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On the existence and stability of solutions of a mixed general type of variational relation problems
Journal of Inequalities and Applications volume 2014, Article number: 337 (2014)
Abstract
In this paper, we introduce a mixed general type of variational relation problems, and establish the existence theorem of solutions of mixed general types of variational relation problems. Moreover, we study the stability of a solution set of mixed general types of variational relation problems. We prove that most of mixed general types of variational relation problems (in the sense of Baire category) are essential and, for any mixed general type of variational relation problems, there exists at least one essential connected component of its solution set.
MSC:49J53, 49J40.
1 Introduction
It is well known that the equilibrium problems are unified models of several problems, namely optimization problems, saddle point problems, variational inequalities, fixed point problems, Nash equilibrium problems etc. Recently, Luc [1] introduced a more general model of equilibrium problems, which is called a variational relation problem (VR). The stability of the solution set of variational relation problems was studied in [2, 3]. Further studies of variational relation problems have been done (see [4–12]). Recently, Agarwal et al. [13] presented a unified approach in studying the existence of solutions for two types of variational relation problems, which encompass several generalized equilibrium problems, variational inequalities and variational inclusions investigated in the recent literature. Balaj and Lin [14] established existence criteria for the solutions of two very general types of variational relation problems.
Motivated and inspired by research works mentioned above, we introduce mixed general types of variational relation problems, which is a mixed structure of two general types of variational relation problems in [14]. Moreover, we study the stability of a solution set of mixed general types of variational relation problems.
2 Mixed general types of variational relation problems
In [14], let X, Y be convex sets in two Hausdorff topological vector spaces, Z be a topological space, , , be set-valued mappings with nonempty values, and be a relation linking elements , and . Balaj and Lin [14] established existence criteria for the solutions of the following variational relation problems:
(VRP1) Find such that , and, , for which holds.
(VRP2) Find such that , and holds and .
In this paper, we introduce mixed general types of variational relation problems. Let X, Y be convex sets in two Hausdorff topological vector spaces, Z be a topological space, , , , , be set-valued mappings with nonempty values, and , be two relations linking elements , and . A mixed general type of variational relation problems (MGVR) consists in finding such that , and
Remark 2.1 Balaj and Lin [14] established existence criteria for the solutions of two very general types of variational relation problems. The mixed general type of variational relation problems is a combination of (VRP1) and (VRP2), and (VRP1) and (VRP2) are some special cases of (MGVR).
Theorem 2.1 Assume that
-
(i)
X, Y, Z are three nonempty, compact and convex subsets of three Hausdorff linear topological spaces;
-
(ii)
and are closed in ;
-
(iii)
P is continuous with nonempty compact values;
-
(iv)
for any , , , H, G have open fibers, and , are closed;
-
(v)
for any fixed , any finite subset of X and any , there are and such that holds;
-
(vi)
for any fixed , any finite subset of Y and any , there is such that holds for any .
Then (MGVR) has at least one solution.
Proof Define and as follows:
As , are closed for any , and P is continuous with nonempty compact values, by Propositions 3.1, 3.3 of [14], A, B have open fibers.
Suppose that there exists such that , then there is a finite subset of such that . By (v), there are and such that holds, which contradicts the fact that for any , i.e., does not hold for any and any . Hence for any .
Suppose that there exists such that , then there is a finite subset of such that . By (vi), there is such that holds, which contradicts the fact that for any , i.e., there is such that does not hold for any . Hence for any .
Define and as follows:
For any , is open in . Similarly, is open for any . Hence, , are open for any , and , for any . By Theorem 3 of [15], there exists such that and , which implies that , and
□
Theorem 2.2 Assume that
-
(i)
X, Y, Z are three nonempty, compact and convex subsets of three normed linear topological spaces;
-
(ii)
S, T are continuous with nonempty convex compact values;
-
(iii)
P is continuous with nonempty compact values;
-
(iv)
and are closed;
-
(v)
for any fixed , any finite subset of X and any , there are and such that holds;
-
(vi)
for any fixed , any finite subset of Y and any , there is such that holds for any .
Then there exists such that , and
Proof For any n, define , , and by
where , . Since S, T are continuous, , have open fibers, and , are closed in , by Theorem 2.1, there exist , and
Since X, Y are nonempty and compact, without loss of generality, we assume that . Let be arbitrarily fixed and such that . Since S is continuous and , there is such that and for any . Therefore, for any ,
Hence . Similarly, .
Suppose that there exists such that does not hold. Since is closed, there exists such that does not hold. Since S is continuous, there exists a sequence convergent to u with . Since P is continuous, there exists such that, for any , , , , which implies that and does not hold. It is a contradiction.
Suppose that there exist and such that does not hold. Since T, P are continuous, there exist two sequences and convergent to v and z with and . As is closed, there exists such that, for any , does not hold, which is a contradiction. This completes the proof. □
3 Generic stability analysis
Let X, Y, Z be three nonempty, compact and convex subsets of three normed linear topological spaces. Denote by ℳ the collection of all (MGVR) such that all conditions of Theorem 2.2 hold. For each , denote by the solution set of q. Thus, a correspondence is well defined. For each , define the distance on ℳ by
where , , (, ) is the Hausdorff distance defined on X (Y, Z), and H is the Hausdorff distance defined on .
Definition 3.1 Let . An is said to be an essential point of if, for any open neighborhood of in , there is a positive δ such that for any with . q is said to be essential if each is essential.
Definition 3.2 Let . A nonempty closed subset of is said to be an essential set of if, for any open set U, , there is a positive δ such that for any with .
Definition 3.3 Let . An essential subset is said to be a minimal essential set of if it is a minimal element of the family of essential sets in ordered by set inclusion. A connected component of is said to be an essential component of if is essential.
Remark 3.1 (1) It is easy to see that the problem is essential if and only if the mapping is lower semicontinuous at q. (2) For two closed , if is essential, then is also essential.
Lemma 3.1 (17.8 Lemma, 17.11 Closed Graph Theorem of [16])
(i) The image of a compact set under a compact-valued upper semicontinuous set-valued mapping is compact. (ii) A correspondence with compact Hausdorff range space is closed if and only if it is upper hemicontinuous and closed-valued.
Lemma 3.2 ([17])
If X, Y are two metric spaces, X is complete and is upper semicontinuous with nonempty compact values, then the set of points, where F is lower semicontinuous, is a dense residual set in X.
Theorem 3.1 is a complete metric space.
Proof Let be any Cauchy sequence in ℳ, then, for any , there is such that for any , that is, for any ,
-
(1)
Clearly, we consult Proposition 3.1 of [18]. There are , and such that S, T are continuous with nonempty convex compact values, and P is continuous with nonempty compact values.
-
(2)
There exist two closed subsets A, B of such that and . Denote , where
Clearly and are closed.
-
(3)
Suppose the existence of , finite subset of X and such that does not hold for any , which implies , . Since for enough large m, , , i.e., does not hold for any , which is a contradiction.
Suppose the existence of , finite subset of Y, and such that does not hold for any , which implies , . Since , there exists a sequence convergent to z with . Hence, for enough large m, , , i.e., does not hold for any , which is a contradiction. Hence and is complete. □
Theorem 3.2 The mapping is upper semicontinuous with nonempty compact values.
Proof The desired conclusion follows from Lemma 3.1 as soon as we show that is closed. Denote and . Let be a sequence converging to such that for any n. Then and ,
Clearly, and .
Suppose the existence of such that does not hold, then . Since S, P are continuous, and
then and . Thus, there exists a sequence convergent to u with such that, for enough large n, , i.e., and does not hold, which is a contradiction.
Suppose the existence of and such that does not hold, then . Similarly, and . Thus, there exist two sequences and convergent to u and z with and . Hence, for enough large n, , i.e., , and does not hold, which is a contradiction. Hence . □
Theorem 3.3 (i) There exists a dense residual subset of ℳ such that q is essential for each . (ii) For any , there exists at least one minimal essential subset of .
Proof The proofs are similar to those of Theorems 3.3 and 3.4 of [19]. Here, we do not repeat the process. □
4 Existence of essential connected components
In this section, let be fixed. Assume that (i) is continuous with nonempty compact values; (ii) if , ; (iii) for any . Denote by the collection of mixed general types of variational relation problems such that all conditions of Theorem 2.2 hold. Clearly . For convenience in the later presentation, for any subset A of X, denote .
Lemma 4.1 ([20])
Let C, D be two nonempty, convex and compact subsets of a linear normed space E. Then , where h is the Hausdorff distance defined on E, and , .
Lemma 4.2 ([21])
Let be a metric space, and be two nonempty compact subsets of Y, and be two nonempty disjoint open subsets of Y. If , then
where h is the Hausdorff metric defined on Y.
Theorem 4.1 For any , every minimal essential subset of is connected.
Proof For each fixed , let be a minimal essential subset of . If is not connected, then there are two nonempty compact subsets , and two disjoint open subsets , of such that and , . Since is a minimal essential set of , neither nor is essential. There exist two open sets , such that, for any , there exist with
Denote , , we know that , are open, , and we may assume that , . Denote
Since is continuous with nonempty compact values, and , are nonempty compact in , by Lemma 3.1, , are nonempty compact in Z. Thus , are open in .
To prove by contraposition that , suppose the existence of such that , which implies that and , i.e., , . It follows that and , which contradicts the fact that .
Denote . Since is essential, and , there exists such that for any with . Since is a minimal essential set of , neither nor is essential. Thus, for , there exist two such that
Thus . Next, define as follows:
where
Easily, we check that (i) , are continuous with nonempty compact convex values. (ii) Since and are closed in , A is closed in , which implies that is closed. Similarly, is closed. (iii) Suppose the existence of , finite subset and such that does not hold for any , i.e., , . Since
and , without loss of generality, we may assume that , which implies . Since for all , that is,
then , , i.e., does not hold for any , which is a contradiction.
(iv) Suppose the existence of , finite subset , and such that does not hold for any , i.e., , . Since
and , without loss of generality, we may assume that , which implies . Since for all , that is, for all , then
that is, there is such that does not hold for any , which is a contradiction. Hence .
(v) By Lemmas 4.1, 4.2,
Thus and .
Since , without loss of generality, we assume . Then there exists such that , , , and
It follows from that , , and . Therefore,
Then , which is a contradiction. This completes the proof. □
Theorem 4.2 For any , there exists at least one essential connected component of .
Proof By Theorem 4.1, there exists at least one connected minimal essential subset of . Thus, there is a component C of such that . It is obvious that C is essential by Remark 3.1(2). This completes the proof. □
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Acknowledgements
This research is supported the Chen Guang Project sponsored by the Shanghai Municipal Education Commission and Shanghai Education Development Foundation (no. 13CG35), and open project of Key Laboratory of Mathematical Economics (SUFE), Ministry of Education (no. 201309KF02).
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Yang, Z. On the existence and stability of solutions of a mixed general type of variational relation problems. J Inequal Appl 2014, 337 (2014). https://doi.org/10.1186/1029-242X-2014-337
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DOI: https://doi.org/10.1186/1029-242X-2014-337