Triple Hierarchical Variational Inequalities, Systems of Variational Inequalities, and Fixed Point Problems

Abstract

In this paper, we introduce a multiple hybrid implicit iteration method for finding a solution for a monotone variational inequality with a variational inequality constraint over the common solution set of a general system of variational inequalities, and a common fixed point problem of a countable family of uniformly Lipschitzian pseudocontractive mappings and an asymptotically nonexpansive mapping in Hilbert spaces. Strong convergence of the proposed method to the unique solution of the problem is established under some suitable assumptions.

1. Introduction

We suppose that H is a real Hilbert space. We use $\langle ·,·\rangle$ to stand for the inner product and $\parallel ·\parallel$ the norm. We suppose that C is a convex closed nonempty set in the Hilbert space H, and $P_C$ is the well-known metric projection from the space H onto the set C. Here, we also suppose that T is a nonlinear self mapping defined in C. Let $\mathrm{Fix}\left(T\right)$ be the set of all fixed points of T, that is, $\mathrm{Fix}\left(T\right)=\{x\in C:x=Tx\}$. We use the notations → and ⇀ to indicate the norm convergence and the weak convergence, respectively. Now, we suppose that $A:C\to H$ is a nonlinear nonself mapping in C to H. The well-known classical variational inequality (VI), whose set of all solutions denoted by VI$(C,A)$, is to find $x^{*}\in C$ such that

$$\langle A{x}^{*},x-x^{*}\rangle \ge 0,\forall x\in C.$$

(1)

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

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

(2)

This mapping is Lipschitz continuous with the Lipschitz constant $L>1$. Fixed points of Lipschitz continuous mappings are a hot topic and have a lot of applications both in theoretical research, such as in differential equations, control theory, equilibrium problems, and in engineering applications; see References [1,2,3,4,5,6] and the references therein. In particular, T is said to be nonexpansive if $|Tx-Ty\parallel \le \parallel x-y\parallel ,\forall x,y\in C$, that is, $\theta \equiv 1$ for all $n.$ Recently, the variational inequality problem (1) has been extensively studied via the iterative methods of Lipschitz continuous mappings, in particular, (asymptotically) nonexpansve mappings; see References [7,8,9,10,11,12] and the references therein.

We suppose that $B_1,B_2:C\to H$ are two nonlinear monotone mappings. We also suppose that ${\mu }_1$ and ${\mu }_2$ are two positive real constants. We consider the problem of finding $(x^{*},y^{*})\in C\times C$ such that

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

(3)

Problem (3) is called a general system of variational inequalities (GSVI). From Reference [8], the GSVI (3) can be translated into a fixed point problem of a Lipschitz continuous nonlinear operator in the following way.

Lemma 1

([8]). We suppose that C is a convex subset in a Hilbert space H. Fix two elements $x^{*}$ and $y^{*}$ in C, $(x^{*},y^{*})$ is a solution of GSVI (3) 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 $y^{*}=P_C(I-{\mu }_2{B}_2)x^{*}$.

The GSVI (3), which includes the variational inequality (1) as a special case, has been investigated via fixed-point algorithms recently in real or complex Hilbert spaces; see References [13,14,15,16,17,18] and the references therein.

A self mapping $f:C\to C$ is said to be a strict contraction on C if there is a number $\delta \in [0,1)$ such that $\parallel f\left(x\right)-f\left(y\right)\parallel \le \delta \parallel x-y\parallel$ for all $x,y\in C$. A nonself mapping $F:C\to H$ is called monotone if $\langle Fx-Fy,x-y\rangle \ge 0\forall x,y\in C$. It is called $\eta$-strongly monotone if there is $\eta >0$ such that

$${\eta \parallel x-y\parallel }^2\le \langle Fx-Fy,x-y\rangle ,\forall x,y\in C.$$

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

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

The class of inverse-strongly monotone operators or $\alpha$-cocoercive operators has been in the spotlight of theoretical research and studied from the viewpoint of numerical computation and many results were obtained in Hilbert (and more generally, in Banach) spaces; see References [19,20,21,22,23,24] and the references therein.

Let X be a real Banach space whose dual space is denoted by $X^{*}$. The well-known normalized duality operator $J:X\to 2^{X^{*}}$ is defined by

$$J\left(x\right)=\{\psi \in X^{*}:\langle x,\psi \rangle =\parallel x{\parallel }^2=\parallel \psi {\parallel }^2\},\forall x\in X,$$

where $\langle ·,·\rangle$ is the duality pairing between E and $E^{*}$. A mapping T with domain $D\left(T\right)$ and range $R\left(T\right)$ in X is called pseudocontractive if the inequality holds

$$\parallel x-y+r\left(\right(I-T)x-(I-T\left)y\right)\parallel \ge \parallel x-y\parallel ,\forall x,y\in D\left(T\right),\forall r>0.$$

Kato’s results [25] told us that the notion of pseudocontraction is equivalent to the one that for each $x,y\in D\left(T\right)$, there exists $j(x-y)\in J(x-y)$ such that

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

The purpose of this paper is act as a continuation of Reference [26], that is to introduce and analyze a multiple hybrid implicit iteration method for solving a monotone variational inequality with a variational inequality constraint for two inverse-strongly monotone mappings and a common fixed point problem (CFPP) of a countable family of uniformly Lipschitzian pseudocontractive mappings and an asymptotically nonexpansive mapping in Hilbert spaces, which is called the triple hierarchical constrained variational inequality (THCVI). Here, the multiple hybrid implicit iteration method is based on the Moudafi’s viscosity approximation method, Korpelevich’s extragradient method, Mann’s mean method, and the hybrid steepest-descent method. Under some suitable assumptions, strong convergence of the proposed method to the unique solution of the THCVI is derived.

2. Preliminaries

Let ${\left\{T_n\right\}}_{n=0}^{\infty }$ be a sequence of continuous pseudocontractive self-mappings on C. Then, ${\left\{T_n\right\}}_{n=0}^{\infty }$ is said to be a countable family of ℓ-uniformly Lipschitzian pseudocontractive self-mappings on C if there exists a constant $\ell >0$ such that each $T_n$ is ℓ-Lipschitz continuous. We fix an element x in H to see that there exists a unique nearest point in C, denoted by $P_{C}x$, such that

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

$P_C$ is called a metric projection of H onto C. It may be a set-valued operator. Further, C is assumed to be convex and closed, and X is assumed to be Hilbert, $P_C$ is, in such a situation, a single-valued operator.

We need the following propositions and lemmas to prove our main results.

Proposition 1

([27]). We suppose C is a convex closed subset of a Banach space X. Let $S_0,S_1,\dots$ be a self-mapping sequence on C. Let ${\sum }_{n=1}^{\infty }sup\{\parallel S_{n}x-S_{n-1}x\parallel :x\in C\}<\infty$. We conclude $\left\{S_{n}y\right\}$, where $y\in C,$ converges strongly to some point in C. Moreover, we assume S is a self mapping on C generated by $Sy={lim}_{n\to \infty }{S}_{n}y$ for all $y\in C$. Therefore, ${lim}_{n\to \infty }sup\{\parallel Sx-S_{n}x\parallel :x\in C\}=0$.

Proposition 2

([28]). We suppose C is a convex closed subset of a Banach space X and T is a continuous strong pseudocontraction self-mapping. Therefore, T enjoys fixed points. Indeed, it has a unique fixed point.

The following lemma is trivial.

Lemma 2.

In a real Hilbert space H, there holds the inequality

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

Lemma 3

([29]). We suppose that $\left\{a_n\right\}$ is a nonnegative number sequence satisfying the restrictions

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

where $\left\{{\lambda }_n\right\}$ and $\left\{{\gamma }_n\right\}$ are sequences of real sequences such that

(i) ${lim\; sup}_{n\to \infty }{\gamma }_n\le 0$ or ${\sum }_{n=0}^{\infty }\left|{\lambda }_n{\gamma }_n\right|<\infty$;

(ii) $\left\{{\lambda }_n\right\}\subset [0,1]$ and ${\sum }_{n=0}^{\infty }{\lambda }_n=\infty$, or equivalently,

$$\underset{n\to \infty }{lim}\prod _{k=0}^n(1-{\lambda }_k)=0.$$

Hence, $a_n\to 0$ as $n\to \infty$.

The following lemma is a direct consequence of Yamada [30].

Lemma 4.

Let $F:H\to H$ be a κ-Lipschitzian and η-strongly monotone. We suppose λ is a positive real number in $(0,1]$ and $T:C\to H$ is a nonexpansive nonself mapping, and we define the mapping $T^{\lambda }:C\to H$ by

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

If $0<\mu <\frac{2\eta }{{\kappa }^2}$, then $T^{\lambda }$ is a contraction operator, that is,

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

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

Lemma 5

([31]). We suppose that the nonself mapping $A:C\to H$ is α-inverse-strongly 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. Further, we suppose $A:C\to H$ is a monotone and hemicontinuous mapping. Then, the following hold:

(i) $\mathrm{VI}(C,A)=\{x^{*}\in C:\langle Ay,y-x^{*}\rangle \ge 0,\forall y\in C\}$;

(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.

Lemma 6

([8]). We suppose the nonself operators $B_1,B_2:C\to H$ are α-inverse-strongly monotone and β-inverse-strongly monotone, respectively. Let the self operator $G:C\to C$ be defined in $G:=P_C(I-{\mu }_1{B}_1)P_C(I-{\mu }_2{B}_2)$. $G:C\to C$ is nonexpansive if $0\le {\mu }_1\le 2\alpha$ and $0\le {\mu }_2\le 2\beta$.

Lemma 7

([32]). We suppose the Banach space X enjoys a weakly continuous duality mapping, and C is a convex closed set in X. Let $T:C\to C$ be an asymptotically nonexpansive self mapping on C with a nonempty fixed point set. Then, $I-T$ is demiclosed at zero, i.e., if $\left\{x_n\right\}$ is a sequence in C converging weakly to some $x\in C$ and the sequence $\left\{(I-T)x_n\right\}$ converges strongly to zero, then $(I-T)x=0$, where I is the identity mapping of X.

Lemma 8

([33]). Let both $\left\{x_n\right\}$ and $\left\{h_n\right\}$ be a bounded sequence in a Banach space X. Let $\left\{{\beta }_n\right\}\subset (0,1)$ be a number sequence such that

$$0<\underset{n\to \infty }{lim\; inf}{\beta }_n\le \underset{n\to \infty }{lim\; sup}{\beta }_n<1.$$

Suppose that $x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)h_n\forall n\ge 0$ and ${lim\; sup}_{n\to \infty }(\parallel h_{n+1}-h_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0$. So, ${lim}_{n\to \infty }\parallel h_n-x_n\parallel =0$.

3. Main Results

Let C be a convex closed subset of a real Hilbert space H. Let $B_1,B_2:C\to H$ be monotone mappings, $A-g:C\to H$ be a monotone mapping with $A,g:C\to H$, $T:C\to C$ be an asymptotically nonexpansive mapping, and ${\left\{S_n\right\}}_{n=0}^{\infty }$ be a countable family of ℓ-uniformly Lipschitzian pseudocontractive self-mappings defined on C. We suppose $\mathsf{\Omega }:={\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)\cap \mathrm{GSVI}(C,B_1,B_2)\cap \mathrm{Fix}\left(T\right)\ne \varnothing$ and studied the variational inequality for monotone mapping $A-g$ over the common solution set $\mathsf{\Omega }$ of the GSVI (3) and the CFPP of ${\left\{S_n\right\}}_{n=0}^{\infty }$ and T:

$$\mathrm{Find}\overline{x}\in \mathrm{VI}(\mathsf{\Omega },A-g):=\{\overline{x}\in \mathsf{\Omega }:\langle (A-g)\overline{x},y-\overline{x}\rangle \ge 0\forall y\in \mathsf{\Omega }\}.$$

This section introduces the following monotone variational inequality problem with the inequality constraint over the common solution set of the GSVI (2) and the CFPP of T and ${\left\{S_n\right\}}_{n=0}^{\infty }$, which is named the triple hierarchical constrained variational inequality:

Assume that

(C1) $T:C\to C$ is an asymptotically nonexpansive mapping with a sequence $\left\{{\theta }_n\right\}$;

(C2) ${\left\{S_n\right\}}_{n=0}^{\infty }$ is a countable family of ℓ-uniformly Lipschitzian pseudocontractive self-mappings on C;

(C3) $B_1,B_2:C\to H$ are $\alpha$-inverse-strongly monotone and $\beta$-inverse-strongly monotone, respectively;

(C4) $\mathrm{GSVI}(C,B_1,B_2):=\mathrm{Fix}\left(G\right)$ where $G:=P_C(I-{\mu }_1{B}_1)P_C(I-{\mu }_2{B}_2)$ for ${\mu }_1,{\mu }_2>0$;

(C5) $\mathsf{\Omega }:={\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)\cap \mathrm{GSVI}(C,B_1,B_2)\cap \mathrm{Fix}\left(T\right)\ne \varnothing$;

(C6) ${\sum }_{n=1}^{\infty }{sup}_{x\in D}\parallel S_{n}x-S_{n-1}x\parallel <\infty$ for any bounded subset D of C;

(C7) $S:C\to C$ is the mapping defined by $Sx={lim}_{n\to \infty }{S}_{n}x\forall x\in C$, such that $\mathrm{Fix}\left(S\right)={\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)$;

(C8) $g:C\to H$ is l-Lipschitzian and $A:C\to H$ is $\zeta$-inverse-strongly monotone such that $A-g$ is monotone;

(C9) $f:C\to C$ is a contraction mapping with coefficient $\delta \in [0,1)$ and $F:C\to H$ is $\kappa$-Lipschitzian and $\eta$-strongly monotone;

(C10) $\mathrm{VI}(\mathsf{\Omega },A-g)\ne \varnothing$.

Problem 1.

The objective is to

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

Since the original problem is a variational inequality, in this paper, we call it a triple hierarchical constrained variational inequality. Since the mapping f is a contractive, we easily get that the solution of the problem is unique. Inspired by the results announced recently, we introduce the following multiple hybrid implicit iterative algorithm to find the solution of such a problem.

Algorithm 1: Multiple hybrid implicit iterative algorithm.

Step 0. Take ${\left\{{\alpha }_n\right\}}_{n=0}^{\infty },{\left\{{\beta }_n\right\}}_{n=0}^{\infty },{\left\{{\gamma }_n\right\}}_{n=0}^{\infty }\subset (0,\infty )$, 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 $$\left\{\begin{array}{c}{u}_n={\gamma }_n{x}_n+(1-{\gamma }_n)S_n{u}_n,\hfill \\ v_n=P_C(u_n-{\mu }_2{B}_2{u}_n),\hfill \\ z_n=P_C(v_n-{\mu }_1{B}_1{v}_n),\hfill \\ y_n=P_C[{\alpha }_{n}g\left(x_n\right)+(I-{\alpha }_{n}A)z_n],\hfill \\ w_n=P_C[{\alpha }_n{x}_n+(I-{\alpha }_n\mu F)T^n{y}_n],\hfill \\ x_{n+1}={\alpha }_{n}f\left(x_n\right)+{\beta }_n{x}_n+(1-{\alpha }_n-{\beta }_n)w_n.\hfill \end{array}\right.$$ (4)

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

We remark here that our algorithm is quite general. It includes mean-valued techniques, gradient techniques, and implicit iteration techniques. Our algorithm can also generate a strong convergence without any compact assumptions in infinite dimensional spaces.

We now state and prove the main result of this paper, that is, the following convergence analysis is presented for our Algorithm 1.

Theorem 1.

We suppose ${\mu }_1\in (0,2\alpha ),{\mu }_2\in (0,2\beta )$, and $l+\delta <\tau :=1-\sqrt{1-\mu (2\eta -\mu {\kappa }^2)}\in (0,1]$ for $\mu \in (0,\frac{2\eta }{{\kappa }^2})$. Let number sequences $\left\{{\alpha }_n\right\},\left\{{\beta }_n\right\}$ and $\left\{{\gamma }_n\right\}$ lie in $(0,1]$ such that

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

(ii) ${lim}_{n\to \infty }\frac{{\theta }_n}{{\alpha }_n}=0$;

(iii) $0<{lim\; inf}_{n\to \infty }{\beta }_n\le {lim\; sup}_{n\to \infty }{\beta }_n<1$ and ${\alpha }_n+{\beta }_n\le 1\forall n\ge 0$;

(iv) $0<{lim\; inf}_{n\to \infty }{\gamma }_n\le {lim\; sup}_{n\to \infty }{\gamma }_n<1$ 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, we have the following conclusions:

(a) ${\left\{x_n\right\}}_{n=0}^{\infty }$ is bounded;

(b) ${lim}_{n\to \infty }\parallel x_n-y_n\parallel =0$ and ${lim}_{n\to \infty }\parallel x_n-w_n\parallel =0$;

(c) ${lim}_{n\to \infty }\parallel x_n-G{x}_n\parallel =0,{lim}_{n\to \infty }\parallel x_n-T{x}_n\parallel =0$ and ${lim}_{n\to \infty }\parallel x_n-S{x}_n\parallel =0$;

(d) If $(\parallel x_n-y_n\parallel +\parallel x_n-w_n\parallel )=o\left({\alpha }_n\right)$, then ${\left\{x_n\right\}}_{n=0}^{\infty }$ converges strongly to the unique solution of the Problem 1.

Proof. 

Observe that the metric projection $P_{\mathrm{VI}(\mathsf{\Omega },A-g)}$ is nonexpansive. Indeed, it is firmly nonexpansive. The mapping f is contractive. Thus, the composition mapping $P_{\mathrm{VI}(\mathsf{\Omega },A-g)}f$ is a contraction mapping and hence $P_{\mathrm{VI}(\mathsf{\Omega },A-g)}f$ has a unique fixed point. Say $x^{*}\in C$, that is, $x^{*}=P_{\mathrm{VI}(\mathsf{\Omega },A-g)}f\left(x^{*}\right)$. By Lemma 5,

$$\left\{x^{*}\right\}=\mathrm{Fix}\left(P_{\mathrm{VI}(\mathsf{\Omega },A-g)}f\right)=\mathrm{VI}(\mathrm{VI}(\mathsf{\Omega },A-g),I-f).$$

Therefore, Problem 1 has a unique solution. Without loss of the generality, we can assume that $\left\{{\alpha }_n\right\}\subset (0,2\zeta ]$ and $\left\{{\gamma }_n\right\}\subset [a,b]\subset (0,1)$ for some $a,b\in (0,1)$. By Lemma 6, we know that G is nonexpansive. It is easy to see that for each $n\ge 0$ there exists a unique element $u_n\in C$ such that

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

(5)

Therefore, it can be seen that the multiple hybrid implicit iterative scheme (4) can be rewritten as

$$\left\{\begin{array}{c}{u}_n={\gamma }_n{x}_n+(1-{\gamma }_n)S_n{u}_n,\hfill \\ z_n=G{u}_n,\hfill \\ y_n=P_C[{\alpha }_{n}g\left(x_n\right)+(I-{\alpha }_{n}A)z_n],\hfill \\ x_{n+1}={\alpha }_{n}f\left(x_n\right)+{\beta }_n{x}_n+(1-{\alpha }_n-{\beta }_n)P_C[{\alpha }_n{x}_n+(I-{\alpha }_n\mu F)T^n{y}_n],\forall n\ge 0.\hfill \end{array}\right.$$

(6)

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

Step 1. We prove $\left\{x_n\right\},\left\{y_n\right\},\left\{z_n\right\},\left\{u_n\right\},\left\{v_n\right\},\left\{T^n{y}_n\right\}$, and $\left\{F\left(T^n{y}_n\right)\right\}$ are bounded. Indeed, We can take an element $p\in \mathsf{\Omega }={\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)\cap \mathrm{GSVI}(C,B_1,B_2)\cap \mathrm{Fix}\left(T\right)$ arbitrarily. Then, we have $S_{n}p=p$, $Gp=p$ and $Tp=p$. Since $S_n:C\to C$ is a pseudocontraction self mapping, one can show that

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

(7)

Hence, we get

$$\parallel z_n-p\parallel =\parallel G{u}_n-p\parallel \le \parallel u_n-p\parallel \le \parallel x_n-p\parallel .$$

(8)

Since ${lim}_{n\to \infty }{\alpha }_n=0$ and $1>{lim\; sup}_{n\to \infty }{\beta }_n\ge {lim\; inf}_{n\to \infty }{\beta }_n>0$, we may assume that $\{{\alpha }_n+{\beta }_n\}$ is a set in $[c,d]$. Here, $c,d\in (0,1)$. In addition, since ${lim}_{n\to \infty }\frac{{\theta }_n}{{\alpha }_n}=0$, we may further assume that

$${\theta }_n(1+{\alpha }_{n}l)\le \frac{{\alpha }_n(\tau -l-\delta )}{2}(\le {\alpha }_n(\tau -l-\delta )).$$

From Lemma 5 and (8), we can prove that

$$\begin{array}{cc}\parallel y_n-p\parallel \hfill & \le \parallel (I-{\alpha }_{n}A)z_n-(I-{\alpha }_{n}A)p+{\alpha }_n(g\left(x_n\right)-Ap)\parallel \hfill \\ & \le \parallel z_n-p\parallel +{\alpha }_n\parallel g\left(x_n\right)-Ap\parallel \hfill \\ & \le (1+{\alpha }_{n}l)\parallel x_n-p\parallel +{\alpha }_n\parallel g\left(p\right)-Ap\parallel .\hfill \end{array}$$

(9)

We have from (6) and using Lemma 4 and (9) that

$$\begin{array}{c}\parallel x_{n+1}-p\parallel \hfill \\ \le {\alpha }_n\parallel f\left(x_n\right)-p\parallel +{\beta }_n\parallel x_n-p\parallel +(1-{\alpha }_n-{\beta }_n)\parallel P_C[{\alpha }_n{x}_n+(I-{\alpha }_n\mu F)T^n{y}_n]-p\parallel \hfill \\ \le {\alpha }_n\parallel f\left(x_n\right)-f\left(p\right)+f\left(p\right)-p\parallel +{\beta }_n\parallel x_n-p\parallel \hfill \\ +(1-{\alpha }_n-{\beta }_n)\parallel {\alpha }_n(x_n-p)+(I-{\alpha }_n\mu F)T^n{y}_n-(I-{\alpha }_n\mu F)p+{\alpha }_n(I-\mu F)p\parallel \hfill \\ \le ({\alpha }_n\delta +{\beta }_n)\parallel x_n-p\parallel +{\alpha }_n\parallel f\left(p\right)-p\parallel \hfill \\ +(1-{\alpha }_n-{\beta }_n)[{\alpha }_n\parallel x_n-p\parallel +(1-{\alpha }_n\tau )(1+{\theta }_n)\parallel y_n-p\parallel +{\alpha }_n\parallel (I-\mu F)p\parallel ]\hfill \\ \le ({\alpha }_n\delta +{\beta }_n)\parallel x_n-p\parallel +{\alpha }_n\parallel f\left(p\right)-p\parallel \hfill \\ +(1-{\alpha }_n-{\beta }_n)[{\alpha }_n\parallel x_n-p\parallel +(1-{\alpha }_n\tau +{\theta }_n)\parallel y_n-p\parallel +{\alpha }_n\parallel (I-\mu F)p\parallel ]\hfill \\ \le ({\alpha }_n\delta +{\beta }_n)\parallel x_n-p\parallel +{\alpha }_n\parallel f\left(p\right)-p\parallel +(1-{\alpha }_n-{\beta }_n)\{{\alpha }_n\parallel x_n-p\parallel \hfill \\ +(1-{\alpha }_n\tau )[(1+{\alpha }_{n}l)\parallel x_n-p\parallel +{\alpha }_n\parallel g\left(p\right)-Ap\parallel ]+{\theta }_n[(1+{\alpha }_{n}l)\parallel x_n-p\parallel \hfill \\ +{\alpha }_n\parallel g\left(p\right)-Ap\parallel ]+{\alpha }_n\parallel (I-\mu F)p\parallel \}\hfill \\ \le ({\alpha }_n\delta +{\beta }_n)\parallel x_n-p\parallel +{\alpha }_n\parallel f\left(p\right)-p\parallel +(1-{\alpha }_n-{\beta }_n)\{{\alpha }_n\parallel x_n-p\parallel \hfill \\ +[1-{\alpha }_n(\tau -l)]\parallel x_n-p\parallel +(1-{\alpha }_n\tau ){\alpha }_n\parallel g\left(p\right)-Ap\parallel +{\theta }_n(1+{\alpha }_{n}l)\parallel x_n-p\parallel \hfill \\ +{\alpha }_n^2\tau \parallel g\left(p\right)-Ap\parallel \}+{\alpha }_n\parallel (I-\mu F)p\parallel \hfill \\ \le ({\alpha }_n\delta +{\beta }_n)\parallel x_n-p\parallel +{\alpha }_n\parallel f\left(p\right)-p\parallel +(1-{\alpha }_n-{\beta }_n)[{\alpha }_n+1-{\alpha }_n(\tau -l)]\parallel x_n-p\parallel \hfill \\ +{\alpha }_n\parallel g\left(p\right)-Ap\parallel +{\theta }_n(1+{\alpha }_{n}l)\parallel x_n-p\parallel +{\alpha }_n\parallel (I-\mu F)p\parallel \hfill \\ =\{1-{\alpha }_n(\tau -l-\delta )-({\alpha }_n+{\beta }_n){\alpha }_n[1-(\tau -l)]\}\parallel x_n-p\parallel \hfill \\ +{\theta }_n(1+{\alpha }_{n}l)\parallel x_n-p\parallel +{\alpha }_n(\parallel f\left(p\right)-p\parallel +\parallel g\left(p\right)-Ap\parallel +\parallel (I-\mu F)p\parallel )\hfill \\ \le [1-{\alpha }_n(\tau -l-\delta )]\parallel x_n-p\parallel +{\displaystyle \frac{{\alpha }_n(\tau -l-\delta )}{2}}\parallel x_n-p\parallel \hfill \\ +{\alpha }_n(\parallel f\left(p\right)-p\parallel +\parallel g\left(p\right)-Ap\parallel +\parallel (I-\mu F)p\parallel )\hfill \\ =[1-{\displaystyle \frac{{\alpha }_n(\tau -l-\delta )}{2}}]\parallel x_n-p\parallel +{\displaystyle \frac{{\alpha }_n(\tau -l-\delta )}{2}}·{\displaystyle \frac{2(\parallel f(p)-p\parallel +\parallel g(p)-Ap\parallel +\parallel (I-\mu F)p\parallel )}{\tau -l-\delta }}\hfill \\ \le max\{\parallel x_n-p\parallel ,{\displaystyle \frac{2(\parallel f(p)-p\parallel +\parallel g(p)-Ap\parallel +\parallel (I-\mu F)p\parallel )}{\tau -l-\delta }}\}.\hfill \end{array}$$

By induction, we have

$$\parallel x_{n+1}-p\parallel \le max\{\frac{2(\parallel p-f(p)\parallel +\parallel Ap-g(p)\parallel +\parallel (I-\mu F)p\parallel )}{\tau -l-\delta },\parallel p-x_0\parallel \},\forall n\ge 0.$$

Thus, $\left\{x_n\right\}$ is a bounded sequence, and so are the sequences $\left\{y_n\right\},\left\{z_n\right\},\left\{u_n\right\},\left\{T^n{y}_n\right\}$, and $\left\{F\left(T^n{y}_n\right)\right\}$. Since $\left\{S_n\right\}$ is ℓ-uniformly Lipschitzian on C, we know that

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

which implies that the set $\left\{S_n{u}_n\right\}$ is bounded. Additionally, from Lemma 1 and $p\in \mathsf{\Omega }\subset \mathrm{GSVI}(C,B_1,B_2)$, it follows that $(p,q)$ is a solution of the GSVI (3), where $q=P_C(I-{\mu }_2{B}_2)p$. Noting that $v_n=P_C(I-{\mu }_2{B}_2)u_n$ for all $n\ge 0$, by Lemma 5, we have

$$\begin{array}{cc}\parallel v_n\parallel \hfill & \le \parallel P_C(I-{\mu }_2{B}_2)u_n-q\parallel +\parallel q\parallel \hfill \\ & =\parallel P_C(I-{\mu }_2{B}_2)u_n-P_C(I-{\mu }_2{B}_2)q\parallel +\parallel q\parallel \hfill \\ & \le \parallel u_n-q\parallel +\parallel q\parallel ,\hfill \end{array}$$

which shows that $\left\{v_n\right\}$ also is bounded.

Step 2. We prove that $\parallel x_{n+1}-x_n\parallel \to 0$ and $\parallel y_{n+1}-y_n\parallel \to 0$ as $n\to \infty$. Indeed, we set

$$x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)h_n$$

and notice

$$w_n=P_C[(I-{\alpha }_n\mu F)T^n{y}_n+{\alpha }_n{x}_n].$$

Then,

$$h_n=\frac{{\alpha }_n}{1-{\beta }_n}f\left(x_n\right)+(1-\frac{{\alpha }_n}{1-{\beta }_n})P_C[{\alpha }_n{x}_n+(I-{\alpha }_n\mu F)T^n{y}_n].$$

Simple calculations show that

$$\begin{array}{c}{h}_{n+1}-h_n\hfill \\ =\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}(f\left(x_{n+1}\right)-f\left(x_n\right))+(1-\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}})\{P_C[{\alpha }_{n+1}{x}_{n+1}+(I-{\alpha }_{n+1}\mu F)T^{n+1}{y}_{n+1}]\hfill \\ -P_C[{\alpha }_n{x}_n+(I-{\alpha }_n\mu F)T^n{y}_n)\}+(\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\alpha }_n}{1-{\beta }_n})(f\left(x_n\right)-w_n).\hfill \end{array}$$

It follows from (6) that

$$\begin{array}{c}\parallel h_{n+1}-h_n\parallel \hfill \\ \le \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}\parallel f\left(x_n\right)-f\left(x_{n+1}\right)\parallel +(1-\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}})\parallel P_C[{\alpha }_{n+1}{x}_{n+1}+(I-{\alpha }_{n+1}\mu F)T^{n+1}{y}_{n+1}]\hfill \\ -P_C[{\alpha }_n{x}_n+(I-{\alpha }_n\mu F)T^n{y}_n]\parallel +|\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\alpha }_n}{1-{\beta }_n}|\parallel w_n-f\left(x_n\right)\parallel \hfill \\ \le \frac{{\alpha }_{n+1}\delta }{1-{\beta }_{n+1}}\parallel x_n-x_{n+1}\parallel +(1-\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}})\parallel T^{n+1}{y}_{n+1}-T^{n+1}{y}_n+T^{n+1}{y}_n-T^n{y}_n\hfill \\ +{\alpha }_{n+1}(x_{n+1}-\mu F\left(T^{n+1}{y}_{n+1}\right))-{\alpha }_n(x_n-\mu F\left(T^n{y}_n\right))\parallel \hfill \\ +|\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\alpha }_n}{1-{\beta }_n}|\parallel f\left(x_n\right)-T^n{y}_n-{\alpha }_n(x_n-\mu F\left(T^n{y}_n\right))\parallel \hfill \\ \le \frac{{\alpha }_{n+1}\delta }{1-{\beta }_{n+1}}\parallel x_{n+1}-x_n\parallel +(1-\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}})(1+{\theta }_{n+1})\parallel y_{n+1}-y_n\parallel +\parallel T^n{y}_n-T^{n+1}{y}_n\parallel \hfill \\ +{\alpha }_{n+1}\parallel x_{n+1}-\mu F\left(T^{n+1}{y}_{n+1}\right)\parallel +{\alpha }_n\parallel x_n-\mu F\left(T^n{y}_n\right)\parallel \hfill \\ +|\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\alpha }_n}{1-{\beta }_n}|\parallel f\left(x_n\right)-T^n{y}_n-{\alpha }_n(x_n-\mu F\left(T^n{y}_n\right))\parallel \hfill \\ \le \parallel y_{n+1}-y_n\parallel +{\theta }_{n+1}\parallel y_n-y_{n+1}\parallel +\frac{{\alpha }_{n+1}\delta }{1-{\beta }_{n+1}}\parallel x_{n+1}-x_n\parallel +\parallel T^{n+1}{y}_n-T^n{y}_n\parallel \hfill \\ +{\alpha }_{n+1}\parallel x_{n+1}-\mu F\left(T^{n+1}{y}_{n+1}\right)\parallel +{\alpha }_n\parallel x_n-\mu F\left(T^n{y}_n\right)\parallel \hfill \\ +|\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\alpha }_n}{1-{\beta }_n}|\parallel f\left(x_n\right)-T^n{y}_n-{\alpha }_n(x_n-\mu F\left(T^n{y}_n\right))\parallel .\hfill \end{array}$$

(10)

Since $\left\{{\alpha }_n\right\}\subset (0,2\zeta ]$ and A is $\zeta$-inverse-strongly monotone, by Lemma 5, we obtain

$$\begin{array}{c}\parallel y_{n+1}-y_n\parallel \hfill \\ =\parallel P_C[{\alpha }_{n}g\left(x_n\right)+(I-{\alpha }_{n}A)z_n]-P_C[{\alpha }_{n+1}g\left(x_{n+1}\right)+(I-{\alpha }_{n+1}A)z_{n+1}]\parallel \hfill \\ \le \parallel (I-{\alpha }_{n+1}A)z_{n+1}-(I-{\alpha }_{n}A)z_n+{\alpha }_{n+1}g\left(x_{n+1}\right)-{\alpha }_{n}g\left(x_n\right)\parallel \hfill \\ =\parallel (I-{\alpha }_{n+1}A)z_{n+1}-(I-{\alpha }_{n+1}A)z_n+({\alpha }_n-{\alpha }_{n+1})A{z}_n+{\alpha }_{n+1}g\left(x_{n+1}\right)-{\alpha }_{n}g\left(x_n\right)\parallel \hfill \\ \le \parallel (I-{\alpha }_{n+1}A)z_{n+1}-(I-{\alpha }_{n+1}A)z_n\parallel +|{\alpha }_n-{\alpha }_{n+1}|\parallel A{z}_n\parallel +\parallel {\alpha }_{n+1}g\left(x_{n+1}\right)-{\alpha }_{n}g\left(x_n\right)\parallel \hfill \\ \le \parallel z_{n+1}-z_n\parallel +|{\alpha }_n-{\alpha }_{n+1}|\parallel A{z}_n\parallel +{\alpha }_{n+1}\parallel g\left(x_{n+1}\right)\parallel +{\alpha }_n\parallel g\left(x_n\right)\parallel \hfill \\ \le \parallel u_{n+1}-u_n\parallel +\parallel A{z}_n\parallel |{\alpha }_n-{\alpha }_{n+1}|+{\alpha }_{n+1}\parallel g\left(x_{n+1}\right)\parallel +{\alpha }_n\parallel g\left(x_n\right)\parallel .\hfill \end{array}$$

(11)

Furthermore, simple calculations show that

$$u_{n+1}-u_n={\gamma }_{n+1}(x_{n+1}-x_n)+(1-{\gamma }_{n+1})(S_{n+1}{u}_{n+1}-S_n{u}_n)+({\gamma }_{n+1}-{\gamma }_n)(x_n-S_n{u}_n),$$

which hence yields

$$\begin{array}{c}\parallel u_{n+1}-u_n{\parallel }^2\hfill \\ ={\gamma }_{n+1}\langle x_{n+1}-x_n,u_{n+1}-u_n\rangle +(1-{\gamma }_{n+1})\langle S_{n+1}{u}_{n+1}-S_n{u}_n,u_{n+1}-u_n\rangle \hfill \\ +({\gamma }_{n+1}-{\gamma }_n)\langle x_n-S_n{u}_n,u_{n+1}-u_n\rangle \hfill \\ ={\gamma }_{n+1}\langle x_{n+1}-x_n,u_{n+1}-u_n\rangle +(1-{\gamma }_{n+1})[\langle S_{n+1}{u}_{n+1}-S_n{u}_{n+1},u_{n+1}-u_n\rangle \hfill \\ +\langle S_n{u}_{n+1}-S_n{u}_n,u_{n+1}-u_n\rangle ]+({\gamma }_{n+1}-{\gamma }_n)\langle x_n-S_n{u}_n,u_{n+1}-u_n\rangle \hfill \\ \le {\gamma }_{n+1}\parallel x_{n+1}-x_n\parallel \parallel u_{n+1}-u_n\parallel +(1-{\gamma }_{n+1})[\parallel S_{n+1}{u}_{n+1}-S_n{u}_{n+1}\parallel \parallel u_{n+1}-u_n\parallel \hfill \\ +\parallel u_{n+1}-u_n{\parallel }^2]+|{\gamma }_{n+1}-{\gamma }_n|\parallel x_n-S_n{u}_n\parallel \parallel u_{n+1}-u_n\parallel .\hfill \end{array}$$

So it follows that

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

which immediately leads to

$$\begin{array}{cc}\parallel u_{n+1}-u_n\parallel \hfill & \le \parallel x_{n+1}-x_n\parallel +\frac{1-{\gamma }_{n+1}}{{\gamma }_{n+1}}\parallel S_{n+1}{u}_{n+1}-S_n{u}_{n+1}\parallel +|{\gamma }_{n+1}-{\gamma }_n|\frac{\parallel x_n-S_n{u}_n\parallel }{{\gamma }_{n+1}}\hfill \\ & \le \parallel x_{n+1}-x_n\parallel +\frac{1}{a}\parallel S_{n+1}{u}_{n+1}-S_n{u}_{n+1}\parallel +|{\gamma }_{n+1}-{\gamma }_n|\frac{\parallel x_n-S_n{u}_n\parallel }{a}.\hfill \end{array}$$

(12)

Put $D=\{u_n:n\ge 0\}$. Since $\left\{u_n\right\}$ is a bounded sequence, we know that D is a bounded set. Then, by the assumption of this theorem, we get

$$\sum _{n=0}^{\infty }\underset{x\in D}{sup}\parallel S_{n+1}x-S_{n}x\parallel <\infty .$$

Noticing that

$$\parallel S_{n+1}{u}_{n+1}-S_n{u}_{n+1}\parallel \le \underset{x\in D}{sup}\parallel S_{n+1}x-S_{n}x\parallel ,\forall n\ge 0,$$

we have

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

(13)

Therefore, from (10)–(12) we deduce that

$$\begin{array}{c}\parallel h_{n+1}-h_n\parallel \hfill \\ \le {\theta }_{n+1}\parallel y_{n+1}-y_n\parallel +{\displaystyle \frac{{\alpha }_{n+1}\delta }{1-{\beta }_{n+1}}}\parallel x_{n+1}-x_n\parallel +\parallel T^{n+1}{y}_n-T^n{y}_n\parallel +\parallel y_{n+1}-y_n\parallel \hfill \\ +{\alpha }_{n+1}\parallel x_{n+1}-\mu F\left(T^{n+1}{y}_{n+1}\right)\parallel +{\alpha }_n\parallel x_n-\mu F\left(T^n{y}_n\right)\parallel \hfill \\ +|{\displaystyle \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}}-{\displaystyle \frac{{\alpha }_n}{1-{\beta }_n}}|\parallel f\left(x_n\right)-T^n{y}_n-{\alpha }_n(x_n-\mu F\left(T^n{y}_n\right))\parallel \hfill \\ \le \parallel u_{n+1}-u_n\parallel +|{\alpha }_n-{\alpha }_{n+1}|\parallel A{z}_n\parallel +{\alpha }_{n+1}\parallel g\left(x_{n+1}\right)\parallel +{\alpha }_n\parallel g\left(x_n\right)\parallel \hfill \\ +{\theta }_{n+1}\parallel y_{n+1}-y_n\parallel +{\displaystyle \frac{{\alpha }_{n+1}\delta }{1-{\beta }_{n+1}}}\parallel x_{n+1}-x_n\parallel +\parallel T^{n+1}{y}_n-T^n{y}_n\parallel \hfill \\ +{\alpha }_{n+1}\parallel x_{n+1}-\mu F\left(T^{n+1}{y}_{n+1}\right)\parallel +{\alpha }_n\parallel x_n-\mu F\left(T^n{y}_n\right)\parallel \hfill \\ +|{\displaystyle \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}}-{\displaystyle \frac{{\alpha }_n}{1-{\beta }_n}}|\parallel f\left(x_n\right)-T^n{y}_n-{\alpha }_n(x_n-\mu F\left(T^n{y}_n\right))\parallel \hfill \\ \le \parallel x_{n+1}-x_n\parallel +{\displaystyle \frac{1}{a}}\parallel S_{n+1}{u}_{n+1}-S_n{u}_{n+1}\parallel +|{\gamma }_{n+1}-{\gamma }_n|{\displaystyle \frac{\parallel x_n-S_n{u}_n\parallel }{a}}\hfill \\ +|{\alpha }_n-{\alpha }_{n+1}|\parallel A{z}_n\parallel +{\alpha }_{n+1}\parallel g\left(x_{n+1}\right)\parallel +{\alpha }_n\parallel g\left(x_n\right)\parallel +{\theta }_{n+1}\parallel y_{n+1}-y_n\parallel \hfill \\ +{\displaystyle \frac{{\alpha }_{n+1}\delta }{1-{\beta }_{n+1}}}\parallel x_{n+1}-x_n\parallel +\parallel T^{n+1}{y}_n-T^n{y}_n\parallel +{\alpha }_{n+1}\parallel x_{n+1}-\mu F\left(T^{n+1}{y}_{n+1}\right)\parallel \hfill \\ +{\alpha }_n\parallel x_n-\mu F\left(T^n{y}_n\right)\parallel +|{\displaystyle \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}}-{\displaystyle \frac{{\alpha }_n}{1-{\beta }_n}}|\parallel f\left(x_n\right)-T^n{y}_n-{\alpha }_n(x_n-\mu F\left(T^n{y}_n\right))\parallel ,\hfill \end{array}$$

which immediately attains

$$\begin{array}{c}\parallel h_{n+1}-h_n\parallel -\parallel x_{n+1}-x_n\parallel \le \frac{1}{a}\parallel S_{n+1}{u}_{n+1}-S_n{u}_{n+1}\parallel +|{\gamma }_{n+1}-{\gamma }_n|\frac{\parallel x_n-S_n{u}_n\parallel }{a}\hfill \\ +|{\alpha }_n-{\alpha }_{n+1}|\parallel A{z}_n\parallel +{\alpha }_{n+1}\parallel g\left(x_{n+1}\right)\parallel +{\alpha }_n\parallel g\left(x_n\right)\parallel +{\theta }_{n+1}\parallel y_{n+1}-y_n\parallel \hfill \\ +\frac{{\alpha }_{n+1}\delta }{1-{\beta }_{n+1}}\parallel x_{n+1}-x_n\parallel +\parallel T^{n+1}{y}_n-T^n{y}_n\parallel +{\alpha }_{n+1}\parallel x_{n+1}-\mu F\left(T^{n+1}{y}_{n+1}\right)\parallel \hfill \\ +{\alpha }_n\parallel x_n-\mu F\left(T^n{y}_n\right)\parallel +|\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\alpha }_n}{1-{\beta }_n}|\parallel f\left(x_n\right)-T^n{y}_n-{\alpha }_n(x_n-\mu F\left(T^n{y}_n\right))\parallel .\hfill \end{array}$$

(14)

Since ${lim}_{n\to \infty }{\theta }_n=0$ and ${lim}_{n\to \infty }\parallel T^{n+1}{y}_n-T^n{y}_n\parallel =0$ (due to condition (v)), from (13) and conditions (i), (iii), (iv), it follows that

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

Hence, by condition (iii) and Lemma 8, we get ${lim}_{n\to \infty }\parallel h_n-x_n\parallel =0$. Consequently,

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

(15)

Again from (11) and (12), we conclude that

$$\begin{array}{c}\parallel y_{n+1}-y_n\parallel \hfill \\ \le \parallel u_{n+1}-u_n\parallel +|{\alpha }_n-{\alpha }_{n+1}|\parallel A{z}_n\parallel +{\alpha }_{n+1}\parallel g\left(x_{n+1}\right)\parallel +{\alpha }_n\parallel g\left(x_n\right)\parallel \hfill \\ \le \parallel x_{n+1}-x_n\parallel +\frac{1}{a}\parallel S_{n+1}{u}_{n+1}-S_n{u}_{n+1}\parallel +|{\gamma }_{n+1}-{\gamma }_n|\frac{\parallel x_n-S_n{u}_n\parallel }{a}+|{\alpha }_n-{\alpha }_{n+1}|\parallel A{z}_n\parallel \hfill \\ +{\alpha }_{n+1}\parallel g\left(x_{n+1}\right)\parallel +{\alpha }_n\parallel g\left(x_n\right)\parallel \to 0(n\to \infty ),\hfill \end{array}$$

and

$$\parallel z_n-z_{n+1}\parallel =\parallel G{u}_n-G{u}_{n+1}\parallel \le \parallel u_n-u_{n+1}\parallel \to 0(n\to \infty ).$$

Thus,

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

Step 3. We prove $\parallel x_n-G{x}_n\parallel \to 0$ as $n\to \infty$. Indeed, noticing $w_n=P_C[(I-{\alpha }_n\mu F)T^n{y}_n+{\alpha }_n{x}_n]$ for all $n\ge 0$, we have

$$\langle (I-{\alpha }_n\mu F)T^n{y}_n+{\alpha }_n{x}_n-P_C[{\alpha }_n{x}_n+(I-{\alpha }_n\mu F)T^n{y}_n],p-w_n\rangle \le 0.$$

(16)

From (16), we have

$$\begin{array}{cc}\parallel w_n{-p\parallel }^2\hfill & =\langle P_C[(I-{\alpha }_n\mu F)T^n{y}_n+{\alpha }_n{x}_n]-p,w_n-p\rangle \hfill \\ & =\langle P_C[(I-{\alpha }_n\mu F)T^n{y}_n+{\alpha }_n{x}_n]-{\alpha }_n{x}_n-(I-{\alpha }_n\mu F)T^n{y}_n,w_n-p\rangle \hfill \\ & +\langle {\alpha }_n{x}_n+(I-{\alpha }_n\mu F)T^n{y}_n-p,w_n-p\rangle \hfill \\ & \le \langle {\alpha }_n{x}_n+(I-{\alpha }_n\mu F)T^n{y}_n-p,w_n-p\rangle \hfill \\ & =\langle (I-{\alpha }_n\mu F)T^n{y}_n-(I-{\alpha }_n\mu F)p,w_n-p\rangle +{\alpha }_n\langle x_n-\mu Fp,w_n-p\rangle \hfill \\ & \le (1-{\alpha }_n\tau )\parallel T^n{y}_n-p\parallel \parallel w_n-p\parallel +{\alpha }_n\langle x_n-\mu Fp,w_n-p\rangle \hfill \\ & \le {\displaystyle \frac{1}{2}}{(1-{\alpha }_n\tau )}^2\parallel T^n{y}_n{-p\parallel }^2+{\displaystyle \frac{1}{2}}{\parallel w_n-p\parallel }^2+{\alpha }_n\langle x_n-\mu Fp,w_n-p\rangle .\hfill \end{array}$$

Hence, we have

$$\begin{array}{cc}\parallel w_n{-p\parallel }^2\hfill & \le {(1-{\alpha }_n\tau )}^2{\parallel T^n{y}_n-p\parallel }^2+2{\alpha }_n\langle x_n-\mu Fp,w_n-p\rangle \hfill \\ & \le (1-{\alpha }_n\tau ){(1+{\theta }_n)}^2{\parallel y_n-p\parallel }^2+2{\alpha }_n\langle x_n-\mu Fp,w_n-p\rangle \hfill \\ & =(1-{\alpha }_n\tau )[\parallel y_n{-p\parallel }^2+{\theta }_n(2+{\theta }_n)\parallel y_n-p{\parallel }^2]+2{\alpha }_n\langle x_n-\mu Fp,w_n-p\rangle \hfill \\ & \le (1-{\alpha }_n\tau )\parallel y_n{-p\parallel }^2+{\theta }_n(2+{\theta }_n){\parallel y_n-p\parallel }^2+2{\alpha }_n\langle x_n-\mu Fp,w_n-p\rangle .\hfill \end{array}$$

(17)

From (9) and (17), we get

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

(18)

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

$$\begin{array}{cc}\parallel v_n{-q\parallel }^2\hfill & =\parallel P_C(u_n-{\mu }_2{B}_2{u}_n)-P_C(p-{\mu }_2{B}_{2}p){\parallel }^2\hfill \\ & \le \parallel u_n-p-{\mu }_2(B_2{u}_n-B_{2}p){\parallel }^2\hfill \\ & \le \parallel u_n{-p\parallel }^2-{\mu }_2(2\beta -{\mu }_2){\parallel B_2{u}_n-B_{2}p\parallel }^2\hfill \end{array}$$

(19)

and

$$\begin{array}{cc}\parallel z_n{-p\parallel }^2\hfill & =\parallel P_C(v_n-{\mu }_1{B}_1{v}_n)-P_C(q-{\mu }_1{B}_{1}q){\parallel }^2\hfill \\ & \le \parallel v_n-q-{\mu }_1(B_1{v}_n-B_{1}q){\parallel }^2\hfill \\ & \le \parallel v_n{-q\parallel }^2-{\mu }_1(2\alpha -{\mu }_1){\parallel B_1{v}_n-B_{1}q\parallel }^2.\hfill \end{array}$$

(20)

Substituting (19) for (20), we obtain from (7) that

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

(21)

Combining (18) and (21), we get

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

which immediately yields

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

$(1-{\alpha }_n-{\beta }_n)(1-{\alpha }_n\tau )[{\mu }_2(2\beta -{\mu }_2)\parallel B_2{u}_n-B_2{p\parallel }^2+{\mu }_1(2\alpha -{\mu }_1)\parallel B_1{v}_n-B_{1}q{\parallel }^2]\to 0$ as $n\to \infty .$ Since ${lim\; inf}_{n\to \infty }(1-{\alpha }_n-{\beta }_n)>0$ (due to condition (iii)), ${\mu }_1\in (0,2\alpha ),{\mu }_2\in (0,2\beta ),{lim}_{n\to \infty }{\theta }_n=0$ and ${lim}_{n\to \infty }{\alpha }_n=0$, we obtain from (15) that

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

(22)

On the other hand, we have

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

which implies that

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

(23)

In the same way, we derive

$$\begin{array}{cc}\parallel z_n{-p\parallel }^2\hfill & =\parallel P_C(v_n-{\mu }_1{B}_1{v}_n)-P_C(q-{\mu }_1{B}_{1}q){\parallel }^2\hfill \\ & \le \langle v_n-{\mu }_1{B}_1{v}_n-(q-{\mu }_1{B}_{1}q),z_n-p\rangle \hfill \\ & =\langle v_n-q,z_n-p\rangle +{\mu }_1\langle B_{1}q-B_1{v}_n,z_n-p\rangle \hfill \\ & \le \frac{1}{2}[\parallel v_n{-q\parallel }^2+\parallel z_n{-p\parallel }^2-\parallel v_n-z_n{+(p-q)\parallel }^2]+{\mu }_1\parallel B_{1}q-B_1{v}_n\parallel \parallel z_n-p\parallel ,\hfill \end{array}$$

which implies that

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

(24)

Substituting (23) for (24), we deduce from (7) that

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

(25)

Combining (18) and (25), we have

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

which hence yields

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

Since ${lim\; inf}_{n\to \infty }(1-{\alpha }_n-{\beta }_n)>0$ (due to condition (iii)), ${lim}_{n\to \infty }{\theta }_n=0$ and ${lim}_{n\to \infty }{\alpha }_n=0$, we conclude from (15) and (22) that

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

(26)

It follows that

$$\parallel u_n-z_n\parallel \le \parallel u_n-v_n-(p-q)\parallel +\parallel v_n-z_n+(p-q)\parallel \to 0(n\to \infty ).$$

That is,

$$\underset{n\to \infty }{lim}\parallel u_n-G{u}_n\parallel =\underset{n\to \infty }{lim}\parallel u_n-z_n\parallel =0.$$

(27)

Additionally, according to (6), we have

$$\begin{array}{cc}\parallel u_n{-p\parallel }^2\hfill & ={\gamma }_n\langle x_n-p,u_n-p\rangle +(1-{\gamma }_n)\langle S_n{u}_n-p,u_n-p\rangle \hfill \\ & \le {\gamma }_n\langle x_n-p,u_n-p\rangle +(1-{\gamma }_n){\parallel u_n-p\parallel }^2,\hfill \end{array}$$

which implicitly yields that

$$\begin{array}{cc}2\parallel u_n{-p\parallel }^2\hfill & \le 2\langle x_n-p,u_n-p\rangle \hfill \\ & =\parallel x_n{-p\parallel }^2-\parallel x_n-u_n{\parallel }^2+{\parallel u_n-p\parallel }^2.\hfill \end{array}$$

This immediately implies that

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

which together with (3.16), yields

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

Hence, we have

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

Since ${lim\; inf}_{n\to \infty }(1-{\alpha }_n-{\beta }_n)>0$, ${lim}_{n\to \infty }{\theta }_n=0$ and ${lim}_{n\to \infty }{\alpha }_n=0$, we obtain from (15) that

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

(28)

Moreover, observe that

$$\parallel x_n-z_n\parallel \le \parallel x_n-u_n\parallel +\parallel G{u}_n-u_n\parallel ,$$

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

and

$$\parallel x_n-y_n\parallel \le \parallel x_n-{\alpha }_{n}g\left(x_n\right)-(I-{\alpha }_{n}A)z_n\parallel \le \parallel x_n-z_n\parallel +{\alpha }_n\parallel g\left(x_n\right)-A{z}_n\parallel .$$

Then, from (27) and (28), it follows that

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

(29)

Step 4. Let us prove $\parallel x_n-S_n{x}_n\parallel \to 0,\parallel x_n-w_n\parallel \to 0$ and $\parallel x_n-T{x}_n\parallel \to 0$ as $n\to \infty$. Indeed, combining (5) and (8), we obtain that

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

That is,

$$\underset{n\to \infty }{lim}\parallel S_n{u}_n-u_n\parallel =0.$$

(30)

Observe ${\left\{S_n\right\}}_{n=0}^{\infty }$ is ℓ-uniformly Lipschitzian. We further get from (28) and (30) that

$$\begin{array}{cc}\parallel S_n{x}_n-x_n\parallel \hfill & \le \parallel S_n{x}_n-S_n{u}_n\parallel +\parallel S_n{u}_n-u_n\parallel +\parallel u_n-x_n\parallel \hfill \\ & \le \ell \parallel x_n-u_n\parallel +\parallel S_n{u}_n-u_n\parallel +\parallel u_n-x_n\parallel \hfill \\ & =(\ell +1)\parallel x_n-u_n\parallel +\parallel S_n{u}_n-u_n\parallel \to 0(n\to \infty ).\hfill \end{array}$$

That is,

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

(31)

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

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

Then,

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

Hence, we get

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

Consequently, from (15), (29) and ${lim}_{n\to \infty }{\alpha }_n=0$, we obtain that

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

(32)

So it follows that

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

That is,

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

(33)

We also note that

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

By the condition (v) and (32), we get

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

Further, 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 (29) that

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

(34)

Step 5. Set $\overline{S}:={(2I-S)}^{-1}$. We aim to prove $\parallel x_n-\overline{S}{x}_n\parallel \to 0$ as $n\to \infty$. We show that $S:C\to C$ is pseudocontractive and ℓ-Lipschitzian such that ${lim}_{n\to \infty }\parallel S{x}_n-x_n\parallel =0$, where $Sx={lim}_{n\to \infty }{S}_{n}x\forall x\in C$. Observe that for all $x,y\in C$, ${lim}_{n\to \infty }\parallel S_{n}x-Sx\parallel =0$ and ${lim}_{n\to \infty }\parallel S_{n}y-Sy\parallel =0$. Since each $S_n$ is a pseudocontractive operator, we get

$$\langle Sx-Sy,x-y\rangle =\underset{n\to \infty }{lim}\langle S_{n}x-S_{n}y,x-y\rangle \le {\parallel x-y\parallel }^2.$$

This presents that S is pseudocontractive. Note that ${\left\{S_n\right\}}_{n=0}^{\infty }$ is ℓ-uniformly Lipschitzian

$$\parallel Sx-Sy\parallel =\underset{n\to \infty }{lim}\parallel S_{n}x-S_{n}y\parallel \le \ell \parallel x-y\parallel ,\forall x,y\in C.$$

This means that S is ℓ-Lipschitzian. Since the boundedness of $\left\{x_n\right\}$ and putting $D=\overline{\mathrm{conv}}\{x_n:n\ge 0\}$ (the closure of convex hull of the set $\{x_n:n\ge 0\}$), we have ${\sum }_{n=1}^{\infty }{sup}_{x\in D}\parallel S_{n}x-S_{n-1}x\parallel <\infty$. Hence, by Proposition 1, we get

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

(35)

Thus, combining (31) with (35) we have

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

That is,

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

(36)

Define $\overline{S}:={(2I-S)}^{-1}$. $\overline{S}:C\to C$ is nonexpansive, $\mathrm{Fix}\left(\overline{S}\right)=\mathrm{Fix}\left(S\right)={\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)$ and ${lim}_{n\to \infty }\parallel x_n-\overline{S}{x}_n\parallel =0$. Indeed, put $\overline{S}:={(2I-S)}^{-1}$, where I is the identity mapping of H. Then, $\overline{S}$ is nonexpansive and the fixed point set $\mathrm{Fix}\left(\overline{S}\right)=\mathrm{Fix}\left(S\right)={\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)$. Observe that

$$\begin{array}{cc}\parallel x_n-\overline{S}{x}_n\parallel \hfill & =\parallel \overline{S}{\overline{S}}^{-1}{x}_n-\overline{S}{x}_n\parallel \hfill \\ & \le \parallel {\overline{S}}^{-1}{x}_n-x_n\parallel \hfill \\ & =\parallel x_n-S{x}_n\parallel .\hfill \end{array}$$

From (36), it follows that

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

(37)

Step 6. We aim to present

$$\underset{n\to \infty }{lim\; sup}\langle (I-f)x^{*},x^{*}-x_n\rangle \le 0,$$

(38)

where $\left\{x^{*}\right\}=\mathrm{VI}(\mathrm{VI}(\mathsf{\Omega },A-g),I-f)$. Indeed, we choose a subsequence $\left\{x_{n_i}\right\}$ of $\left\{x_n\right\}$ such that

$$\underset{i\to \infty }{lim}\langle (I-f)x^{*},x^{*}-x_{n_i}\rangle =\underset{n\to \infty }{lim\; sup}\langle (I-f)x^{*},x^{*}-x_n\rangle .$$

We suppose a subsequence $x_{n_i}\rightharpoonup \overline{x}\in C$. Observe that G and $\overline{S}$ have the nonexpansivity and that T has the asymptotically nonexpansivity. Since $(I-G)x_n\to 0,(I-T)x_n\to 0$ and $(I-\overline{S})x_n\to 0$, by Lemma 7, we have that $\overline{x}\in \mathrm{Fix}\left(G\right)=\mathrm{GSVI}(C,B_1,B_2),\overline{x}\in \mathrm{Fix}\left(T\right)$ and $\overline{x}\in \mathrm{Fix}\left(\overline{S}\right)={\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)$. Then, $\overline{x}\in \mathsf{\Omega }={\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)\cap \mathrm{GSVI}(C,B_1,B_2)\cap \mathrm{Fix}\left(T\right)$. We present that $\overline{x}\in \mathrm{VI}(\mathsf{\Omega },A-g)$. As a fact, let $y\in \mathsf{\Omega }$ be a arbitrarily fixed point. Then, it follows from (6), (8), and the monotonicity of $A-g$ that

$$\begin{array}{c}\parallel y_n{-y\parallel }^2\le {\parallel (z_n-y)-{\alpha }_n(A{z}_n-g\left(x_n\right))\parallel }^2\hfill \\ =\parallel z_n{-y\parallel }^2+2{\alpha }_n\langle A{z}_n-g\left(x_n\right),y-z_n\rangle +{\alpha }_n^2{\parallel A{z}_n-g\left(x_n\right)\parallel }^2\hfill \\ \le \parallel x_n{-y\parallel }^2+2{\alpha }_n\langle A{z}_n-g\left(z_n\right),y-z_n\rangle +2{\alpha }_{n}l\parallel z_n-x_n\parallel \parallel y-z_n\parallel +{\alpha }_n^2{\parallel A{z}_n-g\left(x_n\right)\parallel }^2\hfill \\ \le \parallel x_n{-y\parallel }^2+2{\alpha }_n\langle Ay-g\left(y\right),y-z_n\rangle +2{\alpha }_{n}l\parallel z_n-x_n\parallel \parallel y-z_n\parallel +{\alpha }_n^2{\parallel A{z}_n-g\left(x_n\right)\parallel }^2,\hfill \end{array}$$

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

$$\begin{array}{c}0\le \frac{\parallel x_n{-y\parallel }^2-{\parallel y_n-y\parallel }^2}{{\alpha }_n}+2\langle (A-g)y,y-z_n\rangle +2l\parallel z_n-x_n\parallel \parallel y-z_n\parallel +{\alpha }_n{\parallel A{z}_n-g\left(x_n\right)\parallel }^2\hfill \\ \le \frac{(\parallel x_n-y\parallel +\parallel y_n-y\parallel )\parallel x_n-y_n\parallel }{{\alpha }_n}+2\langle (A-g)y,y-z_n\rangle +2l\parallel z_n-x_n\parallel \parallel y-z_n\parallel +{\alpha }_n{\parallel A{z}_n-g\left(x_n\right)\parallel }^2.\hfill \end{array}$$

From (29), it is easy to see that $x_{n_i}\rightharpoonup \overline{x}$ leads to $z_{n_i}\rightharpoonup \overline{x}$. Since ${lim}_{n\to \infty }{\alpha }_n=0$ and $\parallel x_n-y_n\parallel =o\left({\alpha }_n\right)$ (due to the assumption), we have

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

It follows that

$$\langle (A-g)y,y-\overline{x}\rangle \ge 0,\forall y\in \mathsf{\Omega }.$$

Accordingly, Lemma 5 and the Lipschitz continuity and monotonicity of $A-g$ grant that

$$\langle (A-g)\overline{x},y-\overline{x}\rangle \ge 0,\forall y\in \mathsf{\Omega };$$

that is, $\overline{x}\in \mathrm{VI}(\mathsf{\Omega },A-g)$. Consequently, from $\left\{x^{*}\right\}=\mathrm{VI}(\mathrm{VI}(\mathsf{\Omega },A-g),I-f)$, we have

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

(39)

Step 7. Finally, we prove $x_n\to x^{*}$ as $n\to \infty$. Indeed, from (4) we get

$$\begin{array}{cc}\parallel x_{n+1}-x^{*}{\parallel }^2\hfill & ={\beta }_n\langle x_n-x^{*},x_{n+1}-x^{*}\rangle +{\alpha }_n\langle f\left(x_n\right)-x^{*},x_{n+1}-x^{*}\rangle \hfill \\ & +(1-{\alpha }_n-{\beta }_n)\langle w_n-x^{*},x_{n+1}-x^{*}\rangle \hfill \\ & ={\alpha }_n[\langle f\left(x_n\right)-f\left(x^{*}\right),x_{n+1}-x^{*}\rangle +\langle f\left(x^{*}\right)-x^{*},x_{n+1}-x_n\rangle \hfill \\ & +\langle f\left(x^{*}\right)-x^{*},x_n-x^{*}\rangle ]+{\beta }_n\langle x_n-x^{*},x_{n+1}-x^{*}\rangle \hfill \\ & +(1-{\alpha }_n-{\beta }_n)[\langle w_n-x_n,x_{n+1}-x^{*}\rangle +\langle x_n-x^{*},x_{n+1}-x^{*}\rangle ]\hfill \\ & \le {\alpha }_n[\delta \parallel x_n-x^{*}\parallel \parallel x_{n+1}-x^{*}\parallel +\parallel f\left(x^{*}\right)-x^{*}\parallel \parallel x_{n+1}-x_n\parallel \hfill \\ & +\langle f\left(x^{*}\right)-x^{*},x_n-x^{*}\rangle ]+{\beta }_n\parallel x_n-x^{*}\parallel \parallel x_{n+1}-x^{*}\parallel \hfill \\ & +\parallel w_n-x_n\parallel \parallel x_{n+1}-x^{*}\parallel +(1-{\alpha }_n-{\beta }_n)\parallel x_n-x^{*}\parallel \parallel x_{n+1}-x^{*}\parallel \hfill \\ & \le [1-{\alpha }_n(1-\delta )]\parallel x_n-x^{*}\parallel \parallel x_{n+1}-x^{*}\parallel +{\alpha }_n\parallel f\left(x^{*}\right)-x^{*}\parallel \parallel x_{n+1}-x_n\parallel \hfill \\ & +{\alpha }_n\langle f\left(x^{*}\right)-x^{*},x_n-x^{*}\rangle +\parallel w_n-x_n\parallel \parallel x_{n+1}-x^{*}\parallel \hfill \\ & \le {\displaystyle \frac{{[1-{\alpha }_n(1-\delta )]}^2}{2}}\parallel x_n-x^{*}{\parallel }^2+{\displaystyle \frac{1}{2}}\parallel x_{n+1}-x^{*}{\parallel }^2+{\alpha }_n\parallel f\left(x^{*}\right)-x^{*}\parallel \parallel x_{n+1}-x_n\parallel \hfill \\ & +{\alpha }_n\langle f\left(x^{*}\right)-x^{*},x_n-x^{*}\rangle +\parallel w_n-x_n\parallel \parallel x_{n+1}-x^{*}\parallel ,\hfill \end{array}$$

which immediately yields

$$\begin{array}{cc}\parallel x_{n+1}-x^{*}{\parallel }^2\hfill & \le 2{\alpha }_n\parallel f\left(x^{*}\right)-x^{*}\parallel \parallel x_{n+1}-x_n\parallel +{[1-{\alpha }_n(1-\delta )]}^2{\parallel x_n-x^{*}\parallel }^2\hfill \\ & +2{\alpha }_n\langle f\left(x^{*}\right)-x^{*},x_n-x^{*}\rangle +2\parallel w_n-x_n\parallel \parallel x_{n+1}-x^{*}\parallel \hfill \\ & \le [1-{\alpha }_n(1-\delta )]\parallel x_n-x^{*}{\parallel }^2+{\alpha }_n(1-\delta )·\frac{2}{1-\delta }\{\parallel f\left(x^{*}\right)-x^{*}\parallel \parallel x_{n+1}-x_n\parallel \hfill \\ & +\langle f\left(x^{*}\right)-x^{*},x_n-x^{*}\rangle +\frac{\parallel w_n-x_n\parallel }{{\alpha }_n}·\parallel x_{n+1}-x^{*}\parallel \}.\hfill \end{array}$$

(40)

Since $\parallel w_n-x_n\parallel =o\left({\alpha }_n\right)$, ${\sum }_{n=0}^{\infty }{\alpha }_n=\infty ,$ and ${lim}_{n\to \infty }{\alpha }_n=0$, we deduce from (15), (38), and (39) that ${\sum }_{n=0}^{\infty }{\alpha }_n(1-\delta )=\infty$ and

$$\underset{n\to \infty }{lim\; sup}\{\parallel f\left(x^{*}\right)-x^{*}\parallel \parallel x_{n+1}-x_n\parallel +\langle f\left(x^{*}\right)-x^{*},x_n-x^{*}\rangle +\frac{\parallel w_n-x_n\parallel }{{\alpha }_n}·\parallel x_{n+1}-x^{*}\parallel \}\le 0.$$

Therefore, applying Lemma 3 to relation (40), we conclude that $x_n\to x^{*}$ as $n\to \infty$. This completes the proof. □