)find such that ;(II)find such that .
Proof. Suppose that (I) holds. Then for every , we have
Since is --pseudomonotone, from (12) we have
Also from assumptions (iii) and (13), we get
that is, (II) holds.
Conversely, assume that (II) holds for all . Then there exists such that
For a fixed , set , for . Obviously, and it follows that
Multiplying (16) by , we have
Since is affine and , we have
That is,
Since , adding on both sides of (19), we obtain
Combining (17) and (20), we get
Since is affine in the second argument and , (21) implies that
Since is affine and , then from (22) we deduce that
Dividing (23) by , we have
Using -hemicontinuity of , we get
and hence (II) holds.
Proof of Theorem 9. For each , consider the sets
Then and are nonempty sets, since and .
First, we prove that is a KKM mapping. Indeed, assume that is not a KKM mapping. Then, there exists finite subset of , for each with and such that
That is,
As is convex, therefore
Since is affine in the second argument and is affine, from (29) we have
By assumptions (ii) and (iii), we know . Then (30) implies that , which contradicts the pointedness of and hence is a KKM mapping.
Further, we prove that
Let , so that
Since is --pseudomonotone and , then (32) implies that
and so for each ; that is, and hence
Conversely, suppose that . Then
It follows from Proposition 10 that
that is, and so
Combining (34) and (37), we obtain
Now, since is a KKM mapping, for any finite subset of , we have
This implies that is also a KKM mapping.
In order to show that is closed for all , let us assume that is a net in such that . Then
Since is continuous in the first argument, is continuous in the second argument, and is continuous, we have
As is upper semicontinuous, we obtain
and thus, we have
Therefore , for all and hence is closed. In view of assumption (vii), has compactly closed values in .
By assumption (vii), we see that the family satisfies the condition which is for all there exists such that
and consequently, it is a coercing family for .
Finally, we conclude that satisfies all the hypotheses of Theorem 7 and thus we have
Hence, there exists such that for all
This completes the proof.
Now, we prove an existence result for extended strong mixed vector equilibrium problem (2).
Theorem 11. Let and satisfy the assumptions (i)–(iv) of Theorem 9. In addition, assume that the following conditions are satisfied:(v)′for each , the set is open in ;(vi)′there exists a nonempty compact and convex subset of and, for each , there exists such that (vii)′there exists a family satisfying conditions (i) and (ii) of Definition 4 and the following condition which is for each there exists such that Then, there exists a point such that for all
Proof. Let be defined by
Obviously, for all
As is closed subset of and is compact, therefore is compactly closed.
Now, we show that, for any finite set of , . For this, let . Then, by Lemma 8, is a compact and convex subset of .
Let be defined by
First, we prove that is a KKM mapping. On contrary, suppose that is not a KKM mapping; then there exists such that, for with , we have
which implies
Since and are affine in the second argument, (54) implies that
Since , (55) implies that , which is a contradiction. Hence, is a KKM mapping.
As is closed subset of , therefore it is compactly closed. From assumption (vii)′, it is clear that the family satisfies the condition and therefore it is a coercing family for . Applying Theorem 7, we obtain
Thus we conclude that there exists .
To show that , on contrary suppose that . Then condition (vi)′ implies that there exists such that
which contradicts the fact that , and hence . Since , for each , it follows that ; that is, , for finite subset . As is closed and compact, it follows that, for each , there exists such that . Hence, there exists such that, for all ,
This completes the proof.
Theorem 12. Let the assumptions (i)–(iv) of Theorem 9 hold. In addition, we assume that is strongly --pseudomonotone and -hemicontinuous. Then the following problems are equivalent:(I)find such that ;(II)find such that .
Proof. Suppose (I) holds. By using the definition of strong --pseudomonotonicity of , (II) follows directly.
Conversely, suppose (II) holds for all . Then we can find such that
By substituting , for , in (59), we obtain
As is affine and , (60) implies that
Since is affine in the second argument and , from (61) we get
As is a cone, therefore
On contrary suppose that
As is -hemicontinuous, we have
for sufficiently small , which contradicts (63). Therefore we have
and hence (I) holds. This completes the proof.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgment
The authors gratefully acknowledge that this research was partially supported by the Universiti Putra Malaysia under GP-IBT Grant Scheme having Project no. GP-IBT/2013/9420100.