`Mathematical Problems in EngineeringVolume 2010, Article ID 572451, 13 pageshttp://dx.doi.org/10.1155/2010/572451`
Research Article

## A Leontief-Type Input-Output Inclusion

1College of Science, Nanjing University of Posts and Telecommunications, Nanjing 210003, China
2Department of Mathematics and Computer Science, Northern Michigan University, Marquette,MI 49855, USA

Received 13 January 2010; Accepted 10 April 2010

Copyright © 2010 Yingfan Liu and Qinghong Zhang. 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

A Leontief-type input-output inclusion problem based on a set-valued consuming map is studied. By applying nonlinear analysis approach, in particular using the surjection and continuity technique with respect to set-valued maps, solvability and stability results with and without continuity assumption concerning this inclusion are obtained.

#### 1. Introduction

In this paper, we study the solvability and stability of the following input-output inclusion: where is the set of nonnegative vectors of the -dimensional Euclidean space ,

The present study is essentially a continuation of the investigation initiated in [15] where the classical Leontief input-output model was briefly reviewed, several generalized Leontief input-output models were introduced, numerous key references were cited, and some arguments about the assumptions on and were made. For the necessary background material and preliminaries, the reader is referred to [15]. Here we will make use of the Rogalski-Cornet Theorem in [6] and the Rogalski-Cornet-type Theorem proved in [5] to prove several solvability and stability theorems with and without continuous conditions concerning for model (1.1). Obviously, for some , inclusion (1.1) may not have solutions. Specifically, if makes (1.1) solvable, then While gives us an expression for all possible for which (1.1) has solutions, it is required that all the information regarding is available. It is our intention in this paper to discover some conditions under which (1.1) has solutions for the situation that the information of is only available near the boundary of . We also provide a stability analysis for the solution set in terms of closeness, upper semicontinuity and, upper hemicontinuity of certain related set-valued maps.

The paper is organized as follows. In the rest of this section, we review some necessary concepts and several useful results, which are used throughout this paper. In Section 2, we study (1.1) under the assumption that is upper semicontinuous. In Section 3, we recall a Rogalski-Cornet-type theorem appearing in [5], and use it to obtain three solvability and stability results. We give our concluding remarks in Section 4.

In the sequel, we use several classes of maps, including upper and lower semicontinuous (in short, u.s.c. and l.s.c.), upper hemicontinuous (in short, u.h.c.), continuous, and closed set-valued maps between Hausdorff topological (or Hausdorff locally convex) spaces, whose definitions and some other related concepts are given below and can also be found in [69].

Definition 1.1. Let and be two Hausdorff topological spaces and a set-valued map from to . The domain of is the set denoted by , and the graph of is the set denoted by .(1)We say that is strict if , and is closed if is closed in .(2)We say that is u.s.c. at if for any neighborhood of , there exists a neighborhood of such that . is said to be u.s.c. if is u.s.c. at every point .(3)We say that is l.s.c. at if for any and any neighborhood of , there exists a neighborhood of such that for any . is said to be l.s.c. if is l.s.c. at every point .(4)If is a Hausdorff locally convex vector space, is its dual and is the duality paring on . We say that is u.h.c. at if for any , the function is upper semicontinuous (in short, u.s.c.) at . is said to be u.h.c. if it is u.h.c. at every point of .

We now recall a number of auxiliary results that will be needed in proving our main theorems. They are stated below as lemmas.

Lemma 1.2 (see [8]). Let be an u.s.c. set-valued map from a Hausdorff topological space to a Hausdorff topological space with closed values. Then is closed.

Lemma 1.3 (see [8]). Let be a closed set-valued map from a Hausdorff topological space to a compact Hausdorff topological space . Then is u.s.c..

Lemma 1.4 (see [8]). Suppose that is a Hausdorff topological space, a Hausdorff locally convex vector space equipped with the weak topology , and a set-valued map that is u.s.c. at . Then is u.h.c. at .

Remark 1.5. If , then the weak topology on coincides with the norm topology.

Lemma 1.6 (see [8]). Suppose that is a complete metric space, and a compact metric space. If is a closed and strict set-valued map, then the subset of points at which is continuous is residual, that is, the interior of the discontinuous point set of is empty.

In order to use the Rogalski-Cornet Theorem and Rogalski-Cornet type Theorem to discuss the solvability and stability of (1.1), we need some further concepts.

Let be a Hausdorff locally convex vector space ( its dual, the duality paring on ), a subset of , int the interior of , the boundary of , and a set-valued map from to . Let . The normal cone to at , the supporting set of , and the upper and lower supporting functions and on are defined by We say that

With the help of these concepts, the Rogalski-Cornet theorem can be stated as follows.

Theorem 1.7 (see [6] Rogalski-Cornet). Suppose that is a convex compact subset of supplied with the weak topology, and is an u.h.c. set-valued map from to with nonempty closed convex values. If is either outward or strongly inward, then for any , there exists a solution to , that is, .

We use the following notations throughout this paper:

#### 2. Theorems With u.h.c. Condition

In this section, we assume that of (1.1) is u.s.c. on .

Associating this assumption with (1.2), we can show that and are compact. Therefore, is a strict and u.s.c. set-valued map from (a convex compact subset of ) to (a compact subset of ) with convex compact values, and defined by (1.6) is finite. Moreover, by Lemmas 1.2 and 1.4 and Remark 1.5, is also closed and u.h.c. on . Suppose that is a set-valued map from to defined by Then we have the following results.

Theorem 2.1. If , then is compact and is closed, u.s.c., and u.h.c., and the subset of points at which is continuous is residual.

Proof. Since and are compact, we know that is bounded, so is . Suppose that and as . Then there exist such that . Since has a convergent subsequence, we may assume that as . As is closed, we then obtain , which shows that is closed, and hence, also compact.
For the continuity results of , according to Lemmas 1.31.6 and Remark 1.5, we only need to prove that is closed because and are compact. Suppose that such that as . Then . Since is closed, and , we get . Hence, is closed and has all the continuity results stated in the theorem.

Next, we use Theorem 1.7 to obtain two solvability and stability results by means of the interior and exterior approximation methods used in [2] (three approximation methods have been used to study the single-valued input-output equation in [2]). Besides the assumptions that is convex compact and is strict and u.s.c. with convex compact values, we further assume that in this section. We have the following lemma. The proof is straightforward and hence omitted.

Lemma 2.2. Let be a nonempty convex set, , and the boundary of . If and (or ), then .

##### 2.1. Interior Approximation Method

Define a subset of and a set-valued map from to by Then we have the first solvability and stability result for (1.1) as follows.

Theorem 2.3. Suppose and . Then there exist , such that and . Moreover, defined by (2.2) is closed, u.s.c., and u.h.c., and the subset of points at which is continuous is residual.

Proof. Since is compact for and is u.s.c., it is easy to see that for , is an open neighborhood of , and that for each and , there exists a neighborhood of with . As is compact and , there exist such that where . Let . Then it is a closed subset of with . Let . Then we have . Otherwise, there exist sequences and , such that as . Since is compact, so are and , which imply that there exist convergent subsequences of and . Without loss of generality, we may assume that and are convergent to the same point as . Since and are closed, we obtain , a contradiction. Therefore, . Let . Then it is easy to see that is an open subset of , and
For each , by the truncation technique of generalized functions in partial differential equations [10], there is a continuous function from to such that , int Here denotes the closure of . Define by
We claim that for each , is a closed, u.s.c., and u.h.c. set-valued map with nonempty convex compact values. Indeed, is clearly a convex compact subset of for each . Assume that (i.e., ) such that as . Then , and for each there exists with . Since is compact, we may suppose that as . By letting and using the fact that is closed, we get and , that is, . Hence, is closed. On the other hand, by (2.5), it is easy to see that is compact. Hence by Lemmas 1.3 and 1.4 and Remark 1.5, is also u.s.c. and u.h.c..
Since for and , it follows that for each , the function is u.s.c. on . And so, by Definition 1.1, is u.h.c.
Assume that and . If , then . If , then by Lemma 2.2, , which further implies by that and hence . Therefore, satisfies the outward condition stated in (1.5). In view of Theorem 1.7, we have . Hence for each , there exists such that If , by (2.5) and , we have . Set , then we get If , by (2.5) and (2.6) there exists such that . Let . Then and . By virtue of (2.4), (2.5), and (1.6), we see that (2.7) is also true.
Now, we obtain two sequences and with . Since is compact and is bounded, we may assume that and as . Combining this with (2.7) and also using the fact that is closed, we obtain that
Next, we prove the second part of the theorem. Since and are compact, also by Lemmas 1.31.6 and Remark 1.5, we only need to prove that is closed. Assume that with . Then for each there exists such that We may suppose that as . This implies by (2.8) and the closeness of that and . Hence, is closed and has all the continuity properties stated in the theorem. This completes the proof.

Remark 2.4. In the proof of the theorem, the condition is not used. We impose this requirement in order to make sure that .

##### 2.2. Exterior Approximation Method

Define a function on and a set-valued map from to by where is defined as in (2.2). Then we have the following result.

Theorem 2.5. Suppose that and . Then there exist , such that , and . Furthermore, defined by (2.9) is closed, u.s.c. and u.h.c., and the subset of points at which is continuous is residual.

Proof. Let for . For each , let and let be the projection from to , that is, such that for . It is easy to see that is a convex compact set with nonempty interior, if and if , and for all .
As in the proof of Theorem 2.3, we assume that is a continuous function from to with compact support set such that Let be a set-valued map from to defined by
Utilizing the similar method as in the proof of Theorem 2.3, we can show that is a closed, u.s.c., and u.h.c. set-valued map with nonempty convex compact values and satisfies In fact, if with , then , , and for each , there exists such that . Since and is compact, we may suppose that . By letting , from the closeness of , Lemmas 1.3 and 1.4, and Remark 1.5, we conclude that , , that is, , and thus is closed, u.s.c. and u.h.c..
Also as in the proof of Theorem 2.3, we can prove that satisfies the outward condition stated in (1.5). (In fact, assume that and . If , then . If , then by Lemma 2.2, , which implies by (2.12) that , and hence .)By Theorem 1.7,we obtain Combining this with (2.11), we know that for each , there exists such that
Set . Then . If there is such that , then from (2.10), (2.12), and (2.14) we have and Set and . Then If for all , , then and . By (2.14), there exists such that . Let . Then we obtain
Since is compact and is bounded, we may assume that and as . This implies by that . On the other hand, if and is a -neighborhood of , then there exists a neighborhood of such that because is u.s.c., and thus for any , . Hence the function By letting and using the closeness of , from (2.16) and (2.17), we get Combining this with (2.15), we conclude that the first part of the theorem is true.
As in the proof of Theorem 2.3, we also need to verify that is closed. Suppose that and as . Then for each there exists such that We may suppose as . By (2.18)(a) we have . On the other hand, by (2.9) and (2.17), it is easy to see that the function is also u.s.c. on . Combining this with (2.18)(b), we obtain Therefore, is closed and has the continuity results stated in the theorem.

Remark 2.6. As the discussion in Remark 2.4, the assumption that is used to make sure that is a nonnegative net output vector.

Remark 2.7. Another approach to obtain Theorem 2.5 is to use the so-called neighborhood approximation method discussed in [2]. Indeed, if we define the set-valued map from to by , then we can prove that is also an upper hemicontinuous set-valued map with convex compact values and satisfies the outward condition stated in (1.5). Hence, by Theorem 1.7, we obtain that , which can be used to prove Theorem 2.5.

#### 3. Theorems Without Continuity Assumption

In Theorem 1.7 the associated set-valued map is assumed to be u.h.c.. Recently, a Rogalski-Cornet-type theorem without any continuity conditions was proved in [5]. As a simple application, it is briefly applied to (1.1) in [5]. In this section, we develop more solvability and stability results for (1.1). We first review this theorem in the framework of -dimensional Euclidean space.

Let be a nonempty convex compact subset of , a set-valued map from to such that is a nonempty closed convex subset of for each , and its range. For each , , and , we set Let be the set of all such that Let be a subset of and a set-valued map from to defined by Then the following Rogalski-Cornet-type theorem was proved in [5].

In the sequel, we assume that in Lemma 3.1 is precisely the same subset of as stated in (1.2) and use Lemma 3.1 to obtain three results for (1.1).

Lemma 3.1. (i) (see [5, Theorem ]) If , then .
(ii) (see [5, Theorem ]) If , then is compact. Moreover, defined by (3.3) is closed, u.s.c. and u.h.c., and the subset of points at which is continuous is residual.

Remark 3.2. (i) Two counter-examples were presented in [5] to show that the set-valued map in Lemma 3.1 does not need the u.s.c., l.s.c., and u.h.c. conditions.
(ii) In case is u.h.c., then for each , , and , the function is u.s.c. on , and thus the upper section is closed. This proves that is closed for all , , and provided is u.h.c..

##### 3.1. The First Result

The first result is similar to Theorem 2.5. However, the u.s.c. assumption concerning introduced in Section 2 has been removed. Let , and be defined as in (1.6), (2.2), and (2.9), respectively. Define Applying Lemma 3.1 to , we have the following results.

Theorem 3.3. Assume that is finite, and .(i)Let . If (3.2)(a) holds for , then , that is, (ii)If defined by (3.1) is closed for all , , and , then defined by (2.9) is closed, u.s.c., and u.h.c., and the subset of points at which is continuous is residual.(iii)If is u.s.c., then has the same properties as stated in (ii).

Proof. (i) Since has convex compact values, is compact and is finite; by (3.4) it is easy to see that is a strict set-valued map from to with convex compact values. We shall prove by showing satisfying (3.2). By the assumption of (i), it is sufficient to prove that (3.2)(b) holds for . Indeed, we can prove that (3.2)(b) holds for any . Suppose that , and . If , then . If , then by Lemma 2.2. Hence by (2.9), , , , and thus that is, . So (3.2)(b) holds for all . By Lemma 3.1(i) and , we have . This implies that (3.5) is true, and thus (i) follows.
(ii) By the proof of (i), we have known that (3.2)(b) holds for all . Combining this with the assumption of (ii) and using (3.3)(a), we obtain that . So is nonempty and also compact by Lemma 3.1(ii). In view of Lemmas 1.31.6 and Remark 1.5, to complete the proof of (ii), it is enough to show that is closed since and are compact. Suppose that with . Then , and for each there exists such that . By and (3.4), we can see that , that is, where is defined by (3.3) for . By Lemma 3.1(ii), is closed, and thus , which implies that and . So we can select such that , that is, . Hence is closed.
(iii) Since is u.s.c, by the proof of Theorem 2.5, we have known that the function is u.s.c. on . Combining this with (3.4), it is easy to see that the set-valued map is u.s.c., and, by Lemma 1.4, also u.h.c.. Thus for each , the function is u.s.c. on . This implies that is u.h.c. on . Hence by Remark 3.2 and , the condition of (ii) is true. This completes the proof.

##### 3.2. The Second Result

Next we use a new number , a new set , and a new set-valued map from to to obtain further solvability and stability result for (1.1). Set and define and by Applying Lemma 3.1 to , we have the following results.

Theorem 3.4. Assume that is finite, , and .(i)Let . If (3.2)(a) holds for , then , that is, (ii)If defined by (3.1) is closed for all , and , then defined by (3.7) is closed, u.s.c., and u.h.c., and the subset of points at which is continuous is residual.(iii)If is u.s.c. on and also l.s.c. on , then the result of (ii) for is also true.

Proof. (i) It is easy to see that is a strict set-valued map with convex compact values. As in the proof of Theorem 3.3, we can prove that (3.2)(b) holds for all . Thus we can follow the proof of Theorem 3.3(i) to obtain the conclusion of (i). In fact, let , , and . If , then . If , then . So , and . This implies that is, . Hence, (3.2)(b) holds for all , and by Lemma 3.1(i), . Thus (3.8) is true.
(ii) The proof of (ii) is similar to that of Theorem 3.3(ii). Indeed, by the proof of (i) and the assumption of (ii), we can see that , and thus is compact by Lemma 3.1(ii). If such that as , then , and there exists with . Hence by (3.3) and (3.7), , where is defined by (3.3) for . Also by Lemma 3.1(ii), we have , that is, . So we can select such that . Therefore, and is closed.
(iii) Similar to the proof of Theorem 3.3(iii), it suffices to show that is u.h.c.. For and , we have . In this case, since holds for any and is u.h.c., we obtain that is u.h.c. in . Now, we prove that is also u.h.c. at any point . Because holds for any and , it is sufficient to show that is u.h.c. at . Since is compact, there is such that . As is l.s.c. at by the assumption, for , there exists an open neighborhood of such that for all . This implies that for each and hence for all . So we conclude that is u.s.c. at and, by Lemma 1.4, also u.h.c.. Therefore, is u.h.c. on , and the conclusion of (iii) follows. This completes the proof.

Remark 3.5. As in the previous theorems, the assumptions in Theorem 3.3 and in Theorem 3.4 are used to make sure that the net output vectors are nonnegative.

##### 3.3. The Third Result

Finally, we use the values of at the boundary of to consider the solvability and stability of (1.1). Define by Applying Lemma 3.1 to , we have the following results.

Theorem 3.6. Assume that and .(i)If (3.2)(a) holds for all , then , and is compact. In particular, if , then for any , (1.1) is solvable.(ii)If defined by (3.1) is closed for all , , and , then defined by (3.9) is closed, u.s.c., and u.h.c., and the subset of points at which is continuous is residual.(iii)If is u.s.c., then both results of (i) and (ii) are also true.

Proof. (i) Let , , and . If , then . If , then . From (3.9), we know that . This implies that .Hence (3.2)(b) holds for all , and then . By Lemma 3.1(i), it follows that
To prove that is compact, we only need to prove that is closed. Suppose that such that as . Let . Then by (3.9), for Hence there exists a sequence such that . Therefore, . Because is compact, there exists a convergent subsequence of . Since as , we may assume that , which implies . Hence and is closed. The rest of (i) is clear because .
(ii) By assumption of (ii), the proof of (i), and Lemma 3.1, we can see that , is compact, and defined by (3.3) for is closed. With the same method as in proving Theorem 3.3(ii), we can show that is closed and has all the continuous properties stated in this theorem (ii).
(iii) By assumption of (iii) and , we can see that is u.h.c., hence is closed for all , and . Hence both conditions of (i) and (ii) are satisfied, and therefore the result of (iii) follows. This completes the proof.

#### 4. Conclusions

In this paper, the Leontief-type input-output inclusion has been studied. First applying Rogalski-Cornet theorem [6], we have proved two solvability and stability theorems (Theorems 2.3 and 2.5) under the assumption that the set-valued map in this inclusion is upper semicontinuous. Then utilizing a Rogalski-Cornet-type theorem proved in [5], we have also proved three solvability and stability theorems (Theorems 3.33.6) in which the continuity assumption regarding the set-valued map in this inclusion is no longer needed.

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