Iterative Design for the Common Solution of Monotone Inclusions and Variational Inequalities

Abstract

Some new forward–backward multi-choice iterative algorithms with superposition perturbations are presented in a real Hilbert space for approximating common solution of monotone inclusions and variational inequalities. Some new ideas of constructing iterative elements can be found and strong convergence theorems are proved under mild restrictions, which extend and complement some already existing work.

1. Introduction and Preliminaries

Let C be a non-empty closed and convex subset of a real Hilbert space $H.$ Symbols $\parallel ·\parallel$ and $\langle ·,·\rangle$ denote the norm and inner-product of H, respectively. Symbols → and ⇀ denote the strong and weak convergence in $H,$ respectively.

The classical variational inequality [1] is to find $u\in C$ such that for any $v\in C$,

$$\langle v-u,Tu\rangle \ge 0,$$

(1)

where $T:C\to H$ is a nonlinear mapping. We use $VI(C,T)$ to denote the set of solutions of the variational inequality (1).

The theory of variational inequality draws much attention of mathematicians due to its wide application in several branches of pure and applied sciences [1]. Until now, it is still a hot topic (see [2,3,4,5,6] and the references therein).

An operator $A:D(A)\subset H\to 2^H$ is called monotone ([7]) if for each $x,y\in D(A),$ there exist $u\in Ax$ and $v\in Ay$ such that $\langle x-y,u-v\rangle \ge 0.$ The monotone operator A is called maximal monotone if $R(I+kA)=H,$ for any $k>0.$ In a Hilbert space, a maximal monotone operator can also be called an m-accretive mapping.

A mapping $B:D(B)\subset H\to H$ is called a $\theta$-inversely strongly monotone operator ([8]) if for each $x,y\in D(B)$ and $\theta >0,$ $\langle x-y,Bx-By\rangle \ge {\theta \parallel Bx-By\parallel }^2.$

Let $U:D(U)\subset H\to H$ be a mapping. If $x\in D(U)$ and $Ux=0,$ then x is called a zero point of $U.$ The set of zero points of U is denoted by $N(U).$ If $x\in D(U)$ satisfies that $Ux=x,$ then x is called a fixed point of U. The set of fixed points of U is denoted by $F(U).$

The monotone inclusion problem is to find $u\in H$ such that

$$0\in Au+Bu,$$

(2)

where $A:H\to 2^H$ is maximal monotone and $B:H\to H$ is $\theta$-inversely strongly monotone. The study of monotone inclusions is a hot topic since quite a lot problems appear in minimization problem, convex programming, split feasibility problems, variational inequalities, inverse problem, and image processing can be modeled by it. The construction of iterative algorithms for approximating the solution of (2) has been considered (see [8,9,10,11,12,13,14] and the references therein). The forward–backward splitting iterative method is one of them, which means an iteration involves only A as the forward step and B as the backward step, not the sum $A+B.$ The classical forward–backward splitting iterative method is as follows:

$$\left\{\begin{array}{c}{x}_1\in Hchosenarbitrarily,\hfill \\ x_{n+1}={(I+r_{n}A)}^{-1}(x_n-r_{n}B{x}_n),n\in N.\hfill \end{array}\right.$$

Some of the related work can be seen in [9,10] (and the references therein).

Recall that $f:H\to H$ is called a contraction with contractive constant k ([15]) if $k\in (0,1)$ is that $\parallel f(x)-f(y)\parallel \le k\parallel x-y\parallel$ for $x,y\in H.$

A mapping $S:H\to H$ is called non-expansive ([15]) if $\parallel Sx-Sy\parallel \le \parallel x-y\parallel ,$ for $x,y\in H.$

A mapping $F:H\to H$ is called a strongly positive mapping with $\xi$ ([15]) if $\xi >0$ such that $\langle x,Fx\rangle \ge {\xi \parallel x\parallel }^2$ for $x\in H.$ Furthermore,

$$\parallel aI-bF\parallel =su{p}_{\parallel x\parallel \le 1}|\langle (aI-bF)x,x\rangle |,$$

where I is the identity mapping, $a\in [0,1]$, and $b\in [-1,1].$

In [15], the study of monotone inclusion (2) is extended to the system of monotone inclusions:

$$0\in A_{i}u+B_{i}u,$$

(3)

for $i\in N,$ where $A_i:H\to H$ is maximal monotone and $B_i:H\to H$ is ${\theta }_i$-inversely strongly monotone, for $i\in N.$

Moreover, the iterative algorithm presented in [15] is proved to be strongly convergent to not only the solution of monotone inclusions (3) but also the solution of one kind variational inequality. Specially, the authors constructed the following one by combining the ideas of the splitting method and the midpoint method:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{x}_0\in C\subset Hchosenarbitrarily,\hfill \\ y_n=P_C\left[(1-{\alpha }_n)(x_n+e_n^{\prime })\right],\hfill \\ z_n={\delta }_n{y}_n+{\beta }_n\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)(\frac{y_n+z_n}{2})+{\zeta }_n{e}_n^{″},\hfill \\ x_{n+1}={\gamma }_n\eta f(x_n)+(I-{\gamma }_{n}F)z_n+e_n^{‴},n\in N,\hfill \end{array}\right.\end{array}$$

(4)

where f is a contraction, F is a strongly positive linear bounded mapping, and $P_C$ is the metric projection. Under some conditions, $x_n\to p_0\in {\bigcap }_{i=1}^{\infty }N(A_i+B_i)$ and $p_0$ solves the following variational inequality:

$$\langle F{p}_0-\eta f(p_0),p_0-z\rangle ,\forall z\in \bigcap _{i=1}^{\infty }N(A_i+B_i).$$

(5)

Recall that $T:D(T)\subset H\to H$ is called $\vartheta$-strongly monotone ([16]) if for each $x,y\in D(T)$,

$$\langle Tx-Ty,x-y\rangle \ge {\vartheta \parallel x-y\parallel }^2,$$

for some $\vartheta \in (0,+\infty )$. Furthermore, $T:D(T)\subset H\to H$ is called $\mu$-strictly pseudo-contractive ([16]) if for each $x,y\in D(T)$,

$$\langle Tx-Ty,x-y\rangle \le {\parallel x-y\parallel }^2-\mu {\parallel x-y-(Tx-Ty)\parallel }^2,$$

for some $\mu \in (0,1).$

In 2012, Ceng et al. proposed an iterative algorithm with a perturbed operator for approximating a zero point of the maximal monotone operator A in a Hilbert space ([16]).

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{x}_1\in Hchosenarbitrarily,\hfill \\ y_n={\alpha }_n{x}_n+(1-{\alpha }_n){(I+r_{n}A)}^{-1}{x}_n,\hfill \\ x_{n+1}={\beta }_{n}f(x_n)+(1-{\beta }_n)[{(I+r_{n}A)}^{-1}{y}_n-{\lambda }_n{\mu }_{n}T({(I+r_{n}A)}^{-1}{y}_n)],n\in N,\hfill \end{array}\right.\end{array}$$

(6)

where $T:H\to H$ is a $\vartheta$-strongly monotone and $\mu$-strictly pseudo-contractive mapping with $\vartheta +\mu >1,$ $f:H\to H$ is a contraction, and $A:H\to H$ is maximal monotone. Under some assumptions, $\{x_n\}$ is proved to be convergent strongly to the unique element $p_0\in N(A),$ which solves the following variational inequality:

$$\langle p_0-f(p_0),p_0-u\rangle \le 0,\forall u\in N(A).$$

(7)

The mapping T, which is called a perturbed operator, only plays a role in the construction of the iterative algorithm (6) for selecting a particular zero point of A, but it is not involved in the variational inequality (7).

Later, in 2017, the work in (6) is extended to approximate the solutions of the systems of monotone inclusions (3). The following is a special case in Hilbert space presented in [17]:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{x}_1\in C,\hfill \\ u_n=P_C({\alpha }_n{x}_n+{\beta }_n{a}_n),\hfill \\ v_n={\tau }_n{u}_n+{\nu }_n\sum _{i=1}^{\infty }{\omega }_i^{(2)}{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)(\frac{u_n+v_n}{2})+{\xi }_n{b}_n,\hfill \\ x_{n+1}={\delta }_{n}f(x_n)\hfill \\ +(1-{\delta }_n)(I-{\zeta }_n\sum _{i=1}^{\infty }{\omega }_i^{(1)}{T}_i)\sum _{i=1}^{\infty }{\omega }_i^{(2)}{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)(\frac{u_n+v_n}{2}),n\in N.\hfill \end{array}\right.\end{array}$$

(8)

In (8), ${\sum }_{i=1}^{\infty }{\omega }_i^{(1)}{T}_i$ is called a superposition perturbation, where $T_i:H\to H$ is perturbed operator in the sense of (6); that is, $T_i:H\to H$ is a ${\vartheta }_i$-strongly monotone and ${\mu }_i$-strictly pseudo-contractive mapping, for each $i\in N.$

The iterative sequence $\{x_n\}$ generated by (8) is proved to be strongly convergent to $p_0\in {\bigcap }_{i=1}^{\infty }N(A_i+B_i),$ which solves the variational inequality:

$$\langle p_0-f(p_0),p_0-u\rangle \le 0,\forall u\in \bigcap _{i=1}^{\infty }N(A_i+B_i).$$

(9)

In 2019, Wei et al., proposed some new iterative algorithms. The inertial forward–backward iterative algorithm for approximating the solution of monotone inclusions (3) in [18] is as follows:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{x}_0,x_1\in Hchosenarbitrarily,e_1\in Hischosenarbitrarily,\hfill \\ y_n=x_n+k_n(x_n-x_{n-1}),\hfill \\ w_n={\alpha }_n{x}_n+{\beta }_n\sum _{i=1}^{\infty }{\omega }_{n,i}{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)y_n+{\gamma }_n{e}_n,\hfill \\ C_1=H=Q_1,\hfill \\ C_{n+1}=\{p\in C_n:\parallel w_n{-p\parallel }^2\le (1-{\gamma }_n)\parallel x_n{-p\parallel }^2+{\gamma }_n{\parallel e_n-p\parallel }^2\hfill \\ +k_n^2\parallel x_n-x_{n-1}{\parallel }^2-2{\beta }_n{k}_n\langle x_n-p,x_{n-1}-x_n\rangle \},\hfill \\ Q_{n+1}=\{p\in C_{n+1}:\parallel x_1{-p\parallel }^2\le \parallel x_1-P_{C_{n+1}}(x_1){\parallel }^2+{\sigma }_{n+1}\},\hfill \\ x_{n+1}\in Q_{n+1},n\in N.\hfill \end{array}\right.\end{array}$$

(10)

The result that $x_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(x_1)\in {\bigcap }_{i=1}^{\infty }N(A_i+B_i),$ as $n\to \infty ,$ is proved under some conditions.

The mid-point inertial forward–backward iterative algorithm in [18] is presented as follows:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{x}_0,x_1\in Hchosenarbitrarily,e_1\in Hchosenarbitrarily,\hfill \\ z_0=x_0,\hfill \\ z_n={\delta }_n\lambda f(x_n)+(I-{\delta }_{n}F)x_n,\hfill \\ v_n=z_n+k_n(z_n-z_{n-1}),\hfill \\ w_n={\alpha }_n{v}_n+{\beta }_n\sum _{i=1}^{\infty }{\omega }_{n,i}{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)(\frac{v_n+w_n}{2})+{\gamma }_n{e}_n,\hfill \\ C_1=H=Q_1,\hfill \\ C_{n+1}=\{p\in C_n:\parallel w_n{-p\parallel }^2\le \frac{2{\alpha }_n+{\beta }_n}{2-{\beta }_n}\parallel z_n{-p\parallel }^2+\frac{2{\gamma }_n}{2-{\beta }_n}{\parallel e_n-p\parallel }^2,\hfill \\ +\frac{2{\alpha }_n+{\beta }_n}{2-{\beta }_n}{k}_n\parallel z_n-z_{n-1}{\parallel }^2-2\frac{2{\alpha }_n+{\beta }_n}{2-{\beta }_n}{k}_n\langle z_n-p,z_{n-1}-z_n\rangle \},\hfill \\ Q_{n+1}=\{p\in C_{n+1}:\parallel x_1{-p\parallel }^2\le \parallel x_1-P_{C_{n+1}}(x_1){\parallel }^2+{\sigma }_{n+1}\},\hfill \\ x_{n+1}\in Q_{n+1},n\in N,\hfill \end{array}\right.\end{array}$$

(11)

where f is a contraction and F is a strongly positive linear bounded mapping. The result that $x_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(x_1)=P_{{\bigcap }_{i=1}^{\infty }N(A_i+B_i)}(x_1),$ as $n\to \infty ,$ is proved under some conditions. Furthermore, under the additional assumptions that $\tilde{x}=P_{{\bigcap }_{i=1}^{\infty }N(A_i+B_i)}(x_1)$ and $\tilde{x}=P_{{\bigcap }_{i=1}^{\infty }N(A_i+B_i)}(x_1)[\lambda f(\tilde{x})-F(\tilde{x})+\tilde{x}]$, one has that $\tilde{x}$ solves the variational inequality

$$\langle F\tilde{x}-\lambda f(\tilde{x}),\tilde{x}-z\rangle \le 0,\forall z\in \bigcap _{i=1}^{\infty }N(A_i+B_i).$$

The inertial forward–backward iterative algorithm for approximating common solution of monotone inclusions and one kind variational inequalities, where $T:C\to H$ is maximal monotone and $\tau$-Lipschitz continuous, is presented as follows in [18]:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{u}_0,u_1\in Cchosenarbitrarily,e_1\in Hchosenarbitrarily,\hfill \\ y_0=P_C(u_0-{\lambda }_{0}T{u}_0),\hfill \\ y_n=P_C(u_n-{\lambda }_{n}T{u}_n),\hfill \\ v_n=y_n+k_n(y_n-y_{n-1}),\hfill \\ w_n={\alpha }_n{v}_n+{\beta }_n\sum _{i=1}^{\infty }{\omega }_{n,i}{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)P_C(u_n-{\lambda }_{n}T{u}_n)+{\gamma }_n{e}_n,\hfill \\ C_1=C=Q_1,\hfill \\ C_{n+1}=\{p\in C_n:\parallel w_n{-p\parallel }^2\le {\alpha }_n\parallel y_n{-p\parallel }^2+{\beta }_n\parallel u_n{-p\parallel }^2+{\gamma }_n{\parallel e_n-p\parallel }^2\hfill \\ +k_n^2\parallel y_n-y_{n-1}{\parallel }^2-2{\alpha }_n{k}_n\langle y_n-p,y_{n-1}-y_n\rangle \},\hfill \\ Q_{n+1}=\{p\in C_{n+1}:\parallel u_1{-p\parallel }^2\le \parallel u_1-P_{C_{n+1}}(u_1){\parallel }^2+{\sigma }_{n+1}\},\hfill \\ u_{n+1}\in Q_{n+1},n\in N,\hfill \end{array}\right.\end{array}$$

(12)

The result that $u_n\to P_{{\bigcap }_{i=1}^{\infty }N(A_i+B_i)\bigcap VI(C,T)}(u_1)$, as $n\to \infty$, is proved.

Although two sets $C_n$ and $Q_n$ are needed in (10)–(12), infinite choices of iterative sequences can be made from them whose idea is totally different from that in (4) or (8).

Motivated by the above work, in this paper, we construct some new forward–backward multi-choice iterative algorithms with superposition perturbations in a Hilbert space. Furthermore, some strong convergence theorems for approximating common solution of monotone inclusions and variational inequalities are proved under mild conditions.

To begin our study, the following preliminaries are needed.

Definition 1 

([19]). There exists a unique element $x_0\in C$ such that $\parallel x-x_0\parallel =inf\{\parallel x-y\parallel :y\in C\},$ for each $x\in H.$ Define the metric projection mapping $P_C:H\to C$ by $P_{C}x=x_0$, for any $x\in H.$

Lemma 1 

([20]). For a contraction $f:H\to H$, there is a unique element $x\in H$ that satisfies $f(x)=x.$

Lemma 2 

([19]). For a monotone operator $A:H\to H$ and $r>0,$ one has that ${(I+rA)}^{-1}:H\to H$ is non-expansive.

Lemma 3 

([21]). If $T_i:C\to C$ is non-expansive for $i\in N$ and ${\sum }_{i=1}^{\infty }{a}_i=1$ for $\{a_i\}\subset (0,1),$ then ${\sum }_{i=1}^{\infty }{a}_i{T}_i$ is non-expansive with $F({\sum }_{i=1}^{\infty }{a}_i{T}_i)={\bigcap }_{i=1}^{\infty }F(T_i)$ under the assumption that ${\bigcap }_{i=1}^{\infty }F(T_i)\ne \varnothing .$

Lemma 4 

([15]). If $S:C\to H$ is a single-valued mapping and $T:H\to 2^H$ is maximal monotone, then

$$F({(I+rT)}^{-1}(I-rS))=N(T+S),$$

for $\forall r>0.$

Definition 2 

([22]). Suppose $\{K_n\}$ is a sequence of non-empty closed and convex subsets of H. One has:

(1) The strong lower limit of $\{K_n\}$, $s-liminf{K}_n,$ is defined as the set of all $x\in H$ such that there exists $x_n\in K_n$ for almost all n and it tends to x as $n\to \infty$ in the norm.

(2) The weak upper limit of $\{K_n\},$ $w-limsup{K}_n,$ is defined as the set of all $x\in H$ such that there exists a subsequence $\{K_{n_m}\}$ of $\{K_n\}$ and $x_{n_m}\in K_{n_m}$ for every $n_m$ and it tends to x as $n_m\to \infty$ in the weak topology;.

(3) The limit of $\{K_n\},$ $lim{K}_n,$ is the common value when $s-liminf{K}_n=w-limsup{K}_n$.

Lemma 5 

([22]). Let $\{K_n\}$ be a decreasing sequence of closed and convex subsets of H, i.e., $K_n\subset K_m$ if $n\ge m.$ Then, $\{K_n\}$ converges in H and $lim{K}_n={\bigcap }_{n=1}^{\infty }{K}_n.$

Lemma 6 

([23]). If $lim{K}_n$ exists and is not empty, then $P_{K_n}x\to P_{lim{K}_n}x$ for every $x\in H,$ as $n\to \infty .$

Lemma 7 

([24]). Let $r\in (0,+\infty )$. Then, there exists a continuous, strictly increasing and convex function $g:[0,2r]\to [0,+\infty )$ with $g(0)=0$ such that ${\parallel kx+(1-k)y\parallel }^2\le {k\parallel x\parallel }^2+{(1-k)\parallel y\parallel }^2-k(1-k)g(\parallel x-y\parallel ),$ for $k\in [0,1],x,y\in H$ with $\parallel x\parallel \le r$ and $\parallel y\parallel \le r$.

Lemma 8 

([25]). Let $B:H\to H$ be a ϑ-strongly monotone and μ-strictly pseudo-contractive mapping with $\vartheta +\mu >1$. Then, for any fixed number $\delta \in (0,1)$, $I-\delta B$ is a contraction with contractive constant $1-\delta (1-\sqrt{\frac{1-\vartheta }{\mu }})$.

Lemma 9 

([15]). Suppose $F:H\to H$ is strongly positive bounded mapping with coefficient $\xi >0$ and $0<\rho \le {\parallel F\parallel }^{-1},$ then $\parallel I-\rho F\parallel \le 1-\rho \xi .$

Lemma 10 

([15]). Let $f:H\to H$ be a contraction with contractive constant $k\in (0,1)$, $F:H\to H$ be strongly positive bounded mapping with coefficient $\xi >0$ and $U:H\to H$ be a non-expansive mapping. Suppose $0<\eta \le \frac{\xi }{2k}$ and $F(U)\ne \varnothing$. If for each $t\in (0,1),$ define $T_t:H\to H$ by

$$T_{t}x:=t\eta f(x)+(I-tF)Ux,$$

then $T_t$ has a fixed point $x_t$, for each $t\in (0,\parallel F{\parallel }^{-1}].$ Moreover, $x_t\to q_0,$ as $t\to 0,$ where $q_0\in F(U)$ which satisfies the variational inequality:

$$\langle F{q}_0-\eta f(q_0),q_0-z\rangle \le 0,\forall z\in F(U).$$

Lemma 11 

([15]). In a real Hilbert space H, the following inequality holds:

$${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2\langle y,x+y\rangle ,\forall x,y\in H.$$

Lemma 12 

([26]). Let $\{x_n\}$ and $\{b_n\}$ be two sequences of non-negative real number sequences satisfying

$$x_{n+1}\le (1-t_n)x_n+b_n,\forall n\in N,$$

where $\{t_n\}\subset (0,1)$ with ${\sum }_{n=1}^{\infty }{t}_n=+\infty$ and $t_n\to 0$, as $n\to \infty$. If $limsu{p}_{n\to \infty }\frac{b_n}{t_n}\le 0$, then $li{m}_{n\to \infty }{x}_n=0.$

Lemma 13 

([17]). Let H be a real Hilbert space, $A_i:H\to H$ be maximal monotone, $B_i:H\to H$ be ${\theta }_i$-inversely strongly monotone, and $W_i:H\to H$ be ${\vartheta }_i$-strongly monotone and ${\mu }_i$-strictly pseudo-contractive with ${\vartheta }_i+{\mu }_i>1$ for $i\in N.$ Suppose $0<r_{n,i}\le 2{\theta }_i$ for $i\in N$ and $n\in N,$$k_t\in (0,1)$ for $t\in (0,1), {\sum }_{n=1}^{\infty }{c}_n\parallel W_n\parallel <+\infty ,$ ${\sum }_{i=1}^{\infty }{a}_i=1={\sum }_{i=1}^{\infty }{c}_i$ and ${\bigcap }_{i=1}^{\infty }N(A_i+B_i)\ne \varnothing .$ If, for each $t\in (0,1),$$Z_t^n:H\to H$ is defined by

$$Z_t^{n}u=tf(u)+(1-t)(I-k_t\sum _{i=1}^{\infty }{c}_i{W}_i)\sum _{i=1}^{\infty }{a}_i{(I-r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)u,$$

then $Z_t^n$ has a fixed point $u_t^n.$ That is,

$$u_t^n=tf(u_t^n)+(1-t)(I-k_t\sum _{i=1}^{\infty }{c}_i{W}_i)\sum _{i=1}^{\infty }{a}_i{(I-r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)u_t^n.$$

Moreover, if $\frac{k_t}{t}\to 0,$ then $u_t^n\to p_0,$ as $t\to 0,$ where $p_0$ is the solution of variational inequality:

$$\langle p_0-f(p_0),p_0-z\rangle \le 0,\forall z\in \bigcap _{i=1}^{\infty }N(A_i+B_i).$$

2. Strong Convergence Theorems

Our discussion is based on the following assumptions in this section:

(a)

H is a real Hilbert space.

(b)

$A_i:H\to H$ is maximal monotone and $B_i:H\to H$ is ${\theta }_i$-inversely strongly monotone, for each $i\in N$.

(c)

$f:H\to H$ is a contraction with contractive constant $k\in (0,\frac{1}{2}]$. Furthermore, if $\langle f(x)-x,y-x\rangle =0,$ then $x=0$ or $y=x,$ for $x,y\in H$.

(d)

$F:H\to H$ is a strongly positive linear bounded mapping with $\xi >0$ and $\langle F(x)-\eta f(x)+f(y)-y,x-y\rangle \ge 0,$ for $x,y\in H$.

(e)

$W_i:H\to H$ is ${\vartheta }_i$-strongly monotone and ${\mu }_i$-strictly pseudo-contractive, for $i\in N$;.

(f)

$\{e_n\}\subset H$ and $\{{\epsilon }_n\}\subset H$ are the computational errors.

(g)

$\{a_i\}$ and $\{c_i\}$ are two real number sequences in $(0,1)$ with ${\sum }_{i=1}^{\infty }{a}_i={\sum }_{i=1}^{\infty }{c}_i=1,$ for $n\in N.$

(h)

$\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\delta }_n\}$, $\{{\zeta }_n\}$, $\{{\omega }_n\}$ and $\{{\lambda }_n\}$ are real number sequences in $(0,1),$ for $n\in N.$

(i)

$\{{\sigma }_n\}$ and $\{r_{n,i}\}$ are real number sequences in $(0,+\infty )$, for $n,i\in N.$

Theorem 1. 

Let $\{x_n\}$ be generated by the following iterative algorithm:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{x}_1,y_1\in Hchosenarbitrarily,e_1,{\epsilon }_1\in Hchosenarbitrarily,\hfill \\ C_1=H=Q_1,\hfill \\ u_n={\omega }_n{x}_n+{\epsilon }_n,\hfill \\ v_n={\beta }_n{u}_n+(1-{\beta }_n)\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)v_n,\hfill \\ z_n={\delta }_{n}f(x_n)+(1-{\delta }_n)(I-{\zeta }_n\sum _{i=1}^{\infty }{c}_i{W}_i)\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)v_n,\hfill \\ w_n={\alpha }_n{x}_n+(1-{\alpha }_n)\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)z_n,\hfill \\ x_{n+1}={\lambda }_n\eta f(x_n)+(I-{\lambda }_{n}F)w_n+e_n,\hfill \\ C_{n+1}=\{p\in C_n:2\langle {\alpha }_n{x}_n+(1-{\alpha }_n)z_n-w_n,p\rangle \hfill \\ \le {\alpha }_n\parallel x_n{\parallel }^2+(1-{\alpha }_n)\parallel z_n{\parallel }^2-\parallel w_n{\parallel }^2\}\hfill \\ Q_{n+1}=\{p\in C_{n+1}:\parallel x_1{-p\parallel }^2\le \parallel P_{C_{n+1}}(x_1)-x_1{\parallel }^2+{\sigma }_{n+1}\},n\in N,\hfill \\ y_{n+1}\in Q_{n+1},n\in N,\hfill \end{array}\right.\end{array}$$

(13)

Under the assumptions that:

(i)

$0\in {\bigcap }_{i=1}^{\infty }N(A_i+B_i).$

(ii)

${\mu }_i+{\vartheta }_i>1,$${\mu }_i\in (0,1)$ and ${\vartheta }_i\in (0,1),$ for $i\in N.$

(iii)

$0<r_{n,i}\le 2{\theta }_i,$ for $i,n\in N.$

(iv)

${\sigma }_n\to 0,{\alpha }_n\to 0,{\beta }_n\to 0,{\delta }_n\to 0,$ and ${\zeta }_n\to 0$, as $n\to \infty$.

(v)

$0<\eta <\frac{\xi }{2k}$.

(vi)

${\sum }_{i=1}^{\infty }{c}_i\parallel W_i\parallel <+\infty ;$${\sum }_{n=1}^{\infty }\parallel e_n\parallel <+\infty ,$${\sum }_{n=1}^{\infty }\parallel {\epsilon }_n\parallel <+\infty ,$ and ${\sum }_{n=1}^{\infty }(1-{\omega }_n)<+\infty$.

(vii)

$\frac{{\delta }_n}{{\lambda }_n}\to 0,\frac{\parallel e_n\parallel }{{\lambda }_n}\to 0,\frac{\parallel {\epsilon }_n\parallel }{{\lambda }_n}\to 0,\frac{1-{\omega }_n}{{\lambda }_n}\to 0,\frac{{\zeta }_n}{{\lambda }_n}\to 0,$ as $n\to \infty$.

(viii)

${\sum }_{n=1}^{\infty }{\lambda }_n=+\infty$ and ${\lambda }_n\to 0$,

one has $x_n\to q_0\in {\bigcap }_{i=1}^{\infty }N(A_i+B_i),$ as $n\to \infty ,$ where $q_0$ satisfies the following variational inequalities:

$$\begin{array}{c}\hfill \langle F{q}_0-\eta f(q_0),q_0-y\rangle \le 0,\forall y\in \bigcap _{i=1}^{\infty }N(A_i+B_i),\end{array}$$

(14)

and

$$\begin{array}{c}\hfill \langle q_0-f(q_0),q_0-y\rangle \le 0,\forall y\in \bigcap _{i=1}^{\infty }N(A_i+B_i).\end{array}$$

(15)

Moreover, $y_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(x_1)\in {\bigcap }_{i=1}^{\infty }N(A_i+B_i),$ as $n\to \infty$, which means

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\langle F{q}_0-\eta f(q_0), q_0-P_{{\bigcap }_{m=1}^{\infty }{C}_m}(x_1)\rangle \le 0,\hfill \\ \langle q_0-f(q_0), q_0-P_{{\bigcap }_{m=1}^{\infty }{C}_m}(x_1)\rangle \le 0.\hfill \end{array}\right.\end{array}$$

Proof. 

We split the proof into eleven steps.

• Step 1. $\{v_n\}$ is well-defined.

For $s\in (0,1),$ define $U_s:H\to H$ by

$$\begin{array}{c}\hfill U_{s}x:=su+(1-s)Tx,\end{array}$$

(16)

for any $x\in H$ and for fixed element $u\in H$, where $T:H\to H$ is any fixed non-expansive mapping.

It is easy to check that $\parallel U_{s}x-U_{s}y\parallel =(1-s)\parallel Tx-Ty\parallel \le (1-s)\parallel x-y\parallel .$ Thus, $U_s$ is a contraction, which ensures from Lemma 1 that there exists $x_s\in H$ such that $U_s{x}_s=x_s.$ That is, $x_s=su+(1-s)T{x}_s.$

Since $0<r_{n,i}\le 2{\theta }_i,$ for $i,n\in N,$ for any $x,y\in H,$

$$\begin{array}{ccc}& \parallel (I-r_{n,i}{B}_i)x-(I-r_{n,i}{B}_i){y\parallel }^2={\parallel (x-y)-r_{n,i}(B_{i}x-B_{i}y)\parallel }^2& \\ & ={\parallel x-y\parallel }^2-2r_{n,i}\langle x-y,B_{i}x-B_{i}y\rangle +r_{n,i}^2{\parallel B_{i}x-B_{i}y\parallel }^2& \\ & \le {\parallel x-y\parallel }^2+r_{n,i}(r_{n,i}-2{\theta }_i)\parallel B_{i}x-B_i{y\parallel }^2\le {\parallel x-y\parallel }^2.\end{array}$$

This ensures that $(I-r_{n,i}{B}_i):H\to H$ is non-expansive, for $i,n\in N.$ Since ${\sum }_{i=1}^{\infty }{a}_i=1$, from Lemmas 2–4, one has ${\sum }_{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i):H\to H$ is non-expansive, for $n\in N.$ Moreover, $F({\sum }_{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i))={\bigcap }_{i=1}^{\infty }N(A_i+B_i).$

Considering T in (16) as ${\sum }_{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i),$ one can see that $\{v_n\}$ is well-defined.

• Step 2. $C_n$ is non-empty closed and convex subset of H, for any $n\in N.$

We can easily know from the construction of $C_n$ that $C_n$ is closed and convex subset of H, for any $n\in N.$ We are left to show that $C_n\ne \varnothing .$ For this, it suffices to show that ${\bigcap }_{i=1}^{\infty }N(A_i+B_i)\subset C_n,$ for $n\ge 2.$

In fact, for any $p\in {\bigcap }_{i=1}^{\infty }N(A_i+B_i),$ one has

$$\begin{array}{ccc}& \parallel w_n{-p\parallel }^2\le {\alpha }_n\parallel x_n{-p\parallel }^2+(1-{\alpha }_n)\sum _{i=1}^{\infty }{a}_i{\parallel {(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)z_n-p\parallel }^2& \\ & \le {\alpha }_n\parallel x_n{-p\parallel }^2+(1-{\alpha }_n){\parallel z_n-p\parallel }^2.\end{array}$$

Then,

$$\begin{array}{c}\hfill 2\langle {\alpha }_n{x}_n+(1-{\alpha }_n)z_n-w_n,p\rangle \le {\alpha }_n\parallel x_n{\parallel }^2+(1-{\alpha }_n)\parallel z_n{\parallel }^2-{\parallel w_n\parallel }^2,\end{array}$$

which implies that $p\in C_n,$ for $n\ge 2.$ Therefore, ${\bigcap }_{i=1}^{\infty }N(A_i+B_i)\subset C_n,$ for all $n\in N,$ and then $C_n\ne \varnothing ,$ for all $n\in N.$

• Step 3. $Q_n$ is a non-empty subset of H, for each $n\in N,$ which ensures that $\{y_n\}$ is well-defined.

It follows from Step 2 and Definition 1 that, for ${\sigma }_{n+1},$ there exists $b_{n+1}\in C_{n+1}$ such that $\parallel x_1-b_{n+1}{\parallel }^2\le (in{f}_{z\in C_{n+1}}\parallel x_1-{z\parallel )}^2+{\sigma }_{n+1}={\parallel P_{C_{n+1}}(x_1)-x_1\parallel }^2+{\sigma }_{n+1}.$ Thus, $Q_{n+1}\ne \varnothing ,$ for $n\in N.$ Then, $\{y_n\}$ is well-defined.

• Step 4. $P_{C_n}(x_1)\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(x_1),$ as $n\to \infty$.

It follows from Lemma 5 that $lim{C}_n$ exists and $lim{C}_n={\bigcap }_{n=1}^{\infty }{C}_n\ne \varnothing$. Then, Lemma 6 implies that $P_{C_n}(x_1)\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(x_1),$ as $n\to \infty$.

• Step 5. $y_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(x_1),$ as $n\to \infty$.

Since $y_{n+1}\in Q_{n+1}\subset C_{n+1}$ and $C_n$ is a convex subset of H, for $\forall t\in (0,1),$ $t{P}_{C_{n+1}}(x_1)+(1-t)y_{n+1}\in C_{n+1},$ which implies that

$$\begin{array}{c}\hfill \parallel P_{C_{n+1}}(x_1)-x_1\parallel \le \parallel t{P}_{C_{n+1}}(x_1)+(1-t)y_{n+1}-x_1\parallel .\end{array}$$

(17)

Using Lemma 7, one has:

$$\begin{array}{c}\hfill \begin{array}{c}\parallel t{P}_{C_{n+1}}(x_1)+(1-t)y_{n+1}-x_1{\parallel }^2={\parallel t(P_{C_{n+1}}(x_1)-x_1)+(1-t)(y_{n+1}-x_1)\parallel }^2\hfill \\ \le t\parallel P_{C_{n+1}}(x_1)-x_1{\parallel }^2+(1-t)\parallel y_{n+1}-x_1{\parallel }^2-t(1-t)g(\parallel P_{C_{n+1}}(x_1)-y_{n+1}\parallel ).\hfill \end{array}\end{array}$$

(18)

From (17) and (18), we have $tg(\parallel P_{C_{n+1}}(x_1)-y_{n+1}\parallel )\le \parallel y_{n+1}-x_1{\parallel }^2-{\parallel P_{C_{n+1}}(x_1)-x_1\parallel }^2\le {\sigma }_{n+1}.$ Letting $t\to 1$ first and then $n\to \infty$, one has $P_{C_{n+1}}(x_1)-y_{n+1}\to 0,$ as $n\to \infty .$ Combining with Step 4, $y_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(x_1),$ as $n\to \infty$.

• Step 6. $\{u_n\},$$\{v_n\},$$\{z_n\},$$\{w_n\}$ and $\{x_n\}$ are all bounded.

For $p\in {\bigcap }_{i=1}^{\infty }N(A_i+B_i),$ one has for any $n\in N,$

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \parallel u_n-p\parallel \le {\omega }_n\parallel x_n-p\parallel +(1-{\omega }_n)\parallel p\parallel +\parallel {\epsilon }_n\parallel .\end{array}\end{array}$$

(19)

Furthermore, $\parallel v_n-p\parallel \le {\beta }_n\parallel u_n-p\parallel +(1-{\beta }_n)\parallel v_n-p\parallel$ implies that for any $n\in N,$

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \parallel v_n-p\parallel \le \parallel u_n-p\parallel .\end{array}\end{array}$$

(20)

In view of Lemma 8 and (20), one has

$$\begin{array}{c}\hfill \begin{array}{cc}& \parallel z_n-p\parallel \le {\delta }_n\parallel f(x_n)-f(p)\parallel +{\delta }_n\parallel f(p)-p\parallel \hfill \\ & +(1-{\delta }_n)\parallel (I-{\zeta }_n\sum _{i=1}^{\infty }{c}_i{W}_i)\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)v_n-p\parallel \hfill \\ & \le {\delta }_{n}k\parallel x_n-p\parallel +{\delta }_n\parallel f(p)-p\parallel \hfill \\ & +(1-{\delta }_n)\parallel (I-{\zeta }_n\sum _{i=1}^{\infty }{c}_i{W}_i)\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)(v_n-p)\parallel \hfill \\ & +(1-{\delta }_n)\parallel (I-{\zeta }_n\sum _{i=1}^{\infty }{c}_i{W}_i)\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)p-p\parallel \hfill \\ & \le {\delta }_{n}k\parallel x_n-p\parallel +{\delta }_n\parallel f(p)-p\parallel +(1-{\delta }_n)[1-{\zeta }_n(1-\sum _{i=1}^{\infty }{c}_i\sqrt{\frac{1-{\vartheta }_i}{{\mu }_i}})]\parallel u_n-p\parallel \hfill \\ & +(1-{\delta }_n){\zeta }_n\sum _{i=1}^{\infty }{c}_i\parallel W_i\parallel \parallel p\parallel \hfill \end{array}\end{array}$$

(21)

Note that, for any $n\in N,$

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \parallel w_n-p\parallel \le {\alpha }_n\parallel x_n-p\parallel +(1-{\alpha }_n)\parallel z_n-p\parallel .\end{array}\end{array}$$

(22)

Now, in view of Lemma 9, one has

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \parallel x_{n+1}-p\parallel \le {\lambda }_n\parallel \eta f(x_n)-Fp\parallel +\parallel (I-{\lambda }_{n}F)(w_n-p)\parallel +\parallel e_n\parallel \\ \hfill \le {\lambda }_n\eta k\parallel x_n-p\parallel +{\lambda }_n\parallel \eta f(p)-Fp\parallel +(1-{\lambda }_n\xi )\parallel w_n-p\parallel +\parallel e_n\parallel \end{array}\end{array}$$

(23)

Combing with inequalities (19)–(23), by induction, one has

$$\begin{array}{c}\parallel x_{n+1}-p\parallel \hfill \\ \le \{{\lambda }_n\eta k+(1-{\lambda }_n\xi ){\alpha }_n+(1-{\lambda }_n\xi )(1-{\alpha }_n){\delta }_{n}k\hfill \\ +(1-{\lambda }_n\xi )(1-{\alpha }_n)(1-{\delta }_n)[1-{\zeta }_n(1-\sum _{i=1}^{\infty }{c}_i\sqrt{\frac{1-{\vartheta }_i}{{\mu }_i}})]{\omega }_n\}\parallel x_n-p\parallel \hfill \\ +{\lambda }_n\parallel \eta f(p)-F(p)\parallel +\parallel e_n\parallel \hfill \\ +(1-{\lambda }_n\xi )(1-{\alpha }_n){\delta }_n\parallel f(p)-p\parallel +(1-{\lambda }_n\xi )(1-{\alpha }_n)(1-{\delta }_n){\zeta }_n\sum _{i=1}^{\infty }{c}_i\parallel W_i\parallel \parallel p\parallel \hfill \\ +(1-{\lambda }_n\xi )(1-{\alpha }_n)(1-{\omega }_n)(1-{\delta }_n)[1-{\zeta }_n(1-\sum _{i=1}^{\infty }{c}_i\sqrt{\frac{1-{\vartheta }_i}{{\mu }_i}})]\parallel p\parallel \hfill \\ +(1-{\lambda }_n\xi )(1-{\alpha }_n)(1-{\delta }_n)[1-{\zeta }_n(1-\sum _{i=1}^{\infty }{c}_i\sqrt{\frac{1-{\vartheta }_i}{{\mu }_i}})]\parallel {\epsilon }_n\parallel \hfill \\ \le \{{\lambda }_n\eta k+(1-{\lambda }_n\xi ){\alpha }_n+(1-{\lambda }_n\xi )(1-{\alpha }_n){\delta }_{n}k\hfill \\ +(1-{\lambda }_n\xi )(1-{\alpha }_n)(1-{\delta }_n)[1-{\zeta }_n(1-\sum _{i=1}^{\infty }{c}_i\sqrt{\frac{1-{\vartheta }_i}{{\mu }_i}})]\}\parallel x_n-p\parallel \hfill \\ +{\lambda }_n(\xi -\eta k)\frac{\parallel \eta f(p)-F(p)\parallel }{\xi -\eta k}+\parallel e_n\parallel \hfill \\ +(1-{\lambda }_n\xi )(1-{\alpha }_n){\delta }_n(1-k)\frac{\parallel f(p)-p\parallel }{1-k}\hfill \\ +(1-{\lambda }_n\xi )(1-{\alpha }_n)(1-{\delta }_n){\zeta }_n(1-\sum _{i=1}^{\infty }{c}_i\sqrt{\frac{1-{\vartheta }_i}{{\mu }_i}})\frac{{\sum }_{i=1}^{\infty }{c}_i\parallel W_i\parallel \parallel p\parallel }{1-{\sum }_{i=1}^{\infty }{c}_i\sqrt{\frac{1-{\vartheta }_i}{{\mu }_i}}}\hfill \\ +(1-{\omega }_n)\parallel p\parallel +\parallel {\epsilon }_n\parallel \hfill \\ \le max\{\parallel x_n-p\parallel ,\frac{\parallel \eta f(p)-F(p)\parallel }{\xi -\eta k},\frac{\parallel f(p)-p\parallel }{1-k},\frac{{\sum }_{i=1}^{\infty }{c}_i\parallel W_i\parallel \parallel p\parallel }{1-{\sum }_{i=1}^{\infty }{c}_i\sqrt{\frac{1-{\vartheta }_i}{{\mu }_i}}}\}+\parallel e_n\parallel +(1-{\omega }_n)\parallel p\parallel +\parallel {\epsilon }_n\parallel \hfill \\ \cdots \hfill \\ \le max\{\parallel x_1-p\parallel ,\frac{\parallel \eta f(p)-F(p)\parallel }{\xi -\eta k},\frac{\parallel f(p)-p\parallel }{1-k},\frac{{\sum }_{i=1}^{\infty }{c}_i\parallel W_i\parallel \parallel p\parallel }{1-{\sum }_{i=1}^{\infty }{c}_i\sqrt{\frac{1-{\vartheta }_i}{{\mu }_i}}}\}\hfill \\ +\sum _{i=1}^n\parallel e_i\parallel +\sum _{i=1}^n(1-{\omega }_i)\parallel p\parallel +\sum _{i=1}^n\parallel {\epsilon }_i\parallel \hfill \end{array}$$

Based on the assumptions, one has $\{x_n\}$ is bounded. Following (19)–(22), it is easy to see that $\{u_n\},$$\{v_n\},$$\{z_n\}$ and $\{w_n\}$ are all bounded.

Note that, for $p\in {\bigcap }_{i=1}^{\infty }N(A_i+B_i),$ $\parallel {\sum }_{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)v_n-p\parallel \le \parallel v_n-p\parallel .$ Then, $\{{\sum }_{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)v_n\}$ is bounded. Similarly, $\{{\sum }_{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}$$(I-r_{n,i}{B}_i)z_n\}$ is bounded.

Since $\parallel {\sum }_{i=1}^{\infty }{c}_i{W}_i{\sum }_{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)v_n\parallel \le {\sum }_{i=1}^{\infty }{c}_i\parallel W_i\parallel {\sum }_{i=1}^{\infty }\parallel a_i{(I+r_{n,i}{A}_i)}^{-1}$$(I-r_{n,i}{B}_i)v_n\parallel ,$ then $\{{\sum }_{i=1}^{\infty }{c}_i{W}_i{\sum }_{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)v_n\}$ is bounded.

• Step 7. There exists $q_0\in {\bigcap }_{i=1}^{\infty }N(A_i+B_i),$ which is the solution of variational inclusion (14).

It follows from Lemma 10 that there exists $z_t$ such that

$$z_t=t\eta f(z_t)+(I-tF)\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)z_t$$

and $z_t\to q_0,$ as $t\to 0,$ where $q_0$ is the solution of (14).

• Step 8. $limsu{p}_{n\to \infty }\langle \eta f(q_0)-F{q}_0,x_{n+1}-q_0\rangle \le 0,$ where $q_0$ is the same as that in Step 7.

Note that

$$\begin{array}{c}\hfill \begin{array}{ccc}& \parallel w_n-z_n\parallel \hfill & \\ & =\parallel w_n-\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)z_n\hfill & \\ & +\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)(z_n-v_n)\hfill & \\ & +\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)v_n-v_n+v_n-z_n\parallel \hfill & \\ & =\parallel {\alpha }_n[x_n-\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)z_n]\hfill & \\ & +\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)(z_n-v_n)\hfill & \\ & +{\beta }_n[\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)v_n-u_n]+v_n-z_n\parallel \hfill & \\ & \le {\alpha }_n\parallel x_n-\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)z_n\parallel +2\parallel z_n-v_n\parallel \hfill & \\ & +{\beta }_n\parallel \sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)v_n-u_n\parallel \hfill & \end{array}\end{array}$$

(24)

Furthermore,

$$\begin{array}{c}\hfill \begin{array}{ccc}& \parallel z_n-v_n\parallel \hfill & \\ & \le {\delta }_n\parallel f(x_n)\parallel +{\zeta }_n(1-{\delta }_n)\parallel \sum _{i=1}^{\infty }{c}_i{W}_i\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)v_n\parallel \hfill & \\ & +{\beta }_n\parallel u_n-\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)v_n\parallel \hfill & \\ & +{\delta }_n\parallel \sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)v_n\parallel \hfill \end{array}\end{array}$$

(25)

Since $\{x_n\},$$\{u_n\}$,$\{v_n\}$,$\{z_n\}$, $\{{\sum }_{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}$$(I-r_{n,i}{B}_i)v_n\}$, $\{{\sum }_{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}$$(I-r_{n,i}{B}_i)z_n\}$ and $\{{\sum }_{i=1}^{\infty }{c}_i{W}_i{\sum }_{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)v_n\}$ are bounded, then, based on the assumptions and (24) and (25), $w_n-z_n\to 0,$ as $n\to \infty .$

Let $z_t$ be the same as that in Step 7, then $\parallel z_t\parallel \le \parallel z_t-q_0\parallel +\parallel q_0\parallel ,$ which implies that $\{z_t\}$ is bounded.

Note that

$$\begin{array}{c}\hfill \begin{array}{ccc}& \parallel \sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)w_n-w_n\parallel \hfill & \\ & \le \parallel \sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)(w_n-z_n)\parallel \hfill & \\ & +\parallel \sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)z_n-w_n\parallel \hfill & \\ & \le \parallel w_n-z_n\parallel +\parallel \sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)z_n-w_n\parallel \hfill & \\ & =\parallel w_n-z_n\parallel +{\alpha }_n\parallel x_n-\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)z_n\parallel \hfill & \end{array}\end{array}$$

(26)

Since ${\alpha }_n\to 0$ and $w_n-z_n\to 0,$ as $n\to \infty ,$ then ${\sum }_{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)w_n-w_n\to 0,$ as $n\to \infty .$ In view of Lemma 11,

$$\begin{array}{c}\parallel z_t-w_n{\parallel }^2\hfill \\ =\parallel z_t-\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)w_n+\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)w_n-w_n{\parallel }^2\hfill \\ \le \parallel z_t-\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)w_n{\parallel }^2\hfill \\ +2\langle z_t-w_n,\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)w_n-w_n\rangle \hfill \\ \le \parallel z_t-w_n{\parallel }^2\hfill \\ +2\langle z_t-\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)w_n,t\eta f(z_t)-tF(\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)z_t)\rangle \hfill \\ +2\parallel z_t-w_n\parallel \parallel \sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)w_n-w_n\parallel \hfill \end{array}$$

Therefore,

$$\begin{array}{c}t\langle z_t-\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)w_n,F(\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)z_t)-\eta f(z_t)\rangle \hfill \\ \le \parallel z_t-w_n\parallel \parallel \sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)w_n-w_n\parallel ,\hfill \end{array}$$

which implies that

$$li{m}_{t\to 0}limsu{p}_{n\to \infty }\langle z_t-\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)w_n,F(\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)z_t)-\eta f(z_t)\rangle \le 0.$$

Since $z_t\to q_0$ as $t\to 0,$ then

$$\begin{array}{c}limsu{p}_{n\to \infty }\langle q_0-\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)w_n,F{q}_0-\eta f(q_0)\rangle \le 0.\hfill \end{array}$$

Since ${\sum }_{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)w_n-w_n\to 0,$ $w_n-z_n\to 0$ and $x_{n+1}-w_n={\lambda }_n(\eta f(x_n)-F{w}_n)+e_n\to 0,$ then $limsu{p}_{n\to \infty }\langle q_0-x_{n+1},F{q}_0-\eta f(q_0)\rangle \le 0.$

• Step 9. $x_n\to q_0,$ as $n\to \infty$, where $q_0$ is the same as that in Steps 7 and 8.

In fact, using Lemma 11 again, one has

$$\begin{array}{c}\hfill \begin{array}{ccc}& \parallel u_n-q_0{\parallel }^2={\parallel {\omega }_n(x_n-q_0)+({\omega }_n-1)q_0+{\epsilon }_n\parallel }^2\hfill & \\ & \le {\omega }_n{\parallel x_n-q_0\parallel }^2+2\langle {\epsilon }_n,u_n-q_0\rangle +2(1-{\omega }_n)\langle q_0,q_0-u_n\rangle \hfill & \\ & \le {\omega }_n\parallel x_n-q_0{\parallel }^2+2\parallel {\epsilon }_n\parallel \parallel u_n-q_0\parallel +2(1-{\omega }_n)\parallel q_0\parallel \parallel u_n-q_0\parallel .\hfill \end{array}\end{array}$$

(27)

Furthermore, $\parallel v_n-q_0{\parallel }^2\le {\beta }_n\parallel u_n-q_0{\parallel }^2+(1-{\beta }_n){\parallel v_n-q_0\parallel }^2$ ensures that

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \parallel v_n-q_0{\parallel }^2\le {\parallel u_n-q_0\parallel }^2.\end{array}\end{array}$$

(28)

In view of Lemma 11 again, one has

$$\begin{array}{c}\hfill \begin{array}{ccc}& \parallel z_n-q_0{\parallel }^2\hfill & \\ & =\parallel {\delta }_n(f(x_n)-q_0)+(1-{\delta }_n)[(I-{\zeta }_n\sum _{i=1}^{\infty }{c}_i{W}_i)\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)v_n-q_0]{\parallel }^2\hfill & \\ & \le (1-{\delta }_n){\parallel v_n-q_0\parallel }^2+2{\delta }_n\langle z_n-q_0,f(x_n)-q_0\rangle \hfill \\ & -2(1-{\delta }_n){\zeta }_n\langle \sum _{i=1}^{\infty }{c}_i{W}_i\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)v_n,z_n-q_0\rangle \hfill & \\ & \le (1-{\delta }_n){\parallel v_n-q_0\parallel }^2+2{\delta }_n\langle z_n-q_0,f(x_n)-f(q_0)\rangle +2{\delta }_n\langle z_n-q_0,f(q_0)-q_0\rangle \hfill & \\ & +2(1-{\delta }_n){\zeta }_n\parallel \sum _{i=1}^{\infty }{c}_i{W}_i\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)v_n\parallel \parallel z_n-q_0\parallel \hfill & \\ & \le (1-{\delta }_n)\parallel v_n-q_0{\parallel }^2+2{\delta }_{n}k\parallel z_n-x_n\parallel \parallel x_n-q_0\parallel +2{\delta }_{n}k{\parallel x_n-q_0\parallel }^2+2{\delta }_n\langle z_n-q_0,f(q_0)-q_0\rangle \hfill & \\ & +2{\zeta }_n\parallel \sum _{i=1}^{\infty }{c}_i{W}_i\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)v_n\parallel \parallel z_n-q_0\parallel \hfill \end{array}\end{array}$$

(29)

Note that

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \parallel w_n-q_0{\parallel }^2\le {\alpha }_n\parallel x_n-q_0{\parallel }^2+(1-{\alpha }_n){\parallel z_n-q_0\parallel }^2.\end{array}\end{array}$$

(30)

Now, in view of Lemma 9 and using (27)–(30), one has

$$\begin{array}{c}\hfill \begin{array}{ccc}& \parallel x_{n+1}-q_0{\parallel }^2={\parallel {\lambda }_n\eta f(x_n)+(I-{\lambda }_{n}F)w_n+e_n-q_0\parallel }^2\hfill & \\ & =\parallel {\lambda }_n(\eta f(x_n)-F(q_0))+(I-{\lambda }_{n}F)(w_n-q_0)+e_n{\parallel }^2\hfill & \\ & \le (1-{\lambda }_n\xi ){\parallel w_n-q_0\parallel }^2+2\langle e_n,x_{n+1}-q_0\rangle +2{\lambda }_n\langle \eta f(x_n)-F{q}_0,x_{n+1}-q_0\rangle \hfill & \\ & \le (1-{\lambda }_n\xi )\parallel w_n-q_0{\parallel }^2+2\parallel e_n\parallel \parallel x_{n+1}-q_0\parallel +2{\lambda }_n\eta \langle f(x_n)-f(q_0),x_{n+1}-q_0\rangle \hfill & \\ & +2{\lambda }_n\langle \eta f(q_0)-F{q}_0,x_{n+1}-q_0\rangle \hfill & \\ & \le (1-{\lambda }_n\xi )\parallel w_n-q_0{\parallel }^2+2\parallel e_n\parallel \parallel x_{n+1}-q_0\parallel +2{\lambda }_n\eta k\parallel x_n-q_0\parallel \parallel x_{n+1}-q_0\parallel \hfill & \\ & +2{\lambda }_n\langle \eta f(q_0)-F(q_0),x_{n+1}-q_0\rangle \hfill & \\ & \le \{(1-{\lambda }_n\xi ){\alpha }_n+{\lambda }_n\eta k+2(1-{\lambda }_n\xi )(1-{\alpha }_n){\delta }_{n}k\hfill & \\ & +(1-{\lambda }_n\xi )(1-{\alpha }_n)(1-{\delta }_n){\omega }_n\}\parallel x_n-q_0{\parallel }^2\hfill & \\ & +2\parallel e_n\parallel \parallel x_{n+1}-q_0\parallel +2(1-{\lambda }_n\xi )(1-{\alpha }_n)(1-{\delta }_n)\parallel {\epsilon }_n\parallel \parallel u_n-q_0\parallel \hfill & \\ & +2(1-{\lambda }_n\xi )(1-{\alpha }_n)(1-{\delta }_n)(1-{\omega }_n)\parallel q_0\parallel \parallel u_n-q_0\parallel \hfill & \\ & +2{\delta }_{n}k(1-{\lambda }_n\xi )(1-{\alpha }_n)\parallel x_n-q_0\parallel \parallel x_n-z_n\parallel \hfill & \\ & +2(1-{\lambda }_n\xi )(1-{\alpha }_n){\delta }_n\parallel z_n-q_0\parallel \parallel f(q_0)-q_0\parallel \hfill & \\ & +2(1-{\lambda }_n\xi )(1-{\alpha }_n){\zeta }_n\parallel z_n-q_0\parallel \parallel \sum _{i=1}^{\infty }{c}_i{W}_i\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)v_n\parallel \hfill & \\ & +{\lambda }_n\eta k{\parallel x_{n+1}-q_0\parallel }^2+2{\lambda }_n\langle \eta f(q_0)-F(q_0),x_{n+1}-q_0\rangle .\hfill \end{array}\end{array}$$

(31)

Let $M_1=su{p}_n\{2\parallel x_{n+1}-q_0\parallel ,2\parallel u_n-q_0\parallel ,2k\parallel x_n-q_0\parallel \parallel x_n-z_n\parallel ,2\parallel u_n-q_0\parallel \parallel q_0\parallel ,$$2\parallel f(q_0)-q_0\parallel \parallel z_n-q_0\parallel , 2\parallel z_n-q_0\parallel \parallel {\sum }_{i=1}^{\infty }{c}_i{W}_i{\sum }_{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)v_n\parallel :n\in N\}$. Then, from Step 6, one has $M_1<+\infty .$

Therefore, it follows from (31) that

$$\begin{array}{c}\hfill \begin{array}{ccc}& (1-{\lambda }_n\eta k){\parallel x_{n+1}-q_0\parallel }^2\hfill & \\ & \le \{(1-{\lambda }_n\xi ){\alpha }_n+{\lambda }_n\eta k+2(1-{\lambda }_n\xi )(1-{\alpha }_n){\delta }_{n}k+(1-{\lambda }_n\xi )(1-{\alpha }_n)(1-{\delta }_n)\}{\parallel x_n-q_0\parallel }^2\hfill & \\ & +[\parallel e_n\parallel +\parallel {\epsilon }_n\parallel +(1-{\omega }_n)+2{\delta }_n+{\zeta }_n]M_1\hfill & \\ & +2{\lambda }_n\langle \eta f(q_0)-F(q_0),x_{n+1}-q_0\rangle \hfill & \end{array}\end{array}$$

(32)

If we set $b_n^{(1)}=\frac{{\lambda }_n(\xi -2\eta k)}{1-{\lambda }_n\eta k},b_n^{(2)}=\frac{M_1}{1-{\lambda }_n\eta k}[\parallel e_n\parallel +\parallel {\epsilon }_n\parallel +(1-{\omega }_n)+2{\delta }_n+{\zeta }_n]+$$\frac{2{\lambda }_n}{1-{\lambda }_n\eta k}\langle \eta f(q_0)-F{q}_0,x_{n+1}-q_0\rangle$, then (32) can be reduced as follows:

$$\parallel x_{n+1}-q_0{\parallel }^2\le (1-b_n^{(1)}){\parallel x_n-q_0\parallel }^2+b_n^{(2)}.$$

Based on the assumptions and Step 8, we know that $b_n^{(1)}\to 0,$ ${\sum }_{n=1}^{\infty }{b}_n^{(1)}=+\infty$ and $limsu{p}_{n\to \infty }\frac{b_n^{(2)}}{b_n^{(1)}}\le 0.$ Then, from Lemma 12, $x_n\to q_0,$ as $n\to \infty$.

• Step 10. There exists $p_0\in {\bigcap }_{i=1}^{\infty }N(A_i+B_i),$ which is the solution of the variational inclusion

  $$\begin{array}{c}\hfill \begin{array}{c}\hfill \langle p_0-f(p_0),p_0-y\rangle \le 0,\forall y\in \bigcap _{i=1}^{\infty }N(A_i+B_i).\end{array}\end{array}$$

  (33)

In fact, it follows from Lemma 13 that there exists $u_t^n$ such that

$$u_t^n=tf(u_t^n)+(1-t)(I-k_t\sum _{i=1}^{\infty }{c}_i{W}_i)\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)u_t^n$$

and $u_t^n\to p_0,$ as $t\to 0,$ where $p_0$ is the solution of (33).

• Step 11. $x_n\to p_0,$ as $n\to \infty$, where $p_0$ is the same as that in Step 10.

It suffices to show that $p_0=q_0$.

Since $p_0\in {\bigcap }_{i=1}^{\infty }N(A_i+B_i)$,

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \langle F{q}_0-\eta f(q_0),q_0-p_0\rangle \le 0.\end{array}\end{array}$$

(34)

Since F is strongly positive linear bounded, f is a contraction, and $0<\eta <\frac{\xi }{2k}$,

$$\begin{array}{c}\hfill \begin{array}{ccc}& \langle (F{q}_0-\eta f(q_0))-(F{p}_0-\eta f(p_0)),q_0-p_0\rangle \hfill & \\ & =\langle F(q_0-p_0),q_0-p_0\rangle +\eta \langle f(p_0)-f(q_0),q_0-p_0\rangle \hfill & \\ & \ge \xi \parallel q_0-p_0{\parallel }^2-\eta k{\parallel q_0-p_0\parallel }^2\ge 0.\hfill \end{array}\end{array}$$

(35)

Therefore, (34) ensures that

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \langle F{p}_0-\eta f(p_0),q_0-p_0\rangle \le \langle F{q}_0-\eta f(q_0),q_0-p_0\rangle \le 0.\end{array}\end{array}$$

(36)

On the other hand, it follows from (33) that

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \langle f(p_0)-p_0,q_0-p_0\rangle \le 0.\end{array}\end{array}$$

(37)

Combining with (34), one has $\langle F{q}_0-\eta f(q_0)+f(p_0)-p_0,q_0-p_0\rangle \le 0.$ Following Condition (d), we know that $\langle F{q}_0-\eta f(q_0)+f(p_0)-p_0,q_0-p_0\rangle =0.$ Then, (34) and (37) ensure that $\langle F{q}_0-\eta f(q_0),q_0-p_0\rangle =\langle f(p_0)-p_0,q_0-p_0\rangle =0.$

Since $0\in {\bigcap }_{i=1}^{\infty }N(A_i+B_i),$ from Condition (c), we know that $p_0=0$ or $p_0=q_0.$ If $p_0=q_0,$ then the result follows. If $p_0=0,$ then $\langle F{q}_0-\eta f(q_0),q_0-p_0\rangle =0$ implies that $\langle F{q}_0-\eta f(q_0),q_0\rangle =0.$ Therefore, $\xi \parallel q_0{\parallel }^2\le \langle F{q}_0,q_0\rangle =\eta \langle f(q_0),q_0\rangle \le \eta k{\parallel q_0\parallel }^2.$ Since $\xi >2\eta k,$ then $q_0=0$, which means that $p_0=q_0=0.$ Therefore, $x_n\to p_0=q_0,$ as $n\to \infty .$

This completes the proof. □

Theorem 2. 

Let $\{x_n\}$ be generated by the following iterative algorithm:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{x}_1,y_1\in Hchosenarbitrarily,{\epsilon }_1,e_1\in Hchosenarbitrarily,\hfill \\ C_1=H=Q_1,\hfill \\ u_n={\omega }_n{x}_n+{\epsilon }_n,\hfill \\ v_n={\beta }_n{u}_n+(1-{\beta }_n)\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)(\frac{u_n+v_n}{2}),\hfill \\ z_n={\delta }_{n}f(x_n)+(1-{\delta }_n)(I-{\zeta }_n\sum _{i=1}^{\infty }{c}_i{W}_i)\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)(\frac{u_n+v_n}{2}),\hfill \\ w_n={\alpha }_n{x}_n+(1-{\alpha }_n)\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)(\frac{u_n+z_n}{2}),\hfill \\ x_{n+1}={\lambda }_n\eta f(x_n)+(I-{\lambda }_{n}F)w_n+e_n,\hfill \\ C_{n+1}=\{p\in C_n:2\langle {\alpha }_n{x}_n+(1-{\alpha }_n)z_n-w_n,p\rangle \hfill \\ \le {\alpha }_n\parallel x_n{\parallel }^2+(1-{\alpha }_n)\parallel z_n{\parallel }^2-\parallel w_n{\parallel }^2\}\hfill \\ Q_{n+1}=\{p\in C_{n+1}:\parallel x_1{-p\parallel }^2\le \parallel P_{C_{n+1}}(x_1)-x_1{\parallel }^2+{\sigma }_{n+1}\},n\in N,\hfill \\ y_{n+1}\in Q_{n+1},n\in N.\hfill \end{array}\right.\end{array}$$

(38)

Under the assumptions of Theorem 1, one has

$x_n\to q_0\in {\bigcap }_{i=1}^{\infty }N(A_i+B_i),$ as $n\to \infty ,$ where $q_0$ is the unique solution of the system of variational inclusions (14) and (15). Moreover, $y_n\to P_{{\bigcap }_{m=1}^{\infty }{C}_m}(x_1)\in {\bigcap }_{i=1}^{\infty }N(A_i+B_i),$ as $n\to \infty$.

Proof. 

The proof is split into eleven steps. Copy Steps 2–5, 7, 10 and 11 in Theorem 1. Furthermore, modify the other steps in Theorem 1 as follows

Step 1. $\{v_n\}$ is well-defined.

For $s\in (0,1),$ define $U_s:H\to H$ by $U_{s}x:=su+(1-s)T(\frac{u+x}{2}),$ for any $x\in H$ and for fixed $u\in H$, where $T:H\to H$ is any fixed non-expansive mapping.

It is easy to check that $\parallel U_{s}x-U_{s}y\parallel =(1-s)\parallel T(\frac{u+x}{2})-T(\frac{u+y}{2})\parallel \le \frac{1-s}{2}\parallel x-y\parallel .$ Thus, $U_s$ is a contraction, which ensures from Lemma 1 that there exists $x_s\in H$ such that $U_s{x}_s=x_s.$ That is, $x_s=su+(1-s)T(\frac{u+x_s}{2}).$

Considering T here as ${\sum }_{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i),$ similar to Step 1 of Theorem 1, one can see that $\{v_n\}$ is well-defined.

Step 6. $\{u_n\},$$\{v_n\},$$\{z_n\},$$\{w_n\}$, and $\{x_n\}$ are all bounded.

For $p\in {\bigcap }_{i=1}^{\infty }N(A_i+B_i),$ one has

$\parallel v_n-p\parallel \le {\beta }_n\parallel u_n-p\parallel +(1-{\beta }_n)\frac{\parallel u_n-p\parallel }{2}+(1-{\beta }_n)\frac{\parallel v_n-p\parallel }{2}.$ This implies that (20) is still true.

In view of Lemma 8 and (20), one has

$$\begin{array}{c}\parallel z_n-p\parallel \le {\delta }_n\parallel f(x_n)-f(p)\parallel +{\delta }_n\parallel f(p)-p\parallel \hfill \\ +(1-{\delta }_n)\parallel (I-{\zeta }_n\sum _{i=1}^{\infty }{c}_i{W}_i)\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)(\frac{u_n+v_n}{2})-p\parallel \hfill \\ \le {\delta }_{n}k\parallel x_n-p\parallel +{\delta }_n\parallel f(p)-p\parallel \hfill \\ +(1-{\delta }_n)\parallel (I-{\zeta }_n\sum _{i=1}^{\infty }{c}_i{W}_i)\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)(\frac{u_n+v_n}{2}-p)\parallel \hfill \\ +(1-{\delta }_n)\parallel (I-{\zeta }_n\sum _{i=1}^{\infty }{c}_i{W}_i)\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)p-p\parallel \hfill \\ \le {\delta }_{n}k\parallel x_n-p\parallel +{\delta }_n\parallel f(p)-p\parallel +(1-{\delta }_n)[1-{\zeta }_n(1-\sum _{i=1}^{\infty }{c}_i\sqrt{\frac{1-{\vartheta }_i}{{\mu }_i}})]\parallel \frac{u_n-p}{2}+\frac{v_n-p}{2}\parallel \hfill \\ +(1-{\delta }_n){\zeta }_n\sum _{i=1}^{\infty }{c}_i\parallel W_i\parallel \parallel p\parallel \hfill \\ \le {\delta }_{n}k\parallel x_n-p\parallel +{\delta }_n\parallel f(p)-p\parallel +(1-{\delta }_n)[1-{\zeta }_n(1-\sum _{i=1}^{\infty }{c}_i\sqrt{\frac{1-{\vartheta }_i}{{\mu }_i}})]\parallel u_n-p\parallel \hfill \\ +(1-{\delta }_n){\zeta }_n\sum _{i=1}^{\infty }{c}_i\parallel W_i\parallel \parallel p\parallel ,\hfill \end{array}$$

which ensures that (21) is still true.

Note that

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \parallel w_n-p\parallel \le {\alpha }_n\parallel x_n-p\parallel +(1-{\alpha }_n)\frac{\parallel u_n-p\parallel }{2}+(1-{\alpha }_n)\frac{\parallel z_n-p\parallel }{2}.\end{array}\end{array}$$

(39)

Combining with inequalities (19)–(21) and (39), one has

$$\begin{array}{c}\parallel x_{n+1}-p\parallel \hfill \\ \le \{{\lambda }_n\eta k+(1-{\lambda }_n\xi ){\alpha }_n+\frac{(1-{\lambda }_n\xi )(1-{\alpha }_n){\delta }_{n}k}{2}\hfill \\ +\frac{(1-{\lambda }_n\xi )(1-{\alpha }_n)(1-{\delta }_n)[1-{\zeta }_n(1-{\sum }_{i=1}^{\infty }{c}_i\sqrt{\frac{1-{\vartheta }_i}{{\mu }_i}})]{\omega }_n}{2}+\frac{(1-{\lambda }_n\xi )(1-{\alpha }_n){\omega }_n}{2}\}\parallel x_n-p\parallel \hfill \\ +{\lambda }_n\parallel \eta f(p)-F(p)\parallel +\parallel e_n\parallel +\frac{(1-{\lambda }_n\xi )(1-{\alpha }_n)(1-{\omega }_n)}{2}\parallel p\parallel +\frac{(1-{\lambda }_n\xi )(1-{\alpha }_n)\parallel {\epsilon }_n\parallel }{2}\hfill \\ +\frac{(1-{\lambda }_n\xi )(1-{\alpha }_n){\delta }_n}{2}\parallel f(p)-p\parallel +\frac{(1-{\lambda }_n\xi )(1-{\alpha }_n)(1-{\delta }_n){\zeta }_n{\sum }_{i=1}^{\infty }{c}_i\parallel W_i\parallel \parallel p\parallel }{2}\hfill \\ +\frac{(1-{\lambda }_n\xi )(1-{\alpha }_n)(1-{\omega }_n)(1-{\delta }_n)[1-{\zeta }_n(1-{\sum }_{i=1}^{\infty }{c}_i\sqrt{\frac{1-{\vartheta }_i}{{\mu }_i}})]\parallel p\parallel }{2}\hfill \\ +\frac{(1-{\lambda }_n\xi )(1-{\alpha }_n)(1-{\delta }_n)[1-{\zeta }_n(1-{\sum }_{i=1}^{\infty }{c}_i\sqrt{\frac{1-{\vartheta }_i}{{\mu }_i}})]\parallel {\epsilon }_n\parallel }{2}\hfill \\ \le \{{\lambda }_n\eta k+(1-{\lambda }_n\xi ){\alpha }_n+\frac{(1-{\lambda }_n\xi )(1-{\alpha }_n)}{2}+\frac{(1-{\lambda }_n\xi )(1-{\alpha }_n){\delta }_{n}k}{2}\hfill \\ +\frac{(1-{\lambda }_n\xi )(1-{\alpha }_n)(1-{\delta }_n)[1-{\zeta }_n(1-{\sum }_{i=1}^{\infty }{c}_i\sqrt{\frac{1-{\vartheta }_i}{{\mu }_i}})]}{2}\}\parallel x_n-p\parallel \hfill \\ +{\lambda }_n(\xi -\eta k)\frac{\parallel \eta f(p)-F(p)\parallel }{\xi -\eta k}+\parallel e_n\parallel \hfill \\ +(1-{\lambda }_n\xi )\frac{(1-{\alpha }_n)}{2}{\delta }_n(1-k)\frac{\parallel f(p)-p\parallel }{1-k}\hfill \\ +(1-{\lambda }_n\xi )\frac{(1-{\alpha }_n)}{2}(1-{\delta }_n){\zeta }_n(1-\sum _{i=1}^{\infty }{c}_i\sqrt{\frac{1-{\vartheta }_i}{{\mu }_i}})\frac{{\sum }_{i=1}^{\infty }{c}_i\parallel W_i\parallel \parallel p\parallel }{1-{\sum }_{i=1}^{\infty }{c}_i\sqrt{\frac{1-{\vartheta }_i}{{\mu }_i}}}\hfill \\ +2(1-{\omega }_n)\parallel p\parallel +2\parallel {\epsilon }_n\parallel \hfill \\ \le max\{\parallel x_n-p\parallel ,\frac{\parallel \eta f(p)-p\parallel }{\xi -\eta k},\frac{\parallel f(p)-p\parallel }{1-k},\frac{{\sum }_{i=1}^{\infty }{c}_i\parallel W_i\parallel \parallel p\parallel }{1-{\sum }_{i=1}^{\infty }{c}_i\sqrt{\frac{1-{\vartheta }_i}{{\mu }_i}}}\}+\parallel e_n\parallel +2(1-{\omega }_n)\parallel p\parallel +2\parallel {\epsilon }_n\parallel \hfill \\ \cdots \hfill \\ \le max\{\parallel x_1-p\parallel ,\frac{\parallel \eta f(p)-p\parallel }{\xi -\eta k},\frac{\parallel f(p)-p\parallel }{1-k},\frac{{\sum }_{i=1}^{\infty }{c}_i\parallel W_i\parallel \parallel p\parallel }{1-{\sum }_{i=1}^{\infty }{c}_i\sqrt{\frac{1-{\vartheta }_i}{{\mu }_i}}}\}\hfill \\ +\sum _{i=1}^n\parallel e_i\parallel +2\sum _{i=1}^n(1-{\omega }_i)\parallel p\parallel +2\sum _{i=1}^n\parallel {\epsilon }_i\parallel \hfill \end{array}$$

Therefore, $\{x_n\}$ is bounded. Similar to Step 6 in Theorem 1, $\{u_n\},$$\{v_n\},$$\{z_n\}$, $\{w_n\}$, $\{{\sum }_{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)(\frac{u_n+v_n}{2})\}$, $\{{\sum }_{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)(\frac{u_n+z_n}{2})\}$, and $\{{\sum }_{i=1}^{\infty }{c}_i{W}_i{\sum }_{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)(\frac{u_n+v_n}{2})\}$ are all bounded.

Step 8. $limsu{p}_{n\to \infty }\langle \eta f(q_0)-F{q}_0,x_{n+1}-q_0\rangle \le 0,$ where $q_0$ is the same as that in Step 7.

Note that

$$\begin{array}{c}\hfill \begin{array}{ccc}& \parallel w_n-z_n\parallel \hfill & \\ & \le \parallel w_n-\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)\frac{u_n+z_n}{2}\parallel \hfill & \\ & +\parallel \sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)(\frac{u_n+z_n}{2}-\frac{u_n+v_n}{2})\parallel \hfill & \\ & +\parallel \sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)\frac{u_n+v_n}{2}-v_n\parallel +\parallel v_n-z_n\parallel \hfill & \\ & \le \parallel {\alpha }_n[x_n-\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)\frac{u_n+z_n}{2}]\parallel \hfill & \\ & +{\beta }_n\parallel \sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)\frac{u_n+v_n}{2}-u_n\parallel +\frac{3}{2}\parallel v_n-z_n\parallel \hfill & \\ & ={\alpha }_n\parallel x_n-\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)\frac{u_n+z_n}{2}\parallel +\frac{3}{2}\parallel z_n-v_n\parallel \hfill & \\ & +{\beta }_n\parallel \sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)\frac{u_n+v_n}{2}-u_n\parallel \hfill & \end{array}\end{array}$$

(40)

Furthermore,

$$\begin{array}{c}\hfill \begin{array}{ccc}& \parallel z_n-v_n\parallel \hfill & \\ & \le {\delta }_n\parallel f(x_n)\parallel +{\zeta }_n(1-{\delta }_n)\parallel \sum _{i=1}^{\infty }{c}_i{W}_i\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)(\frac{u_n+v_n}{2})\parallel \hfill & \\ & +{\beta }_n\parallel u_n-\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)(\frac{u_n+v_n}{2})\parallel \hfill & \\ & +{\delta }_n\parallel \sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)(\frac{u_n+v_n}{2})\parallel \hfill & \end{array}\end{array}$$

(41)

Based on the assumptions, (40) and (41), and Step 6, one has $w_n-z_n\to 0,$ as $n\to \infty .$ Copying the corresponding part of Step 8 in Theorem 1, one can see that $limsu{p}_{n\to \infty }\langle \eta f(q_0)-F{q}_0,x_{n+1}-q_0\rangle \le 0$.

Step 9. $x_n\to q_0,$ as $n\to \infty$, where $q_0$ is the same as that in Steps 7 and 8.

Similar to Step 9 in Theorem 1, we can easily see that both (27) and (28) are still true.

In view of Lemma 11 and (28), one has

$$\begin{array}{c}\hfill \begin{array}{ccc}& \parallel z_n-q_0{\parallel }^2\hfill & \\ & =\parallel {\delta }_n(f(x_n)-q_0)+(1-{\delta }_n)[(I-{\zeta }_n\sum _{i=1}^{\infty }{c}_i{W}_i)\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)\frac{u_n+v_n}{2}-q_0]{\parallel }^2\hfill & \\ & \le (1-{\delta }_n){\parallel \frac{u_n+v_n}{2}-q_0\parallel }^2+2{\delta }_n\langle z_n-q_0,f(x_n)-q_0\rangle \hfill & \\ & -2(1-{\delta }_n){\zeta }_n\langle \sum _{i=1}^{\infty }{c}_i{W}_i\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)\frac{u_n+v_n}{2},z_n-q_0\rangle \hfill & \\ & \le (1-{\delta }_n){\parallel \frac{u_n+v_n}{2}-q_0\parallel }^2+2{\delta }_n\langle z_n-q_0,f(x_n)-f(q_0)\rangle +2{\delta }_n\langle z_n-q_0,f(q_0)-q_0\rangle \hfill & \\ & +2(1-{\delta }_n){\zeta }_n\parallel \sum _{i=1}^{\infty }{c}_i{W}_i\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)\frac{u_n+v_n}{2}\parallel \parallel z_n-q_0\parallel \hfill & \\ & \le (1-{\delta }_n)\parallel u_n-q_0{\parallel }^2+2{\delta }_{n}k\parallel z_n-x_n\parallel \parallel x_n-q_0\parallel +2{\delta }_{n}k{\parallel x_n-q_0\parallel }^2+2{\delta }_n\langle z_n-q_0,f(q_0)-q_0\rangle \hfill & \\ & +2{\zeta }_n\parallel \sum _{i=1}^{\infty }{c}_i{W}_i\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)\frac{u_n+v_n}{2}\parallel \parallel z_n-q_0\parallel \hfill \end{array}\end{array}$$

(42)

Note that

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \parallel w_n-q_0{\parallel }^2\le {\alpha }_n{\parallel x_n-q_0\parallel }^2+(1-{\alpha }_n)\frac{\parallel u_n-q_0{\parallel }^2}{2}+(1-{\alpha }_n)\frac{\parallel z_n-q_0{\parallel }^2}{2}.\end{array}\end{array}$$

(43)

Now, in view of Lemma 9 and using (27), (28), (42), and (43), one has

$$\begin{array}{c}\hfill \begin{array}{ccc}& \parallel x_{n+1}-q_0{\parallel }^2={\parallel {\lambda }_n(\eta f(x_n)-F{q}_0)+(I-{\lambda }_{n}F)(w_n-q_0)+e_n\parallel }^2\hfill & \\ & \le (1-{\lambda }_n\xi )\parallel w_n-q_0{\parallel }^2+2\parallel e_n\parallel \parallel x_{n+1}-q_0\parallel +2{\lambda }_n\eta \langle f(x_n)-f(q_0),x_{n+1}-q_0\rangle \hfill & \\ & +2{\lambda }_n\langle \eta f(q_0)-F{q}_0,x_{n+1}-q_0\rangle \hfill & \\ & \le \{(1-{\lambda }_n\xi ){\alpha }_n+{\lambda }_n\eta k+(1-{\lambda }_n\xi )(1-{\alpha }_n){\delta }_{n}k+(1-{\lambda }_n\xi )(1-{\alpha }_n)(1-\frac{{\delta }_n}{2}){\omega }_n\}{\parallel x_n-q_0\parallel }^2\hfill & \\ & +2\parallel e_n\parallel \parallel x_{n+1}-q_0\parallel +(1-{\lambda }_n\xi )(1-{\alpha }_n)(2-{\delta }_n)\parallel {\epsilon }_n\parallel \parallel u_n-q_0\parallel \hfill & \\ & +(1-{\lambda }_n\xi )(1-{\alpha }_n)(2-{\delta }_n)(1-{\omega }_n)\parallel q_0\parallel \parallel u_n-q_0\parallel \hfill & \\ & +{\delta }_n\eta k(1-{\lambda }_n\xi )(1-{\alpha }_n)\parallel x_n-q_0\parallel \parallel x_n-z_n\parallel \hfill & \\ & +(1-{\lambda }_n\xi )(1-{\alpha }_n){\delta }_n\parallel z_n-q_0\parallel \parallel f(q_0)-q_0\parallel \hfill & \\ & +(1-{\lambda }_n\xi )(1-{\alpha }_n){\zeta }_n\parallel z_n-q_0\parallel \parallel \sum _{i=1}^{\infty }{c}_i{W}_i\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)(\frac{u_n+v_n}{2})-q_0\parallel \hfill & \\ & +{\lambda }_n\eta k{\parallel x_{n+1}-q_0\parallel }^2+2{\lambda }_n\langle \eta f(q_0)-F(q_0),x_{n+1}-q_0\rangle \hfill \end{array}\end{array}$$

(44)

Therefore,

$$\begin{array}{c}\hfill \begin{array}{ccc}& (1-{\lambda }_n\eta k){\parallel x_{n+1}-q_0\parallel }^2\hfill & \\ & \le \{(1-{\lambda }_n\xi ){\alpha }_n+{\lambda }_n\eta k+(1-{\lambda }_n\xi )(1-{\alpha }_n){\delta }_{n}k+(1-{\lambda }_n\xi )(1-{\alpha }_n)(1-\frac{{\delta }_n}{2})\}{\parallel x_n-q_0\parallel }^2\hfill & \\ & +[\parallel e_n\parallel +\parallel {\epsilon }_n\parallel +(1-{\omega }_n)+2{\delta }_n+{\zeta }_n]M_2+2{\lambda }_n\langle \eta f(q_0)-F(q_0),x_{n+1}-q_0\rangle ,\hfill \end{array}\end{array}$$

(45)

where $M_2=su{p}_n\{2\parallel x_{n+1}-q_0\parallel , \parallel x_n-q_0\parallel \parallel z_n-x_n\parallel , \parallel z_n-q_0\parallel \parallel f(q_0)-q_0\parallel , \parallel z_n-q_0\parallel$$\parallel {\sum }_{i=1}^{\infty }{c}_i{W}_i{\sum }_{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)(\frac{u_n+v_n}{2})-q_0\parallel , \parallel u_n-q_0\parallel , 2\parallel q_0\parallel \parallel u_n-q_0\parallel :$$n\in N\}<+\infty .$

If we set $b_n^{(3)}=b_n^{(1)},b_n^{(4)}=\frac{[\parallel e_n\parallel +\parallel {\epsilon }_n\parallel +(1-{\omega }_n)+2{\delta }_n+{\zeta }_n]M_2+2{\lambda }_n\langle \eta f(q_0)-F(q_0),x_{n+1}-q_0\rangle }{1-{\lambda }_n\eta k}$, then

$$\parallel x_{n+1}-q_0{\parallel }^2\le (1-b_n^{(3)}){\parallel x_n-q_0\parallel }^2+b_n^{(4)}.$$

Similar to Step 9 in Theorem 1, in view of Lemma 12, we have $x_n\to q_0,$ as $n\to \infty$.

This completes the proof. □

Remark 1. 

The restrictions imposed on the mappings $f(x)$ and $F(x)$ are available. For example, take $F(x)=\frac{3}{4}x$ and $f(x)=\frac{x}{2},$ for $x\in (-\infty ,+\infty ).$ Take $\eta =\frac{1}{2},$ $k=\frac{1}{2}, \xi =\frac{3}{4}$. Then, we can easily see that F is a strongly positive linear bounded mapping with ξ, f is a contraction, and $\xi >2\eta k$. Moreover, $\langle F(x)-\eta f(x)+f(y)-y,x-y\rangle =\frac{1}{2}{\parallel x-y\parallel }^2\ge 0,$ for $x,y\in (-\infty ,+\infty ).$ Furthermore, if $\langle f(x)-x,y-x\rangle =0,$ then $\frac{x}{2}(y-x)=0,$ which implies that $x=0$ or $y=x.$

Remark 2. 

In both (13) and (38), the idea of forward–backward splitting method is embodied, the superposition perturbation is considered and multi-choice sets are constructed, which extends and complements the corresponding studies.

Remark 3. 

From Theorems 1 and 2, we may find that the limit $q_0$ of the iterative sequence $\{x_n\}$ is not only the solution of the system of monotone inclusions (3) but also the solution of variational inequalities (14) and (15). That is, the study on iterative construction of the solution of (14) in [18] and the solution of (15) in [17] are unified in our paper.

Remark 4. 

From Theorems 1 and 2, we may find that the relationship between the metric projection $P_{{\bigcap }_{m=1}^{\infty }{C}_m}(x_1)$ and the common solution of variational inequalities and monotone inclusions $q_0$ is set up in our paper.

3. Applications

In this section, one kind capillarity system discussed in [18] is employed again to demonstrate the application of Theorems 1 and 2.

The discussion begins under the following assumptions:

(1)

$\Omega$ is a bounded conical domain in $R^n$ ($n\in N$) with its boundary $\Gamma \in C^1.$

(2)

$\vartheta$ is the exterior normal derivative of $\Gamma .$

(3)

${\lambda }_i$ is a positive number, for $i\in N.$

(4)

$p_i\in (\frac{2n}{n+1},+\infty )$, for $i\in N.$ Moreover, if $p_i\ge n,$ then suppose $1\le q_i,r_i<+\infty$, for $i\in N$. If $p_i<n,$ then suppose $1\le q_i,r_i\le \frac{n{p}_i}{n-p_i}$, for $i\in N.$

(5)

$|·|$ denotes the norm in $R^n$ and $<·,·>$ the inner-product.

Now, examine the capillarity systems:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}-div[(1+\frac{|\nabla u^{(i)}{|}^{p_i}}{\sqrt{1+|\nabla u^{(i)}{|}^{2p_i}}})|\nabla u^{(i)}{|}^{p_i-2}\nabla u^{(i)}]\hfill \\ \hfill \\ +{\lambda }_i(|u^{(i)}{|}^{q_i-2}{u}^{(i)}+|u^{(i)}{|}^{r_i-2}{u}^{(i)})+u^{(i)}(x)=f_i(x),x\in \Omega \hfill \\ \hfill \\ -<\vartheta ,(1+\frac{|\nabla u^{(i)}{|}^{p_i}}{\sqrt{1+|\nabla u^{(i)}{|}^{2p_i}}}){|\nabla u^{(i)}|}^{p_i-2}\nabla u^{(i)}>=0,x\in \Gamma ,i\in N.\hfill \end{array}\right.\end{array}$$

(46)

Lemma 14. 

(see [18]) For $i\in N,$ define $A_i:L^2(\Omega )\to L^2(\Omega )$ by

(1)

$D(A_i)=\{u\in L^2(\Omega )|\exists f\in L^2(\Omega )suchthatf\in \widetilde{A_i}u\},$ where $\widetilde{A_i}:W^{1,p_i}(\Omega )\to {(W^{1,p_i}(\Omega ))}^{*}$ is defined by

$$\begin{array}{cc}& \langle w,\widetilde{A_i}u\rangle ={\int }_{\Omega }<(1+\frac{{|\nabla u|}^{p_i}}{\sqrt{1+{|\nabla u|}^{2p_i}}}){|\nabla u|}^{p_i-2}\nabla u,\nabla v>dx\hfill \\ & +{\lambda }_i{\int }_{\Omega }{|u(x)|}^{q_i-2}u(x)v(x)dx+{\lambda }_i{\int }_{\Omega }{|u(x)|}^{r_i-2}u(x)v(x)dx,\hfill \end{array}$$

for any $u,w\in W^{1,p_i}(\Omega );$

(2)

$A_{i}u=\{f\in L^2(\Omega )|f\in \widetilde{A_i}u\}$.

Then, $A_i:L^2(\Omega )\to L^2(\Omega )$ is maximal monotone, for each $i\in N$.

Lemma 15. 

(see [18]) Define $B_i:L^2(\Omega )\to L^2(\Omega )$ by

$$(B_{i}u)(x)=u(x)-f_i(x),forallu(x)\in D(B_i),$$

and then $B_i$ is ${\theta }_i$-inversely strongly accretive, for ${\theta }_i\in (0,1]$ and $i\in N.$

Lemma 16. 

(see [18]) If, in (46), $f_i(x)\equiv {\lambda }_i{(|k|}^{q_i-1}+{|k|}^{r_i-1})sgnk+k,$ where k is a constant, then $\{u^{(i)}\equiv k:i\in N\}$ is the solution of capillarity system (46). Furthermore, $\{k\}={\bigcap }_{i=1}^{\infty }N(A_i+B_i).$

Theorem 3. 

Suppose $f_i(x)\equiv {\lambda }_i{(|k|}^{q_i-1}+{|k|}^{r_i-1})sgnk+k,$$A_i$ and $B_i$ are the same as those in Lemmas 14 and 15, $F:L^2(\Omega )\to L^2(\Omega )$ is a strongly positive linear bounded operator with coefficient $\xi >0$, $f:L^2(\Omega )\to L^2(\Omega )$ is a contraction with coefficient $k\in (0,1)$ and $W_i:L^2(\Omega )\to L^2(\Omega )$ is ${\vartheta }_i$-strongly monotone and ${\mu }_i$-strictly pseudo-contractive mapping, for $i\in N.$

Two iterative algorithms are constructed as follows:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{x}_1,y_1\in L^2(\Omega )chosenarbitrarily,e_1,{\epsilon }_1\in L^2(\Omega )chosenarbitrarily,\hfill \\ C_1=L^2(\Omega )=Q_1,\hfill \\ u_n={\omega }_n{x}_n+{\epsilon }_n,\hfill \\ v_n={\beta }_n{u}_n+(1-{\beta }_n)\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)v_n,\hfill \\ z_n={\delta }_{n}f(x_n)+(1-{\delta }_n)(I-{\zeta }_n\sum _{i=1}^{\infty }{c}_i{W}_i)\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)v_n,\hfill \\ w_n={\alpha }_n{x}_n+(1-{\alpha }_n)\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)z_n,\hfill \\ x_{n+1}={\lambda }_n\eta f(x_n)+(I-{\lambda }_{n}F)w_n+e_n,\hfill \\ C_{n+1}=\{p\in C_n:2\langle {\alpha }_n{x}_n+(1-{\alpha }_n)z_n-w_n,p\rangle \hfill \\ \le {\alpha }_n\parallel x_n{\parallel }^2+(1-{\alpha }_n)\parallel z_n{\parallel }^2-\parallel w_n{\parallel }^2\}\hfill \\ Q_{n+1}=\{p\in C_{n+1}:\parallel x_1{-p\parallel }^2\le \parallel P_{C_{n+1}}(x_1)-x_1{\parallel }^2+{\sigma }_{n+1}\},n\in N,\hfill \\ y_{n+1}\in Q_{n+1},n\in N,\hfill \end{array}\right.\end{array}$$

(47)

and

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{x}_1,y_1\in L^2(\Omega )chosenarbitrarily,{\epsilon }_1,e_1\in L^2(\Omega )chosenarbitrarily,\hfill \\ C_1=L^2(\Omega )=Q_1,\hfill \\ u_n={\omega }_n{x}_n+{\epsilon }_n,\hfill \\ v_n={\beta }_n{u}_n+(1-{\beta }_n)\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)(\frac{u_n+v_n}{2}),\hfill \\ z_n={\delta }_{n}f(x_n)+(1-{\delta }_n)(I-{\zeta }_n\sum _{i=1}^{\infty }{c}_i{W}_i)\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)(\frac{u_n+v_n}{2}),\hfill \\ w_n={\alpha }_n{x}_n+(1-{\alpha }_n)\sum _{i=1}^{\infty }{a}_i{(I+r_{n,i}{A}_i)}^{-1}(I-r_{n,i}{B}_i)(\frac{u_n+z_n}{2}),\hfill \\ x_{n+1}={\lambda }_n\eta f(x_n)+(I-{\lambda }_{n}F)w_n+e_n,\hfill \\ C_{n+1}=\{p\in C_n:2\langle {\alpha }_n{x}_n+(1-{\alpha }_n)z_n-w_n,p\rangle \hfill \\ \le {\alpha }_n\parallel x_n{\parallel }^2+(1-{\alpha }_n)\parallel z_n{\parallel }^2-\parallel w_n{\parallel }^2\}\hfill \\ Q_{n+1}=\{p\in C_{n+1}:\parallel x_1{-p\parallel }^2\le \parallel P_{C_{n+1}}(x_1)-x_1{\parallel }^2+{\sigma }_{n+1}\},n\in N,\hfill \\ y_{n+1}\in Q_{n+1},n\in N.\hfill \end{array}\right.\end{array}$$

(48)

Under the assumptions of Theorems 1 and 2, one has $x_n\to q_0(x)\in {\bigcap }_{i=1}^{\infty }N(A_i+B_i),$ where $q_0(x)$ is common solution of the capillarity system (46) and the system of variational inclusions (14) and (15).

4. Conclusions

Some new forward–backward multi-choice iterative algorithm with superposition perturbations are presented in a real Hilbert space. The iterative sequences are proved to be strongly convergent to not only the solution of monotone inclusions but also the solution of variational inequalities. In the near future, more work can be done to weaken the restrictions imposed on the contraction f and the strongly positive linear bounded mapping F.