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Journal of Applied Mathematics
Volume 2012 (2012), Article ID 602513, 17 pages
http://dx.doi.org/10.1155/2012/602513
Research Article

General Iterative Methods for Equilibrium Problems and Infinitely Many Strict Pseudocontractions in Hilbert Spaces

College of Science, Civil Aviation University of China, Tianjin 300300, China

Received 11 January 2012; Revised 24 February 2012; Accepted 25 February 2012

Academic Editor: Rudong Chen

Copyright © 2012 Peichao Duan and Aihong Wang. 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 propose an implicit iterative scheme and an explicit iterative scheme for finding a common element of the set of fixed point of infinitely many strict pseudocontractive mappings and the set of solutions of an equilibrium problem by the general iterative methods. In the setting of real Hilbert spaces, strong convergence theorems are proved. Our results improve and extend the corresponding results reported by many others.

1. Introduction

Let 𝐻 be a real Hilbert space and let 𝐶 be a nonempty closed convex subset of 𝐻. Let 𝐹 be a bifunction from 𝐶×𝐶 to , where is the set of real numbers.

The equilibrium problem for 𝐹𝐶×𝐶 is to find 𝑥𝐶 such that𝐹(𝑥,𝑦)0(1.1) for all 𝑦𝐶. The set of such solutions is denoted by EP(𝐹).

A mapping 𝑆 of 𝐶 is said to be a 𝜅-strict pseudocontraction if there exists a constant 𝜅[0,1) such that𝑆𝑥𝑆𝑦2𝑥𝑦2+𝜅(𝐼𝑆)𝑥(𝐼𝑆)𝑦2(1.2) for all 𝑥,𝑦𝐶; see [1]. We denote the set of fixed points of 𝑆 by 𝐹(𝑆) (i.e., 𝐹(𝑆)={𝑥𝐶𝑆𝑥=𝑥}).

Note that the class of strict pseudocontractions strictly includes the class of nonexpansive mappings which are mapping 𝑆on𝐶 such that𝑆𝑥𝑆𝑦𝑥𝑦(1.3) for all 𝑥,𝑦𝐶. That is, 𝑆 is nonexpansive if and only if 𝑆 is a 0-strict pseudocontraction.

Numerous problems in physics, optimization, and economics reduce to finding a solution of the equilibrium problem. Some methods have been proposed to solve the equilibrium problem (1.1); see, for instance, [24]. In particular, Combettes and Hirstoaga [5] proposed several methods for solving the equilibrium problem. On the other hand, Mann [6], Shimoji and Takahashi [7] considered iterative schemes for finding a fixed point of a nonexpansive mapping. Further, Acedo and Xu [8] projected new iterative methods for finding a fixed point of strict pseudocontractions.

In 2006, Marino and Xu [3] introduced the general iterative method and proved that the algorithm converged strongly. Recently, Liu [2] considered a general iterative method for equilibrium problems and strict pseudocontractions. Tian [9] proposed a new general iterative algorithm combining an 𝐿-Lipschitzian and 𝜂-strong monotone operator. Very recently, Wang [10] considered a general composite iterative method for infinite family strict pseudocontractions.

In this paper, motivated by the above facts, we introduce two iterative schemes and obtain strong convergence theorems for finding a common element of the set of fixed points of a infinite family of strict pseudocontractions and the set of solutions of the equilibrium problem (1.1).

2. Preliminaries

Throughout this paper, we always write for weak convergence and for strong convergence. We need some facts and tools in a real Hilbert space 𝐻 which are listed as below.

Lemma 2.1. Let 𝐻 be a real Hilbert space. There hold the following identities:(i)𝑥𝑦2=𝑥2𝑦22𝑥𝑦,𝑦, 𝑥,𝑦𝐻;(ii)𝑡𝑥+(1𝑡)𝑦2=𝑡𝑥2+(1𝑡)𝑦2𝑡(1𝑡)𝑥𝑦2,𝑡[0,1],𝑥,𝑦𝐻.

Lemma 2.2 (see [11]). Assume that {𝛼𝑛} is a sequence of nonnegative real numbers such that 𝛼𝑛+11𝛾𝑛𝛼𝑛+𝛿𝑛,(2.1) where {𝛾𝑛} is a sequence in (0,1) and {𝛿𝑛} is a sequence such that(i)𝑛=1𝛾𝑛=;(ii)lim𝑛sup(𝛿𝑛/𝛾𝑛)0or𝑛=1|𝛿𝑛|<.Then, lim𝑛𝛼𝑛=0.
Recall that given a nonempty closed convex subset 𝐶 of a real Hilbert space 𝐻, for any 𝑥𝐻, there exists a unique nearest point in 𝐶, denoted by 𝑃𝐶𝑥, such that 𝑥𝑃𝐶𝑥𝑥𝑦(2.2) for all 𝑦𝐶. Such a 𝑃𝐶 is called the metric (or the nearest point) projection of 𝐻 onto 𝐶. As known, 𝑦=𝑃𝐶𝑥 if and only if there holds the relation: 𝑥𝑦,𝑦𝑧0𝑧𝐶.(2.3)

Lemma 2.3 (see [10]). Let 𝐴𝐻𝐻 be a 𝐿-Lipschitzian and 𝜂-strongly monotone operator on a Hilbert space 𝐻 with 𝐿>0, 𝜂>0, 0<𝜇<2𝜂/𝐿2, and 0<𝑡<1. Then, 𝑆=(𝐼𝑡𝜇𝐴)𝐻𝐻 is a contraction with contractive coefficient 1𝑡𝜏 and 𝜏=(1/2)𝜇(2𝜂𝜇𝐿2).

Lemma 2.4 (see [1]). Let 𝑆𝐶𝐶 be a 𝜅-strict pseudocontraction. Define 𝑇𝐶𝐶 by 𝑇𝑥=𝜆𝑥+(1𝜆)𝑆𝑥 for each 𝑥𝐶. Then, as 𝜆[𝜅,1), 𝑇 is a nonexpansive mapping such that 𝐹(𝑇)=𝐹(𝑆).

Lemma 2.5 (see [9]). Let 𝐻 be a Hilbert space and 𝑓𝐻𝐻 be a contraction with coefficient 0<𝛼<1, and 𝐴𝐻𝐻 an 𝐿-Lipschitzian continuous operator and 𝜂-strongly monotone with 𝐿>0, 𝜂>0. Then for 0<𝛾<𝜇𝜂/𝛼: 𝑥𝑦,(𝜇𝐴𝛾𝑓)𝑥(𝜇𝐴𝛾𝑓)𝑦(𝜇𝜂𝛾𝛼)𝑥𝑦2,𝑥,𝑦𝐻.(2.4) That is, 𝜇𝐴𝛾𝑓 is strongly monotone with coefficient 𝜇𝜂𝛾𝛼.
Let {𝑆𝑛} be a sequence of 𝜅𝑛-strict pseudo-contractions. Define 𝑆𝑛=𝜃𝑛𝐼+(1𝜃𝑛)𝑆𝑛,𝜃𝑛[𝜅𝑛,1). Then, by Lemma 2.4, 𝑆𝑛 is nonexpansive. In this paper, consider the mapping 𝑊𝑛 defined by 𝑈𝑛,𝑛+1𝑈=𝐼,𝑛,𝑛=𝑡𝑛𝑆𝑛𝑈𝑛,𝑛+1+1𝑡𝑛𝑈𝐼,𝑛,𝑛1=𝑡𝑛1𝑆𝑛1𝑈𝑛,𝑛+1𝑡𝑛1𝑈𝐼,,𝑛,𝑖=𝑡𝑖𝑆𝑖𝑈𝑛,𝑖+1+1𝑡𝑖𝑈𝐼,,𝑛,2=𝑡2𝑆2𝑈𝑛,3+1𝑡2𝑊𝐼,𝑛=𝑈𝑛,1=𝑡1𝑆1𝑈𝑛,2+1𝑡1𝐼,(2.5) where 𝑡1,𝑡2, are real numbers such that 0𝑡𝑛<1. Such a mapping 𝑊𝑛 is called a 𝑊-mapping generated by 𝑆1,𝑆2, and 𝑡1,𝑡2,. It is easy to see 𝑊𝑛 is nonexpansive.

Lemma 2.6 (see [7]). Let 𝐶 be a nonempty closed convex subset of a strictly convex Banach space 𝐸, let 𝑆1,𝑆2, be nonexpansive mappings of 𝐶 into itself such that 𝑖=1𝐹(𝑆𝑖) and let 𝑡1,𝑡2, be real numbers such that 0<𝑡𝑖𝑏<1, for every 𝑖=1,2,. Then, for any 𝑥𝐶 and 𝑘𝑁, the limit lim𝑛𝑈𝑛,𝑘𝑥 exists.
Using Lemma 2.6, one can define the mapping 𝑊 of 𝐶 into itself as follows: 𝑊𝑥=lim𝑛𝑊𝑛𝑥=lim𝑛𝑈𝑛,1𝑥,𝑥𝐶.(2.6)

Lemma 2.7 (see [7]). Let 𝐶 be a nonempty closed convex subset of a strictly convex Banach space 𝐸. Let 𝑆1,𝑆2, be nonexpansive mappings of 𝐶 into itself such that 𝑖=1𝐹(𝑆𝑖) and let 𝑡1,𝑡2, be real numbers such that 0<𝑡𝑖𝑏<1,for all 𝑖1. If 𝐾 is any bounded subset of 𝐶, then lim𝑛sup𝑥𝐾𝑊𝑥𝑊𝑛𝑥=0.(2.7)

Lemma 2.8 (see [12]). Let 𝐶 be a nonempty closed convex subset of a Hilbert space 𝐻,let{𝑆𝑖𝐶𝐶} be a family of infinite nonexpansive mappings with 𝑖=1𝐹(𝑆𝑖), let 𝑡1,𝑡2, be real numbers such that 0<𝑡𝑖𝑏<1, for every 𝑖=1,2,. Then 𝐹(𝑊)=𝑖=1𝐹(𝑆𝑖).
For solving the equilibrium problem, assume that the bifunction 𝐹 satisfies the following conditions:(A1)𝐹(𝑥,𝑥)=0 for all 𝑥𝐶;(A2)𝐹 is monotone, that is, 𝐹(𝑥,𝑦)+𝐹(𝑦,𝑥)0 for any 𝑥,𝑦𝐶;(A3) for each 𝑥,𝑦,𝑧𝐶,limsup𝑡0𝐹(𝑡𝑧+(1𝑡)𝑥,𝑦)𝐹(𝑥,𝑦);(A4)𝐹(𝑥,) is convex and lower semicontinuous for each 𝑥𝐶.
Recall some lemmas which will be needed in the rest of this paper.

Lemma 2.9 (see [13]). Let 𝐶 be a nonempty closed convex subset of 𝐻, let 𝐹 be bifunction from 𝐶×𝐶 to satisfying (A1)–(A4), and let 𝑟>0 and 𝑥𝐻. Then, there exists 𝑧𝐶 such that 1𝐹(𝑧,𝑦)+𝑟𝑦𝑧,𝑧𝑥0,𝑦𝐶.(2.8)

Lemma 2.10 (see [5]). For 𝑟>0,𝑥𝐻, define a mapping 𝑇𝑟𝐻𝐶 as follows: 𝑇𝑟1(𝑥)=𝑧𝐶𝐹(𝑧,𝑦)+𝑟𝑦𝑧,𝑧𝑥0,𝑦𝐶(2.9) for all 𝑥𝐻. Then, the following statements hold:(i)𝑇𝑟 is single-valued;(ii)𝑇𝑟 is firmly nonexpansive, that is, for any 𝑥,𝑦𝐻,𝑇𝑟𝑥𝑇𝑟𝑦2𝑇𝑟𝑥𝑇𝑟𝑦,𝑥𝑦;(2.10)(iii)𝐹(𝑇𝑟)=𝐸𝑃(𝐹); (iv)𝐸𝑃(𝐹) is closed and convex.

Lemma 2.11 (see [14]). Let {𝑥𝑛} and {𝑧𝑛} be bounded sequences in a Banach space and let {𝛽𝑛} be a sequence of real numbers such that 0<liminf𝑛𝛽𝑛limsup𝑛𝛽𝑛<1 for all 𝑛=0,1,2,. Suppose that 𝑥𝑛+1=(1𝛽𝑛)𝑧𝑛+𝛽𝑛𝑥𝑛 for all 𝑛=0,1,2, and limsup𝑛𝑧𝑛+1𝑧𝑛𝑥𝑛+1𝑥𝑛0. Then lim𝑛𝑧𝑛𝑥𝑛=0.

Lemma 2.12 (see [4]). Let 𝐶,𝐻,𝐹, and 𝑇𝑟𝑥 be as in Lemma 2.10. Then, the following holds: 𝑇𝑠𝑥𝑇𝑡𝑥2𝑠𝑡𝑠𝑇𝑠𝑥𝑇𝑡𝑥,𝑇𝑠𝑥𝑥(2.11) for all 𝑠,𝑡>0 and 𝑥𝐻.

Lemma 2.13 (see [10]). Let 𝐻 be a Hilbert space and let 𝐶 be a nonempty closed convex subset of 𝐻, and 𝑇𝐶𝐶 a nonexpansive mapping with 𝐹(𝑇). If {𝑥𝑛} is a sequence in 𝐶 weakly converging to 𝑥 and if {(𝐼𝑇)𝑥𝑛} converges strongly to 𝑦, then (𝐼𝑇)𝑥=𝑦.

3. Main Result

Throughout the rest of this paper, we always assume that 𝑓 is a contraction of 𝐻 into itself with coefficient 𝛼(0,1), and 𝐴 is a 𝐿-Lipschitzian continuous operator and 𝜂-strongly monotone on 𝐻 with 𝐿>0, 𝜂>0. Assume that 0<𝜇<2𝜂/𝐿2 and 0<𝛾<𝜇(𝜂(𝜇𝐿2/2))/𝛼=𝜏/𝛼.

Define a mapping 𝑉𝑛=𝛽𝑛𝐼+(1𝛽𝑛)𝑊𝑛𝑇𝑟𝑛. Since both 𝑊𝑛 and 𝑇𝑟𝑛 are nonexpansive, it is easy to get 𝑉𝑛 is also nonexpansive. Consider the following mapping 𝐺𝑛 on 𝐻 defined by𝐺𝑛𝑥=𝛼𝑛𝛾𝑓(𝑥)+𝐼𝛼𝑛𝑉𝜇𝐴𝑛𝑥,𝑥𝐻,𝑛𝑁,(3.1) where 𝛼𝑛(0,1). By Lemmas 2.3 and 2.10, we have𝐺𝑛𝑥𝐺𝑛𝑦𝛼𝑛𝛾𝑓(𝑥)𝑓(𝑦)+1𝛼𝑛𝜏𝑉𝑛𝑥𝑉𝑛𝑦𝛼𝑛𝛾𝛼𝑥𝑦+1𝛼𝑛𝜏=𝑥𝑦1𝛼𝑛(𝜏𝛾𝛼)𝑥𝑦.(3.2) Since 0<1𝛼𝑛(𝜏𝛾𝛼)<1, it follows that 𝐺𝑛 is a contraction. Therefore, by the Banach contraction principle, 𝐺𝑛 has a unique fixed pointed 𝑥𝑓𝑛𝐻 such that𝑥𝑓𝑛=𝛼𝑛𝑥𝛾𝑓𝑓𝑛+𝐼𝛼𝑛𝑉𝜇𝐴𝑛𝑥𝑓𝑛.(3.3)

For simplicity, we will write 𝑥𝑛 for 𝑥𝑓𝑛 provided no confusion occurs. Next we prove the sequences {𝑥𝑛} converges strongly to a 𝑥Ω=𝑖=1𝐹(𝑆𝑖)𝐸𝑃(𝐹) which solves the variational inequality:(𝛾𝑓𝜇𝐴)𝑥,𝑝𝑥0,𝑝Ω.(3.4) Equivalently, 𝑥=𝑃Ω(𝐼𝜇𝐴+𝛾𝑓)𝑥.

Theorem 3.1. Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻 and 𝐹 a bifunction from 𝐶×𝐶 to satisfying (A1)–(A4). Let 𝑆𝑖𝐶𝐶 be a family 𝜅𝑖-strict pseudocontractions for some 0𝜅𝑖<1. Assume the set Ω=𝑖=1𝐹(𝑆𝑖)𝐸𝑃(𝐹). Let 𝑓 be a contraction of 𝐻 into itself with 𝛼(0,1) and let 𝐴 be a 𝐿-Lipschitzian continuous operator and 𝜂-strongly monotone with 𝐿>0,𝜂>0,0<𝜇<2𝜂/𝐿2 and 0<𝛾<𝜇(𝜂(𝜇𝐿2/2))/𝛼=𝜏/𝛼. For every 𝑛, let 𝑊𝑛 be the mapping generated by 𝑆𝑖 and 𝑡𝑖 as in (2.5). Let {𝑥𝑛} and {𝑢𝑛} be sequences generated by the following algorithm: 𝑢𝑛=𝑇𝑟𝑛𝑥𝑛,𝑦𝑛=𝛽𝑛𝑥𝑛+1𝛽𝑛𝑊𝑛𝑢𝑛,𝑥𝑛=𝛼𝑛𝑥𝛾𝑓𝑛+𝐼𝜇𝛼𝑛𝐴𝑦𝑛.(3.5) If {𝛼𝑛}, {𝛽𝑛}, and {𝑟𝑛} satisfy the following conditions:(i){𝛼𝑛}(0,1), lim𝑛𝛼𝑛=0;(ii)0<liminf𝑛𝛽𝑛limsup𝑛𝛽𝑛<1; (iii){𝑟𝑛}(0,), liminf𝑛𝑟𝑛>0.Then, {𝑥𝑛} converges strongly to a point 𝑥Ω, which solves the variational inequality (3.4).

Proof. The proof is divided into several steps.
Step 1. Show first that {𝑥𝑛} is bounded.
Take any 𝑝Ω, by (3.5) and Lemma 2.3, we derive that 𝑥𝑛=𝛼𝑝𝑛𝑥𝛾𝑓𝑛+𝜇𝐴𝑝𝐼𝜇𝛼𝑛𝐴𝑦𝑛𝐼𝜇𝛼𝑛𝐴𝑝𝛼𝑛𝑥𝛼𝛾𝑛𝑝+𝛼𝑛(𝛾𝑓𝑝)𝜇𝐴𝑝+1𝛼𝑛𝜏𝑦𝑛𝑝1𝛼𝑛𝑥(𝜏𝛾𝛼)𝑛𝑝+𝛼𝑛𝛾𝑓(𝑝)𝜇𝐴𝑝.(3.6) It follows that 𝑥𝑛𝑝(𝛾𝑓(𝑝)𝜇𝐴𝑝)/(𝜏𝛾𝛼).
Hence, {𝑥𝑛} is bounded, so are {𝑢𝑛} and {𝑦𝑛}. It follows from the Lipschitz continuity of 𝐴 that {𝐴𝑥𝑛} and {𝐴𝑢𝑛} are also bounded. From the nonexpansivity of 𝑓 and 𝑊𝑛, it follows that {𝑓(𝑥𝑛)} and {𝑊𝑛𝑥𝑛} are also bounded.
Step 2. Show that lim𝑛𝑢𝑛𝑥𝑛=0,lim𝑛𝑢𝑛𝑦𝑛=0.(3.7) Notice that 𝑢𝑛𝑦𝑛𝑢𝑛𝑥𝑛+𝑥𝑛𝑦𝑛=𝑢𝑛𝑥𝑛+𝛼𝑛𝑥𝛾𝑓𝑛𝜇𝐴𝑦𝑛.(3.8)
By Lemma 2.10, we have 𝑢𝑛𝑝2=𝑇𝑟𝑛𝑥𝑛𝑇𝑟𝑛𝑝2𝑥𝑛𝑝,𝑢𝑛1𝑝=2𝑢𝑛𝑝2+𝑥𝑛𝑝2𝑥𝑛𝑢𝑛2.(3.9) It follows that 𝑢𝑛𝑝2𝑥𝑛𝑝2𝑥𝑛𝑢𝑛2.(3.10)
Thus, from Lemma 2.1 and (3.10), we get 𝑥𝑛𝑝2=𝛼𝑛𝑥𝛾𝑓𝑛+𝜇𝐴𝑝𝐼𝜇𝛼𝑛𝐴𝑦𝑛𝐼𝜇𝛼𝑛𝐴𝑝21𝛼𝑛𝜏2𝑦𝑛𝑝2+2𝛼𝑛𝑥𝛾𝑓𝑛𝛾𝑓(𝑝)+𝛾𝑓(𝑝)𝜇𝐴𝑝,𝑥𝑛𝑝1𝛼𝑛𝜏2𝑢𝑛𝑝2+2𝛼𝑛𝑥𝛾𝑓𝑛𝛾𝑓(𝑝)+𝛾𝑓(𝑝)𝜇𝐴𝑝,𝑥𝑛𝑝1𝛼𝑛𝜏2𝑥𝑛𝑝2𝑥𝑛𝑢𝑛2+2𝛼𝑛𝑥𝛾𝛼𝑛𝑝2+2𝛼𝑛𝑥𝛾𝑓(𝑝)𝜇𝐴𝑝𝑛=𝑝12𝛼𝑛𝛼(𝜏𝛾𝛼)+𝑛𝜏2𝑥𝑛𝑝21𝛼𝑛𝜏2𝑥𝑛𝑢𝑛2+2𝛼𝑛𝑥𝛾𝑓(𝑝)𝜇𝐴𝑝𝑛𝑥𝑝𝑛𝑝2+𝛼𝑛𝜏2𝑥𝑛𝑝21𝛼𝑛𝜏2𝑥𝑛𝑢𝑛2+2𝛼𝑛𝑥𝛾𝑓(𝑝)𝜇𝐴𝑝𝑛.𝑝(3.11) It follows that 1𝛼𝑛𝜏2𝑥𝑛𝑢𝑛2𝛼𝑛𝜏2𝑥𝑛𝑝2+2𝛼𝑛𝑥𝛾𝑓(𝑝)𝜇𝐴𝑝𝑛.𝑝(3.12) Since 𝛼𝑛0, we have lim𝑛𝑢𝑛𝑥𝑛=0.(3.13) From (3.8), it is easy to get lim𝑛𝑢𝑛𝑦𝑛=0.(3.14)
Step 3. Show that lim𝑛𝑢𝑛𝑊𝑢𝑛𝑢=0,(3.15)𝑛𝑊𝑛𝑢𝑛𝑢𝑛𝑦𝑛+𝑦𝑛𝑊𝑛𝑢𝑛=𝑢𝑛𝑦𝑛+𝛽𝑛𝑥𝑛𝑢𝑛+𝑢𝑛𝑊𝑛𝑢𝑛.(3.16) This implies that 1𝛽𝑛𝑢𝑛𝑊𝑛𝑢𝑛𝑢𝑛𝑦𝑛+𝛽𝑛𝑥𝑛𝑢𝑛.(3.17) From condition (ii), (3.13), and (3.14), we have 𝑢𝑛𝑊𝑛𝑢𝑛0.(3.18) Notice that 𝑢𝑛𝑊𝑢𝑛𝑢𝑛𝑊𝑛𝑢𝑛+𝑊𝑛𝑢𝑛𝑊𝑢𝑛.(3.19) By Lemma 2.7 and (3.18), we get (3.15).
Since {𝑢𝑛} is bounded, so there exists a subsequence {𝑢𝑛𝑗} which converges weakly to 𝑥.
Step 4. Show that 𝑥Ω.
Since 𝐶 is closed and convex, 𝐶 is weakly closed. So, we have 𝑥𝐶.
From (3.15), we obtain 𝑊𝑢𝑛𝑗𝑥. From Lemmas 2.8, 2.4, and 2.13, we have 𝑥𝐹(𝑊)=𝑖=1𝐹(𝑆𝑖)=𝑖=1𝐹(𝑆𝑖).
By 𝑢𝑛=𝑇𝑟𝑛𝑥𝑛, for all 𝑛1, we have 𝐹𝑢𝑛+1,𝑦𝑟𝑛𝑦𝑢𝑛,𝑢𝑛𝑥𝑛0,𝑦𝐶.(3.20) It follows from (A2) that 1𝑟𝑛𝑦𝑢𝑛,𝑢𝑛𝑥𝑛𝐹𝑦,𝑢𝑛,𝑦𝐶.(3.21) Hence, we get 1𝑟𝑛𝑗𝑦𝑢𝑛𝑗,𝑢𝑛𝑗𝑥𝑛𝑗𝐹𝑦,𝑢𝑛𝑗,𝑦𝐶.(3.22) It follows from condition (iii), (3.13), and (A4) that 0𝐹𝑦,𝑥,y𝐶.(3.23) For 𝑠 with 0<𝑠1 and 𝑦𝐶, let 𝑦𝑠=𝑠𝑦+(1𝑠)𝑥. Since 𝑦𝐶 and 𝑥𝐶, we obtain 𝑦𝑠𝐶 and hence 𝐹(𝑦𝑠,𝑥)0. So, we have 𝑦0=𝑓𝑠,𝑦𝑠𝑦𝑠𝐹𝑠+𝑦,𝑦(1𝑠)𝐹𝑠,𝑥𝑦𝑠𝐹𝑠,𝑦.(3.24) Dividing by 𝑠, we get 𝐹𝑦𝑠,y0,𝑦𝐶.(3.25) Letting 𝑠0 and from (A3), we get 𝐹𝑥,𝑦0(3.26) for all 𝑦𝐶and𝑥EP(𝐹).Hence𝑥Ω.
Step 5. Show that 𝑥𝑛𝑥,where𝑥=𝑃Ω(𝐼𝜇𝐴+𝛾𝑓)𝑥: 𝑥𝑛𝑥=𝛼𝑛𝑥𝛾𝑓𝑛𝜇𝐴𝑥+𝐼𝜇𝛼𝑛𝐴𝑦𝑛𝐼𝜇𝛼𝑛𝐴𝑥.(3.27) Hence, we obtain 𝑥𝑛𝑥2=𝛼𝑛𝑥𝛾𝑓𝑛𝜇𝐴𝑥,𝑥𝑛𝑥+𝐼𝜇𝛼𝑛𝐴𝑦𝑛𝐼𝜇𝛼𝑛𝐴𝑥,𝑥𝑛𝑥;𝛼𝑛𝑥𝛾𝑓𝑛𝜇𝐴𝑥,𝑥𝑛𝑥+1𝛼𝑛𝜏𝑥𝑛𝑥2.(3.28) It follows that 𝑥𝑛𝑥21𝜏𝑥𝛾𝑓𝑛𝜇𝐴𝑥,𝑥𝑛𝑥=1𝜏𝛾𝑓𝑥𝑛𝑥𝑓,𝑥𝑛𝑥+𝑥𝛾𝑓𝜇𝐴𝑥,𝑥𝑛𝑥1𝜏𝑥𝛾𝛼𝑛𝑥2+𝑥𝛾𝑓𝜇𝐴𝑥,𝑥𝑛𝑥.(3.29) This implies that 𝑥𝑛𝑥2𝑥𝛾𝑓𝜇𝐴𝑥,𝑥𝑛𝑥.𝜏𝛾𝛼(3.30) In particular, 𝑥𝑛𝑗𝑥2𝑥𝛾𝑓𝜇𝐴𝑥,𝑥𝑛𝑗𝑥.𝜏𝛾𝛼(3.31)
Since 𝑥𝑛𝑗𝑥, it follows from (3.31) that 𝑥𝑛𝑗𝑥 as 𝑗. Next, we show that 𝑥 solves the variational inequality (3.4).
By the iterative algorithm (3.5), we have 𝑥𝑛=𝛼𝑛𝑥𝛾𝑓𝑛+𝐼𝜇𝛼𝑛𝐴𝑦𝑛=𝛼𝑛𝑥𝛾𝑓𝑛+𝐼𝜇𝛼𝑛𝐴𝑉𝑛𝑥𝑛.(3.32) Therefore, we have 𝜇𝛼𝑛𝐴𝑥𝑛𝛼𝑛𝑥𝛾𝑓𝑛=𝜇𝛼𝑛𝐴𝑥𝑛𝑥𝑛+𝐼𝜇𝛼𝑛𝐴𝑉𝑛𝑥𝑛,(3.33) that is, (𝜇𝐴𝛾𝑓)𝑥𝑛1=𝛼𝑛𝐼𝑉𝑛𝑥𝑛𝜇𝛼𝑛𝐴𝑥𝑛𝐴𝑉𝑛𝑥𝑛.(3.34) Hence, for 𝑝Ω, (𝜇𝐴𝛾𝑓)𝑥𝑛,𝑥𝑛1𝑝=𝛼𝑛𝐼𝑉𝑛𝑥𝑛𝜇𝛼𝑛𝐴𝑥𝑛𝐴𝑉𝑛𝑥𝑛,𝑥𝑛1𝑝=𝛼𝑛𝐼𝑉𝑛𝑥𝑛𝐼𝑉𝑛𝑝,𝑥𝑛𝑝+𝜇𝐴𝑥𝑛𝐴𝑉𝑛𝑥𝑛,𝑥𝑛𝑝𝜇𝐴𝑥𝑛𝐴𝑉𝑛𝑥𝑛,𝑥𝑛𝑝.(3.35)
Since 𝐼𝑉𝑛 is monotone (i.e., 𝑥𝑦,(𝐼𝑉𝑛)𝑥(𝐼𝑉𝑛)𝑦0, for all 𝑥,𝑦𝐻). This is due to the nonexpansivity of 𝑉𝑛.
Now replacing 𝑛 in (3.35) with 𝑛𝑗 and letting 𝑗, we obtain (𝜇𝐴𝛾𝑓)𝑥,𝑥𝑝=lim𝑗(𝜇𝐴𝛾𝑓)𝑥𝑛𝑗,𝑥𝑛𝑗𝑝lim𝑗𝜇𝐴𝑥𝑛𝑗𝐴𝑉𝑛𝑥𝑛𝑗,𝑥𝑛𝑗𝑝=0.(3.36)
That is, 𝑥Ω is a solution of (3.4). To show that the sequence {𝑥𝑛} converges strongly to 𝑥, we assume that 𝑥𝑛𝑘̂𝑥. By the same processing as the proof above, we derive ̂𝑥Ω. Moreover, it follows from the inequality (3.36) that (𝜇𝐴𝛾𝑓)𝑥,𝑥̂𝑥0.(3.37) Interchanging 𝑥 and ̂𝑥, we get (𝜇𝐴𝛾𝑓)̂𝑥,̂𝑥𝑥0.(3.38) By Lemma 2.5, adding up (3.37) and (3.38) yields (𝜇𝜂𝛾𝛼)𝑥̂𝑥2(𝜇𝐴𝛾𝑓)𝑥(𝜇𝐴𝛾𝑓)̂𝑥,𝑥̂𝑥0.(3.39) Hence 𝑥=̂𝑥 and, therefore, 𝑥𝑛𝑥 as 𝑛, (𝐼𝜇𝐴+𝛾𝑓)𝑥𝑥,𝑥𝑝0,𝑝Ω.(3.40) This is equivalent to the fixed point equation: 𝑃Ω(𝐼𝜇𝐴+𝛾𝑓)𝑥=𝑥.(3.41)

Theorem 3.2. Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻 and 𝐹 a bifunction from 𝐶×𝐶 to satisfying (A1)–(A4). Let 𝑆𝑖𝐶𝐶 be a family 𝜅𝑖-strict pseudocontractions for some 0𝜅𝑖<1. Assume the set Ω=𝑖=1𝐹(𝑆𝑖)𝐸𝑃(𝐹). Let 𝑓 be a contraction of 𝐻 into itself with 𝛼(0,1) and let 𝐴 be a 𝐿-Lipschitzian continuous operator and 𝜂-strongly monotone with 𝐿>0, 𝜂>0,0<𝜇<2𝜂/𝐿2, and 0<𝛾<𝜇(𝜂(𝜇𝐿2/2))/𝛼=𝜏/𝛼. For every 𝑛, let 𝑊𝑛 be the mapping generated by 𝑆𝑖 and 0<𝑡𝑖𝑏<1. Given 𝑥1𝐻, let {𝑥𝑛} and {𝑢𝑛} be sequences generated by the following algorithm: 𝑢𝑛=𝑇𝑟𝑛𝑥𝑛,𝑦𝑛=𝛽𝑛𝑥𝑛+1𝛽𝑛𝑊𝑛𝑢𝑛,𝑥𝑛+1=𝛼𝑛𝑥𝛾𝑓𝑛+𝐼𝜇𝛼𝑛𝐴𝑦𝑛.(3.42) If {𝛼𝑛}, {𝛽𝑛} and {𝑟𝑛} satisfy the following conditions:(i){𝛼𝑛}(0,1), lim𝑛𝛼𝑛=0 and 𝑛=1𝛼𝑛=;(ii)0<liminf𝑛𝛽𝑛limsup𝑛𝛽𝑛<1; (iii){𝑟𝑛}(0,), liminf𝑛𝑟𝑛>0 and lim𝑛|𝑟𝑛+1𝑟𝑛|=0.Then, {𝑥𝑛} converges strongly to 𝑥Ω, which solves the variational inequality (3.4).

Proof. The proof is divided into several steps.
Step 1. Show first that {𝑥𝑛} is bounded.
Taking any 𝑝Ω, we have 𝑥𝑛+1=𝛼𝑝𝑛𝑥𝛾𝑓𝑛+𝜇𝐴𝑝𝐼𝜇𝛼𝑛𝐴𝑦𝑛𝐼𝜇𝛼𝑛𝐴𝑝𝛼𝑛𝑥𝛾𝑓𝑛+𝛾𝑓(𝑝)+𝛾𝑓(𝑝)𝜇𝐴𝑝1𝛼𝑛𝜏𝑦𝑛𝑝𝛼𝑛𝑥𝛼𝛾𝑛𝑝+𝛼𝑛𝛾𝑓(𝑝)𝜇𝐴𝑝+1𝛼𝑛𝜏𝑦𝑛=𝑝1𝛼𝑛𝑥(𝜏𝛼𝛾)𝑛𝑝+𝛼𝑛(𝜏𝛼𝛾)𝛾𝑓(𝑝)𝜇𝐴𝑝𝑥𝜏𝛼𝛾max𝑛,𝑝𝛾𝑓(𝑝)𝜇𝐴𝑝.𝜏𝛼𝛾(3.43) By induction, we obtain 𝑥𝑛𝑝max{𝑥1𝑝,𝛾𝑓(𝑝)𝜇𝐴𝑝)/(𝜏𝛼𝛾)},𝑛1.Hence, {𝑥𝑛} is bounded, so are {𝑢𝑛} and {𝑦𝑛}. It follows from the Lipschitz continuity of 𝐴 that {𝐴𝑥𝑛} and {𝐴𝑢𝑛} are also bounded. From the nonexpansivity of 𝑓 and 𝑊𝑛, it follows that {𝑓(𝑥𝑛)} and {𝑊𝑛𝑥𝑛} are also bounded.
Step 2. Show that 𝑥𝑛+1𝑥𝑛0.(3.44)
Observe that 𝑢𝑛+1𝑢𝑛=𝑇𝑟𝑛+1𝑥𝑛+1𝑇𝑟𝑛𝑥𝑛𝑇𝑟𝑛+1𝑥𝑛+1𝑇𝑟𝑛+1𝑥𝑛+𝑇𝑟𝑛+1𝑥𝑛𝑇𝑟𝑛𝑥𝑛𝑥𝑛+1𝑥𝑛+𝑇𝑟𝑛+1𝑥𝑛𝑇𝑟𝑛𝑥𝑛,(3.45) and from (2.5), we have 𝑊𝑛+1𝑢𝑛𝑊𝑛𝑢𝑛=𝑡1𝑆1𝑈𝑛+1,2𝑢𝑛𝑡1𝑆1𝑈𝑛,2𝑢𝑛𝑡1𝑈𝑛+1,2𝑢𝑛𝑈𝑛,2𝑢𝑛=𝑡1𝑡2𝑆2𝑈𝑛+1,3𝑢𝑛𝑡2𝑆2𝑈𝑛,3𝑢𝑛𝑡1𝑡2𝑈𝑛+1,3𝑢𝑛𝑈𝑛,3𝑢𝑛𝑛𝑖=1𝑡𝑖𝑈𝑛+1,𝑛+1𝑢𝑛𝑈𝑛,𝑛+1𝑢𝑛𝑀1𝑛𝑖=1𝑡𝑖,(3.46) where 𝑀1=sup𝑛{𝑈𝑛+1,𝑛+1𝑢𝑛𝑈𝑛,𝑛+1𝑢𝑛}.
Suppose 𝑥𝑛+1=𝛽𝑛𝑥𝑛+(1𝛽𝑛)𝑧𝑛, then 𝑧𝑛=(𝑥𝑛+1𝛽𝑛𝑥𝑛)/(1𝛽𝑛)=(𝛼𝑛𝛾𝑓(𝑥𝑛)+(𝐼𝜇𝛼𝑛𝐴)𝑦𝑛𝛽𝑛𝑥𝑛)/(1𝛽𝑛).
Hence, we have 𝑧𝑛+1𝑧𝑛=𝛼𝑛+1𝑥𝛾𝑓𝑛+1+𝐼𝜇𝛼𝑛+1𝐴𝑦𝑛+1𝛽𝑛+1𝑥𝑛+11𝛽𝑛+1𝛼𝑛𝑥𝛾𝑓𝑛+𝐼𝜇𝛼𝑛𝐴𝑦𝑛𝛽𝑛𝑥𝑛1𝛽𝑛=𝛼𝑛+1𝑥𝛾𝑓𝑛+1𝜇𝐴𝑦𝑛+11𝛽𝑛+1+𝑦𝑛+1𝛽𝑛+1𝑥𝑛+11𝛽𝑛+1𝛼𝑛𝑥𝛾𝑓𝑛𝜇𝐴𝑦𝑛1𝛽𝑛𝑦𝑛𝛽𝑛𝑥𝑛1𝛽𝑛=𝛼𝑛+1𝑥𝛾𝑓𝑛+1𝜇𝐴𝑦𝑛+11𝛽𝑛+1+𝛽𝑛+1𝑥𝑛+1+1𝛽𝑛+1𝑊𝑛+1𝑢𝑛+1𝛽𝑛+1𝑥𝑛+11𝛽𝑛+1𝛼𝑛𝑥𝛾𝑓𝑛𝜇𝐴𝑦𝑛1𝛽𝑛𝛽𝑛𝑥𝑛+1𝛽𝑛𝑊𝑛𝑢𝑛𝛽𝑛𝑥𝑛1𝛽𝑛𝛼𝑛+1𝑥𝛾𝑓𝑛+1𝜇𝐴𝑦𝑛+11𝛽𝑛+1𝛼𝑛𝑥𝛾𝑓𝑛𝜇𝐴𝑦𝑛1𝛽𝑛+𝑊𝑛+1𝑢𝑛+1𝑊𝑛𝑢𝑛.(3.47)
It follows from (3.45), (3.46), and the above result that 𝑧𝑛+1𝑧𝑛𝛼𝑛+11𝛽𝑛+1𝑥𝛾𝑓𝑛+1+𝜇𝐴𝑦𝑛+1+𝛼𝑛1𝛽𝑛𝑥𝛾𝑓𝑛+𝜇𝐴𝑦𝑛+𝑊𝑛+1𝑢𝑛+1𝑊𝑛𝑢𝑛𝛼𝑛+11𝛽𝑛+1+𝛼𝑛1𝛽𝑛𝑀2+𝑊𝑛+1𝑢𝑛+1𝑊𝑛+1𝑢𝑛+𝑊𝑛+1𝑢𝑛𝑊𝑛𝑢𝑛𝛼𝑛+11𝛽𝑛+1+𝛼𝑛1𝛽𝑛𝑀2+𝑢𝑛+1𝑢𝑛+𝑊𝑛+1𝑢𝑛𝑊𝑛𝑢𝑛𝑥𝑛+1𝑥𝑛+𝑇𝑟𝑛+1𝑥𝑛𝑇𝑟𝑛𝑥𝑛+𝛼𝑛+11𝛽𝑛+1+𝛼𝑛1𝛽𝑛𝑀2+𝑀1𝑛𝑖=1𝑡𝑖,(3.48) where 𝑀2=sup𝑛{𝛾𝑓(𝑥𝑛)+𝜇𝐴𝑦𝑛}. Hence, we get 𝑧𝑛+1𝑧𝑛𝑥𝑛+1𝑥𝑛𝑇𝑟𝑛+1𝑥𝑛𝑇𝑟𝑛𝑥𝑛+𝛼𝑛+11𝛽𝑛+1+𝛼𝑛1𝛽𝑛𝑀2+𝑀1𝑛𝑖=1𝑡𝑖.(3.49) From condition (i), (iii), 0<𝑡𝑛𝑏<1, and Lemma 2.12, we obtain limsup𝑛𝑧𝑛+1𝑧𝑛𝑥𝑛+1𝑥𝑛0.(3.50) By Lemma 2.11,we have lim𝑛𝑧𝑛𝑥𝑛=0. Thus, lim𝑛𝑥𝑛+1𝑥𝑛=lim𝑛1𝛽𝑛𝑧𝑛𝑥𝑛=0.(3.51)
By Lemma 2.12, (3.45) and (3.44), we obtain 𝑢𝑛+1𝑢𝑛0.(3.52)
Step 3. Show that 𝑥𝑛𝑊𝑥𝑛0.(3.53) Observe that 𝑥𝑛𝑊𝑛𝑥𝑛𝑥𝑛𝑊𝑛𝑢𝑛+𝑊𝑛𝑢𝑛𝑊𝑛𝑥𝑛𝑥𝑛𝑊𝑛𝑢𝑛+𝑢𝑛𝑥𝑛,𝑥𝑛𝑊𝑛𝑢𝑛𝑥𝑛𝑥𝑛+1+𝑥𝑛+1𝑦𝑛+𝑦𝑛𝑊𝑛𝑢𝑛=𝑥𝑛𝑥𝑛+1+𝑥𝑛+1𝑦𝑛+𝛽𝑛𝑢𝑛𝑥𝑛+𝑥𝑛𝑊𝑛𝑢𝑛.(3.54) From condition (i) and (3.5),we can obtain 1𝛽𝑛𝑥𝑛𝑊𝑛𝑢𝑛𝑥𝑛𝑥𝑛+1+𝑥𝑛+1𝑦𝑛+𝛽𝑛𝑢𝑛𝑥𝑛𝑥𝑛𝑥𝑛+1+𝛼𝑛𝑥𝛾𝑓𝑛𝜇𝐴𝑦𝑛+𝛽𝑛𝑢𝑛𝑥𝑛.(3.55)
By Lemma 2.10, we get 𝑢𝑛𝑝2=𝑇𝑟𝑛𝑥𝑛𝑇𝑟𝑛𝑝2𝑇𝑟𝑛𝑥𝑛𝑇𝑟𝑛𝑝,𝑥𝑛=1𝑝2𝑢𝑛𝑝2+𝑥𝑛𝑝2+𝑥𝑛𝑢𝑛2.(3.56) This implies that 𝑢𝑛𝑝2𝑥𝑛𝑝2𝑥𝑛𝑢𝑛2.(3.57) By nonexpansivity of 𝑊𝑛, we have 𝑦𝑛𝑝2𝛽𝑛𝑥𝑛𝑝2+1𝛽𝑛𝑢𝑛𝑝2𝑥𝑛𝑝21𝛽𝑛𝑥𝑛𝑢𝑛2.(3.58) It follows from (3.42) that 𝑥𝑛+1𝑝2=𝛼𝑛𝑥𝛾𝑓𝑛+𝑝𝐼𝜇𝛼𝑛𝐴𝑦𝑛𝐼𝜇𝛼𝑛𝐴𝑝+𝛼𝑛(𝑝𝜇𝐴𝑝)2𝛼𝑛𝑥𝛾𝑓𝑛𝑝2+1𝛼𝑛𝜏𝑦𝑛𝑝2+𝛼𝑛𝑝𝜇𝐴𝑝2𝛼𝑛𝑥𝛾𝑓𝑛𝑝2+1𝛼𝑛𝜏𝑥𝑛𝑝21𝛽𝑛𝑥𝑛𝑢𝑛2+𝛼𝑛𝑝𝜇𝐴𝑝2𝛼𝑛𝑥𝛾𝑓𝑛𝑝2+𝑥𝑛𝑝21𝛽𝑛𝑥𝑛𝑢𝑛2+𝛼𝑛𝑝𝜇𝐴𝑝2.(3.59) This implies that 1𝛽𝑛𝑥𝑛𝑢𝑛2𝛼𝑛𝑥𝛾𝑓𝑛𝑝2+𝑝𝜇𝐴𝑝2+𝑥𝑛𝑝2𝑥𝑛+1𝑝2𝛼𝑛𝑥𝛾𝑓𝑛𝑝2+𝑝𝜇𝐴𝑝2+𝑥𝑛+𝑥𝑝𝑛+1𝑥𝑝𝑛+1𝑥𝑛.(3.60) From condition (i), (ii), and (3.44), we have 𝑥𝑛𝑢𝑛0.(3.61) Further we have 𝑥𝑛𝑊𝑛𝑢𝑛0. Thus we get 𝑥𝑛𝑊𝑛𝑥𝑛0.(3.62)
On the other hand, we have 𝑥𝑛𝑊𝑥𝑛𝑥𝑛𝑊𝑛𝑥𝑛+𝑊𝑛𝑥𝑛𝑊𝑥𝑛𝑥𝑛𝑊𝑛𝑥𝑛+sup𝑥𝑛𝐶𝑊𝑛𝑥𝑛𝑊𝑥𝑛.(3.63) Combining (3.62), the last inequality, and Lemma 2.7, we obtain (3.53).
Step 4. Show that limsup𝑛(𝛾𝑓𝜇𝐴)𝑥,𝑥𝑛𝑥0,(3.64) where 𝑥=𝑃Ω(𝐼𝜇𝐴+𝛾𝑓)𝑥 is a unique solution of the variational inequality (3.4). Indeed, take a subsequence {𝑥𝑛𝑗} of {𝑥𝑛} such that limsup𝑛(𝛾𝑓𝜇𝐴)𝑥,𝑥𝑛𝑥=lim𝑗(𝛾𝑓𝜇𝐴)𝑥,𝑥𝑛𝑗𝑥.(3.65)
Since {𝑥𝑛𝑗} is bounded, there exists a subsequence {𝑥𝑛𝑗𝑘} of {𝑥𝑛𝑗} which converges weakly to 𝑞. Without loss of generality, we can assume 𝑥𝑛𝑗𝑞. From (3.53), we obtain 𝑊𝑥𝑛𝑗𝑞.
By the same argument as in the proof of Theorem 3.1, we have 𝑞Ω. Since 𝑥=𝑃Ω(𝐼𝜇𝐴+𝛾𝑓)𝑥, it follows that limsup𝑛(𝛾𝑓𝜇𝐴)𝑥,𝑥𝑛𝑥=lim𝑗(𝛾𝑓𝜇𝐴)𝑥,𝑥𝑛𝑗𝑥=(𝛾𝑓𝜇𝐴)𝑥,𝑞𝑥0.(3.66)
Step 5. Show that 𝑥𝑛𝑥.(3.67) Since (𝛾𝑓𝜇𝐴)𝑥,𝑥𝑛+1𝑥=(𝛾𝑓𝜇𝐴)𝑥,𝑥𝑛+1𝑥𝑛+(𝛾𝑓𝜇𝐴)𝑥,𝑥𝑛𝑥(𝛾𝑓𝜇𝐴)𝑥𝑥𝑛+1𝑥𝑛+(𝛾f𝜇𝐴)𝑥,𝑥𝑛𝑥.(3.68) It follows from (3.44) and (3.66) that limsup𝑛(𝛾𝑓𝜇𝐴)𝑥,𝑥𝑛+1𝑥𝑥0.𝑛+1𝑥2=𝛼𝑛𝑥𝛾𝑓𝑛+𝐼𝜇𝛼𝑛𝐴𝑦𝑛𝑥2=𝐼𝜇𝛼𝑛𝐴𝑦𝑛𝐼𝜇𝛼𝑛𝐴𝑥+𝛼𝑛𝑥𝛾𝑓𝑛𝜇𝐴𝑥2𝐼𝜇𝛼𝑛𝐴𝑦𝑛𝐼𝜇𝛼𝑛𝐴𝑥2+2𝛼𝑛𝑥𝛾𝑓𝑛𝜇𝐴𝑥,𝑥𝑛+1𝑥1𝛼𝑛𝜏2𝑦𝑛𝑥2+2𝛼𝑛𝑥𝛾𝑓𝑛𝑥𝛾𝑓,𝑥𝑛+1𝑥+2𝛼𝑛(𝛾𝑓𝜇𝐴)𝑥,𝑥𝑛+1𝑥1𝛼𝑛𝜏2𝑥𝑛𝑥2+𝛼𝑛𝑥𝛼𝛾𝑛𝑥2+𝑥𝑛+1𝑥2+2𝛼𝑛(𝛾𝑓𝜇𝐴)𝑥,𝑥𝑛+1𝑥.(3.69) This implies that 𝑥𝑛+1𝑥21𝛼𝑛𝜏2+𝛼𝑛𝛼𝛾1𝛼𝑛𝑥𝛼𝛾𝑛𝑥2+2𝛼𝑛1𝛼𝑛𝛼𝛾(𝛾𝑓𝜇𝐴)𝑥,𝑥𝑛+1𝑥12𝛼𝑛(𝜏𝛼𝛾)1𝛼𝑛𝑥𝛼𝛾𝑛𝑥2+2𝛼𝑛1𝛼𝑛𝛼𝛾(𝛾𝑓𝜇𝐴)𝑥,𝑥𝑛+1𝑥+𝛼𝑛𝜏21𝛼𝑛𝑀𝛼𝛾3,(3.70) where 𝑀3=sup𝑛𝑥𝑛𝑥2, 𝑛1. It is easily to see that 𝛾𝑛=2𝛼𝑛(𝜏𝛼𝛾)/(1𝛼𝑛𝛼𝛾). Hence, by Lemma 2.2, the sequence {𝑥𝑛} converges strongly to 𝑥.

Remark 3.3. If 𝐹0, then Theorem 3.2 reduces to Theorem 3.1 of Wang [10].

Acknowledgments

The authors would like to thank the referee for valuable suggestions to improve the manuscript NSFC Tianyuan Youth Foundation of Mathematics of China (no. 11126136), and the Fundamental Research Funds for the Central Universities (GRANT: ZXH2011C002).

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