Hybrid Mann Viscosity Implicit Iteration Methods for Triple Hierarchical Variational Inequalities, Systems of Variational Inequalities and Fixed Point Problems

Abstract

In the present work, we introduce a hybrid Mann viscosity-like implicit iteration to find solutions of a monotone classical variational inequality with a variational inequality constraint over the common solution set of a general system of variational inequalities and a problem of common fixed points of an asymptotically nonexpansive mapping and a countable of uniformly Lipschitzian pseudocontractive mappings in Hilbert spaces, which is called the triple hierarchical constrained variational inequality. Strong convergence of the proposed method to the unique solution of the problem is guaranteed under some suitable assumptions. As a sub-result, we provide an algorithm to solve problem of common fixed points of pseudocontractive, nonexpansive mappings, variational inequality problems and generalized mixed bifunction equilibrium problems in Hilbert spaces.

1. Introduction

We suppose that H is a real or complex Hilbert space and let H be with inner product $\langle ·,·\rangle$ and norm $\parallel ·\parallel$. We suppose that C is a convex nonempty closed set of H. We also suppose that $P_C$ is the metric projection from H onto C. Since C is a convex nonempty closed set, we conclude that $P_C$ is defined. Let T be a mapping on convex nonempty closed set C. Denote by $\mathrm{Fix}\left(T\right)$ the set of fixed points of T, i.e., $\mathrm{Fix}\left(T\right)=\{x\in C:(I-T)x=0\}$. → and ⇀ present strong convergence and weak convergence, respectively. A mapping $T:C\to C$ is named to be asymptotically nonexpansive if there exists a sequence $\{{\theta }_n\}\subset [0,+\infty )$ with ${\lim }_{n\to \infty }{\theta }_n=0$ such that

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

(1)

If $\theta \equiv 0,$ then T is named to be nonexpansive, that is,

$$\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,\forall x,y\in C.$$

(2)

Suppose that A is a nonself mapping from convex nonempty closed set C to entire space H. The classical variational inequality (VI) is to find $x^{*}\in C$ such that

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

(3)

where $\mu$ is some positive real number. We denote by VI$(C,A)$ the set of solutions of VI (3).

Assume that $B_1$ is a nonself mapping from convex nonempty closed set C to entire space H and $B_2$ is a nonself mapping from convex nonempty closed set C to entire space H, respectively. we study the system of approximating $(x^{*},y^{*})\in C\times C$ such that

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

(4)

Here, ${\mu }_1$ and ${\mu }_2$ are two real numbers. The system (4) is named to be a general system of variational inequalities (GSVI). We note that the system (4) can be transformed into a problem of zero points $(I-T)x=0$, that is, the fixed point of T as following

Lemma 1 

([1]). Fix $x^{*},y^{*}\in C$, where $(x^{*},y^{*})$ satisfies the system (4) if and only if

$$x^{*}\in \mathrm{GSVI}(C,B_1,B_2),$$

where $\mathrm{GSVI}(C,B_1,B_2)$ is the set of solutions of the mapping $G:=P_C(I-{\mu }_1{B}_1)P_C(I-{\mu }_2{B}_2)$, and $y^{*}=P_C(I-{\mu }_2{B}_2)x^{*}$.

Recently, the variational inequality (3) and the system (4) have been intensively investigated by many authors via fixed-point methods; see [2,3,4,5,6,7,8,9,10,11] and the references therein. A mapping $f:C\to C$ is said to be a contraction on C if there exists a constant $\delta \in [0,1)$ 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 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 exists a constant $\eta >0$ such that $\langle Fx-Fy,x-y\rangle \ge {\eta \parallel x-y\parallel }^2\forall x,y\in C$. Moreover, it is called $\alpha$-inverse-strongly monotone (or $\alpha$-cocoercive), if there exists a constant $\alpha >0$ such that

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

Furthermore, let X be a real Banach space whose topological dual space is denoted with $X^{*}$. The normalized duality $J:X\to 2^{X^{*}}$ is defined through

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

where $\langle ·,·\rangle$ denotes the generalized duality pairing. We suppose that T is a mapping. Its domain and range are denoted by $D\left(T\right)$ and range $R\left(T\right)$, respectively. It called pseudocontractive if

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

From a result of Kato [12], we know that the notion of pseudocontraction is equivalent to the following definition: There exists $j(x-y)\in J(x-y)$ such that

$$\langle j(x-y),Tx-Ty\rangle \le {\parallel x-y\parallel }^2,\forall x,y\in D\left(T\right).$$

It is well known that the class of pseudocontractive mappings, whose complementary operators are accretive, is an important and significant generation of nonexpansive mappings (see [13,14,15,16,17,18,19]). In 2011, Ceng et al. [20] introduced an implicit viscosity approximation method for computing approximate fixed points of pseudocontractive mapping T, and obtained the norm convergence of sequence $\{x_n\}$ generated by their implicit method to a fixed-point of T.

The main aim of this paper is to introduce and analyze a hybrid Mann viscosity implicit iteration method for solving a monotone variational inequality with a variational inequality constraint over the common solution set of the GSVI (4) 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 hybrid Mann viscosity implicit iteration method is based on the viscosity approximation method, Korpelevich extragradient method, Mann iteration method and hybrid steepest-descent method. With relatively weak assumptions, the authors prove the strong convergence analysis of the their method to the unique solution of the THCVI. As an application, we list an algorithm to solve problems of common fixed point of pseudocontractive and nonexpansive mappings, classical variational inequalities and generalized mixed equilibrium problems in Hilbert setting.

2. Preliminaries

In this subsection, we suppose H is a Hilbert space. Its inner product denoted by $\langle ·,·\rangle$. We also suppose C is a convex nonempty closed set of H. Here, we list some basic concepts and facts. A nonself mapping F from convex nonempty closed set C to entire space H is said to be $\kappa$-Lipschitzian if there is a number $\kappa >0$ with $\parallel F\left(x\right)-F\left(y\right)\parallel \le \kappa \parallel x-y\parallel \forall x,y\in C$. In particular, if $\kappa =1$, then the nonself mapping F is named to be a nonexpansive operator. A self mapping A on entire space H is name to be a strongly positive bounded linear operator if we have a number $\gamma >0$ with

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

It is easy to see that the self mapping A is a $\gamma$-strongly monotone $\parallel A\parallel$-Lipschitzian operator. Recall that a self mapping T on convex nonempty closed set C is named to be

(a)

a contraction if we have a number $\alpha \in (0,1)$ with

$$\parallel Tx-Ty\parallel \le \alpha \parallel x-y\parallel ,\forall x,y\in C;$$

(b)

a pseudocontraction if

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

(c)

strong pseudocontraction if we have a number $\alpha \in (0,1)$ with

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

We use the following concept in the sequel.

Definition 1.

Let ${\{T_n\}}_{n=0}^{\infty }$ be a mapping sequence of continuous self pseudocontractions on C. Then, ${\{T_n\}}_{n=0}^{\infty }$ is said to be a countable family of ℓ-uniformly Lipschitzian pseudocontractive self-mappings on C if we have a number $\ell >0$ such that each $T_n$ is ℓ-Lipschitz continuous.

Fix $x\in H$, there is a unique element in C, denoted by $P_{C}x$, with

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

(5)

where $P_C$ stands for a metric projection of entire space H onto convex nonempty closed set C. It is well known that $P_C$ is a nonexpansive mapping with

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

(6)

Nevertheless, $P_{C}x$ has the functions: $P_{C}x\in C$ and

$$\langle x-P_{C}x,y-P_{C}x\rangle \le 0,$$

(7)

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

(8)

We also have

$${2\langle x-y,y\rangle +\parallel x-y\parallel }^2={\parallel x\parallel }^2-{\parallel y\parallel }^2.$$

(9)

We need the following propositions and lemmas for our main presentation.

Proposition 1 

([21]). We suppose C is a convex nonempty closed set of a Banach space X. We suppose $S_0,S_1,\dots$ is an operator sequence on convex nonempty closed C. Let

$$\sum _{n=1}^{\infty }\sup \{\parallel S_{n}x-S_{n-1}x\parallel :x\in C\}<\infty .$$

It follows that $\{S_{n}y\}$ converges strongly to some point of C for each $y\in C$. Nevertheless, we let S be a mapping on convex nonempty closed C defined through $Sy={\lim }_{n\to \infty }{S}_{n}y$ for all $y\in C$. Then ${\lim }_{n\to \infty }\sup \{\parallel Sx-S_{n}x\parallel :x\in C\}=0$.

Proposition 2 

([22]). We suppose C is a convex nonempty closed set of a Banach space X. We also suppose T is a continuous and strong pseudocontraction on convex nonempty closed C. This shows the fact that T has a fixed point in C. Indeed, it is also unique.

The following lemma is trivial. In fact, it an immediate consequence of the subdifferential of $\frac{1}{2}{\parallel ·\parallel }^2$.

Lemma 2.

We suppose H is a Hilbert space. In H, we have

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

Lemma 3 

([23]). We suppose $\{a_n\}$ is a number sequence such that

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

where $\{{\lambda }_n\}$ and $\{{\gamma }_n\}$ are real numbers such that

(i) 

$\{{\lambda }_n\}\subset [0,1]$ and ${\sum }_{n=0}^{\infty }{\lambda }_n=\infty$; or, equivalently,

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

(ii) 

${\lim \hspace{0.17em}\sup }_{n\to \infty }{\gamma }_n\le 0$ or ${\sum }_{n=0}^{\infty }|{\lambda }_n{\gamma }_n|<\infty$.

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

Lemma 4 

([24]). We suppose T is a nonexpansive mapping defined on a convex nonempty subset C of a Hilbert space H. Let λ be a number in $(0,1]$. We suppose F is a self κ-Lipschitzian and η-strongly monotone mapping on entire space H. Define the mapping $T^{\lambda }:C\to H$ through

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

Then, $T^{\lambda }$ is a contraction if $0<\mu <\frac{2\eta }{{\kappa }^2}$; 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.

Let the mapping $A:C\to H$ be α-inverse-strongly monotone. Then, for a given $\lambda \ge 0$,

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

In particular, if $0\le \lambda \le 2\alpha$, then $I-\lambda A$ is nonexpansive.

Proof. 

$$\begin{array}{c}{\parallel (I-\lambda A)y-(I-\lambda A)x\parallel }^2\hfill \\ ={\parallel \lambda (Ay-Ax)\parallel }^2-2\langle \lambda (Ay-Ax),y-x\rangle +{\parallel y-x\parallel }^2\hfill \\ \le {\lambda }^2{\parallel Ay-Ax\parallel }^2{-2\lambda \alpha \parallel Ay-Ax\parallel }^2+{\parallel y-x\parallel }^2\hfill \\ ={\lambda (\lambda -2\alpha )\parallel Ax-Ay\parallel }^2+{\parallel y-x\parallel }^2.\hfill \end{array}$$

 □

Utilizing Lemma 5, we immediately obtain the following lemma.

Lemma 6.

We suppose the nonself mappings $B_1,B_2$ is α-inverse-strongly monotone and β-inverse-strongly monotone defined on convex nonempty closed subset C of entire space H, respectively. Let the self mapping G 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 

([25]). We suppose that X is a real Banach space with a weakly continuous duality and C is a convex nonempty closed set in X. Let T be a self mapping defined the set C and we also suppose it is asymptotically nonexpansive with a empty fixed-point set. Then, $T-I$ is demiclosed at zero, i.e., let $\{x_n\}$ be a sequence in set C converging weakly to some x, where x in C and the sequence $\{(I-T)x_n\}$ converges strongly to 0, then $(T-I)x=0$, where I is the identity mapping of X.

Lemma 8 

([26]). We suppose C is a convex nonempty closed set in a Hilbert space H and A is a monotone and hemicontinuous nonself mapping defined on convex nonempty closed set C to H. Then, we have

(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 $\lambda >0$; and

(iii) 

$\mathrm{VI}(C,A)$ is singlton, if A is Lipschitz continuous strongly monotone.

3. Main Results

We suppose C is a convex nonempty closed set. Let the mappings $A_1,B_i$ be nonself monotone mappings for $i=1,2$ from C to H. We also let T be a self asymptotically nonexpansive mapping. Suppose ${\{S_n\}}_{n=0}^{\infty }$ is a countable family of self mapping. We also assume it is ℓ-uniformly Lipschitzian pseudocontractive on set C. Consider the variational inequality for monotone mapping $A_1$ over the common solution set $\mathsf{\Omega }$ of the GSVI (4) and the CFPP of ${\{S_n\}}_{n=0}^{\infty }$ and T:

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

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

This section introduces the following monotone variational inequality with the variational inequality constraint over the common solution set of the GSVI (4) and the CFPP of ${\{S_n\}}_{n=0}^{\infty }$ and T, which is called the triple hierarchical constrained variational inequality (THCVI):

Problem 1.

Assume that

(C1) 

$T:C\to C$ is an asymptotically nonexpansive mapping with a sequence $\{{\theta }_n\}$.

(C2) 

${\{S_n\}}_{n=0}^{\infty }$ is a countable family of ℓ-uniformly Lipschitzian pseudocontractive self-mappings on C.

(C3) 

$B_1:C\to H$ is an α-inverse-strongly monotone operator and $B_2:C\to H$ is a β-inverse-strongly monotone operator.

(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) 

$A_1:C\to H$ is an ζ-inverse-strongly monotone operator and $A_2:C\to H$ is a κ-Lipschitzian and η-strongly monotone operator.

(C9) 

$f:C\to C$ is a contraction mapping with coefficient $\delta \in [0,1)$.

(C10) 

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

Then, the objective is to

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

Since the original problem is a variational inequality problem, we therefore call it a triple hierarchical constrained variational inequality (THCVI). We introduce the following hybrid Mann viscosity implicit iteration method to find the solution of such a problem.

We show the main result of this paper, that is, the strong convergence analysis for Algorithm 1.

Algorithm 1: Hybrid Mann viscosity-like implicit iterative algorithm.

Step 0. Take ${\{{\alpha }_n\}}_{n=0}^{\infty },{\{{\beta }_n\}}_{n=0}^{\infty },{\{{\gamma }_n\}}_{n=0}^{\infty },{\{{\delta }_n\}}_{n=0}^{\infty },{\{{\sigma }_n\}}_{n=0}^{\infty }\subset (0,\infty )$, and $\mu >0$; arbitrarily choose $x_0\in C$; and let $n:=0$.   Step 1. Given $x_n\in C$, compute $x_{n+1}\in C$ as $$\left\{\begin{array}{c}\hspace{1em}{u}_n={\gamma }_n{x}_n+(1-{\gamma }_n)S_n{u}_n,\hfill \\ \hspace{1em}{v}_n=P_C(u_n-{\mu }_2{B}_2{u}_n),\hfill \\ \hspace{1em}{z}_n=P_C(v_n-{\mu }_1{B}_1{v}_n),\hfill \\ \hspace{1em}{y}_n={\sigma }_n{x}_n+(1-{\sigma }_n)P_C(I-{\delta }_n{A}_1)z_n,\hfill \\ \hspace{1em}{x}_{n+1}={\beta }_{n}f\left(y_n\right)+(1-{\beta }_n)P_C(I-{\alpha }_n\mu A_2)T^n{y}_n.\hfill \end{array}\right.$$ (10)   Update $n:=n+1$ and go to Step 1.

Theorem 1.

Assume that ${\mu }_1\in (0,2\alpha ),{\mu }_2\in (0,2\beta )$, and $\delta <\tau :=1-\sqrt{1-\mu (2\eta -\mu {\kappa }^2)}\in (0,1]$ for $\mu \in (0,\frac{2\eta }{{\kappa }^2})$. Suppose that $\{{\alpha }_n\},\{{\beta }_n\},\{{\gamma }_n\},\{{\sigma }_n\}\subset (0,1]$ and $\{{\delta }_n\}\subset (0,2\zeta ]$ are the sequences such that

(i) 

${\lim }_{n\to \infty }{\alpha }_n=0,{\sum }_{n=0}^{\infty }{\alpha }_n=\infty$ and ${\sum }_{n=0}^{\infty }|{\alpha }_{n+1}-{\alpha }_n|<\infty$.

(ii) 

${\lim }_{n\to \infty }\frac{{\theta }_n}{{\alpha }_n}=0,{\lim }_{n\to \infty }\frac{{\beta }_n}{{\alpha }_n}=0,{\sum }_{n=0}^{\infty }|{\beta }_{n+1}-{\beta }_n|<\infty$ and ${\sum }_{n=0}^{\infty }|{\delta }_{n+1}-{\delta }_n|<\infty$.

(iii) 

$0<{\lim \hspace{0.17em}\inf }_{n\to \infty }{\sigma }_n\le {\lim \hspace{0.17em}\sup }_{n\to \infty }{\sigma }_n<1$ and ${\sum }_{n=0}^{\infty }|{\sigma }_{n+1}-{\sigma }_n|<\infty$.

(iv) 

$0<{\lim \hspace{0.17em}\inf }_{n\to \infty }{\gamma }_n\le {\lim \hspace{0.17em}\sup }_{n\to \infty }{\gamma }_n<1$ and ${\sum }_{n=0}^{\infty }|{\gamma }_{n+1}-{\gamma }_n|<\infty$.

(v) 

${\delta }_n\le {\alpha }_n$ and ${\sum }_{n=0}^{\infty }\parallel T^{n+1}{y}_n-T^n{y}_n\parallel <\infty$.

Then, the sequence ${\{x_n\}}_{n=0}^{\infty }$ generated by Algorithm 1 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

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

(c) 

${\{x_n\}}_{n=0}^{\infty }$ converges to the unique solution of Problem 1 if $\frac{\parallel x_n-y_n\parallel }{{\delta }_n}\to 0$ as $n\to \infty$.

Proof. 

First, let us show that $P_{\mathrm{VI}(\mathsf{\Omega },A_1)}(I-\mu A_2)$ is a contractive mapping. Indeed, by Lemma 4, we have

$$\begin{array}{cc}(1-\tau )\parallel x-y\parallel \hfill & \ge \parallel (I-\mu A_2)x-(I-\mu A_2)y\parallel \hfill \\ & \ge \parallel P_{\mathrm{VI}(\mathsf{\Omega },A_1)}(I-\mu A_2)x-P_{\mathrm{VI}(\mathsf{\Omega },A_1)}(I-\mu A_2)y\parallel ,\hfill \end{array}$$

for any $x,y\in C$, which implies that $P_{\mathrm{VI}(\mathsf{\Omega },A_1)}(I-\mu A_2)$ is a contraction mapping. Banach’s Contraction Mapping Principle tell us that $P_{\mathrm{VI}(\mathsf{\Omega },A_1)}(I-\mu A_2)$ has a fixed point and further it is unique. For example, $x^{*}\in C$, that is, $x^{*}=P_{\mathrm{VI}(\mathsf{\Omega },A_1)}(I-\mu A_2)x^{*}$. Hence, by Lemma 8, we get

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

That is, Problem 1 has a unique solution. Taking into account that

$$0<\underset{n\to \infty }{\lim \hspace{0.17em}\inf }\hspace{0.17em}{\gamma }_n\le \underset{n\to \infty }{\lim \hspace{0.17em}\sup }\hspace{0.17em}{\gamma }_n<1,$$

we usually suppose $\{{\gamma }_n\}\subset [a,b]\subset (0,1)$ for some $a,b\in (0,1)$. Note that the mapping $G:C\to C$ is defined as $G:=P_C(I-{\mu }_1{B}_1)P_C(I-{\mu }_2{B}_2)$, where ${\mu }_1\in (0,2\alpha )$ and ${\mu }_2\in (0,2\beta )$. Thus, by Lemma 6, we know that G is nonexpansive. It is easy to see that there exists an element $u_n\in C$ such that

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

(11)

In fact, it is a unique element. Thus, we can consider the mapping

$$F_{n}x={\gamma }_n{x}_n+(1-{\gamma }_n)S_{n}x,\forall x\in C.$$

Since $S_n:C\to C$ is a continuous pseudocontraction mapping, we deduce that all $x,y\in C$,

$$\langle F_{n}x-F_{n}y,x-y\rangle =(1-{\gamma }_n)\langle S_{n}x-S_{n}y,x-y\rangle \le (1-{\gamma }_n){\parallel x-y\parallel }^2.$$

In addition, from $\{{\gamma }_n\}\subset [a,b]\subset (0,1)$ we get $0<1-{\gamma }_n<1$ for all $n\ge 0$. Thus, $F_n$ is a continuous and strong pseudocontraction mapping of C into itself. By Proposition 2, we know that there exists a unique element $u_n\in C$, for each $n\ge 0$, satisfying (11). Thus, it can be readily seen that the hybrid Mann viscosity implicit iterative scheme (10) can be rewritten as

$$\left\{\begin{array}{c} u_n={\gamma }_n{x}_n+(1-{\gamma }_n)S_n{u}_n,\hfill \\ z_n=G{u}_n,\hfill \\ y_n={\sigma }_n{x}_n+(1-{\sigma }_n)P_C(I-{\delta }_n{A}_1)z_n,\hfill \\ x_{n+1}=(1-{\beta }_n)P_C(I-{\alpha }_n\mu A_2)T^n{y}_n+{\beta }_{n}f\left(y_n\right),\forall n\ge 0.\hfill \end{array}\right.$$

(12)

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

Step 1. We claim that $\{x_n\},\{y_n\},\{z_n\},\{u_n\},\{v_n\},\{T^n{y}_n\}$ and $\{A_2\left(T^n{y}_n\right)\}$ are bounded. Indeed, 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 each $S_n:C\to C$ is a pseudocontraction mapping, it follows that

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

which hence yields

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

(13)

Then, we get

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

(14)

Since $1>{\lim \hspace{0.17em}\sup }_{n\to \infty }{\sigma }_n\ge {\lim \hspace{0.17em}\inf }_{n\to \infty }{\sigma }_n>0$, we reach $\{{\sigma }_n\}\subset [c,d]$ for some $c,d\in (0,1)$. In addition, since ${\lim }_{n\to \infty }\frac{{\theta }_n}{{\alpha }_n}=0$ and ${\lim }_{n\to \infty }\frac{{\beta }_n}{{\alpha }_n}=0$, we may assume, without loss of generality, that

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

and ${\beta }_n\le {\alpha }_n$ for all $n\ge 0$. Taking into account the $\zeta$-inverse-strong monotonicity of $A_1$ with $\{{\delta }_n\}\subset (0,2\zeta ]$, we deduce from Lemma 5 and (14) that

$$\begin{array}{cc}\parallel y_n-p\parallel \hfill & \le (1-{\sigma }_n)\parallel P_C(I-{\delta }_n{A}_1)z_n-p\parallel +{\sigma }_n\parallel p-x_n\parallel \hfill \\ & \le (1-{\sigma }_n)\parallel (I-{\delta }_n{A}_1)z_n-(I-{\delta }_n{A}_1)p-{\delta }_n{A}_{1}p\parallel +{\sigma }_n\parallel p-x_n\parallel \hfill \\ & \le (1-{\sigma }_n)(\parallel z_n-p\parallel +{\delta }_n\parallel A_{1}p\parallel )+{\sigma }_n\parallel p-x_n\parallel \hfill \\ & \le (1-{\sigma }_n)\parallel x_n-p\parallel +{\delta }_n\parallel A_{1}p\parallel +{\sigma }_n\parallel p-x_n\parallel \hfill \\ & =\parallel x_n-p\parallel +{\delta }_n\parallel A_{1}p\parallel .\hfill \end{array}$$

(15)

Utilizing Lemma 4 and (15), we obtain from (12) that

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

By induction, we have

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

It immediately follows that $\{x_n\}$ is bounded, and so are the sequences $\{y_n\},\{z_n\},\{u_n\},\{T^n{y}_n\}$ and $\{A_2\left(T^n{y}_n\right)\}$ (due to (13)–(15) and the Lipschitz continuity of T and $A_2$). Taking into account that $\{S_n\}$ 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 $\{S_n{u}_n\}$ is bounded. In addition, from Lemma 1 and $p\in \mathsf{\Omega }\subset \mathrm{GSVI}(C,B_1,B_2)$, it also follows that $(p,q)$ is a solution of GSVI (4) 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 obtain

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

This shows that $\{v_n\}$ is bounded.

Step 2. We claim 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 $p_n=P_C(I-{\delta }_n{A}_1)z_n$ and $q_n=P_C(I-{\alpha }_n\mu A_2)T^n{y}_n$. Then, from (12), we have

$$\left\{\begin{array}{c} u_n={\gamma }_n{x}_n+(1-{\gamma }_n)S_n{u}_n,\hfill \\ y_n={\sigma }_n{x}_n+(1-{\sigma }_n)p_n,\hfill \\ x_{n+1}={\beta }_{n}f\left(y_n\right)+(1-{\beta }_n)q_n.\hfill \end{array}\right.$$

Simple calculations show that

$$\left\{\begin{array}{c} u_n-u_{n-1}={\gamma }_n(x_n-x_{n-1})+({\gamma }_n-{\gamma }_{n-1})(x_{n-1}-S_{n-1}{u}_{n-1})+(1-{\gamma }_n)(S_n{u}_n-S_{n-1}{u}_{n-1}),\hfill \\ y_n-y_{n-1}={\sigma }_n(x_n-x_{n-1})+({\sigma }_n-{\sigma }_{n-1})(x_{n-1}-p_{n-1})+(1-{\sigma }_n)(p_n-p_{n-1}),\hfill \\ x_{n+1}-x_n={\beta }_n(f\left(y_n\right)-f\left(y_{n-1}\right))+({\beta }_n-{\beta }_{n-1})(f\left(y_{n-1}\right)-q_{n-1})+(1-{\beta }_n)(q_n-q_{n-1}).\hfill \end{array}\right.$$

(16)

It follows that

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

which hence yields

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

This immediately leads to

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

(17)

Putting $D=\{u_n:n\ge 0\}$, we know that D is a bounded subset of C. Then, by the assumption, we get ${\sum }_{n=1}^{\infty }{\sup }_{x\in D}\parallel S_{n}x-S_{n-1}x\parallel <\infty$. Noticing $\parallel S_n{u}_n-S_{n-1}{u}_n\parallel \le {\sup }_{x\in D}\parallel S_{n}x-S_{n-1}x\parallel \forall n\ge 1$, we have

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

(18)

In addition, from $p_n=P_C(I-{\delta }_n{A}_1)z_n$ and $\{{\delta }_n\}\subset (0,2\zeta ]$, we observe that

$$\begin{array}{cc}\parallel p_n-p_{n-1}\parallel \hfill & \le \parallel (I-{\delta }_n{A}_1)z_n-(I-{\delta }_{n-1}{A}_1)z_{n-1}\parallel \hfill \\ & =\parallel (I-{\delta }_n{A}_1)z_n-(I-{\delta }_n{A}_1)z_{n-1}-({\delta }_n-{\delta }_{n-1})A_1{z}_{n-1}\parallel \hfill \\ & \le \parallel (I-{\delta }_n{A}_1)z_n-(I-{\delta }_n{A}_1)z_{n-1}\parallel +|{\delta }_n-{\delta }_{n-1}|\parallel A_1{z}_{n-1}\parallel \hfill \\ & \le \parallel z_n-z_{n-1}\parallel +|{\delta }_n-{\delta }_{n-1}|\parallel A_1{z}_{n-1}\parallel \hfill \\ & \le \parallel u_n-u_{n-1}\parallel +M_0|{\delta }_n-{\delta }_{n-1}|,\hfill \end{array}$$

(19)

where ${\sup }_{n\ge 0}\parallel A_1{z}_n\parallel \le M_0$ for some $M_0>0$. Thus, from (16), (17) and (19), we get

$$\begin{array}{c}\parallel y_n-y_{n-1}\parallel \hfill \\ \le {\sigma }_n\parallel x_n-x_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel x_{n-1}-p_{n-1}\parallel +(1-{\sigma }_n)\parallel p_n-p_{n-1}\parallel \hfill \\ \le {\sigma }_n\parallel x_n-x_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel x_{n-1}-p_{n-1}\parallel +(1-{\sigma }_n)(\parallel u_n-u_{n-1}\parallel +M_0|{\delta }_n-{\delta }_{n-1}|)\hfill \\ \le {\sigma }_n\parallel x_n-x_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel x_{n-1}-p_{n-1}\parallel +(1-{\sigma }_n)(\parallel x_n-x_{n-1}\parallel \hfill \\ +\frac{1}{a}\parallel S_n{u}_n-S_{n-1}{u}_n\parallel +|{\gamma }_n-{\gamma }_{n-1}|\frac{\parallel x_{n-1}-S_{n-1}{u}_{n-1}\parallel }{a}+M_0|{\delta }_n-{\delta }_{n-1}|)\hfill \\ \le \parallel x_n-x_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel x_{n-1}-p_{n-1}\parallel +\frac{1}{a}\parallel S_n{u}_n-S_{n-1}{u}_n\parallel \hfill \\ +|{\gamma }_n-{\gamma }_{n-1}|\frac{\parallel x_{n-1}-S_{n-1}{u}_{n-1}\parallel }{a}+M_0|{\delta }_n-{\delta }_{n-1}|\hfill \\ \le \parallel x_n-x_{n-1}\parallel +M(|{\sigma }_n-{\sigma }_{n-1}|+\parallel S_n{u}_n-S_{n-1}{u}_n\parallel +|{\gamma }_n-{\gamma }_{n-1}|+|{\delta }_n-{\delta }_{n-1}|),\hfill \end{array}$$

(20)

where ${\sup }_{n\ge 0}\{\parallel x_n-p_n\parallel +\frac{1}{a}+\frac{\parallel x_n-S_n{u}_n\parallel }{a}+M_0\}\le M$ for some $M>0$.

Furthermore, from $q_n=P_C(I-{\alpha }_n\mu A_2)T^n{y}_n$ and Lemma 4, we note that

$$\begin{array}{c}\parallel q_n-q_{n-1}\parallel \hfill \\ \le \parallel (I-{\alpha }_n\mu A_2)T^n{y}_n-(I-{\alpha }_{n-1}\mu A_2)T^{n-1}{y}_{n-1}\parallel \hfill \\ =\parallel (I-{\alpha }_n\mu A_2)T^n{y}_n-(I-{\alpha }_n\mu A_2)T^{n-1}{y}_{n-1}-({\alpha }_n-{\alpha }_{n-1})\mu A_2{T}^{n-1}{y}_{n-1}\parallel \hfill \\ \le (1-{\alpha }_n\tau )\parallel T^n{y}_n-T^n{y}_{n-1}+T^n{y}_{n-1}-T^{n-1}{y}_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|\parallel \mu A_2{T}^{n-1}{y}_{n-1}\parallel \hfill \\ \le (1-{\alpha }_n\tau )[(1+{\theta }_n)\parallel y_n-y_{n-1}\parallel +\parallel T^n{y}_{n-1}-T^{n-1}{y}_{n-1}\parallel ]+|{\alpha }_n-{\alpha }_{n-1}|\parallel \mu A_2{T}^{n-1}{y}_{n-1}\parallel \hfill \\ \le (1-{\alpha }_n\tau +{\theta }_n)\parallel y_n-y_{n-1}\parallel +\parallel T^n{y}_{n-1}-T^{n-1}{y}_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|\parallel \mu A_2{T}^{n-1}{y}_{n-1}\parallel .\hfill \end{array}$$

(21)

Hence, from (16), (20) and (21), we get

$$\begin{array}{c}\parallel x_{n+1}-x_n\parallel \hfill \\ \le {\beta }_n\parallel f\left(y_n\right)-f\left(y_{n-1}\right)\parallel +|{\beta }_n-{\beta }_{n-1}|\parallel f\left(y_{n-1}\right)-q_{n-1}\parallel +(1-{\beta }_n)\parallel q_n-q_{n-1}\parallel \hfill \\ \le {\alpha }_n\delta \parallel y_n-y_{n-1}\parallel +|{\beta }_n-{\beta }_{n-1}|\parallel f\left(y_{n-1}\right)-q_{n-1}\parallel +(1-{\alpha }_n\tau +{\theta }_n)\parallel y_n-y_{n-1}\parallel \hfill \\ +\parallel T^n{y}_{n-1}-T^{n-1}{y}_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|\parallel \mu A_2{T}^{n-1}{y}_{n-1}\parallel \hfill \\ \le [1-{\alpha }_n(\tau -\delta )+\frac{{\alpha }_n(\tau -\delta )}{2}]\parallel y_n-y_{n-1}\parallel +|{\beta }_n-{\beta }_{n-1}|\parallel f\left(y_{n-1}\right)-q_{n-1}\parallel \hfill \\ +\parallel T^n{y}_{n-1}-T^{n-1}{y}_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|\parallel \mu A_2{T}^{n-1}{y}_{n-1}\parallel \hfill \\ \le (1-\frac{{\alpha }_n(\tau -\delta )}{2})[\parallel x_n-x_{n-1}\parallel +M(|{\sigma }_n-{\sigma }_{n-1}|+\parallel S_n{u}_n-S_{n-1}{u}_n\parallel \hfill \\ +|{\gamma }_n-{\gamma }_{n-1}|+|{\delta }_n-{\delta }_{n-1}|)]+|{\beta }_n-{\beta }_{n-1}|\parallel f\left(y_{n-1}\right)-q_{n-1}\parallel \hfill \\ +\parallel T^n{y}_{n-1}-T^{n-1}{y}_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|\parallel \mu A_2{T}^{n-1}{y}_{n-1}\parallel \hfill \\ \le (1-\frac{{\alpha }_n(\tau -\delta )}{2})\parallel x_n-x_{n-1}\parallel +M_1(|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|+|{\gamma }_n-{\gamma }_{n-1}|\hfill \\ +|{\delta }_n-{\delta }_{n-1}|+|{\sigma }_n-{\sigma }_{n-1}|+\parallel S_n{u}_n-S_{n-1}{u}_n\parallel )+\parallel T^n{y}_{n-1}-T^{n-1}{y}_{n-1}\parallel ,\hfill \end{array}$$

(22)

where ${\sup }_{n\ge 1}\{M+\parallel f\left(y_{n-1}\right)-q_{n-1}\parallel +\parallel \mu A_2{T}^{n-1}{y}_{n-1}\parallel \}\le M_1$ for some $M_1>0$. From (18) and Conditions (i)–(v), we know that ${\sum }_{n=0}^{\infty }\frac{{\alpha }_n(\tau -\delta )}{2}=\infty$ and

$$\begin{array}{c}{\displaystyle \sum _{n=1}^{\infty }}\{M_1(|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|+|{\gamma }_n-{\gamma }_{n-1}|+|{\delta }_n-{\delta }_{n-1}|\hfill \\ +|{\sigma }_n-{\sigma }_{n-1}|+\parallel S_n{u}_n-S_{n-1}{u}_n\parallel )+\parallel T^n{y}_{n-1}-T^{n-1}{y}_{n-1}\parallel \}<\infty .\hfill \end{array}$$

Consequently, applying Lemma 3 to (22), we obtain that

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

(23)

In terms of (18), (23) and Conditions (ii)–(iv), we deduce from (20) that

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

Step 3. We claim that $\parallel x_n-G{x}_n\parallel \to 0$ as $n\to \infty$. Indeed, noticing $q_n=P_C(I-{\alpha }_n\mu A_2)T^n{y}_n$ for all $n\ge 0$, we obtain from (7) that for each $p\in \mathsf{\Omega }$,

$$\langle (I-{\alpha }_n\mu A_2)T^n{y}_n-P_C(I-{\alpha }_n\mu A_2)T^n{y}_n,p-q_n\rangle \le 0,$$

which hence leads to

$$\begin{array}{cc}\parallel q_n{-p\parallel }^2\hfill & =\langle P_C(I-{\alpha }_n\mu A_2)T^n{y}_n-(I-{\alpha }_n\mu A_2)T^n{y}_n,q_n-p\rangle \hfill \\ & +\langle (I-{\alpha }_n\mu A_2)T^n{y}_n-p,q_n-p\rangle \hfill \\ & \le \langle (I-{\alpha }_n\mu A_2)T^n{y}_n-p,q_n-p\rangle \hfill \\ & =\langle (I-{\alpha }_n\mu A_2)T^n{y}_n-(I-{\alpha }_n\mu A_2)p,q_n-p\rangle -{\alpha }_n\langle \mu A_{2}p,q_n-p\rangle \hfill \\ & \le \frac{1}{2}{(1-{\alpha }_n\tau )}^2\parallel T^n{y}_n{-p\parallel }^2+\frac{1}{2}{\parallel q_n-p\parallel }^2-{\alpha }_n\langle \mu A_{2}p,q_n-p\rangle .\hfill \end{array}$$

It follows from (1) that

$$\begin{array}{cc}\parallel q_n{-p\parallel }^2\hfill & \le (1-{\alpha }_n\tau ){(1+{\theta }_n)}^2{\parallel y_n-p\parallel }^2-2{\alpha }_n\langle \mu A_{2}p,q_n-p\rangle \hfill \\ & \le (1-{\alpha }_n\tau )\parallel y_n{-p\parallel }^2+{\theta }_n(2+{\theta }_n){\parallel y_n-p\parallel }^2-2{\alpha }_n\langle \mu A_{2}p,q_n-p\rangle .\hfill \end{array}$$

(24)

From (15) and (24), we get

$$\begin{array}{c}\parallel x_{n+1}{-p\parallel }^2\hfill \\ \le {\beta }_n\parallel f\left(y_n\right)-f\left(p\right){\parallel }^2+(1-{\beta }_n){\parallel q_n-p\parallel }^2+2{\beta }_n\langle f\left(p\right)-p,x_{n+1}-p\rangle \hfill \\ \le {\alpha }_n\delta \parallel y_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 \mu A_{2}p,q_n-p\rangle +2{\beta }_n\langle f\left(p\right)-p,x_{n+1}-p\rangle \hfill \\ =[1-{\alpha }_n(\tau -\delta )]\parallel y_n{-p\parallel }^2+{\theta }_n(2+{\theta }_n){\parallel y_n-p\parallel }^2-2{\alpha }_n\langle \mu A_{2}p,q_n-p\rangle \hfill \\ +2{\beta }_n\langle f\left(p\right)-p,x_{n+1}-p\rangle \hfill \\ \le [1-{\alpha }_n(\tau -\delta )][{\sigma }_n\parallel x_n-p\parallel +(1-{\sigma }_n)(\parallel z_n-p\parallel +{\delta }_n\parallel A_1{p\parallel \left)\right]}^2\hfill \\ +{\theta }_n(2+{\theta }_n){\parallel y_n-p\parallel }^2-2{\alpha }_n\langle \mu A_{2}p,q_n-p\rangle +2{\beta }_n\langle f\left(p\right)-p,x_{n+1}-p\rangle \hfill \\ \le [1-{\alpha }_n(\tau -\delta )][{\sigma }_n\parallel x_n{-p\parallel }^2+(1-{\sigma }_n)(\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 \mu A_{2}p,q_n-p\rangle +2{\beta }_n\langle f\left(p\right)-p,x_{n+1}-p\rangle \hfill \\ \le [1-{\alpha }_n(\tau -\delta )][{\sigma }_n\parallel x_n{-p\parallel }^2+(1-{\sigma }_n)\parallel z_n{-p\parallel }^2+{\delta }_n\parallel A_{1}p\parallel (2\parallel z_n-p\parallel +{\delta }_n\parallel A_{1}p\parallel \left)\right]\hfill \\ +{\theta }_n(2+{\theta }_n){\parallel y_n-p\parallel }^2-2{\alpha }_n\langle \mu A_{2}p,q_n-p\rangle +2{\beta }_n\langle f\left(p\right)-p,x_{n+1}-p\rangle \hfill \\ \le [1-{\alpha }_n(\tau -\delta )][{\sigma }_n\parallel x_n{-p\parallel }^2+(1-{\sigma }_n)\parallel z_n{-p\parallel }^2]+{\alpha }_n\parallel A_{1}p\parallel (2\parallel z_n-p\parallel +{\alpha }_n\parallel A_{1}p\parallel )\hfill \\ +{\theta }_n(2+{\theta }_n)\parallel y_n{-p\parallel }^2+2{\alpha }_n(\parallel \mu A_{2}p\parallel \parallel q_n-p\parallel +\parallel f\left(p\right)-p\parallel \parallel x_{n+1}-p\parallel ).\hfill \end{array}$$

(25)

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

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

(26)

and

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

(27)

Substituting (26) for (27), we obtain from (13) that

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

(28)

Combining (25) and (28), we get

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

which immediately yields

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

Since ${\lim \hspace{0.17em}\inf }_{n\to \infty }(1-{\sigma }_n)>0$ (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 (23) that

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

(29)

Additionally, from (6) and (9), 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

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

(30)

In the same way, we derive

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

which implies that

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

(31)

Substituting (3) for (31), we deduce from (13) that

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

(32)

Combining (25) and (32), we have

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

which hence yields

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

Since ${\lim \hspace{0.17em}\inf }_{n\to \infty }(1-{\sigma }_n)>0$ (due to Condition (iii)), ${\lim }_{n\to \infty }{\theta }_n=0$ and ${\lim }_{n\to \infty }{\alpha }_n=0$, we conclude from (23) and (29) that

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

(33)

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,

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

(34)

In addition, according to (12), we have

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

which, together with (9), yields

$$\begin{array}{cc}\parallel p-u_n{\parallel }^2\hfill & \le \langle x_n-p,u_n-p\rangle \hfill \\ & =\frac{1}{2}[\parallel x_n{-p\parallel }^2+\parallel u_n{-p\parallel }^2-\parallel x_n-u_n{\parallel }^2].\hfill \end{array}$$

This immediately implies that

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

which together with (14) and (26), yields

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

Hence, we have

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

Since ${\lim \hspace{0.17em}\inf }_{n\to \infty }(1-{\sigma }_n)>0$ (due to Condition (iii)), ${\lim }_{n\to \infty }{\theta }_n=0$ and ${\lim }_{n\to \infty }{\alpha }_n=0$, we obtain from (23) that

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

(35)

In addition, observe that

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

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

and

$$\parallel x_n-y_n\parallel \le (1-{\sigma }_n)\parallel x_n-(I-{\delta }_n{A}_1)z_n\parallel \le \parallel x_n-z_n\parallel +{\alpha }_n\parallel A_1{z}_n\parallel .$$

Then, from (34) and (35), it follows that

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

(36)

Step 4. We claim that $\parallel x_n-S_n{x}_n\parallel \to 0,\parallel x_n-q_n\parallel \to 0$ and $\parallel x_n-T{x}_n\parallel \to 0$ as $n\to \infty$. Indeed, combining (11) and (25), 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,

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

(37)

Since ${\{S_n\}}_{n=0}^{\infty }$ is ℓ-uniformly Lipschitzian on C, we deduce from (35) and (37) 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,

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

(38)

In addition, we observe that

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

Hence, we get

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

Consequently, from (23), (36) and ${\lim }_{n\to \infty }{\alpha }_n=0$, we obtain that

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

(39)

Thus, it follows that

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

That is,

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

(40)

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 Condition (v) and (39), we get

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

In addition, noticing that

$$\begin{array}{cc}\parallel x_n-T{x}_n\parallel \hfill & \le \parallel x_n-y_n\parallel +\parallel y_n-T{y}_n\parallel +\parallel T{y}_n-T{x}_n\parallel \hfill \\ & \le \parallel y_n-T{y}_n\parallel +(2+{\theta }_1)\parallel x_n-y_n\parallel ,\hfill \end{array}$$

we deduce from (36) that

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

(41)

Step 5. We claim that $\parallel x_n-\overline{S}{x}_n\parallel \to 0$ as $n\to \infty$ where $\overline{S}:={(2I-S)}^{-1}$. Indeed, first, let us 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 pseudocontractive, 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 means that S is pseudocontractive. Noting that ${\{S_n\}}_{n=0}^{\infty }$ is ℓ-uniformly Lipschitzian on C, we have

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

This means that S is ℓ-Lipschitzian. Taking into account the boundedness of $\{x_n\}$ and putting $D=\overline{\mathrm{conv}}\{x_n:n\ge 0\}$ (the closure of convex hull of the set $\{x_n:n\ge 0\}$), by Assumption (C6) 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

$$\underset{n\to \infty }{\lim }\underset{x\in D}{\sup }\parallel S_{n}x-Sx\parallel =0,$$

which immediately yields

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

(42)

Thus, combining (38) with (42), 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,

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

(43)

Now, let us show that. if we define $\overline{S}:={(2I-S)}^{-1}$, then $\overline{S}:C\to C$ 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, it is known that $\overline{S}$ is nonexpansive and the fixed point set $\mathrm{Fix}\left(\overline{S}\right)=\mathrm{Fix}\left(S\right)={\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)$. From (43), it follows that

$$\begin{array}{cc}\parallel x_n-\overline{S}{x}_n\parallel \hfill & =\parallel \overline{S}{\overline{S}}^{-1}{x}_n-\overline{S}{x}_n\parallel \hfill \\ & \le \parallel {\overline{S}}^{-1}{x}_n-x_n\parallel \hfill \\ & =\parallel (2I-S)x_n-x_n\parallel =\parallel x_n-S{x}_n\parallel \to 0(n\to \infty ).\hfill \end{array}$$

That is,

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

(44)

Step 6. We claim that

$$\underset{n\to \infty }{\lim \hspace{0.17em}\sup }\langle A_2{x}^{*},x^{*}-q_n\rangle \le 0\mathrm{and}\underset{n\to \infty }{\lim \hspace{0.17em}\sup }\langle A_1{x}^{*},x^{*}-z_n\rangle \le 0,$$

(45)

where $\{x^{*}\}=\mathrm{VI}(\mathrm{VI}(\mathsf{\Omega },A_1),A_2)$. Indeed, we fix sequence $\{q_{n_i}\}$ of $\{q_n\}$ such that

$$\underset{n\to \infty }{\lim \hspace{0.17em}\sup }\langle A_2{x}^{*},x^{*}-q_n\rangle =\underset{i\to \infty }{\lim }\langle A_2{x}^{*},x^{*}-q_{n_i}\rangle .$$

Since $\{q_n\}$ is a bounded sequence in C, we may assume, without loss of generality, that $q_{n_i}\rightharpoonup \overline{x}\in C$. Since ${\lim }_{n\to \infty }\parallel x_n-q_n\parallel =0$ (due to (40)), it follows from $q_{n_i}\rightharpoonup \overline{x}$ that $x_{n_i}\rightharpoonup \overline{x}$.

Note that G and $\overline{S}$ are nonexpansive and that T is asymptotically nonexpansive. Since $(I-G)x_n\to 0,(I-T)x_n\to 0$ and $(I-\overline{S})x_n\to 0$ (due to (36), (41) and (44)), 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 claim that $\overline{x}\in \mathrm{VI}(\mathsf{\Omega },A_1)$. In fact, let $y\in \mathsf{\Omega }$ be fixed arbitrarily. Then, it follows from (12), (14) and the $\zeta$-inverse-strong monotonicity of $A_1$ that

$$\begin{array}{cc}\parallel y_n{-y\parallel }^2\hfill & \le {\sigma }_n\parallel x_n{-y\parallel }^2+(1-{\sigma }_n){\parallel P_C(z_n-{\delta }_n{A}_1{z}_n)-P_{C}y\parallel }^2\hfill \\ & \le {\sigma }_n\parallel x_n{-y\parallel }^2+(1-{\sigma }_n){\parallel (z_n-y)-{\delta }_n{A}_1{z}_n\parallel }^2\hfill \\ & ={\sigma }_n\parallel x_n{-y\parallel }^2+(1-{\sigma }_n)[\parallel z_n{-y\parallel }^2+2{\delta }_n\langle A_1{z}_n,y-z_n\rangle +{\delta }_n^2\parallel A_1{z}_n{\parallel }^2]\hfill \\ & \le \parallel x_n{-y\parallel }^2+(1-{\sigma }_n)[2{\delta }_n\langle A_{1}y,y-z_n\rangle +{\delta }_n^2\parallel A_1{z}_n{\parallel }^2],\hfill \end{array}$$

which together with $\{{\sigma }_n\}\subset [c,d]$, implies that for all $n\ge 0$,

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

From (36) it is easy to see that $x_{n_i}\rightharpoonup \overline{x}$ leads to $z_{n_i}\rightharpoonup \overline{x}$. Since ${\lim }_{n\to \infty }{\delta }_n=0$ and $\parallel x_n-y_n\parallel =o\left({\delta }_n\right)$ (due to the assumption), we have

$$\begin{array}{cc}0\hfill & \le {\displaystyle \underset{n\to \infty }{\lim \hspace{0.17em}\inf }}\{(\parallel x_n-y\parallel +\parallel y_n-y\parallel )\frac{\parallel x_n-y_n\parallel }{{\delta }_n(1-d)}+2\langle A_{1}y,y-z_n\rangle +\frac{{\delta }_n}{1-d}\parallel A_1{z}_n{\parallel }^2\}\hfill \\ & ={\displaystyle \underset{n\to \infty }{\lim \hspace{0.17em}\inf }}2\langle A_{1}y,y-z_n\rangle \le {\displaystyle \underset{i\to \infty }{\lim }}2\langle A_{1}y,y-z_{n_i}\rangle =2\langle A_{1}y,y-\overline{x}\rangle .\hfill \end{array}$$

It follows that

$$\langle A_{1}y,y-\overline{x}\rangle \ge 0,\forall y\in \mathsf{\Omega }.$$

Accordingly, Lemma 8 and the $\zeta$-inverse-strong monotonicity of $A_1$ ensure that

$$\langle A_1\overline{x},y-\overline{x}\rangle \ge 0,\forall y\in \mathsf{\Omega };$$

that is, $\overline{x}\in \mathrm{VI}(\mathsf{\Omega },A_1)$. Consequently, from $\{x^{*}\}=\mathrm{VI}(\mathrm{VI}(\mathsf{\Omega },A_1),A_2)$, we have

$$\underset{n\to \infty }{\lim \hspace{0.17em}\sup }\langle A_2{x}^{*},x^{*}-q_n\rangle =\underset{i\to \infty }{\lim }\langle A_2{x}^{*},x^{*}-q_{n_i}\rangle =\langle A_2{x}^{*},x^{*}-\overline{x}\rangle \le 0.$$

On the other hand, we choose a subsequence $\{z_{n_k}\}$ of $\{z_n\}$ such that

$$\underset{n\to \infty }{\lim \hspace{0.17em}\sup }\langle A_1{x}^{*},x^{*}-z_n\rangle =\underset{k\to \infty }{\lim }\langle A_1{x}^{*},x^{*}-z_{n_k}\rangle .$$

Since $\{z_n\}$ is a bounded sequence in C, we may assume, without loss of generality, that $z_{n_k}\rightharpoonup \widehat{x}\in C$. From (36), it is easy to see that $z_{n_k}\rightharpoonup \widehat{x}$ yields $x_{n_k}\rightharpoonup \widehat{x}$. By the same arguments as in the proof of $\overline{x}\in \mathsf{\Omega }$, we have $\widehat{x}\in \mathsf{\Omega }$. From $x^{*}\in \mathrm{VI}(\mathsf{\Omega },A_1)$, we get

$$\underset{n\to \infty }{\lim \hspace{0.17em}\sup }\langle A_1{x}^{*},x^{*}-z_n\rangle =\underset{k\to \infty }{\lim }\langle A_1{x}^{*},x^{*}-z_{n_k}\rangle =\langle A_1{x}^{*},x^{*}-\widehat{x}\rangle \le 0.$$

(46)

Therefore, the inequalities in (45) hold.

Step 7. We claim that $x_n\to x^{*}$ as $n\to \infty$. Indeed, putting $p=x^{*}$ in (14) and at Lines 5–6 in (25), we obtain that $\parallel z_n-x^{*}\parallel \le \parallel x_n-x^{*}\parallel$ and

$$\begin{array}{cc}\parallel x_{n+1}-x^{*}{\parallel }^2\hfill & \le [1-{\alpha }_n(\tau -\delta )]\parallel y_n-x^{*}{\parallel }^2+{\theta }_n(2+{\theta }_n){\parallel y_n-x^{*}\parallel }^2\hfill \\ & -2{\alpha }_n\langle \mu A_2{x}^{*},q_n-x^{*}\rangle +2{\beta }_n\langle f\left(x^{*}\right)-x^{*},x_{n+1}-x^{*}\rangle .\hfill \end{array}$$

(47)

From (12) and the $\zeta$-inverse-strong monotonicity of $A_1$, it follows that

$$\begin{array}{cc}\parallel y_n-x^{*}{\parallel }^2\hfill & \le {\sigma }_n\parallel x_n-x^{*}{\parallel }^2+(1-{\sigma }_n){\parallel P_C(z_n-{\delta }_n{A}_1{z}_n)-x^{*}\parallel }^2\hfill \\ & \le {\sigma }_n\parallel x_n-x^{*}{\parallel }^2+(1-{\sigma }_n){\parallel (z_n-x^{*})-{\delta }_n{A}_1{z}_n\parallel }^2\hfill \\ & ={\sigma }_n\parallel x_n-x^{*}{\parallel }^2+(1-{\sigma }_n)[\parallel z_n-x^{*}{\parallel }^2+2{\delta }_n\langle A_1{z}_n,x^{*}-z_n\rangle +{\delta }_n^2\parallel A_1{z}_n{\parallel }^2]\hfill \\ & \le {\sigma }_n\parallel x_n-x^{*}{\parallel }^2+(1-{\sigma }_n)[\parallel x_n-x^{*}{\parallel }^2+2{\delta }_n\langle A_1{x}^{*},x^{*}-z_n\rangle +{\delta }_n^2\parallel A_1{z}_n{\parallel }^2]\hfill \\ & =\parallel x_n-x^{*}{\parallel }^2+(1-{\sigma }_n)[2{\delta }_n\langle A_1{x}^{*},x^{*}-z_n\rangle +{\delta }_n^2\parallel A_1{z}_n{\parallel }^2].\hfill \end{array}$$

(48)

Thus, in terms of (47) and (48), we get

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

(49)

It can be readily seen that (45) guarantees that

$$\underset{n\to \infty }{\lim \hspace{0.17em}\sup }\frac{(1-{\alpha }_n(\tau -\delta ))(1-{\sigma }_n)2{\delta }_n}{{\alpha }_n(\tau -\delta )}\langle A_1{x}^{*},x^{*}-z_n\rangle \le 0$$

and

$$\underset{n\to \infty }{\lim \hspace{0.17em}\sup }\frac{2}{\tau -\delta }\langle \mu A_2{x}^{*},x^{*}-q_n\rangle \le 0.$$

In fact, from ${\lim \hspace{0.17em}\sup }_{n\to \infty }\langle A_1{x}^{*},x^{*}-z_n\rangle \le 0$, 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

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

which hence yields

$$\underset{n\to \infty }{\lim \hspace{0.17em}\sup }\frac{(1-{\alpha }_n(\tau -\delta ))(1-{\sigma }_n)2{\delta }_n}{{\alpha }_n(\tau -\delta )}\langle A_1{x}^{*},x^{*}-z_n\rangle \le \frac{2}{\tau -\delta }\epsilon .$$

Letting $\epsilon \to 0$, we get ${\lim \hspace{0.17em}\sup }_{n\to \infty }\frac{(1-{\alpha }_n(\tau -\delta ))(1-{\sigma }_n)2{\delta }_n}{{\alpha }_n(\tau -\delta )}\langle A_1{x}^{*},x^{*}-z_n\rangle \le 0$.

Since ${\sum }_{n=0}^{\infty }{\alpha }_n=\infty ,{\lim }_{n\to \infty }\frac{{\theta }_n}{{\alpha }_n}=0$ and ${\lim }_{n\to \infty }\frac{{\beta }_n}{{\alpha }_n}=0$ (due to Conditions (i) and (ii)), we deduce that ${\sum }_{n=0}^{\infty }{\alpha }_n(\tau -\delta )=\infty$ and

$$\begin{array}{c}{\displaystyle \underset{n\to \infty }{\lim \hspace{0.17em}\sup }}\{\frac{(1-{\alpha }_n(\tau -\delta ))(1-{\sigma }_n)2{\delta }_n}{{\alpha }_n(\tau -\delta )}\langle A_1{x}^{*},x^{*}-z_n\rangle +\frac{{\alpha }_n{\parallel A_1{z}_n\parallel }^2}{\tau -\delta }\hfill \\ +\frac{{\theta }_n(2+{\theta }_n){\parallel y_n-x^{*}\parallel }^2}{{\alpha }_n(\tau -\delta )}+\frac{2}{\tau -\delta }\langle \mu A_2{x}^{*},x^{*}-q_n\rangle +\frac{2{\beta }_n\parallel f\left(x^{*}\right)-x^{*}\parallel \parallel x_{n+1}-x^{*}\parallel }{{\alpha }_n(\tau -\delta )}\}\le 0.\hfill \end{array}$$

We can apply Lemma 3 to the relation (49) and conclude that $x_n\to x^{*}$ as $n\to \infty$. This completes the proof. □

The following results can be obtained by Theorem 1 easily, and hence we omit the details.

Corollary 1.

We suppose C is a convex nonempty closed set of a real Hilbert space H and $f:C\to C$ is a contraction with the parameter $\delta \in [0,1)$. Let $A_1$ be a ζ-inverse-strongly monotone nonself mapping on C and $A_2$ be a strongly positive bounded linear self operator one H with the parameter $\gamma >0$, where $\delta <\tau :=1-\sqrt{1-\mu (2\gamma -\mu \parallel A_2{\parallel }^2)}\in (0,1]$, $0<\mu <\frac{2\gamma }{\parallel A_2{\parallel }^2}$. Let the mappings $B_1,B_2:C\to H$ be α-inverse-strongly monotone and β-inverse-strongly monotone, respectively. Let T be an asymptotically nonexpansive self mapping on set C with a sequence $\{{\theta }_n\}$. Let ${\{S_n\}}_{n=0}^{\infty }$ be a countable family of ℓ-uniformly Lipschitzian pseudocontractive self-mappings on C satisfying the assumptions in Problem 1. For any given $x_0\in C$, we suppose $\{x_n\}$ is a vector sequence through

$$\left\{\begin{array}{c} u_n=\frac{x_n+S_n{u}_n}{2},\hfill \\ v_n=P_C(u_n-{\mu }_2{B}_2{u}_n),\hfill \\ z_n=P_C(v_n-{\mu }_1{B}_1{v}_n),\hfill \\ y_n={\sigma }_n{x}_n+(1-{\sigma }_n)P_C(I-{\delta }_n{A}_1)z_n,\hfill \\ x_{n+1}={\beta }_{n}f\left(y_n\right)+(1-{\beta }_n)P_C(I-{\alpha }_n\mu A_2)T^n{y}_n,\forall n\ge 0,\hfill \end{array}\right.$$

(50)

where ${\mu }_1\in (0,2\alpha )$ and ${\mu }_2\in (0,2\beta )$. Suppose that $\{{\alpha }_n\},\{{\beta }_n\},\{{\sigma }_n\}\subset (0,1]$ and $\{{\delta }_n\}\subset (0,2\zeta ]$ are the sequences as in Theorem 1. Then, the sequence ${\{x_n\}}_{n=0}^{\infty }$ generated by (50) 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

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

(c) 

${\{x_n\}}_{n=0}^{\infty }$ reaches to the unique solution of Problem 1 if $\frac{\parallel x_n-y_n\parallel }{{\delta }_n}\to 0$ as $n\to \infty$.

Proof. 

Since the linear bounded operator $A_2:H\to H$ is positive and strong with the parameter $\gamma >0$, we know that $A_2$ is $\kappa$-Lipschitzian and $\eta$-strongly monotone where $\kappa =\parallel A_2\parallel$ and $\eta =\gamma$. In this case, we obtain that $0<\mu <\frac{2\eta }{{\kappa }^2}=\frac{2\gamma }{\parallel A_2{\parallel }^2}$, and

$$\delta <\tau :=1-\sqrt{1-\mu (2\eta -\mu {\kappa }^2)}=1-\sqrt{1-\mu (2\gamma -\mu \parallel A_2{\parallel }^2)}\in (0,1].$$

Therefore, utilizing Theorem 1, we derive the desired result. □

Corollary 2.

We suppose C is a convex nonempty closed set of a real Hilbert space H. Let $f:C\to C$ be a contraction with the parameter $\delta \in [0,1)$. Let $A_1:C\to H$ be a ζ-inverse-strongly monotone mapping and $A_2:C\to H$ be κ-Lipschitzian and η-strongly monotone with the parameters $\kappa ,\eta >0$, where $\delta <\tau :=1-\sqrt{1-\mu (2\eta -\mu {\kappa }^2)}\in (0,1]$, $0<\mu <\frac{2\eta }{{\kappa }^2}$. We suppose the nonself mappings $B_1,B_2:C\to H$ are α-inverse-strongly monotone and β-inverse-strongly monotone, respectively. Let $T:C\to C$ be a nonexpansive mapping and ${\{S_n\}}_{n=0}^{\infty }$ be a countable family of ℓ-uniformly Lipschitzian pseudocontractive self-mappings on C satisfying the assumptions in Problem 1. For any given $x_0\in C$, let $\{x_n\}$ be the sequence generated by

$$\left\{\begin{array}{c} u_n={\gamma }_n{x}_n+(1-{\gamma }_n)S_n{u}_n,\hfill \\ v_n=P_C(u_n-{\mu }_2{B}_2{u}_n),\hfill \\ z_n=P_C(v_n-{\mu }_1{B}_1{v}_n),\hfill \\ y_n={\sigma }_n{x}_n+(1-{\sigma }_n)P_C(I-{\delta }_n{A}_1)z_n,\hfill \\ x_{n+1}={\beta }_{n}f\left(y_n\right)+(1-{\beta }_n)P_C(I-{\alpha }_n\mu A_2)T{y}_n,\forall n\ge 0,\hfill \end{array}\right.$$

(51)

where ${\mu }_1\in (0,2\alpha )$ and ${\mu }_2\in (0,2\beta )$. Suppose that $\{{\alpha }_n\},\{{\beta }_n\},\{{\gamma }_n\},\{{\sigma }_n\}\subset (0,1]$ and $\{{\delta }_n\}\subset (0,2\zeta ]$ are the sequences as in Theorem 1. Then, the sequence ${\{x_n\}}_{n=0}^{\infty }$ generated by (51) 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 ${\lim }_{n\to \infty }\parallel x_n-S{x}_n\parallel =0$.

(c) 

${\{x_n\}}_{n=0}^{\infty }$ reaches to the unique solution of Problem 1 if $\frac{\parallel x_n-y_n\parallel }{{\delta }_n}\to 0$ as $n\to \infty$.

Proof. 

Since T is a nonexpansive self mapping defined on set C, T is, of course, an asymptotically nonexpansive mapping with the parameter sequence $\{{\theta }_n\}$, where ${\theta }_n=0$$\forall n\ge 0$. Therefore, utilizing the similar argument process to that of Theorem 1, we obtain the desired result. □

4. Applications to Finite Generalized Mixed Equilibria

We suppose set C is convex nonempty closed and a mapping T with fixed points is named as a attracting nonexpansive mapping if it is nonexpansive and satisfies:

$$\parallel Tx-p\parallel <\parallel x-p\parallel \mathrm{for}\mathrm{all}x\notin \mathrm{Fix}\left(T\right)\mathrm{and}p\in \mathrm{Fix}\left(T\right).$$

Lemma 9 

([27]). Let X be a strictly convex space, $T_1$ be an attracting nonexpansive mapping and $T_2$ be a nonexpansive mapping. We suppose they have common fixed points. Then, $\mathrm{Fix}\left(T_1{T}_2\right)=\mathrm{Fix}\left(T_2{T}_1\right)=\mathrm{Fix}\left(T_1\right)\cap \mathrm{Fix}\left(T_2\right)$.

Let $A:C\to H$ be nonself mapping, $\phi :C\to \mathbf{R}$ be a single-valued real function, and $\mathsf{\Theta }:C\times C\to \mathbf{R}$ be a bifunction to R. The mixed generalized equilibrium problem (MGEP) is to find $x\in C$ such that

$$\mathsf{\Theta }(x,y)+\langle Ax,y-x\rangle \ge 0+\phi \left(y\right)-\phi \left(x\right),\forall y\in C.$$

(52)

We borrow the collection of solutions of MGEP (52) by MGEP($\mathsf{\Theta },\phi ,A$). The GMEP (52) is quite useful in the sense that it includes many problems, namely, vector optimization problems, minimax problems, classical variational inequalities, Nash equilibrium problems in noncooperative games and others. For different aspects and solution methods, we refer to [28,29,30,31,32,33,34,35,36,37,38] and the references therein.

In particular, if $\phi =0$, then MGEP (52) become the generalized equilibrium problem (GEP) of finding $x\in C$ such that

$$\mathsf{\Theta }(x,y)+\langle Ax,y-x\rangle \ge 0,\forall y\in C.$$

(53)

The collection of solutions of GEP is used by GEP($\mathsf{\Theta },A$).

If $A=0$, then MGEP (52) become the mixed equilibrium problem (MEP). which is to find $x\in C$ such that

$$\mathsf{\Theta }(x,y)+\phi \left(y\right)-\phi \left(x\right)\ge 0,\forall y\in C.$$

The collection of solutions of MEP is used by MEP($\mathsf{\Theta },\phi$).

If $\phi =0$ and $A=0$, then MGEP (52) become to the equilibrium problem (EP) (see Blum and Oettli [30]), which will approximate $x\in C$ with

$$\mathsf{\Theta }(x,y)\ge 0,\forall y\in C.$$

The collection of solutions of EP is used by EP(Θ).

Here, we list some elementary conclusions for the MEP.

It is first used in [38] that $\mathsf{\Theta }:C\times C\to \mathbf{R}$ is a bifunction and $\phi :C\to \mathbf{R}$ is a convex lower semicontinuous function restricted to the following items

(A1) 

$\forall x\in C$, $\mathsf{\Theta }(x,x)\equiv 0$.

(A2) 

Θ has the monotonicity, i.e., $\forall x,y\in C$, $\mathsf{\Theta }(x,y)+\mathsf{\Theta }(y,x)\le 0$.

(A3) 

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

(A4) 

$\forall x\in C$, $\mathsf{\Theta }(x,·)$ is lower semicontinuous convex.

(B1) 

$\forall x\in H$ and $\forall r>0$, we fix a set $D_x\subset C$ and $y_x\in C$ with

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

$\forall z\in C\backslash D_x$.

(B2) 

C acts as a bounded set.

Lemma 10 

([38]). We suppose that $\mathsf{\Theta }:C\times C\to \mathbf{R}$ has conditions (A1)–(A4) and $\phi :C\to \mathbf{R}$ has the properties proper lower semicontinuous and convex, if either condition (B1) or condition (B2) is true. For $r>0$ and $x\in H$, generate an operator $T_r^{(\mathsf{\Theta },\phi )}:H\to C$ through

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

for all $x\in H$. Then,

(i) 

Set $T_r^{(\mathsf{\Theta },\phi )}\left(x\right)$ is a singleton set.

(ii) 

$\forall x,y\in H$,

$$\parallel T_r^{(\mathsf{\Theta },\phi )}x-T_r^{(\mathsf{\Theta },\phi )}{y\parallel }^2\le \langle T_r^{(\mathsf{\Theta },\phi )}x-T_r^{(\mathsf{\Theta },\phi )}y,x-y\rangle .$$

(iii) 

$\mathrm{MEP}(\mathsf{\Theta },\phi )=\mathrm{Fix}\left(T_r^{(\mathsf{\Theta },\phi )}\right)$.

(iv) 

$\mathrm{MEP}(\mathsf{\Theta },\phi )$ is convex closed.

(v) 

$\parallel T_s^{(\mathsf{\Theta },\phi )}x-T_t^{(\mathsf{\Theta },\phi )}{x\parallel }^2\le \frac{s-t}{s}\langle T_s^{(\mathsf{\Theta },\phi )}x-T_t^{(\mathsf{\Theta },\phi )}x,T_s^{(\mathsf{\Theta },\phi )}x-x\rangle$, $\forall s,t>0$ and $\forall x\in H$.

Next, under some mild control conditions, we establish the strong convergence of the proposed algorithm to the unique element $\{x^{*}\}=\mathrm{VI}(\mathrm{VI}(\mathsf{\Omega },{\mathcal{A}}_1),{\mathcal{A}}_2)$ (i.e., the unique solution of a THCVI), where $\mathsf{\Omega }:={\bigcap }_{i=1}^N\mathrm{GMEP}({\mathsf{\Theta }}_i,{\phi }_i,A_i)\cap \mathrm{GSVI}(C,B_1,B_2)\cap \mathrm{Fix}\left(S\right)\cap \mathrm{Fix}\left(T\right)$.

Theorem 2.

We suppose C is a convex nonempty closed set. Assume that, $\forall i=1,2,\dots ,N$, ${\mathsf{\Theta }}_i:C\times C\to \mathbf{R}$ a bifunction has Conditions (A1)–(A4), ${\phi }_i:C\to \mathbf{R}\cup \{+\infty \}$ is a lower semicontinuous, convex proper function with Condition (B1) or Condition (B2), and $A_i:C\to H$ is an ${\eta }_i$-inverse-strongly monotone nonself mapping. Let $f:C\to C$ be a contraction with the parameter $\delta \in [0,1)$, ${\mathcal{A}}_1:C\to H$ be a ζ-inverse-strongly monotone nonself mapping and ${\mathcal{A}}_2:C\to H$ be κ-Lipschitzian and η-strongly monotone with parameters $\kappa ,\eta >0$, where $\delta <\tau :=1-\sqrt{1-\mu (2\eta -\mu {\kappa }^2)}\in (0,1]$, $0<\mu <\frac{2\eta }{{\kappa }^2}$. Let the nonself mappings $B_1,B_2:C\to H$ be α-inverse-strongly monotone and β-inverse-strongly monotone, respectively. Let self mapping T, defined on C, be a nonexpansive mapping and self mapping S, also defined on $C,$ be an ℓ-Lipschitzian pseudocontractive mapping such that $\mathsf{\Omega }:={\bigcap }_{i=1}^N\mathrm{GMEP}({\mathsf{\Theta }}_i,{\phi }_i,A_i)\cap \mathrm{GSVI}(C,B_1,B_2)\cap \mathrm{Fix}\left(S\right)\cap \mathrm{Fix}\left(T\right)\ne \varnothing$ and $\mathrm{VI}(\mathsf{\Omega },{\mathcal{A}}_1)\ne \varnothing$, where $\mathrm{GSVI}(C,B_1,B_2)$ is the fixed point set of the mapping $G:=P_C(I-{\mu }_1{B}_1)P_C(I-{\mu }_2{B}_2)$ with ${\mu }_1\in (0,2\alpha )$ and ${\mu }_2\in (0,2\beta )$. For any given $x_0\in C$, let $\{x_n\}$ be the sequence generated by

$$\left\{\begin{array}{c} u_n={\gamma }_n{x}_n+(1-{\gamma }_n)S{u}_n,\hfill \\ v_n=P_C(u_n-{\mu }_2{B}_2{u}_n),\hfill \\ z_n=P_C(v_n-{\mu }_1{B}_1{v}_n),\hfill \\ y_n={\sigma }_n{x}_n+(1-{\sigma }_n)P_C(I-{\delta }_n{\mathcal{A}}_1)z_n,\hfill \\ x_{n+1}={\beta }_{n}f\left(y_n\right)+(1-{\beta }_n)P_C(I-{\alpha }_n\mu {\mathcal{A}}_2)T{\Lambda }^N{y}_n,\forall n\ge 0,\hfill \end{array}\right.$$

(54)

where ${\mathsf{\Delta }}^N=T_{r_1}^{({\mathsf{\Theta }}_1,{\phi }_1)}(I-r_1{A}_1)\cdots T_{r_N}^{({\mathsf{\Theta }}_N,{\phi }_N)}(I-r_N{A}_N)$ with $r_i\in (0,2{\eta }_i)$ for each $i=1,2,\dots ,N$. Suppose that $\{{\alpha }_n\},\{{\beta }_n\},\{{\gamma }_n\},\{{\sigma }_n\}\subset (0,1]$ and $\{{\delta }_n\}\subset (0,2\zeta ]$ are the sequences such that

(i) 

${\alpha }_n\to 0$ as $n\to \infty$, ${\sum }_{n=0}^{\infty }{\alpha }_n=\infty$ and ${\sum }_{n=0}^{\infty }|{\alpha }_{n+1}-{\alpha }_n|<\infty$.

(ii) 

$\frac{{\beta }_n}{{\alpha }_n}\to 0$ as $n\to \infty$ and ${\sum }_{n=0}^{\infty }|{\beta }_{n+1}-{\beta }_n|<\infty$.

(iii) 

$0<{\lim \hspace{0.17em}\inf }_{n\to \infty }{\sigma }_n\le {\lim \hspace{0.17em}\sup }_{n\to \infty }{\sigma }_n<1$ and ${\sum }_{n=0}^{\infty }|{\sigma }_{n+1}-{\sigma }_n|<\infty$.

(iv) 

$0<{\lim \hspace{0.17em}\inf }_{n\to \infty }{\gamma }_n\le {\lim \hspace{0.17em}\sup }_{n\to \infty }{\gamma }_n<1$ and ${\sum }_{n=0}^{\infty }|{\gamma }_{n+1}-{\gamma }_n|<\infty$.

(v) 

${\delta }_n\le {\alpha }_n\forall n\ge 0$ and ${\sum }_{n=0}^{\infty }|{\delta }_{n+1}-{\delta }_n|<\infty$.

Then, the sequence ${\{x_n\}}_{n=0}^{\infty }$ generated by (54) satisfies the following properties:

(a) 

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

(b) 

${\lim }_{n\to \infty }\parallel x_n-y_n\parallel =0,{\lim }_{n\to \infty }\parallel x_n-G{x}_n\parallel =0,{\lim }_{n\to \infty }\parallel x_n-T{\mathsf{\Delta }}^N{x}_n\parallel =0$ and ${\lim }_{n\to \infty }\parallel x_n-S{x}_n\parallel =0$.

(c) 

${\{x_n\}}_{n=0}^{\infty }$ converges strongly to the unique element $\{x^{*}\}=\mathrm{VI}(\mathrm{VI}(\mathsf{\Omega },{\mathcal{A}}_1),{\mathcal{A}}_2)$ (i.e., the unique solution of a THCVI), provided $\parallel x_n-y_n\parallel =o\left({\delta }_n\right)$.

Proof. 

First, let us show that for each $i=1,2,\dots ,N$, the composite mapping $T_{r_i}^{({\mathsf{\Theta }}_i,{\phi }_i)}(I-r_i{A}_i)$ with $r_i\in (0,2{\eta }_i)$ is nonexpansive. Indeed, from Lemma 10 (iii), it is not difficult to obtain

$$\mathrm{GMEP}({\mathsf{\Theta }}_i,{\phi }_i,A_i)=\mathrm{Fix}\left(T_{r_i}^{({\mathsf{\Theta }}_i,{\phi }_i)}(I-r_i{A}_i)\right).$$

Utilizing Lemma 5 and Lemma 10 (ii), we have

$$\begin{array}{c}\parallel T_{r_i}^{({\mathsf{\Theta }}_i,{\phi }_i)}(I-r_i{A}_i)x-T_{r_i}^{({\mathsf{\Theta }}_i,{\phi }_i)}(I-r_i{A}_i){y\parallel }^2\hfill \\ \le \parallel (I-r_i{A}_i)x-(I-r_i{A}_i){y\parallel }^2\hfill \\ \le {\parallel x-y\parallel }^2+r_i(r_i-2{\eta }_i){\parallel A_i-A_{i}y\parallel }^2\hfill \\ \le {\parallel x-y\parallel }^2.\hfill \end{array}$$

Thus, each composite mapping $T_{r_i}^{({\mathsf{\Theta }}_i,{\phi }_i)}(I-r_i{A}_i)$ is nonexpansive. Moreover, we claim that $T_{r_i}^{({\mathsf{\Theta }}_i,{\phi }_i)}(I-r_i{A}_i)$ is also attracting nonexpansive for each $i=1,2,\dots ,N$. In fact, for all $x\notin \mathrm{GMEP}({\mathsf{\Theta }}_i,{\phi }_i,A_i)$ and $p\in \mathrm{GMEP}({\mathsf{\Theta }}_i,{\phi }_i,A_i)$, by the firm nonexpansivity of $T_{r_i}^{({\mathsf{\Theta }}_i,{\phi }_i)}$ (due to Lemma 10 (ii)), we obtain

$$\begin{array}{c}\parallel T_{r_i}^{({\mathsf{\Theta }}_i,{\phi }_i)}(I-r_i{A}_i){x-p\parallel }^2={\parallel T_{r_i}^{({\mathsf{\Theta }}_i,{\phi }_i)}(I-r_i{A}_i)x-T_{r_i}^{({\mathsf{\Theta }}_i,{\phi }_i)}(I-r_i{A}_i)p\parallel }^2\hfill \\ \le \langle T_{r_i}^{({\mathsf{\Theta }}_i,{\phi }_i)}(I-r_i{A}_i)x-T_{r_i}^{({\mathsf{\Theta }}_i,{\phi }_i)}(I-r_i{A}_i)p,(I-r_i{A}_i)x-(I-r_i{A}_i)p\rangle \hfill \\ =\frac{1}{2}\{\parallel T_{r_i}^{({\mathsf{\Theta }}_i,{\phi }_i)}(I-r_i{A}_i)x-T_{r_i}^{({\mathsf{\Theta }}_i,{\phi }_i)}(I-r_i{A}_i){p\parallel }^2+{\parallel (I-r_i{A}_i)x-(I-r_i{A}_i)p\parallel }^2\hfill \\ -\parallel T_{r_i}^{({\mathsf{\Theta }}_i,{\phi }_i)}(I-r_i{A}_i)x-T_{r_i}^{({\mathsf{\Theta }}_i,{\phi }_i)}(I-r_i{A}_i)p-(I-r_i{A}_i)x+(I-r_i{A}_i)p{\parallel }^2\}\hfill \\ =\frac{1}{2}\{\parallel T_{r_i}^{({\mathsf{\Theta }}_i,{\phi }_i)}(I-r_i{A}_i){x-p\parallel }^2+{\parallel (I-r_i{A}_i)x-(I-r_i{A}_i)p\parallel }^2\hfill \\ -\parallel T_{r_i}^{({\mathsf{\Theta }}_i,{\phi }_i)}(I-r_i{A}_i)x-p-(I-r_i{A}_i)x+(I-r_i{A}_i)p{\parallel }^2\},\hfill \end{array}$$

which immediately implies that

$$\begin{array}{c}\parallel T_{r_i}^{({\mathsf{\Theta }}_i,{\phi }_i)}(I-r_i{A}_i){x-p\parallel }^2\hfill \\ \le \parallel (I-r_i{A}_i)x-(I-r_i{A}_i){p\parallel }^2-{\parallel T_{r_i}^{({\mathsf{\Theta }}_i,{\phi }_i)}(I-r_i{A}_i)x-p-(I-r_i{A}_i)x+(I-r_i{A}_i)p\parallel }^2.\hfill \end{array}$$

(55)

Next, we discuss two cases.

Case 1. If $A_{i}x=A_{i}p$, then from (55) we have

$$\begin{array}{c}\parallel T_{r_i}^{({\mathsf{\Theta }}_i,{\phi }_i)}(I-r_i{A}_i){x-p\parallel }^2\hfill \\ \le \parallel (I-r_i{A}_i)x-(I-r_i{A}_i){p\parallel }^2-{\parallel T_{r_i}^{({\mathsf{\Theta }}_i,{\phi }_i)}(I-r_i{A}_i)x-p-(I-r_i{A}_i)x+(I-r_i{A}_i)p\parallel }^2\hfill \\ ={\parallel x-p\parallel }^2-{\parallel T_{r_i}^{({\mathsf{\Theta }}_i,{\phi }_i)}(I-r_i{A}_i)x-x\parallel }^2\hfill \\ {<\parallel x-p\parallel }^2.\hfill \end{array}$$

Case 2. If $A_{i}x\ne A_{i}p$, then from (55) we get

$$\begin{array}{c}\parallel T_{r_i}^{({\mathsf{\Theta }}_i,{\phi }_i)}(I-r_i{A}_i){x-p\parallel }^2\hfill \\ \le \parallel (x-r_i{A}_{i}x)-(p-r_i{A}_{i}p){\parallel }^2-{\parallel T_{r_i}^{({\mathsf{\Theta }}_i,{\phi }_i)}(I-r_i{A}_i)x-p-(I-r_i{A}_i)x+(I-r_i{A}_i)p\parallel }^2\hfill \\ \le \parallel (x-r_i{A}_{i}x)-(p-r_i{A}_{i}p){\parallel }^2\hfill \\ \le {\parallel x-p\parallel }^2+r_i(r_i-2{\eta }_i){\parallel A_{i}x-A_{i}p\parallel }^2\hfill \\ {<\parallel x-p\parallel }^2\hfill \end{array}$$

(due to $r_i\in (0,2{\eta }_i)$). Summing up the above two cases, we know that each composite mapping $T_{r_i}^{({\mathsf{\Theta }}_i,{\phi }_i)}(I-r_i{A}_i)$ is also attracting nonexpansive. Therefore, by Lemma 9, we conclude that $\mathrm{Fix}\left(T{\mathsf{\Delta }}^N\right)={\bigcap }_{i=1}^N\mathrm{GMEP}({\mathsf{\Theta }}_i,{\phi }_i,A_i)\cap \mathrm{Fix}\left(T\right)$. Then, we get the desired result by Theorem 1 easily. □