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Journal of Applied Mathematics
Volume 2012 (2012), Article ID 401960, 20 pages
http://dx.doi.org/10.1155/2012/401960
Research Article

Hybrid Algorithm for Common Fixed Points of Uniformly Closed Countable Families of Hemirelatively Nonexpansive Mappings and Applications

1Department of Mathematics, Cangzhou Normal University, Cangzhou 061001, China
2Department of Mathematics, Tianjin Polytechnic University, Tianjin 300387, China

Received 27 November 2011; Accepted 15 December 2011

Academic Editor: Yonghong Yao

Copyright © 2012 Sumei Ai and Yongfu Su. 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

The authors have obtained the following results: (1) the definition of uniformly closed countable family of nonlinear mappings, (2) strong convergence theorem by the monotone hybrid algorithm for two countable families of hemirelatively nonexpansive mappings in a Banach space with new method of proof, (3) two examples of uniformly closed countable families of nonlinear mappings and applications, (4) an example which is hemirelatively nonexpansive mapping but not weak relatively nonexpansive mapping, and (5) an example which is weak relatively nonexpansive mapping but not relatively nonexpansive mapping. Therefore, the results of this paper improve and extend the results of Plubtieng and Ungchittrakool (2010) and many others.

1. Introduction and Preliminaries

Let 𝐸 be a Banach space with the dual 𝐸. We denote by 𝐽 the normalized duality mapping from 𝐸 to 2𝐸 defined by 𝐽𝑥=𝑓𝐸𝑥,𝑓=𝑥2=𝑓2,(1.1)where , denotes the generalized duality pairing. It is well known that the normalized duality 𝐽 has the following properties: (1) if 𝐸 is smooth, then 𝐽 is single valued; (2) if 𝐸 is strictly convex, then 𝐽 is one-to-one (i.e., 𝐽𝑥𝐽𝑦= for all 𝑥𝑦); (3) if 𝐸 is reflexive, then 𝐽 is surjective; (4) if 𝐸 has Frchet differentiable norm, then 𝐽 is uniformly norm-to-norm continuous; (5) if 𝐸 is uniformly smooth, then 𝐽 is uniformly norm-to-norm continuous on each bounded subset of 𝐸; (7) if 𝐸 is a Hilbert space, then 𝐽 is the identity operator.

As we all know that if 𝐶 is a nonempty closed convex subset of a Hilbert space 𝐻 and 𝑃𝐶𝐻𝐶 is the metric projection of 𝐻 onto 𝐶, then 𝑃𝐶 is nonexpansive. This fact actually characterizes Hilbert spaces, and, consequently, it is not available in more general Banach spaces. In this connection, Alber [1] has recently introduced a generalized projection operator Π𝐶 in a Banach space 𝐸 which is an analogue of the metric projection in Hilbert spaces.

Let 𝐸 be a smooth Banach space. Consider the function defined by 𝜙(𝑥,𝑦)=𝑥22𝑥,𝐽𝑦+𝑦2for𝑥,𝑦𝐸.(1.2) Observe that, in a Hilbert space 𝐻, (1.2) reduces to 𝜙(𝑥,𝑦)=𝑥𝑦2,𝑥,𝑦𝐻.

The generalized projection Π𝐶𝐸𝐶 is a map that assigns to an arbitrary point 𝑥𝐸 the minimum point of the functional 𝜙(𝑥,𝑦), that is, Π𝐶𝑥=𝑥, where 𝑥 is the solution to the minimization problem 𝜙𝑥,𝑥=min𝑦𝐶𝜙(𝑦,𝑥),(1.3) existence and uniqueness of the operator Π𝐶 follow from the properties of the functional 𝜙(𝑥,𝑦) and strict monotonicity of the mapping 𝐽 (see, e.g., [1, 2]). In Hilbert space, Π𝐶=𝑃𝐶. It is obvious from the definition of function 𝜙 that ()𝑦𝑥2)𝜙(𝑦,𝑥)(𝑦+𝑥2𝑥,𝑦𝐸.(1.4)

If 𝐸 is a reflexive strictly convex and smooth Banach space, then for 𝑥,𝑦𝐸, 𝜙(𝑥,𝑦)=0 if and only if 𝑥=𝑦. It is sufficient to show that if 𝜙(𝑥,𝑦)=0, then 𝑥=𝑦. From (1.4), we have 𝑥=𝑦. This implies that 𝑥,𝐽𝑦=𝑥2=𝐽𝑦2. From the definitions of 𝑗, we have 𝐽𝑥=𝐽𝑦. That is, 𝑥=𝑦; see [3, 4] for more details.

Let 𝐶 be a closed convex subset of 𝐸, and let 𝑇 be a mapping from 𝐶 into itself with nonempty set of fixed points. We denote by 𝐹(𝑇) the set of fixed points of 𝑇. 𝑇 is called hemi-relatively nonexpansive if 𝜙(𝑝,𝑇𝑥)𝜙(𝑝,𝑥) for all 𝑥𝐶 and 𝑝𝐹(𝑇).

A point 𝑝 in 𝐶 is said to be an asymptotic fixed point of 𝑇 if 𝐶 contains a sequence {𝑥𝑛} which converges weakly to 𝑝 such that the strong lim𝑛(𝑇𝑥𝑛𝑥𝑛)=0. The set of asymptotic fixed points of 𝑇 will be denoted by 𝐹(𝑇). A hemi-relatively nonexpansive mapping 𝑇 from 𝐶 into itself is called relatively nonexpansive if 𝐹(𝑇)=𝐹(𝑇) (see, [5]).

A point 𝑝 in 𝐶 is said to be a strong asymptotic fixed point of 𝑇 if 𝐶 contains a sequence {𝑥𝑛} which converges strongly to 𝑝 such that the strong lim𝑛(𝑇𝑥𝑛𝑥𝑛)=0. The set of strong asymptotic fixed points of 𝑇 will be denoted by 𝐹(𝑇). A hemi-relatively nonexpansive mapping 𝑇 from 𝐶 into itself is called weak relatively nonexpansive if 𝐹(𝑇)=𝐹(𝑇) (see, [6]).

The following conclusions are obvious: (1) relatively nonexpansive mapping must be weak relatively nonexpansive mapping; (2) weak relatively nonexpansive mapping must be hemi-relatively nonexpansive mapping.

In this paper, we will give two examples to show that the inverses of above two conclusions are not hold.

In an infinite-dimensional Hilbert space, Mann's iterative algorithm has only weak convergence, in general, even for nonexpansive mappings. Hence in order to have strong convergence, in recent years, the hybrid iteration methods for approximating fixed points of nonlinear mappings has been introduced and studied by various authors.

In 2003, Nakajo and Takahashi [7] proposed the following modification of Mann iteration method for a single nonexpansive mapping 𝑇 in a Hilbert space 𝐻: 𝑥0𝑦𝐶chosenonlyarbitrarily,𝑛=𝛼𝑛𝑥𝑛+1𝛼𝑛𝑇𝑥𝑛,𝐶𝑛=𝑦𝑧𝐶𝑛𝑥𝑧𝑛,𝑄𝑧𝑛=𝑧𝐶𝑥𝑛𝑧,𝑥0𝑥𝑛,𝑥0𝑛+1=𝑃𝐶𝑛𝑄𝑛𝑥0,(1.5) where 𝐶 is a closed convex subset of 𝐻, and 𝑃𝐾 denotes the metric projection from 𝐻 onto a closed convex subset 𝐾 of 𝐻. They proved that if the sequence {𝛼𝑛} is bounded above from one, then the sequence {𝑥𝑛} generated by (1.5) converges strongly to 𝑃𝐹(𝑇)(𝑥0), where 𝐹(𝑇) denote the fixed points set of 𝑇.

The ideas to generalize the process (1.5) from Hilbert space to Banach space have recently been made. By using available properties on uniformly convex and uniformly smooth Banach space, Matsushita and Takahashi [8] presented their ideas as the following method for a single relatively nonexpansive mapping 𝑇 in a Banach space 𝐸: 𝑥0𝑦𝐶chosenonlyarbitrarily,𝑛=𝐽1𝛼𝑛𝐽𝑥0+1𝛼𝑛𝐽𝑇𝑥𝑛,𝐶𝑛=𝑧𝐶𝜙𝑧,𝑦𝑛𝜙𝑧,𝑥𝑛,𝑄𝑛=𝑧𝐶𝑥𝑛𝑧,𝐽𝑥0𝐽𝑥𝑛,𝑥0𝑛+1=Π𝐶𝑛𝑄𝑛𝑥0,(1.6) where 𝐽 is the duality mapping on 𝐸, and Π𝐾() is the generalized projection from 𝐸 onto a nonempty closed convex subset 𝐾. They proved the following convergence theorem.

Theorem MT. Let 𝐸 be a uniformly convex and uniformly smooth Banach space, let 𝐶 be a nonempty closed convex subset of 𝐸, let 𝑇 be a relatively nonexpansive mapping from 𝐶 into itself, and let {𝛼𝑛} be a sequence of real numbers such that 0𝛼𝑛<1 and limsup𝑛𝛼𝑛<1. Suppose that {𝑥𝑛} is given by (1.6), where 𝐽 is the duality mapping on 𝐸. If 𝐹(𝑇) is nonempty, then {𝑥𝑛} converges strongly to Π𝐹(𝑇)𝑥0, where Π𝐹(𝑇)() is the generalized projection from 𝐶 onto 𝐹(𝑇).

Recently, Plubtieng and Ungchittrakool [9] here proposed the following hybrid iteration method for a countable family of relatively nonexpansive mappings in a Banach space and proved the convergence theorem.

Theorem PU. Let 𝐸 be a uniformly smooth and uniformly convex Banach space and let 𝐶 and 𝐶 be two nonempty closed convex subsets of 𝐸 such that 𝐶𝐶. Let {𝑇𝑛} be a sequence of relatively nonexpansive mappings from 𝐶 into 𝐸 such that 𝑛=1𝐹(𝑇𝑛) is nonempty, and let {𝑥𝑛} be a sequence defined as follows: 𝑥0𝐶𝐶,1𝑥=𝐶,1=Π𝐶1𝑥0,𝑦𝑛=𝐽1𝛼𝑛𝐽𝑥𝑛+1𝛼𝑛𝐽𝑇𝑛𝑥𝑛,𝐶𝑛+1=𝑧𝐶𝑛𝜙𝑧,𝑦𝑛𝜙𝑧,𝑥𝑛𝑥,𝑛1,𝑛+1=Π𝐶𝑛+1𝑥0,(1.7) where 𝛼𝑛[0,1] satisfies either(a)0𝛼𝑛<1 for all 𝑛1 and limsup𝑛𝛼𝑛<1 or(b)liminf𝑛𝛼𝑛(1𝛼𝑛)>0. Suppose that for any bounded subset 𝐵 of 𝐶 there exists an increasing, continuous and convex function 𝐵 from 𝑅+ into 𝑅+ such that 𝐵(0)=0, and lim𝑙,𝑘sup𝐵𝑇𝑙𝑧𝑇𝑘𝑧𝑧𝐵=0.(1.8) Let 𝑇 be a mapping from 𝐶 into 𝐸 defined by 𝑇𝑥=lim𝑛𝑇𝑛𝑥 for all 𝑥𝐶 and suppose that 𝐹(𝑇)=𝑛=1𝐹𝑇𝑛=𝑛=1𝐹𝑇𝑛=𝐹(𝑇).(1.9) Then {𝑥𝑛}, {𝑇𝑛𝑥𝑛}, and {𝑦𝑛} converge strongly to Π𝐹(𝑇)𝑥0.

In this paper, the authors have obtained the following results: (1) the definition of uniformly closed countable family of nonlinear mappings, (2) strong convergence theorem by the monotone hybrid algorithm for a countable family of hemi-relatively nonexpansive mappings in a Banach space with new method of proof, (3) two examples of uniformly closed countable families of nonlinear mappings and applications, (4) an example which is hemi-relatively nonexpansive mapping but not weak relatively nonexpansive mapping, and (5) an example which is weak relatively nonexpansive mapping but not relatively nonexpansive mapping. Therefore, the results of this paper improve and extend the results of Plubtieng and Ungchittrakool [9] and many others.

We need the following definitions and lemmas.

Lemma 1.1 (Kamimura and Takahashi [10]). Let 𝐸 be a uniformly convex and smooth Banach space, and let {𝑥𝑛} and {𝑦𝑛} be two sequences of 𝐸. If 𝜙(𝑥𝑛,𝑦𝑛)0 and either {𝑥𝑛} or {𝑦𝑛} is bounded, then 𝑥𝑛𝑦𝑛0.

Lemma 1.2 (Alber [1]). Let 𝐶 be a nonempty closed convex subset of a smooth Banach space 𝐸 and 𝑥𝐸. Then, 𝑥0=Π𝐶𝑥 if and only if 𝑥0𝑦,𝐽𝑥𝐽𝑥00for𝑦𝐶.(1.10)

Lemma 1.3 (Alber [1]). Let 𝐸 be a reflexive, strictly convex, and smooth Banach space, let 𝐶 be a nonempty closed convex subset of 𝐸, and let 𝑥𝐸. Then 𝜙𝑦,Π𝐶𝑥Π+𝜙𝐶𝑥,𝑥𝜙(𝑦,𝑥)𝑦𝐶.(1.11)

The following lemma is not hard to prove.

Lemma 1.4. Let 𝐸 be a strictly convex and smooth Banach space, let 𝐶 be a closed convex subset of 𝐸, and let 𝑇 be a hemi-relatively nonexpansive mapping from 𝐶 into itself. Then 𝐹(𝑇) is closed and convex.

In this paper, we present the definition of uniformly closed for a sequence of mappings as follows.

Definition 1.5. Let 𝐸 be a Banach space, 𝐶 be a closed convex subset of 𝐸, let {𝑇𝑛}𝑛=1 be a sequence of mappings of 𝐶 into 𝐸 such that 𝑛=1𝐹(𝑇𝑛) is nonempty. We say that {𝑇𝑛}𝑛=1 is uniformly closed if 𝑝𝑛=1𝐹(𝑇𝑛) whenever {𝑥𝑛}𝐶 converges strongly to 𝑝 and 𝑥𝑛𝑇𝑛𝑥𝑛0 as 𝑛.

Lemma 1.6 (see [11, 12]). Let 𝐸 be a 𝑝-uniformly convex Banach space with 𝑝2. Then, for all 𝑥,𝑦𝐸, 𝑗(𝑥)𝐽𝑝(𝑥), and 𝑗(𝑦)𝐽𝑝(𝑦), 𝑐𝑥𝑦,𝑗(𝑥)𝑗(𝑦)𝑝𝑐𝑝2𝑝𝑥𝑦𝑝,(1.12) where 𝐽𝑝 is the generalized duality mapping from 𝐸 into 𝐸, and 1/𝑐 is the 𝑝-uniformly convexity constant of 𝐸.

2. Main Results

Definition 2.1. Let 𝐸 be a Banach space, let 𝐶 be a nonempty closed convex subset of 𝐸, and let {𝑇𝑛}𝑛=1𝐶𝐸, {𝑆𝑛}𝑛=1𝐶𝐸 be two sequences of mappings. If for any convergence sequence {𝑥𝑛}𝐶, the following holds: lim𝑛+𝑇𝑛𝑥𝑛𝑆𝑛𝑥𝑛=0.(2.1) Then {𝑇𝑛} and {𝑆𝑛} are said to satisfy the only asymptotically condition.

Theorem 2.2. Let 𝐸 be a uniformly convex and uniformly smooth Banach space, let 𝐶 be a nonempty closed convex subset of 𝐸, let {𝑇𝑛}𝑛=1𝐶𝐸, {𝑆𝑛}𝑛=1𝐶𝐸 be two uniformly closed sequences of hemi-relatively nonexpansive mappings satisfying the asymptotically condition such that 𝐹1=𝑛=1𝐹(𝑇𝑛), 𝐹2=𝑛=1𝐹(𝑆𝑛), and 𝐹=𝐹1𝐹2. Assume that {𝛼𝑛}𝑛=0, {𝛽𝑛}𝑛=0, {𝛾𝑛}𝑛=0 and {𝛿𝑛}𝑛=0 are four sequences in [0,1] such that 𝛼𝑛+𝛽𝑛+𝛾𝑛+𝛿𝑛=1, lim𝑛𝛼𝑛=0 and 0<𝛾𝛾𝑛1, 𝛿𝑛0 for some constant 𝛾(0,1). Define a sequence {𝑥𝑛} in 𝐶 by the following algorithm: 𝑥0𝑦𝐶arbitrarily,𝑛=𝐽1𝛼𝑛𝐽𝑥0+𝛽𝑛𝐽𝑥𝑛+𝛾𝑛𝐽𝑇𝑛𝑥𝑛+𝛿𝑛𝐽𝑆𝑛𝑥𝑛𝐶,𝑛1,𝑛=𝑧𝐶𝑛1𝜙𝑧,𝑦𝑛1𝛼𝑛𝜙𝑧,𝑥𝑛+𝛼𝑛𝜙𝑧,𝑥0𝐶,𝑛1,0𝑥=𝐶,𝑛+1=Π𝐶𝑛𝑥0,𝑛0.(2.2) Then {𝑥𝑛} converges strongly to 𝑞=Π𝐹𝑥0.

Proof . We first show that 𝐶𝑛 is closed and convex for all 𝑛0. From the definitions of 𝐶𝑛, it is obvious that 𝐶𝑛 is closed for all 𝑛0. Next, we prove that 𝐶𝑛 is convex for all 𝑛0. Since 𝜙𝑧,𝑦𝑛1𝛼𝑛𝜙𝑧,𝑥𝑛+𝛼𝑛𝜙𝑧,𝑥0(2.3) is equivalent to 2𝑧,1𝛼𝑛𝐽𝑥𝑛+𝛼𝑛𝐽𝑥0𝐽𝑦𝑛1𝛼𝑛𝑥𝑛2+𝛼𝑛𝑥02𝑦𝑛2,(2.4) it is easy to get that 𝐶𝑛 is convex for all 𝑛0.
Next, we show that 𝐹𝐶𝑛 for all 𝑛1. Indeed, for each 𝑝𝐹, we have 𝜙𝑝,𝑦𝑛=𝜙𝑝,𝐽1𝛼𝑛𝐽𝑥0+𝛽𝑛𝐽𝑥𝑛+𝛾𝑛𝐽𝑇𝑛𝑥𝑛+𝛿𝑛𝐽𝑆𝑛𝑥𝑛=𝑝2𝛼2𝑝,𝑛𝐽𝑥0+𝛽𝑛𝐽𝑥𝑛+𝛾𝑛𝐽𝑇𝑛𝑥𝑛+𝛿𝑛𝐽𝑆𝑛𝑥𝑛+𝛼𝑛𝐽𝑥0+𝛽𝑛𝐽𝑥𝑛+𝛾𝑛𝐽𝑇𝑛𝑥𝑛+𝛿𝑛𝐽𝑆𝑛𝑥𝑛2𝑝22𝛼𝑛𝑝,𝐽𝑥02𝛽𝑛𝑝,𝐽𝑥𝑛2𝛾𝑛𝑝,𝐽𝑇𝑛𝑥𝑛2𝛿𝑛𝑝,𝐽𝑆𝑛𝑥𝑛+𝛼𝑛𝑥02+𝛽𝑛𝑥𝑛+𝛾𝑛𝑇𝑛𝑥𝑛2+𝛿𝑛𝑆𝑛𝑥𝑛2𝛼𝑛𝜙𝑝,𝑥0+𝛽𝑛𝜙𝑝,𝑥𝑛+𝛾𝑛𝜙𝑝,𝑇𝑛𝑥𝑛+𝛿𝑛𝜙𝑝,𝑆𝑛𝑥𝑛𝛼𝑛𝜙𝑝,𝑥0+1𝛼𝑛𝜙𝑝,𝑥𝑛.(2.5) So, 𝑝𝐶𝑛, which implies that 𝐹𝐶𝑛 for all 𝑛1.
Since 𝑥𝑛+1=Π𝐶𝑛𝑥0 and 𝐶𝑛𝐶𝑛1, then we get the following: 𝜙𝑥𝑛,𝑥0𝑥𝜙𝑛+1,𝑥0,𝑛0.(2.6) Therefore, {𝜙(𝑥𝑛,𝑥0)} is nondecreasing. On the other hand, by Lemma 1.3, we have 𝜙𝑥𝑛,𝑥0Π=𝜙𝐶𝑛1𝑥0,𝑥0𝜙𝑝,𝑥0𝜙𝑝,𝑥𝑛𝜙𝑝,𝑥0,(2.7) for all 𝑝𝐹(𝑇)𝐶𝑛1 and for all 𝑛1. Therefore, 𝜙(𝑥𝑛,𝑥0) is also bounded. This together with (3.1) implies that the limit of {𝜙(𝑥𝑛,𝑥0)} exists. Put the following: lim𝑛𝜙𝑥𝑛,𝑥0=𝑑.(2.8) From Lemma 1.3, we have, for any positive integer 𝑚, that 𝜙𝑥𝑛+𝑚,𝑥𝑛+1𝑥=𝜙𝑛+𝑚,Π𝐶𝑛𝑥0𝑥𝜙𝑛+𝑚,𝑥0Π𝜙𝐶𝑛𝑥0,𝑥0𝑥=𝜙𝑛+𝑚,𝑥0𝑥𝜙𝑛+1,𝑥0,(2.9) for all 𝑛0. This together with (3.6) implies that lim𝑛𝜙𝑥𝑛+𝑚,𝑥𝑛+1=0(2.10) is, uniformly for all 𝑚, holds. By using Lemma 1.1, we get that lim𝑛𝑥𝑛+𝑚𝑥𝑛+1=0(2.11) is, uniformly for all 𝑚, holds. Then {𝑥𝑛} is a Cauchy sequence; therefore, there exists a point 𝑝𝐶 such that 𝑥𝑛𝑝.
Since 𝑥𝑛+1=Π𝐶𝑛𝑥0𝐶𝑛, from the definition of 𝐶𝑛, we have the following: 𝜙𝑥𝑛+1,𝑦𝑛1𝛼𝑛𝜙𝑥𝑛+1,𝑥𝑛+𝛼𝑛𝜙𝑥𝑛+1,𝑥0.(2.12) This together with (2.10) and lim𝑛𝛼𝑛=0 implies that lim𝑛𝜙𝑥𝑛+1,𝑦𝑛=0.(2.13) Therefore, by using Lemma 1.1, we obtain the following: lim𝑛𝑥𝑛+1𝑦𝑛=0.(2.14) Since 𝐽 is uniformly norm-to-norm continuous on bounded sets, then we have the following: lim𝑛𝐽𝑥𝑛+1𝐽𝑦𝑛=lim𝑛𝐽𝑥𝑛+1𝐽𝑥𝑛=0.(2.15) Noticing that 𝐽𝑥𝑛+1𝐽𝑦𝑛=𝐽𝑥𝑛+1𝛼𝑛𝐽𝑥0+𝛽𝑛𝐽𝑥𝑛+𝛾𝑛𝐽𝑇𝑛𝑥𝑛+𝛿𝑛𝐽𝑆𝑛𝑥𝑛=𝛼𝑛𝐽𝑥𝑛+1𝐽𝑥0+𝛽𝑛𝐽𝑥𝑛+1𝐽𝑥𝑛+𝛾𝑛𝐽𝑥𝑛+1𝐽𝑇𝑛𝑥𝑛+𝛿𝑛𝐽𝑥𝑛+1𝐽𝑇𝑛𝑥𝑛𝛾𝑛𝐽𝑥𝑛+1𝐽𝑇𝑛𝑥𝑛𝛿𝑛𝐽𝑥𝑛+1𝐽𝑆𝑛𝑥𝑛𝛼𝑛𝐽𝑥𝑛+1𝐽𝑥0𝛽𝑛𝐽𝑥𝑛+1𝐽𝑥𝑛,(2.16) which leads to 𝛾𝑛𝐽𝑥𝑛+1𝐽𝑇𝑛𝑥𝑛𝐽𝑥𝑛+1𝐽𝑦𝑛+𝛼𝑛𝐽𝑥0𝐽𝑥𝑛+1+𝛽𝑛𝐽𝑥𝑛+1𝐽𝑥𝑛+𝛿𝑛𝐽𝑥𝑛+1𝐽𝑆𝑛𝑥𝑛.(2.17) From (2.15) andlim𝑛𝛼𝑛=0, lim𝑛𝛿𝑛=0, 0<𝛾𝛾𝑛1 we obtain that lim𝑛𝐽𝑥𝑛+1𝐽𝑇𝑛𝑥𝑛=0.(2.18) Since 𝐽1 is also uniformly norm-to-norm continuous on bounded sets, then we obtain that lim𝑛𝑥𝑛+1𝑇𝑛𝑥𝑛=0.(2.19) This together with (2.11) implies that lim𝑛𝑥𝑛𝑇𝑛𝑥𝑛=0.(2.20) Sine {𝑇𝑛} and {𝑆𝑛} satisfy the asymptotically condition, we also have lim𝑛𝑥𝑛𝑆𝑛𝑥𝑛=0.(2.21) Since 𝑥𝑛𝑝 and {𝑇𝑛}𝑛=1, {𝑆𝑛}𝑛=1 are uniformly closed, we have 𝑝𝐹=𝑛=1𝐹𝑇𝑛𝑛=1𝐹𝑆𝑛.(2.22)
Finally, we prove that 𝑝=Π𝐹𝑥0, from Lemma 1.3, we have 𝜙𝑝,Π𝐹𝑥0Π+𝜙𝐹𝑥0,𝑥0𝜙𝑝,𝑥0.(2.23) On the other hand, 𝑥𝑛+1=Π𝐶𝑛𝑥0 and 𝐹𝐶𝑛, for all 𝑛. Also from Lemma 1.3, we have 𝜙Π𝐹𝑥0,𝑥𝑛+1𝑥+𝜙𝑛+1,𝑥0Π𝜙𝐹𝑥0,𝑥0.(2.24) By the definition of 𝜙(𝑥,𝑦), we know that lim𝑛𝜙𝑥𝑛+1,𝑥0=𝜙𝑝,𝑥0.(2.25) Combining (2.24) and (2.25), we know that 𝜙(𝑝,𝑥0)=𝜙(Π𝐹𝑥0,𝑥0). Therefore, it follows from the uniqueness of Π𝐹𝑥0 that 𝑝=Π𝐹𝑥0. This completes the proof.

When 𝛼𝑛0, 𝛿𝑛0 in Theorem 2.2, we obtain the following result.

Theorem 2.3. Let 𝐸 be a uniformly convex and uniformly smooth Banach space, let 𝐶 be a nonempty closed convex subset of 𝐸, and let {𝑇𝑛}𝑛=1𝐶𝐸 be a uniformly closed sequence of hemi-relatively nonexpansive mappings such that 𝐹=𝑛=1𝐹(𝑇𝑛). Assume that {𝛼𝑛}𝑛=0 is a sequences in [0,1] such that 0𝛼𝑛𝛼<1 for some constant 𝛼(0,1). Define a sequence {𝑥𝑛} in 𝐶 by the following algorithm: 𝑥0𝑦𝐶arbitrarily,𝑛=𝐽1𝛼𝑛𝐽𝑥𝑛+1𝛼𝑛𝐽𝑇𝑛𝑥𝑛𝐶,𝑛1,𝑛=𝑧𝐶𝑛1𝜙𝑧,𝑦𝑛𝜙𝑧,𝑥𝑛𝐶,𝑛1,0𝑥=𝐶,𝑛+1=Π𝐶𝑛𝑥0,𝑛0.(2.26) Then {𝑥𝑛} converges strongly to 𝑞=Π𝐹𝑥0.

3. Applications for Equilibrium Problem

Let 𝐸 be a real Banach space, and let 𝐸 be the dual space of 𝐸. Let 𝐶 be a closed convex subset of 𝐸. Let 𝑓 be a bifunction from 𝐶×𝐶 to 𝑅=(,+). The equilibrium problem is to find 𝑥𝐶 such that 𝑓(𝑥,𝑦)0,𝑦𝐶.(3.1) The set of solutions of (1.2) is denoted by EP(𝑓). Given a mapping 𝑇𝐶𝐸 let 𝑓(𝑥,𝑦)=𝑇𝑥,𝑦𝑥 for all 𝑥,𝑦𝐶. Then, 𝑝EP(𝑓) if and only if 𝑇𝑝,𝑦𝑝0 for all 𝑦𝐶, that is, 𝑝 is a solution of the variational inequality. Numerous problems in physics, optimization, and economics reduce to find a solution of (1.2). Some methods have been proposed to solve the equilibrium problem in Hilbert spaces, see, for instance, [1315].

For solving the equilibrium problem, let us assume that a bifunction 𝑓 satisfies the following conditions:(A1)𝑓(𝑥,𝑥)=0,forall𝑥𝐸,(A2)𝑓 is monotone, that is, 𝑓(𝑥,𝑦)+𝑓(𝑦,𝑥)0,forall𝑥,𝑦𝐸,(A3) for all 𝑥,𝑦,𝑧𝐸, only limsup𝑡0𝑓(𝑡𝑧+(1𝑡)𝑥,𝑦)𝑓(𝑥,𝑦),(A4) for all 𝑥𝐶, 𝑓(𝑥,) is convex and lower semicontinuous.

Lemma 3.1 (Blum and Oettli [13]). Let 𝐶 be a closed convex subset of a smooth, strictly convex, and reflexive Banach space 𝐸, let 𝑓 be a bifunction from 𝐶×𝐶 to 𝑅=(,+) satisfying (A1)–(A4), and let 𝑟>0 and 𝑥𝐸. Then, there exists 𝑧𝐶 such that 1𝑓(𝑧,𝑦)+𝑟𝑦𝑧,𝐽𝑧𝐽𝑥0,𝑦𝐶.(3.2)

Lemma 3.2 (Takahashi and Zembayashi [15]). Let 𝐶 be a closed convex subset of a uniformly smooth, strictly convex, and reflexive Banach space 𝐸, and let 𝑓 be a bifunction from 𝐶×𝐶 to 𝑅=(,+) satisfying (A1)–(A4). For 𝑟>0, define a mapping 𝑇𝑟𝐸𝐶 as follows: 𝑇𝑟1(𝑥)=𝑧𝐶𝑓(𝑧,𝑦)+𝑟𝑦𝑧,𝐽𝑧𝐽𝑥0,𝑦𝐶(3.3) for all 𝑥𝐸. Then, the following hold:(1)𝑇𝑟 is single-valued;(2)𝑇𝑟 is a firmly nonexpansive-type mapping, that is, for all 𝑥,𝑦𝐸, 𝑇𝑟𝑥𝑇𝑟𝑦,𝐽𝑇𝑟𝑥𝐽𝑇𝑟𝑦𝑇𝑟𝑥𝑇𝑟𝑦,𝐽𝑥𝐽𝑦,(3.4)(3)𝐹(𝑇𝑟)=EP(𝑓);(4)EP(𝑓) is closed and convex;(5)𝑇𝑟 is also a relatively nonexpansive mapping.

Lemma 3.3 (Takahashi and Zembayashi [15]). Let 𝐶 be a closed convex subset of a smooth, strictly convex, and reflexive Banach space 𝐸, let 𝑓 be a bifunction from 𝐶×𝐶 to 𝑅=(,+) satisfying (A1)–(A4), let 𝑟>0, and let 𝑥𝐸, 𝑞𝐹(𝑇𝑟), then the following holds: 𝜙𝑞,𝑇𝑟𝑥𝑇+𝜙𝑟𝑥,𝑥𝜙(𝑞,𝑥).(3.5)

Lemma 3.4. Let 𝐸 be a 𝑝-uniformly convex with 𝑝2 and uniformly smooth Banach space, and let 𝐶 be a nonempty closed convex subset of 𝐸. Let 𝑓 be a bifunction from 𝐶×𝐶 to 𝑅=(,+) satisfying (A1)–(A4). Let {𝑟𝑛} be a positive real sequence such that lim𝑛𝑟𝑛=𝑟>0. Then the sequence of mappings {𝑇𝑟𝑛} is uniformly closed.

Proof. (1) Let {𝑥𝑛} be a convergent sequence in 𝐶. Let 𝑧𝑛=𝑇𝑟𝑛𝑥𝑛 for all 𝑛, then 𝑓𝑧𝑛+1,𝑦𝑟𝑛𝑦𝑧𝑛,𝐽𝑧𝑛𝐽𝑥𝑛𝑓𝑧0,𝑦𝐶,(3.6)𝑛+𝑚+1,𝑦𝑟𝑛+𝑚𝑦𝑧𝑛+𝑚,𝐽𝑧𝑛+𝑚𝐽𝑥𝑛+𝑚0,𝑦𝐶.(3.7) Putting 𝑦=𝑧𝑛+𝑚 in (3.6) and 𝑦=𝑧𝑛 in (3.7), we have 𝑓𝑧𝑛,𝑧𝑛+𝑚+1𝑟𝑛𝑧𝑛+𝑚𝑧𝑛,𝐽𝑧𝑛𝐽𝑥𝑛𝑓𝑧0,𝑦𝐶,𝑛+𝑚,𝑧𝑛+1𝑟𝑛+𝑚𝑧𝑛𝑧𝑛+𝑚,𝐽𝑧𝑛+𝑚𝐽𝑥𝑛+𝑚0,𝑦𝐶.(3.8) So, from (A2), we have 𝑧𝑛+𝑚𝑧𝑛,𝐽𝑧𝑛𝐽𝑥𝑛𝑟𝑛𝐽𝑧𝑛+𝑚𝐽𝑥𝑛+𝑚𝑟𝑛+𝑚0,(3.9) and hence 𝑧𝑛+𝑚𝑧𝑛,𝐽𝑧𝑛𝐽𝑥𝑛𝑟𝑛𝑟𝑛+𝑚𝐽𝑧𝑛+𝑚𝐽𝑥𝑛+𝑚0.(3.10) Thus, we have 𝑧𝑛+𝑚𝑧𝑛,𝐽𝑧𝑛𝐽𝑧𝑛+𝑚+𝐽𝑧𝑛+𝑚𝐽𝑥𝑛𝑟𝑛𝑟𝑛+𝑚𝐽𝑧𝑛+𝑚𝐽𝑥𝑛+𝑚0,(3.11) which implies that 𝑧𝑛+𝑚𝑧𝑛,𝐽𝑧𝑛+𝑚𝐽𝑧𝑛𝑧𝑛+𝑚𝑧𝑛,𝐽𝑧𝑛+𝑚𝐽𝑥𝑛𝑟𝑛𝑟𝑛+𝑚𝐽𝑧𝑛+𝑚𝐽𝑥𝑛+𝑚0.(3.12) By using Lemma 1.6, we obtain the following: 𝑐𝑝𝑐𝑝2𝑝𝑧𝑛+𝑚𝑧𝑛𝑝𝑧𝑛+𝑚𝑧𝑛,𝐽𝑧𝑛+𝑚𝐽𝑥𝑛𝑟𝑛𝑟𝑛+𝑚𝐽𝑧𝑛+𝑚𝐽𝑥𝑛+𝑚=𝑧0𝑛+𝑚𝑧𝑛,𝑟1𝑛𝑟𝑛+𝑚𝐽𝑧𝑛+𝑚+𝑟𝑛𝑟𝑛+𝑚𝐽𝑥𝑛+𝑚𝐽𝑥𝑛.(3.13) Therefore, we get the following: 𝑐𝑝𝑐𝑝2𝑝𝑧𝑛+𝑚𝑧𝑛𝑝1||||𝑟1𝑛𝑟𝑛+𝑚||||𝐽𝑧𝑛+𝑚+𝑟𝑛𝑟𝑛+𝑚𝐽𝑥𝑛+𝑚𝐽𝑥𝑛.(3.14) On the other hand, for any 𝑝EP(𝑓), from 𝑧𝑛=𝑇𝑟𝑛𝑥𝑛, we have 𝑧𝑛=𝑇𝑝𝑟𝑛𝑥𝑛𝑥𝑝𝑛,𝑝(3.15) so that {𝑧𝑛} is bounded. Since lim𝑛𝑟𝑛=𝑟>0, this together with (3.14) implies that {𝑧𝑛} is a Cauchy sequence. Hence 𝑇𝑟𝑛𝑥𝑛=𝑧𝑛 is convergent.
(2) By using Lemma 3.2, we know that 𝑛=1𝐹𝑇𝑟𝑛=EP(𝑓).(3.16)
(3) From (1) we know that, lim𝑛𝑇𝑟𝑛𝑥 exists for all 𝑥𝐶. So, we can define a mapping 𝑇 from 𝐶 into itself by 𝑇𝑥=lim𝑛𝑇𝑟𝑛𝑥,𝑥𝐶.(3.17) It is obvious that 𝑇 is nonexpansive. It is easy to see that EP(𝑓)=𝑛=1𝐹𝑇𝑟𝑛𝐹(𝑇).(3.18) On the other hand, let 𝑤𝐹(𝑇) and 𝑤𝑛=𝑇𝑟𝑛𝑤, we have 𝑓𝑤𝑛+1,𝑦𝑟𝑛𝑦𝑤𝑛,𝐽𝑤𝑛𝐽𝑤0,𝑦𝐶.(3.19) By (A2) we know that 1𝑟𝑛𝑦𝑤𝑛,𝐽𝑤𝑛𝐽𝑤𝑓𝑦,𝑤𝑛,𝑦𝐶.(3.20) Since 𝑤𝑛𝑇𝑤=𝑤 and from (A4), we have 𝑓(𝑦,𝑤)0, for all 𝑦𝐶. Then, for 𝑡(0,1] and 𝑦𝐶, 0=𝑓(𝑡𝑦+(1𝑡)𝑤,𝑡𝑦+(1𝑡)𝑤)𝑡𝑓(𝑡𝑦+(1𝑡)𝑤,𝑦)+(1𝑡)𝑓(𝑡𝑦+(1𝑡)𝑤,𝑤)𝑡𝑓(𝑡𝑦+(1𝑡)𝑤,𝑦).(3.21) Therefore, we have 𝑓(𝑡𝑦+(1𝑡)𝑤,𝑦)0.(3.22) Letting 𝑡0 and using (A3), we get the following: 𝑓(𝑤,𝑦)0,𝑦𝐶,(3.23) and hence 𝑤EP(𝑓). From above two respects, we know that 𝐹(𝑇)=𝑛=0𝐹(𝑇𝑟𝑛).
Next we show {𝑇𝑟𝑛} is uniformly closed. Assume 𝑥𝑛𝑥 and 𝑥𝑛𝑇𝑟𝑛𝑥𝑛0, from above results, we know that 𝑇𝑥=lim𝑛𝑇𝑟𝑛𝑥. On the other hand, from 𝑥𝑛𝑇𝑟𝑛𝑥𝑛0, we also get lim𝑛𝑇𝑟𝑛𝑥=𝑥, so that 𝑥𝐹(𝑇)=𝑛=1𝐹(𝑇𝑟𝑛). That is, the sequence of mappings {𝑇𝑟𝑛} is uniformly closed. This completes the proof.

Theorem 3.5. Let 𝐸 be a 𝑝-uniformly convex with 𝑝2 and uniformly smooth Banach space, and let 𝐶 be a nonempty closed convex subset of 𝐸. Let 𝑓 and 𝑔 be two bifunctions from 𝐶×𝐶 to 𝑅=(,+) satisfying (A1)–(A4). Assume that {𝛼𝑛}𝑛=0, {𝛽𝑛}𝑛=0, {𝛾𝑛}𝑛=0, and {𝛿𝑛}𝑛=1 are four sequences in [0,1] such that 𝛼𝑛+𝛽𝑛+𝛾𝑛+𝛿𝑛=1, lim𝑛𝛼𝑛=0,lim𝑛𝛿𝑛=0, and 0<𝛾𝛾𝑛1 for some constant 𝛾(0,1). Let {𝑥𝑛} be a sequence generated by 𝑥0𝑦𝐶arbitrarily,𝑛=𝐽1𝛼𝑛𝐽𝑥0+𝛽𝑛𝐽𝑥𝑛+𝛾𝑛𝐽𝑇𝑟𝑛𝑥𝑛+𝛿𝑛𝐽𝑆𝑟𝑛𝑥𝑛𝐶,𝑛1,𝑛=𝑧𝐶𝑛1𝜙z,𝑦𝑛1𝛼𝑛𝜙𝑧,𝑥𝑛+𝛼𝑛𝜙𝑧,𝑥0𝐶,𝑛1,0𝑥=𝐶,𝑛+1=Π𝐶𝑛𝑥0,𝑛0,(3.24) where 𝑇𝑟1(𝑥)=𝑧𝐶𝑓(𝑧,𝑦)+𝑟𝑆𝑦𝑧,𝐽𝑧𝐽𝑥0,𝑦𝐶,𝑥𝐸,𝑟1(𝑥)=𝑧𝐶𝑔(𝑧,𝑦)+𝑟𝑦𝑧,𝐽𝑧𝐽𝑥0,𝑦𝐶,𝑥𝐸,(3.25) and lim𝑛𝑟𝑛=𝑟>0. Assume that the mappings 𝑇𝑛 and 𝑆𝑛 satisfy the asymptotically condition. Then {𝑥𝑛} converges strongly to 𝑞=ΠEP(𝑓)EP(𝑔)𝑥0.

Proof. By Lemma 3.4, {𝑇𝑟𝑛}𝑛=1, {𝑆𝑟𝑛}𝑛=1 are uniformly closed; therefore, by using Theorem 2.2 and Lemma 3.2, we can obtain the conclusion of Theorem 3.5. This completes the proof.

When 𝛼𝑛0, 𝛿𝑛0 in the Theorem 3.5, we obtain the following result.

Theorem 3.6. Let 𝐸 be a 𝑝uniformly convex with 𝑝2 and uniformly smooth Banach space, and let 𝐶 be a nonempty closed convex subset of 𝐸. Let 𝑓 be a bifunction from 𝐶×𝐶 to 𝑅=(,+) satisfying (A1)–(A4). Assume that {𝛼𝑛}𝑛=0 is a sequences in [0,1] such that 0𝛼𝑛𝛼<1 for some constant 𝛼(0,1). Define a sequence {𝑥𝑛} in 𝐶 by the following algorithm: 𝑥0𝑦𝐶arbitrarily,𝑛=𝐽1𝛼𝑛𝐽𝑥𝑛+1𝛼𝑛𝐽𝑇𝑟𝑛𝑥𝑛𝐶,𝑛1,𝑛=𝑧𝐶𝑛1𝜙𝑧,𝑦𝑛𝜙𝑧,𝑥𝑛𝐶,𝑛1,0𝑥=𝐶,𝑛+1=Π𝐶𝑛𝑥0,𝑛0,(3.26) where lim𝑛𝑟𝑛=𝑟>0. Then {𝑥𝑛} converges strongly to 𝑞=ΠEP(𝑓)𝑥0.

4. Applications for Maximal Monotone Operators

In this section, we apply our above results to prove some strong convergence theorem concerning maximal monotone operators in a Banach space 𝐸.

Let 𝐴 be a multivalued operator from 𝐸 to 𝐸 with domain 𝐷(𝐴)={𝑧𝐸𝐴𝑧} and range 𝑅(𝐴)={𝑧𝐸𝑧𝐷(𝐴)}. An operator 𝐴 is said to be monotone if 𝑥1𝑥2,𝑦1𝑦20(4.1) for each 𝑥1,𝑥2𝐷(𝐴) and 𝑦1𝐴𝑥1, 𝑦2𝐴𝑥2. A monotone operator 𝐴 is said to be maximal if it's graph 𝐺(𝐴)={(𝑥,𝑦)𝑦𝐴𝑥} is not properly contained in the graph of any other monotone operator. We know that if 𝐴 is a maximal monotone operator, then 𝐴10 is closed and convex. The following result is also wellknown.

Theorem 4.1. Let 𝐸 be a reflexive, strictly convex, and smooth Banach space, and let 𝐴 be a monotone operator from 𝐸 to 𝐸. Then 𝐴 is maximal if and only if 𝑅(𝐽+𝑟𝐴)=𝐸. for all 𝑟>0.

Let 𝐸 be a reflexive, strictly convex, and smooth Banach space, and let 𝐴 be a maximal monotone operator from 𝐸 to 𝐸. Using Theorem 4.1 and strict convexity of 𝐸, we obtain that for every 𝑟>0 and 𝑥𝐸, there exists a unique 𝑥𝑟 such that 𝐽𝑥𝐽𝑥𝑟+𝑟𝐴𝑥𝑟.(4.2)

Then we can define a single-valued mapping 𝐽𝑟𝐸𝐷(𝐴) by 𝐽𝑟=(𝐽+𝑟𝐴)1𝐽 and such a 𝐽𝑟 is called the resolvent of 𝐴, we know that 𝐴10=𝐹(𝐽𝑟) for all 𝑟>0.

Theorem 4.2. Let 𝐸 be a uniformly convex and uniformly smooth Banach space, let 𝐴 be a maximal monotone operator from 𝐸 to 𝐸, and let 𝐽𝑟 be a resolvent of 𝐴 for 𝑟>0. Then for any sequence {𝑟𝑛}𝑛=1 such that liminf𝑛𝑟𝑛>0, {𝐽𝑟𝑛}𝑛=1 is a uniformly closed sequence of hemi-relatively nonespansive mappings.

Proof. Firstly, we show that {𝐽𝑟𝑛}𝑛=1 is uniformly closed. Let {𝑧𝑛}𝐸 be a sequence such that 𝑧𝑛𝑝 and lim𝑛𝑧𝑛𝐽𝑟𝑛𝑧𝑛=0. Since 𝐽 is uniformly norm-to-norm continuous on bounded sets, we obtain that 1𝑟𝑛𝐽𝑧𝑛𝐽𝐽𝑟𝑛𝑧𝑛0.(4.3) It follows from 1𝑟𝑛𝐽𝑧𝑛𝐽𝐽𝑟𝑛𝑧𝑛𝐴𝐽𝑟𝑛𝑧𝑛(4.4) and the monotonicity of 𝐴 that 𝑤𝐽𝑟𝑛𝑧𝑛,𝑤1𝑟𝑛𝐽𝑧𝑛𝐽𝐽𝑟𝑛𝑧𝑛0(4.5) for all 𝑤𝐷(𝐴) and 𝑤𝐴𝑤. Letting 𝑛, we have 𝑤𝑝,𝑤0 for all 𝑤𝐷(𝐴) and 𝑤𝐴𝑤. Therefore, from the maximality of 𝐴, we obtain that 𝑝𝐴10=𝐹(𝐽𝑟𝑛) for all 𝑛1, that is, 𝑝𝑛=1𝐹(𝐽𝑟𝑛).
Next we show that 𝐽𝑟𝑛 is a hemi-relatively nonexpansive mapping for all 𝑛1. For any 𝑤𝐸 and 𝑝𝐹(𝐽𝑟𝑛)=𝐴10, from the monotonicity of 𝐴, we have 𝜙𝑝,𝐽𝑟𝑛𝑤=𝑝22𝑝,𝐽𝐽𝑟𝑛𝑤+𝐽𝑟𝑛𝑤2=𝑝2+2𝑝,𝐽𝑤𝐽𝐽𝑟𝑛+𝐽𝑤𝐽𝑤𝑟𝑛𝑤2=𝑝2+2𝑝,𝐽𝑤𝐽𝐽𝑟𝑛𝑤𝐽2𝑝,𝐽𝑤+𝑟𝑛𝑤2=𝑝2𝐽2𝑟𝑤𝑝𝐽𝑟𝑛𝑤,𝐽𝑤𝐽𝐽𝑟𝑛𝑤𝐽2𝑝,𝐽𝑤+𝑟𝑛𝑤2=𝑝2𝐽2𝑟𝑛𝑤𝑝,𝐽𝑤𝐽𝐽𝑟𝑛𝑤𝐽+2𝑟𝑛𝑤,𝐽𝑤𝐽𝐽𝑟𝑛𝑤𝐽2𝑝,𝐽𝑤+𝑟𝑛𝑤2𝑝2𝐽+2𝑟𝑛𝑤,𝐽𝑤𝐽𝐽𝑟𝑛𝑤𝐽2𝑝,𝐽𝑤+𝑟𝑛𝑤2=𝑝22𝑝,𝐽𝑤+𝑤2𝐽𝑟𝑛𝑤2𝐽+2𝑟𝑛𝑤,𝐽𝑤𝑤2𝐽=𝜙(𝑝,𝑤)𝜙𝑟𝑛𝑤,𝑤𝜙(𝑝,𝑤).(4.6) This implies that 𝐽𝑟𝑛 is a hemi-relatively nonexpansive mapping for all 𝑛1. This completes the proof.

Theorem 4.3. Let 𝐸 be a uniformly convex and uniformly smooth Banach space, let 𝐴 and 𝐵 be two maximal monotone operators from 𝐸 to 𝐸 with nonempty common zero point set 𝐴1(0)𝐵1(0), let 𝐽𝐴𝑟 be a resolvent of 𝐴 and 𝐽𝐵𝑟 a resolvent of 𝐵 for 𝑟>0. Assume that {𝛼𝑛}𝑛=0, {𝛽𝑛}𝑛=0, {𝛾𝑛}𝑛=0, and {𝛿𝑛}𝑛=1 are four sequences in [0,1] such that 𝛼𝑛+𝛽𝑛+𝛾𝑛=1, lim𝑛𝛼𝑛=0, lim𝑛𝛿𝑛=0, and 0<𝛾𝛾𝑛1 for some constant 𝛾(0,1). Let {𝑥𝑛} be a sequence generated by 𝑥0𝑦𝐶arbitrarily,𝑛=𝐽1𝛼𝑛𝐽𝑥0+𝛽𝑛𝐽𝑥𝑛+𝛾𝑛𝐽𝐽𝐴𝑟𝑛𝑥𝑛+𝛿𝑛𝐽𝐽𝐵𝑟𝑛𝑥𝑛𝐶,𝑛1,𝑛=𝑧𝐶𝑛1𝜙𝑧,𝑦𝑛1𝛼𝑛𝜙𝑧,𝑥𝑛+𝛼𝑛𝜙𝑧,𝑥0𝐶,𝑛1,0𝑥=𝐶,𝑛+1=Π𝐶𝑛𝑥0,𝑛0,(4.7) where liminf𝑛𝑟𝑛>0. Assume that the mappings 𝐽𝐴𝑟𝑛 and 𝐽𝐵𝑟𝑛 satisfy the asymptotically condition. Then {𝑥𝑛} converges strongly to 𝑞=Π𝐴1(0)𝐵1(0)𝑥0.

Proof. From Theorem 4.2, {𝐽𝑟𝑛}𝑛=1 is uniformly closed countable family of hemi-relatively nonexpansive mappings, on the other hand, 𝐴1(0)=𝑛=1𝐹(𝐽𝑟𝑛), by using Theorem 2.2, we can obtain the conclusion of Theorem 4.3.

When 𝛼𝑛0,𝛿𝑛0 in Theorem 4.3, we obtain the following result.

Theorem 4.4. Let 𝐸 be a uniformly convex and uniformly smooth Banach space, let 𝐴 be a maximal monotone operator from 𝐸 to 𝐸 with nonempty zero point set 𝐴1(0), and let 𝐽𝑟 be a resolvent of 𝐴 for 𝑟>0. Assume that {𝛼𝑛}𝑛=0 is a sequence in [0,1] such that 0𝛼𝑛𝛼<1 for some constant 𝛼(0,1). Define a sequence {𝑥𝑛} in 𝐶 by the following algorithm: 𝑥0𝑦𝐶arbitrarily,𝑛=𝐽1𝛼𝑛𝐽𝑥𝑛+1𝛼𝑛𝐽𝐽𝑟𝑛𝑥𝑛𝐶,𝑛1,𝑛=𝑧𝐶𝑛1𝜙𝑧,𝑦𝑛𝜙𝑧,𝑥𝑛𝐶,𝑛1,0𝑥=𝐶,𝑛+1=Π𝐶𝑛𝑥0,𝑛0,(4.8) where liminf𝑛𝑟𝑛>0. Then {𝑥𝑛} converges strongly to 𝑞=Π𝐴1(0)𝑥0.

5. Examples

Firstly, we give an example which is hemi-relatively nonexpansive mapping but not weak relatively nonexpansive mapping.

Example 5.1. Let 𝐸=𝑅𝑛 and 𝑥00 be a any element of 𝐸. We define a mapping 𝑇𝐸𝐸 as follows: 1𝑇(𝑥)=2+12𝑛+1𝑥01if𝑥=2+12𝑛𝑥0,1𝑥if𝑥2+12𝑛𝑥0,(5.1) for 𝑛=1,2,3,. Next we show that 𝑇 is a hemi-relatively nonexpansive mapping but no weak relatively nonexpansive mapping. First, it is obvious that 𝐹(𝑇)={0}. In addition, it is easy to see that 𝑇𝑥𝑥,𝑥𝐸.(5.2) This implies that 𝑇𝑥2𝑥220,𝐽𝑇𝑥𝐽𝑥=2𝑝,𝐽𝑇𝑥𝐽𝑥(5.3) for all 𝑝𝐹(𝑇). It follows from above inequality that 𝑝22𝑝,𝐽𝑇𝑥+𝑇𝑥2𝑝22𝑝,𝐽𝑥+𝑥2,(5.4) for all 𝑝𝐹(𝑇) and 𝑥𝐸. That is 𝜙(𝑝,𝑇𝑥)𝜙(𝑝,𝑥),(5.5) for all 𝑝𝐹(𝑇) and 𝑥𝐸; hence 𝑇 is a hemi-relatively nonexpansive mapping. Finally, we show that 𝑇 is not weak relatively nonexpansive mapping. In fact that, letting 𝑥𝑛=12+12𝑛𝑥0,𝑛=1,2,3,(5.6) from the definition of 𝑇, we have 𝑇𝑥𝑛=12+12𝑛+1𝑥0,𝑛=1,2,3,(5.7) which implies that 𝑥𝑛𝑇𝑥𝑛0 and 𝑥𝑛𝑥0 (𝑥𝑛𝑥0) as 𝑛. That is 𝑥0𝐹(𝑇) but 𝑥0𝐹(𝑇).
Next, we give an example which is weak relatively nonexpansive mapping but not relatively nonexpansive mapping.

Example 5.2. Let 𝐸=𝑙2, where 𝑙2=𝜉𝜉=1,𝜉2,𝜉3,,𝜉𝑛,𝑛=1||𝜉𝑛||2,<𝜉=𝑛=1||𝜉𝑛||21/2,𝜉𝑙2,𝜉,𝜂=𝑛=1𝜉𝑛𝜂𝑛𝜉,𝜉=1,𝜉2,𝜉3,,𝜉𝑛𝜂,,𝜂=1,𝜂2,𝜂3,,𝜂𝑛,.𝑙2.(5.8) It is well known that 𝑙2 is a Hilbert space, so that (𝑙2)=𝑙2. Let {𝑥𝑛}𝐸 be a sequence defined by 𝑥0𝑥=(1,0,0,0,),1=𝑥(1,1,0,0,),2𝑥=(1,0,1,0,0,),3𝑥=(1,0,0,1,0,0,),𝑛=𝜉𝑛,1,𝜉𝑛,2,𝜉𝑛,3,,𝜉𝑛,𝑘,,(5.9) where 𝜉𝑛,𝑘=1if𝑘=1,𝑛+1,0if𝑘1,𝑘𝑛+1,(5.10) for all 𝑛1. Define a mapping 𝑇𝐸𝐸 as follows: 𝑛𝑇(𝑥)=𝑥𝑛+1𝑛if𝑥=𝑥𝑛(𝑛1),𝑥if𝑥𝑥𝑛(𝑛1).(5.11)

Conclusion 1. {𝑥𝑛} converges weakly to 𝑥0.

Proof. For any 𝑓=(𝜁1,𝜁2,𝜁3,,𝜁𝑘,)𝑙2=(𝑙2), we have 𝑓𝑥𝑛𝑥0=𝑓,𝑥𝑛𝑥0=𝑘=2𝜁𝑘𝜉𝑛,𝑘=𝜁𝑛+10,(5.12) as 𝑛. That is, {𝑥𝑛} converges weakly to 𝑥0.

Conclusion 2. {𝑥𝑛} is not a Cauchy sequence, so that, it does not converge strongly to any element of 𝑙2.

Proof. In fact, we have 𝑥𝑛𝑥𝑚=2 for any 𝑛𝑚. Then {𝑥𝑛} is not a Cauchy sequence.

Conclusion 3. 𝑇 has a unique fixed point 0, that is, 𝐹(𝑇)={0}.

Proof . The conclusion is obvious.

Conclusion 4. 𝑥0 is an asymptotic fixed point of 𝑇.

Proof. Since {𝑥𝑛} converges weakly to 𝑥0 and 𝑇𝑥𝑛𝑥𝑛=𝑛𝑥𝑛+1𝑛𝑥𝑛=1𝑥𝑛+1𝑛0(5.13) as 𝑛, so that, 𝑥0 is an asymptotic fixed point of 𝑇.

Conclusion 5. 𝑇 has a unique strong asymptotic fixed point 0, so that 𝐹(𝑇)=𝐹(𝑇).

Proof . In fact that, for any strong convergent sequence {𝑧𝑛}𝐸 such that 𝑧𝑛𝑧0 and 𝑧𝑛𝑇𝑧𝑛0 as 𝑛, from Conclusion 2, there exists sufficiently large nature number 𝑁 such that 𝑧𝑛𝑥𝑚, for any 𝑛,𝑚>𝑁. Then 𝑇𝑧𝑛=𝑧𝑛 for 𝑛>𝑁, and it follows from 𝑧𝑛𝑇𝑧𝑛0 that 2𝑧𝑛0 and hence 𝑧𝑛𝑧0=0.

Conclusion 6. 𝑇 is a weak relatively nonexpansive mapping.

Proof. Since 𝐸=𝑙2 is a Hilbert space, we have 𝜙(0,𝑇𝑥)=0𝑇𝑥2=𝑇𝑥2𝑥2=𝑥02=𝜙(0,𝑥),𝑥𝐸.(5.14) From Conclusion 2, we have 𝐹(𝑇)=𝐹(𝑇), then 𝑇 is a weak relatively nonexpansive mapping.

Conclusion 7. 𝑇 is not a relatively nonexpansive mapping.

Proof. From Conclusions 3 and 4, we have 𝐹(𝑇)𝐹(𝑇), so that 𝑇 is not a relatively nonexpansive mapping.

Acknowledgment

This project is supported by the National Natural Science Foundation of China under Grant (11071279).

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