An approximation of a common fixed point of nonexpansive mappings on convex metric spaces

Abstract

Sokhuma and Kaewkhao (2011) introduced an iteration scheme to compute a common fixed point of a single-valued nonexpansive mapping and a multivalued nonexpansive mapping on a uniformly convex Banach space. In this paper, we extend the above result of Sokhuma and Kaewkhao from a single-valued mapping to a countable number of mappings and, at the same time, we extend the underlying spaces to strictly convex Banach spaces. The corresponding results are also obtained for the $CAT(0)$ space setting.

MSC:47H09, 47H10.

1 Introduction

Let X be a complete metric space, and E a nonempty subset of X. We will denote by $2^E$ the family of nonempty subsets of E and by $FB(E)$ the family of nonempty bounded closed subsets of E. Let $H(\cdot ,\cdot )$ be theHausdorff distance on $FB(X)$, that is,

$$\begin{array}{r}H(A,B)=max\{\underset{a\in A}{sup}dist(a,B),\underset{b\in B}{sup}dist(b,A)\},A,B\in FB(X),\end{array}$$

where $dist(a,B)=inf\{d(a,b):b\in B\}$ is the distance from the point a to the subset B.

A mapping $t:E\to E$ and a multivalued mapping $T:E\to FB(X)$ are said to be nonexpansive if for each $x,y\in E$,

$$\begin{array}{r}d(tx,ty)\le d(x,y),\text{and}\\ H(Tx,Ty)\le d(x,y),\end{array}$$

respectively. If $tx=x$, we call x a fixed point of a single-valued mapping t. Moreover, if $x\in Tx$, we call x a fixed point of a multivalued mapping T. We use the notation $Fix(S)$ to stand for the set of fixed points of a mapping S. Thus $Fix(t)\cap Fix(T)$ is the set of common fixed points of t and T, i.e., $x\in Fix(t)\cap Fix(T)$ if and only if $x=tx\in Tx$.

Following [8], a bounded closed and convex subset E of a Banach space X has the fixed point property for nonexpansive mappings (FPP) (respectively, for multivalued nonexpansive mappings (MFPP)) if every nonexpansive mapping of E into E has a fixed point (respectively, every nonexpansive mapping of E into $2^E$ with compact convex values has a fixed point). For a bounded closed and convex subset E of a Banach space X, a mapping $t:E\to X$ is said to satisfy the conditional fixed point property (CFP) if either t has no fixed points, or t has a fixed point in each nonempty bounded closed convex set that leaves t invariant. A set E is said to have the conditional fixed point property for nonexpansive mappings (CFPP) if every nonexpansive $t:E\to E$ satisfies (CFP). For commuting family of nonexpansive mappings, the following is a remarkable common fixed point property due to Bruck [6].

Theorem 1.1 ([6])

Let X be a Banach space and E a nonempty closed convex subset of X. If E has both the (FPP) and the (CFPP) for nonexpansive mappings, then for any commuting family$\mathcal{S}$of nonexpansive mappings of E into E, there is a common fixed point for$\mathcal{S}$.

Theorem 1.1 was proved by Belluce and Kirk [1] when $\mathcal{S}$ is finite and E is weakly compact and has a normal structure; by Belluce and Kirk [2] when E is weakly compact and has a complete normal structure; by Browder [4] when X is uniformly convex and E is bounded; by Lau and Holmes [11] when $\mathcal{S}$ is left reversible and E is compact; and finally, by Lim [14] when $\mathcal{S}$ is left reversible and E is weakly compact and has a normal structure.

Open Problem (Bruck [6]). Can commutativity of $\mathcal{S}$ be replaced by left reversibility?

The answer to this Problem is not known even when the semigroup is left amenable (see [13] for more details).

In 2011, Sokhuma and Kaewkhao [17] introduced a new iteration method for approximating a common fixed point of a pair of a single-valued and a multivalued nonexpansive mappings and proved the following strong convergence theorem:

Theorem 1.2 ([17], Theorem 3.5])

Let E be a nonempty compact convex subset of a uniformly convex Banach space X, and let$t:E\to E$and$T:E\to FB(E)$be a single-valued and a multivalued nonexpansive mappings respectively, and$Fix(t)\cap Fix(T)\ne \mathrm{\varnothing }$satisfying$Tw=\{w\}$for all$w\in Fix(t)\cap Fix(T)$. Let$\{x_n\}$be the sequence of the modified Ishikawa iteration defined by

$$\begin{array}{r}{y}_n=(1-{\beta }_n)x_n+{\beta }_n{z}_n,\\ x_{n+1}=(1-{\alpha }_n)x_n+{\alpha }_{n}t{y}_n,\end{array}$$

where$x_1\in E$, $z_n\in T{x}_n$and$0<a\le {\alpha }_n$, ${\beta }_n\le b<1$. Then$\{x_n\}$converges strongly to a common fixed point of t and T.

For a single-valued nonexpansive mapping $t:E\to E$ with $Fix(t)\ne \mathrm{\varnothing }$, where E is a convex nonexpansive retract of a real uniformly smooth Banach space, Reich and Shemen [15], Theorem 3.4] obtained a strong convergence to a fixed point of t of a sequence $\{x_n\}$ of the form

$$\begin{array}{r}{y}_n=R_E[(1-{\beta }_n)x_n],\\ x_{n+1}=(1-{\alpha }_n)x_n+{\alpha }_{n}t{y}_n,\end{array}$$

where $R_E$ is a retraction on the subset E and the sequences $\{{\alpha }_n\}$$\{{\beta }_n\}$ satisfy conditions: (i) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n<1$, (ii) ${lim}_{n\to \mathrm{\infty }}{\beta }_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\beta }_n=\mathrm{\infty }$. Clearly, conditions (i) and (ii) on the sequences $\{{\alpha }_n\}$$\{{\beta }_n\}$ are different from the ones in Theorem 1.2.

In 2003, Suzuki [18] proved the following result.

Theorem 1.3 ([18], Theorem 2])

Let E be a compact convex subset of a strictly convex Banach space X. Let$\{t_n:n\in \mathbb{N}\}$be a sequence of nonexpansive mappings on E with${\bigcap }_{n=1}^{\mathrm{\infty }}Fix(t_n)\ne \mathrm{\varnothing }$. Let$\{{\gamma }_n\}$be a sequence of positive numbers such that${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n<1$, and let$\{I_n\}$be a sequence of subsets of$\mathbb{N}$satisfying$I_n\subset I_{n+1}$for$n\in \mathbb{N}$and${\bigcup }_{n=1}^{\mathrm{\infty }}{I}_n=\mathbb{N}$. Define a sequence$\{x_n\}$in E by$x_1\in C$and

$$x_{n+1}=(1-\sum _{i\in I_n}{\gamma }_i)x_n+\sum _{i\in I_n}{\gamma }_i{t}_i{x}_n$$

for$n\in \mathbb{N}$. Then$\{x_n\}$converges strongly to a common fixed point of$\{t_n:n\in \mathbb{N}\}$.

The purpose of this paper is to extend Theorem 1.2 to countably many numbers of single-valued nonexpansive mappings on strictly convex Banach spaces, thereby the result in Theorem 1.3 is covered. The results for $CAT(0)$ spaces are also derived. Our main discoveries are Theorem 3.2 and Theorem 3.6.

2 Preliminaries

We recall that the graph $G(U)$ of a multivalued mapping $U:E\to 2^X$ is $G(U)=\{(x,y)\in X\times X;x\in E,y\in Ux\}$. The following theorem is essentially proved by Dozo [10].

Theorem 2.1 ([10], Theorem 3.1])

Let X be a Banach space which satisfies Opial’s condition, E be a weakly compact convex subset of X. Let$T:E\to K(X)$, where$K(X)$is a family of nonempty compact subsets of X. Then the graph of$U=I-T$is closed in$(X,\sigma (X,X^{\ast }))\times (X,\parallel \cdot \parallel )$, where I denotes the identity on X, $\sigma (X,X^{\ast })$the weak topology and$\parallel \cdot \parallel$the norm (or strong) topology.

We will use the theorem in the following form: If $\{x_n\}$ is a sequence in E such that $\{x_n\}$ converges weakly to some $z\in E$ and $\{dist(x_n,T{x}_n)\}$ converges to 0, then $z\in Tz$.

Let $\{t_n:n\in \mathbb{N}\}$ be a family of nonexpansive mappings from E to E. The following lemma proved by Bruck [5] plays a very important role to our proof of the main result.

Lemma 2.2 ([5], Lemma 3])

Let E be a nonempty closed convex subset of a strictly convex Banach space X, let$\{t_n:n\in \mathbb{N}\}$be a family of single-valued nonexpansive mappings on E. Suppose${\bigcap }_{n=1}^{\mathrm{\infty }}Fix(t_n)$is nonempty. Given$\{{\lambda }_n\}$a sequence of positive numbers with${\sum }_{n=1}^{\mathrm{\infty }}{\lambda }_n=1$. Then a mapping t on E defined by

$$tx=\sum _{n=1}^{\mathrm{\infty }}{\lambda }_n{t}_{n}x$$

for all$x\in E$is well defined, nonexpansive and$Fix(t)={\bigcap }_{n=1}^{\mathrm{\infty }}Fix(t_n)$.

The following results show examples when the required condition on the nonemptiness of the common fixed point set always satisfies:

Theorem 2.3 ([8], Theorem 3.1])

Let E be a weakly compact convex subset of a Banach space X. Suppose E has (MFPP) and (CFPP). Let$\mathcal{S}$be any commuting family of nonexpansive self-mappings of E. If$T:E\to KC(E)$is a multivalued nonexpansive mapping which commutes with every member of$\mathcal{S}$, where$KC(E)$is the family of nonempty compact convex subsets of E. Then$F(\mathcal{S})\cap Fix(T)\ne \mathrm{\varnothing }$where$F(\mathcal{S})={\bigcap }_{t\in \mathcal{S}}Fix(t)$.

Theorem 2.4 ([8], Theorem 3.2])

Let X be a Banach space satisfying the Kirk-Massa condition, i.e., the asymptotic center of each bounded sequence of X in each bounded closed and convex subset is nonempty and compact. Let E be a weakly compact convex subset of X and let$\mathcal{S}$be any commuting family of nonexpansive self-mappings of E. Suppose$T:E\to KC(E)$is a multivalued mapping satisfying condition$(C_{\lambda })$for some$\lambda \in (0,1)$which commutes with every member of$\mathcal{S}$. If T is upper semi-continuous, then$F(\mathcal{S})\cap Fix(T)\ne \mathrm{\varnothing }$.

Note that strictly convex Banach spaces satisfy the condition in the above theorems.

Remark 2.5 In our main theorems (Theorem 3.2 and Theorem 3.6), we assume the following conditions:

$$\begin{array}{r}F(\mathcal{S})\cap Fix(T)\ne \mathrm{\varnothing }\text{and}Tw=\{w\}\text{for all}w\in F(\mathcal{S})\cap Fix(T).\end{array}$$

(2.1)

It is an open problem to find a sufficient condition to assure that the condition (2.1) is satisfied.

Let $(X,d)$ be a metric space. A geodesic joining $x\in X$ 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 $t,t^{\prime }\in [0,l]$. Thus c is an isometry and $d(x,y)=l$. The image of c is called a geodesic (or metric) segment joining x and y. We denote $[x,y]$ for this geodesic if it is unique. Write $c(\alpha 0+(1-\alpha )l)=\alpha x\oplus (1-\alpha )y$ for $\alpha \in (0,1)$. The space X is said to be a geodesic space if every two points of X are joined by a geodesic. It is said to be of hyperbolic type [12] if it satisfies:

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

(2.2)

for all $p\in X$. Let $\{v_1,v_2,\dots ,v_n\}\subset X$ and $\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n\}\subset (0,1)$ with ${\sum }_{i=1}^n{\lambda }_i=1$. It had been defined, by induction, in [7] that

$$\underset{i=1}{\overset{n}{⨁}}{\lambda }_i{v}_i:=(1-{\lambda }_n)(\frac{{\lambda }_1}{1-{\lambda }_n}{v}_1\oplus \frac{{\lambda }_2}{1-{\lambda }_n}{v}_2\oplus \cdots \oplus \frac{{\lambda }_{n-1}}{1-{\lambda }_n}{v}_{n-1})\oplus {\lambda }_n{v}_n.$$

(2.3)

The definition of ⊕ in (2.3) is an ordered one in the sense that it depends on the order of points $v_1,\dots ,v_n$. Under (2.2) we can see that

$$d(\underset{i=1}{\overset{n}{⨁}}{\lambda }_i{v}_i,x)\le \sum _{n=1}^n{\lambda }_{i}d(v_i,x)$$

(2.4)

for each $x\in X$.

Following [3], a metric space X is said to be a $CAT(0)$space if it is geodesically connected and if every geodesic triangle in X is at least as thin as its comparison triangle in the Euclidean plane ${\mathbb{E}}^2$. In fact (cf. [3] p.163), the following are equivalent for a geodesic space X:

• (i) X is a $CAT(0)$ space.

• (ii) X satisfies the (CN) inequality: If $x_0,x_1\in X$ and $\frac{x_0\oplus x_1}{2}$ is the midpoint of $x_0$ and $x_1$, then

  $$d^2(y,\frac{x_0\oplus x_1}{2})\le \frac{1}{2}{d}^2(y,x_0)+\frac{1}{2}{d}^2(y,x_1)-\frac{1}{4}{d}^2(x_0,x_1),\text{for all}y\in X.$$

Lemma 2.6 ([3], Proposition 2.2]) Let X be a$CAT(0)$space. Then for each$p,q,r,s\in X$and$\alpha \in [0,1]$

$$\begin{array}{r}d(\alpha p\oplus (1-\alpha )q,\alpha r\oplus (1-\alpha )s)\le \alpha d(p,r)+(1-\alpha )d(q,s).\end{array}$$

(2.5)

In particular, (2.2) holds in$CAT(0)$spaces.

In [9] the element $x={⨁}_{n=1}^{\mathrm{\infty }}{\lambda }_n{v}_n$ has been defined. Let $\{{\lambda }_n\}$ be a given sequence in $(0,1)$ such that ${\sum }_{n=1}^{\mathrm{\infty }}{\lambda }_n=1$, let $\{v_n\}$ be a bounded sequence in X, and let $v_0$ be an arbitrary point in X. Let ${\lambda }_n^{\prime }={\sum }_{i=n+1}^{\mathrm{\infty }}{\lambda }_i$ and assume that ${\sum }_{i=n}^{\mathrm{\infty }}{\lambda }_i^{\prime }\to 0$ as $n\to \mathrm{\infty }$. Set

$$s_n:={\lambda }_1{v}_1\oplus {\lambda }_2{v}_2\oplus \cdots \oplus {\lambda }_n{v}_n\oplus {\lambda }_n^{\prime }{v}_0.$$

Thus, by (2.3),

$$s_n=\left(\sum _{i=1}^n{\lambda }_i\right)w_n\oplus {\lambda }_n^{\prime }{v}_0,$$

(2.6)

where $w_1=v_1$ and for each $n\ge 2$

$$w_n=\frac{{\lambda }_1}{{\sum }_{i=1}^n{\lambda }_i}{v}_1\oplus \frac{{\lambda }_2}{{\sum }_{i=1}^n{\lambda }_i}{v}_2\oplus \cdots \oplus \frac{{\lambda }_n}{{\sum }_{i=1}^n{\lambda }_i}{v}_n.$$

We know that $\{s_n\}$ is a Cauchy sequence (see [9]). Thus $s_n\to x$ as $n\to \mathrm{\infty }$ for some $x\in X$. Write

$$x=\underset{n=1}{\overset{\mathrm{\infty }}{⨁}}{\lambda }_n{v}_n.$$

By (2.6), $d(s_n,w_n)\le {\lambda }_n^{\prime }d(w_n,v_0)$, it is seen that ${lim}_{n\to \mathrm{\infty }}{s}_n={lim}_{n\to \mathrm{\infty }}{w}_n$. Thus the limit x is independent of the choice of $v_0$.

Lemma 2.7 ([9], Lemma 3.8])

Let C be a nonempty closed convex subset of a complete$CAT(0)$space X, let$\{t_n:n\in \mathbb{N}\}$be a family of single-valued nonexpansive mappings on C. Suppose${\bigcap }_{n=1}^{\mathrm{\infty }}Fix(t_n)$is nonempty. Define$t:C\to C$by$t(x)={⨁}_{n=1}^{\mathrm{\infty }}{\lambda }_n{t}_n(x)$for all$x\in C$where$\{{\lambda }_n\}\subset (0,1)$with${\sum }_{n=1}^{\mathrm{\infty }}{\lambda }_n=1$and${\sum }_{i=n}^{\mathrm{\infty }}{\lambda }_i^{\prime }\to 0$as$n\to \mathrm{\infty }$ . Then t is nonexpansive and$Fix(t)={\bigcap }_{n=1}^{\mathrm{\infty }}Fix(t_n)$.

3 Main results

3.1 Strictly convex Banach spaces

The following result is a generalization of the result of [16], Lemma 1.3].

Lemma 3.1 Let E be a compact subset of a strictly convex Banach space X, let$\{{\alpha }_n\}$be a sequence of real numbers such that$0<a\le {\alpha }_n\le b<1$for all$n\in \mathbb{N}$, and let$\{u_n\}$, $\{v_n\}$be sequences of E satisfying, for some$c\ge 0$,

• (i) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\parallel u_n\parallel \le c$,

• (ii) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\parallel v_n\parallel \le c$ and

• (iii) ${lim}_{n\to \mathrm{\infty }}\parallel {\alpha }_n{u}_n+(1-{\alpha }_n)v_n\parallel =c$.

Then, ${lim}_{n\to \mathrm{\infty }}\parallel u_n-v_n\parallel =0$.

Proof We suppose on the contrary that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\parallel u_n-v_n\parallel \ne 0$. Since E and $[a,b]$ are compact, there exist subsequences $\{u_{n_k}\}$ of $\{u_n\}$, $\{v_{n_k}\}$ of $\{v_n\}$ and $\{{\alpha }_{n_k}\}$ of $\{{\alpha }_n\}$ such that ${lim}_{k\to \mathrm{\infty }}{u}_{n_k}=u$, ${lim}_{k\to \mathrm{\infty }}{v}_{n_k}=v$, ${lim}_{k\to \mathrm{\infty }}{\alpha }_{n_k}=\alpha$ for some $u,v\in E$ with $u\ne v$ and for some $\alpha \in [0,1]$. From (i) and (ii) we have $\parallel u\parallel ={lim}_{k\to \mathrm{\infty }}\parallel u_{n_k}\parallel \le c$ and $\parallel v\parallel ={lim}_{k\to \mathrm{\infty }}\parallel v_{n_k}\parallel \le c$. Using the strict convexity of X and (iii), we have $c={lim}_{k\to \mathrm{\infty }}\parallel {\alpha }_{n_k}{u}_{n_k}+(1-{\alpha }_{n_k})v_{n_k}\parallel =\parallel \alpha u+(1-\alpha )v\parallel <\alpha \parallel u\parallel +(1-\alpha )\parallel v\parallel \le c$, a contradiction. Hence ${lim}_{n\to \mathrm{\infty }}\parallel u_n-v_n\parallel =0$. □

Now we introduce a new iteration method for a family of single-valued nonexpansive mappings and a multivalued nonexpansive mapping. Let E be a nonempty bounded closed convex subset of a Banach space X, let $\{t_n:n\in \mathbb{N}\}$ be a family of single-valued nonexpansive mappings on E, and let $T:E\to FB(E)$ be a multivalued nonexpansive mapping. Given a sequence of positive numbers $\{{\gamma }_n\}$ with ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n<1$. The sequence $\{x_n\}$ of the modified Ishikawa iteration is defined by $x_1\in E$, and

$$\begin{array}{r}{y}_n=(1-{\beta }_n)x_n+{\beta }_n{z}_n,\\ x_{n+1}=(1-\sum _{i=1}^n{\gamma }_i)x_n+\sum _{i=1}^n{\gamma }_i{t}_i{y}_n,\end{array}$$

(3.1)

where $z_n\in T{x}_n$, and $0<a\le {\beta }_n\le b<1$. Put $F:=({\bigcap }_n^{\mathrm{\infty }}Fix(t_n))\cap Fix(T)$.

Theorem 3.2 Let E be a nonempty compact convex subset of a strictly convex Banach space X, let$\{t_n:n\in \mathbb{N}\}$be a family of single-valued nonexpansive mappings on E, and let$T:E\to FB(E)$be a multivalued nonexpansive mapping. Suppose$F\ne \mathrm{\varnothing }$and$Tw=\{w\}$for all$w\in F$. Given a sequence of positive numbers$\{{\gamma }_n\}$with${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n<1$and$\{{\beta }_n\}$with$0<a\le {\beta }_n\le b<1$. Then the sequence$\{x_n\}$defined by (3.1) converges strongly to some$v\in F$.

Proof We follow the proof of [17], Theorem 3.6] and split the proof into five steps.

Step 1. ${lim}_{n\to \mathrm{\infty }}\parallel x_n-w\parallel$ exists for all $w\in F$:

We first note that, since $Tw=\{w\}$,

$$\parallel z_n-w\parallel =dist(z_n,Tw)\le H(T{x}_n,Tw)\le \parallel x_n-w\parallel .$$

Consider the following estimates:

$$\begin{array}{rcl}\parallel x_{n+1}-w\parallel & \le & (1-\sum _{i=1}^n{\gamma }_i)\parallel x_n-w\parallel +\sum _{i=1}^n{\gamma }_i\parallel t_i{y}_n-w\parallel \\ \le & (1-\sum _{i=1}^n{\gamma }_i)\parallel x_n-w\parallel +\sum _{i=1}^n{\gamma }_i\parallel y_n-w\parallel \\ \le & (1-\sum _{i=1}^n{\gamma }_i)\parallel x_n-w\parallel +\left(\sum _{i=1}^n{\gamma }_i\right)((1-{\beta }_n)\parallel x_n-w\parallel +{\beta }_n\parallel z_n-w\parallel )\\ \le & \parallel x_n-w\parallel .\end{array}$$

Therefore, $\{\parallel x_n-w\parallel \}$ is a bounded decreasing sequence in $\mathbb{R}$, and hence ${lim}_{n\to \mathrm{\infty }}\parallel x_n-w\parallel$ exists.

Step 2. ${lim}_{n\to \mathrm{\infty }}\parallel x_n-\frac{{\sum }_{i=1}^n{\gamma }_i{t}_i{y}_n}{{\sum }_{i=1}^n{\gamma }_i}\parallel =0$:

From Step 1, suppose ${lim}_{n\to \mathrm{\infty }}\parallel x_n-w\parallel =c$. We have

$$\begin{array}{rcl}\parallel \frac{{\sum }_{i=1}^n{\gamma }_i{t}_i{y}_n}{{\sum }_{i=1}^n{\gamma }_i}-w\parallel & \le & \frac{1}{{\sum }_{i=1}^n{\gamma }_i}\parallel \sum _{i=1}^n{\gamma }_i{t}_i{y}_n-\sum _{i=1}^n{\gamma }_{i}w\parallel \\ \le & \parallel y_n-w\parallel \le \parallel x_n-w\parallel .\end{array}$$

Thus

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel \frac{{\sum }_{i=1}^n{\gamma }_i{t}_i{y}_n}{{\sum }_{i=1}^n{\gamma }_i}-w\parallel \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel y_n-w\parallel \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_n-w\parallel =c.$$

(3.2)

We also have

$$\begin{array}{rcl}c& =& \underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-w\parallel \\ =& \underset{n\to \mathrm{\infty }}{lim}\parallel (1-\sum _{i=1}^n{\gamma }_i)x_n+\sum _{i=1}^n{\gamma }_i{t}_i{y}_n-w\parallel \\ =& \underset{n\to \mathrm{\infty }}{lim}\parallel (1-\sum _{i=1}^n{\gamma }_i)(x_n-w)+\sum _{i=1}^n{\gamma }_i(\frac{{\sum }_{i=1}^n{\gamma }_i{t}_i{t}_n}{{\sum }_{i=1}^n{\gamma }_i}-w)\parallel .\end{array}$$

By Lemma 3.1, since $0<{\gamma }_1<{\sum }_{i=1}^n{\gamma }_i\le {\sum }_{i=1}^{\mathrm{\infty }}{\gamma }_i<1$, .

Step 3. ${lim}_{n\to \mathrm{\infty }}\parallel x_n-z_n\parallel =0$:

From (3.1), we can see that

$$\parallel x_{n+1}-w\parallel \le (1-\sum _{i=1}^n{\gamma }_i)\parallel x_n-w\parallel +\sum _{i=1}^n{\gamma }_i\parallel y_n-w\parallel ,$$

and hence $\parallel x_{n+1}-w\parallel -\parallel x_n-w\parallel \le {\sum }_{i=1}^n{\gamma }_i(\parallel y_n-w\parallel -\parallel x_n-w\parallel )$. Therefore, $(\frac{\parallel x_{n+1}-w\parallel -\parallel x_n-w\parallel }{{\sum }_{i=1}^n{\gamma }_i})+\parallel x_n-w\parallel \le \parallel y_n-w\parallel$ and by (3.2) we obtain

$$\begin{array}{rcl}c& =& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\{\left(\frac{\parallel x_{n+1}-w\parallel -\parallel x_n-w\parallel }{{\sum }_{i=1}^n{\gamma }_i}\right)+\parallel x_n-w\parallel \}\\ \le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel y_n-w\parallel \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel y_n-w\parallel \le c.\end{array}$$

Thus $c={lim}_{n\to \mathrm{\infty }}\parallel y_n-w\parallel ={lim}_{n\to \mathrm{\infty }}\parallel (1-{\beta }_n)(x_n-w)+{\beta }_n(z_n-w)\parallel$. By Lemma 3.1, since $0<a\le {\beta }_n\le b<1$, ${lim}_{n\to \mathrm{\infty }}\parallel x_n-z_n\parallel =0$.

Step 4. ${lim}_{n\to \mathrm{\infty }}\parallel x_n-\frac{{\sum }_{i=1}^{\mathrm{\infty }}{\gamma }_i{t}_i{x}_n}{{\sum }_{i=1}^{\mathrm{\infty }}{\gamma }_i}\parallel =0$:

We note from Step 3 that

$$\parallel \frac{{\sum }_{i=1}^n{\gamma }_i{t}_i{x}_n}{{\sum }_{i=1}^n{\gamma }_i}-\frac{{\sum }_{i=1}^n{\gamma }_i{t}_i{y}_n}{{\sum }_{i=1}^n{\gamma }_i}\parallel \le \parallel x_n-y_n\parallel ={\beta }_n\parallel x_n-z_n\parallel \to 0\text{as}n\to \mathrm{\infty }$$

(3.3)

and

$$\begin{array}{rcl}\parallel t_i{x}_n\parallel & \le & \parallel t_i{x}_n-w\parallel +\parallel w\parallel \le \parallel x_n-w\parallel +\parallel w\parallel \\ \le & \parallel x_1-w\parallel +\parallel w\parallel :=M\end{array}$$

for all $i\in \mathbb{N}$. Therefore,

$$\begin{array}{rcl}\parallel x_n-\frac{{\sum }_{i=1}^{\mathrm{\infty }}{\gamma }_i{t}_i{x}_n}{{\sum }_{i=1}^{\mathrm{\infty }}{\gamma }_i}\parallel & \le & \parallel x_n-\frac{{\sum }_{i=1}^n{\gamma }_i{t}_i{x}_n}{{\sum }_{i=1}^n{\gamma }_i}\parallel +\parallel \frac{{\sum }_{i=1}^n{\gamma }_i{t}_i{x}_n}{{\sum }_{i=1}^n{\gamma }_i}-\frac{{\sum }_{i=1}^{\mathrm{\infty }}{\gamma }_i{t}_i{x}_n}{{\sum }_{i=1}^{\mathrm{\infty }}{\gamma }_i}\parallel \\ \le & \parallel x_n-\frac{{\sum }_{i=1}^n{\gamma }_i{t}_i{x}_n}{{\sum }_{i=1}^n{\gamma }_i}\parallel +\parallel \frac{{\sum }_{i=1}^n{\gamma }_i{t}_i{x}_n}{{\sum }_{i=1}^n{\gamma }_i}-\frac{{\sum }_{i=1}^n{\gamma }_i{t}_i{x}_n}{{\sum }_{i=1}^{\mathrm{\infty }}{\gamma }_i}\parallel \\ +\frac{1}{{\sum }_{i=1}^{\mathrm{\infty }}{\gamma }_i}\sum _{i=n+1}^{\mathrm{\infty }}{\gamma }_i\parallel t_i{x}_n\parallel \\ \le & \parallel x_n-\frac{{\sum }_{i=1}^n{\gamma }_i{t}_i{x}_n}{{\sum }_{i=1}^n{\gamma }_i}\parallel +\frac{{\sum }_{i=n+1}^{\mathrm{\infty }}{\gamma }_i}{({\sum }_{i=1}^n{\gamma }_i)({\sum }_{i=1}^{\mathrm{\infty }}{\gamma }_i)}\sum _{i=1}^n{\gamma }_{i}M\\ +\frac{{\sum }_{i=n+1}^{\mathrm{\infty }}{\gamma }_i}{{\sum }_{i=1}^{\mathrm{\infty }}{\gamma }_i}M\\ =& \parallel x_n-\frac{{\sum }_{i=1}^n{\gamma }_i{t}_i{x}_n}{{\sum }_{i=1}^n{\gamma }_i}\parallel +\frac{2{\sum }_{i=n+1}^{\mathrm{\infty }}{\gamma }_i}{{\sum }_{i=1}^{\mathrm{\infty }}{\gamma }_i}M\\ \le & \parallel x_n-\frac{{\sum }_{i=1}^n{\gamma }_i{t}_i{y}_n}{{\sum }_{i=1}^n{\gamma }_i}\parallel +\parallel \frac{{\sum }_{i=1}^n{\gamma }_i{t}_i{y}_n}{{\sum }_{i=1}^n{\gamma }_i}-\frac{{\sum }_{i=1}^n{\gamma }_i{t}_i{x}_n}{{\sum }_{i=1}^n{\gamma }_i}\parallel \\ +\frac{2{\sum }_{i=n+1}^{\mathrm{\infty }}{\gamma }_i}{{\sum }_{i=1}^{\mathrm{\infty }}{\gamma }_i}M.\end{array}$$

From Step 2 and (3.3), we obtain ${lim}_{n\to \mathrm{\infty }}\parallel x_n-\frac{{\sum }_{i=1}^{\mathrm{\infty }}{\gamma }_i{t}_i{x}_n}{{\sum }_{i=1}^{\mathrm{\infty }}{\gamma }_i}\parallel =0$.

Step 5. ${lim}_{n\to \mathrm{\infty }}{x}_n=v\in F$:

Define a mapping $t:E\to E$ by

$$tx=\frac{{\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n{t}_{n}x}{{\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n}$$

for any $x\in E$. By Lemma 2.2, t is well defined, nonexpansive and $Fix(t)={\bigcap }_{n=1}^{\mathrm{\infty }}Fix(t_n)$. Since E is compact, there exists a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ which converges to v for some $v\in E$. Using Step 3 and Step 4, we have

$$\begin{array}{rcl}\parallel tv-v\parallel & \le & \underset{k\to \mathrm{\infty }}{lim}(\parallel tv-t{x}_{n_k}\parallel +\parallel t{x}_{n_k}-x_{n_k}\parallel +\parallel x_{n_k}-v\parallel )\\ \le & \underset{k\to \mathrm{\infty }}{lim}(\parallel t{x}_{n_k}-x_{n_k}\parallel +2\parallel x_{n_k}-v\parallel )=0\end{array}$$

and

$$\begin{array}{rcl}dist(v,Tv)& \le & \parallel v-x_{n_k}\parallel +dist(x_{n_k},T{x}_{n_k})+H(T{x}_{n_k},Tv)\\ \le & \parallel v-x_{n_k}\parallel +\parallel x_{n_k}-z_{n_k}\parallel +\parallel x_{n_k}-v\parallel \to 0\text{as}k\to \mathrm{\infty }.\end{array}$$

It follows that $v\in Fix(T)\cap Fix(t)=F$. Since ${lim}_{n\to \mathrm{\infty }}\parallel x_n-v\parallel$ exists by Step 1, ${lim}_{n\to \mathrm{\infty }}\parallel x_n-v\parallel ={lim}_{k\to \mathrm{\infty }}\parallel x_{n_k}-v\parallel =0$. □

The following example shows that the condition ‘$Tw=\{w\}$ for all $w\in F$’ in Theorem 3.2 is necessary.

Example 3.3 We consider the space X of Example 3.9 in [8]. Let X be the Hilbert space ${\mathbb{R}}^2$ with the usual norm, and let $f:[0,1]\to [0,1]$ be a continuous strictly concave function such that $f(0)=\frac{1}{2}$$f(1)=1$ and $f^{\prime }(x)\le 1$ for all $x\in [0,1]$. Let ${\epsilon }_n={\sum }_{i=1}^n{(\frac{1}{2})}^{i+1}$$T:{[0,1]}^2\to FB({[0,1]}^2)$ be defined by $T(a,b)=[0,1]\times [f(a),1]$ and $t_n:{[0,1]}^2\to {[0,1]}^2$ be defined by

$$t_n(a,b)=\{\begin{array}{ll}(a,{\epsilon }_n),& b<{\epsilon }_n,\\ (a,b),& \text{otherwise.}\end{array}$$

It is straightforward showing that T and each $t_n$ are nonexpansive. Set $x_1=(1,0)\in {[0,1]}^2$ and for a subsequence $\{{\gamma }_n\}$ in $(0,1)$ with ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n<1$. Let $\{x_n=(a_n,b_n)\}$ be a sequence in ${[0,1]}^2$ defined as

$$\begin{array}{r}{y}_n=\frac{1}{2}{x}_n+\frac{1}{2}{z}_n,\\ x_{n+1}=(1-\sum _{i=1}^n{\gamma }_i)x_n+\sum _{i=1}^n{\gamma }_i{t}_i{y}_n,\end{array}$$

(3.4)

where

$$z_n=\{\begin{array}{ll}(0,f(a_n)),& n\text{is odd},\\ (1,f(a_n)),& n\text{is even}.\end{array}$$

We will show that $\{x_n\}$ does not converge to a common fixed point of T and $\{t_n\}$.

Proof Clearly, $\{z_n\}$ is a divergent sequence. We note that ${\epsilon }_n\uparrow \frac{1}{2}$ and for each $y=(a,b)\in {[0,1]}^2$ with $b\ge \frac{1}{2}$, we have $t_{i}y=y$ for all i. If we put $y_n=(c_n,d_n)$, then $d_n\ge \frac{1}{2}$ for all n. Since ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n<1$, we must have $d(x_n,z_n)\to 0$ as $n\to \mathrm{\infty }$. Suppose $\{x_n\}$ converges to z for some $z\in F=\{(a,b)\in {[0,1]}^2:b\ge f(a)\}$. Thus $\{z_n\}$ also converges to z, a contradiction. □

It is noticed that F is not convex. Thus it is not a nonexpansive retract of any convex set. It can be also observed that if we redefine the mapping T as $T(a,b)=\{a\}\times [\frac{1+b}{2},1]$ we can easily verify that T is nonexpansive and the condition (2.1) is satisfied.

Remark 3.4 With the same proof, Theorem 3.2 is valid when $\{x_n\}$ is of the following form: For a permutation π on $\mathbb{N}$, define $\{x_n\}$ in E by $x_1\in E$ and

$$\begin{array}{r}{y}_n=(1-{\beta }_n)x_n+{\beta }_n{z}_n,\\ x_{n+1}=(1-\sum _{i=1}^n{\gamma }_{\pi (i)})x_n+\sum _{i=1}^n{\gamma }_{\pi (i)}{t}_{\pi (i)}{y}_n,\end{array}$$

$z_n\in T{x}_n$, and $0<a\le {\beta }_n\le b<1$.

Note also that the above result is equivalent to:

Let $\{I_n\}$ be a sequence of subsets of $\mathbb{N}$ satisfying $I_n\subset I_{n+1}$ for $n\in \mathbb{N}$ and ${\bigcup }_{n=1}^{\mathrm{\infty }}{I}_n=\mathbb{N}$. Define $\{x_n\}$ in E by $x_1\in E$ and

$$\begin{array}{r}{y}_n=(1-{\beta }_n)x_n+{\beta }_n{z}_n,\\ x_{n+1}=(1-\sum _{i\in I_n}{\gamma }_i)x_n+\sum _{i\in I_n}{\gamma }_i{t}_i{y}_n,\end{array}$$

$z_n\in T{x}_n$, and $0<a\le {\beta }_n\le b<1$. Then the sequence $\{x_n\}$ converges strongly to some $v\in F$.

Thus Theorem 3.2 contains Theorem 1.3.

With the application of the demiclosedness principle (Theorem 2.1), a weak convergence version of Theorem 3.2 also holds:

Theorem 3.5 Let X be a strictly convex Banach space satisfying the Opial’s condition, E be a weakly compact convex subset of X, let$\{t_n:n\in \mathbb{N}\}$be a family of single-valued nonexpansive mappings on E, and let$T:E\to FB(E)$be a multivalued nonexpansive mapping. Suppose$F\ne \mathrm{\varnothing }$and$Tw=\{w\}$for all$w\in F$. Given a sequence of positive numbers$\{{\gamma }_n\}$with$0<{\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n<1$and$\{{\beta }_n\}$with$0<a\le {\beta }_n\le b<1$. Then the sequence$\{x_n\}$defined by (3.1) converges weakly to some$v\in F$.

Proof In the proof of Theorem 3.2, by applying the Opial’s condition, it follows from a standard argument that $\{x_n\}$ converges weakly to some $v\in E$. Then Theorem 2.1 implies that v is a point in F. □

3.2 $\mathrm{CAT}(0)$ spaces

Let E be a nonempty bounded closed convex subset of a complete $CAT(0)$ space X, let $\{t_n:n\in \mathbb{N}\}$ be a family of single-valued nonexpansive mappings on E, and $T:E\to FB(E)$ be a multivalued nonexpansive mapping. Given $\{{\gamma }_n\}$ a sequence of positive numbers with ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n<1$ and ${\sum }_{i=n}^{\mathrm{\infty }}{\gamma }_i^{\prime }\to 0$ as $n\to \mathrm{\infty }$ where ${\gamma }_n^{\prime }={\sum }_{i=n+1}^{\mathrm{\infty }}{\gamma }_i$. The sequence $\{x_n\}$ of the modified Ishikawa iteration is defined by

$$\begin{array}{r}{y}_n=(1-{\beta }_n)x_n\oplus {\beta }_n{z}_n,\\ x_{n+1}=(1-\sum _{i=1}^n{\gamma }_i)x_n\oplus \left(\sum _{i=1}^n{\gamma }_i\right)\underset{i=1}{\overset{n}{⨁}}\frac{{\gamma }_i}{{\sum }_{i=1}^n{\gamma }_i}{t}_i{y}_n,\end{array}$$

(3.5)

where $x_1\in E$, $z_n\in T{x}_n$, and $0<a\le {\beta }_n\le b<1$. Put $F:={\bigcap }_{n=1}^{\mathrm{\infty }}Fix(t_n)\cap Fix(T)$.

Theorem 3.6 Let E be a compact convex subset of a complete$CAT(0)$space X. Let$\{t_n:n\in \mathbb{N}\}$be a family of single-valued nonexpansive mappings on E, and let$T:E\to FB(E)$be a multivalued nonexpansive mapping. Suppose$F\ne \mathrm{\varnothing }$and$Tw=\{w\}$for all$w\in F$. Given$\{{\gamma }_n\}$a sequence of positive numbers with${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n<1$and${\sum }_{i=n}^{\mathrm{\infty }}{\gamma }_i^{\prime }\to 0$as$n\to \mathrm{\infty }$where${\gamma }_n^{\prime }={\sum }_{i=n+1}^{\mathrm{\infty }}{\gamma }_i$. If$0<a\le {\beta }_n\le b<1$, then the sequence$\{x_n\}$defined by (3.5) converges strongly to some$v\in F$.

Proof The proof follows along the lines with the proof of Theorem 3.2. Recall that $w_{1}x=t_{1}x$ and $w_{n}x={⨁}_{i=1}^n\frac{{\gamma }_i}{{\sum }_{i=1}^n{\gamma }_i}{t}_{i}x$ for all $n\ge 2$. Thus, by (3.5),

$$x_{n+1}=(1-\sum _{i=1}^n{\gamma }_i)x_n\oplus \left(\sum _{i=1}^n{\gamma }_i\right)w_n{y}_n.$$

As before, we consider the proof in 5 steps. Because of the same details in some cases, we only present proofs for Step 2 to Step 4.

Step 2. ${lim}_{n\to \mathrm{\infty }}d(x_n,w_n{y}_n)=0$:

Let $w\in F$, we have $w_{n}w=w$ for all n. Using the nonexpansiveness of $w_n$, we see that

$$\begin{array}{r}d(w_n{y}_n,w)\le d(y_n,w)\le (1-{\beta }_n)d(x_n,w)+{\beta }_{n}d(z_n,w)\le d(x_n,w).\end{array}$$

(3.6)

By (3.6) and using (CN) inequality,

$$\begin{array}{rcl}{d}^2(x_{n+1},w)& \le & (1-\sum _{i=1}^n{\gamma }_i)d^2(x_n,w)+\left(\sum _{i=1}^n{\gamma }_i\right)d^2(w_n{y}_n,w)\\ -\sum _{i=1}^n{\gamma }_i(1-\sum _{i=1}^n{\gamma }_i)d^2(x_n,w_n{y}_n)\\ \le & d^2(x_n,w)-\sum _{i=1}^n{\gamma }_i(1-\sum _{i=1}^n{\gamma }_i)d^2(x_n,w_n{y}_n).\end{array}$$

Let $\gamma ={\sum }_{i=1}^{\mathrm{\infty }}{\gamma }_i$. Since $0<{\gamma }_1\le {\sum }_{i=1}^n{\gamma }_i\le \gamma <1$,

$${\gamma }_1(1-\gamma )d^2(x_n,w_n{y}_n)\le \sum _{i=1}^n{\gamma }_i(1-\sum _{i=1}^n{\gamma }_i)d^2(x_n,w_n{y}_n)\le d^2(x_n,w)-d^2(x_{n+1},w).$$

This implies that

$$\sum _{n=1}^{\mathrm{\infty }}[{\gamma }_1(1-\gamma )d^2(x_n,w_n{y}_n)]\le d^2(x_1,w)<\mathrm{\infty },$$

and hence ${lim}_{n\to \mathrm{\infty }}d(x_n,w_n{y}_n)=0$.

Step 3. ${lim}_{n\to \mathrm{\infty }}d(x_n,z_n)=0$:

Using (3.6) and (CN) inequality, we have

$$\begin{array}{rcl}{d}^2(w_n{y}_n,w)& \le & d^2(y_n,w)\le (1-{\beta }_n)d^2(x_n,w)+{\beta }_n{d}^2(z_n,w)-{\beta }_n(1-{\beta }_n)d^2(x_n,z_n)\\ \le & d^2(x_n,w)-{\beta }_n(1-{\beta }_n)d^2(x_n,z_n),\end{array}$$

and thus

$$\begin{array}{rcl}{d}^2(x_{n+1},u)& \le & (1-\sum _{i=1}^n{\gamma }_i)d^2(x_n,w)+\sum _{i=1}^n{\gamma }_i{d}^2(w_n{y}_n,w)\\ \le & d^2(x_n,w)-{\beta }_n\left(\sum _{i=1}^n{\gamma }_i\right)(1-{\beta }_n)d^2(x_n,z_n).\end{array}$$

As before,

$$a{\gamma }_1(1-b)d^2(x_n,z_n)\le {\beta }_n\left(\sum _{i=1}^n{\gamma }_i\right)(1-{\beta }_n)d^2(x_n,z_n)\le d^2(x_n,w)-d^2(x_{n+1},w).$$

This also implies that ${lim}_{n\to \mathrm{\infty }}d(x_n,z_n)=0$.

Step 4. ${lim}_{n\to \mathrm{\infty }}d(x_n,t{x}_n)=0$, where $t={⨁}_{i=1}^{\mathrm{\infty }}\frac{{\gamma }_i}{{\sum }_{i=1}^{\mathrm{\infty }}{\gamma }_i}{t}_i$:

Since E is compact, there exists a subsequence $\{y_{n^{\prime }}\}$ of $\{y_n\}$ such that $y_{n^{\prime }}\to y$ as $n^{\prime }\to \mathrm{\infty }$ for some $y\in E$. Using the nonexpansiveness of $w_{n^{\prime }}$ and t, we have

$$\begin{array}{rcl}d(w_{n^{\prime }}{y}_{n^{\prime }},t{y}_{n^{\prime }})& \le & d(w_{n^{\prime }}{y}_{n^{\prime }},w_{n^{\prime }}y)+d(w_{n^{\prime }}y,ty)+d(ty,t{y}_{n^{\prime }})\\ \le & 2d(y_{n^{\prime }},y)+d(w_{n^{\prime }}y,ty)\to 0\text{as}{n}^{\prime }\to \mathrm{\infty }.\end{array}$$

Therefore, ${lim}_{n\to \mathrm{\infty }}d(w_n{y}_n,t{y}_n)=0$. From Step 2 and Step 3 we have

$$\begin{array}{rcl}d(x_n,t{x}_n)& \le & d(x_n,t{y}_n)+d(t{y}_n,t{x}_n)\\ \le & d(x_n,t{y}_n)+d(y_n,x_n)\\ \le & d(x_n,w_n{y}_n)+d(w_n{y}_n,t{y}_n)+{\beta }_{n}d(x_n,z_n)\to 0\text{as}n\to \mathrm{\infty }.\end{array}$$

 □