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Journal of Applied Mathematics
Volume 2012, Article ID 174318, 20 pages
http://dx.doi.org/10.1155/2012/174318
Research Article

General Iterative Algorithms for Hierarchical Fixed Points Approach to Variational Inequalities

Department of Mathematics, Faculty of Science, King Mongkut’s University of Technology Thonburi (KMUTT), Bangmod, Bangkok 10140, Thailand

Received 24 March 2012; Accepted 16 May 2012

Academic Editor: Zhenyu Huang

Copyright © 2012 Nopparat Wairojjana 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

This paper deals with new methods for approximating a solution to the fixed point problem; find ̃𝑥𝐹(𝑇), where 𝐻 is a Hilbert space, 𝐶 is a closed convex subset of 𝐻, 𝑓 is a 𝜌-contraction from 𝐶 into 𝐻, 0<𝜌<1, 𝐴 is a strongly positive linear-bounded operator with coefficient 𝛾>0, 0<𝛾<𝛾/𝜌, 𝑇 is a nonexpansive mapping on 𝐶, and 𝑃𝐹(𝑇) denotes the metric projection on the set of fixed point of 𝑇. Under a suitable different parameter, we obtain strong convergence theorems by using the projection method which solves the variational inequality (𝐴𝛾𝑓)̃𝑥+𝜏(𝐼𝑆)̃𝑥,𝑥̃𝑥0 for 𝑥𝐹(𝑇), where 𝜏[0,). Our results generalize and improve the corresponding results of Yao et al. (2010) and some authors. Furthermore, we give an example which supports our main theorem in the last part.

1. Introduction

Throughout this paper, we assume that 𝐻 is a real Hilbert space where inner product and norm are denoted by , and , respectively, and let 𝐶 be a nonempty closed convex subset of 𝐻. A mapping 𝑇𝐶𝐶 is called nonexpansive if 𝑇𝑥𝑇𝑦𝑥𝑦,𝑥,𝑦𝐶.(1.1)

We use 𝐹(𝑇) to denote the set of fixed points of 𝑇, that is, 𝐹(𝑇)={𝑥𝐶𝑇𝑥=𝑥}. It is assumed throughout the paper that 𝑇 is a nonexpansive mapping such that 𝐹(𝑇).

Recall that a mapping 𝑓𝐶𝐻 is a contraction on 𝐶 if there exists a constant 𝜌(0,1) such that 𝑓(𝑥)𝑓(𝑦)𝜌𝑥𝑦,𝑥,𝑦𝐶.(1.2)

A mapping 𝐴𝐻𝐻 is called a strongly positive linear bounded operator on 𝐻 if there exists a constant 𝛾>0 with property 𝐴𝑥,𝑥𝛾𝑥2,𝑥𝐻.(1.3)

A mapping 𝑀𝐻𝐻 is called a strongly monotone operator with 𝛼 if 𝑥𝑦,𝑀𝑥𝑀𝑦𝛼𝑥𝑦2,𝑥,𝑦𝐻,(1.4) and 𝑀 is called a monotone operator if 𝑥𝑦,𝑀𝑥𝑀𝑦0,𝑥,𝑦𝐻.(1.5) We easily prove that the mapping (𝐼𝑇) is monotone operator, if 𝑇 is nonexpansive mapping.

The metric (or nearest point) projection from 𝐻 onto 𝐶 is mapping 𝑃𝐶[]𝐻𝐶 which assigns to each point 𝑥𝐶 the unique point 𝑃𝐶[𝑥]𝐶 satisfying the property 𝑥𝑃𝐶[𝑥]=inf𝑦𝐶𝑥𝑦=𝑑(𝑥,𝐶).(1.6)

The variational inequality for a monotone operator, 𝑀𝐻𝐻 over 𝐶, is to find a point in VI(𝐶,𝑀)={̃𝑥𝐶𝑥̃𝑥,𝑀̃𝑥0,𝑥𝐶}.(1.7)

A hierarchical fixed point problem is equivalent to the variational inequality for a monotone operator over the fixed point set. Moreover, to find a hierarchically fixed point of a nonexpansive mapping 𝑇 with respect to another nonexpansive mapping 𝑆, namely, we find ̃𝑥𝐹(𝑇) such that 𝑥̃𝑥,(𝐼𝑆)̃𝑥0,𝑥𝐹(𝑇).(1.8)

Iterative methods for nonexpansive mappings have recently been applied to solve a convex minimization problem; see, for example, [15] and the references therein. A typical problem is to minimize a quadratic function over the set of the fixed points of a nonexpansive mapping on a real Hilbert space 𝐻: min𝑥𝐹(𝑇)12𝐴𝑥,𝑥𝑥,𝑏,(1.9) where 𝑏 is a given point in 𝐻. In [5], it is proved that the sequence {𝑥𝑛} defined by the iterative method below, with the initial guess 𝑥0𝐻 chosen arbitrarily, 𝑥𝑛+1=𝐼𝛼𝑛𝐴𝑇𝑥𝑛+𝛼𝑛𝑏,𝑛0,(1.10) converges strongly to the unique solution of the minimization problem (1.9) provided the sequence {𝛼𝑛} of parameters satisfies certain appropriate conditions.

On the other hand, Moudafi [6] introduced the viscosity approximation method for nonexpansive mappings (see [7] for further developments in both Hilbert and Banach spaces). Starting with an arbitrary initial 𝑥0𝐻, define a sequence {𝑥𝑛} recursively by 𝑥𝑛+1=𝜎𝑛𝑓𝑥𝑛+1𝜎𝑛𝑇𝑥𝑛,𝑛0,(1.11) where {𝜎𝑛} is a sequence in (0,1). It is proved in [6, 7] that under certain appropriate conditions imposed on {𝜎𝑛}, the sequence {𝑥𝑛} generated by (1.11) strongly converges to the unique solution 𝑥 in 𝐶 of the variational inequality (𝐼𝑓)𝑥,𝑥𝑥0,𝑥𝐶.(1.12)

In 2006, Marino and Xu [8] introduced a general iterative method for nonexpansive mapping. Starting with an arbitrary initial 𝑥0𝐻, define a sequence {𝑥𝑛} recursively by 𝑥𝑛+1=𝜖𝑛𝑥𝛾𝑓𝑛+𝐼𝜖𝑛𝐴𝑇𝑥𝑛,𝑛0.(1.13) They proved that if the sequence {𝜖𝑛} of parameters satisfies appropriate conditions, then the sequence {𝑥𝑛} generated by (1.13) strongly converges to the unique solution ̃𝑥=𝑃𝐹(𝑇)(𝐼𝐴+𝛾𝑓)̃𝑥 of the variational inequality (𝐴𝛾𝑓)̃𝑥,𝑥̃𝑥0,𝑥𝐹(𝑇),(1.14) which is the optimality condition for the minimization problem min𝑥𝐹(𝑇)12𝐴𝑥,𝑥(𝑥),(1.15) where is a potential function for 𝛾𝑓 (i.e., (𝑥)=𝛾𝑓(𝑥) for 𝑥𝐻).

In 2010, Yao et al. [9] introduced an iterative algorithm for solving some hierarchical fixed point problem, starting with an arbitrary initial guess 𝑥0𝐶, define a sequence {𝑥𝑛} iteratively by 𝑦𝑛=𝛽𝑛𝑆𝑥𝑛+1𝛽𝑛𝑥𝑛,𝑥𝑛+1=𝑃𝐶𝛼𝑛𝑓𝑥𝑛+1𝛼𝑛𝑇𝑦𝑛,𝑛1.(1.16) They proved that if the sequences {𝛼𝑛} and {𝛽𝑛} of parameters satisfies appropriate conditions, then the sequence {𝑥𝑛} generated by (1.16) strongly converges to the unique solution 𝑧 in 𝐻 of the variational inequality 𝑧𝐹(𝑇),(𝐼𝑓)𝑧,𝑥𝑧0,𝑥𝐹(𝑇).(1.17)

In this paper we will combine the general iterative method (1.13) with the iterative algorithm (1.16) and consider the following iterative algorithm: 𝑦𝑛=𝛽𝑛𝑆𝑥𝑛+1𝛽𝑛𝑥𝑛,𝑥𝑛+1=𝑃𝐶𝛼𝑛𝑥𝛾𝑓𝑛+𝐼𝛼𝑛𝐴𝑇𝑦𝑛,𝑛1.(1.18) We will prove in Section 3 that if the sequences {𝛼𝑛} and {𝛽𝑛} of parameters satisfy appropriate conditions and lim𝑛(𝛽𝑛/𝛼𝑛)=𝜏(0,) then the sequence {𝑥𝑛} generated by (1.18) converges strongly to the unique solution ̃𝑥 in 𝐻 of the following variational inequality 1̃𝑥𝐹(𝑇),𝜏(𝐴𝛾𝑓)̃𝑥+(𝐼𝑆)̃𝑥,𝑥̃𝑥0,𝑥𝐹(𝑇).(1.19) In particular, if we take 𝜏=0, under suitable difference assumptions on parameter, then the sequence {𝑥𝑛} generated by (1.18) converges strongly to the unique solution ̃𝑥 in 𝐻 of the following variational inequality ̃𝑥𝐹(𝑇),(𝐴𝛾𝑓)̃𝑥,𝑥̃𝑥0,𝑥𝐹(𝑇).(1.20) Our results improve and extend the recent results of Yao et al. [9] and some authors. Furthermore, we give an example which supports our main theorem in the last part.

2. Preliminaries

This section collects some lemma which can be used in the proofs for the main results in the next section. Some of them are known, others are not hard to derive.

Lemma 2.1 (Browder [10]). Let 𝐻 be a Hilbert space, 𝐶 be a closed convex subset of 𝐻, and 𝑇𝐶𝐶 be 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; in particular, if 𝑦=0 then 𝑥𝐹(𝑇).

Lemma 2.2. Let 𝑥𝐻 and 𝑧𝐶 be any points. Then one has the following:
(1) That 𝑧=𝑃𝐶[𝑥] if and only if there holds the relation: 𝑥𝑧,𝑦𝑧0,𝑦𝐶.(2.1)(2) That 𝑧=𝑃𝐶[𝑥] if and only if there holds the relation: 𝑥𝑧2𝑥𝑦2𝑦𝑧2,𝑦𝐶.(2.2)(3) There holds the relation: 𝑃𝐶[𝑥]𝑃𝐶[𝑦]𝑃,𝑥𝑦𝐶[𝑥]𝑃𝐶[𝑦]2,𝑥,𝑦𝐻.(2.3) Consequently, 𝑃𝐶 is nonexpansive and monotone.

Lemma 2.3 (Marino and Xu [8]). Let 𝐻 be a Hilbert space, 𝐶 be a closed convex subset of 𝐻, 𝑓𝐶𝐻 be a contraction with coefficient 0<𝜌<1, and 𝑇𝐶𝐶 be nonexpansive mapping. Let 𝐴 be a strongly positive linear bounded operator on a Hilbert space 𝐻 with coefficient 𝛾>0. Then, for 0<𝛾<𝛾/𝜌, for 𝑥,𝑦𝐶, (1)the mapping (𝐼𝑓) is strongly monotone with coefficient (1𝜌), that is, 𝑥𝑦,(𝐼𝑓)𝑥(𝐼𝑓)𝑦(1𝜌)𝑥𝑦2,(2.4)(2)the mapping (𝐴𝛾𝑓) is strongly monotone with coefficient 𝛾𝛾𝜌 that is 𝑥𝑦,(𝐴𝛾𝑓)𝑥(𝐴𝛾𝑓)𝑦𝛾𝛾𝜌𝑥𝑦2.(2.5)

Lemma 2.4 (Xu [4]). Assume that {𝑎𝑛} is a sequence of nonnegative numbers such that 𝑎𝑛+11𝛾𝑛𝑎𝑛+𝛿𝑛,𝑛0,(2.6) where {𝛾𝑛} is a sequence in (0,1) and {𝛿𝑛} is a sequence in such that (1)𝑛=1𝛾𝑛=, (2)limsup𝑛(𝛿𝑛/𝛾𝑛)0 or 𝑛=1|𝛿𝑛|<. Then lim𝑛𝑎𝑛=0.

Lemma 2.5 (Marino and Xu [8]). Assume 𝐴 is a strongly positive linear bounded operator on a Hilbert space 𝐻 with coefficient 𝛾>0 and 0<𝛼𝐴1. Then 𝐼𝛼𝐴1𝛼𝛾.

Lemma 2.6 (Acedo and Xu [11]). Let 𝐶 be a closed convex subset of 𝐻. Let {𝑥𝑛} be a bounded sequence in 𝐻. Assume that (1)The weak 𝜔-limit set 𝜔𝑤(𝑥𝑛)𝐶,(2)For each 𝑧𝐶, lim𝑛𝑥𝑛𝑧 exists. Then {𝑥𝑛} is weakly convergent to a point in 𝐶.

NotationWe use for strong convergence and for weak convergence.

3. Main Results

Theorem 3.1. Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻. Let 𝑓𝐶𝐻 be a 𝜌-contraction with 𝜌(0,1). Let 𝑆,𝑇𝐶𝐶 be two nonexpansive mappings with 𝐹(𝑇). Let 𝐴 be a strongly positive linear bounded operator on 𝐻 with coefficient 𝛾>0. {𝛼𝑛} and {𝛽𝑛} are two sequences in (0,1) and 0<𝛾<𝛾/𝜌. Starting with an arbitrary initial guess 𝑥0𝐶 and {𝑥𝑛} is a sequence generated by 𝑦𝑛=𝛽𝑛𝑆𝑥𝑛+1𝛽𝑛𝑥𝑛,𝑥𝑛+1=𝑃𝐶𝛼𝑛𝑥𝛾𝑓𝑛+𝐼𝛼𝑛𝐴𝑇𝑦𝑛,𝑛1.(3.1)
Suppose that the following conditions are satisfied: (C1)  lim𝑛𝛼𝑛=0 and 𝑛=1𝛼𝑛=, (C2)  lim𝑛(𝛽𝑛/𝛼𝑛)=𝜏=0, (C3)  lim𝑛(|𝛼𝑛𝛼𝑛1|/𝛼𝑛)=0 and lim𝑛(|𝛽𝑛𝛽𝑛1|/𝛽𝑛)=0, or (C4)  𝑛=1|𝛼𝑛𝛼𝑛1|< and 𝑛=1|𝛽𝑛𝛽𝑛1|<. Then the sequence {𝑥𝑛} converges strongly to a point ̃𝑥𝐻, which is the unique solution of the variational inequality: ̃𝑥𝐹(𝑇),(𝐴𝛾𝑓)̃𝑥,𝑥̃𝑥0,𝑥𝐹(𝑇).(3.2) Equivalently, one has 𝑃𝐹(𝑇)(𝐼𝐴+𝛾𝑓)̃𝑥=̃𝑥.

Proof. We first show the uniqueness of a solution of the variational inequality (3.2), which is indeed a consequence of the strong monotonicity of 𝐴𝛾𝑓. Suppose 𝑥𝐹(𝑇) and ̃𝑥𝐹(𝑇) both are solutions to (3.2), then (𝐴𝛾𝑓)𝑥,𝑥̃𝑥0 and (𝐴𝛾𝑓)̃𝑥,̃𝑥𝑥0. It follows that (𝐴𝛾𝑓)𝑥,𝑥+̃𝑥(𝐴𝛾𝑓)̃𝑥,̃𝑥𝑥=(𝐴𝛾𝑓)𝑥,𝑥̃𝑥(𝐴𝛾𝑓)̃𝑥,𝑥̃𝑥=(𝐴𝛾𝑓)𝑥(𝐴𝛾𝑓)̃𝑥,𝑥̃𝑥0.(3.3) The strong monotonicity of 𝐴𝛾𝑓 (Lemma 2.3) implies that 𝑥=̃𝑥 and the uniqueness is proved.
Next, we prove that the sequence {𝑥𝑛} is bounded. Since 𝛼𝑛0 and lim𝑛(𝛽𝑛/𝛼𝑛)=0 by condition (C1) and (C2), respectively, we can assume, without loss of generality, that 𝛼𝑛<𝐴1 and 𝛽𝑛<𝛼𝑛 for all 𝑛1. Take 𝑢𝐹(𝑇) and from (3.1), we have 𝑥𝑛+1=𝑃𝑢𝐶𝛼𝑛𝑥𝛾𝑓𝑛+𝐼𝛼𝑛𝐴𝑇𝑦𝑛𝑃𝐶[𝑢]𝛼𝑛𝑥𝛾𝑓𝑛+𝐼𝛼𝑛𝐴𝑇𝑦𝑛𝑢𝛼𝑛𝛾𝑓𝑥𝑛𝑓(𝑢)+𝛼𝑛𝛾𝑓(𝑢)𝐴𝑢+𝐼𝛼𝑛𝐴𝑇𝑦𝑛.𝑢(3.4) Since 𝐼𝛼𝑛𝐴1𝛼𝑛𝛾 and by Lemma 2.5, we note that 𝑥𝑛+1𝑢𝛼𝑛𝛾𝑓𝑥𝑛𝑓(𝑢)+𝛼𝑛𝛾𝑓(𝑢)𝐴𝑢+1𝛼𝑛𝛾𝑇𝑦𝑛𝑢𝛼𝑛𝑥𝛾𝜌𝑛𝑢+𝛼𝑛𝛾𝑓(𝑢)𝐴𝑢+1𝛼𝑛𝛾𝑇𝑦𝑛𝑇𝑢𝛼𝑛𝑥𝛾𝜌𝑛𝑢+𝛼𝑛𝛾𝑓(𝑢)𝐴𝑢+1𝛼𝑛𝛾𝑦𝑛𝑢𝛼𝑛𝑥𝛾𝜌𝑛𝑢+𝛼𝑛+𝛾𝑓(𝑢)𝐴𝑢1𝛼𝑛𝛾𝛽𝑛𝑆𝑥𝑛𝑆𝑢+𝛽𝑛𝑆𝑢𝑢+1𝛽𝑛𝑥𝑛𝑢𝛼𝑛𝑥𝛾𝜌𝑛𝑢+𝛼𝑛+𝛾𝑓(𝑢)𝐴𝑢1𝛼𝑛𝛾𝛽𝑛𝑥𝑛𝑢+𝛽𝑛𝑆𝑢𝑢+1𝛽𝑛𝑥𝑛=𝑢1𝛼𝑛𝑥𝛾𝛾𝜌𝑛𝑢+𝛼𝑛𝛾𝑓(𝑢)𝐴𝑢+1𝛼𝑛𝛾𝛽𝑛𝑆𝑢𝑢1𝛼𝑛𝑥𝛾𝛾𝜌𝑛𝑢+𝛼𝑛𝛾𝑓(𝑢)𝐴𝑢+𝛽𝑛𝑆𝑢𝑢1𝛼𝑛𝑥𝛾𝛾𝜌𝑛𝑢+𝛼𝑛𝛾𝑓(𝑢)𝐴𝑢+𝛼𝑛=𝑆𝑢𝑢1𝛼𝑛𝑥𝛾𝛾𝜌𝑛𝑢+𝛼𝑛[(]=𝛾𝑓𝑢)𝐴𝑢+𝑆𝑢𝑢1𝛼𝑛𝑥𝛾𝛾𝜌𝑛𝑢+𝛼𝑛𝛾𝛾𝜌𝛾𝑓(𝑢)𝐴𝑢+𝑆𝑢𝑢.𝛾𝛾𝜌(3.5) By induction, we can obtain 𝑥𝑛+1𝑥𝑢max0,𝑢𝛾𝑓(𝑢)𝐴𝑢+𝑆𝑢𝑢,𝛾𝛾𝜌(3.6) which implies that the sequence {𝑥𝑛} is bounded and so are the sequences {𝑓(𝑥𝑛)}, {𝑆𝑥𝑛}, and {𝐴𝑇𝑦𝑛}.
Set 𝑤𝑛=𝛼𝑛𝛾𝑓(𝑥𝑛)+(𝐼𝛼𝑛𝐴)𝑇𝑦𝑛,𝑛1. We get 𝑥𝑛+1𝑥𝑛=𝑃𝐶𝑤𝑛+1𝑃𝐶𝑤𝑛𝑤𝑛+1𝑤𝑛.(3.7) It follows that 𝑥𝑛+1𝑥𝑛𝛼𝑛𝑥𝛾𝑓𝑛+𝐼𝛼𝑛𝐴𝑇𝑦𝑛𝛼𝑛1𝑥𝛾𝑓𝑛1+𝐼𝛼𝑛1𝐴𝑇𝑦𝑛1𝛼𝑛𝛾𝑓𝑥𝑛𝑥𝑓𝑛1+||𝛼𝑛𝛼𝑛1||𝑥𝛾𝑓𝑛1𝐴𝑇𝑦𝑛1+1𝛼𝑛𝛾𝑇𝑦𝑛𝑇𝑦𝑛1𝛼𝑛𝑥𝛾𝜌𝑛𝑥𝑛1+||𝛼𝑛𝛼𝑛1||𝑥𝛾𝑓𝑛1𝐴𝑇𝑦𝑛1+1𝛼𝑛𝛾𝑦𝑛𝑦𝑛1.(3.8) By (3.7) and (3.8), we get 𝑥𝑛+1𝑥𝑛𝛼𝑛𝑤𝛾𝜌𝑛𝑤𝑛1+||𝛼𝑛𝛼𝑛1||𝑥𝛾𝑓𝑛1𝐴𝑇𝑦𝑛1+1𝛼𝑛𝛾𝑦𝑛𝑦𝑛1.(3.9) From (3.1), we obtain 𝑦𝑛𝑦𝑛1=𝛽𝑛𝑆𝑥𝑛+1𝛽𝑛𝑥𝑛𝛽𝑛1𝑆𝑥𝑛1+1𝛽𝑛1𝑥𝑛1=𝛽𝑛𝑆𝑥𝑛𝑆𝑥𝑛1+𝛽𝑛𝛽𝑛1𝑆𝑥𝑛1𝑥𝑛1+1𝛽𝑛𝑥𝑛𝑥𝑛1𝑥𝑛𝑥𝑛1+||𝛽𝑛𝛽𝑛1||𝑆𝑥𝑛1𝑥𝑛1𝑥𝑛𝑥𝑛1+||𝛽𝑛𝛽𝑛1||𝑀,(3.10) where 𝑀 is a constant such that sup𝑛𝑥𝛾𝑓𝑛1𝐴𝑇𝑦𝑛1+𝑆𝑥𝑛1𝑥𝑛1𝑀.(3.11) Substituting (3.10) into (3.8) to obtain 𝑥𝑛+1𝑥𝑛𝛼𝑛𝑥𝛾𝜌𝑛𝑥𝑛1+||𝛼𝑛𝛼𝑛1||𝑥𝛾𝑓𝑛1𝐴𝑇𝑦𝑛1+1𝛼𝑛𝛾𝑥𝑛𝑥𝑛1+||𝛽𝑛𝛽𝑛1||𝑀𝛼𝑛𝑥𝛾𝜌𝑛𝑥𝑛1+||𝛼𝑛𝛼𝑛1||𝑀+1𝛼𝑛𝛾𝑥𝑛𝑥𝑛1+||𝛽𝑛𝛽𝑛1||𝑀=1𝛼𝑛𝑥𝛾𝛾𝜌𝑛𝑥𝑛1||𝛼+𝑀𝑛𝛼𝑛1||+||𝛽𝑛𝛽𝑛1||1𝛼𝑛𝑤𝛾𝛾𝜌𝑛𝑤𝑛1||𝛼+𝑀𝑛𝛼𝑛1||+||𝛽𝑛𝛽𝑛1||.(3.12) At the same time, we can write (3.12) as 𝑥𝑛+1𝑥𝑛1𝛼𝑛𝑤𝛾𝛾𝜌𝑛𝑤𝑛1+𝑀𝛼𝑛||𝛼𝑛𝛼𝑛1||𝛼𝑛+||𝛽𝑛𝛽𝑛1||𝛼𝑛1𝛼𝑛𝑤𝛾𝛾𝜌𝑛𝑤𝑛1+𝑀𝛼𝑛||𝛼𝑛𝛼𝑛1||𝛽𝑛+||𝛽𝑛𝛽𝑛1||𝛽𝑛.(3.13) From (3.12), (C4), and Lemma 2.5 or from (3.13), (C3), and Lemma 2.5, we can deduce that 𝑥𝑛+1𝑥𝑛0, respectively.
From (3.1), we have 𝑥𝑛𝑇𝑥𝑛𝑥𝑛𝑥𝑛+1+𝑥𝑛+1𝑇𝑥𝑛=𝑥𝑛𝑥𝑛+1+𝑃𝐶𝑤𝑛𝑃𝐶𝑇𝑥𝑛𝑥𝑛𝑥𝑛+1+𝑤𝑛𝑇𝑥𝑛=𝑥𝑛𝑥𝑛+1+𝛼𝑛𝑥𝛾𝑓𝑛+𝐼𝛼𝑛𝐴𝑇𝑦𝑛𝑇𝑥𝑛𝑥𝑛𝑥𝑛+1+𝛼𝑛𝑥𝛾𝑓𝑛𝐴𝑇𝑥𝑛+1𝛼𝑛𝛾𝑇𝑦𝑛𝑇𝑥𝑛𝑥𝑛𝑥𝑛+1+𝛼𝑛𝑥𝛾𝑓𝑛𝐴𝑇𝑥𝑛+1𝛼𝑛𝛾𝑦𝑛𝑥𝑛=𝑥𝑛𝑥𝑛+1+𝛼𝑛𝑥𝛾𝑓𝑛𝐴𝑇𝑥𝑛+1𝛼𝑛𝛾𝛽𝑛𝑆𝑥𝑛𝑥𝑛.(3.14) Notice that 𝛼𝑛0, 𝛽𝑛0, and 𝑥𝑛+1𝑥𝑛0, so we obtain 𝑥𝑛𝑇𝑥𝑛0.(3.15) Next, we prove limsup𝑛𝛾𝑓(𝑧)𝐴𝑧,𝑥𝑛𝑧0,(3.16) where 𝑧=𝑃𝐹(𝑇)(𝐼𝐴+𝛾𝑓)𝑧. Since the sequence {𝑥𝑛} is bounded we can take a subsequence {𝑥𝑛𝑘} of {𝑥𝑛} such that limsup𝑛𝛾𝑓(𝑧)𝐴𝑧,𝑥𝑛𝑧=lim𝑘𝛾𝑓(𝑧)𝐴𝑧,𝑥𝑛𝑘,𝑧(3.17) and 𝑥𝑛𝑘0𝑥00086̃𝑥. From (3.15) and by Lemma 2.1, it follows that ̃𝑥𝐹(𝑇). Hence, by Lemma 2.2(1) that limsup𝑛𝛾𝑓(𝑧)𝐴𝑧,𝑥𝑛𝑧=lim𝑘𝛾𝑓(𝑧)𝐴𝑧,𝑥𝑛𝑘𝑧=𝛾𝑓(𝑧)𝐴𝑧,̃𝑥𝑧=(𝐼𝐴+𝛾𝑓)𝑧𝑧,̃𝑥𝑧0.(3.18) Now, by Lemma 2.2(1), we observe that 𝑃𝐶𝑤𝑛𝑤𝑛,𝑃𝐶𝑤𝑛𝑧0,(3.19) and so 𝑥𝑛+1𝑧2=𝑃𝐶𝑤𝑛𝑧,𝑃𝐶𝑤𝑛=𝑃𝑧𝐶𝑤𝑛𝑤𝑛,𝑃𝐶𝑤𝑛+𝑤𝑧𝑛𝑧,𝑃𝐶𝑤𝑛𝑤𝑧𝑛𝑧,𝑃𝐶𝑤𝑛=𝛼𝑧𝑛𝑥𝛾𝑓𝑛+𝐼𝛼𝑛𝐴𝑇𝑦𝑛𝑧,𝑥𝑛+1𝑧𝛼𝑛𝛾𝑓𝑥𝑛𝑥𝑓(𝑧)𝑛+1𝑧+𝛼𝑛𝛾𝑓(𝑧)𝐴𝑧,𝑥𝑛+1+𝑧1𝛼𝑛𝛾𝑇𝑦𝑛𝑥𝑧𝑛+1𝑧𝛼𝑛𝑥𝛾𝜌𝑛𝑥𝑧𝑛+1𝑧+𝛼𝑛𝛾𝑓(𝑧)𝐴𝑧,𝑥𝑛+1+𝑧1𝛼𝑛𝛾𝑦𝑛𝑥𝑧𝑛+1𝑧=𝛼𝑛𝑥𝛾𝜌𝑛𝑥𝑧𝑛+1𝑧+𝛼𝑛𝛾𝑓(𝑧)𝐴𝑧,𝑥𝑛+1+𝑧1𝛼𝑛𝛾𝛽𝑛𝑆𝑥𝑛+1𝛽𝑛𝑥𝑛𝑥𝑧𝑛+1𝑧𝛼𝑛𝑥𝛾𝜌𝑛𝑥𝑧𝑛+1𝑧+𝛼𝑛𝛾𝑓(𝑧)𝐴𝑧,𝑥𝑛+1+𝑧1𝛼𝑛𝛾𝛽𝑛𝑆𝑥𝑛𝑆𝑧+𝛽𝑛𝑆𝑧𝑧+1𝛽𝑛𝑥𝑛𝑧𝑥𝑛+1𝑧𝛼𝑛𝑥𝛾𝜌𝑛𝑥𝑧𝑛+1𝑧+𝛼𝑛𝛾𝑓(𝑧)𝐴𝑧,𝑥𝑛+1+𝑧1𝛼𝑛𝛾𝛽𝑛𝑥𝑛𝑧+𝛽𝑛𝑆𝑧𝑧+1𝛽𝑛𝑥𝑛𝑥𝑧𝑛+1=𝑧1𝛼𝑛𝑥𝛾𝛾𝜌𝑛𝑥𝑧𝑛+1𝑧+𝛼𝑛𝛾𝑓(𝑧)𝐴𝑧,𝑥𝑛+1+𝑧1𝛼𝑛𝛾𝛽𝑛𝑥𝑆𝑧𝑧𝑛+1𝑧1𝛼𝑛𝛾𝛾𝜌2𝑥𝑛𝑧2+𝑥𝑛+1𝑧2+𝛼𝑛𝛾𝑓(𝑧)𝐴𝑧,𝑥𝑛+1+𝑧1𝛼𝑛𝛾𝛽𝑛𝑥𝑆𝑧𝑧𝑛+1.𝑧(3.20) Hence, it follows that 𝑥𝑛+1𝑧21𝛼𝑛𝛾𝛾𝜌1+𝛼𝑛𝑥𝛾𝛾𝜌𝑛𝑧2+2𝛼𝑛1+𝛼𝑛𝛾𝛾𝜌𝛾𝑓(𝑧)𝐴𝑧,𝑥𝑛+1+2𝑧1𝛼𝑛𝛾𝛽𝑛1+𝛼𝑛𝑥𝛾𝛾𝜌𝑆𝑧𝑧𝑛+1=𝑧2𝛼𝑛𝛾𝛾𝜌1+𝛼𝑛1𝛾𝛾𝜌𝛼𝑛𝛾𝛾𝜌𝛾𝑓(𝑧)𝐴𝑧,𝑥𝑛+1+𝛽𝑧𝑛1𝛼𝑛𝛾𝛼𝑛𝑥𝛾𝛾𝜌𝑆𝑧𝑧𝑛+1×𝑧12𝛼𝑛𝛾𝛾𝜌1+𝛼𝑛𝑥𝛾𝛾𝜌𝑛𝑧2.(3.21) We observe that limsup𝑛1𝛼𝑛𝛾𝛾𝜌𝛾𝑓(𝑧)𝐴𝑧,𝑥𝑛+1+𝛽𝑧𝑛1𝛼𝑛𝛾𝛼𝑛𝑥𝛾𝛾𝜌𝑆𝑧𝑧𝑛+1𝑧0.(3.22) Thus, by Lemma 2.4, 𝑥𝑛𝑧 as 𝑛. This is completes.

From Theorem 3.1, we can deduce the following interesting corollary.

Corollary 3.2 (Yao et al. [9]). Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻. Let 𝑓𝐶𝐻 be a 𝜌-contraction (possibly nonself) with 𝜌(0,1). Let 𝑆,𝑇𝐶𝐶 be two nonexpansive mappings with 𝐹(𝑇).  {𝛼𝑛} and {𝛽𝑛} are two sequences in (0,1). Starting with an arbitrary initial guess 𝑥0𝐶 and {𝑥𝑛} is a sequence generated by 𝑦𝑛=𝛽𝑛𝑆𝑥𝑛+1𝛽𝑛𝑥𝑛,𝑥𝑛+1=𝑃𝐶𝛼𝑛𝑓𝑥𝑛+1𝛼𝑛𝑇𝑦𝑛,𝑛1.(3.23) Suppose that the following conditions are satisfied: (C1)  lim𝑛𝛼𝑛=0 and 𝑛=1𝛼𝑛=, (C2)  lim𝑛(𝛽𝑛/𝛼𝑛)=0, (C3)  lim𝑛(|𝛼𝑛𝛼𝑛1|/𝛼𝑛)=0 and lim𝑛(|𝛽𝑛𝛽𝑛1|/𝛽𝑛)=0, or (C4)  𝑛=1|𝛼𝑛𝛼𝑛1|< and 𝑛=1|𝛽𝑛𝛽𝑛1|<. Then the sequence {𝑥𝑛} converges strongly to a point ̃𝑥𝐻, which is the unique solution of the variational inequality: ̃𝑥𝐹(𝑇),(𝐼𝑓)̃𝑥,𝑥̃𝑥0,𝑥𝐹(𝑇).(3.24) Equivalently, one has 𝑃𝐹(𝑇)(𝑓)̃𝑥=̃𝑥. In particular, if one takes 𝑓=0, then the sequence {𝑥𝑛} converges in norm to the Minimum norm fixed point ̃𝑥 of 𝑇, namely, the point ̃𝑥 is the unique solution to the quadratic minimization problem: 𝑧=argmin𝑥𝐹(𝑇)𝑥2.(3.25)

Proof. As a matter of fact, if we take 𝐴=𝐼 and 𝛾=1 in Theorem 3.1. This completes the proof.

Under different conditions on data we obtain the following result.

Theorem 3.3. Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻. Let 𝑓𝐶𝐻 be a 𝜌-contraction (possibly nonself) with 𝜌(0,1). Let 𝑆,𝑇𝐶𝐶 be two nonexpansive mappings with 𝐹(𝑇). Let 𝐴 be a strongly positive linear bounded operator on a Hilbert space 𝐻 with coefficient 𝛾>0 and 0<𝛾<𝛾/𝜌. {𝛼𝑛} and {𝛽𝑛} are two sequences in (0,1). Starting with an arbitrary initial guess 𝑥0𝐶 and {𝑥𝑛} is a sequence generated by 𝑦𝑛=𝛽𝑛𝑆𝑥𝑛+1𝛽𝑛𝑥𝑛,𝑥𝑛+1=𝑃𝐶𝛼𝑛𝑥𝛾𝑓𝑛+𝐼𝛼𝑛𝐴𝑇𝑦𝑛,𝑛1.(3.26) Suppose that the following conditions are satisfied: (C1)  lim𝑛𝛼𝑛=0 and 𝑛=1𝛼𝑛=, (C2)  lim𝑛(𝛽𝑛/𝛼𝑛)=𝜏(0,), (C5)  lim𝑛((|𝛼𝑛𝛼𝑛1|+|𝛽𝑛𝛽𝑛1|)/𝛼𝑛𝛽𝑛)=0, (C6)  there exists a constant 𝐾>0 such that (1/𝛼𝑛)|1/𝛽𝑛1/𝛽𝑛1|𝐾. Then the sequence {𝑥𝑛} converges strongly to a point ̃𝑥𝐻, which is the unique solution of the variational inequality: 1̃𝑥𝐹(𝑇),𝜏(𝐴𝛾𝑓)̃𝑥+(𝐼𝑆)̃𝑥,𝑥̃𝑥0,𝑥𝐹(𝑇).(3.27)

Proof. First of all, we show that (3.27) has the unique solution. Indeed, let 𝑥 and ̃𝑥 be two solutions. Then (𝐴𝛾𝑓)̃𝑥,̃𝑥𝑥𝜏(𝐼𝑆)̃𝑥,𝑥.̃𝑥(3.28) Analogously, we have (𝐴𝛾𝑓)𝑥,𝑥̃𝑥𝜏(𝐼𝑆)𝑥,̃𝑥𝑥.(3.29) Adding (3.28) and (3.29), by Lemma 2.3, we obtain 𝛾𝛾𝜌̃𝑥𝑥2(𝐴𝛾𝑓)̃𝑥(𝐴𝛾𝑓)𝑥,̃𝑥𝑥𝜏(𝐼𝑆)̃𝑥(𝐼𝑆)𝑥,̃𝑥𝑥0,(3.30) and so ̃𝑥=𝑥. From (C2), we can assume, without loss of generality, that 𝛽𝑛(𝜏+1)𝛼𝑛 for all 𝑛1. By a similar argument in Theorem 3.1, we have 𝑥𝑛+1𝑢𝛼𝑛𝑥𝛾𝜌𝑛𝑢+𝛼𝑛+𝛾𝑓(𝑢)𝐴𝑢1𝛼𝑛𝛾𝑥𝑛𝑢+𝛽𝑛𝑆𝑢𝑢+1𝛽𝑛𝑥𝑛=𝑢1𝛼𝑛𝑥𝛾𝛾𝜌𝑛𝑢+𝛼𝑛𝛾𝑓(𝑢)𝐴𝑢+1𝛼𝑛𝛾𝛽𝑛𝑆𝑢𝑢1𝛼𝑛𝑥𝛾𝛾𝜌𝑛𝑢+𝛼𝑛(𝛾𝑓𝑢)𝐴𝑢+𝛽𝑛𝑆𝑢𝑢1𝛼𝑛𝑥𝛾𝛾𝜌𝑛𝑢+𝛼𝑛𝛾𝑓(𝑢)𝐴𝑢+(𝜏+1)𝛼𝑛=𝑆𝑢𝑢1𝛼𝑛𝑥𝛾𝛾𝜌𝑛𝑢+𝛼𝑛[(]=𝛾𝑓𝑢)𝐴𝑢+(𝜏+1)𝑆𝑢𝑢1𝛼𝑛𝑥𝛾𝛾𝜌𝑛𝑢+𝛼𝑛𝛾𝛾𝜌𝛾𝑓(𝑢)𝐴𝑢+(𝜏+1)𝑆𝑢𝑢.𝛾𝛾𝜌(3.31) By induction, we obtain 𝑥𝑛𝑥𝑢max0,1𝑢[],𝛾𝛾𝜌𝛾𝑓(𝑢)𝐴𝑢+(𝜏+1)𝑆𝑢𝑢(3.32) which implies that the sequence {𝑥𝑛} is bounded. Since (C5) implies (C4) then, from Theorem 3.1, we can deduce 𝑥𝑛+1𝑥𝑛0.
From (3.1), we note that 𝑥𝑛+1=𝑃𝐶𝑤𝑛𝑤𝑛+𝑤𝑛+𝑦𝑛𝑦𝑛=𝑃𝐶𝑤𝑛𝑤𝑛+𝛼𝑛𝑥𝛾𝑓𝑛+𝑇𝑦𝑛𝑦𝑛+𝑦𝑛𝛼𝑛𝐴𝑇𝑦𝑛.(3.33) Hence, it follows that 𝑥𝑛𝑥𝑛+1=𝑤𝑛𝑃𝐶𝑤𝑛+𝛼𝑛𝐴𝑥𝑛𝑥𝛾𝑓𝑛+𝑦𝑛𝑇𝑦𝑛+𝑥𝑛𝑦𝑛+𝛼𝑛𝐴𝑇𝑦𝑛𝐴𝑥𝑛=𝑤𝑛𝑃𝐶𝑤𝑛+𝛼𝑛(𝐴𝛾𝑓)𝑥𝑛+(𝐼𝑇)𝑦𝑛+𝛽𝑛(𝐼𝑆)𝑥𝑛+𝛼𝑛𝐴𝑇𝑦𝑛𝑥𝑛,(3.34) and so 𝑥𝑛𝑥𝑛+11𝛼𝑛𝛽𝑛=11𝛼𝑛𝛽𝑛𝑤𝑛𝑃𝐶𝑤𝑛+𝛼𝑛1𝛼𝑛𝛽𝑛(𝐴𝛾𝑓)𝑥𝑛+11𝛼𝑛𝛽𝑛(𝐼𝑇)𝑦𝑛+11𝛼𝑛(𝐼𝑆)𝑥𝑛+𝛼𝑛1𝛼𝑛𝛽𝑛𝐴𝑇𝑦𝑛𝑥𝑛.(3.35) Set 𝑣𝑛=(𝑥𝑛𝑥𝑛+1)/(1𝛼𝑛)𝛽𝑛. Then, we have 𝑣𝑛=11𝛼𝑛𝛽𝑛𝑤𝑛𝑃𝐶𝑤𝑛+𝛼𝑛1𝛼𝑛𝛽𝑛(𝐴𝛾𝑓)𝑥𝑛+11𝛼𝑛𝛽𝑛(𝐼𝑇)𝑦𝑛+11𝛼𝑛(𝐼𝑆)𝑥𝑛+𝛼𝑛1𝛼𝑛𝛽𝑛𝐴𝑇𝑦𝑛𝑥𝑛.(3.36) From (3.12) in Theorem 3.1 and (C6), we obtain 𝑥𝑛+1𝑥𝑛𝛽𝑛1𝛼𝑛𝑥𝛾𝛾𝜌𝑛𝑥𝑛1𝛽𝑛||𝛼+𝑀𝑛𝛼𝑛1||𝛽𝑛+||𝛽𝑛𝛽𝑛1||𝛽𝑛=1𝛼𝑛𝑥𝛾𝛾𝜌𝑛𝑥𝑛1𝛽𝑛+1𝛼𝑛𝑥𝛾𝛾𝜌𝑛𝑥𝑛1𝛽𝑛11𝛼𝑛𝑥𝛾𝛾𝜌𝑛𝑥𝑛1𝛽𝑛1||𝛼+𝑀𝑛𝛼𝑛1||𝛽𝑛+||𝛽𝑛𝛽𝑛1||𝛽𝑛=1𝛼𝑛𝑥𝛾𝛾𝜌𝑛𝑥𝑛1𝛽𝑛1+1𝛼𝑛𝑥𝛾𝛾𝜌𝑛𝑥𝑛11𝛽𝑛1𝛽𝑛1||𝛼+𝑀𝑛𝛼𝑛1||𝛽𝑛+||𝛽𝑛𝛽𝑛1||𝛽𝑛1𝛼𝑛𝑥𝛾𝛾𝜌𝑛𝑥𝑛1𝛽𝑛1+𝑥𝑛𝑥𝑛1||||1𝛽𝑛1𝛽𝑛1||||||𝛼+𝑀𝑛𝛼𝑛1||𝛽𝑛+||𝛽𝑛𝛽𝑛1||𝛽𝑛1𝛼𝑛𝑥𝛾𝛾𝜌𝑛𝑥𝑛1𝛽𝑛1+𝛼𝑛𝐾𝑥𝑛𝑥𝑛1||𝛼+𝑀𝑛𝛼𝑛1||𝛽𝑛+||𝛽𝑛𝛽𝑛1||𝛽𝑛1𝛼𝑛𝑤𝛾𝛾𝜌𝑛𝑤𝑛1𝛽𝑛1+𝛼𝑛𝐾𝑥𝑛𝑥𝑛1||𝛼+𝑀𝑛𝛼𝑛1||𝛽𝑛+||𝛽𝑛𝛽𝑛1||𝛽𝑛.(3.37) This together with Lemma 2.4 and (C2) imply that lim𝑛𝑥𝑛+1𝑥𝑛𝛽𝑛=lim𝑛𝑤𝑛+1𝑤𝑛𝛽𝑛=lim𝑛𝑤𝑛+1𝑤𝑛𝛼𝑛=0.(3.38) From (3.36), for 𝑧𝐹(𝑇), we have 𝑣𝑛,𝑥𝑛1𝑧=1𝛼𝑛𝛽𝑛𝑤𝑛𝑃𝐶𝑤𝑛,𝑃𝐶𝑤𝑛1+𝛼𝑧𝑛1𝛼𝑛𝛽𝑛(𝐴𝛾𝑓)𝑥𝑛,𝑥𝑛+1𝑧1𝛼𝑛𝛽𝑛(𝐼𝑇)𝑦𝑛,𝑥𝑛1𝑧+1𝛼𝑛(𝐼𝑆)𝑥𝑛,𝑥𝑛+𝛼𝑧𝑛1𝛼𝑛𝛽𝑛𝐴𝑇𝑦𝑛𝑥𝑛,𝑥𝑛=1𝑧1𝛼𝑛𝛽𝑛𝑤𝑛𝑃𝐶𝑤𝑛,𝑃𝐶𝑤𝑛+1𝑧1𝛼𝑛𝛽𝑛𝑤𝑛𝑃𝐶𝑤𝑛,𝑃𝐶𝑤𝑛1𝑃𝐶𝑤𝑛+𝛼𝑛1𝛼𝑛𝛽𝑛(𝐴𝛾𝑓)𝑥𝑛(𝐴𝛾𝑓)𝑧,𝑥𝑛𝛼𝑧+𝑛1𝛼𝑛𝛽𝑛(𝐴𝛾𝑓)𝑧,𝑥𝑛+1𝑧1𝛼𝑛(𝐼𝑆)𝑥𝑛(𝐼𝑆)𝑧,𝑥𝑛1𝑧+1𝛼𝑛(𝐼𝑆)𝑧,𝑥𝑛+1𝑧1𝛼𝑛𝛽𝑛(𝐼𝑇)𝑦𝑛,𝑥𝑛𝛼𝑧+𝑛1𝛼𝑛𝛽𝑛𝐴𝑇𝑦𝑛𝑥𝑛,𝑥𝑛.𝑧(3.39) By Lemmas 2.2 and 2.3, we obtain 𝑣𝑛,𝑥𝑛1𝑧1𝛼𝑛𝛽𝑛𝑤𝑛𝑃𝐶𝑤𝑛,𝑃𝐶𝑤𝑛1𝑃𝐶𝑤𝑛+𝛼𝛾𝛾𝜌𝑛1𝛼𝑛𝛽𝑛𝑥𝑛𝑧2+𝛼𝑛1𝛼𝑛𝛽𝑛(𝐴𝛾𝑓)𝑧,𝑥𝑛1𝑧+1𝛼𝑛(𝐼𝑆)𝑧,𝑥𝑛+1𝑧1𝛼𝑛𝛽𝑛(𝐼𝑇)𝑦𝑛,𝑥𝑛𝛼𝑧+𝑛1𝛼𝑛𝛽𝑛𝐴𝑇𝑦𝑛𝑥𝑛,𝑥𝑛.𝑧(3.40) Now, we observe that 𝑥𝑛𝑧21𝛼𝑛𝛽𝑛𝛼𝛾𝛾𝜌𝑛𝑣,𝑥𝑛𝛽𝑧𝑛𝛼𝛾𝛾𝜌𝑛(𝐼𝑆)𝑧,𝑥𝑛1𝑧𝛾𝛾𝜌(𝐴𝛾𝑓)𝑧,𝑥𝑛1𝑧𝛼𝛾𝛾𝜌𝑛(𝐼𝑇)𝑦𝑛,𝑥𝑛1𝑧𝐴𝛾𝛾𝜌𝑇𝑦𝑛𝑥𝑛,𝑥𝑛1𝑧𝛼𝛾𝛾𝜌𝑛𝑤𝑛𝑃𝐶𝑤𝑛,𝑃𝐶𝑤𝑛1𝑃𝐶𝑤𝑛1𝛼𝑛𝛽𝑛𝛼𝛾𝛾𝜌𝑛𝑣,𝑥𝑛𝛽𝑧𝑛𝛼𝛾𝛾𝜌𝑛(𝐼𝑆)𝑧,𝑥𝑛1𝑧𝛾𝛾𝜌(𝐴𝛾𝑓)𝑧,𝑥𝑛1𝑧𝛼𝛾𝛾𝜌𝑛(𝐼𝑇)𝑦𝑛,𝑥𝑛1𝑧𝐴𝛾𝛾𝜌𝑇𝑦𝑛𝑥𝑛,𝑥𝑛+𝑤𝑧𝑛𝑤𝑛1𝑤𝛾𝛾𝜌𝑛𝑃𝐶𝑤𝑛.(3.41) From (C1) and (C2), we have 𝛽𝑛0. Hence, from (3.1), we deduce 𝑦𝑛𝑥𝑛0 and 𝑥𝑛+1𝑇𝑦𝑛0. Therefore, 𝑦𝑛𝑇𝑦𝑛𝑦𝑛𝑥𝑛+𝑥𝑛𝑥𝑛+1+𝑥𝑛+1𝑇𝑦𝑛0.(3.42)
Since 𝑣𝑛0, (𝐼𝑇)𝑦𝑛0, 𝐴(𝑇𝑦𝑛𝑥𝑛)0, and 𝑤𝑛𝑤𝑛1/(𝛾𝛾𝜌)0, every weak cluster point of {𝑥𝑛} is also a strong cluster point. Note that the sequence {𝑥𝑛} is bounded, thus there exists a subsequence {𝑥𝑛𝑘} converging to a point ̃𝑥𝐻. For all 𝑧𝐹(𝑇), it follows from (3.39) that (𝐴𝛾𝑓)𝑥𝑛𝑘,𝑥𝑛𝑘=𝑧1𝛼𝑛𝑘𝛽𝑛𝑘𝛼𝑛𝑘𝑣𝑛𝑘,𝑥𝑛𝑘1𝑧𝛼𝑛𝑘(𝐼𝑇)𝑦𝑛𝑘,𝑥𝑛𝑘𝛽𝑧𝑛𝑘𝛼𝑛𝑘(𝐼𝑆)𝑥𝑛𝑘,𝑥𝑛𝑘𝐴𝑧𝑇𝑦𝑛𝑘𝑥𝑛𝑘,𝑥𝑛𝑘1𝑧𝛼𝑛𝑘𝑤𝑛𝑘𝑃𝐶𝑤𝑛𝑘,𝑃𝐶𝑤𝑛𝑘1𝑧1𝛼𝑛𝑘𝛽𝑛𝑘𝛼𝑛𝑘𝑣𝑛𝑘,𝑥𝑛𝑘1𝑧𝛼𝑛𝑘(𝐼𝑇)𝑦𝑛𝑘,𝑥𝑛𝑘𝛽𝑧𝑛𝑘𝛼𝑛𝑘(𝐼𝑆)𝑥𝑛𝑘,𝑥𝑛𝑘𝐴𝑧𝑇𝑦𝑛𝑘𝑥𝑛𝑘,𝑥𝑛𝑘1𝑧𝛼𝑛𝑘𝑤𝑛𝑘𝑃𝐶𝑤𝑛𝑘,𝑃𝐶𝑤𝑛𝑘1𝑃𝐶𝑤𝑛𝑘𝐴𝑇𝑦𝑛𝑘𝑥𝑛𝑘,𝑥𝑛𝑘1𝑧𝛼𝑛𝑘𝑤𝑛𝑘𝑃𝐶𝑤𝑛𝑘,𝑃𝐶𝑤𝑛𝑘1𝑧1𝛼𝑛𝑘𝛽𝑛𝑘𝛼𝑛𝑘𝑣𝑛𝑘,𝑥𝑛𝑘1𝑧𝛼𝑛𝑘(𝐼𝑇)𝑦𝑛𝑘,𝑥𝑛𝑘𝛽𝑧𝑛𝑘𝛼𝑛𝑘(𝐼𝑆)𝑥𝑛𝑘,𝑥𝑛𝑘𝐴𝑧𝑇𝑦𝑛𝑘𝑥𝑛𝑘,𝑥𝑛𝑘+𝑤𝑧𝑛𝑘𝑤𝑛𝑘1𝛼𝑛𝑘𝑤𝑛𝑘𝑃𝐶𝑤𝑛𝑘.(3.43) Letting 𝑘, we obtain (𝐴𝛾𝑓)̃𝑥,̃𝑥𝑧𝜏(𝐼𝑆)̃𝑥,̃𝑥𝑧,𝑧𝐹(𝑇).(3.44) By Lemma 2.6 and (3.27) having the unique solution, it follows that 𝜔𝑤(𝑥𝑛)={̃𝑥}. Therefore, 𝑥𝑛̃𝑥 as 𝑛. This completes the proof.

From Theorem 3.3, we can deduce the following interesting corollary.

Corollary 3.4 (Yao et al. [9]). Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻. Let 𝑓𝐶𝐻 be a 𝜌-contraction (possibly nonself) with 𝜌(0,1). Let 𝑆,𝑇𝐶𝐶 be two nonexpansive mappings with 𝐹(𝑇). {𝛼𝑛} and {𝛽𝑛} are two sequences in (0,1) Starting with an arbitrary initial guess 𝑥0𝐶 and {𝑥𝑛} is a sequence generated by 𝑦𝑛=𝛽𝑛𝑆𝑥𝑛+1𝛽𝑛𝑥𝑛,𝑥𝑛+1=𝑃𝐶𝛼𝑛𝑓𝑥𝑛+1𝛼𝑛𝑇𝑦𝑛,𝑛1.(3.45) Suppose that the following conditions are satisfied: (C1)  lim𝑛𝛼𝑛=0 and 𝑛=1𝛼𝑛=, (C2) lim𝑛(𝛽𝑛/𝛼𝑛)=𝜏(0,), (C5)  lim𝑛((|𝛼𝑛𝛼𝑛1|+|𝛽𝑛𝛽𝑛1|)/𝛼𝑛𝛽𝑛)=0, (C6)  there exists a constant 𝐾>0 such that (1/𝛼𝑛)|1/𝛽𝑛1/𝛽𝑛1|𝐾. Then the sequence {𝑥𝑛} converges strongly to a point ̃𝑥𝐻, which is the unique solution of the variational inequality: 1̃𝑥𝐹(𝑇),𝜏(𝐼𝑓)̃𝑥+(𝐼𝑆)̃𝑥,𝑥̃𝑥0,𝑥𝐹(𝑇).(3.46)

Proof. As a matter of fact, if we take 𝐴=𝐼 and 𝛾=1 in Theorem 3.3 then this completes the proof.

Corollary 3.5 (Yao et al. [9]). Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻. Let 𝑆,𝑇𝐶𝐶 be two nonexpansive mappings with 𝐹(𝑇). {𝛼𝑛} and {𝛽𝑛} are two sequences in (0,1). Starting with an arbitrary initial guess 𝑥0𝐶 and suppose {𝑥𝑛} is a sequence generated by 𝑦𝑛=𝛽𝑛𝑆𝑥𝑛+1𝛽𝑛𝑥𝑛,𝑥𝑛+1=𝑃𝐶1𝛼𝑛𝑇𝑦𝑛,𝑛1.(3.47) Suppose that the following conditions are satisfied: (C1)  lim𝑛𝛼𝑛=0 and 𝑛=1𝛼𝑛=, (C2)  lim𝑛(𝛽𝑛/𝛼𝑛)=1, (C5)  lim𝑛((|𝛼𝑛𝛼𝑛1|+|𝛽𝑛𝛽𝑛1|)/𝛼𝑛𝛽𝑛)=0, (C6)  there exists a constant 𝐾>0 such that (1/𝛼𝑛)|1/𝛽𝑛1/𝛽𝑛1|𝐾. Then the sequence {𝑥𝑛} converges strongly to a point ̃𝑥𝐻, which is the unique solution of the variational inequality: 𝑆̃𝑥𝐹(𝑇),𝐼2̃𝑥,𝑥̃𝑥0,𝑥𝐹(𝑇).(3.48)

Proof. As a matter of fact, if we take 𝐴=𝐼, 𝑓=0, and 𝛾=1 in Theorem 3.3 then this is completes the proof.

Remark 3.6. Prototypes for the iterative parameters are, for example, 𝛼𝑛=𝑛𝜃 and 𝛽𝑛=𝑛𝜔 (with 𝜃,𝜔>0). Since |𝛼𝑛𝛼𝑛1|𝑛𝜃 and |𝛽𝑛𝛽𝑛1|𝑛𝜔, it is not difficult to prove that (C5) is satisfied for 0<𝜃,𝜔<1 and (C6) is satisfied if 𝜃+𝜔1.

Remark 3.7. Our results improve and extend the results of Yao et al. [9] by taking 𝐴=𝐼 and 𝛾=1 in Theorems 3.1 and 3.3.

The following is an example to support Theorem 3.3.

Example 3.8. Let 𝐻=,𝐶=[1/4,1/4],𝑇=𝐼,𝑆=𝐼,𝐴=𝐼, 𝑓(𝑥)=𝑥2, 𝑃𝐶=𝐼,𝛽𝑛=1/𝑛, 𝛼𝑛=1/𝑛 for every 𝑛, we have 𝜏=1 and choose 𝛾=1/2, 𝜌=1/3 and 𝛾=1. Then {𝑥𝑛} is the sequence 𝑥𝑛+1=𝑥2𝑛𝑛+11𝑛21𝑛𝑥𝑛,(3.49) and 𝑥𝑛̃𝑥=0 as 𝑛, where ̃𝑥=0 is the unique solution of the variational inequality 1̃𝑥𝐹(𝑇)=4,14,3̃𝑥̃𝑥21,𝑥̃𝑥0,𝑥𝐹(𝑇)=4,14.(3.50)

Acknowledgments

The authors would like to thank the National Research University Project of Thailand’s Office of the Higher Education Commission under the Project NRU-CSEC no. 55000613 for financial support.

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