Abstract

Let 𝑋 be a uniformly convex Banach space and 𝒮={𝑇(𝑠)∶0≤𝑠<∞} be a nonexpansive semigroup such that ⋂𝐹(𝒮)=𝑠>0𝐹(𝑇(𝑠))≠∅. Consider the iterative method that generates the sequence {𝑥𝑛} by the algorithm 𝑥𝑛+1=𝛼𝑛𝑓(𝑥𝑛)+𝛽𝑛𝑥𝑛+(1−𝛼𝑛−𝛽𝑛)(1/𝑠𝑛)∫𝑠𝑛0𝑇(𝑠)𝑥𝑛𝑑𝑠,𝑛≥0, where {𝛼𝑛}, {𝛽𝑛}, and {𝑠𝑛} are three sequences satisfying certain conditions, 𝑓∶𝐶→𝐶 is a contraction mapping. Strong convergence of the algorithm {𝑥𝑛} is proved assuming 𝑋 either has a weakly continuous duality map or has a uniformly Gâteaux differentiable norm.

1. Introduction

Let 𝑋 be a real Banach space and let 𝐶 be a nonempty closed convex subset of 𝑋. A mapping 𝑇 of 𝐶 into itself is said to be nonexpansive if ‖𝑇𝑥−𝑇𝑦‖≤‖𝑥−𝑦‖ for each 𝑥,𝑦∈𝐶. We denote by 𝐹(𝑇) the set of fixed points of 𝑇. One classical way to study nonexpansive mappings is to use contractions to approximate a nonexpansive mapping (Browder [1] and Reich [2]). More precisely, take 𝑡∈(0,1) and define a contraction 𝑇𝑡∶𝐶→𝐶 by𝑇𝑡𝑥=𝑡𝑢+(1−𝑡)𝑇𝑥,𝑥∈𝐶,(1.1)

where 𝑢∈𝐶 is a fixed point. Banach’s contraction mapping principle guarantees that 𝑇𝑡 has a unique fixed point 𝑥𝑡 in 𝐶. It is unclear, in general, what is the behavior of {𝑥𝑡} as 𝑡→0, even if 𝑇 has a fixed point. In 1967, in the case of 𝑇 having a fixed point, Browder [3] proved that if 𝑋 is a Hilbert space, then 𝑥𝑡 converges strongly to the element of 𝐹(𝑇) which is nearest to 𝑢 in 𝐹(𝑇) as 𝑡↓0. Song and Xu [4] extended Browder’s result to the setting of Banach spaces and proved that if 𝑋 is a uniformly smooth Banach space, then 𝑥𝑡 converges strongly to a fixed point of 𝑇 and the limit defines the (unique) sunny nonexpansive retraction from 𝐶 onto 𝐹(𝑇).

Let 𝑓 be a contraction on 𝐻 such that ‖𝑓𝑥−𝑓𝑦‖≤𝛼‖𝑥−𝑦‖, where 𝛼∈[0,1) is a constant. Let 𝑥∈𝐶, 𝑡∈(0,1) and 𝑥𝑡∈𝐶 be the unique fixed point of the contraction 𝑆𝑡𝑥=𝑡𝑓(𝑥)+(1−𝑡)𝑇𝑥, that is,𝑥𝑡𝑥=𝑡𝑓𝑡+(1−𝑡)𝑇𝑥𝑡.(1.2)

Concerning the convergence problem of the net {𝑥𝑡}, Moudafi [5] and Xu [6] by using the viscosity approximation method proved that the net {𝑥𝑡} converges strongly to a fixed point ̃𝑥 of T in C which is the unique solution to the following variational inequality:⟨(𝐼−𝑓)̃𝑥,𝑥−̃𝑥⟩≥0,∀𝑥∈𝐹(𝑇).(1.3)

Moreover, Xu [6] also studied the strong convergence of the following iterative sequence generated by𝑥𝑛+1=𝛽𝑛𝑓𝑥𝑛+1−𝛽𝑛𝑇𝑥𝑛,𝑛≥0,(1.4)

where 𝑥0∈𝐶 is arbitrary, the sequence {𝛽𝑛} in (0,1) satisfies the certain appropriate conditions.

A family {𝑇(𝑠)∶0≤𝑠<∞} of mappings of 𝐶 into itself is called a nonexpansive semigroup if it satisfies the following conditions:(i)𝑇(0)𝑥=𝑥 for all 𝑥∈𝐶;(ii)𝑇(𝑠+𝑡)=𝑇(𝑠)𝑇(𝑡) for all 𝑥,𝑦∈𝐶 and 𝑠,𝑡≥0;(iii)‖𝑇(𝑠)𝑥−𝑇(𝑠)𝑦‖≤‖𝑥−𝑦‖ for all 𝑥,𝑦∈𝐶 and 𝑠≥0;(iv)for all 𝑥∈𝐶, 𝑠↦𝑇(𝑠)𝑥 is continuous.

We denote by 𝐹(𝒮) the set of all common fixed points of 𝒮, that is, 𝐹(𝒮)={𝑥∈𝐶∶𝑇(𝑠)𝑥=𝑥,0≤𝑠<∞}. It is known that 𝐹(𝒮) is closed and convex.

It is an interesting problem to extend above (Moudafi’s [5], Xu’s [6], and so on) results to the nonexpansive semigroup case. Recently, for the nonexpansive semigroups 𝒮={𝑇(𝑠)∶0≤𝑠<∞}, Plubtieng and Punpaeng [7] studied the continuous scheme {𝑥𝑡} defined by𝑥𝑡𝑥=𝑡𝑓𝑡1+(1−𝑡)𝜆𝑡𝜆𝑡0𝑇(𝑠)𝑥𝑡𝑑𝑠,(1.5)

where 𝑡∈(0,1) and {𝜆𝑡} is a positive real divergent net, and the iterative scheme {𝑥𝑛} defined by𝑥𝑛+1=𝛼𝑛𝑓𝑥𝑛+𝛽𝑛𝑥𝑛+1−𝛼𝑛−𝛽𝑛1𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛𝑑𝑠,𝑛≥0,(1.6)

where 𝑥0∈𝐶, {𝛼𝑛},{𝛽𝑛} are a sequence in (0,1) and {𝑠𝑛} is a positive real divergent real sequence in the setting of a real Hilbert space. They proved the continuous scheme {𝑥𝑡} defined by (1.5) and the iterative scheme {𝑥𝑛} defined by (1.6) converge strongly to a fixed point ̃𝑥 of 𝒮 which is the unique solution of the variational inequality (1.3). At this stage, the following question arises naturally.

Question 1. Do Plubtieng and Punpaeng’s results hold for the nonexpansive semigroups in a Banach space?

The purpose of this paper is to give affirmative answers of Question 1. One result of this paper says that Plubtieng and Punpaeng’s results hold in a uniformly convex Banach space which has a weakly continuous duality map.

On the other hand, Chen and Song [8] proved the following implicit and explicit viscosity iteration processes defined by (1.7) to nonexpansive semigroup case,𝑥𝑛=𝛼𝑛𝑓𝑥𝑛+1−𝛼𝑛1𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛𝑥𝑑𝑠,𝑛≥0,𝑛+1=𝛼𝑛𝑓𝑥𝑛+1−𝛼𝑛1𝑠𝑛s𝑛0𝑇(𝑠)𝑥𝑛𝑑𝑠,𝑛≥0.(1.7)

And they proved that {𝑥𝑛} converges strongly to a common fixed point of 𝐹(𝒮) in a uniformly convex Banach space with a uniformly Gâteaux differentiable norm.

Motivated by the above results, the other result of this paper says that Plubtieng and Punpaeng’s results hold in the framework of uniformly convex Banach space with a uniformly Gâteaux differentiable norm. The results improve and extend the corresponding results of Plubtieng and Punpaeng [7], Chen and Song [8], Moudafi’s [5], Xu’s [6], and others.

2. Preliminaries

Let 𝑋 be a real Banach space with inner product ⟨⋅,⋅⟩ and norm ‖⋅‖, respectively. Let 𝐽 denote the normalized duality mapping from 𝑋 into the dual space 2𝑋∗ given by𝑥𝐽(𝑥)=∗∈𝑋∗∶⟨𝑥,𝑥∗⟩=‖𝑥‖2=‖𝑥∗‖2,𝑥∈𝑋.(2.1)

In the sequel, we will denote the single valued duality mapping by 𝑗. When {𝑥𝑛} is a sequence in 𝑋, then 𝑥𝑛→𝑥(𝑥𝑛⇀𝑥) will denote strong (weak) convergence of the sequence {𝑥𝑛} to 𝑥.

Let 𝑆(𝑋)={𝑥∈𝑋∶‖𝑥‖=1}. Then the norm of 𝑋 is said to be Gâteaux differentiable iflim𝑡→0‖𝑥+𝑡𝑦‖−‖𝑥‖𝑡(2.2) exists for each 𝑥,𝑦∈𝑆(𝑋). In this case, 𝑋 is called smooth. The norm of 𝑋 is said to be uniformly Gâteaux differentiable if for each 𝑦∈𝑆(𝑋), the limit (2.2) is attained uniformly for 𝑥∈𝑆(𝑋). It is well known that 𝑋 is smooth if and only if any duality mapping on 𝑋 is sigle valued. Also if 𝑋 has a uniformly Gâteaux differentiable norm, then the duality mapping is norm-to-weak* uniformly continuous on bounded sets. The norm of E is called Fréchet differentiable, if for each 𝑥∈𝑆(𝑋), the limit (2.2) is attained uniformly for 𝑦∈𝑆(𝑋). The norm of 𝑋 is called uniformly Fréchet differentiable, if the limit (2.2) is attained uniformly for 𝑥,𝑦∈𝑆(𝑋). It is well known that (uniformly) Fréchet differentiability of the norm of 𝑋 implies (uniformly) Gâteaux differentiability of the norm of 𝑋 and 𝑋 is uniformly smooth if and only if the norm of 𝑋 is uniformly Fréchet differentiable.

A Banach space 𝑋 is said to be strictly convex if‖𝑥‖=‖𝑦‖=1,𝑥≠𝑦implies‖𝑥+𝑦‖2<1.(2.3)

A Banach space 𝑋 is said to be uniformly convex if 𝛿𝑋(𝜀)>0 for all 𝜀>0, where 𝛿𝑋(𝜀) is modulus of convexity of 𝐸 defined by𝛿𝐸(𝜀)=inf1−‖𝑥+𝑦‖2∶[]‖𝑥‖≤1,‖𝑦‖≤1,‖𝑥−𝑦‖≥𝜀,𝜀∈0,2.(2.4)

A uniformly convex Banach space 𝐸 is reflexive and strictly convex [9, Theorem 4.1.6, Theorem 4.1.2].

Lemma 2.1 (Goebel and Reich [10], Proposition 5.3). Let 𝐶 be a nonempty closed convex subset of a strictly convex Banach space 𝑋 and 𝑇∶𝐶→𝐶 a nonexpansive mapping with 𝐹(𝑇)≠∅. Then 𝐹(𝑇) is closed and convex.

Lemma 2.2 (see Xu [11]). In a smooth Banach space 𝑋 there holds the inequality ‖𝑥+𝑦‖2≤‖𝑥‖2+2⟨𝑦,𝐽(𝑥+𝑦)⟩,𝑥,𝑦,∈𝑋.(2.5)

Lemma 2.3 (Browder [12]). Let 𝐸 be a uniformly convex Banach space, 𝐾 a nonempty closed convex subset of 𝐸, and 𝑇∶𝐾→𝐸 a nonexpansive mapping. Then 𝐼−𝑇 is demi closed at zero.

Lemma 2.4 (see [8, Lemma 2.7]). Let 𝐶 be a nonempty bounded closed convex subset of a uniformly convex Banach space 𝑋, and let 𝒮={𝑇(𝑠)∶0≤𝑠<∞} be a nonexpansive semigroup on 𝐶 such that 𝐹(𝒮)≠∅. For 𝑥∈𝐶 and 𝑡>0. Then, for any 0≤ℎ<∞, lim𝑡→∞sup𝑥∈𝐶‖‖‖1𝑡𝑡01𝑇(𝑠)𝑥𝑑𝑠−𝑇(ℎ)𝑡𝑡0‖‖‖𝑇(𝑠)𝑥𝑑𝑠=0.(2.6)

Recall that a gauge is a continuous strictly increasing function 𝜑∶[0,∞)→[0,∞) such that 𝜑(0)=0 and 𝜑(𝑡)→∞ as 𝑡→∞. Associated to a gauge 𝜑 is the duality map 𝐽𝜑∶𝑋→𝑋∗ defined by𝐽𝜑𝑥(𝑥)=∗∈𝑋∗∶⟨𝑥,𝑥∗(⟩=‖𝑥‖𝜑‖𝑥‖),‖𝑥∗()‖=𝜑‖𝑥‖,𝑥∈𝑋.(2.7)

Following Browder [13], we say that a Banach space 𝑋 has a weakly continuous duality map if there exists a gauge 𝜑 for which the duality map 𝐽𝜑 is single valued and weak-to-weak* sequentially continuous (i.e., if {𝑥𝑛} is a sequence in 𝑋 weakly convergent to a point 𝑥, then the sequence 𝐽𝜑(𝑥𝑛) converges weakly* to 𝐽𝜑(𝑥)). It is known that 𝑙𝑝 has a weakly continuous duality map for all 1<𝑝<∞. SetΦ(𝑡)=𝑡0𝜑(𝜏)𝑑𝜏,𝑡≥0.(2.8)

Then𝐽𝜑(𝑥)=𝜕Φ(‖𝑥‖),𝑥∈𝑋,(2.9)

where 𝜕 denotes the subdifferential in the sense of convex analysis. The next lemma is an immediate consequence of the subdifferential inequality.

Lemma 2.5 (Xu [11, Lemma 2.6]). Assume that 𝑋 has a weakly continuous duality map 𝐽𝜑 with gauge 𝜑, for all 𝑥,𝑦∈𝑋, there holds the inequality Φ(‖𝑥+𝑦‖)≤Φ(‖𝑥‖)+𝑦,𝐽𝜑.(𝑥+𝑦)(2.10)

Lemma 2.6 (Xu [6]). Assume {𝛼𝑛} is a sequence of nonnegative real numbers such that 𝛼𝑛+1≤1−𝛾𝑛𝛼𝑛+𝛿𝑛,𝑛≥0,(2.11)where {𝛾𝑛} is a sequence in (0,1) and {𝛿𝑛} is a sequence in 𝐑 such that(i)∑∞𝑛=1𝛾𝑛=∞; (ii)limsup𝑛→∞𝛿𝑛/𝛾𝑛≤0 or ∑∞𝑛=1|𝛿𝑛|<∞.Then lim𝑛→∞𝛼𝑛=0.

Finally, we also need the following definitions and results [9, 14]. Let 𝜇 be a continuous linear functional on 𝑙∞ satisfying ‖𝜇‖=1=𝜇(1). Then we know that 𝜇 is a mean on 𝑁 if and only if𝑎inf𝑛𝑎;𝑛∈𝑁≤𝜇(𝑎)≤sup𝑛;𝑛∈𝑁,(2.12) for every 𝑎=(𝑎1,𝑎2,…)∈𝑙∞. Occasionally, we will use 𝜇𝑛(𝑎𝑛) instead of 𝜇(𝑎). A mean 𝜇 on 𝑁 is called a Banach limit if𝜇𝑛𝑎𝑛=𝜇𝑛𝑎𝑛+1,(2.13) for every 𝑎=(𝑎1,𝑎2,…)∈𝑙∞. Using the Hahn-Banach theorem, or the Tychonoff fixed point theorem, we can prove the existence of a Banach limit. We know that if 𝜇 is a Banach limit, thenliminf𝑛→∞𝑎𝑛≤𝜇𝑛𝑎𝑛≤limsup𝑛→∞𝑎𝑛,(2.14) for every 𝑎=(𝑎1,𝑎2,…)∈𝑙∞. So, if 𝑎=(𝑎1,𝑎2,…), 𝑏=(𝑏1,𝑏2,…)∈𝑙∞, and 𝑎𝑛→𝑐 (resp., 𝑎𝑛−𝑏𝑛→0), as 𝑛→∞, we have𝜇𝑛𝑎𝑛=𝜇(𝑎)=𝑐resp.,𝜇𝑛𝑎𝑛=𝜇𝑛𝑏𝑛.(2.15)

Subsequently, the following result was showed in [14, Lemma 1] and [9, Lemma 4.5.4].

Lemma 2.7 (see [14, Lemma 1]). Let 𝐶 be a nonempty closed convex subset of a Banach space 𝑋 with a uniformly Gâteaux differentiable norm and {𝑥𝑛} a bounded sequence of 𝐸. If 𝑧0∈𝐶, then 𝜇𝑛‖‖𝑥𝑛−𝑧0‖‖2=min𝑥∈𝐶𝜇𝑛‖‖𝑥𝑛‖‖−𝑥2,(2.16) if and only if 𝜇𝑛𝑥−𝑧0𝑥,𝐽𝑛−𝑧0≤0,∀𝑥∈𝐶.(2.17)

Lemma 2.8 (Song and Xu [4, Proposition 3.1]). Let 𝑋 be a reflexive strictly convex Banach space with a uniformly Gâteaux differentiable norm, and 𝐶 a nonempty closed convex subset of 𝑋. Suppose {𝑥𝑛} is a bounded sequence in 𝐶 such that lim𝑛→∞‖𝑥𝑛−𝑇𝑥𝑛‖=0, an approximate fixed point of nonexpansive self-mapping 𝑇 on 𝐶. Define the set 𝐶∗=𝑦∈𝐶∶𝜇𝑛‖‖𝑥𝑛‖‖−𝑦2=inf𝑥∈𝐶𝜇𝑛‖‖𝑥𝑛‖‖−𝑥2.(2.18) If 𝐹(𝑇)≠∅, then 𝐶∗∩𝐹(𝑇)≠∅.

3. Implicit Iteration Scheme

Theorem 3.1. Let 𝑋 be a uniformly convex Banach space that has a weakly continuous duality map 𝐽𝜑 with gauge 𝜑, and let 𝐶 be a nonempty closed convex subset of 𝑋. Let 𝒮={𝑇(𝑠)∶0≤𝑠<∞} be a nonexpansive semigroup from 𝐶 into itself such that ⋂𝐹(𝒮)=𝑠>0𝐹(𝑇(𝑠))≠∅ and 𝑓∶𝐶→𝐶 a contraction mapping with the contractive coefficient 𝛼∈[0,1). Suppose {𝜆𝑡}0<𝑡<1 is a net of positive real numbers such that lim𝑡→0+𝜆𝑡=∞, the sequence {𝑥𝑡} is given by the following equation: 𝑥𝑡𝑥=𝑡𝑓𝑡1+(1−𝑡)𝜆𝑡𝜆𝑡0𝑇(s)𝑥𝑡𝑑𝑠.(3.1) Then {𝑥𝑡} converges strongly to ̃𝑥 as 𝑡→0+, where ̃𝑥 is the unique solution in 𝐹(𝒮) of the variational inequality ⟨(𝐼−𝑓)̃𝑥,𝐽(𝑥−̃𝑥)⟩≥0,𝑥∈𝐹(𝒮).(3.2)

Proof. Note that 𝐹(𝒮) is a nonempty closed convex set by Lemma 2.1. We first show that {𝑥𝑡} is bounded. Indeed, for any fixed 𝑝∈𝐹(𝒮), we have ‖‖𝑥𝑡‖‖‖‖𝑓𝑥−𝑝≤𝑡𝑡‖‖‖‖‖1−𝑝+(1−𝑡)𝜆𝑡𝜆𝑡0𝑇(𝑠)𝑥𝑡‖‖‖‖‖𝑓𝑥𝑑𝑠−𝑝≤𝑡𝑡‖‖1−𝑓(𝑝)+‖𝑓(𝑝)−𝑝‖+(1−𝑡)𝜆𝑡𝜆𝑡0‖‖𝑇(𝑠)𝑥𝑡‖‖𝛼‖‖𝑥−𝑝𝑑𝑠≤𝑡𝑡‖‖‖+‖‖𝑥−𝑝+‖𝑓(𝑝)−𝑝(1−𝑡)𝑡‖‖=‖‖𝑥−𝑝𝑡‖‖‖‖𝑥−𝑝−𝑡(1−𝛼)𝑡‖‖−𝑝+𝑡‖𝑓(𝑝)−𝑝‖.(3.3) It follows that ‖‖𝑥𝑡‖‖≤1−𝑝1−𝛼‖𝑓(𝑝)−𝑝‖.(3.4) Thus {𝑥𝑡} is bounded, so are {𝑓(𝑥𝑡)} and {𝑇(𝑠)𝑥𝑡} for every 0≤𝑠<∞. Furthermore, we note that ‖‖𝑥𝑡−𝑇(𝑠)𝑥𝑡‖‖≤‖‖‖𝑥𝑡−1𝜆𝑡𝜆𝑡0𝑇(𝑠)𝑥𝑡‖‖‖+‖‖‖1𝑑𝑠𝜆𝑡𝜆𝑡0𝑇(𝑠)𝑥𝑡1𝑑𝑠−𝑇(𝑠)𝜆𝑡𝜆𝑡0𝑇(𝑠)𝑥𝑡‖‖‖+‖‖‖1𝑑𝑠𝑇(𝑠)𝜆𝑡𝜆𝑡0𝑇(𝑠)𝑥𝑡𝑑𝑠−𝑇(𝑠)𝑥𝑡‖‖‖‖‖‖𝑥≤2𝑡−1𝜆𝑡𝜆𝑡0𝑇(𝑠)𝑥𝑡‖‖‖+‖‖‖1d𝑠𝜆𝑡𝜆𝑡0𝑇(𝑠)𝑥𝑡1𝑑𝑠−𝑇(𝑠)𝜆𝑡𝜆𝑡0𝑇(𝑠)𝑥𝑡‖‖‖,𝑑𝑠(3.5) for every 0≤𝑠<∞. On the one hand, we observe that ‖‖‖𝑥𝑡−1𝜆𝑡𝜆𝑡0𝑇(𝑠)𝑥𝑡‖‖‖=𝑡𝑑𝑠‖‖𝑥1−𝑡𝑡𝑥−𝑓𝑡‖‖,(3.6) for every 𝑡>0. On the other hand, let 𝑧0∈𝐹(𝑆) and 𝐷={𝑧∈𝐶∶‖𝑧−𝑧0‖≤‖𝑓(𝑧0)−𝑧0‖}, then 𝐷 is a nonempty closed bounded convex subset of 𝐶 which is 𝑇(𝑠)-invariant for each 0≤𝑠<∞ and contains {𝑥𝑡}. It follows by Lemma 2.4 that lim𝜆𝑡→∞‖‖‖1𝜆𝑡𝜆𝑡0𝑇(𝑠)𝑥𝑡1𝑑𝑠−𝑇(s)𝜆𝑡𝜆𝑡0𝑇(𝑠)𝑥𝑡‖‖‖𝑑𝑠≤lim𝜆𝑡→∞sup𝑥∈𝐷‖‖‖1𝜆𝑡𝜆𝑡0𝑇(𝑠)𝑥𝑡1𝑑𝑠−𝑇(𝑠)𝜆𝑡𝜆𝑡0𝑇(𝑠)𝑥𝑡‖‖‖𝑑𝑠=0.(3.7) Hence, by (3.5)–(3.7), we obtain ‖‖𝑥𝑡−𝑇(𝑠)𝑥𝑡‖‖⟶0as𝑡⟶0,(3.8) for every 0≤𝑠<∞. Assume {𝑡𝑛}∞𝑛=1⊂(0,1) is such that 𝑡𝑛→0 as 𝑛→∞. Put 𝑥𝑛∶=𝑥𝑡𝑛, 𝜆𝑛∶=𝜆𝑡𝑛, we will show that {𝑥𝑛} contains s subsequence converging strongly to ̃𝑥, where ̃𝑥∈𝐹(𝒮). Since {𝑥𝑛} is a bounded sequence, there is a subsequence {𝑥𝑛𝑗} of {𝑥𝑛} which converges weakly to ̃𝑥∈𝐶. By Lemma 2.3, we have ̃𝑥∈𝐹(𝒮). For each 𝑛≥1, we have 𝑥𝑛−̃𝑥=𝑡𝑛𝑓𝑥𝑛+−̃𝑥1−𝑡𝑛1𝜆𝑛𝜆𝑛0𝑇(𝑠)𝑥𝑛.𝑑𝑠−̃𝑥(3.9) Thus, by Lemma 2.5, we obtain Φ‖‖𝑥𝑛‖‖‖‖‖𝑡−̃𝑥=Φ𝑛𝑓𝑥𝑛+−̃𝑥1−𝑡𝑛1𝜆𝑛𝜆𝑛0𝑇(𝑠)𝑥𝑛‖‖‖‖‖‖𝑑𝑠−̃𝑥≤Φ1−𝑡𝑛1𝜆𝑛𝜆𝑛0𝑇(𝑠)𝑥𝑛‖‖‖𝑑𝑠−̃𝑥+𝑡𝑛𝑓𝑥𝑛−̃𝑥,𝐽𝜑𝑥𝑛≤−̃𝑥1−𝑡𝑛Φ‖‖‖1𝜆𝑛𝜆𝑛0𝑇(𝑠)𝑥𝑛‖‖‖𝑑𝑠−̃𝑥+𝑡𝑛𝑓𝑥𝑛−̃𝑥,𝐽𝜑𝑥𝑛≤−̃𝑥1−𝑡𝑛Φ‖‖𝑥𝑛‖‖−̃𝑥+𝑡𝑛𝑓𝑥𝑛−̃𝑥,𝐽𝜑𝑥𝑛.−̃𝑥(3.10) This implies that Φ‖‖𝑥𝑛‖‖≤𝑓𝑥−̃𝑥𝑛−̃𝑥,𝐽𝜑𝑥𝑛.−̃𝑥(3.11) In particular, we have Φ‖‖𝑥𝑛𝑗‖‖≤𝑓𝑥−̃𝑥𝑛𝑗−̃𝑥,𝐽𝜑𝑥𝑛𝑗−̃𝑥.(3.12) Now observing that {𝑥𝑛𝑗}⇀̃𝑥 implies 𝐽𝜑(𝑥𝑛𝑗−̃𝑥)⇀0. And since 𝑓(𝑥𝑛𝑗) is bounded, it follows from (3.12) that Φ‖‖𝑥𝑛𝑗‖‖−̃𝑥⟶0as𝑗⟶∞.(3.13) Hence 𝑥𝑛𝑗→̃𝑥.
Next, we show that ̃𝑥∈𝐹(𝒮) solves the variational inequality (3.2). Indeed, for 𝑞∈𝐹(𝒮), it is easy to see that 𝑥𝑡−1𝜆𝑡𝜆𝑡0𝑇(𝑠)𝑥𝑡𝑑𝑠,𝐽𝜑𝑥𝑡‖‖𝑥−𝑞=Φ𝑡‖‖+1−𝑞𝑞−𝜆𝑡𝜆𝑡0𝑇(𝑠)𝑥𝑡𝑑𝑠,𝐽𝜑𝑥𝑡‖‖𝑥−𝑞≥Φ𝑡‖‖−‖‖‖1−𝑞𝑞−𝜆𝑡𝜆𝑡0𝑇(𝑠)𝑥𝑡‖‖‖‖‖𝐽𝑑𝑠𝜑𝑥𝑡‖‖‖‖𝑥−𝑞≥Φ𝑡‖‖‖‖𝑥−𝑞−Φ𝑡‖‖−𝑞=0.(3.14) However, we note that 𝑥𝑡−1𝜆𝑡𝜆𝑡0𝑇(𝑠)𝑥𝑡𝑡𝑑𝑠=𝑓𝑥1−𝑡𝑡−𝑥𝑡.(3.15) Thus, we get that for 𝑡∈(0,1) and 𝑞∈𝐹(𝒮)𝑥𝑡𝑥−𝑓𝑡,𝐽𝜑𝑥𝑡−𝑞≤0.(3.16) Taking the limit through 𝑡∶=𝑡𝑛𝑗→0, we obtain (𝐼−𝑓)̃𝑥,𝐽𝜑(̃𝑥−𝑞)≤0,∀𝑞∈𝐹(𝒮).(3.17) This implies that ⟨(𝐼−𝑓)̃𝑥,𝐽(̃𝑥−𝑞)⟩≤0,∀𝑞∈𝐹(𝒮),(3.18) since 𝐽𝜑(𝑥)=(𝜑(‖𝑥‖)/‖𝑥‖)𝐽(𝑥) for 𝑥≠0.
Finally, we show that the net {𝑥𝑡} convergence strong to ̃𝑥. Assume that there is a sequence {𝑠𝑛}⊂(0,1) such that 𝑥𝑠𝑛→𝑥, where 𝑠𝑛→0. we note by Lemma 2.3 that 𝑥∈𝐹(𝒮). It follows from the inequality (3.18) that (𝐼−𝑓)̃𝑥,𝐽̃𝑥−𝑥≤0.(3.19) Interchange ̃𝑥 and 𝑥 to obtain (𝐼−𝑓)𝑥,𝐽𝑥−̃𝑥≤0.(3.20) Adding (3.19) and (3.20) yields ‖‖(1−𝛼)̃𝑥−𝑥‖‖2≤𝑥−̃𝑥−𝑓(̃𝑥)−𝑓𝑥,𝐽𝑥−̃𝑥≤0.(3.21) We must have ̃𝑥=𝑥 and the uniqueness is proved. In a summary, we have shown that each cluster point of {𝑥𝑡} as 𝑡→0 equals ̃𝑥. Therefore 𝑥𝑡→̃𝑥 as 𝑡→0.

Theorem 3.2. Let 𝑋 be a uniformly convex Banach space with a uniformly Gâteaux differentiable norm and 𝐶 be a nonempty closed convex subset of 𝑋. Let 𝒮={𝑇(𝑠)∶0≤𝑠<∞} be a nonexpansive semigroup from 𝐶 into itself such that ⋂𝐹(𝒮)=𝑠>0𝐹(𝑇(𝑠))≠∅ and 𝑓∶𝐶→𝐶 a contraction mapping with the contractive coefficient 𝛼∈[0,1). Suppose {𝜆𝑡}0<𝑡<1 is a net of positive real numbers such that lim𝑡→0+𝜆𝑡=∞, the sequence {𝑥𝑡} is given by the following equation: 𝑥𝑡𝑥=𝑡𝑓𝑡1+(1−𝑡)𝜆𝑡𝜆𝑡0𝑇(𝑠)𝑥𝑡𝑑𝑠.(3.22) Then {𝑥𝑡} converges strongly to ̃𝑥 as 𝑡→0+, where ̃𝑥 is the unique solution in 𝐹(𝒮) of the variational inequality ⟨(𝐼−𝑓)̃𝑥,𝐽(𝑥−̃𝑥)⟩≥0,𝑥∈𝐹(𝒮).(3.23)

Proof. We include only those points in this proof which are different from those already presented in the proof of Theorem 3.1. As in the proof of Theorem 3.1, we obtain that there is a subsequence {𝑥𝑛𝑗} of {𝑥𝑛} which converges weakly to ̃𝑥∈𝐹(𝒮). For each 𝑛≥1, we have 𝑥𝑛−̃𝑥=𝑡𝑛𝑓𝑥𝑛+−̃𝑥1−𝑡𝑛1𝜆𝑛𝜆𝑛0𝑇(𝑠)𝑥𝑛.𝑑𝑠−̃𝑥(3.24) Thus, we have ‖‖𝑥𝑛‖‖−̃𝑥2=𝑡𝑛𝑓𝑥𝑛+−̃𝑥1−𝑡𝑛1𝜆𝑛𝜆𝑛0𝑇(𝑠)𝑥𝑛𝑥𝑑𝑠−̃𝑥,𝐽𝑛−̃𝑥=𝑡𝑛𝑓𝑥𝑛𝑥−𝑓(̃𝑥)+𝑓(̃𝑥)−̃𝑥,𝐽𝑛+−̃𝑥1−𝑡𝑛1𝜆𝑛𝜆𝑛0𝑇(𝑠)𝑥𝑛𝑥𝑑𝑠−̃𝑥,𝐽𝑛−̃𝑥≤𝑡𝑛‖‖𝑓𝑥𝑛‖‖‖‖𝐽𝑥−𝑓(̃𝑥)𝑛‖‖−̃𝑥+𝑡𝑛𝑥𝑓(̃𝑥)−̃𝑥,𝐽𝑛+−̃𝑥1−𝑡𝑛‖‖‖1𝜆𝑛𝜆𝑛0𝑇(𝑠)𝑥𝑛‖‖‖‖‖𝐽𝑥𝑑𝑠−̃𝑥𝑛‖‖≤−̃𝑥1−(1−𝛼)𝑡𝑛‖‖𝑥𝑛‖‖−̃𝑥2+𝑡𝑛𝑥𝑓(̃𝑥)−̃𝑥,𝐽𝑛.−̃𝑥(3.25) Therefore, ‖‖𝑥𝑛‖‖−̃𝑥2≤1𝑥1−𝛼𝑓(̃𝑥)−̃𝑥,𝐽𝑛.−̃𝑥(3.26)
We claim that the set {𝑥𝑛} is sequentially compact. Indeed, define the set 𝐶∗=𝑦∈𝐶∶𝜇𝑛‖‖𝑥𝑛‖‖−𝑦2=inf𝑥∈𝐶𝜇𝑛‖‖𝑥𝑛‖‖−𝑥2.(3.27) By Lemma 2.8, we found ̃𝑥∈𝐶∗. Using Lemma 2.7 we get that 𝜇𝑛𝑥𝑥−̃𝑥,𝐽𝑛−̃𝑥≤0,∀𝑥∈𝐶.(3.28) From (3.26), we get 𝜇𝑛‖‖𝑥𝑛‖‖−̃𝑥2≤1𝜇1−𝛼𝑛𝑥𝑓(̃𝑥)−̃𝑥,𝐽𝑛−̃𝑥≤0,(3.29) that is 𝜇𝑛‖‖𝑥𝑛‖‖−̃𝑥=0.(3.30) Hence, there exists a subsequence {𝑥𝑛𝑘} of {𝑥𝑛} converges strongly to ̃𝑥∈𝐹(𝒮) as 𝑘→∞.
Next we show that ̃𝑥 is a solution in 𝐹(𝒮) to the variational inequality (3.23). In fact, for any fixed 𝑥∈𝐹(𝒮), there exists a constant 𝑀>0 such that ‖𝑥𝑛−𝑥‖≤𝑀, then ‖‖𝑥𝑛‖‖−𝑥2=𝑡𝑛𝑓𝑥𝑛−𝑓(̃𝑥)+̃𝑥−𝑥𝑛𝑥,𝐽𝑛−𝑥+𝑡𝑛𝑥𝑓(̃𝑥)−̃𝑥,𝐽𝑛−𝑥+𝑡𝑛𝑥𝑛𝑥−𝑥,𝐽𝑛+−𝑥1−𝑡𝑛1𝜆𝑛𝜆𝑛0𝑇(𝑠)𝑥𝑛𝑥𝑑𝑠−𝑥,𝐽𝑛−𝑥≤(1+𝛼)𝑡𝑛𝑀‖‖𝑥𝑛‖‖−̃𝑥+𝑡𝑛𝑥𝑓(̃𝑥)−̃𝑥,𝐽𝑛+‖‖𝑥−𝑥𝑛‖‖−𝑥2.(3.31) Therefore, 𝑓(̃𝑥)−̃𝑥,𝐽𝑥−𝑥𝑛≤‖‖𝑥(1+𝛼)𝑀𝑛‖‖−̃𝑥.(3.32) Since the duality mapping 𝐽 is single valued and norm topology to weak* topology uniformly continuous on any bounded subset of a Banach space 𝑋 with a uniformly Gâteaux differentiable norm, we have 𝑓(̃𝑥)−̃𝑥,𝐽𝑥−𝑥𝑛𝑘⟶⟨𝑓(̃𝑥)−̃𝑥,𝐽(𝑥−̃𝑥)⟩.(3.33) Taking limit as 𝑗→∞ in two sides of (3.32), we get ⟨𝑓(̃𝑥)−̃𝑥,𝐽(𝑥−̃𝑥)⟩≤0,∀𝑥∈𝐹(𝒮).(3.34)
Finally we will show that the net {𝑥𝑡} convergence strong to ̃𝑥. This section is similar to that of Theorem 3.1.

4. Explicit Iterative Scheme

Theorem 4.1. Let 𝑋 be a uniformly convex Banach space that has a weakly continuous duality map 𝐽𝜑 with gauge 𝜑 and 𝐶 be a nonempty closed convex subset of 𝑋. Let 𝒮={𝑇(𝑠)∶0≤𝑠<∞} be a nonexpansive semigroup from 𝐶 into itself such that ⋂𝐹(𝒮)=𝑠>0𝐹(𝑇(𝑠))≠∅ and 𝑓∶𝐶→𝐶 a contraction mapping with the contractive coefficient 𝛼∈[0,1). Let {𝛼𝑛} and {𝛽𝑛} be the sequence in (0,1) which satisfies 𝛼𝑛+𝛽𝑛<1, lim𝑛→∞𝛼𝑛→0, lim𝑛→∞𝛽𝑛→0 and ∑∞𝑛=1𝛼𝑛=∞, and {𝑠𝑛} is a positive real divergent sequence such that lim𝑛→∞𝑠𝑛→∞. If the sequence {𝑥𝑛} defined by 𝑥0∈𝐶 and 𝑥𝑛+1=𝛼𝑛𝑓𝑥𝑛+𝛽𝑛𝑥𝑛+1−𝛼𝑛−𝛽𝑛1𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛𝑑𝑠,𝑛≥0.(4.1) Then {𝑥𝑛} converges strongly to ̃𝑥 as 𝑛→∞, where ̃𝑥 is the unique solution in 𝐹(𝒮) of the variational inequality ⟨(𝐼−𝑓)̃𝑥,𝐽(𝑥−̃𝑥)⟩≥0,𝑥∈𝐹(𝒮).(4.2)

Proof. Note that 𝐹(𝒮) is a nonempty closed convex set. We first show that {𝑥𝑛} is bounded. Let 𝑞∈𝐹(𝒮). Thus, we compute that ‖‖𝑥𝑛+1‖‖=‖‖‖𝛼−𝑞𝑛𝑓𝑥𝑛+𝛽𝑛𝑥𝑛+1−𝛼𝑛−𝛽𝑛1𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛‖‖‖𝑑𝑠−𝑞≤𝛼𝑛‖‖𝑓𝑥𝑛‖‖−𝑞+𝛽𝑛‖‖𝑥𝑛‖‖+−𝑞1−𝛼𝑛−𝛽𝑛‖‖‖1𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛‖‖‖𝑑𝑠−𝑞≤𝛼𝑛‖‖𝑓𝑥𝑛‖‖−𝑓(𝑞)+‖𝑓(𝑞)−𝑞‖+𝛽𝑛‖‖𝑥𝑛‖‖+−𝑞1−𝛼𝑛−𝛽𝑛1𝑠𝑛𝑠𝑛0‖‖𝑇(𝑠)𝑥𝑛‖‖−𝑞𝑑𝑠≤𝛼𝑛𝛼‖‖𝑥𝑛‖‖−𝑞+𝛼𝑛‖𝑓(𝑞)−𝑞‖+1−𝛼𝑛‖‖𝑥𝑛‖‖=−𝑞1−𝛼𝑛‖‖𝑥(1−𝛼)𝑛‖‖−𝑞+𝛼𝑛‖‖𝑥‖𝑓(𝑞)−𝑞‖≤max𝑛‖‖,1−𝑞⋮‖‖𝑥1−𝛼‖𝑓(𝑞)−𝑞‖≤max0‖‖,1−𝑞‖.1−𝛼‖𝑓(𝑞)−𝑞(4.3) Therefore, {𝑥𝑛} is bounded, {𝑓(𝑥𝑛)} and {𝑇(𝑠)𝑥𝑛} for every 0≤𝑠<∞ are also bounded.
Next we show ‖𝑥𝑛−𝑇(ℎ)𝑥𝑛‖→0 as 𝑛→∞. Notice that ‖‖𝑥𝑛+1−𝑇(ℎ)𝑥𝑛+1‖‖≤‖‖‖𝑥𝑛+1−1𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛‖‖‖+‖‖‖1𝑑𝑠𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛1𝑑𝑠−𝑇(ℎ)𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛‖‖‖+‖‖‖1𝑑𝑠𝑇(ℎ)𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛𝑑𝑠−𝑇(ℎ)𝑥𝑛+1‖‖‖‖‖‖𝑥≤2𝑛+1−1𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛‖‖‖+‖‖‖1𝑑𝑠𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛1𝑑𝑠−𝑇(ℎ)𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛‖‖‖𝑑𝑠≤2𝛼𝑛‖‖‖𝑓𝑥𝑛−1𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛‖‖‖𝑑𝑠+2𝛽𝑛‖‖‖𝑥𝑛−1𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛‖‖‖+‖‖‖1𝑑𝑠𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛1𝑑𝑠−𝑇(ℎ)𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛‖‖‖.𝑑𝑠(4.4)
Put 𝑧0=𝑃𝐹(𝒮)𝑥0 and 𝐷={𝑧∈𝐶∶‖𝑧−𝑧0‖≤‖𝑥0−𝑧0‖+1/(1−𝛼)‖𝑓(𝑧0)−𝑧0‖}. Then 𝐷 is a nonempty closed bounded convex subset of 𝐶 which is 𝑇(𝑠)-invariant for each 𝑠∈[0,∞) and contains {𝑥𝑛}. So without loss of generality, we may assume that 𝒮={𝑇(𝑠)∶0≤𝑠<∞} is a nonexpansive semigroup on 𝐷. By Lemma 2.4, we get lim𝑛→∞‖‖‖1𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛1𝑑𝑠−𝑇(ℎ)𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛‖‖‖𝑑𝑠=0,(4.5) for every ℎ∈[0,∞). On the other hand, since {𝑥𝑛}, {𝑓(𝑥𝑛)}, and {𝑇(𝑠)𝑥𝑛} are bounded, using the assumption that lim𝑛→∞𝛼𝑛→0, lim𝑛→∞𝛽𝑛→0, and (4.5) into (4.4), we get that ‖‖𝑥𝑛+1−𝑇(ℎ)𝑥𝑛+1‖‖⟶0as𝑛⟶∞,(4.6) and hence ‖‖𝑥𝑛−𝑇(ℎ)𝑥𝑛‖‖⟶0as𝑛⟶∞.(4.7)
We now show that 𝑓(̃𝑥)−̃𝑥,𝐽𝜑𝑥𝑛−̃𝑥≤0.(4.8) Let 𝑥𝑡=𝑡𝑓(𝑥𝑡)+(1−𝑡)(1/𝜆𝑡)∫𝜆𝑡0𝑇(𝑠)𝑥𝑡𝑑𝑠, where 𝑡 and 𝜆𝑡 satisfies the condition of Theorem 3.1. Then it follows from Theorem 3.1 that ̃𝑥=lim𝑡→0𝑥𝑡 and ̃𝑥 be the unique solution in 𝐹(𝒮) of the variational inequality (3.2). Clearly ̃𝑥 is a unique solution of (4.2). Take a subsequence {𝑥𝑛𝑘} of {𝑥𝑛} such that limsup𝑛→∞𝑓(̃𝑥)−̃𝑥,𝐽𝜑𝑥𝑛−̃𝑥=lim𝑘→∞𝑓(̃𝑥)−̃𝑥,𝐽𝜑𝑥𝑛𝑘−̃𝑥.(4.9) Since 𝑋 is uniformly convex and hence it is reflexive, we may further assume that 𝑥𝑛𝑘⇀𝑝. Moreover, we note that 𝑝∈𝐹(𝒮) by Lemma 2.3 and (4.7). Therefore, from (4.9) and (3.17), we have limsup𝑛→∞𝑓(̃𝑥)−̃𝑥,𝐽𝜑𝑥𝑛=𝑓−̃𝑥(̃𝑥)−̃𝑥,𝐽𝜑(𝑝−̃𝑥)≤0.(4.10) That is (4.8) holds.
Finally we will show that 𝑥𝑛→̃𝑥. For each 𝑛≥0, we have Φ‖‖𝑥𝑛+1‖‖‖‖‖𝛼−̃𝑥=Φ𝑛𝑓𝑥𝑛−̃𝑥+𝛽𝑛𝑥𝑛+−̃𝑥1−𝛼𝑛−𝛽𝑛1𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛‖‖‖‖‖𝛼𝑑𝑠−̃𝑥≤Φ𝑛𝑓𝑥𝑛−𝑓(̃𝑥)+𝛼𝑛(𝑓(̃𝑥)−̃𝑥)+𝛽𝑛𝑥𝑛+−̃𝑥1−𝛼𝑛−𝛽𝑛1𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛‖‖‖‖‖‖𝛼𝑑𝑠−̃𝑥≤Φ𝑛𝑓𝑥𝑛−𝑓(̃𝑥)+𝛽𝑛𝑥𝑛+−̃𝑥1−𝛼𝑛−𝛽𝑛1𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛‖‖‖𝑑𝑠−̃𝑥+𝛼𝑛𝑓(̃𝑥)−̃𝑥,𝐽𝜑𝑥𝑛+1𝛼−̃𝑥≤Φ𝑛𝛼‖‖𝑥𝑛‖‖−̃𝑥+𝛽𝑛‖‖𝑥𝑛‖‖+−̃𝑥1−𝛼𝑛−𝛽𝑛‖‖‖1𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛‖‖‖𝑑𝑠−̃𝑥+𝛼𝑛𝑓(̃𝑥)−̃𝑥,𝐽𝜑𝑥𝑛+1−̃𝑥≤Φ1−𝛼𝑛‖‖𝑥(1−𝛼)𝑛‖‖−̃𝑥+𝛼𝑛𝑓(̃𝑥)−̃𝑥,𝐽𝜑𝑥𝑛+1≤−̃𝑥1−𝛼𝑛Φ‖‖𝑥(1−𝛼)𝑛‖‖−̃𝑥+𝛼𝑛𝑓(̃𝑥)−̃𝑥,𝐽𝜑𝑥𝑛+1.−̃𝑥(4.11) An application of Lemma 2.6, we can obtain Φ(‖𝑥𝑛−̃𝑥‖)→0, hence ‖𝑥𝑛−̃𝑥‖→0. That is, {𝑥𝑛} converges strongly to a fixed point ̃𝑥 of 𝒮. This completes the proof.

Theorem 4.2. Let 𝑋 be a uniformly convex Banach space with a uniformly Gâteaux differentiable norm and 𝐶 be a nonempty closed convex subset of 𝑋. Let 𝒮={𝑇(𝑠)∶0≤𝑠<∞} be a nonexpansive semigroup from 𝐶 into itself such that ⋂𝐹(𝒮)=𝑠>0𝐹(𝑇(𝑠))≠∅ and 𝑓∶𝐶→𝐶 a contraction mapping with the contractive coefficient 𝛼∈[0,1). Let {𝛼𝑛} and {𝛽𝑛} be the sequence in (0,1) which satisfies 𝛼𝑛+𝛽𝑛<1, lim𝑛→∞𝛼𝑛→0, lim𝑛→∞𝛽𝑛→0, and ∑∞𝑛=1𝛼𝑛=∞, and {𝑠𝑛} is a positive real divergent sequence such that lim𝑛→∞𝑠𝑛→∞. If the sequence {𝑥𝑛} defined by 𝑥0∈𝐶 and 𝑥𝑛+1=𝛼𝑛𝑓𝑥𝑛+𝛽𝑛𝑥𝑛+1−𝛼𝑛−𝛽𝑛1𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛𝑑𝑠,𝑛≥0.(4.12) Then {𝑥𝑛} converges strongly to ̃𝑥 as 𝑛→∞, where ̃𝑥 is the unique solution in 𝐹(𝒮) of the variational inequality ⟨(𝐼−𝑓)̃𝑥,𝐽(𝑥−̃𝑥)⟩≥0,𝑥∈𝐹(𝒮).(4.13)

Proof. We also show only those points in this proof which are different from that already presented in the proof of Theorem 4.1. We now show that 𝑓𝑥(̃𝑥)−̃𝑥,𝐽𝑛−̃𝑥≤0.(4.14) Let 𝑥𝑡=𝑡𝑓(𝑥𝑡)+(1−𝑡)(1/𝜆𝑡)∫𝜆𝑡0𝑇(𝑠)𝑥𝑡𝑑𝑠, where 𝑡 and 𝜆𝑡 satisfies the condition of Theorem 3.2. Then it follows from Theorem 3.2 that ̃𝑥=lim𝑡→0𝑥𝑡 and ̃𝑥 is the unique solution in 𝐹(𝒮) of the variational inequality (3.23). Clearly ̃𝑥 is a unique solution of (4.13). Take a subsequence {𝑥𝑛𝑘} of {𝑥𝑛} such that limsup𝑛→∞𝑓𝑥(̃𝑥)−̃𝑥,𝐽𝑛−̃𝑥=lim𝑘→∞𝑓𝑥(̃𝑥)−̃𝑥,𝐽𝑛𝑘−̃𝑥.(4.15) Since 𝑋 is uniformly convex and hence it is reflexive, we may further assume that 𝑥𝑛𝑘⇀𝑝. Moreover, we note that 𝑝∈𝐹(𝒮) by Lemma 2.3 and (4.7). Therefore, from (4.15) and (3.23), we have limsup𝑛→∞𝑓𝑥(̃𝑥)−̃𝑥,𝐽𝑛−̃𝑥=⟨𝑓(̃𝑥)−̃𝑥,𝐽(𝑝−̃x)⟩≤0.(4.16) That is, (4.14) holds.
Finally we will show that 𝑥𝑛→̃𝑥. For each 𝑛≥0, by Lemma 2.2, we have ‖‖𝑥𝑛+1‖‖−̃𝑥2=‖‖‖𝛼𝑛𝑓𝑥𝑛−̃𝑥+𝛽𝑛𝑥𝑛+−̃𝑥1−𝛼𝑛−𝛽𝑛1𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛‖‖‖𝑑𝑠−̃𝑥2≤‖‖‖1−𝛼𝑛−𝛽𝑛1𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛𝑑𝑠−̃𝑥+𝛽𝑛𝑥𝑛‖‖‖−̃𝑥2+2𝛼𝑛𝑓𝑥𝑛𝑥−̃𝑥,𝐽𝑛+1≤−̃𝑥1−𝛼𝑛−𝛽𝑛‖‖‖1𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛‖‖‖𝑑𝑠−̃𝑥+𝛽𝑛‖‖𝑥𝑛‖‖−̃𝑥2+2𝛼𝑛𝑓𝑥𝑛𝑥−̃𝑥,𝐽𝑛+1≤−̃𝑥1−𝛼𝑛‖‖𝑥𝑛‖‖−̃𝑥2+2𝛼𝑛𝑓𝑥𝑛𝑥−𝑓(̃𝑥),𝐽𝑛+1−̃𝑥+2𝛼𝑛𝑥𝑓(̃𝑥)−̃𝑥,𝐽𝑛+1≤−̃𝑥1−𝛼𝑛2‖‖𝑥𝑛‖‖−̃𝑥2+2𝛼𝑛‖‖𝑓𝑥𝑛‖‖‖‖𝐽𝑥−𝑓(̃𝑥)𝑛+1‖‖−̃𝑥+2𝛼𝑛𝑓𝑥(̃𝑥)−̃𝑥,𝐽𝑛+1≤−̃𝑥1−𝛼𝑛2‖‖𝑥𝑛‖‖−̃𝑥2+2𝛼𝑛𝛼‖‖𝑥𝑛‖‖‖‖𝑥−̃𝑥𝑛+1‖‖−̃𝑥+2𝛼𝑛𝑥𝑓(̃𝑥)−̃𝑥,𝐽𝑛+1≤−̃𝑥1−𝛼𝑛2‖‖𝑥𝑛‖‖−̃𝑥2+𝛼𝑛𝛼‖‖𝑥𝑛‖‖−̃𝑥2+‖‖𝑥𝑛+1‖‖−̃𝑥2+2𝛼𝑛𝑥𝑓(̃𝑥)−̃𝑥,𝐽𝑛+1=−̃𝑥1−𝛼𝑛2+𝛼𝑛𝛼‖‖𝑥𝑛‖‖−̃𝑥2+𝛼𝑛𝛼‖‖𝑥𝑛+1‖‖−̃𝑥2+2𝛼𝑛𝑥𝑓(̃𝑥)−̃𝑥,𝐽𝑛+1,−̃𝑥(4.17) which implies that ‖‖𝑥𝑛+1‖‖−̃𝑥2≤1−2𝛼𝑛+𝛼2𝑛+𝛼𝑛𝛼1−𝛼𝑛𝛼‖‖𝑥𝑛‖‖−̃𝑥2+2𝛼𝑛1−𝛼𝑛𝛼𝑥𝑓(̃𝑥)−̃𝑥,𝐽𝑛+1=−̃𝑥1−2(1−𝛼)𝛼𝑛1−𝛼𝑛𝛼‖‖𝑥𝑛‖‖−̃𝑥2+𝛼2𝑛1−𝛼𝑛𝛼‖‖𝑥𝑛‖‖−̃𝑥2+2𝛼𝑛1−𝛼𝑛𝛼𝑥𝑓(̃𝑥)−̃𝑥,𝐽𝑛+1≤−̃𝑥1−2(1−𝛼)𝛼𝑛1−𝛼𝑛𝛼‖‖𝑥𝑛‖‖−̃𝑥2+2(1−𝛼)𝛼𝑛1−𝛼𝑛𝛼𝛼𝑛𝑀+12(1−𝛼)𝑥1−𝛼𝑓(̃𝑥)−̃𝑥,𝐽𝑛+1=−̃𝑥1−𝛿𝑛‖‖𝑥𝑛‖‖−̃𝑥2+𝛿𝑛𝛾𝑛,(4.18)where 𝑀=sup{‖𝑥𝑛−̃𝑥‖2∶𝑛∈ℕ}, 𝛿𝑛∶=2(1−𝛼)𝛼𝑛/(1−𝛼𝑛𝛼), and 𝛾𝑛∶=(𝛼𝑛𝑀/2(1−𝛼))+(1/(1−𝛼))⟨𝑓(̃𝑥)−̃𝑥,𝐽(𝑥𝑛+1−̃𝑥)⟩. It is easily to see that 𝛿𝑛→0, ∑∞𝑛=1𝛿𝑛=∞ and limsup𝑛→∞𝛾𝑛≤0 by (4.14). Finally by using Lemma 2.6, we can obtain {𝑥𝑛} converges strongly to a fixed point ̃𝑥∈𝐹(𝒮). This completes the proof.

5. Applications

Theorem 5.1. Let 𝑋 be a uniformly convex Banach space that has a weakly continuous duality map 𝐽𝜑 with gauge 𝜑 and 𝐶 be a nonempty closed convex subset of 𝑋. Let 𝒮={𝑇(𝑠)∶0≤𝑠<∞} be a nonexpansive semigroup from 𝐶 into itself such that ⋂𝐹(𝒮)=𝑠>0𝐹(𝑇(𝑠))≠∅ and 𝑓∶𝐶→𝐶 a contraction mapping with the contractive coefficient 𝛼∈[0,1). Let {𝛼𝑛} be the sequence in (0,1) which satisfies lim𝑛→∞𝛼𝑛→0 and ∑∞𝑛=1𝛼𝑛=∞, and {𝑠𝑛} is a positive real divergent sequence such that lim𝑛→∞𝑠𝑛→∞. If the sequence {𝑥𝑛} defined by 𝑥0∈𝐶 and 𝑥𝑛+1=𝛼𝑛𝑓𝑥𝑛+1−𝛼𝑛1𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛𝑑𝑠,𝑛≥0.(5.1) Then {𝑥𝑛} converges strongly to ̃𝑥 as 𝑛→∞, where ̃𝑥 is the unique solution in 𝐹(𝒮) of the variational inequality ⟨(𝐼−𝑓)̃𝑥,𝐽(𝑥−̃𝑥)⟩≥0,𝑥∈𝐹(𝒮).(5.2)

Proof. Taking 𝛽𝑛=0 in the in Theorem 4.1, we get the desired conclusion easily.

Theorem 5.2. Let 𝑋 be a uniformly convex Banach space with a uniformly Gâteaux differentiable norm and 𝐶 be a nonempty closed convex subset of 𝑋. Let 𝒮={𝑇(𝑠)∶0≤𝑠<∞} be a nonexpansive semigroup from 𝐶 into itself such that ⋂𝐹(𝒮)=𝑠>0𝐹(𝑇(𝑠))≠∅ and 𝑓∶𝐶→𝐶 a contraction mapping with the contractive coefficient 𝛼∈[0,1). Let {𝛼𝑛} be the sequence in (0,1) which satisfies lim𝑛→∞𝛼𝑛→0 and ∑∞𝑛=1𝛼𝑛=∞, and {𝑠𝑛} is a positive real divergent sequence such that lim𝑛→∞𝑠𝑛→∞. If the sequence {𝑥𝑛} defined by 𝑥0∈𝐶 and 𝑥𝑛+1=𝛼𝑛𝑓𝑥𝑛+1−𝛼𝑛1𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛𝑑𝑠,𝑛≥0.(5.3) Then {𝑥𝑛} converges strongly to ̃𝑥 as 𝑛→∞, where ̃𝑥 is the unique solution in 𝐹(𝒮) of the variational inequality ⟨(𝐼−𝑓)̃𝑥,𝐽(𝑥−̃𝑥)⟩≥0,𝑥∈𝐹(𝒮).(5.4)

Proof. Taking 𝛽𝑛=0 in the in Theorem 4.2, we get the desired conclusion easily.

When 𝑋 is a Hilbert space, we can get the following corollary easily.

Corollary 5.3 (Reich [2]). Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻. Let 𝒮={𝑇(𝑠)∶0≤𝑠<∞} be a strongly continuous semigroup of nonexpansive mapping on 𝐶 such that 𝐹(𝒮) is nonempty. Let {𝛼𝑛} and {𝛽𝑛} be sequences of real numbers in (0,1) which satisfies 𝛼𝑛+𝛽𝑛<1, lim𝑛→∞𝛼𝑛→0, lim𝑛→∞𝛽𝑛→0, and ∑∞𝑛=1𝛼𝑛=∞. Let 𝑓 be a contraction of 𝐶 into itself with a coefficient 𝛼∈[0,1) and {𝑠𝑛} be a positive real divergent sequence such that lim𝑛→∞𝑠𝑛→∞. Then the sequence {𝑥𝑛} defined by 𝑥0∈𝐶 and 𝑥𝑛+1=𝛼𝑛𝑓𝑥𝑛+𝛽𝑛𝑥𝑛+1−𝛼𝑛−𝛽𝑛1𝑠𝑛𝑠𝑛0𝑇(𝑠)𝑥𝑛𝑑𝑠,𝑛≥0.(5.5)
Then {𝑥𝑛} converges strongly to ̃𝑥, where ̃𝑥 is the unique solution in 𝐹(𝒮) of the variational inequality ⟨(𝐼−𝑓)̃𝑥,𝑥−̃𝑥⟩≥0,𝑥∈𝐹(𝒮),(5.6) or equivalent ̃𝑥=𝑃𝐹(𝒮)(𝑓)(̃𝑥), where 𝑃 is a metric projection mapping from 𝐻 into 𝐹(𝒮).

Funding

This paper is supported by the National Science Foundation of China under Grants (10771050 and 11101305).