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Abstract and Applied Analysis
Volume 2013 (2013), Article ID 323126, 9 pages
http://dx.doi.org/10.1155/2013/323126
Research Article

Ordered Variational Inequalities and Ordered Complementarity Problems in Banach Lattices

1Department of Mathematics, Shawnee State University, Portsmouth, OH 45662, USA
2Center for Fundamental Science, Kaohsiung Medical University, Kaohsiung 807, Taiwan

Received 18 November 2012; Accepted 11 December 2012

Academic Editor: Jen-Chih Yao

Copyright © 2013 Jinlu Li and Ching-Feng Wen. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We introduce the concepts of ordered variational inequalities and ordered complementarity problems with both domain and range in Banach lattices. Then we apply the Fan-KKM theorem and KKM mappings to study the solvability of these problems.

1. Introduction

Let be a real Banach space with its norm dual . Let be a nonempty convex subset of and a single-valued mapping. The variational inequality problem associated with and , simply denoted as , is to find an such that A nonempty convex subset of a Banach space is called a convex cone in whenever the following two conditions hold:(1) and , for any nonnegative number ;(2).Let be a convex cone in ; the complementarity problem associated with and , simply denoted as CP, is to find an such that The variational inequality problem VI and complementarity problem CP have been extensively studied by many authors. This theory has been recognized as an important branch in nonlinear analysis and has been widely applied to optimization theory, game theory, economic equilibrium, mechanics, and so forth. During the last five decades, many researchers have studied the existence of solutions of these problems and their applications to applied mathematical fields from finite-dimensional Euclidean spaces to infinite-dimensional general Banach spaces (see, e.g., [111]).

Since most classical Banach spaces are Banach lattices equipped with some lattice orders on which the positive operators appear naturally, the domain of an ordinal variational inequality defined in (1) and the complementarity problem defined in (2) may be in a Banach lattice (in particular, a Hilbert lattice). In this case, to investigate the properties of the solution set of (1) related to the partial order may be an important topic in economics theory and other applied-mathematics fields. In 2011, Li and Yao [8] and Nishimura and Ok [12] studied the solvability and the existence of order-maximum and order-minimum solutions to general variational inequalities defined in Hilbert lattices. In 2012, Li and Ok [13] extended these results to Banach lattices as the domain of the variational inequalities.

The ranges of the pairing in the variational inequality (1) and the complementarity problem (2) are both the set of real numbers, where the inequalities in (1) and (2) are the usual inequality for real numbers, which is the ordinal order of real numbers that is a complete order. In some economic circumstances, the preferences of a certain type of outcomes may not be totally ordered; that is, it may be a partial order, in particular a lattice order. In this case, any preference optimal problem must be defined under the given partial order that describes the preferences. To further demonstrate this aspect, we consider the following example.

For any positive integer , let denote the -dimensional Hilbert lattice where is the -dimensional Euclidean space equipped with the coordinate partial order , which is defined as whenever , for , for , and . Now let and be two finite-dimensional Hilbert lattices and a nonempty closed convex subset of . Let be a mapping. Then a new and more general (than (1) and (2)) problem is to find an such that where, without causing any confusion, 0 is the origin of . Taking , this problem turns to be the VI defined in (1), and hence it is an obvious generalization of (1).

Based on this motivation, we consider two Banach lattices and , where is considered as the domain and as the range for the values of a mapping that is from a subset of to . Then we extend the variational inequality problem VI and the complementarity problem CP to more general cases which are called the ordered variational inequalities and ordered complementarity problems defined by (9) and (10) in Section 3.

This paper is organized as follows. In Section 2, we recall some concepts and provide some properties of Banach lattices; in Section 3, we introduce the concepts of ordered variational inequalities, ordered complementarity problems and prove some solution existence theorems; in Section 4, the properties of the solution set of ordered variational inequalities, such as the order optimal solutions, will be provided; in Section 5, we give an example as an application of the main theorem (Theorem 10) in this paper.

2. Preliminaries

In this section, we recall some definitions of Banach lattices and provide some properties that are useful in this paper. Here, we adopt the notions from [14].

Let be a vector lattice with a partial order . As usual, the origin of is denoted by 0, and , and , for all . A Banach space equipped with a lattice order is called a Banach lattice, which is written as , if the following properties hold:(1) implies , for all ;(2) implies , for all and ;(3) implies , for every .It is well known that, in a Banach lattice , the distributive properties hold: and , for all , and in . We could not find the extension of these distributive properties to infinite cases, so we include them below as a lemma. They will be used in the content of this paper.

Lemma 1. Let be an element and let be subsets of a Banach lattice satisfying that , and exist; then the following distributive properties hold:(1) and ;(2) and ;(3) and ;(4) and .

Proof. The proofs of Parts 1, 2, and 3 are straightforward (e.g., see page  4 in [14] for the first equality in Part 1). We only prove Part 4. For every , it is clear that and . From the order linearity of Banach lattices, it yields . It implies that is a lower bound of . On the other hand, suppose that is an arbitrary lower bound of . Then for all , from Part 1, we have . Applying Part 1 again, it implies that . We obtain . This completes the proof of this lemma.

A net in a Banach lattice is said to be decreasing (it is denoted by ) whenever implies , where is the partial order on the index net. If a net satisfies and , then we denote it by . The meanings ↑ and are analogously defined. A net in a Banach lattice is said to be order convergent to a vector , which is denoted by , whenever there exists another net with the same index net satisfying and , for each . A subset of a Banach lattice is said to be order closed whenever for any satisfying implies .

The positive cone of a Banach lattice is denoted by which is defined as . It is well known that the positive cone of a Banach lattice is norm closed. In the next two lemmas, we show that the positive cone also has the order closeness and weak closeness.

Lemma 2. Let be an arbitrary Banach lattice. Then the positive cone is order closed.

Proof. Let be a net in , which order converges to . That is, . So there exists another net with the same index net satisfying and , for each . Then for every , we have It implies that . Since , from Lemma 1, we have . Substituting this to the above order inequality and noticing , we get . So , and hence is order closed.

Lemma 3. Let be a Banach lattice. Then the positive cone is weakly closed.

Proof. It is clear that the positive cone of the Banach lattice is convex. We have mentioned that the positive cone of the Banach lattice is norm closed. Applying Mazur’s lemma (see [12] or [15]), we have in a Banach space, a convex set is norm closed if and only if it is weakly closed. This completes the proof of this lemma.

Let and be two Banach lattices. A linear operator is said to be order bounded if it maps every order bounded subset of to order-bounded subset of . Let denote the collection of all order bounded linear operators from to . is also a vector space. A natural partial order on is induced by the positive cone as follows: for any , we say that if and only if , for all . Then is a partially ordered vector space with respect to .

A linear operator between two Banach lattices is said to be order continuous if in implies in . It is known that all order-continuous linear operators between two Banach lattices are order-bounded linear operators. The following lemma is useful in the content of this paper.

Lemma 4. Let and be two Banach lattices with being Dedekind complete. Then one has

Proof. A linear operator between two Banach lattices is said to be positive whenever . It is known that any positive linear operator between two Banach lattices is an order-bounded linear operator. Let denote the collection of all linear operators from to which can be represented as a difference between two positive operators. Then as a consequence, we have . Furthermore, if is Dedekind complete, then . From Theorem  4.3 in [14], we have that every positive linear operator between two Banach lattices is (strongly) continuous. It implies that if is Dedekind complete, then we have . This lemma is proved.

If and are two Banach lattices with Dedekind complete, then for any in , we have ; therefore, from Lemma 4, holds. Hence we can define a norm on by , for all . This norm is called the regular norm on that satisfies the following inequality By applying the Riesz-Kantorovich theorem, under the regular norm and with the partial order becomes a Dedekind-complete Banach lattice. In addition, for any net in , we have Let be a Banach lattice. The norm of is said to be an order-continuous norm, if for any net in , in implies . A Banach lattice with order-continuous norm has many useful properties. We list some below.(1)Every Banach lattice with order-continuous norm is Dedekind complete. (2)Every reflexive Banach lattice has order-continuous norm (Nakano theorem); therefore, every reflexive Banach lattice is Dedekind complete.

The class of Banach lattices with order-continuous norms is pretty large and includes many useful Banach spaces. For example, the classical , where , are Banach lattices with order-continuous norms. The following result is a consequence of order-continuous norm. We list it as a lemma which is useful in the following sections.

Lemma 5. If the norm of a Banach lattice is order continuous, then the -order convergence implies norm convergence, that is,

Proof. Suppose that is a sequence in satisfying . It is equivalent to . Since is a Banach lattice with order-continuous norm, it implies . This completes the proof of this lemma.

3. The Solvability of Ordered Variational Inequalities in Banach Lattices

In this section, we introduce the concepts of ordered variational inequalities and ordered complementarity problems on suitable Banach lattices. Then we extend some already known solvability results about variational inequalities and complementarity problems (see [38, 10, 12, 13]) to the cases of ordered variational inequalities and ordered complementarity problems.

Definition 6. Let and be two Banach lattices. Let be a nonempty convex subset of and a mapping. The ordered variational inequality problem associated with and , denoted by VOI, is to find an such that where, as usual, 0 denotes the origin of . If is linear, then the problem VOI is called a linear ordered variational inequality problem; otherwise, it is called a nonlinear ordered variational inequality problem.

Definition 7. Let and be two Banach lattices. Let be a convex cone of and a mapping. The ordered complementarity problem associated with and , denoted by OCP, is to find an such that If is linear, then the problem OCP is called a linear ordered complementarity problem; otherwise, it is called a nonlinear ordered complementarity problem.

For a given Banach lattice , let denote the norm dual of , that is, . The order dual of is denoted by that is defined to be , where is the Banach lattice of the set of real numbers with the ordinal topology and the standard order , which is complete. From Garrett Birkhoff Theorem, the norm dual of a Banach lattice coincides with its order dual , that is,

In Definitions 6 and 7, if we take , then holds; therefore, in this case, we have where is the pairing between and . Hence, the ordered variational inequality VOI and the ordered complementarity problem OCP turn to be an ordinary variational inequality VI and an ordinal complementarity problem CP, respectively. Thus the ordered variational inequality problems and the ordered complementarity problems in Banach lattices are generalizations of the variational inequality problems and complementarity problems in Banach spaces that extend the ranges from the real numbers to more general Banach lattices.

There are close connections between variational inequality problems and complementarity problems in Banach spaces (e.g., see [35, 911]). In the next lemma, we show the similar connections between ordered variational inequality problems and ordered complementarity problems in Banach lattices.

Lemma 8. Let and be two Banach lattices. Let be a convex cone of and a mapping. Then is a solution to VOI if and only if is a solution to OCP.

Proof. It can be seen that is a solution to OCP that implies that is a solution to VOI. Conversely, we will show that (9) implies (10). Suppose is a solution to VOI satisfying (9). In the case, if , then (10) obviously follows from (9). So we assume . Since is a convex cone, then and are both in . From (9), we have They imply The last order inequality is equivalent to . From the antisymmetric property of , we obtain . Since , then from the linearity of and by substituting into (9), it yields , for all . This lemma is proved.

Definition 9. Let and be two Banach lattices. Let be a nonempty convex subset of . A mapping is said to be linearly order comparable on whenever for any given and for every , and are -comparable in , that is, either or
Now we prove the main theorem of this paper.

Theorem 10. Let and be two Banach lattices. Let be a nonempty convex closed subset of and let be a linearly order comparable and continuous mapping (see Remark 11 below). If there exists a point such that then the problem VOI is solvable, that is, there exists such that

Remark 11. That a mapping is continuous means that, for any sequence , whenever in , the following conditions hold:

Proof of Theorem 10. In the proof and the following contents, not causing confusion, we drop the foot marks for the norms of the Banach spaces , and . Define a set valued mapping as follows: It is clear that , and hence , for all . Next, we show that for any is a closed subset of . To this end, take any sequence satisfying in . Since is closed, then . On the other hand, the condition clearly implies in . We have Since is a convergent sequence in , then is bounded. Applying (20) and (19), inequality (22) implies The assumption that implies , that is, for all . In Section 2, we recalled that for any Banach lattice , its positive cone is -closed and from the limit (23), it yields that , that is, . Hence , and, therefore, is closed in .
Now we show that is a KKM mapping. Assume, by way of contradiction, that is not a KKM mapping, that is, there is a finite subset of , for some positive integer , and a finite set of positive numbers satisfying , such that . Set . It implies , for all , that is, Notice that for any fixed can be rewritten as where . Since is convex, then , for any fixed . We have Since , for all , from the linearly order comparable property of , we must have From (26), we obtain Multiplying by the above order inequality and summing up from 1 to , we get It is a contradiction to the fact that , which must satisfy . Hence we must have ; therefore, is a KKM mapping. Condition (17) implies that there exists a point such that is compact. Applying the Fan-KKM Theorem, we obtain Taking , then , for all . Hence is a solution to VOI. This completes the proof of this theorem.

In particular, if is compact, as a special case of Theorem 10, we get the following corollary.

Corollary 12. Let and be two Banach lattices. Let be a nonempty convex compact subset of and let be a linearly order comparable and continuous mapping. Then the problem VOI is solvable.

In the following result, we apply Theorem 10 to the case that has order-continuous norm.

Corollary 13. Let and be two Banach lattices with having order-continuous norm. Let be a nonempty convex closed subset of and let be a linearly order comparable and continuous mapping (with respect to the regular norm on ). If there exists a point such that then the problem VOI is solvable.

Proof. From the properties of Banach lattices with order-continuous norms, is Dedekind complete. Then under the regular norm is a Dedekind-complete Banach lattice. From Lemma 4, we have . From (6), holds, for all . Since is continuous in with respect to the regular norm on , it implies that is continuous in with respect to the norm in . Then this corollary immediately follows from Theorem 10.

It is well known that every reflexive Banach lattice has order-continuous norm. As reflexive Banach lattices have been widely used in many mathematics fields, we list the following result as a special case of Corollary 13.

Corollary 14. Let be a Banach lattice and a reflexive Banach lattice. Let be a nonempty convex closed subset of and let be a linearly order comparable and continuous mapping (with respect to the regular norm on ). If there exists a point such that then the problem VOI is solvable.

Taking into account Lemma 8, an easy application of Theorem 10 to ordered complementarity problem yields the following result that provides a solvability of ordered complementarity problem in Banach lattices. Similarly, the solvability results for problem VOI provided in Corollaries 13 and 14 can be extended to solvability of ordered complementarity problems.

Corollary 15. Let and be two Banach lattices. Let be a nonempty convex closed cone of , and let be a linearly order comparable and continuous mapping. If there exists a point such that then the problem OCP is solvable; that is, there exists such that

Recall that for a given Banach lattice , the partial order in its order dual is induced by its positive cone , that is, for any , if and only if , for all . In particular, if is reflexive, applying the Garrett Birkhoff Theorem, we have the following result. It describes the connections between ordinal variational inequalities and ordered variational inequalities.

Proposition 16. Let be a Banach lattice and a reflexive Banach lattice. Let be a nonempty convex closed subset of , and let be a mapping. Then is a solution to the problem VOI if and only if the following inequality holds:

4. The Existence of Order-Optimal Solutions of Ordered Variational Inequalities

As mentioned in the introduction, Li and Yao [8], Nishimura and Ok [12], and Li and Ok [13] have studied the existence of order-maximum and order-minimum solutions to general variational inequalities defined in Banach lattices with real values. After we studied the solvability of ordered variational inequalities in Section 3, in this section, we investigate the existence of order-optimal solutions and the convexity of the solution set of ordered variational inequalities defined in Banach lattices.

Definition 17. A linear operator from a Banach lattice to a Banach lattice is said to be(1)a positive operator whenever it maps positive element to positive element, that is, whenever implies ; (2)a negative operator whenever it maps positive element to negative element, that is, whenever implies .

The collection of all positive (negative) operators between and is denoted by . It is clear that a linear operator is negative if and only if is positive. It is well known that every positive operator between two Banach lattices and is an order-bounded linear operator, so is every negative operator; therefore we have The theory of positive operators between two Banach lattices has become a major theme in the field of Banach lattices. It has been widely applied to many fields. In this section, we apply it to study the existence of order-optimal solutions and the order-preserving properties of solutions to ordered variational inequalities defined in Banach lattices.

The following results are easy consequences of positive and negative operators. We state them as a lemma without proof.

Lemma 18. Let and be two Banach lattices and a nonempty convex subset of . Let be a mapping. Let denote the set of solutions to VOI. Then has the following order preserving properties.(1)If , exists, and , then and . (2)If , exists, and , then and .

Definition 19. Let and be two Banach lattices and a nonempty subset of . Let be a mapping. is said to be(1)totally order increasing on whenever implies , for all ;(2)totally order decreasing on whenever implies , for all .

Noticing that if and only if , for all , we immediately obtain that a mapping is totally order increasing (decreasing) on implying that is -increasing (decreasing) on .

Lemma 20. Let and be two Banach lattices and a nonempty convex subset of . Let be a mapping. Let denote the set of solutions to VOI. Then has the following order preserving properties.(1)If is totally order decreasing on , then , and imply .(2)If is totally order increasing on , then , and imply .

Proof. At first we prove Part 1. Suppose that is totally order decreasing on . For any and for any satisfying , we have It implies . To show Part 2, suppose that is totally order increasing on . For any and for any satisfying , we have It implies . This proves the lemma.

As an immediate consequence, we have the following result.

Corollary 21. Let and be two Banach lattices and a nonempty convex Dedekind-complete subset of . Let be a mapping. Let denote the set of solutions to VOI. Then has the following order preserving properties.(1)If is totally order decreasing on and exists, then .(2)If is totally order increasing on and exists, then .

Proof. Since is Dedekind complete, if exists, then . The rest of the proof immediately follows from Lemma 20.

Lemma 20 and Corollary 21 do not claim the convexity of the solution set of the problem VOI. However, there are some conditions on the mapping to guarantee the convexity of . To this end, we need some concepts.

Definition 22. Let and be two Banach lattices. Let be a nonempty convex subset of and a mapping. is said to be(1)order monotone if , for any ;(2)order pseudomonotone if implies , for any .

From the above definition, it is clear that every order-monotone mapping is order-pseudomonotone. It is well-known that, in the special case , there are examples of pseudomonotone mappings which are not monotonic.

Definition 23. Let and be two Banach lattices. Let be a nonempty convex subset of and a mapping. is said to be hemicontinuous on whenever, for any , the following limit holds:

Lemma 24. Let and be two Banach lattices. Let be a nonempty convex order-closed subset of and an order-pseudomonotone and hemicontinuous mapping. Then is a solution to the problem VOI, for some , if and only if the following order inequality holds:

Proof. If is a solution to the problem VOI, then The order inequality (40) immediately follows from the order pseudomonotony of . Conversely, for any and , we define . Since is convex, then for all . From (40), we have Noticing that , for , and , the above order inequality implies Since is hemicontinuous on , we have From (42), and since is norm closed, (44) implies . This completes the proof of this lemma.

Theorem 25. Let and be two Banach lattices both with order-continuous norms. Let be a nonempty convex, closed, Dedekind-complete subset of , and let be a linearly order comparable and continuous mapping. Suppose that there exists a point such that If is also order-pseudomonotone and hemicontinuous mapping, then the solution set to the problem VOI is a nonempty closed convex subset of .

Proof. From Theorem 10, we have . We need to show that is convex. To this end, for any , from Lemma 24, we have Then for any , it implies From Lemma 24 again, we obtain It implies that , and hence is convex. The closeness of follows from the closeness of the positive cone of a Banach lattice. The proof is finished.

5. An Application

In this section, we give an example of ordered variational inequality problem in finite-dimensional cases as an application of Theorem 10, which can be considered as an application to economics theory.

Example 26. In an economy, we consider two finite-dimensional Hilbert lattices , equipped with the coordinate partial orders. Let be a bounded closed convex subset of , which is the capital resources set and is the outcome set. Suppose that a mapping is the plan-product function in this economy which assigns every point in to a producing plan. This plan is a production function with as the capital resources and with as the outcome space. Suppose that there is a fixed matrix satisfying , for , and the plan-product is a linearly weighted distribution defined by where is an square with every column as . Then there exists such that the economy takes the least production at the capital under the producing plan : That is, the problem VOI is solvable. Furthermore, there is such that the economy takes the most production at the capital under the producing plan :

Remark 27. The capitals and obtained in the previous example can be, respectively, viewed as the worst-case scenario and the best-case scenario in this economy.

Proof. At first, we show that is linearly order comparable on . To this end, for any given and for every , we have Calculating gets We obtain Since and are both real numbers and , for , it yields that and are -comparable in . Hence is linearly order comparable on . To apply Theorem 10, we have to check that and satisfies Conditions (19) and (20). For this purpose, suppose that satisfying in . Since is closed, it implies . For every fixed , ,  is an matrix. Then immediately following from in , we have Hence, satisfies Condition (19) in Theorem 10. Similarly, and are also matrices and from in , we obtain This is Condition (20) in Theorem 10. Since is compact, all conditions for in Theorem 10 are satisfied. Hence the problem VOI is solvable, that is, there exists such that The proof of the second part can be reduced to Part 1 by considering a new function . This completes the proof of this example.

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