Abstract

We introduce a new iterative method for finding a common element of the set of solutions of an equilibrium problem and the set of zeros of the sum of maximal monotone operators, and we obtain strong convergence theorems in Hilbert spaces. We also apply our results to the variational inequality and convex minimization problems. Our results extend and improve the recent result of Takahashi et al. (2012).

1. Introduction

Let be a real Hilbert space and let be a nonempty closed convex subset of . Let be a bifunction from to , where is the set of real numbers. The equilibrium problem for is to find such that The set of solutions of (1) is denoted by . Numerous problems in physics, optimization, and economics are reduced to find the solution of an equilibrium problem (e.g., see [1]). For solving an equilibrium problem, we assume that the bifunction satisfies the following conditions:(A1) for all ; (A2) is monotone; that is, for all ; (A3) for every , ; (A4) is convex and lower semicontinuous for each . Equilibrium problems have been studied extensively (see [16]).

Let be a mapping of into . The effective domain of is denoted by ; that is, . A multivalued mapping is said to be monotone if A monotone operator is said to be maximal if its graph is not properly contained in the graph of any other monotone operator. For a maximal monotone operator on and , we may define a single-valued operator , which is called the resolvent of for . It is known that is firmly nonexpansive. A basic problem for maximal monotone operator is to The classical method for solving problem (3) is the proximal point algorithm. Rockafellar [7] established the weak convergence of the proximal point algorithm for maximal monotone operators. The question whether the strong convergence of the proximal point algorithm holds was answered in the negative by Güler [8]. In order to obtain the strong convergence theorem, some researchers have been devoted to modifying the proximal point algorithm (see [9, 10]). Given a positive constant , a mapping is said to be -inverse strongly monotone if Generally, monotone inclusion problem is to where the mapping is inverse strongly monotone and is maximal monotone. We write for the set of solutions of problem (5); that is, . It is well known that (see [11]). Takahashi et al. [12] constructed the following iterative sequence. Let , and let be a sequence generated by Under appropriate conditions, they proved that the sequence converges strongly to a point . Lin and Takahashi [13] introduced an iterative sequence that converges strongly to an element of , where is another maximal monotone operator.

Motivated by the above results, especially by Chuang et al. [2] and Takahashi et al. [12], we introduce a new iterative method for finding a common element of the set of solutions of an equilibrium problem and the set of zeros of the sum of maximal monotone operators, and we obtain strong convergence theorems in Hilbert spaces. We also apply our results to the variational inequality and convex minimization problems. Our results extend and improve the recent result of [12].

2. Preliminaries

Throughout this paper, let be a real Hilbert space with inner product and norm and let be a nonempty closed convex subset of . A mapping is nonexpansive if for all . The set of fixed points of is denoted by ; that is, . It is well known that is closed and convex. We denote the strong convergence and the weak convergence of to by and , respectively. For any , there exists a unique point such that is called the metric projection of onto . Note that is a nonexpansive mapping. For and , we have

Now consider inverse strongly monotone operators. For , let be an -inverse strongly monotone operator. If , then is a nonexpansive mapping. Indeed, for   and , we get Therefore, the operator is a nonexpansive mapping of into .

We collect some important facts and tools.

Lemma 1 (see [1]). Let be a nonempty closed convex subset of and let be a bifunction from to satisfying (A1)–(A4). If and , then there exists such that

Lemma 2 (see [4]). Let be a nonempty closed convex subset of , and let be a bifunction from to satisfying (A1)–(A4). For , define a mapping as follows: Then, the following hold: (i) is single valued;(ii) is firmly nonexpansive; that is, for any , ;(iii);(iv) is closed and convex.

Lemma 3 (see [6]). Suppose that (A1)–(A4) hold. If and , then

Lemma 4 (see [14]). Let be a real Hilbert space and let be a maximal monotone operator on . Then, the following holds: for all and .

Lemma 5 (see [15]). Let and be bounded sequences in a Banach space and let be a sequence in with . Suppose for all integers and . Then, .

The following lemma is an immediate consequence of the inner product on .

Lemma 6. For all , the inequality holds.

Lemma 7 (demiclosedness principle [16]). Let be a nonempty closed convex subset of and a nonexpansive mapping. Let be a point in and let be a sequence in . Suppose that and that . Then, .

Lemma 8 (see [17]). Let be a sequence of nonnegative real numbers satisfying , where(i), ; (ii). Then, .

3. Strong Convergence Theorems

In this section, we introduce a new iterative method for finding a common element of the set of zeros of the sum of maximal monotone operators and the set of solutions of an equilibrium problem, and we prove the strong convergence theorems in Hilbert spaces.

Theorem 9. Let be a nonempty closed convex subset of a real Hilbert space and an -inverse strongly monotone operator of into with . Suppose that is a maximal monotone operator on and is a bifunction from to with (A1)–(A4). Assume that . Let and be sequences in and let be a sequence in . Let be a sequence generated by Assume the following conditions are satisfied:(c1) for some ;(c2) and ; (c3); (c4) and ; (c5) and . Then, the sequence converges strongly to .

Proof. Note that the set is closed and convex since and are closed and convex.
With the help of Lemmas 1 and 2, we have , and, for any , Notice that It follows that By condition (c1), the sequence is bounded. Hence, there exists a positive number such that . From a simple inductive process, we have which yields that is bounded. So are and .
We have from Lemmas 3 and 4 that Thus, conditions (c2), (c4), and (c5) yield that Recall that . It follows from Lemma 5 that Consequently,
For any , according to Lemma 2, we have which yields that By (14), one has which implies that This together with (22) deduces that
Next, we prove that where . In order to show this inequality, we can choose a subsequence of such that Owing to the boundedness of , without loss of generality, we assume that . Now we show that . As is contained in and is a closed convex set, one has .
We firstly prove that . By (14), we have The monotonicity of yields that Replacing by , we obtain It follows from (21), (27), and (A4) that For , , set . Then, and . Thus, Dividing by , we see that Letting , we get namely, .
Secondly, we prove that . Thanks to (c5), there is a subsequence of such that . Without loss of generality, we assume that . Observe that Note that expressions (21) and (27) imply that Therefore, Since is nonexpansive and , Lemma 7 gives ; that is, . Thus, we have .
By (29) and the property of metric projection, we have Inequality (28) arrives.
Finally, we prove that . Using (14) again, we obtain Note that It follows from Lemma 8 that converges strongly to .

The following results are direct consequences of Theorem 9.

Corollary 10. Let be a nonempty closed convex subset of a real Hilbert space and an -inverse strongly monotone operator of into with . Suppose that is a maximal monotone operator on with . Let and be sequences in and let be a sequence in . Let be a sequence generated by If conditions (c1)–(c3) and (c5) are satisfied, then the sequence converges strongly to .

Proof. Letting for all and in Theorem 9, the desired result follows.

Remark 11. When , the algorithm in Corollary 10 corresponds to the one in Theorem 10 of Takahashi et al. [12]. Thus, Corollary 10 includes, as a special case, Theorem 10 of [12].

Corollary 12. Let be a nonempty closed convex subset of a real Hilbert space and let be a bifunction from to with (A1)–(A4). Assume that the set is nonempty. Let and be sequences in and let be a sequence in . Let be a sequence generated by If conditions (c1)–(c4) are satisfied, then the sequence converges strongly to .

Proof. Putting , (i.e., the subdifferential of the indicator function of  ) in Theorem 9, we get the result.

4. Applications

In this section, we apply the results in the preceding section to variational inequality and optimization problems.

Now we consider the variational inequality problem. Let be a real Hilbert space and let be a proper lower semicontinuous convex function of into . Then, the subdifferential of is defined as follows: for all . Rockafellar [18] claimed that is a maximal monotone operator. Let be the indicator function of ; that is, Since is a proper lower semicontinuous convex function on , the subdifferential of is a maximal monotone operator. The resolvent of for is defined by We have where , .

Let be a nonlinear mapping. The variational inequality problem for is to find such that The set of its solutions is denoted by . It is easy to see that (for more details, see [13]). Using Corollary 10, we obtain the strong convergence theorem for the variational inequality problem.

Theorem 13. Let be a nonempty closed convex subset of a real Hilbert space and an -inverse strongly monotone operator of into with . Assume that the set is nonempty. Let and be sequences in and let be a sequence in . Let be a sequence generated by If conditions (c1)–(c3) and (c5) are satisfied, then the sequence converges strongly to .

Proof. Notice that . Corollary 10 yields the conclusion.

Next we study the optimization problem where is a proper lower semicontinuous convex function of into such that is contained in . We denote by the set of solutions of problem (51). Let be a bifunction defined by It is clear that satisfies (A1)–(A4) and . Therefore, by Corollary 12, the following result is obtained.

Theorem 14. Let be a proper lower semicontinuous convex function of into and let be a nonempty closed convex subset of such that is contained in . Suppose that the set is nonempty. Let and be sequences in and let be a sequence in . Let be a sequence generated by If conditions (c1)–(c4) are satisfied, then the sequence converges strongly to .

Proof. Letting in Corollary 12, we get the required result.

Remark 15. Letting in our results, we obtain the algorithms for minimal-norm solutions of corresponding problems.

Conflict of Interests

The authors declare that there is no conflict of interests.

Acknowledgments

The authors would like to thank the referees and editors for their valuable comments and suggestions. The work was supported by graduate funds of Beijing University of Technology (no. ykj-2013-9422).