An Iterative Algorithm for Mixed Equilibrium Problems and Variational Inclusions Approach to Variational Inequalities

Abstract

We present an iterative algorithm for finding a common element of the set of solutions of a mixed equilibrium problem and the set of a variational inclusion in a real Hilbert space. Furthermore, we prove that the proposed iterative algorithms strongly converge to which solves some variational inequality.

1. Introduction

Let be a nonempty closed convex subset of a real Hilbert space . Let be a nonlinear mapping, let be a function, and let be a bifunction of into . Now we consider the following mixed equilibrium problem:

(1.1)

The set of solution of problem (1.1) is denoted by EP.

If , then the mixed equilibrium problem (1.1) becomes the following mixed equilibrium problem:

(1.2)

which was considered by Ceng and Yao [1]. If , then the mixed equilibrium problem (1.1) becomes the following equilibrium problem:

(1.3)

which was studied by S. Takahashi and W. Takahashi [2]. If and , then the mixed equilibrium problem (1.1) becomes the following equilibrium problem:

(1.4)

If for all , the mixed equilibrium problem (1.1) becomes the following variational inequality problem:

(1.5)

The mixed equilibrium problems include fixed point problems, optimization problems, variational inequality problems, Nash equilibrium problems, and the equilibrium problems as special cases; see, for example, [3–8]. Some methods have been proposed to solve the mixed equilibrium problem and the equilibrium problem. In 1997, Flaim and Antipen [4] introduced an iterative method of finding the best approximation to the initial data and proved a strong convergence theorem. Subsequently, S. Takahashi and W. Takahashi [9] introduced another iterative scheme for finding a common element of the set of solutions of the equilibrium problem (1.2) and the set of fixed point points of a nonexpansive mapping. Furthermore, Yao et al. [10] introduced some new iterative schemes for finding a common element of the set of solutions of the equilibrium problem (1.2) and the set of common fixed points of finitely (infinitely) nonexpansive mappings. Very recently, Ceng and Yao [1] considered a new iterative scheme for finding a common element of the set of solutions of the mixed equilibrium problem and the set of common fixed points of finitely many nonexpansive mappings. Peng and Yao [11] developed a CQ method. They obtained some strong convergence results for finding a common element of the set of solutions of the mixed equilibrium problem (1.1) and the set of the variational inequality and the set of fixed points of a nonexpansive mapping. Their results extend and improve the corresponding results in [1, 9, 12].

Recall that a mapping is said to be -inverse strongly monotone if there exists a constant such that for all . A mapping is strongly positive on if there exists a constant such that .

Let be a single-valued nonlinear mapping and let be a set-valued mapping. Now we concern the following variational inclusion, which is to find a point such that

(1.6)

where is the zero vector in . The set of solutions of problem (1.6) is denoted by . If , then problem (1.6) becomes the generalized equation introduced by Robinson [13]. If , then problem (1.6) becomes the inclusion problem introduced by Rockafellar [14]. It is known that (1.6) provides a convenient framework for the unified study of optimal solutions in many optimization related areas including mathematical programming, complementarity, variational inequalities, optimal control, mathematical economics, equilibria, and game theory. Also various types of variational inclusions problems have been extended and generalized. Recently, Zhang et al. [15] introduced a new iterative scheme for finding a common element of the set of solutions to the problem (1.6) and the set of fixed points of nonexpansive mappings in Hilbert spaces. Peng et al. [16] introduced another iterative scheme by the viscosity approximate method for finding a common element of the set of solutions of a variational inclusion with set-valued maximal monotone mapping and inverse strongly monotone mappings, the set of solutions of an equilibrium problem, and the set of fixed points of a nonexpansive mapping. For some related works, please see [1, 2, 9–11, 13–34] and the references therein.

Inspired and motivated by the works in the literature, in this paper, we present an iterative algorithm for finding a common element of the set of solutions of a mixed equilibrium problem and the set of a variational inclusion in a real Hilbert space. Furthermore, we prove that the proposed iterative algorithms strongly converge to which solves some variational inequality.

2. Preliminaries

Let be a real Hilbert space with inner product and norm . Let be a nonempty closed convex subset of . Then, for any , there exists a unique nearest point in , denoted by , such that

(2.1)

Such a is called the metric projection of onto . We know that is nonexpansive. Further, for and ,

(2.2)

A set-valued mapping is called monotone if, for all , and imply . A monotone mapping is maximal if its graph is not properly contained in the graph of any other monotone mapping. It is known that a monotone mapping is maximal if and only if, for , for every implies .

Let the set-valued mapping be maximal monotone. We define the resolvent operator associated with and as follows:

(2.3)

where is a positive number. It is worth mentioning that the resolvent operator is single-valued, nonexpansive, and 1-inverse strongly monotone and that a solution of problem (1.6) is a fixed point of the operator for all , see, for instance, [25].

Throughout this paper, we assume that a bifunction and a convex function satisfy the following conditions:

(H1) for all ;

(H2) is monotone, that is, for all ;

(H3)for each , is weakly upper semicontinuous;

(H4)for each , is convex and lower semicontinuous;

(H5)for each and , there exists a bounded subset and such that for any ,

(2.4)

Lemma (see [11]).

Let be a nonempty closed convex subset of a real Hilbert space . Let be a bifunction and let be a proper lower semicontinuous and convex function. For and , define a mapping as follows:

(2.5)

for all . Assume that the conditions (H1)–(H5) hold. Then one has the following results:

(1)for each , and is single-valued;

(2) is firmly nonexpansive, that is, for any ,

(2.6)

(3);

(4) is closed and convex.

Lemma (see [24]).

Let be a maximal monotone mapping and let be a Lipschitz-continuous mapping. Then the mapping is maximal monotone.

Lemma (see [34]).

Assume taht is a sequence of nonnegative real numbers such that where is a sequence in and is a sequence such that

(1);

(2) or .

Then .

3. Main Results

In this section, we will prove our main result. First, we give some assumptions on the operators and the parameters. Subsequently, we introduce our iterative algorithm for finding a common element of the set of solutions of a mixed equilibrium problem and the set of a variational inclusion. Finally, we will show that the proposed algorithm has strong convergence.

Let be a nonempty closed convex subset of a real Hilbert space . Let be a lower semicontinuous and convex function and let be a bifunction satisfying conditions (H1)–(H5). Let be a strongly positive bounded linear operator with coefficient and let be a maximal monotone mapping. Let the mappings be -inverse strongly monotone and -inverse strongly monotone, respectively. Let and be two constants such that and .

Now we introduce the following iteration algorithm.

Algorithm 3.1.

For given arbitrarily, compute the sequences and as follows:

(3.1)

where is a real sequence in .

Now we study the strong convergence of the algorithm (3.1).

Theorem 3.2.

Suppose that . Assume the following conditions are satisfied:

(i);

(ii);

(iii).

Then the sequence generated by (3.1) converges strongly to which solves the following variational inequality:

(3.2)

Proof.

Take . It is clear that

(3.3)

We divide our proofs into the following five steps:

(1)the sequences and are bounded;

(2);

(3);

(4);

(5).

Proof.

Since is -inverse strongly monotone and is -inverse strongly monotone, we have

(3.4)

(3.5)

It is clear that if and , then and are all nonexpansive. Set . It follows that

(3.6)

By Lemma 2.1, we have for all . Then, we have

(3.7)

Hence, we have

(3.8)

Since is linear bounded self-adjoint operator on , then

(3.9)

Observe that

(3.10)

that is to say is positive. It follows that

(3.11)

From (3.1), we deduce that

(3.12)

Therefore, is bounded. Hence, , , and are all bounded.

Proof.

From (3.1), we have

(3.13)

Note that

(3.14)

Substituting (3.14) into (3.13), we get

(3.15)

Notice that This together with the last inequality and Lemma 2.3 implies that

(3.16)

Proof.

From (3.5) and (3.7), we get

(3.17)

By (3.1), we obtain

(3.18)

where is some constant satisfying . From (3.17) and (3.18), we have

(3.19)

Thus,

(3.20)

which imply that

(3.21)

Proof.

Since is firmly nonexpansive, we have

(3.22)

Hence, we have

(3.23)

Since is 1-inverse strongly monotone, we have

(3.24)

which implies that

(3.25)

Thus, by (3.23) and (3.25), we obtain

(3.26)

Substitute (3.26) into (3.18) to get

(3.27)

Then we derive

(3.28)

So, we have

(3.29)

Proof.

We note that is a contraction. As a matter of fact,

(3.30)

for all . Hence has a unique fixed point, say . That is, . This implies that for all . Next, we prove that

(3.31)

First, we note that there exists a subsequence of such that

(3.32)

Since is bounded, there exists a subsequence of which converges weakly to . Without loss of generality, we can assume that .

We next show that . By , we know that

(3.33)

It follows from (H2) that

(3.34)

Hence,

(3.35)

For and , let . From (3.35) we have

(3.36)

Since , we have . Further, from the inverse strongly monotonicity of , we have . So, from (H4) and the weakly lower semicontinuity of , and weakly, we have

(3.37)

From (H1), (H4), and (3.37), we also have

(3.38)

and hence

(3.39)

Letting , we have, for each ,

(3.40)

This implies that .

Next, we show that . In fact, since is -inverse strongly monotone, is Lipschitz continuous monotone mapping. It follows from Lemma 2.2 that is maximal monotone. Let , that is, . Again since , we have , that is, . By virtue of the maximal monotonicity of , we have

(3.41)

and so

(3.42)

It follows from , and that

(3.43)

It follows from the maximal monotonicity of that , that is, . Therefore, . It follows that

(3.44)

Proof.

First, we note that ; then for all , we have .

From (3.1), we have

(3.45)

that is,

(3.46)

where and . It is easy to see that and . Hence, by Lemma 2.3, we conclude that the sequence converges strongly to . This completes the proof.