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

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.

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 → ℝ 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 $\bar{\gamma }$ if there exists a constant $\bar{\gamma }>0$ with the property

$$〈Ax,x〉\ge \bar{\gamma }\parallel x{\parallel }^2.$$

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

$$〈v-u,Au〉\ge 0forallv\in C.$$

(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., [5–8]).

In 2008, Qin et al. [9] introduced the following iterative scheme:

$$\left\{\begin{array}{c}{y}_n=P_C\left(I-s_{n}A\right)x_n\\ x_{n+1}={\alpha }_n\gamma f\left(W_n{x}_n\right)+\left(I-{\alpha }_{n}B\right)W_n{P}_C\left(I-r_{n}A\right)y_n,\forall n\in \mathbb{N},\end{array}\right.$$

(1.2)

where W _n is the W-mapping generated by a finite family of nonexpansive mappings and real numbers, A : C → H 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:

$$x_0\in H,x_{n+1}=\left(I-{\alpha }_{n}A\right)S{x}_n+{\alpha }_n\gamma f\left(x_n\right),\forall n\ge 0,$$

(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 $\bar{\gamma }$, 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 : H → H 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 : H → H, M : H → 2^H are two nonlinear mappings, and θ is zero vector in H (see, for instance, [11–16]). 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 : C → C as follows:

$$\begin{array}{lll}\hfill U_{n,0}& =I& \hfill \text{(1)}\\ \hfill U_{n,1}& ={\lambda }_1{T}_1{U}_{n,0}+\left(1-{\lambda }_1\right)U_{n,0},& \hfill \text{(2)}\\ \hfill U_{n,2}& ={\lambda }_2{T}_2{U}_{n,1}+\left(1-{\lambda }_2\right)U_{n,1},& \hfill \text{(3)}\\ \hfill U_{n,3}& ={\lambda }_3{T}_3{U}_{n,2}+\left(1-{\lambda }_3\right)U_{n,2},& \hfill \text{(4)}\\ ⋮& \hfill \text{(5)}\\ \hfill U_{n,k}& ={\lambda }_k{T}_k{U}_{n,k-1}+\left(1-{\lambda }_k\right)U_{n,k-1}& \hfill \text{(6)}\\ \hfill U_{n,k+1}& ={\lambda }_{k+1}{T}_{k+1}{U}_{n,k}+\left(1-{\lambda }_{k+1}\right)U_{n,k}& \hfill \text{(7)}\\ ⋮& \hfill \text{(8)}\\ \hfill U_{n,n-1}& ={\lambda }_{n-1}{T}_{n-1}{U}_{n,n-2}+\left(1-{\lambda }_{n-1}\right)U_{n,n-2}& \hfill \text{(9)}\\ \hfill K_n& =U_{n,n}={\lambda }_n{T}_n{U}_{n,n-1}+\left(1-{\lambda }_n\right)U_{n,n-1}.& \hfill \text{(10)}\\ \hfill \text{(11)}\end{array}$$

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 : C → C is the S - mapping generated by G₁, G₂,..., G _N and ν₁, ν₂,..., ν _N , where G _i : H → H 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 : H → H 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:

$$\parallel x-P_{C}x\parallel =\underset{y\in C}{\min }\parallel x-y\parallel .$$

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

$$〈x-y,y-z〉\ge 0\forall z\in C.$$

Lemma 2.2. (see [18]) Let {s _n } be a sequence of nonnegative real number satisfying

$$s_{n+1}=\left(1-{\alpha }_n\right)s_n+{\alpha }_n{\beta }_n,\forall n\ge 0$$

where {α _n }, {β _n } satisfy the conditions:

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

• (2) $\underset{n\to \infty }{{\displaystyle \text{lim}\text{sup}}}{\beta }_n\le 0or{\displaystyle \sum _{n=1}^{\infty }}\left|{\alpha }_n{\beta }_n\right|<\infty$.

Then lim_n→∞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 ∈ 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 : C → C 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 sup_n→∞β _n < 1.

Suppose x_n+1= β _n x _n +(1 - β _n )z _n for all integer n ≥ 0 and lim sup_n→∞(||z_n+1- z _n || - ||x_n+1- x _n ||) ≤ 0. Then lim_n→∞||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 ${\alpha }_j=\left({\alpha }_1^j,{\alpha }_2^j,{\alpha }_3^j\right)\in I\times I\times I$ where I ∈ [0, 1] and ${\alpha }_1^j+{\alpha }_2^j+{\alpha }_3^j=1$, define the mapping S : C → C as follows:

$$\begin{array}{lll}\hfill U_0& =I& \hfill \text{(1)}\\ \hfill U_1& ={\alpha }_1^1{T}_1{U}_0+{\alpha }_2^1{U}_0+{\alpha }_3^{1}I& \hfill \text{(2)}\\ \hfill U_2& ={\alpha }_1^2{T}_2{U}_1+{\alpha }_2^2{U}_1+{\alpha }_3^{2}I& \hfill \text{(3)}\\ \hfill U_3& ={\alpha }_1^3{T}_3{U}_2+{\alpha }_2^3{U}_2+{\alpha }_3^{3}I& \hfill \text{(4)}\\ ⋮& \hfill \text{(5)}\\ \hfill U_{N-1}& ={\alpha }_1^{N-1}{T}_{N-1}{U}_{N-2}+{\alpha }_2^{N-1}{U}_{N-2}+{\alpha }_3^{N-1}I& \hfill \text{(6)}\\ \hfill S& =U_N={\alpha }_1^N{T}_N{U}_{N-1}+{\alpha }_2^N{U}_{N-1}+{\alpha }_3^{N}I.& \hfill \text{(7)}\\ \hfill \text{(8)}\end{array}$$

(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 ${\alpha }_j=\left({\alpha }_1^j,{\alpha }_2^j,{\alpha }_3^j\right)\in I\times I\times I$, 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 ${\alpha }_j^{\left(n\right)}=\left({\alpha }_1^{n,j},{\alpha }_2^{n,j},{\alpha }_3^{n,j}\right)$, ${\alpha }_j=\left({\alpha }_1^j,{\alpha }_2^j,{\alpha }_3^j\right)\in I\times I\times I$, 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 ∈ 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 lim_n→∞||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 : H → H defined by

$$J_{M\mathrm{,}\lambda }(u)=(I+\lambda M{)}^{-1}(u)\text{,\hspace{1em}}\forall u\in H\mathrm{,}$$

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 = J_{M, λ}(u - λBu), ∀λ > 0, i.e.,

$$VI\left(H,B,M\right)=F\left(J_{M,\lambda }\left(I-\lambda B\right)\right),\forall \lambda >0.$$

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

$$\left|\right|x\left|\right|=\left|\right|y\left|\right|=\left|\right|\lambda x+\left(1-\lambda \right)y\left|\right|$$

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 ∈ 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 ∈ ℕ, lim_n→∞K _n x exits.

Proof. Let x ∈ C. Then for k, n ∈ ℕ, we have

$$\begin{array}{lll}\hfill \parallel U_{n+1,k}x-U_{n,k}x\parallel & =\parallel {\lambda }_k{T}_k{U}_{n+1,k-1}x+\left(1-{\lambda }_k\right)U_{n+1,k-1}x-{\lambda }_k{T}_k{U}_{n,k-1}x-\left(1-{\lambda }_k\right)U_{n,k-1}x\parallel & \hfill \text{(1)}\\ =\parallel {\lambda }_k\left(T_k{U}_{n+1,k-1}x-T_k{U}_{n,k-1}x\right)+\left(1-{\lambda }_k\right)\left(U_{n+1,k-1}x-U_{n,k-1}x\right)\parallel & \hfill \text{(2)}\\ \le {\lambda }_k\parallel T_k{U}_{n+1,k-1}x-T_k{U}_{n,k-1}x\parallel +\left(1-{\lambda }_k\right)\parallel U_{n+1,k-1}x-U_{n,k-1}x\parallel & \hfill \text{(3)}\\ \le {\lambda }_k\parallel U_{n+1,k-1}x-U_{n,k-1}x\parallel +\left(1-{\lambda }_k\right)\parallel U_{n+1,k-1}x-U_{n,k-1}x\parallel & \hfill \text{(4)}\\ =\parallel U_{n+1,k-1}x-U_{n,k-1}x\parallel & \hfill \text{(5)}\\ =\parallel {\lambda }_{k-1}{T}_{k-1}{U}_{n+1,k-2}x+\left(1-{\lambda }_{k-1}\right)U_{n+1,k-2}x-{\lambda }_{k-1}{T}_{k-1}{U}_{n,k-2}x-\left(1-{\lambda }_{k-1}\right)U_{n,k-2}x\parallel & \hfill \text{(6)}\\ \le {\lambda }_{k-1}\parallel T_{k-1}{U}_{n+1,k-2}x-T_{k-1}{U}_{n,k-2}x\parallel +\left(1-{\lambda }_{k-1}\right)\parallel U_{n+1,k-2}x-U_{n,k-2}x\parallel & \hfill \text{(7)}\\ \le \parallel U_{n+1,k-2}x-U_{n,k-2}x\parallel & \hfill \text{(8)}\\ ⋮& \hfill \text{(9)}\\ \le \parallel U_{n+1,1}x-U_{n,1}x\parallel & \hfill \text{(10)}\\ =\parallel {\lambda }_1{T}_1{U}_{n+1,0}x+\left(1-{\lambda }_1\right)U_{n+1,0}x-{\lambda }_1{T}_1{U}_{n,0}x-\left(1-{\lambda }_1\right)U_{n,0}x\parallel & \hfill \text{(11)}\\ =\parallel {\lambda }_1{T}_{1}x+\left(1-{\lambda }_1\right)x-{\lambda }_1{T}_{1}x-\left(1-{\lambda }_1\right)x\parallel & \hfill \text{(12)}\\ =0,& \hfill \text{(13)}\\ \hfill \text{(14) }\end{array}$$

(2.2)

which implies that U_n+1,k= U_{n, k}for every k, n ∈ ℕ. Hence, K _n = U_{n, n}= U_n+1,n. Since K_n+1x = U_n+1,n+1x = λ_n+1T_n+1K _n x + (1 - λ_n+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 ∈ ℕ, we have

$$\begin{array}{lll}\hfill \parallel K_{n}x-x^{*}\parallel & =\parallel {\lambda }_n{T}_n{U}_{n,n-1}x+\left(1-{\lambda }_n\right)U_{n,n-1}x-x^{*}\parallel & \hfill \text{(1)}\\ \le {\lambda }_n\parallel T_n{U}_{n,n-1}x-x^{*}\parallel +\left(1-{\lambda }_n\right)\parallel U_{n,n-1}x-x^{*}\parallel & \hfill \text{(2)}\\ \le \parallel U_{n,n-1}x-x^{*}\parallel & \hfill \text{(3)}\\ =\parallel {\lambda }_{n-1}{T}_{n-1}{U}_{n,n-2}x+\left(1-{\lambda }_{n-1}\right)U_{n,n-2}x-x^{*}\parallel & \hfill \text{(4)}\\ \le {\lambda }_{n-1}\parallel T_{n-1}{U}_{n,n-2}x-x^{*}\parallel +\left(1-{\lambda }_{n-1}\right)\parallel U_{n,n-2}x-x^{*}\parallel & \hfill \text{(5)}\\ \le \parallel U_{n,n-2}x-x^{*}\parallel & \hfill \text{(6)}\\ .& \hfill \text{(7)}\\ .& \hfill \text{(8)}\\ .& \hfill \text{(9)}\\ \le \parallel U_{n,1}x-x^{*}\parallel & \hfill \text{(10)}\\ =\parallel {\lambda }_1{T}_1{U}_{n,0}x+\left(1-{\lambda }_1\right)U_{n,0}x-x^{*}\parallel & \hfill \text{(11)}\\ \le {\lambda }_1\parallel T_{1}x-x^{*}\parallel +\left(1-{\lambda }_1\right)\parallel x-x^{*}\parallel & \hfill \text{(12)}\\ =\parallel x-x^{*}\parallel ,& \hfill \text{(13)}\\ \hfill \text{(14)}\end{array}$$

(2.4)

which implies that {K _n x} is bounded, and so is {T _n K _n x}. For m ≥ n, by (2.3) we have

$$\begin{array}{lll}\hfill \parallel K_{m}x-K_{n}x\parallel & =\parallel K_{m}x-K_{m-1}x+K_{m-1}x-K_{m-2}x+K_{m-2}x-+\cdot \cdot \cdot & \hfill \text{(1)}\\ -K_{n+1}x+K_{n+1}x-K_{n}x\parallel & \hfill \text{(2)}\\ \le \parallel K_{m}x-K_{m-1}x\parallel +\parallel K_{m-1}x-K_{m-2}x\parallel +\parallel K_{m-2}x-K_{m-3}x\parallel +\cdot \cdot \cdot & \hfill \text{(3)}\\ +\parallel K_{n+2}-K_{n+1}x\parallel +\parallel K_{n+1}x-K_{n}x\parallel & \hfill \text{(4)}\\ ={\lambda }_m\parallel T_m{K}_{m-1}x-K_{m-1}x\parallel +{\lambda }_{m-1}\parallel T_{m-1}{K}_{m-2}x-K_{m-2}x\parallel +\cdot \cdot \cdot & \hfill \text{(5)}\\ +{\lambda }_{n+1}\parallel T_{n+1}{K}_{n}x-K_{n}x\parallel & \hfill \text{(6)}\\ \le M\sum _{k=n+1}^m{\lambda }_k,& \hfill \text{(7)}\\ \hfill \text{(8) }\end{array}$$

(2.5)

where M = sup_n∈ℕ{||T_n+1K _n x - K _n x||}. This implies that {K _n x} is Cauchy sequence. Hence lim_n→∞K _n x exists.

From Lemma 2.11, we can define a mapping K : C → C as follows:

$$Kx=\underset{n\to \infty }{\lim }{K}_{n}x,x\in C.$$

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 ∈ ℕ, K _n is nonexpansive mappings. Let x, y ∈ C, then we have

$$\parallel Kx-Ky\parallel =\underset{n\Rightarrow \infty }{\lim }\parallel K_{n}x-K_{n}y\parallel \le \parallel x-y\parallel .$$

(2.6)

By (2.6), we have K : C → C is nonexpansive mapping. Next, we will show that lim_n→∞sup_x∈D||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 m ≥ n, 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

$$\underset{n\to \infty }{\lim }\underset{x\in D}{\sup }\parallel Kx-K_{n}x\parallel =0.$$

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 ∈ ℕ be fixed. Since

$$\begin{array}{lll}\hfill \parallel K_n{x}_0-x^{*}\parallel & =\parallel {\lambda }_n{T}_n{U}_{n,n-1}{x}_0+\left(1-{\lambda }_n\right)U_{n,n-1}{x}_0-x^{*}\parallel & \hfill \text{(1) }\\ =\parallel {\lambda }_n\left(T_n{U}_{n,n-1}{x}_0-x^{*}\right)+\left(1-{\lambda }_n\right)\left(U_{n,n-1}{x}_0-x^{*}\right)\parallel & \hfill \text{(2) }\\ \le {\lambda }_n\parallel T_n{U}_{n,n-1}{x}_0-x^{*}\parallel +\left(1-{\lambda }_n\right)\parallel U_{n,n-1}{x}_0-x^{*}\parallel & \hfill \text{(3) }\\ \le \parallel U_{n,n-1}{x}_0-x^{*}\parallel & \hfill \text{(4) }\\ =\parallel {\lambda }_{n-1}\left(T_{n-1}{U}_{n,n-2}{x}_0-x^{*}\right)+\left(1-{\lambda }_{n-1}\right)U_{n,n-2}\left(x_0-x^{*}\right)\parallel & \hfill \text{(5) }\\ \le {\lambda }_{n-1}\parallel T_{n-1}{U}_{n,n-2}{x}_0-x^{*}\parallel +\left(1-{\lambda }_{n-1}\right)\parallel U_{n,n-2}{x}_0-x^{*}\parallel & \hfill \text{(6) }\\ \le \parallel U_{n,n-2}{x}_0-x^{*}\parallel & \hfill \text{(7) }\\ ⋮& \hfill \text{(8) }\\ .& \hfill \text{(9) }\\ .& \hfill \text{(10) }\\ \le \parallel U_{n,k}{x}_0-x^{*}\parallel & \hfill \text{(11) }\\ =\parallel {\lambda }_k\left(T_k{U}_{n,k-1}{x}_0-x^{*}\right)+\left(1-{\lambda }_k\right)\left(U_{n,k-1}{x}_0-x^{*}\right)\parallel & \hfill \text{(12) }\\ \le {\lambda }_k\parallel T_k{U}_{n,k-1}{x}_0-x^{*}\parallel +\left(1-{\lambda }_k\right)\parallel U_{n,k-1}{x}_0-x^{*}\parallel & \hfill \text{(13) }\\ \le \parallel U_{n,k-1}{x}_0-x^{*}\parallel & \hfill \text{(14) }\\ .& \hfill \text{(15) }\\ .& \hfill \text{(16) }\\ .& \hfill \text{(17) }\\ \le \parallel U_{n,1}{x}_0-x^{*}\parallel & \hfill \text{(18) }\\ =\parallel {\lambda }_1\left(T_1{x}_0-x^{*}\right)+\left(1-{\lambda }_1\right)\left(x_0-x^{*}\right)\parallel & \hfill \text{(19) }\\ \le {\lambda }_1\parallel T_1{x}_0-x^{*}\parallel +\left(1-{\lambda }_1\right)\parallel x_0-x^{*}\parallel & \hfill \text{(20) }\\ \le \parallel x_0-x^{*}\parallel ,& \hfill \text{(21) }\\ \hfill \text{(22) }\end{array}$$

(2.7)

we have

$$\begin{array}{lll}\hfill \parallel x_0-x^{*}\parallel & =\underset{n\to \infty }{\lim }\parallel K_n{x}_0-x^{*}\parallel \le \parallel {\lambda }_1\left(T_1{x}_0-x^{*}\right)+\left(1-{\lambda }_1\right)\left(x_0-x^{*}\right)\parallel & \hfill \text{(1) }\\ \le {\lambda }_1\parallel T_1{x}_0-x^{*}\parallel +\left(1-{\lambda }_1\right)\parallel x_0-x^{*}\parallel & \hfill \text{(2) }\\ \le \parallel x_0-x^{*}\parallel ,& \hfill \text{(3) }\\ \hfill \text{(4) }\end{array}$$

this implies that

$$\parallel x_0-x^{*}\parallel =\parallel T_1{x}_0-x^{*}\parallel =\parallel {\lambda }_1\left(T_1{x}_0-x^{*}\right)+\left(1-{\lambda }_1\right)\left(x_0-x^{*}\right)\parallel .$$

By Lemma 2.10, we have T₁x₀ = x₀, that is x₀ ∈ F (T₁). It follows that U_n,1x₀ = x₀. By (2.7), we have

$$\begin{array}{lll}\hfill \parallel K_n{x}_0-x^{*}\parallel & \le \parallel U_{n,2}{x}_0-x^{*}\parallel =\parallel {\lambda }_2\left(T_2{U}_{n,1}{x}_0-x^{*}\right)+\left(1-{\lambda }_2\right)\left(U_{n,1}{x}_0-x^{*}\right)\parallel & \hfill \text{(1)}\\ =\parallel {\lambda }_2\left(T_2{x}_0-x^{*}\right)+\left(1-{\lambda }_2\right)\left(x_0-x^{*}\right)\parallel & \hfill \text{(2)}\\ \le {\lambda }_2\parallel T_2{x}_0-x^{*}\parallel +\left(1-{\lambda }_2\right)\parallel x_0-x^{*}\parallel & \hfill \text{(3)}\\ \le \parallel x_0-x^{*}\parallel .& \hfill \text{(4)}\\ \hfill \text{(5) }\end{array}$$

It follows that

$$\begin{array}{lll}\hfill \parallel x_0-x^{*}\parallel & =\underset{n\to \infty }{\lim }\parallel K_n{x}_0-x^{*}\parallel & \hfill \text{(1)}\\ \le \parallel {\lambda }_2\left(T_2{x}_0-x^{*}\right)+\left(1-{\lambda }_2\right)\left(x_0-x^{*}\right)\parallel & \hfill \text{(2)}\\ \le {\lambda }_2\parallel T_2{x}_0-x^{*}\parallel +\left(1-{\lambda }_2\right)\parallel x_0-x^{*}\parallel & \hfill \text{(3)}\\ \le \parallel x_0-x^{*}\parallel ,& \hfill \text{(4)}\\ \hfill \text{(5)}\end{array}$$

which implies

$$\parallel x_0-x^{*}\parallel =\parallel T_2{x}_0-x^{*}\parallel =\parallel {\lambda }_2\left(T_2{x}_0-x^{*}\right)+\left(1-{\lambda }_2\right)\left(x_0-x^{*}\right)\parallel .$$

By Lemma 2.10, we obtain that T₂x₀ = x₀, that is x₀ ∈ F (T₂). It follows that U_n,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

$$\begin{array}{lll}\hfill \parallel K_n{x}_0-x^{*}\parallel & \le \parallel U_{n,k}{x}_0-x^{*}\parallel & \hfill \text{(1)}\\ =\parallel {\lambda }_k\left(T_k{U}_{n,k-1}{x}_0-x^{*}\right)+\left(1-{\lambda }_k\right)\left(U_{n,k-1}{x}_0-x^{*}\right)\parallel & \hfill \text{(2)}\\ =\parallel {\lambda }_k\left(T_k{x}_0-x^{*}\right)+\left(1-{\lambda }_k\right)\left(x_0-x^{*}\right)\parallel & \hfill \text{(3)}\\ \le {\lambda }_k\parallel T_k{x}_0-x^{*}\parallel +\left(1-{\lambda }_k\right)\parallel x_0-x^{*}\parallel & \hfill \text{(4)}\\ \le \parallel x_0-x^{*}\parallel .& \hfill \text{(5)}\\ \hfill \text{(6)}\end{array}$$

It follows that

$$\begin{array}{lll}\hfill \parallel x_0-x^{*}\parallel & =\underset{n\to \infty }{\lim }\parallel K_n{x}_0-x^{*}\parallel & \hfill \text{(1)}\\ =\parallel {\lambda }_k\left(T_k{x}_0-x^{*}\right)+\left(1-{\lambda }_k\right)\left(x_0-x^{*}\right)\parallel & \hfill \text{(2)}\\ \le {\lambda }_k\parallel T_k{x}_0-x^{*}\parallel +\left(1-{\lambda }_k\right)\parallel x_0-x^{*}\parallel & \hfill \text{(3)}\\ \le \parallel x_0-x^{*}\parallel ,& \hfill \text{(4)}\\ \hfill \text{(5)}\end{array}$$

(2.8)

which implies

$$\parallel x_0-x^{*}\parallel =\parallel T_k{x}_0-x^{*}\parallel =\parallel {\lambda }_k\left(T_k{x}_0-x^{*}\right)+\left(1-{\lambda }_k\right)\left(x_0-x^{*}\right)\parallel .$$

(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 : H → H 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 : H → H 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 ${\nu }_j=\left({\alpha }_1^j,{\alpha }_2^j,{\alpha }_3^j\right)\in I\times I\times I$, 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 : C → C 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.,

$$Kx=\underset{n\to \infty }{\lim }{K}_{n}x$$

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 ∈ 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 : H → H 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) $\underset{n\to \infty }{\lim }{\alpha }_n=0$ 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) $\underset{n\to \infty }{\lim }{\gamma }_n=c\in \left(0,1\right)$

Then {x _n } converges strongly to $z\in \mathfrak{F}$, which solves uniquely the following variational inequality:

$$〈\left(A-\gamma f\right)z,z-x^{*}〉\le 0,\forall x^{*}\in \mathfrak{F}.$$

(3.2)

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

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

$$\begin{array}{lll}\hfill \parallel \left(I-\eta B_i\right)x-\left(I-\eta B_i\right)y{\parallel }^2& =\parallel x-y-\eta \left(B_{i}x-B_{i}y\right){\parallel }^2& \hfill \text{(1)}\\ =\parallel x-y{\parallel }^2-2\eta 〈x-y,B_{i}x-B_{i}y〉+{\eta }^2\parallel B_{i}x-B_{i}y{\parallel }^2& \hfill \text{(2)}\\ \le \parallel x-y{\parallel }^2-2{\delta }_i\eta \parallel B_{i}x-B_{i}y{\parallel }^2+{\eta }^2\parallel B_{i}x-B_{i}y{\parallel }^2& \hfill \text{(3)}\\ =\parallel x-y{\parallel }^2+\eta \left(\eta -2{\delta }_i\right)\parallel B_{i}x-B_{i}y{\parallel }^2& \hfill \text{(4)}\\ \le \parallel x-y{\parallel }^2.& \hfill \text{(5)}\\ \hfill \text{(6) }\end{array}$$

(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

$$\begin{array}{lll}\hfill \parallel e_n-x^{*}\parallel & =\parallel {\gamma }_n\left(K_n{x}_n-x^{*}\right)+\left(1-{\gamma }_n\right)\left(S{x}_n-x^{*}\right)\parallel & \hfill \text{(1)}\\ \le {\gamma }_n\parallel K_n{x}_n-x^{*}\parallel +\left(1-{\gamma }_n\right)\parallel S{x}_n-x^{*}\parallel & \hfill \text{(2)}\\ \le {\gamma }_n\parallel x_n-x^{*}\parallel +\left(1-{\gamma }_n\right)\parallel x_n-x^{*}\parallel & \hfill \text{(3)}\\ \le \parallel x_n-x^{*}\parallel .& \hfill \text{(4)}\\ \hfill \text{(5)}\end{array}$$

(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

$$\parallel A\parallel =sup\left\{\left|〈Ax,x〉\right|:x\in H,\parallel x\parallel =1\right\}.$$

(3.6)

For each x ∈ C with ||x|| = 1, we have

$$〈\left(\left(1-{\beta }_n\right)I-{\alpha }_{n}A\right)x,x〉=1-{\beta }_n-{\alpha }_n〈Ax,x〉\ge 1-{\beta }_n-{\alpha }_n\parallel A\parallel \ge 0,$$

(3.7)

then (1 - β _n )I -α _n A is positive. By (3.6) and (3.7), we have

$$\begin{array}{c}\parallel (1-{\beta }_n)I-{\alpha }_{n}A\parallel =\sup \{〈((1-{\beta }_n)I-{\alpha }_{n}A)x\text{,}x〉:x\in C\text{,}\parallel x\parallel =1\}\\ =\sup \{〈1-{\beta }_n-{\alpha }_n〈Ax\text{,}x〉:x\in C\text{,}\parallel x\parallel =1\}\\ \le 1-{\beta }_n-{\alpha }_n〈Ax\text{,}x〉\\ \le 1-{\beta }_n-{\alpha }_n\overline{\gamma }\text{.}\end{array}$$

(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

$$\begin{array}{lll}\hfill \parallel x_{n+1}-x^{*}\parallel & =\parallel {\alpha }_n\gamma f\left(x_n\right)+{\beta }_n{x}_n+\left(\left(1-{\beta }_n\right)I-{\alpha }_{n}A\right)e_n-x^{*}\parallel & \hfill \text{(1)}\\ =\parallel {\alpha }_n\gamma f\left(x_n\right)-{\alpha }_{n}A{x}^{*}+{\alpha }_{n}A{x}^{*}-{\beta }_n{x}^{*}+{\beta }_n{x}^{*}+{\beta }_n{x}_n& \hfill \text{(2)}\\ +\left(\left(1-{\beta }_n\right)I-{\alpha }_{n}A\right)e_n-x^{*}\parallel & \hfill \text{(3)}\\ \le {\alpha }_n\parallel \gamma f\left(x_n\right)-A{x}^{*}\parallel +{\beta }_n\parallel x_n-x^{*}\parallel +\parallel \left(\left(1-{\beta }_n\right)I-{\alpha }_{n}A\right)\left(e_n-x^{*}\right)\parallel & \hfill \text{(4)}\\ \le {\alpha }_n\parallel \gamma f\left(x_n\right)-A{x}^{*}\parallel +{\beta }_n\parallel x_n-x^{*}\parallel +\left(\left(1-{\beta }_n\right)I-{\alpha }_n\stackrel{}{\gamma }\right)\parallel x_n-x^{*}\parallel & \hfill \text{(5)}\\ \le {\alpha }_n\left(\parallel \gamma f\left(x_n\right)-\gamma f\left(x^{*}\right)\parallel +\parallel \gamma f\left(x^{*}\right)-A{x}^{*}\parallel \right)+\left(1-{\alpha }_n\stackrel{}{\gamma }\right)\parallel x_n-x^{*}\parallel & \hfill \text{(6)}\\ \le {\alpha }_n\gamma \theta \parallel x_n-x^{*}\parallel +{\alpha }_n\parallel \gamma f\left(x^{*}\right)-A{x}^{*}\parallel +\left(1-{\alpha }_n\stackrel{}{\gamma }\right)\parallel x_n-x^{*}\parallel & \hfill \text{(7)}\\ ={\alpha }_n\parallel \gamma f\left(x^{*}\right)-A{x}^{*}\parallel +\left(1-{\alpha }_n\left(\stackrel{}{\gamma }-\gamma \theta \right)\right)\parallel x_n-x^{*}\parallel & \hfill \text{(8)}\\ \le max\left\{\parallel x_n-x^{*}\parallel ,\frac{\parallel \gamma f\left(x^{*}\right)-A{x}^{*}\parallel }{\stackrel{}{\gamma }-\gamma \theta }\right\}.& \hfill \text{(9)}\\ \hfill \text{(10) }\end{array}$$

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

$$e_n,x_n,S{x}_n,K_n{x}_n,G_i{x}_n\in D,\forall n\in \mathbb{N}andi=1,2,\dots ,N.$$

(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 x_n+1= β _n x _n + (1 - β _n )z _n .

Since {x _n } is bounded, we have

$$\begin{array}{lll}\hfill \left|\right|z_{n+1}-z_n\left|\right|& =\left|\right|\frac{x_{n+2}-{\beta }_{n+1}{x}_{n+1}}{1-{\beta }_{n+1}}-\left(\frac{x_{n+1}-{\beta }_n{x}_n}{1-{\beta }_n}\right)\left|\right|& \hfill \text{(1)}\\ =\left|\right|\frac{{\alpha }_{n+1}\gamma f\left(x_{n+1}\right)+\left(\left(1-{\beta }_{n+1}\right)I-{\alpha }_{n+1}A\right)e_{n+1}}{1-{\beta }_{n+1}}& \hfill \text{(2)}\\ -\left(\frac{{\alpha }_n\gamma f\left(x_n\right)+\left(\left(1-{\beta }_n\right)I-{\alpha }_{n}A\right)e_n}{1-{\beta }_n}\right)\left|\right|& \hfill \text{(3)}\\ \le {\alpha }_{n+1}\left|\right|\frac{\gamma f\left(x_{n+1}\right)-A{e}_{n+1}}{1-{\beta }_{n+1}}\left|\right|+\left|\right|e_{n+1}-e_n\left|\right|& \hfill \text{(4)}\\ +{\alpha }_n\left|\right|\frac{\gamma f\left(x_n\right)-A{e}_n}{1-{\beta }_n}\left|\right|.& \hfill \text{(5)}\\ \hfill \text{(6)}\end{array}$$

(3.10)

By definition of e _n and nonexpansiveness of S, we have

$$\begin{array}{lll}\hfill ∥e_{n+1}-e_n∥& =∥{\gamma }_{n+1}{K}_{n+1}{x}_{n+1}+\left(1-{\gamma }_{n+1}\right)S{x}_{n+1}-{\gamma }_n{K}_n{x}_n-\left(1-{\gamma }_n\right)S{x}_n∥& \hfill \text{(1) }\\ =∥{\gamma }_{n+1}{K}_{n+1}{x}_{n+1}+\left(1-{\gamma }_{n+1}\right)S{x}_{n+1}-{\gamma }_{n+1}{K}_n{x}_n+{\gamma }_{n+1}{K}_n{x}_n& \hfill \text{(2) }\\ -\left(1-{\gamma }_{n+1}\right)S{x}_n+\left(1-{\gamma }_{n+1}\right)S{x}_n-{\gamma }_n{K}_n{x}_n-\left(1-{\gamma }_n\right)S{x}_n∥& \hfill \text{(3) }\\ =∥{\gamma }_{n+1}\left(K_{n+1}{x}_{n+1}-K_n{x}_n\right)+\left(1-{\gamma }_{n+1}\right)\left(S{x}_{n+1}-S{x}_n\right)& \hfill \text{(4) }\\ +\left({\gamma }_{n+1}-{\gamma }_n\right)K_n{x}_n+\left({\gamma }_n-{\gamma }_{n+1}\right)S{x}_n∥& \hfill \text{(5) }\\ \le {\gamma }_{n+1}∥K_{n+1}{x}_{n+1}-K_n{x}_n∥+\left(1-{\gamma }_{n+1}\right)∥x_{n+1}-x_n∥& \hfill \text{(6) }\\ +\left|{\gamma }_{n+1}-{\gamma }_n\right|∥K_n{x}_n∥+\left|{\gamma }_n-{\gamma }_{n+1}\right|∥S{x}_n∥& \hfill \text{(7) }\\ \le {\gamma }_{n+1}∥K_{n+1}{x}_{n+1}-K_n{x}_n∥+\left(1-{\gamma }_{n+1}\right)∥x_{n+1}-x_n∥& \hfill \text{(8) }\\ +2|{\gamma }_{n+1}-{\gamma }_n|M,& \hfill \text{(9)}\\ \hfill \text{(10) }\end{array}$$

where M = max_n∈ℕ{||K _n x _n ||, ||Sx _n ||}. Substituting (3.11) into (3.10), we have

$$\begin{array}{lll}\hfill ∥z_{n+1}-z_n∥& \le {\alpha }_{n+1}\left|\right|\frac{\gamma f\left(x_{n+1}\right)-A{e}_{n+1}}{1-{\beta }_{n+1}}\left|\right|+{\alpha }_n\left|\right|\frac{\gamma f\left(x_n\right)-A{e}_n}{1-{\beta }_n}\left|\right|+∥e_{n+1}-e_n∥& \hfill \text{(1)}\\ \le {\alpha }_{n+1}\left|\right|\frac{\gamma f\left(x_{n+1}\right)-A{e}_{n+1}}{1-{\beta }_{n+1}}\left|\right|+{\alpha }_n\left|\right|\frac{\gamma f\left(x_n\right)-A{e}_n}{1-{\beta }_n}\left|\right|& \hfill \text{(2)}\\ +{\gamma }_{n+1}∥K_{n+1}{x}_{n+1}-K_n{x}_n∥+\left(1-{\gamma }_{n+1}\right)∥x_{n+1}-x_n∥& \hfill \text{(3)}\\ +2|{\gamma }_{n+1}-{\gamma }_n|M& \hfill \text{(4)}\\ \le {\alpha }_{n+1}\left|\right|\frac{\gamma f\left(x_{n+1}\right)-A{e}_{n+1}}{1-{\beta }_{n+1}}\left|\right|+{\alpha }_n\left|\right|\frac{\gamma f\left(x_n\right)-A{e}_n}{1-{\beta }_n}\left|\right|& \hfill \text{(5)}\\ +{\gamma }_{n+1}\left(∥K_{n+1}{x}_{n+1}-K_{n+1}{x}_n∥+∥K_{n+1}{x}_n-K_n{x}_n∥\right)& \hfill \text{(6)}\\ +\left(1-{\gamma }_{n+1}\right)∥x_{n+1}-x_n∥+2|{\gamma }_{n+1}-{\gamma }_n|M& \hfill \text{(7)}\\ \le {\alpha }_{n+1}\left|\right|\frac{\gamma f\left(x_{n+1}\right)-A{e}_{n+1}}{1-{\beta }_{n+1}}\left|\right|+{\alpha }_n\left|\right|\frac{\gamma f\left(x_n\right)-A{e}_n}{1-{\beta }_n}\left|\right|+∥x_{n+1}-x_n∥& \hfill \text{(8)}\\ +∥K_{n+1}{x}_n-K_n{x}_n∥+2|{\gamma }_{n+1}-{\gamma }_n|M.& \hfill \text{(9)}\\ \hfill \text{(10) }\end{array}$$

(3.12)

It implies that

$$\begin{array}{lll}\hfill ∥z_{n+1}-z_n∥-∥x_{n+1}-x_n∥& \le {\alpha }_{n+1}\left|\right|\frac{\gamma f\left(x_{n+1}\right)-A{e}_{n+1}}{1-{\beta }_{n+1}}\left|\right|+{\alpha }_n\left|\right|\frac{\gamma f\left(x_n\right)-A{e}_n}{1-{\beta }_n}\left|\right|& \hfill \text{(1)}\\ +∥K_{n+1}{x}_n-K_n{x}_n∥+2|{\gamma }_{n+1}-{\gamma }_n|M.& \hfill \text{(2)}\\ \hfill \text{(3) }\end{array}$$

(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

$$\underset{n\to \infty }{\lim }∥K_{n+1}{x}_n-K_n{x}_n∥=0.$$

(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

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

(3.16)

By condition (ii) and (3.16)

$$\underset{n\to \infty }{\lim }∥x_{n+1}-x_n∥=\underset{n\to \infty }{\lim }\left(1-{\beta }_n\right)∥z_n-x_n∥=0.$$

(3.17)

Step 3. We will show that

$$\underset{n\to \infty }{\lim }∥e_n-x_n∥=0.$$

(3.18)

Since x_n+1= α _n γf + β _n x _n + ((1 - β _n )I - α _n A) e _n , we have

$$\begin{array}{lll}\hfill ∥x_{n+1}-e_n∥& =∥{\alpha }_n\left(\gamma f\left(x_n\right)-A{e}_n\right)+{\beta }_n\left(x_n-e_n\right)∥& \hfill \text{(1)}\\ \le {\alpha }_n∥\gamma f\left(x_n\right)-A{e}_n∥+{\beta }_n\left(∥x_n-x_{n+1}∥+∥x_{n+1}-e_n∥\right),& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$$

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

$$\underset{n\to \infty }{\lim }∥x_{n+1}-e_n∥=0.$$

(3.19)

Since ||e _n - x _n || ≤ ||e _n - x_n+1|| + ||x_n+1- x _n ||, by (3.17) and (3.19), we have

$$\underset{n\to \infty }{\lim }∥e_n-x_n∥=0.$$

Step 4. Define a mapping Q : H → H by

$$Qx=cKx+\left(1-c\right)Sx,\forall x\in H.$$

(3.20)

We will show that

$$\underset{n\to \infty }{\lim }\left|\right|Q{x}_n-x_n\left|\right|=0.$$

(3.21)

Since

$$\begin{array}{lll}\hfill ∥Q{x}_n-e_n∥& =∥cK{x}_n+\left(1-c\right)S{x}_n-{\gamma }_n{K}_n{x}_n-\left(1-{\gamma }_n\right)S{x}_n∥& \hfill \text{(1)}\\ \le ∥cK{x}_n-{\gamma }_n{K}_n{x}_n∥+|{\gamma }_n-c|∥S{x}_n∥& \hfill \text{(2)}\\ =∥cK{x}_n-{\gamma }_{n}K{x}_n+{\gamma }_{n}K{x}_n-{\gamma }_n{K}_n{x}_n∥+|{\gamma }_n-c|∥S{x}_n∥& \hfill \text{(3)}\\ \le |c-{\gamma }_n|∥K{x}_n∥+{\gamma }_n∥K{x}_n-K_n{x}_n∥+|{\gamma }_n-c|∥S{x}_n∥& \hfill \text{(4)}\\ \le |c-{\gamma }_n|∥K{x}_n∥+\underset{x\in D}{\mathrm{sup>}}\left\{∥Kx-K_{n}x∥\right\}+|{\gamma }_n-c|∥S{x}_n∥.& \hfill \text{(5)}\\ \hfill \text{(6)}\end{array}$$

By remark 2.12 and condition (iii), we have

$$\underset{n\to \infty }{\lim }∥Q{x}_n-e_n∥=0.$$

(3.22)

Since ||Qx _n - x _n || ≤ ||Qx _n - e _n || + ||e _n - x _n ||, from (3.22) and (3.18), we have

$$\underset{n\to \infty }{\lim }\left|\right|Q{x}_n-x_n\left|\right|=0.$$

Step 5. We will show that

$$\underset{n\to \infty }{\text{lim}\text{sup}}〈(\gamma f-A)z,x_n-z〉\le 0,$$

(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

$$\underset{n\to \infty }{\mathrm{limsup}}〈\left(\gamma f-A\right)z,x_n-z〉=\underset{j\to \infty }{\lim }〈\left(\gamma f-A\right)z,x_{n_j}-z〉.$$

(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

$$F\left(Q\right)=F\left(K\right)\bigcap F\left(S\right).$$

(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

$$VI\left(H,B_i,M_i\right)=F\left(J_{M_{i,\eta }}\left(I-\eta B_i\right)\right)=F\left(G_i\right),\forall i=1,2,...N.$$

(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 $x_{n_j}\rightharpoonup q$ 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

$$\underset{n\to \infty }{\text{lim}\text{sup}}〈(\gamma f-A)z,x_n-z〉=\underset{n\to \infty }{\text{lim}}〈(\gamma f-A)z,x_{n_j}-z〉=〈(\gamma f-A)z,q-z〉\le 0.$$

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

$$\begin{array}{lll}\hfill {∥x_{n+1}-z∥}^2& ={∥{\alpha }_n\gamma f\left(x_n\right)+{\beta }_n{x}_n+\left(\left(1-{\beta }_n\right)I-{\alpha }_{n}A\right)e_n-z∥}^2& \hfill \text{(1)}\\ ={∥{\alpha }_n\left(\gamma f\left(x_n\right)-Az\right)+{\beta }_n\left(x_n-z\right)+\left(\left(1-{\beta }_n\right)I-{\alpha }_{n}A\right)\left(e_n-z\right)∥}^2& \hfill \text{(2)}\\ \le {∥{\beta }_n\left(x_n-z\right)+\left(\left(1-{\beta }_n\right)I-{\alpha }_{n}A\right)\left(e_n-z\right)∥}^2+2{\alpha }_n〈\gamma f\left(x_n\right)-Az,x_{n+1}-z〉& \hfill \text{(3)}\\ ={\left(∥{\beta }_n\left(x_n-z\right)+\left(\left(1-{\beta }_n\right)I-{\alpha }_{n}A\right)\left(e_n-z\right)∥\right)}^2+2{\alpha }_n〈\gamma f\left(x_n\right)-Az,x_{n+1}-z〉& \hfill \text{(4)}\\ \le {\left(∥{\beta }_n\left(x_n-z\right)∥+∥\left(1-{\beta }_n\right)I-{\alpha }_{n}A∥∥e_n-z∥\right)}^2& \hfill \text{(5)}\\ +2{\alpha }_n〈\gamma f\left(x_n\right)-\gamma f\left(z\right),x_{n+1}-z〉+2{\alpha }_n〈\gamma f\left(z\right)-Az,x_{n+1}-z〉& \hfill \text{(6)}\\ \le {\left(∥{\beta }_n\left(x_n-z\right)∥+\left(1-{\beta }_n-{\alpha }_n\bar{\gamma }\right)∥e_n-z∥\right)}^2& \hfill \text{(7)}\\ +2{\alpha }_n\gamma \theta ∥x_n-z∥∥x_{n+1}-z∥+2{\alpha }_n〈\gamma f\left(z\right)-Az,x_{n+1}-z〉& \hfill \text{(8)}\\ \le {\left(∥{\beta }_n\left(x_n-z\right)∥+\left(1-{\beta }_n-{\alpha }_n\bar{\gamma }\right)∥x_n-z∥\right)}^2& \hfill \text{(9)}\\ +2{\alpha }_n\gamma \theta ∥x_n-z∥∥x_{n+1}-z∥+2{\alpha }_n〈\gamma f\left(z\right)-Az,x_{n+1}-z〉& \hfill \text{(10)}\\ \le {\left(\left(1-{\alpha }_n\bar{\gamma }\right)∥x_n-z∥\right)}^2+{\alpha }_n\gamma \theta \left({∥x_n-z∥}^2+{∥x_{n+1}-z∥}^2\right)& \hfill \text{(11)}\\ +2{\alpha }_n〈\gamma f\left(z\right)-Az,x_{n+1}-z〉& \hfill \text{(12)}\\ \le \left(1-2{\alpha }_n\bar{\gamma }+{\alpha }_n\gamma \theta \right){∥x_n-z∥}^2+{\alpha }_n^2{\bar{\gamma }}^2{∥x_n-z∥}^2+{\alpha }_n\gamma \theta {∥x_{n+1}-z∥}^2& \hfill \text{(13)}\\ +2{\alpha }_n〈\gamma f\left(z\right)-Az,x_{n+1}-z〉,& \hfill \text{(14)}\\ \hfill \text{(15) }\end{array}$$

it implies that

$$\begin{array}{c}{\Vert x_{n+1}-z‱}^2\le \frac{(1-2{\alpha }_n\overline{\gamma }+{\alpha }_n\gamma \theta )}{1-{\alpha }_n\gamma \theta }{\Vert x_n-z‱}^2+\frac{{\alpha }_n^2{\overline{\gamma }}^2}{1-{\alpha }_n\gamma \theta }{\Vert x_n-z‱}^2\\ +\frac{2{\alpha }_n}{1-{\alpha }_n\gamma \theta }〈\gamma f(z)-Az,x_{n+1}-z〉\\ =\frac{(1-{\alpha }_n\gamma \theta +{\alpha }_n\gamma \theta -2{\alpha }_n\overline{\gamma }+{\alpha }_n\gamma \theta )}{1-{\alpha }_n\gamma \theta }{\Vert x_n-z‱}^2+\frac{{\alpha }_n^2{\overline{\gamma }}^2}{1-{\alpha }_n\gamma \theta }{\Vert x_n-z‱}^2\\ +\frac{2{\alpha }_n}{1-{\alpha }_n\gamma \theta }〈\gamma f(z)-Az,x_{n+1}-z〉\\ =\frac{\left(1-{\alpha }_n\gamma \theta -2{\alpha }_n(\overline{\gamma }-\gamma \theta )\right)}{1-{\alpha }_n\gamma \theta }{\Vert x_n-z‱}^2+\frac{{\alpha }_n^2{\overline{\gamma }}^2}{1-{\alpha }_n\gamma \theta }{\Vert x_n-z‱}^2\\ +\frac{2{\alpha }_n}{1-{\alpha }_n\gamma \theta }〈\gamma f(z)-Az,x_{n+1}-z〉\\ =(1-\frac{2{\alpha }_n(\overline{\gamma }-\gamma \theta ))}{1-{\alpha }_n\gamma \theta }{\Vert x_n-z‱}^2+\frac{{\alpha }_n^2{\overline{\gamma }}^2}{1-{\alpha }_n\gamma \theta }{\Vert x_n-z‱}^2\\ +\frac{2{\alpha }_n}{1-{\alpha }_n\gamma \theta }〈\gamma f(z)-Az,x_{n+1}-z〉\\ =(1-\frac{2{\alpha }_n(\overline{\gamma }-\gamma \theta ))}{1-{\alpha }_n\gamma \theta }{\Vert x_n-z‱}^2+\frac{{\alpha }_n}{1-{\alpha }_n\gamma \theta }({\alpha }_n{\overline{\gamma }}^2{\Vert x_n-z‱}^2\\ +2〈\gamma f(z)-Az,x_{n+1}-z〉)\\ =(1-\frac{2{\alpha }_n(\overline{\gamma }-\gamma \theta ))}{1-{\alpha }_n\gamma \theta }{\Vert x_n-z‱}^2+\frac{2(\overline{\gamma }-\gamma \theta )}{2(\overline{\gamma }-\gamma \theta )}\frac{{\alpha }_n}{1-{\alpha }_n\gamma \theta }({\alpha }_n{\overline{\gamma }}^2{\Vert x_n-z‱}^2\\ +2〈\gamma f(z)-Az,x_{n+1}-z〉)\\ =(1-\frac{2{\alpha }_n(\overline{\gamma }-\gamma \theta ))}{1-{\alpha }_n\gamma \theta }{\Vert x_n-z‱}^2+\frac{2{\alpha }_n(\overline{\gamma }-\gamma \theta )}{1-{\alpha }_n\gamma \theta }(\frac{{\alpha }_n{\overline{\gamma }}^2}{2(\overline{\gamma }-\gamma \theta )}{\Vert x_n-z‱}^2\\ +\frac{2}{2(\overline{\gamma }-\gamma \theta )}〈\gamma f(z)-Az,x_{n+1}-z〉),\end{array}$$

from condition i, step 5 and Lemma 2.2, we can conclude that x _n → z as n → ∞, where z = P_F (I -(A - γf))z. This completes the proof.

By means of our main result, we have the following results in the framework of Hilbert space. To prove these results, we need definition and lemma as follows:

Definition 3.1. A mapping T : C → C is said to be a κ-strict pseudo-contraction mapping, if there exists κ ∈ [0, 1) such that

$${∥Tx-Ty∥}^2\le {∥x-y∥}^2+\kappa {∥\left(I-T\right)x-\left(I-T\right)y∥}^2,\forall x,y\in C.$$

Lemma 3.2. (see [23]) Let C be a nonempty closed convex subset of a real Hilbert space H and T : C → C a κ-strict pseudo-contraction. Define S : C → C by Sx = αx + (1 - α)Tx, for each x ∈ C. Then, as α ∈ [κ, 1), S is nonexpansive such that F(S) = F(T).

Corollary 3.3. Let H be a real Hilbert space and let M _i : H → 2^Hbe maximal monotone mappings for every i = 1, 2,... N. Let B _i : H → H 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 κ _i - strictly pseudo-contractive mappings from H into itself. Define a mapping $T_{{\kappa }_i}$ by $T_{{\kappa }_i}={\kappa }_{i}x+\left(1-{\kappa }_i\right)T_{i}x$, ∀x ∈ H, i ∈ ℕ. Let A be a strongly positive linear-bounded self-adjoint operator with the coefficient $0<\stackrel{}{\gamma }<1$. Let G _i : H → H 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 ${\nu }_j=\left({\alpha }_1^j,{\alpha }_2^j,{\alpha }_3^j\right)\in I\times I\times I$, 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 : C → C be the S - mapping generated by G₁, G₂,..., G _N and ν₁, ν₂,..., ν _N . Let λ₁, λ ₂,..., 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 $T_{{\kappa }_1},T_{{\kappa }_2},\dots ,T_{{\kappa }_n}$, and λ₁, λ₂,..., λ _n , and let K be the K-mapping generated by $T_{{\kappa }_1},T_{{\kappa }_2},\dots ,$ and λ₁, λ₂,..., i.e.,

$$Kx=\underset{n\to \infty }{\lim }{K}_{n}x$$

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 ∈ ℕ, i = 1, 2,..., N, let x₁ ∈ H and {x _n } be the sequence generated by

$$x_{n+1}={\alpha }_n\gamma f\left(x_n\right)+{\beta }_n{x}_n+\left(\left(1-{\beta }_n\right)I-{\alpha }_{n}A\right)\left({\gamma }_n{K}_n{x}_n+\left(1-{\gamma }_n\right)S{x}_n\right),$$

(3.31)

where f : H → H 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) $\underset{n\to \infty }{\lim }{\alpha }_n=0$ 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) $\underset{n\to \infty }{\lim }{\gamma }_n=c\in \left(0,1\right)$.

Then {x _n } converges strongly to $z\in \mathfrak{F}$, which solves uniquely the following variational inequality:

$$〈\left(A-\gamma f\right)z,z-x^{*}〉\le 0,\forall x^{*}\in \mathfrak{F}.$$

(3.32)

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

Proof. For every i ∈ ℕ, by Lemma 3.2, we have that $T_{{\kappa }_i}$ is a nonexpansive mapping and ${\bigcap }_{i=1}^{\infty }F\left(T_{{\kappa }_i}\right)={\bigcap }_{i=1}^{\infty }F\left(T_i\right)$. From Theorem 3.1 and Lemma 2.13, we can reach the desired conclusion.

Corollary 3.4. Let H be a real Hilbert space and let M : H → 2^H be maximal monotone mappings. Let B : H → H be a δ - inverse strongly monotone mapping and ${\left\{T_i\right\}}_{i=1}^{\infty }$ an infinite family of κ _i - strictly pseudo-contractive map-pings from H into itself. Define a mapping $T_{{\kappa }_i}$ by $T_{{\kappa }_i}={\kappa }_{i}x+\left(1-{\kappa }_i\right)T_{i}x$, ∀x ∈ H, i ∈ ℕ. Let A be a strongly positive linear-bounded self-adjoint operator with the coefficient $0<\stackrel{}{\gamma }<1$. 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 $T_{{\kappa }_1},T_{{\kappa }_2},\dots ,T_{{\kappa }_n}$ and λ₁, λ₂,..., λ _n , and let K be the K-mapping generated by $T_{{\kappa }_1},T_{{\kappa }_2},\dots ,$ and λ₁, λ₂,..., i.e.,

$$Kx=\underset{n\to \infty }{\lim }{K}_{n}x$$

for every x ∈ C. Assume that $\mathfrak{F}={\bigcap }_{i=1}^{\infty }F\left(T_i\right)\bigcap V\left(H,B,M\right)\ne \varnothing$. For every n ∈ ℕ, let x₁ ∈ H and {x _n } be the sequence generated by

$$x_{n+1}={\alpha }_n\gamma f\left(x_n\right)+{\beta }_n{x}_n+\left(\left(1-{\beta }_n\right)I-{\alpha }_{n}A\right)\left({\gamma }_n{K}_n{x}_n+\left(1-{\gamma }_n\right)J_{M,\eta }\left(I-\eta B\right)x_n\right),$$

(3.33)

where f : H → H is contractive mapping with coefficient θ ∈ (0, 1) and $0<\gamma <\frac{\stackrel{}{\gamma }}{\theta }$, η ∈ (0, 2δ), {α _n }, {β _n }, {γ _n } are sequences in [0, 1], satisfying the following conditions:

(i) $\underset{n\to \infty }{\lim }{\alpha }_n=0$ 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) $\underset{n\to \infty }{\lim }{\gamma }_n=c\in \left(0,1\right)$.

Then {x _n } converges strongly to $z\in \mathfrak{F}$, which solves uniquely the following variational inequality

$$〈\left(A-\gamma f\right)z,z-x^{*}〉\le 0,\forall x^{*}\in \mathfrak{F}.$$

(3.34)

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

Proof. Putting N = 1 in Corollary 3.3, we can reach the desired conclusion.