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

Fixed Point Theory and Applications20132013:248

https://doi.org/10.1186/1687-1812-2013-248

• Received: 3 June 2013
• Accepted: 17 September 2013
• Published:

## 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.

## Keywords

• nonexpansive semigroup
• fixed point
• reflexive and strictly convex Banach space
• uniformly smooth
• Opial’s condition
• sunny nonexpansive retraction

## 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
$\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in K.$
(1.1)
One parameter family $S:=\left\{T\left(s\right):0\le s<\mathrm{\infty }\right\}$ is said to be a nonexpansive semigroup from K into K if the following conditions are satisfied:
1. (1)

$T\left(0\right)x=x$ for all $x\in K$;

2. (2)

$T\left(s+t\right)=T\left(s\right)T\left(t\right)$ for all $s,t\ge 0$;

3. (3)

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

4. (4)

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

Let $F\left(S\right)$ denote the common fixed point set of the semigroup S, i.e., $F\left(S\right):=\left\{x\in K:T\left(s\right)x=x,\mathrm{\forall }s>0\right\}$. It is known that $F\left(S\right)$ 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, ${lim}_{s\to \mathrm{\infty }}{sup}_{x\in C}\parallel T\left(h\right)T\left(s\right)x-T\left(s\right)x\parallel =0$ (see ).

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

Theorem 1.1 

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $S={\left\{T\left(s\right)\right\}}_{s\ge 0}:C\to C$ be a nonexpansive semigroup with $Fix\left(S\right)\ne \mathrm{\varnothing }$. Let ${\left\{{\gamma }_{t}\right\}}_{0 and ${\left\{{\lambda }_{t}\right\}}_{0 be two continuous nets of positive real numbers such that ${\gamma }_{t}\in \left(0,1\right)$, ${lim}_{t\to 0}{\gamma }_{t}=1$ and ${lim}_{t\to 0}{\lambda }_{t}=+\mathrm{\infty }$. Let $\left\{{x}_{t}\right\}$ be the net defined in the following implicit manner:
${x}_{t}={P}_{C}\left[t\left({\gamma }_{t}{x}_{t}\right)+\left(1-t\right)\frac{1}{{\lambda }_{t}}{\int }_{0}^{{\lambda }_{t}}T\left(s\right){x}_{t}\phantom{\rule{0.2em}{0ex}}ds\right],\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in \left(0,1\right).$
(1.2)

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

Theorem 1.2 

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $S={\left\{T\left(s\right)\right\}}_{s\ge 0}:C\to C$ be a nonexpansive semigroup with $Fix\left(S\right)\ne \mathrm{\varnothing }$. Let $\left\{{x}_{n}\right\}$ be the sequence generated iteratively by the following explicit algorithm:
${x}_{n+1}=\left(1-{\beta }_{n}\right){x}_{n}+{\beta }_{n}{P}_{C}\left[{\alpha }_{n}\left({\gamma }_{n}{x}_{n}\right)+\left(1-{\alpha }_{n}\right)\frac{1}{{\lambda }_{n}}{\int }_{0}^{{\lambda }_{n}}T\left(s\right){x}_{n}\phantom{\rule{0.2em}{0ex}}ds\right],\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 0,$
(1.3)
where $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$ and $\left\{{\gamma }_{n}\right\}$ are sequences of real numbers in $\left[0,1\right]$ and $\left\{{\lambda }_{n}\right\}$ is a sequence of positive real numbers. Suppose that the following conditions are satisfied:
1. (i)

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

2. (ii)

$0<{lim inf}_{n\to \mathrm{\infty }}{\beta }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\beta }_{n}<1$;

3. (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 $\left\{{x}_{n}\right\}$ generated by (1.3) strongly converges to a point ${x}^{\ast }\in Fix\left(s\right)$.

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:
$\begin{array}{c}{x}_{t}={Q}_{K}\left[t\left({\gamma }_{t}{x}_{t}\right)+\left(1-t\right)T\left({s}_{t}\right){x}_{t}\right],\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in \left(0,1\right),\hfill \\ {x}_{n+1}=\left(1-{\beta }_{n}\right){x}_{n}+{\beta }_{n}{Q}_{K}\left[{\alpha }_{n}\left({\gamma }_{n}{x}_{n}\right)+\left(1-{\alpha }_{n}\right)T\left({s}_{n}\right){x}_{n}\right],\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 0.\hfill \end{array}$

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

## 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
$\left(x\right)=\left\{f\in {E}^{\ast }:〈x,f〉={\parallel x\parallel }^{2}={\parallel f\parallel }^{2}\right\}.$
(2.1)

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

Let $dimE\ge 2$. The modulus of convexity of E is the function ${\delta }_{E}:\left(0,2\right]\to \left[0,1\right]$ defined by
${\delta }_{E}\left(ϵ\right):=inf\left\{1-\parallel \frac{x-y}{2}\parallel :\parallel x\parallel =\parallel y\parallel =1;ϵ=\parallel x-y\parallel \right\}.$
(2.2)

E is uniformly convex if $\mathrm{\forall }ϵ\in \left(0,2\right]$, there exists $\delta =\delta \left(ϵ\right)>0$ 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 ${\delta }_{E}\left(ϵ\right)>0$, $\mathrm{\forall }ϵ\in \left(0,2\right]$. E is strictly convex if for all $x,y\in E$, $x\ne y$, $\parallel x\parallel =\parallel y\parallel =1$, we have $\parallel \lambda x+\left(1-\lambda \right)y\parallel <1$, $\mathrm{\forall }\lambda \in \left(0,1\right)$.

Let $S\left(E\right)=\left\{x\in E:\parallel x\parallel =1\right\}$. The space E is said to be smooth if
$\underset{t\to 0}{lim}\left(\parallel x+ty\parallel -\parallel x\parallel \right)/t$
(2.3)

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

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 $\left\{{x}_{n}\right\}\in E$ with ${x}_{n}⇀x$, $J\left({x}_{n}\right){⇀}^{\ast }J\left(x\right)$. Gossez and Lami Dozo  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
$Q\left(Qx+t\left(x-Qx\right)\right)=Qx,$

where $Qx+t\left(x-Qx\right)\in C$ 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 

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. (1)

Q is sunny and nonexpansive.

2. (2)

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

3. (3)

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

Proposition 2.2 

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 $F\left(T\right)\ne \mathrm{\varnothing }$. Then the set $F\left(S\right)$ is a sunny nonexpansive retraction of C.

Lemma 2.3 

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, ${x}_{n}⇀x$, ${x}_{n}-T{x}_{n}\to 0$ implies $x=Tx$.

Lemma 2.4 

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

Lemma 2.5 

Let $\left\{{a}_{n}\right\}$ be a sequence of nonnegative real numbers satisfying the following relation:
${a}_{n+1}\le \left(1-{\rho }_{n}\right){a}_{n}+{\rho }_{n}{\sigma }_{n},\phantom{\rule{1em}{0ex}}n\ge 0,$
where $\left\{{\rho }_{n}\right\}$ and $\left\{{\sigma }_{n}\right\}$ are sequences of real numbers such that
1. (i)

$0<{\rho }_{n}<1$;

2. (ii)

${\sum }_{n=1}^{\mathrm{\infty }}{\rho }_{n}=\mathrm{\infty }$;

3. (iii)

${lim sup}_{n\to \mathrm{\infty }}{\sigma }_{n}\le 0$ 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 $S=\left\{T\left(s\right):s\ge 0\right\}:K\to K$ be a uniformly asymptotically regular nonexpansive semigroup such that $F\left(S\right)\ne \mathrm{\varnothing }$. Let ${\left\{{\gamma }_{t}\right\}}_{0 and ${\left\{{s}_{t}\right\}}_{0 be two continuous nets of positive real numbers such that ${\gamma }_{t}\in \left(0,1\right)$, ${lim}_{t\to 0}{\gamma }_{t}=1$ and ${lim}_{t\to 0}{s}_{t}=+\mathrm{\infty }$. Let $\left\{{x}_{t}\right\}$ be the net defined by
${x}_{t}={Q}_{K}\left[t\left({\gamma }_{t}{x}_{t}\right)+\left(1-t\right)T\left({s}_{t}\right){x}_{t}\right],\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in \left(0,1\right).$
(3.1)

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

Proof Consider a mapping W on K defined by
$Wx:={Q}_{K}\left[t\left({\gamma }_{t}x\right)+\left(1-t\right)T\left({s}_{t}\right)x\right],\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in \left(0,1\right).$
$\mathrm{\forall }x,y\in K$, we have
$\begin{array}{rl}\parallel Wx-Wy\parallel & \le \parallel t{\gamma }_{t}\left(x-y\right)+\left(1-t\right)\left(T\left({s}_{t}\right)x-T\left({s}_{t}\right)y\right)\parallel \\ \le t{\gamma }_{t}\parallel x-y\parallel +\left(1-t\right)\parallel x-y\parallel \\ =\left[1-\left(1-{\gamma }_{t}\right)t\right]\parallel x-y\parallel .\end{array}$
Hence, W is a contraction. So, it has a unique fixed point, denoted by ${x}_{t}$. That is,
${x}_{t}={Q}_{K}\left[t\left({\gamma }_{t}{x}_{t}\right)+\left(1-t\right)T\left({s}_{t}\right){x}_{t}\right].$

Therefore, the sequence $\left\{{x}_{t}\right\}$ defined by (3.1) is well defined.

Let $p\in F\left(S\right)$, then
$\begin{array}{rl}\parallel {x}_{t}-p\parallel & =\parallel {Q}_{K}\left[t\left({\gamma }_{t}{x}_{t}\right)+\left(1-t\right)T\left({s}_{t}\right){x}_{t}\right]-p\parallel \\ \le \parallel t{\gamma }_{t}\left({x}_{t}-p\right)-t\left(1-{\gamma }_{t}\right)p+\left(1-t\right)\left(T\left({s}_{t}\right){x}_{t}-p\right)\parallel \\ \le t{\gamma }_{t}\parallel {x}_{t}-p\parallel +t\left(1-{\gamma }_{t}\right)\parallel p\parallel +\left(1-t\right)\parallel {x}_{t}-p\parallel \\ =\left[1-\left(1-{\gamma }_{t}\right)t\right]\parallel {x}_{t}-p\parallel +t\left(1-{\gamma }_{t}\right)\parallel p\parallel .\end{array}$
It follows that
$\parallel {x}_{t}-p\parallel \le \parallel p\parallel .$

Thus, $\left\{{x}_{t}\right\}$ is bounded, so is $\left\{T\left({s}_{t}\right){u}_{n}\right\}$.

Let $R=\parallel p\parallel$. It is clear that $\left\{{x}_{t}\right\}\subset B\left(p,R\right)$. Then $B\left(p,R\right)\cap K$ is a nonempty bounded closed convex subset of K and $T\left(s\right)$-invariant. Since $\left\{T\left(s\right)\right\}$ is u.a.r. nonexpansive semigroup and ${lim}_{t\to 0}{s}_{t}=\mathrm{\infty }$, then for all $s>0$,
$\underset{t\to 0}{lim}\parallel T\left(s\right)\left(T\left({s}_{t}\right){x}_{t}\right)-T\left({s}_{t}\right){x}_{t}\parallel \le \underset{n\to \mathrm{\infty }}{lim}\underset{x\in D}{sup}\parallel T\left(s\right)\left(T\left({s}_{t}\right)x\right)-T\left({s}_{t}\right)x\parallel =0,$
where D is any bounded subset of K containing $\left\{{u}_{n}\right\}$. Since
$\parallel {x}_{t}-T\left({s}_{t}\right){x}_{t}\parallel \le t\parallel {\gamma }_{t}{x}_{t}-T\left({s}_{t}\right){x}_{t}\parallel \to 0,$
and
$\begin{array}{rl}\parallel {x}_{t}-T\left(s\right){x}_{t}\parallel & \le \parallel {x}_{t}-T\left({s}_{t}\right){x}_{t}\parallel +\parallel T\left({s}_{t}\right){x}_{t}-T\left(s\right)\left(T\left({s}_{t}\right){x}_{t}\right)\parallel +\parallel T\left(s\right)\left(T\left({s}_{t}\right){x}_{t}\right)-T\left(s\right){x}_{t}\parallel \\ \le 2\parallel {x}_{t}-T\left({s}_{t}\right){x}_{t}\parallel +\parallel T\left({s}_{t}\right){x}_{t}-T\left(s\right)\left(T\left({s}_{t}\right){x}_{t}\right)\parallel .\end{array}$
Thus, for all $s>0$, we have
$\underset{t\to 0}{lim}\parallel {x}_{t}-T\left(s\right){x}_{t}\parallel =0.$
(3.2)
Set ${y}_{t}=t\left({\gamma }_{t}{x}_{t}\right)+\left(1-t\right)T\left({s}_{t}\right){x}_{t}$. Then ${x}_{t}={Q}_{K}{y}_{t}$. By Proposition 2.1(2), we can get that
$\begin{array}{rl}{\parallel {x}_{t}-p\parallel }^{2}& ={\parallel {Q}_{K}{y}_{t}-{Q}_{K}p\parallel }^{2}\\ \le 〈{y}_{t}-p,j\left({x}_{t}-p\right)〉\\ =t{\gamma }_{t}〈{x}_{t}-p,j\left({x}_{t}-p\right)〉-t\left(1-{\gamma }_{t}\right)〈p,j\left({x}_{t}-p\right)〉+\left(1-t\right)〈T\left({s}_{t}\right){x}_{t}-p,j\left({x}_{t}-p\right)〉\\ \le \left[1-\left(1-{\gamma }_{t}\right)t\right]{\parallel {x}_{t}-p\parallel }^{2}-t\left(1-{\gamma }_{t}\right)〈p,j\left({x}_{t}-p\right)〉.\end{array}$
Thus
${\parallel {x}_{t}-p\parallel }^{2}\le -〈p,j\left({x}_{t}-p\right)〉,\phantom{\rule{1em}{0ex}}\mathrm{\forall }p\in F\left(S\right).$
(3.3)
Since $\left\{{x}_{t}\right\}$ is bounded and E is reflexive, there exists a subsequence $\left\{{x}_{{t}_{n}}\right\}$ of $\left\{{x}_{t}\right\}$ such that ${x}_{{t}_{n}}⇀{x}^{\ast }$. From (3.2), we have ${x}_{{t}_{n}}-T\left(s\right){x}_{{t}_{n}}\to 0$ as $n\to \mathrm{\infty }$. Since E satisfies Opial’s condition, it follows from Lemma 2.3 that ${x}^{\ast }\in F\left(S\right)$. From (3.3), we have
${\parallel {x}_{{t}_{n}}-p\parallel }^{2}\le -〈p,j\left({x}_{{t}_{n}}-p\right)〉,\phantom{\rule{1em}{0ex}}\mathrm{\forall }p\in F\left(S\right).$
(3.4)
In particular, if we substitute ${x}^{\ast }$ for p in (3.4), then we have
${\parallel {x}_{{t}_{n}}-{x}^{\ast }\parallel }^{2}\le -〈{x}^{\ast },j\left({x}_{{t}_{n}}-{x}^{\ast }\right)〉.$
(3.5)
Since j is weakly sequentially continuous from E to ${E}^{\ast }$, it follows from (3.5) that
$\underset{n\to \mathrm{\infty }}{lim}{\parallel {x}_{{t}_{n}}-{x}^{\ast }\parallel }^{2}\le \underset{n\to \mathrm{\infty }}{lim}-〈{x}^{\ast },j\left({x}_{{t}_{n}}-{x}^{\ast }\right)〉=0.$
Suppose that there exists a subsequence $\left\{{x}_{{t}_{m}}\right\}$ of $\left\{{x}_{t}\right\}$ such that ${x}_{{t}_{m}}⇀\stackrel{˜}{x}$. Then we have $\stackrel{˜}{x}\in F\left(S\right)$ and
${\parallel {x}_{{t}_{m}}-p\parallel }^{2}\le -〈p,j\left({x}_{{t}_{m}}-p\right)〉,\phantom{\rule{1em}{0ex}}\mathrm{\forall }p\in F\left(S\right).$
(3.6)
Since ${x}^{\ast },\stackrel{˜}{x}\in F\left(S\right)$, from (3.4) and (3.6), we have
${\parallel {x}_{{t}_{n}}-\stackrel{˜}{x}\parallel }^{2}\le -〈\stackrel{˜}{x},j\left({x}_{{t}_{n}}-\stackrel{˜}{x}\right)〉,$
(3.7)
and
${\parallel {x}_{{t}_{m}}-{x}^{\ast }\parallel }^{2}\le -〈{x}^{\ast },j\left({x}_{{t}_{m}}-{x}^{\ast }\right)〉.$
(3.8)
Now, in (3.7) and (3.8), taking $n\to \mathrm{\infty }$ and $m\to \mathrm{\infty }$, respectively. We get
${\parallel {x}^{\ast }-\stackrel{˜}{x}\parallel }^{2}\le -〈\stackrel{˜}{x},j\left({x}^{\ast }-\stackrel{˜}{x}\right)〉,$
(3.9)
and
${\parallel \stackrel{˜}{x}-{x}^{\ast }\parallel }^{2}\le -〈{x}^{\ast },j\left(\stackrel{˜}{x}-{x}^{\ast }\right)〉.$
(3.10)
Adding up (3.9) and (3.10), we have
${\parallel {x}^{\ast }-\stackrel{˜}{x}\parallel }^{2}\le 0.$

We have proved that each cluster point of $\left\{{x}_{t}\right\}$ (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.  in the following aspects.
1. (1)

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

2. (2)

$\frac{1}{{\lambda }_{t}}{\int }_{0}^{{\lambda }_{t}}T\left(s\right){x}_{t}\phantom{\rule{0.2em}{0ex}}ds$ is replaced by $T\left({s}_{t}\right){x}_{t}$.

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 $S=\left\{T\left(s\right):s\ge 0\right\}:K\to K$ be a uniformly asymptotically regular nonexpansive semigroup such that $F\left(S\right)\ne \mathrm{\varnothing }$. Let $\left\{{x}_{n}\right\}$ be a sequence generated in the following iterative process:
${x}_{n+1}=\left(1-{\beta }_{n}\right){x}_{n}+{\beta }_{n}{Q}_{K}\left[{\alpha }_{n}\left({\gamma }_{n}{x}_{n}\right)+\left(1-{\alpha }_{n}\right)T\left({s}_{n}\right){x}_{n}\right],\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 0,$
(3.11)
where $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$ and $\left\{{\gamma }_{n}\right\}$ are sequences of real numbers in $\left[0,1\right]$ satisfying the following conditions:
1. (1)

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

2. (2)

$0<{lim inf}_{n\to \mathrm{\infty }}{\beta }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\beta }_{n}<1$.

3. (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 $\left\{{x}_{n}\right\}$ converges strongly to ${x}^{\ast }\in F\left(S\right)$.

Proof Let $p\in F\left(S\right)$, we can get
$\begin{array}{rl}\parallel {x}_{n+1}-p\parallel & =\parallel \left(1-{\beta }_{n}\right){x}_{n}+{\beta }_{n}{Q}_{K}\left[{\alpha }_{n}\left({\gamma }_{n}{x}_{n}\right)+\left(1-{\alpha }_{n}\right)T\left({s}_{n}\right){x}_{n}\right]-p\parallel \\ \le \left(1-{\beta }_{n}\right)\parallel {x}_{n}-p\parallel +{\beta }_{n}\parallel {Q}_{K}\left[{\alpha }_{n}\left({\gamma }_{n}{x}_{n}\right)+\left(1-{\alpha }_{n}\right)T\left({s}_{n}\right){x}_{n}\right]-p\parallel \\ \le \left(1-{\beta }_{n}\right)\parallel {x}_{n}-p\parallel +{\beta }_{n}\parallel {\alpha }_{n}{\gamma }_{n}\left({x}_{n}-p\right)-{\alpha }_{n}\left(1-{\gamma }_{n}\right)p+\left(1-{\alpha }_{n}\right)\left(T\left({s}_{n}\right){x}_{n}-p\right)\parallel \\ \le \left(1-{\beta }_{n}\right)\parallel {x}_{n}-p\parallel +{\beta }_{n}\left({\alpha }_{n}{\gamma }_{n}\parallel {x}_{n}-p\parallel -{\alpha }_{n}\left(1-{\gamma }_{n}\right)\parallel p\parallel +\left(1-{\alpha }_{n}\right)\parallel {x}_{n}-p\parallel \right)\\ =\left[1-\left(1-{\gamma }_{n}\right){\alpha }_{n}{\beta }_{n}\right]\parallel {x}_{n}-p\parallel +\left(1-{\gamma }_{n}\right){\alpha }_{n}{\beta }_{n}\parallel p\parallel \\ \le max\left\{\parallel {x}_{n}-p\parallel ,\parallel p\parallel \right\}\\ \le max\left\{\parallel {x}_{0}-p\parallel ,\parallel p\parallel \right\}.\end{array}$

Hence, $\left\{{x}_{n}\right\}$ is bounded, so is $\left\{T\left({s}_{n}\right){x}_{n}\right\}$.

Set ${y}_{n}={Q}_{K}\left[{\alpha }_{n}\left({\gamma }_{n}{x}_{n}\right)+\left(1-{\alpha }_{n}\right)T\left({s}_{n}\right){x}_{n}\right]$ for all $n\ge 0$. Then ${x}_{n+1}=\left(1-{\beta }_{n}\right){x}_{n}+{\beta }_{n}{y}_{n}$.
$\begin{array}{rl}\parallel {y}_{n+1}-{y}_{n}\parallel =& \parallel {Q}_{K}\left[{\alpha }_{n+1}\left({\gamma }_{n+1}{x}_{n+1}\right)+\left(1-{\alpha }_{n+1}\right)T\left({s}_{n+1}\right){x}_{n+1}\right]\\ -{Q}_{K}\left[{\alpha }_{n}\left({\gamma }_{n}{x}_{n}\right)+\left(1-{\alpha }_{n}\right)T\left({s}_{n}\right){x}_{n}\right]\parallel \\ \le & \parallel \left[{\alpha }_{n+1}\left({\gamma }_{n+1}{x}_{n+1}\right)+\left(1-{\alpha }_{n+1}\right)T\left({s}_{n+1}\right){x}_{n+1}\right]-\left[{\alpha }_{n}\left({\gamma }_{n}{x}_{n}\right)+\left(1-{\alpha }_{n}\right)T\left({s}_{n}\right){x}_{n}\right]\parallel \\ =& \parallel {\alpha }_{n+1}{\gamma }_{n+1}\left({x}_{n+1}-{x}_{n}\right)+\left({\alpha }_{n+1}{\gamma }_{n+1}-{\alpha }_{n}{\gamma }_{n}\right){x}_{n}+\left(1-{\alpha }_{n+1}\right)\\ ×\left(T\left({s}_{n+1}\right){x}_{n+1}-T\left({s}_{n+1}\right){x}_{n}+T\left({s}_{n+1}\right){x}_{n}-T\left({s}_{n}\right){x}_{n}\right)+\left({\alpha }_{n+1}-{\alpha }_{n}\right)T\left({s}_{n}\right){x}_{n}\parallel \\ \le & {\alpha }_{n+1}{\gamma }_{n+1}\parallel {x}_{n+1}-{x}_{n}\parallel +|{\alpha }_{n+1}{\gamma }_{n+1}-{\alpha }_{n}{\gamma }_{n}|\parallel {x}_{n}\parallel \\ +\left(1-{\alpha }_{n+1}\right)\left(\parallel {x}_{n+1}-{x}_{n}\parallel +\parallel T\left(h\right)T\left({s}_{n}\right){x}_{n}-T\left({s}_{n}\right){x}_{n}\parallel \right)\\ +|{\alpha }_{n+1}-{\alpha }_{n}|\parallel T\left({s}_{n}\right){x}_{n}\parallel \\ =& \left[1-\left(1-{\gamma }_{n+1}\right){\alpha }_{n+1}\right]\parallel {x}_{n+1}-{x}_{n}\parallel +|{\alpha }_{n+1}{\gamma }_{n+1}-{\alpha }_{n}{\gamma }_{n}|\parallel {x}_{n}\parallel \\ +\left(1-{\alpha }_{n+1}\right)\parallel T\left(h\right)\left({s}_{n}\right){x}_{n}-T\left({s}_{n}\right){x}_{n}\parallel +|{\alpha }_{n+1}-{\alpha }_{n}|\parallel T\left({s}_{n}\right){x}_{n}\parallel .\end{array}$
So,
$\begin{array}{r}\parallel {y}_{n+1}-{y}_{n}\parallel -\parallel {x}_{n+1}-{x}_{n}\parallel \\ \phantom{\rule{1em}{0ex}}\le -\left(1-{\gamma }_{n+1}\right){\alpha }_{n+1}\parallel {x}_{n+1}-{x}_{n}\parallel +|{\alpha }_{n+1}{\gamma }_{n+1}-{\alpha }_{n}{\gamma }_{n}|\parallel {x}_{n}\parallel \\ \phantom{\rule{2em}{0ex}}+\left(1-{\alpha }_{n+1}\right)\parallel T\left(h\right)\left({s}_{n}\right){x}_{n}-T\left({s}_{n}\right){x}_{n}\parallel +|{\alpha }_{n+1}-{\alpha }_{n}|\parallel T\left({s}_{n}\right){x}_{n}\parallel .\end{array}$
(3.12)
Since $\left\{T\left(s\right):s\ge 0\right\}$ is uniformly asymptotically regular and ${lim}_{n\to \mathrm{\infty }}{s}_{n}=\mathrm{\infty }$, it follows that
$\underset{n\to \mathrm{\infty }}{lim}\parallel T\left(h\right)T\left({s}_{n}\right){x}_{n}-T\left({s}_{n}\right){x}_{n}\parallel \le \underset{n\to \mathrm{\infty }}{lim}\underset{x\in B}{sup}\parallel T\left(h\right)T\left({s}_{n}\right)x-T\left({s}_{n}\right)x\parallel =0,$
(3.13)
where B is any bounded set containing $\left\{{x}_{n}\right\}$. Moreover, since $\left\{{x}_{n}\right\}$, $\left\{T\left({s}_{n}\right){x}_{n}\right\}$ are bounded, and ${\alpha }_{n}\to 0$ as $n\to \mathrm{\infty }$, (3.12) implies that
$\underset{n\to \mathrm{\infty }}{lim sup}\left(\parallel {y}_{n+1}-{y}_{n}\parallel -\parallel {x}_{n+1}-{x}_{n}\parallel \right)\le 0.$

Hence, by Lemma 2.4 we have ${lim}_{n\to \mathrm{\infty }}\parallel {y}_{n}-{x}_{n}\parallel =0$ since ${x}_{n+1}-{x}_{n}={\beta }_{n}\left({y}_{n}-{x}_{n}\right)$. Consequently, ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n+1}-{x}_{n}\parallel =0$.

It follows from (3.11) that
$\begin{array}{rl}\parallel {x}_{n}-T\left({s}_{n}\right){x}_{n}\parallel \le & \parallel {x}_{n}-{x}_{n+1}\parallel +\parallel {x}_{n+1}-T\left({s}_{n}\right){x}_{n}\parallel \\ \le & \parallel {x}_{n}-{x}_{n+1}\parallel +\parallel \left(1-{\beta }_{n}\right)\left({x}_{n}-T\left({s}_{n}\right){x}_{n}\right)\\ +{\beta }_{n}\left({Q}_{K}\left[{\alpha }_{n}\left({\gamma }_{n}{x}_{n}\right)+\left(1-{\alpha }_{n}\right)T\left({s}_{n}\right){x}_{n}\right]-T\left({s}_{n}\right){x}_{n}\right)\parallel \\ \le & \parallel {x}_{n}-{x}_{n+1}\parallel +\left(1-{\beta }_{n}\right)\parallel {x}_{n}-T\left({s}_{n}\right){x}_{n}\parallel +{\alpha }_{n}{\gamma }_{n}\parallel {x}_{n}-T\left({s}_{n}\right){x}_{n}\parallel \\ +{\alpha }_{n}\left(1-{\gamma }_{n}\right)\parallel T\left({s}_{n}\right){x}_{n}\parallel \\ =& \parallel {x}_{n}-{x}_{n+1}\parallel +\left(1-{\beta }_{n}+{\alpha }_{n}{\gamma }_{n}\right)\parallel {x}_{n}-T\left({s}_{n}\right){x}_{n}\parallel +{\alpha }_{n}\left(1-{\gamma }_{n}\right)\parallel T\left({s}_{n}\right){x}_{n}\parallel .\end{array}$
So,
$\parallel {x}_{n}-T\left({s}_{n}\right){x}_{n}\parallel \le \frac{1}{{\beta }_{n}-{\alpha }_{n}{\gamma }_{n}}\left(\parallel {x}_{n}-{x}_{n+1}\parallel +{\alpha }_{n}\left(1-{\gamma }_{n}\right)\parallel T\left({s}_{n}\right){x}_{n}\parallel \right)\to 0.$
(3.14)
Since
$\begin{array}{r}\parallel {x}_{n}-T\left(h\right){x}_{n}\parallel \\ \phantom{\rule{1em}{0ex}}\le \parallel {x}_{n}-T\left({s}_{n}\right){x}_{n}\parallel +\parallel T\left({s}_{n}\right){x}_{n}-T\left(h\right)T\left({s}_{n}\right){x}_{n}\parallel +\parallel T\left(h\right)T\left({s}_{n}\right){x}_{n}-T\left(h\right){x}_{n}\parallel \\ \phantom{\rule{1em}{0ex}}\le 2\parallel {x}_{n}-T\left({s}_{n}\right){x}_{n}\parallel +\parallel T\left({s}_{n}\right){x}_{n}-T\left(h\right)T\left({s}_{n}\right){x}_{n}\parallel ,\end{array}$
from (3.13) and (3.14), we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-T\left(h\right){x}_{n}\parallel =0.$
(3.15)
Notice that $\left\{{x}_{n}\right\}$ is bounded. Put ${x}^{\ast }={Q}_{F\left(S\right)}\left(0\right)$. Then there exists a positive number R such that $B\left({x}^{\ast },R\right)\cap K$ contains $\left\{{x}_{n}\right\}$. Moreover, $B\left({x}^{\ast },R\right)\cap K$ is $T\left(s\right)$-invariant for all $s\ge 0$ and so, without loss of generality, we can assume that $\left\{T\left(s\right):s\ge 0\right\}$ is a nonexpansive semigroup on $B\left({x}^{\ast },R\right)\cap K$. We take a subsequence $\left\{{x}_{{n}_{k}}\right\}$ of $\left\{{x}_{n}\right\}$ such that
$\underset{n\to \mathrm{\infty }}{lim sup}〈-{x}^{\ast },j\left({x}_{n}-{x}^{\ast }\right)〉=\underset{k\to \mathrm{\infty }}{lim}〈{x}^{\ast },j\left({x}_{{n}_{k}}-{x}^{\ast }\right)〉.$
We may also assume that ${x}_{{n}_{k}}⇀\stackrel{˜}{x}$. It follows from Lemma 2.3 and (3.15) that $\stackrel{˜}{x}\in F\left(S\right)$ and hence
$〈-{x}^{\ast },j\left(\stackrel{˜}{x}-{x}^{\ast }\right)〉\le 0.$
Since j is weakly sequentially continuous, we have
$\underset{n\to \mathrm{\infty }}{lim sup}〈-{x}^{\ast },j\left({x}_{n}-{x}^{\ast }\right)〉=\underset{k\to \mathrm{\infty }}{lim}〈-{x}^{\ast },j\left({x}_{{n}_{k}}-{x}^{\ast }\right)〉=〈-{x}^{\ast },j\left(\stackrel{˜}{x}-{x}^{\ast }\right)〉\le 0.$
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
$\underset{n\to \mathrm{\infty }}{lim sup}〈-{x}^{\ast },j\left({y}_{n}-{x}^{\ast }\right)〉=\underset{n\to \mathrm{\infty }}{lim sup}〈-{x}^{\ast },j\left({x}_{n}-{x}^{\ast }\right)〉\le 0.$
Set ${u}_{n}={\alpha }_{n}\left({\gamma }_{n}{x}_{n}\right)+\left(1-{\alpha }_{n}\right)T\left({s}_{n}\right){x}_{n}$. It follows that ${y}_{n}={Q}_{K}{u}_{n}$ for all $n\ge 0$. By Proposition 2.1(3), we have
$〈{y}_{n}-{u}_{n},j\left({y}_{n}-{x}^{\ast }\right)〉\le 0,$
and so
$\begin{array}{rl}{\parallel {y}_{n}-{x}^{\ast }\parallel }^{2}=& 〈{y}_{n}-{x}^{\ast },j\left({y}_{n}-{x}^{\ast }\right)〉\\ =& 〈{y}_{n}-{u}_{n},j\left({y}_{n}-{x}^{\ast }\right)〉+〈{u}_{n}-{x}^{\ast },j\left({y}_{n}-{x}^{\ast }\right)〉\\ \le & 〈{u}_{n}-{x}^{\ast },j\left({y}_{n}-{x}^{\ast }\right)〉\\ =& {\alpha }_{n}{\gamma }_{n}〈{x}_{n}-{x}^{\ast },j\left({y}_{n}-{x}^{\ast }\right)〉-{\alpha }_{n}\left(1-{\gamma }_{n}\right)〈{x}^{\ast },j\left({y}_{n}-{x}^{\ast }\right)〉\\ +\left(1-{\alpha }_{n}\right)〈T\left({s}_{n}\right){x}_{n}-{x}^{\ast },j\left({y}_{n}-{x}^{\ast }\right)〉\\ \le & {\alpha }_{n}{\gamma }_{n}\parallel {x}_{n}-{x}^{\ast }\parallel \parallel j\left({y}_{n}-{x}^{\ast }\right)\parallel -{\alpha }_{n}\left(1-{\gamma }_{n}\right)〈{x}^{\ast },j\left({y}_{n}-{x}^{\ast }\right)〉\\ +\left(1-{\alpha }_{n}\right)\parallel T\left({s}_{n}\right){x}_{n}-{x}^{\ast }\parallel \parallel j\left({y}_{n}-{x}^{\ast }\right)\parallel \\ \le & \left[1-\left(1-{\gamma }_{n}\right){\alpha }_{n}\right]\parallel {x}_{n}-{x}^{\ast }\parallel \parallel {y}_{n}-{x}^{\ast }\parallel -{\alpha }_{n}\left(1-{\gamma }_{n}\right)〈{x}^{\ast },j\left({y}_{n}-{x}^{\ast }\right)〉\\ \le & \frac{1-\left(1-{\gamma }_{n}\right){\alpha }_{n}}{2}{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+\frac{1}{2}{\parallel {y}_{n}-{x}^{\ast }\parallel }^{2}-{\alpha }_{n}\left(1-{\gamma }_{n}\right)〈{x}^{\ast },j\left({y}_{n}-{x}^{\ast }\right)〉,\end{array}$
that is,
${\parallel {y}_{n}-{x}^{\ast }\parallel }^{2}\le \left[1-\left(1-{\gamma }_{n}\right){\alpha }_{n}\right]{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}-2{\alpha }_{n}\left(1-{\gamma }_{n}\right)〈{x}^{\ast },j\left({y}_{n}-{x}^{\ast }\right)〉.$
By the convexity of ${\parallel \cdot \parallel }^{2}$, we have
$\begin{array}{rl}{\parallel {x}_{n+1}-{x}^{\ast }\parallel }^{2}& \le \left(1-{\beta }_{n}\right){\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+{\beta }_{n}{\parallel {y}_{n}-{x}^{\ast }\parallel }^{2}\\ \le \left[1-\left(1-{\gamma }_{n}\right){\alpha }_{n}{\beta }_{n}\right]{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}-2\left(1-{\gamma }_{n}\right){\alpha }_{n}{\beta }_{n}〈{x}^{\ast },j\left({y}_{n}-{x}^{\ast }\right)〉.\end{array}$

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.  in the following aspects.
1. (1)

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

2. (2)

$\frac{1}{{\lambda }_{n}}{\int }_{0}^{{\lambda }_{n}}T\left(s\right){x}_{n}\phantom{\rule{0.2em}{0ex}}ds$ is replaced by $T\left({s}_{n}\right){x}_{n}$.

## Authors’ Affiliations

(1)
School of Mathematics and Statistics, Hubei Normal University, Huangshi, 435002, China

## References

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