## 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 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

1. (i)

$$\sum_{n=1}^{\infty} \alpha_{n}=\infty$$;

2. (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

1. (i)

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

2. (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 as $$t\to0$$. Moreover, $$\tilde{x}=P_{\operatorname{Fix}(T)}(f(\tilde{x}))$$ and 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:

1. (i)

$$\lim_{n\to\infty} \alpha_{n} =0$$;

2. (ii)

$$\sum^{\infty}_{n=1} \alpha_{n} = \infty$$;

3. (iii)

$$0<\liminf_{n} \beta_{n} \leq\limsup_{n} \beta_{n} <1$$.

Then $$\{x_{n}\}$$ converges strongly to such that $$\tilde{x}=P_{\operatorname{Fix}(T)}(f(\tilde{x}))$$ and 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 , 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 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. (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. (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 $$k\in [\frac{1}{2}, 1 )$$.