A modified Picard-Mann hybrid iterative algorithm for common fixed points of countable families of nonexpansive mappings

Abstract

An up-to-date method is used for approximating common fixed points of countable families of nonlinear mappings. A modified Picard-Mann hybrid iterative algorithm is introduced with the help of our method for the class of nonexpansive mappings. Strong convergence and weak convergence theorems are established in the framework of uniformly convex Banach spaces. Our results extend the corresponding ones announced by Khan (Fixed Point Theory Appl. 2013:69, 2013, doi:10.1186/1687-1812-2013-69) to the case of countable families of nonexpansive mappings.

MSC:47H09, 47J25.

1 Introduction

Let K be a nonempty closed convex subset of a real uniformly convex Banach space E. A mapping $T:K\to K$ is said to be nonexpansive if

$$\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,\mathrm{\forall }x,y\in K.$$

Iterative techniques for approximating fixed points of nonexpansive mappings have been studied by various authors (see, e.g., [1–5]) who used the Mann iteration process or the Ishikawa process. In 2013, Khan [6] introduced the following Picard-Mann hybrid iterative process for a single nonexpansive mapping T. For any initial point $x_0\in K$:

$$\{\begin{array}{l}{x}_{n+1}=T{y}_n,\\ y_n=(1-{\alpha }_n)x_n+{\alpha }_{n}T{x}_n,n\in \mathbb{N},\end{array}$$

(1.1)

where $\{{\alpha }_n\}$ is a real sequence in $(0,1)$. He showed that the new process converges faster than all of Picard, Mann and Ishikawa iterative processes in the sense of Berinde [7] for contractions. He also proved strong convergence and weak convergence theorems with the help of his process for the class of nonexpansive mappings in general Banach spaces and apply it to obtain a result in uniformly convex Banach spaces.

Inspired and motivated by the studies mentioned above, in this paper, we use an up-to-date method for the approximation of common fixed points of countable families of nonlinear operators. We introduce a modified Picard-Mann hybrid iterative algorithm with the help of our method for the class of nonexpansive mappings. We prove strong convergence and weak convergence theorems in the framework of Banach spaces. Our results extend the corresponding ones for one map in [6].

2 Preliminaries

Throughout this paper we assume that E is a real Banach space with its dual $E^{\ast }$, K is a nonempty closed convex subset of E and $J:E\to 2^{E^{\ast }}$ is the normalized duality mapping defined by

$$Jx=\{f\in E^{\ast }:〈x,f〉={\parallel x\parallel }^2={\parallel f\parallel }^2\},\mathrm{\forall }x\in E.$$

In the sequel, we use $F(T)$ to denote the set of fixed points of a mapping T.

We say that E is strictly convex if the following implication holds for $x,y\in E$:

$$\parallel x\parallel =\parallel y\parallel =1,x\ne y\Rightarrow \parallel \frac{x+y}{2}\parallel <1.$$

(2.1)

It is also said to be uniformly convex if, for any $ϵ>0$, there exists a $\delta >0$ such that

$$\parallel x\parallel =\parallel y\parallel =1,\parallel x-y\parallel \ge ϵ\Rightarrow \parallel \frac{x+y}{2}\parallel \le 1-\delta .$$

(2.2)

It is well known that if E is a uniformly convex Banach space, then E is reflexive and strictly convex. A Banach space E is said to be smooth if

$$\underset{t\to 0}{lim}\frac{\parallel x+ty\parallel -\parallel x\parallel }{t}$$

(2.3)

exists for each $x,y\in S(E):=\{x\in E:\parallel x\parallel =1\}$. In this case, the norm of E is said to be Gâteaux differentiable. The space E is said to have uniformly Gâteaux differentiable norm if for each $y\in S(E)$, the limit (2.3) is attained uniformly for $x\in S(E)$. The norm of E is said to be Fréchet differentiable if for each $x\in S(E)$, the limit (2.3) is attained uniformly for $y\in S(E)$. The norm of E is said to be uniformly Fréchet differentiable (and E is said to be uniformly smooth) if the limit (2.3) is attained uniformly for $x,y\in S(E)$.

Note The readers can find all the definitions and concepts mentioned above in [8].

A Banach space E is said to satisfy Opial’s condition if, for any sequence $\{x_n\}$ in E, $x_n\rightharpoonup x$ implies that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_n-x\parallel <\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_n-y\parallel$$

(2.4)

for all $y\in E$ with $y\ne x$, where $x_n\rightharpoonup x$ denotes that $\{x_n\}$ converges weakly to x.

A mapping T with domain $D(T)$ and range $R(T)$ in E is said to be demi-closed at p if whenever $\{x_n\}$ is a sequence in $D(T)$ such that $\{x_n\}$ converges weakly to $x^{\ast }\in D(T)$ and $\{T{x}_n\}$ converges strongly to p, then $T{x}^{\ast }=p$.

Remark 2.1 The following basic properties for a Banach space E can be found in [9].

• (i) If E is uniformly smooth, then J is uniformly continuous on each bounded subset of E.

• (ii) If E is reflexive and strictly convex, then $J^{-1}$ is norm-weak-continuous.

• (iii) If E is a smooth, strictly convex and reflexive Banach space, then the normalized duality mapping $J:E\to 2^{E^{\ast }}$ is single valued, one-to-one and onto.

• (iv) A Banach space E is uniformly smooth if and only if $E^{\ast }$ is uniformly convex.

• (v) Each uniformly convex Banach space E has the Kadec-Klee property, i.e., for any sequence $\{x_n\}\subset E$, if $x_n\rightharpoonup x\in E$ and $\parallel x_n\parallel \to \parallel x\parallel$, then $x_n\to x$ as $n\to \mathrm{\infty }$.

We need the following lemmas for our main results.

Lemma 2.2 [10]

Let E be a real uniformly convex Banach space and let a, b be two constant with $0<a<b<1$. Suppose that $\{t_n\}\subset [a,b]$ is a real sequence and $\{x_n\}$, $\{y_n\}$ are two sequences in E. Then the conditions

$$\underset{n\to \mathrm{\infty }}{lim}\parallel t_n{x}_n+(1-t_n)y_n\parallel =d,\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_n\parallel \le d,\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel y_n\parallel \le d$$

(2.5)

imply that $li{m}_{n\to \mathrm{\infty }}\parallel x_n-y_n\parallel =0$, where $d\ge 0$ is a constant.

Lemma 2.3 [11]

Let E be a real uniformly convex Banach space, let K be a nonempty closed convex subset of E, and let $T:K\to K$ be a nonexpansive mapping. Then $I-T$ is demi-closed at zero.

Lemma 2.4 [12]

The unique solutions to the positive integer equation

$$n=i+\frac{(m-1)m}{2},m\ge i,n=1,2,3,\dots$$

(2.6)

are

$$i=n-\frac{(m-1)m}{2},m=-[\frac{1}{2}-\sqrt{2n+\frac{1}{4}}],n=1,2,3,\dots ,$$

(2.7)

where $[x]$ denotes the maximal integer that is not larger than x.

3 Main results

Lemma 3.1 Let E be a real uniformly convex Banach space and K a nonempty closed convex subset of E. Let ${\{T_i\}}_{i=1}^{\mathrm{\infty }}$ be a sequence of nonexpansive mappings from K to itself. For an arbitrary initial point $x_1\in K$, the modified Picard-Mann hybrid iterative scheme $\{x_n\}$ is defined as follows:

$$\{\begin{array}{l}{x}_{n+1}=T_{i_n}{y}_n,\\ y_n=(1-{\alpha }_n)x_n+{\alpha }_n{T}_{i_n}{x}_n,n\in \mathbb{N},\end{array}$$

(3.1)

where $\{{\alpha }_n\}$ is a sequence in $[ϵ,1-ϵ]$ for some $ϵ\in (0,1)$ and $i_n$ is the solution to the positive integer equation: $n=i+\frac{(m-1)m}{2}$ ($m\ge i$, $n=1,2,\dots$), that is, for each $n\ge 1$, there exists a unique $i_n$ such that

$$\begin{array}{c}{i}_1=1,i_2=1,i_3=2,i_4=1,i_5=2,i_6=3,i_7=1,\hfill \\ i_8=2,i_9=3,i_{10}=4,i_{11}=1,\dots .\hfill \end{array}$$

If $F:=\{x\in K:T_{i}x=x,\mathrm{\forall }i\ge 1\}\ne \mathrm{\varnothing }$, then

• (1) ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q\parallel$ exists, $\mathrm{\forall }q\in F$;

• (2) ${lim}_{n\to \mathrm{\infty }}d(x_n,F)$ exists, where $d(x_n,F)={inf}_{q\in F}\parallel x_n-q\parallel$;

• (3) ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T_i{x}_n\parallel =0$, $\mathrm{\forall }i\ge 1$.

Proof (1) For any $q\in F$, by (3.1), we have

$$\begin{array}{rcl}\parallel y_n-q\parallel & =& \parallel (1-{\alpha }_n)(x_n-q)+{\alpha }_n(T_{i_n}{x}_n-q)\parallel \\ \le & (1-{\alpha }_n)\parallel x_n-q\parallel +{\alpha }_n\parallel T_{i_n}{x}_n-T_{i_n}q\parallel \\ \le & \parallel x_n-q\parallel ,\end{array}$$

(3.2)

and hence

$$\parallel x_{n+1}-q\parallel =\parallel T_{i_n}{y}_n-q\parallel \le \parallel y_n-q\parallel \le \parallel x_n-q\parallel .$$

(3.3)

This shows that $\{\parallel x_n-q\parallel \}$ is decreasing and hence ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q\parallel$ exists.

(2) This conclusion can easily be shown by taking the infimum in (3.3) for all $q\in F$.

(3) Assume, by the conclusion of (1), ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q\parallel =d$. We then claim that ${lim}_{n\to \mathrm{\infty }}\parallel y_n-q\parallel =d$, that is,

$$\underset{n\to \mathrm{\infty }}{lim}\parallel (1-{\alpha }_n)(x_n-q)+{\alpha }_n(T_{i_n}{x}_n-q)\parallel =d.$$

(3.4)

In fact, noting that $\parallel x_{n+1}-q\parallel <\parallel y_n-q\parallel$, we have

$$d=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_{n+1}-q\parallel \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel y_n-q\parallel ;$$

on the other hand, it follows from (3.2) that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel y_n-q\parallel \le d,$$

which implies that ${lim}_{n\to \mathrm{\infty }}\parallel y_n-q\parallel =d$.

Next, $\parallel T_{i_n}{x}_n-q\parallel \le \parallel x_n-q\parallel$ implies that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel T_{i_n}{x}_n-q\parallel \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_n-q\parallel =d,$$

(3.5)

and hence, it follows from (3.4), (3.5), and Lemma 2.2 that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T_{i_n}{x}_n\parallel =0.$$

(3.6)

On the other hand, since

$$\begin{array}{rcl}\parallel x_{n+1}-x_n\parallel & \le & \parallel x_{n+1}-T_{i_n}{x}_n\parallel +\parallel T_{i_n}{x}_n-x_n\parallel \\ =& \parallel T_{i_n}{y}_n-T_{i_n}{x}_n\parallel +\parallel T_{i_n}{x}_n-x_n\parallel \\ \le & \parallel y_n-x_n\parallel +\parallel T_{i_n}{x}_n-x_n\parallel \\ \le & {\alpha }_n\parallel T_{i_n}{x}_n-x_n\parallel +\parallel T_{i_n}{x}_n-x_n\parallel ,\end{array}$$

we have, from (3.6),

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-x_n\parallel =0.$$

(3.7)

By induction, for any nonnegative integer p, we also have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+p}-x_n\parallel =0.$$

(3.8)

For each $p\ge 0$, since

$$\begin{array}{rcl}\parallel x_n-T_{i_{n+p}}{x}_n\parallel & \le & \parallel x_n-x_{n+p}\parallel +\parallel x_{n+p}-T_{i_{n+p}}{x}_n\parallel \\ \le & \parallel x_n-x_{n+p}\parallel +\parallel x_{n+p}-T_{i_{n+p}}{x}_{n+p}\parallel \\ +\parallel T_{i_{n+p}}{x}_{n+p}-T_{i_{n+p}}{x}_n\parallel \\ \le & 2\parallel x_n-x_{n+p}\parallel +\parallel x_{n+p}-T_{i_{n+p}}{x}_{n+p}\parallel ,\end{array}$$

(3.9)

it follows from (3.6) and (3.8) that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T_{i_{n+p}}{x}_n\parallel =0.$$

(3.10)

Now, for each $i\ge 1$, we claim that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T_i{x}_n\parallel =0.$$

(3.11)

As a matter of fact, setting $n=k_m+i$, where $k_m=\frac{(m-1)m}{2}$, $m\ge i$, we obtain

$$\begin{array}{rcl}\parallel x_n-T_i{x}_n\parallel & \le & \parallel x_n-x_{k_m}\parallel +\parallel x_{k_m}-T_i{x}_n\parallel \\ \le & \parallel x_n-x_{k_m}\parallel +\parallel x_{k_m}-T_{i_{k_m+i}}{x}_{k_m}\parallel \\ +\parallel T_{i_{k_m+i}}{x}_{k_m}-T_i{x}_n\parallel \\ =& \parallel x_n-x_{k_m}\parallel +\parallel x_{k_m}-T_{i_{k_m+i}}{x}_{k_m}\parallel \\ +\parallel T_i{x}_{k_m}-T_i{x}_n\parallel \\ \le & 2\parallel x_n-x_{k_m}\parallel +\parallel x_{k_m}-T_{i_{k_m+i}}{x}_{k_m}\parallel \\ =& 2\parallel x_n-x_{n-i}\parallel +\parallel x_{k_m}-T_{i_{k_m+i}}{x}_{k_m}\parallel .\end{array}$$

(3.12)

Note that $k_m\to \mathrm{\infty }$ as $n\to \mathrm{\infty }$. It then follows from (3.8) and (3.10) that (3.11) holds obviously. This completes the proof. □

Remark 3.2 The key point of the proof of Lemma 3.1 lies in the use of a special way of choosing the indices of involved mappings, which makes the generalization of finite families of nonlinear mappings to infinite ones possible. Moreover, with the help of our method, some known results on the common fixed points of countable families of nonexpansive mappings have been improved. We now give an example to show why our work, compared with that of others, is an improvement.

In 2011, for the approximation of common fixed points of a countable family of nonexpansive mappings $\{T_n\}$, Zhang et al. [13] introduced in his iterative algorithm a mapping T defined by a convex linear combination of $\{T_n\}$, i.e., $T={\sum }_{n=1}^{\mathrm{\infty }}{\lambda }_n{T}_n$, ${\lambda }_n\ge 0$ ($n=1,2,\dots$) with ${\sum }_{n=1}^{\mathrm{\infty }}{\lambda }_n=1$. However, it is easy to see that the accurate computation of $T{x}_n$ at each step of the iteration process is not easily attainable, which will leads to gradually increasing errors. By using a special way of choosing the indices of involved mappings, Deng [14] recently improved the corresponding results announced by Zhang et al. [13]. Since the strong convergence theorems for solving some variational inequality problems and hierarchical fixed point problems are obtained without the aid of the convex linear combination of a countable family of nonexpansive mappings, our results are more applicable than those of other authors with related research interest.

Theorem 3.3 Let E be a real uniformly convex Banach space and K a nonempty closed convex subset of E. Let ${\{T_i\}}_{i=1}^{\mathrm{\infty }}$ be a sequence of nonexpansive mappings from K to itself. Suppose that $\{x_n\}$ is a sequence defined by (3.1). If $F:=\{x\in K:T_{i}x=x,\mathrm{\forall }i\ge 1\}\ne \mathrm{\varnothing }$ and there exist $T_{i_0}\in {\{T_i\}}_{i=1}^{\mathrm{\infty }}$ and a nondecreasing function $f:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ with $f(0)=0$ and $f(r)>0$ for all $r\in (0,\mathrm{\infty })$ such that $f(d(x_n,F))\le \parallel x_n-T_{i_0}{x}_n\parallel$ for all $n\ge 1$, then $\{x_n\}$ converges strongly to some common fixed point of ${\{T_i\}}_{i=1}^{\mathrm{\infty }}$.

Proof Since

$$f(d(x_n,F))\le \parallel x_n-T_{i_0}{x}_n\parallel ,$$

by taking lim sup as $n\to \mathrm{\infty }$ on both sides in the inequality above, we obtain

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

which implies ${lim}_{n\to \mathrm{\infty }}d(x_n,F)=0$ by the definition of the function f.

Now we show that $\{x_n\}$ is a Cauchy sequence. Since ${lim}_{n\to \mathrm{\infty }}d(x_n,F)=0$, then for any $ϵ>0$, there exists a positive integer N such that $d(x_n,F)<\frac{ϵ}{2}$ for all $n\ge N$. On the other hand, there exists a $p\in F$ such that $\parallel x_N-p\parallel =d(x_N,F)<\frac{ϵ}{2}$, because $d(x_N,F)={inf}_{q\in F}\parallel x_N-q\parallel$ and F is closed.

Thus, for any $n,m\ge N$, it follows from (3.3) that

$$\parallel x_n-x_m\parallel \le \parallel x_n-p\parallel +\parallel x_m-p\parallel \le 2\parallel x_N-p\parallel <ϵ.$$

This implies that $\{x_n\}$ is a Cauchy sequence, and hence there exists an $x\in K$ such that $x_n\to x$ as $n\to \mathrm{\infty }$. Then ${lim}_{n\to \mathrm{\infty }}d(x_n,F)=0$ yields $d(x,F)=0$. Further, it follows from the closedness of F that $x\in F$. This completes the proof. □

Theorem 3.4 Let E be a real uniformly convex Banach space satisfying Opial’s condition and K a nonempty closed convex subset of E. Let ${\{T_i\}}_{i=1}^{\mathrm{\infty }}$ be a sequence of nonexpansive mappings from K to itself. Suppose that $\{x_n\}$ is a sequence defined by (3.1). If $F:=\{x\in K:T_{i}x=x,\mathrm{\forall }i\ge 1\}\ne \mathrm{\varnothing }$, then $\{x_n\}$ converges weakly to some common fixed point of ${\{T_i\}}_{i=1}^{\mathrm{\infty }}$.

Proof For any $q\in F$, by Lemma 3.1, we know that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q\parallel$ exists. We now prove that $\{x_n\}$ has a unique weakly subsequential limit in F. First of all, Lemmas 2.3 and 3.1 guarantee that each weakly subsequential limit of $\{x_n\}$ is a common fixed point of ${\{T_i\}}_{i=1}^{\mathrm{\infty }}$. Secondly, Opial’s condition guarantees that the weakly subsequential limit of $x_n$ is unique. Consequently, $\{x_n\}$ converges weakly to a common fixed point of ${\{T_i\}}_{i=1}^{\mathrm{\infty }}$. This completes the proof. □

Remark 3.5 The results presented in this paper extend those of Khan [6], whose research areas are limited to the situation of a single nonexpansive mapping.