# Strong Convergence Theorems for Nonexpansive Semigroups and Variational Inequalities in Banach Spaces

Authors: Haiqing Wang; Yongfu Su; Hong Zhang
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

LetX 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)∫0snT(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.


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## Body

## 1. Introduction

LetX 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 by(1.1)Ttx=tu+(1-t)Tx,x∈C,whereu∈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).Letf 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,(1.2)xt=tf(xt)+(1-t)Txt.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 x̃ of T in C which is the unique solution to the following variational inequality:(1.3)〈(I-f)x̃,x-x̃〉≥0,∀x∈F(T).Moreover, Xu [6] also studied the strong convergence of the following iterative sequence generated by(1.4)xn+1=βnf(xn)+(1-βn)Txn,n≥0,wherex0∈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 allx∈C, s↦T(s)x is continuous.We denote byF(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 by(1.5)xt=tf(xt)+(1-t)1λt∫0λtT(s)xtds,wheret∈(0,1) and {λt} is a positive real divergent net, and the iterative scheme {xn} defined by(1.6)xn+1=αnf(xn)+βnxn+(1-αn-βn)1sn∫0snT(s)xnds,n≥0,wherex0∈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 x̃ 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 Question1. 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,(1.7)xn=αnf(xn)+(1-αn)1sn∫0snT(s)xnds,n≥0,xn+1=αnf(xn)+(1-αn)1sn∫0snT(s)xnds,n≥0.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

LetX 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 by(2.1)J(x)={x*∈X*:〈x,x*〉=‖x‖2=‖x*‖2},x∈X.In the sequel, we will denote the single valued duality mapping byj. When {xn} is a sequence in X, then xn→x(xn⇀x) will denote strong (weak) convergence of the sequence {xn} to x.LetS(X)={x∈X:∥x∥=1}. Then the norm of X is said to be Gâteaux differentiable if(2.2)lim⁡t→0‖x+ty‖-‖x‖t
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 spaceX is said to be strictly convex if(2.3)‖x‖=‖y‖=1,x≠yimplies‖x+y‖2<1.A Banach spaceX is said to be uniformly convex if δX(ɛ)>0 for all ɛ>0, where δX(ɛ) is modulus of convexity of E defined by(2.4)δE(ɛ)=inf⁡{1-‖x+y‖2:‖x‖≤1,‖y‖≤1,‖x-y‖≥ɛ},ɛ∈[0,2].A uniformly convex Banach spaceE is reflexive and strictly convex [9, Theorem 4.1.6, Theorem 4.1.2].Lemma 2.1 (Goebel and Reich [10], Proposition 5.3).
LetC 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 spaceX there holds the inequality
(2.5)‖x+y‖2≤‖x‖2+2〈y,J(x+y)〉,x,y,∈X.Lemma 2.3 (Browder [12]).
LetE 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]).
LetC 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<∞,
(2.6)lim⁡t→∞sup⁡x∈C‖1t∫0tT(s)xds-T(h)(1t∫0tT(s)xds)‖=0.Recall that a gauge is a continuous strictly increasing functionφ:[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 by(2.7)Jφ(x)={x*∈X*:〈x,x*〉=‖x‖φ(‖x‖),‖x*‖=φ(‖x‖)},x∈X.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<∞. Set(2.8)Φ(t)=∫0tφ(τ)dτ,t≥0.Then(2.9)Jφ(x)=∂Φ(‖x‖),x∈X,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 thatX has a weakly continuous duality map  Jφ with gauge  φ, for all  x,y∈X, there holds the inequality
(2.10)Φ(‖x+y‖)≤Φ(‖x‖)+〈y,Jφ(x+y)〉.Lemma 2.6 (Xu [6]).
Assume{αn} is a sequence of nonnegative real numbers such that
(2.11)αn+1≤(1-γn)αn+δn,n≥0,where {γn} is a sequence in (0,1) and {δn} is a sequence in R such that(i)
∑n=1∞γn=∞;(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 if(2.12)inf⁡{an;n∈N}≤μ(a)≤sup⁡{an;n∈N},
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 if(2.13)μn(an)=μn(an+1),
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, then(2.14)liminf⁡n→∞an≤μn(an)≤limsup⁡n→∞an,
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 have(2.15)μn(an)=μ(a)=c(resp.,μn(an)=μn(bn)).Subsequently, the following result was showed in [14, Lemma 1] and [9, Lemma 4.5.4].Lemma 2.7 (see [14, Lemma 1]).
LetC 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
(2.16)μn‖xn-z0‖2=min⁡x∈Cμn‖xn-x‖2,
if and only if
(2.17)μn〈x-z0,J(xn-z0)〉≤0,∀x∈C.Lemma 2.8 (Song and Xu [4, Proposition 3.1]).
LetX 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
(2.18)C*={y∈C:μn‖xn-y‖2=inf⁡x∈Cμn‖xn-x‖2}.
If F(T)≠∅, then C*∩F(T)≠∅.


## 3. Implicit Iteration Scheme

Theorem 3.1.
LetX 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:
(3.1)xt=tf(xt)+(1-t)1λt∫0λtT(s)xtds.
Then {xt} converges strongly to x̃ as t→0+, where x̃ is the unique solution in F(𝒮) of the variational inequality
(3.2)〈(I-f)x̃,J(x-x̃)〉≥0,x∈F(S).Proof.
Note thatF(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
(3.3)‖xt-p‖≤t‖f(xt)-p‖+(1-t)‖1λt∫0λtT(s)xtds-p‖≤t(‖f(xt)-f(p)‖+‖f(p)-p‖)+(1-t)1λt∫0λt‖T(s)xt-p‖ds≤t(α‖xt-p‖+‖f(p)-p‖)+(1-t)‖xt-p‖=‖xt-p‖-t(1-α)‖xt-p‖+t‖f(p)-p‖.
It follows that
(3.4)‖xt-p‖≤11-α‖f(p)-p‖.
Thus {xt} is bounded, so are {f(xt)} and {T(s)xt} for every 0≤s<∞. Furthermore, we note that
(3.5)‖xt-T(s)xt‖≤‖xt-1λt∫0λtT(s)xtds‖+‖1λt∫0λtT(s)xtds-T(s)(1λt∫0λtT(s)xtds)‖+‖T(s)(1λt∫0λtT(s)xtds)-T(s)xt‖≤2‖xt-1λt∫0λtT(s)xtds‖+‖1λt∫0λtT(s)xtds-T(s)(1λt∫0λtT(s)xtds)‖,
for every 0≤s<∞. On the one hand, we observe that
(3.6)‖xt-1λt∫0λtT(s)xtds‖=t1-t‖xt-f(xt)‖,
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
(3.7)lim⁡λt→∞‖1λt∫0λtT(s)xtds-T(s)(1λt∫0λtT(s)xtds)‖≤lim⁡λt→∞sup⁡x∈D‖1λt∫0λtT(s)xtds-T(s)(1λt∫0λtT(s)xtds)‖=0.
Hence, by (3.5)–(3.7), we obtain
(3.8)‖xt-T(s)xt‖⟶0ast⟶0,
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 x̃, where x̃∈F(𝒮). Since {xn} is a bounded sequence, there is a subsequence {xnj} of {xn} which converges weakly to x̃∈C. By Lemma 2.3, we have x̃∈F(𝒮). For each n≥1, we have
(3.9)xn-x̃=tn(f(xn)-x̃)+(1-tn)(1λn∫0λnT(s)xnds-x̃).
Thus, by Lemma 2.5, we obtain
(3.10)Φ(‖xn-x̃‖)=Φ(‖tn(f(xn)-x̃)+(1-tn)(1λn∫0λnT(s)xnds-x̃)‖)≤Φ(‖(1-tn)(1λn∫0λnT(s)xnds-x̃)‖)+tn〈f(xn)-x̃,Jφ(xn-x̃)〉≤(1-tn)Φ(‖1λn∫0λnT(s)xnds-x̃‖)+tn〈f(xn)-x̃,Jφ(xn-x̃)〉≤(1-tn)Φ(‖xn-x̃‖)+tn〈f(xn)-x̃,Jφ(xn-x̃)〉.
This implies that
(3.11)Φ(‖xn-x̃‖)≤〈f(xn)-x̃,Jφ(xn-x̃)〉.
In particular, we have
(3.12)Φ(‖xnj-x̃‖)≤〈f(xnj)-x̃,Jφ(xnj-x̃)〉.
Now observing that {xnj}⇀x̃ implies Jφ(xnj-x̃)⇀0. And since f(xnj) is bounded, it follows from (3.12) that
(3.13)Φ(‖xnj-x̃‖)⟶0asj⟶∞.
Hence xnj→x̃.
Next, we show thatx̃∈F(𝒮) solves the variational inequality (3.2). Indeed, for q∈F(𝒮), it is easy to see that
(3.14)〈xt-1λt∫0λtT(s)xtds,Jφ(xt-q)〉=Φ(‖xt-q‖)+〈q-1λt∫0λtT(s)xtds,Jφ(xt-q)〉≥Φ(‖xt-q‖)-‖q-1λt∫0λtT(s)xtds‖‖Jφ(xt-q)‖≥Φ(‖xt-q‖)-Φ(‖xt-q‖)=0.
However, we note that
(3.15)xt-1λt∫0λtT(s)xtds=t1-t(f(xt)-xt).
Thus, we get that for t∈(0,1) and q∈F(𝒮)(3.16)〈xt-f(xt),Jφ(xt-q)〉≤0.
Taking the limit through t:=tnj→0, we obtain
(3.17)〈(I-f)x̃,Jφ(x̃-q)〉≤0,∀q∈F(S).
This implies that
(3.18)〈(I-f)x̃,J(x̃-q)〉≤0,∀q∈F(S),
since Jφ(x)=(φ(∥x∥)/∥x∥)J(x) for x≠0.
Finally, we show that the net{xt} convergence strong to x̃. 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
(3.19)〈(I-f)x̃,J(x̃-x¯)〉≤0.
Interchange x̃ and x¯ to obtain
(3.20)〈(I-f)x¯,J(x¯-x̃)〉≤0.
Adding (3.19) and (3.20) yields
(3.21)(1-α)‖x̃-x¯‖2≤〈x¯-x̃-(f(x̃)-f(x¯)),J(x¯-x̃)〉≤0.
We must have x̃=x¯ and the uniqueness is proved. In a summary, we have shown that each cluster point of {xt} as t→0 equals x̃. Therefore xt→x̃ as t→0.Theorem 3.2.
LetX 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:
(3.22)xt=tf(xt)+(1-t)1λt∫0λtT(s)xtds.
Then {xt} converges strongly to x̃ as t→0+, where x̃ is the unique solution in F(𝒮) of the variational inequality
(3.23)〈(I-f)x̃,J(x-x̃)〉≥0,x∈F(S).Proof.
We include only those points in this proof which are different from those already presented in the proof of Theorem3.1. As in the proof of Theorem 3.1, we obtain that there is a subsequence {xnj} of {xn} which converges weakly to x̃∈F(𝒮). For each n≥1, we have
(3.24)xn-x̃=tn(f(xn)-x̃)+(1-tn)(1λn∫0λnT(s)xnds-x̃).
Thus, we have
(3.25)‖xn-x̃‖2=〈tn(f(xn)-x̃)+(1-tn)(1λn∫0λnT(s)xnds-x̃),J(xn-x̃)〉=tn〈f(xn)-f(x̃)+f(x̃)-x̃,J(xn-x̃)〉+(1-tn)〈(1λn∫0λnT(s)xnds-x̃),J(xn-x̃)〉≤tn‖f(xn)-f(x̃)‖‖J(xn-x̃)‖+tn〈f(x̃)-x̃,J(xn-x̃)〉+(1-tn)‖1λn∫0λnT(s)xnds-x̃‖‖J(xn-x̃)‖≤(1-(1-α)tn)‖xn-x̃‖2+tn〈f(x̃)-x̃,J(xn-x̃)〉.
Therefore,
(3.26)‖xn-x̃‖2≤11-α〈f(x̃)-x̃,J(xn-x̃)〉.
We claim that the set{xn} is sequentially compact. Indeed, define the set
(3.27)C*={y∈C:μn‖xn-y‖2=inf⁡x∈Cμn‖xn-x‖2}.
By Lemma 2.8, we found x̃∈C*. Using Lemma 2.7 we get that
(3.28)μn〈x-x̃,J(xn-x̃)〉≤0,∀x∈C.
From (3.26), we get
(3.29)μn‖xn-x̃‖2≤11-αμn〈f(x̃)-x̃,J(xn-x̃)〉≤0,
that is
(3.30)μn‖xn-x̃‖=0.
Hence, there exists a subsequence {xnk} of {xn} converges strongly to x̃∈F(𝒮) as k→∞.
Next we show thatx̃ 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
(3.31)‖xn-x‖2=tn〈f(xn)-f(x̃)+x̃-xn,J(xn-x)〉+tn〈f(x̃)-x̃,J(xn-x)〉+tn〈xn-x,J(xn-x)〉+(1-tn)〈1λn∫0λnT(s)xnds-x,J(xn-x)〉≤(1+α)tnM‖xn-x̃‖+tn〈f(x̃)-x̃,J(xn-x)〉+‖xn-x‖2.
Therefore,
(3.32)〈f(x̃)-x̃,J(x-xn)〉≤(1+α)M‖xn-x̃‖.
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
(3.33)〈f(x̃)-x̃,J(x-xnk)〉⟶〈f(x̃)-x̃,J(x-x̃)〉.
Taking limit as j→∞ in two sides of (3.32), we get
(3.34)〈f(x̃)-x̃,J(x-x̃)〉≤0,∀x∈F(S).
Finally we will show that the net{xt} convergence strong to x̃. This section is similar to that of Theorem 3.1.


## 4. Explicit Iterative Scheme

Theorem 4.1.
LetX 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=1∞αn=∞, and  {sn} is a positive real divergent sequence such that  limn→∞sn→∞. If the sequence  {xn} defined by  x0∈C and
(4.1)xn+1=αnf(xn)+βnxn+(1-αn-βn)1sn∫0snT(s)xnds,n≥0.
Then {xn} converges strongly to x̃ as n→∞, where x̃ is the unique solution in F(𝒮) of the variational inequality
(4.2)〈(I-f)x̃,J(x-x̃)〉≥0,x∈F(S).Proof.
Note thatF(𝒮) is a nonempty closed convex set. We first show that {xn} is bounded. Let q∈F(𝒮). Thus, we compute that
(4.3)‖xn+1-q‖=‖αnf(xn)+βnxn+(1-αn-βn)1sn∫0snT(s)xnds-q‖≤αn‖f(xn)-q‖+βn‖xn-q‖+(1-αn-βn)‖1sn∫0snT(s)xnds-q‖≤αn(‖f(xn)-f(q)‖+‖f(q)-q‖)+βn‖xn-q‖+(1-αn-βn)1sn∫0sn‖T(s)xn-q‖ds≤αnα‖xn-q‖+αn‖f(q)-q‖+(1-αn)‖xn-q‖=(1-αn(1-α))‖xn-q‖+αn‖f(q)-q‖≤max⁡{‖xn-q‖,11-α‖f(q)-q‖}⋮≤max⁡{‖x0-q‖,11-α‖f(q)-q‖}.
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
(4.4)‖xn+1-T(h)xn+1‖≤‖xn+1-1sn∫0snT(s)xnds‖+‖1sn∫0snT(s)xnds-T(h)(1sn∫0snT(s)xnds)‖+‖T(h)(1sn∫0snT(s)xnds)-T(h)xn+1‖≤2‖xn+1-1sn∫0snT(s)xnds‖+‖1sn∫0snT(s)xnds-T(h)(1sn∫0snT(s)xnds)‖≤2αn‖f(xn)-1sn∫0snT(s)xnds‖+2βn‖xn-1sn∫0snT(s)xnds‖+‖1sn∫0snT(s)xnds-T(h)(1sn∫0snT(s)xnds)‖.
Putz0=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
(4.5)lim⁡n→∞‖1sn∫0snT(s)xnds-T(h)(1sn∫0snT(s)xnds)‖=0,
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
(4.6)‖xn+1-T(h)xn+1‖⟶0asn⟶∞,
and hence
(4.7)‖xn-T(h)xn‖⟶0asn⟶∞.
We now show that(4.8)〈f(x̃)-x̃,Jφ(xn-x̃)〉≤0.
Let xt=tf(xt)+(1-t)(1/λt)∫0λtT(s)xtds, where t and λt satisfies the condition of Theorem 3.1. Then it follows from Theorem 3.1 that x̃=limt→0xt and x̃ be the unique solution in F(𝒮) of the variational inequality (3.2). Clearly x̃ is a unique solution of (4.2). Take a subsequence {xnk} of {xn} such that
(4.9)limsup⁡n→∞〈f(x̃)-x̃,Jφ(xn-x̃)〉=lim⁡k→∞〈f(x̃)-x̃,Jφ(xnk-x̃)〉.
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
(4.10)limsup⁡n→∞〈f(x̃)-x̃,Jφ(xn-x̃)〉=〈f(x̃)-x̃,Jφ(p-x̃)〉≤0.
That is (4.8) holds.
Finally we will show thatxn→x̃. For each n≥0, we have
(4.11)Φ(‖xn+1-x̃‖)=Φ(‖αn(f(xn)-x̃)+βn(xn-x̃)+(1-αn-βn)(1sn∫0snT(s)xnds-x̃)‖)≤Φ((1sn∫0snT(s)xnds-x̃)‖αn(f(xn)-f(x̃))+αn(f(x̃)-x̃)+βn(xn-x̃)+(1-αn-βn)(1sn∫0snT(s)xnds-x̃)‖)≤Φ(‖αn(f(xn)-f(x̃))+βn(xn-x̃)+(1-αn-βn)(1sn∫0snT(s)xnds-x̃)‖)+αn〈f(x̃)-x̃,Jφ(xn+1-x̃)〉≤Φ(αnα‖xn-x̃‖+βn‖xn-x̃‖+(1-αn-βn)‖1sn∫0snT(s)xnds-x̃‖)+αn〈f(x̃)-x̃,Jφ(xn+1-x̃)〉≤Φ((1-αn(1-α))‖xn-x̃‖)+αn〈f(x̃)-x̃,Jφ(xn+1-x̃)〉≤(1-αn(1-α))Φ(‖xn-x̃‖)+αn〈f(x̃)-x̃,Jφ(xn+1-x̃)〉.
An application of Lemma 2.6, we can obtain Φ(∥xn-x̃∥)→0, hence ∥xn-x̃∥→0. That is, {xn} converges strongly to a fixed point x̃ of 𝒮. This completes the proof.Theorem 4.2.
LetX 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=1∞αn=∞, and  {sn} is a positive real divergent sequence such that  limn→∞sn→∞. If the sequence  {xn} defined by  x0∈C and
(4.12)xn+1=αnf(xn)+βnxn+(1-αn-βn)1sn∫0snT(s)xnds,n≥0.
Then {xn} converges strongly to x̃ as n→∞, where x̃ is the unique solution in F(𝒮) of the variational inequality
(4.13)〈(I-f)x̃,J(x-x̃)〉≥0,x∈F(S).Proof.
We also show only those points in this proof which are different from that already presented in the proof of Theorem4.1. We now show that
(4.14)〈f(x̃)-x̃,J(xn-x̃)〉≤0.
Let xt=tf(xt)+(1-t)(1/λt)∫0λtT(s)xtds, where t and λt satisfies the condition of Theorem 3.2. Then it follows from Theorem 3.2 that x̃=limt→0xt and x̃ is the unique solution in F(𝒮) of the variational inequality (3.23). Clearly x̃ is a unique solution of (4.13). Take a subsequence {xnk} of {xn} such that
(4.15)limsup⁡n→∞〈f(x̃)-x̃,J(xn-x̃)〉=lim⁡k→∞〈f(x̃)-x̃,J(xnk-x̃)〉.
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
(4.16)limsup⁡n→∞〈f(x̃)-x̃,J(xn-x̃)〉=〈f(x̃)-x̃,J(p-x̃)〉≤0.
That is, (4.14) holds.
Finally we will show thatxn→x̃. For each n≥0, by Lemma 2.2, we have
(4.17)‖xn+1-x̃‖2=‖αn(f(xn)-x̃)+βn(xn-x̃)+(1-αn-βn)(1sn∫0snT(s)xnds-x̃)‖2≤‖(1-αn-βn)(1sn∫0snT(s)xnds-x̃)+βn(xn-x̃)‖2+2αn〈f(xn)-x̃,J(xn+1-x̃)〉≤((1-αn-βn)‖1sn∫0snT(s)xnds-x̃‖+βn‖xn-x̃‖)2+2αn〈f(xn)-x̃,J(xn+1-x̃)〉≤((1-αn)‖xn-x̃‖)2+2αn〈f(xn)-f(x̃),J(xn+1-x̃)〉+2αn〈f(x̃)-x̃,J(xn+1-x̃)〉≤(1-αn)2‖xn-x̃‖2+2αn‖f(xn)-f(x̃)‖‖J(xn+1-x̃)‖+2αn〈f(x̃)-x̃,J(xn+1-x̃)〉≤(1-αn)2‖xn-x̃‖2+2αnα‖xn-x̃‖‖xn+1-x̃‖+2αn〈f(x̃)-x̃,J(xn+1-x̃)〉≤(1-αn)2‖xn-x̃‖2+αnα(‖xn-x̃‖2+‖xn+1-x̃‖2)+2αn〈f(x̃)-x̃,J(xn+1-x̃)〉=((1-αn)2+αnα)‖xn-x̃‖2+αnα‖xn+1-x̃‖2+2αn〈f(x̃)-x̃,J(xn+1-x̃)〉,
which implies that
(4.18)‖xn+1-x̃‖2≤1-2αn+αn2+αnα1-αnα‖xn-x̃‖2+2αn1-αnα〈f(x̃)-x̃,J(xn+1-x̃)〉=[1-2(1-α)αn1-αnα]‖xn-x̃‖2+αn21-αnα‖xn-x̃‖2+2αn1-αnα〈f(x̃)-x̃,J(xn+1-x̃)〉≤[1-2(1-α)αn1-αnα]‖xn-x̃‖2+2(1-α)αn1-αnα{αnM2(1-α)+11-α〈f(x̃)-x̃,J(xn+1-x̃)〉}=(1-δn)‖xn-x̃‖2+δnγn,where M=sup{∥xn-x̃∥2:n∈ℕ}, δn:=2(1-α)αn/(1-αnα), and γn:=(αnM/2(1-α))+(1/(1-α))〈f(x̃)-x̃,J(xn+1-x̃)〉. It is easily to see that δn→0, ∑n=1∞δn=∞ and limsupn→∞γn≤0 by (4.14). Finally by using Lemma 2.6, we can obtain {xn} converges strongly to a fixed point x̃∈F(𝒮). This completes the proof.


## 5. Applications

Theorem 5.1.
LetX 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=1∞αn=∞, and  {sn} is a positive  real divergent sequence such that  limn→∞sn→∞. If the sequence  {xn} defined by  x0∈C and
(5.1)xn+1=αnf(xn)+(1-αn)1sn∫0snT(s)xnds,n≥0.
Then {xn} converges strongly to x̃ as n→∞, where x̃ is the unique solution in F(𝒮) of the variational inequality
(5.2)〈(I-f)x̃,J(x-x̃)〉≥0,x∈F(S).Proof.
Takingβn=0 in the in Theorem 4.1, we get the desired conclusion easily.Theorem 5.2.
LetX 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=1∞αn=∞, and  {sn} is a positive real divergent sequence such that  limn→∞sn→∞. If the sequence  {xn} defined by  x0∈C and
(5.3)xn+1=αnf(xn)+(1-αn)1sn∫0snT(s)xnds,n≥0.
Then {xn} converges strongly to x̃ as n→∞, where x̃ is the unique solution in F(𝒮) of the variational inequality
(5.4)〈(I-f)x̃,J(x-x̃)〉≥0,x∈F(S).Proof.
Takingβn=0 in the in Theorem 4.2, we get the desired conclusion easily.WhenX is a Hilbert space, we can get the following corollary easily.Corollary 5.3 (Reich [2]).
LetC 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=1∞αn=∞. 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
(5.5)xn+1=αnf(xn)+βnxn+(1-αn-βn)1sn∫0snT(s)xnds,n≥0.
Then{xn} converges strongly to x̃, where x̃ is the unique solution in F(𝒮) of the variational inequality
(5.6)〈(I-f)x̃,x-x̃〉≥0,x∈F(S),
or equivalent x̃=PF(𝒮)(f)(x̃), where P is a metric projection mapping from H into F(𝒮).



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*Source: 641479-2012-02-22.xml*