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Journal of Applied Mathematics
Volume 2012 (2012), Article ID 572326, 18 pages
http://dx.doi.org/10.1155/2012/572326
Research Article

Hybrid Iterative Scheme by a Relaxed Extragradient Method for Equilibrium Problems, a General System of Variational Inequalities and Fixed-Point Problems of a Countable Family of Nonexpansive Mappings

1College of Science, Civil Aviation University of China, Tianjin 300300, China
2College of Science, National University of Defense Technology, Changsha 410073, China

Received 29 November 2011; Accepted 5 December 2011

Academic Editor: Yonghong Yao

Copyright © 2012 Qiao-Li Dong 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

Based on the relaxed extragradient method and viscosity method, we introduce a new iterative method for finding a common element of solution of equilibrium problems, the solution set of a general system of variational inequalities, and the set of fixed points of a countable family of nonexpansive mappings in a real Hilbert space. Furthermore, we prove the strong convergence theorem of the studied iterative method. The results of this paper extend and improve the results of Ceng et al., (2008), W. Kumam and P. Kumam, (2009), Yao et al., (2010) and many others.

1. Introduction

Let 𝐻 be a real Hilbert space with the inner product , and the norm . Let 𝐶 be a closed convex subset of 𝐻. Let 𝐹 be a bifunction of 𝐶×𝐶 into , where is the set of real numbers. The equilibrium problem for 𝐹𝐶×𝐶 is to find 𝑥𝐶 such that 𝐹(𝑥,𝑦)0,𝑦𝐶.(1.1) The set of solutions of (1.1) is denoted by EP(𝐹). The equilibrium problems covers, as special cases, monotone inclusion problems, saddle point problems, minimization problems, optimization problems, variational inequality problems, Nash equilibria in noncooperative games, and various forms of feasibility problems (see [14] and the references therein).

A mapping 𝐴𝐶𝐻 is called 𝛼-inverse-strongly monotone if there exists a positive real number 𝛼 such that 𝐴𝑢𝐴𝑣,𝑢𝑣𝛼𝐴𝑢𝐴𝑣2,𝑢,𝑣𝐶.(1.2) It is obvious that any 𝛼-inverse-strongly monotone mapping 𝐴 is monotone and Lipschitz continuous. A mapping 𝑇𝐶𝐶 is said to be nonexpansive if 𝑇𝑥𝑇𝑦𝑥𝑦,𝑥,𝑦𝐶.(1.3) We denote by 𝐹(𝑇) the set of fixed points of 𝑇. Recently, Wang and Guo [5] introduced an iterative scheme for a countable family of nonexpansive mappings.

Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻. For a given nonlinear operator 𝐴𝐶𝐻, consider the following variational inequality problem of finding 𝑥𝐶 such that 𝐴𝑥,𝑥𝑥0,𝑥𝐶.(1.4) The set of solutions of the variational inequality (1.4) is denoted by VI(𝐶,𝐴) (see [69] and the references therein).

Let 𝐴,𝐵𝐶𝐻 be two mappings. Consider the following problem of finding (𝑥,𝑦)𝐶×𝐶 such that 𝜆𝐴𝑦+𝑥𝑦,𝑥𝑥0,𝑥𝐶,𝜇𝐵𝑥+𝑦𝑥,𝑥𝑦0,𝑥𝐶,(1.5) which is called a general system of variational inequalities, where 𝜆>0 and 𝜇>0 are two constants. The set of solutions of (1.5) is denoted by GSVI(𝐴,𝐵,and𝐶). In particular, if 𝐴=𝐵, then problem (1.5) reduces to finding (𝑥,𝑦)𝐶×𝐶 such that 𝜆𝐴𝑦+𝑥𝑦,𝑥𝑥0,𝑥𝐶,𝜇𝐴𝑥+𝑦𝑥,𝑥𝑦0,𝑥𝐶,(1.6) which is defined by Verma [7] (see also [10]) and is called the new system of variational inequalities. Further, if we add up the requirement that 𝑥=𝑦, then problem (1.6) reduces to the classical variational inequality problem (1.4). Recently, Yao et al. [11] presented system of variational inequalities in Banach space. For solving problem (1.5), recently, Ceng et al. [12] introduced and studied a relaxed extragradient method. Based on the relaxed extragradient method and the viscosity approximation method, W. Kumam and P. Kumam [13] constructed a new viscosity-relaxed extragradient approximation method. Very recently, based on the extragradient method, Yao et al. [14] proposed an iterative method for finding a common element of the set of a general system of variational inequalities and the set of fixed points of a strictly pseudocontractive mapping in a real Hilbert space.

Motivated and inspired by the above works, in this paper, we introduce an iterative method based on the extragradient method and viscosity method for finding a common element of solution of equilibrium problems, the solution set of a general system of variational inequalities, and the set of fixed points of a countable family of nonexpansive mappings in a real Hilbert space. Furthermore, we prove the strong convergence theorem of the proposed iterative method.

2. Preliminaries

Let 𝐶 be a closed convex subset of 𝐻, and let 𝑇𝐶𝐶 be nonexpansive such that 𝐹(𝑇). For all ̂𝑥𝐹(𝑇) and all 𝑥𝐶, we have 𝑥̂𝑥2𝑇𝑥𝑇̂𝑥2=𝑇𝑥̂𝑥2=(𝑇𝑥𝑥)+(𝑥̂𝑥)2=𝑇𝑥𝑥2+𝑥̂𝑥2+2𝑇𝑥𝑥,𝑥̂𝑥,(2.1) and hence 𝑇𝑥𝑥22𝑥𝑇𝑥,𝑥̂𝑥,̂𝑥𝐹(𝑇),𝑥𝐶.(2.2)

Remark 2.1. Let A be 𝛼-inverse-strongly monotone. For all 𝑥,𝑦𝐶, 𝜆>0, we have (𝐼𝜆𝐴)𝑥(𝐼𝜆𝐴)𝑦2=𝑥𝑦22𝜆𝐴𝑥𝐴𝑦,𝑥𝑦+𝜆2𝐴𝑥𝐴𝑦2𝑥𝑦2+𝜆(𝜆2𝛼)𝐴𝑥𝐴𝑦2.(2.3)So, if 𝜆2𝛼, then 𝐼𝜆𝐴 is a nonexpansive mapping from 𝐶 to 𝐻.

Recall that the (nearest point) projection 𝑃𝐶 from 𝐻 onto 𝐶 assigns to each 𝑥𝐻 the unique point 𝑃𝐶𝑥𝐶 satisfying the property 𝑥𝑃𝐶𝑥=min𝑦𝐶𝑥𝑦.(2.4) The following characterizes the projection 𝑃𝐶.

Lemma 2.2. Given that 𝑥𝐻 and 𝑦𝐶, then 𝑃𝐶𝑥=𝑦 if and only if there holds the inequality 𝑥𝑦,𝑦𝑧0𝑧𝐶.(2.5)

Lemma 2.3 (see [15]). Let 𝐻 be a Hilbert space, 𝐶 a closed convex subset of 𝐻, and 𝑇𝐶𝐶 a nonexpansive mapping with 𝐹(𝑇). If {𝑥𝑛} is a sequence in 𝐶 weakly converging to 𝑥𝐶 and if {(1𝑇)𝑥𝑛} converges strongly to 𝑦, then(1𝑇)𝑥=𝑦.

Lemma 2.4 (see [16]). Assume that {𝑎𝑛} is a sequence of nonnegative real numbers such that 𝛼𝑛+11𝛾𝑛𝛼𝑛+𝛿𝑛,𝑛0,(2.6) where two sequences {𝛾𝑛}(0,1) and {𝛿𝑛} satisfy(1)𝑛=1𝛾𝑛=; (2)limsup𝑛𝛿𝑛/𝛾𝑛0𝑜𝑟𝑛=1|𝛿𝑛|<. Then lim𝑛𝛼𝑛=0.

Lemma 2.5 (see [12]). For given 𝑥,𝑦𝐶, (𝑥,𝑦) is a solution of problem (1.5) if and only if 𝑥 is a fixed point of the mapping 𝐺𝐶𝐶 defined by 𝐺(𝑥)=𝑃𝐶𝑃𝐶(𝑥𝜇𝐵𝑥)𝜆𝐴𝑃𝐶(𝑥𝜇𝐵𝑥),𝑥𝐶,(2.7) where 𝑦=𝑃𝐶(𝑥𝜇𝐵𝑥).

Note that the mapping 𝐺 is nonexpansive provided that 𝜆(0,2𝛼) and 𝜇(0,2𝛽).

Throughout this paper, the set of fixed points of the mapping 𝐺 is denoted by Γ.

Lemma 2.6 (see [1]). Let 𝐶 be a nonempty closed convex subset of 𝐻 and 𝐹𝐶×𝐶 satisfy following conditions: (A1)𝐹(𝑥,𝑥)=0, 𝑥𝐶; (A2)𝐹 is monotone, that is, 𝐹(𝑥,𝑦)+𝐹(𝑦,𝑥)0, 𝑥,𝑦𝐶;(A3)limsup𝑡0+𝐹(𝑡𝑧+(1𝑡)𝑥,𝑦)𝐹(𝑥,𝑦), 𝑥,𝑦,𝑧𝐶;(A4)for each 𝑥𝐶, 𝐹(𝑥,) is convex and lower semicontinuous.For 𝑥𝐶 and 𝑟>0, set 𝑇𝐹𝑟𝐻𝐶 to be 𝑇𝐹𝑟1(𝑥)=𝑧𝐶𝐹(𝑧,𝑦)+𝑟𝑦𝑧,𝑧𝑥0,𝑦𝐶.(2.8) Then 𝑇𝐹𝑟 is well defined and the following holds: (1)𝑇𝐹𝑟 is single valued;(2)𝑇𝐹𝑟 is firmly nonexpansive [17], that is, for any 𝑥,𝑦𝐸, 𝑇𝐹𝑟𝑥𝑇𝐹𝑟𝑦2𝑇𝐹𝑟𝑥𝑇𝐹𝑟𝑦,𝑥𝑦;(2.9)(3)𝐹(𝑇𝐹𝑟)=𝐸𝑃(𝐹); (4)𝐸𝑃(𝐹) is closed and convex.

By the proof of Lemma  5 in [2] (also see [3]), we have following lemma.

Lemma 2.7. Let 𝐶 be a nonempty closed convex subset of a Hilbert space 𝐻 and 𝐹𝐶×𝐶 be a bifunction. Let 𝑥𝐶 and 𝑟1,𝑟2(0,). Then 𝑇𝑟1𝑥𝑇𝑟2𝑥||||𝑟12𝑟1||||𝑇𝑟1𝑥+𝑥.(2.10)

3. Main Results

Theorem 3.1. Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻. Let the mappings 𝐴,𝐵𝐶𝐻 be 𝛼-inverse strongly monotone and 𝛽-inverse strongly monotone, respectively. Let 𝐹 be a bifunction from 𝐶×𝐶 to satisfying (A1)–(A4) and {𝑇𝑛}𝑛=1𝐶𝐶 be a countable family of nonexpansive mappings such that Ω=𝑛=1𝐹(𝑇𝑛)𝐸𝑃(𝐹)Γ. Let 𝑓𝐶𝐶 be a contraction with coefficient 𝜌(0,1/2). Set 𝛽0=1. For given 𝑥1𝐶 arbitrarily, let the sequences {𝑥𝑛}, {𝑦𝑛}, {𝑧𝑛}, and {𝑢𝑛} be generated by 𝐹𝑢𝑛+1,𝑦𝑟𝑛𝑦𝑢𝑛,𝑢𝑛𝑥𝑛𝑧0,𝑦𝐶,𝑛=𝑃𝐶𝑢𝑛𝜇𝐵𝑢𝑛,𝑦𝑛=𝛼𝑛𝑓𝑥𝑛+1𝛼𝑛𝑃𝐶𝑧𝑛𝜆𝐴𝑧𝑛,𝑥𝑛+1=𝛽𝑛𝑥𝑛+𝜎𝑛𝑛𝑖=1𝛽𝑖1𝛽𝑖𝑇𝑖𝑦𝑛+1𝛽𝑛1𝜎𝑛𝑃𝐶𝑧𝑛𝜆𝐴𝑧𝑛,(3.1) where 𝜆(0,2𝛼), 𝜇(0,2𝛽), and sequences {𝛼𝑛}[0,1], {𝛽𝑛}[0,1], {𝜎𝑛}[0,1], and {𝑟𝑛}(𝑟,), 𝑟>0, are such that
(i){𝛽𝑛} is strictly decreasing, (ii)0<liminf𝑛𝛽nlimsup𝑛𝛽𝑛<1, (iii)lim𝑛𝛼𝑛=0 and 𝑛=1𝛼𝑛=,(iv)𝜎𝑛>1/2(1𝜌),𝑛=1|𝜎𝑛𝜎𝑛1|<, (v)𝑛=1|𝑟𝑛𝑟𝑛1|<.Then the sequence {𝑥𝑛} generated by (3.1) converges strongly to 𝑥=𝑃Ω𝑓(𝑥), and (𝑥,𝑦) is a solution of the general system of variational inequalities (1.5), where 𝑦=𝑃𝐶(𝑥𝜇𝐵𝑥).

Proof. The proof is divided into several steps.
Step 1. The sequence {𝑥𝑛} defined by (3.1) is bounded.
For each 𝑥Ω, from Lemma 2.6, we have 𝑢𝑛=𝑇𝑟𝑛𝑥𝑛 and hence 𝑢𝑛𝑥=𝑇𝑟𝑛𝑥𝑛𝑇𝑟𝑛𝑥𝑥𝑛𝑥.(3.2) Since 𝑓 is a 𝜌-contraction mapping, using (2.3) and (3.2), we have 𝑧𝑛𝑦2=𝑃𝐶𝑢𝑛𝜇𝐵𝑢𝑛𝑃𝐶𝑥𝜇𝐵𝑥2𝑢𝑛𝜇𝐵𝑢𝑛𝑥𝜇𝐵𝑥2𝑢𝑛𝑥2+𝜇(𝜇2𝛽)𝐵𝑢𝑛𝐵𝑥2𝑥𝑛𝑥2+𝜇(𝜇2𝛽)𝐵𝑢𝑛𝐵𝑥2𝑥𝑛𝑥2.(3.3) Set 𝑣𝑛=𝑃𝐶(𝑧𝑛𝜆𝐴𝑧𝑛). Since 𝑃𝐶 is nonexpansive, from (2.3), we have 𝑣𝑛𝑥2=𝑃𝐶𝑧𝑛𝜆𝐴𝑧𝑛𝑃𝐶𝑦𝜆𝐴𝑦2𝑧𝑛𝑦2+𝜆(𝜆2𝛼)𝐴𝑧𝑛𝐴𝑦2𝑧𝑛𝑦2.(3.4) Hence we get 𝑦𝑛𝑥=𝛼𝑛𝑓𝑥𝑛+1𝛼𝑛𝑣𝑛𝑥𝛼𝑛𝑓𝑥𝑛𝑥+1𝛼𝑛𝑧𝑛𝑦𝛼𝑛𝑓𝑥𝑛𝑥𝑓+𝑓𝑥𝑥+1𝛼𝑛𝑥𝑛𝑥1𝛼𝑛(𝑥1𝜌)𝑛𝑥+𝛼𝑛𝑓𝑥𝑥.(3.5) From (3.1) and (3.3)–(3.5), we get 𝑥𝑛+1𝑥=𝛽𝑛𝑥𝑛+𝜎𝑛𝑛𝑖=1𝛽𝑖1𝛽𝑖𝑇𝑖𝑦𝑛+1𝛽𝑛1𝜎𝑛𝑣𝑛𝑥𝛽𝑛𝑥𝑛𝑥+𝜎𝑛𝑛𝑖=1𝛽𝑖1𝛽𝑖𝑇𝑖𝑦𝑛𝑥+1𝛽𝑛1𝜎𝑛𝑣𝑛𝑥𝛽𝑛𝑥𝑛𝑥+𝜎𝑛𝑛𝑖=1𝛽𝑖1𝛽𝑖𝑦𝑛𝑥+1𝛽𝑛1𝜎𝑛𝑧𝑛𝑦1𝜎𝑛1𝛽𝑛𝑥𝑛𝑥+𝜎𝑛1𝛽𝑛𝑦𝑛𝑦1𝜎𝑛1𝛽𝑛𝑥𝑛𝑥+𝜎𝑛1𝛽𝑛1𝛼𝑛𝑥(1𝜌)𝑛𝑥+𝛼𝑛𝑓𝑥𝑥1𝜎𝑛1𝛽𝑛𝛼𝑛𝑥(1𝜌)𝑛𝑥+𝜎𝑛1𝛽𝑛𝛼𝑛1(1𝜌)𝑓𝑥1𝜌𝑥𝑀,(3.6) where 𝑀=max{𝑥1𝑥,1/(1𝜌)𝑓(𝑥)𝑥}. Hence, {𝑥𝑛} is bounded and therefore {𝑢𝑛}, {𝑧𝑛}, {𝑣𝑛}, and {𝑦𝑛} are also bounded.
Step 2. lim𝑛𝑥𝑛+1𝑥𝑛=0. Since 𝑢𝑛=𝑇𝑟𝑛𝑥𝑛 and 𝑢𝑛1=𝑇𝑟𝑛1𝑥𝑛1, using Lemma 2.7, we have 𝑢𝑛𝑢𝑛1=𝑇𝑟𝑛𝑥𝑛𝑇𝑟𝑛1𝑥𝑛1𝑇𝑟𝑛𝑥𝑛𝑇𝑟𝑛𝑥𝑛1+𝑇𝑟𝑛𝑥𝑛1𝑇𝑟𝑛1𝑥𝑛1𝑥𝑛𝑥𝑛1+||||𝑟1𝑛𝑟𝑛1||||𝑢𝑛1+𝑥𝑛1𝑥𝑛𝑥𝑛1+1𝑟||𝑟𝑛1𝑟𝑛||𝐿,(3.7) where 𝐿=sup{𝑢𝑛+𝑥𝑛𝑛=1,2,}. From (3.1) and (3.7), it follows that 𝑧𝑛𝑧𝑛1=𝑃𝐶(𝐼𝜇𝐵)𝑢𝑛𝑃𝐶(𝐼𝜇𝐵)𝑢𝑛1𝑢𝑛𝑢𝑛1𝑥𝑛𝑥𝑛1+1𝑟||𝑟𝑛1𝑟𝑛||𝐿.(3.8) From (3.1) and (3.8), we have 𝑦𝑛𝑦𝑛1𝛼𝑛𝑓𝑥𝑛𝑥𝑓𝑛1+1𝛼𝑛𝑃𝐶(𝐼𝜆𝐴)𝑧𝑛𝑃𝐶(𝐼𝜆𝐴)𝑧𝑛1𝛼𝑛𝜌𝑥𝑛𝑥𝑛1+1𝛼𝑛𝑧𝑛𝑧𝑛1𝛼𝑛𝜌𝑥𝑛𝑥𝑛1+1𝛼𝑛𝑥𝑛𝑥𝑛1+1𝑟||𝑟𝑛1𝑟𝑛||𝐿1𝛼𝑛𝑥(1𝜌)𝑛𝑥𝑛1+1𝛼𝑛1𝑟||𝑟𝑛1𝑟𝑛||𝐿.(3.9) By definition of scheme (3.1), we have 𝑥𝑛+1𝑥𝑛=𝛽𝑛𝑥𝑛𝑥𝑛1+𝛽𝑛𝛽𝑛1𝑥𝑛1+𝜎𝑛𝑛𝑖=1𝛽𝑖1𝛽𝑖𝑇𝑖𝑦𝑛𝑇𝑖𝑦𝑛1+𝜎𝑛𝜎𝑛1𝑛1𝑖=1𝛽𝑖1𝛽𝑖𝑇𝑖𝑦𝑛1+𝜎𝑛𝛽𝑛1𝛽𝑛𝑇𝑛𝑦𝑛1+1𝛽𝑛1𝜎𝑛𝑃𝐶𝑧𝑛𝜆𝐴𝑧𝑛𝑃𝐶𝑧𝑛1𝜆𝐴𝑧𝑛11𝛽𝑛11𝜎𝑛1𝑃𝐶𝑧𝑛1𝜆𝐴𝑧𝑛1+1𝛽𝑛1𝜎𝑛𝑃𝐶𝑧𝑛1𝜆𝐴𝑧𝑛1.(3.10) Thus, from (3.8)–(3.10), 𝑥𝑛+1𝑥𝑛𝛽𝑛𝑥𝑛𝑥𝑛1+𝛽𝑛1𝛽𝑛𝑥𝑛1+𝜎𝑛𝑛𝑖=1𝛽𝑖1𝛽𝑖𝑇𝑖𝑦𝑛𝑇𝑖𝑦𝑛1+||𝜎𝑛𝜎𝑛1||𝑛1𝑖=1𝛽𝑖1𝛽𝑖𝑇𝑖𝑦𝑛1+𝜎𝑛𝛽𝑛1𝛽𝑛𝑇𝑛𝑦𝑛1+1𝛽𝑛1𝜎𝑛𝑃𝐶𝑧𝑛𝜆𝐴𝑧𝑛𝑃𝐶𝑧𝑛1𝜆𝐴𝑧𝑛1+||1𝛽𝑛1𝜎𝑛1𝛽𝑛11𝜎𝑛1||𝑃𝐶𝑧𝑛1𝜆𝐴𝑧𝑛1𝛽𝑛𝑥𝑛𝑥𝑛1+𝛽𝑛1𝛽𝑛𝑥𝑛1+𝜎𝑛𝑛𝑖=1𝛽𝑖1𝛽𝑖𝑦𝑛𝑦𝑛1+||𝜎𝑛𝜎𝑛1||𝑛1𝑖=1𝛽𝑖1𝛽𝑖𝑇𝑖𝑦𝑛1+𝛽𝑛1𝛽𝑛𝑇𝑛𝑦𝑛1+1𝛽𝑛1𝜎𝑛𝑧𝑛𝑧𝑛1+𝛽𝑛1𝛽𝑛1𝜎𝑛𝑃𝐶(𝐼𝜆𝐴)𝑧𝑛1+1𝛽𝑛1||𝜎𝑛𝜎𝑛1||𝑃𝐶(𝐼𝜆𝐴)𝑧𝑛1𝛽𝑛𝑥𝑛𝑥𝑛1+𝛽𝑛1𝛽𝑛𝑥𝑛1+𝜎𝑛1𝛽𝑛×1𝛼𝑛𝑥(1𝜌)𝑛𝑥𝑛1+1𝛼𝑛1𝑟||𝑟𝑛1𝑟𝑛||𝐿+||𝜎𝑛𝜎𝑛1||𝑛1𝑖=1𝛽𝑖1𝛽𝑖𝑇𝑖𝑦𝑛1+𝛽𝑛1𝛽𝑛𝑇𝑛𝑦𝑛1+1𝛽𝑛1𝜎𝑛𝑥𝑛𝑥𝑛1+1𝑟||𝑟𝑛1𝑟𝑛||𝐿+𝛽𝑛1𝛽𝑛𝑃𝐶(𝐼𝜆𝐴)𝑧𝑛1+||𝜎𝑛𝜎𝑛1||𝑃𝐶(𝐼𝜆𝐴)𝑧𝑛11𝜎𝑛1𝛽𝑛𝛼𝑛(𝑥1𝜌)𝑛𝑥𝑛1+𝛽𝑛1𝛽𝑛𝑀+1𝑟||𝑟𝑛1𝑟𝑛||||𝜎𝐿+𝑛𝜎𝑛1||𝑛1𝑖=1𝛽𝑖1𝛽𝑖𝛽𝑀+𝑛1𝛽𝑛𝑀+𝛽𝑛1𝛽𝑛+||𝜎𝑛𝜎𝑛1||𝑀1𝜎𝑛1𝛽𝑛𝛼𝑛𝑥(1𝜌)𝑛𝑥𝑛1+1𝑟||𝑟𝑛1𝑟𝑛||𝐿𝛽+3𝑛1𝛽𝑛||𝜎𝑀+2𝑛𝜎𝑛1||𝑀,(3.11) where 𝑀=max{sup𝑛1𝑥𝑛,sup𝑖1,𝑛1𝑇𝑖𝑦𝑛,sup𝑛1𝑃𝐶(𝐼𝜆𝐴)𝑧𝑛}. Since {𝛽𝑛} is strictly decreasing, we have 𝑛=2(𝛽𝑛1𝛽𝑛)=𝛽1<. Further, by assumption conditions (iv)-(v), we have 𝑛=21𝑟||𝑟𝑛+1𝑟𝑛||𝛽𝐿+3𝑛1𝛽𝑛||𝜎𝑀+2𝑛𝜎𝑛1||𝑀<.(3.12) Thus, using Lemma 2.4, we have lim𝑛𝑥𝑛+1𝑥𝑛=0.Step 3. lim𝑛𝑥𝑛𝑢𝑛=0.
For any 𝑥Ω, it follows from Lemma 2.6 that 𝑢𝑛𝑥2=𝑇𝑟𝑛𝑥𝑛𝑇𝑟𝑛𝑥2𝑇𝑟𝑛𝑥𝑛𝑇𝑟𝑛𝑥,𝑥𝑛𝑥=𝑢𝑛𝑥,𝑥𝑛𝑥=12𝑢𝑛𝑥2+𝑥𝑛𝑥2𝑢𝑛𝑥𝑛2.(3.13) From (2.3), (3.1)–(3.4), we can get 𝑦𝑛𝑥2𝛼𝑛𝑓𝑥𝑛𝑥2+1𝛼𝑛𝑣𝑛𝑥2𝛼𝑛𝑓𝑥𝑛𝑥2+1𝛼𝑛𝑧𝑛𝑦2+𝜆(𝜆2𝛼)𝐴𝑧𝑛𝐴𝑦2𝛼𝑛𝑓𝑥𝑛𝑥2+1𝛼𝑛𝑧𝑛𝑦2𝛼𝑛𝑓𝑥𝑛𝑥2+1𝛼𝑛𝑢𝑛𝑥2.(3.14) From (3.1), (3.3)-(3.4), and (3.13)-(3.14), it follows that 𝑥𝑛+1𝑥2=𝛽𝑛𝑥𝑛+𝜎𝑛𝑛𝑖=1𝛽𝑖1𝛽𝑖𝑇𝑖𝑦𝑛+1𝛽𝑛1𝜎𝑛𝑣𝑛𝑥2𝛽𝑛𝑥𝑛𝑥2+1𝛽𝑛𝜎𝑛11𝛽𝑛𝑛𝑖=1𝛽𝑖1𝛽𝑖𝑇𝑖𝑦𝑛+1𝜎𝑛𝑣𝑛𝑥2𝛽𝑛𝑥𝑛𝑥2+1𝛽𝑛𝜎𝑛11𝛽𝑛𝑛𝑖=1𝛽𝑖1𝛽𝑖𝑇𝑖𝑦𝑛𝑥2+1𝛽𝑛1𝜎𝑛𝑣𝑛𝑥2𝛽𝑛𝑥𝑛𝑥2+1𝛽𝑛𝜎𝑛11𝛽𝑛𝑛𝑖=1𝛽𝑖1𝛽𝑖𝑇𝑖𝑦𝑛𝑥2+1𝛽𝑛1𝜎𝑛𝑣𝑛𝑥2𝛽𝑛𝑥𝑛𝑥2+1𝛽𝑛𝜎𝑛𝑦𝑛𝑥2+1𝛽𝑛1𝜎𝑛𝑧𝑛𝑦2𝛽𝑛𝑥𝑛𝑥2+1𝛽𝑛𝜎𝑛1𝛼𝑛𝑢𝑛𝑥2+𝛼𝑛𝑓𝑥𝑛𝑥2+1𝛽𝑛1𝜎𝑛𝑢𝑛𝑥2𝑥𝑛𝑥2+1𝛽𝑛𝜎𝑛𝛼𝑛𝑓𝑥𝑛𝑥2𝑥𝑛𝑥21𝛽𝑛1𝜎𝑛𝛼𝑛𝑢𝑛𝑥𝑛2(3.15) which implies that 1𝛽𝑛1𝜎𝑛𝛼𝑛𝑢𝑛𝑥𝑛2𝑥𝑛𝑥+𝑥𝑛+1𝑥𝑥𝑛𝑥𝑛+1+𝛼𝑛𝑓𝑥𝑛𝑥2.(3.16) Since {𝑥𝑛} is both bounded, using Step 2 and conditions (ii)–(iv), we conclude the result.
Step 4. lim𝑛𝐴𝑧𝑛𝐴𝑦=0 and lim𝑛𝐵𝑥𝑛𝐵𝑥=0.Using (3.3), (3.14), and (3.15), we have 𝑥𝑛+1𝑥2𝛽𝑛𝑥𝑛𝑥2+1𝛽𝑛𝜎𝑛𝑦𝑛𝑥2+1𝛽𝑛1𝜎𝑛𝑧𝑛𝑦2𝛽𝑛𝑥𝑛𝑥2+1𝛽𝑛𝜎𝑛×𝛼𝑛𝑓𝑥𝑛𝑥2+1𝛼𝑛𝑧𝑛𝑦2+𝜆(𝜆2𝛼)𝐴𝑧𝑛𝐴𝑦2+1𝛽𝑛1𝜎𝑛𝑧𝑛𝑦2𝛽𝑛𝑥𝑛𝑥2+1𝛽𝑛𝜎𝑛1𝛼𝑛𝜆(𝜆2𝛼)𝐴𝑧𝑛𝐴𝑦2+𝛼𝑛𝑓𝑥𝑛𝑥2+1𝛽𝑛𝑧𝑛𝑦2𝛽𝑛𝑥𝑛𝑥2+𝜆(𝜆2𝛼)𝐴𝑧𝑛𝐴𝑦2+𝛼𝑛𝑓𝑥𝑛𝑥2+1𝛽𝑛𝑥𝑛𝑥2+𝜇(𝜇2𝛽)𝐵𝑢𝑛𝐵𝑥2𝑥𝑛𝑥2+𝜆(𝜆2𝛼)𝐴𝑧𝑛𝐴𝑦2+𝛼𝑛𝑓𝑥𝑛𝑥2+𝜇(𝜇2𝛽)𝐵𝑢𝑛𝐵𝑥2.(3.17) Therefore, 𝜆(2𝛼𝜆)𝐴𝑧𝑛𝐴𝑦2+𝜇(2𝛽𝜇)𝐵𝑢𝑛𝐵𝑥2𝑥𝑛𝑥+𝑥𝑛+1𝑥𝑥𝑛𝑥𝑛+1+𝛼𝑛𝑓𝑥𝑛𝑦2.(3.18) From Step 2, using condition (iii), we get lim𝑛𝐴𝑧𝑛𝐴𝑦=0 and lim𝑛𝐵𝑢𝑛𝐵𝑥=0. From the fact that the 𝐵 is 𝛽-inverse strongly monotone operator, it follows that 𝐵𝑥𝑛𝐵𝑥𝐵𝑥𝑛𝐵𝑢𝑛+𝐵𝑢𝑛𝐵𝑥1𝛽𝑥𝑛𝑢𝑛+𝐵𝑢𝑛𝐵𝑥.(3.19) Applying Step 3, we have lim𝑛𝐵𝑥𝑛𝐵𝑥=0.Step 5. lim𝑛𝑥𝑛𝑇𝑖𝑥𝑛=0,𝑓𝑜𝑟𝑎𝑙𝑙𝑖=1,2,.
Noting that 𝑃𝐶 is firmly nonexpansive, we have𝑧𝑛𝑦2=𝑃𝐶𝑢𝑛𝜇𝐵𝑢𝑛𝑃𝐶𝑥𝜇𝐵𝑥2𝑢𝑛𝜇𝐵𝑢𝑛𝑥𝜇𝐵𝑥,𝑧𝑛𝑦=12𝑢𝑛𝑥𝜇𝐵𝑢𝑛𝐵𝑥2+𝑧𝑛𝑦2𝑢𝑛𝑧𝑛𝜇𝐵𝑢𝑛𝐵𝑥𝑥𝑦212𝑢𝑛𝑥2+𝑧𝑛𝑦2𝑢𝑛𝑧𝑛𝑥𝑦2𝑢+2𝜇𝑛𝑧𝑛𝑥𝑦,𝐵𝑢𝑛𝐵𝑥𝜇2𝐵𝑢𝑛𝐵𝑥212𝑥𝑛𝑥2+𝑧𝑛𝑦2𝑢𝑛𝑧𝑛𝑥𝑦2𝑢+2𝜇𝑛𝑧𝑛𝑥𝑦,𝐵𝑢𝑛𝐵𝑥,𝑣𝑛𝑥2=𝑃𝐶𝑧𝑛𝜆𝐴𝑧𝑛𝑃𝐶𝑦𝜆𝐴𝑦2𝑧𝑛𝜆𝐴𝑧𝑛𝑦𝜆𝐴𝑦,𝑣𝑛𝑥=12𝑧𝑛𝑦𝜆𝐴𝑧𝑛𝐴𝑦2+𝑣𝑛𝑥2𝑧𝑛𝑣𝑛𝜆𝐴𝑧𝑛𝐴𝑦𝑦𝑥212𝑧𝑛𝑦2+𝑣𝑛𝑥2𝑧𝑛𝑣𝑛𝑦𝑥2𝑧+2𝜆𝑛𝑣𝑛𝑦𝑥,𝐴𝑧𝑛𝐴𝑦𝜆2𝐴𝑧𝑛𝐴𝑦212𝑥𝑛𝑥2+𝑣𝑛𝑥2𝑧𝑛𝑣𝑛𝑦𝑥2𝑧+2𝜆𝑛𝑣𝑛𝑦𝑥𝐴𝑧𝑛𝐴𝑦.(3.20) It follows that 𝑧𝑛𝑦2𝑥𝑛𝑥2𝑢𝑛𝑧𝑛𝑥𝑦2𝑢+2𝜇𝑛𝑧𝑛𝑥𝑦,𝐵𝑢𝑛𝐵𝑥,𝑣(3.21)𝑛𝑥2𝑥𝑛𝑥2𝑧𝑛𝑣𝑛𝑦𝑥2𝑧+2𝜆𝑛𝑣𝑛𝑦𝑥,𝐴𝑧𝑛𝐴𝑦.(3.22) By (3.3), (3.14)-(3.15), and (3.22), we have 𝑥𝑛+1𝑥211𝛽𝑛𝜎𝑛𝑥𝑛𝑥2+1𝛽𝑛𝜎𝑛𝑦𝑛𝑥211𝛽𝑛𝜎𝑛𝑥𝑛𝑥2+1𝛽𝑛𝜎𝑛𝛼𝑛𝑓𝑥𝑛𝑥2+𝑣𝑛𝑥211𝛽𝑛𝜎𝑛𝑥𝑛𝑥2+1𝛽𝑛𝜎𝑛𝛼𝑛𝑓𝑥𝑛𝑥2+𝑥𝑛𝑥2𝑧𝑛𝑣𝑛𝑦𝑥2𝑧+2𝜆𝑛𝑣𝑛𝑦𝑥𝐴𝑧𝑛𝐴𝑦=𝑥𝑛𝑥2+𝛼𝑛𝑓𝑥𝑛𝑥2+1𝛽𝑛1𝜎𝑛×𝑧𝑛𝑣𝑛𝑦𝑥2𝑧+2𝜆𝑛𝑣𝑛𝑦𝑥,𝐴𝑧𝑛𝐴𝑦.(3.23) It follows that 1𝛽𝑛𝜎𝑛𝑣𝑛𝑥𝑧𝑦2𝑥𝑛𝑥+𝑥𝑛+1𝑥𝑥𝑛𝑥𝑛+1+𝛼𝑛𝑓𝑥𝑛𝑥2+2𝜆1𝛽𝑛𝜎𝑛𝑧𝑛𝑣𝑛𝑦𝑥,𝐴𝑧𝑛𝐴𝑦.(3.24) From conditions (ii)–(iv), Steps 2 and 4, we get the following: lim𝑛𝑣𝑛𝑧𝑛𝑥𝑦=0.(3.25) On the other hand, from (3.14)–(3.21), we have 𝑥𝑛+1𝑥2𝛽𝑛𝑥𝑛𝑥2+1𝛽𝑛𝜎𝑛𝛼𝑛𝑓𝑥𝑛𝑥2+𝑧𝑛𝑦2+1𝛽𝑛1𝜎𝑛𝑧𝑛𝑦2𝛽𝑛𝑥𝑛𝑥2+𝛼𝑛𝑓𝑥𝑛𝑥2+1𝛽𝑛𝑧𝑛𝑦2𝛽𝑛𝑥𝑛𝑥2+𝛼𝑛𝑓𝑥𝑛𝑥2+1𝛽𝑛×𝑥𝑛𝑥2𝑢𝑛𝑧𝑛𝑥𝑦2𝑢+2𝜇𝑛𝑧𝑛𝑥𝑦,𝐵𝑢𝑛𝐵𝑥=𝑥𝑛𝑥2+𝛼𝑛𝑓𝑥𝑛𝑥2+1𝛽𝑛𝑢𝑛𝑧𝑛𝑥𝑦2𝑢+2𝜇𝑛𝑧𝑛𝑥𝑦,𝐵𝑢𝑛𝐵𝑥,(3.26) which implies that 1𝛽𝑛𝑢𝑛𝑧𝑛𝑥𝑦2𝑥𝑛𝑥+𝑥𝑛+1𝑥𝑥𝑛𝑥𝑛+1+𝛼𝑛𝑓𝑥𝑛𝑥2+2𝜇1𝛽𝑛×𝑢𝑛𝑧𝑛𝑥𝑦,𝐵𝑢𝑛𝐵𝑥.(3.27) From (ii)-(iii), Steps 2 and 4, we get the following: lim𝑛𝑢𝑛𝑧𝑛𝑥𝑦=0.(3.28) Combining (3.25) and (3.28), we get the following: lim𝑛𝑢𝑛𝑣𝑛=0.(3.29) Using Step 3, we obtain the following: lim𝑛𝑥𝑛𝑣𝑛=0.(3.30) This together with 𝑦𝑛𝑣𝑛=𝛼𝑛𝑓(𝑥𝑛)𝑣𝑛0 implies that lim𝑛𝑥𝑛𝑦𝑛=0.(3.31) For any 𝑥Ω, we have from (3.1) that 𝜎𝑛𝑛𝑖=1𝛽𝑖1𝛽𝑖𝑥𝑛𝑇𝑖𝑥𝑛,𝑥𝑛𝑥+𝑇𝑖𝑥𝑛𝑇𝑖𝑦𝑛,𝑥𝑛𝑥=𝜎𝑛𝑛𝑖=1𝛽𝑖1𝛽𝑖𝑥𝑛𝑇𝑖𝑦𝑛,𝑥𝑛𝑥=𝑥𝑛𝑥𝑛+1,𝑥𝑛𝑥+1𝛽𝑛1𝜎𝑛𝑣𝑛𝑥𝑛,𝑥𝑛𝑥.(3.32) Since each 𝑇𝑖 is nonexpansive, from (2.2), we have 𝑇𝑖𝑥𝑛𝑥𝑛22𝑥𝑛𝑇𝑖𝑥𝑛,𝑥𝑛𝑥.(3.33) Hence, combining this inequality with (3.32), we get the following: 12𝜎𝑛𝑛𝑖=1𝛽𝑖1𝛽𝑖𝑇𝑖𝑥𝑛𝑥𝑛2𝜎𝑛𝑛𝑖=1𝛽𝑖1𝛽𝑖𝑇𝑖𝑥𝑛𝑇𝑖𝑦𝑛,𝑥𝑛𝑥+𝑥𝑛𝑥𝑛+1,𝑥𝑛𝑥+1𝛽𝑛1𝜎𝑛×𝑣𝑛𝑥𝑛,𝑥𝑛𝑥,(3.34) that is (noting that 𝛽𝑛 is strictly decreasing), 𝑇𝑖𝑥𝑛𝑥𝑛221𝛽𝑛𝛽𝑖1𝛽𝑖𝑥𝑛𝑦𝑛𝑥𝑛𝑥+2𝜎𝑛𝛽𝑖1𝛽𝑖𝑥𝑛𝑥𝑛+1𝑥𝑛𝑥+21𝛽𝑛1𝜎𝑛𝜎𝑛𝛽𝑖1𝛽𝑖𝑣𝑛𝑥𝑛𝑥𝑛𝑥.(3.35) Now from (3.30)-(3.31) and Step 2, we conclude that lim𝑛𝑇𝑖𝑥𝑛𝑥𝑛=0,𝑖=1,2,,(3.36) which completes the proof.
Step 6. limsup𝑛𝑓(𝑥)𝑥,𝑥𝑛𝑥0, where 𝑥=𝑃Ω𝑓(𝑥).
As {𝑥𝑛} is bounded, there exists a subsequence {𝑥𝑛𝑖} of {𝑥𝑛} such that {𝑥𝑛𝑖}̃𝑥 weakly. First, it is clear from Step 5 and Lemma 2.3 that ̃𝑥𝑛=1𝐹(𝑇𝑛). Next, we prove that ̃𝑥Γ. From (iii), Step 3 and (3.31), we note that 𝑦𝑛𝑦𝐺𝑛𝛼𝑛𝑓𝑥𝑛𝑦𝐺𝑛+1𝛼𝑛𝑃𝐶𝑃C𝑢𝑛𝜇𝐵𝑢𝑛𝜆𝐴𝑃𝐶𝑢𝑛𝜇𝐵𝑢𝑛𝑦𝐺𝑛=𝛼𝑛𝑓𝑥𝑛𝑦𝐺𝑛+1𝛼𝑛𝐺𝑢𝑛𝑦𝐺𝑛𝛼𝑛𝑓𝑥𝑛𝑦𝐺𝑛+1𝛼𝑛𝑢𝑛𝑦𝑛0.(3.37) According to (3.31) and Lemma 2.3, we obtain that ̃𝑥Γ. Next, we show that ̃𝑥EP(𝐹). Indeed, by 𝑢𝑛=𝑇𝑟𝑛𝑥𝑛, we have 𝐹𝑢𝑛+1,𝑦𝑟𝑛𝑦𝑢𝑛,𝑢𝑛𝑥𝑛0,𝑦𝐶.(3.38) From (A2), we have also 1𝑟𝑛𝑦𝑢𝑛,𝑢𝑛𝑥𝑛𝑢𝐹𝑛,𝑦𝐹𝑦,𝑢𝑛,(3.39) and hence, 𝑦𝑢𝑛𝑘,𝑢𝑛𝑘𝑥𝑛𝑘𝑟𝑛𝑘𝐹𝑦,𝑢𝑛𝑘.(3.40) According to 𝑟𝑛>𝑟>0 and 𝑢𝑛𝑥𝑛0, we conclude that (𝑢𝑛𝑖𝑥𝑛𝑖)/𝑟𝑛𝑖0 and 𝑢𝑛𝑘̃𝑥. From (A4), we obtain the following: 𝐹(𝑦,̃𝑥)0,𝑦𝐶.(3.41) For 𝑡 with 0<𝑡1 and 𝑦𝐶, let 𝑦𝑡=𝑡𝑦+(1𝑡)̃𝑥. Since 𝑦𝐶 and ̃𝑥𝐶 (due to 𝑢𝑛𝑖̃𝑥 as 𝑖), we have 𝑦𝑡𝐶 and hence 𝐹(𝑦𝑡,̃𝑥)0. So, from (A1) and (A4), we have 𝑦𝑡𝐹𝑡𝑦,𝑦𝑡𝐹𝑡+𝑦,𝑦(1𝑡)𝐹𝑡𝑦,̃𝑥𝑡𝐹𝑡,𝑦𝑡=0(3.42) and 𝐹(𝑦𝑡,𝑦)0, for all 𝑡(0,1) and 𝑦𝐶. From (A3), we obtain the following: 𝐹(̃𝑥,𝑦)0𝑦𝐶,(3.43) and hence ̃𝑥EP(𝐹). Therefore, we obtain that ̃𝑥Ω. Hence, it follows from Lemma 2.2 that limsup𝑛𝑓𝑥𝑥,𝑥𝑛𝑥=lim𝑖𝑓𝑥𝑥,𝑥𝑛𝑖𝑥=𝑓𝑥𝑥,̃𝑥𝑥0.(3.44)
Step 7. lim𝑛𝑥𝑛=𝑥.
From (3.1), (3.3)-(3.4), and the convexity of , we have 𝑦𝑛𝑥2=𝛼𝑛𝑓𝑥𝑛+1𝛼𝑛𝑣𝑛𝑥21𝛼𝑛𝑣𝑛𝑥2+2𝛼𝑛𝑓𝑥𝑛𝑥,𝑦𝑛𝑥1𝛼𝑛𝑥𝑛𝑥2+2𝛼𝑛𝑓𝑥𝑛𝑥,𝑦𝑛𝑥.(3.45) We can also get 𝑥𝑛+1𝑥2=𝛽𝑛𝑥𝑛𝑥+𝜎𝑛𝑛𝑖=1𝛽𝑖1𝛽𝑖𝑇𝑖𝑦𝑛𝑥+1𝛽𝑛1𝜎𝑛𝑦𝑛𝑥+1𝛽𝑛1𝜎𝑛𝛼𝑛𝑣𝑛𝑥𝑓𝑛2𝛽𝑛𝑥𝑛𝑥+𝜎𝑛𝑛𝑖=1𝛽𝑖1𝛽𝑖𝑇𝑖𝑦𝑛𝑥+1𝛽𝑛1𝜎𝑛𝑦𝑛𝑥2+21𝛽𝑛1𝜎𝑛𝛼𝑛𝑣𝑛𝑥𝑓𝑛,𝑥𝑛+1𝑥𝛽𝑛𝑥𝑛𝑥2+1𝛽𝑛𝑦𝑛𝑥2+21𝛽𝑛1𝜎𝑛𝛼𝑛𝑣𝑛𝑥,𝑥𝑛+1𝑥+21𝛽𝑛1𝜎𝑛𝛼𝑛𝑥𝑥𝑓𝑛,𝑥𝑛+1𝑥.(3.46) By (3.45), we have 𝑥𝑛+1𝑥2𝛽𝑛𝑥𝑛𝑥2+1𝛽𝑛1𝛼𝑛𝑥𝑛𝑥2+2𝛼𝑛𝑓𝑥𝑛𝑥,𝑦𝑛𝑥+21𝛽𝑛1𝜎𝑛𝛼𝑛𝑣𝑛𝑥𝑥𝑛+1𝑥+21𝛽𝑛𝛼𝑛1𝜎𝑛𝑥𝑥𝑓𝑛,𝑥𝑛+1𝑥11𝛽𝑛𝛼𝑛𝑥𝑛𝑥2+21𝛽𝑛1𝜎𝑛𝛼𝑛𝑓𝑥𝑛𝑥,𝑦𝑛𝑥𝑛+1+21𝛽𝑛1𝜎𝑛𝛼𝑛𝑧𝑛𝑦𝑥𝑛+1𝑥+21𝛽𝑛𝛼𝑛𝜎𝑛𝑓𝑥𝑛𝑥,𝑦𝑛𝑥11𝛽𝑛𝛼𝑛𝑥𝑛𝑥2+21𝛽𝑛1𝜎𝑛𝛼𝑛𝑓𝑥𝑛𝑥𝑥𝑛+1𝑦𝑛+21𝛽𝑛1𝜎𝑛𝛼𝑛𝑥𝑛𝑥𝑥𝑛+1𝑥+21𝛽𝑛𝛼𝑛𝜎𝑛𝑓𝑥𝑛𝑥,𝑦𝑛𝑥𝑛+21𝛽𝑛𝛼𝑛𝜎𝑛𝑓𝑥𝑛𝑥𝑓,𝑥𝑛𝑥+21𝛽𝑛𝛼𝑛𝜎𝑛𝑓𝑥𝑥,𝑥𝑛𝑥11𝛽𝑛𝛼𝑛𝑥𝑛𝑥2+21𝛽𝑛1𝜎𝑛𝛼𝑛𝑓𝑥𝑛𝑥𝑥𝑛+1𝑦𝑛+1𝛽𝑛1𝜎𝑛𝛼𝑛𝑥𝑛𝑥2+𝑥𝑛+1𝑥2+21𝛽𝑛𝛼𝑛𝜎𝑛𝑓𝑥𝑛𝑥𝑦𝑛𝑥𝑛+21𝛽𝑛𝛼𝑛𝜎𝑛𝜌𝑥𝑛𝑥2+21𝛽𝑛𝛼𝑛𝜎𝑛𝑓𝑥𝑥,𝑥𝑛𝑥,(3.47) which implies that 𝑥𝑛+1𝑥212𝜎𝑛(1𝜌)11𝛽𝑛11𝛽𝑛1𝜎𝑛𝛼𝑛𝛼𝑛𝑥𝑛𝑥2+2𝜎𝑛(1𝜌)11𝛽𝑛𝛼𝑛11𝛽𝑛1𝜎𝑛𝛼𝑛×21𝜎𝑛2𝜎𝑛𝑓𝑥(1𝜌)1𝑛𝑥𝑥𝑛+1𝑦𝑛+2𝜎𝑛2𝜎𝑛𝑓𝑥(1𝜌)1𝑛𝑥×𝑦𝑛𝑥𝑛+2𝜎𝑛2𝜎𝑛𝑓𝑥(1𝜌)1𝑛𝑥,𝑥𝑛𝑥.(3.48) From liminf𝑛(2𝜎𝑛(1𝜌)1)(1𝛽𝑛)/(1(1𝛽𝑛)(1𝜎𝑛))>0, it follows that 𝑛=0((2𝜎𝑛(1𝜌)1)(1𝛽𝑛)/(1(1𝛽𝑛)(1𝜎𝑛)𝛼𝑛))𝛼𝑛=. It is clear that limsup𝑛21𝜎𝑛2𝜎𝑛𝑓𝑥(1𝜌)1𝑛𝑥𝑥𝑛+1𝑦𝑛+2𝜎𝑛2𝜎𝑛𝑓𝑥(1𝜌)1𝑛𝑥×𝑦𝑛𝑥𝑛+2𝜎𝑛2𝜎𝑛(𝑓𝑥1𝜌)1𝑛𝑥,𝑥𝑛𝑥0.(3.49) Therefore, all conditions of Lemma 2.4 are satisfied. Therefore, we immediately deduce that 𝑥𝑛𝑥. This completes the proof.

Corollary 3.2. Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻. Let the mappings 𝐴,𝐵𝐶𝐻 be 𝛼-inverse strongly monotone and 𝛽-inverse strongly monotone, respectively. Let 𝐹 be a bifunction from 𝐶×𝐶 to satisfying (A1)–(A4) and 𝑇𝐶𝐶 be a nonexpansive mapping such that Ω=𝐹(𝑇)𝐸𝑃(𝐹)Γ.Let 𝑓𝐶𝐶 be a contraction with coefficient 𝜌(0,1/2). For given 𝑥0𝐶 arbitrarily, let the sequences {𝑥𝑛}, {𝑦𝑛}, {𝑧𝑛} and, {𝑢𝑛} be generated By 𝐹𝑢𝑛+1,𝑦𝑟𝑛𝑦𝑢𝑛,𝑢𝑛𝑥𝑛𝑧0,𝑦𝐶,𝑛=𝑃𝐶𝑢𝑛𝜇𝐵𝑢𝑛,𝑦𝑛=𝛼𝑛𝑓𝑥𝑛+1𝛼𝑛𝑃𝐶𝑧𝑛𝜆𝐴𝑧𝑛,𝑥𝑛+1=𝛽𝑛𝑥𝑛+𝜎𝑛1𝛽𝑛𝑇𝑦𝑛+1𝛽𝑛1𝜎𝑛𝑃𝐶𝑧𝑛𝜆𝐴𝑧𝑛,(3.50) where 𝜆(0,2𝛼), 𝜇(0,2𝛽), and sequences {𝛼𝑛}[0,1], {𝛽𝑛}[0,1], {𝜎𝑛}[0,1], and {𝑟𝑛}(𝑟,), 𝑟>0, are such that (i)0<liminf𝑛𝛽𝑛limsup𝑛𝛽𝑛<1, (ii)lim𝑛𝛼𝑛=0and 𝑛=1𝛼𝑛=,(iii)𝜎𝑛>1/2(1𝜌),𝑛=1|𝜎𝑛𝜎𝑛1|<, (iv)𝑛=1|𝑟𝑛𝑟𝑛1|<. Then the sequence {𝑥𝑛} generated by (3.50) converges strongly to 𝑥=𝑃Ω𝑓(𝑥), and (𝑥,𝑦) is a solution of the general system of variational inequalities (1.5), where 𝑦=𝑃𝐶(𝑥𝜇𝐵𝑥).

Acknowledgments

The authors would like to express their thanks to the referee for the valuable comments and suggestions for improving this article. This paper was supported by the NSFC Tianyuan Youth Foundation of Mathematics of China (no. 11126136) and Fundamental Research Funds for the Central Universities (no. ZXH2011C002).

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