Convergence to Common Fixed Point for Generalized Asymptotically Nonexpansive Semigroup in Banach Spaces

Abstract

Let 𝐾 be a nonempty closed convex subset of a reflexive and strictly convex Banach space 𝐸 with a uniformly Gâteaux differentiable norm, ℱ={𝑇(ℎ)∶ℎ≥0} a generalized asymptotically nonexpansive self-mapping semigroup of 𝐾, and 𝑓∶𝐾→𝐾 a fixed contractive mapping with contractive coefficient 𝛽∈(0,1). We prove that the following implicit and modified implicit viscosity iterative schemes {𝑥_𝑛} defined by 𝑥_𝑛=𝛼_𝑛𝑓(𝑥_𝑛)+(1−𝛼_𝑛)𝑇(𝑡_𝑛)𝑥_𝑛 and 𝑥_𝑛=𝛼_𝑛𝑦_𝑛+(1−𝛼_𝑛)𝑇(𝑡_𝑛)𝑥_𝑛,𝑦_𝑛=𝛽_𝑛𝑓(𝑥_𝑛−1)+(1−𝛽_𝑛)𝑥_𝑛−1 strongly converge to 𝑝∈𝐹 as 𝑛→∞ and 𝑝 is the unique solution to the following variational inequality: ⟨𝑓(𝑝)−𝑝,𝑗(𝑦−𝑝)⟩≤0 for all 𝑦∈𝐹.

1. Introduction

Let 𝐶 be a closed convex subset of a Hilbert space 𝐻 and 𝑇 a nonexpansive mapping from 𝐶 into itself. We denote by 𝐹(𝑇) the set of fixed points of 𝑇. Let 𝐹(𝑇) be nonempty and 𝑢 an element of 𝐶. For each 𝑡 with 0<𝑡<1, let  𝑥_𝑡∈𝐶 be the unique fixed point of the contraction 𝑥↦𝑡𝑢+(1−𝑡)𝑇𝑥. Browder [1] showed that {𝑥_𝑡} defined by 𝑥_𝑡=𝑡𝑢+(1−𝑡)𝑇𝑥_𝑡 converges strongly to the element of 𝐹(𝑇) which is nearest to 𝑢 in 𝐹(𝑇) as 𝑡→0.

In 2004, for a contraction 𝑓∶𝐶→𝐶 and a nonexpansive mapping 𝑇∶𝐶→𝐶, Xu [2] proposed the following viscosity approximation method in Banach space:𝑥_𝑡=𝑡𝑓(𝑥_𝑡)+(1−𝑡)𝑇𝑥_𝑡,𝑡∈(0,1),𝑡→0, (1.1)and Song and Xu [3] studied the convergence of the following implicit viscosity iterative scheme:𝑥_𝑛=𝛼_𝑛𝑓(𝑥_𝑛)+(1−𝛼_𝑛)𝑇(𝑡_𝑛)𝑥_𝑛, (1.2)where {𝛼_𝑛}⊂(0,1).

On the other hand, for a fixed Lipschitz strongly pseudocontractive mapping 𝑓 and a continuous pseudocontractive mapping 𝑇, Song and Chen [4] proposed the following motivated implicit viscosity iterative scheme:𝑥_𝑛=𝛼_𝑛𝑦_𝑛+(1−𝛼_𝑛)𝑇𝑥_𝑛,𝑦_𝑛=𝛽_𝑛𝑓(𝑥_𝑛−1)+(1−𝛽_𝑛)𝑥_𝑛−1. (1.3)

In this paper, we will still study the implicit viscosity iterative scheme (1.2) and propose the following iterative scheme:𝑥_𝑛=𝛼_𝑛𝑦_𝑛+(1−𝛼_𝑛)𝑇(𝑡_𝑛)𝑥_𝑛,𝑦_𝑛=𝛽_𝑛𝑓(𝑥_𝑛−1)+(1−𝛽_𝑛)𝑥_𝑛−1, (1.4)where {𝑇(ℎ)∶ℎ≥0} is a generalized asymptotically nonexpansive self-mappings semigroup and 𝑓 a fixed contractive mapping with contractive coefficient 𝛽∈(0,1).

2. Preliminaries

Throughout this paper, we assume that 𝐸 is a Banach space and 𝐾 a nonempty closed convex subset of 𝐸. Let 𝐸^∗ be a dual space of 𝐸, 𝐽∶𝐸→2^𝐸∗ the normalized duality mapping defined by𝐽(𝑥)={𝑓∈𝐸^∗,⟨𝑥,𝑓⟩=‖𝑥‖⋅‖𝑓‖,‖𝑥‖=‖𝑓‖}, (2.1)where ⟨⋅,⋅⟩ denotes the generalized duality pairing.

Definition 2.1 (see {[5]}). A mapping 𝑇∶𝐸→𝐸 is said to be total asymptotically nonexpansive if there exist nonnegative real sequences {𝑘_𝑛⁽¹⁾} and {𝑘_𝑛⁽²⁾},𝑛>0, with 𝑘_𝑛⁽¹⁾ and 𝑘_𝑛⁽²⁾→0 as 𝑛→∞, and strictly increasing and continuous functions 𝜙∶𝑅⁺→𝑅⁺ with 𝜙(0)=0 such that‖𝑇^𝑛𝑥−𝑇^𝑛𝑦‖≤‖𝑥−𝑦‖+𝑘_𝑛⁽¹⁾𝜙(‖𝑥−𝑦‖)+𝑘_𝑛⁽²⁾∀𝑥,𝑦∈𝐾. (2.2)

Remark 2.2. If 𝜙(𝜆)=𝜆, the total asymptotically nonexpansive mapping coincides with generalized asymptotically nonexpansive mapping. In addition, for all 𝑛∈ℕ, if 𝑘_𝑛⁽²⁾=0, then generalized asymptotically nonexpansive mapping coincide with asymptotically nonexpansive mapping; if 𝑘_𝑛⁽¹⁾=0,𝑘_𝑛⁽²⁾=max{0,𝑝_𝑛}, where 𝑝_𝑛∶=sup_{𝑥,𝑦∈𝐾}(‖𝑇_𝑛𝑥−𝑇_𝑛𝑦‖−‖𝑥−𝑦‖), then generalized asymptotically nonexpansive mapping coincide with asymptotically nonexpansive mapping in the intermediate sense; if 𝑘_𝑛⁽¹⁾=0 and 𝑘_𝑛⁽²⁾=0, then we obtain from (2.2) the class of nonexpansive mapping.

Remark 2.3. In [5], for the total asymptotically nonexpansive mapping, the authors assume that there exist 𝑀,𝑀^∗>0 such that 𝜙(𝜆)≤𝑀^∗𝜆 for all 𝜆≥𝑀, so for 𝑀₀=max{𝜙(𝑀),𝑀^∗}, 𝜙(𝜆)≤𝑀₀(1+𝜆) for all 𝜆≥0, then the total asymptotically nonexpansive mapping studied by [5] coincides with generalized asymptotically nonexpansive mapping.

A (one-parameter) generalized asymptotically nonexpansive semigroup is a family ℱ={𝑇(ℎ)∶ℎ≥0} of self-mapping of 𝐾 such that

(i)𝑇(0)𝑥=𝑥 for 𝑥∈𝐾;(ii)𝑇(𝑠+𝑡)𝑥=𝑇(𝑠)𝑇(𝑡)𝑥 for 𝑡,𝑠≥0 and 𝑥∈𝐾;(iii)lim_𝑡→0𝑇(𝑡)𝑥=𝑥 for 𝑥∈𝐾;(iv)for each ℎ≥0,𝑇(ℎ) is generalized asymptotically nonexpansive, that is,‖𝑇(ℎ)𝑥−𝑇(ℎ)𝑦‖≤(1+𝑘_ℎ⁽¹⁾)‖𝑥−𝑦‖+𝑘_ℎ⁽²⁾∀𝑥,𝑦∈𝐾. (2.3)

We will denote by 𝐹 the common fixed point set of ℱ, that is,𝐹∶=Fix(ℱ)={𝑥∈𝐾∶𝑇(ℎ)𝑥=𝑥,ℎ≥0}=_ℎ≥0Fix(𝑇(ℎ)). (2.4)

Definition 2.4. A Banach space 𝐸 is said to be strictly convex if ‖𝑥+𝑦‖/2<1 for ‖𝑥‖=‖𝑦‖=1 and 𝑥≠𝑦.

Definition 2.5. Let 𝑈={𝑥∈𝐸∶‖𝑥‖=1}, the norm of 𝐸 is said to be uniformly 𝐺𝑎𝑡𝑒𝑎𝑢𝑥 differentiable, if for each 𝑦∈𝑈, lim_𝑡→0((‖𝑥+𝑡𝑦‖−‖𝑥‖)/𝑡) exists uniformly for 𝑥∈𝑈.

Definition 2.6. Let 𝜇 be a continuous liner functional on 𝑙^∞ and let (𝑎₀,𝑎₁,…)∈𝑙^∞. One writes 𝜇_𝑛(𝑎_𝑛) instead of 𝜇((𝑎₀,𝑎₁,…)). One calls 𝜇 a Banach limit when 𝜇 satisfies ‖𝜇‖=𝜇_𝑛(1)=1 and 𝜇_𝑛(𝑎_𝑛+1)=𝜇_𝑛(𝑎_𝑛) for each (𝑎₀,𝑎₁,…)∈𝑙^∞.

For a Banach limit 𝜇, one knows that lim_𝑛→∞𝑎_𝑛≤𝜇_𝑛(𝑎_𝑛)≤lim_𝑛→∞𝑎_𝑛 for every 𝑎=(𝑎₀,𝑎₁,…)∈𝑙^∞. So if 𝑎=(𝑎₀,𝑎₁,…)∈𝑙^∞,𝑏=(𝑏₀,𝑏₁,…)∈𝑙^∞ and 𝑎_𝑛−𝑏_𝑛→0 as 𝑛→∞, one has 𝜇_𝑛(𝑎_𝑛)=𝜇_𝑛(𝑏_𝑛).

Definition 2.7. Let 𝐾 be a nonempty closed convex subset of a Banach space 𝐸, ℱ={𝑇(ℎ)∶ℎ≥0} a continuous operator semigroup on 𝐾. Then ℱ is said to be uniformly asymptotically regular (in short, u.a.r.) on 𝐾 if for all ℎ≥0 and any bounded subset 𝐶 of 𝐾, lim_𝑡→∞sup_𝑥∈𝐶‖𝑇(ℎ)(𝑇(𝑡)𝑥)−𝑇(𝑡)𝑥‖=0.

Lemma 2.8 (see {[6]}). Let 𝐸 be a Banach space with a uniformly Gâteaux differentiable norm, then the normalized duality mapping 𝐽∶𝐸→2^𝐸∗ defined by (2.1) is single-valued and uniformly continuous from the norm topology of 𝐸 to the weak∗ topology of 𝐸^∗ on each bounded subset of 𝐸.

The single-valued normalized duality mapping is denoted by 𝑗.

Lemma 2.9. Let 𝐸 be a reflexive and strictly convex Banach space with a uniformly Gâteaux differentiable norm, 𝐾 a nonempty closed convex subset of 𝐸. Suppose that {𝑥_𝑛} is a bounded sequence in 𝐾, {𝑇(ℎ)∶ℎ≥0} a continuous generalized asymptotically nonexpansive semigroup from 𝐾 into itself such that lim_𝑛→∞‖𝑥_𝑛−𝑇(ℎ)𝑥_𝑛‖=0 for all ℎ≥0. Define the set𝐾^∗={𝑥∈𝐾∶𝜇_𝑛‖𝑥_𝑛−𝑥‖²=min_𝑦∈𝐾𝜇_𝑛‖𝑥_𝑛−𝑦‖²}. (2.5)If 𝐹≠∅, then 𝐾^∗∩𝐹≠∅.

Proof. Set 𝑔(𝑦)=𝜇_𝑛‖𝑥_𝑛−𝑦‖², then 𝑔(𝑦) is a convex and continuous function, and 𝑔(𝑦)→∞ as ‖𝑦‖→∞. Using [7, Theorem 1.3.11], there exists 𝑥∈𝐾 such that 𝑔(𝑥)=inf_𝑦∈𝐾𝑔(𝑦) by the reflexivity of 𝐸, that is, 𝐾^∗ is nonempty. Clearly, 𝐾^∗ is closed convex by the convexity and continuity of 𝑔(𝑦).
Since lim_𝑛→∞‖𝑥_𝑛−𝑇(ℎ)𝑥_𝑛‖=0,lim_ℎ→∞𝑘_ℎ^(𝑖)=0(𝑖=1,2), and 𝑔(𝑦) is continuous for all 𝑧∈𝐾^∗, we have𝑔(lim_ℎ→∞𝑇(ℎ)𝑧)=lim_ℎ→∞𝑔(𝑇(ℎ)𝑧)=lim_ℎ→∞𝜇_𝑛‖𝑥_𝑛−𝑇(ℎ)𝑧‖²≤lim_ℎ→∞𝜇_𝑛‖𝑇(ℎ)𝑥_𝑛−𝑇(ℎ)𝑧‖²≤lim_ℎ→∞𝜇_𝑛((1+𝑘_ℎ⁽¹⁾)‖𝑥_𝑛−𝑧‖+𝑘_ℎ⁽²⁾)²=𝜇_𝑛‖𝑥_𝑛−𝑧‖². (2.6)Hence lim_ℎ→∞𝑇(ℎ)𝑧∈𝐾^∗.
Let 𝑝∈𝐹. Since 𝐾^∗ is closed convex set, there exists a unique 𝑣∈𝐾^∗ such that‖𝑝−𝑣‖=min_{𝑥∈𝐾∗}‖𝑝−𝑥‖. (2.7)
Since 𝑝=lim_ℎ→∞𝑇(ℎ)𝑝 and lim_ℎ→∞𝑇(ℎ)𝑣∈𝐾^∗,‖𝑝−lim_ℎ→∞𝑇(ℎ)𝑣‖=‖lim_ℎ→∞𝑇(ℎ)𝑝−lim_ℎ→∞𝑇(ℎ)𝑣‖=lim_ℎ→∞‖𝑇(ℎ)𝑝−𝑇(ℎ)𝑣‖≤lim_ℎ→∞(1+𝑘_ℎ⁽¹⁾)‖𝑝−𝑣‖+𝑘_ℎ⁽²⁾=‖𝑝−𝑣‖. (2.8)Therefore, lim_ℎ→∞𝑇(ℎ)𝑣=𝑣. Since 𝑇(𝑠+𝑡)𝑥=𝑇(𝑠)𝑇(𝑡)𝑥 for all 𝑥∈𝐾, then we have𝑣=lim_𝑡→∞𝑇(𝑡)𝑣=lim_𝑡→∞𝑇(𝑠+𝑡)𝑣=lim_𝑡→∞𝑇(𝑠)𝑇(𝑡)𝑣=𝑇(𝑠)lim_𝑡→∞𝑇(𝑡)𝑣=𝑇(𝑠)𝑣 (2.9)for all 𝑠≥0. Therefore 𝑣∈𝐹 and the proof is complete.

Lemma 2.10 (see {[8]}). Let 𝐾 be a nonempty convex subset of a Banach space 𝐸 with a uniformly Gâteaux differentiable norm, and {𝑥_𝑛} a bounded sequence of 𝐸. If 𝑧₀∈𝐾, then𝜇_𝑛‖𝑥_𝑛−𝑧₀‖²=min_𝑦∈𝐾𝜇_𝑛‖𝑥_𝑛−𝑦‖² (2.10)if and only if𝜇_𝑛⟨𝑦−𝑧₀,𝐽(𝑥_𝑛−𝑧₀)⟩≤0∀𝑦∈𝐾. (2.11)

Lemma 2.11 (see {[9]}). Let {𝑎_𝑛} be a sequence of nonnegative real numbers satisfying the following conditions:𝑎_𝑛+1≤(1−𝜆_𝑛)𝑎_𝑛+𝑏_𝑛+𝑐_𝑛∀𝑛≥𝑛₀, (2.12)where 𝑛₀ is some nonnegative integer, 𝜆_𝑛∈[0,1] with ∑^∞_𝑛=1𝜆_𝑛=∞,limsup_𝑛→∞(𝑏_𝑛/𝜆_𝑛)≤0, and ∑^∞_𝑛=1𝑐_𝑛<∞. Then 𝑎_𝑛→0 as 𝑛→∞.

3. Implicit Iteration Scheme

Theorem 3.1. Let 𝐸 be a real reflexive and strictly convex Banach space with a uniformly Gâteaux differentiable norm, 𝐾 a nonempty closed convex subset of 𝐸, ℱ={𝑇(ℎ)∶ℎ≥0} a u.a.r generalized asymptotically nonexpansive semigroup from 𝐾 into itself with sequences {𝑘_ℎ⁽¹⁾}{𝑘_ℎ⁽²⁾},ℎ≥0, such that 𝐹≠∅, and 𝑓∶𝐾→𝐾 a fixed contractive mapping with contractive coefficient 𝛽∈(0,1). If {𝑥_𝑛} is given by (1.2), where lim_𝑛→∞𝑡_𝑛=∞,𝛼_𝑛∈(0,1),lim_𝑛→∞𝛼_𝑛=0 and lim_𝑛→∞(𝑘_𝑡^(𝑖)_𝑛/𝛼_𝑛)=0(𝑖=1,2), then {𝑥_𝑛} converges strongly to some common fixed point 𝑝 of 𝐹 such that 𝑝 is the unique solution in 𝐹 to variational inequality:⟨𝑓(𝑝)−𝑝,𝑗(𝑦−𝑝)⟩≤0∀𝑦∈𝐹. (3.1)

Proof. For any fixed 𝑦∈𝐹,‖𝑥_𝑛−𝑦‖²=⟨𝛼_𝑛(𝑓(𝑥_𝑛)−𝑦)+(1−𝛼_𝑛)(𝑇(𝑡_𝑛)𝑥_𝑛−𝑦),𝑗(𝑥_𝑛−𝑦)⟩=𝛼_𝑛⟨𝑓(𝑥_𝑛)−𝑓(𝑦),𝑗(𝑥_𝑛−𝑦)⟩+𝛼_𝑛⟨𝑓(𝑦)−𝑦,𝑗(𝑥_𝑛−𝑦)⟩+(1−𝛼_𝑛)⟨𝑇(𝑡_𝑛)𝑥_𝑛−𝑇(𝑡_𝑛)𝑦,𝑗(𝑥_𝑛−𝑦)⟩≤𝛼_𝑛𝛽‖𝑥_𝑛−𝑦‖²+𝛼_𝑛⟨𝑓(𝑦)−𝑦,𝑗(𝑥_𝑛−𝑦)⟩+(1−𝛼_𝑛)‖𝑥_𝑛−𝑦‖[(1+𝑘_𝑡⁽¹⁾_𝑛)‖𝑥_𝑛−𝑦‖+𝑘_𝑡⁽²⁾_𝑛]=(1−𝛼_𝑛(1−𝛽)+(1−𝛼_𝑛)𝑘_𝑡⁽¹⁾_𝑛)‖𝑥_𝑛−𝑦‖²+𝛼_𝑛⟨𝑓(𝑦)−𝑦,𝑗(𝑥_𝑛−𝑦)⟩+(1−𝛼_𝑛)𝑘_𝑡⁽²⁾_𝑛‖𝑥_𝑛−𝑦‖. (3.2) Let 𝑑_𝑛^(𝑖)=(𝑘_𝑡^(𝑖)_𝑛/𝛼_𝑛)(𝑖=1,2). Since lim_𝑛→∞(𝑘_𝑡^(𝑖)_𝑛/𝛼_𝑛)=0 for all 𝜀∈(0,1−𝛽), there exists 𝑁∈ℕ such that 𝑘_𝑡^(𝑖)_𝑛/𝛼_𝑛<𝜀<1−𝛽<(1−𝛽)/(1−𝛼_𝑛) for all 𝑛≥𝑁.
Furthermore,‖𝑥_𝑛−𝑦‖²≤⟨𝑓(𝑦)−𝑦,𝑗(𝑥_𝑛−𝑦)⟩1−𝛽−(1−𝛼_𝑛)𝑑_𝑛⁽¹⁾+(1−𝛼_𝑛)𝑑_𝑛⁽²⁾‖𝑥_𝑛−𝑦‖1−𝛽−(1−𝛼_𝑛)𝑑_𝑛⁽¹⁾ (3.3)for all 𝑛≥𝑁. That is, ‖𝑥_𝑛−𝑦‖≤(‖𝑓(𝑦)−𝑦‖+(1−𝛼_𝑛)𝑑_𝑛⁽²⁾)/(1−𝛽−(1−𝛼_𝑛)𝑑_𝑛⁽¹⁾) for all 𝑛≥𝑁. Thus {𝑥_𝑛} is bounded, so are {𝑇(𝑡_𝑛)𝑥_𝑛} and {𝑓(𝑥_𝑛)}. This imply thatlim_𝑛→∞‖𝑥_𝑛−𝑇(𝑡_𝑛)𝑥_𝑛‖=lim_𝑛→∞𝛼_𝑛‖𝑇(𝑡_𝑛)𝑥_𝑛−𝑓(𝑥_𝑛)‖=0. (3.4)Since {𝑇(ℎ)} is u.a.r and lim_𝑛→∞𝑡_𝑛=∞, then for all ℎ≥0,lim_𝑛→∞‖𝑇(ℎ)𝑇(𝑡_𝑛)𝑥_𝑛−𝑇(𝑡_𝑛)𝑥_𝑛‖≤lim_𝑛→∞sup_𝑥∈𝐶‖𝑇(ℎ)𝑇(𝑡_𝑛)𝑥−𝑇(𝑡_𝑛)𝑥‖=0, (3.5)where 𝐶 is any bounded subset of 𝐾 containing {𝑥_𝑛}. Since {𝑇(ℎ)} is continuous, hence‖𝑥_𝑛−𝑇(ℎ)𝑥_𝑛‖≤‖𝑥_𝑛−𝑇(𝑡_𝑛)𝑥_𝑛‖+‖𝑇(𝑡_𝑛)𝑥_𝑛−𝑇(ℎ)(𝑇(𝑡_𝑛)𝑥_𝑛)‖+‖𝑇(ℎ)(𝑇(𝑡_𝑛)𝑥_𝑛)−𝑇(ℎ)𝑥_𝑛‖→0. (3.6)That is, for all ℎ≥0,lim_𝑛→∞‖𝑥_𝑛−𝑇(ℎ)𝑥_𝑛‖=0. We claim that the set {𝑥_𝑛} is sequentially compact. Indeed, define the set𝐾^∗={𝑥∈𝐾∶𝜇_𝑛‖𝑥_𝑛−𝑥‖²=min_𝑦∈𝐾𝜇_𝑛‖𝑥_𝑛−𝑦‖²}. (3.7)By Lemma 2.9, we can found 𝑝∈𝐾^∗∩𝐹. Using Lemma 2.10, we get that𝜇_𝑛⟨𝑦−𝑝,𝑗(𝑥_𝑛−𝑝)⟩≤0∀𝑦∈𝐾. (3.8)It follows from (3.3) that
𝜇_𝑛‖𝑥_𝑛−𝑝‖²≤𝜇_𝑛⟨𝑓(𝑝)−𝑝,𝑗(𝑥_𝑛−𝑝)⟩1−𝛽−(1−𝛼_𝑛)𝑑_𝑛⁽¹⁾+𝜇_𝑛(1−𝛼_𝑛)𝑑_𝑛⁽²⁾‖𝑥_𝑛−𝑝‖1−𝛽−(1−𝛼_𝑛)𝑑_𝑛⁽¹⁾→0. (3.9)Then we have 𝜇_𝑛‖𝑥_𝑛−𝑝‖=0.
Hence, there exists a subsequence {𝑥_𝑛𝑘} of {𝑥_𝑛} which strongly converges to 𝑝∈𝐹 as 𝑘→∞.
Next we show that 𝑝 is a solution in 𝐹 to the variational inequality (3.1). In fact, for any fixed 𝑦∈𝐹, there exists a constant 𝑄>0 such that ‖𝑥_𝑛−𝑦‖≤𝑄, then‖𝑥_𝑛−𝑦‖²=⟨𝛼_𝑛(𝑓(𝑥_𝑛)−𝑦)+(1−𝛼_𝑛)(𝑇(𝑡_𝑛)𝑥_𝑛−𝑦),𝑗(𝑥_𝑛−𝑦)⟩=𝛼_𝑛⟨𝑓(𝑥_𝑛)−𝑓(𝑝)+𝑝−𝑥_𝑛,𝑗(𝑥_𝑛−𝑦)⟩+𝛼_𝑛⟨𝑓(𝑝)−𝑝,𝑗(𝑥_𝑛−𝑦)⟩+𝛼_𝑛⟨𝑥_𝑛−𝑦,𝑗(𝑥_𝑛−𝑦)⟩+(1−𝛼_𝑛)⟨𝑇(𝑡_𝑛)𝑥_𝑛−𝑇(𝑡_𝑛)𝑦,𝑗(𝑥_𝑛−𝑦)⟩≤𝛼_𝑛(𝛽+1)‖𝑥_𝑛−𝑝‖𝑄+𝛼_𝑛⟨𝑓(𝑝)−𝑝,𝑗(𝑥_𝑛−𝑦)⟩+‖𝑥_𝑛−𝑦‖²+(1−𝛼_𝑛)𝑘_𝑡⁽¹⁾_𝑛𝑄²+(1−𝛼_𝑛)𝑘_𝑡⁽²⁾_𝑛𝑄. (3.10)Therefore,⟨𝑓(𝑝)−𝑝,𝑗(𝑦−𝑥_𝑛)⟩≤(𝛽+1)‖𝑥_𝑛−𝑝‖𝑄+(1−𝛼_𝑛)𝑑_𝑛⁽¹⁾𝑄²+(1−𝛼_𝑛)𝑑_𝑛⁽²⁾𝑄. (3.11)Taking limit as 𝑛_𝑘→∞ in two sides of (3.11), by Lemma 2.8 and {𝑥_𝑛𝑘}→𝑝 as 𝑘→∞, we obtain⟨𝑓(𝑝)−𝑝,𝑗(𝑦−𝑝)⟩≤0∀𝑦∈𝐹. (3.12)That is, 𝑝∈𝐹 is a solution of variational inequality (3.1).
Suppose that 𝑝,𝑞∈𝐹 satisfy (3.1), we have ⟨𝑓(𝑝)−𝑝,𝑗(𝑞−𝑝)⟩≤0,⟨𝑓(𝑞)−𝑞,𝑗(𝑝−𝑞)⟩≤0. (3.13)(3.14) Combining (3.13) and (3.14), it follows that(1−𝛽)‖𝑝−𝑞‖²≤⟨(𝑝−𝑞)−𝑓(𝑝)+𝑓(𝑞),𝑗(𝑝−𝑞)⟩≤0. (3.15)Hence 𝑝=𝑞, that is, 𝑝∈𝐹 is the unique solution of variational inequality (3.1), so each cluster point of sequence {𝑥_𝑛} is equal to 𝑝. Therefore, {𝑥_𝑛} converges to 𝑝 and the proof is complete.

Remark 3.2. Let 𝐸,𝐾,𝐹,𝑓,{𝛼_𝑛}, and {𝑡_𝑛} be as in Theorem 3.1, ℱ={𝑇(ℎ)∶ℎ≥0} a u.a.r nonexpansive semigroup from 𝐾 into itself, then our result coincides with Theorem 3.2 in [3].

4. Modified Implicit Iteration Scheme

Theorem 4.1. Let 𝐸 be a real reflexive and strictly convex Banach space with a uniformly Gâteaux differentiable norm, 𝐾 a nonempty closed convex subset of 𝐸, ℱ={𝑇(ℎ)∶ℎ≥0} a u.a.r generalized asymptotically nonexpansive semigroup from 𝐾 into itself with sequences {𝑘_ℎ⁽¹⁾}{𝑘_ℎ⁽²⁾}, ℎ≥0, such that 𝐹≠∅, and 𝑓∶𝐾→𝐾 a fixed contractive mapping with contractive coefficient 𝛽∈(0,1). If {𝑥_𝑛} is given by (1.4), where lim_𝑛→∞𝑡_𝑛=∞, 𝛼_𝑛,𝛽_𝑛∈(0,1], lim_𝑛→∞𝛼_𝑛=0, ∑^∞_𝑛=1𝛽_𝑛=∞, and ∑^∞_𝑛=1(𝑘_𝑡⁽¹⁾_𝑛/𝛼_𝑛)<∞,∑^∞_𝑛=1(𝑘_𝑡⁽²⁾_𝑛/𝛼_𝑛)<∞, then {𝑥_𝑛} converges strongly to some common fixed point 𝑝 of 𝐹 such that 𝑝 is the unique solution in 𝐹 to variational inequality (3.1).

Proof. For any fixed 𝑦∈𝐹,‖𝑥_𝑛−𝑦‖=‖𝛼_𝑛(𝑦_𝑛−𝑦)+(1−𝛼_𝑛)(𝑇(𝑡_𝑛)𝑥_𝑛−𝑦)‖≤(1−𝛼_𝑛)‖𝑇(𝑡_𝑛)𝑥_𝑛−𝑦‖+𝛼_𝑛‖𝑦_𝑛−𝑦‖≤(1−𝛼_𝑛)[(1+𝑘_𝑡⁽¹⁾_𝑛)‖𝑥_𝑛−𝑦‖+𝑘_𝑡⁽²⁾_𝑛]+𝛼_𝑛‖𝑦_𝑛−𝑦‖. (4.1)Let 𝑑_𝑛^(𝑖)=(𝑘_𝑡^(𝑖)_𝑛/𝛼_𝑛)(𝑖=1,2). Hence,‖𝑥_𝑛−𝑦‖≤(1−𝛼_𝑛)𝑑_𝑛⁽²⁾1−(1−𝛼_𝑛)𝑑_𝑛⁽¹⁾+‖𝑦_𝑛−𝑦‖1−(1−𝛼_𝑛)𝑑_𝑛⁽¹⁾≤𝑑_𝑛⁽²⁾1−𝑑_𝑛⁽¹⁾+𝛽_𝑛‖𝑓(𝑥_𝑛−1)−𝑦‖+(1−𝛽_𝑛)‖𝑥_𝑛−1−𝑦‖1−𝑑_𝑛⁽¹⁾≤𝑑_𝑛⁽²⁾1−𝑑_𝑛⁽¹⁾+𝛽_𝑛‖𝑓(𝑥_𝑛−1)−𝑓(𝑦)‖1−𝑑_𝑛⁽¹⁾+𝛽_𝑛‖𝑓(𝑦)−𝑦‖1−𝑑_𝑛⁽¹⁾+(1−𝛽_𝑛)‖𝑥_𝑛−1−𝑦‖1−𝑑_𝑛⁽¹⁾≤(1−𝛽_𝑛(1−𝛽))(‖𝑥_𝑛−1−𝑦‖+𝑑_𝑛⁽²⁾)1−𝑑_𝑛⁽¹⁾+𝛽_𝑛(‖𝑓(𝑦)−𝑦‖+𝑑_𝑛⁽²⁾)1−𝑑_𝑛⁽¹⁾≤11−𝑑_𝑛⁽¹⁾max{‖𝑥_𝑛−1−𝑦‖+𝑑_𝑛⁽²⁾,‖𝑓(𝑦)−𝑦‖+𝑑_𝑛⁽²⁾1−𝛽}. (4.2)By induction, we get that‖𝑥_𝑛1−𝑦‖≤1−𝑑_𝑛⁽¹⁾max{‖𝑥_𝑛−2−𝑦‖+𝑑⁽²⁾_𝑛−11−𝑑⁽¹⁾_𝑛−1+𝑑_𝑛⁽²⁾,‖𝑓(𝑦)−𝑦‖+𝑑⁽²⁾_𝑛−1(1−𝑑⁽¹⁾_𝑛−1,)(1−𝛽)‖𝑓(𝑦)−𝑦‖+𝑑_𝑛⁽²⁾}⋯≤11−𝛽(1−𝑑_𝑛⁽¹⁾)⋯(1−𝑑₁⁽¹⁾)×max{‖𝑥₁−𝑦‖+^𝑛_𝑖=1𝑑_𝑖⁽²⁾,‖𝑓(𝑦)−𝑦‖+𝑑₁⁽²⁾1−𝛽,…,‖𝑓(𝑦)−𝑦‖+𝑑_𝑛⁽²⁾1−𝛽}. (4.3)Since ∑^∞_𝑛=1𝑑_𝑛^(𝑖)<∞(𝑖=1,2), we know from Abel–Dini theorem that there exists 𝑟>0 such that lim_𝑛→∞(1−𝑑_𝑛⁽¹⁾)⋯(1−𝑑₁⁽¹⁾)=𝑟. Thus {𝑥_𝑛} is bounded, so are {𝑇(𝑡_𝑛)𝑥_𝑛}, {𝑓(𝑥_𝑛)} and {𝑦_𝑛}. This imply thatlim_𝑛→∞‖𝑥_𝑛−𝑇(𝑡_𝑛)𝑥_𝑛‖=lim_𝑛→∞𝛼_𝑛‖𝑦_𝑛−𝑇(𝑡_𝑛)𝑥_𝑛‖=0. (4.4)Since {𝑇(ℎ)∶ℎ≥0} is u.a.r. and lim_𝑛→∞𝑡_𝑛=∞ for all ℎ≥0,lim_𝑛→∞‖𝑇(ℎ)𝑇(𝑡_𝑛)𝑥_𝑛−𝑇(𝑡_𝑛)𝑥_𝑛‖≤lim_𝑛→∞sup_𝑥∈𝐶‖𝑇(ℎ)𝑇(𝑡_𝑛)𝑥−𝑇(𝑡_𝑛)𝑥‖=0, (4.5)where 𝐶 is any bounded subset of 𝐾 containing {𝑥_𝑛}. Since 𝑇(ℎ) is continuous, hence‖𝑥_𝑛−𝑇(ℎ)𝑥_𝑛‖≤‖𝑥_𝑛−𝑇(𝑡_𝑛)𝑥_𝑛‖+‖𝑇(𝑡_𝑛)𝑥_𝑛−𝑇(ℎ)(𝑇(𝑡_𝑛)𝑥_𝑛)‖+‖𝑇(ℎ)(𝑇(𝑡_𝑛)𝑥_𝑛)−𝑇(ℎ)𝑥_𝑛‖→0. (4.6)That is, for all ℎ≥0,lim_𝑛→∞‖𝑥_𝑛−𝑇(ℎ)𝑥_𝑛‖=0. (4.7)From Theorem 3.1, there exists the unique solution 𝑝∈𝐹 to the variational inequality (3.1). Since 𝑝=𝑇(ℎ)𝑝 for all ℎ≥0, we have‖𝑥_𝑛+1−𝑝‖²=𝛼_𝑛+1⟨𝑦_𝑛+1−𝑝,𝑗(𝑥_𝑛+1−𝑝)⟩+(1−𝛼_𝑛+1)⟨𝑇(𝑡_𝑛+1)𝑥_𝑛+1−𝑝,𝑗(𝑥_𝑛+1−𝑝)⟩=𝛼_𝑛+1⟨𝛽_𝑛+1𝑓(𝑥_𝑛)+(1−𝛽_𝑛+1)𝑥_𝑛−𝑝,𝑗(𝑥_𝑛+1−𝑝)⟩+(1−𝛼_𝑛+1)⟨𝑇(𝑡_𝑛+1)𝑥_𝑛+1−𝑝,𝑗(𝑥_𝑛+1−𝑝)⟩≤𝛼_𝑛+1𝛽_𝑛+1⟨𝑓(𝑥_𝑛)−𝑓(𝑝),𝑗(𝑥_𝑛+1−𝑝)⟩+𝛼_𝑛+1𝛽_𝑛+1⟨𝑓(𝑝)−𝑝,𝑗(𝑥_𝑛+1−𝑝)⟩+𝛼_𝑛+1(1−𝛽_𝑛+1)⟨𝑥_𝑛−𝑝,𝑗(𝑥_𝑛+1−𝑝)⟩+(1−𝛼_𝑛+1)‖𝑥_𝑛+1−𝑝‖[(1+𝑘_𝑡⁽¹⁾_𝑛+1)‖𝑥_𝑛+1−𝑝‖+𝑘_𝑡⁽²⁾_𝑛+1]≤𝛼_𝑛+1𝛽_𝑛+1𝛽²‖𝑥_𝑛−𝑝‖²+‖𝑥_𝑛+1−𝑝‖²2+𝛼_𝑛+1(1−𝛽_𝑛+1)‖𝑥_𝑛−𝑝‖²+‖𝑥_𝑛+1−𝑝‖²2+(1−𝛼_𝑛+1)‖𝑥_𝑛+1−𝑝‖²+(1−𝛼_𝑛+1)𝑘_𝑡⁽¹⁾_𝑛+1‖𝑥_𝑛+1−𝑝‖²+(1−𝛼_𝑛+1)𝑘_𝑡⁽²⁾_𝑛+1‖𝑥_𝑛+1−𝑝‖+𝛼_𝑛+1𝛽_𝑛+1⟨𝑓(𝑝)−𝑝,𝑗(𝑥_𝑛+1=𝛼−𝑝)⟩_𝑛+12‖𝑥_𝑛+1−𝑝‖²+(1−𝛼_𝑛+1)‖𝑥_𝑛+1−𝑝‖²+𝛼_𝑛+12(1+𝛽²𝛽_𝑛+1−𝛽_𝑛+1)‖𝑥_𝑛−𝑝‖²+(1−𝛼_𝑛+1)𝑘_𝑡⁽¹⁾_𝑛+1‖𝑥_𝑛+1−𝑝‖²+(1−𝛼_𝑛+1)𝑘_𝑡⁽²⁾_𝑛+1‖𝑥_𝑛+1−𝑝‖+𝛼_𝑛+1𝛽_𝑛+1⟨𝑓(𝑝)−𝑝,𝑗(𝑥_𝑛+1−𝑝)⟩. (4.8)Therefore‖𝑥_𝑛+1−𝑝‖²≤(1−(1−𝛽²)𝛽_𝑛+1)‖𝑥_𝑛−𝑝‖²+2𝛽_𝑛+1⟨𝑓(𝑝)−𝑝,𝑗(𝑥_𝑛+1−𝑝)⟩+2(1−𝛼_𝑛+1)(𝑑⁽¹⁾_𝑛+1‖𝑥_𝑛+1−𝑝‖+𝑑⁽²⁾_𝑛+1)‖𝑥_𝑛+1−𝑝‖. (4.9)That is,‖𝑥_𝑛+1−𝑝‖²≤(1−𝜆_𝑛)‖𝑥_𝑛−𝑝‖²+𝑏_𝑛+𝑐_𝑛, (4.10)where 𝜆_𝑛=(1−𝛽²)𝛽_𝑛+1,𝑏_𝑛=2𝛽_𝑛+1⟨𝑓(𝑝)−𝑝,𝑗(𝑥_𝑛+1−𝑝)⟩ and 𝑐_𝑛=2(1−𝛼_𝑛+1)(𝑑⁽¹⁾_𝑛+1‖𝑥_𝑛+1−𝑝‖+𝑑⁽²⁾_𝑛+1)‖𝑥_𝑛+1−𝑝‖. Since ∑^∞_𝑛=1𝜆_𝑛=∞, ∑^∞_𝑛=1𝑑_𝑛^(𝑖)<∞(𝑖=1,2),‖𝑥_𝑛+1−𝑝‖ is bounded, we have ∑^∞_𝑛=1𝑐_𝑛<∞. So we only need to show that limsup_𝑛→∞(𝑏_𝑛/𝜆_𝑛)≤0, that is,lim_𝑛→∞⟨𝑓(𝑝)−𝑝,𝑗(𝑥_𝑛+1−𝑝)⟩≤0. (4.11)Let 𝑧_𝑚=𝑎_𝑚𝑓(𝑧_𝑚)+(1−𝑎_𝑚)𝑇(𝑡_𝑚)𝑧_𝑚, where 𝑡_𝑚 and 𝛼_𝑚 satisfy the condition of Theorem 3.1. Then it follows from Theorem 3.1 that 𝑝=lim_𝑚→∞𝑧_𝑚.
Since‖𝑥_𝑛+1−𝑧_𝑚‖²=(1−𝛼_𝑚)⟨𝑇(𝑡_𝑚)𝑧_𝑚−𝑥_𝑛+1,𝑗(𝑧_𝑚−𝑥_𝑛+1)⟩+𝛼_𝑚⟨𝑓(𝑧_𝑚)−𝑥_𝑛+1,𝑗(𝑧_𝑚−𝑥_𝑛+1)⟩=(1−𝛼_𝑚)(⟨𝑇(𝑡_𝑚)𝑧_𝑚−𝑇(𝑡_𝑚)𝑥_𝑛+1,𝑗(𝑧_𝑚−𝑥_𝑛+1)⟩+⟨𝑇(𝑡_𝑚)𝑥_𝑛+1−𝑥_𝑛+1,𝑗(𝑧_𝑚−𝑥_𝑛+1)⟩)+𝛼_𝑚⟨𝑓(𝑧_𝑚)−𝑧_𝑚−(𝑓(𝑝)−𝑝),𝑗(𝑧_𝑚−𝑥_𝑛+1)⟩+𝛼_𝑚⟨𝑓(𝑝)−𝑝,𝑗(𝑧_𝑚−𝑥_𝑛+1)⟩+𝛼_𝑚⟨𝑧_𝑚−𝑥_𝑛+1,𝑗(𝑧_𝑚−𝑥_𝑛+1)⟩≤‖𝑥_𝑛+1−𝑧_𝑚‖²+(1−𝛼_𝑚)(𝑘_𝑡⁽¹⁾_𝑚𝑄+𝑘_𝑡⁽²⁾_𝑚)𝑄+(1−𝛼_𝑚)‖𝑇(𝑡_𝑚)𝑥_𝑛+1−𝑥_𝑛+1‖𝑄+𝛼_𝑚⟨𝑓(𝑝)−𝑝,𝑗(𝑧_𝑚−𝑥_𝑛+1)⟩+𝛼_𝑚(1+𝛽)‖𝑧_𝑚−𝑝‖𝑄. (4.12)Furthermore,⟨𝑓(𝑝)−𝑝,𝑗(𝑥_𝑛+1−𝑧_𝑚)⟩≤1−𝛼_𝑚𝛼_𝑚(𝑘_𝑡⁽¹⁾_𝑚𝑄+𝑘_𝑡⁽²⁾_𝑚)𝑄+‖𝑇(𝑡_𝑚)𝑥_𝑛+1−𝑥_𝑛+1‖𝛼_𝑚𝑄+(1+𝛽)‖𝑧_𝑚−𝑝‖𝑄, (4.13)where 𝑄 is a constant such that 𝑄≥‖𝑧_𝑚−𝑥_𝑛+1‖. Hence, taking upper limit as 𝑛→∞ firstly, and then as 𝑚→∞ in (4.13), we havelim_𝑚→∞lim_𝑛→∞⟨𝑓(𝑝)−𝑝,𝑗(𝑥_𝑛+1−𝑧_𝑚)⟩≤0. (4.14)On the other hand, since 𝑝=lim_𝑚→∞𝑧_𝑚 and by Lemma 2.8, we have⟨𝑓(𝑝)−𝑝,𝑗(𝑥_𝑛+1−𝑧_𝑚)⟩→⟨𝑓(𝑝)−𝑝,𝑗(𝑥_𝑛+1−𝑝)⟩uniformly. (4.15)Thus given 𝜀>0, there exists 𝑁≥1 such that if 𝑚>𝑁 for all 𝑛 we have⟨𝑓(𝑝)−𝑝,𝑗(𝑥_𝑛+1−𝑝)⟩<⟨𝑓(𝑝)−𝑝,𝑗(𝑥_𝑛+1−𝑧_𝑚)⟩+𝜀. (4.16)Hence, taking upper limit as 𝑛→∞ firstly, and then as 𝑚→∞ in two sides of (4.16), we get thatlim_𝑛→∞⟨𝑓(𝑝)−𝑝,𝑗(𝑥_𝑛+1−𝑝)⟩≤lim_𝑚→∞lim_𝑛→∞⟨𝑓(𝑝)−𝑝,𝑗(𝑥_𝑛+1−𝑧_𝑚)⟩+𝜀≤𝜀. (4.17)For the arbitrariness of 𝜀, (4.11) holds. By Lemma 2.11, 𝑥_𝑛→𝑝 and the proof is complete.

Acknowledgments

This work was supported by National Natural Science Foundation of China (10771173) and Natural Science Foundation Project of Henan (2008B110012).