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# Generalized viscosity approximation methods for nonexpansive mappings

Fixed Point Theory and Applications20142014:68

https://doi.org/10.1186/1687-1812-2014-68

• Received: 30 October 2013
• Accepted: 6 March 2014
• Published:

## Abstract

We combine a sequence of contractive mappings $\left\{{f}_{n}\right\}$ and propose a generalized viscosity approximation method. One side, we consider a nonexpansive mapping S with the nonempty fixed point set defined on a nonempty closed convex subset C of a real Hilbert space H and design a new iterative method to approximate some fixed point of S, which is also a unique solution of the variational inequality. On the other hand, using similar ideas, we consider N nonexpansive mappings ${\left\{{S}_{i}\right\}}_{i=1}^{N}$ with the nonempty common fixed point set defined on a nonempty closed convex subset C. Under reasonable conditions, strong convergence theorems are proven. The results presented in this paper improve and extend the corresponding results reported by some authors recently.

MSC:47H09, 47H10, 47J20, 47J25.

## Keywords

• nonexpansive mapping
• contractive mapping
• variational inequality
• fixed point
• viscosity approximation method

## 1 Introduction

Let H be a real Hilbert space with an inner product $〈\phantom{\rule{0.2em}{0ex}},\phantom{\rule{0.2em}{0ex}}〉$ and norm $\parallel \cdot \parallel$, and C be a nonempty closed convex subset of H.

Let $S:C\to C$ be a nonlinear mapping, we use $Fix\left(S\right)$ to denote the set of fixed points of S (i.e., $Fix\left(S\right)=\left\{x\in C:Sx=x\right\}$). A mapping is called nonexpansive if the following inequality holds:
$\parallel Sx-Sy\parallel \le \parallel x-y\parallel$

for all $x,y\in C$.

In 1967, Halpern  used contractions to approximate a nonexpansive mapping and considered the following explicit iterative process:
${x}_{0}\in C,\phantom{\rule{1em}{0ex}}{x}_{n+1}={\alpha }_{n}u+\left(1-{\alpha }_{n}\right)S{x}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 0,$
where u is a given point and $S:C\to C$ is nonexpansive. He proved the strong convergence of $\left\{{x}_{n}\right\}$ to a fixed point of S provided that ${\alpha }_{n}={n}^{-\theta }$ with $\theta \in \left(0,1\right)$. In 2000, Moudafi  introduced the viscosity approximation method for nonexpansive mappings. Until now, in many references, viscosity approximation methods still are used and studied, which formally generates the sequence $\left\{{x}_{n}\right\}$ by the recursive formula:
${x}_{n+1}={\alpha }_{n}f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right)S{x}_{n},$

where f is a contraction and ${\alpha }_{n}\subset \left(0,1\right)$ is a slowly vanishing sequence. See, for instance, . In fact, Yamada’s hybrid steepest descent algorithm is also a kind of viscosity approximation method (see ).

The variational inequality problem is to find a point ${x}^{\ast }\in C$ such that
$〈F{x}^{\ast },x-{x}^{\ast }〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in C.$
(1.1)

In recent years, the theory of variational inequality has been extended to the study of a large variety of problems arising in structural analysis, economics, engineering sciences, and so on. See  and the references cited therein.

Recently, Zhou and Wang  proposed a simpler explicit iterative algorithm for finding a solution of variational inequality over the set of common fixed points of a finite family nonexpansive mappings. They introduced an explicit scheme as follows.

Theorem 1.1 Let H be a real Hilbert space and $F:H\to H$ be an L-Lipschitz continuous and η-strongly monotone mapping. Let ${\left\{{S}_{i}\right\}}_{i=1}^{N}$ be N nonexpansive self-mappings of H such that $C={\bigcap }_{i=1}^{N}Fix\left({S}_{i}\right)\ne \mathrm{\varnothing }$. For any point ${x}_{0}\in H$, define a sequence $\left\{{x}_{n}\right\}$ in the following manner:
${x}_{n+1}=\left(I-{\lambda }_{n}\mu F\right){S}_{N}^{n}{S}_{N-1}^{n}\cdots {S}_{1}^{n}{x}_{n},\phantom{\rule{1em}{0ex}}n\ge 0,$
(1.2)

where $\mu \in \left(0,2\eta /{L}^{2}\right)$ and ${S}_{i}^{n}:=\left(1-{\beta }_{n}^{i}\right)I+{\beta }_{n}^{i}{S}_{i}$ for $i=1,2,\dots ,N$. When the parameters satisfy appropriate conditions, the sequence converges strongly to the unique solution of the variational inequality (1.1).

In this paper, motivated by the above works, we introduce a more generalized iterative method like viscosity approximation. In Section 3, we combine a sequence of contractive mappings and obtain strong convergence theorem for approximating fixed point of a nonexpansive mapping. In Section 4, we propose a new iterative algorithm for finding some common fixed point of a finite family nonexpansive mappings, which is also a unique solution for the variational inequality over the set of fixed point of these mappings on Hilbert spaces.

## 2 Preliminaries

In order to prove our results, we collect some facts and tools in a real Hilbert space H, which are listed as below.

Lemma 2.1 Let H be a real Hilbert space. We have the following inequalities:
1. (i)

${\parallel x+y\parallel }^{2}\le {\parallel x\parallel }^{2}+2〈x+y,y〉$, $\mathrm{\forall }x,y\in H$.

2. (ii)

${\parallel tx+\left(1-t\right)y\parallel }^{2}\le t{\parallel x\parallel }^{2}+\left(1-t\right){\parallel y\parallel }^{2}$, $\mathrm{\forall }t\in \left[0,1\right]$, $\mathrm{\forall }x,y\in H$.

Lemma 2.2 

Let ${\left\{{S}_{i}\right\}}_{i=1}^{2}$ be ${\gamma }_{i}$-averaged on C such that $Fix\left({S}_{1}\right)\cap Fix\left({S}_{2}\right)\ne \mathrm{\varnothing }$. Then the following conclusions hold:
1. (i)

both ${S}_{1}{S}_{2}$ and ${S}_{2}{S}_{1}$ are γ-averaged, where $\gamma ={\gamma }_{1}+{\gamma }_{2}-{\gamma }_{1}{\gamma }_{2}$;

2. (ii)

$Fix\left({S}_{1}\right)\cap Fix\left({S}_{2}\right)=Fix\left({S}_{1}{S}_{2}\right)=Fix\left({S}_{2}{S}_{1}\right)$.

Recall that given a nonempty closed convex subset C of a real Hilbert space H, for any $x\in H$, there exists a unique nearest point in C, denoted by ${P}_{C}x$, such that
$\parallel x-{P}_{C}x\parallel \le \parallel x-y\parallel$

for all $y\in C$. Such a ${P}_{C}$ is called the metric (or the nearest point) projection of H onto C.

Lemma 2.3 

Let C be a nonempty closed convex subset of a real Hilbert space H. Given $x\in H$ and $z\in C$, then $y={P}_{C}x$ if and only if we have the relation

Lemma 2.4 

Let H be a Hilbert space and C be a nonempty closed convex subset of H, and $T:C\to C$ a nonexpansive mapping with $Fix\left(T\right)\ne \mathrm{\varnothing }$. If $\left\{{x}_{n}\right\}$ is a sequence in C weakly converging to x and if $\left\{\left(I-T\right){x}_{n}\right\}$ converges strongly to y, then $\left(I-T\right)x=y$.

Lemma 2.5 

Assume that $\left\{{a}_{n}\right\}$ is a sequence of nonnegative real numbers such that
${a}_{n+1}\le \left(1-{\gamma }_{n}\right){a}_{n}+{\delta }_{n},$
where $\left\{{\gamma }_{n}\right\}$ is a sequence in $\left(0,1\right)$ and $\left\{{\delta }_{n}\right\}$ is a sequence such that
$\begin{array}{r}\phantom{i}\left(\mathrm{i}\right)\phantom{\rule{1em}{0ex}}\sum _{n=1}^{\mathrm{\infty }}{\gamma }_{n}=\mathrm{\infty };\\ \left(\mathrm{ii}\right)\phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty }}{lim sup}\frac{{\delta }_{n}}{{\gamma }_{n}}\le 0\phantom{\rule{1em}{0ex}}\mathit{\text{or}}\phantom{\rule{1em}{0ex}}\sum _{n=1}^{\mathrm{\infty }}|{\delta }_{n}|<\mathrm{\infty }.\end{array}$

Then, ${lim}_{n\to \mathrm{\infty }}{a}_{n}=0$.

Lemma 2.6 

Let $\left\{{x}_{n}\right\}$ and $\left\{{z}_{n}\right\}$ be bounded sequences in a Banach space and $\left\{{\beta }_{n}\right\}$ be a sequence of real numbers such that $0<{lim inf}_{n\to \mathrm{\infty }}{\beta }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\beta }_{n}<1$ for all $n=0,1,2,\dots$ . Suppose that ${x}_{n+1}=\left(1-{\beta }_{n}\right){z}_{n}+{\beta }_{n}{x}_{n}$ for all $n=0,1,2,\dots$ and ${lim sup}_{n\to \mathrm{\infty }}\left(\parallel {z}_{n+1}-{z}_{n}\parallel -\parallel {x}_{n+1}-{x}_{n}\parallel \right)\le 0$. Then ${lim}_{n\to \mathrm{\infty }}\parallel {z}_{n}-{x}_{n}\parallel =0$.

## 3 Generalized viscosity approximation method combining with a nonexpansive mapping

In this section, we combine a sequence of contractive mappings and apply a more generalized iterative method like viscosity approximation to approximate some fixed point of a nonexpansive mapping defined on a closed convex subset C of a Hilbert space H, which is also the solution of the variational inequality
$〈f\left({x}^{\ast }\right)-{x}^{\ast },p-{x}^{\ast }〉\le 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }p\in Fix\left(S\right).$
(3.1)

Suppose the contractive mapping sequence $\left\{{f}_{n}\left(x\right)\right\}$ is uniformly convergent for any $x\in D$, where D is any bounded subset of C. The uniform convergence of $\left\{{f}_{n}\left(x\right)\right\}$ on D is denoted by ${f}_{n}\left(x\right)⇉f\left(x\right)$ ($n\to \mathrm{\infty }$), $x\in D$.

Theorem 3.1 Let C be a nonempty closed convex subset of a real Hilbert space H and let $\left\{{f}_{n}\right\}$ be a sequence of ${\rho }_{n}$-contractive self-maps of C with $0\le {\rho }_{l}={lim inf}_{n\to \mathrm{\infty }}{\rho }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\rho }_{n}={\rho }_{u}<1$. Let $S:C\to C$ be a nonexpansive mapping. Assume the set $Fix\left(S\right)\ne \mathrm{\varnothing }$ and $\left\{{f}_{n}\left(x\right)\right\}$ is uniformly convergent for any $x\in D$, where D is any bounded subset of C. Given ${x}_{1}\in C$, let {${x}_{n}$} be generated by the following algorithm:
${x}_{n+1}={\alpha }_{n}{f}_{n}\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right)S{x}_{n}.$
(3.2)
If the sequence $\left\{{\alpha }_{n}\right\}\subset \left(0,1\right)$ satisfies the following conditions:
1. (i)

${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$;

2. (ii)

${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$;

3. (iii)

${\sum }_{n=1}^{\mathrm{\infty }}|{\alpha }_{n+1}-{\alpha }_{n}|<\mathrm{\infty }$,

then the sequence $\left\{{x}_{n}\right\}$ converges strongly to a point ${x}^{\ast }\in Fix\left(S\right)$, which is also the unique solution of the variational inequality (3.1).

Proof The proof is divided into several steps.

Step 1. Show first that $\left\{{x}_{n}\right\}$ is bounded.

For any $q\in Fix\left(S\right)$, we have
$\begin{array}{rl}\parallel {x}_{n+1}-q\parallel & =\parallel {\alpha }_{n}{f}_{n}\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right)S{x}_{n}-q\parallel \\ \le {\alpha }_{n}\parallel {f}_{n}\left({x}_{n}\right)-q\parallel +\left(1-{\alpha }_{n}\right)\parallel S{x}_{n}-Sq\parallel \\ \le {\alpha }_{n}{\rho }_{n}\parallel {x}_{n}-q\parallel +\left(1-{\alpha }_{n}\right)\parallel {x}_{n}-q\parallel +{\alpha }_{n}\parallel {f}_{n}\left(q\right)-q\parallel \\ \le \left(1-{\alpha }_{n}\left(1-{\rho }_{n}\right)\right)\parallel {x}_{n}-q\parallel +{\alpha }_{n}\left(1-{\rho }_{n}\right)\frac{\parallel {f}_{n}\left(q\right)-q\parallel }{1-{\rho }_{n}}\\ \le max\left\{\parallel {x}_{n}-q\parallel ,\frac{\parallel {f}_{n}\left(q\right)-q\parallel }{1-{\rho }_{n}}\right\}.\end{array}$

From the uniform convergence of $\left\{{f}_{n}\right\}$ on D, it is easy to get the boundedness of $\left\{{f}_{n}\left(q\right)\right\}$. Thus there exists a positive constant ${M}_{1}$, such that $\parallel {f}_{n}\left(q\right)-q\parallel \le {M}_{1}$. By induction, we obtain $\parallel {x}_{n}-p\parallel \le max\left\{\parallel {x}_{1}-p\parallel ,\frac{{M}_{1}}{1-{\rho }_{u}}\right\}$. Hence, $\left\{{x}_{n}\right\}$ is bounded, so are $\left\{S{x}_{n}\right\}$ and $\left\{{f}_{n}\left({x}_{n}\right)\right\}$.

Step 2. Show that
(3.3)
Indeed, observe that
$\begin{array}{rl}\parallel {x}_{n+1}-{x}_{n}\parallel =& \parallel {\alpha }_{n}{f}_{n}\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right)S{x}_{n}-{\alpha }_{n-1}{f}_{n-1}\left({x}_{n-1}\right)-\left(1-{\alpha }_{n-1}\right)S{x}_{n-1}\parallel \\ =& \parallel {\alpha }_{n}\left({f}_{n}\left({x}_{n}\right)-{f}_{n}\left({x}_{n-1}\right)\right)+{\alpha }_{n}\left({f}_{n}\left({x}_{n-1}\right)-{f}_{n-1}\left({x}_{n-1}\right)\right)\\ +\left({\alpha }_{n}-{\alpha }_{n-1}\right)\left({f}_{n-1}\left({x}_{n-1}\right)-S{x}_{n-1}\right)+\left(1-{\alpha }_{n}\right)\left(S{x}_{n}-S{x}_{n-1}\right)\parallel \\ \le & {\alpha }_{n}{\rho }_{n}\parallel {x}_{n}-{x}_{n-1}\parallel +{\alpha }_{n}\parallel {f}_{n}\left({x}_{n-1}\right)-{f}_{n-1}\left({x}_{n-1}\right)\parallel \\ +|{\alpha }_{n}-{\alpha }_{n-1}|\left(\parallel S{x}_{n}\parallel +\parallel {f}_{n-1}\left({x}_{n-1}\right)\parallel \right)+\left(1-{\alpha }_{n}\right)\parallel {x}_{n}-{x}_{n-1}\parallel \\ =& \left(1-{\alpha }_{n}\left(1-{\rho }_{n}\right)\right)\parallel {x}_{n}-{x}_{n-1}\parallel +{\alpha }_{n}\parallel {f}_{n}\left({x}_{n-1}\right)-{f}_{n-1}\left({x}_{n-1}\right)\parallel \\ +|{\alpha }_{n}-{\alpha }_{n-1}|\left(\parallel S{x}_{n}\parallel +\parallel {f}_{n-1}\left({x}_{n-1}\right)\parallel \right).\end{array}$
By the conditions (i)-(iii) and the uniform convergence of ${f}_{n}\left(x\right)$, we have
$\frac{{\alpha }_{n}\parallel {f}_{n}\left({x}_{n-1}\right)-{f}_{n-1}{x}_{n-1}\parallel +|{\alpha }_{n}-{\alpha }_{n-1}|\left(\parallel S{x}_{n}\parallel +\parallel {f}_{n-1}{x}_{n-1}\parallel \right)}{{\alpha }_{n}\left(1-{\rho }_{n}\right)}\to 0$

as $n\to \mathrm{\infty }$. By Lemma 2.5, (3.3) holds.

Step 3. Show that
$\parallel S{x}_{n}-{x}_{n}\parallel \to 0.$
(3.4)
Since
$\parallel S{x}_{n}-{x}_{n}\parallel \le \parallel {x}_{n+1}-{x}_{n}\parallel +\parallel {x}_{n+1}-S{x}_{n}\parallel .$

By the condition (i), we have $\parallel {x}_{n+1}-S{x}_{n}\parallel ={\alpha }_{n}\parallel {f}_{n}\left({x}_{n}\right)-S{x}_{n}\parallel \to 0$. Combining with (3.3), it is easy to get (3.4).

Step 4.
$\underset{n\to \mathrm{\infty }}{lim sup}〈f\left({x}^{\ast }\right)-{x}^{\ast },{x}_{n}-{x}^{\ast }〉\le 0,$
(3.5)

where ${x}^{\ast }={P}_{Fix\left(S\right)}f\left({x}^{\ast }\right)$ is a unique solution of the variational inequality (3.1).

Since ${f}_{n}\left(x\right)$ is uniformly convergent on D, we have ${lim}_{n\to \mathrm{\infty }}\left({f}_{n}\left({x}^{\ast }\right)-{x}^{\ast }\right)=f\left({x}^{\ast }\right)-{x}^{\ast }$.

Indeed, take a subsequence $\left\{{x}_{{n}_{j}}\right\}$ of $\left\{{x}_{n}\right\}$ such that
$\underset{n\to \mathrm{\infty }}{lim sup}〈f\left({x}^{\ast }\right)-{x}^{\ast },{x}_{n}-{x}^{\ast }〉=\underset{j\to \mathrm{\infty }}{lim}〈f\left({x}^{\ast }\right)-{x}^{\ast },{x}_{{n}_{j}}-{x}^{\ast }〉.$
(3.6)
Since $\left\{{x}_{{n}_{j}}\right\}$ is bounded, there exists a subsequence $\left\{{x}_{{n}_{{j}_{k}}}\right\}$ of $\left\{{x}_{{n}_{j}}\right\}$ which converges weakly to $\stackrel{ˆ}{x}$. Without loss of generality, we can assume ${x}_{{n}_{j}}⇀\stackrel{ˆ}{x}$. From (3.4), we obtain $S{x}_{{n}_{j}}⇀\stackrel{ˆ}{x}$. Using Lemma 2.4, we have $\stackrel{ˆ}{x}\in Fix\left(S\right)$. Since ${x}^{\ast }={P}_{Fix\left(S\right)}f\left({x}^{\ast }\right)$, we get
$\underset{j\to \mathrm{\infty }}{lim}〈f\left({x}^{\ast }\right)-{x}^{\ast },{x}_{{n}_{j}}-{x}^{\ast }〉=〈f\left({x}^{\ast }\right)-{x}^{\ast },\stackrel{ˆ}{x}-{x}^{\ast }〉\le 0.$

Combining with (3.6), the inequality (3.5) holds.

Step 5. Show that
$\begin{array}{r}{x}_{n}\to {x}^{\ast },\\ {\parallel {x}_{n+1}-{x}^{\ast }\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}={\parallel {\alpha }_{n}{f}_{n}\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right)S{x}_{n}-{x}^{\ast }\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\left(1-{\alpha }_{n}\right)}^{2}{\parallel S{x}_{n}-{x}^{\ast }\parallel }^{2}+2{\alpha }_{n}〈{x}_{n+1}-{x}^{\ast },{f}_{n}\left({x}_{n}\right)-{x}^{\ast }〉\\ \phantom{\rule{1em}{0ex}}\le {\left(1-{\alpha }_{n}\right)}^{2}{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+2{\alpha }_{n}〈{x}_{n+1}-{x}^{\ast },{f}_{n}\left({x}_{n}\right)-{f}_{n}\left({x}^{\ast }\right)〉+2{\alpha }_{n}〈{x}_{n+1}-{x}^{\ast },{f}_{n}\left({x}^{\ast }\right)-{x}^{\ast }〉\\ \phantom{\rule{1em}{0ex}}\le {\left(1-{\alpha }_{n}\right)}^{2}{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+{\alpha }_{n}{\rho }_{n}\left({\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+{\parallel {x}_{n+1}-{x}^{\ast }\parallel }^{2}\right)\\ \phantom{\rule{2em}{0ex}}+2{\alpha }_{n}〈{x}_{n+1}-{x}^{\ast },{f}_{n}\left({x}^{\ast }\right)-{x}^{\ast }〉.\end{array}$
(3.7)
Transform the inequality into another form, we obtain
${\parallel {x}_{n+1}-{x}^{\ast }\parallel }^{2}\le \left(1-\frac{{\alpha }_{n}\left(2-{\alpha }_{n}-2{\rho }_{n}\right)}{1-{\alpha }_{n}{\rho }_{n}}\right){\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+\frac{2{\alpha }_{n}}{1-{\alpha }_{n}{\rho }_{n}}〈{x}_{n+1}-{x}^{\ast },{f}_{n}\left({x}^{\ast }\right)-{x}^{\ast }〉.$
By Schwartz’s inequality, we have
$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim sup}〈{x}_{n+1}-{x}^{\ast },{f}_{n}\left({x}^{\ast }\right)-{x}^{\ast }〉\\ \phantom{\rule{1em}{0ex}}\le \underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n+1}-{x}^{\ast }\parallel \parallel {f}_{n}\left({x}^{\ast }\right)-f\left({x}^{\ast }\right)\parallel +\underset{n\to \mathrm{\infty }}{lim sup}〈{x}_{n+1}-{x}^{\ast },f\left({x}^{\ast }\right)-{x}^{\ast }〉.\end{array}$
By the boundedness of $\left\{{x}_{n}\right\}$, ${f}_{n}\left(x\right)⇉f\left(x\right)$, (3.3) and (3.5), we have
$\underset{n\to \mathrm{\infty }}{lim sup}〈{x}_{n+1}-{x}^{\ast },{f}_{n}\left({x}^{\ast }\right)-{x}^{\ast }〉\le 0.$

It follows from Lemma 2.5 that (3.7) holds. □

Remark 3.2 In , Moudafi proposed the viscosity iterative algorithm as follows:
${x}_{n+1}={\alpha }_{n}f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right)S{x}_{n},$
(3.8)

where f is a contraction on H. It is a special case of (3.2) in this paper when ${f}_{1}={f}_{2}=\cdots ={f}_{n}=\cdots =f$, $\mathrm{\forall }n\in \mathbb{N}$ and $C=H$. Of course, Halpern’s iteration method is also a special case of (3.2) when ${f}_{1}={f}_{2}=\cdots ={f}_{n}=\cdots =u$, $\mathrm{\forall }n\in \mathbb{N}$.

Remark 3.3 In , the following iterative process was introduced:
${x}_{n+1}=S{x}_{n}-\mu {\alpha }_{n}F\left(S{x}_{n}\right).$
Rewriting the equation, we get
$\begin{array}{rl}{x}_{n+1}& ={\alpha }_{n}\left(I-\mu F\right)S{x}_{n}+\left(1-{\alpha }_{n}\right)S{x}_{n}\\ ={\alpha }_{n}f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right)S{x}_{n}.\end{array}$
(3.9)

It is easily to verify $f:=\left(I-\mu F\right)S$ is a contractive mapping on H when $0<\mu <2\eta /{L}^{2}$. That is, Yamada’s method is a kind of viscosity approximation method. Of course it is also a special case of Theorem 3.1.

## 4 Generalized viscosity approximation method combining with a finite family of nonexpansive mappings

In this section, we apply a more generalized iterative method like viscosity approximation to approximate a common element of the set of fixed points of a finite family of nonexpansive mappings on Hilbert spaces.

Let $\left\{{f}_{n}\right\}$ be a sequence of ${\rho }_{n}$-contractive self-maps of C with $0<{\rho }_{l}={lim inf}_{n\to \mathrm{\infty }}{\rho }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\rho }_{n}={\rho }_{u}<1$ and ${\left\{{S}_{i}\right\}}_{i=1}^{N}$ be N nonexpansive self-mapping of C. Assume the common fixed point set $F={\bigcap }_{i=1}^{N}Fix\left({S}_{i}\right)\ne \mathrm{\varnothing }$ and $\left\{{f}_{n}\left(q\right)\right\}$ is convergent for any $q\in F$. Put $f\left(q\right):={lim}_{n\to \mathrm{\infty }}{f}_{n}\left(q\right)$, since every ${f}_{n}$ is ${\rho }_{n}$-contractive, we have
$\parallel {f}_{n}\left(p\right)-{f}_{n}\left(q\right)\parallel \le {\rho }_{n}\parallel p-q\parallel \le {\rho }_{u}\parallel p-q\parallel$
for any $p,q\in F$. Further we obtain $\parallel f\left(p\right)-f\left(q\right)\parallel \le {\rho }_{u}\parallel p-q\parallel$. Next we prove the sequence $\left\{{x}_{n}\right\}$ converges strongly to a point ${x}^{\ast }\in F={\bigcap }_{i=1}^{N}Fix\left({S}_{i}\right)$, which also solves the variational inequality
$〈f\left({x}^{\ast }\right)-{x}^{\ast },p-{x}^{\ast }〉\le 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }p\in F.$
(4.1)

As we know, it is equivalent to the fixed point equation ${x}^{\ast }={P}_{F}f\left({x}^{\ast }\right)$.

Theorem 4.1 Let C be a nonempty closed convex subset of a real Hilbert space H and let $\left\{{f}_{n}\right\}$ be a sequence of ${\rho }_{n}$-contractive self-maps of C with $0\le {\rho }_{l}={lim inf}_{n\to \mathrm{\infty }}{\rho }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\rho }_{n}={\rho }_{u}<1$. Let, for each $1\le i\le N$ ($N\ge 1$ be an integer), ${S}_{i}:C\to C$ be a nonexpansive mapping. Assume the set $F={\bigcap }_{i=1}^{N}Fix\left({S}_{i}\right)\ne \mathrm{\varnothing }$ and $\left\{{f}_{n}\left(q\right)\right\}$ is convergent for any $q\in F$. Given ${x}_{1}\in C$, let {${x}_{n}$} be generated by the following algorithm:
$\left\{\begin{array}{l}{x}_{n+1}={\alpha }_{n}{f}_{n}\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){S}_{N}^{n}{S}_{N-1}^{n}\cdots {S}_{1}^{n}{x}_{n},\\ {S}_{i}^{n}=\left(1-{\lambda }_{i}^{n}\right)I+{\lambda }_{i}^{n}{S}_{i},\phantom{\rule{1em}{0ex}}i=1,2,\dots ,N.\end{array}$
(4.2)
If the parameters $\left\{{\alpha }_{n}\right\}$ and $\left\{{\lambda }_{i}^{n}\right\}$ satisfy the following conditions:
1. (i)

$\left\{{\alpha }_{n}\right\}\subset \left(0,1\right)$, ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$;

2. (ii)

${\lambda }_{i}^{n}\in \left({\lambda }_{l},{\lambda }_{u}\right)$ for some ${\lambda }_{l},{\lambda }_{u}\in \left(0,1\right)$ and ${lim}_{n\to \mathrm{\infty }}|{\lambda }_{i}^{n}-{\lambda }_{i}^{n+1}|=0$, $\mathrm{\forall }i=1,2,\dots ,N$,

then the sequence $\left\{{x}_{n}\right\}$ converges strongly to a point ${x}^{\ast }\in F$, which is also the unique solution of the variational inequality (4.1).

Proof We will prove the theorem in the case of $N=2$. The proof is divided into several steps.

Step 1. We show first that $\left\{{x}_{n}\right\}$ is bounded.

For any $q\in F$, we have
$\begin{array}{rl}\parallel {x}_{n+1}-q\parallel & =\parallel {\alpha }_{n}{f}_{n}\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){S}_{2}^{n}{S}_{1}^{n}{x}_{n}-q\parallel \\ \le {\alpha }_{n}\parallel {f}_{n}\left({x}_{n}\right)-q\parallel +\left(1-{\alpha }_{n}\right)\parallel {S}_{2}^{n}{S}_{1}^{n}{x}_{n}-{S}_{2}^{n}{S}_{1}^{n}q\parallel \\ \le {\alpha }_{n}{\rho }_{n}\parallel {x}_{n}-q\parallel +\left(1-{\alpha }_{n}\right)\parallel {x}_{n}-q\parallel +{\alpha }_{n}\parallel {f}_{n}\left(q\right)-q\parallel \\ \le \left(1-{\alpha }_{n}\left(1-{\rho }_{n}\right)\right)\parallel {x}_{n}-q\parallel +{\alpha }_{n}\left(1-{\rho }_{n}\right)\frac{\parallel {f}_{n}\left(q\right)-q\parallel }{1-{\rho }_{n}}\\ \le max\left\{\parallel {x}_{n}-q\parallel ,\frac{\parallel {f}_{n}\left(q\right)-q\parallel }{1-{\rho }_{n}}\right\}.\end{array}$

From the convergence of $\left\{{f}_{n}\left(q\right)\right\}$, it is easy to get the boundness of $\left\{{f}_{n}\left(q\right)\right\}$. Thus there exists a positive constant ${M}_{1}$, such that $\parallel {f}_{n}\left(q\right)-q\parallel \le {M}_{1}$. By induction, we obtain $\parallel {x}_{n}-p\parallel \le max\left\{\parallel {x}_{1}-p\parallel ,\frac{{M}_{1}}{1-{\rho }_{u}}\right\}$. Hence, $\left\{{x}_{n}\right\}$ is bounded, and so are $\left\{{S}_{1}{x}_{n}\right\}$ and $\left\{{S}_{2}^{n}{S}_{1}^{n}{x}_{n}\right\}$.

Step 2. We show that
(4.3)
Since both ${S}_{2}^{n}$ and ${S}_{1}^{n}$ are averaged nonexpansive mappings, by Lemma 2.2, ${S}_{2}^{n}{S}_{1}^{n}$ is also averaged. Rewrite ${S}_{2}^{n}{S}_{1}^{n}=\left(1-{\beta }_{n}\right)I+{\beta }_{n}{V}_{n}$, where ${\beta }_{n}={\lambda }_{1}^{n}+{\lambda }_{2}^{n}-{\lambda }_{1}^{n}{\lambda }_{2}^{n}$. Then we have
$\begin{array}{rl}{x}_{n+1}& ={\alpha }_{n}{f}_{n}\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right)\left[\left(1-{\beta }_{n}\right)I+{\beta }_{n}{V}_{n}\right]{x}_{n}\\ ={\alpha }_{n}{f}_{n}\left({x}_{n}\right)+\left(1-{\beta }_{n}\right){x}_{n}-{\alpha }_{n}\left(1-{\beta }_{n}\right){x}_{n}+\left(1-{\alpha }_{n}\right){\beta }_{n}{V}_{n}{x}_{n}\\ =\left(1-{\beta }_{n}\right){x}_{n}+{\beta }_{n}\left[{\alpha }_{n}\frac{{f}_{n}\left({x}_{n}\right)-\left(1-{\beta }_{n}\right){x}_{n}}{{\beta }_{n}}+\left(1-{\alpha }_{n}\right){V}_{n}{x}_{n}\right]\\ =\left(1-{\beta }_{n}\right){x}_{n}+{\beta }_{n}{z}_{n}.\end{array}$
Further we obtain
$\begin{array}{r}\parallel {z}_{n+1}-{z}_{n}\parallel \\ \phantom{\rule{1em}{0ex}}=\parallel \frac{{\alpha }_{n+1}}{{\beta }_{n+1}}\left[{f}_{n+1}\left({x}_{n+1}\right)-\left(1-{\beta }_{n+1}\right){x}_{n+1}\right]+\left(1-{\alpha }_{n+1}\right){V}_{n+1}{x}_{n+1}\\ \phantom{\rule{2em}{0ex}}-\frac{{\alpha }_{n}}{{\beta }_{n}}\left[{f}_{n}\left({x}_{n}\right)-\left(1-{\beta }_{n}\right){x}_{n}\right]-\left(1-{\alpha }_{n}\right){V}_{n}{x}_{n}\parallel \\ \phantom{\rule{1em}{0ex}}=\parallel {V}_{n+1}{x}_{n+1}-{V}_{n}{x}_{n}\parallel +\parallel \left[\frac{{\alpha }_{n+1}}{{\beta }_{n+1}}{f}_{n+1}\left({x}_{n+1}\right)-\frac{{\alpha }_{n}}{{\beta }_{n}}{f}_{n}\left({x}_{n}\right)\right]\\ \phantom{\rule{2em}{0ex}}-\left[\frac{{\alpha }_{n+1}\left(1-{\beta }_{n+1}\right)}{{\beta }_{n+1}}{x}_{n+1}-\frac{{\alpha }_{n}\left(1-{\beta }_{n}\right)}{{\beta }_{n}}{x}_{n}\right]-{\alpha }_{n+1}{V}_{n+1}{x}_{n+1}+{\alpha }_{n}{V}_{n}{x}_{n}\parallel \\ \phantom{\rule{1em}{0ex}}\le \parallel {x}_{n+1}-{x}_{n}\parallel +\parallel {V}_{n+1}{x}_{n}-{V}_{n}{x}_{n}\parallel +|\frac{{\alpha }_{n+1}}{{\beta }_{n+1}}{f}_{n+1}\left({x}_{n+1}\right)-\frac{{\alpha }_{n}}{{\beta }_{n}}{f}_{n}\left({x}_{n}\right)|\\ \phantom{\rule{2em}{0ex}}+\parallel \frac{{\alpha }_{n+1}\left(1-{\beta }_{n+1}\right)}{{\beta }_{n+1}}{x}_{n+1}-\frac{{\alpha }_{n}\left(1-{\beta }_{n}\right)}{{\beta }_{n}}{x}_{n}\parallel \\ \phantom{\rule{2em}{0ex}}+\parallel {\alpha }_{n+1}{V}_{n+1}{x}_{n+1}-{\alpha }_{n}{V}_{n}{x}_{n}\parallel .\end{array}$
(4.4)
Write ${\lambda }_{1}=2{\lambda }_{l}-{\lambda }_{l}^{2}$, ${\lambda }_{2}=2{\lambda }_{u}-{\lambda }_{u}^{2}$. From the condition (iii), it is easily to get $0<{\lambda }_{1}\le {\beta }_{n}\le {\lambda }_{2}$ and ${\beta }_{n+1}-{\beta }_{n}\to 0$ as $n\to \mathrm{\infty }$. We have
$\begin{array}{rl}\parallel {V}_{n+1}{x}_{n}-{V}_{n}{x}_{n}\parallel =& \parallel \frac{{S}_{2}^{n+1}{S}_{1}^{n+1}-\left(1-{\beta }_{n+1}\right)I}{{\beta }_{n+1}}{x}_{n}-\frac{{S}_{2}^{n}{S}_{1}^{n}-\left(1-{\beta }_{n}\right)I}{{\beta }_{n}}{x}_{n}\parallel \\ \le & \parallel \frac{{S}_{2}^{n+1}{S}_{1}^{n+1}}{{\beta }_{n+1}}{x}_{n}-\frac{{S}_{2}^{n}{S}_{1}^{n}}{{\beta }_{n}}{x}_{n}\parallel +|\frac{1}{{\beta }_{n}}-\frac{1}{{\beta }_{n+1}}|\parallel {x}_{n}\parallel \\ \le & \frac{1}{{\beta }_{n}}\parallel {S}_{2}^{n+1}{S}_{1}^{n+1}{x}_{n}-{S}_{2}^{n}{S}_{1}^{n}{x}_{n}\parallel +|\frac{1}{{\beta }_{n}}-\frac{1}{{\beta }_{n+1}}|\left(\parallel {S}_{2}^{n+1}{S}_{1}^{n+1}{x}_{n}\parallel +\parallel {x}_{n}\parallel \right)\\ \le & \frac{1}{{\lambda }_{1}}\left(\parallel {S}_{1}^{n+1}{x}_{n}-{S}_{1}^{n}{x}_{n}\parallel +\parallel {S}_{2}^{n+1}{S}_{1}^{n}{x}_{n}-{S}_{2}^{n}{S}_{1}^{n}{x}_{n}\parallel \right)\\ +|\frac{1}{{\beta }_{n}}-\frac{1}{{\beta }_{n+1}}|\left(\parallel {S}_{2}^{n+1}{S}_{1}^{n+1}{x}_{n}\parallel +\parallel {x}_{n}\parallel \right)\\ \le & \frac{1}{{\lambda }_{1}}\left(\parallel {S}_{1}^{n+1}{x}_{n}-{S}_{1}^{n}{x}_{n}\parallel +\parallel {S}_{2}^{n+1}{S}_{1}^{n}{x}_{n}-{S}_{2}^{n}{S}_{1}^{n}{x}_{n}\parallel \right)\\ +|\frac{1}{{\beta }_{n}}-\frac{1}{{\beta }_{n+1}}|{M}_{2},\end{array}$
(4.5)
where ${M}_{2}={sup}_{n}\left\{\parallel {S}_{2}^{n+1}{S}_{1}^{n+1}{x}_{n}\parallel +\parallel {x}_{n}\parallel \right\}$. Since $|{\lambda }_{i}^{n+1}-{\lambda }_{i}^{n}|\to 0$, $i=1,2$, we can deduce
$\parallel {S}_{1}^{n+1}{x}_{n}-{S}_{1}^{n}{x}_{n}\parallel \le |{\lambda }_{1}^{n+1}-{\lambda }_{1}^{n}|\left(\parallel {x}_{n}\parallel +\parallel {S}_{1}{x}_{n}\parallel \right)\to 0$
(4.6)
and
$\parallel {S}_{2}^{n+1}{S}_{1}^{n}{x}_{n}-{S}_{2}^{n}{S}_{1}^{n}{x}_{n}\parallel \le |{\lambda }_{2}^{n+1}-{\lambda }_{2}^{n}|\left(\parallel {S}_{1}^{n}{x}_{n}\parallel +\parallel {S}_{2}{S}_{1}^{n}{x}_{n}\parallel \right)\to 0.$
(4.7)
Substituting (4.5) into (4.4), we have
$\begin{array}{r}\parallel {z}_{n+1}-{z}_{n}\parallel -\parallel {x}_{n+1}-{x}_{n}\parallel \\ \phantom{\rule{1em}{0ex}}\le \frac{1}{{\lambda }_{1}}\left(\parallel {S}_{1}^{n+1}{x}_{n}-{S}_{1}^{n}{x}_{n}\parallel +\parallel {S}_{2}^{n+1}{S}_{1}^{n}{x}_{n}-{S}_{2}^{n}{S}_{1}^{n}{x}_{n}\parallel \right)+\frac{|{\beta }_{n}-{\beta }_{n+1}|}{{\beta }_{n}{\beta }_{n+1}}{M}_{2}\\ \phantom{\rule{2em}{0ex}}+\parallel \frac{{\alpha }_{n+1}}{{\beta }_{n+1}}{f}_{n+1}\left({x}_{n+1}\right)-\frac{{\alpha }_{n}}{{\beta }_{n}}{f}_{n}\left({x}_{n}\right)\parallel +\parallel \frac{{\alpha }_{n+1}\left(1-{\beta }_{n+1}\right)}{{\beta }_{n+1}}{x}_{n+1}-\frac{{\alpha }_{n}\left(1-{\beta }_{n}\right)}{{\beta }_{n}}{x}_{n}\parallel \\ \phantom{\rule{2em}{0ex}}+\parallel {\alpha }_{n+1}{V}_{n+1}{x}_{n+1}-{\alpha }_{n}{V}_{n}{x}_{n}\parallel .\end{array}$
Combining (4.6), (4.7), and condition (i), we get
$\underset{n\to \mathrm{\infty }}{lim sup}\left(\parallel {z}_{n+1}-{z}_{n}\parallel -\parallel {x}_{n+1}-{x}_{n}\parallel \right)\le 0.$
By Lemma 2.6, we conclude that ${lim}_{n\to \mathrm{\infty }}\parallel {z}_{n}-{x}_{n}\parallel \to 0$. Further we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n+1}-{x}_{n}\parallel =\underset{n\to \mathrm{\infty }}{lim}{\beta }_{n}\parallel {z}_{n}-{x}_{n}\parallel \to 0.$
Step 3. We show that
$\parallel {S}_{2}^{n}{S}_{1}^{n}{x}_{n}-{x}_{n}\parallel \to 0.$
(4.8)
By (4.2), we get
$\parallel {x}_{n+1}-{S}_{2}^{n}{S}_{1}^{n}{x}_{n}\parallel ={\alpha }_{n}\parallel {f}_{n}\left({x}_{n}\right)-{S}_{2}^{n}{S}_{1}^{n}{x}_{n}\parallel \to 0.$
We have
$\parallel {x}_{n}-{S}_{2}^{n}{S}_{1}^{n}{x}_{n}\parallel \le \parallel {x}_{n+1}-{S}_{2}^{n}{S}_{1}^{n}{x}_{n}\parallel +\parallel {x}_{n}-{x}_{n+1}\parallel .$

Combining with (4.3), (4.8) holds.

Since $\left\{{\lambda }_{i}^{n}\right\}\subset \left({\lambda }_{l},{\lambda }_{u}\right)$, we can assume that ${\lambda }_{i}^{{n}_{j}}\to {\lambda }_{i}^{0}$ as $n\to \mathrm{\infty }$. It is easy to get $0<{\lambda }_{i}^{0}<1$ for $i=1,2$. Write ${S}_{i}^{0}=\left(1-{\lambda }_{i}^{0}\right)I+{\lambda }_{i}^{0}{S}_{i}$, $i=1,2$. Then we have $Fix\left({S}_{i}^{0}\right)=Fix\left({S}_{i}\right)$, $i=1,2$ and
$\underset{j\to \mathrm{\infty }}{lim}\underset{x\in D}{sup}\parallel {S}_{i}^{{n}_{j}}x-{S}_{i}^{0}x\parallel =0,$
(4.9)

where D is an arbitrary bounded subset including $\left\{{x}_{{n}_{j}}\right\}$. By using (4.8) and (4.9), we obtain $\parallel {S}_{2}^{0}{S}_{1}^{0}{x}_{n}-{x}_{n}\parallel \to 0$.

Step 4. We have
$\underset{n\to \mathrm{\infty }}{lim sup}〈f\left({x}^{\ast }\right)-{x}^{\ast },{x}_{n}-{x}^{\ast }〉\le 0,$
(4.10)

where ${x}^{\ast }={P}_{F}f\left({x}^{\ast }\right)$ is a unique solution of the variational inequality (4.1).

Since ${f}_{n}\left(q\right)$ is convergent, we have ${lim}_{n\to \mathrm{\infty }}\left({f}_{n}\left({x}^{\ast }\right)-{x}^{\ast }\right)=f\left({x}^{\ast }\right)-{x}^{\ast }$.

The proof of Step 4 is similar to that of Theorem 3.1.

Step 5. We show that
${x}_{n}\to {x}^{\ast }.$
(4.11)

The proof of Step 5 is similar to that of Theorem 3.1. □

Remark 4.2 In , put ${S}_{n}={S}_{N}^{n}{S}_{N-1}^{n}\cdots {S}_{1}^{n}$, and we rewrite Zhou and Wang’s iterative algorithm as follows:
$\begin{array}{rl}{x}_{n+1}& =\left(I-{\alpha }_{n}\mu F\right){S}_{n}{x}_{n}\\ ={\alpha }_{n}\left(I-\mu F\right){S}_{n}{x}_{n}+\left(1-{\alpha }_{n}\right){S}_{n}{x}_{n}\\ ={\alpha }_{n}{f}_{n}\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){S}_{n}{x}_{n}.\end{array}$
(4.12)

It is easily to verify $\left(I-\mu F\right){S}_{n}$ is a contractive mapping on H when $0<\mu <2\eta /{L}^{2}$. Thus it is a special case of Theorem 4.1 when ${f}_{n}:=\left(I-\mu F\right){S}_{n}$, $\mathrm{\forall }n\in N$ and $C=H$.

## Declarations

### Acknowledgements

The authors would like to thank the referee for valuable suggestions to improve the manuscript and the Fundamental Research Funds for the Central Universities (GRANT: 3122013k004).

## Authors’ Affiliations

(1)
College of Science, Civil Aviation University of China, Tianjin, 300300, P.R. China

## References

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