Nonexpansive semigroups in $CAT(\kappa )$ spaces

Abstract

The main purpose of this paper is to study Browder type convergence theorems for a nonexpansive semigroup with geometric approaches in a $CAT(\kappa )$ space. Besides, we determine a necessary and sufficient condition for convergence of a Browder type iteration associated to a uniformly asymptotically regular nonexpansive semigroup on the unit sphere in an infinite-dimensional Hilbert space.

MSC:47H20, 47H10.

1 Introduction

Let $(X,d)$ be a metric space, C a closed convex subset of X and $T:C\to C$ a mapping. Recall that T is nonexpansive on C if $d(Tx,Ty)\le d(x,y)$, for all $x,y\in C$. We denote by $\mathfrak{F}(T)$ the fixed point set of the mapping T. A one-parameter family $\mathfrak{S}=\{T(t):t\ge 0\}$ of self-mappings of C is called a strongly continuous nonexpansive semigroup on C if the following conditions are satisfied:

• (i) for each $t\ge 0$, $T(t)$ is a nonexpansive mapping on C;

• (ii) $T(0)x=x$, for all $x\in C$;

• (iii) $T(s+t)=T(s)\circ T(t)$, for all $s,t\ge 0$;

• (iv) for each $x\in C$, the mapping $t\mapsto T(t)x$ from $[0,\mathrm{\infty })$ into C is continuous.

Let $\mathfrak{F}(\mathfrak{S})$ denote the common fixed point set of all mappings in $\mathfrak{S}$.

There have been considerably many interesting results of iterative methods for approximating fixed points of nonexpansive mappings, nonexpansive semigroups, and their generalizations which solve some variational inequalities problems due to their various applications in several physical problems, such as in operations research, economics, and engineering design; see, e.g., [1–4] and the references therein.

Suppose that X is a real Hilbert space and u is an arbitrary point of X. If T is nonexpansive on C, then for each $\alpha \in (0,1)$ there exists a unique $x_{\alpha }\in C$ such that $x_{\alpha }=\alpha u+(1-\alpha )T{x}_{\alpha }$ because the mapping $x\mapsto \alpha u+(1-\alpha )Tx$ is a contraction. In 1967, Browder [5] was the pioneer to consider an implicit scheme and prove the following strong convergence theorem of this algorithm in a Hilbert space.

Theorem 1.1 Let C be a bounded closed convex subset of a Hilbert space and T a nonexpansive mapping on C. Let u be an arbitrary point of C and define $x_{\alpha }\in C$ by

$$x_{\alpha }=\alpha u+(1-\alpha )T{x}_{\alpha },\mathit{\text{for }}\alpha \in (0,1).$$

Then as $\alpha \to 0$, $\{x_{\alpha }\}$ converges strongly to a point of $\mathfrak{F}(T)$ nearest to u.

The extension work of Browder’s type convergence theorems has been tremendously studied for not only one single nonexpansive mapping, but, most significantly, a semigroup of nonexpansive mappings; see, e.g., [6, 7] and the references therein.

This paper is devoted to studying Browder’s type iterations for a nonexpansive semigroup in a $CAT(\kappa )$ space, where $\kappa \in \mathbb{R}$, which is a specific type of metric space. Intuitively, triangles in a $CAT(\kappa )$ space are ‘slimmer’ than corresponding ‘model triangles’ in a standard space of constant curvature κ (see Section 2). Complete $CAT(0)$ spaces are often called Hadamard spaces. A few recent new convergence results of classical iterations on $CAT(\kappa )$ spaces with $\kappa >0$ are obtained; see, e.g., [8–10] and the references therein.

In [11], Dhompongsa et al. extended Suzuki’s result [7], Theorem 3] on common fixed points of a nonexpansive semigroup in a Hilbert space to a complete $CAT(0)$ space.

Theorem 1.2 ([11], Theorem 4.4])

Let C be a bounded closed convex subset of a $CAT(0)$ space, $\mathfrak{S}=\{T(t):t\ge 0\}$ a strongly continuous nonexpansive semigroup on C, and two sequences $\{{\alpha }_n\}\subset (0,1)$, $\{t_n\}\subset (0,\mathrm{\infty })$ such that ${lim}_{n\to \mathrm{\infty }}{t}_n={lim}_{n\to \mathrm{\infty }}{\alpha }_n/t_n=0$. Choose arbitrarily a point $x_0\in C$ and for each $n\in \mathbb{N}$ let $x_n$ be the fixed point of the mapping $x\mapsto {\alpha }_n{x}_0\oplus (1-{\alpha }_n)T(t_n)x$. Then $\mathfrak{F}(\mathfrak{S})\ne \mathrm{\varnothing }$ and $\{x_n\}$ converges strongly to a point of $\mathfrak{F}(\mathfrak{S})$ nearest to $x_0$.

In 2010, Acedo and Suzuki [6] proved a Browder type convergence theorem for uniformly asymptotically regular ($UAR$ for short) nonexpansive semigroups (see Section 2) in Hilbert spaces under a weaker condition on $\{{\alpha }_n\}$ and $\{t_n\}$ than that in Theorem 1.2.

Theorem 1.3 ([6], Theorem 2.3])

Let C be a closed convex subset of a Hilbert space, $\mathfrak{S}=\{T(t):t\ge 0\}$ a $UAR$ and strongly continuous nonexpansive semigroup on C such that $\mathfrak{F}(\mathfrak{S})\ne \mathrm{\varnothing }$, and two sequences $\{{\alpha }_n\}\subset (0,1)$, $\{t_n\}\subset [0,\mathrm{\infty })$ such that ${lim}_{n\to \mathrm{\infty }}{\alpha }_n={lim}_{n\to \mathrm{\infty }}{\alpha }_n/t_n=0$. Fix $x_0\in C$ and define a sequence $\{x_n\}$ in C by

$$x_n={\alpha }_n{x}_0+(1-{\alpha }_n)T(t_n)x_n.$$

Then $\{x_n\}$ converges strongly to a point of $\mathfrak{F}(\mathfrak{S})$ nearest to $x_0$.

The preceding two theorems lead naturally to the question of whether or not they can be extended to a $CAT(\kappa )$ space with $\kappa >0$. The purpose of this article is to investigate this question with the geometric approaches in $CAT(\kappa )$ spaces.

This paper is organized as follows. In Section 2 we recall the definition of geodesic metric spaces and summarize some useful lemmas and the main properties of $CAT(\kappa )$ spaces. In Section 3 we present some technical results about Δ-convergence of a sequence in a complete $CAT(1)$ space. In Section 4 we establish our main results (Theorems 4.2, 4.3, 4.4) of Browder’s iterations for nonexpansive semigroups in $CAT(\kappa )$ spaces and conclude that Theorems 1.2 and 1.3 can be generalized to $CAT(\kappa )$ spaces under the same respective conditions on the coefficients $\{{\alpha }_n\}$ and $\{t_n\}$. It is noteworthy that, however, without the $UAR$ assumption, the sequence $\{x_n\}$ established in Theorem 1.2 is not necessarily convergent to the nearest point of $\mathfrak{F}(\mathfrak{S})$ to $x_0$ if ${lim}_{n\to \mathrm{\infty }}{t}_n=\hat{t}\in (0,\mathrm{\infty }]$ even when ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ (and therefore ${lim}_{n\to \mathrm{\infty }}{\alpha }_n/t_n=0$); see Example 4.5. Furthermore, we determine a necessary and sufficient condition for a Browder type convergence theorem associated to a nonexpansive semigroup on the unit sphere in an infinite-dimensional Hilbert space. We also propose an open problem in Section 5.

2 Preliminaries

Let $(X,d)$ be a metric space. For any subset E of X and $x\in X$, the diameter of E and the distance from x to E are defined, respectively, by

$$\begin{array}{r}diamE=sup\{d(x,y):x,y\in E\},\\ d(x,E)=inf\{d(x,y):y\in E\}.\end{array}$$

We always denote the open ball and the closed ball centered at x with radius $r>0$ by $B(x,r)$ and $\overline{B}(x,r)$, respectively.

Let C be a closed convex subset of X and let $\mathfrak{S}=\{T(t):t\ge 0\}$ be a family of self-mappings of C. Then $\mathfrak{S}$ is called

• (i) asymptotically regular on C if for any $h\ge 0$ and any $x\in C$,

  $$\underset{t\to \mathrm{\infty }}{lim}d(T(h)T(t)x,T(t)x)=0;$$

• (ii) uniformly asymptotically regular (in short $UAR$) on C if for any $h\ge 0$ and any bounded subset D of C,

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

For $x,y\in X$, a geodesic path joining x to y (or a geodesic from x to y) is an isometric mapping $\gamma :[a,b]\subset \mathbb{R}\to X$ such that $\gamma (a)=x$, $\gamma (b)=y$, i.e., $d(\gamma (t),\gamma (t^{\prime }))=|t-t^{\prime }|$, for all $t,t^{\prime }\in [a,b]$. Therefore $d(x,y)=b-a$. The image of γ is called a geodesic (segment) from x to y and we shall denote a definite choice of this geodesic segment by $[x,y]$. We remark that composing γ with a translation (this is still an isometry), one can always choose the interval $[a,b]$ to be $[0,\ell ]$, where $\ell =b-a$. A point $z=\gamma (t)$ in the geodesic $[x,y]$ will be written as $z=(1-\lambda )x\oplus \lambda y$, where $\lambda =(t-a)/(b-a)$, and so $d(z,x)=\lambda d(x,y)$ and $d(z,y)=(1-\lambda )d(x,y)$.

Let $r>0$. The metric space $(X,d)$ is said to be

• (i) a geodesic (metric) space if any two points in X are joined by a geodesic;

• (ii) uniquely geodesic if there is exactly one geodesic joining x to y for all $x,y\in X$;

• (iii) r-geodesic space if any two points $x,y\in X$ with $d(x,y)<r$ are joined by a geodesic;

• (iv) r-uniquely geodesic if any two points $x,y\in X$ with $d(x,y)<r$ are joined by a unique geodesic in X.

A subset C of X is convex if every pair of points $x,y\in C$ can be joined by a geodesic in X and the image of every such geodesic is contained in C. If this condition holds for all points $x,y\in C$ with $d(x,y)<r$, then C is said to be r-convex.

The n-dimensional sphere ${\mathbb{S}}^n$ is the set $\{x=(x_1,\dots ,x_{n+1})\in {\mathbb{R}}^{n+1}:(x\mid x)=1\}$, where $(\cdot \mid \cdot )$ denote the Euclidean scalar product. It is endowed with the following metric: Let $d_{{\mathbb{S}}^n}:{\mathbb{S}}^n\times {\mathbb{S}}^n\to \mathbb{R}$ be the function that assigns to each pair $(A,B)\in {\mathbb{S}}^n\times {\mathbb{S}}^n$ the unique number $d(A,B)\in [0,\pi ]$ such that

$$cos{d}_{{\mathbb{S}}^n}(A,B)=(A\mid B).$$

Then $d_{{\mathbb{S}}^n}$ is a metric [12], I.2.1].

Definition 2.1 Given a real number κ, we denote by $M_{\kappa }^n$ the following metric spaces:

• (i) if $\kappa =0$, then $M_0^n$ is the Euclidean space ${\mathbb{R}}^n$;

• (ii) if $\kappa >0$, then $M_{\kappa }^n$ is obtained from the sphere ${\mathbb{S}}^n$ by multiplying the distance function by $1/\sqrt{\kappa }$;

• (iii) if $\kappa <0$, then $M_{\kappa }^n$ is obtained from the sphere ${\mathbb{H}}^n$ by multiplying the distance function by $1/\sqrt{-\kappa }$, where ${\mathbb{H}}^n$ is the hyperbolic n-space.

It is well known that $M_{\kappa }^n$ is a geodesic metric space. If $\kappa \le 0$, then $M_{\kappa }^n$ is uniquely geodesic. If $\kappa >0$, then $M_{\kappa }^n$ is $\pi /\sqrt{\kappa }$-uniquely geodesic, and any open ball (respectively, closed) ball of radius $\le \pi /(2\sqrt{\kappa })$ (respectively, $<\pi /(2\sqrt{\kappa })$) in X is convex [12], I.2.11]. The diameter of $M_{\kappa }^n$ will be denoted $D_{\kappa }$ and thus $D_{\kappa }$ is $\pi /\sqrt{\kappa }$ if $\kappa >0$, and ∞ otherwise.

Given two distinct points $A,B\in {\mathbb{S}}^n$ with $d(A,B)=\ell <\pi$ there is a natural way to parameterize a unique geodesic joining A to B: consider the path $c(t)=(cost)A+(sint)u$, $t\in [0,\ell ]$, where the initial vector $u\in {\mathbb{R}}^{n+1}$ is the unit vector in the direction of $B-(A\mid B)A$. We shall refer to the image of c as a minimal great arc joining A to B.

The spherical angle between two minimal great arcs issuing from a point of ${\mathbb{S}}^n$, with the initial vectors u and v, say, is the unique number $\alpha \in [0,\pi ]$ such that $cos\alpha =(u\mid v)$. A spherical triangle △ in ${\mathbb{S}}^n$ consists of a choice of three distinct points (its vertices) $A,B,C\in {\mathbb{S}}^n$, and three minimal great arcs (its sides), one joining each other of vertices. The vertex angle at C is defined to be the spherical angle between the sides of △ joining C to A and C to B.

Proposition 2.2 (The Spherical Law of Cosines [12], I.2.13])

Let a geodesic triangle in $M_{\kappa }^n$ ($\kappa >0$) have sides a, b, c and angles α, β, γ at the vertices opposite to the sides of length a, b, c, respectively. Then

$$cos(\sqrt{\kappa }c)=cos(\sqrt{\kappa }a)cos(\sqrt{\kappa }b)+sin(\sqrt{\kappa }a)sin(\sqrt{\kappa }b)cos\gamma .$$

In particular, fixing a, b and κ, c is a strictly increasing function of γ, varying from $|a-b|$ to $a+b$ as γ varies from 0 to π.

A geodesic triangle △ in a metric space X consists of three points $p,q,r\in X$, its vertices, and a choice of three geodesic segments $[p,q]$, $[q,r]$, $[r,p]$ joining them, its sides. Such a geodesic triangle will be denoted $\mathrm{△}([p,q],[q,r],[r,p])$ or (less accurately if X is not uniquely geodesic) $\mathrm{△}(p,q,r)$. If a point $x\in X$ lies in the union of $[p,q]$, $[q,r]$ and $[r,p]$, then we write $x\in \mathrm{△}$.

A triangle $\overline{\mathrm{△}}=\mathrm{△}(\overline{p},\overline{q},\overline{r})$ in $M_{\kappa }^2$ is called a comparison triangle for $\mathrm{△}=\mathrm{△}([p,q],[q,r],[r,p])$ in X if $d_{M_{\kappa }^2}(\overline{p},\overline{q})=d(p,q)$, $d_{M_{\kappa }^2}(\overline{q},\overline{r})=d(q,r)$ and $d_{M_{\kappa }^2}(\overline{r},\overline{p})=d(r,p)$. Such a triangle $\overline{\mathrm{△}}\subset M_{\kappa }^2$ always exists if the perimeter $d(p,q)+d(q,r)+d(r,p)$ of △ is less than $2D_{\kappa }$; it is unique up to an isometry of $M_{\kappa }^2$ [12], I.2.14]. A point $\overline{x}\in [\overline{q},\overline{r}]$ is called a comparison point for $x\in [q,r]$ if $d_{M_{\kappa }^2}(\overline{q},\overline{x})=d(q,x)$.

A geodesic triangle △ in X is said to satisfy the $CAT(\kappa )$ inequality if, given a comparison triangle $\overline{\mathrm{△}}\subset M_{\kappa }^2$ for △, for all $x,y\in \mathrm{△}$,

$$d(x,y)\le d_{M_{\kappa }^2}(\overline{x},\overline{y}),$$

where $\overline{x},\overline{y}\in \overline{\mathrm{△}}$ are respective comparison points of x, y.

Definition 2.3 If $\kappa \le 0$, then X is called a $CAT(\kappa )$ space if X is a geodesic space all of whose geodesic triangles satisfy the $CAT(\kappa )$ inequality.

If $\kappa >0$, then X is called a $CAT(\kappa )$ space if X is $D_{\kappa }$-geodesic and all geodesic triangles in X of perimeter less than $2D_{\kappa }$ satisfy the $CAT(\kappa )$ inequality.

In particular, Hilbert spaces are $CAT(0)$. A $CAT(\kappa )$ space is a $CAT({\kappa }^{\prime })$ space for every ${\kappa }^{\prime }\ge \kappa$. A $CAT(\kappa )$ space X is $(D_{\kappa }\text{-})$uniquely geodesic (if $\kappa >0$) and any open (respectively, closed) ball of radius $\le D_{\kappa }/2$ (respectively, $<D_{\kappa }/2$) in X is convex [12], II.1.4].

Lemma 2.4 Let $(X,d)$ be a $CAT(\kappa )$ space and let $\alpha ,\beta \in [0,1]$. Then

• (i) [12], II.2. Exercise 2.3(1)] for $p\in X$ and $x,y\in B(p,D_{\kappa }/2)$, we have

  $$d(\alpha x\oplus (1-\alpha )y,p)\le \alpha d(x,p)+(1-\alpha )d(y,p);$$

• (ii) for $x,y\in X$, we have

  $$d(\alpha x\oplus (1-\alpha )y,\beta x\oplus (1-\beta )y)=|\alpha -\beta |d(x,y);$$

• (iii) [10], Lemma 3.3] if $\kappa >0$, for $x,y,z\in X$ with $max\{d(x,y),d(y,z),d(x,z)\}<M\le D_{\kappa }/2$, we have

  $$d(\alpha x\oplus (1-\alpha )y,\alpha x\oplus (1-\alpha )z)\le \frac{sin[(1-\alpha )M]}{sinM}d(y,z).$$

Let p, q, r be three distinct points of X with $d(p,q)+d(q,r)+d(r,p)<2D_{\kappa }$. The κ-comparison angle between q and r at p, denoted ${\mathrm{\angle }}_p^{(\kappa )}(q,r)$, is the angle at $\overline{p}$ in a comparison triangle $\mathrm{△}(\overline{p},\overline{q},\overline{r})\subset M_{\kappa }^2$ for $\mathrm{△}(p,q,r)$.

Let $\gamma :[0,a]\to X$ and let ${\gamma }^{\prime }:[0,a^{\prime }]\to X$ be two geodesic paths with $\gamma (0)={\gamma }^{\prime }(0)$. Given $t\in (0,a]$ and $t^{\prime }\in (0,a^{\prime }]$, we consider the comparison triangle $\overline{\mathrm{△}}(\gamma (0),\gamma (t),{\gamma }^{\prime }(t^{\prime }))$ and the κ-comparison angle ${\mathrm{\angle }}_{\gamma (0)}^{(\kappa )}(\gamma (t),{\gamma }^{\prime }(t^{\prime }))$. The (Alexandrov) angle or the upper angle between the geodesic paths γ and ${\gamma }^{\prime }$ is the number $\mathrm{\angle }(\gamma ,{\gamma }^{\prime })\in [0,\pi ]$ defined by

$$\mathrm{\angle }(\gamma ,{\gamma }^{\prime })=\underset{t,t^{\prime }\to 0}{lim\hspace{0.17em}sup}{\mathrm{\angle }}_{\gamma (0)}^{(\kappa )}(\gamma (t),{\gamma }^{\prime }\left(t^{\prime }\right))=\underset{ϵ\to 0}{lim}\underset{0<t,t^{\prime }<ϵ}{sup}{\mathrm{\angle }}_{\gamma (0)}^{(\kappa )}(\gamma (t),{\gamma }^{\prime }\left(t^{\prime }\right)).$$

If X is uniquely geodesic, $p\ne q$ and $p\ne r$, the angle of $\mathrm{△}(p,q,r)$ in X at p is the (Alexandrov) angle between the geodesic segments $[p,q]$ and $[p,r]$ issuing from p and is denoted ${\mathrm{\angle }}_p(q,r)$.

Proposition 2.5 ([12], I.1.13, 1.14 and II.1.8])

Let X be a metric space and let γ, ${\gamma }^{\prime }$, ${\gamma }^{″}$ be three geodesics issuing from a common point. Then

• (i) $\mathrm{\angle }(\gamma ,{\gamma }^{\prime })=\mathrm{\angle }({\gamma }^{\prime },\gamma )$;

• (ii) $\mathrm{\angle }(\gamma ,{\gamma }^{\prime })\le \mathrm{\angle }(\gamma ,{\gamma }^{″})+\mathrm{\angle }({\gamma }^{\prime },{\gamma }^{″})$;

• (iii) if $\gamma :[-a,a]\to X$ is a geodesic with $a>0$, and if ${\gamma }^{\prime },{\gamma }^{″}:[0,a]\to X$ are defined by ${\gamma }^{\prime }(t)=\gamma (-t)$ and ${\gamma }^{″}(t)=\gamma (t)$, then $\mathrm{\angle }({\gamma }^{\prime },{\gamma }^{″})=\pi$.

3 Basic properties and Δ-convergence

This section contains a number of primary results in [8] which are crucial to the study of our problem. The following proposition states very useful properties of the metric projection in a complete $CAT(1)$ space.

Proposition 3.1 ([8], Proposition 3.5])

Let X be a complete $CAT(1)$ space and let $C\subset X$ be nonempty closed and π-convex. Suppose that $x\in X$ such that $d(x,C)<\pi /2$. Then the following are satisfied:

• (i) There exists a unique point $P_{C}x\in C$ such that $d(x,P_{C}x)=d(x,C)$.

• (ii) If $x\notin C$ and $y\in C$ with $y\ne P_{C}x$, then ${\mathrm{\angle }}_{P_{C}x}(x,y)\ge \pi /2$.

• (iii) If $diam(X)\le \pi$, then for any $y\in C$,

  $$d(P_{C}x,P_{C}y)=d(P_{C}x,y)\le d(x,y).$$

The mapping $P_C$ of X onto C in Proposition 3.1 is called the metric projection.

The next result shows the existence property of fixed points for a nonexpansive mapping.

Proposition 3.2 ([8], Theorem 3.9])

Let X be a complete $CAT(1)$ space such that $diamX<\pi /2$. Then every nonexpansive mapping $T:X\to X$ has at least one fixed point.

The rest of this section is devoted to presenting several closely related characterizations of Δ-convergence. For this purpose, we start with some basic definitions of an asymptotic radius and an asymptotic center. Let $\{x_n\}$ be a bounded sequence in a complete $CAT(1)$ space X. For $x\in X$ and $C\subset X$, let $r(x,\{x_n\})={lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}d(x,x_n)$. The asymptotic radius $r(\{x_n\})$ of $\{x_n\}$ is given by

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

the asymptotic radius $r_C(\{x_n\})$ with respective to C of $\{x_n\}$ is given by

$$r_C(\{x_n\})=inf\{r(x,\{x_n\}):x\in C\};$$

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

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

the asymptotic center $A_C(\{x_n\})$ with respective to C of $\{x_n\}$ is given by the set

$$A_C(\{x_n\})=\{x\in C:r(x,\{x_n\})=r_C(\{x_n\})\}.$$

Proposition 3.3 ([8], Proposition 4.1])

Let X be a complete $CAT(1)$ space and $C\subset X$ nonempty closed and π-convex. If $\{x_n\}$ be a sequence in X such that $r_C(\{x_n\})<\pi /2$, then $A_C(\{x_n\})$ consists of exactly one point.

Definition 3.4 A sequence $\{x_n\}$ in X is said to Δ-converge to $x\in X$ if x is the unique asymptotic center of every subsequence of $\{x_n\}$. In this case, we write $\mathrm{\Delta }\text{-}{lim}_{n\to \mathrm{\infty }}{x}_n=x$ and x is called the Δ-limit of $\{x_n\}$.

The next result is an immediate consequence of the preceding proposition.

Proposition 3.5 ([8], Corollary 4.4])

Let X be a complete $CAT(1)$ space and $\{x_n\}$ a sequence in X. If $r(\{x_n\})<\pi /2$, then $\{x_n\}$ has a Δ-convergent subsequence.

4 Main results

Since the validity of all our results, including the proofs as well, on $CAT(1)$ spaces can be restored on any $CAT(\kappa )$ space with $\kappa >0$ by rescaling without major changes, we will pay our attention to $CAT(1)$ spaces. In addition, when we deal with a $CAT(\kappa )$ space, the hypothesis $d(x_0,\mathfrak{F}(T))<\pi /4$ for each theorem in this section is replaced by $d(x_0,\mathfrak{F}(T))<D_{\kappa }/4$ and so can be dropped if $\kappa \le 0$ (in this case, $D_{\kappa }=\mathrm{\infty }$; refer to Section 2 for the definition).

To verify our result, the following basic property of asymptotical regularity is required; also cf. [6], Proposition 2.1] in a topological vector space.

Lemma 4.1 Let C be a subset of a metric space $(X,d)$ and let $\mathfrak{S}=\{T(t):t\ge 0\}$ be a family of self-mappings of C such that $T(s+t)=T(s)\circ T(t)$, for all $s,t\ge 0$. If $\mathfrak{S}$ is asymptotically regular, then

$$\mathfrak{F}(T(t))=\mathfrak{F}(\mathfrak{S}),\mathit{\text{for all }}t>0.$$

Proof Fix $t>0$. Then $\mathfrak{F}(\mathfrak{S})\subset \mathfrak{F}(T(t))$. To prove $\mathfrak{F}(T(t))\subset \mathfrak{F}(\mathfrak{S})$, let x be a fixed point of $T(t)$. For $h\ge 0$, we obtain

$$\begin{array}{rl}d(T(h)x,x)& =\underset{k\to \mathrm{\infty }}{lim}d(T(h)\circ T{(t)}^{k}x,T{(t)}^{k}x)\\ =\underset{k\to \mathrm{\infty }}{lim}d(T(h+tk)x,T(tk)x)\\ =\underset{s\to \mathrm{\infty }}{lim}d(T(h+s)x,T(s)x)\\ =0\end{array}$$

and therefore $x\in \mathfrak{F}(\mathfrak{S})$. □

Let $(X,d)$ be a complete $CAT(1)$ space, C a closed π-convex subset of X and $T:C\to C$ a nonexpansive mapping. Then $\mathfrak{F}(T)$ is closed. Also it is seen that $\mathfrak{F}(T)$ is π-convex. Indeed, for $x,y\in \mathfrak{F}(T)$ with $d(x,y)<\pi$, we have

$$d(x,T(\lambda x\oplus (1-\lambda )y))\le d(x,\lambda x\oplus (1-\lambda )y)=(1-\lambda )d(x,y)$$

and similarly $d(y,T(\lambda x\oplus (1-\lambda )y))\le \lambda d(x,y)$. Thus both equalities must hold and hence $\lambda x\oplus (1-\lambda )y$ and $T(\lambda x\oplus (1-\lambda )y)$ belong to the unique geodesic $[x,y]$. This implies that $T(\lambda x\oplus (1-\lambda )y)=\lambda x\oplus (1-\lambda )y$.

Now consider a family $\mathfrak{S}=\{T(t):t\ge 0\}$ of nonexpansive self-mappings of C such that $\mathfrak{F}(\mathfrak{S})\ne \mathrm{\varnothing }$. From the previous remark, each fixed point set $\mathfrak{F}(T(t))$ is closed and convex, hence so is $\mathfrak{F}(\mathfrak{S})$. Fix $x_0\in C$ with $0<d(x_0,\mathfrak{F}(\mathfrak{S}))<\pi /4$. Let $p=P_{\mathfrak{F}(\mathfrak{S})}{x}_0$ (the uniqueness follows from Proposition 3.1(i)) and $ϵ=d(x_0,p)$. Then for each $t\ge 0$, $T(t)$ maps the closed π-convex set $C\cap \overline{B}(p,ϵ)$ into itself. For any $\alpha \in (0,1]$ and $t\ge 0$, define the mapping $S_{\alpha ,t}:C\cap \overline{B}(p,ϵ)\to C\cap \overline{B}(p,ϵ)$ by $S_{\alpha ,t}x=\alpha x_0\oplus (1-\alpha )T(t)x$. Observe that $S_{\alpha ,t}$ is a contraction. In fact, for $x,y\in C\cap \overline{B}(p,ϵ)$, we apply Lemma 2.4(iii) to get

$$d(S_{\alpha ,t}x,S_{\alpha ,t}y)\le sin[(1-\alpha )\frac{\pi }{2}]d(T(t)x,T(t)y)\le sin[(1-\alpha )\frac{\pi }{2}]d(x,y).$$

Then $S_{\alpha ,t}$ has a unique fixed point in $C\cap \overline{B}(p,ϵ)$.

Let $[t]$ denote the maximum integer no greater than t. We now extend Theorem 1.3 in Hilbert spaces to $CAT(\kappa )$ spaces as the following result.

Theorem 4.2 Let X be a complete $CAT(1)$ space, C a closed π-convex subset of X, $\mathfrak{S}=\{T(t):t\ge 0\}$ a $UAR$ and strongly continuous nonexpansive semigroup on C with $\mathfrak{F}(\mathfrak{S})\ne \mathrm{\varnothing }$, and two sequences $\{{\alpha }_n\}\subset (0,1]$, $\{t_n\}\subset (0,\mathrm{\infty })$ such that ${lim}_{n\to \mathrm{\infty }}{\alpha }_n={lim}_{n\to \mathrm{\infty }}{\alpha }_n/t_n=0$. Choose arbitrarily a point $x_0\in C$ such that $d(x_0,\mathfrak{F}(\mathfrak{S}))<\pi /4$. Let $p=P_{\mathfrak{F}(\mathfrak{S})}{x}_0$ and $ϵ=d(x_0,p)$. Define a sequence $\{x_n\}$ in $C\cap \overline{B}(p,ϵ)$ by the implicit iteration

$$x_n={\alpha }_n{x}_0\oplus (1-{\alpha }_n)T(t_n)x_n.$$

(4.1)

Then $\{x_n\}$ converges strongly to the point p of $\mathfrak{F}(\mathfrak{S})$ nearest to $x_0$.

Proof If $x_0\in \mathfrak{F}(\mathfrak{S})$, then $\{x_n\}$ reduces to the constant sequence $\{x_0,x_0,\dots \}$. Suppose that $x_0\notin \mathfrak{F}(\mathfrak{S})$. We shall prove that any subsequence of $\{x_n\}$ contains a subsequence converging strongly to p from which it follows that $\{x_n\}$ also converges strongly to the point p. Let $\{y_n\}$ be any subsequence of $\{x_n\}$. Proposition 3.5 asserts that $\{y_n\}$ has a Δ-convergent subsequence. By passing to a subsequence we may assume that $\{y_n\}$ Δ-converges to a point y. Let $\{{\beta }_n\}$ and $\{s_n\}$ be the respective corresponding subsequences of $\{{\alpha }_n\}$ and $\{t_n\}$.

Observe that $y\in \mathfrak{F}(\mathfrak{S})$. To see this, take $\hat{s}={lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{s}_n$. There are three cases:

• (i) $\hat{s}=0$,

• (ii) $0<\hat{s}<\mathrm{\infty }$,

• (iii) $\hat{s}=\mathrm{\infty }$.

By passing to a subsequence if necessary it may be assumed that $\hat{s}={lim}_{n\to \mathrm{\infty }}{s}_n$. We discuss each case as follows.

Suppose that $\hat{s}=0$. Fix $t\ge 0$ and we obtain

$$\begin{array}{rl}d(T(t)y,y_n)\le & d(T(t)y,T\left(\left[\frac{t}{s_n}\right]s_n\right)y)+d(T\left(\left[\frac{t}{s_n}\right]s_n\right)y,T\left(\left[\frac{t}{s_n}\right]s_n\right)y_n)\\ +\sum _{j=1}^{[t/s_n]}d(T((j-1)s_n)y_n,T(j{s}_n)y_n)\\ \le & d(T(t-\left[\frac{t}{s_n}\right]s_n)y,y)+d(y,y_n)+\left[\frac{t}{s_n}\right]d(y_n,T(s_n)y_n)\\ \le & \underset{0\le s\le s_n}{max}d(T(s)y,y)+d(y,y_n)+\frac{t{\beta }_n}{s_n}d(x_0,T(s_n)y_n).\end{array}$$

This implies that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(T(t)y,y_n)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(y,y_n);$$

hence $T(t)y=y$ for all $t\ge 0$, that is, $y\in \mathfrak{F}(\mathfrak{S})$.

If $0<\hat{s}<\mathrm{\infty }$, then

$$\begin{array}{rl}d(T(\hat{s})y,y_n)& \le d(T(\hat{s})y,T(s_n)y)+d(T(s_n)y,T(s_n)y_n)+d(T(s_n)y_n,y_n)\\ \le d(y,T(|s_n-\hat{s}|)y)+d(y,y_n)+d(T(s_n)y_n,y_n)\\ \le d(y,T(|s_n-\hat{s}|)y)+d(y,y_n)+{\beta }_{n}d(x_0,T(s_n)y_n)\end{array}$$

and thus

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(T(\hat{s})y,y_n)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(y,y_n).$$

Hence $T(\hat{s})y=y$ from which it follows that $y\in \mathfrak{F}(\mathfrak{S})$ by Lemma 4.1.

If $\hat{s}=\mathrm{\infty }$, fix $t\ge 0$. For all sufficiently large n we get

$$\begin{array}{rl}d(T(t)y,y_n)\le & d(T(t)y,T(t)y_n)+d(T(t)y_n,y_n)\\ \le & d(y,y_n)+{\beta }_{n}d(T(t)y_n,x_0)+(1-{\beta }_n)d(T(t)y_n,T(s_n)y_n)\\ \le & d(y,y_n)+{\beta }_{n}d(T(t)y_n,x_0)+(1-{\beta }_n)d(y_n,T(s_n-t)y_n)\\ \le & d(y,y_n)+{\beta }_{n}d(T(t)y_n,x_0)+{\beta }_n(1-{\beta }_n)d(x_0,T(s_n-t)y_n)\\ +{(1-{\beta }_n)}^{2}d(T(s_n)y_n,T(s_n-t)y_n)\\ \le & d(y,y_n)+{\beta }_{n}d(T(t)y_n,x_0)+{\beta }_n(1-{\beta }_n)d(x_0,T(s_n-t)y_n)\\ +{(1-{\beta }_n)}^{2}d(T(s_n-t+t)y_n,T(s_n-t)y_n)\end{array}$$

which yields

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(T(t)y,y_n)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(y,y_n)$$

since

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(T(s_n-t+t)y_n,T(s_n-t)y_n)\le \underset{n\to \mathrm{\infty }}{lim}\underset{x\in C\cap \overline{B}(p,ϵ)}{sup}d(T(h_n+t)x,T(h_n)x)=0,$$

where $h_n=s_n-t$. It follows that $T(t)y=y$ and therefore $y\in \mathfrak{F}(\mathfrak{S})$. This finishes the proof that y is a common fixed point of all mappings in $\mathfrak{S}$.

Next, we claim that $\{y_n\}$ converges strongly to y. We suppose on the contrary that

$$r(\{y_n\})=\sigma >0.$$

From $y=T(s_n)y$, we obtain

$$\begin{array}{rl}d(y,y_n)& \le {\beta }_{n}d(y,x_0)+(1-{\beta }_n)d(y,T(s_n)y_n)\\ \le {\beta }_{n}d(y,x_0)+(1-{\beta }_n)d(y,y_n),\end{array}$$

and hence taking the limit superior as $n\to \mathrm{\infty }$ yields

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(y,T(s_n)y_n)=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(y,y_n)=\sigma ,$$

(4.2)

since ${lim}_{n\to \mathrm{\infty }}{\beta }_n=0$. Recall that

$$d(x_0,y_n)=(1-{\beta }_n)d(x_0,T(s_n)y_n)<d(x_0,T(s_n)y_n),\text{for all }n.$$

Then

$$d(y_n,T(s_n)y_n)=d(x_0,T(s_n)y_n)-d(x_0,y_n)>0,\text{for all }n.$$

Recall that $x_0\notin \mathfrak{F}(\mathfrak{S})$. We have $y\ne x_0$ and hence

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_0,y_n)>\sigma .$$

(4.3)

According to (4.2) and (4.3), by passing to a subsequence again we may assume that

$$d(x_0,y_n)>0,d(y,y_n)>0,d(y,T(s_n)y_n)>0,\text{for all }n.$$

Let $\mathrm{△}({\overline{x}}_0,\overline{y},\overline{T(s_n)y_n})$ be a comparison triangle for $\mathrm{△}(x_0,y,T(s_n)y_n)$ in ${\mathbb{S}}^2$. Observe that

$${\mathrm{\angle }}_{{\overline{y}}_n}({\overline{x}}_0,\overline{y})\ge \pi /2.$$

(4.4)

For, contrarily, if ${\mathrm{\angle }}_{{\overline{y}}_n}({\overline{x}}_0,\overline{y})<\pi /2$, since ${\mathrm{\angle }}_{{\overline{y}}_n}({\overline{x}}_0,\overline{T(s_n)y_n})=\pi$, then ${\mathrm{\angle }}_{{\overline{y}}_n}(\overline{y},\overline{T(s_n)y_n})>\pi /2$; see Proposition 2.5. By the law of cosines in ${\mathbb{S}}^2$ (Proposition 2.2), since $d(y_n,T(s_n)y_n)>0$, it follows that

$$cos{d}_{{\mathbb{S}}^2}(\overline{y},\overline{T(s_n)y_n})<cos{d}_{{\mathbb{S}}^2}(\overline{y},{\overline{y}}_n),$$

which is a contradiction because $\overline{T(s_n)}$ is nonexpansive. Notice that

$$0<d(y,y_n)\le d_{{\mathbb{S}}^2}(\overline{y},{\overline{y}}_n).$$

Hence (4.4) implies that

$$cos{d}_{{\mathbb{S}}^2}({\overline{x}}_0,\overline{y})\le cos{d}_{{\mathbb{S}}^2}({\overline{x}}_0,{\overline{y}}_n)cos{d}_{{\mathbb{S}}^2}(\overline{y},{\overline{y}}_n)<cos{d}_{{\mathbb{S}}^2}({\overline{x}}_0,{\overline{y}}_n),$$

or equivalently,

$$d_{{\mathbb{S}}^2}({\overline{x}}_0,{\overline{y}}_n)<d_{{\mathbb{S}}^2}({\overline{x}}_0,\overline{y}).$$

Similarly,

$$d_{{\mathbb{S}}^2}(\overline{y},{\overline{y}}_n)<d_{{\mathbb{S}}^2}({\overline{x}}_0,\overline{y}).$$

The previous two inequalities, together with the $CAT(\kappa )$ inequality, yield

$$d(x_0,y_n)<d(x_0,y),d(y,y_n)<d(x_0,y),\text{for all }n.$$

(4.5)

Let $\mathrm{△}({\overline{x}}_0,\overline{y},{\overline{y}}_n)$ be a comparison triangle for $\mathrm{△}(x_0,y,y_n)$ in ${\mathbb{S}}^2$. Since $d_{{\mathbb{S}}^2}({\overline{y}}_n,[{\overline{x}}_0,\overline{y}])<\pi /2$, Proposition 3.1(i) assures that there is a unique point ${\overline{u}}_n\in [{\overline{x}}_0,\overline{y}]$ nearest to ${\overline{y}}_n$. Let $u_n\in [x_0,y]$ such that $d(x_0,u_n)=d_{{\mathbb{S}}^2}({\overline{x}}_0,{\overline{u}}_n)$ and $d(u_n,y)=d_{{\mathbb{S}}^2}({\overline{u}}_n,\overline{y})$. By passing to a subsequence again it may be assumed that $\{{\overline{u}}_n\}$ and $\{u_n\}$ converge, respectively, to $\overline{u}\in [{\overline{x}}_0,\overline{y}]$ and $u\in [x_0,y]$. Since

$$\begin{array}{rl}\sigma & =\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(y,y_n)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(u,y_n)\\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{d}_{{\mathbb{S}}^2}(\overline{u},{\overline{u}}_n)+\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{d}_{{\mathbb{S}}^2}({\overline{u}}_n,{\overline{y}}_n)\\ =\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{d}_{{\mathbb{S}}^2}({\overline{u}}_n,{\overline{y}}_n)\\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{d}_{{\mathbb{S}}^2}(\overline{y},{\overline{y}}_n)\\ =\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(y,y_n),\end{array}$$

(4.6)

this guarantees that $u=y$.

On the other hand, according to (4.6), by passing to a subsequence we may suppose that

$$d_{{\mathbb{S}}^2}({\overline{u}}_n,{\overline{y}}_n)>\frac{\sigma }{2},\text{for all }n.$$

Let $a_n=d_{{\mathbb{S}}^2}({\overline{x}}_0,{\overline{u}}_n)$, $b_n=d_{{\mathbb{S}}^2}({\overline{u}}_n,{\overline{y}}_n)$, $c_n=d_{{\mathbb{S}}^2}({\overline{x}}_0,{\overline{y}}_n)$ and ${\gamma }_n={\mathrm{\angle }}_{{\overline{u}}_n}({\overline{x}}_0,{\overline{y}}_n)$. Then by (4.5), $b_n>\sigma /2$ and $c_n<d_{{\mathbb{S}}^2}({\overline{x}}_0,\overline{y})$. Observe that $u_n\ne x_0$. Otherwise, since $y\ne x_0$, we have ${\mathrm{\angle }}_{{\overline{x}}_0}(\overline{y},{\overline{y}}_n)={\mathrm{\angle }}_{{\overline{u}}_n}(\overline{y},{\overline{y}}_n)\ge \pi /2$ which implies that

$$d(y,y_n)=d_{{\mathbb{S}}^2}(\overline{y},{\overline{y}}_n)\ge d_{{\mathbb{S}}^2}({\overline{x}}_0,\overline{y})=d(x_0,y).$$

But this contradicts to (4.5). Therefore $u_n\ne x_0$ and so ${\gamma }_n\ge \pi /2$ by Proposition 3.1(ii). The law of cosines in ${\mathbb{S}}^2$ yields

$$\begin{array}{rl}cos{c}_n& =cos{a}_{n}cos{b}_n+sin{a}_{n}sin{b}_{n}cos{\gamma }_n\\ \le cos{a}_{n}cos{b}_n.\end{array}$$

Since $\sigma /2<b_n\le c_n<d_{{\mathbb{S}}^2}({\overline{x}}_0,\overline{y})<\pi /2$, this implies that

$$cos{a}_n\ge \frac{cos{c}_n}{cos{b}_n}>\frac{cos{d}_{{\mathbb{S}}^2}({\overline{x}}_0,\overline{y})}{cos(\sigma /2)}>cos{d}_{{\mathbb{S}}^2}({\overline{x}}_0,\overline{y}).$$

Hence

$$d_{{\mathbb{S}}^2}({\overline{x}}_0,{\overline{u}}_n)=a_n<{cos}^{-1}\left(\frac{cos{d}_{{\mathbb{S}}^2}({\overline{x}}_0,\overline{y})}{cos(\sigma /2)}\right)<d_{{\mathbb{S}}^2}({\overline{x}}_0,\overline{y}).$$

It follows that

$$d(y,u_n)=d_{{\mathbb{S}}^2}(\overline{y},{\overline{u}}_n)=d_{{\mathbb{S}}^2}({\overline{x}}_0,\overline{y})-d_{{\mathbb{S}}^2}({\overline{x}}_0,{\overline{u}}_n)>d_{{\mathbb{S}}^2}({\overline{x}}_0,\overline{y})-\delta >0,$$

where $\delta ={cos}^{-1}[cos{d}_{{\mathbb{S}}^2}({\overline{x}}_0,\overline{y})/cos(\sigma /2)]$. Taking the limit as $n\to \mathrm{\infty }$ yields $u\ne y$, which is a contradiction. Thus $\sigma =0$ and so $\{y_n\}$ converges strongly to y, as claimed.

It remains to prove that $y=P_{\mathfrak{F}(\mathfrak{S})}{x}_0$. Let $q\in \mathfrak{F}(\mathfrak{S})$ and consider a comparison triangle $\mathrm{△}({\overline{x}}_0,\overline{q},\overline{T(s_n)y_n})$ for $\mathrm{△}(x_0,q,T(s_n)y_n)$. From

$$d_{{\mathbb{S}}^2}(\overline{q},\overline{T(s_n)y_n})\le d_{{\mathbb{S}}^2}(\overline{q},{\overline{y}}_n),$$

it is seen that ${\mathrm{\angle }}_{{\overline{y}}_n}(\overline{q},\overline{T(s_n)y_n})\le \pi /2$. Hence ${\mathrm{\angle }}_{{\overline{y}}_n}({\overline{x}}_0,\overline{q})\ge \pi /2$ and so

$$d(x_0,y_n)=d_{{\mathbb{S}}^2}({\overline{x}}_0,{\overline{y}}_n)\le d_{{\mathbb{S}}^2}({\overline{x}}_0,\overline{q})=d(x_0,q).$$

We then take the limit as $n\to \mathrm{\infty }$ and obtain

$$d(x_0,y)\le d(x_0,q),$$

which shows that y is the nearest point in $\mathfrak{F}(\mathfrak{S})$ to $x_0$. Consequently, we conclude that $\{x_n\}$ converges strongly to $P_{\mathfrak{F}(\mathfrak{S})}{x}_0$, which completes the proof. □

It is worthy emphasizing that a uniformly asymptotical regularity hypothesis of Theorem 4.2 is superfluous when ${lim}_{n\to \mathrm{\infty }}{t}_n={lim}_{n\to \mathrm{\infty }}{\alpha }_n/t_n=0$; see case (i) in the proof of Theorem 4.2. Thus, we state the result as follows.

Theorem 4.3 Let X be a complete $CAT(1)$ space, C a closed π-convex subset of X, $\mathfrak{S}=\{T(t):t\ge 0\}$ a strongly continuous nonexpansive semigroup on C with $\mathfrak{F}(\mathfrak{S})\ne \mathrm{\varnothing }$, and two sequences $\{{\alpha }_n\}\subset (0,1]$, $\{t_n\}\subset (0,\mathrm{\infty })$ such that ${lim}_{n\to \mathrm{\infty }}{t}_n={lim}_{n\to \mathrm{\infty }}{\alpha }_n/t_n=0$. Choose arbitrarily a point $x_0\in C$ such that $d(x_0,\mathfrak{F}(\mathfrak{S}))<\pi /4$. Let $p=P_{\mathfrak{F}(\mathfrak{S})}{x}_0$ and $ϵ=d(x_0,p)$. Then the sequence $\{x_n\}$ in $C\cap \overline{B}(p,ϵ)$ defined by (4.1) converges strongly to the point p.

Furthermore, if X is a $CAT(0)$ space (recall that $D_0=\mathrm{\infty }$) in Theorem 4.3, then the corresponding assumption $d(x_0,\mathfrak{F}(\mathfrak{S}))<\pi /4$ is $d(x_0,\mathfrak{F}(\mathfrak{S}))<\mathrm{\infty }$, which can be dropped. Moreover, the sequence $\{x_n\}$ defined by (4.1) is bounded. In fact, since

$$\begin{array}{rl}d(p,x_n)& \le {\alpha }_{n}d(p,x_0)+(1-{\alpha }_n)d(p,T(t_n)x_n)\\ \le {\alpha }_{n}d(p,x_0)+(1-{\alpha }_n)d(p,x_n),\end{array}$$

this shows that $d(p,x_n)\le d(p,x_0)$. Then $\{x_n\}$ is bounded and so is $\{T(t)x_n:t\ge 0,n\in \mathbb{N}\}$. We restate Theorem 1.2 without assuming boundedness of C as follows.

Theorem 4.4 Let X be a complete $CAT(0)$ space, C a closed convex subset of X, $\mathfrak{S}=\{T(t):t\ge 0\}$ a strongly continuous nonexpansive semigroup on C with $\mathfrak{F}(\mathfrak{S})\ne \mathrm{\varnothing }$, and two sequences $\{{\alpha }_n\}\subset (0,1]$, $\{t_n\}\subset (0,\mathrm{\infty })$ such that ${lim}_{n\to \mathrm{\infty }}{t}_n={lim}_{n\to \mathrm{\infty }}{\alpha }_n/t_n=0$. Choose arbitrarily a point $x_0\in C$. Then the sequence $\{x_n\}$ in C defined by (4.1) converges strongly to a point of $\mathfrak{F}(\mathfrak{S})$ nearest to $x_0$.

Nonetheless, Theorem 4.2 has a uniformly asymptotical regularity hypothesis that cannot in general be removed when ${lim}_{n\to \mathrm{\infty }}{t}_n=\hat{t}\in (0,\mathrm{\infty }]$ as illustrated in the following example.

Example 4.5 Consider the complex plane $(\mathbb{C},|\cdot |)$, where $|\cdot |$ is the absolute value, so that it is a complete $CAT(0)$ space. For $t\ge 0$, define a self-mapping $T(t)$ of ℂ by $T(t)(z)=z{e}^{it}$, $z\in \mathbb{C}$. Then $\mathfrak{S}=\{T(t):t\ge 0\}$ is a strongly continuous nonexpansive semigroup with $\mathfrak{F}(\mathfrak{S})=\{0\}$. The family $\mathfrak{S}$ is not $UAR$ on ℂ because

$$\underset{t\to \mathrm{\infty }}{lim}\underset{|z|=1}{sup}|T(\pi )T(t)(z)-T(t)(z)|=\underset{t\to \mathrm{\infty }}{lim}\underset{|z|=1}{sup}|z{e}^{i(t+\pi )}-z{e}^{it}|=2.$$

Let ${\alpha }_n=1/n$, $t_n=n\pi$, $n\in \mathbb{N}$, so that ${lim}_{n\to \mathrm{\infty }}{t}_n=\mathrm{\infty }$ and ${lim}_{n\to \mathrm{\infty }}{\alpha }_n/t_n=0$. Choose $z_0=1$. Define a sequence $\{z_n\}$ in ℂ by (4.1), that is,

$$z_n=\{\begin{array}{ll}\frac{1}{2n-1},& n\text{ is odd},\\ 1,& n\text{ is even}.\end{array}$$

Hence the sequence $\{z_n\}$ is divergent.

Next if we take ${\alpha }_n=1/n$, $t_n=(1/n)+2\pi$, $n\in \mathbb{N}$, then ${lim}_{n\to \mathrm{\infty }}{t}_n=2\pi$ and ${lim}_{n\to \mathrm{\infty }}{\alpha }_n/t_n=0$. Choose $z_0=1$ and define a sequence $\{z_n\}$ in ℂ by (4.1) to get

$$z_n=\frac{1/n}{1-[1-(1/n)]e^{i/n}}.$$

However, the sequence $\{z_n\}$ converges to $(1+i)/2\notin \mathfrak{F}(\mathfrak{S})$.

These two examples explain that strong convergence of the sequence defined by (4.1) in Theorem 4.2 ceases to be true without the $UAR$ assumption if ${lim}_{n\to \mathrm{\infty }}{t}_n=\hat{t}\in (0,\mathrm{\infty }]$.

The next example shows that the condition ${lim}_{n\to \mathrm{\infty }}{\alpha }_n={lim}_{n\to \mathrm{\infty }}{\alpha }_n/t_n=0$ in Theorem 4.2 is also necessary when X is the unit sphere in ${\ell }_2$, the infinite-dimensional Hilbert space of square-summable sequences.

Example 4.6 Let ${\mathbb{S}}_{{\ell }_2}$ be the unit sphere in ${\ell }_2$ endowed with the following intrinsic metric $d:{\mathbb{S}}_{{\ell }_2}\times {\mathbb{S}}_{{\ell }_2}\to \mathbb{R}$: for $x,y\in {\mathbb{S}}_{{\ell }_2}$, $d(x,y)\in [0,\pi ]$ such that

$$cosd(x,y)=(x\mid y)=\sum _{k=1}^{\mathrm{\infty }}x(k)y(k).$$

This space $({\mathbb{S}}_{{\ell }_2},d)$ is a $CAT(1)$ space.

Take one element $v=(1,0,\dots ,0,\dots )$ of the canonical basis of ${\mathbb{S}}_{{\ell }_2}$. Let $C=\{x\in {\mathbb{S}}_{{\ell }_2}:d(x,v)\le r\}$ be a closed ball centered at v in ${\mathbb{S}}_{{\ell }_2}$, where $0<r<\pi /4$. For $t\ge 0$, define a mapping $T(t):C\to C$ by

$$T(t)(x)=(1-e^{-t})v\oplus e^{-t}x,$$

that is,

$$(T(t)(x))(1)=cos[e^{-t}{cos}^{-1}x(1)],$$

(4.7)

and for $k=2,3,\dots$ ,

$$(T(t)(x))(k)=\{\begin{array}{ll}0,& \text{if }x(1)=1,\\ \frac{x(k)sin[e^{-t}{cos}^{-1}x(1)]}{\sqrt{1-x{(1)}^2}},& \text{if }x(1)\ne 1.\end{array}$$

(4.8)

For $x,y\in C$, from Lemma 2.4(iii), we obtain

$$\begin{array}{rl}d(T(t)x,T(t)y)& =d((1-e^{-t})v\oplus e^{-t}x,(1-e^{-t})v\oplus e^{-t}y)\\ \le \frac{sin(e^{-t}\pi /2)}{sin\pi /2}d(x,y)\\ \le d(x,y).\end{array}$$

Then $T(t)$ is nonexpansive. In fact, v is the unique fixed point of $T(t)$ for $t>0$. Furthermore, it is seen that $\mathfrak{S}=\{T(t):t\ge 0\}$ is a $UAR$ and strongly continuous nonexpansive semigroup on C.

Choose $u=(u_k)\in C$ with $0<u_1<1$ so that $0<{cos}^{-1}{u}_1=d(u,v)\le r$. For $\alpha \in (0,1)$ and $t\in (0,\mathrm{\infty })$, define $x(\alpha ,t)$ by

$$\begin{array}{rl}x(\alpha ,t)& =\alpha u\oplus (1-\alpha )T(t)x(\alpha ,t)\\ =cos[(1-\alpha )d(u,T(t)x(\alpha ,t))]u+sin[(1-\alpha )d(u,T(t)x(\alpha ,t))]\frac{A(\alpha ,t)}{\parallel A(\alpha ,t)\parallel },\end{array}$$

(4.9)

where

$$A(\alpha ,t)=T(t)x(\alpha ,t)-(u\mid T(t)x(\alpha ,t))u.$$

We remark that ${\parallel A(\alpha ,t)\parallel }^2=1-{(u\mid T(t)x(\alpha ,t))}^2$. Let $\{{\alpha }_n\}\subset (0,1]$, $\{t_n\}\subset (0,\mathrm{\infty })$ be two sequences and define a sequence $\{x_n\}$ in $\overline{B}(v,{cos}^{-1}{u}_1)\subset C$ by

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

We shall prove that if the sequence $\{x_n\}$ converges to the common fixed point v of $\mathfrak{S}$, then ${lim}_{n\to \mathrm{\infty }}{\alpha }_n={lim}_{n\to \mathrm{\infty }}{\alpha }_n/t_n=0$.

First, assume that $\beta ={lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n$. Since ${lim}_{n\to \mathrm{\infty }}{x}_n(1)=1$ and $|x_n(k)|\le \sqrt{1-x_n{(1)}^2}$ for $k\ge 2$, we obtain from (4.7) and (4.8) that

$$\underset{n\to \mathrm{\infty }}{lim}(T(t_n)(x_n))(1)=1,\underset{n\to \mathrm{\infty }}{lim}(T(t_n)(x_n))(k)=0.$$

Therefore ${lim}_{n\to \mathrm{\infty }}T(t_n)(x_n)=v$, which implies that

$$\underset{n\to \mathrm{\infty }}{lim}(u\mid T(t_n)(x_n))=u_1,\underset{n\to \mathrm{\infty }}{lim}d(u,T(t_n)x_n)={cos}^{-1}{u}_1;$$

(4.10)

hence

$$\underset{n\to \mathrm{\infty }}{lim}A({\alpha }_n,t_n)(1)=1-u_1^2,\underset{n\to \mathrm{\infty }}{lim}\parallel A({\alpha }_n,t_n)\parallel =\sqrt{1-u_1^2}.$$

(4.11)

Let $\theta =d(u,v)={cos}^{-1}{u}_1$ so that $sin\theta =\sqrt{1-u_1^2}$. Using (4.9), (4.10), and (4.11), we have

$$\begin{array}{rl}1=& \underset{n\to \mathrm{\infty }}{lim}{x}_n(1)\\ =& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\{u_{1}cos[(1-{\alpha }_n)d(u,T(t_n)x_n)]+\frac{A({\alpha }_n,t_n)(1)}{\parallel A({\alpha }_n,t_n)\parallel }sin[(1-{\alpha }_n)d(u,T(t_n)x_n)]\}\\ \le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\{u_{1}cos[(1-{\alpha }_n)d(u,T(t_n)x_n)]\}\\ +\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\{\frac{A({\alpha }_n,t_n)(1)}{\parallel A({\alpha }_n,t_n)\parallel }sin[(1-{\alpha }_n)d(u,T(t_n)x_n)]\}\\ =& cos\theta cos[(1-\beta )\theta ]+sin\theta sin[(1-\beta )\theta ]\\ =& cos(\beta \theta ).\end{array}$$

This shows that $\beta =0$ because $\theta >0$ and so ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$.

To prove that ${lim}_{n\to \mathrm{\infty }}{\alpha }_n/t_n=0$, we recall the inequality $t\ge 1-e^{-t}$ for all $t\ge 0$. Since

$$t_{n}d(v,x_n)\ge (1-e^{-t_n})d(v,x_n)=d(x_n,T(t_n)(x_n))={\alpha }_{n}d(u,T(t_n)(x_n)),$$

it follows that

$$\begin{array}{rl}0& =\underset{n\to \mathrm{\infty }}{lim}d(v,x_n)\\ \ge \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\left[\frac{{\alpha }_n}{t_n}d(u,T(t_n)(x_n))\right]\\ =\theta \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{{\alpha }_n}{t_n}.\end{array}$$

Consequently, ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n/t_n=0$ and therefore ${lim}_{n\to \mathrm{\infty }}{\alpha }_n/t_n=0$ as desired.

The following result is an immediate consequence of Theorem 4.2 and Example 4.6.

Theorem 4.7 Let X be the unit sphere in an infinite-dimensional Hilbert space, and let $\{{\alpha }_n\}\subset (0,1]$, $\{t_n\}\subset (0,\mathrm{\infty })$ be two sequences. Then the following conditions are equivalent:

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

• (ii) Let C be a closed convex subset of X with $diamC<\pi /2$, $\mathfrak{S}=\{T(t):t\ge 0\}$ a $UAR$ and strongly continuous nonexpansive semigroup on C with $\mathfrak{F}(\mathfrak{S})\ne \mathrm{\varnothing }$, $x_0\in C$ such that $d(x_0,\mathfrak{F}(\mathfrak{S}))<\pi /4$, $p=P_{\mathfrak{F}(\mathfrak{S})}{x}_0$, $ϵ=d(x_0,p)$ and $\{x_n\}$ is a sequence in $C\cap \overline{B}(p,ϵ)$ defined by (4.1).

Then $\{x_n\}$ converges strongly to p.

5 Remark

A semitopological semigroup S is a semigroup equipped with a Hausdorff topology such that for each $t\in S$ the mappings $s\mapsto ts$ and $s\mapsto st$ from S into S are continuous. A semitopological semigroup S is left (respectively, right) reversible if any two closed right (respectively, left) ideals of S have nonvoid intersection, i.e., $\overline{aS}\cap \overline{bS}\ne \mathrm{\varnothing }$ (respectively, $\overline{Sa}\cap \overline{Sb}\ne \mathrm{\varnothing }$), for $a,b\in S$, where $\overline{E}$ denotes the closure of a set E in a topological space. The class S of all left reversible semitopological semigroups includes all groups, all commuting semigroups, and all normal left amenable semitopological semigroups, i.e., the space $CB(S)$ of bounded continuous functions on S has a left invariant mean; see [2, 3, 13–15]. If S is left (respectively, right) reversible, $(S,⪯)$ is a directed system when the binary relation ⪯ on S is defined by $a⪯b$ if and only if $\{a\}\cup \overline{aS}\supset \{b\}\cup \overline{bS}$ (respectively, $\{a\}\cup \overline{Sa}\supset \{b\}\cup \overline{Sb}$), $a,b\in S$.

Let S be a semitopological semigroup and C a closed convex subset of a metric space $(X,d)$. A family $\mathfrak{S}=\{T(s):s\in S\}$ is called a representation of S on C if for each $s\in S$, $T(s)$ is a mapping from C into C and $T(st)=T(s)\circ T(t)$, for all $s,t\in S$. The representation is called a strongly continuous nonexpansive semigroup on C (or a continuous representation of S as nonexpansive mappings on C) if the following conditions are satisfied (see [3]):

• (i) for each $t\in S$, $T(t)$ is a nonexpansive mapping on C;

• (ii) for each $x\in C$, the mapping $t\mapsto T(t)x$ from S into C is continuous.

The representation $\mathfrak{S}$ of S is called

• (i) asymptotically regular on C if for any $h\in S$ and any $x\in C$,

  $$\underset{t\in S}{lim}d(T(h)T(t)x,T(t)x)=0;$$

• (ii) uniformly asymptotically regular (in short $UAR$) on C if for any $h\in S$ and any bounded subset D of C,

  $$\underset{t\in S}{lim}\underset{x\in D}{sup}d(T(h)T(t)x,T(t)x)=0.$$

Problem 5.1 Let X be a complete $CAT(1)$ space, C a closed π-convex subset of X, S a commutative (or left reversible) semitopological semigroup, and $\mathfrak{S}=\{T(t):t\in S\}$ a continuous representation of S as nonexpansive mappings on C. Can we obtain the result analogous to Theorem 4.2 corresponding to a continuous representation $\mathfrak{S}=\{T(t):t\in S\}$ of S as nonexpansive mappings on C?

To answer this problem, the following conjecture is required and hence needs to be verified.

Conjecture 5.2 Let S be a commutative (or left reversible) semitopological semigroup, C a subset of a metric space $(X,d)$ and $\mathfrak{S}=\{T(t):t\in S\}$ a representation of S on C. If $\mathfrak{S}$ is asymptotically regular, then

$$\mathfrak{F}(T(t))=\mathfrak{F}(\mathfrak{S}),\mathit{\text{for all }}t\in S.$$