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# Viscosity approximation methods for nonexpansive semigroups in $CAT\left(0\right)$ spaces

Fixed Point Theory and Applications20132013:160

https://doi.org/10.1186/1687-1812-2013-160

• Accepted: 14 May 2013
• Published:

## Abstract

In this paper, we study the strong convergence of Moudafi’s viscosity approximation methods for approximating a common fixed point of a one-parameter continuous semigroup of nonexpansive mappings in $CAT\left(0\right)$ spaces. We prove that the proposed iterative scheme converges strongly to a common fixed point of a one-parameter continuous semigroup of nonexpansive mappings which is also a unique solution of the variational inequality. The results presented in this paper extend and enrich the existing literature.

## Keywords

• viscosity approximation method
• nonexpansive semigroup
• variational inequality
• $CAT\left(0\right)$ space
• common fixed point

## 1 Introduction

Let $\left(X,d\right)$ be a metric space. A geodesic path joining $x\in X$ to $y\in X$ (or, more briefly, a geodesic from x to y) is a map c from a closed interval $\left[0,l\right]\subset \mathbb{R}$ to X such that $c\left(0\right)=x$, $c\left(l\right)=y$, and $d\left(c\left(t\right),c\left({t}^{\prime }\right)\right)=|t-{t}^{\prime }|$ for all $t,{t}^{\prime }\in \left[0,l\right]$. In particular, c is an isometry and $d\left(x,y\right)=l$. The image α of c is called a geodesic (or metric) segment joining x and y. When it is unique, this geodesic segment is denoted by $\left[x,y\right]$. The space $\left(X,d\right)$ is said to be a geodesic space if every two points of 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 $Y\subseteq X$ is said to be convex if Y includes every geodesic segment joining any two of its points. A geodesic triangle $\mathrm{△}\left({x}_{1},{x}_{2},{x}_{3}\right)$ in a geodesic metric space $\left(X,d\right)$ consists of three points ${x}_{1}$, ${x}_{2}$, ${x}_{3}$ in X (the vertices of ) and a geodesic segment between each pair of vertices (the edges of ). A comparison triangle for the geodesic triangle $\mathrm{△}\left({x}_{1},{x}_{2},{x}_{3}\right)$ in $\left(X,d\right)$ is a triangle $\overline{\mathrm{△}}\left({x}_{1},{x}_{2},{x}_{3}\right):=\mathrm{△}\left({\overline{x}}_{1},{\overline{x}}_{2},{\overline{x}}_{3}\right)$ in the Euclidean plane ${\mathbb{E}}^{2}$ such that ${d}_{{\mathbb{E}}_{2}}\left({\overline{x}}_{i},{\overline{x}}_{j}\right)=d\left({x}_{i},{x}_{j}\right)$ for all $i,j\in \left\{1,2,3\right\}$.

A geodesic space is said to be a $CAT\left(0\right)$ space if all geodesic triangles of appropriate size satisfy the following comparison axiom.

$CAT\left(0\right)$: Let be a geodesic triangle in X and let $\overline{\mathrm{△}}$ be a comparison triangle for . Then is said to satisfy the $CAT\left(0\right)$ inequality if for all $x,y\in \mathrm{△}$ and all comparison points $\overline{x},\overline{y}\in \overline{\mathrm{△}}$,
$d\left(x,y\right)\le {d}_{{\mathbb{E}}^{2}}\left(\overline{x},\overline{y}\right).$
If x, ${y}_{1}$, ${y}_{2}$ are points in a $CAT\left(0\right)$ space and if ${y}_{0}$ is the midpoint of the segment $\left[{y}_{1},{y}_{2}\right]$, then the $CAT\left(0\right)$ inequality implies
${d}^{2}\left(x,{y}_{0}\right)\le \frac{1}{2}{d}^{2}\left(x,{y}_{1}\right)+\frac{1}{2}{d}^{2}\left(x,{y}_{2}\right)-\frac{1}{4}{d}^{2}\left({y}_{1},{y}_{2}\right).$
(1.1)

This is the (CN)-inequality of Bruhat and Tits [1]. In fact (cf. [2], p.163), a geodesic space is a $CAT\left(0\right)$ space if and only if it satisfies the (CN)-inequality.

It is well known that any complete, simply connected Riemannian manifold having nonpositive sectional curvature is a $CAT\left(0\right)$ space. Other examples include pre-Hilbert spaces, -trees (see [2]), Euclidean buildings (see [3]), the complex Hilbert ball with a hyperbolic metric (see [4]), and many others. Complete $CAT\left(0\right)$ spaces are often called Hadamard spaces.

It is proved in [2] that a normed linear space satisfies the (CN)-inequality if and only if it satisfies the parallelogram identity, i.e., is a pre-Hilbert space; hence it is not so unusual to have an inner product-like notion in Hadamard spaces. Berg and Nikolaev [5] introduced the concept of quasilinearization as follows.

Let us formally denote a pair $\left(a,b\right)\in X×X$ by $\stackrel{\to }{ab}$ and call it a vector. Then quasilinearization is defined as a map $〈\cdot ,\cdot 〉:\left(X×X\right)×\left(X×X\right)\to \mathbb{R}$ defined by
$〈\stackrel{\to }{ab},\stackrel{\to }{cd}〉=\frac{1}{2}\left({d}^{2}\left(a,d\right)+{d}^{2}\left(b,c\right)-{d}^{2}\left(a,c\right)-{d}^{2}\left(b,d\right)\right)\phantom{\rule{1em}{0ex}}\left(a,b,c,d\in X\right).$
(1.2)
It is easily seen that $〈\stackrel{\to }{ab},\stackrel{\to }{cd}〉=〈\stackrel{\to }{cd},\stackrel{\to }{ab}〉$, $〈\stackrel{\to }{ab},\stackrel{\to }{cd}〉=-〈\stackrel{\to }{ba},\stackrel{\to }{cd}〉$ and $〈\stackrel{\to }{ax},\stackrel{\to }{cd}〉+〈\stackrel{\to }{xb},\stackrel{\to }{cd}〉=〈\stackrel{\to }{ab},\stackrel{\to }{cd}〉$ for all $a,b,c,d,x\in X$. We say that X satisfies the Cauchy-Schwarz inequality if
$〈\stackrel{\to }{ab},\stackrel{\to }{cd}〉\le d\left(a,b\right)d\left(c,d\right)$
(1.3)

for all $a,b,c,d\in X$. It is known [[5], Corollary 3] that a geodesically connected metric space is a $CAT\left(0\right)$ space if and only if it satisfies the Cauchy-Schwarz inequality.

In 2010, Kakavandi and Amini [6] introduced the concept of a dual space for $CAT\left(0\right)$ spaces as follows. Consider the map $\mathrm{\Theta }:\mathbb{R}×X×X\to C\left(X\right)$ defined by
$\mathrm{\Theta }\left(t,a,b\right)\left(x\right)=t〈\stackrel{\to }{ab},\stackrel{\to }{ax}〉,$
(1.4)
where $C\left(X\right)$ is the space of all continuous real-valued functions on X. Then the Cauchy-Schwarz inequality implies that $\mathrm{\Theta }\left(t,a,b\right)$ is a Lipschitz function with a Lipschitz semi-norm $L\left(\mathrm{\Theta }\left(t,a,b\right)\right)=|t|d\left(a,b\right)$ for all $t\in \mathbb{R}$ and $a,b\in X$, where
$L\left(f\right)=sup\left\{\frac{f\left(x\right)-f\left(y\right)}{d\left(x,y\right)}:x,y\in X,x\ne y\right\}$
is the Lipschitz semi-norm of the function $f:X\to \mathbb{R}$. Now, define the pseudometric D on $\mathbb{R}×X×X$ by
$D\left(\left(t,a,b\right),\left(s,c,d\right)\right)=L\left(\mathrm{\Theta }\left(t,a,b\right)-\mathrm{\Theta }\left(s,c,d\right)\right).$

Lemma 1.1 [[6], Lemma 2.1]

$D\left(\left(t,a,b\right),\left(s,c,d\right)\right)=0$ if and only if $t〈\stackrel{\to }{ab},\stackrel{\to }{xy}〉=s〈\stackrel{\to }{cd},\stackrel{\to }{xy}〉$ for all $x,y\in X$.

For a complete $CAT\left(0\right)$ space $\left(X,d\right)$, the pseudometric space $\left(\mathbb{R}×X×X,D\right)$ can be considered as a subspace of the pseudometric space $\left(Lip\left(X,R\right),L\right)$ of all real-valued Lipschitz functions. Also, D defines an equivalence relation on $\mathbb{R}×X×X$, where the equivalence class of $t\stackrel{\to }{ab}:=\left(t,a,b\right)$ is
$\left[t\stackrel{\to }{ab}\right]=\left\{s\stackrel{\to }{cd}:t〈\stackrel{\to }{ab},\stackrel{\to }{xy}〉=s〈\stackrel{\to }{cd},\stackrel{\to }{xy}〉\phantom{\rule{0.25em}{0ex}}\mathrm{\forall }x,y\in X\right\}.$

The set ${X}^{\ast }:=\left\{\left[t\stackrel{\to }{ab}\right]:\left(t,a,b\right)\in \mathbb{R}×X×X\right\}$ is a metric space with metric D, which is called the dual metric space of $\left(X,d\right)$.

Recently, Dehghan and Rooin [7] introduced the duality mapping in $CAT\left(0\right)$ spaces and studied its relation with subdifferential, by using the concept of quasilinearization. Then they presented a characterization of metric projection in $CAT\left(0\right)$ spaces as follows.

Theorem 1.2 [[7], Theorem 2.4]

Let C be a nonempty convex subset of a complete $CAT\left(0\right)$ space X, $x\in X$ and $u\in C$. Then
$u={P}_{C}x\phantom{\rule{1em}{0ex}}\mathit{\text{if and only if}}\phantom{\rule{1em}{0ex}}〈\stackrel{\to }{yu},\stackrel{\to }{ux}〉\ge 0\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}y\in C.$
From now on, let be the set of positive integers, let be the set of real numbers, and let ${\mathbb{R}}^{+}$ be the set of nonnegative real numbers. Let C be a nonempty, closed and convex subset of a complete $CAT\left(0\right)$ space X. A family $\mathcal{S}:=\left\{T\left(t\right):t\in {\mathbb{R}}^{+}\right\}$ of self-mappings of C is called a one-parameter continuous semigroup of nonexpansive mappings if the following conditions hold:
1. (i)
for each $t\in {\mathbb{R}}^{+}$, $T\left(t\right)$ is a nonexpansive mapping on C, i.e.,
$d\left(T\left(t\right)x,T\left(t\right)y\right)\le d\left(x,y\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C;$

2. (ii)

$T\left(s+t\right)=T\left(t\right)\circ T\left(s\right)$ for all $t,s\in {\mathbb{R}}^{+}$;

3. (iii)

for each $x\in X$, the mapping $T\left(\cdot \right)x$ from ${\mathbb{R}}^{+}$ into C is continuous.

A family $\mathcal{S}:=\left\{T\left(t\right):t\in {\mathbb{R}}^{+}\right\}$ of mappings is called a one-parameter strongly continuous semigroup of nonexpansive mappings if conditions (i), (ii) and (iii) and the following condition are satisfied:
1. (iv)

$T\left(0\right)x=x$ for all $x\in C$.

We shall denote by the common fixed point set of $\mathcal{S}$, that is,
$\mathcal{F}:=F\left(\mathcal{S}\right)=\left\{x\in C:T\left(t\right)x=x,t\in {\mathbb{R}}^{+}\right\}=\bigcap _{t\in {\mathbb{R}}^{+}}F\left(T\left(t\right)\right).$
One classical way to study nonexpansive mappings is to use contractions to approximate nonexpansive mappings. More precisely, take $t\in \left(0,1\right)$ and define a contraction ${T}_{t}:C\to C$ by
${T}_{t}=tu+\left(1-t\right)Tx,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in C,$

where $u\in C$ is an arbitrary fixed element. Banach’s contraction mapping principle guarantees that ${T}_{t}$ has a unique fixed point ${x}_{t}$ in C. It is unclear, in general, what the behavior of ${x}_{t}$ is as $t\to 0$, even if T has a fixed point. However, in the case of T having a fixed point, Browder [8] proved that ${x}_{t}$ converges strongly to a fixed point of T that is nearest to u in the framework of Hilbert spaces. Reich [9] extended Browder’s result to the setting of Banach spaces and proved, in a uniformly smooth Banach space, that ${x}_{t}$ converges strongly to a fixed point of T and the limit defines the (unique) sunny nonexpansive retraction from C onto $F\left(T\right)$.

Halpern [10] introduced the following explicit iterative scheme (1.5) for a nonexpansive mapping T on a subset C of a Hilbert space by taking any points $u,{x}_{1}\in C$ and defined the iterative sequence $\left\{{x}_{n}\right\}$ by
${x}_{n+1}={\alpha }_{n}u+\left(1-{\alpha }_{n}\right)T{x}_{n}.$
(1.5)

He proved that the sequence $\left\{{x}_{n}\right\}$ generated by (1.5) converges to a fixed point of T.

It is an interesting problem to extend the above (Browder’s [8] and Halpern’s [10]) results to the nonexpansive semigroup case. In [11], Shioji and Takahashi introduced the following implicit iteration in a Hilbert space:
${x}_{n}={\alpha }_{n}u+\left(1-{\alpha }_{n}\right)\frac{1}{{t}_{n}}{\int }_{0}^{{t}_{n}}T\left(s\right){x}_{n}\phantom{\rule{0.2em}{0ex}}ds,$
(1.6)

where C is a nonempty closed convex subset of a real Hilbert space H, $u\in C$, $\left\{{\alpha }_{n}\right\}$ is a sequence in $\left(0,1\right)$, $\left\{{t}_{n}\right\}$ is a sequence of positive real numbers divergent to ∞. Under suitable conditions, they proved strong convergence of $\left\{{x}_{n}\right\}$ to a member of .

Later, Suzuki [12] was the first to introduce in a Hilbert space the following iteration process:
${x}_{n}={\alpha }_{n}u+\left(1-{\alpha }_{n}\right)T\left({t}_{n}\right){x}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,$
(1.7)
where $\left\{T\left(t\right):t\ge 0\right\}$ is a strongly continuous semigroup of nonexpansive mappings on C such that $\mathcal{F}\ne \mathrm{\varnothing }$ and $\left\{{\alpha }_{n}\right\}$ and $\left\{{t}_{n}\right\}$ are appropriate sequences of real numbers. He proved that $\left\{{x}_{n}\right\}$ generated by (1.7) converges strongly to the element of nearest to u. Using Moudafi’s viscosity approximation methods, Song and Xu [13] introduced the following iteration process:
${x}_{n}={\alpha }_{n}f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right)T\left({t}_{n}\right){x}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,$
(1.8)
and
${x}_{n+1}={\alpha }_{n}f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right)T\left({t}_{n}\right){x}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1.$
(1.9)

They proved that $\left\{{x}_{n}\right\}$ converges to the same point of in a reflexive strictly Banach space with a uniformly Gâteaux differentiable norm.

In the similar way, Dhompongsa et al. [14] extended Browder’s iteration to a strongly continuous semigroup of nonexpansive mappings $\left\{T\left(t\right):t\ge 0\right\}$ in a complete $CAT\left(0\right)$ space X as follows:
${x}_{n}={\alpha }_{n}{x}_{0}\oplus T\left({t}_{n}\right){x}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,$

where C is a nonempty closed convex subset of a complete $CAT\left(0\right)$ space X, ${x}_{0}\in C$, $\left\{{\alpha }_{n}\right\}$ and $\left\{{t}_{n}\right\}$ are sequences of real numbers satisfying $0<{\alpha }_{n}<1$, ${t}_{n}>0$, and ${lim}_{n\to \mathrm{\infty }}{t}_{n}={lim}_{n\to \mathrm{\infty }}{\alpha }_{n}/{t}_{n}=0$. The proved that $\mathcal{F}\ne \mathrm{\varnothing }$ and $\left\{{x}_{n}\right\}$ converges to the element of nearest to u. For other related results, see [15, 16].

In 2012, Shi and Chen [17], studied the convergence theorems of the following Moudafi’s viscosity iterations for a nonexpansive mapping T: for a contraction f on C and $t\in \left(0,1\right)$, let ${x}_{t}\in C$ be a unique fixed point of the contraction $x↦tf\left(x\right)\oplus \left(1-t\right)Tx$; i.e.,
${x}_{t}=tf\left({x}_{t}\right)\oplus \left(1-t\right)T{x}_{t},$
(1.10)
and ${x}_{0}\in C$ is arbitrarily chosen and
${x}_{n+1}={\alpha }_{n}f\left({x}_{n}\right)\oplus \left(1-{\alpha }_{n}\right)T{x}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 0,$
(1.11)
where $\left\{{\alpha }_{n}\right\}\subset \left(0,1\right)$. They proved $\left\{{x}_{t}\right\}$ defined by (1.10) converges strongly as $t\to 0$ to $\stackrel{˜}{x}\in F\left(T\right)$ such that $\stackrel{˜}{x}={P}_{F\left(T\right)}f\left(\stackrel{˜}{x}\right)$ in the framework of $CAT\left(0\right)$ space satisfying property $\mathcal{P}$, i.e., if for $x,u,{y}_{1},{y}_{2}\in X$,
$d\left(x,{P}_{\left[x,{y}_{1}\right]}u\right)d\left(x,{y}_{1}\right)\le d\left(x,{P}_{\left[x,{y}_{2}\right]}u\right)d\left(x,{y}_{2}\right)+d\left(x,u\right)d\left({y}_{1},{y}_{2}\right).$

Furthermore, they also obtained that $\left\{{x}_{n}\right\}$ defined by (1.11) converges strongly as $n\to \mathrm{\infty }$ to $\stackrel{˜}{x}\in F\left(T\right)$ under certain appropriate conditions imposed on $\left\{{\alpha }_{n}\right\}$.

By using the concept of quasilinearization, Wangkeeree and Preechasilp [18] improved Shi and Chen’s results. In fact, they proved the strong convergence theorems for two given iterative schemes (1.10) and (1.11) in a complete $CAT\left(0\right)$ space without the property $\mathcal{P}$.

Motivated and inspired by Song and Xu [13], Dhompongsa et al. [14], and Wangkeeree and Preechasilp [18], in this paper we aim to study the strong convergence theorems of Moudafi’s viscosity approximation methods for a one-parameter continuous semigroup of nonexpansive mappings $\mathcal{S}:=\left\{T\left(t\right):t\in {\mathbb{R}}^{+}\right\}$ in $CAT\left(0\right)$ spaces. Let C be a nonempty, closed and convex subset of a $CAT\left(0\right)$ space X. For a given contraction f on C and ${\alpha }_{n}\in \left(0,1\right)$, let ${x}_{n}\in C$ be a unique fixed point of the contraction $x↦{\alpha }_{n}f\left(x\right)\oplus \left(1-{\alpha }_{n}\right)T\left({t}_{n}\right)x$; i.e.,
${x}_{n}={\alpha }_{n}f\left({x}_{n}\right)\oplus \left(1-{\alpha }_{n}\right)T\left({t}_{n}\right){x}_{n},\phantom{\rule{1em}{0ex}}n\ge 0,$
(1.12)
and
${x}_{n+1}={\alpha }_{n}f\left({x}_{n}\right)\oplus \left(1-{\alpha }_{n}\right)T\left({t}_{n}\right){x}_{n},\phantom{\rule{1em}{0ex}}n\ge 0.$
(1.13)
We prove that the iterative schemes $\left\{{x}_{n}\right\}$ defined by (1.12) and $\left\{{x}_{n}\right\}$ defined by (1.13) converge strongly to the same point $\stackrel{˜}{x}$ such that $\stackrel{˜}{x}={P}_{\mathcal{F}}f\left(\stackrel{˜}{x}\right)$, which is the unique solution of the variational inequality
$〈\stackrel{\to }{\stackrel{˜}{x}f\stackrel{˜}{x}},\stackrel{\to }{x\stackrel{˜}{x}}〉\ge 0,\phantom{\rule{1em}{0ex}}x\in \mathcal{F},$
where is the common fixed point set of $\mathcal{S}$, that is,
$\mathcal{F}:=F\left(\mathcal{S}\right)=\left\{x\in C:T\left(t\right)x=x,t\in {\mathbb{R}}^{+}\right\}=\bigcap _{t\in {\mathbb{R}}^{+}}F\left(T\left(t\right)\right).$

## 2 Preliminaries

In this paper, we write $\left(1-t\right)x\oplus ty$ for the unique point z in the geodesic segment joining from x to y such that
$d\left(z,x\right)=td\left(x,y\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}d\left(z,y\right)=\left(1-t\right)d\left(x,y\right).$

We also denote by $\left[x,y\right]$ the geodesic segment joining from x to y, that is, $\left[x,y\right]=\left\{\left(1-t\right)x\oplus ty:t\in \left[0,1\right]\right\}$. A subset C of a $CAT\left(0\right)$ space is convex if $\left[x,y\right]\subseteq C$ for all $x,y\in C$.

The following lemmas play an important role in our paper.

Lemma 2.1 [[2], Proposition 2.2]

Let X be a $CAT\left(0\right)$ space, $p,q,r,s\in X$ and $\lambda \in \left[0,1\right]$. Then
$d\left(\lambda p\oplus \left(1-\lambda \right)q,\lambda r\oplus \left(1-\lambda \right)s\right)\le \lambda d\left(p,r\right)+\left(1-\lambda \right)d\left(q,s\right).$

Lemma 2.2 [[19], Lemma 2.4]

Let X be a $CAT\left(0\right)$ space, $x,y,z\in X$ and $\lambda \in \left[0,1\right]$. Then
$d\left(\lambda x\oplus \left(1-\lambda \right)y,z\right)\le \lambda d\left(x,z\right)+\left(1-\lambda \right)d\left(y,z\right).$

Lemma 2.3 [[19], Lemma 2.5]

Let X be a $CAT\left(0\right)$ space, $x,y,z\in X$ and $\lambda \in \left[0,1\right]$. Then
${d}^{2}\left(\lambda x\oplus \left(1-\lambda \right)y,z\right)\le \lambda {d}^{2}\left(x,z\right)+\left(1-\lambda \right){d}^{2}\left(y,z\right)-\lambda \left(1-\lambda \right){d}^{2}\left(x,y\right).$

The concept of Δ-convergence introduced by Lim [20] in 1976 was shown by Kirk and Panyanak [21] in $CAT\left(0\right)$ spaces to be very similar to the weak convergence in Banach space setting. Next, we give the concept of Δ-convergence and collect some basic properties.

Let $\left\{{x}_{n}\right\}$ be a bounded sequence in a $CAT\left(0\right)$ space X. For $x\in X$, we set
$r\left(x,\left\{{x}_{n}\right\}\right)=\underset{n\to \mathrm{\infty }}{lim sup}d\left(x,{x}_{n}\right).$
The asymptotic radius $r\left(\left\{{x}_{n}\right\}\right)$ of $\left\{{x}_{n}\right\}$ is given by
$r\left(\left\{{x}_{n}\right\}\right)=inf\left\{r\left(x,\left\{{x}_{n}\right\}\right):x\in X\right\},$
and the asymptotic center $A\left(\left\{{x}_{n}\right\}\right)$ of $\left\{{x}_{n}\right\}$ is the set
$A\left(\left\{{x}_{n}\right\}\right)=\left\{x\in X:r\left(x,\left\{{x}_{n}\right\}\right)=r\left(\left\{{x}_{n}\right\}\right)\right\}.$
It is known from Proposition 7 of [22] that in a complete $CAT\left(0\right)$ space, $A\left(\left\{{x}_{n}\right\}\right)$ consists of exactly one point. A sequence $\left\{{x}_{n}\right\}\subset X$ is said to Δ-converge to $x\in X$ if $A\left(\left\{{x}_{{n}_{k}}\right\}\right)=\left\{x\right\}$ for every subsequence $\left\{{x}_{{n}_{k}}\right\}$ of $\left\{{x}_{n}\right\}$. The uniqueness of an asymptotic center implies that a $CAT\left(0\right)$ space X satisfies Opial’s property, i.e., for given $\left\{{x}_{n}\right\}\subset X$ such that $\left\{{x}_{n}\right\}\mathrm{\Delta }$-converges to x and given $y\in X$ with $y\ne x$,
$\underset{n\to \mathrm{\infty }}{lim sup}d\left({x}_{n},x\right)<\underset{n\to \mathrm{\infty }}{lim sup}d\left({x}_{n},y\right).$

Since it is not possible to formulate the concept of demiclosedness in a $CAT\left(0\right)$ setting, as stated in linear spaces, let us formally say that ‘$I-T$ is demiclosed at zero’ if the conditions $\left\{{x}_{n}\right\}\subseteq C$ Δ-converges to x and $d\left({x}_{n},T{x}_{n}\right)\to 0$ imply $x\in F\left(T\right)$.

Lemma 2.4 [21]

Every bounded sequence in a complete $CAT\left(0\right)$ space always has a Δ-convergent subsequence.

Lemma 2.5 [23]

If C is a closed convex subset of a complete $CAT\left(0\right)$ space and if $\left\{{x}_{n}\right\}$ is a bounded sequence in C, then the asymptotic center of $\left\{{x}_{n}\right\}$ is in C.

Lemma 2.6 [23]

If C is a closed convex subset of X and $T:C\to X$ is a nonexpansive mapping, then the conditions $\left\{{x}_{n}\right\}$ Δ-converges to x and $d\left({x}_{n},T{x}_{n}\right)\to 0$ imply $x\in C$ and $Tx=x$.

Having the notion of quasilinearization, Kakavandi and Amini [6] introduced the following notion of convergence.

A sequence $\left\{{x}_{n}\right\}$ in the complete $CAT\left(0\right)$ space $\left(X,d\right)$ w-converges to $x\in X$ if
$\underset{n\to \mathrm{\infty }}{lim}〈\stackrel{\to }{x{x}_{n}},\stackrel{\to }{xy}〉=0,$

i.e., ${lim}_{n\to \mathrm{\infty }}\left({d}^{2}\left({x}_{n},x\right)-{d}^{2}\left({x}_{n},y\right)+{d}^{2}\left(x,y\right)\right)=0$ for all $y\in X$.

It is obvious that convergence in the metric implies w-convergence, and it is easy to check that w-convergence implies Δ-convergence [[6], Proposition 2.5], but it is showed in [[24], Example 4.7] that the converse is not valid. However, the following lemma shows another characterization of Δ-convergence as well as, more explicitly, a relation between w-convergence and Δ-convergence.

Lemma 2.7 [[24], Theorem 2.6]

Let X be a complete $CAT\left(0\right)$ space, $\left\{{x}_{n}\right\}$ be a sequence in X and $x\in X$. Then $\left\{{x}_{n}\right\}\mathrm{\Delta }$-converges to x if and only if ${lim sup}_{n\to \mathrm{\infty }}〈\stackrel{\to }{x{x}_{n}},\stackrel{\to }{xy}〉\le 0$ for all $y\in X$.

Lemma 2.8 [[25], Lemma 2.1]

Let $\left\{{a}_{n}\right\}$ be a sequence of non-negative real numbers satisfying the property
${a}_{n+1}\le \left(1-{\alpha }_{n}\right){a}_{n}+{\alpha }_{n}{\beta }_{n},\phantom{\rule{1em}{0ex}}n\ge 0,$
where $\left\{{\alpha }_{n}\right\}\subseteq \left(0,1\right)$ and $\left\{{\beta }_{n}\right\}\subseteq \mathbb{R}$ such that
1. (i)

${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$;

2. (ii)

${lim sup}_{n\to \mathrm{\infty }}{\beta }_{n}\le 0$ or ${\sum }_{n=0}^{\mathrm{\infty }}|{\alpha }_{n}{\beta }_{n}|<\mathrm{\infty }$.

Then $\left\{{a}_{n}\right\}$ converges to zero as $n\to \mathrm{\infty }$.

## 3 Viscosity approximation methods

In this section, we present the strong convergence theorems of Moudafi’s viscosity approximation methods for a one-parameter continuous semigroup of nonexpansive mappings $\mathcal{S}:=\left\{T\left(t\right):t\in {\mathbb{R}}^{+}\right\}$ in $CAT\left(0\right)$ spaces. Before proving main results, we need the following two vital lemmas.

Lemma 3.1 Let X be a complete $CAT\left(0\right)$ space. Then, for all $u,x,y\in X$, the following inequality holds:
${d}^{2}\left(x,u\right)\le {d}^{2}\left(y,u\right)+2〈\stackrel{\to }{xy},\stackrel{\to }{xu}〉.$
Proof Using (1.2), we have that
$\begin{array}{rcl}{d}^{2}\left(y,u\right)-{d}^{2}\left(x,u\right)-2〈\stackrel{\to }{yx},\stackrel{\to }{xu}〉& =& {d}^{2}\left(y,u\right)-{d}^{2}\left(x,u\right)-2〈\stackrel{\to }{yu},\stackrel{\to }{xu}〉-2〈\stackrel{\to }{ux},\stackrel{\to }{xu}〉\\ =& {d}^{2}\left(y,u\right)-{d}^{2}\left(x,u\right)-2〈\stackrel{\to }{yu},\stackrel{\to }{xu}〉+2{d}^{2}\left(x,u\right)\\ =& {d}^{2}\left(y,u\right)+{d}^{2}\left(x,u\right)-2〈\stackrel{\to }{yu},\stackrel{\to }{xu}〉\\ \ge & {d}^{2}\left(y,u\right)+{d}^{2}\left(x,u\right)-2d\left(y,u\right)d\left(x,u\right)\\ =& {\left({d}^{2}\left(y,u\right)-{d}^{2}\left(x,u\right)\right)}^{2}\ge 0.\end{array}$
Therefore we obtain that
${d}^{2}\left(x,u\right)\le {d}^{2}\left(y,u\right)+2〈\stackrel{\to }{xy},\stackrel{\to }{xu}〉,$

which is the desired result. □

Lemma 3.2 Let X be a $CAT\left(0\right)$ space. For any $t\in \left[0,1\right]$ and $u,v\in X$, let ${u}_{t}=tu\oplus \left(1-t\right)v$. Then, for all $x,y\in X$,
1. (i)

$〈\stackrel{\to }{{u}_{t}x},\stackrel{\to }{{u}_{t}y}〉\le t〈\stackrel{\to }{ux},\stackrel{\to }{{u}_{t}y}〉+\left(1-t\right)〈\stackrel{\to }{vx},\stackrel{\to }{{u}_{t}y}〉$;

2. (ii)

$〈\stackrel{\to }{{u}_{t}x},\stackrel{\to }{uy}〉\le t〈\stackrel{\to }{ux},\stackrel{\to }{uy}〉+\left(1-t\right)〈\stackrel{\to }{vx},\stackrel{\to }{uy}〉$ and $〈\stackrel{\to }{{u}_{t}x},\stackrel{\to }{vy}〉\le t〈\stackrel{\to }{ux},\stackrel{\to }{vy}〉+\left(1-t\right)〈\stackrel{\to }{vx},\stackrel{\to }{vy}〉$.

Proof (i) It follows from (CN)-inequality (1.1) that
$\begin{array}{rcl}2〈\stackrel{\to }{{u}_{t}x},\stackrel{\to }{{u}_{t}y}〉& =& {d}^{2}\left({u}_{t},y\right)+{d}^{2}\left(x,{u}_{t}\right)-{d}^{2}\left(x,y\right)\\ \le & t{d}^{2}\left(u,y\right)+\left(1-t\right){d}^{2}\left(v,y\right)-t\left(1-t\right){d}^{2}\left(u,v\right)+{d}^{2}\left(x,{u}_{t}\right)-{d}^{2}\left(x,y\right)\\ =& t{d}^{2}\left(u,y\right)+t{d}^{2}\left(x,{u}_{t}\right)-t{d}^{2}\left(u,{u}_{t}\right)-t{d}^{2}\left(x,y\right)\\ +\left(1-t\right){d}^{2}\left(v,y\right)+\left(1-t\right){d}^{2}\left(x,{u}_{t}\right)-\left(1-t\right){d}^{2}\left(v,{u}_{t}\right)-\left(1-t\right){d}^{2}\left(x,y\right)\\ +t{d}^{2}\left(u,{u}_{t}\right)+\left(1-t\right){d}^{2}\left(v,{u}_{t}\right)-t\left(1-t\right){d}^{2}\left(u,v\right)\\ =& t\left[{d}^{2}\left(u,y\right)+{d}^{2}\left(x,{u}_{t}\right)-{d}^{2}\left(u,{u}_{t}\right)-{d}^{2}\left(x,y\right)\right]\\ +\left(1-t\right)\left[{d}^{2}\left(v,y\right)+{d}^{2}\left(x,{u}_{t}\right)-{d}^{2}\left(v,{u}_{t}\right)-{d}^{2}\left(x,y\right)\right]\\ +t{\left(1-t\right)}^{2}{d}^{2}\left(u,v\right)+\left(1-t\right){t}^{2}{d}^{2}\left(u,v\right)-t\left(1-t\right){d}^{2}\left(u,v\right)\\ =& t〈\stackrel{\to }{ux},\stackrel{\to }{{u}_{t}y}〉+\left(1-t\right)〈\stackrel{\to }{vx},\stackrel{\to }{{u}_{t}y}〉.\end{array}$

(ii) The proof is similar to (i). □

For any ${\alpha }_{n}\in \left(0,1\right)$, ${t}_{n}\in \left[0,\mathrm{\infty }\right)$ and a contraction f with coefficient $\alpha \in \left(0,1\right)$, define the mapping ${G}_{n}:C\to C$ by
${G}_{n}\left(x\right)={\alpha }_{n}f\left(x\right)\oplus \left(1-{\alpha }_{n}\right)T\left({t}_{n}\right)x,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in C.$
(3.1)
It is not hard to see that ${G}_{n}$ is a contraction on C. Indeed, for $x,y\in C$, we have
$\begin{array}{rcl}d\left({G}_{n}\left(x\right),{G}_{n}\left(y\right)\right)& =& d\left({\alpha }_{n}f\left(x\right)\oplus \left(1-{\alpha }_{n}\right)T\left({t}_{n}\right)x,{\alpha }_{n}f\left(y\right)\oplus \left(1-{\alpha }_{n}\right)T\left({t}_{n}\right)y\right)\\ \le & d\left({\alpha }_{n}f\left(x\right)\oplus \left(1-{\alpha }_{n}\right)T\left({t}_{n}\right)x,{\alpha }_{n}f\left(y\right)\oplus \left(1-{\alpha }_{n}\right)T\left({t}_{n}\right)x\right)\\ +d\left({\alpha }_{n}f\left(y\right)\oplus \left(1-{\alpha }_{n}\right)T\left({t}_{n}\right)x,{\alpha }_{n}f\left(y\right)\oplus \left(1-{\alpha }_{n}\right)T\left({t}_{n}\right)y\right)\\ \le & {\alpha }_{n}d\left(f\left(x\right),f\left(y\right)\right)+\left(1-{\alpha }_{n}\right)d\left(T\left({t}_{n}\right)x,T\left({t}_{n}\right)y\right)\\ \le & {\alpha }_{n}\alpha d\left(x,y\right)+\left(1-{\alpha }_{n}\right)d\left(x,y\right)\\ =& \left(1-{\alpha }_{n}\left(1-\alpha \right)\right)d\left(x,y\right).\end{array}$
Therefore we have that ${G}_{n}$ is a contraction mapping. Let ${x}_{n}\in C$ be the unique fixed point of ${G}_{n}$; that is,
(3.2)

Now we are in a position to state and prove our main results.

Theorem 3.3 Let C be a closed convex subset of a complete $CAT\left(0\right)$ space X, and let $\left\{T\left(t\right)\right\}$ be a one-parameter continuous semigroup of nonexpansive mappings on C satisfying $\mathcal{F}\ne \mathrm{\varnothing }$ and uniformly asymptotically regular (in short, u.a.r.) on C, that is, for all $h\ge 0$ and any bounded subset B of C,
$\underset{t\to \mathrm{\infty }}{lim}\underset{x\in B}{sup}d\left(T\left(h\right)\left(T\left(t\right)x\right),T\left(t\right)x\right)=0.$
Let f be a contraction on C with coefficient $0<\alpha <1$. Suppose that ${t}_{n}\in \left[0,\mathrm{\infty }\right)$, ${\alpha }_{n}\in \left(0,1\right)$ such that ${lim}_{n\to \mathrm{\infty }}{t}_{n}=\mathrm{\infty }$, ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$ and let $\left\{{x}_{n}\right\}$ be given by (3.2). Then $\left\{{x}_{n}\right\}$ converges strongly as $n\to \mathrm{\infty }$ to $\stackrel{˜}{x}$ such that $\stackrel{˜}{x}={P}_{\mathcal{F}}f\left(\stackrel{˜}{x}\right)$, which is equivalent to the following variational inequality:
$〈\stackrel{\to }{\stackrel{˜}{x}f\stackrel{˜}{x}},\stackrel{\to }{x\stackrel{˜}{x}}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in \mathcal{F}.$
(3.3)
Proof We first show that $\left\{{x}_{n}\right\}$ is bounded. For any $p\in \mathcal{F}$, we have that
$\begin{array}{rcl}d\left({x}_{n},p\right)& =& d\left({\alpha }_{n}f\left({x}_{n}\right)\oplus \left(1-{\alpha }_{n}\right)T\left({t}_{n}\right){x}_{n},p\right)\le {\alpha }_{n}d\left(f\left({x}_{n}\right),p\right)+\left(1-{\alpha }_{n}\right)d\left(T\left({t}_{n}\right){x}_{n},p\right)\\ \le & {\alpha }_{n}d\left(f\left({x}_{n}\right),p\right)+\left(1-{\alpha }_{n}\right)d\left({x}_{n},p\right).\end{array}$
Then
$d\left({x}_{n},p\right)\le d\left(f\left({x}_{n}\right),p\right)\le d\left(f\left({x}_{n}\right),f\left(p\right)\right)+d\left(f\left(p\right),p\right)\le \alpha d\left({x}_{n},p\right)+d\left(f\left(p\right),p\right).$
This implies that
$d\left({x}_{n},p\right)\le \frac{1}{1-\alpha }d\left(f\left(p\right),p\right).$
Hence $\left\{{x}_{n}\right\}$ is bounded, so are $\left\{T\left({t}_{n}\right){x}_{n}\right\}$ and $\left\{f\left({x}_{n}\right)\right\}$. We get that
Since $\left\{T\left(t\right)\right\}$ is u.a.r. and ${lim}_{n\to \mathrm{\infty }}{t}_{n}=\mathrm{\infty }$, then for all $h>0$,
$\underset{n\to \mathrm{\infty }}{lim}d\left(T\left(h\right)\left(T\left({t}_{n}\right){x}_{n}\right),T\left({t}_{n}\right){x}_{n}\right)\le \underset{n\to \mathrm{\infty }}{lim}\underset{x\in B}{sup}d\left(T\left(h\right)\left(T\left({t}_{n}\right)x\right),T\left({t}_{n}\right)x\right)=0,$
where B is any bounded subset of C containing $\left\{{x}_{n}\right\}$. Hence
(3.4)
We will show that $\left\{{x}_{n}\right\}$ contains a subsequence converging strongly to $\stackrel{˜}{x}$ such that $\stackrel{˜}{x}={P}_{F\left(T\right)}f\left(\stackrel{˜}{x}\right)$, which is equivalent to the following variational inequality:
$〈\stackrel{\to }{\stackrel{˜}{x}f\stackrel{˜}{x}},\stackrel{\to }{x\stackrel{˜}{x}}〉\ge 0,\phantom{\rule{1em}{0ex}}x\in \mathcal{F}.$
(3.5)
Since $\left\{{x}_{n}\right\}$ is bounded, by Lemma 2.4, there exists a subsequence $\left\{{x}_{{n}_{j}}\right\}$ of $\left\{{x}_{n}\right\}$ which Δ-converges to a point $\stackrel{˜}{x}$, denoted by $\left\{{x}_{j}\right\}$. We claim that $\stackrel{˜}{x}\in \mathcal{F}$. Since every $CAT\left(0\right)$ space has Opial’s property, for any $h\ge 0$, if $T\left(h\right)\stackrel{˜}{x}\ne \stackrel{˜}{x}$, we have
$\begin{array}{rcl}\underset{j\to \mathrm{\infty }}{lim sup}d\left({x}_{j},T\left(h\right)\stackrel{˜}{x}\right)& \le & \underset{j\to \mathrm{\infty }}{lim sup}\left\{d\left({x}_{j},T\left(h\right){x}_{j}\right)+d\left(T\left(h\right){x}_{j},T\left(h\right)\stackrel{˜}{x}\right)\right\}\\ \le & \underset{j\to \mathrm{\infty }}{lim sup}\left\{d\left({x}_{j},T\left(h\right){x}_{j}\right)+d\left({x}_{j},\stackrel{˜}{x}\right)\right\}\\ =& \underset{j\to \mathrm{\infty }}{lim sup}d\left({x}_{j},\stackrel{˜}{x}\right)\\ <& \underset{j\to \mathrm{\infty }}{lim sup}d\left({x}_{j},T\left(h\right)\stackrel{˜}{x}\right).\end{array}$
This is a contradiction, and hence $\stackrel{˜}{x}\in \mathcal{F}$. So we have the claim. It follows from Lemma 3.2(i) that
$\begin{array}{rcl}{d}^{2}\left({x}_{j},\stackrel{˜}{x}\right)& =& 〈\stackrel{\to }{{x}_{j}\stackrel{˜}{x}},\stackrel{\to }{{x}_{j}\stackrel{˜}{x}}〉\\ \le & {\alpha }_{j}〈\stackrel{\to }{f\left({x}_{j}\right)\stackrel{˜}{x}},\stackrel{\to }{{x}_{j}\stackrel{˜}{x}}〉+\left(1-{\alpha }_{j}\right)〈\stackrel{\to }{T\left({t}_{j}\right){x}_{j}\stackrel{˜}{x}},\stackrel{\to }{{x}_{j}\stackrel{˜}{x}}〉\\ \le & {\alpha }_{j}〈\stackrel{\to }{f\left({x}_{j}\right)\stackrel{˜}{x}},\stackrel{\to }{{x}_{j}\stackrel{˜}{x}}〉+\left(1-{\alpha }_{j}\right)d\left(T\left({t}_{j}\right){x}_{j},\stackrel{˜}{x}\right)d\left({x}_{j},\stackrel{˜}{x}\right)\\ \le & {\alpha }_{j}〈\stackrel{\to }{f\left({x}_{j}\right)\stackrel{˜}{x}},\stackrel{\to }{{x}_{j}\stackrel{˜}{x}}〉+\left(1-{\alpha }_{j}\right){d}^{2}\left({x}_{j},\stackrel{˜}{x}\right).\end{array}$
It follows that
$\begin{array}{rcl}{d}^{2}\left({x}_{j},\stackrel{˜}{x}\right)& \le & 〈\stackrel{\to }{f\left({x}_{j}\right)\stackrel{˜}{x}},\stackrel{\to }{{x}_{j}\stackrel{˜}{x}}〉\\ =& 〈\stackrel{\to }{f\left({x}_{j}\right)f\left(\stackrel{˜}{x}\right)},\stackrel{\to }{{x}_{j}\stackrel{˜}{x}}〉+〈\stackrel{\to }{f\left(\stackrel{˜}{x}\right)\stackrel{˜}{x}},\stackrel{\to }{{x}_{j}\stackrel{˜}{x}}〉\\ \le & d\left(f\left({x}_{j}\right),f\left(\stackrel{˜}{x}\right)\right)d\left({x}_{j},\stackrel{˜}{x}\right)+〈\stackrel{\to }{f\left(\stackrel{˜}{x}\right)\stackrel{˜}{x}},\stackrel{\to }{{x}_{j}\stackrel{˜}{x}}〉\\ \le & \alpha {d}^{2}\left({x}_{j},\stackrel{˜}{x}\right)+〈\stackrel{\to }{f\left(\stackrel{˜}{x}\right)\stackrel{˜}{x}},\stackrel{\to }{{x}_{j}\stackrel{˜}{x}}〉,\end{array}$
and thus
${d}^{2}\left({x}_{j},\stackrel{˜}{x}\right)\le \frac{1}{1-\alpha }〈\stackrel{\to }{f\left(\stackrel{˜}{x}\right)\stackrel{˜}{x}},\stackrel{\to }{{x}_{j}\stackrel{˜}{x}}〉.$
(3.6)
Since $\left\{{x}_{j}\right\}$ Δ-converges to $\stackrel{˜}{x}$, by Lemma 2.7, we have
$\underset{n\to \mathrm{\infty }}{lim sup}〈\stackrel{\to }{f\left(\stackrel{˜}{x}\right)\stackrel{˜}{x}},\stackrel{\to }{{x}_{j}\stackrel{˜}{x}}〉\le 0.$
It follows from (3.6) that $\left\{{x}_{j}\right\}$ converges strongly to $\stackrel{˜}{x}$. Next, we show that $\stackrel{˜}{x}$ solves the variational inequality (3.3). Applying Lemma 2.3, for any $q\in \mathcal{F}$,
$\begin{array}{rcl}{d}^{2}\left({x}_{j},q\right)& =& {d}^{2}\left({\alpha }_{j}f\left({x}_{j}\right)\oplus \left(1-{\alpha }_{j}\right)T\left({t}_{j}\right){x}_{j},q\right)\\ \le & {\alpha }_{j}{d}^{2}\left(f\left({x}_{j}\right),q\right)+\left(1-{\alpha }_{j}\right){d}^{2}\left(T\left({t}_{j}\right){x}_{j},q\right)-{\alpha }_{j}\left(1-{\alpha }_{j}\right){d}^{2}\left(f\left({x}_{j}\right),T\left({t}_{j}\right){x}_{j}\right)\\ \le & {\alpha }_{j}{d}^{2}\left(f\left({x}_{j}\right),q\right)+\left(1-{\alpha }_{j}\right){d}^{2}\left({x}_{j},q\right)-{\alpha }_{j}\left(1-{\alpha }_{j}\right){d}^{2}\left(f\left({x}_{j}\right),T\left({t}_{j}\right){x}_{j}\right).\end{array}$
It implies that
${d}^{2}\left({x}_{j},q\right)\le {d}^{2}\left(f\left({x}_{j}\right),q\right)-\left(1-{\alpha }_{j}\right){d}^{2}\left(f\left({x}_{j}\right),T\left({t}_{j}\right){x}_{j}\right).$
Taking the limit through $j\to \mathrm{\infty }$, we can get that
${d}^{2}\left(\stackrel{˜}{x},q\right)\le {d}^{2}\left(f\left(\stackrel{˜}{x}\right),q\right)-{d}^{2}\left(f\left(\stackrel{˜}{x}\right),\stackrel{˜}{x}\right).$
Hence
$0\le \frac{1}{2}\left[{d}^{2}\left(\stackrel{˜}{x},\stackrel{˜}{x}\right)+{d}^{2}\left(f\left(\stackrel{˜}{x}\right),q\right)-{d}^{2}\left(\stackrel{˜}{x},q\right)-{d}^{2}\left(f\left(\stackrel{˜}{x}\right),\stackrel{˜}{x}\right)\right]=〈\stackrel{\to }{\stackrel{˜}{x}f\left(\stackrel{˜}{x}\right)},\stackrel{\to }{q\stackrel{˜}{x}}〉,\phantom{\rule{1em}{0ex}}\mathrm{\forall }q\in \mathcal{F}.$
That is, $\stackrel{˜}{x}$ solves the inequality (3.3). Finally, we show that the sequence $\left\{{x}_{n}\right\}$ converges to $\stackrel{˜}{x}$. Assume that ${x}_{{n}_{i}}\to \stackrel{ˆ}{x}$, where $i\to \mathrm{\infty }$. By the same argument, we get that $\stackrel{ˆ}{x}\in \mathcal{F}$ and solves the variational inequality (3.3), i.e.,
$〈\stackrel{\to }{\stackrel{˜}{x}f\stackrel{˜}{x}},\stackrel{\to }{\stackrel{˜}{x}\stackrel{ˆ}{x}}〉\le 0,$
(3.7)
and
$〈\stackrel{\to }{\stackrel{ˆ}{x}f\stackrel{ˆ}{x}},\stackrel{\to }{\stackrel{ˆ}{x}\stackrel{˜}{x}}〉\le 0.$
(3.8)
Adding up (3.7) and (3.8), we get that
$\begin{array}{rcl}0& \ge & 〈\stackrel{\to }{\stackrel{˜}{x}f\left(\stackrel{˜}{x}\right)},\stackrel{\to }{\stackrel{˜}{x}\stackrel{ˆ}{x}}〉-〈\stackrel{\to }{\stackrel{ˆ}{x}f\left(\stackrel{ˆ}{x}\right)},\stackrel{\to }{\stackrel{˜}{x}\stackrel{ˆ}{x}}〉\\ =& 〈\stackrel{\to }{\stackrel{˜}{x}f\left(\stackrel{ˆ}{x}\right)},\stackrel{\to }{\stackrel{˜}{x}\stackrel{ˆ}{x}}〉+〈\stackrel{\to }{f\left(\stackrel{ˆ}{x}\right)f\left(\stackrel{˜}{x}\right)},\stackrel{\to }{\stackrel{˜}{x}\stackrel{ˆ}{x}}〉-〈\stackrel{\to }{\stackrel{ˆ}{x}\stackrel{˜}{x}},\stackrel{\to }{\stackrel{˜}{x}\stackrel{ˆ}{x}}〉-〈\stackrel{\to }{\stackrel{˜}{x}f\left(\stackrel{ˆ}{x}\right)},\stackrel{\to }{\stackrel{˜}{x}\stackrel{ˆ}{x}}〉\\ =& 〈\stackrel{\to }{\stackrel{˜}{x}\stackrel{ˆ}{x}},\stackrel{\to }{\stackrel{˜}{x}\stackrel{ˆ}{x}}〉-〈\stackrel{\to }{f\left(\stackrel{ˆ}{x}\right)f\left(\stackrel{˜}{x}\right)},\stackrel{\to }{\stackrel{ˆ}{x}\stackrel{˜}{x}}〉\\ \ge & 〈\stackrel{\to }{\stackrel{˜}{x}\stackrel{ˆ}{x}},\stackrel{\to }{\stackrel{˜}{x}\stackrel{ˆ}{x}}〉-d\left(f\left(\stackrel{ˆ}{x}\right),f\left(\stackrel{˜}{x}\right)\right)d\left(\stackrel{ˆ}{x},\stackrel{˜}{x}\right)\\ \ge & {d}^{2}\left(\stackrel{˜}{x},\stackrel{ˆ}{x}\right)-\alpha d\left(\stackrel{ˆ}{x},\stackrel{˜}{x}\right)d\left(\stackrel{ˆ}{x},\stackrel{˜}{x}\right)\\ \ge & {d}^{2}\left(\stackrel{˜}{x},\stackrel{ˆ}{x}\right)-\alpha {d}^{2}\left(\stackrel{ˆ}{x},\stackrel{˜}{x}\right)\\ \ge & \left(1-\alpha \right){d}^{2}\left(\stackrel{˜}{x},\stackrel{ˆ}{x}\right).\end{array}$

Since $0<\alpha <1$, we have that $d\left(\stackrel{˜}{x},\stackrel{ˆ}{x}\right)=0$, and so $\stackrel{˜}{x}=\stackrel{ˆ}{x}$. Hence the sequence ${x}_{n}$ converges strongly to $\stackrel{˜}{x}$, which is the unique solution to the variational inequality (3.3). This completes the proof. □

If $f\equiv u$, then the following result can be obtained directly from Theorem 3.3.

Corollary 3.4 Let C be a closed convex subset of a complete $CAT\left(0\right)$ space X, and let $\left\{T\left(t\right)\right\}$ be a one-parameter continuous semigroup of nonexpansive mappings on C satisfying $\mathcal{F}\ne \mathrm{\varnothing }$ and uniformly asymptotically regular (in short, u.a.r.) on C, that is, for all $h\ge 0$ and any bounded subset B of C,
$\underset{t\to \mathrm{\infty }}{lim}\underset{x\in B}{sup}d\left(T\left(h\right)\left(T\left(t\right)x\right),T\left(t\right)x\right)=0.$
Let u be any element in C. Suppose ${t}_{n}\in \left[0,\mathrm{\infty }\right)$, ${\alpha }_{n}\in \left(0,1\right)$ such that ${lim}_{n\to \mathrm{\infty }}{t}_{n}=\mathrm{\infty }$ and ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$ and let $\left\{{x}_{n}\right\}$ be given by
${x}_{n}={\alpha }_{n}u\oplus \left(1-{\alpha }_{n}\right)T\left({t}_{n}\right){x}_{n}.$
Then $\left\{{x}_{n}\right\}$ converges strongly as $n\to \mathrm{\infty }$ to $\stackrel{˜}{x}$ such that $\stackrel{˜}{x}={P}_{\mathcal{F}}\stackrel{˜}{x}$, which is equivalent to the following variational inequality:
$〈\stackrel{\to }{\stackrel{˜}{x}u},\stackrel{\to }{x\stackrel{˜}{x}}〉\ge 0,\phantom{\rule{1em}{0ex}}x\in \mathcal{F}.$
(3.9)
Theorem 3.5 Let C be a closed convex subset of a complete $CAT\left(0\right)$ space X, and let $\left\{T\left(t\right)\right\}$ be a one-parameter continuous semigroup of nonexpansive mappings on C satisfying $\mathcal{F}\ne \mathrm{\varnothing }$ and uniformly asymptotically regular (in short, u.a.r.) on C, that is, for all $h\ge 0$ and any bounded subset B of C,
$\underset{t\to \mathrm{\infty }}{lim}\underset{x\in B}{sup}d\left(T\left(h\right)\left(T\left(t\right)x\right),T\left(t\right)x\right)=0.$
Let f be a contraction on C with coefficient $0<\alpha <1$. Suppose that ${t}_{n}\in \left[0,\mathrm{\infty }\right)$, ${\alpha }_{n}\in \left(0,1\right)$, ${x}_{0}\in C$, and $\left\{{x}_{n}\right\}$ is given by
${x}_{n+1}={\alpha }_{n}f\left({x}_{n}\right)\oplus \left(1-{\alpha }_{n}\right)T\left({t}_{n}\right){x}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 0,$
(3.10)
where $\left\{{\alpha }_{n}\right\}\subset \left(0,1\right)$ satisfies the following conditions:
1. (i)

${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$;

2. (ii)

${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$ and

3. (iii)

${lim}_{n\to \mathrm{\infty }}{t}_{n}=\mathrm{\infty }$.

Then $\left\{{x}_{n}\right\}$ converges strongly as $n\to \mathrm{\infty }$ to $\stackrel{˜}{x}$ such that $\stackrel{˜}{x}={P}_{\mathcal{F}}f\left(\stackrel{˜}{x}\right)$, which is equivalent to the variational inequality (3.3).

Proof We first show that the sequence $\left\{{x}_{n}\right\}$ is bounded. For any $p\in \mathcal{F}$, we have that
$\begin{array}{rcl}d\left({x}_{n+1},p\right)& =& d\left({\alpha }_{n}f\left({x}_{n}\right)\oplus \left(1-{\alpha }_{n}\right)T\left({t}_{n}\right){x}_{n},p\right)\\ \le & {\alpha }_{n}d\left(f\left({x}_{n}\right),p\right)+\left(1-{\alpha }_{n}\right)d\left(T\left({t}_{n}\right){x}_{n},p\right)\\ \le & {\alpha }_{n}\left(d\left(f\left({x}_{n}\right),f\left(p\right)\right)+d\left(f\left(p\right),p\right)\right)+\left(1-{\alpha }_{n}\right)d\left(T\left({t}_{n}\right){x}_{n},p\right)\\ \le & max\left\{d\left({x}_{n},p\right),\frac{1}{1-\alpha }d\left(f\left(p\right),p\right)\right\}.\end{array}$
By induction, we have
$d\left({x}_{n},p\right)\le max\left\{d\left({x}_{0},p\right),\frac{1}{1-\alpha }d\left(f\left(p\right),p\right)\right\}$
for all $n\in \mathbb{N}$. Hence $\left\{{x}_{n}\right\}$ is bounded, so are $\left\{T\left({t}_{n}\right){x}_{n}\right\}$ and $\left\{f\left({x}_{n}\right)\right\}$. Using the assumption that ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$, we get that
Since $\left\{T\left(t\right)\right\}$ is u.a.r. and ${lim}_{n\to \mathrm{\infty }}{t}_{n}=\mathrm{\infty }$, then for all $h\ge 0$,
$\underset{n\to \mathrm{\infty }}{lim}d\left(T\left(h\right)\left(T\left({t}_{n}\right){x}_{n}\right),T\left({t}_{n}\right){x}_{n}\right)\le \underset{n\to \mathrm{\infty }}{lim}\underset{x\in B}{sup}d\left(T\left(h\right)\left(T\left({t}_{n}\right)x\right),T\left({t}_{n}\right)x\right)=0,$
where B is any bounded subset of C containing $\left\{{x}_{n}\right\}$. Hence
(3.11)
Let $\left\{{z}_{m}\right\}$ be a sequence in C such that
${z}_{m}={\alpha }_{m}f\left({z}_{m}\right)\oplus \left(1-{\alpha }_{m}\right)T\left({t}_{m}\right){z}_{m}.$
It follows from Theorem 3.3 that $\left\{{z}_{m}\right\}$ converges strongly as $m\to \mathrm{\infty }$ to a fixed point $\stackrel{˜}{x}\in \mathcal{F}$, which solves the variational inequality (3.3). Now, we claim that
$\underset{n\to \mathrm{\infty }}{lim sup}〈\stackrel{\to }{f\left(\stackrel{˜}{x}\right)\stackrel{˜}{x}},\stackrel{\to }{{x}_{n+1}\stackrel{˜}{x}}〉\le 0.$
It follows from Lemma 3.2(i) that
$\begin{array}{rcl}{d}^{2}\left({z}_{m},{x}_{n+1}\right)& =& 〈\stackrel{\to }{{z}_{m}{x}_{n+1}},\stackrel{\to }{{z}_{m}{x}_{n+1}}〉\\ \le & {\alpha }_{m}〈\stackrel{\to }{f\left({z}_{m}\right){x}_{n+1}},\stackrel{\to }{{z}_{m}{x}_{n+1}}〉+\left(1-{\alpha }_{m}\right)〈\stackrel{\to }{T\left({t}_{m}\right){z}_{m}{x}_{n+1}},\stackrel{\to }{{z}_{m}{x}_{n+1}}〉\\ =& {\alpha }_{m}〈\stackrel{\to }{f\left({z}_{m}\right)f\left(\stackrel{˜}{x}\right)},\stackrel{\to }{{z}_{m}{x}_{n+1}}〉+{\alpha }_{m}〈\stackrel{\to }{f\left(\stackrel{˜}{x}\right)\stackrel{˜}{x}},\stackrel{\to }{{z}_{m}{x}_{n+1}}〉+{\alpha }_{m}〈\stackrel{\to }{\stackrel{˜}{x}{z}_{m}},\stackrel{\to }{{z}_{m}{x}_{n+1}}〉\\ +{\alpha }_{m}〈\stackrel{\to }{{z}_{m}{x}_{n+1}},\stackrel{\to }{{z}_{m}{x}_{n+1}}〉+\left(1-{\alpha }_{m}\right)〈\stackrel{\to }{T\left({t}_{m}\right){z}_{m}T\left({t}_{m}\right){x}_{n+1}},\stackrel{\to }{{z}_{m}{x}_{n+1}}〉\\ +\left(1-{\alpha }_{m}\right)〈\stackrel{\to }{T\left({t}_{m}\right){x}_{n+1}{x}_{n+1}},\stackrel{\to }{{z}_{m}{x}_{n+1}}〉\\ \le & {\alpha }_{m}\alpha d\left({z}_{m},\stackrel{˜}{x}\right)d\left({z}_{m},{x}_{n+1}\right)+{\alpha }_{m}〈\stackrel{\to }{f\left(\stackrel{˜}{x}\right)\stackrel{˜}{x}},\stackrel{\to }{{z}_{m}{x}_{n+1}}〉+{\alpha }_{m}d\left(\stackrel{˜}{x},{z}_{m}\right)d\left({z}_{m},{x}_{n+1}\right)\\ +{\alpha }_{m}{d}^{2}\left({z}_{m},{x}_{n+1}\right)+\left(1-{\alpha }_{m}\right){d}^{2}\left({z}_{m},{x}_{n+1}\right)\\ +\left(1-{\alpha }_{m}\right)d\left(T\left({t}_{m}\right){x}_{n+1},{x}_{n+1}\right)d\left({z}_{m},{x}_{n+1}\right)\\ \le & {\alpha }_{m}\alpha d\left({z}_{m},\stackrel{˜}{x}\right)M+{\alpha }_{m}〈\stackrel{\to }{f\left(\stackrel{˜}{x}\right)\stackrel{˜}{x}},\stackrel{\to }{{z}_{m}{x}_{n+1}}〉+{\alpha }_{m}d\left(\stackrel{˜}{x},{z}_{m}\right)M+{\alpha }_{m}{d}^{2}\left({z}_{m},{x}_{n+1}\right)\\ +\left(1-{\alpha }_{m}\right){d}^{2}\left({z}_{m},{x}_{n+1}\right)+\left(1-{\alpha }_{m}\right)d\left(T\left({t}_{m}\right){x}_{n+1},{x}_{n+1}\right)M\\ \le & {d}^{2}\left({z}_{m},{x}_{n+1}\right)+{\alpha }_{m}\alpha d\left({z}_{m},\stackrel{˜}{x}\right)M+{\alpha }_{m}d\left(\stackrel{˜}{x},{z}_{m}\right)M+d\left(T\left({t}_{m}\right){x}_{n+1},{x}_{n+1}\right)M\\ +{\alpha }_{m}〈\stackrel{\to }{f\left(\stackrel{˜}{x}\right)\stackrel{˜}{x}},\stackrel{\to }{{z}_{m}{x}_{n+1}}〉,\end{array}$
where $M\ge {sup}_{m,n\ge 1}\left\{d\left({z}_{m},{x}_{n}\right)\right\}$. This implies that
$〈\stackrel{\to }{f\left(\stackrel{˜}{x}\right)\stackrel{˜}{x}},\stackrel{\to }{{x}_{n+1}{z}_{m}}〉\le \left(1+\alpha \right)d\left({z}_{m},\stackrel{˜}{x}\right)M+\frac{d\left(T\left({t}_{m}\right){x}_{n+1},{x}_{n+1}\right)}{{\alpha }_{m}}M.$
(3.12)
Taking the upper limit as $n\to \mathrm{\infty }$ first, and then $m\to \mathrm{\infty }$, inequality (3.12) yields that
$\underset{m\to \mathrm{\infty }}{lim sup}\underset{n\to \mathrm{\infty }}{lim sup}〈\stackrel{\to }{f\left(\stackrel{˜}{x}\right)\stackrel{˜}{x}},\stackrel{\to }{{x}_{n+1}{z}_{m}}〉\le 0.$
(3.13)
Since
$\begin{array}{rcl}〈\stackrel{\to }{f\left(\stackrel{˜}{x}\right)\stackrel{˜}{x}},\stackrel{\to }{{x}_{n+1}\stackrel{˜}{x}}〉& =& 〈\stackrel{\to }{f\left(\stackrel{˜}{x}\right)\stackrel{˜}{x}},\stackrel{\to }{{x}_{n+1}{z}_{m}}〉+〈\stackrel{\to }{f\left(\stackrel{˜}{x}\right)\stackrel{˜}{x}},\overline{{z}_{m}\stackrel{˜}{x}}〉\\ \le & 〈\stackrel{\to }{f\left(\stackrel{˜}{x}\right)\stackrel{˜}{x}},\stackrel{\to }{{x}_{n+1}{z}_{m}}〉+d\left(f\left(\stackrel{˜}{x}\right),\stackrel{˜}{x}\right)d\left({z}_{m},\stackrel{˜}{x}\right).\end{array}$
Thus, by taking the upper limit as $n\to \mathrm{\infty }$ first, and then $m\to \mathrm{\infty }$ the last inequality, it follows from ${z}_{m}\to \stackrel{˜}{x}$ and (3.13) that
$\underset{n\to \mathrm{\infty }}{lim sup}〈\stackrel{\to }{f\left(\stackrel{˜}{x}\right)\stackrel{˜}{x}},\stackrel{\to }{{x}_{n+1}\stackrel{˜}{x}}〉\le 0.$
Finally, we prove that ${x}_{n}\to \stackrel{˜}{x}$ as $n\to \mathrm{\infty }$. For any $n\in \mathbb{N}$, we set ${y}_{n}={\alpha }_{n}\stackrel{˜}{x}\oplus \left(1-{\alpha }_{n}\right)T\left({t}_{n}\right){x}_{n}$. It follows from Lemma 3.1 and Lemma 3.2(i), (ii) that
$\begin{array}{rcl}{d}^{2}\left({x}_{n+1},\stackrel{˜}{x}\right)& \le & {d}^{2}\left({y}_{n},\stackrel{˜}{x}\right)+2〈\stackrel{\to }{{x}_{n+1}{y}_{n}},\stackrel{\to }{{x}_{n+1}\stackrel{˜}{x}}〉\\ \le & {\left({\alpha }_{n}d\left(\stackrel{˜}{x},\stackrel{˜}{x}\right)+\left(1-{\alpha }_{n}\right)d\left(T\left({t}_{n}\right){x}_{n},\stackrel{˜}{x}\right)\right)}^{2}\\ +2\left[{\alpha }_{n}〈\stackrel{\to }{f\left({x}_{n}\right){y}_{n}},\stackrel{\to }{{x}_{n+1}\stackrel{˜}{x}}〉+\left(1-{\alpha }_{n}\right)〈\stackrel{\to }{T\left({t}_{n}\right){x}_{n}{y}_{n}},\stackrel{\to }{{x}_{n+1}\stackrel{˜}{x}}〉\right]\\ \le & {\left(1-{\alpha }_{n}\right)}^{2}{d}^{2}\left({x}_{n},\stackrel{˜}{x}\right)+2\left[{\alpha }_{n}{\alpha }_{n}〈\stackrel{\to }{f\left({x}_{n}\right)\stackrel{˜}{x}},\stackrel{\to }{{x}_{n+1}\stackrel{˜}{x}}〉+{\alpha }_{n}\left(1-{\alpha }_{n}\right)〈\stackrel{\to }{f\left({x}_{n}\right)T\left({t}_{n}\right){x}_{n}},\stackrel{\to }{{x}_{n+1}\stackrel{˜}{x}}〉\\ +\left(1-{\alpha }_{n}\right){\alpha }_{n}〈\stackrel{\to }{T\left({t}_{n}\right){x}_{n}\stackrel{˜}{x}},\stackrel{\to }{{x}_{n+1}\stackrel{˜}{x}}〉+\left(1-{\alpha }_{n}\right)\left(1-{\alpha }_{n}\right)〈\stackrel{\to }{T\left({t}_{n}\right){x}_{n}T\left({t}_{n}\right){x}_{n}},\stackrel{\to }{{x}_{n+1}\stackrel{˜}{x}}〉\right]\\ \le & {\left(1-{\alpha }_{n}\right)}^{2}{d}^{2}\left({x}_{n},\stackrel{˜}{x}\right)+2\left[{\alpha }_{n}{\alpha }_{n}〈\stackrel{\to }{f\left({x}_{n}\right)\stackrel{˜}{x}},\stackrel{\to }{{x}_{n+1}\stackrel{˜}{x}}〉+{\alpha }_{n}\left(1-{\alpha }_{n}\right)〈\stackrel{\to }{f\left({x}_{n}\right)T\left({t}_{n}\right){x}_{n}},\stackrel{\to }{{x}_{n+1}\stackrel{˜}{x}}〉\\ +\left(1-{\alpha }_{n}\right){\alpha }_{n}〈\stackrel{\to }{T\left({t}_{n}\right){x}_{n}\stackrel{˜}{x}},\stackrel{\to }{{x}_{n+1}\stackrel{˜}{x}}〉+{\left(1-{\alpha }_{n}\right)}^{2}d\left(T\left({t}_{n}\right){x}_{n},T\left({t}_{n}\right){x}_{n}\right)d\left({x}_{n+1}\stackrel{˜}{x}\right)\right]\\ =& {\left(1-{\alpha }_{n}\right)}^{2}{d}^{2}\left({x}_{n},\stackrel{˜}{x}\right)+2\left[{\alpha }_{n}^{2}〈\stackrel{\to }{f\left({x}_{n}\right)\stackrel{˜}{x}},\stackrel{\to }{{x}_{n+1}\stackrel{˜}{x}}〉+{\alpha }_{n}\left(1-{\alpha }_{n}\right)〈\stackrel{\to }{f\left({x}_{n}\right)\stackrel{˜}{x}},\stackrel{\to }{{x}_{n+1}\stackrel{˜}{x}}〉\right]\\ =& {\left(1-{\alpha }_{n}\right)}^{2}{d}^{2}\left({x}_{n},\stackrel{˜}{x}\right)+2{\alpha }_{n}〈\stackrel{\to }{f\left({x}_{n}\right)\stackrel{˜}{x}},\stackrel{\to }{{x}_{n+1}\stackrel{˜}{x}}〉\\ =& {\left(1-{\alpha }_{n}\right)}^{2}{d}^{2}\left({x}_{n},\stackrel{˜}{x}\right)+2{\alpha }_{n}〈\stackrel{\to }{f\left({x}_{n}\right)f\left(\stackrel{˜}{x}\right)},\stackrel{\to }{{x}_{n+1}\stackrel{˜}{x}}〉+2{\alpha }_{n}〈\stackrel{\to }{f\left(\stackrel{˜}{x}\right)\stackrel{˜}{x}},\stackrel{\to }{{x}_{n+1}\stackrel{˜}{x}}〉\\ \le & {\left(1-{\alpha }_{n}\right)}^{2}{d}^{2}\left({x}_{n},\stackrel{˜}{x}\right)+2{\alpha }_{n}\alpha d\left({x}_{n},\stackrel{˜}{x}\right)d\left({x}_{n+1},\stackrel{˜}{x}\right)+2{\alpha }_{n}〈\stackrel{\to }{f\left(\stackrel{˜}{x}\right)\stackrel{˜}{x}},\stackrel{\to }{{x}_{n+1}\stackrel{˜}{x}}〉\\ \le & {\left(1-{\alpha }_{n}\right)}^{2}{d}^{2}\left({x}_{n},\stackrel{˜}{x}\right)+{\alpha }_{n}\alpha \left({d}^{2}\left({x}_{n},\stackrel{˜}{x}\right)+{d}^{2}\left({x}_{n+1},\stackrel{˜}{x}\right)\right)+2{\alpha }_{n}〈\stackrel{\to }{f\left(\stackrel{˜}{x}\right)\stackrel{˜}{x}},\stackrel{\to }{{x}_{n+1}\stackrel{˜}{x}}〉,\end{array}$
which implies that
$\begin{array}{rcl}{d}^{2}\left({x}_{n+1},\stackrel{˜}{x}\right)& \le & \frac{1-\left(2-\alpha \right){\alpha }_{n}+{\alpha }_{n}^{2}}{1-\alpha {\alpha }_{n}}{d}^{2}\left({x}_{n},\stackrel{˜}{x}\right)+\frac{2{\alpha }_{n}}{1-\alpha {\alpha }_{n}}〈\stackrel{\to }{f\left(\stackrel{˜}{x}\right)\stackrel{˜}{x}},\stackrel{\to }{{x}_{n+1}\stackrel{˜}{x}}〉\\ \le & \frac{1-\left(2-\alpha \right){\alpha }_{n}}{1-\alpha {\alpha }_{n}}{d}^{2}\left({x}_{n},\stackrel{˜}{x}\right)+\frac{2{\alpha }_{n}}{1-\alpha {\alpha }_{n}}〈\stackrel{\to }{f\left(\stackrel{˜}{x}\right)\stackrel{˜}{x}},\stackrel{\to }{{x}_{n+1}\stackrel{˜}{x}}〉+{\alpha }_{n}^{2}M,\end{array}$
where $M\ge {sup}_{n\ge 0}\left\{{d}^{2}\left({x}_{n},\stackrel{˜}{x}\right)\right\}$. It then follows that
${d}^{2}\left({x}_{n+1},\stackrel{˜}{x}\right)\le \left(1-{\alpha }_{n}^{\prime }\right){d}^{2}\left({x}_{n},\stackrel{˜}{x}\right)+{\alpha }_{n}^{\prime }{\beta }_{n}^{\prime },$
where
${\alpha }_{n}^{\prime }=\frac{2\left(1-\alpha \right){\alpha }_{n}}{1-\alpha {\alpha }_{n}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\beta }_{n}^{\prime }=\frac{\left(1-\alpha {\alpha }_{n}\right){\alpha }_{n}}{2\left(1-\alpha \right)}M+\frac{1}{\left(1-\alpha \right)}〈\stackrel{\to }{f\left(\stackrel{˜}{x}\right)\stackrel{˜}{x}},\stackrel{\to }{{x}_{n+1}\stackrel{˜}{x}}〉.$

Applying Lemma 2.8, we can conclude that ${x}_{n}\to \stackrel{˜}{x}$. This completes the proof. □

If $f\equiv u$, then the following corollary can be obtained directly from Theorem 3.5.

Corollary 3.6 Let C be a closed convex subset of a complete $CAT\left(0\right)$ space X, and let $\left\{T\left(t\right)\right\}$ be a one-parameter continuous semigroup of nonexpansive mappings on C satisfying $\mathcal{F}\ne \mathrm{\varnothing }$ and uniformly asymptotically regular (in short, u.a.r.) on C, that is, for all $h\ge 0$ and any bounded subset B of C,
$\underset{t\to \mathrm{\infty }}{lim}\underset{x\in B}{sup}d\left(T\left(h\right)\left(T\left(t\right)x\right),T\left(t\right)x\right)=0.$
Suppose that ${t}_{n}\in \left[0,\mathrm{\infty }\right)$, ${\alpha }_{n}\in \left(0,1\right)$, ${x}_{0}\in C$ and $\left\{{x}_{n}\right\}$ is given by
${x}_{n+1}={\alpha }_{n}u\oplus \left(1-{\alpha }_{n}\right)T\left({t}_{n}\right){x}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 0,$
(3.14)
where $\left\{{\alpha }_{n}\right\}\subset \left(0,1\right)$ satisfies the following conditions:
1. (i)

${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$;

2. (ii)

${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$ and

3. (iii)

${lim}_{n\to \mathrm{\infty }}{t}_{n}=\mathrm{\infty }$.

Then $\left\{{x}_{n}\right\}$ converges strongly as $n\to \mathrm{\infty }$ to $\stackrel{˜}{x}$ such that $\stackrel{˜}{x}={P}_{\mathcal{F}}\stackrel{˜}{x}$, which is equivalent to the variational inequality (3.9).

## Declarations

### Acknowledgements

The first author is supported by the Centre of Excellence in Mathematics under the Commission on Higher Education, Ministry of Education, Thailand.

## Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, Naresuan University, Phitsanulok, 65000, Thailand
(2)
Centre of Excellence in Mathematics (CHE), Si Ayutthaya Road, Bangkok, 10400, Thailand

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