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Discrete Dynamics in Nature and Society
Volume 2011 (2011), Article ID 976505, 23 pages
http://dx.doi.org/10.1155/2011/976505
Research Article

Generalized Systems of Variational Inequalities and Projection Methods for Inverse-Strongly Monotone Mappings

1Department of Mathematics, Faculty of Science and Technology, Rajamangala University of Technology Thanyaburi (RMUTT), Thanyaburi, Pathumthani 12110, Thailand
2Department of Mathematics, Faculty of Science, King Mongkut's University of Technology Thonburi (KMUTT), Bangmod, Bangkok 10140, Thailand
3Centre of Excellence in Mathematics, CHE, Si Ayutthaya Road, Bangkok 10400, Thailand

Received 20 February 2011; Accepted 3 May 2011

Academic Editor: Jianshe Yu

Copyright © 2011 Wiyada Kumam 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 introduce an iterative sequence for finding a common element of the set of fixed points of a nonexpansive mapping and the solutions of the variational inequality problem for three inverse-strongly monotone mappings. Under suitable conditions, some strong convergence theorems for approximating a common element of the above two sets are obtained. Moreover, using the above theorem, we also apply to find solutions of a general system of variational inequality and a zero of a maximal monotone operator in a real Hilbert space. As applications, at the end of the paper we utilize our results to study some convergence problem for strictly pseudocontractive mappings. Our results include the previous results as special cases extend and improve the results of Ceng et al., (2008) and many others.

1. Introduction

Variational inequalities are known to play a crucial role in mathematics as a unified framework for studying a large variety of problems arising, for instance, in structural analysis, engineering sciences and others. Roughly speaking, they can be recast as fixed-point problems, and most of the numerical methods related to this topic are based on projection methods. Let 𝐻 be a real Hilbert space with inner product , and , and let 𝐸 be a nonempty, closed, convex subset of 𝐻. A mapping 𝐴𝐸𝐻 is called 𝛼-inverse-strongly monotone if there exists a positive real number 𝛼>0 such that 𝐴𝑥𝐴𝑦,𝑥𝑦𝛼𝐴𝑥𝐴𝑦2,𝑥,𝑦𝐸(1.1) (see [1, 2]). It is obvious that every 𝛼-inverse-strongly monotone mapping 𝐴 is monotone and Lipschitz continuous. A mapping 𝑆𝐸𝐸 is called nonexpansive if 𝑆𝑥𝑆𝑦𝑥𝑦,𝑥,𝑦𝐸.(1.2) We denote by 𝐹(𝑆) the set of fixed points of 𝑆 and by 𝑃𝐸 the metric projection of 𝐻 onto 𝐸. Recall that the classical variational inequality, denoted by 𝑉𝐼(𝐴,𝐸), is to find an 𝑥𝐸 such that 𝐴𝑥,𝑥𝑥0,𝑥𝐸.(1.3) The set of solutions of 𝑉𝐼(𝐴,𝐸) is denoted by Γ. The variational inequality has been widely studied in the literature; see, for example, [36] and the references therein.

For finding an element of 𝐹(𝑆)Γ, Takahashi and Toyoda [7] introduced the following iterative scheme:𝑥𝑛+1=𝛼𝑛𝑥𝑛+1𝛼𝑛𝑆𝑃𝐸𝑥𝑛𝜆𝑛𝐴𝑥𝑛,(1.4) for every 𝑛=0,1,2,, where 𝑥0=𝑥𝐸,{𝛼𝑛} is a sequence in (0,1), and {𝜆𝑛} is a sequence in (0,2𝛼). On the other hand, for solving the variational inequality problem in the finite-dimensional Euclidean space 𝐑𝑛, Korpelevich (1976) [8] introduced the following so-called extragradient method:𝑥0𝑦=𝑥𝐸,𝑛=𝑃𝐸𝑥𝑛𝜆𝑛𝐴𝑥𝑛,𝑥𝑛+1=𝑃𝐸𝑥n𝜆𝑛𝐴𝑦𝑛,(1.5) for every 𝑛=0,1,2,, where 𝜆𝑛(0,1/𝑘). Many authors using extragradient method for approximating common fixed points and variational inequality problems (see also [9, 10]). Recently, Nadezhkina and Takahashi [11] and Zeng et al. [12] proposed some iterative schemes for finding elements in 𝐹(𝑆)Γ by combining (1.4) and (1.5). Further, these iterative schemes are extended in Y. Yao and J. C. Yao [13] to develop a new iterative scheme for finding elements in 𝐹(𝑆)Γ.

Consider the following problem of finding (𝑥,𝑦)𝐸×𝐸 such that (see cf. Ceng et al. [14]):𝜆𝐴𝑦+𝑥𝑦,𝑥𝑥0,𝑥𝐸,𝜇𝐵𝑥+𝑦𝑥,𝑥𝑦0,𝑥𝐸,(1.6) which is called general system of variational inequalities (GSVI), where 𝜆>0 and 𝜇>0 are two constants. In particular, if 𝐴=𝐵, then problem (1.6) reduces to finding (𝑥,𝑦)𝐸×𝐸 such that𝜆𝐴𝑦+𝑥𝑦,𝑥𝑥0,𝑥𝐸,𝜇𝐴𝑥+𝑦𝑥,𝑥𝑦0,𝑥𝐸,(1.7) which is defined by [15, 16], and is called the new system of variational inequalities. Further, if 𝑥=𝑦, then problem (1.7) reduces to the classical variational inequality VI(𝐴,𝐸), that is, find 𝑥𝐸 such that 𝐴𝑥,𝑥𝑥0, forall𝑥𝐸.

We can characteristic problem, if 𝑥𝐹(𝑆)𝑉𝐼(𝐴,𝐸), then it follows that 𝑥=𝑆𝑥=𝑃𝐸[𝑥𝜌𝐴𝑥], where 𝜌>0 is a constant.

In 2008, Ceng et al. [14] introduced a relaxed extragradient method for finding solutions of problem (1.6). Let the mappings 𝐴,𝐵𝐸𝐻 be 𝛼-inverse-strongly monotone and 𝛽-inverse-strongly monotone, respectively. Let 𝑆𝐸𝐸 be a nonexpansive mapping. Suppose 𝑥1=𝑢𝐸 and {𝑥𝑛} is generated by𝑦𝑛=𝑃𝐸𝑥𝑛𝜇𝐵𝑥𝑛,𝑥𝑛+1=𝛼𝑛𝑢+𝛽𝑛𝑥𝑛+𝛾𝑛𝑆𝑃𝐸𝑦𝑛𝜆𝑛A𝑦𝑛,(1.8) where 𝜆(0,2𝛼),𝜇(0,2𝛽), and {𝛼𝑛},{𝛽𝑛},{𝛾𝑛} are three sequences in [0,1] such that 𝛼𝑛+𝛽𝑛+𝛾𝑛=1,forall𝑛1. First, problem (1.6) is proven to be equivalent to a fixed point problem of a nonexpansive mapping.

In this paper, motivated by what is mentioned above, we consider generalized system of variational inequalities as follows.

Let 𝐸 be a nonempty, closed, convex subset of a real Hilbert space 𝐻. Let 𝐴,𝐵,𝐶𝐸𝐻 be three mappings. We consider the following problem of finding (𝑥,𝑦,𝑧)𝐸×𝐸×𝐸 such that𝜆𝐴𝑦+𝑥𝑦,𝑥𝑥0,𝑥𝐸,𝜇𝐵𝑧+𝑦𝑧,𝑥𝑦0,𝑥𝐸,𝜏𝐶𝑥+𝑧𝑥,𝑥𝑧0,𝑥𝐸,(1.9) which is called a general system of variational inequalities where 𝜆>0,𝜇>0 and 𝜏>0 are three constants.

In particular, if 𝐴=𝐵=𝐶, then problem (1.9) reduces to finding (𝑥,𝑦,𝑧)𝐸×𝐸×𝐸 such that𝜆𝐴𝑦+𝑥𝑦,𝑥𝑥0,𝑥𝐸,𝜇𝐴𝑧+𝑦𝑧,𝑥𝑦0,𝑥𝐸,𝜏𝐴𝑥+𝑧𝑥,𝑥𝑧0,𝑥𝐸.(1.10) Next, we consider some special classes of the GSVI problem (1.9) reduce to the following GSVI.(i)If 𝜏=0, then the GSVI problems (1.9) reduce to GSVI problem (1.6).(ii)If 𝜏=𝜇=0, then the GSVI problems (1.9) reduce to classical variational inequality VI(A,E) problem.

The above system enters a class of more general problems which originated mainly from the Nash equilibrium points and was treated from a theoretical viewpoint in [17, 18]. Observe at the same time that, to construct a mathematical model which is as close as possible to a real complex problem, we often have to use constraints which can be expressed as one several subproblems of a general problem. These constrains can be given, for instance, by variational inequalities, by fixed point problems, or by problems of different types.

This paper deals with a relaxed extragradient approximation method for solving a system of variational inequalities over the fixed-point sets of nonexpansive mapping. Under classical conditions, we prove a strong convergence theorem for this method. Moreover, the proposed algorithm can be applied for instance to solving the classical variational inequality problems.

2. Preliminaries

Let 𝐸 be a nonempty, closed, convex subset of a real Hilbert space 𝐻. For every point 𝑥𝐻, there exists a unique nearest point in 𝐸, denoted by 𝑃𝐸𝑥, such that 𝑥𝑃𝐸𝑥𝑥𝑦,𝑦𝐸.(2.1)𝑃𝐸 is call the metric projection of 𝐻 onto 𝐸.

Recall that, 𝑃𝐸𝑥 is characterized by following properties: 𝑃𝐸𝑥𝐸 and 𝑥𝑃𝐸𝑥,𝑦𝑃𝐸𝑥0,𝑥𝑦2𝑥𝑃𝐸𝑥2+𝑃𝐸𝑥𝑦2,(2.2) for all 𝑥𝐻 and 𝑦𝐸.

Lemma 2.1 (see cf. Zhang et al. [19]). The metric projection 𝑃𝐸 has the following properties: (i)𝑃𝐸𝐻𝐸 is nonexpansive;(ii)𝑃𝐸𝐻𝐸 is firmly nonexpansive, that is, 𝑃𝐸𝑥𝑃𝐸𝑦2𝑃𝐸𝑥𝑃𝐸𝑦,𝑥𝑦,𝑥,𝑦𝐻;(2.3)(iii)for each 𝑥𝐻, 𝑧=𝑃𝐸(𝑥)𝑥𝑧,𝑧𝑦0,𝑦𝐸.(2.4)

Lemma 2.2 (see Osilike and Igbokwe [20]). Let (𝐸,,) be an inner product space. Then for all 𝑥,𝑦,𝑧𝐸 and 𝛼,𝛽,𝛾[0,1] with 𝛼+𝛽+𝛾=1, one has 𝛼𝑥+𝛽𝑦+𝛾𝑧2=𝛼𝑥2+𝛽𝑦2+𝛾𝑧2𝛼𝛽𝑥𝑦2𝛼𝛾𝑥𝑧2𝛽𝛾𝑦𝑧2.(2.5)

Lemma 2.3 (see Suzuki [21]). Let {𝑥𝑛} and {𝑦𝑛} be bounded sequences in a Banach space 𝑋 and let {𝛽𝑛} be a sequence in [0,1] with 0<liminf𝑛𝛽𝑛limsup𝑛𝛽𝑛<1. Suppose 𝑥𝑛+1=(1𝛽𝑛)𝑦𝑛+𝛽𝑛𝑥𝑛 for all integers 𝑛0 and limsup𝑛(𝑦𝑛+1y𝑛𝑥𝑛+1𝑥𝑛)0. Then, lim𝑛𝑦𝑛𝑥𝑛=0.

Lemma 2.4 (see Xu [22]). Assume {𝑎𝑛} is a sequence of nonnegative real numbers such that 𝑎𝑛+11𝛼𝑛𝑎𝑛+𝛿𝑛,𝑛0,(2.6) where {𝛼𝑛} is a sequence in (0,1) and {𝛿𝑛} is a sequence in 𝐑 such that (i)𝑛=1𝛼𝑛=, (ii)limsup𝑛(𝛿𝑛/𝛼𝑛)0 or 𝑛=1|𝛿𝑛|<. Then, lim𝑛𝑎𝑛=0.

Lemma 2.5 (Goebel and Kirk [23]). Demiclosedness Principle. Assume that 𝑇 is a nonexpansive self-mapping of a nonempty, closed, convex subset 𝐸 of a real Hilbert space 𝐻. If 𝑇 has a fixed point, then 𝐼𝑇 is demiclosed; that is, whenever {𝑥𝑛} is a sequence in 𝐸 converging weakly to some 𝑥𝐸 (for short, 𝑥𝑛𝑥𝐸), and the sequence {(𝐼𝑇)𝑥𝑛} converges strongly to some 𝑦 (for short, (𝐼𝑇)𝑥𝑛𝑦), it follows that (𝐼𝑇)𝑥=𝑦. Here, 𝐼 is the identity operator of 𝐻.

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

Lemma 2.6. In a real Hilbert space 𝐻, there holds the inequality 𝑥+𝑦2𝑥2+2𝑦,𝑥+𝑦,𝑥,𝑦𝐻.(2.7)

Remark 2.7. We also have that, for all 𝑢,𝑣𝐸 and 𝜆>0, (𝐼𝜆𝐴)𝑢(𝐼𝜆𝐴)𝑣2=(𝑢𝑣)𝜆(𝐴𝑢𝐴𝑣)2=𝑢𝑣22𝜆𝑢𝑣,𝐴𝑢𝐴𝑣+𝜆2𝐴𝑢𝐴𝑣2𝑢𝑣2+𝜆(𝜆2𝛼)𝐴𝑢𝐴𝑣2.(2.8) So, if 𝜆2𝛼, then 𝐼𝜆𝐴 is a nonexpansive mapping from 𝐸 to 𝐻.

3. Main Results

In this section, we introduce an iterative precess by the relaxed extragradient approximation method for finding a common element of the set of fixed points of a nonexpansive mapping and the solution set of the variational inequality problem for three inverse-strongly monotone mappings in a real Hilbert space. We prove that the iterative sequence converges strongly to a common element of the above two sets.

In order to prove our main result, the following lemmas are needed.

Lemma 3.1. For given 𝑥,𝑦,𝑧𝐸×𝐸×𝐸,(𝑥,y,𝑧) is a solution of problem (1.9) if and only if 𝑥 is a fixed point of the mapping 𝐺𝐸𝐸 defined by 𝐺(𝑥)=𝑃𝐸𝑃𝐸𝑃𝐸(𝑥𝜏𝐶𝑥)𝜇𝐵𝑃𝐸(𝑥𝜏𝐶𝑥)𝜆𝐴𝑃𝐸𝑃𝐸(𝑥𝜏𝐶𝑥)𝜇𝐵𝑃𝐸(𝑥𝜏𝐶𝑥),𝑥𝐸,(3.1) where 𝑦=𝑃𝐸(𝑧𝜇𝐵𝑧) and 𝑧=𝑃𝐸(𝑥𝜏𝐶𝑥).

Proof. 𝜆𝐴𝑦+𝑥𝑦,𝑥𝑥0,𝑥𝐸,𝜇𝐵𝑧+𝑦𝑧,𝑥𝑦0,𝑥𝐸,𝜏𝐶𝑥+𝑧𝑥,𝑥𝑧0,𝑥𝐸,(3.2)𝑦+𝜆𝐴𝑦+𝑥,𝑥𝑥0,𝑥𝐸,𝑧+𝜇𝐵𝑧+𝑦,𝑥𝑦0,𝑥𝐸,𝑥+𝜏𝐶𝑥+𝑧,𝑥𝑧0,𝑥𝐸,(3.3)𝑦𝜆𝐴𝑦𝑥,𝑥𝑧𝑥0,𝑥𝐸,𝜇𝐵𝑧𝑦,𝑦𝑥𝑥0,𝑥𝐸,𝜏𝐶𝑥𝑧,𝑧𝑥0,𝑥𝐸,(3.4)x=𝑃𝐸𝑦𝜆𝐴𝑦,𝑦=𝑃𝐸𝑧𝜇𝐵𝑧,𝑧=𝑃𝐸𝑥𝜏𝐶𝑥,(3.5)𝑥=𝑃𝐸[𝑃𝐸(𝑧𝜇𝐵𝑧)𝜆𝐴𝑃𝐸(𝑧𝜇𝐵𝑧)].
Thus, 𝑥=𝑃𝐸𝑃𝐸𝑃𝐸𝑥𝜏𝐶𝑥𝜇𝐵𝑃𝐸𝑥𝜏𝐶𝑥𝜆𝐴𝑃𝐸𝑃𝐸𝑥𝜏𝐶𝑥𝜇𝐵𝑃𝐸𝑥𝜏𝐶𝑥.(3.6)

Lemma 3.2. The mapping 𝐺 defined by Lemma 3.1 is nonexpansive mappings.

Proof. For all 𝑥,𝑦𝐸, 𝑃𝐺(𝑥)𝐺(𝑦)=𝐸𝑃𝐸𝑃𝐸(𝑥𝜏𝐶𝑥)𝜇𝐵𝑃𝐸(𝑥𝜏𝐶𝑥)𝜆𝐴𝑃𝐸𝑃𝐸(𝑥𝜏𝐶𝑥)𝜇𝐵𝑃𝐸(𝑥𝜏𝐶𝑥)𝑃𝐸𝑃𝐸𝑃𝐸(𝑦𝜏𝐶𝑦)𝜇𝐵𝑃𝐸(𝑦𝜏𝐶𝑦)𝜆𝐴𝑃𝐸𝑃𝐸(𝑦𝜏𝐶𝑦)𝜇𝐵𝑃𝐸(𝑃𝑦𝜏𝐶𝑦)𝐸(𝑥𝜏𝐶𝑥)𝜇𝐵𝑃𝐸(𝑥𝜏𝐶𝑥)𝜆𝐴𝑃𝐸𝑃𝐸(𝑥𝜏𝐶𝑥)𝜇𝐵𝑃𝐸(𝑃𝑥𝜏𝐶𝑥)𝐸(𝑦𝜏𝐶𝑦)𝜇𝐵𝑃𝐸(𝑦𝜏𝐶𝑦)𝜆𝐴𝑃𝐸𝑃𝐸(𝑦𝜏𝐶𝑦)𝜇𝐵𝑃𝐸=𝑃(𝑦𝜏𝐶𝑦)(𝐼𝜆𝐴)𝐸(𝑥𝜏𝐶𝑥)𝜇𝐵𝑃𝐸𝑃(𝑥𝜏𝐶𝑥)(𝐼𝜆𝐴)𝐸(𝑦𝜏𝐶𝑦)𝜇𝐵𝑃𝐸𝑃(𝑦𝜏𝐶𝑦)𝐸(𝑥𝜏𝐶𝑥)𝜇𝐵𝑃𝐸𝑃(𝑥𝜏𝐶𝑥)𝐸(𝑦𝜏𝐶𝑦)𝜇𝐵𝑃𝐸=𝑃(𝑦𝜏𝐶𝑦)(𝐼𝜇𝐵)𝐸𝑃(𝑥𝜏𝐶𝑥)(𝐼𝜇𝐵)𝐸𝑃(𝑦𝜏𝐶𝑦)𝐸(𝑥𝜏𝐶𝑥)𝑃𝐸(𝑦𝜏𝐶𝑦)(𝑥𝜏𝐶𝑥)(𝑦𝜏𝐶𝑦)=(𝐼𝜏𝐶)(𝑥)(𝐼𝜏𝐶)(𝑦)𝑥𝑦.(3.7) This shows that 𝐺𝐸𝐸 is a nonexpansive mapping.

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

Now, we are ready to proof our main results in this paper.

Theorem 3.3. Let 𝐸 be a nonempty, closed, convex subset of a real Hilbert space 𝐻. Let the mapping 𝐴,𝐵,𝐶𝐸𝐻 be 𝛼-inverse-strongly monotone, 𝛽-inverse-strongly monotone, and 𝛾-inverse-strongly monotone, respectively. Let 𝑆 be a nonexpansive mapping of 𝐸 into itself such that 𝐹(𝑆)Υ. Let 𝑓 be a contraction of 𝐻 into itself and given 𝑥1𝐻 arbitrarily and {𝑥𝑛} is generated by 𝑧𝑛=𝑃𝐸𝑥𝑛𝜏𝐶𝑥𝑛,𝑦𝑛=𝑃𝐸𝑧𝑛𝜇𝐵𝑧𝑛,𝑥𝑛+1=𝛼𝑛𝑓𝑥𝑛+𝛽𝑛𝑥𝑛+𝛾𝑛𝑆𝑃𝐸𝑦𝑛𝜆𝐴𝑦𝑛,𝑛0,(3.8) where 𝜆(0,2𝛼),𝜇(0,2𝛽),𝜏(0,2𝛾), and {𝛼𝑛},{𝛽𝑛},{𝛾𝑛} are three sequences in [0,1] such that (i)𝛼𝑛+𝛽𝑛+𝛾𝑛=1,(ii)lim𝑛𝛼𝑛=0 and 𝑛=1𝛼𝑛=,(iii)0<liminf𝑛𝛽𝑛limsup𝑛𝛽𝑛<1. Then, {𝑥𝑛} converges strongly to 𝑥𝐹(𝑆)Γ, where 𝑥=𝑃𝐹(𝑆)Γ𝑓(𝑥) and (𝑥,𝑦,𝑧) is a solution of problem (1.9), where 𝑦=𝑃𝐸𝑧𝜇𝐵𝑧,𝑧=𝑃𝐸𝑥𝜏𝐶𝑥.(3.9)

Proof. Let 𝑥𝐹(𝑆)Γ. Then, 𝑥=𝑆𝑥 and 𝑥=𝐺𝑥, that is, 𝑥=𝑃𝐸𝑃𝐸𝑃𝐸𝑥𝜏𝐶𝑥𝜇𝐵𝑃𝐸𝑥𝜏𝐶𝑥𝜆𝐴𝑃𝐸𝑃𝐸𝑥𝜏𝐶𝑥𝜇𝐵𝑃𝐸𝑥𝜏𝐶𝑥.(3.10) Put 𝑥=𝑃𝐸(y𝜆𝐴𝑦) and 𝑡𝑛=𝑃𝐸(𝑦𝑛𝜆𝐴𝑦𝑛). Then, 𝑥=𝑃𝐸[𝑃𝐸(𝑧𝜇𝐵𝑧)𝜆𝐴𝑃𝐸(𝑧𝜇𝐵𝑧)] implies that 𝑦=𝑃𝐸(𝑧𝜇𝐵𝑧), where 𝑧=𝑃𝐸(𝑥𝜏𝐶𝑥). Since 𝐼𝜆𝐴, 𝐼𝜇𝐵 and 𝐼𝜏𝐶 are nonexpansive mappings. We obtain that 𝑡𝑛𝑥=𝑃𝐸𝑦𝑛𝜆𝐴𝑦𝑛𝑥=𝑃𝐸𝑦𝑛𝜆𝐴𝑦𝑛𝑃𝐸𝑦𝜆𝐴𝑦𝑦𝑛𝜆𝐴𝑦𝑛𝑦𝜆𝐴𝑦=(𝐼𝜆𝐴)𝑦𝑛(𝐼𝜆𝐴)𝑦𝑦𝑛𝑦=𝑦(3.11)𝑛𝑃𝐸𝑧𝜇𝐵𝑧=𝑃𝐸𝑧𝑛𝜇𝐵𝑧𝑛𝑃𝐸𝑧𝜇𝐵𝑧(𝐼𝜇𝐵)𝑧𝑛(𝐼𝜇𝐵)𝑧𝑧𝑛𝑧,𝑧(3.12)𝑛𝑧=𝑃𝐸𝑥𝑛𝜏𝐶𝑥𝑛𝑃𝐸𝑥𝜏𝐶𝑥𝑥𝑛𝜏𝐶𝑥𝑛𝑥𝜏𝐶𝑥=(𝐼𝜏𝐶)𝑥𝑛(𝐼𝜏𝐶)𝑥𝑥𝑛𝑥.(3.13) Substituting (3.13) into (3.12), we have 𝑡𝑛𝑥𝑥𝑛𝑥,(3.14) and by (3.11) we also have 𝑦𝑛𝑦𝑥𝑛𝑥.(3.15) Since 𝑥𝑛+1=𝛼𝑛𝑓(𝑥𝑛)+𝛽𝑛𝑥𝑛+𝛾𝑛𝑆𝑡𝑛 and by Lemma 2.2, we compute 𝑥𝑛+1𝑥=𝛼𝑛𝑓𝑥𝑛+𝛽𝑛𝑥𝑛+𝛾𝑛𝑆𝑡𝑛𝑥=𝛼𝑛𝑓𝑥𝑛𝑥+𝛽𝑛𝑥𝑛𝑥+𝛾𝑛𝑆𝑡𝑛𝑥𝛼𝑛𝑓𝑥𝑛𝑥+𝛽𝑛𝑥𝑛𝑥+𝛾𝑛𝑆𝑡𝑛𝑥𝛼𝑛𝑓𝑥𝑛𝑥+𝛽𝑛𝑥𝑛𝑥+𝛾𝑛𝑡𝑛𝑥𝛼𝑛𝑓𝑥𝑛𝑥+𝛽𝑛𝑥𝑛𝑥+𝛾𝑛𝑥𝑛𝑥=𝛼𝑛𝑓𝑥𝑛𝑥+1𝛼𝑛𝑥𝑛𝑥=𝛼𝑛𝑓𝑥𝑛𝑥𝑓𝑥+𝑓𝑥+1𝛼𝑛𝑥𝑛𝑥𝛼𝑛𝑓𝑥𝑛𝑥𝑓+𝛼𝑛𝑓𝑥𝑥+1𝛼𝑛𝑥𝑛𝑥𝛼𝑛𝑘𝑥𝑛𝑥+𝛼𝑛𝑓𝑥𝑥+1𝛼𝑛𝑥𝑛𝑥=𝛼𝑛𝑘+1𝛼𝑛𝑥𝑛𝑥+𝛼𝑛𝑓𝑥𝑥=1𝛼𝑛𝑥(1𝑘)𝑛𝑥+𝛼𝑛𝑓𝑥𝑥=1𝛼𝑛(𝑥1𝑘)𝑛𝑥+𝛼𝑛(𝑓𝑥1𝑘)𝑥.(1𝑘)(3.16) By induction, we get 𝑥𝑛+1𝑥𝑀,(3.17) where 𝑀=max{𝑥0𝑥+(1/(1𝑘))𝑓(𝑥)𝑥},𝑛0. Therefore, {𝑥𝑛}is bounded. Consequently, by (3.11), (3.12) and (3.13), the sequences {𝑡𝑛},{𝑆𝑡𝑛},{𝑦𝑛},{𝐴𝑦𝑛},{𝑧𝑛},{𝐵𝑧𝑛},{𝐶𝑥𝑛}, and {𝑓(𝑥𝑛)} are also bounded. Also, we observe that 𝑧𝑛+1𝑧𝑛=𝑃𝐸𝑥𝑛+1𝜏𝐶𝑥𝑛+1𝑃𝐸𝑥𝑛𝜏𝐶𝑥𝑛(𝐼𝜏𝐶)𝑥𝑛+1(𝐼𝜏𝐶)𝑥𝑛𝑥𝑛+1𝑥𝑛,𝑡(3.18)𝑛+1𝑡𝑛=𝑃𝐸𝑦𝑛+1𝜆𝐴𝑦𝑛+1𝑃𝐸𝑦𝑛𝜆𝐴𝑦𝑛𝑦𝑛+1𝜆𝐴𝑦𝑛+1𝑦𝑛𝜆𝐴𝑦𝑛=(𝐼𝜆𝐴)𝑦𝑛+1(𝐼𝜆𝐴)𝑦𝑛𝑦𝑛+1𝑦𝑛=𝑃𝐸𝑧𝑛+1𝜇𝐵𝑧𝑛+1𝑃𝐸𝑧𝑛𝜇𝐵𝑧𝑛𝑧𝑛+1𝑧𝑛𝑥𝑛+1𝑥𝑛.(3.19) Let 𝑥𝑛+1=(1𝛽𝑛)𝑤𝑛+𝛽𝑛𝑥𝑛. Thus, we get 𝑤𝑛=𝑥𝑛+1𝛽𝑛𝑥𝑛1𝛽𝑛=𝛼𝑛𝑓𝑥𝑛+𝛾𝑛𝑆𝑃𝐶𝑦𝑛𝜆𝑛𝐴𝑦𝑛1𝛽𝑛=𝛼𝑛𝑢+𝛾𝑛𝑆𝑡𝑛1𝛽𝑛(3.20) It follows that 𝑤𝑛+1𝑤𝑛=𝛼𝑛+1𝑓𝑥𝑛+1+𝛾𝑛+1𝑆𝑡𝑛+11𝛽𝑛+1𝛼𝑛𝑓𝑥𝑛+𝛾𝑛𝑆𝑡𝑛1𝛽𝑛=𝛼𝑛+1𝑓𝑥𝑛+11𝛽𝑛+1+𝛾𝑛+1𝑆𝑡𝑛+11𝛽𝑛+1𝛼𝑛+1𝑓𝑥𝑛1𝛽𝑛+1+𝛼𝑛+1𝑓𝑥𝑛1𝛽𝑛+1𝛼𝑛𝑓𝑥𝑛1𝛽𝑛𝛾𝑛𝑆𝑡𝑛1𝛽𝑛=𝛼𝑛+11𝛽𝑛+1𝑓𝑥𝑛+1𝑥𝑓𝑛+𝛼𝑛+11𝛽𝑛+1𝛼𝑛1𝛽𝑛𝑓𝑥𝑛+𝛾𝑛+1𝑆𝑡𝑛+11𝛽𝑛+1𝛾𝑛𝑆𝑡𝑛1𝛽𝑛=𝛼𝑛+11𝛽𝑛+1𝑓𝑥𝑛+1𝑥𝑓𝑛+𝛼𝑛+11𝛽𝑛+1𝛼𝑛1𝛽𝑛𝑓𝑥𝑛+𝛾𝑛+1𝑆𝑡𝑛+11𝛽𝑛+1𝛾𝑛+1𝑆𝑡𝑛1𝛽𝑛+1+𝛾𝑛+1𝑆𝑡𝑛1𝛽𝑛+1𝛾𝑛𝑆𝑡𝑛1𝛽𝑛=𝛼𝑛+11𝛽𝑛+1𝑓𝑥𝑛+1𝑥𝑓𝑛+𝛼𝑛+11𝛽𝑛+1𝛼𝑛1𝛽𝑛𝑓𝑥𝑛+𝛾𝑛+11𝛽𝑛+1𝑆𝑡𝑛+1𝑆𝑡𝑛+𝛾𝑛+11𝛽𝑛+1𝛾𝑛1𝛽𝑛𝑆𝑡𝑛=𝛼𝑛+11𝛽𝑛+1𝑓𝑥𝑛+1𝑥𝑓𝑛+𝛼𝑛+11𝛽𝑛+1𝛼𝑛1𝛽𝑛𝑓𝑥𝑛+𝛼𝑛+11𝛽𝑛+1𝛼𝑛1𝛽𝑛𝑆𝑡𝑛+𝛾𝑛+11𝛽𝑛+1𝑆𝑡𝑛+1𝑆𝑡𝑛=𝛼𝑛+11𝛽𝑛+1𝑓𝑥𝑛+1𝑥𝑓𝑛+𝛼𝑛+11𝛽𝑛+1𝛼𝑛1𝛽𝑛𝑓𝑥𝑛+𝑆𝑡𝑛+𝛾𝑛+11𝛽𝑛+1𝑆𝑡𝑛+1𝑆𝑡𝑛.(3.21) Combining (3.19) and (3.21), we obtain 𝑤𝑛+1𝑤𝑛𝑥𝑛+1𝑥𝑛||||𝛼𝑛+11𝛽𝑛+1||||𝑓𝑥𝑛+1𝑥𝑓𝑛+||||𝛼𝑛+11𝛽𝑛+1𝛼𝑛1𝛽𝑛||||𝑓𝑥𝑛+𝑆𝑡𝑛+||||𝛾𝑛+11𝛽𝑛+1||||𝑆𝑡𝑛+1𝑆𝑡𝑛𝑥𝑛+1𝑥𝑛||||𝛼𝑛+11𝛽𝑛+1||||𝑘𝑥𝑛+1𝑥𝑛+||||𝛼𝑛+11𝛽𝑛+1𝛼𝑛1𝛽𝑛||||𝑓𝑥𝑛+𝑆𝑡𝑛+||||𝛾𝑛+11𝛽𝑛+1||||𝑡𝑛+1𝑡𝑛𝑥𝑛+1𝑥𝑛||||𝛼𝑛+11𝛽𝑛+1||||𝑘𝑥𝑛+1𝑥𝑛+||||𝛼𝑛+11𝛽𝑛+1𝛼𝑛1𝛽𝑛||||𝑓𝑥𝑛+𝑆𝑡𝑛+||||𝛾𝑛+11𝛽𝑛+1||||𝑥𝑛+1𝑥𝑛𝑥𝑛+1𝑥𝑛=||||𝛼𝑛+11𝛽𝑛+1||||𝑘𝑥𝑛+1𝑥𝑛+||||𝛼𝑛+11𝛽𝑛+1𝛼𝑛1𝛽𝑛||||𝑓𝑥𝑛+𝑆𝑡𝑛+||||𝛾𝑛+11+𝛽𝑛+11𝛽𝑛+1||||𝑥𝑛+1𝑥𝑛=||||𝛼𝑛+11𝛽𝑛+1||||𝑘𝑥𝑛+1𝑥𝑛+||||𝛼𝑛+11𝛽𝑛+1𝛼𝑛1𝛽𝑛||||𝑓𝑥𝑛+𝑆𝑡𝑛+||||𝛼𝑛+11𝛽𝑛+1||||𝑥𝑛+1𝑥𝑛.(3.22) This together with (i), (ii), and (iii) implies that limsup𝑛𝑤𝑛+1𝑤𝑛𝑥𝑛+1𝑥𝑛0.(3.23) Hence, by Lemma 2.3, we have lim𝑛𝑤𝑛𝑥𝑛=0.(3.24) Consequently, lim𝑛𝑥𝑛+1𝑥𝑛=lim𝑛1𝛽𝑛𝑤𝑛𝑥𝑛=0.(3.25) From (3.18) and (3.19), we also have 𝑧𝑛+1𝑧𝑛0𝑡𝑛+1𝑡𝑛0 and 𝑦𝑛+1𝑦𝑛0 as 𝑛. Since 𝑥𝑛+1𝑥𝑛=𝛼𝑛𝑓𝑥𝑛+𝛽𝑛𝑥𝑛+𝛾𝑛𝑆𝑡𝑛𝑥𝑛=𝛼𝑛𝑓𝑥𝑛𝑥𝑛+𝛾𝑛𝑆𝑡𝑛𝑥𝑛,(3.26) it follows by (ii) and (3.25) that lim𝑛𝑥𝑛𝑆𝑡𝑛=0.(3.27) Since 𝑥𝐹(𝑆)Γ, from (3.15) and Lemma 2.2, we get 𝑥𝑛+1𝑥2=𝛼𝑛𝑓𝑥𝑛+𝛽𝑛𝑥𝑛+𝛾𝑛𝑆𝑡𝑛𝑥2𝛼𝑛𝑓𝑥𝑛𝑥2+𝛽𝑛𝑥𝑛𝑥2+𝛾𝑛𝑆𝑡𝑛𝑥2𝛼𝑛𝑓𝑥𝑛𝑥2+𝛽𝑛𝑥𝑛𝑥2+𝛾𝑛𝑡𝑛𝑥2=𝛼𝑛𝑓𝑥𝑛𝑥2+𝛽𝑛𝑥𝑛𝑥2+𝛾𝑛𝑃𝐸𝑦𝑛𝜆𝐴𝑦𝑛𝑃𝐸𝑦𝜆𝐴𝑦2𝛼𝑛𝑓𝑥𝑛𝑥2+𝛽𝑛𝑥𝑛𝑥2+𝛾𝑛𝑦𝑛𝜆𝐴𝑦𝑛𝑦𝜆𝐴𝑦2=𝛼𝑛𝑓𝑥𝑛𝑥2+𝛽𝑛𝑥𝑛𝑥2+𝛾𝑛𝑦𝑛𝑦𝜆𝐴𝑦𝑛𝐴𝑦2=𝛼𝑛𝑓𝑥𝑛𝑥2+𝛽𝑛𝑥𝑛𝑥2+𝛾𝑛𝑦𝑛𝑦22𝜆𝑦𝑛𝑦,𝐴𝑦𝑛𝐴𝑦+𝜆2𝐴𝑦𝑛𝐴𝑦2𝛼𝑛𝑓𝑥𝑛𝑥2+𝛽𝑛𝑥𝑛𝑥2+𝛾𝑛𝑦𝑛𝑦22𝜆𝛼𝐴𝑦𝑛𝐴𝑦2+𝜆2𝐴𝑦𝑛𝐴𝑦2=𝛼𝑛𝑓𝑥𝑛𝑥2+𝛽𝑛𝑥𝑛𝑥2+𝛾𝑛𝑦𝑛𝑦2+𝜆(𝜆2𝛼)𝐴𝑦𝑛𝐴𝑦2𝛼𝑛𝑓𝑥𝑛𝑥2+𝛽𝑛𝑥𝑛𝑥2+𝛾𝑛𝑥𝑛𝑥2+𝜆(𝜆2𝛼)𝐴𝑦𝑛𝐴𝑦2=𝛼𝑛𝑓𝑥𝑛𝑥2+𝛽𝑛𝑥𝑛𝑥2+𝛾𝑛𝑥𝑛𝑥2+𝛾𝑛𝜆(𝜆2𝛼)𝐴𝑦𝑛𝐴𝑦2=𝛼𝑛𝑓𝑥𝑛𝑥2+𝛽𝑛+𝛾𝑛𝑥𝑛𝑥2+𝛾𝑛𝜆(𝜆2𝛼)𝐴𝑦𝑛𝐴𝑦2=𝛼𝑛𝑓𝑥𝑛𝑥2+1𝛼𝑛𝑥𝑛𝑥2+𝛾𝑛𝜆(𝜆2𝛼)𝐴𝑦𝑛𝐴𝑦2𝛼𝑛𝑓𝑥𝑛𝑥2+𝑥𝑛𝑥2+𝛾𝑛𝜆(𝜆2𝛼)𝐴𝑦𝑛𝐴𝑦2.(3.28) Therefore, we have 𝛾𝑛𝜆(𝜆2𝛼)𝐴𝑦𝑛𝐴𝑦2𝛼𝑛𝑓𝑥𝑛𝑥2+𝑥𝑛𝑥2𝑥𝑛+1𝑥2=𝛼𝑛𝑓𝑥𝑛𝑥2+𝑥𝑛𝑥+𝑥𝑛+1𝑥×𝑥𝑛𝑥𝑥𝑛+1𝑥𝛼𝑛𝑓𝑥𝑛𝑥2+𝑥𝑛𝑥+𝑥𝑛+1𝑥𝑥𝑛𝑥𝑛+1.(3.29) From (ii), (iii), and 𝑥𝑛+1𝑥𝑛0, as 𝑛, we get 𝐴𝑦𝑛𝐴𝑦0 as 𝑛.
Since 𝑥𝐹(𝑆)Υ, from (3.11) and Lemma 2.2, we get 𝑥𝑛+1𝑥2=𝛼𝑛𝑓𝑥𝑛+𝛽𝑛𝑥𝑛+𝛾𝑛𝑆𝑡𝑛𝑥2𝛼𝑛𝑓𝑥𝑛𝑥2+𝛽𝑛𝑥𝑛𝑥2+𝛾𝑛𝑡𝑛𝑥2𝛼𝑛𝑓𝑥𝑛𝑥2+𝛽𝑛𝑥𝑛𝑥2+𝛾𝑛𝑦𝑛𝑦2=𝛼𝑛𝑓𝑥𝑛𝑥2+𝛽𝑛𝑥𝑛𝑥2+𝛾𝑛𝑃𝐸𝑧𝑛𝜇𝐵𝑧𝑛𝑃𝐸𝑧𝜇𝐵𝑧2𝛼𝑛𝑓𝑥𝑛𝑥2+𝛽𝑛𝑥𝑛𝑥2+𝛾𝑛𝑧𝑛𝜇𝐵𝑧𝑛𝑧𝜇𝐵𝑧2=𝛼𝑛𝑓𝑥𝑛𝑥2+𝛽𝑛𝑥𝑛𝑥2+𝛾𝑛𝑧𝑛𝑧𝜇𝐵𝑧𝑛𝜇𝐵𝑧2𝛼𝑛𝑓𝑥𝑛𝑥2+𝛽𝑛𝑥𝑛𝑥2+𝛾𝑛𝑧𝑛𝑧2+𝜇(𝜇2𝛽)𝐵𝑧𝑛𝐵𝑧2𝛼𝑛𝑓𝑥𝑛𝑥2+𝑥𝑛𝑥2+𝛾𝑛𝜇(𝜇2𝛽)𝐵𝑧𝑛𝐵𝑧2.(3.30)
Thus, we also have 𝛾𝑛𝜇(𝜇2𝛽)𝐵𝑧𝑛𝐵𝑧2𝛼𝑛𝑓𝑥𝑛𝑥2+𝑥𝑛𝑥2𝑥𝑛+1𝑥2=𝛼𝑛𝑓𝑥𝑛𝑥2+𝑥𝑛𝑥+𝑥𝑛+1𝑥×𝑥𝑛𝑥𝑥𝑛+1𝑥𝛼𝑛𝑓𝑥𝑛𝑥2+𝑥𝑛𝑥+𝑥𝑛+1𝑥𝑥𝑛𝑥𝑛+1.(3.31) By again (ii), (iii), and (3.25), we also get 𝐵𝑧𝑛𝐵𝑧0 as 𝑛.
Let 𝑥𝐹(𝑆)Υ; again from (3.12), (3.13) and Lemma 2.2, we get 𝑥𝑛+1𝑥2=𝛼𝑛𝑓𝑥𝑛+𝛽𝑛𝑥𝑛+𝛾𝑛𝑆𝑡𝑛𝑥2𝛼𝑛𝑓𝑥𝑛𝑥2+𝛽𝑛𝑥𝑛𝑥2+𝛾𝑛𝑡𝑛𝑥2𝛼𝑛𝑓𝑥𝑛𝑥2+𝛽𝑛𝑥𝑛𝑥2+𝛾𝑛𝑧𝑛𝑧2𝛼𝑛𝑓𝑥𝑛𝑥2+𝛽𝑛𝑥𝑛𝑥2+𝛾𝑛𝑥𝑛𝜏𝐶𝑥𝑛𝑥𝜏𝐶𝑥2𝛼𝑛𝑓𝑥𝑛𝑥2+𝛽𝑛𝑥𝑛𝑥2+𝛾𝑛𝑥𝑛𝑥2+𝜏(𝜏2𝛾)𝐶𝑥𝑛𝐶𝑥2𝛼𝑛𝑓𝑥𝑛𝑥2+𝑥𝑛𝑥2+𝛾𝑛𝜏(𝜏2𝛾)𝐶𝑥𝑛𝐶𝑥2.(3.32)
Again, we have 𝛾𝑛𝜏(𝜏2𝛾)𝐶𝑥𝑛𝐶𝑥2𝛼𝑛𝑓𝑥𝑛𝑥2+𝑥𝑛𝑥2𝑥𝑛+1𝑥2=𝛼𝑛𝑓𝑥𝑛𝑥2+𝑥𝑛𝑥+𝑥𝑛+1𝑥×𝑥𝑛𝑥𝑥𝑛+1𝑥𝛼𝑛𝑓𝑥𝑛𝑥2+𝑥𝑛𝑥+𝑥𝑛+1𝑥𝑥𝑛𝑥𝑛+1.(3.33) Similarly again by (ii), (iii), and 𝑥𝑛𝑥𝑛+10 as 𝑛, and from (3.33), we also that 𝐶𝑥𝑛𝐶𝑥0.
On the other hand, we compute that 𝑧𝑛𝑧2=𝑃𝐸𝑥𝑛𝜏𝐶𝑥𝑛𝑃𝐸𝑥𝜏𝐶𝑥2𝑥𝑛𝜏𝐶𝑥𝑛𝑥𝜏𝐶𝑥,𝑃𝐸𝑥𝑛𝜏𝐶𝑥𝑛𝑃𝐸𝑥𝜏𝐶𝑥=𝑥𝑛𝜏𝐶𝑥𝑛𝑥𝜏𝐶𝑥,𝑧𝑛𝑧=12𝑥𝑛𝜏𝐶𝑥𝑛𝑥𝜏𝐶𝑥2+𝑧𝑛𝑧2𝑥𝑛𝜏𝐶𝑥𝑛𝑥𝜏𝐶𝑥𝑧𝑛𝑧2=12(𝐼𝜏𝐶)𝑥𝑛(𝐼𝜏𝐶)𝑥2+𝑧𝑛𝑧2𝑥𝑛𝜏𝐶𝑥𝑛𝑥𝜏𝐶𝑥𝑧𝑛𝑧212𝑥𝑛𝑥2+𝑧𝑛𝑧2𝑥𝑛𝑧𝑛𝜏𝐶𝑥𝑛𝐶𝑥𝑥𝑧2=12𝑥𝑛𝑥2+𝑧𝑛𝑧2𝑥𝑛𝑧𝑛𝑥𝑧𝜏𝐶𝑥𝑛𝐶𝑥2=12𝑥𝑛𝑥2+𝑧𝑛𝑧2𝑥𝑛𝑧𝑛𝑥𝑧2𝑥+2𝜏𝑛𝑧𝑛𝑥𝑧,𝐶𝑥𝑛𝐶𝑥𝜏2𝐶𝑥𝑛𝐶𝑥2.(3.34)
So, we obtain 𝑧𝑛𝑧2𝑥𝑛𝑥2𝑥𝑛z𝑛𝑥𝑧2𝑥+2𝜏𝑛𝑧𝑛𝑥𝑧,𝐶𝑥𝑛𝐶𝑥𝜏2𝐶𝑥𝑛𝐶𝑥2.(3.35) Hence, by (3.12), it follows that 𝑥𝑛+1𝑥2=𝛼𝑛𝑓𝑥𝑛+𝛽𝑛𝑥𝑛+𝛾𝑛𝑆𝑡𝑛𝑥2𝛼𝑛𝑓𝑥𝑛𝑥2+𝛽𝑛𝑥𝑛𝑥2+𝛾𝑛𝑆𝑡𝑛𝑥2𝛼𝑛𝑘𝑥𝑛𝑥2+𝛽𝑛𝑥𝑛𝑥2+𝛾𝑛𝑡𝑛𝑥2𝛼𝑛𝑘𝑥𝑛𝑥2+𝛽𝑛𝑥𝑛𝑥2+𝛾𝑛𝑧𝑛𝑧2𝛼𝑛𝑘𝑥𝑛𝑥2+𝛽𝑛𝑥𝑛𝑥2+𝛾𝑛𝑥𝑛𝑥2𝛾𝑛𝑥𝑛𝑧𝑛𝑥𝑧2+2𝜏𝛾𝑛𝑥𝑛𝑧𝑛𝑥𝑧,𝐶𝑥𝑛𝐶𝑥𝜏2𝛾𝑛𝐶𝑥𝑛𝐶𝑥2=𝛼𝑛𝑘𝑥𝑛𝑥2+1𝛼𝑛𝑥𝑛𝑥2𝛾𝑛𝑥𝑛𝑧𝑛𝑥𝑧2+2𝜏𝛾𝑛𝑥𝑛𝑧𝑛𝑥𝑧,𝐶𝑥𝑛𝐶𝑥𝜏2𝛾𝑛𝐶𝑥𝑛𝐶𝑥2𝛼𝑛𝑘𝑥𝑛𝑥2+𝑥𝑛𝑥2𝛾𝑛𝑥𝑛𝑧𝑛𝑥𝑧2+2𝜏𝛾𝑛𝑥𝑛𝑧𝑛𝑥𝑧,𝐶𝑥𝑛𝐶𝑥𝛼𝑛𝑘𝑥𝑛𝑥2+𝑥𝑛𝑥2𝛾𝑛𝑥𝑛𝑧𝑛𝑥𝑧2+2𝜏𝛾𝑛𝑥𝑛𝑧𝑛𝑥𝑧𝐶𝑥𝑛𝐶𝑥,(3.36) which implies that 𝛾𝑛𝑥𝑛𝑧𝑛𝑥𝑧2𝛼𝑛𝑘𝑥𝑛𝑥2+𝑥𝑛𝑥2𝑥𝑛+1𝑥2+2𝜏𝛾𝑛𝑥𝑛𝑧𝑛𝑥𝑧𝐶𝑥𝑛𝐶𝑥𝛼𝑛𝑘𝑥𝑛+1𝑥2+2𝛾𝑛𝜏𝑥𝑛𝑧𝑛𝑥𝑧𝐶𝑥𝑛𝐶𝑥+𝑥𝑛𝑥𝑛+1𝑥𝑛𝑥+𝑥𝑛+1𝑥.(3.37) By (ii), (iii), 𝑥𝑛𝑥𝑛+10, and 𝐶𝑥𝑛𝐶𝑥0 as 𝑛, from (3.37) and we get (𝑥𝑛𝑧𝑛)(𝑥𝑧)0 as 𝑛. Now, observe that 𝑧𝑛𝑡𝑛+𝑥𝑧2=𝑧𝑛𝑃𝐸𝑦𝑛𝜆𝐴𝑦𝑛+𝑃𝐸𝑦𝜆𝐴y𝑧2=𝑧𝑛𝑃𝐸𝑦𝑛𝜆𝐴𝑦𝑛+𝑃𝐸𝑦𝜆𝐴𝑦𝑧+𝜇𝐵𝑧𝑛𝜇𝐵𝑧𝑛+𝜇𝐵𝑧𝜇𝐵𝑧2=𝑧𝑛𝜇𝐵𝑧𝑛𝑧𝜇𝐵𝑧𝑃𝐸𝑦𝑛𝜆𝐴𝑦𝑛𝑃𝐸𝑦𝜆𝐴𝑦+𝜇𝐵𝑧𝑛𝐵𝑧2𝑧𝑛𝜇𝐵𝑧𝑛𝑧𝜇𝐵𝑧𝑃𝐸𝑦𝑛𝜆𝐴𝑦𝑛𝑃𝐸𝑦𝜆𝐴𝑦2+2𝜇𝐵𝑧𝑛𝐵𝑧,𝑧𝑛𝜇𝐵𝑧𝑛𝑧𝜇𝐵𝑧𝑃𝐸𝑦𝑛𝜆𝐴𝑦𝑛𝑃𝐸𝑦𝜆𝐴𝑦+𝜇𝐵𝑧𝑛𝐵𝑧=𝑧𝑛𝜇𝐵𝑧𝑛𝑧𝜇𝐵𝑧𝑃𝐸𝑦𝑛𝜆𝐴𝑦𝑛𝑃𝐸𝑦𝜆𝐴𝑦2+2𝜇𝐵𝑧𝑛𝐵𝑧,𝑧𝑛𝑡𝑛+𝑥𝑧𝑧𝑛𝜇𝐵𝑧𝑛𝑧𝜇𝐵𝑧2𝑃𝐸𝑦𝑛𝜆𝑛𝐴𝑦𝑛𝑃𝐸𝑦𝜆𝑛𝐴𝑦2+2𝜇𝐵𝑧𝑛𝐵𝑧𝑧𝑛𝑡𝑛+𝑥𝑧𝑧𝑛𝜇𝐵𝑧𝑛𝑧𝜇𝐵𝑧2𝑆𝑃𝐸𝑦𝑛𝜆𝑛𝐴𝑦𝑛𝑆𝑃𝐸𝑦𝜆𝑛𝐴𝑦2+2𝜇𝐵𝑧𝑛𝐵𝑧𝑧𝑛𝑡𝑛+𝑥𝑧=𝑧𝑛𝜇𝐵𝑧𝑛𝑧𝜇𝐵𝑧2𝑆𝑡𝑛𝑆𝑥2+2𝜇𝐵𝑧𝑛𝐵𝑧𝑧𝑛𝑡𝑛+𝑥𝑧𝑧𝑛𝜇𝐵𝑧𝑛𝑧𝜇𝐵𝑧𝑆𝑡𝑛𝑥×𝑧𝑛𝜇𝐵𝑧𝑛𝑧𝜇𝐵𝑧+𝑆𝑡𝑛𝑥+2𝜇𝐵𝑧𝑛𝐵𝑧𝑧𝑛𝑡𝑛+𝑥𝑧.(3.38) Since 𝑆𝑡𝑛𝑥𝑛0, (𝑥𝑛𝑧𝑛)(𝑥𝑧)0, and 𝐵𝑧𝑛𝐵𝑧0,as𝑛, it follows that 𝑧𝑛𝑡𝑛+𝑥𝑧0,as𝑛.(3.39) Since 𝑆𝑡𝑛𝑡𝑛𝑆𝑡𝑛𝑥𝑛+𝑥𝑛𝑧𝑛𝑥𝑧+𝑧𝑛𝑡𝑛+𝑥𝑧,(3.40) we obtain lim𝑛𝑆𝑡𝑛𝑡𝑛=0.(3.41) Next, we show that limsup𝑛𝑓𝑥𝑥,𝑥𝑛𝑥0,(3.42) where 𝑥=𝑃𝐹(𝑆)Υ𝑓(𝑥).
Indeed, since {𝑡𝑛} and {𝑆𝑡𝑛} are two bounded sequence in 𝐸, we can choose a subsequence {𝑡𝑛𝑖} of {𝑡𝑛} such that 𝑡𝑛𝑖 of 𝑡𝑛 such that 𝑡𝑛𝑖𝑧𝐸 and limsup𝑛𝑓𝑥𝑥,𝑆𝑡𝑛𝑥=lim𝑖𝑓𝑥𝑥,𝑆𝑡𝑛𝑖𝑥.(3.43) Since lim𝑛𝑆𝑡𝑛𝑡𝑛=0, we obtain 𝑆𝑡𝑛𝑖𝑧 as 𝑖. Now, we claim that 𝑧𝐹(𝑆)Υ. First by Lemma 2.5, it is easy to see that 𝑧𝐹(𝑆).
Since 𝑆𝑡𝑛𝑡𝑛0, 𝑆𝑡𝑛𝑥𝑛0, and 𝑡𝑛𝑥𝑛=𝑡𝑛𝑆𝑡𝑛+𝑆𝑡𝑛𝑥𝑛𝑡𝑛𝑆𝑡𝑛+𝑆𝑡𝑛𝑥𝑛=𝑆𝑡𝑛𝑡𝑛+𝑆𝑡𝑛𝑥𝑛,(3.44)
we conclude that 𝑡𝑛𝑥𝑛0 as 𝑛. Furthermore, by Lemma 3.2 that 𝐺 is nonexpansive, then 𝑡𝑛𝑡𝐺𝑛=𝐺𝑥𝑛𝑡𝐺𝑛𝑥𝑛𝑡𝑛.(3.45) Thus lim𝑛𝑡𝑛𝐺(𝑡𝑛)=0. According to Lemma 2.5, we obtain 𝑧Υ. Therefore, there holds 𝑧𝐹(𝑆)Υ.
On the other hand, it follows from (2.2) that limsup𝑛𝑓𝑥𝑥,𝑥𝑛𝑥=limsup𝑛𝑓𝑥𝑥,𝑆𝑡𝑛𝑥=lim𝑛𝑓𝑥𝑥,𝑆𝑡𝑛𝑖𝑥=𝑓𝑥𝑥,𝑧