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Journal of Function Spaces and Applications

Volume 2013 (2013), Article ID 267585, 9 pages

http://dx.doi.org/10.1155/2013/267585

## Ekeland Variational Principle for Generalized Vector Equilibrium Problems with Equivalences and Applications

^{1}College of Applied Science, Beijing University of Technology, Beijing 100124, China^{2}College of Mathematics, Jilin Normal University, Siping, Jilin 136000, China

Received 2 April 2013; Accepted 12 July 2013

Academic Editor: Yongsheng S. Han

Copyright © 2013 De-ning Qu and Cao-zong Cheng. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

The aim of this paper is to introduce Ekeland variational principle with variants for generalized vector equilibrium problems and to establish some existence results of solutions of generalized vector equilibrium problems with compact or noncompact domain as applications. Finally, some equivalent results of the established Ekeland variational principle are presented.

#### 1. Introduction

An Ekeland variational principle [1] (also [2, 3]) appeared first as an existence result of approximate minimizer for a lower semicontinuous and bounded below function on complete metric spaces. It subsequently developed an important tool of many subjects, such as in nonlinear analysis (e.g., [4]), optimization theory (e.g., [5–11]), game theory (e.g., [12]), dynamical systems (e.g., [13]), and others (e.g., [14, 15]). By reason of the fact that equilibrium problems contain many problems as their special cases, such as optimization problems, fixed-point problems, variational inequality problems, complementary problems, and Nash equilibrium problems (see [16]), a new direction of research on variational principle for equilibrium problems has arisen. The variational principle for equilibrium problems (e.g., [17]) and for vector equilibrium problems (e.g., [18–23]) and/or their applications or equivalent results were discussed. In terms of set-valued objective mappings, variational principle for vector optimization problems was first introduced by Chen and Huang [5] and was reported in many literatures (e.g., see [7, 9–11]) in the sequel. In 2009, Zeng and Li [24] discussed Ekeland variational principle for vector equilibrium problems with set-valued objective mappings (generalized vector equilibrium problems).

One of the most important tools of obtaining variational principle for vector problems is the scalarization functions (see [5, 6, 21–24], for instance). While these scalarization functions always involve single-valued mappings. Motivated by the works mentioned above, we establish the Ekeland variational principle for generalized vector equilibrium problems by applying a nonlinear scalarization function involving set-valued mappings. It is worth noting that the generalized vector equilibrium problems considered in this paper are rather than those in [24].

Let , and be denoted by the sets of real numbers, nonnegative real numbers and positive integers, respectively, and let be the collection of open neighborhoods of , where is a point or a set. A subset of a real topological vector space is called a *cone* if for all and . Let be a cone in and . is called *proper* if . is said to be -*closed* [25] if is closed; to be -*bounded* [25] if, for each neighborhood , there exists such that .

Throughout this paper, unless otherwise specified, let be a Hausdorff topological space and a real Hausdorff topological vector space, let be a proper, closed, and convex cone with nonempty interior and , and let be a strict, set-valued mapping, for each and for each . A set-valued mapping is said to be strict if it has nonempty values. Consider the following generalized vector equilibrium problems:

This paper is divided into five sections. In Section 2, some preliminaries are provided. In Section 3, Ekeland variational principle with variants for is argued in complete quasimetric spaces. In Section 4, some existence results of solutions of with compact or noncompact domain are established as applications. Finally, in Section 5, some equivalent results of the established Ekeland variational principle are presented.

#### 2. Preliminaries

Let be a nonempty set. “” is called a quasiorder on if it is of(o1) reflexivity: , for all ;(o2) transitivity: .

Here is called a *quasiorder space*. An element is said to be a *maximal element *of a quasiorder space if there is no element , other than , such that , in other words, for some implies .

Let and be topological spaces. A real-valued function is said to be *upper semicontinuous on * if is open for each ; to be *lower semicontinuous on *, is open for each . The following conceptions of continuity for a set-valued mapping can be found in [4]. A set-valued mapping is said to be *upper semicontinuous at * if, for any , there exists such that for all ; to be *lower semicontinuous at *, if for any and any , there exists such that for all ; to be *upper semicontinuous* (resp., *lower semicontinuous*) *on *, if is upper semicontinuous (resp., lower semicontinuous) at each ; to be *continuous at * (resp., on ), if is both upper semicontinuous and lower semicontinuous at (resp., on ); to be *closed*, if its graph is closed in .

Lemma 1 (see [26]). *Let and be Hausdorff topological spaces and a strict set-valued mapping. If is compact and is upper semicontinuous with compact values, then is compact.*

*Definition 2. *Let be a nonempty set. A function is said to be a *quasimetric on * if (d1) if and only if ;(d2).

equipped a quasimetric is said to be a *quasimetric space*, denoted by . Furthermore, if also satisfies(d3),

then is called a metric space, where is a metric on .

Let be a quasimetric space. means . is called a Cauchy sequence on if, for any , there exists such that . is said to be complete if, for each Cauchy sequence , there exists such that as .

*Definition 3 (see [27]). *Let be a quasimetric space. A function is said to be a -*distance* on if (w1);
(w2) for each fixed , is lower semicontinuous;(w3) for any , there exists such that and imply that .

The -distance includes metric and quasimetric as its special cases. But the converse fails to be true. Moreover, -distance is not necessary to be symmetric. Some examples and properties of a -distance in metric spaces are provided by Kada et al. [27].

Lemma 4 (see [27]). *Let be a -distance on a quasimetric space , let and with as , and let . Then the following assertions are true:*(i)*if and , then . In particular, if and , then ;*(ii)*if , then is a Cauchy sequence;*(iii)*if , then is also a -distance on .*

Let be a topological space and be a topological vector space in the rest of this section.

Since each compact subset in is both -closed and -bounded by Definition 3.1 and Proposition 3.1 in [25], the following general nonlinear scalarization function is well defined in view of Lemma 3.1 in [28].

*Definition 5. *Let be a strict and compact-valued mapping. A generalized nonlinear scalarization function of is defined by

According to Proposition 3.1 in [28], we have the following.

Lemma 6. *Let be a strict and compact-valued mapping. The following assertions are true for each and :*(i)*.
*(ii)*.
*

Lemma 7. *Let be a strict and compact-valued mapping.*(i)*If is lower semicontinuous on , then is upper semicontinuous on ;*(ii)*If is upper semicontinuous on , then is lower semicontinuous on .*

*Proof. *This proof is completed by letting , and for all in Corollary 3.1 in [28] since upper (resp., lower) semicontinuity implies the -upper (resp., lower) semicontinuity (see [29]).

Lemma 8. *If*(a1)* is compact valued;*(a2)*,
**then
*

*Proof. *For each , it follows from Definition 5 that and can be chosen. Then by (a2). Select to satisfy . This deduces that
Therefore, by Lemma 6(ii) and .

#### 3. Ekeland Variational Principle for

From now on, unless otherwise specified, suppose that is a Hausdorff complete quasimetric space and that is a -distance on .

Lemma 9. * (i) If (a1)-(a2) and*(a3)

*hold, then*

*where is defined by*

*(a4)*

*(ii)*Besides (a1)–(a3), if*for each is upper semicontinuous on ;*(a5)

*for each for some ,*

*then, for each with , there exists such that .*

*Proof. *Obviously, by (a1) and Lemma 6(ii),
(i) For each and , if , then the conclusion holds trivially. Otherwise, for each ,
and so
according to Lemma 8. Moreover, by and ,
Now claim that . Otherwise, by (a3) and Lemma 6(i). Thus,
since
by (2) and (9). Similarly, , and so by (w1). This, together with , implies that by Lemma 4(i), which contradicts with . Thus , and so .

(ii) For given in (a5), for some by Lemma 2.1 in [30], which leads to . This, together with (a5), implies that . It follows form Lemma 4(ii) that
Then, for each ,
For each , let satisfy that and take such that
If for some , this conclusion holds by letting . Now consider . The rest proof is divided into four steps as follows.

(a) Show that as for some . Indeed, for each , by (2),
which implies that
and so
Therefore, is a Cauchy sequence by Lemma 4(ii), and so there exists such that as by the completeness of . In addition, according to (w2)
(b) Show that . In fact, for each ,
by (2). Since, for each , is lower semicontinuous by (a1), (a4), and Lemma 7(ii),
by letting in (19). Hence,
Now it is enough to prove that by (20). Indeed, if for some , then in view of (a3) and Lemma 6(i). By applying (21) and adopting the same argument of the proof of (10),
Thus,
It follows form that . Hence, by (21) and the similar argument of (10), which, together with , implies in virtue of (22) and Lemma 4(i). This contradicts with . Thus, . Clearly, .

(c) Show that . Indeed, for any , , and . As a result,
which, together with (18), implies that by Lemma 4(i).

(d) Show that and . As a matter of fact, by (b). If , then in view of (c). This leads to a contradiction by the definition of .

Theorem 10. *Define as
**
If (a1)–(a5) hold, then, for any and for any , there exists such that *(i)*;*(ii)* with ;*(iii)* if, further, , where defined as
*

*Proof. *Since is another -distance by Lemma 4(iii), without loss of generality, we set . Then
and is identical to (5). Take if . Otherwise, there exists such that by Lemma 9(ii). All in all, . The conclusion (i) is true. Also, is equivalent to , and so the conclusion (ii) holds by Lemma 6(ii). If, further, , (i) implies that
by Lemma 6(ii).

Corollary 11. *If (a5) is replaced by*(a6)* is compact**in Theorem 10, then the conclusions still hold.*

*Proof. *It is easy to see that is compact for each by (a1), (a4), (a6), and Lemma 1. The rest proof is divided into three steps.

(a) For any with ,
(b) For any , there exists such that . In fact, if , then by (a), which implies
Since by Lemma 2.1 in [30], for each , there exists and such that . Then . Thus and a contradiction with properness if arises.

(c) For some , and (a5) holds. Indeed, for each , there exists such that , and so for some . By the completeness of , there exists such that . Taking , we have .

Apparently, we have the following by applying Theorem 10.

Corollary 12. *If*(a7)*, where is the topological bound of **instead of (a3), hold in Theorem 10, then the conclusions are true as well.*

*Remark 13. *Under all the hypotheses in Theorem 10, Corollary 11, or Corollary 12, for each , there exists such that

In fact, it is obvious that with . For , if then , which contradicts with (a3) or (a7).

*Remark 14. *(a) The condition: (a8) there exists such that ,often appeared in the results on Ekeland variational principle for (generalized) vector equilibrium problems, such as Theorem 2.1 in [22]. Actually, (a8) is equivalent to (a5). As the case stands, it is apparent that (a5) implies (a8). On the other hand, taking , then , which guarantees (a5).

(b) The condition:(a9) for each , or ,instead of (a3) or (a7), is always required to prove variational principle for equilibrium problems in many literatures, such as in [20–24]. But this condition is unnecessary in Theorem 10. See the following examples.

*Example 15. *Let , and , and let . Define as
By computing simply, (a1)–(a5) hold. But (a9) is false since unless .

*Example 16. *Let , and , and let , and define as . It is clear that (a1)-(a2), (a4)-(a5), and (a7) are satisfied, but (a9) does not hold.

It is effortless to obtain the following variant of vectorial form of Ekeland variational principle by Theorem 10.

Theorem 17. *Let be given. If, besides (a1)–(a5),**
for some with ,
**
where is defined as (26), then there exists such that *(i)* if, further, ;*(ii)*;
*(iii)* with .*

*Proof. *Let be the point provided by Theorem 10 with instead of . Then (iii) is obtained by the conclusion (ii) in Theorem 10. In addition, in view of the conclusion (i) in Theorem 10,
It follows from (34)-(35), (a10), and Lemma 6(ii) that and so the conclusion (ii) holds. Moreover, according to the conclusion (iii) in Theorem 10, there exists such that
Taking and in (a2), we have
Then
Thereby, (i) holds.

*Remark 18. *(a) If the solution set of or is nonempty, then there must exist a point satisfying (34).

(b) If, further, is a point cone (that is, ) in Theorems 17, then the condition (a10) can be weakened as follows: for some with ,
where is defined as (5).

In fact, we only prove (ii) by the proof of Theorem 17. In view of Theorem 10(iii),
Then there exist and such that . By (39),
or equivalently,
If , then and so , which implies that . Thereupon, by the pointedness of , which contradicts with .

#### 4. Applications of Vectorial Form of Ekelend Variational Principle for

In this section, some existence results of solutions of are established as applications of Ekelend variational principle for .

Theorem 19. *Let admit a topology (possibly different from the initial topology induced by ). Besides (a1)-(a2), (a4)-(a5) and (a7) in the sense of , if the following conditions hold:*(b1)* is -compact; namely, is compact with respect to ;*(b2)* for each , as implies the boundedness of ;*(b3)* is -lower semicontinuous on for each ,**then has a solution.*

*Proof. *If and in (31), then there exists such that
Thus by Lemma 6(i). Without loss of generality, let by (b1). Since is -upper semicontinuous for each by (b3) and Lemma 7(i),
due to (b2). This deduces that by Lemma 6(i); that is, has a solution.

*Remark 20. *It is worth noting that the condition (b2) cannot be guaranteed by (b1). For instance, let and . For the trivial topology , is -compact. For , it is clear that , but is unbounded.

Corollary 21. *If (a1)-(a2), (a4), and (a6)-(a7) and*(a11)* is lower semicontinuous on for each ,**hold, then has a solution.*

*Proof. *Clearly, (a6) implies (a5). Taking in Theorem 19, (b1) and (b3) hold by (a6) and (a11), respectively, and (b2) holds trivially since when . Thus this conclusion is true by Theorem 19.

When the existence of solutions of equilibrium problems on a noncompact domain is discussed, some sufficient assumptions, such as coercivity condition [31] and condition [20], and so forth, must be required to be substitute for compact domain. Motivated by these ideas, we obtain the following existence results of solutions of by applying Theorem 19.

Theorem 22. *Let (a1)-(a2), (a4)-(a5), and (a7) hold in the sense of induced by , and let admit a topological (possibly different from the initial topology induced by ). If (b3) is satisfied and the following hold:*(b4)* each -closed bounded ball on is -compact,*(b5)* for given point , there exists a -compact set satisfying that, for each and , implies the boundedness of and for each ,
* *for some ,*(b6)* for each , is -closed,*(b7)* for each , is -lower semicontinuous,* *then has a solution.*

*Proof. *Define as
We see the following. (a) For each , since by (a7). (b) For each , implies that by the similar argument of the proof in Lemma 9. (c) is -compact for each by (b4) and (b6).

By regarding the -compact set as in Theorem 19, there exists such that
Now argue the conclusion by contradiction. Assume that there exists such that
Let

Now assert that
In fact, if , then , which, together with (48), deduces that in view of (a2) and Lemmas 6 and 8. This contradicts with (47).

The minimum in (49) is attained since is nonempty -compact by (a) and (c), and is -lower semicontinuous. Take such that . Since by (50), can be chosen to satisfy that and by (b5). Accordingly, the assertion is absurd by the definition of . Thereby, , or in other words, is a solution of .

Theorem 23. *Suppose that is a real reflexive Banach space. Besides (a1)-(a2), (a4)-(a5) and (a7), if the following conditions hold:*(b8)* for each , is weakly lower semicontinuous,*(b9)* there exists such that for each , can be selected such that and , where ,**then has a solution.*

*Proof. *Denote and by the strong topology and the weak topology on , respectively, and define as
Letting , , and , we see that is just equal to (46). (b3) and (b5) are satisfied by (b8) and (b9), respectively. Also, (b4) and (b7) hold thanks to weak compactness of each closed bounded ball and weak lower semicontinuity of in a real reflexive Banach space, respectively. Besides, (b6) guarantees the weak compactness of in the proof of Theorem 22. While this property can be guaranteed by the assertions that is weakly compact and is closed by (a4) and Lemma 7(ii), this rest proof completes by the similar argument of the proof in Theorem 22.

#### 5. Several Equivalent Results of Vectorial Form of Ekelend Variational Principle

In this section, we shall present several equivalent results of the established Ekelend variational principle. Now Define as follows: Then, under the conditions (a1)–(a3), is a quasiorder on by Lemma 8(i). Also, has a maximal element under all assumptions of Theorem 10. Indeed, , where is defined as (27). Then there exists such that by Theorem 10(i), which implies that is a maximal element of .

Theorem 24. *Let (a1)–(a3) hold and equip a quasiorder defined as (52). Then the following assertions are equivalent.*(i)* (Existence result of maximal element) has a maximal element .*(ii)* (Ekelend variational principle for ) There exists such that
*(iii)* (Caristi-kirk fixe-point theorem) Let be a strict, set-valued mapping such that, for any , . Then there exists such that .*

*Proof. *(i)(iii) Let be a maximal element on . If , then there exists such that since is strict. By the assumption (iii), , which is absurd. Thus .

(iii)(ii) Let , where is defined as (27). Then for any , ; that is, the assumption of (iii) is satisfied. Consequently, for some and so (ii) holds.

(ii)(i) Define as (27). Then , and there exists such that by (ii). This implies that is a maximal element on .

Theorem 25. *Let be defined as (5) and be any given point with . Then the results below are equivalent.*(i)* (Ekelend variational principle for ) There exists such that (53) holds.*(ii)* (Existence result of solutions of ) If, for any ,
* *implies that there exists with such that
* *then has a solution .*(iii)* (Caristi-kirk type fixe-point Theorem) Let be a strict, set-valued mapping such that, for any , there exists satisfying (54). Then there exists such that .*(iv)* (Oettli-Théra type Theorem) Let . If, for every , there exists with such that (54) holds, then .*

*Proof. *(i)(iv) Let satisfy (53). If , then there exists with such that holds by the hypothesis of (iv). This is absurd by (53) and so .

(iv)(i) Let (iv) hold and define . If , then ; that is, there exists with such that (54) holds; that is, the assumption of (iv) is satisfied. Thus, . By taking , satisfies (53); that is, (i) is true.

(ii)(iv) This conclusion is argued by contradiction. Assume that . If , then and so there exists with such that (54) holds by the hypothesis of (iv). Thus the assumption (ii) is satisfied and so there exists such that
which implies that
by and . This contradicts with (54).

(iv)(ii) Let . By the assumption of (ii), for any , there exists with such that (54) holds. This is just the assumption of (iv). Choosing , we have that and .

(iii)(iv) Define as . Argue this by contradiction. If , then the assumption of (iii) holds by the assumption of (iv). Hence, has a fixed point on , which is absurd by the definition of .

(iv)(iii) Let . The assumption of (iv) is deduced straight by the assumption of (iii). Resultingly, ; that is, for some .

*Remark 26. *Suppose that is a single-valued mapping. Theorem 25 reduces Theorem 5.1 in [21] if hold instead of (a3), and if, further, (a4) hold and is -lower semicontinuous for each (namely, is closed for each and each ).

#### Acknowledgments

This research was supported by the Doctoral Fund of Innovation of Beijing University of Technology; the Science and Technology Development Project of Siping (2011005).

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