On an open problem of Kyung Soo Kim

Abstract

We prove a convergence theorem of the Mann iteration scheme for a uniformly L-Lipschitzian asymptotically demicontractive mapping in a $\operatorname{CAT}(\kappa)$ space with $\kappa>0$. We also obtain a convergence theorem of the Ishikawa iteration scheme for a uniformly L-Lipschitzian asymptotically hemicontractive mapping. Our results provide a complete solution to an open problem raised by Kim (Abstr. Appl. Anal. 2013:381715, 2013).

1 Introduction

Roughly speaking, $\operatorname{CAT}(\kappa)$ spaces are geodesic spaces of bounded curvature and generalizations of Riemannian manifolds of sectional curvature bounded above. The precise definition is given below. The letters C, A, and T stand for Cartan, Alexandrov, and Toponogov, who have made important contributions to the understanding of curvature via inequalities for the distance function, and κ is a real number that we impose it as the curvature bound of the space.

Fixed point theory in $\operatorname{CAT}(\kappa)$ spaces was first studied by Kirk [1, 2]. His work was followed by a series of new works by many authors, mainly focusing on $\operatorname{CAT}(0)$ spaces (see e.g., [3–25]). Since any $\operatorname{CAT}(\kappa )$ space is a $\operatorname{CAT}(\kappa')$ space for $\kappa' \geq\kappa$, all results for $\operatorname{CAT}(0)$ spaces immediately apply to any $\operatorname {CAT}(\kappa)$ space with $\kappa\leq0$. However, there are only a few articles that contain fixed point results in the setting of $\operatorname{CAT}(\kappa)$ spaces with $\kappa>0$, because in this case the proof seems to be more complicated.

The notion of uniformly L-Lipschitzian mappings, which is more general than the notion of asymptotically nonexpansive mappings, was introduced by Goebel and Kirk [26]. In 1991, Schu [27] proved the strong convergence of Mann iteration for asymptotically nonexpansive mappings in Hilbert spaces. Qihou [28] extended Schu’s result to the general setting of asymptotically demicontractive mappings and also obtained the strong convergence of Ishikawa iteration for asymptotically hemicontractive mappings. Recently, Kim [29] proved the analogous results of Qihou in the framework of the so-called $\operatorname{CAT}(0)$ spaces. Precisely, Kim obtained the following theorems.

Theorem A

Let $(X, \rho)$ be a complete $\operatorname {CAT}(0)$ space, C be a nonempty bounded closed convex subset of X, and $T:C\to C$ be a completely continuous and uniformly L-Lipschitzian asymptotically demicontractive mapping with constant $k\in[0,1)$ and sequence $\{a_{n}\}$ in $[1,\infty)$ such that $\sum^{\infty}_{n=1}(a_{n}^{2}-1)<\infty$. Let $\{\alpha_{n}\}$ be a sequence in $[\varepsilon, 1-k-\varepsilon]$ for some $\varepsilon>0$. Given $x_{1}\in C$, define the iteration scheme $\{x_{n}\}$ by

$$x_{n+1}=(1-\alpha_{n})x_{n}\oplus \alpha_{n} T^{n}x_{n}, \quad n\geq1. $$

Then $\{x_{n}\}$ converges strongly to a fixed point of T.

Theorem B

Let $(X, \rho)$ be a complete $\operatorname {CAT}(0)$ space, let C be a nonempty bounded closed convex subset of X, and let $T:C\to C$ be a completely continuous and uniformly L-Lipschitzian asymptotically hemicontractive mapping with sequence $\{a_{n}\}$ in $[1,\infty)$ such that $\sum_{n=1}^{\infty}(a_{n}-1)<\infty$. Let $\{\alpha_{n}\}, \{\beta_{n}\}\subset[0,1]$ be such that $\varepsilon \leq\alpha_{n}\leq\beta_{n}\leq b$ for some $\varepsilon>0$ and $b\in (0,\frac{\sqrt{1+L^{2}}-1}{L^{2}} )$. Given $x_{1}\in C$, define the iteration scheme $\{x_{n}\}$ by

$$\begin{aligned}& x_{n+1}=(1-\alpha_{n})x_{n}\oplus \alpha_{n}T^{n}y_{n}, \\& y_{n}=(1-\beta_{n})x_{n}\oplus\beta_{n} T^{n}x_{n},\quad n\geq1. \end{aligned}$$

Then $\{x_{n}\}$ converges strongly to a fixed point of T.

In [29], the author raised the following problem.

Problem

Can we extend Theorems A and B to the general setting of $\operatorname{CAT}(\kappa)$ spaces with $\kappa>0$?

The purpose of the paper is to solve this problem. Our main discoveries are Theorems 3.2 and 3.6.

2 Preliminaries

Let $(X,\rho)$ be a metric space. A geodesic path joining $x\in X$ to $y\in X$ (or, more briefly, a geodesic from x to y) is a map c from a closed interval $[0,l]\subset \mathbb{R}$ to X such that $c(0)=x$, $c(l)=y$, and $\rho(c(t),c(t^{\prime}))=|t-t^{\prime}|$ for all $t,t^{\prime}\in[0,l]$. In particular, c is an isometry and $\rho(x,y)=l$. The image $c([0,l])$ of c is called a geodesic segment joining x and y. When it is unique this geodesic segment is denoted by $[x,y]$. This means that $z\in[x, y]$ if and only if there exists $\alpha\in[0, 1]$ such that

$$\rho(x,z)=(1-\alpha)\rho(x,y) \quad \mbox{and} \quad \rho(y,z)=\alpha\rho(x,y). $$

In this case, we write $z=\alpha x\oplus(1-\alpha)y$. The space $(X,\rho)$ is said to be a geodesic space (D-geodesic space) if every two points of X (every two points of distance smaller than D) are joined by a geodesic, and X is said to be uniquely geodesic (D-uniquely geodesic) if there is exactly one geodesic joining x and y for each $x, y\in X$ (for $x, y \in X$ with $\rho(x, y) < D$). A subset C of X is said to be convex if C includes every geodesic segment joining any two of its points. The set C is said to be bounded if

$$\operatorname{diam}(C) := \sup\bigl\{ \rho(x,y) : x, y\in C\bigr\} < \infty. $$

Now we introduce the model spaces $M^{n}_{\kappa}$, for more details on these spaces the reader is referred to [30, 31]. Let $n\in\mathbb{N}$. We denote by $\mathbb{E}^{n}$ the metric space $\mathbb{R}^{n}$ endowed with the usual Euclidean distance. We denote by $(\cdot|\cdot)$ the Euclidean scalar product in $\mathbb{R}^{n}$, that is,

$$( x|y )=x_{1}y_{1}+\cdots+x_{n}y_{n},\quad \mbox{where } x=(x_{1},\ldots,x_{n}), y=(y_{1}, \ldots,y_{n}). $$

Let $\mathbb{S}^{n}$ denote the n-dimensional sphere defined by

$$\mathbb{S}^{n}= \bigl\{ x=(x_{1},\ldots,x_{n+1})\in \mathbb{R}^{n+1} : ( x|x )=1 \bigr\} , $$

with metric $d_{\mathbb{S}^{n}}(x,y)=\arccos( x|y )$, $x,y\in \mathbb{S}^{n}$.

Let $\mathbb{E}^{n,1}$ denote the vector space $\mathbb{R}^{n+1}$ endowed with the symmetric bilinear form which associates to vectors $u = (u_{1},\ldots, u_{n+1})$ and $v = (v_{1},\ldots, v_{n+1})$ the real number $\langle u|v\rangle$ defined by

$$\langle u|v\rangle= - u_{n+1} v_{n+1}+\sum _{i=1}^{n}u_{i} v_{i}. $$

Let $\mathbb{H}^{n}$ denote the hyperbolic n-space defined by

$$\mathbb{H}^{n}= \bigl\{ u=(u_{1},\ldots,u_{n+1})\in \mathbb{E}^{n,1} : \langle u|u \rangle= -1, u_{n+1}> 0 \bigr\} , $$

with metric $d_{\mathbb{H}^{n}}$ such that

$$\cosh d_{\mathbb{H}^{n}}(x,y)= -\langle x|y \rangle,\quad x,y\in \mathbb{H}^{n}. $$

Definition 2.1

Given $\kappa\in\mathbb{R}$, we denote by $M^{n}_{\kappa}$ the following metric spaces:

• (i) if $\kappa= 0$ then $M^{n}_{0}$ is the Euclidean space $\mathbb{E}^{n}$;

• (ii) if $\kappa> 0$ then $M^{n}_{\kappa}$ is obtained from the spherical space $\mathbb{S}^{n}$ by multiplying the distance function by the constant $1/\sqrt{\kappa}$;

• (iii) if $\kappa< 0$ then $M^{n}_{\kappa}$ is obtained from the hyperbolic space $\mathbb{H}^{n}$ by multiplying the distance function by the constant $1/\sqrt{-\kappa}$.

A geodesic triangle $\triangle(x, y, z)$ in a geodesic space $(X,\rho)$ consists of three points x, y, z in X (the vertices of △) and three geodesic segments between each pair of vertices (the edges of △). A comparison triangle for a geodesic triangle $\triangle(x, y, z)$ in $(X,\rho)$ is a triangle $\overline{\triangle}(\bar{x}, \bar{y}, \bar{z})$ in $M^{2}_{\kappa}$ such that

$$\rho(x,y)=d_{M_{\kappa}^{2}}(\bar{x},\bar{y}), \qquad \rho(y,z)=d_{M_{\kappa }^{2}}( \bar{y},\bar{z})\quad \mbox{and} \quad \rho(z,x)=d_{M_{\kappa}^{2}}(\bar {z}, \bar{x}). $$

If $\kappa\leq0$ then such a comparison triangle always exists in $M^{2}_{\kappa}$. If $\kappa> 0$ then such a triangle exists whenever $\rho(x, y) + \rho(y, z) + \rho(z, x) < 2D_{\kappa}$, where $D_{\kappa}=\pi/\sqrt{\kappa}$. A point $\bar{p}\in[\bar{x}, \bar{y}]$ is called a comparison point for $p\in[x, y]$ if $\rho(x, p) = d_{M_{\kappa}^{2}}(\bar{x}, \bar{p})$.

A geodesic triangle $\triangle(x, y, z)$ in X is said to satisfy the $\operatorname{CAT}(\kappa)$ inequality if for any $p,q\in \triangle(x, y, z)$ and for their comparison points $\bar{p}, \bar{q}\in \overline{\triangle}(\bar{x}, \bar{y}, \bar{z})$, one has

$$\rho(p,q)\leq d_{M_{\kappa}^{2}}(\bar{p}, \bar{q}). $$

Definition 2.2

If $\kappa\leq0$, then X is called a $\operatorname{CAT}(\kappa)$ space if X is a geodesic space such that all of its geodesic triangles satisfy the $\operatorname{CAT}(\kappa)$ inequality.

If $\kappa> 0$, then X is called a $\operatorname {CAT}(\kappa)$ space if X is $D_{\kappa}$-geodesic and any geodesic triangle $\triangle(x, y, z)$ in X with $\rho(x, y) + \rho(y, z) + \rho(z, x) < 2D_{\kappa}$ satisfies the $\operatorname{CAT}(\kappa)$ inequality.

Notice that in a $\operatorname{CAT}(0)$ space $(X,\rho)$, if $x,y,z\in X$ then the $\operatorname{CAT}(0)$ inequality implies

$$(\mathrm{CN})\quad \rho^{2} \biggl(x,\frac{1}{2}y\oplus \frac{1}{2}z \biggr)\leq\frac{1}{2}\rho ^{2}(x,y)+ \frac{1}{2} \rho^{2}(x,z)-\frac{1}{4}\rho^{2}(y,z). $$

This is the (CN) inequality of Bruhat and Tits [32]. This inequality is extended by Dhompongsa and Panyanak [9] as

$$(\mathrm{CN}^{*})\quad \rho^{2}\bigl(x,(1-\alpha)y\oplus\alpha z\bigr)\leq(1-\alpha)\rho^{2}(x,y)+\alpha \rho^{2}(x,z)- \alpha(1-\alpha)\rho^{2}(y,z) $$

for all $\alpha\in[0,1]$ and $x, y, z\in X$. In fact, if X is a geodesic space then the following statements are equivalent:

• (i) X is a $\operatorname{CAT}(0)$ space;

• (ii) X satisfies (CN);

• (iii) X satisfies (CN^∗).

Let $R\in(0,2]$. Recall that a geodesic space $(X, \rho)$ is said to be R-convex for R [33] if for any three points $x, y, z \in X$, we have

$$ \rho^{2}\bigl(x,(1-\alpha)y\oplus\alpha z\bigr)\leq (1- \alpha)\rho^{2}(x,y)+\alpha\rho^{2}(x,z)-\frac{R}{2} \alpha(1-\alpha)\rho^{2}(y,z). $$

(1)

It follows from (CN^∗) that a geodesic space $(X, \rho)$ is a $\operatorname{CAT}(0)$ space if and only if $(X, \rho)$ is R-convex for $R=2$. The following lemma generalizes Proposition 3.1 of Ohta [33].

Lemma 2.3

Let κ be an arbitrary positive real number and $(X, \rho)$ be a $\operatorname{CAT}(\kappa)$ space with $\operatorname{diam}(X)\leq\frac{\pi/2-\eta}{\sqrt{\kappa}}$ for some $\eta\in(0,\pi/2)$. Then $(X, \rho)$ is R-convex for $R=(\pi-2\eta)\tan(\eta)$.

Proof

Let $x,y,z\in X$. Since $\operatorname{diam}(X) < \frac{\pi}{2\sqrt {\kappa}}$, $\rho(x, y) + \rho(x, z) + \rho(y, z) < 2D_{\kappa}$ where $D_{\kappa}=\frac{\pi}{\sqrt{\kappa}}$. Let $\triangle(x, y, z)$ be the geodesic triangle constructed from x, y, z and $\overline{\triangle}(\bar{x}, \bar{y}, \bar{z})$ its comparison triangle. Then

$$ \rho(x,y)=d_{M_{\kappa}^{2}}(\bar{x},\bar {y}), \qquad \rho(x,z)=d_{M_{\kappa}^{2}}(\bar{x},\bar{z})\quad \mbox{and}\quad \rho(y,z)=d_{M_{\kappa}^{2}}(\bar{y},\bar{z}). $$

(2)

It is sufficient to prove (1) only the case of $\alpha=1/2$. Let $a=d_{\mathbb{S}^{2}}(\bar{x},\bar{y})$, $b=d_{\mathbb{S}^{2}}(\bar{x},\bar{z})$, $c=d_{\mathbb{S}^{2}}(\bar{y},\bar{z})/2$, and $d=d_{\mathbb{S}^{2}} (\bar{x},\frac{1}{2}\bar{y}\oplus\frac{1}{2}\bar {z} )$ and define

$$f(a,b,c):=\frac{2}{c^{2}} \biggl(\frac{1}{2}a^{2}+ \frac{1}{2}b^{2}-d^{2} \biggr). $$

By using the same argument in the proof of Proposition 3.1 in [33], we obtain

$$d^{2}_{\mathbb{S}^{2}} \biggl(\bar{x},\frac{1}{2}\bar{y}\oplus \frac{1}{2}\bar {z} \biggr)\leq \frac{1}{2}d^{2}_{\mathbb{S}^{2}}( \bar{x},\bar{y})+\frac{1}{2} d^{2}_{\mathbb{S}^{2}}(\bar{x}, \bar{z})- \biggl(\frac{R}{2} \biggr) \biggl(\frac {1}{4} \biggr)d^{2}_{\mathbb{S}^{2}}(\bar{y},\bar{z}), $$

where $R=(\pi-2\eta)\tan(\eta)$. This implies that

$$ d^{2}_{M_{\kappa}^{2}} \biggl(\bar{x},\frac {1}{2} \bar{y}\oplus\frac{1}{2}\bar{z} \biggr)\leq \frac{1}{2}d^{2}_{M_{\kappa}^{2}}( \bar{x},\bar{y})+\frac{1}{2} d^{2}_{M_{\kappa}^{2}}(\bar{x}, \bar{z})- \biggl(\frac{R}{2} \biggr) \biggl(\frac {1}{4} \biggr)d^{2}_{M_{\kappa}^{2}}(\bar{y},\bar{z}). $$

(3)

By (2) and (3), we get

$$\rho^{2} \biggl(x,\frac{1}{2}y\oplus \frac{1}{2}z \biggr) \leq\frac{1}{2}\rho^{2}(x,y)+\frac{1}{2} \rho^{2}(x,z)- \biggl(\frac{R}{2} \biggr) \biggl(\frac{1}{4} \biggr)\rho^{2}(y,z). $$

This completes the proof. □

The following lemma is also needed.

Lemma 2.4

Let $\{s_{n}\}$ and $\{t_{n}\}$ be sequences of nonnegative real numbers satisfying

$$s_{n+1}\leq s_{n} + t_{n} \quad \textit{for all } n \in\mathbb{N}. $$

If $\sum_{n=1}^{\infty} t_{n}<\infty$ and $\{s_{n}\}$ has a subsequence converging to 0, then $\lim_{n\to\infty} s_{n}=0$.

Definition 2.5

Let C be a nonempty subset of a $\operatorname{CAT}(\kappa)$ space $(X,\rho)$ and $T:C\to C$ be a mapping. We denote by $F(T)$ the set of all fixed points of T, i.e., $F(T)=\{x\in C: x=Tx\}$. Then T is said to

• (i) be completely continuous if T is continuous and for any bounded sequence $\{x_{n}\}$ in C, $\{Tx_{n}\}$ has a convergent subsequence in C;

• (ii) be uniformly L-Lipschitzian if there exists a constant $L>0$ such that

  $$\rho\bigl(T^{n}x,T^{n}y\bigr)\leq L \rho(x,y) \quad \text{for all } x, y\in C \text{ and all } n\in\mathbb{N}; $$

• (iii) be asymptotically demicontractive if $F(T)\neq \emptyset$ and there exist $k\in[0,1)$ and a sequence $\{a_{n}\}$ with $\lim_{n\to\infty}a_{n}=1$ such that

  $$\rho^{2}\bigl(T^{n}x, p\bigr)\leq a^{2}_{n} \rho^{2}(x,p)+k \rho^{2}\bigl(x,T^{n}x\bigr) \quad \text{for all } x\in C, p\in F(T) \text{ and } n\in\mathbb{N}; $$

• (iv) be asymptotically hemicontractive if $F(T)\neq \emptyset$ and there exists a sequence $\{a_{n}\}$ with $\lim_{n\to\infty}a_{n}=1$ such that

  $$\rho^{2}\bigl(T^{n}x, p\bigr)\leq a_{n} \rho^{2}(x,p)+ \rho^{2}\bigl(x,T^{n}x\bigr) \quad \text{for all } x\in C, p\in F(T) \text{ and } n\in\mathbb{N}. $$

It follows from the definition that every asymptotically demicontractive mapping is asymptotically hemicontractive. For more details as regards these classes of mappings the reader is referred to [27, 28].

Let C be a nonempty convex subset of a $\operatorname{CAT}(\kappa)$ space $(X,\rho)$ and $T:C\to C$ be a mapping. Given $x_{1}\in C$.

Algorithm 1

The sequence $\{x_{n}\}$ defined by

$$\begin{aligned}& x_{n+1}=(1-\alpha_{n})x_{n}\oplus \alpha_{n}T^{n}y_{n}, \\& y_{n}=(1-\beta_{n})x_{n}\oplus\beta_{n} T^{n}x_{n},\quad n\geq1, \end{aligned}$$

is called an Ishikawa iterative sequence (see [34]).

If $\beta_{n} = 0$ for all $n\in\mathbb{N}$, then Algorithm 1 reduces to the following.

Algorithm 2

The sequence $\{x_{n}\}$ defined by

$$x_{n+1}=(1-\alpha_{n})x_{n}\oplus \alpha_{n} T^{n}x_{n}, \quad n\geq1, $$

is called a Mann iterative sequence (see [35]).

3 Main results

We first discuss the strong convergence of Mann iteration for uniformly L-Lipschitzian asymptotically demicontractive mappings. The following lemma follows immediately from Lemma 6 of [29] and [30], p.176.

Lemma 3.1

Let $\kappa>0$ and $(X, \rho)$ be a $\operatorname{CAT}(\kappa)$ space with $\operatorname{diam}(X)\leq \frac{\pi/2-\eta}{\sqrt{\kappa}}$ for some $\eta\in(0,\pi/2)$. Let C be a nonempty convex subset of X, $T:C\to C$ be a uniformly L-Lipschitzian mapping, and $\{\alpha _{n}\}$, $\{\beta_{n}\}$ be sequences in $[0, 1]$. Given $x_{1}\in C$, define the iteration scheme $\{x_{n}\}$ by

$$\begin{aligned}& x_{n+1}=(1-\alpha_{n})x_{n}\oplus \alpha_{n}T^{n}y_{n}, \\& y_{n}=(1-\beta_{n})x_{n}\oplus\beta_{n} T^{n}x_{n}, \quad n\geq1. \end{aligned}$$

Then

$$\rho(x_{n},Tx_{n})\leq\rho\bigl(x_{n},T^{n}x_{n} \bigr)+ L\bigl(1+2L+L^{2}\bigr)\rho\bigl(x_{n-1},T^{n-1}x_{n-1} \bigr) $$

for all $n\geq1$.

The following theorem is one of our main results.

Theorem 3.2

Let $\kappa>0$ and $(X, \rho)$ be a $\operatorname{CAT}(\kappa)$ space with $\operatorname{diam}(X)\leq \frac{\pi/2-\eta}{\sqrt{\kappa}}$ for some $\eta\in(0,\pi/2)$. Let C be a nonempty closed convex subset of X, and $T:C\to C$ be a completely continuous and uniformly L-Lipschitzian asymptotically demicontractive mapping with constant $k\in[0,1)$ and sequence $\{a_{n}\}$ in $[1,\infty)$ such that $\sum^{\infty}_{n=1}(a_{n}^{2}-1)<\infty$. Let $\{\alpha_{n}\}$ be a sequence in $[\varepsilon, R/2-k-\varepsilon]$ for some $\varepsilon>0$ where $R=(\pi-2\eta)\tan(\eta)$. Given $x_{1}\in C$, define the iteration scheme $\{x_{n}\}$ by

$$x_{n+1}=(1-\alpha_{n})x_{n}\oplus \alpha_{n} T^{n}x_{n},\quad n\geq1. $$

Then $\{x_{n}\}$ converges strongly to a fixed point of T.

Proof

Let $p\in F(T)$. By (1), we have

$$\rho^{2}(x_{n+1},p)\leq(1-\alpha_{n}) \rho^{2}(x_{n},p)+\alpha_{n}\rho ^{2} \bigl(T^{n}x_{n},p\bigr)-\frac{R}{2} \alpha_{n}(1-\alpha_{n})\rho^{2} \bigl(x_{n},T^{n}x_{n}\bigr). $$

It follows from the asymptotically demicontractiveness of T that

$$\begin{aligned} \rho^{2}(x_{n+1},p) \leq&(1- \alpha_{n})\rho^{2}(x_{n},p)+\alpha_{n} \bigl[a_{n}^{2}\rho^{2}(x_{n},p)+k \rho^{2}\bigl(x_{n},T^{n}x_{n}\bigr) \bigr] \\ &{}-\frac{R}{2}\alpha _{n}(1-\alpha_{n}) \rho^{2}\bigl(x_{n},T^{n}x_{n}\bigr) \\ =& \rho^{2}(x_{n},p)+\alpha_{n} \bigl(a^{2}_{n}-1\bigr) \rho^{2}(x_{n},p)- \alpha_{n} \biggl(\frac{R}{2}-\frac{R}{2} \alpha_{n}-k \biggr)\rho^{2}\bigl(x_{n},T^{n}x_{n} \bigr) \\ \leq&\rho^{2}(x_{n},p)+\alpha_{n} \bigl(a^{2}_{n}-1\bigr) \rho^{2}(x_{n},p)- \alpha_{n} \biggl(\frac{R}{2}-\alpha_{n}-k \biggr)\rho ^{2}\bigl(x_{n},T^{n}x_{n}\bigr). \end{aligned}$$

(4)

Since $\varepsilon\leq\alpha_{n}\leq R/2-k-\varepsilon$, we have $\varepsilon\leq R/2-\alpha_{n}-k$. Thus,

$$ \varepsilon^{2}\leq\alpha_{n} (R/2- \alpha_{n}-k). $$

(5)

By (4) and (5), we have

$$\begin{aligned} \rho^{2}(x_{n+1},p) \leq&\rho^{2}(x_{n},p)+ \alpha_{n} \bigl(a^{2}_{n}-1\bigr) \rho ^{2}(x_{n},p)- \varepsilon^{2}\rho^{2} \bigl(x_{n},T^{n}x_{n}\bigr) \\ \leq&\rho^{2}(x_{n},p)+ \frac{\pi^{2}(a^{2}_{n}-1)}{4\kappa} - \varepsilon^{2}\rho^{2}\bigl(x_{n},T^{n}x_{n} \bigr). \end{aligned}$$

(6)

Therefore,

$$\varepsilon^{2}\rho^{2}\bigl(x_{n},T^{n}x_{n} \bigr)\leq\rho^{2}(x_{n},p)-\rho^{2}(x_{n+1},p) + \frac{\pi^{2}(a^{2}_{n}-1)}{4\kappa}. $$

Since $\sum^{\infty}_{n=1}(a_{n}^{2}-1)<\infty$, $\sum^{\infty}_{n=1}\rho^{2}(x_{n},T^{n}x_{n})<\infty$, which implies that $\lim_{n\to\infty}\rho(x_{n}, T^{n}x_{n})=0$. By Lemma 3.1, we have

$$ \lim_{n\to\infty}\rho(x_{n},Tx_{n})=0. $$

(7)

Since T is completely continuous, $\{Tx_{n}\}$ has a convergent subsequence in C. By (7), $\{x_{n}\}$ has a convergent subsequence, say $x_{n_{k}}\to q\in C$. Moreover,

$$\rho(q,Tq)\leq\rho(q,x_{n_{k}})+\rho(x_{n_{k}},Tx_{n_{k}})+ \rho (Tx_{n_{k}},Tq)\to0 \quad \text{as } k\to\infty. $$

That is $q\in F(T)$. It follows from (6) that

$$\rho^{2}(x_{n+1},p)\leq\rho^{2}(x_{n},p)+ \frac{\pi^{2}(a^{2}_{n}-1)}{4\kappa}. $$

Since $\sum^{\infty}_{n=1}(a_{n}^{2}-1)<\infty$, by Lemma 2.4 we have $x_{n}\to q$. This completes the proof. □

Corollary 3.3

(Theorem 7 of [29])

Let $(X, \rho)$ be a $\operatorname{CAT}(0)$ space, C be a nonempty bounded closed convex subset of X, and $T:C\to C$ be a completely continuous and uniformly L-Lipschitzian asymptotically demicontractive mapping with constant $k\in[0,1)$ and sequence $\{a_{n}\}$ in $[1,\infty)$ such that $\sum^{\infty}_{n=1}(a_{n}^{2}-1)<\infty$. Let $\{\alpha_{n}\}$ be a sequence in $[\varepsilon, 1-k-\varepsilon]$ for some $\varepsilon>0$. Given $x_{1}\in C$, define the iteration scheme $\{x_{n}\}$ by

$$x_{n+1}=(1-\alpha_{n})x_{n}\oplus \alpha_{n} T^{n}x_{n}, \quad n\geq1. $$

Then $\{x_{n}\}$ converges strongly to a fixed point of T.

Proof

It is well known that every convex subset of a $\operatorname {CAT}(0)$ space, equipped with the induced metric, is a $\operatorname{CAT}(0)$ space. Then $(C,\rho)$ is a $\operatorname{CAT}(0)$ space and hence it is a $\operatorname{CAT}(\kappa)$ space for all $\kappa>0$. Notice also that C is R-convex for $R=2$. Since C is bounded, we can choose $\eta\in(0,\pi /2)$ and $\kappa>0$ so that $\operatorname{diam}(C)\leq \frac{\pi/2-\eta}{\sqrt{\kappa}}$. The conclusion follows from Theorem 3.2. □

Next, we prove the strong convergence of Ishikawa iteration for uniformly L-Lipschitzian asymptotically hemicontractive mappings. The following lemmas are also needed.

Lemma 3.4

Let $\kappa>0$ and $(X, \rho )$ be a $\operatorname{CAT}(\kappa)$ space with $\operatorname{diam}(X)\leq \frac{\pi/2-\eta}{\sqrt{\kappa}}$ for some $\eta\in(0,\pi/2)$. Let $R=(\pi-2\eta)\tan(\eta)$, C be a nonempty convex subset of X, and $T:C\to C$ be a uniformly L-Lipschitzian and asymptotically hemicontractive mapping with sequence $\{a_{n}\}$ in $[1,\infty)$. Given $x_{1}\in C$, define the iteration scheme $\{x_{n}\}$ by

$$\begin{aligned}& x_{n+1}=(1-\alpha_{n})x_{n}\oplus \alpha_{n}T^{n}y_{n}, \\& y_{n}=(1-\beta_{n})x_{n}\oplus\beta_{n} T^{n}x_{n},\quad n\geq1, \end{aligned}$$

where $\{\alpha_{n}\}$ and $\{\beta_{n}\}$ are sequences in $[0,1]$. Then the following inequality holds:

$$\begin{aligned} \rho^{2}(x_{n+1},p) \leq& \bigl[1+\alpha_{n}(a_{n}-1) (1+a_{n}\beta_{n}) \bigr]\rho ^{2}(x_{n},p) \\ &{} -\alpha_{n}\beta_{n} \biggl[\frac{R}{2}(1-\beta _{n}) (1+a_{n})-\bigl(a_{n}+L^{2} \beta_{n}^{2}\bigr) \biggr]\rho^{2} \bigl(x_{n},T^{n}x_{n}\bigr) \\ &{} -\alpha_{n} \biggl[\frac{R}{2}(1-\alpha_{n})-(1- \beta_{n}) \biggr]\rho^{2}\bigl(x_{n},T^{n}y_{n} \bigr) \end{aligned}$$

for all $p\in F(T)$.

Proof

Let $p\in F(T)$. By (1), we have

$$ \rho^{2}(x_{n+1},p)\leq(1-\alpha_{n}) \rho ^{2}(x_{n},p)+\alpha_{n}\rho^{2} \bigl(T^{n}y_{n},p\bigr)-\frac{R}{2} \alpha_{n}(1-\alpha _{n})\rho^{2} \bigl(x_{n},T^{n}y_{n}\bigr) $$

(8)

and

$$ \rho^{2}(y_{n},p)\leq(1-\beta_{n}) \rho ^{2}(x_{n},p)+\beta_{n}\rho^{2} \bigl(T^{n}x_{n},p\bigr)-\frac{R}{2} \beta_{n}(1-\beta_{n})\rho ^{2}\bigl(x_{n},T^{n}x_{n} \bigr). $$

(9)

Since T is asymptotically hemicontractive,

$$ \rho^{2}\bigl(T^{n}y_{n},p\bigr) \leq a_{n}\rho ^{2}(y_{n},p)+\rho^{2} \bigl(y_{n},T^{n}y_{n}\bigr) $$

(10)

and

$$ \rho^{2}\bigl(T^{n}x_{n},p\bigr) \leq a_{n}\rho ^{2}(x_{n},p)+\rho^{2} \bigl(x_{n},T^{n}x_{n}\bigr). $$

(11)

It follows from (9) and (11) that

$$\begin{aligned} \begin{aligned}[b] \rho^{2}(y_{n},p) \leq{}&(1- \beta_{n})\rho^{2}(x_{n},p)+\beta_{n} \bigl[a_{n}\rho ^{2}(x_{n},p)+\rho^{2} \bigl(x_{n},T^{n}x_{n}\bigr) \bigr] \\ &{}- \frac{R}{2}\beta_{n}(1-\beta _{n})\rho^{2} \bigl(x_{n},T^{n}x_{n}\bigr) \\ ={}& \bigl(1+\beta_{n}(a_{n}-1) \bigr)\rho ^{2}(x_{n},p)+\beta_{n} \biggl(1- \frac{R}{2}(1-\beta_{n}) \biggr)\rho ^{2} \bigl(x_{n},T^{n}x_{n}\bigr). \end{aligned} \end{aligned}$$

(12)

Substituting (12) into (10) and using (1), we get

$$\begin{aligned} \rho^{2}\bigl(T^{n}y_{n},p\bigr) \leq& a_{n} \bigl(1+\beta_{n}(a_{n}-1) \bigr)\rho ^{2}(x_{n},p) \\ &{}+a_{n}\beta_{n} \biggl(1- \frac{R}{2}(1-\beta_{n}) \biggr)\rho ^{2} \bigl(x_{n},T^{n}x_{n}\bigr)+\rho^{2} \bigl(y_{n}, T^{n}y_{n}\bigr) \\ \leq& a_{n} \bigl(1+\beta_{n}(a_{n}-1) \bigr) \rho^{2}(x_{n},p)+a_{n}\beta_{n} \biggl(1- \frac{R}{2}(1-\beta_{n}) \biggr)\rho^{2} \bigl(x_{n},T^{n}x_{n}\bigr) \\ &{} +(1-\beta_{n})\rho^{2}\bigl(x_{n},T^{n}y_{n} \bigr)+\beta_{n}\rho ^{2}\bigl(T^{n}x_{n},T^{n}y_{n} \bigr) \\ &{}-\frac{R}{2}\beta_{n}(1-\beta_{n}) \rho^{2}\bigl(x_{n},T^{n}x_{n}\bigr) \\ \leq& a_{n} \bigl(1+\beta_{n}(a_{n}-1) \bigr) \rho^{2}(x_{n},p) \\ &{}+ \biggl[a_{n}\beta _{n}-a_{n}\beta_{n}\frac{R}{2}(1- \beta_{n})-\beta_{n}\frac{R}{2}(1-\beta_{n}) \biggr]\rho^{2}\bigl(x_{n},T^{n}x_{n} \bigr) \\ &{} +(1-\beta_{n})\rho^{2}\bigl(x_{n},T^{n}y_{n} \bigr)+\beta_{n}L^{2}\rho ^{2}(x_{n},y_{n}) \\ \leq& a_{n} \bigl(1+\beta_{n}(a_{n}-1) \bigr) \rho^{2}(x_{n},p) \\ &{} + \biggl[a_{n}\beta_{n}-a_{n} \beta_{n}\frac{R}{2}(1-\beta_{n})-\beta _{n} \frac{R}{2}(1-\beta_{n})+\beta^{3}L^{2} \biggr]\rho^{2}\bigl(x_{n},T^{n}x_{n} \bigr) \\ &{} +(1-\beta_{n})\rho^{2}\bigl(x_{n},T^{n}y_{n} \bigr). \end{aligned}$$

(13)

Substituting (13) into (8), we obtain

$$\begin{aligned} \rho^{2}(x_{n+1},p) \leq&(1-\alpha_{n}) \rho^{2}(x_{n},p)+\alpha_{n}a_{n} \bigl(1+\beta_{n}(a_{n}-1) \bigr)\rho^{2}(x_{n}, p) \\ &{}+\alpha_{n} \biggl[a_{n}\beta_{n}-a_{n} \beta_{n}\frac{R}{2}(1-\beta _{n})-\beta_{n} \frac{R}{2}(1-\beta_{n})+\beta^{3}L^{2} \biggr]\rho^{2}\bigl(x_{n},T^{n}x_{n} \bigr) \\ &{}+\alpha_{n}(1-\beta_{n})\rho^{2} \bigl(x_{n},T^{n}y_{n}\bigr)-\frac{R}{2} \alpha _{n}(1-\alpha_{n})\rho^{2} \bigl(x_{n},T^{n}y_{n}\bigr) \\ =& \bigl[1+\alpha_{n}(a_{n}-1) (1+a_{n} \beta_{n}) \bigr]\rho^{2}(x_{n},p) \\ &{} -\alpha_{n}\beta_{n} \biggl[\frac{R}{2}(1-\beta _{n}) (1+a_{n})-\bigl(a_{n}+L^{2} \beta_{n}^{2}\bigr) \biggr]\rho^{2} \bigl(x_{n},T^{n}x_{n}\bigr) \\ &{} -\alpha_{n} \biggl[\frac{R}{2}(1-\alpha_{n})-(1- \beta_{n}) \biggr]\rho^{2}\bigl(x_{n},T^{n}y_{n} \bigr). \end{aligned}$$

This completes the proof. □

Lemma 3.5

Let $\kappa>0$ and $(X, \rho)$ be a $\operatorname{CAT}(\kappa)$ space with $\operatorname{diam}(X)\leq\frac{\pi/2-\eta}{\sqrt{\kappa}}$ for some $\eta\in(0,\pi/2)$. Let C be a nonempty convex subset of X, and $T:C\to C$ be a uniformly L-Lipschitzian and asymptotically hemicontractive mapping with sequence $\{a_{n}\}$ in $[1,\infty)$ such that $\sum_{n=1}^{\infty}(a_{n}-1)<\infty$. Let $\{\alpha_{n}\}, \{\beta_{n}\}\subset[0,1]$ be such that $\frac{1-\beta_{n}}{1-\alpha_{n}}\leq\frac{R}{2}$ where $R=(\pi-2\eta)\tan(\eta)$ and $\alpha_{n}, \beta_{n}\in[\varepsilon, b]$ for some $\varepsilon>0$ and $b\in (0,\frac{\sqrt{R^{2}+4RL^{2}-4L^{2}}-R}{2L^{2}} )$. Given $x_{1}\in C$, define the iteration scheme $\{x_{n}\}$ by

$$\begin{aligned}& x_{n+1}=(1-\alpha_{n})x_{n}\oplus \alpha_{n}T^{n}y_{n}, \\& y_{n}=(1-\beta_{n})x_{n}\oplus\beta_{n} T^{n}x_{n},\quad n\geq1. \end{aligned}$$

Then

$$ \lim_{n\to\infty}\rho(x_{n},Tx_{n})=0. $$

(14)

Proof

First, we prove that $\lim_{n\to\infty}\rho(x_{n},T^{n}x_{n})=0$. Since $\frac{1-\beta_{n}}{1-\alpha_{n}}\leq\frac{R}{2}$, by Lemma 3.4 we have

$$\begin{aligned} \rho^{2}(x_{n+1},p)-\rho^{2}(x_{n},p) \leq&\alpha_{n} (a_{n}-1) (1+a_{n}\beta _{n})\rho^{2}(x_{n},p) \\ &{} -\alpha_{n}\beta_{n} \biggl[\frac{R}{2}(1-\beta _{n}) (1+a_{n})-\bigl(a_{n}+L^{2} \beta_{n}^{2}\bigr) \biggr]\rho^{2} \bigl(x_{n},T^{n}x_{n}\bigr). \end{aligned}$$

Since $\{\alpha_{n}(1+a_{n}\beta_{n})\rho^{2}(x_{n},p) \}^{\infty }_{n=1}$ is a bounded sequence, there exists $M>0$ such that

$$\begin{aligned} \rho^{2}(x_{n+1},p)-\rho^{2}(x_{n},p) \leq& (a_{n}-1) M \\ &{}-\alpha_{n}\beta_{n} \biggl[\frac {R}{2}(1- \beta_{n}) (1+a_{n})-\bigl(a_{n}+L^{2} \beta_{n}^{2}\bigr) \biggr]\rho^{2} \bigl(x_{n},T^{n}x_{n}\bigr). \end{aligned}$$

(15)

Let $D=R(1-b)-(1+L^{2}b^{2})>0$. Since $\lim_{n\to\infty}a_{n}=1$, there exists a natural number N such that

$$ \frac{R}{2}(1-\beta_{n}) (1+a_{n})- \bigl(a_{n}+L^{2}\beta _{n}^{2}\bigr)\geq \frac{R}{2}(1-b) (1+a_{n})-\bigl(a_{n}+L^{2}b^{2} \bigr)\geq\frac{D}{2}>0 $$

(16)

for all $n\geq N$. Suppose that $\lim_{n\to\infty}\rho(x_{n},T^{n}x_{n})\neq0$. Then there exist $\varepsilon_{0}>0$ and a subsequence $\{x_{n_{i}}\}$ of $\{x_{n}\}$ such that

$$ \rho^{2}\bigl(x_{n_{i}},T^{n_{i}}x_{n_{i}} \bigr)\geq \varepsilon_{0}. $$

(17)

Without loss of generality, we let $n_{1}\geq N$. From (15), we have

$$\alpha_{n}\beta_{n} \biggl[\frac{R}{2}(1- \beta_{n}) (1+a_{n})-\bigl(a_{n}+L^{2} \beta _{n}^{2}\bigr) \biggr]\rho^{2} \bigl(x_{n},T^{n}x_{n}\bigr)\leq(a_{n}-1)M+ \rho^{2}(x_{n},p)-\rho^{2}(x_{n+1},p). $$

Then

$$\begin{aligned}& \sum_{l=1}^{i} \alpha_{n_{l}} \beta_{n_{l}} \biggl[\frac{R}{2}(1-\beta _{n_{l}}) (1+a_{n_{l}})-\bigl(a_{n_{l}}+L^{2}\beta_{n_{l}}^{2} \bigr) \biggr]\rho ^{2}\bigl(x_{n_{l}},T^{n_{l}}x_{n_{l}} \bigr) \\& \quad =\sum_{m=n_{1}}^{n_{i}} \alpha_{m}\beta_{m} \biggl[\frac{R}{2}(1-\beta _{m}) (1+a_{m})-\bigl(a_{m}+L^{2} \beta_{m}^{2}\bigr) \biggr]\rho^{2} \bigl(x_{m},T^{m}x_{m}\bigr) \\& \quad \leq\sum_{m=n_{1}}^{n_{i}} (a_{m}-1)M+\rho^{2}(x_{n_{1}},p)- \rho^{2}(x_{n_{i}+1},p). \end{aligned}$$

From this, together with (16), (17) and the fact that $\varepsilon\leq\alpha_{n}\leq\beta_{n}$, we obtain

$$ i\cdot\varepsilon^{2}\cdot\frac{D}{2}\cdot \varepsilon_{0} \leq\sum_{m=n_{1}}^{n_{i}} (a_{m}-1)M+\rho^{2}(x_{n_{1}},p)-\rho ^{2}(x_{n_{i}+1},p). $$

(18)

If we take $i\to\infty$, the right side of (18) is bounded while the left side is unbounded. This is a contradiction. Therefore $\lim_{n\to\infty}\rho(x_{n},T^{n}x_{n})=0$, and hence $\lim_{n\to\infty}\rho (x_{n},Tx_{n})=0$ by Lemma 3.1. □

Theorem 3.6

Let $\kappa>0$ and $(X, \rho)$ be a $\operatorname{CAT}(\kappa)$ space with $\operatorname{diam}(X)\leq\frac{\pi/2-\eta}{\sqrt{\kappa}}$ for some $\eta\in(0,\pi/2)$. Let C be a nonempty closed convex subset of X, and $T:C\to C$ be a completely continuous and uniformly L-Lipschitzian asymptotically hemicontractive mapping with sequence $\{a_{n}\}$ in $[1,\infty)$ such that $\sum_{n=1}^{\infty}(a_{n}-1)<\infty$. Let $\{\alpha_{n}\}, \{\beta_{n}\}\subset[0,1]$ be such that $\frac{1-\beta_{n}}{1-\alpha_{n}}\leq\frac{R}{2}$ where $R=(\pi-2\eta)\tan(\eta)$ and $\alpha_{n}, \beta_{n}\in[\varepsilon, b]$ for some $\varepsilon>0$ and $b\in (0,\frac{\sqrt{R^{2}+4RL^{2}-4L^{2}}-R}{2L^{2}} )$. Given $x_{1}\in C$, define the iteration scheme $\{x_{n}\}$ by

$$\begin{aligned}& x_{n+1}=(1-\alpha_{n})x_{n}\oplus \alpha_{n}T^{n}y_{n}, \\& y_{n}=(1-\beta_{n})x_{n}\oplus\beta_{n} T^{n}x_{n}, \quad n\geq1. \end{aligned}$$

Then $\{x_{n}\}$ converges strongly to a fixed point of T.

Proof

Since T is completely continuous, $\{Tx_{n}\}$ has a convergent subsequence in C. By using Lemma 3.5, we can show that $\{x_{n}\}$ has a convergent subsequence, say $x_{n_{k}}\to q\in C$. Hence $q\in F(T)$ by (14) and the continuity of T. It follows from (15) and (16) that

$$\rho^{2}(x_{n+1},p)\leq\rho^{2}(x_{n},p)+ (a_{n}-1)M. $$

Since $\sum^{\infty}_{n=1}(a_{n}-1)<\infty$, by Lemma 2.4 we have $x_{n}\to q$. This completes the proof. □

As consequences of Theorem 3.6, we obtain the following.

Corollary 3.7

Let $\kappa>0$ and $(X, \rho)$ be a $\operatorname {CAT}(\kappa)$ space with $\operatorname{diam}(X)\leq\frac{\pi/2-\eta}{\sqrt{\kappa}}$ for some $\eta\in(0,\pi/2)$. Let C be a nonempty closed convex subset of X, and $T:C\to C$ be a completely continuous and uniformly L-Lipschitzian asymptotically demicontractive mapping with sequence $\{a_{n}\}$ in $[1,\infty)$ such that $\sum_{n=1}^{\infty}(a_{n}^{2}-1)<\infty$. Let $\{\alpha_{n}\}, \{\beta_{n}\}\subset[0,1]$ be such that $\frac{1-\beta_{n}}{1-\alpha_{n}}\leq\frac{R}{2}$ where $R=(\pi-2\eta)\tan(\eta)$ and $\alpha_{n}, \beta_{n}\in[\varepsilon, b]$ for some $\varepsilon>0$ and $b\in (0,\frac{\sqrt{R^{2}+4RL^{2}-4L^{2}}-R}{2L^{2}} )$. Given $x_{1}\in C$, define the iteration scheme $\{x_{n}\}$ by

$$\begin{aligned}& x_{n+1}=(1-\alpha_{n})x_{n}\oplus \alpha_{n}T^{n}y_{n}, \\& y_{n}=(1-\beta_{n})x_{n}\oplus\beta_{n} T^{n}x_{n}, \quad n\geq1. \end{aligned}$$

Then $\{x_{n}\}$ converges strongly to a fixed point of T.

Corollary 3.8

(Theorem 11 of [29])

Let $(X, \rho)$ be a $\operatorname{CAT}(0)$ space, let C be a nonempty bounded closed convex subset of X, and let $T:C\to C$ be a completely continuous and uniformly L-Lipschitzian asymptotically hemicontractive mapping with sequence $\{a_{n}\}$ in $[1,\infty)$ such that $\sum_{n=1}^{\infty}(a_{n}-1)<\infty$. Let $\{\alpha_{n}\}, \{\beta_{n}\}\subset[0,1]$ be such that $\varepsilon \leq\alpha_{n}\leq\beta_{n}\leq b$ for some $\varepsilon>0$ and $b\in (0,\frac{\sqrt{1+L^{2}}-1}{L^{2}} )$. Given $x_{1}\in C$, define the iteration scheme $\{x_{n}\}$ by

$$\begin{aligned}& x_{n+1}=(1-\alpha_{n})x_{n}\oplus \alpha_{n}T^{n}y_{n}, \\& y_{n}=(1-\beta_{n})x_{n}\oplus\beta_{n} T^{n}x_{n},\quad n\geq1. \end{aligned}$$

Then $\{x_{n}\}$ converges strongly to a fixed point of T.