Viscosity approximation methods for nonexpansive semigroups in $CAT(0)$ spaces

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 $CAT(0)$ 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.

1 Introduction

Let $(X,d)$ 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 $[0,l]\subset \mathbb{R}$ to X such that $c(0)=x$, $c(l)=y$, and $d(c(t),c(t^{\prime }))=|t-t^{\prime }|$ for all $t,t^{\prime }\in [0,l]$. In particular, c is an isometry and $d(x,y)=l$. 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 $[x,y]$. The space $(X,d)$ 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 $\mathrm{△}(x_1,x_2,x_3)$ in a geodesic metric space $(X,d)$ 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 $\mathrm{△}(x_1,x_2,x_3)$ in $(X,d)$ is a triangle $\overline{\mathrm{△}}(x_1,x_2,x_3):=\mathrm{△}({\overline{x}}_1,{\overline{x}}_2,{\overline{x}}_3)$ in the Euclidean plane ${\mathbb{E}}^2$ such that $d_{{\mathbb{E}}_2}({\overline{x}}_i,{\overline{x}}_j)=d(x_i,x_j)$ for all $i,j\in \{1,2,3\}$.

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

$CAT(0)$: Let △ be a geodesic triangle in X and let $\overline{\mathrm{△}}$ be a comparison triangle for △. Then △ is said to satisfy the $CAT(0)$ inequality if for all $x,y\in \mathrm{△}$ and all comparison points $\overline{x},\overline{y}\in \overline{\mathrm{△}}$,

$$d(x,y)\le d_{{\mathbb{E}}^2}(\overline{x},\overline{y}).$$

If x, $y_1$, $y_2$ are points in a $CAT(0)$ space and if $y_0$ is the midpoint of the segment $[y_1,y_2]$, then the $CAT(0)$ inequality implies

$$d^2(x,y_0)\le \frac{1}{2}{d}^2(x,y_1)+\frac{1}{2}{d}^2(x,y_2)-\frac{1}{4}{d}^2(y_1,y_2).$$

(1.1)

This is the (CN)-inequality of Bruhat and Tits [1]. In fact (cf. [2], p.163), a geodesic space is a $CAT(0)$ 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 $CAT(0)$ space. Other examples include pre-Hilbert spaces, ℝ-trees (see [2]), Euclidean buildings (see [3]), the complex Hilbert ball with a hyperbolic metric (see [4]), and many others. Complete $CAT(0)$ 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 $(a,b)\in X\times X$ by $\overrightarrow{ab}$ and call it a vector. Then quasilinearization is defined as a map $〈\cdot ,\cdot 〉:(X\times X)\times (X\times X)\to \mathbb{R}$ defined by

$$〈\overrightarrow{ab},\overrightarrow{cd}〉=\frac{1}{2}(d^2(a,d)+d^2(b,c)-d^2(a,c)-d^2(b,d))(a,b,c,d\in X).$$

(1.2)

It is easily seen that $〈\overrightarrow{ab},\overrightarrow{cd}〉=〈\overrightarrow{cd},\overrightarrow{ab}〉$, $〈\overrightarrow{ab},\overrightarrow{cd}〉=-〈\overrightarrow{ba},\overrightarrow{cd}〉$ and $〈\overrightarrow{ax},\overrightarrow{cd}〉+〈\overrightarrow{xb},\overrightarrow{cd}〉=〈\overrightarrow{ab},\overrightarrow{cd}〉$ for all $a,b,c,d,x\in X$. We say that X satisfies the Cauchy-Schwarz inequality if

$$〈\overrightarrow{ab},\overrightarrow{cd}〉\le d(a,b)d(c,d)$$

(1.3)

for all $a,b,c,d\in X$. It is known [[5], Corollary 3] that a geodesically connected metric space is a $CAT(0)$ 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 $CAT(0)$ spaces as follows. Consider the map $\mathrm{\Theta }:\mathbb{R}\times X\times X\to C(X)$ defined by

$$\mathrm{\Theta }(t,a,b)(x)=t〈\overrightarrow{ab},\overrightarrow{ax}〉,$$

(1.4)

where $C(X)$ is the space of all continuous real-valued functions on X. Then the Cauchy-Schwarz inequality implies that $\mathrm{\Theta }(t,a,b)$ is a Lipschitz function with a Lipschitz semi-norm $L(\mathrm{\Theta }(t,a,b))=|t|d(a,b)$ for all $t\in \mathbb{R}$ and $a,b\in X$, where

$$L(f)=sup\{\frac{f(x)-f(y)}{d(x,y)}:x,y\in X,x\ne y\}$$

is the Lipschitz semi-norm of the function $f:X\to \mathbb{R}$. Now, define the pseudometric D on $\mathbb{R}\times X\times X$ by

$$D((t,a,b),(s,c,d))=L(\mathrm{\Theta }(t,a,b)-\mathrm{\Theta }(s,c,d)).$$

Lemma 1.1 [[6], Lemma 2.1]

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

For a complete $CAT(0)$ space $(X,d)$, the pseudometric space $(\mathbb{R}\times X\times X,D)$ can be considered as a subspace of the pseudometric space $(Lip(X,R),L)$ of all real-valued Lipschitz functions. Also, D defines an equivalence relation on $\mathbb{R}\times X\times X$, where the equivalence class of $t\overrightarrow{ab}:=(t,a,b)$ is

$$[t\overrightarrow{ab}]=\{s\overrightarrow{cd}:t〈\overrightarrow{ab},\overrightarrow{xy}〉=s〈\overrightarrow{cd},\overrightarrow{xy}〉\mathrm{\forall }x,y\in X\}.$$

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

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

Theorem 1.2 [[7], Theorem 2.4]

Let C be a nonempty convex subset of a complete $CAT(0)$ space X, $x\in X$ and $u\in C$. Then

$$u=P_{C}x\mathit{\text{if and only if}}〈\overrightarrow{yu},\overrightarrow{ux}〉\ge 0\mathit{\text{for all}}y\in C.$$

From now on, let ℕ be the set of positive integers, let ℝ 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 $CAT(0)$ space X. A family $\mathcal{S}:=\{T(t):t\in {\mathbb{R}}^{+}\}$ of self-mappings of C is called a one-parameter continuous semigroup of nonexpansive mappings if the following conditions hold:

• (i) for each $t\in {\mathbb{R}}^{+}$, $T(t)$ is a nonexpansive mapping on C, i.e.,

  $$d(T(t)x,T(t)y)\le d(x,y),\mathrm{\forall }x,y\in C;$$

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

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

A family $\mathcal{S}:=\{T(t):t\in {\mathbb{R}}^{+}\}$ 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:

• (iv) $T(0)x=x$ for all $x\in C$.

We shall denote by ℱ the common fixed point set of $\mathcal{S}$, that is,

$$\mathcal{F}:=F(\mathcal{S})=\{x\in C:T(t)x=x,t\in {\mathbb{R}}^{+}\}=\bigcap _{t\in {\mathbb{R}}^{+}}F(T(t)).$$

One classical way to study nonexpansive mappings is to use contractions to approximate nonexpansive mappings. More precisely, take $t\in (0,1)$ and define a contraction $T_t:C\to C$ by

$$T_t=tu+(1-t)Tx,\mathrm{\forall }x\in C,$$

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 $F(T)$.

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 $\{x_n\}$ by

$$x_{n+1}={\alpha }_{n}u+(1-{\alpha }_n)T{x}_n.$$

(1.5)

He proved that the sequence $\{x_n\}$ 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:

$$x_n={\alpha }_{n}u+(1-{\alpha }_n)\frac{1}{t_n}{\int }_0^{t_n}T(s)x_{n}ds,$$

(1.6)

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

Later, Suzuki [12] was the first to introduce in a Hilbert space the following iteration process:

$$x_n={\alpha }_{n}u+(1-{\alpha }_n)T(t_n)x_n,\mathrm{\forall }n\ge 1,$$

(1.7)

where $\{T(t):t\ge 0\}$ is a strongly continuous semigroup of nonexpansive mappings on C such that $\mathcal{F}\ne \mathrm{\varnothing }$ and $\{{\alpha }_n\}$ and $\{t_n\}$ are appropriate sequences of real numbers. He proved that $\{x_n\}$ generated by (1.7) converges strongly to the element of ℱ nearest to u. Using Moudafi’s viscosity approximation methods, Song and Xu [13] introduced the following iteration process:

$$x_n={\alpha }_{n}f(x_n)+(1-{\alpha }_n)T(t_n)x_n,\mathrm{\forall }n\ge 1,$$

(1.8)

and

$$x_{n+1}={\alpha }_{n}f(x_n)+(1-{\alpha }_n)T(t_n)x_n,\mathrm{\forall }n\ge 1.$$

(1.9)

They proved that $\{x_n\}$ converges to the same point of ℱ 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 $\{T(t):t\ge 0\}$ in a complete $CAT(0)$ space X as follows:

$$x_n={\alpha }_n{x}_0\oplus T(t_n)x_n,\mathrm{\forall }n\ge 1,$$

where C is a nonempty closed convex subset of a complete $CAT(0)$ space X, $x_0\in C$, $\{{\alpha }_n\}$ and $\{t_n\}$ 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 $\{x_n\}$ converges to the element of ℱ 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 $t\in (0,1)$, let $x_t\in C$ be a unique fixed point of the contraction $x\mapsto tf(x)\oplus (1-t)Tx$; i.e.,

$$x_t=tf(x_t)\oplus (1-t)T{x}_t,$$

(1.10)

and $x_0\in C$ is arbitrarily chosen and

$$x_{n+1}={\alpha }_{n}f(x_n)\oplus (1-{\alpha }_n)T{x}_n,\mathrm{\forall }n\ge 0,$$

(1.11)

where $\{{\alpha }_n\}\subset (0,1)$. They proved $\{x_t\}$ defined by (1.10) converges strongly as $t\to 0$ to $\tilde{x}\in F(T)$ such that $\tilde{x}=P_{F(T)}f(\tilde{x})$ in the framework of $CAT(0)$ space satisfying property $\mathcal{P}$, i.e., if for $x,u,y_1,y_2\in X$,

$$d(x,P_{[x,y_1]}u)d(x,y_1)\le d(x,P_{[x,y_2]}u)d(x,y_2)+d(x,u)d(y_1,y_2).$$

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

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 $CAT(0)$ 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 $\mathcal{S}:=\{T(t):t\in {\mathbb{R}}^{+}\}$ in $CAT(0)$ spaces. Let C be a nonempty, closed and convex subset of a $CAT(0)$ space X. For a given contraction f on C and ${\alpha }_n\in (0,1)$, let $x_n\in C$ be a unique fixed point of the contraction $x\mapsto {\alpha }_{n}f(x)\oplus (1-{\alpha }_n)T(t_n)x$; i.e.,

$$x_n={\alpha }_{n}f(x_n)\oplus (1-{\alpha }_n)T(t_n)x_n,n\ge 0,$$

(1.12)

and

$$x_{n+1}={\alpha }_{n}f(x_n)\oplus (1-{\alpha }_n)T(t_n)x_n,n\ge 0.$$

(1.13)

We prove that the iterative schemes $\{x_n\}$ defined by (1.12) and $\{x_n\}$ defined by (1.13) converge strongly to the same point $\tilde{x}$ such that $\tilde{x}=P_{\mathcal{F}}f(\tilde{x})$, which is the unique solution of the variational inequality

$$〈\overrightarrow{\tilde{x}f\tilde{x}},\overrightarrow{x\tilde{x}}〉\ge 0,x\in \mathcal{F},$$

where ℱ is the common fixed point set of $\mathcal{S}$, that is,

$$\mathcal{F}:=F(\mathcal{S})=\{x\in C:T(t)x=x,t\in {\mathbb{R}}^{+}\}=\bigcap _{t\in {\mathbb{R}}^{+}}F(T(t)).$$

2 Preliminaries

In this paper, we write $(1-t)x\oplus ty$ for the unique point z in the geodesic segment joining from x to y such that

$$d(z,x)=td(x,y)\text{and}d(z,y)=(1-t)d(x,y).$$

We also denote by $[x,y]$ the geodesic segment joining from x to y, that is, $[x,y]=\{(1-t)x\oplus ty:t\in [0,1]\}$. A subset C of a $CAT(0)$ space is convex if $[x,y]\subseteq C$ 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 $CAT(0)$ space, $p,q,r,s\in X$ and $\lambda \in [0,1]$. Then

$$d(\lambda p\oplus (1-\lambda )q,\lambda r\oplus (1-\lambda )s)\le \lambda d(p,r)+(1-\lambda )d(q,s).$$

Lemma 2.2 [[19], Lemma 2.4]

Let X be a $CAT(0)$ space, $x,y,z\in X$ and $\lambda \in [0,1]$. Then

$$d(\lambda x\oplus (1-\lambda )y,z)\le \lambda d(x,z)+(1-\lambda )d(y,z).$$

Lemma 2.3 [[19], Lemma 2.5]

Let X be a $CAT(0)$ space, $x,y,z\in X$ and $\lambda \in [0,1]$. Then

$$d^2(\lambda x\oplus (1-\lambda )y,z)\le \lambda d^2(x,z)+(1-\lambda )d^2(y,z)-\lambda (1-\lambda )d^2(x,y).$$

The concept of Δ-convergence introduced by Lim [20] in 1976 was shown by Kirk and Panyanak [21] in $CAT(0)$ 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 $\{x_n\}$ be a bounded sequence in a $CAT(0)$ space X. For $x\in X$, we set

$$r(x,\{x_n\})=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x,x_n).$$

The asymptotic radius $r(\{x_n\})$ of $\{x_n\}$ is given by

$$r(\{x_n\})=inf\{r(x,\{x_n\}):x\in X\},$$

and the asymptotic center $A(\{x_n\})$ of $\{x_n\}$ is the set

$$A(\{x_n\})=\{x\in X:r(x,\{x_n\})=r(\{x_n\})\}.$$

It is known from Proposition 7 of [22] that in a complete $CAT(0)$ space, $A(\{x_n\})$ consists of exactly one point. A sequence $\{x_n\}\subset X$ is said to Δ-converge to $x\in X$ if $A(\{x_{n_k}\})=\{x\}$ for every subsequence $\{x_{n_k}\}$ of $\{x_n\}$. The uniqueness of an asymptotic center implies that a $CAT(0)$ space X satisfies Opial’s property, i.e., for given $\{x_n\}\subset X$ such that $\{x_n\}\mathrm{\Delta }$-converges to x and given $y\in X$ with $y\ne x$,

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_n,x)<\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_n,y).$$

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

Lemma 2.4 [21]

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

Lemma 2.5 [23]

If C is a closed convex subset of a complete $CAT(0)$ space and if $\{x_n\}$ is a bounded sequence in C, then the asymptotic center of $\{x_n\}$ 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 $\{x_n\}$ Δ-converges to x and $d(x_n,T{x}_n)\to 0$ imply $x\in C$ and $Tx=x$.

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

A sequence $\{x_n\}$ in the complete $CAT(0)$ space $(X,d)$ w-converges to $x\in X$ if

$$\underset{n\to \mathrm{\infty }}{lim}〈\overrightarrow{x{x}_n},\overrightarrow{xy}〉=0,$$

i.e., ${lim}_{n\to \mathrm{\infty }}(d^2(x_n,x)-d^2(x_n,y)+d^2(x,y))=0$ 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 $CAT(0)$ space, $\{x_n\}$ be a sequence in X and $x\in X$. Then $\{x_n\}\mathrm{\Delta }$-converges to x if and only if ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈\overrightarrow{x{x}_n},\overrightarrow{xy}〉\le 0$ for all $y\in X$.

Lemma 2.8 [[25], Lemma 2.1]

Let $\{a_n\}$ be a sequence of non-negative real numbers satisfying the property

$$a_{n+1}\le (1-{\alpha }_n)a_n+{\alpha }_n{\beta }_n,n\ge 0,$$

where $\{{\alpha }_n\}\subseteq (0,1)$ and $\{{\beta }_n\}\subseteq \mathbb{R}$ such that

• (i) ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (ii) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n\le 0$ or ${\sum }_{n=0}^{\mathrm{\infty }}|{\alpha }_n{\beta }_n|<\mathrm{\infty }$.

Then $\{a_n\}$ 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 $\mathcal{S}:=\{T(t):t\in {\mathbb{R}}^{+}\}$ in $CAT(0)$ spaces. Before proving main results, we need the following two vital lemmas.

Lemma 3.1 Let X be a complete $CAT(0)$ space. Then, for all $u,x,y\in X$, the following inequality holds:

$$d^2(x,u)\le d^2(y,u)+2〈\overrightarrow{xy},\overrightarrow{xu}〉.$$

Proof Using (1.2), we have that

$$\begin{array}{rcl}{d}^2(y,u)-d^2(x,u)-2〈\overrightarrow{yx},\overrightarrow{xu}〉& =& d^2(y,u)-d^2(x,u)-2〈\overrightarrow{yu},\overrightarrow{xu}〉-2〈\overrightarrow{ux},\overrightarrow{xu}〉\\ =& d^2(y,u)-d^2(x,u)-2〈\overrightarrow{yu},\overrightarrow{xu}〉+2d^2(x,u)\\ =& d^2(y,u)+d^2(x,u)-2〈\overrightarrow{yu},\overrightarrow{xu}〉\\ \ge & d^2(y,u)+d^2(x,u)-2d(y,u)d(x,u)\\ =& {(d^2(y,u)-d^2(x,u))}^2\ge 0.\end{array}$$

Therefore we obtain that

$$d^2(x,u)\le d^2(y,u)+2〈\overrightarrow{xy},\overrightarrow{xu}〉,$$

which is the desired result. □

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

• (i) $〈\overrightarrow{u_{t}x},\overrightarrow{u_{t}y}〉\le t〈\overrightarrow{ux},\overrightarrow{u_{t}y}〉+(1-t)〈\overrightarrow{vx},\overrightarrow{u_{t}y}〉$;

• (ii) $〈\overrightarrow{u_{t}x},\overrightarrow{uy}〉\le t〈\overrightarrow{ux},\overrightarrow{uy}〉+(1-t)〈\overrightarrow{vx},\overrightarrow{uy}〉$ and $〈\overrightarrow{u_{t}x},\overrightarrow{vy}〉\le t〈\overrightarrow{ux},\overrightarrow{vy}〉+(1-t)〈\overrightarrow{vx},\overrightarrow{vy}〉$.

Proof (i) It follows from (CN)-inequality (1.1) that

$$\begin{array}{rcl}2〈\overrightarrow{u_{t}x},\overrightarrow{u_{t}y}〉& =& d^2(u_t,y)+d^2(x,u_t)-d^2(x,y)\\ \le & t{d}^2(u,y)+(1-t)d^2(v,y)-t(1-t)d^2(u,v)+d^2(x,u_t)-d^2(x,y)\\ =& t{d}^2(u,y)+t{d}^2(x,u_t)-t{d}^2(u,u_t)-t{d}^2(x,y)\\ +(1-t)d^2(v,y)+(1-t)d^2(x,u_t)-(1-t)d^2(v,u_t)-(1-t)d^2(x,y)\\ +t{d}^2(u,u_t)+(1-t)d^2(v,u_t)-t(1-t)d^2(u,v)\\ =& t[d^2(u,y)+d^2(x,u_t)-d^2(u,u_t)-d^2(x,y)]\\ +(1-t)[d^2(v,y)+d^2(x,u_t)-d^2(v,u_t)-d^2(x,y)]\\ +t{(1-t)}^2{d}^2(u,v)+(1-t)t^2{d}^2(u,v)-t(1-t)d^2(u,v)\\ =& t〈\overrightarrow{ux},\overrightarrow{u_{t}y}〉+(1-t)〈\overrightarrow{vx},\overrightarrow{u_{t}y}〉.\end{array}$$

(ii) The proof is similar to (i). □

For any ${\alpha }_n\in (0,1)$, $t_n\in [0,\mathrm{\infty })$ and a contraction f with coefficient $\alpha \in (0,1)$, define the mapping $G_n:C\to C$ by

$$G_n(x)={\alpha }_{n}f(x)\oplus (1-{\alpha }_n)T(t_n)x,\mathrm{\forall }x\in C.$$

(3.1)

It is not hard to see that $G_n$ is a contraction on C. Indeed, for $x,y\in C$, we have

$$\begin{array}{rcl}d(G_n(x),G_n(y))& =& d({\alpha }_{n}f(x)\oplus (1-{\alpha }_n)T(t_n)x,{\alpha }_{n}f(y)\oplus (1-{\alpha }_n)T(t_n)y)\\ \le & d({\alpha }_{n}f(x)\oplus (1-{\alpha }_n)T(t_n)x,{\alpha }_{n}f(y)\oplus (1-{\alpha }_n)T(t_n)x)\\ +d({\alpha }_{n}f(y)\oplus (1-{\alpha }_n)T(t_n)x,{\alpha }_{n}f(y)\oplus (1-{\alpha }_n)T(t_n)y)\\ \le & {\alpha }_{n}d(f(x),f(y))+(1-{\alpha }_n)d(T(t_n)x,T(t_n)y)\\ \le & {\alpha }_n\alpha d(x,y)+(1-{\alpha }_n)d(x,y)\\ =& (1-{\alpha }_n(1-\alpha ))d(x,y).\end{array}$$

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,

$$x_n={\alpha }_{n}f(x_n)\oplus (1-{\alpha }_n)T(t_n)x_n\text{for all }n\ge 0.$$

(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 $CAT(0)$ space X, and let $\{T(t)\}$ 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,

$$\underset{t\to \mathrm{\infty }}{lim}\underset{x\in B}{sup}d(T(h)(T(t)x),T(t)x)=0.$$

Let f be a contraction on C with coefficient $0<\alpha <1$. Suppose that $t_n\in [0,\mathrm{\infty })$, ${\alpha }_n\in (0,1)$ such that ${lim}_{n\to \mathrm{\infty }}{t}_n=\mathrm{\infty }$, ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and let $\{x_n\}$ be given by (3.2). Then $\{x_n\}$ converges strongly as $n\to \mathrm{\infty }$ to $\tilde{x}$ such that $\tilde{x}=P_{\mathcal{F}}f(\tilde{x})$, which is equivalent to the following variational inequality:

$$〈\overrightarrow{\tilde{x}f\tilde{x}},\overrightarrow{x\tilde{x}}〉\ge 0,\mathrm{\forall }x\in \mathcal{F}.$$

(3.3)

Proof We first show that $\{x_n\}$ is bounded. For any $p\in \mathcal{F}$, we have that

$$\begin{array}{rcl}d(x_n,p)& =& d({\alpha }_{n}f(x_n)\oplus (1-{\alpha }_n)T(t_n)x_n,p)\le {\alpha }_{n}d(f(x_n),p)+(1-{\alpha }_n)d(T(t_n)x_n,p)\\ \le & {\alpha }_{n}d(f(x_n),p)+(1-{\alpha }_n)d(x_n,p).\end{array}$$

Then

$$d(x_n,p)\le d(f(x_n),p)\le d(f(x_n),f(p))+d(f(p),p)\le \alpha d(x_n,p)+d(f(p),p).$$

This implies that

$$d(x_n,p)\le \frac{1}{1-\alpha }d(f(p),p).$$

Hence $\{x_n\}$ is bounded, so are $\{T(t_n)x_n\}$ and $\{f(x_n)\}$. We get that

$$\begin{array}{rcl}d(x_n,T(t_n)x_n)& =& d({\alpha }_{n}f(x_n)\oplus (1-{\alpha }_n)T(t_n)x_n,T(t_n)x_n)\\ \le & {\alpha }_{n}d(f(x_n),T(t_n)x_n)+(1-{\alpha }_n)d(T(t_n)x_n,T(t_n)x_n)\\ \le & {\alpha }_{n}d(f(x_n),T(t_n)x_n)\to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

Since $\{T(t)\}$ is u.a.r. and ${lim}_{n\to \mathrm{\infty }}{t}_n=\mathrm{\infty }$, then for all $h>0$,

$$\underset{n\to \mathrm{\infty }}{lim}d(T(h)(T(t_n)x_n),T(t_n)x_n)\le \underset{n\to \mathrm{\infty }}{lim}\underset{x\in B}{sup}d(T(h)(T(t_n)x),T(t_n)x)=0,$$

where B is any bounded subset of C containing $\{x_n\}$. Hence

$$\begin{array}{rcl}d(x_n,T(h)x_n)& \le & d(x_n,T(t_n)x_n)+d(T(t_n)x_n,T(h)(T(t_n)x_n))\\ +d(T(h)(T(t_n)x_n),T(h)x_n)\\ \le & 2d(x_n,T(t_n)x_n)+d(T(t_n)x_n,T(h)(T(t_n)x_n))\to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

(3.4)

We will show that $\{x_n\}$ contains a subsequence converging strongly to $\tilde{x}$ such that $\tilde{x}=P_{F(T)}f(\tilde{x})$, which is equivalent to the following variational inequality:

$$〈\overrightarrow{\tilde{x}f\tilde{x}},\overrightarrow{x\tilde{x}}〉\ge 0,x\in \mathcal{F}.$$

(3.5)

Since $\{x_n\}$ is bounded, by Lemma 2.4, there exists a subsequence $\{x_{n_j}\}$ of $\{x_n\}$ which Δ-converges to a point $\tilde{x}$, denoted by $\{x_j\}$. We claim that $\tilde{x}\in \mathcal{F}$. Since every $CAT(0)$ space has Opial’s property, for any $h\ge 0$, if $T(h)\tilde{x}\ne \tilde{x}$, we have

$$\begin{array}{rcl}\underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_j,T(h)\tilde{x})& \le & \underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\{d(x_j,T(h)x_j)+d(T(h)x_j,T(h)\tilde{x})\}\\ \le & \underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\{d(x_j,T(h)x_j)+d(x_j,\tilde{x})\}\\ =& \underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_j,\tilde{x})\\ <& \underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_j,T(h)\tilde{x}).\end{array}$$

This is a contradiction, and hence $\tilde{x}\in \mathcal{F}$. So we have the claim. It follows from Lemma 3.2(i) that

$$\begin{array}{rcl}{d}^2(x_j,\tilde{x})& =& 〈\overrightarrow{x_j\tilde{x}},\overrightarrow{x_j\tilde{x}}〉\\ \le & {\alpha }_j〈\overrightarrow{f(x_j)\tilde{x}},\overrightarrow{x_j\tilde{x}}〉+(1-{\alpha }_j)〈\overrightarrow{T(t_j)x_j\tilde{x}},\overrightarrow{x_j\tilde{x}}〉\\ \le & {\alpha }_j〈\overrightarrow{f(x_j)\tilde{x}},\overrightarrow{x_j\tilde{x}}〉+(1-{\alpha }_j)d(T(t_j)x_j,\tilde{x})d(x_j,\tilde{x})\\ \le & {\alpha }_j〈\overrightarrow{f(x_j)\tilde{x}},\overrightarrow{x_j\tilde{x}}〉+(1-{\alpha }_j)d^2(x_j,\tilde{x}).\end{array}$$

It follows that

$$\begin{array}{rcl}{d}^2(x_j,\tilde{x})& \le & 〈\overrightarrow{f(x_j)\tilde{x}},\overrightarrow{x_j\tilde{x}}〉\\ =& 〈\overrightarrow{f(x_j)f(\tilde{x})},\overrightarrow{x_j\tilde{x}}〉+〈\overrightarrow{f(\tilde{x})\tilde{x}},\overrightarrow{x_j\tilde{x}}〉\\ \le & d(f(x_j),f(\tilde{x}))d(x_j,\tilde{x})+〈\overrightarrow{f(\tilde{x})\tilde{x}},\overrightarrow{x_j\tilde{x}}〉\\ \le & \alpha d^2(x_j,\tilde{x})+〈\overrightarrow{f(\tilde{x})\tilde{x}},\overrightarrow{x_j\tilde{x}}〉,\end{array}$$

and thus

$$d^2(x_j,\tilde{x})\le \frac{1}{1-\alpha }〈\overrightarrow{f(\tilde{x})\tilde{x}},\overrightarrow{x_j\tilde{x}}〉.$$

(3.6)

Since $\{x_j\}$ Δ-converges to $\tilde{x}$, by Lemma 2.7, we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈\overrightarrow{f(\tilde{x})\tilde{x}},\overrightarrow{x_j\tilde{x}}〉\le 0.$$

It follows from (3.6) that $\{x_j\}$ converges strongly to $\tilde{x}$. Next, we show that $\tilde{x}$ solves the variational inequality (3.3). Applying Lemma 2.3, for any $q\in \mathcal{F}$,

$$\begin{array}{rcl}{d}^2(x_j,q)& =& d^2({\alpha }_{j}f(x_j)\oplus (1-{\alpha }_j)T(t_j)x_j,q)\\ \le & {\alpha }_j{d}^2(f(x_j),q)+(1-{\alpha }_j)d^2(T(t_j)x_j,q)-{\alpha }_j(1-{\alpha }_j)d^2(f(x_j),T(t_j)x_j)\\ \le & {\alpha }_j{d}^2(f(x_j),q)+(1-{\alpha }_j)d^2(x_j,q)-{\alpha }_j(1-{\alpha }_j)d^2(f(x_j),T(t_j)x_j).\end{array}$$

It implies that

$$d^2(x_j,q)\le d^2(f(x_j),q)-(1-{\alpha }_j)d^2(f(x_j),T(t_j)x_j).$$

Taking the limit through $j\to \mathrm{\infty }$, we can get that

$$d^2(\tilde{x},q)\le d^2(f(\tilde{x}),q)-d^2(f(\tilde{x}),\tilde{x}).$$

Hence

$$0\le \frac{1}{2}[d^2(\tilde{x},\tilde{x})+d^2(f(\tilde{x}),q)-d^2(\tilde{x},q)-d^2(f(\tilde{x}),\tilde{x})]=〈\overrightarrow{\tilde{x}f(\tilde{x})},\overrightarrow{q\tilde{x}}〉,\mathrm{\forall }q\in \mathcal{F}.$$

That is, $\tilde{x}$ solves the inequality (3.3). Finally, we show that the sequence $\{x_n\}$ converges to $\tilde{x}$. Assume that $x_{n_i}\to \hat{x}$, where $i\to \mathrm{\infty }$. By the same argument, we get that $\hat{x}\in \mathcal{F}$ and solves the variational inequality (3.3), i.e.,

$$〈\overrightarrow{\tilde{x}f\tilde{x}},\overrightarrow{\tilde{x}\hat{x}}〉\le 0,$$

(3.7)

and

$$〈\overrightarrow{\hat{x}f\hat{x}},\overrightarrow{\hat{x}\tilde{x}}〉\le 0.$$

(3.8)

Adding up (3.7) and (3.8), we get that

$$\begin{array}{rcl}0& \ge & 〈\overrightarrow{\tilde{x}f(\tilde{x})},\overrightarrow{\tilde{x}\hat{x}}〉-〈\overrightarrow{\hat{x}f(\hat{x})},\overrightarrow{\tilde{x}\hat{x}}〉\\ =& 〈\overrightarrow{\tilde{x}f(\hat{x})},\overrightarrow{\tilde{x}\hat{x}}〉+〈\overrightarrow{f(\hat{x})f(\tilde{x})},\overrightarrow{\tilde{x}\hat{x}}〉-〈\overrightarrow{\hat{x}\tilde{x}},\overrightarrow{\tilde{x}\hat{x}}〉-〈\overrightarrow{\tilde{x}f(\hat{x})},\overrightarrow{\tilde{x}\hat{x}}〉\\ =& 〈\overrightarrow{\tilde{x}\hat{x}},\overrightarrow{\tilde{x}\hat{x}}〉-〈\overrightarrow{f(\hat{x})f(\tilde{x})},\overrightarrow{\hat{x}\tilde{x}}〉\\ \ge & 〈\overrightarrow{\tilde{x}\hat{x}},\overrightarrow{\tilde{x}\hat{x}}〉-d(f(\hat{x}),f(\tilde{x}))d(\hat{x},\tilde{x})\\ \ge & d^2(\tilde{x},\hat{x})-\alpha d(\hat{x},\tilde{x})d(\hat{x},\tilde{x})\\ \ge & d^2(\tilde{x},\hat{x})-\alpha d^2(\hat{x},\tilde{x})\\ \ge & (1-\alpha )d^2(\tilde{x},\hat{x}).\end{array}$$

Since $0<\alpha <1$, we have that $d(\tilde{x},\hat{x})=0$, and so $\tilde{x}=\hat{x}$. Hence the sequence $x_n$ converges strongly to $\tilde{x}$, 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 $CAT(0)$ space X, and let $\{T(t)\}$ 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,

$$\underset{t\to \mathrm{\infty }}{lim}\underset{x\in B}{sup}d(T(h)(T(t)x),T(t)x)=0.$$

Let u be any element in C. Suppose $t_n\in [0,\mathrm{\infty })$, ${\alpha }_n\in (0,1)$ such that ${lim}_{n\to \mathrm{\infty }}{t}_n=\mathrm{\infty }$ and ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and let $\{x_n\}$ be given by

$$x_n={\alpha }_{n}u\oplus (1-{\alpha }_n)T(t_n)x_n.$$

Then $\{x_n\}$ converges strongly as $n\to \mathrm{\infty }$ to $\tilde{x}$ such that $\tilde{x}=P_{\mathcal{F}}\tilde{x}$, which is equivalent to the following variational inequality:

$$〈\overrightarrow{\tilde{x}u},\overrightarrow{x\tilde{x}}〉\ge 0,x\in \mathcal{F}.$$

(3.9)

Theorem 3.5 Let C be a closed convex subset of a complete $CAT(0)$ space X, and let $\{T(t)\}$ 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,

$$\underset{t\to \mathrm{\infty }}{lim}\underset{x\in B}{sup}d(T(h)(T(t)x),T(t)x)=0.$$

Let f be a contraction on C with coefficient $0<\alpha <1$. Suppose that $t_n\in [0,\mathrm{\infty })$, ${\alpha }_n\in (0,1)$, $x_0\in C$, and $\{x_n\}$ is given by

$$x_{n+1}={\alpha }_{n}f(x_n)\oplus (1-{\alpha }_n)T(t_n)x_n,\mathrm{\forall }n\ge 0,$$

(3.10)

where $\{{\alpha }_n\}\subset (0,1)$ satisfies the following conditions:

• (i) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$;

• (ii) ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$ and

• (iii) ${lim}_{n\to \mathrm{\infty }}{t}_n=\mathrm{\infty }$.

Then $\{x_n\}$ converges strongly as $n\to \mathrm{\infty }$ to $\tilde{x}$ such that $\tilde{x}=P_{\mathcal{F}}f(\tilde{x})$, which is equivalent to the variational inequality (3.3).

Proof We first show that the sequence $\{x_n\}$ is bounded. For any $p\in \mathcal{F}$, we have that

$$\begin{array}{rcl}d(x_{n+1},p)& =& d({\alpha }_{n}f(x_n)\oplus (1-{\alpha }_n)T(t_n)x_n,p)\\ \le & {\alpha }_{n}d(f(x_n),p)+(1-{\alpha }_n)d(T(t_n)x_n,p)\\ \le & {\alpha }_n(d(f(x_n),f(p))+d(f(p),p))+(1-{\alpha }_n)d(T(t_n)x_n,p)\\ \le & max\{d(x_n,p),\frac{1}{1-\alpha }d(f(p),p)\}.\end{array}$$

By induction, we have

$$d(x_n,p)\le max\{d(x_0,p),\frac{1}{1-\alpha }d(f(p),p)\}$$

for all $n\in \mathbb{N}$. Hence $\{x_n\}$ is bounded, so are $\{T(t_n)x_n\}$ and $\{f(x_n)\}$. Using the assumption that ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, we get that

$$d(x_{n+1},T(t_n)x_n)\le {\alpha }_{n}d(f(x_n),T(t_n)x_n)\to 0\text{as }n\to \mathrm{\infty }.$$

Since $\{T(t)\}$ is u.a.r. and ${lim}_{n\to \mathrm{\infty }}{t}_n=\mathrm{\infty }$, then for all $h\ge 0$,

$$\underset{n\to \mathrm{\infty }}{lim}d(T(h)(T(t_n)x_n),T(t_n)x_n)\le \underset{n\to \mathrm{\infty }}{lim}\underset{x\in B}{sup}d(T(h)(T(t_n)x),T(t_n)x)=0,$$

where B is any bounded subset of C containing $\{x_n\}$. Hence

$$\begin{array}{r}d(x_{n+1},T(h)x_{n+1})\\ \le d(x_{n+1},T(t_n)x_n)+d(T(t_n)x_n,T(h)(T(t_n)x_n))\\ +d(T(h)(T(t_n)x_n),T(h)x_{n+1})\\ \le 2d(x_{n+1},T(t_n)x_n)+d(T(t_n)x_n,T(h)(T(t_n)x_n))\to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

(3.11)

Let $\{z_m\}$ be a sequence in C such that

$$z_m={\alpha }_{m}f(z_m)\oplus (1-{\alpha }_m)T(t_m)z_m.$$

It follows from Theorem 3.3 that $\{z_m\}$ converges strongly as $m\to \mathrm{\infty }$ to a fixed point $\tilde{x}\in \mathcal{F}$, which solves the variational inequality (3.3). Now, we claim that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈\overrightarrow{f(\tilde{x})\tilde{x}},\overrightarrow{x_{n+1}\tilde{x}}〉\le 0.$$

It follows from Lemma 3.2(i) that

$$\begin{array}{rcl}{d}^2(z_m,x_{n+1})& =& 〈\overrightarrow{z_m{x}_{n+1}},\overrightarrow{z_m{x}_{n+1}}〉\\ \le & {\alpha }_m〈\overrightarrow{f(z_m)x_{n+1}},\overrightarrow{z_m{x}_{n+1}}〉+(1-{\alpha }_m)〈\overrightarrow{T(t_m)z_m{x}_{n+1}},\overrightarrow{z_m{x}_{n+1}}〉\\ =& {\alpha }_m〈\overrightarrow{f(z_m)f(\tilde{x})},\overrightarrow{z_m{x}_{n+1}}〉+{\alpha }_m〈\overrightarrow{f(\tilde{x})\tilde{x}},\overrightarrow{z_m{x}_{n+1}}〉+{\alpha }_m〈\overrightarrow{\tilde{x}{z}_m},\overrightarrow{z_m{x}_{n+1}}〉\\ +{\alpha }_m〈\overrightarrow{z_m{x}_{n+1}},\overrightarrow{z_m{x}_{n+1}}〉+(1-{\alpha }_m)〈\overrightarrow{T(t_m)z_{m}T(t_m)x_{n+1}},\overrightarrow{z_m{x}_{n+1}}〉\\ +(1-{\alpha }_m)〈\overrightarrow{T(t_m)x_{n+1}{x}_{n+1}},\overrightarrow{z_m{x}_{n+1}}〉\\ \le & {\alpha }_m\alpha d(z_m,\tilde{x})d(z_m,x_{n+1})+{\alpha }_m〈\overrightarrow{f(\tilde{x})\tilde{x}},\overrightarrow{z_m{x}_{n+1}}〉+{\alpha }_{m}d(\tilde{x},z_m)d(z_m,x_{n+1})\\ +{\alpha }_m{d}^2(z_m,x_{n+1})+(1-{\alpha }_m)d^2(z_m,x_{n+1})\\ +(1-{\alpha }_m)d(T(t_m)x_{n+1},x_{n+1})d(z_m,x_{n+1})\\ \le & {\alpha }_m\alpha d(z_m,\tilde{x})M+{\alpha }_m〈\overrightarrow{f(\tilde{x})\tilde{x}},\overrightarrow{z_m{x}_{n+1}}〉+{\alpha }_{m}d(\tilde{x},z_m)M+{\alpha }_m{d}^2(z_m,x_{n+1})\\ +(1-{\alpha }_m)d^2(z_m,x_{n+1})+(1-{\alpha }_m)d(T(t_m)x_{n+1},x_{n+1})M\\ \le & d^2(z_m,x_{n+1})+{\alpha }_m\alpha d(z_m,\tilde{x})M+{\alpha }_{m}d(\tilde{x},z_m)M+d(T(t_m)x_{n+1},x_{n+1})M\\ +{\alpha }_m〈\overrightarrow{f(\tilde{x})\tilde{x}},\overrightarrow{z_m{x}_{n+1}}〉,\end{array}$$

where $M\ge {sup}_{m,n\ge 1}\{d(z_m,x_n)\}$. This implies that

$$〈\overrightarrow{f(\tilde{x})\tilde{x}},\overrightarrow{x_{n+1}{z}_m}〉\le (1+\alpha )d(z_m,\tilde{x})M+\frac{d(T(t_m)x_{n+1},x_{n+1})}{{\alpha }_m}M.$$

(3.12)

Taking the upper limit as $n\to \mathrm{\infty }$ first, and then $m\to \mathrm{\infty }$, inequality (3.12) yields that

$$\underset{m\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈\overrightarrow{f(\tilde{x})\tilde{x}},\overrightarrow{x_{n+1}{z}_m}〉\le 0.$$

(3.13)

Since

$$\begin{array}{rcl}〈\overrightarrow{f(\tilde{x})\tilde{x}},\overrightarrow{x_{n+1}\tilde{x}}〉& =& 〈\overrightarrow{f(\tilde{x})\tilde{x}},\overrightarrow{x_{n+1}{z}_m}〉+〈\overrightarrow{f(\tilde{x})\tilde{x}},\overline{z_m\tilde{x}}〉\\ \le & 〈\overrightarrow{f(\tilde{x})\tilde{x}},\overrightarrow{x_{n+1}{z}_m}〉+d(f(\tilde{x}),\tilde{x})d(z_m,\tilde{x}).\end{array}$$

Thus, by taking the upper limit as $n\to \mathrm{\infty }$ first, and then $m\to \mathrm{\infty }$ the last inequality, it follows from $z_m\to \tilde{x}$ and (3.13) that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈\overrightarrow{f(\tilde{x})\tilde{x}},\overrightarrow{x_{n+1}\tilde{x}}〉\le 0.$$

Finally, we prove that $x_n\to \tilde{x}$ as $n\to \mathrm{\infty }$. For any $n\in \mathbb{N}$, we set $y_n={\alpha }_n\tilde{x}\oplus (1-{\alpha }_n)T(t_n)x_n$. It follows from Lemma 3.1 and Lemma 3.2(i), (ii) that

$$\begin{array}{rcl}{d}^2(x_{n+1},\tilde{x})& \le & d^2(y_n,\tilde{x})+2〈\overrightarrow{x_{n+1}{y}_n},\overrightarrow{x_{n+1}\tilde{x}}〉\\ \le & {({\alpha }_{n}d(\tilde{x},\tilde{x})+(1-{\alpha }_n)d(T(t_n)x_n,\tilde{x}))}^2\\ +2[{\alpha }_n〈\overrightarrow{f(x_n)y_n},\overrightarrow{x_{n+1}\tilde{x}}〉+(1-{\alpha }_n)〈\overrightarrow{T(t_n)x_n{y}_n},\overrightarrow{x_{n+1}\tilde{x}}〉]\\ \le & {(1-{\alpha }_n)}^2{d}^2(x_n,\tilde{x})+2[{\alpha }_n{\alpha }_n〈\overrightarrow{f(x_n)\tilde{x}},\overrightarrow{x_{n+1}\tilde{x}}〉+{\alpha }_n(1-{\alpha }_n)〈\overrightarrow{f(x_n)T(t_n)x_n},\overrightarrow{x_{n+1}\tilde{x}}〉\\ +(1-{\alpha }_n){\alpha }_n〈\overrightarrow{T(t_n)x_n\tilde{x}},\overrightarrow{x_{n+1}\tilde{x}}〉+(1-{\alpha }_n)(1-{\alpha }_n)〈\overrightarrow{T(t_n)x_{n}T(t_n)x_n},\overrightarrow{x_{n+1}\tilde{x}}〉]\\ \le & {(1-{\alpha }_n)}^2{d}^2(x_n,\tilde{x})+2[{\alpha }_n{\alpha }_n〈\overrightarrow{f(x_n)\tilde{x}},\overrightarrow{x_{n+1}\tilde{x}}〉+{\alpha }_n(1-{\alpha }_n)〈\overrightarrow{f(x_n)T(t_n)x_n},\overrightarrow{x_{n+1}\tilde{x}}〉\\ +(1-{\alpha }_n){\alpha }_n〈\overrightarrow{T(t_n)x_n\tilde{x}},\overrightarrow{x_{n+1}\tilde{x}}〉+{(1-{\alpha }_n)}^{2}d(T(t_n)x_n,T(t_n)x_n)d(x_{n+1}\tilde{x})]\\ =& {(1-{\alpha }_n)}^2{d}^2(x_n,\tilde{x})+2[{\alpha }_n^2〈\overrightarrow{f(x_n)\tilde{x}},\overrightarrow{x_{n+1}\tilde{x}}〉+{\alpha }_n(1-{\alpha }_n)〈\overrightarrow{f(x_n)\tilde{x}},\overrightarrow{x_{n+1}\tilde{x}}〉]\\ =& {(1-{\alpha }_n)}^2{d}^2(x_n,\tilde{x})+2{\alpha }_n〈\overrightarrow{f(x_n)\tilde{x}},\overrightarrow{x_{n+1}\tilde{x}}〉\\ =& {(1-{\alpha }_n)}^2{d}^2(x_n,\tilde{x})+2{\alpha }_n〈\overrightarrow{f(x_n)f(\tilde{x})},\overrightarrow{x_{n+1}\tilde{x}}〉+2{\alpha }_n〈\overrightarrow{f(\tilde{x})\tilde{x}},\overrightarrow{x_{n+1}\tilde{x}}〉\\ \le & {(1-{\alpha }_n)}^2{d}^2(x_n,\tilde{x})+2{\alpha }_n\alpha d(x_n,\tilde{x})d(x_{n+1},\tilde{x})+2{\alpha }_n〈\overrightarrow{f(\tilde{x})\tilde{x}},\overrightarrow{x_{n+1}\tilde{x}}〉\\ \le & {(1-{\alpha }_n)}^2{d}^2(x_n,\tilde{x})+{\alpha }_n\alpha (d^2(x_n,\tilde{x})+d^2(x_{n+1},\tilde{x}))+2{\alpha }_n〈\overrightarrow{f(\tilde{x})\tilde{x}},\overrightarrow{x_{n+1}\tilde{x}}〉,\end{array}$$

which implies that

$$\begin{array}{rcl}{d}^2(x_{n+1},\tilde{x})& \le & \frac{1-(2-\alpha ){\alpha }_n+{\alpha }_n^2}{1-\alpha {\alpha }_n}{d}^2(x_n,\tilde{x})+\frac{2{\alpha }_n}{1-\alpha {\alpha }_n}〈\overrightarrow{f(\tilde{x})\tilde{x}},\overrightarrow{x_{n+1}\tilde{x}}〉\\ \le & \frac{1-(2-\alpha ){\alpha }_n}{1-\alpha {\alpha }_n}{d}^2(x_n,\tilde{x})+\frac{2{\alpha }_n}{1-\alpha {\alpha }_n}〈\overrightarrow{f(\tilde{x})\tilde{x}},\overrightarrow{x_{n+1}\tilde{x}}〉+{\alpha }_n^{2}M,\end{array}$$

where $M\ge {sup}_{n\ge 0}\{d^2(x_n,\tilde{x})\}$. It then follows that

$$d^2(x_{n+1},\tilde{x})\le (1-{\alpha }_n^{\prime })d^2(x_n,\tilde{x})+{\alpha }_n^{\prime }{\beta }_n^{\prime },$$

where

$${\alpha }_n^{\prime }=\frac{2(1-\alpha ){\alpha }_n}{1-\alpha {\alpha }_n}\text{and}{\beta }_n^{\prime }=\frac{(1-\alpha {\alpha }_n){\alpha }_n}{2(1-\alpha )}M+\frac{1}{(1-\alpha )}〈\overrightarrow{f(\tilde{x})\tilde{x}},\overrightarrow{x_{n+1}\tilde{x}}〉.$$

Applying Lemma 2.8, we can conclude that $x_n\to \tilde{x}$. 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 $CAT(0)$ space X, and let $\{T(t)\}$ 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,

$$\underset{t\to \mathrm{\infty }}{lim}\underset{x\in B}{sup}d(T(h)(T(t)x),T(t)x)=0.$$

Suppose that $t_n\in [0,\mathrm{\infty })$, ${\alpha }_n\in (0,1)$, $x_0\in C$ and $\{x_n\}$ is given by

$$x_{n+1}={\alpha }_{n}u\oplus (1-{\alpha }_n)T(t_n)x_n,\mathrm{\forall }n\ge 0,$$

(3.14)

where $\{{\alpha }_n\}\subset (0,1)$ satisfies the following conditions:

• (i) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$;

• (ii) ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$ and

• (iii) ${lim}_{n\to \mathrm{\infty }}{t}_n=\mathrm{\infty }$.

Then $\{x_n\}$ converges strongly as $n\to \mathrm{\infty }$ to $\tilde{x}$ such that $\tilde{x}=P_{\mathcal{F}}\tilde{x}$, which is equivalent to the variational inequality (3.9).