- Research Article
- Open Access

# Solvability Criteria for Some Set-Valued Inequality Systems

- Yingfan Liu
^{1}Email author

**2010**:543061

https://doi.org/10.1155/2010/543061

© Yingfan Liu. 2010

**Received:**23 May 2010**Accepted:**9 July 2010**Published:**28 July 2010

## Abstract

Arising from studying some multivalued von Neumann model, three set-valued inequality systems are introduced, and two solvability questions are considered. By constructing some auxiliary functions and studying their minimax and saddle-point properties, solvability criteria composed of necessary and sufficient conditions regarding these inequality systems are obtained.

## Keywords

- Auxiliary Function
- Convex Compact
- Duality Pairing
- Point Property
- Dimensional Euclidean Space

## 1. Introduction

Arising from considering some multivalued von Neumann model, this paper aims to study three set-valued inequality systems and try to find their solvability criteria. Before starting with this subject, we need to review some necessary backgrounds as follows.

We denote by the -dimensional Euclidean space, its dual, and the duality pairing on ; moreover, we denote that and int is its interior. We also define (or ) in by (or by int ).

and obtained several necessary and sufficient conditions for its solvability, where and is a class of set-valued maps from to . Along the way, three further set-valued inequality systems that we will study in the sequel can be stated as follows.

- (1)
For (1.2) and (1.3), it is possible that only (1.2) has solution for some . Indeed, if is compact and is continuous, compact valued with , then is compact and there is with for . Hence solves (1.2) but does not solve (1.3).

- (2)
It seems that the solvability criteria (namely, necessary and sufficient results concerning existence) to (1.4) can be obtained immediately by [8] with the replacement . However, this type of result is trivial because it depends only on the property of but not on the respective information of and . This opinion is also applicable to (1.3) and (1.5).

- (3)
Clearly, (1.3) (or (1.5)) is more fine and more useful than (1.2) (or (1.4)). However, the method used for in [8] (or the possible idea for (1.4)) to obtain solvability criteria fails to be applied to find the similar characteristic results for (1.3) (or (1.5)) because there are some examples (see Examples 3.5 and 4.4) to show that, without any additional restrictions, no necessary and sufficient conditions concerning existence for them can be obtained. This is also a main cause that the author did not consider and in [8].

So some new methods should be introduced if we want to search out the solvability criteria to (1.3)–(1.5). In the sections below, we are devoted to study (1.3)–(1.5) by considering two questions under two assumptions as follows:

Question 1.

Whether there exist any criteria that solves (1.3) in some proper way?

Question 2.

Like Question 1, whether there exist any solvability criteria to (1.4) or (1.5) that depend on the respective information of and ?

Assumption.

is a fixed point and is a convex compact subset.

Assumption.

Consider the following: and are upper semicontinuous and convex set-valued maps with nonempty convex compact values.

By constructing some functions and studying their minimax properties, some progress concerning both questions has been made. The paper is arranged as follows. We review some concepts and known results in Section 2 and prove three Theorems composed of necessary and sufficient conditions regarding the solvability of (1.3)–(1.5) in Sections 3 and 4. Then we present the conclusion in Section 5.

## 2. Terminology

Let , , and ( ). Let , , and be functions and a set-valued map. We need some concepts concerning , and and such as convex or concave and upper or lower semicontinuous (in short, u.s.c. or l.s.c.) and continuous (i.e., both u.s.c. and l.s.c.), whose definitions can be found in [9–11], therefore, the details are omitted here. We also need some further concepts to , , and as follows.

Definition 2.1.

(1) is said to be closed if its graph defined by graph is closed in . Moreover, is said to be upper semicontinuous (in short, u.s.c.) if, for each and each neighborhood of , there exists a neighborhood of such that .

- (2)
Assume that ( ), and define , ( ). Then is said to be upper hemicontinuous (in short, u.h.c.) if is u.s.c. on for any .

(3) is said to be convex if is convex and for any and .

(4)

(a)If , then one claims that the minimax equality of holds. Denoting by the value of the preceding equality, one also says that the minimax value of exists. If such that , then one calls a saddle point of . Denote by the set of all saddle points of (i.e., ), and define , the restriction of to if is nonempty.

(b)Replacing by and by , with the similar method one can also define (the minimax value of ), (the saddle-point set of ), and (the restriction of to ).

- (5)
If is a convex set and a subset of , one claims that is an extremal subset of if and for some entails . is an extremal point of if is an extremal subset of , and the set of all extremal points of is denoted by ext .

- (1)
Since and , we can see that is u.h.c. if and only if is l.s.c. on for any .

- (2)

We also need three known results as follows.

- (1)
(see [9]) If is u.s.c., then is u.h.c.

- (2)
(see [9]) If is u.s.c. with closed values, then is closed.

- (3)
(see [9]) If (the closure of ) is compact and is closed, then is u.s.c.

- (4)
If is compact and is u.s.c. with compact values, then is compact in .

- (5)
If is convex (or compact) and are convex (or u.s.c. with compact values), then are also convex (or u.s.c.) for all .

Proof.

- (b)
Now we assume that is compact.

In case is u.s.c. with compact values and , then by (2), (4), is closed and the range of is compact. If , then for any ; hence, is u.s.c. If , supposing that graph with , then graph such that as , which implies that . Hence, is closed and also u.s.c. because of (3).

In case are u.s.c. with compact values, if graph with , then and there exist , such that for all By (4), and are compact, so we can suppose and as . By (2), both are closed, this implies that , and thus is closed. Hence by (3), is u.s.c. because and is compact.

Lemma 2.4 (see [8, Theorems and ]).

Let be convex compact with , , and . Assume that is convex and u.s.c. with nonempty convex compact values, and define on by for . Then

(1) exists and is a convex compact subset of ,

(2) solves (1.2) for .

In particular, both (1) and (2) are also true if .

- (1)
(see [10, 11]) If is convex or l.s.c. (resp., concave or u.s.c.) on for and (resp., ) is finite for , then (resp., ) is also convex or l.s.c. (resp., concave or u.s.c.) on .

- (2)
(see [11]) If is l.s.c. (or u.s.c.) and is compact, then defined by (or defined by ) is also l.s.c. (or u.s.c.).

- (3)
(see [9–11], Minimax Theorem) Let , be convex compact, and let be defined on . If, for each is convex and l.s.c. and, for each is concave and u.s.c., then and there exists such that .

## 3. Solvability Theorem to (1.3)

Remark 3.1.

By both Assumptions in Section 1, Definition 2.1, and Lemmas 2.4 and 2.5, we can see that

(1) for all ,

(2) and exist, and and are nonempty,

(3) solves (1.2) if and only if if and only if with .

Hence, and are nonempty. Moreover, we have the following.

Theorem 3.2.

For (1.3), the following three statements are equivalent to each other:

(1) ,

(2)for all , for all , , ,

(3) , , , .

Remark 3.3.

Clearly, each of (2) and (3) implies that solves (1.3) because and . So we conclude from Theorem 3.2 that solves (1.3) in the way of (2) or in the way of (3) if and only if .

Proof of Theorem 3.2.

We only need to prove (1) (2) and (3) (1).

By taking , it follows that . Hence, . On the other hand, it is easy to verify that is a nonempty extremal subset of . The Crain-Milmann Theorem (see [12]) shows that ext is nonempty with ext ext . So there exists such that ext . This implies by (3.3) that , and therefore (2) follows.

where . Hence, and the theorem follows.

Remark 3.4.

From the Theorem, we know that implies that solves (1.3). However, without any additional restricting conditions, the inverse may not be true.

Example 3.5.

This implies that (or ) may not be the necessary (or the sufficient) condition that solves (1.3).

## 4. Solvability Theorems to (1.4) and (1.5)

By both Assumptions, we know that and are finite with and for and , so the functions and defined by (4.1) are well defined.

In view of Definition 2.1, we denote by (or ) the minimax value of (or ) if it exists, (or ) the saddle point set if it is nonempty, and (or ) the restriction of to (or to ). Then we have the solvability result to (1.4) and (1.5) as follows.

Theorem 4.1.

(i) exists if and only if .

(ii)

(1) solves (1.4) if and only if exists with if and only if with for .

(2)In particular, if exists with , then for each , there exists such that

Theorem 4.2.

For (1.5), the following three statements are equivalent to each other:

(1) ,

(2) , , and for all , for all , , ,

(3) , , and , , ,

That is, solves (1.5) in the way of (2) or in the way of (3) if and only if .

Remark 4.3.

It is also needed to point out that is not the necessary condition of making (1.5) solvable without any other restricting conditions.

Example 4.4.

This implies that, for each , solves (1.5) but .

The proof of both Theorems 4.1 and 4.2 can be divided into eight lemmas.

denote by the minimax value of if it exists, and denote by the saddle point set if it is nonempty. Then we have the following.

- (1)
For each , exists and is nonempty. Moreover, solves (4.3) if and only if if and only if for .

- (2)
The function is continuous and strictly decreasing on with .

- (1)
By both Assumptions and Lemmas 2.3(4) and 2.3(5), is convex and u.s.c. with nonempty convex compact values for each . Since for , applying Lemma 2.4 to and substituting for , we know that (1) is true.

- (2)
We prove (2) in three steps as follows.

So we conclude from (4.5), (4.8), and Statement (1) that is continuous on .

which implies that , and hence is strict decreasing on .

This implies that . Therefore, . This completes the proof.

Lemma 4.6.

(1) and are l.s.c. on .

(2) ) and are u.s.c. on .

(3) exists if and only if is nonempty.

- (1)
Since, for each and , the function is continuous on , from Lemma 2.5(1), we can see that and are l.s.c., hence (1) is true.

- (2)

- (3)
Assume that exists. By (1) and (2), there exist and such that

By Remark 2.2(2), . Hence is nonempty. The inverse is obvious. This completes the proof.

- (1)
If solves(1.4), then exists with .

(2)If exists with , then and for .

- (1)

then . It is needed to show that .

- (2)
follows immediately from Lemma 4.6(3) and Remark 2.2(2). The third lemma follows.

Lemma 4.8.

If with for , then solves (1.4). Moreover, for each , there exists such that

Proof.

Applying Lemma 4.5(1) to , we then conclude that solves (4.3). So there exist , , and such that . Hence solves (1.4) because , , and .

By taking , we get , and therefore satisfies because . This completes the proof.

Proof of Theorem 4.1.

By Lemmas 4.6(3), 4.7, and 4.8, we know that Theorem 4.1 is true.

We denote by the minimax values of if it exists, the saddle point set if it is nonempty, and its restriction to .

Lemma 4.9.

Let and be fixed. Then one has the following.

(1)For each exists and is nonempty.

(2) solves (4.25) if and only if if and only if implies that

(3) is continuous and strict decreasing on with

Proof.

Then is a single-valued continuous map with the convex condition defined by Definition 2.1(3) because for all and .

Hence by (1), is continuous on .

Let , , and be defined as in the proof of Lemma 4.5(2).

If , also by (4.26), we can see that for . It follows that and thus is strict decreasing on .

If , then for . This implies that with .) Hence the fifth lemma follows.

Lemma 4.10.

- (2)
If solves (4.25) for and , then exists with , is nonempty with for , and for .

Proof.

(1)By (4.29) and with the same method as in proving Lemma 4.6(3), we can show that (1) is true. (Indeed, we only need to prove the necessary part. If exists, then by (4.29), there exists such that . Hence is nonempty.)

(2)If solves (4.25) for and , then from Lemma 4.9 we know that exists with , and there is a unique such that . In particular, is the biggest number that makes (4.25) solvable for and .

Then . We need to show that .

Since solves (4.25) for and , there exist and such that . It follows that for any , hence On the other hand, by the definition of , we have for any . By (4.29)(a) and (4.1)(b), there exists such that which implies by (4.26) that for any . Hence from Lemma 4.9, solves (4.25), and . Therefore, exist with .) So we conclude from (1) and Remark 2.2 that (2) is true. This completes the proof.

Lemma 4.11.

If , then Theorem 4.2(2) is true.

By using the same method as in proving (1) (2) of Theorem 3.2, we conclude that and there exists such that . Hence Theorem 4.2(2) is true.

Lemma 4.12.

If Theorem 4.2(3) holds,then .

Proof.

If Theorem 4.2(3) holds, then solves both (1.4) and (1.5), and by Lemma 4.7(1), exists with .

Hence, .

Proof of Theorem 4.2.

Since (2) (3) of is clear, Theorem 4.2 follows immediately from Lemmas 4.11 and 4.12.

## 5. Conclusion

Based on the generalized and multivalued input-output inequality models, in this paper we have considered three types of set-valued inequality systems (namely, (1.3)–(1.5)) and two corresponding solvability questions. By constructing some auxiliary functions and studying their minimax and saddle point properties with the nonlinear analysis approaches, three solvability theorems (i.e., Theorems 3.2, 4.1, and 4.2) composed of necessary and sufficient conditions regarding these inequality systems have been obtained.

## Authors’ Affiliations

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