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A set-valued type inequality system
Nonlinear Analysis 69 (2008) 4131–4142 www.elsevier.com/locate/na
A set-valued type inequality system Yingfan Liu College of Mathematics and Physics, Nanjing University of Post and Telecommunications, Nanjing 210009, PR China Received 29 August 2007; accepted 24 October 2007
Abstract A set-valued type inequality system is introduced along with two solvability questions composed of existence and perturbation, and main theorems are obtained, which include three necessary and sufficient conditions concerning the existence and a continuity property concerning the perturbation. As applications, two existence criteria with respect to a single-valued type inequality system have also been obtained. c 2007 Elsevier Ltd. All rights reserved.
MSC: 28B15; 54C60 Keywords: Input–output; Inequality; Minimax; Saddle point; Nonlinear analysis
1. Introduction In this paper, we shall study a class of set-valued type inequality systems arising from considering some models of the input–output analysis. Before introducing the inequality and corresponding solvability questions, we need to review some background material. Denote by R n the n-dimensional Euclidean space supplied with the Euclidean norm k · k, R n∗ = R n its dual, and n the set of all nonnegative vectors of R n , we need h·, ·i the duality pairing on space pair hR n∗ , R n i. Also denote as R+ n , ⇐⇒ n to use the semi-ordering in R defined by x = (x1 , x2 , . . . , xn ) ≤ y = (y1 , y2 , . . . , yn ) ⇐⇒ y − x ∈ R+ xi ≤ yi (1 ≤ i ≤ n). In the classical input–output analysis found by Leontief [1,2], there are two types of economic models. The first is Leontief’s matrix type input–output equation n x ∈ R+ s.t. (I − A)x = x − Ax = c, n is an expected final demand of the market, R n is some enterprise’s admission output vector set, where c ∈ R+ + n to R n called the enterprise’s consuming coefficient and A = (ai j )n×n is a nonnegative linear transformation from R+ + matrix. This linear equation has been studied systematically in Ref. [1,2]. Since a series of relative nonlinear extensions have been considered by [3–6], we do not consider this here.
E-mail address: [email protected]. c 2007 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2007.10.043
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The second is the classical von Neumann’s matrix type input–output inequality n ∃ x ∈ R+ s.t. Bx − Ax = (B − A)x ≥ c,
(1.1)
m is also an expected final demand vector of the market, while R n is not the enterprise’s admission where c ∈ R+ + m×n output vector set, but the admission raw material bundle set. B =(b ˆ i j )m×n , A=(a ˆ i j )m×n ∈ R+ are two linear n to R m called the output and consuming coefficient matrices, respectively, which means that transformations from R+ + n , Bx ∈ R m is the output, Ax ∈ R m the consuming, and thus (B − A)x ∈ R m the for each raw material bundle x ∈ R+ + + enterprise’s pure output vector. The economic meaning of (1.1) is that if it has a solution x to c, then the enterprise’s pure output (B − A)x satisfies sufficiently the demand of the market c. With the matrix analysis approach, (1.1) has already been studied by [1,2]. Applying a similar method, without any difficulty one can consider the following input–output inequality which is more precise than (1.1): n ∃ x ∈ R+ , ∃ i 0 ∈ {1, 2, . . . , m} (1.2) s.t. (B − A)x ≥ c and (B − A)i0 x = ci0 .
As indicated by (1.1) and (1.2), no restrictions on the positive solution set have been made as long as each component of x is positive, and both output and consuming maps are strictly based on the linear assumptions. However, an economic system may be much more complicated than what (1.1) implies. Indeed, with the enterprise’s n due to the limitation of the material resources, and any raw material bundle set it is impossible to fill up the whole R+ enterprise’s output and consuming models can hardly be described by linear maps. So some general models might be considered and some nonlinear extensions should be made. The single-valued nonlinear extensions to both (1.1) and (1.2) are the following two nonlinear inequalities: ∃ x ∈ X, ∃ x ∈ X, ∃ i 0 ∈ {1, 2, . . . , m} (a) (b) (1.3) s.t. (g − f )(x) ≥ c, s.t. (g − f )(x) ≥ c and (g − f )i0 (x) = ci0 , m is the same as that of (1.1), while g =(g where c ∈ R+ ˆ 1 , g2 , . . . , gm ) and f =( ˆ f 1 , f 2 , . . . , f m ) are single-valued m , respectively. It means that for each raw material x ∈ X , nonlinear output and consuming maps from X to R+ (g − f )(x)=g(x) ˆ − f (x) is the pure output vector produced by x. With some convexity and continuity assumptions, both (a) and (b) of (1.3) have been considered by [7–10] and several results making (1.3) solvable have also been presented. n is some enterprise’s admission raw material bundle set. For each x ∈ X , in the process from Suppose X ⊂ R+ x to admission outputs, the enterprise can design different types of production plans, which give rise to varieties of possible outputs, and so produces a technique output subset Gx. On the other hand, a lot of expected and unexpected factors can also bring out many types of consuming vectors, thus causing a consuming subset F x. This implies that each raw material x may produce a pure output subset (G − F)x = Gx − F x = {y = u − v | u ∈ Gx, v ∈ F x}, and therefore, yields the following set-valued type input–output inequality: ∃x ∈X (1.4) s.t. ∃ y ∈ (G − F)(x) H⇒ y ≥ c, m
where G and F, from X to 2 R+ , are said to be the set-valued type output and consuming maps, respectively. If we take h = g − f in (1.3) and T = G − F in (1.4), then we obtain two inequality systems as follows: ∃x ∈X ∃ x ∈ X, ∃ i 0 ∈ {1, 2, . . . , m} (a) (b) s.t. h(x) ≥ c, s.t. h(x) ≥ c and h i0 (x) = ci0
(1.5)
and
∃x ∈X m s.t. ∃ y ∈ T (x) H⇒ y ≥ c, i.e., T x ∩ (c + R+ ) 6= ∅
(1.6)
n is a nonempty subset, c ∈ R m a nonnegative vector, h in (1.5) is a single-valued nonlinear map from where X ⊂ R+ + m m ), T in (1.6) a set-valued map from X to 2 R m (not necessarily to 2 R+ m X to R (not necessarily to R+ ) with nonempty
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m solves (1.6) (resp. (1.5)) if c makes (1.6) (resp. (1.5)) values, and n may not be equal to m. We claim that c ∈ R+ solvable. Taking (1.5) as the special instance of (1.6), we shall devote our consideration to studying the solvability of (1.6) m ) 6= ∅}. Since for each c ∈ R m , c solves (1.6) and x ∈ X is the in the sequel. Define E(c) = {x ∈ X | T x ∩ (c + R+ + corresponding solution if and only if x ∈ E(c), we introduce the solvability questions concerning (1.6) composed of existence and perturbation as follows. m ; does it solve (1.6)? Can we present any sufficient conditions, or if possible, any Question 1 (Existence). Let c ∈ R+ necessary and sufficient conditions?
Question 2 (Perturbation). If c solves (1.6) and is made by some small perturbations, what kinds of changes will be made for E(c)? Just as in [9,10], in order to analyse the solvability questions for both (1.5) and (1.6), some proper conditions are needed. From the viewpoint of economics, the raw material bundle set X and the pure output value T x (x ∈ X ) are generally bounded subsets due to both restrictions of the enterprise’s production ability and the demand of the market. According to the mixed strategy idea of game theory [11], we may consider X and T x (x ∈ X ) to be self-closed with respect to convex combinations and limits. So we introduce two assumptions as follows. n is a nonempty convex compact subset. Assumption 1. X ⊂ R+ m
Assumption 2. T : X −→ 2 R is a set-valued map with nonempty convex compact values. With both assumptions and some additional continuity conditions, and by applying the nonlinear analysis approaches attributed to [11–13], we have made some tentative progress as regards (1.6) and obtained two solvability theorems, which include three necessary and sufficient conditions concerning Question 1, and a continuity property concerning Question 2. As applications, two existence criteria concerning (1.5) have also been obtained. The paper is organized as follows. We introduce some concepts, definitions and some known results described by lemmas in Section 2, present and prove six lemmas in Section 3, then use these lemmas to obtain the main theorems concerning (1.6) in Section 4, and as applications, to show the solvability results concerning (1.5) in Section 5. Finally, we present the conclusion in Section 6. 2. Terminology 2.1. Some concepts Let U and V be two sets, X ⊆ U a nonempty subset. Let f : X −→ R be a real function and S : X −→ 2V be a set-valued map, namely, a correspondence, whose graph is the subset Graph S = {(x, y) ∈ X × V | y ∈ Sx}. In the sequel, besides some well known concepts such as f being convex, concave, upper semi-continuous (for short, u.s.c.), lower semi-continuous (for short, l.s.c.) and continuous when U is a Hausdorff locally convex vector space and X the closed convex subset of U , we need to use a number of concepts concerning S, whose definitions are given below for convenience of reference. For more information pertaining to these and some other related concepts, the reader may consult [11–13]. Definition 2.1 ([11–13]). (i) S is said to be strict if Sx 6= ∅ for each x ∈ X . (ii) S is said to be convex if both U and V are linear vector spaces, X is a convex subset of U and αSx (1) + (1 − α)Sx (2) ⊂ S[αx (1) + (1 − α)x (2) ] hold for all α ∈ [0, 1] and x (i) ∈ X (i = 1, 2). (iii) When U and V are Hausdorff topological spaces: (1) S is said to be closed if Graph S is closed in U × V (in particular, if X is a closed subset of U and Graph S is a closed subset of X × V ). (2) S is said to be u.s.c. at x0 ∈ X if for any neighborhood N (Sx0 ) of Sx0 , there exists a neighborhood N (x0 ) of x0 such that ∀ x ∈ N (x0 ), Sx ⊂ N (Sx0 ). (3) S is said to be l.s.c. at x0 ∈ X if for any y0 ∈ Sx0 and any neighborhood N (y0 ) of y0 , there exists a neighborhood N (x0 ) of x0 such that ∀x ∈ N (x0 ), Sx ∩ N (y0 ) 6= ∅.
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(4) S is said to be continuous at x0 ∈ X if it is both u.s.c. and l.s.c. at x0 . (5) S is said to be u.s.c., l.s.c. or continuous on X if it is u.s.c., l.s.c. or continuous at all points x ∈ X . (6) In the case where V is a Hausdorff locally convex vector space (V ∗ its dual) supplied with the weak topology σ (V, V ∗ ), then S : X −→ 2V is said to be u.h.c. at x0 ∈ X if for each p ∈ V ∗ , the upper supporting function X −→ R : x 7−→ σ ] (Sx, p) = sup y∈Sx h p, yi is u.s.c. at x0 , and S to be u.h.c. on X if it is u.h.c. at all points x ∈ X . Definition 2.2 ([11–13]). Let U, V be two Hausdorff locally convex vector spaces, P ⊂ U , X ⊂ V , and let ϕ( p, x) be a function defined on P × X . (i) If inf sup ϕ( p, x) = sup inf ϕ( p, x),
p∈P x∈X
x∈X p∈P
(2.1)
and we let v(ϕ)= ˆ inf p∈P supx∈X ϕ( p, x) = supx∈X inf p∈P ϕ( p, x), then we say that the balanced value v(ϕ) of ϕ exists. (ii) If sup ϕ( p, ¯ x) = inf ϕ( p, x), ¯ x∈X
p∈P
(2.2)
then we call ( p, ¯ x) ¯ the saddle point of ϕ, and denote as S(ϕ) the set of all saddle points. Remark 2.1. From the definition, it is easy to see that: (1) (2.1) holds (i.e., v(ϕ) exists) iff inf p∈P supx∈X ϕ( p, x) ≤ supx∈X inf p∈P ϕ( p, x). (2) (2.2) holds for ( p, ¯ x) ¯ (i.e., ( p, ¯ x) ¯ ∈ S(ϕ)) iff supx∈X ϕ( p, ¯ x) ≤ inf p∈P ϕ( p, x). ¯ (3) If S(ϕ) 6= φ, then (2.1) holds, and for each ( p, ¯ x) ¯ ∈ S(ϕ), v(ϕ) = ϕ( p, ¯ x) ¯ = supx∈X ϕ( p, ¯ x) = inf p∈P ϕ( p, x). ¯ Definition 2.3 ([14]). Let U be a real vector space, A ⊂ U a convex set and E a subset of A, we claim that E is an extremal subset of A if x, y ∈ A and t x + (1 − t)y ∈ E for some t ∈ (0, 1) entails x, y ∈ E, x0 ∈ A is an extremal point of A if E = {x0 } is an extremal subset of A, and the set of all extremal points of A is denoted by ExtA. 2.2. Some known results To prove the main theorems in this paper, we also need some known results as follows. Lemma 2.1 ([14]). For each finite dimensional normed linear space U , the norm topology coincides with the weak topology. Lemma 2.2 ([11]). Let U and V be two Hausdorff topological spaces and S be a strict set-valued map. (1) If S is u.s.c. with closed values, then S is closed. (2) If V is compact and S is closed, then S is u.s.c. Lemma 2.3 ([11]). Let U be a Hausdorff topological space, V a Hausdorff locally convex vector space supplied with the weak topology σ (V, V ∗ ), and S be a strict u.s.c. set-valued map. Then S is u.h.c. Lemma 2.4 ([11]). Let U be a complete metric space, V a compact metric space, and let S : U → 2V be a strict closed set-valued map. Then the subset of points at which S is continuous is residual, i.e., the interior of the discontinuous point set of S is empty. Lemma 2.5 ([12]). Let U be a Hausdorff locally convex vector space, X ⊂ U a closed convex subset, and let f α : X −→ R(α ∈ Λ) be a family of functions. (1) If X (2) If X
x 7−→ −→ R : x 7−→ −→ R :
f α (x) is convex (resp., l.s.c.) for each α ∈ Λ and supα∈Λ f α (x) is finite for each x ∈ X , then x 7−→ supα∈Λ f α (x) is also a convex (resp., a l.s.c.) function. f α (x) is concave (resp., u.s.c.) for each α ∈ Λ and infα∈Λ f α (x) is finite for each x ∈ X , then x 7−→ infα∈Λ f α (x) is also a concave (resp., an u.s.c.) function.
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Lemma 2.6 ([13]). Let U be a metric space, V a compact metric space and let g : U × V −→ R be a l.s.c. (or an u.s.c.) function; then f : U −→ R defined by f (x) = inf y∈V g(x, y) (or h : U −→ R defined by h(x) = sup y∈V g(x, y)) is also a l.s.c. (or an u.s.c.) function. Lemma 2.7 ([11,12]). Let U, V be two Hausdorff locally convex vector spaces, P ⊂ U, X ⊂ V be convex compact subsets, and let ϕ( p, x) be a function defined on P × X . If ∀x ∈ X, p 7−→ ϕ( p, x) is convex and l.s.c., and ∀ p ∈ P, x 7−→ ϕ( p, x) is concave and u.s.c., then inf p∈P supx∈X ϕ( p, x) = supx∈X inf p∈P ϕ( p, x) and there exists ( p, ¯ x) ¯ ∈ P × X such that supx∈X ϕ( p, ¯ x) = inf p∈P ϕ( p, x). ¯ Lemma 2.8 ([14]). Let U be a Hausdorff locally convex vector space and X be its nonempty convex compact subset. Then Ext X 6= ∅ and co(E xt X ) = X (i.e., the closed convex hull of E xt X is equal to X ). 3. Some lemmas Let X, T be given as in (1.6). Assume: (1) both assumptions in section 1 are satisfied, m (2) T : X −→ 2 R is convex and u.s.c., m m (3) P ⊂ R+ is convex compact with R+ P = R+ ,
(3.1)
m × P × X by and define a function (c, p, x) 7−→ φc ( p, x) on R+
φc ( p, x) = sup h p, y − ci,
m c ∈ R+ , p ∈ P, x ∈ X,
(3.2)
y∈T x m , φ =φ then for each c ∈ R+ c ˆ c (·, ·) is a function on P × X . Just as stated in Definition 2.2, we claim v(φc ) (or S(φc )) the balanced value (or the saddle point set) of φc if (2.1) (or if there exists ( p, ¯ x) ¯ ∈ P × X such that (2.2)) holds for ϕ = φc . For the following six lemmas except for Lemma 3.6, we always suppose (3.1) holds and φc ( p, x) is defined by (3.2).
Lemma 3.1. (i) T is closed and u.h.c. (ii) T X is a convex compact set. m , sup (iii) For each c ∈ R+ p∈P,x∈X |φc ( p, x)| < +∞. Thus φc ( p, x)(( p, x) ∈ P × X ), supx∈X φc ( p, x)( p ∈ P) and inf p∈P φc ( p, x)(x ∈ X ) are all finite. Proof. (i) can be deduced from Lemmas 2.1 to 2.3, and both (ii) and (iii) can be proved by the definitions.
m be fixed, then: Lemma 3.2. Let c ∈ R+
(i) For each x ∈ X , P −→ R : p 7−→ φc ( p, x) is l.s.c. and convex. (ii) For each p ∈ P, X −→ R : x − 7 → φc ( p, x) is u.s.c. and concave. Proof. (i) Since P −→ R : p 7−→ h p, y − ci is continuous, affine and φc ( p, x) = sup y∈T x h p, y − ci, then (i) follows immediately from the Lemma 3.1(iii) and Lemma 2.5(1). (ii) For each α ∈ [0, 1], x (1) , x (2) ∈ X , by the convexity of T , we have T [αx (1) +(1−α)x (2) ] ⊃ αT x (1) +(1−α)T x (2) , which implies that φc [ p, αx (1) + (1 − α)x (2) ] = =
sup y (1) ∈T x (1) ,y (2) ∈T x (2)
sup
h p, y − ci ≥
y∈T [αx (1) +(1−α)x (2) ]
sup
h p, y − ci
y∈αT x (1) +(1−α)T x (2)
h p, α(y (1) − c) + (1 − α)(y (2) − c)i = αφc ( p, x (1) ) + (1 − α)φc ( p, x (2) ).
Hence, x 7−→ φc ( p, x) is concave.
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Suppose x (k) −→ x (0) ∈ X as k −→ ∞. Since T x (k) is compact, there exists y (k) ∈ T x (k) such that φc ( p, x (k) ) = sup y∈T x (k) h p, y − ci = h p, y (k) − ci. As T X is compact and T is closed by Lemma 3.1(i) (ii), choosing a subsequence and rebelling as necessary, we may suppose lim supk−→∞ h p, y (k) − ci = limk−→∞ h p, y (k) − ci and y (k) −→ y (0) ∈ T x (0) . It follows that lim supk−→∞ φc ( p, x (k) ) = h p, y (0) − ci ≤ sup y∈T x (0) h p, y − ci = φc ( p, x (0) ). Hence x −→ φc ( p, x) is u.s.c.. This completes the proof. m , we have Lemma 3.3. For each c ∈ R+
(i) P −→ R : p 7−→ sup φc ( p, x) is l.s.c. and convex, x∈X
(3.3)
X −→ R : x 7−→ inf φc ( p, x) is u.s.c. and concave. p∈P
(ii) The balanced value v(φc ) (of φc ) exists. (iii) The saddle point set S(φc ) (of φc ) is a nonempty convex compact subset of P × X . Proof. By Lemma 3.1(iii), supx∈X φc ( p, x)( p ∈ P) and inf p∈P φc ( p, x) (x ∈ X ) are all finite. Hence, (i) follows immediately from Lemmas 2.5 and 3.2. Associating Lemma 3.2 with Lemma 2.7, we conclude that (ii) holds and S(φc ) is nonempty. It remains to show that S(φc ) is closed and convex. Assume α ∈ [0, 1], ( p (i) , x (i) ) ∈ S(φc ) (i = 1, 2), then sup φc ( p (i) , x) = inf φc ( p, x (i) ).
(3.4)
p∈P
x∈X
Associating (3.3) with (3.4) we have sup φc [αp (1) + (1 − α) p (2) , x] ≤ α sup φc ( p (1) , x) + (1 − α) sup φc ( p (2) , x) x∈X
x∈X
= α inf φc ( p, x p∈P
(1)
) + (1 − α) inf φc ( p, x p∈P
(2)
x∈X (1)
) ≤ inf φc [ p, αx p∈P
+ (1 − α)x (2) ].
Combining this with Remark 2.1 (2), we conclude that α( p (1) , x (1) ) + (1 − α)( p (2) , x (2) ) ∈ S(φc ), and thus S(φc ) is convex. Suppose ( p (k) , x (k) ) ∈ S(φc ) such that ( p (k) , x (k) ) −→ ( p (0) , x (0) ) ∈ P × X as k −→ ∞, then also by (3.3), (3.4) (where we replace i by k) and Remark 2.1 (2) we obtain sup φc ( p (0) , x) ≤ lim inf[sup φc ( p (k) , x)] ≤ lim sup[ inf φc ( p, x (k) )] ≤ inf φc ( p, x (0) ) k−→∞ x∈X
x∈X
k−→∞ p∈P
and ( p (0) , x (0) ) ∈ S(φc ). Thus S(φc ) is closed and the proof is completed.
p∈P
m ), i.e., c 7−→ v(φ ) is continuous on R m . Lemma 3.4. v(φc ) ∈ C(R+ c +
Proof. First we prove m (a) R+ × P −→ R : (c, p) 7−→ sup φc ( p, x) is l.s.c.; x∈X
m × X −→ R : (c, x) 7−→ inf φc ( p, x) is u.s.c. (b) R+
(3.5)
p∈P
m × P, sup (a) Since for each (c, p) ∈ R+ x∈X φc ( p, x) = sup y∈T X h p, y − ci is finite, and for each y ∈ T x, (c, p) 7−→ h p, y − ci is continuous (and hence, l.s.c.), then from Lemma 2.5(1) we know that (3.5)(a) is true. (b) By (3.2), we have inf p∈P φc ( p, x) = inf p∈P sup y∈T x h p, y − ci. Since p 7−→ h p, y − ci is l.s.c. convex on P, y 7−→ h p, y − ci is u.s.c. concave on T x, and both P and T x are convex compact, Lemma 2.7 implies
inf φc ( p, x) = sup inf h p, y − ci.
p∈P
y∈T x p∈P
(3.6)
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m × X , and denote by B m (0, ε) the ε-open neighborhood of 0 in R m ; then for each ε > 0, Suppose (c(0) , x (0) ) ∈ R+ R there exists δ (> 0), an open neighborhood B R n (0, δ) of 0 in R n , such that T x ⊂ T x (0) + B R m (0, ε) holds for all x ∈ x (0) + B R n (0, δ). Associating this with (3.6), it follows that
∀c ∈ c(0) + B R m (0, ε), ∀x ∈ x (0) + B R n (0, δ), inf φc ( p, x) = sup inf h p, y − ci ≤ sup p∈P
y∈T x p∈P
=
sup
inf h p, y − ci
y∈T x (0) +B R m (0,ε) p∈P
sup
inf [h p, y − ci + h p, zi]
y∈T x (0) z∈B R m (0,ε) p∈P
≤
sup y∈T x (0)
≤
inf [h p, y − c(0) i + h p, c(0) − ci] +
p∈P
sup
inf h p, y − c(0) i + 2 sup k pkε
y∈T x (0) p∈P
= inf φc(0) ( p, x p∈P
sup
sup |h p, zi|
z∈B R m (0,ε) p∈P
p∈P (0)
) + 2 sup k pkε. p∈P
Hence, (3.5)(b) holds also. By (3.5) and applying Lemma 2.6, we get m R+ −→ R : c 7−→ inf sup φc ( p, x) is l.s.c., p∈P x∈X
m −→ R : c 7−→ sup inf φc ( p, x) is u.s.c. R+ x∈X p∈P
m. So we conclude from Lemma 3.3(ii) that the function c 7−→ v(φc ) is continuous on R+
m −→ R : c −→ v(φ ) is a bounded above and continuous Lemma 3.5. (i) v = ˆ supc∈R+m v(φc ) is finite (and hence, R+ c function), and there exists a > 0 with v = supc∈[0,a]m v(φc ). m is a nonempty closed subset, then S : C −→ 2 P×X : c 7−→ S(φ ) is a strict u.s.c. and (ii) If C ⊂ R+ c u.h.c. correspondence with convex compact values, and the set of points at which S is continuous is residual. m , then for each i = 1, 2, . . . , m there exists d (i) = Proof. (i) Since R+ P = R+ i > 0 such that d (0, . . . , 0, di , 0, . . . , 0) ∈ P. Define y = (y1 , y2 , . . . , ym ), c = (c1 , c2 , . . . , cm ), d = (d1 , d2 , . . . , dm ), b = kdk · sup y∈T X kyk and d0 = min1≤i≤m di , we have
m inf sup h p, y − ci ≤ min sup hd (i) , y − ci ∀c ∈ R+ , v(φc ) = p∈P 1≤i≤m y∈T X y∈T X = min sup (di yi − di ci ) ≤ sup kdk · kyk − max di ci 1≤i≤m y∈T X 1≤i≤m y∈T X kck d 0 ≤ b − d0 max ci ≤ b − √ . 1≤i≤m m
(3.7)
m \ {0} be fixed, then from (3.7) we obtain that v(φ ) < v(φ m Let c(0) ∈ R+ c c(0) ) if c ∈ R+ satisfies kck > a = √ m[b−v(φ
(0) )]
c max{ , kc(0) k}, which implies that v = ˆ supc∈R+m v(φc ) is finite and v = supc∈[0,a]m v(φc ). Combining this d0 with Lemma 3.4, we know that (i) follows. (ii) By Lemma 3.3(iii), S is clearly a strict correspondence with convex compact values. In order to obtain the continuous properties of S stated in (ii), it is sufficient to prove that S is closed thanks to Lemmas 2.1–2.4. m × (P × X ) as Suppose (c(k) , ( p (k) , x (k) )) ∈ graph S is such that (c(k) , ( p (k) , x (k) )) −→ (c(0) , ( p (0) , x (0) )) ∈ R+ (0) (k) (k) k −→ ∞; then c ∈ C and for all k ≥ 1, ( p , x ) ∈ S(φc(k) ), i.e.,
sup φc(k) ( p (k) , x) = inf φc(k) ( p, x (k) ),
x∈X
p∈P
k ≥ 1.
(3.8)
Since φc(k) ( p (k) , x) = sup y∈T x h p (k) , y − c(k) i = φc(0) ( p (k) , x) + h p (k) , c(0) − c(k) i and φc(k) ( p, x (k) ) = sup y∈T x (k) h p, y − c(k) i = φc(0) ( p, x (k) ) + h p, c(0) − c(k) i, then
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(a) sup φc(k) ( p (k) , x) = sup φc(0) ( p (k) , x) + h p (k) , c(0) − c(k) i, x∈X x∈X (b) inf φc(k) ( p, x (k) ) = inf [φc(0) ( p, x (k) ) + h p, c(0) − c(k) i] p∈P p∈P ≤ inf φc(0) ( p, x (k) ) + sup k pk · kc(0) − c(k) k p∈P
p∈P
with limk−→∞ h p (k) , c(0) − c(k) i = 0 and limk−→∞ kc(0) − c(k) k = 0. It follows that (k) (k) lim inf[sup φc(k) ( p , x)] = lim inf[sup φc(0) ( p , x)], k−→∞ x∈X
k−→∞ x∈X
(3.9)
lim sup[ inf φc(k) ( p, x (k) )] ≤ lim sup[ inf φc(0) ( p, x (k) )]. k−→∞ p∈P
k−→∞ p∈P
Let k −→ ∞. By (3.8), (3.9) and applying Lemma 3.3(i), we obtain sup φc(0) ( p (0) , x) ≤ lim inf[sup φc(0) ( p (k) , x)] = lim inf[sup φc(k) ( p (k) , x)] k−→∞ x∈X
x∈X
k−→∞ x∈X
≤ lim sup[ inf φc(k) ( p, x k−→∞ p∈P
(k)
)] ≤ lim sup[ inf φc(0) ( p, x (k) )] ≤ inf φc(0) ( p, x (0) ). k−→∞ p∈P
p∈P
This implies that supx∈X φc(0) ( p (0) , x) = inf p∈P φc(0) ( p, x (0) ), and thus ( p (0) , x (0) ) ∈ S(φc(0) ). Therefore, S is closed and the statement (ii) follows. m such that Z − R m is a closed convex set and R Q = R m , then Z ∩ R m 6= φ Lemma 3.6. If Z ⊂ R m and Q ⊂ R+ + + + + m (i.e., 0 ∈ Z − R+ ) if and only if
∀q ∈ Q,
suphq, zi ≥ 0,
i.e., inf suph p, zi ≥ 0.
z∈Z
q∈Q z∈Z
(3.10)
m and Z ∩ R m 6= ∅. So we only need to Proof. The necessary part follows immediately from the conditions Q ⊂ R+ + m m prove that 0 ∈ Z − R+ if (3.10) holds. As Z − R+ is closed and convex, by the Hahn–Banach separation theorem, it is enough to show that
∀q ∈ R m \ {0},
sup h p, ui = suphq, zi − infm hq, yi ≥ 0.
m u∈Z −R+
z∈Z
(3.11)
y∈R+
m , then inf m hq, yi = −∞, and hence (3.11) is true. If q 6∈ R+ y∈R+ m \ {0}, then inf m hq, yi = 0. Since R+ Q = R m , we can find t > 0 such that t −1 q ∈ Q. Hence (3.10) If q ∈ R+ y∈R+ + implies supz∈Z hq, zi = t supz∈Z ht −1 q, zi ≥ 0, and thus (3.11) is also true. This completes the proof.
4. The solvability theorems of (1.6) Before state the main theorem of (1.6), we summarize the main results included in Lemmas 3.3–3.5 with respect to v(φc ) and S(φc ) as follows. m , v(φ ) exists, and c 7−→ v(φ ) is continuous on R m with v = sup m v(φc ) < Theorem 4.1. (1) For each c ∈ R+ c c c∈R+ + +∞. Furthermore, there exists a > 0 such that v = supc∈[0,a]m v(φc ). m , S(φ ) is a nonempty convex compact subset of P × X , and for each nonempty closed subset (2) For each c ∈ R+ c m C ⊂ R+ , S : C −→ 2 P×X : c 7−→ S(φc ) is a strict u.s.c. and u.h.c. correspondence, and the set of points at which S is continuous is residual.
Then the solvability result of (1.6) is the following. m solves (1.6) iff v(φ ) ≥ 0 iff ( p ∗ , x ∗ ) ∈ S(φ ) with φ ( p ∗ , x ∗ ) ≥ 0 iff ( p ∗ , x ∗ ) ∈ S(φ ) Theorem 4.2. (1) c ∈ R+ c c c c ∗ with inf p∈P φc ( p, x ) ≥ 0 or supx∈X φc ( p ∗ , x) ≥ 0. m that solves (1.6) iff v = (2) There exists at least one c ∈ R+ ˆ supc∈R+m v(φc ) ≥ 0. m ∗ (3) Suppose c ∈ R+ , then x ∈ E(c) (i.e., c solves (1.6) and x ∗ ∈ X is the corresponding solution) iff x ∗ ∈ X with inf p∈P φc ( p, x ∗ ) ≥ 0. In particular, if ( p ∗ , x ∗ ) ∈ S(φc ) with φc ( p ∗ , x ∗ ) ≥ 0, then x ∗ ∈ E(c).
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m with v(φ ) > 0, then there exists an open or closed neighborhood O(c) of c in R m such that (4) Suppose c ∈ R+ c m solves (1.6). each element of O(c) ∩ R+ m is a closed subset such that each element of C (0) solves (1.6), then E : C (0) −→ 2 X : c 7−→ (5) Suppose C (0) ⊂ R+ m 6= ∅} is a strict u.s.c. and u.h.c. correspondence with convex compact values, and E(c) = {x ∈ X |(T x − c) ∩ R+ its continuous point set is residual. m , by Lemma 3.1(ii) we can see that A =T Proof. First we prove (1). For each c ∈ R+ c ˆ X − c is convex compact, m is closed and convex. Since c solves (1.6) if and only if A ∩ R m 6= ∅ and v(φ ) = and hence Ac − R+ c c + inf p∈P supx∈X φc ( p, x) = inf p∈P supz∈T X −c h p, zi, then from Lemma 3.6 (where we take Z = Ac and Q = P), Theorem 4.1 and Remark 2.1(3), we conclude that c solves (1.6) ⇐⇒ v(φc ) ≥ 0 ⇐⇒ ( p ∗ , x ∗ ) ∈ S(φc ) with φc ( p ∗ , x ∗ ) ≥ 0 ⇐⇒ ( p ∗ , x ∗ ) ∈ S(φc ) with inf p∈P φc ( p, x ∗ ) ≥ 0 or supx∈X φc ( p ∗ , x) ≥ 0. Hence (1) follows. Next we prove (2). The necessary part follows from (1), so we prove the sufficient part. By Theorem 4.1(1), we can find a > 0 such that v = supc∈[0,a]m v(φc ) ≥ 0, so there exists a sequence {c(k) } ⊂ [0, a]m satisfying v(φc(k) ) > v − k1 ≥ − k1 . Up to the subsequence, we may assume c(k) −→ c(0) ∈ [0, a]m (k −→ ∞). As the function c 7−→ v(φc ) is continuous on [0, a]m , by taking k −→ ∞ and using statement (1), we have v(φc(0) ) ≥ 0, and hence c(0) solves (1.6). Then we prove (3). m ), H⇒ Assume x ∗ ∈ E(c), then there exists y ∗ ∈ T x ∗ such that y ∗ − c ≥ 0. This implies that for each p ∈ P(⊂ R+ ∗ ∗ ∗ φc ( p, x ) = sup y∈T x ∗ h p, y − ci ≥ h p, y − ci ≥ 0; hence inf p∈P φc ( p, x ) ≥ 0. ⇐H Suppose x ∗ ∈ X with inf p∈P φc ( p, x ∗ ) ≥ 0, by (3.2) we have
inf sup h p, y − ci = inf φc ( p, x ∗ ) ≥ 0.
p∈P y∈T x ∗
p∈P
(4.1)
Since both P and T x ∗ are convex compact, and ∀y ∈ T x ∗ , p 7−→ h p, y − ci is l.s.c. convex on P, ∀ p ∈ P, y 7−→ h p, y − ci is u.s.c. concave on T x ∗ , by (4.1) and Lemma 2.7 we obtain sup inf h p, y − ci = inf sup h p, y − ci ≥ 0.
y∈T x ∗ p∈P
p∈P y∈T x ∗
On the other hand, by Lemma 2.5(2), we know that y 7−→ inf p∈P h p, y − ci is u.s.c. on T x ∗ . So there exists y ∗ = (y1∗ , y2∗ , . . . , ym∗ ) ∈ T x ∗ such that inf h p, y ∗ − ci = sup inf h p, y − ci ≥ 0.
p∈P
y∈T x ∗ p∈P
(4.2)
As in the proof of Lemma 3.5(i), for each 1 ≤ i ≤ m, assume d (i) = (0, . . . , 0, di , 0, . . . , 0) ∈ P with di > 0. Take p = d (i) in (4.2), we obtain di (yi∗ − ci ) ≥ inf p∈P h p, y ∗ − ci ≥ 0. Therefore, yi∗ ≥ ci (1 ≤ i ≤ m 6= ∅ and hence x ∗ ∈ E(c). m), (T x ∗ − c) ∩ R+ In particular, if ( p ∗ , x ∗ ) ∈ S(φc ) with φc ( p ∗ , x ∗ ) ≥ 0, by Remark 2.1(3) we have inf p∈P φc ( p, x ∗ ) ≥ 0, and thus (3) follows. (4) can be proved with Theorem 4.1(1) and the above (1) or (2). Finally we prove (5). As in the proof of Lemma 3.5(ii), it is sufficient to prove that E(c) is a nonempty convex compact set for each c ∈ C (0) and E is closed. m ) 6= ∅} is nonempty. If Assume c ∈ C (0) , by the definition of C (0) , we know that E(c) = {x ∈ X | T x ∩ (c + R+ (i) (i) (i) (i) α ∈ [0, 1] and x ∈ E(c) (i = 1, 2), then there exists y ∈ T x such that y ≥ c (i = 1, 2). By the convexity of T , we have αy (1) +(1−α)y (2) −c ≥ 0 and αy (1) +(1−α)y (2) −c ∈ αT x (1) +(1−α)T x (2) −c ⊂ T (αx (1) +(1−α)x (2) )−c. m 6= ∅ and thus E(c) is convex. If x (k) ∈ E(c) with This implies that (T (αx (1) + (1 − α)x (2) ) − c) ∩ R+ m ) 6= ∅ and there exists y (k) ∈ T x (k) such that y (k) ≥ c. x (k) −→ x (0) ∈ X (k −→ ∞), then T x (k) ∩ (c + R+ (k) (0) Since T X is compact, we may suppose y −→ y . By taking k −→ ∞ and using the fact that T is closed, we get m ), hence E(c) is closed. Therefore, E(c) is convex and compact. y (0) ≥ c and y (0) ∈ T x (0) , i.e.,y (0) ∈ T x 0 ∩ (c + R+ (k) (k) (k) (k) Suppose (c , x ) ∈ graph E satisfy (c , x ) −→ (c(0) , x (0) )(k −→ ∞), then (c(0) , x (0) ) ∈ C (0) × X , m ) 6= ∅, and for each k ≥ 1 there exists y (k) ∈ T x (k) ∩ (c(k) + R m ). We may also suppose T x (k) ∩ (c(k) + R+ + m ). Hence, (k) (0) y −→ y (k −→ ∞). Up to the limit, we obtain y (0) ≥ c(0) , y (0) ∈ T x (0) and y (0) ∈ T x (0) ∩ (c(0) + R+ (c(0) , x (0) ) ∈ graph E, and E is closed. This completes the proof.
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m | Σ m p = 1}, then P = S m−1 satisfies the condition Remark 4.1. Suppose S m−1 = { p = ( p1 , p2 , . . . , pm ) ∈ R+ i=1 i (3) of (3.1). Hence, both Theorems 4.1 and 4.2 hold true if we take P = S m−1 in (3.1). If G, F are defined as in (1.4) and T = G − F satisfies (3.1)(1)(2), then Theorem 4.1 holds for T = G − F.
5. Solvability results for (1.5) In this section, we shall use both theorems concerning (1.6) in Section 4 to study the solvability of (1.5). Assume (a)∀i = 1, 2, . . . , m, |h i | : X → R : x 7−→ |h i (x)| is u.s.c. and concave, m (b)P m−1 ⊂ R+ is a convex compact polyhedron with ext (P m−1 ) (5.1) (i) = {d | i = 1, 2, . . . , m}, where d i = (0, . . . , 0, di , 0, . . . , 0) such that di > 0, define |h|(x) = (|h 1 (x)|, |h 2 (x)|, . . . , |h m (x)|) and define |h|c : P m−1 × X −→ R by |h|c ( p, x) = h p, |h|(x) − ci =
m X
pi (|h i (x)| − ci ), ( p, x) ∈ P m−1 × X.
(5.2)
i=1
Then we have Theorem 5.1. Let (5.1) hold and |h|c be defined by (5.2), then: m , the balanced value v(|h| ) exists and the saddle point set S(|h| ) is a nonempty convex compact (1) For each c ∈ R+ c c subset of P × X . (2) Define S(|h|c )| X = {x ∈ X | ( p, x) ∈ S(|h|c )}, then: m solves (1.5) (a) and x¯ ∈ S(|h| )| is its solution iff v(|h| ) ≥ 0 and h( x) (a) c ∈ R+ ¯ ≥ 0. c X c m solves (1.5)(b) and x ∗ ∈ S(|h| )| is its solution iff v(|h| ) = 0 and h(x ∗ ) ≥ 0. (b) c ∈ R+ c X c |h|
|h|
|h|
Proof. First we prove (1). Define Tc : X −→ R m and φc = φc (·, ·) : P m−1 × X −→ R by m Y [−|h i (x)|, |h i (x)|], x ∈ X, (a) T |h| (x) = i=1
|h| (b) φc ( p, x) =
sup h p, y − ci, y∈T |h| (x)
( p, x) ∈ P m−1 × X.
(5.3)
In view of (5.1)–(5.3), we have |h|c ( p, x) = φc|h| ( p, x),
( p, x) ∈ P m−1 × X. (i)
(5.4) (i)
(i)
Suppose α ∈ [0, 1], x (i) ∈ X (i = 1, 2) and y (i) = (y1 , y2 , . . . , ym ) ∈ T |h| (x (i) ) (i = 1, 2). By (5.1)(a) we obtain (2)
(1)
|αy j + (1 − α)y j | ≤ α|h j (x (1) )| + (1 − α)|h j (x (2) )| ≤ |h j (αx (1) + (1 − α)x (2) )|, which implies αT |h| (x (1) ) + (1 − α)T |h| (x (2) ) ⊂ T |h| (αx (1) + (1 − α)x (2) ), i.e., T |h| is convex. Suppose x ∈ X and ε > 0. Also by (5.1)(a) we can see that for each i = 1, 2, . . . , m, there exists an open neighborhood O(x) of x in R n such that |h i (x 0 )| < |h i (x)| + ε(x 0 ∈ O(x) ∩ X ) hold for all i = 1, 2, . . . , m. Hence we have m m m Y Y Y T |h| (x 0 ) = [−|h i (x 0 )|, |h i (x 0 )|] ⊂ (−|h i (x)| − ε, |h i (x)| + ε) ⊂ T |h| (x) + (−ε, ε). i=1
T |h|
i=1
i=1
Therefore, is a strict u.s.c. convex correspondence with convex compact values, and both Theorems 4.1 and 4.2 |h| |h| |h| hold for φc = φc . In particular, Theorem 4.1 and (5.4) imply that v(|h|c ) = v(φc ) exists and S(|h|c ) = S(φc ) is a m . Thus (1) follows. nonempty convex compact subset of P × X for each c ∈ R+ Then we prove (2). Since the proof of (a) is similar to but simpler than that of (b), we only prove (b). m solves (1.5)(b) and x ∗ ∈ S(|h| )| ∗ H⇒ If c ∈ R+ c X is its solution, then h(x ) ≥ c(≥ 0) (and hence, |h| ∗ m |h| ∗ Tc (x ) ∩ (c + R+ ) 6= ∅, which means that c solves (1.6) for T = T and x is its solution), and there exists
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i 0 ∈ {1, 2, . . . , m} such that h i0 (x ∗ ) = ci0 . Applying Theorem 4.2(1) to φc = φc and using (5.1), (5.4), Remark 2.1 and d (i0 ) = (0, . . . , 0, di0 , 0, . . . , 0) ∈ P m−1 (see (5.1)(b)), we obtain 0 ≤ v(φc|h| ) = v(|h|c ) =
inf h p, |h|(x ∗ ) − ci =
p∈P m−1
inf h p, h(x ∗ ) − ci ≤ di0 (h i0 (x ∗ ) − ci0 ) = 0.
p∈P m−1
Hence, v(|h|c ) = 0, and the necessary part follows. ⇐H If v(|h|c ) = 0 and ( p ∗ , x ∗ ) ∈ S(|h|c ) with h(x ∗ ) ≥ 0, by taking d (i) ∈ ext(P m−1 ) in (5.1), we have ∀i = 1, 2, . . . , m, di (h i (x ∗ ) − ci ) = |h|c (d (i) , x ∗ ) ≥
inf
p∈P m−1
|h|c ( p, x ∗ ) = v(|h|c ) = 0.
Combining this with ( p ∗ , x ∗ ) ∈ S(|h|c ), it implies that ( h i (x ∗ ) ≥ ci , i = 1, 2, . . . , m, i.e.,h(x ∗ ) ≥ c, inf h p, h(x ∗ ) − ci = inf h p, |h|(x ∗ ) − ci = v(|h|c ) = 0. p∈P m−1
(5.5)
p∈P m−1
Let P ∗ = { p ∈ P m−1 | h p, h(x ∗ ) − ci = 0}. As P m−1 → R : p 7−→ h p, h(x ∗ ) − ci be a continuous affine function, from (5.5) we know that P ∗ is a nonempty convex compact subset. Furthermore, if α ∈ (0, 1) and p (i) ∈ P m−1 (i = 1, 2) are such that αp (1) + (1 − α) p (2) ∈ P ∗ , also from (5.5) we have (i) h p , h(x ∗ ) − ci ≥ 0 (i = 1, 2), αh p (1) , h(x ∗ ) − ci + (1 − α)h p (2) , h(x ∗ ) − ci = 0, which shows that p (i) ∈ P ∗ (i = 1, 2) and P ∗ is an extremal subset of P m−1 . By Lemma 2.8, we obtain ext(P ∗ ) ⊂ ext(P m−1 ) = {d (1) , d (2) , . . . , d (m) }, and thus min1≤i≤m hd (i) , h(x ∗ ) − ci = inf p∈P m−1 h p, h(x ∗ ) − ci = 0. So there exists i 0 ∈ {1, 2, . . . , m} such that h i0 (x ∗ ) = ci0 . This completes the proof. If we consider h = (h 1 , h 2 , . . . , h m ) as some enterprise’s pure output map and this enterprise’s management is m . Thus (5.1) becomes efficient, we may suppose that h is a map from X to R+ (a) ∀i = 1, 2, . . . , m, h i : X → R+ : x 7−→ h i (x) is u.s.c. and concave , (5.6) m (b) P m−1 ⊂ R+ is a convex compact polyhedron with ext(P m−1 ) = {d (i) | i = 1, 2, . . . , m}, Define h c = h c (·, ·) : P m−1 × X −→ R by h c ( p, x) = h p, h(x) − ci =
m X
pi (h i (x) − ci ),
( p, x) ∈ P m−1 × X,
(5.7)
i=1
and m Y [−h i (x), h i (x)], (a) T h (x) = i=1
h (b) φc ( p, x) =
sup h p, y − ci, y∈T h (x)
x ∈ X, ( p, x) ∈ P m−1 × X
(5.8)
or m Y (a) T h (x) = [0, h i (x)],
x ∈ X,
i=1
h (b) φc ( p, x) =
sup h p, y − ci, y∈T h (x)
( p, x) ∈ P m−1 × X.
(5.9)
By (5.6)–(5.9) and using the same method as that of Theorem 5.1, we have Theorem 5.2. Let (5.6) hold and h c = h c (·, ·) be defined by (5.7), then: (1) The balanced value v(h c ) exists and the saddle point set S(h c ) is a nonempty convex compact subset for each m. c ∈ R+
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(2) Define S(h c )| X = {x ∈ X | ( p, x) ∈ S(h c )}, then: m solves (1.5)(a) in S(h )| iff v(h ) ≥ 0. (a) c ∈ R+ c X c m makes (1.5)(b) solvable in S(h )| iff v(h ) = 0. (b) c ∈ R+ c X c
6. Conclusion In this article, two classes of nonlinear inequality systems (i.e., (1.6) and (1.5)) are introduced along with two solvability questions, namely those of the existence and perturbation. By constructing some auxiliary functions such as |h| φc ((3.2)), |h|c ((5.2)), φc ((5.3)(b)) etc. and applying the nonlinear analysis approach, especially the minimax, saddle point theorems, the solvability results of (1.6) and (1.5) have been obtained. The main results can be summarized as follows. (i) Theorems 4.1 and 4.2, composed mainly of the necessary and sufficient criteria, have answered both questions concerning (1.6). (ii) As applications of (i), both Theorems 5.1 and 5.2 have presented the necessary and sufficient conditions as regards the existence question for (1.5), which are different from those of Ref. [9,10]. References [1] E.W. Leontief, Input–Output Analysis, Pergamon Press, Oxford, 1985 (Reprinted from the Encyclopedia of Materials Science and Engineering); Input–Output Analysis, 2nd edition, Oxford University Press, New York, 1986. [2] R. Miller, P. Blair, Input–Output Analysis: Foundations and Extensions, Prentice-Hall, Englewood Cliffs, NJ, 1985. [3] T. Fujimoto, Nonlinear model in abstract spaces, Journal of Mathematical Economics 15 (1986) 151–156. [4] Yingfan Liu, Xiaohong Chen, Some results on continuous type conditional input–output equation, Applied Mathematics and Mechanics (English edition) 25 (3) (2004) 358–366. [5] Yingfan Liu, Tianping Chen, A solvability theorem on a class of conditional input–output equation, Chinese Annals of Mathematics 25A (6) (2004) 791–798 (in Chinese). [6] Yingfan Liu, Qinghong Zhang, Rogalski–Cornet theorem and a Leontief type input–output inclusion, Nonlinear Analysis–Theory, Methods and Applications 69 (2) (2008) 425–433. [7] P. Medvegyev, A generalized existence theorem for von Neumann economic growth model, Econometrica 52 (4) (1974) 963–974. [8] C. Bidard, E. Hosoda, On consumption basket in a generalized von Neumann model, International Economic Review 28 (2) (1987) 509–519. [9] Yingfan Liu, Some results on a class of generalized Leontief conditional input output inequalities, Acta Mathematicae Applicatae Sinica 28 (3) (2005) 358–366 (in Chinese). [10] Yingfan Liu, Qinghong Zhang, Generalized input–output inequality system, Applied Mathematics and Optimization 54 (2) (2006) 189–204. [11] J.P. Aubin, I. Ekeland, Applied Nonlinear Analysis, John Wiley and Sons, New York, 1984. [12] J.P. Aubin, Mathematical Methods of Game and Economic Theory, North-Holland, Amsterdam, 1979. [13] J.P. Aubin, Optima and Equilibria, Springer-Verlag, New York, Berlin, Heidelberg, 1991. [14] R.B. Holmes, Geometric Functional Analysis and Applications, Springer-Verlag, New York, Berlin, 1974.
A set-valued type inequality system Yingfan Liu College of Mathematics and Physics, Nanjing University of Post and Telecommunications, Nanjing 210009, PR China Received 29 August 2007; accepted 24 October 2007
Abstract A set-valued type inequality system is introduced along with two solvability questions composed of existence and perturbation, and main theorems are obtained, which include three necessary and sufficient conditions concerning the existence and a continuity property concerning the perturbation. As applications, two existence criteria with respect to a single-valued type inequality system have also been obtained. c 2007 Elsevier Ltd. All rights reserved.
MSC: 28B15; 54C60 Keywords: Input–output; Inequality; Minimax; Saddle point; Nonlinear analysis
1. Introduction In this paper, we shall study a class of set-valued type inequality systems arising from considering some models of the input–output analysis. Before introducing the inequality and corresponding solvability questions, we need to review some background material. Denote by R n the n-dimensional Euclidean space supplied with the Euclidean norm k · k, R n∗ = R n its dual, and n the set of all nonnegative vectors of R n , we need h·, ·i the duality pairing on space pair hR n∗ , R n i. Also denote as R+ n , ⇐⇒ n to use the semi-ordering in R defined by x = (x1 , x2 , . . . , xn ) ≤ y = (y1 , y2 , . . . , yn ) ⇐⇒ y − x ∈ R+ xi ≤ yi (1 ≤ i ≤ n). In the classical input–output analysis found by Leontief [1,2], there are two types of economic models. The first is Leontief’s matrix type input–output equation n x ∈ R+ s.t. (I − A)x = x − Ax = c, n is an expected final demand of the market, R n is some enterprise’s admission output vector set, where c ∈ R+ + n to R n called the enterprise’s consuming coefficient and A = (ai j )n×n is a nonnegative linear transformation from R+ + matrix. This linear equation has been studied systematically in Ref. [1,2]. Since a series of relative nonlinear extensions have been considered by [3–6], we do not consider this here.
E-mail address: [email protected]. c 2007 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2007.10.043
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The second is the classical von Neumann’s matrix type input–output inequality n ∃ x ∈ R+ s.t. Bx − Ax = (B − A)x ≥ c,
(1.1)
m is also an expected final demand vector of the market, while R n is not the enterprise’s admission where c ∈ R+ + m×n output vector set, but the admission raw material bundle set. B =(b ˆ i j )m×n , A=(a ˆ i j )m×n ∈ R+ are two linear n to R m called the output and consuming coefficient matrices, respectively, which means that transformations from R+ + n , Bx ∈ R m is the output, Ax ∈ R m the consuming, and thus (B − A)x ∈ R m the for each raw material bundle x ∈ R+ + + enterprise’s pure output vector. The economic meaning of (1.1) is that if it has a solution x to c, then the enterprise’s pure output (B − A)x satisfies sufficiently the demand of the market c. With the matrix analysis approach, (1.1) has already been studied by [1,2]. Applying a similar method, without any difficulty one can consider the following input–output inequality which is more precise than (1.1): n ∃ x ∈ R+ , ∃ i 0 ∈ {1, 2, . . . , m} (1.2) s.t. (B − A)x ≥ c and (B − A)i0 x = ci0 .
As indicated by (1.1) and (1.2), no restrictions on the positive solution set have been made as long as each component of x is positive, and both output and consuming maps are strictly based on the linear assumptions. However, an economic system may be much more complicated than what (1.1) implies. Indeed, with the enterprise’s n due to the limitation of the material resources, and any raw material bundle set it is impossible to fill up the whole R+ enterprise’s output and consuming models can hardly be described by linear maps. So some general models might be considered and some nonlinear extensions should be made. The single-valued nonlinear extensions to both (1.1) and (1.2) are the following two nonlinear inequalities: ∃ x ∈ X, ∃ x ∈ X, ∃ i 0 ∈ {1, 2, . . . , m} (a) (b) (1.3) s.t. (g − f )(x) ≥ c, s.t. (g − f )(x) ≥ c and (g − f )i0 (x) = ci0 , m is the same as that of (1.1), while g =(g where c ∈ R+ ˆ 1 , g2 , . . . , gm ) and f =( ˆ f 1 , f 2 , . . . , f m ) are single-valued m , respectively. It means that for each raw material x ∈ X , nonlinear output and consuming maps from X to R+ (g − f )(x)=g(x) ˆ − f (x) is the pure output vector produced by x. With some convexity and continuity assumptions, both (a) and (b) of (1.3) have been considered by [7–10] and several results making (1.3) solvable have also been presented. n is some enterprise’s admission raw material bundle set. For each x ∈ X , in the process from Suppose X ⊂ R+ x to admission outputs, the enterprise can design different types of production plans, which give rise to varieties of possible outputs, and so produces a technique output subset Gx. On the other hand, a lot of expected and unexpected factors can also bring out many types of consuming vectors, thus causing a consuming subset F x. This implies that each raw material x may produce a pure output subset (G − F)x = Gx − F x = {y = u − v | u ∈ Gx, v ∈ F x}, and therefore, yields the following set-valued type input–output inequality: ∃x ∈X (1.4) s.t. ∃ y ∈ (G − F)(x) H⇒ y ≥ c, m
where G and F, from X to 2 R+ , are said to be the set-valued type output and consuming maps, respectively. If we take h = g − f in (1.3) and T = G − F in (1.4), then we obtain two inequality systems as follows: ∃x ∈X ∃ x ∈ X, ∃ i 0 ∈ {1, 2, . . . , m} (a) (b) s.t. h(x) ≥ c, s.t. h(x) ≥ c and h i0 (x) = ci0
(1.5)
and
∃x ∈X m s.t. ∃ y ∈ T (x) H⇒ y ≥ c, i.e., T x ∩ (c + R+ ) 6= ∅
(1.6)
n is a nonempty subset, c ∈ R m a nonnegative vector, h in (1.5) is a single-valued nonlinear map from where X ⊂ R+ + m m ), T in (1.6) a set-valued map from X to 2 R m (not necessarily to 2 R+ m X to R (not necessarily to R+ ) with nonempty
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m solves (1.6) (resp. (1.5)) if c makes (1.6) (resp. (1.5)) values, and n may not be equal to m. We claim that c ∈ R+ solvable. Taking (1.5) as the special instance of (1.6), we shall devote our consideration to studying the solvability of (1.6) m ) 6= ∅}. Since for each c ∈ R m , c solves (1.6) and x ∈ X is the in the sequel. Define E(c) = {x ∈ X | T x ∩ (c + R+ + corresponding solution if and only if x ∈ E(c), we introduce the solvability questions concerning (1.6) composed of existence and perturbation as follows. m ; does it solve (1.6)? Can we present any sufficient conditions, or if possible, any Question 1 (Existence). Let c ∈ R+ necessary and sufficient conditions?
Question 2 (Perturbation). If c solves (1.6) and is made by some small perturbations, what kinds of changes will be made for E(c)? Just as in [9,10], in order to analyse the solvability questions for both (1.5) and (1.6), some proper conditions are needed. From the viewpoint of economics, the raw material bundle set X and the pure output value T x (x ∈ X ) are generally bounded subsets due to both restrictions of the enterprise’s production ability and the demand of the market. According to the mixed strategy idea of game theory [11], we may consider X and T x (x ∈ X ) to be self-closed with respect to convex combinations and limits. So we introduce two assumptions as follows. n is a nonempty convex compact subset. Assumption 1. X ⊂ R+ m
Assumption 2. T : X −→ 2 R is a set-valued map with nonempty convex compact values. With both assumptions and some additional continuity conditions, and by applying the nonlinear analysis approaches attributed to [11–13], we have made some tentative progress as regards (1.6) and obtained two solvability theorems, which include three necessary and sufficient conditions concerning Question 1, and a continuity property concerning Question 2. As applications, two existence criteria concerning (1.5) have also been obtained. The paper is organized as follows. We introduce some concepts, definitions and some known results described by lemmas in Section 2, present and prove six lemmas in Section 3, then use these lemmas to obtain the main theorems concerning (1.6) in Section 4, and as applications, to show the solvability results concerning (1.5) in Section 5. Finally, we present the conclusion in Section 6. 2. Terminology 2.1. Some concepts Let U and V be two sets, X ⊆ U a nonempty subset. Let f : X −→ R be a real function and S : X −→ 2V be a set-valued map, namely, a correspondence, whose graph is the subset Graph S = {(x, y) ∈ X × V | y ∈ Sx}. In the sequel, besides some well known concepts such as f being convex, concave, upper semi-continuous (for short, u.s.c.), lower semi-continuous (for short, l.s.c.) and continuous when U is a Hausdorff locally convex vector space and X the closed convex subset of U , we need to use a number of concepts concerning S, whose definitions are given below for convenience of reference. For more information pertaining to these and some other related concepts, the reader may consult [11–13]. Definition 2.1 ([11–13]). (i) S is said to be strict if Sx 6= ∅ for each x ∈ X . (ii) S is said to be convex if both U and V are linear vector spaces, X is a convex subset of U and αSx (1) + (1 − α)Sx (2) ⊂ S[αx (1) + (1 − α)x (2) ] hold for all α ∈ [0, 1] and x (i) ∈ X (i = 1, 2). (iii) When U and V are Hausdorff topological spaces: (1) S is said to be closed if Graph S is closed in U × V (in particular, if X is a closed subset of U and Graph S is a closed subset of X × V ). (2) S is said to be u.s.c. at x0 ∈ X if for any neighborhood N (Sx0 ) of Sx0 , there exists a neighborhood N (x0 ) of x0 such that ∀ x ∈ N (x0 ), Sx ⊂ N (Sx0 ). (3) S is said to be l.s.c. at x0 ∈ X if for any y0 ∈ Sx0 and any neighborhood N (y0 ) of y0 , there exists a neighborhood N (x0 ) of x0 such that ∀x ∈ N (x0 ), Sx ∩ N (y0 ) 6= ∅.
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(4) S is said to be continuous at x0 ∈ X if it is both u.s.c. and l.s.c. at x0 . (5) S is said to be u.s.c., l.s.c. or continuous on X if it is u.s.c., l.s.c. or continuous at all points x ∈ X . (6) In the case where V is a Hausdorff locally convex vector space (V ∗ its dual) supplied with the weak topology σ (V, V ∗ ), then S : X −→ 2V is said to be u.h.c. at x0 ∈ X if for each p ∈ V ∗ , the upper supporting function X −→ R : x 7−→ σ ] (Sx, p) = sup y∈Sx h p, yi is u.s.c. at x0 , and S to be u.h.c. on X if it is u.h.c. at all points x ∈ X . Definition 2.2 ([11–13]). Let U, V be two Hausdorff locally convex vector spaces, P ⊂ U , X ⊂ V , and let ϕ( p, x) be a function defined on P × X . (i) If inf sup ϕ( p, x) = sup inf ϕ( p, x),
p∈P x∈X
x∈X p∈P
(2.1)
and we let v(ϕ)= ˆ inf p∈P supx∈X ϕ( p, x) = supx∈X inf p∈P ϕ( p, x), then we say that the balanced value v(ϕ) of ϕ exists. (ii) If sup ϕ( p, ¯ x) = inf ϕ( p, x), ¯ x∈X
p∈P
(2.2)
then we call ( p, ¯ x) ¯ the saddle point of ϕ, and denote as S(ϕ) the set of all saddle points. Remark 2.1. From the definition, it is easy to see that: (1) (2.1) holds (i.e., v(ϕ) exists) iff inf p∈P supx∈X ϕ( p, x) ≤ supx∈X inf p∈P ϕ( p, x). (2) (2.2) holds for ( p, ¯ x) ¯ (i.e., ( p, ¯ x) ¯ ∈ S(ϕ)) iff supx∈X ϕ( p, ¯ x) ≤ inf p∈P ϕ( p, x). ¯ (3) If S(ϕ) 6= φ, then (2.1) holds, and for each ( p, ¯ x) ¯ ∈ S(ϕ), v(ϕ) = ϕ( p, ¯ x) ¯ = supx∈X ϕ( p, ¯ x) = inf p∈P ϕ( p, x). ¯ Definition 2.3 ([14]). Let U be a real vector space, A ⊂ U a convex set and E a subset of A, we claim that E is an extremal subset of A if x, y ∈ A and t x + (1 − t)y ∈ E for some t ∈ (0, 1) entails x, y ∈ E, x0 ∈ A is an extremal point of A if E = {x0 } is an extremal subset of A, and the set of all extremal points of A is denoted by ExtA. 2.2. Some known results To prove the main theorems in this paper, we also need some known results as follows. Lemma 2.1 ([14]). For each finite dimensional normed linear space U , the norm topology coincides with the weak topology. Lemma 2.2 ([11]). Let U and V be two Hausdorff topological spaces and S be a strict set-valued map. (1) If S is u.s.c. with closed values, then S is closed. (2) If V is compact and S is closed, then S is u.s.c. Lemma 2.3 ([11]). Let U be a Hausdorff topological space, V a Hausdorff locally convex vector space supplied with the weak topology σ (V, V ∗ ), and S be a strict u.s.c. set-valued map. Then S is u.h.c. Lemma 2.4 ([11]). Let U be a complete metric space, V a compact metric space, and let S : U → 2V be a strict closed set-valued map. Then the subset of points at which S is continuous is residual, i.e., the interior of the discontinuous point set of S is empty. Lemma 2.5 ([12]). Let U be a Hausdorff locally convex vector space, X ⊂ U a closed convex subset, and let f α : X −→ R(α ∈ Λ) be a family of functions. (1) If X (2) If X
x 7−→ −→ R : x 7−→ −→ R :
f α (x) is convex (resp., l.s.c.) for each α ∈ Λ and supα∈Λ f α (x) is finite for each x ∈ X , then x 7−→ supα∈Λ f α (x) is also a convex (resp., a l.s.c.) function. f α (x) is concave (resp., u.s.c.) for each α ∈ Λ and infα∈Λ f α (x) is finite for each x ∈ X , then x 7−→ infα∈Λ f α (x) is also a concave (resp., an u.s.c.) function.
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Lemma 2.6 ([13]). Let U be a metric space, V a compact metric space and let g : U × V −→ R be a l.s.c. (or an u.s.c.) function; then f : U −→ R defined by f (x) = inf y∈V g(x, y) (or h : U −→ R defined by h(x) = sup y∈V g(x, y)) is also a l.s.c. (or an u.s.c.) function. Lemma 2.7 ([11,12]). Let U, V be two Hausdorff locally convex vector spaces, P ⊂ U, X ⊂ V be convex compact subsets, and let ϕ( p, x) be a function defined on P × X . If ∀x ∈ X, p 7−→ ϕ( p, x) is convex and l.s.c., and ∀ p ∈ P, x 7−→ ϕ( p, x) is concave and u.s.c., then inf p∈P supx∈X ϕ( p, x) = supx∈X inf p∈P ϕ( p, x) and there exists ( p, ¯ x) ¯ ∈ P × X such that supx∈X ϕ( p, ¯ x) = inf p∈P ϕ( p, x). ¯ Lemma 2.8 ([14]). Let U be a Hausdorff locally convex vector space and X be its nonempty convex compact subset. Then Ext X 6= ∅ and co(E xt X ) = X (i.e., the closed convex hull of E xt X is equal to X ). 3. Some lemmas Let X, T be given as in (1.6). Assume: (1) both assumptions in section 1 are satisfied, m (2) T : X −→ 2 R is convex and u.s.c., m m (3) P ⊂ R+ is convex compact with R+ P = R+ ,
(3.1)
m × P × X by and define a function (c, p, x) 7−→ φc ( p, x) on R+
φc ( p, x) = sup h p, y − ci,
m c ∈ R+ , p ∈ P, x ∈ X,
(3.2)
y∈T x m , φ =φ then for each c ∈ R+ c ˆ c (·, ·) is a function on P × X . Just as stated in Definition 2.2, we claim v(φc ) (or S(φc )) the balanced value (or the saddle point set) of φc if (2.1) (or if there exists ( p, ¯ x) ¯ ∈ P × X such that (2.2)) holds for ϕ = φc . For the following six lemmas except for Lemma 3.6, we always suppose (3.1) holds and φc ( p, x) is defined by (3.2).
Lemma 3.1. (i) T is closed and u.h.c. (ii) T X is a convex compact set. m , sup (iii) For each c ∈ R+ p∈P,x∈X |φc ( p, x)| < +∞. Thus φc ( p, x)(( p, x) ∈ P × X ), supx∈X φc ( p, x)( p ∈ P) and inf p∈P φc ( p, x)(x ∈ X ) are all finite. Proof. (i) can be deduced from Lemmas 2.1 to 2.3, and both (ii) and (iii) can be proved by the definitions.
m be fixed, then: Lemma 3.2. Let c ∈ R+
(i) For each x ∈ X , P −→ R : p 7−→ φc ( p, x) is l.s.c. and convex. (ii) For each p ∈ P, X −→ R : x − 7 → φc ( p, x) is u.s.c. and concave. Proof. (i) Since P −→ R : p 7−→ h p, y − ci is continuous, affine and φc ( p, x) = sup y∈T x h p, y − ci, then (i) follows immediately from the Lemma 3.1(iii) and Lemma 2.5(1). (ii) For each α ∈ [0, 1], x (1) , x (2) ∈ X , by the convexity of T , we have T [αx (1) +(1−α)x (2) ] ⊃ αT x (1) +(1−α)T x (2) , which implies that φc [ p, αx (1) + (1 − α)x (2) ] = =
sup y (1) ∈T x (1) ,y (2) ∈T x (2)
sup
h p, y − ci ≥
y∈T [αx (1) +(1−α)x (2) ]
sup
h p, y − ci
y∈αT x (1) +(1−α)T x (2)
h p, α(y (1) − c) + (1 − α)(y (2) − c)i = αφc ( p, x (1) ) + (1 − α)φc ( p, x (2) ).
Hence, x 7−→ φc ( p, x) is concave.
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Suppose x (k) −→ x (0) ∈ X as k −→ ∞. Since T x (k) is compact, there exists y (k) ∈ T x (k) such that φc ( p, x (k) ) = sup y∈T x (k) h p, y − ci = h p, y (k) − ci. As T X is compact and T is closed by Lemma 3.1(i) (ii), choosing a subsequence and rebelling as necessary, we may suppose lim supk−→∞ h p, y (k) − ci = limk−→∞ h p, y (k) − ci and y (k) −→ y (0) ∈ T x (0) . It follows that lim supk−→∞ φc ( p, x (k) ) = h p, y (0) − ci ≤ sup y∈T x (0) h p, y − ci = φc ( p, x (0) ). Hence x −→ φc ( p, x) is u.s.c.. This completes the proof. m , we have Lemma 3.3. For each c ∈ R+
(i) P −→ R : p 7−→ sup φc ( p, x) is l.s.c. and convex, x∈X
(3.3)
X −→ R : x 7−→ inf φc ( p, x) is u.s.c. and concave. p∈P
(ii) The balanced value v(φc ) (of φc ) exists. (iii) The saddle point set S(φc ) (of φc ) is a nonempty convex compact subset of P × X . Proof. By Lemma 3.1(iii), supx∈X φc ( p, x)( p ∈ P) and inf p∈P φc ( p, x) (x ∈ X ) are all finite. Hence, (i) follows immediately from Lemmas 2.5 and 3.2. Associating Lemma 3.2 with Lemma 2.7, we conclude that (ii) holds and S(φc ) is nonempty. It remains to show that S(φc ) is closed and convex. Assume α ∈ [0, 1], ( p (i) , x (i) ) ∈ S(φc ) (i = 1, 2), then sup φc ( p (i) , x) = inf φc ( p, x (i) ).
(3.4)
p∈P
x∈X
Associating (3.3) with (3.4) we have sup φc [αp (1) + (1 − α) p (2) , x] ≤ α sup φc ( p (1) , x) + (1 − α) sup φc ( p (2) , x) x∈X
x∈X
= α inf φc ( p, x p∈P
(1)
) + (1 − α) inf φc ( p, x p∈P
(2)
x∈X (1)
) ≤ inf φc [ p, αx p∈P
+ (1 − α)x (2) ].
Combining this with Remark 2.1 (2), we conclude that α( p (1) , x (1) ) + (1 − α)( p (2) , x (2) ) ∈ S(φc ), and thus S(φc ) is convex. Suppose ( p (k) , x (k) ) ∈ S(φc ) such that ( p (k) , x (k) ) −→ ( p (0) , x (0) ) ∈ P × X as k −→ ∞, then also by (3.3), (3.4) (where we replace i by k) and Remark 2.1 (2) we obtain sup φc ( p (0) , x) ≤ lim inf[sup φc ( p (k) , x)] ≤ lim sup[ inf φc ( p, x (k) )] ≤ inf φc ( p, x (0) ) k−→∞ x∈X
x∈X
k−→∞ p∈P
and ( p (0) , x (0) ) ∈ S(φc ). Thus S(φc ) is closed and the proof is completed.
p∈P
m ), i.e., c 7−→ v(φ ) is continuous on R m . Lemma 3.4. v(φc ) ∈ C(R+ c +
Proof. First we prove m (a) R+ × P −→ R : (c, p) 7−→ sup φc ( p, x) is l.s.c.; x∈X
m × X −→ R : (c, x) 7−→ inf φc ( p, x) is u.s.c. (b) R+
(3.5)
p∈P
m × P, sup (a) Since for each (c, p) ∈ R+ x∈X φc ( p, x) = sup y∈T X h p, y − ci is finite, and for each y ∈ T x, (c, p) 7−→ h p, y − ci is continuous (and hence, l.s.c.), then from Lemma 2.5(1) we know that (3.5)(a) is true. (b) By (3.2), we have inf p∈P φc ( p, x) = inf p∈P sup y∈T x h p, y − ci. Since p 7−→ h p, y − ci is l.s.c. convex on P, y 7−→ h p, y − ci is u.s.c. concave on T x, and both P and T x are convex compact, Lemma 2.7 implies
inf φc ( p, x) = sup inf h p, y − ci.
p∈P
y∈T x p∈P
(3.6)
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m × X , and denote by B m (0, ε) the ε-open neighborhood of 0 in R m ; then for each ε > 0, Suppose (c(0) , x (0) ) ∈ R+ R there exists δ (> 0), an open neighborhood B R n (0, δ) of 0 in R n , such that T x ⊂ T x (0) + B R m (0, ε) holds for all x ∈ x (0) + B R n (0, δ). Associating this with (3.6), it follows that
∀c ∈ c(0) + B R m (0, ε), ∀x ∈ x (0) + B R n (0, δ), inf φc ( p, x) = sup inf h p, y − ci ≤ sup p∈P
y∈T x p∈P
=
sup
inf h p, y − ci
y∈T x (0) +B R m (0,ε) p∈P
sup
inf [h p, y − ci + h p, zi]
y∈T x (0) z∈B R m (0,ε) p∈P
≤
sup y∈T x (0)
≤
inf [h p, y − c(0) i + h p, c(0) − ci] +
p∈P
sup
inf h p, y − c(0) i + 2 sup k pkε
y∈T x (0) p∈P
= inf φc(0) ( p, x p∈P
sup
sup |h p, zi|
z∈B R m (0,ε) p∈P
p∈P (0)
) + 2 sup k pkε. p∈P
Hence, (3.5)(b) holds also. By (3.5) and applying Lemma 2.6, we get m R+ −→ R : c 7−→ inf sup φc ( p, x) is l.s.c., p∈P x∈X
m −→ R : c 7−→ sup inf φc ( p, x) is u.s.c. R+ x∈X p∈P
m. So we conclude from Lemma 3.3(ii) that the function c 7−→ v(φc ) is continuous on R+
m −→ R : c −→ v(φ ) is a bounded above and continuous Lemma 3.5. (i) v = ˆ supc∈R+m v(φc ) is finite (and hence, R+ c function), and there exists a > 0 with v = supc∈[0,a]m v(φc ). m is a nonempty closed subset, then S : C −→ 2 P×X : c 7−→ S(φ ) is a strict u.s.c. and (ii) If C ⊂ R+ c u.h.c. correspondence with convex compact values, and the set of points at which S is continuous is residual. m , then for each i = 1, 2, . . . , m there exists d (i) = Proof. (i) Since R+ P = R+ i > 0 such that d (0, . . . , 0, di , 0, . . . , 0) ∈ P. Define y = (y1 , y2 , . . . , ym ), c = (c1 , c2 , . . . , cm ), d = (d1 , d2 , . . . , dm ), b = kdk · sup y∈T X kyk and d0 = min1≤i≤m di , we have
m inf sup h p, y − ci ≤ min sup hd (i) , y − ci ∀c ∈ R+ , v(φc ) = p∈P 1≤i≤m y∈T X y∈T X = min sup (di yi − di ci ) ≤ sup kdk · kyk − max di ci 1≤i≤m y∈T X 1≤i≤m y∈T X kck d 0 ≤ b − d0 max ci ≤ b − √ . 1≤i≤m m
(3.7)
m \ {0} be fixed, then from (3.7) we obtain that v(φ ) < v(φ m Let c(0) ∈ R+ c c(0) ) if c ∈ R+ satisfies kck > a = √ m[b−v(φ
(0) )]
c max{ , kc(0) k}, which implies that v = ˆ supc∈R+m v(φc ) is finite and v = supc∈[0,a]m v(φc ). Combining this d0 with Lemma 3.4, we know that (i) follows. (ii) By Lemma 3.3(iii), S is clearly a strict correspondence with convex compact values. In order to obtain the continuous properties of S stated in (ii), it is sufficient to prove that S is closed thanks to Lemmas 2.1–2.4. m × (P × X ) as Suppose (c(k) , ( p (k) , x (k) )) ∈ graph S is such that (c(k) , ( p (k) , x (k) )) −→ (c(0) , ( p (0) , x (0) )) ∈ R+ (0) (k) (k) k −→ ∞; then c ∈ C and for all k ≥ 1, ( p , x ) ∈ S(φc(k) ), i.e.,
sup φc(k) ( p (k) , x) = inf φc(k) ( p, x (k) ),
x∈X
p∈P
k ≥ 1.
(3.8)
Since φc(k) ( p (k) , x) = sup y∈T x h p (k) , y − c(k) i = φc(0) ( p (k) , x) + h p (k) , c(0) − c(k) i and φc(k) ( p, x (k) ) = sup y∈T x (k) h p, y − c(k) i = φc(0) ( p, x (k) ) + h p, c(0) − c(k) i, then
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(a) sup φc(k) ( p (k) , x) = sup φc(0) ( p (k) , x) + h p (k) , c(0) − c(k) i, x∈X x∈X (b) inf φc(k) ( p, x (k) ) = inf [φc(0) ( p, x (k) ) + h p, c(0) − c(k) i] p∈P p∈P ≤ inf φc(0) ( p, x (k) ) + sup k pk · kc(0) − c(k) k p∈P
p∈P
with limk−→∞ h p (k) , c(0) − c(k) i = 0 and limk−→∞ kc(0) − c(k) k = 0. It follows that (k) (k) lim inf[sup φc(k) ( p , x)] = lim inf[sup φc(0) ( p , x)], k−→∞ x∈X
k−→∞ x∈X
(3.9)
lim sup[ inf φc(k) ( p, x (k) )] ≤ lim sup[ inf φc(0) ( p, x (k) )]. k−→∞ p∈P
k−→∞ p∈P
Let k −→ ∞. By (3.8), (3.9) and applying Lemma 3.3(i), we obtain sup φc(0) ( p (0) , x) ≤ lim inf[sup φc(0) ( p (k) , x)] = lim inf[sup φc(k) ( p (k) , x)] k−→∞ x∈X
x∈X
k−→∞ x∈X
≤ lim sup[ inf φc(k) ( p, x k−→∞ p∈P
(k)
)] ≤ lim sup[ inf φc(0) ( p, x (k) )] ≤ inf φc(0) ( p, x (0) ). k−→∞ p∈P
p∈P
This implies that supx∈X φc(0) ( p (0) , x) = inf p∈P φc(0) ( p, x (0) ), and thus ( p (0) , x (0) ) ∈ S(φc(0) ). Therefore, S is closed and the statement (ii) follows. m such that Z − R m is a closed convex set and R Q = R m , then Z ∩ R m 6= φ Lemma 3.6. If Z ⊂ R m and Q ⊂ R+ + + + + m (i.e., 0 ∈ Z − R+ ) if and only if
∀q ∈ Q,
suphq, zi ≥ 0,
i.e., inf suph p, zi ≥ 0.
z∈Z
q∈Q z∈Z
(3.10)
m and Z ∩ R m 6= ∅. So we only need to Proof. The necessary part follows immediately from the conditions Q ⊂ R+ + m m prove that 0 ∈ Z − R+ if (3.10) holds. As Z − R+ is closed and convex, by the Hahn–Banach separation theorem, it is enough to show that
∀q ∈ R m \ {0},
sup h p, ui = suphq, zi − infm hq, yi ≥ 0.
m u∈Z −R+
z∈Z
(3.11)
y∈R+
m , then inf m hq, yi = −∞, and hence (3.11) is true. If q 6∈ R+ y∈R+ m \ {0}, then inf m hq, yi = 0. Since R+ Q = R m , we can find t > 0 such that t −1 q ∈ Q. Hence (3.10) If q ∈ R+ y∈R+ + implies supz∈Z hq, zi = t supz∈Z ht −1 q, zi ≥ 0, and thus (3.11) is also true. This completes the proof.
4. The solvability theorems of (1.6) Before state the main theorem of (1.6), we summarize the main results included in Lemmas 3.3–3.5 with respect to v(φc ) and S(φc ) as follows. m , v(φ ) exists, and c 7−→ v(φ ) is continuous on R m with v = sup m v(φc ) < Theorem 4.1. (1) For each c ∈ R+ c c c∈R+ + +∞. Furthermore, there exists a > 0 such that v = supc∈[0,a]m v(φc ). m , S(φ ) is a nonempty convex compact subset of P × X , and for each nonempty closed subset (2) For each c ∈ R+ c m C ⊂ R+ , S : C −→ 2 P×X : c 7−→ S(φc ) is a strict u.s.c. and u.h.c. correspondence, and the set of points at which S is continuous is residual.
Then the solvability result of (1.6) is the following. m solves (1.6) iff v(φ ) ≥ 0 iff ( p ∗ , x ∗ ) ∈ S(φ ) with φ ( p ∗ , x ∗ ) ≥ 0 iff ( p ∗ , x ∗ ) ∈ S(φ ) Theorem 4.2. (1) c ∈ R+ c c c c ∗ with inf p∈P φc ( p, x ) ≥ 0 or supx∈X φc ( p ∗ , x) ≥ 0. m that solves (1.6) iff v = (2) There exists at least one c ∈ R+ ˆ supc∈R+m v(φc ) ≥ 0. m ∗ (3) Suppose c ∈ R+ , then x ∈ E(c) (i.e., c solves (1.6) and x ∗ ∈ X is the corresponding solution) iff x ∗ ∈ X with inf p∈P φc ( p, x ∗ ) ≥ 0. In particular, if ( p ∗ , x ∗ ) ∈ S(φc ) with φc ( p ∗ , x ∗ ) ≥ 0, then x ∗ ∈ E(c).
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m with v(φ ) > 0, then there exists an open or closed neighborhood O(c) of c in R m such that (4) Suppose c ∈ R+ c m solves (1.6). each element of O(c) ∩ R+ m is a closed subset such that each element of C (0) solves (1.6), then E : C (0) −→ 2 X : c 7−→ (5) Suppose C (0) ⊂ R+ m 6= ∅} is a strict u.s.c. and u.h.c. correspondence with convex compact values, and E(c) = {x ∈ X |(T x − c) ∩ R+ its continuous point set is residual. m , by Lemma 3.1(ii) we can see that A =T Proof. First we prove (1). For each c ∈ R+ c ˆ X − c is convex compact, m is closed and convex. Since c solves (1.6) if and only if A ∩ R m 6= ∅ and v(φ ) = and hence Ac − R+ c c + inf p∈P supx∈X φc ( p, x) = inf p∈P supz∈T X −c h p, zi, then from Lemma 3.6 (where we take Z = Ac and Q = P), Theorem 4.1 and Remark 2.1(3), we conclude that c solves (1.6) ⇐⇒ v(φc ) ≥ 0 ⇐⇒ ( p ∗ , x ∗ ) ∈ S(φc ) with φc ( p ∗ , x ∗ ) ≥ 0 ⇐⇒ ( p ∗ , x ∗ ) ∈ S(φc ) with inf p∈P φc ( p, x ∗ ) ≥ 0 or supx∈X φc ( p ∗ , x) ≥ 0. Hence (1) follows. Next we prove (2). The necessary part follows from (1), so we prove the sufficient part. By Theorem 4.1(1), we can find a > 0 such that v = supc∈[0,a]m v(φc ) ≥ 0, so there exists a sequence {c(k) } ⊂ [0, a]m satisfying v(φc(k) ) > v − k1 ≥ − k1 . Up to the subsequence, we may assume c(k) −→ c(0) ∈ [0, a]m (k −→ ∞). As the function c 7−→ v(φc ) is continuous on [0, a]m , by taking k −→ ∞ and using statement (1), we have v(φc(0) ) ≥ 0, and hence c(0) solves (1.6). Then we prove (3). m ), H⇒ Assume x ∗ ∈ E(c), then there exists y ∗ ∈ T x ∗ such that y ∗ − c ≥ 0. This implies that for each p ∈ P(⊂ R+ ∗ ∗ ∗ φc ( p, x ) = sup y∈T x ∗ h p, y − ci ≥ h p, y − ci ≥ 0; hence inf p∈P φc ( p, x ) ≥ 0. ⇐H Suppose x ∗ ∈ X with inf p∈P φc ( p, x ∗ ) ≥ 0, by (3.2) we have
inf sup h p, y − ci = inf φc ( p, x ∗ ) ≥ 0.
p∈P y∈T x ∗
p∈P
(4.1)
Since both P and T x ∗ are convex compact, and ∀y ∈ T x ∗ , p 7−→ h p, y − ci is l.s.c. convex on P, ∀ p ∈ P, y 7−→ h p, y − ci is u.s.c. concave on T x ∗ , by (4.1) and Lemma 2.7 we obtain sup inf h p, y − ci = inf sup h p, y − ci ≥ 0.
y∈T x ∗ p∈P
p∈P y∈T x ∗
On the other hand, by Lemma 2.5(2), we know that y 7−→ inf p∈P h p, y − ci is u.s.c. on T x ∗ . So there exists y ∗ = (y1∗ , y2∗ , . . . , ym∗ ) ∈ T x ∗ such that inf h p, y ∗ − ci = sup inf h p, y − ci ≥ 0.
p∈P
y∈T x ∗ p∈P
(4.2)
As in the proof of Lemma 3.5(i), for each 1 ≤ i ≤ m, assume d (i) = (0, . . . , 0, di , 0, . . . , 0) ∈ P with di > 0. Take p = d (i) in (4.2), we obtain di (yi∗ − ci ) ≥ inf p∈P h p, y ∗ − ci ≥ 0. Therefore, yi∗ ≥ ci (1 ≤ i ≤ m 6= ∅ and hence x ∗ ∈ E(c). m), (T x ∗ − c) ∩ R+ In particular, if ( p ∗ , x ∗ ) ∈ S(φc ) with φc ( p ∗ , x ∗ ) ≥ 0, by Remark 2.1(3) we have inf p∈P φc ( p, x ∗ ) ≥ 0, and thus (3) follows. (4) can be proved with Theorem 4.1(1) and the above (1) or (2). Finally we prove (5). As in the proof of Lemma 3.5(ii), it is sufficient to prove that E(c) is a nonempty convex compact set for each c ∈ C (0) and E is closed. m ) 6= ∅} is nonempty. If Assume c ∈ C (0) , by the definition of C (0) , we know that E(c) = {x ∈ X | T x ∩ (c + R+ (i) (i) (i) (i) α ∈ [0, 1] and x ∈ E(c) (i = 1, 2), then there exists y ∈ T x such that y ≥ c (i = 1, 2). By the convexity of T , we have αy (1) +(1−α)y (2) −c ≥ 0 and αy (1) +(1−α)y (2) −c ∈ αT x (1) +(1−α)T x (2) −c ⊂ T (αx (1) +(1−α)x (2) )−c. m 6= ∅ and thus E(c) is convex. If x (k) ∈ E(c) with This implies that (T (αx (1) + (1 − α)x (2) ) − c) ∩ R+ m ) 6= ∅ and there exists y (k) ∈ T x (k) such that y (k) ≥ c. x (k) −→ x (0) ∈ X (k −→ ∞), then T x (k) ∩ (c + R+ (k) (0) Since T X is compact, we may suppose y −→ y . By taking k −→ ∞ and using the fact that T is closed, we get m ), hence E(c) is closed. Therefore, E(c) is convex and compact. y (0) ≥ c and y (0) ∈ T x (0) , i.e.,y (0) ∈ T x 0 ∩ (c + R+ (k) (k) (k) (k) Suppose (c , x ) ∈ graph E satisfy (c , x ) −→ (c(0) , x (0) )(k −→ ∞), then (c(0) , x (0) ) ∈ C (0) × X , m ) 6= ∅, and for each k ≥ 1 there exists y (k) ∈ T x (k) ∩ (c(k) + R m ). We may also suppose T x (k) ∩ (c(k) + R+ + m ). Hence, (k) (0) y −→ y (k −→ ∞). Up to the limit, we obtain y (0) ≥ c(0) , y (0) ∈ T x (0) and y (0) ∈ T x (0) ∩ (c(0) + R+ (c(0) , x (0) ) ∈ graph E, and E is closed. This completes the proof.
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m | Σ m p = 1}, then P = S m−1 satisfies the condition Remark 4.1. Suppose S m−1 = { p = ( p1 , p2 , . . . , pm ) ∈ R+ i=1 i (3) of (3.1). Hence, both Theorems 4.1 and 4.2 hold true if we take P = S m−1 in (3.1). If G, F are defined as in (1.4) and T = G − F satisfies (3.1)(1)(2), then Theorem 4.1 holds for T = G − F.
5. Solvability results for (1.5) In this section, we shall use both theorems concerning (1.6) in Section 4 to study the solvability of (1.5). Assume (a)∀i = 1, 2, . . . , m, |h i | : X → R : x 7−→ |h i (x)| is u.s.c. and concave, m (b)P m−1 ⊂ R+ is a convex compact polyhedron with ext (P m−1 ) (5.1) (i) = {d | i = 1, 2, . . . , m}, where d i = (0, . . . , 0, di , 0, . . . , 0) such that di > 0, define |h|(x) = (|h 1 (x)|, |h 2 (x)|, . . . , |h m (x)|) and define |h|c : P m−1 × X −→ R by |h|c ( p, x) = h p, |h|(x) − ci =
m X
pi (|h i (x)| − ci ), ( p, x) ∈ P m−1 × X.
(5.2)
i=1
Then we have Theorem 5.1. Let (5.1) hold and |h|c be defined by (5.2), then: m , the balanced value v(|h| ) exists and the saddle point set S(|h| ) is a nonempty convex compact (1) For each c ∈ R+ c c subset of P × X . (2) Define S(|h|c )| X = {x ∈ X | ( p, x) ∈ S(|h|c )}, then: m solves (1.5) (a) and x¯ ∈ S(|h| )| is its solution iff v(|h| ) ≥ 0 and h( x) (a) c ∈ R+ ¯ ≥ 0. c X c m solves (1.5)(b) and x ∗ ∈ S(|h| )| is its solution iff v(|h| ) = 0 and h(x ∗ ) ≥ 0. (b) c ∈ R+ c X c |h|
|h|
|h|
Proof. First we prove (1). Define Tc : X −→ R m and φc = φc (·, ·) : P m−1 × X −→ R by m Y [−|h i (x)|, |h i (x)|], x ∈ X, (a) T |h| (x) = i=1
|h| (b) φc ( p, x) =
sup h p, y − ci, y∈T |h| (x)
( p, x) ∈ P m−1 × X.
(5.3)
In view of (5.1)–(5.3), we have |h|c ( p, x) = φc|h| ( p, x),
( p, x) ∈ P m−1 × X. (i)
(5.4) (i)
(i)
Suppose α ∈ [0, 1], x (i) ∈ X (i = 1, 2) and y (i) = (y1 , y2 , . . . , ym ) ∈ T |h| (x (i) ) (i = 1, 2). By (5.1)(a) we obtain (2)
(1)
|αy j + (1 − α)y j | ≤ α|h j (x (1) )| + (1 − α)|h j (x (2) )| ≤ |h j (αx (1) + (1 − α)x (2) )|, which implies αT |h| (x (1) ) + (1 − α)T |h| (x (2) ) ⊂ T |h| (αx (1) + (1 − α)x (2) ), i.e., T |h| is convex. Suppose x ∈ X and ε > 0. Also by (5.1)(a) we can see that for each i = 1, 2, . . . , m, there exists an open neighborhood O(x) of x in R n such that |h i (x 0 )| < |h i (x)| + ε(x 0 ∈ O(x) ∩ X ) hold for all i = 1, 2, . . . , m. Hence we have m m m Y Y Y T |h| (x 0 ) = [−|h i (x 0 )|, |h i (x 0 )|] ⊂ (−|h i (x)| − ε, |h i (x)| + ε) ⊂ T |h| (x) + (−ε, ε). i=1
T |h|
i=1
i=1
Therefore, is a strict u.s.c. convex correspondence with convex compact values, and both Theorems 4.1 and 4.2 |h| |h| |h| hold for φc = φc . In particular, Theorem 4.1 and (5.4) imply that v(|h|c ) = v(φc ) exists and S(|h|c ) = S(φc ) is a m . Thus (1) follows. nonempty convex compact subset of P × X for each c ∈ R+ Then we prove (2). Since the proof of (a) is similar to but simpler than that of (b), we only prove (b). m solves (1.5)(b) and x ∗ ∈ S(|h| )| ∗ H⇒ If c ∈ R+ c X is its solution, then h(x ) ≥ c(≥ 0) (and hence, |h| ∗ m |h| ∗ Tc (x ) ∩ (c + R+ ) 6= ∅, which means that c solves (1.6) for T = T and x is its solution), and there exists
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i 0 ∈ {1, 2, . . . , m} such that h i0 (x ∗ ) = ci0 . Applying Theorem 4.2(1) to φc = φc and using (5.1), (5.4), Remark 2.1 and d (i0 ) = (0, . . . , 0, di0 , 0, . . . , 0) ∈ P m−1 (see (5.1)(b)), we obtain 0 ≤ v(φc|h| ) = v(|h|c ) =
inf h p, |h|(x ∗ ) − ci =
p∈P m−1
inf h p, h(x ∗ ) − ci ≤ di0 (h i0 (x ∗ ) − ci0 ) = 0.
p∈P m−1
Hence, v(|h|c ) = 0, and the necessary part follows. ⇐H If v(|h|c ) = 0 and ( p ∗ , x ∗ ) ∈ S(|h|c ) with h(x ∗ ) ≥ 0, by taking d (i) ∈ ext(P m−1 ) in (5.1), we have ∀i = 1, 2, . . . , m, di (h i (x ∗ ) − ci ) = |h|c (d (i) , x ∗ ) ≥
inf
p∈P m−1
|h|c ( p, x ∗ ) = v(|h|c ) = 0.
Combining this with ( p ∗ , x ∗ ) ∈ S(|h|c ), it implies that ( h i (x ∗ ) ≥ ci , i = 1, 2, . . . , m, i.e.,h(x ∗ ) ≥ c, inf h p, h(x ∗ ) − ci = inf h p, |h|(x ∗ ) − ci = v(|h|c ) = 0. p∈P m−1
(5.5)
p∈P m−1
Let P ∗ = { p ∈ P m−1 | h p, h(x ∗ ) − ci = 0}. As P m−1 → R : p 7−→ h p, h(x ∗ ) − ci be a continuous affine function, from (5.5) we know that P ∗ is a nonempty convex compact subset. Furthermore, if α ∈ (0, 1) and p (i) ∈ P m−1 (i = 1, 2) are such that αp (1) + (1 − α) p (2) ∈ P ∗ , also from (5.5) we have (i) h p , h(x ∗ ) − ci ≥ 0 (i = 1, 2), αh p (1) , h(x ∗ ) − ci + (1 − α)h p (2) , h(x ∗ ) − ci = 0, which shows that p (i) ∈ P ∗ (i = 1, 2) and P ∗ is an extremal subset of P m−1 . By Lemma 2.8, we obtain ext(P ∗ ) ⊂ ext(P m−1 ) = {d (1) , d (2) , . . . , d (m) }, and thus min1≤i≤m hd (i) , h(x ∗ ) − ci = inf p∈P m−1 h p, h(x ∗ ) − ci = 0. So there exists i 0 ∈ {1, 2, . . . , m} such that h i0 (x ∗ ) = ci0 . This completes the proof. If we consider h = (h 1 , h 2 , . . . , h m ) as some enterprise’s pure output map and this enterprise’s management is m . Thus (5.1) becomes efficient, we may suppose that h is a map from X to R+ (a) ∀i = 1, 2, . . . , m, h i : X → R+ : x 7−→ h i (x) is u.s.c. and concave , (5.6) m (b) P m−1 ⊂ R+ is a convex compact polyhedron with ext(P m−1 ) = {d (i) | i = 1, 2, . . . , m}, Define h c = h c (·, ·) : P m−1 × X −→ R by h c ( p, x) = h p, h(x) − ci =
m X
pi (h i (x) − ci ),
( p, x) ∈ P m−1 × X,
(5.7)
i=1
and m Y [−h i (x), h i (x)], (a) T h (x) = i=1
h (b) φc ( p, x) =
sup h p, y − ci, y∈T h (x)
x ∈ X, ( p, x) ∈ P m−1 × X
(5.8)
or m Y (a) T h (x) = [0, h i (x)],
x ∈ X,
i=1
h (b) φc ( p, x) =
sup h p, y − ci, y∈T h (x)
( p, x) ∈ P m−1 × X.
(5.9)
By (5.6)–(5.9) and using the same method as that of Theorem 5.1, we have Theorem 5.2. Let (5.6) hold and h c = h c (·, ·) be defined by (5.7), then: (1) The balanced value v(h c ) exists and the saddle point set S(h c ) is a nonempty convex compact subset for each m. c ∈ R+
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(2) Define S(h c )| X = {x ∈ X | ( p, x) ∈ S(h c )}, then: m solves (1.5)(a) in S(h )| iff v(h ) ≥ 0. (a) c ∈ R+ c X c m makes (1.5)(b) solvable in S(h )| iff v(h ) = 0. (b) c ∈ R+ c X c
6. Conclusion In this article, two classes of nonlinear inequality systems (i.e., (1.6) and (1.5)) are introduced along with two solvability questions, namely those of the existence and perturbation. By constructing some auxiliary functions such as |h| φc ((3.2)), |h|c ((5.2)), φc ((5.3)(b)) etc. and applying the nonlinear analysis approach, especially the minimax, saddle point theorems, the solvability results of (1.6) and (1.5) have been obtained. The main results can be summarized as follows. (i) Theorems 4.1 and 4.2, composed mainly of the necessary and sufficient criteria, have answered both questions concerning (1.6). (ii) As applications of (i), both Theorems 5.1 and 5.2 have presented the necessary and sufficient conditions as regards the existence question for (1.5), which are different from those of Ref. [9,10]. References [1] E.W. Leontief, Input–Output Analysis, Pergamon Press, Oxford, 1985 (Reprinted from the Encyclopedia of Materials Science and Engineering); Input–Output Analysis, 2nd edition, Oxford University Press, New York, 1986. [2] R. Miller, P. Blair, Input–Output Analysis: Foundations and Extensions, Prentice-Hall, Englewood Cliffs, NJ, 1985. [3] T. Fujimoto, Nonlinear model in abstract spaces, Journal of Mathematical Economics 15 (1986) 151–156. [4] Yingfan Liu, Xiaohong Chen, Some results on continuous type conditional input–output equation, Applied Mathematics and Mechanics (English edition) 25 (3) (2004) 358–366. [5] Yingfan Liu, Tianping Chen, A solvability theorem on a class of conditional input–output equation, Chinese Annals of Mathematics 25A (6) (2004) 791–798 (in Chinese). [6] Yingfan Liu, Qinghong Zhang, Rogalski–Cornet theorem and a Leontief type input–output inclusion, Nonlinear Analysis–Theory, Methods and Applications 69 (2) (2008) 425–433. [7] P. Medvegyev, A generalized existence theorem for von Neumann economic growth model, Econometrica 52 (4) (1974) 963–974. [8] C. Bidard, E. Hosoda, On consumption basket in a generalized von Neumann model, International Economic Review 28 (2) (1987) 509–519. [9] Yingfan Liu, Some results on a class of generalized Leontief conditional input output inequalities, Acta Mathematicae Applicatae Sinica 28 (3) (2005) 358–366 (in Chinese). [10] Yingfan Liu, Qinghong Zhang, Generalized input–output inequality system, Applied Mathematics and Optimization 54 (2) (2006) 189–204. [11] J.P. Aubin, I. Ekeland, Applied Nonlinear Analysis, John Wiley and Sons, New York, 1984. [12] J.P. Aubin, Mathematical Methods of Game and Economic Theory, North-Holland, Amsterdam, 1979. [13] J.P. Aubin, Optima and Equilibria, Springer-Verlag, New York, Berlin, Heidelberg, 1991. [14] R.B. Holmes, Geometric Functional Analysis and Applications, Springer-Verlag, New York, Berlin, 1974.