Approximating common fixed points in hyperbolic spaces

Abstract

We establish strong convergence and Δ-convergence theorems of an iteration scheme associated to a pair of nonexpansive mappings on a nonlinear domain. In particular we prove that such a scheme converges to a common fixed point of both mappings. Our results are a generalization of well-known similar results in the linear setting. In particular, we avoid assumptions such as smoothness of the norm, necessary in the linear case.

MSC:47H09, 46B20, 47H10, 47E10.

1 Introduction

Let C be a nonempty subset of a metric space $(X,d)$ and $T:C\to C$ be a mapping. Denote the set of fixed points of T by $F(T)$. Then T is (i) nonexpansive if $d(Tx,Ty)\le d(x,y)$ for $x,y\in C$ (ii) quasi-nonexpansive if $F(T)\ne \mathrm{\varnothing }$ and $d(Tx,y)\le d(x,y)$ for $x\in C$ and $y\in F(T)$. For an initial value $x_1\in C$, Das and Debata [1] studied the strong convergence of Ishikawa iterates $\{x_n\}$ defined by

$$x_{n+1}={\alpha }_{n}S({\beta }_{n}T{x}_n+(1-{\beta }_n)x_n)+(1-{\alpha }_n)x_n$$

(1.1)

for two quasi-nonexpansive mappings S, T on a nonempty closed and convex subset of a strictly convex Banach space. Takahashi and Tamura [2] proved weak convergence of (1.1) to a common fixed point of two nonexpansive mappings in a uniformly convex Banach space which satisfies Opial’s condition or whose norm is Fréchet differentiable and strong convergence in a strictly convex Banach space (see also [3, 4]). Mann and Ishikawa iterative procedures are well-defined in a vector space through its built-in convexity. In the literature, several mathematicians have introduced the notion of convexity in metric spaces; for example [5–8]. In this work, we follow the original metric convexity introduced by Menger [9] and used by many authors like Kirk [5, 6] and Takahashi [8]. Note that Mann iterative procedures were also investigated in hyperbolic metric spaces [10, 11].

In this paper we investigate the results published in [2] and generalize them to uniformly convex hyperbolic spaces. A particular example of such metric spaces is the class of $CAT(0)$-spaces (in the sense of Gromov) and ℝ-trees (in the sense of Tits). Heavy use of the linear structure of Banach spaces in [2] presents some difficulties when extending these results to metric spaces. For example, a key assumption in many of their results is the smoothness of the norm which is hard to extend to metric spaces.

2 Menger convexity in metric spaces

Let $(X,d)$ be a metric space. Assume that for any x and y in X, there exists a unique metric segment $[x,y]$, which is an isometric copy of the real line interval $[0,d(x,y)]$. Note by ℱ the family of the metric segments in X. For any $\beta \in [0,1]$, there exists a unique point $z\in [x,y]$ such that

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

Throughout this paper we will denote such point by $\beta x\oplus (1-\beta )y$. Such metric spaces are usually called convex metric spaces [9]. Moreover, if we have

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

for all p, q, x, y in X and $\alpha \in [0,1]$, then X is said to be a hyperbolic metric space (see [11–13]). For $q=y$, the hyperbolic inequality reduces to the convex structure inequality [8]. Throughout this paper, we will assume

$$\alpha x\oplus (1-\alpha )y=(1-\alpha )y\oplus \alpha x$$

for any $\alpha \in [0,1]$ and any $x,y\in X$.

An example of hyperbolic spaces is the family of Banach vector spaces or any normed vector spaces. Hadamard manifolds [14], the Hilbert open unit ball equipped with the hyperbolic metric [15], and the $CAT(0)$ spaces [6, 16–20] (see Example 2.1) are examples of nonlinear structures which play a major role in recent research in metric fixed point theory. A subset C of a hyperbolic space X is said to be convex if $[x,y]\subset C$, whenever $x,y\in C$ (see also [21]).

Definition 2.1 [22, 23]

Let $(X,d)$ be a hyperbolic metric space. For any $r>0$ and $\epsilon >0$, set

$$\delta (r,\epsilon )=inf\{1-\frac{1}{r}d(\frac{1}{2}x\oplus \frac{1}{2}y,a);d(x,a)\le r,d(y,a)\le r,d(x,y)\ge r\epsilon \}$$

for any $a\in X$. X is said to be uniformly convex whenever $\delta (r,\epsilon )>0$, for any $r>0$ and $\epsilon >0$.

Throughout this paper we assume that if X is a uniformly convex hyperbolic space, then for every $s\ge 0$ and $\epsilon >0$, there exists $\eta (s,\epsilon )>0$ such that

$$\delta (r,\epsilon )>\eta (s,\epsilon )>0\text{for any }r>s.$$

Remark 2.1

• (i) We have $\delta (r,0)=0$. Moreover, $\delta (r,\epsilon )$ is an increasing function of ε.

• (ii) For $r_1\le r_2$, we have

  $$1-\frac{r_2}{r_1}(1-\delta (r_2,\epsilon \frac{r_1}{r_2}))\le \delta (r_1,\epsilon ).$$

Next we give a very important example of uniformly convex hyperbolic metric space.

Example 2.1 [16]

Let $(X,d)$ be a metric space. A geodesic from x to y in X is a mapping 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. 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 $x,y\in X$, which will be denoted by $[x,y]$, and called the segment joining x to y.

A geodesic triangle $\mathrm{\Delta }(x_1,x_2,x_3)$ in a geodesic metric space X consists of three points $x_1$, $x_2$, $x_3$ in X (the vertices of Δ) and a geodesic segment between each pair of vertices (the edges of Δ). A comparison triangle for geodesic triangle $\mathrm{\Delta }(x_1,x_2,x_3)$ in $(X,d)$ is a triangle $\overline{\mathrm{\Delta }}(x_1,x_2,x_3):=\mathrm{\Delta }({\overline{x}}_1,{\overline{x}}_2,{\overline{x}}_3)$ in ${\mathbb{R}}^2$ such that $d_{{\mathbb{R}}^2}({\overline{x}}_i,{\overline{x}}_j)=d(x_i,x_j)$ for $i,j\in \{1,2,3\}$. Such a triangle always exists (see [18]).

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

Let Δ be a geodesic triangle in X and let $\overline{\mathrm{\Delta }}\subset {\mathbb{R}}^2$ be a comparison triangle for Δ. Then Δ is said to satisfy the $CAT(0)$ inequality if for all $x,y\in \mathrm{\Delta }$ and all comparison points $\overline{x},\overline{y}\in \overline{\mathrm{\Delta }}$,

$$d(x,y)\le d(\overline{x},\overline{y}).$$

Complete $CAT(0)$ spaces are often called Hadamard spaces (see [16]). If x, $y_1$, $y_2$ are points of a $CAT(0)$ space, then the $CAT(0)$ inequality implies

$$d^2(x,\frac{1}{2}{y}_1\oplus \frac{1}{2}{y}_2)\le \frac{1}{2}{d}^2(x,y_1)+\frac{1}{2}{d}^2(x,y_2)-\frac{1}{4}{d}^2(y_1,y_2).$$

The above inequality is known as the (CN) inequality of Bruhat and Tits [24]. The (CN) inequality implies that $CAT(0)$ spaces are uniformly convex with

$$\delta (r,\epsilon )=1-\sqrt{1-\frac{{\epsilon }^2}{4}}.$$

In a hyperbolic space X, (1.1) is written as

$$x_{n+1}={\alpha }_{n}S({\beta }_{n}T{x}_n\oplus (1-{\beta }_n)x_n)\oplus (1-{\alpha }_n)x_n,$$

(2.1)

where ${\alpha }_n,{\beta }_n\in [0,1]$. If $S=T$ in (2.1), it reduces to the following Ishikawa iteration process of one mapping:

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

(2.2)

where ${\alpha }_n,{\beta }_n\in [0,1]$. Let $\{x_n\}$ be a bounded sequence in a metric space X and C be a nonempty subset. Define $r(\cdot ,\{x_n\}):C\to [0,\mathrm{\infty })$, by

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

The asymptotic radius ${\rho }_C$ of $\{x_n\}$ with respect to C is given by

$${\rho }_C=inf\{r(x,\{x_n\}):x\in C\}.$$

ρ will denote the asymptotic radius of $\{x_n\}$ with respect to X. A point $\xi \in C$ is said to be an asymptotic center of $\{x_n\}$ with respect to C if $r(\xi ,\{x_n\})=r(C,\{x_n\})=min\{r(x,\{x_n\}):x\in C\}$. We denote with $A(C,\{x_n\})$, the set of asymptotic centers of $\{x_n\}$ with respect to C. When $C=X$, we call ξ an asymptotic center of $\{x_n\}$ and we use the notation $A(\{x_n\})$ instead of $A(X,\{x_n\})$. In general, the set $A(C,\{x_n\})$ of asymptotic centers of a bounded sequence $\{x_n\}$ may be empty or may even contain infinitely many points. Note that in the study of the geometry of Banach spaces, the function $r(\cdot ,\{x_n\})$ is also known as a type. For more on types in metric spaces, we refer to [25].

The Δ-convergence, introduced independently several years ago by Kuczumow [26] and Lim [27], is shown in $CAT(0)$ spaces to behave similarly as the weak convergence in Banach spaces.

Definition 2.2 A bounded sequence $\{x_n\}$ in X is said to Δ-converge to $x\in X$ if x is the unique asymptotic center of every subsequence $\{u_n\}$ of $\{x_n\}$. We write $x_n\stackrel{\mathrm{\Delta }}{\to }x$ ($\{x_n\}$ Δ-converges to x).

In this paper, we study the iteration schemes (2.1)-(2.2) for nonexpansive mappings. We study strong convergence of these iterates in strictly convex hyperbolic spaces and prove Δ-convergence results in uniformly convex hyperbolic spaces without requiring any condition similar to norm Fréchet differentiability.

In the sequel, the following results will be needed.

Lemma 2.1 [25, 28]

Let X be a hyperbolic metric spaces. Assume that X is uniformly convex. Let C be a nonempty, closed and convex subset of X. Then every bounded sequence $\{x_n\}\in X$ has a unique asymptotic center with respect to C.

Lemma 2.2 [25, 28]

Let X be a hyperbolic metric spaces. Assume that X is uniformly convex. Let C be a nonempty, closed and convex subset of X. Let C be a nonempty closed and convex subset of X, and $\{x_n\}$ be a bounded sequence in C such that $A(\{x_n\})=\{y\}$ and $r(\{x_n\})=\rho$. If $\{y_m\}$ is a sequence in C such that ${lim}_{m\to \mathrm{\infty }}r(y_m,\{x_n\})=\rho$, then ${lim}_{m\to \mathrm{\infty }}{y}_m=y$.

The following lemma [29] will be useful in studying the sequence generated by (2.1) in uniformly convex metric spaces. Here we give a proof based on the ideas developed in [25].

Lemma 2.3 Let X be a uniformly convex hyperbolic space. Then for arbitrary positive numbers $\epsilon >0$ and $r>0$, and $\alpha \in [0,1]$, we have

$$d(a,\alpha x\oplus (1-\alpha )y)\le r(1-\delta (r,2min\{\alpha ,1-\alpha \}\epsilon ))$$

for all $a,x,y\in X$, such that $d(z,x)\le r$, $d(z,y)\le r$, and $d(x,y)\ge r\epsilon$.

Proof Without loss of generality, we may assume $\alpha <\frac{1}{2}$. In this case, we have $min\{\alpha ,1-\alpha \}=\alpha$. Let $a\in X$ be fixed and $x,y\in X$. Set $\overline{x}=2\alpha x\oplus (1-2\alpha )y$. Since

$$d(\frac{1}{2}\overline{x}\oplus \frac{1}{2}y,x)\le (1-\alpha )d(x,y)\text{and}d(\frac{1}{2}\overline{x}\oplus \frac{1}{2}y,y)=\alpha d(x,y),$$

the uniform convexity of X will imply $\frac{1}{2}\overline{x}\oplus \frac{1}{2}y=\alpha x\oplus (1-\alpha )y$. Using the uniform convexity of X, we get

$$d(a,\frac{1}{2}\overline{x}\oplus \frac{1}{2}y)\le r(1-\delta (r,2\alpha \epsilon )).$$

Hence

$$d(a,\alpha x\oplus (1-\alpha )y)\le r(1-\delta (r,2min\{\alpha ,1-\alpha \}\epsilon )).$$

 □

Remark 2.2 If $(X,d)$ is uniformly convex, then $(X,d)$ is strictly convex, i.e., whenever

$$d(\alpha x\oplus (1-\alpha )y,a)=d(x,a)=d(y,a)$$

for $\alpha \in (0,1)$ and any $x,y,a\in X$, then we must have $x=y$.

The following result is very useful.

Lemma 2.4 [25]

Let $(X,d)$ be a uniformly convex hyperbolic space. Let $R\in [0,+\mathrm{\infty })$ be such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_n,a)\le R,\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(y_n,a)\le R\mathit{\text{and}}\underset{n\to \mathrm{\infty }}{lim}d(a,\frac{1}{2}{x}_n\oplus \frac{1}{2}{y}_n)=R.$$

Then we have

$$\underset{n\to \mathrm{\infty }}{lim}d(x_n,y_n)=0.$$

But since we use convex combinations other than the middle point, we will need the following generalization obtained by using Lemma 2.3.

Lemma 2.5 Let $(X,d)$ be a uniformly convex hyperbolic space. Let $R\in [0,+\mathrm{\infty })$ be such that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}d(x_n,a)\le R$, ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}d(y_n,a)\le R$, and

$$\underset{n\to \mathrm{\infty }}{lim}d(a,{\alpha }_n{x}_n\oplus (1-{\alpha }_n)y_n)=R,$$

where ${\alpha }_n\in [a,b]$, with $0<a\le b<1$. Then we have

$$\underset{n\to \mathrm{\infty }}{lim}d(x_n,y_n)=0.$$

A subset C of a metric space X is Chebyshev if for every $x\in X$, there exists $c_0\in C$ such that $d(c_0,x)<d(c,x)$ for all $c\in C$, $c\ne c_0$. In other words, for each point of the space, there is a well-defined nearest point of C. We can then define the nearest point projection $P:X\to C$ by sending x to $c_0$. We have the following result.

Lemma 2.6 [25]

Let $(X,d)$ be a complete uniformly convex hyperbolic space. Let C be nonempty, convex and closed subset of X. Let $x\in X$ be such that $d(x,C)<\mathrm{\infty }$. Then there exists a unique best approximant of x in C, i.e., there exists a unique $c_0\in C$ such that

$$d(x,c_0)=d(x,C)=inf\{d(x,c);c\in C\},$$

i.e., C is Chebyshev.

3 Convergence in strictly convex hyperbolic space

Let $(X,d)$ be a hyperbolic metric space. Let C be a nonempty closed convex subset of X. Let $S,T:C\to C$ be two nonexpansive mappings. Throughout the paper, assume that $F=F(S)\cap F(T)$. Let $x_1\in C$ and $p\in F$ (assuming F is not empty). Set $r=d(x_1,p)$. Then

$$C(x_1)=C\cap B(p,r)=\{x\in C;d(p,x)\le r\}$$

is nonempty and invariant by both S and T. Therefore one may always assume that C is bounded provided that S and T have a common fixed point. Moreover, if $\{x_n\}$ is the sequence generated by (2.1), then we have

$$\begin{array}{rcl}d(x_{n+1},p)& =& d({\alpha }_{n}S{y}_n\oplus (1-{\alpha }_n)x_n,p)\\ \le & {\alpha }_{n}d(S{y}_n,p)+(1-{\alpha }_n)d(x_n,p)\\ \le & {\alpha }_{n}d(y_n,p)+(1-{\alpha }_n)d(x_n,p)\\ =& {\alpha }_{n}d({\beta }_{n}T{x}_n\oplus (1-{\beta }_n)x_n,p)+(1-{\alpha }_n)d(x_n,p)\\ \le & {\alpha }_n[{\beta }_{n}d(T{x}_n,p)+(1-{\beta }_n)d(x_n,p)]+(1-{\alpha }_n)d(x_n,p)\\ \le & d(x_n,p),\end{array}$$

where $y_n={\beta }_{n}T{x}_n\oplus (1-{\beta }_n)x_n$. This proves that $\{d(x_n,p)\}$ is decreasing, which implies that ${lim}_{n\to \mathrm{\infty }}d(x_n,p)$ exists. Using the above inequalities, we get

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim}d(x_n,p)& =& \underset{n\to \mathrm{\infty }}{lim}d({\alpha }_{n}S{y}_n\oplus (1-{\alpha }_n)x_n,p)\\ =& \underset{n\to \mathrm{\infty }}{lim}{\alpha }_{n}d(S{y}_n,p)+(1-{\alpha }_n)d(x_n,p)\\ =& \underset{n\to \mathrm{\infty }}{lim}{\alpha }_{n}d(y_n,p)+(1-{\alpha }_n)d(x_n,p)\\ =& \underset{n\to \mathrm{\infty }}{lim}{\alpha }_{n}d({\beta }_{n}T{x}_n\oplus (1-{\beta }_n)x_n,p)+(1-{\alpha }_n)d(x_n,p)\\ =& \underset{n\to \mathrm{\infty }}{lim}{\alpha }_n[{\beta }_{n}d(T{x}_n,p)+(1-{\beta }_n)d(x_n,p)]+(1-{\alpha }_n)d(x_n,p).\end{array}$$

The first result of this work discusses the convergence behavior of the sequence generated by (2.1).

Theorem 3.1 Let X be a strictly convex hyperbolic space. Let C be a nonempty bounded, closed and convex subset of X. Let $S,T:C\to C$ be two nonexpansive mappings. Assume that $F\ne \mathrm{\varnothing }$. Let $x_1\in C$ and $\{x_n\}$ be given by (2.1). Then the following hold:

• (i) if ${\alpha }_n\in [a,b]$ and ${\beta }_n\in [0,b]$, with $0<a\le b<1$, then $x_{n_i}\to y$ implies $y\in F(S)$;

• (ii) if ${\alpha }_n\in [a,1]$ and ${\beta }_n\in [a,b]$, with $0<a\le b<1$, then $x_{n_i}\to y$ implies $y\in F(T)$;

• (iii) if ${\alpha }_n,{\beta }_n\in [a,b]$, with $0<a\le b<1$, then $x_{n_i}\to y$ implies $y\in F$. In this case, we have $x_n\to y$.

Proof Assume that $x_{n_i}\to y$. Let $p\in F$. Without loss of generality, we may assume ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n_i}=\alpha$ and ${lim}_{n\to \mathrm{\infty }}{\beta }_{n_i}=\beta$. Since $\{d(x_n,p)\}$ is decreasing, we get

$$\underset{n_i\to \mathrm{\infty }}{lim}d(x_n,p)=\underset{n_i\to \mathrm{\infty }}{lim}d(x_{n_i},p)=d(y,p).$$

The above inequalities imply

$$\begin{array}{rcl}d(y,p)& =& d(\alpha S(\beta Ty\oplus (1-\beta )y)\oplus (1-\alpha )y,p)\\ =& \alpha d(S(\beta Ty\oplus (1-\beta )y),p)+(1-\alpha )d(y,p)\\ =& \alpha d(\beta Ty\oplus (1-\beta )y,p)+(1-\alpha )d(y,p)\\ =& \alpha [\beta d(Ty,p)+(1-\beta )d(y,p)]+(1-\alpha )d(y,p).\end{array}$$

Set $r=d(y,p)$. Without loss of generality we may assume $r>0$ otherwise most of the conclusions in the theorem are trivial. Assume that ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n>0$. Then $\alpha \ne 0$. Hence

$$d(S(\beta Ty\oplus (1-\beta )y),p)=d(\beta Ty\oplus (1-\beta )y,p)=\beta d(Ty,p)+(1-\beta )r=r,$$

which implies $\beta d(Ty,p)=\beta r$. If we assume that ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n>0$, then $\beta \ne 0$, which implies $d(Ty,p)=r$.

• (1) If $\alpha \in (0,1)$ and $\beta >0$, then

  $$d(p,S(\beta Ty\oplus (1-\beta )y))=d(\alpha S(\beta Ty\oplus (1-\beta )y)\oplus (1-\alpha )y,p)=r.$$

  The strict convexity of X will imply $S(\beta Ty\oplus (1-\beta )y)=y$.

• (2) If $\alpha \in (0,1)$ and $\beta =0$, then

  $$d(p,y)=d(p,S(y))=d(\alpha S(y)\oplus (1-\alpha )y,p).$$

The strict convexity of X will imply $S(y)=y$.

• (3) If $\beta \in (0,1)$ and $\alpha >0$, then

  $$d(p,y)=d(p,T(y))=d(p,\beta Ty\oplus (1-\beta )y).$$

The strict convexity of X will imply $T(y)=y$.

• (4) If $\alpha ,\beta \in (0,1)$, then $T(y)=y$ and $S(\beta Ty\oplus (1-\beta )y)=y$. Hence $S(y)=y$.

Let us finish the proof of Theorem 3.1. Note that (i) implies $\alpha \in [a,b]$ and $\beta \in [0,b]$. If $\beta =0$, then the conclusion (2) above implies $y\in F(S)$. Otherwise the conclusion (4) will imply $y\in F$. This proves (i).

For (ii), notice that $\alpha \in [a,1]$ and $\beta \in [a,b]$. Hence the conclusion (3) will imply $y\in F(T)$ which proves (ii).

For (iii), notice that $\alpha ,\beta \in [a,b]$. Hence the conclusion (4) will imply $y\in F(T)$. Since

$$\underset{n\to \mathrm{\infty }}{lim}d(y_n,y)=\underset{n\to \mathrm{\infty }}{lim}d(y_{n_i},y)=0,$$

we get $x_n\to y$, which completes the proof of (iii). □

If we assume compactness, Theorem 3.1 will imply the following result.

Theorem 3.2 Let X be a strictly convex hyperbolic space. Let C be a nonempty bounded, closed and convex subset of X. Let $S,T:C\to C$ be two nonexpansive mappings. Assume that $F\ne \mathrm{\varnothing }$. Fix $x_1\in C$. Assume that $\overline{co}\{\{x_1\}\cup S(C)\cup T(C)\}$ is a compact subset of C. Define $\{x_n\}$ as in (2.1) where ${\alpha }_n,{\beta }_n\in [a,b]$, with $0<a\le b<1$, and $x_1$ is the initial element of the sequence. Then $\{x_n\}$ converges strongly to a common fixed point of S and T.

Proof We have $x_n\in \overline{co}\{\{x_1\}\cup S(C)\cup T(C)\}$. Since $\overline{co}\{\{x_1\}\cup S(C)\cup T(C)\}$ is compact, $\{x_n\}$ has a convergent subsequence $\{x_{n_i}\}$, i.e., $x_{n_i}\to z$. By Theorem 3.1, we have $z\in F$ and $x_n\to z$. □

The existence of a common fixed point T and S is crucial. If one assumes that T and S commute, i.e., $S\circ T=T\circ S$, then a common fixed point exists under the assumptions of Theorem 3.2. Indeed, fix $x_0\in C$ and define

$$T_{n}x=\frac{1}{n}{x}_0\oplus (1-\frac{1}{n})Tx$$

for $x\in C$ and $n\ge 1$. Then

$$\begin{array}{rcl}d(T_{n}x,T_{n}y)& =& d(\frac{1}{n}{x}_0\oplus (1-\frac{1}{n})Tx,\frac{1}{n}{x}_0\oplus (1-\frac{1}{n})Ty)\\ \le & (1-\frac{1}{n})d(Tx,Ty)\\ \le & (1-\frac{1}{n})d(x,y).\end{array}$$

That is, $T_n$ is a contraction. By the Banach contraction principle (BCP), $T_n$ has a unique fixed point $u_n$ in C. Since the closure of $T(C)$ is compact, there exists a subsequence $\{T{u}_{n_i}\}$ of $\{T{u}_n\}$ such that $T{u}_{n_i}\to u$. Since $T(C)$ is bounded and

$$d(u_n,T{u}_n)=d(\frac{1}{n}{x}_0\oplus (1-\frac{1}{n})T{u}_n,T{u}_n)\le \frac{1}{n}d(x_0,T{u}_n),$$

we have $d(u_n,T{u}_n)\to 0$. In particular, we have $u_{n_i}\to u$. Continuity of T implies $Tu=u$. Since X is strictly convex, then $F(T)$ is a nonempty convex subset of X. Since T and S commute, we have $S(F(T))\subset F(T)$. Moreover, since the closure of $T(C)$ is compact, we see that $F(T)$ is compact. The above proof shows that S has a fixed point in $F(T)$, i.e., $F=F(S)\cap F(T)\ne \mathrm{\varnothing }$.

The case $S=T$ gives the following result.

Theorem 3.3 Let C be a nonempty closed and convex subset of a complete strictly convex hyperbolic space X. Let $T:C\to C$ be a nonexpansive mapping such that $\overline{co}\{\{c_0\}\cup T(C)\}$ is a compact subset of C, where $c_0\in C$. Define $\{x_n\}$ by (2.2), where $x_1=c_0$, ${\alpha }_n\in [a,b]$ and ${\beta }_n\in [0,b]$ or ${\alpha }_n\in [a,1]$ and ${\beta }_n\in [a,b]$, with $0<a\le b<1$. Then $\{x_n\}$ converges strongly to a fixed point of T.

Proof We saw that in this case, we have $F(T)\ne \mathrm{\varnothing }$. Since $x_n\in \overline{co}\{\{x_1\}\cup T(C)\}$. Then there exists a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ such that $x_{n_i}\to z\in C$. By Theorem 3.1, we have $Tz=z$ and $x_n\to z$. □

4 Convergence in uniformly convex hyperbolic spaces

The following result is similar to the well-known demi-closedness principle discovered by Göhde in uniformly convex Banach spaces [30].

Lemma 4.1 Let C be a nonempty, closed and convex subset of a complete uniformly convex hyperbolic space X. Let $T:C\to C$ be a nonexpansive mapping. Let $\{x_n\}\in C$ be an approximate fixed point sequence of T, i.e., ${lim}_{n\to \mathrm{\infty }}d(x_n,T{x}_n)=0$. If $x\in C$ is the asymptotic center of $\{x_n\}$ with respect to C, then x is a fixed point of T, i.e., $x\in F(T)$. In particular, if $\{x_n\}\in C$ is an approximate fixed point sequence of T, such that $x_n\stackrel{\mathrm{\Delta }}{\to }x$, then $x\in F(T)$.

Proof Let $\{x_n\}$ be an approximate fixed point sequence of T. Let $x\in C$ be the unique asymptotic center of $\{x_n\}$ with respect to C. Since

$$d(Tx,x_n)\le d(Tx,T{x}_n)+d(T{x}_n,x_n)\le d(x,x_n)+d(T{x}_n,x_n),$$

we get

$$\begin{array}{rcl}r(Tx,\{x_n\})& =& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(Tx,x_n)\\ \le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}[d(x,x_n)+d(T{x}_n,x_n)]=r(x,\{x_n\}).\end{array}$$

By the uniqueness of the asymptotic center, we get $Tx=x$. □

The following theorem is necessary to discuss the behavior of the iterates in (2.1).

Theorem 4.1 Let C be a nonempty, closed and convex subset of a complete uniformly convex hyperbolic space X. Let $S,T:C\to C$ be nonexpansive mappings such that $F\ne \mathrm{\varnothing }$. Fix $x_1\in C$ and generate $\{x_n\}$ by (2.1). Set

$$y_n={\beta }_{n}T{x}_n\oplus (1-{\beta }_n)x_n$$

for any $n\ge 1$.

• (i) If ${\alpha }_n\in [a,b]$, where $0<a\le b<1$, then

  $$\underset{n\to \mathrm{\infty }}{lim}d(x_n,S{y}_n)=0.$$

• (ii) If ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n>0$ and ${\beta }_n\in [a,b]$, with $0<a\le b<1$, then

  $$\underset{n\to \mathrm{\infty }}{lim}d(x_n,T{x}_n)=0.$$

• (iii) If ${\alpha }_n,{\beta }_n\in [a,b]$, with $0<a\le b<1$, then

  $$\underset{n\to \mathrm{\infty }}{lim}d(x_n,S{x}_n)=0\mathit{\text{and}}\underset{n\to \mathrm{\infty }}{lim}d(x_n,T{x}_n)=0.$$

Proof Let $p\in F$. Then the sequence $\{d(x_n,p)\}$ is decreasing. Set $c={lim}_{n\to \mathrm{\infty }}d(x_n,p)$. If $c=0$, then all the conclusions are trivial. Therefore we will assume that $c>0$. Note that we have

$$d(x_{n+1},p)\le {\alpha }_{n}d(S{y}_n,p)+(1-{\alpha }_n)d(x_n,p)$$

(4.1)

and

$$d(S{y}_n,p)\le d(y_n,p)\le {\beta }_{n}d(T{x}_n,p)+(1-{\beta }_n)d(x_n,p)\le d(x_n,p)$$

(4.2)

for any $n\ge 1$. In order to prove (i), assume that ${\alpha }_n\in [a,b]$, where $0<a\le b<1$. From the inequalities (4.1) and (4.2), we get

$$d(x_{n+1},p)=d({\alpha }_{n}S{y}_n\oplus (1-{\alpha }_n)x_n,p)\le {\alpha }_{n}d(S{y}_n,p)+(1-{\alpha }_n)d(x_n,p)\le d(x_n,p),$$

which implies ${lim}_{n\to \mathrm{\infty }}d(S{y}_n,p)=c$. Indeed, let $\mathcal{U}$ be an ultrafilter over ℕ. Then we have ${lim}_{\mathcal{U}}{\alpha }_n=\alpha \in [a,b]$ and ${lim}_{\mathcal{U}}d(x_n,p)={lim}_{\mathcal{U}}d(x_{n+1},p)=c$. Hence

$$c\le \alpha \underset{\mathcal{U}}{lim}d(S{y}_n,p)+(1-\alpha )c\le c.$$

Since $\alpha \ne 0$, we get ${lim}_{\mathcal{U}}d(S{y}_n,p)=c$. Since $\mathcal{U}$ was an arbitrary ultrafilter, we get ${lim}_{n\to \mathrm{\infty }}d(S{y}_n,p)=c$ as claimed. Therefore we have

$$\underset{n\to \mathrm{\infty }}{lim}d(x_n,p)=\underset{n\to \mathrm{\infty }}{lim}d(S{y}_n,p)=\underset{n\to \mathrm{\infty }}{lim}d({\alpha }_{n}S{y}_n\oplus (1-{\alpha }_n)x_n,p)=c.$$

Using Lemma 2.5, we get ${lim}_{n\to \mathrm{\infty }}d(S{y}_n,x_n)=0$.

Next we prove (ii). Assume ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n>0$ and ${\beta }_n\in [a,b]$, with $0<a\le b<1$. First note that from (4.1) and (4.2), we get

$$d(x_{n+1},p)\le {\alpha }_{n}d(y_n,p)+(1-{\alpha }_n)d(x_n,p)\le d(x_n,p),$$

which implies ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}d(y_n,p)+(1-{\alpha }_n)d(x_n,p)=c$. Since ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n>0$, we conclude that ${lim}_{n\to \mathrm{\infty }}d(y_n,p)=c$. Since ${\beta }_n\ge a>0$, we get in a similar fashion ${lim}_{n\to \mathrm{\infty }}d(T{x}_n,p)=c$. Therefore we have

$$\underset{n\to \mathrm{\infty }}{lim}d(x_n,p)=\underset{n\to \mathrm{\infty }}{lim}d(T{x}_n,p)=\underset{n\to \mathrm{\infty }}{lim}d({\beta }_{n}T{x}_n\oplus (1-{\beta }_n)x_n,p)=c.$$

Using Lemma 2.5, we get ${lim}_{n\to \mathrm{\infty }}d(T{x}_n,x_n)=0$.

Finally let us prove (iii). Assume that ${\alpha }_n,{\beta }_n\in [a,b]$, with $0<a\le b<1$. Then from (i) and (ii), we know that

$$\underset{n\to \mathrm{\infty }}{lim}d(x_n,S{y}_n)=0\text{and}\underset{n\to \mathrm{\infty }}{lim}d(x_n,T{x}_n)=0.$$

Since

$$\begin{array}{rcl}d(x_n,S{x}_n)& \le & d(x_n,S{y}_n)+d(S{y}_n,S{x}_n)\\ \le & d(x_n,S{y}_n)+d(y_n,x_n)\\ =& d(x_n,S{y}_n)+{\beta }_{n}d(T{x}_n,x_n)\\ \le & d(x_n,S{y}_n)+d(T{x}_n,x_n),\end{array}$$

we conclude that ${lim}_{n\to \mathrm{\infty }}d(x_n,S{x}_n)=0$. □

The conclusion of Theorem 4.1(iii) is amazing because the sequence generated by (2.1) gives an approximate fixed point sequence for both S and T without assuming that these mappings commute.

Remark 4.1 If we assume that $0\le {\beta }_n\le b<1$ and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n>0$, then

$$\underset{n\to \mathrm{\infty }}{lim}{\beta }_{n}d(x_n,T{x}_n)=0.$$

Indeed, if we assume this not to be so, then there exists a subsequence $\{{\beta }_{n^{\prime }}\}$ and $\delta >0$ such that

$${\beta }_{n^{\prime }}d(x_{n^{\prime }},T{x}_{n^{\prime }})\ge \delta$$

for any $n\ge 1$. In particular, it is clear, since $\{d(x_n,T{x}_n)\}$ is bounded, that ${lim}_{n\to \mathrm{\infty }}{\beta }_{n^{\prime }}\ne 0$. Without loss of generality, we may assume that ${\beta }_{n^{\prime }}\ge a>0$, for $n\ge 1$. The proof of (ii) will imply

$$\underset{n\to \mathrm{\infty }}{lim}d(x_{n^{\prime }},T{x}_{n^{\prime }})=0,$$

which is a contradiction since $\{{\beta }_n\}$ is a bounded sequence. Therefore we must have

$$\underset{n\to \mathrm{\infty }}{lim}{\beta }_{n}d(x_n,T{x}_n)=0.$$

In particular, if we assume ${\alpha }_n\in [a,b]$, then we get

$$\underset{n\to \mathrm{\infty }}{lim}d(x_n,S{x}_n)=0.$$

As a direct consequence to Theorem 4.1 and Remark 4.1, we get the following result which discusses the Δ-convergence of the iterative sequence defined by (2.1).

Theorem 4.2 Let C be a nonempty, closed and convex subset of a complete uniformly convex hyperbolic space X. Let $S,T:C\to C$ be two nonexpansive mappings such that $F\ne \mathrm{\varnothing }$. Fix $x_1\in C$ and generate $\{x_n\}$ by (2.1). Then

• (i) if ${\alpha }_n\in [a,b]$ and ${\beta }_n\in [0,b]$, with $0<a\le b<1$, then $x_n\stackrel{\mathrm{\Delta }}{\to }y$ and $y\in F(S)$;

• (ii) if ${\alpha }_n\in [a,1]$ and ${\beta }_n\in [a,b]$, with $0<a\le b<1$, then $x_n\stackrel{\mathrm{\Delta }}{\to }y$ and $y\in F(T)$;

• (iii) if ${\alpha }_n,{\beta }_n\in [a,b]$, with $0<a\le b<1$, then $x_n\stackrel{\mathrm{\Delta }}{\to }y$ and $y\in F$.

Proof Let us prove (i). Assume ${\alpha }_n\in [a,b]$ and ${\beta }_n\in [0,b]$, with $0<a\le b<1$. Theorem 4.1 and Remark 4.1 imply that $\{x_n\}$ is an approximate fixed point sequence of S, i.e.,

$$\underset{n\to \mathrm{\infty }}{lim}d(x_n,S{x}_n)=0.$$

Let y be the unique asymptotic center of $\{x_n\}$. Then Lemma 4.1 implies that $y\in F(S)$. Let us prove that in fact $\{x_n\}$ Δ-converges to y. Let $\{x_{n_i}\}$ be a subsequence of $\{x_n\}$. Let z be the unique asymptotic center of $\{x_{n_i}\}$. Again since $\{x_{n_i}\}$ is an approximate fixed point sequence of S, we get $z\in F(S)$. Hence

$$\underset{n_i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_{n_i},z)\le \underset{n_i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_{n_i},y).$$

Since $y,z\in F(S)$, we get

$$\underset{n_i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_{n_i},z)=\underset{n\to \mathrm{\infty }}{lim}d(x_n,z)\text{and}\underset{n_i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_{n_i},y)=\underset{n\to \mathrm{\infty }}{lim}d(x_n,y).$$

Since y is the unique asymptotic center of $\{x_n\}$, we get $y=z$. This proves that $\{x_n\}$ Δ-converges to y.

Next we prove (ii). Assume ${\alpha }_n\in [a,1]$ and ${\beta }_n\in [a,b]$, with $0<a\le b<1$. Then Theorem 4.1 implies that $\{x_n\}$ is an approximate fixed point sequence of T, i.e.,

$$\underset{n\to \mathrm{\infty }}{lim}d(x_n,T{x}_n)=0.$$

Following the same proof as given above for (i), we get $\{x_n\}$ Δ-converges to its unique asymptotic center which is a fixed point of T.

The conclusion (iii) follows easily from (i) and (ii). □

As a corollary to Theorem 4.2, we get the following result when $S=T$.

Corollary 4.1 Let C be a nonempty, closed and convex subset of a complete uniformly convex hyperbolic space X. Let $T:C\to C$ be a nonexpansive mapping with a fixed point. Suppose that $\{x_n\}$ is given by (2.2), where ${\alpha }_n\in [a,b]$ and ${\beta }_n\in [0,b]$ or ${\alpha }_n\in [a,1]$ and ${\beta }_n\in [a,b]$, with $0<a\le b<1$. Then $x_n\stackrel{\mathrm{\Delta }}{\to }p$, with $p\in F(T)$.

Using the concept of near point projection, we establish the following amazing convergence result.

Theorem 4.3 Let C be a nonempty, closed and convex subset of a complete uniformly convex hyperbolic space X. Let $S,T:C\to C$ be nonexpansive mappings such that $F\ne \mathrm{\varnothing }$. Let P be the nearest point projection of C onto F. For an initial value $x_1\in C$, define $\{x_n\}$ as given in (2.1), where ${\alpha }_n,{\beta }_n\in [a,b]$, with $0<a\le b<1$. Then $\{P{x}_n\}$ converges strongly to the asymptotic center of $\{x_n\}$.

Proof First, we claim that

$$d(P{x}_n,x_{n+m})\le d(P{x}_n,x_n)\text{for }m\ge 1,n\ge 1.$$

(4.3)

We prove (4.3) by induction on $m\ge 1$. For $m=1$, we have

$$\begin{array}{rcl}d(P{x}_n,x_{n+1})& =& d(P{x}_n,{\alpha }_{n}S{y}_n\oplus (1-{\alpha }_n)x_n)\\ \le & {\alpha }_{n}d(P{x}_n,S{y}_n)+(1-{\alpha }_n)d(P{x}_n,x_n)\\ \le & {\alpha }_{n}d(P{x}_n,y_n)+(1-{\alpha }_n)d(P{x}_n,x_n)\\ =& {\alpha }_{n}d(P{x}_n,{\beta }_{n}T{x}_n\oplus (1-{\beta }_n)x_n)+(1-{\alpha }_n)d(P{x}_n,x_n)\\ \le & {\alpha }_n[{\beta }_{n}d(P{x}_n,T{x}_n)+(1-{\beta }_n)d(P{x}_n,x_n)]+(1-{\alpha }_n)d(P{x}_n,x_n)\\ \le & {\alpha }_n[{\beta }_{n}d(P{x}_n,x_n)+(1-{\beta }_n)d(P{x}_n,x_n)]+(1-{\alpha }_n)d(P{x}_n,x_n)\\ =& d(P{x}_n,x_n).\end{array}$$

That is,

$$d(P{x}_n,x_{n+1})\le d(P{x}_n,x_n)$$

for $n\ge 1$. Assume that (4.3) is true for $m=k$. That is,

$$d(P{x}_n,x_{n+k})\le d(P{x}_n,x_n)$$

for $n\ge 1$. Hence

$$\begin{array}{rcl}d(P{x}_n,x_{n+k+1})& =& d(P{x}_n,{\alpha }_{n+k}S{y}_{n+k}\oplus (1-{\alpha }_{n+k})x_{n+k})\\ \le & {\alpha }_{n+k}d(P{x}_n,S{y}_{n+k})+(1-{\alpha }_{n+k})d(P{x}_n,x_{n+k})\\ \le & {\alpha }_{n+k}d(P{x}_n,y_{n+k})+(1-{\alpha }_{n+k})d(P{x}_n,x_{n+k})\\ =& {\alpha }_{n+k}d(P{x}_n,{\beta }_{n+k}T{x}_{n+k}\oplus (1-{\beta }_{n+k})x_{n+k})\\ +(1-{\alpha }_{n+k})d(P{x}_n,x_{n+k})\\ \le & {\alpha }_{n+k}[{\beta }_{n+k}d(P{x}_n,T{x}_{n+k})+(1-{\beta }_{n+k})d(P{x}_n,x_{n+k})]\\ +(1-{\alpha }_{n+k})d(P{x}_n,x_{n+k})\\ \le & {\alpha }_{n+k}[{\beta }_{n+k}d(P{x}_n,x_{n+k})+(1-{\beta }_{n+k})d(P{x}_n,x_{n+k})]\\ +(1-{\alpha }_{n+k})d(P{x}_n,x_{n+k})\\ =& d(P{x}_n,x_{n+k})\\ \le & d(P{x}_n,x_n).\end{array}$$

This completes the proof of (4.3). Let us complete the proof of Theorem 4.3. We know from Theorem 4.2(iii) that $\{x_n\}$ Δ-converges to its unique asymptotic center y, which is in F. Let us prove that $\{P{x}_n\}$ converges strongly to y. Assume not, i.e., there exist $\epsilon >0$ and a subsequence $\{P{x}_{n_i}\}$ such that $d(P{x}_{n_i},y)\ge \epsilon$, for any $n_i\ge 1$. It is clear that we must have $R=d(x_1,y)>0$, otherwise $\{x_n\}$ is a constant sequence. Since

$$\{\begin{array}{l}d(x_{n_i},y)\le d(x_{n_i},y),\\ d(x_{n_i},P{x}_{n_i})\le d(x_{n_i},y),\\ d(P{x}_{n_i},y)\ge \epsilon =d(x_{n_i},y)\frac{\epsilon }{d(x_{n_i},y)}\ge d(x_{n_i},y)\frac{\epsilon }{R}\end{array}$$

we get

$$d(x_{n_i},\frac{1}{2}P{x}_{n_i}\oplus \frac{1}{2}y)\le d(x_{n_i},y)(1-\delta (d(x_{n_i},y),\frac{\epsilon }{R}))$$

for any $n_i\ge 1$. Using the properties of the modulus of uniform convexity, there exists $\eta >0$ such that

$$\delta (d(x_{n_i},y),\frac{\epsilon }{R})\ge \eta$$

for any $n_i\ge 1$. Hence

$$d(x_{n_i},\frac{1}{2}P{x}_{n_i}\oplus \frac{1}{2}y)\le d(x_{n_i},y)(1-\eta )$$

for any $n_i\ge 1$. Using the definition of the nearest point projection P, we get

$$d(x_{n_i},P{x}_{n_i})\le d(x_{n_i},y)(1-\eta )$$

for any $n_i\ge 1$. Using the inequality (4.3) above, we get

$$d(x_{n_i+m},P{x}_{n_i})\le d(x_{n_i},y)(1-\eta )$$

for any $n_i\ge 1$ and $m\ge 1$. Since $P{x}_{n_i}\in F$, we know that $\{d(x_n,P{x}_{n_i})\}$ is decreasing (in n and fixed $n_i$). Hence

$$\underset{m\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_{n_i+m},P{x}_{n_i})=\underset{n\to \mathrm{\infty }}{lim}d(x_n,P{x}_{n_i})\le d(x_{n_i},y)(1-\eta )$$

for any $n_i\ge 1$. Since y is the asymptotic center of $\{x_n\}$, we get

$$\underset{n\to \mathrm{\infty }}{lim}d(x_n,y)\le \underset{n\to \mathrm{\infty }}{lim}d(x_n,P{x}_{n_i})\le d(x_{n_i},y)(1-\eta )$$

for any $n_i\ge 1$. Finally since $y\in F$, if we let $n_i\to \mathrm{\infty }$, we get

$$\underset{n\to \mathrm{\infty }}{lim}d(x_n,y)\le \underset{n\to \mathrm{\infty }}{lim}d(x_n,y)(1-\eta ).$$

Since $\epsilon \le d(x_{n_i},P{x}_{n_i})\le d(x_{n_i},y)$, we conclude that $\epsilon \le {lim}_{n\to \mathrm{\infty }}d(x_n,y)$, which implies $1\le 1-\eta$ which is our desired contradiction. Therefore $\{P{x}_n\}$ converges strongly to y. □