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ISRN Applied Mathematics
Volume 2013 (2013), Article ID 708548, 11 pages
http://dx.doi.org/10.1155/2013/708548
Research Article

Strong Convergence Theorems for Maximal Monotone Operators, Fixed-Point Problems, and Equilibrium Problems

College of Applied Science, Beijing University of Technology, Beijing 100124, China

Received 31 May 2013; Accepted 19 June 2013

Academic Editors: C. Lu, E. Skubalska-Rafajlowicz, Q. Song, and F. Zirilli

Copyright © 2013 Huan-chun Wu et al. 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 present a new iterative method for finding a common element of the set of fixed points of a nonexpansive mapping, the set of solutions to an equilibrium problem, and the set of zeros of the sum of maximal monotone operators and prove the strong convergence theorems in the Hilbert spaces. We also apply our results to variational inequality and optimization problems.

1. Introduction

Let be a nonempty closed convex subset of a real Hilbert space . A mapping is nonexpansive if . The set of fixed points of is denoted by . It is well known that is closed and convex. There are two iterative methods for approximating fixed points of a nonexpansive mapping. One is introduced by Mann in [1] and the other by Halpern in [2]. The iteration procedure of Mann's type for approximating fixed points of a nonexpansive mapping is the following: and where is a sequence in . The iteration procedure of Halpern's type is the following: and where is a sequence in .

Let be a bifunction from to , where is the set of real numbers. The equilibrium problem is to find such that for all . The set of such solutions is denoted by . Numerous problems in physics, optimization, and economics reduce to finding a solution to the equilibrium problem (e.g., see [3]). For solving the 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 [39].

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 , the operator is called the resolvent of for . It is known that is firmly nonexpansive. Given a positive constant , a mapping is said to be -inverse strongly monotone if Some authors have paid more attention to finding an element in the set of zeros of . For a mapping from into , we know that ; see [10]. Takahashi et al. [11] 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 [12] introduced an iterative sequence that converges strongly to an element of , where is another maximal monotone operator. Takahashi et al. [13] presented a new iterative sequence converging strongly to an element of .

Motivated by the above results, in this paper, we introduce a new iterative algorithm for finding a common element of the set of fixed points of a nonexpansive mapping, the set of solutions to an equilibrium problem, and the set of zeros of the sum of maximal monotone operators and prove the strong convergence theorems in the Hilbert spaces. Finally, we give the applications to the variational inequality and optimization problems.

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 . We write to indicate that the sequence converges strongly to . Similarly, will mean weak convergence. It is well known that satisfies Opial's condition; that is, for any sequence with , we have For any , there exists a unique point such that   is called the metric projection of onto . Note that   is a nonexpansive mapping of onto . For and , we have

For , a mapping on is called -strongly monotone if Taking , a mapping on is said to be -Lipschitzian continuous if It is easy to see that is -inverse strongly monotone whenever is -strongly monotone and -Lipschitzian continuous. Now we consider inverse strongly monotone. Let , and 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 need the following lemmas.

Lemma 1 (see [3]). Let C 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 [7]). Let C 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. Let be a -strongly monotone and -Lipschitzian continuous operator on a real Hilbert space with and . Suppose that is a sequence in . For all , one has where .

Proof. Observe that Since the sequence and , we obtain

Lemma 4 (see [8]). Suppose that (A1)–(A4) hold. If and , then

Lemma 5 (see [13]). Let be a real Hilbert space, and let be a maximal monotone operator on . Then the following holds: for all and .

Lemma 6 (see [14, 15]). Let be a sequence of nonnegative real numbers satisfying where , , and satisfy the following conditions: (i),  , (ii), (iii). Then .

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

Lemma 7. For all , the inequality holds.

Lemma 8 (see [16] (demiclosedness principle)). Let be a nonempty closed convex subset of , a nonexpansive mapping, and a point in , the sequence in . Suppose that and that . Then .

3. Strong Convergence Theorems

In this section, we present a new iterative method for finding a common element of the set of fixed points of a nonexpansive mapping, the set of solutions to an equilibrium problem, and the set of zeros of the sum of maximal monotone operators.

Theorem 9. Let be a nonempty closed convex subset of a real Hilbert space and an -inverse strongly monotone operator of into . Let be a maximal monotone operator on such that the domain of is included in . Let be the resolvent of for , and let be a nonexpansive mapping of into itself. Suppose that is a -strongly monotone and -Lipschitzian continuous operator on with and . Assume that satisfies (A1)–(A4). Suppose that . Let and , and let be a sequence generated by where the sequences ,  ,  ,  and   satisfy the following conditions: (1)  and  , (2),  and  , (3)  and  ,(4)  and  . Then the sequence converges strongly to an element of .

Proof. The proof will be completed by eight steps.
Step  1. Show that the sequences and are bounded.
Note that is a closed convex subset of since , , and   are closed and convex. For simplicity, we write
From Lemmas 1 and 2, we have , and for any , Set . It follows that Lemma 3 implies that From a simple inductive process, it follows that which yields that is bounded, so is the sequence .
Step  2. Show that
Since it follows from Lemmas 4 and 5 that
Set  . We have By the assumptions ,  ,  ,  ,   and ,   it follows from Lemma 6 that
Step  3. Show that .
For any , we have which implies that
With the help of Lemma 7, we get Consequently, Hence, Since and , we get
Step  4. Show that , for all .
For , we get This together with (33) deduces that Thus, Since   and  , the sequence is a Cauchy sequence. Assume that . It follows that
Step  5. Show that .
Set . For , we have
Therefore, Using (33) again, we obtain that Thus, It follows from (30), (40), and that
Step  6. Show that .
Since equality (45) implies that As it follows from (30), (36), and (47) that
Step  7. Show that , where .
Observe that the mapping is a contraction. Indeed, for any , As , we have . The Banach contraction mapping principle guarantees that the mapping has a unique fixed point ; that is, .
In order to show this inequality, we can choose a subsequence of such that
In view of the boundedness of , there exists a subsequence of such that . Without loss of generality, we assume that . It follows from (36) that . Since and is closed and convex, we get . Now we show that .
First we prove that . By (21), The monotonicity of implies that Replacing by , we obtain Applying (36) and (A4), we have For , ,   set  . Then   and  . Thus, Dividing by , we see that Letting , we get That is, .
Now we prove that . Otherwise, assume that , that is, . Opial's condition and (49) imply that This is a contradiction. Thus, .
Next we will show that .
In fact, let , and let . It follows from Lemma 5 that Thus, Since equalities (45) and (61) imply that Therefore, It follows from and (49) that . As is nonexpansive, Lemma 8 implies that . That is, . Hence, . By (51) and the property of metric projection, we have
Step  8. Show that , where .
According to (21), we get It follows from (65) and Lemma 6 that converges strongly to .

Remark 10. By an examination of the proof of Theorem 9, the conclusion still holds in the case that .

Remark 11. Consider the following quadratic optimization problem: where is a real Hilbert space, is a self-adjoint bounded linear operator on such that Letting  ,   (i.e., the subdifferential of the indicator function of ),  ,   and  , algorithm (21) reduces to Xu [17] showed that the sequence in algorithm (69) converges strongly to the solution of problem (67).

Remark 12. Consider the setting of Theorem 9 with ,  and  . Then algorithm (21) corresponds to the algorithm in [11, Theorem 9].

The corollaries below are the direct consequences of Theorem 9.

Corollary 13. Let be a nonempty closed convex subset of a real Hilbert space and an -inverse strongly monotone operator of into . Let be a maximal monotone operator on such that the domain of is included in . Let be the resolvent of for , and let be a nonexpansive mapping of into itself. Suppose that is a -strongly monotone and -Lipschitzian continuous operator on with and . Suppose that . Let and , and let be a sequence generated by where the sequences , , and satisfy the following conditions: (1)  and  , (2),  and  , (3)  and  . Then the sequence converges strongly to an element of  .

Proof. Letting for all and in Theorem 9, we get the result.

Corollary 14. Let be a nonempty closed convex subset of a real Hilbert space and an -inverse strongly monotone operator of into . Let be a maximal monotone operator on such that the domain of is included in . Let be the resolvent of for . Assume that satisfies (A1)–(A4). Suppose that . Let and , and let be a sequence generated by where the sequences , , , and satisfy the following conditions: (1)  and  , (2),  and  , (3)  and  , (4)  and  . Then the sequence converges strongly to an element of .

Proof. Putting in Theorem 9, we can obtain the desired 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 for all . Rockafellar [18] claimed that is a maximal monotone operator. Let be a nonempty closed convex subset of , and 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 . The variational inequality problem for nonlinear operator is to find such that The set of its solutions is denoted by . Then we have Using Theorem 9, we obtain the strong convergence theorem for the variational inequality problem.

Theorem 15. Let be a nonempty closed convex subset of a real Hilbert space and an -inverse strongly monotone operator of into , and let be a -strongly monotone and -Lipschitzian continuous operator on with and . Suppose that . Let and , and let be a sequence generated by where the sequences , , and satisfy the following conditions: (1)  and  (2),  and  , (3)  and  . Then the sequence converges strongly to an element of .

Proof. Notice that . Letting for all ,  , and , Theorem 9 yields that the sequence converges strongly to an element of .

Next we study the optimization problem where is a proper lower semicontinuous convex function of into such that is included in . We denote by the set of solutions to problem (79). Let