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Solvability Criteria for Some SetValued Inequality Systems
Journal of Inequalities and Applicationsvolume 2010, Article number: 543061 (2010)
Abstract
Arising from studying some multivalued von Neumann model, three setvalued inequality systems are introduced, and two solvability questions are considered. By constructing some auxiliary functions and studying their minimax and saddlepoint properties, solvability criteria composed of necessary and sufficient conditions regarding these inequality systems are obtained.
1. Introduction
Arising from considering some multivalued von Neumann model, this paper aims to study three setvalued 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 ).
It is known that the generalized (linear or nonlinear) von Neumann model, which is composed of an inequality system and a growth factor problem described by
is one of the most important issues in the inputoutput analysis [1–3], where , ( may not be equal to ), and are two nonnegative or positive maps from to . A series of researches on (1.1) have been made by the authors of [1–5] for the linear case (i.e., are matrices) and by the authors of [6, 7] for the nonlinear case (i.e., are some types of nonlinear maps). Since (a) or (b) of (1.1) is precisely a special example of the inequality if we restrict or , it is enough for (1.1) to consider the inequality system. This idea can be extended to the setvalued version. Indeed, if and are replaced by setvalued maps and , respectively, then (1.1) yields a class of multivalued von Neumann model, and it solves a proper setvalued inequality system to study. With this idea, by [8] (as a setvalued extension to [6, 7]) we have considered the following multivalued inequality system:
and obtained several necessary and sufficient conditions for its solvability, where and is a class of setvalued maps from to . Along the way, three further setvalued inequality systems that we will study in the sequel can be stated as follows.
Let , be as above, and let be setvalued maps from to , then we try to find the solvability criteria (i.e., the necessary and sufficient conditions) that solves
When and are singlevalued maps, then (1.3)–(1.5) return to the models of [6, 7]. When and , are setvalued maps, there are three troubles if we try to obtain some meaningful solvability criteria regarding (1.3)–(1.5) just like what we did in [8].

(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 setvalued 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 setvalued 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 saddlepoint 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 .
Remark 2.2.

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

(2)
For the function on , exists if and only if , and if and only if if and only if for any . If , then exists, and for any . The same properties are also true for on . Moreover, we have
(2.1)
We also need three known results as follows.
Lemma 2.3.

(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.
We only need to prove (5).

(a)
If are convex, , , and , then
(2.2)
Hence is convex.

(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 .
Lemma 2.5.

(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)
Let be introduced as in Lemma 2.4, and define the functions on and on by
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).
(1)(2). Assume that (1) holds. By (3.1) and Remarks 2.2 and 3.1, it is easy to see that
Then for each and each , we have
By taking , it follows that . Hence, . On the other hand, it is easy to verify that is a nonempty extremal subset of . The CrainMilmann 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.
(3)(1). Let , , and let be presented in (3). Since solves (1.3) and also solves (1.2), by (3.1) and Remarks 2.2 and 3.1, we obtain
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.
Let , and let be defined by for Then is an u.s.c. and convex setvalued map with convex compact values, and for each , and , . Hence, for all and therefore,
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)
Taking , then from both Assumptions, Lemma 2.4, and Theorem 3.2, we immediately obtain the necessary and sufficient conditions to the solvability of (1.4) and (1.5). However, just as indicated in Section 1, this type of result is only concerned with . To get some further solvability criteria to (1.4) and (1.5) depending on the respective information of and , we define the functions on and on by
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.
Let , , and be defined by , for Then both and are u.s.c. convex setvalued maps with convex compact values, and for any , , and , we have , , , and Therefore,
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.
Let , , and . Consider the auxiliary inequality system
Then solves (4.3) if and only if solves (1.2) for , and in particular, solves (4.3) if and only if solves (1.4). Define on by
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.
Lemma 4.5.

(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 .
Proof.

(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.
(a)By Lemma 2.3(1), and are u.h.c., which implies by Definition 2.1(2) and Remark 2.2(1) that for each , is u.s.c. on . Then from Lemma 2.5(1)(2), we know that both functions
Since and are compact by both Assumptions and Lemma 2.3(4), and are finite. Then for any , and , we have
This implies that, for each ,
Hence for each , is continuous on . Also from Lemmas 2.5(1) and 2.5(2), it follows that both functions
So we conclude from (4.5), (4.8), and Statement (1) that is continuous on .
(b)Assume that . Since int is compact, it is easy to see that . Thus for any , we have
which implies that , and hence is strict decreasing on .
(c)Let and . By both Assumptions, and are finite. Thus for any and , we have
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.
Proof.

(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)
Assume that is a sequence with , then for each , there exist and such that , . Since are compact and , , we may choose and such that
(4.11)
By Lemma 2.3(2), both and are closed. Hence, and , which in turn imply that and
Combining this with for , it follows that
Hence by (4.1), is u.s.c. on , so is on thanks to Lemma 2.5(1).

(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.
Lemma 4.7.

(1)
If solves(1.4), then exists with .
(2)If exists with , then and for .
Proof.

(1)
If solves (1.4), then solves (4.3). From Lemma 4.5, we know that , and there is a unique such that . Moreover, also from Lemma 4.5, is the biggest number that makes (4.3) solvable, and thus solves (4.3) if and only if . We will prove that exists with . Let
(4.15)
then . It is needed to show that .
Since solves (4.3), there exist , , and such that . Hence for each As for , it follows from (4.1) that and thus
On the other hand, by (4.15), for each we have . By (4.1) and Lemma 4.6(2), there exists such that
It deduces from (4.4) that for each
Hence by Lemma 4.5(1), , and solves (4.3). Since is the biggest number that makes (4.3) solvable, we have . Combining this with (4.16), we obtain . Therefore, exists and

(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.
By (4.1) and Remark 2.2(2), we know that, for each ,
Combining this with the definition of (i.e., (4.4) for ), it follows that, for each and each ,
Hence by Definition 2.1(4) and Remark 2.2(2),
It follows that implies that with and so
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 .
For each . Since by (4.4), applying Lemma 2.5(3) to the function on and associating with (4.22), we obtain
Since is u.s.c. on , from (4.23) there exist , such that satisfies
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.
To prove Theorem 4.2, besides using Lemmas 4.5–4.8, for and , we also need to study the condition that solves
Define on by
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.
Define from to by
Then is a singlevalued continuous map with the convex condition defined by Definition 2.1(3) because for all and .
Since is convex and compact in , replacing by , by , and by , from Lemma 2.4, we know that both (1) and (2) are true. Moreover, with the same method as in proving Lemma 4.5(2), we can show that (3) is also true. (In fact, since is continuous on and and are compact, by (4.26) and Lemmas 2.5(1) and 2.5(2), we can see that
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.
(1) exists if and only if is nonempty, where is defined by (4.1)(b).

(2)
If solves (4.25) for and , then exists with , is nonempty with for , and for .
Proof.
Since is continuous on , by Lemma 2.5(1), it is easy to see that
(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 .
Applying the same method as in proving Lemma 4.7(1), we can show that exists with . (In fact, let
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.
Proof.

(i)
If , then by Lemma 4.6(3) and Remark 2.2, and
(4.31)
By the same proof of (4.21) we can show that
Combining this with Lemma 4.5(1) and using Remark 2.2(2), we have
As by (4.4), applying Lemma 2.5(3) to the function on , we obtain that, for each ,
Since is u.s.c. on , from (4.34) there exists such that
Hence, This implies that solves (4.25) for and any . So we conclude from Lemma 4.10 and Remark 2.2(2) that
On the other hand, by (4.1), . Combining this with (4.36), it follows that, for each and each ,
Hence, also by (4.1), . This implies that, for each and there exists such that So we obtain
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 .
Now we let , , and satisfy and , then we have for and (where This implies by (4.1) that
Combining this with the fact that and using Remark 2.2 and (4.1), we obtain that
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 inputoutput inequality models, in this paper we have considered three types of setvalued 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.
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Keywords
 Auxiliary Function
 Convex Compact
 Duality Pairing
 Point Property
 Dimensional Euclidean Space