Iterative Algorithms for a System of Variational Inclusions in Banach Spaces

Abstract

A system of variational inclusions (GSVI) is considered in Banach spaces. An implicit iterative procedure is proposed for solving the GSVI. Strong convergence of the proposed algorithm is given.

1. Introduction

Let X be a smooth Banach space and $\varnothing \ne C\subset X$ a closed convex set. Let $A_1,A_2:C\to X$ and $M_1,M_2:C\to 2^X$ be nonlinear mappings. In the present article, we consider the following system of variational inclusions (GSVI, for short) which aims to seek $(u^{*},v^{*})\in C\times C$ verifying

$$\left\{\begin{array}{l}0\in {\varsigma }_1(A_1{v}^{*}+M_1{u}^{*})+u^{*}-v^{*},\hfill \\ 0\in {\varsigma }_2(A_2{u}^{*}+M_2{v}^{*})+v^{*}-u^{*},\hfill \end{array}\right.$$

(1)

where ${\varsigma }_1$ and ${\varsigma }_2$ are two positive constants.

Special cases: If $A_1=A_2=A$ and $M_1=M_2=M$, then the relation (1) reduces to seek $(u^{*},v^{*})\in C\times C$ verifying

$$\left\{\begin{array}{l}0\in {\varsigma }_1(A{v}^{*}+M{u}^{*})+u^{*}-v^{*},\hfill \\ 0\in {\varsigma }_2(A{u}^{*}+M{v}^{*})+v^{*}-u^{*}.\hfill \end{array}\right.$$

(2)

If $u^{*}=v^{*}$ in (2), then the relation (2) reduces to seek $(u^{*},v^{*})\in C\times C$ verifying

$$0\in A{u}^{*}+M{u}^{*}.$$

(3)

Especially, if $M=\partial \varphi$, where $\varphi :H\to R\cup +\infty$ is a proper convex lower semi-continuous function, then we have the following mixed quasi-variational inequality

$$\langle Au,y-u\rangle +\varphi \left(y\right)-\varphi \left(u\right)\rangle \ge 0,\forall y\in H.$$

Variational inequalities and variational inclusions have played vital roles in practical applications. Numerous iterative procedures for approaching variational inequalities and variational inclusions have been computed by the researchers [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29].

In [4], the authors introduced an iterative procedure for approaching GSVI (1). Qin et al. [30] suggested an extragradient algorithm for solving GSVI (1), and demonstrated the strong convergence analysis of the presented algorithm. Lan et al. [28], Buong et al. [11], Zhang et al. [13] studied iterative procedures for approaching variational inclusion (3).

On the other hand, iterative computation of zeros or fixed points of nonlinear operators has been studied extensively in the literature [14,31,32,33,34,35,36]. Zhang et al. [37] introduced an iterative procedure for approaching a solution of the inclusion problem (3) and a fixed point of a nonexpansive mapping in Hilbert spaces. Peng et al. [38] presented a viscosity algorithm for finding a solution of a variational inclusion with set-valued maximal monotone mapping and inverse strongly monotone mappings, the set of solutions of an equilibrium problem and a fixed point of a nonexpansive mapping.

Motivated by the above work, in the present paper, we consider the GSVI (1) with the hierarchical variational inequality constraint for a strict pseudocontraction T in Banach spaces. We suggest an implicit iterative procedure for solving the GSVI (1) with the HVI constraint for strict pseudocontraction T. We show the strong convergence of the suggested procedure to a solution of the GSVI (1).

2. Preliminaries

Let X be a real Banach space and $\varnothing \ne C\subset X$ a closed convex set. A mapping $f:C\to C$ is said to be k-Lipschitz if $\parallel f\left(u\right)-f\left(v\right)\parallel \le k\parallel u-v\parallel ,\forall u,v\in C$ for some $k\ge 0$. If $k<1$, then f is said to be a k-contraction. If $k=1$, then f is said to be nonexpansive.

Recall that an operator $T:C\to X$ is called

(i)

accretive if

$$\langle Tu-Tv,j(u-v)\rangle \ge 0,\forall u,v\in C,$$

where $j(u-v)\in J(u-v)$.

(ii)

$\alpha$-inverse-strongly accretive if

$$\langle Tu-Tv,j(u-v)\rangle \ge {\alpha \parallel Tu-Tv\parallel }^2,\forall u,v\in C,$$

where $j(u-v)\in J(u-v)$ and $\alpha >0$.

(iii)

strictly pseudocontractive if

$$\langle Tu-Tv,j(u-v)\rangle \le {\parallel u-v\parallel }^2-\beta {\parallel u-v-(Tu-Tv)\parallel }^2,\forall u,v\in C,$$

where $j(u-v)\in J(u-v)$ and $\beta >0$.

If X is q-uniformly smooth with $1<q\le 2$, then

$${\parallel u+v\parallel }^q{+\parallel u-v\parallel }^q\le {2(\parallel u\parallel }^q+{\parallel cv\parallel }^q),\forall u,v\in X,$$

where $c>0$ is some constant.

Proposition 1

([32]). In a smooth and uniformly convex Banach space X, for all $u,v\in B_r=\{u\in X:\parallel u\parallel \le r\}$, there holds

$$g(\parallel u-v\parallel )\le {\parallel u\parallel }^2-2\langle u,j\left(v\right)\rangle +{\parallel v\parallel }^2,$$

where $g:[0,2r]\to \mathbf{R}$ is a strictly increasing, continuous, and convex function satisfying $g\left(0\right)=0$.

Proposition 2

([35]). In a 2-uniformly smooth Banach space X, there holds

$${\parallel u+v\parallel }^2\le {\parallel u\parallel }^2+2\langle v,j\left(u\right)\rangle +2{\parallel cv\parallel }^2,\forall u,v\in X.$$

Let $D\subset C$ and $\Pi :C\to D$ be an operator. If $\Pi \left[\right(1-s)\Pi (u)+su]=\Pi \left(u\right),$ whenever $(1-s)\Pi \left(u\right)+su\in C$ for $u\in C$ and $s\ge 0$, we call $\Pi$ is sunny.

Proposition 3

([26]). Let X be a smooth Banach space and $\varnothing \ne C\subset X$ a closed convex set. Let $\varnothing \ne D\subset C$ be a set and $\Pi :C\to D$ be a retraction. Then the following conclusions are equivalent:

(i) 

$\langle \Pi \left(u\right)-u,j(v-\Pi (u\left)\right)\rangle \ge 0,\forall u\in C$ and $\forall v\in D$;

(ii) 

${\parallel \Pi \left(u\right)-\Pi \left(v\right)\parallel }^2\le \langle u-v,j(\Pi \left(u\right)-\Pi \left(v\right))\rangle ,\forall u,v\in C$;

(iii) 

Π is sunny nonexpansive operator.

If an accretive operator M satisfies $R(I+rM)=X$ for each $r>0$, then M is said to be m-accretive. Assume that an accretive M satisfies the range condition $\overline{D\left(M\right)}\subset R(I+rM)$. Define the resolvent $J_r^M:R(I+rM)\to D\left(M\right)$ of M by $J_r^M={(I+rM)}^{-1}$. Note that $J_r^M$ is nonexpansive and $F\left(J_r^M\right)=M^{-1}0=\{x\in D\left(M\right):0\in Mx\}$ [31]. If $M^{-1}0\ne \varnothing$, then the inclusion $0\in Mx$ is solvable.

Lemma 1.

Let X be a smooth Banach space and $\varnothing \ne C\subset X$ a closed convex set. Let $M:C\to 2^X$ be an m-accretive operator. Then, for any given $r>0$,

$$\parallel J_r^{M}x-J_r^M{y\parallel }^2\le \langle x-y,j(J_r^{M}x-J_r^{M}y)\rangle ,\forall x,y\in X.$$

This means that $J_r^M:X\to C$ is nonexpansive.

Proof. 

Put $u=J_r^{M}x$ and $v=J_r^{M}y$. Then we have $x\in (I+rM)u$ and $y\in (I+rM)v$. Hence, there exist $\tilde{u}\in Mu$ and $\tilde{v}\in Mv$ such that $x=u+r\tilde{u}$ and $y=v+r\tilde{v}$. Utilizing the accretiveness of M, we obtain

$$\begin{array}{ll}\langle x-y,j(J_r^{M}x-J_r^{M}y)\rangle & =\langle u+r\tilde{u}-(v+r\tilde{v}),j(u-v)\rangle \hfill \\ & =\langle u-v,j(u-v)\rangle +r\langle \tilde{u}-\tilde{v},j(u-v)\rangle \hfill \\ & ={\parallel u-v\parallel }^2+r\langle \tilde{u}-\tilde{v},j(u-v)\rangle \hfill \\ & \ge {\parallel u-v\parallel }^2\hfill \\ & =\parallel J_r^{M}x-J_r^M{y\parallel }^2.\hfill \end{array}$$

 □

Lemma 2.

Let $M_1,M_2:C\to 2^X$ be two m-accretive operators and $A_1,A_2:C\to X$ be two operators. $(x^{*},y^{*})$ is a solution of the GSVI (1) iff $Q{x}^{*}=J_{{\varsigma }_1}^{M_1}(I-{\varsigma }_1{A}_1)J_{{\varsigma }_2}^{M_2}(I-{\varsigma }_2{A}_2)x^{*}$, where $y^{*}=J_{{\varsigma }_2}^{M_2}(I-{\varsigma }_2{A}_2)x^{*}$.

Proof. 

Observe that

$$\left\{\begin{array}{l}0\in x^{*}-y^{*}+{\varsigma }_1(A_1{y}^{*}+M_1{x}^{*})\hfill \\ 0\in y^{*}-x^{*}+{\varsigma }_2(A_2{x}^{*}+M_2{y}^{*})\hfill \end{array}\right.\iff \left\{\begin{array}{l}{x}^{*}=J_{{\varsigma }_1}^{M_1}(I-{\varsigma }_1{A}_1)y^{*},\hfill \\ y^{*}=J_{{\varsigma }_2}^{M_2}(I-{\varsigma }_2{A}_2)x^{*}\hfill \end{array}\right.\iff x^{*}=Q{x}^{*}.$$

 □

Lemma 3

([3]). Let X be a strictly convex Banach space and $\varnothing \ne C\subset X$ a closed convex set. Let $\mu \in (0,1)$ be a constant. Define an operator $S:C\to X$ by $Sx=\mu T_{1}x+(1-\mu )T_{2}x,\forall x\in C$, where $T_1,T_2:C\to X$ be two nonexpansive mappings with $F\left(T_1\right)\cap F\left(T_2\right)\ne \varnothing$. Then S is nonexpansive and $F\left(S\right)=F\left(T_1\right)\cap F\left(T_2\right)$.

Lemma 4

([3]). Let X be a 2-uniformly smooth Banach space and $\varnothing \ne C\subset X$ a closed convex set. If the operator $A:C\to X$ is α-inverse-strongly accretive, then

$${\parallel (I-\zeta A)u-(I-\zeta A)v\parallel }^2\le {\parallel u-v\parallel }^2+2\zeta (c^2\zeta -\alpha ){\parallel Au-Av\parallel }^2,\forall u,v\in C.$$

Lemma 5

([3]). Let X be a 2-uniformly smooth Banach space and $\varnothing \ne C\subset X$ a closed convex set. Let $M_1,M_2:C\to 2^X$ be two m-accretive operators and $A_i:C\to X(i=1,2)$ be ${\zeta }_i$-inverse-strongly accretive operator. Define an operator $Q:C\to C$ by $Q:=J_{{\varsigma }_1}^{M_1}(I-{\varsigma }_1{A}_1)J_{{\varsigma }_2}^{M_2}(I-{\varsigma }_2{A}_2)$. If $0\le {\varsigma }_i\le {\displaystyle \frac{{\zeta }_i}{c^2}}(i=1,2)$, then $Q:C\to C$ is nonexpansive.

Lemma 6

([36]). Let X be a uniformly smooth Banach space and $\varnothing \ne C\subset X$ a closed convex set. Let $A:C\to C$ be a nonexpansive mapping with $F\left(A\right)\ne \varnothing$, and $f:C\to X$ be a contraction. Let $t\in (0,1)$. Define a net $z_t$ by $z_t=tf\left(z_t\right)+(1-t)A{z}_t$. Then $z_t\to x^{*}\in F\left(A\right)$ and

$$\langle (I-f)x^{*},j(x^{*}-x)\rangle \le 0,\forall x\in F\left(A\right).$$

Lemma 7

([36]). Assume the sequence $\{a_n\}\subset [0,\infty )$ satisfies $a_{n+1}\le (1-{\lambda }_n)a_n+{\lambda }_n{\sigma }_n(\forall n\ge 0),$ where the sequences $\{{\lambda }_n\}\subset (0,1)$ and $\{{\sigma }_n\}$ satisfy

(i) 

${\sum }_{n=0}^{\infty }{\lambda }_n=\infty$;

(ii) 

either ${\sum }_{n=0}^{\infty }\left|{\lambda }_n{\sigma }_n\right|<\infty$ or ${lim\; sup}_{n\to \infty }{\sigma }_n\le 0$.

Then ${lim}_{n\to \infty }{a}_n=0$.

Lemma 8

([33]). Let X be a 2-uniformly smooth Banach space and $\varnothing \ne C\subset X$ a closed convex set. Let $T:C\to C$ be a λ-strict pseudocontraction. Define an operator $T_{\alpha }$ by $T_{\alpha }x=(1-\alpha )x+\alpha Tx,\alpha \in (0,1)$. Then, $T_{\alpha }:C\to C$ is nonexpansive with $F\left(T_{\alpha }\right)=F\left(T\right)$ provided $\alpha \in (0,{\displaystyle \frac{\lambda }{c^2}}]$.

3. Main Results

Theorem 1.

Let X be a uniformly convex and 2-uniformly smooth Banach space and $\varnothing \ne C\subset X$ a closed convex set. Let $M_1,M_2:C\to 2^X$ be two m-accretive operators and $A_i:C\to X(i=1,2)$ be ${\zeta }_i$-inverse-strongly accretive operator. Let $f:C\to C$ be a contraction with coefficient $k\in [0,1)$. Let $V:C\to C$ be a nonexpansive operator and $T:C\to C$ be a λ-strict pseudocontraction with $\Omega :=F\left(T\right)\cap F\left(Q\right)\ne \varnothing$, where the operator Q is defined as in Lemma 5. Assume that the sequences $\{{\alpha }_n\}\subset (0,1),\{{\beta }_n\}\subset (0,1)$, $\{{\delta }_n\}\subset (0,1)$ and $\{{\gamma }_n\}\subset (0,1)$ satisfy

(i) 

${\alpha }_n+{\delta }_n+{\beta }_n+{\gamma }_n=1(\forall n\ge 1)$;

(ii) 

${\alpha }_n\to 0$ and ${\displaystyle \frac{{\beta }_n}{{\alpha }_n}}\to 0$;

(iii) 

${\gamma }_n\to 1$;

(iv) 

${\sum }_{n=1}^{\infty }{\alpha }_n=\infty$.

Given $x_0\in C$, compute the sequences $\{x_n\}$ and $\{y_n\}$ such that

$$\left\{\begin{array}{ll}{y}_n& =J_{{\varsigma }_2}^{M_2}(x_n-{\varsigma }_2{A}_2{x}_n),\hfill \\ x_n& ={\alpha }_{n}f\left(x_{n-1}\right)+{\delta }_n{x}_{n-1}+{\beta }_{n}V{x}_{n-1}+{\gamma }_n[\mu S{x}_n+(1-\mu )J_{{\varsigma }_1}^{M_1}(y_n-{\varsigma }_1{A}_1{y}_n)],\forall n\ge 1,\hfill \end{array}\right.$$

(4)

where $Sx=(1-\alpha )x+\alpha Tx,\forall x\in C$ with $0<\alpha <min\{1,{\displaystyle \frac{\lambda }{c^2}}\}$ and $\mu \in (0,1)$. Then $x_n\to x^{*}$, $y_n\to y^{*}$ and

(a) 

$(x^{*},y^{*})$ solves the GSVI (1);

(b) 

$x^{*}$ solves the variational inequality: $\langle (I-f)x^{*},j(u-x^{*})\rangle \ge 0,\forall u\in \Omega$.

Proof. 

By Lemmas 5 and 8, Q and S are nonexpansive and $F\left(S\right)=F\left(T\right)$. Put $A:=\mu S+(1-\mu )Q$ with $\mu \in (0,1)$. It is easy to see that the implicit iterative scheme (4) can be rewritten as

$$x_n={\alpha }_{n}f\left(x_{n-1}\right)+{\delta }_n{x}_{n-1}+{\beta }_{n}V{x}_{n-1}+{\gamma }_{n}A{x}_n,\forall n\ge 1.$$

(5)

Consider the mapping $F_{n}u={\alpha }_{n}f\left(x_{n-1}\right)+{\delta }_n{x}_{n-1}+{\beta }_{n}V{x}_{n-1}+{\gamma }_{n}Au,\forall u\in C$. According to Lemma 3, we have

$$\parallel F_{n}u-F_{n}v\parallel ={\gamma }_n\parallel Au-Av\parallel \le {\gamma }_n\parallel u-v\parallel ,\forall u,v\in C.$$

Hence $F_n$ is a contraction. Thus, (5) and hence (4) are all well-posed.

Let $u^{†}\in \Omega$. Thus, $T{u}^{†}=u^{†}$ and $Q{u}^{†}=u^{†}$. It is clear that

$$\begin{array}{ll}{x}_n-u^{†}& ={\alpha }_{n}f\left(x_{n-1}\right)+{\delta }_n{x}_{n-1}+{\beta }_{n}V{x}_{n-1}+{\gamma }_{n}A{x}_n-u^{†}\hfill \\ & ={\alpha }_n(f\left(x_{n-1}\right)-u^{†})+{\delta }_n(x_{n-1}-u^{†})+{\beta }_n(V{x}_{n-1}-u^{†})+{\gamma }_n(A{x}_n-u^{†}).\hfill \end{array}$$

By Lemma 3, we get

$$\begin{array}{ll}\parallel x_n-u^{†}\parallel & \le {\alpha }_n\parallel f\left(x_{n-1}\right)-u^{†}\parallel +{\delta }_n\parallel x_{n-1}-u^{†}\parallel +{\beta }_n\parallel V{x}_{n-1}-u^{†}\parallel +{\gamma }_n\parallel A{x}_n-u^{†}\parallel \hfill \\ & \le {\alpha }_n(\parallel f\left(x_{n-1}\right)-f\left(u^{†}\right)\parallel +\parallel f\left(u^{†}\right)-u^{†}\parallel )+{\delta }_n\parallel x_{n-1}-u^{†}\parallel \hfill \\ & +{\beta }_n(\parallel V{x}_{n-1}-V{u}^{†}\parallel +\parallel V{u}^{†}-u^{†}\parallel )+{\gamma }_n\parallel x_n-u^{†}\parallel \hfill \\ & \le {\alpha }_n(k\parallel x_{n-1}-u^{†}\parallel +\parallel f\left(u^{†}\right)-u^{†}\parallel )+{\delta }_n\parallel x_{n-1}-u^{†}\parallel \hfill \\ & +{\beta }_n(\parallel x_{n-1}-u^{†}\parallel +\parallel V{u}^{†}-u^{†}\parallel )+{\gamma }_n\parallel x_n-u^{†}\parallel \hfill \\ & ={\alpha }_n\parallel f\left(u^{†}\right)-u^{†}\parallel +(1-(1-k){\alpha }_n-{\gamma }_n)\parallel x_{n-1}-u^{†}\parallel +{\beta }_n\parallel V{u}^{†}-u^{†}\parallel +{\gamma }_n\parallel x_n-u^{†}\parallel .\hfill \end{array}$$

By condition (ii), without loss of generality, we assume that ${\beta }_n\le {\alpha }_n$ for all $n\ge 1$. Hence,

$$\begin{array}{ll}\parallel x_n-u^{†}\parallel & \le [1-(1-k){\displaystyle \frac{{\alpha }_n}{1-{\gamma }_n}}]\parallel x_{n-1}-u^{†}\parallel +{\displaystyle \frac{{\alpha }_n}{1-{\gamma }_n}}\parallel f\left(u^{†}\right)-u^{†}\parallel +{\displaystyle \frac{{\beta }_n}{1-{\gamma }_n}}\parallel V{u}^{†}-u^{†}\parallel \hfill \\ & \le [1-(1-k){\displaystyle \frac{{\alpha }_n}{1-{\gamma }_n}}]\parallel x_{n-1}-u^{†}\parallel +{\displaystyle \frac{{\alpha }_n}{1-{\gamma }_n}}\parallel f\left(u^{†}\right)-u^{†}\parallel +{\displaystyle \frac{{\alpha }_n}{1-{\gamma }_n}}\parallel V{u}^{†}-u^{†}\parallel \hfill \\ & =[1-(1-k){\displaystyle \frac{{\alpha }_n}{1-{\gamma }_n}}]\parallel x_{n-1}-u^{†}\parallel +{\displaystyle \frac{{\alpha }_n}{1-{\gamma }_n}}(\parallel f\left(u^{†}\right)-u^{†}\parallel +\parallel V{u}^{†}-u^{†}\parallel )\hfill \\ & \le max\{\parallel x_{n-1}-u^{†}\parallel ,(\parallel f\left(u^{†}\right)-u^{†}\parallel +\parallel V{u}^{†}-u^{†}\parallel )/(1-k)\}.\hfill \end{array}$$

(6)

Thus, $\{x_n\}$, $\{T{x}_n\},\{S{x}_n\},\{y_n\},\{Q{x}_n\}$ and $\{A{x}_n\}$ are all bounded.

Set $q=J_{{\varsigma }_2}^{M_2}(u^{†}-{\varsigma }_2{A}_2{u}^{†})$ and $z_n=J_{{\varsigma }_1}^{M_1}(y_n-{\varsigma }_1{A}_1{y}_n)$. Then $z_n=Q{x}_n,\forall n\ge 1$. By virtue of Lemma 4, we get

$$\begin{array}{ll}\parallel y_n{-q\parallel }^2& =\parallel J_{{\varsigma }_2}^{M_2}(x_n-{\varsigma }_2{A}_2{x}_n)-J_{{\varsigma }_2}^{M_2}(u^{†}-{\varsigma }_2{A}_2{u}^{†}){\parallel }^2\hfill \\ & \le \parallel x_n-u^{†}-{\varsigma }_2(A_2{x}_n-A_2{u}^{†}){\parallel }^2\hfill \\ & \le \parallel x_n-u^{†}{\parallel }^2-2{\varsigma }_2({\zeta }_2-c^2{\varsigma }_2){\parallel A_2{x}_n-A_2{u}^{†}\parallel }^2,\hfill \end{array}$$

(7)

and

$$\begin{array}{ll}\parallel z_n-u^{†}{\parallel }^2& =\parallel J_{{\varsigma }_1}^{M_1}(y_n-{\varsigma }_1{A}_1{y}_n)-J_{{\varsigma }_1}^{M_1}(q-{\varsigma }_1{A}_{1}q){\parallel }^2\hfill \\ & \le \parallel y_n-q-{\varsigma }_1(A_1{y}_n-A_{1}q){\parallel }^2\hfill \\ & \le \parallel y_n{-q\parallel }^2-2{\varsigma }_1({\zeta }_1-c^2{\varsigma }_1){\parallel A_1{y}_n-A_{1}q\parallel }^2.\hfill \end{array}$$

(8)

Substituting (7) for (8), we derive

$$\parallel z_n-u^{†}{\parallel }^2\le \parallel x_n-u^{†}{\parallel }^2-2{\varsigma }_2({\zeta }_2-c^2{\varsigma }_2)\parallel A_2{x}_n-A_2{u}^{†}{\parallel }^2-2{\varsigma }_1({\zeta }_1-c^2{\varsigma }_1){\parallel A_1{y}_n-A_{1}q\parallel }^2.$$

(9)

In view of (5) and (9), we obtain

$$\begin{array}{lll}\parallel x_n-u^{†}{\parallel }^2& =\hfill & {\alpha }_n\langle f\left(x_{n-1}\right)-u^{†},j(x_n-u^{†})\rangle +{\delta }_n\langle x_{n-1}-u^{†},j(x_n-u^{†})\rangle \hfill \\ & & +{\beta }_n\langle V{x}_{n-1}-u^{†},j(x_n-u^{†})\rangle +{\gamma }_n\langle (1-\mu )S{x}_n+\mu z_n-u^{†},j(x_n-u^{†})\rangle \hfill \\ & \le \hfill & {\alpha }_n\langle f\left(x_{n-1}\right)-u^{†},j(x_n-u^{†})\rangle +{\delta }_n\parallel x_{n-1}-u^{†}\parallel \parallel x_n-u^{†}\parallel \hfill \\ & & +{\beta }_n\parallel V{x}_{n-1}-u^{†}\parallel \parallel x_n-u^{†}\parallel +{\gamma }_n\parallel (1-\mu )S{x}_n+\mu z_n-u^{†}\parallel \parallel x_n-u^{†}\parallel \hfill \\ & \le \hfill & {\delta }_n\parallel x_{n-1}-u^{†}\parallel \parallel x_n-u^{†}\parallel +{\gamma }_n[(1-\mu )\parallel x_n-u^{†}\parallel +\mu \parallel z_n-u^{†}\parallel ]\parallel x_n-u^{†}\parallel \hfill \\ & & +{\alpha }_n[\langle f\left(x_{n-1}\right)-f\left(u^{†}\right),j(x_n-u^{†})\rangle +\langle f\left(u^{†}\right)-u^{†},j(x_n-u^{†})\rangle ]\hfill \\ & & +{\beta }_n(\parallel V{x}_{n-1}-V{u}^{†}\parallel +\parallel V{u}^{†}-u^{†}\parallel )\parallel x_n-u^{†}\parallel \hfill \\ & \le \hfill & {\delta }_n\parallel x_{n-1}-u^{†}\parallel \parallel x_n-u^{†}\parallel +{\gamma }_n[(1-\mu )\parallel x_n-u^{†}\parallel +\mu \parallel z_n-u^{†}\parallel ]\parallel x_n-u^{†}\parallel \hfill \\ & & +{\alpha }_n[k\parallel x_{n-1}-u^{†}\parallel \parallel x_n-u^{†}\parallel +\langle f\left(u^{†}\right)-u^{†},j(x_n-u^{†})\rangle ]\hfill \\ & & +{\beta }_n(\parallel x_{n-1}-u^{†}\parallel +\parallel V{u}^{†}-u^{†}\parallel )\parallel x_n-u^{†}\parallel \hfill \\ & \le \hfill & {\alpha }_n\langle f\left(u^{†}\right)-u^{†},j(x_n-u^{†})\rangle +[1-(1-k){\alpha }_n-{\gamma }_n]/2(\parallel x_{n-1}-u^{†}{\parallel }^2+\parallel x_n-u^{†}{\parallel }^2)\hfill \\ & & +{\beta }_n\parallel V{u}^{†}-u^{†}\parallel \parallel x_n-u^{†}\parallel +{\gamma }_n\parallel x_n-u^{†}{\parallel }^2-{\gamma }_n\mu [{\varsigma }_2({\zeta }_2-c^2{\varsigma }_2){\parallel A_2{x}_n-A_2{u}^{†}\parallel }^2\hfill \\ & & +{\varsigma }_1({\zeta }_1-c^2{\varsigma }_1)\parallel A_1{y}_n-A_{1}q{\parallel }^2].\hfill \end{array}$$

(10)

It follows that

$$\begin{array}{l}{\gamma }_n\mu [{\varsigma }_2({\zeta }_2-c^2{\varsigma }_2)\parallel A_2{x}_n-A_2{u}^{†}{\parallel }^2+{\varsigma }_1({\zeta }_1-c^2{\varsigma }_1)\parallel A_1{y}_n-A_{1}q{\parallel }^2]\hfill \\ \le [1-(1-k){\alpha }_n-{\gamma }_n]/2(\parallel x_{n-1}-u^{†}{\parallel }^2+\parallel x_n-u^{†}{\parallel }^2)+{\alpha }_n\langle f\left(u^{†}\right)-u^{†},j(x_n-u^{†})\rangle \hfill \\ +{\beta }_n\parallel V{u}^{†}-u^{†}\parallel \parallel x_n-u^{†}\parallel -(1-{\gamma }_n){\parallel x_n-u^{†}\parallel }^2\hfill \\ \le {\alpha }_n\parallel f\left(u^{†}\right)-u^{†}\parallel \parallel x_n-u^{†}\parallel +[1-(1-k){\alpha }_n-{\gamma }_n]/2(\parallel x_{n-1}-u^{†}{\parallel }^2+\parallel x_n-u^{†}{\parallel }^2)\hfill \\ +{\beta }_n\parallel V{u}^{†}-u^{†}\parallel \parallel x_n-u^{†}\parallel .\hfill \end{array}$$

By the assumptions (ii) and (iii), we conclude

$$\underset{n\to \infty }{lim}\parallel A_2{x}_n-A_2{u}^{†}\parallel =0\mathrm{and}\underset{n\to \infty }{lim}\parallel A_1{y}_n-A_{1}q\parallel =0.$$

(11)

Utilizing Lemma 1 and Proposition 1, we have

$$\begin{array}{ll}\parallel y_n{-q\parallel }^2& =\parallel J_{{\varsigma }_2}^{M_2}(x_n-{\varsigma }_2{A}_2{x}_n)-J_{{\varsigma }_2}^{M_2}(u^{†}-{\varsigma }_2{A}_2{u}^{†}){\parallel }^2\hfill \\ & \le \langle (x_n-{\varsigma }_2{A}_2{x}_n)-(u^{†}-{\varsigma }_2{A}_2{u}^{†}),j(y_n-q)\rangle \hfill \\ & =\langle x_n-u^{†},j(y_n-q)\rangle +{\varsigma }_2\langle A_2{u}^{†}-A_2{x}_n,j(y_n-q)\rangle \hfill \\ & \le [\parallel x_n-u^{†}{\parallel }^2+\parallel y_n-u^{†}{\parallel }^2-g_1(\parallel x_n-y_n-(u^{†}-q)\parallel \left)\right]/2+{\varsigma }_2\parallel A_2{u}^{†}-A_2{x}_n\parallel \parallel y_n-q\parallel .\hfill \end{array}$$

It follows that

$$\parallel y_n{-q\parallel }^2\le \parallel x_n-u^{†}{\parallel }^2-g_1(\parallel x_n-y_n-(u^{†}-q)\parallel )+2{\varsigma }_2\parallel A_2{u}^{†}-A_2{x}_n\parallel \parallel y_n-q\parallel .$$

(12)

Similarly,

$$\begin{array}{ll}\parallel z_n-u^{†}{\parallel }^2& =\parallel J_{{\varsigma }_1}^{M_1}(y_n-{\varsigma }_1{A}_1{y}_n)-J_{{\varsigma }_1}^{M_1}(q-{\varsigma }_1{A}_{1}q){\parallel }^2\hfill \\ & \le \langle (y_n-{\varsigma }_1{A}_1{y}_n)-(q-{\varsigma }_1{A}_{1}q),j(z_n-u^{†})\rangle \hfill \\ & =\langle y_n-q,j(z_n-q)\rangle +{\varsigma }_1\langle A_{1}q-A_1{y}_n,j(z_n-u^{†})\rangle \hfill \\ & \le {\displaystyle \frac{1}{2}}[\parallel y_n{-q\parallel }^2+\parallel z_n-u^{†}{\parallel }^2-g_2(\parallel y_n-z_n+(u^{†}-q)\parallel \left)\right]+{\varsigma }_1\parallel A_{1}q-A_1{y}_n\parallel \parallel z_n-u^{†}\parallel ,\hfill \end{array}$$

which implies that

$$\parallel z_n-u^{†}{\parallel }^2\le \parallel y_n{-q\parallel }^2-g_2(\parallel y_n-z_n+(u^{†}-q)\parallel )+2{\varsigma }_1\parallel A_{1}q-A_1{y}_n\parallel \parallel z_n-u^{†}\parallel .$$

(13)

Substituting (12) for (13), we get

$$\begin{array}{ll}\parallel z_n-u^{†}{\parallel }^2\le & \parallel x_n-u^{†}{\parallel }^2-g_1(\parallel x_n-y_n-(u^{†}-q)\parallel )-g_2(\parallel y_n-z_n+(u^{†}-q)\parallel )\hfill \\ & +2{\varsigma }_2\parallel A_2{u}^{†}-A_2{x}_n\parallel \parallel y_n-q\parallel +2{\varsigma }_1\parallel A_{1}q-A_1{y}_n\parallel \parallel z_n-u^{†}\parallel .\hfill \end{array}$$

(14)

From (10) and (14), we have

$$\begin{array}{lll}\parallel x_n-u^{†}{\parallel }^2& \le \hfill & {\alpha }_n\langle f\left(u^{†}\right)-u^{†},j(x_n-u^{†})\rangle +[1-(1-k){\alpha }_n-{\gamma }_n](\parallel x_{n-1}-u^{†}{\parallel }^2+\parallel x_n-u^{†}{\parallel }^2)/2\hfill \\ & & +{\beta }_n\parallel V{u}^{†}-u^{†}\parallel \parallel x_n-u^{†}\parallel +{\gamma }_n(\parallel x_n-u^{†}{\parallel }^2+(1-\mu )\parallel x_n-u^{†}{\parallel }^2+\mu \parallel z_n-u^{†}{\parallel }^2)/2\hfill \\ & \le \hfill & {\alpha }_n\langle f\left(u^{†}\right)-u^{†},j(x_n-u^{†})\rangle +[1-(1-k){\alpha }_n-{\gamma }_n](\parallel x_{n-1}-u^{†}{\parallel }^2+\parallel x_n-u^{†}{\parallel }^2)/2\hfill \\ & & +{\beta }_n\parallel V{u}^{†}-u^{†}\parallel \parallel x_n-u^{†}\parallel +{\gamma }_n\parallel x_n-u^{†}{\parallel }^2-{\displaystyle \frac{{\gamma }_n\mu }{2}}[g_1(\parallel x_n-y_n-(u^{†}-q)\parallel )\hfill \\ & & +g_2(\parallel y_n-z_n+(u^{†}-q)\parallel \left)\right]+{\gamma }_n\mu ({\varsigma }_2\parallel A_2{u}^{†}-A_2{x}_n\parallel \parallel y_n-q\parallel \hfill \\ & & +{\varsigma }_1\parallel A_{1}q-A_1{y}_n\parallel \parallel z_n-u^{†}\parallel ).\hfill \end{array}$$

It follows that

$$\begin{array}{l}{\displaystyle \frac{{\gamma }_n\mu }{2}}[g_1(\parallel x_n-y_n-(u^{†}-q)\parallel )+g_2(\parallel y_n-z_n+(u^{†}-q)\parallel \left)\right]\hfill \\ \le {\alpha }_n\langle f\left(u^{†}\right)-u^{†},j(x_n-u^{†})\rangle +[1-(1-k){\alpha }_n-{\gamma }_n](\parallel x_{n-1}-u^{†}{\parallel }^2+\parallel x_n-u^{†}{\parallel }^2)/2\hfill \\ +{\beta }_n\parallel V{u}^{†}-u^{†}\parallel \parallel x_n-u^{†}\parallel -(1-{\gamma }_n)\parallel x_n-u^{†}{\parallel }^2+{\gamma }_n\mu ({\varsigma }_2\parallel A_2{u}^{†}-A_2{x}_n\parallel \parallel y_n-q\parallel \hfill \\ +{\varsigma }_1\parallel A_{1}q-A_1{y}_n\parallel \parallel z_n-u^{†}\parallel )\hfill \\ \le {\alpha }_n\parallel f\left(u^{†}\right)-u^{†}\parallel \parallel x_n-u^{†}\parallel +[1-(1-k){\alpha }_n-{\gamma }_n](\parallel x_{n-1}-u^{†}{\parallel }^2+\parallel x_n-u^{†}{\parallel }^2)/2\hfill \\ +{\beta }_n\parallel V{u}^{†}-u^{†}\parallel \parallel x_n-u^{†}\parallel +{\varsigma }_2\parallel A_2{u}^{†}-A_2{x}_n\parallel \parallel y_n-q\parallel +{\varsigma }_1\parallel A_{1}q-A_1{y}_n\parallel \parallel z_n-u^{†}\parallel .\hfill \end{array}$$

This together with conditions (ii) and (iii) implies that

$$\underset{n\to \infty }{lim}{g}_1(\parallel x_n-y_n-(u^{†}-q)\parallel )=0\mathrm{and}\underset{n\to \infty }{lim}{g}_2(\parallel y_n-z_n+(u^{†}-q)\parallel )=0.$$

Hence,

$$\underset{n\to \infty }{lim}\parallel x_n-y_n-(u^{†}-q)\parallel =0\mathrm{and}\underset{n\to \infty }{lim}\parallel y_n-z_n+(u^{†}-q)\parallel =0.$$

(15)

In light of (15), we have

$$\parallel x_n-z_n\parallel \le \parallel x_n-y_n-(u^{†}-q)\parallel +\parallel y_n-z_n+(u^{†}-q)\parallel \to 0,$$

which means that

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

(16)

Note that

$$\begin{array}{ll}{\gamma }_n\parallel x_n-A{x}_n\parallel & =\parallel {\alpha }_n(f\left(x_{n-1}\right)-x_n)+{\delta }_n(x_{n-1}-x_n)+{\beta }_n(V{x}_{n-1}-x_n)\parallel \hfill \\ & \le {\alpha }_n\parallel f\left(x_{n-1}\right)-x_n\parallel +{\delta }_n\parallel x_{n-1}-x_n\parallel +{\beta }_n\parallel V{x}_{n-1}-x_n\parallel .\hfill \end{array}$$

Thus,

$$\underset{n\to \infty }{lim}\parallel x_n-A{x}_n\parallel =0.$$

(17)

Also, observe that

$$\mu \parallel S{x}_n-x_n\parallel =\parallel A{x}_n-x_n-(1-\mu )(Q{x}_n-x_n)\parallel \le \parallel A{x}_n-x_n\parallel +\parallel Q{x}_n-x_n\parallel .$$

In terms of (16) and (17), we obtain

$$\underset{n\to \infty }{lim}\parallel x_n-S{x}_n\parallel =0.$$

(18)

Since $S=(1-\alpha )I+\alpha T$ with $0<\alpha <min\{1,{\displaystyle \frac{\lambda }{c^2}}\}$, it is easy from (3.15) that

$$\underset{n\to \infty }{lim}\parallel x_n-T{x}_n\parallel =0.$$

Define a net $\{u_t\}$ by $u_t=(1-t)A{u}_t+tf\left(u_t\right)$. So,

$$u_t-x_n=t(f\left(u_t\right)-x_n)+(1-t)(A{u}_t-x_n).$$

(19)

It follows that

$$\begin{array}{ll}\parallel u_t-x_n{\parallel }^2& \le 2t\langle f\left(u_t\right)-x_n,j(u_t-x_n)\rangle +{(1-t)}^2{\parallel A{u}_t-x_n\parallel }^2\hfill \\ & \le 2t\langle f\left(u_t\right)-x_n,j(u_t-x_n)\rangle +{(1-t)}^2[\parallel A{u}_t-A{x}_n\parallel +\parallel A{x}_n-x_n{\parallel ]}^2\hfill \\ & \le 2t\langle f\left(u_t\right)-x_n,j(u_t-x_n)\rangle +{(1-t)}^2[\parallel u_t-x_n\parallel +\parallel A{x}_n-x_n{\parallel ]}^2\hfill \\ & ={(1-t)}^2[\parallel u_t-x_n{\parallel }^2+\parallel A{x}_n-x_n{\parallel }^2+2\parallel u_t-x_n\parallel \parallel A{x}_n-x_n\parallel ]\hfill \\ & +2t\langle f\left(u_t\right)-x_n,j(u_t-x_n)\rangle ,\hfill \end{array}$$

that is,

$$\begin{array}{lll}\parallel u_t-x_n{\parallel }^2& \le \hfill & \parallel A{x}_n-x_n\parallel (2\parallel u_t-x_n\parallel +\parallel A{x}_n-x_n\parallel )+2t\parallel u_t-x_n{\parallel }^2\hfill \\ & & +2t\langle f\left(u_t\right)-u_t,j(u_t-x_n)\rangle +{(1-t)}^2{\parallel u_t-x_n\parallel }^2\hfill \\ & =\hfill & (1+t^2){\parallel u_t-x_n\parallel }^2+2t\langle f\left(u_t\right)-u_t,j(u_t-x_n)\rangle \hfill \\ & & +\parallel A{x}_n-x_n\parallel (2\parallel u_t-x_n\parallel +\parallel A{x}_n-x_n\parallel ).\hfill \end{array}$$

It follows that

$$\langle u_t-f\left(u_t\right),j(u_t-x_n)\rangle \le {\displaystyle \frac{t}{2}}\parallel u_t-x_n{\parallel }^2+{\displaystyle \frac{1}{2t}}(2\parallel u_t-x_n\parallel +\parallel A{x}_n-x_n\parallel )\parallel A{x}_n-x_n\parallel .$$

(20)

Letting $n\to \infty$ in (20), from (17), we have

$$\overline{{lim}_{n\to \infty }}\langle u_t-f\left(u_t\right),j(u_t-x_n)\rangle \le {\displaystyle \frac{t{M}_0}{2}}$$

(21)

where $M_0$ is a constant such that $\parallel u_t-x_n{\parallel }^2\le M_0,\forall n\ge 0,\forall t\in (0,1)$. By Lemma 6, $u_t\to x^{*}\in \Omega$, which solves $\langle (I-f)x^{*},j(x^{*}-x)\rangle \le 0,\forall x\in \Omega$. Letting $t\to 0^{+}$ in (21), we deduce

$$\overline{{lim}_{n\to \infty }}\langle x^{*}-f\left(x^{*}\right),j(x^{*}-x_n)\rangle \le 0.$$

Putting $u^{†}=x^{*}$ in (10), we obtain

$$\begin{array}{ll}\parallel x_n-x^{*}{\parallel }^2& \le {\alpha }_n\langle f\left(x^{*}\right)-x^{*},j(x_n-x^{*})\rangle +[1-(1-k){\alpha }_n-{\gamma }_n](\parallel x_{n-1}-x^{*}{\parallel }^2+\parallel x_n-x^{*}{\parallel }^2)/2\hfill \\ & +{\beta }_n\parallel V{x}^{*}-x^{*}\parallel \parallel x_n-x^{*}\parallel +{\gamma }_n{\parallel x_n-x^{*}\parallel }^2.\hfill \end{array}$$

Consequently, we have

$$\begin{array}{ll}\parallel x_n-x^{*}{\parallel }^2\le & {\displaystyle \frac{2{\alpha }_n}{1+(1-k){\alpha }_n-{\gamma }_n}}\langle f\left(x^{*}\right)-x^{*},j(x_n-x^{*})\rangle +{\displaystyle \frac{1-(1-k){\alpha }_n-{\gamma }_n}{1+(1-k){\alpha }_n-{\gamma }_n}}{\parallel x_{n-1}-x^{*}\parallel }^2\hfill \\ & +{\displaystyle \frac{2{\beta }_n}{1+(1-k){\alpha }_n-{\gamma }_n}}\parallel V{x}^{*}-x^{*}\parallel \parallel x_n-x^{*}\parallel \hfill \\ \hfill =& (1-{\upsilon }_n){\parallel x_{n-1}-x^{*}\parallel }^2+{\upsilon }_n{\varrho }_n,\hfill \end{array}$$

(22)

where ${\upsilon }_n={\displaystyle \frac{2(1-k){\alpha }_n}{1+(1-k){\alpha }_n-{\gamma }_n}}$ and

$${\varrho }_n={\displaystyle \frac{{\beta }_n}{(1-k){\alpha }_n}}\parallel V{x}^{*}-x^{*}\parallel \parallel x_n-x^{*}\parallel +{\displaystyle \frac{1}{1-k}}\langle f\left(x^{*}\right)-x^{*},j(x_n-x^{*})\rangle .$$

Now, observe that

$$(1-k){\alpha }_n={\displaystyle \frac{2(1-k){\alpha }_n}{2}}\le {\displaystyle \frac{2(1-k){\alpha }_n}{1-{\gamma }_n+(1-k){\alpha }_n}}={\upsilon }_n.$$

Observe that $\overline{{lim}_{n\to \infty }}{\varrho }_n\le 0$. With the help of Lemma 7, we get $x_n\to x^{*}$. Moreover, putting $u^{†}=x^{*}$ and $q=y^{*}=J_{{\varsigma }_2}^{M_2}(I-{\varsigma }_2{A}_2)x^{*}$ in (15), we obtain

$$\underset{n\to \infty }{lim}\parallel x_n-y_n-(x^{*}-y^{*})\parallel =0.$$

Note that

$$\parallel y_n-y^{*}\parallel \le \parallel x_n-y_n-(x^{*}-y^{*})\parallel +\parallel x^{*}-x_n\parallel .$$

So, it follows that $y_n\to y^{*}$ as $n\to \infty$. Consequently, $(x^{*},y^{*})$ is a solution of (1) by Lemma 2. □

Corollary 1.

Let X be a uniformly convex and 2-uniformly smooth Banach space and $\varnothing \ne C\subset X$ a closed convex set. Let $M:C\to 2^X$ be an m-accretive operator and $A:C\to X$ be a ζ-inverse-strongly accretive operator. Let $f:C\to C$ be a contraction with coefficient $k\in [0,1)$. Let $V:C\to C$ be a nonexpansive operator and $T:C\to C$ be a λ-strict pseudocontraction with $\Omega :=F\left(T\right)\cap F\left(Q\right)\ne \varnothing$, where the operator $Q=J_{{\varsigma }_1}^M(I-{\varsigma }_{1}A)J_{{\varsigma }_2}^M(I-{\varsigma }_{2}A)$ and $0<{\varsigma }_i<{\displaystyle \frac{\zeta }{c^2}}(i=1,2)$. Assume that the sequences $\{{\alpha }_n\}\subset (0,1),\{{\beta }_n\}\subset (0,1)$ and $\{{\gamma }_n\}\subset (0,1)$ satisfy

(i) 

${\alpha }_n+{\delta }_n+{\beta }_n+{\gamma }_n=1(\forall n\ge 1)$;

(ii) 

${\alpha }_n\to 0$ and ${\displaystyle \frac{{\beta }_n}{{\alpha }_n}}\to 0$;

(iii) 

${\gamma }_n\to 1$;

(iv) 

${\sum }_{n=1}^{\infty }{\alpha }_n=\infty$.

Given $x_0\in C$, compute the sequences $\{x_n\}$ and $\{y_n\}$ such that

$$\left\{\begin{array}{c}{y}_n=J_{{\varsigma }_2}^M(x_n-{\varsigma }_{2}A{x}_n),\hfill \\ x_n={\delta }_n{x}_{n-1}+{\alpha }_{n}f\left(x_{n-1}\right)+{\beta }_{n}V{x}_{n-1}+{\gamma }_n[\mu S{x}_n+(1-\mu )J_{{\varsigma }_1}^M(y_n-{\varsigma }_{1}A{y}_n)],\forall n\ge 1,\hfill \end{array}\right.$$

where $Sx=(1-\alpha )x+\alpha Tx,\forall x\in C$ with $0<\alpha <min\{1,{\displaystyle \frac{\lambda }{c^2}}\}$ and $\mu \in (0,1)$. Then $x_n\to x^{*}$, $y_n\to y^{*}$ and

(a) 

$(x^{*},y^{*})$ solves the GSVI (2);

(b) 

$x^{*}$ solves the variational inequality: $\langle (I-f)x^{*},j(u-x^{*})\rangle \ge 0,\forall u\in \Omega$.

Corollary 2.

Let H be a Hilbert space and $\varnothing \ne C\subset H$ a closed convex set. Let $M:C\to 2^H$ be a maximal monotone operator and $A:C\to H$ be a ζ-inverse-strongly monotone operator. Let $f:C\to C$ be a contraction with coefficient $k\in [0,1)$. Let $V:C\to C$ be a nonexpansive operator and $T:C\to C$ be a λ-strict pseudocontraction with $\Omega :=F\left(T\right)\cap F\left(Q\right)\ne \varnothing$, where the operator $Q=J_{{\varsigma }_1}^M(I-{\varsigma }_{1}A)J_{{\varsigma }_2}^M(I-{\varsigma }_{2}A)$ and $0<{\varsigma }_i<{\displaystyle \frac{\zeta }{c^2}}(i=1,2)$. Assume that the sequences $\{{\alpha }_n\}\subset (0,1),\{{\beta }_n\}\subset (0,1)$ and $\{{\gamma }_n\}\subset (0,1)$ satisfy

(i) 

${\alpha }_n+{\delta }_n+{\beta }_n+{\gamma }_n=1(\forall n\ge 1)$;

(ii) 

${\alpha }_n\to 0$ and ${\displaystyle \frac{{\beta }_n}{{\alpha }_n}}\to 0$;

(iii) 

${\gamma }_n\to 1$;

(iv) 

${\sum }_{n=1}^{\infty }{\alpha }_n=\infty$.

Given $x_0\in C$, compute the sequences $\{x_n\}$ and $\{y_n\}$ such that

$$\left\{\begin{array}{c}{y}_n=J_{{\varsigma }_2}^M(x_n-{\varsigma }_{2}A{x}_n),\hfill \\ x_n={\delta }_n{x}_{n-1}+{\alpha }_{n}f\left(x_{n-1}\right)+{\beta }_{n}V{x}_{n-1}+{\gamma }_n[\mu S{x}_n+(1-\mu )J_{{\varsigma }_1}^M(y_n-{\varsigma }_{1}A{y}_n)],\forall n\ge 1,\hfill \end{array}\right.$$

where $Sx=(1-\alpha )x+\alpha Tx,\forall x\in C$ with $0<\alpha <min\{1,{\displaystyle \frac{\lambda }{c^2}}\}$ and $\mu \in (0,1)$. Then $x_n\to x^{*}$, $y_n\to y^{*}$ and

(a) 

$(x^{*},y^{*})$ solves the GSVI (2);

(b) 

$x^{*}$ solves the variational inequality: $\langle (I-f)x^{*},j(u-x^{*})\rangle \ge 0,\forall u\in \Omega$.

4. Conclusions

In this paper, we consider the GSVI (1) with the hierarchical variational inequality (HVI) constraint for a strict pseudocontraction in a uniformly convex and 2-uniformly smooth Banach space. By utilizing the equivalence between the GSVI (1) and the fixed point problem, we construct an implicit composite viscosity approximation method for solving the GSVI (1) with the HVI constraint for strict pseudocontractions. We prove the strong convergence of the proposed algorithm to a solution of the GSVI (1) with the HVI constraint for strict pseudocontraction under very mild conditions. Note that our algorithm (4) is an implicit manner. This brings us a natural question: could we construct an explicit algorithm with strong convergence?