`Journal of Applied MathematicsVolumeΒ 2012, Article IDΒ 638632, 11 pageshttp://dx.doi.org/10.1155/2012/638632`
Research Article

## Iterative Methods for the Sum of Two Monotone Operators

Department of Information Management, Cheng Shiu University, Kaohsiung 833, Taiwan

Received 3 October 2011; Accepted 7 October 2011

Copyright Β© 2012 Yeong-Cheng Liou. 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 introduce an iterative for finding the zeros point of the sum of two monotone operators. We prove that the suggested method converges strongly to the zeros point of the sum of two monotone operators.

#### 1. Introduction

Let be a nonempty closed convex subset of a real Hilbert space . Let be a single-valued nonlinear mapping and let be a multivalued mapping. The βso-calledβ quasi-variational inclusion problem is to find a such that The set of solutions of (1.1) is denoted by . A number of problems arising in structural analysis, mechanics, and economics can be studied in the framework of this kind of variational inclusions; see, for instance, [1β4]. The problem (1.1) includes many problems as special cases.(1)If , where is a proper convex lower semicontinuous function and is the subdif and if onlyerential of , then the variational inclusion problem (1.1) is equivalent to find such that which is called the mixed quasi-variational inequality (see, Noor [5]).(2)If , where is a nonempty closed convex subset of and is the indicator function of , that is, then the variational inclusion problem (1.1) is equivalent to find such that

This problem is called Hartman-Stampacchia variational inequality (see, e.g., [6]).

Recently, Zhang et al. [7] introduced a new iterative scheme for finding a common element of the set of solutions to the inclusion problem, and the set of fixed points of nonexpansive mappings in Hilbert spaces. Peng et al. [8] 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 [9β27] and the references therein.

Inspired and motivated by the works in the literature, in this paper, we introduce an iterative for solving the problem (1.1). We prove that the suggested method converges strongly to the zeros point of the sum of two monotone operators .

#### 2. Preliminaries

Let be a real Hilbert space with inner product and norm , respectively. Let be a nonempty closed convex subset of . Recall that a mapping is said to be -inverse strongly-monotone if and if only for some and for all . It is known that if is -inverse strongly monotone, then for all .

Let be a mapping of into . The effective domain of is denoted by , that is, A multivalued mapping is said to be a monotone operator on if and if only for all , , and . A monotone operator on is said to be maximal if and if only its graph is not strictly contained in the graph of any other monotone operator on . Let be a maximal monotone operator on and let .

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 the resolvent is firmly nonexpansive, that is, for all and for all .

The following resolvent identity is well known: for and , there holds the following identity:

We use the following notation: (i) stands for the weak convergence of to ; (ii) stands for the strong convergence of to .

We need the following lemmas for the next section.

Lemma 2.1 (see [28]). Let be a nonempty closed convex subset of a real Hilbert space . Let the mapping be -inverse strongly monotone and let be a constant. Then, one has In particular, if , then is nonexpansive.

Lemma 2.2 (see [29]). Let and be bounded sequences in a Banach space and let be a sequence in with Suppose that for all and Then, .

Lemma 2.3 (see [30]). Assume that 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.

Theorem 3.1. Let be a nonempty closed and convex subset of a real Hilbert space . Let be an -inverse strongly monotone mapping of into H and let be a maximal monotone operator on , such that the domain of is included in . Let be the resolvent of for . Suppose that . For and given , let be a sequence generated by for all , where , , and satisfy (i) and ; (ii); (iii) where and . Then generated by (3.1) converges strongly to .

Proof. First, we choose any . Note that for all . Since is nonexpansive for all , we have Since is -inverse strongly monotone, we get By (3.3) and (3.4), we obtain It follows from (3.1) and (3.5) that By induction, we have Therefore, is bounded. We deduce immediately that is also bounded. Set for all . Then and are bounded.
Next, we estimate . In fact, we have
Since is nonexpansive for , we have . By the resolvent identity (2.7), we have It follows that So, Thus, From Lemma 2.2, we get Consequently, we obtain From (3.5) and (3.6), we have It follows that Since , , and , we have Put . Set for all . Take in (3.17) 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, also converges weakly to because of . By the similar argument as that in [31], we can show that . This implies that Note that . Then, . Therefore, Finally, we prove that . From (3.1), we have It is clear that and . We can therefore apply Lemma 2.3 to conclude that . This completes the proof.

#### 4. Applications

Next, we consider the problem for finding the minimum norm solution of a mathematical model related to equilibrium problems. Let be a nonempty, closed, and convex subset of a Hilbert space and let be a bifunction satisfying the following conditions:(E1) for all ;(E2) is monotone, that is, for all ;(E3)for all , ;(E4)for all , is convex and lower semicontinuous.

Then, the mathematical model related to equilibrium problems (with respect to ) is to find such that for all . The set of such solutions is denoted by . The following lemma appears implicitly in Blum and Oettli [32].

Lemma 4.1. Let be a nonempty, closed, and convex subset of and let be a bifunction of into satisfying (E1)β(E4). Let and . Then, there exists such that

The following lemma was given by Combettes and Hirstoaga [33].

Lemma 4.2. Assume that satisfies (E1)β(E4). For and , define a mapping as follows: for all . Then, the following holds: (1) is single valued; (2) is a firmly nonexpansive mapping, that is, for all , (3); (4) is closed and convex.

We call such the resolvent of for . Using Lemmas 4.1 and 4.2, we have the following lemma. See [34] for a more general result.

Lemma 4.3. Let be a Hilbert space and let be a nonempty, closed, and convex subset of . Let satisfy (E1)β(E4). Let be a multivalued mapping of into itself defined by Then, and is a maximal monotone operator with . Further, for any and , the resolvent of coincides with the resolvent of ; that is,

Form Lemma 4.3 and Theorems 3.1, we have the following result.

Theorem 4.4. Let be a nonempty, closed, and convex subset of a real Hilbert space . Let be a bifunction from satisfying (E1)β(E4) and let be the resolvent of for . Suppose . For and given , let be a sequence generated by for all , where , , and satisfy (i) and ;(ii);(iii) where and .Then converges strongly to a point .

#### Acknowledgment

The author was supported in part by NSC 100-2221-E-230-012.

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