# A Leontief-Type Input-Output Inclusion

Authors: Yingfan Liu; Qinghong Zhang
Journal: Mathematical Problems in Engineering
 (2010)

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## 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.


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## Body

## 1. Introduction

In this paper, we study the solvability and stability of the following input-output inclusion:(1.1)x∈X⊂R+n,c∈(I-A)x=x-Ax,
where R+n is the set of nonnegative vectors of the n-dimensional Euclidean space Rn,
(1.2)X⊂R+nisanonemptyconvexcompactsubset,c∈R+nisanetoutputvector,IistheidentitymapfromRntoitself,andA:X→2R+nisaset-valuedmapfromXtoR+nwithnonemptyconvexcompactvalues(i.e.,foreachx∈X,AxisanonemptyandconvexcompactsubsetofR+n).The present study is essentially a continuation of the investigation initiated in [1–5] 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 X and A were made. For the necessary background material and preliminaries, the reader is referred to [1–5]. 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 A for model (1.1). Obviously, for some c∈R+n, inclusion (1.1) may not have solutions. Specifically, if c makes (1.1) solvable, then (1.3)c∈C(A)=̂R+n⋂(I-A)X,where(I-A)X=⋃x∈X(x-Ax).
While C(A) gives us an expression for all possible c for which (1.1) has solutions, it is required that all the information regarding A 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 A is only available near the boundary of X. 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 Section2, we study (1.1) under the assumption that A 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 [6–9].Definition 1.1.
LetU  and V  be two Hausdorff topological spaces and F:U→2V  a set-valued map from U  to V.  The domain of F  is the set {x∈U:Fx≠∅}  denoted by domF,  and the graph of F  is the set {(x,y)∈U×V:x∈U,y∈Fx}  denoted by graphF.(1)
We say thatF  is strict if domF=U,  and F  is closed if graphF  is closed in U×V.(2)
We say thatF  is u.s.c. at x0∈U if for any neighborhood N(Fx0)  of Fx0, there exists a neighborhood N(x0) of x0 such that F(N(x0))⊆N(Fx0). F  is said to be u.s.c. if F is u.s.c. at every point x∈U.(3)
We say thatF  is l.s.c. at x0∈U  if for any y0∈Fx0  and any neighborhood N(y0)  of y0,  there exists a neighborhood N(x0)  of x0  such that Fx⋂N(y0)≠∅  for any x∈N(x0). F  is said to be l.s.c. if F  is l.s.c. at every point x∈U.(4)
IfV  is a Hausdorff locally convex vector space, V*  is its dual and 〈·,·〉 is the duality paring on V*×V.  We say that F  is u.h.c. at x0∈domF  if for any p∈V*,  the function x↦σ#(Fx,p)=supy∈Fx〈p,y〉  is upper semicontinuous (in short, u.s.c.) at x0. F  is said to be u.h.c. if it is u.h.c. at every point of domF.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]).
LetF be an u.s.c. set-valued map from a Hausdorff topological space U to a Hausdorff topological space V with closed values. Then F is closed.Lemma 1.3 (see [8]).
LetF be a closed set-valued map from a Hausdorff topological space U to a compact Hausdorff topological space V.  Then F is u.s.c..Lemma 1.4 (see [8]).
Suppose thatU is a Hausdorff topological space, V a Hausdorff locally convex vector space equipped with the weak topology σ(V,V*),  and F:U→2(V,σ(V,V*)) a set-valued map that is u.s.c. at x0∈U.  Then F is u.h.c. at x0∈U.Remark 1.5.
IfV=Rn,  then the weak topology σ(Rn,Rn*)  on Rn  coincides with the norm topology.Lemma 1.6 (see [8]).
Suppose thatU is a complete metric space, and V a compact metric space. If F:U→2V is a closed  and  strict set-valued map, then the subset of points at which F is continuous is residual, that is, the interior of the discontinuous point set of F 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.LetU be a Hausdorff locally convex vector space (U* its dual, 〈·,·〉 the duality paring on 〈U*,U〉), X a subset of U, intX the interior of X, ∂X=X∖intX the boundary of X, and S:X→2U a set-valued map from X to U. Let p∈U*. The normal cone NX(x) to X at x∈X, the supporting set ∂(X,p) of X, and the upper and lower supporting functions x↦σ#(Sx,p) and x↦σ♭(Sx,p) on X are defined by (1.4)NX(x)={p∈U*:〈p,x〉=σ#(X,p)=̂supy∈X〈p,y〉},∂(X,p)={x∈X:〈p,x〉=σ♭(X,p)=̂infy∈X〈p,y〉},σ#(Sx,p)=supy∈Sx〈p,y〉,σ♭(Sx,p)=infy∈Sx〈p,y〉forx∈X.
We say that(1.5)Sisoutwardif∀p∈U*,∀x∈∂(X,p),〈p,x〉≥σb(Sx,p),Sisinwardif∀p∈U*,∀x∈∂(X,p),〈p,x〉≤σ#(Sx,p),Sisstronglyinwardif∀p∈U*,∀x∈∂(X,p),σ#(Sx,p)≥σ#(X,p).With the help of these concepts, the Rogalski-Cornet theorem can be stated as follows.Theorem 1.7 (see [6] Rogalski-Cornet).
Suppose  thatX is a convex compact subset of U supplied with the weak topology, and S is an u.h.c. set-valued map from X to U with nonempty closed convex values.  If S is either outward or strongly inward, then for any y∈X,  there exists a solution x∈X to y∈Sx,  that is, SX=⋃x∈XSx⊇X.We use the following notations throughout this paper:(1.6)|Ax|∞=supz∈Ax∥z∥,μ∞=supx∈∂X|Ax|∞,d(x,F)=infz∈F∥x-z∥foreachx∈Rn,F⊆RnwithF≠∅.


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

In this section, we assume thatA of (1.1) is u.s.c. on X.Associating this assumption with (1.2), we can show that AX=⋃x∈XAx and A(∂X)=⋃x∈∂XAx are compact. Therefore, A is a strict and u.s.c. set-valued map from X (a convex compact subset of R+n) to AX (a compact subset of R+n) with convex compact values, and μ∞ defined by (1.6) is finite. Moreover, by Lemmas 1.2 and 1.4 and Remark 1.5, A is also closed and u.h.c. on X. Suppose that GA is a set-valued map from C(A) to X defined by
(2.1)C(A)=R+n⋂(I-A)X,GAc={x∈X:c∈x-Ax}forc∈C(A).
Then we have the following results.Theorem 2.1.
IfC(A)≠∅, then C(A) is compact and GA is closed, u.s.c., and u.h.c., and the subset of points at which GA is continuous is residual.Proof.
SinceX and AX are compact, we know that (I-A)X is bounded, so is C(A). Suppose that {ck:k≥1}⊆C(A) and ck→c0∈R+n as k→∞. Then there exist {xk:k≥1}⊂X such that ck∈xk-Axk. Since {xk:k≥1} has a convergent subsequence, we may assume that  xk→x0∈X as k→∞. As A is closed, we then obtain c0∈x0-Ax0, which shows that C(A) is closed, and hence, also compact.
For the continuity results ofGA, according to Lemmas 1.3–1.6 and Remark 1.5, we only need to prove that GA is closed because C(A) and X are compact. Suppose that {(ck,xk):k≥1}⊆graphGA such that (ck,xk)→(c,x) as k→∞. Then ck∈xk-Axk(k≥1). Since A is closed, ck→c∈C(A) and xk→x∈X, we get c∈x-Ax. Hence, GA is closed and has all the continuity results stated in the theorem.Next, we use Theorem1.7 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 X is convex compact and A is strict and u.s.c. with convex compact values, we further assume that intX≠∅ in this section. We have the following lemma. The proof is straightforward and hence omitted.Lemma 2.2.
LetY⊆Rn  be a nonempty convex set, y∈Y,  and ∂Y=Y∖intY  the boundary of Y.  If p≠0  and y∈∂(Y,p) (or p∈NY(y)),  then y∈∂Y.


### 2.1. Interior Approximation Method

Define a subsetX∞ of X and a set-valued map F1 from X∞ to X by(2.2)X∞=X⋂[μ∞,+∞)n,where[μ∞,+∞)n={(x1,x2,…,xn)∈R+n:xi≥μ∞},F1c={x∈X:∃Δc∈Rns.t.c+Δc∈x-Ax,∥Δc∥≤μ∞},forc∈X∞.
Then we have the first solvability and stability result for (1.1) as follows.Theorem 2.3.
SupposethatX∞≠∅ and c∈X∞. Then there exist x∈X, Δc∈Rn such that c+Δc∈x-Ax and ∥Δc∥≤μ∞. Moreover, F1 defined by (2.2) is closed, u.s.c., and u.h.c., and the subset of points at which F1 is continuous is residual.Proof.
SinceAx is compact for x∈X and A is u.s.c., it is easy to see that for ε>0, B(Ax,ε)={y∈Rn:d(y,Ax)<ε} is an open neighborhood of Ax, and that for each k≥1 and x∈∂X, there exists a neighborhood U(x) of x with A(U(x)∩X)⊆B(Ax,1/k). As ∂X is compact and ∂X⊆⋃x∈∂XU(x), there exist {zi:1≤i≤m}⊆∂X such that
(2.3)∂X⊆⋃i=1mUi,A(Ui∩X)⊆B(Azi,1k),fori=1,2,…,m,
where Ui=̂U(zi). Let Y=X∖(⋃i=1mUi). Then it is a closed subset of X with ∂X∩Y=∅. Let d=d(∂X,Y)=infx∈∂X,y∈Yd(x,y). Then we have d>0. Otherwise, there exist sequences {xj:j≥1}⊆∂X and {yj:j≥1}⊆Y, such that d(xj,yj)→0 as j→∞. Since X is compact, so are ∂X and Y, which imply that there exist convergent subsequences of {xj:j≥1} and {yj:j≥1}. Without loss of generality, we may assume that {xj:j≥1} and {yj:j≥1} are convergent to the same point x as j→∞. Since ∂X and Y are closed, we obtain x∈∂X∩Y, a contradiction. Therefore, d>0. Let U0=⋃y∈YB(y,d/2). Then it is easy to see that U0 is an open subset of X, X∖U0⊆⋃i=1mUi and
(2.4)A(X∖U0)⊆⋃i=1mB(Azi,1k).
For eachk≥1, by the truncation technique of generalized functions in partial differential equations [10], there is a continuous function gk from X to R+ such that 0≤gk(x)≤1(x∈X),gk|U0=1, suppgk≡{x∈X:gk(x)≠0}¯⊆ intX,andhencegk|∂X=0. Here M¯ denotes the closure of M. Define Ak:X→2Rn by
(2.5)Akx=gk(x)Ax,forx∈X.
We claim that for eachk≥1, Ak is a closed, u.s.c., and u.h.c. set-valued map with nonempty convex compact values. Indeed, Akx is clearly a convex compact subset of Rn for each x∈X. Assume that (xj,yj)∈graphAk (i.e., yj∈gk(xj)Axj) such that (xj,yj)→(x0,y0) as j→∞. Then xj→x0,yj→y0(j→∞), and for each j≥1 there exists wj∈Axj with yj=gk(xj)wj. Since AX is compact, we may suppose that  wj→w0 as j→∞. By letting j→∞ and using the fact that A is closed, we get w0∈Ax0 and y0=gk(x0)w0∈gk(x0)Ax0, that is, (x0,y0)∈graphAk. Hence, Ak is closed. On the other hand, by (2.5), it is easy to see that Ak(X)¯ is compact. Hence by Lemmas 1.3 and 1.4 and Remark 1.5, Ak is also u.s.c. and u.h.c..
Sinceσ#((I-Ak)x,p)=〈p,x〉+σ#(Akx,-p) for p∈Rn and x∈X, it follows that for each p∈Rn, the function x→σ#((I-Ak)x,p) is u.s.c. on X. And so, by Definition 1.1, S=I-Ak is u.h.c.
Assume thatp∈Rn and x∈∂(X,p). If p=0, then 〈p,x〉=0=σb(x-Akx,p). If p≠0, then by Lemma 2.2, x∈∂X, which further implies by gk|∂X=0 that Akx={0} and hence 〈p,x〉=σb(x-Akx,p). Therefore, S=I-Ak satisfies the outward condition stated in (1.5). In view of Theorem 1.7, we have (I-Ak)X⊇X. Hence for each c∈X, there exists xk∈X such that
(2.6)c∈xk-Akxk.
If xk∈U0, by (2.5) and gk|U0=1, we have c∈xk-Axk. Set Δck=0, then we get
(2.7)c+Δck∈xk-Axk,∥Δck∥≤μ∞+1k.
If xk∈X∖U0, by (2.5) and (2.6) there exists wk∈Axk such that c=xk-gk(xk)wk. Let Δck=(gk(xk)-1)wk. Then c+Δck=xk-wk∈xk-Axk and Δck∈(gk(xk)-1)Axk. By virtue of (2.4), (2.5), and (1.6), we see that (2.7) is also true.
Now, we obtain two sequences{xk:k≥1}⊆X and {Δck:k≥1}⊆Rn with ∥Δck∥≤μ∞+(1/k). Since X is compact and {Δck:k≥1} is bounded, we may assume that xk→x∈X and Δck→Δc as k→∞. Combining this with (2.7) and also using the fact that A is closed, we obtain that c+Δc∈x-Axand∥Δc∥≤μ∞.
Next, we prove the second part of the theorem. SinceX∞ and X are compact, also by Lemmas 1.3–1.6 and Remark 1.5, we only need to prove that F1 is closed. Assume that  {(ck,xk):k≥1}⊆graphF1 with (ck,xk)→(c,x)(k→∞). Then for each k≥1 there exists Δck∈Rn such that
(2.8)ck+Δck∈xk-Axk,∥Δck∥≤μ∞.
We may suppose that  Δck→Δc as k→∞. This implies by (2.8) and the closeness of A that c+Δc∈x-Ax and ∥Δc∥≤μ∞. Hence, F1 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 conditionc∈X∞  is not used. We impose this requirement in order to make sure that c+Δc∈R+n.


### 2.2. Exterior Approximation Method

Define a functionδ∞(x) on X and a set-valued map F2 from X∞ to X by(2.9)δ∞(x)={0ifx∈intX,|Ax|∞ifx∈∂X,F2c={x∈X:∃Δc∈Rns.t.c+Δc∈x-Ax,∥Δc∥≤δ∞(x)}forc∈X∞,
where X∞ is defined as in (2.2). Then we have the following result.Theorem 2.5.
Suppose thatX∞≠∅ and c∈X∞.  Then there exist x∈X, Δc∈Rn such that c+Δc∈x-Ax,  and ∥Δc∥≤δ∞(x).  Furthermore, F2 defined by (2.9) is closed, u.s.c. and u.h.c., and the subset of points at which F2 is continuous is residual.Proof.
LetB¯(X,ε)={x∈Rn:d(x,X)≤ε} for ε>0. For each k≥1, let B¯k(X)=̂B¯(X,1/k) and let Pk be the projection from B¯k(X) to X, that is, Pkx∈X such that ∥Pkx-x∥=infy∈X∥y-x∥ for x∈B¯k(X). It is easy to see that B¯k(X) is a convex compact set with nonempty interior, Pkx=x if x∈X and Pkx∈∂X if x∈B¯k(X)∖X, and ∥Pkx-x∥≤1/k for all x∈B¯k(X).
As in the proof of Theorem2.3, we assume that gk is a continuous function from B¯k(X) to R+ with compact support set such that
(2.10)0≤gk(x)≤1forx∈B¯k(X),gk|X=1,gk|∂(B¯k(X))=0.
Let Ak be a set-valued map from B¯k(X) to Rn defined by
(2.11)Akx=gk(x)A(Pkx)forx∈B¯k(X).
Utilizing the similar method as in the proof of Theorem2.3, we can show that Ak is a closed, u.s.c., and u.h.c. set-valued map with nonempty convex compact values and satisfies
(2.12)Ak|X=A,Ak|∂(B¯k(X))={0}.
In fact, if (xm,ym)∈graphAk with (xm,ym)→(x0,y0)(m→∞), then x0∈B¯kX, Pkxm→Pkx0∈X(m→∞), and for each m≥1, there exists wm∈A(Pkxm) such that ym=gk(xm)wm. Since A(Pk(B¯k(X)))⊆AX and AX is compact, we may suppose that wm→w0(m→∞). By letting m→∞, from the closeness of A, Lemmas 1.3 and 1.4, and Remark 1.5, we conclude that w0∈A(Pkx0), y0=gk(x0)w0∈Akx0, that is, (x0,y0)∈graphAk, and thus Ak is closed, u.s.c. and u.h.c..
Also as in the proof of Theorem2.3, we can prove that I-Ak satisfies the outward condition stated in (1.5). (In fact, assume that p∈Rn and x∈∂(B¯k(X),p). If p=0, then 〈p,x〉=0=σb(x-Akx,p). If p≠0, then by Lemma 2.2, x∈∂(B¯k(X)), which implies by (2.12) that Akx={0}, and hence 〈p,x〉=σb(x-Akx,p).)By Theorem 1.7,we obtain
(2.13)(I-Ak)B¯k(X)⊇B¯k(X)⊇X(⊇X∞).
Combining this with (2.11), we know that for each c∈X, there exists x¯k∈B¯k(X) such that
(2.14)c∈x¯k-gk(x¯k)A(Pkx¯k).
Setxk=Pkx¯k. Then xk∈X. If there is k≥1 such that x¯k∈intX, then from (2.10), (2.12), and (2.14) we have x¯k=xk and c∈xk-Axk. Set x=xk and Δc=0. Then
(2.15)c+Δc∈x-Ax,∥Δc∥=0≤δ∞(x).
If for all k≥1, x¯k∈B¯k(X)∖intX, then xk=Pkx¯k∈∂X and ∥xk-x¯k∥≤1/k. By (2.14), there exists zk∈Axk such that c=x¯k-g(x¯k)zk. Let Δck=xk-x¯k+[g(x¯k)-1]zk. Then we obtain
(2.16)(a)c+Δck=xk-zk∈xk-Axk,(b)∥Δck∥≤∥xk-x¯k∥+|g(x¯k)-1|∥zk∥≤1k+|Axk|∞.
SinceX is compact and {Δck:k≥1} is bounded, we may assume that xk→x and Δck→Δc as k→∞. This implies by xk∈∂X(k≥1) that x∈∂X. On the other hand, if ε>0 and B(Ax,ε) is a ε-neighborhood of Ax, then there exists a neighborhood U(x) of x such that A(U(x)⋂X)⊆B(Ax,ε) because A is u.s.c., and thus for any x′∈U(x)∩X, |Ax′|∞≤|Ax|∞+ε. Hence the function
(2.17)x↦|Ax|∞isuppersemicontinuous(i.e.,u.s.c.)onX.
By letting k→∞ and using the closeness of A, from (2.16) and (2.17), we get c+Δc∈x-Axand∥Δc∥≤|Ax|∞=δ∞(x). Combining this with (2.15), we conclude that the first part of the theorem is true.
As in the proof of Theorem2.3, we also need to verify that graphF2 is closed. Suppose that  {(ck,xk):k≥1}⊆graphF2 and (ck,xk)→(c,x) as k→∞. Then for each k≥1 there exists Δck∈Rn such that
(2.18)(a)ck+Δck∈xk-Axk,(b)∥Δck∥≤δ∞(xk).
We may suppose Δck→Δc as k→∞. By (2.18)(a) we have c+Δc∈x-Ax. On the other hand, by (2.9) and (2.17), it is easy to see that the function x↦δ∞(x) is also u.s.c. on X. Combining this with (2.18)(b), we obtain ∥Δc∥≤δ∞(x). Therefore, F2 is closed and has the continuity results stated in the theorem.Remark 2.6.
As the discussion in Remark2.4, the assumption that c∈X∞  is used to make sure that c+Δc  is a nonnegative net output vector.Remark 2.7.
Another approach to obtain Theorem2.5 is to use the so-called neighborhood approximation method discussed in [2].  Indeed, if we define the set-valued map T  from X  to Rn  by Tx=Ax+B¯(x,δ∞(x)),  then we can prove that T  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 T(X)⊇X⊇X∞,  which can be used to prove Theorem 2.5.


## 2.1. Interior Approximation Method

Define a subsetX∞ of X and a set-valued map F1 from X∞ to X by(2.2)X∞=X⋂[μ∞,+∞)n,where[μ∞,+∞)n={(x1,x2,…,xn)∈R+n:xi≥μ∞},F1c={x∈X:∃Δc∈Rns.t.c+Δc∈x-Ax,∥Δc∥≤μ∞},forc∈X∞.
Then we have the first solvability and stability result for (1.1) as follows.Theorem 2.3.
SupposethatX∞≠∅ and c∈X∞. Then there exist x∈X, Δc∈Rn such that c+Δc∈x-Ax and ∥Δc∥≤μ∞. Moreover, F1 defined by (2.2) is closed, u.s.c., and u.h.c., and the subset of points at which F1 is continuous is residual.Proof.
SinceAx is compact for x∈X and A is u.s.c., it is easy to see that for ε>0, B(Ax,ε)={y∈Rn:d(y,Ax)<ε} is an open neighborhood of Ax, and that for each k≥1 and x∈∂X, there exists a neighborhood U(x) of x with A(U(x)∩X)⊆B(Ax,1/k). As ∂X is compact and ∂X⊆⋃x∈∂XU(x), there exist {zi:1≤i≤m}⊆∂X such that
(2.3)∂X⊆⋃i=1mUi,A(Ui∩X)⊆B(Azi,1k),fori=1,2,…,m,
where Ui=̂U(zi). Let Y=X∖(⋃i=1mUi). Then it is a closed subset of X with ∂X∩Y=∅. Let d=d(∂X,Y)=infx∈∂X,y∈Yd(x,y). Then we have d>0. Otherwise, there exist sequences {xj:j≥1}⊆∂X and {yj:j≥1}⊆Y, such that d(xj,yj)→0 as j→∞. Since X is compact, so are ∂X and Y, which imply that there exist convergent subsequences of {xj:j≥1} and {yj:j≥1}. Without loss of generality, we may assume that {xj:j≥1} and {yj:j≥1} are convergent to the same point x as j→∞. Since ∂X and Y are closed, we obtain x∈∂X∩Y, a contradiction. Therefore, d>0. Let U0=⋃y∈YB(y,d/2). Then it is easy to see that U0 is an open subset of X, X∖U0⊆⋃i=1mUi and
(2.4)A(X∖U0)⊆⋃i=1mB(Azi,1k).
For eachk≥1, by the truncation technique of generalized functions in partial differential equations [10], there is a continuous function gk from X to R+ such that 0≤gk(x)≤1(x∈X),gk|U0=1, suppgk≡{x∈X:gk(x)≠0}¯⊆ intX,andhencegk|∂X=0. Here M¯ denotes the closure of M. Define Ak:X→2Rn by
(2.5)Akx=gk(x)Ax,forx∈X.
We claim that for eachk≥1, Ak is a closed, u.s.c., and u.h.c. set-valued map with nonempty convex compact values. Indeed, Akx is clearly a convex compact subset of Rn for each x∈X. Assume that (xj,yj)∈graphAk (i.e., yj∈gk(xj)Axj) such that (xj,yj)→(x0,y0) as j→∞. Then xj→x0,yj→y0(j→∞), and for each j≥1 there exists wj∈Axj with yj=gk(xj)wj. Since AX is compact, we may suppose that  wj→w0 as j→∞. By letting j→∞ and using the fact that A is closed, we get w0∈Ax0 and y0=gk(x0)w0∈gk(x0)Ax0, that is, (x0,y0)∈graphAk. Hence, Ak is closed. On the other hand, by (2.5), it is easy to see that Ak(X)¯ is compact. Hence by Lemmas 1.3 and 1.4 and Remark 1.5, Ak is also u.s.c. and u.h.c..
Sinceσ#((I-Ak)x,p)=〈p,x〉+σ#(Akx,-p) for p∈Rn and x∈X, it follows that for each p∈Rn, the function x→σ#((I-Ak)x,p) is u.s.c. on X. And so, by Definition 1.1, S=I-Ak is u.h.c.
Assume thatp∈Rn and x∈∂(X,p). If p=0, then 〈p,x〉=0=σb(x-Akx,p). If p≠0, then by Lemma 2.2, x∈∂X, which further implies by gk|∂X=0 that Akx={0} and hence 〈p,x〉=σb(x-Akx,p). Therefore, S=I-Ak satisfies the outward condition stated in (1.5). In view of Theorem 1.7, we have (I-Ak)X⊇X. Hence for each c∈X, there exists xk∈X such that
(2.6)c∈xk-Akxk.
If xk∈U0, by (2.5) and gk|U0=1, we have c∈xk-Axk. Set Δck=0, then we get
(2.7)c+Δck∈xk-Axk,∥Δck∥≤μ∞+1k.
If xk∈X∖U0, by (2.5) and (2.6) there exists wk∈Axk such that c=xk-gk(xk)wk. Let Δck=(gk(xk)-1)wk. Then c+Δck=xk-wk∈xk-Axk and Δck∈(gk(xk)-1)Axk. By virtue of (2.4), (2.5), and (1.6), we see that (2.7) is also true.
Now, we obtain two sequences{xk:k≥1}⊆X and {Δck:k≥1}⊆Rn with ∥Δck∥≤μ∞+(1/k). Since X is compact and {Δck:k≥1} is bounded, we may assume that xk→x∈X and Δck→Δc as k→∞. Combining this with (2.7) and also using the fact that A is closed, we obtain that c+Δc∈x-Axand∥Δc∥≤μ∞.
Next, we prove the second part of the theorem. SinceX∞ and X are compact, also by Lemmas 1.3–1.6 and Remark 1.5, we only need to prove that F1 is closed. Assume that  {(ck,xk):k≥1}⊆graphF1 with (ck,xk)→(c,x)(k→∞). Then for each k≥1 there exists Δck∈Rn such that
(2.8)ck+Δck∈xk-Axk,∥Δck∥≤μ∞.
We may suppose that  Δck→Δc as k→∞. This implies by (2.8) and the closeness of A that c+Δc∈x-Ax and ∥Δc∥≤μ∞. Hence, F1 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 conditionc∈X∞  is not used. We impose this requirement in order to make sure that c+Δc∈R+n.


## 2.2. Exterior Approximation Method

Define a functionδ∞(x) on X and a set-valued map F2 from X∞ to X by(2.9)δ∞(x)={0ifx∈intX,|Ax|∞ifx∈∂X,F2c={x∈X:∃Δc∈Rns.t.c+Δc∈x-Ax,∥Δc∥≤δ∞(x)}forc∈X∞,
where X∞ is defined as in (2.2). Then we have the following result.Theorem 2.5.
Suppose thatX∞≠∅ and c∈X∞.  Then there exist x∈X, Δc∈Rn such that c+Δc∈x-Ax,  and ∥Δc∥≤δ∞(x).  Furthermore, F2 defined by (2.9) is closed, u.s.c. and u.h.c., and the subset of points at which F2 is continuous is residual.Proof.
LetB¯(X,ε)={x∈Rn:d(x,X)≤ε} for ε>0. For each k≥1, let B¯k(X)=̂B¯(X,1/k) and let Pk be the projection from B¯k(X) to X, that is, Pkx∈X such that ∥Pkx-x∥=infy∈X∥y-x∥ for x∈B¯k(X). It is easy to see that B¯k(X) is a convex compact set with nonempty interior, Pkx=x if x∈X and Pkx∈∂X if x∈B¯k(X)∖X, and ∥Pkx-x∥≤1/k for all x∈B¯k(X).
As in the proof of Theorem2.3, we assume that gk is a continuous function from B¯k(X) to R+ with compact support set such that
(2.10)0≤gk(x)≤1forx∈B¯k(X),gk|X=1,gk|∂(B¯k(X))=0.
Let Ak be a set-valued map from B¯k(X) to Rn defined by
(2.11)Akx=gk(x)A(Pkx)forx∈B¯k(X).
Utilizing the similar method as in the proof of Theorem2.3, we can show that Ak is a closed, u.s.c., and u.h.c. set-valued map with nonempty convex compact values and satisfies
(2.12)Ak|X=A,Ak|∂(B¯k(X))={0}.
In fact, if (xm,ym)∈graphAk with (xm,ym)→(x0,y0)(m→∞), then x0∈B¯kX, Pkxm→Pkx0∈X(m→∞), and for each m≥1, there exists wm∈A(Pkxm) such that ym=gk(xm)wm. Since A(Pk(B¯k(X)))⊆AX and AX is compact, we may suppose that wm→w0(m→∞). By letting m→∞, from the closeness of A, Lemmas 1.3 and 1.4, and Remark 1.5, we conclude that w0∈A(Pkx0), y0=gk(x0)w0∈Akx0, that is, (x0,y0)∈graphAk, and thus Ak is closed, u.s.c. and u.h.c..
Also as in the proof of Theorem2.3, we can prove that I-Ak satisfies the outward condition stated in (1.5). (In fact, assume that p∈Rn and x∈∂(B¯k(X),p). If p=0, then 〈p,x〉=0=σb(x-Akx,p). If p≠0, then by Lemma 2.2, x∈∂(B¯k(X)), which implies by (2.12) that Akx={0}, and hence 〈p,x〉=σb(x-Akx,p).)By Theorem 1.7,we obtain
(2.13)(I-Ak)B¯k(X)⊇B¯k(X)⊇X(⊇X∞).
Combining this with (2.11), we know that for each c∈X, there exists x¯k∈B¯k(X) such that
(2.14)c∈x¯k-gk(x¯k)A(Pkx¯k).
Setxk=Pkx¯k. Then xk∈X. If there is k≥1 such that x¯k∈intX, then from (2.10), (2.12), and (2.14) we have x¯k=xk and c∈xk-Axk. Set x=xk and Δc=0. Then
(2.15)c+Δc∈x-Ax,∥Δc∥=0≤δ∞(x).
If for all k≥1, x¯k∈B¯k(X)∖intX, then xk=Pkx¯k∈∂X and ∥xk-x¯k∥≤1/k. By (2.14), there exists zk∈Axk such that c=x¯k-g(x¯k)zk. Let Δck=xk-x¯k+[g(x¯k)-1]zk. Then we obtain
(2.16)(a)c+Δck=xk-zk∈xk-Axk,(b)∥Δck∥≤∥xk-x¯k∥+|g(x¯k)-1|∥zk∥≤1k+|Axk|∞.
SinceX is compact and {Δck:k≥1} is bounded, we may assume that xk→x and Δck→Δc as k→∞. This implies by xk∈∂X(k≥1) that x∈∂X. On the other hand, if ε>0 and B(Ax,ε) is a ε-neighborhood of Ax, then there exists a neighborhood U(x) of x such that A(U(x)⋂X)⊆B(Ax,ε) because A is u.s.c., and thus for any x′∈U(x)∩X, |Ax′|∞≤|Ax|∞+ε. Hence the function
(2.17)x↦|Ax|∞isuppersemicontinuous(i.e.,u.s.c.)onX.
By letting k→∞ and using the closeness of A, from (2.16) and (2.17), we get c+Δc∈x-Axand∥Δc∥≤|Ax|∞=δ∞(x). Combining this with (2.15), we conclude that the first part of the theorem is true.
As in the proof of Theorem2.3, we also need to verify that graphF2 is closed. Suppose that  {(ck,xk):k≥1}⊆graphF2 and (ck,xk)→(c,x) as k→∞. Then for each k≥1 there exists Δck∈Rn such that
(2.18)(a)ck+Δck∈xk-Axk,(b)∥Δck∥≤δ∞(xk).
We may suppose Δck→Δc as k→∞. By (2.18)(a) we have c+Δc∈x-Ax. On the other hand, by (2.9) and (2.17), it is easy to see that the function x↦δ∞(x) is also u.s.c. on X. Combining this with (2.18)(b), we obtain ∥Δc∥≤δ∞(x). Therefore, F2 is closed and has the continuity results stated in the theorem.Remark 2.6.
As the discussion in Remark2.4, the assumption that c∈X∞  is used to make sure that c+Δc  is a nonnegative net output vector.Remark 2.7.
Another approach to obtain Theorem2.5 is to use the so-called neighborhood approximation method discussed in [2].  Indeed, if we define the set-valued map T  from X  to Rn  by Tx=Ax+B¯(x,δ∞(x)),  then we can prove that T  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 T(X)⊇X⊇X∞,  which can be used to prove Theorem 2.5.


## 3. Theorems Without Continuity Assumption

In Theorem1.7 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 n-dimensional Euclidean space.LetX be a nonempty convex compact subset of Rn, S:X→2Rn a set-valued map from X to Rn such that Sx is a nonempty closed convex subset of Rn for each x∈X, and SX=⋃x∈XSx its range. For each p∈Rn, c∈X, and ε≥0, we set(3.1)Y(S,p,c,ε)={x∈X:σ#(Sx-c,p)+ε≥0},Y(S,p,c)=Y(S,p,c,0).
Let Y(S) be the set of all c∈X such that(3.2)(a)Y(S,p,c)isclosed,∀p∈Rn,(b)x∈Y(S,p,c)ifx∈X,p∈NX(x).
Let Ỹ(S) be a subset of Y(S) and FS a set-valued map from Ỹ(S) to X defined by(3.3)(a)Ỹ(S)={c∈Y(S):∀p∈Rn,∀ε≥0,Y(S,p,c,ε)isclosed},(b)FSc={x∈X:c∈Sx}=X∩S-1c,forc∈Ỹ(S).
Then the following Rogalski-Cornet-type theorem was proved in [5].In the sequel, we assume thatX in Lemma 3.1 is precisely the same subset X of R+n as stated in (1.2) and use Lemma 3.1 to obtain three results for (1.1).Lemma 3.1.
(i) (see [5, Theorem 2.8])  If Y(S)≠∅,  then Y(S)⊆SX.
(ii) (see [5, Theorem 2.12])   If Ỹ(S)≠∅,  then Ỹ(S) is compact. Moreover, FS defined by (3.3) is closed, u.s.c. and u.h.c., and the subset of points at which FS is continuous is residual.Remark 3.2.
(i) Two counter-examples were presented in[5] to show that the set-valued map S  in Lemma 3.1 does not need the u.s.c., l.s.c., and u.h.c. conditions.
(ii) In caseS  is u.h.c., then for each p∈Rn, c∈Rn,  and ε≥0,  the function f(x)=σ#(Sx-c,p)+ε=σ#(Sx,p)-〈p,c〉+ε  is u.s.c. on X,  and thus the upper section Y(S,p,c,ε)={x∈X∣f(x)≥0}  is closed. This proves that Y(S,p,c,ε)  is closed for all p∈Rn, c∈Rn,  and ε≥0  provided S  is u.h.c..


### 3.1. The First Result

The first result is similar to Theorem2.5. However, the u.s.c. assumption concerning A introduced in Section 2 has been removed. Let μ∞,X∞,δ∞(x), and F2 be defined as in (1.6), (2.2), and (2.9), respectively. Define(3.4)A∞:X→2RnbyA∞x=B¯(0,δ∞(x))forx∈X.
Applying Lemma 3.1 to S=I-(A+A∞), we have the following results.Theorem 3.3.
Assume  thatμ∞ is finite, X∞≠∅ and S=I-(A+A∞).(i)
Letc0∈X∞.  If (3.2)(a) holds for c0,  then F2c0≠∅, that is,
(3.5)∃x0∈X,∃Δc0∈Rns.t.c0+Δc0∈x0-Ax0,∥Δc∥≤σ∞(x0).(ii)
IfY(S,p,c,ε) defined by (3.1) is closed for all p∈Rn, c∈X∞,  and ε≥0,  then F2 defined by (2.9) is closed, u.s.c., and u.h.c., and the subset of points at which F2 is continuous is residual.(iii)
IfA is u.s.c., then F2 has the same properties as stated in (ii).Proof.
(i) SinceA has convex compact values, X is compact and μ∞ is finite; by (3.4) it is easy to see that S=I-(A+A∞) is a strict set-valued map from X to Rn with convex compact values. We shall prove c0∈Y(S) by showing c0 satisfying (3.2). By the assumption of (i), it is sufficient to prove that (3.2)(b) holds for c0. Indeed, we can prove that (3.2)(b) holds for any c∈X. Suppose that x∈X,p∈NX(x), and c∈X. If p=0, then σ#(Sx-c,0)=0. If p≠0, then x∈∂X by Lemma 2.2. Hence by (2.9), δ∞(x)=|Ax|∞, 0∈Ax+B¯(0,|Ax|∞), σb(Ax+A∞x,p)≤0, and thus 〈p,x〉=supy∈X〈p,y〉≥〈p,c〉≥σb(Ax+A∞x,p)+〈p,c〉, that is, σ#(Sx-c,p)=σ#(x-Ax-A∞x-c,p)≥0. So (3.2)(b) holds for all c∈X. By Lemma 3.1(i) and X∞⊆X, we have c0∈Y(S)⊆SX=⋃x∈X(x-Ax-A∞x). 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 c∈X. Combining this with the assumption of (ii) and using (3.3)(a), we obtain that X∞⊆Ỹ(S). So Ỹ(S) is nonempty and also compact by Lemma 3.1(ii). In view of Lemmas 1.3–1.6 and Remark 1.5, to complete the proof of (ii), it is enough to show that F2 is closed since X∞ and X are compact. Suppose that {(cm,xm):m≥1}⊆graphF2 with (cm,xm)→(c0,x0)(m→∞). Then c0∈X∞⊂Ỹ(S), and for each m≥1 there exists Δcm∈B¯(0,δ∞(xm)) such that cm+Δcm∈xm-Axm. By graphF2⊆X∞×X⊆Ỹ(S)×X and (3.4), we can see that cm∈Sxm, that is, (cm,xm)∈graphFS, where FS is defined by (3.3) for S=I-(A+A∞). By Lemma 3.1(ii), FS is closed, and thus (c0,x0)∈graphFS, which implies that x0∈X and c0∈Sx0=(I-A-A∞)x0. So we can select Δc0∈B¯(0,δ∞(x0)) such that c0+Δc0∈x0-Ax0, that is, (c0,x0)∈graphF2. Hence F2 is closed.
(iii) SinceA is u.s.c, by the proof of Theorem 2.5, we have known that the function x↦δ∞(x) is u.s.c. on X. Combining this with (3.4), it is easy to see that the set-valued map A∞ is u.s.c., and, by Lemma 1.4, also u.h.c.. Thus for each p∈Rn, the function x↦σ#(Sx,p)=〈p,x〉+σ#(Ax,-p)+σ#(A∞x,-p) is u.s.c. on X. This implies that S=I-(A+A∞) is u.h.c. on X. Hence by Remark 3.2 and X∞⊆X, the condition of (ii) is true. This completes the proof.


### 3.2. The Second Result

Next we use a new numberμ0, a new set X0, and a new set-valued map F0 from X0 to X to obtain further solvability and stability result for (1.1). Set(3.6)|Ax|0=infz∈Ax∥z∥,(x∈X),μ0=supx∈∂X|Ax|0,δ0(x)={0ifx∈intX,|Ax|0ifx∈∂X,X0=X⋂[μ0,+∞)n,
and define A0:X→2Rn and F0:X0→2X by(3.7)A0x=B¯(0,δ0(x))forx∈X,F0c={x∈X:∃Δc∈Rns.t.c+Δc∈x-Ax,∥Δc∥≤δ0(x)}forc∈X0.
Applying Lemma 3.1 to S=I-(A+A0), we have the following results.Theorem 3.4.
Assume thatμ0 is finite, X0≠∅,  and S=I-(A+A0).(i)
Letc0∈X0.  If (3.2)(a) holds for c0,  then F0c0≠∅,  that is,
(3.8)∃x0∈X,∃Δc0∈Rns.t.c0+Δc0∈x0-Ax0,∥Δc0∥≤δ0(x).(ii)
IfY(S,p,c,ε) defined by (3.1) is closed for all p∈Rn, c∈X0 and ε≥0,  then F0 defined by (3.7) is closed, u.s.c., and u.h.c., and the subset of points at which F0 is continuous is residual.(iii)
IfA is u.s.c. on X and also l.s.c. on ∂X,  then the result of (ii) for F0 is also true.Proof.
(i) It is easy to see thatS 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 c∈X. Thus we can follow the proof of Theorem 3.3(i) to obtain the conclusion of (i). In fact, let x∈X, p∈NX(x), and c∈X. If p=0, then σ#(Sx-c,p)=0. If p≠0, then x∈∂X. So δ0(x)=|Ax|0, 0∈Ax+B¯(0,δ0(x)) and σb(Ax+A0x,p)≤0. This implies 〈p,x〉=supy∈X〈p,y〉≥〈p,c〉≥σb(Ax+A0x,p)+〈p,c〉, that is, σ#(Sx-c,p)=σ#(x-Ax-A0x-c,p)≥0. Hence, (3.2)(b) holds for all c∈X, and by Lemma 3.1(i), c0∈Y(S)⊆SX=⋃x∈X(x-Ax-A0x). Thus (3.8) is true.
(ii) The proof of (ii) is similar to that of Theorem3.3(ii). Indeed, by the proof of (i) and the assumption of (ii), we can see that X0⊆Ỹ(S), and thus Ỹ(S) is compact by Lemma 3.1(ii). If {(cm,xm):m≥1}⊆graphF0 such that (cm,xm)→(c0,x0) as m→∞, then c0∈X0⊂Ỹ(S),x0∈X, and there exists Δcm∈B¯(0,δ0(xm)) with cm+Δcm∈xm-Axm. Hence by (3.3) and (3.7), (cm,xm)∈graphFS, where FS is defined by (3.3) for S=I-(A+A0). Also by Lemma 3.1(ii), we have (c0,x0)∈graphFS, that is, c0∈(I-A-A0)x0. So we can select Δc0∈B¯(0,δ0(x0)) such that c0+Δc0∈x0-Ax0. Therefore, (c0,x0)∈graphF0 and F0 is closed.
(iii) Similar to the proof of Theorem3.3(iii), it suffices to show that S=I-(A+A0) is u.h.c.. For x∈intX and p∈Rn, we have A0x={0}. In this case, since σ#(Sx,p)=〈x,p〉+σ#(Ax,-p) holds for any x∈intX and A is u.h.c., we obtain that S is u.h.c. in intX. Now, we prove that S is also u.h.c. at any point x0∈∂X. Because σ#(Sx,p)=〈x,p〉+σ#(Ax,-p)+σ#(A0x,-p) holds for any x∈X and p∈Rn, it is sufficient to show that A0 is u.h.c. at x0∈∂X. Since Ax0 is compact, there is z0∈Ax0 such that ∥z0∥=|Ax0|0=δ0(x0). As A is l.s.c. at x0 by the assumption, for ε>0, there exists an open neighborhood U(x0) of x0 such that Ax∩B(z0,ε)≠∅ for all x∈X∩U(x0). This implies that for each x∈X∩U(x0),δ0(x)≤|Ax|0≤infy∈Ax∩B(z0,ε)∥y∥≤supy∈B(z0,ε)∥y∥≤δ0(x0)+ε, and hence B¯(0,δ0(x))⊆B¯(B¯(0,δ0(x0)),ε) for all x∈X∩U(x0). So we conclude that A0 is u.s.c. at x0 and, by Lemma 1.4, also u.h.c.. Therefore, S=I-(A+A0) is u.h.c. on X, and the conclusion of (iii) follows. This completes the proof.Remark 3.5.
As in the previous theorems, the assumptionsc0∈X∞  in Theorem 3.3 and c0∈X0  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 ofA at the boundary ∂X of X to consider the solvability and stability of (1.1). Define F̂:X̂→2X by(3.9)X̂={c∈X:∀x∈∂X,(Ax+c)∩X≠∅},F̂c={x∈X:c∈x-Ax}forc∈X̂.
Applying Lemma 3.1 to S=I-A, we have the following results.Theorem 3.6.
Assume thatX̂≠∅ and S=I-A.(i)
If (3.2)(a) holds for all c∈X̂,  then X̂⊆Y(S)⊆(I-A)X,  and X̂ is compact. In particular, if X̂∩R+n≠∅,  then for any c∈X̂∩R+n,  (1.1) is solvable.(ii)
IfY(S,p,z,ε) defined by (3.1) is closed for all p∈Rn, c∈X̂,  and ε≥0,  then F̂ defined by (3.9) is closed, u.s.c., and u.h.c., and the subset of points at which F̂ is continuous is residual.(iii)
IfA is u.s.c., then both results of (i) and (ii) are also true.Proof.
(i) Letc∈X̂, x∈X, and p∈NX(x). If p=0, then σ#(x-Ax-c,0)=〈0,x〉. If p≠0, then x∈∂X. From (3.9), we know that (Ax+c)∩X≠∅. This implies that σb(Ax+c,p)≤σb((Ax+c)∩X,p)≤σ#(X,p)=〈p,x〉.Hence (3.2)(b) holds for all c∈X̂, and then c∈Y(S). By Lemma 3.1(i), it follows that X̂⊆Y(S)⊆SX=(I-A)X.
To prove thatX̂ is compact, we only need to prove that X̂ is closed. Suppose that {cm:m≥1}⊆X̂ such that cm→c0∈X as m→∞. Let x∈∂X. Then by (3.9), (Ax+cm)∩X≠∅ for m≥1. Hence there exists a sequence {ym:m≥1}⊆X such that ym∈Ax+cm. Therefore, ym-cm∈Ax. Because Ax is compact, there exists a convergent subsequence of {ym-cm:m≥1}. Since cm→c0 as m→∞, we may assume that  ym→y0∈X, which implies y0∈(Ax+c0)∩X. Hence c0∈X̂ and X̂ is closed. The rest of (i) is clear because X̂∩R+n⊂(I-A)X.
(ii) By assumption of (ii), the proof of (i), and Lemma3.1, we can see that X̂⊂Ỹ(S), Ỹ(S) is compact, and FS defined by (3.3) for S=I-A is closed. With the same method as in proving Theorem 3.3(ii), we can show that F̂ is closed and has all the continuous properties stated in this theorem (ii).
(iii) By assumption of (iii) andσ#((I-A)x,p)=〈p,x〉+σ#(Ax,-p), we can see that S=I-A is u.h.c., hence Y(S,p,c,ε) is closed for all p∈Rn, c∈X̂ and ε≥0. Hence both conditions of (i) and (ii) are satisfied, and therefore the result of (iii) follows. This completes the proof.


## 3.1. The First Result

The first result is similar to Theorem2.5. However, the u.s.c. assumption concerning A introduced in Section 2 has been removed. Let μ∞,X∞,δ∞(x), and F2 be defined as in (1.6), (2.2), and (2.9), respectively. Define(3.4)A∞:X→2RnbyA∞x=B¯(0,δ∞(x))forx∈X.
Applying Lemma 3.1 to S=I-(A+A∞), we have the following results.Theorem 3.3.
Assume  thatμ∞ is finite, X∞≠∅ and S=I-(A+A∞).(i)
Letc0∈X∞.  If (3.2)(a) holds for c0,  then F2c0≠∅, that is,
(3.5)∃x0∈X,∃Δc0∈Rns.t.c0+Δc0∈x0-Ax0,∥Δc∥≤σ∞(x0).(ii)
IfY(S,p,c,ε) defined by (3.1) is closed for all p∈Rn, c∈X∞,  and ε≥0,  then F2 defined by (2.9) is closed, u.s.c., and u.h.c., and the subset of points at which F2 is continuous is residual.(iii)
IfA is u.s.c., then F2 has the same properties as stated in (ii).Proof.
(i) SinceA has convex compact values, X is compact and μ∞ is finite; by (3.4) it is easy to see that S=I-(A+A∞) is a strict set-valued map from X to Rn with convex compact values. We shall prove c0∈Y(S) by showing c0 satisfying (3.2). By the assumption of (i), it is sufficient to prove that (3.2)(b) holds for c0. Indeed, we can prove that (3.2)(b) holds for any c∈X. Suppose that x∈X,p∈NX(x), and c∈X. If p=0, then σ#(Sx-c,0)=0. If p≠0, then x∈∂X by Lemma 2.2. Hence by (2.9), δ∞(x)=|Ax|∞, 0∈Ax+B¯(0,|Ax|∞), σb(Ax+A∞x,p)≤0, and thus 〈p,x〉=supy∈X〈p,y〉≥〈p,c〉≥σb(Ax+A∞x,p)+〈p,c〉, that is, σ#(Sx-c,p)=σ#(x-Ax-A∞x-c,p)≥0. So (3.2)(b) holds for all c∈X. By Lemma 3.1(i) and X∞⊆X, we have c0∈Y(S)⊆SX=⋃x∈X(x-Ax-A∞x). 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 c∈X. Combining this with the assumption of (ii) and using (3.3)(a), we obtain that X∞⊆Ỹ(S). So Ỹ(S) is nonempty and also compact by Lemma 3.1(ii). In view of Lemmas 1.3–1.6 and Remark 1.5, to complete the proof of (ii), it is enough to show that F2 is closed since X∞ and X are compact. Suppose that {(cm,xm):m≥1}⊆graphF2 with (cm,xm)→(c0,x0)(m→∞). Then c0∈X∞⊂Ỹ(S), and for each m≥1 there exists Δcm∈B¯(0,δ∞(xm)) such that cm+Δcm∈xm-Axm. By graphF2⊆X∞×X⊆Ỹ(S)×X and (3.4), we can see that cm∈Sxm, that is, (cm,xm)∈graphFS, where FS is defined by (3.3) for S=I-(A+A∞). By Lemma 3.1(ii), FS is closed, and thus (c0,x0)∈graphFS, which implies that x0∈X and c0∈Sx0=(I-A-A∞)x0. So we can select Δc0∈B¯(0,δ∞(x0)) such that c0+Δc0∈x0-Ax0, that is, (c0,x0)∈graphF2. Hence F2 is closed.
(iii) SinceA is u.s.c, by the proof of Theorem 2.5, we have known that the function x↦δ∞(x) is u.s.c. on X. Combining this with (3.4), it is easy to see that the set-valued map A∞ is u.s.c., and, by Lemma 1.4, also u.h.c.. Thus for each p∈Rn, the function x↦σ#(Sx,p)=〈p,x〉+σ#(Ax,-p)+σ#(A∞x,-p) is u.s.c. on X. This implies that S=I-(A+A∞) is u.h.c. on X. Hence by Remark 3.2 and X∞⊆X, the condition of (ii) is true. This completes the proof.


## 3.2. The Second Result

Next we use a new numberμ0, a new set X0, and a new set-valued map F0 from X0 to X to obtain further solvability and stability result for (1.1). Set(3.6)|Ax|0=infz∈Ax∥z∥,(x∈X),μ0=supx∈∂X|Ax|0,δ0(x)={0ifx∈intX,|Ax|0ifx∈∂X,X0=X⋂[μ0,+∞)n,
and define A0:X→2Rn and F0:X0→2X by(3.7)A0x=B¯(0,δ0(x))forx∈X,F0c={x∈X:∃Δc∈Rns.t.c+Δc∈x-Ax,∥Δc∥≤δ0(x)}forc∈X0.
Applying Lemma 3.1 to S=I-(A+A0), we have the following results.Theorem 3.4.
Assume thatμ0 is finite, X0≠∅,  and S=I-(A+A0).(i)
Letc0∈X0.  If (3.2)(a) holds for c0,  then F0c0≠∅,  that is,
(3.8)∃x0∈X,∃Δc0∈Rns.t.c0+Δc0∈x0-Ax0,∥Δc0∥≤δ0(x).(ii)
IfY(S,p,c,ε) defined by (3.1) is closed for all p∈Rn, c∈X0 and ε≥0,  then F0 defined by (3.7) is closed, u.s.c., and u.h.c., and the subset of points at which F0 is continuous is residual.(iii)
IfA is u.s.c. on X and also l.s.c. on ∂X,  then the result of (ii) for F0 is also true.Proof.
(i) It is easy to see thatS 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 c∈X. Thus we can follow the proof of Theorem 3.3(i) to obtain the conclusion of (i). In fact, let x∈X, p∈NX(x), and c∈X. If p=0, then σ#(Sx-c,p)=0. If p≠0, then x∈∂X. So δ0(x)=|Ax|0, 0∈Ax+B¯(0,δ0(x)) and σb(Ax+A0x,p)≤0. This implies 〈p,x〉=supy∈X〈p,y〉≥〈p,c〉≥σb(Ax+A0x,p)+〈p,c〉, that is, σ#(Sx-c,p)=σ#(x-Ax-A0x-c,p)≥0. Hence, (3.2)(b) holds for all c∈X, and by Lemma 3.1(i), c0∈Y(S)⊆SX=⋃x∈X(x-Ax-A0x). Thus (3.8) is true.
(ii) The proof of (ii) is similar to that of Theorem3.3(ii). Indeed, by the proof of (i) and the assumption of (ii), we can see that X0⊆Ỹ(S), and thus Ỹ(S) is compact by Lemma 3.1(ii). If {(cm,xm):m≥1}⊆graphF0 such that (cm,xm)→(c0,x0) as m→∞, then c0∈X0⊂Ỹ(S),x0∈X, and there exists Δcm∈B¯(0,δ0(xm)) with cm+Δcm∈xm-Axm. Hence by (3.3) and (3.7), (cm,xm)∈graphFS, where FS is defined by (3.3) for S=I-(A+A0). Also by Lemma 3.1(ii), we have (c0,x0)∈graphFS, that is, c0∈(I-A-A0)x0. So we can select Δc0∈B¯(0,δ0(x0)) such that c0+Δc0∈x0-Ax0. Therefore, (c0,x0)∈graphF0 and F0 is closed.
(iii) Similar to the proof of Theorem3.3(iii), it suffices to show that S=I-(A+A0) is u.h.c.. For x∈intX and p∈Rn, we have A0x={0}. In this case, since σ#(Sx,p)=〈x,p〉+σ#(Ax,-p) holds for any x∈intX and A is u.h.c., we obtain that S is u.h.c. in intX. Now, we prove that S is also u.h.c. at any point x0∈∂X. Because σ#(Sx,p)=〈x,p〉+σ#(Ax,-p)+σ#(A0x,-p) holds for any x∈X and p∈Rn, it is sufficient to show that A0 is u.h.c. at x0∈∂X. Since Ax0 is compact, there is z0∈Ax0 such that ∥z0∥=|Ax0|0=δ0(x0). As A is l.s.c. at x0 by the assumption, for ε>0, there exists an open neighborhood U(x0) of x0 such that Ax∩B(z0,ε)≠∅ for all x∈X∩U(x0). This implies that for each x∈X∩U(x0),δ0(x)≤|Ax|0≤infy∈Ax∩B(z0,ε)∥y∥≤supy∈B(z0,ε)∥y∥≤δ0(x0)+ε, and hence B¯(0,δ0(x))⊆B¯(B¯(0,δ0(x0)),ε) for all x∈X∩U(x0). So we conclude that A0 is u.s.c. at x0 and, by Lemma 1.4, also u.h.c.. Therefore, S=I-(A+A0) is u.h.c. on X, and the conclusion of (iii) follows. This completes the proof.Remark 3.5.
As in the previous theorems, the assumptionsc0∈X∞  in Theorem 3.3 and c0∈X0  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 ofA at the boundary ∂X of X to consider the solvability and stability of (1.1). Define F̂:X̂→2X by(3.9)X̂={c∈X:∀x∈∂X,(Ax+c)∩X≠∅},F̂c={x∈X:c∈x-Ax}forc∈X̂.
Applying Lemma 3.1 to S=I-A, we have the following results.Theorem 3.6.
Assume thatX̂≠∅ and S=I-A.(i)
If (3.2)(a) holds for all c∈X̂,  then X̂⊆Y(S)⊆(I-A)X,  and X̂ is compact. In particular, if X̂∩R+n≠∅,  then for any c∈X̂∩R+n,  (1.1) is solvable.(ii)
IfY(S,p,z,ε) defined by (3.1) is closed for all p∈Rn, c∈X̂,  and ε≥0,  then F̂ defined by (3.9) is closed, u.s.c., and u.h.c., and the subset of points at which F̂ is continuous is residual.(iii)
IfA is u.s.c., then both results of (i) and (ii) are also true.Proof.
(i) Letc∈X̂, x∈X, and p∈NX(x). If p=0, then σ#(x-Ax-c,0)=〈0,x〉. If p≠0, then x∈∂X. From (3.9), we know that (Ax+c)∩X≠∅. This implies that σb(Ax+c,p)≤σb((Ax+c)∩X,p)≤σ#(X,p)=〈p,x〉.Hence (3.2)(b) holds for all c∈X̂, and then c∈Y(S). By Lemma 3.1(i), it follows that X̂⊆Y(S)⊆SX=(I-A)X.
To prove thatX̂ is compact, we only need to prove that X̂ is closed. Suppose that {cm:m≥1}⊆X̂ such that cm→c0∈X as m→∞. Let x∈∂X. Then by (3.9), (Ax+cm)∩X≠∅ for m≥1. Hence there exists a sequence {ym:m≥1}⊆X such that ym∈Ax+cm. Therefore, ym-cm∈Ax. Because Ax is compact, there exists a convergent subsequence of {ym-cm:m≥1}. Since cm→c0 as m→∞, we may assume that  ym→y0∈X, which implies y0∈(Ax+c0)∩X. Hence c0∈X̂ and X̂ is closed. The rest of (i) is clear because X̂∩R+n⊂(I-A)X.
(ii) By assumption of (ii), the proof of (i), and Lemma3.1, we can see that X̂⊂Ỹ(S), Ỹ(S) is compact, and FS defined by (3.3) for S=I-A is closed. With the same method as in proving Theorem 3.3(ii), we can show that F̂ is closed and has all the continuous properties stated in this theorem (ii).
(iii) By assumption of (iii) andσ#((I-A)x,p)=〈p,x〉+σ#(Ax,-p), we can see that S=I-A is u.h.c., hence Y(S,p,c,ε) is closed for all p∈Rn, c∈X̂ and ε≥0. 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 3.3–3.6) in which the continuity assumption regarding the set-valued map in this inclusion is no longer needed.



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*Source: 572451-2010-06-06.xml*