Some Mann-Type Implicit Iteration Methods for Triple Hierarchical Variational Inequalities, Systems Variational Inequalities and Fixed Point Problems

Abstract

This paper discusses a monotone variational inequality problem with a variational inequality constraint over the common solution set of a general system of variational inequalities (GSVI) and a common fixed point (CFP) of a countable family of nonexpansive mappings and an asymptotically nonexpansive mapping in Hilbert spaces, which is called the triple hierarchical constrained variational inequality (THCVI), and introduces some Mann-type implicit iteration methods for solving it. Norm convergence of the proposed methods of the iteration methods is guaranteed under some suitable assumptions.

1. Introduction

Let C be a convex closed nonempty subset of a real Hilbert space H with norm $\parallel ·\parallel$ and inner product $\langle ·,·\rangle$. Let $P_C$ be the metric (or nearest point) projection from H onto C, that is, for all $x\in H,P_{C}x\in C$ and $\parallel x-P_{C}x\parallel ={inf}_{y\in C}\parallel x-y\parallel$. Let $T:C\to C$ be a possible nonlinear mapping. Denote by $\mathrm{Fix}\left(T\right)$ the set of fixed points of T, i.e., $\mathrm{Fix}\left(T\right)=\{x\in C:x=Tx\}$. We use the notations $\mathbf{R}$, ⇀ and → to indicate the set of real numbers, weak convergence and strong convergence, respectively.

A mapping $T:C\to C$ is said to be asymptotically nonexpansive (see [1]), if there exists a sequence $\{{\theta }_n\}\subset [0,+\infty )$ with ${lim}_{n\to \infty }{\theta }_n=0$ such that

$$\parallel T^{n}x-T^{n}y\parallel \le (1+{\theta }_n)\parallel x-y\parallel \forall n\ge 0,x,y\in C.$$

In particular, T is said to be nonexpansive if $\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,$$\forall x,y\in C$, that is, $\theta \equiv 0$. If C is also a bounded set, then the fixed-point set of T is nonempty, that is $\mathrm{Fix}\left(T\right)\ne \varnothing$. Via iterative techniques, fixed points of (asymptotically) nonexpansive mappings have been studied because of their applications in convex optimization problems; see [2,3,4,5,6,7,8,9,10] and the references therein.

Let $B_1,B_2:C\to H$ be two nonlinear single-valued mappings. We consider the following problem of finding $(x^{*},y^{*})\in C\times C$ such that

$$\left\{\begin{array}{c}\langle x-x^{*},{\mu }_1{B}_1{y}^{*}+x^{*}-y^{*}\rangle \ge 0,\forall x\in C,\hfill \\ \langle x-y^{*},{\mu }_2{B}_2{x}^{*}+y^{*}-x^{*}\rangle \ge 0,\forall x\in C,\hfill \end{array}\right.$$

(1)

which is called a general system of variational inequalities (GSVI) with real number constants ${\mu }_1$ and ${\mu }_2>0$, which covers as special subcases the problems arising, especially from nonlinear complementarity problems, quadratic mathematical programming and other variational problems. The reader is referred to [11,12,13,14,15,16,17,18] and the references therein. Particularly, if both $B_1$ and $B_2$ are equal to A and $x^{*}=y^{*}$, then problem (1) become the classical variational inequality (VI), that set of solutions is stated by VI($C,A$). Note that, problem (1) can be transformed into a fixed-point problem in the following way.

Lemma 1

([19]). Let both $x^{*}$ and $y^{*}$ be points in C. $(x^{*},y^{*})$ is a solution of GSVI (1) if and only if $x^{*}\in \mathrm{GSVI}(C,B_1,B_2)$, where $\mathrm{GSVI}(C,B_1,B_2)$ is the fixed point set of the mapping $G:=P_C(I-{\mu }_1{B}_1)P_C(I-{\mu }_2{B}_2)$, and $y^{*}=P_C(I-{\mu }_2{B}_2)x^{*}$.

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

$$\langle Ax-Ay,x-y\rangle \ge 0,\forall x,y\in C.$$

It is called $\eta$-strongly monotone if there exists a constant $\eta >0$ such that

$$\langle Ax-Ay,x-y\rangle \ge {\eta \parallel x-y\parallel }^2,\forall x,y\in C.$$

Moreover, it is called $\alpha$-inverse-strongly monotone (or $\alpha$-cocoercive), if there exists a constant $\alpha >0$ such that

$$\langle Ax-Ay,x-y\rangle \ge {\alpha \parallel Ax-Ay\parallel }^2,\forall x,y\in C.$$

Obviously, each inverse-strongly monotone mapping is monotone and Lipschitzian, and each strongly monotone and Lipschitzian mapping is inverse-strongly monotone but the converse is not true.

Recently, Cai et al. [20] proposed a new implicit-rule for obtaining a common element of the solution set of GSVI (1) and the fixed point set of an asymptotically nonexpansive mapping T, and presented norm convergence of the sequence generated by the proposed rule to an element of $\mathrm{GSVI}(C,B_1,B_2)\cap \mathrm{Fix}\left(T\right)$, which also solves certain VI.

On the other hand, Iiduka [21] considered a monotone variational inequality linked to a inequality constraint over the set of fixed points of a nonexpansive mapping. Iiduka’s problem is a triple mathematical programming in contrast with bilevel mathematical programming problems or hierarchical constrained optimization problems or nonlinear hierarchical problem, it is referred as triple hierarchical constrained optimization problem (THCOP). Since the THCOP is a general variational inequality, we also call it a triple hierarchical variational inequality (THVI). This kind of problems play an important role in nonlinear minimizer problems and nonlinear operator equations; see [22,23,24,25,26] and the references therein.

To begin with, let us recall the variational inequality for a monotone mapping, $A_1:H\to H$, over the fixed point set of a nonexpansive mapping, $T:H\to H$:

$$\begin{array}{c}\mathrm{Find}\overline{x}\in \mathrm{VI}(\mathrm{Fix}\left(T\right),A_1)\hfill \\ :=\{\overline{x}\in \mathrm{Fix}\left(T\right):\langle A_1\overline{x},y-\overline{x}\rangle \ge 0\forall y\in \mathrm{Fix}\left(T\right)\},\hfill \end{array}$$

where $\mathrm{Fix}\left(T\right):=\{x\in H:Tx=x\}\ne \varnothing$. Iiduka’s THCOP and its algorithm (Algorithm 1) are stated below.

Problem 1. 

(see [21], Problem 3.1) Assume that

(C1) 

$T:H\to H$ is a nonexpansive mapping such that $\mathrm{Fix}\left(T\right)\ne \varnothing$;

(C2) 

$A_2:H\to H$ is κ-Lipschitz continuous η-strongly monotone;

(C3) 

$A_1:H\to H$ is ζ-inverse-strongly monotone;

(C4) 

$\mathrm{VI}(\mathrm{Fix}\left(T\right),A_1)\ne \varnothing$.

Then the objective is to

$$\begin{array}{c}\mathrm{Find}{x}^{*}\in \mathrm{VI}(\mathrm{VI}(\mathrm{Fix}\left(T\right),A_1),A_2)\hfill \\ :=\{x^{*}\in \mathrm{VI}(\mathrm{Fix}\left(T\right),A_1):\langle v-x^{*},A_2{x}^{*}\rangle \ge 0\forall v\in \mathrm{VI}(\mathrm{Fix}\left(T\right),A_1)\}.\hfill \end{array}$$

Algorithm 1.

(see [21], Algorithm 4.1)

Step 0. Take ${\{{\alpha }_n\}}_{n=0}^{\infty },{\{{\delta }_n\}}_{n=0}^{\infty }\subset (0,\infty )$, and $\mu >0$, choose $x_0\in H$ arbitrarily, and let $n:=0$.

Step 1. Given$x_n\in H$, compute$x_{n+1}\in H$as

$$\left\{\begin{array}{c}{y}_n=T(x_n-{\delta }_n{A}_1{x}_n),\hfill \\ x_{n+1}=y_n-{\alpha }_n\mu A_2{y}_n.\hfill \end{array}\right.$$

Update$n:=n+1$and go to Step 1.

The purpose of this paper is to introduce and analyze some Mann-type implicit iteration methods for treating a monotone variational inequality with a inequality constraint over the common solution set of the GSVI (1) for two inverse-strongly monotone mappings and a common fixed point problem (CFPP) of a countable family of nonexpansive mappings and an asymptotically nonexpansive mapping in Hilbert spaces, which is called the triple hierarchical constrained variational inequality (THCVI). Here the Mann-type implicit iteration methods are based on the Mann iteration method, viscosity approximation method, Korpelevich’s extragradient method and hybrid steepest-descent method. Under some suitable assumptions, we prove strong convergence of the proposed methods to the unique solution of the THCVI.

2. Preliminaries

Now we recall some necessary concepts and facts. A mapping $F:C\to H$ is named to be $\kappa$-Lipschitzian if there is a real number $\kappa >0$ with

$$\kappa \parallel x-y\parallel \ge \parallel F\left(x\right)-F\left(y\right)\parallel ,\forall x,y\in C.$$

Particularly, if $\kappa \in (0,1)$, then F is said to be contractive. If $\kappa =1$, then F is said to be a nonexpansivity. A mapping $A:H\to H$ is named to be a strongly positive bounded linear operator if there is a real number $\gamma >0$ with

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

For a fixed $x\in H$, we know that there is a unique point in C, presented by $P_{C}x$, with

$$\parallel x-y\parallel \ge \parallel x-P_{C}x\parallel ,\forall y\in C.$$

$P_C$ is called a metric projection of H onto C.

Lemma 2.

There hold the following important relations for metric projection $P_C$:

(i) 

$\langle x-y,P_{C}x-P_{C}y\rangle \ge {\parallel P_{C}x-P_{C}y\parallel }^2,$$\forall x,y\in H$;

(ii) 

$0\ge \langle x-P_{C}x,y-P_{C}x\rangle ,$$\forall x\in H,y\in C$;

(iii) 

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

(iv) 

${\parallel x-y\parallel }^2\ge \parallel x-P_C{x\parallel }^2+{\parallel y-P_{C}x\parallel }^2,$$\forall x\in H,y\in C$.

Lemma 3

([27]). Let $\{a_n\}$ be a sequence of real numbers with the conditions:

$$a_{n+1}\le (1-{\lambda }_n)a_n+{\lambda }_n{\gamma }_n,\forall n\ge 0,$$

where $\{{\lambda }_n\}$ and $\{{\gamma }_n\}$ are sequences of real numbers such that (i) $\{{\lambda }_n\}\subset [0,1]$ and ${\sum }_{n=0}^{\infty }{\lambda }_n=\infty$, and (ii) ${\sum }_{n=0}^{\infty }|{\gamma }_n{\lambda }_n|<\infty$ or ${lim\; sup}_{n\to \infty }{\gamma }_n\le 0$. Then ${lim}_{n\to \infty }{a}_n=0$.

Lemma 4

([27]). Let λ be real number in $(0,1]$. Let $T:C\to H$ be a nonexpansive nonself mapping. Let $T^{\lambda }:C\to H$ be a nonself mapping defined by

$$T^{\lambda }x:=Tx-\lambda \mu F\left(Tx\right),\forall x\in C.$$

Here $F:H\to H$ is κ-Lipschitzian and η-strongly monotone. So, $T^{\lambda }$ is a contraction if $0<\mu <\frac{2\eta }{{\kappa }^2}$, i.e.,

$$\parallel T^{\lambda }x-T^{\lambda }y\parallel \le (1-\lambda \tau )\parallel x-y\parallel ,\forall x,y\in C,$$

where $\tau =1-\sqrt{1-\mu (2\eta -\mu {\kappa }^2)}\in (0,1]$.

Lemma 5

([17]). Let the mapping $A:C\to H$ be α-inverse-strongly nonself monotone. Then, for a given $\lambda \ge 0$, ${\parallel (I-\lambda A)x-(I-\lambda A)y\parallel }^2\le {\parallel x-y\parallel }^2+\lambda (\lambda -2\alpha ){\parallel Ax-Ay\parallel }^2$. In particular, if $0\le \lambda \le 2\alpha$, then $I-\lambda A$ is nonexpansive.

Lemma 6

([17]). Let the mappings $B_1,B_2:C\to H$ be α-inverse-strongly monotone and β-inverse-strongly monotone, respectively. Let the mapping $G:C\to C$ be defined as $G:=P_C(I-{\mu }_1{B}_1)P_C(I-{\mu }_2{B}_2)$. If $0\le {\mu }_1\le 2\alpha$ and $0\le {\mu }_2\le 2\beta$, then $G:C\to C$ is nonexpansive.

Lemma 7

([28]). Let H be a Hilbert space. We suppose that C is a convex closed nonempty set in H, and $T:C\to C$ is an asymptotically nonexpansive nonself mapping with a nonempty fixed point set, that is, $\mathrm{Fix}\left(T\right)\ne \varnothing$. Then $I-T$ is demiclosed at zero, i.e., if $\{x_n\}\subset C$ converges weakly to some $x\in C$, and $\{(I-T)x_n\}$ converges strongly to zero, then $(I-T)x=0$, where I is the identity mapping on H.

Lemma 8

([29]). Let H be a Hilbert space. We suppose that $\{x_n\}$ and $\{w_n\}$ are bounded vector sequences in H and $\{{\beta }_n\}$ is a real number sequence in $(0,1)$ such that ${lim\; sup}_{n\to \infty }{\beta }_n\le 1$ and ${lim\; inf}_{n\to \infty }{\beta }_n>0$. We also suppose that $x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)w_n$, $\forall n\ge 0$ and

$$\underset{n\to \infty }{lim\; sup}(\parallel w_{n+1}-w_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0.$$

Then ${lim}_{n\to \infty }\parallel w_n-x_n\parallel =0$.

Let C be a convex closed nonempty set. Let ${\{S_n\}}_{n=0}^{\infty }$ be a countable family of nonexpansive self mappings defined on C, and ${\{{\lambda }_n\}}_{n=0}^{\infty }$ be a sequence of real numbers in $[0,1]$. On C, we define a self mapping $W_n$:

$$\left\{\begin{array}{c}{U}_{n,n+1}=I,\hfill \\ U_{n,n}=(1-{\lambda }_n)I+{\lambda }_n{S}_n{U}_{n,n+1},\hfill \\ U_{n,n-1}=(1-{\lambda }_{n-1})I+{\lambda }_{n-1}{S}_{n-1}{U}_{n,n},\hfill \\ \cdots ,\hfill \\ U_{n,k}=(1-{\lambda }_k)I+{\lambda }_k{S}_k{U}_{n,k+1},\hfill \\ U_{n,k-1}=(1-{\lambda }_{k-1})I+{\lambda }_{k-1}{S}_{k-1}{U}_{n,k},\hfill \\ \cdots ,\hfill \\ U_{n,1}=(1-{\lambda }_1)I+{\lambda }_1{S}_1{U}_{n,2},\hfill \\ W_n=U_{n,0}=(1-{\lambda }_0)I+{\lambda }_0{S}_0{U}_{n,1}.\hfill \end{array}\right.$$

Such a $W_n$ is named the W-mapping generated by $S_n,S_{n-1},...,S_0$ and ${\lambda }_n,{\lambda }_{n-1},...,{\lambda }_0$; see [30].

Lemma 9

([30]). Let C be a convex closed nonempty set in a Hilbert space H. Let ${\{S_n\}}_{n=0}^{\infty }$ be a mapping sequence of nonexpansivity on C with ${\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)\ne \varnothing$. Let ${\{{\lambda }_n\}}_{n=0}^{\infty }$ be a number sequence in $(0,b]$ for some $b\in (0,1)$. Then ${lim}_{n\to \infty }{U}_{n,k}x$ exists for every $x\in C$ and $k\ge 0$.

Using Lemma 9, $W:C\to C$ is defined by $Wx={lim}_{n\to \infty }{W}_{n}x={lim}_{n\to \infty }{U}_{n,0}x,$$\forall x\in C$. We call W is the W-mapping defined by ${\{S_n\}}_{n=0}^{\infty }$ and ${\{{\lambda }_n\}}_{n=0}^{\infty }$. Next, we assume that ${\{{\lambda }_n\}}_{n=0}^{\infty }$ is a sequence of positive numbers in $(0,b]$ for some $b\in (0,1)$.

Lemma 10

([30]). Let C be a convex closed nonempty set of a Hilbert space H. Let ${\{S_n\}}_{n=0}^{\infty }$ be a mapping sequence of nonexpansivity on C with ${\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)\ne \varnothing$. Let ${\{{\lambda }_n\}}_{n=0}^{\infty }$ be a number sequence in $(0,b]$ for some $b\in (0,1)$. Then ${\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)=\mathrm{Fix}\left(W\right)$.

Lemma 11

([30]). Let C be a convex closed nonempty set of a Hilbert space H. Let ${\{S_n\}}_{n=0}^{\infty }$ be a sequence of nonexpansive self-mappings on C with ${\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)\ne \varnothing$, and ${\{{\lambda }_n\}}_{n=0}^{\infty }$ be a real sequence in $(0,b]$ for some $b\in (0,1)$. If D is any bounded subset of C, then ${lim}_{n\to \infty }{sup}_{x\in D}\parallel W_{n}x-Wx\parallel =0$.

Lemma 12

([21]). Let C be a convex closed nonempty set of a Hilbert space H. Let $A:C\to H$ be a hemicontinuous nonself monotone mapping. Then the following hold: (i) $\mathrm{VI}(C,A)=\{x^{*}\in C:\langle x^{*}-y,Ay\rangle \le 0\forall y\in C\}$; (ii) $\mathrm{VI}(C,A)=\mathrm{Fix}\left(P_C(I-\lambda A)\right)$ for all $\lambda >0$; (iii) $\mathrm{VI}(C,A)$ consists of one point, if A is strongly monotone and Lipschitz continuous.

3. Main Results

Let C be a convex closed nonempty set of a real Hilbert space H. Let the mappings $A_1,B_i:C\to H$ be monotone for $i=1,2$. Let $T:C\to C$ be an asymptotically nonexpansive self mapping and ${\{S_n\}}_{n=0}^{\infty }$ be a countable family of nonexpansive self mappings on C. We now consider the variational inequality for mapping $A_1$ over the common solution set $\Omega$ of the GSVI (1) and the CFPP of ${\{S_n\}}_{n=0}^{\infty }$ and T:

$$\begin{array}{c}\mathrm{Find}\overline{x}\in \mathrm{VI}(\Omega ,A_1)\hfill \\ :=\{\overline{x}\in \Omega :\langle A_1\overline{x},y-\overline{x}\rangle \ge 0\forall y\in \Omega \},\hfill \end{array}$$

where $\Omega :={\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)\cap \mathrm{GSVI}(C,B_1,B_2)\cap \mathrm{Fix}\left(T\right)\ne \varnothing$. This section introduces the following general monotone variational inequality with the variational inequality constraint on the common solution set of the GSVI (1) and the CFPP of ${\{S_n\}}_{n=0}^{\infty }$ and T, which is named as the triple hierarchical constrained variational inequality (THCVI):

Problem 2.

Assume that

(C1) 

$T:C\to C$ is an asymptotically nonexpansive self mapping with a sequence $\{{\theta }_n\}\subset [0,+\infty )$;

(C2) 

${\{S_n\}}_{n=0}^{\infty }$ is a countable family of nonexpansive self mappings on C;

(C3) 

$B_1,B_2:C\to H$ are α-inverse-strongly monotone and β-inverse-strongly monotone, respectively;

(C4) 

$\mathrm{GSVI}(C,B_1,B_2):=\mathrm{Fix}\left(G\right)$ where $G:=P_C(P_C(I-{\mu }_2{B}_2)-{\mu }_1{B}_1{P}_C(I-{\mu }_2{B}_2))$ for real numbers ${\mu }_1,{\mu }_2>0$;

(C5) 

$\Omega :={\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)\cap \mathrm{GSVI}(C,B_1,B_2)\cap \mathrm{Fix}\left(T\right)\ne \varnothing$;

(C6) 

$W_n$ is the W-mapping defined by $S_n,S_{n-1},...,S_0$ and ${\lambda }_n,{\lambda }_{n-1},...,{\lambda }_0$, where ${\{{\lambda }_n\}}_{n=0}^{\infty }\subset (0,1)$;

(C7) 

$A_1:C\to H$ is ζ-inverse-strongly monotone;

(C8) 

$A_2:C\to H$ is η-strongly monotone and κ-Lipschitzian;

(C9) 

$f:C\to C$ is a δ-contraction mapping with real coefficient $\delta \in [0,1)$;

(C10) 

$\mathrm{VI}(\Omega ,A_1)\ne \varnothing$.

Then the objective is to

$$\begin{array}{c}\mathrm{find}{x}^{*}\in \mathrm{VI}(\mathrm{VI}(\Omega ,A_1),\mu A_2-f)\hfill \\ :=\{x^{*}\in \mathrm{VI}(\Omega ,A_1):\langle x^{*}-v,(\mu A_2-f)x^{*}\rangle \le 0\forall v\in \mathrm{VI}(\Omega ,A_1)\},\hfill \end{array}$$

for some $\mu >0$.

Problem 3.

If we put $f=0$ in Problem 2, then the objective is to

$$\begin{array}{c}\mathrm{find}{x}^{*}\in \mathrm{VI}(\mathrm{VI}(\Omega ,A_1),A_2)\hfill \\ :=\{x^{*}\in \mathrm{VI}(\Omega ,A_1):\langle A_2{x}^{*},v-x^{*}\rangle \ge 0\forall v\in \mathrm{VI}(\Omega ,A_1)\}.\hfill \end{array}$$

Here we propose the following implicit Mann-type iteration algorithms (Algorithms 2 and 3) for solving Problems 2 and 3, respectively.

Algorithm 2.

 

Step 0. Take${\{{\alpha }_n\}}_{n=0}^{\infty },{\{{\beta }_n\}}_{n=0}^{\infty },{\{{\gamma }_n\}}_{n=0}^{\infty },{\{{\delta }_n\}}_{n=0}^{\infty }\subset (0,\infty )$, and$\mu >0$, choose$x_0\in C$arbitrarily, and let$n:=0$.

Step 1. Given$x_n\in C$, compute$x_{n+1}\in C$as

$$\left\{\begin{array}{c}{u}_n=(1-{\gamma }_n)W_n{u}_n+{\gamma }_n{x}_n,\hfill \\ y_n=P_C(I-{\delta }_n{A}_1)G{u}_n,\hfill \\ x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)P_C[{\alpha }_{n}f\left(x_n\right)+(I-{\alpha }_n\mu A_2)T^n{y}_n].\hfill \end{array}\right.$$

(2)

Update$n:=n+1$and go to Step 1.

Algorithm 3.

 

Step 0. Take${\{{\alpha }_n\}}_{n=0}^{\infty },{\{{\beta }_n\}}_{n=0}^{\infty },{\{{\gamma }_n\}}_{n=0}^{\infty },{\{{\delta }_n\}}_{n=0}^{\infty }\subset (0,\infty )$, and$\mu >0$, choose$x_0\in C$arbitrarily, and let$n:=0$.

Step 1. Given$x_n\in C$, compute$x_{n+1}\in C$as

$$\left\{\begin{array}{c}{u}_n=(1-{\gamma }_n)W_n{z}_n+{\gamma }_n{x}_n,\hfill \\ v_n=P_C(u_n-{\mu }_2{B}_2{u}_n),\hfill \\ z_n=P_C(v_n-{\mu }_1{B}_1{v}_n),\hfill \\ y_n=P_C(z_n-{\delta }_n{A}_1{z}_n),\hfill \\ x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)P_C(I-{\alpha }_n\mu A_2)T^n{y}_n.\hfill \end{array}\right.$$

Update$n:=n+1$and go to Step 1.

We are now able to state and prove the main results of this paper: the following convergence analysis is presented for our Algorithms 2 and 3.

Theorem 1.

Assume that ${\mu }_1$ is a real number in $(0,2\alpha ),$ and ${\mu }_2$ is a real number in $(0,2\beta )$. Let $\delta <\tau :=1-\sqrt{1-\mu (2\eta -\mu {\kappa }^2)}\in (0,1]$ for $\mu \in (0,\frac{2\eta }{{\kappa }^2})$. We suppose ${\{{\lambda }_n\}}_{n=0}^{\infty }$ is a real sequence in $(0,b]$ for some real number b in $(0,1)$. We also suppose that $\{{\alpha }_n\},\{{\beta }_n\},\{{\gamma }_n\}\subset (0,1]$ and $\{{\delta }_n\}\subset (0,2\zeta ]$ such that

(i) 

${\sum }_{n=0}^{\infty }{\alpha }_n=\infty$ and ${lim}_{n\to \infty }{\alpha }_n=0$;

(ii) 

${\delta }_n\le {\alpha }_n\forall n\ge 0$ and ${lim}_{n\to \infty }\frac{{\theta }_n}{{\alpha }_n}=0$;

(iii) 

${lim\; inf}_{n\to \infty }{\beta }_n>0$ and ${lim\; sup}_{n\to \infty }{\beta }_n<1$;

(iv) 

${lim\; inf}_{n\to \infty }{\gamma }_n>0$, ${lim\; sup}_{n\to \infty }{\gamma }_n<1$ and ${lim}_{n\to \infty }|{\gamma }_{n+1}-{\gamma }_n|=0$;

(v) 

${lim}_{n\to \infty }\parallel T^{n+1}{y}_n-T^n{y}_n\parallel =0$.

Then the sequence ${\{x_n\}}_{n=0}^{\infty }$ generated by Algorithm 2 satisfies the following properties:

(a) 

${\{x_n\}}_{n=0}^{\infty }$ is bounded;

(b) 

${lim}_{n\to \infty }\parallel x_n-y_n\parallel =0,$${lim}_{n\to \infty }\parallel x_n-G{x}_n\parallel =0,$${lim}_{n\to \infty }\parallel x_n-T{x}_n\parallel =0$ and ${lim}_{n\to \infty }\parallel x_n-W{x}_n\parallel =0$;

(c) 

if ${lim}_{n\to }\frac{\parallel x_n-y_n\parallel }{{\delta }_n}=0$, then $x_n\to x^{*}\in \mathrm{VI}(\Omega ,A_1)$.

Proof. 

First of all, for any $x,y\in C$, by Lemma 4, we have

$$\parallel P_{\mathrm{VI}(\Omega ,A_1)}(f+I-\mu A_2)x-P_{\mathrm{VI}(\Omega ,A_1)}(f+I-\mu A_2)y\parallel \le \delta \parallel x-y\parallel +(1-\tau )\parallel x-y\parallel =[1-(\tau -\delta )]\parallel x-y\parallel ,$$

which implies that $P_{\mathrm{VI}(\Omega ,A_1)}(f+I-\mu A_2)$ is a contraction. Banach’s Contraction Principle tells us that $P_{\mathrm{VI}(\Omega ,A_1)}(f+I-\mu A_2)$ has a fixed point. Indeed, it is also unique, say $x^{*}\in C$, that is, $x^{*}=P_{\mathrm{VI}(\Omega ,A_1)}(f+I-\mu A_2)x^{*}$. Utilizing Lemma 12, we get

$$\{x^{*}\}=\mathrm{Fix}\left(P_{\mathrm{VI}(\Omega ,A_1)}(f+I-\mu A_2)\right)=\mathrm{VI}(\mathrm{VI}(\Omega ,A_1),\mu A_2-f).$$

That is, the Problem 2 has the unique solution. Since ${lim\; inf}_{n\to \infty }{\gamma }_n>0$ and ${lim\; sup}_{n\to \infty }{\gamma }_n<1$, we can suppose that $\{{\gamma }_n\}\subset [a_0,b_0]$ is subset of $(0,1)$ for some $a_0,b_0\in (0,1)$. Since $G:$ is defined from C to C as $G:=P_C(P_C(I-{\mu }_2{B}_2)-{\mu }_1{B}_1{P}_C(I-{\mu }_2{B}_2))$. Here ${\mu }_1\in (0,2\alpha )$ and ${\mu }_2\in (0,2\beta )$, G is nonexpansive by Lemma 6. It is easy to see that for each $n\ge 0$ there exists a unique element $u_n\in C$ such that

$$u_n={\gamma }_n{x}_n+(1-{\gamma }_n)W_n{u}_n.$$

(3)

As a matter of fact, we utilize $F_{n}x:={\gamma }_n{x}_n+(1-{\gamma }_n)W_{n}x\forall x\in C$. Since each $W_n:C\to C$ is a nonexpansive mapping, we get

$$\parallel F_{n}x-F_{n}y\parallel =(1-{\gamma }_n)\parallel W_{n}x-W_{n}y\parallel \le (1-{\gamma }_n)\parallel x-y\parallel ,\forall x,y\in C.$$

Also, from $\{{\gamma }_n\}\subset [a_0,b_0]$ and $[a_0,b_0]\subset (0,1)$ we have $0<1-{\gamma }_n<1$, $\forall n\ge 0$. Thus, $F_n:C\to C$ is a contraction. Banach’s Contraction Principle infers there exists a unique element $u_n$ in set C satisfying (3).

Here, we are able to divide the rest of the proof into several steps.

Step 1. We claim that all the vector sequences $\{x_n\},\{y_n\},\{z_n\},\{u_n\},\{v_n\},\{T^n{y}_n\}$ and $\{A_2\left(T^n{y}_n\right)\}$ are bounded, where $v_n=P_C(u_n-{\mu }_2{B}_2{u}_n)$ and $z_n=P_C(v_n-{\mu }_1{B}_1{v}_n)$ for all $n\ge 0$. Indeed, it is clear that (2) can be rewritten as

$$\left\{\begin{array}{c}{u}_n=(1-{\gamma }_n)W_n{u}_n+{\gamma }_n{x}_n,\hfill \\ z_n=G{u}_n,\hfill \\ y_n=P_C(I-{\delta }_n{A}_1)z_n,\hfill \\ x_{n+1}=(1-{\beta }_n)P_C[(I-{\alpha }_n\mu A_2)T^n{y}_n+{\alpha }_{n}f\left(x_n\right)]+{\beta }_n{x}_n.\hfill \end{array}\right.$$

(4)

Take an arbitrary

$$p\in \Omega =\bigcap _{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)\cap \mathrm{GSVI}(C,B_1,B_2)\cap \mathrm{Fix}\left(T\right).$$

Then $p=W_{n}p$, $p=Tp$ and $p=Gp$. Since each $W_n:C\to C$ is nonexpansive, (4) infers to

$$\parallel u_n-p\parallel \le (1-{\gamma }_n)\parallel u_n-p\parallel +{\gamma }_n\parallel x_n-p\parallel ,$$

which hence yields

$$\parallel u_n-p\parallel \le \parallel x_n-p\parallel ,\forall n\ge 0.$$

(5)

It is easy to infer from (4) that

$$\parallel z_n-p\parallel =\parallel G{u}_n-p\parallel \le \parallel u_n-p\parallel \le \parallel x_n-p\parallel ,\forall n\ge 0.$$

(6)

Since ${lim\; inf}_{n\to \infty }{\beta }_n>0$ and ${lim\; sup}_{n\to \infty }{\beta }_n<1$, we suppose that $\{{\beta }_n\}\subset [c,d]$. Since ${lim}_{n\to \infty }\frac{{\theta }_n}{{\alpha }_n}=0$, we can also suppose that

$${\theta }_n\le \frac{{\alpha }_n(\tau -\delta )(1-d)}{2}\le {\alpha }_n(\tau -\delta ).$$

Note that ${\delta }_n\le {\alpha }_n$, $\forall n\ge 0$. $\zeta$-inverse-strong monotonicity of $A_1$ and Lemma 5 yield

$$\parallel y_n-p\parallel \le \parallel (I-{\delta }_n{A}_1)z_n-(I-{\delta }_n{A}_1)p-{\delta }_n{A}_{1}p\parallel \le {\delta }_n\parallel A_{1}p\parallel +\parallel z_n-p\parallel \le \parallel x_n-p\parallel +{\delta }_n\parallel A_{1}p\parallel .$$

(7)

Utilizing Lemma 4 and (7), we obtain from (4) that

$$\begin{array}{c}\parallel x_{n+1}-p\parallel \hfill \\ \le (1-{\beta }_n)\parallel {\alpha }_n(f\left(x_n\right)-\mu A_{2}p)+(I-{\alpha }_n\mu A_2)T^n{y}_n-(I-{\alpha }_n\mu A_2)p\parallel +{\beta }_n\parallel x_n-p\parallel \hfill \\ \le (1-{\beta }_n)[{\alpha }_n\delta \parallel x_n-p\parallel +{\alpha }_n\parallel f\left(p\right)-\mu A_{2}p\parallel +(1-{\alpha }_n\tau )(1+{\theta }_n)\parallel y_n-p\parallel ]+{\beta }_n\parallel x_n-p\parallel \hfill \\ \le {\beta }_n\parallel x_n-p\parallel +(1-{\beta }_n)\{{\alpha }_n\delta \parallel x_n-p\parallel +{\alpha }_n\parallel \mu A_{2}p-f\left(p\right)\parallel \hfill \\ +(1-{\alpha }_n\tau )[\parallel x_n-p\parallel +{\delta }_n\parallel A_{1}p\parallel ]+{\theta }_n[{\delta }_n\parallel A_{1}p\parallel +\parallel x_n-p\parallel ]\}\hfill \\ \le {\beta }_n\parallel x_n-p\parallel +(1-{\beta }_n)\{{\alpha }_n\delta \parallel x_n-p\parallel +{\alpha }_n\parallel \mu A_{2}p-f\left(p\right)\parallel \hfill \\ +(1-{\alpha }_n\tau )[\parallel x_n-p\parallel +{\delta }_n\parallel A_{1}p\parallel ]+{\theta }_n\parallel x_n-p\parallel +(\tau -\delta ){\alpha }_n{\delta }_n\parallel A_{1}p\parallel \}\hfill \\ \le [1-{\alpha }_n(1-{\beta }_n)(\tau -\delta )]\parallel x_n-p\parallel +{\theta }_n\parallel x_n-p\parallel +{\alpha }_n\parallel \mu A_{2}p-f\left(p\right)\parallel +{\alpha }_n\parallel A_{1}p\parallel \hfill \\ \le [1-\frac{{\alpha }_n(1-d)(\tau -\delta )}{2}]\parallel x_n-p\parallel +\frac{{\alpha }_n(1-d)(\tau -\delta )}{2}·\frac{2(\parallel A_{1}p\parallel +\parallel \mu A_{2}p-f\left(p\right)\parallel )}{(1-d)(\tau -\delta )}\hfill \\ \le max\{\frac{2(\parallel A_{1}p\parallel +\parallel \mu A_{2}p-f\left(p\right)\parallel )}{(1-d)(\tau -\delta )},\parallel x_n-p\parallel \}.\hfill \end{array}$$

By simple induction, we have

$$\parallel x_{n+1}-p\parallel \le max\{\frac{2(\parallel f\left(p\right)-\mu A_{2}p\parallel )+\parallel A_{1}p\parallel }{(1-d)(\tau -\delta )},\parallel x_0-p\parallel \},\forall n\ge 0.$$

Therefore, $\{x_n\}$ is a bounded vector sequence, and so are all the other sequences $\{y_n\},\{z_n\},\{u_n\},\{T^n{y}_n\}$ and $\{A_2\left(T^n{y}_n\right)\}$ (due to the Lipschitz continuity of T and $A_2$). Since each $W_n$ enjoys the nonexpansivity on C, we get

$$\parallel W_n{u}_n\parallel \le \parallel W_n{u}_n-p\parallel +\parallel p\parallel \le \parallel u_n-p\parallel +\parallel p\parallel ,$$

which yields that $\{W_n{u}_n\}$ is bounded too. In addition, from Lemma 2 and p is a element in $\Omega \subset \mathrm{GSVI}(C,B_1,B_2)$, it also follows that $(p,q)$ is a solution of GSVI (1) where $q=P_C(I-{\mu }_2{B}_2)p$. Note that $v_n=P_C(I-{\mu }_2{B}_2)u_n$ for all $n\ge 0$. Then by Lemma 5, we get

$$\parallel v_n\parallel \le \parallel P_C(I-{\mu }_2{B}_2)u_n-P_C(I-{\mu }_2{B}_2)p\parallel +\parallel q\parallel \le \parallel u_n-p\parallel +\parallel q\parallel .$$

This yields vector sequence $\{v_n\}$ is bounded.

Step 2. We claim that $\parallel x_n-x_{n+1}\parallel \to 0$ and $\parallel y_n-y_{n+1}\parallel \to 0$ as $n\to \infty$. Indeed, we set $x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)w_n$, $\forall n\ge 0$. Then $w_n=P_C[(I-{\alpha }_n\mu A_2)T^n{y}_n+{\alpha }_{n}f\left(x_n\right)]$. It follows from (4) that

$$\begin{array}{cc}\parallel w_{n+1}-w_n\parallel \hfill & \le \parallel {\alpha }_{n+1}f\left(x_{n+1}\right)+(I-{\alpha }_{n+1}\mu A_2)T^{n+1}{y}_{n+1}-{\alpha }_{n}f\left(x_n\right)-(I-{\alpha }_n\mu A_2)T^n{y}_n\parallel \hfill \\ & \le \parallel T^{n+1}{y}_{n+1}-T^{n+1}{y}_n\parallel +\parallel T^{n+1}{y}_n-T^n{y}_n\parallel +{\alpha }_{n+1}\parallel \mu A_2\left(T^{n+1}{y}_{n+1}\right)\parallel \hfill \\ & +{\alpha }_n\parallel \mu A_2\left(T^n{y}_n\right)\parallel +{\alpha }_{n+1}\parallel f\left(x_{n+1}\right)\parallel +{\alpha }_n\parallel f\left(x_n\right)\parallel \hfill \\ & \le (1+{\theta }_{n+1})\parallel y_{n+1}-y_n\parallel +\parallel T^{n+1}{y}_n-T^n{y}_n\parallel +{\alpha }_{n+1}(\parallel f\left(x_{n+1}\right)\parallel \hfill \\ & +\parallel \mu A_2\left(T^{n+1}{y}_{n+1}\right)\parallel )+{\alpha }_n(\parallel f\left(x_n\right)\parallel +\parallel \mu A_2\left(T^n{y}_n\right)\parallel ).\hfill \end{array}$$

(8)

Since vector sequence $\{{\delta }_n\}$ falls into $(0,2\zeta ]$ and $A_1$ is $\zeta$-inverse-strongly monotone, by Lemma 5 we obtain

$$\begin{array}{cc}\parallel y_{n+1}-y_n\parallel \hfill & \le \parallel (z_{n+1}-{\delta }_{n+1}{A}_1{z}_{n+1})-(z_n-{\delta }_n{A}_1{z}_n)\parallel \hfill \\ & \le \parallel (z_{n+1}-{\delta }_{n+1}{A}_1{z}_{n+1})-(z_n-{\delta }_{n+1}{A}_1{z}_n)\parallel +|{\delta }_n-{\delta }_{n+1}|\parallel A_1{z}_n\parallel \hfill \\ & \le \parallel u_{n+1}-u_n\parallel +|{\delta }_n-{\delta }_{n+1}|\parallel A_1{z}_n\parallel .\hfill \end{array}$$

(9)

Since simple calculations show that

$$\begin{array}{cc}\parallel u_n{u}_{n+1}\parallel \hfill & \le {\gamma }_{n+1}\parallel x_n-x_{n+1}\parallel +(1-{\gamma }_{n+1})\parallel W_n{u}_n-W_{n+1}{u}_{n+1}\parallel \hfill \\ & +|{\gamma }_n-{\gamma }_{n+1}|\parallel W_n{u}_n-x_n\parallel \hfill \\ & \le {\gamma }_{n+1}\parallel x_n-x_{n+1}\parallel +(1-{\gamma }_{n+1})[\parallel W_n{u}_{n+1}-W_{n+1}{u}_{n+1}\parallel \hfill \\ & +\parallel W_n{u}_n-W_n{u}_{n+1}\parallel ]+|{\gamma }_n-{\gamma }_{n+1}|\parallel W_n{u}_n-x_n\parallel \hfill \\ & \le (1-{\gamma }_{n+1})[\parallel W_n{u}_{n+1}-W_{n+1}{u}_{n+1}\parallel +{\gamma }_{n+1}\parallel x_n-x_{n+1}\parallel \hfill \\ & +\parallel u_n-u_{n+1}\parallel ]+|{\gamma }_n-{\gamma }_{n+1}|\parallel W_n{u}_n-x_n\parallel ,\hfill \end{array}$$

it follows that

$$a_0\parallel u_n-u_{n+1}\parallel \le a_0\parallel x_{n+1}-x_n\parallel +\parallel W_{n+1}{u}_{n+1}-W_n{u}_{n+1}\parallel +a_0\parallel x_n-W_n{u}_n\parallel |{\gamma }_{n+1}-{\gamma }_n|.$$

(10)

Since $D:=\{u_n:n\ge 0\}\subset C$ is bounded subset, by the argument process in Lemma 11 we get ${\sum }_{n=0}^{\infty }{sup}_{x\in D}\parallel W_{n+1}x-W_{n}x\parallel <\infty$. Thus we have

$$\sum _{n=0}^{\infty }\parallel W_{n+1}{u}_{n+1}-W_n{u}_{n+1}\parallel <\infty .$$

(11)

Therefore, from (8)–(10) we deduce that

$$\begin{array}{c}\parallel w_n-w_{n+1}\parallel \hfill \\ \le |{\delta }_n-{\delta }_{n+1}|\parallel A_1{z}_n\parallel +{\theta }_{n+1}\parallel y_n-y_{n+1}\parallel +\parallel T^n{y}_n-T^{n+1}{y}_n\parallel +\parallel u_n-u_{n+1}\parallel \hfill \\ +{\alpha }_{n+1}(\parallel f\left(x_{n+1}\right)\parallel +\parallel \mu A_2\left(T^{n+1}{y}_{n+1}\right)\parallel )+{\alpha }_n(\parallel f\left(x_n\right)\parallel +\parallel \mu A_2\left(T^n{y}_n\right)\parallel )\hfill \\ \le \frac{1}{a_0}\parallel W_n{u}_{n+1}-W_{n+1}{u}_{n+1}\parallel +\parallel x_{n+1}-x_n\parallel +|{\gamma }_{n+1}-{\gamma }_n|\frac{\parallel W_n{u}_n-x_n\parallel }{a_0}+|{\delta }_n-{\delta }_{n+1}|\parallel A_1{z}_n\parallel \hfill \\ +{\theta }_{n+1}\parallel y_n-y_{n+1}\parallel +\parallel T^n{y}_n-T^{n+1}{y}_n\parallel +{\alpha }_{n+1}(\parallel f\left(x_{n+1}\right)\parallel +\parallel \mu A_2\left(T^{n+1}{y}_{n+1}\right)\parallel \hfill \\ +{\alpha }_n\parallel (\parallel f\left(x_n\right)\parallel +\mu A_2\left(T^n{y}_n\right)\parallel ),\hfill \end{array}$$

which immediately attains

$$\begin{array}{c}\parallel w_n-w_{n+1}\parallel -\parallel x_n-x_{n+1}\parallel \hfill \\ \le \frac{1}{a_0}\parallel W_n{u}_{n+1}-W_{n+1}{u}_{n+1}\parallel +|{\gamma }_n-{\gamma }_{n+1}|\frac{\parallel W_n{u}_n-x_n\parallel }{a_0}+|{\delta }_n-{\delta }_{n+1}|\parallel A_1{z}_n\parallel \hfill \\ +{\theta }_{n+1}\parallel y_n-y_{n+1}\parallel +\parallel T^n{y}_n-T^{n+1}{y}_n\parallel +{\alpha }_{n+1}(\parallel f\left(x_{n+1}\right)\parallel +\parallel \mu A_2\left(T^{n+1}{y}_{n+1}\right)\parallel )\hfill \\ +{\alpha }_n(\parallel f\left(x_n\right)\parallel +\parallel \mu A_2\left(T^n{y}_n\right)\parallel ).\hfill \end{array}$$

(12)

Since

$$\underset{n\to \infty }{lim}\parallel T^n{y}_n-T^{n+1}{y}_n\parallel =\underset{n\to \infty }{lim}{\theta }_n=0,$$

from (11) and conditions (i), (ii), (iv) we get ${lim\; sup}_{n\to \infty }(\parallel w_n-w_{n+1}\parallel -\parallel x_n-x_{n+1}\parallel )\le 0$. Hence, by condition (iii) and Lemma 8, we get ${lim}_{n\to \infty }\parallel w_n-x_n\parallel =0$. Consequently,

$$\underset{n\to \infty }{lim}(1-{\beta }_n)\parallel w_n-x_n\parallel =\underset{n\to \infty }{lim}\parallel x_n-x_{n+1}\parallel =0.$$

(13)

Again from (9) and (10) we conclude that

$$\begin{array}{c}{a}_0\parallel y_n-y_{n+1}\parallel \hfill \\ \le a_0\parallel x_n-x_{n+1}\parallel +\parallel W_n{u}_{n+1}-W_{n+1}{u}_{n+1}\parallel +a_0|{\gamma }_n-{\gamma }_{n+1}|\hfill \\ \parallel W_n{u}_n-x_n\parallel +|{\delta }_n-{\delta }_{n+1}|a_0\parallel A_1{z}_n\parallel \to 0\hfill \end{array}$$

and $\parallel z_{n+1}-z_n\parallel =\parallel G{u}_{n+1}-G{u}_n\parallel \le \parallel u_{n+1}-u_n\parallel \to 0$. Thus,

$$\underset{n\to \infty }{lim}\parallel y_n-y_{n+1}\parallel =\underset{n\to \infty }{lim}\parallel u_n-u_{n+1}\parallel =\underset{n\to \infty }{lim}\parallel z_n-z_{n+1}\parallel =0.$$

(14)

Step 3. We claim that ${lim}_{n\to \infty }\parallel G{x}_n-x_n\parallel =0$ as $n\to \infty$. Indeed, noticing $w_n=P_C[(I-{\alpha }_n\mu A_2)T^n{y}_n+{\alpha }_{n}f\left(x_n\right)]\forall n\ge 0$, we obtain from Lemma 2 that for each $p\in \Omega$,

$$\langle p-w_n,(I-{\alpha }_n\mu A_2)T^n{y}_n+{\alpha }_{n}f\left(x_n\right)-P_C[{\alpha }_{n}f\left(x_n\right)+(I-{\alpha }_n\mu A_2)T^n{y}_n]\rangle \le 0.$$

(15)

From (15), we have

$$\begin{array}{cc}\parallel w_n{-p\parallel }^2\hfill & =\langle P_C[(I-{\alpha }_n\mu A_2)T^n{y}_n+{\alpha }_{n}f\left(x_n\right)]-{\alpha }_{n}f\left(x_n\right)-(I-{\alpha }_n\mu A_2)T^n{y}_n,w_n-p\rangle \hfill \\ & +\langle (I-{\alpha }_n\mu A_2)T^n{y}_n+{\alpha }_{n}f\left(x_n\right)-p,w_n-p\rangle \hfill \\ & \le \langle (I-{\alpha }_n\mu A_2)T^n{y}_n+{\alpha }_{n}f\left(x_n\right)-p,w_n-p\rangle \hfill \\ & =\langle w_n-p,(I-{\alpha }_n\mu A_2)T^n{y}_n-(I-{\alpha }_n\mu A_2)p\rangle +{\alpha }_n\langle f\left(x_n\right)-\mu A_{2}p,w_n-p\rangle \hfill \\ & \le [(1-{\alpha }_n\tau )\parallel T^n{y}_n-p\parallel +\delta {\alpha }_n\parallel x_n-p\parallel ]\parallel w_n-p\parallel +{\alpha }_n\langle w_n-p,f\left(p\right)-\mu A_{2}p\rangle \hfill \\ & \le \frac{[(1-{\alpha }_n\tau )\parallel T^n{y}_n-p\parallel +{\alpha }_n\delta \parallel x_n-p{\parallel ]}^2}{2}+\frac{1}{2}{\parallel w_n-p\parallel }^2+{\alpha }_n\langle w_n-p,f\left(p\right)-\mu A_{2}p\rangle ,\hfill \end{array}$$

which leads to

$$\begin{array}{c}\parallel w_n{-p\parallel }^2\hfill \\ \le (1-{\alpha }_n\tau )\parallel T^n{y}_n{-p\parallel }^2+\delta {\alpha }_n{\parallel x_n-p\parallel }^2-2{\alpha }_n\langle w_n-p,\mu A_{2}p-f\left(p\right)\rangle \hfill \\ \le (1-{\alpha }_n\tau ){(1+{\theta }_n)}^2\parallel y_n{-p\parallel }^2+{\alpha }_n\delta {\parallel x_n-p\parallel }^2-2{\alpha }_n\langle w_n-p,\mu A_{2}p-f\left(p\right)\rangle \hfill \\ \le (1-{\alpha }_n\tau )\parallel y_n{-p\parallel }^2+{\alpha }_n\delta \parallel x_n{-p\parallel }^2+{\theta }_n(2+{\theta }_n){\parallel y_n-p\parallel }^2-2{\alpha }_n\langle w_n-p,\mu A_{2}p-f\left(p\right)\rangle .\hfill \end{array}$$

(16)

From (7) and (16), we get

$$\begin{array}{c}\parallel x_{n+1}{-p\parallel }^2\hfill \\ \le {\beta }_n\parallel x_n{-p\parallel }^2+(1-{\beta }_n)[{\alpha }_n\delta \parallel x_n{-p\parallel }^2+(1-{\alpha }_n\tau )\parallel y_n{-p\parallel }^2+{\theta }_n(2+{\theta }_n){\parallel y_n-p\parallel }^2\hfill \\ +2{\alpha }_n\langle f\left(p\right)-\mu A_{2}p,w_n-p\rangle ]\hfill \\ \le {\beta }_n\parallel x_n{-p\parallel }^2+(1-{\beta }_n)\{{\alpha }_n\delta \parallel x_n{-p\parallel }^2+(1-{\alpha }_n\tau )(\parallel z_n-p\parallel +{\delta }_n\parallel A_{1}p{\parallel )}^2\hfill \\ +{\theta }_n(2+{\theta }_n)\parallel y_n-p{\parallel }^2+2{\alpha }_n\langle f\left(p\right)-\mu A_{2}p,w_n-p\rangle \}\hfill \\ \le {\beta }_n\parallel x_n{-p\parallel }^2+(1-{\beta }_n)[{\alpha }_n\delta \parallel x_n{-p\parallel }^2+(1-{\alpha }_n\tau )\parallel z_n-p{\parallel }^2]\hfill \\ +{\delta }_n\parallel A_{1}p\parallel (2\parallel z_n-p\parallel +{\delta }_n\parallel A_{1}p\parallel )+{\theta }_n(2+{\theta }_n)\parallel y_n{-p\parallel }^2+2{\alpha }_n\parallel f\left(p\right)-\mu A_{2}p\parallel \parallel p-w_n\parallel .\hfill \end{array}$$

(17)

We now note that $q=P_C(p-{\mu }_2{B}_{2}p),v_n=P_C(u_n-{\mu }_2{B}_2{u}_n)$ and $z_n=P_C(v_n-{\mu }_1{B}_1{v}_n)$. Then $z_n=G{u}_n$. By Lemma 5 we have

$$\parallel v_n{-q\parallel }^2\le \parallel u_n-p-{\mu }_2(B_2{u}_n-B_{2}p){\parallel }^2\le \parallel u_n{-p\parallel }^2-{\mu }_2(2\beta -{\mu }_2){\parallel B_2{u}_n-B_{2}p\parallel }^2$$

(18)

and

$$\parallel z_n{-p\parallel }^2\le \parallel v_n-q-{\mu }_1(B_1{v}_n-B_{1}q){\parallel }^2\le \parallel v_n{-q\parallel }^2-{\mu }_1(2\alpha -{\mu }_1){\parallel B_1{v}_n-B_{1}q\parallel }^2.$$

(19)

Substituting (18) for (19), we obtain from (5) that

$$\parallel z_n{-p\parallel }^2\le {\mu }_2({\mu }_2-2\beta )\parallel B_2{u}_n-B_2{p\parallel }^2+{\mu }_1({\mu }_1+2\alpha )\parallel B_1{v}_n-B_1{q\parallel }^2+{\parallel x_n-p\parallel }^2.$$

(20)

Combining (17) and (20), we get

$$\begin{array}{c}\parallel x_{n+1}{-p\parallel }^2\hfill \\ \le {\beta }_n\parallel x_n{-p\parallel }^2+(1-{\beta }_n)\{{\alpha }_n\delta \parallel x_n{-p\parallel }^2+(1-\tau {\alpha }_n)[\parallel x_n{-p\parallel }^2\hfill \\ -(2\beta -{\mu }_2){\mu }_2\parallel B_{2}p-B_2{u}_n{\parallel }^2-(2\alpha -{\mu }_1){\mu }_1\parallel B_{1}q-B_1{v}_n{\parallel }^2]\}\hfill \\ +{\delta }_n\parallel A_{1}p\parallel (2\parallel z_n-p\parallel +{\delta }_n\parallel A_{1}p\parallel )+(2+{\theta }_n){\theta }_n\parallel y_n{-p\parallel }^2+2{\alpha }_n\parallel \mu A_{2}p-f\left(p\right)\parallel \parallel p-w_n\parallel \hfill \\ =[1-(\tau -\delta ){\alpha }_n(1-{\beta }_n)]\parallel x_n{-p\parallel }^2-(1-{\alpha }_n\tau )(1-{\beta }_n)[{\mu }_2(2\beta -{\mu }_2){\parallel B_{2}p-B_2{u}_n\parallel }^2\hfill \\ +(2\alpha -{\mu }_1){\mu }_1\parallel B_{1}q-B_1{v}_n{\parallel }^2]+{\delta }_n\parallel A_{1}p\parallel (2\parallel z_n-p\parallel +{\delta }_n\parallel A_{1}p\parallel )\hfill \\ +(2+{\theta }_n){\theta }_n\parallel y_n{-p\parallel }^2+2{\alpha }_n\parallel \mu A_{2}p-f\left(p\right)\parallel \parallel p-w_n\parallel ,\hfill \end{array}$$

which immediately yields

$$\begin{array}{c}(1-{\alpha }_n\tau )(1-{\beta }_n)[{\mu }_2(2\beta -{\mu }_2)\parallel B_{2}p-B_2{u}_n{\parallel }^2+(2\alpha -{\mu }_1){\mu }_1\parallel B_{1}q-B_1{v}_n{\parallel }^2]\hfill \\ \le \parallel x_n{-p\parallel }^2-\parallel x_{n+1}{-p\parallel }^2+{\delta }_n\parallel A_{1}p\parallel ({\delta }_n\parallel A_{1}p\parallel +2\parallel z_n-p\parallel )\hfill \\ +(2+{\theta }_n){\theta }_n\parallel y_n{-p\parallel }^2+2{\alpha }_n\parallel \mu A_{2}p-f\left(p\right)\parallel \parallel p-w_n\parallel \hfill \\ \le \parallel x_n-x_{n+1}\parallel (\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )+{\delta }_n\parallel A_{1}p\parallel ({\delta }_n\parallel A_{1}p\parallel +2\parallel z_n-p\parallel )\hfill \\ +(2+{\theta }_n){\theta }_n\parallel y_n{-p\parallel }^2+2{\alpha }_n\parallel p-w_n\parallel \parallel f\left(p\right)-\mu A_{2}p\parallel .\hfill \end{array}$$

Due to condition (iii), ${lim\; inf}_{n\to \infty }(1-{\beta }_n)>0$, ${\mu }_1\in (0,2\alpha ),$${\mu }_2\in (0,2\beta ),$${lim}_{n\to \infty }{\theta }_n=0,$${lim}_{n\to \infty }{\alpha }_n=0$ and ${lim}_{n\to \infty }{\delta }_n=0$, we obtain from (13) that

$$\underset{n\to \infty }{lim}\parallel B_2{u}_n-B_{2}p\parallel =0\mathrm{and}\underset{n\to \infty }{lim}\parallel B_1{v}_n-B_{1}q\parallel =0.$$

(21)

On the other hand, from Lemma 2 we have

$$\begin{array}{cc}\parallel v_n{-q\parallel }^2\hfill & \le \langle v_n-q,u_n-(p-{\mu }_2{B}_{2}p)-{\mu }_2{B}_2{u}_n\rangle \hfill \\ & \le \frac{1}{2}[\parallel u_n{-p\parallel }^2+\parallel v_n{-q\parallel }^2-\parallel u_n-v_n{-(p-q)\parallel }^2]+{\mu }_2\parallel v_n-q\parallel \parallel B_2{u}_n-B_{2}p\parallel ,\hfill \end{array}$$

which implies that

$$\parallel v_n{-q\parallel }^2\le \parallel u_n{-p\parallel }^2-\parallel (p-q)-u_n+v_n{\parallel }^2+2{\mu }_2\parallel v_n-q\parallel \parallel B_2{u}_n-B_{2}p\parallel .$$

(22)

In the same way, we derive

$$\parallel z_n{-p\parallel }^2\le \parallel v_n{-q\parallel }^2-\parallel (p-q)-v_n+z_n{\parallel }^2+2{\mu }_1\parallel z_n-p\parallel \parallel B_1{v}_n-B_{1}q\parallel .$$

(23)

Substituting (22) for (23), we deduce from (5) that

$$\begin{array}{cc}\parallel z_n{-p\parallel }^2\hfill & \le \parallel x_n{-p\parallel }^2-\parallel u_n-v_n{-(p-q)\parallel }^2-{\parallel v_n-z_n+(p-q)\parallel }^2\hfill \\ & +2{\mu }_2\parallel B_{2}p-B_2{u}_n\parallel \parallel v_n-q\parallel +2{\mu }_1\parallel B_{1}q-B_1{v}_n\parallel \parallel z_n-p\parallel .\hfill \end{array}$$

(24)

Combining (17) and (24), we have

$$\begin{array}{c}\parallel x_{n+1}{-p\parallel }^2\hfill \\ \le {\beta }_n\parallel x_n{-p\parallel }^2+(1-{\beta }_n)\{{\alpha }_n\delta \parallel x_n{-p\parallel }^2+(1-{\alpha }_n\tau )[\parallel x_n{-p\parallel }^2-{\parallel p-q-u_n+v_n\parallel }^2\hfill \\ -\parallel p-q+v_n-z_n{\parallel }^2+2{\mu }_1\parallel z_n-p\parallel \parallel B_1{v}_n-B_{1}q\parallel +2{\mu }_2\parallel v_n-q\parallel \parallel B_2{u}_n-B_{2}p\parallel ]\}\hfill \\ +{\delta }_n\parallel A_{1}p\parallel (2\parallel z_n-p\parallel +{\delta }_n\parallel A_{1}p\parallel )+(2+{\theta }_n){\theta }_n\parallel y_n{-p\parallel }^2+2{\alpha }_n\parallel \mu A_{2}p-f\left(p\right)\parallel \parallel w_n-p\parallel \hfill \\ \le [1-(\tau -\delta ){\alpha }_n(1-{\beta }_n)]\parallel x_n{-p\parallel }^2-(1-{\alpha }_n\tau )(1-{\beta }_n)[\parallel p-q-u_n+v_n{\parallel }^2\hfill \\ +\parallel p-q+v_n-z_n{\parallel }^2]+2{\mu }_1\parallel B_1{v}_n-B_{1}q\parallel \parallel z_n-p\parallel +2{\mu }_2\parallel v_n-q\parallel \parallel B_{2}p-B_2{u}_n\parallel \hfill \\ +{\delta }_n\parallel A_{1}p\parallel ({\delta }_n\parallel A_{1}p\parallel +2\parallel z_n-p\parallel )+(2+{\theta }_n){\theta }_n\parallel y_n{-p\parallel }^2+2{\alpha }_n\parallel w_n-p\parallel \parallel f\left(p\right)-\mu A_{2}p\parallel ,\hfill \end{array}$$

which hence yields

$$\begin{array}{c}(1-{\alpha }_n\tau )(1-{\beta }_n)[\parallel p-q-u_n+v_n{\parallel }^2+\parallel p-q+v_n-z_n{\parallel }^2]\hfill \\ \le \parallel x_n{-p\parallel }^2-\parallel x_{n+1}{-p\parallel }^2+2{\mu }_2\parallel v_n-q\parallel \parallel B_{2}p-B_2{u}_n\parallel \hfill \\ +2{\mu }_1\parallel z_n-p\parallel \parallel B_{1}q-B_1{v}_n\parallel +{\delta }_n\parallel A_{1}p\parallel ({\delta }_n\parallel A_{1}p\parallel +2\parallel z_n-p\parallel )\hfill \\ +(2+{\theta }_n){\theta }_n\parallel y_n{-p\parallel }^2+2{\alpha }_n\parallel p-w_n\parallel \parallel f\left(p\right)-\mu A_{2}p\parallel \hfill \\ \le \parallel x_n-x_{n+1}\parallel (\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )+2{\mu }_2\parallel z_n-q\parallel \parallel B_{2}p-B_2{u}_n\parallel \hfill \\ +2{\mu }_1\parallel y_n-p\parallel \parallel B_{1}q-B_1{z}_n\parallel +{\delta }_n\parallel A_{1}p\parallel ({\delta }_n\parallel A_{1}p\parallel +2\parallel z_n-p\parallel )\hfill \\ +(2+{\theta }_n){\theta }_n\parallel y_n{-p\parallel }^2+2{\alpha }_n\parallel p-w_n\parallel \parallel f\left(p\right)-\mu A_{2}p\parallel .\hfill \end{array}$$

Since ${lim\; inf}_{n\to \infty }(1-{\beta }_n)>0$, ${lim}_{n\to \infty }{\theta }_n=0,$${lim}_{n\to \infty }{\alpha }_n=0$ and ${lim}_{n\to \infty }{\delta }_n=0$, we conclude from (13) and (21) that

$$\underset{n\to \infty }{lim}\parallel u_n-v_n-(p-q)\parallel =0\mathrm{and}\underset{n\to \infty }{lim}\parallel v_n-z_n+(p-q)\parallel =0.$$

(25)

It follows that

$$\parallel u_n-G{u}_n\parallel =\parallel u_n-z_n\parallel \le \parallel u_n-v_n-(p-q)\parallel +\parallel v_n-z_n+(p-q)\parallel \to 0(n\to \infty ).$$

(26)

Also, from (4) we have $\parallel u_n{-p\parallel }^2\le (1-{\gamma }_n){\parallel u_n-p\parallel }^2+{\gamma }_n\langle u_n-p,x_n-p\rangle$, which together with Lemma 2, yields $\parallel u_n{-p\parallel }^2\le \langle u_n-p,x_n-p\rangle =\frac{1}{2}[\parallel x_n{-p\parallel }^2+\parallel u_n{-p\parallel }^2-\parallel x_n-u_n{\parallel }^2]$. Thus, we get

$$\parallel u_n{-p\parallel }^2\le \parallel x_n{-p\parallel }^2-{\parallel x_n-u_n\parallel }^2,$$

which together with (17), yields

$$\begin{array}{c}\parallel x_{n+1}{-p\parallel }^2\hfill \\ \le {\beta }_n\parallel x_n{-p\parallel }^2+(1-{\beta }_n)[(1-{\alpha }_n\tau )\parallel u_n{-p\parallel }^2+\delta {\alpha }_n\parallel x_n-p{\parallel }^2]\hfill \\ +{\delta }_n\parallel A_{1}p\parallel ({\delta }_n\parallel A_{1}p\parallel +2\parallel z_n-p\parallel )+(2+{\theta }_n){\theta }_n\parallel y_n{-p\parallel }^2+2{\alpha }_n\parallel p-w_n\parallel \parallel f\left(p\right)-\mu A_{2}p\parallel \hfill \\ \le {\beta }_n\parallel x_n{-p\parallel }^2+(1-{\beta }_n)\{{\alpha }_n\delta \parallel x_n{-p\parallel }^2+(1-{\alpha }_n\tau )[\parallel x_n{-p\parallel }^2-\parallel x_n-u_n{\parallel }^2]\}\hfill \\ +{\delta }_n\parallel A_{1}p\parallel ({\delta }_n\parallel A_{1}p\parallel +2\parallel z_n-p\parallel )+(2+{\theta }_n){\theta }_n\parallel y_n{-p\parallel }^2+2{\alpha }_n\parallel p-w_n\parallel \parallel f\left(p\right)-\mu A_{2}p\parallel \hfill \\ =[1-{\alpha }_n(\tau -\delta )(1-{\beta }_n)]\parallel x_n{-p\parallel }^2-(1-{\alpha }_n\tau )(1-{\beta }_n){\parallel x_n-u_n\parallel }^2\hfill \\ +{\delta }_n\parallel A_{1}p\parallel ({\delta }_n\parallel A_{1}p\parallel +2\parallel z_n-p\parallel )+(2+{\theta }_n){\theta }_n\parallel y_n{-p\parallel }^2+2{\alpha }_n\parallel p-w_n\parallel \parallel f\left(p\right)-\mu A_{2}p\parallel .\hfill \end{array}$$

Hence we have

$$\begin{array}{c}(1-{\alpha }_n\tau )(1-{\beta }_n){\parallel x_n-u_n\parallel }^2\hfill \\ \le \parallel x_n{-p\parallel }^2-\parallel x_{n+1}{-p\parallel }^2+{\delta }_n\parallel A_{1}p\parallel ({\delta }_n\parallel A_{1}p\parallel +2\parallel z_n-p\parallel )\hfill \\ +(2+{\theta }_n){\theta }_n\parallel y_n{-p\parallel }^2+2{\alpha }_n\parallel \mu A_{2}p-f\left(p\right)\parallel \parallel p-w_n\parallel \hfill \\ \le \parallel x_n-x_{n+1}\parallel (\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )+{\delta }_n\parallel A_{1}p\parallel ({\delta }_n\parallel A_{1}p\parallel +2\parallel z_n-p\parallel )\hfill \\ +(2+{\theta }_n){\theta }_n\parallel y_n{-p\parallel }^2+2{\alpha }_n\parallel f\left(p\right)-\mu A_{2}p\parallel \parallel p-w_n\parallel .\hfill \end{array}$$

Since ${lim\; inf}_{n\to \infty }(1-{\beta }_n)>0$, ${lim}_{n\to \infty }{\theta }_n=0,$ ${lim}_{n\to \infty }{\alpha }_n=0$ and ${lim}_{n\to \infty }{\delta }_n=0$, we obtain from (13) that

$$\underset{n\to \infty }{lim}\parallel x_n-u_n\parallel =0.$$

(27)

Also, observe that $\parallel x_n-z_n\parallel \le \parallel x_n-u_n\parallel +\parallel G{u}_n-u_n\parallel$, $\parallel x_n-G{x}_n\parallel \le \parallel x_n-z_n\parallel +\parallel u_n-x_n\parallel ,$ and

$$\parallel x_n-y_n\parallel \le \parallel x_n-(z_n-{\delta }_n{A}_1{z}_n)\parallel \le \parallel x_n-z_n\parallel +{\delta }_n\parallel A_1{z}_n\parallel .$$

Then from (26) and (27) it follows that

$$\underset{n\to \infty }{lim}\parallel x_n-z_n\parallel =0,\underset{n\to \infty }{lim}\parallel x_n-G{x}_n\parallel =0\mathrm{and}\underset{n\to \infty }{lim}\parallel x_n-y_n\parallel =0.$$

(28)

Step 4. We claim that ${lim}_{n\to \infty }\parallel T{x}_n-x_n\parallel =0$ and ${lim}_{n\to \infty }\parallel W_n{x}_n-x_n\parallel =0$. Indeed, combining (4) and (27), we obtain

$$\parallel W_n{u}_n-u_n\parallel =\frac{{\gamma }_n}{1-{\gamma }_n}\parallel x_n-u_n\parallel \le \frac{b_0}{1-b_0}\parallel x_n-u_n\parallel \to 0(n\to \infty ).$$

(29)

Since each $W_n$ is nonexpansive on C, from (27) and (29) we get

$$\begin{array}{cc}\parallel W_n{x}_n-x_n\parallel \hfill & \le \parallel W_n{u}_n-u_n\parallel +\parallel u_n-x_n\parallel +\parallel W_n{x}_n-W_n{u}_n\parallel \hfill \\ & \le \parallel W_n{u}_n-u_n\parallel +2\parallel u_n-x_n\parallel \to 0(n\to \infty ).\hfill \end{array}$$

(30)

We note that $\{{\beta }_n\}\subset [c,d]$ and $[c,d]\subset (0,1)$ for some $c,d\in (0,1)$, and observe that

$$\begin{array}{c}\parallel x_n-T^n{y}_n\parallel \le \parallel x_n-x_{n+1}\parallel +\parallel T^n{y}_n-x_{n+1}\parallel \hfill \\ \le \parallel x_n-x_{n+1}\parallel +{\beta }_n\parallel x_n-T^n{y}_n\parallel +(1-{\beta }_n)\parallel T^n{y}_n-P_C[(I-{\alpha }_n\mu A_2)T^n{y}_n+{\alpha }_{n}f\left(x_n\right)]\parallel \hfill \\ \le \parallel x_n-x_{n+1}\parallel +{\beta }_n\parallel x_n-T^n{y}_n\parallel +(1-{\beta }_n){\alpha }_n(\parallel \mu A_2\left(T^n{y}_n\right)\parallel +\parallel f\left(x_n\right)\parallel ).\hfill \end{array}$$

Then we have

$$(1-d)\parallel x_n-T^n{y}_n\parallel \le \parallel x_n-x_{n+1}\parallel +(1-d){\alpha }_n(\parallel f\left(x_n\right)\parallel +\parallel \mu A_2\left(T^n{y}_n\right)\parallel ).$$

Hence we get

$$\begin{array}{cc}(1-d)\parallel y_n-T^n{y}_n\parallel \hfill & \le (1-d)\parallel y_n-x_n\parallel +(1-d)\parallel x_n-T^n{y}_n\parallel \hfill \\ & \le (1-d)\parallel y_n-x_n\parallel +\parallel x_n-x_{n+1}\parallel +{\alpha }_n(1-d)(\parallel f\left(x_n\right)\parallel +\parallel \mu A_2\left(T^n{y}_n\right)\parallel ).\hfill \end{array}$$

Consequently, from (13), (28) and ${lim}_{n\to \infty }{\alpha }_n=0$, it follows that

$$\underset{n\to \infty }{lim}\parallel x_n-T^n{y}_n\parallel =0\mathrm{and}\underset{n\to \infty }{lim}\parallel y_n-T^n{y}_n\parallel =0.$$

(31)

We also note that

$$\begin{array}{cc}\parallel y_n-T{y}_n\parallel \hfill & \le \parallel y_n-T^n{y}_n\parallel +\parallel T^n{y}_n-T^{n+1}{y}_n\parallel +\parallel T^{n+1}{y}_n-T{y}_n\parallel \hfill \\ & \le (2+{\theta }_1)\parallel T^n{y}_n-y_n\parallel +\parallel T^{n+1}{y}_n-T^n{y}_n\parallel .\hfill \end{array}$$

From ${lim}_{n\to \infty }\parallel T^n{y}_n-T^{n+1}{y}_n\parallel =0$ and (31), we get

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

(32)

In addition, noticing that

$$\parallel x_n-T{x}_n\parallel \le \parallel x_n-y_n\parallel +\parallel y_n-T{y}_n\parallel +\parallel T{y}_n-T{x}_n\parallel \le \parallel y_n-T{y}_n\parallel +(2+{\theta }_1)\parallel x_n-y_n\parallel ,$$

we deduce from (28) and (32) that

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

(33)

Step 5. We claim that $W:C\to C$ is nonexpansive, $\mathrm{Fix}\left(W\right)={\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)$ and ${lim}_{n\to \infty }\parallel W{x}_n-x_n\parallel =0$ where $Wx:={lim}_{n\to \infty }{W}_{n}x$ for all $x\in C$. Indeed, we observe that for all $x,y\in C$, ${lim}_{n\to \infty }\parallel W_{n}x-Wx\parallel =0$ and ${lim}_{n\to \infty }\parallel W_{n}y-Wy\parallel =0$. Since each $W_n$ enjoys the nonexpansivity, we get

$$\parallel Wx-Wy\parallel =\underset{n\to \infty }{lim}\parallel W_{n}x-W_{n}y\parallel \le \parallel x-y\parallel .$$

This means that W is nonexpansive. Also, noticing the boundedness of $\{x_n\}$ and putting $D:=\{x_n:n\ge 0\}$, we obtain from Lemma 11 that ${lim}_{n\to \infty }{sup}_{x\in D}\parallel W_{n}x-Wx\parallel =0$, which immediately sends to

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

(34)

Thus, combining (30) with (34) we have

$$\parallel x_n-W{x}_n\parallel \le \parallel x_n-W_n{x}_n\parallel +\parallel W_n{x}_n-W{x}_n\parallel \to 0(n\to \infty ).$$

(35)

In addition, utilizing Lemma 10 we get

$$\mathrm{Fix}\left(W\right)=\bigcap _{n=0}^{\infty }\mathrm{Fix}\left(S_n\right).$$

(36)

Step 6. We prove that

$$\underset{n\to \infty }{lim\; sup}\langle A_2{x}^{*},x^{*}-w_n\rangle \le 0\mathrm{and}\underset{n\to \infty }{lim\; sup}\langle A_1{x}^{*},x^{*}-z_n\rangle \le 0,$$

(37)

where $\{x^{*}\}=\mathrm{VI}(\mathrm{VI}(\Omega ,A_1),\mu A_2-f)$. Indeed, we choose a subsequence $\{w_{n_i}\}$ of $\{w_n\}$ such that

$$\underset{n\to \infty }{lim\; sup}\langle x^{*}-w_n,A_2{x}^{*}\rangle =\underset{i\to \infty }{lim}\langle x^{*}-w_{n_i},A_2{x}^{*}\rangle .$$

Utilizing the boundedness of $\{w_n\}\subset C$, we suppose that $w_{n_i}\rightharpoonup \overline{x}\in C$. Since ${lim}_{n\to \infty }\parallel x_n-T^n{y}_n\parallel =0$ (due to (31)) and ${lim}_{n\to \infty }{\alpha }_n=0$, it follows that

$$\begin{array}{cc}\parallel x_n-w_n\parallel \hfill & \le \parallel x_n-T^n{y}_n\parallel +\parallel T^n{y}_n-{\alpha }_{n}f\left(x_n\right)-(I-{\alpha }_n\mu A_2)T^n{y}_n\parallel \hfill \\ & \le \parallel T^n{y}_n{x}_n\parallel +{\alpha }_n(\parallel \mu A_2\left(T^n{y}_n\right)\parallel +\parallel f\left(x_n\right)\parallel )\to 0(n\to \infty ).\hfill \end{array}$$

(38)

Hence, from $w_{n_i}\rightharpoonup \overline{x}$, we get $x_{n_i}\rightharpoonup \overline{x}$.

Note that G and W are nonexpansive and T is asymptotical. Since $(I-G)x_n\to 0,(I-T)x_n\to 0$ and $(I-W)x_n\to 0$ (due to (28), (33) and (35)), by Lemma 7 we get $\overline{x}\in \mathrm{Fix}\left(G\right)=\mathrm{GSVI}(C,B_1,B_2)$, $\overline{x}\in \mathrm{Fix}\left(T\right)$ and $\overline{x}\in \mathrm{Fix}\left(W\right)={\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)$. So,

$$\overline{x}\in \Omega =\bigcap _{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)\cap \mathrm{GSVI}(C,B_1,B_2)\cap \mathrm{Fix}\left(T\right).$$

We show $\overline{x}\in \mathrm{VI}(\Omega ,A_1)$. Actually, let $y\in \Omega$ be fixed arbitrarily. From (4), (6) and $\zeta$-inverse strong monotonicity of $A_1$, we get

$$\parallel y_n{-y\parallel }^2\le \parallel (z_n-y)-{\delta }_n{A}_1{z}_n{\parallel }^2\le \parallel x_n{-y\parallel }^2+2{\delta }_n\langle y-z_n,A_{1}y\rangle +{\delta }_n^2{\parallel A_1{z}_n\parallel }^2,$$

which implies that, for all $n\ge 0$,

$$\begin{array}{cc}0\hfill & \le \frac{1}{{\delta }_n}(\parallel x_n{-y\parallel }^2-\parallel y_n{-y\parallel }^2)+2\langle A_{1}y,y-z_n\rangle +{\delta }_n{\parallel A_1{z}_n\parallel }^2\hfill \\ & \le (\parallel x_n-y\parallel +\parallel y_n-y\parallel )\frac{\parallel x_n-y_n\parallel }{{\delta }_n}+2\langle A_{1}y,y-z_n\rangle +{\delta }_n{\parallel A_1{z}_n\parallel }^2.\hfill \end{array}$$

From (28) it is easy to see $x_{n_i}\rightharpoonup \overline{x}$ leads to $z_{n_i}\rightharpoonup \overline{x}$. Since ${lim}_{n\to \infty }{\delta }_n=0$ and $\parallel x_n-y_n\parallel =o\left({\delta }_n\right)$, we have

$$\begin{array}{cc}0\hfill & \le {\displaystyle \underset{n\to \infty }{lim\; inf}}\{(\parallel x_n-y\parallel +\parallel y_n-y\parallel )\frac{\parallel x_n-y_n\parallel }{{\delta }_n}+2\langle A_{1}y,y-z_n\rangle +{\delta }_n\parallel A_1{z}_n{\parallel }^2\}\hfill \\ & ={\displaystyle \underset{n\to \infty }{lim\; inf}}2\langle y-z_n,A_{1}y\rangle \le {\displaystyle \underset{i\to \infty }{lim}}2\langle y-z_{n_i},A_{1}y\rangle =2\langle y-\overline{x},A_{1}y\rangle .\hfill \end{array}$$

It follows that $\langle A_{1}y,y-\overline{x}\rangle \ge 0,$$\forall y\in \Omega$. So, Lemma 12 and the $\zeta$-inverse-strong monotonicity of $A_1$ ensure that $\langle y-\overline{x},A_1\overline{x}\rangle \ge 0,\forall y\in \Omega$, that is, $\overline{x}\in \mathrm{VI}(\Omega ,A_1)$. Consequently, from $\{x^{*}\}=\mathrm{VI}(\mathrm{VI}(\Omega ,A_1),\mu A_2-f)$, we have

$$\underset{n\to \infty }{lim\; sup}\langle x^{*}-w_n,(\mu A_2-f)x^{*}\rangle =\underset{i\to \infty }{lim}\langle x^{*}-w_{n_i},(\mu A_2-f)x^{*}\rangle =\langle x^{*}-\overline{x},(\mu A_2-f)x^{*}\rangle \le 0.$$

Also, we pick a subsequence $\{z_{n_k}\}\subset \{z_n\}$ such that

$$\underset{n\to \infty }{lim\; sup}\langle x^{*}-z_n,A_1{x}^{*}\rangle =\underset{k\to \infty }{lim}\langle x^{*}-z_{n_k},A_1{x}^{*}\rangle .$$

Since vector sequence $\{z_n\}$ is bounded in C, we suppose that $z_{n_k}\rightharpoonup \widehat{x}\in C$. From (28) it is clear that $z_{n_k}\rightharpoonup \widehat{x}$ yields $x_{n_k}\rightharpoonup \widehat{x}$. By the same arguments as in the proof of $\overline{x}\in \Omega$, we have $\widehat{x}\in \Omega$. From $x^{*}\in \mathrm{VI}(\Omega ,A_1)$, we get

$$\underset{n\to \infty }{lim\; sup}\langle x^{*}-z_n,A_1{x}^{*}\rangle =\underset{k\to \infty }{lim}\langle x^{*}-z_{n_k},A_1{x}^{*}\rangle =\langle x^{*}-\widehat{x},A_1{x}^{*}\rangle \le 0.$$

Therefore, the inequalities in (37) hold.

Step 7. We propose $x_n\to x^{*}$ as $n\to \infty$. Indeed, putting $p=x^{*}$ in (6) and (16) we obtain that $\parallel z_n-x^{*}\parallel \le \parallel x_n-x^{*}\parallel$ and

$$\parallel w_n-x^{*}{\parallel }^2\le {\alpha }_n\delta \parallel x_n-x^{*}{\parallel }^2+(1-{\alpha }_n\tau )\parallel y_n-x^{*}{\parallel }^2+{\theta }_n(2+{\theta }_n){\parallel y_n-x^{*}\parallel }^2+2{\alpha }_n\langle (\mu A_2-f)x^{*},x^{*}-w_n\rangle .$$

(39)

From (4) and the $\zeta$-inverse-strong monotonicity of $A_1$ it follows that

$$\parallel y_n-x^{*}{\parallel }^2\le \parallel (z_n-x^{*})-{\delta }_n{A}_1{z}_n{\parallel }^2\le \parallel x_n-x^{*}{\parallel }^2+2{\delta }_n\langle A_1{x}^{*},x^{*}-z_n\rangle +{\delta }_n^2{\parallel A_1{z}_n\parallel }^2.$$

(40)

Thus, in terms of (4), (39) and (40), we get

$$\begin{array}{c}\parallel x_{n+1}-x^{*}{\parallel }^2\le (1-{\beta }_n)\parallel w_n-x^{*}{\parallel }^2+{\beta }_n{\parallel x_n-x^{*}\parallel }^2\hfill \\ \le (1-{\beta }_n)[{\alpha }_n\delta \parallel x_n-x^{*}{\parallel }^2+(1-{\alpha }_n\tau )\parallel y_n-x^{*}{\parallel }^2+{\theta }_n(2+{\theta }_n)\parallel y_n-x^{*}{\parallel }^2+{\beta }_n{\parallel x_n-x^{*}\parallel }^2\hfill \\ +2{\alpha }_n\langle (\mu A_2-f)x^{*},x^{*}-w_n\rangle ]\hfill \\ \le {\beta }_n\parallel x_n-x^{*}{\parallel }^2+(1-{\beta }_n)\{{\alpha }_n\delta \parallel x_n-x^{*}{\parallel }^2+(1-{\alpha }_n\tau )[2{\delta }_n\langle A_1{x}^{*},x^{*}-z_n\rangle +{\parallel x_n-x^{*}\parallel }^2\hfill \\ +{\delta }_n^2\parallel A_1{z}_n{\parallel }^2]+{\theta }_n(2+{\theta }_n)\parallel y_n-x^{*}{\parallel }^2+2{\alpha }_n\langle (\mu A_2-f)x^{*},x^{*}-w_n\rangle \}\hfill \\ \le [1-{\alpha }_n(\tau -\delta )(1-{\beta }_n)]\parallel x_n-x^{*}{\parallel }^2+{\alpha }_n(\tau -\delta )(1-{\beta }_n)\{\frac{(1-{\alpha }_n\tau )2{\delta }_n}{(\tau -\delta ){\alpha }_n}\langle x^{*}-z_n,A_1{x}^{*}\rangle \hfill \\ +{\alpha }_n\frac{\parallel A_1{z}_n{\parallel }^2}{\tau -\delta }+\frac{(2+{\theta }_n){\theta }_n{\parallel y_n-x^{*}\parallel }^2}{(\tau -\delta ){\alpha }_n}+\frac{2}{\tau -\delta }\langle (\mu A_2-f)x^{*},x^{*}-w_n\rangle \}.\hfill \end{array}$$

(41)

Obviously, (37) yields

$$\underset{n\to \infty }{lim\; sup}\frac{(1-{\alpha }_n\tau )2{\delta }_n}{(\tau -\delta ){\alpha }_n}\langle x^{*}-z_n,A_1{x}^{*}\rangle \le 0$$

and

$$\underset{n\to \infty }{lim\; sup}\frac{2}{\tau -\delta }·\langle x^{*}-w_n,(\mu A_2-f)x^{*}\rangle \le 0.$$

Actually, from ${lim\; sup}_{n\to \infty }\langle A_1{x}^{*},x^{*}-z_n\rangle \le 0$ it follows that for any given $\epsilon >0$ there exists an integer $n_0\ge 1$ such that $\langle A_1{x}^{*},x^{*}-z_n\rangle \le \epsilon$, $\forall n\ge n_0$. Then from ${\delta }_n\le {\alpha }_n$ we get

$$\frac{2{\delta }_n(1-{\alpha }_n\tau )}{(\tau -\delta ){\alpha }_n}\langle A_1{x}^{*},x^{*}-z_n\rangle \le \frac{2{\delta }_n(1-{\alpha }_n\tau )}{(\tau -\delta ){\alpha }_n}\epsilon \le \frac{2}{\tau -\delta }\epsilon ,\forall n\ge n_0,$$

which hence yields

$$\underset{n\to \infty }{lim\; sup}\frac{2{\delta }_n(1-{\alpha }_n\tau )\langle A_1{x}^{*},x^{*}-z_n\rangle }{(\tau -\delta ){\alpha }_n}\le \frac{2}{\tau -\delta }\epsilon .$$

Letting $\epsilon \to 0$, we get

$$\underset{n\to \infty }{lim\; sup}\frac{2{\delta }_n(1-{\alpha }_n\tau )\langle x^{*}-z_n,A_1{x}^{*}\rangle }{(\tau -\delta ){\alpha }_n}\le 0.$$

Since ${\sum }_{n=0}^{\infty }{\alpha }_n=\infty ,$${lim\; inf}_{n\to \infty }(1-{\beta }_n)>0$ and ${lim}_{n\to \infty }\frac{{\theta }_n}{{\alpha }_n}=0$, we deduce that

$$\sum _{n=0}^{\infty }{\alpha }_n(\tau -\delta )(1-{\beta }_n)=\infty$$

and

$$\begin{array}{c}{\displaystyle \underset{n\to \infty }{lim\; sup}}\{\frac{(1-{\alpha }_n\tau )2{\delta }_n}{(\tau -\delta ){\alpha }_n}\langle A_1{x}^{*},x^{*}-z_n\rangle +{\alpha }_n\frac{\parallel A_1{z}_n{\parallel }^2}{\tau -\delta }\hfill \\ +\frac{{\theta }_n(2+{\theta }_n){\parallel y_n-x^{*}\parallel }^2}{(\tau -\delta ){\alpha }_n}+\frac{2}{\tau -\delta }\langle x^{*}-w_n,(\mu A_2-f)x^{*}\rangle \}\le 0.\hfill \end{array}$$

We can infer Lemma 3 to the relation (41) and conclude that $x_n\to x^{*}$ as $n\to \infty$. This completes the proof. □

From Theorem 1, we have the following sub-result.

Corollary 1.

Assume that ${\mu }_1$ is a real number in $(0,2\alpha ),$ and ${\mu }_2$ is a real number in $(0,2\beta )$. Let $\delta <\tau :=1-\sqrt{1-\mu (2\eta -\mu {\kappa }^2)}\in (0,1]$ for $\mu \in (0,\frac{2\eta }{{\kappa }^2})$. We suppose ${\{{\lambda }_n\}}_{n=0}^{\infty }$ is a real sequence in $(0,b]$ for some real number b in $(0,1)$. We also suppose that $\{{\alpha }_n\},\{{\beta }_n\},\{{\gamma }_n\}\subset (0,1]$ and $\{{\delta }_n\}\subset (0,2\zeta ]$ such that

(i) 

${\sum }_{n=0}^{\infty }{\alpha }_n=\infty$ and ${lim}_{n\to \infty }{\alpha }_n=0$;

(ii) 

${\delta }_n\le {\alpha }_n\forall n\ge 0$ and ${lim}_{n\to \infty }\frac{{\theta }_n}{{\alpha }_n}=0$;

(iii) 

${lim\; inf}_{n\to \infty }{\beta }_n>0$ and ${lim\; sup}_{n\to \infty }{\beta }_n<1$;

(iv) 

${lim\; inf}_{n\to \infty }{\gamma }_n>0$, ${lim\; sup}_{n\to \infty }{\gamma }_n<1$ and ${lim}_{n\to \infty }|{\gamma }_{n+1}-{\gamma }_n|=0$;

(v) 

${lim}_{n\to \infty }\parallel T^{n+1}{y}_n-T^n{y}_n\parallel =0$.

Let ${\{x_n\}}_{n=0}^{\infty }$ be a sequence defined by

$$\left\{\begin{array}{c}{y}_n=(1-{\gamma }_n)W_n{y}_n+{\gamma }_n{x}_n,\hfill \\ x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)P_C[{\alpha }_{n}f\left(x_n\right)+(I-{\alpha }_n\mu A_2)T^n{y}_n].\hfill \end{array}\right.$$

Then we have

(a) 

${\{x_n\}}_{n=0}^{\infty }$ is bounded;

(b) 

${lim}_{n\to \infty }\parallel x_n-y_n\parallel =0,$${lim}_{n\to \infty }\parallel x_n-T{x}_n\parallel =0$ and ${lim}_{n\to \infty }\parallel x_n-W{x}_n\parallel =0$;

(c) 

if ${lim}_{n\to }\frac{\parallel x_n-y_n\parallel }{{\delta }_n}=0$, then $\{x_n\}$ converges to a common fixed point of the asymptotically nonexpansive and nonexpansive mappings.

Theorem 2.

Assume that ${\mu }_1$ is a real number in $(0,2\alpha ),$ and ${\mu }_2$ is a real number in $(0,2\beta )$. Let $\tau =1-\sqrt{1-\mu (2\eta -\mu {\kappa }^2)}\in (0,1]$ for μ in $(0,\frac{2\eta }{{\kappa }^2})$, and let ${\{{\lambda }_n\}}_{n=0}^{\infty }$ be a real sequence in $(0,b]$ for some b in $(0,1)$. Suppose that $\{{\alpha }_n\},\{{\beta }_n\},\{{\gamma }_n\}\subset (0,1]$ and $\{{\delta }_n\}\subset (0,2\zeta ]$ such that

(i) 

${\sum }_{n=0}^{\infty }{\alpha }_n=\infty$ and ${lim}_{n\to \infty }{\alpha }_n=0$;

(ii) 

${\delta }_n\le {\alpha }_n\forall n\ge 0$ and ${lim}_{n\to \infty }\frac{{\theta }_n}{{\alpha }_n}=0$;

(iii) 

${lim\; inf}_{n\to \infty }{\beta }_n>0$ and ${lim\; sup}_{n\to \infty }{\beta }_n<1$;

(iv) 

${lim\; inf}_{n\to \infty }{\gamma }_n>0$, ${lim\; sup}_{n\to \infty }{\gamma }_n<1$ and ${lim}_{n\to \infty }|{\gamma }_{n+1}-{\gamma }_n|=0$;

(v) 

${lim}_{n\to \infty }\parallel T^{n+1}{y}_n-T^n{y}_n\parallel =0$.

Then the sequence ${\{x_n\}}_{n=0}^{\infty }$ generated by Algorithm 3 satisfies the following properties:

(a) 

${\{x_n\}}_{n=0}^{\infty }$ is bounded;

(b) 

${lim}_{n\to \infty }\parallel x_n-y_n\parallel =0,{lim}_{n\to \infty }\parallel x_n-G{x}_n\parallel =0,{lim}_{n\to \infty }\parallel x_n-T{x}_n\parallel =0$ and ${lim}_{n\to \infty }\parallel x_n-W{x}_n\parallel =0$;

(c) 

If $\frac{\parallel x_n-y_n\parallel }{{\delta }_n}=0$, $x_n\to x^{*}\in \mathrm{VI}(\Omega ,A_1)$.

Proof. 

Since $A_2:C\to H$ is $\kappa$-Lipschitzian and $\eta$-strongly monotone, by Lemma 12 we know that the Problem 2 has the unique solution. We let $\{x^{*}\}=\mathrm{VI}(\mathrm{VI}(\Omega ,A_1),A_2)$. For each $n\ge 0$, we consider the mapping $F_{n}x:=G({\gamma }_n{x}_n+(1-{\gamma }_n)W_{n}x)$, $\forall x\in C$. Utilizing the same argument as in the proof of Theorem 1, we can deduce from Banach’s contraction principle that for each $n\ge 0$ there exists a unique element $z_n\in C$ such that $z_n=G({\gamma }_n{x}_n+(1-{\gamma }_n)W_n{z}_n)$. Thus, the iterative scheme in Algorithm 3 can be rewritten as

$$\left\{\begin{array}{c}{u}_n={\gamma }_n{x}_n+(1-{\gamma }_n)W_n{z}_n,\hfill \\ z_n=G{u}_n,\hfill \\ y_n=P_C(z_n-{\delta }_n{A}_1{z}_n),\hfill \\ x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)P_C(I-\mu {\alpha }_n{A}_2)T^n{y}_n.\hfill \end{array}\right.$$

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

Step 1. We prove $\{x_n\},\{y_n\},\{z_n\},\{u_n\},\{v_n\},\{T^n{y}_n\}$ and $\{A_2\left(T^n{y}_n\right)\}$ are bounded vector sequences, where $v_n=P_C(u_n-{\mu }_2{B}_2{u}_n)$ and $z_n=P_C(v_n-{\mu }_1{B}_1{v}_n)$ for all $n\ge 0$. Indeed, utilizing the similar argument to that of Step 1 in the proof of Theorem 1, we obtain the desired assertion.

Step 2. We prove $\parallel x_{n+1}-x_n\parallel \to 0$ and $\parallel y_{n+1}-y_n\parallel \to 0$ as $n\to \infty$. Indeed, utilizing the similar argument to that of Step 2 in the proof of Theorem 1, we obtain the desired assertion.

Step 3. We prove $\parallel x_n-G{x}_n\parallel \to 0$ as $n\to \infty$. Indeed, utilizing the similar argument to that of Step 3 in the proof of Theorem 1, we obtain the desired assertion.

Step 4. We prove $\parallel T{x}_n-x_n\parallel \to 0$ and $\parallel W_n{x}_n-x_n\parallel \to 0$ as $n\to \infty$. Indeed, utilizing the similar argument to that of Step 4 in the proof of Theorem 1, we obtain the desired assertion.

Step 5. We prove $W:C\to C$ enjoys the nonexpansivity, $\mathrm{Fix}\left(W\right)={\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)$ and ${lim}_{n\to \infty }\parallel W{x}_n-x_n\parallel =0$ where $Wx:={lim}_{n\to \infty }{W}_{n}x$ for all $x\in C$. Indeed, utilizing the similar argument to that of Step 5 in the proof of Theorem 1, we obtain the desired assertion.

Step 6. We prove ${lim\; sup}_{n\to \infty }\langle A_2{x}^{*},x^{*}-w_n\rangle \le 0$ and ${lim\; sup}_{n\to \infty }\langle A_1{x}^{*},x^{*}-z_n\rangle \le 0$, where $\{x^{*}\}=\mathrm{VI}(\mathrm{VI}(\Omega ,A_1),A_2)$. Indeed, utilizing the similar argument to that of Step 6 in the proof of Theorem 1, we obtain the desired assertion.

Step 7. We prove $x_n\to x^{*}$ as $n\to \infty$. Indeed, utilizing the similar argument to that of Step 7 in the proof of Theorem 1, we obtain the desired assertion.

This completes the entire proof. □

Corollary 2.

Assume that ${\mu }_1$ is a real number in $(0,2\alpha ),$ and ${\mu }_2$ is a real number in $(0,2\beta )$. Let $\tau =1-\sqrt{1-\mu (2\eta -\mu {\kappa }^2)}\in (0,1]$ for μ in $(0,\frac{2\eta }{{\kappa }^2})$, and let ${\{{\lambda }_n\}}_{n=0}^{\infty }$ be a real sequence in $(0,b]$ for some b in $(0,1)$. Suppose that $\{{\alpha }_n\},\{{\beta }_n\},\{{\gamma }_n\}\subset (0,1]$ and $\{{\delta }_n\}\subset (0,2\zeta ]$ such that

(i) 

${\sum }_{n=0}^{\infty }{\alpha }_n=\infty$ and ${lim}_{n\to \infty }{\alpha }_n=0$;

(ii) 

${\delta }_n\le {\alpha }_n\forall n\ge 0$ and ${lim}_{n\to \infty }\frac{{\theta }_n}{{\alpha }_n}=0$;

(iii) 

${lim\; inf}_{n\to \infty }{\beta }_n>0$ and ${lim\; sup}_{n\to \infty }{\beta }_n<1$;

(iv) 

${lim\; inf}_{n\to \infty }{\gamma }_n>0$, ${lim\; sup}_{n\to \infty }{\gamma }_n<1$ and ${lim}_{n\to \infty }|{\gamma }_{n+1}-{\gamma }_n|=0$;

(v) 

${lim}_{n\to \infty }\parallel T^{n+1}{y}_n-T^n{y}_n\parallel =0$.

Let ${\{x_n\}}_{n=0}^{\infty }$ be a sequence defined by

$$\left\{\begin{array}{c}{u}_n=(1-{\gamma }_n)W_n{u}_n+{\gamma }_n{x}_n,\hfill \\ x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)P_C(I-{\alpha }_n\mu A_2)T^n{u}_n.\hfill \end{array}\right.$$

Then we have

(a) 

${\{x_n\}}_{n=0}^{\infty }$ is bounded;

(b) 

${lim}_{n\to \infty }\parallel x_n-u_n\parallel =0,$${lim}_{n\to \infty }\parallel x_n-T{x}_n\parallel =0$ and ${lim}_{n\to \infty }\parallel x_n-W{x}_n\parallel =0$;

(c) 

If $\frac{\parallel x_n-u_n\parallel }{{\delta }_n}=0$, $\{x_n\}$

converges to a common fixed point of the asymptotically nonexpansive and nonexpansive mappings.

4. Concluding Remark

This paper discussed a monotone variational inequality problem with a variational inequality constraint over the common solution set of a general system of variational inequalities and a common fixed point of a countable family of nonexpansive mappings and an asymptotically nonexpansive mapping in Hilbert spaces, which is called the triple hierarchical constrained variational inequality, and introduced some Mann-type implicit iteration methods for solving it. Norm convergence of the proposed methods of the iteration methods is guaranteed under some suitable assumptions.