# A projection method for approximating fixed points of quasinonexpansive mappings in Hadamard spaces

## Abstract

This work is devoted to analyzing the feasibility study of a Moudafi viscosity projection method with a weak contraction for a finite family of quasinonexpansive mappings in a Hadamard space. To this end, we need to construct a countable family of nonexpansive mappings satisfying AKTT condition with a weak contraction by choosing an appropriate control sequence under certain conditions.

## Introduction

Let C be a nonempty subset of a metric space $$(X,d)$$. Suppose that, for each $$x \in X$$, there exists a unique point $$P_{C}x \in C$$ such that $$d(x,P_{C}x)=d(x,C)=\inf_{y \in C}d(x,y)$$. Then, the mapping $$P_{C}$$ of X onto C is called the metric projection.

The well-known Banach contraction principle is an important tool in the theory of metric spaces; it guarantees the existence and uniqueness of fixed points of certain self-mappings of metric spaces. One generalization of the contraction principle for weak contractions is obtained by Alber and Guerre-Delabriere  in Hilbert spaces. A mapping $$f:X \to X$$ is called a φ-weak contraction if

$$d\bigl(f(x),f(y)\bigr) \le d(x,y)-\varphi\bigl(d(x,y)\bigr),\quad x,y \in X,$$

where $$\varphi:[0,\infty) \to[0,\infty)$$ is a continuous and nondecreasing function with $$\varphi(t)=0$$ if and only if $$t=0$$.

Let $$T:C \to X$$ be a mapping. If $$d(Tx,Ty) \le d(x,y)$$ for all $$x,y \in C$$, then T is nonexpansive. We denote by $$\mathfrak{F}(T)$$ the set o fixed points of T. The mapping T is quasinonexpansive if $$\mathfrak{F}(T)$$ is nonempty and

$$d(Tx,y) \le d(x,y), \quad x \in C, y \in\mathfrak{F}(T).$$

A point $$p \in C$$ is said to be a strongly asymptotic fixed point  of T if there exists a sequence $$\{x_{n}\}$$ in C that converges strongly to p and $$\lim_{n \to\infty}d(x_{n},Tx_{n})=0$$. We denote by $$\mathfrak{\widetilde{F}}(T)$$ the set of strongly asymptotic fixed points of T. It is known that the fixed point set of a quasinonexpansive mapping defined on a $$\operatorname{CAT}(0)$$ space (see Section 2 for the definition) is closed and convex.

Approximation methods for finding specific fixed points of a family of nonexpansive mappings in Hilbert, Banach, and geodesic metric spaces have been studied by many researchers; see, e.g.,  and the references therein. One well-known method, called the shrinking projection method, was first proposed by Takahashi et al.  and has been applied to a variety of approximation problems; see, e.g., [11, 12]. In particular, Kimura and Takahashi  applied this method to the zero-point problem for a maximal monotone operator defined in a Banach space and obtained strong convergence theorems. To generate the iterative sequence by the shrinking projection method, they use the metric projection onto a closed convex set $$C_{n}$$ for each $$n \in\mathbb{N}$$. It is noticeable that the larger the integer n, the more complicated the shape of $$C_{n}$$. Hence, the calculation of the projection is tedious as n gets larger. In 2011, Kimura et al.  overcome this difficulty and introduce the so-called averaged projection method of Halpern type for a family of quasinonexpansive mappings by combining the Halpern iteration. They still use the metric projection approach; nevertheless, the subsets corresponding to these projections have simpler shapes than the classical ones. Let us denote by $$\mathfrak{F}(\mathfrak{T})$$ the common fixed point set of all mappings in a family $$\mathfrak{T}$$. Their theorem is stated as follows.

### Theorem 1.1

(Kimura et al. , Theorem 3.1)

Let C be a closed convex subset of a Hilbert space H, $$\mathfrak{T}=\{T_{j}:j=1,\ldots,N\}$$ a finite family of quasinonexpansive mappings of C into H with $$\mathfrak{F}(\mathfrak{T}) \ne\emptyset$$ and $$\widetilde{\mathfrak{F}}(T_{j})=\mathfrak{F}(T_{j})$$ for $$j=1,\ldots,N$$. Let $$u,x_{1} \in C$$ and define the sequence $$\{x_{n}\}$$ by

\begin{aligned}& y^{j}_{n} =\alpha_{n}x_{n}+(1- \alpha_{n})T_{j}x_{n}, \\& C^{j}_{n} =\bigl\{ z \in C:\bigl\Vert y^{j}_{n}-z \bigr\Vert \le \Vert x_{n}-z\Vert \bigr\} , \quad j=1,\ldots,N, \\& v^{j}_{n,k} =P_{C^{j}_{k}}x_{n}, \quad k=1, \ldots,n, j=1,\ldots,N, \\& w_{n,k} =\sum_{j=1}^{N} \beta^{j}_{k}v^{j}_{n,k}, \quad k=1, \ldots,n, \\& x_{n+1} = \delta_{n}u+(1-\delta_{n})\sum _{k=1}^{n}\gamma_{n,k}w_{n,k}, \end{aligned}

where $$\{\alpha_{n}\}$$, $$\{\beta^{j}_{n}:j=1,\ldots,N\}$$, $$\{\gamma_{n,k}:k \le n\}$$, and $$\{\delta_{n}\}$$ are sequences in $$[0,1]$$ satisfying the following conditions:

1. (i)

$$\liminf_{n \to\infty}\alpha_{n}<1$$,

2. (ii)

$$\beta^{j}_{n}>0$$ for $$j=1,\ldots,N$$, and $$\sum_{j=1}^{N}\beta^{j}_{n}=1$$ for $$n \in\mathbb{N}$$,

3. (iii)

$$\sum_{k=1}^{n}\gamma_{n,k}=1$$ for $$n \in\mathbb{N}$$, $$\lim_{n \to\infty}\gamma_{n,k}>0$$ for $$k \in\mathbb{N}$$, and $$\sum_{n=1}^{\infty}\sum_{k=1}^{n}|\gamma_{n+1,k}-\gamma _{n,k}|<\infty$$,

4. (iv)

$$\lim_{n \to\infty}\delta_{n}=0$$, $$\sum_{n=1}^{\infty}\delta_{n}=\infty$$, and $$\sum_{n=1}^{\infty}|\delta_{n+1}-\delta_{n}|<\infty$$.

Then $$\{x_{n}\}$$ converges strongly to the point $$P_{\mathfrak{F}(\mathfrak{T})}u$$.

The problem of whether or not we can construct a shrinking projection method analogous to that given in Theorem 1.1 for solving a common fixed point problem for a finite family of quasinonexpansive mappings in a geodesic metric space is still open. The purpose of this paper is to analyze the feasibility study of Moudafi viscosity type of projection method with a weak contraction for a finite family of quasinonexpansive mappings in a complete $$\operatorname{CAT}(0)$$ space, also known as a Hadamard space.

This paper is organized as follows. In Section 2 we recall the definition of geodesic metric spaces and summarize some useful lemmas and the main properties of $$\operatorname{CAT}(0)$$ spaces. Besides, without vector addition as in a Banach space, we present an inequality to estimate the distance between two elements defined by finite convex combination ‘’ in a $$\operatorname{CAT}(0)$$ space; see Lemma 2.2. In Section 3 we construct a sequence of nonexpansive mappings satisfying AKTT condition by choosing an appropriate control sequence under certain conditions; see Theorem 3.2. Therefore, a convergence theorem of a new Moudafi viscosity approximation follows from Theorem 3.2; see Theorem 3.3. Using Theorem 3.3, we also derive a strong convergence theorem by a Moudafi type viscosity approximation with a weak contraction for a family of quasinonexpansive mappings; see Theorem 3.4. As a particular case where a weak contraction is constant in Theorem 3.4, a strong convergence theorem by the averaged projection method of Halpern type is then obtained; see Theorem 3.5.

## Preliminaries

Let $$(X,d)$$ be a metric space. For $$x,y \in X$$, a geodesic path joining x to y (or a geodesic from x to y) is an isometric mapping $$c:[0,\ell] \subset\mathbb{R} \to X$$ such that $$c(0)=x$$, $$c(\ell)=y$$, that is, $$d(c(t),c(t'))=|t-t'|$$ for all $$t,t' \in[0,\ell]$$. Therefore, $$d(x,y)=\ell$$. The image of c is called a geodesic (segment) from x to y, and we shall denote a definite choice of this geodesic segment by $$[x,y]$$. A point $$z=c(t)$$ in the geodesic $$[x,y]$$ will be written as $$z=(1-\lambda)x \oplus\lambda y$$, where $$\lambda=t/\ell$$, and so $$d(z,x)=\lambda d(x,y)$$ and $$d(z,y)=(1-\lambda)d(x,y)$$. A subset C of X is 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 $$\triangle(x_{1},x_{2},x_{3})$$ in $$(X,d)$$ consists of three points $$x_{i} \in X$$ ($$i=1,2,3$$), its vertices, and a geodesic segment between each pair of vertices, its sides. If a point $$x \in X$$ lies in the union of $$[x_{i},x_{j}]$$, $$i,j \in\{ 1,2,3\}$$, then we write $$x \in\triangle(x_{1},x_{2},x_{3})$$. A comparison triangle for the geodesic triangle $$\triangle (x_{1},x_{2},x_{3})$$ in X is a triangle $$\triangle(\bar{x}_{1},\bar{x}_{2},\bar{x}_{3})$$ in the Euclidean plane $$\mathbb{E}^{2}$$ such that $$d_{\mathbb{E}^{2}}(\bar{x}_{i},\bar{x}_{j})=d(x_{i},x_{j})$$ for $$i,j \in\{1,2,3\}$$.

A geodesic triangle in X is said to satisfy the $$\operatorname{CAT}(0)$$ inequality if, given a comparison triangle ̅ in $$\mathbb{E}^{2}$$ for ,

$$d(x,y) \le d_{\mathbb{E}^{2}}(\bar{x},\bar{y})\quad \mbox{for }x,y \in \triangle,$$

where $$\bar{x},\bar{y} \in\overline{\triangle}$$ are the corresponding comparison points of x, y. The geodesic metric space X is called a $$\operatorname{CAT}(0)$$ space if all geodesic triangles in X satisfy the $$\operatorname{CAT}(0)$$ inequality. Note that Hilbert spaces are $$\operatorname{CAT}(0)$$.

### Lemma 2.1

Let $$(X,d)$$ be a $$\operatorname{CAT}(0)$$ space, and let $$\alpha,\beta\in [0,1]$$. Then:

1. (i)

For $$x,y \in X$$, we have

$$d\bigl(\alpha x \oplus(1-\alpha)y,\beta x \oplus(1-\beta)y\bigr)=|\alpha- \beta|d(x,y).$$
2. (ii)

(, Chapter II.2. Proposition 2.2) For $$x,y,p,q \in X$$, we have

$$d\bigl(\alpha x \oplus(1-\alpha)y,\alpha p \oplus(1-\alpha)q\bigr) \le\alpha d(x,p)+(1-\alpha) d(y,q).$$

In particular, if $$p=q$$, this reduces to

$$d\bigl(\alpha x \oplus(1-\alpha)y,p\bigr) \le\alpha d(x,p)+(1-\alpha)d(y,p).$$
3. (iii)

(, Lemma 2.5) For $$x,y,z \in X$$, we have

$$d\bigl(\alpha x \oplus(1-\alpha)y,z\bigr)^{2} \le\alpha d(x,z)^{2}+(1-\alpha )d(y,z)^{2}-\alpha(1- \alpha)d(x,y)^{2}.$$

We will extend the equality in Lemma 2.1(i) to any finitely many elements in X. First, we recall the notion of a finite sum ‘’ defined by Butsan et al. . Fix $$n \in\mathbb{N}$$ with $$n \ge2$$ and let $$\{\alpha_{1},\ldots,\alpha _{n}\} \subset(0,1)$$ with $$\sum_{k=1}^{n}\alpha_{k}=1$$ and $$\{x_{1},\ldots,x_{n}\} \subset X$$. By induction we define

$$\bigoplus_{k=1}^{n}\alpha_{k}x_{k} =(1-\alpha_{n}) \biggl(\frac{\alpha_{1}}{1-\alpha_{n}}x_{1}\oplus\cdots \oplus\frac {\alpha_{n-1}}{1-\alpha_{n}}x_{n-1} \biggr) \oplus\alpha_{n}x_{n}.$$
(2.1)

The definition of in (2.1) is an ordered one in the sense that it depends on the order of points $$x_{1},\ldots,x_{n}$$. However, we occasionally use the notation $$\alpha_{1}x_{1} \oplus\alpha_{2}x_{2} \oplus\cdots\oplus\alpha_{n}x_{n}$$ for such a point. Lemma 2.1(ii) assures that, for $$y \in X$$,

$$d \Biggl(\bigoplus_{k=1}^{n} \alpha_{k}x_{k},y \Biggr) \le\sum _{k=1}^{n}\alpha _{k}d(x_{k},y).$$
(2.2)

### Lemma 2.2

Let $$(X,d)$$ be a $$\operatorname{CAT}(0)$$ space, and for $$n \in\mathbb{N}$$ with $$n \ge2$$, let $$\{\alpha_{k}\}_{k=1}^{n}$$ and $$\{\beta_{k}\}_{k=1}^{n} \subset(0,1)$$ be two sequences such that $$\sum_{k=1}^{n}\alpha_{k}=\sum_{k=1}^{n}\beta_{k}=1$$. Then, for $$x_{1},\ldots,x_{n} \in X$$, we have

\begin{aligned}& d \Biggl(\bigoplus_{k=1}^{n} \alpha_{k}x_{k},\bigoplus_{k=1}^{n} \beta_{k}x_{k} \Biggr) \\& \quad \le\biggl\vert \frac{\alpha_{2}}{\alpha_{1}+\alpha_{2}}-\frac{\beta_{2}}{\beta _{1}+\beta_{2}}\biggr\vert ( \alpha_{1}+\alpha_{2})d(x_{1},x_{2}) \\& \qquad {}+\biggl\vert \frac{\alpha_{3}}{\sum_{k=1}^{3}\alpha_{k}}-\frac{\beta_{3}}{\sum_{k=1}^{3}\beta_{k}}\biggr\vert \sum _{k=1}^{3}\alpha_{k}\cdot\sum _{k=1}^{2}\frac{\beta_{k}}{\beta_{1}+\beta _{2}}d(x_{k},x_{3}) \\& \qquad {}+\cdots+\biggl\vert \frac{\alpha_{j}}{\sum_{k=1}^{j}\alpha_{k}}-\frac{\beta _{j}}{\sum_{k=1}^{j}\beta_{j}}\biggr\vert \sum_{k=1}^{j}\alpha_{j}\cdot\sum _{k=1}^{j-1}\frac{\beta_{k}}{\beta_{1}+\cdots +\beta_{j-1}}d(x_{k},x_{j}) \\& \qquad {}+\cdots+\vert \alpha_{n}-\beta_{n}\vert \sum _{k=1}^{n-1}\frac{\beta_{k}}{1-\beta _{n}}d(x_{k},x_{n}). \end{aligned}

### Proof

We will prove the result by induction.

Step 1. According to Lemma 2.1(ii), (2.1), and (2.2), we derive

\begin{aligned}& d \Biggl(\bigoplus_{k=1}^{n} \alpha_{k}x_{k},\bigoplus_{k=1}^{n} \beta_{k}x_{k} \Biggr) \\& \quad \le d \Biggl((1-\alpha_{n}) \Biggl(\bigoplus _{k=1}^{n-1}\frac{\alpha _{k}}{1-\alpha_{n}}x_{k} \Biggr) \oplus\alpha_{n}x_{n}, (1-\alpha_{n}) \Biggl( \bigoplus_{k=1}^{n-1}\frac{\beta_{k}}{1-\beta _{n}}x_{k} \Biggr)\oplus\alpha_{n}x_{n} \Biggr) \\& \qquad {}+d \Biggl((1-\alpha_{n}) \Biggl(\bigoplus _{k=1}^{n-1}\frac{\beta_{k}}{1-\beta _{n}}x_{k} \Biggr) \oplus\alpha_{n}x_{n}, (1-\beta_{n}) \Biggl( \bigoplus_{k=1}^{n-1}\frac{\beta_{k}}{1-\beta _{n}}x_{k} \Biggr)\oplus\beta_{n}x_{n} \Biggr) \\& \quad \le (1-\alpha_{n})d \Biggl(\bigoplus _{k=1}^{n-1}\frac{\alpha_{k}}{1-\alpha _{n}}x_{k},\bigoplus _{k=1}^{n-1}\frac{\beta_{k}}{1-\beta_{n}}x_{k} \Biggr) + |\alpha_{n}-\beta_{n}|d \Biggl(\bigoplus _{k=1}^{n-1}\frac{\beta_{k}}{1-\beta _{n}}x_{k},x_{n} \Biggr) \\& \quad \le (1-\alpha_{n})d \Biggl(\bigoplus _{k=1}^{n-1}\frac{\alpha_{k}}{1-\alpha _{n}}x_{k},\bigoplus _{k=1}^{n-1}\frac{\beta_{k}}{1-\beta_{n}}x_{k} \Biggr) \\& \qquad {}+ |\alpha_{n}-\beta_{n}|\sum _{k=1}^{n-1}\frac{\beta_{k}}{1-\beta_{n}}d(x_{k},x_{n}). \end{aligned}

Step 2. Apply the inequality in Step 1 for the case $$n-1$$ to obtain

\begin{aligned}& d \Biggl(\bigoplus_{k=1}^{n-1} \frac{\alpha_{k}}{1-\alpha_{n}}x_{k},\bigoplus_{k=1}^{n-1} \frac{\beta_{k}}{1-\beta_{n}}x_{k} \Biggr) \\& \quad \le\frac{1-\alpha_{n-1}-\alpha_{n}}{1-\alpha_{n}} d \Biggl(\bigoplus_{k=1}^{n-2} \frac{\alpha_{k}}{1-\alpha_{n-1}-\alpha _{n}}x_{k},\bigoplus_{k=1}^{n-2} \frac{\beta_{k}}{1-\beta_{n-1}-\beta _{n}}x_{k} \Biggr) \\& \qquad {}+\biggl\vert \frac{\alpha_{n-1}}{1-\alpha_{n}}-\frac{\beta _{n-1}}{1-\beta_{n}}\biggr\vert \sum _{k=1}^{n-2}\frac{\beta_{k}}{1-\beta _{n-1}-\beta_{n}}d(x_{k},x_{n-1}). \end{aligned}

Step 3. Recall that $$\sum_{k=1}^{n}\alpha_{k}=\sum_{k=1}^{n}\beta_{k}=1$$. Hence, the two inequalities in Step 1 and Step 2 imply that

\begin{aligned}& d \Biggl(\bigoplus_{k=1}^{n} \alpha_{k}x_{k},\bigoplus_{k=1}^{n} \beta_{k}x_{k} \Biggr) \\& \quad \le(1-\alpha_{n-1}-\alpha_{n})d \Biggl(\bigoplus _{k=1}^{n-2}\frac{\alpha _{k}}{1-\alpha_{n-1}-\alpha_{n}}x_{k}, \bigoplus_{k=1}^{n-2}\frac{\beta _{k}}{1-\beta_{n-1}-\beta_{n}}x_{k} \Biggr) \\& \qquad {}+\biggl\vert \frac{\alpha_{n-1}}{\sum_{k=1}^{n-1}\alpha_{k}}-\frac{\beta _{n-1}}{\sum_{k=1}^{n-1}\beta_{k}}\biggr\vert \sum _{k=1}^{n-1}\alpha_{k}\cdot\sum _{k=1}^{n-2}\frac{\beta_{k}}{1-\beta _{n-1}-\beta_{n}}d(x_{k},x_{n-1}) \\& \qquad {}+|\alpha_{n}-\beta_{n}|\sum _{k=1}^{n-1}\frac{\beta_{k}}{1-\beta_{n}}d(x_{k},x_{n}). \end{aligned}

Continuing the process in Step 1 to estimate the first term of this inequality on the right-hand side, after $$n-2$$ steps, we have

\begin{aligned}& d \Biggl(\bigoplus_{k=1}^{n} \alpha_{k}x_{k},\bigoplus_{k=1}^{n} \beta_{k}x_{k} \Biggr) \\& \quad \le\biggl\vert \frac{\alpha_{2}}{\alpha_{1}+\alpha_{2}}-\frac{\beta_{2}}{\beta _{1}+\beta_{2}}\biggr\vert ( \alpha_{1}+\alpha_{2})d(x_{1},x_{2}) \\& \qquad {}+\biggl\vert \frac{\alpha_{3}}{\sum_{k=1}^{3}\alpha_{k}}-\frac{\beta_{3}}{\sum_{k=1}^{3}\beta_{k}}\biggr\vert \sum _{k=1}^{3}\alpha_{k}\cdot\sum _{k=1}^{2}\frac{\beta_{k}}{\beta_{1}+\beta _{2}}d(x_{k},x_{3}) \\& \qquad {}+\cdots+\biggl\vert \frac{\alpha_{j}}{\sum_{k=1}^{j}\alpha_{k}}-\frac{\beta _{j}}{\sum_{k=1}^{j}\beta_{j}}\biggr\vert \sum_{k=1}^{j}\alpha_{j}\cdot\sum _{k=1}^{j-1}\frac{\beta_{k}}{\beta_{1}+\cdots +\beta_{j-1}}d(x_{k},x_{j}) \\& \qquad {}+\cdots+|\alpha_{n}-\beta_{n}|\sum _{k=1}^{n-1}\frac{\beta_{k}}{1-\beta_{n}}d(x_{k},x_{n}). \end{aligned}

□

Let $$\{\alpha_{n}\}_{n=1}^{\infty}$$ be a sequence in $$(0,1)$$ such that $$\sum_{n=1}^{\infty}\alpha_{n}=1$$. For notational convenience, let

$$\bar{\alpha}_{k}=\frac{\alpha_{k}}{\sum_{j=1}^{k}\alpha_{j}},\qquad \alpha'_{k}= \sum_{j=k+1}^{\infty}\alpha_{j} \quad \text{for }k \in \mathbb{N},$$

The following result is an immediate consequence of Lemma 2.2.

### Lemma 2.3

Let $$(X,d)$$ be a $$\operatorname{CAT}(0)$$ space, and for $$n \in\mathbb{N}$$ ($$n \ge2$$), let $$\{\alpha_{k}\}_{k=1}^{n},\{\beta _{k}\}_{k=1}^{n} \subset(0,1)$$ be such that $$\sum_{k=1}^{n}\alpha_{k}=\sum_{k=1}^{n}\beta_{k}=1$$. Then for $$x_{1},\ldots,x_{n} \in X$$, we have

$$d \Biggl(\bigoplus_{k=1}^{n} \alpha_{k}x_{k},\bigoplus_{k=1}^{n} \beta_{k}x_{k} \Biggr) \le M\sum_{k=1}^{n}| \bar{\alpha}_{k}-\bar{\beta}_{k}|,$$

where $$M=\max\{d(x_{i},x_{j}):i,j=1,\ldots,n\}$$.

It is remarkable that Dhompongsa et al.  define an infinite sum ‘’ as follows. Let $$\{\alpha_{n}\} \subset(0,1)$$ with $$\sum_{n=1}^{\infty}\alpha_{n}=1$$, and let $$\{x_{n}\}$$ be a bounded sequence in a complete metric space X. Choose arbitrary $$u \in X$$. Suppose that $$\lim_{n \to\infty}\sum_{k=n}^{\infty}\alpha'_{k}=0$$. Define the sequence $$\{y_{n}\}$$ in X by

$$y_{n}=\alpha_{1}x_{1} \oplus \alpha_{2}x_{2} \oplus\cdots\oplus\alpha_{n}x_{n} \oplus\alpha'_{n}u.$$

Then, according to (2.1),

$$y_{n}= \Biggl(\sum_{k=1}^{n} \alpha_{k} \Biggr)z_{n} \oplus\alpha'_{n}u,$$
(2.3)

where

$$z_{n}=\frac{\alpha_{1}}{\sum_{k=1}^{n}\alpha_{k}}x_{1} \oplus\cdots\oplus \frac {\alpha_{n}}{\sum_{k=1}^{n}\alpha_{k}}x_{n}.$$

Recall that $$\{y_{n}\}$$ is a Cauchy sequence  and therefore converges to some point $$x \in X$$. We can write

$$x=\bigoplus_{n=1}^{\infty}\alpha_{n}x_{n}.$$

By (2.3), $$d(y_{n},z_{n})=\alpha'_{n}d(z_{n},u)$$. Hence, $$\{z_{n}\}$$ also converges to x, and the limit x is independent of the choice of u.

To verify our main results in Section 3, the following property is required and crucial.

### Lemma 2.4

(Dhompongsa et al. , Lemma 3.8)

Let C be a closed convex subset of a complete $$\operatorname{CAT}(0)$$ space X, $$\{T_{n}\}$$ a sequence of nonexpansive mappings on C with $$\bigcap_{n=1}^{\infty}\mathfrak{F}(T_{n}) \ne\emptyset$$, and $$\{\alpha_{n}\}$$ a sequence in $$(0,1)$$ such that $$\sum_{n=1}^{\infty}\alpha_{n}=1$$ and $$\lim_{n \to\infty}\sum_{k=n}^{\infty}\alpha'_{k}=0$$. Define the mapping $$S:C \to C$$ by $$Sx=\bigoplus_{n=1}^{\infty}\alpha _{n}T_{n}x$$, $$x \in C$$. Then S is nonexpansive, and $$\mathfrak{F}(S)=\bigcap_{n=1}^{\infty}\mathfrak{F}(T_{n})$$.

## Projection method

Let C be a closed convex subset of a complete metric space X. A family $$\{T_{n}\}$$ of nonexpansive self-mappings of C is said to satisfy AKTT condition  if for every bounded subset B of C,

$$\sum_{n=1}^{\infty}\sup\bigl\{ d(T_{n+1}x,T_{n}x):x \in B\bigr\} < \infty.$$

In this case, the sequence $$\{T_{n}x\}$$ is Cauchy for each $$x \in C$$ and so converges in X. We recall the following convergence theorem with a weak contraction for a sequence of nonexpansive mappings with AKTT condition.

### Theorem 3.1

(Huang , Theorem 4.11)

Let X be a complete $$\operatorname{CAT}(0)$$ space, C a closed convex subset of X, $$\{T_{n}\}$$ a family of nonexpansive mappings on C satisfying AKTT condition such that $$\bigcap_{n=1}^{\infty}\mathfrak{F}(T_{n})\ne\emptyset$$, f a φ-weak contraction on C, where φ is strictly increasing, and $$\{\alpha_{n}\}$$ is a sequence in $$(0,1]$$ satisfying

1. (C1)

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

2. (C2)

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

3. (C3)

either $$\sum_{n=1}^{\infty}|\alpha_{n+1}-\alpha_{n}|<\infty$$, or $$\lim_{n \to\infty}(\alpha_{n+1}/\alpha_{n})=1$$.

Define the mapping $$S:C \to C$$ by $$Sx=\lim_{n \to\infty}T_{n}x$$ for $$x \in C$$. Suppose that $$\mathfrak{F}(S)=\bigcap_{n=1}^{\infty}\mathfrak{F}(T_{n})$$. Then the sequence $$\{x_{n}\}$$ defined by $$x_{1} \in C$$ and

$$x_{n+1}=\alpha_{n}f(x_{n}) \oplus(1- \alpha_{n})T_{n}x_{n}$$

converges strongly to a point $$\hat{x} \in C$$ such that $$\hat{x}=P_{\mathfrak{F}(S)}f(\hat{x})$$.

We now construct a sequence of nonexpansive mappings satisfying AKTT condition by choosing an appropriate control sequence under certain conditions.

### Theorem 3.2

Let C be a closed convex subset of a complete $$\operatorname{CAT}(0)$$ space X, $$\mathfrak{T}=\{T_{n}\}$$ a family of nonexpansive mappings on C with $$\mathfrak{F}(\mathfrak{T}) \ne\emptyset$$, and $$\{\gamma_{n,k}:k \le n\} \subset(0,1)$$ a sequence satisfying

1. (D1)

$$\sum_{k=1}^{n}\gamma_{n,k}=1$$, $$\forall n \in\mathbb{N}$$;

2. (D2)

$$\lambda_{k}=\lim_{n \to\infty}\gamma_{n,k}>0$$, $$\forall k \in\mathbb {N}$$, and $$\lim_{n \to\infty}\sum_{k=n}^{\infty}\lambda'_{k}=0$$;

3. (D3)

$$\sum_{n=1}^{\infty}\sum_{k=1}^{n+1}|\bar{\gamma}_{n+1,k}-\bar{\gamma }_{n,k}|<\infty$$, where $$\gamma_{n,n+1}=0$$ and

$$\bar{\gamma}_{n,k}=\frac{\gamma_{n,k}}{\gamma_{n,1}+\cdots+\gamma _{n,k}}, \quad k=1,\ldots,n+1.$$

For each $$n \in\mathbb{N}$$, define the mapping $$S_{n}:C \to C$$ by

$$S_{n}x=\bigoplus_{k=1}^{n} \gamma_{n,k}T_{k}x.$$

Then $$\{S_{n}\}$$ is a family of nonexpansive mappings satisfying AKTT condition and

$$\bigcap_{n=1}^{\infty}\mathfrak{F}(S_{n})=\bigcap_{n=1}^{\infty}\mathfrak {F}(T_{n}).$$

Moreover, the mapping $$S:C \to C$$ defined by $$Sx=\lim_{n \to\infty}S_{n}x$$ is also nonexpansive, and $$\mathfrak{F}(S)=\bigcap_{n=1}^{\infty}\mathfrak {F}(S_{n})$$.

### Proof

Fix any $$n \in\mathbb{N}$$. We may assume that $$\gamma_{n,k}=0$$ for all $$k>n$$. Then Lemma 2.4 states that $$S_{n}$$ is nonexpansive and $$\mathfrak {F}(S_{n})=\bigcap_{k=1}^{n}\mathfrak{F}(T_{k})$$. Thus,

$$\bigcap_{n=1}^{\infty}\mathfrak{F}(S_{n})= \bigcap_{k=1}^{\infty}\mathfrak {F}(T_{k}) \ne\emptyset.$$

For every bounded subset B of C, the set $$\{T_{k}x:x \in B, k \in \mathbb{N}\}$$ is bounded since $$\bigcap_{k=1}^{\infty}\mathfrak{F}(T_{k}) \ne\emptyset$$. Let

$$M=\operatorname{diam}\{T_{k}x:x \in B, k \in\mathbb{N}\},$$

so that by Lemma 2.3, for $$x \in B$$ and $$n \in\mathbb{N}$$, we have

\begin{aligned} d(S_{n+1}x,S_{n}x) \le& d \Biggl(\bigoplus _{k=1}^{n}\gamma_{n+1,k}T_{k}x, \bigoplus_{k=1}^{n}\gamma _{n,k}T_{k}x \Biggr) +\gamma_{n+1,n+1}d \Biggl(T_{n+1}x,\bigoplus _{k=1}^{n}\gamma _{n,k}T_{k}x \Biggr) \\ \le& M\sum_{k=1}^{n}|\bar{ \gamma}_{n+1,k}-\bar{\gamma}_{n,k}|+M\gamma _{n+1,n+1}\sum _{k=1}^{n}\gamma_{n,k} \\ =& M\sum_{k=1}^{n}|\bar{ \gamma}_{n+1,k}-\bar{\gamma}_{n,k}|+M\bar{\gamma }_{n+1,n+1} \\ =& M\sum_{k=1}^{n+1}|\bar{ \gamma}_{n+1,k}-\bar{\gamma}_{n,k}|. \end{aligned}

It follows that

$$\sum_{n=1}^{\infty}\sup\bigl\{ d(S_{n+1}x,S_{n}x):x \in B\bigr\} \le M\sum _{n=1}^{\infty}\sum_{k=1}^{n+1}| \bar{\gamma}_{n+1,k}-\bar{\gamma }_{n,k}|< \infty.$$

Therefore, $$\{S_{n}\}$$ is a family of nonexpansive mappings on C satisfying AKTT condition such that $$\bigcap_{n=1}^{\infty}\mathfrak {F}(S_{n}) \ne\emptyset$$. It follows that $$\{S_{n}x\}$$ converges for all $$x \in C$$, and thus S is well defined.

If $$m,n \in\mathbb{N}$$ and $$m > n$$, then we get

\begin{aligned} \begin{aligned} \sum_{k=1}^{n}|\bar{\gamma}_{m,k}- \bar{\gamma}_{n,k}| & \le\sum_{k=1}^{n} \bigl(\vert \bar{\gamma}_{n+1,k}-\bar{\gamma}_{n,k}\vert +| \bar {\gamma}_{n+2,k}-\bar{\gamma}_{n+1,k}|+\cdots +|\bar{\gamma}_{m,k}-\bar{\gamma}_{m-1,k}|\bigr) \\ & = \sum_{j=n}^{m-1}\sum _{k=1}^{n}|\bar{\gamma}_{j+1,k}-\bar{\gamma }_{j,k}| \\ & \le\sum_{j=n}^{m-1}\sum _{k=1}^{j+1}|\bar{\gamma}_{j+1,k}-\bar{ \gamma}_{j,k}|. \end{aligned} \end{aligned}

Recall that $$\bar{\lambda}_{k}=\lim_{n \to\infty}\bar{\gamma}_{n,k}$$ for $$k \in\mathbb{N}$$. We take the limit as $$m \to\infty$$ to obtain

$$\sum_{k=1}^{n}|\bar{\lambda}_{k}- \bar{\gamma}_{n,k}| \le\sum_{j=n}^{\infty}\sum_{k=1}^{j+1}|\bar{\gamma}_{j+1,k}- \bar{\gamma}_{j,k}|$$

and then take the limit as $$n \to\infty$$ to obtain

$$\lim_{n \to\infty}\sum_{k=1}^{n}| \bar{\lambda}_{k}-\bar{\gamma}_{n,k}|=0.$$
(3.1)

On the other hand, the absolute convergence of the series

$$\sum_{n=1}^{\infty}\sum _{k=1}^{n+1}(\bar{\gamma}_{n+1,k}-\bar{ \gamma}_{n,k})$$

implies the convergence of its partial sums

$$\sum_{n=1}^{m}\sum _{k=1}^{n+1}(\bar{\gamma}_{n+1,k}-\bar{ \gamma}_{n,k}) = \Biggl(\sum_{k=1}^{m+1} \bar{\gamma}_{m+1,k} \Biggr)-\bar{\gamma}_{1,1} = \Biggl(\sum _{k=1}^{m+1}\bar{\gamma}_{m+1,k} \Biggr)-1.$$

Hence, by (3.1), $$\sum_{k=1}^{\infty}\bar{\lambda}_{k}$$ converges (in fact, to $$\sum_{k=1}^{\infty}\bar{\gamma}_{n,k}$$), and so does $$\sum_{k=1}^{\infty}\lambda_{k}$$ because $$\lambda_{k} \le\bar {\lambda}_{k}$$. Let $$\lambda=\sum_{k=1}^{\infty}\lambda_{k}$$. Define the mapping $$W:C \to C$$ by

$$Wx=\bigoplus_{n=1}^{\infty}\frac{\lambda_{n}}{\lambda}T_{n}x.$$

Then by (D2) Lemma 2.4 guarantees that W is nonexpansive and $$\mathfrak{F}(W)=\bigcap_{n=1}^{\infty}\mathfrak{F}(T_{n})$$. If

$$W_{n}x=\bigoplus_{k=1}^{n} \frac{\lambda_{k}}{\sum_{j=1}^{n}\lambda_{j}}T_{k}x,\quad x \in C,$$

then $$\{W_{n}x\}$$ converges to Wx. Recall that

$$\overline{ \biggl(\frac{\lambda_{k}}{\sum_{j=1}^{n}\lambda_{j}} \biggr)}=\bar {\lambda}_{k}\quad \text{for }k=1,\ldots,n.$$

Fix any $$x \in C$$. Then by Lemma 2.3 and (3.1) we get

$$d(S_{n}x,W_{n}x) \le K\sum_{k=1}^{n}| \bar{\gamma}_{n,k}-\bar{\lambda}_{k}| \to0\quad \text{as }n \to \infty,$$

where $$K=\max\{d(T_{i}x,T_{j}x):i,j=1,\ldots,n\}$$. This shows that $$Wx=Sx$$ for all $$x \in C$$, as required. □

The following result follows immediately from Theorems 3.1 and 3.2.

### Theorem 3.3

Let C be a closed convex subset of a complete $$\operatorname{CAT}(0)$$ space X, $$\mathfrak{T}=\{T_{n}\}$$ a family of nonexpansive mappings on C such that $$\mathfrak{F}(\mathfrak{T}) \ne\emptyset$$, and f a φ-weak contraction on C, where φ is strictly increasing. Let $$\{\alpha_{n}\} \subset(0,1]$$ and $$\{\gamma_{n,k}:k \le n\} \subset (0,1)$$ be two sequences such that $$\{\alpha_{n}\}$$ satisfies (C1)-(C3) and $$\{\gamma_{n,k}:k \le n\}$$ satisfies (D1)-(D3). Let $$x_{1} \in C$$ and define the sequence $$\{x_{n}\}$$ by

$$x_{n+1}=\alpha_{n}f(x_{n}) \oplus(1- \alpha_{n})\bigoplus_{k=1}^{n} \gamma_{n,k}T_{k}x_{n}.$$

Then $$\{x_{n}\}$$ converges strongly to a point $$\hat{x} \in C$$ such that $$\hat{x}=P_{\mathfrak{F}(\mathfrak{T})}f(\hat{x})$$.

### Proof

For each $$n \in\mathbb{N}$$, let $$S_{n}:C \to C$$ be the mapping defined by

$$S_{n}x=\bigoplus_{k=1}^{n} \gamma_{n,k}T_{k}x.$$

Then by Theorem 3.2, $$\{S_{n}\}$$ is a family of nonexpansive mappings satisfying the AKTT condition and $$\bigcap_{n=1}^{\infty}\mathfrak{F}(S_{n})=\bigcap_{n=1}^{\infty}\mathfrak {F}(T_{n})$$. We can write

$$x_{n+1}=\alpha_{n}f(x_{n}) \oplus(1- \alpha_{n})S_{n}x_{n}.$$

Define the mapping $$S:C \to C$$ by $$Sx=\lim_{n \to\infty}S_{n}x$$ for $$x \in C$$, so that S is nonexpansive and $$\mathfrak{F}(S)=\bigcap_{n=1}^{\infty}\mathfrak{F}(S_{n})$$. Consequently, Theorem 3.1 assures the strong convergence of $$\{ x_{n}\}$$ with limit , say, such that $$\hat{x}=P_{\mathfrak{F}(S)}f(\hat{x})$$. □

Using Theorem 3.3, we establish a strong convergence theorem by a Moudafi type of shrinking projection method for a family of quasinonexpansive mappings as follows.

### Theorem 3.4

Let C be a closed convex subset of a complete $$\operatorname{CAT}(0)$$ space X such that $$\{z \in C: d(u, z) \leq d(v,z)\}$$ is a convex subset of C for every $$u, v \in C$$. Let $$\mathfrak{T}=\{T_{j}:j=1,\ldots,N\}$$ be a finite family of quasinonexpansive mappings of C into X with $$\mathfrak{F}(\mathfrak{T}) \ne\emptyset$$ and $$\widetilde{\mathfrak{F}}(T_{j})=\mathfrak{F}(T_{j})$$ for $$j=1,\ldots,N$$, and f a φ-weak contraction on C, where φ is strictly increasing. Let $$\{\alpha_{n}\}$$, $$\{\delta_{n}\}$$ be sequences in $$(0,1]$$, and $$\{\beta^{j}_{n}:j=1,\ldots,N\}$$ and $$\{\gamma_{n,k}:k \le n\}$$ be sequences in $$(0,1)$$. Let $$x_{1} \in C$$ and define the sequence $$\{x_{n}\}$$ by

\begin{aligned}& y^{j}_{n} =\delta_{n}x_{n} \oplus(1- \delta_{n})T_{j}x_{n}, \\& C^{j}_{n} =\bigl\{ z \in C:d\bigl(y^{j}_{n},z \bigr) \le d(x_{n},z)\bigr\} ,\quad j=1,\ldots,N, \\& v^{j}_{n,k} =P_{C^{j}_{k}}x_{n}, \quad k=1, \ldots,n, j=1,\ldots,N, \\& w_{n,k} =\bigoplus_{j=1}^{N} \beta^{j}_{k}v^{j}_{n,k}, \quad k=1, \ldots,n, \\& x_{n+1} = \alpha_{n}f(x_{n}) \oplus(1- \alpha_{n})\bigoplus_{k=1}^{n} \gamma _{n,k}w_{n,k}, \end{aligned}

where $$\{\alpha_{n}\}$$ satisfies (C1)-(C3), $$\{\gamma_{n,k}:k \le n\}$$ satisfies (D1)-(D3), and $$\{\delta_{n}\}$$, $$\{\beta^{j}_{n}\}$$ satisfy the following conditions:

1. (i)

$$\liminf_{n \to\infty}\delta_{n}<1$$;

2. (ii)

$$\sum_{j=1}^{N}\beta^{j}_{n}=1$$ for $$n \in\mathbb{N}$$.

Then $$\{x_{n}\}$$ converges strongly to a point $$\hat{x} \in C$$ such that $$\hat{x}=P_{\mathfrak{F}(\mathfrak{T})}f(\hat{x})$$.

### Proof

First, we can see that every $$C^{j}_{n}$$ is closed and convex by the assumption on the space. To prove that the metric projection $$P_{C^{j}_{k}}$$ is well defined, let $$z \in\mathfrak{F}(\mathfrak{T})$$. Since $$T_{j}$$ is quasinonexpansive, we have

$$d\bigl(y^{j}_{n},z\bigr) \le\delta_{n}d(x_{n},z)+(1- \delta_{n})d(T_{j}x_{n},z) \le d(x_{n},z),$$

and so $$z \in C^{j}_{n}$$. This implies that

$$\emptyset\ne\mathfrak{F}(\mathfrak{T}) \subset C^{j}_{n}, \quad j=1,\ldots,N, n \in \mathbb{N}.$$

Thus, the metric projection onto $$C^{j}_{n}$$ is well defined. For $$n \in\mathbb{N}$$, define $$Q_{n}:C \to C$$ by

$$Q_{n}x=\bigoplus_{j=1}^{N} \beta^{j}_{n}P_{C^{j}_{n}}x, \quad x \in C.$$

It follows from Lemma 2.4 and condition (ii) that $$Q_{n}$$ is nonexpansive and $$\mathfrak{F}(Q_{n})=\bigcap_{j=1}^{N}C^{j}_{n}$$. According to our construction, we can write

\begin{aligned}& w_{n,k}=Q_{k}x_{n}, \quad k=1,\ldots,n, \\& x_{n+1}=\alpha_{n}f(x_{n}) \oplus(1- \alpha_{n})\bigoplus_{j=1}^{n} \gamma _{n,k}Q_{k}x_{n}, \quad n \in\mathbb{N}. \end{aligned}

Hence, Theorem 3.3 and conditions (C1)-(C3) and (D1)-(D3) assure the strong convergence of $$\{x_{n}\}$$ to a point $$\hat{x} \in C$$ such that $$\hat{x}=P_{F}f(\hat{x})$$, where

$$F=\bigcap_{n=1}^{\infty}\mathfrak{F}(Q_{n})= \bigcap_{n=1}^{\infty}\bigcap _{j=1}^{N}C^{j}_{n}=\bigcap _{j=1}^{N}\bigcap _{n=1}^{\infty}C^{j}_{n}.$$

Notice that $$\mathfrak{F}(\mathfrak{T}) \subset F$$. Condition (i) asserts that there exists a convergent subsequence $$\{ \delta_{n_{i}}\}$$ of $$\{\delta_{n}\}$$ such that $$\lim_{i \to\infty}\delta_{n_{i}}<1$$. Since $$\hat{x} \in C^{j}_{n}$$ for all $$j=1,\ldots,N$$ and $$n \in\mathbb{N}$$, we obtain

\begin{aligned} d(x_{n_{i}},\hat{x}) & \ge d(y_{n_{i}},\hat{x}) \\ & =d\bigl(\delta_{n_{i}}x_{n_{i}} \oplus(1-\delta_{n_{i}})T_{j}x_{n_{i}}, \hat{x}\bigr) \\ & \ge d\bigl(x_{n_{i}},\delta_{n_{i}}x_{n_{i}} \oplus(1- \delta _{n_{i}})T_{j}x_{n_{i}}\bigr)-d(x_{n_{i}}, \hat{x}) \\ & =(1-\delta_{n_{i}})d(x_{n_{i}},T_{j}x_{n_{i}})-d(x_{n_{i}}, \hat{x}), \end{aligned}

which yields

$$\frac{2}{1-\delta_{n_{i}}}d(x_{n_{i}},\hat{x}) \ge d(x_{n_{i}},T_{j}x_{n_{i}}).$$

We then take the limit as $$i \to\infty$$ and get

$$\lim_{i \to\infty}d(x_{n_{i}},T_{j}x_{n_{i}})=0, \quad j=1,\ldots,N.$$

This shows that $$\hat{x} \in\widetilde{\mathfrak{F}}(T_{j})=\mathfrak {F}(T_{j})$$ for $$j=1,\ldots,N$$, that is, $$\hat{x} \in\mathfrak{F}(\mathfrak{T})$$. Since $$\mathfrak{F}(\mathfrak{T}) \subset F$$, we then have $$\hat{x}=P_{F}f(\hat{x})=P_{\mathfrak{F}(\mathfrak{T})}f(\hat{x})$$, which completes the proof. □

Consequently, when f is constant in Theorem 3.4, we obtain the following strong convergence theorem by a new Halpern type of shrinking projection method.

### Theorem 3.5

Let X, C, $$\mathfrak{T}=\{T_{j}:j=1, \ldots,N\}$$, and the sequences $$\{\alpha_{n}\}$$, $$\{\delta_{n}\}$$, $$\{\beta^{j}_{n}:j=1,\ldots ,N\}$$, $$\{\gamma_{n,k}:k \le n\}$$ be as in Theorem  3.4. Let $$u,x_{1} \in C$$ and define the sequence $$\{x_{n}\}$$ by

\begin{aligned}& y^{j}_{n} =\delta_{n}x_{n} \oplus(1- \delta_{n})T_{j}x_{n}, \\& C^{j}_{n} =\bigl\{ z \in C:d\bigl(y^{j}_{n},z \bigr) \le d(x_{n},z)\bigr\} , \quad j=1,\ldots,N, \\& v^{j}_{n,k} =P_{C^{j}_{k}}x_{n}, \quad k=1, \ldots,n, j=1,\ldots,N, \\& w_{n,k} =\bigoplus_{j=1}^{N} \beta^{j}_{k}v^{j}_{n,k},\quad k=1, \ldots,n, \\& x_{n+1} = \alpha_{n}u \oplus(1-\alpha_{n})\bigoplus _{k=1}^{n}\gamma_{n,k}w_{n,k}. \end{aligned}

Then $$\{x_{n}\}$$ converges strongly to the point $$P_{\mathfrak{F}(\mathfrak{T})}u$$.

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## Acknowledgements

The first author is supported by a grant MOST 104-2115-M-259-004 from the Ministry of Science and Technology of Taiwan and therefore thanks the MOST financial support. The second author is supported by JSPS KAKENHI Grant Number 15K05007 from Japan Society for the Promotion of Science. The authors would like to express the most sincere thanks to Professor Qamrul Hasan Ansari, Editor of Fixed Point Theory and Applications, and the anonymous referees for their careful reading of the manuscript and for giving insightful comments and the citation of .

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Both authors contributed equally to the writing of this paper. Both authors read and approved the final manuscript.

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