Strong convergence theorems for fixed point problems of a nonexpansive semigroup in a Banach space

Abstract

In this paper, we study the implicit and explicit viscosity iteration schemes for a nonexpansive semigroup in a reflexive, strictly convex and uniformly smooth Banach space which satisfies Opial’s condition. Our results improve and generalize the corresponding results given by Yao et al. (Fixed Point Theory Appl. 2013, doi:10.1186/1687-1812-2013-31) and many others.

MSC:47H05, 47H10, 47H17.

1 Introduction

Let E be a real Banach space, and let K be a nonempty, closed and convex subset of E. A mapping $T:K\to K$ is called nonexpansive if

$$\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,\mathrm{\forall }x,y\in K.$$

(1.1)

One parameter family $S:=\{T(s):0\le s<\mathrm{\infty }\}$ is said to be a nonexpansive semigroup from K into K if the following conditions are satisfied:

• (1) $T(0)x=x$ for all $x\in K$;

• (2) $T(s+t)=T(s)T(t)$ for all $s,t\ge 0$;

• (3) $\parallel T(t)x-T(t)y\parallel \le \parallel x-y\parallel$, $\mathrm{\forall }x,y\in K$ and $t\ge 0$;

• (4) for each $x\in K$, the mapping $T(\cdot )x$ from $[0,\mathrm{\infty })$ into K is continuous.

Let $F(S)$ denote the common fixed point set of the semigroup S, i.e., $F(S):=\{x\in K:T(s)x=x,\mathrm{\forall }s>0\}$. It is known that $F(S)$ is closed and convex.

A continuous operator of the semigroup S is said to be uniformly asymptotically regular (u.a.r.) on K if for all $h\ge 0$ and any bounded subset C of K, ${lim}_{s\to \mathrm{\infty }}{sup}_{x\in C}\parallel T(h)T(s)x-T(s)x\parallel =0$ (see [1]).

Approximation of fixed points of nonexpansive mappings by a sequence of finite means has been considered by many authors (see [2–6]). In 2013, Yao et al. [7] introduced two new algorithms for finding a common fixed point of a nonexpansive semigroup ${\{T(s)\}}_{s\ge 0}$ in Hilbert spaces and proved that both approaches converge strongly to a common fixed point of ${\{T(s)\}}_{s\ge 0}$.

Theorem 1.1 [7]

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $S={\{T(s)\}}_{s\ge 0}:C\to C$ be a nonexpansive semigroup with $Fix(S)\ne \mathrm{\varnothing }$. Let ${\{{\gamma }_t\}}_{0<t<1}$ and ${\{{\lambda }_t\}}_{0<t<1}$ be two continuous nets of positive real numbers such that ${\gamma }_t\in (0,1)$, ${lim}_{t\to 0}{\gamma }_t=1$ and ${lim}_{t\to 0}{\lambda }_t=+\mathrm{\infty }$. Let $\{x_t\}$ be the net defined in the following implicit manner:

$$x_t=P_C[t({\gamma }_t{x}_t)+(1-t)\frac{1}{{\lambda }_t}{\int }_0^{{\lambda }_t}T(s)x_{t}ds],\mathrm{\forall }t\in (0,1).$$

(1.2)

Then, as $t\to 0^{+}$, the net $\{x_t\}$ strongly converges to $x^{\ast }\in Fix(s)$.

Theorem 1.2 [7]

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $S={\{T(s)\}}_{s\ge 0}:C\to C$ be a nonexpansive semigroup with $Fix(S)\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be the sequence generated iteratively by the following explicit algorithm:

$$x_{n+1}=(1-{\beta }_n)x_n+{\beta }_n{P}_C[{\alpha }_n({\gamma }_n{x}_n)+(1-{\alpha }_n)\frac{1}{{\lambda }_n}{\int }_0^{{\lambda }_n}T(s)x_{n}ds],\mathrm{\forall }n\ge 0,$$

(1.3)

where $\{{\alpha }_n\}$, $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ are sequences of real numbers in $[0,1]$ and $\{{\lambda }_n\}$ is a sequence of positive real numbers. Suppose that the following conditions are satisfied:

• (i) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$ and ${lim}_{n\to \mathrm{\infty }}{\gamma }_n=1$;

• (ii) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$;

• (iii) ${lim}_{n\to \mathrm{\infty }}{\lambda }_n=\mathrm{\infty }$ and ${lim}_{n\to \mathrm{\infty }}\frac{{\lambda }_{n-1}}{{\lambda }_n}=1$.

Then the sequence $\{x_n\}$ generated by (1.3) strongly converges to a point $x^{\ast }\in Fix(s)$.

In this paper, we study the convergence of the following iterative schemes in a reflexive, strictly convex and uniformly smooth Banach space which satisfies Opial’s condition:

$$\begin{array}{c}{x}_t=Q_K[t({\gamma }_t{x}_t)+(1-t)T(s_t)x_t],\mathrm{\forall }t\in (0,1),\hfill \\ x_{n+1}=(1-{\beta }_n)x_n+{\beta }_n{Q}_K[{\alpha }_n({\gamma }_n{x}_n)+(1-{\alpha }_n)T(s_n)x_n],\mathrm{\forall }n\ge 0.\hfill \end{array}$$

Our work improves and generalizes many others. In particular, our results extend the main results of Yao et al. [7].

2 Preliminaries

Let E be a real Banach space and $E^{\ast }$ be the dual space of E. The duality mapping $J:E\to 2^{E^{\ast }}$ is defined by

$$(x)=\{f\in E^{\ast }:〈x,f〉={\parallel x\parallel }^2={\parallel f\parallel }^2\}.$$

(2.1)

By the Hahn-Banach theorem, $J(x)$ is nonempty.

Let $dimE\ge 2$. The modulus of convexity of E is the function ${\delta }_E:(0,2]\to [0,1]$ defined by

$${\delta }_E(ϵ):=inf\{1-\parallel \frac{x-y}{2}\parallel :\parallel x\parallel =\parallel y\parallel =1;ϵ=\parallel x-y\parallel \}.$$

(2.2)

E is uniformly convex if $\mathrm{\forall }ϵ\in (0,2]$, there exists $\delta =\delta (ϵ)>0$ such that if $x,y\in E$ with $\parallel x\parallel \le 1$, $\parallel y\parallel \le 1$ and $\parallel x-y\parallel \ge ϵ$, then $\parallel \frac{x+y}{2}\parallel \le 1-\delta$. Equivalently, E is uniformly convex if and only if ${\delta }_E(ϵ)>0$, $\mathrm{\forall }ϵ\in (0,2]$. E is strictly convex if for all $x,y\in E$, $x\ne y$, $\parallel x\parallel =\parallel y\parallel =1$, we have $\parallel \lambda x+(1-\lambda )y\parallel <1$, $\mathrm{\forall }\lambda \in (0,1)$.

Let $S(E)=\{x\in E:\parallel x\parallel =1\}$. The space E is said to be smooth if

$$\underset{t\to 0}{lim}(\parallel x+ty\parallel -\parallel x\parallel )/t$$

(2.3)

exists for all $x,y\in S(E)$. The norm of E is said to be Fréchet differentiable if for all $x\in S(E)$, the limit (2.3) exists uniformly for all $y\in S(E)$. E is said to have a uniformly Gâteaux differentiable norm if for all $y\in S(E)$, the limit (2.3) is attained uniformly for all $x\in S(E)$. The norm of E is said to be uniformly Fréchet differentiable (or uniformly smooth) if the limit (2.3) is attained uniformly for $x,y\in S(E)\times S(E)$.

It is well known that if E is smooth, then J is single-valued, which is denoted by j. And if E has a uniformly Gâteaux differentiable norm, then J is norm-to-weak^∗ uniformly continuous on each bounded subset of E. The duality mapping J is said to be weakly sequentially continuous if J is single-valued and for any $\{x_n\}\in E$ with $x_n\rightharpoonup x$, $J(x_n){\rightharpoonup }^{\ast }J(x)$. Gossez and Lami Dozo [8] proved that a space with a weakly continuous duality mappings satisfies Opial’s condition. Conversely, if a space satisfies Opial’s condition and has a uniformly Gâteaux differentiable norm, then it has a weakly continuous duality mapping.

Recall that if C and D are nonempty subsets of a Banach space E such that C is nonempty closed convex and $D\subset C$, the mapping $Q:C\to D$ is said to be sunny if

$$Q(Qx+t(x-Qx))=Qx,$$

where $Qx+t(x-Qx)\in C$ for all $x\in C$ and $t\ge 0$.

A mapping $Q:C\to D$ is called a retraction if $Qx=x$ for all $x\in D$.

A subset D of C is called a sunny nonexpansive retraction of C if there exists a sunny nonexpansive retraction from C into D (see [9, 10]). It is well known that if E is a Hilbert space, then a sunny nonexpansive retraction is coincident with the metric projection from E onto C.

Proposition 2.1 [11]

Let C be a closed convex subset of a smooth Banach space E. Let D be a nonempty subset of C. Let $Q:C\to D$ be a retraction, and let J be the normalized duality mapping on E. Then the following are equivalent:

• (1) Q is sunny and nonexpansive.

• (2) ${\parallel Qx-Qy\parallel }^2\le 〈x-y,J(Qx-Qy)〉$, $\mathrm{\forall }x,y\in C$.

• (3) $〈x-Qx,J(y-Qx)〉\le 0$, $\mathrm{\forall }x\in C$, $y\in D$.

Proposition 2.2 [12]

Let C be a nonempty closed convex subset of a strictly convex and uniformly smooth Banach space E, and let T be a nonexpansive mapping of C into itself with $F(T)\ne \mathrm{\varnothing }$. Then the set $F(S)$ is a sunny nonexpansive retraction of C.

Lemma 2.3 [13]

Let K be a nonempty closed convex subset of a reflexive Banach space E which satisfies Opial’s condition, and suppose that $T:K\to E$ is nonexpansive. Then the mapping $I-T$ is demiclosed at zero, that is, $x_n\rightharpoonup x$, $x_n-T{x}_n\to 0$ implies $x=Tx$.

Lemma 2.4 [14]

Let $\{x_n\}$, $\{y_n\}$ be two bounded sequences in a Banach space E and ${\beta }_n\in (0,1)$ with $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$. Suppose $x_{n+1}={\beta }_n{y}_n+(1-{\beta }_n)x_n$ for all integers $n\ge 0$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}(\parallel y_{n+1}-y_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0$. Then ${lim}_{n\to \mathrm{\infty }}\parallel x_n-y_n\parallel =0$.

Lemma 2.5 [15]

Let $\{a_n\}$ be a sequence of nonnegative real numbers satisfying the following relation:

$$a_{n+1}\le (1-{\rho }_n)a_n+{\rho }_n{\sigma }_n,n\ge 0,$$

where $\{{\rho }_n\}$ and $\{{\sigma }_n\}$ are sequences of real numbers such that

• (i) $0<{\rho }_n<1$;

• (ii) ${\sum }_{n=1}^{\mathrm{\infty }}{\rho }_n=\mathrm{\infty }$;

• (iii) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\sigma }_n\le 0$ or ${\sum }_{n=1}^{\mathrm{\infty }}|{\rho }_n{\sigma }_n|$ is convergent.

Then ${lim}_{n\to \mathrm{\infty }}{a}_n=0$.

3 Main result

Theorem 3.1 Let E be a reflexive, strictly convex and uniformly smooth Banach space which satisfies Opial’s condition, and let K be a nonempty closed convex subset of E. Let $S=\{T(s):s\ge 0\}:K\to K$ be a uniformly asymptotically regular nonexpansive semigroup such that $F(S)\ne \mathrm{\varnothing }$. Let ${\{{\gamma }_t\}}_{0<t<1}$ and ${\{s_t\}}_{0<t<1}$ be two continuous nets of positive real numbers such that ${\gamma }_t\in (0,1)$, ${lim}_{t\to 0}{\gamma }_t=1$ and ${lim}_{t\to 0}{s}_t=+\mathrm{\infty }$. Let $\{x_t\}$ be the net defined by

$$x_t=Q_K[t({\gamma }_t{x}_t)+(1-t)T(s_t)x_t],\mathrm{\forall }t\in (0,1).$$

(3.1)

Then, as $t\to 0^{+}$, the net $\{x_t\}$ converges strongly to a point $x^{\ast }\in F(S)$.

Proof Consider a mapping W on K defined by

$$Wx:=Q_K[t({\gamma }_{t}x)+(1-t)T(s_t)x],\mathrm{\forall }t\in (0,1).$$

$\mathrm{\forall }x,y\in K$, we have

$$\begin{array}{rl}\parallel Wx-Wy\parallel & \le \parallel t{\gamma }_t(x-y)+(1-t)(T(s_t)x-T(s_t)y)\parallel \\ \le t{\gamma }_t\parallel x-y\parallel +(1-t)\parallel x-y\parallel \\ =[1-(1-{\gamma }_t)t]\parallel x-y\parallel .\end{array}$$

Hence, W is a contraction. So, it has a unique fixed point, denoted by $x_t$. That is,

$$x_t=Q_K[t({\gamma }_t{x}_t)+(1-t)T(s_t)x_t].$$

Therefore, the sequence $\{x_t\}$ defined by (3.1) is well defined.

Let $p\in F(S)$, then

$$\begin{array}{rl}\parallel x_t-p\parallel & =\parallel Q_K[t({\gamma }_t{x}_t)+(1-t)T(s_t)x_t]-p\parallel \\ \le \parallel t{\gamma }_t(x_t-p)-t(1-{\gamma }_t)p+(1-t)(T(s_t)x_t-p)\parallel \\ \le t{\gamma }_t\parallel x_t-p\parallel +t(1-{\gamma }_t)\parallel p\parallel +(1-t)\parallel x_t-p\parallel \\ =[1-(1-{\gamma }_t)t]\parallel x_t-p\parallel +t(1-{\gamma }_t)\parallel p\parallel .\end{array}$$

It follows that

$$\parallel x_t-p\parallel \le \parallel p\parallel .$$

Thus, $\{x_t\}$ is bounded, so is $\{T(s_t)u_n\}$.

Let $R=\parallel p\parallel$. It is clear that $\{x_t\}\subset B(p,R)$. Then $B(p,R)\cap K$ is a nonempty bounded closed convex subset of K and $T(s)$-invariant. Since $\{T(s)\}$ is u.a.r. nonexpansive semigroup and ${lim}_{t\to 0}{s}_t=\mathrm{\infty }$, then for all $s>0$,

$$\underset{t\to 0}{lim}\parallel T(s)(T(s_t)x_t)-T(s_t)x_t\parallel \le \underset{n\to \mathrm{\infty }}{lim}\underset{x\in D}{sup}\parallel T(s)(T(s_t)x)-T(s_t)x\parallel =0,$$

where D is any bounded subset of K containing $\{u_n\}$. Since

$$\parallel x_t-T(s_t)x_t\parallel \le t\parallel {\gamma }_t{x}_t-T(s_t)x_t\parallel \to 0,$$

and

$$\begin{array}{rl}\parallel x_t-T(s)x_t\parallel & \le \parallel x_t-T(s_t)x_t\parallel +\parallel T(s_t)x_t-T(s)(T(s_t)x_t)\parallel +\parallel T(s)(T(s_t)x_t)-T(s)x_t\parallel \\ \le 2\parallel x_t-T(s_t)x_t\parallel +\parallel T(s_t)x_t-T(s)(T(s_t)x_t)\parallel .\end{array}$$

Thus, for all $s>0$, we have

$$\underset{t\to 0}{lim}\parallel x_t-T(s)x_t\parallel =0.$$

(3.2)

Set $y_t=t({\gamma }_t{x}_t)+(1-t)T(s_t)x_t$. Then $x_t=Q_K{y}_t$. By Proposition 2.1(2), we can get that

$$\begin{array}{rl}{\parallel x_t-p\parallel }^2& ={\parallel Q_K{y}_t-Q_{K}p\parallel }^2\\ \le 〈y_t-p,j(x_t-p)〉\\ =t{\gamma }_t〈x_t-p,j(x_t-p)〉-t(1-{\gamma }_t)〈p,j(x_t-p)〉+(1-t)〈T(s_t)x_t-p,j(x_t-p)〉\\ \le [1-(1-{\gamma }_t)t]{\parallel x_t-p\parallel }^2-t(1-{\gamma }_t)〈p,j(x_t-p)〉.\end{array}$$

Thus

$${\parallel x_t-p\parallel }^2\le -〈p,j(x_t-p)〉,\mathrm{\forall }p\in F(S).$$

(3.3)

Since $\{x_t\}$ is bounded and E is reflexive, there exists a subsequence $\{x_{t_n}\}$ of $\{x_t\}$ such that $x_{t_n}\rightharpoonup x^{\ast }$. From (3.2), we have $x_{t_n}-T(s)x_{t_n}\to 0$ as $n\to \mathrm{\infty }$. Since E satisfies Opial’s condition, it follows from Lemma 2.3 that $x^{\ast }\in F(S)$. From (3.3), we have

$${\parallel x_{t_n}-p\parallel }^2\le -〈p,j(x_{t_n}-p)〉,\mathrm{\forall }p\in F(S).$$

(3.4)

In particular, if we substitute $x^{\ast }$ for p in (3.4), then we have

$${\parallel x_{t_n}-x^{\ast }\parallel }^2\le -〈x^{\ast },j(x_{t_n}-x^{\ast })〉.$$

(3.5)

Since j is weakly sequentially continuous from E to $E^{\ast }$, it follows from (3.5) that

$$\underset{n\to \mathrm{\infty }}{lim}{\parallel x_{t_n}-x^{\ast }\parallel }^2\le \underset{n\to \mathrm{\infty }}{lim}-〈x^{\ast },j(x_{t_n}-x^{\ast })〉=0.$$

Suppose that there exists a subsequence $\{x_{t_m}\}$ of $\{x_t\}$ such that $x_{t_m}\rightharpoonup \tilde{x}$. Then we have $\tilde{x}\in F(S)$ and

$${\parallel x_{t_m}-p\parallel }^2\le -〈p,j(x_{t_m}-p)〉,\mathrm{\forall }p\in F(S).$$

(3.6)

Since $x^{\ast },\tilde{x}\in F(S)$, from (3.4) and (3.6), we have

$${\parallel x_{t_n}-\tilde{x}\parallel }^2\le -〈\tilde{x},j(x_{t_n}-\tilde{x})〉,$$

(3.7)

and

$${\parallel x_{t_m}-x^{\ast }\parallel }^2\le -〈x^{\ast },j(x_{t_m}-x^{\ast })〉.$$

(3.8)

Now, in (3.7) and (3.8), taking $n\to \mathrm{\infty }$ and $m\to \mathrm{\infty }$, respectively. We get

$${\parallel x^{\ast }-\tilde{x}\parallel }^2\le -〈\tilde{x},j(x^{\ast }-\tilde{x})〉,$$

(3.9)

and

$${\parallel \tilde{x}-x^{\ast }\parallel }^2\le -〈x^{\ast },j(\tilde{x}-x^{\ast })〉.$$

(3.10)

Adding up (3.9) and (3.10), we have

$${\parallel x^{\ast }-\tilde{x}\parallel }^2\le 0.$$

We have proved that each cluster point of $\{x_t\}$ (as $t\to 0$) equals $x^{\ast }$. Thus $x_t\to x^{\ast }$ as $t\to 0$. □

Remark 3.2 Theorem 3.1 improves and extends Theorem 3.1 of Yao et al. [7] in the following aspects.

• (1) From a real Hilbert space to a reflexive, strictly convex and uniformly smooth Banach space which satisfies Opial’s condition.

• (2) $\frac{1}{{\lambda }_t}{\int }_0^{{\lambda }_t}T(s)x_{t}ds$ is replaced by $T(s_t)x_t$.

Theorem 3.3 Let E be a reflexive, strictly convex and uniformly smooth Banach space which satisfies Opial’s condition, and let K be a nonempty closed convex subset of E. Let $S=\{T(s):s\ge 0\}:K\to K$ be a uniformly asymptotically regular nonexpansive semigroup such that $F(S)\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be a sequence generated in the following iterative process:

$$x_{n+1}=(1-{\beta }_n)x_n+{\beta }_n{Q}_K[{\alpha }_n({\gamma }_n{x}_n)+(1-{\alpha }_n)T(s_n)x_n],\mathrm{\forall }n\ge 0,$$

(3.11)

where $\{{\alpha }_n\}$, $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ are sequences of real numbers in $[0,1]$ satisfying the following conditions:

• (1) ${lim}_{n\to \mathrm{\infty }}{\gamma }_n=1$, ${\sum }_{n=1}^{\mathrm{\infty }}(1-{\gamma }_n){\alpha }_n=\mathrm{\infty }$, ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$.

• (2) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$.

• (3) $h,s_n\ge 0$ such that $s_{n+1}=h+s_n$ and ${lim}_{n\to \mathrm{\infty }}{s}_n=\mathrm{\infty }$.

Then $\{x_n\}$ converges strongly to $x^{\ast }\in F(S)$.

Proof Let $p\in F(S)$, we can get

$$\begin{array}{rl}\parallel x_{n+1}-p\parallel & =\parallel (1-{\beta }_n)x_n+{\beta }_n{Q}_K[{\alpha }_n({\gamma }_n{x}_n)+(1-{\alpha }_n)T(s_n)x_n]-p\parallel \\ \le (1-{\beta }_n)\parallel x_n-p\parallel +{\beta }_n\parallel Q_K[{\alpha }_n({\gamma }_n{x}_n)+(1-{\alpha }_n)T(s_n)x_n]-p\parallel \\ \le (1-{\beta }_n)\parallel x_n-p\parallel +{\beta }_n\parallel {\alpha }_n{\gamma }_n(x_n-p)-{\alpha }_n(1-{\gamma }_n)p+(1-{\alpha }_n)(T(s_n)x_n-p)\parallel \\ \le (1-{\beta }_n)\parallel x_n-p\parallel +{\beta }_n({\alpha }_n{\gamma }_n\parallel x_n-p\parallel -{\alpha }_n(1-{\gamma }_n)\parallel p\parallel +(1-{\alpha }_n)\parallel x_n-p\parallel )\\ =[1-(1-{\gamma }_n){\alpha }_n{\beta }_n]\parallel x_n-p\parallel +(1-{\gamma }_n){\alpha }_n{\beta }_n\parallel p\parallel \\ \le max\{\parallel x_n-p\parallel ,\parallel p\parallel \}\\ \le max\{\parallel x_0-p\parallel ,\parallel p\parallel \}.\end{array}$$

Hence, $\{x_n\}$ is bounded, so is $\{T(s_n)x_n\}$.

Set $y_n=Q_K[{\alpha }_n({\gamma }_n{x}_n)+(1-{\alpha }_n)T(s_n)x_n]$ for all $n\ge 0$. Then $x_{n+1}=(1-{\beta }_n)x_n+{\beta }_n{y}_n$.

$$\begin{array}{rl}\parallel y_{n+1}-y_n\parallel =& \parallel Q_K[{\alpha }_{n+1}({\gamma }_{n+1}{x}_{n+1})+(1-{\alpha }_{n+1})T(s_{n+1})x_{n+1}]\\ -Q_K[{\alpha }_n({\gamma }_n{x}_n)+(1-{\alpha }_n)T(s_n)x_n]\parallel \\ \le & \parallel [{\alpha }_{n+1}({\gamma }_{n+1}{x}_{n+1})+(1-{\alpha }_{n+1})T(s_{n+1})x_{n+1}]-[{\alpha }_n({\gamma }_n{x}_n)+(1-{\alpha }_n)T(s_n)x_n]\parallel \\ =& \parallel {\alpha }_{n+1}{\gamma }_{n+1}(x_{n+1}-x_n)+({\alpha }_{n+1}{\gamma }_{n+1}-{\alpha }_n{\gamma }_n)x_n+(1-{\alpha }_{n+1})\\ \times (T(s_{n+1})x_{n+1}-T(s_{n+1})x_n+T(s_{n+1})x_n-T(s_n)x_n)+({\alpha }_{n+1}-{\alpha }_n)T(s_n)x_n\parallel \\ \le & {\alpha }_{n+1}{\gamma }_{n+1}\parallel x_{n+1}-x_n\parallel +|{\alpha }_{n+1}{\gamma }_{n+1}-{\alpha }_n{\gamma }_n|\parallel x_n\parallel \\ +(1-{\alpha }_{n+1})(\parallel x_{n+1}-x_n\parallel +\parallel T(h)T(s_n)x_n-T(s_n)x_n\parallel )\\ +|{\alpha }_{n+1}-{\alpha }_n|\parallel T(s_n)x_n\parallel \\ =& [1-(1-{\gamma }_{n+1}){\alpha }_{n+1}]\parallel x_{n+1}-x_n\parallel +|{\alpha }_{n+1}{\gamma }_{n+1}-{\alpha }_n{\gamma }_n|\parallel x_n\parallel \\ +(1-{\alpha }_{n+1})\parallel T(h)(s_n)x_n-T(s_n)x_n\parallel +|{\alpha }_{n+1}-{\alpha }_n|\parallel T(s_n)x_n\parallel .\end{array}$$

So,

$$\begin{array}{r}\parallel y_{n+1}-y_n\parallel -\parallel x_{n+1}-x_n\parallel \\ \le -(1-{\gamma }_{n+1}){\alpha }_{n+1}\parallel x_{n+1}-x_n\parallel +|{\alpha }_{n+1}{\gamma }_{n+1}-{\alpha }_n{\gamma }_n|\parallel x_n\parallel \\ +(1-{\alpha }_{n+1})\parallel T(h)(s_n)x_n-T(s_n)x_n\parallel +|{\alpha }_{n+1}-{\alpha }_n|\parallel T(s_n)x_n\parallel .\end{array}$$

(3.12)

Since $\{T(s):s\ge 0\}$ is uniformly asymptotically regular and ${lim}_{n\to \mathrm{\infty }}{s}_n=\mathrm{\infty }$, it follows that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T(h)T(s_n)x_n-T(s_n)x_n\parallel \le \underset{n\to \mathrm{\infty }}{lim}\underset{x\in B}{sup}\parallel T(h)T(s_n)x-T(s_n)x\parallel =0,$$

(3.13)

where B is any bounded set containing $\{x_n\}$. Moreover, since $\{x_n\}$, $\{T(s_n)x_n\}$ are bounded, and ${\alpha }_n\to 0$ as $n\to \mathrm{\infty }$, (3.12) implies that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(\parallel y_{n+1}-y_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0.$$

Hence, by Lemma 2.4 we have ${lim}_{n\to \mathrm{\infty }}\parallel y_n-x_n\parallel =0$ since $x_{n+1}-x_n={\beta }_n(y_n-x_n)$. Consequently, ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$.

It follows from (3.11) that

$$\begin{array}{rl}\parallel x_n-T(s_n)x_n\parallel \le & \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-T(s_n)x_n\parallel \\ \le & \parallel x_n-x_{n+1}\parallel +\parallel (1-{\beta }_n)(x_n-T(s_n)x_n)\\ +{\beta }_n(Q_K[{\alpha }_n({\gamma }_n{x}_n)+(1-{\alpha }_n)T(s_n)x_n]-T(s_n)x_n)\parallel \\ \le & \parallel x_n-x_{n+1}\parallel +(1-{\beta }_n)\parallel x_n-T(s_n)x_n\parallel +{\alpha }_n{\gamma }_n\parallel x_n-T(s_n)x_n\parallel \\ +{\alpha }_n(1-{\gamma }_n)\parallel T(s_n)x_n\parallel \\ =& \parallel x_n-x_{n+1}\parallel +(1-{\beta }_n+{\alpha }_n{\gamma }_n)\parallel x_n-T(s_n)x_n\parallel +{\alpha }_n(1-{\gamma }_n)\parallel T(s_n)x_n\parallel .\end{array}$$

So,

$$\parallel x_n-T(s_n)x_n\parallel \le \frac{1}{{\beta }_n-{\alpha }_n{\gamma }_n}(\parallel x_n-x_{n+1}\parallel +{\alpha }_n(1-{\gamma }_n)\parallel T(s_n)x_n\parallel )\to 0.$$

(3.14)

Since

$$\begin{array}{r}\parallel x_n-T(h)x_n\parallel \\ \le \parallel x_n-T(s_n)x_n\parallel +\parallel T(s_n)x_n-T(h)T(s_n)x_n\parallel +\parallel T(h)T(s_n)x_n-T(h)x_n\parallel \\ \le 2\parallel x_n-T(s_n)x_n\parallel +\parallel T(s_n)x_n-T(h)T(s_n)x_n\parallel ,\end{array}$$

from (3.13) and (3.14), we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T(h)x_n\parallel =0.$$

(3.15)

Notice that $\{x_n\}$ is bounded. Put $x^{\ast }=Q_{F(S)}(0)$. Then there exists a positive number R such that $B(x^{\ast },R)\cap K$ contains $\{x_n\}$. Moreover, $B(x^{\ast },R)\cap K$ is $T(s)$-invariant for all $s\ge 0$ and so, without loss of generality, we can assume that $\{T(s):s\ge 0\}$ is a nonexpansive semigroup on $B(x^{\ast },R)\cap K$. We take a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈-x^{\ast },j(x_n-x^{\ast })〉=\underset{k\to \mathrm{\infty }}{lim}〈x^{\ast },j(x_{n_k}-x^{\ast })〉.$$

We may also assume that $x_{n_k}\rightharpoonup \tilde{x}$. It follows from Lemma 2.3 and (3.15) that $\tilde{x}\in F(S)$ and hence

$$〈-x^{\ast },j(\tilde{x}-x^{\ast })〉\le 0.$$

Since j is weakly sequentially continuous, we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈-x^{\ast },j(x_n-x^{\ast })〉=\underset{k\to \mathrm{\infty }}{lim}〈-x^{\ast },j(x_{n_k}-x^{\ast })〉=〈-x^{\ast },j(\tilde{x}-x^{\ast })〉\le 0.$$

Since ${lim}_{n\to \mathrm{\infty }}\parallel y_n-x_n\parallel =0$, we have $y_n-x^{\ast }\to x_n-x^{\ast }$, so

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈-x^{\ast },j(y_n-x^{\ast })〉=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈-x^{\ast },j(x_n-x^{\ast })〉\le 0.$$

Set $u_n={\alpha }_n({\gamma }_n{x}_n)+(1-{\alpha }_n)T(s_n)x_n$. It follows that $y_n=Q_K{u}_n$ for all $n\ge 0$. By Proposition 2.1(3), we have

$$〈y_n-u_n,j(y_n-x^{\ast })〉\le 0,$$

and so

$$\begin{array}{rl}{\parallel y_n-x^{\ast }\parallel }^2=& 〈y_n-x^{\ast },j(y_n-x^{\ast })〉\\ =& 〈y_n-u_n,j(y_n-x^{\ast })〉+〈u_n-x^{\ast },j(y_n-x^{\ast })〉\\ \le & 〈u_n-x^{\ast },j(y_n-x^{\ast })〉\\ =& {\alpha }_n{\gamma }_n〈x_n-x^{\ast },j(y_n-x^{\ast })〉-{\alpha }_n(1-{\gamma }_n)〈x^{\ast },j(y_n-x^{\ast })〉\\ +(1-{\alpha }_n)〈T(s_n)x_n-x^{\ast },j(y_n-x^{\ast })〉\\ \le & {\alpha }_n{\gamma }_n\parallel x_n-x^{\ast }\parallel \parallel j(y_n-x^{\ast })\parallel -{\alpha }_n(1-{\gamma }_n)〈x^{\ast },j(y_n-x^{\ast })〉\\ +(1-{\alpha }_n)\parallel T(s_n)x_n-x^{\ast }\parallel \parallel j(y_n-x^{\ast })\parallel \\ \le & [1-(1-{\gamma }_n){\alpha }_n]\parallel x_n-x^{\ast }\parallel \parallel y_n-x^{\ast }\parallel -{\alpha }_n(1-{\gamma }_n)〈x^{\ast },j(y_n-x^{\ast })〉\\ \le & \frac{1-(1-{\gamma }_n){\alpha }_n}{2}{\parallel x_n-x^{\ast }\parallel }^2+\frac{1}{2}{\parallel y_n-x^{\ast }\parallel }^2-{\alpha }_n(1-{\gamma }_n)〈x^{\ast },j(y_n-x^{\ast })〉,\end{array}$$

that is,

$${\parallel y_n-x^{\ast }\parallel }^2\le [1-(1-{\gamma }_n){\alpha }_n]{\parallel x_n-x^{\ast }\parallel }^2-2{\alpha }_n(1-{\gamma }_n)〈x^{\ast },j(y_n-x^{\ast })〉.$$

By the convexity of ${\parallel \cdot \parallel }^2$, we have

$$\begin{array}{rl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& \le (1-{\beta }_n){\parallel x_n-x^{\ast }\parallel }^2+{\beta }_n{\parallel y_n-x^{\ast }\parallel }^2\\ \le [1-(1-{\gamma }_n){\alpha }_n{\beta }_n]{\parallel x_n-x^{\ast }\parallel }^2-2(1-{\gamma }_n){\alpha }_n{\beta }_n〈x^{\ast },j(y_n-x^{\ast })〉.\end{array}$$

By Lemma 2.5, we conclude that $x_n\to x^{\ast }$. □

Remark 3.4 Theorem 3.3 improves and extends Theorem 3.3 of Yao et al. [7] in the following aspects.

• (1) From a real Hilbert space to a reflexive, strictly convex and uniformly smooth Banach space which satisfies Opial’s condition.

• (2) $\frac{1}{{\lambda }_n}{\int }_0^{{\lambda }_n}T(s)x_{n}ds$ is replaced by $T(s_n)x_n$.