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# Saddle Point Problems, Bilevel Problems, and Mathematical Program with Equilibrium Constraint on Complete Metric Spaces

## Abstract

We apply an existence theorem of variational inclusion problem on metric spaces to study optimization problems, set-valued vector saddle point problems, bilevel problems, and mathematical programs with equilibrium constraint on metric spaces. We study these problems without any convexity and compactness assumptions. Our results are different from any existence results of these types of problems in topological vector spaces.

## 1. Introduction

Let and be two metric spaces, let be a real Hausdorff topological vector space ordered by a nonempty pointed closed convex in with nonempty interior, and let be a real Banach space ordered by a nonempty pointed closed convex with nonempty interior. Let , , and be multivalued maps. Throughout this paper, we use these notations unless specified otherwise. In this paper, the following vector mathematical programs with equilibrium constraint on metric spaces are considered.

(MPEC-1) is subject to , and for all .

(MPEC-2) is subject to , and for all .

If , , and is a real function, then problems (MPEC-1) and (MPEC-2) are reduced to the following problem:

(MPEC-3) is subject to , and for all .

We also study the following bilevel problems.

(BL-1) is subject to , for , is a solution of problem : (i.e., for , ), for , and is a solution of problem : . (i.e., for , ).

(BL-2) is subject to , for all , and for all

(BL-3) is subject to , for , is a solution of problem : and (i.e., for , ), for , is a solution of problem : . (i.e., for , ).

If is a single valued function, then (BL-2) is reduced to the following problem.

(BL-4) is subject to , for , and is a solution of problem and : ; and for , is a solution of problem : .

Problem (BL-1) has applications in real world. Let be the set of government's agricultural policies and let be the set of government's industrial policies. For each and , let be the amount of money that the government uses to promote agriculture and industrial developments, and let be the degree of industrial development. We suppose that as the industry becomes more and more developed, the losses from the agriculture sector rise accordingly. Therefore, the solution of problem (BL-1) represents the government's best policy to promote the development of the industry so that the losses from agriculture sector will be as minimal as possible, while the amount of money that the government uses to promote the policies can be the lowest possible.

Mathematical program with equilibrium constraint and bilevel problem represent two important classes of optimization problems which have been investigated in a large number of articles and books. We find in the literatures that Luo et al. [1], Stein and Still [2], Stein [3], Birbil et al. [4], Liou et al. [5], Lin and Still [6],Lin [7], as well asLin and Hsu [8] have studied mathematical program with equilibrium constraint and bilevel problem on topological vector spaces. As usual in linear and nonlinear optimization, these studies mainly deal with optimality conditions and numerical methods to solve these problems and typically the existence of feasible points is tacitly assumed. Besides, the domains of the functions they consider are subsets of topological vector spaces and certain convexity assumptions on the functions are needed. In this paper, we study these problems with functions defined on metric spaces, so we do not need any convexity assumptions on the functions we consider. We study the existence theorems of solutions for (MPEC-1), (MPEC-2), (BL-1), (BL-2), and (BL-3). To the best of our knowledge, there is no result of these types of problems on metric space.

In this paper, we also study the following loose set-valued vector saddle point problems.

(LSP-1)For each , find such that , , and .

(LSP-2)For each , find such that , and for all and all .

If is a map, then problems (LSP-1) and (LSP-2) are reduced to the following vector saddle point problems (VSP).

(i)Find such that .

There are many results on loose saddle point problems, vector saddle point, and saddle point problems (see, e.g., [9â€“17]). But to the best of our knowledge, there are no existence theorems in the literatures for loose saddle point and vector saddle point problems for functions defined on the product of metric spaces. We study the loose saddle point problems and vector saddle point problems for functions defined on the product of metric spaces. We do not assume any compact assumptions on the spaces and convexity assumptions on the maps we consider. As for applications of our existence theorems on saddle point problems, we study bilevel problems on metric spaces. We also study mathematical programs with equilibrium constraint. Our results on mathematical programs with equilibrium constraint, bilevel problems, loose saddle point problems, and vector saddle point problems are different from any existence results of these types of problems in the literatures.

## 2. Preliminaries

Let and be topological spaces (in short t.s.), be a multivalued map. is said to be u.s.c. (resp., l.s.c.) at if for every open set in with (resp., ), there exists an open neighborhood of such that (resp., ) for all ; is said to be u.s.c. (resp., l.s.c.) on if is u.s.c. (resp., l.s.c.) at every point of ; is continuous at if is both u.s.c. and l.s.c. at ; is said to be closed if = is a closed set in ; is said to be open if is an open set in . For a subset of topological space , let denote the closure of .

Lemma (see [18]).

Let and be topological spaces, and let be a multivalued map. Then is l.s.c. at if and only if for any and any net in converges to , there exists a net such that for all and .

Lemma (see [19]).

Let and be Hausdorff topological spaces, and let be a multivalued map. (i) If is an u.s.c. multivalued map with nonempty closed values, then is closed; (ii) if is a compact set and is an u.s.c. multivalued map with nonempty compact values, then is compact.

Definition.

Let be a nonempty subset of a t.v.s. ordered by a nonempty pointed closed convex . An element is said to be a minimal (resp., maximal) point of if (resp., ). Here, and denote the sets of minimal point of and maximal point of , respectively.

Definition.

Let be a nonempty subset of a t.v.s. ordered by a nonempty pointed closed convex with nonempty interior. An element is said to be a weakly minimal (resp., weakly maximal) point of if (resp., ). Here, and denote the sets of weakly minimal point of and weakly maximal point of , respectively.

Theorem (see [20]).

Let be a Hausdorff t.v.s. ordered by a nonempty pointed closed convex with nonempty interior. If is a nonempty compact subset of , then and .

Definition (see [21]).

Let be pointed closed convex with nonempty interior in a Banach space . Then is called normal if there exists such that if and , then . Here, it is called -normal pointed closed convex .

Theorem (see [22]).

Let be a complete metric space, and let be a Hausdorff t.v.s.. Let be a multivalued map. Assume that

(i) for each

(ii)for each , is a closed subset of

(iii)for each , if and , then ,

(iv)for each sequence in , if , then as .

Then there exists such that for all .

Theorem.

Let be a complete metric space. Let be a multivalued map with nonempty values. Assume that

(i) is closed,

(ii)for each sequence in , if and for all , then and as .

Then there exists such that .

Proof.

Let be fixed. Clearly, is a complete metric space. Let be defined by for each . Clearly, for each . For each , we know that

(2.1)

If is a sequence in and as , then and for all . Clearly, and . By (ii), condition (iv) of Theorem 2.7 is satisfied. By Theorem 2.7, there exists such that for all . Clearly, . If and , then there exists such that and . Then and this implies that . This leads to a contradiction. Hence, . This implies that .

Example.

Let , , , and . Let be defined by for each . Then by Theorem 2.8, there exists such that .

Example.

Let , , and . Let be defined by for each . Clearly, is closed. But condition (ii) of Theorem 2.8 does not hold. Indeed, let . Then for all , and and for all . Furthermore, there is no such that . Hence, condition (ii) of Theorem 2.8 is essential in Theorem 2.8.

Theorem.

Let be a metric space. Let be a multivalued map with nonempty values. Assume that

(i) is a nonempty closed subset of

(ii)for each sequence in , if , then as .

Then there exists such that .

Proof.

Let be fixed. Clearly, is a complete metric space. Let be defined by for each . Clearly, for each . Besides, for each , let . Then is a closed set. Next, for each , if and , then . By (ii), for each sequence in with , we have as . By Theorem 2.7, there exists such that .

Example.

Let , , , and . Let be defined by

(2.2)

Thus by Theorem 2.11, there exists such that . Note that is not closed, not open, and not convex.

Example.

Let , , and . Let be defined by

(2.3)

By Theorem 2.11, there exists such that . Indeed, .

Remark.

Example 2.10 also shows that condition (ii) of Theorem 2.11 is essential in Theorem 2.11. Next, the following result is a special case of Theorem 2.11. Note that it is different from Theorem 2.5 since we do not assume that is a compact set, but we assume that is a Banach space.

Corollary.

Let be a nonempty closed subset of . Suppose that for each sequence in if , then as . Then .

Proof.

Let be a singleton subset of . Then is a complete metric space. Let be defined by for each . By Theorem 2.11, there exists such that and this implies that .

Example.

Let and . It is easy to see that all conditions of Corollary 2.15 are satisfied. Hence, . Indeed, . Note that is not a compact subset of .

Corollary.

Let be a Banach space, and let be a -normal pointed closed convex with nonempty interior . If is a nonempty bounded closed subset of , then .

Proof.

Take any sequence in with . This implies that By assumption, we get

(2.4)

Since is bounded, there exists such that This implies that as and Corollary 2.17 follows from Corollary 2.15.

Remark.

Theorem 2.11 and Corollary 2.15 are equivalent.

Remark.

From the above results and examples, we observe that if is a closed set and there exists such that , then . Indeed, for each sequence in with , it is easy to see that, for each , is a decreasing sequence in and bounded from below. Then as and .

Theorem.

Let and be two complete metric spaces. Assume that

(i)for each , is a closed set,

(ii)for each , if and , then ,

(iii)for each sequence in , if , then and as .

Then, for each , there exists such that , , and .

Proof.

Let . Then is a complete metric space. Take any . Define the set Clearly, is a complete metric space. Let be defined by . By Theorem 2.7, there exists such that for all . By (ii) and the definition of , it is easy to see that for all . Hence,

(3.1)

Since is a nonempty compact set, by Theorem 2.5, there exist such that and . Next, it is easy to see that and . Therefore, and .

Remark.

If is an u.s.c. multivalued map with nonempty compact values, then condition (i) of Theorem 3.1 holds.

Example.

Let , , , and . Let be defined by for each . By Theorem 3.1, for each , there exists such that , , and . Indeed, for each , .

In Theorem 3.1, if is singleton, then we have the following result.

Corollary.

Let be a multivalued map with nonempty compact values. Assume that

(i)for each , is a closed set,

(ii)for each , if and , then ,

(iii)for each sequence in , if , then as .

Then, for each , there exists such that and .

Theorem.

Assume that

(i)for each , is closed,

(ii)for each , if and , then ,

(iii)for each sequence in , if , then and as .

Then, for each , there exists such that , and for all .

Proof.

Let . Take any . Define the set . Clearly, is a complete metric space. Let be defined by . For each , by (i), is a closed set. Then by Theorem 2.7, there exists such that for all .

Now, we want to show that for all . In fact, we only need to consider that . Suppose that . By (ii) and , and this is a contradiction. Therefore, for all .

Remark.

Condition (i) of Theorem 3.5 can be replaced by being l.s.c. and being an u.s.c. multivalued map with nonempty compact values.

Example.

Let , , , and , and let be defined by

(3.2)

This is an example for Theorem 3.5.

Example.

Let , , , and . Let be defined by for each . By Theorem 3.5, for each , there exists such that , and for all . Indeed, for each .

## 4. Bilevel Problems

Proposition.

Let be defined by

(4.1)

If is a continuous multivalued map with nonempty compact values, then is closed, and is a closed set.

Proof.

Here, we only need to show that is closed. If and , for each , we have then

(1),

(2),

(3).

By (1), for each , there exists and such that . Let , , , and . Then , , , and are compact sets. By Lemma 2.2, and are compact sets. Hence, we may assume that and . By Lemma 2.2, and . Clearly, and . By (2), for each , there exists . By Lemma 2.2, is a compact set. Hence, we may assume that . By Lemma 2.2, .

For each , Take any . Then there exists such that . There exists a net in such that , Therefore, for all . That is, . Hence, . Similarly, we can prove that . So, and is closed.

Lemma.

Let and be topological spaces, and let be a multivalued map. Let be defined by for each . If is an u.s.c. multivalued map with nonempty compact values and is a nonempty closed subset of , then is a closed subset of .

Proof.

If , then there exist a net such that and a net in such that for all . Hence, for all . Let . Then is a compact set and is a compact set. We may assume that . Since is a closed set, . By Lemma 2.2, and . Therefore, is a closed subset of .

Theorem.

In Theorem 3.1, let be a multivalued map with nonempty values, and further assume that

(a) is an u.s.c. multivalued map with nonempty compact values,

(b) is a continuous multivalued map with nonempty compact values,

(c)for each sequence in , if , then as .

Then there is a solution of problem (BL-1).

Proof.

Let and be defined as in Proposition 4.1. By Theorem 3.1 and Proposition 4.1, is a nonempty closed subset of . By Lemma 4.2, is a nonempty closed subset of . By Theorem 2.11, there exists such that .

Example.

In Example 3.3, let , , and for each . Clearly, is a closed subset of . Besides, we have

(4.2)

and is an u.s.c. multivalued map with nonempty compact values. By Theorem 4.3, there is a solution of problem (BL-1). Indeed, the solution set is .

The following theorem is similar to Theorem 4.3. Note that the conditions of Theorems 4.3 and 4.5 are different.

Theorem.

In Theorem 3.1, let be a multivalued map, and further assume that

(a) is an u.s.c. multivalued map with nonempty compact values,

(b)for each , if and , then ,

(c) is a continuous multivalued map with nonempty compact values,

(d)for each in , if , then as .

Then there is a solution of problem (BL-1).

Proof.

Let and be defined as in Proposition 4.1. By Theorem 3.1 and Proposition 4.1, is a nonempty closed subset of . Hence, is a complete metric space. Now, for each , let . If , then there exists a net in such that . Then for each , there exists . Since is a compact set, we may assume that . There exists such that . Let . Clearly, and are compact sets. Hence, we may assume that . By Lemma 2.2, . Clearly, and this implies that . Therefore, and is a closed set. By Corollary 3.4, there exists such that .

Furthermore, we have the following result which is different from Theorems 4.3 and 4.5. In Theorem 4.6, is a Hausdorff t.v.s., and and are compact metric spaces. In Theorems 4.3 and 4.5, is a Banach space, and and are complete metric spaces.

Theorem.

In Theorem 3.1, let be a multivalued map, and further assume that

(a) is an u.s.c. multivalued map with nonempty compact values,

(b) is a continuous multivalued map with nonempty compact values,

(c) and are compact.

Then there is a solution of problem (BL-1).

Proposition.

Let be defined by

(4.3)

If is a continuous multivalued map with nonempty compact values, then is closed, and is a closed set.

Proof.

Here, we only need to show that is closed. If and , then we have

(1),

(2) for all ,

(3) for all .

Take any ; there exists a net such that for all and . There exists such that . Let , , , and . Then , , , and are compact sets and is a compact set. Hence, we may assume that . By Lemma 2.2, . Clearly, . Hence, , and .

Take any there exists such that . is a compact set, and we may assume that ; By Lemma 2.2, . Take any , there exists a net such that for all and , for all . Clearly, for all . Hence, for all . Similarly, for all . Therefore, is closed.

Applying Proposition 4.7 and following the similar argument as in the proof of Theorems 4.3â€“4.6, we can get the following similar results.

Theorem.

In Theorem 3.5, let be a multivalued map with nonempty values. Further assume that conditions (a)â€“(c) of Theorem 4.3 (resp., conditions (a)â€“(d) of Theorem 4.5) are satisfied. Then there is a solution of problem (BL-2).

Proposition.

Let be a map with . Let be defined by

(4.4)

Suppose that, for each , is continuous, is one to one, and is one to one. Then is closed, and is a closed set.

Proof.

Let and as . Then , for all , and for all . Since is continuous, . Furthermore, for each , we have

(4.5)

Take any . There exists such that for all . Furthermore, for each with , there exists such that Indeed, if not, there exists with such that for all . Since and is one to one, for all . Hence, . This leads to a contradiction.

Therefore, there exist and , of and , respectively, such that Suppose that Then , Hence, . This leads to a contradiction since is one to one. Therefore, Similarly, we have Therefore, is closed.

Theorem.

Let be a map with , . Let be a multivalued map with nonempty values. Further assume that

(i)for each , is continuous, is one to one, and is one to one,

(ii)for each , if and , then ,

(iii)for each sequence in with , and as ,

(iv) is an u.s.c. multivalued map with nonempty compact values,

(v)for each sequence in , if , then as .

Then there is a solution of problem (BL-3).

Proof.

Applying Proposition 4.9 and following the similar argument as in the proof of Theorem 4.3, we can get the proof of Theorem 4.10.

Remark.

The conditions of Theorem 4.10 and Theorem 3.5 in Liou et al. [5] are different. Note that Liou et al. [5] assumed that the feasible set is nonempty, and let the considered multivalued map be proper, lower semicontinuous, and weakly coercive on .

## 5. Equilibrium Problems and Mathematical Program with Equilibrium Constraint on Complete Metric Spaces

Theorem.

Let and be a complete metric spaces, and let be a multivalued map. Assume that

(i)for each , ,

(ii)for each , is a closed subset of ,

(iii)for each and , if and , then ,

(iv)for each sequence in with , and as .

Then there exists such that for all .

Proof.

Let be defined by for each . Then Theorem 5.1 follows from Theorem 2.7.

Theorem.

Let and be a complete metric spaces, and let be a multivalued map. Assume that

(i) is a closed set; and for each , ,

(ii)for each , is a closed subset of ,

(iii)for each and , if and , then ,

(iv)for each sequence in with , and as ,

(v) is closed; and for each , is l.s.c.,

(vi)for each sequence in with , and as .

Then there is a solution of problem (MPEC-1).

Proof.

By Theorem 5.1, there exists such that for all . Since , for all .

Let for all . Clearly, . If , then there exists a net in such that . Then, for each , for all . That is, Take any and any ; there exists a net such that for all and . Clearly, . Hence, for all . Then is a closed subset of a complete metric space . Furthermore, is a complete metric space. Let . Then is a closed subset of . By Theorem 2.8, there is a solution of problem (MPEC-1).

Theorem.

Let and be a complete metric spaces, and let be a multivalued map. Assume that

(i)for each , ,

(ii)for each , is a closed subset of ,

(iii)for each and , if and , then ,

(iv)for each sequence in with , and as .,

(v) is a closed set,

(vi)for each , is an u.s.c. multivalued map with nonempty compact values,

(vii) is closed,

(viii)for each sequence in with , and as .

Then there is a solution of problem (MPEC-2).

Proof.

Let be defined by for each . By Theorem 5.1 and condition (i), there exists such that for all .

Let for all . If , then there exists a sequence in such that . Let . Then is a compact set. For each , since , For each and , there exists such that and . Then . By (vi), we may assume that . Since is a closed set, . Hence, and is closed. Following the similar argument as in the last part of the proof of Theorem 5.2, we get the proof of Theorem 5.3.

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Lin, LJ., Chuang, CS. Saddle Point Problems, Bilevel Problems, and Mathematical Program with Equilibrium Constraint on Complete Metric Spaces. J Inequal Appl 2010, 306403 (2010). https://doi.org/10.1155/2010/306403