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Journal of Applied Mathematics
Volume 2012 (2012), Article ID 638632, 11 pages
http://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

Academic Editor: Yonghong Yao

Copyright © 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 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻. Let 𝐴𝐶𝐻 be a single-valued nonlinear mapping and let 𝐵𝐻2𝐻 be a multivalued mapping. The “so-called” quasi-variational inclusion problem is to find a 𝑢2𝐻 such that0𝐴𝑥+𝐵𝑥.(1.1) The set of solutions of (1.1) is denoted by (𝐴+𝐵)1(0). 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, [14]. The problem (1.1) includes many problems as special cases.(1)If 𝐵=𝜕𝜙𝐻2𝐻, where 𝜙𝐻𝑅+ is a proper convex lower semicontinuous function and 𝜕𝜙 is the subdif and if onlyerential of 𝜙, then the variational inclusion problem (1.1) is equivalent to find 𝑢𝐻 such that 𝐴𝑢,𝑦𝑢+𝜙(𝑦)𝜙(𝑢)0,𝑦𝐻,(1.2) which is called the mixed quasi-variational inequality (see, Noor [5]).(2)If 𝐵=𝜕𝛿𝐶, where 𝐶 is a nonempty closed convex subset of 𝐻 and 𝛿𝐶𝐻[0,] is the indicator function of 𝐶, that is, 𝛿𝐶=0,𝑥𝐶,+,𝑥𝐶,(1.3) then the variational inclusion problem (1.1) is equivalent to find 𝑢𝐶 such that 𝐴𝑢,𝑣𝑢0,𝑣𝐶.(1.4)

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 [927] 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 𝐴+𝐵.

2. Preliminaries

Let 𝐻 be a real Hilbert space with inner product , and norm , respectively. Let 𝐶 be a nonempty closed convex subset of 𝐻. Recall that a mapping 𝐴𝐶𝐻 is said to be 𝛼-inverse strongly-monotone if and if only 𝐴𝑥𝐴𝑦,𝑥𝑦𝛼𝐴𝑥𝐴𝑦2(2.1) for some 𝛼>0 and for all 𝑥,𝑦𝐶. It is known that if 𝐴 is 𝛼-inverse strongly monotone, then 1𝐴𝑥𝐴𝑦𝛼𝑥𝑦(2.2) for all 𝑥,𝑦𝐶.

Let 𝐵 be a mapping of 𝐻 into 2𝐻. The effective domain of 𝐵 is denoted by dom(𝐵), that is,dom(𝐵)={𝑥𝐻𝐵𝑥}.(2.3) A multivalued mapping 𝐵 is said to be a monotone operator on 𝐻 if and if only 𝑥𝑦,𝑢𝑣0(2.4) for all 𝑥,𝑦dom(𝐵), 𝑢𝐵𝑥, and 𝑣𝐵𝑦. A monotone operator 𝐵 on 𝐻 is said to be maximal if and if only its graph is not strictly contained in the graph of any other monotone operator on 𝐻. Let 𝐵 be a maximal monotone operator on 𝐻 and let 𝐵10={𝑥𝐻0𝐵𝑥}.

For a maximal monotone operator 𝐵 on 𝐻 and 𝜆>0, we may define a single-valued operator: 𝐽𝐵𝜆=(𝐼+𝜆𝐵)1𝐻dom(𝐵),(2.5) which is called the resolvent of 𝐵 for 𝜆. It is known that the resolvent 𝐽𝐵𝜆 is firmly nonexpansive, that is, 𝐽𝐵𝜆𝑥𝐽𝐵𝜆𝑦2𝐽𝐵𝜆𝑥𝐽𝐵𝜆𝑦,𝑥𝑦(2.6) for all 𝑥,𝑦𝐶 and 𝐵10=𝐹(𝐽𝐵𝜆) for all 𝜆>0.

The following resolvent identity is well known: for 𝜆>0 and 𝜇>0, there holds the following identity:𝐽𝐵𝜆𝑥=𝐽𝐵𝜇𝜇𝜆𝜇𝑥+1𝜆𝐽𝐵𝜆𝑥,𝑥𝐻.(2.7)

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

We need the following lemmas for the next section.

Lemma 2.1 (see [28]). Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻. Let the mapping 𝐴𝐶𝐻 be 𝛼-inverse strongly monotone and let 𝜆>0 be a constant. Then, one has (𝐼𝜆𝐴)𝑥(𝐼𝜆𝐴)𝑦2𝑥𝑦2+𝜆(𝜆2𝛼)𝐴𝑥𝐴𝑦2,𝑥,𝑦𝐶.(2.8) In particular, if 0𝜆2𝛼, then 𝐼𝜆𝐴 is nonexpansive.

Lemma 2.2 (see [29]). Let {𝑥𝑛} and {𝑦𝑛} be bounded sequences in a Banach space 𝑋 and let {𝛽𝑛} be a sequence in [0,1] with 0<liminf𝑛𝛽𝑛limsup𝑛𝛽𝑛<1.(2.9) Suppose that 𝑥𝑛+1=1𝛽𝑛𝑦𝑛+𝛽𝑛𝑥𝑛(2.10) for all 𝑛0 and limsup𝑛𝑦𝑛+1𝑦𝑛𝑥𝑛+1𝑥𝑛0.(2.11) Then, lim𝑛𝑦𝑛𝑥𝑛=0.

Lemma 2.3 (see [30]). Assume that {𝑎𝑛} is a sequence of nonnegative real numbers such that 𝑎𝑛+11𝛾𝑛𝑎𝑛+𝛿𝑛𝛾𝑛,(2.12) where {𝛾𝑛} is a sequence in (0,1) and {𝛿𝑛} is a sequence such that (1)𝑛=1𝛾𝑛=; (2)limsup𝑛𝛿𝑛0 or 𝑛=1|𝛿𝑛𝛾𝑛|<. Then lim𝑛𝑎𝑛=0.

3. Main Results

In this section, we will prove our main result.

Theorem 3.1. Let 𝐶 be a nonempty closed and convex subset of a real Hilbert space 𝐻. Let 𝐴 be an 𝛼-inverse strongly monotone mapping of 𝐶 into H and let 𝐵 be a maximal monotone operator on 𝐻, such that the domain of 𝐵 is included in 𝐶. Let 𝐽𝐵𝜆=(𝐼+𝜆𝐵)1 be the resolvent of 𝐵 for 𝜆>0. Suppose that (𝐴+𝐵)10. For 𝑢𝐶 and given 𝑥0𝐶, let {𝑥𝑛}𝐶 be a sequence generated by 𝑥𝑛+1=𝛽𝑛𝑥𝑛+1𝛽𝑛𝐽𝐵𝜆𝑛𝛼𝑛𝑢+1𝛼𝑛𝑥𝑛𝜆𝑛𝐴𝑥𝑛(3.1) for all 𝑛0, where {𝜆𝑛}(0,2𝛼), {𝛼𝑛}(0,1), and {𝛽𝑛}(0,1) satisfy (i)lim𝑛𝛼𝑛=0 and 𝑛𝛼𝑛=; (ii)0<liminf𝑛𝛽𝑛limsup𝑛𝛽𝑛<1; (iii)𝑎𝜆𝑛𝑏 where [𝑎,𝑏](0,2𝛼) and lim𝑛(𝜆𝑛+1𝜆𝑛)=0. Then {𝑥𝑛} generated by (3.1) converges strongly to ̃𝑥=𝑃(𝐴+𝐵)10(𝑢).

Proof. First, we choose any 𝑧(𝐴+𝐵)10. Note that 𝑧=𝐽𝐵𝜆𝑛𝑧𝜆𝑛1𝛼𝑛𝐴𝑧=𝐽𝐵𝜆𝑛𝛼𝑛𝑧+1𝛼𝑛𝑧𝜆𝑛𝐴𝑧(3.2) for all 𝑛0. Since 𝐽𝐵𝜆 is nonexpansive for all 𝜆>0, we have 𝐽𝐵𝜆𝑛𝛼𝑛𝑢+1𝛼𝑛𝑥𝑛𝜆𝑛𝐴𝑥𝑛𝑧2=𝐽𝐵𝜆𝑛𝛼𝑛𝑢+1𝛼𝑛𝑥𝑛𝜆𝑛𝐴𝑥𝑛𝐽𝐵𝜆𝑛𝛼𝑛𝑧+1𝛼𝑛𝑧𝜆𝑛𝐴𝑧2𝛼𝑛𝑢+1𝛼𝑛𝑥𝑛𝜆𝑛𝐴𝑥𝑛𝛼𝑛𝑧+1𝛼𝑛𝑧𝜆𝑛𝐴𝑧2=1𝛼𝑛𝑥𝑛𝜆𝑛𝐴𝑥𝑛𝑧𝜆𝑛𝐴𝑧+𝛼𝑛(𝑢𝑧)2.(3.3) Since 𝐴 is 𝛼-inverse strongly monotone, we get 1𝛼𝑛𝑥𝑛𝜆𝑛𝐴𝑥𝑛𝑧𝜆𝑛𝐴𝑧+𝛼𝑛(𝑢𝑧)21𝛼𝑛𝑥𝑛𝜆𝑛𝐴𝑥𝑛𝑧𝜆𝑛𝐴𝑧2+𝛼𝑛𝑢𝑧2=1𝛼𝑛𝑥𝑛𝑧𝜆𝑛𝐴𝑥𝑛𝐴𝑧2+𝛼𝑛𝑢𝑧2=1𝛼𝑛𝑥𝑛𝑧22𝜆𝑛𝐴𝑥𝑛𝐴𝑧,𝑥𝑛𝑧+𝜆2𝑛𝐴𝑥𝑛𝐴𝑧2+𝛼𝑛𝑢𝑧21𝛼𝑛𝑥𝑛𝑧22𝛼𝜆𝑛𝐴𝑥𝑛𝐴𝑧2+𝜆2𝑛𝐴𝑥𝑛𝐴𝑧2+𝛼𝑛𝑢𝑧2=1𝛼𝑛𝑥𝑛𝑧2+𝜆𝑛𝜆𝑛2𝛼𝐴𝑥𝑛𝐴𝑧2+𝛼𝑛𝑢𝑧2.(3.4) By (3.3) and (3.4), we obtain 𝐽𝐵𝜆𝑛𝛼𝑛𝑢+1𝛼𝑛𝑥𝑛𝜆𝑛𝐴𝑥𝑛𝑧21𝛼𝑛𝑥𝑛𝑧2+𝜆𝑛𝜆𝑛2𝛼𝐴𝑥𝑛𝐴𝑧2+𝛼𝑛𝑢𝑧21𝛼𝑛𝑥𝑛𝑧2+𝛼𝑛𝑢𝑧2.(3.5) It follows from (3.1) and (3.5) that 𝑥𝑛+1𝑧2=𝛽𝑛𝑥𝑛+𝑧1𝛽𝑛𝐽𝐵𝜆𝑛𝛼𝑛𝑢+1𝛼𝑛𝑥𝑛𝜆𝑛𝐴𝑥𝑛𝑧2𝛽𝑛𝑥𝑛𝑧2+1𝛽𝑛𝐽𝐵𝜆𝑛𝛼𝑛𝑢+1𝛼𝑛𝑥𝑛𝜆𝑛𝐴𝑥𝑛𝑧2𝛽𝑛𝑥𝑛𝑧2+1𝛽𝑛1𝛼𝑛𝑥𝑛𝑧2+𝛼𝑛𝑢𝑧2=11𝛽𝑛𝛼𝑛𝑥𝑛𝑧2+1𝛽𝑛𝛼𝑛𝑢𝑧2𝑥max𝑛𝑧2,𝑢𝑧2.(3.6) By induction, we have 𝑥𝑛+1𝑥𝑧max0𝑧,𝑢𝑧.(3.7) Therefore, {𝑥𝑛} is bounded. We deduce immediately that {𝐴𝑥𝑛} is also bounded. Set 𝑢𝑛=𝛼𝑛𝑢+(1𝛼𝑛)(𝑥𝑛𝜆𝑛𝐴𝑥𝑛) for all 𝑛. Then {𝑢𝑛} and {𝐽𝐵𝜆𝑛𝑢𝑛} are bounded.
Next, we estimate 𝐽𝐵𝜆𝑛+1𝑢𝑛+1𝐽𝐵𝜆𝑛𝑢𝑛. In fact, we have 𝐽𝐵𝜆𝑛+1𝑢𝑛+1𝐽𝐵𝜆𝑛𝑢𝑛=𝐽𝐵𝜆𝑛+1𝛼𝑛+1𝑢+1𝛼𝑛+1𝑥𝑛+1𝜆𝑛+1𝐴𝑥𝑛+1𝐽𝐵𝜆𝑛𝛼𝑛𝑢+1𝛼𝑛𝑥𝑛𝜆𝑛𝐴𝑥𝑛𝐽𝐵𝜆𝑛+1𝛼𝑛+1𝑢+1𝛼𝑛+1𝑥𝑛+1𝜆𝑛+1𝐴𝑥𝑛+1𝐽𝐵𝜆𝑛+1𝛼𝑛𝑢+1𝛼𝑛𝑥𝑛𝜆𝑛𝐴𝑥𝑛+𝐽𝐵𝜆𝑛+1𝛼𝑛𝑢+1𝛼𝑛𝑥𝑛𝜆𝑛𝐴𝑥𝑛𝐽𝐵𝜆𝑛𝛼𝑛𝑢+1𝛼𝑛𝑥𝑛𝜆𝑛𝐴𝑥𝑛𝛼𝑛+1𝑢+1𝛼𝑛+1𝑥𝑛+1𝜆𝑛+1𝐴𝑥𝑛+1𝛼𝑛𝑢+1𝛼𝑛𝑥𝑛𝜆𝑛𝐴𝑥𝑛+𝐽𝐵𝜆𝑛+1𝑢𝑛𝐽𝐵𝜆𝑛𝑢𝑛𝐼𝜆𝑛+1𝐴𝑥𝑛+1𝐼𝜆𝑛+1𝐴𝑥𝑛+||𝜆𝑛+1𝜆𝑛||𝐴𝑥𝑛+𝛼𝑛+1𝑥𝑢+𝑛+1+𝜆𝑛+1𝐴𝑥𝑛+1+𝛼𝑛𝑥𝑢+𝑛+𝜆𝑛𝐴𝑥𝑛+𝐽𝐵𝜆𝑛+1𝑢𝑛𝐽𝐵𝜆𝑛𝑢𝑛.(3.8)
Since 𝐼𝜆𝑛+1𝐴 is nonexpansive for 𝜆𝑛+1(0,2𝛼), we have (𝐼𝜆𝑛+1𝐴)𝑥𝑛+1(𝐼𝜆𝑛+1𝐴)𝑥𝑛𝑥𝑛+1𝑥𝑛. By the resolvent identity (2.7), we have 𝐽𝐵𝜆𝑛+1𝑢𝑛=𝐽𝐵𝜆𝑛𝜆𝑛𝜆𝑛+1𝑢𝑛+𝜆1𝑛𝜆𝑛+1𝐽𝐵𝜆𝑛+1𝑢𝑛.(3.9) It follows that 𝐽𝐵𝜆𝑛+1𝑢𝑛𝐽𝐵𝜆𝑛𝑢𝑛=𝐽𝐵𝜆𝑛𝜆𝑛𝜆𝑛+1𝑢𝑛+𝜆1𝑛𝜆𝑛+1𝐽𝐵𝜆𝑛+1𝑢𝑛𝐽𝐵𝜆𝑛𝑢𝑛𝜆𝑛𝜆𝑛+1𝑢𝑛+𝜆1𝑛𝜆𝑛+1𝐽𝐵𝜆𝑛+1𝑢𝑛𝑢𝑛||𝜆𝑛+1𝜆𝑛||𝜆𝑛+1𝑢𝑛𝐽𝐵𝜆𝑛+1𝑢𝑛.(3.10) So, 𝐽𝐵𝜆𝑛+1𝑢𝑛+1𝐽𝐵𝜆𝑛𝑢𝑛𝑥𝑛+1𝑥𝑛+||𝜆𝑛+1𝜆𝑛||𝐴𝑥𝑛+𝛼𝑛+1𝑥𝑢+𝑛+1+𝜆𝑛+1𝐴𝑥𝑛+1+𝛼𝑛𝑥𝑢+𝑛+𝜆𝑛𝐴𝑥𝑛+||𝜆𝑛+1𝜆𝑛||𝜆𝑛+1𝑢𝑛𝐽𝐵𝜆𝑛+1𝑢𝑛.(3.11) Thus, limsup𝑛𝐽𝐵𝜆𝑛+1𝑢𝑛+1𝐽𝐵𝜆𝑛𝑢𝑛𝑥𝑛+1𝑥𝑛0.(3.12) From Lemma 2.2, we get lim𝑛𝐽𝐵𝜆𝑛𝑢𝑛𝑥𝑛=0.(3.13) Consequently, we obtain lim𝑛𝑥𝑛+1𝑥𝑛=lim𝑛1𝛽𝑛𝐽𝐵𝜆𝑛𝑢𝑛𝑥𝑛=0.(3.14) From (3.5) and (3.6), we have 𝑥𝑛+1𝑧2𝛽𝑛𝑥𝑛𝑧2+1𝛽𝑛𝐽𝐵𝜆𝑛𝛼𝑛𝑢+1𝛼𝑛𝑥𝑛𝜆𝑛𝐴𝑥𝑛𝑧21𝛽𝑛1𝛼𝑛𝑥𝑛𝑧2+𝜆𝑛𝜆𝑛2𝛼𝐴𝑥𝑛𝐴𝑧2+𝛼𝑛𝑢𝑧2+𝛽𝑛𝑥𝑛𝑧2=11𝛽𝑛𝛼𝑛𝑥𝑛𝑧2+1𝛽𝑛𝜆𝑛𝜆𝑛2𝛼𝐴𝑥𝑛𝐴𝑧2+1𝛽𝑛𝛼𝑛𝑢𝑧2𝑥𝑛𝑧2+1𝛽𝑛𝜆𝑛𝜆𝑛2𝛼𝐴𝑥𝑛𝐴𝑧2+1𝛽𝑛𝛼𝑛𝑢𝑧2.(3.15) It follows that 1𝛽𝑛𝜆𝑛2𝛼𝜆𝑛𝐴𝑥𝑛𝐴𝑧2𝑥𝑛𝑧2𝑥𝑛+1𝑧2+1𝛽𝑛𝛼𝑛𝑢𝑧2𝑥𝑛𝑥𝑧𝑛+1𝑥𝑧𝑛+1𝑥𝑛+1𝛽𝑛𝛼𝑛𝑢𝑧2.(3.16) Since lim𝑛𝛼𝑛=0, lim𝑛𝑥𝑛+1𝑥𝑛=0, and liminf𝑛(1𝛽𝑛)𝜆𝑛(2𝛼𝜆𝑛)>0, we have lim𝑛𝐴𝑥𝑛𝐴𝑧=0.(3.17) Put ̃𝑥=𝑃(𝐴+𝐵)10(𝑢). Set 𝑣𝑛=𝑥𝑛(𝜆𝑛/(1𝛼𝑛))(𝐴𝑥𝑛𝐴̃𝑥) for all 𝑛. Take 𝑧=̃𝑥 in (3.17) to get 𝐴𝑥𝑛𝐴̃𝑥0. First, we prove limsup𝑛𝑢̃𝑥,𝑣𝑛̃𝑥0. We take a subsequence {𝑣𝑛𝑖} of {𝑣𝑛} such that limsup𝑛𝑢̃𝑥,𝑣𝑛̃𝑥=lim𝑖𝑢̃𝑥,𝑣𝑛𝑖.̃𝑥(3.18) It is clear that {𝑣𝑛𝑖} is bounded due to the boundedness of {𝑥𝑛} and 𝐴𝑥𝑛𝐴̃𝑥0. Then, there exists a subsequence {𝑣𝑛𝑖𝑗} of {𝑣𝑛𝑖} which converges weakly to some point 𝑤𝐶. Hence, {𝑥𝑛𝑖𝑗} also converges weakly to 𝑤 because of 𝑣𝑛𝑖𝑗𝑥𝑛𝑖𝑗0. By the similar argument as that in [31], we can show that 𝑤(𝐴+𝐵)10. This implies that limsup𝑛𝑢̃𝑥,𝑣𝑛̃𝑥=lim𝑗𝑢̃𝑥,𝑣𝑛𝑖𝑗̃𝑥=𝑢̃𝑥,𝑤̃𝑥.(3.19) Note that ̃𝑥=𝑃(𝐴+𝐵)10(𝑢). Then, 𝑢̃𝑥,𝑤̃𝑥0,𝑤(𝐴+𝐵)10. Therefore, limsup𝑛𝑢̃𝑥,𝑣𝑛̃𝑥0.(3.20) Finally, we prove that 𝑥𝑛̃𝑥. From (3.1), we have 𝑥𝑛+1̃𝑥2𝛽𝑛𝑥𝑛̃𝑥2+1𝛽𝑛𝐽𝐵𝜆𝑛𝑢𝑛̃𝑥2=𝛽𝑛𝑥𝑛̃𝑥2+1𝛽𝑛𝐽𝐵𝜆𝑛𝑢𝑛𝐽𝐵𝜆𝑛̃𝑥1𝛼𝑛𝜆𝑛𝐴̃𝑥2𝛽𝑛𝑥𝑛̃𝑥2+1𝛽𝑛𝑢𝑛̃𝑥1𝛼𝑛𝜆𝑛𝐴̃𝑥2=𝛽𝑛𝑥𝑛̃𝑥2+1𝛽𝑛𝛼𝑛𝑢+1𝛼𝑛𝑥𝑛𝜆𝑛𝐴𝑥𝑛̃𝑥1𝛼𝑛𝜆𝑛𝐴̃𝑥2=𝛽𝑛𝑥𝑛̃𝑥2+1𝛽𝑛1𝛼𝑛𝑥𝑛𝜆𝑛𝐴𝑥𝑛̃𝑥𝜆𝑛𝐴̃𝑥+𝛼𝑛(𝑢̃𝑥)2=𝛽𝑛𝑥𝑛̃𝑥2+1𝛽𝑛×1𝛼𝑛2𝑥𝑛𝜆𝑛𝐴𝑥𝑛̃𝑥𝜆𝑛𝐴̃𝑥2+2𝛼𝑛1𝛼𝑛𝑥𝑢̃𝑥,𝑛𝜆𝑛𝐴𝑥𝑛̃𝑥𝜆𝑛𝐴̃𝑥+𝛼2𝑛𝑢̃𝑥2𝛽𝑛𝑥𝑛̃𝑥2+1𝛽𝑛×1𝛼𝑛𝑥𝑛̃𝑥2+2𝛼𝑛1𝛼𝑛𝑢̃𝑥,𝑥𝑛𝜆𝑛𝐴𝑥𝑛𝐴̃𝑥̃𝑥+𝛼2𝑛𝑢̃𝑥211𝛽𝑛𝛼𝑛𝑥𝑛̃𝑥2+1𝛽𝑛𝛼𝑛21𝛼𝑛𝑢̃𝑥,𝑣𝑛̃𝑥+𝛼𝑛𝑢̃𝑥2.(3.21) It is clear that 𝑛(1𝛽𝑛)𝛼𝑛= and limsup𝑛(2(1𝛼𝑛)𝑢̃𝑥,𝑣𝑛̃𝑥+𝛼𝑛𝑢̃𝑥2)0. We can therefore apply Lemma 2.3 to conclude that 𝑥𝑛̃𝑥. 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 𝐶 be a nonempty, closed, and convex subset of a Hilbert space and let 𝐺𝐶×𝐶𝑅 be a bifunction satisfying the following conditions:(E1)𝐺(𝑥,𝑥)=0 for all 𝑥𝐶;(E2)𝐺 is monotone, that is, 𝐺(𝑥,𝑦)+𝐺(𝑦,𝑥)0 for all 𝑥,𝑦𝐶;(E3)for all 𝑥,𝑦,𝑧𝐶, limsup𝑡0𝐺(𝑡𝑧+(1𝑡)𝑥,𝑦)𝐺(𝑥,𝑦);(E4)for all 𝑥𝐶, 𝐺(𝑥,) is convex and lower semicontinuous.

Then, the mathematical model related to equilibrium problems (with respect to 𝐶) is to find ̃𝑥𝐶 such that𝐺(̃𝑥,𝑦)0(4.1) for all 𝑦𝐶. The set of such solutions ̃𝑥 is denoted by 𝐸𝑃(𝐺). The following lemma appears implicitly in Blum and Oettli [32].

Lemma 4.1. Let 𝐶 be a nonempty, closed, and convex subset of 𝐻 and let 𝐺 be a bifunction of 𝐶×𝐶 into 𝑅 satisfying (E1)–(E4). Let 𝑟>0 and 𝑥𝐻. Then, there exists 𝑧𝐶 such that 1𝐺(𝑧,𝑦)+𝑟𝑦𝑧,𝑧𝑥0,𝑦C.(4.2)

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

Lemma 4.2. Assume that 𝐺𝐶×𝐶𝑅 satisfies (E1)–(E4). For 𝑟>0 and 𝑥𝐻, define a mapping 𝑇𝑟𝐻𝐶 as follows: 𝑇𝑟1(𝑥)=𝑧𝐶𝐺(𝑧,𝑦)+𝑟𝑦𝑧,𝑧𝑥0,𝑦𝐶(4.3) for all 𝑥𝐻. Then, the following holds: (1)𝑇𝑟 is single valued; (2)𝑇𝑟 is a firmly nonexpansive mapping, that is, for all 𝑥,𝑦𝐻, 𝑇𝑟𝑥𝑇𝑟𝑦2𝑇𝑟𝑥𝑇𝑟𝑦,𝑥𝑦;(4.4)(3)𝐹(𝑇𝑟)=𝐸𝑃(𝐺); (4)𝐸𝑃(𝐺) is closed and convex.

We call such 𝑇𝑟 the resolvent of 𝐺 for 𝑟>0. Using Lemmas 4.1 and 4.2, we have the following lemma. See [34] for a more general result.

Lemma 4.3. Let 𝐻 be a Hilbert space and let 𝐶 be a nonempty, closed, and convex subset of 𝐻. Let 𝐺𝐶×𝐶𝑅 satisfy (E1)–(E4). Let 𝐴𝐺 be a multivalued mapping of 𝐻 into itself defined by 𝐴𝐺𝑥={𝑧𝐻𝐺(𝑥,𝑦)𝑦𝑥,𝑧,𝑦𝐶},,𝑥𝐶,𝑥𝐶.(4.5) Then, 𝐸𝑃(𝐺)=𝐴𝐺1(0) and 𝐴𝐺 is a maximal monotone operator with dom(𝐴𝐺)𝐶. Further, for any 𝑥𝐻 and 𝑟>0, the resolvent 𝑇𝑟 of 𝐺 coincides with the resolvent of 𝐴𝐺; that is, 𝑇𝑟𝑥=(𝐼+𝑟𝐴𝐺)1𝑥.(4.6)

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

Theorem 4.4. Let 𝐶 be a nonempty, closed, and convex subset of a real Hilbert space 𝐻. Let 𝐺 be a bifunction from 𝐶×𝐶𝑅 satisfying (E1)–(E4) and let 𝑇𝜆 be the resolvent of 𝐺 for 𝜆>0. Suppose 𝐸𝑃(𝐺). For 𝑢𝐶 and given 𝑥0𝐶, let {𝑥𝑛}𝐶 be a sequence generated by 𝑥𝑛+1=𝛽𝑛𝑥𝑛+1𝛽𝑛𝑇𝜆𝑛𝛼𝑛𝑢+1𝛼𝑛𝑥𝑛(4.7) for all 𝑛0, where {𝜆𝑛}(0,), {𝛼𝑛}(0,1), and {𝛽𝑛}(0,1) satisfy (i)lim𝑛𝛼𝑛=0 and 𝑛𝛼𝑛=;(ii)0<liminf𝑛𝛽𝑛limsup𝑛𝛽𝑛<1;(iii)𝑎𝜆𝑛𝑏 where [𝑎,𝑏](0,) and lim𝑛(𝜆𝑛+1𝜆𝑛)=0.Then {𝑥𝑛} converges strongly to a point ̃𝑥=𝑃𝐸𝑃(𝐺)(𝑢).

Acknowledgment

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

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