About this Journal Submit a Manuscript Table of Contents
Abstract and Applied Analysis
Volume 2010 (2010), Article ID 390972, 39 pages
http://dx.doi.org/10.1155/2010/390972
Research Article

A Viscosity Hybrid Steepest Descent Method for Generalized Mixed Equilibrium Problems and Variational Inequalities for Relaxed Cocoercive Mapping in Hilbert Spaces

1Department of Mathematics, Faculty of Science, Kasetsart University (KU), Bangkok 10900, Thailand
2Department of Mathematics, Faculty of Liberal Arts, Rajamangala University of Technology Rattanakosin (RMUTR), Bangkok 10100, Thailand
3Department of Mathematics, Faculty of Science, King Mongkut's University of Technology Thonburi (KMUTT), Bangkok 10140, Thailand
4Centre of Excellence in Mathematics, CHE, Si Ayuthaya Road, Bangkok 10400, Thailand

Received 5 March 2010; Revised 24 May 2010; Accepted 30 May 2010

Academic Editor: S. Reich

Copyright © 2010 Wanpen Chantarangsi 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 an iterative method for fixed point problems, generalized mixed equilibrium problems, and variational inequality problems. Our method is based on the so-called viscosity hybrid steepest descent method. Using this method, we can find the common element of the set of fixed points of a nonexpansive mapping, the set of solutions of generalized mixed equilibrium problems, and the set of solutions of variational inequality problems for a relaxed cocoercive mapping in a real Hilbert space. Then, we prove the strong convergence of the proposed iterative scheme to the unique solution of variational inequality. The results presented in this paper generalize and extend some well-known strong convergence theorems in the literature.

1. Introduction

Throughout this paper, unless otherwise specified, we consider 𝐻 to be a real Hilbert space with inner product , and its induced norm . Let 𝐶 be a nonempty closed convex subset of 𝐻 and let 𝑃𝐶 be the metric projection of 𝐻 onto the closed convex subset 𝐶. Let 𝑆𝐶𝐶 be a nonexpansive mapping, that is, 𝑆𝑥𝑆𝑦𝑥𝑦 for all 𝑥,𝑦𝐶. The fixed point set of 𝑆 is defined by 𝐹(𝑆)={𝑥𝐶𝑆𝑥=𝑥}.(1.1) If 𝐶𝐻 is nonempty, bounded, closed, and convex and 𝑆 is a nonexpansive mapping of 𝐶 into itself, then 𝐹(𝑆) is nonempty; see, for example, [1, 2]. A mapping 𝑓𝐶𝐶 is a contraction on 𝐶 if there exists a constant 𝜂(0,1) such that 𝑓(𝑥)𝑓(𝑦)𝜂𝑥𝑦 for all 𝑥,𝑦𝐶. In addition, let Ψ𝐶𝐻 be a nonlinear mapping. Let 𝜑𝐶{+} be a real-valued function and let Θ𝐶×𝐶 be a bifunction such that 𝐶dom𝜑, where is the set of real numbers and dom𝜑={𝑥𝐶𝜑(𝑥)<+}.

The generalized mixed equilibrium problem for finding 𝑥𝐶Θ(𝑥,𝑦)+Ψ𝑥,𝑦𝑥+𝜑(𝑦)𝜑(𝑥)0,𝑦𝐶.(1.2) The set of solutions of (1.2) is denoted by GMEP(Θ,𝜑,Ψ), that is, GMEP(Θ,𝜑,Ψ)={𝑥𝐶Θ(𝑥,𝑦)+Ψ𝑥,𝑦𝑥+𝜑(𝑦)𝜑(𝑥)0,𝑦𝐶}.(1.3) We see that if 𝑥 is a solution of a problem (1.2), then 𝑥dom𝜑.

Special Examples
(1)If Ψ=0, then the problem (1.2) is reduced into the mixed equilibrium problem for finding 𝑥𝐶 such that Θ(𝑥,𝑦)+𝜑(𝑦)𝜑(𝑥)0,𝑦𝐶.(1.4) The set of solutions of (1.4) is denoted by MEP(Θ,𝜑).(2)If 𝜑=0, then the problem (1.2) is reduced into the generalized equilibrium problem for finding 𝑥𝐶 such that Θ(𝑥,𝑦)+Ψ𝑥,𝑦𝑥0,𝑦𝐶.(1.5) The set of solutions of (1.5) is denoted by GEP(Θ,Ψ).(3)If Ψ=0 and 𝜑=0, then the problem (1.2) is reduced into the equilibrium problem for finding 𝑥𝐶 such that Θ(𝑥,𝑦)0,𝑦𝐶.(1.6) The set of solutions of (1.6) is denoted by EP(Θ).(4)If Θ=0,𝜑=0, and Ψ=𝐵, then the problem (1.2) is reduced into the variational inequality problem for finding 𝑥𝐶 such that 𝐵𝑥,𝑦𝑥0,𝑦𝐶.(1.7) The set of solutions of (1.7) is denoted by VI(𝐶,𝐵).

The generalized mixed equilibrium problem is very general in the sense that it includes, as special cases, fixed point problems, variational inequality problems, optimization problems, Nash equilibrium problems in noncooperative games, the equilibrium problem, and Numerous problems in physics, economics, and others. Some methods have been proposed to solve problem (1.2); see, for instance, [3, 4] and the references therein.

Let 𝐵𝐶𝐻 be a nonlinear mapping. Now, we recall the following definitions.(d1)𝐵 is said to be monotone if for each 𝑥,𝑦𝐶𝐵𝑥𝐵𝑦,𝑥𝑦0.(1.8)(d2)𝐵 is said to be 𝜌-strongly monotone if there exists a positive real number 𝜌 such that 𝐵𝑥𝐵𝑦,𝑥𝑦𝜌𝑥𝑦2,𝑥,𝑦𝐶.(1.9)(d3)𝐵 is said to be 𝜔-Lipschitz continuous if there exists a positive real number 𝜔 such that 𝐵𝑥𝐵𝑦𝜔𝑥𝑦,𝑥,𝑦𝐶.(1.10)(d4)𝐵 is said to be 𝜉-inverse-strongly monotone if there exists a constant 𝜉>0 such that 𝐵𝑥𝐵𝑦,𝑥𝑦𝜉𝐵𝑥𝐵𝑦2,𝑥,𝑦𝐶.(1.11)(d5)𝐵 is said to be relaxed (𝑢,𝑣)-cocoercive if there exist positive real numbers 𝑢,𝑣 such that 𝐵𝑥𝐵𝑦,𝑥𝑦(𝑢)𝐵𝑥𝐵𝑦2+𝑣𝑥𝑦2,𝑥,𝑦𝐶.(1.12)(d6) A set-valued mapping 𝑄𝐻2𝐻 is called monotone if for all 𝑥,𝑦𝐻, 𝑓𝑄𝑥 and 𝑔𝑄𝑦 imply 𝑥𝑦,𝑓𝑔0.(d7) A monotone mapping 𝑄𝐻2𝐻 is called maximal if the graph 𝐺(𝑄) of 𝑄 is not properly contained in the graph of any other monotone mapping. It is well known that a monotone mapping 𝑄 is maximal if and only if for (𝑥,𝑓)𝐻×𝐻, 𝑥𝑦,𝑓𝑔0 for every (𝑦,𝑔)𝐺(𝑄) implies 𝑓𝑄𝑥.

For finding a common element of the set of fixed points of a nonexpansive mapping and the set of solutions of variational inequalities for a 𝜉-inverse-strongly monotone mapping, Takahashi and Toyoda [5] introduced the following iterative scheme: 𝑥0𝑥𝐶chosenarbitrary,𝑛+1=𝛾𝑛𝑥𝑛+1𝛾𝑛𝑆𝑃𝐶𝑥𝑛𝛼𝑛𝐵𝑥𝑛,𝑛0,(1.13) where 𝐵 is a 𝜉-inverse-strongly monotone mapping, {𝛾𝑛} is a sequence in (0,1), and {𝛼𝑛} is a sequence in (0,2𝜉). They showed that if 𝐹(𝑆)VI(𝐶,𝐵) is nonempty, then the sequence {𝑥𝑛} generated by (1.13) converges weakly to some 𝑧𝐹(𝑆)VI(𝐶,𝐵).

For finding an element of VI(𝐶,𝐵), Iiduka et al. [6] introduced the following iterative scheme:𝑥0𝑥𝐶chosenarbitrary,𝑛+1=𝑃𝐶𝛾𝑛𝑥𝑛+1𝛾𝑛𝑃𝐶𝑥𝑛𝛼𝑛𝐵𝑥𝑛,𝑛0,(1.14) where 𝐵 is a 𝜉-inverse-strongly monotone mapping, {𝛾𝑛} is a sequence in (1,1), and {𝛼𝑛} is a sequence in (0,2𝜉). They showed that if VI(𝐶,𝐵) is nonempty, then the sequence {𝑥𝑛} generated by (1.14) converges weakly to some 𝑧VI(𝐶,𝐵).

For finding a common element of 𝐹(𝑆)VI(𝐶,𝐵), let 𝑆𝐻𝐻 be a nonexpansive mapping. Yamada [7] introduced the following iterative scheme called the hybrid steepest descent method: 𝑥𝑛+1=𝑆𝑥𝑛𝛼𝑛𝜇𝐵𝑆𝑥𝑛,𝑛1,(1.15) where 𝑥1=𝑥𝐻,{𝛼𝑛}(0,1),𝐵𝐻𝐻 is a strongly monotone and Lipschitz continuous mapping, and 𝜇 is a positive real number. He proved that the sequence {𝑥𝑛} generated by (1.15) converges strongly to the unique solution of the 𝐹(𝑆)VI(𝐶,𝐵).

The hybrid steepest descent method is constructed by blending important ideas in the steepest descent method and in the fixed point theory. The remarkable applicability of this method to the convexly constrained generalized pseudoinverse problem as well as to the convex feasibility problem is demonstrated by constructing nonexpansive mappings whose fixed point sets are the feasible sets of the problems.

On the other hand, Shang et al. [8] introduced a new iterative process for finding a common element of the set of fixed points of a nonexpansive mapping and the set of solutions of the variational inequalities for relaxed (𝑢,𝑣)-cocoercive mappings in a real Hilbert space by using viscosity approximation method. Let 𝑆𝐶𝐶 be a nonexpansive mapping and let 𝑓𝐶𝐶 be a contraction mapping. Starting with arbitrary initial 𝑥1𝐶 and define sequences {𝑥𝑛} recursively by 𝑥𝑛+1=𝜖𝑛𝑓𝑥𝑛+𝛽𝑛𝑥𝑛+𝛾𝑛𝑆𝑃𝐶𝑥𝑛𝛼𝑛𝐵𝑥𝑛,𝑛1.(1.16) They proved that under certain appropriate conditions imposed on {𝜖𝑛},{𝛽𝑛},{𝛾𝑛}, and {𝛼𝑛}, the sequence {𝑥𝑛} converges strongly to 𝑧𝐹(𝑆)VI(𝐶,𝐵), where 𝑧=𝑃𝐹(𝑆)VI(𝐶,𝐵)𝑓(𝑧).

For finding a common element of 𝐹(𝑆)GEF(Θ,Ψ), let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻. Let Ψ be a 𝜉-inverse-strongly monotone mapping of 𝐶 into 𝐻 and let 𝑆 be a nonexpansive mapping of 𝐶 into itself. S. Takahashi and W. Takahashi [9] introduced the following iterative scheme: Θ𝑢𝑛,𝑦+Ψ𝑥𝑛,𝑦𝑢𝑛1+𝑟𝑛𝑦𝑢𝑛,𝑢𝑛𝑥𝑛𝑦0,𝑦𝐶,𝑛=𝛼𝑛𝑥+1𝛼𝑛𝑢𝑛,𝑥𝑛+1=𝛾𝑛𝑥𝑛+1𝛾𝑛𝑆𝑦𝑛,(1.17) where {𝛼𝑛}[0,1],{𝛾𝑛}[0,1], and {𝑟𝑛}[0,2𝜉] satisfy some parameters controlling conditions. They proved that the sequence {𝑥𝑛} defined by (1.17) converges strongly to a common element of 𝐹(𝑆)GEF(Θ,Ψ).

Iterative methods for nonexpansive mappings have recently been applied to solve convex minimization problems; see, for example, [7, 1012] and the references therein. Convex minimization problems have a great impact and influence in the development of almost all branches of pure and applied sciences.

A typical problem is to minimize a quadratic function over the set of fixed points of a nonexpansive mapping defined on a real Hilbert space 𝐻: min𝑥𝐹12𝐴𝑥,𝑥𝑥,𝑏,(1.18) where 𝐹 is the fixed point set of a nonexpansive mapping 𝑆 defined on 𝐻 and 𝑏 is a given point in 𝐻.

A linear bounded operator 𝐴 is strongly positive if there exists a constant 𝛾>0 with the property 𝐴𝑥,𝑥𝛾𝑥2,𝑥𝐻.(1.19)

Recently, Marino and Xu [13] introduced a new iterative scheme by the viscosity approximation method: 𝑥𝑛+1=𝜖𝑛𝑥𝛾𝑓𝑛+1𝜖𝑛𝐴𝑆𝑥𝑛.(1.20) They proved that the sequence {𝑥𝑛} generated by (1.20) converges strongly to the unique solution of the variational inequality: 𝛾𝑓𝑧𝐴𝑧,𝑥𝑧0,𝑥𝐹(𝑆),(1.21) which is the optimality condition for the minimization problem: min𝑥𝐹(𝑆)12𝐴𝑥,𝑥(𝑥),(1.22) where is a potential function for 𝛾𝑓.

In 2008, Qin et al. [14] proposed the following iterative algorithm: Θ𝑢𝑛+1,𝑦𝑟𝑛𝑦𝑢𝑛,𝑢𝑛𝑥𝑛𝑥0,𝑦𝐻,𝑛+1=𝜖𝑛𝑥𝛾𝑓𝑛+𝐼𝜖𝑛𝐴𝑆𝑃𝐶𝑢𝑛𝛼𝑛𝐵𝑢𝑛,(1.23) where 𝐴 is a strongly positive linear bounded operator and 𝐵 is a relaxed cocoercive mapping of 𝐶 into 𝐻. They proved that if the sequences {𝜖𝑛},{𝛼𝑛}, and {𝑟𝑛} of parameters satisfy appropriate condition, then the sequence {𝑥𝑛} defined by (1.23) converges strongly to the unique solution 𝑧 of the variational inequality: 𝛾𝑓𝑧𝐴𝑧,𝑥𝑧0,𝑥𝐹(𝑆)VI(𝐶,𝐵)EP(Θ),(1.24) which is the optimality condition for the minimization problem: min𝑥𝐹(𝑆)VI(𝐶,𝐵)EP(Θ)12𝐴𝑥,𝑥(𝑥),(1.25) where is a potential function for 𝛾𝑓.

In this paper, we introduce an iterative scheme by using a viscosity hybrid steepest descent method for finding a common element of the set of solutions of a generalized mixed equilibrium problem, the set of fixed points of a nonexpansive mapping, and the set of solutions of variational inequality problem for a relaxed cocoercive mapping in a real Hilbert space. The results shown in this paper improve and extend the recent ones announced by many others.

2. Preliminaries

Throughout this paper, we always assume that 𝐻 is a real Hilbert space and 𝐶 is a nonempty closed convex subset of 𝐻. For a sequence {𝑥𝑛}, the notation of 𝑥𝑛𝑥 and 𝑥𝑛𝑥 means that the sequence {𝑥𝑛} converges weakly and strongly to 𝑥, respectively.

The following lemmata give some characterizations and useful properties of the metric projection 𝑃𝐶 in a real Hilbert space. The metric (or nearest point) projection from 𝐻 onto 𝐶 is the mapping 𝑃𝐶𝐻𝐶 which assigns to each point 𝑥𝐻 the unique point 𝑃𝐶𝑥𝐶 satisfying the following property: 𝑥𝑃𝐶𝑥=inf𝑦𝐶𝑥𝑦.(2.1)

Lemma 2.1. It is well known that the metric projection 𝑃𝐶 has the following properties: (m1) for each 𝑥𝐻 and 𝑧𝐶, 𝑧=𝑃𝐶𝑥𝑥𝑧,𝑦𝑧0,𝑦𝐶;(2.2)(m2)𝑃𝐶𝐻𝐶 is nonexpansive, that is, 𝑃𝐶𝑥𝑃𝐶𝑦𝑥𝑦,𝑥,𝑦𝐻;(2.3)(m3)𝑃𝐶 is firmly nonexpansive, that is, 𝑃𝐶𝑥𝑃𝐶𝑦2𝑃𝐶𝑥𝑃𝐶𝑦,𝑥𝑦𝑥,𝑦𝐻.(2.4)

In order to prove our main results, we also need the following lemmata.

Lemma 2.2 (see [2]). Let 𝐻 be a Hilbert space, let 𝐶 be a nonempty closed convex subset of 𝐻, and let 𝐵 be a mapping of 𝐶 into 𝐻. Let 𝑥𝐶. Then, for 𝜆>0, 𝑥VI(𝐶,𝐵)𝑥=𝑃𝐶𝑥𝜆𝐵𝑥,(2.5) that is, 𝑥VI(𝐶,𝐵)𝑥𝑃𝐹𝐶(𝐼𝜆𝐵),(2.6) where 𝑃𝐶 is the metric projection of 𝐻 onto 𝐶.

Lemma 2.3 (see [15]). Let 𝐵 be a monotone mapping of 𝐶 into 𝐻 and let 𝑁𝐶𝑤1 be the normal cone to 𝐶 at 𝑤1𝐶, that is, 𝑁𝐶𝑤1=𝑤𝐻𝑤1𝑤2,𝑤0,𝑤2𝐶,(2.7) and define a mapping 𝑄 on 𝐶 by 𝑄𝑤1=𝐵𝑤1+𝑁𝐶𝑤1,𝑤1𝐶,,𝑤1𝐶.(2.8) Then 𝑄 is maximal monotone and 0𝑄𝑤1 if and only if 𝐵𝑤1,𝑤2𝑤10 for all 𝑤2𝐶.

Lemma 2.4 (see [16]). Each Hilbert space 𝐻 satisfies Opials condition; that is, for any sequence {𝑥𝑛}𝐻 with 𝑥𝑛𝑥, the inequality liminf𝑛𝑥𝑛𝑥<liminf𝑛𝑥𝑛𝑦(2.9) holds for each 𝑦𝐻 with 𝑦𝑥.

Lemma 2.5 (see [13]). Let 𝐶 be a nonempty closed convex subset of 𝐻, let 𝑓 be a contraction of 𝐻 into itself with coefficient 𝜂(0,1), and let 𝐴 be a strongly positive linear bounded operator on 𝐻 with coefficient 𝛾>0. Then, for 0<𝛾<𝛾/𝜂, 𝑥𝑦,(𝐴𝛾𝑓)𝑥(𝐴𝛾𝑓)𝑦𝛾𝜂𝛾𝑥𝑦2,𝑥,𝑦𝐻.(2.10) That is, 𝐴𝛾𝑓 is strongly monotone with coefficient 𝛾𝜂𝛾.

Lemma 2.6 (see [13]). Assume that 𝐴 is a strongly positive linear bounded operator on 𝐻 with coefficient 𝛾>0 and 0<𝜌𝐴1. Then 𝐼𝜌𝐴1𝜌𝛾.

Lemma 2.7 (see [17]). Let {𝑥𝑛} and {𝑦𝑛} be bounded sequences in a Banach space 𝑋 and let {𝛾𝑛} be a sequence in [0,1] with 0<liminf𝑛𝛾𝑛limsup𝑛𝛾𝑛<1.(2.11) Suppose 𝑥𝑛+1=1𝛾𝑛𝑦𝑛+𝛾𝑛𝑥𝑛,𝑛0,limsup𝑛𝑦𝑛+1𝑦𝑛𝑥𝑛+1𝑥𝑛0.(2.12) Then, lim𝑛𝑦𝑛𝑥𝑛=0.

Lemma 2.8 (see [18]). Assume that {𝑎𝑛} is a sequence of nonnegative real numbers such that 𝑎𝑛+11𝜚𝑛𝑎𝑛+𝜎𝑛,𝑛0,(2.13) where {𝜚𝑛} is a sequence in (0,1) and {𝜎𝑛} is a sequence in such that (1)𝑛=1𝜚𝑛=, (2)limsup𝑛(𝜎𝑛/𝜚𝑛)0 or 𝑛=1|𝜎𝑛|<.Then lim𝑛𝑎𝑛=0.

For solving the generalized mixed equilibrium problem and the mixed equilibrium problem, let us give the following assumptions for the bifunction Θ, the function 𝜑, and the set 𝐶: (H1)Θ(𝑥,𝑥)=0,𝑥𝐶;(H2)Θ is monotone, that is, Θ(𝑥,𝑦)+Θ(𝑦,𝑥)0,𝑥,𝑦𝐶;(H3) for each 𝑦𝐶,𝑥Θ(𝑥,𝑦) is weakly upper semicontinuous; (H4) for each 𝑥𝐶,𝑦Θ(𝑥,𝑦) is convex; (H5) for each 𝑥𝐶,𝑦Θ(𝑥,𝑦) is lower semicontinuous; (B1) for each 𝑥𝐻 and 𝑟>0, there exist abounded subsets 𝐷𝑥𝐶 and 𝑦𝑥𝐶 such that for any 𝑧𝐶𝐷𝑥, Θ𝑧,𝑦𝑥𝑦+𝜑𝑥1𝜑(𝑧)+𝑟𝑦𝑥𝑧,𝑧𝑥<0;(2.14)(B2)𝐶 is a bounded set.

Lemma 2.9 (see [19]). Let 𝐶 be a nonempty closed convex subset of 𝐻. Let Θ𝐶×𝐶 be a bifunction that satisfies (H1)–(H5) and let 𝜑𝐶{+} be a proper lower semicontinuous and convex function. Assume that either (B1) or (B2) holds. For 𝑟>0 and 𝑥𝐻, define a mapping 𝑇𝑟(Θ,𝜑)𝐻𝐶 as follows: 𝑇𝑟(Θ,𝜑)1(𝑥)=𝑧𝐶Θ(𝑧,𝑦)+𝜑(𝑦)𝜑(𝑧)+𝑟𝑦𝑧,𝑧𝑥0,𝑦𝐶(2.15) for all 𝑧𝐻. Then, the following properties hold: (i)for each 𝑥𝐻,𝑇𝑟(Θ,𝜑)(𝑥); (ii)𝑇𝑟(Θ,𝜑) is single-valued; (iii)𝑇𝑟(Θ,𝜑) is firmly nonexpansive; that is, for any 𝑥,𝑦𝐻,𝑇𝑟(Θ,𝜑)𝑥𝑇𝑟(Θ,𝜑)𝑦2𝑇𝑟(Θ,𝜑)𝑥𝑇𝑟(Θ,𝜑)𝑦,𝑥𝑦;(2.16)(iv)𝐹(𝑇𝑟(Θ,𝜑))=MEP(Θ,𝜑);(v)MEP(Θ,𝜑) is closed and convex.

Remark 2.10. If 𝜑=0, then 𝑇𝑟(Θ,𝜑) is rewritten as 𝑇Θ𝑟.

Lemma 2.11 (see [9]). Let 𝐶,𝐻,Θ, and 𝑇Θ𝑟 be as in Remark 2.10. Then the following holds: 𝑇Θ𝑠𝑥𝑇Θ𝑡𝑥2𝑠𝑡𝑠𝑇Θ𝑠𝑥𝑇Θ𝑡𝑥,𝑇Θ𝑠𝑥𝑥(2.17) for all 𝑠,𝑡>0 and 𝑥𝐻.

The following lemma is an immediate consequence of an inner product.

Lemma 2.12. Let 𝐻 be a real Hilbert space, let 𝑥 and 𝑦 be elements in 𝐻, and let 𝜆[0,1]. Then (1)𝜆𝑥+(1𝜆)𝑦2=𝜆𝑥2+(1𝜆)𝑦2𝜆(1𝜆)𝑥𝑦2, (2)𝑥+𝑦2𝑥2+2𝑦,𝑥+𝑦.

3. Main Results

In this section, we will introduce an iterative scheme by using a viscosity hybrid steepest descent method for finding a common element of the set of fixed points for nonexpansive mappings, the set of solutions of a generalized mixed equilibrium problem, and the set of solutions of variational inequality problem for a relaxed cocoercive mapping in a real Hilbert space. We show that the iterative sequence converges strongly to a common element of the three sets.

In order to prove our main results, we first prove the following lemmata.

Lemma 3.1. Let 𝐶,𝐻,Θ,𝜑, and 𝑇𝑟(Θ,𝜑) be as in Lemma 2.9. Then the following holds: 𝑇𝑠(Θ,𝜑)𝑥T𝑡(Θ,𝜑)𝑥2𝑠𝑡𝑠𝑇𝑠(Θ,𝜑)𝑥𝑇𝑡(Θ,𝜑)𝑥,𝑇𝑠(Θ,𝜑)𝑥𝑥(3.1) for all 𝑠,𝑡>0 and 𝑥𝐻.

Proof. By similar argument as in the proof of Lemma  2.11 in [9], for 𝑠,𝑡>0 and 𝑥𝐻. Observing that =𝑇𝑠(Θ,𝜑)𝑥 and 𝜗=𝑇𝑡(Θ,𝜑)𝑥, we have 1Θ(,𝑦)+𝜑(𝑦)𝜑()+𝑠1𝑦,𝑥0,𝑦𝐶,(3.2)Θ(𝜗,𝑦)+𝜑(𝑦)𝜑(𝜗)+𝑡𝑦𝜗,𝜗𝑥0,𝑦𝐶.(3.3) Putting 𝑦=𝜗 in (3.2) and 𝑦= in (3.3), we obtain 1Θ(,𝜗)+𝜑(𝜗)𝜑()+𝑠1𝜗,𝑥0,Θ(𝜗,)+𝜑()𝜑(𝜗)+𝑡𝜗,𝜗𝑥0.(3.4) So, summing up these two equalities and using the monotonicity of Θ (H2), we get 1𝑠1𝜗,𝑥+𝑡𝜗,𝜗𝑥0,(3.5) and hence 𝜗,𝜗𝑥𝑡𝑥𝑠0.(3.6) We derive from (3.6) that 𝑡𝜗,𝜗𝑥𝑠(𝑥)0,(3.7) and so 𝜗2+𝑡𝜗,1𝑠(𝑥)0.(3.8) This indicates that 𝑡1𝑠𝜗,𝑥𝜗2.(3.9) In other words, 𝜗2𝑠𝑡𝑠𝜗,𝑥,(3.10) and thus the claim holds.

Lemma 3.2. Let 𝐻 be a real Hilbert space, let 𝐶 be a nonempty closed convex subset of 𝐻, let 𝑆𝐶𝐶 be a nonexpansive mapping, and let 𝐵𝐶𝐻 be an 𝜔-Lipschitz continuous and relaxed (𝑢,𝑣)-cocoercive mappings with 𝑣>𝑢𝜔2. If 0𝛼𝑛2(𝑣𝑢𝜔2)/𝜔2, then 𝑆𝛼𝑛𝐵𝑆 is a nonexpansive mapping in 𝐻.

Proof. Let 𝛼𝑛2(𝑣𝑢𝜔2)/𝜔2,𝑣>𝑢𝜔2. Then, for every 𝑥,𝑦𝐶, we have 𝑆𝛼𝑛𝐵𝑆𝑥𝑆𝛼𝑛𝑦𝐵𝑆2=(𝑆𝑥𝑆𝑦)𝛼𝑛(𝐵𝑆𝑥𝐵𝑆𝑦)2=𝑆𝑥𝑆𝑦22𝛼𝑛𝑆𝑥𝑆𝑦,𝐵𝑆𝑥𝐵𝑆𝑦+𝛼2𝑛𝐵𝑆𝑥𝐵𝑆𝑦2𝑆𝑥𝑆𝑦22𝛼𝑛𝑢𝐵𝑆𝑥𝐵𝑆𝑦2+𝑣𝑆𝑥𝑆𝑦2+𝛼2𝑛𝐵𝑆𝑥𝐵𝑆𝑦2𝑆𝑥𝑆𝑦2+2𝛼𝑛𝑢𝜔2𝑆𝑥𝑆𝑦22𝛼𝑛𝑣𝑆𝑥𝑆𝑦2+𝛼2𝑛𝜔2𝑆𝑥𝑆𝑦2=1+2𝛼𝑛𝑢𝜔22𝛼𝑛𝑣+𝛼2𝑛𝜔2𝑆𝑥𝑆𝑦21𝛼𝑛𝜔22𝑣𝑢𝜔2𝜔2𝛼𝑛𝑥𝑦2.(3.11) Now, since (1𝛼𝑛𝜔2[2(𝑣𝑢𝜔2)/𝜔2𝛼𝑛])<1, thus (𝑆𝛼𝑛𝐵𝑆)𝑥(𝑆𝛼𝑛𝐵𝑆)𝑦𝑥𝑦.
Thus, 𝑆𝛼𝑛𝐵𝑆 is a nonexpansive mapping of 𝐶 into 𝐻.

Now we can prove that a strong convergence theorem is a real Hilbert space.

Theorem 3.3. Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻. Let Θ1 and Θ2 be two bifunctions from 𝐶×𝐶 to satisfying (H1)–(H5) and let 𝜑𝐶{+} be a proper lower semicontinuous and convex function with assumption (B1) or (B2). Let (i)Ψ1𝐶𝐻 be a 𝜉-inverse-strongly monotone mapping, (ii)Ψ2𝐶𝐻 be a 𝛽-inverse-strongly monotone mapping, (iii)𝐵𝐶𝐻 be an 𝜔-Lipschitz continuous and relaxed (𝑢,𝑣)-cocoercive mappings, (iv)𝑓𝐶𝐶 be a contraction mapping with coefficient 𝜂(0,1) and let 𝐴 be a strongly positive linear bounded self-adjoint operator with the coefficient 𝛾>0 and 0<𝛾<𝛾/𝜂. Let 𝑆𝐶𝐶 be a nonexpansive mapping with 𝐹(𝑆).
Assume that Θ=𝐹(𝑆)GMEP1,𝜑,Ψ1ΘGMEP2,𝜑,Ψ2VI(𝐶,𝐵).(3.12) Let {𝑥𝑛}, {𝑦𝑛}, {𝑧𝑛}, {𝑣𝑛}, and {𝑢𝑛} be the sequences generated by 𝑢𝑛=𝑇(Θ1𝑟,𝜑)𝑛𝑥𝑛𝑟𝑛Ψ1𝑥𝑛,𝑣𝑛=𝑇(Θ2𝑠,𝜑)𝑛𝑢𝑛𝑠𝑛Ψ2𝑢𝑛,𝑧𝑛=𝑃𝐶𝑆𝑣𝑛𝛼𝑛𝐵𝑆𝑣𝑛,𝑦𝑛=𝜖𝑛𝑥𝛾𝑓𝑛+𝛽𝑛𝑥𝑛+1𝛽𝑛𝐼𝜖𝑛𝐴𝑧𝑛,𝑥𝑛+1=𝛾𝑛𝑥𝑛+1𝛾𝑛𝑦𝑛,𝑛1,(3.13) where {𝑟𝑛}[𝑎,𝑏][0,2𝜉],{𝑠𝑛}[𝑐,𝑑][0,2𝛽],{𝛾𝑛}[,𝑗](0,1), and {𝛾𝑛},{𝜖𝑛}, and {𝛽𝑛} are three sequences in (0,1) satisfying the following conditions: (C1)lim𝑛𝜖𝑛=0 and 𝑛=1𝜖𝑛=, (C2)0<liminf𝑛𝛽𝑛limsup𝑛𝛽𝑛<1, and lim𝑛𝛽𝑛=0, (C3)0<liminf𝑛𝑟𝑛limsup𝑛𝑟𝑛<2𝜉, and lim𝑛|𝑟𝑛+1𝑟𝑛|=0, (C4)0<liminf𝑛𝑠𝑛limsup𝑛𝑠𝑛<2𝛽, and lim𝑛|𝑠𝑛+1𝑠𝑛|=0, (C5){𝛼𝑛}[𝑒,𝑔](0,2(𝑣𝑢𝜔2)/𝜔2),𝑣>𝑢𝜔2, and lim𝑛|𝛼𝑛+1𝛼𝑛|=0. Then, {𝑥𝑛} converges strongly to 𝑧=𝑃(𝛾𝑓+(𝐼𝐴))(𝑧), which is the unique solution of the variational inequality: 𝛾𝑓(𝑧)𝐴𝑧,𝑥𝑧0,𝑥.(3.14)

Proof. From the restrictions on control sequences, we may assume, without loss of generality, that 𝜖𝑛(1𝛽𝑛)𝐴1 for all 𝑛1. Since 𝐴 is a strongly positive linear bounded self-adjoint operator on 𝐻, we have ||||𝐴=sup𝐴𝑥,𝑥𝑥𝐻,𝑥=1.(3.15) Observe that 1𝛽𝑛𝐼𝜖𝑛𝐴𝑥,𝑥=1𝛽𝑛𝜖𝑛𝐴𝑥,𝑥1𝛽𝑛𝜖𝑛𝐴0.(3.16) That is, (1𝛽𝑛)𝐼𝜖𝑛𝐴 is positive. It follows that 1𝛽𝑛𝐼𝜖𝑛𝐴||=sup1𝛽𝑛𝐼𝜖𝑛𝐴||𝑥,𝑥𝑥𝐻,𝑥=1=sup1𝛽𝑛𝜖𝑛𝐴𝑥,𝑥𝑥𝐻,𝑥=11𝛽𝑛𝜖𝑛𝛾.(3.17) We will split the proof of Theorem 3.3 into six steps.Step 1. We claim that the sequence {𝑥𝑛} is bounded.
Indeed, let 𝑥, by Lemmas 2.2 and 2.9, we obtain 𝑥=𝑆𝑥=𝑃𝐶𝑥𝛼𝑛𝐵𝑥=𝑇(Θ1𝑟,𝜑)𝑛𝑥𝑟𝑛Ψ1𝑥=𝑇(Θ2𝑠,𝜑)𝑛𝑥𝑠𝑛Ψ2𝑥.(3.18) Since 𝑢𝑛=𝑇(Θ1𝑟,𝜑)𝑛(𝑥𝑛𝑟𝑛Ψ1𝑥𝑛)dom𝜑,Ψ1 is 𝜉-inverse-strongly monotone, and 0𝑟𝑛2𝜉, we know that, for any 𝑛, we have 𝑢𝑛𝑥2=𝑇(Θ1𝑟,𝜑)𝑛𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑇(Θ1𝑟,𝜑)𝑛𝑥𝑟𝑛Ψ1𝑥2𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑥𝑟𝑛Ψ1𝑥2=𝑥𝑛𝑥𝑟𝑛Ψ1𝑥𝑛Ψ1𝑥2=𝑥𝑛𝑥22𝑟𝑛𝑥𝑛𝑥,Ψ1𝑥𝑛Ψ1𝑥+𝑟2𝑛Ψ1𝑥𝑛Ψ1𝑥2𝑥𝑛𝑥22𝑟𝑛𝜉Ψ1𝑥𝑛Ψ1𝑥2+𝑟2𝑛Ψ1𝑥𝑛Ψ1𝑥2=𝑥𝑛𝑥2+𝑟𝑛𝑟𝑛Ψ2𝜉1𝑥𝑛Ψ1𝑥2𝑥𝑛𝑥2.(3.19) Similarly, from (3.19), 𝑣𝑛=𝑇(Θ2𝑠,𝜑)𝑛(𝑢𝑛𝑠𝑛Ψ2𝑢𝑛)dom𝜑, and 0𝑠𝑛2𝛽, we can prove that 𝑣𝑛𝑥2=𝑇(Θ2𝑠,𝜑)𝑛𝑢𝑛𝑠𝑛Ψ2𝑢𝑛𝑇(Θ2𝑠,𝜑)𝑛𝑥𝑠𝑛Ψ2𝑥2𝑢𝑛𝑥2𝑥𝑛𝑥2,(3.20) and hence 𝑣𝑛𝑥𝑢n𝑥𝑥𝑛𝑥.(3.21) Let 𝑥, and form Lemma 3.2  𝑆𝛼𝑛𝐵𝑆 is a nonexpansive mapping and from Lemma  2.2 𝑥=𝑃𝐶(𝑥𝛼𝑛𝐵𝑥), we have 𝑧𝑛𝑥=𝑃𝐶𝑆𝑣𝑛𝛼𝑛𝐵𝑆𝑣𝑛𝑃𝐶𝑥𝛼𝑛𝐵𝑥𝑆𝑣𝑛𝛼𝑛𝐵𝑆𝑣𝑛𝑥𝛼𝑛𝐵𝑥=𝑆𝑣𝑛𝛼𝑛𝐵𝑆𝑣𝑛𝑆𝑥𝛼𝑛𝐵𝑆𝑥=𝑆𝛼𝑛𝑣𝐵𝑆𝑛𝑆𝛼𝑛𝑥𝐵𝑆𝑣𝑛𝑥𝑥𝑛𝑥,𝑦𝑛𝑥=𝜖𝑛𝑥𝛾𝑓𝑛𝐴𝑥+𝛽𝑛𝑥𝑛𝑥+1𝛽𝑛𝐼𝜖𝑛𝐴𝑧𝑛𝑥1𝛽𝑛𝜖𝑛𝛾𝑧𝑛𝑥+𝛽𝑛𝑥𝑛𝑥+𝜖𝑛𝑥𝛾𝑓𝑛𝐴𝑥1𝛽𝑛𝜖𝑛𝛾𝑥𝑛𝑥+𝛽𝑛𝑥𝑛𝑥+𝜖𝑛𝑥𝛾𝑓𝑛𝐴𝑥1𝜖𝑛𝛾𝑥𝑛𝑥+𝜖𝑛𝛾𝑓𝑥𝑛𝑥𝑓+𝜖𝑛𝑥𝛾𝑓𝐴𝑥1𝜖𝑛𝛾𝑥𝑛𝑥+𝜖𝑛𝑥𝛾𝜂𝑛𝑥+𝜖𝑛𝑥𝛾𝑓𝐴𝑥=1𝜖𝛾𝜂𝛾𝑛𝑥𝑛𝑥+𝜖𝑛𝑥𝛾𝑓𝐴𝑥,(3.22) which yields that 𝑥𝑛+1𝑥𝛾𝑛𝑥𝑛𝑥+1𝛾𝑛𝑦𝑛𝑥𝛾𝑛𝑥𝑛𝑥+1𝛾𝑛1𝜖𝛾𝜂𝛾𝑛𝑥𝑛𝑥+𝜖𝑛𝑥𝛾𝑓𝐴𝑥=𝛾𝑛𝑥𝑛𝑥+1𝛾𝑛𝑥𝑛𝑥1𝛾𝑛𝜖𝛾𝜂𝛾𝑛𝑥𝑛𝑥+1𝛾𝑛𝜖𝑛𝑥𝛾𝑓𝐴𝑥=11𝛾𝑛𝜖𝛾𝜂𝛾𝑛𝑥𝑛𝑥+1𝛾𝑛𝜖𝑛𝑥𝛾𝑓𝐴𝑥.(3.23) By mathematical induction, putting 𝐷=max{𝑥1𝑥,𝛾𝑓(𝑥)𝐴𝑥/𝛾𝜂𝛾}, we have that 𝑥𝑛𝑥𝐷 for all 𝑛1. Indeed, we can easily see that 𝑥1𝑥𝐷. Suppose that 𝑥𝑘𝑥𝐷 for some positive integral 𝑘. Then we have that 𝑥𝑘+1𝑥11𝛾𝑘𝜖𝛾𝜂𝛾𝑘𝑥𝑘𝑥+1𝛾𝑘𝜖𝑘𝑥𝛾𝑓𝐴𝑥11𝛾𝑘𝜖𝛾𝜂𝛾𝑘𝐷+1𝛾𝑘𝜖𝑘𝑥𝛾𝑓𝐴𝑥=11𝛾𝑘𝜖𝛾𝜂𝛾𝑘𝐷+1𝛾𝑘𝜖𝛾𝜂𝛾𝑘𝑥𝛾𝑓𝐴𝑥𝛾𝜂𝛾11𝛾𝑘𝜖𝛾𝜂𝛾𝑘𝐷+1𝛾𝑘𝜖𝛾𝜂𝛾𝑘𝐷=𝐷.(3.24) This shows that {𝑥𝑛} is bounded in 𝐻. From (3.21), we know that {𝑢𝑛} and {𝑣𝑛} are bounded in 𝐶 and so {𝑦𝑛}, {Ψ1𝑢𝑛}, {Ψ2𝑥𝑛}, {𝑆𝑣𝑛}, {𝐵𝑆𝑣𝑛}, and {𝑓(𝑥𝑛)} are bounded sequence in 𝐻.
Step 2. We claim that lim𝑛𝑥𝑛+1𝑥𝑛=0.
Since 𝑆𝛼𝑛𝐵𝑆 is nonexpansive, we have 𝑧𝑛+1𝑧𝑛=𝑃𝐶𝑆𝑣𝑛+1𝛼𝑛+1𝐵𝑆𝑣𝑛+1𝑃𝐶𝑆𝑣𝑛𝛼𝑛𝐵𝑆𝑣𝑛𝑆𝑣𝑛+1𝛼𝑛+1𝐵𝑆𝑣𝑛+1𝑆𝑣𝑛𝛼𝑛𝐵𝑆𝑣𝑛=𝑆𝑣𝑛+1𝛼𝑛+1𝐵𝑆𝑣𝑛+1𝑆𝑣𝑛𝛼𝑛+1𝐵𝑆𝑣𝑛+𝛼𝑛𝛼𝑛+1𝐵𝑆𝑣𝑛𝑆𝑣𝑛+1𝛼𝑛+1𝐵𝑆𝑣𝑛+1𝑆𝑣𝑛𝛼𝑛+1𝐵𝑆𝑣𝑛+||𝛼𝑛𝛼𝑛+1||𝐵𝑆𝑣𝑛𝑣𝑛+1𝑣𝑛+||𝛼𝑛𝛼𝑛+1||𝐵𝑆𝑣𝑛.(3.25) Next, we estimate 𝑢𝑛+1𝑢𝑛. Observing that 𝑢𝑛=𝑇(Θ1𝑟,𝜑)𝑛(𝑥𝑛𝑟𝑛Ψ1𝑥𝑛) and 𝑢𝑛+1=𝑇(Θ1𝑟,𝜑)𝑛+1(𝑥𝑛+1𝑟𝑛+1Ψ1𝑥𝑛+1), we have 𝑢𝑛+1𝑢𝑛=𝑇(Θ1𝑟,𝜑)𝑛+1𝑥𝑛+1𝑟𝑛+1Ψ1𝑥𝑛+1𝑇(Θ1𝑟,𝜑)𝑛𝑥𝑛𝑟𝑛Ψ1𝑥𝑛=𝑇(Θ1𝑟,𝜑)𝑛+1𝑥𝑛+1𝑟𝑛+1Ψ1𝑥𝑛+1𝑇(Θ1𝑟,𝜑)𝑛+1𝑥𝑛𝑟𝑛Ψ1𝑥𝑛+𝑇(Θ1𝑟,𝜑)𝑛+1𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑇(Θ1𝑟,𝜑)𝑛𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑇(Θ1𝑟,𝜑)𝑛+1𝑥𝑛+1𝑟𝑛+1Ψ1𝑥𝑛+1𝑇(Θ1𝑟,𝜑)𝑛+1𝑥𝑛𝑟𝑛Ψ1𝑥𝑛+𝑇(Θ1𝑟,𝜑)𝑛+1𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑇(Θ1𝑟,𝜑)𝑛𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑥𝑛+1𝑟𝑛+1Ψ1𝑥𝑛+1𝑥𝑛𝑟𝑛Ψ1𝑥𝑛+𝑇(Θ1𝑟,𝜑)𝑛+1𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑇(Θ1𝑟,𝜑)𝑛𝑥𝑛𝑟𝑛Ψ1𝑥𝑛=𝑥𝑛+1𝑥𝑛𝑟𝑛+1Ψ1𝑥𝑛+1Ψ1𝑥𝑛+𝑟𝑛+1𝑟𝑛Ψ1𝑥𝑛+𝑇(Θ1𝑟,𝜑)𝑛+1𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑇(Θ1𝑟,𝜑)𝑛𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑥𝑛+1𝑥𝑛𝑟𝑛+1Ψ1𝑥𝑛+1Ψ1𝑥𝑛+||𝑟𝑛+1𝑟𝑛||Ψ1𝑥𝑛+𝑇(Θ1𝑟,𝜑)𝑛+1𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑇(Θ1𝑟,𝜑)𝑛𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑥𝑛+1𝑥𝑛+||𝑟𝑛+1𝑟𝑛||Ψ1𝑥𝑛+𝑇(Θ1𝑟,𝜑)𝑛+1𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑇(Θ1𝑟,𝜑)𝑛𝑥𝑛𝑟𝑛Ψ1𝑥𝑛.(3.26) Similarly, we can prove that 𝑣𝑛+1𝑣𝑛𝑢𝑛+1𝑢𝑛+||𝑠𝑛+1𝑠𝑛||Ψ2𝑢𝑛+𝑇(Θ2𝑠,𝜑)𝑛+1𝑢𝑛𝑠𝑛Ψ2𝑢𝑛𝑇(Θ2𝑠,𝜑)𝑛𝑢𝑛𝑠𝑛Ψ2𝑢𝑛.(3.27) Substitution (3.26) into (3.27), we derive 𝑣𝑛+1𝑣𝑛𝑥𝑛+1𝑥𝑛+||𝑟𝑛+1𝑟𝑛||Ψ1𝑥𝑛+||𝑠𝑛+1𝑠𝑛||Ψ2𝑢𝑛+𝑇(Θ1𝑟,𝜑)𝑛+1𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑇(Θ1𝑟,𝜑)𝑛𝑥𝑛𝑟𝑛Ψ1𝑥𝑛+𝑇(Θ2𝑠,𝜑)𝑛+1𝑢𝑛𝑠𝑛Ψ2𝑢𝑛𝑇(Θ2𝑠,𝜑)𝑛𝑢𝑛𝑠𝑛Ψ2𝑢𝑛.(3.28) Since {Ψ1𝑥𝑛},{Ψ2𝑢𝑛}, and {𝐵𝑆𝑣𝑛} are bounded, 𝐾 is an appropriate constant such that 𝐾maxsup𝑛1Ψ1𝑥𝑛,sup𝑛1Ψ2𝑢𝑛,sup𝑛1𝐵𝑆𝑣𝑛.(3.29) Substitution (3.28) into (3.25), we obtain 𝑧𝑛+1𝑧𝑛𝑣𝑛+1𝑣𝑛+||𝛼𝑛𝛼𝑛+1||𝐵𝑆𝑣𝑛𝑥𝑛+1𝑥𝑛+||𝑟𝑛+1𝑟𝑛||Ψ1𝑥𝑛+||𝑠𝑛+1𝑠𝑛||Ψ2𝑢𝑛+||𝛼𝑛𝛼𝑛+1||𝐵𝑆𝑣𝑛+𝑇(Θ1𝑟,𝜑)𝑛+1𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑇(Θ1𝑟,𝜑)𝑛𝑥𝑛𝑟𝑛Ψ1𝑥𝑛+𝑇(Θ2𝑠,𝜑)𝑛+1𝑢𝑛𝑠𝑛Ψ2𝑢𝑛𝑇(Θ2𝑠,𝜑)𝑛𝑢𝑛𝑠𝑛Ψ2𝑢𝑛𝑥𝑛+1𝑥𝑛||𝑟+𝐾𝑛+1𝑟𝑛||+||𝑠𝑛+1𝑠𝑛||+||𝛼𝑛𝛼𝑛+1||+𝑇(Θ1𝑟,𝜑)𝑛+1𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑇(Θ1𝑟,𝜑)𝑛𝑥𝑛𝑟𝑛Ψ1𝑥𝑛+𝑇(Θ2𝑠,𝜑)𝑛+1𝑢𝑛𝑠𝑛Ψ2𝑢𝑛𝑇(Θ2𝑠,𝜑)𝑛𝑢𝑛𝑠𝑛Ψ2𝑢𝑛.(3.30) From (3.13), we have 𝑦𝑛=𝜖𝑛𝑥𝛾𝑓𝑛+𝛽𝑛𝑥𝑛+1𝛽𝑛𝐼𝜖𝑛𝐴𝑧𝑛,𝑦𝑛+1=𝜖𝑛+1𝑥𝛾𝑓𝑛+1+𝛽𝑛+1𝑥𝑛+1+1𝛽𝑛+1𝐼𝜖𝑛+1𝐴𝑧𝑛+1.(3.31) Simple calculations show that 𝑦𝑛+1𝑦𝑛=𝜖𝑛+1𝛾𝑓𝑥𝑛+1𝑥𝑓𝑛+𝜖𝑛+1𝜖𝑛𝑥𝛾𝑓𝑛+𝛽𝑛+1𝑥𝑛+1𝑥𝑛+𝛽𝑛+1𝛽𝑛𝑥𝑛+1𝛽𝑛+1𝐼𝜖𝑛+1𝐴𝑧𝑛+1𝑧𝑛𝛽𝑛+1𝛽𝑛𝑧𝑛𝜖𝑛+1𝜖𝑛𝐴𝑧𝑛=𝜖𝑛+1𝛾𝑓𝑥𝑛+1𝑥𝑓𝑛+𝜖𝑛+1𝜖𝑛𝑥𝛾𝑓𝑛𝐴𝑧𝑛+𝛽𝑛+1𝑥𝑛+1𝑥𝑛+𝛽𝑛+1𝛽𝑛𝑥𝑛𝑧𝑛+1𝛽𝑛+1𝐼𝜖𝑛+1𝐴𝑧𝑛+1𝑧𝑛,(3.32) which yields that 𝑦𝑛+1𝑦𝑛𝜖𝑛+1𝛾𝑓𝑥𝑛+1𝑥𝑓𝑛+||𝜖𝑛+1𝜖𝑛||𝑥𝛾𝑓𝑛𝐴𝑧𝑛+𝛽𝑛+1𝑥𝑛+1𝑥𝑛+||𝛽𝑛+1𝛽𝑛||𝑥𝑛𝑧𝑛+1𝛽𝑛+1𝜖𝑛+1𝛾𝑧𝑛+1𝑧𝑛𝜖𝑛+1𝑥𝛾𝜂𝑛+1𝑥𝑛+||𝜖𝑛+1𝜖𝑛||𝑥𝛾𝑓𝑛𝐴𝑧𝑛+𝛽𝑛+1𝑥𝑛+1𝑥𝑛+||𝛽𝑛+1𝛽𝑛||𝑥𝑛𝑧𝑛+1𝛽𝑛+1𝜖𝑛+1𝛾𝑧𝑛+1𝑧𝑛.(3.33) Substitution (3.30) into (3.33) yields that 𝑦𝑛+1𝑦𝑛𝜖𝑛+1𝑥𝛾𝜂𝑛+1𝑥𝑛+||𝜖𝑛+1𝜖𝑛||𝑥𝛾𝑓𝑛𝐴𝑧𝑛+𝛽𝑛+1𝑥𝑛+1𝑥𝑛+||𝛽𝑛+1𝛽𝑛||𝑥𝑛𝑧𝑛+1𝛽𝑛+1𝜖𝑛+1𝛾×𝑥𝑛+1𝑥𝑛||𝑟+𝐾𝑛+1𝑟𝑛||+||𝑠𝑛+1𝑠𝑛||+||𝛼𝑛𝛼𝑛+1||+𝑇(Θ1𝑟,𝜑)𝑛+1𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑇(Θ1𝑟,𝜑)𝑛𝑥𝑛𝑟𝑛Ψ1𝑥𝑛+𝑇(Θ2𝑠,𝜑)𝑛+1𝑢𝑛𝑠𝑛Ψ2𝑢𝑛𝑇(Θ2𝑠,𝜑)𝑛𝑢𝑛𝑠𝑛Ψ2𝑢𝑛=1𝜖𝑛+1𝑥𝛾𝛾𝜂𝑛+1𝑥𝑛+||𝜖𝑛+1𝜖𝑛||𝑥𝛾𝑓𝑛𝐴𝑧𝑛+||𝛽𝑛+1𝛽𝑛||𝑥𝑛𝑧𝑛+1𝛽𝑛+1𝜖𝑛+1𝛾𝐾×||𝑟𝑛+1𝑟𝑛||+||𝑠𝑛+1𝑠𝑛||+||𝛼𝑛𝛼𝑛+1||+1𝛽𝑛+1𝜖𝑛+1𝛾×𝑇(Θ1𝑟,𝜑)𝑛+1𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑇(Θ1𝑟,𝜑)𝑛𝑥𝑛𝑟𝑛Ψ1𝑥𝑛+1𝛽𝑛+1𝜖𝑛+1𝛾𝑇(Θ2𝑠,𝜑)𝑛+1𝑢𝑛𝑠𝑛Ψ2𝑢𝑛𝑇(Θ2𝑠,𝜑)𝑛𝑢𝑛𝑠𝑛Ψ2𝑢𝑛1𝜖𝑛+1𝑥𝛾𝛾𝜂𝑛+1𝑥𝑛||𝜖+𝑀𝑛+1𝜖𝑛||+||𝛽𝑛+1𝛽𝑛||+||𝑟𝑛+1𝑟𝑛||+||𝑠𝑛+1𝑠𝑛||+||𝛼𝑛𝛼𝑛+1||+𝑇(Θ1𝑟,𝜑)𝑛+1𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑇(Θ1𝑟,𝜑)𝑛𝑥𝑛𝑟𝑛Ψ1𝑥𝑛+𝑇(Θ2𝑠,𝜑)𝑛+1𝑢𝑛𝑠𝑛Ψ2𝑢𝑛𝑇(Θ2𝑠,𝜑)𝑛𝑢𝑛𝑠𝑛Ψ2𝑢𝑛,(3.34) where 𝑀 is an appropriate constant such that 𝑀maxsup𝑛1𝑥𝛾𝑓𝑛𝐴𝑧𝑛,sup𝑛1𝑥𝑛𝑧𝑛.,𝐾(3.35) Since 𝑥𝑛+1=𝛾𝑛𝑥𝑛+(1𝛾𝑛)𝑦𝑛 and 𝑦𝑛+1𝑦𝑛𝑥𝑛+1𝑥𝑛||𝜖𝑀𝑛+1𝜖𝑛||+||𝛽𝑛+1𝛽𝑛||+||𝑟𝑛+1𝑟𝑛||+||𝑠𝑛+1𝑠𝑛||+||𝛼𝑛𝛼𝑛+1||+𝑇(Θ1𝑟,𝜑)𝑛+1𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑇(Θ1𝑟,𝜑)𝑛𝑥𝑛𝑟𝑛Ψ1𝑥𝑛+𝑇(Θ2𝑠,𝜑)𝑛+1𝑢𝑛𝑠𝑛Ψ2𝑢𝑛𝑇(Θ2𝑠,𝜑)𝑛𝑢𝑛𝑠𝑛Ψ2𝑢𝑛,(3.36) next, we estimate 𝑇(Θ1𝑟,𝜑)𝑛+1𝑥n𝑟𝑛Ψ1𝑥𝑛𝑇(Θ1𝑟,𝜑)𝑛𝑥𝑛𝑟𝑛Ψ1𝑥𝑛.(3.37) Note that liminf𝑛𝑟𝑛>0; there exists a constant ̃𝑟>0 such that 𝑟𝑛̃𝑟>0 for all 𝑛1. From Lemma 3.1, we get 𝑇(Θ1𝑟,𝜑)𝑛+1𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑇(Θ1𝑟,𝜑)𝑛𝑥𝑛𝑟𝑛Ψ1𝑥𝑛2𝑟𝑛+1𝑟𝑛𝑟𝑛+1𝑇(Θ1𝑟,𝜑)𝑛+1𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑇(Θ1𝑟,𝜑)𝑛𝑥𝑛𝑟𝑛Ψ1𝑥𝑛,𝑇(Θ1𝑟,𝜑)𝑛+1𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑥𝑛𝑟𝑛Ψ1𝑥𝑛=𝑟𝑛+1𝑟𝑛𝑟𝑛+1𝑇(Θ1𝑟,𝜑)𝑛+1𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑇(Θ1𝑟,𝜑)𝑛𝑥𝑛𝑟𝑛Ψ1𝑥𝑛,𝑇(Θ1𝑟,𝜑)𝑛+1𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑥𝑛𝑟𝑛Ψ1𝑥𝑛||𝑟𝑛+1𝑟𝑛||𝑟𝑛+1𝑇(Θ1𝑟,𝜑)𝑛+1𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑇(Θ1𝑟,𝜑)𝑛𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑇(Θ1𝑟,𝜑)𝑛+1𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑥𝑛𝑟𝑛Ψ1𝑥𝑛.(3.38) It follows that 𝑇(Θ1𝑟,𝜑)𝑛+1𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑇(Θ1𝑟,𝜑)𝑛𝑥𝑛𝑟𝑛Ψ1𝑥𝑛||𝑟𝑛+1𝑟𝑛||𝑇̃𝑟(Θ1𝑟,𝜑)𝑛+1𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑥𝑛𝑟𝑛Ψ1𝑥𝑛.(3.39) Since lim𝑛|𝑟𝑛+1𝑟𝑛|=0, we obtain lim𝑛𝑇(Θ1𝑟,𝜑)𝑛+1𝑥𝑛𝑟𝑛Ψ1𝑥𝑛𝑇(Θ1𝑟,𝜑)𝑛𝑥𝑛𝑟𝑛Ψ1𝑥𝑛=0.(3.40) Similarly, we can prove that lim𝑛𝑇(Θ2𝑠,𝜑)𝑛+1𝑢𝑛𝑠𝑛Ψ2𝑢𝑛𝑇(Θ2𝑠,𝜑)𝑛𝑢𝑛𝑠𝑛Ψ2𝑢𝑛=0.(3.41) Consequently, from (3.40), (3.41), and conditions in Theorem 3.3, we obtain lim𝑛𝑦𝑛+1𝑦𝑛𝑥𝑛+1𝑥𝑛0.(3.42) It follows from Lemma 2.7 that lim𝑛𝑦𝑛𝑥𝑛=0.(3.43) In view of (3.13), we see that 𝑥𝑛+1𝑥𝑛=1𝛾𝑛𝑥𝑛𝑦𝑛,𝑛1,(3.44) which, combining with (3.43) and 0<𝛾𝑛𝑗<1, yields that lim𝑛𝑥𝑛+1𝑥𝑛=0.(3.45)
Step 3. We claim that lim𝑛𝑆𝑧𝑛𝑧𝑛=0.
Observing that 𝑦𝑛=𝜖𝑛𝛾𝑓(𝑥𝑛)+𝛽𝑛𝑥𝑛+((1𝛽𝑛)𝐼𝜖𝑛𝐴)𝑧𝑛, we have 𝑦𝑛