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)∫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.

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,

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,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:〈(I-f)x̃,x-x̃〉≥0,∀x∈F(T).

Moreover, Xu [6] also studied the strong convergence of the following iterative sequence generated byxn+1=βnf(xn)+(1-βn)Txn,n≥0,

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:

T(0)x=x for all x∈C;

T(s+t)=T(s)T(t) for all x,y∈C and s,t≥0;

∥T(s)x-T(s)y∥≤∥x-y∥ for all x,y∈C and s≥0;

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 byxt=tf(xt)+(1-t)1λt∫0λtT(s)xtds,

where t∈(0,1) and {λt} is a positive real divergent net, and the iterative scheme {xn} defined byxn+1=αnf(xn)+βnxn+(1-αn-βn)1sn∫0snT(s)xnds,n≥0,

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 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 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=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

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 byJ(x)={x*∈X*:〈x,x*〉=‖x‖2=‖x*‖2},x∈X.

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
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.

A Banach space X is said to be uniformly convex if δX(ɛ)>0 for all ɛ>0, where δX(ɛ) is modulus of convexity of E defined byδE(ɛ)=inf{1-‖x+y‖2:‖x‖≤1,‖y‖≤1,‖x-y‖≥ɛ},ɛ∈[0,2].

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 [<xref ref-type="bibr" rid="B6">10</xref>], 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 [<xref ref-type="bibr" rid="B14">11</xref>]).

In a smooth Banach space X there holds the inequality
‖x+y‖2≤‖x‖2+2〈y,J(x+y)〉,x,y,∈X.

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 [<xref ref-type="bibr" rid="B5">8</xref>, 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‖1t∫0tT(s)xds-T(h)(1t∫0tT(s)xds)‖=0.

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 byJφ(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<∞. SetΦ(t)=∫0tφ(τ)dτ,t≥0.

ThenJφ(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.

Assume {αn} is a sequence of nonnegative real numbers such that
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

∑n=1∞γn=∞;

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 ifinf{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 ifμn(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, thenliminfn→∞an≤μn(an)≤limsupn→∞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 haveμn(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 [<xref ref-type="bibr" rid="B11">14</xref>, 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-x‖2,
if and only if
μn〈x-z0,J(xn-z0)〉≤0,∀x∈C.

Lemma 2.8 (Song and Xu [<xref ref-type="bibr" rid="B10">4</xref>, 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-y‖2=infx∈Cμn‖xn-x‖2}.
If F(T)≠∅, then C*∩F(T)≠∅.

3. Implicit Iteration SchemeTheorem 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:
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
〈(I-f)x̃,J(x-x̃)〉≥0,x∈F(S).

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-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
‖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
‖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
‖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
limλt→∞‖1λt∫0λtT(s)xtds-T(s)(1λt∫0λtT(s)xtds)‖≤limλt→∞supx∈D‖1λt∫0λtT(s)xtds-T(s)(1λt∫0λtT(s)xtds)‖=0.
Hence, by (3.5)–(3.7), we obtain
‖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
xn-x̃=tn(f(xn)-x̃)+(1-tn)(1λn∫0λnT(s)xnds-x̃).
Thus, by Lemma 2.5, we obtain
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
Φ(‖xn-x̃‖)≤〈f(xn)-x̃,Jφ(xn-x̃)〉.
In particular, we have
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
Φ(‖xnj-x̃‖)⟶0asj⟶∞.
Hence xnj→x̃.

Next, we show that x̃∈F(𝒮) solves the variational inequality (3.2). Indeed, for q∈F(𝒮), it is easy to see that
〈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
xt-1λt∫0λtT(s)xtds=t1-t(f(xt)-xt).
Thus, we get that for t∈(0,1) and q∈F(𝒮)〈xt-f(xt),Jφ(xt-q)〉≤0.
Taking the limit through t:=tnj→0, we obtain
〈(I-f)x̃,Jφ(x̃-q)〉≤0,∀q∈F(S).
This implies that
〈(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
〈(I-f)x̃,J(x̃-x¯)〉≤0.
Interchange x̃ and x¯ to obtain
〈(I-f)x¯,J(x¯-x̃)〉≤0.
Adding (3.19) and (3.20) yields
(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.

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:
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
〈(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 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 x̃∈F(𝒮). For each n≥1, we have
xn-x̃=tn(f(xn)-x̃)+(1-tn)(1λn∫0λnT(s)xnds-x̃).
Thus, we have
‖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,
‖xn-x̃‖2≤11-α〈f(x̃)-x̃,J(xn-x̃)〉.

We claim that the set {xn} is sequentially compact. Indeed, define the set
C*={y∈C:μn‖xn-y‖2=infx∈Cμn‖xn-x‖2}.
By Lemma 2.8, we found x̃∈C*. Using Lemma 2.7 we get that
μn〈x-x̃,J(xn-x̃)〉≤0,∀x∈C.
From (3.26), we get
μn‖xn-x̃‖2≤11-αμn〈f(x̃)-x̃,J(xn-x̃)〉≤0,
that is
μn‖xn-x̃‖=0.
Hence, there exists a subsequence {xnk} of {xn} converges strongly to x̃∈F(𝒮) as k→∞.

Next we show that x̃ 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-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,
〈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
〈f(x̃)-x̃,J(x-xnk)〉⟶〈f(x̃)-x̃,J(x-x̃)〉.
Taking limit as j→∞ in two sides of (3.32), we get
〈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 SchemeTheorem 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=1∞αn=∞, and {sn} is a positive real divergent sequence such that limn→∞sn→∞. If the sequence {xn} defined by x0∈C and
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
〈(I-f)x̃,J(x-x̃)〉≥0,x∈F(S).

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-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
‖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)‖.

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→∞‖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
‖xn+1-T(h)xn+1‖⟶0asn⟶∞,
and hence
‖xn-T(h)xn‖⟶0asn⟶∞.

We now show that
〈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
limsupn→∞〈f(x̃)-x̃,Jφ(xn-x̃)〉=limk→∞〈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
limsupn→∞〈f(x̃)-x̃,Jφ(xn-x̃)〉=〈f(x̃)-x̃,Jφ(p-x̃)〉≤0.
That is (4.8) holds.

Finally we will show that xn→x̃. For each n≥0, we have
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.

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=1∞αn=∞, and {sn} is a positive real divergent sequence such that limn→∞sn→∞. If the sequence {xn} defined by x0∈C and
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
〈(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 Theorem 4.1. We now show that
〈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
limsupn→∞〈f(x̃)-x̃,J(xn-x̃)〉=limk→∞〈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
limsupn→∞〈f(x̃)-x̃,J(xn-x̃)〉=〈f(x̃)-x̃,J(p-x̃)〉≤0.
That is, (4.14) holds.

Finally we will show that xn→x̃. For each n≥0, by Lemma 2.2, we have
‖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
‖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. ApplicationsTheorem 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=1∞αn=∞, and {sn} is a positive real divergent sequence such that limn→∞sn→∞. If the sequence {xn} defined by x0∈C and
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
〈(I-f)x̃,J(x-x̃)〉≥0,x∈F(S).

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=1∞αn=∞, and {sn} is a positive real divergent sequence such that limn→∞sn→∞. If the sequence {xn} defined by x0∈C and
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
〈(I-f)x̃,J(x-x̃)〉≥0,x∈F(S).

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.

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=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
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
〈(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).

Funding

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

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