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# Strong convergence to a fixed point of a total asymptotically nonexpansive mapping

Fixed Point Theory and Applications20132013:302

https://doi.org/10.1186/1687-1812-2013-302

• Received: 16 May 2013
• Accepted: 22 October 2013
• Published:

## Abstract

In this paper, we prove strong convergence for the modified Ishikawa iteration process of a total asymptotically nonexpansive mapping satisfying condition (A) in a real uniformly convex Banach space. Our result generalizes the results due to Rhoades (J. Math. Anal. Appl. 183:118-120, 1994).

MSC:47H05, 47H10.

## Keywords

• strong convergence
• fixed point
• modified Ishikawa iteration process
• total asymptotically nonexpansive mapping

## 1 Introduction

Let X be a real Banach space, let C be a nonempty closed convex subset of X, and let T be a mapping of C into itself. Then T is said to be asymptotically nonexpansive  if there exists a sequence $\left\{{k}_{n}\right\}$, ${k}_{n}\ge 1$, with ${lim}_{n\to \mathrm{\infty }}{k}_{n}=1$, such that
$\parallel {T}^{n}x-{T}^{n}y\parallel \le {k}_{n}\parallel x-y\parallel$
(1.1)
for all $x,y\in C$ and $n\ge 1$. T is said to be uniformly L-Lipschitzian if there exists a constant $L>0$ such that
$\parallel {T}^{n}x-{T}^{n}y\parallel \le L\parallel x-y\parallel$
for all $x,y\in C$ and $n\ge 1$. If T is asymptotically nonexpansive, then it is uniformly L-Lipschitzian. We denote by the set of all positive integers. T is said to be total asymptotically nonexpansive (in brief, TAN)  if there exist two nonnegative real sequences $\left\{{c}_{n}\right\}$ and $\left\{{d}_{n}\right\}$ with ${c}_{n},{d}_{n}\to 0$ as $n\to \mathrm{\infty }$, $\varphi \in \mathrm{\Gamma }\left({R}^{+}\right)$ such that
$\parallel {T}^{n}x-{T}^{n}y\parallel \le \parallel x-y\parallel +{c}_{n}\varphi \left(\parallel x-y\parallel \right)+{d}_{n},$
(1.2)
for all $x,y\in C$ and $n\ge 1$, where ${R}^{+}:=\left[0,\mathrm{\infty }\right)$ and $\varphi \in \mathrm{\Gamma }\left({R}^{+}\right)$ if and only if ϕ is strictly increasing, continuous on ${R}^{+}$ and $\varphi \left(0\right)=0$. It is clear that if we take $\varphi \left(t\right)=t$ for all $t\ge 0$ and ${d}_{n}=0$ for all $n\ge 1$ in (1.2), it is reduced to (1.1). Approximating fixed points of the modified Ishikawa iterative scheme under total asymptotically nonexpansive mappings has been investigated by several authors; see, for example, Chidume and Ofoedu [4, 5], Kim , Kim and Kim  and others. For a mapping T of C into itself in a Hilbert space, Schu  considered the following modified Ishikawa iteration process (cf. Ishikawa ) in C defined by
$\begin{array}{c}{x}_{1}\in C,\hfill \\ {x}_{n+1}=\left(1-{\alpha }_{n}\right){x}_{n}+{\alpha }_{n}{T}^{n}{y}_{n},\hfill \\ {y}_{n}=\left(1-{\beta }_{n}\right){x}_{n}+{\beta }_{n}{T}^{n}{x}_{n},\hfill \end{array}$
(1.3)
where $\left\{{\alpha }_{n}\right\}$ and $\left\{{\beta }_{n}\right\}$ are two real sequences in $\left[0,1\right]$. If ${\beta }_{n}=0$ for all $n\ge 1$, then iteration process (1.3) becomes the following modified Mann iteration process (cf. Mann ):
$\begin{array}{r}{x}_{1}\in C,\\ {x}_{n+1}=\left(1-{\alpha }_{n}\right){x}_{n}+{\alpha }_{n}{T}^{n}{x}_{n},\end{array}$
(1.4)

where $\left\{{\alpha }_{n}\right\}$ is a real sequence in $\left[0,1\right]$.

Rhoades  proved the following results which extended Theorems 1.5 and 2.3 of Schu  to uniformly convex Banach spaces.

Theorem 1.1 Let X be a uniformly convex Banach space, let C be a nonempty bounded closed convex subset of X, and let $T:C\to C$ be a completely continuous asymptotically nonexpansive mapping with $\left\{{k}_{n}\right\}$ satisfying ${k}_{n}\ge 1$, ${\sum }_{n=1}^{\mathrm{\infty }}\left({k}_{n}^{r}-1\right)<\mathrm{\infty }$, $r=max\left\{2,p\right\}$. Then, for any ${x}_{1}\in C$, the sequence $\left\{{x}_{n}\right\}$ defined by (1.4), where $\left\{{\alpha }_{n}\right\}$ satisfies $a\le {\alpha }_{n}\le 1-a$ for all $n\ge 1$ and some $a>0$, converges strongly to some fixed point of T.

Theorem 1.2 Let X be a uniformly convex Banach space, let C be a nonempty bounded closed convex subset of E, and let $T:C\to C$ be a completely continuous asymptotically nonexpansive mapping with $\left\{{k}_{n}\right\}$ satisfying ${k}_{n}\ge 1$, ${\sum }_{n=1}^{\mathrm{\infty }}\left({k}_{n}^{r}-1\right)<\mathrm{\infty }$, $r=max\left\{2,p\right\}$. Then, for any ${x}_{1}\in C$, the sequence $\left\{{x}_{n}\right\}$ defined by (1.3), where $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$ satisfy $a\le \left(1-{\alpha }_{n}\right),\left(1-{\beta }_{n}\right)\le 1-a$ for all $n\ge 1$ and some $a>0$, converges strongly to some fixed point of T.

On the other hand, Kim  proved the following result which generalized Theorem 1 of Senter and Dotson .

Theorem 1.3 Let X be a real uniformly convex Banach space, let C be a nonempty closed convex subset of X, and let T be a nonexpansive mapping of C into itself satisfying condition (A) with $F\left(T\right)\ne \mathrm{\varnothing }$. Suppose that for any ${x}_{1}$ in C, the sequence $\left\{{x}_{n}\right\}$ is defined by ${x}_{n+1}=\left(1-{\alpha }_{n}\right){x}_{n}+{\alpha }_{n}\left[{\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right)T{x}_{n}\right]$, for all $n\ge 1$, where $\left\{{\alpha }_{n}\right\}$ and $\left\{{\beta }_{n}\right\}$ are sequences in $\left[0,1\right]$ such that ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}\left(1-{\alpha }_{n}\right)=\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\beta }_{n}<\mathrm{\infty }$. Then $\left\{{x}_{n}\right\}$ converges strongly to some fixed point of T.

In this paper, we prove that if T is a total asymptotically nonexpansive self-mapping satisfying condition (A), the iteration $\left\{{x}_{n}\right\}$ defined by (1.3) converges strongly to some fixed point of T, which generalizes the results due to Rhoades .

## 2 Preliminaries

Throughout this paper, we denote by X a real Banach space. Let C be a nonempty closed convex subset of X, and let T be a mapping from C into itself. Then we denote by $F\left(T\right)$ the set of all fixed points of T, i.e., $F\left(T\right)=\left\{x\in C:Tx=x\right\}$. We also denote by $a\vee b:=max\left\{a,b\right\}$. A Banach space X is said to be uniformly convex if the modulus of convexity ${\delta }_{X}={\delta }_{X}\left(ϵ\right)$, $0<ϵ\le 2$, of X defined by
${\delta }_{X}\left(ϵ\right)=inf\left\{1-\frac{\parallel x+y\parallel }{2}:x,y\in X,\parallel x\parallel \le 1,\parallel y\parallel \le 1,\parallel x-y\parallel \ge ϵ\right\}$

satisfies the inequality ${\delta }_{X}\left(ϵ\right)>0$ for every $ϵ\in \left(0,2\right]$. When $\left\{{x}_{n}\right\}$ is a sequence in X, then ${x}_{n}\to x$ will denote strong convergence of the sequence $\left\{{x}_{n}\right\}$ to x.

Definition 2.1 

A mapping $T:C\to C$ with $F\left(T\right)\ne \mathrm{\varnothing }$ is said to satisfy condition (A) if there exists a nondecreasing function $f:\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ with $f\left(0\right)=0$ and $f\left(r\right)>0$ for all $r\in \left(0,\mathrm{\infty }\right)$ such that
$\parallel x-Tx\parallel \ge f\left(d\left(x,F\left(T\right)\right)\right)$

for all $x\in C$, where $d\left(x,F\left(T\right)\right)={inf}_{z\in F\left(T\right)}\parallel x-z\parallel$.

## 3 Strong convergence theorem

We first begin with the following lemma.

Lemma 3.1 

Let $\left\{{a}_{n}\right\}$, $\left\{{b}_{n}\right\}$ and $\left\{{c}_{n}\right\}$ be sequences of nonnegative real numbers such that ${\sum }_{n=1}^{\mathrm{\infty }}{b}_{n}<\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}{c}_{n}<\mathrm{\infty }$ and
${a}_{n+1}\le \left(1+{b}_{n}\right){a}_{n}+{c}_{n}$

for all $n\ge 1$. Then ${lim}_{n\to \mathrm{\infty }}{a}_{n}$ exists.

Lemma 3.2 

Let X be a uniformly convex Banach space. Let $x,y\in X$. If $\parallel x\parallel \le 1$, $\parallel y\parallel \le 1$ and $\parallel x-y\parallel \ge ϵ>0$, then $\parallel \lambda x+\left(1-\lambda \right)y\parallel \le 1-2\lambda \left(1-\lambda \right)\delta \left(ϵ\right)$ for $0\le \lambda \le 1$.

Lemma 3.3 Let C be a nonempty closed convex subset of a uniformly convex Banach space X, and let $T:C\to C$ be a TAN mapping with $F\left(T\right)\ne \mathrm{\varnothing }$. Suppose that $\left\{{c}_{n}\right\}$, $\left\{{d}_{n}\right\}$ and ϕ satisfy the following two conditions:
1. (I)

$\mathrm{\exists }\alpha ,\beta >0$ such that $\varphi \left(t\right)\le \alpha t$ for all $t\ge \beta$.

2. (II)

${\sum }_{n=1}^{\mathrm{\infty }}{c}_{n}<\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}{d}_{n}<\mathrm{\infty }$.

Suppose that the sequence $\left\{{x}_{n}\right\}$ is defined by (1.3). Then ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-z\parallel$ exists for any $z\in F\left(T\right)$.

Proof For any $z\in F\left(T\right)$, we set
$M:=1\vee \varphi \left(\beta \right)<\mathrm{\infty }.$
From (I) and strict increasing of ϕ, we obtain
$\varphi \left(t\right)\le \varphi \left(\beta \right)+\alpha t,\phantom{\rule{1em}{0ex}}t\ge 0.$
(3.1)
By using (3.1), we have
$\begin{array}{rcl}\parallel {T}^{n}{x}_{n}-z\parallel & \le & \parallel {x}_{n}-z\parallel +{c}_{n}\varphi \left(\parallel {x}_{n}-z\parallel \right)+{d}_{n}\\ \le & \parallel {x}_{n}-z\parallel +{c}_{n}\left\{\varphi \left(\beta \right)+\alpha \parallel {x}_{n}-z\parallel \right\}+{d}_{n}\\ \le & \left(1+\alpha {c}_{n}\right)\parallel {x}_{n}-z\parallel +{\kappa }_{n}M,\end{array}$
where ${\kappa }_{n}={c}_{n}+{d}_{n}$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\kappa }_{n}<\mathrm{\infty }$. Since
$\begin{array}{rcl}\parallel {y}_{n}-z\parallel & =& \parallel {\beta }_{n}{T}^{n}{x}_{n}+\left(1-{\beta }_{n}\right){x}_{n}-z\parallel \\ \le & {\beta }_{n}\parallel {T}^{n}{x}_{n}-z\parallel +\left(1-{\beta }_{n}\right)\parallel {x}_{n}-z\parallel \\ \le & {\beta }_{n}\left\{\left(1+\alpha {c}_{n}\right)\parallel {x}_{n}-z\parallel +{\kappa }_{n}M\right\}+\left(1-{\beta }_{n}\right)\parallel {x}_{n}-z\parallel \\ \le & \left(1+\alpha {c}_{n}\right)\parallel {x}_{n}-z\parallel +{\kappa }_{n}M,\end{array}$
and thus
$\begin{array}{c}\parallel {y}_{n}-z\parallel +{c}_{n}\varphi \left(\parallel {y}_{n}-z\parallel \right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \left(1+\alpha {c}_{n}\right)\parallel {x}_{n}-z\parallel +{\kappa }_{n}M+{c}_{n}\left\{\varphi \left(\beta \right)+\alpha \parallel {y}_{n}-z\parallel \right\}\hfill \\ \phantom{\rule{1em}{0ex}}\le \left(1+\alpha {c}_{n}\right)\parallel {x}_{n}-z\parallel +{\kappa }_{n}M+{c}_{n}\varphi \left(\beta \right)+\alpha {c}_{n}\left(1+\alpha {c}_{n}\right)\parallel {x}_{n}-z\parallel +\alpha {c}_{n}{\kappa }_{n}M\hfill \\ \phantom{\rule{1em}{0ex}}\le \left(1+{\sigma }_{n}\right)\parallel {x}_{n}-z\parallel +{\delta }_{n}M,\hfill \end{array}$
where ${\sigma }_{n}=2\alpha {c}_{n}+{\alpha }^{2}{c}_{n}^{2}$, ${\delta }_{n}={\kappa }_{n}+{c}_{n}+\alpha {c}_{n}{\kappa }_{n}$, ${\sum }_{n=1}^{\mathrm{\infty }}{\sigma }_{n}<\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\delta }_{n}<\mathrm{\infty }$. So, we have
$\begin{array}{rcl}\parallel {T}^{n}{y}_{n}-z\parallel & \le & \parallel {y}_{n}-z\parallel +{c}_{n}\varphi \left(\parallel {y}_{n}-z\parallel \right)+{d}_{n}\\ \le & \left(1+{\sigma }_{n}\right)\parallel {x}_{n}-z\parallel +{\delta }_{n}M+{d}_{n}\\ \le & \left(1+{\sigma }_{n}\right)\parallel {x}_{n}-z\parallel +{\eta }_{n}M,\end{array}$
where ${\eta }_{n}={\delta }_{n}+{d}_{n}$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\eta }_{n}<\mathrm{\infty }$. Hence
$\begin{array}{rcl}\parallel {x}_{n+1}-z\parallel & =& \parallel \left(1-{\alpha }_{n}\right){x}_{n}+{\alpha }_{n}{T}^{n}{y}_{n}-z\parallel \\ \le & \left(1-{\alpha }_{n}\right)\parallel {x}_{n}-z\parallel +{\alpha }_{n}\parallel {T}^{n}{y}_{n}-z\parallel \\ \le & \left(1-{\alpha }_{n}\right)\parallel {x}_{n}-z\parallel +{\alpha }_{n}\left\{\left(1+{\sigma }_{n}\right)\parallel {x}_{n}-z\parallel +{\eta }_{n}M\right\}\\ \le & \left(1+{\sigma }_{n}\right)\parallel {x}_{n}-z\parallel +{\eta }_{n}M.\end{array}$

By Lemma 3.1, we see that ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-z\parallel$ exists. □

Theorem 3.4 Let X be a uniformly convex Banach space, and let C be a nonempty closed convex subset of X. Let $T:C\to C$ be a uniformly continuous and TAN mapping with $F\left(T\right)\ne \mathrm{\varnothing }$. Suppose that $\left\{{c}_{n}\right\}$, $\left\{{d}_{n}\right\}$ and ϕ satisfy the following two conditions:
1. (I)

$\mathrm{\exists }\alpha ,\beta >0$ such that $\varphi \left(t\right)\le \alpha t$ for all $t\ge \beta$.

2. (II)

${\sum }_{n=1}^{\mathrm{\infty }}{c}_{n}<\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}{d}_{n}<\mathrm{\infty }$.

Suppose that for any ${x}_{1}$ in C, the sequence $\left\{{x}_{n}\right\}$ defined by (1.3) satisfies ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}\left(1-{\alpha }_{n}\right)=\mathrm{\infty }$ and $lim{\beta }_{n}=0$. Then $\left\{{x}_{n}\right\}$ converges strongly to some fixed point of T.

Proof For any $z\in F\left(T\right)$, by Lemma 3.3, $\left\{{x}_{n}\right\}$ is bounded. We set
$M:=1\vee \varphi \left(\beta \right)\vee \underset{n\ge 1}{sup}\parallel {x}_{n}-z\parallel <\mathrm{\infty }.$
By Lemma 3.3, we see that ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-z\parallel \phantom{\rule{0.25em}{0ex}}\left(\equiv r\right)$ exists. Without loss of generality, we assume $r>0$. As in the proof of Lemma 3.3, we obtain
$\begin{array}{rcl}\parallel {T}^{n}{y}_{n}-z\parallel & \le & \left(1+{\sigma }_{n}\right)\parallel {x}_{n}-z\parallel +{\eta }_{n}M\\ \le & \parallel {x}_{n}-z\parallel +{\nu }_{n}M,\end{array}$
where ${\nu }_{n}={\sigma }_{n}+{\eta }_{n}$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\nu }_{n}<\mathrm{\infty }$. By using Lemma 3.2 and Takahashi , we obtain
$\begin{array}{rcl}\parallel {x}_{n+1}-z\parallel & =& \parallel \left(1-{\alpha }_{n}\right){x}_{n}+{\alpha }_{n}{T}^{n}{y}_{n}-z\parallel \\ =& \parallel \left(1-{\alpha }_{n}\right)\left({x}_{n}-z\right)+{\alpha }_{n}\left({T}^{n}{y}_{n}-z\right)\parallel \\ \le & \left(\parallel {x}_{n}-z\parallel +{\nu }_{n}M\right)\left[1-2{\alpha }_{n}\left(1-{\alpha }_{n}\right){\delta }_{X}\left(\frac{\parallel {T}^{n}{y}_{n}-{x}_{n}\parallel }{\parallel {x}_{n}-z\parallel +{\nu }_{n}M}\right)\right].\end{array}$
Hence we obtain
$\begin{array}{c}2{\alpha }_{n}\left(1-{\alpha }_{n}\right)\left(\parallel {x}_{n}-z\parallel +{\nu }_{n}M\right){\delta }_{X}\left(\frac{\parallel {T}^{n}{y}_{n}-{x}_{n}\parallel }{\parallel {x}_{n}-z\parallel +{\nu }_{n}M}\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \parallel {x}_{n}-z\parallel -\parallel {x}_{n+1}-z\parallel +{\nu }_{n}M.\hfill \end{array}$
Thus
$2{\alpha }_{n}\left(1-{\alpha }_{n}\right)\left(\parallel {x}_{n}-z\parallel +{\nu }_{n}M\right){\delta }_{X}\left(\frac{\parallel {T}^{n}{y}_{n}-{x}_{n}\parallel }{\parallel {x}_{n}-z\parallel +{\nu }_{n}M}\right)<\mathrm{\infty }.$
Since ${\delta }_{X}$ is strictly increasing, continuous and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}\left(1-{\alpha }_{n}\right)=\mathrm{\infty }$, we obtain
$\underset{n\to \mathrm{\infty }}{lim inf}\parallel {T}^{n}{y}_{n}-{x}_{n}\parallel =0.$
(3.2)
By using (3.1) in the proof of Lemma 3.3, we have
$\begin{array}{rcl}\parallel {T}^{n-1}{x}_{n-1}-z\parallel & \le & \parallel {x}_{n-1}-z\parallel +{c}_{n-1}\varphi \left(\parallel {x}_{n-1}-z\parallel \right)+{d}_{n-1}\\ \le & \parallel {x}_{n-1}-z\parallel +{c}_{n-1}\left\{\varphi \left(\beta \right)+\alpha \parallel {x}_{n-1}-z\parallel \right\}+{d}_{n-1}\\ \le & \left(1+\alpha {c}_{n-1}\right)\parallel {x}_{n-1}-z\parallel +{\rho }_{n-1}M,\end{array}$
where ${\rho }_{n-1}={c}_{n-1}+{d}_{n-1}$ and ${\sum }_{n=2}^{\mathrm{\infty }}{\rho }_{n-1}<\mathrm{\infty }$. Thus
$\begin{array}{rcl}\parallel {y}_{n-1}-z\parallel & =& \parallel {\beta }_{n-1}{T}^{n-1}{x}_{n-1}+\left(1-{\beta }_{n-1}\right){x}_{n-1}-z\parallel \\ \le & {\beta }_{n-1}\parallel {T}^{n-1}{x}_{n-1}-z\parallel +\left(1-{\beta }_{n-1}\right)\parallel {x}_{n-1}-z\parallel \\ \le & {\beta }_{n-1}\left\{\left(1+\alpha {c}_{n-1}\right)\parallel {x}_{n-1}-z\parallel +{\rho }_{n-1}M\right\}+\left(1-{\beta }_{n-1}\right)\parallel {x}_{n-1}-z\parallel \\ \le & \left(1+\alpha {c}_{n-1}\right)\parallel {x}_{n-1}-z\parallel +{\rho }_{n-1}M,\end{array}$
and hence
$\begin{array}{c}\parallel {y}_{n-1}-z\parallel +{c}_{n-1}\varphi \left(\parallel {y}_{n-1}-z\parallel \right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \left(1+\alpha {c}_{n-1}\right)\parallel {x}_{n-1}-z\parallel +{\rho }_{n-1}M+{c}_{n-1}\left\{\varphi \left(\beta \right)+\alpha \parallel {y}_{n-1}-z\parallel \right\}\hfill \\ \phantom{\rule{1em}{0ex}}\le \left(1+\alpha {c}_{n-1}\right)\parallel {x}_{n-1}-z\parallel +{\rho }_{n-1}M+{c}_{n-1}\varphi \left(\beta \right)+\alpha {c}_{n-1}\left(1+\alpha {c}_{n-1}\right)\parallel {x}_{n-1}-z\parallel \hfill \\ \phantom{\rule{2em}{0ex}}+\alpha {c}_{n-1}{\rho }_{n-1}M\hfill \\ \phantom{\rule{1em}{0ex}}\le \left(1+{\mu }_{n-1}\right)\parallel {x}_{n-1}-z\parallel +{\phi }_{n-1}M,\hfill \end{array}$
where ${\mu }_{n-1}=2\alpha {c}_{n-1}+{\alpha }^{2}{c}_{n-1}^{2}$, ${\phi }_{n-1}={\rho }_{n-1}+{c}_{n-1}+\alpha {c}_{n-1}{\rho }_{n-1}$, ${\sum }_{n=2}^{\mathrm{\infty }}{\mu }_{n-1}<\mathrm{\infty }$ and ${\sum }_{n=2}^{\mathrm{\infty }}{\phi }_{n-1}<\mathrm{\infty }$. So, we have
$\begin{array}{rcl}\parallel {T}^{n-1}{y}_{n-1}-z\parallel & \le & \parallel {y}_{n-1}-z\parallel +{c}_{n-1}\varphi \left(\parallel {y}_{n-1}-z\parallel \right)+{d}_{n-1}\\ \le & \left(1+{\mu }_{n-1}\right)\parallel {x}_{n-1}-z\parallel +{\phi }_{n-1}M+{d}_{n-1}\\ \le & \parallel {x}_{n-1}-z\parallel +{\omega }_{n-1}M,\end{array}$
where ${\omega }_{n-1}={\mu }_{n-1}+{\phi }_{n-1}+{d}_{n-1}$ and ${\sum }_{n=2}^{\mathrm{\infty }}{\omega }_{n-1}<\mathrm{\infty }$. By using Lemma 3.2 and Takahashi , we obtain
$\begin{array}{rcl}\parallel {x}_{n}-z\parallel & =& \parallel \left(1-{\alpha }_{n-1}\right){x}_{n-1}+{\alpha }_{n-1}{T}^{n-1}{y}_{n-1}-z\parallel \\ =& \parallel \left(1-{\alpha }_{n-1}\right)\left({x}_{n-1}-z\right)+{\alpha }_{n-1}\left({T}^{n-1}{y}_{n-1}-z\right)\parallel \\ \le & \left(\parallel {x}_{n-1}-z\parallel +{\omega }_{n-1}M\right)\left[1-2{\alpha }_{n}\left(1-{\alpha }_{n}\right){\delta }_{X}\left(\frac{\parallel {T}^{n-1}{y}_{n-1}-{x}_{n-1}\parallel }{\parallel {x}_{n-1}-z\parallel +{\omega }_{n-1}M}\right)\right].\end{array}$
By the same method as above, we obtain
$\underset{n\to \mathrm{\infty }}{lim inf}\parallel {T}^{n-1}{y}_{n-1}-{x}_{n-1}\parallel =0.$
(3.3)
Since $\left\{{x}_{n}\right\}$ is bounded and T is a TAN mapping, we obtain
$\begin{array}{rcl}\parallel {y}_{n}-{x}_{n}\parallel & =& \parallel {\beta }_{n}{T}^{n}{x}_{n}+\left(1-{\beta }_{n}\right){x}_{n}-{x}_{n}\parallel \\ \le & {\beta }_{n}\parallel {T}^{n}{x}_{n}-{x}_{n}\parallel \\ \le & {\beta }_{n}{M}^{\prime },\end{array}$
where ${M}^{\prime }={sup}_{n\ge 1}\parallel {T}^{n}{x}_{n}-{x}_{n}\parallel <\mathrm{\infty }$. By using $lim{\beta }_{n}=0$, we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-{y}_{n}\parallel =0.$
(3.4)
Since
$\parallel {T}^{n}{y}_{n}-{y}_{n}\parallel \le \parallel {T}^{n}{y}_{n}-{x}_{n}\parallel +\parallel {x}_{n}-{y}_{n}\parallel ,$
by (3.2) and (3.4), we obtain
$\underset{n\to \mathrm{\infty }}{lim inf}\parallel {T}^{n}{y}_{n}-{y}_{n}\parallel =0.$
(3.5)
By using (3.3) and (3.4), we obtain
$\underset{n\to \mathrm{\infty }}{lim inf}\parallel {T}^{n-1}{y}_{n-1}-{y}_{n-1}\parallel =0.$
(3.6)
Since
$\begin{array}{rcl}\parallel {T}^{n-1}{x}_{n-1}-{x}_{n-1}\parallel & \le & \parallel {T}^{n-1}{x}_{n-1}-{T}^{n-1}{y}_{n-1}\parallel +\parallel {T}^{n-1}{y}_{n-1}-{x}_{n-1}\parallel \\ \le & \parallel {x}_{n-1}-{y}_{n-1}\parallel +{c}_{n-1}\varphi \left(\parallel {x}_{n-1}-{y}_{n-1}\parallel \right)+{d}_{n-1}\\ +\parallel {T}^{n-1}{y}_{n-1}-{x}_{n-1}\parallel ,\end{array}$
by using (3.3) and (3.4), we have
$\underset{n\to \mathrm{\infty }}{lim inf}\parallel {T}^{n-1}{x}_{n-1}-{x}_{n-1}\parallel =0.$
(3.7)
Since
$\begin{array}{rcl}\parallel {x}_{n}-{x}_{n-1}\parallel & =& \parallel \left(1-{\alpha }_{n-1}\right){x}_{n-1}+{\alpha }_{n-1}{T}^{n-1}{y}_{n-1}-{x}_{n-1}\parallel \\ =& {\alpha }_{n-1}\parallel {T}^{n-1}{y}_{n-1}-{x}_{n-1}\parallel \\ \le & \parallel {T}^{n-1}{y}_{n-1}-{y}_{n-1}\parallel +\parallel {y}_{n-1}-{x}_{n-1}\parallel ,\end{array}$
by (3.4) and (3.6), we get
$\underset{n\to \mathrm{\infty }}{lim inf}\parallel {x}_{n}-{x}_{n-1}\parallel =0.$
(3.8)
From
$\begin{array}{rcl}\parallel {T}^{n-1}{x}_{n}-{x}_{n}\parallel & \le & \parallel {T}^{n-1}{x}_{n}-{T}^{n-1}{x}_{n-1}\parallel +\parallel {T}^{n-1}{x}_{n-1}-{x}_{n-1}\parallel +\parallel {x}_{n-1}-{x}_{n}\parallel \\ \le & 2\parallel {x}_{n}-{x}_{n-1}\parallel +{c}_{n-1}\varphi \left(\parallel {x}_{n}-{x}_{n-1}\parallel \right)+{d}_{n-1}+\parallel {T}^{n-1}{x}_{n-1}-{x}_{n-1}\parallel ,\end{array}$
by (3.7) and (3.8), we obtain
$\underset{n\to \mathrm{\infty }}{lim inf}\parallel {T}^{n-1}{x}_{n}-{x}_{n}\parallel =0.$
(3.9)
Since
$\begin{array}{c}\parallel {x}_{n}-T{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}\le \parallel {x}_{n}-{y}_{n}\parallel +\parallel {y}_{n}-{T}^{n}{y}_{n}\parallel +\parallel {T}^{n}{y}_{n}-{T}^{n}{x}_{n}\parallel +\parallel {T}^{n}{x}_{n}-T{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}\le \parallel {y}_{n}-{T}^{n}{y}_{n}\parallel +2\parallel {x}_{n}-{y}_{n}\parallel +{c}_{n}\varphi \left(\parallel {x}_{n}-{y}_{n}\parallel \right)+{d}_{n}+\parallel {T}^{n}{x}_{n}-T{x}_{n}\parallel \hfill \end{array}$
and by the uniform continuity of T, (3.4), (3.5) and (3.9), we have
$\underset{n\to \mathrm{\infty }}{lim inf}\parallel {x}_{n}-T{x}_{n}\parallel =0.$
(3.10)
By using condition (A), we obtain
$f\left(d\left({x}_{n},F\left(T\right)\right)\right)\le \parallel {x}_{n}-T{x}_{n}\parallel$
(3.11)
for all $n\ge 1$. As in the proof of Lemma 3.3, we obtain
$\parallel {x}_{n+1}-z\parallel \le \left(1+{\sigma }_{n}\right)\parallel {x}_{n}-z\parallel +{\eta }_{n}M.$
(3.12)
Thus
$\underset{z\in F\left(T\right)}{inf}\parallel {x}_{n+1}-z\parallel \le \left(1+{\sigma }_{n}\right)\underset{z\in F\left(T\right)}{inf}\parallel {x}_{n}-z\parallel +{\eta }_{n}M.$
By using Lemma 3.1, we see that ${lim}_{n\to \mathrm{\infty }}d\left({x}_{n},F\left(T\right)\right)\phantom{\rule{0.25em}{0ex}}\left(\equiv c\right)$ exists. We first claim that ${lim}_{n\to \mathrm{\infty }}d\left({x}_{n},F\left(T\right)\right)=0$. In fact, assume that $c={lim}_{n\to \mathrm{\infty }}d\left({x}_{n},F\left(T\right)\right)>0$. Then we can choose ${n}_{0}\in \mathbb{N}$ such that $0<\frac{c}{2} for all $n\ge {n}_{0}$. By using condition (A), (3.10) and (3.11), we obtain
$0
as $i\to \mathrm{\infty }$. This is a contradiction. So, we obtain $c=0$. Next, we claim that $\left\{{x}_{n}\right\}$ is a Cauchy sequence. Since ${\sum }_{n=1}^{\mathrm{\infty }}{\sigma }_{n}<\mathrm{\infty }$, we obtain ${\prod }_{n=1}^{\mathrm{\infty }}\left(1+{\sigma }_{n}\right):=U<\mathrm{\infty }$. Let $ϵ>0$ be given. Since ${lim}_{n\to \mathrm{\infty }}d\left({x}_{n},F\left(T\right)\right)=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\eta }_{n}<\mathrm{\infty }$, there exists ${n}_{0}\in \mathbb{N}$ such that for all $n\ge {n}_{0}$, we obtain
$d\left({x}_{n},F\left(T\right)\right)<\frac{ϵ}{4U+4}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\sum _{i={n}_{0}}^{\mathrm{\infty }}{\eta }_{i}<\frac{ϵ}{4M}.$
(3.13)
Let $n,m\ge {n}_{0}$ and $p\in F\left(T\right)$. Then, by (3.12), we obtain
$\begin{array}{rcl}\parallel {x}_{n}-{x}_{m}\parallel & \le & \parallel {x}_{n}-p\parallel +\parallel {x}_{m}-p\parallel \\ \le & \prod _{i={n}_{0}}^{n-1}\left(1+{\sigma }_{i}\right)\parallel {x}_{{n}_{0}}-p\parallel +M\sum _{i={n}_{0}}^{n-1}{\eta }_{i}+\prod _{i={n}_{0}}^{m-1}\left(1+{\sigma }_{i}\right)\parallel {x}_{{n}_{0}}-p\parallel +M\sum _{i={n}_{0}}^{m-1}{\eta }_{i}\\ \le & 2\left[\prod _{i={n}_{0}}^{\mathrm{\infty }}\left(1+{\sigma }_{i}\right)\parallel {x}_{{n}_{0}}-p\parallel +M\sum _{i={n}_{0}}^{\mathrm{\infty }}{\eta }_{i}\right].\end{array}$
Taking the infimum over all $p\in F\left(T\right)$ on both sides and by (3.13), we obtain
$\begin{array}{rcl}\parallel {x}_{n}-{x}_{m}\parallel & \le & 2\left[\prod _{i={n}_{0}}^{\mathrm{\infty }}\left(1+{\sigma }_{i}\right)d\left({x}_{{n}_{0}},F\left(T\right)\right)+M\sum _{i={n}_{0}}^{\mathrm{\infty }}{\eta }_{i}\right]\\ <& 2\left[\left(U+1\right)\frac{ϵ}{4U+4}+M\frac{ϵ}{4M}\right]=ϵ\end{array}$

for all $n,m\ge {n}_{0}$. This implies that $\left\{{x}_{n}\right\}$ is a Cauchy sequence. Let ${lim}_{n\to \mathrm{\infty }}{x}_{n}=q$. Then $d\left(q,F\left(T\right)\right)=0$. Since $F\left(T\right)$ is closed, we obtain $q\in F\left(T\right)$. Hence $\left\{{x}_{n}\right\}$ converges strongly to some fixed point of T. □

Remark 3.5 If $T:C\to C$ is completely continuous, then it satisfies demicompact and, if T is continuous and demicompact, it satisfies condition (A); see Senter and Dotson .

Remark 3.6 If $\left\{{\alpha }_{n}\right\}$ is bounded away from both 0 and 1, i.e., $a\le {\alpha }_{n}\le b$ for all $n\ge 1$ and some $a,b\in \left(0,1\right)$, then ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}\left(1-{\alpha }_{n}\right)=\mathrm{\infty }$ and ${lim}_{n\to \mathrm{\infty }}{\beta }_{n}=0$ hold. However, the converse is not true. For example, consider ${\alpha }_{n}=\frac{1}{n}$.

We give an example of a mapping $T:C\to C$ which satisfies all the assumptions of T in Theorem 3.4, i.e., $T:C\to C$ is a uniformly continuous mapping with $F\left(T\right)\ne \mathrm{\varnothing }$ which is TAN on C, not Lipschitzian and hence not asymptotically nonexpansive.

Example 3.7 Let $X:=\mathbb{R}$ and $C:=\left[0,2\right]$. Define $T:C\to C$ by
$Tx=\left\{\begin{array}{cc}1,\hfill & x\in \left[0,1\right];\hfill \\ \frac{1}{\sqrt{3}}\sqrt{4-{x}^{2}},\hfill & x\in \left[1,2\right].\hfill \end{array}$
Note that ${T}^{n}x=1$ for all $x\in C$ and $n\ge 2$ and $F\left(T\right)=\left\{1\right\}$. Clearly, T is both uniformly continuous and TAN on C. We show that T satisfies condition (A). In fact, if $x\in \left[0,1\right]$, then $|x-1|=|x-Tx|$. Similarly, if $x\in \left[1,2\right]$, then
$|x-1|=x-1\le x-\frac{1}{\sqrt{3}}\sqrt{4-{x}^{2}}=|x-Tx|.$
So, we get $d\left(x,F\left(T\right)\right)=|x-1|\le |x-Tx|$ for all $x\in C$. But T is not Lipschitzian. Indeed, suppose not, i.e., there exists $L>0$ such that
$|Tx-Ty|\le L|x-y|$
for all $x,y\in C$. If we take $x=2-\frac{1}{3{\left(L+1\right)}^{2}}>1$ and $y=2$, then
$\frac{1}{\sqrt{3}}\sqrt{4-{x}^{2}}\le L\left(2-x\right)\phantom{\rule{1em}{0ex}}⇔\phantom{\rule{1em}{0ex}}\frac{1}{3{L}^{2}}\le \frac{2-x}{2+x}=\frac{1}{12{L}^{2}+24L+1}.$

This is a contradiction.

## Declarations

### Acknowledgements

The author would like to express their sincere appreciation to the anonymous referees for useful suggestions which improved the contents of this manuscript.

## Authors’ Affiliations

(1)
Department of Applied Mathematics, Pukyong National University, Busan, 608-737, Korea

## References

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