Recent Contributions to Fixed Point Theory and Its ApplicationsView this Special Issue
Noncompact Equilibrium Points for Set-Valued Maps
We prove a generalized result on the existence of equilibria for a monotone set-valued map defined on noncompact domain and take its values in an order of topological vector space. As consequence, we give a new variational inequality.
In the literature, the notion of an equilibrium point (or equilibrium problem) has been firstly introduced by Karamardian in  and Allen in . By using the well-known KKM principle, they proved that for a real valued function defined on a product of two sets and , there exists an element of , which will be called an equilibrium point, satisfying for all :
The classical hypothesis used to prove this type of equilibrium result concerns the convexity and compactness of the domain , the monotonicity, the convexity, and the continuity of and all extensions of this result obtained in the literature are about these hypotheses. In a recent work (see ), this result was extended to the noncompact case by using a coercivity type condition on a bifunction . In this context the function is supposed to take its values in a topological vector space endowed with an order defined by a cone in the same way that has been used by [4–7]. Note that the result on the existence of equilibrium points proved in  was obtained via a result on the existence of what we called weak equilibrium points, that is, a point , satisfying the following condition: where denotes the interior of the cone in .
In this paper, we investigate the extension of equilibrium points to set-valued maps in the same context. Generally, we have many choices to formulate the notion of equilibrium point. In fact, if is a closed convex cone of a topological vector space with nonempty interior, (, and is a set-valued map, then the equilibrium point for a set-valued map can be extended in several possible ways (see [8, 9]) as follows: ; ; ; . In this paper, we select the one that will be more adapted technically to our arguments. We will put a “moving” order on by a cone and the notions of convexity and continuity are naturally extended in our setting. We will use the pseudomonotonicity condition on borrowed from . As an application, we prove a variational inequality. The results obtained in this paper generalize the corresponding one in [9, 10].
We extend the notions of convexity, monotonicity, and continuity given previously to set-valued maps. If and are two sets, a set-valued map , where denotes the family of all subsets of , is a map that is assigned to each , a subset . Note that for the notation of set-valued maps, we will simply write instead of .
We firstly need to define an order on the codomain of set-valued maps as it has done for single valued maps. If is a subset of some real topological vector space , let be another real topological vector space, and let be a closed convex cone (not necessarily pointed) with nonempty interior and . Then defines an ordering “” on by means of We extend this notation to arbitrary subset by setting
By using this order, we naturally extend the notion of convexity for set-valued maps as follows.
Definition 1. Given a set-valued map defined on a vector space with values in a vector space endowed with an order defined by a convex cone , we say that is convex with respect to if for all and : which means that
Note that in particular, if and , we obtain the standard definition of convex set-valued maps.
As in the case of single valued maps, we can find many kinds of monotonicity for set-valued maps in the literature. We will use the notion of pseudomonotonicity defined in  which in turn extends the corresponding one defined in  for single valued maps.
Definition 2. Let and be two real topological vector spaces, let be a nonempty closed and convex set, and let be a set-valued map such that for every , is a closed and convex cone in with . Consider a set-valued map . is said to be pseudomonotone if, for any given ,
We recall the classical notions of continuity for set-valued maps as follows.
Definition 3. Given a set-valued map defined on a vector space with values in a vector space . Then(1)is said to be lower semicontinuous (l.s.c) at if, for every open set with , there exists a neighborhood of with for all . is said to be l.s.c. on if is l.s.c. at every .(2) is said to be upper semicontinuous (u.s.c) at if, for every open set with , there exists a neighborhood set of with for all . is said to be u.s.c. on if is u.s.c. at every .(3)A set-valued map which is both lower and upper semicontinuous is called continuous.
In this paper, we will use the definition of coercing family borrowed from .
Definition 4. 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 set-valued map 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. Definition 1 can be reformulated by using the “dual” set-valued map defined for all by . Indeed, a family is coercing for if and only if it satisfies conditions (i), (ii) of Definition 4, and the following one:
Note that in the case where the family is reduced to one element, condition (iii) of Definition 4 and in the sense of Remark 5 appeared first in this generality (with two sets and ) in  and generalized condition of Karamardian  and Allen . Condition (iii) is also an extension of the coercivity condition given by Fan . For other examples of set-valued maps admitting a coercing family that is not necessarily reduced to one element, see .
The following generalization of KKM principle obtained in  will be used in the proof of the main result of this paper.
Proposition 6. Let be a Hausdorff topological vector space, a convex subset of , a nonempty subset of , and a KKM map with compactly closed values in (i.e., for all , is closed for every compact set of ). If admits a coercing family, then .
3. The Main Result
As it is mentioned in the introduction, at an abstract level all possible extension of equilibria can be handled equally well. But there are great practical differences if we try to replace the resulting abstract conditions by simpler, verifiable hypotheses like convexity or semicontinuity. This is even more so if we admit a “moving” ordering cone (see ). For these reasons we choose to consider here the following generalized equilibrium problem.
Definition 7. Let be a nonempty convex subset of some real topological vector space , a real topological vector space, and a set-valued map such that for any , is a closed convex cone with and . Let be a set-valued map. The generalized equilibrium problem is to find such that in this case, is said to be an equilibrium point.
Theorem 8. Let and be real topological vector spaces (not necessarily Hausdorff). Let a nonempty, convex set and three set-valued mappings , , and be given. Suppose that the following conditions are satisfied.(1)For all , implies (pseudomonotonicity).(2)For all , is closed in .(3)For all , is convex.(4)For all , .(5)There exists a family satisfying conditions (i) and (ii) of Definition 4 and the following one: for each , there exists such that Then there exists such that for all .
Proof. Let us consider a set-valued map defined for every by
Then we can see firstly that is a KKM map; that is, for every finite subset of there holds
In fact, let and assume by contradiction that ; it means that with , and for all . Then for all , hence from condition for all . It follows from condition that , and then , which contradicts condition ; thus is a KKM map.
It is also clear from condition that, for all , is closed.
In addition, we can verify that condition implies that the family satisfies the following condition: for all there exists with
We deduce that satisfies all hypothesis of Proposition 6, so we have Therefore there exists such that for any , . Hence
Theorem 9. Let , , , , , and satisfy the assumptions of Theorem 8 and the additional following conditions.(6)For all with and if and , then for all .(7)For all with , is open in .
Then there exists such that for all .
Proof. By Theorem 8, there exists with for all . Assume that for some ; then by and from there exists such that . Since , we deduce that , but this contradicts and the theorem is proved.
The following result, which corresponds to Theorem 1 in , can be deduced from the two previous theorems.
Corollary 10. Let , , satisfy hypothesis of Theorem 8, of Theorem 9 and the following condition.There exists a nonempty compact set and a compact convex set such that for every there exists with .
Then there exists such that for all .
Proof. By taking for all , , which is convex compact set, and , which is compact set, and by using hypothesis , we can see that admits a coercing family in the sense of Remark 5; that is, for all , . Suppose, per absurdum, that there exists with . Hence for all , . This means that for all , and so . Therefore, there exists such that for all , we have
Then by Theorem 9, we deduce that there exists such that for all
but this contradicts hypothesis (5′).
Corollary 11. Let be a set-valued map satisfy the following conditions. (1)For all , implies .(2)For all , is lower semicontinuous.(3)For all , is convex with respect to .(4)The map has open graph in .(5)For all , is upper semicontinuous and compact valued on .(6)For all , .(7)There exists a family satisfying conditions and of Definition 4 coercing and the following one. For each , there exists such that Then there exists such that for all .
Proof. Following , if the map is lower semicontinuous and is closed, then condition of Theorem 9 is satisfied. Furthermore and also by , condition of Theorem 9 is fulfilled, if for all , the map is upper semicontinuous along line segments with compact values, and the map has open graph in .
Now let denote the space of all continuous linear operators . For , we write and for , we write . The following result is a variational inequality formulation of our main result.
Corollary 12. Let a map be given such that for all is nonempty. Suppose the following. (1)For all , implies .(2)The map has open graph in .(3)For all , is upper semicontinuous on and compact valued.(4)There exists a family satisfying conditions and of Definition 4 and the following one: for each , there exists such that Then there exists such that for all .
Proof. Take , , and . Then conditions and of Theorem 9 are clearly satisfied. holds since each member of is continuous and is closed. is satisfied since and . and hold since for all :
To verify hypothesis , we have to show that is closed in . Let be a net in converging to ; we may assume , since , and we may assume for all as well. Thus with and with . For every , there exists with ; then . We conclude as above that there is a subnet converging to some . The corresponding converges to , since has open graph; we obtain ; hence .
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
The authors extend their appreciation to the Deanship of Scientific Research at King Saud University for funding the work through the research group Project (RGP-VPP 237).