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# Strong convergence of the hybrid method for a finite family of nonspreading mappings and variational inequality problems

Fixed Point Theory and Applications20122012:188

https://doi.org/10.1186/1687-1812-2012-188

• Accepted: 5 October 2012
• Published:

## Abstract

In this paper, we prove a strong convergence theorem by the hybrid method for finding a common element of the set of fixed points of a finite family of nonspreading mappings and the set of solutions of a finite family of variational inequality problems.

## Keywords

• quasi-nonexpansive mapping
• S-mapping

## 1 Introduction

Let C be a nonempty closed convex subset of a real Hilbert space H. Then a mapping $T:C\to C$ is said to be nonexpansive if $\parallel Tx-Ty\parallel \le \parallel x-y\parallel$ for all $x,y\in C$. Recall that the mapping $T:C\to C$ is said to be quasi-nonexpansive if $\parallel Tx-p\parallel \le \parallel x-p\parallel$, $\mathrm{\forall }x\in C$ and $\mathrm{\forall }p\in F\left(T\right)$, where $F\left(T\right)$ denotes the set of fixed points of T. In 2008, Kohsaka and Takahashi [1] introduced the mapping T called the nonspreading mapping in Hilbert spaces H and defined it as follows: $2{\parallel Tx-Ty\parallel }^{2}\le {\parallel Tx-y\parallel }^{2}+{\parallel x-Ty\parallel }^{2}$, $\mathrm{\forall }x,y\in C$.

Let $A:C\to H$. The variational inequality problem is to find a point $u\in C$ such that
$〈Au,v-u〉\ge 0$
(1.1)

for all $v\in C$. The set of solutions of (1.1) is denoted by $VI\left(C,A\right)$.

The variational inequality has emerged as a fascinating and interesting branch of mathematical and engineering sciences with a wide range of applications in industry, finance, economics, social, ecology, regional, pure and applied sciences; see, e.g., [25].

A mapping A of C into H is called inverse-strongly monotone (see [6]) if there exists a positive real number α such that
$〈x-y,Ax-Ay〉\ge \alpha {\parallel Ax-Ay\parallel }^{2}$
for all $x,y\in C$. Throughout this paper, we will use the following notation:
1. 1.

for weak convergence and → for strong convergence.

2. 2.

$\omega \left({x}_{n}\right)=\left\{x:\mathrm{\exists }{x}_{{n}_{i}}⇀x\right\}$ denotes the weak ω-limit set of $\left\{{x}_{n}\right\}$.

In 2008, Takahashi, Takeuchi and Kubota [7] proved the following strong convergence theorems by using the hybrid method for nonexpansive mappings in Hilbert spaces.

Theorem 1.1 Let H be a Hilbert space and C be a nonempty closed convex subset of H. Let T be a nonexpansive mapping of C into H such that $F\left(T\right)\ne \mathrm{\varnothing }$ and let ${x}_{0}\in H$. For ${C}_{1}=C$ and ${u}_{1}\in {P}_{{C}_{1}}{x}_{0}$, define a sequence $\left\{{u}_{n}\right\}$ of C as follows:
$\left\{\begin{array}{c}{y}_{n}={\alpha }_{n}{u}_{n}+\left(1-{\alpha }_{n}\right){u}_{n},\hfill \\ {C}_{n+1}=\left\{z\in {C}_{n}:\parallel {y}_{n}-z\parallel \le \parallel {u}_{n}-z\parallel \right\},\hfill \\ {u}_{n+1}={P}_{{C}_{n+1}}{x}_{0},\phantom{\rule{1em}{0ex}}n\in \mathbb{N},\hfill \end{array}$

where $0\le {\alpha }_{n}\le a<1$ for all $n\in \mathbb{N}$. Then $\left\{{u}_{n}\right\}$ converges strongly to ${z}_{0}={P}_{F\left(T\right)}{x}_{0}$.

In 2009, Iemoto and Takahashi [8] proved the convergence theorem of nonexpansive and nonspreading mappings as follows.

Theorem 1.2 Let H be a Hilbert space, and let C be a nonempty closed convex subset of H. Let S be a nonspreading mapping of C into itself, and let T be a nonexpansive mapping of C into itself such that $F\left(S\right)\cap F\left(T\right)\ne \mathrm{\varnothing }$. Define a sequence $\left\{{x}_{n}\right\}$ as follows.
$\left\{\begin{array}{c}{x}_{1}\in C,\hfill \\ {x}_{n+1}={\alpha }_{n}{x}_{n}+\left(1-{\alpha }_{n}\right)\left({\beta }_{n}S{x}_{n}+\left(1-{\beta }_{n}\right)T{x}_{n}\right)\hfill \end{array}$
for all $n\in \mathbb{N}$, where $\left\{{\alpha }_{n}\right\},\left\{{\beta }_{n}\right\}\subset \left[0,1\right]$. Then the following hold:
1. (i)

If ${lim inf}_{n\to \mathrm{\infty }}{\alpha }_{n}\left(1-{\alpha }_{n}\right)>0$ and ${\sum }_{n=1}^{\mathrm{\infty }}\left(1-{\beta }_{n}\right)<\mathrm{\infty }$, then $\left\{{x}_{n}\right\}$ converges weakly to $v\in F\left(S\right)$.

2. (ii)

If ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}\left(1-{\alpha }_{n}\right)=\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\beta }_{n}<\mathrm{\infty }$, then $\left\{{x}_{n}\right\}$ converges weakly to $v\in F\left(T\right)$.

3. (iii)

If ${lim inf}_{n\to \mathrm{\infty }}{\alpha }_{n}\left(1-{\alpha }_{n}\right)>0$ and ${lim inf}_{n\to \mathrm{\infty }}{\beta }_{n}\left(1-{\beta }_{n}\right)>0$, then $\left\{{x}_{n}\right\}$ converges weakly to $v\in F\left(S\right)\cap F\left(T\right)$.

Inspired and motivated by these facts and the research in this direction, we prove the strong convergence theorem by the hybrid method for finding a common element of the set of fixed points of a finite family of nonspreading mappings and the set of solutions of a finite family of variational inequality problems.

## 2 Preliminaries

In this section, we collect and give some useful lemmas that will be used for our main result in the next section.

Let C be a closed convex subset of a real Hilbert space H, let ${P}_{C}$ be the metric projection of H onto C, i.e., for $x\in H$, ${P}_{C}x$ satisfies the property
$\parallel x-{P}_{C}x\parallel =\underset{y\in C}{min}\parallel x-y\parallel .$

The following characterizes the projection ${P}_{C}$.

Lemma 2.1 (See [9])

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 [8])

Let C be a nonempty closed convex subset of H. Then a mapping $S:C\to C$ is nonspreading if and only if
${\parallel Sx-Sy\parallel }^{2}\le {\parallel x-y\parallel }^{2}+2〈x-Sx,y-Sy〉$

for all $x,y\in C$.

Example 2.3 Let denote the reals with the usual norm. Let $T:\mathcal{R}\to \mathcal{R}$ be defined by

for all $x\in \mathcal{R}$.

To see that T is a nonspreading mapping, if $x,y\in \left(0,\mathrm{\infty }\right)$, then we have $Tx=-\left(x+1\right)$ and $Ty=-\left(y+1\right)$. From the definition of the mapping T, we have
$\begin{array}{rcl}{|Tx-Ty|}^{2}& =& {|-\left(x+1\right)-\left(-\left(y+1\right)\right)|}^{2}\\ =& {|y-x|}^{2}={|x-y|}^{2}\end{array}$
and
The above implies that
${|Tx-Ty|}^{2}={|x-y|}^{2}<{|x-y|}^{2}+2〈x-Tx,y-Ty〉.$
For every $x,y\in \left(-\mathrm{\infty },0\right]$, we have $Tx=x-1$ and $Ty=y-1$. From the definition of T, we have
$\begin{array}{rcl}{|Tx-Ty|}^{2}& =& {|x-1-\left(y-1\right)|}^{2}\\ =& {|x-y|}^{2},\end{array}$
and
$2〈x-Tx,y-Ty〉=2〈x-\left(x-1\right),y-\left(y-1\right)〉=2.$
From above, we have
${|Tx-Ty|}^{2}={|x-y|}^{2}<{|x-y|}^{2}+2〈x-Tx,y-Ty〉.$
Finally, for every $x\in \left(-\mathrm{\infty },0\right]$ and $y\in \left(0,\mathrm{\infty }\right)$, we have $Tx=x-1$ and $Ty=-\left(y+1\right)$. From the definition of T, we have
and
From above, we have
$\begin{array}{rcl}{|Tx-Ty|}^{2}& =& {|x+y|}^{2}={\left(x+y\right)}^{2}\\ \le & {|x-y|}^{2}\\ <& {|x-y|}^{2}+2〈x-Tx,y-Ty〉.\end{array}$
Hence, for all $x,y\in \mathcal{R}$, we have
${|Tx-Ty|}^{2}<{|x-y|}^{2}+2〈x-Tx,y-Ty〉.$

Then T is a nonspreading mapping.

Lemma 2.4 (See [1])

Let H be a Hilbert space, let C be a nonempty closed convex subset of H, and let S be a nonspreading mapping of C into itself. Then $F\left(S\right)$ is closed and convex.

Lemma 2.5 (See [9])

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 $\lambda >0$,
$u={P}_{C}\left(I-\lambda A\right)u\phantom{\rule{1em}{0ex}}⇔\phantom{\rule{1em}{0ex}}u\in VI\left(C,A\right),$

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

Lemma 2.6 (See [10])

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

Lemma 2.7 (See [11])

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

Lemma 2.8 (See [12])

Let C be a closed convex subset of H. Let $\left\{{x}_{n}\right\}$ be a sequence in H and $u\in H$. Let $q={P}_{C}u$. If $\left\{{x}_{n}\right\}$ is such that $\omega \left({x}_{n}\right)\subset C$ and satisfies the condition
$\parallel {x}_{n}-u\parallel \le \parallel u-q\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\in \mathbb{N},$

then ${x}_{n}\to q$, as $n\to \mathrm{\infty }$.

In 2009, Kangtunyakarn and Suantai [13] introduced an S-mapping generated by ${T}_{1},\dots ,{T}_{N}$ and ${\lambda }_{1},\dots ,{\lambda }_{N}$ as follows.

Definition 2.1 Let C be a nonempty convex subset of a real Banach space. Let ${\left\{{T}_{i}\right\}}_{i=1}^{N}$ be a finite family of (nonexpansive) mappings of C into itself. For each $j=1,2,\dots ,N$, let ${\alpha }_{j}=\left({\alpha }_{1}^{j},{\alpha }_{2}^{j},{\alpha }_{3}^{j}\right)\in I×I×I$, where $I\in \left[0,1\right]$ and ${\alpha }_{1}^{j}+{\alpha }_{2}^{j}+{\alpha }_{3}^{j}=1$. Define the mapping $S:C\to C$ as follows:
(2.1)
(2.2)

This mapping is called an S-mapping generated by ${T}_{1},\dots ,{T}_{N}$ and ${\alpha }_{1},{\alpha }_{2},\dots ,{\alpha }_{N}$.

The next lemma is very useful for our consideration.

Lemma 2.9 Let C be a nonempty closed convex subset of a real Hilbert space. Let ${\left\{{T}_{i}\right\}}_{i=1}^{N}$ be a finite family of nonspreading mappings of C into C with ${\bigcap }_{i=1}^{N}F\left({T}_{i}\right)\ne \mathrm{\varnothing }$, and let ${\alpha }_{j}=\left({\alpha }_{1}^{j},{\alpha }_{2}^{j},{\alpha }_{3}^{j}\right)\in I×I×I$, $j=1,2,3,\dots ,N$, where $I=\left[0,1\right]$, ${\alpha }_{1}^{j}+{\alpha }_{2}^{j}+{\alpha }_{3}^{j}=1$, ${\alpha }_{1}^{j},{\alpha }_{3}^{j}\in \left(0,1\right)$ for all $j=1,2,\dots ,N-1$ and ${\alpha }_{1}^{N}\in \left(0,1\right]$, ${\alpha }_{3}^{N}\in \left[0,1\right)$, ${\alpha }_{2}^{j}\in \left[0,1\right)$ for all $j=1,2,\dots ,N$. Let S be the mapping generated by ${T}_{1},\dots ,{T}_{N}$ and ${\alpha }_{1},{\alpha }_{2},\dots ,{\alpha }_{N}$. Then $F\left(S\right)={\bigcap }_{i=1}^{N}F\left({T}_{i}\right)$ and S is a quasi-nonexpansive mapping.

Proof It easy to see that ${\bigcap }_{i=1}^{N}F\left({T}_{i}\right)\subseteq F\left(S\right)$. Let ${x}_{0}\in F\left(S\right)$ and ${x}^{\ast }\in {\bigcap }_{i=1}^{N}F\left({T}_{i}\right)$. Since ${\left\{{T}_{i}\right\}}_{i=1}^{N}$ is a finite family of nonspreading mappings of C into itself, for every $y\in C$, we have
${\parallel {T}_{i}y-{x}^{\ast }\parallel }^{2}\le \frac{1}{2}\left({\parallel {T}_{i}y-{x}^{\ast }\parallel }^{2}+{\parallel y-{x}^{\ast }\parallel }^{2}\right).$
(2.3)
This implies that
(2.4)
From the definition of S and (2.4),
$\begin{array}{rcl}{\parallel S{x}_{0}-{x}^{\ast }\parallel }^{2}& =& {\parallel {\alpha }_{1}^{N}{T}_{N}{U}_{N-1}{x}_{0}+{\alpha }_{2}^{N}{U}_{N-1}{x}_{0}+{\alpha }_{3}^{N}{x}_{0}-{x}^{\ast }\parallel }^{2}\\ =& {\parallel {\alpha }_{1}^{N}\left({T}_{N}{U}_{N-1}{x}_{0}-{x}^{\ast }\right)+{\alpha }_{2}^{N}\left({U}_{N-1}{x}_{0}-{x}^{\ast }\right)+{\alpha }_{3}^{N}\left({x}_{0}-{x}^{\ast }\right)\parallel }^{2}\\ \le & {\alpha }_{1}^{N}{\parallel {T}_{N}{U}_{N-1}{x}_{0}-{x}^{\ast }\parallel }^{2}+{\alpha }_{2}^{N}{\parallel {U}_{N-1}{x}_{0}-{x}^{\ast }\parallel }^{2}+{\alpha }_{3}^{N}{\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}\\ \le & \left(1-{\alpha }_{3}^{N}\right){\parallel {U}_{N-1}{x}_{0}-{x}^{\ast }\parallel }^{2}+{\alpha }_{3}^{N}{\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}\\ =& \left(1-{\alpha }_{3}^{N}\right)\parallel {\alpha }_{1}^{N-1}\left({T}_{N-1}{U}_{N-2}{x}_{0}-{x}^{\ast }\right)+{\alpha }_{2}^{N-1}\left({U}_{N-2}{x}_{0}-{x}^{\ast }\right)\\ +{\alpha }_{3}^{N-1}\left({x}_{0}-{x}^{\ast }\right){\parallel }^{2}+{\alpha }_{3}^{N}{\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}\\ \le & \left(1-{\alpha }_{3}^{N}\right)\left({\alpha }_{1}^{N-1}{\parallel {T}_{N-1}{U}_{N-2}{x}_{0}-{x}^{\ast }\parallel }^{2}+{\alpha }_{2}^{N-1}{\parallel {U}_{N-2}{x}_{0}-{x}^{\ast }\parallel }^{2}\\ +{\alpha }_{3}^{N-1}{\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}\right)+{\alpha }_{3}^{N}{\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}\\ \le & \left(1-{\alpha }_{3}^{N}\right)\left(\left(1-{\alpha }_{3}^{N-1}\right){\parallel {U}_{N-2}{x}_{0}-{x}^{\ast }\parallel }^{2}+{\alpha }_{3}^{N-1}{\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}\right)\\ +{\alpha }_{3}^{N}{\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}\\ =& \left(1-{\alpha }_{3}^{N}\right)\left(1-{\alpha }_{3}^{N-1}\right){\parallel {U}_{N-2}{x}_{0}-{x}^{\ast }\parallel }^{2}+{\alpha }_{3}^{N-1}\left(1-{\alpha }_{3}^{N}\right){\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}\\ +{\alpha }_{3}^{N}{\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}\\ =& \prod _{j=N-1}^{N}\left(1-{\alpha }_{3}^{j}\right){\parallel {U}_{N-2}{x}_{0}-{x}^{\ast }\parallel }^{2}+\left(1-\prod _{j=N-1}^{N}\left(1-{\alpha }_{3}^{j}\right)\right){\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}\\ ⋮\\ \le & \prod _{j=3}^{N}\left(1-{\alpha }_{3}^{j}\right){\parallel {U}_{2}{x}_{0}-{x}^{\ast }\parallel }^{2}+\left(1-\prod _{j=3}^{N}\left(1-{\alpha }_{3}^{j}\right)\right){\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}\\ \le & \prod _{j=2}^{N}\left(1-{\alpha }_{3}^{j}\right){\parallel {U}_{1}{x}_{0}-{x}^{\ast }\parallel }^{2}+\left(1-\prod _{j=2}^{N}\left(1-{\alpha }_{3}^{j}\right)\right){\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}\\ =& \prod _{j=2}^{N}\left(1-{\alpha }_{3}^{j}\right){\parallel {\alpha }_{1}^{1}\left({T}_{1}{x}_{0}-{x}^{\ast }\right)+\left(1-{\alpha }_{1}^{1}\right)\left({x}_{0}-{x}^{\ast }\right)\parallel }^{2}\\ +\left(1-\prod _{j=2}^{N}\left(1-{\alpha }_{3}^{j}\right)\right){\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}\\ =& \prod _{j=2}^{N}\left(1-{\alpha }_{3}^{j}\right)\left({\alpha }_{1}^{1}{\parallel {T}_{1}{x}_{0}-{x}^{\ast }\parallel }^{2}+\left(1-{\alpha }_{1}^{1}\right){\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}\\ -{\alpha }_{1}^{1}\left(1-{\alpha }_{1}^{1}\right)\parallel {T}_{1}{x}_{0}-{x}_{0}\parallel \right)+\left(1-\prod _{j=2}^{N}\left(1-{\alpha }_{3}^{j}\right)\right){\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}\\ \le & \prod _{j=2}^{N}\left(1-{\alpha }_{3}^{j}\right)\left({\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}-{\alpha }_{1}^{1}\left(1-{\alpha }_{1}^{1}\right){\parallel {T}_{1}{x}_{0}-{x}_{0}\parallel }^{2}\right)\\ +\left(1-\prod _{j=2}^{N}\left(1-{\alpha }_{3}^{j}\right)\right){\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}.\end{array}$
(2.5)
From (2.5), we have
$\begin{array}{rcl}{\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}& \le & \prod _{j=2}^{N}\left(1-{\alpha }_{3}^{j}\right)\left({\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}-{\alpha }_{1}^{1}\left(1-{\alpha }_{1}^{1}\right){\parallel {T}_{1}{x}_{0}-{x}_{0}\parallel }^{2}\right)\\ +\left(1-\prod _{j=2}^{N}\left(1-{\alpha }_{3}^{j}\right)\right){\parallel {x}_{0}-{x}^{\ast }\parallel }^{2},\end{array}$
which implies that
${\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}\le {\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}-{\alpha }_{1}^{1}\left(1-{\alpha }_{1}^{1}\right){\parallel {T}_{1}{x}_{0}-{x}_{0}\parallel }^{2}.$
(2.6)
Since ${\alpha }_{1}^{j}\in \left(0,1\right)$ for all $j=1,2,\dots ,N-1$ and (2.6), we have ${x}_{0}\in F\left({T}_{1}\right)$. From ${x}_{0}={T}_{1}{x}_{0}$ and the definition of S, we have
${U}_{1}{x}_{0}={\alpha }_{1}^{1}{T}_{1}{x}_{0}+{\alpha }_{2}^{1}{x}_{0}+{\alpha }_{3}^{1}{x}_{0}={x}_{0}.$
From (2.5) and ${x}_{0}\in F\left({U}_{1}\right)$, we have
$\begin{array}{rcl}{\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}& \le & \prod _{j=3}^{N}\left(1-{\alpha }_{3}^{j}\right){\parallel {U}_{2}{x}_{0}-{x}^{\ast }\parallel }^{2}+\left(1-\prod _{j=3}^{N}\left(1-{\alpha }_{3}^{j}\right)\right){\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}\\ =& \prod _{j=3}^{N}\left(1-{\alpha }_{3}^{j}\right){\parallel {\alpha }_{1}^{2}{T}_{2}{U}_{1}{x}_{0}+{\alpha }_{2}^{2}{U}_{1}{x}_{0}+{\alpha }_{3}^{2}{x}_{0}-{x}^{\ast }\parallel }^{2}\\ +\left(1-\prod _{j=3}^{N}\left(1-{\alpha }_{3}^{j}\right)\right){\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}\\ =& \prod _{j=3}^{N}\left(1-{\alpha }_{3}^{j}\right){\parallel {\alpha }_{1}^{2}\left({T}_{2}{x}_{0}-{x}^{\ast }\right)+\left(1-{\alpha }_{1}^{2}\right)\left({x}_{0}-{x}^{\ast }\right)\parallel }^{2}\\ +\left(1-\prod _{j=3}^{N}\left(1-{\alpha }_{3}^{j}\right)\right){\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}\\ =& \prod _{j=3}^{N}\left(1-{\alpha }_{3}^{j}\right)\left({\alpha }_{1}^{2}{\parallel {T}_{2}{x}_{0}-{x}^{\ast }\parallel }^{2}+\left(1-{\alpha }_{1}^{2}\right){\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}\\ -{\alpha }_{1}^{2}\left(1-{\alpha }_{1}^{2}\right){\parallel {T}_{2}{x}_{0}-{x}_{0}\parallel }^{2}\right)\\ +\left(1-\prod _{j=3}^{N}\left(1-{\alpha }_{3}^{j}\right)\right){\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}\\ \le & \prod _{j=3}^{N}\left(1-{\alpha }_{3}^{j}\right)\left({\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}-{\alpha }_{1}^{2}\left(1-{\alpha }_{1}^{2}\right){\parallel {T}_{2}{x}_{0}-{x}_{0}\parallel }^{2}\right)\\ +\left(1-\prod _{j=3}^{N}\left(1-{\alpha }_{3}^{j}\right)\right){\parallel {x}_{0}-{x}^{\ast }\parallel }^{2},\end{array}$
which implies that
${\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}\le {\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}-{\alpha }_{1}^{2}\left(1-{\alpha }_{1}^{2}\right){\parallel {T}_{2}{x}_{0}-{x}_{0}\parallel }^{2}.$
(2.7)
Since ${\alpha }_{1}^{j}\in \left(0,1\right)$ for all $j=1,2,\dots ,N-1$ and (2.7), we have ${x}_{0}\in F\left({T}_{2}\right)$. From the definition of S and ${x}_{0}={T}_{2}{x}_{0}$, we have
${U}_{2}{x}_{0}={\alpha }_{1}^{2}{T}_{2}{U}_{1}{x}_{0}+{\alpha }_{2}^{2}{U}_{1}{x}_{0}+{\alpha }_{3}^{2}{x}_{0}={x}_{0}.$

By continuing in this way, we can show that ${x}_{0}\in F\left({T}_{i}\right)$ and ${x}_{0}\in F\left({U}_{i}\right)$ for all $i=1,2,\dots ,N-1$.

Finally, we shall show that ${x}_{0}\in F\left({T}_{N}\right)$.

Since
$\begin{array}{rcl}0& =& S{x}_{0}-{x}_{0}={\alpha }_{1}^{N}{T}_{N}{U}_{N-1}{x}_{0}+{\alpha }_{2}^{N}{U}_{N-1}{x}_{0}+{\alpha }_{3}^{N}{x}_{0}-{x}_{0}\\ =& {\alpha }_{1}^{N}\left({T}_{N}{x}_{0}-{x}_{0}\right),\end{array}$

and ${\alpha }_{1}^{N}\in \left(0,1\right]$, we obtain ${T}_{N}{x}_{0}={x}_{0}$ so that ${x}_{0}\in F\left({T}_{N}\right)$. Then we have ${x}_{0}\in {\bigcap }_{i=1}^{N}F\left({T}_{i}\right)$. Hence, $F\left(S\right)\subseteq {\bigcap }_{i=1}^{N}F\left({T}_{i}\right)$.

Next, we show that S is a quasi-nonexpansive mapping. Let $x\in C$ and $y\in F\left(S\right)$. From (2.5), we can imply that
$\begin{array}{rcl}{\parallel Sx-y\parallel }^{2}& \le & \prod _{j=2}^{N}\left(1-{\alpha }_{3}^{j}\right)\left({\parallel x-y\parallel }^{2}-{\alpha }_{1}^{1}\left(1-{\alpha }_{1}^{1}\right)\parallel {T}_{1}x-x\parallel \right)\\ +\left(1-\prod _{j=2}^{N}\left(1-{\alpha }_{3}^{j}\right)\right){\parallel x-y\parallel }^{2}\\ \le & {\parallel x-y\parallel }^{2}.\end{array}$

Then we have the S-mapping is quasi-nonexpansive. □

Example 2.10 Let ${T}_{1}:\left[-1,1\right]\to \left[-1,1\right]$ be a mapping defined by

for all $x\in \left[-1,1\right]$.

Let ${T}_{2}:\left[-1,1\right]\to \left[-1,1\right]$ be a mapping defined by

for all $x\in \left[-1,1\right]$.

To see that ${T}_{1}$ is a nonspreading mapping, observe that if $x,y\in \left(0,1\right]$, we have ${T}_{1}x=\frac{x+1}{2}$ and ${T}_{1}y=\frac{y+1}{2}$. Then we have
$\begin{array}{rcl}{|{T}_{1}x-{T}_{1}y|}^{2}& =& {|\frac{x+1}{2}-\frac{y+1}{2}|}^{2}\\ =& \frac{1}{4}{|x-y|}^{2}\end{array}$
and
From above, we have
$\begin{array}{rcl}{|x-y|}^{2}+2〈x-{T}_{1}x,y-{T}_{1}y〉& \ge & {|x-y|}^{2}\\ \ge & \frac{1}{4}{|x-y|}^{2}\\ =& {|{T}_{1}x-{T}_{1}y|}^{2}.\end{array}$
For every $x,y\in \left[-1,0\right]$, we have ${T}_{1}x=\frac{-x+1}{2}$ and ${T}_{1}y=\frac{-y+1}{2}$. From the definition of ${T}_{1}$, we have
$\begin{array}{rcl}{|{T}_{1}x-{T}_{1}y|}^{2}& =& {|\frac{-x+1}{2}-\left(\frac{-y+1}{2}\right)|}^{2}\\ =& {|\frac{y-x}{2}|}^{2}\\ =& \frac{1}{4}{|x-y|}^{2}\end{array}$
and
From above, we have
$\begin{array}{rcl}{|x-y|}^{2}+2〈x-{T}_{1}x,y-{T}_{1}y〉& >& {|x-y|}^{2}\\ \ge & \frac{1}{4}{|x-y|}^{2}\\ =& {|{T}_{1}x-{T}_{1}y|}^{2}.\end{array}$
Finally, for every $x\in \left(0,1\right]$ and $y\in \left[-1,0\right]$, we have ${T}_{1}x=\frac{x+1}{2}$ and ${T}_{1}y=\frac{-y+1}{2}$. From the definition of ${T}_{1}$, we have
$\begin{array}{rcl}{|{T}_{1}x-{T}_{1}y|}^{2}& =& {|\frac{x+1}{2}-\frac{-y+1}{2}|}^{2}\\ =& \frac{1}{4}{|x+y|}^{2}\end{array}$
and
From above, we have
Then for all $x,y\in \left[-1,1\right]$, we have
${|{T}_{1}x-{T}_{1}y|}^{2}\le {|x-y|}^{2}+〈x-{T}_{1}x,y-{T}_{1}y〉.$

Hence, we have ${T}_{1}$ is a nonspreading mapping.

Next, we show that ${T}_{2}$ is a nonspreading mapping. Let $x,y\in \left(0,1\right]$, then we have ${T}_{2}x=\frac{x+2}{3}$ and ${T}_{2}y=\frac{y+2}{3}$. From the definition of ${T}_{2}$, we have
$\begin{array}{rcl}{|{T}_{2}x-{T}_{2}y|}^{2}& =& {|\frac{x+2}{3}-\frac{y+2}{3}|}^{2}\\ =& \frac{1}{9}{|x-y|}^{2}\end{array}$
and
From above, we have
$\begin{array}{rcl}{|x-y|}^{2}+2〈x-{T}_{2}x,y-{T}_{2}y〉& \ge & {|x-y|}^{2}\\ \ge & \frac{1}{9}{|x-y|}^{2}\\ =& {|{T}_{2}x-{T}_{2}y|}^{2}.\end{array}$
For every $x,y\in \left[-1,0\right]$, we have ${T}_{2}x=\frac{2-x}{3}$ and ${T}_{2}y=\frac{2-y}{3}$. From the definition of ${T}_{2}$, we have
$\begin{array}{rcl}{|{T}_{2}x-{T}_{2}y|}^{2}& =& {|\frac{2-x}{3}-\frac{2-y}{3}|}^{2}\\ =& {|\frac{y-x}{3}|}^{2}\\ =& \frac{1}{9}{|x-y|}^{2}\end{array}$
and
From above, we have
$\begin{array}{rcl}{|x-y|}^{2}+2〈x-{T}_{2}x,y-{T}_{2}y〉& >& {|x-y|}^{2}\\ \ge & \frac{1}{9}{|x-y|}^{2}\\ =& {|{T}_{2}x-{T}_{2}y|}^{2}.\end{array}$
Finally, for every $x\in \left(0,1\right]$ and $y\in \left[-1,0\right]$, we have ${T}_{2}x=\frac{x+2}{3}$ and ${T}_{2}y=\frac{2-y}{3}$. From the definition of ${T}_{2}$, we have
$\begin{array}{rcl}{|{T}_{2}x-{T}_{2}y|}^{2}& =& {|\frac{x+2}{3}-\frac{2-y}{3}|}^{2}\\ =& \frac{1}{9}{|x+y|}^{2}\end{array}$
and
From above, we have
Then for every $x,y\in \left[-1,1\right]$, we have
${|{T}_{2}x-{T}_{2}y|}^{2}\le {|x-y|}^{2}+2〈x-{T}_{2}x,y-{T}_{2}y〉.$

Hence, we have ${T}_{2}$ is a nonspreading mapping. Observe that $1\in F\left({T}_{1}\right)\cap F\left({T}_{2}\right)$. Let the mapping $S:\left[-1,1\right]\to \left[-1,1\right]$ be the S-mapping generated by ${T}_{1}$, ${T}_{2}$ and ${\alpha }_{1}$, ${\alpha }_{2}$, where ${\alpha }_{1}=\left(\frac{1}{6},\frac{2}{6},\frac{3}{6}\right)$ and $\left(\frac{4}{15},\frac{5}{15},\frac{6}{15}\right)$. From Lemma 2.9, we have $1\in F\left(S\right)$.

## 3 Main result

Theorem 3.1 Let C be a nonempty closed convex subset of a Hilbert space H. For every $i=1,2,\dots ,N$, let ${A}_{i}:C\to H$ be an ${\alpha }_{i}$-inverse strongly monotone mapping, and let ${\left\{{T}_{i}\right\}}_{i=1}^{N}$ be a finite family of nonspreading mappings with $\mathfrak{F}={\bigcap }_{i=1}^{N}F\left({T}_{i}\right)\cap {\bigcap }_{i=1}^{N}VI\left(C,{A}_{i}\right)\ne \mathrm{\varnothing }$. For every $i=1,2,\dots ,N$, define the mapping ${G}_{i}:C\to C$ by ${G}_{i}x={P}_{C}\left(I-\lambda {A}_{i}\right)x$ $\mathrm{\forall }x\in C$ and $\lambda \in \left[c,d\right]\subset \left(0,2{\alpha }_{i}\right)$. Let ${\rho }_{j}=\left({\alpha }_{1}^{j},{\alpha }_{2}^{j},{\alpha }_{3}^{j}\right)\in I×I×I$, $j=1,2,3,\dots ,N$, where $I=\left[0,1\right]$, ${\alpha }_{1}^{j}+{\alpha }_{2}^{j}+{\alpha }_{3}^{j}=1$, ${\alpha }_{1}^{j},{\alpha }_{3}^{j}\in \left(0,1\right)$ for all $j=1,2,\dots ,N-1$ and ${\alpha }_{1}^{N}\in \left(0,1\right]$, ${\alpha }_{3}^{N}\in \left[0,1\right)$ ${\alpha }_{2}^{j}\in \left(0,1\right)$ for all $j=1,2,\dots ,N$, and let S be the S-mapping generated by ${T}_{1},{T}_{2},\dots ,{T}_{N}$ and ${\rho }_{1},{\rho }_{2},\dots ,{\rho }_{N}$. Let $\left\{{x}_{n}\right\}$ be a sequence generated by ${x}_{1}\in {C}_{1}=C$ and
$\left\{\begin{array}{c}{z}_{n}={\sum }_{i=1}^{N}{\delta }_{n}^{i}{G}_{i}{x}_{n},\hfill \\ {y}_{n}={\alpha }_{n}{x}_{n}+{\beta }_{n}S{x}_{n}+{\gamma }_{n}{z}_{n},\hfill \\ {C}_{n+1}=\left\{z\in {C}_{n}:\parallel {y}_{n}-z\parallel \le \parallel {x}_{n}-z\parallel \right\},\hfill \\ {x}_{n+1}={P}_{{C}_{n+1}}{x}_{1},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,\hfill \end{array}$
(3.1)
where $\left\{{\alpha }_{n}\right\},\left\{{\beta }_{n}\right\},\left\{{\gamma }_{n}\right\}\subseteq \left[0,1\right]$, ${\alpha }_{n}+{\beta }_{n}+{\gamma }_{n}=1$ and suppose the following conditions hold:

Then the sequence $\left\{{x}_{n}\right\}$ converges strongly to ${P}_{\mathfrak{F}}{x}_{1}$.

Proof First, we show that $\left(I-\lambda {A}_{i}\right)$ is a nonexpansive mapping for every $i=1,2,\dots ,N$. Let $x,y\in C$. Since A is an ${\alpha }_{i}$-inverse strongly monotone and $\lambda <2{\alpha }_{i}$, we have
$\begin{array}{rcl}{\parallel \left(I-\lambda {A}_{i}\right)x-\left(I-\lambda {A}_{i}\right)y\parallel }^{2}& =& {\parallel x-y-\lambda \left({A}_{i}x-{A}_{i}y\right)\parallel }^{2}\\ =& {\parallel x-y\parallel }^{2}-2\lambda 〈x-y,{A}_{i}x-{A}_{i}y〉+{\lambda }^{2}{\parallel {A}_{i}x-{A}_{i}y\parallel }^{2}\\ \le & {\parallel x-y\parallel }^{2}-2{\alpha }_{i}\lambda {\parallel {A}_{i}x-{A}_{i}y\parallel }^{2}+{\lambda }^{2}{\parallel {A}_{i}x-{A}_{i}y\parallel }^{2}\\ =& {\parallel x-y\parallel }^{2}+\lambda \left(\lambda -2{\alpha }_{i}\right){\parallel {A}_{i}x-{A}_{i}y\parallel }^{2}\\ \le & {\parallel x-y\parallel }^{2}.\end{array}$
Thus $\left(I-\lambda {A}_{i}\right)$ is a nonexpansive mapping for every $i=1,2,\dots ,N$. Since ${P}_{C}$ is a nonexpansive mapping, we have ${G}_{i}$ is a nonexpansive mapping for every $i=1,2,\dots ,N$. From Lemma 2.5, we have
$F\left({G}_{i}\right)=F\left({P}_{C}\left(I-\lambda {A}_{i}\right)\right)=VI\left(C,{A}_{i}\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }i=1,2,\dots ,N.$
(3.2)
From (3.2), $VI\left(C,{A}_{i}\right)$ is closed and convex. Let $z\in \mathfrak{F}$. From (3.2), we have $z\in F\left({P}_{C}\left(I-\lambda {A}_{i}\right)\right)$ for every $i=1,2,\dots ,N$. By nonexpansiveness of ${G}_{i}$, we have
$\parallel {z}_{n}-z\parallel =\parallel \sum _{i=1}^{N}{\delta }_{n}^{i}\left({G}_{i}{x}_{n}-z\right)\parallel \le \sum _{i=1}^{N}{\delta }_{n}^{i}\parallel {x}_{n}-z\parallel =\parallel {x}_{n}-z\parallel .$
(3.3)
Next, we show that ${C}_{n}$ is closed and convex for every $n\in \mathbb{N}$. It is obvious that ${C}_{n}$ is closed. In fact, we know that for $z\in {C}_{n}$,
$\parallel {y}_{n}-z\parallel \le \parallel {x}_{n}-z\parallel \phantom{\rule{1em}{0ex}}\text{is equivalent to}\phantom{\rule{1em}{0ex}}{\parallel {y}_{n}-{x}_{n}\parallel }^{2}+2〈{y}_{n}-{x}_{n},{x}_{n}-z〉\le 0.$
So, for every ${z}_{1},{z}_{2}\in {C}_{n}$ and $t\in \left(0,1\right)$, it follows that
then, we have ${C}_{n}$ is convex. Since $VI\left(C,{A}_{i}\right)$ is closed and convex for every $i=1,2,\dots ,N$, we have ${\bigcap }_{i=1}^{N}VI\left(C,{A}_{i}\right)$ is closed and convex. From Lemma 2.4, we have ${\bigcap }_{i=1}^{N}F\left({T}_{i}\right)$ is closed and convex. Hence, we have $\mathfrak{F}$ is closed and convex. This implies that ${P}_{\mathfrak{F}}$ is well defined. Next, we show that $\mathfrak{F}\subset {C}_{n}$ for every $n\in \mathbb{N}$. Let $z\in \mathfrak{F}$, then we have
$\begin{array}{rcl}\parallel {y}_{n}-z\parallel & =& \parallel {\alpha }_{n}\left({x}_{n}-z\right)+{\beta }_{n}\left(S{x}_{n}-z\right)+{\gamma }_{n}\left({z}_{n}-z\right)\parallel \\ \le & {\alpha }_{n}\parallel {x}_{n}-z\parallel +{\beta }_{n}\parallel S{x}_{n}-z\parallel +{\gamma }_{n}\parallel {z}_{n}-z\parallel \\ \le & \parallel {x}_{n}-z\parallel .\end{array}$
It follows that $z\in {C}_{n}$. Hence, we have $\mathfrak{F}\subset {C}_{n}$ for every $n\in \mathbb{N}$. This implies that $\left\{{x}_{n}\right\}$ is well defined. Since ${x}_{n}={P}_{{C}_{n}}{x}_{1}$, for every $w\in {C}_{n}$, we have
$\parallel {x}_{n}-{x}_{1}\parallel \le \parallel w-{x}_{1}\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\in \mathbb{N}.$
(3.4)
In particular, we have
$\parallel {x}_{n}-{x}_{1}\parallel \le \parallel {P}_{\mathfrak{F}}{x}_{1}-{x}_{1}\parallel .$
(3.5)
By (3.4) we have $\left\{{x}_{n}\right\}$ is bounded, so are $\left\{{G}_{i}{x}_{n}\right\}$, $\left\{{T}_{i}{x}_{n}\right\}$ for every $i=1,2,\dots ,N$, $\left\{{z}_{n}\right\}$, $\left\{{y}_{n}\right\}$ and $\left\{S{x}_{n}\right\}$. Since ${x}_{n+1}={P}_{{C}_{n+1}}{x}_{1}\in {C}_{n+1}\subset {C}_{n}$ and ${x}_{n}={P}_{{C}_{n}}{x}_{1}$, we have
$\begin{array}{rcl}0& \le & 〈{x}_{1}-{x}_{n},{x}_{n}-{x}_{n+1}〉\\ =& 〈{x}_{1}-{x}_{n},{x}_{n}-{x}_{1}+{x}_{1}-{x}_{n+1}〉\\ \le & -{\parallel {x}_{n}-{x}_{1}\parallel }^{2}+\parallel {x}_{n}-{x}_{1}\parallel \parallel {x}_{1}-{x}_{n+1}\parallel ,\end{array}$
which implies that
$\parallel {x}_{n}-{x}_{1}\parallel \le \parallel {x}_{n+1}-{x}_{1}\parallel .$
Hence, we have ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-{x}_{1}\parallel$ exists. Since
$\begin{array}{rcl}{\parallel {x}_{n}-{x}_{n+1}\parallel }^{2}& =& {\parallel {x}_{n}-{x}_{1}+{x}_{1}-{x}_{n+1}\parallel }^{2}\\ =& {\parallel {x}_{n}-{x}_{1}\parallel }^{2}+2〈{x}_{n}-{x}_{1},{x}_{1}-{x}_{n+1}〉+{\parallel {x}_{1}-{x}_{n+1}\parallel }^{2}\\ =& {\parallel {x}_{n}-{x}_{1}\parallel }^{2}+2〈{x}_{n}-{x}_{1},{x}_{1}-{x}_{n}+{x}_{n}-{x}_{n+1}〉+{\parallel {x}_{1}-{x}_{n+1}\parallel }^{2}\\ =& {\parallel {x}_{n}-{x}_{1}\parallel }^{2}-2{\parallel {x}_{n}-{x}_{1}\parallel }^{2}+2〈{x}_{n}-{x}_{1},{x}_{n}-{x}_{n+1}〉+{\parallel {x}_{1}-{x}_{n+1}\parallel }^{2}\\ \le & {\parallel {x}_{1}-{x}_{n+1}\parallel }^{2}-{\parallel {x}_{n}-{x}_{1}\parallel }^{2},\end{array}$
(3.6)
it implies that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-{x}_{n+1}\parallel =0.$
(3.7)
Since ${x}_{n+1}={P}_{{C}_{n+1}}{x}_{1}\in {C}_{n+1}$, we have
$\parallel {y}_{n}-{x}_{n+1}\parallel \le \parallel {x}_{n}-{x}_{n+1}\parallel .$
By (3.7) we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {y}_{n}-{x}_{n+1}\parallel =0.$
(3.8)
Since
$\parallel {y}_{n}-{x}_{n}\parallel \le \parallel {y}_{n}-{x}_{n+1}\parallel +\parallel {x}_{n+1}-{x}_{n}\parallel ,$
by (3.7) and (3.8), we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {y}_{n}-{x}_{n}\parallel =0.$
(3.9)
Next, we will show that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-S{x}_{n}\parallel =0.$
(3.10)
For every $i=1,2,\dots ,N$, we have
(3.11)
From the definition of ${y}_{n}$ and (3.11), we have
$\begin{array}{rcl}{\parallel {y}_{n}-z\parallel }^{2}& \le & {\alpha }_{n}{\parallel {x}_{n}-z\parallel }^{2}+{\beta }_{n}{\parallel S{x}_{n}-z\parallel }^{2}+{\gamma }_{n}{\parallel {z}_{n}-z\parallel }^{2}\\ \le & {\alpha }_{n}{\parallel {x}_{n}-z\parallel }^{2}+{\beta }_{n}{\parallel S{x}_{n}-z\parallel }^{2}+{\gamma }_{n}\sum _{i=1}^{N}{\delta }_{n}^{i}{\parallel {P}_{C}\left(I-\lambda {A}_{i}\right){x}_{n}-z\parallel }^{2}\\ \le & {\alpha }_{n}{\parallel {x}_{n}-z\parallel }^{2}+{\beta }_{n}{\parallel S{x}_{n}-z\parallel }^{2}\\ +{\gamma }_{n}\sum _{i=1}^{N}{\delta }_{n}^{i}\left({\parallel {x}_{n}-z\parallel }^{2}-\lambda \left(2{\alpha }_{i}-\lambda \right){\parallel {A}_{i}{x}_{n}-{A}_{i}z\parallel }^{2}\right)\\ =& {\alpha }_{n}{\parallel {x}_{n}-z\parallel }^{2}+{\beta }_{n}{\parallel S{x}_{n}-z\parallel }^{2}+{\gamma }_{n}{\parallel {x}_{n}-z\parallel }^{2}\\ -{\gamma }_{n}\sum _{i=1}^{N}{\delta }_{n}^{i}\lambda \left(2{\alpha }_{i}-\lambda \right){\parallel {A}_{i}{x}_{n}-{A}_{i}z\parallel }^{2}\\ \le & {\parallel {x}_{n}-z\parallel }^{2}-{\gamma }_{n}\sum _{i=1}^{N}{\delta }_{n}^{i}\lambda \left(2{\alpha }_{i}-\lambda \right){\parallel {A}_{i}{x}_{n}-{A}_{i}z\parallel }^{2}.\end{array}$
It follows that
$\begin{array}{rcl}{\gamma }_{n}\sum _{i=1}^{N}{\delta }_{n}^{i}\lambda \left(2{\alpha }_{i}-\lambda \right){\parallel {A}_{i}{x}_{n}-{A}_{i}z\parallel }^{2}& \le & {\parallel {x}_{n}-z\parallel }^{2}-{\parallel {y}_{n}-z\parallel }^{2}\\ \le & \left(\parallel {x}_{n}-z\parallel +\parallel {y}_{n}-z\parallel \right)\parallel {y}_{n}-{x}_{n}\parallel .\end{array}$
From conditions (i), (ii) and (3.9), it implies that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {A}_{i}{x}_{n}-{A}_{i}z\parallel =0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }i=1,2,\dots ,N.$
(3.12)
Since
$\begin{array}{rcl}{\parallel {P}_{C}\left(I-\lambda {A}_{i}\right){x}_{n}-z\parallel }^{2}& \le & 〈\left(I-\lambda {A}_{i}\right){x}_{n}-\left(I-\lambda {A}_{i}\right)z,{P}_{C}\left(I-\lambda {A}_{i}\right){x}_{n}-z〉\\ =& \frac{1}{2}\left({\parallel \left(I-\lambda {A}_{i}\right){x}_{n}-\left(I-\lambda {A}_{i}\right)z\parallel }^{2}+{\parallel {P}_{C}\left(I-\lambda {A}_{i}\right){x}_{n}-z\parallel }^{2}\\ -{\parallel \left(I-\lambda {A}_{i}\right){x}_{n}-\left(I-\lambda {A}_{i}\right)z-{P}_{C}\left(I-\lambda {A}_{i}\right){x}_{n}+z\parallel }^{2}\right)\\ \le & \frac{1}{2}\left({\parallel {x}_{n}-z\parallel }^{2}+{\parallel {P}_{C}\left(I-\lambda {A}_{i}\right){x}_{n}-z\parallel }^{2}\\ -{\parallel {x}_{n}-{P}_{C}\left(I-\lambda {A}_{i}\right){x}_{n}-\lambda \left({A}_{i}{x}_{n}-{A}_{i}z\right)\parallel }^{2}\right)\\ =& \frac{1}{2}\left({\parallel {x}_{n}-z\parallel }^{2}+{\parallel {P}_{C}\left(I-\lambda {A}_{i}\right){x}_{n}-z\parallel }^{2}\\ -{\parallel {x}_{n}-{P}_{C}\left(I-\lambda {A}_{i}\right){x}_{n}\parallel }^{2}-{\parallel \lambda \left({A}_{i}{x}_{n}-{A}_{i}z\right)\parallel }^{2}\\ +2\lambda 〈{x}_{n}-{P}_{C}\left(I-\lambda {A}_{i}\right){x}_{n},{A}_{i}{x}_{n}-{A}_{i}z〉\right),\end{array}$
it implies that
$\begin{array}{rcl}{\parallel {P}_{C}\left(I-\lambda {A}_{i}\right){x}_{n}-z\parallel }^{2}& \le & {\parallel {x}_{n}-z\parallel }^{2}-{\parallel {x}_{n}-{P}_{C}\left(I-\lambda {A}_{i}\right){x}_{n}\parallel }^{2}\\ +2\lambda \parallel {x}_{n}-{P}_{C}\left(I-\lambda {A}_{i}\right){x}_{n}\parallel \parallel {A}_{i}{x}_{n}-{A}_{i}z\parallel .\end{array}$
(3.13)
From the definition of ${y}_{n}$ and (3.13), we have
$\begin{array}{rcl}{\parallel {y}_{n}-z\parallel }^{2}& \le & {\alpha }_{n}{\parallel {x}_{n}-z\parallel }^{2}+{\beta }_{n}{\parallel S{x}_{n}-z\parallel }^{2}+{\gamma }_{n}{\parallel {z}_{n}-z\parallel }^{2}\\ \le & \left(1-{\gamma }_{n}\right){\parallel {x}_{n}-z\parallel }^{2}+{\gamma }_{n}\sum _{i=1}^{N}{\delta }_{n}^{i}{\parallel {P}_{C}\left(I-\lambda {A}_{i}\right){x}_{n}-z\parallel }^{2}\\ \le & \left(1-{\gamma }_{n}\right){\parallel {x}_{n}-z\parallel }^{2}+{\gamma }_{n}\sum _{i=1}^{N}{\delta }_{n}^{i}\left({\parallel {x}_{n}-z\parallel }^{2}-{\parallel {x}_{n}-{P}_{C}\left(I-\lambda {A}_{i}\right){x}_{n}\parallel }^{2}\\ +2\lambda \parallel {x}_{n}-{P}_{C}\left(I-\lambda {A}_{i}\right){x}_{n}\parallel \parallel {A}_{i}{x}_{n}-{A}_{i}z\parallel \right)\\ =& {\parallel {x}_{n}-z\parallel }^{2}-{\gamma }_{n}\sum _{i=1}^{N}{\delta }_{n}^{i}{\parallel {x}_{n}-{P}_{C}\left(I-\lambda {A}_{i}\right){x}_{n}\parallel }^{2}\\ +2{\gamma }_{n}\sum _{i=1}^{N}{\delta }_{n}^{i}\lambda \parallel {x}_{n}-{P}_{C}\left(I-\lambda {A}_{i}\right){x}_{n}\parallel \parallel {A}_{i}{x}_{n}-{A}_{i}z\parallel ,\end{array}$
which implies that
$\begin{array}{rcl}{\gamma }_{n}\sum _{i=1}^{N}{\delta }_{n}^{i}{\parallel {x}_{n}-{P}_{C}\left(I-\lambda {A}_{i}\right){x}_{n}\parallel }^{2}& \le & {\parallel {x}_{n}-z\parallel }^{2}-{\parallel {y}_{n}-z\parallel }^{2}\\ +2{\gamma }_{n}\sum _{i=1}^{N}{\delta }_{n}^{i}\lambda \parallel {x}_{n}-{P}_{C}\left(I-\lambda {A}_{i}\right){x}_{n}\parallel \parallel {A}_{i}{x}_{n}-{A}_{i}z\parallel \\ \le & \left(\parallel {x}_{n}-z\parallel +\parallel {y}_{n}-z\parallel \right)\parallel {y}_{n}-{x}_{n}\parallel \\ +2{\gamma }_{n}\sum _{i=1}^{N}{\delta }_{n}^{i}\lambda \parallel {x}_{n}-{P}_{C}\left(I-\lambda {A}_{i}\right){x}_{n}\parallel \parallel {A}_{i}{x}_{n}-{A}_{i}z\parallel .\end{array}$
From conditions (i), (ii), (3.9) and (3.12), we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {P}_{C}\left(I-\lambda {A}_{i}\right){x}_{n}-{x}_{n}\parallel =0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }i=1,2,\dots ,N.$
(3.14)
Since
$\parallel {z}_{n}-{x}_{n}\parallel \le \sum _{i=1}^{N}{\delta }_{n}^{i}\parallel {P}_{C}\left(I-\lambda {A}_{i}\right){x}_{n}-{x}_{n}\parallel ,$
from (3.14), we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {z}_{n}-{x}_{n}\parallel =0.$
(3.15)
Since
${y}_{n}-{x}_{n}={\beta }_{n}\left(S{x}_{n}-{x}_{n}\right)+{\gamma }_{n}\left({z}_{n}-{x}_{n}\right)$
from (3.9) and (3.15), we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel S{x}_{n}-{x}_{n}\parallel =0.$
Next, we will show that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {T}_{i}{U}_{i-1}{x}_{n}-{U}_{i-1}{x}_{n}\parallel =0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }i=1,2,\dots ,N.$
(3.16)
From the definition of ${y}_{n}$, we have
$\begin{array}{rcl}{\parallel {y}_{n}-z\parallel }^{2}& \le & {\alpha }_{n}{\parallel {x}_{n}-z\parallel }^{2}+{\beta }_{n}{\parallel S{x}_{n}-z\parallel }^{2}+{\gamma }_{n}{\parallel {z}_{n}-z\parallel }^{2}\\ \le & \left(1-{\beta }_{n}\right){\parallel {x}_{n}-z\parallel }^{2}+{\beta }_{n}\parallel {\alpha }_{1}^{N}\left({T}_{N}{U}_{N-1}{x}_{n}-z\right)\\ +{\alpha }_{2}^{N}\left({U}_{N-1}{x}_{n}-z\right)+{\alpha }_{3}^{N}\left({x}_{n}-z\right)\parallel \\ \le & \left(1-{\beta }_{n}\right){\parallel {x}_{n}-z\parallel }^{2}+{\beta }_{n}\left({\alpha }_{1}^{N}{\parallel {T}_{N}{U}_{N-1}{x}_{n}-z\parallel }^{2}+{\alpha }_{2}^{N}{\parallel {U}_{N-1}{x}_{n}-z\parallel }^{2}\\ +{\alpha }_{3}^{N}{\parallel {x}_{n}-z\parallel }^{2}-{\alpha }_{1}^{N}{\alpha }_{2}^{N}{\parallel {T}_{N}{U}_{N-1}{x}_{n}-{U}_{N-1}{x}_{n}\parallel }^{2}\right)\\ \le & \left(1-{\beta }_{n}\right){\parallel {x}_{n}-z\parallel }^{2}+{\beta }_{n}\left(\left(1-{\alpha }_{3}^{N}\right){\parallel {U}_{N-1}{x}_{n}-z\parallel }^{2}\\ +{\alpha }_{3}^{N}{\parallel {x}_{n}-z\parallel }^{2}-{\alpha }_{1}^{N}{\alpha }_{2}^{N}{\parallel {T}_{N}{U}_{N-1}{x}_{n}-{U}_{N-1}{x}_{n}\parallel }^{2}\right)\\ =& \left(1-{\beta }_{n}\right){\parallel {x}_{n}-z\parallel }^{2}+{\beta }_{n}\left(\left(1-{\alpha }_{3}^{N}\right)\parallel {\alpha }_{1}^{N-1}\left({T}_{N-1}{U}_{N-2}{x}_{n}-z\right)\\ +{\alpha }_{2}^{N-1}\left({U}_{N-2}{x}_{n}-z\right)+{\alpha }_{3}^{N-1}\left({x}_{n}-z\right){\parallel }^{2}\\ +{\alpha }_{3}^{N}{\parallel {x}_{n}-z\parallel }^{2}-{\alpha }_{1}^{N}{\alpha }_{2}^{N}{\parallel {T}_{N}{U}_{N-1}{x}_{n}-{U}_{N-1}{x}_{n}\parallel }^{2}\right)\\ \le & \left(1-{\beta }_{n}\right){\parallel {x}_{n}-z\parallel }^{2}+{\beta }_{n}\left(\left(1-{\alpha }_{3}^{N}\right)\left({\alpha }_{1}^{N-1}{\parallel {T}_{N-1}{U}_{N-2}{x}_{n}-z\parallel }^{2}\\ +{\alpha }_{2}^{N-1}{\parallel {U}_{N-2}{x}_{n}-z\parallel }^{2}+{\alpha }_{3}^{N-1}{\parallel {x}_{n}-z\parallel }^{2}\\ -{\alpha }_{1}^{N-1}{\alpha }_{2}^{N-1}{\parallel {T}_{N-1}{U}_{N-2}{x}_{n}-{U}_{N-2}{x}_{n}\parallel }^{2}\right)\\ +{\alpha }_{3}^{N}{\parallel {x}_{n}-z\parallel }^{2}-{\alpha }_{1}^{N}{\alpha }_{2}^{N}{\parallel {T}_{N}{U}_{N-1}{x}_{n}-{U}_{N-1}{x}_{n}\parallel }^{2}\right)\\ \le & \left(1-{\beta }_{n}\right){\parallel {x}_{n}-z\parallel }^{2}+{\beta }_{n}\left(\left(1-{\alpha }_{3}^{N}\right)\left(\left(1-{\alpha }_{3}^{N-1}\right){\parallel {U}_{N-2}{x}_{n}-z\parallel }^{2}\\ +{\alpha }_{3}^{N-1}{\parallel {x}_{n}-z\parallel }^{2}-{\alpha }_{1}^{N-1}{\alpha }_{2}^{N-1}{\parallel {T}_{N-1}{U}_{N-2}{x}_{n}-{U}_{N-2}{x}_{n}\parallel }^{2}\right)\\ +{\alpha }_{3}^{N}{\parallel {x}_{n}-z\parallel }^{2}-{\alpha }_{1}^{N}{\alpha }_{2}^{N}{\parallel {T}_{N}{U}_{N-1}{x}_{n}-{U}_{N-1}{x}_{n}\parallel }^{2}\right)\\ =& \left(1-{\beta }_{n}\right){\parallel {x}_{n}-z\parallel }^{2}+{\beta }_{n}\left(\left(1-{\alpha }_{3}^{N}\right)\left(1-{\alpha }_{3}^{N-1}\right){\parallel {U}_{N-2}{x}_{n}-z\parallel }^{2}\\ +\left(1-{\alpha }_{3}^{N}\right){\alpha }_{3}^{N-1}{\parallel {x}_{n}-z\parallel }^{2}-{\alpha }_{1}^{N-1}{\alpha }_{2}^{N-1}\left(1-{\alpha }_{3}^{N}\right){\parallel {T}_{N-1}{U}_{N-2}{x}_{n}-{U}_{N-2}{x}_{n}\parallel }^{2}\\ +{\alpha }_{3}^{N}{\parallel {x}_{n}-z\parallel }^{2}-{\alpha }_{1}^{N}{\alpha }_{2}^{N}{\parallel {T}_{N}{U}_{N-1}{x}_{n}-{U}_{N-1}{x}_{n}\parallel }^{2}\right)\\ =& \left(1-{\beta }_{n}\right){\parallel {x}_{n}-z\parallel }^{2}+{\beta }_{n}\left(\prod _{j=N-1}^{N}\left(1-{\alpha }_{3}^{j}\right){\parallel {U}_{N-2}{x}_{n}-z\parallel }^{2}\\ +\left(1-\prod _{j=N-1}^{N}\left(1-{\alpha }_{3}^{j}\right)\right){\parallel {x}_{n}-z\parallel }^{2}\\ -{\alpha }_{1}^{N-1}{\alpha }_{2}^{N-1}\left(1-{\alpha }_{3}^{N}\right){\parallel {T}_{N-1}{U}_{N-2}{x}_{n}-{U}_{N-2}{x}_{n}\parallel }^{2}\\ -{\alpha }_{1}^{N}{\alpha }_{2}^{N}{\parallel {T}_{N}{U}_{N-1}{x}_{n}-{U}_{N-1}{x}_{n}\parallel }^{2}\right)\\ =& \left(1-{\beta }_{n}\right){\parallel {x}_{n}-z\parallel }^{2}+{\beta }_{n}\left(\prod _{j=N-1}^{N}\left(1-{\alpha }_{3}^{j}\right)\parallel {\alpha }_{1}^{N-2}\left({T}_{N-2}{U}_{N-3}{x}_{n}-z\right)\\ +{\alpha }_{2}^{N-2}\left({U}_{N-3}{x}_{n}-z\right)+{\alpha }_{3}^{N-2}\left({x}_{n}-z\right){\parallel }^{2}\\ +\left(1-\prod _{j=N-1}^{N}\left(1-{\alpha }_{3}^{j}\right)\right){\parallel {x}_{n}-z\parallel }^{2}\\ -{\alpha }_{1}^{N-1}{\alpha }_{2}^{N-1}\left(1-{\alpha }_{3}^{N}\right){\parallel {T}_{N-1}{U}_{N-2}{x}_{n}-{U}_{N-2}{x}_{n}\parallel }^{2}\\ -{\alpha }_{1}^{N}{\alpha }_{2}^{N}{\parallel {T}_{N}{U}_{N-1}{x}_{n}-{U}_{N-1}{x}_{n}\parallel }^{2}\right)\\ \le & \left(1-{\beta }_{n}\right){\parallel {x}_{n}-z\parallel }^{2}+{\beta }_{n}\left(\prod _{j=N-1}^{N}\left(1-{\alpha }_{3}^{j}\right)\left({\alpha }_{1}^{N-2}{\parallel {T}_{N-2}{U}_{N-3}{x}_{n}-z\parallel }^{2}\\ +{\alpha }_{2}^{N-2}{\parallel {U}_{N-3}{x}_{n}-z\parallel }^{2}+{\alpha }_{3}^{N-2}{\parallel {x}_{n}-z\parallel }^{2}\\ -{\alpha }_{1}^{N-2}{\alpha }_{2}^{N-2}{\parallel {T}_{N-2}{U}_{N-3}{x}_{n}-{U}_{N-3}{x}_{n}\parallel }^{2}\right)\\ +\left(1-\prod _{j=N-1}^{N}\left(1-{\alpha }_{3}^{j}\right)\right){\parallel {x}_{n}-z\parallel }^{2}\\ -{\alpha }_{1}^{N-1}{\alpha }_{2}^{N-1}\left(1-{\alpha }_{3}^{N}\right){\parallel {T}_{N-1}{U}_{N-2}{x}_{n}-{U}_{N-2}{x}_{n}\parallel }^{2}\\ -{\alpha }_{1}^{N}{\alpha }_{2}^{N}{\parallel {T}_{N}{U}_{N-1}{x}_{n}-{U}_{N-1}{x}_{n}\parallel }^{2}\right)\\ \le & \left(1-{\beta }_{n}\right){\parallel {x}_{n}-z\parallel }^{2}+{\beta }_{n}\left(\prod _{j=N-1}^{N}\left(1-{\alpha }_{3}^{j}\right)\left(\left(1-{\alpha }_{3}^{N-2}\right){\parallel {U}_{N-3}{x}_{n}-z\parallel }^{2}\\ +{\alpha }_{3}^{N-2}{\parallel {x}_{n}-z\parallel }^{2}-{\alpha }_{1}^{N-2}{\alpha }_{2}^{N-2}{\parallel {T}_{N-2}{U}_{N-3}{x}_{n}-{U}_{N-3}{x}_{n}\parallel }^{2}\right)\\ +\left(1-\prod _{j=N-1}^{N}\left(1-{\alpha }_{3}^{j}\right)\right){\parallel {x}_{n}-z\parallel }^{2}\\ -{\alpha }_{1}^{N-1}{\alpha }_{2}^{N-1}\left(1-{\alpha }_{3}^{N}\right){\parallel {T}_{N-1}{U}_{N-2}{x}_{n}-{U}_{N-2}{x}_{n}\parallel }^{2}\\ -{\alpha }_{1}^{N}{\alpha }_{2}^{N}{\parallel {T}_{N}{U}_{N-1}{x}_{n}-{U}_{N-1}{x}_{n}\parallel }^{2}\right)\\ =& \left(1-{\beta }_{n}\right){\parallel {x}_{n}-z\parallel }^{2}+{\beta }_{n}\left(\prod _{j=N-2}^{N}\left(1-{\alpha }_{3}^{j}\right){\parallel {U}_{N-3}{x}_{n}-z\parallel }^{2}\\ +\left(1-\prod _{j=N-2}^{N}\left(1-{\alpha }_{3}^{j}\right)\right){\parallel {x}_{n}-z\parallel }^{2}\\ -{\alpha }_{1}^{N-2}{\alpha }_{2}^{N-2}\prod _{j=N-1}^{N}\left(1-{\alpha }_{3}^{j}\right){\parallel {T}_{N-2}{U}_{N-3}{x}_{n}-{U}_{N-3}{x}_{n}\parallel }^{2}\\ -{\alpha }_{1}^{N-1}{\alpha }_{2}^{N-1}\left(1-{\alpha }_{3}^{N}\right){\parallel {T}_{N-1}{U}_{N-2}{x}_{n}-{U}_{N-2}{x}_{n}\parallel }^{2}\\ -{\alpha }_{1}^{N}{\alpha }_{2}^{N}{\parallel {T}_{N}{U}_{N-1}{x}_{n}-{U}_{N-1}{x}_{n}\parallel }^{2}\right)\\ \le \\ ⋮\\ \le & \left(1-{\beta }_{n}\right){\parallel {x}_{n}-z\parallel }^{2}+{\beta }_{n}\left(\prod _{j=1}^{N}\left(1-{\alpha }_{3}^{j}\right){\parallel {U}_{0}{x}_{n}-z\parallel }^{2}\\ +\left(1-\prod _{j=1}^{N}\left(1-{\alpha }_{3}^{j}\right)\right){\parallel {x}_{n}-z\parallel }^{2}\\ -{\alpha }_{1}^{1}{\alpha }_{2}^{1}\prod _{j=2}^{N}\left(1-{\alpha }_{3}^{j}\right){\parallel {T}_{1}{U}_{0}{x}_{n}-{U}_{0}{x}_{n}\parallel }^{2}\\ ⋮\\ -{\alpha }_{1}^{N-2}{\alpha }_{2}^{N-2}\prod _{j=N-1}^{N}\left(1-{\alpha }_{3}^{j}\right){\parallel {T}_{N-2}{U}_{N-3}{x}_{n}-{U}_{N-3}{x}_{n}\parallel }^{2}\\ -{\alpha }_{1}^{N-1}{\alpha }_{2}^{N-1}\left(1-{\alpha }_{3}^{N}\right){\parallel {T}_{N-1}{U}_{N-2}{x}_{n}-{U}_{N-2}{x}_{n}\parallel }^{2}\\ -{\alpha }_{1}^{N}{\alpha }_{2}^{N}{\parallel {T}_{N}{U}_{N-1}{x}_{n}-{U}_{N-1}{x}_{n}\parallel }^{2}\right)\\ =& {\parallel {x}_{n}-z\parallel }^{2}\\ -{\beta }_{n}{\alpha }_{1}^{1}{\alpha }_{2}^{1}\prod _{j=2}^{N}\left(1-{\alpha }_{3}^{j}\right){\parallel {T}_{1}{x}_{n}-{x}_{n}\parallel }^{2}\\ ⋮\\ -{\beta }_{n}{\alpha }_{1}^{N-2}{\alpha }_{2}^{N-2}\prod _{j=N-1}^{N}\left(1-{\alpha }_{3}^{j}\right){\parallel {T}_{N-2}{U}_{N-3}{x}_{n}-{U}_{N-3}{x}_{n}\parallel }^{2}\\ -{\beta }_{n}{\alpha }_{1}^{N-1}{\alpha }_{2}^{N-1}\left(1-{\alpha }_{3}^{N}\right){\parallel {T}_{N-1}{U}_{N-2}{x}_{n}-{U}_{N-2}{x}_{n}\parallel }^{2}\\ -{\beta }_{n}{\alpha }_{1}^{N}{\alpha }_{2}^{N}{\parallel {T}_{N}{U}_{N-1}{x}_{n}-{U}_{N-1}{x}_{n}\parallel }^{2}.\end{array}$
(3.17)
From (3.17) and condition (ii), we have
$\begin{array}{rcl}{\beta }_{n}{\alpha }_{1}^{1}{\alpha }_{2}^{1}\prod _{j=2}^{N}\left(1-{\alpha }_{3}^{j}\right){\parallel {T}_{1}{x}_{n}-{x}_{n}\parallel }^{2}& \le & {\parallel {x}_{n}-z\parallel }^{2}-{\parallel {y}_{n}-z\parallel }^{2}\\ \le & \left(\parallel {x}_{n}-z\parallel +\parallel {y}_{n}-z\parallel \right)\parallel {y}_{n}-{x}_{n}\parallel .\end{array}$
Form (3.9), we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {T}_{1}{x}_{n}-{x}_{n}\parallel =0.$
(3.18)
By using the same method as (3.18), we can conclude that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {T}_{i}{U}_{i-1}{x}_{n}-{U}_{i-1}{x}_{n}\parallel =0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }i=1,2,\dots ,N.$

Let $\omega \left({x}_{n}\right)$ be the set of all weakly ω-limit of $\left\{{x}_{n}\right\}$. We shall show that $\omega \left({x}_{n}\right)\subset \mathfrak{F}$. Since $\left\{{x}_{n}\right\}$ is bounded, then $\omega \left({x}_{n}\right)\ne \mathrm{\varnothing }$. Let $q\in \omega \left({x}_{n}\right)$, there exists a subsequence $\left\{{x}_{{n}_{i}}\right\}$ of $\left\{{x}_{n}\right\}$ which converges weakly to q.

Put $Q:C\to C$ defined by
$Qx=\sum _{i=1}^{N}{\delta }^{i}{G}_{i}x,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in C.$
(3.19)
Since ${G}_{i}={P}_{C}\left(I-\lambda {A}_{i}\right)$ is a nonexpansive mapping, for every $i=1,2,\dots ,N$, from Lemma 2.6 and 2.5, we have
$F\left(Q\right)=\bigcap _{i=1}^{N}F\left({G}_{i}\right)=\bigcap _{i=1}^{N}VI\left(C,{A}_{i}\right).$
(3.20)
Since
$\begin{array}{rcl}\parallel {x}_{n}-Q{x}_{n}\parallel & \le & \parallel {x}_{n}-{z}_{n}\parallel +\parallel {z}_{n}-Q{x}_{n}\parallel \\ =& \parallel {x}_{n}-{z}_{n}\parallel +\parallel \sum _{i=1}^{N}{\delta }_{n}^{i}{G}_{i}{x}_{n}-\sum _{i=1}^{N}{\delta }^{i}{G}_{i}{x}_{n}\parallel \\ =& \parallel {x}_{n}-{z}_{n}\parallel +\parallel \sum _{i=1}^{N}\left({\delta }_{n}^{i}-{\delta }^{i}\right){G}_{i}{x}_{n}\parallel \\ \le & \parallel {x}_{n}-{z}_{n}\parallel +\sum _{i=1}^{N}|{\delta }_{n}^{i}-{\delta }^{i}|\parallel {G}_{i}{x}_{n}\parallel ,\end{array}$
from the condition (i) and (3.15), we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-Q{x}_{n}\parallel =0.$
(3.21)
From (3.21), we have
$\underset{i\to \mathrm{\infty }}{lim}\parallel {x}_{{n}_{i}}-Q{x}_{{n}_{i}}\parallel =0.$
From (3.19), it is easy to see that Q is a nonexpansive mapping. By Lemma 2.7 and ${x}_{{n}_{i}}⇀q$ as $i\to \mathrm{\infty }$, we have $q\in F\left(Q\right)={\bigcap }_{i=1}^{N}F\left({G}_{i}\right)$ From (3.2), we have
$q\in \bigcap _{i=1}^{N}VI\left(C,{A}_{i}\right).$
(3.22)
Next, we will show that $q\in F\left(S\right)$. Assume that $q\ne Sq$. From the Opial property, (3.10) and (3.16), we have
$\begin{array}{rcl}\underset{i\to \mathrm{\infty }}{lim inf}{\parallel {x}_{{n}_{i}}-q\parallel }^{2}& <& \underset{i\to \mathrm{\infty }}{lim inf}{\parallel {x}_{{n}_{i}}-Sq\parallel }^{2}\\ =& \underset{i\to \mathrm{\infty }}{lim inf}{\parallel {x}_{{n}_{i}}-S{x}_{{n}_{i}}+\left(S{x}_{{n}_{i}}-Sq\right)\parallel }^{2}\\ =& \underset{i\to \mathrm{\infty }}{lim inf}\left({\parallel {x}_{{n}_{i}}-S{x}_{{n}_{i}}\parallel }^{2}+{\parallel S{x}_{{n}_{i}}-Sq\parallel }^{2}+2〈{x}_{{n}_{i}}-S{x}_{{n}_{i}},S{x}_{{n}_{i}}-Sq〉\right)\\ =& \underset{i\to \mathrm{\infty }}{lim inf}{\parallel S{x}_{{n}_{i}}-Sq\parallel }^{2}\\ =& \underset{i\to \mathrm{\infty }}{lim inf}\parallel {\alpha }_{1}^{N}{T}_{N}{U}_{N-1}{x}_{{n}_{i}}+{\alpha }_{2}^{N}{U}_{N-1}{x}_{{n}_{i}}+{\alpha }_{3}^{N}{x}_{{n}_{i}}\\ -{\alpha }_{1}^{N}{T}_{N}{U}_{N-1}q-{\alpha }_{2}^{N}{U}_{N-1}q-{\alpha }_{3}^{N}q{\parallel }^{2}\\ =& \underset{i\to \mathrm{\infty }}{lim inf}\parallel {\alpha }_{1}^{N}\left({T}_{N}{U}_{N-1}{x}_{{n}_{i}}-{T}_{N}{U}_{N-1}q\right)\\ +{\alpha }_{2}^{N}\left({U}_{N-1}{x}_{{n}_{i}}-{U}_{N-1}q\right)+{\alpha }_{3}^{N}\left({x}_{{n}_{i}}-q\right){\parallel }^{2}\\ \le & \underset{i\to \mathrm{\infty }}{lim inf}\left({\alpha }_{1}^{N}{\parallel {T}_{N}{U}_{N-1}{x}_{{n}_{i}}-{T}_{N}{U}_{N-1}q\parallel }^{2}\\ +{\alpha }_{2}^{N}{\parallel {U}_{N-1}{x}_{{n}_{i}}-{U}_{N-1}q\parallel }^{2}+{\alpha }_{3}^{N}{\parallel {x}_{{n}_{i}}-q\parallel }^{2}\right)\\ \le & \underset{i\to \mathrm{\infty }}{lim inf}\left({\alpha }_{1}^{N}\left({\parallel {U}_{N-1}{x}_{{n}_{i}}-{U}_{N-1}q\parallel }^{2}\\ +2〈{U}_{N-1}{x}_{{n}_{i}}-{T}_{N}{U}_{N-1}{x}_{{n}_{i}},{U}_{N-1}q-{T}_{N}{U}_{N-1}q〉\right)\\ +{\alpha }_{2}^{N}{\parallel {U}_{N-1}{x}_{{n}_{i}}-{U}_{N-1}q\parallel }^{2}+{\alpha }_{3}^{N}{\parallel {x}_{{n}_{i}}-q\parallel }^{2}\right)\\ =& \underset{i\to \mathrm{\infty }}{lim inf}\left(\left(1-{\alpha }_{3}^{N}\right){\parallel {U}_{N-1}{x}_{{n}_{i}}-{U}_{N-1}q\parallel }^{2}+{\alpha }_{3}^{N}{\parallel {x}_{{n}_{i}}-q\parallel }^{2}\right)\\ =& \underset{i\to \mathrm{\infty }}{lim inf}\left(\left(1-{\alpha }_{3}^{N}\right)\parallel {\alpha }_{1}^{N-1}\left({T}_{N-1}{U}_{N-2}{x}_{{n}_{i}}-{T}_{N-1}{U}_{N-2}q\right)\\ +{\alpha }_{2}^{N-1}\left({U}_{N-2}{x}_{{n}_{i}}-{U}_{N-2}q\right)+{\alpha }_{3}^{N-1}\left({x}_{{n}_{i}}-q\right){\parallel }^{2}+{\alpha }_{3}^{N}{\parallel {x}_{{n}_{i}}-q\parallel }^{2}\right)\\ \le & \underset{i\to \mathrm{\infty }}{lim inf}\left(\left(1-{\alpha }_{3}^{N}\right)\left({\alpha }_{1}^{N-1}{\parallel {T}_{N-1}{U}_{N-2}{x}_{{n}_{i}}-{T}_{N-1}{U}_{N-2}q\parallel }^{2}\\ +{\alpha }_{2}^{N-1}{\parallel {U}_{N-2}{x}_{{n}_{i}}-{U}_{N-2}q\parallel }^{2}+{\alpha }_{3}^{N-1}{\parallel {x}_{{n}_{i}}-q\parallel }^{2}\right)+{\alpha }_{3}^{N}{\parallel {x}_{{n}_{i}}-q\parallel }^{2}\right)\\ \le & \underset{i\to \mathrm{\infty }}{lim inf}\left(\left(1-{\alpha }_{3}^{N}\right)\left({\alpha }_{1}^{N-1}\left({\parallel {U}_{N-2}{x}_{{n}_{i}}-{U}_{N-2}q\parallel }^{2}\\ +2〈{U}_{N-2}{x}_{{n}_{i}}-{T}_{N-1}{U}_{N-2}{x}_{{n}_{i}},{U}_{N-2}q-{T}_{N-1}{U}_{N-2}q〉\right)\\ +{\alpha }_{2}^{N-1}{\parallel {U}_{N-2}{x}_{{n}_{i}}-{U}_{N-2}q\parallel }^{2}+{\alpha }_{3}^{N-1}{\parallel {x}_{{n}_{i}}-q\parallel }^{2}\right)+{\alpha }_{3}^{N}{\parallel {x}_{{n}_{i}}-q\parallel }^{2}\right)\\ =& \underset{i\to \mathrm{\infty }}{lim inf}\left(\left(1-{\alpha }_{3}^{N}\right)\left(\left(1-{\alpha }_{3}^{N-1}\right){\parallel {U}_{N-2}{x}_{{n}_{i}}-{U}_{N-2}q\parallel }^{2}\\ +{\alpha }_{3}^{N-1}{\parallel {x}_{{n}_{i}}-q\parallel }^{2}\right)+{\alpha }_{3}^{N}{\parallel {x}_{{n}_{i}}-q\parallel }^{2}\right)\\ =& \underset{i\to \mathrm{\infty }}{lim inf}\left(\prod _{j=N-1}^{N}\left(1-{\alpha }_{3}^{j}\right){\parallel {U}_{N-2}{x}_{{n}_{i}}-{U}_{N-2}q\parallel }^{2}\\ +\left(1-\prod _{j=N-1}^{N}\left(1-{\alpha }_{3}^{j}\right)\right){\parallel {x}_{{n}_{i}}-q\parallel }^{2}\right)\\ \le \\ ⋮\\ \le & \underset{i\to \mathrm{\infty }}{lim inf}\left(\prod _{j=1}^{N}\left(1-{\alpha }_{3}^{j}\right){\parallel {U}_{0}{x}_{{n}_{i}}-{U}_{0}q\parallel }^{2}\\ +\left(1-\prod _{j=1}^{N}\left(1-{\alpha }_{3}^{j}\right)\right){\parallel {x}_{{n}_{i}}-q\parallel }^{2}\right)\\ =& \underset{i\to \mathrm{\infty }}{lim inf}\left(\prod _{j=1}^{N}\left(1-{\alpha }_{3}^{j}\right){\parallel {x}_{{n}_{i}}-q\parallel }^{2}\\ +\left(1-\prod _{j=1}^{N}\left(1-{\alpha }_{3}^{j}\right)\right){\parallel {x}_{{n}_{i}}-q\parallel }^{2}\right)\\ =& \underset{i\to \mathrm{\infty }}{lim inf}{\parallel {x}_{{n}_{i}}-q\parallel }^{2}.\end{array}$
This is a contradiction. Then, we have $q\in F\left(S\right)$. From Lemma 2.9, we have
$q\in \bigcap _{i=1}^{N}F\left({T}_{i}\right).$
(3.23)

From (3.22) and (3.23), we have $q\in \mathfrak{F}$. Hence, $\omega \left({x}_{n}\right)\subset \mathfrak{F}$. Therefore, by (3.5) and Lemma 2.8, we have $\left\{{x}_{n}\right\}$ converges strongly to ${P}_{\mathfrak{F}}{x}_{1}$. This completes the proof. □

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

Corollary 3.2 Let C be a nonempty closed convex subset of a Hilbert space H. For every $i=1,2,\dots ,N$, let ${A}_{i}:C\to H$ be an ${\alpha }_{i}$-inverse strongly monotone mapping, and let $T:C\to C$ be a nonspreading mapping with $\mathfrak{F}=F\left(T\right)\cap {\bigcap }_{i=1}^{N}VI\left(C,{A}_{i}\right)\ne \mathrm{\varnothing }$. For every $i=1,2,\dots ,N$, define the mapping ${G}_{i}:C\to C$ by ${G}_{i}x={P}_{C}\left(I-\lambda {A}_{i}\right)x$ $\mathrm{\forall }x\in C$ and $\lambda \in \left[c,d\right]\subset \left(0,2{\alpha }_{i}\right)$. Let $\left\{{x}_{n}\right\}$ be a sequence generated by ${x}_{1}\in {C}_{1}=C$ and
$\left\{\begin{array}{c}{z}_{n}={\sum }_{i=1}^{N}{\delta }_{n}^{i}{G}_{i}{x}_{n},\hfill \\ {y}_{n}={\alpha }_{n}{x}_{n}+{\beta }_{n}T{x}_{n}+{\gamma }_{n}{z}_{n},\hfill \\ {C}_{n+1}=\left\{z\in {C}_{n}:\parallel {y}_{n}-z\parallel \le \parallel {x}_{n}-z\parallel \right\},\hfill \\ {x}_{n+1}={P}_{{C}_{n+1}}{x}_{1},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,\hfill \end{array}$
(3.24)
where $\left\{{\alpha }_{n}\right\},\left\{{\beta }_{n}\right\},\left\{{\gamma }_{n}\right\}\subseteq \left[0,1\right]$, ${\alpha }_{n}+{\beta }_{n}+{\gamma }_{n}=1$ and suppose the following conditions hold:

Then the sequence $\left\{{x}_{n}\right\}$ converges strongly to ${P}_{\mathfrak{F}}{x}_{1}$.

Corollary 3.3 Let C be a nonempty closed convex subset of a Hilbert space H. Let $A:C\to H$ be an α-inverse strongly monotone mapping, and let ${\left\{{T}_{i}\right\}}_{i=1}^{N}$ be a finite family of nonspreading mappings with $\mathfrak{F}={\bigcap }_{i=1}^{N}F\left({T}_{i}\right)\cap VI\left(C,A\right)\ne \mathrm{\varnothing }$. Let ${\rho }_{j}=\left({\alpha }_{1}^{j},{\alpha }_{2}^{j},{\alpha }_{3}^{j}\right)\in I×I×I$, $j=1,2,3,\dots ,N$, where $I=\left[0,1\right]$, ${\alpha }_{1}^{j}+{\alpha }_{2}^{j}+{\alpha }_{3}^{j}=1$, ${\alpha }_{1}^{j},{\alpha }_{3}^{j}\in \left(0,1\right)$ for all $j=1,2,\dots ,N-1$ and ${\alpha }_{1}^{N}\in \left(0,1\right]$, ${\alpha }_{3}^{N}\in \left[0,1\right)$, ${\alpha }_{2}^{j}\in \left(0,1\right)$ for all $j=1,2,\dots ,N$, and let S be the S-mapping generated by ${T}_{1},{T}_{2},\dots ,{T}_{N}$ and ${\rho }_{1},{\rho }_{2},\dots ,{\rho }_{N}$. Let $\left\{{x}_{n}\right\}$ be a sequence generated by ${x}_{1}\in {C}_{1}=C$ and
$\left\{\begin{array}{c}{y}_{n}={\alpha }_{n}{x}_{n}+{\beta }_{n}S{x}_{n}+{\gamma }_{n}{P}_{C}\left(I-\lambda A\right){x}_{n},\hfill \\ {C}_{n+1}=\left\{z\in {C}_{n}:\parallel {y}_{n}-z\parallel \le \parallel {x}_{n}-z\parallel \right\},\hfill \\ {x}_{n+1}={P}_{{C}_{n+1}}{x}_{1},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,\hfill \end{array}$
(3.25)

where $\left\{{\alpha }_{n}\right\},\left\{{\beta }_{n}\right\},\left\{{\gamma }_{n}\right\}\subseteq \left[a,b\right]\subset \left(0,1\right)$, ${\alpha }_{n}+{\beta }_{n}+{\gamma }_{n}=1$ and $\lambda \subseteq \left[c,d\right]\subset \left(0,2\alpha \right)$. Then the sequence $\left\{{x}_{n}\right\}$ converges strongly to ${P}_{\mathfrak{F}}{x}_{1}$.

## 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

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