Abstract and Applied Analysis

Volume 2015, Article ID 451320, 10 pages

http://dx.doi.org/10.1155/2015/451320

## Approximating Iterations for Nonexpansive and Maximal Monotone Operators

^{1}School of Mathematics & Information Technology, Nanjing Xiaozhuang University, Nanjing 211171, China^{2}Department of Mathematics, Gyeongsang National University, Jinju 660-701, Republic of Korea^{3}Department of Mathematics and RINS, Gyeongsang National University, Jinju 660-701, Republic of Korea^{4}School of Mathematics and Information Science, Beifang University of Nationalities, Yinchuan 750021, China

Received 29 May 2014; Accepted 3 August 2014

Academic Editor: Poom Kumam

Copyright © 2015 Zhangsong Yao 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 two algorithms for finding a zero of the sum of two monotone operators and a fixed point of a nonexpansive operator in Hilbert spaces. We show that these two algorithms converge strongly to the minimum norm common element of the zero of the sum of two monotone operators and the fixed point of a nonexpansive operator.

#### 1. Introduction

Throughout, we assume that is a real Hilbert space with inner product and norm , respectively. Let be a nonempty closed convex set.

*Definition 1. *An operator is said to be* nonexpansive* iffor all .

We denote by the set of fixed points of .

*Definition 2. *An operator is said to be *-inverse strong monotone* if for some and for all .

It is known that if is -inverse strong monotone, then is -lipschitz, that is, for all . Furthermore,In particular, if , then is nonexpansive.

Let be a set-valued operator. The effective domain of is denoted by , that is, .

*Definition 3. *A multivalued operator is said to be a* monotone* on if and only if for all , , and .

A monotone operator on is said to be* maximal* if and only if its graph is not strictly contained in the graph of any other monotone operator on . We denote by the set of the zero points of , that is, .

For , we define a single-valued operator which is called the resolvent of for . It is known that the resolvent is firmly nonexpansive, that is,for all and for all .

In the present paper, we consider the variational inclusion of finding a zero of the sum of two monotone operators and such thatwhere is a single-valued operator and is a set-valued operator. The set of solutions of problem (8) is denoted by .

*Special Cases*. (i) If , then problem (8) becomes the generalized equation introduced by Robinson [1].

(ii) If , then problem (8) becomes the inclusion problem introduced by Rockafellar [2].

It is known that (8) 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. For related work, please see [3–20].

Zhang et al. [21] introduced the following iterative algorithm for finding a common element of the set of solutions to the problem (8) and the set of fixed points of a nonexpansive operator:where is a nonexpansive operator. Under some mild conditions, they prove that the sequence converges strongly to .

Recently, Takahashi et al. [22] introduced another iterative algorithm for finding a zero of the sum of two monotone operators and a fixed point of a nonexpansive operatorfor all . Under some assumptions, they proved that the sequence converges strongly to a point of .

Motivated and inspired by (9) and (10), in the present paper, we suggest two algorithmsIt is obvious that (12) is very different from (9) and (10). Furthermore, we prove that both (11) and (12) converge strongly to the minimum norm element in . It should be pointed out that we do not use the metric projection in (11) and (12).

#### 2. Lemmas

In this section, we collect several useful lemmas for our next section.

First, the following resolvent equality is well known.

Lemma 4. *For and , one has*

*Lemma 5 (see [23]). Let be a closed convex set. Let be a nonexpansive operator. Then is a closed convex subset of and the operator is demiclosed at .*

*Lemma 6 (see [24]). Let be a Banach space. Let and be two bounded sequences. Let the sequence satisfy . Suppose for all and . Then .*

*Lemma 7 (see [25]). Let , , and be three sequences satisfying If and (or ), then .*

*3. Strong Convergence Results*

*Let be a nonempty closed convex set. Let be a -inverse strong monotone operator. Let be a maximal monotone operator on such that . Let be a nonexpansive operator.*

*Pick up a constant . For any , we define an operator for all .*

*Since , , and are nonexpansive, we havefor any . Hence is a contraction on . We use to denote the unique fixed point of in . Thus, satisfies the fixed point equationNext, we give the convergence analysis of (17).*

*Theorem 8. Assume that . Then defined by (17) converges strongly, as , to the minimum norm element in .*

*Proof. *Choose any . It is obvious that for all . So, we havefor all .

From (17), we haveHence, we getThus, is bounded.

By (4) and (19), we deriveSo,Since for all , we obtainUsing the firm nonexpansivity of , we haveNote thatThus,It follows thatHence,This together with (23) implies that So,By (19), we haveIt follows thatwhere is some constant such thatNow we show that is relatively norm-compact as . Assume such that as . Put . From (32), we haveSince is bounded, without loss of generality, we may assume that . Hence, because of by (23). From (30), we haveBy Lemma 5 and (35), we deduce .

Next, we show that . Let . Note that for all . Then, we haveSo,Since is monotone, we have, for ,It follows thatSince, and , we have . We also observe that and . Then, from (39), we deriveThat is, . Since is maximal monotone, we have . This shows that . Hence, we have . Therefore, we can substitute for in (34) to getConsequently, the weak convergence of to actually implies that . This has proved the relative norm-compactness of the net as .

From (34), we getThat is,It follows thatIt is obvious that by (44). This denotes that the entire net converges to . This completes the proof.

*Next, we present another algorithm.*

*Algorithm 9. *For given , define a sequence iteratively bywhere , , and .

*Theorem 10. Suppose that . Assume that the following conditions are satisfied:(i) and ;(ii);(iii), where and .Then generated by (46) converges strongly to a point which is the minimum norm element in .*

*Proof. *Let . We have for all . Since , , and are nonexpansive, we haveThus,By induction, we haveTherefore, is bounded.

From (4) and (47), we deriveSet for all Since for all , we obtainFrom (46), we haveSet for all . Then for all . Next, we estimate . In fact, we haveSince is nonexpansive for , we haveFrom (13), we have It follows thatSo,Then,Since , and , we obtain By Lemma 6, we getConsequently, we obtainFrom (51) and (52), we haveThen, we obtainSince , , and , we haveNext, we show . By using the firm nonexpansivity of , we haveObserve thatHence,It follows thatThis together with (52) implies thatHence,Since , , , and (by (60)), we deduce This indicates thatCombining (60) and (72), we getPut , where is the net defined by (17). We will finally show that .

Set for all . Take in (64) to get . First, we prove . We take a subsequence of such that It is clear that is bounded due to the boundedness of and . Then, there exists a subsequence of which converges weakly to some point . Hence, and also converge weakly to because of and . By the demiclosedness principle of the nonexpansive mapping (see Lemma 5) and (73), we deduce . Furthermore, by similar argument as that of Theorem 8, we can show that is also in . Hence, we have . This implies thatNote that . Then, , . Therefore,From (46), we have