Strong convergence for asymptotically nonexpansive mappings in the intermediate sense

Abstract

In this paper, let C be a nonempty closed convex subset of a strictly convex Banach space. Then we prove strong convergence of the modified Ishikawa iteration process when T is an ANI self-mapping such that $T(C)$ is contained in a compact subset of C, which generalizes the result due to Takahashi and Kim (Math. Jpn. 48:1-9, 1998).

MSC:47H05, 47H10.

1 Introduction

Let C be a nonempty closed convex subset of a Banach space E, and let T be a mapping of C into itself. Then T is said to be asymptotically nonexpansive [1] if there exists a sequence $\{k_n\}$, $k_n\ge 1$, with ${lim}_{n\to \mathrm{\infty }}{k}_n=1$, such that

$$\parallel T^{n}x-T^{n}y\parallel \le k_n\parallel x-y\parallel$$

for all $x,y\in C$ and $n\ge 1$. In particular, if $k_n=1$ for all $n\ge 1$, T is said to be nonexpansive. 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 $n\ge 1$. T is said to be asymptotically nonexpansive in the intermediate sense (in brief, ANI) [2] provided T is uniformly continuous and

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{x,y\in C}{sup}(\parallel T^{n}x-T^{n}y\parallel -\parallel x-y\parallel )\le 0.$$

We denote by $F(T)$ the set of all fixed points of T, i.e., $F(T)=\{x\in C:Tx=x\}$. We define the modulus of convexity for a convex subset of a Banach space; see also [3]. Let C be a nonempty bounded convex subset of a Banach space E with $d(C)>0$, where $d(C)$ is the diameter of C. Then we define $\delta (C,ϵ)$ with $0\le ϵ\le 1$ as follows:

$$\delta (C,ϵ)=\frac{1}{r}inf\{max(\parallel x-z\parallel ,\parallel y-z\parallel )-\parallel z-\frac{x+y}{2}\parallel :x,y,z\in C,\parallel x-y\parallel \ge rϵ\},$$

where $r=d(C)$. When $\{x_n\}$ is a sequence in E, then $x_n\to x$ will denote strong convergence of the sequence $\{x_n\}$ to x. For a mappings T of C into itself, Rhoades [4] considered the following modified Ishikawa iteration process (cf. Ishikawa [5]) in C defined by $x_1\in C$:

$$x_{n+1}={\alpha }_n{T}^n{y}_n+(1-{\alpha }_n)x_n,y_n={\beta }_n{T}^n{x}_n+(1-{\beta }_n)x_n,$$

(1.1)

where $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ are two real sequences in $[0,1]$. If ${\beta }_n=0$ for all $n\ge 1$, then the iteration process (1.1) reduces to the modified Mann iteration process [6] (cf. Mann [7]).

Takahashi and Kim [8] proved the following result: Let E be a strictly convex Banach space and C be a nonempty closed convex subset of E and $T:C\to C$ be a nonexpansive mapping such that $T(C)$ is contained in a compact subset of C. Suppose $x_1\in C$, and the sequence $\{x_n\}$ is defined by $x_{n+1}={\alpha }_{n}T[{\beta }_{n}T{x}_n+(1-{\beta }_n)x_n]+(1-{\alpha }_n)x_n$, where $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ are chosen so that ${\alpha }_n\in [a,b]$ and ${\beta }_n\in [0,b]$ or ${\alpha }_n\in [a,1]$ and ${\beta }_n\in [a,b]$ for some a, b with $0<a\le b<1$. Then $\{x_n\}$ converges strongly to a fixed point of T. In 2000, Tsukiyama and Takahashi [9] generalized the result due to Takahashi and Kim [8] to a nonexpansive mapping under much less restrictions on the iterative parameters $\{{\alpha }_n\}$ and $\{{\beta }_n\}$.

In this paper, let C be a nonempty closed convex subset of a strictly convex Banach space. We prove that if $T:C\to C$ is an ANI mapping such that $T(C)$ is contained in a compact subset of C, then the iteration $\{x_n\}$ defined by (1.1) converges strongly to a fixed point of T, which generalizes the result due to Takahashi and Kim [8].

2 Strong convergence theorem

We first begin with the following lemma.

Lemma 2.1 [9]

Let C be a nonempty compact convex subset of a Banach space E with $r=d(C)>0$. Let $x,y,z\in C$ and suppose $\parallel x-y\parallel \ge ϵr$ for some ϵ with $0\le ϵ\le 1$. Then, for all λ with $0\le \lambda \le 1$,

$$\parallel \lambda (x-z)+(1-\lambda )(y-z)\parallel \le max(\parallel x-z\parallel ,\parallel y-z\parallel )-2\lambda (1-\lambda )r\delta (C,ϵ).$$

Lemma 2.2 [9]

Let C be a nonempty compact convex subset of a strictly convex Banach space E with $r=d(C)>0$. If ${lim}_{n\to \mathrm{\infty }}\delta (C,{ϵ}_n)=0$, then ${lim}_{n\to \mathrm{\infty }}{ϵ}_n=0$.

Lemma 2.3 [10]

Let $\{a_n\}$ and $\{b_n\}$ be two sequences of nonnegative real numbers such that ${\sum }_{n=1}^{\mathrm{\infty }}{b}_n<\mathrm{\infty }$ and

$$a_{n+1}\le a_n+b_n$$

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

Lemma 2.4 Let C be a nonempty compact convex subset of a Banach space E, and let $T:C\to C$ be an ANI mapping. Put

$$c_n=\underset{x,y\in C}{sup}(\parallel T^{n}x-T^{n}y\parallel -\parallel x-y\parallel )\vee 0,$$

so that ${\sum }_{n=1}^{\mathrm{\infty }}{c}_n<\mathrm{\infty }$. Suppose that the sequence $\{x_n\}$ is defined by (1.1). Then ${lim}_{n\to \mathrm{\infty }}\parallel x_n-z\parallel$ exists for any $z\in F(T)$.

Proof The existence of a fixed point of T follows from Schauder’s fixed theorem [11]. For a fixed $z\in F(T)$, since

$$\begin{array}{rcl}\parallel T^n{y}_n-z\parallel & \le & \parallel y_n-z\parallel +c_n\\ =& \parallel {\beta }_n{T}^n{x}_n+(1-{\beta }_n)x_n-z\parallel +c_n\\ \le & {\beta }_n\parallel T^n{x}_n-z\parallel +(1-{\beta }_n)\parallel x_n-z\parallel +c_n\\ \le & {\beta }_n\parallel x_n-z\parallel +c_n+(1-{\beta }_n)\parallel x_n-z\parallel +c_n\\ \le & \parallel x_n-z\parallel +2c_n,\end{array}$$

we obtain

$$\begin{array}{rcl}\parallel x_{n+1}-z\parallel & =& \parallel {\alpha }_n{T}^n{y}_n+(1-{\alpha }_n)x_n-z\parallel \\ \le & {\alpha }_n\parallel T^n{y}_n-z\parallel +(1-{\alpha }_n)\parallel x_n-z\parallel \\ \le & {\alpha }_n(\parallel x_n-z\parallel +2c_n)+(1-{\alpha }_n)\parallel x_n-z\parallel \\ \le & \parallel x_n-z\parallel +2c_n.\end{array}$$

By Lemma 2.3, we readily see that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-z\parallel$ exists. □

Theorem 2.5 Let C be a nonempty compact convex subset of a strictly convex Banach space E with $r=d(C)>0$. Let $T:C\to C$ be an ANI mapping. Put

$$c_n=\underset{x,y\in C}{sup}(\parallel T^{n}x-T^{n}y\parallel -\parallel x-y\parallel )\vee 0,$$

so that ${\sum }_{n=1}^{\mathrm{\infty }}{c}_n<\mathrm{\infty }$. Suppose $x_1\in C$, and the sequence $\{x_n\}$ defined by (1.1) satisfies ${\alpha }_n\in [a,b]$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n=b<1$ or ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n>0$ and ${\beta }_n\in [a,b]$ for some a, b with $0<a\le b<1$. Then ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel =0$.

Proof The existence of a fixed point of T follows from Schauder’s fixed theorem [11]. For any fixed $z\in F(T)$, we first show that if ${\alpha }_n\in [a,b]$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n=b<1$ for some $a,b\in (0,1)$, then we obtain ${lim}_{n\to \mathrm{\infty }}\parallel T{x}_n-x_n\parallel =0$. In fact, let ${ϵ}_n=\frac{\parallel T^n{y}_n-x_n\parallel }{r}$. Then we have $0\le {ϵ}_n\le 1$ since $\parallel T^n{y}_n-x_n\parallel \le r$. As in the proof of Lemma 2.4, we obtain

$$\parallel T^n{y}_n-z\parallel \le \parallel x_n-z\parallel +2c_n.$$

(2.1)

Since

$$\parallel T^n{y}_n-x_n\parallel =r{ϵ}_n,$$

and by (2.1) and Lemma 2.1, we have

$$\begin{array}{rcl}\parallel x_{n+1}-z\parallel & =& \parallel {\alpha }_n(T^n{y}_n-z)+(1-{\alpha }_n)(x_n-z)\parallel \\ \le & \parallel x_n-z\parallel +2c_n-2{\alpha }_n(1-{\alpha }_n)r\delta (C,{ϵ}_n).\end{array}$$

Thus

$$2{\alpha }_n(1-{\alpha }_n)r\delta (C,{ϵ}_n)\le \parallel x_n-z\parallel -\parallel x_{n+1}-z\parallel +2c_n.$$

Since

$$2r\sum _{n=1}^{\mathrm{\infty }}a(1-b)\delta (C,\frac{\parallel T^n{y}_n-x_n\parallel }{r})<\mathrm{\infty },$$

we obtain

$$\underset{n\to \mathrm{\infty }}{lim}\delta (C,\frac{\parallel T^n{y}_n-x_n\parallel }{r})=0.$$

By using Lemma 2.2, we obtain

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

(2.2)

Since

$$\begin{array}{rcl}\parallel T^n{x}_n-x_n\parallel & \le & \parallel T^n{x}_n-T^n{y}_n\parallel +\parallel T^n{y}_n-x_n\parallel \\ \le & \parallel x_n-y_n\parallel +c_n+\parallel T^n{y}_n-x_n\parallel \\ =& {\beta }_n\parallel T^n{x}_n-x_n\parallel +c_n+\parallel T^n{y}_n-x_n\parallel ,\end{array}$$

we obtain

$$(1-{\beta }_n)\parallel T^n{x}_n-x_n\parallel \le c_n+\parallel T^n{y}_n-x_n\parallel .$$

(2.3)

Since ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n=b<1$, we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}(1-{\beta }_n)=1-b>0.$$

(2.4)

From (2.2), (2.3) and (2.4), we obtain

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

(2.5)

Since

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

and by (2.2), we obtain

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-x_n\parallel =0.$$

(2.6)

Since

$$\begin{array}{r}\parallel x_n-T{x}_n\parallel \\ \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-T^{n+1}{x}_{n+1}\parallel +\parallel T^{n+1}{x}_{n+1}-T^{n+1}{x}_n\parallel +\parallel T^{n+1}{x}_n-T{x}_n\parallel \\ \le 2\parallel x_n-x_{n+1}\parallel +c_{n+1}+\parallel x_{n+1}-T^{n+1}{x}_{n+1}\parallel +\parallel T^{n+1}{x}_n-T{x}_n\parallel \end{array}$$

and by the uniform continuity of T, (2.5) and (2.6), we have

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

(2.7)

Next, we show that if ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n>0$ and ${\beta }_n\in [a,b]$, then we also obtain (2.7). In fact, let ${ϵ}_n=\frac{\parallel T^n{x}_n-x_n\parallel }{r}$. Then we have $0\le {ϵ}_n\le 1$. From ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n>0$, there are some positive integer $n_0$ and a positive number a such that ${\alpha }_n>a>0$ for all $n\ge n_0$. Since

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

and hence

$$\frac{\parallel x_{n+1}-z\parallel -\parallel x_n-z\parallel }{{\alpha }_n}\le \parallel y_n-z\parallel -\parallel x_n-z\parallel +c_n.$$

So, we obtain

$$\begin{array}{rcl}\parallel x_n-z\parallel -\parallel y_n-z\parallel & \le & \frac{\parallel x_n-z\parallel -\parallel x_{n+1}-z\parallel }{{\alpha }_n}+c_n\\ \le & \frac{\parallel x_n-z\parallel -\parallel x_{n+1}-z\parallel }{a}+c_n.\end{array}$$

(2.8)

Since

$$\parallel T^n{x}_n-z\parallel \le \parallel x_n-z\parallel +c_n,$$

from Lemma 2.1, we obtain

$$\begin{array}{rcl}\parallel y_n-z\parallel & =& \parallel {\beta }_n{T}^n{x}_n+(1-{\beta }_n)x_n-z\parallel \\ =& \parallel {\beta }_n(T^n{x}_n-z)+(1-{\beta }_n)(x_n-z)\parallel \\ \le & \parallel x_n-z\parallel +c_n-2{\beta }_n(1-{\beta }_n)r\delta (C,{ϵ}_n).\end{array}$$

(2.9)

By using (2.8) and (2.9), we obtain

$$\begin{array}{rcl}2{\beta }_n(1-{\beta }_n)r\delta (C,{ϵ}_n)& \le & \parallel x_n-z\parallel -\parallel y_n-z\parallel +c_n\\ \le & \frac{\parallel x_n-z\parallel -\parallel x_{n+1}-z\parallel }{a}+2c_n.\end{array}$$

Hence

$$2r\sum _{n=1}^{\mathrm{\infty }}a(1-b)\delta (C,\frac{\parallel T^n{x}_n-x_n\parallel }{r})<\mathrm{\infty }.$$

We also obtain

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T^n{x}_n\parallel =0$$

(2.10)

similarly to the argument above. Since

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

and by using (2.10), we obtain

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-y_n\parallel =0.$$

(2.11)

Since

$$\begin{array}{rcl}\parallel T^n{y}_n-x_n\parallel & \le & \parallel T^n{y}_n-T^n{x}_n\parallel +\parallel T^n{x}_n-x_n\parallel \\ \le & \parallel y_n-x_n\parallel +c_n+\parallel T^n{x}_n-x_n\parallel ,\end{array}$$

by using (2.10) and (2.11), we obtain

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

(2.12)

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 using (2.11) and (2.12), we obtain

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

(2.13)

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 (2.11) and (2.13), we get

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

(2.14)

From

$$\begin{array}{r}\parallel T^{n-1}{x}_n-x_n\parallel \\ \le \parallel T^{n-1}{x}_n-T^{n-1}{x}_{n-1}\parallel +\parallel T^{n-1}{x}_{n-1}-x_{n-1}\parallel +\parallel x_{n-1}-x_n\parallel \\ \le 2\parallel x_n-x_{n-1}\parallel +c_{n-1}+\parallel T^{n-1}{x}_{n-1}-x_{n-1}\parallel \end{array}$$

and by (2.10) and (2.14), we obtain

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

(2.15)

Since

$$\begin{array}{r}\parallel x_n-T{x}_n\parallel \\ \le \parallel x_n-y_n\parallel +\parallel y_n-T^n{y}_n\parallel +\parallel T^n{y}_n-T^n{x}_n\parallel +\parallel T^n{x}_n-T{x}_n\parallel \\ \le \parallel y_n-T^n{y}_n\parallel +2\parallel x_n-y_n\parallel +c_n+\parallel T^n{x}_n-T{x}_n\parallel \end{array}$$

and by the uniform continuity of T, (2.11), (2.13) and (2.15), we have

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

 □

Our Theorem 2.6 carries over Theorem 3 of Takahashi and Kim [8] to an ANI mapping.

Theorem 2.6 Let C be a nonempty closed convex subset of a strictly convex Banach space E, and let $T:C\to C$ be an ANI mapping, and let $T(C)$ be contained in a compact subset of C. Put

$$c_n=\underset{x,y\in C}{sup}(\parallel T^{n}x-T^{n}y\parallel -\parallel x-y\parallel )\vee 0,$$

so that ${\sum }_{n=1}^{\mathrm{\infty }}{c}_n<\mathrm{\infty }$. Suppose $x_1\in C$, and the sequence $\{x_n\}$ defined by (1.1) satisfies ${\alpha }_n\in [a,b]$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n=b<1$ or ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n>0$ and ${\beta }_n\in [a,b]$ for some a, b with $0<a\le b<1$. Then $\{x_n\}$ converges strongly to a fixed point of T.

Proof By Mazur’s theorem [12], $A:=\overline{co}(\{x_1\}\cup T(C))$ is a compact subset of C containing $\{x_n\}$ which is invariant under T. So, without loss of generality, we may assume that C is compact and $\{x_n\}$ is well defined. The existence of a fixed point of T follows from Schauder’s fixed theorem [11]. If $d(C)=0$, then the conclusion is obvious. So, we assume $d(C)>0$. From Theorem 2.5, we obtain

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

(2.16)

Since C is compact, there exist a subsequence $\{x_{n_k}\}$ of the sequence $\{x_n\}$ and a point $p\in C$ such that $x_{n_k}\to p$. Thus we obtain $p\in F(T)$ by the continuity of T and (2.16). Hence we obtain ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel =0$ by Lemma 2.4. □

Corollary 2.7 Let C be a nonempty closed convex subset of a strictly convex Banach space E, and let $T:C\to C$ be an asymptotically nonexpansive mapping with $\{k_n\}$ satisfying $k_n\ge 1$, ${\sum }_{n=1}^{\mathrm{\infty }}(k_n-1)<\mathrm{\infty }$, and let $T(C)$ be contained in a compact subset of C. Suppose $x_1\in C$, and the sequence $\{x_n\}$ defined by (1.1) satisfies ${\alpha }_n\in [a,b]$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n=b<1$ or ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n>0$ and ${\beta }_n\in [a,b]$ for some a, b with $0<a\le b<1$. Then $\{x_n\}$ converges strongly to a fixed point of T.

Proof Note that

$$\sum _{n=1}^{\mathrm{\infty }}{c}_n=\sum _{n=1}^{\mathrm{\infty }}(k_n-1)diam(C)<\mathrm{\infty },$$

where $diam(C)={sup}_{x,y\in C}\parallel x-y\parallel <\mathrm{\infty }$. The conclusion now follows easily from Theorem 2.6. □

We give an example which satisfies all assumptions of T in Theorem 2.6, i.e., $T:C\to C$ is an ANI mapping which is not Lipschitzian and hence not asymptotically nonexpansive.

Example 2.8 Let $E:=\mathbb{R}$ and $C:=[0,2]$. Define $T:C\to C$ by

$$Tx=\{\begin{array}{ll}1,& x\in [0,1];\\ \sqrt{2-x},& x\in [1,2].\end{array}$$

Note that $T^{n}x=1$ for all $x\in C$ and $n\ge 2$ and $F(T)=\{1\}$. Clearly, T is uniformly continuous, ANI on 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 $y:=2$ and $x:=2-\frac{1}{{(L+1)}^2}>1$, then

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

This is a contradiction.

We also give an example of an ANI mapping which is not a Lipschitz function.

Example 2.9 Let $E=\mathbb{R}$ and $C=[-3\pi ,3\pi ]$ and let $|h|<1$. Let $T:C\to C$ be defined by

$$Tx=hxsinnx$$

for each $x\in C$ and for all $n\in \mathbb{N}$, where ℕ denotes the set of all positive integers. Clearly $F(T)=\{0\}$. Since

$$\begin{array}{c}T(x)=hxsinnx,\hfill \\ T^{2}x=h^{2}xsinnxsinnhxsinn(sinnx)\cdots ,\hfill \end{array}$$

we obtain $\{T^{n}x\}\to 0$ uniformly on C as $n\to \mathrm{\infty }$. Thus

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\{\parallel T^{n}x-T^{n}y\parallel -\parallel x-y\parallel \vee 0\}=0$$

for all $x,y\in C$. Hence T is an ANI mapping, but it is not a Lipschitz function. In fact, suppose that there exists $h>0$ such that $|Tx-Ty|\le h|x-y|$ for all $x,y\in C$. If we take $x=\frac{5\pi }{2n}$ and $y=\frac{3\pi }{2n}$, then

$$|Tx-Ty|=|h\frac{5\pi }{2n}sinn\frac{5\pi }{2n}-h\frac{3\pi }{2n}sinn\frac{3\pi }{2n}|=\frac{4h\pi }{n},$$

whereas

$$h|x-y|=h|\frac{5\pi }{2n}-\frac{3\pi }{2n}|=\frac{h\pi }{n}.$$