Another type of Mann iterative scheme for two mappings in a complete geodesic space

Abstract

In this paper, we show a Δ-convergence theorem for a Mann iteration procedure in a complete geodesic space with two quasinonexpansive and Δ-demiclosed mappings. The proposed method is different from known procedures with respect to the order of taking the convex combination.

1 Introduction

The fixed point approximation has been studied in a variety of ways and its results are useful for the other studies. In 1953, Mann [1] introduced an iteration procedure for approximating fixed points of a nonexpansive mapping T in a Hilbert space. Later, Reich [2] discussed this iteration procedure in a uniformly convex Banach space whose norm is Fréchet differentiable. In 1998, Takahashi and Tamura [3] considered an iteration procedure with two nonexpansive mappings and obtained weak convergence theorems for this procedure in a uniformly convex Banach space which satisfies Opial’s condition or whose norm is Fréchet differentiable. On the other hand, in 2008, Dhompongsa and Panyanak [4] proved the following theorem.

Theorem 1.1 Let C be a bounded closed convex subset of a complete $CAT(0)$ space and $T:C\to C$ a nonexpansive mapping. For any initial point $x_0$ in C, define the Mann iterative sequence $\{x_n\}$ by

$$x_{n+1}=(1-t_n)x_n\oplus t_{n}T{x}_n,n=0,1,2,\dots ,$$

where $\{t_n\}$ is a sequence in $[0,1]$, with the restrictions that ${\sum }_{n=0}^{\mathrm{\infty }}{t}_n$ diverges and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{t}_n<1$. Then $\{x_n\}$ Δ-converges to a fixed point of T.

Further, in a $CAT(1)$ space, Kimura et al. [5] proved the Δ-convergence theorem for a family of nonexpansive mappings including the following scheme:

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

In a Hilbert space H, the following equality holds for any $x,y,z\in H$:

$$\alpha x+(1-\alpha )(\beta y+(1-\beta )z)=\gamma (\delta x+(1-\delta )y)+(1-\gamma )z,$$

where $\alpha ,\beta ,\gamma ,\delta \in ]0,1[$ such that $\alpha =\gamma \delta$ and $\beta =\gamma (1-\delta )/(1-\gamma \delta )$. However, in $CAT(\kappa )$ spaces with $\kappa >0$, it does not hold in general, that is, the value of the convex combination taken twice depends on their order. Thus, the following formulas are different in general:

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

(1)

In this paper, we show an analogous result to Theorem 1.1 using the iterative scheme (1) in a complete $CAT(1)$ space with two quasinonexpansive and Δ-demiclosed mappings. We also deal with the image recovery problem for two closed convex sets.

2 Preliminaries

Let X be a metric space. For $x,y\in X$, a mapping $c:[0,l]\to X$ is said to be a geodesic if c satisfies $c(0)=x$, $c(l)=y$ and $d(c(s),c(t))=|s-t|$ for all $s,t\in [0,l]$. An image $[x,y]$ of c is called a geodesic segment joining x and y. For $r>0$, X is said to be an r-geodesic metric space if, for any $x,y\in X$ with $d(x,y)<r$, there exists a geodesic segment $[x,y]$. In particular, if a segment $[x,y]$ is unique for any $x,y\in X$ with $d(x,y)<r$, then X is said to be a uniquely r-geodesic metric space. In what follows, we always assume $d(x,y)<\pi /2$ for any $x,y\in X$. Thus, we say X is a geodesic metric space instead of a $\pi /2$-geodesic metric space. For the more general case, see [6].

Let X be a uniquely geodesic metric space. A geodesic triangle is defined by $\mathrm{△}(x,y,z)=[x,y]\cup [y,z]\cup [z,x]$. Let M be the two-dimensional unit sphere in ${\mathbb{R}}^3$. For $\overline{x},\overline{y},\overline{z}\in M$, a triangle $\mathrm{△}(\overline{x},\overline{y},\overline{z})\subset M$ is called a comparison triangle of $\mathrm{△}(x,y,z)$ if $d(x,y)=d_M(\overline{x},\overline{y})$, $d(y,z)=d_M(\overline{y},\overline{z})$, $d(z,x)=d_M(\overline{z},\overline{x})$. Further, for any $x,y\in X$ and $t\in ]0,1[$, if $z\in [x,y]$ satisfies $d(x,z)=(1-t)d(x,y)$ and $d(z,y)=td(x,y)$, then z is denoted by $z=tx\oplus (1-t)y$. A point $\overline{z}\in [\overline{x},\overline{y}]$ is called a comparison point of $z\in [x,y]$ if $d(x,z)=d_M(\overline{x},\overline{z})$. X is said to be a $CAT(1)$ space if, for any $p,q\in \mathrm{△}(x,y,z)\subset X$ and its comparison points $\overline{p},\overline{q}\in \mathrm{△}(\overline{x},\overline{y},\overline{z})\subset M$, the inequality $d(p,q)\le d_M(\overline{p},\overline{q})$ holds.

Let X be a geodesic metric space and $\{x_n\}$ a bounded sequence of X. For $x\in X$, we put $r(x,\{x_n\})={lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}d(x,x_n)$. The asymptotic radius of $\{x_n\}$ is defined by $r(\{x_n\})={inf}_{x\in X}r(x,\{x_n\})$. Further, the asymptotic center of $\{x_n\}$ is defined by $AC(\{x_n\})=\{x\in X:r(x,\{x_n\})=r(\{x_n\})\}$. If, for any subsequences $\{x_{n_k}\}$ of $\{x_n\}$, $AC(\{x_{n_k}\})=\{x_0\}$, i.e., their asymptotic center consists of the unique element $x_0$, then we say $\{x_n\}$ Δ-converges to $x_0$ and we denote it by $x_n\stackrel{\mathrm{\Delta }}{\rightharpoonup }{x}_0$.

Let X be a metric space. A mapping $T:X\to X$ is said to be a nonexpansive if T satisfies $d(Tx,Ty)\le d(x,y)$ for any $x,y\in X$. The set of fixed points of T is denoted by $F(T)=\{z\in X:Tz=z\}$. Further, a mapping $T:X\to X$ with $F(T)\ne \mathrm{\varnothing }$ is said to be a quasinonexpansive if T satisfies $d(Tx,z)\le d(x,z)$ for any $x\in X$ and $z\in F(T)$. Moreover, T is said to be Δ-demiclosed if, for any bounded sequence $\{x_n\}\subset X$ and $x_0\in X$ satisfying $d(x_n,T{x}_n)\to 0$ and $x_n\stackrel{\mathrm{\Delta }}{\rightharpoonup }{x}_0$, we have $x_0\in F(T)$.

3 Tools for the main results

In this section, we introduce some tools for using the main theorem.

Theorem 3.1 (Kimura and Satô [7])

Let $\mathrm{△}(x,y,z)$ be a geodesic triangle in a $CAT(1)$ space such that $d(x,y)+d(y,z)+d(z,x)<2\pi$. Let $u=tx\oplus (1-t)y$ for some $t\in [0,1]$. Then

$$cosd(u,z)sind(x,y)\ge cosd(x,z)sintd(x,y)+cosd(y,z)sin(1-t)d(x,y).$$

Corollary 3.2 (Kimura and Satô [8])

Let $\mathrm{△}(x,y,z)$ be a geodesic triangle in a $CAT(1)$ space such that $d(x,y)+d(y,z)+d(z,x)<2\pi$. Let $u=tx\oplus (1-t)y$ for some $t\in [0,1]$. Then

$$cosd(u,z)\ge tcosd(x,z)+(1-t)cosd(y,z).$$

Theorem 3.3 (Espínola and Fernández-León [9])

Let X be a complete $CAT(1)$ space and $\{x_n\}$ a sequence in X. If $r(\{x_n\})<\pi /2$, then the following hold.

• (i) $AC(\{x_n\})$ consists of exactly one point;

• (ii) $\{x_n\}$ has a Δ-convergent subsequence.

Theorem 3.4 (Kimura and Satô [8])

Let X be a metric space and T a mapping from X into itself. If T is a nonexpansive with $F(T)\ne \mathrm{\varnothing }$, then T is quasinonexpansive and Δ-demiclosed.

The following lemmas are important properties of real numbers and they are easy to show. Thus, we omit the proofs.

Lemma 3.5 Let δ be a real number such that $-1<\delta <0$ and $\{b_n\}$, $\{c_n\}$ real sequences satisfying $\delta \le b_n\le 1$, $\delta \le c_n\le 1$ and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{b}_n{c}_n\ge 1$. Then ${lim}_{n\to \mathrm{\infty }}{b}_n={lim}_{n\to \mathrm{\infty }}{c}_n=1$.

Lemma 3.6 Let $s\in ]0,\mathrm{\infty }[$ and $\{b_n\}$, $\{c_n\}$ bounded real sequences satisfying $b_n\le 0$, $s<c_n$ and ${lim}_{n\to \mathrm{\infty }}{b}_n/c_n=0$. Then ${lim}_{n\to \mathrm{\infty }}{b}_n=0$.

Lemma 3.7 Let $\{b_n\}$ and $\{c_n\}$ be bounded real sequences satisfying ${lim}_{n\to \mathrm{\infty }}(b_n-c_n)=0$. Then ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{b}_n={lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{c}_n$.

4 The main result

In this section, we show the main result.

Theorem 4.1 Let X be a complete $CAT(1)$ space such that for any $u,v\in X$, $d(u,v)<\pi /2$. Let S and T be quasinonexpansive and Δ-demiclosed mappings from X into itself with $F(S)\cap F(T)\ne \mathrm{\varnothing }$. Let $\{{\alpha }_n\}$, $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ be sequences of $[a,b]\subset ]0,1[$. Define a sequence $\{x_n\}\subset X$ by the following recurrence formula: $x_1\in X$ and

$$\{\begin{array}{l}{u}_n=(1-{\beta }_n)x_n\oplus {\beta }_{n}S{x}_n,\\ v_n=(1-{\gamma }_n)x_n\oplus {\gamma }_{n}T{x}_n,\\ x_{n+1}=(1-{\alpha }_n)u_n\oplus {\alpha }_n{v}_n\end{array}$$

for $n\in \mathbb{N}$. Then $\{x_n\}$ Δ-converges to a common fixed point of S and T.

Proof Let $z\in F(S)\cap F(T)$. By Corollary 3.2, we have

$$\begin{array}{c}\begin{array}{rl}cosd(u_n,z)& \ge (1-{\beta }_n)cosd(x_n,z)+{\beta }_{n}cosd(S{x}_n,z)\\ \ge (1-{\beta }_n)cosd(x_n,z)+{\beta }_{n}cosd(x_n,z)\\ =cosd(x_n,z),\end{array}\hfill \\ \begin{array}{rl}cosd(v_n,z)& \ge (1-{\gamma }_n)cosd(x_n,z)+{\gamma }_{n}cosd(T{x}_n,z)\\ \ge (1-{\gamma }_n)cosd(x_n,z)+{\gamma }_{n}cosd(x_n,z)\\ =cosd(x_n,z).\end{array}\hfill \end{array}$$

Then, by Corollary 3.2 again, we have

$$\begin{array}{rl}cosd(x_{n+1},z)& \ge (1-{\alpha }_n)cosd(u_n,z)+{\alpha }_{n}cosd(v_n,z)\\ \ge (1-{\alpha }_n)cosd(x_n,z)+{\alpha }_{n}cosd(x_n,z)\\ \ge cosd(x_n,z).\end{array}$$

So, we get $d(x_{n+1},z)\le d(x_n,z)$ for all $n\in \mathbb{N}$ and there exists $d_0={lim}_{n\to \mathrm{\infty }}d(x_n,z)\le d(x_1,z)<\pi /2$.

Furthermore, by Theorem 3.1, we have

$$\begin{array}{r}cosd(u_n,z)sind(x_n,S{x}_n)\\ \ge cosd(x_n,z)sin(1-{\beta }_n)d(x_n,S{x}_n)+cosd(S{x}_n,z)sin{\beta }_{n}d(x_n,S{x}_n)\\ \ge 2cosd(x_n,z)sin\frac{d(x_n,S{x}_n)}{2}cos\frac{(1-2{\beta }_n)d(x_n,S{x}_n)}{2}\end{array}$$

(2)

and

$$\begin{array}{r}cosd(v_n,z)sind(x_n,T{x}_n)\\ \ge cosd(x_n,z)sin(1-{\gamma }_n)d(x_n,T{x}_n)+cosd(T{x}_n,z)sin{\gamma }_{n}d(x_n,T{x}_n)\\ \ge 2cosd(x_n,z)sin\frac{d(x_n,T{x}_n)}{2}cos\frac{(1-2{\gamma }_n)d(x_n,T{x}_n)}{2}.\end{array}$$

(3)

Let $d_n=d(x_n,z)$, $s_n=d(x_n,S{x}_n)/2$ and $t_n=d(x_n,T{x}_n)/2$ for $n\in \mathbb{N}$. If there exists $n_0\in \mathbb{N}$ such that $s_{n_0}=t_{n_0}=0$, then we have $x_{n_0}\in F(S)\cap F(T)$ and since

$$\begin{array}{rl}{x}_{n_0+1}& =(1-{\alpha }_{n_0})((1-{\beta }_{n_0})x_{n_0}\oplus {\beta }_{n_0}S{x}_{n_0})\oplus {\alpha }_{n_0}((1-{\gamma }_{n_0})x_{n_0}\oplus {\gamma }_{n_0}T{x}_{n_0})\\ =(1-{\alpha }_{n_0})x_{n_0}\oplus {\alpha }_{n_0}{x}_{n_0}\\ =x_{n_0},\end{array}$$

and the proof is finished. So, we may assume $s_n\ne 0$ or $t_n\ne 0$ for all $n\in \mathbb{N}$.

If $s_n=0$ and $t_n\ne 0$, then we have $u_n=x_n$. From (2), (3), and Corollary 3.2, we get

$$\begin{array}{r}2cos{d}_{n+1}sin{t}_{n}cos{t}_n\\ =cos{d}_{n+1}sin2t_n\\ \ge (1-{\alpha }_n)cosd(u_n,z)sin2t_n+{\alpha }_{n}cosd(v_n,z)sin2t_n\\ \ge 2(1-{\alpha }_n)cos{d}_{n}sin{t}_{n}cos{t}_n+2{\alpha }_{n}cos{d}_{n}sin{t}_{n}cos(1-2{\gamma }_n)t_n.\end{array}$$

Dividing by $2sin{t}_n>0$, we get

$$cos{d}_{n+1}cos{t}_n\ge (1-{\alpha }_n)cos{d}_{n}cos{t}_n+{\alpha }_{n}cos{d}_{n}cos(1-2{\gamma }_n)t_n.$$

(4)

If $t_n=0$ and $s_n\ne 0$, then we have $v_n=x_n$. In a similar way as above, we get

$$cos{d}_{n+1}cos{s}_n\ge (1-{\alpha }_n)cos{d}_{n}cos(1-2{\beta }_n)s_n+{\alpha }_{n}cos{d}_{n}cos{s}_n.$$

(5)

If $s_n\ne 0$ and $t_n\ne 0$, then from (2), (3), and Corollary 3.2, we get

$$\begin{array}{r}cos{d}_{n+1}sin2s_{n}sin2t_n\\ \ge (1-{\alpha }_n)cosd(u_n,z)sin2s_{n}sin2t_n+{\alpha }_{n}cosd(v_n,z)sin2s_{n}sin2t_n\\ \ge 4cos{d}_{n}sin{s}_{n}sin{t}_n((1-{\alpha }_n)cos{t}_{n}cos(1-2{\beta }_n)s_n+{\alpha }_{n}cos{s}_{n}cos(1-2{\gamma }_n)t_n).\end{array}$$

Dividing by $4sin{s}_{n}sin{t}_n>0$, we get

$$\begin{array}{r}cos{d}_{n+1}cos{s}_{n}cos{t}_n\\ \ge (1-{\alpha }_n)cos{d}_{n}cos{t}_{n}cos(1-2{\beta }_n)s_n+{\alpha }_{n}cos{d}_{n}cos{s}_{n}cos(1-2{\gamma }_n)t_n.\end{array}$$

(6)

Therefore, (4) and (5) can be reduced to the inequality (6) and it is equivalent to

$$(\frac{{\epsilon }_{n}cos{s}_n}{{\alpha }_{n}cos(1-2{\beta }_n)s_n}-\frac{1-{\alpha }_n}{{\alpha }_n})(\frac{{\epsilon }_{n}cos{t}_n}{(1-{\alpha }_n)cos(1-2{\gamma }_n)t_n}-\frac{{\alpha }_n}{1-{\alpha }_n})\ge 1,$$

where ${\epsilon }_n=cos{d}_{n+1}/cos{d}_n$ for $n\in \mathbb{N}$. It follows that ${lim}_{n\to \mathrm{\infty }}{\epsilon }_n=cos{d}_0/cos{d}_0=1$. Since $\{{\alpha }_n\}\subset [a,b]\subset ]0,1[$ for $n\in \mathbb{N}$, we get

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}(\frac{cos{s}_n}{{\alpha }_{n}cos(1-2{\beta }_n)s_n}-\frac{1-{\alpha }_n}{{\alpha }_n})(\frac{cos{t}_n}{(1-{\alpha }_n)cos(1-2{\gamma }_n)t_n}-\frac{{\alpha }_n}{1-{\alpha }_n})\ge 1.$$

(7)

Then we show that there exists $n_0\in \mathbb{N}$ such that for all $n\ge n_0$, the following hold:

$$-\frac{1}{2}\le \frac{cos{s}_n}{{\alpha }_{n}cos(1-2{\beta }_n)s_n}-\frac{1-{\alpha }_n}{{\alpha }_n}\le 1$$

(8)

and

$$-\frac{1}{2}\le \frac{cos{t}_n}{(1-{\alpha }_n)cos(1-2{\gamma }_n)t_n}-\frac{{\alpha }_n}{1-{\alpha }_n}\le 1.$$

(9)

First, we show the right inequality of (8). Since $\{{\beta }_n\}\subset [a,b]\subset ]0,1[$ for $n\in \mathbb{N}$, we get $cos{s}_n\le cos|1-2{\beta }_n|s_n=cos(1-2{\beta }_n)s_n$. Hence we get

$$\frac{cos{s}_n}{{\alpha }_{n}cos(1-2{\beta }_n)s_n}-\frac{1-{\alpha }_n}{{\alpha }_n}\le \frac{1}{{\alpha }_n}-\frac{1-{\alpha }_n}{{\alpha }_n}=1.$$

By the same method as above, the right inequality of (9) also holds. Next, let us show the left inequality of (8). If it does not hold, then letting

$${\sigma }_n=\frac{cos{s}_n}{{\alpha }_{n}cos(1-2{\beta }_n)s_n}-\frac{1-{\alpha }_n}{{\alpha }_n}\text{and}{\tau }_n=\frac{cos{t}_n}{(1-{\alpha }_n)cos(1-2{\gamma }_n)t_n}-\frac{{\alpha }_n}{1-{\alpha }_n},$$

we can find a subsequence $\{{\sigma }_{n_i}\}\subset \{{\sigma }_n\}$ such that ${\sigma }_{n_i}<-1/2$ for $i\in \mathbb{N}$ and ${lim}_{i\to \mathrm{\infty }}{\sigma }_{n_i}=\sigma \le -1/2$. Since $\{{\alpha }_n\},\{{\gamma }_n\}\subset [a,b]\subset ]0,1[$ and $\{t_n\}\subset [0,\pi /4[\subset [0,\pi /2[$, we have $\{{\tau }_n\}$ is bounded. Therefore, by taking a subsequence again if necessary, we may assume that $\{{\tau }_{n_i}\}$ converges to $\tau \in \mathbb{R}$. Then, by (7), we get $\sigma \tau ={lim}_{i\to \mathrm{\infty }}{\sigma }_{n_i}{\tau }_{n_i}\ge {lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\sigma }_n{\tau }_n\ge 1$. Hence we may assume that ${\tau }_{n_i}<0$ for all $i\in \mathbb{N}$. Since $\sqrt{2}/2<cos{s}_n$, $\sqrt{2}/2<cos{t}_n$, $0<cos(1-2{\beta }_n)s_n\le 1$, $0<cos(1-2{\gamma }_n)t_n\le 1$ and $\{{\alpha }_n\}\subset [a,b]\subset ]0,1[$, we also have

$$0<\frac{\sqrt{2}}{2b}\le \frac{cos{s}_n}{{\alpha }_{n}cos(1-2{\beta }_n)s_n}\text{and}0<\frac{\sqrt{2}}{2(1-a)}\le \frac{cos{t}_n}{(1-{\alpha }_n)cos(1-2{\gamma }_n)t_n}.$$

(10)

Let ρ be a real number such that

$$0<\rho <min\{\frac{\sqrt{2}}{2b},\frac{\sqrt{2}}{2(1-a)},\frac{1-b}{b}+\frac{a}{1-a}\}.$$

(11)

Then, by (10), we get

$$\rho -\frac{1-{\alpha }_{n_i}}{{\alpha }_{n_i}}\le {\sigma }_{n_i}<0\text{and}\rho -\frac{{\alpha }_{n_i}}{1-{\alpha }_{n_i}}\le {\tau }_{n_i}<0.$$

(12)

Then, by (11) and (12), we have

$$\begin{array}{rl}{\sigma }_{n_i}{\tau }_{n_i}& \le (\rho -\frac{1-{\alpha }_{n_i}}{{\alpha }_{n_i}})(\rho -\frac{{\alpha }_{n_i}}{1-{\alpha }_{n_i}})\\ ={\rho }^2-(\frac{1-{\alpha }_{n_i}}{{\alpha }_{n_i}}+\frac{{\alpha }_{n_i}}{1-{\alpha }_{n_i}})\rho +1\\ \le {\rho }^2-(\frac{1-b}{b}+\frac{a}{1-a})\rho +1\\ =\rho (\rho -(\frac{1-b}{b}+\frac{a}{1-a}))+1.\end{array}$$

Thus, as $i\to \mathrm{\infty }$, we have

$$1\le \sigma \tau \le \rho (\rho -(\frac{1-b}{b}+\frac{a}{1-a}))+1<1.$$

(13)

This is a contradiction. We also obtain the left inequality of (9) in a similar way. Hence we get

$$\underset{n\to \mathrm{\infty }}{lim}(\frac{cos{s}_n}{{\alpha }_{n}cos(1-2{\beta }_n)s_n}-\frac{1-{\alpha }_n}{{\alpha }_n})=\underset{n\to \mathrm{\infty }}{lim}(\frac{cos{t}_n}{(1-{\alpha }_n)cos(1-2{\gamma }_n)t_n}-\frac{{\alpha }_n}{1-{\alpha }_n})=1$$

(14)

by Lemma 3.5, (8), and (9). Furthermore, from (14), we get

$$\underset{n\to \mathrm{\infty }}{lim}\frac{cos{s}_n-cos(1-2{\beta }_n)s_n}{{\alpha }_{n}cos(1-2{\beta }_n)s_n}=0.$$

(15)

By Lemma 3.6 and (15), we get

$$\underset{n\to \mathrm{\infty }}{lim}(cos{s}_n-cos(1-2{\beta }_n)s_n)=0.$$

(16)

Moreover, by Lemma 3.7 and (16), we get

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}cos{s}_n=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}cos(1-2{\beta }_n)s_n=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}cos|1-2{\beta }_n|s_n.$$

Hence we get

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{s}_n=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(|1-2{\beta }_n|s_n)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}|1-2{\beta }_n|\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{s}_n,$$

and we have

$$0\ge (1-\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}|1-2{\beta }_n|)\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{s}_n=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}(1-|1-2{\beta }_n|)\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{s}_n.$$

Since $\{{\beta }_n\}\subset [a,b]\subset ]0,1[$ for $n\in \mathbb{N}$, we have ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}(1-|1-2{\beta }_n|)>0$ and thus, ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{s}_n\le 0$. Therefore, we get ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{s}_n=0$ and we also get ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{t}_n=0$. It implies $d(x_n,S{x}_n)\to 0$ and $d(x_n,T{x}_n)\to 0$.

Next, let $\{y_k\}$ be a subsequence of $\{x_n\}$. Since $r(\{x_n\})\le d_0<\pi /2$, by Theorem 3.3(i), there exists a unique asymptotic center $x_0$ of $\{y_k\}$. Moreover, since $r(\{y_k\})<\pi /2$, by Theorem 3.3(ii), there exists a subsequence $\{z_l\}$ of $\{y_k\}$ such that $z_l\stackrel{\mathrm{\Delta }}{\rightharpoonup }{z}_0\in X$. Further, since $d(z_l,S{z}_l)\to 0$, $d(z_l,T{z}_l)\to 0$ and S, T are Δ-demiclosed, we have $z_0\in F(S)\cap F(T)$. Then we can show that $x_0=z_0$, i.e., $x_0\in F(S)\cap F(T)$. If not, from the uniqueness of the asymptotic centers $x_0$, $z_0$ of $\{y_k\}$, $\{z_l\}$, respectively, due to Theorem 3.3(i), we have

$$\begin{array}{rl}\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(y_k,x_0)& <\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(y_k,z_0)\\ =\underset{n\to \mathrm{\infty }}{lim}d(x_n,z_0)\\ =\underset{l\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(z_l,z_0)\\ <\underset{l\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(z_l,x_0)\\ \le \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(y_k,x_0).\end{array}$$

This is a contradiction. Hence we get $x_0\in F(S)\cap F(T)$. Next, we show that for any subsequences of $\{x_n\}$, their asymptotic center consists of the unique element. Let $\{u_k\}$, $\{v_k\}$ be subsequences of $\{x_n\}$, $x_0\in AC(\{u_k\})$ and $x_0^{\prime }\in AC(\{v_k\})$. We show $x_0=x_0^{\prime }$ by using contradiction. Assume $x_0\ne x_0^{\prime }$. Then $x_0^{\prime }\notin AC(\{u_k\})$ and $x_0\notin AC(\{v_k\})$ by Theorem 3.3(i). It follows that

$$\begin{array}{rl}\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(u_k,x_0)& <\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(u_k,x_0^{\prime })\\ =\underset{n\to \mathrm{\infty }}{lim}d(x_n,x_0^{\prime })\\ =\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(v_k,x_0^{\prime })\\ <\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(v_k,x_0)\\ =\underset{n\to \mathrm{\infty }}{lim}d(x_n,x_0)\\ =\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(u_k,x_0).\end{array}$$

This is a contradiction. Hence we get $x_0=x_0^{\prime }$. Therefore, we have $\{x_n\}$ Δ-converges to a common fixed point of S and T. □

By Theorem 3.4, we know that a nonexpansive mapping having a fixed point satisfies the assumptions in Theorem 4.1. Thus, we get the following result.

Corollary 4.2 Let X be a complete $CAT(1)$ space such that for any $u,v\in X$, $d(u,v)<\pi /2$. Let S and T be nonexpansive mappings of X into itself such that $F(S)\cap F(T)\ne \mathrm{\varnothing }$. Let $\{{\alpha }_n\}$, $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ be sequences in $[a,b]\subset ]0,1[$. Define a sequence $\{x_n\}\subset X$ as the following recurrence formula: $x_1\in X$ and

$$\{\begin{array}{l}{u}_n=(1-{\beta }_n)x_n\oplus {\beta }_{n}S{x}_n,\\ v_n=(1-{\gamma }_n)x_n\oplus {\gamma }_{n}T{x}_n,\\ x_{n+1}=(1-{\alpha }_n)u_n\oplus {\alpha }_n{v}_n\end{array}$$

for $n\in \mathbb{N}$. Then $\{x_n\}$ Δ-converges to a common fixed point of S and T.

5 An application to the image recovery

The image recovery problem is formulated as to find the nearest point in the intersection of family of closed convex subsets from a given point by using corresponding metric projection of each subset. In this section, we consider this problem for two subsets of a complete $CAT(1)$ space.

Theorem 5.1 Let X be a complete $CAT(1)$ space such that for any $u,v\in X$, $d(u,v)<\pi /2$. Let $C_1$ and $C_2$ be nonempty closed convex subsets of X such that $C_1\cap C_2\ne \mathrm{\varnothing }$. Let $P_1$ and $P_2$ be metric projections onto $C_1$ and $C_2$, respectively. Let $\{{\alpha }_n\}$, $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ be real sequences in $[a,b]\subset ]0,1[$. Define a sequence $\{x_n\}\subset X$ by the following recurrence formula: $x_1\in X$ and

$$\{\begin{array}{l}{u}_n=(1-{\beta }_n)x_n\oplus {\beta }_n{P}_1{x}_n,\\ v_n=(1-{\gamma }_n)x_n\oplus {\gamma }_n{P}_2{x}_n,\\ x_{n+1}=(1-{\alpha }_n)u_n\oplus {\alpha }_n{v}_n\end{array}$$

for $n\in \mathbb{N}$. Then $\{x_n\}$ Δ-converges to a fixed point of the intersection of $C_1$ and $C_2$.

Proof We see that $P_1$ and $P_2$ are quasinonexpansive [9] and Δ-demiclosed [8]. Further, we also get $F(P_1)=C_1$ and $F(P_2)=C_2$. Thus, letting $S=P_1$ and $T=P_2$ in Theorem 4.1, we obtain the desired result. □