Viscosity iteration method in CAT(0) spaces without the nice projection property

Abstract

A complete CAT(0) space X is said to have the nice projection property (property $\mathcal{N}$ for short) if its metric projection onto a geodesic segment preserves points on each geodesic segment, that is, for any geodesic segment L in X and $x,y\in X$, $m\in[x,y]$ implies $P_{L}(m)\in[P_{L}(x), P_{L}(y)]$, where $P_{L}$ denotes the metric projection from X onto L. In this paper, we prove a strong convergence theorem of a two-step viscosity iteration method for nonexpansive mappings in CAT(0) spaces without the condition on the property $\mathcal{N}$. Our result gives an affirmative answer to a problem raised by Piatek (Numer. Funct. Anal. Optim. 34:1245-1264, 2013).

1 Introduction

A mapping T on a metric space $(X,\rho)$ is said to be a contraction if there exists a constant $k\in[0,1)$ such that

$$ \rho\bigl(T(x),T(y)\bigr)\leq k \rho(x,y) \quad \mbox{for all } x, y\in X. $$

(1)

If (1) is valid when $k=1$, then T is called nonexpansive. A point $x\in X$ is called a fixed point of T if $x=T(x)$. We shall denote by $\operatorname{Fix}(T)$ the set of all fixed points of T.

One of the powerful iteration methods for finding fixed points of nonexpansive mappings was given by Moudafi [1]. More precisely, let C be a nonempty, closed, and convex subset of a Hilbert space H and $T : C \to C$ be a nonexpansive mapping with $\operatorname{Fix}(T)\neq\emptyset$, the following scheme is known as the viscosity iteration method:

$$\begin{aligned} &x_{1}=u\in C \mbox{ arbitrarily chosen}, \\ &x_{n+1}=\frac{\alpha_{n}}{1+\alpha _{n}}f(x_{n})+\frac{1}{1+\alpha_{n}}T(x_{n}), \end{aligned}$$

(2)

where $f : C \to C$ is a contraction and $\{\alpha_{n}\}$ is a sequence in $(0, 1)$ satisfying (i) $\lim_{n\to\infty} \alpha_{n} =0$, (ii) $\sum^{\infty}_{n=1} \alpha_{n} = \infty$, and (iii) $\lim_{n\to\infty} (1/\alpha_{n} - 1/\alpha_{n+1})=0$. In [1], the author proved that the sequence $\{x_{n}\}$ defined by (2) converges strongly to a fixed point z of T. The point z also satisfies the following variational inequality:

$$\bigl\langle f(z)-z, z-x \bigr\rangle \geq0,\quad x\in \operatorname{Fix}(T). $$

The first extension of Moudafi’s result to the so-called CAT(0) space was proved by Shi and Chen [2]. They assumed that the space $(X, \rho)$ must satisfy the property $\mathcal{P}$, i.e., for $x, u, y_{1}, y_{2}\in X$, one has

$$\rho(x, m_{1})\rho(x, y_{1})\leq\rho(x, m_{2}) \rho(x, y_{2}) + \rho(x, u)\rho (y_{1}, y_{2}), $$

where $m_{1}$ and $m_{2}$ are the unique nearest points of u on the segments $[x, y_{1}]$ and $[x, y_{2}]$, respectively. By using the concept of quasi-linearization introduced by Berg and Nikolaev [3], Wangkeeree and Preechasilp [4] could omit the property $\mathcal{P}$ from Shi and Chen’s result as the following theorem.

Theorem A

([4], Theorem 3.4)

Let C be a nonempty, closed, and convex subset of a complete CAT(0) space X, $T : C \to C$ be a nonexpansive mapping with $\operatorname{Fix}(T)\neq\emptyset$, and $f:C\to C$ be a contraction with $k\in[0,1)$. For $x_{1}\in C$, let $\{x_{n}\}$ be generated by

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

where $\{\alpha_{n}\}\subset(0,1)$ satisfies the conditions: (i) $\lim_{n\to\infty} \alpha_{n} =0$, (ii) $\sum^{\infty}_{n=1} \alpha_{n} = \infty$, (iii) either $\sum^{\infty}_{n=1} |\alpha_{n+1}-\alpha_{n}|<\infty$ or $\lim_{n\to\infty} (\alpha_{n+1}/\alpha_{n}) =1$. Then $\{x_{n}\}$ converges strongly to x̃ such that $\tilde{x}=P_{\operatorname{Fix}(T)}(f(\tilde{x}))$ which is equivalent to the variational inequality:

$$\bigl\langle \overrightarrow{\tilde{x}f(\tilde{x})}, \overrightarrow{x \tilde {x}}\bigr\rangle \geq0, \quad x\in \operatorname{Fix}(T). $$

Among other things, by using the geometric properties of CAT(0) spaces, Piatek [5] proved the strong convergence of a two-step viscosity iteration method as the following result.

Theorem B

([5], Theorem 4.3)

Let X be a complete CAT(0) space with the property $\mathcal{N}$. Let $T: X \to X$ be a nonexpansive mapping with $\operatorname{Fix}(T)\neq\emptyset$ and $f : X\to X$ be a contraction with $k\in [0,\frac{1}{2} )$. Then there is a unique point $q\in \operatorname{Fix}(T)$ such that $q=P_{\operatorname{Fix}(T)}(f(q))$. Moreover, for each $u\in X$ and for each couple of sequences $\{\alpha_{n}\}$ and $\{\beta_{n}\}$ in $(0,1)$ satisfying (i) $\lim_{n\to\infty} \alpha_{n} =0$, (ii) $\sum^{\infty}_{n=1} \alpha_{n} = \infty$, and (iii) $0<\liminf_{n} \beta_{n} \leq\limsup_{n} \beta_{n} <1$, the viscosity iterative sequence defined by $x_{1}=u$,

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

converges to q.

In [5], the author provided an example of a CAT(0) space lacking property $\mathcal{N}$ and also raised the following open problem.

Problem

Can we omit the property $\mathcal{N}$ in Theorem B?

In this paper, by combining the ideas of [4] and [5] intensively, we can omit the property $\mathcal{N}$ from Theorem B. This gives a complete solution to the problem mentioned above.

2 Preliminaries

Let $[0,l]$ be a closed interval in $\mathbb{R}$ and x, y be two points in a metric space $(X,\rho)$. A geodesic joining x to y is a map $\xi:[0,l]\to X$ such that $\xi(0)=x$, $\xi(l)=y$, and $\rho(\xi(s),\xi(t))=|s-t|$ for all $s, t\in[0,l]$. The image of ξ is called a geodesic segment joining x and y which when unique is denoted by $[x,y]$. The space $(X,\rho)$ is said to be a geodesic space if every two points in X are joined by a geodesic, and X is said to be uniquely geodesic if there is exactly one geodesic joining x and y for each $x,y\in X$. A subset C of X is said to be convex if every pair of points $x,y\in C$ can be joined by a geodesic in X and the image of every such geodesic is contained in C.

A geodesic triangle $\triangle(p, q, r)$ in a geodesic space $(X,\rho)$ consists of three points p, q, r in X and a choice of three geodesic segments $[p, q]$, $[q, r]$, $[r, p]$ joining them. A comparison triangle for the geodesic triangle $\triangle(p, q, r)$ in X is a triangle $\overline{\triangle}(\bar{p}, \bar{q}, \bar{r})$ in the Euclidean plane $\mathbb{R}^{2}$ such that $d_{\mathbb{R}^{2}} ( \bar{p},\bar{q} ) =\rho(p, q)$, $d_{\mathbb{R}^{2}} ( \bar{q},\bar{r} ) =\rho(q, r)$, and $d_{\mathbb{R}^{2}} ( \bar{r},\bar{p} ) =\rho(r, p)$. A point $\bar{u}\in[\bar{p}, \bar{q}]$ is called a comparison point for $u\in[p, q]$ if $\rho(p, u)=d_{\mathbb{R}^{2}}(\bar{p},\bar{u})$. Comparison points on $[\bar{q}, \bar{r}]$ and $[\bar{r}, \bar{p}]$ are defined in the same way.

Definition 2.1

A geodesic triangle $\triangle(p, q, r)$ in $(X,\rho)$ is said to satisfy the CAT(0) inequality if for any $u,v\in\triangle(p, q, r)$ and for their comparison points $\bar{u}, \bar{v}\in \overline{\triangle}(\bar{p}, \bar{q}, \bar{r})$, one has

$$\rho(u,v)\leq d_{\mathbb{R}^{2}}(\bar{u}, \bar{v}). $$

A geodesic space X is said to be a CAT(0) space if all of its geodesic triangles satisfy the CAT(0) inequality. For other equivalent definitions and basic properties of CAT(0) spaces, we refer the reader to standard texts, such as [6, 7]. It is well known that every CAT(0) space is uniquely geodesic. Notice also that pre-Hilbert spaces, $\mathbb{R}$-trees, Euclidean buildings are examples of CAT(0) spaces (see [6, 8]). Let C be a nonempty, closed, and convex subset of a complete CAT(0) space $(X,\rho)$. It follows from Proposition 2.4 of [6] that for each $x\in X$, there exists a unique point $x_{0}\in C$ such that

$$\rho(x, x_{0})=\inf\bigl\{ \rho(x,y) : y\in C\bigr\} . $$

In this case, $x_{0}$ is called the unique nearest point of x in C. The metric projection of X onto C is the mapping $P_{C}:X\to C$ defined by

$$P_{C}(x):= \mbox{the unique nearest point of } x \mbox{ in } C. $$

Definition 2.2

A complete CAT(0) space X is said to have the nice projection property [9] if for any geodesic segment L in X, it is the case that $P_{L}(m)\in[P_{L}(x), P_{L}(y)]$ for any $x,y\in X$ and $m\in[x,y]$.

Let $(X,\rho)$ be a CAT(0) space. For each $x,y\in X$ and $t\in[0,1]$, there exists a unique point $z\in[x,y]$ such that

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

(3)

We shall denote by $tx\oplus(1-t)y$ the unique point z satisfying (3). Now, we collect some elementary facts about CAT(0) spaces which will be used in the proof of our main theorem.

Lemma 2.3

([10], Lemma 2.4)

Let $(X,\rho)$ be a CAT(0) space. Then

$$\rho\bigl(tx\oplus(1-t)y,z\bigr)\leq t\rho(x,z)+(1-t)\rho(y,z) $$

for all $x,y,z\in X$ and $t\in[0,1]$.

Lemma 2.4

([10], Lemma 2.5)

Let $(X,\rho)$ be a CAT(0) space. Then

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

for all $x,y,z\in X$ and $t\in[0,1]$.

Lemma 2.5

([11], Lemma 3)

Let $(X,\rho)$ be a CAT(0) space. Then

$$\rho\bigl(tx\oplus(1-t)z,ty\oplus(1-t)z\bigr)\leq t \rho(x,y) $$

for all $x,y,z\in X$ and $t\in[0,1]$.

Lemma 2.6

(cf. [12, 13])

Let $\{x_{n}\}$ and $\{y_{n}\}$ be bounded sequences in a CAT(0) space $(X,\rho)$ and let $\{\beta_{n}\}$ be a sequence in $[0,1]$ with $0<\liminf_{n}\beta_{n}\leq\limsup_{n}\beta_{n}<1$. Suppose that $x_{n+1}=\beta_{n}x_{n}\oplus(1-\beta_{n})y_{n}$ for all $n\in \mathbb{N}$ and

$$\limsup_{n\to\infty} \bigl(\rho(y_{n+1},y_{n})- \rho(x_{n+1},x_{n}) \bigr)\leq0. $$

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

Lemma 2.7

([14], Lemma 2.1)

Let $\{s_{n}\}$ be a sequence of non-negative real numbers satisfying

$$s_{n+1}\leq(1-\alpha_{n})s_{n}+\alpha_{n} \beta_{n}, \quad \forall n\geq1, $$

where $\{\alpha_{n}\}\subset(0,1)$ and $\{\beta_{n}\}\subset \mathbb{R}$ such that

• (i) $\sum_{n=1}^{\infty} \alpha_{n}=\infty$;

• (ii) $\limsup_{n\to\infty}\beta_{n} \leq0$ or $\sum_{n=1}^{\infty} |\alpha_{n}\beta_{n}| < \infty$.

Then $\{s_{n}\}$ converges to zero as $n\to\infty$.

We finish this section by recalling an important concept of quasi-linearization introduced by Berg and Nikolaev [3]. Let us denote a pair $(a, b) \in X\times X$ by $\overrightarrow{ab}$ and call it a vector. The quasi-linearization is a map $\langle\cdot, \cdot\rangle: (X \times X)\times(X\times X)\to\mathbb{R}$ defined by

$$\langle\overrightarrow{ab}, \overrightarrow{cd}\rangle=\frac {1}{2} \bigl( \rho^{2}(a, d) + \rho^{2}(b, c) -\rho^{2}(a, c)- \rho^{2}(b, d) \bigr) \quad\mbox{for all } a, b, c, d \in X. $$

It is easy to see that $\langle\overrightarrow{ab}, \overrightarrow{cd}\rangle=\langle\overrightarrow{cd}, \overrightarrow{ab}\rangle$, $\langle\overrightarrow{ab}, \overrightarrow{cd}\rangle=- \langle\overrightarrow{ba}, \overrightarrow{cd}\rangle$, and $\langle\overrightarrow{ax}, \overrightarrow{cd}\rangle+\langle\overrightarrow{xb}, \overrightarrow{cd}\rangle=\langle\overrightarrow{ab}, \overrightarrow{cd}\rangle$ for all $a, b, c, d, x \in X$. We say that $(X,\rho)$ satisfies the Cauchy-Schwarz inequality if

$$\bigl\vert \langle\overrightarrow{ab}, \overrightarrow{cd}\rangle\bigr\vert \leq\rho(a, b)\rho(c, d) \quad \mbox{for all } a, b, c, d \in X. $$

It is known from [3], Corollary 3, that a geodesic space X is a CAT(0) space if and only if X satisfies the Cauchy-Schwarz inequality. Some other properties of quasi-linearization are included as follows.

Lemma 2.8

([4], Lemma 2.9)

Let X be a CAT(0) space. Then

$$\rho^{2}(x,u) \leq\rho^{2}(y,u) + 2 \langle \overrightarrow{xy}, \overrightarrow{xu} \rangle $$

for all $u, x, y \in X$.

Lemma 2.9

([4], Lemma 2.10)

Let u and v be two points in a CAT(0) space X. For each $t\in[0, 1]$, we set $u_{t} = t u\oplus (1-t)v$. Then, for each $x, y\in X$, we have

• (i) $\langle\overrightarrow{u_{t} x}, \overrightarrow{u_{t} y}\rangle\leq t\langle\overrightarrow{ux}, \overrightarrow{u_{t} y}\rangle+(1-t)\langle\overrightarrow{vx}, \overrightarrow{u_{t} y}\rangle$;

• (ii) $\langle\overrightarrow{u_{t} x}, \overrightarrow{uy}\rangle\leq t\langle\overrightarrow{ux}, \overrightarrow{uy}\rangle+(1-t)\langle\overrightarrow{vx}, \overrightarrow{uy}\rangle$ and $\langle\overrightarrow{u_{t} x}, \overrightarrow{vy}\rangle\leq t\langle\overrightarrow{ux}, \overrightarrow{vy}\rangle+(1-t)\langle\overrightarrow{vx}, \overrightarrow{vy}\rangle$.

The following fact, which can be found in [15], is an immediate consequence of Lemma 2.4.

Lemma 2.10

Let X be a CAT(0) space. Then

$$\rho^{2}\bigl(t x\oplus(1-t)y, z\bigr)\leq t^{2} \rho^{2}(x,z)+(1-t)^{2} \rho ^{2}(y,z)+2t(1-t) \langle\overrightarrow{xz}, \overrightarrow{yz}\rangle $$

for all $x,y,z\in X$ and $t\in[0,1]$.

3 Main theorem

Before proving our main theorem, we need one more lemma, which is proved by Wangkeeree and Preechasilp (see [4], Theorem 3.1).

Lemma 3.1

Let C be a nonempty, closed, and convex subset of a complete CAT(0) space X, $T : C \to C$ be a nonexpansive mapping with $\operatorname{Fix}(T)\neq\emptyset$, and $f:C\to C$ be a contraction with $k\in[0,1)$. For each $t \in (0, 1)$, let $\{z_{t}\}$ be given by

$$z_{t}=t f(z_{t})\oplus(1-t) T(z_{t}). $$

Then $\{z_{t}\}$ converges strongly to x̃ as $t\to0$. Moreover, $\tilde{x}=P_{\operatorname{Fix}(T)}(f(\tilde{x}))$ and x̃ also satisfies the following variational inequality:

$$ \bigl\langle \overrightarrow{\tilde{x}f(\tilde{x})}, \overrightarrow{x \tilde{x}}\bigr\rangle \geq0, \quad x\in \operatorname{Fix}(T). $$

(4)

Now, we are ready to prove our main theorem.

Theorem 3.2

Let C be a nonempty, closed, and convex subset of a complete CAT(0) space X, $T : C \to C$ be a nonexpansive mapping with $\operatorname{Fix}(T)\neq\emptyset$, and $f:C\to C$ be a contraction with $k\in [0,\frac{1}{2} )$. For the arbitrary initial point $u\in C$, let $\{x_{n}\}$ be generated by

$$\begin{aligned}[b] &x_{1}=u, \\ &y_{n}=\alpha_{n} f(x_{n})\oplus(1- \alpha_{n}) T(x_{n}), \\ &x_{n+1}=\beta_{n} x_{n}\oplus(1- \beta_{n}) y_{n}, \quad \forall n\geq1, \end{aligned} $$

where $\{\alpha_{n}\}$ and $\{\beta_{n}\}$ are sequences in $(0,1)$ satisfying the following conditions:

• (i) $\lim_{n\to\infty} \alpha_{n} =0$;

• (ii) $\sum^{\infty}_{n=1} \alpha_{n} = \infty$;

• (iii) $0<\liminf_{n} \beta_{n} \leq\limsup_{n} \beta_{n} <1$.

Then $\{x_{n}\}$ converges strongly to x̃ such that $\tilde{x}=P_{\operatorname{Fix}(T)}(f(\tilde{x}))$ and x̃ also satisfies

$$\bigl\langle \overrightarrow{\tilde{x}f(\tilde{x})}, \overrightarrow{x \tilde {x}}\bigr\rangle \geq0, \quad x\in \operatorname{Fix}(T). $$

Proof

We divide the proof into three steps.

Step 1. We show that $\{x_{n}\}$, $\{y_{n}\}$, $\{T(x_{n})\}$, and $\{f(x_{n})\}$ are bounded sequences. Let $p\in \operatorname{Fix}(T)$. By Lemma 2.3, we have

$$\begin{aligned} \rho(x_{n+1}, p) \leq& \beta_{n} \rho(x_{n}, p)+(1-\beta _{n}) \rho(y_{n}, p) \\ \leq& \beta_{n} \rho(x_{n}, p)+(1-\beta_{n}) \bigl[\alpha_{n} \rho\bigl(f(x_{n}), p\bigr)+(1- \alpha_{n})\rho\bigl(T(x_{n}), p\bigr)\bigr] \\ \leq& \bigl[\beta_{n}+(1-\beta_{n}) (1- \alpha_{n})\bigr]\rho(x_{n}, p)+(1-\beta_{n})\alpha _{n} \rho\bigl(f(x_{n}), f(p)\bigr)\\ &{}+(1-\beta_{n}) \alpha_{n} \rho\bigl(f(p), p\bigr) \\ \leq& \bigl[1-(1-k)\alpha_{n}+(1-k)\alpha_{n} \beta_{n}\bigr]\rho(x_{n}, p)+(1-\beta _{n}) \alpha_{n} \rho\bigl(f(p), p\bigr) \\ \leq& \max \biggl\{ \rho(x_{n},p), \frac{\rho(f(p), p)}{1-k} \biggr\} . \end{aligned}$$

By induction, we also have

$$\rho(x_{n},p)\leq\max \biggl\{ \rho(x_{1},p), \frac{\rho(f(p), p)}{1-k} \biggr\} . $$

Hence, $\{x_{n}\}$ is bounded and so are $\{y_{n}\}$, $\{f(x_{n})\}$, and $\{T(x_{n})\}$.

Step 2. We show that $\lim_{n\to\infty} \rho(x_{n},T(x_{n}))=0$. By applying Lemma 2.5 twice for geodesic triangles $\triangle(f(x_{n}), T(x_{n}), T(x_{n+1}))$ and $\triangle(f(x_{n}), f(x_{n+1}), T(x_{n+1}))$, respectively, we obtain

$$\begin{aligned} \rho(y_{n}, y_{n+1}) \leq& (1-\alpha_{n}) \rho \bigl(T(x_{n}), T(x_{n+1})\bigr) + |\alpha_{n}- \alpha_{n+1}| \rho\bigl(f(x_{n}), T(x_{n+1})\bigr) \\ &{}+\alpha_{n+1}\rho\bigl(f(x_{n}), f(x_{n+1}) \bigr) \\ \leq& (1-\alpha_{n}) \rho(x_{n}, x_{n+1}) + | \alpha_{n}-\alpha_{n+1}| \rho\bigl(f(x_{n}), T(x_{n+1})\bigr) \\ &{} +\alpha_{n+1} k \rho(x_{n}, x_{n+1}), \end{aligned}$$

which implies

$$\rho(y_{n}, y_{n+1})-\rho(x_{n}, x_{n+1}) \leq(\alpha_{n+1} k-\alpha_{n}) \rho (x_{n}, x_{n+1})+|\alpha_{n}-\alpha_{n+1}| \rho \bigl(f(x_{n}), T(x_{n+1})\bigr). $$

Since $\lim_{n\to\infty} \alpha_{n} =0$, $\limsup_{n\to\infty} (\rho(y_{n+1},y_{n})-\rho(x_{n+1},x_{n}) )\leq0$. By Lemma 2.6 we have $\lim_{n\to\infty} \rho(x_{n}, y_{n})=0$. Thus,

$$\begin{aligned} \rho\bigl(x_{n}, T(x_{n})\bigr)&\leq\rho(x_{n}, y_{n}) +\rho\bigl(y_{n}, T(x_{n})\bigr)\\ &= \rho (x_{n},y_{n})+\alpha_{n} \rho\bigl(f(x_{n}), T(x_{n})\bigr)\to0 \quad \mbox{as } n\to\infty. \end{aligned}$$

Step 3. We show that $\{x_{n}\}$ converges to x̃, which satisfies $\tilde{x}=P_{\operatorname{Fix}(T)}(f(\tilde{x}))$ and

$$\bigl\langle \overrightarrow{\tilde{x}f(\tilde{x})}, \overrightarrow{x \tilde {x}}\bigr\rangle \geq0, \quad x\in \operatorname{Fix}(T). $$

Let $\{z_{m}\}$ be a sequence in C defined by

$$z_{m}=\alpha_{m} f(z_{m})\oplus(1- \alpha_{m}) T(z_{m}), \quad \forall m\in\mathbb{N}. $$

By Lemma 3.1, $\{z_{m}\}$ converges strongly as $m\to \infty$ to x̃ which satisfies (4) and $\tilde{x}=P_{\operatorname{Fix}(T)}(f(\tilde{x}))$. We claim that

$$\limsup_{n\to\infty} \bigl\langle \overrightarrow{f(\tilde{x}) \tilde{x}}, \overrightarrow{x_{n}\tilde{x}}\bigr\rangle \leq0. $$

It follows from Lemma 2.9(i) that

$$\begin{aligned} \rho^{2}(z_{m},x_{n}) = & \langle \overrightarrow{z_{m} x_{n}}, \overrightarrow{z_{m} x_{n}}\rangle \\ \leq& \alpha_{m}\bigl\langle \overrightarrow{f(z_{m}) x_{n}}, \overrightarrow{z_{m} x_{n}}\bigr\rangle +(1-\alpha_{m})\bigl\langle \overrightarrow{T(z_{m}) x_{n}}, \overrightarrow{z_{m} x_{n}}\bigr\rangle \\ = & \alpha_{m}\bigl\langle \overrightarrow{f(z_{m}) f( \tilde{x})}, \overrightarrow{z_{m} x_{n}}\bigr\rangle + \alpha_{m}\bigl\langle \overrightarrow{f(\tilde{x}) \tilde{x}}, \overrightarrow{z_{m} x_{n}}\bigr\rangle + \alpha_{m}\langle\overrightarrow{\tilde{x} z_{m}}, \overrightarrow{z_{m} x_{n}}\rangle+\alpha_{m} \langle\overrightarrow{z_{m} x_{n}}, \overrightarrow{z_{m} x_{n}}\rangle \\ &{}+(1-\alpha_{m})\bigl\langle \overrightarrow{T(z_{m}) T(x_{n})}, \overrightarrow{z_{m} x_{n}}\bigr\rangle +(1-\alpha_{m})\bigl\langle \overrightarrow{T(x_{n}) x_{n}}, \overrightarrow{z_{m} x_{n}}\bigr\rangle \\ \leq& \alpha_{m} k \rho(z_{m},\tilde{x}) \rho(z_{m},x_{n})+ \alpha_{m}\bigl\langle \overrightarrow{f(\tilde{x}) \tilde{x}}, \overrightarrow{z_{m} x_{n}}\bigr\rangle +\alpha_{m}\rho(\tilde{x},z_{m}) \rho(z_{m},x_{n}) \\ &{}+\alpha_{m}\rho ^{2}(z_{m},x_{n})+(1-\alpha_{m})\rho^{2}(z_{m},x_{n})+(1- \alpha_{m})\rho\bigl(T(x_{n}),x_{n}\bigr)\rho (z_{m},x_{n}) \\ \leq& \alpha_{m} (k+1) \rho(z_{m},\tilde{x})M+\rho \bigl(T(x_{n}),x_{n}\bigr)M+\rho^{2}(z_{m},x_{n})+ \alpha_{m}\bigl\langle \overrightarrow{f(\tilde{x}) \tilde{x}}, \overrightarrow{z_{m} x_{n}}\bigr\rangle , \end{aligned}$$

for some $M>0$. This implies

$$ \bigl\langle \overrightarrow{f(\tilde{x}) \tilde{x}}, \overrightarrow{x_{n} z_{m}}\bigr\rangle \leq (k+1) \rho(z_{m},\tilde{x})M+\frac{\rho(x_{n},T(x_{n}))}{\alpha_{m}}M. $$

(5)

Taking the upper limit as $n\to\infty$ first and then $m\to\infty$, the inequality (5) yields

$$ \limsup_{m\to\infty}\limsup_{n\to\infty } \bigl\langle \overrightarrow{f(\tilde{x}) \tilde{x}}, \overrightarrow{x_{n} z_{m}}\bigr\rangle \leq0. $$

(6)

Notice also that

$$\begin{aligned} \bigl\langle \overrightarrow{f(\tilde{x}) \tilde{x}}, \overrightarrow{x_{n} \tilde{x}}\bigr\rangle = \bigl\langle \overrightarrow{f(\tilde{x}) \tilde{x}}, \overrightarrow{x_{n} z_{m}}\bigr\rangle + \bigl\langle \overrightarrow{f(\tilde{x}) \tilde{x}}, \overrightarrow{z_{m} \tilde{x}}\bigr\rangle \leq \bigl\langle \overrightarrow{f(\tilde{x}) \tilde{x}}, \overrightarrow{x_{n} z_{m}}\bigr\rangle + \rho\bigl(f( \tilde{x}), \tilde{x}\bigr) \rho(z_{m}, \tilde{x}). \end{aligned}$$

This, together with (6), implies that

$$\limsup_{n\to\infty} \bigl\langle \overrightarrow{f(\tilde{x}) \tilde{x}}, \overrightarrow{x_{n}\tilde{x}}\bigr\rangle \leq0. $$

Finally, we show that $x_{n}\to\tilde{x}$ as $n\to\infty$. It follows from Lemmas 2.4, 2.8, 2.9, and 2.10 that

$$\begin{aligned} \rho^{2}(x_{n+1},\tilde{x}) \leq& \beta_{n} \rho^{2}(x_{n},\tilde{x})+(1-\beta_{n}) \rho^{2}(y_{n},\tilde{x}) \\ \leq& \beta_{n} \rho^{2}(x_{n},\tilde{x})+(1- \beta_{n}) \bigl[\alpha_{n}^{2}\rho^{2} \bigl(f(x_{n}),\tilde {x}\bigr)+(1-\alpha_{n})^{2} \rho^{2}\bigl(T(x_{n}),\tilde{x}\bigr) \bigr] \\ &{}+2\alpha_{n}(1-\alpha_{n}) (1-\beta_{n})\bigl\langle \overrightarrow{f(x_{n})\tilde {x}},\overrightarrow{T(x_{n}) \tilde{x}}\bigr\rangle \\ \leq& \beta_{n}\rho^{2}(x_{n},\tilde{x})+(1- \beta_{n}) (1-\alpha_{n})^{2}\rho ^{2}(x_{n},\tilde{x}) \\ &{}+\alpha^{2}_{n}(1-\beta_{n}) \bigl[ \rho^{2}\bigl(x_{n+1},f(x_{n})\bigr)+2\bigl\langle \overrightarrow{\tilde{x}x_{n+1}},\overrightarrow{\tilde {x}f(x_{n})}\bigr\rangle \bigr] \\ &{}+2\alpha_{n}(1-\alpha_{n}) (1-\beta_{n}) \bigl[\bigl\langle \overrightarrow {f(x_{n})\tilde{x}}, \overrightarrow{T(x_{n})x_{n}}\bigr\rangle +\bigl\langle \overrightarrow{f(x_{n})\tilde{x}},\overrightarrow{x_{n} \tilde{x}}\bigr\rangle \bigr] \\ \leq& \bigl[\beta_{n}+(1-\beta_{n}) (1- \alpha_{n}) \bigr]\rho^{2}(x_{n},\tilde {x})+ \alpha^{2}_{n}(1-\beta_{n})\rho^{2} \bigl(x_{n+1},f(x_{n})\bigr) \\ &{}+2\alpha^{2}_{n}(1-\beta_{n}) \bigl[\bigl\langle \overrightarrow{f(x_{n})f(\tilde {x})},\overrightarrow{x_{n+1} \tilde{x}}\bigr\rangle +\bigl\langle \overrightarrow {f(\tilde{x})\tilde{x}}, \overrightarrow{x_{n+1}\tilde{x}}\bigr\rangle \bigr] \\ &{}+2\alpha_{n}(1-\alpha_{n}) (1-\beta_{n})\bigl\langle \overrightarrow{f(x_{n})\tilde {x}},\overrightarrow{T(x_{n})x_{n}} \bigr\rangle \\ &{}+2\alpha_{n}(1-\alpha_{n}) (1-\beta_{n}) \bigl[\bigl\langle \overrightarrow {f(x_{n})f(\tilde{x})}, \overrightarrow{x_{n}\tilde{x}}\bigr\rangle +\bigl\langle \overrightarrow{f(\tilde{x})\tilde{x}},\overrightarrow{x_{n}\tilde {x}} \bigr\rangle \bigr] \\ \leq& \bigl[\beta_{n}+(1-\beta_{n}) (1- \alpha_{n}) \bigr]\rho^{2}(x_{n},\tilde {x})+ \alpha^{2}_{n}(1-\beta_{n})\rho^{2} \bigl(x_{n+1},f(x_{n})\bigr) \\ &{}+2\alpha^{2}_{n}(1-\beta_{n})\rho \bigl(f(x_{n}),f(\tilde{x})\bigr)\rho(x_{n+1},\tilde {x})+2 \alpha^{2}_{n}(1-\beta_{n})\bigl\langle \overrightarrow{f(\tilde{x})\tilde {x}},\overrightarrow{x_{n+1}\tilde{x}} \bigr\rangle \\ &{}+2\alpha_{n}(1-\alpha_{n}) (1-\beta_{n})\rho \bigl(f(x_{n}),\tilde{x}\bigr)\rho \bigl(T(x_{n}),x_{n} \bigr) \\ &{}+2\alpha_{n}(1-\alpha_{n}) (1-\beta_{n})\rho \bigl(f(x_{n}),f(\tilde{x})\bigr)\rho (x_{n},\tilde{x})\\ &{}+2 \alpha_{n}(1-\alpha_{n}) (1-\beta_{n})\bigl\langle \overrightarrow {f(\tilde{x})\tilde{x}},\overrightarrow{x_{n}\tilde{x}} \bigr\rangle \\ \leq& \bigl[\beta_{n}+(1-\beta_{n}) (1- \alpha_{n}) \bigr]\rho^{2}(x_{n},\tilde {x})+ \alpha^{2}_{n}(1-\beta_{n})\rho^{2} \bigl(x_{n+1},f(x_{n})\bigr) \\ &{}+2k\alpha^{2}_{n}(1-\beta_{n}) \rho(x_{n},\tilde{x})\rho(x_{n+1},\tilde {x})+2 \alpha^{2}_{n}(1-\beta_{n})\bigl\langle \overrightarrow{f(\tilde{x})\tilde {x}},\overrightarrow{x_{n+1}\tilde{x}} \bigr\rangle \\ &{}+2\alpha_{n}(1-\alpha_{n}) (1-\beta_{n})\rho \bigl(f(x_{n}),\tilde{x}\bigr)\rho \bigl(x_{n},T(x_{n}) \bigr) \\ &{}+2k\alpha_{n}(1-\alpha_{n}) (1-\beta_{n}) \rho^{2}(x_{n},\tilde{x})+2\alpha _{n}(1- \alpha_{n}) (1-\beta_{n})\bigl\langle \overrightarrow{f( \tilde{x})\tilde {x}},\overrightarrow{x_{n}\tilde{x}}\bigr\rangle \\ \leq& \bigl[\beta_{n}+(1-\beta_{n}) (1- \alpha_{n}) \bigr]\rho^{2}(x_{n},\tilde {x})+ \alpha^{2}_{n}(1-\beta_{n})\rho^{2} \bigl(x_{n+1},f(x_{n})\bigr) \\ &{}+k\alpha^{2}_{n}(1-\beta_{n}) \bigl[ \rho^{2}(x_{n},\tilde{x})+\rho ^{2}(x_{n+1}, \tilde{x}) \bigr]+2\alpha^{2}_{n}(1-\beta_{n})\bigl\langle \overrightarrow{f(\tilde{x})\tilde{x}},\overrightarrow{x_{n+1} \tilde {x}}\bigr\rangle \\ &{}+2\alpha_{n}(1-\alpha_{n}) (1-\beta_{n})\rho \bigl(f(x_{n}),\tilde{x}\bigr)\rho \bigl(x_{n},T(x_{n}) \bigr) \\ &{}+2k\alpha_{n}(1-\alpha_{n}) (1-\beta_{n}) \rho^{2}(x_{n},\tilde{x})+2\alpha _{n}(1- \alpha_{n}) (1-\beta_{n})\bigl\langle \overrightarrow{f( \tilde{x})\tilde {x}},\overrightarrow{x_{n}\tilde{x}}\bigr\rangle . \end{aligned}$$

This implies that

$$\begin{aligned} \rho^{2}(x_{n+1},\tilde{x}) \leq& \biggl[\frac{\beta_{n}+(1-\beta_{n})(1-\alpha_{n})+2k\alpha_{n}(1-\alpha _{n})(1-\beta_{n})}{1-k\alpha^{2}_{n}(1-\beta_{n})} \biggr]\rho^{2}(x_{n},\tilde{x}) \\ &{}+\frac{k\alpha^{2}_{n}(1-\beta_{n})}{1-k\alpha^{2}_{n}(1-\beta_{n})}\rho ^{2}(x_{n},\tilde{x})+ \frac{\alpha^{2}_{n}(1-\beta_{n})}{1-k\alpha^{2}_{n}(1-\beta _{n})}\rho^{2}\bigl(x_{n+1},f(x_{n})\bigr) \\ &{}+\frac{2\alpha_{n}(1-\alpha_{n})(1-\beta_{n})}{1-k\alpha^{2}_{n}(1-\beta_{n})}\rho \bigl(f(x_{n}),\tilde{x}\bigr)\rho \bigl(x_{n},T(x_{n})\bigr) \\ &{}+\frac{2\alpha^{2}_{n}(1-\beta_{n})}{1-k\alpha^{2}_{n}(1-\beta_{n})}\bigl\langle \overrightarrow{f(\tilde{x})\tilde{x}}, \overrightarrow{x_{n+1}\tilde {x}}\bigr\rangle +\frac{2\alpha_{n}(1-\alpha_{n})(1-\beta_{n})}{1-k\alpha^{2}_{n}(1-\beta _{n})}\bigl\langle \overrightarrow{f(\tilde{x})\tilde{x}},\overrightarrow {x_{n} \tilde{x}}\bigr\rangle . \end{aligned}$$

Thus,

$$ \rho^{2}(x_{n+1},\tilde{x})\leq \bigl(1- \alpha_{n}'\bigr)\rho^{2}(x_{n}, \tilde{x})+\alpha_{n}'\beta_{n}', $$

(7)

where $\alpha_{n}'=\frac{\alpha_{n}(1-\beta_{n})(1-k(2-\alpha_{n}))}{1-k\alpha ^{2}_{n}(1-\beta_{n})}$ and

$$\begin{aligned} \beta_{n}' = & \frac{k\alpha_{n}}{1-k(2-\alpha_{n})}\rho ^{2}(x_{n},\tilde{x})+\frac{\alpha_{n}}{1-k(2-\alpha_{n})}\rho ^{2} \bigl(x_{n+1},f(x_{n})\bigr) \\ &{}+\frac{2(1-\alpha_{n})}{1-k(2-\alpha_{n})}\rho\bigl(f(x_{n}),\tilde{x}\bigr)\rho \bigl(x_{n},T(x_{n})\bigr) \\ &{}+\frac{2\alpha_{n}}{1-k(2-\alpha_{n})}\bigl\langle \overrightarrow{f(\tilde {x})\tilde{x}}, \overrightarrow{x_{n+1}\tilde{x}}\bigr\rangle +\frac{2(1-\alpha_{n})}{1-k(2-\alpha_{n})}\bigl\langle \overrightarrow{f(\tilde {x})\tilde{x}},\overrightarrow{x_{n} \tilde{x}}\bigr\rangle . \end{aligned}$$

Since $k\in [0,\frac{1}{2} )$, $\alpha_{n}'\in(0,1)$. Applying Lemma 2.7 to the inequality (7), we can conclude that $x_{n}\to\tilde{x}$ as $n\to\infty$. This completes the proof. □

4 Concluding remarks and open problems

• (1) Our main theorem can be applied to $\operatorname{CAT}(\kappa)$ spaces with $\kappa\leq0$ since any $\operatorname{CAT}(\kappa)$ space is a $\operatorname{CAT}(\kappa')$ space for $\kappa' \geq\kappa$ (see [6]). However, the result for $\kappa> 0$ is still unknown (see [5], p.1264).

• (2) Our main theorem can be viewed as an extension of Corollary 8 in [16] for a contraction f with $k\in [0,\frac{1}{2} )$. It remains an open problem whether Theorem 3.2 holds for $k\in [\frac{1}{2}, 1 )$.