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

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.

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 [2] if there exists a sequence $\{k_n\}$, $k_n\ge 1$, 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 ℕ the set of all positive integers. T is said to be total asymptotically nonexpansive (in brief, TAN) [3] if there exist two nonnegative real sequences $\{c_n\}$ and $\{d_n\}$ with $c_n,d_n\to 0$ as $n\to \mathrm{\infty }$, $\varphi \in \mathrm{\Gamma }(R^{+})$ such that

$$\parallel T^{n}x-T^{n}y\parallel \le \parallel x-y\parallel +c_n\varphi (\parallel x-y\parallel )+d_n,$$

(1.2)

for all $x,y\in C$ and $n\ge 1$, where $R^{+}:=[0,\mathrm{\infty })$ and $\varphi \in \mathrm{\Gamma }(R^{+})$ if and only if ϕ is strictly increasing, continuous on $R^{+}$ and $\varphi (0)=0$. It is clear that if we take $\varphi (t)=t$ 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 [6], Kim and Kim [7] and others. For a mapping T of C into itself in a Hilbert space, Schu [8] considered the following modified Ishikawa iteration process (cf. Ishikawa [9]) in C defined by

$$\begin{array}{c}{x}_1\in C,\hfill \\ x_{n+1}=(1-{\alpha }_n)x_n+{\alpha }_n{T}^n{y}_n,\hfill \\ y_n=(1-{\beta }_n)x_n+{\beta }_n{T}^n{x}_n,\hfill \end{array}$$

(1.3)

where $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ are two real sequences in $[0,1]$. If ${\beta }_n=0$ for all $n\ge 1$, then iteration process (1.3) becomes the following modified Mann iteration process (cf. Mann [10]):

$$\begin{array}{r}{x}_1\in C,\\ x_{n+1}=(1-{\alpha }_n)x_n+{\alpha }_n{T}^n{x}_n,\end{array}$$

(1.4)

where $\{{\alpha }_n\}$ is a real sequence in $[0,1]$.

Rhoades [1] proved the following results which extended Theorems 1.5 and 2.3 of Schu [8] 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 $\{k_n\}$ satisfying $k_n\ge 1$, ${\sum }_{n=1}^{\mathrm{\infty }}(k_n^r-1)<\mathrm{\infty }$, $r=max\{2,p\}$. Then, for any $x_1\in C$, the sequence $\{x_n\}$ defined by (1.4), where $\{{\alpha }_n\}$ 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 $\{k_n\}$ satisfying $k_n\ge 1$, ${\sum }_{n=1}^{\mathrm{\infty }}(k_n^r-1)<\mathrm{\infty }$, $r=max\{2,p\}$. Then, for any $x_1\in C$, the sequence $\{x_n\}$ defined by (1.3), where $\{{\alpha }_n\}$, $\{{\beta }_n\}$ satisfy $a\le (1-{\alpha }_n),(1-{\beta }_n)\le 1-a$ for all $n\ge 1$ and some $a>0$, converges strongly to some fixed point of T.

On the other hand, Kim [11] proved the following result which generalized Theorem 1 of Senter and Dotson [12].

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 $F(T)\ne \mathrm{\varnothing }$. Suppose that for any $x_1$ in C, the sequence $\{x_n\}$ is defined by $x_{n+1}=(1-{\alpha }_n)x_n+{\alpha }_n[{\beta }_n{x}_n+(1-{\beta }_n)T{x}_n]$, for all $n\ge 1$, where $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ are sequences in $[0,1]$ such that ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)=\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\beta }_n<\mathrm{\infty }$. Then $\{x_n\}$ 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 $\{x_n\}$ defined by (1.3) converges strongly to some fixed point of T, which generalizes the results due to Rhoades [1].

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 $F(T)$ the set of all fixed points of T, i.e., $F(T)=\{x\in C:Tx=x\}$. We also denote by $a\vee b:=max\{a,b\}$. A Banach space X is said to be uniformly convex if the modulus of convexity ${\delta }_X={\delta }_X(ϵ)$, $0<ϵ\le 2$, of X defined by

$${\delta }_X(ϵ)=inf\{1-\frac{\parallel x+y\parallel }{2}:x,y\in X,\parallel x\parallel \le 1,\parallel y\parallel \le 1,\parallel x-y\parallel \ge ϵ\}$$

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

Definition 2.1 [12]

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

$$\parallel x-Tx\parallel \ge f\left(d(x,F(T))\right)$$

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

3 Strong convergence theorem

We first begin with the following lemma.

Lemma 3.1 [13]

Let $\{a_n\}$, $\{b_n\}$ and $\{c_n\}$ 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

$$a_{n+1}\le (1+b_n)a_n+c_n$$

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

Lemma 3.2 [14]

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 $\parallel \lambda x+(1-\lambda )y\parallel \le 1-2\lambda (1-\lambda )\delta (ϵ)$ 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 $F(T)\ne \mathrm{\varnothing }$. Suppose that $\{c_n\}$, $\{d_n\}$ and ϕ satisfy the following two conditions:

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

• (II) ${\sum }_{n=1}^{\mathrm{\infty }}{c}_n<\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}{d}_n<\mathrm{\infty }$.

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

Proof For any $z\in F(T)$, we set

$$M:=1\vee \varphi (\beta )<\mathrm{\infty }.$$

From (I) and strict increasing of ϕ, we obtain

$$\varphi (t)\le \varphi (\beta )+\alpha t,t\ge 0.$$

(3.1)

By using (3.1), we have

$$\begin{array}{rcl}\parallel T^n{x}_n-z\parallel & \le & \parallel x_n-z\parallel +c_n\varphi (\parallel x_n-z\parallel )+d_n\\ \le & \parallel x_n-z\parallel +c_n\{\varphi (\beta )+\alpha \parallel x_n-z\parallel \}+d_n\\ \le & (1+\alpha c_n)\parallel x_n-z\parallel +{\kappa }_{n}M,\end{array}$$

where ${\kappa }_n=c_n+d_n$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\kappa }_n<\mathrm{\infty }$. Since

$$\begin{array}{rcl}\parallel y_n-z\parallel & =& \parallel {\beta }_n{T}^n{x}_n+(1-{\beta }_n)x_n-z\parallel \\ \le & {\beta }_n\parallel T^n{x}_n-z\parallel +(1-{\beta }_n)\parallel x_n-z\parallel \\ \le & {\beta }_n\{(1+\alpha c_n)\parallel x_n-z\parallel +{\kappa }_{n}M\}+(1-{\beta }_n)\parallel x_n-z\parallel \\ \le & (1+\alpha c_n)\parallel x_n-z\parallel +{\kappa }_{n}M,\end{array}$$

and thus

$$\begin{array}{c}\parallel y_n-z\parallel +c_n\varphi (\parallel y_n-z\parallel )\hfill \\ \le (1+\alpha c_n)\parallel x_n-z\parallel +{\kappa }_{n}M+c_n\{\varphi (\beta )+\alpha \parallel y_n-z\parallel \}\hfill \\ \le (1+\alpha c_n)\parallel x_n-z\parallel +{\kappa }_{n}M+c_n\varphi (\beta )+\alpha c_n(1+\alpha c_n)\parallel x_n-z\parallel +\alpha c_n{\kappa }_{n}M\hfill \\ \le (1+{\sigma }_n)\parallel x_n-z\parallel +{\delta }_{n}M,\hfill \end{array}$$

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

$$\begin{array}{rcl}\parallel T^n{y}_n-z\parallel & \le & \parallel y_n-z\parallel +c_n\varphi (\parallel y_n-z\parallel )+d_n\\ \le & (1+{\sigma }_n)\parallel x_n-z\parallel +{\delta }_{n}M+d_n\\ \le & (1+{\sigma }_n)\parallel x_n-z\parallel +{\eta }_{n}M,\end{array}$$

where ${\eta }_n={\delta }_n+d_n$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\eta }_n<\mathrm{\infty }$. Hence

$$\begin{array}{rcl}\parallel x_{n+1}-z\parallel & =& \parallel (1-{\alpha }_n)x_n+{\alpha }_n{T}^n{y}_n-z\parallel \\ \le & (1-{\alpha }_n)\parallel x_n-z\parallel +{\alpha }_n\parallel T^n{y}_n-z\parallel \\ \le & (1-{\alpha }_n)\parallel x_n-z\parallel +{\alpha }_n\{(1+{\sigma }_n)\parallel x_n-z\parallel +{\eta }_{n}M\}\\ \le & (1+{\sigma }_n)\parallel x_n-z\parallel +{\eta }_{n}M.\end{array}$$

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 $F(T)\ne \mathrm{\varnothing }$. Suppose that $\{c_n\}$, $\{d_n\}$ and ϕ satisfy the following two conditions:

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

• (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 $\{x_n\}$ defined by (1.3) satisfies ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)=\mathrm{\infty }$ and $lim{\beta }_n=0$. Then $\{x_n\}$ converges strongly to some fixed point of T.

Proof For any $z\in F(T)$, by Lemma 3.3, $\{x_n\}$ is bounded. We set

$$M:=1\vee \varphi (\beta )\vee \underset{n\ge 1}{sup}\parallel x_n-z\parallel <\mathrm{\infty }.$$

By Lemma 3.3, we see that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-z\parallel (\equiv r)$ exists. Without loss of generality, we assume $r>0$. As in the proof of Lemma 3.3, we obtain

$$\begin{array}{rcl}\parallel T^n{y}_n-z\parallel & \le & (1+{\sigma }_n)\parallel x_n-z\parallel +{\eta }_{n}M\\ \le & \parallel x_n-z\parallel +{\nu }_{n}M,\end{array}$$

where ${\nu }_n={\sigma }_n+{\eta }_n$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\nu }_n<\mathrm{\infty }$. By using Lemma 3.2 and Takahashi [15], we obtain

$$\begin{array}{rcl}\parallel x_{n+1}-z\parallel & =& \parallel (1-{\alpha }_n)x_n+{\alpha }_n{T}^n{y}_n-z\parallel \\ =& \parallel (1-{\alpha }_n)(x_n-z)+{\alpha }_n(T^n{y}_n-z)\parallel \\ \le & (\parallel x_n-z\parallel +{\nu }_{n}M)[1-2{\alpha }_n(1-{\alpha }_n){\delta }_X\left(\frac{\parallel T^n{y}_n-x_n\parallel }{\parallel x_n-z\parallel +{\nu }_{n}M}\right)].\end{array}$$

Hence we obtain

$$\begin{array}{c}2{\alpha }_n(1-{\alpha }_n)(\parallel x_n-z\parallel +{\nu }_{n}M){\delta }_X\left(\frac{\parallel T^n{y}_n-x_n\parallel }{\parallel x_n-z\parallel +{\nu }_{n}M}\right)\hfill \\ \le \parallel x_n-z\parallel -\parallel x_{n+1}-z\parallel +{\nu }_{n}M.\hfill \end{array}$$

Thus

$$2{\alpha }_n(1-{\alpha }_n)(\parallel x_n-z\parallel +{\nu }_{n}M){\delta }_X\left(\frac{\parallel T^n{y}_n-x_n\parallel }{\parallel x_n-z\parallel +{\nu }_{n}M}\right)<\mathrm{\infty }.$$

Since ${\delta }_X$ is strictly increasing, continuous and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)=\mathrm{\infty }$, we obtain

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel T^n{y}_n-x_n\parallel =0.$$

(3.2)

By using (3.1) in the proof of Lemma 3.3, we have

$$\begin{array}{rcl}\parallel T^{n-1}{x}_{n-1}-z\parallel & \le & \parallel x_{n-1}-z\parallel +c_{n-1}\varphi (\parallel x_{n-1}-z\parallel )+d_{n-1}\\ \le & \parallel x_{n-1}-z\parallel +c_{n-1}\{\varphi (\beta )+\alpha \parallel x_{n-1}-z\parallel \}+d_{n-1}\\ \le & (1+\alpha c_{n-1})\parallel x_{n-1}-z\parallel +{\rho }_{n-1}M,\end{array}$$

where ${\rho }_{n-1}=c_{n-1}+d_{n-1}$ and ${\sum }_{n=2}^{\mathrm{\infty }}{\rho }_{n-1}<\mathrm{\infty }$. Thus

$$\begin{array}{rcl}\parallel y_{n-1}-z\parallel & =& \parallel {\beta }_{n-1}{T}^{n-1}{x}_{n-1}+(1-{\beta }_{n-1})x_{n-1}-z\parallel \\ \le & {\beta }_{n-1}\parallel T^{n-1}{x}_{n-1}-z\parallel +(1-{\beta }_{n-1})\parallel x_{n-1}-z\parallel \\ \le & {\beta }_{n-1}\{(1+\alpha c_{n-1})\parallel x_{n-1}-z\parallel +{\rho }_{n-1}M\}+(1-{\beta }_{n-1})\parallel x_{n-1}-z\parallel \\ \le & (1+\alpha c_{n-1})\parallel x_{n-1}-z\parallel +{\rho }_{n-1}M,\end{array}$$

and hence

$$\begin{array}{c}\parallel y_{n-1}-z\parallel +c_{n-1}\varphi (\parallel y_{n-1}-z\parallel )\hfill \\ \le (1+\alpha c_{n-1})\parallel x_{n-1}-z\parallel +{\rho }_{n-1}M+c_{n-1}\{\varphi (\beta )+\alpha \parallel y_{n-1}-z\parallel \}\hfill \\ \le (1+\alpha c_{n-1})\parallel x_{n-1}-z\parallel +{\rho }_{n-1}M+c_{n-1}\varphi (\beta )+\alpha c_{n-1}(1+\alpha c_{n-1})\parallel x_{n-1}-z\parallel \hfill \\ +\alpha c_{n-1}{\rho }_{n-1}M\hfill \\ \le (1+{\mu }_{n-1})\parallel x_{n-1}-z\parallel +{\phi }_{n-1}M,\hfill \end{array}$$

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

$$\begin{array}{rcl}\parallel T^{n-1}{y}_{n-1}-z\parallel & \le & \parallel y_{n-1}-z\parallel +c_{n-1}\varphi (\parallel y_{n-1}-z\parallel )+d_{n-1}\\ \le & (1+{\mu }_{n-1})\parallel x_{n-1}-z\parallel +{\phi }_{n-1}M+d_{n-1}\\ \le & \parallel x_{n-1}-z\parallel +{\omega }_{n-1}M,\end{array}$$

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 [15], we obtain

$$\begin{array}{rcl}\parallel x_n-z\parallel & =& \parallel (1-{\alpha }_{n-1})x_{n-1}+{\alpha }_{n-1}{T}^{n-1}{y}_{n-1}-z\parallel \\ =& \parallel (1-{\alpha }_{n-1})(x_{n-1}-z)+{\alpha }_{n-1}(T^{n-1}{y}_{n-1}-z)\parallel \\ \le & (\parallel x_{n-1}-z\parallel +{\omega }_{n-1}M)[1-2{\alpha }_n(1-{\alpha }_n){\delta }_X\left(\frac{\parallel T^{n-1}{y}_{n-1}-x_{n-1}\parallel }{\parallel x_{n-1}-z\parallel +{\omega }_{n-1}M}\right)].\end{array}$$

By the same method as above, we obtain

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel T^{n-1}{y}_{n-1}-x_{n-1}\parallel =0.$$

(3.3)

Since $\{x_n\}$ is bounded and T is a TAN mapping, we obtain

$$\begin{array}{rcl}\parallel y_n-x_n\parallel & =& \parallel {\beta }_n{T}^n{x}_n+(1-{\beta }_n)x_n-x_n\parallel \\ \le & {\beta }_n\parallel T^n{x}_n-x_n\parallel \\ \le & {\beta }_n{M}^{\prime },\end{array}$$

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

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel T^n{y}_n-y_n\parallel =0.$$

(3.5)

By using (3.3) and (3.4), we obtain

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel T^{n-1}{y}_{n-1}-y_{n-1}\parallel =0.$$

(3.6)

Since

$$\begin{array}{rcl}\parallel T^{n-1}{x}_{n-1}-x_{n-1}\parallel & \le & \parallel T^{n-1}{x}_{n-1}-T^{n-1}{y}_{n-1}\parallel +\parallel T^{n-1}{y}_{n-1}-x_{n-1}\parallel \\ \le & \parallel x_{n-1}-y_{n-1}\parallel +c_{n-1}\varphi (\parallel x_{n-1}-y_{n-1}\parallel )+d_{n-1}\\ +\parallel T^{n-1}{y}_{n-1}-x_{n-1}\parallel ,\end{array}$$

by using (3.3) and (3.4), we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel T^{n-1}{x}_{n-1}-x_{n-1}\parallel =0.$$

(3.7)

Since

$$\begin{array}{rcl}\parallel x_n-x_{n-1}\parallel & =& \parallel (1-{\alpha }_{n-1})x_{n-1}+{\alpha }_{n-1}{T}^{n-1}{y}_{n-1}-x_{n-1}\parallel \\ =& {\alpha }_{n-1}\parallel T^{n-1}{y}_{n-1}-x_{n-1}\parallel \\ \le & \parallel T^{n-1}{y}_{n-1}-y_{n-1}\parallel +\parallel y_{n-1}-x_{n-1}\parallel ,\end{array}$$

by (3.4) and (3.6), we get

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_n-x_{n-1}\parallel =0.$$

(3.8)

From

$$\begin{array}{rcl}\parallel T^{n-1}{x}_n-x_n\parallel & \le & \parallel T^{n-1}{x}_n-T^{n-1}{x}_{n-1}\parallel +\parallel T^{n-1}{x}_{n-1}-x_{n-1}\parallel +\parallel x_{n-1}-x_n\parallel \\ \le & 2\parallel x_n-x_{n-1}\parallel +c_{n-1}\varphi (\parallel x_n-x_{n-1}\parallel )+d_{n-1}+\parallel T^{n-1}{x}_{n-1}-x_{n-1}\parallel ,\end{array}$$

by (3.7) and (3.8), we obtain

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel T^{n-1}{x}_n-x_n\parallel =0.$$

(3.9)

Since

$$\begin{array}{c}\parallel x_n-T{x}_n\parallel \hfill \\ \le \parallel x_n-y_n\parallel +\parallel y_n-T^n{y}_n\parallel +\parallel T^n{y}_n-T^n{x}_n\parallel +\parallel T^n{x}_n-T{x}_n\parallel \hfill \\ \le \parallel y_n-T^n{y}_n\parallel +2\parallel x_n-y_n\parallel +c_n\varphi (\parallel x_n-y_n\parallel )+d_n+\parallel T^n{x}_n-T{x}_n\parallel \hfill \end{array}$$

and by the uniform continuity of T, (3.4), (3.5) and (3.9), we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_n-T{x}_n\parallel =0.$$

(3.10)

By using condition (A), we obtain

$$f\left(d(x_n,F(T))\right)\le \parallel x_n-T{x}_n\parallel$$

(3.11)

for all $n\ge 1$. As in the proof of Lemma 3.3, we obtain

$$\parallel x_{n+1}-z\parallel \le (1+{\sigma }_n)\parallel x_n-z\parallel +{\eta }_{n}M.$$

(3.12)

Thus

$$\underset{z\in F(T)}{inf}\parallel x_{n+1}-z\parallel \le (1+{\sigma }_n)\underset{z\in F(T)}{inf}\parallel x_n-z\parallel +{\eta }_{n}M.$$

By using Lemma 3.1, we see that ${lim}_{n\to \mathrm{\infty }}d(x_n,F(T))(\equiv c)$ exists. We first claim that ${lim}_{n\to \mathrm{\infty }}d(x_n,F(T))=0$. In fact, assume that $c={lim}_{n\to \mathrm{\infty }}d(x_n,F(T))>0$. Then we can choose $n_0\in \mathbb{N}$ such that $0<\frac{c}{2}<d(x_n,F(T))$ for all $n\ge n_0$. By using condition (A), (3.10) and (3.11), we obtain

$$0<f\left(\frac{c}{2}\right)\le f\left(d(x_{n_i},F(T))\right)\le \parallel x_{n_i}-T{x}_{n_i}\parallel \to 0$$

as $i\to \mathrm{\infty }$. This is a contradiction. So, we obtain $c=0$. Next, we claim that $\{x_n\}$ is a Cauchy sequence. Since ${\sum }_{n=1}^{\mathrm{\infty }}{\sigma }_n<\mathrm{\infty }$, we obtain ${\prod }_{n=1}^{\mathrm{\infty }}(1+{\sigma }_n):=U<\mathrm{\infty }$. Let $ϵ>0$ be given. Since ${lim}_{n\to \mathrm{\infty }}d(x_n,F(T))=0$ 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

$$d(x_n,F(T))<\frac{ϵ}{4U+4}\text{and}\sum _{i=n_0}^{\mathrm{\infty }}{\eta }_i<\frac{ϵ}{4M}.$$

(3.13)

Let $n,m\ge n_0$ and $p\in F(T)$. Then, by (3.12), we obtain

$$\begin{array}{rcl}\parallel x_n-x_m\parallel & \le & \parallel x_n-p\parallel +\parallel x_m-p\parallel \\ \le & \prod _{i=n_0}^{n-1}(1+{\sigma }_i)\parallel x_{n_0}-p\parallel +M\sum _{i=n_0}^{n-1}{\eta }_i+\prod _{i=n_0}^{m-1}(1+{\sigma }_i)\parallel x_{n_0}-p\parallel +M\sum _{i=n_0}^{m-1}{\eta }_i\\ \le & 2[\prod _{i=n_0}^{\mathrm{\infty }}(1+{\sigma }_i)\parallel x_{n_0}-p\parallel +M\sum _{i=n_0}^{\mathrm{\infty }}{\eta }_i].\end{array}$$

Taking the infimum over all $p\in F(T)$ on both sides and by (3.13), we obtain

$$\begin{array}{rcl}\parallel x_n-x_m\parallel & \le & 2[\prod _{i=n_0}^{\mathrm{\infty }}(1+{\sigma }_i)d(x_{n_0},F(T))+M\sum _{i=n_0}^{\mathrm{\infty }}{\eta }_i]\\ <& 2[(U+1)\frac{ϵ}{4U+4}+M\frac{ϵ}{4M}]=ϵ\end{array}$$

for all $n,m\ge n_0$. This implies that $\{x_n\}$ is a Cauchy sequence. Let ${lim}_{n\to \mathrm{\infty }}{x}_n=q$. Then $d(q,F(T))=0$. Since $F(T)$ is closed, we obtain $q\in F(T)$. Hence $\{x_n\}$ 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 [12].

Remark 3.6 If $\{{\alpha }_n\}$ is bounded away from both 0 and 1, i.e., $a\le {\alpha }_n\le b$ for all $n\ge 1$ and some $a,b\in (0,1)$, then ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)=\mathrm{\infty }$ 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 $F(T)\ne \mathrm{\varnothing }$ which is TAN on C, not Lipschitzian and hence not asymptotically nonexpansive.

Example 3.7 Let $X:=\mathbb{R}$ and $C:=[0,2]$. Define $T:C\to C$ by

$$Tx=\{\begin{array}{cc}1,\hfill & x\in [0,1];\hfill \\ \frac{1}{\sqrt{3}}\sqrt{4-x^2},\hfill & x\in [1,2].\hfill \end{array}$$

Note that $T^{n}x=1$ for all $x\in C$ and $n\ge 2$ and $F(T)=\{1\}$. Clearly, T is both uniformly continuous and TAN on C. We show that T satisfies condition (A). In fact, if $x\in [0,1]$, then $|x-1|=|x-Tx|$. Similarly, if $x\in [1,2]$, then

$$|x-1|=x-1\le x-\frac{1}{\sqrt{3}}\sqrt{4-x^2}=|x-Tx|.$$

So, we get $d(x,F(T))=|x-1|\le |x-Tx|$ 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 $x=2-\frac{1}{3{(L+1)}^2}>1$ and $y=2$, then

$$\frac{1}{\sqrt{3}}\sqrt{4-x^2}\le L(2-x)\iff \frac{1}{3L^2}\le \frac{2-x}{2+x}=\frac{1}{12L^2+24L+1}.$$

This is a contradiction.