• Research
• Open Access

# Convergence theorems for a system of equilibrium problems and fixed point problems of a strongly nonexpansive sequence

Fixed Point Theory and Applications20132013:193

https://doi.org/10.1186/1687-1812-2013-193

• Accepted: 9 July 2013
• Published:

## Abstract

The purpose of this paper is to prove a strong convergence theorem of an iterative scheme associated to a strongly nonexpansive sequence for finding a common element of the set of equilibrium problems and the set of fixed point problems of a pair of sequences of nonexpansive mappings where one of them is a strongly nonexpansive sequence. Moreover, in the last section, by using our main result, we obtain a strong convergence theorem of an iterative scheme associated to a strongly nonexpansive sequence for finding a common element of the set of a finite family of equilibrium problems and the set of fixed point problems of a pair of sequences of nonexpansive mappings where one of them is a strongly nonexpansive sequence in a Hilbert space, and we also give some examples to support our main result.

## Keywords

• nonexpansive mappings
• strongly nonexpansive sequence
• equilibrium problem
• fixed point

## 1 Introduction

Throughout this paper, we assume that H is a real Hilbert space with the inner product $〈\cdot ,\cdot 〉$ and the norm $\parallel \cdot \parallel$. A mapping T of C into itself is called nonexpansive if $\parallel Tx-Ty\parallel \le \parallel x-y\parallel$ for all $x,y\in H$. The set of fixed points of T is denoted by $F\left(T\right)$, i.e., $F\left(T\right)=\left\{x\in H:Tx=x\right\}$. It is known that $F\left(T\right)$ is closed and convex if T is nonexpansive. Let ${P}_{C}$ be a metric projection of H onto C, i.e., for $x\in H$, ${P}_{C}x$ satisfies the property
$\parallel x-{P}_{C}x\parallel =\underset{y\in C}{min}\parallel x-y\parallel .$
We use ${}^{\mathrm{‵}\mathrm{‵}}⇀^{\mathrm{\prime }\mathrm{\prime }}$ and ${}^{\mathrm{‵}\mathrm{‵}}\to ^{\mathrm{\prime }\mathrm{\prime }}$ to denote weak and strong convergence, respectively. Let $\left\{{T}_{n}\right\}$ be a sequence of mappings of C into H. The set of common fixed points of $\left\{{T}_{n}\right\}$ is denoted by $F\left(\left\{{T}_{n}\right\}\right)={\bigcap }_{n=1}^{\mathrm{\infty }}F\left({T}_{n}\right)$. Recall the main concepts as follows:
1. (1)

A sequence $\left\{{z}_{n}\right\}$ in C is said to be an approximate fixed point sequence of $\left\{{T}_{n}\right\}$ if ${z}_{n}-{T}_{n}{z}_{n}\to 0$. The set of all bounded approximate fixed point sequences of $\left\{{T}_{n}\right\}$ is denoted by $\stackrel{˜}{F}\left(\left\{{T}_{n}\right\}\right)$; see [1]. It is clear that if $\left\{{T}_{n}\right\}$ has a common fixed point, then $\stackrel{˜}{F}\left(\left\{{T}_{n}\right\}\right)$ is nonempty.

2. (2)
A sequence $\left\{{T}_{n}\right\}$ is said to be a strongly nonexpansive sequence if each ${T}_{n}$ is nonexpansive and
${x}_{n}-{y}_{n}-\left({T}_{n}{x}_{n}-{T}_{n}{y}_{n}\right)\to 0,$

whenever $\left\{{x}_{n}\right\}$ and $\left\{{y}_{n}\right\}$ are sequences in C such that $\left\{{x}_{n}-{y}_{n}\right\}$ is bounded and $\parallel {x}_{n}-{y}_{n}\parallel -\parallel {T}_{n}{x}_{n}-{T}_{n}{y}_{n}\parallel \to 0$.

3. (3)

A sequence $\left\{{T}_{n}\right\}$ having a common fixed point is said to satisfy the condition (Z) if every weak cluster point of $\left\{{x}_{n}\right\}$ is a common fixed point whenever $\left\{{x}_{n}\right\}\in \stackrel{˜}{F}\left(\left\{{T}_{n}\right\}\right)$.

4. (4)
A sequence $\left\{{T}_{n}\right\}$ of nonexpansive mappings of C into H is said to satisfy the condition (R) if
$\underset{n\to \mathrm{\infty }}{lim}\underset{y\in D}{sup}\parallel {T}_{n+1}y-{T}_{n}y\parallel =0$

for every nonempty bounded subset D of C; see [2].

Example 1.1 Let be a set of real numbers. For every $n\in \mathbb{N}$, the mapping ${T}_{n}:\mathbb{R}\to \mathbb{R}$ is defined by ${T}_{n}x=\frac{1}{n}x$ for all $x\in \mathbb{R}$.

Then $\left\{{T}_{n}\right\}$ is a nonexpansive sequence, but it is not a strongly nonexpansive sequence.

Example 1.2 For every $n\in \mathbb{N}$, the mapping ${T}_{n}:\left[0,1\right]\to \left[0,1\right]$ is defined by ${T}_{n}x=\left(1-\frac{1}{n}\right)x$ for all $x\in \left[0,1\right]$.

Then $\left\{{T}_{n}\right\}$ is a strongly nonexpansive sequence.

Solution It is easy to see that ${T}_{n}$ is a nonexpansive mapping for all $n\in \mathbb{N}$.

Let $\left\{{x}_{n}\right\}$ and $\left\{{y}_{n}\right\}$ be sequences in $\left[0,1\right]$ with $\left\{{x}_{n}-{y}_{n}\right\}$ being bounded and $|{x}_{n}-{y}_{n}|-|{T}_{n}{x}_{n}-{T}_{n}{y}_{n}|\to 0$ as $n\to \mathrm{\infty }$.

Since ${x}_{n}-{y}_{n}-\left({T}_{n}{x}_{n}-{T}_{n}{y}_{n}\right)=\frac{1}{n}\left({x}_{n}-{y}_{n}\right)$, for all $n\in \mathbb{N}$, then we have

Then $\left\{{T}_{n}\right\}$ is a strongly nonexpansive sequence.

Let $G:C×C\to \mathbb{R}$ be a bifunction. The equilibrium problem for G is to determine its equilibrium points, i.e., the set
$\mathit{EP}\left(G\right)=\left\{x\in G:G\left(x,y\right)\ge 0,\mathrm{\forall }y\in C\right\}.$

It is a unified model of several problems, namely, variational inequality problem, complementary problem, saddle point problem, optimization problem, fixed point problem, etc.; see [35]. Several iterative methods have been proposed to solve the equilibrium problem; see, for instance, [68]. In 2005, Combettes and Hirstoaga [4] introduced some iterative schemes of finding the best approximation to the initial data when $\mathit{EP}\left(G\right)$ is nonempty and proved a strong convergence theorem.

Also in [4], Combettes and Hiratoaga, following [3], defined ${S}_{r}:H\to C$ by
${S}_{r}\left(x\right)=\left\{z\in C:G\left(z,y\right)+\frac{1}{r}〈y-z,z-x〉\ge 0,\mathrm{\forall }y\phantom{\rule{0.25em}{0ex}}\in C\right\}.$

They proved that under suitable hypotheses ${S}_{r}$ is single-valued and firmly nonexpansive with $F\left({S}_{r}\right)=\mathit{EP}\left(G\right)$.

In 2007, Takahashi and Takahashi [9] proved the following theorem.

Theorem 1.3 Let C be a nonempty closed convex subset of H. Let G be a bifunction from $C×C$ to satisfying
1. (A1)

$G\left(x,x\right)=0$, $\mathrm{\forall }x\in C$;

2. (A2)

G is monotone, i.e., $G\left(x,y\right)+G\left(y,x\right)\le 0$, $\mathrm{\forall }x,y\in C$;

3. (A3)

$\mathrm{\forall }x,y,z\in C$, ${lim}_{t\to {0}^{+}}G\left(tz+\left(1-t\right)x,y\right)\le G\left(x,y\right)$;

4. (A4)

$\mathrm{\forall }x\in C$, $y↦G\left(x,y\right)$ is convex and lower semicontinuous;

and let S be a nonexpansive mapping of C into H such that $F\left(S\right)\cap \mathit{EP}\left(G\right)\ne \mathrm{\varnothing }$. Let f be a contraction of H into itself, and let $\left\{{x}_{n}\right\}$ and $\left\{{u}_{n}\right\}$ be sequences generated by ${x}_{1}\in H$ and
$\begin{array}{c}G\left({u}_{n},y\right)+\frac{1}{{r}_{n}}〈y-{u}_{n},{u}_{n}-{x}_{n}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in C,\hfill \\ {x}_{n+1}={\alpha }_{n}f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right)S{u}_{n}\hfill \end{array}$
for all $n\in \mathbb{N}$, where $\left\{{\alpha }_{n}\right\}\subset \left[0,1\right]$ and $\left\{{r}_{n}\right\}\subset \left(0,1\right)$ satisfy
1. (C1)

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

2. (C2)

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

3. (C3)

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

and ${lim inf}_{n\to \mathrm{\infty }}{r}_{n}>0$ and ${\sum }_{n=1}^{\mathrm{\infty }}|{r}_{n+1}-{r}_{n}|<\mathrm{\infty }$.

Then $\left\{{x}_{n}\right\}$ and $\left\{{u}_{n}\right\}$ converge strongly to $z\in F\left(S\right)\cap \mathit{EP}\left(G\right)$, where $z={P}_{F\left(S\right)\cap \mathit{EP}\left(G\right)}f\left(z\right)$.

Very recently, in 2011, Aoyama and Kimura [10] proved a strong convergence theorem of the iterative scheme of $\left\{{x}_{n}\right\}$ associated to a strongly nonexpansive sequence as follows.

Theorem 1.4 Let H be a Hilbert space, let C be a nonempty closed convex subset of H, and let $\left\{{S}_{n}\right\}$ and $\left\{{T}_{n}\right\}$ be sequences of nonexpansive self-mappings of C. Suppose that $F=F\left(\left\{{S}_{n}\right\}\right)\cap F\left(\left\{{T}_{n}\right\}\right)$ is nonempty, both $\left\{{S}_{n}\right\}$ and $\left\{{T}_{n}\right\}$ satisfy the conditions (R) and (Z), and $\left\{{S}_{n}\right\}$ or $\left\{{T}_{n}\right\}$ is a strongly nonexpansive sequence. Let $\left\{{\alpha }_{n}\right\}$ and $\left\{{\beta }_{n}\right\}$ be sequences in [0, 1] such that
$\underset{n\to \mathrm{\infty }}{lim}{\alpha }_{n}=0,\phantom{\rule{2em}{0ex}}\sum _{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }\phantom{\rule{1em}{0ex}}\phantom{\rule{0.25em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}0<\underset{n\to \mathrm{\infty }}{lim inf}{\beta }_{n}\le \underset{n\to \mathrm{\infty }}{lim sup}{\beta }_{n}<1.$
Let $x,u\in C$ and let $\left\{{x}_{n}\right\}$ be a sequence in C defined by ${x}_{1}=x\in C$ and
${x}_{n+1}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right){S}_{n}\left({\alpha }_{n}u+\left(1-{\alpha }_{n}\right){T}_{n}{x}_{n}\right)$

for all $n\in \mathbb{N}$. Then $\left\{{x}_{n}\right\}$ converges strongly to ${P}_{F}u$.

For ${x}_{1},u,v\in C$, let $\left\{{u}_{n}\right\}$, $\left\{{v}_{n}\right\}$, $\left\{{y}_{n}\right\}$ and $\left\{{x}_{n}\right\}$ be the sequences defined by
$\left\{\begin{array}{c}{F}_{1}\left({u}_{n},u\right)+\frac{1}{{r}_{n}}〈u-{u}_{n},{u}_{n}-{x}_{n}〉\ge 0,\hfill \\ {F}_{2}\left({v}_{n},v\right)+\frac{1}{{s}_{n}}〈v-{v}_{n},{v}_{n}-{x}_{n}〉\ge 0,\hfill \\ {y}_{n}={\delta }_{n}{u}_{n}+\left(1-{\delta }_{n}\right){v}_{n},\hfill \\ {x}_{n+1}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right){S}_{n}\left({\alpha }_{n}f\left({T}_{n}{y}_{n}\right)+\left(1-{\alpha }_{n}\right){T}_{n}{y}_{n}\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,\hfill \end{array}$
(1.1)

where $f:C\to C$ is a contractive mapping with $\alpha \in \left(0,\frac{1}{2}\right)$ and $\left\{{S}_{n}\right\}$, $\left\{{T}_{n}\right\}$ are sequences of nonexpansive mappings, one of them is a strongly nonexpansive sequence.

In this paper, inspired and motivated by [10] and [9], we prove that a strong convergence theorem of the iterative scheme $\left\{{x}_{n}\right\}$ defined by (1.1) converges strongly to $z={P}_{\mathbb{F}}f\left(z\right)$, where $\mathbb{F}=\mathit{EP}\left({F}_{1}\right)\cap \mathit{EP}\left({F}_{2}\right)\cap F\left(\left\{{S}_{n}\right\}\right)\cap F\left(\left\{{T}_{n}\right\}\right)$, under the conditions (R) and (Z) and suitable conditions of $\left\{{r}_{n}\right\}$, $\left\{{s}_{n}\right\}$, $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$ and $\left\{{\delta }_{n}\right\}$.

## 2 Preliminaries

In this section, we need the following lemmas to prove our main result in the next section.

Lemma 2.1 (See [11])

Given $x\in H$ and $y\in C$. Then ${P}_{C}x=y$ if and only if the following inequality holds:
$〈x-y,y-z〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }z\in C.$

Lemma 2.2 (See [12])

Let $\left\{{s}_{n}\right\}$ be a sequence of nonnegative real numbers satisfying
${s}_{n+1}=\left(1-{\alpha }_{n}\right){s}_{n}+{\alpha }_{n}{\beta }_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 0,$
where $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$ satisfy the conditions
1. (1)

$\left\{{\alpha }_{n}\right\}\subset \left[0,1\right]$, ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$;

2. (2)

${lim sup}_{n\to \mathrm{\infty }}{\beta }_{n}\le 0$ or ${\sum }_{n=1}^{\mathrm{\infty }}|{\alpha }_{n}{\beta }_{n}|<\mathrm{\infty }$.

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

Lemma 2.3 (See [13])

Let $\left\{{x}_{n}\right\}$ and $\left\{{z}_{n}\right\}$ be bounded sequences in a Banach space X, and let $\left\{{\beta }_{n}\right\}$ be a sequence in $\left[0,1\right]$ with $0<{lim inf}_{n\to \mathrm{\infty }}{\beta }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\beta }_{n}<1$. Suppose that
${x}_{n+1}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right){z}_{n}$
for all integers $n\ge 0$ and
$\underset{n\to \mathrm{\infty }}{lim sup}\left(\parallel {z}_{n+1}-{z}_{n}\parallel -\parallel {x}_{n+1}-{x}_{n}\parallel \right)\le 0.$

Then ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-{z}_{n}\parallel =0$.

Lemma 2.4 (See [14])

Let C be a closed convex subset of a strictly convex Banach space E. Let $\left\{{T}_{n}:n\in \mathbb{N}\right\}$ be a sequence of nonexpansive mappings on C. Suppose that ${\bigcap }_{n=1}^{\mathrm{\infty }}F\left({T}_{n}\right)$ is nonempty. Let $\left\{{\lambda }_{n}\right\}$ be a sequence of positive numbers with ${\sum }_{n=1}^{\mathrm{\infty }}{\lambda }_{n}=1$. Then a mapping S on C defined by
$S\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}{\lambda }_{n}{T}_{n}x$

for all $x\in C$ is well defined, nonexpansive and $F\left(S\right)={\bigcap }_{n=1}^{\mathrm{\infty }}F\left({T}_{n}\right)$ holds.

Lemma 2.5 (See [4])

Let C be a nonempty closed convex subset of a Hilbert space H, and let $G:C×C\to \mathbb{R}$ satisfy
1. (A1)

$G\left(x,x\right)=0$, $\mathrm{\forall }x\in C$;

2. (A2)

G is monotone, i.e., $G\left(x,y\right)+G\left(y,x\right)\le 0$, $\mathrm{\forall }x,y\in C$;

3. (A3)

$\mathrm{\forall }x,y,z\in C$, ${lim}_{t\to {0}^{+}}G\left(tz+\left(1-t\right)x,y\right)\le G\left(x,y\right)$;

4. (A4)

$\mathrm{\forall }x\in C$, $y↦G\left(x,y\right)$ is convex and lower semicontinuous.

For $x\in H$ and $r>0$, define a mapping ${S}_{r}:H\to C$ as follows:
${S}_{r}\left(x\right)=\left\{z\in C:G\left(z,y\right)+\frac{1}{r}〈y-z,z-x〉\ge 0,\mathrm{\forall }y\in C\right\}.$
Then ${S}_{r}$ is well defined and the following hold:
1. (1)

${S}_{r}$ is single-valued;

2. (2)

${S}_{r}$ is firmly nonexpansive, i.e., ${\parallel {S}_{r}\left(x\right)-{S}_{r}\left(y\right)\parallel }^{2}\le 〈{S}_{r}\left(x\right)-{S}_{r}\left(y\right),x-y〉$, $\mathrm{\forall }x,y\in H$;

3. (3)

$F\left({S}_{r}\right)=\mathit{EP}\left(G\right)$;

4. (4)

$\mathit{EP}\left(G\right)$ is closed and convex.

Lemma 2.6 (See [11]) (Demiclosedness principle)

Assume that T is a nonexpansive self-mapping of a closed convex subset C of a Hilbert space H. If T has a fixed point, then $I-T$ is demiclosed. That is, whenever $\left\{{x}_{n}\right\}$ is a sequence in C weakly converging to some $x\in C$ and the sequence $\left\{\left(I-T\right){x}_{n}\right\}$ converges strongly to some y, it follows that $\left(I-T\right)x=y$. Here, I is the identity mapping of H.

Lemma 2.7 Let H be a real Hilbert space. Then, for all $x,y\in H$,
${\parallel x+y\parallel }^{2}\le {\parallel x\parallel }^{2}+2〈y,x+y〉.$

Lemma 2.8 (See [10])

Let H be a Hilbert space, let C be a nonempty subset of H, and let $\left\{{S}_{n}\right\}$ and $\left\{{T}_{n}\right\}$ be the sequences of nonexpansive self-mappings of C. Suppose that $\left\{{S}_{n}\right\}$ and $\left\{{T}_{n}\right\}$ satisfy the condition (R) and that $\left\{{T}_{n}y:n\in \mathbb{N},y\in D\right\}$ is bounded for any bounded subset D of C. Then $\left\{{S}_{n}{T}_{n}\right\}$ satisfies the condition (R).

Lemma 2.9 (See [1])

Let H be a Hilbert space, let C be a nonempty subset of H, and let $\left\{{S}_{n}\right\}$ and $\left\{{T}_{n}\right\}$ be the sequences of nonexpansive self-mappings of C. Suppose that $\left\{{S}_{n}\right\}$ or $\left\{{T}_{n}\right\}$ is a strongly nonexpansive sequence and that $\stackrel{˜}{F}\left(\left\{{S}_{n}\right\}\right)\cap \stackrel{˜}{F}\left(\left\{{T}_{n}\right\}\right)$ is nonempty. Then $\stackrel{˜}{F}\left(\left\{{S}_{n}\right\}\right)\cap \stackrel{˜}{F}\left(\left\{{T}_{n}\right\}\right)=\stackrel{˜}{F}\left(\left\{{S}_{n}{T}_{n}\right\}\right)$.

## 3 Main result

Theorem 3.1 Let H be a Hilbert space, let C be a nonempty closed convex subset of H. Let ${F}_{1}$ and ${F}_{2}$ be two bifunctions from $C×C$ into satisfying (A1)-(A4), respectively, and let $\left\{{S}_{n}\right\}$ and $\left\{{T}_{n}\right\}$ be sequences of nonexpansive self-mappings of C with $\mathbb{F}=\mathit{EP}\left({F}_{1}\right)\cap \mathit{EP}\left({F}_{2}\right)\cap F\left(\left\{{S}_{n}\right\}\right)\cap F\left(\left\{{T}_{n}\right\}\right)\ne \mathrm{\varnothing }$. Let $\left\{{T}_{n}\right\}$ or $\left\{{S}_{n}\right\}$ be a sequence of strongly nonexpansive mappings, and let $f:C\to C$ be a contractive mapping with $\alpha \in \left(0,\frac{1}{2}\right)$. Let $\left\{{x}_{n}\right\}$, $\left\{{u}_{n}\right\}$, $\left\{{v}_{n}\right\}$ be sequences generated by ${x}_{1},u,v\in C$ and
$\left\{\begin{array}{c}{F}_{1}\left({u}_{n},u\right)+\frac{1}{{r}_{n}}〈u-{u}_{n},{u}_{n}-{x}_{n}〉\ge 0,\hfill \\ {F}_{2}\left({v}_{n},v\right)+\frac{1}{{s}_{n}}〈v-{v}_{n},{v}_{n}-{x}_{n}〉\ge 0,\hfill \\ {y}_{n}={\delta }_{n}{u}_{n}+\left(1-{\delta }_{n}\right){v}_{n},\hfill \\ {x}_{n+1}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right){S}_{n}\left({\alpha }_{n}f\left({T}_{n}{y}_{n}\right)+\left(1-{\alpha }_{n}\right){T}_{n}{y}_{n}\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,\hfill \end{array}$
(3.1)
where $\left\{{\alpha }_{n}\right\},\left\{{\beta }_{n}\right\}\in \left[0,1\right]$, $\left\{{r}_{n}\right\},\left\{{s}_{n}\right\}\in \left(a,b\right)\in \left[0,1\right]$. Assume that the following conditions hold:
1. (i)

${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$;

2. (ii)

$0<{lim inf}_{n\to \mathrm{\infty }}{\beta }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\beta }_{n}<1$;

3. (iii)

${\sum }_{n=0}^{\mathrm{\infty }}|{r}_{n+1}-{r}_{n}|$, ${\sum }_{n=0}^{\mathrm{\infty }}|{s}_{n+1}-{s}_{n}|<\mathrm{\infty }$;

4. (iv)

${lim}_{n\to \mathrm{\infty }}{\delta }_{n}=\delta \in \left(0,1\right)$;

5. (v)

$\left\{{S}_{n}\right\}$ and $\left\{{T}_{n}\right\}$ satisfy the conditions R and Z.

Then the sequences $\left\{{x}_{n}\right\}$, $\left\{{u}_{n}\right\}$, $\left\{{v}_{n}\right\}$, $\left\{{y}_{n}\right\}$ converge strongly to $z={P}_{\mathbb{F}}f\left(z\right)$.

Proof Let $v\in \mathbb{F}$. From the definition of ${x}_{n}$, we have
$\begin{array}{rcl}\parallel {x}_{n+1}-v\parallel & =& \parallel {\beta }_{n}\left({x}_{n}-v\right)+\left(1-{\beta }_{n}\right)\left({S}_{n}\left({\alpha }_{n}f\left({T}_{n}{y}_{n}\right)+\left(1-{\alpha }_{n}\right){T}_{n}{y}_{n}\right)-v\right)\parallel \\ \le & {\beta }_{n}\parallel {x}_{n}-v\parallel +\left(1-{\beta }_{n}\right)\parallel {\alpha }_{n}f\left({T}_{n}{y}_{n}\right)+\left(1-{\alpha }_{n}\right){T}_{n}{y}_{n}-v\parallel \\ \le & {\beta }_{n}\parallel {x}_{n}-v\parallel +\left(1-{\beta }_{n}\right)\left({\alpha }_{n}\parallel f\left({T}_{n}{y}_{n}\right)-v\parallel +\left(1-{\alpha }_{n}\right)\parallel {T}_{n}{y}_{n}-v\parallel \right)\\ \le & {\beta }_{n}\parallel {x}_{n}-v\parallel +\left(1-{\beta }_{n}\right)\left({\alpha }_{n}\parallel f\left({T}_{n}{y}_{n}\right)-f\left(v\right)\parallel +{\alpha }_{n}\parallel f\left(v\right)-v\parallel \\ +\left(1-{\alpha }_{n}\right)\parallel {T}_{n}{y}_{n}-v\parallel \right)\\ \le & {\beta }_{n}\parallel {x}_{n}-v\parallel +\left(1-{\beta }_{n}\right)\left({\alpha }_{n}\alpha \parallel {y}_{n}-v\parallel +{\alpha }_{n}\parallel f\left(v\right)-v\parallel \\ +\left(1-{\alpha }_{n}\right)\parallel {y}_{n}-v\parallel \right)\\ =& {\beta }_{n}\parallel {x}_{n}-v\parallel +\left(1-{\beta }_{n}\right)\left({\alpha }_{n}\parallel f\left(v\right)-v\parallel \\ +\left(1-{\alpha }_{n}\left(1-\alpha \right)\right)\parallel {y}_{n}-v\parallel \right).\end{array}$
(3.2)
From Lemma 2.5 and (3.1), we have $\mathit{EP}\left({F}_{1}\right)=F\left({S}_{{r}_{n}}\right)$, $\mathit{EP}\left({F}_{2}\right)=F\left({S}_{{s}_{n}}\right)$, ${S}_{{r}_{n}}{x}_{n}={u}_{n}$ and ${S}_{{s}_{n}}{x}_{n}={v}_{n}$. By $v\in \mathbb{F}$ and the nonexpansiveness of ${S}_{{r}_{n}}$ and ${S}_{{s}_{n}}$, we have
$\begin{array}{rcl}\parallel {y}_{n}-v\parallel & =& \parallel {\delta }_{n}\left({u}_{n}-v\right)+\left(1-{\delta }_{n}\right)\left({v}_{n}-v\right)\parallel \\ \le & {\delta }_{n}\parallel {u}_{n}-v\parallel +\left(1-{\delta }_{n}\right)\parallel {v}_{n}-v\parallel \\ =& {\delta }_{n}\parallel {S}_{{r}_{n}}{x}_{n}-v\parallel +\left(1-{\delta }_{n}\right)\parallel {S}_{{s}_{n}}{x}_{n}-v\parallel \\ \le & \parallel {x}_{n}-v\parallel .\end{array}$
(3.3)
Substituting (3.3) into (3.2), we have
$\begin{array}{rcl}\parallel {x}_{n+1}-v\parallel & \le & {\beta }_{n}\parallel {x}_{n}-v\parallel +\left(1-{\beta }_{n}\right)\left({\alpha }_{n}\parallel f\left(v\right)-v\parallel \\ +\left(1-{\alpha }_{n}\left(1-\alpha \right)\right)\parallel {y}_{n}-v\parallel \right)\\ \le & {\beta }_{n}\parallel {x}_{n}-v\parallel +\left(1-{\beta }_{n}\right)\left({\alpha }_{n}\parallel f\left(v\right)-v\parallel \\ +\left(1-{\alpha }_{n}\left(1-\alpha \right)\right)\parallel {x}_{n}-v\parallel \right)\\ =& {\beta }_{n}\parallel {x}_{n}-v\parallel +\left(1-{\beta }_{n}\right){\alpha }_{n}\parallel f\left(v\right)-v\parallel \\ +\left(1-{\beta }_{n}\right)\left(1-{\alpha }_{n}\left(1-\alpha \right)\right)\parallel {x}_{n}-v\parallel \\ =& \left(1-{\beta }_{n}\right){\alpha }_{n}\parallel f\left(v\right)-v\parallel +\left(1-{\alpha }_{n}\left(1-{\beta }_{n}\right)\left(1-\alpha \right)\right)\parallel {x}_{n}-v\parallel \\ \le & max\left\{\parallel {x}_{n}-v\parallel ,\frac{\parallel f\left(v\right)-v\parallel }{1-\alpha }\right\}.\end{array}$

By induction we can conclude that $\left\{{x}_{n}\right\}$ is bounded and so are $\left\{{u}_{n}\right\}$, $\left\{{v}_{n}\right\}$, $\left\{{y}_{n}\right\}$. Next, we show that $\stackrel{˜}{F}\left(\left\{{S}_{n}{A}_{n}\right\}\right)=\stackrel{˜}{F}\left(\left\{{S}_{n}\right\}\right)$ and $\stackrel{˜}{F}\left(\left\{{A}_{n}{T}_{n}\right\}\right)=\stackrel{˜}{F}\left(\left\{{T}_{n}\right\}\right)$, where ${A}_{n}={\alpha }_{n}f+\left(1-{\alpha }_{n}\right)I$.

Let $\left\{{z}_{n}\right\}$ be a bounded sequence in C. From the nonexpansiveness of ${S}_{n}$, we have
$\parallel {S}_{n}{A}_{n}{z}_{n}-{S}_{n}{z}_{n}\parallel \le \parallel {A}_{n}{z}_{n}-{z}_{n}\parallel ={\alpha }_{n}\parallel f\left({z}_{n}\right)-{z}_{n}\parallel .$
(3.4)
From (3.4) and ${\alpha }_{n}\to 0$ as $n\to \mathrm{\infty }$, we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {S}_{n}{A}_{n}{z}_{n}-{S}_{n}{z}_{n}\parallel =0.$
(3.5)
Let $\left\{{z}_{n}\right\}\in \stackrel{˜}{F}\left(\left\{{S}_{n}{A}_{n}\right\}\right)$, then we have
$\parallel {z}_{n}-{S}_{n}{z}_{n}\parallel \le \parallel {z}_{n}-{S}_{n}{A}_{n}{z}_{n}\parallel +\parallel {S}_{n}{A}_{n}{z}_{n}-{S}_{n}{z}_{n}\parallel .$
From (3.5), we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {z}_{n}-{S}_{n}{z}_{n}\parallel =0,$
which implies that $\left\{{z}_{n}\right\}\in \stackrel{˜}{F}\left(\left\{{S}_{n}\right\}\right)$. It follows that
$\stackrel{˜}{F}\left(\left\{{S}_{n}{A}_{n}\right\}\right)\subseteq \stackrel{˜}{F}\left(\left\{{S}_{n}\right\}\right).$
(3.6)
Let $\left\{{z}_{n}\right\}\in \stackrel{˜}{F}\left(\left\{{S}_{n}\right\}\right)$, then we have
$\parallel {z}_{n}-{S}_{n}{A}_{n}{z}_{n}\parallel \le \parallel {z}_{n}-{S}_{n}{z}_{n}\parallel +\parallel {S}_{n}{z}_{n}-{S}_{n}{A}_{n}{z}_{n}\parallel .$
From (3.5), we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {z}_{n}-{S}_{n}{A}_{n}{z}_{n}\parallel =0,$
which implies that $\left\{{z}_{n}\right\}\in \stackrel{˜}{F}\left(\left\{{S}_{n}{A}_{n}\right\}\right)$. It follows that
$\stackrel{˜}{F}\left(\left\{{S}_{n}\right\}\right)\subseteq \stackrel{˜}{F}\left(\left\{{S}_{n}{A}_{n}\right\}\right).$
(3.7)
From (3.6) and (3.7), we have
$\stackrel{˜}{F}\left(\left\{{S}_{n}\right\}\right)=\stackrel{˜}{F}\left(\left\{{S}_{n}{A}_{n}\right\}\right).$
(3.8)
Let $\left\{{z}_{n}\right\}$ be a bounded sequence in C, then we have $\left\{{T}_{n}{z}_{n}\right\}$ is bounded and so is $\left\{f\left({T}_{n}{z}_{n}\right)\right\}$. Since
$\parallel {A}_{n}{T}_{n}{z}_{n}-{T}_{n}{z}_{n}\parallel ={\alpha }_{n}\parallel f\left({T}_{n}{z}_{n}\right)-{T}_{n}{z}_{n}\parallel$
and ${\alpha }_{n}\to 0$ as $n\to \mathrm{\infty }$, we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {A}_{n}{T}_{n}{z}_{n}-{T}_{n}{z}_{n}\parallel =0.$
(3.9)
Let $\left\{{z}_{n}\right\}\in \stackrel{˜}{F}\left(\left\{{A}_{n}{T}_{n}\right\}\right)$, then we have
$\parallel {z}_{n}-{T}_{n}{z}_{n}\parallel \le \parallel {z}_{n}-{A}_{n}{T}_{n}{z}_{n}\parallel +\parallel {A}_{n}{T}_{n}{z}_{n}-{T}_{n}{z}_{n}\parallel .$
From (3.9), we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {z}_{n}-{T}_{n}{z}_{n}\parallel =0,$
which implies that
$\left\{{z}_{n}\right\}\in \stackrel{˜}{F}\left(\left\{{T}_{n}\right\}\right).$
It follows that
$\stackrel{˜}{F}\left(\left\{{A}_{n}{T}_{n}\right\}\right)\subseteq \stackrel{˜}{F}\left(\left\{{T}_{n}\right\}\right).$
(3.10)
Let $\left\{{z}_{n}\right\}\in \stackrel{˜}{F}\left(\left\{{T}_{n}\right\}\right)$, then we have
$\parallel {z}_{n}-{A}_{n}{T}_{n}{z}_{n}\parallel \le \parallel {z}_{n}-{T}_{n}{z}_{n}\parallel +\parallel {T}_{n}{z}_{n}-{A}_{n}{T}_{n}{z}_{n}\parallel .$
From (3.9), we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {z}_{n}-{A}_{n}{T}_{n}{z}_{n}\parallel =0,$
which implies that
$\left\{{z}_{n}\right\}\in \stackrel{˜}{F}\left(\left\{{A}_{n}{T}_{n}\right\}\right).$
It follows that
$\stackrel{˜}{F}\left(\left\{{T}_{n}\right\}\right)\subseteq \stackrel{˜}{F}\left(\left\{{A}_{n}{T}_{n}\right\}\right).$
(3.11)
From (3.10) and (3.11), we have
$\stackrel{˜}{F}\left(\left\{{T}_{n}\right\}\right)=\stackrel{˜}{F}\left(\left\{{A}_{n}{T}_{n}\right\}\right).$
(3.12)
Next, we show that
$\stackrel{˜}{F}\left(\left\{{S}_{n}{A}_{n}{T}_{n}\right\}\right)=\stackrel{˜}{F}\left(\left\{{S}_{n}\right\}\right)\cap \stackrel{˜}{F}\left(\left\{{T}_{n}\right\}\right).$
Since is nonempty, from (3.8), (3.12), we have
$\stackrel{˜}{F}\left(\left\{{S}_{n}{A}_{n}\right\}\right)\cap \stackrel{˜}{F}\left(\left\{{T}_{n}\right\}\right)=\stackrel{˜}{F}\left(\left\{{S}_{n}\right\}\right)\cap \stackrel{˜}{F}\left(\left\{{T}_{n}\right\}\right)\ne \mathrm{\varnothing }$
(3.13)
and
$\stackrel{˜}{F}\left(\left\{{S}_{n}\right\}\right)\cap \stackrel{˜}{F}\left(\left\{{A}_{n}{T}_{n}\right\}\right)=\stackrel{˜}{F}\left(\left\{{S}_{n}\right\}\right)\cap \stackrel{˜}{F}\left(\left\{{T}_{n}\right\}\right)\ne \mathrm{\varnothing }.$
(3.14)
Suppose that $\left\{{S}_{n}\right\}$ is a strongly nonexpansive sequence. From (3.14) and Lemma 2.9, we have
$\stackrel{˜}{F}\left(\left\{{S}_{n}{A}_{n}{T}_{n}\right\}\right)=\stackrel{˜}{F}\left(\left\{{S}_{n}\right\}\right)\cap \stackrel{˜}{F}\left(\left\{{A}_{n}{T}_{n}\right\}\right)=\stackrel{˜}{F}\left(\left\{{S}_{n}\right\}\right)\cap \stackrel{˜}{F}\left(\left\{{T}_{n}\right\}\right).$
(3.15)
On the other hand, suppose that $\left\{{T}_{n}\right\}$ is a strongly nonexpansive sequence. From (3.13) and Lemma 2.9, we have
$\stackrel{˜}{F}\left(\left\{{S}_{n}{A}_{n}{T}_{n}\right\}\right)=\stackrel{˜}{F}\left(\left\{{S}_{n}{A}_{n}\right\}\right)\cap \stackrel{˜}{F}\left(\left\{{T}_{n}\right\}\right)=\stackrel{˜}{F}\left(\left\{{S}_{n}\right\}\right)\cap \stackrel{˜}{F}\left(\left\{{T}_{n}\right\}\right).$
(3.16)
From (3.16) and (3.15), we have $\stackrel{˜}{F}\left(\left\{{S}_{n}{A}_{n}{T}_{n}\right\}\right)=\stackrel{˜}{F}\left(\left\{{S}_{n}\right\}\right)\cap \stackrel{˜}{F}\left(\left\{{T}_{n}\right\}\right)$. Next, we show that $\left\{{A}_{n}\right\}$ and $\left\{{S}_{n}{A}_{n}{T}_{n}\right\}$ satisfy the condition (R). It is easy to see that ${A}_{n}$ is a nonexpansive mapping for every $n\in \mathbb{N}$ and that $\left\{{A}_{n}y:n\in \mathbb{N},y\in D\right\}$ is bounded, where D is a bounded subset of C. Let $y\in D$, then we have
$\begin{array}{rcl}\parallel {A}_{n+1}y-{A}_{n}y\parallel & =& \parallel {\alpha }_{n+1}f\left(y\right)+\left(1-{\alpha }_{n+1}\right)y-{\alpha }_{n}f\left(y\right)-\left(1-{\alpha }_{n}\right)y\parallel \\ \le & |{\alpha }_{n+1}-{\alpha }_{n}|\parallel f\left(y\right)\parallel +|{\alpha }_{n+1}-{\alpha }_{n}|\parallel y\parallel .\end{array}$
From the condition (i), we have
$\underset{n\to \mathrm{\infty }}{lim}\underset{y\in D}{sup}\parallel {A}_{n+1}y-{A}_{n}y\parallel =0.$

It follows that $\left\{{A}_{n}\right\}$ satisfies the condition (R). From Lemma 2.8, we have that $\left\{{S}_{n}{A}_{n}\right\}$ satisfies the condition (R). From the nonexpansiveness of ${T}_{n}$ and $\mathbb{F}\ne \mathrm{\varnothing }$, we have $\left\{{T}_{n}y:n\in \mathbb{N},y\in D\right\}$ is bounded for any bounded subset D of C. From Lemma 2.8, we have that $\left\{{S}_{n}{A}_{n}{T}_{n}\right\}$ satisfies the condition (R).

Next, we show that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n+1}-{x}_{n}\parallel =0.$
(3.17)
Put
${x}_{n+1}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right){w}_{n},$
(3.18)
where ${w}_{n}={S}_{n}\left({\alpha }_{n}f\left({T}_{n}{y}_{n}\right)+\left(1-{\alpha }_{n}\right){T}_{n}{y}_{n}\right)$. From the definition of ${w}_{n}$, we have
$\begin{array}{rcl}\parallel {w}_{n+1}-{w}_{n}\parallel & =& \parallel {S}_{n+1}{A}_{n+1}{T}_{n+1}{y}_{n+1}-{S}_{n}{A}_{n}{T}_{n}{y}_{n}\parallel \\ \le & \parallel {S}_{n+1}{A}_{n+1}{T}_{n+1}{y}_{n+1}-{S}_{n}{A}_{n}{T}_{n}{y}_{n+1}\parallel +\parallel {S}_{n}{A}_{n}{T}_{n}{y}_{n+1}-{S}_{n}{A}_{n}{T}_{n}{y}_{n}\parallel \\ \le & \underset{y\in D}{sup}\parallel {S}_{n+1}{A}_{n+1}{T}_{n+1}y-{S}_{n}{A}_{n}{T}_{n}y\parallel +\parallel {y}_{n+1}-{y}_{n}\parallel ,\end{array}$
(3.19)
where D is a bounded subset of C. Besides, we have
$\begin{array}{rcl}\parallel {y}_{n+1}-{y}_{n}\parallel & =& \parallel {\delta }_{n+1}{u}_{n+1}+\left(1-{\delta }_{n+1}\right){v}_{n+1}-{\delta }_{n}{u}_{n}-\left(1-{\delta }_{n}\right){v}_{n}\parallel \\ =& \parallel {\delta }_{n+1}{u}_{n+1}-{\delta }_{n+1}{u}_{n}+{\delta }_{n+1}{u}_{n}-\left(1-{\delta }_{n+1}\right){v}_{n}+\left(1-{\delta }_{n+1}\right){v}_{n}\\ +\left(1-{\delta }_{n+1}\right){v}_{n+1}-{\delta }_{n}{u}_{n}-\left(1-{\delta }_{n}\right){v}_{n}\parallel \\ =& \parallel {\delta }_{n+1}\left({u}_{n+1}-{u}_{n}\right)+\left({\delta }_{n+1}-{\delta }_{n}\right){u}_{n}+\left(1-{\delta }_{n+1}\right)\left({v}_{n+1}-{v}_{n}\right)+\left({\delta }_{n}-{\delta }_{n+1}\right){v}_{n}\parallel \\ \le & {\delta }_{n+1}\parallel {u}_{n+1}-{u}_{n}\parallel +|{\delta }_{n+1}-{\delta }_{n}|\parallel {u}_{n}\parallel +\left(1-{\delta }_{n+1}\right)\parallel {v}_{n+1}-{v}_{n}\parallel \\ +|{\delta }_{n}-{\delta }_{n+1}|\parallel {v}_{n}\parallel .\end{array}$
(3.20)
From (3.1) and Lemma 2.5, we have ${u}_{n}={S}_{{r}_{n}}{x}_{n}$. This implies that
(3.21)
and
(3.22)
Putting $u={u}_{n+1}$ in (3.21) and $u={u}_{n}$ in (3.22), we have
${F}_{1}\left({u}_{n},{u}_{n+1}\right)+\frac{1}{{r}_{n}}〈{u}_{n+1}-{u}_{n},{u}_{n}-{x}_{n}〉\ge 0$
(3.23)
and
${F}_{1}\left({u}_{n+1},{u}_{n}\right)+\frac{1}{{r}_{n+1}}〈{u}_{n}-{u}_{n+1},{u}_{n+1}-{x}_{n+1}〉\ge 0.$
(3.24)
Summing up the last two inequalities and using (A2), we obtain
$〈{u}_{n+1}-{u}_{n},\frac{{u}_{n}-{x}_{n}}{{r}_{n}}-\frac{{u}_{n+1}-{x}_{n+1}}{{r}_{n+1}}〉\ge 0.$
This implies that
$〈{u}_{n+1}-{u}_{n},{u}_{n}-{u}_{n+1}+{u}_{n+1}-{x}_{n}-\frac{{r}_{n}}{{r}_{n+1}}\left({u}_{n+1}-{x}_{n+1}\right)〉\ge 0.$
Hence,
$\begin{array}{rcl}{\parallel {u}_{n+1}-{u}_{n}\parallel }^{2}& \le & 〈{u}_{n+1}-{u}_{n},{u}_{n+1}-{x}_{n}-\frac{{r}_{n}}{{r}_{n+1}}\left({u}_{n+1}-{x}_{n+1}\right)〉\\ =& 〈{u}_{n+1}-{u}_{n},{u}_{n+1}-{x}_{n+1}+{x}_{n+1}-{x}_{n}-\frac{{r}_{n}}{{r}_{n+1}}\left({u}_{n+1}-{x}_{n+1}\right)〉\\ =& 〈{u}_{n+1}-{u}_{n},{x}_{n+1}-{x}_{n}+\left(1-\frac{{r}_{n}}{{r}_{n+1}}\right)\left({u}_{n+1}-{x}_{n+1}\right)〉\\ \le & \parallel {u}_{n+1}-{u}_{n}\parallel \left(\parallel {x}_{n+1}-{x}_{n}\parallel +\frac{1}{{r}_{n+1}}|{r}_{n+1}-{r}_{n}|\parallel {u}_{n+1}-{x}_{n+1}\parallel \right)\\ \le & \parallel {u}_{n+1}-{u}_{n}\parallel \left(\parallel {x}_{n+1}-{x}_{n}\parallel +\frac{1}{a}|{r}_{n+1}-{r}_{n}|\parallel {u}_{n+1}-{x}_{n+1}\parallel \right).\end{array}$
Then we have
$\parallel {u}_{n+1}-{u}_{n}\parallel \le \parallel {x}_{n+1}-{x}_{n}\parallel +\frac{1}{a}|{r}_{n+1}-{r}_{n}|\parallel {u}_{n+1}-{x}_{n+1}\parallel .$
(3.25)
From (3.1) and Lemma 2.5, we have ${v}_{n}={S}_{{s}_{n}}{x}_{n}$. This implies that
By using the same method as (3.25), we have
$\parallel {v}_{n+1}-{v}_{n}\parallel \le \parallel {x}_{n+1}-{x}_{n}\parallel +\frac{1}{a}|{s}_{n+1}-{s}_{n}|\parallel {v}_{n+1}-{x}_{n+1}\parallel .$
(3.26)
Substituting (3.25) and (3.26) into (3.20), we have
$\begin{array}{rcl}\parallel {y}_{n+1}-{y}_{n}\parallel & \le & {\delta }_{n+1}\parallel {u}_{n+1}-{u}_{n}\parallel +|{\delta }_{n+1}-{\delta }_{n}|\parallel {u}_{n}\parallel +\left(1-{\delta }_{n+1}\right)\parallel {v}_{n+1}-{v}_{n}\parallel \\ +|{\delta }_{n}-{\delta }_{n+1}|\parallel {v}_{n}\parallel \\ \le & {\delta }_{n+1}\left(\parallel {x}_{n+1}-{x}_{n}\parallel +\frac{1}{a}|{r}_{n+1}-{r}_{n}|\parallel {u}_{n+1}-{x}_{n+1}\parallel \right)\\ +\left(1-{\delta }_{n+1}\right)\left(\parallel {x}_{n+1}-{x}_{n}\parallel +\frac{1}{a}|{s}_{n+1}-{s}_{n}|\parallel {v}_{n+1}-{x}_{n+1}\parallel \right)\\ +2M|{\delta }_{n}-{\delta }_{n+1}|\\ \le & \parallel {x}_{n+1}-{x}_{n}\parallel +\frac{1}{a}|{r}_{n+1}-{r}_{n}|\parallel {u}_{n+1}-{x}_{n+1}\parallel \\ +\frac{1}{a}|{s}_{n+1}-{s}_{n}|\parallel {v}_{n+1}-{x}_{n+1}\parallel +2M|{\delta }_{n}-{\delta }_{n+1}|,\end{array}$
(3.27)
where $M={sup}_{n\in \mathbb{N}}\left\{\parallel {u}_{n}\parallel ,\parallel {v}_{n}\parallel \right\}$. Substituting (3.27) into (3.19), we have
$\begin{array}{rcl}\parallel {w}_{n+1}-{w}_{n}\parallel & \le & \underset{y\in D}{sup}\parallel {S}_{n+1}{A}_{n+1}{T}_{n+1}y-{S}_{n}{A}_{n}{T}_{n}y\parallel +\parallel {y}_{n+1}-{y}_{n}\parallel \\ \le & \underset{y\in D}{sup}\parallel {S}_{n+1}{A}_{n+1}{T}_{n+1}y-{S}_{n}{A}_{n}{T}_{n}y\parallel +\parallel {x}_{n+1}-{x}_{n}\parallel \\ +\frac{1}{a}|{r}_{n+1}-{r}_{n}|\parallel {u}_{n+1}-{x}_{n+1}\parallel \\ +\frac{1}{a}|{s}_{n+1}-{s}_{n}|\parallel {v}_{n+1}-{x}_{n+1}\parallel +2M|{\delta }_{n}-{\delta }_{n+1}|.\end{array}$
(3.28)
From (3.28), the conditions (iii), (iv) and $\left\{{S}_{n}{A}_{n}{T}_{n}\right\}$ satisfying the condition (R), we have
$\underset{n\to \mathrm{\infty }}{lim sup}\left(\parallel {w}_{n+1}-{w}_{n}\parallel -\parallel {x}_{n+1}-{x}_{n}\parallel \right)\le 0.$
(3.29)
From Lemma 2.3 and the definition of ${x}_{n}$, we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-{w}_{n}\parallel =0.$
(3.30)
From the definition of ${x}_{n}$, we have
${x}_{n+1}-{x}_{n}=\left(1-{\beta }_{n}\right)\left({w}_{n}-{x}_{n}\right).$
(3.31)
From (3.30), (3.31) and the condition (ii), we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n+1}-{x}_{n}\parallel =0.$
Next, we show that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {y}_{n}-{x}_{n}\parallel =0.$
From the definition of ${y}_{n}$, we have
$\parallel {y}_{n}-{x}_{n}\parallel \le {\delta }_{n}\parallel {u}_{n}-{x}_{n}\parallel +\left(1-{\delta }_{n}\right)\parallel {v}_{n}-{x}_{n}\parallel .$
(3.32)
Next, we show that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {u}_{n}-{x}_{n}\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel {v}_{n}-{x}_{n}\parallel =0.$
Let $v\in \mathbb{F}$. From the definition of ${x}_{n}$, we have
$\begin{array}{rcl}{\parallel {x}_{n+1}-v\parallel }^{2}& \le & {\beta }_{n}{\parallel {x}_{n}-v\parallel }^{2}+\left(1-{\beta }_{n}\right){\parallel {S}_{n}\left({\alpha }_{n}f\left({T}_{n}{y}_{n}\right)+\left(1-{\alpha }_{n}\right){T}_{n}{y}_{n}\right)-v\parallel }^{2}\\ \le & {\beta }_{n}{\parallel {x}_{n}-v\parallel }^{2}+\left(1-{\beta }_{n}\right){\parallel {\alpha }_{n}\left(f\left({T}_{n}{y}_{n}\right)-v\right)+\left(1-{\alpha }_{n}\right)\left({T}_{n}{y}_{n}-v\right)\parallel }^{2}\\ \le & {\beta }_{n}{\parallel {x}_{n}-v\parallel }^{2}+\left(1-{\beta }_{n}\right)\left({\alpha }_{n}{\parallel f\left({T}_{n}{y}_{n}\right)-v\parallel }^{2}+\left(1-{\alpha }_{n}\right){\parallel {T}_{n}{y}_{n}-v\parallel }^{2}\right)\\ \le & {\beta }_{n}{\parallel {x}_{n}-v\parallel }^{2}+\left(1-{\beta }_{n}\right)\left({\alpha }_{n}{\parallel f\left({T}_{n}{y}_{n}\right)-v\parallel }^{2}+\left(1-{\alpha }_{n}\right){\parallel {y}_{n}-v\parallel }^{2}\right)\\ \le & {\beta }_{n}{\parallel {x}_{n}-v\parallel }^{2}+\left(1-{\beta }_{n}\right)\left({\alpha }_{n}{\parallel f\left({T}_{n}{y}_{n}\right)-v\parallel }^{2}\\ +\left(1-{\alpha }_{n}\right)\left({\delta }_{n}{\parallel {u}_{n}-v\parallel }^{2}+\left(1-{\delta }_{n}\right){\parallel {v}_{n}-v\parallel }^{2}\right)\right).\end{array}$
(3.33)
From the firm nonexpansiveness of ${S}_{{r}_{n}}$ and ${u}_{n}={S}_{{r}_{n}}{x}_{n}$, we have
$\begin{array}{rcl}{\parallel {u}_{n}-v\parallel }^{2}& =& {\parallel {S}_{{r}_{n}}{x}_{n}-{S}_{{r}_{n}}v\parallel }^{2}\\ \le & 〈{u}_{n}-v,{x}_{n}-v〉\\ =& \frac{1}{2}\left({\parallel {u}_{n}-v\parallel }^{2}+{\parallel {x}_{n}-v\parallel }^{2}-{\parallel {u}_{n}-{x}_{n}\parallel }^{2}\right).\end{array}$
It implies that
${\parallel {u}_{n}-v\parallel }^{2}\le {\parallel {x}_{n}-v\parallel }^{2}-{\parallel {u}_{n}-{x}_{n}\parallel }^{2}.$
(3.34)
Since ${S}_{{s}_{n}}$ is a firmly nonexpansive mapping and ${v}_{n}={S}_{{s}_{n}}{x}_{n}$, by using the same method as (3.34), we have
${\parallel {v}_{n}-v\parallel }^{2}\le {\parallel {x}_{n}-v\parallel }^{2}-{\parallel {v}_{n}-{x}_{n}\parallel }^{2}.$
(3.35)
Substituting (3.34), (3.35) into (3.33), we have
$\begin{array}{rcl}{\parallel {x}_{n+1}-v\parallel }^{2}& \le & {\beta }_{n}{\parallel {x}_{n}-v\parallel }^{2}+\left(1-{\beta }_{n}\right)\left({\alpha }_{n}{\parallel f\left({T}_{n}{y}_{n}\right)-v\parallel }^{2}\\ +\left(1-{\alpha }_{n}\right)\left({\delta }_{n}{\parallel {u}_{n}-v\parallel }^{2}+\left(1-{\delta }_{n}\right){\parallel {v}_{n}-v\parallel }^{2}\right)\right)\\ \le & {\beta }_{n}{\parallel {x}_{n}-v\parallel }^{2}+\left(1-{\beta }_{n}\right)\left({\alpha }_{n}{\parallel f\left({T}_{n}{y}_{n}\right)-v\parallel }^{2}\\ +\left(1-{\alpha }_{n}\right)\left({\delta }_{n}\left({\parallel {x}_{n}-v\parallel }^{2}-{\parallel {u}_{n}-{x}_{n}\parallel }^{2}\right)\\ +\left(1-{\delta }_{n}\right)\left({\parallel {x}_{n}-v\parallel }^{2}-{\parallel {v}_{n}-{x}_{n}\parallel }^{2}\right)\right)\right)\\ =& {\beta }_{n}{\parallel {x}_{n}-v\parallel }^{2}+\left(1-{\beta }_{n}\right)\left({\alpha }_{n}{\parallel f\left({T}_{n}{y}_{n}\right)-v\parallel }^{2}\\ +\left(1-{\alpha }_{n}\right)\left({\delta }_{n}{\parallel {x}_{n}-v\parallel }^{2}-{\delta }_{n}{\parallel {u}_{n}-{x}_{n}\parallel }^{2}\\ +\left(1-{\delta }_{n}\right){\parallel {x}_{n}-v\parallel }^{2}-\left(1-{\delta }_{n}\right){\parallel {v}_{n}-{x}_{n}\parallel }^{2}\right)\right)\\ =& {\beta }_{n}{\parallel {x}_{n}-v\parallel }^{2}+\left(1-{\beta }_{n}\right)\left({\alpha }_{n}{\parallel f\left({T}_{n}{y}_{n}\right)-v\parallel }^{2}\\ +\left(1-{\alpha }_{n}\right)\left({\parallel {x}_{n}-v\parallel }^{2}-{\delta }_{n}{\parallel {u}_{n}-{x}_{n}\parallel }^{2}\\ -\left(1-{\delta }_{n}\right){\parallel {v}_{n}-{x}_{n}\parallel }^{2}\right)\right)\\ =& {\beta }_{n}{\parallel {x}_{n}-v\parallel }^{2}+\left(1-{\beta }_{n}\right){\alpha }_{n}{\parallel f\left({T}_{n}{y}_{n}\right)-v\parallel }^{2}\\ +\left(1-{\alpha }_{n}\right)\left(1-{\beta }_{n}\right)\left({\parallel {x}_{n}-v\parallel }^{2}-{\delta }_{n}{\parallel {u}_{n}-{x}_{n}\parallel }^{2}\\ -\left(1-{\delta }_{n}\right){\parallel {v}_{n}-{x}_{n}\parallel }^{2}\right)\\ =& {\beta }_{n}{\parallel {x}_{n}-v\parallel }^{2}+\left(1-{\beta }_{n}\right){\alpha }_{n}{\parallel f\left({T}_{n}{y}_{n}\right)-v\parallel }^{2}\\ +\left(1-{\alpha }_{n}\right)\left(1-{\beta }_{n}\right){\parallel {x}_{n}-v\parallel }^{2}-{\delta }_{n}\left(1-{\alpha }_{n}\right)\left(1-{\beta }_{n}\right){\parallel {u}_{n}-{x}_{n}\parallel }^{2}\\ -\left(1-{\delta }_{n}\right)\left(1-{\alpha }_{n}\right)\left(1-{\beta }_{n}\right){\parallel {v}_{n}-{x}_{n}\parallel }^{2}\\ \le & {\parallel {x}_{n}-v\parallel }^{2}+{\alpha }_{n}{\parallel f\left({T}_{n}{y}_{n}\right)-v\parallel }^{2}-{\delta }_{n}\left(1-{\alpha }_{n}\right)\left(1-{\beta }_{n}\right){\parallel {u}_{n}-{x}_{n}\parallel }^{2}\\ -\left(1-{\delta }_{n}\right)\left(1-{\alpha }_{n}\right)\left(1-{\beta }_{n}\right){\parallel {v}_{n}-{x}_{n}\parallel }^{2}.\end{array}$
(3.36)
From (3.36), we have
$\begin{array}{rcl}{\delta }_{n}\left(1-{\alpha }_{n}\right)\left(1-{\beta }_{n}\right){\parallel {u}_{n}-{x}_{n}\parallel }^{2}& \le & {\parallel {x}_{n}-v\parallel }^{2}-{\parallel {x}_{n+1}-v\parallel }^{2}+{\alpha }_{n}{\parallel f\left({T}_{n}{y}_{n}\right)-v\parallel }^{2}\\ -\left(1-{\delta }_{n}\right)\left(1-{\alpha }_{n}\right)\left(1-{\beta }_{n}\right){\parallel {v}_{n}-{x}_{n}\parallel }^{2}\\ \le & \left(\parallel {x}_{n}-v\parallel +\parallel {x}_{n+1}-v\parallel \right)\parallel {x}_{n+1}-{x}_{n}\parallel +{\alpha }_{n}{\parallel f\left({T}_{n}{y}_{n}\right)-v\parallel }^{2}\\ -\left(1-{\delta }_{n}\right)\left(1-{\alpha }_{n}\right)\left(1-{\beta }_{n}\right){\parallel {v}_{n}-{x}_{n}\parallel }^{2}\\ \le & \left(\parallel {x}_{n}-v\parallel +\parallel {x}_{n+1}-v\parallel \right)\parallel {x}_{n+1}-{x}_{n}\parallel \\ +{\alpha }_{n}{\parallel f\left({T}_{n}{y}_{n}\right)-v\parallel }^{2}.\end{array}$
From the conditions (i), (ii), (iv) and (3.17), we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {u}_{n}-{x}_{n}\parallel =0.$
(3.37)
By using the method as (3.37), we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {v}_{n}-{x}_{n}\parallel =0.$
(3.38)
From (3.32), (3.37) and (3.38), we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {y}_{n}-{x}_{n}\parallel =0.$
(3.39)
Next, we show that
$\left\{{y}_{n}\right\}\in \stackrel{˜}{F}\left(\left\{{S}_{n}\right\}\right)\cap \stackrel{˜}{F}\left(\left\{{T}_{n}\right\}\right).$
(3.40)
Since
$\begin{array}{rcl}\parallel {S}_{n}{A}_{n}{T}_{n}{y}_{n}-{y}_{n}\parallel & \le & \parallel {S}_{n}{A}_{n}{T}_{n}{y}_{n}-{x}_{n}\parallel +\parallel {x}_{n}-{y}_{n}\parallel \\ =& \parallel {w}_{n}-{x}_{n}\parallel +\parallel {x}_{n}-{y}_{n}\parallel ,\end{array}$
from (3.30) and (3.39), we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {S}_{n}{A}_{n}{T}_{n}{y}_{n}-{y}_{n}\parallel =0.$
Since $\left\{{y}_{n}\right\}$ is bounded, we have
$\left\{{y}_{n}\right\}\in \stackrel{˜}{F}\left(\left\{{S}_{n}{A}_{n}{T}_{n}\right\}\right).$
(3.41)

Since $\stackrel{˜}{F}\left(\left\{{S}_{n}{A}_{n}{T}_{n}\right\}\right)=\stackrel{˜}{F}\left(\left\{{S}_{n}\right\}\right)\cap \stackrel{˜}{F}\left(\left\{{T}_{n}\right\}\right)$ and (3.41), we have (3.40).

Next, we show that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {S}_{n}{m}_{n}-{m}_{n}\parallel =0,$
where ${m}_{n}={\alpha }_{n}f\left({T}_{n}{y}_{n}\right)+\left(1-{\alpha }_{n}\right){T}_{n}{y}_{n}$. From the definition of ${m}_{n}$, we have
$\begin{array}{rcl}\parallel {S}_{n}{m}_{n}-{m}_{n}\parallel & \le & \parallel {S}_{n}{m}_{n}-{x}_{n}\parallel +\parallel {m}_{n}-{x}_{n}\parallel \\ =& \parallel {S}_{n}{m}_{n}-{x}_{n}\parallel +\parallel {\alpha }_{n}\left(f\left({T}_{n}{y}_{n}\right)-{x}_{n}\right)+\left(1-{\alpha }_{n}\right)\left({T}_{n}{y}_{n}-{x}_{n}\right)\parallel \\ \le & \parallel {w}_{n}-{x}_{n}\parallel +{\alpha }_{n}\parallel f\left({T}_{n}{y}_{n}\right)-{x}_{n}\parallel +\left(1-{\alpha }_{n}\right)\parallel {T}_{n}{y}_{n}-{x}_{n}\parallel \\ \le & \parallel {w}_{n}-{x}_{n}\parallel +{\alpha }_{n}\parallel f\left({T}_{n}{y}_{n}\right)-{x}_{n}\parallel \\ +\parallel {T}_{n}{y}_{n}-{y}_{n}\parallel +\parallel {y}_{n}-{x}_{n}\parallel .\end{array}$
From (3.39), (3.40), (3.30) and the condition (i), we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {S}_{n}{m}_{n}-{m}_{n}\parallel =0.$
Next, we show that
$\underset{n\to \mathrm{\infty }}{lim sup}〈f\left(z\right)-z,{m}_{n}-z〉\le 0,$

where $z={P}_{\mathbb{F}}f\left(z\right)$. Since $\left\{{y}_{n}\right\}$ is bounded, there exists a subsequence $\left\{{y}_{{n}_{i}}\right\}$ of $\left\{{y}_{n}\right\}$ converging weakly to v, that is, ${y}_{{n}_{i}}⇀v$ as $i\to \mathrm{\infty }$. From (3.40), $\left\{{S}_{n}\right\}$ and $\left\{{T}_{n}\right\}$ satisfying the condition (Z), we have $v\in F\left(\left\{{S}_{n}\right\}\right)\cap F\left(\left\{{T}_{n}\right\}\right)$.

Define the mapping $Q:C\to C$ by
where ${lim}_{n\to \mathrm{\infty }}{\delta }_{n}=\delta \in \left(0,1\right)$. From the nonexpansiveness of ${S}_{{r}_{n}}$, ${S}_{{s}_{n}}$ and Lemma 2.4, we have
$F\left(Q\right)=F\left({S}_{{r}_{n}}\right)\cap F\left({S}_{{s}_{n}}\right)=\mathit{EP}\left({F}_{1}\right)\cap \mathit{EP}\left({F}_{2}\right).$
From the definitions of ${y}_{n}$ and Q, we have
$\begin{array}{rcl}\parallel {x}_{n}-Q{x}_{n}\parallel & \le & \parallel {x}_{n}-{y}_{n}\parallel +\parallel {y}_{n}-Q{x}_{n}\parallel \\ \le & \parallel {x}_{n}-{y}_{n}\parallel +\parallel {\delta }_{n}{u}_{n}+\left(1-{\delta }_{n}\right){v}_{n}-\delta {S}_{{r}_{n}}{x}_{n}-\left(1-\delta \right){S}_{{s}_{n}}{x}_{n}\parallel \\ \le & \parallel {x}_{n}-{y}_{n}\parallel +|{\delta }_{n}-\delta |\parallel {u}_{n}\parallel +|{\delta }_{n}-\delta |\parallel {v}_{n}\parallel .\end{array}$
(3.42)
From (3.39), (3.42) and the condition (iv), we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-Q{x}_{n}\parallel =0.$
(3.43)
From (3.39) and ${y}_{{n}_{i}}⇀v$ as $i\to \mathrm{\infty }$, we have ${x}_{{n}_{i}}⇀v$ as $i\to \mathrm{\infty }$. By (3.43), ${x}_{{n}_{i}}⇀v$ as $i\to \mathrm{\infty }$ and Lemma 2.6, we have
$v\in F\left(Q\right)=\mathit{EP}\left({F}_{1}\right)\cap \mathit{EP}\left({F}_{2}\right).$
Hence,
$v\in \mathit{EP}\left({F}_{1}\right)\cap \mathit{EP}\left({F}_{2}\right)\cap F\left(\left\{{S}_{n}\right\}\right)\cap F\left(\left\{{T}_{n}\right\}\right)=\mathbb{F}.$
(3.44)
By (3.40), (3.44) and the condition (i), we have
$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim sup}〈f\left(z\right)-z,{m}_{n}-z〉& =& \underset{n\to \mathrm{\infty }}{lim sup}\left({\alpha }_{n}〈f\left(z\right)-z,f\left({T}_{n}{y}_{n}\right)-{T}_{n}{y}_{n}〉\\ +〈f\left(z\right)-z,{T}_{n}{y}_{n}-z〉\right)\\ =& \underset{i\to \mathrm{\infty }}{lim}\left({\alpha }_{{n}_{i}}〈f\left(z\right)-z,f\left({T}_{{n}_{i}}{y}_{{n}_{i}}\right)-{T}_{{n}_{i}}{y}_{{n}_{i}}〉\\ +〈f\left(z\right)-z,{T}_{{n}_{i}}{y}_{{n}_{i}}-z〉\right)\\ =& \underset{i\to \mathrm{\infty }}{lim}\left({\alpha }_{{n}_{i}}〈f\left(z\right)-z,f\left({T}_{{n}_{i}}{y}_{{n}_{i}}\right)-{T}_{{n}_{i}}{y}_{{n}_{i}}〉\\ +〈f\left(z\right)-z,{T}_{{n}_{i}}{y}_{{n}_{i}}-{y}_{{n}_{i}}〉+〈f\left(z\right)-z,{y}_{{n}_{i}}-z〉\right)\\ =& 〈f\left(z\right)-z,v-z〉\le 0.\end{array}$
Finally, we show that the sequence $\left\{{x}_{n}\right\}$ converges strongly to $z={P}_{\mathbb{F}}f\left(z\right)$. From the definition of $\left\{{x}_{n}\right\}$, we have
$\begin{array}{rcl}{\parallel {x}_{n+1}-z\parallel }^{2}& =& {\parallel {\beta }_{n}\left({x}_{n}-z\right)+\left(1-{\beta }_{n}\right)\left({S}_{n}{m}_{n}-z\right)\parallel }^{2}\\ \le & {\beta }_{n}{\parallel {x}_{n}-z\parallel }^{2}+\left(1-{\beta }_{n}\right){\parallel {S}_{n}{m}_{n}-z\parallel }^{2}\\ \le & {\beta }_{n}{\parallel {x}_{n}-z\parallel }^{2}+\left(1-{\beta }_{n}\right){\parallel {m}_{n}-z\parallel }^{2}.\end{array}$
(3.45)
Since ${m}_{n}={\alpha }_{n}f\left({T}_{n}{y}_{n}\right)+\left(1-{\alpha }_{n}\right){T}_{n}{y}_{n}$, we have
$\begin{array}{rcl}{\parallel {m}_{n}-z\parallel }^{2}& =& {\parallel {\alpha }_{n}\left(f\left({T}_{n}{y}_{n}\right)-z\right)+\left(1-{\alpha }_{n}\right)\left({T}_{n}{y}_{n}-z\right)\parallel }^{2}\\ \le & {\left(1-{\alpha }_{n}\right)}^{2}{\parallel {T}_{n}{y}_{n}-z\parallel }^{2}+2{\alpha }_{n}〈f\left({T}_{n}{y}_{n}\right)-z,{m}_{n}-z〉\\ \le & \left(1-{\alpha }_{n}\right){\parallel {x}_{n}-z\parallel }^{2}+2{\alpha }_{n}〈f\left({T}_{n}{y}_{n}\right)-f\left(z\right),{m}_{n}-z〉\\ +2{\alpha }_{n}〈f\left(z\right)-z,{m}_{n}-z〉\\ \le & \left(1-{\alpha }_{n}\right){\parallel {x}_{n}-z\parallel }^{2}+2{\alpha }_{n}\alpha \parallel {x}_{n}-z\parallel \parallel {m}_{n}-z\parallel \\ +2{\alpha }_{n}〈f\left(z\right)-z,{m}_{n}-z〉\\ \le & \left(1-{\alpha }_{n}\right){\parallel {x}_{n}-z\parallel }^{2}+{\alpha }_{n}\alpha \left({\parallel {x}_{n}-z\parallel }^{2}+{\parallel {m}_{n}-z\parallel }^{2}\right)\\ +2{\alpha }_{n}〈f\left(z\right)-z,{m}_{n}-z〉\\ =& \left(1-{\alpha }_{n}\right){\parallel {x}_{n}-z\parallel }^{2}+{\alpha }_{n}\alpha {\parallel {x}_{n}-z\parallel }^{2}+{\alpha }_{n}\alpha {\parallel {m}_{n}-z\parallel }^{2}\\ +2{\alpha }_{n}〈f\left(z\right)-z,{m}_{n}-z〉\\ =& \left(1-{\alpha }_{n}\left(1-\alpha \right)\right){\parallel {x}_{n}-z\parallel }^{2}+{\alpha }_{n}\alpha {\parallel {m}_{n}-z\parallel }^{2}\\ +2{\alpha }_{n}〈f\left(z\right)-z,{m}_{n}-z〉.\end{array}$
This implies that
$\begin{array}{rcl}{\parallel {m}_{n}-z\parallel }^{2}& \le & \frac{1-{\alpha }_{n}\left(1-\alpha \right)}{1-{\alpha }_{n}\alpha }{\parallel {x}_{n}-z\parallel }^{2}+\frac{2{\alpha }_{n}}{1-{\alpha }_{n}\alpha }〈f\left(z\right)-z,{m}_{n}-z〉\\ =& \frac{1-{\alpha }_{n}\alpha +{\alpha }_{n}\alpha -{\alpha }_{n}\left(1-\alpha \right)}{1-{\alpha }_{n}\alpha }{\parallel {x}_{n}-z\parallel }^{2}+\frac{2{\alpha }_{n}}{1-{\alpha }_{n}\alpha }〈f\left(z\right)-z,{m}_{n}-z〉\\ =& \left(1-\frac{{\alpha }_{n}\left(1-2\alpha \right)}{1-{\alpha }_{n}\alpha }\right){\parallel {x}_{n}-z\parallel }^{2}+\frac{2{\alpha }_{n}}{1-{\alpha }_{n}\alpha }〈f\left(z\right)-z,{m}_{n}-z〉.\end{array}$
(3.46)
Substituting (3.46) into (3.45), we have
$\begin{array}{rcl}{\parallel {x}_{n+1}-z\parallel }^{2}& \le & {\beta }_{n}{\parallel {x}_{n}-z\parallel }^{2}+\left(1-{\beta }_{n}\right){\parallel {m}_{n}-z\parallel }^{2}\\ \le & {\beta }_{n}{\parallel {x}_{n}-z\parallel }^{2}+\left(1-{\beta }_{n}\right)\left(\left(1-\frac{{\alpha }_{n}\left(1-2\alpha \right)}{1-{\alpha }_{n}\alpha }\right){\parallel {x}_{n}-z\parallel }^{2}\\ +\frac{2{\alpha }_{n}}{1-{\alpha }_{n}\alpha }〈f\left(z\right)-z,{m}_{n}-z〉\right)\\ \le & {\beta }_{n}{\parallel {x}_{n}-z\parallel }^{2}+\left(1-{\beta }_{n}\right)\left(1-\frac{{\alpha }_{n}\left(1-2\alpha \right)}{1-{\alpha }_{n}\alpha }\right){\parallel {x}_{n}-z\parallel }^{2}\\ +\frac{2{\alpha }_{n}\left(1-{\beta }_{n}\right)}{1-{\alpha }_{n}\alpha }〈f\left(z\right)-z,{m}_{n}-z〉\\ =& {\beta }_{n}{\parallel {x}_{n}-z\parallel }^{2}+\left(\left(1-{\beta }_{n}\right)-\frac{{\alpha }_{n}\left(1-2\alpha \right)\left(1-{\beta }_{n}\right)}{1-{\alpha }_{n}\alpha }\right){\parallel {x}_{n}-z\parallel }^{2}\\ +\frac{2{\alpha }_{n}\left(1-{\beta }_{n}\right)}{1-{\alpha }_{n}\alpha }〈f\left(z\right)-z,{m}_{n}-z〉\\ =& \left(1-\frac{{\alpha }_{n}\left(1-2\alpha \right)\left(1-{\beta }_{n}\right)}{1-{\alpha }_{n}\alpha }\right){\parallel {x}_{n}-z\parallel }^{2}\\ +\frac{{\alpha }_{n}\left(1-{\beta }_{n}\right)\left(1-2\alpha \right)}{1-{\alpha }_{n}\alpha }\frac{2〈f\left(z\right)-z,{m}_{n}-z〉}{\left(1-2\alpha \right)}.\end{array}$
(3.47)

Applying (3.47), the conditions (i), (ii) and Lemma 2.2, we have $\left\{{x}_{n}\right\}$ converges strongly to $z={P}_{\mathbb{F}}f\left(z\right)$. From (3.39), (3.37) and (3.38), it is easy to see that $\left\{{y}_{n}\right\}$, $\left\{{u}_{n}\right\}$, $\left\{{v}_{n}\right\}$ converge strongly to $z={P}_{\mathbb{F}}f\left(z\right)$. This completes the proof. □

## 4 Applications

In this section, we give three examples for a strongly nonexpansive sequence and prove a strong convergence theorem associated to the variational inequality problem.

Before we give three examples, we need the following definition and lemmas.

Definition 4.1 Let C be a nonempty closed convex subset of a real Hilbert space H. A mapping $A:C\to H$ is called an α-inverse strongly monotone mapping if there exists an $\alpha >0$ such that
$〈x-y,ax-Ay〉\ge \alpha {\parallel Ax-Ay\parallel }^{2}$

for all $x,y\in C$.

A mapping $A:C\to H$ is called α-strongly monotone if there exists $\alpha >0$ such that
$〈x-y,ax-Ay〉\ge \alpha {\parallel x-y\parallel }^{2}$

for all $x,y\in C$.

A mapping $T:C\to C$ is called a κ-strictly pseudo-contractive mapping if there is $\kappa \in \left[0,1\right)$ such that
${\parallel Tx-Ty\parallel }^{2}\le {\parallel x-y\parallel }^{2}+\kappa {\parallel \left(I-T\right)x-\left(I-T\right)y\parallel }^{2}$
(4.1)

for all $x,y\in C$.

Then (4.1) is equivalent to
$〈x-y,\left(I-T\right)x-\left(I-T\right)y〉\ge \frac{1-\kappa }{2}{\parallel \left(I-T\right)x-\left(I-T\right)y\parallel }^{2}$

for all $x,y\in C$.

The set of solutions of the variational inequality problem of the mapping $A:C\to H$ is denoted by $\mathit{VI}\left(C,A\right)$, that is,
$\mathit{VI}\left(C,A\right)=\left\{x\in C:〈y-x,Ax〉\ge 0,\mathrm{\forall }y\in C\right\}.$
Let $A,B:C\to H$ be two mappings. In 2013, Kangtunyakarn [15] modified $\mathit{VI}\left(C,A\right)$ as follows:

Remark 4.1 If $T:C\to C$ is a κ-strictly pseudo-contractive mapping with $F\left(T\right)\ne \mathrm{\varnothing }$, then $\left(I-T\right)$ is a $\frac{1-\kappa }{2}$-inverse strongly monotone mapping and $F\left(T\right)=\mathit{VI}\left(C,I-T\right)$.

Lemma 4.2 (See [16])

Let H be a Hilbert space, let C be a nonempty closed convex subset of H, and let A be a mapping of C into H. Let $u\in C$. Then, for $\lambda >0$,
$u={P}_{C}\left(I-\lambda A\right)u\phantom{\rule{1em}{0ex}}⇔\phantom{\rule{1em}{0ex}}u\in \mathit{VI}\left(C,A\right),$

where ${P}_{C}$ is the metric projection of H onto C.

Lemma 4.3 (See [15])

Let C be a nonempty closed convex subset of a real Hilbert space H, and let $A,B:C\to H$ be α and β-inverse strongly monotone mappings, respectively, with $\alpha ,\beta >0$ and $\mathit{VI}\left(C,A\right)\cap \mathit{VI}\left(C,B\right)\ne \mathrm{\varnothing }$. Then
$\mathit{VI}\left(C,aA+\left(1-a\right)B\right)=\mathit{VI}\left(C,A\right)\cap \mathit{VI}\left(C,B\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }a\in \left(0,1\right).$
(4.2)

Furthermore, if $0<\gamma <2\eta$, where $\eta =min\left\{\alpha ,\beta \right\}$, we have $I-\gamma \left(aA+\left(1-a\right)B\right)$ is a nonexpansive mapping.

Example 4.4 Let $T:C\to C$ be a κ-strictly pseudo-contractive mapping with $F\left(T\right)\ne \mathrm{\varnothing }$. Let $\left\{{\lambda }_{n}\right\}$ be a sequence of positive real numbers such that
$0<\underset{n\in \mathbb{N}}{inf}{\lambda }_{n}\le \underset{n\in \mathbb{N}}{sup}{\lambda }_{n}<1-\kappa \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty }}{lim}\left({\lambda }_{n+1}-{\lambda }_{n}\right)=0,$

and let $\left\{{T}_{n}\right\}$ be a sequence of mappings defined by ${T}_{n}={P}_{C}\left(I-{\lambda }_{n}\left(I-T\right)\right)$. Then $\left\{{T}_{n}\right\}$ is a strongly nonexpansive sequence satisfying the conditions (R) and (Z).

Proof Since T is a κ-strictly pseudo-contractive mapping, then $I-T$ is $\frac{1-\kappa }{2}$-inverse strongly monotone. From Example 4.3 in [10], we have $\left\{{T}_{n}\right\}$ is a strongly nonexpansive sequence satisfying the conditions (R) and (Z). □

Example 4.5 Let $A,B:C\to H$ be $\alpha ,\beta$-inverse strongly monotone mappings, respectively, with $\overline{\gamma }=min\left\{\alpha ,\beta \right\}$ and $\mathit{VI}\left(C,A\right)\cap \mathit{VI}\left(C,B\right)\ne \mathrm{\varnothing }$. Let $\left\{{\lambda }_{n}\right\}$ be a sequence of positive real numbers such that
$0<\underset{n\in \mathbb{N}}{inf}{\lambda }_{n}\le \underset{n\in \mathbb{N}}{sup}{\lambda }_{n}<2\overline{\gamma }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty }}{lim}\left({\lambda }_{n+1}-{\lambda }_{n}\right)=0,$

and let $\left\{{T}_{n}\right\}$ be a sequence of mappings defined by ${T}_{n}={P}_{C}\left(I-{\lambda }_{n}D\right)$, where $D=aA+\left(1-a\right)B$ for all $a\in \left(0,1\right)$. Then $\left\{{T}_{n}\right\}$ is a strongly nonexpansive sequence satisfying the conditions (R) and (Z).

Proof Let $x,y\in C$, then we have
$\begin{array}{rcl}〈x-y,Dx-Dy〉& =& 〈x-y,\left(aA+\left(1-a\right)B\right)x-\left(aA+\left(1-a\right)B\right)y〉\\ \ge & a〈x-y,Ax-Ay〉+\left(1-a\right)〈x-y,Bx-By〉\\ \ge & a\alpha {\parallel Ax-Ay\parallel }^{2}+\left(1-a\right)\beta {\parallel Bx-By\parallel }^{2}\\ \ge & \overline{\gamma }\left({\parallel aAx+\left(1-a\right)Bx-aAy-\left(1-a\right)By\parallel }^{2}\right)\\ \ge & \overline{\gamma }{\parallel Dx-Dy\parallel }^{2}.\end{array}$

Then D is a $\overline{\gamma }$-inverse strongly monotone mapping. From Example 4.3 in [10], we have that $\left\{{T}_{n}\right\}$ is a strongly nonexpansive sequence satisfying the conditions (R) and (Z). □

Example 4.6 Let $A:C\to H$ be an α-strongly monotone and L-Lipschitzian mapping with $\mathit{VI}\left(C,A\right)\ne \mathrm{\varnothing }$. Let $\left\{{\lambda }_{n}\right\}$ be a sequence of positive real numbers such that
$0<\underset{n\in \mathbb{N}}{inf}{\lambda }_{n}\le \underset{n\in \mathbb{N}}{sup}{\lambda }_{n}<\frac{2\alpha }{{L}^{2}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty }}{lim}\left({\lambda }_{n+1}-{\lambda }_{n}\right)=0,$

and let $\left\{{T}_{n}\right\}$ be a sequence of mappings defined by ${T}_{n}={P}_{C}\left(I-{\lambda }_{n}A\right)$. Then $\left\{{T}_{n}\right\}$ is a strongly nonexpansive sequence satisfying the conditions (R) and (Z).

Proof Let $x,y\in C$, then we have
$\begin{array}{rcl}〈x-y,Ax-Ay〉& \ge & \alpha {\parallel x-y\parallel }^{2}\\ \ge & \frac{\alpha }{{L}^{2}}{\parallel Ax-Ay\parallel }^{2}.\end{array}$

Then A is an $\frac{\alpha }{{L}^{2}}$-inverse strongly monotone mapping. From Example 4.3 in [10], we have that $\left\{{T}_{n}\right\}$ is a strongly nonexpansive sequence satisfying the conditions (R) and (Z). □

Example 4.7 (See [10])

Let $\left\{{R}_{n}\right\}$ be a sequence of nonexpansive mappings of C into itself having a common fixed point, and let $\left\{{\mu }_{n}\right\}$ be a sequence in $\left[0,1\right]$. For each $n\in \mathbb{N}$, a W-mapping [17]${T}_{n}$ generated by ${R}_{n},{R}_{n-1},\dots ,{R}_{1}$ and ${\mu }_{n},{\mu }_{n-1},\dots ,{\mu }_{1}$ is defined as follows:
$\begin{array}{c}{U}_{n,n}={\mu }_{n}{R}_{n}+\left(1-{\mu }_{n}\right)I,\hfill \\ {U}_{n,n-1}={\mu }_{n-1}{R}_{n-1}{U}_{n,n}+\left(1-{\mu }_{n-1}\right)I,\hfill \\ {U}_{n,n-2}={\mu }_{n-2}{R}_{n-2}{U}_{n,n-1}+\left(1-{\mu }_{n-2}\right)I,\hfill \\ ⋮\hfill \\ {U}_{n,k}={\mu }_{k}{R}_{k}{U}_{n,k+1}+\left(1-{\mu }_{k}\right)I,\hfill \\ ⋮\hfill \\ {U}_{n,2}={\mu }_{2}{R}_{2}{U}_{n,3}+\left(1-{\mu }_{2}\right)I,\hfill \\ {T}_{n}={U}_{n,1}={\mu }_{1}{R}_{1}{U}_{n,2}+\left(1-{\mu }_{1}\right)I.\hfill \end{array}$

If $0<{\mu }_{1}\le 1$ and $0<{\mu }_{n}\le b$, for all $n\ge 2$ and $0, then $\left\{{T}_{n}\right\}$ satisfies the conditions (R) and (Z).

By using our main result and these three examples, we obtain the following results.

Theorem 4.8 Let H be a Hilbert space, let C be a nonempty closed convex subset of H. Let ${F}_{1}$ and ${F}_{2}$ be two bifunctions from $C×C$ into satisfying (A1)-(A4), respectively. Let $T:C\to C$ be a κ-strictly pseudo-contractive mapping with $F\left(T\right)\ne \mathrm{\varnothing }$. Let $\left\{{\lambda }_{n}\right\}$ be a sequence of positive real numbers such that
$0<\underset{n\in \mathbb{N}}{inf}{\lambda }_{n}\le \underset{n\in \mathbb{N}}{sup}{\lambda }_{n}<1-\kappa \phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty }}{lim}\left({\lambda }_{n+1}-{\lambda }_{n}\right)=0,$
and let $\left\{{T}_{n}\right\}$ be a sequence of mappings defined by ${T}_{n}={P}_{C}\left(I-{\lambda }_{n}\left(I-T\right)\right)$. Let $\left\{{R}_{n}\right\}$ be a sequence of nonexpansive mappings of C into itself having a common fixed point, and let $\left\{{\mu }_{n}\right\}$ be a sequence in $\left[0,1\right]$. For each $n\in \mathbb{N}$, ${W}_{n}$ is a W-mapping generated by ${R}_{n},{R}_{n-1},\dots ,{R}_{1}$ and ${\mu }_{n},{\mu }_{n-1},\dots ,{\mu }_{1}$. Assume that $\mathbb{F}=\mathit{EP}\left({F}_{1}\right)\cap \mathit{EP}\left({F}_{2}\right)\cap F\left(\left\{{R}_{n}\right\}\right)\cap F\left(T\right)\ne \mathrm{\varnothing }$. Let $f:C\to C$ be a contractive mapping with $\alpha \in \left(0,\frac{1}{2}\right)$. Let $\left\{{x}_{n}\right\}$, $\left\{{u}_{n}\right\}$, $\left\{{v}_{n}\right\}$ be sequences generated by ${x}_{1},u,v\in C$ and
$\left\{\begin{array}{c}{F}_{1}\left({u}_{n},u\right)+\frac{1}{{r}_{n}}〈u-{u}_{n},{u}_{n}-{x}_{n}〉\ge 0,\hfill \\ {F}_{2}\left({v}_{n},v\right)+\frac{1}{{s}_{n}}〈v-{v}_{n},{v}_{n}-{x}_{n}〉\ge 0,\hfill \\ {y}_{n}={\delta }_{n}{u}_{n}+\left(1-{\delta }_{n}\right){v}_{n},\hfill \\ {x}_{n+1}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right){W}_{n}\left({\alpha }_{n}f\left({T}_{n}{y}_{n}\right)+\left(1-{\alpha }_{n}\right){T}_{n}{y}_{n}\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,\hfill \end{array}$
(4.3)
where $\left\{{\alpha }_{n}\right\},\left\{{\beta }_{n}\right\}\in \left[0,1\right]$, $\left\{{r}_{n}\right\},\left\{{s}_{n}\right\}\in \left(a,b\right)\in \left[0,1\right]$. Assume that the following conditions hold:
1. (i)

${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$;

2. (ii)

$0<{lim inf}_{n\to \mathrm{\infty }}{\beta }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\beta }_{n}<1$;

3. (iii)

${\sum }_{n=0}^{\mathrm{\infty }}|{r}_{n+1}-{r}_{n}|$, ${\sum }_{n=0}^{\mathrm{\infty }}|{s}_{n+1}-{s}_{n}|<\mathrm{\infty }$;

4. (iv)

${lim}_{n\to \mathrm{\infty }}{\delta }_{n}=\delta \in \left(0,1\right)$.

Then the sequences $\left\{{x}_{n}\right\}$, $\left\{{u}_{n}\right\}$, $\left\{{v}_{n}\right\}$, $\left\{{y}_{n}\right\}$ converge strongly to $z={P}_{\mathbb{F}}f\left(z\right)$.

Proof From Example 4.4, we have $\left\{{T}_{n}\right\}$ is a strongly nonexpansive sequence satisfying the conditions (R) and (Z). From Lemma 4.2, we have $F\left({T}_{n}\right)=F\left({P}_{C}\left(I-{\lambda }_{n}\left(I-T\right)\right)\right)=\mathit{VI}\left(C,I-T\right)=F\left(T\right)$ for all $n\in \mathbb{N}$. It implies that $F\left(\left\{{T}_{n}\right\}\right)=F\left(T\right)$. From [18], we have $F\left(\left\{{W}_{n}\right\}\right)=F\left(\left\{{R}_{n}\right\}\right)$. It follows that $\mathbb{F}=\mathit{EP}\left({F}_{1}\right)\cap \mathit{EP}\left({F}_{2}\right)\cap F\left(\left\{{W}_{n}\right\}\right)\cap F\left(\left\{{T}_{n}\right\}\right)\ne \mathrm{\varnothing }$. From Example 4.7, we have $\left\{{W}_{n}\right\}$ is a nonexpansive sequence satisfying the conditions (R) and (Z). By Theorem 3.1, we can conclude the desired result. □

Theorem 4.9 Let H be a Hilbert space, let C be a nonempty closed convex subset of H. Let ${F}_{1}$ and ${F}_{2}$ be two bifunctions from $C×C$ into satisfying (A1)-(A4), respectively. Let $A,B:C\to H$ be $\alpha ,\beta$-inverse strongly monotone mappings, respectively, with $\overline{\gamma }=min\left\{\alpha ,\beta \right\}$ and $\mathit{VI}\left(C,A\right)\cap \mathit{VI}\left(C,B\right)\ne \mathrm{\varnothing }$. Let $\left\{{\lambda }_{n}\right\}$ be a sequence of positive real numbers such that
$0<\underset{n\in \mathbb{N}}{inf}{\lambda }_{n}\le \underset{n\in \mathbb{N}}{sup}{\lambda }_{n}<2\overline{\gamma }\phantom{\rule{1em}{0ex}}\phantom{\rule{0.25em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty }}{lim}\left({\lambda }_{n+1}-{\lambda }_{n}\right)=0,$
and let $\left\{{T}_{n}\right\}$ be a sequence of mappings defined by ${T}_{n}={P}_{C}\left(I-{\lambda }_{n}D\right)$, where $D=aA+\left(1-a\right)B$ for all $a\in \left(0,1\right)$. Let $\left\{{R}_{n}\right\}$ be a sequence of nonexpansive mappings of C into itself having a common fixed point, and let $\left\{{\mu }_{n}\right\}$ be a sequence in $\left[0,1\right]$. For each $n\in \mathbb{N}$, ${W}_{n}$ is a W-mapping generated by ${R}_{n},{R}_{n-1},\dots ,{R}_{1}$ and ${\mu }_{n},{\mu }_{n-1},\dots ,{\mu }_{1}$. Assume that $\mathbb{F}=\mathit{EP}\left({F}_{1}\right)\cap \mathit{EP}\left({F}_{2}\right)\cap F\left(\left\{{R}_{n}\right\}\right)\cap \mathit{VI}\left(C,A\right)\cap \mathit{VI}\left(C,B\right)\ne \mathrm{\varnothing }$. Let $f:C\to C$ be a contractive mapping with $\alpha \in \left(0,\frac{1}{2}\right)$. Let $\left\{{x}_{n}\right\}$, $\left\{{u}_{n}\right\}$, $\left\{{v}_{n}\right\}$ be sequences generated by ${x}_{1},u,v\in C$ and
$\left\{\begin{array}{c}{F}_{1}\left({u}_{n},u\right)+\frac{1}{{r}_{n}}〈u-{u}_{n},{u}_{n}-{x}_{n}〉\ge 0,\hfill \\ {F}_{2}\left({v}_{n},v\right)+\frac{1}{{s}_{n}}〈v-{v}_{n},{v}_{n}-{x}_{n}〉\ge 0,\hfill \\ {y}_{n}={\delta }_{n}{u}_{n}+\left(1-{\delta }_{n}\right){v}_{n},\hfill \\ {x}_{n+1}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right){W}_{n}\left({\alpha }_{n}f\left({T}_{n}{y}_{n}\right)+\left(1-{\alpha }_{n}\right){T}_{n}{y}_{n}\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,\hfill \end{array}$
(4.4)
where $\left\{{\alpha }_{n}\right\},\left\{{\beta }_{n}\right\}\in \left[0,1\right]$, $\left\{{r}_{n}\right\},\left\{{s}_{n}\right\}\in \left(a,b\right)\in \left[0,1\right]$. Assume that the following conditions hold:
1. (i)

${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$;

2. (ii)

$0<{lim inf}_{n\to \mathrm{\infty }}{\beta }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\beta }_{n}<1$;

3. (iii)

${\sum }_{n=0}^{\mathrm{\infty }}|{r}_{n+1}-{r}_{n}|$, ${\sum }_{n=0}^{\mathrm{\infty }}|{s}_{n+1}-{s}_{n}|<\mathrm{\infty }$;

4. (iv)

${lim}_{n\to \mathrm{\infty }}{\delta }_{n}=\delta \in \left(0,1\right)$.

Then the sequences $\left\{{x}_{n}\right\}$, $\left\{{u}_{n}\right\}$, $\left\{{v}_{n}\right\}$, $\left\{{y}_{n}\right\}$ converge strongly to $z={P}_{\mathbb{F}}f\left(z\right)$.

Proof From Example 4.5, we have $\left\{{T}_{n}\right\}$ is a strongly nonexpansive sequence satisfying the conditions (R) and (Z). From Lemmas 4.2 and 4.3, we have $F\left({T}_{n}\right)=F\left({P}_{C}\left(I-{\lambda }_{n}D\right)\right)=\mathit{VI}\left(C,D\right)=\mathit{VI}\left(C,A\right)\cap \mathit{VI}\left(C,B\right)$ for all $n\in \mathbb{N}$. It implies that $F\left(\left\{{T}_{n}\right\}\right)=\mathit{VI}\left(C,A\right)\cap \mathit{VI}\left(C,B\right)$. From [18], we have $F\left(\left\{{W}_{n}\right\}\right)=F\left(\left\{{R}_{n}\right\}\right)$. It follows that $\mathbb{F}=\mathit{EP}\left({F}_{1}\right)\cap \mathit{EP}\left({F}_{2}\right)\cap F\left(\left\{{W}_{n}\right\}\right)\cap F\left(\left\{{T}_{n}\right\}\right)\ne \mathrm{\varnothing }$. From Example 4.7, we have $\left\{{W}_{n}\right\}$ is a nonexpansive sequence satisfying the conditions (R) and (Z). By Theorem 3.1, we can conclude the desired result. □

Theorem 4.10 Let H be a Hilbert space, let C be a nonempty closed convex subset of H. Let ${F}_{1}$ and ${F}_{2}$ be two bifunctions from $C×C$ into satisfying (A1)-(A4), respectively. Let $A:C\to H$ be an α-strongly monotone and L-Lipschitzian mapping with $\mathit{VI}\left(C,A\right)\ne \mathrm{\varnothing }$. Let $\left\{{\lambda }_{n}\right\}$ be a sequence of positive real numbers such that
$0<\underset{n\in \mathbb{N}}{inf}{\lambda }_{n}\le \underset{n\in \mathbb{N}}{sup}{\lambda }_{n}<\frac{2\alpha }{{L}^{2}}\phantom{\rule{1em}{0ex}}\phantom{\rule{0.25em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty }}{lim}\left({\lambda }_{n+1}-{\lambda }_{n}\right)=0,$
and let $\left\{{T}_{n}\right\}$ be a sequence of mappings defined by ${T}_{n}={P}_{C}\left(I-{\lambda }_{n}A\right)$. Let $\left\{{R}_{n}\right\}$ be a sequence of nonexpansive mappings of C into itself having a common fixed point, and let $\left\{{\mu }_{n}\right\}$ be a sequence in $\left[0,1\right]$. For each $n\in \mathbb{N}$, ${W}_{n}$ is a W-mapping generated by ${R}_{n},{R}_{n-1},\dots ,{R}_{1}$ and ${\mu }_{n},{\mu }_{n-1},\dots ,{\mu }_{1}$. Assume that $\mathbb{F}=\mathit{EP}\left({F}_{1}\right)\cap \mathit{EP}\left({F}_{2}\right)\cap F\left(\left\{{R}_{n}\right\}\right)\cap \mathit{VI}\left(C,A\right)\ne \mathrm{\varnothing }$. Let $f:C\to C$ be a contractive mapping with $\alpha \in \left(0,\frac{1}{2}\right)$. Let $\left\{{x}_{n}\right\}$, $\left\{{u}_{n}\right\}$, $\left\{{v}_{n}\right\}$ be sequences generated by ${x}_{1},u,v\in C$ and
$\left\{\begin{array}{c}{F}_{1}\left({u}_{n},u\right)+\frac{1}{{r}_{n}}〈u-{u}_{n},{u}_{n}-{x}_{n}〉\ge 0,\hfill \\ {F}_{2}\left({v}_{n},v\right)+\frac{1}{{s}_{n}}〈v-{v}_{n},{v}_{n}-{x}_{n}〉\ge 0,\hfill \\ {y}_{n}={\delta }_{n}{u}_{n}+\left(1-{\delta }_{n}\right){v}_{n},\hfill \\ {x}_{n+1}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right){W}_{n}\left({\alpha }_{n}f\left({T}_{n}{y}_{n}\right)+\left(1-{\alpha }_{n}\right){T}_{n}{y}_{n}\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,\hfill \end{array}$
(4.5)
where $\left\{{\alpha }_{n}\right\},\left\{{\beta }_{n}\right\}\in \left[0,1\right]$, $\left\{{r}_{n}\right\},\left\{{s}_{n}\right\}\in \left(a,b\right)\in \left[0,1\right]$. Assume that the following conditions hold:
1. (i)

${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$;

2. (ii)

$0<{lim inf}_{n\to \mathrm{\infty }}{\beta }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\beta }_{n}<1$;

3. (iii)

${\sum }_{n=0}^{\mathrm{\infty }}|{r}_{n+1}-{r}_{n}|$, ${\sum }_{n=0}^{\mathrm{\infty }}|{s}_{n+1}-{s}_{n}|<\mathrm{\infty }$;

4. (iv)

${lim}_{n\to \mathrm{\infty }}{\delta }_{n}=\delta \in \left(0,1\right)$.

Then the sequences $\left\{{x}_{n}\right\}$, $\left\{{u}_{n}\right\}$, $\left\{{v}_{n}\right\}$, $\left\{{y}_{n}\right\}$ converge strongly to $z={P}_{\mathbb{F}}f\left(z\right)$.

Proof From Example 4.6, we have $\left\{{T}_{n}\right\}$ is a strongly nonexpansive sequence satisfying the conditions (R) and (Z). From Lemma 4.2, we have $F\left({T}_{n}\right)=F\left({P}_{C}\left(I-{\lambda }_{n}A\right)\right)=\mathit{VI}\left(C,A\right)$ for all $n\in \mathbb{N}$. It implies that $F\left(\left\{{T}_{n}\right\}\right)=\mathit{VI}\left(C,A\right)$. From [18], we have $F\left(\left\{{W}_{n}\right\}\right)=F\left(\left\{{R}_{n}\right\}\right)$. It follows that $\mathbb{F}=\mathit{EP}\left({F}_{1}\right)\cap \mathit{EP}\left({F}_{2}\right)\cap F\left(\left\{{W}_{n}\right\}\right)\cap F\left(\left\{{T}_{n}\right\}\right)\ne \mathrm{\varnothing }$. From Example 4.7, we have $\left\{{W}_{n}\right\}$ is a nonexpansive sequence satisfying the conditions (R) and (Z). By Theorem 3.1, we can conclude the desired result. □

Theorem 4.11 Let H be a Hilbert space, let C be a nonempty closed convex subset of H. Let ${F}_{1}$ be a bifunction from $C×C$ into satisfying (A1)-(A4), and let $\left\{{S}_{n}\right\}$ and $\left\{{T}_{n}\right\}$ be sequences of nonexpansive self-mappings of C with $\mathbb{F}=\mathit{EP}\left({F}_{1}\right)\cap F\left(\left\{{S}_{n}\right\}\right)\cap F\left(\left\{{T}_{n}\right\}\right)\ne \mathrm{\varnothing }$. Let $\left\{{T}_{n}\right\}$ or $\left\{{S}_{n}\right\}$ be a sequence of strongly nonexpansive mappings, and let $f:C\to C$ be a contractive mapping with $\alpha \in \left(0,\frac{1}{2}\right)$. Let $\left\{{x}_{n}\right\}$, $\left\{{u}_{n}\right\}$ be sequences generated by ${x}_{1},u\in C$ and
$\left\{\begin{array}{c}{F}_{1}\left({u}_{n},u\right)+\frac{1}{{r}_{n}}〈u-{u}_{n},{u}_{n}-{x}_{n}〉\ge 0,\hfill \\ {x}_{n+1}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right){S}_{n}\left({\alpha }_{n}f\left({T}_{n}{u}_{n}\right)+\left(1-{\alpha }_{n}\right){T}_{n}{u}_{n}\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,\hfill \end{array}$
(4.6)
where $\left\{{\alpha }_{n}\right\},\left\{{\beta }_{n}\right\}\in \left[0,1\right]$, $\left\{{r}_{n}\right\},\left\{{s}_{n}\right\}\in \left(a,b\right)\in \left[0,1\right]$. Assume that the following conditions hold:
1. (i)

${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$;

2. (ii)

$0<{lim inf}_{n\to \mathrm{\infty }}{\beta }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\beta }_{n}<1$;

3. (iii)

${\sum }_{n=0}^{\mathrm{\infty }}|{r}_{n+1}-{r}_{n}|<\mathrm{\infty }$;

4. (iv)

$\left\{{S}_{n}\right\}$ and $\left\{{T}_{n}\right\}$ satisfy the conditions R and Z.

Then the sequences $\left\{{x}_{n}\right\}$, $\left\{{u}_{n}\right\}$ converge strongly to $z={P}_{\mathbb{F}}f\left(z\right)$.

Proof Put ${F}_{1}\equiv {F}_{2}$, ${s}_{n}={r}_{n}$ and ${u}_{n}={v}_{n}$. From Theorem 3.1, we can conclude the desired conclusion. □

The following result can be obtained from Theorem 3.1. We, therefore, omit the proof.

Theorem 4.12 Let H be a Hilbert space, let C be a nonempty closed convex subset of H. Let ${F}_{i}$ be bifunctions from $C×C$ into , for every $i=1,2,\dots ,N$, satisfying (A1)-(A4), and let $\left\{{S}_{n}\right\}$ and $\left\{{T}_{n}\right\}$ be sequences of nonexpansive self-mappings of C with $\mathbb{F}={\bigcap }_{i=1}^{N}\mathit{EP}\left({F}_{i}\right)\cap F\left(\left\{{S}_{n}\right\}\right)\cap F\left(\left\{{T}_{n}\right\}\right)\ne \mathrm{\varnothing }$. Let $\left\{{T}_{n}\right\}$ or $\left\{{S}_{n}\right\}$ be a sequence of strongly nonexpansive mappings, and let $f:C\to C$ be a contractive mapping with $\alpha \in \left(0,\frac{1}{2}\right)$. Let $\left\{{x}_{n}\right\}$, $\left\{{u}_{n}\right\}$, $\left\{{v}_{n}\right\}$ be sequences generated by ${x}_{1},{u}^{i}\in C$, for every $i\in 1,2,\dots ,N$, and
$\left\{\begin{array}{c}{F}_{i}\left({u}_{n}^{i},{u}^{i}\right)+\frac{1}{{r}_{n}^{i}}〈u-{u}_{n}^{i},{u}_{n}^{i}-{x}_{n}〉\ge 0,\hfill \\ {y}_{n}={\sum }_{i=1}^{N}{\delta }_{n}^{i}{u}_{n}^{i},\hfill \\ {x}_{n+1}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right){S}_{n}\left({\alpha }_{n}f\left({T}_{n}{y}_{n}\right)+\left(1-{\alpha }_{n}\right){T}_{n}{y}_{n}\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,\hfill \end{array}$
(4.7)
where $\left\{{\alpha }_{n}\right\},\left\{{\beta }_{n}\right\}\in \left[0,1\right]$, $\left\{{r}_{n}\right\},\left\{{s}_{n}\right\}\in \left(a,b\right)\in \left[0,1\right]$. Assume that the following conditions hold:
1. (i)

${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$;

2. (ii)

$0<{lim inf}_{n\to \mathrm{\infty }}{\beta }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\beta }_{n}<1$;

3. (iii)

${\sum }_{n=0}^{\mathrm{\infty }}|{r}_{n+1}^{i}-{r}_{n}^{i}|<\mathrm{\infty }$, $\mathrm{\forall }i=1,2,\dots ,N$;

4. (iv)

${\sum }_{i=1}^{N}{\delta }_{n}^{i}=1$;

5. (v)

${lim}_{n\to \mathrm{\infty }}{\delta }_{n}^{i}={\delta }^{i}\in \left(0,1\right)$, $\mathrm{\forall }i=1,2,\dots ,N$;

6. (vi)

$\left\{{S}_{n}\right\}$ and $\left\{{T}_{n}\right\}$ satisfy the conditions R and Z.

Then the sequences $\left\{{x}_{n}\right\}$, $\left\{{y}_{n}\right\}$ and $\left\{{u}_{n}^{i}\right\}$, for every $i=1,2,\dots ,N$, converge strongly to $z={P}_{\mathbb{F}}f\left(z\right)$.

## Declarations

### Acknowledgements

This research was supported by the Research Administration Division of King Mongkut’s Institute of Technology Ladkrabang.

## Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, King Mongkut’s Institute of Technology Ladkrabang, Bangkok, 10520, Thailand

## References

1. Aoyama K, Kimura Y, Takahashi W, Toyoda M: On a strongly nonexpansive sequence in Hilbert spaces. J. Nonlinear Convex Anal. 2007, 8: 471–489.
2. Aoyama K: An iterative method for fixed point problems for sequences of nonexpansive mappings. In Fixed Point Theory and Applications. Yokohama Publ., Yokohama; 2010:1–7.Google Scholar
3. Blum E, Oettli W: From optimization and variational inequalities to equilibrium problems. Math. Stud. 1994, 63(1–4):123–145.
4. Combettes PL, Hirstoaga SA: Equilibrium programming in Hilbert spaces. J. Nonlinear Convex Anal. 2005, 6(1):117–136.
5. Kangtunyakarn A: Iterative methods for finding common solution of generalized equilibrium problems and variational inequality problems and fixed point problems of a finite family of nonexpansive mappings. Fixed Point Theory Appl. 2010., 2010: Article ID 836714 10.1155/2010/836714Google Scholar
6. Kangtunyakarn A: Hybrid algorithm for finding common elements of the set of generalized equilibrium problems and the set of fixed point problems of strictly pseudocontractive mapping. Fixed Point Theory Appl. 2011., 2011: Article ID 274820 10.1155/2011/274820Google Scholar
7. Cholamjiak W, Suantai S: A hybrid method for a countable family of multivalued maps, equilibrium problems, and variational inequality problems. Discrete Dyn. Nat. Soc. 2010., 2010: Article ID 349158 10.1155/2010/349158Google Scholar
8. Takahashi W, Zembayashi K: Strong convergence theorem by a new hybrid method for equilibrium problems and relatively nonexpansive mappings. Fixed Point Theory Appl. 2008., 2008: Article ID 528476 10.1155/2008/528476Google Scholar
9. Takahashi S, Takahashi W: Viscosity approximation methods for equilibrium problems and fixed point problems in Hilbert spaces. J. Math. Anal. Appl. 2007, 331(1):506–515. 10.1016/j.jmaa.2006.08.036
10. Aoyama K, Kimura Y: Strong convergence theorems for strongly nonexpansive sequences. Appl. Math. Comput. 2011, 217: 7537–7545. 10.1016/j.amc.2011.01.092
11. Browder FE: Convergence of approximants to fixed points of nonexpansive nonlinear mappings in Banach space. Arch. Ration. Mech. Anal. 1967, 24: 82–89.
12. Xu HK: An iterative approach to quadratic optimization. J. Optim. Theory Appl. 2003, 116(3):659–678. 10.1023/A:1023073621589
13. Suzuki T: Strong convergence of Krasnoselskii and Mann’s type sequences for one-parameter nonexpansive semigroups without Bochner integrals. J. Math. Anal. Appl. 2005, 305(1):227–239. 10.1016/j.jmaa.2004.11.017
14. Bruck RE: Properties of fixed point sets of nonexpansive mappings in Banach spaces. Trans. Am. Math. Soc. 1973, 179: 251–262.
15. Kangtunyakarn A: Convergence theorem of κ -strictly pseudocontractive mapping and a modification of generalized equilibrium problems. Fixed Point Theory Appl. 2012., 2012: Article ID 89Google Scholar
16. Takahashi W: Nonlinear Functional Analysis. Yokohama Publ., Yokohama; 2000.
17. Takahashi W: Weak and strong convergence theorems for families of nonexpansive mappings and their applications. Proceedings of Workshop on Fixed Point Theory Kazimierz Dolny 1997, 277–292.Google Scholar
18. Atsushiba S, Takahashi W: Strong convergence theorems for a finite family of nonexpansive mappings and applications. Indian J. Math. 1999, 41: 435–453.MATH