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# Iterative algorithms for finding a common solution of system of the set of variational inclusion problems and the set of fixed point problems

Fixed Point Theory and Applications20112011:38

https://doi.org/10.1186/1687-1812-2011-38

• Accepted: 18 August 2011
• Published:

## Abstract

In this article, we introduce a new mapping generated by infinite family of nonexpansive mapping and infinite real numbers. By means of the new mapping, we prove a strong convergence theorem for finding a common element of the set of fixed point problems of infinite family of nonexpansive mappings and the set of a finite family of variational inclusion problems in Hilbert space. In the last section, we apply our main result to prove a strong convergence theorem for finding a common element of the set of fixed point problems of infinite family of strictly pseudo-contractive mappings and the set of finite family of variational inclusion problems.

## Keywords

• nonexpansive mapping
• strict pseudo contraction
• strongly positive operator
• variational inclusion problem
• fixed point

## 1 Introduction

Let H be a real Hilbert space and let C be a nonempty closed convex subset of H. Let A : C → H be a nonlinear mapping and let F : C × C be a bifunction. A mapping T of H into itself is called nonexpansive if ||Tx - Ty|| ≤ ||x - y|| for all x, y H. We denote by F(T) the set of fixed points of T (i.e. F(T) = {x H : Tx = x}). Goebel and Kirk [1] showed that F(T) is always closed convex and also nonempty provided T has a bounded trajectory.

The problem for finding a common fixed point of a family of nonexpansive mappings has been studied by many authors. The well-known convex feasibility problem reduces to finding a point in the intersection of the fixed point sets of a family of nonexpansive mappings (see, e.g., [2, 3]).

A bounded linear operator A on H is called strongly positive with coefficient $\stackrel{̄}{\gamma }$ if there exists a constant $\stackrel{̄}{\gamma }>0$ with the property
$〈Ax,x〉\ge \stackrel{̄}{\gamma }\parallel x{\parallel }^{2}.$
A mapping A of C into H is called inverse-strongly monotone, see [4], 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 C.

The variational inequality problem is to find a point u C such that
$〈v-u,Au〉\ge 0\phantom{\rule{1em}{0ex}}forallv\in C.$
(1.1)

The set of solutions of (1.1) is denoted by V I(C, A). Many authors have studied methods for finding solution of variational inequality problems (see, e.g., [58]).

In 2008, Qin et al. [9] introduced the following iterative scheme:
$\left\{\begin{array}{c}{y}_{n}={P}_{C}\left(I-{s}_{n}A\right){x}_{n}\\ {x}_{n+1}={\alpha }_{n}\gamma f\left({W}_{n}{x}_{n}\right)+\left(I-{\alpha }_{n}B\right){W}_{n}{P}_{C}\left(I-{r}_{n}A\right){y}_{n},\phantom{\rule{1em}{0ex}}\forall n\in ℕ,\end{array}\right\$
(1.2)

where W n is the W-mapping generated by a finite family of nonexpansive mappings and real numbers, A : CH is relaxed (u,v) cocoercive and μ-Lipschitz continuous, and P C is a metric projection H onto C. Under suitable conditions of {s n }, {r n }{α n }, γ, they proved that {x n } converges strongly to an element of the set of variational inequality problem and the set of a common fixed point of a finite family of nonexpansive mappings.

In 2006, Marino and Xu [10] introduced the iterative scheme as follows:
${x}_{0}\in H,{x}_{n+1}=\left(I-{\alpha }_{n}A\right)S{x}_{n}+{\alpha }_{n}\gamma f\left({x}_{n}\right),\phantom{\rule{1em}{0ex}}\forall n\ge 0,$
(1.3)
where S is a nonexpansive mapping, f is a contraction with the coefficient a (0, 1), A is a strongly positive bounded linear self-adjoint operator with the coefficient $\stackrel{̄}{\gamma }$, and γ is a constant such that $0<\gamma <\frac{\stackrel{}{\gamma }}{a}$. They proved that {x n } generated by the above iterative scheme converges strongly to the unique solution of the variational inequality:
$〈\left(A-\gamma f\right){x}^{*},x-{x}^{*}〉\ge 0,x\in F\left(S\right).$
We know that a mapping B : HH is said to be monotone, if for each x, y H, we have
$〈Bx-By,x-y〉\ge 0.$

A set-valued mapping M : H → 2 H is called monotone if for all x, y H, f Mx and g My imply 〈x - y, f - g〉 ≥ 0. A monotone mapping M : H → 2 H is maximal if the graph of Graph(M) of M is not properly contained in the graph of any other monotone mapping. It is known that a monotone mapping M is maximal if and only if for (x, f) H × H, 〈x - y, f - g〉 ≥ 0 for every (y, g) Graph(M) implies f Mx.

Next, we consider the following so-called variational inclusion problem:

Find a u H such that
$\theta \in Bu+Mu$
(1.4)

where B : HH, M : H → 2 H are two nonlinear mappings, and θ is zero vector in H (see, for instance, [1116]). The set of the solution of (1.4) is denoted by V I(H, B, M).

Let C be a nonempty closed convex subset of Banach space X. Let ${\left\{{T}_{n}\right\}}_{n=1}^{\infty }$ be an infinite family of nonexpansive mappings of C into itself, and let λ1, λ2,..., be real numbers in [0, 1]; then we define the mapping K n : CC as follows:
$\begin{array}{lll}\hfill {U}_{n,0}& =I\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ \hfill {U}_{n,1}& ={\lambda }_{1}{T}_{1}{U}_{n,0}+\left(1-{\lambda }_{1}\right){U}_{n,0},\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ \hfill {U}_{n,2}& ={\lambda }_{2}{T}_{2}{U}_{n,1}+\left(1-{\lambda }_{2}\right){U}_{n,1},\phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ \hfill {U}_{n,3}& ={\lambda }_{3}{T}_{3}{U}_{n,2}+\left(1-{\lambda }_{3}\right){U}_{n,2},\phantom{\rule{2em}{0ex}}& \hfill \text{(4)}\\ \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}⋮\phantom{\rule{2em}{0ex}}& \hfill \text{(5)}\\ \hfill {U}_{n,k}& ={\lambda }_{k}{T}_{k}{U}_{n,k-1}+\left(1-{\lambda }_{k}\right){U}_{n,k-1}\phantom{\rule{2em}{0ex}}& \hfill \text{(6)}\\ \hfill {U}_{n,k+1}& ={\lambda }_{k+1}{T}_{k+1}{U}_{n,k}+\left(1-{\lambda }_{k+1}\right){U}_{n,k}\phantom{\rule{2em}{0ex}}& \hfill \text{(7)}\\ \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}⋮\phantom{\rule{2em}{0ex}}& \hfill \text{(8)}\\ \hfill {U}_{n,n-1}& ={\lambda }_{n-1}{T}_{n-1}{U}_{n,n-2}+\left(1-{\lambda }_{n-1}\right){U}_{n,n-2}\phantom{\rule{2em}{0ex}}& \hfill \text{(9)}\\ \hfill {K}_{n}& ={U}_{n,n}={\lambda }_{n}{T}_{n}{U}_{n,n-1}+\left(1-{\lambda }_{n}\right){U}_{n,n-1}.\phantom{\rule{2em}{0ex}}& \hfill \text{(10)}\\ \hfill \text{(11)}\end{array}$

Such a mapping K n is called the K-mapping generated by T1, T2,..., T n and λ1, λ2,..., λ n .

Let x1 H and {x n } be the sequence generated by
${x}_{n+1}={\alpha }_{n}\gamma f\left({x}_{n}\right)+{\beta }_{n}{x}_{n}+\left(\left(1-{\beta }_{n}\right)I-{\alpha }_{n}A\right)\left({\gamma }_{n}{K}_{n}{x}_{n}+\left(1-{\gamma }_{n}\right)S{x}_{n}\right),$
(1.5)

where A is a strongly positive linear-bounded self-adjoint operator with the coefficient $0<\stackrel{}{\gamma }<1$, S : CC is the S - mapping generated by G1, G2,..., G N and ν1, ν2,..., ν N , where G i : HH is a mapping defined by ${J}_{{M}_{i,\eta }}\left(I-\eta {B}_{i}\right)x={G}_{i}x$ for every x H, and η (0, 2δ i ) for every i = 1, 2,..., N, f : HH is contractive mapping with coefficient θ (0, 1) and $0<\gamma <\frac{\stackrel{}{\gamma }}{\theta }$, {α n }, {β n }, {γ n } are sequences in [0, 1].

In this article, by motivation of (1.3), we prove a strong convergence theorem of the proposed algorithm scheme (1.5) to an element $z\in {\bigcap }_{i=1}^{\infty }F\left({T}_{i}\right)\bigcap {\bigcap }_{i=1}^{N}V\left(H,{B}_{i},{M}_{i}\right)$, under suitable conditions of {α n }, {β n }, {γ n }.

## 2 Preliminaries

In this section, we provide 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, and let P C be the metric projection of H onto C, i.e., for x H, P C x satisfies the property:
$\parallel x-{P}_{C}x\parallel =\underset{y\in C}{\mathrm{min}}\parallel x-y\parallel .$

The following characterizes the projection P C .

Lemma 2.1. (see [17]) Given x H and y C. Then P C x = y if and only if there holds the inequality
$〈x-y,y-z〉\ge 0\phantom{\rule{1em}{0ex}}\forall z\in C.$
Lemma 2.2. (see [18]) Let {s n } be a sequence of nonnegative real number satisfying
${s}_{n+1}=\left(1-{\alpha }_{n}\right){s}_{n}+{\alpha }_{n}{\beta }_{n},\phantom{\rule{1em}{0ex}}\forall n\ge 0$
where {α n }, {β n } satisfy the conditions:
1. (1)

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

2. (2)

$\underset{n\to \infty }{\text{lim}\phantom{\rule{0.1em}{0ex}}\text{sup}}{\beta }_{n}\le 0\phantom{\rule{0.1em}{0ex}}or\phantom{\rule{0.1em}{0ex}}\sum _{n=1}^{\infty }|{\alpha }_{n}{\beta }_{n}|\phantom{\rule{0.1em}{0ex}}<\infty$.

Then limn→∞s n = 0.

Lemma 2.3. (see [19]) Let C be a closed convex subset of a strictly convex Banach space E. Let {T n : n } be a sequence of nonexpansive mappings on C. Suppose ${\bigcap }_{n=1}^{\infty }F\left({T}_{n}\right)$ is nonempty. Let {λ n } be a sequence of positive numbers with ${\Sigma }_{n=1}^{\infty }{\lambda }_{n}=1$. Then a mapping S on C defined by
$S\left(x\right)={\Sigma }_{n=1}^{\infty }{\lambda }_{n}{T}_{n}{x}_{n}$

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

Lemma 2.4. (see [20]) Let E be a uniformly convex Banach space, C be a nonempty closed convex subset of E, and S : CC be a nonexpansive mapping. Then I - S is demi-closed at zero.

Lemma 2.5. (see [21]) Let {x n } and {z n } be bounded sequences in a Banach space X and let {β n } be a sequence in [0,1] with 0 < lim inf n→∞ β n ≤ lim supn→∞β n < 1.

Suppose xn+1= β n x n +(1 - β n )z n for all integer n ≥ 0 and lim supn→∞(||zn+1- z n || - ||xn+1- x n ||) ≤ 0. Then limn→∞||x n - z n || = 0.

In 2009, Kangtunykarn and Suantai [5] introduced the S-mapping generated by a finite family of nonexpansive mappings and real numbers as follows:

Definition 2.1. Let C be a nonempty convex subset of real Banach space. Let ${\left\{{T}_{i}\right\}}_{i=1}^{N}$ be a finite family of nonexpanxive mappings of C into itself. For each j = 1, 2,..., N, let ${\alpha }_{j}=\left({\alpha }_{1}^{j},{\alpha }_{2}^{j},{\alpha }_{3}^{j}\right)\in I×I×I$ where I [0, 1] and ${\alpha }_{1}^{j}+{\alpha }_{2}^{j}+{\alpha }_{3}^{j}=1$, define the mapping S : C → C as follows:
$\begin{array}{lll}\hfill {U}_{0}& =I\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ \hfill {U}_{1}& ={\alpha }_{1}^{1}{T}_{1}{U}_{0}+{\alpha }_{2}^{1}{U}_{0}+{\alpha }_{3}^{1}I\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ \hfill {U}_{2}& ={\alpha }_{1}^{2}{T}_{2}{U}_{1}+{\alpha }_{2}^{2}{U}_{1}+{\alpha }_{3}^{2}I\phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ \hfill {U}_{3}& ={\alpha }_{1}^{3}{T}_{3}{U}_{2}+{\alpha }_{2}^{3}{U}_{2}+{\alpha }_{3}^{3}I\phantom{\rule{2em}{0ex}}& \hfill \text{(4)}\\ \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}⋮\phantom{\rule{2em}{0ex}}& \hfill \text{(5)}\\ \hfill {U}_{N-1}& ={\alpha }_{1}^{N-1}{T}_{N-1}{U}_{N-2}+{\alpha }_{2}^{N-1}{U}_{N-2}+{\alpha }_{3}^{N-1}I\phantom{\rule{2em}{0ex}}& \hfill \text{(6)}\\ \hfill S& ={U}_{N}={\alpha }_{1}^{N}{T}_{N}{U}_{N-1}+{\alpha }_{2}^{N}{U}_{N-1}+{\alpha }_{3}^{N}I.\phantom{\rule{2em}{0ex}}& \hfill \text{(7)}\\ \hfill \text{(8)}\end{array}$
(2.1)

This mapping is called the S-mapping generated by T1,..., T N and α1, α2,..., α N .

Lemma 2.6. (see [5]) Let C be a nonempty closed convex subset of strictly convex. Let ${\left\{{T}_{i}\right\}}_{i=1}^{N}$ be a finite family of nonexpanxive mappings of C into itself with ${\bigcap }_{i=1}^{N}F\left({T}_{i}\right)\ne \varnothing$ and let ${\alpha }_{j}=\left({\alpha }_{1}^{j},{\alpha }_{2}^{j},{\alpha }_{3}^{j}\right)\in I×I×I$, j = 1,2,3,..., N, where I = [0, 1], ${\alpha }_{1}^{j}+{\alpha }_{2}^{j}+{\alpha }_{3}^{j}=1$, ${\alpha }_{1}^{j}\in \left(0,1\right)$ for all j = 1,2,..., N-1, ${\alpha }_{1}^{N}\in \left(0,1\right]{\alpha }_{2}^{j}$, ${\alpha }_{3}^{j}\in \left[0,1\right)$ for all j = 1,2,..., N. Let S be the mapping generated by T1,..., T N and α1, α2,..., α N . Then $F\left(S\right)={\bigcap }_{i=1}^{N}F\left({T}_{i}\right)$.

Lemma 2.7. (see [5]) Let C be a nonempty closed convex subset of Banach space. Let ${\left\{{T}_{i}\right\}}_{i=1}^{N}$ be a finite family of nonexpansive mappings of C into itself and ${\alpha }_{j}^{\left(n\right)}=\left({\alpha }_{1}^{n,j},{\alpha }_{2}^{n,j},{\alpha }_{3}^{n,j}\right)$, ${\alpha }_{j}=\left({\alpha }_{1}^{j},{\alpha }_{2}^{j},{\alpha }_{3}^{j}\right)\in I×I×I$, where I = [0,1], ${\alpha }_{1}^{n,j}+{\alpha }_{2}^{n,j}+{\alpha }_{3}^{n,j}=1$ and ${\alpha }_{1}^{j}+{\alpha }_{2}^{j}+{\alpha }_{3}^{j}=1$ such that ${\alpha }_{i}^{n,j}\to {\alpha }_{i}^{j}\in \left[0,1\right]$ as n → ∞ for i = 1,3 and j = 1,2,3,..., N. Moreover, for every n , let S and S n be the S-mappings generated by T1, T2,..., T N and α1, α2,..., α N and T1, T2,..., T N and ${\alpha }_{1}^{\left(n\right)},{\alpha }_{2}^{\left(n\right)},...,{\alpha }_{N}^{\left(n\right)}$, respectively. Then limn→∞||S n x - Sx|| = 0 for every x C.

Definition 2.2. (see [11]) Let M : H → 2 H be a multi-valued maximal monotone mapping, then the single-valued mapping J M,λ : HH defined by
${J}_{M,\lambda }\left(u\right)=\left(I+\lambda M{\right)}^{-1}\left(u\right)\text{, }\forall u\in H,$

is called the resolvent operator associated with M, where λ is any positive number and I is identity mapping.

Lemma 2.8. (see [11]) u H is a solution of variational inclusion (1.4) if and only if u = JM, λ(u - λBu), λ > 0, i.e.,
$VI\left(H,B,M\right)=F\left({J}_{M,\lambda }\left(I-\lambda B\right)\right),\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\forall \lambda >0.$

Further, if λ (0, 2α], then V I(H, B, M) is closed convex subset in H.

Lemma 2.9. (see [22]) The resolvent operator J M,λ associated with M is single-valued, nonexpansive for all λ > 0 and 1-inverse-strongly monotone.

Lemma 2.10. In a strictly convex Banach space E, if
$||x||\phantom{\rule{2.77695pt}{0ex}}=\phantom{\rule{2.77695pt}{0ex}}||y||\phantom{\rule{2.77695pt}{0ex}}=\phantom{\rule{2.77695pt}{0ex}}||\lambda x+\left(1-\lambda \right)y||$

for all x, y E and λ (0, 1), then x = y.

Lemma 2.11. Let C be a nonempty closed convex subset of a strictly convex Banach space. Let ${\left\{{T}_{i}\right\}}_{i=1}^{\infty }$ be an infinite family of nonexpanxive mappings of C into itself with ${\bigcap }_{i=1}^{\infty }F\left({T}_{i}\right)\ne \varnothing$ and let λ1, λ2,..., be real numbers such that 0 < λ i < 1 for every i = 1, 2,..., and ${\sum }_{i=1}^{\infty }{\lambda }_{i}<\infty$. For every n , let K n be the K-mapping generated by T1, T2,..., T n and λ1, λ2,..., λ n . Then for every x C and k , limn→∞K n x exits.

Proof. Let x C. Then for k, n , we have
(2.2)
which implies that Un+1,k= Un, kfor every k, n . Hence, K n = Un, n= Un+1,n. Since Kn+1x = Un+1,n+1x = λn+1Tn+1K n x + (1 - λn+1)K n x, we have
${K}_{n+1}x-{K}_{n}x={\lambda }_{n+1}\left({T}_{n+1}{K}_{n}x-{K}_{n}x\right).$
(2.3)
Let ${x}^{*}\in {\bigcap }_{i=1}^{\infty }F\left({T}_{i}\right)$ and x C. For each n , we have
$\begin{array}{lll}\hfill \parallel {K}_{n}x-{x}^{*}\parallel & =\parallel {\lambda }_{n}{T}_{n}{U}_{n,n-1}x+\left(1-{\lambda }_{n}\right){U}_{n,n-1}x-{x}^{*}\parallel \phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ \le {\lambda }_{n}\parallel {T}_{n}{U}_{n,n-1}x-{x}^{*}\parallel +\left(1-{\lambda }_{n}\right)\parallel {U}_{n,n-1}x-{x}^{*}\parallel \phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ \le \parallel {U}_{n,n-1}x-{x}^{*}\parallel \phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ =\parallel {\lambda }_{n-1}{T}_{n-1}{U}_{n,n-2}x+\left(1-{\lambda }_{n-1}\right){U}_{n,n-2}x-{x}^{*}\parallel \phantom{\rule{2em}{0ex}}& \hfill \text{(4)}\\ \le {\lambda }_{n-1}\parallel {T}_{n-1}{U}_{n,n-2}x-{x}^{*}\parallel +\left(1-{\lambda }_{n-1}\right)\parallel {U}_{n,n-2}x-{x}^{*}\parallel \phantom{\rule{2em}{0ex}}& \hfill \text{(5)}\\ \le \parallel {U}_{n,n-2}x-{x}^{*}\parallel \phantom{\rule{2em}{0ex}}& \hfill \text{(6)}\\ \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}.\phantom{\rule{2em}{0ex}}& \hfill \text{(7)}\\ \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}.\phantom{\rule{2em}{0ex}}& \hfill \text{(8)}\\ \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}.\phantom{\rule{2em}{0ex}}& \hfill \text{(9)}\\ \le \phantom{\rule{1em}{0ex}}\parallel {U}_{n,1}x-{x}^{*}\parallel \phantom{\rule{2em}{0ex}}& \hfill \text{(10)}\\ =\phantom{\rule{1em}{0ex}}\parallel {\lambda }_{1}{T}_{1}{U}_{n,0}x+\left(1-{\lambda }_{1}\right){U}_{n,0}x-{x}^{*}\parallel \phantom{\rule{2em}{0ex}}& \hfill \text{(11)}\\ \le \phantom{\rule{1em}{0ex}}{\lambda }_{1}\parallel {T}_{1}x-{x}^{*}\parallel +\left(1-{\lambda }_{1}\right)\parallel x-{x}^{*}\parallel \phantom{\rule{2em}{0ex}}& \hfill \text{(12)}\\ =\phantom{\rule{1em}{0ex}}\parallel x-{x}^{*}\parallel ,\phantom{\rule{2em}{0ex}}& \hfill \text{(13)}\\ \hfill \text{(14)}\end{array}$
(2.4)
which implies that {K n x} is bounded, and so is {T n K n x}. For mn, by (2.3) we have
(2.5)

where M = supn{||Tn+1K n x - K n x||}. This implies that {K n x} is Cauchy sequence. Hence limn→∞K n x exists.

From Lemma 2.11, we can define a mapping K : CC as follows:
$Kx=\underset{n\to \infty }{\mathrm{lim}}{K}_{n}x,\phantom{\rule{1em}{0ex}}x\in C.$

Such a mapping K is called the K-mapping generated by T1, T2,..., and λ1, λ2,.....

Remark 2.12. It is easy to see that for each n , K n is nonexpansive mappings. Let x, y C, then we have
$\parallel Kx-Ky\parallel =\underset{n⇒\infty }{\mathrm{lim}}\parallel {K}_{n}x-{K}_{n}y\parallel \le \phantom{\rule{2.77695pt}{0ex}}\parallel x-y\parallel .$
(2.6)
By (2.6), we have K : CC is nonexpansive mapping. Next, we will show that limn→∞supxD||K n x - Kx|| = 0 for every bounded subset D of C. To show this, let x, y C and D be a bounded subset of C. By (2.5), for mn, we have
$\parallel {K}_{m}x-{K}_{n}x\parallel \le M\sum _{k=n+1}^{m}{\lambda }_{k}.$
By letting m → ∞, for any x D, we have
$\parallel Kx-{K}_{n}x\parallel \le M\sum _{k=n+1}^{\infty }{\lambda }_{k}.$
Since ${\sum }_{n=1}^{\infty }{\lambda }_{n}<\infty$, we have
$\underset{n\to \infty }{\mathrm{lim}}\phantom{\rule{2.77695pt}{0ex}}\underset{x\in D}{\mathrm{sup}}\parallel Kx-{K}_{n}x\parallel =0.$

By the next lemma, we will show that $F\left(K\right)={\bigcap }_{i=1}^{\infty }F\left({T}_{i}\right)$

Lemma 2.13. Let C be a nonempty closed convex subset of a strictly convex Banach space. Let ${\left\{{T}_{i}\right\}}_{i=1}^{\infty }$ be an infinite family of nonexpansive mappings of C into itself with ${\bigcap }_{i=1}^{\infty }F\left({T}_{i}\right)\ne \varnothing$, and let λ1, λ2,..., be real numbers such that 0 < λ i < 1 for every i = 1, 2,... with ${\sum }_{i=1}^{\infty }{\lambda }_{i}<\infty$. Let K n and K be the K-mapping generated by T1, T2,... T n and λ1, λ2,... λ n and T1, T2,... and λ1, λ2,..., respectively. Then $F\left(K\right)={\bigcap }_{i=1}^{\infty }F\left({T}_{i}\right)$.

Proof. It is easy to see that ${\bigcap }_{i=1}^{\infty }F\left({T}_{i}\right)\subseteq F\left(K\right)$. Next, we show that $F\left(K\right)\subseteq {\bigcap }_{i=1}^{\infty }F\left({T}_{i}\right)$. Let x0 F (K) and ${x}^{*}\in {\bigcap }_{i=1}^{\infty }F\left({T}_{i}\right)$. Let k be fixed. Since
(2.7)
we have
this implies that
$\parallel {x}_{0}-{x}^{*}\parallel \phantom{\rule{2.77695pt}{0ex}}=\phantom{\rule{2.77695pt}{0ex}}\parallel {T}_{1}{x}_{0}-{x}^{*}\parallel \phantom{\rule{2.77695pt}{0ex}}=\phantom{\rule{2.77695pt}{0ex}}\parallel {\lambda }_{1}\left({T}_{1}{x}_{0}-{x}^{*}\right)+\left(1-{\lambda }_{1}\right)\left({x}_{0}-{x}^{*}\right)\parallel .$
By Lemma 2.10, we have T1x0 = x0, that is x0 F (T1). It follows that Un,1x0 = x0. By (2.7), we have
It follows that
$\begin{array}{lll}\hfill \parallel {x}_{0}-{x}^{*}\parallel & =\underset{n\to \infty }{\mathrm{lim}}\parallel {K}_{n}{x}_{0}-{x}^{*}\parallel \phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ \le \parallel {\lambda }_{2}\left({T}_{2}{x}_{0}-{x}^{*}\right)+\left(1-{\lambda }_{2}\right)\left({x}_{0}-{x}^{*}\right)\parallel \phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ \le {\lambda }_{2}\parallel {T}_{2}{x}_{0}-{x}^{*}\parallel +\left(1-{\lambda }_{2}\right)\parallel {x}_{0}-{x}^{*}\parallel \phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ \le \parallel {x}_{0}-{x}^{*}\parallel ,\phantom{\rule{2em}{0ex}}& \hfill \text{(4)}\\ \hfill \text{(5)}\end{array}$
which implies
$\parallel {x}_{0}-{x}^{*}\parallel =\phantom{\rule{2.77695pt}{0ex}}\parallel {T}_{2}{x}_{0}-{x}^{*}\parallel =\phantom{\rule{2.77695pt}{0ex}}\parallel {\lambda }_{2}\left({T}_{2}{x}_{0}-{x}^{*}\right)+\left(1-{\lambda }_{2}\right)\left({x}_{0}-{x}^{*}\right)\parallel .$
By Lemma 2.10, we obtain that T2x0 = x0, that is x0 F (T2). It follows that Un,2x0 = x0. By using the same argument, we can conclude that T i x0 = x0 and U i x0 = x0 for i = 1, 2,..., k - 1. By (2.7), we have
$\begin{array}{lll}\hfill \parallel {K}_{n}{x}_{0}-{x}^{*}\parallel & \le \parallel {U}_{n,k}{x}_{0}-{x}^{*}\parallel \phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ =\parallel {\lambda }_{k}\left({T}_{k}{U}_{n,k-1}{x}_{0}-{x}^{*}\right)+\left(1-{\lambda }_{k}\right)\left({U}_{n,k-1}{x}_{0}-{x}^{*}\right)\parallel \phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ =\parallel {\lambda }_{k}\left({T}_{k}{x}_{0}-{x}^{*}\right)+\left(1-{\lambda }_{k}\right)\left({x}_{0}-{x}^{*}\right)\parallel \phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ \le {\lambda }_{k}\parallel {T}_{k}{x}_{0}-{x}^{*}\parallel +\left(1-{\lambda }_{k}\right)\parallel {x}_{0}-{x}^{*}\parallel \phantom{\rule{2em}{0ex}}& \hfill \text{(4)}\\ \le \parallel {x}_{0}-{x}^{*}\parallel .\phantom{\rule{2em}{0ex}}& \hfill \text{(5)}\\ \hfill \text{(6)}\end{array}$
It follows that
$\begin{array}{lll}\hfill \parallel {x}_{0}-{x}^{*}\parallel & =\underset{n\to \infty }{\mathrm{lim}}\parallel {K}_{n}{x}_{0}-{x}^{*}\parallel \phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ =\parallel {\lambda }_{k}\left({T}_{k}{x}_{0}-{x}^{*}\right)+\left(1-{\lambda }_{k}\right)\left({x}_{0}-{x}^{*}\right)\parallel \phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ \le {\lambda }_{k}\parallel {T}_{k}{x}_{0}-{x}^{*}\parallel +\left(1-{\lambda }_{k}\right)\parallel {x}_{0}-{x}^{*}\parallel \phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ \le \parallel {x}_{0}-{x}^{*}\parallel ,\phantom{\rule{2em}{0ex}}& \hfill \text{(4)}\\ \hfill \text{(5)}\end{array}$
(2.8)
which implies
$\parallel {x}_{0}-{x}^{*}\parallel =\phantom{\rule{2.77695pt}{0ex}}\parallel {T}_{k}{x}_{0}-{x}^{*}\parallel =\phantom{\rule{2.77695pt}{0ex}}\parallel {\lambda }_{k}\left({T}_{k}{x}_{0}-{x}^{*}\right)+\left(1-{\lambda }_{k}\right)\left({x}_{0}-{x}^{*}\right)\parallel .$
(2.9)

By Lemma 2.10, we have T k x0 = x0, that is x0 F (T k ). This implies that ${x}_{0}\in {\bigcap }_{i=1}^{\infty }F\left({T}_{i}\right)$.

## 3 Main result

Theorem 3.1. Let H be a real Hilbert space, and let M i : H → 2 H be maximal monotone mappings for every i = 1, 2,..., N. Let B i : HH be a δ i - inverse strongly monotone mapping for every i = 1, 2,..., N and ${\left\{{T}_{i}\right\}}_{i=1}^{\infty }$ an infinite family of nonexpansive mappings from H into itself. Let A be a strongly positive linear-bounded self-adjoint operator with the coefficient $0<\stackrel{}{\gamma }<1$. Let G i : HH be defined by ${J}_{{M}_{i,\eta }}\left(I-\eta {B}_{i}\right)x={G}_{i}x$ for every x H and η (0, 2δ i ) for every i = 1, 2,..., N and let ${\nu }_{j}=\left({\alpha }_{1}^{j},{\alpha }_{2}^{j},{\alpha }_{3}^{j}\right)\in I×I×I$, j = 1, 2, 3,..., N, where I = [0, 1], ${\alpha }_{1}^{j}+{\alpha }_{2}^{j}+{\alpha }_{3}^{j}=1$, ${\alpha }_{1}^{j}\in \left(0,1\right)$ for all j = 1,2,..., N-1, ${\alpha }_{1}^{N}\in \left(0,1\right]{\alpha }_{2}^{j}$, ${\alpha }_{3}^{j}\in \left[0,1\right)$ for all j = 1, 2,..., N.. Let S : CC be the S-mapping generated by G1, G2,..., G N and ν1, ν2,..., ν N . Let λ1, λ2,..., be real numbers such that 0 < λ i < 1 for every i = 1, 2,..., with ${\sum }_{i=1}^{\infty }{\lambda }_{i}<\infty$, and let K n be the K-mapping generated by T1, T2,..., T n and λ1, λ2,..., λ n , and let K be the K-mapping generated by T1, T2,..., and λ1, λ2,..., i.e.,
$Kx=\underset{n\to \infty }{\mathrm{lim}}{K}_{n}x$
for every x C. Assume that $\mathfrak{F}={\bigcap }_{i=1}^{\infty }F\left({T}_{i}\right)\bigcap {\bigcap }_{i=1}^{N}V\left(H,{B}_{i},{M}_{i}\right)\ne \varnothing$. For every n , i = 1, 2,..., N, let x1 H and {x n } be the sequence generated by
${x}_{n+1}={\alpha }_{n}\gamma f\left({x}_{n}\right)+{\beta }_{n}{x}_{n}+\left(\left(1-{\beta }_{n}\right)I-{\alpha }_{n}A\right)\left({\gamma }_{n}{K}_{n}{x}_{n}+\left(1-{\gamma }_{n}\right)S{x}_{n}\right),$
(3.1)

where f : HH is contractive mapping with coefficient θ (0, 1) and $0<\gamma <\frac{\stackrel{}{\gamma }}{\theta }$. Let {α n }, {β n }, {γ n } be sequences in [0, 1], satisfying the following conditions:

(i) $\underset{n\to \infty }{\mathrm{lim}}{\alpha }_{n}=0$ and ${\Sigma }_{n=0}^{\infty }{\alpha }_{n}=\infty$

(ii) $0<\underset{n\to \infty }{\mathrm{liminf}}{\beta }_{n}\le \underset{n\to \infty }{\mathrm{limsup}}{\beta }_{n}<1,$

(iii) $\underset{n\to \infty }{\mathrm{lim}}{\gamma }_{n}=c\in \left(0,1\right)$

Then {x n } converges strongly to $z\in \mathfrak{F}$, which solves uniquely the following variational inequality:
$〈\left(A-\gamma f\right)z,z-{x}^{*}〉\le 0,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\forall {x}^{*}\in \mathfrak{F}.$
(3.2)

Equivalently, we have ${P}_{\mathfrak{F}}\left(I-A+\gamma f\right)z=z$.

Proof. Let z be the unique solution of (3.2). First, we will show that the mapping G i is a nonexpansive mapping for every i = 1, 2,..., N. Let x, y H, since B i is δ i - inverse strongly monotone mapping and 0 < η < 2δ i , for every i = 1, 2,..., N, we have
(3.3)
Thus, (I - ηB i ) is a nonexpansive mapping for every i = 1, 2,..., N . By Lemma 2.9, we have ${G}_{i}={J}_{{M}_{i,\eta }}\left(I-\eta {B}_{i}\right)$ is a nonexpansive mappings for every i = 1, 2,..., N. Let ${x}^{*}\in \mathfrak{F}$; by Lemma 2.8, we have
${x}^{*}={G}_{i}{x}^{*}={J}_{{M}_{i},\eta }\left(I-\eta {B}_{i}\right){x}^{*},\forall i=1,2,...N.$
(3.4)
Let e n = γ n K n x n + (1 - γ n )Sx n . Since G i is a nonexpansive mapping for every i = 1, 2,..., N, we have that S is a nonexpansive mapping. By nonexpansiveness of K n we have
$\begin{array}{lll}\hfill \parallel {e}_{n}-{x}^{*}\parallel & =\parallel {\gamma }_{n}\left({K}_{n}{x}_{n}-{x}^{*}\right)+\left(1-{\gamma }_{n}\right)\left(S{x}_{n}-{x}^{*}\right)\parallel \phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ \le {\gamma }_{n}\parallel {K}_{n}{x}_{n}-{x}^{*}\parallel +\left(1-{\gamma }_{n}\right)\parallel S{x}_{n}-{x}^{*}\parallel \phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ \le {\gamma }_{n}\parallel {x}_{n}-{x}^{*}\parallel +\left(1-{\gamma }_{n}\right)\parallel {x}_{n}-{x}^{*}\parallel \phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ \le \parallel {x}_{n}-{x}^{*}\parallel .\phantom{\rule{2em}{0ex}}& \hfill \text{(4)}\\ \hfill \text{(5)}\end{array}$
(3.5)
Without loss of generality, by conditions (i) and (ii), we have α n ≤ (1 - β n )||A||-1. Since A is a strongly positive linear-bounded self-adjoint operator, we have
$\parallel A\parallel \phantom{\rule{2.77695pt}{0ex}}=sup\left\{|〈Ax,x〉|:x\in H,\parallel x\parallel \phantom{\rule{2.77695pt}{0ex}}=1\right\}.$
(3.6)
For each x C with ||x|| = 1, we have
$〈\left(\left(1-{\beta }_{n}\right)I-{\alpha }_{n}A\right)x,x〉=1-{\beta }_{n}-{\alpha }_{n}〈Ax,x〉\ge 1-{\beta }_{n}-{\alpha }_{n}\parallel A\parallel \phantom{\rule{2.77695pt}{0ex}}\ge 0,$
(3.7)
then (1 - β n )I -α n A is positive. By (3.6) and (3.7), we have
$\begin{array}{c}\parallel \left(1-{\beta }_{n}\right)I-{\alpha }_{n}A\parallel =\mathrm{sup}\left\{〈\left(\left(1-{\beta }_{n}\right)I-{\alpha }_{n}A\right)x\text{,}x〉:x\in C\text{,}\parallel x\parallel \phantom{\rule{0.1em}{0ex}}=1\right\}\\ =\mathrm{sup}\left\{〈1-{\beta }_{n}-{\alpha }_{n}〈Ax\text{,}x〉:x\in C\text{,}\parallel x\parallel \phantom{\rule{0.1em}{0ex}}=1\right\}\\ \le 1-{\beta }_{n}-{\alpha }_{n}〈Ax\text{,}x〉\\ \le 1-{\beta }_{n}-{\alpha }_{n}\overline{\gamma }\text{.}\end{array}$
(3.8)

We shall divide our proof into six steps.

Step 1. We will show that the sequence {x n } is bounded. Let ${x}^{*}\in \mathfrak{F}$, by (3.5) and (3.8), we have
By induction, we can prove that {x n } is bounded, and so are {e n }, {K n x n }, {Sx n } and {G i (x n )} for every i = 1, 2,..., N. Without loss of generality, we can assume that there exists a bounded set D H such that
${e}_{n},{x}_{n},S{x}_{n},{K}_{n}{x}_{n},{G}_{i}{x}_{n}\in D,\forall n\in ℕandi=1,2,\dots ,N.$
(3.9)

Step 2. We will show that $li{m}_{n\to \infty }∥{x}_{n+1}-{x}_{n}∥=0$.

Define sequence {z n } by ${z}_{n}=\frac{1}{1-{\beta }_{n}}\left({x}_{n+1}-{\beta }_{n}{x}_{n}\right)$.

Then xn+1= β n x n + (1 - β n )z n .

Since {x n } is bounded, we have
$\begin{array}{lll}\hfill ||{z}_{n+1}-{z}_{n}||\phantom{\rule{2.77695pt}{0ex}}& =\phantom{\rule{2.77695pt}{0ex}}||\frac{{x}_{n+2}-{\beta }_{n+1}{x}_{n+1}}{1-{\beta }_{n+1}}-\left(\frac{{x}_{n+1}-{\beta }_{n}{x}_{n}}{1-{\beta }_{n}}\right)||\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ =\phantom{\rule{2.77695pt}{0ex}}||\frac{{\alpha }_{n+1}\gamma f\left({x}_{n+1}\right)+\left(\left(1-{\beta }_{n+1}\right)I-{\alpha }_{n+1}A\right){e}_{n+1}}{1-{\beta }_{n+1}}\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ \phantom{\rule{1em}{0ex}}-\left(\frac{{\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(\left(1-{\beta }_{n}\right)I-{\alpha }_{n}A\right){e}_{n}}{1-{\beta }_{n}}\right)||\phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ \le {\alpha }_{n+1}||\frac{\gamma f\left({x}_{n+1}\right)-A{e}_{n+1}}{1-{\beta }_{n+1}}||+||{e}_{n+1}-{e}_{n}||\phantom{\rule{2em}{0ex}}& \hfill \text{(4)}\\ \phantom{\rule{1em}{0ex}}+{\alpha }_{n}||\frac{\gamma f\left({x}_{n}\right)-A{e}_{n}}{1-{\beta }_{n}}||.\phantom{\rule{2em}{0ex}}& \hfill \text{(5)}\\ \hfill \text{(6)}\end{array}$
(3.10)
By definition of e n and nonexpansiveness of S, we have
where M = maxn{||K n x n ||, ||Sx n ||}. Substituting (3.11) into (3.10), we have
(3.12)
It implies that
(3.13)
By (2.3), it implies that
${K}_{n+1}{x}_{n}-{K}_{n}{x}_{n}={\lambda }_{n+1}\left({T}_{n+1}{K}_{n}{x}_{n}-{K}_{n}{x}_{n}\right),$
since λ n 0 as n → ∞, we have
$\underset{n\to \infty }{\mathrm{lim}}∥{K}_{n+1}{x}_{n}-{K}_{n}{x}_{n}∥=0.$
(3.14)
By (3.13), (3.14) and conditions (i), (iii), we have
$\underset{n\to \infty }{\mathrm{limsup}}\left(∥{z}_{n+1}-{z}_{n}∥-∥{x}_{n+1}-{x}_{n}∥\right)\le 0.$
(3.15)
By Lemma 2.5, we have
$\underset{n\to \infty }{\mathrm{lim}}∥{z}_{n}-{x}_{n}∥=0.$
(3.16)
By condition (ii) and (3.16)
$\underset{n\to \infty }{\mathrm{lim}}∥{x}_{n+1}-{x}_{n}∥=\underset{n\to \infty }{\mathrm{lim}}\left(1-{\beta }_{n}\right)∥{z}_{n}-{x}_{n}∥=0.$
(3.17)
Step 3. We will show that
$\underset{n\to \infty }{\mathrm{lim}}∥{e}_{n}-{x}_{n}∥=0.$
(3.18)
Since xn+1= α n γf + β n x n + ((1 - β n )I - α n A) e n , we have
$\begin{array}{lll}\hfill ∥{x}_{n+1}-{e}_{n}∥& =∥{\alpha }_{n}\left(\gamma f\left({x}_{n}\right)-A{e}_{n}\right)+{\beta }_{n}\left({x}_{n}-{e}_{n}\right)∥\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ \le {\alpha }_{n}∥\gamma f\left({x}_{n}\right)-A{e}_{n}∥+{\beta }_{n}\left(∥{x}_{n}-{x}_{n+1}∥+∥{x}_{n+1}-{e}_{n}∥\right),\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$
it implies that
$\left(1-{\beta }_{n}\right)∥{x}_{n+1}-{e}_{n}∥\le {\alpha }_{n}∥\gamma f\left({x}_{n}\right)-A{e}_{n}∥+{\beta }_{n}∥{x}_{n}-{x}_{n+1}∥,$
and it follows that
$∥{x}_{n+1}-{e}_{n}∥\le \frac{{\alpha }_{n}}{\left(1-{\beta }_{n}\right)}∥\gamma f\left({x}_{n}\right)-A{e}_{n}∥+\frac{{\beta }_{n}}{\left(1-{\beta }_{n}\right)}∥{x}_{n}-{x}_{n+1}∥.$
By conditions (i), (ii) and (3.17), we have
$\underset{n\to \infty }{\mathrm{lim}}∥{x}_{n+1}-{e}_{n}∥=0.$
(3.19)
Since ||e n - x n || ≤ ||e n - xn+1|| + ||xn+1- x n ||, by (3.17) and (3.19), we have
$\underset{n\to \infty }{\mathrm{lim}}∥{e}_{n}-{x}_{n}∥=0.$
Step 4. Define a mapping Q : HH by
$Qx=cKx+\left(1-c\right)Sx,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\forall x\in H.$
(3.20)
We will show that
$\underset{n\to \infty }{\mathrm{lim}}||Q{x}_{n}-{x}_{n}||=0.$
(3.21)
Since
$\begin{array}{lll}\hfill ∥Q{x}_{n}-{e}_{n}∥& =∥cK{x}_{n}+\left(1-c\right)S{x}_{n}-{\gamma }_{n}{K}_{n}{x}_{n}-\left(1-{\gamma }_{n}\right)S{x}_{n}∥\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ \le ∥cK{x}_{n}-{\gamma }_{n}{K}_{n}{x}_{n}∥+|{\gamma }_{n}-c|∥S{x}_{n}∥\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ =∥cK{x}_{n}-{\gamma }_{n}K{x}_{n}+{\gamma }_{n}K{x}_{n}-{\gamma }_{n}{K}_{n}{x}_{n}∥+|{\gamma }_{n}-c|∥S{x}_{n}∥\phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ \le \phantom{\rule{2.77695pt}{0ex}}|c-{\gamma }_{n}|∥K{x}_{n}∥+{\gamma }_{n}∥K{x}_{n}-{K}_{n}{x}_{n}∥+|{\gamma }_{n}-c|∥S{x}_{n}∥\phantom{\rule{2em}{0ex}}& \hfill \text{(4)}\\ \le \phantom{\rule{2.77695pt}{0ex}}|c-{\gamma }_{n}|∥K{x}_{n}∥+\underset{x\in D}{\mathrm{sup>}}\left\{∥Kx-{K}_{n}x∥\right\}+|{\gamma }_{n}-c|∥S{x}_{n}∥.\phantom{\rule{2em}{0ex}}& \hfill \text{(5)}\\ \hfill \text{(6)}\end{array}$
By remark 2.12 and condition (iii), we have
$\underset{n\to \infty }{\mathrm{lim}}∥Q{x}_{n}-{e}_{n}∥=0.$
(3.22)
Since ||Qx n - x n || ≤ ||Qx n - e n || + ||e n - x n ||, from (3.22) and (3.18), we have
$\underset{n\to \infty }{\mathrm{lim}}\phantom{\rule{2.77695pt}{0ex}}||Q{x}_{n}-{x}_{n}||\phantom{\rule{2.77695pt}{0ex}}=0.$
Step 5. We will show that
$\underset{n\to \infty }{\text{lim}\phantom{\rule{0.1em}{0ex}}\text{sup}}〈\left(\gamma f-A\right)z,{x}_{n}-z〉\le 0,$
(3.23)
where $z=P\mathfrak{F}\left(I-\left(A-\gamma f\right)\right)z$. Let $\left\{{{x}_{n}}_{{}_{j}}\right\}$ be subsequence of {x n } such that
$\underset{n\to \infty }{\mathrm{limsup}}〈\left(\gamma f-A\right)z,{x}_{n}-z〉=\underset{j\to \infty }{\mathrm{lim}}〈\left(\gamma f-A\right)z,{x}_{{n}_{j}}-z〉.$
(3.24)
Without loss of generality, we may assume that $\left\{{{x}_{n}}_{{}_{j}}\right\}$ converses weakly to some q H. By nonexpansiveness of S and K, (3.20) and Lemma 2.3, we have that Q is nonexpansive mapping and
$F\left(Q\right)=F\left(K\right)\phantom{\rule{2.77695pt}{0ex}}\bigcap F\left(S\right).$
(3.25)
Since ${J}_{{M}_{i,\eta }}\left(I-\eta {B}_{i}\right)x={G}_{i}x$ for every x H and i = 1, 2,... N , by Lemma 2.8, we have
$VI\left(H,{B}_{i},{M}_{i}\right)=F\left({J}_{{M}_{i,\eta }}\left(I-\eta {B}_{i}\right)\right)=F\left({G}_{i}\right),\phantom{\rule{1em}{0ex}}\forall i=1,2,...\phantom{\rule{0.3em}{0ex}}N.$
(3.26)
By Lemma 2.6 and Lemma 2.9, we have
$F\left(S\right)=\bigcap _{i=1}^{N}F\left({G}_{i}\right)=\bigcap _{i=1}^{N}VI\left(H,{B}_{i},{M}_{i}\right)$
(3.27)
By Lemma 2.13, we have
$F\left(K\right)=\bigcap _{i=1}^{\infty }F\left({T}_{i}\right).$
(3.28)
By (3.25), (3.27), and (3.28), we have
$F\left(Q\right)=F\left(K\right)\bigcap F\left(S\right)=\bigcap _{i=1}^{\infty }F\left({T}_{i}\right)\bigcap \bigcap _{i=1}^{N}VI\left(H,{B}_{i},{M}_{i}\right).$
(3.29)
Since ${x}_{{n}_{j}}⇀q$ as j → ∞, nonexpansiveness of Q, (3.21) and Lemma 2.4, we have
$q\in F\left(Q\right)=\bigcap _{i=1}^{\infty }F\left({T}_{i}\right)\bigcap \bigcap _{i=1}^{N}VI\left(H,{B}_{i},{M}_{i}\right)=\mathfrak{F}.$
(3.30)
By (3.24) and (3.30), we have
$\underset{n\to \infty }{\text{lim}\phantom{\rule{0.1em}{0ex}}\text{sup}}〈\left(\gamma f-A\right)z,{x}_{n}-z〉=\underset{n\to \infty }{\text{lim}}〈\left(\gamma f-A\right)z,{x}_{{n}_{j}}-z〉=〈\left(\gamma f-A\right)z,q-z〉\le 0.$
Step 6. Finally, we will show that x n z as n → ∞, where $z=P\mathfrak{F}\left(I-\left(A-\gamma f\right)\right)z$. Since
it implies that
$\begin{array}{c}{‖{x}_{n+1}-z‱}^{2}\le \frac{\left(1-2{\alpha }_{n}\overline{\gamma }+{\alpha }_{n}\gamma \theta \right)}{1-{\alpha }_{n}\gamma \theta }{‖{x}_{n}-z‱}^{2}+\frac{{\alpha }_{n}^{2}{\overline{\gamma }}^{2}}{1-{\alpha }_{n}\gamma \theta }{‖{x}_{n}-z‱}^{2}\\ \phantom{\rule{0.1em}{0ex}}+\frac{2{\alpha }_{n}}{1-{\alpha }_{n}\gamma \theta }〈\gamma f\left(z\right)-Az,{x}_{n+1}-z〉\\ =\frac{\left(1-{\alpha }_{n}\gamma \theta +{\alpha }_{n}\gamma \theta -2{\alpha }_{n}\overline{\gamma }+{\alpha }_{n}\gamma \theta \right)}{1-{\alpha }_{n}\gamma \theta }{‖{x}_{n}-z‱}^{2}+\frac{{\alpha }_{n}^{2}{\overline{\gamma }}^{2}}{1-{\alpha }_{n}\gamma \theta }{‖{x}_{n}-z‱}^{2}\\ \phantom{\rule{0.1em}{0ex}}+\frac{2{\alpha }_{n}}{1-{\alpha }_{n}\gamma \theta }〈\gamma f\left(z\right)-Az,{x}_{n+1}-z〉\\ =\frac{\left(1-{\alpha }_{n}\gamma \theta -2{\alpha }_{n}\left(\overline{\gamma }-\gamma \theta \right)\right)}{1-{\alpha }_{n}\gamma \theta }{‖{x}_{n}-z‱}^{2}+\frac{{\alpha }_{n}^{2}{\overline{\gamma }}^{2}}{1-{\alpha }_{n}\gamma \theta }{‖{x}_{n}-z‱}^{2}\\ \phantom{\rule{0.1em}{0ex}}+\frac{2{\alpha }_{n}}{1-{\alpha }_{n}\gamma \theta }〈\gamma f\left(z\right)-Az,{x}_{n+1}-z〉\\ =\left(1-\frac{2{\alpha }_{n}\left(\overline{\gamma }-\gamma \theta \right)\right)}{1-{\alpha }_{n}\gamma \theta }{‖{x}_{n}-z‱}^{2}+\frac{{\alpha }_{n}^{2}{\overline{\gamma }}^{2}}{1-{\alpha }_{n}\gamma \theta }{‖{x}_{n}-z‱}^{2}\\ \phantom{\rule{0.1em}{0ex}}+\frac{2{\alpha }_{n}}{1-{\alpha }_{n}\gamma \theta }〈\gamma f\left(z\right)-Az,{x}_{n+1}-z〉\\ =\left(1-\frac{2{\alpha }_{n}\left(\overline{\gamma }-\gamma \theta \right)\right)}{1-{\alpha }_{n}\gamma \theta }{‖{x}_{n}-z‱}^{2}+\frac{{\alpha }_{n}}{1-{\alpha }_{n}\gamma \theta }\left({\alpha }_{n}{\overline{\gamma }}^{2}{‖{x}_{n}-z‱}^{2}\\ \phantom{\rule{0.1em}{0ex}}+2〈\gamma f\left(z\right)-Az,{x}_{n+1}-z〉\right)\\ =\left(1-\frac{2{\alpha }_{n}\left(\overline{\gamma }-\gamma \theta \right)}{}\end{array}$