General iterative methods for systems of variational inequalities with the constraints of generalized mixed equilibria and fixed point problem of pseudocontractions

Abstract

In this paper, we introduce two general iterative methods (one implicit method and one explicit method) for finding a solution of a general system of variational inequalities (GSVI) with the constraints of finitely many generalized mixed equilibrium problems and a fixed point problem of a continuous pseudocontractive mapping in a Hilbert space. Then we establish strong convergence of the proposed implicit and explicit iterative methods to a solution of the GSVI with the above constraints, which is the unique solution of a certain variational inequality. The results presented in this paper improve, extend, and develop the corresponding results in the earlier and recent literature.

Introduction

Let C be a nonempty closed convex subset of a real Hilbert space H with inner product $\langle \cdot ,\cdot \rangle$ and induced norm $\parallel \cdot \parallel$. We denote by $P_C$ the metric projection of H onto C and by $Fix(S)$ the set of fixed points of the mapping S. Recall that a mapping $T:C\to H$ is nonexpansive if $\parallel Tx-Ty\parallel \le \parallel x-y\parallel$, $\mathrm{\forall }x,y\in C$. A mapping $T:C\to H$ is called pseudocontractive if

$$\langle Tx-Ty,x-y\rangle \le {\parallel x-y\parallel }^2,\mathrm{\forall }x,y\in C.$$

This inequality can be equivalently rewritten as

$${\parallel Tx-Ty\parallel }^2\le {\parallel x-y\parallel }^2+{\parallel (I-T)x-(I-T)y\parallel }^2,\mathrm{\forall }x,y\in C,$$

where I is the identity mapping.

$T:C\to H$ is said to be k-strictly pseudocontractive if there exists a constant $k\in [0,1)$ such that

$${\parallel Tx-Ty\parallel }^2\le {\parallel x-y\parallel }^2+k{\parallel (I-T)x-(I-T)y\parallel }^2,\mathrm{\forall }x,y\in C.$$

A mapping $V:C\to H$ is said to be l-Lipschitzian if there exists a constant $l\ge 0$ such that

$$\parallel Vx-Vy\parallel \le l\parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

A mapping $F:C\to H$ is called monotone if

$$\langle x-y,Fx-Fy\rangle \ge 0,\mathrm{\forall }x,y\in C,$$

and F is called α-inverse-strongly monotone if there exists a constant $\alpha >0$ such that

$$\langle x-y,Fx-Fy\rangle \ge \alpha {\parallel Fx-Fy\parallel }^2,\mathrm{\forall }x,y\in C.$$

If F is an α-inverse-strongly monotone mapping, then it is obvious that F is $\frac{1}{\alpha }$-Lipschitz continuous, that is, $\parallel Fx-Fy\parallel \le \frac{1}{\alpha }\parallel x-y\parallel$ for all $x,y\in C$.

A mapping $F:C\to H$ is called β-strongly monotone if there exists a constant $\beta >0$ such that

$$\langle x-y,Fx-Fy\rangle \ge \beta {\parallel x-y\parallel }^2,\mathrm{\forall }x,y\in C.$$

A linear operator $A:H\to H$ is said to be strongly positive on H if there exists a constant $\overline{\gamma }>0$ such that

$$\langle Ax,x\rangle \ge \overline{\gamma }{\parallel x\parallel }^2,\mathrm{\forall }x\in H.$$

Let $F:C\to H$ be a mapping. The classical variational inequality problem (VIP) is to find $x^{\ast }\in C$ such that

$$\langle F{x}^{\ast },x-x^{\ast }\rangle \ge 0,\mathrm{\forall }x\in C.$$ 1.1

We denote the set of solutions of VIP (1.1) by $VI(C,F)$.

In 2008, Ceng et al. [1] considered the following general system of variational inequalities (GSVI):

$$\{\begin{array}{c}\langle \lambda F_1{y}^{\ast }+x^{\ast }-y^{\ast },x-x^{\ast }\rangle \ge 0,\mathrm{\forall }x\in C,\hfill \\ \langle \nu F_2{x}^{\ast }+y^{\ast }-x^{\ast },x-y^{\ast }\rangle \ge 0,\mathrm{\forall }x\in C,\hfill \end{array}$$ 1.2

where $F_1$, $F_2$ are α-inverse-strongly monotone and β-inverse-strongly monotone, respectively, and $\lambda \in (0,2\alpha )$ and $\nu \in (0,2\beta )$ are two constants. Many iterative methods have been developed for solving GSVI (1.2); see [2–7] and the references therein.

Subsequently, Alofi et al. [8] also introduced two composite iterative algorithms based on the composite iterative methods in Ceng et al. [9] and Jung [10] for solving the problem of GSVI (1.2). Moreover, they showed strong convergence of the proposed algorithms to a common solution of these two problems.

Very recently, Kong et al. [11] established the strong convergence of two hybrid steepest-descent schemes to the same solution of GSVI (1.2), which is also a common solution of finitely many variational inclusions and a minimization problem.

Lemma 1.1

(see [12, Proposition 3.1])

Let C be a nonempty closed convex subset of a real Hilbert space H. For given $x^{\ast },y^{\ast }\in C$, $(x^{\ast },y^{\ast })$ is a solution of GSVI (1.3) for continuous monotone mappings $F_1$ and $F_2$ if and only if $x^{\ast }$ is a fixed point of the composite $R=F_{1,\lambda }{F}_{2,\nu }:H\to C$ of nonexpansive mappings $F_{1,\lambda }:H\to C$ and $F_{2,\nu }:H\to C$, where $y^{\ast }=F_{2,\nu }{x}^{\ast }$,

$$F_{1,\lambda }x=\{z\in C:\langle y-z,F_{1}z\rangle +\frac{1}{\lambda }\langle y-z,z-x\rangle \ge 0,\mathrm{\forall }y\in C\},$$

and

$$F_{2,\nu }x=\{z\in C:\langle y-z,F_{2}z\rangle +\frac{1}{\nu }\langle y-z,z-x\rangle \ge 0,\mathrm{\forall }y\in C\}.$$

For simplicity, we denote by $GSVI(C,F_1,F_2)$ the fixed point set of mapping R.

In the meantime, inspired by Ceng et al. [1], Jung [12] introduced a general system of variational inequalities (GSVI) for two continuous monotone mappings $F_1$ and $F_2$ of finding $(x^{\ast },y^{\ast })\in C\times C$ such that

$$\{\begin{array}{c}\langle \lambda F_1{x}^{\ast }+x^{\ast }-y^{\ast },x-x^{\ast }\rangle \ge 0,\mathrm{\forall }x\in C,\hfill \\ \langle \nu F_2{y}^{\ast }+y^{\ast }-x^{\ast },x-y^{\ast }\rangle \ge 0,\mathrm{\forall }x\in C,\hfill \end{array}$$ 1.3

where $\lambda ,\nu >0$ are two constants. In order to find an element of $Fix(R)\cap Fix(T)$, he proposed one implicit algorithm generating a net $\{x_t\}$:

$$x_t=(I-{\theta }_{t}A)T_{r_t}R{x}_t+{\theta }_t[t\gamma V{x}_t+(I-t\mu G)T_{r_t}R{x}_t],$$ 1.4

with $t\in (0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma l}\})$ and ${\theta }_t\in (0,min\{\frac{1}{2},{\parallel A\parallel }^{-1}\})$, and an explicit algorithm generating a sequence $\{x_n\}$:

$$\{\begin{array}{c}{y}_n={\alpha }_n\gamma V{x}_n+(I-{\alpha }_n\mu G)T_{r_n}R{x}_n,\hfill \\ x_{n+1}=(I-{\beta }_{n}A)T_{r_n}R{x}_n+{\beta }_n{y}_n,\mathrm{\forall }n\ge 0,\hfill \end{array}$$ 1.5

with $\{{\alpha }_n\}\subset [0,1]$, $\{{\beta }_n\}\subset (0,1]$, $\{r_n\}\subset (0,\mathrm{\infty })$, and $x_0\in C$ any initial guess, where $T_{r_t}x=\{z\in C:\langle y-z,Tz\rangle -\frac{1}{r_t}\langle y-z,(1+r_t)z-x\rangle \le 0,\mathrm{\forall }y\in C\}$ for $r_t\in (0,\mathrm{\infty })$, and $T_{r_n}x=\{z\in C:\langle y-z,Tz\rangle -\frac{1}{r_n}\langle y-z,(1+r_n)z-x\rangle \le 0,\mathrm{\forall }y\in C\}$ for $r_n\in (0,\mathrm{\infty })$. Moreover, he established strong convergence of the proposed iterative algorithms to an element $\tilde{x}\in Fix(R)\cap Fix(T)$, which uniquely solves the variational inequality

$$\langle (A-I)\tilde{x},\tilde{x}-p\rangle \le 0,\mathrm{\forall }p\in Fix(R)\cap Fix(T).$$

On the other hand, the generalized mixed equilibrium problem (GMEP) is to find $x\in C$ such that

$$\Theta (x,y)+\phi (y)-\phi (x)+\langle Bx,y-x\rangle \ge 0,\mathrm{\forall }y\in C.$$ 1.6

We denote the set of solutions of GMEP (1.6) by $GMEP(\Theta ,\phi ,B)$. GMEP (1.6) is very general in the sense that it includes many problems as special cases, namely optimization problems, variational inequalities, minimax problems, Nash equilibrium problems in noncooperative games, and others. For different aspects and solution methods, we refer to [13–18] and the references therein.

In this paper, we introduce implicit and explicit iterative methods for finding a solution of GSVI (1.3) with solutions belonging also to the common solution set ${\bigcap }_{i=1}^{N}GMEP({\Theta }_i,{\phi }_i,B_i)$ of finitely many generalized mixed equilibrium problems and the fixed point set of a continuous pseudocontractive mapping T. First, GSVI (1.3) and each generalized mixed equilibrium problem both are transformed into fixed point problems of nonexpansive mappings. Then we establish strong convergence of the proposed iterative methods to an element of ${\bigcap }_{i=1}^{N}GMEP({\Theta }_i,{\phi }_i,B_i)\cap GSVI(C,F_1,F_2)\cap Fix(T)$, which is the unique solution of a certain variational inequality.

Preliminaries and lemmas

Let H be a real Hilbert space, and let C be a nonempty closed convex subset of H. We write $x_n\to x$ and $x_n\rightharpoonup x$ to indicate the strong convergence of the sequence $\{x_n\}$ to x and the weak convergence of the sequence $\{x_n\}$ to x, respectively.

For every point $x\in H$, there exists a unique nearest point in C, denoted by $P_C(x)$, such that

$$\parallel x-P_C(x)\parallel \le \parallel x-y\parallel ,\mathrm{\forall }y\in C.$$

$P_C$ is called the metric projection of H onto C. It is well known that $P_C$ is nonexpansive and is characterized by the property

$$u=P_C(x)\iff \langle x-u,u-y\rangle \ge 0,\mathrm{\forall }x\in H,y\in C.$$ 2.1

In a Hilbert space H, the following equality holds:

$${\parallel x-y\parallel }^2={\parallel x\parallel }^2+{\parallel y\parallel }^2-2\langle x,y\rangle ,\mathrm{\forall }x,y\in H.$$ 2.2

The following lemma is an immediate consequence of an inner product.

Lemma 2.1

In a real Hilbert space H, there holds the following inequality:

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

Next we list some elementary conclusions for the MEP.

It is first assumed as in [19] that $\Theta :C\times C\to \mathbf{R}$ is a bifunction satisfying conditions (A1)–(A4) and $\phi :C\to \mathbf{R}$ is a lower semicontinuous and convex function with restriction (B1) or (B2), where

1. $\Theta (x,x)=0$ for all $x\in C$;

2. Θ is monotone, i.e., $\Theta (x,y)+\Theta (y,x)\le 0$ for any $x,y\in C$;

3. Θ is upper-hemicontinuous, i.e., for each $x,y,z\in C$,

   $$\underset{t\to 0^{+}}{lim\hspace{0.17em}sup}\Theta (tz+(1-t)x,y)\le \Theta (x,y);$$

4. $\Theta (x,\cdot )$ is convex and lower semicontinuous for each $x\in C$;

5. for $\mathrm{\forall }x\in H$ and $r>0$, there exists a bounded subset $D_x\subset C$ and $y_x\in C$ such that, for $\mathrm{\forall }z\in C\setminus D_x$,

   $$\Theta (z,y_x)+\phi (y_x)-\phi (z)+\frac{1}{r}\langle y_x-z,z-x\rangle <0;$$

6. C is a bounded set.

Proposition 2.1

([19])

Assume that $\Theta :C\times C\to \mathbf{R}$ satisfies (A1)–(A4), and let $\phi :C\to \mathbf{R}$ be a proper lower semicontinuous and convex function. Assume that either (B1) or (B2) holds. For $r>0$ and $x\in H$, define a mapping $T_r^{(\Theta ,\phi )}:H\to C$ as follows:

$$T_r^{(\Theta ,\phi )}(x):=\{z\in C:\Theta (z,y)+\phi (y)-\phi (z)+\frac{1}{r}\langle y-z,z-x\rangle \ge 0,\mathrm{\forall }y\in C\}$$

for all $x\in H$. Then the following hold:

• (i)for each $x\in H$, $T_r^{(\Theta ,\phi )}(x)$ is nonempty and single-valued;
• (ii)
  $T_r^{(\Theta ,\phi )}$ is firmly nonexpansive, that is, for any $x,y\in H$,

  $${\parallel T_r^{(\Theta ,\phi )}x-T_r^{(\Theta ,\phi )}y\parallel }^2\le \langle T_r^{(\Theta ,\phi )}x-T_r^{(\Theta ,\phi )}y,x-y\rangle ;$$
• (iii)$Fix(T_r^{(\Theta ,\phi )})=MEP(\Theta ,\phi )$;
• (iv)$MEP(\Theta ,\phi )$ is closed and convex;
• (v)${\parallel T_s^{(\Theta ,\phi )}x-T_t^{(\Theta ,\phi )}x\parallel }^2\le \frac{s-t}{s}\langle T_s^{(\Theta ,\phi )}x-T_t^{(\Theta ,\phi )}x,T_s^{(\Theta ,\phi )}x-x\rangle$ for all $s,t>0$ and $x\in H$.

Proposition 2.2

Let $F:C\to H$ be an α-inverse-strongly monotone mapping. Then, for all $x,y\in C$ and $\lambda >0$, one has

$${\parallel (I-\lambda F)x-(I-\lambda F)y\parallel }^2\le {\parallel x-y\parallel }^2+\lambda (\lambda -2\alpha ){\parallel Fx-Fy\parallel }^2.$$

In particular, if $\lambda \in (0,2\alpha ]$, $I-\lambda F:C\to H$ is a nonexpansive mapping.

We will use the following lemmas for the proof of our main results in the sequel.

Lemma 2.2

([20])

Let $\{s_n\}$ be a sequence of nonnegative real numbers satisfying

$$s_{n+1}\le (1-{\omega }_n)s_n+{\omega }_n{\delta }_n+{\gamma }_n,\mathrm{\forall }n\ge 0,$$

where $\{{\omega }_n\}$, $\{{\delta }_n\}$, and $\{{\gamma }_n\}$ satisfy the following conditions:

• (i)$\{{\omega }_n\}\subset [0,1]$ and ${\sum }_{n=0}^{\mathrm{\infty }}{\omega }_n=\mathrm{\infty }$ or, equivalently, ${\prod }_{n=0}^{\mathrm{\infty }}(1-{\omega }_n)=0$;
• (ii)${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\delta }_n\le 0$ or ${\sum }_{n=0}^{\mathrm{\infty }}{\omega }_n|{\delta }_n|<\mathrm{\infty }$;
• (iii)${\gamma }_n\ge 0$ ($n\ge 0$), ${\sum }_{n=0}^{\mathrm{\infty }}{\gamma }_n<\mathrm{\infty }$.

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

Lemma 2.3

(Demiclosedness principle [21])

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $S:C\to C$ be a nonexpansive mapping with $Fix(S)\ne \mathrm{\varnothing }$. Then the mapping $I-S$ is demiclosed. That is, if $\{x_n\}$ is a sequence in C such that $x_n\rightharpoonup x^{\ast }$ and $(I-S)x_n\to y$, then $(I-S)x^{\ast }=y$. Here I is the identity mapping of H.

Lemma 2.4

([22])

Let H be a real Hilbert space. Let $A:H\to H$ be a strongly positive bounded linear operator with a constant $\overline{\gamma }>1$. Then

$$\langle (A-I)x-(A-I)y,x-y\rangle \ge (\overline{\gamma }-1){\parallel x-y\parallel }^2,\mathrm{\forall }x,y\in C.$$

That is, $A-I$ is strongly monotone with a constant $\overline{\gamma }-1$.

Lemma 2.5

([22])

Assume that $A:H\to H$ is a strongly positive bounded linear operator with a coefficient $\overline{\gamma }>0$ and $0<\zeta \le {\parallel A\parallel }^{-1}$. Then $\parallel I-\zeta A\parallel \le 1-\zeta \overline{\gamma }$.

Lemma 2.6

([23])

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $G:C\to H$ be a ρ-Lipschitzian and η-strongly monotone mapping with constants $\rho ,\eta >0$. Let $0<\mu <\frac{2\eta }{{\rho }^2}$ and $0<t<\sigma \le 1$. Then $S:=\sigma I-t\mu G:C\to H$ is a contractive mapping with constant $\sigma -t\tau$, where $\tau =1-\sqrt{1-\mu (2\eta -\mu {\rho }^2)}$.

Lemma 2.7

([24])

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $F:C\to H$ be a continuous monotone mapping. Then, for $r>0$ and $x\in H$, there exists $z\in C$ such that

$$\langle y-z,Fz\rangle +\frac{1}{r}\langle y-z,z-x\rangle \ge 0,\mathrm{\forall }y\in C.$$

For $r>0$ and $x\in H$, define $F_r:H\to C$ by

$$F_{r}x=\{z\in C:\langle y-z,Fz\rangle +\frac{1}{r}\langle y-z,z-x\rangle \ge 0,\mathrm{\forall }y\in C\}.$$

Then the following hold:

• (i)$F_r$ is single-valued;
• (ii)
  $F_r$ is firmly nonexpansive, that is,

  $${\parallel F_{r}x-F_{r}y\parallel }^2\le \langle x-y,F_{r}x-F_{r}y\rangle ,\mathrm{\forall }x,y\in H;$$
• (iii)$Fix(F_r)=VI(C,F)$;
• (iv)$VI(C,F)$ is a closed convex subset of C.

Lemma 2.8

([24])

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $T:C\to H$ be a continuous pseudocontractive mapping. Then, for $r>0$ and $x\in H$, there exists $z\in C$ such that

$$\langle y-z,Tz\rangle -\frac{1}{r}\langle y-z,(1+r)z-x\rangle \le 0,\mathrm{\forall }y\in C.$$

For $r>0$ and $x\in H$, define $T_r:H\to C$ by

$$T_{r}x=\{z\in C:\langle y-z,Tz\rangle -\frac{1}{r}\langle y-z,(1+r)z-x\rangle \le 0,\mathrm{\forall }y\in C\}.$$

Then the following hold:

• (i)$T_r$ is single-valued;
• (ii)
  $T_r$ is firmly nonexpansive, that is,

  $${\parallel T_{r}x-T_{r}y\parallel }^2\le \langle x-y,T_{r}x-T_{r}y\rangle ,\mathrm{\forall }x,y\in H;$$
• (iii)$Fix(T_r)=Fix(T)$;
• (iv)$Fix(T)$ is a closed convex subset of C.

Main results

Throughout this section, we always assume the following:

• $B_i:C\to H$ is a ${\mu }_i$-inverse-strongly monotone mapping for each $i=1,2,\dots ,N$;

• ${\Theta }_i:C\times C\to \mathbf{R}$ is a bifunction satisfying conditions (A1)–(A4) for each $i=1,2,\dots ,N$;

• ${\phi }_i:C\to \mathbf{R}$ is a proper lower semicontinuous and convex function with restriction (B1) or (B2) for each $i=1,2,\dots ,N$;

• $A:H\to H$ is a strongly positive linear bounded self-adjoint operator with a constant $\overline{\gamma }\in (1,2)$;

• $V:C\to C$ is l-Lipschitzian with constant $l\in [0,\mathrm{\infty })$;

• $G:C\to C$ is a ρ-Lipschitzian and η-strongly monotone mapping with constants $\rho >0$ and $\eta >0$;

• constants μ, l, τ, and γ satisfy $0<\mu <\frac{2\eta }{{\rho }^2}$ and $0\le \gamma l<\tau$, where $\tau =1-\sqrt{1-\mu (2\eta -\mu {\rho }^2)}$;

• $F_1,F_2:C\to H$ are continuous monotone mappings and $T:C\to C$ is a continuous pseudocontractive mapping such that $\Omega :={\bigcap }_{i=1}^{N}GMEP({\Theta }_i,{\phi }_i,B_i)\cap GSVI(C,F_1,F_2)\cap Fix(T)\ne \mathrm{\varnothing }$;

• $R_t=F_{1,{\lambda }_t}{F}_{2,{\nu }_t}:H\to C$, where $F_{1,{\lambda }_t},F_{2,{\nu }_t}:H\to C$ are defined as follows:

  $$\begin{array}{c}{F}_{1,{\lambda }_t}x=\{z\in C:\langle y-z,F_{1}z\rangle +\frac{1}{{\lambda }_t}\langle y-z,z-x\rangle \ge 0,\mathrm{\forall }y\in C\},\hfill \\ F_{2,{\nu }_t}x=\{z\in C:\langle y-z,F_{2}z\rangle +\frac{1}{{\nu }_t}\langle y-z,z-x\rangle \ge 0,\mathrm{\forall }y\in C\},\hfill \end{array}$$

  for ${\lambda }_t,{\nu }_t\in (0,\mathrm{\infty })$, $t\in (0,1)$, ${lim}_{t\to 0}{\lambda }_t=\lambda >0$, and ${lim}_{t\to 0}{\nu }_t=\nu >0$;
• $R_n=F_{1,{\lambda }_n}{F}_{2,{\nu }_n}:H\to C$, where $F_{1,{\lambda }_n},F_{2,{\nu }_n}:H\to C$ are defined as follows:

  $$\begin{array}{c}{F}_{1,{\lambda }_n}x=\{z\in C:\langle y-z,F_{1}z\rangle +\frac{1}{{\lambda }_n}\langle y-z,z-x\rangle \ge 0,\mathrm{\forall }y\in C\},\hfill \\ F_{2,{\nu }_n}x=\{z\in C:\langle y-z,F_{2}z\rangle +\frac{1}{{\nu }_n}\langle y-z,z-x\rangle \ge 0,\mathrm{\forall }y\in C\},\hfill \end{array}$$

  for ${\lambda }_n,{\nu }_n\in (0,\mathrm{\infty })$, ${lim}_{n\to \mathrm{\infty }}{\lambda }_n=\lambda >0$, and ${lim}_{n\to \mathrm{\infty }}{\nu }_n=\nu >0$;
• $T_{r_t}:H\to C$ is a mapping defined by

  $$T_{r_t}x=\{z\in C:\langle y-z,Tz\rangle -\frac{1}{r_t}\langle y-z,(1+r_t)z-x\rangle \ge 0,\mathrm{\forall }y\in C\}$$

  for $r_t\in (0,\mathrm{\infty })$, $t\in (0,1)$, and ${lim\hspace{0.17em}inf}_{t\to 0}{r}_t>0$;
• $T_{r_n}:H\to C$ is a mapping defined by

  $$T_{r_n}x=\{z\in C:\langle y-z,Tz\rangle -\frac{1}{r_n}\langle y-z,(1+r_n)z-x\rangle \ge 0,\mathrm{\forall }y\in C\}$$

  for $r_n\in (0,\mathrm{\infty })$, and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$;
• $T_{r_{i,t}}^{({\Theta }_i,{\phi }_i)}:H\to C$ is a mapping defined by

  $$T_{r_{i,t}}^{({\Theta }_i,{\phi }_i)}x=\{z\in C:{\Theta }_i(z,y)+{\phi }_i(y)-{\phi }_i(z)+\frac{1}{r_{i,t}}\langle y-z,z-x\rangle \ge 0,\mathrm{\forall }y\in C\}$$

  for ${\{r_{i,t}\}}_{t\in (0,1)}\subset [c_i,d_i]\subset (0,2{\mu }_i)$ and $i\in \{1,2,\dots ,N\}$;
• $T_{r_{i,n}}^{({\Theta }_i,{\phi }_i)}:H\to C$ is a mapping defined by

  $$T_{r_{i,n}}^{({\Theta }_i,{\phi }_i)}x=\{z\in C:{\Theta }_i(z,y)+{\phi }_i(y)-{\phi }_i(z)+\frac{1}{r_{i,n}}\langle y-z,z-x\rangle \ge 0,\mathrm{\forall }y\in C\}$$

  for ${\{r_{i,n}\}}_{n=1}^{\mathrm{\infty }}\subset [c_i,d_i]\subset (0,2{\mu }_i)$ and $i\in \{1,2,\dots ,N\}$.

By Proposition 2.1 and Lemmas 2.7 and 2.8, we note that $T_{r_{i,t}}^{({\Theta }_i,{\phi }_i)}$, $T_{r_{i,n}}^{({\Theta }_i,{\phi }_i)}$, $F_{1,{\lambda }_t}$, $F_{1,{\lambda }_n}$, $F_{2,{\nu }_t}$, $F_{2,{\nu }_n}$, $T_{r_t}$, and $T_{r_n}$ are nonexpansive, $GMEP({\Theta }_i,{\phi }_i,B_i)=Fix(T_{r_{i,t}}^{({\Theta }_i,{\phi }_i)}(I-r_{i,t}{B}_i))=Fix(T_{r_{i,n}}^{({\Theta }_i,{\phi }_i)}(I-r_{i,n}{B}_i))$, and $Fix(T)=Fix(T_{r_t})=Fix(T_{r_n})$. So it is known that the composite mappings $R_t=F_{1,{\lambda }_t}{F}_{2,{\nu }_t}$ and $R_n=F_{1,{\lambda }_n}{F}_{2,{\nu }_n}$ are nonexpansive. Also, we note that $GSVI(C,F_1,F_2)=Fix(R_t)=Fix(R_n)$ by Lemma 1.1.

In this section, for $t\in (0,1)$, $n\ge 1$ and $i\in \{1,2,\dots ,N\}$, we put

$$\begin{array}{c}{\Delta }_t^i=T_{r_{i,t}}^{({\Theta }_i,{\phi }_i)}(I-r_{i,t}{B}_i)T_{r_{i-1,t}}^{({\Theta }_{i-1},{\phi }_{i-1})}(I-r_{i-1,t}{B}_{i-1})\cdots T_{r_{1,t}}^{({\Theta }_1,{\phi }_1)}(I-r_{1,t}{B}_1),\hfill \\ {\Delta }_n^i=T_{r_{i,n}}^{({\Theta }_i,{\phi }_i)}(I-r_{i,n}{B}_i)T_{r_{i-1,n}}^{({\Theta }_{i-1},{\phi }_{i-1})}(I-r_{i-1,n}{B}_{i-1})\cdots T_{r_{1,n}}^{({\Theta }_1,{\phi }_1)}(I-r_{1,n}{B}_1),\hfill \end{array}$$

and ${\Delta }_t^0={\Delta }_n^0=I$.

We now introduce the first general iterative scheme that generates a net $\{x_t\}$ in an implicit way:

$$x_t=P_C[(I-{\theta }_{t}A)T_{r_t}{\Delta }_t^N{R}_t{x}_t+{\theta }_t(t\gamma V{x}_t+(I-t\mu G)T_{r_t}{\Delta }_t^N{R}_t{x}_t)],$$ 3.1

where $t\in (0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma l}\})$ and ${\theta }_t\in (0,min\{\frac{1}{2},{\parallel A\parallel }^{-1}\})$.

We prove the strong convergence of $\{x_t\}$ as $t\to 0$ to a point $\tilde{x}\in \Omega$, which is a unique solution to the VI

$$\langle (A-I)\tilde{x},p-\tilde{x}\rangle \ge 0,\mathrm{\forall }p\in \Omega .$$ 3.2

In the meantime, we also propose the second general iterative scheme that generates a sequence $\{x_n\}$ in an explicit way:

$$\{\begin{array}{c}{w}_n={\alpha }_n\gamma V{x}_n+(I-{\alpha }_n\mu G)T_{r_n}{\Delta }_n^N{R}_n{x}_n,\hfill \\ x_{n+1}=P_C[(I-{\beta }_{n}A)T_{r_n}{\Delta }_n^N{R}_n{x}_n+{\beta }_n{w}_n],\mathrm{\forall }n\ge 1,\hfill \end{array}$$ 3.3

where $\{{\alpha }_n\},\{{\beta }_n\}\subset [0,1]$ and $x_0\in C$ is an arbitrary initial guess, and establish the strong convergence of $\{x_n\}$ as $n\to \mathrm{\infty }$ to the same point $\tilde{x}\in \Omega$, which is the unique solution to VI (3.2).

Next, for $t\in (0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma l}\})$ and ${\theta }_t\in (0,min\{\frac{1}{2},{\parallel A\parallel }^{-1}\})$, consider a mapping $Q_t:C\to C$ defined by

$$Q_{t}x=P_C[(I-{\theta }_{t}A)T_{r_t}{\Delta }_t^N{R}_{t}x+{\theta }_t(t\gamma Vx+(I-t\mu G)T_{r_t}{\Delta }_t^N{R}_{t}x)],\mathrm{\forall }x\in C.$$

It is easy to see that $Q_t$ is a contractive mapping with constant $1-{\theta }_t(\overline{\gamma }-1+t(\tau -\gamma l))$. Indeed, by Propositions 2.1 and 2.2 and Lemmas 2.5 and 2.6, we have

$$\begin{array}{rl}\parallel Q_{t}x-Q_{t}y\parallel & \le \parallel (I-{\theta }_{t}A)T_{r_t}{\Delta }_t^N{R}_{t}x+{\theta }_t(t\gamma Vx+(I-t\mu G)T_{r_t}{\Delta }_t^N{R}_{t}x)\\ & -(I-{\theta }_{t}A)T_{r_t}{\Delta }_t^N{R}_{t}y-{\theta }_t(t\gamma Vy+(I-t\mu G)T_{r_t}{\Delta }_t^N{R}_{t}y)\parallel \\ & \le \parallel (I-{\theta }_{t}A)T_{r_t}{\Delta }_t^N{R}_{t}x-(I-{\theta }_{t}A)T_{r_t}{\Delta }_t^N{R}_{t}y\parallel \\ & +{\theta }_t\parallel (t\gamma Vx+(I-t\mu G)T_{r_t}{\Delta }_t^N{R}_{t}x)-(t\gamma Vy+(I-t\mu G)T_{r_t}{\Delta }_t^N{R}_{t}y)\parallel \\ & \le (1-{\theta }_t\overline{\gamma })\parallel T_{r_t}{\Delta }_t^N{R}_{t}x-T_{r_t}{\Delta }_t^N{R}_{t}y\parallel +{\theta }_t[t\gamma \parallel Vx-Vy\parallel \\ & +\parallel (I-t\mu G)T_{r_t}{\Delta }_t^N{R}_{t}x-(I-t\mu G)T_{r_t}{\Delta }_t^N{R}_{t}y\parallel ]\\ & \le (1-{\theta }_t\overline{\gamma })\parallel x-y\parallel +{\theta }_t[t\gamma l\parallel x-y\parallel +(1-t\tau )\parallel x-y\parallel ]\\ & =[1-{\theta }_t(\overline{\gamma }-1+t(\tau -\gamma l))]\parallel x-y\parallel .\end{array}$$

Since $\overline{\gamma }\in (1,2)$, $\tau -\gamma l>0$ and $0<t<min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma l}\}\le \frac{2-\overline{\gamma }}{\tau -\gamma l}$, it follows that $0<\overline{\gamma }-1+t(\tau -\gamma l)<1$, which together with $0<{\theta }_t<min\{\frac{1}{2},{\parallel A\parallel }^{-1}\}<1$ yields $0<1-{\theta }_t(\overline{\gamma }-1+t(\tau -\gamma l))<1$. Hence $Q_t$ is a contractive mapping. By the Banach contraction principle, $Q_t$ has a unique fixed point, denoted by $x_t$, which uniquely solves the fixed point equation (3.1).

We summarize the basic properties of $\{x_t\}$.

Theorem 3.1

Let $\{x_t\}$ be defined via (3.1). Then

• (i)$\{x_t\}$ is bounded for $t\in (0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma l}\})$;
• (ii)${lim}_{t\to 0}\parallel x_t-R_t{x}_t\parallel =0$, ${lim}_{t\to 0}\parallel x_t-{\Delta }_t^N{x}_t\parallel =0$, and ${lim}_{t\to 0}\parallel x_t-T_{r_t}{x}_t\parallel =0$ provided ${lim}_{t\to 0}{\theta }_t=0$;
• (iii)$x_t:(0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma l}\})\to H$ is locally Lipschitzian provided ${\theta }_t:(0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma l}\})\to (0,min\{\frac{1}{2},{\parallel A\parallel }^{-1}\})$ is locally Lipschitzian, $r_t,{\lambda }_t,{\nu }_t:(0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma l}\})\to (0,\mathrm{\infty })$ are locally Lipschitzian, and $r_{i,t}:(0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma l}\})\to [c_i,d_i]$ is locally Lipschitzian for each $i=1,2,\dots ,N$;
• (iv)$x_t$ defines a continuous path from $(0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma l}\})$ into H provided ${\theta }_t:(0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma l}\})\to (0,min\{\frac{1}{2},{\parallel A\parallel }^{-1}\})$ is continuous, $r_t,{\lambda }_t,{\nu }_t:(0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma l}\})\to (0,\mathrm{\infty })$ are continuous, and $r_{i,t}:(0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma l}\})\to [c_i,d_i]$ is continuous for each $i=1,2,\dots ,N$.

Proof

Let $z_t=R_t{x}_t$, $u_t={\Delta }_t^N{z}_t$, and $v_t=T_{r_t}{u}_t$. Take $p\in \Omega$. Then $p=T_{r_t}p$ by Lemma 2.8(iii), $p={\Delta }_t^{i}p$ ($=T_{r_{i,t}}^{({\Theta }_i,{\phi }_i)}(I-r_{i,t}{B}_i)p$) by Proposition 2.1(iii), and $p=R_{t}p$ by Lemma 1.1.

(i) Utilizing Proposition 2.1(ii) and Proposition 2.2, we have

$$\begin{array}{rl}\parallel u_t-p\parallel & =\parallel T_{r_{N,t}}^{({\Theta }_N,{\phi }_N)}(I-r_{N,t}{B}_N){\Delta }_t^{N-1}{z}_t-T_{r_{N,t}}^{({\Theta }_N,{\phi }_N)}(I-r_{N,t}{B}_N){\Delta }_t^{N-1}p\parallel \\ & \le \parallel (I-r_{N,t}{B}_N){\Delta }_t^{N-1}{z}_t-(I-r_{N,t}{B}_N){\Delta }_t^{N-1}p\parallel \\ & \le \parallel {\Delta }_t^{N-1}{z}_t-{\Delta }_t^{N-1}p\parallel \\ & \le \cdots \\ & \le \parallel {\Delta }_t^0{z}_t-{\Delta }_t^{0}p\parallel =\parallel z_t-p\parallel .\end{array}$$ 3.4

Moreover, it is easy from the nonexpansivity of $R_t$ to see that

$$\parallel z_t-p\parallel =\parallel R_t{x}_t-R_{t}p\parallel \le \parallel x_t-p\parallel ,$$

which together with the nonexpansivity of $T_{r_t}$ and (3.4) implies that

$$\parallel v_t-p\parallel =\parallel T_{r_t}{u}_t-T_{r_t}p\parallel \le \parallel u_t-p\parallel \le \parallel z_t-p\parallel \le \parallel x_t-p\parallel .$$ 3.5

By (3.5), we have

$$\begin{array}{rl}\parallel x_t-p\parallel & \le \parallel (I-{\theta }_{t}A)v_t+{\theta }_t(t\gamma V{x}_t+(I-t\mu G)v_t)-p\parallel \\ & =\parallel (I-{\theta }_{t}A)v_t-(I-{\theta }_{t}A)p+{\theta }_t(t\gamma V{x}_t+(I-t\mu G)v_t-p)+{\theta }_t(I-A)p\parallel \\ & \le \parallel (I-{\theta }_{t}A)v_t-(I-{\theta }_{t}A)p\parallel +{\theta }_t\parallel t\gamma V{x}_t+(I-t\mu G)v_t-p\parallel +{\theta }_t\parallel (I-A)p\parallel \\ & =\parallel (I-{\theta }_{t}A)v_t-(I-{\theta }_{t}A)p\parallel \\ & +{\theta }_t\parallel (I-t\mu G)v_t-(I-t\mu G)p+t(\gamma V{x}_t-\mu Gp)\parallel +{\theta }_t\parallel (I-A)p\parallel \\ & \le (1-{\theta }_t\overline{\gamma })\parallel v_t-p\parallel +{\theta }_t[\parallel (I-t\mu G)v_t-(I-t\mu G)p\parallel \\ & +t(\gamma \parallel V{x}_t-Vp\parallel +\parallel \gamma Vp-\mu Gp\parallel )]+{\theta }_t\parallel (I-A)p\parallel \\ & \le (1-{\theta }_t\overline{\gamma })\parallel x_t-p\parallel +{\theta }_t[(1-t\tau )\parallel x_t-p\parallel +t(\gamma l\parallel x_t-p\parallel +\parallel (\gamma V-\mu G)p\parallel )]\\ & +{\theta }_t\parallel I-A\parallel \parallel p\parallel \\ & =[1-{\theta }_t(\overline{\gamma }-1+t(\tau -\gamma l))]\parallel x_t-p\parallel +{\theta }_t[\parallel I-A\parallel \parallel p\parallel +t\parallel (\gamma V-\mu G)p\parallel ].\end{array}$$

So, it follows that

$$\parallel x_t-p\parallel \le \frac{\parallel I-A\parallel \parallel p\parallel +t\parallel (\gamma V-\mu G)p\parallel }{\overline{\gamma }-1+t(\tau -\gamma l)}\le \frac{\parallel I-A\parallel \parallel p\parallel +\parallel (\gamma V-\mu G)p\parallel }{\overline{\gamma }-1}.$$

Hence $\{x_t\}$ is bounded and so are $\{V{x}_t\}$, $\{u_t\}$, $\{v_t\}$, $\{z_t\}$, and $\{G{v}_t\}$.

(ii) By the definition of $\{x_t\}$, we have

$$\begin{array}{rl}\parallel x_t-v_t\parallel & =\parallel P_C[(I-{\theta }_{t}A)v_t+{\theta }_t(t\gamma V{x}_t+(I-t\mu G)v_t)]-v_t\parallel \\ & \le \parallel (I-{\theta }_{t}A)v_t+{\theta }_t(t\gamma V{x}_t+(I-t\mu G)v_t)-v_t\parallel \\ & =\parallel {\theta }_t[(I-A)v_t+t(\gamma V{x}_t-\mu G{v}_t)]\parallel \\ & ={\theta }_t\parallel (I-A)v_t+t(\gamma V{x}_t-\mu G{v}_t)\parallel \\ & \le {\theta }_t\parallel I-A\parallel \parallel v_t\parallel +t\parallel \gamma V{x}_t-\mu G{v}_t\parallel \to 0\text{as }t\to 0,\end{array}$$

using the boundedness of $\{V{x}_t\}$, $\{v_t\}$, and $\{G{v}_t\}$ in the proof of assertion (i). That is,

$$\underset{t\to 0}{lim}\parallel x_t-v_t\parallel =0.$$ 3.6

In view of (3.5) and Lemma 2.7(ii), we get

$$\begin{array}{rl}{\parallel v_t-p\parallel }^2& \le {\parallel z_t-p\parallel }^2={\parallel R_t{x}_t-R_{t}p\parallel }^2\\ & ={\parallel F_{1,{\lambda }_t}{F}_{2,{\nu }_t}{x}_t-F_{1,{\lambda }_t}{F}_{2,{\nu }_t}p\parallel }^2\\ & \le \langle F_{2,{\nu }_t}{x}_t-F_{2,{\nu }_t}p,F_{1,{\lambda }_t}{F}_{2,{\nu }_t}{x}_t-F_{1,{\lambda }_t}{F}_{2,{\nu }_t}p\rangle \\ & =\langle F_{2,{\nu }_t}{x}_t-F_{2,{\nu }_t}p,z_t-p\rangle \\ & \le \frac{1}{2}[{\parallel F_{2,{\nu }_t}{x}_t-F_{2,{\nu }_t}p\parallel }^2+{\parallel z_t-p\parallel }^2-{\parallel (F_{2,{\nu }_t}{x}_t-F_{2,{\nu }_t}p)-(z_t-p)\parallel }^2]\\ & \le \frac{1}{2}[{\parallel x_t-p\parallel }^2+{\parallel x_t-p\parallel }^2-{\parallel (F_{2,{\nu }_t}{x}_t-F_{2,{\nu }_t}p)-(z_t-p)\parallel }^2]\\ & ={\parallel x_t-p\parallel }^2-\frac{1}{2}{\parallel (F_{2,{\nu }_t}{x}_t-F_{2,{\nu }_t}p)-(z_t-p)\parallel }^2,\end{array}$$

which immediately yields

$$\frac{1}{2}{\parallel (F_{2,{\nu }_t}{x}_t-F_{2,{\nu }_t}p)-(z_t-p)\parallel }^2\le {\parallel x_t-p\parallel }^2-{\parallel v_t-p\parallel }^2\le (\parallel x_t-p\parallel +\parallel v_t-p\parallel )\parallel x_t-v_t\parallel .$$

From (3.6) and the boundedness of $\{x_t\}$ and $\{v_t\}$, we have

$$\underset{t\to 0}{lim}\parallel (F_{2,{\nu }_t}{x}_t-F_{2,{\nu }_t}p)-(z_t-p)\parallel =0.$$ 3.7

Again from (3.5) and Lemma 2.7(ii), we obtain

$$\begin{array}{rl}{\parallel v_t-p\parallel }^2& \le {\parallel z_t-p\parallel }^2={\parallel R_t{x}_t-R_{t}p\parallel }^2\\ & \le {\parallel F_{2,{\nu }_t}{x}_t-F_{2,{\nu }_t}p\parallel }^2\\ & \le \langle x_t-p,F_{2,{\nu }_t}{x}_t-F_{2,{\nu }_t}p\rangle \\ & \le \frac{1}{2}[{\parallel x_t-p\parallel }^2+{\parallel F_{2,{\nu }_t}{x}_t-F_{2,{\nu }_t}p\parallel }^2-{\parallel (x_t-p)-(F_{2,{\nu }_t}{x}_t-F_{2,{\nu }_t}p)\parallel }^2]\\ & \le \frac{1}{2}[{\parallel x_t-p\parallel }^2+{\parallel x_t-p\parallel }^2-{\parallel (x_t-p)-(F_{2,{\nu }_t}{x}_t-F_{2,{\nu }_t}p)\parallel }^2]\\ & ={\parallel x_t-p\parallel }^2-\frac{1}{2}{\parallel (x_t-p)-(F_{2,{\nu }_t}{x}_t-F_{2,{\nu }_t}p)\parallel }^2,\end{array}$$

which hence leads to

$$\frac{1}{2}{\parallel (x_t-p)-(F_{2,{\nu }_t}{x}_t-F_{2,{\nu }_t}p)\parallel }^2\le {\parallel x_t-p\parallel }^2-{\parallel v_t-p\parallel }^2\le (\parallel x_t-p\parallel +\parallel v_t-p\parallel )\parallel x_t-v_t\parallel .$$

Again from (3.6) and the boundedness of $\{x_t\}$ and $\{v_t\}$, we have

$$\underset{t\to 0}{lim}\parallel (x_t-p)-(F_{2,{\nu }_t}{x}_t-F_{2,{\nu }_t}p)\parallel =0.$$ 3.8

So it follows from (3.7) and (3.8) that

$$\parallel x_t-z_t\parallel \le \parallel (x_t-p)-(F_{2,{\nu }_t}{x}_t-F_{2,{\nu }_t}p)\parallel +\parallel (F_{2,{\nu }_t}{x}_t-F_{2,{\nu }_t}p)-(z_t-p)\parallel \to 0\text{as }t\to 0.$$

That is,

$$\underset{t\to 0}{lim}\parallel x_t-z_t\parallel =0.$$ 3.9

Furthermore, from (3.5) and Proposition 2.1(ii) and Proposition 2.2, it follows that

$$\begin{array}{rl}{\parallel v_t-p\parallel }^2& \le {\parallel u_t-p\parallel }^2={\parallel {\Delta }_t^N{z}_t-p\parallel }^2\\ & \le {\parallel {\Delta }_t^i{z}_t-p\parallel }^2\\ & ={\parallel T_{r_{i,t}}^{({\Theta }_i,{\phi }_i)}(I-r_{i,t}{B}_i){\Delta }_t^{i-1}{z}_t-T_{r_{i,t}}^{({\Theta }_i,{\phi }_i)}(I-r_{i,t}{B}_i)p\parallel }^2\\ & \le {\parallel (I-r_{i,t}{B}_i){\Delta }_t^{i-1}{z}_t-(I-r_{i,t}{B}_i)p\parallel }^2\\ & \le {\parallel {\Delta }_t^{i-1}{z}_t-p\parallel }^2+r_{i,t}(r_{i,t}-2{\mu }_i){\parallel B_i{\Delta }_t^{i-1}{z}_t-B_{i}p\parallel }^2\\ & \le {\parallel z_t-p\parallel }^2+r_{i,t}(r_{i,t}-2{\mu }_i){\parallel B_i{\Delta }_t^{i-1}{z}_t-B_{i}p\parallel }^2\\ & \le {\parallel x_t-p\parallel }^2+r_{i,t}(r_{i,t}-2{\mu }_i){\parallel B_i{\Delta }_t^{i-1}{z}_t-B_{i}p\parallel }^2,\end{array}$$

which together with ${\{r_{i,t}\}}_{t\in (0,1)}\subset [c_i,d_i]\subset (0,2{\mu }_i)$ for $i\in \{1,2,\dots ,N\}$ implies that

$$\begin{array}{rl}{c}_i(2{\mu }_i-d_i){\parallel B_i{\Delta }_t^{i-1}{z}_t-B_{i}p\parallel }^2& \le r_{i,t}(2{\mu }_i-r_{i,t}){\parallel B_i{\Delta }_t^{i-1}{z}_t-B_{i}p\parallel }^2\\ & \le {\parallel x_t-p\parallel }^2-{\parallel v_t-p\parallel }^2\le (\parallel x_t-p\parallel +\parallel v_t-p\parallel )\parallel x_t-v_t\parallel .\end{array}$$

From (3.6) and the boundedness of $\{x_t\}$ and $\{v_t\}$, we have

$$\underset{t\to 0}{lim}\parallel B_i{\Delta }_t^{i-1}{z}_t-B_{i}p\parallel =0.$$ 3.10

Also, by Proposition 2.1(ii), we obtain that, for each $i=1,2,\dots ,N$,

$$\begin{array}{c}{\parallel {\Delta }_t^i{z}_t-p\parallel }^2\hfill \\ ={\parallel T_{r_{i,t}}^{({\Theta }_i,{\phi }_i)}(I-r_{i,t}{B}_i){\Delta }_t^{i-1}{z}_t-T_{r_{i,t}}^{({\Theta }_i,{\phi }_i)}(I-r_{i,t}{B}_i)p\parallel }^2\hfill \\ \le \langle (I-r_{i,t}{B}_i){\Delta }_t^{i-1}{z}_t-(I-r_{i,t}{B}_i)p,{\Delta }_t^i{z}_t-p\rangle \hfill \\ =\frac{1}{2}[{\parallel (I-r_{i,t}{B}_i){\Delta }_t^{i-1}{z}_t-(I-r_{i,t}{B}_i)p\parallel }^2+{\parallel {\Delta }_t^i{z}_t-p\parallel }^2\hfill \\ -{\parallel (I-r_{i,t}{B}_i){\Delta }_t^{i-1}{z}_t-(I-r_{i,t}{B}_i)p-({\Delta }_t^i{z}_t-p)\parallel }^2]\hfill \\ \le \frac{1}{2}[{\parallel {\Delta }_t^{i-1}{z}_t-p\parallel }^2+{\parallel {\Delta }_t^i{z}_t-p\parallel }^2-{\parallel {\Delta }_t^{i-1}{z}_t-{\Delta }_t^i{z}_t-r_{i,t}(B_i{\Delta }_t^{i-1}{z}_t-B_{i}p)\parallel }^2]\hfill \\ \le \frac{1}{2}[{\parallel x_t-p\parallel }^2+{\parallel {\Delta }_t^i{z}_t-p\parallel }^2-{\parallel {\Delta }_t^{i-1}{z}_t-{\Delta }_t^i{z}_t-r_{i,t}(B_i{\Delta }_t^{i-1}{z}_t-B_{i}p)\parallel }^2],\hfill \end{array}$$

which immediately implies that

$${\parallel {\Delta }_t^i{z}_t-p\parallel }^2\le {\parallel x_t-p\parallel }^2-{\parallel {\Delta }_t^{i-1}{z}_t-{\Delta }_t^i{z}_t-r_{i,t}(B_i{\Delta }_t^{i-1}{z}_t-B_{i}p)\parallel }^2.$$

This together with (3.5) leads to

$$\begin{array}{rl}{\parallel v_t-p\parallel }^2& \le {\parallel u_t-p\parallel }^2={\parallel {\Delta }_t^N{z}_t-p\parallel }^2\le {\parallel {\Delta }_t^i{z}_t-p\parallel }^2\\ & \le {\parallel x_t-p\parallel }^2-{\parallel {\Delta }_t^{i-1}{z}_t-{\Delta }_t^i{z}_t-r_{i,t}(B_i{\Delta }_t^{i-1}{z}_t-B_{i}p)\parallel }^2,\end{array}$$

which hence implies

$$\begin{array}{rl}{\parallel {\Delta }_t^{i-1}{z}_t-{\Delta }_t^i{z}_t-r_{i,t}(B_i{\Delta }_t^{i-1}{z}_t-B_{i}p)\parallel }^2& \le {\parallel x_t-p\parallel }^2-{\parallel v_t-p\parallel }^2\\ & \le (\parallel x_t-p\parallel +\parallel v_t-p\parallel )\parallel x_t-v_t\parallel .\end{array}$$

From (3.6) and the boundedness of $\{x_t\}$ and $\{v_t\}$, we have

$$\underset{t\to 0}{lim}\parallel {\Delta }_t^{i-1}{z}_t-{\Delta }_t^i{z}_t-r_{i,t}(B_i{\Delta }_t^{i-1}{z}_t-B_{i}p)\parallel =0,$$

which together with (3.10) implies that, for each $i=1,2,\dots ,N$,

$$\underset{t\to 0}{lim}\parallel {\Delta }_t^{i-1}{z}_t-{\Delta }_t^i{z}_t\parallel =0.$$ 3.11

Note that

$$\parallel z_t-u_t\parallel \le \sum _{i=1}^N\parallel {\Delta }_t^{i-1}{z}_t-{\Delta }_t^i{z}_t\parallel .$$

From (3.11), it is easy to see that

$$\underset{t\to 0}{lim}\parallel z_t-u_t\parallel =0.$$ 3.12

Also, observe that

$$\begin{array}{rl}\parallel x_t-{\Delta }_t^N{x}_t\parallel & \le \parallel x_t-z_t\parallel +\parallel z_t-{\Delta }_t^N{z}_t\parallel +\parallel {\Delta }_t^N{z}_t-{\Delta }_t^N{x}_t\parallel \\ & \le \parallel x_t-z_t\parallel +\parallel z_t-{\Delta }_t^N{z}_t\parallel +\parallel z_t-x_t\parallel \\ & =2\parallel x_t-z_t\parallel +\parallel z_t-u_t\parallel .\end{array}$$

From (3.9) and (3.12), it is easy to see that

$$\underset{t\to 0}{lim}\parallel x_t-{\Delta }_t^N{x}_t\parallel =0.$$ 3.13

In the meantime, again from (3.5) and Lemma 2.7(ii), we obtain

$$\begin{array}{rl}{\parallel v_t-p\parallel }^2& ={\parallel T_{r_t}{u}_t-T_{r_t}p\parallel }^2\\ & \le \langle u_t-p,T_{r_t}{u}_t-T_{r_t}p\rangle =\langle u_t-p,v_t-p\rangle \\ & =\frac{1}{2}[{\parallel u_t-p\parallel }^2+{\parallel v_t-p\parallel }^2-{\parallel u_t-p-(v_t-p)\parallel }^2]\\ & \le \frac{1}{2}[{\parallel x_t-p\parallel }^2+{\parallel x_t-p\parallel }^2-{\parallel u_t-v_t\parallel }^2]\\ & ={\parallel x_t-p\parallel }^2-\frac{1}{2}{\parallel u_t-v_t\parallel }^2,\end{array}$$

which immediately yields

$$\frac{1}{2}{\parallel u_t-v_t\parallel }^2\le {\parallel x_t-p\parallel }^2-{\parallel v_t-p\parallel }^2\le (\parallel x_t-p\parallel +\parallel v_t-p\parallel )\parallel x_t-v_t\parallel .$$

From (3.6) and the boundedness of $\{x_t\}$ and $\{v_t\}$, we have

$$\underset{t\to 0}{lim}\parallel u_t-v_t\parallel =0.$$ 3.14

Taking into account that

$$\begin{array}{rl}\parallel x_t-T_{r_t}{x}_t\parallel & \le \parallel x_t-u_t\parallel +\parallel u_t-T_{r_t}{u}_t\parallel +\parallel T_{r_t}{u}_t-T_{r_t}{x}_t\parallel \\ & \le \parallel x_t-u_t\parallel +\parallel u_t-T_{r_t}{u}_t\parallel +\parallel u_t-x_t\parallel \\ & =2\parallel x_t-u_t\parallel +\parallel u_t-v_t\parallel \\ & \le 2(\parallel x_t-z_t\parallel +\parallel z_t-u_t\parallel )+\parallel u_t-v_t\parallel ,\end{array}$$

we deduce from (3.9), (3.12), and (3.14) that

$$\underset{t\to 0}{lim}\parallel x_t-T_{r_t}{x}_t\parallel =0.$$ 3.15

(iii) Let $t,t_0\in (0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma l}\})$. Since $v_t=T_{r_t}{u}_t$ and $v_{t_0}=T_{r_{t_0}}{u}_{t_0}$, we get

$$\langle y-v_t,(I-T)v_t\rangle +\frac{1}{r_t}\langle y-v_t,v_t-u_t\rangle \ge 0,\mathrm{\forall }y\in C,$$ 3.16

and

$$\langle y-v_{t_0},(I-T)v_{t_0}\rangle +\frac{1}{r_{t_0}}\langle y-v_{t_0},v_{t_0}-u_{t_0}\rangle \ge 0,\mathrm{\forall }y\in C.$$ 3.17

Putting $y=v_{t_0}$ in (3.16) and $y=v_t$ in (3.17), we obtain

$$\langle v_{t_0}-v_t,(I-T)v_t\rangle +\frac{1}{r_t}\langle v_{t_0}-v_t,v_t-u_t\rangle \ge 0$$ 3.18

and

$$\langle v_t-v_{t_0},(I-T)v_{t_0}\rangle +\frac{1}{r_{t_0}}\langle v_t-v_{t_0},v_{t_0}-u_{t_0}\rangle \ge 0.$$ 3.19

Adding up (3.18) and (3.19), we have

$$-\langle v_t-v_{t_0},(I-T)v_t-(I-T)v_{t_0}\rangle +\langle v_{t_0}-v_t,\frac{v_t-u_t}{r_t}-\frac{v_{t_0}-u_{t_0}}{r_{t_0}}\rangle \ge 0.$$

Since T is pseudocontractive, we know that $I-T$ is a monotone mapping such that

$$\langle v_{t_0}-v_t,\frac{v_t-u_t}{r_t}-\frac{v_{t_0}-u_{t_0}}{r_{t_0}}\rangle \ge 0,$$

and hence

$$\langle v_t-v_{t_0},v_{t_0}-v_t+v_t-u_{t_0}-\frac{r_{t_0}}{r_t}(v_t-u_t)\rangle \ge 0.$$ 3.20

Taking into account that ${lim\hspace{0.17em}inf}_{t\to 0}{r}_t>0$, without loss of generality, we may assume that $r_t>b>0$ $\mathrm{\forall }t\in (0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma l}\})$ for some $b>0$. Then from (3.20) we have

$$\begin{array}{rl}{\parallel v_t-v_{t_0}\parallel }^2& \le \langle v_t-v_{t_0},v_t-u_t+u_t-u_{t_0}-\frac{r_{t_0}}{r_t}(v_t-u_t)\rangle \\ & =\langle v_t-v_{t_0},u_t-u_{t_0}+(1-\frac{r_{t_0}}{r_t})(v_t-u_t)\rangle \\ & \le \parallel v_t-v_{t_0}\parallel \parallel u_t-u_{t_0}+(1-\frac{r_{t_0}}{r_t})(v_t-u_t)\parallel \\ & \le \parallel v_t-v_{t_0}\parallel \{\parallel u_t-u_{t_0}\parallel +|1-\frac{r_{t_0}}{r_t}|\parallel v_t-u_t\parallel \},\end{array}$$

which immediately yields

$$\begin{array}{rl}\parallel v_t-v_{t_0}\parallel & \le \parallel u_t-u_{t_0}\parallel +\frac{1}{r_t}|r_t-r_{t_0}|\parallel v_t-u_t\parallel \\ & \le \parallel u_t-u_{t_0}\parallel +\frac{{\tilde{L}}_1}{b}|r_t-r_{t_0}|,\end{array}$$ 3.21

where ${\tilde{L}}_1=sup\{\parallel v_t-u_t\parallel :t\in (0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma l}\})\}$.

Also, taking into account that ${lim}_{t\to 0}{\lambda }_t=\lambda >0$ and ${lim}_{t\to 0}{\nu }_t=\nu >0$, without loss of generality, we may assume that $min\{{\lambda }_t,{\nu }_t\}>a>0$ $\mathrm{\forall }t\in (0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma l}\})$ for some $a>0$. Since $z_t=F_{1,{\lambda }_t}{y}_t$ and $z_{t_0}=F_{1,{\lambda }_{t_0}}{y}_{t_0}$, where $y_t=F_{2,{\nu }_t}{x}_t$ and $y_{t_0}=F_{2,{\nu }_{t_0}}{x}_{t_0}$ for $t,t_0\in (0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma l}\})$, by using arguments similar to those of (3.21), we get

$$\parallel z_t-z_{t_0}\parallel \le \parallel y_t-y_{t_0}\parallel +\frac{1}{a}|{\lambda }_t-{\lambda }_{t_0}|{\tilde{L}}_2$$ 3.22

and

$$\parallel y_t-y_{t_0}\parallel \le \parallel x_t-x_{t_0}\parallel +\frac{1}{a}|{\nu }_t-{\nu }_{t_0}|{\tilde{L}}_2,$$ 3.23

where ${\tilde{L}}_2=sup\{\parallel z_t-y_t\parallel +\parallel y_t-x_t\parallel :t\in (0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma l}\})\}$. Substituting (3.23) for (3.22), we obtain

$$\parallel z_t-z_{t_0}\parallel \le \parallel x_t-x_{t_0}\parallel +\frac{{\tilde{L}}_2}{a}(|{\lambda }_t-{\lambda }_{t_0}|+|{\nu }_t-{\nu }_{t_0}|).$$ 3.24

In the meantime, by Proposition 2.1(ii), (v) and Proposition 2.2, we deduce that

$$\begin{array}{rl}\parallel u_t-u_{t_0}\parallel & =\parallel {\Delta }_t^N{z}_t-{\Delta }_{t_0}^N{z}_{t_0}\parallel \\ & =\parallel T_{r_{N,t}}^{({\Theta }_N,{\phi }_N)}(I-r_{N,t}{B}_N){\Delta }_t^{N-1}{z}_t-T_{r_{N,t_0}}^{({\Theta }_N,{\phi }_N)}(I-r_{N,t_0}{B}_N){\Delta }_{t_0}^{N-1}{z}_{t_0}\parallel \\ & \le \parallel T_{r_{N,t}}^{({\Theta }_N,{\phi }_N)}(I-r_{N,t}{B}_N){\Delta }_t^{N-1}{z}_t-T_{r_{N,t_0}}^{({\Theta }_N,{\phi }_N)}(I-r_{N,t_0}{B}_N){\Delta }_t^{N-1}{z}_t\parallel \\ & +\parallel T_{r_{N,t_0}}^{({\Theta }_N,{\phi }_N)}(I-r_{N,t_0}{B}_N){\Delta }_t^{N-1}{z}_t-T_{r_{N,t_0}}^{({\Theta }_N,{\phi }_N)}(I-r_{N,t_0}{B}_N){\Delta }_{t_0}^{N-1}{z}_{t_0}\parallel \\ & \le \parallel T_{r_{N,t}}^{({\Theta }_N,{\phi }_N)}(I-r_{N,t}{B}_N){\Delta }_t^{N-1}{z}_t-T_{r_{N,t_0}}^{({\Theta }_N,{\phi }_N)}(I-r_{N,t}{B}_N){\Delta }_t^{N-1}{z}_t\parallel \\ & +\parallel T_{r_{N,t_0}}^{({\Theta }_N,{\phi }_N)}(I-r_{N,t}{B}_N){\Delta }_t^{N-1}{z}_t-T_{r_{N,t_0}}^{({\Theta }_N,{\phi }_N)}(I-r_{N,t_0}{B}_N){\Delta }_t^{N-1}{z}_t\parallel \\ & +\parallel (I-r_{N,t_0}{B}_N){\Delta }_t^{N-1}{z}_t-(I-r_{N,t_0}{B}_N){\Delta }_{t_0}^{N-1}{z}_{t_0}\parallel \\ & \le \frac{|r_{N,t}-r_{N,t_0}|}{r_{N,t}}\parallel T_{r_{N,t}}^{({\Theta }_N,{\phi }_N)}(I-r_{N,t}{B}_N){\Delta }_t^{N-1}{z}_t-(I-r_{N,t}{B}_N){\Delta }_t^{N-1}{z}_t\parallel \\ & +|r_{N,t}-r_{N,t_0}|\parallel B_N{\Delta }_t^{N-1}{z}_t\parallel +\parallel {\Delta }_t^{N-1}{z}_t-{\Delta }_{t_0}^{N-1}{z}_{t_0}\parallel \\ & =|r_{N,t}-r_{N,t_0}|[\parallel B_N{\Delta }_t^{N-1}{z}_t\parallel +\frac{1}{r_{N,t}}\parallel T_{r_{N,t}}^{({\Theta }_N,{\phi }_N)}(I-r_{N,t}{B}_N){\Delta }_t^{N-1}{z}_t\\ & -(I-r_{N,t}{B}_N){\Delta }_t^{N-1}{z}_t\parallel ]+\parallel {\Delta }_t^{N-1}{z}_t-{\Delta }_{t_0}^{N-1}{z}_{t_0}\parallel \\ & \le \cdots \\ & \le |r_{N,t}-r_{N,t_0}|[\parallel B_N{\Delta }_t^{N-1}{z}_t\parallel +\frac{1}{r_{N,t}}\parallel T_{r_{N,t}}^{({\Theta }_N,{\phi }_N)}(I-r_{N,t}{B}_N){\Delta }_t^{N-1}{z}_t\\ & -(I-r_{N,t}{B}_N){\Delta }_t^{N-1}{z}_t\parallel ]+\cdots +|r_{1,t}-r_{1,t_0}|[\parallel B_1{\Delta }_t^0{z}_t\parallel \\ & +\frac{1}{r_{1,t}}\parallel T_{r_{1,t}}^{({\Theta }_1,{\phi }_1)}(I-r_{1,t}{B}_1){\Delta }_t^0{z}_t-(I-r_{1,t}{B}_1){\Delta }_t^0{z}_t\parallel ]+\parallel {\Delta }_t^0{z}_t-{\Delta }_{t_0}^0{z}_{t_0}\parallel \\ & \le {\tilde{L}}_3\sum _{i=1}^N|r_{i,t}-r_{i,t_0}|+\parallel z_t-z_{t_0}\parallel ,\end{array}$$ 3.25

where

$$\begin{array}{rl}& \underset{t\in (0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma l}\})}{sup}\left\{\sum _{i=1}^N[\parallel B_i{\Delta }_t^{i-1}{z}_t\parallel +\frac{1}{r_{i,t}}\parallel T_{r_{i,t}}^{({\Theta }_i,{\phi }_i)}(I-r_{i,t}{B}_i){\Delta }_t^{i-1}{z}_t-(I-r_{i,t}{B}_i){\Delta }_t^{i-1}{z}_t\parallel ]\right\}\\ & \le {\tilde{L}}_3\end{array}$$

for some ${\tilde{L}}_3>0$. This together with (3.21) and (3.24) implies that

$$\begin{array}{rl}\parallel v_t-v_{t_0}\parallel & \le \parallel u_t-u_{t_0}\parallel +\frac{{\tilde{L}}_1}{b}|r_t-r_{t_0}|\\ & \le {\tilde{L}}_3\sum _{i=1}^N|r_{i,t}-r_{i,t_0}|+\parallel z_t-z_{t_0}\parallel +\frac{{\tilde{L}}_1}{b}|r_t-r_{t_0}|\\ & \le {\tilde{L}}_3\sum _{i=1}^N|r_{i,t}-r_{i,t_0}|+\parallel x_t-x_{t_0}\parallel +\frac{{\tilde{L}}_2}{a}(|{\lambda }_t-{\lambda }_{t_0}|+|{\nu }_t-{\nu }_{t_0}|)+\frac{{\tilde{L}}_1}{b}|r_t-r_{t_0}|\\ & \le \parallel x_t-x_{t_0}\parallel +(\frac{{\tilde{L}}_1}{b}+\frac{{\tilde{L}}_2}{a})(|{\lambda }_t-{\lambda }_{t_0}|+|{\nu }_t-{\nu }_{t_0}|+|r_t-r_{t_0}|)\\ & +{\tilde{L}}_3\sum _{i=1}^N|r_{i,t}-r_{i,t_0}|.\end{array}$$

Taking into account that both ${\theta }_{t_0}\in (0,min\{\frac{1}{2},{\parallel A\parallel }^{-1}\})$ and $0\le \gamma l<\tau =1-\sqrt{1-\mu (2\eta -\mu {\rho }^2)}$ imply

$$0<1-{\theta }_{t_0}(\overline{\gamma }-1+t_0\tau )<1,$$

we calculate from (3.1)

$$\begin{array}{rl}\parallel x_t-x_{t_0}\parallel & \le \parallel (I-{\theta }_{t}A)T_{r_t}{\Delta }_t^N{R}_t{x}_t+{\theta }_t(t\gamma V{x}_t+(I-t\mu G)T_{r_t}{\Delta }_t^N{R}_t{x}_t)\\ & -(I-{\theta }_{t_0}A)T_{r_{t_0}}{\Delta }_{t_0}^N{R}_{t_0}{x}_{t_0}+{\theta }_{t_0}(t_0\gamma V{x}_{t_0}+(I-t_0\mu G)T_{r_{t_0}}{\Delta }_{t_0}^N{R}_{t_0}{x}_{t_0})\parallel \\ & =\parallel (I-{\theta }_{t}A)v_t+{\theta }_t(t\gamma V{x}_t+(I-t\mu G)v_t)-(I-{\theta }_{t_0}A)v_{t_0}\\ & +{\theta }_{t_0}(t_0\gamma V{x}_{t_0}+(I-t_0\mu G)v_{t_0})\parallel \\ & \le \parallel (I-{\theta }_{t}A)v_t-(I-{\theta }_{t_0}A)v_t\parallel +\parallel (I-{\theta }_{t_0}A)v_t-(I-{\theta }_{t_0}A)v_{t_0}\parallel \\ & +|{\theta }_t-{\theta }_{t_0}|\parallel t\gamma V{x}_t+(I-t\mu G)v_t\parallel +{\theta }_{t_0}\parallel [t\gamma V{x}_t+(I-t\mu G)v_t]\\ & -[t_0\gamma V{x}_{t_0}+(I-t_0\mu G)v_{t_0}]\parallel \\ & \le |{\theta }_t-{\theta }_{t_0}|\parallel A\parallel \parallel v_t\parallel +(1-{\theta }_{t_0}\overline{\gamma })\parallel v_t-v_{t_0}\parallel +|{\theta }_t-{\theta }_{t_0}|\parallel t\gamma V{x}_t+(I-t\mu G)v_t\parallel \\ & +{\theta }_{t_0}\parallel (t-t_0)\gamma V{x}_t+t_0\gamma (V{x}_t-V{x}_{t_0})-(t-t_0)\mu G{v}_t+(I-t_0\mu G)v_t\\ & -(I-t_0\mu G)v_{t_0}\parallel \\ & \le |{\theta }_t-{\theta }_{t_0}|\parallel A\parallel \parallel v_t\parallel +(1-{\theta }_{t_0}\overline{\gamma })\parallel v_t-v_{t_0}\parallel +|{\theta }_t-{\theta }_{t_0}|[\parallel v_t\parallel \\ & +t(\gamma \parallel V{x}_t\parallel +\mu \parallel G{v}_t\parallel )]\\ & +{\theta }_{t_0}[(\gamma \parallel V{x}_t\parallel +\mu \parallel G{v}_t\parallel )|t-t_0|+t_0\gamma l\parallel x_t-x_{t_0}\parallel +(1-t_0\tau )\parallel v_t-v_{t_0}\parallel ]\\ & \le |{\theta }_t-{\theta }_{t_0}|[\parallel v_t\parallel +\parallel A\parallel \parallel v_t\parallel +\gamma \parallel V{x}_t\parallel +\mu \parallel G{v}_t\parallel ]+{\theta }_{t_0}{t}_0\gamma l\parallel x_t-x_{t_0}\parallel \\ & +[1-{\theta }_{t_0}(\overline{\gamma }-1+t_0\tau )]\\ & \times \{\parallel x_t-x_{t_0}\parallel +(\frac{{\tilde{L}}_1}{b}+\frac{{\tilde{L}}_2}{a})(|{\lambda }_t-{\lambda }_{t_0}|+|{\nu }_t-{\nu }_{t_0}|+|r_t-r_{t_0}|)\\ & +{\tilde{L}}_3\sum _{i=1}^N|r_{i,t}-r_{i,t_0}|\}+{\theta }_{t_0}(\gamma \parallel V{x}_t\parallel +\mu \parallel G{v}_t\parallel )|t-t_0|\\ & =|{\theta }_t-{\theta }_{t_0}|[\parallel v_t\parallel +\parallel A\parallel \parallel v_t\parallel +\gamma \parallel V{x}_t\parallel +\mu \parallel G{v}_t\parallel ]\\ & +[1-{\theta }_{t_0}(\overline{\gamma }-1+t_0(\tau -\gamma l)]\parallel x_t-x_{t_0}\parallel \\ & +[1-{\theta }_{t_0}(\overline{\gamma }-1+t_0\tau )]\{(\frac{{\tilde{L}}_1}{b}+\frac{{\tilde{L}}_2}{a})(|{\lambda }_t-{\lambda }_{t_0}|+|{\nu }_t-{\nu }_{t_0}|+|r_t-r_{t_0}|)\\ & +{\tilde{L}}_3\sum _{i=1}^N|r_{i,t}-r_{i,t_0}|\}+{\theta }_{t_0}(\gamma \parallel V{x}_t\parallel +\mu \parallel G{v}_t\parallel )|t-t_0|.\end{array}$$

This immediately implies that

$$\begin{array}{rl}\parallel x_t-x_{t_0}\parallel & \le \frac{\parallel v_t\parallel +\parallel A\parallel \parallel v_t\parallel +\gamma \parallel V{x}_t\parallel +\mu \parallel G{v}_t\parallel }{{\theta }_{t_0}(\overline{\gamma }-1+t_0(\tau -\gamma l)}|{\theta }_t-{\theta }_{t_0}|+\frac{\gamma \parallel V{x}_t\parallel +\mu \parallel G{v}_t\parallel }{\overline{\gamma }-1+t_0(\tau -\gamma l)}|t-t_0|\\ & +\frac{1-{\theta }_{t_0}(\overline{\gamma }-1+t_0\tau )}{{\theta }_{t_0}(\overline{\gamma }-1+t_0(\tau -\gamma l)}\{(\frac{{\tilde{L}}_1}{b}+\frac{{\tilde{L}}_2}{a})(|{\lambda }_t-{\lambda }_{t_0}|+|{\nu }_t-{\nu }_{t_0}|+|r_t-r_{t_0}|)\\ & +{\tilde{L}}_3\sum _{i=1}^N|r_{i,t}-r_{i,t_0}|\}.\end{array}$$

Since ${\theta }_t:(0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma l}\})\to (0,min\{\frac{1}{2},{\parallel A\parallel }^{-1}\})$ is locally Lipschitzian, $r_t,{\lambda }_t,{\nu }_t:(0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma l}\})\to (0,\mathrm{\infty })$ are locally Lipschitzian, and $r_{i,t}:(0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma l}\})\to [c_i,d_i]$ is locally Lipschitzian for each $i=1,2,\dots ,N$, we deduce that $x_t:(0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma l}\})\to H$ is locally Lipschitzian.

(iv) From the last inequality in (iii), the desired result follows immediately. □

We prove the following strong convergence theorem for the net $\{x_t\}$ as $t\to 0$, which guarantees the existence of solutions of the variational inequality (3.2).

Theorem 3.2

Let the net $\{x_t\}$ be defined via (3.1). If ${lim}_{t\to 0}{\theta }_t=0$, then $x_t$ converges strongly to $\tilde{x}\in \Omega$ as $t\to 0$, which solves VI (3.2). Equivalently, we have $P_{\Omega }(2I-A)\tilde{x}=\tilde{x}$.

Proof

We first note that the uniqueness of a solution of VI (3.2) is a consequence of the strong monotonicity of $A-I$ (due to Lemma 2.4). See [2, 4, 5] for this fact.

Next, we prove that $x_t\to \tilde{x}$ as $t\to 0$. For simplicity, let $v_t=T_{r_t}{u}_t$, $u_t={\Delta }_t^N{z}_t$, $y_t=F_{2,{\nu }_t}{x}_t$, and $z_t=R_t{x}_t=F_{1,{\lambda }_t}{y}_t$. For any given $p\in \Omega$, we observe that $T_{r_t}p=p$, ${\Delta }_t^{N}p=p$, and $R_{t}p=p$. From (3.1), we write

$$\begin{array}{rl}{x}_t-p& =x_t-w_t+w_t-p=x_t-w_t+(I-{\theta }_{t}A)v_t+{\theta }_t(t\gamma V{x}_t+(I-t\mu G)v_t)-p\\ & =x_t-w_t+(I-{\theta }_{t}A)(v_t-p)+{\theta }_t[t(\gamma V{x}_t-\mu Gp)+(I-t\mu G)v_t-(I-t\mu G)p]\\ & +{\theta }_t(I-A)p,\end{array}$$

where $w_t=(I-{\theta }_{t}A)v_t+{\theta }_t(t\gamma V{x}_t+(I-t\mu G)v_t)$. In terms of (2.1) and (3.5), we have

$$\begin{array}{rl}{\parallel x_t-p\parallel }^2& =\langle x_t-w_t,x_t-p\rangle +\langle (I-{\theta }_{t}A)(v_t-p),x_t-p\rangle +{\theta }_t[t\langle \gamma V{x}_t-\mu Gp,x_t-p\rangle \\ & +\langle (I-t\mu G)v_t-(I-t\mu G)p,x_t-p\rangle ]+{\theta }_t\langle (I-A)p,x_t-p\rangle \\ & \le (1-{\theta }_t\overline{\gamma }){\parallel x_t-p\parallel }^2+{\theta }_t[(1-t\tau ){\parallel x_t-p\parallel }^2+t\gamma l{\parallel x_t-p\parallel }^2\\ & +t\langle (\gamma V-\mu G)p,x_t-p\rangle ]+{\theta }_t\langle (I-A)p,x_t-p\rangle \\ & =[1-{\theta }_t(\overline{\gamma }-1+t(\tau -\gamma l))]{\parallel x_t-p\parallel }^2+{\theta }_t(t\langle (\gamma V-\mu G)p,x_t-p\rangle \\ & +\langle (I-A)p,x_t-p\rangle ).\end{array}$$

Therefore,

$${\parallel x_t-p\parallel }^2\le \frac{1}{\overline{\gamma }-1+t(\tau -\gamma l)}(t\langle (\gamma V-\mu G)p,x_t-p\rangle +\langle (I-A)p,x_t-p\rangle ).$$ 3.26

Since $\{x_t\}$ is bounded as $t\to 0$ (due to Theorem 3.1(i)), there exists a subsequence $\{t_n\}$ in $(0,min\{1,\frac{2-\overline{\gamma }}{\tau -\gamma l}\})$ such that $t_n\to 0$ and $x_{t_n}\rightharpoonup x^{\ast }$. We first show that $x^{\ast }\in \Omega$. To this end, we divide its proof into four steps.

Step 1. We claim that ${lim}_{n\to \mathrm{\infty }}\parallel x_{t_n}-z_{t_n}\parallel =0$, ${lim}_{n\to \mathrm{\infty }}\parallel z_{t_n}-u_{t_n}\parallel =0$, and ${lim}_{n\to \mathrm{\infty }}\parallel u_{t_n}-v_{t_n}\parallel =0$, where $z_{t_n}=R_{t_n}{x}_{t_n}$, $u_{t_n}={\Delta }_{t_n}^N{z}_{t_n}$, and $v_{t_n}=T_{r_{t_n}}{u}_{t_n}$. Indeed, according to (3.9), (3.12), and (3.14) in the proof of Theorem 3.1, we obtain the assertion.

Step 2. We claim that $x^{\ast }\in Fix(T)$. In fact, from the definition of $v_{t_n}=T_{r_{t_n}}{u}_{t_n}$, we have

$$\langle y-v_{t_n},(I-T)v_{t_n}\rangle +\langle y-v_{t_n},\frac{v_{t_n}-u_{t_n}}{r_{t_n}}\rangle \ge 0,\mathrm{\forall }y\in C.$$ 3.27

Set $w_t=tv+(1-t)x^{\ast }$ for all $t\in (0,1]$ and $v\in C$. Then $w_t\in C$. From (3.27) it follows that

$$\begin{array}{rl}\langle w_t-v_{t_n},(I-T)w_t\rangle & \ge \langle w_t-v_{t_n},(I-T)w_t\rangle -\langle w_t-v_{t_n},(I-T)v_{t_n}\rangle -\langle w_t-v_{t_n},\frac{v_{t_n}-u_{t_n}}{r_{t_n}}\rangle \\ & =\langle w_t-v_{t_n},(I-T)w_t-(I-T)v_{t_n}\rangle -\langle w_t-v_{t_n},\frac{v_{t_n}-u_{t_n}}{r_{t_n}}\rangle .\end{array}$$ 3.28

By Step 1, we have $\frac{v_{t_n}-u_{t_n}}{r_{t_n}}\to 0$ as $n\to \mathrm{\infty }$. Moreover, since $x_{t_n}\rightharpoonup x^{\ast }$, by Step 1 we have $v_{t_n}\rightharpoonup x^{\ast }$. Since $I-T$ is monotone, we also have that $\langle w_t-v_{t_n},(I-T)w_t-(I-T)v_{t_n}\rangle \ge 0$. Thus, from (3.28) it follows that

$$0\le \underset{n\to \mathrm{\infty }}{lim}\langle w_t-v_{t_n},(I-T)w_t\rangle =\langle w_t-x^{\ast },(I-T)w_t\rangle ,$$

and hence

$$\langle v-x^{\ast },(I-T)w_t\rangle \ge 0,\mathrm{\forall }v\in C.$$

Letting $t\to 0$, we know from the continuity of $I-T$ that

$$\langle v-x^{\ast },(I-T)x^{\ast }\rangle \ge 0,\mathrm{\forall }v\in C.$$

Putting $v=T{x}^{\ast }$, we get ${\parallel (I-T)x^{\ast }\parallel }^2=0$, which leads to $x^{\ast }\in Fix(T)$.

Step 3. We claim that $x^{\ast }\in GSVI(C,F_1,F_2)$. Indeed, note that ${lim}_{t\to 0}{\lambda }_t=\lambda >0$ and ${lim}_{t\to 0}{\nu }_t=\nu >0$. For each $x\in C$, we put $x(t):=F_{1,{\lambda }_t}x$, $x(0):=F_{1,\lambda }x$, $y(t):=F_{2,{\nu }_t}x$, and $y(0):=F_{2,\nu }x$. Then, by Lemma 1.1, we have $GSVI(C,F_1,F_2)=Fix(R)$, where $R=F_{1,\lambda }{F}_{2,\nu }$ and R is nonexpansive. Moreover, it is easy to see that

$$\langle y-x(t),F_{1}x(t)\rangle +\frac{1}{{\lambda }_t}\langle y-x(t),x(t)-x\rangle \ge 0,\mathrm{\forall }y\in C,$$ 3.29

and

$$\langle y-x(0),F_{1}x(0)\rangle +\frac{1}{\lambda }\langle y-x(0),x(0)-x\rangle \ge 0,\mathrm{\forall }y\in C.$$ 3.30

Putting $y=x(0)$ in (3.29) and $y=x(t)$ in (3.30), we obtain

$$\langle x(0)-x(t),F_{1}x(t)\rangle +\frac{1}{{\lambda }_t}\langle x(0)-x(t),x(t)-x\rangle \ge 0$$ 3.31

and

$$\langle x(t)-x(0),F_{1}x(0)\rangle +\frac{1}{\lambda }\langle x(t)-x(0),x(0)-x\rangle \ge 0.$$ 3.32

Adding up (3.31) and (3.32), we have

$$-\langle x(t)-x(0),F_{1}x(t)-F_{1}x(0)\rangle +\langle x(0)-x(t),\frac{x(t)-x}{{\lambda }_t}-\frac{x(0)-x}{\lambda }\rangle \ge 0.$$

Since $F_1$ is a monotone mapping, we know that

$$\langle x(0)-x(t),\frac{x(t)-x}{{\lambda }_t}-\frac{x(0)-x}{\lambda }\rangle \ge 0,$$

and hence

$$\langle x(t)-x(0),x(0)-x(t)+x(t)-x-\frac{\lambda }{{\lambda }_t}(x(t)-x)\rangle \ge 0.$$

So it follows that

$$\begin{array}{rl}{\parallel x(t)-x(0)\parallel }^2& \le \langle x(t)-x(0),x(t)-x-\frac{\lambda }{{\lambda }_t}(x(t)-x)\rangle \\ & =\langle x(t)-x(0),(1-\frac{\lambda }{{\lambda }_t})(x(t)-x)\rangle \\ & \le \parallel x(t)-x(0)\parallel \cdot \frac{|{\lambda }_t-\lambda |}{{\lambda }_t}\parallel x(t)-x\parallel ,\end{array}$$

which immediately yields

$$\parallel F_{1,{\lambda }_t}x-F_{1,\lambda }x\parallel \le \frac{|{\lambda }_t-\lambda |}{{\lambda }_t}\parallel F_{1,{\lambda }_t}x-x\parallel .$$ 3.33

By using arguments similar to those of (3.33), we have

$$\parallel F_{2,{\nu }_t}x-F_{2,\nu }x\parallel \le \frac{|{\nu }_t-\nu |}{{\nu }_t}\parallel F_{2,{\nu }_t}x-x\parallel .$$ 3.34

Now, putting $t=t_n$, $x=F_{2,\nu }{x}_{t_n}$ in (3.33), and $t=t_n$, $x=x_{t_n}$ in (3.34), respectively, we deduce that

$$\parallel F_{1,{\lambda }_{t_n}}{F}_{2,\nu }{x}_{t_n}-F_{1,\lambda }{F}_{2,\nu }{x}_{t_n}\parallel \le \frac{|{\lambda }_{t_n}-\lambda |}{{\lambda }_{t_n}}\parallel F_{1,{\lambda }_{t_n}}{F}_{2,\nu }{x}_{t_n}-F_{2,\nu }{x}_{t_n}\parallel$$

and

$$\parallel F_{2,{\nu }_{t_n}}{x}_{t_n}-F_{2,\nu }{x}_{t_n}\parallel \le \frac{|{\nu }_{t_n}-\nu |}{{\nu }_{t_n}}\parallel F_{2,{\nu }_{t_n}}{x}_{t_n}-x_{t_n}\parallel .$$

Since ${lim}_{n\to \mathrm{\infty }}{\lambda }_{t_n}=\lambda >0$ and ${lim}_{n\to \mathrm{\infty }}{\nu }_{t_n}=\nu >0$, it follows from the last two inequalities that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel F_{1,{\lambda }_{t_n}}{F}_{2,\nu }{x}_{t_n}-F_{1,\lambda }{F}_{2,\nu }{x}_{t_n}\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel F_{2,{\nu }_{t_n}}{x}_{t_n}-F_{2,\nu }{x}_{t_n}\parallel =0.$$ 3.35

Also, we observe that

$$\begin{array}{rl}& \parallel R{x}_{t_n}-x_{t_n}\parallel \\ & \le \parallel F_{1,\lambda }{F}_{2,\nu }{x}_{t_n}-F_{1,{\lambda }_{t_n}}{F}_{2,\nu }{x}_{t_n}\parallel +\parallel F_{1,{\lambda }_{t_n}}{F}_{2,\nu }{x}_{t_n}-F_{1,{\lambda }_{t_n}}{F}_{2,{\nu }_{t_n}}{x}_{t_n}\parallel \\ & +\parallel F_{1,{\lambda }_{t_n}}{F}_{2,{\nu }_{t_n}}{x}_{t_n}-x_{t_n}\parallel \\ & \le \parallel F_{1,\lambda }{F}_{2,\nu }{x}_{t_n}-F_{1,{\lambda }_{t_n}}{F}_{2,\nu }{x}_{t_n}\parallel +\parallel F_{2,\nu }{x}_{t_n}-F_{2,{\nu }_{t_n}}{x}_{t_n}\parallel +\parallel F_{1,{\lambda }_{t_n}}{F}_{2,{\nu }_{t_n}}{x}_{t_n}-x_{t_n}\parallel \\ & =\parallel F_{1,\lambda }{F}_{2,\nu }{x}_{t_n}-F_{1,{\lambda }_{t_n}}{F}_{2,\nu }{x}_{t_n}\parallel +\parallel F_{2,\nu }{x}_{t_n}-F_{2,{\nu }_{t_n}}{x}_{t_n}\parallel +\parallel R_{t_n}{x}_{t_n}-x_{t_n}\parallel .\end{array}$$ 3.36

Since $R_{t_n}{x}_{t_n}-x_{t_n}\to 0$ (due to Step 1), from (3.35) and (3.36) we get

$$\underset{n\to \mathrm{\infty }}{lim}\parallel R{x}_{t_n}-x_{t_n}\parallel =0.$$ 3.37

Taking into account that $x_{t_n}\rightharpoonup x^{\ast }$ and $x_{t_n}-R{x}_{t_n}\to 0$ (due to (3.37)), from Lemma 2.3 we get $x^{\ast }=R{x}^{\ast }$, that is, $x^{\ast }\in Fix(R)=GSVI(C,F_1,F_2)$.

Step 4. We claim that $x^{\ast }\in {\bigcap }_{i=1}^{N}GMEP({\Theta }_i,{\phi }_i,B_i)$. In fact, since ${\Delta }_{t_n}^i{z}_{t_n}=T_{r_{i,t_n}}^{({\Theta }_i,{\phi }_i)}(I-r_{i,t_n}{B}_i){\Delta }_{t_n}^{i-1}{z}_{t_n}$, for each $i=1,2,\dots ,N$, we have

$$\begin{array}{rl}0& \le {\Theta }_i({\Delta }_{t_n}^i{z}_{t_n},y)+{\phi }_i(y)-{\phi }_i\left({\Delta }_{t_n}^i{z}_{t_n}\right)\\ & +\langle B_i{\Delta }_{t_n}^{i-1}{z}_{t_n},y-{\Delta }_{t_n}^i{z}_{t_n}\rangle +\frac{1}{r_{i,t_n}}\langle y-{\Delta }_{t_n}^i{z}_{t_n},{\Delta }_{t_n}^i{z}_{t_n}-{\Delta }_{t_n}^{i-1}{z}_{t_n}\rangle .\end{array}$$

By (A2), we have

$$\begin{array}{rl}{\Theta }_i(y,{\Delta }_{t_n}^i{z}_{t_n})& \le {\phi }_i(y)-{\phi }_i\left({\Delta }_{t_n}^i{z}_{t_n}\right)+\langle B_i{\Delta }_{t_n}^{i-1}{z}_{t_n},y-{\Delta }_{t_n}^i{z}_{t_n}\rangle \\ & +\frac{1}{r_{i,t_n}}\langle y-{\Delta }_{t_n}^i{z}_{t_n},{\Delta }_{t_n}^i{z}_{t_n}-{\Delta }_{t_n}^{i-1}{z}_{t_n}\rangle .\end{array}$$

Let $w_t=tv+(1-t)x^{\ast }$ for all $t\in (0,1]$ and $v\in C$. This implies that $w_t\in C$. Then we have

$$\begin{array}{rl}& \langle w_t-{\Delta }_{t_n}^i{z}_{t_n},B_i{w}_t\rangle \\ & \ge {\phi }_i\left({\Delta }_{t_n}^i{z}_{t_n}\right)-{\phi }_i(w_t)+\langle w_t-{\Delta }_{t_n}^i{z}_{t_n},B_i{w}_t\rangle -\langle w_t-{\Delta }_{t_n}^i{z}_{t_n},B_i{\Delta }_{t_n}^{i-1}{z}_{t_n}\rangle \\ & -\langle w_t-{\Delta }_{t_n}^i{z}_{t_n},\frac{{\Delta }_{t_n}^i{z}_{t_n}-{\Delta }_{t_n}^{i-1}{z}_{t_n}}{r_{i,t_n}}\rangle +{\Theta }_i(w_t,{\Delta }_{t_n}^i{z}_{t_n})\\ & ={\phi }_i\left({\Delta }_{t_n}^i{z}_{t_n}\right)-{\phi }_i(w_t)+\langle w_t-{\Delta }_{t_n}^i{z}_{t_n},B_i{w}_t-B_i{\Delta }_{t_n}^i{z}_{t_n}\rangle \\ & +\langle w_t-{\Delta }_{t_n}^i{z}_{t_n},B_j{\Delta }_{t_n}^i{z}_{t_n}-B_i{\Delta }_{t_n}^{i-1}{z}_{t_n}\rangle \\ & -\langle w_t-{\Delta }_{t_n}^i{z}_{t_n},\frac{{\Delta }_{t_n}^i{z}_{t_n}-{\Delta }_{t_n}^{i-1}{z}_{t_n}}{r_{i,t_n}}\rangle +{\Theta }_i(w_t,{\Delta }_{t_n}^i{z}_{t_n}).\end{array}$$

By the same arguments as in the proof of Theorem 3.1, we have $\parallel B_i{\Delta }_{t_n}^i{z}_{t_n}-B_i{\Delta }_{t_n}^{i-1}{z}_{t_n}\parallel \to 0$ as $n\to \mathrm{\infty }$. In the meantime, by the monotonicity of $B_i$, we obtain $\langle w_t-{\Delta }_{t_n}^i{z}_{t_n},B_i{w}_t-B_i{\Delta }_{t_n}^i{z}_{t_n}\rangle \ge 0$. Then by (A4) we get

$$\langle w_t-x^{\ast },B_i{w}_t\rangle \ge {\phi }_i\left(x^{\ast }\right)-{\phi }_i(w_t)+{\Theta }_i(w_t,x^{\ast }).$$

Utilizing (A1), (A4), and the last inequality, we obtain

$$\begin{array}{rl}0& ={\Theta }_i(w_t,w_t)+{\phi }_i(w_t)-{\phi }_i(w_t)\\ & \le t{\Theta }_i(w_t,v)+(1-t){\Theta }_i(w_t,x^{\ast })+t{\phi }_i(v)+(1-t){\phi }_i\left(x^{\ast }\right)-{\phi }_i(w_t)\\ & \le t[{\Theta }_i(w_t,v)+{\phi }_i(v)-{\phi }_i(w_t)]+(1-t)\langle w_t-x^{\ast },B_i{w}_t\rangle \\ & =t[{\Theta }_i(w_t,v)+{\phi }_i(v)-{\phi }_i(w_t)]+(1-t)t\langle v-x^{\ast },B_i{w}_t\rangle ,\end{array}$$

and hence

$$0\le {\Theta }_i(w_t,v)+{\phi }_i(v)-{\phi }_i(w_t)+(1-t)\langle v-x^{\ast },B_i{w}_t\rangle .$$

Letting $t\to 0$, we have, for each $v\in C$,

$$0\le {\Theta }_i(x^{\ast },v)+{\phi }_i(v)-{\phi }_i\left(x^{\ast }\right)+\langle v-x^{\ast },B_i{x}^{\ast }\rangle .$$

This implies that $x^{\ast }\in GMEP({\Theta }_i,{\phi }_i,B_i)$ and hence $x^{\ast }\in {\bigcap }_{i=1}^{N}GMEP({\Theta }_i,{\phi }_i,B_i)$. This together with Steps 2 and 3 attains $x^{\ast }\in \Omega$.

Finally, we show that $x^{\ast }$ is a solution of VI (3.2). In fact, putting $x_{t_n}$ in place of $x_t$ in (3.26) and taking the limit as $t_n\to 0$, we obtain

$${\parallel x^{\ast }-p\parallel }^2\le \frac{1}{\overline{\gamma }-1}\langle (I-A)p,x^{\ast }-p\rangle ,\mathrm{\forall }p\in \Omega .$$

In particular, $x^{\ast }$ solves the following VI:

$$x^{\ast }\in \Omega ,\langle (A-I)p,x^{\ast }-p\rangle \le 0,\mathrm{\forall }p\in \Omega ,$$

or the equivalent dual variational inequality

$$x^{\ast }\in \Omega ,\langle (A-I)x^{\ast },x^{\ast }-p\rangle \le 0,\mathrm{\forall }p\in \Omega .$$

That is, $x^{\ast }\in \Omega$ is a solution of VI (3.2). Hence $x^{\ast }=\tilde{x}$ by uniqueness. In a summary, we have proven that each cluster point of $\{x_t\}$ (as $t\to 0$) equals x̃. Therefore $x_t\to \tilde{x}$ as $t\to 0$. VI (3.2) can be rewritten as

$$\langle (2I-A)\tilde{x}-\tilde{x},\tilde{x}-p\rangle \ge 0,\mathrm{\forall }p\in \Omega .$$

So, in terms of (2.1), this is equivalent to the fixed point equation

$$P_{\Omega }(2I-A)\tilde{x}=\tilde{x}.$$

This completes the proof. □

Taking $T\equiv I$, $G\equiv I$, $\mu =1$, and $\gamma =1$ in Theorem 3.2, we have the following corollary.

Corollary 3.1

Let $\{x_t\}$ be defined by

$$x_t=P_C[(I-{\theta }_{t}A){\Delta }_t^N{R}_t{x}_t+{\theta }_t(tV{x}_t+(1-t){\Delta }_t^N{R}_t{x}_t)].$$

If ${lim}_{t\to 0}{\theta }_t=0$, then $x_t$ converges strongly as $t\to 0$ to $\tilde{x}\in \Omega :={\bigcap }_{i=1}^{N}GMEP({\Theta }_i,{\phi }_i,B_i)\cap GSVI(C,B_1,B_2)$, which is the unique solution of the VI

$$\langle (A-I)\tilde{x},\tilde{x}-p\rangle \le 0,\mathrm{\forall }p\in \Omega .$$ 3.38

Proof

If $T\equiv I$, then $T_r$ in Lemma 2.8 is the identity mapping. Thus the result follows from Theorem 3.2. □

We are now in a position to prove the strong convergence of the sequence $\{x_n\}$ generated by the general explicit iterative scheme (3.3) to $\tilde{x}\in \Omega$, which is the unique solution to VI (3.2).

Theorem 3.3

Let $\{x_n\}$ be the sequence generated by the explicit algorithm (3.3). Let $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{r_n\}$, $\{{\lambda }_n\}$, $\{{\nu }_n\}$, and ${\{r_{i,n}\}}_{i=1}^N$ satisfy the following conditions:

1. $\{{\alpha }_n\}\subset [0,1]$ and $\{{\beta }_n\}\subset (0,1]$, ${\alpha }_n\to 0$ and ${\beta }_n\to 0$ as $n\to \mathrm{\infty }$;

2. ${\sum }_{n=0}^{\mathrm{\infty }}{\beta }_n=\mathrm{\infty }$;

3. ${\sum }_{n=0}^{\mathrm{\infty }}|{\alpha }_{n+1}-{\alpha }_n|<\mathrm{\infty }$, and $|{\beta }_{n+1}-{\beta }_n|\le o({\beta }_{n+1})+{\sigma }_n$, ${\sum }_{n=0}^{\mathrm{\infty }}{\sigma }_n<\mathrm{\infty }$ (the perturbed control condition);

4. $\{r_n\}\subset (0,\mathrm{\infty })$, ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$, and ${\sum }_{n=0}^{\mathrm{\infty }}|r_{n+1}-r_n|<\mathrm{\infty }$;

5. $\{{\lambda }_n\}\subset (0,\mathrm{\infty })$, ${lim}_{n\to \mathrm{\infty }}{\lambda }_n=\lambda >0$, and ${\sum }_{n=0}^{\mathrm{\infty }}|{\lambda }_{n+1}-{\lambda }_n|<\mathrm{\infty }$;

6. $\{{\nu }_n\}\subset (0,\mathrm{\infty })$, ${lim}_{n\to \mathrm{\infty }}{\nu }_n=\nu >0$, and ${\sum }_{n=0}^{\mathrm{\infty }}|{\nu }_{n+1}-{\nu }_n|<\mathrm{\infty }$;

7. $\{r_{i,n}\}\subset [c_i,d_i]\subset (0,2{\mu }_i)$ $\mathrm{\forall }i\in \{1,2,\dots ,N\}$, and ${\sum }_{n=0}^{\mathrm{\infty }}({\sum }_{i=1}^N|r_{i,n+1}-r_{i,n}|)<\mathrm{\infty }$.

Then $\{x_n\}$ converges strongly to $\tilde{x}\in \Omega :={\bigcap }_{i=1}^{N}GMEP({\Theta }_i,{\phi }_i,B_i)\cap GSVI(C,F_1,F_2)\cap Fix(T)$, which is the unique solution of VI (3.2).

Proof

First, note that from condition (C1), without loss of generality, we assume that ${\alpha }_n\tau <1$, ${\beta }_n\overline{\gamma }<1$ and $\frac{2{\beta }_n(\overline{\gamma }-1)}{1-{\beta }_n}<1$ for all $n\ge 0$. Let $\tilde{x}\in \Omega$ be the unique solution of VI (3.2). (The existence of x̃ follows from Theorem 3.2.)

From now, we put $z_n=R_n{x}_n$, $u_n={\Delta }_n^N{z}_n$, and $v_n=T_{r_n}{u}_n$. Take $p\in \Omega$. Then $p=T_{r_n}p$ by Lemma 2.8(iii), $p={\Delta }_n^{i}p$ ($=T_{r_{i,n}}^{({\Theta }_i,{\phi }_i)}(I-r_{i,n}{B}_i)p$) by Proposition 2.1(iii), and $p=R_{n}p$ by Lemma 1.1.

We divide the proof into several steps as follows.

Step 1. We show that $\{x_n\}$ is bounded. Indeed, utilizing Proposition 2.1(ii) and Proposition 2.2, we have

$$\begin{array}{rl}\parallel u_n-p\parallel & =\parallel T_{r_{N,n}}^{({\Theta }_N,{\phi }_N)}(I-r_{N,n}{B}_N){\Delta }_n^{N-1}{z}_n-T_{r_{N,n}}^{({\Theta }_N,{\phi }_N)}(I-r_{N,n}{B}_N){\Delta }_n^{N-1}p\parallel \\ & \le \parallel (I-r_{N,n}{B}_N){\Delta }_n^{N-1}{z}_n-(I-r_{N,n}{B}_N){\Delta }_n^{N-1}p\parallel \\ & \le \parallel {\Delta }_n^{N-1}{z}_n-{\Delta }_n^{N-1}p\parallel \\ & \le \cdots \\ & \le \parallel {\Delta }_n^0{z}_n-{\Delta }_n^{0}p\parallel =\parallel z_n-p\parallel .\end{array}$$ 3.39

It is easy from the nonexpansion of $R_n$ to see that

$$\parallel z_n-p\parallel =\parallel R_n{x}_n-R_{n}p\parallel \le \parallel x_n-p\parallel ,$$

which together with the nonexpansion of $T_{r_n}$ and (3.39) implies that

$$\parallel v_n-p\parallel =\parallel T_{r_n}{u}_n-T_{r_n}p\parallel \le \parallel u_n-p\parallel \le \parallel z_n-p\parallel \le \parallel x_n-p\parallel .$$ 3.40

From (3.3) and (3.40), we get

$$\begin{array}{rl}& \parallel x_{n+1}-p\parallel \\ & \le \parallel (I-{\beta }_{n}A)v_n+{\beta }_n({\alpha }_n\gamma V{x}_n+(I-{\alpha }_n\mu G)v_n)-p\parallel \\ & =\parallel (I-{\beta }_{n}A)v_n-(I-{\beta }_{n}A)p+{\beta }_n({\alpha }_n\gamma V{x}_n+(I-{\alpha }_n\mu G)v_n-p)+{\beta }_n(I-A)p\parallel \\ & \le \parallel (I-{\beta }_{n}A)v_n-(I-{\beta }_{n}A)p\parallel +{\beta }_n\parallel {\alpha }_n\gamma V{x}_n+(I-{\alpha }_n\mu G)v_n-p\parallel +{\beta }_n\parallel (I-A)p\parallel \\ & =\parallel (I-{\beta }_{n}A)v_n-(I-{\beta }_{n}A)p\parallel \\ & +{\beta }_n\parallel (I-{\alpha }_n\mu G)v_n-(I-{\alpha }_n\mu G)p+{\alpha }_n(\gamma V{x}_n-\mu Gp)\parallel +{\beta }_n\parallel (I-A)p\parallel \\ & \le (1-{\beta }_n\overline{\gamma })\parallel v_n-p\parallel +{\beta }_n[\parallel (I-{\alpha }_n\mu G)v_n-(I-{\alpha }_n\mu G)p\parallel \\ & +{\alpha }_n(\gamma \parallel V{x}_n-Vp\parallel +\parallel \gamma Vp-\mu Gp\parallel )]+{\beta }_n\parallel (I-A)p\parallel \\ & \le (1-{\beta }_n\overline{\gamma })\parallel x_n-p\parallel +{\beta }_n[(1-{\alpha }_n\tau )\parallel x_n-p\parallel +{\alpha }_n(\gamma l\parallel x_n-p\parallel +\parallel (\gamma V-\mu G)p\parallel )]\\ & +{\beta }_n\parallel I-A\parallel \parallel p\parallel \\ & =[1-{\beta }_n(\overline{\gamma }-1+{\alpha }_n(\tau -\gamma l))]\parallel x_n-p\parallel +{\beta }_n[\parallel I-A\parallel \parallel p\parallel +{\alpha }_n\parallel (\gamma V-\mu G)p\parallel ]\\ & \le [1-{\beta }_n(\overline{\gamma }-1)]\parallel x_n-p\parallel +{\beta }_n[\parallel I-A\parallel \parallel p\parallel +\parallel (\gamma V-\mu G)p\parallel ]\\ & =[1-{\beta }_n(\overline{\gamma }-1)]\parallel x_n-p\parallel +{\beta }_n(\overline{\gamma }-1)\frac{\parallel I-A\parallel \parallel p\parallel +\parallel (\gamma V-\mu G)p\parallel }{\overline{\gamma }-1}\\ & \le max\{\parallel x_n-p\parallel ,\frac{\parallel I-A\parallel \parallel p\parallel +\parallel (\gamma V-\mu G)p\parallel }{\overline{\gamma }-1}\}.\end{array}$$

By induction, we derive

$$\parallel x_n-p\parallel \le max\{\parallel x_0-p\parallel ,\frac{\parallel I-A\parallel \parallel p\parallel +\parallel (\gamma V-\mu G)p\parallel }{\overline{\gamma }-1}\},\mathrm{\forall }n\ge 0.$$

This implies that $\{x_n\}$ is bounded and so are $\{V{x}_n\}$, $\{u_n\}$, $\{v_n\}$, $\{w_n\}$, $\{z_n\}$, and $\{G{v}_n\}$. As a consequence, with the control condition (C1), we get

$$\parallel x_{n+1}-v_n\parallel \le {\beta }_n\parallel w_n-A{v}_n\parallel \to 0(n\to \mathrm{\infty }).$$ 3.41

Step 2. We show that ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$. To this end, let $y_n=F_{2,{\nu }_n}{x}_n$, $y_{n-1}=F_{2,{\nu }_{n-1}}{x}_{n-1}$, $z_n=F_{1,{\lambda }_n}{y}_n$, and $z_{n-1}=F_{1,{\lambda }_{n-1}}{y}_{n-1}$. Then we derive

$$\langle y-y_{n-1},F_2{y}_{n-1}\rangle +\frac{1}{{\nu }_{n-1}}\langle y-y_{n-1},y_{n-1}-x_{n-1}\rangle \ge 0,\mathrm{\forall }y\in C,$$ 3.42

and

$$\langle y-y_n,F_2{y}_n\rangle +\frac{1}{{\nu }_n}\langle y-y_n,y_n-x_n\rangle \ge 0,\mathrm{\forall }y\in C.$$ 3.43

Putting $y=y_n$ in (3.42) and $y=y_{n-1}$ in (3.43), we obtain

$$\langle y_n-y_{n-1},F_2{y}_{n-1}\rangle +\frac{1}{{\nu }_{n-1}}\langle y_n-y_{n-1},y_{n-1}-x_{n-1}\rangle \ge 0$$ 3.44

and

$$\langle y_{n-1}-y_n,F_2{y}_n\rangle +\frac{1}{{\nu }_n}\langle y_{n-1}-y_n,y_n-x_n\rangle \ge 0.$$ 3.45

Adding up (3.44) and (3.45), we have

$$\langle y_n-y_{n-1},F_2{y}_{n-1}-F_2{y}_n\rangle +\langle y_n-y_{n-1},\frac{y_{n-1}-x_{n-1}}{{\nu }_{n-1}}-\frac{y_n-x_n}{{\nu }_n}\rangle \ge 0,$$

which together with the monotonicity of $F_2$ implies that

$$\langle y_n-y_{n-1},\frac{y_{n-1}-x_{n-1}}{{\nu }_{n-1}}-\frac{y_n-x_n}{{\nu }_n}\rangle \ge 0,$$

and hence

$$\langle y_n-y_{n-1},y_{n-1}-y_n+y_n-x_{n-1}-\frac{{\nu }_{n-1}}{{\nu }_n}(y_n-x_n)\rangle \ge 0.$$

It follows that

$$\begin{array}{rl}{\parallel y_n-y_{n-1}\parallel }^2& \le \langle y_n-y_{n-1},x_n-x_{n-1}+(1-\frac{{\nu }_{n-1}}{{\nu }_n})(y_n-x_n)\rangle \\ & \le \parallel y_n-y_{n-1}\parallel (\parallel x_n-x_{n-1}\parallel +\frac{1}{{\nu }_n}|{\nu }_n-{\nu }_{n-1}|\parallel y_n-x_n\parallel ),\end{array}$$

which immediately yields

$$\parallel y_n-y_{n-1}\parallel \le \parallel x_n-x_{n-1}\parallel +\frac{1}{{\nu }_n}|{\nu }_n-{\nu }_{n-1}|\parallel y_n-x_n\parallel .$$ 3.46

By using arguments similar to those of (3.46), we get

$$\parallel z_n-z_{n-1}\parallel \le \parallel y_n-y_{n-1}\parallel +\frac{1}{{\lambda }_n}|{\lambda }_n-{\lambda }_{n-1}|\parallel z_n-y_n\parallel .$$ 3.47

Substituting (3.46) for (3.47), we have

$$\begin{array}{rl}\parallel z_n-z_{n-1}\parallel & \le \parallel y_n-y_{n-1}\parallel +\frac{1}{{\lambda }_n}|{\lambda }_n-{\lambda }_{n-1}|\parallel z_n-y_n\parallel \\ & \le \parallel x_n-x_{n-1}\parallel +\frac{1}{{\nu }_n}|{\nu }_n-{\nu }_{n-1}|\parallel y_n-x_n\parallel +\frac{1}{{\lambda }_n}|{\lambda }_n-{\lambda }_{n-1}|\parallel z_n-y_n\parallel .\end{array}$$ 3.48

Note that $v_n=T_{r_n}{u}_n$ and $v_{n-1}=T_{r_{n-1}}{u}_{n-1}$. By using arguments similar to those of (3.46), we obtain

$$\parallel v_n-v_{n-1}\parallel \le \parallel u_n-u_{n-1}\parallel +\frac{1}{r_n}|r_n-r_{n-1}|\parallel v_n-u_n\parallel .$$ 3.49

Also, utilizing arguments similar to those of (3.25) in the proof of Theorem 3.1, we have

$$\begin{array}{rl}\parallel u_n-u_{n-1}\parallel & =\parallel {\Delta }_n^N{z}_n-{\Delta }_{n-1}^N{z}_{n-1}\parallel \\ & \le |r_{N,n}-r_{N,n-1}|[\parallel B_N{\Delta }_n^{N-1}{z}_n\parallel +\frac{1}{r_{N,n}}\parallel T_{r_{N,n}}^{({\Theta }_N,{\phi }_N)}(I-r_{N,n}{B}_N){\Delta }_n^{N-1}{z}_n\\ & -(I-r_{N,n}{B}_N){\Delta }_n^{N-1}{z}_n\parallel ]+\cdots +|r_{1,n}-r_{1,n-1}|[\parallel B_1{\Delta }_n^0{z}_n\parallel \\ & +\frac{1}{r_{1,n}}\parallel T_{r_{1,n}}^{({\Theta }_1,{\phi }_1)}(I-r_{1,n}{B}_1){\Delta }_n^0{z}_n-(I-r_{1,n}{B}_1){\Delta }_n^0{z}_n\parallel ]\\ & +\parallel {\Delta }_n^0{z}_n-{\Delta }_{n-1}^0{z}_{n-1}\parallel \\ & \le {\tilde{M}}_1\sum _{i=1}^N|r_{i,n}-r_{i,n-1}|+\parallel z_n-z_{n-1}\parallel ,\end{array}$$ 3.50

where ${\tilde{M}}_1>0$ is a constant such that, for each $n\ge 0$,

$$\sum _{i=1}^N[\parallel B_i{\Delta }_n^{i-1}{z}_n\parallel +\frac{1}{r_{i,n}}\parallel T_{r_{i,n}}^{({\Theta }_i,{\phi }_i)}(I-r_{i,n}{B}_i){\Delta }_n^{i-1}{z}_n-(I-r_{i,n}{B}_i){\Delta }_n^{i-1}{z}_n\parallel ]\}\le {\tilde{M}}_1.$$

So it follows from (3.48), (3.49), and (3.50) that

$$\begin{array}{rl}\parallel v_n-v_{n-1}\parallel & \le \parallel u_n-u_{n-1}\parallel +\frac{1}{r_n}|r_n-r_{n-1}|\parallel v_n-u_n\parallel \\ & \le {\tilde{M}}_1\sum _{i=1}^N|r_{i,n}-r_{i,n-1}|+\parallel z_n-z_{n-1}\parallel +\frac{1}{r_n}|r_n-r_{n-1}|\parallel v_n-u_n\parallel \\ & \le {\tilde{M}}_1\sum _{i=1}^N|r_{i,n}-r_{i,n-1}|+\parallel x_n-x_{n-1}\parallel +\frac{1}{{\nu }_n}|{\nu }_n-{\nu }_{n-1}|\parallel y_n-x_n\parallel \\ & +\frac{1}{{\lambda }_n}|{\lambda }_n-{\lambda }_{n-1}|\parallel z_n-y_n\parallel +\frac{1}{r_n}|r_n-r_{n-1}|\parallel v_n-u_n\parallel .\end{array}$$ 3.51

Since ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$, ${lim}_{n\to \mathrm{\infty }}{\lambda }_n=\lambda >0$, and ${lim}_{n\to \mathrm{\infty }}{\nu }_n=\nu >0$, it is easy to see from (3.51) that, for each $n\ge 0$,

$$\begin{array}{rl}\parallel v_n-v_{n-1}\parallel \le & \parallel x_n-x_{n-1}\parallel +\tilde{M}[\sum _{i=1}^N|r_{i,n}-r_{i,n-1}|+|{\nu }_n-{\nu }_{n-1}|\\ & +|{\lambda }_n-{\lambda }_{n-1}|+|r_n-r_{n-1}|],\end{array}$$ 3.52

where $\tilde{M}>0$ is a constant such that

$$\underset{n\ge 0}{sup}\{{\tilde{M}}_1+\frac{1}{{\nu }_n}\parallel y_n-x_n\parallel +\frac{1}{{\lambda }_n}\parallel z_n-y_n\parallel +\frac{1}{r_n}\parallel v_n-u_n\parallel \}\le \tilde{M}.$$

Now, simple calculations yield that

$$\begin{array}{rl}{w}_n-w_{n-1}& ={\alpha }_n\gamma V{x}_n+(I-{\alpha }_n\mu G)v_n-{\alpha }_{n-1}\gamma V{x}_{n-1}-(I-{\alpha }_{n-1}\mu G)v_{n-1}\\ & =({\alpha }_n-{\alpha }_{n-1})(\gamma V{x}_{n-1}-\mu G{v}_{n-1})+{\alpha }_n\gamma (V{x}_n-V{x}_{n-1})\\ & +(I-{\alpha }_n\mu G)v_n-(I-{\alpha }_n\mu G)v_{n-1}.\end{array}$$

In terms of (3.52) and Lemma 2.6, we obtain

$$\begin{array}{rl}\parallel w_n-w_{n-1}\parallel & \le |{\alpha }_n-{\alpha }_{n-1}|(\gamma \parallel V{x}_{n-1}\parallel +\mu \parallel G{v}_{n-1}\parallel )+{\alpha }_n\gamma l\parallel x_n-x_{n-1}\parallel \\ & +(1-\tau {\alpha }_n)\parallel v_n-v_{n-1}\parallel \\ & \le |{\alpha }_n-{\alpha }_{n-1}|(\gamma \parallel V{x}_{n-1}\parallel +\mu \parallel G{v}_{n-1}\parallel )+{\alpha }_n\gamma l\parallel x_n-x_{n-1}\parallel \\ & +(1-\tau {\alpha }_n)\parallel x_n-x_{n-1}\parallel +\tilde{M}[\sum _{i=1}^N|r_{i,n}-r_{i,n-1}|\\ & +|{\nu }_n-{\nu }_{n-1}|+|{\lambda }_n-{\lambda }_{n-1}|+|r_n-r_{n-1}|]\\ & =|{\alpha }_n-{\alpha }_{n-1}|(\gamma \parallel V{x}_{n-1}\parallel +\mu \parallel G{v}_{n-1}\parallel )+(1-{\alpha }_n(\tau -\gamma l))\parallel x_n-x_{n-1}\parallel \\ & +\tilde{M}[\sum _{i=1}^N|r_{i,n}-r_{i,n-1}|+|{\nu }_n-{\nu }_{n-1}|+|{\lambda }_n-{\lambda }_{n-1}|+|r_n-r_{n-1}|]\\ & \le \parallel x_n-x_{n-1}\parallel +{\tilde{M}}_2[\sum _{i=1}^N|r_{i,n}-r_{i,n-1}|+|{\alpha }_n-{\alpha }_{n-1}|\\ & +|{\nu }_n-{\nu }_{n-1}|+|{\lambda }_n-{\lambda }_{n-1}|+|r_n-r_{n-1}|],\end{array}$$ 3.53

where ${\tilde{M}}_2={sup}_{n\ge 0}\{\gamma \parallel V{x}_n\parallel +\mu \parallel G{v}_n\parallel +\tilde{M}\}$. By (3.53) and Lemma 2.5, we derive

$$\begin{array}{rl}\parallel x_{n+1}-x_n\parallel & \le \parallel (I-{\beta }_{n}A)v_n+{\beta }_n{w}_n-(I-{\beta }_{n-1}A)v_{n-1}-{\beta }_{n-1}{w}_{n-1}\parallel \\ & \le \parallel (I-{\beta }_{n}A)(v_n-v_{n-1})\parallel +|{\beta }_n-{\beta }_{n-1}|\parallel A\parallel \parallel v_{n-1}\parallel \\ & +{\beta }_n\parallel w_n-w_{n-1}\parallel +|{\beta }_n-{\beta }_{n-1}|\parallel w_{n-1}\parallel \\ & \le (1-{\beta }_n\overline{\gamma })\parallel v_n-v_{n-1}\parallel +{\beta }_n[\parallel x_n-x_{n-1}\parallel +{\tilde{M}}_2(\sum _{i=1}^N|r_{i,n}-r_{i,n-1}|\\ & +|{\alpha }_n-{\alpha }_{n-1}|+|{\nu }_n-{\nu }_{n-1}|+|{\lambda }_n-{\lambda }_{n-1}|+|r_n-r_{n-1}|\left)\right]+|{\beta }_n-{\beta }_{n-1}|{\tilde{M}}_3\\ & \le (1-{\beta }_n\overline{\gamma })[\parallel x_n-x_{n-1}\parallel +\tilde{M}(\sum _{i=1}^N|r_{i,n}-r_{i,n-1}|+|{\nu }_n-{\nu }_{n-1}|+|{\lambda }_n-{\lambda }_{n-1}|\\ & +|r_n-r_{n-1}|\left)\right]+{\beta }_n[\parallel x_n-x_{n-1}\parallel +{\tilde{M}}_2(\sum _{i=1}^N|r_{i,n}-r_{i,n-1}|+|{\alpha }_n-{\alpha }_{n-1}|\\ & +|{\nu }_n-{\nu }_{n-1}|+|{\lambda }_n-{\lambda }_{n-1}|+|r_n-r_{n-1}|\left)\right]+|{\beta }_n-{\beta }_{n-1}|{\tilde{M}}_3\\ & \le (1-{\beta }_n(\overline{\gamma }-1))\parallel x_n-x_{n-1}\parallel +(1-{\beta }_n(\overline{\gamma }-1)){\tilde{M}}_2(\sum _{i=1}^N|r_{i,n}-r_{i,n-1}|\\ & +|{\alpha }_n-{\alpha }_{n-1}|+|{\nu }_n-{\nu }_{n-1}|+|{\lambda }_n-{\lambda }_{n-1}|+|r_n-r_{n-1}|)+|{\beta }_n-{\beta }_{n-1}|{\tilde{M}}_3\\ & \le (1-{\beta }_n(\overline{\gamma }-1))\parallel x_n-x_{n-1}\parallel +{\tilde{M}}_2(\sum _{i=1}^N|r_{i,n}-r_{i,n-1}|+|{\alpha }_n-{\alpha }_{n-1}|\\ & +|{\nu }_n-{\nu }_{n-1}|+|{\lambda }_n-{\lambda }_{n-1}|+|r_n-r_{n-1}|)+|{\beta }_n-{\beta }_{n-1}|{\tilde{M}}_3\\ & \le (1-{\beta }_n(\overline{\gamma }-1))\parallel x_n-x_{n-1}\parallel +{\tilde{M}}_2(\sum _{i=1}^N|r_{i,n}-r_{i,n-1}|+|{\alpha }_n-{\alpha }_{n-1}|\\ & +|{\nu }_n-{\nu }_{n-1}|+|{\lambda }_n-{\lambda }_{n-1}|+|r_n-r_{n-1}|)+(o({\beta }_n)+{\sigma }_{n-1}){\tilde{M}}_3,\end{array}$$ 3.54

where ${\tilde{M}}_3={sup}_{n\ge 0}\{\parallel A\parallel \parallel v_n\parallel +\parallel w_n\parallel \}$. By taking $s_{n+1}=\parallel x_{n+1}-x_n\parallel$, ${\omega }_n={\beta }_n(\overline{\gamma }-1)$, ${\omega }_n{\delta }_n={\tilde{M}}_{3}o({\beta }_n)$, and

$${\gamma }_n={\sigma }_{n-1}{\tilde{M}}_3+{\tilde{M}}_2(\sum _{i=1}^N|r_{i,n}-r_{i,n-1}|+|{\alpha }_n-{\alpha }_{n-1}|+|{\nu }_n-{\nu }_{n-1}|+|{\lambda }_n-{\lambda }_{n-1}|+|r_n-r_{n-1}|),$$

we deduce from (3.54) that

$$s_{n+1}\le (1-{\omega }_n)s_n+{\omega }_n{\delta }_n+{\gamma }_n.$$

Hence, by conditions (C2)–(C7) and Lemma 2.2, we obtain

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-x_n\parallel =0.$$

Step 3. We show that ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-w_n\parallel =0$. Indeed, from (3.41) and condition (C1), we derive

$$\begin{array}{rl}\parallel x_{n+1}-w_n\parallel & \le \parallel x_{n+1}-v_n\parallel +\parallel v_n-w_n\parallel \\ & \le {\beta }_n\parallel w_n-A{v}_n\parallel +{\alpha }_n\parallel \gamma V{x}_n-\mu G{v}_n\parallel \to 0(n\to \mathrm{\infty }).\end{array}$$

Step 4. We show that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-w_n\parallel =0$. In fact, by Step 2 and Step 3, we get

$$\parallel x_n-w_n\parallel \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-w_n\parallel \to 0(n\to \mathrm{\infty }).$$

Step 5. We show that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-z_n\parallel =0$ and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-R{x}_n\parallel =0$. In fact, we first derive ${lim}_{n\to \mathrm{\infty }}\parallel x_n-z_n\parallel =0$ by using arguments similar to those of (3.9) in the proof of Theorem 3.1, and then we obtain ${lim}_{n\to \mathrm{\infty }}\parallel x_n-R{x}_n\parallel =0$ by using arguments similar to those of (3.37) in the proof of Theorem 3.2.

Step 6. We show that ${lim}_{n\to \mathrm{\infty }}\parallel z_n-u_n\parallel =0$ and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-{\Delta }_n^N{x}_n\parallel =0$. In fact, by using arguments similar to those of (3.12) and (3.13) in the proof of Theorem 3.1, we obtain the desired conclusions.

Step 7. We show that ${lim}_{n\to \mathrm{\infty }}\parallel u_n-v_n\parallel =0$ and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T_{r_n}{x}_n\parallel =0$. In fact, by using arguments similar to those of (3.14) and (3.15) in the proof of Theorem 3.1, we obtain the desired conclusions.

Step 8. We show that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\langle (I-A)\tilde{x},x_n-\tilde{x}\rangle \le 0$. To this end, take a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\langle (I-A)\tilde{x},x_n-\tilde{x}\rangle =\underset{k\to \mathrm{\infty }}{lim}\langle (I-A)\tilde{x},x_{n_k}-\tilde{x}\rangle .$$

Without loss of generality, we may assume that $x_{n_k}\rightharpoonup \hat{x}$. Utilizing Steps 5, 6, and 7 and arguments similar to those of Steps 2, 3, and 4 in the proof of Theorem 3.2, we derive $\hat{x}\in \Omega$. Thus, from VI (3.2), we conclude

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\langle (I-A)\tilde{x},x_n-\tilde{x}\rangle =\underset{k\to \mathrm{\infty }}{lim}\langle (I-A)\tilde{x},x_{n_k}-\tilde{x}\rangle =\langle (I-A)\tilde{x},\hat{x}-\tilde{x}\rangle \le 0.$$

Step 9. We show that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-\tilde{x}\parallel =0$. Note that $\tilde{x}\in \Omega$. From (3.3), $\tilde{x}=R_n\tilde{x}$, $\tilde{x}={\Delta }_n^N\tilde{x}$, and $\tilde{x}=T_{r_n}\tilde{x}$, we obtain

$$w_n-\tilde{x}=(I-{\alpha }_n\mu G)v_n-(I-{\alpha }_n\mu G)\tilde{x}+{\alpha }_n(\gamma V{x}_n-\mu G\tilde{x})$$

and

$$\begin{array}{rl}{x}_{n+1}-\tilde{x}& =x_{n+1}-(I-{\beta }_{n}A)v_n-{\beta }_n{w}_n+(I-{\beta }_{n}A)v_n+{\beta }_n{w}_n-\tilde{x}\\ & =x_{n+1}-(I-{\beta }_{n}A)v_n-{\beta }_n{w}_n+(I-{\beta }_{n}A)(v_n-\tilde{x})+{\beta }_n(w_n-\tilde{x})+{\beta }_n(I-A)\tilde{x}.\end{array}$$

Applying (2.1), (3.40) and Lemmas 2.1, 2.5, and 2.6, we deduce that

$$\begin{array}{rl}{\parallel w_n-\tilde{x}\parallel }^2& ={\parallel (I-{\alpha }_n\mu G)v_n-(I-{\alpha }_n\mu G)\tilde{x}+{\alpha }_n(\gamma V{x}_n-\mu G\tilde{x})\parallel }^2\\ & \le {\parallel (I-{\alpha }_n\mu G)v_n-(I-{\alpha }_n\mu G)\tilde{x}\parallel }^2+2{\alpha }_n\langle \gamma V{x}_n-\mu G\tilde{x},w_n-\tilde{x}\rangle \\ & \le {(1-{\alpha }_n\tau )}^2{\parallel v_n-\tilde{x}\parallel }^2+2{\alpha }_n\parallel \gamma V{x}_n-\mu G\tilde{x}\parallel \parallel w_n-\tilde{x}\parallel \\ & \le {\parallel x_n-\tilde{x}\parallel }^2+2{\alpha }_n\parallel \gamma V{x}_n-\mu G\tilde{x}\parallel \parallel w_n-\tilde{x}\parallel ,\end{array}$$

and hence

$$\begin{array}{rl}& {\parallel x_{n+1}-\tilde{x}\parallel }^2\\ & ={\parallel (I-{\beta }_{n}A)(v_n-\tilde{x})+{\beta }_n(w_n-\tilde{x})+{\beta }_n(I-A)\tilde{x}+x_{n+1}-(I-{\beta }_{n}A)v_n-{\beta }_n{w}_n\parallel }^2\\ & \le {\parallel (I-{\beta }_{n}A)(v_n-\tilde{x})\parallel }^2+2{\beta }_n\langle w_n-\tilde{x},x_{n+1}-\tilde{x}\rangle \\ & +2{\beta }_n\langle (I-A)\tilde{x},x_{n+1}-\tilde{x}\rangle +2\langle x_{n+1}-(I-{\beta }_{n}A)v_n-{\beta }_n{w}_n,x_{n+1}-\tilde{x}\rangle \\ & \le {\parallel (I-{\beta }_{n}A)(v_n-\tilde{x})\parallel }^2+2{\beta }_n\langle w_n-\tilde{x},x_{n+1}-\tilde{x}\rangle +2{\beta }_n\langle (I-A)\tilde{x},x_{n+1}-\tilde{x}\rangle \\ & \le {(1-{\beta }_n\overline{\gamma })}^2{\parallel v_n-\tilde{x}\parallel }^2+2{\beta }_n\parallel w_n-\tilde{x}\parallel \parallel x_{n+1}-\tilde{x}\parallel +2{\beta }_n\langle (I-A)\tilde{x},x_{n+1}-\tilde{x}\rangle \\ & \le {(1-{\beta }_n\overline{\gamma })}^2{\parallel x_n-\tilde{x}\parallel }^2+{\beta }_n({\parallel w_n-\tilde{x}\parallel }^2+{\parallel x_{n+1}-\tilde{x}\parallel }^2)+2{\beta }_n\langle (I-A)\tilde{x},x_{n+1}-\tilde{x}\rangle \\ & \le {(1-{\beta }_n\overline{\gamma })}^2{\parallel x_n-\tilde{x}\parallel }^2+{\beta }_n[{\parallel x_n-\tilde{x}\parallel }^2+2{\alpha }_n\parallel \gamma V{x}_n-\mu G\tilde{x}\parallel \parallel w_n-\tilde{x}\parallel ]\\ & +{\beta }_n{\parallel x_{n+1}-\tilde{x}\parallel }^2+2{\beta }_n\langle (I-A)\tilde{x},x_{n+1}-\tilde{x}\rangle \\ & =[{(1-{\beta }_n\overline{\gamma })}^2+{\beta }_n]{\parallel x_n-\tilde{x}\parallel }^2+2{\alpha }_n{\beta }_n\parallel \gamma V{x}_n-\mu G\tilde{x}\parallel \parallel w_n-\tilde{x}\parallel +{\beta }_n{\parallel x_{n+1}-\tilde{x}\parallel }^2\\ & +2{\beta }_n\langle (I-A)\tilde{x},x_{n+1}-\tilde{x}\rangle .\end{array}$$ 3.55

It then follows from (3.55) that

$$\begin{array}{rl}& {\parallel x_{n+1}-\tilde{x}\parallel }^2\\ & \le \frac{{(1-{\beta }_n\overline{\gamma })}^2+{\beta }_n}{1-{\beta }_n}{\parallel x_n-\tilde{x}\parallel }^2+\frac{{\beta }_n}{1-{\beta }_n}[2{\alpha }_n\parallel \gamma V{x}_n-\mu G\tilde{x}\parallel \parallel w_n-\tilde{x}\parallel \\ & +2\langle (I-A)\tilde{x},x_{n+1}-\tilde{x}\rangle ]\\ & =(1-\frac{2{\beta }_n(\overline{\gamma }-1)}{1-{\beta }_n}){\parallel x_n-\tilde{x}\parallel }^2+\frac{2{\beta }_n(\overline{\gamma }-1)}{1-{\beta }_n}\cdot \frac{1}{2(\overline{\gamma }-1)}[2{\alpha }_n\parallel \gamma V{x}_n-\mu G\tilde{x}\parallel \parallel w_n-\tilde{x}\parallel \\ & +{\beta }_n{\overline{\gamma }}^2{\parallel x_n-\tilde{x}\parallel }^2+2\langle (I-A)\tilde{x},x_{n+1}-\tilde{x}\rangle ]\\ & =(1-{\xi }_n){\parallel x_n-\tilde{x}\parallel }^2+{\xi }_n{\delta }_n,\end{array}$$

where ${\xi }_n=\frac{2{\beta }_n(\overline{\gamma }-1)}{1-{\beta }_n}$, ${\delta }_n=\frac{1}{2(\overline{\gamma }-1)}[2{\alpha }_n\parallel \gamma V{x}_n-\mu G\tilde{x}\parallel \parallel w_n-\tilde{x}\parallel +{\beta }_n{\overline{\gamma }}^2{\parallel x_n-\tilde{x}\parallel }^2+2\langle (I-A)\tilde{x},x_{n+1}-\tilde{x}\rangle ]$. It can be readily seen from Step 2 and conditions (C1) and (C2) that ${\xi }_n\to 0$, ${\sum }_{n=0}^{\mathrm{\infty }}{\xi }_n=\mathrm{\infty }$, and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\delta }_n\le 0$. By Lemma 2.2, we conclude that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-\tilde{x}\parallel =0$. This completes the proof. □

Taking $T\equiv I$, $G\equiv I$, $\mu =1$, and $\gamma =1$ in Theorem 3.3, we have the following corollary.

Corollary 3.2

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

$$\{\begin{array}{c}{w}_n={\alpha }_{n}V{x}_n+(1-{\alpha }_n){\Delta }_n^N{R}_n{x}_n,\hfill \\ x_{n+1}=P_C[(I-{\beta }_{n}A){\Delta }_n^N{R}_n{x}_n+{\beta }_n{w}_n],\mathrm{\forall }n\ge 1.\hfill \end{array}$$

Assume that the sequences $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\lambda }_n\}$, $\{{\nu }_n\}$, and ${\{r_{i,n}\}}_{i=1}^N$ satisfy conditions (C1)–(C3) and (C5)–(C7) in Theorem 3.3. Then $\{x_n\}$ converges strongly to $\tilde{x}\in \Omega :={\bigcap }_{i=1}^{N}GMEP({\Theta }_i,{\phi }_i,B_i)\cap GSVI(C,F_1,F_2)$, which is the unique solution of VI (3.38).

Remark 3.1

Compared with Proposition 3.3, Theorem 3.4, and Theorem 3.7 in [11], respectively, our Theorems 3.1, 3.2, and 3.3 improve and develop them in the following aspects:

• (i)GSVI (1.3) with solutions being also fixed points of a continuous pseudocontinuous mapping in [12, Proposition 3.3, Theorem 3.4, and Theorem 3.7] is extended to GSVI (1.3) with solutions being also common solutions of a finite family of generalized mixed equilibrium problems (GMEPs) and fixed points of a continuous pseudocontinuous mapping in our Theorems 3.1, 3.2, and 3.3;
• (ii)in the argument process of our Theorems 3.1, 3.2, and 3.3, we use the variable parameters ${\lambda }_t$ and ${\nu }_t$ (resp., ${\lambda }_n$ and ${\nu }_n$) in place of the fixed parameters λ and ν in the proof of [12, Proposition 3.3, Theorem 3.4, and Theorem 3.7], and additionally deal with a pool of variable parameters ${\{r_{i,t}\}}_{i=1}^N$ (resp., ${\{r_{i,n}\}}_{i=1}^N$) involving a finite family of GMEPs;
• (iii)the iterative schemes in our Theorems 3.1, 3.2, and 3.3 are more advantageous and more flexible than the iterative schemes in [12, Proposition 3.3, Theorem 3.4, and Theorem 3.7], because they can be applied to solving three problems (i.e., GSVI (1.3), a finite family of GMEPs, and the fixed point problem of a continuous pseudocontractive mapping) and involve much more parameter sequences;
• (iv)it is worth emphasizing that our general implicit iterative scheme (3.1) is very different from Jung’s composite implicit iterative scheme in [12], because the term “$T_{r_t}R{x}_t$” in Jung’s implicit scheme is replaced by the term “$T_{r_t}{\Delta }_t^N{R}_t{x}_t$” in our implicit scheme (3.1). Moreover, the term “$T_{r_n}R{x}_n$” in Jung’s explicit scheme is replaced by the term “$T_{r_n}{\Delta }_n^N{R}_n{x}_n$” in our explicit scheme (3.3).

Numerical examples

The purpose of this section is to give two examples and numerical results to illustrate the applicability, effectiveness, and stability of our algorithm.

Example 4.1

(Example of Theorem 3.3)

Let $H=\mathbf{R}$ and $C=[0,100]$. Let the inner product $\langle \cdot ,\cdot \rangle :\mathbf{R}\times \mathbf{R}\to \mathbf{R}$ be defined by $\langle x,y\rangle =xy$. Let $N=2$, $Vx=2x$, $Gx=\frac{1}{2}x$, $Tx=x$, $B_{1}x=\frac{1}{2}x$, $B_{2}x=\frac{1}{3}x$, $F_{1}x=\frac{1}{2}x$, $F_{2}x=x$, ${\Theta }_1(x,y)=y^2-x^2$, ${\Theta }_2(x,y)=-3x^2+xy+2y^2$, ${\phi }_{1}x=x^2$, ${\phi }_{2}x=0$, and $Ax=\frac{3}{2}x$. Let ${\alpha }_n=\frac{1}{n}$, ${\beta }_n=\frac{1}{3(n+1)}$, $r_n=1$, $r_{1,n}=\frac{1}{2}$, $r_{2,n}=1$, ${\lambda }_n=1$, ${\nu }_n=\frac{1}{2}$, $\gamma =\frac{1}{8}$, $\mu =\frac{2}{3}$. It is easy to calculate that $T_{r_{1,n}}^{({\Theta }_1,{\phi }_1)}x=\frac{1}{3}x$, $T_{r_{2,n}}^{({\Theta }_2,{\phi }_2)}x=\frac{1}{6}x$, $T_{r_n}x=x$, $F_{1,{\lambda }_n}x=\frac{1}{2}x$, and $F_{2,{\nu }_n}x=\frac{1}{2}x$. Choose an arbitrary initial guess $x_1=4$. We get the numerical results of Algorithm (3.3).

Table 1 shows the value of the sequence $\{x_n\}$.

Table 1.

The values of $x_n$

n	$x_n$
1	4.0000
2	1.8261 × 10−1
3	3.3191 × 10−3
4	3.7633 × 10−5
5	3.2426 × 10−7
6	2.3546 × 10−9
7	1.5285 × 10−11
8	9.1892 × 10−14
9	5.2325 × 10−16
10	2.8636 × 10−18
11	1.5212 × 10−20
12	7.8994 × 10−23

Figure 1 shows the convergence of the iterative sequence of Algorithm (3.3).

Figure 1.

The convergence of $\{x_n\}$ with initial $x_1=4$

Solution: We can see from both Table 1 and Fig. 1 that the sequence $\{x_n\}$ converges to 0, that is, 0 is the solution in Example 4.1. In addition, it is also easy to check from Example 4.1 that ${\bigcap }_{i=1}^{2}GMEP({\Theta }_i,{\phi }_i,B_i)\cap GSVI(C,F_1,F_2)\cap Fix(T)=\{0\}$. Therefore, the iterative algorithm of Theorem 3.3 is efficient.

Example 4.2

(Example of Theorem 3.7 in [12])

Let $H=\mathbf{R}$ and $C=[0,100]$. Let the inner product $\langle \cdot ,\cdot \rangle :\mathbf{R}\times \mathbf{R}\to \mathbf{R}$ be defined by $\langle x,y\rangle =xy$. Let $Vx=2x$, $Gx=\frac{1}{2}x$, $Tx=x$, $F_{1}x=\frac{1}{2}x$, $F_{2}x=x$, and $Ax=\frac{3}{2}x$. Let ${\alpha }_n=\frac{1}{n}$, ${\beta }_n=\frac{1}{3(n+1)}$, $r_n=1$, $\lambda =1$, $\nu =\frac{1}{2}$, $\gamma =\frac{1}{8}$, $\mu =\frac{2}{3}$. Choose an arbitrary initial guess $x_1=4$. We get the numerical results of Algorithm (1.5) (Algorithm (3.10) of [12]).

Table 2 shows the value of the sequence $\{x_n\}$.

Table 2.

The values of $x_n$

n	$x_n$
1	4.0000
2	1.0278
3	2.5219 × 10−1
4	6.1587 × 10−2
5	1.5055 × 10−2
6	3.6870 × 10−3
7	9.0468 × 10−4
8	2.2236 × 10−4
9	5.4731 × 10−5
10	1.3488 × 10−5
11	3.3278 × 10−6
12	8.2181 × 10−7

The Fig. 2 shows the convergence of the iterative sequence of Algorithm (1.5).

Figure 2.

The convergence of $\{x_n\}$ with initial $x_1=4$

Solution: We can see from both Table 2 and Fig. 2 that the sequence $\{x_n\}$ converges to 0, that is, 0 is the solution in Example 4.2. In addition, it is also easy to check from Example 4.2 that $GSVI(C,F_1,F_2)\cap Fix(T)=\{0\}$.

Remark 4.1

From Tables 1, 2 and Figs. 1, 2, it is readily seen that the convergence of $\{x_n\}$ to 0 in Example 4.1 is faster than the one of $\{x_n\}$ to 0 in Example 4.2. Therefore, our algorithm is more applicable, efficient, and stable than the algorithm in [12].

Application

In this section, applying our main result Theorem 3.3, we can prove strong convergence theorems for approximating the solution of the standard constrained convex optimization problem.

Let C be a closed convex subset of H. The standard constrained convex optimization problem is to find $x^{\ast }\in C$ such that

$$f\left(x^{\ast }\right)=\underset{x\in C}{min}f(x),$$ 5.1

where $f:C\to \mathbf{R}$ is a convex, Fréchet differentiable function. The set of the solutions of (5.1) is denoted by ${\Phi }_f$.

Lemma 5.1

(Optimality condition, [25])

A necessary condition of optimality for a point $x^{\ast }\in C$ to be a solution of the minimization problem (5.1) is that $x^{\ast }$ solves the variational inequality

$$\langle \mathrm{\nabla }f\left(x^{\ast }\right),x-x^{\ast }\rangle \ge 0$$ 5.2

for all $x\in C$. Equivalently, $x^{\ast }\in C$ solves the fixed point equation

$$x^{\ast }=P_C(I-\lambda \mathrm{\nabla }f)x^{\ast }$$

for every $\lambda >0$. If, in addition, f is convex, then the optimality condition (5.2) is also sufficient.

Theorem 5.1

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $f_i$ ($i=1,2,\dots ,N$): $C\to \mathbf{R}$ be a real-valued convex function with the gradient $\mathrm{\nabla }{f}_i$ being $\frac{1}{L_{f_i}}$-inverse strongly monotone and continuous with $L_{f_i}>0$. Let ${\Theta }_i$, ${\phi }_i$, A, V, G, $F_1$, $F_2$, $R_n$, $F_{1,{\lambda }_n}$, $F_{2,{\nu }_n}$, $T_{r_n}$, and $T_{r_{i,n}}^{({\Theta }_i,{\phi }_i)}$ be defined as in Theorem 3.3. Given $x_1\in C$ and let $\{x_n\}$ be the sequence generated by the following explicit algorithm:

$$\{\begin{array}{c}{w}_n={\alpha }_n\gamma V{x}_n+(I-{\alpha }_n\mu G)T_{r_n}{\Lambda }_n^N{R}_n{x}_n,\hfill \\ x_{n+1}=P_C[(I-{\beta }_{n}A)T_{r_n}{\Lambda }_n^N{R}_n{x}_n+{\beta }_n{w}_n],\mathrm{\forall }n\ge 1,\hfill \end{array}$$ 5.3

where ${\Lambda }_n^i=T_{r_{i,n}}^{({\Theta }_i,{\phi }_i)}(I-r_{i,n}\mathrm{\nabla }{f}_i)T_{r_{i-1,n}}^{({\Theta }_{i-1},{\phi }_{i-1})}(I-r_{i-1,n}\mathrm{\nabla }{f}_{i-1})\cdots T_{r_{1,n}}^{({\Theta }_1,{\phi }_1)}(I-r_{1,n}\mathrm{\nabla }{f}_1)$ and ${\Lambda }_n^0=I$. Assume that $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{r_n\}$, $\{{\lambda }_n\}$, $\{{\nu }_n\}$, and ${\{r_{i,n}\}}_{i=1}^N$ satisfy conditions (C1)–(C7) in Theorem 3.3. Then $\{x_n\}$ converges strongly to $\tilde{x}\in \Omega :={\bigcap }_{i=1}^{N}MEP({\Theta }_i,{\phi }_i)\cap {\bigcap }_{i=1}^N{\Phi }_{f_i}\cap GSVI(C,F_1,F_2)\cap Fix(T)$, which is the unique solution of VI (3.2).

Proof

By using Lemma 5.1 and Theorem 3.3, we obtain the desired conclusion directly. □

Conclusions

We introduced and analyzed one general implicit iterative scheme and another general explicit iterative scheme for finding a solution of a general system of variational inequalities (GSVI) with the constraints of finitely many generalized mixed equilibrium problems and a fixed point problem of a continuous pseudocontractive mapping in a Hilbert space. Moreover, we established strong convergence of the proposed implicit and explicit iterative schemes to a solution of the GSVI, which is the unique solution of a certain variational inequality. Our Theorems 3.1–3.3 not only improve and develop the main results of [1] and [12] but also improve and develop Theorems 3.1 and 3.2 of [9], Theorems 3.1 and 3.2 of [10], and Proposition 3.1, Theorems 3.2 and 3.5 of [11].

Authors’ contributions

All authors read and approved the final manuscript.

Funding

L.-C. Ceng was partially supported by the Innovation Program of Shanghai Municipal Education Commission (15ZZ068), Ph.D. Program Foundation of the Ministry of Education of China (20123127110002), and Program for Outstanding Academic Leaders in Shanghai City (15XD1503100).

Competing interests

The authors declare that they have no competing interests.