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Journal of Applied Mathematics
Volume 2012 (2012), Article ID 787419, 13 pages
http://dx.doi.org/10.1155/2012/787419
Research Article

Iterative Algorithm for Common Fixed Points of Infinite Family of Nonexpansive Mappings in Banach Spaces

College of Science, Civil Aviation University of China, Tianjin 300300, China

Received 9 January 2012; Accepted 18 January 2012

Academic Editor: Yonghong Yao

Copyright © 2012 Songnian He and Jun Guo. 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

Let 𝐶 be a nonempty closed convex subset of a real uniformly smooth Banach space 𝑋, {𝑇𝑘}𝑘=1𝐶𝐶 an infinite family of nonexpansive mappings with the nonempty set of common fixed points 𝑘=1Fix(𝑇𝑘), and 𝑓𝐶𝐶 a contraction. We introduce an explicit iterative algorithm 𝑥𝑛+1=𝛼𝑛𝑓(𝑥𝑛)+(1𝛼𝑛)𝐿𝑛𝑥𝑛, where 𝐿𝑛=𝑛𝑘=1𝜔𝑘/s𝑛𝑇𝑘,𝑆𝑛=𝑛𝑘=1𝜔𝑘, and 𝑤𝑘>0 with 𝑘=1𝜔𝑘=1. Under certain appropriate conditions on {𝛼𝑛}, we prove that {𝑥𝑛} converges strongly to a common fixed point 𝑥 of {𝑇𝑘}𝑘=1, which solves the following variational inequality: 𝑥𝑓(𝑥),𝐽(𝑥𝑝)0,𝑝𝑘=1Fix(𝑇𝑘), where 𝐽 is the (normalized) duality mapping of 𝑋. This algorithm is brief and needs less computational work, since it does not involve 𝑊-mapping.

1. Introduction

Let 𝑋 be a real Banach space, 𝐶 a nonempty closed convex subset of 𝑋, and 𝑋 the dual space of 𝑋. The (normalized) duality mapping 𝐽𝑋2𝑋 is defined by 𝐽𝑥(𝑥)=𝑋𝑥,𝑥=𝑥2,𝑥=𝑥,𝑥𝑋.(1.1) If 𝑋 is a Hilbert space, then 𝐽=𝐼, where 𝐼 is the identity mapping. It is well known that if 𝑋 is smooth, then 𝐽 is single valued.

Recall that a mapping 𝑓𝐶𝐶 is a contraction, if there exists a constant 𝛼[0,1) such that 𝑓(𝑥)𝑓(𝑦)𝛼𝑥𝑦,𝑥,𝑦𝐶.(1.2) We use Π𝐶 to denote the collection of all contractions on 𝐶, that is, Π𝐶={𝑓𝑓isacontractionon𝐶}.(1.3) A mapping 𝑇𝐶𝐶 is said to be nonexpansive, if 𝑇𝑥𝑇𝑦𝑥𝑦,𝑥,𝑦𝐶.(1.4) We use Fix(𝑇) to denote the set of fixed points of 𝑇, namely, Fix(𝑇)={𝑥𝐶𝑇𝑥=𝑥}. One classical way to study nonexpansive mappings is to use contractions to approximate a nonexpansive mapping ([111]). Browder [1] first considered the following approximation in a Hilbert space. Fix 𝑢𝐶 and define a contraction 𝐹𝑡 from 𝐶 into itself by𝐹𝑡𝑥=𝑡𝑢+(1𝑡)𝑇𝑥,𝑥𝐶,(1.5) where 𝑡(0,1). Banach contraction mapping principle guarantees that 𝐹𝑡 has a unique fixed point in 𝐶. Denote by 𝑧𝑡𝐶 the unique fixed point of 𝐹𝑡, that is,𝑧𝑡=𝑡𝑢+(1𝑡)𝑇𝑧𝑡.(1.6) In the case of 𝑇 having fixed points, Browder [1] proved the following.

Theorem 1.1. In a Hilbert space, as 𝑡0, 𝑧𝑡 defined in (1.6) converges strongly to a fixed point of 𝑇 that is closest to 𝑢, that is, the nearest point projection of 𝑢 onto Fix(𝑇).

Halpern [3] introduced an iteration process (discretization of (1.6)) in a Hilbert as follows:𝑧𝑛+1=𝛼𝑛𝑢+1𝛼𝑛𝑇𝑧𝑛,𝑛0,(1.7) where 𝑢,𝑧0𝐶 are arbitrary (but fixed) and {𝛼𝑛} is a sequence in (0,1). Lions [4] proved the following.

Theorem 1.2. In a Hilbert space, if {𝛼𝑛} satisfies the following conditions: (K1)lim𝑛𝛼𝑛=0;(K2)𝑛=0𝛼𝑛=;(K3)lim𝑛|𝛼𝑛𝛼𝑛1|/𝛼2𝑛+1=0. Then {𝑧𝑛} converges strongly to the nearest point projection of 𝑢 onto Fix(𝑇).

The Banach space versions of Theorems 1.1 and 1.2 were obtained by Reich [5]. He proved the following.

Theorem 1.3. In a uniformly smooth Banach space 𝑋, both 𝑧𝑡 defined in (1.6) and {𝑧𝑛} defined in (1.7) converge strongly to a same fixed point of T. If one defines 𝑄𝐶Fix(𝑇) by 𝑄(𝑢)=lim𝑡0𝑧𝑡,(1.8) then 𝑄 is the sunny nonexpansive retraction from 𝐶 onto Fix(𝑇). Namely, Q satisfies the property: 𝑄𝑥𝑄𝑦2𝑥𝑦,𝐽(𝑄𝑥𝑄𝑦),𝑥,𝑦𝐶,(1.9) where 𝐽 is the duality mapping of 𝑋.

Moudafi [6] introduced a viscosity approximation method and proved the strong convergence of both the implicit and explicit methods in Hilbert spaces. Xu [7] extended Moudafi's results in Hilbert spaces. Given a real number 𝑡(0,1) and a contraction 𝑓Π𝐶, define a contraction 𝑇𝑓𝑡𝐶𝐶 by𝑇𝑓𝑡𝑥=𝑡𝑓(𝑥)+(1𝑡)𝑇𝑥,𝑥𝐶.(1.10) Let 𝑥𝑡𝐶 be the unique fixed point of 𝑇𝑓𝑡. Thus,𝑥𝑡𝑥=𝑡𝑓𝑡+(1𝑡)𝑇𝑥𝑡.(1.11) Corresponding explicit iterative process is defined by𝑥𝑛+1=𝛼𝑛𝑓𝑥𝑛+1𝛼𝑛𝑇𝑥𝑛,(1.12) where 𝑥0𝐶 is arbitrary (but fixed) and {𝛼𝑛} is a sequence in (0,1). It was proved by Xu [7] that under certain appropriate conditions on {𝛼𝑛}, both 𝑥𝑡 defined in (1.11) and {𝑥𝑛} defined in (1.12) converged strongly to 𝑥𝐶, which is the unique solution of the variational inequality: (𝐼𝑓)𝑥,𝑥𝑥0,𝑥Fix(𝑇).(1.13) Xu [7] also extended Moudafi's results to the setting of Banach spaces and proved the strong convergence of both the implicit method (1.11) and explicit method (1.12) in uniformly smooth Banach spaces.

In order to deal with some problems involving the common fixed points of infinite family of nonexpansive mappings, 𝑊-mapping is often used, see [1220]. Let {𝑇𝑘}𝑘=1𝐶𝐶 be an infinite family of nonexpansive mappings and let {𝜉𝑘}𝑘=1 be a real number sequence such that 0<𝜉𝑘<1 for every 𝑘. For any 𝑛, we define a mapping 𝑊𝑛 of 𝐶 into itself as follows:𝑈𝑛,𝑛+1𝑈=𝐼,𝑛,𝑛=𝜉𝑛𝑇𝑛𝑈𝑛,𝑛+1+1𝜉𝑛𝑈𝐼,𝑛,𝑛1=𝜉𝑛1𝑇𝑛1𝑈𝑛,𝑛+1𝜉𝑛1𝑈𝐼,𝑛,𝑘=𝜉𝑘𝑇𝑘𝑈𝑛,𝑘+1+1𝜉𝑘𝑈𝐼,𝑛,𝑘1=𝜉𝑘1𝑇𝑘1𝑈𝑛,𝑘+1𝜉𝑘1𝑈𝐼,𝑛,2=𝜉2𝑇2𝑈𝑛,3+1𝜉2𝑊𝐼,𝑛=𝑈𝑛,1=𝜉1𝑇1𝑈𝑛,2+1𝜉1𝐼.(1.14) Such 𝑊𝑛 is called the 𝑊-mapping generated by {𝑇𝑘}𝑘=1 and {𝜉𝑘}𝑘=1, see [12, 13].

Yao et al. [10] introduced the following iterative algorithm for infinite family of nonexpansive mappings. Let 𝑋 be a uniformly convex Banach space with a uniformly Gâteaux differentiable norm and 𝐶 a nonempty closed convex subset of 𝑋. Sequence {𝑥𝑛} is defined by𝑥𝑛+1=𝛼𝑛𝑢+1𝛼𝑛𝑊𝑛𝑥𝑛,𝑛0,(1.15) where 𝑢,𝑥0𝐶 are arbitrary (but fixed) and {𝛼𝑛}(0,1). It was proved that under certain appropriate conditions on {𝛼𝑛}, the sequence {𝑥𝑛} generated by (1.15) converges strongly to a common fixed point of {𝑇𝑘}𝑘=1 [13].

Since 𝑊-mapping contains many composite operations of {𝑇𝑘}, it is complicated and needs large computational work. In this paper, we introduce a new iterative algorithm for solving the common fixed point problem of infinite family of nonexpansive mappings. Let 𝑋 be a real uniformly smooth Banach space, 𝐶 a nonempty closed convex subset of 𝑋, {𝑇𝑘}𝑘=1𝐶𝐶 an infinite family of nonexpansive mappings with the nonempty set of common fixed points 𝑘=1Fix(𝑇𝑘), and 𝑓Π𝐶. Given any 𝑥0𝐶, define a sequence {𝑥𝑛} by𝑥𝑛+1=𝛼𝑛𝑓𝑥𝑛+1𝛼𝑛𝐿𝑛𝑥𝑛,𝑛0,(1.16) where {𝛼𝑛}(0,1), 𝐿𝑛=𝑛𝑘=1(𝜔𝑘/𝑠𝑛)𝑇𝑘, 𝑆𝑛=𝑛𝑘=1𝜔𝑘 and 𝑤𝑘>0 with 𝑘=1𝜔𝑘=1. Under certain appropriate conditions on {𝛼𝑛}, we prove that {𝑥𝑛} converges strongly to 𝑥𝑘=1Fix(𝑇𝑘), which solves the following variational inequality: 𝑥𝑥𝑓𝑥,𝐽𝑝0,𝑝𝑘=1𝑇Fix𝑘,(1.17) where 𝐽 is the duality mapping of 𝑋. Because 𝐿𝑛 doesn't contain many composite operations of {𝑇𝑘}, this algorithm is brief and needs less computational wok.

We will use 𝑀 to denote a constant, which may be different in different places.

2. Preliminaries

Let 𝐵={𝑥𝑋𝑥=1} denotes the unit sphere of 𝑋. A Banach space 𝑋 is said to be strictly convex, if (𝑥+𝑦)/2<1 holds for all 𝑥,𝑦𝐵, 𝑥𝑦. A Banach space 𝑋 is said to be uniformly convex if for each 𝜀(0,2], there exists a constant 𝛿>0 such that for any x,𝑦𝐵, 𝑥𝑦𝜀 implies (𝑥+𝑦)/21𝛿. It is known that a uniformly convex Banach space is reflexive and strictly convex, see [21].

The norm of 𝑋 is said to be Gâteaux differentiable iflim𝑡0𝑥+𝑡𝑦𝑥𝑡(2.1) exists for each 𝑥,𝑦𝐵 and in this case 𝑋 is said to be smooth. The norm of 𝑋 is said to be uniformly Gâteaux differentiable if for each 𝑦𝐵, the limit (2.1) is attained uniformly for 𝑥𝐵. The norm of 𝑋 is said to be Frêchet differentiable, if for each 𝑥𝐵, the limit (2.1) is attained uniformly for 𝑦𝐵. The norm of 𝑋 is said to be uniformly Frêchet differentiable, if the limit (2.1) is attained uniformly for 𝑥,𝑦𝐵 and in this case 𝑋 is said to be uniformly smooth.

Let 𝐷 be a nonempty subset of 𝐶. A mapping 𝑄𝐶𝐷 is said to be sunny [22] if𝑄(𝑥+𝑡(𝑥𝑄(𝑥)))=𝑄(𝑥),𝑥𝐶,𝑡0,(2.2) whenever 𝑥+𝑡(𝑥𝑄(𝑥))𝐶. A mapping 𝑄𝐶𝐷 is called a retraction if 𝑄𝑥=𝑥 for all 𝑥𝐷. Furthermore, 𝑄 is sunny nonexpansive retraction from 𝐶 onto 𝐷 if 𝑄 is a retraction from 𝐶 onto 𝐷 which is also sunny and nonexpansive.

A subset 𝐷 of 𝐶 is called a sunny nonexpansive retraction of 𝐶 if there exits a sunny nonexpansive retraction from 𝐶 onto 𝐷.

Lemma 2.1 (see [22]). Let 𝐶 be a closed convex subset of a smooth Banach space 𝑋. Let 𝐷 be a nonempty subset of 𝐶 and 𝑄𝐶𝐷 be a retraction. Then the following are equivalent.(a)𝑄 is sunny and nonexpansive.(b)𝑄𝑥𝑄𝑦2𝑥𝑦,𝐽(𝑄𝑥𝑄𝑦), for all 𝑥,𝑦𝐶.(c)𝑥𝑄𝑥,𝐽(𝑦𝑄𝑥)0, for all 𝑥𝐶,𝑦𝐷.

Lemma 2.2 (see [23]). Let {𝑠𝑛} be a sequence of nonnegative real numbers satisfying 𝑠𝑛+11𝛾𝑛𝑠𝑛+𝛾𝑛𝛽𝑛+𝛿𝑛,𝑛0,(2.3) where {𝛾𝑛}(0,1), {𝛽𝑛} and {𝛿𝑛} satisfy the following conditions: (A1)𝑛=0𝛾𝑛=;(A2)limsup𝑛𝛽𝑛0;(A3)𝛿𝑛0 (𝑛0), 𝑛=1𝛿𝑛<. Then lim𝑛𝑠𝑛=0.

Lemma 2.3 (see [24]). In a Banach space 𝑋, the following inequality holds: 𝑥+𝑦2𝑥2+2𝑦,𝑗(𝑥+𝑦),𝑥,𝑦𝑋,(2.4) where 𝑗(𝑥+𝑦)𝐽(𝑥+𝑦).

Lemma 2.4 (see [25]). Let 𝐶 be a closed convex subset of a strictly convex Banach space 𝑋. Let {𝑇𝑛𝑛} be a sequence of nonexpansive mappings on 𝐶. Suppose 𝑛=1Fix(𝑇𝑛) is nonempty. Let {𝜆𝑛} be a sequence of positive numbers with 𝑛=1𝜆𝑛=1. Then a mapping 𝑆 on 𝐶 defined by 𝑆𝑥=𝑛=1𝜆𝑛𝑇𝑛𝑥 for 𝑥𝐶 is well defined, nonexpansive and Fix(𝑆)=𝑛=1Fix(𝑇𝑛) holds.

Lemma 2.5 (see [7]). Let 𝑋 be a uniformly smooth Banach space, 𝐶 a closed convex subset of 𝑋, 𝑇𝐶𝐶 a nonexpansive mapping with Fix(𝑇), and 𝑓Π𝐶. Then {𝑥𝑡} defined by 𝑥𝑡𝑥=𝑡𝑓𝑡+(1𝑡)𝑇𝑥𝑡(2.5) converges strongly to a point in Fix(𝑇). If we define a mapping 𝑄Π𝐶Fix(𝑇) by 𝑄(𝑓)=lim𝑡0𝑥𝑡,𝑓Π𝐶,(2.6) then 𝑄(𝑓) solves the variational inequality: (𝐼𝑓)𝑄(𝑓),𝐽(𝑄(𝑓)𝑝)0,𝑓Π𝐶,𝑝Fix(𝑇).(2.7)

Lemma 2.6. Let 𝑋 be a Banach space, {𝑥𝑘} a bounded sequence of 𝑋, and {𝜔𝑘} a sequence of positive numbers with 𝑘=1𝜔𝑘=1. Then 𝑘=1𝜔𝑘𝑥𝑘 is convergent in 𝑋.

Lemma 2.7. Let X be Banach space, {𝑇𝑘𝑘} a sequence of nonexpansive mappings on X with 𝑘=1Fix(𝑇𝑘), and {𝜔𝑘} a sequence of positive numbers with 𝑘=1𝜔𝑘=1. Let 𝑇=𝑘=1𝜔𝑘𝑇𝑘, 𝐿𝑚=𝑚𝑘=1(𝜔𝑘/𝑆𝑚)𝑇𝑘, and 𝑆𝑚=𝑚𝑘=1𝜔𝑘. Then 𝐿𝑚 uniformly converges to 𝑇 in each bounded subset 𝑆 of 𝑋.

Proof. Forall𝑥𝑆, we observe that 𝐿𝑚=𝑥𝑇𝑥𝑚𝑘=1𝜔𝑘𝑆𝑚𝑇𝑘𝑥𝑘=1𝜔𝑘𝑇𝑘𝑥=𝑚𝑘=1𝜔𝑘𝜔𝑘𝑆𝑚𝑆𝑚𝑇𝑘𝑥𝑘=𝑚+1𝜔𝑘𝑇𝑘𝑥𝑚𝑘=11𝑆𝑚𝑆𝑚𝜔𝑘𝑇𝑘𝑥+𝑘=𝑚+1𝜔𝑘𝑇𝑘𝑥1𝑆𝑚𝑆𝑚𝑚𝑘=1𝜔𝑘𝑇𝑘𝑥+𝑘=𝑚+1𝜔𝑘𝑇𝑘𝑥1𝑆𝑚𝑆𝑚𝑀+𝑀𝑘=𝑚+1𝜔𝑘,(2.8) where 𝑀=sup𝑥𝑆,𝑘1𝑇𝑘𝑥<. Taking 𝑚 in above last inequality, we have that lim𝑚𝐿𝑚𝑥𝑇𝑥=0(2.9) holds uniformly for 𝑥𝑆 and this completes the proof.

3. Main Results

Theorem 3.1. Let 𝐶 be a nonempty closed convex subset of a real uniformly smooth Banach space 𝑋, {𝑇𝑘}𝑘=1𝐶𝐶 an infinite family of nonexpansive mappings with 𝑘=1Fix(𝑇𝑘), and {𝜔𝑘} a sequence of positive numbers with 𝑘=1𝜔𝑘=1. Let 𝐿𝑛=𝑛𝑘=1(𝜔𝑘/𝑆𝑛)𝑇𝑘, 𝑆𝑛=𝑛𝑘=1𝜔𝑘, and 𝑓Π𝐶 with coefficient 𝛼[0,1). Given any 𝑥0𝐶, let {𝑥𝑛} be a sequence generated by 𝑥𝑛+1=𝛼𝑛𝑓𝑥𝑛+1𝛼𝑛𝐿𝑛𝑥𝑛,𝑛0,(3.1) where {𝛼𝑛}(0,1) satisfies the following conditions: (A1)lim𝑛𝛼𝑛=0;(A2)𝑛=0𝛼𝑛=;(A3)either 𝑛=0|𝛼𝑛+1𝛼𝑛|< or lim𝑛(𝛼𝑛+1/𝛼𝑛)=1. Then {𝑥𝑛} converges strongly to 𝑥𝑘=1Fix(𝑇𝑘), which solves the following variational inequality: 𝑥𝑥𝑓𝑥,𝐽𝑝0,𝑝𝑘=1𝑇Fix𝑘.(3.2)

Proof. Step 1. We show that {𝑥𝑛} is bounded.
Noticing nonexpansiveness of 𝐿𝑛, take a 𝑝𝑘=1Fix(𝑇𝑘) to derive that 𝑥𝑛+1=𝛼p𝑛𝑓𝑥𝑛+1𝛼𝑛𝐿𝑛𝑥𝑛𝑝𝛼𝑛𝑓𝑥𝑛+𝑝1𝛼𝑛𝐿𝑛𝑥𝑛𝑝=𝛼𝑛𝑓𝑥𝑛+𝑓(𝑝)+𝑓(𝑝)𝑝1𝛼𝑛𝐿𝑛𝑥𝑛𝑝𝛼𝑛𝑓𝑥𝑛𝑓(𝑝)+𝛼𝑛(𝑓𝑝)𝑝+1𝛼𝑛𝐿𝑛𝑥𝑛𝑝𝛼𝛼𝑛𝑥𝑛𝑝+𝛼𝑛𝑓(𝑝)𝑝+1𝛼𝑛𝑥𝑛=𝑝1(1𝛼)𝛼𝑛𝑥𝑛𝑝+𝛼𝑛(=𝑓(𝑝)𝑝)1(1𝛼)𝛼𝑛𝑥𝑛+𝑝(1𝛼)𝛼𝑛(𝑓(𝑝)𝑝)(1𝛼max𝑓(𝑝)𝑝),𝑥1𝛼𝑛.𝑝(3.3) By induction, we obtain 𝑥𝑛+1(𝑝max𝑓(𝑝)𝑝),𝑥1𝛼0𝑝,𝑛0,(3.4) and {𝑥𝑛} is bounded, so are {𝑇𝑘𝑥𝑛}, {𝐿𝑛𝑥𝑛}, and {𝑓(𝑥𝑛)}.
Step 2. We prove that lim𝑛𝑥𝑛+1𝑥𝑛=0.
By (3.1), We have 𝑥𝑛+1𝑥𝑛=𝛼𝑛𝑓𝑥𝑛+1𝛼𝑛𝐿𝑛𝑥𝑛𝛼𝑛1𝑓𝑥𝑛11𝛼𝑛1𝐿𝑛1𝑥𝑛1=𝛼𝑛𝑓𝑥𝑛𝛼𝑛𝑓𝑥𝑛1+𝛼𝑛𝑓𝑥𝑛1𝛼𝑛1𝑓𝑥𝑛1+1𝛼𝑛𝐿𝑛𝑥𝑛1𝛼𝑛𝐿𝑛1𝑥𝑛1+1𝛼𝑛𝐿𝑛1𝑥𝑛11𝛼𝑛1𝐿𝑛1𝑥𝑛1𝛼𝛼𝑛𝑥𝑛𝑥𝑛1+||𝛼𝑛𝛼𝑛1||𝑓𝑥𝑛1+1𝛼𝑛𝐿𝑛𝑥𝑛𝐿𝑛1𝑥𝑛1+||𝛼𝑛𝛼𝑛1||𝐿𝑛1𝑥𝑛1=𝛼𝛼𝑛𝑥𝑛𝑥𝑛1+||𝛼𝑛𝛼𝑛1||𝑓𝑥𝑛1+𝐿𝑛1𝑥𝑛1+1𝛼𝑛𝐿𝑛𝑥𝑛𝐿𝑛1𝑥𝑛1𝛼𝛼𝑛𝑥𝑛𝑥𝑛1+||𝛼𝑛𝛼𝑛1||𝑓𝑥𝑛1+𝐿𝑛1𝑥𝑛1+1𝛼𝑛𝐿𝑛𝑥𝑛𝐿𝑛𝑥𝑛1+1𝛼𝑛𝐿𝑛𝑥𝑛1𝐿𝑛1𝑥𝑛11(1𝛼)𝛼𝑛𝑥𝑛𝑥𝑛1+1𝛼𝑛𝐿𝑛𝑥𝑛1𝐿𝑛1𝑥𝑛1+||𝛼𝑛𝛼𝑛1||𝑓𝑥𝑛1+𝐿𝑛1𝑥𝑛11(1𝛼)𝛼𝑛𝑥𝑛𝑥𝑛1+1𝛼𝑛𝐿𝑛𝑥𝑛1𝐿𝑛1𝑥𝑛1+||𝛼𝑛𝛼𝑛1||𝑀,(3.5) where 𝑀=sup𝑛1(𝑓(𝑥𝑛1)+𝐿𝑛1𝑥𝑛1). At the same time, we observe that 𝑛=1𝐿𝑛𝑥𝑛1𝐿𝑛1𝑥𝑛1=𝑛=1𝑛𝑘=1𝜔𝑘𝑆𝑛𝑇𝑘𝑥𝑛1𝑛1𝑘=1𝜔𝑘𝑆𝑛1𝑇𝑘𝑥𝑛1=𝑛=1𝜔𝑛𝑆𝑛𝑇𝑛𝑥𝑛1+𝑛1𝑘=1𝜔𝑛𝜔𝑘𝑆𝑛𝑆𝑛1𝑇𝑘𝑥𝑛1𝑛=1𝜔𝑛𝑆𝑛𝑇𝑛𝑥𝑛1+𝑛1𝑘=1𝜔𝑛𝜔𝑘𝑆𝑛𝑆𝑛1𝑇𝑘𝑥𝑛1𝑛=1𝜔𝑛𝑆𝑛𝑇𝑛𝑥𝑛1+𝑛=1𝑛1𝑘=1𝜔𝑛𝜔𝑘𝑆𝑛𝑆𝑛1𝑇𝑘𝑥𝑛1𝑛=1𝜔𝑛𝑆𝑛𝑀+𝑛=1𝜔𝑛𝑆𝑛𝑀=𝑛=12𝑀𝑆𝑛𝜔𝑛,(3.6) where 𝑀=sup𝑘1,𝑛1𝑇𝑘𝑥𝑛1. Applying Lemma 2.6 and compatibility test of series, we have 𝑛=1𝐿𝑛𝑥𝑛1𝐿𝑛1𝑥𝑛1<.(3.7) Put 𝛾𝑛=(1𝛼)𝛼𝑛,𝛽𝑛=||𝛼𝑛𝛼𝑛1||𝑀(1𝛼)𝛼𝑛,𝛿𝑛=1𝛼𝑛𝐿𝑛𝑥𝑛1𝐿𝑛1𝑥𝑛1.(3.8) It follows that 𝑥𝑛+1𝑥𝑛1𝛾𝑛𝑥𝑛𝑥𝑛1+𝛾𝑛𝛽𝑛+𝛿𝑛.(3.9) It is easily seen from (A2), (A3), and (3.7) that 𝑛=1𝛾𝑛=,limsup𝑛𝛽𝑛0,𝑛=1𝛿𝑛<.(3.10) Applying Lemma 2.2 to (3.9), we obtain lim𝑛𝑥𝑛+1𝑥𝑛=0.(3.11)
Step 3. We show that lim𝑛𝑥𝑛𝑇𝑥𝑛=0.
Indeed we observe that 𝑥𝑛𝑇𝑥𝑛𝑥𝑛𝑥𝑛+1+𝑥𝑛+1𝑇𝑥𝑛=𝑥𝑛𝑥𝑛+1+𝛼𝑛𝑓𝑥𝑛+1𝛼𝑛𝐿𝑛𝑥𝑛𝑇𝑥𝑛𝑥𝑛𝑥𝑛+1+𝛼𝑛𝑓𝑥𝑛𝑇𝑥𝑛+1𝛼𝑛𝐿𝑛𝑥𝑛𝑇𝑥𝑛.(3.12) Hence, by (3.11), (A1), and Lemma 2.7, we have lim𝑛𝑥𝑛𝑇𝑥𝑛=0.(3.13)
Step 4. We prove that limsup𝑛𝑓𝑥𝑥𝑥,𝐽𝑛𝑥0,(3.14) where 𝑥=lim𝑡0𝑥𝑡 with 𝑥𝑡 being the fixed point of the contraction 𝑥𝑡𝑓(𝑥)+(1𝑡)𝑇𝑥.(3.15) From Lemma 2.5, we have 𝑥Fix(𝑇) and (𝐼𝑓)𝑥𝑥,𝐽𝑝0,𝑝Fix(𝑇).(3.16) By Lemma 2.4, we have 𝑥𝑘=1Fix(𝑇𝑘) and (𝐼𝑓)𝑥𝑥,𝐽𝑝0,𝑝𝑘=1𝑇Fix𝑘.(3.17) By 𝑥𝑡=𝑡𝑓(𝑥𝑡)+(1𝑡)𝑇𝑥𝑡, we have 𝑥𝑡𝑥𝑛𝑓𝑥=𝑡𝑡𝑥𝑛+(1𝑡)𝑇𝑥𝑡𝑥𝑛.(3.18) It follows from Lemma 2.3 that 𝑥𝑡𝑥𝑛2=𝑡(𝑓(𝑥𝑡)𝑥𝑛)+(1𝑡)(𝑇𝑥𝑡𝑥𝑛)2(1𝑡)2𝑇𝑥𝑡𝑥𝑛2𝑥+2𝑡𝑓𝑡𝑥𝑛𝑥,𝐽𝑡𝑥𝑛(1𝑡)2𝑇𝑥𝑡𝑇𝑥𝑛+𝑇𝑥𝑛𝑥𝑛2𝑥+2𝑡𝑓𝑡𝑥𝑡𝑥,𝐽𝑡𝑥𝑛+2𝑡𝑥𝑡𝑥𝑛𝑥,𝐽𝑡𝑥𝑛(1𝑡)2𝑥𝑡𝑥𝑛+𝑇𝑥𝑛𝑥𝑛2𝑥+2𝑡𝑓𝑡𝑥𝑡𝑥,𝐽𝑡𝑥𝑛+2𝑡𝑥𝑡𝑥𝑛𝑥,𝐽𝑡𝑥𝑛(1𝑡)2𝑥𝑡𝑥𝑛2+𝑏𝑛𝑥(𝑡)+2𝑡𝑓𝑡𝑥𝑡𝑥,𝐽𝑡𝑥𝑛𝑥+2𝑡𝑡𝑥𝑛2,(3.19) where 𝑏𝑛(𝑡)=𝑇𝑥𝑛𝑥𝑛(2𝑥𝑡𝑥𝑛+𝑇𝑥𝑛𝑥𝑛)0 (𝑛). It follows from above last inequality that 𝑥𝑡𝑥𝑓𝑡𝑥,𝐽𝑡𝑥𝑛𝑡2𝑥𝑡𝑥𝑛2+1𝑏2𝑡𝑛(𝑡).(3.20) Taking 𝑛 in (3.20) yields limsup𝑛𝑥𝑡𝑥𝑓𝑡𝑥,𝐽𝑡𝑥𝑛𝑡2𝑀,(3.21) where 𝑀𝑥𝑡𝑥𝑛2 for all 𝑛1 and 𝑡(0,1). Taking 𝑡0 in (3.21), we have limsup𝑡0limsup𝑛𝑥𝑡𝑥𝑓𝑡𝑥,𝐽𝑡𝑥𝑛0.(3.22) Noticing the fact that two limits are interchangeable due to the fact the duality mapping 𝐽 is norm-to-norm uniformly continuous on bounded sets, it follows from (3.22), we have limsup𝑛𝑓𝑥𝑥𝑥,𝐽𝑛𝑥=limsup𝑛limsup𝑡0𝑥𝑡𝑥𝑓𝑡𝑥,𝐽𝑡𝑥𝑛=limsup𝑡0limsup𝑛𝑥𝑡𝑥𝑓𝑡𝑥,𝐽𝑡𝑥𝑛0.(3.23) Hence (3.14) holds.Step 5. Finally, we prove that 𝑥𝑛𝑥 (𝑛).
Indeed we observe that 𝑥𝑛+1𝑥2=𝛼𝑛𝑓(𝑥𝑛)+(1𝛼𝑛)𝐿𝑛𝑥𝑛𝑥2=𝛼𝑛(𝑓(𝑥𝑛)𝑓(𝑥))+(1𝛼𝑛)(𝐿𝑛𝑥𝑛𝑥)+𝛼𝑛(𝑓(𝑥)𝑥)2𝛼𝑛(𝑓(𝑥𝑛)𝑓(𝑥))+(1𝛼𝑛)(𝐿𝑛𝑥𝑛𝑥)2+2𝛼𝑛𝑓𝑥𝑥𝑥,𝐽𝑛+1𝑥𝛼𝑛𝑓𝑥𝑛𝑥𝑓+1𝛼𝑛𝐿𝑛𝑥𝑛𝑥2+2𝛼𝑛𝑥𝑓𝑥𝑥,𝐽𝑛+1𝑥𝛼𝛼𝑛𝑥𝑛𝑥+1𝛼𝑛𝑥𝑛𝑥2+2𝛼𝑛𝑥𝑓𝑥𝑥,𝐽𝑛+1𝑥1(1𝛼)𝛼𝑛2𝑥𝑛𝑥2+2𝛼𝑛𝑥𝑓𝑥𝑥,𝐽𝑛+1𝑥1(1𝛼)𝛼𝑛𝑥𝑛𝑥2+2𝛼𝑛𝑥𝑓𝑥𝑥,𝐽𝑛+1𝑥.(3.24) By view of (3.14) and condition (A2), it follows from Lemma 2.2 that 𝑥𝑛𝑥(𝑛). This completes the proof.

Corollary 3.2. Let 𝐶 be a nonempty closed convex subset of a real uniformly smooth Banach space 𝑋, {𝑇𝑘}𝑘=1𝐶𝐶 an infinite family of nonexpansive mappings with 𝑘=1Fix(𝑇𝑘), {𝜔𝑘} a sequence of positive numbers with 𝑘=1𝜔𝑘=1. Let 𝐿𝑛=𝑛𝑘=1(𝜔𝑘/𝑆𝑛)𝑇𝑘, 𝑆𝑛=𝑛𝑘=1𝜔𝑘, and 𝑢𝐶. Given any 𝑥0𝐶, let {𝑥𝑛} be a sequence generated by 𝑥𝑛+1=𝛼𝑛𝑢+1𝛼𝑛𝐿𝑛𝑥𝑛,𝑛0,(3.25) where {𝛼𝑛}(0,1) satisfies the following conditions: (A1)lim𝑛𝛼𝑛=0;(A2)𝑛=0𝛼𝑛=;(A3)either𝑛=0|𝛼𝑛+1𝛼𝑛|< or lim𝑛(𝛼𝑛+1/𝛼𝑛=)1. Then {𝑥𝑛} converges strongly to 𝑥𝑘=1Fix(𝑇𝑘), which solves the following variational inequality: 𝑥𝑥𝑢,𝐽𝑝0,𝑝𝑘=1𝑇Fix𝑘.(3.26)

Acknowledgment

This paper is supported by the Fundamental Research Funds for the Central Universities (ZXH2011D005).

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