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

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.

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 $\mathrm{\forall }p\in F(T)$, where $F(T)$ 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 $VI(C,A)$.

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., [2–5].

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. ⇀ for weak convergence and → for strong convergence.

• 2. $\omega (x_n)=\{x:\mathrm{\exists }{x}_{n_i}\rightharpoonup x\}$ denotes the weak ω-limit set of $\{x_n\}$.

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 $F(T)\ne \mathrm{\varnothing }$ and let $x_0\in H$. For $C_1=C$ and $u_1\in P_{C_1}{x}_0$, define a sequence $\{u_n\}$ of C as follows:

$$\{\begin{array}{c}{y}_n={\alpha }_n{u}_n+(1-{\alpha }_n)u_n,\hfill \\ C_{n+1}=\{z\in C_n:\parallel y_n-z\parallel \le \parallel u_n-z\parallel \},\hfill \\ u_{n+1}=P_{C_{n+1}}{x}_0,n\in \mathbb{N},\hfill \end{array}$$

where $0\le {\alpha }_n\le a<1$ for all $n\in \mathbb{N}$. Then $\{u_n\}$ converges strongly to $z_0=P_{F(T)}{x}_0$.

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 $F(S)\cap F(T)\ne \mathrm{\varnothing }$. Define a sequence $\{x_n\}$ as follows.

$$\{\begin{array}{c}{x}_1\in C,\hfill \\ x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)({\beta }_{n}S{x}_n+(1-{\beta }_n)T{x}_n)\hfill \end{array}$$

for all $n\in \mathbb{N}$, where $\{{\alpha }_n\},\{{\beta }_n\}\subset [0,1]$. Then the following hold:

• (i) If ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$ and ${\sum }_{n=1}^{\mathrm{\infty }}(1-{\beta }_n)<\mathrm{\infty }$, then $\{x_n\}$ converges weakly to $v\in F(S)$.

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

• (iii) If ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$ and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n(1-{\beta }_n)>0$, then $\{x_n\}$ converges weakly to $v\in F(S)\cap F(T)$.

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:

$$〈x-y,y-z〉\ge 0\mathrm{\forall }z\in C.$$

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

$$Tx=\{\begin{array}{c}x-1\text{if }x\in (-\mathrm{\infty },0],\hfill \\ -(x+1)\text{if }x\in (0,\mathrm{\infty })\hfill \end{array}$$

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

To see that T is a nonspreading mapping, if $x,y\in (0,\mathrm{\infty })$, then we have $Tx=-(x+1)$ and $Ty=-(y+1)$. From the definition of the mapping T, we have

$$\begin{array}{rcl}{|Tx-Ty|}^2& =& {|-(x+1)-(-(y+1))|}^2\\ =& {|y-x|}^2={|x-y|}^2\end{array}$$

and

$$\begin{array}{rcl}2〈x-Tx,y-Ty〉& =& 2〈x+x+1,y+y+1〉\\ =& 2〈2x+1,2y+1〉\\ =& 2(2x+1)(2y+1)>0(\text{since }x,y>0).\end{array}$$

The above implies that

$${|Tx-Ty|}^2={|x-y|}^2<{|x-y|}^2+2〈x-Tx,y-Ty〉.$$

For every $x,y\in (-\mathrm{\infty },0]$, we have $Tx=x-1$ and $Ty=y-1$. From the definition of T, we have

$$\begin{array}{rcl}{|Tx-Ty|}^2& =& {|x-1-(y-1)|}^2\\ =& {|x-y|}^2,\end{array}$$

and

$$2〈x-Tx,y-Ty〉=2〈x-(x-1),y-(y-1)〉=2.$$

From above, we have

$${|Tx-Ty|}^2={|x-y|}^2<{|x-y|}^2+2〈x-Tx,y-Ty〉.$$

Finally, for every $x\in (-\mathrm{\infty },0]$ and $y\in (0,\mathrm{\infty })$, we have $Tx=x-1$ and $Ty=-(y+1)$. From the definition of T, we have

and

$$\begin{array}{rcl}2〈x-Tx,y-Ty〉& =& 2〈x-(x-1),y+(y+1)〉\\ =& 2〈1,2y+1〉\\ =& 2(2y+1)>0(\text{since }y>0).\end{array}$$

From above, we have

$$\begin{array}{rcl}{|Tx-Ty|}^2& =& {|x+y|}^2={(x+y)}^2\\ \le & {|x-y|}^2\\ <& {|x-y|}^2+2〈x-Tx,y-Ty〉.\end{array}$$

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 $F(S)$ 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 $\lambda >0$,

$$u=P_C(I-\lambda A)u\iff u\in VI(C,A),$$

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 $\{T_n:n\in \mathbb{N}\}$ be a sequence of nonexpansive mappings on C. Suppose ${\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)$ is nonempty. Let $\{{\lambda }_n\}$ be a sequence of positive numbers with ${\sum }_{n=1}^{\mathrm{\infty }}{\lambda }_n=1$. Then a mapping S on C defined by

$$S(x)=\sum _{n=1}^{\mathrm{\infty }}{\lambda }_n{T}_{n}x$$

for $x\in C$ is well defined, nonexpansive and $F(S)={\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)$ 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 $\{x_n\}$ be a sequence in H and $u\in H$. Let $q=P_{C}u$. If $\{x_n\}$ is such that $\omega (x_n)\subset C$ and satisfies the condition

$$\parallel x_n-u\parallel \le \parallel u-q\parallel ,\mathrm{\forall }n\in \mathbb{N},$$

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 ${\{T_i\}}_{i=1}^N$ be a finite family of (nonexpansive) mappings of C into itself. For each $j=1,2,\dots ,N$, let ${\alpha }_j=({\alpha }_1^j,{\alpha }_2^j,{\alpha }_3^j)\in I\times I\times I$, where $I\in [0,1]$ 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 ${\{T_i\}}_{i=1}^N$ be a finite family of nonspreading mappings of C into C with ${\bigcap }_{i=1}^{N}F(T_i)\ne \mathrm{\varnothing }$, and let ${\alpha }_j=({\alpha }_1^j,{\alpha }_2^j,{\alpha }_3^j)\in I\times I\times I$, $j=1,2,3,\dots ,N$, where $I=[0,1]$, ${\alpha }_1^j+{\alpha }_2^j+{\alpha }_3^j=1$, ${\alpha }_1^j,{\alpha }_3^j\in (0,1)$ for all $j=1,2,\dots ,N-1$ and ${\alpha }_1^N\in (0,1]$, ${\alpha }_3^N\in [0,1)$, ${\alpha }_2^j\in [0,1)$ 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 $F(S)={\bigcap }_{i=1}^{N}F(T_i)$ and S is a quasi-nonexpansive mapping.

Proof It easy to see that ${\bigcap }_{i=1}^{N}F(T_i)\subseteq F(S)$. Let $x_0\in F(S)$ and $x^{\ast }\in {\bigcap }_{i=1}^{N}F(T_i)$. Since ${\{T_i\}}_{i=1}^N$ is a finite family of nonspreading mappings of C into itself, for every $y\in C$, we have

$${\parallel T_{i}y-x^{\ast }\parallel }^2\le \frac{1}{2}({\parallel T_{i}y-x^{\ast }\parallel }^2+{\parallel y-x^{\ast }\parallel }^2).$$

(2.3)

This implies that

$${\parallel T_{i}y-x^{\ast }\parallel }^2\le {\parallel y-x^{\ast }\parallel }^2,\mathrm{\forall }y\in C\text{ and }i=1,2,\dots ,N.$$

(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(T_N{U}_{N-1}{x}_0-x^{\ast })+{\alpha }_2^N(U_{N-1}{x}_0-x^{\ast })+{\alpha }_3^N(x_0-x^{\ast })\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 & (1-{\alpha }_3^N){\parallel U_{N-1}{x}_0-x^{\ast }\parallel }^2+{\alpha }_3^N{\parallel x_0-x^{\ast }\parallel }^2\\ =& (1-{\alpha }_3^N)\parallel {\alpha }_1^{N-1}(T_{N-1}{U}_{N-2}{x}_0-x^{\ast })+{\alpha }_2^{N-1}(U_{N-2}{x}_0-x^{\ast })\\ +{\alpha }_3^{N-1}(x_0-x^{\ast }){\parallel }^2+{\alpha }_3^N{\parallel x_0-x^{\ast }\parallel }^2\\ \le & (1-{\alpha }_3^N)({\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)+{\alpha }_3^N{\parallel x_0-x^{\ast }\parallel }^2\\ \le & (1-{\alpha }_3^N)((1-{\alpha }_3^{N-1}){\parallel U_{N-2}{x}_0-x^{\ast }\parallel }^2+{\alpha }_3^{N-1}{\parallel x_0-x^{\ast }\parallel }^2)\\ +{\alpha }_3^N{\parallel x_0-x^{\ast }\parallel }^2\\ =& (1-{\alpha }_3^N)(1-{\alpha }_3^{N-1}){\parallel U_{N-2}{x}_0-x^{\ast }\parallel }^2+{\alpha }_3^{N-1}(1-{\alpha }_3^N){\parallel x_0-x^{\ast }\parallel }^2\\ +{\alpha }_3^N{\parallel x_0-x^{\ast }\parallel }^2\\ =& \prod _{j=N-1}^N(1-{\alpha }_3^j){\parallel U_{N-2}{x}_0-x^{\ast }\parallel }^2+(1-\prod _{j=N-1}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ ⋮\\ \le & \prod _{j=3}^N(1-{\alpha }_3^j){\parallel U_2{x}_0-x^{\ast }\parallel }^2+(1-\prod _{j=3}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ \le & \prod _{j=2}^N(1-{\alpha }_3^j){\parallel U_1{x}_0-x^{\ast }\parallel }^2+(1-\prod _{j=2}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ =& \prod _{j=2}^N(1-{\alpha }_3^j){\parallel {\alpha }_1^1(T_1{x}_0-x^{\ast })+(1-{\alpha }_1^1)(x_0-x^{\ast })\parallel }^2\\ +(1-\prod _{j=2}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ =& \prod _{j=2}^N(1-{\alpha }_3^j)({\alpha }_1^1{\parallel T_1{x}_0-x^{\ast }\parallel }^2+(1-{\alpha }_1^1){\parallel x_0-x^{\ast }\parallel }^2\\ -{\alpha }_1^1(1-{\alpha }_1^1)\parallel T_1{x}_0-x_0\parallel )+(1-\prod _{j=2}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ \le & \prod _{j=2}^N(1-{\alpha }_3^j)({\parallel x_0-x^{\ast }\parallel }^2-{\alpha }_1^1(1-{\alpha }_1^1){\parallel T_1{x}_0-x_0\parallel }^2)\\ +(1-\prod _{j=2}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2.\end{array}$$

(2.5)

From (2.5), we have

$$\begin{array}{rcl}{\parallel x_0-x^{\ast }\parallel }^2& \le & \prod _{j=2}^N(1-{\alpha }_3^j)({\parallel x_0-x^{\ast }\parallel }^2-{\alpha }_1^1(1-{\alpha }_1^1){\parallel T_1{x}_0-x_0\parallel }^2)\\ +(1-\prod _{j=2}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2,\end{array}$$

which implies that

$${\parallel x_0-x^{\ast }\parallel }^2\le {\parallel x_0-x^{\ast }\parallel }^2-{\alpha }_1^1(1-{\alpha }_1^1){\parallel T_1{x}_0-x_0\parallel }^2.$$

(2.6)

Since ${\alpha }_1^j\in (0,1)$ for all $j=1,2,\dots ,N-1$ and (2.6), we have $x_0\in F(T_1)$. 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 $x_0\in F(U_1)$, we have

$$\begin{array}{rcl}{\parallel x_0-x^{\ast }\parallel }^2& \le & \prod _{j=3}^N(1-{\alpha }_3^j){\parallel U_2{x}_0-x^{\ast }\parallel }^2+(1-\prod _{j=3}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ =& \prod _{j=3}^N(1-{\alpha }_3^j){\parallel {\alpha }_1^2{T}_2{U}_1{x}_0+{\alpha }_2^2{U}_1{x}_0+{\alpha }_3^2{x}_0-x^{\ast }\parallel }^2\\ +(1-\prod _{j=3}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ =& \prod _{j=3}^N(1-{\alpha }_3^j){\parallel {\alpha }_1^2(T_2{x}_0-x^{\ast })+(1-{\alpha }_1^2)(x_0-x^{\ast })\parallel }^2\\ +(1-\prod _{j=3}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ =& \prod _{j=3}^N(1-{\alpha }_3^j)({\alpha }_1^2{\parallel T_2{x}_0-x^{\ast }\parallel }^2+(1-{\alpha }_1^2){\parallel x_0-x^{\ast }\parallel }^2\\ -{\alpha }_1^2(1-{\alpha }_1^2){\parallel T_2{x}_0-x_0\parallel }^2)\\ +(1-\prod _{j=3}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ \le & \prod _{j=3}^N(1-{\alpha }_3^j)({\parallel x_0-x^{\ast }\parallel }^2-{\alpha }_1^2(1-{\alpha }_1^2){\parallel T_2{x}_0-x_0\parallel }^2)\\ +(1-\prod _{j=3}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2,\end{array}$$

which implies that

$${\parallel x_0-x^{\ast }\parallel }^2\le {\parallel x_0-x^{\ast }\parallel }^2-{\alpha }_1^2(1-{\alpha }_1^2){\parallel T_2{x}_0-x_0\parallel }^2.$$

(2.7)

Since ${\alpha }_1^j\in (0,1)$ for all $j=1,2,\dots ,N-1$ and (2.7), we have $x_0\in F(T_2)$. 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 $x_0\in F(T_i)$ and $x_0\in F(U_i)$ for all $i=1,2,\dots ,N-1$.

Finally, we shall show that $x_0\in F(T_N)$.

Since

$$\begin{array}{rcl}0& =& S{x}_0-x_0={\alpha }_1^N{T}_N{U}_{N-1}{x}_0+{\alpha }_2^N{U}_{N-1}{x}_0+{\alpha }_3^N{x}_0-x_0\\ =& {\alpha }_1^N(T_N{x}_0-x_0),\end{array}$$

and ${\alpha }_1^N\in (0,1]$, we obtain $T_N{x}_0=x_0$ so that $x_0\in F(T_N)$. Then we have $x_0\in {\bigcap }_{i=1}^{N}F(T_i)$. Hence, $F(S)\subseteq {\bigcap }_{i=1}^{N}F(T_i)$.

Next, we show that S is a quasi-nonexpansive mapping. Let $x\in C$ and $y\in F(S)$. From (2.5), we can imply that

$$\begin{array}{rcl}{\parallel Sx-y\parallel }^2& \le & \prod _{j=2}^N(1-{\alpha }_3^j)({\parallel x-y\parallel }^2-{\alpha }_1^1(1-{\alpha }_1^1)\parallel T_{1}x-x\parallel )\\ +(1-\prod _{j=2}^N(1-{\alpha }_3^j)){\parallel x-y\parallel }^2\\ \le & {\parallel x-y\parallel }^2.\end{array}$$

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

Example 2.10 Let $T_1:[-1,1]\to [-1,1]$ be a mapping defined by

$$T_{1}x=\{\begin{array}{cc}\frac{x+1}{2}\hfill & \text{if }x\in (0,1],\hfill \\ \frac{-x+1}{2}\hfill & \text{if }x\in [-1,0]\hfill \end{array}$$

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

Let $T_2:[-1,1]\to [-1,1]$ be a mapping defined by

$$T_{2}x=\{\begin{array}{cc}\frac{x+2}{3}\hfill & \text{if }x\in (0,1],\hfill \\ \frac{-x+2}{3}\hfill & \text{if }x\in [-1,0]\hfill \end{array}$$

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

To see that $T_1$ is a nonspreading mapping, observe that if $x,y\in (0,1]$, 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

$$\begin{array}{rcl}2〈x-T_{1}x,y-T_{1}y〉& =& 2〈x-\left(\frac{x+1}{2}\right),y-\left(\frac{y+1}{2}\right)〉\\ =& 2〈\frac{x-1}{2},\frac{y-1}{2}〉\\ =& \frac{1}{2}(x-1)(y-1)\\ \ge & 0(\text{since }x\le 1,y\le 1,\text{ then }(x-1)(y-1)\ge 0).\end{array}$$

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 $x,y\in [-1,0]$, 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

$$\begin{array}{rcl}{|T_{1}x-T_{1}y|}^2& =& {|\frac{-x+1}{2}-\left(\frac{-y+1}{2}\right)|}^2\\ =& {\left|\frac{y-x}{2}\right|}^2\\ =& \frac{1}{4}{|x-y|}^2\end{array}$$

and

$$\begin{array}{rcl}2〈x-T_{1}x,y-T_{1}y〉& =& 2〈x-\left(\frac{1-x}{2}\right),y-\left(\frac{1-y}{2}\right)〉\\ =& 2〈\frac{3x-1}{2},\frac{3y-1}{2}〉\\ =& \frac{1}{2}(3x-1)(3y-1)\\ =& \frac{1}{2}(3x(3y-1)-(3y-1))\\ =& \frac{1}{2}(9xy-3x-3y+1)\\ >& 0(\text{since }-1\le x,y\le 0,\text{ then }9xy,-3x,-3y\ge 0).\end{array}$$

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 $x\in (0,1]$ and $y\in [-1,0]$, 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

$$\begin{array}{rcl}2〈x-T_{1}x,y-T_{1}y〉& =& 2〈x-\left(\frac{x+1}{2}\right),y-\left(\frac{-y+1}{2}\right)〉\\ =& 2〈\frac{x-1}{2},\frac{3y-1}{2}〉\\ =& \frac{1}{2}(x-1)(3y-1)\\ =& \frac{1}{2}(x(3y-1)-(3y-1))\\ =& \frac{1}{2}(3xy-x-3y+1)\\ =& \frac{1}{2}(3y(x-1)+(1-x))\\ \ge & 0(\text{since }0<x\le 1\text{ and }-1\le y\le 0,\text{ then }3y(x-1),(1-x)\ge 0).\end{array}$$

From above, we have

$$\begin{array}{rcl}{|x-y|}^2+2〈x-T_{1}x,y-T_{1}y〉& \ge & {|x-y|}^2\\ =& x^2-2xy+y^2\\ =& x^2+2xy+y^2-4xy\\ \ge & x^2+2xy+y^2(\text{since }-4xy\ge 0)\\ =& {(x+y)}^2\\ \ge & \frac{1}{4}{(x+y)}^2\\ =& {|T_{1}x-T_{1}y|}^2.\end{array}$$

Then for all $x,y\in [-1,1]$, 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 $x,y\in (0,1]$, 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

$$\begin{array}{rcl}2〈x-T_{2}x,y-T_{2}y〉& =& 2〈x-\left(\frac{x+2}{3}\right),y-\left(\frac{y+2}{3}\right)〉\\ =& 2〈\frac{2x-2}{3},\frac{2y-2}{3}〉\\ =& \frac{8}{9}(x-1)(y-1)\\ \ge & 0(\text{since }0<x,y\le 1,\text{ then }(x-1)(y-1)\ge 0).\end{array}$$

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 $x,y\in [-1,0]$, 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

$$\begin{array}{rcl}{|T_{2}x-T_{2}y|}^2& =& {|\frac{2-x}{3}-\frac{2-y}{3}|}^2\\ =& {\left|\frac{y-x}{3}\right|}^2\\ =& \frac{1}{9}{|x-y|}^2\end{array}$$

and

$$\begin{array}{rcl}2〈x-T_{2}x,y-T_{2}y〉& =& 2〈x-\left(\frac{2-x}{3}\right),y-\left(\frac{2-y}{3}\right)〉\\ =& 2〈\frac{4x-2}{3},\frac{4y-2}{3}〉\\ =& \frac{8}{9}(2x-1)(2y-1)\\ =& \frac{8}{9}(2x(2y-1)-(2y-1))\\ =& \frac{8}{9}(4xy-2x-2y+1)\\ >& 0(\text{since }-1\le x,y\le 0,\text{ then }4xy,-2x,-2y\ge 0).\end{array}$$

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 $x\in (0,1]$ and $y\in [-1,0]$, 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

$$\begin{array}{rcl}2〈x-T_{2}x,y-T_{2}y〉& =& 2〈x-\left(\frac{x+2}{3}\right),y-\left(\frac{2-y}{3}\right)〉\\ =& 2〈\frac{2x-2}{3},\frac{4y-2}{3}〉\\ =& \frac{8}{9}(x-1)(2y-1)\\ =& \frac{8}{9}(x(2y-1)-(2y-1))\\ =& \frac{8}{9}(2xy-x-2y+1)\\ =& \frac{8}{9}(2y(x-1)+(1-x))\\ \ge & 0(\text{since }0<x\le 1\text{ and }-1\le y\le 0,\text{ then }2y(x-1),(1-x)\ge 0).\end{array}$$

From above, we have

$$\begin{array}{rcl}{|x-y|}^2+2〈x-T_{2}x,y-T_{2}y〉& \ge & {|x-y|}^2\\ =& x^2-2xy+y^2\\ =& x^2+2xy+y^2-4xy\\ \ge & {(x+y)}^2(\text{since }-4xy\ge 0)\\ \ge & \frac{1}{9}{|x+y|}^2\\ =& {|T_{2}x-T_{2}y|}^2.\end{array}$$

Then for every $x,y\in [-1,1]$, 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 $1\in F(T_1)\cap F(T_2)$. Let the mapping $S:[-1,1]\to [-1,1]$ be the S-mapping generated by $T_1$, $T_2$ and ${\alpha }_1$, ${\alpha }_2$, where ${\alpha }_1=(\frac{1}{6},\frac{2}{6},\frac{3}{6})$ and $(\frac{4}{15},\frac{5}{15},\frac{6}{15})$. From Lemma 2.9, we have $1\in F(S)$.

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 ${\{T_i\}}_{i=1}^N$ be a finite family of nonspreading mappings with $\mathfrak{F}={\bigcap }_{i=1}^{N}F(T_i)\cap {\bigcap }_{i=1}^{N}VI(C,A_i)\ne \mathrm{\varnothing }$. For every $i=1,2,\dots ,N$, define the mapping $G_i:C\to C$ by $G_{i}x=P_C(I-\lambda A_i)x$ $\mathrm{\forall }x\in C$ and $\lambda \in [c,d]\subset (0,2{\alpha }_i)$. Let ${\rho }_j=({\alpha }_1^j,{\alpha }_2^j,{\alpha }_3^j)\in I\times I\times I$, $j=1,2,3,\dots ,N$, where $I=[0,1]$, ${\alpha }_1^j+{\alpha }_2^j+{\alpha }_3^j=1$, ${\alpha }_1^j,{\alpha }_3^j\in (0,1)$ for all $j=1,2,\dots ,N-1$ and ${\alpha }_1^N\in (0,1]$, ${\alpha }_3^N\in [0,1)$ ${\alpha }_2^j\in (0,1)$ 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 $\{x_n\}$ be a sequence generated by $x_1\in C_1=C$ and

$$\{\begin{array}{c}{z}_n={\sum }_{i=1}^N{\delta }_n^i{G}_i{x}_n,\hfill \\ y_n={\alpha }_n{x}_n+{\beta }_{n}S{x}_n+{\gamma }_n{z}_n,\hfill \\ C_{n+1}=\{z\in C_n:\parallel y_n-z\parallel \le \parallel x_n-z\parallel \},\hfill \\ x_{n+1}=P_{C_{n+1}}{x}_1,\mathrm{\forall }n\ge 1,\hfill \end{array}$$

(3.1)

where $\{{\alpha }_n\},\{{\beta }_n\},\{{\gamma }_n\}\subseteq [0,1]$, ${\alpha }_n+{\beta }_n+{\gamma }_n=1$ and suppose the following conditions hold:

Then the sequence $\{x_n\}$ converges strongly to $P_{\mathfrak{F}}{x}_1$.

Proof First, we show that $(I-\lambda A_i)$ 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

$$\begin{array}{rcl}{\parallel (I-\lambda A_i)x-(I-\lambda A_i)y\parallel }^2& =& {\parallel x-y-\lambda (A_{i}x-A_{i}y)\parallel }^2\\ =& {\parallel x-y\parallel }^2-2\lambda 〈x-y,A_{i}x-A_{i}y〉+{\lambda }^2{\parallel A_{i}x-A_{i}y\parallel }^2\\ \le & {\parallel x-y\parallel }^2-2{\alpha }_i\lambda {\parallel A_{i}x-A_{i}y\parallel }^2+{\lambda }^2{\parallel A_{i}x-A_{i}y\parallel }^2\\ =& {\parallel x-y\parallel }^2+\lambda (\lambda -2{\alpha }_i){\parallel A_{i}x-A_{i}y\parallel }^2\\ \le & {\parallel x-y\parallel }^2.\end{array}$$

Thus $(I-\lambda A_i)$ 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

$$F(G_i)=F(P_C(I-\lambda A_i))=VI(C,A_i),\mathrm{\forall }i=1,2,\dots ,N.$$

(3.2)

From (3.2), $VI(C,A_i)$ is closed and convex. Let $z\in \mathfrak{F}$. From (3.2), we have $z\in F(P_C(I-\lambda A_i))$ for every $i=1,2,\dots ,N$. By nonexpansiveness of $G_i$, we have

$$\parallel z_n-z\parallel =\parallel \sum _{i=1}^N{\delta }_n^i(G_i{x}_n-z)\parallel \le \sum _{i=1}^N{\delta }_n^i\parallel x_n-z\parallel =\parallel x_n-z\parallel .$$

(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 $z\in C_n$,

$$\parallel y_n-z\parallel \le \parallel x_n-z\parallel \text{is equivalent to}{\parallel y_n-x_n\parallel }^2+2〈y_n-x_n,x_n-z〉\le 0.$$

So, for every $z_1,z_2\in C_n$ and $t\in (0,1)$, it follows that

then, we have $C_n$ is convex. Since $VI(C,A_i)$ is closed and convex for every $i=1,2,\dots ,N$, we have ${\bigcap }_{i=1}^{N}VI(C,A_i)$ is closed and convex. From Lemma 2.4, we have ${\bigcap }_{i=1}^{N}F(T_i)$ 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

$$\begin{array}{rcl}\parallel y_n-z\parallel & =& \parallel {\alpha }_n(x_n-z)+{\beta }_n(S{x}_n-z)+{\gamma }_n(z_n-z)\parallel \\ \le & {\alpha }_n\parallel x_n-z\parallel +{\beta }_n\parallel S{x}_n-z\parallel +{\gamma }_n\parallel z_n-z\parallel \\ \le & \parallel x_n-z\parallel .\end{array}$$

It follows that $z\in C_n$. Hence, we have $\mathfrak{F}\subset C_n$ for every $n\in \mathbb{N}$. This implies that $\{x_n\}$ is well defined. Since $x_n=P_{C_n}{x}_1$, for every $w\in C_n$, we have

$$\parallel x_n-x_1\parallel \le \parallel w-x_1\parallel ,\mathrm{\forall }n\in \mathbb{N}.$$

(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 $\{x_n\}$ is bounded, so are $\{G_i{x}_n\}$, $\{T_i{x}_n\}$ for every $i=1,2,\dots ,N$, $\{z_n\}$, $\{y_n\}$ and $\{S{x}_n\}$. 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

$$\begin{array}{rcl}{\parallel y_n-z\parallel }^2& \le & {\alpha }_n{\parallel x_n-z\parallel }^2+{\beta }_n{\parallel S{x}_n-z\parallel }^2+{\gamma }_n{\parallel z_n-z\parallel }^2\\ \le & {\alpha }_n{\parallel x_n-z\parallel }^2+{\beta }_n{\parallel S{x}_n-z\parallel }^2+{\gamma }_n\sum _{i=1}^N{\delta }_n^i{\parallel P_C(I-\lambda A_i)x_n-z\parallel }^2\\ \le & {\alpha }_n{\parallel x_n-z\parallel }^2+{\beta }_n{\parallel S{x}_n-z\parallel }^2\\ +{\gamma }_n\sum _{i=1}^N{\delta }_n^i({\parallel x_n-z\parallel }^2-\lambda (2{\alpha }_i-\lambda ){\parallel A_i{x}_n-A_{i}z\parallel }^2)\\ =& {\alpha }_n{\parallel x_n-z\parallel }^2+{\beta }_n{\parallel S{x}_n-z\parallel }^2+{\gamma }_n{\parallel x_n-z\parallel }^2\\ -{\gamma }_n\sum _{i=1}^N{\delta }_n^i\lambda (2{\alpha }_i-\lambda ){\parallel A_i{x}_n-A_{i}z\parallel }^2\\ \le & {\parallel x_n-z\parallel }^2-{\gamma }_n\sum _{i=1}^N{\delta }_n^i\lambda (2{\alpha }_i-\lambda ){\parallel A_i{x}_n-A_{i}z\parallel }^2.\end{array}$$

It follows that

$$\begin{array}{rcl}{\gamma }_n\sum _{i=1}^N{\delta }_n^i\lambda (2{\alpha }_i-\lambda ){\parallel A_i{x}_n-A_{i}z\parallel }^2& \le & {\parallel x_n-z\parallel }^2-{\parallel y_n-z\parallel }^2\\ \le & (\parallel x_n-z\parallel +\parallel y_n-z\parallel )\parallel y_n-x_n\parallel .\end{array}$$

From conditions (i), (ii) and (3.9), it implies that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel A_i{x}_n-A_{i}z\parallel =0,\mathrm{\forall }i=1,2,\dots ,N.$$

(3.12)

Since

$$\begin{array}{rcl}{\parallel P_C(I-\lambda A_i)x_n-z\parallel }^2& \le & 〈(I-\lambda A_i)x_n-(I-\lambda A_i)z,P_C(I-\lambda A_i)x_n-z〉\\ =& \frac{1}{2}({\parallel (I-\lambda A_i)x_n-(I-\lambda A_i)z\parallel }^2+{\parallel P_C(I-\lambda A_i)x_n-z\parallel }^2\\ -{\parallel (I-\lambda A_i)x_n-(I-\lambda A_i)z-P_C(I-\lambda A_i)x_n+z\parallel }^2)\\ \le & \frac{1}{2}({\parallel x_n-z\parallel }^2+{\parallel P_C(I-\lambda A_i)x_n-z\parallel }^2\\ -{\parallel x_n-P_C(I-\lambda A_i)x_n-\lambda (A_i{x}_n-A_{i}z)\parallel }^2)\\ =& \frac{1}{2}({\parallel x_n-z\parallel }^2+{\parallel P_C(I-\lambda A_i)x_n-z\parallel }^2\\ -{\parallel x_n-P_C(I-\lambda A_i)x_n\parallel }^2-{\parallel \lambda (A_i{x}_n-A_{i}z)\parallel }^2\\ +2\lambda 〈x_n-P_C(I-\lambda A_i)x_n,A_i{x}_n-A_{i}z〉),\end{array}$$

it implies that

$$\begin{array}{rcl}{\parallel P_C(I-\lambda A_i)x_n-z\parallel }^2& \le & {\parallel x_n-z\parallel }^2-{\parallel x_n-P_C(I-\lambda A_i)x_n\parallel }^2\\ +2\lambda \parallel x_n-P_C(I-\lambda A_i)x_n\parallel \parallel A_i{x}_n-A_{i}z\parallel .\end{array}$$

(3.13)

From the definition of $y_n$ and (3.13), we have

$$\begin{array}{rcl}{\parallel y_n-z\parallel }^2& \le & {\alpha }_n{\parallel x_n-z\parallel }^2+{\beta }_n{\parallel S{x}_n-z\parallel }^2+{\gamma }_n{\parallel z_n-z\parallel }^2\\ \le & (1-{\gamma }_n){\parallel x_n-z\parallel }^2+{\gamma }_n\sum _{i=1}^N{\delta }_n^i{\parallel P_C(I-\lambda A_i)x_n-z\parallel }^2\\ \le & (1-{\gamma }_n){\parallel x_n-z\parallel }^2+{\gamma }_n\sum _{i=1}^N{\delta }_n^i({\parallel x_n-z\parallel }^2-{\parallel x_n-P_C(I-\lambda A_i)x_n\parallel }^2\\ +2\lambda \parallel x_n-P_C(I-\lambda A_i)x_n\parallel \parallel A_i{x}_n-A_{i}z\parallel )\\ =& {\parallel x_n-z\parallel }^2-{\gamma }_n\sum _{i=1}^N{\delta }_n^i{\parallel x_n-P_C(I-\lambda A_i)x_n\parallel }^2\\ +2{\gamma }_n\sum _{i=1}^N{\delta }_n^i\lambda \parallel x_n-P_C(I-\lambda A_i)x_n\parallel \parallel A_i{x}_n-A_{i}z\parallel ,\end{array}$$

which implies that

$$\begin{array}{rcl}{\gamma }_n\sum _{i=1}^N{\delta }_n^i{\parallel x_n-P_C(I-\lambda A_i)x_n\parallel }^2& \le & {\parallel x_n-z\parallel }^2-{\parallel y_n-z\parallel }^2\\ +2{\gamma }_n\sum _{i=1}^N{\delta }_n^i\lambda \parallel x_n-P_C(I-\lambda A_i)x_n\parallel \parallel A_i{x}_n-A_{i}z\parallel \\ \le & (\parallel x_n-z\parallel +\parallel y_n-z\parallel )\parallel y_n-x_n\parallel \\ +2{\gamma }_n\sum _{i=1}^N{\delta }_n^i\lambda \parallel x_n-P_C(I-\lambda A_i)x_n\parallel \parallel A_i{x}_n-A_{i}z\parallel .\end{array}$$

From conditions (i), (ii), (3.9) and (3.12), we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel P_C(I-\lambda A_i)x_n-x_n\parallel =0,\mathrm{\forall }i=1,2,\dots ,N.$$

(3.14)

Since

$$\parallel z_n-x_n\parallel \le \sum _{i=1}^N{\delta }_n^i\parallel P_C(I-\lambda A_i)x_n-x_n\parallel ,$$

from (3.14), we have

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

(3.15)

Since

$$y_n-x_n={\beta }_n(S{x}_n-x_n)+{\gamma }_n(z_n-x_n)$$

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

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T_i{U}_{i-1}{x}_n-U_{i-1}{x}_n\parallel =0,\mathrm{\forall }i=1,2,\dots ,N.$$

(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 & (1-{\beta }_n){\parallel x_n-z\parallel }^2+{\beta }_n\parallel {\alpha }_1^N(T_N{U}_{N-1}{x}_n-z)\\ +{\alpha }_2^N(U_{N-1}{x}_n-z)+{\alpha }_3^N(x_n-z)\parallel \\ \le & (1-{\beta }_n){\parallel x_n-z\parallel }^2+{\beta }_n({\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)\\ \le & (1-{\beta }_n){\parallel x_n-z\parallel }^2+{\beta }_n((1-{\alpha }_3^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)\\ =& (1-{\beta }_n){\parallel x_n-z\parallel }^2+{\beta }_n((1-{\alpha }_3^N)\parallel {\alpha }_1^{N-1}(T_{N-1}{U}_{N-2}{x}_n-z)\\ +{\alpha }_2^{N-1}(U_{N-2}{x}_n-z)+{\alpha }_3^{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)\\ \le & (1-{\beta }_n){\parallel x_n-z\parallel }^2+{\beta }_n\left((1-{\alpha }_3^N)\right({\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)\\ +{\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)\\ \le & (1-{\beta }_n){\parallel x_n-z\parallel }^2+{\beta }_n\left((1-{\alpha }_3^N)\right((1-{\alpha }_3^{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)\\ +{\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)\\ =& (1-{\beta }_n){\parallel x_n-z\parallel }^2+{\beta }_n((1-{\alpha }_3^N)(1-{\alpha }_3^{N-1}){\parallel U_{N-2}{x}_n-z\parallel }^2\\ +(1-{\alpha }_3^N){\alpha }_3^{N-1}{\parallel x_n-z\parallel }^2-{\alpha }_1^{N-1}{\alpha }_2^{N-1}(1-{\alpha }_3^N){\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)\\ =& (1-{\beta }_n){\parallel x_n-z\parallel }^2+{\beta }_n(\prod _{j=N-1}^N(1-{\alpha }_3^j){\parallel U_{N-2}{x}_n-z\parallel }^2\\ +(1-\prod _{j=N-1}^N(1-{\alpha }_3^j)){\parallel x_n-z\parallel }^2\\ -{\alpha }_1^{N-1}{\alpha }_2^{N-1}(1-{\alpha }_3^N){\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)\\ =& (1-{\beta }_n){\parallel x_n-z\parallel }^2+{\beta }_n(\prod _{j=N-1}^N(1-{\alpha }_3^j)\parallel {\alpha }_1^{N-2}(T_{N-2}{U}_{N-3}{x}_n-z)\\ +{\alpha }_2^{N-2}(U_{N-3}{x}_n-z)+{\alpha }_3^{N-2}(x_n-z){\parallel }^2\\ +(1-\prod _{j=N-1}^N(1-{\alpha }_3^j)){\parallel x_n-z\parallel }^2\\ -{\alpha }_1^{N-1}{\alpha }_2^{N-1}(1-{\alpha }_3^N){\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)\\ \le & (1-{\beta }_n){\parallel x_n-z\parallel }^2+{\beta }_n\left(\prod _{j=N-1}^N(1-{\alpha }_3^j)\right({\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)\\ +(1-\prod _{j=N-1}^N(1-{\alpha }_3^j)){\parallel x_n-z\parallel }^2\\ -{\alpha }_1^{N-1}{\alpha }_2^{N-1}(1-{\alpha }_3^N){\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)\\ \le & (1-{\beta }_n){\parallel x_n-z\parallel }^2+{\beta }_n\left(\prod _{j=N-1}^N(1-{\alpha }_3^j)\right((1-{\alpha }_3^{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)\\ +(1-\prod _{j=N-1}^N(1-{\alpha }_3^j)){\parallel x_n-z\parallel }^2\\ -{\alpha }_1^{N-1}{\alpha }_2^{N-1}(1-{\alpha }_3^N){\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)\\ =& (1-{\beta }_n){\parallel x_n-z\parallel }^2+{\beta }_n(\prod _{j=N-2}^N(1-{\alpha }_3^j){\parallel U_{N-3}{x}_n-z\parallel }^2\\ +(1-\prod _{j=N-2}^N(1-{\alpha }_3^j)){\parallel x_n-z\parallel }^2\\ -{\alpha }_1^{N-2}{\alpha }_2^{N-2}\prod _{j=N-1}^N(1-{\alpha }_3^j){\parallel T_{N-2}{U}_{N-3}{x}_n-U_{N-3}{x}_n\parallel }^2\\ -{\alpha }_1^{N-1}{\alpha }_2^{N-1}(1-{\alpha }_3^N){\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)\\ \le \\ ⋮\\ \le & (1-{\beta }_n){\parallel x_n-z\parallel }^2+{\beta }_n(\prod _{j=1}^N(1-{\alpha }_3^j){\parallel U_0{x}_n-z\parallel }^2\\ +(1-\prod _{j=1}^N(1-{\alpha }_3^j)){\parallel x_n-z\parallel }^2\\ -{\alpha }_1^1{\alpha }_2^1\prod _{j=2}^N(1-{\alpha }_3^j){\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(1-{\alpha }_3^j){\parallel T_{N-2}{U}_{N-3}{x}_n-U_{N-3}{x}_n\parallel }^2\\ -{\alpha }_1^{N-1}{\alpha }_2^{N-1}(1-{\alpha }_3^N){\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)\\ =& {\parallel x_n-z\parallel }^2\\ -{\beta }_n{\alpha }_1^1{\alpha }_2^1\prod _{j=2}^N(1-{\alpha }_3^j){\parallel T_1{x}_n-x_n\parallel }^2\\ ⋮\\ -{\beta }_n{\alpha }_1^{N-2}{\alpha }_2^{N-2}\prod _{j=N-1}^N(1-{\alpha }_3^j){\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}(1-{\alpha }_3^N){\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

$$\begin{array}{rcl}{\beta }_n{\alpha }_1^1{\alpha }_2^1\prod _{j=2}^N(1-{\alpha }_3^j){\parallel T_1{x}_n-x_n\parallel }^2& \le & {\parallel x_n-z\parallel }^2-{\parallel y_n-z\parallel }^2\\ \le & (\parallel x_n-z\parallel +\parallel y_n-z\parallel )\parallel y_n-x_n\parallel .\end{array}$$

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

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T_i{U}_{i-1}{x}_n-U_{i-1}{x}_n\parallel =0,\mathrm{\forall }i=1,2,\dots ,N.$$

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

Put $Q:C\to C$ defined by

$$Qx=\sum _{i=1}^N{\delta }^i{G}_{i}x,\mathrm{\forall }x\in C.$$

(3.19)

Since $G_i=P_C(I-\lambda A_i)$ is a nonexpansive mapping, for every $i=1,2,\dots ,N$, from Lemma 2.6 and 2.5, we have

$$F(Q)=\bigcap _{i=1}^{N}F(G_i)=\bigcap _{i=1}^{N}VI(C,A_i).$$

(3.20)

Since

$$\begin{array}{rcl}\parallel x_n-Q{x}_n\parallel & \le & \parallel x_n-z_n\parallel +\parallel z_n-Q{x}_n\parallel \\ =& \parallel x_n-z_n\parallel +\parallel \sum _{i=1}^N{\delta }_n^i{G}_i{x}_n-\sum _{i=1}^N{\delta }^i{G}_i{x}_n\parallel \\ =& \parallel x_n-z_n\parallel +\parallel \sum _{i=1}^N({\delta }_n^i-{\delta }^i)G_i{x}_n\parallel \\ \le & \parallel x_n-z_n\parallel +\sum _{i=1}^N|{\delta }_n^i-{\delta }^i|\parallel G_i{x}_n\parallel ,\end{array}$$

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 $x_{n_i}\rightharpoonup q$ as $i\to \mathrm{\infty }$, we have $q\in F(Q)={\bigcap }_{i=1}^{N}F(G_i)$ From (3.2), we have

$$q\in \bigcap _{i=1}^{N}VI(C,A_i).$$

(3.22)

Next, we will show that $q\in F(S)$. 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\hspace{0.17em}inf}{\parallel x_{n_i}-q\parallel }^2& <& \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\parallel x_{n_i}-Sq\parallel }^2\\ =& \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\parallel x_{n_i}-S{x}_{n_i}+(S{x}_{n_i}-Sq)\parallel }^2\\ =& \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}({\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〉)\\ =& \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\parallel S{x}_{n_i}-Sq\parallel }^2\\ =& \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}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\hspace{0.17em}inf}\parallel {\alpha }_1^N(T_N{U}_{N-1}{x}_{n_i}-T_N{U}_{N-1}q)\\ +{\alpha }_2^N(U_{N-1}{x}_{n_i}-U_{N-1}q)+{\alpha }_3^N(x_{n_i}-q){\parallel }^2\\ \le & \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}({\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)\\ \le & \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\left({\alpha }_1^N\right({\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〉)\\ +{\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)\\ =& \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}((1-{\alpha }_3^N){\parallel U_{N-1}{x}_{n_i}-U_{N-1}q\parallel }^2+{\alpha }_3^N{\parallel x_{n_i}-q\parallel }^2)\\ =& \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}((1-{\alpha }_3^N)\parallel {\alpha }_1^{N-1}(T_{N-1}{U}_{N-2}{x}_{n_i}-T_{N-1}{U}_{N-2}q)\\ +{\alpha }_2^{N-1}(U_{N-2}{x}_{n_i}-U_{N-2}q)+{\alpha }_3^{N-1}(x_{n_i}-q){\parallel }^2+{\alpha }_3^N{\parallel x_{n_i}-q\parallel }^2)\\ \le & \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\left((1-{\alpha }_3^N)\right({\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)+{\alpha }_3^N{\parallel x_{n_i}-q\parallel }^2)\\ \le & \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\left((1-{\alpha }_3^N)\right({\alpha }_1^{N-1}({\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〉)\\ +{\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)+{\alpha }_3^N{\parallel x_{n_i}-q\parallel }^2)\\ =& \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\left((1-{\alpha }_3^N)\right((1-{\alpha }_3^{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)+{\alpha }_3^N{\parallel x_{n_i}-q\parallel }^2)\\ =& \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}(\prod _{j=N-1}^N(1-{\alpha }_3^j){\parallel U_{N-2}{x}_{n_i}-U_{N-2}q\parallel }^2\\ +(1-\prod _{j=N-1}^N(1-{\alpha }_3^j)){\parallel x_{n_i}-q\parallel }^2)\\ \le \\ ⋮\\ \le & \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}(\prod _{j=1}^N(1-{\alpha }_3^j){\parallel U_0{x}_{n_i}-U_{0}q\parallel }^2\\ +(1-\prod _{j=1}^N(1-{\alpha }_3^j)){\parallel x_{n_i}-q\parallel }^2)\\ =& \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}(\prod _{j=1}^N(1-{\alpha }_3^j){\parallel x_{n_i}-q\parallel }^2\\ +(1-\prod _{j=1}^N(1-{\alpha }_3^j)){\parallel x_{n_i}-q\parallel }^2)\\ =& \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\parallel x_{n_i}-q\parallel }^2.\end{array}$$

This is a contradiction. Then, we have $q\in F(S)$. From Lemma 2.9, we have

$$q\in \bigcap _{i=1}^{N}F(T_i).$$

(3.23)

From (3.22) and (3.23), we have $q\in \mathfrak{F}$. Hence, $\omega (x_n)\subset \mathfrak{F}$. Therefore, by (3.5) and Lemma 2.8, we have $\{x_n\}$ 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 $\mathfrak{F}=F(T)\cap {\bigcap }_{i=1}^{N}VI(C,A_i)\ne \mathrm{\varnothing }$. For every $i=1,2,\dots ,N$, define the mapping $G_i:C\to C$ by $G_{i}x=P_C(I-\lambda A_i)x$ $\mathrm{\forall }x\in C$ and $\lambda \in [c,d]\subset (0,2{\alpha }_i)$. Let $\{x_n\}$ be a sequence generated by $x_1\in C_1=C$ and

$$\{\begin{array}{c}{z}_n={\sum }_{i=1}^N{\delta }_n^i{G}_i{x}_n,\hfill \\ y_n={\alpha }_n{x}_n+{\beta }_{n}T{x}_n+{\gamma }_n{z}_n,\hfill \\ C_{n+1}=\{z\in C_n:\parallel y_n-z\parallel \le \parallel x_n-z\parallel \},\hfill \\ x_{n+1}=P_{C_{n+1}}{x}_1,\mathrm{\forall }n\ge 1,\hfill \end{array}$$

(3.24)

where $\{{\alpha }_n\},\{{\beta }_n\},\{{\gamma }_n\}\subseteq [0,1]$, ${\alpha }_n+{\beta }_n+{\gamma }_n=1$ and suppose the following conditions hold:

Then the sequence $\{x_n\}$ 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 ${\{T_i\}}_{i=1}^N$ be a finite family of nonspreading mappings with $\mathfrak{F}={\bigcap }_{i=1}^{N}F(T_i)\cap VI(C,A)\ne \mathrm{\varnothing }$. Let ${\rho }_j=({\alpha }_1^j,{\alpha }_2^j,{\alpha }_3^j)\in I\times I\times I$, $j=1,2,3,\dots ,N$, where $I=[0,1]$, ${\alpha }_1^j+{\alpha }_2^j+{\alpha }_3^j=1$, ${\alpha }_1^j,{\alpha }_3^j\in (0,1)$ for all $j=1,2,\dots ,N-1$ and ${\alpha }_1^N\in (0,1]$, ${\alpha }_3^N\in [0,1)$, ${\alpha }_2^j\in (0,1)$ 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 $\{x_n\}$ be a sequence generated by $x_1\in C_1=C$ and

$$\{\begin{array}{c}{y}_n={\alpha }_n{x}_n+{\beta }_{n}S{x}_n+{\gamma }_n{P}_C(I-\lambda A)x_n,\hfill \\ C_{n+1}=\{z\in C_n:\parallel y_n-z\parallel \le \parallel x_n-z\parallel \},\hfill \\ x_{n+1}=P_{C_{n+1}}{x}_1,\mathrm{\forall }n\ge 1,\hfill \end{array}$$

(3.25)

where $\{{\alpha }_n\},\{{\beta }_n\},\{{\gamma }_n\}\subseteq [a,b]\subset (0,1)$, ${\alpha }_n+{\beta }_n+{\gamma }_n=1$ and $\lambda \subseteq [c,d]\subset (0,2\alpha )$. Then the sequence $\{x_n\}$ converges strongly to $P_{\mathfrak{F}}{x}_1$.