System of Variational Inclusions and Fixed Points of Pseudocontractive Mappings in Banach Spaces

Abstract

The purpose of this paper is to solve the general system of variational inclusions (GSVI) with hierarchical variational inequality (HVI) constraint, for an infinite family of continuous pseudocontractive mappings in Banach spaces. By utilizing the equivalence between the GSVI and the fixed point problem, we construct an implicit multiple-viscosity approximation method for solving the GSVI. Under very mild conditions, we prove the strong convergence of the proposed method to a solution of the GSVI with the HVI constraint, for infinitely many pseudocontractions.

1. Introduction

Let C be a nonempty closed convex subset of a real Hilbert space H equipped with inner product $⟨·,·⟩$ and norm $\parallel ·\parallel$. Let $S:C\to C$ be nonexpansive, with its fixed point set $F\left(S\right)\ne \varnothing$. Let $A,B:C\to H$ be $\alpha$-inverse-strongly monotone and $\beta$-inverse-strongly monotone, respectively. In 2008, Ceng et al. [1] introduced and considered the following general system of variational inequality problems of finding $(x^{\ast },y^{\ast })\in C\times C$, such that

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{c}⟨\rho A{y}^{\ast }+x^{\ast }-y^{\ast },x-x^{\ast }⟩\ge 0,\forall x\in C,\hfill \\ ⟨\eta B{x}^{\ast }+y^{\ast }-x^{\ast },x-y^{\ast }⟩\ge 0,\forall x\in C.\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(1)

They first transformed problem (1) into a fixed point problem for the mapping $G=P_C(I-\rho A)P_C(I-\eta B)$, and then proved strong convergence of the following relaxed extragradient method for solving the problem (1), and the fixed point problem of S:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{cc}\hfill y_n& =P_C(x_n-\eta B{x}_n),n\ge 0,\hfill \\ \hfill x_{n+1}& ={\alpha }_{n}u+{\beta }_n{x}_n+{\gamma }_{n}S{P}_C(y_n-\rho A{y}_n),\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(2)

where $\rho \in (0,2\alpha ),\eta \in (0,2\beta )$, and $\left\{{\alpha }_n\right\}$, $\left\{{\beta }_n\right\}$, $\left\{{\gamma }_n\right\}$ are sequences in $[0,1]$.

Let E be a real Banach space with the dual $E^{\ast }$ and C a nonempty closed convex subset of E. A self-mapping $f:C\to C$ is said to be k-Lipschitz on C if $k\in {\mathbf{R}}_{+}=[0,+\infty )$ and $\parallel f\left(x\right)-f\left(y\right)\parallel \le k\parallel x-y\parallel$ for all $x,y\in C$. If f is k-Lipschitz with $k<1$, then f is called a k-contraction mapping (or a contraction mapping with coefficient k). A self-mapping $f:C\to C$ is said to be nonexpansive if it is Lipschitz with $k=1$. Also, recall that a mapping T with domain $D\left(T\right)$ and range $R\left(T\right)$ in E is called pseudocontractive if the following inequality holds

$$\begin{array}{c}\hfill \parallel x-y\parallel \le \parallel x-y+r\left(\right(I-T)x-(I-T\left)y\right)\parallel ,\forall x,y\in D\left(T\right),r>0,\end{array}$$

which is equivalent to the inequality (see [2]) that for each $x,y\in D\left(T\right)$ there exists $j(x-y)\in J(x-y)$ such that

$$\begin{array}{c}\hfill ⟨Tx-Ty,j(x-y)⟩\le {\parallel x-y\parallel }^2.\end{array}$$

(3)

It is known that the class of pseudocontractive mappings is an important and significant generator of nonexpansive mappings [3]. Moreover, interest in pseudocontractive mappings stems mainly from their firm connection with the class of accretive mappings.

Let $A_1,A_2:C\to E$ and $M_1,M_2:C\to 2^E$ be nonlinear mappings. The general system of variational inclusions (GSVI) is to find $(x^{\ast },y^{\ast })\in C\times C$ such that

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{c}0\in x^{\ast }-y^{\ast }+{\rho }_1(A_1{y}^{\ast }+M_1{x}^{\ast }),\hfill \\ 0\in y^{\ast }-x^{\ast }+{\rho }_2(A_2{x}^{\ast }+M_2{y}^{\ast }),\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(4)

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

In 2010, Qin et al. [4] introduced a relaxed extragradient-type method for solving GSVI (4), and proved a strong convergence theorem for the proposed method (for its related results in the literature, see, e.g., [1,5,6,7,8,9,10,11,12,13,14,15,16,17,18]). Furthermore, Aoyama et al. [19] considered the following variational inequality: Find $x^{\ast }\in C$, such that

$$\begin{array}{c}\hfill ⟨A{x}^{\ast },j(x-x^{\ast })⟩\ge 0,\forall x\in C.\end{array}$$

(5)

They proved that the problem (5) is equivalent to a fixed point problem; that is, the element $x^{\ast }\in C$ is a solution of problem (5) if and only if $x^{\ast }\in C$ satisfies the following equation:

$$\begin{array}{c}\hfill x^{\ast }={\mathsf{\Pi }}_C(x^{\ast }-\eta A{x}^{\ast }),\end{array}$$

(6)

where $\eta >0$ is a constant and ${\mathsf{\Pi }}_C$ is a sunny nonexpansive retraction from E onto C. In particular, if $E=H$ a Hilbert space, then ${\mathsf{\Pi }}_C$ coincides with the metric projection $P_C$ from H onto C. Recently, many authors have studied the problem of finding a common element of the set of fixed points of nonlinear mappings and the set of solutions to variational inequalities by iterative methods (see, e.g., [1,2,3,5,6,8,9,10,12,14,15,16,18,19,20,21,22,23,24]). In particular, Ceng et al. [22] introduced an implicit viscosity approximation method for computing approximate fixed points of a pseudocontractive mapping T, and derived strong convergence of the proposed implicit method to a point in $F\left(T\right)$, which solves a certain variational inequality.

The purpose of this paper is to solve the GSVI (4) with the hierarchical variational inequality (HVI) constraint, for an infinite family of continuous pseudocontractive mappings ${\left\{T_n\right\}}_{n=1}^{\infty }$ in a uniformly convex and two-uniformly smooth Banach space E. By utilizing the equivalence between the GSVI (4) and the fixed point problem, we construct an implicit multiple-viscosity approximation method for solving the GSVI (4) with the HVI constraint, for infinitely many pseudocontractions ${\left\{T_n\right\}}_{n=1}^{\infty }$. Under very mild conditions, we prove the strong convergence of the proposed method to a solution of the GSVI (4) with the HVI constraint, for infinitely many pseudocontractions ${\left\{T_n\right\}}_{n=1}^{\infty }$. Our results improve and extend the corresponding results announced by some others; for example, Yao et al. [13] and Ceng et al. [22].

2. Preliminaries

Let E be a real Banach space with dual $E^{\ast }$. Throughout this paper, we write $x_n\rightharpoonup x$ (respectively, $x_n\to x$) to indicate that the sequence $\left\{x_n\right\}$ converges weakly (respectively, strongly) to x. Let C be a nonempty closed convex subset of E. Recall that a mapping $T:C\to E$ is said to be

(a)

accretive if, for each $x,y\in C$, $\exists j(x-y)\in J(x-y)$ such that $⟨Tx-Ty,j(x-y)⟩\ge 0$, where J is the normalized duality mapping;

(b)

$\alpha$-strongly accretive if, for each $x,y\in C$, $\exists j(x-y)\in J(x-y)$ such that $⟨Tx-Ty,j(x-y)⟩\ge {\alpha \parallel x-y\parallel }^2\mathrm{for}\mathrm{some}\alpha \in (0,1)$;

(c)

$\beta$-inverse-strongly accretive if, for each $x,y\in C$, $\exists j(x-y)\in J(x-y)$ such that $⟨Tx-Ty,j(x-y)⟩\ge {\beta \parallel Tx-Ty\parallel }^2\mathrm{for}\mathrm{some}\beta >0$.

Let $U=\{x\in E:\parallel x\parallel =1\}$ be the unit sphere of E. Then E is said to be strictly convex if for any $x,y\in U$, ${\displaystyle x\ne y\Rightarrow \parallel \frac{x+y}{2}\parallel <1}$. It is also said to be uniformly convex if for each $ϵ\in (0,2]$, there exists $\delta >0$ such that for any $x,y\in U$, ${\displaystyle \parallel x-y\parallel \ge ϵ\Rightarrow \parallel \frac{x+y}{2}\parallel \le 1-\delta }$. It is known that a uniformly convex Banach space is reflexive and strictly convex. Also, it is known that if a Banach space E is reflexive, then E is strictly convex if and only if $E^{\ast }$ is smooth, as well as that E is smooth if and only if $E^{\ast }$ is strictly convex. A Banach space E is said to have a Gateaux differentiable norm if the limit ${\displaystyle \underset{t\to 0}{\lim }\frac{\parallel x+ty\parallel -\parallel x\parallel }{t}}$ exists for each $x,y\in U$ and, in this case, we call E smooth. E is said to have a uniformly Fréchet differentiable norm if the limit is attained uniformly for $x,y\in U$ and, in this case, we call E uniformly smooth. E is also said to have a Fréchet differentiable norm if for each $x\in U$, the limit is attained uniformly for $y\in U$ and, in this case, we call E strongly smooth. The modulus of smoothness of E is defined by

$$\begin{array}{c}\hfill \varrho \left(\tau \right)=\sup \{\frac{1}{2}(\parallel x+y\parallel +\parallel x-y\parallel )-1:x,y\in E,\parallel x\parallel =1,\parallel y\parallel =\tau \},\end{array}$$

where $\varrho :[0,\infty )\to [0,\infty )$ is a function. It is known that E is uniformly smooth if and only if ${\displaystyle \underset{\tau \to 0}{\lim }\frac{\varrho \left(\tau \right)}{\tau }=0}$. Let q be a fixed real number with $1<q\le 2$. A Banach space E is said to be q-uniformly smooth if there exists a constant $\kappa >0$, such that $\varrho \left(\tau \right)\le \kappa {\tau }^q$ for all $\tau >0$.

Let q be a real number with $1<q\le 2$ and let E be a Banach space. Then E is q-uniformly smooth if and only if there exists a constant $c>0$ such that

$$\begin{array}{c}\hfill {\parallel x+y\parallel }^q{+\parallel x-y\parallel }^q\le {2(\parallel x\parallel }^q+{\parallel cy\parallel }^q),\forall x,y\in E.\end{array}$$

The best constant c in the above inequality is called the q-uniformly smooth constant of E; see [11] for more details. Note that no Banach space is q-uniformly smooth for $q>2$; see [25].

For $q>1$, the generalized duality mapping $J_q:E\to 2^{E^{\ast }}$ is defined by

$$\begin{array}{c}\hfill J_q\left(x\right)=\{\phi \in E^{\ast }:⟨x,\phi ⟩={\parallel x\parallel }^q,\parallel \phi \parallel ={\parallel x\parallel }^{q-1}\},\forall x\in E.\end{array}$$

In particular, $J=J_2$ is called the normalized duality mapping. It is known that $J_q\left(x\right)={\parallel x\parallel }^{q-2}J\left(x\right),\forall x\in E$. If E is a Hilbert space, then $J=I$ the identity mapping. Recall that

(1)

if E is smooth, then J is single-valued and norm-to-weak${}^{\ast }$ continuous on E;

(2)

if E is uniformly smooth, then J is single-valued and norm-to-norm uniformly continuous on bounded subsets of E;

(3)

all Hilbert spaces, $L^p$ (or $l^p$) spaces ($p\ge 2$) and the Sobolev spaces $W_m^p(p\ge 2)$, are two-uniformly smooth, while $L^p$ (or $l^p$) and $W_m^p$ spaces ($1<p\le 2$) are p-uniformly smooth;

(4)

typical examples of both uniformly convex and uniformly smooth Banach space are $L^p$, where $p>1$. More precisely, $L^p$ is $\min \{p,2\}$-uniformly smooth for any $p>1$.

Proposition 1

([26]). Let E be a smooth and uniformly convex Banach space, and let $r>0$. Then there exists a strictly increasing, continuous and convex function $g:[0,2r]\to \mathbf{R},g\left(0\right)=0$ such that $g(\parallel x-y\parallel )\le {\parallel x\parallel }^2-2⟨x,j\left(y\right)⟩+{\parallel y\parallel }^2,\forall x,y\in B_r$, where $B_r=\{x\in E:\parallel x\parallel \le r\}$.

Proposition 2

([27]). If E is a two-uniformly smooth Banach space, then ${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2⟨y,j\left(x\right)⟩+2{\parallel cy\parallel }^2,\forall x,y\in E$, where c is the two-uniformly smooth constant of E. In particular, if E is a Hilbert space, then the duality pairing $⟨·,·⟩$ reduces to the inner product, $j=I$ the identity mapping, and $c=1/\sqrt{2}$.

Let D be a subset of C and let $\mathsf{\Pi }$ be a mapping of C into D. Then $\mathsf{\Pi }$ is said to be sunny if $\mathsf{\Pi }\left[\mathsf{\Pi }\right(x)+t(x-\mathsf{\Pi }\left(x\right)\left)\right]=\mathsf{\Pi }\left(x\right)$, whenever $\mathsf{\Pi }\left(x\right)+t(x-\mathsf{\Pi }(x\left)\right)\in C$ for $x\in C$ and $t\ge 0$. A mapping $\mathsf{\Pi }$ of C into itself is called a retraction if ${\mathsf{\Pi }}^2=\mathsf{\Pi }$. If a mapping $\mathsf{\Pi }$ of C into itself is a retraction, then $\mathsf{\Pi }\left(z\right)=z$ for each $z\in R\left(\mathsf{\Pi }\right)$, where $R\left(\mathsf{\Pi }\right)$ is the range of $\mathsf{\Pi }$. A subset D of C is called a sunny nonexpansive retract of C if there exists a sunny nonexpansive retraction from C onto D.

Proposition 3

([28]). Let C be a nonempty closed convex subset of a smooth Banach space E, D be a nonempty subset of C and Π be a retraction of C onto D. Then the following are equivalent:

(i) 

Π is sunny and nonexpansive;

(ii) 

${\parallel \mathsf{\Pi }\left(x\right)-\mathsf{\Pi }\left(y\right)\parallel }^2\le ⟨x-y,j(\mathsf{\Pi }\left(x\right)-\mathsf{\Pi }\left(y\right))⟩,\forall x,y\in C$;

(iii) 

$⟨x-\mathsf{\Pi }\left(x\right),j(y-\mathsf{\Pi }(x\left)\right)⟩\le 0,\forall x\in C,y\in D$.

Let C be a nonempty closed convex subset of a uniformly convex and uniformly smooth Banach space E and let T be a nonexpansive mapping of C into itself with the fixed point set $F\left(T\right)\ne \varnothing$. Then the set $F\left(T\right)$ is a sunny nonexpansive retract of C; see [29].

Proposition 4

([30]). Let C be a nonempty closed convex subset of a Banach space E and $T:C\to C$ be a continuous and strong pseudocontraction mapping. Then, T has a unique fixed point in C.

Recall that a possibly multivalued operator $M\subset E\times E$ with domain $D\left(M\right)$ and range $R\left(M\right)$ in E is accretive if, for each $x_i\in D\left(M\right)$ and $y_i\in M{x}_i(i=1,2)$, there exists $j(x_1-x_2)\in J(x_1-x_2)$ such that $⟨y_1-y_2,j(x_1-x_2)⟩\ge 0$. An accretive operator M is said to satisfy the range condition if $\overline{D\left(M\right)}\subset R(I+rM)$ for all $r>0$. An accretive operator M is m-accretive if $R(I+rM)=E$ for each $r>0$. If M is an accretive operator which satisfies the range condition, then we can define, for each $r>0$, a mapping $J_r^M:R(I+rM)\to D\left(M\right)$ by $J_r^M={(I+rM)}^{-1}$, which is called the resolvent of M. It is well known that $J_r^M$ is nonexpansive and $F\left(J_r^M\right)=M^{-1}0,\forall r>0$; see [31]. Hence, $F\left(J_r^M\right)=M^{-1}0=\{x\in D\left(M\right):0\in Mx\}$. If $M^{-1}0\ne \varnothing$, then the inclusion $0\in Mx$ is solvable. We below present some lemmas which will be used in the sequel. Some of them are known, and others are not hard to prove.

Lemma 1

([24]). Let C be a nonempty closed convex subset of a smooth Banach space E and $M:C\to 2^E$ be an m-accretive mapping. Then for any given $r>0$, the inequality holds: $\parallel J_r^{M}x-J_r^M{y\parallel }^2\le ⟨x-y,j(J_r^{M}x-J_r^{M}y)⟩,\forall x,y\in E$. This means that $J_r^M:E\to C$ is nonexpansive.

Lemma 2

([24]). Let $M:C\to 2^E$ be an m-accretive mapping and $A:C\to E$ is a mapping. Then $x^{\ast }\in C$ is a solution of the variational inclusion $0\in Ax+Mx$ if and only if $x^{\ast }=J_{\rho }^M(x^{\ast }-\rho A{x}^{\ast })$, for all $\rho >0$, that is, $\mathrm{VI}(C,A,M)=F\left(J_{\rho }^M(I-\rho A)\right),\forall \rho >0$, where $\mathrm{VI}(C,A,M)$ denotes the set of solutions to this variational inclusion.

Lemma 3

([24]). Let $M_1,M_2:C\to 2^E$ be two m-accretive mappings and $A_1,A_2:C\to E$ be two mappings. For given $x^{\ast },y^{\ast }\in C,(x^{\ast },y^{\ast })$ is a solution of the GSVI (1.4) if and only if $x^{\ast }$ is a fixed point of the mapping $Q:=J_{{\rho }_1}^{M_1}(I-{\rho }_1{A}_1)J_{{\rho }_2}^{M_2}(I-{\rho }_2{A}_2)$, where $y^{\ast }=J_{{\rho }_2}^{M_2}(I-{\rho }_2{A}_2)x^{\ast }$.

Lemma 4

([32]). Let C be a nonempty closed convex subset of a strictly convex Banach space E. Let $T_1,T_2:C\to E$ be nonexpansive mappings with $F\left(T_1\right)\cap F\left(T_2\right)\ne \varnothing$. Define a mapping $S:C\to E$ by $Sx=\nu T_{1}x+(1-\nu )T_{2}x,$ $\forall x\in C$, where ν is a constant in $(0,1)$. Then S is nonexpansive and $F\left(S\right)=F\left(T_1\right)\cap F\left(T_2\right)$.

Lemma 5

([24]). Let C be a nonempty closed convex subset of a two-uniformly smooth Banach space E. Let the mapping $A:C\to E$ be α-inverse-strongly accretive. Then,

$${\parallel (I-\lambda A)x-(I-\lambda A)y\parallel }^2\le {\parallel x-y\parallel }^2+2\lambda (c^2\lambda -\alpha ){\parallel Ax-Ay\parallel }^2.$$

In particular, if $0\le \lambda \le \frac{\alpha }{c^2}$, then $I-\lambda A$ is nonexpansive.

Lemma 6

([24]). Let C be a nonempty closed convex subset of a two-uniformly smooth Banach space E. Let $M_1,M_2:C\to 2^E$ be two m-accretive mappings and $A_i:C\to E$ be ${\zeta }_i$-inverse-strongly accretive for $i=1,2$. Let the mapping $Q:C\to C$ be defined as $Q:=J_{{\rho }_1}^{M_1}(I-{\rho }_1{A}_1)J_{{\rho }_2}^{M_2}(I-{\rho }_2{A}_2)$. If $0\le {\rho }_i\le \frac{{\zeta }_i}{c^2}$ for $i=1,2$, then $Q:C\to C$ is nonexpansive.

Lemma 7

([33]). Let J be the normalized duality mapping on a real Banach space E. Then for all $x,y\in E$, the inequality holds: ${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2⟨y,j(x+y)⟩,\forall j(x+y)\in J(x+y)$.

Lemma 8

([33]). Let C be a nonempty closed convex subset of a uniformly smooth Banach space E, $A:C\to C$ be a nonexpansive mapping with $F\left(A\right)\ne \varnothing$, and $f:C\to C$ be a fixed contraction mapping. For each $t\in (0,1)$, let $z_t\in C$ be the unique fixed point of the contraction $C\ni z\mapsto tf\left(z\right)+(1-t)Az$ on C, that is, $z_t=tf\left(z_t\right)+(1-t)A{z}_t$. Then $\left\{z_t\right\}$ converges strongly to a point $x^{\ast }\in F\left(A\right)$, which solves the variational inequality: $⟨(I-f)x^{\ast },j(x^{\ast }-x)⟩\le 0,\forall x\in F\left(A\right)$.

Lemma 9

([33]). Let ${\left\{a_n\right\}}_{n=0}^{\infty }$ be a sequence of nonnegative real numbers satisfying $a_{n+1}\le (1-{\lambda }_n)a_n+{\lambda }_n{\sigma }_n,\forall n\ge 0$, where ${\left\{{\lambda }_n\right\}}_{n=0}^{\infty }$ and ${\left\{{\sigma }_n\right\}}_{n=0}^{\infty }$ are real sequences such that (i) ${\left\{{\lambda }_n\right\}}_{n=0}^{\infty }\subset (0,1),{\sum }_{n=0}^{\infty }{\lambda }_n=\infty$, and (ii) either ${\lim \sup }_{n\to \infty }{\sigma }_n\le 0$ or ${\sum }_{n=0}^{\infty }\left|{\lambda }_n{\sigma }_n\right|<\infty$. Then ${\lim }_{n\to \infty }{a}_n=0$.

Lemma 10

([34]). Let C be a nonempty closed convex subset of a Banach space E. Let $T_0,T_1,...$ be a sequence of mappings of C into itself. Suppose that ${\sum }_{n=1}^{\infty }\sup \{\parallel T_{n}x-T_{n-1}x\parallel :x\in C\}<\infty$. Then for each $y\in C,\left\{T_{n}y\right\}$ converges strongly to some point of C. Moreover, let T be a mapping of C into itself defined by $Ty={\lim }_{n\to \infty }{T}_{n}y$ for all $y\in C$. Then ${\lim }_{n\to \infty }\sup \{\parallel Tx-T_{n}x\parallel :x\in C\}=0$.

3. Main Results

Now, we are in a position to state and prove our main result.

Theorem 1.

Let C be a nonempty closed convex subset of a uniformly convex and two-uniformly smooth Banach space E. Let $M_1,M_2:C\to 2^E$ be two m-accretive mappings and $A_i:C\to E$ be ${\zeta }_i$-inverse-strongly accretive for $i=1,2$. Let the mapping $Q:C\to C$ be defined as $Q:=J_{{\rho }_1}^{M_1}(I-{\rho }_1{A}_1)J_{{\rho }_2}^{M_2}(I-{\rho }_2{A}_2)$, where $0<{\rho }_i<\frac{{\zeta }_i}{c^2},i=1,2$, for c the 2-uniformly smooth constant of E. Let $f:C\to C$ be a fixed contraction mapping with coefficient $k\in [0,1)$, $S:C\to C$ be a nonexpansive mapping, and ${\left\{T_n\right\}}_{n=1}^{\infty }$ be an infinite family of continuous pseudocontractive mappings of C into itself, such that $\mathsf{\Omega }:={\bigcap }_{n=1}^{\infty }F\left(T_n\right)\cap F\left(Q\right)\ne \varnothing$. Let $\left\{{\alpha }_n\right\},\left\{{\beta }_n\right\}$ and $\left\{{\gamma }_n\right\}$ be three real sequences in $(0,1)$ satisfying the following conditions:

(i) 

${\alpha }_n+{\beta }_n+{\gamma }_n\le 1,\forall n\ge 1$;

(ii) 

${\lim }_{n\to \infty }{\alpha }_n=0$ and ${\displaystyle \underset{n\to \infty }{\lim }\frac{{\beta }_n}{{\alpha }_n}=0}$;

(iii) 

${\lim }_{n\to \infty }{\gamma }_n=1$;

(iv) 

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

For arbitrary initial value $x_0\in C$, compute the sequences $\left\{x_n\right\}$ and $\left\{y_n\right\}$ such that

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

(7)

where $\mu \in (0,1)$, and $J_{{\rho }_i}^{M_i}$ is the resolvent of $M_i$ for $i=1,2$. Assume that ${\sum }_{n=1}^{\infty }{\sup }_{x\in D}\parallel T_{n+1}x-T_{n}x\parallel <\infty$ for any bounded subset D of C, let T be a mapping of C into itself defined by $Tx={\lim }_{n\to \infty }{T}_{n}x$ for all $x\in C$, and suppose that $F\left(T\right)={\bigcap }_{n=1}^{\infty }F\left(T_n\right)$. Then $\left\{x_n\right\}$ and $\left\{y_n\right\}$ converge strongly to $x^{\ast }(\in \mathsf{\Omega })$ and $y^{\ast }$, respectively, where

(a) 

$(x^{\ast },y^{\ast })$ solves the GSVI (4);

(b) 

$x^{\ast }$ solves the variational inequality: $⟨(I-f)x^{\ast },j(x^{\ast }-x)⟩\le 0,\forall x\in \mathsf{\Omega }$ (i.e., $x^{\ast }={\mathsf{\Pi }}_{\mathsf{\Omega }}f\left(x^{\ast }\right)$ where ${\mathsf{\Pi }}_{\mathsf{\Omega }}$ is a sunny nonexpansive retraction from C onto Ω).

Proof. 

Note that the mapping $Q:C\to C$ is defined as $Q:=J_{{\rho }_1}^{M_1}(I-{\rho }_1{A}_1)J_{{\rho }_2}^{M_2}(I-{\rho }_2{A}_2)$, where $0<{\rho }_i<\frac{{\zeta }_i}{c^2},i=1,2$, for c the two-uniformly smooth constant of E. So, by Lemma 6, we know that Q is nonexpansive. It is easy to see that the implicit iterative scheme (7) can be rewritten as

$$\begin{array}{c}\hfill x_n=(1-{\alpha }_n-{\beta }_n-{\gamma }_n)x_{n-1}+{\alpha }_{n}f\left(x_{n-1}\right)+{\beta }_{n}S{x}_{n-1}+{\gamma }_n(\mu T_n{x}_n+(1-\mu )Q{x}_n),\forall n\ge 1.\end{array}$$

(8)

Consider the mapping

$$\begin{array}{c}\hfill F_{n}x=(1-{\alpha }_n-{\beta }_n-{\gamma }_n)x_{n-1}+{\alpha }_{n}f\left(x_{n-1}\right)+{\beta }_{n}S{x}_{n-1}+{\gamma }_n(\mu T_{n}x+(1-\mu )Qx),\forall x\in C.\end{array}$$

Since $Q:C\to C$ is a nonexpansive mapping and $T_n:C\to C$ is a continuous pseudocontraction mapping, we deduce that all $x,y\in C$,

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill ⟨F_{n}x-F_{n}y,j(x-y)⟩& ={\gamma }_n[\mu ⟨T_{n}x-T_{n}y,j(x-y)⟩+(1-\mu )⟨Qx-Qy,j(x-y)⟩]\hfill \\ & \le {\gamma }_n{[\mu \parallel x-y\parallel }^2+(1-\mu )\parallel Qx-Qy\parallel \parallel x-y\parallel ]\le {\gamma }_n{\parallel x-y\parallel }^2.\hfill \end{array}\end{array}$$

Hence $F_n$ is a continuous and strong pseudocontraction mapping of C into itself (due to ${\gamma }_n\in (0,1)$). By Proposition 4, we know that for each $n\ge 1$ there exists a unique element $x_n\in C$, satisfying (8).

Next, we divide the rest of the proof into several steps.

Step 1. We claim that $\left\{x_n\right\},\left\{y_n\right\},\left\{f\left(x_n\right)\right\},\left\{S{x}_n\right\},\left\{T_n{x}_n\right\}$ and $\left\{Q{x}_n\right\}$ are bounded. Indeed, take an arbitrarily given $p\in \mathsf{\Omega }$. Then we have $T_{n}p=p$ and $Qp=p$. Putting $W_n:=\mu T_n+(1-\mu )Q$, we know that $W_n$ is a continuous pseudocontraction mapping of C into itself. Then it follows that $W_{n}p=p$ and

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel x_n{-p\parallel }^2& =⟨x_n-p,j(x_n-p)⟩\hfill \\ & \le (1-{\alpha }_n-{\beta }_n-{\gamma }_n)\parallel x_{n-1}-p\parallel \parallel x_n-p\parallel +{\alpha }_n\parallel f\left(x_{n-1}\right)-p\parallel \parallel x_n-p\parallel \hfill \\ & +{\beta }_n\parallel S{x}_{n-1}-p\parallel \parallel x_n-p\parallel +{\gamma }_n{\parallel x_n-p\parallel }^2,\hfill \end{array}\end{array}$$

which hence implies that

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel x_n-p\parallel & \le (1-{\alpha }_n-{\beta }_n-{\gamma }_n)\parallel x_{n-1}-p\parallel +{\alpha }_n(\parallel f\left(x_{n-1}\right)-f\left(p\right)\parallel +\parallel f\left(p\right)-p\parallel )\hfill \\ & +{\beta }_n(\parallel S{x}_{n-1}-Sp\parallel +\parallel Sp-p\parallel )+{\gamma }_n\parallel x_n-p\parallel \hfill \\ & \le (1-(1-k){\alpha }_n-{\gamma }_n)\parallel x_{n-1}-p\parallel +{\alpha }_n\parallel f\left(p\right)-p\parallel +{\beta }_n\parallel Sp-p\parallel +{\gamma }_n\parallel x_n-p\parallel .\hfill \end{array}\end{array}$$

Since ${\displaystyle \underset{n\to \infty }{\lim }\frac{{\beta }_n}{{\alpha }_n}=0}$, we may assume, without loss of generality, that ${\beta }_n\le {\alpha }_n$ for all $n\ge 1$. This implies that

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel x_n-p\parallel & \le [1-(1-k)\frac{{\alpha }_n}{1-{\gamma }_n}]\parallel x_{n-1}-p\parallel +\frac{{\alpha }_n}{1-{\gamma }_n}\parallel f\left(p\right)-p\parallel +\frac{{\beta }_n}{1-{\gamma }_n}\parallel Sp-p\parallel \hfill \\ & \le [1-(1-k)\frac{{\alpha }_n}{1-{\gamma }_n}]\parallel x_{n-1}-p\parallel +\frac{{\alpha }_n}{1-{\gamma }_n}(\parallel f\left(p\right)-p\parallel +\parallel Sp-p\parallel )\hfill \\ & \le \max \{\parallel x_{n-1}-p\parallel ,\frac{1}{1-k}(\parallel f\left(p\right)-p\parallel +\parallel Sp-p\parallel )\}.\hfill \end{array}\end{array}$$

(9)

By induction, we derive $\parallel x_n-p\parallel \le \max \{\parallel x_0-p\parallel ,\frac{1}{1-k}(\parallel f\left(p\right)-p\parallel +\parallel Sp-p\parallel )\},\forall n\ge 0$. So, $\left\{x_n\right\}$ is bounded. Observe that $\parallel f\left(x_n\right)\parallel \le \parallel f\left(x_n\right)-f\left(p\right)\parallel +\parallel f\left(p\right)\parallel \le k\parallel x_n-p\parallel +\parallel f\left(p\right)\parallel$ and $\parallel S{x}_n\parallel \le \parallel S{x}_n-Sp\parallel +\parallel Sp\parallel \le \parallel x_n-p\parallel +\parallel Sp\parallel$. This implies that $\left\{f\left(x_n\right)\right\}$ and $\left\{S{x}_n\right\}$ are bounded. Similarly, by the nonexpansivity of Q we know that $\left\{Q{x}_n\right\}$ is bounded. Note that ${\lim }_{n\to \infty }{\gamma }_n=1$. Hence there exist $n_0\ge 1$ and ${\epsilon }_0\in (0,1)$, such that ${\gamma }_n\ge {\epsilon }_0$ for all $n\ge n_0$. Consequently, we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\epsilon }_0\parallel W_n{x}_n\parallel & \le {\gamma }_n\parallel W_n{x}_n\parallel =\parallel x_n-(1-{\alpha }_n-{\beta }_n-{\gamma }_n)x_{n-1}-{\alpha }_{n}f\left(x_{n-1}\right)-{\beta }_{n}S{x}_{n-1}\parallel \hfill \\ & \le \parallel x_n\parallel +\parallel x_{n-1}\parallel +\parallel f\left(x_{n-1}\right)\parallel +\parallel S{x}_{n-1}\parallel .\hfill \end{array}\end{array}$$

This means that $\left\{W_n{x}_n\right\}$ is bounded. Since $W_n{x}_n=\mu T_n{x}_n+(1-\mu )Q{x}_n$, we get

$$\mu \parallel T_n{x}_n\parallel =\parallel W_n{x}_n-(1-\mu )Q{x}_n\parallel \le \parallel W_n{x}_n\parallel +(1-\mu )\parallel Q{x}_n\parallel \le \parallel W_n{x}_n\parallel +\parallel Q{x}_n\parallel .$$

Hence $\left\{T_n{x}_n\right\}$ is bounded. In addition, from Lemma 3 and $p\in \mathsf{\Omega }\subset F\left(Q\right)$, it follows that $(p,q)$ is a solution of GSVI (4) where $q=J_{{\rho }_2}^{M_2}(I-{\rho }_2{A}_2)p$. So, by Lemmas 1 and 5 we get $\parallel y_n\parallel \le \parallel J_{{\rho }_2}^{M_2}(I-{\rho }_2{A}_2)x_n-q\parallel +\parallel q\parallel \le \parallel x_n-p\parallel +\parallel q\parallel$. That is, $\left\{y_n\right\}$ is bounded.

Step 2. We show that $\parallel x_n-Q{x}_n\parallel \to 0$ and $\parallel x_n-T{x}_n\parallel \to 0$ as $n\to \infty$, where $T:C\to C$ is defined as $Tx={\lim }_{n\to \infty }{T}_{n}x,\forall x\in C$. For simplicity, put $q=J_{{\rho }_2}^{M_2}(p-{\rho }_2{A}_{2}p)$ and $z_n=J_{{\rho }_1}^{M_1}(y_n-{\rho }_1{A}_1{y}_n)$. Then $z_n=Q{x}_n,\forall n\ge 1$. From Lemmas 1 and 5, we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel y_n{-q\parallel }^2& =\parallel J_{{\rho }_2}^{M_2}(x_n-{\rho }_2{A}_2{x}_n)-J_{{\rho }_2}^{M_2}(p-{\rho }_2{A}_{2}p){\parallel }^2\hfill \\ & \le \parallel x_n{-p\parallel }^2-2{\rho }_2({\zeta }_2-c^2{\rho }_2){\parallel A_2{x}_n-A_{2}p\parallel }^2.\hfill \end{array}\end{array}$$

(10)

Similarly, we get

$$\begin{array}{c}\hfill \parallel z_n{-p\parallel }^2\le \parallel y_n{-q\parallel }^2-2{\rho }_1({\zeta }_1-c^2{\rho }_1){\parallel A_1{y}_n-A_{1}q\parallel }^2.\end{array}$$

(11)

Substituting (10) into (11), we obtain

$$\begin{array}{c}\hfill \parallel z_n{-p\parallel }^2\le \parallel x_n{-p\parallel }^2-2{\rho }_2({\zeta }_2-c^2{\rho }_2)\parallel A_2{x}_n-A_2{p\parallel }^2-2{\rho }_1({\zeta }_1-c^2{\rho }_1){\parallel A_1{y}_n-A_{1}q\parallel }^2.\end{array}$$

(12)

From (8) and (12), we conclude

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel x_n{-p\parallel }^2& \le (1-{\alpha }_n-{\beta }_n-{\gamma }_n)\parallel x_{n-1}-p\parallel \parallel x_n-p\parallel \hfill \\ & +{\alpha }_n[⟨f\left(x_{n-1}\right)-f\left(p\right),j(x_n-p)⟩+⟨f\left(p\right)-p,j(x_n-p)⟩]\hfill \\ & +{\beta }_n(\parallel S{x}_{n-1}-Sp\parallel +\parallel Sp-p\parallel )\parallel x_n-p\parallel \hfill \\ & +{\gamma }_n(\mu \parallel x_n-p\parallel +(1-\mu )\parallel z_n-p\parallel )\parallel x_n-p\parallel \hfill \\ & \le (1-{\alpha }_n-{\beta }_n-{\gamma }_n)\parallel x_{n-1}-p\parallel \parallel x_n-p\parallel \hfill \\ & +{\alpha }_n[k\parallel x_{n-1}-p\parallel \parallel x_n-p\parallel +⟨f\left(p\right)-p,j(x_n-p)⟩]+{\beta }_n(\parallel x_{n-1}-p\parallel \hfill \\ & +\parallel Sp-p\parallel )\parallel x_n-p\parallel +{\gamma }_n\frac{(\mu \parallel x_n-p\parallel +(1-\mu )\parallel z_n-{p\parallel )}^2+{\parallel x_n-p\parallel }^2}{2}\hfill \\ & \le [1-(1-k){\alpha }_n-{\gamma }_n]\frac{\parallel x_{n-1}{-p\parallel }^2+{\parallel x_n-p\parallel }^2}{2}+{\alpha }_n⟨f\left(p\right)-p,j(x_n-p)⟩\hfill \\ & +{\beta }_n\parallel Sp-p\parallel \parallel x_n-p\parallel +{\gamma }_n\parallel x_n{-p\parallel }^2-{\gamma }_n(1-\mu )[{\rho }_2({\zeta }_2-c^2{\rho }_2){\parallel A_2{x}_n-A_{2}p\parallel }^2\hfill \\ & +{\rho }_1({\zeta }_1-c^2{\rho }_1)\parallel A_1{y}_n-A_{1}q{\parallel }^2],\hfill \end{array}\end{array}$$

(13)

which together with ${\alpha }_n+{\beta }_n+{\gamma }_n\le 1$, immediately yields

$$\begin{array}{c}{\gamma }_n(1-\mu )[{\rho }_2({\zeta }_2-c^2{\rho }_2)\parallel A_2{x}_n-A_2{p\parallel }^2+{\rho }_1({\zeta }_1-c^2{\rho }_1)\parallel A_1{y}_n-A_{1}q{\parallel }^2]\\ {\displaystyle \le [1-(1-k){\alpha }_n-{\gamma }_n]\frac{\parallel x_{n-1}{-p\parallel }^2+{\parallel x_n-p\parallel }^2}{2}+{\alpha }_n\parallel f\left(p\right)-p\parallel \parallel x_n-p\parallel +{\beta }_n\parallel Sp-p\parallel \parallel x_n-p\parallel .}\end{array}$$

From ${\alpha }_n\to 0$, ${\displaystyle \frac{{\beta }_n}{{\alpha }_n}\to 0}$, ${\gamma }_n\to 1$, ${\displaystyle {\rho }_i\in (0,\frac{{\zeta }_i}{c^2})}$ and the boundedness of $\left\{x_n\right\}$, we deduce that

$$\begin{array}{c}\hfill \underset{n\to \infty }{\lim }\parallel A_2{x}_n-A_{2}p\parallel =0\mathrm{and}\underset{n\to \infty }{\lim }\parallel A_1{y}_n-A_{1}q\parallel =0.\end{array}$$

(14)

Also, utilizing Lemma 1 and Proposition 1, we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel y_n{-q\parallel }^2& \le ⟨x_n-p,j(y_n-q)⟩+{\rho }_2⟨A_{2}p-A_2{x}_n,j(y_n-q)⟩\hfill \\ & \le \frac{1}{2}[\parallel x_n{-p\parallel }^2+\parallel y_n{-p\parallel }^2-g_1(\parallel x_n-y_n-(p-q)\parallel \left)\right]+{\rho }_2\parallel A_{2}p-A_2{x}_n\parallel \parallel y_n-q\parallel ,\hfill \end{array}\end{array}$$

which implies that

$$\begin{array}{c}\hfill \parallel y_n{-q\parallel }^2\le \parallel x_n{-p\parallel }^2-g_1(\parallel x_n-y_n-(p-q)\parallel )+2{\rho }_2\parallel A_{2}p-A_2{x}_n\parallel \parallel y_n-q\parallel .\end{array}$$

(15)

Similarly, we get

$$\begin{array}{c}\hfill \parallel z_n{-p\parallel }^2\le \parallel y_n{-q\parallel }^2-g_2(\parallel y_n-z_n+(p-q)\parallel )+2{\rho }_1\parallel A_{1}q-A_1{y}_n\parallel \parallel z_n-p\parallel .\end{array}$$

(16)

Substituting (15) into (16), we get

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

(17)

From (13) and (17), we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel x_n{-p\parallel }^2& \le [1-(1-k){\alpha }_n-{\gamma }_n]\frac{\parallel x_{n-1}{-p\parallel }^2+{\parallel x_n-p\parallel }^2}{2}+{\alpha }_n⟨f\left(p\right)-p,j(x_n-p)⟩\hfill \\ & +{\beta }_n\parallel Sp-p\parallel \parallel x_n-p\parallel +\frac{{\gamma }_n}{2}\{\parallel x_n{-p\parallel }^2+\mu \parallel x_n{-p\parallel }^2+(1-\mu )[\parallel x_n{-p\parallel }^2\hfill \\ & -g_1(\parallel x_n-y_n-(p-q)\parallel )-g_2(\parallel y_n-z_n+(p-q)\parallel )+2{\rho }_2\parallel A_{2}p-A_2{x}_n\parallel \parallel y_n-q\parallel \hfill \\ & +2{\rho }_1\parallel A_{1}q-A_1{y}_n\parallel \parallel z_n-p\parallel \left]\right\}\hfill \\ & =[1-(1-k){\alpha }_n-{\gamma }_n]\frac{\parallel x_{n-1}{-p\parallel }^2+{\parallel x_n-p\parallel }^2}{2}+{\alpha }_n⟨f\left(p\right)-p,j(x_n-p)⟩\hfill \\ & +{\beta }_n\parallel Sp-p\parallel \parallel x_n-p\parallel +{\gamma }_n\parallel x_n{-p\parallel }^2-\frac{{\gamma }_n(1-\mu )}{2}[g_1(\parallel x_n-y_n-(p-q)\parallel )\hfill \\ & +g_2(\parallel y_n-z_n+(p-q)\parallel \left)\right]+{\gamma }_n(1-\mu )({\rho }_2\parallel A_{2}p-A_2{x}_n\parallel \parallel y_n-q\parallel \hfill \\ & +{\rho }_1\parallel A_{1}q-A_1{y}_n\parallel \parallel z_n-p\parallel ),\hfill \end{array}\end{array}$$

which together with ${\alpha }_n+{\beta }_n+{\gamma }_n\le 1$, leads to

$$\begin{array}{c}\hfill \begin{array}{c}{\displaystyle \frac{{\gamma }_n(1-\mu )}{2}[g_1(\parallel x_n-y_n-(p-q)\parallel )+g_2(\parallel y_n-z_n+(p-q)\parallel \left)\right]}\hfill \\ {\displaystyle \le [1-(1-k){\alpha }_n-{\gamma }_n]\frac{\parallel x_{n-1}{-p\parallel }^2+{\parallel x_n-p\parallel }^2}{2}+{\alpha }_n\parallel f\left(p\right)-p\parallel \parallel x_n-p\parallel +{\beta }_n\parallel Sp-p\parallel \parallel x_n-p\parallel }\hfill \\ +{\rho }_2\parallel A_{2}p-A_2{x}_n\parallel \parallel y_n-q\parallel +{\rho }_1\parallel A_{1}q-A_1{y}_n\parallel \parallel z_n-p\parallel .\hfill \end{array}\end{array}$$

Since ${\alpha }_n\to 0$, ${\displaystyle \frac{{\beta }_n}{{\alpha }_n}\to 0}$ and ${\gamma }_n\to 1$ as $n\to \infty$ (from (14)), and by the boundedness of $\left\{y_n\right\}$ and $\left\{z_n\right\}$, we deduce that ${\lim }_{n\to \infty }{g}_1(\parallel x_n-y_n-(p-q)\parallel )=0$ and ${\lim }_{n\to \infty }{g}_2(\parallel y_n-z_n+(p-q)\parallel )=0$. Utilizing the properties of $g_1$ and $g_2$, we conclude that

$$\begin{array}{c}\hfill \underset{n\to \infty }{\lim }\parallel x_n-y_n-(p-q)\parallel =0\mathrm{and}\underset{n\to \infty }{\lim }\parallel y_n-z_n+(p-q)\parallel =0.\end{array}$$

(18)

From (18), we get $\parallel x_n-z_n\parallel \le \parallel x_n-y_n-(p-q)\parallel +\parallel y_n-z_n+(p-q)\parallel \to 0\mathrm{as}n\to \infty$. That is,

$$\begin{array}{c}\hfill \underset{n\to \infty }{\lim }\parallel x_n-Q{x}_n\parallel =0.\end{array}$$

(19)

Note that

$$\begin{array}{c}\hfill {\gamma }_n\parallel x_n-W_n{x}_n\parallel \le (1-{\alpha }_n-{\beta }_n-{\gamma }_n)\parallel x_{n-1}-x_n\parallel +{\alpha }_n\parallel f\left(x_{n-1}\right)-x_n\parallel +{\beta }_n\parallel S{x}_{n-1}-x_n\parallel .\end{array}$$

From ${\alpha }_n\to 0$, ${\displaystyle \frac{{\beta }_n}{{\alpha }_n}\to 0}$, ${\gamma }_n\to 1$, and the boundedness of $\left\{x_n\right\},\left\{f\left(x_n\right)\right\}$, and $\left\{S{x}_n\right\}$, we know that

$$\begin{array}{c}\hfill \underset{n\to \infty }{\lim }\parallel x_n-W_n{x}_n\parallel =0.\end{array}$$

(20)

Observe that $\mu \parallel T_n{x}_n-x_n\parallel =\parallel W_n{x}_n-x_n-(1-\mu )(Q{x}_n-x_n)\parallel \le \parallel W_n{x}_n-x_n\parallel +\parallel Q{x}_n-x_n\parallel$. In terms of (19) and (20), we obtain

$$\begin{array}{c}\hfill \underset{n\to \infty }{\lim }\parallel x_n-T_n{x}_n\parallel =0.\end{array}$$

(21)

It is easy to see that $\overline{\mathrm{conv}}\left\{x_n\right\}$ is a nonempty bounded closed convex subset of C, where $\overline{\mathrm{conv}}\left\{x_n\right\}$ is the closed convex hull of the set $\left\{x_n\right\}$. By assumption, we get ${\sum }_{n=1}^{\infty }{\sup }_{x\in \overline{\mathrm{conv}}\left\{x_n\right\}}$ $\parallel T_{n+1}x-T_{n}x\parallel <\infty$. By Lemma 10, we have ${\lim }_{n\to \infty }{\sup }_{x\in \overline{\mathrm{conv}}\left\{x_n\right\}}\parallel T_{n}x-Tx\parallel =0$. Therefore, by (21), we conclude that

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \underset{n\to \infty }{\lim \sup }\parallel x_n-T{x}_n\parallel & \le \underset{n\to \infty }{\lim \sup }(\parallel x_n-T_n{x}_n\parallel +\parallel T_n{x}_n-T{x}_n\parallel )\hfill \\ & \le \underset{n\to \infty }{\lim \sup }\parallel x_n-T_n{x}_n\parallel +\underset{n\to \infty }{\lim \sup }\underset{x\in \overline{\mathrm{conv}}\left\{x_n\right\}}{\sup }\parallel T_{n}x-Tx\parallel =0.\hfill \end{array}\end{array}$$

That is, ${\lim }_{n\to \infty }\parallel x_n-T{x}_n\parallel =0$.

Step 3. We claim that

$$\begin{array}{c}\hfill \underset{n\to \infty }{\lim \sup }⟨x^{\ast }-f\left(x^{\ast }\right),j(x^{\ast }-x_n)⟩\le 0,x^{\ast }\in \mathsf{\Omega },\end{array}$$

(22)

where $z_t$ is the fixed point of the mapping $z\mapsto tf\left(z\right)+(1-t)(\mu A+(1-\mu \left)Q\right)z$ with $A:={(2I-T)}^{-1}$, $x^{\ast }={\lim }_{t\to 0^{+}}{z}_t$ and $x^{\ast }$ solves the VI: $⟨(I-f)x^{\ast },j(x^{\ast }-x)⟩\le 0,\forall x\in \mathsf{\Omega }$.

Indeed, note that the mapping $T:C\to C$ is defined as $Tx:={\lim }_{n\to \infty }{T}_{n}x$ for all $x\in C$. By assumption, we have that $F\left(T\right)={\bigcap }_{n=1}^{\infty }F\left(T_n\right)$. Let us show that $T:C\to C$ is pseudocontractive and continuous. As a matter of fact, observe that for all $x,y\in C$, ${\lim }_{n\to \infty }\parallel T_{n}x-Tx\parallel =0$ and ${\lim }_{n\to \infty }\parallel T_{n}y-Ty\parallel =0$. Since each $T_n$ is pseudocontractive, we get

$$\begin{array}{c}\hfill ⟨Tx-Ty,j(x-y)⟩=\underset{n\to \infty }{\lim }⟨T_{n}x-T_{n}y,j(x-y)⟩\le {\parallel x-y\parallel }^2.\end{array}$$

This means that T is pseudocontractive. In order to show the continuity of T on C, we suppose that $u_n\to u$ as $n\to \infty$ with $u\in C$ and $\left\{u_n\right\}\subset C$. Let $D:=\overline{\mathrm{conv}}(\left\{u_n\right\}\cup \left\{u\right\})$, where $\overline{\mathrm{conv}}(\left\{u_n\right\}\cup \left\{u\right\})$ is the closed convex hull of the set $\left\{u_n\right\}\cup \left\{u\right\}$. Then D is a nonempty bounded closed convex subset of C. By assumption, we obtain ${\sum }_{n=1}^{\infty }{\sup }_{x\in D}\parallel T_{n+1}x-T_{n}x\parallel <\infty$ for such a subset D of C. So, by Lemma 10 we deduce that ${\lim }_{m\to \infty }{\sup }_{x\in D}\parallel T_{m}x-Tx\parallel =0$.

We now observe that, for each given $m,n\ge 1$,

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel T{u}_n-Tu\parallel & \le \parallel T{u}_n-T_m{u}_n\parallel +\parallel T_m{u}_n-T_{m}u\parallel +\parallel T_{m}u-Tu\parallel \hfill \\ & \le {\displaystyle \underset{x\in D}{\sup }}\parallel Tx-T_{m}x\parallel +\parallel T_m{u}_n-T_{m}u\parallel +\parallel T_{m}u-Tu\parallel .\hfill \end{array}\end{array}$$

Since each $T_m$ is continuous and $u_n\to u$ as $n\to \infty$, we have ${\lim }_{n\to \infty }\parallel T_m{u}_n-T_{m}u\parallel =0$, which together with the last inequality, implies that for each given $m\ge 1$

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \underset{n\to \infty }{\lim \sup }\parallel T{u}_n-Tu\parallel & \le \underset{n\to \infty }{\lim \sup }(\underset{x\in D}{\sup }\parallel Tx-T_{m}x\parallel +\parallel T_m{u}_n-T_{m}u\parallel +\parallel T_{m}u-Tu\parallel )\hfill \\ & \le \underset{x\in D}{\sup }\parallel Tx-T_{m}x\parallel +\underset{n\to \infty }{\lim \sup }\parallel T_m{u}_n-T_{m}u\parallel +\parallel T_{m}u-Tu\parallel \hfill \\ & =\underset{x\in D}{\sup }\parallel Tx-T_{m}x\parallel +\parallel T_{m}u-Tu\parallel .\hfill \end{array}\end{array}$$

Since ${\lim }_{m\to \infty }{\sup }_{x\in D}\parallel Tx-T_{m}x\parallel =0$ and ${\lim }_{m\to \infty }\parallel T_{m}u-Tu\parallel =0$, we obtain

$$\begin{array}{c}\hfill \underset{n\to \infty }{\lim \sup }\parallel T{u}_n-Tu\parallel \le \underset{m\to \infty }{\lim }\underset{x\in D}{\sup }\parallel Tx-T_{m}x\parallel +\underset{m\to \infty }{\lim }\parallel T_{m}u-Tu\parallel =0,\end{array}$$

that is, ${\lim }_{n\to \infty }\parallel T{u}_n-Tu\parallel =0$. This means that T is continuous on C.

Suppose $A:={(2I-T)}^{-1}$. Then A is nonexpansive and $F\left(A\right)=F\left(T\right)$ as a consequence of Theorem 6 of [35]. So it follows that $F\left(A\right)\cap F\left(Q\right)=F\left(T\right)\cap F\left(Q\right)={\bigcap }_{n=1}^{\infty }F\left(T_n\right)\cap F\left(Q\right)(=:\mathsf{\Omega })\ne \varnothing$. Also, we observe that

$$\begin{array}{c}\hfill \parallel x_n-A{x}_n\parallel =\parallel A{A}^{-1}{x}_n-A{x}_n\parallel \le \parallel A^{-1}{x}_n-x_n\parallel =\parallel (2I-T)x_n-x_n\parallel =\parallel x_n-T{x}_n\parallel .\end{array}$$

Since ${\lim }_{n\to \infty }\parallel x_n-T{x}_n\parallel =0$, we have ${\lim }_{n\to \infty }\parallel x_n-A{x}_n\parallel =0$. Meanwhile, from Lemma 4 it is easy to see that $\mu A+(1-\mu )Q$ is nonexpansive and $F(\mu A+(1-\mu )Q)=F\left(A\right)\cap F\left(Q\right)={\bigcap }_{n=1}^{\infty }F\left(T_n\right)\cap F\left(Q\right)(=:\mathsf{\Omega })\ne \varnothing$. Obviously, the mapping $z\mapsto tf\left(z\right)+(1-t)(\mu A+(1-\mu \left)Q\right)z$ is a contraction of C into itself for each $t\in (0,1)$. So, $z_t$ solves the fixed point equation $z_t=tf\left(z_t\right)+(1-t)(\mu A+(1-\mu )Q)z_t$. Then, we have

$$\begin{array}{c}\hfill z_t-x_n=(1-t)[\mu (A{z}_t-x_n)+(1-\mu )(Q{z}_t-x_n)]+t(f\left(z_t\right)-x_n).\end{array}$$

(23)

Thus, from Lemma 7 and (23), we obtain

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel z_t-x_n{\parallel }^2& \le {(1-t)}^2{\parallel \mu (A{z}_t-x_n)+(1-\mu )(Q{z}_t-x_n)\parallel }^2+2t⟨f\left(z_t\right)-x_n,j(z_t-x_n)⟩\hfill \\ & \le {(1-t)}^2\left\{\mu \right(\parallel z_t-x_n{\parallel }^2+2\parallel z_t-x_n\parallel \parallel A{x}_n-x_n\parallel +\parallel A{x}_n-x_n{\parallel }^2)\hfill \\ & +(1-\mu )(\parallel z_t-x_n{\parallel }^2+2\parallel z_t-x_n\parallel \parallel Q{x}_n-x_n\parallel +\parallel Q{x}_n-x_n{\parallel }^2)\}\hfill \\ & +2t⟨f\left(z_t\right)-x_n,j(z_t-x_n)⟩\hfill \\ & \le {(1-t)}^2\{\parallel z_t-x_n{\parallel }^2+\parallel A{x}_n-x_n\parallel (2\parallel z_t-x_n\parallel +\parallel A{x}_n-x_n\parallel )\hfill \\ & +\parallel Q{x}_n-x_n\parallel (2\parallel z_t-x_n\parallel +\parallel Q{x}_n-x_n\parallel \left)\right\}+2t⟨f\left(z_t\right)-x_n,j(z_t-x_n)⟩,\hfill \end{array}\end{array}$$

that is,

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel z_t-x_n{\parallel }^2& \le (1+t^2)\parallel z_t-x_n{\parallel }^2+\parallel A{x}_n-x_n\parallel (2\parallel z_t-x_n\parallel +\parallel A{x}_n-x_n\parallel )\hfill \\ & +\parallel Q{x}_n-x_n\parallel (2\parallel z_t-x_n\parallel +\parallel Q{x}_n-x_n\parallel )+2t⟨f\left(z_t\right)-z_t,j(z_t-x_n)⟩.\hfill \end{array}\end{array}$$

It follows that

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill ⟨z_t-f\left(z_t\right),j(z_t-x_n)⟩& \le \frac{t}{2}\parallel z_t-x_n{\parallel }^2+\frac{1}{2t}[\parallel A{x}_n-x_n\parallel (2\parallel z_t-x_n\parallel +\parallel A{x}_n-x_n\parallel )\hfill \\ & +\parallel Q{x}_n-x_n\parallel (2\parallel z_t-x_n\parallel +\parallel Q{x}_n-x_n\parallel \left)\right].\hfill \end{array}\end{array}$$

(24)

Letting $n\to \infty$ in (24), from $\parallel x_n-A{x}_n\parallel \to 0$ and $\parallel x_n-Q{x}_n\parallel \to 0$ as $n\to 0$, we have

$$\begin{array}{c}\hfill \underset{n\to \infty }{\lim \sup }⟨z_t-f\left(z_t\right),j(z_t-x_n)⟩\le \frac{t}{2}\ell \end{array}$$

(25)

where ℓ is a constant such that $\parallel z_t-x_n{\parallel }^2\le \ell$ for all $n\ge 0$ and $t\in (0,1)$. Utilizing Lemma 8 we deduce that $\left\{z_t\right\}$ converges strongly to a fixed point $x^{\ast }\in F(\mu A+(1-\mu )Q)=F\left(A\right)\cap F\left(Q\right)$ ($=\mathsf{\Omega }$), which solves the variational inequality: $⟨(I-f)x^{\ast },j(x^{\ast }-x)⟩\le 0,\forall x\in \mathsf{\Omega }$. Since j is norm-to-norm uniformly continuous on bounded sets of E, as $t\to 0^{+}$ in (25), we get (22).

Step 4. We claim that $x_n\to x^{\ast }$ and $y_n\to y^{\ast }$ as $n\to \infty$, where $(x^{\ast },y^{\ast })$ solves the GSVI (4). Indeed, putting $p=x^{\ast }$ in (13), we obtain

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel x_n-x^{\ast }{\parallel }^2& \le [1-(1-k){\alpha }_n-{\gamma }_n]\frac{\parallel x_{n-1}-x^{\ast }{\parallel }^2+{\parallel x_n-x^{\ast }\parallel }^2}{2}+{\alpha }_n⟨f\left(x^{\ast }\right)-x^{\ast },j(x_n-x^{\ast })⟩\hfill \\ & +{\beta }_n\parallel S{x}^{\ast }-x^{\ast }\parallel \parallel x_n-x^{\ast }\parallel +{\gamma }_n{\parallel x_n-x^{\ast }\parallel }^2,\hfill \end{array}\end{array}$$

which hence implies that

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel x_n-x^{\ast }{\parallel }^2& \le \frac{1-(1-k){\alpha }_n-{\gamma }_n}{1+(1-k){\alpha }_n-{\gamma }_n}{\parallel x_{n-1}-x^{\ast }\parallel }^2+\frac{2{\alpha }_n}{1+(1-k){\alpha }_n-{\gamma }_n}⟨f\left(x^{\ast }\right)-x^{\ast },j(x_n-x^{\ast })⟩\hfill \\ & +\frac{2{\beta }_n}{1+(1-k){\alpha }_n-{\gamma }_n}\parallel S{x}^{\ast }-x^{\ast }\parallel \parallel x_n-x^{\ast }\parallel \hfill \\ & =(1-{\lambda }_n){\parallel x_{n-1}-x^{\ast }\parallel }^2+{\lambda }_n{\sigma }_n,\hfill \end{array}\end{array}$$

(26)

where ${\displaystyle {\lambda }_n=\frac{2(1-k){\alpha }_n}{1+(1-k){\alpha }_n-{\gamma }_n}}$ and ${\sigma }_n=\frac{1}{1-k}⟨f\left(x^{\ast }\right)-x^{\ast },j(x_n-x^{\ast })⟩+\frac{{\beta }_n}{(1-k){\alpha }_n}\parallel S{x}^{\ast }-x^{\ast }\parallel \parallel x_n-x^{\ast }\parallel$.

Now, observe that ${\displaystyle (1-k){\alpha }_n=\frac{2(1-k){\alpha }_n}{2}\le \frac{2(1-k){\alpha }_n}{1-{\gamma }_n+(1-k){\alpha }_n}={\lambda }_n}$. Since ${\sum }_{n=1}^{\infty }{\alpha }_n=\infty$, we infer that ${\sum }_{n=1}^{\infty }{\lambda }_n=\infty$. Note that ${\displaystyle \underset{n\to \infty }{\lim }\frac{{\beta }_n}{{\alpha }_n}=0}$ and ${\lim \sup }_{n\to \infty }⟨f\left(x^{\ast }\right)-x^{\ast },j(x_n-x^{\ast })⟩\le 0$, due to (22). Thus, in terms of the boundedness of $\{\parallel x_n-x^{\ast }\parallel \}$, we have ${\lim \sup }_{n\to \infty }{\sigma }_n\le 0$. Therefore, applying Lemma 9 to (26) implies that $x_n\to x^{\ast }$ as $n\to \infty$. Moreover, putting $p=x^{\ast }$ and $q=y^{\ast }=J_{{\rho }_2}^{M_2}(I-{\rho }_2{A}_2)x^{\ast }$ in (18), we have ${\lim }_{n\to \infty }\parallel x_n-y_n-(x^{\ast }-y^{\ast })\parallel =0$. Also, since $\parallel y_n-y^{\ast }\parallel \le \parallel x_n-y_n-(x^{\ast }-y^{\ast })\parallel +\parallel x^{\ast }-x_n\parallel$, we know that $y_n\to y^{\ast }$ as $n\to \infty$. In addition, in terms of Lemma 3 and as $x^{\ast }\in \mathsf{\Omega }\subset F\left(Q\right)$, $(x^{\ast },y^{\ast })$ solves the GSVI (4). ☐