# 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  introduced an iteration procedure for approximating fixed points of a nonexpansive mapping T in a Hilbert space. Later, Reich  discussed this iteration procedure in a uniformly convex Banach space whose norm is Fréchet differentiable. In 1998, Takahashi and Tamura  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  proved the following theorem.

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

${x}_{n+1}=\left(1-{t}_{n}\right){x}_{n}\oplus {t}_{n}T{x}_{n},\phantom{\rule{1em}{0ex}}n=0,1,2,\dots ,$

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

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

${x}_{n+1}=\left(1-{\alpha }_{n}\right){x}_{n}\oplus {\alpha }_{n}\left(\left(1-{\beta }_{n}\right)S{x}_{n}\oplus {\beta }_{n}T{x}_{n}\right).$

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

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

where $\alpha ,\beta ,\gamma ,\delta \in \phantom{\rule{0.2em}{0ex}}\right]0,1\left[$ such that $\alpha =\gamma \delta$ and $\beta =\gamma \left(1-\delta \right)/\left(1-\gamma \delta \right)$. However, in $CAT\left(\kappa \right)$ 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:

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

In this paper, we show an analogous result to Theorem 1.1 using the iterative scheme (1) in a complete $CAT\left(1\right)$ 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 $c:\left[0,l\right]\to X$ is said to be a geodesic if c satisfies $c\left(0\right)=x$, $c\left(l\right)=y$ and $d\left(c\left(s\right),c\left(t\right)\right)=|s-t|$ for all $s,t\in \left[0,l\right]$. An image $\left[x,y\right]$ 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\left(x,y\right), there exists a geodesic segment $\left[x,y\right]$. In particular, if a segment $\left[x,y\right]$ is unique for any $x,y\in X$ with $d\left(x,y\right), then X is said to be a uniquely r-geodesic metric space. In what follows, we always assume $d\left(x,y\right)<\pi /2$ 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 .

Let X be a uniquely geodesic metric space. A geodesic triangle is defined by $\mathrm{△}\left(x,y,z\right)=\left[x,y\right]\cup \left[y,z\right]\cup \left[z,x\right]$. Let M be the two-dimensional unit sphere in ${\mathbb{R}}^{3}$. For $\overline{x},\overline{y},\overline{z}\in M$, a triangle $\mathrm{△}\left(\overline{x},\overline{y},\overline{z}\right)\subset M$ is called a comparison triangle of $\mathrm{△}\left(x,y,z\right)$ if $d\left(x,y\right)={d}_{M}\left(\overline{x},\overline{y}\right)$, $d\left(y,z\right)={d}_{M}\left(\overline{y},\overline{z}\right)$, $d\left(z,x\right)={d}_{M}\left(\overline{z},\overline{x}\right)$. Further, for any $x,y\in X$ and $t\in \phantom{\rule{0.2em}{0ex}}\right]0,1\left[$, if $z\in \left[x,y\right]$ satisfies $d\left(x,z\right)=\left(1-t\right)d\left(x,y\right)$ and $d\left(z,y\right)=td\left(x,y\right)$, then z is denoted by $z=tx\oplus \left(1-t\right)y$. A point $\overline{z}\in \left[\overline{x},\overline{y}\right]$ is called a comparison point of $z\in \left[x,y\right]$ if $d\left(x,z\right)={d}_{M}\left(\overline{x},\overline{z}\right)$. X is said to be a $CAT\left(1\right)$ space if, for any $p,q\in \mathrm{△}\left(x,y,z\right)\subset X$ and its comparison points $\overline{p},\overline{q}\in \mathrm{△}\left(\overline{x},\overline{y},\overline{z}\right)\subset M$, the inequality $d\left(p,q\right)\le {d}_{M}\left(\overline{p},\overline{q}\right)$ holds.

Let X be a geodesic metric space and $\left\{{x}_{n}\right\}$ a bounded sequence of X. For $x\in X$, we put $r\left(x,\left\{{x}_{n}\right\}\right)={lim sup}_{n\to \mathrm{\infty }}d\left(x,{x}_{n}\right)$. The asymptotic radius of $\left\{{x}_{n}\right\}$ is defined by $r\left(\left\{{x}_{n}\right\}\right)={inf}_{x\in X}r\left(x,\left\{{x}_{n}\right\}\right)$. Further, the asymptotic center of $\left\{{x}_{n}\right\}$ is defined by $AC\left(\left\{{x}_{n}\right\}\right)=\left\{x\in X:r\left(x,\left\{{x}_{n}\right\}\right)=r\left(\left\{{x}_{n}\right\}\right)\right\}$. If, for any subsequences $\left\{{x}_{{n}_{k}}\right\}$ of $\left\{{x}_{n}\right\}$, $AC\left(\left\{{x}_{{n}_{k}}\right\}\right)=\left\{{x}_{0}\right\}$, i.e., their asymptotic center consists of the unique element ${x}_{0}$, then we say $\left\{{x}_{n}\right\}$ Δ-converges to ${x}_{0}$ and we denote it by ${x}_{n}\stackrel{\mathrm{\Delta }}{⇀}{x}_{0}$.

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

## 3 Tools for the main results

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

Theorem 3.1 (Kimura and Satô )

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

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

Corollary 3.2 (Kimura and Satô )

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

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

Theorem 3.3 (Espínola and Fernández-León )

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

1. (i)

$AC\left(\left\{{x}_{n}\right\}\right)$ consists of exactly one point;

2. (ii)

$\left\{{x}_{n}\right\}$ has a Δ-convergent subsequence.

Theorem 3.4 (Kimura and Satô )

Let X be a metric space and T a mapping from X into itself. If T is a nonexpansive with $F\left(T\right)\ne \mathrm{\varnothing }$, 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 $\left\{{b}_{n}\right\}$, $\left\{{c}_{n}\right\}$ real sequences satisfying $\delta \le {b}_{n}\le 1$, $\delta \le {c}_{n}\le 1$ and ${lim inf}_{n\to \mathrm{\infty }}{b}_{n}{c}_{n}\ge 1$. Then ${lim}_{n\to \mathrm{\infty }}{b}_{n}={lim}_{n\to \mathrm{\infty }}{c}_{n}=1$.

Lemma 3.6 Let $s\in \phantom{\rule{0.2em}{0ex}}\right]0,\mathrm{\infty }\left[$ and $\left\{{b}_{n}\right\}$, $\left\{{c}_{n}\right\}$ 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 $\left\{{b}_{n}\right\}$ and $\left\{{c}_{n}\right\}$ be bounded real sequences satisfying ${lim}_{n\to \mathrm{\infty }}\left({b}_{n}-{c}_{n}\right)=0$. Then ${lim inf}_{n\to \mathrm{\infty }}{b}_{n}={lim inf}_{n\to \mathrm{\infty }}{c}_{n}$.

## 4 The main result

In this section, we show the main result.

Theorem 4.1 Let X be a complete $CAT\left(1\right)$ space such that for any $u,v\in X$, $d\left(u,v\right)<\pi /2$. Let S and T be quasinonexpansive and Δ-demiclosed mappings from X into itself with $F\left(S\right)\cap F\left(T\right)\ne \mathrm{\varnothing }$. Let $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$ and $\left\{{\gamma }_{n}\right\}$ be sequences of $\left[a,b\right]\subset \phantom{\rule{0.2em}{0ex}}\right]0,1\left[$. Define a sequence $\left\{{x}_{n}\right\}\subset X$ by the following recurrence formula: ${x}_{1}\in X$ and

$\left\{\begin{array}{l}{u}_{n}=\left(1-{\beta }_{n}\right){x}_{n}\oplus {\beta }_{n}S{x}_{n},\\ {v}_{n}=\left(1-{\gamma }_{n}\right){x}_{n}\oplus {\gamma }_{n}T{x}_{n},\\ {x}_{n+1}=\left(1-{\alpha }_{n}\right){u}_{n}\oplus {\alpha }_{n}{v}_{n}\end{array}$

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

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

$\begin{array}{c}\begin{array}{rl}cosd\left({u}_{n},z\right)& \ge \left(1-{\beta }_{n}\right)cosd\left({x}_{n},z\right)+{\beta }_{n}cosd\left(S{x}_{n},z\right)\\ \ge \left(1-{\beta }_{n}\right)cosd\left({x}_{n},z\right)+{\beta }_{n}cosd\left({x}_{n},z\right)\\ =cosd\left({x}_{n},z\right),\end{array}\hfill \\ \begin{array}{rl}cosd\left({v}_{n},z\right)& \ge \left(1-{\gamma }_{n}\right)cosd\left({x}_{n},z\right)+{\gamma }_{n}cosd\left(T{x}_{n},z\right)\\ \ge \left(1-{\gamma }_{n}\right)cosd\left({x}_{n},z\right)+{\gamma }_{n}cosd\left({x}_{n},z\right)\\ =cosd\left({x}_{n},z\right).\end{array}\hfill \end{array}$

Then, by Corollary 3.2 again, we have

$\begin{array}{rl}cosd\left({x}_{n+1},z\right)& \ge \left(1-{\alpha }_{n}\right)cosd\left({u}_{n},z\right)+{\alpha }_{n}cosd\left({v}_{n},z\right)\\ \ge \left(1-{\alpha }_{n}\right)cosd\left({x}_{n},z\right)+{\alpha }_{n}cosd\left({x}_{n},z\right)\\ \ge cosd\left({x}_{n},z\right).\end{array}$

So, we get $d\left({x}_{n+1},z\right)\le d\left({x}_{n},z\right)$ for all $n\in \mathbb{N}$ and there exists ${d}_{0}={lim}_{n\to \mathrm{\infty }}d\left({x}_{n},z\right)\le d\left({x}_{1},z\right)<\pi /2$.

Furthermore, by Theorem 3.1, we have

$\begin{array}{r}cosd\left({u}_{n},z\right)sind\left({x}_{n},S{x}_{n}\right)\\ \phantom{\rule{1em}{0ex}}\ge cosd\left({x}_{n},z\right)sin\left(1-{\beta }_{n}\right)d\left({x}_{n},S{x}_{n}\right)+cosd\left(S{x}_{n},z\right)sin{\beta }_{n}d\left({x}_{n},S{x}_{n}\right)\\ \phantom{\rule{1em}{0ex}}\ge 2cosd\left({x}_{n},z\right)sin\frac{d\left({x}_{n},S{x}_{n}\right)}{2}cos\frac{\left(1-2{\beta }_{n}\right)d\left({x}_{n},S{x}_{n}\right)}{2}\end{array}$
(2)

and

$\begin{array}{r}cosd\left({v}_{n},z\right)sind\left({x}_{n},T{x}_{n}\right)\\ \phantom{\rule{1em}{0ex}}\ge cosd\left({x}_{n},z\right)sin\left(1-{\gamma }_{n}\right)d\left({x}_{n},T{x}_{n}\right)+cosd\left(T{x}_{n},z\right)sin{\gamma }_{n}d\left({x}_{n},T{x}_{n}\right)\\ \phantom{\rule{1em}{0ex}}\ge 2cosd\left({x}_{n},z\right)sin\frac{d\left({x}_{n},T{x}_{n}\right)}{2}cos\frac{\left(1-2{\gamma }_{n}\right)d\left({x}_{n},T{x}_{n}\right)}{2}.\end{array}$
(3)

Let ${d}_{n}=d\left({x}_{n},z\right)$, ${s}_{n}=d\left({x}_{n},S{x}_{n}\right)/2$ and ${t}_{n}=d\left({x}_{n},T{x}_{n}\right)/2$ 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 ${x}_{{n}_{0}}\in F\left(S\right)\cap F\left(T\right)$ and since

$\begin{array}{rl}{x}_{{n}_{0}+1}& =\left(1-{\alpha }_{{n}_{0}}\right)\left(\left(1-{\beta }_{{n}_{0}}\right){x}_{{n}_{0}}\oplus {\beta }_{{n}_{0}}S{x}_{{n}_{0}}\right)\oplus {\alpha }_{{n}_{0}}\left(\left(1-{\gamma }_{{n}_{0}}\right){x}_{{n}_{0}}\oplus {\gamma }_{{n}_{0}}T{x}_{{n}_{0}}\right)\\ =\left(1-{\alpha }_{{n}_{0}}\right){x}_{{n}_{0}}\oplus {\alpha }_{{n}_{0}}{x}_{{n}_{0}}\\ ={x}_{{n}_{0}},\end{array}$

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

$\begin{array}{r}2cos{d}_{n+1}sin{t}_{n}cos{t}_{n}\\ \phantom{\rule{1em}{0ex}}=cos{d}_{n+1}sin2{t}_{n}\\ \phantom{\rule{1em}{0ex}}\ge \left(1-{\alpha }_{n}\right)cosd\left({u}_{n},z\right)sin2{t}_{n}+{\alpha }_{n}cosd\left({v}_{n},z\right)sin2{t}_{n}\\ \phantom{\rule{1em}{0ex}}\ge 2\left(1-{\alpha }_{n}\right)cos{d}_{n}sin{t}_{n}cos{t}_{n}+2{\alpha }_{n}cos{d}_{n}sin{t}_{n}cos\left(1-2{\gamma }_{n}\right){t}_{n}.\end{array}$

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

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

$cos{d}_{n+1}cos{s}_{n}\ge \left(1-{\alpha }_{n}\right)cos{d}_{n}cos\left(1-2{\beta }_{n}\right){s}_{n}+{\alpha }_{n}cos{d}_{n}cos{s}_{n}.$
(5)

If ${s}_{n}\ne 0$ and ${t}_{n}\ne 0$, then from (2), (3), and Corollary 3.2, we get

$\begin{array}{r}cos{d}_{n+1}sin2{s}_{n}sin2{t}_{n}\\ \phantom{\rule{1em}{0ex}}\ge \left(1-{\alpha }_{n}\right)cosd\left({u}_{n},z\right)sin2{s}_{n}sin2{t}_{n}+{\alpha }_{n}cosd\left({v}_{n},z\right)sin2{s}_{n}sin2{t}_{n}\\ \phantom{\rule{1em}{0ex}}\ge 4cos{d}_{n}sin{s}_{n}sin{t}_{n}\left(\left(1-{\alpha }_{n}\right)cos{t}_{n}cos\left(1-2{\beta }_{n}\right){s}_{n}+{\alpha }_{n}cos{s}_{n}cos\left(1-2{\gamma }_{n}\right){t}_{n}\right).\end{array}$

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

$\begin{array}{r}cos{d}_{n+1}cos{s}_{n}cos{t}_{n}\\ \phantom{\rule{1em}{0ex}}\ge \left(1-{\alpha }_{n}\right)cos{d}_{n}cos{t}_{n}cos\left(1-2{\beta }_{n}\right){s}_{n}+{\alpha }_{n}cos{d}_{n}cos{s}_{n}cos\left(1-2{\gamma }_{n}\right){t}_{n}.\end{array}$
(6)

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

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

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 $\left\{{\alpha }_{n}\right\}\subset \left[a,b\right]\subset \phantom{\rule{0.2em}{0ex}}\right]0,1\left[$ for $n\in \mathbb{N}$, we get

$\underset{n\to \mathrm{\infty }}{lim inf}\left(\frac{cos{s}_{n}}{{\alpha }_{n}cos\left(1-2{\beta }_{n}\right){s}_{n}}-\frac{1-{\alpha }_{n}}{{\alpha }_{n}}\right)\left(\frac{cos{t}_{n}}{\left(1-{\alpha }_{n}\right)cos\left(1-2{\gamma }_{n}\right){t}_{n}}-\frac{{\alpha }_{n}}{1-{\alpha }_{n}}\right)\ge 1.$
(7)

Then we show that there exists ${n}_{0}\in \mathbb{N}$ such that for all $n\ge {n}_{0}$, the following hold:

$-\frac{1}{2}\le \frac{cos{s}_{n}}{{\alpha }_{n}cos\left(1-2{\beta }_{n}\right){s}_{n}}-\frac{1-{\alpha }_{n}}{{\alpha }_{n}}\le 1$
(8)

and

$-\frac{1}{2}\le \frac{cos{t}_{n}}{\left(1-{\alpha }_{n}\right)cos\left(1-2{\gamma }_{n}\right){t}_{n}}-\frac{{\alpha }_{n}}{1-{\alpha }_{n}}\le 1.$
(9)

First, we show the right inequality of (8). Since $\left\{{\beta }_{n}\right\}\subset \left[a,b\right]\subset \phantom{\rule{0.2em}{0ex}}\right]0,1\left[$ for $n\in \mathbb{N}$, we get $cos{s}_{n}\le cos|1-2{\beta }_{n}|{s}_{n}=cos\left(1-2{\beta }_{n}\right){s}_{n}$. Hence we get

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

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

${\sigma }_{n}=\frac{cos{s}_{n}}{{\alpha }_{n}cos\left(1-2{\beta }_{n}\right){s}_{n}}-\frac{1-{\alpha }_{n}}{{\alpha }_{n}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\tau }_{n}=\frac{cos{t}_{n}}{\left(1-{\alpha }_{n}\right)cos\left(1-2{\gamma }_{n}\right){t}_{n}}-\frac{{\alpha }_{n}}{1-{\alpha }_{n}},$

we can find a subsequence $\left\{{\sigma }_{{n}_{i}}\right\}\subset \left\{{\sigma }_{n}\right\}$ 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 $\left\{{\alpha }_{n}\right\},\left\{{\gamma }_{n}\right\}\subset \left[a,b\right]\subset \phantom{\rule{0.2em}{0ex}}\right]0,1\left[$ and $\left\{{t}_{n}\right\}\subset \left[0,\pi /4\left[\phantom{\rule{0.2em}{0ex}}\subset \left[0,\pi /2\left[$, we have $\left\{{\tau }_{n}\right\}$ is bounded. Therefore, by taking a subsequence again if necessary, we may assume that $\left\{{\tau }_{{n}_{i}}\right\}$ converges to $\tau \in \mathbb{R}$. Then, by (7), we get $\sigma \tau ={lim}_{i\to \mathrm{\infty }}{\sigma }_{{n}_{i}}{\tau }_{{n}_{i}}\ge {lim inf}_{n\to \mathrm{\infty }}{\sigma }_{n}{\tau }_{n}\ge 1$. Hence we may assume that ${\tau }_{{n}_{i}}<0$ for all $i\in \mathbb{N}$. Since $\sqrt{2}/2, $\sqrt{2}/2, $0, $0 and $\left\{{\alpha }_{n}\right\}\subset \left[a,b\right]\subset \phantom{\rule{0.2em}{0ex}}\right]0,1\left[$, we also have

$0<\frac{\sqrt{2}}{2b}\le \frac{cos{s}_{n}}{{\alpha }_{n}cos\left(1-2{\beta }_{n}\right){s}_{n}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}0<\frac{\sqrt{2}}{2\left(1-a\right)}\le \frac{cos{t}_{n}}{\left(1-{\alpha }_{n}\right)cos\left(1-2{\gamma }_{n}\right){t}_{n}}.$
(10)

Let ρ be a real number such that

$0<\rho
(11)

Then, by (10), we get

$\rho -\frac{1-{\alpha }_{{n}_{i}}}{{\alpha }_{{n}_{i}}}\le {\sigma }_{{n}_{i}}<0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\rho -\frac{{\alpha }_{{n}_{i}}}{1-{\alpha }_{{n}_{i}}}\le {\tau }_{{n}_{i}}<0.$
(12)

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

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

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

$1\le \sigma \tau \le \rho \left(\rho -\left(\frac{1-b}{b}+\frac{a}{1-a}\right)\right)+1<1.$
(13)

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

$\underset{n\to \mathrm{\infty }}{lim}\left(\frac{cos{s}_{n}}{{\alpha }_{n}cos\left(1-2{\beta }_{n}\right){s}_{n}}-\frac{1-{\alpha }_{n}}{{\alpha }_{n}}\right)=\underset{n\to \mathrm{\infty }}{lim}\left(\frac{cos{t}_{n}}{\left(1-{\alpha }_{n}\right)cos\left(1-2{\gamma }_{n}\right){t}_{n}}-\frac{{\alpha }_{n}}{1-{\alpha }_{n}}\right)=1$
(14)

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

$\underset{n\to \mathrm{\infty }}{lim}\frac{cos{s}_{n}-cos\left(1-2{\beta }_{n}\right){s}_{n}}{{\alpha }_{n}cos\left(1-2{\beta }_{n}\right){s}_{n}}=0.$
(15)

By Lemma 3.6 and (15), we get

$\underset{n\to \mathrm{\infty }}{lim}\left(cos{s}_{n}-cos\left(1-2{\beta }_{n}\right){s}_{n}\right)=0.$
(16)

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

$\underset{n\to \mathrm{\infty }}{lim inf}cos{s}_{n}=\underset{n\to \mathrm{\infty }}{lim inf}cos\left(1-2{\beta }_{n}\right){s}_{n}=\underset{n\to \mathrm{\infty }}{lim inf}cos|1-2{\beta }_{n}|{s}_{n}.$

Hence we get

$\underset{n\to \mathrm{\infty }}{lim sup}{s}_{n}=\underset{n\to \mathrm{\infty }}{lim sup}\left(|1-2{\beta }_{n}|{s}_{n}\right)\le \underset{n\to \mathrm{\infty }}{lim sup}|1-2{\beta }_{n}|\underset{n\to \mathrm{\infty }}{lim sup}{s}_{n},$

and we have

$0\ge \left(1-\underset{n\to \mathrm{\infty }}{lim sup}|1-2{\beta }_{n}|\right)\underset{n\to \mathrm{\infty }}{lim sup}{s}_{n}=\underset{n\to \mathrm{\infty }}{lim inf}\left(1-|1-2{\beta }_{n}|\right)\underset{n\to \mathrm{\infty }}{lim sup}{s}_{n}.$

Since $\left\{{\beta }_{n}\right\}\subset \left[a,b\right]\subset \phantom{\rule{0.2em}{0ex}}\right]0,1\left[$ for $n\in \mathbb{N}$, we have ${lim inf}_{n\to \mathrm{\infty }}\left(1-|1-2{\beta }_{n}|\right)>0$ and thus, ${lim sup}_{n\to \mathrm{\infty }}{s}_{n}\le 0$. Therefore, we get ${lim sup}_{n\to \mathrm{\infty }}{s}_{n}=0$ and we also get ${lim sup}_{n\to \mathrm{\infty }}{t}_{n}=0$. It implies $d\left({x}_{n},S{x}_{n}\right)\to 0$ and $d\left({x}_{n},T{x}_{n}\right)\to 0$.

Next, let $\left\{{y}_{k}\right\}$ be a subsequence of $\left\{{x}_{n}\right\}$. Since $r\left(\left\{{x}_{n}\right\}\right)\le {d}_{0}<\pi /2$, by Theorem 3.3(i), there exists a unique asymptotic center ${x}_{0}$ of $\left\{{y}_{k}\right\}$. Moreover, since $r\left(\left\{{y}_{k}\right\}\right)<\pi /2$, by Theorem 3.3(ii), there exists a subsequence $\left\{{z}_{l}\right\}$ of $\left\{{y}_{k}\right\}$ such that ${z}_{l}\stackrel{\mathrm{\Delta }}{⇀}{z}_{0}\in X$. Further, since $d\left({z}_{l},S{z}_{l}\right)\to 0$, $d\left({z}_{l},T{z}_{l}\right)\to 0$ and S, T are Δ-demiclosed, we have ${z}_{0}\in F\left(S\right)\cap F\left(T\right)$. Then we can show that ${x}_{0}={z}_{0}$, i.e., ${x}_{0}\in F\left(S\right)\cap F\left(T\right)$. If not, from the uniqueness of the asymptotic centers ${x}_{0}$, ${z}_{0}$ of $\left\{{y}_{k}\right\}$, $\left\{{z}_{l}\right\}$, respectively, due to Theorem 3.3(i), we have

$\begin{array}{rl}\underset{k\to \mathrm{\infty }}{lim sup}d\left({y}_{k},{x}_{0}\right)& <\underset{k\to \mathrm{\infty }}{lim sup}d\left({y}_{k},{z}_{0}\right)\\ =\underset{n\to \mathrm{\infty }}{lim}d\left({x}_{n},{z}_{0}\right)\\ =\underset{l\to \mathrm{\infty }}{lim sup}d\left({z}_{l},{z}_{0}\right)\\ <\underset{l\to \mathrm{\infty }}{lim sup}d\left({z}_{l},{x}_{0}\right)\\ \le \underset{k\to \mathrm{\infty }}{lim sup}d\left({y}_{k},{x}_{0}\right).\end{array}$

This is a contradiction. Hence we get ${x}_{0}\in F\left(S\right)\cap F\left(T\right)$. Next, we show that for any subsequences of $\left\{{x}_{n}\right\}$, their asymptotic center consists of the unique element. Let $\left\{{u}_{k}\right\}$, $\left\{{v}_{k}\right\}$ be subsequences of $\left\{{x}_{n}\right\}$, ${x}_{0}\in AC\left(\left\{{u}_{k}\right\}\right)$ and ${x}_{0}^{\prime }\in AC\left(\left\{{v}_{k}\right\}\right)$. We show ${x}_{0}={x}_{0}^{\prime }$ by using contradiction. Assume ${x}_{0}\ne {x}_{0}^{\prime }$. Then ${x}_{0}^{\prime }\notin AC\left(\left\{{u}_{k}\right\}\right)$ and ${x}_{0}\notin AC\left(\left\{{v}_{k}\right\}\right)$ by Theorem 3.3(i). It follows that

$\begin{array}{rl}\underset{k\to \mathrm{\infty }}{lim sup}d\left({u}_{k},{x}_{0}\right)& <\underset{k\to \mathrm{\infty }}{lim sup}d\left({u}_{k},{x}_{0}^{\prime }\right)\\ =\underset{n\to \mathrm{\infty }}{lim}d\left({x}_{n},{x}_{0}^{\prime }\right)\\ =\underset{k\to \mathrm{\infty }}{lim sup}d\left({v}_{k},{x}_{0}^{\prime }\right)\\ <\underset{k\to \mathrm{\infty }}{lim sup}d\left({v}_{k},{x}_{0}\right)\\ =\underset{n\to \mathrm{\infty }}{lim}d\left({x}_{n},{x}_{0}\right)\\ =\underset{k\to \mathrm{\infty }}{lim sup}d\left({u}_{k},{x}_{0}\right).\end{array}$

This is a contradiction. Hence we get ${x}_{0}={x}_{0}^{\prime }$. Therefore, we have $\left\{{x}_{n}\right\}$ 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 $CAT\left(1\right)$ space such that for any $u,v\in X$, $d\left(u,v\right)<\pi /2$. Let S and T be nonexpansive mappings of X into itself such that $F\left(S\right)\cap F\left(T\right)\ne \mathrm{\varnothing }$. Let $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$ and $\left\{{\gamma }_{n}\right\}$ be sequences in $\left[a,b\right]\subset \phantom{\rule{0.2em}{0ex}}\right]0,1\left[$. Define a sequence $\left\{{x}_{n}\right\}\subset X$ as the following recurrence formula: ${x}_{1}\in X$ and

$\left\{\begin{array}{l}{u}_{n}=\left(1-{\beta }_{n}\right){x}_{n}\oplus {\beta }_{n}S{x}_{n},\\ {v}_{n}=\left(1-{\gamma }_{n}\right){x}_{n}\oplus {\gamma }_{n}T{x}_{n},\\ {x}_{n+1}=\left(1-{\alpha }_{n}\right){u}_{n}\oplus {\alpha }_{n}{v}_{n}\end{array}$

for $n\in \mathbb{N}$. Then $\left\{{x}_{n}\right\}$ 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 $CAT\left(1\right)$ space.

Theorem 5.1 Let X be a complete $CAT\left(1\right)$ space such that for any $u,v\in X$, $d\left(u,v\right)<\pi /2$. 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 $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$ and $\left\{{\gamma }_{n}\right\}$ be real sequences in $\left[a,b\right]\subset \phantom{\rule{0.2em}{0ex}}\right]0,1\left[$. Define a sequence $\left\{{x}_{n}\right\}\subset X$ by the following recurrence formula: ${x}_{1}\in X$ and

$\left\{\begin{array}{l}{u}_{n}=\left(1-{\beta }_{n}\right){x}_{n}\oplus {\beta }_{n}{P}_{1}{x}_{n},\\ {v}_{n}=\left(1-{\gamma }_{n}\right){x}_{n}\oplus {\gamma }_{n}{P}_{2}{x}_{n},\\ {x}_{n+1}=\left(1-{\alpha }_{n}\right){u}_{n}\oplus {\alpha }_{n}{v}_{n}\end{array}$

for $n\in \mathbb{N}$. Then $\left\{{x}_{n}\right\}$ 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  and Δ-demiclosed . Further, we also get $F\left({P}_{1}\right)={C}_{1}$ and $F\left({P}_{2}\right)={C}_{2}$. Thus, letting $S={P}_{1}$ and $T={P}_{2}$ in Theorem 4.1, we obtain the desired result. □

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## Acknowledgements

The authors thank the anonymous referees for their valuable comments and suggestions. The first author is supported by Grant-in-Aid for Scientific Research No. 22540175 from the Japan Society for the Promotion of Science.

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Correspondence to Koichi Nakagawa.

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The authors declare that they have no competing interests.

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The authors have contributed in this work on an equal basis. All authors read and approved the final manuscript.

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Kimura, Y., Nakagawa, K. Another type of Mann iterative scheme for two mappings in a complete geodesic space. J Inequal Appl 2014, 72 (2014). https://doi.org/10.1186/1029-242X-2014-72

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• DOI: https://doi.org/10.1186/1029-242X-2014-72

### Keywords

• Nonexpansive Mapping
• Common Fixed Point
• Nonempty Closed Convex Subset
• Iteration Procedure
• Real Sequence 