Abstract

Let {𝑡𝑛}(0,1) be such that 𝑡𝑛1 as 𝑛, let 𝛼 and 𝛽 be two positive numbers such that 𝛼+𝛽=1, and let 𝑓 be a contraction. If 𝑇 be a continuous asymptotically pseudocontractive self-mapping of a nonempty bounded closed convex subset 𝐾 of a real reflexive Banach space with a uniformly Gateaux differentiable norm, under suitable conditions on the sequence {𝑡𝑛}, we show the existence of a sequence {𝑥𝑛}𝑛 satisfying the relation 𝑥𝑛=(1𝑡𝑛/𝑘𝑛)𝑓(𝑥𝑛)+(𝑡𝑛/𝑘𝑛)𝑇𝑛𝑥𝑛 and prove that {𝑥𝑛} converges strongly to the fixed point of 𝑇, which solves some variational inequality provided 𝑇 is uniformly asymptotically regular. As an application, if 𝑇 be an asymptotically nonexpansive self-mapping of a nonempty bounded closed convex subset 𝐾 of a real Banach space with a uniformly Gateaux differentiable norm and which possesses uniform normal structure, we prove that the iterative process defined by 𝑧0𝐾,𝑧𝑛+1=(1𝑡𝑛/𝑘𝑛)𝑓(𝑧𝑛)+(𝛼𝑡𝑛/𝑘𝑛)𝑇𝑛𝑧𝑛+(𝛽𝑡𝑛/𝑘𝑛)𝑧𝑛 converges strongly to the fixed point of 𝑇.

1. Introduction

Let 𝐸 be a real Banach space with dual 𝐸 and 𝐾 a nonempty closed convex subset of 𝐸. Recall that a mapping 𝑇𝐾𝐾 is said to be asymptotically pseudocontractive if, for each 𝑛𝑁 and 𝑥,𝑦𝐾, there exist 𝑗𝐽(𝑥𝑦) and a constant 𝑘𝑛1 with lim𝑛𝑘𝑛=1 such that 𝑇𝑛𝑥𝑇𝑛𝑦𝑘𝑛𝑥𝑦2,(1.1) where 𝐽𝐸2𝐸 denote the normalized duality mapping defined by 𝐽𝑥(𝑥)=𝐸𝑥,𝑥=𝑥2,𝑥.=𝑥,𝑥𝐸(1.2) The class of asymptotically pseudocontractive mappings is essentially wider than the class of asymptotically nonexpansive mappings. A mapping 𝑇 is called asymptotically nonexpansive if there exists a sequence {𝑘𝑛}[1,) with lim𝑛𝑘𝑛=1 such that 𝑇𝑛𝑥𝑇𝑛𝑦𝑘𝑛𝑥𝑦(1.3) for all integers 𝑛0 and all 𝑥,𝑦𝐾. A mapping 𝑓𝐾𝐾 is called a contraction if there exists a constant 𝛾[0,1) such that 𝑓(𝑥)𝑓(𝑦)𝛾𝑥𝑦,𝑥,𝑦𝐾.(1.4) It is clear that every contraction is nonexpansive, every nonexpansive mapping is asymptotically nonexpansive, and every asymptotically nonexpansive mapping is asymptotically pseudocontractive. The converses do not hold. The asymptotically nonexpansive mappings are important generalizations of nonexpansive mappings. For details, you may see [1].

The mapping 𝑇 is called uniformly asymptotically regular (in short u.a.r.) if for each 𝜖>0 there exists 𝑛0𝑁 such that 𝑇𝑛+1𝑥𝑇𝑛𝑥𝜖,(1.5) for all 𝑛𝑛0 and 𝑥𝐾 and it is called uniformly asymptotically regular with sequence {𝜖𝑛} (in short u.a.r.s.) if 𝑇𝑛+1𝑥𝑇𝑛𝑥𝜖𝑛,(1.6) for all integers 𝑛1 and all 𝑥𝐾, where 𝜖𝑛0 as 𝑛.

The viscosity approximation method of selecting a particular fixed point of a given nonexpansive mapping was proposed by Moudafi [2] who proved the strong convergence of both the implicit and explicit methods in Hilbert spaces, see [2, Theorems 2.1 and 2.2]. Xu [3] studied the viscosity approximation methods proposed by Moudafi [2] for a nonexpansive mapping in a uniformly smooth Banach space.

Very recently, Shahzad and Udomene [4] obtained fixed point solutions of variational inequalities for an asymptotically nonexpansive mapping defined on a real Banach space with uniformly Gateaux differentiable norm possessing uniform normal structure. They proved the following theorem.

Theorem 1.1. Let 𝐸 be a real Banach space with a uniformly Gateaux differentiable norm possessing uniform normal structure, let 𝐾 be a nonempty closed convex and bounded subset of 𝐸, let 𝑇𝐾𝐾 be an asymptotically nonexpansive mapping with sequence {𝑘𝑛}𝑛[1,), and let 𝑓𝐾𝐾 be a contraction with constant 𝛼[0,1). Let {𝑡𝑛}𝑛(0,𝜉𝑛) be such that lim𝑛𝑡𝑛=1,  𝑛=1𝑡𝑛(1𝑡𝑛)=, and lim𝑛((𝑘𝑛1)/(𝑘𝑛𝑡𝑛))=0, where 𝜉𝑛=min{(1𝛼)𝑘𝑛/(𝑘𝑛𝛼),1/𝑘𝑛}. For an arbitrary 𝑧0𝐾 let the sequence {𝑧𝑛} be iteratively defined by 𝑧𝑛+1=𝑡1𝑛𝑘𝑛𝑓𝑧𝑛+𝑡𝑛𝑘𝑛𝑇𝑛𝑧𝑛,𝑛𝑁.(1.7) Then(i) for each integer 𝑛0, there is a unique 𝑥𝑛𝐾 such that 𝑥𝑛=𝑡1𝑛𝑘𝑛𝑓𝑥𝑛+𝑡𝑛𝑘𝑛𝑇𝑛𝑥𝑛;(1.8)if in addition lim𝑛𝑥𝑛𝑇𝑥𝑛=0,lim𝑛𝑧𝑛𝑇𝑧𝑛=0,(1.9)then(ii) the sequence {𝑧𝑛}𝑛 converges strongly to the unique solution of the variational inequality: 𝑝𝐹(𝑇)suchthat(𝐼𝑓)𝑝,𝑗𝑝𝑥0,𝑥𝐹(𝑇).(1.10)

Remark 1.2. We note that 𝑇𝑛+1𝑥𝑇𝑛𝑥𝑘𝑛𝑇𝑥𝑥, then the condition (1.9) lim𝑛𝑥𝑛𝑇𝑥𝑛=0 and lim𝑛𝑧𝑛𝑇𝑧𝑛=0 imply that lim𝑛𝑇𝑛+1𝑥𝑛𝑇𝑛𝑥𝑛=0,lim𝑛𝑇𝑛+1𝑧𝑛𝑇𝑛𝑧𝑛=0,(1.11) respectively. In other words, if an asymptotically nonexpansive mapping 𝑇 satisfies the condition (1.9) then 𝑇 must be u.a.r.s.

Inspired by the works in [48], in this paper, we suggest and analyze a modification of the iterative algorithm.

Let {𝑡𝑛}(0,1), let 𝛼 and 𝛽 be two positive numbers such that 𝛼+𝛽=1, and let 𝑓 be a contraction on 𝐾, a sequence {𝑧𝑛} iteratively defined by: 𝑧0𝐾, 𝑧𝑛+1=𝑡1𝑛𝑘𝑛𝑓𝑧𝑛+𝛼𝑡𝑛𝑘𝑛𝑇𝑛𝑧𝑛+𝛽𝑡𝑛𝑘𝑛𝑧𝑛.(1.12)

Remark 1.3. The algorithm (1.12) includes the algorithm (1.7) of Chidume et al. [5] and Shahzad and Udomene [4] as a special case.

We show the convergence of the proposed algorithm (1.12) to the unique solution of some variational inequality (some related works on VI, please see [912]). In this respect, our results can be considered as a refinement and improvement of the known results of Chidume et al. [5], Shahzad and Udomene [4], and Lim and Xu [13].

2. Preliminaries

Let 𝑆={𝑥𝐸𝑥=1} denote the unit sphere of the Banach space 𝐸. The space 𝐸 is said to have a Gateaux differentiable norm if the limit lim𝑛𝑥+𝑡𝑦𝑥𝑡(2.1) exists for each 𝑥,𝑦𝑆, and we call 𝐸 smooth; 𝐸 is said to have a uniformly Gateaux differentiable norm if for each 𝑦𝑆 the limit (2.1) is attained uniformly for 𝑥𝑆. Further, 𝐸 is said to be uniformly smooth if the limit (2.1) exists uniformly for (𝑥,𝑦)𝑆×𝑆. It is well known [14] that if 𝐸 is smooth then any duality mapping on 𝐸 is single-valued, and if 𝐸 has a uniformly Gateaux differentiable norm then the duality mapping is norm-to-weak* uniformly continuous on bounded sets.

Let 𝐾 be a nonempty closed convex and bounded subset of the Banach space 𝐸 and let the diameter of 𝐾 be defined by 𝑑(𝐾)=sup{𝑥𝑦𝑥,𝑦𝐾}. For each 𝑥𝐾, let 𝑟(𝑥,𝐾)=sup{𝑥𝑦𝑦𝐾} and let 𝑟(𝐾)=inf{𝑟(𝑥,𝐾)𝑥𝐾} denote the Chebyshev radius of 𝐾 relative to itself. The normal structure coefficient 𝑁(𝐸) of 𝐸 is defined by 𝑁(𝐸)=inf𝑑(𝐾)𝑟(𝐾)𝐾isaclosedconvexandboundedsubsetof𝐸with𝑑(𝐾)>0.(2.2) A space 𝐸 such that 𝑁(𝐸)>1 is said to have uniform normal structure. It is known that every space with a uniform normal structure is reflexive, and that all uniformly convex and uniformly smooth Banach spaces have uniform normal structure (see [13]).

We will let LIM be a Banach limit. Recall that LIM(𝑙) such that LIM=1, liminf𝑛𝑎𝑛LIM𝑛𝑎𝑛limsup𝑛𝑎𝑛, and LIM𝑛𝑎𝑛=LIM𝑛𝑎𝑛+1 for all {𝑎𝑛}𝑛𝑙. Let {𝑥𝑛} be a bounded sequence of 𝐸. Then we can define the real-valued continuous convex function 𝑔 on 𝐸 by 𝑔(𝑧)=LIM𝑛𝑥𝑛𝑧2 for all 𝑧𝐾.

Let 𝑇𝐾𝐾 be a nonlinear mapping and 𝑀={𝑥𝐾𝑔(𝑥)=min𝑧𝐾𝑔(𝑧)}. 𝑇 is said to satisfy the property (S) if for any bounded sequence {𝑥𝑛} in 𝐾, lim𝑛𝑥𝑛𝑇𝑥𝑛=0 implies 𝑀𝐹(𝑇).

Lemma 2.1 (see [15]). Let 𝐸 be a Banach space with the uniformly Gateaux differentiable norm and 𝑢𝐸. Then 𝑔(𝑢)=inf𝑧𝐸𝑔(𝑧)(2.3) if and only if 𝑥LIM𝑧,𝐽𝑛𝑢0(2.4) for all 𝑧𝐸.

Lemma 2.2 (see [16]). Assume {𝑎𝑛} is a sequence of nonnegative real numbers such that 𝑎𝑛+11𝛾𝑛𝑎𝑛+𝛿𝑛𝛾𝑛,(2.5) where {𝛾𝑛} is a sequence in (0,1) and {𝛿𝑛} is a sequence such that (1)𝑛=1𝛾𝑛=; (2)limsup𝑛𝛿𝑛0 or 𝑛=1|𝛿𝑛𝛾𝑛|<. Then lim𝑛𝑎𝑛=0.

Lemma 2.3 (see [17]). 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 that 𝑥𝑛+1=1𝛽𝑛𝑦𝑛+𝛽𝑛𝑥𝑛(2.6) for all 𝑛0 and limsup𝑛𝑦𝑛+1𝑦𝑛𝑥𝑛+1𝑥𝑛0.(2.7) Then lim𝑛𝑦𝑛𝑥𝑛=0.

Lemma 2.4. Let 𝐸 be an arbitrary real Banach space. Then 𝑥+𝑦2𝑥2+2𝑦,𝑗(𝑥+𝑦),(2.8) for all 𝑥,𝑦𝐸 and for all 𝑗(𝑥+𝑦)𝐽(𝑥+𝑦).

Lemma 2.5 (see [5]). Let 𝐸 be a Banach space with uniform normal structure, 𝐾 a nonempty closed convex and bounded subset of 𝐸, and 𝑇𝐾𝐾 an asymptotically nonexpansive mapping. Then 𝑇 has a fixed point.

3. Main Results

Theorem 3.1. Let 𝐸 be a real reflexive Banach space with a uniformly Gateaux differentiable norm, 𝐾 a nonempty closed convex and bounded subset of 𝐸, 𝑇𝐾𝐾 a continuous asymptotically pseudocontractive mapping with sequence {𝑘𝑛}𝑛[1,), and 𝑓𝐾𝐾 a contraction with constant 𝛾[0,1). Let {𝑡𝑛}(0,(1𝛾)𝑘𝑛/(𝑘𝑛𝛾)) be such that lim𝑛𝑡𝑛=1 and lim𝑛((𝑘𝑛1)/(𝑘𝑛𝑡𝑛))=0. Suppose 𝑇 satisfies the property (S). Then (i) for each integer 𝑛0, there is a unique 𝑥𝑛𝐾 such that 𝑥𝑛=𝑡1𝑛𝑘𝑛𝑓𝑥𝑛+𝑡𝑛𝑘𝑛𝑇𝑛𝑥𝑛;(3.1)if 𝑇 is u.a.r.s., then (ii) the sequence {𝑥𝑛}𝑛 converges strongly to the unique solution of the variational inequality: 𝑝𝐹(𝑇)suchthat(𝐼𝑓)𝑝,𝑗𝑝𝑥0,𝑥𝐹(𝑇).(3.2)

Proof. By the conditions on {𝑡𝑛}, 𝑡𝑛<(1𝛾)𝑘𝑛/(𝑘𝑛𝛾) implies (1𝑡𝑛/𝑘𝑛)𝛾+𝑡𝑛<1 for each integer 𝑛0, then the mapping 𝑆𝑛𝐾𝐾 defined for each 𝑥𝐾 by 𝑆𝑛𝑥=(1𝑡𝑛/𝑘𝑛)𝑓(𝑥)+(𝑡𝑛/𝑘𝑛)𝑇𝑛𝑥 is a strictly pseudocontractive mapping.
Indeed, for all 𝑥,𝑦𝐾, we have 𝑆𝑛𝑥𝑆𝑛𝑡𝑦,𝑗(𝑥𝑦)=1𝑛𝑘𝑛+𝑡𝑓(𝑥)𝑓(𝑦),𝑗(𝑥𝑦)𝑛𝑘𝑛𝑇𝑛𝑥𝑇𝑛𝑡𝑦,𝑗(𝑥𝑦)1𝑛𝑘𝑛𝑓(𝑥)𝑓(𝑦)𝑥𝑦+𝑡𝑛𝑥𝑦2𝑡1𝑛𝑘𝑛𝛾+𝑡𝑛𝑥𝑦2.(3.3) It follows [18, Corollary 1] that 𝑆𝑛 possesses exactly one fixed point 𝑥𝑛 in 𝐾 such that 𝑆𝑛𝑥𝑛=𝑥𝑛.
From (3.1), we have 𝑥𝑛𝑇𝑛𝑥𝑛=𝑡1𝑛𝑘𝑛𝑓𝑥𝑛+𝑡𝑛𝑘𝑛𝑇1𝑛𝑥𝑛=𝑡1𝑛𝑘𝑛𝑓𝑥𝑛𝑇𝑛𝑥𝑛0as𝑛.(3.4) Notice that 𝑥𝑛𝑇𝑥𝑛=𝑡1𝑛𝑘𝑛𝑓𝑥𝑛𝑇𝑥𝑛+𝑡𝑛𝑘𝑛𝑇𝑛𝑥𝑛𝑇𝑥𝑛𝑡1𝑛𝑘𝑛𝑓𝑥𝑛𝑇𝑥𝑛+𝑡𝑛𝑘𝑛𝑇𝑛𝑥𝑛𝑇𝑛+1𝑥𝑛+𝑡𝑛𝑘𝑛𝑇𝑛+1𝑥𝑛𝑇𝑥𝑛𝑡1𝑛𝑘𝑛𝑓𝑥𝑛𝑇𝑥𝑛+𝑡𝑛𝑘𝑛𝑇𝑛𝑥𝑛𝑇𝑛+1𝑥𝑛+𝑡𝑛𝑘𝑛𝑘1𝑥𝑛𝑇𝑛𝑥𝑛.(3.5) Therefore, from (3.4), (3.5), and 𝑇 which is u.a.r.s., we obtain 𝑥𝑛𝑇𝑥𝑛0 as 𝑛.
Define a function 𝑔𝐾𝑅+ by 𝑔(𝑧)=LIM𝑛𝑥𝑛𝑧2(3.6) for all 𝑧𝐾. Since 𝑔 is continuous and convex, 𝑔(𝑧) as 𝑧, and 𝐸 is reflexive, 𝑔 attains it infimum over 𝐾. Let 𝑧0𝐾 such that 𝑔(𝑧0)=min𝑧𝐾𝑔(𝑧) and let 𝑀={𝑥𝐾𝑔(𝑥)=min𝑧𝐾𝑔(𝑧)}. Then 𝑀 is nonempty because 𝑧0𝑀. Since 𝑇 satisfies the property (S), it follows that 𝑀𝐹(𝑇). Suppose that 𝑝𝑀𝐹(𝑇). Then, by Lemma 2.1, we have LIM𝑛𝑥𝑥𝑝,𝑗𝑛𝑝0(3.7) for all 𝑥𝐾. In particular, we have LIM𝑛𝑓𝑥(𝑝)𝑝,𝑗𝑛𝑝0.(3.8) On the other hand, from (3.1), we have 𝑥𝑛𝑇𝑛𝑥𝑛=𝑡1𝑛𝑘𝑛𝑓𝑥𝑛𝑇𝑛𝑥𝑛=1𝑡𝑛/𝑘𝑛𝑡𝑛/𝑘𝑛𝑓𝑥𝑛𝑥𝑛.(3.9) Now, for any 𝑝𝐹(𝑇), we have 𝑥𝑛𝑇𝑛𝑥𝑛𝑥,𝐽𝑛=𝑥𝑝𝑛𝑝+𝑇𝑛𝑝𝑇𝑛𝑥𝑛𝑥,𝐽𝑛𝑘𝑝𝑛𝑥1𝑛𝑝2𝑘𝑛𝐵12(3.10) for some 𝐵>0 and it follows from (3.9) that 𝑥𝑛𝑥𝑓𝑛𝑥,𝑗𝑛𝑡𝑝𝑛𝑘𝑛1𝑘𝑛𝑡𝑛𝐵2,(3.11) which implies that limsup𝑛𝑥𝑛𝑥𝑓𝑛𝑥,𝑗𝑛𝑝0.(3.12) Consequently, similar to the lines of the proof of [4, Theorem 3.1], Theorem 3.1 is easily proved. This completes the proof.

Corollary 3.2. Let 𝐸 be a real Banach space with a uniformly Gateaux differentiable norm possessing uniform normal structure, 𝐾 a nonempty closed convex and bounded subset of 𝐸, 𝑇𝐾𝐾 be an asymptotically nonexpansive mapping with sequence {𝑘𝑛}𝑛[1,), and 𝑓𝐾𝐾 a contraction with constant 𝛾[0,1). Let {𝑡𝑛}(0,(1𝛾)𝑘𝑛/(𝑘𝑛𝛾)) be such that lim𝑛𝑡𝑛=1 and lim𝑛((𝑘𝑛1)/(𝑘𝑛𝑡𝑛))=0. Then (i) for each integer 𝑛0, there is a unique 𝑥𝑛𝐾 such that 𝑥𝑛=𝑡1𝑛𝑘𝑛𝑓𝑥𝑛+𝑡𝑛𝑘𝑛𝑇𝑛𝑥𝑛;(3.13)if 𝑇 is u.a.r.s., then (ii) the sequence {𝑥𝑛}𝑛 converges strongly to the unique solution of the variational inequality: 𝑝𝐹(𝑇)suchthat(𝐼𝑓)𝑝,𝑗𝑝𝑥0,𝑥𝐹(𝑇).(3.14)

Theorem 3.3. Let 𝐸 be a real Banach space with a uniformly Gateaux differentiable norm possessing uniform normal structure, 𝐾 a nonempty closed convex and bounded subset of 𝐸, 𝑇𝐾𝐾 an asymptotically nonexpansive mapping with sequence {𝑘𝑛}𝑛[1,), and 𝑓𝐾𝐾 a contraction with constant 𝛾[0,1). Let {𝑡𝑛}(0,𝜉𝑛) be such that lim𝑛𝑡𝑛=1,𝑛=1𝑡𝑛(1𝑡𝑛)=, and lim𝑛((𝑘𝑛1)/(𝑘𝑛𝑡𝑛))=0, where 𝜉𝑛=min{(1𝛾)𝑘𝑛/(𝑘𝑛𝛾),1/𝑘𝑛}. For an arbitrary 𝑧0𝐾, let the sequence {𝑧𝑛}𝑛 be iteratively defined by (1.12). Then (i) for each integer 𝑛0, there is a unique 𝑥𝑛𝐾 such that 𝑥𝑛=𝑡1𝑛𝑘𝑛𝑓𝑥𝑛+𝑡𝑛𝑘𝑛𝑇𝑛𝑥𝑛;(3.15)if 𝑇 is u.a.r.s., then (ii) the sequence {𝑧𝑛}𝑛 converges strongly to the unique solution of the variational inequality: 𝑝𝐹(𝑇)suchthat(𝐼𝑓)𝑝,𝑗𝑝𝑥0,𝑥𝐹(𝑇).(3.16)

Proof. Set 𝛼𝑛=𝑡𝑛/𝑘𝑛, then 𝛼𝑛1 as 𝑛. Define 𝑧𝑛+1=𝛽𝛼𝑛𝑧𝑛+1𝛽𝛼𝑛𝑦𝑛.(3.17) Observe that 𝑦𝑛+1𝑦𝑛=𝑧𝑛+2𝛽𝛼𝑛+1𝑧𝑛+11𝛽𝛼𝑛+1𝑧𝑛+1𝛽𝛼𝑛𝑧𝑛1𝛽𝛼𝑛=1𝛼𝑛+1𝑓𝑧𝑛+1+𝛼𝛼𝑛+1𝑇𝑛+1𝑧𝑛+11𝛽𝛼𝑛+11𝛼𝑛𝑓𝑧𝑛+𝛼𝛼𝑛𝑇𝑛𝑧𝑛1𝛽𝛼𝑛=1𝛼𝑛+11𝛽𝛼𝑛+1𝑓𝑧𝑛+1𝑧𝑓𝑛+1𝛼𝑛+11𝛽𝛼𝑛+11𝛼𝑛1𝛽𝛼𝑛𝑓𝑧𝑛+𝛼𝛼𝑛+11𝛽𝛼𝑛+1𝑇𝑛+1𝑧𝑛+1𝑇𝑛+1𝑧𝑛+𝛼𝛼𝑛+11𝛽𝛼𝑛+1𝑇𝑛+1𝑧𝑛𝑇𝑛𝑧𝑛+𝛼𝛼𝑛+11𝛽𝛼𝑛+1𝛼𝛼𝑛1𝛽𝛼𝑛𝑇𝑛𝑧𝑛.(3.18) It follows that 𝑦𝑛+1𝑦𝑛𝑧𝑛+1𝑧𝑛1𝛼𝑛+11𝛽𝛼𝑛+1𝛾𝑧𝑛+1𝑧𝑛+||||1𝛼𝑛+11𝛽𝛼𝑛+11𝛼𝑛1𝛽𝛼𝑛||||𝑓𝑧𝑛+𝛼𝛼𝑛+11𝛽𝛼𝑛+1𝑇𝑛+1𝑧𝑛+1𝑇𝑛+1𝑧𝑛+𝛼𝛼𝑛+11𝛽𝛼𝑛+1𝑇𝑛+1𝑧𝑛𝑇𝑛𝑧𝑛+||||𝛼𝛼𝑛+11𝛽𝛼𝑛+1𝛼𝛼𝑛1𝛽𝛼𝑛||||𝑇𝑛𝑧𝑛𝑧𝑛+1𝑧𝑛||||1𝛼𝑛+11𝛽𝛼𝑛+11𝛼𝑛1𝛽𝛼𝑛||||𝑓𝑧𝑛+𝛼𝛼𝑛+11𝛽𝛼𝑛+1𝑇𝑛+1𝑧𝑛𝑇𝑛𝑧𝑛+||||𝛼𝛼𝑛+11𝛽𝛼𝑛+1𝛼𝛼𝑛1𝛽𝛼𝑛||||𝑇𝑛𝑧𝑛+1𝛼𝑛+11𝛽𝛼𝑛+1𝛾+𝛼𝛼𝑛+11𝛽𝛼𝑛+1𝑘𝑛+1𝑧1𝑛+1𝑧𝑛.(3.19) We note that 𝑘𝑛+1𝛾𝛼𝑘𝑛+1=+𝛽𝛾(1𝛼)𝑘𝑛+1𝛽1𝛼𝛽=0.(3.20) It follows that 𝑡𝑛+1(1𝛾)𝑘𝑛+1𝑘𝑛+1𝛾(1𝛾)𝑘𝑛+1𝛼𝑘𝑛+1,+𝛽𝛾(3.21) which implies that 𝑘𝑛+1𝑡𝑛+1𝛼+𝑡𝑛+1𝛽𝑡𝑛+1𝛾(1𝛾)𝑘𝑛+1𝛼𝑘𝑛+1𝛼𝑛+1+𝛼𝑛+1𝛽𝛼𝑛+1𝛾1𝛾𝛼𝑘𝑛+1𝛼𝑛+1+1𝛼𝑛+1𝛾1𝛼𝑛+1𝛽𝛼𝑘𝑛+1𝛼𝑛+1+1𝛼𝑛+1𝛾1𝛼𝑛+1𝛽1.(3.22) From (3.19) and (3.22), we obtain limsup𝑛𝑦𝑛+1𝑦𝑛𝑧𝑛+1𝑧𝑛0.(3.23) Hence, by Lemma 2.3 we know lim𝑛𝑦𝑛𝑧𝑛=0,(3.24) consequently lim𝑛𝑧𝑛+1𝑧𝑛=0.(3.25)
On the other hand, 𝑧𝑛𝑇𝑛𝑧𝑛𝑧𝑛+1𝑧𝑛+𝑧𝑛+1𝑇𝑛𝑧𝑛𝑧𝑛+1𝑧𝑛+1𝛼𝑛𝑓𝑧𝑛𝑇𝑛𝑧𝑛+𝛽𝛼𝑛𝑧𝑛𝑇𝑛𝑧𝑛,(3.26) which implies that lim𝑛𝑧𝑛𝑇𝑛𝑧𝑛=0.(3.27) Hence, we have 𝑧𝑛𝑇𝑧𝑛𝑧𝑛𝑇𝑛𝑧𝑛+𝑇𝑛𝑧𝑛𝑇𝑛+1𝑧𝑛+𝑇𝑛+1𝑧𝑛𝑇𝑧𝑛𝑧𝑛𝑇𝑛𝑧𝑛+𝑇𝑛𝑧𝑛𝑇𝑛+1𝑧𝑛+𝑘1𝑧𝑛𝑇𝑛𝑧𝑛=1+𝑘1𝑧𝑛𝑇𝑛𝑧𝑛+𝑇𝑛𝑧𝑛𝑇𝑛+1𝑧𝑛0(𝑛).(3.28) From (3.15), 𝑥𝑚𝑧𝑛=(1𝛼𝑚)(𝑓(𝑥𝑚)𝑧𝑛)+𝛼𝑚(𝑇𝑚𝑥𝑚𝑧𝑛). Applying Lemma 2.4, we estimate as follows: 𝑥𝑚𝑧𝑛2𝛼2𝑚𝑇𝑚𝑥𝑚𝑧𝑛2+21𝛼𝑚𝑓𝑥𝑚𝑧𝑛𝑥,𝑗𝑚𝑧𝑛𝛼2𝑚𝑇𝑚𝑥𝑚𝑇𝑚𝑧𝑛+𝑇𝑚𝑧𝑛𝑧𝑛2+21𝛼𝑚𝑓𝑥𝑚𝑥𝑚𝑥,𝑗𝑚𝑧𝑛+𝑥𝑚𝑧𝑛2𝛼2𝑚𝑘𝑚𝑥𝑚𝑧𝑛+𝑇𝑚𝑧𝑛𝑧𝑛2+21𝛼𝑚𝑓𝑥𝑚𝑥𝑚𝑥,𝑗𝑚𝑧𝑛+𝑘2𝑚𝑥𝑚𝑧𝑛2=𝛼2𝑚𝑘2𝑚𝑥𝑚𝑧𝑛2+2𝑘𝑚𝑥𝑚𝑧𝑛𝑇𝑚𝑧𝑛𝑧𝑛+𝑇𝑚𝑧𝑛𝑧𝑛2+21𝛼𝑚𝑓𝑥𝑚𝑥𝑚𝑥,𝑗𝑚𝑧𝑛+𝑘2𝑚𝑥𝑚𝑧𝑛2=1(1𝛼𝑚)2𝑘2𝑚𝑥𝑚𝑧𝑛2+𝑇𝑚𝑧𝑛𝑧𝑛2𝑘𝑚𝑥𝑚𝑧𝑛+𝑇𝑚𝑧𝑛𝑧𝑛+21𝛼𝑚𝑓𝑥𝑚𝑥𝑚𝑥,𝑗𝑚𝑧𝑛+𝑘2𝑚𝑥𝑚𝑧𝑛21+1𝛼𝑚2𝑘2𝑚𝑥𝑚𝑧𝑛2+𝑇𝑚𝑧𝑛𝑧𝑛2𝑘𝑚𝑥𝑚𝑧𝑛+𝑇𝑚𝑧𝑛𝑧𝑛+21𝛼𝑚𝑓𝑥𝑚𝑥𝑚𝑥,𝑗𝑚𝑧𝑛.(3.29) Since 𝐾 is bounded, for some constant 𝑀>0, it follows that limsup𝑛𝑓𝑥𝑚𝑥𝑚𝑧,𝑗𝑛𝑥𝑚𝑘2𝑚1+𝑘2𝑚1𝛼𝑚21𝛼𝑚𝑀+limsup𝑛𝑀𝑧𝑛𝑇𝑚z𝑛1𝛼𝑚,(3.30) so that limsup𝑛𝑓𝑥𝑚𝑥𝑚𝑧,𝑗𝑛𝑥𝑚𝑘2𝑚1+𝑘2𝑚1𝛼𝑚21𝛼𝑚𝑀.(3.31) By Corollary 3.2, 𝑥𝑚𝑝𝐹(𝑇), which solve the variational inequality (3.16). Since 𝑗 is norm to weak* continuous on bounded sets, in the limit as 𝑚, we obtain that limsup𝑛𝑓𝑧(𝑝)𝑝,𝑗𝑛𝑝0.(3.32) From Lemma 2.4, we estimate as follows: 𝑧𝑛+1𝑝2=(1𝛼𝑛)(𝑓(𝑧𝑛)𝑝)+𝛼𝛼𝑛(𝑇𝑛𝑧𝑛𝑝)+𝛽𝛼𝑛(𝑧𝑛𝑝)2𝛼𝛼𝑛(𝑇𝑛𝑧𝑛𝑝)+𝛽𝛼𝑛(𝑧𝑛𝑝)2+21𝛼𝑛𝑓𝑧𝑛𝑧𝑝,𝑗𝑛+1𝑝𝛼2𝛼2𝑛𝑇𝑛𝑧𝑛𝑝2+2𝛼𝛽𝛼2𝑛𝑇𝑛𝑧𝑛𝑧𝑝𝑛𝑝+𝛽2𝛼2𝑛𝑧𝑛𝑝2+21𝛼𝑛𝑧𝑓𝑛𝑧𝑓(𝑝),𝑗𝑛+1𝑝+21𝛼𝑛𝑧𝑓(𝑝)𝑝,𝑗𝑛+1𝛼𝑝2𝑘2𝑛+2𝛼𝛽𝑘𝑛+𝛽2𝛼2𝑛𝑧𝑛𝑝2+21𝛼𝑛𝛾𝑧𝑛𝑧𝑝𝑛+1𝑝+21𝛼𝑛𝑧𝑓(𝑝)𝑝,𝑗𝑛+1𝑝𝛼2𝑛𝑘2𝑛𝑧𝑛𝑝2+𝛾1𝛼𝑛𝑧𝑛𝑝2+𝑧𝑛+1𝑝2+21𝛼𝑛𝑧𝑓(𝑝)𝑝,𝑗𝑛+1𝑝,(3.33) so that 𝑧𝑛+1𝑝2𝑡2𝑛+1𝛼𝑛𝛾11𝛼𝑛𝛾𝑧𝑛𝑝2+21𝛼𝑛11𝛼𝑛𝛾𝑓𝑧(𝑝)𝑝,𝑗𝑛+1=𝑝1121𝛼𝑛𝛾𝑡2𝑛11𝛼𝑛𝛾𝑧𝑛𝑝2+21𝛼𝑛11𝛼𝑛𝛾𝑧𝑓(𝑝)𝑝,𝑗𝑛+1.𝑝(3.34) Let 𝜆𝑛=121𝛼𝑛𝛾𝑡2𝑛11𝛼𝑛𝛾.(3.35) Consequently, following the lines of the proof of [4, Theorem 3.3], Theorem 3.3 is easily proved.

From the lines of the proof of Theorem 3.3, we can obtain the following corollary.

Corollary 3.4. Let 𝐸 be a real Banach space with a uniformly Gateaux differentiable norm possessing uniform normal structure, 𝐾 a nonempty closed convex and bounded subset of 𝐸, 𝑇𝐾𝐾 an asymptotically nonexpansive mapping with sequence {𝑘𝑛}𝑛[1,), and 𝑓𝐾𝐾 a contraction with constant 𝛾[0,1). Let {𝑡𝑛}(0,𝜉𝑛) be such that lim𝑛𝑡𝑛=1,𝑛=1𝑡𝑛(1𝑡𝑛)=, and lim𝑛((𝑘𝑛1)/(𝑘𝑛𝑡𝑛))=0, where 𝜉𝑛=min{(1𝛾)𝑘𝑛/(𝑘𝑛𝛾),1/𝑘𝑛}. For an arbitrary 𝑧0𝐾, let the sequence {𝑧𝑛}𝑛 be iteratively defined by (1.12). Then(i) for each integer 𝑛0, there is a unique 𝑥𝑛𝐾 such that 𝑥𝑛=𝑡1𝑛𝑘𝑛𝑓𝑥𝑛+𝑡𝑛𝑘𝑛𝑇𝑛x𝑛;(3.36)if 𝑇 satisfies lim𝑛𝑇𝑛+1𝑥𝑛𝑇𝑛𝑥𝑛=0 and lim𝑛𝑇𝑛+1𝑧𝑛𝑇𝑛𝑧𝑛=0 then(ii) the sequence {𝑧𝑛}𝑛 converges strongly to the unique solution of the variational inequality: 𝑝𝐹(𝑇)suchthat(𝐼𝑓)𝑝,𝑗𝑝𝑥0,𝑥𝐹(𝑇).(3.37)

Remark 3.5. Since every nonexpansive mapping is asymptotically nonexpansive, our theorems hold for the case when 𝑇 is simply nonexpansive. In this case, the boundedness requirement on 𝐾 can be removed from the above results.

Remark 3.6. Our results can be viewed as a refinement and improvement of the corresponding results by Shahzad and Udomene [4], Chidume et al. [5], and Lim and Xu [13].

Example 3.7. Let 𝑇𝐶𝐶 be a nonexpansive mapping. Let the iterative sequence {𝑥𝑛} be defined by 𝑥𝑛+1=1𝑛1𝑢+1𝑛𝑇𝑥𝑛.(3.38) It is easy to see that {𝑥𝑛} converges strongly to some fixed point of 𝑇.
In particular, let 𝐻=𝑅2 and define 𝑇𝑅2𝑅2 by 𝑇𝑟𝑒𝑖𝜃=𝑟𝑒𝑖(𝜃+𝜋/2),(3.39) and take that 𝑢=𝑒𝑖𝜋 is a fix element in 𝐶. It is obvious that 𝑇 is a nonexpansive mapping with a unique fixed point 𝑥=0. In this case, (3.38) becomes 𝑥𝑛+1=1𝑛𝑒𝑖𝜋+11𝑛𝑟𝑛𝑒𝑖(𝜃𝑛+𝜋/2).(3.40) It is clear that the complex number sequence {𝑥𝑛} converges strongly to a fixed point 𝑥=0.

Acknowledgment

Y.-C. Liou was partially supported by NSC 101-2628-E-230-001-MY3.