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Journal of Applied Mathematics
Volume 2012 (2012), Article ID 816529, 26 pages
http://dx.doi.org/10.1155/2012/816529
Research Article

Viscosity Approximation Method for System of Variational Inclusions Problems and Fixed-Point Problems of a Countable Family of Nonexpansive Mappings

1Department of Mathematics, Faculty of Liberal Arts, Rajamangala University of Technology Rattanakosin (RMUTR), Bangkok 10100, Thailand
2Department of Mathematics, Faculty of Science, King Mongkut’s University of Technology Thonburi (KMUTT), Bangkok 10140, Thailand

Received 22 January 2012; Accepted 2 March 2012

Academic Editor: Yonghong Yao

Copyright © 2012 Chaichana Jaiboon and Poom Kumam. 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 new iterative schemes for finding the common element of the set of common fixed points of countable family of nonexpansive mappings, the set of solutions of the variational inequality problem for relaxed cocoercive and Lipschitz continuous, the set of solutions of system of variational inclusions problem, and the set of solutions of equilibrium problems in a real Hilbert space by using the viscosity approximation method. We prove strong convergence theorem under some parameters. The results in this paper unify and generalize some well-known results in the literature.

1. Introduction

Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻. A mapping 𝑆 of 𝐶 into itself is called nonexpansive if 𝑆𝑥𝑆𝑦𝑥𝑦 for all 𝑥,𝑦𝐶. We denote by 𝐹(𝑆) the set of fixed points of 𝑆; that is, 𝐹(𝑆)={𝑥𝐶𝑆𝑥=𝑥}. If 𝐶𝐻 is nonempty, closed and convex and let 𝑆𝐶𝐶 be a nonexpansive mapping, then 𝐹(𝑆) is closed and convex and 𝐹(𝑆), when 𝐶 is bounded; see, for example, [1, 2]. The metric projection, 𝑃𝐶, onto a given nonempty, closed and convex subset 𝐶, satisfies the nonexpansive with 𝐹(𝑃𝐶)=𝐶. A mapping 𝐵𝐶𝐶 is called monotone if 𝐵𝑥𝐵𝑦,𝑥𝑦0 for all 𝑥,𝑦𝐶. A mapping 𝐵𝐶𝐶 is called 𝛽-inverse-strongly monotone if there exists a constant 𝛽>0 such that 𝐵𝑥𝐵𝑦,𝑥𝑦𝛽𝑥𝑦2, for all 𝑥,𝑦𝐶. A mapping 𝐵𝐶𝐶 is called relaxed (𝜙,𝜔)-cocoercive if there exists 𝜙,𝜔>0 such that𝐵𝑥𝐵𝑦,𝑥𝑦(𝜙)𝐵𝑥𝐵𝑦2+𝜔𝑥𝑦2,𝑥,𝑦𝐶.(1.1)

A mapping 𝐵𝐶𝐶 is said to be 𝜉-Lipschitz continuous if there exists 𝜉0 such that𝐵𝑥𝐵𝑦𝜉𝑥𝑦,𝑥,𝑦𝐶.(1.2)

Let 𝐵𝐻𝐻 be a single-valued nonlinear mapping and 𝑀𝐻2𝐻 a multivalued mapping. The variational inclusion problem is to find ̃𝑥𝐻 such that𝜃𝐵(̃𝑥)+𝑀(̃𝑥),(1.3) where 𝜃 is the zero vector in 𝐻. The set of solutions of problem (1.3) is denoted by 𝐼(𝐵,𝑀). If 𝑀=𝜕𝜓𝐶, where 𝐶 is a nonempty closed convex subset of 𝐻 and 𝜕𝜓𝐶𝐻[0,+] is the indicator function of 𝐶; that is,𝜓𝐶(𝑥)=0,𝑥𝐶,+,𝑥𝐶,(1.4)

then, the variational inclusion problem (1.3) is equivalent to the variational inequality problems denoted by VI(𝐶,𝐵) which is to find ̃𝑥𝐶 such that𝐵̃𝑥,𝑦̃𝑥0,𝑦𝐶.(1.5)

In 2003, Takahashi and Toyoda [3] to find 𝑥𝐹(𝑆)VI(𝐶,𝐵) introduced the following iterative scheme:𝑥0𝑥Cchosenarbitrary,𝑛+1=𝛼𝑛𝑥𝑛+1𝛼𝑛𝑆𝑃𝐶𝑥𝑛𝜉𝑛𝐵𝑥𝑛,𝑛0,(1.6) where 𝐵 is a 𝛽-inverse-strongly monotone mapping, {𝛼𝑛} is a sequence in (0, 1), and {𝜉𝑛} is a sequence in (0,2𝛽). They showed that if 𝐹(𝑆)VI(𝐶,𝐵) is nonempty, then the sequence {𝑥𝑛} generated by (1.6) converges weakly to some 𝑥𝐹(𝑆)VI(𝐶,𝐵).

In 2008, Zhang et al. [4] to find 𝑥𝐹(𝑆)𝐼(𝑀,𝐵). They introduced the following new iterative scheme:𝑥0𝑦Cchosenarbitrary,𝑛=𝐽𝑀,𝜆𝑥𝑛𝜆𝐵𝑥𝑛,𝑥𝑛+1=𝛼𝑛𝑥+1𝛼𝑛𝑆𝑦𝑛,𝑛0,(1.7) where 𝐽𝑀,𝜆=(𝐼+𝜆𝑀)1 is the resolvent operator associated with 𝑀 and a positive number 𝜆,{𝛼𝑛} is a sequence in the interval [0,1].

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 that𝐹𝑥,𝑦0,𝑦𝐶.(1.8) The set of solutions of (1.8) is denoted by EP(𝐹). Many problems in applied sciences, such as monotone inclusion problems, variational inequality problems, saddle point problems, Nash equilibria in noncooperative games, as well as certain fixed-point problems reduce to finding some element to EP(𝐹) in Hilbert and Banach spaces (see [514]).

Given any 𝑟>0. The operator 𝑇𝑟𝐻𝐶 defined by𝑇𝑟1(𝑥)=𝑧𝐶𝐹(𝑧,𝑦)+𝑟,𝑦𝑧,𝑧𝑥0,𝑦𝐶(1.9) is called the resolvent of 𝐹 (see [5, 6]).

It is shown in [6] that, under suitable hypotheses on 𝐹 (to be stated precisely in Section 2), 𝑇𝑟𝐻𝐶 is single valued and firmly nonexpansive and satisfies𝐹𝑇𝑟=EP(𝐹),𝑟>0.(1.10)

Using this result, for finding an element of 𝐹(𝑆)VI(𝐶,𝐵)EP(𝐹), Su et al. [15] introduced the following iterative scheme by the viscosity approximation method in Hilbert spaces:𝑥0𝑥𝐶chosenarbitrary,𝑛+1=𝛼𝑛𝑓𝑥𝑛+1𝛼𝑛𝑆𝑃𝐶𝐼𝜉𝑛𝐵𝑇𝑟𝑛𝑥𝑛,𝑛0,(1.11) where 𝑓𝐶𝐶 is a contraction (i.e., 𝑓(𝑥)𝑓(𝑦)𝜓𝑥𝑦,forall𝑥,𝑦𝐶and0𝜓<1) and {𝛼𝑛}(0,1), 𝜉𝑛(0,2𝛽), and 𝑟𝑛(0,) satisfy some appropriate conditions. Furthermore, they prove {𝑥𝑛} converges strongly to the same point 𝑥𝐹(𝑆)VI(𝐶,𝐵)EP(𝐹), where 𝑥=𝑃𝐹(𝑆)VI(𝐶,𝐵)EP(𝐹)𝑓(𝑥).

In this paper, motivated and inspired by the above facts, we introduce a new iterative scheme for finding a common element of the set of solutions of the variational inequalities for 𝜇-Lipschitz continuous and relaxed (𝜙,𝜔)-cocoercive mapping, the set of solutions to the variational inclusion for family of 𝛼-inverse strongly monotone mappings, the set of fixed points of a countable family of nonexpansive mappings, and the set of solutions of an equilibrium problem in a real Hilbert space by using the viscosity approximation method. Strong convergence results are derived under suitable conditions in a real Hilbert space.

2. Preliminaries

In this section, we will recall some basic notations and collect some conclusions that will be used in the next section.

Let 𝐻 be a real Hilbert space whose inner product and norm are denoted by , and , respectively. We denote strong convergence of {𝑥𝑛} to 𝑥𝐻 by 𝑥𝑛𝑥 and weak convergence by 𝑥𝑛𝑥. Let 𝐶 be nonempty closed convex subset of 𝐻. Recall that for all 𝑥𝐻 there exists a unique nearest point in 𝐶 to 𝑥 denoted 𝑃𝐶𝑥; that is, 𝑥𝑃𝐶𝑥𝑥𝑦,forall𝑦𝐶. The mapping 𝑃𝐶 is nonexpansive; that is, 𝑃𝐶𝑥𝑃𝐶𝑦𝑥𝑦,forall𝑥,𝑦𝐻. The mapping 𝑃𝐶 is firmly nonexpansive; that is, 𝑃𝐶𝑥𝑃𝐶𝑦2𝑃𝐶𝑥𝑃𝐶𝑦,𝑥𝑦,forall𝑥,𝑦𝐻. It is well known that̃𝑥VI(𝐶,𝐵)̃𝑥=𝑃𝐶(̃𝑥𝜆𝐵̃𝑥),𝜆>0.(2.1) A set-valued mapping 𝑀𝐻2𝐻 is called monotone if, for all 𝑥,𝑦𝐻, 𝑓𝑀𝑥 and 𝑔𝑀𝑦 imply 𝑥𝑦,𝑓𝑔0. A monotone mapping 𝑀𝐻2𝐻 is called maximal, if its graph of any Graph (𝑀)={(𝑥,𝑓)𝐻×𝐻|𝑓𝑀(𝑥)} of 𝑀 is not properly contained in the graph of any other monotone mapping. It is well known that a monotone mapping 𝑀 is maximal if and only if for all (𝑥,𝑓)𝐻×𝐻,𝑥𝑦,𝑓𝑔0,forall(𝑦,𝑔) Graph (𝑀) (the graph of mapping 𝑀) implies that 𝑓𝑀𝑥.

Definition 2.1. Let 𝑀𝐻2𝐻 be a multivalued maximal monotone mapping; then the set-valued mapping 𝐽𝑀,𝜆𝐻𝐻 defined by 𝐽𝑀,𝜆(̃𝑥)=(𝐼+𝜆𝑀)1(̃𝑥),̃𝑥𝐻,(2.2) is called the resolvent operator associated with 𝑀, where 𝜆 is any positive number and 𝐼 is the identity mapping.

Lemma 2.2 (see [16]). Let 𝑀𝐻2𝐻 be a maximal monotone mapping and let 𝐵𝐻𝐻 be a Lipschitz continuous mapping. Then the mapping 𝑀+𝐵𝐻2𝐻 is a maximal monotone mapping.

Lemma 2.3 (see [16, 17]). (1)The resolvent operator J𝑀,𝜆 is single valued and nonexpansive for all 𝜆>0; that is,𝐽𝑀,𝜆(𝑥)𝐽𝑀,𝜆(𝑦)𝑥𝑦,𝑥,𝑦𝐻,𝜆>0.(2.3)(2)The resolvent operator 𝐽𝑀,𝜆 is 1-inverse-strongly monotone; that is,𝐽𝑀,𝜆(𝑥)𝐽𝑀,𝜆(𝑦)2𝑥𝑦,𝐽𝑀,𝜆(𝑥)𝐽𝑀,𝜆(𝑦),𝑥,𝑦𝐻.(2.4)

Lemma 2.4 (see [17]). (1)Let ̃𝑥𝐻 is a solution of problem (1.3) if and only if ̃𝑥=𝐽𝑀,𝜆(𝐼𝜆𝐵) for all 𝜆>0; that is,𝐼𝐽(𝐵,𝑀)=𝐹𝑀,𝜆(𝐼𝜆𝐵),𝜆>0.(2.5)(2)If 𝜆[0,2𝛽], then 𝐼(𝐵,𝑀) is a closed convex subset in 𝐻.

Lemma 2.5 (see [18]). Each Hilbert space 𝐻 satisfies Opial’s condition; that is, for any sequence {𝑥𝑛}𝐻 with 𝑥𝑛𝑥, the inequality liminf𝑛𝑥𝑛𝑥<liminf𝑛𝑥𝑛𝑦(2.6) holds for each 𝑦𝐻 with 𝑦𝑥.

Lemma 2.6 (see [19]). Let {𝑥𝑛} and {𝑧𝑛} be bounded sequences in a Banach space 𝐸, and let {𝛽𝑛} be a sequence in [0,1] with 0<liminf𝑛𝛽𝑛limsup𝑛𝛽𝑛<1. Suppose 𝑥𝑛+1=(1𝛽𝑛)𝑧𝑛+𝛽𝑛𝑥𝑛 for all integers 𝑛1 and limsup𝑛(𝑧𝑛+1𝑧𝑛𝑥𝑛+1𝑥𝑛)0. Then, lim𝑛𝑧𝑛𝑥𝑛=0.

Lemma 2.7 (see [20]). Assume {𝑎𝑛} is a sequence of nonnegative real numbers such that 𝑎𝑛+11𝑏𝑛𝑎𝑛+𝛿𝑛,𝑛0,(2.7) where {𝑏𝑛} is a sequence in (0,1) and {𝛿𝑛} is a sequence in such that(1)𝑛=1𝑏𝑛=,(2)limsup𝑛𝛿𝑛/𝑏𝑛0 or 𝑛=1|𝛿𝑛|<. Then lim𝑛𝑎𝑛=0.

Lemma 2.8. Let 𝐻 be a real Hilbert space. Then hold the following identities:(i)𝑡𝑥+(1𝑡)𝑦2=𝑡𝑥2+(1𝑡)𝑦2𝑡(1𝑡)𝑥𝑦2,𝑡[0,1],𝑥,𝑦𝐻, (ii)𝑥+𝑦2𝑥2+2𝑦,𝑥+𝑦,𝑥,𝑦𝐻.

Lemma 2.9 (see [21]). Let 𝐶 be a nonempty closed subset of a Banach space, and let {𝑆𝑛} be a sequence of mappings of 𝐶 into itself. Suppose that 𝑛=1sup{𝑆𝑛+1𝑧𝑆𝑛𝑧𝑧𝐶}<. Then, for each 𝑦𝐶, {𝑆𝑛𝑦} converges strongly to some point of 𝐶. Moreover, let 𝑆 be a mapping of 𝐶 into itself defined by 𝑆𝑦=lim𝑛𝑆𝑛𝑦,𝑦𝐶.(2.8) Then lim𝑛sup{𝑆𝑧𝑆𝑛𝑧𝑧𝐶}=0.

For solving the equilibrium problem for a bifunction 𝐹𝐶×𝐶, let us assume that 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,𝑥,𝑦𝐶;(A3) for each 𝑥,𝑦,𝑧𝐶,lim𝑡0𝐹(𝑡𝑧+(1𝑡)𝑥,𝑦)𝐹(𝑥,𝑦);(A4) for each 𝑥𝐶,𝑦𝐹(𝑥,𝑦) is convex and lower semicontinuous.

Lemma 2.10 (see [5]). Let 𝐶 be a nonempty closed convex subset of 𝐻, and let 𝐹 be a bifunction of 𝐶×𝐶 into satisfying (A1)–(A4). Let 𝑟>0 and 𝑥𝐻. Then, there exists 𝑧𝐶 such that 1𝐹(𝑧,𝑦)+𝑟𝑦𝑧,𝑧𝑥0,𝑦𝐶.(2.9)

Lemma 2.11 (see [6]). Assume that 𝐹𝐶×𝐶 satisfies (A1)–(A4). For 𝑟>0 and 𝑥𝐻, define a mapping 𝑇𝑟𝐻𝐶 as follows: 𝑇𝑟1(𝑥)=𝑧𝐶𝐹(𝑧,𝑦)+𝑟,𝑦𝑧,𝑧𝑥0,𝑦𝐶(2.10) for all 𝑥𝐻. Then, the following hold:(i)𝑇𝑟 is single valued;(ii)T𝑟 is firmly nonexpansive; that is, for any 𝑥,𝑦𝐻𝑇𝑟𝑥𝑇𝑟𝑦2𝑇𝑟𝑥𝑇𝑟𝑦,𝑥𝑦;(iii)𝐹(𝑇𝑟)=EP(𝐹);(iv)EP(𝐹) is closed and convex.

Lemma 2.12 (see [22]). Let 𝐻 be a Hilbert space and 𝑀 a maximal monotone on 𝐻. Then, the following holds: 𝐽𝑀,𝑟𝑥𝐽𝑀,𝑠𝑥2𝑟𝑠𝑟𝐽𝑀,𝑟𝑥𝐽𝑀,𝑠𝑥,𝐽𝑀,𝑟𝑥𝑥,𝑠,𝑟>0,𝑥𝐻,(2.11) where 𝐽𝑀,𝑟=(𝐼+𝑟𝑀)1 and 𝐽𝑀,𝑠=(𝐼+𝑠𝑀)1.

3. Main Results

In this section, we will use the viscosity approximation method to prove a strong convergence theorem for finding a common element of the set of fixed points of a countable family of nonexpansive mappings, the set of solutions of the variational inequality problem for relaxed cocoercive and Lipschitz continuous mappings, the set of solutions of system of variational inclusions, and the set of solutions of equilibrium problem in a real Hilbert space.

Theorem 3.1. Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻, and let 𝐵𝐶𝐻 be relaxed (𝜙,𝜔)-cocoercive and 𝜇-Lipschitz continuous with 𝜔>𝜙𝜇2, for some 𝜙,𝜔,𝜇>0. Let 𝒢={𝐺𝑘𝑘=1,2,3,,𝑁} be a finite family of 𝛽-inverse strongly monotone mappings from 𝐶 into 𝐻, and let 𝐹 be a bifunction from 𝐶×𝐶 satisfying (A1)–(A4). Let 𝑓𝐶𝐶 be a contraction with coefficient 𝜓(0𝜓<1), and let {𝑆𝑛} be a sequence of nonexpansive mappings of 𝐶 into itself such that Ω𝑛=1𝐹𝑆𝑛𝑁𝑘=1𝐼𝐺𝑘,𝑀𝑘VI(𝐶,𝐵)EP(𝐹).(3.1) Let the sequences {𝑥𝑛} and {𝑦𝑛} be generated by 𝑥1𝑦=𝑥𝐶chosenarbitrarily,𝑛=𝐽𝑀𝑁,𝜆𝑁,𝑛𝐼𝜆𝑁,𝑛𝐺𝑛𝐽𝑀2,𝜆2,𝑛𝐼𝜆2,𝑛𝐺2𝐽𝑀1,𝜆1,𝑛𝐼𝜆1,𝑛𝐺1𝑇𝑟𝑛𝑥𝑛,𝑥𝑛+1=𝛼𝑛𝑓𝑥𝑛+𝛽𝑛𝑥𝑛+𝛾𝑛𝑆𝑛𝑃𝐶𝑦𝑛𝜉𝑛𝐵𝑦𝑛,𝑛1,(3.2) where {𝛼𝑛},{𝛽𝑛}, {𝛾𝑛}(0,1) and {𝜉𝑛},{𝑟𝑛}(0,) satisfy the following conditions:(C1)𝛼𝑛+𝛽𝑛+𝛾𝑛=1,(C2)lim𝑛𝛼𝑛=0,𝑛=1𝛼𝑛=, (C3)0<liminf𝑛𝛽𝑛limsup𝑛𝛽𝑛<1, (C4){𝜉𝑛}[𝑎,𝑏] for some 𝑎, 𝑏 with 0𝑎𝑏2(𝜔𝜙𝜇2)/𝜇2 and lim𝑛|𝜉𝑛+1𝜉𝑛|=0,(C5){𝜆𝑘,𝑛}𝑁𝑘=1[𝑐,𝑑](0,2𝛽) and lim𝑛|𝜆𝑘,𝑛+1𝜆𝑘,𝑛|=0, for each 𝑘{1,2,,𝑁},(C6)liminf𝑛𝑟𝑛>0 and lim𝑛|𝑟𝑛+1𝑟𝑛|=0.Suppose that 𝑛=1sup{𝑆𝑛+1𝑧𝑆𝑛𝑧𝑧𝐾}< for any bounded subset 𝐾 of 𝐶. Let 𝑆 be a mapping of 𝐶 into itself defined by 𝑆𝑦=lim𝑛𝑆𝑛𝑦forall𝑦𝐶 and suppose that 𝐹(𝑆)=𝑛=1𝐹(𝑆𝑛). Then, the sequences {𝑥𝑛} and {𝑦𝑛} converge strongly to the same point 𝑥Ω, where 𝑥=𝑃Ω𝑓(𝑥).

Proof. First, we prove that the mapping 𝑃Ω𝑓𝐻𝐶 has a unique fixed point.
In fact, since 𝑓𝐶𝐶 is a contraction with 𝜓[0,1) and 𝑃Ω𝑓𝐻Ω is also a contraction, we obtain 𝑃Ω𝑓(𝑥)𝑃Ω𝑓(𝑦)𝑓(𝑥)𝑓(𝑦)𝜓𝑥𝑦,𝑥,𝑦𝐶.(3.3) Therefore, there exists a unique element 𝑥𝐶 such that 𝑥=𝑃Ω𝑓(𝑥), where Ω𝑛=1𝐹𝑆𝑛𝑁𝑘=1𝐼𝐺𝑘,𝑀𝑘VI(𝐶,𝐵)EP(𝐹).(3.4) Now, we prove that (𝐼𝜉𝑛𝐵) is nonexpansive.
Indeed, for any 𝑥,𝑦𝐶, since 𝐵𝐶𝐻 is a 𝜇-Lipschitz continuous and relaxed (𝜙,𝜔)-cocoercive mappings with 𝜔>𝜙𝜇2 and 𝜉𝑛2(𝜔𝜙𝜇2)/𝜇2, we obtain 𝐼𝜉𝑛𝐵𝑥𝐼𝜉𝑛𝐵𝑦2=(𝑥𝑦)𝜉𝑛(𝐵𝑥𝐵𝑦)2=𝑥𝑦22𝜉𝑛𝑥𝑦,𝐵𝑥𝐵𝑦+𝜉2𝑛𝐵𝑥𝐵𝑦2𝑥𝑦22𝜉𝑛𝜙𝐵𝑥𝐵𝑦2+𝜔𝑥𝑦2+𝜉2𝑛𝐵𝑥𝐵𝑦2𝑥𝑦2+2𝜉𝑛𝜙𝜇2𝑥𝑦22𝜉𝑛𝜔𝑥𝑦2+𝜉2𝑛𝜇2𝑥𝑦2=1+2𝜉𝑛𝜙𝜇22𝜉𝑛𝜔+𝜉2𝑛𝜇2𝑥𝑦2=1𝜉𝑛𝜇22𝜔𝜙𝜇2𝜇2𝜉𝑛𝑥𝑦21𝜉𝑛𝜇22𝜔𝜙𝜇2𝜇2𝑏𝑥𝑦2.(3.5) Setting 𝜇𝜁=222𝜔𝜙𝜇2𝜇2𝑏>0,(3.6) thus, 𝐼𝜉𝑛𝐵𝑥𝐼𝜉𝑛𝐵𝑦212𝜉𝑛𝜁𝑥𝑦21𝜉𝑛𝜁2𝑥𝑦2,(3.7) which implies that 𝐼𝜉𝑛𝐵𝑥𝐼𝜉𝑛𝐵𝑦1𝜉𝑛𝜁𝑥𝑦𝑥𝑦.(3.8) Hence (𝐼𝜉𝑛𝐵) is nonexpansive.
We divide the proof of Theorem 3.1 into five steps.
Step 1. We show that the sequence {𝑥𝑛} is bounded.
Now, let ̃𝑥Ω and if {𝑇𝑟𝑛} is a sequence of mappings defined as in Lemma 2.11, then ̃𝑥=𝑃𝐶(̃𝑥𝜆𝑛𝐵̃𝑥)=𝑇𝑟𝑛̃𝑥, and let 𝑢𝑛=𝑇𝑟𝑛𝑥𝑛. So, we have 𝑢𝑛=𝑇̃𝑥𝑟𝑛𝑥𝑛𝑇𝑟𝑛𝑥̃𝑥𝑛̃𝑥.(3.9) For 𝑘{1,2,,𝑁} and for any positive integer number 𝑛, we define the operator Υ𝑘𝑛𝐶𝐻 as follows: Υ0𝑛Υ𝑥=𝑥,𝑘𝑛𝑥=𝐽𝑀𝑘,𝜆𝑘,𝑛𝐼𝜆𝑘,𝑛𝐺𝑘𝐽𝑀2,𝜆2,𝑛𝐼𝜆2,𝑛𝐺2𝐽𝑀1,𝜆1,𝑛𝐼𝜆1,𝑛𝐺1𝑥,(3.10) for all 𝑛, we get 𝑦𝑛=Υ𝑁𝑛𝑢𝑛. On the other hand, since 𝐺𝑘𝐶𝐻 is 𝛽-inverse strongly monotone and 𝜆𝑘,𝑛[𝑐,𝑑](0,2𝛽), then 𝐽𝑀𝑘,𝜆𝑘,𝑛(𝐼𝜆𝑘,𝑛𝐺𝑘) is nonexpansive. Thus Υ𝑘𝑛 is nonexpansive. From Lemma 2.4(1), we have ̃𝑥=Υ𝑁𝑛̃𝑥. It follows that 𝑦𝑛=Υ̃𝑥𝑁𝑛𝑢𝑛Υ𝑁𝑛𝑢̃𝑥𝑛𝑥̃𝑥𝑛.̃𝑥(3.11) Setting 𝑣𝑛=𝑃𝐶(𝑦𝑛𝜉𝑛𝐵𝑦𝑛) and 𝐼𝜉𝑛𝐵 is a nonexpansive mapping, we obtain 𝑣𝑛=𝑃̃𝑥𝐶𝑦𝑛𝜉𝑛𝐵𝑦n𝑃𝐶̃𝑥𝜉𝑛𝑦𝐵̃𝑥𝑛𝜉𝑛𝐵𝑦𝑛̃𝑥𝜉𝑛=𝐵̃𝑥𝐼𝜉𝑛𝐵𝑦𝑛𝐼𝜉𝑛𝐵𝑦̃𝑥𝑛𝑥̃𝑥𝑛.̃𝑥(3.12) From (3.2) and (3.12), we deduce that 𝑥𝑛+1=𝛼̃𝑥𝑛𝑓𝑥𝑛+𝛽𝑛𝑥𝑛+𝛾𝑛𝑆𝑛𝑣𝑛̃𝑥𝛼𝑛𝑓𝑥𝑛̃𝑥+𝛽𝑛𝑥𝑛̃𝑥+𝛾𝑛𝑣𝑛̃𝑥𝛼𝑛𝑓𝑥𝑛𝑓(̃𝑥)+𝛼𝑛𝑓(̃𝑥)̃𝑥+𝛽𝑛𝑥𝑛̃𝑥+𝛾𝑛𝑥𝑛̃𝑥𝛼𝑛𝜓𝑥𝑛̃𝑥+𝛼𝑛(𝑓̃𝑥)̃𝑥+1𝛼𝑛𝑥𝑛̃𝑥1𝛼𝑛𝑥(1𝜓)𝑛̃𝑥+𝛼𝑛=𝑓(̃𝑥)̃𝑥1𝛼𝑛(𝑥1𝜓)𝑛̃𝑥+𝛼𝑛(1𝜓)𝑓(̃𝑥)̃𝑥𝑥(1𝜓)max𝑛,̃𝑥𝑓(̃𝑥)̃𝑥.1𝜓(3.13) It follows from induction that 𝑥𝑛𝑥̃𝑥max1,(̃𝑥𝑓̃𝑥)̃𝑥1𝜓,𝑛1.(3.14) Therefore, {𝑥𝑛} is bounded and hence so are {𝑣𝑛}, {𝑦𝑛}, {𝑢𝑛}, {𝐵𝑦𝑛}, and {𝑆𝑛𝑣𝑛}.
Step 2. We claim that lim𝑛𝑥𝑛+1𝑥𝑛=0.
By the definition of 𝑇𝑟, 𝑢𝑛=𝑇𝑟𝑛𝑥𝑛 and 𝑢𝑛+1=𝑇𝑟𝑛+1𝑥𝑛+1, we get 𝐹𝑢𝑛+1,𝑦𝑟𝑛𝑦𝑢𝑛,𝑢𝑛𝑥𝑛𝐹𝑢0,𝑦𝐻,(3.15)𝑛+1+1,𝑦𝑟𝑛+1𝑦𝑢𝑛+1,𝑢𝑛+1𝑥𝑛+10,𝑦𝐻.(3.16) Taking 𝑦=𝑢𝑛+1 in (3.15) and 𝑦=𝑢𝑛 in (3.16), we have 𝐹𝑢𝑛,𝑢𝑛+1+1𝑟𝑛𝑢𝑛+1𝑢𝑛,𝑢𝑛𝑥𝑛0,(3.17) and hence 𝐹𝑢𝑛+1,𝑢𝑛+1𝑟𝑛+1𝑢𝑛𝑢𝑛+1,𝑢𝑛+1𝑥𝑛+10.(3.18) So, from (A2) we have 𝑢𝑛+1𝑢𝑛,𝑢𝑛𝑥𝑛𝑟𝑛𝑢𝑛+1𝑥𝑛+1𝑟𝑛+10,(3.19) and hence 𝑢𝑛+1𝑢𝑛,𝑢𝑛𝑢𝑛+1+𝑢𝑛+1𝑥𝑛𝑟𝑛𝑟𝑛+1𝑢𝑛+1𝑥𝑛+10.(3.20) Without loss of generality, let us assume that there exists a real number 𝑐 such that 𝑟𝑛>𝑐>0 for all 𝑛. Then, we have 𝑢𝑛+1𝑢𝑛2𝑢𝑛+1𝑢𝑛,𝑥𝑛+1𝑥𝑛+𝑟1𝑛𝑟𝑛+1𝑢𝑛+1𝑥𝑛+1𝑢𝑛+1𝑢𝑛𝑥𝑛+1𝑥𝑛+||||𝑟1𝑛𝑟𝑛+1||||𝑢𝑛+1𝑥𝑛+1,(3.21) and hence 𝑢𝑛+1𝑢𝑛𝑥𝑛+1𝑥𝑛+1𝑟𝑛+1||𝑟𝑛+1𝑟𝑛||𝑢𝑛+1𝑥𝑛+1𝑥𝑛+1𝑥𝑛+𝑀1𝑐||𝑟𝑛+1𝑟𝑛||,(3.22) where 𝑀1=sup{𝑢𝑛𝑥𝑛𝑛}.
Notice from Lemma 2.12 that 𝑦𝑛+1𝑦𝑛=Υ𝑁𝑛+1𝑢𝑛+1Υ𝑁𝑛𝑢𝑛𝑢𝑛+1𝜆𝑘,𝑛+1𝐺𝑘Υ𝑘𝑛+1𝑢𝑛+1𝑢𝑛𝜆𝑘,𝑛𝐺𝑘Υ𝑘𝑛𝑢𝑛+𝐽𝑀𝑘,𝜆𝑘,𝑛+1𝑢𝑛𝜆𝑘,𝑛𝐺𝑘Υ𝑘𝑛𝑢𝑛𝐽𝑀𝑘,𝜆𝑘,𝑛𝑢𝑛𝜆𝑘,𝑛𝐺𝑘Υ𝑘𝑛𝑢𝑛𝑢𝑛+1𝑢𝑛+||𝜆𝑘,𝑛+1𝜆𝑘,𝑛||𝐺𝑘Υ𝑘𝑛𝑢𝑛+||𝜆𝑘,𝑛+1𝜆𝑘,𝑛||𝜆𝑘,𝑛+1𝐽𝑀𝑘,𝜆𝑘,𝑛+1𝑢𝑛𝜆𝑘,𝑛𝐺𝑘Υ𝑘𝑛𝑢𝑛𝑢𝑛𝜆𝑘,𝑛𝐺𝑘Υ𝑘𝑛𝑢𝑛𝑢𝑛+1𝑢𝑛+2𝑀2||𝜆𝑘,𝑛+1𝜆𝑘,𝑛||𝑥𝑛+1𝑥𝑛+𝑀1𝑐||𝑟𝑛+1𝑟𝑛||+2𝑀2||𝜆𝑘,𝑛+1𝜆𝑘,𝑛||,(3.23) where 𝑀2 is an appropriate constant such that 𝑀2=maxsup𝑛1𝐺𝑘Υ𝑘𝑛𝑢𝑛,sup𝑛1𝐽𝑀𝑘,𝜆𝑘,𝑛+1𝑢𝑛𝜆𝑘,𝑛𝐺𝑘Υ𝑘𝑛𝑢𝑛𝑢𝑛𝜆𝑘,𝑛𝐺𝑘Υ𝑘𝑛𝑢𝑛𝐽𝑀𝑘,𝜆𝑘,𝑛+1.(3.24) Since 𝐼𝜉𝑛𝐵 is nonexpansive mappings, we have the following estimates: 𝑣𝑛+1𝑣𝑛𝑃𝐶𝑦𝑛+1𝜉𝑛+1𝐵𝑦𝑛+1𝑃𝐶𝑦𝑛𝜉𝑛𝐵𝑦𝑛𝑦𝑛+1𝜉𝑛+1𝐵𝑦𝑛+1𝑦𝑛𝜉𝑛𝐵𝑦𝑛=𝑦𝑛+1𝜉𝑛+1𝐵𝑦𝑛+1𝑦𝑛𝜉𝑛+1𝐵𝑦𝑛+𝜉𝑛𝜉𝑛+1𝐵𝑦𝑛𝑦𝑛+1𝜉𝑛+1𝐵𝑦𝑛+1𝑦𝑛𝜉𝑛+1𝐵𝑦𝑛+||𝜉𝑛𝜉𝑛+1||𝐵𝑦𝑛=𝐼𝜉𝑛+1𝐵𝑦𝑛+1𝐼𝜉𝑛+1𝐵𝑦𝑛+||𝜉𝑛𝜉𝑛+1||𝐵𝑦𝑛𝑦𝑛+1𝑦𝑛+||𝜉𝑛𝜉𝑛+1||𝐵𝑦𝑛.(3.25) Substituting (3.23) into (3.25), we obtain 𝑣𝑛+1𝑣𝑛𝑥𝑛+1𝑥𝑛+𝑀1𝑐||𝑟𝑛+1𝑟𝑛||+2𝑀2||𝜆𝑘,𝑛+1𝜆𝑘,𝑛||+||𝜉𝑛𝜉𝑛+1||𝐵𝑦𝑛.(3.26) Indeed, define 𝑥𝑛+1=(1𝛽𝑛)𝑧𝑛+𝛽𝑛𝑥𝑛 for all 𝑛. It follows that 𝑧𝑛=𝑥𝑛+1𝛽𝑛𝑥𝑛1𝛽𝑛=𝛼𝑛𝑓𝑥𝑛+𝛾𝑛𝑆𝑛𝑣𝑛1𝛽𝑛.(3.27) Thus, we have 𝑧𝑛+1𝑧𝑛=𝛼𝑛+1𝑓𝑥𝑛+1+𝛾𝑛+1𝑆𝑛+1𝑣𝑛+11𝛽𝑛+1𝛼𝑛𝑓𝑥𝑛+𝛾𝑛𝑆𝑛𝑣𝑛1𝛽𝑛=𝛼𝑛+11𝛽𝑛+1𝑓𝑥𝑛+1𝑥𝑓𝑛+𝛾𝑛+11𝛽𝑛+1𝑆𝑛+1𝑣𝑛+1𝑆𝑛𝑣𝑛+𝛼𝑛+11𝛽𝑛+1𝛼𝑛1𝛽𝑛𝑓𝑥𝑛+𝛾𝑛+11𝛽𝑛+1𝛾𝑛1𝛽𝑛𝑆𝑛𝑣𝑛𝛼𝑛+11𝛽𝑛+1𝑓𝑥𝑛+1𝑥𝑓𝑛+𝛾𝑛+11𝛽𝑛+1𝑆𝑛+1𝑣𝑛+1𝑆𝑛𝑣𝑛+||||𝛼𝑛+11𝛽𝑛+1𝛼𝑛1𝛽𝑛||||𝑓𝑥𝑛𝑆𝑛𝑣𝑛𝜓𝛼𝑛+11𝛽𝑛+1𝑥𝑛+1𝑥𝑛+𝛾𝑛+11𝛽𝑛+1𝑆𝑛+1𝑣𝑛+1𝑆𝑛𝑣𝑛+||||𝛼𝑛+11𝛽𝑛+1𝛼𝑛1𝛽𝑛||||𝑓𝑥𝑛𝑆𝑛𝑣𝑛.(3.28) Now, compute 𝑆𝑛+1𝑣𝑛+1𝑆𝑛𝑣𝑛𝑆𝑛+1𝑣𝑛+1𝑆𝑛+1𝑣𝑛+𝑆𝑛+1𝑣𝑛𝑆𝑛𝑣𝑛𝑣𝑛+1𝑣𝑛+𝑆𝑛+1𝑣𝑛𝑆𝑛𝑣𝑛x𝑛+1𝑥𝑛+𝑀1𝑐||𝑟𝑛+1𝑟𝑛||+||𝜉𝑛𝜉𝑛+1||𝐵𝑦𝑛+2𝑀2||𝜆𝑘,𝑛+1𝜆𝑘,𝑛||+𝑆𝑛+1𝑣𝑛𝑆𝑛𝑣𝑛.(3.29) Combining (3.28) and (3.29), we have 𝑧𝑛+1𝑧𝑛𝜓𝛼𝑛+11𝛽𝑛+1𝑥𝑛+1𝑥𝑛+𝛾𝑛+11𝛽𝑛+1𝑥𝑛+1𝑥𝑛+𝑀1𝑐||𝑟𝑛+1𝑟𝑛||+||𝜉𝑛𝜉𝑛+1||𝐵𝑦𝑛+2𝑀2||𝜆𝑘,𝑛+1𝜆𝑘,𝑛||+𝑆𝑛+1𝑣𝑛𝑆𝑛𝑣𝑛+||||𝛼𝑛+11𝛽𝑛+1𝛼𝑛1𝛽𝑛||||𝑓𝑥𝑛𝑆𝑛𝑣𝑛𝑥𝑛+1𝑥𝑛+𝛾𝑛+11𝛽𝑛+1𝑀1𝑐||𝑟𝑛+1𝑟𝑛||+||𝜉𝑛𝜉𝑛+1||𝐵𝑦𝑛+2𝑀2||𝜆𝑘,𝑛+1𝜆𝑘,𝑛||+𝛾𝑛+11𝛽𝑛+1𝑆𝑛+1𝑣𝑛𝑆𝑛𝑣𝑛+||||𝛼𝑛+11𝛽𝑛+1𝛼𝑛1𝛽𝑛||||𝑓𝑥𝑛𝑆𝑛𝑣𝑛.(3.30) It follows that 𝑧𝑛+1𝑧𝑛𝑥𝑛+1𝑥𝑛𝛾𝑛+11𝛽𝑛+1𝑀1𝑐||𝑟𝑛+1𝑟𝑛||+||𝜉𝑛𝜉𝑛+1||𝐵𝑦𝑛+2𝑀2||𝜆𝑘,𝑛+1𝜆𝑘,𝑛||+𝛾𝑛+11𝛽𝑛+1𝑆𝑛+1𝑣𝑛𝑆𝑛𝑣𝑛+||||𝛼𝑛+11𝛽𝑛+1𝛼𝑛1𝛽𝑛||||𝑓𝑥𝑛𝑆𝑛𝑣𝑛𝛾𝑛+11𝛽𝑛+1𝑀1𝑐||𝑟𝑛+1𝑟𝑛||+||𝜉𝑛𝜉𝑛+1||𝐵𝑦𝑛+2𝑀2||𝜆𝑘,𝑛+1𝜆𝑘,𝑛||+𝛾𝑛+11𝛽𝑛+1𝑆sup𝑛+1𝑧𝑆𝑛𝑧𝑣𝑧𝑛+||||𝛼𝑛+11𝛽𝑛+1𝛼𝑛1𝛽𝑛||||𝑓𝑥𝑛𝑆𝑛𝑣𝑛.(3.31) This together with conditions (C1)–(C6) and lim𝑛sup{𝑆𝑛+1𝑧𝑆𝑛𝑧𝑧{𝑣𝑛}}=0 implies that limsup𝑛𝑧𝑛+1𝑧𝑛𝑥𝑛+1𝑥𝑛0.(3.32) Hence, by Lemma 2.6, we obtain 𝑧𝑛𝑥𝑛0 as 𝑛. It then follows that lim𝑛𝑥𝑛+1𝑥𝑛=lim𝑛1𝛽𝑛𝑧𝑛𝑥𝑛=0.(3.33) By (3.26), we also have lim𝑛𝑣𝑛+1𝑣𝑛=0.(3.34)
Step 3. We claim that lim𝑛𝑆𝑣𝑛𝑣𝑛=0.
Since {𝐺𝑘𝑘=1,2,3,,𝑁} is 𝛽-inverse strongly monotone mappings, by the choice of {𝜆𝑘,𝑛} for given ̃𝑥Ω and 𝑘{0,1,2,,𝑁1}, we also have Υ𝑛𝑘+1𝑢𝑛̃𝑥2=𝐽𝑀𝑘+1,𝜆𝑘+1,𝑛𝐼𝜆𝑘+1,𝑛𝐺𝑘+1Υ𝑘𝑛𝑢𝑛𝐽𝑀𝑘+1,𝜆𝑘+1,𝑛𝐼𝜆𝑘+1,𝑛𝐺𝑘+1̃𝑥2𝐼𝜆𝑘+1,𝑛𝐺𝑘+1Υ𝑘𝑛𝑢𝑛𝐼𝜆𝑘+1,𝑛𝐺𝑘+1̃𝑥2=Υ𝑘𝑛𝑢𝑛𝜆𝑘+1,𝑛𝐺𝑘+1Υ𝑘𝑛𝑢𝑛̃𝑥𝜆𝑘+1,𝑛𝐺𝑘+1̃𝑥2=Υ𝑘𝑛𝑢𝑛̃𝑥𝜆𝑘+1,𝑛𝐺𝑘+1Υ𝑘𝑛𝑢𝑛𝐺𝑘+1̃𝑥2=Υ𝑘𝑛𝑢𝑛̃𝑥22𝜆𝑘+1,𝑛Υ𝑘𝑛𝑢𝑛̃𝑥,𝐺𝑘+1Υ𝑘𝑛𝑢𝑛𝐺𝑘+1̃𝑥+𝜆2𝑘+1,𝑛𝐺𝑘+1Υ𝑘𝑛𝑢𝑛𝐺𝑘+1̃𝑥2Υ𝑘𝑛𝑢𝑛̃𝑥22𝜆𝑘+1,𝑛𝛽𝐺𝑘+1Υ𝑘𝑛𝑢𝑛𝐺𝑘+1̃𝑥+𝜆2𝑘+1,𝑛𝐺𝑘+1Υ𝑘𝑛𝑢𝑛𝐺𝑘+1̃𝑥2𝑢𝑛̃𝑥22𝜆𝑘+1,𝑛𝛽𝐺𝑘+1Υ𝑘𝑛𝑢𝑛𝐺𝑘+1̃𝑥+𝜆2𝑘+1,𝑛𝐺𝑘+1Υ𝑘𝑛𝑢𝑛𝐺𝑘+1̃𝑥2𝑥𝑛̃𝑥2+𝜆𝑘+1,𝑛𝜆𝑘+1,𝑛𝐺2𝛽𝑘+1Υ𝑘𝑛𝑢𝑛𝐺𝑘+1̃𝑥2.(3.35) Form (3.13), we have 𝑥𝑛+1̃𝑥2𝛼𝑛𝑓𝑥𝑛̃𝑥2+𝛽𝑛𝑥𝑛̃𝑥2+𝛾𝑛𝑣𝑛̃𝑥2𝛼𝑛𝑓𝑥𝑛̃𝑥2+𝛽𝑛𝑥𝑛̃𝑥2+𝛾𝑛𝑦𝑛̃𝑥2=𝛼𝑛𝑓𝑥𝑛̃𝑥2+𝛽𝑛𝑥𝑛̃𝑥2+𝛾𝑛Υ𝑁𝑛𝑢𝑛̃𝑥2𝛼𝑛𝑓𝑥𝑛̃𝑥2+𝛽𝑛𝑥𝑛̃𝑥2+𝛾𝑛Υ𝑛𝑘+1𝑢𝑛̃𝑥2𝛼𝑛𝑓𝑥𝑛̃𝑥2+𝛽𝑛𝑥𝑛̃𝑥2+𝛾𝑛𝑥𝑛̃𝑥2+𝜆𝑘+1,𝑛𝜆𝑘+1,𝑛𝐺2𝛽𝑘+1Υ𝑘𝑛𝑢𝑛𝐺𝑘+1̃𝑥2𝛼𝑛𝑓𝑥𝑛̃𝑥2+𝑥𝑛̃𝑥2+𝛾𝑛𝜆𝑘+1,𝑛𝜆𝑘+1,𝑛𝐺2𝛽𝑘+1Υ𝑘𝑛𝑢𝑛𝐺𝑘+1̃𝑥2.(3.36) It follows that 𝛾𝑛𝜆𝑘+1,𝑛2𝛽𝜆𝑘+1,𝑛𝐺𝑘+1Υ𝑘𝑛𝑢𝑛𝐺𝑘+1̃𝑥2𝛾𝑛𝐺𝑐(2𝛽𝑑)𝑘+1Υ𝑘𝑛𝑢𝑛𝐺𝑘+1̃𝑥2𝑥𝑛𝑥𝑛+1𝑥𝑛+𝑥̃𝑥𝑛+1̃𝑥+𝛼𝑛𝑓𝑥𝑛̃𝑥2.(3.37) By condition (C2), (3.33), and liminf𝑛𝛾𝑛>0, we obtain lim𝑛𝐺𝑘+1Υ𝑘𝑛𝑢𝑛𝐺𝑘+1̃𝑥=0.(3.38) From Lemma 2.3(2) and as 𝐼𝜆𝑘+1,𝑛𝐺𝑘+1 is nonexpansive, we have Υ𝑛𝑘+1𝑢𝑛̃𝑥2=𝐽𝑀𝑘+1,𝜆𝑘+1,𝑛𝐼𝜆𝑘+1,𝑛𝐺𝑘+1Υ𝑘𝑛𝑢𝑛𝐽𝑀𝑘+1,𝜆𝑘+1,𝑛𝐼𝜆𝑘+1,𝑛𝐺𝑘+1̃𝑥2𝐼𝜆𝑘+1,𝑛𝐺𝑘+1Υ𝑘𝑛𝑢𝑛𝐼𝜆𝑘+1,𝑛𝐺𝑘+1̃𝑥,Υ𝑛𝑘+1𝑢𝑛=1̃𝑥2𝐼𝜆𝑘+1,𝑛𝐺𝑘+1Υ𝑘𝑛𝑢𝑛𝐼𝜆𝑘+1,𝑛𝐺𝑘+1̃𝑥2+Υ𝑛𝑘+1𝑢𝑛̃𝑥2𝐼𝜆𝑘+1,𝑛𝐺𝑘+1Υ𝑘𝑛𝑢𝑛𝐼𝜆𝑘+1,𝑛𝐺𝑘+1Υ̃𝑥𝑛𝑘+1𝑢𝑛̃𝑥212Υ𝑘𝑛𝑢𝑛̃𝑥2+Υ𝑛𝑘+1𝑢𝑛̃𝑥2Υ𝑘𝑛𝑢𝑛Υ𝑛𝑘+1𝑢𝑛𝜆𝑘+1,𝑛𝐺𝑘+1Υ𝑘𝑛𝑢𝑛𝐺𝑘+1̃𝑥212Υ𝑘𝑛𝑢𝑛̃𝑥2+Υ𝑛𝑘+1𝑢𝑛̃𝑥2Υ𝑘𝑛𝑢𝑛Υ𝑛𝑘+1𝑢𝑛2𝜆2𝑘+1,𝑛𝐺𝑘+1Υ𝑘𝑛𝑢𝑛𝐺𝑘+1̃𝑥2+2𝜆𝑘+1,𝑛Υ𝑘𝑛𝑢𝑛Υ𝑛𝑘+1𝑢𝑛,𝐺𝑘+1Υ𝑘𝑛𝑢𝑛𝐺𝑘+1,̃𝑥(3.39) which yields that Υ𝑛𝑘+1𝑢𝑛̃𝑥2Υ𝑘𝑛𝑢𝑛̃𝑥2Υ𝑘𝑛𝑢𝑛Υ𝑛𝑘+1𝑢𝑛2+2𝜆𝑘+1,𝑛Υ𝑘𝑛𝑢𝑛Υ𝑛𝑘+1𝑢𝑛𝐺𝑘+1Υ𝑘𝑛𝑢𝑛𝐺𝑘+1𝑢̃𝑥𝑛̃𝑥2Υ𝑘𝑛𝑢𝑛Υ𝑛𝑘+1𝑢𝑛2+2𝜆𝑘+1,𝑛Υ𝑘𝑛𝑢𝑛Υ𝑛𝑘+1𝑢𝑛𝐺𝑘+1Υ𝑘𝑛𝑢𝑛𝐺𝑘+1𝑥̃𝑥𝑛̃𝑥2Υ𝑘𝑛𝑢𝑛Υ𝑛𝑘+1𝑢𝑛2+2𝜆𝑘+1,𝑛Υ𝑘𝑛𝑢𝑛Υ𝑛𝑘+1𝑢𝑛𝐺𝑘+1Υ𝑘𝑛𝑢𝑛𝐺𝑘+1.̃𝑥(3.40) Substituting (3.40) into (3.36), we obtain 𝑥𝑛+1̃𝑥2𝛼𝑛𝑓𝑥𝑛̃𝑥2+𝛽𝑛𝑥𝑛̃𝑥2+𝛾𝑛Υ𝑛𝑘+1𝑢𝑛̃𝑥2𝛼𝑛𝑓𝑥𝑛̃𝑥2+𝛽𝑛𝑥𝑛̃𝑥2+𝛾𝑛𝑥𝑛̃𝑥2Υ𝑘𝑛𝑢𝑛Υ𝑛𝑘+1𝑢𝑛2+2𝜆𝑘+1,𝑛Υ𝑘𝑛𝑢𝑛Υ𝑛𝑘+1𝑢𝑛𝐺𝑘+1Υ𝑘𝑛𝑢𝑛𝐺𝑘+1̃𝑥𝛼𝑛𝑓𝑥𝑛̃𝑥2+𝑥𝑛̃𝑥2𝛾𝑛Υ𝑘𝑛𝑢𝑛Υ𝑛𝑘+1𝑢𝑛2+2𝜆𝑘+1,𝑛𝛾𝑛Υ𝑘𝑛𝑢𝑛Υ𝑛𝑘+1𝑢𝑛𝐺𝑘+1Υ𝑘𝑛𝑢𝑛𝐺𝑘+1.̃𝑥(3.41) It follows that 𝛾𝑛Υ𝑘𝑛𝑢𝑛Υ𝑛𝑘+1𝑢𝑛2𝑥𝑛𝑥𝑛+1𝑥𝑛+𝑥̃𝑥𝑛+1̃𝑥+𝛼𝑛𝑓𝑥𝑛̃𝑥2+2𝜆𝑘+1,𝑛𝛾𝑛Υ𝑘𝑛𝑢𝑛Υ𝑛𝑘+1𝑢𝑛𝐺𝑘+1Υ𝑘𝑛𝑢𝑛𝐺𝑘+1.̃𝑥(3.42) By condition (C2), (3.33), (3.38), and liminf𝑛𝛾𝑛>0, we obtain lim𝑛Υ𝑘𝑛𝑢𝑛Υ𝑛𝑘+1𝑢𝑛=0.(3.43) For ̃𝑥Ω, we obtain 𝑣𝑛̃𝑥2=𝑃𝐶𝑦𝑛𝜉𝑛𝐵𝑦𝑛𝑃𝐶̃𝑥𝜉𝑛𝐵̃𝑥2𝑦𝑛𝜉𝑛𝐵𝑦𝑛̃𝑥𝜉𝑛𝐵̃𝑥2=𝑦𝑛̃𝑥𝜉𝑛𝐵𝑦𝑛𝐵̃𝑥2𝑦𝑛̃𝑥22𝜉𝑛𝑦𝑛̃𝑥,𝐵𝑦𝑛𝐵̃𝑥+𝜉2𝑛𝐵𝑦𝑛𝐵̃𝑥2𝑦𝑛̃𝑥22𝜉𝑛𝜙𝐵𝑦𝑛𝐵̃𝑥2𝑦+𝜔𝑛̃𝑥2+𝜉2𝑛𝐵𝑦𝑛𝐵̃𝑥2𝑦𝑛̃𝑥2+2𝜉𝑛𝜙𝐵𝑦𝑛𝐵̃𝑥22𝜉𝑛𝜔𝑦𝑛̃𝑥2+𝜉2𝑛𝐵𝑦𝑛𝐵̃𝑥2𝑦𝑛̃𝑥2+2𝜉𝑛𝜙𝐵𝑦𝑛𝐵̃𝑥22𝜉𝑛𝜔𝜇2𝐵𝑦𝑛𝐵̃𝑥2+𝜉2𝑛𝐵𝑦𝑛𝐵̃𝑥2𝑥𝑛̃𝑥2+2𝜉𝑛𝜙+𝜉2𝑛2𝜉𝑛𝜔𝜇2𝐵𝑦𝑛𝐵̃𝑥2.(3.44) On the other hand, we have 𝑥𝑛+1̃𝑥2=𝛼𝑛𝑓𝑥𝑛+𝛽𝑛𝑥𝑛+𝛾𝑛𝑆𝑛𝑣𝑛̃𝑥2𝛼𝑛𝑓𝑥𝑛̃𝑥2+𝛽𝑛𝑥𝑛̃𝑥2+𝛾𝑛𝑆𝑛𝑣𝑛̃𝑥2𝛼𝑛𝑓𝑥𝑛̃𝑥2+𝛽𝑛𝑥𝑛̃𝑥2+𝛾𝑛𝑣𝑛̃𝑥2𝛼𝑛𝑓𝑥𝑛̃𝑥2+𝛽𝑛𝑥𝑛̃𝑥2+𝛾𝑛𝑥𝑛̃𝑥2+2𝜉𝑛𝜙+𝜉2𝑛2𝜉𝑛𝜔𝜇2𝐵𝑦𝑛𝐵̃𝑥2=𝛼𝑛𝑓𝑥𝑛̃𝑥2+𝛽𝑛𝑥𝑛̃𝑥2+𝛾𝑛𝑥𝑛̃𝑥2+𝛾𝑛2𝜉𝑛𝜙+𝜉2𝑛2𝜉𝑛𝜔𝜇2𝐵𝑦𝑛𝐵̃𝑥2𝛼𝑛𝑓𝑥𝑛̃𝑥2+𝑥𝑛̃𝑥2+𝛾𝑛2𝜉𝑛𝜙+𝜉2𝑛2𝜉𝑛𝜔𝜇2𝐵𝑦𝑛𝐵̃𝑥2.(3.45) It follows that 2𝑎𝜔𝜇2𝑏2𝛾2𝑏𝜙𝑛𝐵𝑦𝑛𝐵̃𝑥22𝜉𝑛𝜔𝜇2𝜉2𝑛2𝜉𝑛𝜙𝛾𝑛𝐵𝑦𝑛𝐵̃𝑥2𝑥𝑛̃𝑥2𝑥𝑛+1̃𝑥2+𝛼𝑛𝑓𝑥𝑛̃𝑥2𝑥𝑛𝑥𝑛+1𝑥𝑛+𝑥̃𝑥𝑛+1̃𝑥+𝛼𝑛𝑓𝑥𝑛̃𝑥2.(3.46) It now follows from the last inequality, conditions (C2), (3.33), and liminf𝑛𝛾𝑛>0 that lim𝑛𝐵𝑦𝑛𝐵̃𝑥=0.(3.47) Since 𝑃𝐶 is firmly nonexpansive, we have 𝑣𝑛̃𝑥2=𝑃𝐶𝑦𝑛𝜉𝑛𝐵𝑦𝑛𝑃𝐶̃𝑥𝜉𝑛𝐵̃𝑥2=𝑃𝐶𝐼𝜉𝑛𝐵𝑦𝑛𝑃𝐶𝐼𝜉𝑛𝐵̃𝑥2𝐼𝜉𝑛𝐵𝑦𝑛𝐼𝜉𝑛𝐵̃𝑥,𝑣𝑛=1̃𝑥2𝐼𝛼𝑛𝐵𝑦𝑛𝐼𝜉𝑛𝐵̃𝑥2+𝑣𝑛̃𝑥2𝐼𝜉𝑛𝐵𝑦𝑛𝐼𝜉𝑛𝐵𝑣̃𝑥𝑛̃𝑥212𝑦𝑛̃𝑥2+𝑣𝑛̃𝑥2𝑦𝑛𝑣𝑛𝜉𝑛𝐵𝑦𝑛𝐵̃𝑥212𝑦𝑛̃𝑥2+𝑣𝑛̃𝑥2𝑦𝑛𝑣𝑛2𝜉2𝑛𝐵𝑦𝑛𝐵̃𝑥2+2𝜉𝑛𝑦𝑛𝑣𝑛,𝐵𝑦𝑛,𝐵̃𝑥(3.48) which yields that 𝑣𝑛̃𝑥2𝑦𝑛̃𝑥2𝑦𝑛𝑣𝑛2+2𝜉𝑛𝑦𝑛𝑣𝑛𝐵𝑦𝑛𝑥𝐵̃𝑥𝑛̃𝑥2𝑦𝑛𝑣𝑛2+2𝜉𝑛𝑦𝑛𝑣𝑛𝐵𝑦𝑛.𝐵̃𝑥(3.49) Substituting (3.49) into (3.45), we obtain 𝑥𝑛+1̃𝑥2𝛼𝑛𝑓𝑥𝑛̃𝑥2+𝛽𝑛𝑥𝑛̃𝑥2+𝛾𝑛𝑣𝑛̃𝑥2𝛼𝑛𝑓𝑥𝑛̃𝑥2+𝛽𝑛𝑥𝑛̃𝑥2+𝛾𝑛𝑥𝑛̃𝑥2𝑦𝑛𝑣𝑛2+2𝜉𝑛𝑦𝑛𝑣𝑛𝐵𝑦𝑛𝐵̃𝑥=𝛼𝑛𝑓𝑥𝑛̃𝑥2+𝛽𝑛𝑥𝑛̃𝑥2+𝛾𝑛𝑥𝑛̃𝑥2𝛾𝑛𝑦𝑛𝑣𝑛2+2𝛾𝑛𝜉𝑛𝑦𝑛𝑣𝑛