Journal of Function Spaces

Volume 2014 (2014), Article ID 853106, 4 pages

http://dx.doi.org/10.1155/2014/853106

## On the Graph of Solution Mappings of Variational Inequalities Problems

College of Science, Guilin University of Technology, Guilin 541004, China

Received 9 December 2013; Revised 27 March 2014; Accepted 27 March 2014; Published 9 April 2014

Academic Editor: Yuming Xing

Copyright © 2014 Qi-Qing Song. 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

This paper shows the property of homeomorphism between the space of monotone variational inequalities problems and the graph of their solution mappings.

#### 1. Introduction and Preliminaries

It is well known that variational inequalities were introduced by Hartman and Stampacchia in the 1960s; see [1, 2]. Variational inequalities have become important methods for the study of many problems such as optimization problems, nonlinear complementary problems, fixed-point problems, saddle-point problems, and Nash equilibrium problems; see [3, 4] and references therein.

Let be compact convex subset of and a mapping. A classic variational inequality problem is to find a point such that

A basic characteristic of monotone variational inequalities is the clarity of the construction of their solution sets, which attracts much attention in theoretical studies. Monotone variational inequalities are also important aspects in algorithm designs. We intend to recall some notions with monotonicity.

*Definition 1. *A function is said to be monotone if , for all ; is said to be strictly monotone if is monotone, and if , then ; is strongly monotone if there exists a constant such that , for all .

Denote by the set as follows: For any two , we adopt the uniform metric between and ; that is, We write as . For each , if a point is a solution of the problem (1) with respect to , we denote it as . Then, we define a set-valued mapping from to . For each , by the finite dimensional case of Theorem 1.4 in Chapter III in [5], it holds that . In fact, is a closed convex set for each , and if is strictly monotone, is a singleton set.

Let be the graph of the set-valued mapping ; that is,

Noting the uniqueness of strictly monotone variational inequalities, the graphs and may show some relations, and this paper aims to illustrate the construction between variational inequalities problems and the graph of solution mappings in detail. The homeomorphism between and is revealed, which is linked closely with the stability of solutions of variational inequalities (1).

For each , let , where denotes the constant mapping such that , for all . Clearly, is monotone; hence, . Then, is a mapping from to .

#### 2. Properties of the Graph Space of Solution Mappings

Theorem 2. *The spaces and are homeomorphic.*

*Proof. *For each , let , where is the unique point in ( is the identity mapping on ); then, is a mapping from to . We need to show the following five steps.(a) maps onto . For each , since , we have , for all ; that is, .(b) is continuous on . In order to show this, firstly, we prove that is closed. Let () and ( and ). Then, ; that is,
As is close to infinity for (5), noting that is continuous, we have
Then, . Consequently, is closed. Hence, the set-valued mapping is upper semicontinuous on .

Next, let () and . Then, for each , , where is the unique point in . We need to show that . Since is upper semicontinuous on , for any open set , there exists a number such that, for each , . Therefore, we can obtain that . For each , as gets close to infinity, since the right side of the following inequality,
tends to zero, we have .(c) is continuous on . Let () and ( and ). For each , as gets close to infinity, it holds that
and, then, we have .(d), where is the identity mapping on . For each , clearly,
where is the unique element of .(e), where is the identity mapping on . For each ,
where satisfies that . From step , we have
Consequently, (11) holds for ; that is,
Since , we have , for all . It follows that
Add (12) and (13); then, we obtain that
that is,
Since is monotone, we have
Thus, (15) and (16) imply that
Noting that is also strictly monotone, (17) can result in the fact that . Therefore, .

From the closedness of , the continuity of is guaranteed in Theorem 2. Conversely, being closed can be induced by Theorem 2.

Corollary 3. *Theorem 2 implies that is closed.*

*Proof. *Since is closed, by the continuity of in Theorem 2, we have that is closed. Additionally, Theorem 2 means that ; then, is closed.

From Theorem 2 or Corollary 3, is closed, noting that is closed for each , which results in the upper semicontinuity of the set-valued mapping . This directly leads to the following stability result of the solution set for each .

Corollary 4. *Let , . For each neighborhood of in , there exists a neighborhood of in such that , for all .*

*Remark 5. *As shown in Theorem 2 or Corollary 3, is upper semicontinuous on . Then, for any , , and with a unique solution , we know that . For any open neighborhood , there is a number such that , . Then, by the arbitrariness of with , we can assert that there exists a convergent subsequence of such that . This is like the Tikhonov regularization method for the ill-posed variational inequalities; see [3].

*Remark 6. *Let . In Theorem 2, for each , let and, for each , let with for the strongly monotone mapping . Then, and are also bijections between and .

From Theorem 2, in a finite dimensional case, the homeomorphism is shown between two spaces in relation to the variational problem (1). This facilitates the generalization of this property into Hilbert spaces.

In the following part, let be a compact convex subset in a Hilbert space , where denotes the inner product on . We denote by the norm space induced by the inner product on . Let be the dual space of which consists of all bounded linear functions on equipped with norm topology. Then, for each , the norm of can be written as . And it is known that is a Banach space using this topology.

A mapping from to is said to be monotone if , for all , where is the pairing of and ; is called strictly monotone if is monotone, and implies .

Let be continuous; then, by Theorem 3 in [6], there is in solving the variational inequality problem This kind of variational inequality problem was introduced by Browder (see [6]). Denote by the set of all (monotone and continuous on ) and by the set of all solutions of the problem (18) for each . Then, we define a set-valued mapping from to . For each , we know that if is strictly monotone, then is a singleton set. For two , we measure the metric between them as Denote by the graph of the set-valued mapping ; that is, To construct a bijection between and , let such that, for each , , where is a constant mapping such that , for all . For each , let such that, for each , Easily, we can check that is strictly monotone. Let for each , where satisfies that and is defined by (21).

Theorem 7. *The spaces and are homeomorphic.*

*Proof. *We will follow the following five steps to complete the proof.(a) maps onto . For each , let ; we need to show that . Since , we have , for all . Hence,
that is, .(b), where is the identity mapping on . For each , by step , it holds that
(c), where is the identity mapping on . For each ,
where ; then, for each ,
On the one hand, by step , . Hence,
Taking a special point in (26), we can obtain that
Then,
On the other hand, since , we have that ; hence,
Inequality (29) implies that
Add (28) and (30); it follows that
Noting that is monotone, it holds that
Therefore, by (31) and (32), we have
From the strict monotonicity of , we can obtain that .(d) is continuous on . Let , , and ( and ). For any , since
we have . That is, .(e) is continuous on . Let , , and with , , and . It is sufficient to show that and . In order to achieve this, we give a proof of upper semicontinuity of by the closedness of .

Let with and . Then, . That is,
By way of contradiction, assume that there exists such that
Let , ; then, .

Since is continuous on , , and , we have
Hence, according to (36), there exists a number such that
Clearly, there is contradiction between (35) and (38). Therefore, we have
that is, . Consequently, the space is closed and is upper semicontinuous on . Similar to step in Theorem 2, the fact that () and the upper semicontinuity of imply that . Since , similar to step , we can obtain that . The proof is completed.

#### Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

The author thanks reviewers for their useful and constructive comments. This project is supported by Guangxi Natural Science Foundation (2012GXNSFBA053013), NNSF (61164020), and Doctoral Research Fund of Guilin University of Technology.

#### References

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