Open Access

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

Journal of Inequalities and Applications20152015:278

https://doi.org/10.1186/s13660-015-0801-6

Received: 8 June 2015

Accepted: 27 August 2015

Published: 17 September 2015

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).

Keywords

viscosity iteration method fixed point strong convergence the nice projection property CAT(0) space

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 \(k\in [\frac{1}{2}, 1 )\).

     

Declarations

Acknowledgements

This research was supported by Chiang Mai University and Thailand Research Fund under Grant RTA5780007.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, Chiang Mai University

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© Kaewkhao et al. 2015