Approximating fixed points of α-nonexpansive mappings in uniformly convex Banach spaces and $\mathrm{CAT}(0)$ spaces

Abstract

An existence theorem for a fixed point of an α-nonexpansive mapping of a nonempty bounded, closed and convex subset of a uniformly convex Banach space has been recently established by Aoyama and Kohsaka with a non-constructive argument. In this paper, we show that appropriate Ishikawa iterate algorithms ensure weak and strong convergence to a fixed point of such a mapping. Our theorems are also extended to $CAT(0)$ spaces.

AMS Subject Classification:54E40, 54H25, 47H10, 37C25.

1 Introduction

The purpose of this paper is to study fixed point theorems of α-nonexpansive mappings of $CAT(0)$ spaces. A metric space X is a $CAT(0)$ space if it is geodesically connected, and if every geodesic triangle in X is at least as ‘thin’ as its comparison triangle in the Euclidean plane (see Section 4 for the precise definition). Our approach is to prove firstly weak and strong convergence theorems for Ishikawa iterations of α-nonexpansive mappings in uniformly convex Banach spaces. Then, we extend the results to $CAT(0)$ spaces.

Here are the details. Let E be a (real) Banach space and let C be a nonempty subset of E. Let $T:C\to E$ be a mapping. Denote by $F(T)$ the set of fixed points of T, i.e., $F(T)=\{x\in C:Tx=x\}$. We say that T is nonexpansive if $\parallel Tx-Ty\parallel \le \parallel x-y\parallel$ for all x, y in C, and that T is quasi-nonexpansive if $F(T)\ne \mathrm{\varnothing }$ and $\parallel Tx-y\parallel \le \parallel x-y\parallel$ for all x in C and y in $F(T)$.

The concept of nonexpansivity of a map T from a convex set C into C plays an important role in the study of the Mann-type iteration given by

$$x_{n+1}={\beta }_{n}T{x}_n+(1-{\beta }_n)x_n,x_1\in C.$$

(1.1)

Here, $\{{\beta }_n\}$ is a real sequence in $[0,1]$ satisfying some appropriate conditions, which is usually called a control sequence. A more general iteration scheme is the Ishikawa iteration given by

$$\{\begin{array}{c}{y}_n={\beta }_{n}T{x}_n+(1-{\beta }_n)x_n,\hfill \\ x_{n+1}={\gamma }_{n}T{y}_n+(1-{\gamma }_n)x_n,\hfill \end{array}$$

(1.2)

where the sequences $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ satisfy some appropriate conditions. In particular, when all ${\beta }_n=0$, the Ishikawa iteration (1.2) becomes the standard Mann iteration (1.1). Let T be nonexpansive and let C be a nonempty closed and convex subset of a uniformly convex Banach space E satisfying the Opial property. Takahashi and Kim [1] proved that, for any initial data $x_1$ in C, the sequence $\{x_n\}$ of iterations defined by the Ishikawa iteration (1.2) converges weakly to a fixed point of T, with appropriate choices of control sequences $\{{\beta }_n\}$ and $\{{\gamma }_n\}$.

Following Aoyama and Kohsaka [2], a mapping $T:C\to E$ is said to be α-nonexpansive for some real number $\alpha <1$ if

$${\parallel Tx-Ty\parallel }^2\le \alpha {\parallel Tx-y\parallel }^2+\alpha {\parallel Ty-x\parallel }^2+(1-2\alpha ){\parallel x-y\parallel }^2,\mathrm{\forall }x,y\in C.$$

Clearly, 0-nonexpansive maps are exactly nonexpansive maps. Moreover, T is Lipschitz continuous whenever $\alpha \le 0$. An example of a discontinuous α-nonexpansive mapping (with $\alpha >0$) has been given in [2]. See also Example 3.6(b).

An existence theorem for a fixed point of an α-nonexpansive mapping T of a nonempty bounded, closed and convex subset C of a uniformly convex Banach space E has been recently established by Aoyama and Kohsaka [2] with a non-constructive argument. In Section 3, we show that, under mild conditions on the control sequences $\{{\beta }_n\}$ and $\{{\gamma }_n\}$, the fixed point set $F(T)$ is nonempty if and only if the sequence $\{x_n\}$ obtained by the Ishikawa iteration (1.2) is bounded and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}\parallel T{x}_n-x_n\parallel =0$. In this case, $\{x_n\}$ converges weakly or strongly to a fixed point of T.

In Section 5, we establish the existence result of an α-nonexpansive mapping in a $CAT(0)$-space in parallel to [2]. We then extend the convergence theorems obtained in Section 3 to the case of $CAT(0)$ spaces, as we planned.

2 Preliminaries

Let E be a (real) Banach space with the norm $\parallel \cdot \parallel$ and the dual space $E^{\ast }$. Denote by $x_n\to x$ the strong convergence of a sequence $\{x_n\}$ to x in E and by $x_n\rightharpoonup x$ the weak convergence. The modulus δ of the convexity of E is defined by

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

for every ϵ with $0\le ϵ\le 2$. A Banach space E is said to be uniformly convex if $\delta (ϵ)>0$ for every $0<ϵ\le 2$. Let $S=\{x\in E:\parallel x\parallel =1\}$. The norm of E is said to be Gâteaux differentiable if for each x, y in S, the limit

$$\underset{t\to 0}{lim}\frac{\parallel x+ty\parallel -\parallel x\parallel }{t}$$

(2.1)

exists. In this case, E is called smooth. If the limit (2.1) is attained uniformly in x, y in S, then E is called uniformly smooth. A Banach space E is said to be strictly convex if $\parallel \frac{x+y}{2}\parallel <1$ whenever $x,y\in S$ and $x\ne y$. It is well-known that E is uniformly convex if and only if $E^{\ast }$ is uniformly smooth. It is also known that if E is reflexive, then E is strictly convex if and only if $E^{\ast }$ is smooth; for more details, see [3].

A Banach space E is said to satisfy the Opial property [4] if, for every weakly convergent sequence $x_n\rightharpoonup x$ in E, we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_n-x\parallel <\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_n-y\parallel$$

for all y in E with $y\ne x$. It is well known that all Hilbert spaces, all finite dimensional Banach spaces and the Banach spaces $l^p$ ($1\le p<\mathrm{\infty }$) satisfy the Opial property, while the uniformly convex spaces $L_p[0,2\pi ]$ ($p\ne 2$) do not; see, for example, [4–6].

Let $\{x_n\}$ be a bounded sequence in a Banach space E. For any x in E, we set

$$r(x,\{x_n\})=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x-x_n\parallel .$$

The asymptotic radius of $\{x_n\}$ relative to a nonempty closed and convex subset C of E is defined by

$$r(C,\{x_n\})=inf\{r(x,\{x_n\}):x\in C\}.$$

The asymptotic center of $\{x_n\}$ relative to C is the set

$$A(C,\{x_n\})=\{x\in C:r(x,\{x_n\})=r(C,\{x_n\})\}.$$

It is well known that if E is uniformly convex, then $A(C,\{x_n\})$ consists of exactly one point; see [7, 8].

Lemma 2.1 Let C be a nonempty subset of a Banach space E. Let $T:C\to E$ be an α-nonexpansive mapping for some $\alpha <1$ such that $F(T)\ne \mathrm{\varnothing }$. Then T is quasi-nonexpansive. Moreover, $F(T)$ is norm closed.

Proof Let $x\in C$ and $z\in F(T)$. Then we have

$$\begin{array}{rcl}{\parallel Tx-z\parallel }^2& =& {\parallel Tx-Tz\parallel }^2\\ \le & \alpha {\parallel Tx-z\parallel }^2+\alpha {\parallel Tz-x\parallel }^2+(1-2\alpha ){\parallel x-z\parallel }^2\\ =& \alpha {\parallel Tx-z\parallel }^2+\alpha {\parallel z-x\parallel }^2+(1-2\alpha ){\parallel x-z\parallel }^2\\ =& \alpha {\parallel Tx-z\parallel }^2+(1-\alpha ){\parallel x-z\parallel }^2.\end{array}$$

Therefore,

$$\parallel Tx-z\parallel \le \parallel x-z\parallel .$$

This inequality ensures the closedness of $F(T)$. □

Lemma 2.2 Let C be a nonempty subset of a Banach space E. Let $T:C\to E$ be an α-nonexpansive mapping for some $\alpha <1$. Then the following assertions hold.

• (i) If $0\le \alpha <1$, then

  $$\begin{array}{c}{\parallel x-Ty\parallel }^2\le \frac{1+\alpha }{1-\alpha }{\parallel x-Tx\parallel }^2+\frac{2}{1-\alpha }(\alpha \parallel x-y\parallel +\parallel Tx-Ty\parallel )\parallel x-Tx\parallel +{\parallel x-y\parallel }^2,\hfill \\ \mathrm{\forall }x,y\in C.\hfill \end{array}$$

• (ii) If $\alpha <0$, then

  $$\begin{array}{c}{\parallel x-Ty\parallel }^2\le {\parallel x-Tx\parallel }^2+\frac{2}{1-\alpha }[(-\alpha )\parallel Tx-y\parallel +\parallel Tx-Ty\parallel ]\parallel x-Tx\parallel +{\parallel x-y\parallel }^2,\hfill \\ \mathrm{\forall }x,y\in C.\hfill \end{array}$$

Proof

• (i) Observe

  $$\begin{array}{rcl}{\parallel x-Ty\parallel }^2& =& {\parallel x-Tx+Tx-Ty\parallel }^2\\ \le & {(\parallel x-Tx\parallel +\parallel Tx-Ty\parallel )}^2\\ =& {\parallel x-Tx\parallel }^2+{\parallel Tx-Ty\parallel }^2+2\parallel x-Tx\parallel \parallel Tx-Ty\parallel \\ \le & {\parallel x-Tx\parallel }^2+\alpha {\parallel Tx-y\parallel }^2+\alpha {\parallel x-Ty\parallel }^2+(1-2\alpha ){\parallel x-y\parallel }^2\\ +2\parallel x-Tx\parallel \parallel Tx-Ty\parallel \\ \le & {\parallel x-Tx\parallel }^2+\alpha {(\parallel Tx-x\parallel +\parallel x-y\parallel )}^2\\ +\alpha {\parallel x-Ty\parallel }^2+(1-2\alpha ){\parallel x-y\parallel }^2+2\parallel x-Tx\parallel \parallel Tx-Ty\parallel \\ \le & {\parallel x-Tx\parallel }^2+\alpha {\parallel Tx-x\parallel }^2+\alpha {\parallel x-y\parallel }^2\\ +2\alpha \parallel Tx-x\parallel \parallel x-y\parallel +\alpha {\parallel x-Ty\parallel }^2\\ +(1-2\alpha ){\parallel x-y\parallel }^2+2\parallel x-Tx\parallel \parallel Tx-Ty\parallel \\ =& (1+\alpha ){\parallel x-Tx\parallel }^2+2\alpha \parallel Tx-x\parallel \parallel x-y\parallel +\alpha {\parallel x-Ty\parallel }^2\\ +(1-\alpha ){\parallel x-y\parallel }^2+2\parallel x-Tx\parallel \parallel Tx-Ty\parallel .\end{array}$$

This implies that

$${\parallel x-Ty\parallel }^2\le \frac{1+\alpha }{1-\alpha }{\parallel x-Tx\parallel }^2+\frac{2}{1-\alpha }(\alpha \parallel x-y\parallel +\parallel Tx-Ty\parallel )\parallel x-Tx\parallel +{\parallel x-y\parallel }^2.$$

(ii) Observe

$$\begin{array}{rcl}{\parallel x-Ty\parallel }^2& =& {\parallel x-Tx+Tx-Ty\parallel }^2\\ \le & {(\parallel x-Tx\parallel +\parallel Tx-Ty\parallel )}^2\\ =& {\parallel x-Tx\parallel }^2+{\parallel Tx-Ty\parallel }^2+2\parallel x-Tx\parallel \parallel Tx-Ty\parallel \\ \le & {\parallel x-Tx\parallel }^2+\alpha {\parallel Tx-y\parallel }^2+\alpha {\parallel x-Ty\parallel }^2+(1-2\alpha ){\parallel x-y\parallel }^2\\ +2\parallel x-Tx\parallel \parallel Tx-Ty\parallel \\ =& {\parallel x-Tx\parallel }^2+\alpha {\parallel Tx-y\parallel }^2+\alpha {\parallel x-Ty\parallel }^2\\ +(1-\alpha ){\parallel x-y\parallel }^2-\alpha {\parallel x-y\parallel }^2+2\parallel x-Tx\parallel \parallel Tx-Ty\parallel \\ \le & {\parallel x-Tx\parallel }^2+\alpha {\parallel Tx-y\parallel }^2+\alpha {\parallel x-Ty\parallel }^2\\ +(1-\alpha ){\parallel x-y\parallel }^2-\alpha [{\parallel x-Tx\parallel }^2+{\parallel Tx-y\parallel }^2+2\parallel x-Tx\parallel \parallel Tx-y\parallel ]\\ +2\parallel x-Tx\parallel \parallel Tx-Ty\parallel \\ =& (1-\alpha ){\parallel x-Tx\parallel }^2+\alpha {\parallel x-Ty\parallel }^2\\ +(1-\alpha ){\parallel x-y\parallel }^2-2\alpha \parallel x-Tx\parallel \parallel Tx-y\parallel +2\parallel x-Tx\parallel \parallel Tx-Ty\parallel \\ =& (1-\alpha ){\parallel x-Tx\parallel }^2+\alpha {\parallel x-Ty\parallel }^2\\ +(1-\alpha ){\parallel x-y\parallel }^2+2[(-\alpha )\parallel Tx-y\parallel +\parallel Tx-Ty\parallel ]\parallel x-Tx\parallel .\end{array}$$

This implies that

$${\parallel x-Ty\parallel }^2\le {\parallel x-Tx\parallel }^2+\frac{2}{1-\alpha }[(-\alpha )\parallel Tx-y\parallel +\parallel Tx-Ty\parallel ]\parallel x-Tx\parallel +{\parallel x-y\parallel }^2.$$

 □

Proposition 2.3 (Demiclosedness principle)

Let C be a subset of a Banach space E with the Opial property. Let $T:C\to C$ be an α-nonexpansive mapping for some $\alpha <1$. If $\{x_n\}$ converges weakly to z and ${lim}_{n\to \mathrm{\infty }}\parallel T{x}_n-x_n\parallel =0$, then $Tz=z$. That is, $I-T$ is demiclosed at zero, where I is the identity mapping on E.

Proof Since $\{x_n\}$ converges weakly to z and ${lim}_{n\to \mathrm{\infty }}\parallel T{x}_n-x_n\parallel =0$, both $\{x_n\}$ and $\{T{x}_n\}$ are bounded. Let $M_1=sup\{\parallel x_n\parallel ,\parallel T{x}_n\parallel ,\parallel z\parallel ,\parallel Tz\parallel :n\in \mathbb{N}\}<\mathrm{\infty }$. If $0\le \alpha <1$, then in view of Lemma 2.2(i),

$$\begin{array}{c}{\parallel x_n-Tz\parallel }^2\hfill \\ \le \frac{1+\alpha }{1-\alpha }{\parallel x_n-T{x}_n\parallel }^2+\frac{2}{1-\alpha }(\alpha \parallel x_n-z\parallel +\parallel T{x}_n-Tz\parallel )\parallel x_n-T{x}_n\parallel +{\parallel x_n-z\parallel }^2\hfill \\ \le \frac{1+\alpha }{1-\alpha }{\parallel x_n-T{x}_n\parallel }^2+\frac{4M_1(1+\alpha )}{1-\alpha }\parallel x_n-T{x}_n\parallel +{\parallel x_n-z\parallel }^2.\hfill \end{array}$$

If $\alpha <0$, then in view of Lemma 2.2(ii),

$$\begin{array}{c}{\parallel x_n-Tz\parallel }^2\hfill \\ \le {\parallel x_n-T{x}_n\parallel }^2+\frac{2}{1-\alpha }[(-\alpha )\parallel T{x}_n-z\parallel +\parallel T{x}_n-Tz\parallel ]\parallel x_n-T{x}_n\parallel +{\parallel x_n-z\parallel }^2\hfill \\ \le {\parallel x_n-T{x}_n\parallel }^2+4M_1\parallel x_n-T{x}_n\parallel +{\parallel x_n-z\parallel }^2.\hfill \end{array}$$

These relations imply

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_n-Tz\parallel \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_n-z\parallel .$$

From the Opial property, we obtain $Tz=z$. □

The following result has been proved in [9].

Lemma 2.4 Let $r>0$ be a fixed real number. If E is a uniformly convex Banach space, then there exists a continuous strictly increasing convex function $g:[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ with $g(0)=0$ such that

$${\parallel \lambda x+(1-\lambda )y\parallel }^2\le \lambda {\parallel x\parallel }^2+(1-\lambda ){\parallel y\parallel }^2-\lambda (1-\lambda )g(\parallel x-y\parallel )$$

for all x, y in $B_r(0)=\{u\in E:\parallel u\parallel \le r\}$ and $\lambda \in [0,1]$.

Recently, Aoyama and Kohsaka [2] proved the following fixed point theorem for α-nonexpansive mappings of Banach spaces.

Lemma 2.5 Let C be a nonempty closed and convex subset of a uniformly convex Banach space E. Let $T:C\to C$ be an α-nonexpansive mapping for some $\alpha <1$. Then the following conditions are equivalent.

• (i) There exists x in C such that ${\{T^{n}x\}}_{n=1}^{\mathrm{\infty }}$ is bounded.

• (ii) $F(T)\ne \mathrm{\varnothing }$.

3 Fixed point and convergence theorems in Banach spaces

Lemma 3.1 Let C be a nonempty closed and convex subset of a Banach space E. Let $T:C\to C$ be an α-nonexpansive mapping for some $\alpha <1$. Let a sequence $\{x_n\}$ with $x_1$ in C be defined by the Ishikawa iteration (1.2) such that $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ are arbitrary sequences in $[0,1]$. Suppose that the fixed point set $F(T)$ contains an element z. Then the following assertions hold.

• (1) $max\{\parallel x_{n+1}-z\parallel ,\parallel y_n-z\parallel \}\le \parallel x_n-z\parallel$ for all $n=1,2,\dots$ .

• (2) ${lim}_{n\to \mathrm{\infty }}\parallel x_n-z\parallel$ exists.

• (3) ${lim}_{n\to \mathrm{\infty }}d(x_n,F(T))$ exists, where $d(x,F(T))$ denotes the distance from x to $F(T)$.

Proof

In view of Lemma 2.1, we conclude that

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

Consequently,

$$\begin{array}{rcl}\parallel x_{n+1}-z\parallel & =& \parallel {\gamma }_{n}T{y}_n+(1-{\gamma }_n)x_n-z\parallel \\ \le & {\gamma }_n\parallel T{y}_n-z\parallel +(1-{\gamma }_n)\parallel x_n-z\parallel \\ \le & {\gamma }_n\parallel y_n-z\parallel +(1-{\gamma }_n)\parallel x_n-z\parallel \\ \le & {\gamma }_n\parallel x_n-z\parallel +(1-{\gamma }_n)\parallel x_n-z\parallel \\ =& \parallel x_n-z\parallel .\end{array}$$

This implies that $\{\parallel x_n-z\parallel \}$ is a bounded and nonincreasing sequence. Thus, ${lim}_{n\to \mathrm{\infty }}\parallel x_n-z\parallel$ exists.

In the same manner, we see that $\{d(x_n,F(T))\}$ is also a bounded nonincreasing real sequence, and thus converges. □

Theorem 3.2 Let C be a nonempty closed and convex subset of a uniformly convex Banach space E. Let $T:C\to C$ be an α-nonexpansive mapping for some $\alpha <1$. Let $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ be sequences in $[0,1]$ and let $\{x_n\}$ be a sequence with $x_1$ in C defined by the Ishikawa iteration (1.2).

• 1. If $\{x_n\}$ is bounded and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}\parallel T{x}_n-x_n\parallel =0$, then the fixed point set $F(T)\ne \mathrm{\varnothing }$.

• 2. Assume $F(T)\ne \mathrm{\varnothing }$. Then $\{x_n\}$ is bounded, and the following hold.

Case 1: $0<\alpha <1$.

• (a) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}\parallel T{x}_n-x_n\parallel =0$ when ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\gamma }_n(1-{\gamma }_n)>0$.

• (b) ${lim}_{n\to \mathrm{\infty }}\parallel T{x}_n-x_n\parallel =0$ when ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\gamma }_n(1-{\gamma }_n)>0$.

Case 2: $\alpha \le 0$.

• (a) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}\parallel T{x}_n-x_n\parallel =0$ when

  $$\left\{\begin{array}{c}{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\gamma }_n(1-{\gamma }_n)>0,\hfill \\ {lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n<1,\hfill \end{array}\mathit{\text{or}}\right\{\begin{array}{c}{lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\gamma }_n(1-{\gamma }_n)>0,\hfill \\ {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1.\hfill \end{array}$$

• (b) ${lim}_{n\to \mathrm{\infty }}\parallel T{x}_n-x_n\parallel =0$ when ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\gamma }_n(1-{\gamma }_n)>0$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$.

Proof Assume that $\{x_n\}$ is bounded and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}\parallel T{x}_n-x_n\parallel =0$. There is a bounded subsequence $\{T{x}_{n_k}\}$ of $\{T{x}_n\}$ such that ${lim}_{k\to \mathrm{\infty }}\parallel T{x}_{n_k}-x_{n_k}\parallel =0$. Suppose $A(C,\{x_{n_k}\})=\{z\}$. Let $M_1=sup\{\parallel x_{n_k}\parallel ,\parallel T{x}_{n_k}\parallel ,\parallel z\parallel ,\parallel Tz\parallel :k\in \mathbb{N}\}<\mathrm{\infty }$. If $0\le \alpha <1$, then, by Lemma 2.2(i), we have

$$\begin{array}{c}{\parallel x_{n_k}-Tz\parallel }^2\hfill \\ \le \frac{1+\alpha }{1-\alpha }{\parallel x_{n_k}-T{x}_{n_k}\parallel }^2+\frac{2}{1-\alpha }(\alpha \parallel x_{n_k}-z\parallel +\parallel T{x}_{n_k}-Tz\parallel )\parallel x_{n_k}-T{x}_{n_k}\parallel +{\parallel x_{n_k}-z\parallel }^2\hfill \\ \le \frac{1+\alpha }{1-\alpha }{\parallel x_{n_k}-T{x}_{n_k}\parallel }^2+\frac{4M_1(1+\alpha )}{1-\alpha }\parallel T{x}_{n_k}-x_{n_k}\parallel +{\parallel x_{n_k}-z\parallel }^2.\hfill \end{array}$$

This implies that

$$\begin{array}{c}\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_{n_k}-Tz\parallel }^2\hfill \\ \le \frac{1+\alpha }{1-\alpha }\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_{n_k}-T{x}_{n_k}\parallel }^2+\frac{4M_1(1+\alpha )}{1-\alpha }\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel T{x}_{n_k}-x_{n_k}\parallel \hfill \\ +\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_{n_k}-z\parallel }^2\hfill \\ =\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_{n_k}-z\parallel }^2.\hfill \end{array}$$

If $\alpha <0$, then, by Lemma 2.2(ii), we have

$$\begin{array}{c}{\parallel x_{n_k}-Tz\parallel }^2\hfill \\ \le {\parallel x_{n_k}-T{x}_{n_k}\parallel }^2+\frac{2}{1-\alpha }((-\alpha )\parallel T{x}_{n_k}-z\parallel +\parallel T{x}_{n_k}-Tz\parallel )\parallel x_{n_k}-T{x}_{n_k}\parallel +{\parallel x_{n_k}-z\parallel }^2\hfill \\ \le \frac{1+\alpha }{1-\alpha }{\parallel x_{n_k}-T{x}_{n_k}\parallel }^2+\frac{4M_1(1+\alpha )}{1-\alpha }\parallel T{x}_{n_k}-x_{n_k}\parallel +{\parallel x_{n_k}-z\parallel }^2.\hfill \end{array}$$

This implies again that

$$\begin{array}{c}\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_{n_k}-Tz\parallel }^2\hfill \\ \le \frac{1+\alpha }{1-\alpha }\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_{n_k}-T{x}_{n_k}\parallel }^2+\frac{4M_1(1+\alpha )}{1-\alpha }\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel T{x}_{n_k}-x_{n_k}\parallel \hfill \\ +\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_{n_k}-z\parallel }^2\hfill \\ =\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_{n_k}-z\parallel }^2.\hfill \end{array}$$

Thus, we have in all cases

$$\begin{array}{rcl}r(Tz,\{x_{n_k}\})& =& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_{n_k}-Tz\parallel \\ \le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_{n_k}-z\parallel \\ =& r(z,\{x_{n_k}\}).\end{array}$$

This means that $Tz\in A(C,\{x_{n_k}\})$. By the uniform convexity of E, we conclude that $Tz=z$.

Conversely, let $F(T)\ne \mathrm{\varnothing }$ and let $z\in F(T)$. It follows from Lemma 3.1 that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-z\parallel$ exists and hence $\{x_n\}$ is bounded. In view of Lemmas 2.1 and 2.4, we obtain a continuous strictly increasing convex function $g:[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ with $g(0)=0$ such that

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

(3.1)

In view of (3.1), we conclude by applying Lemma 3.1 that

$$\begin{array}{rcl}{\gamma }_n(1-{\gamma }_n)g(\parallel T{y}_n-x_n\parallel )& \le & {\parallel x_n-z\parallel }^2-{\parallel x_{n+1}-z\parallel }^2\\ \to & 0,\text{as }n\to \mathrm{\infty }.\end{array}$$

It follows that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}g(\parallel T{y}_n-x_n\parallel )=0\text{whenever }\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\gamma }_n(1-{\gamma }_n)>0.$$

From the property of g, we deduce that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel T{y}_n-x_n\parallel =0\text{in case }\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\gamma }_n(1-{\gamma }_n)>0.$$

(3.2)

In the same manner, we also obtain that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T{y}_n-x_n\parallel =0\text{in case }\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\gamma }_n(1-{\gamma }_n)>0.$$

(3.3)

On the other hand, from (1.2) we get

$$T{x}_n-y_n=(1-{\beta }_n)(T{x}_n-x_n),x_n-y_n={\beta }_n(x_n-T{x}_n).$$

(3.4)

Observing (3.4), we see that the assertions about the case $\alpha \le 0$ follow from (3.2) and (3.3).

In what follows, we discuss the case $0<\alpha <1$. Assume first ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\gamma }_n(1-{\gamma }_n)>0$. By Lemma 2.1 and (3.3), we see that $M_2:=sup\{\parallel T{x}_n\parallel ,\parallel T{y}_n\parallel :n\in \mathbb{N}\}<\mathrm{\infty }$. Since T is α-nonexpansive, in view of (3.4), we obtain

$$\begin{array}{c}{\parallel T{x}_n-x_n\parallel }^2\hfill \\ ={\parallel T{x}_n-T{y}_n+T{y}_n-x_n\parallel }^2\hfill \\ \le {(\parallel T{x}_n-T{y}_n\parallel +\parallel T{y}_n-x_n\parallel )}^2\hfill \\ ={\parallel T{x}_n-T{y}_n\parallel }^2+{\parallel T{y}_n-x_n\parallel }^2+2\parallel T{x}_n-T{y}_n\parallel \parallel T{y}_n-x_n\parallel \hfill \\ \le \alpha {\parallel T{x}_n-y_n\parallel }^2+\alpha {\parallel T{y}_n-x_n\parallel }^2+(1-2\alpha ){\parallel x_n-y_n\parallel }^2+{\parallel T{y}_n-x_n\parallel }^2\hfill \\ +4M_2\parallel T{y}_n-x_n\parallel \hfill \\ \le \alpha {\parallel (1-{\beta }_n)(T{x}_n-x_n)\parallel }^2+(\alpha +1){\parallel T{y}_n-x_n\parallel }^2+(1-2\alpha ){\parallel {\beta }_n(x_n-T{x}_n)\parallel }^2\hfill \\ +4M_2\parallel T{y}_n-x_n\parallel \hfill \\ \le [\alpha {(1-{\beta }_n)}^2+(1-2\alpha ){\beta }_n^2]{\parallel T{x}_n-x_n\parallel }^2+(\alpha +1){\parallel T{y}_n-x_n\parallel }^2\hfill \\ +4M_2\parallel T{y}_n-x_n\parallel .\hfill \end{array}$$

(3.5)

Case (i): If $0<\alpha <\frac{1}{2}$, then (3.5) becomes

$$\begin{array}{c}{\parallel T{x}_n-x_n\parallel }^2\hfill \\ \le [\alpha {(1-{\beta }_n)}^2+(1-2\alpha ){\beta }_n^2]{\parallel T{x}_n-x_n\parallel }^2+(\alpha +1){\parallel T{y}_n-x_n\parallel }^2+4M_2\parallel T{y}_n-x_n\parallel \hfill \\ =(1-\alpha ){\parallel T{x}_n-x_n\parallel }^2+(\alpha +1){\parallel T{y}_n-x_n\parallel }^2+4M_2\parallel T{y}_n-x_n\parallel ,\hfill \end{array}$$

since all ${\beta }_n$ are in $[0,1]$. We then derive from (3.3) that

$${\parallel T{x}_n-x_n\parallel }^2\le \frac{1+\alpha }{\alpha }{\parallel T{y}_n-x_n\parallel }^2+\frac{4M_2}{\alpha }\parallel T{y}_n-x_n\parallel \to 0,\text{as }n\to \mathrm{\infty }.$$

(3.6)

Case (ii): If $\frac{1}{2}\le \alpha <1$, then (3.5) becomes

$$\begin{array}{c}{\parallel T{x}_n-x_n\parallel }^2\hfill \\ \le [\alpha {(1-{\beta }_n)}^2+(1-2\alpha ){\beta }_n^2]{\parallel T{x}_n-x_n\parallel }^2+(\alpha +1){\parallel T{y}_n-x_n\parallel }^2+4M_2\parallel T{y}_n-x_n\parallel \hfill \\ \le \alpha {\parallel T{x}_n-x_n\parallel }^2+(\alpha +1){\parallel T{y}_n-x_n\parallel }^2+4M_2\parallel T{y}_n-x_n\parallel .\hfill \end{array}$$

We then derive from (3.3) again that

$${\parallel T{x}_n-x_n\parallel }^2\le \frac{1+\alpha }{1-\alpha }{\parallel T{y}_n-x_n\parallel }^2+\frac{4M_2}{1-\alpha }\parallel T{y}_n-x_n\parallel \to 0,\text{as }n\to \mathrm{\infty }.$$

(3.7)

Finally, we assume ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\gamma }_n(1-{\gamma }_n)>0$ instead. By (3.2) we have subsequences $\{x_{n_k}\}$ and $\{y_{n_k}\}$ of $\{x_n\}$ and $\{y_n\}$, respectively, such that

$$\underset{k\to \mathrm{\infty }}{lim}\parallel T{y}_{n_k}-x_{n_k}\parallel =0.$$

Replacing $M_2$ by the number $sup\{\parallel T{x}_{n_k}\parallel ,\parallel T{y}_{n_k}\parallel :k\in \mathbb{N}\}<\mathrm{\infty }$ and dealing with the subsequences $\{x_{n_k}\}$ and $\{y_{n_k}\}$ in (3.6) and (3.7), we will arrive at the desired conclusion that ${lim}_{k\to \mathrm{\infty }}\parallel T{x}_{n_k}-x_{n_k}\parallel =0$. This gives ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}\parallel T{x}_n-x_n\parallel =0$. □

Theorem 3.3 Let C be a nonempty closed and convex subset of a uniformly convex Banach space E with the Opial property. Let $T:C\to C$ be an α-nonexpansive mapping with a nonempty fixed point set $F(T)$ for some $\alpha <1$. Let $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ be sequences in $[0,1]$ and let $\{x_n\}$ be a sequence with $x_1$ in C defined by the Ishikawa iteration (1.2).

Assume that ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\gamma }_n(1-{\gamma }_n)>0$, and assume, in addition, ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$ if $\alpha \le 0$. Then $\{x_n\}$ converges weakly to a fixed point of T.

Proof It follows from Theorem 3.2 that $\{x_n\}$ is bounded and ${lim}_{n\to \mathrm{\infty }}\parallel T{x}_n-x_n\parallel =0$. The uniform convexity of E implies that E is reflexive; see, for example, [3]. Then there exists a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ such that $x_{n_i}\rightharpoonup p\in C$ as $i\to \mathrm{\infty }$. In view of Proposition 2.3, we conclude that $p\in F(T)$. We claim that $x_n\rightharpoonup p$ as $n\to \mathrm{\infty }$. Suppose on the contrary that there exists a subsequence $\{x_{n_j}\}$ of $\{x_n\}$ converging weakly to some q in C with $p\ne q$. By Proposition 2.3, we see that $q\in F(T)$. Lemma 3.1 says that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-z\parallel$ exists for all z in $F(T)$. The Opial property then implies

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-p\parallel & =& \underset{i\to \mathrm{\infty }}{lim}\parallel x_{n_i}-p\parallel <\underset{i\to \mathrm{\infty }}{lim}\parallel x_{n_i}-q\parallel \\ =& \underset{n\to \mathrm{\infty }}{lim}\parallel x_n-q\parallel =\underset{j\to \mathrm{\infty }}{lim}\parallel x_{n_j}-q\parallel \\ <& \underset{j\to \mathrm{\infty }}{lim}\parallel x_{n_j}-p\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-p\parallel .\end{array}$$

This is a contradiction. Thus $p=q$, and the desired assertion follows. □

Theorem 3.4 Let C be a nonempty compact and convex subset of a uniformly convex Banach space E. Let $T:C\to C$ be an α-nonexpansive mapping for some $\alpha <1$. Let $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ be sequences in $[0,1]$.

When $0<\alpha <1$, we assume ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\gamma }_n(1-{\gamma }_n)>0$. When $\alpha \le 0$, we assume either

$$\left\{\begin{array}{c}{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\gamma }_n(1-{\gamma }_n)>0,\hfill \\ {lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n<1,\hfill \end{array}\mathit{\text{or}}\right\{\begin{array}{c}{lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\gamma }_n(1-{\gamma }_n)>0,\hfill \\ {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1.\hfill \end{array}$$

Let $\{x_n\}$ be a sequence with $x_1$ in C defined by the Ishikawa iteration (1.2). Then $\{x_n\}$ converges strongly to a fixed point z of T.

Proof Since C is bounded, it follows from Lemma 2.5 that the fixed point set $F(T)$ of T is nonempty. In view of Theorem 3.2, the sequence $\{x_n\}$ is bounded and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}\parallel T{x}_n-x_n\parallel =0$. By the compactness of C, there exists a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ converging strongly to some z in C, and ${lim}_{k\to \mathrm{\infty }}\parallel T{x}_{n_k}-x_{n_k}\parallel =0$. In particular, $\{T{x}_{n_k}\}$ is bounded. Let $M_3=sup\{\parallel x_{n_k}\parallel ,\parallel T{x}_{n_k}\parallel ,\parallel z\parallel ,\parallel Tz\parallel :k\in \mathbb{N}\}<\mathrm{\infty }$. If $0\le \alpha <1$, then, in view of Lemma 2.2(i), we obtain

$$\begin{array}{c}{\parallel x_{n_k}-Tz\parallel }^2\hfill \\ \le \frac{1+\alpha }{1-\alpha }{\parallel x_{n_k}-T{x}_{n_k}\parallel }^2+\frac{2}{1-\alpha }(\alpha \parallel x_{n_k}-z\parallel +\parallel T{x}_{n_k}-Tz\parallel )\parallel x_{n_k}-T{x}_{n_k}\parallel +{\parallel x_{n_k}-z\parallel }^2\hfill \\ \le \frac{1+\alpha }{1-\alpha }{\parallel x_{n_k}-T{x}_{n_k}\parallel }^2+\frac{4M_3(1+\alpha )}{1-\alpha }\parallel T{x}_{n_k}-x_{n_k}\parallel +{\parallel x_{n_k}-z\parallel }^2.\hfill \end{array}$$

Therefore,

$$\begin{array}{c}\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_{n_k}-Tz\parallel }^2\hfill \\ \le \frac{1+\alpha }{1-\alpha }\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_{n_k}-T{x}_{n_k}\parallel }^2+\frac{4M_3(1+\alpha )}{1-\alpha }\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel T{x}_{n_k}-x_{n_k}\parallel \hfill \\ +\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_{n_k}-z\parallel }^2.\hfill \end{array}$$

If $\alpha <0$, then, in view of Lemma 2.2(ii), we obtain

$$\begin{array}{c}{\parallel x_{n_k}-Tz\parallel }^2\hfill \\ \le {\parallel x_{n_k}-T{x}_{n_k}\parallel }^2+\frac{2}{1-\alpha }[(-\alpha )\parallel T{x}_{n_k}-z\parallel +\parallel T{x}_{n_k}-Tz\parallel ]\parallel x_{n_k}-T{x}_{n_k}\parallel +{\parallel x_{n_k}-z\parallel }^2\hfill \\ \le {\parallel x_{n_k}-T{x}_{n_k}\parallel }^2+\frac{4M_3(1-\alpha )}{1-\alpha }\parallel T{x}_{n_k}-x_{n_k}\parallel +{\parallel x_{n_k}-z\parallel }^2.\hfill \end{array}$$

Therefore,

$$\begin{array}{c}\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_{n_k}-Tz\parallel }^2\hfill \\ \le \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_{n_k}-T{x}_{n_k}\parallel }^2+4M_3\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel T{x}_{n_k}-x_{n_k}\parallel +\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_{n_k}-z\parallel }^2.\hfill \end{array}$$

It follows that ${lim}_{k\to \mathrm{\infty }}\parallel x_{n_k}-Tz\parallel =0$. Thus we have $Tz=z$. By Lemma 3.1, ${lim}_{n\to \mathrm{\infty }}\parallel x_n-z\parallel$ exists. Therefore, z is the strong limit of the sequence $\{x_n\}$. □

Let C be a nonempty closed and convex subset of a Banach space E. A mapping $T:C\to C$ is said to satisfy condition (I) [10] if

there exists a nondecreasing function $f:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ with $f(0)=0$ and $f(r)>0$ for all $r>0$ such that

$$d(x,Tx)\ge f\left(d(x,F(T))\right),\mathrm{\forall }x\in C.$$

Using Theorem 3.2, we can prove the following result.

Theorem 3.5 Let C be a nonempty closed and convex subset of a uniformly convex Banach space E. Let $T:C\to C$ be an α-nonexpansive mapping with a nonempty fixed point set $F(T)$ for some $\alpha <1$. Let $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ be sequences in $[0,1]$. When $0<\alpha <1$, we assume ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\gamma }_n(1-{\gamma }_n)>0$. When $\alpha \le 0$, we assume either

$$\left\{\begin{array}{c}{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\gamma }_n(1-{\gamma }_n)>0,\hfill \\ {lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n<1,\hfill \end{array}\mathit{\text{or}}\right\{\begin{array}{c}{lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\gamma }_n(1-{\gamma }_n)>0,\hfill \\ {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1.\hfill \end{array}$$

Let $\{x_n\}$ be a sequence with $x_1$ in C defined by the Ishikawa iteration (1.2). If T satisfies condition (I), then $\{x_n\}$ converges strongly to a fixed point z of T.

Proof

It follows from Theorem 3.2 that

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

Therefore, there is a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that

$$\underset{k\to \mathrm{\infty }}{lim}\parallel T{x}_{n_k}-x_{n_k}\parallel =0.$$

Since T satisfies condition (I), with respect to the sequence $\{x_{n_k}\}$, we obtain

$$\underset{k\to \mathrm{\infty }}{lim}d(x_{n_k},F(T))=0.$$

This implies that, there exist a subsequence of $\{x_{n_k}\}$, denoted also by $\{x_{n_k}\}$, and a sequence $\{z_k\}$ in $F(T)$ such that

$$d(x_{n_k},z_k)<\frac{1}{2^k},\mathrm{\forall }k\in \mathbb{N}.$$

(3.8)

In view of Lemma 3.1, we have

$$\parallel x_{n_{k+1}}-z_k\parallel \le \parallel x_{n_k}-z_k\parallel <\frac{1}{2^k},\mathrm{\forall }k\in \mathbb{N}.$$

This implies

$$\begin{array}{rcl}\parallel z_{k+1}-z_k\parallel & \le & \parallel z_{k+1}-x_{n_{k+1}}\parallel +\parallel x_{n_{k+1}}-z_k\parallel \\ \le & \frac{1}{2^{(k+1)}}+\frac{1}{2^k}\\ <& \frac{1}{2^{(k-1)}},\mathrm{\forall }k=1,2,\dots .\end{array}$$

Consequently, $\{z_k\}$ is a Cauchy sequence in $F(T)$. Due to the closedness of $F(T)$ in E (see Lemma 2.1), we deduce that ${lim}_{k\to \mathrm{\infty }}{z}_k=z$ for some z in $F(T)$. It follows from (3.8) that ${lim}_{k\to \mathrm{\infty }}{x}_{n_k}=z$. By Lemma 3.1, we see that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-z\parallel$ exists. This forces ${lim}_{n\to \mathrm{\infty }}\parallel x_n-z\parallel =0$. □

The following examples explain why we need to impose some conditions on the control sequences in previous theorems.

Examples 3.6 (a) Let $T:[-1,1]\to [-1,1]$ be defined by $Tx=-x$. Then T is a 0-nonexpansive (i.e., nonexpansive) mapping. Setting all ${\beta }_n=1$, the Ishikawa iteration (1.2) provides a sequence

$$x_{n+1}={\gamma }_n{T}^2{x}_n+(1-{\gamma }_n)x_n=x_n,\mathrm{\forall }n=1,2,\dots ,$$

no matter how we choose $\{{\gamma }_n\}$. Unless $x_1=0$, we can never reach the unique fixed point 0 of T via $\{x_n\}$.

(b) Let $T:[0,4]\to [0,4]$ be defined by

$$Tx=\{\begin{array}{cc}0\hfill & \text{if }x\ne 4,\hfill \\ 2\hfill & \text{if }x=4.\hfill \end{array}$$

Then T is a $\frac{1}{2}$-nonexpansive mapping. Indeed, for any x in $[0,4)$ and $y=4$, we have

$${|Tx-Ty|}^2=4\le 8+\frac{1}{2}{|x-2|}^2=\frac{1}{2}{|Tx-y|}^2+\frac{1}{2}{|x-Ty|}^2.$$

The other cases can be verified similarly. It is worth mentioning that T is neither nonexpansive nor continuous. Setting all ${\beta }_n=1$, the Ishikawa iteration (1.2) provides a sequence

$$x_{n+1}={\gamma }_n{T}^2{x}_n+(1-{\gamma }_n)x_n,\mathrm{\forall }n=1,2,\dots .$$

For any arbitrary starting point $x_1$ in $[0,4]$, we have $T^2{x}_n=0$ and

$$\begin{array}{rcl}{x}_{n+1}& =& (1-{\gamma }_n)x_n\\ =& (1-{\gamma }_1)(1-{\gamma }_2)\cdots (1-{\gamma }_n)x_1\\ =& \prod _{k=1}^n(1-{\gamma }_k)x_1,\mathrm{\forall }n=1,2,\dots .\end{array}$$

Consider two possible choices of the values of ${\gamma }_n$:

Case 1. If we set ${\gamma }_n=\frac{1}{2}$, $\mathrm{\forall }n=1,2,\dots$ , then ${lim}_{n\to \mathrm{\infty }}{\gamma }_n(1-{\gamma }_n)=1/4>0$ and $x_n\to 0$, the unique fixed point of T.

Case 2. If we set ${\gamma }_n=\frac{1}{{(n+1)}^2}$, $\mathrm{\forall }n=1,2,\dots$ , then ${lim}_{n\to \mathrm{\infty }}{\gamma }_n(1-{\gamma }_n)=0$ and $x_n=\frac{n+2}{2n+2}{x}_1\to x_1/2$. Unless $x_1=0$, we can never reach the unique fixed point 0 of T via $x_n$.

4 An existence result in $\mathrm{CAT}(0)$ spaces

Let $(X,d)$ be a metric space. A geodesic path joining x to y in X (or briefly, a geodesic from x to y) is a map c from a closed interval $[0,l]\subset \mathbb{R}$ into X such that $c(0)=x$, $c(l)=y$, and $d(c(t),c(t^{\prime }))=|t-t^{\prime }|$ for all t, $t^{\prime }$ in $[0,l]$. In particular, c is an isometry and $d(x,y)=l$. The image α of c is called a geodesic (or metric) segment joining x and y. When it is unique, this geodesic is denoted by $[x,y]$. The space $(X,d)$ is said to be a geodesic space if every two points of X are joined by a geodesic, and X is said to be a uniquely geodesic if there exists exactly one geodesic joining x and y for each x, y in X. A subset Y of X is said to be convex if Y includes every geodesic segment joining any two of its points.

A geodesic triangle $\mathrm{\Delta }(x_1,x_2,x_3)$ in a geodesic space $(X,d)$ consists of three points $x_1$, $x_2$, $x_3$ in X (the vertices of Δ), together with a geodesic segment between each pair of vertices (the edges of Δ). A comparison triangle for a geodesic triangle $\mathrm{\Delta }(x_1,x_2,x_3)$ in a geodesic space $(X,d)$ is a triangle $\overline{\mathrm{\Delta }}(x_1,x_2,x_3):=\mathrm{\Delta }({\overline{x}}_1,{\overline{x}}_2,{\overline{x}}_3)$ in the Euclidean plane ${\mathbb{E}}^2$ together with a one-to-one correspondence $x\mapsto \overline{x}$ from Δ onto $\overline{\mathrm{\Delta }}$ such that it is an isometry on each of the three segments. A geodesic space X is said to be a $CAT(0)$ space if all geodesic triangles Δ satisfy the $CAT(0)$ inequality:

$$d(x,y)\le d_{{\mathbb{E}}^2}(\overline{x},\overline{y}),\mathrm{\forall }x,y\in \mathrm{\Delta }.$$

It is easy to see that a $CAT(0)$ space is uniquely geodesic.

It is well known that any complete, simply connected Riemannian manifold having nonpositive sectional curvature is a $CAT(0)$ space. Other examples include inner product spaces, ℝ-trees (see, for example, [11]), Euclidean building (see, for example, [12]), and the complex Hilbert ball with a hyperbolic metric (see, for example, [8]). For a thorough discussion on other spaces and on the fundamental role they play in geometry, see, for example, [12–14].

We collect some properties of $CAT(0)$ spaces. For more details, we refer the readers to [15–17].

Lemma 4.1 [16]

Let $(X,d)$ be a $CAT(0)$ space. Then the following assertions hold.

• (i) For x, y in X and t in $[0,1]$, there exists a unique point z in $[x,y]$ such that

  $$d(x,z)=td(x,y)\mathit{\text{and}}d(y,z)=(1-t)d(x,y).$$

  (4.1)

  We use the notation $(1-t)x\oplus ty$ for the unique point z satisfying (4.1).

• (ii) For x, y in X and t in $[0,1]$, we have

  $$d((1-t)x\oplus ty,z)\le (1-t)d(x,z)+td(y,z).$$

The notion of asymptotic centers in a Banach space can be extended to a $CAT(0)$ space as well by simply replacing the distance defined by $\parallel \cdot -\cdot \parallel$ with the one defined by the metric $d(\cdot ,\cdot )$. In particular, in a $CAT(0)$ space, $A(C,\{x_n\})$ consists of exactly one point whenever C is a closed and convex set and $\{x_n\}$ is a bounded sequence; see [[18], Proposition 7].

Definition 4.2 [19, 20]

A sequence $\{x_n\}$ in a $CAT(0)$ space X is said to Δ-converge to x in X if x is the unique asymptotic center of $\{x_{n_k}\}$ for every subsequence $\{x_{n_k}\}$ of $\{x_n\}$. In this case, we write $\mathrm{\Delta }\text{-}{lim}_{n\to \mathrm{\infty }}{x}_n=x$, and we call x the Δ-limit of $\{x_n\}$.

Lemma 4.3 [19]

Every bounded sequence in a complete $CAT(0)$ space X has a Δ-convergent subsequence.

Lemma 4.4 [21]

Let C be a closed and convex subset of a complete $CAT(0)$ space X. If $\{x_n\}$ is a bounded sequence in C, then the asymptotic center of $\{x_n\}$ is in C.

Lemma 4.5 [22]

Let X be a complete $CAT(0)$ space and let $x\in X$. Suppose that $0<b\le t_n\le c<1$ and $x_n,y_n\in X$ for $n=1,2,\dots$ . If for some $r\ge 0$ we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_n,x)\le r,\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(y_n,x)\le r,\mathit{\text{and}}\underset{n\to \mathrm{\infty }}{lim}d(t_n{x}_n\oplus (1-t_n)y_n,x)=r,$$

then ${lim}_{n\to \mathrm{\infty }}d(x_n,y_n)=0$.

Recall that the Ishikawa iteration in $CAT(0)$ spaces is described as follows: For any initial point $x_1$ in C, we define the iterates $\{x_n\}$ by

$$\{\begin{array}{c}{y}_n={\beta }_{n}T{x}_n\oplus (1-{\beta }_n)x_n,\hfill \\ x_{n+1}={\gamma }_{n}T{y}_n\oplus (1-{\gamma }_n)x_n,\hfill \end{array}$$

(4.2)

where the sequences $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ satisfy some appropriate conditions.

We introduce the notion of α-nonexpansive mappings of $CAT(0)$ spaces.

Definition 4.6 Let C be a nonempty subset of a $CAT(0)$ space X and let $\alpha <1$. A mapping $T:C\to X$ is said to be α-nonexpansive if

$$d{(Tx,Ty)}^2\le \alpha d{(Tx,y)}^2+\alpha d{(x,Ty)}^2+(1-2\alpha )d{(x,y)}^2,\mathrm{\forall }x,y\in C.$$

The following is the $CAT(0)$ counterpart to Lemma 2.5. However, we do not know if the compactness assumption can be removed from the negative α case.

Lemma 4.7 Let C be a nonempty closed and convex subset of a complete $CAT(0)$ space X. Let $T:C\to C$ be an α-nonexpansive mapping for some $\alpha <1$. In the case $0\le \alpha <1$, we have $F(T)\ne \mathrm{\varnothing }$ if and only if ${\{T^{n}x\}}_{n=1}^{\mathrm{\infty }}$ is bounded for some x in C. If C is compact, we always have $F(T)\ne \mathrm{\varnothing }$.

Proof Assume first that $0\le \alpha <1$. The necessity is obvious. We verify the sufficiency. Suppose that ${\{T^{n}x\}}_{n=1}^{\mathrm{\infty }}$ is bounded for some x in C. Set $x_n:=T^{n}x$ for $n=1,2,\dots$ . By the boundedness of ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$, there exists z in X such that $A(C,\{x_n\})=\{z\}$. It follows from Lemma 4.4 that $z\in C$. Furthermore, we have

$$d{(x_n,Tz)}^2\le \alpha d{(x_n,z)}^2+\alpha d{(x_{n-1},Tz)}^2+(1-2\alpha )d{(x_{n-1},z)}^2,\mathrm{\forall }n=1,2,\dots .$$

This implies

$$\begin{array}{c}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d{(x_n,Tz)}^2\hfill \\ \le \alpha \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d{(x_n,z)}^2+\alpha \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d{(x_{n-1},Tz)}^2+(1-2\alpha )\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d{(x_{n-1},z)}^2.\hfill \end{array}$$

Thus,

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_n,Tz)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_n,z).$$

Consequently, $Tz\in A(\{x_n\})=\{z\}$, ensuring that $F(T)\ne \mathrm{\varnothing }$.

Next, we assume $\alpha <0$ and C is compact. In particular, T is continuous and the sequence of $x_n:=T^{n}x$ for any x in C is bounded. In what follows, we adapt the arguments in [2] with slight modifications.

Let μ be a Banach limit, i.e., μ is a bounded unital positive linear functional of ${\ell }_{\mathrm{\infty }}$ such that $\mu \circ s=\mu$. Here, s is the left shift operator on ${\ell }_{\mathrm{\infty }}$. We write ${\mu }_n{a}_n$ for the value of $\mu (a)$ with $a=(a_n)$ in ${\ell }_{\mathrm{\infty }}$ as usual. In particular, ${\mu }_n{a}_{n+1}=\mu (s(a))=\mu (a)={\mu }_n{a}_n$. As showed in [[2], Lemmas 3.1 and 3.2], we have

$${\mu }_{n}d{(x_n,Ty)}^2\le {\mu }_{n}d{(x_n,y)}^2,\mathrm{\forall }y\in C,$$

(4.3)

and

$$g(y):={\mu }_{n}d{(x_n,y)}^2$$

defines a continuous function from C into ℝ.

By compactness, there exists y in C such that $g(y)=infg(C)$. Suppose that there is another z in C such that $g(z)=g(y)$. Let m be the midpoint in the geodesic segment joining y to z. In view of Lemma 4.1, we see that g is convex. Thus, $g(m)=g(y)$ too. Observing the comparison triangles in ${\mathbb{E}}^2$, we have

$$d{(x_n,y)}^2+d{(x_n,z)}^2\ge 2d{(x_n,m)}^2+\frac{1}{2}d{(y,z)}^2,\mathrm{\forall }n=1,2,\dots .$$

Consequently,

$${\mu }_{n}d{(x_n,y)}^2+{\mu }_{n}d{(x_n,z)}^2\ge 2{\mu }_{n}d{(x_n,m)}^2+\frac{1}{2}{\mu }_{n}d{(y,z)}^2.$$

This amounts to say

$$g(y)+g(z)\ge 2g(m)+\frac{1}{2}d{(y,z)}^2.$$

Since $g(y)=g(z)=g(m)$, we have $y=z$. Finally, it follows from (4.3) that $g(Ty)\le g(y)=infg(C)$. By uniqueness, we have $Ty=y\in F(T)$. □

The proofs of the following results are similar to those in Sections 2 and 3.

Lemma 4.8 Let C be a nonempty subset of a $CAT(0)$ space X. Let $T:C\to X$ be an α-nonexpansive mapping for some $\alpha <1$ such that $F(T)\ne \mathrm{\varnothing }$. Then T is quasi-nonexpansive.

Lemma 4.9 Let C be a nonempty closed and convex subset of a $CAT(0)$ space X. Let $T:C\to X$ be an α-nonexpansive mapping for some $\alpha <1$. Then the following assertions hold.

• (i) If $0\le \alpha <1$, then

  $$d{(x,Ty)}^2\le \frac{1+\alpha }{1-\alpha }d{(x,Tx)}^2+\frac{2}{1-\alpha }(\alpha d(x,y)+d(Tx,Ty))d(x,Tx)+d{(x,y)}^2,\mathrm{\forall }x,y\in C.$$

• (ii) If $\alpha <0$, then

  $$d{(x,Ty)}^2\le d{(x,Tx)}^2+\frac{2}{1-\alpha }[(-\alpha )d(Tx,y)+d(Tx,Ty)]d(x,Tx)+d{(x,y)}^2,\mathrm{\forall }x,y\in C.$$

Lemma 4.10 Let C be a nonempty closed and convex subset of a $CAT(0)$ space X. Let $T:C\to C$ be an α-nonexpansive mapping for some $\alpha <1$. Let a sequence $\{x_n\}$ with $x_1$ in C be defined by (4.2) such that $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ are arbitrary sequences in $[0,1]$. Let $z\in F(T)$. Then the following assertions hold:

• (1) $max\{d(x_{n+1},z),d(y_n,z)\}\le d(x_n,z)$ for $n=1,2,\dots$ .

• (2) ${lim}_{n\to \mathrm{\infty }}d(x_n,z)$ exists.

• (3) ${lim}_{n\to \mathrm{\infty }}d(x_n,F(T))$ exists.

Lemma 4.11 [15]

Let C be a nonempty convex subset of a $CAT(0)$ space X and let $T:C\to C$ be a quasi-nonexpansive map whose fixed point set is nonempty. Then $F(T)$ is closed, convex and hence contractible.

The following result is deduced from Lemmas 4.8 and 4.11.

Lemma 4.12 Let C be a nonempty convex subset of a $CAT(0)$ space X and let $T:C\to C$ be an α-nonexpansive mapping with a nonempty fixed point set $F(T)$ for some $\alpha <1$. Then $F(T)$ is closed, convex, and hence contractible.

Lemma 4.13 Let C be a nonempty closed and convex subset of a complete $CAT(0)$ space X and let $T:C\to C$ be an α-nonexpansive mapping for some $\alpha <1$. If $\{x_n\}$ is a sequence in C such that $d(T{x}_n,x_n)\to 0$ and $\mathrm{\Delta }\text{-}{lim}_{n\to \mathrm{\infty }}{x}_n=z$ for some z in X, then $z\in C$ and $Tz=z$.

Proof It follows from Lemma 4.4 that $z\in C$.

Let $0\le \alpha <1$. By Lemma 4.9(i), we deduce that

$$d{(x_n,Tz)}^2\le \frac{1+\alpha }{1-\alpha }d{(x_n,T{x}_n)}^2+\frac{2}{1-\alpha }(\alpha d(x_n,z)+d(T{x}_n,Tz))d(x_n,T{x}_n)+d{(x_n,z)}^2$$

for all n in ℕ. Thus we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_n,Tz)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_n,z).$$

Let $\alpha <0$. Then, by Lemma 4.9(ii), we have

$$d{(x_n,Tz)}^2\le d{(x_n,T{x}_n)}^2+\frac{2}{1-\alpha }[(-\alpha )d(T{x}_n,z)+d(T{x}_n,Tz)]d(x_n,T{x}_n)+d{(x_n,z)}^2$$

for all n in ℕ. This implies again that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_n,Tz)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_n,z).$$

By the uniqueness of asymptotic centers, $Tz=z$. □

5 Fixed point and convergence theorems in $\mathrm{CAT}(0)$ spaces

In this section, we extend our results in Section 3 to $CAT(0)$ spaces.

Theorem 5.1 Let C be a nonempty closed and convex subset of a complete $CAT(0)$ space X and let $T:C\to C$ be an α-nonexpansive mapping for some $\alpha <1$. Let $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ be sequences in $[0,1]$ such that $0<{lim\hspace{0.17em}inf}_{k\to \mathrm{\infty }}{\gamma }_{n_k}\le {lim\hspace{0.17em}sup}_{k\to \mathrm{\infty }}{\gamma }_{n_k}<1$ for a subsequence $\{{\gamma }_{n_k}\}$ of $\{{\gamma }_n\}$. In the case $\alpha \le 0$, we assume also that ${lim\hspace{0.17em}sup}_{k\to \mathrm{\infty }}{\beta }_{n_k}<1$. Let $\{x_n\}$ be a sequence with $x_1$ in C defined by (4.2). Then the fixed point set $F(T)\ne \mathrm{\varnothing }$ if and only if $\{x_n\}$ is bounded and ${lim}_{k\to \mathrm{\infty }}d(T{x}_{n_k},x_{n_k})=0$.

Proof Suppose that $F(T)\ne \mathrm{\varnothing }$ and z in $F(T)$ is arbitrarily chosen. By Lemma 4.10, ${lim}_{n\to \mathrm{\infty }}d(x_n,z)$ exists and $\{x_n\}$ is bounded. Let

$$\underset{n\to \mathrm{\infty }}{lim}d(x_n,z)=l.$$

(5.1)

It follows from Lemmas 4.8 and 4.1(ii) that

$$\begin{array}{rcl}d(T{y}_n,z)& \le & d(y_n,z)\\ =& d({\beta }_{n}T{x}_n\oplus (1-{\beta }_n)x_n,z)\\ \le & {\beta }_{n}d(T{x}_n,z)+(1-{\beta }_n)d(x_n,z)\\ \le & {\beta }_{n}d(x_n,z)+(1-{\beta }_n)d(x_n,z)\\ =& d(x_n,z).\end{array}$$

Thus, we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(T{y}_n,z)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(y_n,z)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_n,z)=l.$$

(5.2)

On the other hand, it follows from (4.2) and (5.1) that

$$\underset{n\to \mathrm{\infty }}{lim}d({\gamma }_{n}T{y}_n\oplus (1-{\gamma }_n)x_n,z)=\underset{n\to \mathrm{\infty }}{lim}d(x_{n+1},z)=l.$$

(5.3)

In view of (5.1)-(5.3) and Lemma 4.5, we conclude that

$$\underset{k\to \mathrm{\infty }}{lim}d(T{y}_{n_k},x_{n_k})=0.$$

By simply replacing $\parallel \cdot -\cdot \parallel$ with $d(\cdot ,\cdot )$ in the proof of Theorem 3.2, we have the desired result ${lim}_{k\to \mathrm{\infty }}d(T{x}_{n_k},x_{n_k})=0$. The proof in the converse direction follows similarly. □

Theorem 5.2 Let C be a nonempty closed and convex subset of a complete $CAT(0)$ space X and let $T:C\to C$ be an α-nonexpansive mapping for some $\alpha <1$. Let $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ be sequences in $[0,1]$ such that $0<{lim\hspace{0.17em}inf}_{k\to \mathrm{\infty }}{\gamma }_{n_k}\le {lim\hspace{0.17em}sup}_{k\to \mathrm{\infty }}{\gamma }_{n_k}<1$ for a subsequence $\{{\gamma }_{n_k}\}$ of $\{{\gamma }_n\}$. In the case $\alpha \le 0$, we assume also that ${lim\hspace{0.17em}sup}_{k\to \mathrm{\infty }}{\beta }_{n_k}<1$. Let $\{x_n\}$ be a sequence with $x_1$ in C defined by (4.2). If $F(T)\ne \mathrm{\varnothing }$, then $\{x_{n_k}\}$ Δ-converges to a fixed point of T.

Proof It follows from Theorem 5.1 that $\{x_n\}$ is bounded and ${lim}_{k\to \mathrm{\infty }}d(T{x}_{n_k},x_{n_k})=0$. Denote by ${\omega }_w(x_{n_k}):=\bigcup A(C,\{u_n\})$, where the union is taken over all subsequences $\{u_n\}$ of $\{x_{n_k}\}$. We prove that ${\omega }_w(x_{n_k})\subset F(T)$. Let $u\in {\omega }_w(x_{n_k})$. Then there exists a subsequence $\{u_n\}$ of $\{x_{n_k}\}$ such that $A(C,\{u_n\})=\{u\}$. In view of Lemmas 4.3 and 4.4, there exists a subsequence $\{v_n\}$ of $\{u_n\}$ such that $\mathrm{\Delta }\text{-}{lim}_{n\to \mathrm{\infty }}{v}_n=v$ for some v in C. Since ${lim}_{n\to \mathrm{\infty }}d(T{v}_n,v_n)=0$, Lemma 4.13 implies that $v\in F(T)$. By Lemma 4.10, ${lim}_{n\to \mathrm{\infty }}d(x_n,v)$ exists. We claim that $u=v$. For else, the uniqueness of asymptotic centers implies that

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(v_n,v)& <& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(v_n,u)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(u_n,u)\\ <& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(u_n,v)=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_n,v)=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(v_n,v),\end{array}$$

which is a contradiction. Thus, we have $u=v\in F(T)$ and hence ${\omega }_w(x_{n_k})\subset F(T)$.

Now, we prove that $\{x_{n_k}\}$ Δ-converges to a fixed point of T. It suffices to show that ${\omega }_w(x_{n_k})$ consists of exactly one point. Let $\{u_n\}$ be a subsequence of $\{x_{n_k}\}$. In view of Lemmas 4.3 and 4.4, there exists a subsequence $\{v_n\}$ of $\{u_n\}$ such that $\mathrm{\Delta }\text{-}{lim}_{n\to \mathrm{\infty }}{v}_n=v$ for some v in C. Let $A(C,\{u_n\})=\{u\}$ and $A(C,\{x_{n_k}\})=\{x\}$. By the argument mentioned above, we have $u=v$ and $v\in F(T)$. We show that $x=v$. If it is not the case, then the uniqueness of asymptotic centers implies that

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(v_n,v)& <& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(v_n,x)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_n,x)\\ <& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_n,v)=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(v_n,v),\end{array}$$

which is a contradiction. Thus we have the desired result. □

Theorem 5.3 Let C be a nonempty compact convex subset of a complete $CAT(0)$ space X and let $T:C\to C$ be an α-nonexpansive mapping for some $\alpha <1$. Let $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ be sequences in $[0,1]$ such that $0<{lim\hspace{0.17em}inf}_{k\to \mathrm{\infty }}{\gamma }_{n_k}\le {lim\hspace{0.17em}sup}_{k\to \mathrm{\infty }}{\gamma }_{n_k}<1$ for a subsequence $\{{\gamma }_{n_k}\}$ of $\{{\gamma }_n\}$. In the case $\alpha \le 0$, we assume also that ${lim\hspace{0.17em}sup}_{k\to \mathrm{\infty }}{\beta }_{n_k}<1$. Let $\{x_n\}$ be a sequence with $x_1$ in C defined by (4.2). Then $\{x_n\}$ converges in metric to a fixed point of T.

Proof Using Lemmas 4.7 and 4.9 and replacing $\parallel \cdot -\cdot \parallel$ with $d(\cdot ,\cdot )$ in the proof of Theorem 3.4, we conclude the desired result. □

As in the proof of Theorem 3.5, we can verify the following result.

Theorem 5.4 Let C be a nonempty compact convex subset of a complete $CAT(0)$ space X and let $T:C\to C$ be an α-nonexpansive mapping for some $\alpha <1$. Let $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ be sequences in $[0,1]$ such that $0<{lim\hspace{0.17em}inf}_{k\to \mathrm{\infty }}{\gamma }_{n_k}\le {lim\hspace{0.17em}sup}_{k\to \mathrm{\infty }}{\gamma }_{n_k}<1$ for a subsequence $\{{\gamma }_{n_k}\}$ of $\{{\gamma }_n\}$. In the case $\alpha \le 0$, we assume also that ${lim\hspace{0.17em}sup}_{k\to \mathrm{\infty }}{\beta }_{n_k}<1$. Let $\{x_n\}$ be a sequence with $x_1$ in C defined by (4.2). If T satisfies condition (I), then $\{x_n\}$ converges in metric to a fixed point of T.