Abstract and Applied Analysis
Volume 2015 (2015), Article ID 451320, 10 pages
http://dx.doi.org/10.1155/2015/451320

Research Article

Approximating Iterations for Nonexpansive and Maximal Monotone Operators

Zhangsong Yao,¹ Sun Young Cho,² Shin Min Kang,³ and Li-Jun Zhu⁴

¹School of Mathematics & Information Technology, Nanjing Xiaozhuang University, Nanjing 211171, China
²Department of Mathematics, Gyeongsang National University, Jinju 660-701, Republic of Korea
³Department of Mathematics and RINS, Gyeongsang National University, Jinju 660-701, Republic of Korea
⁴School 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 if

for 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 that

where 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 operator

for 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 algorithms

It 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 have

for any . Hence is a contraction on . We use to denote the unique fixed point of in . Thus, satisfies the fixed point equation

Next, 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 have

for all .
From (17), we have

Hence, we get

Thus, is bounded.
By (4) and (19), we derive

So,

Since for all , we obtain

Using the firm nonexpansivity of , we have

Note that

Thus,

It follows that

Hence,

This together with (23) implies that

So,

By (19), we have

It follows that

where is some constant such that

Now we show that is relatively norm-compact as . Assume such that as . Put . From (32), we have

Since is bounded, without loss of generality, we may assume that . Hence, because of by (23). From (30), we have

By Lemma 5 and (35), we deduce .
Next, we show that . Let . Note that for all . Then, we have

So,

Since is monotone, we have, for ,

It follows that

Since

, and , we have . We also observe that and . Then, from (39), we derive

That is, . Since is maximal monotone, we have . This shows that . Hence, we have . Therefore, we can substitute for in (34) to get

Consequently, the weak convergence of to actually implies that . This has proved the relative norm-compactness of the net as .
From (34), we get

That is,

It follows that

It 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 by

where , , 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 have

Thus,

By induction, we have

Therefore, is bounded.
From (4) and (47), we derive

Set for all Since for all , we obtain

From (46), we have

Set for all . Then for all . Next, we estimate . In fact, we have

Since is nonexpansive for , we have

From (13), we have

It follows that

So,

Then,

Since , and , we obtain

By Lemma 6, we get

Consequently, we obtain

From (51) and (52), we have

Then, we obtain

Since , , and , we have

Next, we show . By using the firm nonexpansivity of , we have

Observe that

Hence,

It follows that

This together with (52) implies that

Hence,

Since , , , and (by (60)), we deduce

This indicates that

Combining (60) and (72), we get

Put , 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 that

Note that . Then, , . Therefore,

From (46), we have

It is clear that and

By Lemma 7, we conclude that . This completes the proof.

Corollary 11. Suppose that . Let be a constant satisfying , where . For , let be a net generated by

Then the net converges strongly, as , to a point which is the minimum norm element in .

Corollary 12. Suppose that . For given , let be a sequence generated by

for all , where , , and satisfy

(i) and ;(ii);(iii), where and .

Then converges strongly to a point which is the minimum norm element in .

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors would like to thank the editor and the referees for the useful comments and suggestions. Sun Young Cho was supported by the Basic Science Research Program through the National Research Foundation of Korea funded by the Ministry of Education, Science and Technology (KRF-2013053358). Li-Jun Zhu was supported in part by NNSF of China (61362033 and NZ13087).

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