Multi-step extragradient method with regularization for triple hierarchical variational inequalities with variational inclusion and split feasibility constraints

Abstract

By combining Korpelevich’s extragradient method, the viscosity approximation method, the hybrid steepest-descent method, Mann’s iteration method, and the gradient-projection method with regularization, a hybrid multi-step extragradient algorithm with regularization for finding a solution of triple hierarchical variational inequality problem is introduced and analyzed. It is proven that under appropriate assumptions, the proposed algorithm converges strongly to a unique solution of a triple hierarchical variational inequality problem which is defined over the set of solutions of a hierarchical variational inequality problem defined over the set of common solutions of finitely many generalized mixed equilibrium problems (GMEP), finitely many variational inclusions, fixed point problems, and the split feasibility problem (SFP). We also prove the strong convergence of the proposed algorithm to a common solution of the SFP, finitely many GMEPs, finitely many variational inclusions, and the fixed point problem of a strict pseudocontraction. The results presented in this paper improve and extend the corresponding results announced by several others.

MSC:49J30, 47H09, 47J20, 49M05.

1 Introduction

The following problems have their own importance because of their applications in diverse areas of science, engineering, social sciences, and management:

• Equilibrium problems including variational inequalities.

• Variational inclusion problems.

• Split feasibility problems.

• Fixed point problems.

One way or the other, these problems are related to each other. They are described as follows.

Equilibrium problem

Let C be a nonempty closed convex subset of a real Hilbert space H and $\Theta :C\times C\to \mathbb{R}$ be a real-valued bifunction. The equilibrium problem (EP) is to find an element $x\in C$ such that

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

The set of solutions of EP is denoted by $EP(\Theta )$. It includes several problems, namely, variational inequality problems, optimization problems, saddle point problems, fixed point problems, etc., as special cases. For further details on EP, we refer to [1–6] and the references therein.

Let $A:C\to H$ be a nonlinear operator. If $\Theta (x,y)=〈A(x),y-x〉$, then EP reduces to the variational inequality problem of finding $x\in C$ such that

$$〈A(x),y-x〉\ge 0,\mathrm{\forall }y\in C.$$

For further details on variational inequalities and their generalizations, we refer to [7–13] and the references therein.

During the last two decades, EP has been extended and generalized in several directions. The generalized mixed equilibrium problem (GMEP), one of the generalizations of EP, is to find $x\in C$ such that

$$\Theta (x,y)+\phi (y)-\phi (x)+〈Ax,y-x〉\ge 0,\mathrm{\forall }y\in C,$$

(1.1)

where $\phi :C\to \mathbb{R}$ is a real-valued function. The set of solutions of GMEP is denoted by $GMEP(\Theta ,\phi ,A)$. For different choices of operators/functions Θ, φ, and A, we get different forms of equilibrium problems. For applications of GMEP, we refer to [14, 15] and the references therein.

Variational inclusion problem

Let $B:C\to H$ be a single-valued mapping and $R:C\to 2^H$ be a set-valued mapping with $D(R)=C$, where $D(R)$ denotes the domain of R. The variational inclusion problem is to find $x\in C$ such that

$$0\in Bx+Rx.$$

(1.2)

We denote by $\mathrm{I}(B,R)$ the solution set of the variational inclusion problem (1.2). In particular, if $B=R=0$, then $\mathrm{I}(B,R)=C$. If $B=0$, then problem (1.2) becomes the inclusion problem introduced by Rockafellar [16]. It is well known that problem (1.2) provides a convenient framework for the unified study of optimal solutions in many optimization related areas including mathematical programming, complementarity problems, variational inequalities, optimal control, mathematical economics, equilibria and game theory, etc. Let a set-valued mapping $R:D(R)\subset H\to 2^H$ be maximal monotone. We define the resolvent operator $J_{R,\lambda }:H\to \overline{D(R)}$ associated with R and $\lambda >0$ as follows:

$$J_{R,\lambda }={(I+\lambda R)}^{-1},\mathrm{\forall }x\in H.$$

Huang [17] studied problem (1.2) in the case where R is maximal monotone and B is strongly monotone and Lipschitz continuous with $D(R)=C=H$. Zeng et al. [18] further studied problem (1.2) in a more general setting than in [17]. They gave the geometric convergence rate estimate for approximate solutions. Various types of iterative algorithms for solving variational inclusions have been further studied and developed in the literature; see, for example, [19–22] and the references therein.

Split feasibility problem

Let C and Q be nonempty closed convex subsets of real Hilbert spaces ${\mathcal{H}}_1$ and ${\mathcal{H}}_2$, respectively. The split feasibility problem (SFP) is to find a point x such that

$$x\in C\text{and}Ax\in Q,$$

(1.3)

where $A:{\mathcal{H}}_1\to {\mathcal{H}}_2$ is a bounded linear operator from ${\mathcal{H}}_1$ to ${\mathcal{H}}_2$. We denote by Γ the solution set of the SFP. It is a model of an inverse problem which arises in phase retrievals and in medical image reconstruction. A number of image reconstruction problems can be formulated as SFP; see, for example [23] and the references therein. Recently, it is found that the SFP can also be applied to study intensity-modulated radiation therapy (IMRT); see, for example, [24, 25] and the references therein. In the recent past, a wide variety of iterative methods have been proposed to solve SFP; see, for example, [24–28] the references therein.

Fixed point problem

Let C be a nonempty subset of a H and $T:C\to C$ be a mapping. The fixed point problem is to find an element $x\in C$ such that $T(x)=x$.

It is a well-known problem and has tremendous applications in different branches of science, engineering, social sciences, and management.

The following proposition provides some relations among the above mentioned problems.

Proposition 1.1 Given $x^{\ast }\in H$, the following statements are equivalent:

• (a) $x^{\ast }$ solves the SFP;

• (b) $x^{\ast }$ solves the fixed point equation

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

  where $\lambda >0$, $\mathrm{\nabla }f=A^{\ast }(I-P_Q)A$, $P_Q$ is the projection operator and $A^{\ast }$ is the adjoint of A;

• (c) $x^{\ast }$ solves the variational inequality problem (VIP) of finding $x^{\ast }\in C$ such that

  $$〈\mathrm{\nabla }f\left(x^{\ast }\right),x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in C.$$

A variational inequality problem which is defined over the set of fixed points of a mapping is called hierarchical variational inequality problem; that is, when the set C in variational inequality formulation is equal to the set of fixed points of a mapping. A variational inequality problem which is defined over the set of solutions of a hierarchical variational inequality problem is called a triple hierarchical variational inequality problem. For further details on hierarchical variational inequality problems and triple hierarchical variational inequality problems, we refer to [29], a recent survey on these problems.

Very recently, Kong et al. [30] considered the following triple hierarchical variational inequality problem (THVIP).

Problem 1.1 Let C be a nonempty closed convex subset of a real Hilbert space H and $F:C\to H$ be a κ-Lipschitzian and η-strongly monotone operator, where κ and η are positive constants. Let $A:C\to H$ be a monotone and L-Lipschitzian mapping, $V:C\to H$ be a ρ-contraction with coefficient $\rho \in [0,1)$, $S:C\to C$ be a nonexpansive mapping, and $T:C\to C$ be a ξ-strictly pseudocontractive mapping with $Fix(T)\cap VI(C,A)\ne \mathrm{\varnothing }$, where $Fix(T)$ denotes the set of all fixed points of T. Let $0<\mu <\frac{2\eta }{{\kappa }^2}$ and $0<\gamma \le \tau$, where $\tau =1-\sqrt{1-\mu (2\eta -\mu {\kappa }^2)}$. Then the objective is to find $x^{\ast }\in \Xi$ such that

$$〈(\mu F-\gamma V)x^{\ast },x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in \Xi ,$$

(1.4)

where Ξ denotes the solution set of the hierarchical variational inequality problem (HVIP) of finding $z^{\ast }\in Fix(T)\cap VI(C,A)$ such that

$$〈(\mu F-\gamma S)z^{\ast },z-z^{\ast }〉\ge 0,\mathrm{\forall }z\in Fix(T)\cap VI(C,A).$$

(1.5)

Kong et al. [30] presented an algorithm for finding a solution of Problem 1.1. Under some conditions, they proved that the sequence $\{x_n\}$ generated by the proposed algorithm converges strongly to a point $x^{\ast }\in Fix(T)\cap VI(C,A)$ which is a unique solution of Problem 1.1 provided that $\{S{x}_n\}$ is bounded and $\parallel x_{n+1}-x_n\parallel +\parallel x_n-z_n\parallel =o({ϵ}_n^2)$. They also showed under certain conditions that the sequence $\{x_n\}$ generated by proposed algorithm converges strongly to a unique solution $x^{\ast }$ of the following VIP provided that $\parallel x_{n+1}-x_n\parallel +\parallel x_n-z_n\parallel =o({ϵ}_n^2)$ and the sequence $\{S{x}_n\}$ is bounded:

$$\text{find }{x}^{\ast }\in \Xi \text{ such that }〈F{x}^{\ast },x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in \Xi .$$

In this paper, we consider the following triple hierarchical variational inequality problem (THVIP).

Problem 1.2 Let M, N be two positive integers. Assume that

• (i) $F:H\to H$ is κ-Lipschitzian and η-strongly monotone with positive constants $\kappa ,\eta >0$ such that $0<\gamma \le \tau$ and $0<\mu <\frac{2\eta }{{\kappa }^2}$ where $\tau =1-\sqrt{1-\mu (2\eta -\mu {\kappa }^2)}$;

• (ii) for each $k\in \{1,2,\dots ,M\}$, ${\Theta }_k:C\times C\to \mathbb{R}$ satisfies conditions (A1)-(A4) and ${\phi }_k:C\to \mathbb{R}\cup \{+\mathrm{\infty }\}$ is a proper lower semicontinuous and convex function with restriction (B1) or (B2) (conditions (A1)-(A4) and (B1)-(B2) are given in the next section);

• (iii) for each $k\in \{1,2,\dots ,M\}$ and $i\in \{1,2,\dots ,N\}$, $R_i:C\to 2^H$ is a maximal monotone mapping, and $A_k:H\to H$ and $B_i:C\to H$ are ${\mu }_k$-inverse strongly monotone and ${\eta }_i$-inverse strongly monotone, respectively;

• (iv) $T:H\to H$ is a ξ-strict pseudocontraction, $S:H\to H$ is a nonexpansive mapping and $V:H\to H$ is a ρ-contraction with coefficient $\rho \in [0,1)$;

• (v) $\Omega :=({\bigcap }_{k=1}^{M}GMEP({\Theta }_k,{\phi }_k,A_k))\cap ({\bigcap }_{i=1}^N\mathrm{I}(B_i,R_i))\cap Fix(T)\cap \Gamma \ne \mathrm{\varnothing }$.

Then the objective is to find $x^{\ast }\in \Xi$ such that

$$〈(\mu F-\gamma V)x^{\ast },x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in \Xi ,$$

(1.6)

where Ξ denotes the solution set of the hierarchical variational inequality problem (HVIP) of finding $z^{\ast }\in \Omega$ such that

$$〈(\mu F-\gamma S)z^{\ast },z-z^{\ast }〉\ge 0,\mathrm{\forall }z\in \Omega .$$

(1.7)

By combining Korpelevich’s extragradient method, the viscosity approximation method, the hybrid steepest-descent method, Mann’s iteration method, and the gradient-projection method (GPM) with regularization, we introduce and analyze a hybrid multi-step extragradient algorithm with regularization in the setting of Hilbert spaces. It is proven that under appropriate assumptions, the proposed algorithm converges strongly to a unique solution of THVIP (1.6). The algorithm and convergence result of this paper extend and generalize several existing algorithms and results, respectively, in the literature.

2 Preliminaries

Throughout this paper, unless otherwise specified, we assume that H is a real Hilbert space whose inner product and norm are denoted by $〈\cdot ,\cdot 〉$ and $\parallel \cdot \parallel$, respectively. We write $x_n\to x$ (respectively, $x_n\rightharpoonup x$) to indicate that the sequence $\{x_n\}$ converges (respectively, weakly) to x. Moreover, we use ${\omega }_w(x_n)$ to denote the weak ω-limit set of the sequence $\{x_n\}$, that is,

$${\omega }_w(x_n):=\{x\in H:x_{n_i}\rightharpoonup x\text{ for some subsequence }\{x_{n_i}\}\text{ of }\{x_n\}\}.$$

Definition 2.1 A mapping $T:H\to H$ is said to be

• (a) nonexpansive if

  $$\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,\mathrm{\forall }x,y\in H;$$

• (b) firmly nonexpansive if $2T-I$ is nonexpansive, or equivalently, if T is 1-inverse strongly monotone (1-ism),

  $$〈x-y,Tx-Ty〉\ge {\parallel Tx-Ty\parallel }^2,\mathrm{\forall }x,y\in H;$$

alternatively, T is firmly nonexpansive if and only if T can be expressed as

$$T=\frac{1}{2}(I+S),$$

where $S:H\to H$ is nonexpansive; projections are firmly nonexpansive.

It can easily be seen that if T is nonexpansive, then $I-T$ is monotone.

Definition 2.2 A mapping $T:H\to H$ is said to be an averaged mapping if it can be written as the average of the identity I and a nonexpansive mapping, that is,

$$T\equiv (1-\alpha )I+\alpha S,$$

where $\alpha \in (0,1)$ and $S:H\to H$ is nonexpansive. More precisely, when the last equality holds, we say that T is α-averaged. Thus firmly nonexpansive mappings (in particular, projections) are $\frac{1}{2}$-averaged mappings.

Proposition 2.1 [31]

Let $T:H\to H$ be a given mapping.

• (a) T is nonexpansive if and only if the complement $I-T$ is $\frac{1}{2}$-ism.

• (b) If T is ν-ism, then for $\gamma >0$, γT is $\frac{\nu }{\gamma }$-ism.

• (c) T is averaged if and only if the complement $I-T$ is ν-ism for some $\nu >1/2$. Indeed, for $\alpha \in (0,1)$, T is α-averaged if and only if $I-T$ is $\frac{1}{2\alpha }$-ism.

Proposition 2.2 [31, 32]

Let $S,T,V:H\to H$ be given operators.

• (a) If $T=(1-\alpha )S+\alpha V$ for some $\alpha \in (0,1)$ and if S is averaged and V is nonexpansive, then T is averaged.

• (b) T is firmly nonexpansive if and only if the complement $I-T$ is firmly nonexpansive.

• (c) If $T=(1-\alpha )S+\alpha V$ for some $\alpha \in (0,1)$ and if S is firmly nonexpansive and V is nonexpansive, then T is averaged.

• (d) The composite of finitely many averaged mappings is averaged, that is, if each of the mappings ${\{T_i\}}_{i=1}^N$ is averaged, then so is the composite $T_1\cdots T_N$. In particular, if $T_1$ is ${\alpha }_1$-averaged and $T_2$ is ${\alpha }_2$-averaged, where ${\alpha }_1,{\alpha }_2\in (0,1)$, then the composite $T_1{T}_2$ is α-averaged, where $\alpha ={\alpha }_1+{\alpha }_2-{\alpha }_1{\alpha }_2$.

• (e) If the mappings ${\{T_i\}}_{i=1}^N$ are averaged and have a common fixed point, then

  $$\bigcap _{i=1}^{N}Fix(T_i)=Fix(T_1\cdots T_N).$$

The notation $Fix(T)$ denotes the set of all fixed points of the mapping T, that is, $Fix(T)=\{x\in H:Tx=x\}$.

A mapping $T:C\to C$ is said to be ξ-strictly pseudocontractive if there exists $\xi \in [0,1)$ such that

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

In this case, we also say that T is a ξ-strict pseudocontraction. We denote by $Fix(S)$ the set of fixed points of S. In particular, if $\xi =0$, T is a nonexpansive mapping.

It is clear that, in a real Hilbert space H, $T:C\to C$ is ξ-strictly pseudocontractive if and only if the following inequality holds:

$$〈Tx-Ty,x-y〉\le {\parallel x-y\parallel }^2-\frac{1-\xi }{2}{\parallel (I-T)x-(I-T)y\parallel }^2,\mathrm{\forall }x,y\in C.$$

This immediately implies that if T is a ξ-strictly pseudocontractive mapping, then $I-T$ is $\frac{1-\xi }{2}$-inverse strongly monotone; for further details, we refer to [33] and the references therein. It is well known that the class of strict pseudocontractions strictly includes the class of nonexpansive mappings and that the class of pseudocontractions strictly includes the class of strict pseudocontractions.

Lemma 2.1 [[33], Proposition 2.1]

Let C be a nonempty closed convex subset of a real Hilbert space H and $T:C\to C$ be a mapping.

• (a) If T is a ξ-strictly pseudocontractive mapping, then T satisfies the Lipschitzian condition

  $$\parallel Tx-Ty\parallel \le \frac{1+\xi }{1-\xi }\parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

• (b) If T is a ξ-strictly pseudocontractive mapping, then the mapping $I-T$ is semiclosed at 0, that is, if $\{x_n\}$ is a sequence in C such that $x_n\rightharpoonup \tilde{x}$ and $(I-T)x_n\to 0$, then $(I-T)\tilde{x}=0$.

• (c) If T is ξ-(quasi-)strict pseudocontraction, then the fixed point set $Fix(T)$ of T is closed and convex so that the projection $P_{Fix(T)}$ is well defined.

Lemma 2.2 [34]

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $T:C\to C$ be a ξ-strictly pseudocontractive mapping. Let γ and δ be two nonnegative real numbers such that $(\gamma +\delta )\xi \le \gamma$. Then

$$\parallel \gamma (x-y)+\delta (Tx-Ty)\parallel \le (\gamma +\delta )\parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

Lemma 2.3 (Demiclosedness principle)

Let C be a nonempty closed convex subset of a real Hilbert space H. Let S be a nonexpansive self-mapping on C with $Fix(S)\ne \mathrm{\varnothing }$. Then $I-S$ is demiclosed. That is, whenever $\{x_n\}$ is a sequence in C weakly converging to some $x\in C$ and the sequence $\{(I-S)x_n\}$ strongly converges to some y, it follows that $(I-S)x=y$, where I is the identity operator of H.

Definition 2.3 A nonlinear operator T with the domain $D(T)\subset H$ and the range $R(T)\subset H$ is said to be

• (a) monotone if

  $$〈Tx-Ty,x-y〉\ge 0,\mathrm{\forall }x,y\in D(T);$$

• (b) β-strongly monotone if there exists a constant $\beta >0$ such that

  $$〈Tx-Ty,x-y〉\ge \eta {\parallel x-y\parallel }^2,\mathrm{\forall }x,y\in D(T);$$

• (c) ν-inverse strongly monotone if there exists a constant $\nu >0$ such that

  $$〈Tx-Ty,x-y〉\ge \nu {\parallel Tx-Ty\parallel }^2,\mathrm{\forall }x,y\in D(T).$$

It is easy to see that the projection $P_C$ is 1-inverse strongly monotone. Inverse strongly monotone (also referred to as co-coercive) operators have been applied widely in solving practical problems in various fields, for instance, in traffic assignment problems; see, for example, [35]. It is obvious that if T is ν-inverse strongly monotone, then T is monotone and $\frac{1}{\nu }$-Lipschitz continuous. Moreover, we also have, for all $u,v\in D(T)$ and $\lambda >0$,

$$\begin{array}{rcl}{\parallel (I-\lambda T)u-(I-\lambda T)v\parallel }^2& =& {\parallel (u-v)-\lambda (Tu-Tv)\parallel }^2\\ =& {\parallel u-v\parallel }^2-2\lambda 〈Tu-Tv,u-v〉+{\lambda }^2{\parallel Tu-Tv\parallel }^2\\ \le & {\parallel u-v\parallel }^2+\lambda (\lambda -2\nu ){\parallel Tu-Tv\parallel }^2.\end{array}$$

(2.1)

So, if $\lambda \le 2\nu$, then $I-\lambda T$ is a nonexpansive mapping.

The metric (or nearest point) projection from H onto C is the mapping $P_C:H\to C$ which assigns to each point $x\in H$ the unique point $P_{C}x\in C$ satisfying the property

$$\parallel x-P_{C}x\parallel =\underset{y\in C}{inf}\parallel x-y\parallel =:d(x,C).$$

Some important properties of projections are gathered in the following proposition.

Proposition 2.3 For given $x\in H$ and $z\in C$:

• (a) $z=P_{C}x\iff 〈x-z,y-z〉\le 0$, $\mathrm{\forall }y\in C$;

• (b) $z=P_{C}x\iff {\parallel x-z\parallel }^2\le {\parallel x-y\parallel }^2-{\parallel y-z\parallel }^2$, $\mathrm{\forall }y\in C$;

• (c) $〈P_{C}x-P_{C}y,x-y〉\ge {\parallel P_{C}x-P_{C}y\parallel }^2$, $\mathrm{\forall }y\in H$.

Consequently, $P_C$ is nonexpansive and monotone.

Let λ be a number in $(0,1]$ and let $\mu >0$. Associating with a nonexpansive mapping $T:C\to H$, we define the mapping $T^{\lambda }:C\to H$ by

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

where $F:H\to H$ is an operator such that, for some positive constants $\kappa ,\eta >0$, F is κ-Lipschitzian and η-strongly monotone on H, that is, F satisfies the conditions:

$$\parallel Fx-Fy\parallel \le \kappa \parallel x-y\parallel \text{and}〈Fx-Fy,x-y〉\ge \eta {\parallel x-y\parallel }^2$$

for all $x,y\in H$.

Lemma 2.4 [[36], Lemma 3.1]

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

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

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

Lemma 2.5 Let $A:C\to H$ be a monotone mapping. In the context of the variational inequality problem the characterization of the projection (see Proposition  2.3(a)) implies

$$u\in VI(C,A)\iff u=P_C(u-\lambda Au),\mathrm{\forall }\lambda >0.$$

Let C be a nonempty closed convex subset of H and $\Theta :C\times C\to \mathbb{R}$ satisfy the following conditions.

• (A1) $T(x,x)=0$, $\mathrm{?}x?C$;

• (A2) T is monotone, that is, $T(x,y)+T(y,x)=0$, $\mathrm{?}x,y?C$;

• (A3) T is upper-hemicontinuous, that is, $\mathrm{?}x,y,z?C$,

  $$\underset{t?0^{+}}{lim?sup}T(tz+(1-t)x,y)=T(x,y);$$

• (A4) $T(x,·)$ is convex and lower semicontinuous, for each $x?C$.

Let $\phi :C\to \mathbb{R}$ be a lower semicontinuous and convex function satisfying either (B1) or (B2), where

• (B1) for each $x?H$ and $r>0$, there exist a bounded subset $D_x?C$ and $y_x?C$ such that, for any $z?C\backslash D_x$,

  $$T(z,y_x)+f(y_x)-f(z)+\frac{1}{r}{y}_x-z,z-x><0;$$

• (B2) C is a bounded set.

Given a positive number $r>0$. Let $T_r^{(\Theta ,\phi )}:H\to C$ be the solution set of the auxiliary mixed equilibrium problem, that is, for each $x\in H$,

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

Next we list some elementary conclusions for the MEP.

Proposition 2.4 [37]

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

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

for all $x\in H$. Then

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

• (ii) $T_r^{(\Theta ,\phi )}$ is firmly nonexpansive, that is, for any $x,y\in H$,

  $${\parallel T_r^{(\Theta ,\phi )}x-T_r^{(\Theta ,\phi )}y\parallel }^2\le 〈T_r^{(\Theta ,\phi )}x-T_r^{(\Theta ,\phi )}y,x-y〉;$$

• (iii) $Fix(T_r^{(\Theta ,\phi )})=MEP(\Theta ,\phi )$;

• (iv) $MEP(\Theta ,\phi )$ is closed and convex;

• (v) ${\parallel T_s^{(\Theta ,\phi )}x-T_t^{(\Theta ,\phi )}x\parallel }^2\le \frac{s-t}{s}〈T_s^{(\Theta ,\phi )}x-T_t^{(\Theta ,\phi )}x,T_s^{(\Theta ,\phi )}x-x〉$, for all $s,t>0$ and $x\in H$.

We need some facts and tools in a real Hilbert space H which are listed as lemmas below.

Lemma 2.6 Let X be a real inner product space. Then we have the following inequality:

$${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2〈y,x+y〉,\mathrm{\forall }x,y\in X.$$

Lemma 2.7 Let H be a real Hilbert space. Then the following hold:

• (a) ${\parallel x-y\parallel }^2={\parallel x\parallel }^2-{\parallel y\parallel }^2-2〈x-y,y〉$, for all $x,y\in H$;

• (b) ${\parallel \lambda x+\mu y\parallel }^2=\lambda {\parallel x\parallel }^2+\mu {\parallel y\parallel }^2-\lambda \mu {\parallel x-y\parallel }^2$, for all $x,y\in H$ and $\lambda ,\mu \in [0,1]$ with $\lambda +\mu =1$;

• (c) if $\{x_n\}$ is a sequence in H such that $x_n\rightharpoonup x$, it follows that

  $$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_n-y\parallel }^2=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_n-x\parallel }^2+{\parallel x-y\parallel }^2,\mathrm{\forall }y\in H.$$

Lemma 2.8 [38]

Let $\{a_n\}$ be a sequence of nonnegative real numbers satisfying the property

$$a_{n+1}\le (1-s_n)a_n+s_n{b}_n+t_n,\mathrm{\forall }n\ge 1,$$

where $\{s_n\}\subset (0,1]$ and $\{b_n\}$ are such that:

• (i) ${\sum }_{n=1}^{\mathrm{\infty }}{s}_n=\mathrm{\infty }$;

• (ii) either ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{b}_n\le 0$ or ${\sum }_{n=0}^{\mathrm{\infty }}|s_n{b}_n|<\mathrm{\infty }$;

• (iii) ${\sum }_{n=1}^{\mathrm{\infty }}{t}_n<\mathrm{\infty }$ where $t_n\ge 0$, for all $n\ge 1$.

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

Recall that a set-valued mapping $T:D(T)\subset H\to 2^H$ is called monotone if, for all $x,y\in D(T)$, $f\in Tx$, and $g\in Ty$ imply $〈f-g,x-y〉\ge 0$. A set-valued mapping T is called maximal monotone if T is monotone and $(I+\lambda T)D(T)=H$, for each $\lambda >0$, where I is the identity mapping of H. We denote by $G(T)$ the graph of T. It is well known that a monotone mapping T is maximal if and only if, for $(x,f)\in H\times H$, $〈f-g,x-y〉\ge 0$ for every $(y,g)\in G(T)$ implies $f\in Tx$.

Next we provide an example to illustrate the concept of maximal monotone mapping.

Let $A:C\to H$ be a monotone, k-Lipschitz-continuous mapping and let $N_{C}v$ be the normal cone to C at $v\in C$, that is,

$$N_{C}v=\{u\in H:〈v-p,u〉\ge 0,\mathrm{\forall }p\in C\}.$$

Define

$$\tilde{T}v=\{\begin{array}{ll}Av+N_{C}v,& \text{if }v\in C,\\ \mathrm{\varnothing },& \text{if }v\notin C.\end{array}$$

Then $\tilde{T}$ is maximal monotone (see [16]) such that

$$0\in \tilde{T}v\iff v\in VI(C,A).$$

(2.2)

Let $R:D(R)\subset H\to 2^H$ be a maximal monotone mapping. Let $\lambda ,\mu >0$ be two positive numbers.

Lemma 2.9 [39]

We have the resolvent identity

$$J_{R,\lambda }x=J_{R,\mu }(\frac{\mu }{\lambda }x+(1-\frac{\mu }{\lambda })J_{R,\lambda }x),\mathrm{\forall }x\in H.$$

Remark 2.1 For $\lambda ,\mu >0$, we have the following relation:

$$\parallel J_{R,\lambda }x-J_{R,\mu }y\parallel \le \parallel x-y\parallel +|\lambda -\mu |(\frac{1}{\lambda }\parallel J_{R,\lambda }x-y\parallel +\frac{1}{\mu }\parallel x-J_{R,\mu }y\parallel ),\mathrm{\forall }x,y\in H.$$

(2.3)

The following property for the resolvent operator $J_{R,\lambda }:H\to \overline{D(R)}$ was considered in [17, 18].

Lemma 2.10 $J_{R,\lambda }$ is single-valued and firmly nonexpansive, that is,

$$〈J_{R,\lambda }x-J_{R,\lambda }y,x-y〉\ge {\parallel J_{R,\lambda }x-J_{R,\lambda }y\parallel }^2,\mathrm{\forall }x,y\in H.$$

Consequently, $J_{R,\lambda }$ is nonexpansive and monotone.

Lemma 2.11 [20]

Let R be a maximal monotone mapping with $D(R)=C$. Then, for any given $\lambda >0$, $u\in C$ is a solution of problem (1.6) if and only if $u\in C$ satisfies

$$u=J_{R,\lambda }(u-\lambda Bu).$$

Lemma 2.12 [18]

Let R be a maximal monotone mapping with $D(R)=C$ and let $B:C\to H$ be a strongly monotone, continuous and single-valued mapping. Then, for each $z\in H$, the equation $z\in (B+\lambda R)x$ has a unique solution $x_{\lambda }$ for $\lambda >0$.

Lemma 2.13 [20]

Let R be a maximal monotone mapping with $D(R)=C$ and $B:C\to H$ be a monotone, continuous and single-valued mapping. Then $(I+\lambda (R+B))C=H$, for each $\lambda >0$. In this case, $R+B$ is maximal monotone.

3 Algorithms and convergence results

Let H be a real Hilbert space and $f:H\to \mathbb{R}$ be a function. Then the minimization problem

$$\underset{x\in C}{min}f(x):=\frac{1}{2}{\parallel Ax-P_{Q}Ax\parallel }^2$$

is ill-posed. Xu [40] considered the following Tikhonov’s regularization problem:

$$\underset{x\in C}{min}{f}_{\alpha }(x):=\frac{1}{2}{\parallel Ax-P_{Q}Ax\parallel }^2+\frac{1}{2}\alpha {\parallel x\parallel }^2,$$

where $\alpha >0$ is the regularization parameter. It is clear that the gradient

$$\mathrm{\nabla }{f}_{\alpha }=\mathrm{\nabla }f+\alpha I=A^{\ast }(I-P_Q)A+\alpha I$$

is $(\alpha +{\parallel A\parallel }^2)$-Lipschitz continuous.

Throughout the paper, unless otherwise specified, M, N are positive integers and C be a nonempty closed convex subset of a real Hilbert space H.

Algorithm 3.1 The notations and symbols are the same as in Problem 1.2. Start with a given arbitrary $x_0\in H$, and compute a sequence $\{x_n\}$ by

$$\{\begin{array}{l}{u}_n=T_{r_{M,n}}^{({\Theta }_M,{\phi }_M)}(I-r_{M,n}{A}_M)T_{r_{M-1,n}}^{({\Theta }_{M-1},{\phi }_{M-1})}(I-r_{M-1,n}{A}_{M-1})\cdots T_{r_{1,n}}^{({\Theta }_1,{\phi }_1)}(I-r_{1,n}{A}_1)x_n,\\ v_n=J_{R_N,{\lambda }_{N,n}}(I-{\lambda }_{N,n}{B}_N)J_{R_{N-1},{\lambda }_{N-1,n}}(I-{\lambda }_{N-1,n}{B}_{N-1})\cdots J_{R_1,{\lambda }_{1,n}}(I-{\lambda }_{1,n}{B}_1)u_n,\\ y_n={\beta }_n{x}_n+{\gamma }_n{P}_C(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n+{\sigma }_{n}T{P}_C(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n,\\ x_{n+1}={ϵ}_n\gamma ({\delta }_{n}V{x}_n+(1-{\delta }_n)S{x}_n)+(I-{ϵ}_n\mu F)y_n,\mathrm{\forall }n\ge 0,\end{array}$$

(3.1)

where $\mathrm{\nabla }{f}_{{\alpha }_n}={\alpha }_{n}I+\mathrm{\nabla }f$.

The following result provides the strong convergence of the sequence generated by Algorithm 3.1.

Theorem 3.1 For each $k\in \{1,2,\dots ,M\}$, let ${\Theta }_k:C\times C\to \mathbb{R}$ be a bifunction satisfying conditions (A1)-(A4) and ${\phi }_k:C\to \mathbb{R}\cup \{+\mathrm{\infty }\}$ be a proper lower semicontinuous and convex function with restriction (B1) or (B2). For each $k\in \{1,2,\dots ,M\}$ and $i\in \{1,2,\dots ,N\}$, let $R_i:C\to 2^H$ be a maximal monotone mapping and let $A_k:H\to H$ and $B_i:C\to H$ be ${\mu }_k$-inverse strongly monotone and ${\eta }_i$-inverse strongly monotone, respectively. Let $T:H\to H$ be a ξ-strictly pseudocontractive mapping, $S:H\to H$ be a nonexpansive mapping and $V:H\to H$ be a ρ-contraction with coefficient $\rho \in [0,1)$. Let $F:H\to H$ be κ-Lipschitzian and η-strongly monotone with positive constants $\kappa ,\eta >0$ such that $0<\gamma \le \tau$ and $0<\mu <\frac{2\eta }{{\kappa }^2}$, where $\tau =1-\sqrt{1-\mu (2\eta -\mu {\kappa }^2)}$. Assume that the solution set Ξ of HVIP (1.7) is nonempty where $\Omega :=({\bigcap }_{k=1}^{M}GMEP({\Theta }_k,{\phi }_k,A_k))\cap ({\bigcap }_{i=1}^N\mathrm{I}(B_i,R_i))\cap Fix(T)\cap \Gamma$. Let $\{{\lambda }_n\}\subset [a,b]\subset (0,\frac{2}{{\parallel A\parallel }^2})$, $\{{\alpha }_n\}\subset (0,\mathrm{\infty })$ with ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n<\mathrm{\infty }$, $\{{ϵ}_n\},\{{\delta }_n\},\{{\beta }_n\},\{{\gamma }_n\},\{{\sigma }_n\}\subset (0,1)$ with ${\beta }_n+{\gamma }_n+{\sigma }_n=1$, and $\{{\lambda }_{i,n}\}\subset [a_i,b_i]\subset (0,2{\eta }_i)$, $\{r_{k,n}\}\subset [c_k,d_k]\subset (0,2{\mu }_k)$ where $i\in \{1,2,\dots ,N\}$ and $k\in \{1,2,\dots ,M\}$. Suppose that

• (C1) ${lim}_{n?\mathrm{8}}{d}_n=0$, ${lim}_{n?\mathrm{8}}{?}_n=0$, ${lim}_{n?\mathrm{8}}\frac{|{?}_n-{?}_{n-1}|}{d_n{?}_n^2{?}_{n-1}}=0$ and ${?}_{n=0}^{\mathrm{8}}{?}_n{d}_n=\mathrm{8}$;

• (C2) ${?}_{n=1}^{\mathrm{8}}|d_n-d_{n-1}|<\mathrm{8}$ or ${lim}_{n?\mathrm{8}}|d_n-d_{n-1}|/(d_n{?}_n)=0$;

• (C3) ${?}_{n=1}^{\mathrm{8}}\frac{|{ß}_n-{ß}_{n-1}|}{{?}_n}<\mathrm{8}$ or ${lim}_{n?\mathrm{8}}|{ß}_n-{ß}_{n-1}|/(d_n{?}_n^2)=0$;

• (C4) ${?}_{n=1}^{\mathrm{8}}\frac{|{?}_n-{?}_{n-1}|}{{?}_n}<\mathrm{8}$ or ${lim}_{n?\mathrm{8}}|{?}_n-{?}_{n-1}|/(d_n{?}_n^2)=0$;

• (C5) ${?}_{n=1}^{\mathrm{8}}\frac{|{?}_n-{?}_{n-1}|}{{?}_n}<\mathrm{8}$ or ${lim}_{n?\mathrm{8}}|{?}_n-{?}_{n-1}|/(d_n{?}_n^2)=0$;

• (C6) ${?}_{n=1}^{\mathrm{8}}\frac{|{?}_n{a}_n-{?}_{n-1}{a}_{n-1}|}{{?}_n}<\mathrm{8}$ or ${lim}_{n?\mathrm{8}}|{?}_n{a}_n-{?}_{n-1}{a}_{n-1}|/(d_n{?}_n^2)=0$;

• (C7) $\{{ß}_n\}?[c,d]?(0,1)$, $({?}_n+s_n)?={?}_n$ and ${lim?inf}_{n?\mathrm{8}}{s}_n>0$;

• (C8) for each $i=1,2,\dots ,N$, ${?}_{n=1}^{\mathrm{8}}\frac{|{?}_{i,n}-{?}_{i,n-1}|}{{?}_n}<\mathrm{8}$ or ${lim}_{n?\mathrm{8}}|{?}_{i,n}-{?}_{i,n-1}|/(d_n{?}_n^2)=0$;

• (C9) for each $k=1,2,\dots ,M$, ${?}_{n=1}^{\mathrm{8}}\frac{|r_{k,n}-r_{k,n-1}|}{{?}_n}<\mathrm{8}$ or ${lim}_{n?\mathrm{8}}|r_{k,n}-r_{k,n-1}|/(d_n{?}_n^2)=0$;

• (C10) there exist positive constants $?,\overline{k}>0$ such that ${lim}_{n?\mathrm{8}}{?}_n^{1/?}/d_n=0$ and $?x_n-T{x}_n?=\overline{k}{[d(x_n,O)]}^{?}$, $\mathrm{?}x?C$ for sufficiently large $n=0$.

If $\{x_n\}$ is a sequence generated by Algorithm 3.1 and $\{S{x}_n\}$ is bounded, then

• (a) $\parallel x_{n+1}-x_n\parallel =o({ϵ}_n)$;

• (b) ${\omega }_w(x_n)\subset \Omega$;

• (c) $\{x_n\}$ converges strongly to a point $x^{\ast }\in \Omega$ provided $\parallel x_n-y_n\parallel +{\alpha }_n=o({ϵ}_n^2)$, which is the unique solution of Problem 1.2.

Proof First of all, taking into account $\Xi \ne \mathrm{\varnothing }$, we know that $\Omega \ne \mathrm{\varnothing }$. Observe that

$$\begin{array}{rl}\mu \eta \ge \tau & \iff \mu \eta \ge 1-\sqrt{1-\mu (2\eta -\mu {\kappa }^2)}\\ \iff \sqrt{1-\mu (2\eta -\mu {\kappa }^2)}\ge 1-\mu \eta \\ \iff 1-2\mu \eta +{\mu }^2{\kappa }^2\ge 1-2\mu \eta +{\mu }^2{\eta }^2\\ \iff {\kappa }^2\ge {\eta }^2\\ \iff \kappa \ge \eta \end{array}$$

and

$$\begin{array}{rl}〈(\mu F-\gamma V)x-(\mu F-\gamma V)y,x-y〉& =\mu 〈Fx-Fy,x-y〉-\gamma 〈Vx-Vy,x-y〉\\ \ge \mu \eta {\parallel x-y\parallel }^2-\gamma \rho {\parallel x-y\parallel }^2\\ =(\mu \eta -\gamma \rho ){\parallel x-y\parallel }^2,\mathrm{\forall }x,y\in H.\end{array}$$

Since $\tau \ge \gamma >0$ and $\kappa \ge \eta$, we deduce that $\mu \eta \ge \tau \ge \gamma >\gamma \rho$ and hence the mapping $\mu F-\gamma V$ is $(\mu \eta -\gamma \rho )$-strongly monotone. Moreover, it is clear that the mapping $\mu F-\gamma V$ is $(\mu \kappa +\gamma \rho )$-Lipschitzian. Thus, there exists a unique solution $x^{\ast }$ in Ξ to the VIP

$$〈(\mu F-\gamma V)x^{\ast },p-x^{\ast }〉\ge 0,\mathrm{\forall }p\in \Xi ,$$

that is, $\{x^{\ast }\}=VI(\Xi ,\mu F-\gamma V)$. Now, we put

$${\Delta }_n^k=T_{r_{k,n}}^{({\Theta }_k,{\phi }_k)}(I-r_{k,n}{A}_k)T_{r_{k-1,n}}^{({\Theta }_{k-1},{\phi }_{k-1})}(I-r_{k-1,n}{A}_{k-1})\cdots T_{r_{1,n}}^{({\Theta }_1,{\phi }_1)}(I-r_{1,n}{A}_1)x_n$$

for all $k\in \{1,2,\dots ,M\}$ and $n\ge 0$,

$${\Lambda }_n^i=J_{R_i,{\lambda }_{i,n}}(I-{\lambda }_{i,n}{B}_i)J_{R_{i-1},{\lambda }_{i-1,n}}(I-{\lambda }_{i-1,n}{B}_{i-1})\cdots J_{R_1,{\lambda }_{1,n}}(I-{\lambda }_{1,n}{B}_1)$$

for all $i\in \{1,2,\dots ,N\}$, ${\Delta }_n^0=I$, and ${\Lambda }_n^0=I$, where I is the identity mapping on H. Then we have $u_n={\Delta }_n^M{x}_n$ and $v_n={\Lambda }_n^N{u}_n$.

Now, we show that $P_C(I-\lambda \mathrm{\nabla }{f}_{\alpha })$ is ζ-averaged, for each $\lambda \in (0,\frac{2}{\alpha +{\parallel A\parallel }^2})$, where

$$\zeta =\frac{2+\lambda (\alpha +{\parallel A\parallel }^2)}{4}\in (0,1).$$

Indeed, it is easy to see that $\mathrm{\nabla }f=A^{\ast }(I-P_Q)A$ is $\frac{1}{{\parallel A\parallel }^2}$-ism, that is,

$$〈\mathrm{\nabla }f(x)-\mathrm{\nabla }f(y),x-y〉\ge \frac{1}{{\parallel A\parallel }^2}{\parallel \mathrm{\nabla }f(x)-\mathrm{\nabla }f(y)\parallel }^2.$$

Observe that

$$\begin{array}{c}(\alpha +{\parallel A\parallel }^2)〈\mathrm{\nabla }{f}_{\alpha }(x)-\mathrm{\nabla }{f}_{\alpha }(y),x-y〉\hfill \\ =(\alpha +{\parallel A\parallel }^2)[\alpha {\parallel x-y\parallel }^2+〈\mathrm{\nabla }f(x)-\mathrm{\nabla }f(y),x-y〉]\hfill \\ ={\alpha }^2{\parallel x-y\parallel }^2+\alpha 〈\mathrm{\nabla }f(x)-\mathrm{\nabla }f(y),x-y〉\hfill \\ +\alpha {\parallel A\parallel }^2{\parallel x-y\parallel }^2+{\parallel A\parallel }^2〈\mathrm{\nabla }f(x)-\mathrm{\nabla }f(y),x-y〉\hfill \\ \ge {\alpha }^2{\parallel x-y\parallel }^2+2\alpha 〈\mathrm{\nabla }f(x)-\mathrm{\nabla }f(y),x-y〉+{\parallel \mathrm{\nabla }f(x)-\mathrm{\nabla }f(y)\parallel }^2\hfill \\ ={\parallel \alpha (x-y)+\mathrm{\nabla }f(x)-\mathrm{\nabla }f(y)\parallel }^2\hfill \\ ={\parallel \mathrm{\nabla }{f}_{\alpha }(x)-\mathrm{\nabla }{f}_{\alpha }(y)\parallel }^2.\hfill \end{array}$$

Hence, it follows that $\mathrm{\nabla }{f}_{\alpha }=\alpha I+A^{\ast }(I-P_Q)A$ is $\frac{1}{\alpha +{\parallel A\parallel }^2}$-ism. Thus, by Proposition 2.1(b), $\lambda \mathrm{\nabla }{f}_{\alpha }$ is $\frac{1}{\lambda (\alpha +{\parallel A\parallel }^2)}$-ism. From Proposition 2.1(c), the complement $I-\lambda \mathrm{\nabla }{f}_{\alpha }$ is $\frac{\lambda (\alpha +{\parallel A\parallel }^2)}{2}$-averaged. Therefore, noting that $P_C$ is $\frac{1}{2}$-averaged and utilizing Proposition 2.2(d), we see that, for each $\lambda \in (0,\frac{2}{\alpha +{\parallel A\parallel }^2})$, $P_C(I-\lambda \mathrm{\nabla }{f}_{\alpha })$ is ζ-averaged with

$$\zeta =\frac{1}{2}+\frac{\lambda (\alpha +{\parallel A\parallel }^2)}{2}-\frac{1}{2}\cdot \frac{\lambda (\alpha +{\parallel A\parallel }^2)}{2}=\frac{2+\lambda (\alpha +{\parallel A\parallel }^2)}{4}\in (0,1).$$

This shows that $P_C(I-\lambda \mathrm{\nabla }{f}_{\alpha })$ is nonexpansive. Taking into account that $\{{\lambda }_n\}\subset [a,b]\subset (0,\frac{2}{{\parallel A\parallel }^2})$ and ${\alpha }_n\to 0$, we get

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{2+{\lambda }_n({\alpha }_n+{\parallel A\parallel }^2)}{4}\le \frac{2+b{\parallel A\parallel }^2}{4}<1.$$

Without loss of generality, we may assume that ${\zeta }_n:=\frac{2+{\lambda }_n({\alpha }_n+{\parallel A\parallel }^2)}{4}<1$, for each $n\ge 0$. So, $P_C(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})$ is nonexpansive, for each $n\ge 0$. Similarly, since

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{{\lambda }_n({\alpha }_n+{\parallel A\parallel }^2)}{2}\le \frac{b{\parallel A\parallel }^2}{2}<1,$$

it may be confirmed that $I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n}$ is nonexpansive, for each $n\ge 0$.

We divide the rest of the proof into several steps.

Step 1. We prove that $\{x_n\}$ is bounded.

Indeed, take a fixed $p\in \Omega$ arbitrarily. Utilizing (2.1) and Proposition 2.4(b), we have

$$\begin{array}{rcl}\parallel u_n-p\parallel & =& \parallel T_{r_{M,n}}^{({\Theta }_M,{\phi }_M)}(I-r_{M,n}{B}_M){\Delta }_n^{M-1}{x}_n-T_{r_{M,n}}^{({\Theta }_M,{\phi }_M)}(I-r_{M,n}{B}_M){\Delta }_n^{M-1}p\parallel \\ \le & \parallel (I-r_{M,n}{B}_M){\Delta }_n^{M-1}{x}_n-(I-r_{M,n}{B}_M){\Delta }_n^{M-1}p\parallel \\ \le & \parallel {\Delta }_n^{M-1}{x}_n-{\Delta }_n^{M-1}p\parallel \\ ⋮\\ \le & \parallel {\Delta }_n^0{x}_n-{\Delta }_n^{0}p\parallel \\ =& \parallel x_n-p\parallel .\end{array}$$

(3.2)

Utilizing (2.1) and Lemma 2.10, we have

$$\begin{array}{rcl}\parallel v_n-p\parallel & =& \parallel J_{R_N,{\lambda }_{N,n}}(I-{\lambda }_{N,n}{A}_N){\Lambda }_n^{N-1}{u}_n-J_{R_N,{\lambda }_{N,n}}(I-{\lambda }_{N,n}{A}_N){\Lambda }_n^{N-1}p\parallel \\ \le & \parallel (I-{\lambda }_{N,n}{A}_N){\Lambda }_n^{N-1}{u}_n-(I-{\lambda }_{N,n}{A}_N){\Lambda }_n^{N-1}p\parallel \\ \le & \parallel {\Lambda }_n^{N-1}{u}_n-{\Lambda }_n^{N-1}p\parallel \\ ⋮\\ \le & \parallel {\Lambda }_n^0{u}_n-{\Lambda }_n^{0}p\parallel \\ =& \parallel u_n-p\parallel .\end{array}$$

(3.3)

Combining (3.2) and (3.3), we have

$$\parallel v_n-p\parallel \le \parallel x_n-p\parallel .$$

(3.4)

For simplicity, put $t_n=P_C(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n$, for each $n\ge 0$. Note that $P_C(I-\lambda \mathrm{\nabla }f)p=p$ for $\lambda \in (0,\frac{2}{{\parallel A\parallel }^2})$. Hence, from (3.4), it follows that

$$\begin{array}{rcl}\parallel t_n-p\parallel & =& \parallel P_C(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n-P_C(I-{\lambda }_n\mathrm{\nabla }f)p\parallel \\ \le & \parallel P_C(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n-P_C(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})p\parallel \\ +\parallel P_C(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})p-P_C(I-{\lambda }_n\mathrm{\nabla }f)p\parallel \\ \le & \parallel v_n-p\parallel +\parallel (I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})p-(I-{\lambda }_n\mathrm{\nabla }f)p\parallel \\ =& \parallel v_n-p\parallel +{\lambda }_n{\alpha }_n\parallel p\parallel \\ \le & \parallel x_n-p\parallel +{\lambda }_n{\alpha }_n\parallel p\parallel .\end{array}$$

(3.5)

Since T is a ξ-strictly pseudocontractive mapping and $({\gamma }_n+{\sigma }_n)\xi \le {\gamma }_n$, for all $n\ge 0$, by Lemma 2.2, we obtain from (3.1) and (3.5) that

$$\begin{array}{ll}\parallel y_n-p\parallel & =\parallel {\beta }_n{x}_n+{\gamma }_n{t}_n+{\sigma }_{n}T{t}_n-p\parallel \\ =\parallel {\beta }_n(x_n-p)+{\gamma }_n(t_n-p)+{\sigma }_n(T{t}_n-p)\parallel \\ \le {\beta }_n\parallel x_n-p\parallel +\parallel {\gamma }_n(t_n-p)+{\sigma }_n(T{t}_n-p)\parallel \\ \le {\beta }_n\parallel x_n-p\parallel +({\gamma }_n+{\sigma }_n)\parallel t_n-p\parallel \\ \le {\beta }_n\parallel x_n-p\parallel +({\gamma }_n+{\sigma }_n)[\parallel x_n-p\parallel +{\lambda }_n{\alpha }_n\parallel p\parallel ]\\ \le {\beta }_n\parallel x_n-p\parallel +({\gamma }_n+{\delta }_n)\parallel x_n-p\parallel +{\lambda }_n{\alpha }_n\parallel p\parallel \\ =\parallel x_n-p\parallel +{\lambda }_n{\alpha }_n\parallel p\parallel .\end{array}$$

(3.6)

Noticing the boundedness of $\{S{x}_n\}$, we get ${sup}_{n\ge 0}\parallel \gamma S{x}_n-\mu Fp\parallel \le \hat{M}$ for some $\hat{M}>0$. Moreover, utilizing Lemma 2.4 and (3.1), (3.6), we deduce that $\{{\lambda }_n\}\subset [a,b]\subset (0,\frac{2}{{\parallel A\parallel }^2})$ and $0<\gamma \le \tau$ that, for all $n\ge 0$,

$$\begin{array}{c}\parallel x_{n+1}-p\parallel \hfill \\ =\parallel {ϵ}_n\gamma ({\delta }_{n}V{x}_n+(1-{\delta }_n)S{x}_n)+(I-{ϵ}_n\mu F)y_n-p\parallel \hfill \\ =\parallel {ϵ}_n\gamma ({\delta }_{n}V{x}_n+(1-{\delta }_n)S{x}_n)-{ϵ}_n\mu Fp+(I-{ϵ}_n\mu F)y_n-(I-{ϵ}_n\mu F)p\parallel \hfill \\ \le \parallel {ϵ}_n\gamma ({\delta }_{n}V{x}_n+(1-{\delta }_n)S{x}_n)-{ϵ}_n\mu Fp\parallel +\parallel (I-{ϵ}_n\mu F)y_n-(I-{ϵ}_n\mu F)p\parallel \hfill \\ ={ϵ}_n\parallel {\delta }_n(\gamma V{x}_n-\mu Fp)+(1-{\delta }_n)(\gamma S{x}_n-\mu Fp)\parallel \hfill \\ +\parallel (I-{ϵ}_n\mu F)y_n-(I-{ϵ}_n\mu F)p\parallel \hfill \\ \le {ϵ}_n[{\delta }_n\parallel \gamma V{x}_n-\mu Fp\parallel +(1-{\delta }_n)\parallel \gamma S{x}_n-\mu Fp\parallel ]+(1-{ϵ}_n\tau )\parallel y_n-p\parallel \hfill \\ \le {ϵ}_n[{\delta }_n(\parallel \gamma V{x}_n-\gamma Vp\parallel +\parallel \gamma Vp-\mu Fp\parallel )+(1-{\delta }_n)\hat{M}]+(1-{ϵ}_n\tau )\parallel y_n-p\parallel \hfill \\ \le {ϵ}_n[{\delta }_n\gamma \rho \parallel x_n-p\parallel +{\delta }_n\parallel \gamma Vp-\mu Fp\parallel +(1-{\delta }_n)\hat{M}]\hfill \\ +(1-{ϵ}_n\tau )[\parallel x_n-p\parallel +{\lambda }_n{\alpha }_n\parallel p\parallel ]\hfill \\ \le {ϵ}_n[{\delta }_n\gamma \rho \parallel x_n-p\parallel +max\{\hat{M},\parallel \gamma Vp-\mu Fp\parallel \}]+(1-{ϵ}_n\tau )[\parallel x_n-p\parallel +{\lambda }_n{\alpha }_n\parallel p\parallel ]\hfill \\ \le {ϵ}_n\gamma \rho \parallel x_n-p\parallel +{ϵ}_{n}max\{\hat{M},\parallel \gamma Vp-\mu Fp\parallel \}+(1-{ϵ}_n\tau )\parallel x_n-p\parallel +{\lambda }_n{\alpha }_n\parallel p\parallel \hfill \\ =[1-(\tau -\gamma \rho ){ϵ}_n]\parallel x_n-p\parallel +{ϵ}_{n}max\{\hat{M},\parallel \gamma Vp-\mu Fp\parallel \}+{\lambda }_n{\alpha }_n\parallel p\parallel \hfill \\ =[1-(\tau -\gamma \rho ){ϵ}_n]\parallel x_n-p\parallel +(\tau -\gamma \rho ){ϵ}_{n}max\{\frac{\hat{M}}{\tau -\gamma \rho },\frac{\parallel \gamma Vp-\mu Fp\parallel }{\tau -\gamma \rho }\}+{\lambda }_n{\alpha }_n\parallel p\parallel \hfill \\ \le max\{\parallel x_n-p\parallel ,\frac{\hat{M}}{\tau -\gamma \rho },\frac{\parallel \gamma Vp-\mu Fp\parallel }{\tau -\gamma \rho }\}+{\alpha }_{n}b\parallel p\parallel .\hfill \end{array}$$

By induction, we get

$$\parallel x_{n+1}-p\parallel \le max\{\parallel x_0-p\parallel ,\frac{\hat{M}}{\tau -\gamma \rho },\frac{\parallel \gamma Vp-\mu Fp\parallel }{\tau -\gamma \rho }\}+\sum _{j=0}^n{\alpha }_{j}b\parallel p\parallel ,\mathrm{\forall }n\ge 0.$$

Thus, $\{x_n\}$ is bounded since ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n<\mathrm{\infty }$, and so are the sequences $\{t_n\}$, $\{u_n\}$, $\{v_n\}$, and $\{y_n\}$.

Step 2. We prove that ${lim}_{n\to \mathrm{\infty }}\frac{\parallel x_{n+1}-x_n\parallel }{{ϵ}_n}=0$.

Indeed, utilizing (2.1) and (2.3), we obtain

$$\begin{array}{c}\parallel v_{n+1}-v_n\parallel \hfill \\ =\parallel {\Lambda }_{n+1}^N{u}_{n+1}-{\Lambda }_n^N{u}_n\parallel \hfill \\ =\parallel J_{R_N,{\lambda }_{N,n+1}}(I-{\lambda }_{N,n+1}{B}_N){\Lambda }_{n+1}^{N-1}{u}_{n+1}-J_{R_N,{\lambda }_{N,n}}(I-{\lambda }_{N,n}{B}_N){\Lambda }_n^{N-1}{u}_n\parallel \hfill \\ \le \parallel J_{R_N,{\lambda }_{N,n+1}}(I-{\lambda }_{N,n+1}{B}_N){\Lambda }_{n+1}^{N-1}{u}_{n+1}-J_{R_N,{\lambda }_{N,n+1}}(I-{\lambda }_{N,n}{B}_N){\Lambda }_{n+1}^{N-1}{u}_{n+1}\parallel \hfill \\ +\parallel J_{R_N,{\lambda }_{N,n+1}}(I-{\lambda }_{N,n}{B}_N){\Lambda }_{n+1}^{N-1}{u}_{n+1}-J_{R_N,{\lambda }_{N,n}}(I-{\lambda }_{N,n}{B}_N){\Lambda }_n^{N-1}{u}_n\parallel \hfill \\ \le \parallel (I-{\lambda }_{N,n+1}{B}_N){\Lambda }_{n+1}^{N-1}{u}_{n+1}-(I-{\lambda }_{N,n}{B}_N){\Lambda }_{n+1}^{N-1}{u}_{n+1}\parallel \hfill \\ +\parallel (I-{\lambda }_{N,n}{B}_N){\Lambda }_{n+1}^{N-1}{u}_{n+1}-(I-{\lambda }_{N,n}{B}_N){\Lambda }_n^{N-1}{u}_n\parallel +|{\lambda }_{N,n+1}-{\lambda }_{N,n}|\hfill \\ \times (\frac{1}{{\lambda }_{N,n+1}}\parallel J_{R_N,{\lambda }_{N,n+1}}(I-{\lambda }_{N,n}{B}_N){\Lambda }_{n+1}^{N-1}{u}_{n+1}-(I-{\lambda }_{N,n}{B}_N){\Lambda }_n^{N-1}{u}_n\parallel \hfill \\ +\frac{1}{{\lambda }_{N,n}}\parallel (I-{\lambda }_{N,n}{B}_N){\Lambda }_{n+1}^{N-1}{u}_{n+1}-J_{R_N,{\lambda }_{N,n}}(I-{\lambda }_{N,n}{B}_N){\Lambda }_n^{N-1}{u}_n\parallel )\hfill \\ \le |{\lambda }_{N,n+1}-{\lambda }_{N,n}|(\parallel B_N{\Lambda }_{n+1}^{N-1}{u}_{n+1}\parallel +\tilde{M})+\parallel {\Lambda }_{n+1}^{N-1}{u}_{n+1}-{\Lambda }_n^{N-1}{u}_n\parallel \hfill \\ \le |{\lambda }_{N,n+1}-{\lambda }_{N,n}|(\parallel B_N{\Lambda }_{n+1}^{N-1}{u}_{n+1}\parallel +\tilde{M})\hfill \\ +|{\lambda }_{N-1,n+1}-{\lambda }_{N-1,n}|(\parallel B_{N-1}{\Lambda }_{n+1}^{N-2}{u}_{n+1}\parallel +\tilde{M})+\parallel {\Lambda }_{n+1}^{N-2}{u}_{n+1}-{\Lambda }_n^{N-2}{u}_n\parallel \hfill \\ ⋮\hfill \\ \le |{\lambda }_{N,n+1}-{\lambda }_{N,n}|(\parallel B_N{\Lambda }_{n+1}^{N-1}{u}_{n+1}\parallel +\tilde{M})\hfill \\ +|{\lambda }_{N-1,n+1}-{\lambda }_{N-1,n}|(\parallel B_{N-1}{\Lambda }_{n+1}^{N-2}{u}_{n+1}\parallel +\tilde{M})\hfill \\ +\cdots +|{\lambda }_{1,n+1}-{\lambda }_{1,n}|(\parallel B_1{\Lambda }_{n+1}^0{u}_{n+1}\parallel +\tilde{M})+\parallel {\Lambda }_{n+1}^0{u}_{n+1}-{\Lambda }_n^0{u}_n\parallel \hfill \\ \le {\tilde{M}}_0\sum _{i=1}^N|{\lambda }_{i,n+1}-{\lambda }_{i,n}|+\parallel u_{n+1}-u_n\parallel ,\hfill \end{array}$$

(3.7)

where

$$\begin{array}{c}\underset{n\ge 0}{sup}\{\frac{1}{{\lambda }_{N,n+1}}\parallel J_{R_N,{\lambda }_{N,n+1}}(I-{\lambda }_{N,n}{B}_N){\Lambda }_{n+1}^{N-1}{u}_{n+1}-(I-{\lambda }_{N,n}{B}_N){\Lambda }_n^{N-1}{u}_n\parallel \hfill \\ +\frac{1}{{\lambda }_{N,n}}\parallel (I-{\lambda }_{N,n}{B}_N){\Lambda }_{n+1}^{N-1}{u}_{n+1}-J_{R_N,{\lambda }_{N,n}}(I-{\lambda }_{N,n}{B}_N){\Lambda }_n^{N-1}{u}_n\parallel \}\le \tilde{M}\hfill \end{array}$$

for some $\tilde{M}>0$ and ${sup}_{n\ge 0}\{{\sum }_{i=1}^N\parallel B_i{\Lambda }_{n+1}^{i-1}{u}_{n+1}\parallel +\tilde{M}\}\le {\tilde{M}}_0$ for some ${\tilde{M}}_0>0$.

Utilizing Proposition 2.4(b), (e), we deduce that

$$\begin{array}{c}\parallel u_{n+1}-u_n\parallel \hfill \\ =\parallel {\Delta }_{n+1}^M{x}_{n+1}-{\Delta }_n^M{x}_n\parallel \hfill \\ =\parallel T_{r_{M,n+1}}^{({\Theta }_M,{\phi }_M)}(I-r_{M,n+1}{A}_M){\Delta }_{n+1}^{M-1}{x}_{n+1}-T_{r_{M,n}}^{({\Theta }_M,{\phi }_M)}(I-r_{M,n}{A}_M){\Delta }_n^{M-1}{x}_n\parallel \hfill \\ \le \parallel T_{r_{M,n+1}}^{({\Theta }_M,{\phi }_M)}(I-r_{M,n+1}{A}_M){\Delta }_{n+1}^{M-1}{x}_{n+1}-T_{r_{M,n}}^{({\Theta }_M,{\phi }_M)}(I-r_{M,n}{A}_M){\Delta }_{n+1}^{M-1}{x}_{n+1}\parallel \hfill \\ +\parallel T_{r_{M,n}}^{({\Theta }_M,{\phi }_M)}(I-r_{M,n}{A}_M){\Delta }_{n+1}^{M-1}{x}_{n+1}-T_{r_{M,n}}^{({\Theta }_M,{\phi }_M)}(I-r_{M,n}{A}_M){\Delta }_n^{M-1}{x}_n\parallel \hfill \\ \le \parallel T_{r_{M,n+1}}^{({\Theta }_M,{\phi }_M)}(I-r_{M,n+1}{A}_M){\Delta }_{n+1}^{M-1}{x}_{n+1}-T_{r_{M,n}}^{({\Theta }_M,{\phi }_M)}(I-r_{M,n+1}{A}_M){\Delta }_{n+1}^{M-1}{x}_{n+1}\parallel \hfill \\ +\parallel T_{r_{M,n}}^{({\Theta }_M,{\phi }_M)}(I-r_{M,n+1}{A}_M){\Delta }_{n+1}^{M-1}{x}_{n+1}-T_{r_{M,n}}^{({\Theta }_M,{\phi }_M)}(I-r_{M,n}{A}_M){\Delta }_{n+1}^{M-1}{x}_{n+1}\parallel \hfill \\ +\parallel (I-r_{M,n}{A}_M){\Delta }_{n+1}^{M-1}{x}_{n+1}-(I-r_{M,n}{A}_M){\Delta }_n^{M-1}{x}_n\parallel \hfill \\ \le \frac{|r_{M,n+1}-r_{M,n}|}{r_{M,n+1}}\parallel T_{r_{M,n+1}}^{({\Theta }_M,{\phi }_M)}(I-r_{M,n+1}{A}_M){\Delta }_{n+1}^{M-1}{x}_{n+1}-(I-r_{M,n+1}{A}_M){\Delta }_{n+1}^{M-1}{x}_{n+1}\parallel \hfill \\ +|r_{M,n+1}-r_{M,n}|\parallel A_M{\Delta }_{n+1}^{M-1}{x}_{n+1}\parallel +\parallel {\Delta }_{n+1}^{M-1}{x}_{n+1}-{\Delta }_n^{M-1}{x}_n\parallel \hfill \\ =|r_{M,n+1}-r_{M,n}|[\parallel A_M{\Delta }_{n+1}^{M-1}{x}_{n+1}\parallel +\frac{1}{r_{M,n+1}}\parallel T_{r_{M,n+1}}^{({\Theta }_M,{\phi }_M)}(I-r_{M,n+1}{A}_M){\Delta }_{n+1}^{M-1}{x}_{n+1}\hfill \\ -(I-r_{M,n+1}{A}_M){\Delta }_{n+1}^{M-1}{x}_{n+1}\parallel ]+\parallel {\Delta }_{n+1}^{M-1}{x}_{n+1}-{\Delta }_n^{M-1}{x}_n\parallel \hfill \\ ⋮\hfill \\ \le |r_{M,n+1}-r_{M,n}|[\parallel A_M{\Delta }_{n+1}^{M-1}{x}_{n+1}\parallel +\frac{1}{r_{M,n+1}}\parallel T_{r_{M,n+1}}^{({\Theta }_M,{\phi }_M)}(I-r_{M,n+1}{A}_M){\Delta }_{n+1}^{M-1}{x}_{n+1}\hfill \\ -(I-r_{M,n+1}{A}_M){\Delta }_{n+1}^{M-1}{x}_{n+1}\parallel ]+\cdots +|r_{1,n+1}-r_{1,n}|[\parallel A_1{\Delta }_{n+1}^0{x}_{n+1}\parallel \hfill \\ +\frac{1}{r_{1,n+1}}\parallel T_{r_{1,n+1}}^{({\Theta }_1,{\phi }_1)}(I-r_{1,n+1}{A}_1){\Delta }_{n+1}^0{x}_{n+1}-(I-r_{1,n+1}{A}_1){\Delta }_{n+1}^0{x}_{n+1}\parallel ]\hfill \\ +\parallel {\Delta }_{n+1}^0{x}_{n+1}-{\Delta }_n^0{x}_n\parallel \hfill \\ \le {\tilde{M}}_1\sum _{k=1}^M|r_{k,n+1}-r_{k,n}|+\parallel x_{n+1}-x_n\parallel ,\hfill \end{array}$$

(3.8)

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

$$\begin{array}{c}\sum _{k=1}^M[\parallel A_k{\Delta }_{n+1}^{k-1}{x}_{n+1}\parallel +\frac{1}{r_{k,n+1}}\parallel T_{r_{k,n+1}}^{({\Theta }_k,{\phi }_k)}(I-r_{k,n+1}{A}_k){\Delta }_{n+1}^{k-1}{x}_{n+1}-(I-r_{k,n+1}{A}_k){\Delta }_{n+1}^{k-1}{x}_{n+1}\parallel ]\hfill \\ \le {\tilde{M}}_1.\hfill \end{array}$$

Furthermore, we define $y_n={\beta }_n{x}_n+(1-{\beta }_n)w_n$ for all $n\ge 0$. It follows that

$$\begin{array}{c}{w}_{n+1}-w_n\hfill \\ =\frac{y_{n+1}-{\beta }_{n+1}{x}_{n+1}}{1-{\beta }_{n+1}}-\frac{y_n-{\beta }_n{x}_n}{1-{\beta }_n}\hfill \\ =\frac{{\gamma }_{n+1}{t}_{n+1}+{\sigma }_{n+1}T{t}_{n+1}}{1-{\beta }_{n+1}}-\frac{{\gamma }_n{t}_n+{\sigma }_{n}T{t}_n}{1-{\beta }_n}\hfill \\ =\frac{{\gamma }_{n+1}(t_{n+1}-t_n)+{\sigma }_{n+1}(T{t}_{n+1}-T{t}_n)}{1-{\beta }_{n+1}}+(\frac{{\gamma }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\gamma }_n}{1-{\beta }_n})t_n\hfill \\ +(\frac{{\sigma }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\sigma }_n}{1-{\beta }_n})T{t}_n.\hfill \end{array}$$

(3.9)

Since T is a ξ-strictly pseudocontractive mapping and $({\gamma }_n+{\sigma }_n)\xi \le {\gamma }_n$, for all $n\ge 0$, by Lemma 2.2, we obtain

$$\parallel {\gamma }_{n+1}(t_{n+1}-t_n)+{\sigma }_{n+1}(T{t}_{n+1}-T{t}_n)\parallel \le ({\gamma }_{n+1}+{\sigma }_{n+1})\parallel t_{n+1}-t_n\parallel .$$

(3.10)

Also, utilizing the nonexpansivity of $P_C(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})$, we have

$$\begin{array}{rcl}\parallel t_{n+1}-t_n\parallel & =& \parallel P_C(I-{\lambda }_{n+1}\mathrm{\nabla }{f}_{{\alpha }_{n+1}})v_{n+1}-P_C(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n\parallel \\ \le & \parallel P_C(I-{\lambda }_{n+1}\mathrm{\nabla }{f}_{{\alpha }_{n+1}})v_{n+1}-P_C(I-{\lambda }_{n+1}\mathrm{\nabla }{f}_{{\alpha }_{n+1}})v_n\parallel \\ +\parallel P_C(I-{\lambda }_{n+1}\mathrm{\nabla }{f}_{{\alpha }_{n+1}})v_n-P_C(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n\parallel \\ \le & \parallel v_{n+1}-v_n\parallel +\parallel (I-{\lambda }_{n+1}\mathrm{\nabla }{f}_{{\alpha }_{n+1}})v_n-(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n\parallel \\ \le & \parallel v_{n+1}-v_n\parallel +|{\lambda }_{n+1}{\alpha }_{n+1}-{\lambda }_n{\alpha }_n|\parallel v_n\parallel \\ +|{\lambda }_{n+1}-{\lambda }_n|\parallel \mathrm{\nabla }f(v_n)\parallel .\end{array}$$

(3.11)

Hence, from (3.7)-(3.11), it follows that

$$\begin{array}{c}\parallel w_{n+1}-w_n\parallel \hfill \\ \le \frac{\parallel {\gamma }_{n+1}(t_{n+1}-t_n)+{\sigma }_{n+1}(T{t}_{n+1}-T{t}_n)\parallel }{1-{\beta }_{n+1}}\hfill \\ +|\frac{{\gamma }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\gamma }_n}{1-{\beta }_n}|\parallel t_n\parallel +|\frac{{\sigma }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\sigma }_n}{1-{\beta }_n}|\parallel T{t}_n\parallel \hfill \\ \le \frac{({\gamma }_{n+1}+{\sigma }_{n+1})\parallel t_{n+1}-t_n\parallel }{1-{\beta }_{n+1}}+|\frac{{\gamma }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\gamma }_n}{1-{\beta }_n}|\parallel t_n\parallel +|\frac{{\sigma }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\sigma }_n}{1-{\beta }_n}|\parallel T{t}_n\parallel \hfill \\ =\parallel t_{n+1}-t_n\parallel +|\frac{{\gamma }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\gamma }_n}{1-{\beta }_n}|(\parallel t_n\parallel +\parallel T{t}_n\parallel )\hfill \\ \le \parallel v_{n+1}-v_n\parallel +|{\lambda }_{n+1}{\alpha }_{n+1}-{\lambda }_n{\alpha }_n|\parallel v_n\parallel +|{\lambda }_{n+1}-{\lambda }_n|\parallel \mathrm{\nabla }f(v_n)\parallel \hfill \\ +|\frac{{\gamma }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\gamma }_n}{1-{\beta }_n}|(\parallel t_n\parallel +\parallel T{t}_n\parallel )\hfill \\ \le {\tilde{M}}_0\sum _{i=1}^N|{\lambda }_{i,n+1}-{\lambda }_{i,n}|+\parallel u_{n+1}-u_n\parallel +|{\lambda }_{n+1}{\alpha }_{n+1}-{\lambda }_n{\alpha }_n|\parallel v_n\parallel \hfill \\ +|{\lambda }_{n+1}-{\lambda }_n|\parallel \mathrm{\nabla }f(v_n)\parallel +|\frac{{\gamma }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\gamma }_n}{1-{\beta }_n}|(\parallel t_n\parallel +\parallel T{t}_n\parallel )\hfill \\ \le {\tilde{M}}_0\sum _{i=1}^N|{\lambda }_{i,n+1}-{\lambda }_{i,n}|+{\tilde{M}}_1\sum _{k=1}^M|r_{k,n+1}-r_{k,n}|+\parallel x_{n+1}-x_n\parallel \hfill \\ +|{\lambda }_{n+1}{\alpha }_{n+1}-{\lambda }_n{\alpha }_n|\parallel v_n\parallel +|{\lambda }_{n+1}-{\lambda }_n|\parallel \mathrm{\nabla }f(v_n)\parallel \hfill \\ +|\frac{{\gamma }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\gamma }_n}{1-{\beta }_n}|(\parallel t_n\parallel +\parallel T{t}_n\parallel ).\hfill \end{array}$$

(3.12)

In the meantime, simple calculation shows that

$$y_{n+1}-y_n={\beta }_n(x_{n+1}-x_n)+(1-{\beta }_n)(w_{n+1}-w_n)+({\beta }_{n+1}-{\beta }_n)(x_{n+1}-w_{n+1}).$$

So, it follows from (3.12) that

$$\begin{array}{c}\parallel y_{n+1}-y_n\parallel \hfill \\ \le {\beta }_n\parallel x_{n+1}-x_n\parallel +(1-{\beta }_n)\parallel w_{n+1}-w_n\parallel +|{\beta }_{n+1}-{\beta }_n|\parallel x_{n+1}-w_{n+1}\parallel \hfill \\ \le {\beta }_n\parallel x_{n+1}-x_n\parallel +(1-{\beta }_n)[{\tilde{M}}_0\sum _{i=1}^N|{\lambda }_{i,n+1}-{\lambda }_{i,n}|+{\tilde{M}}_1\sum _{k=1}^M|r_{k,n+1}-r_{k,n}|\hfill \\ +\parallel x_{n+1}-x_n\parallel +|\frac{{\gamma }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\gamma }_n}{1-{\beta }_n}|(\parallel t_n\parallel +\parallel T{t}_n\parallel )+|{\lambda }_{n+1}{\alpha }_{n+1}-{\lambda }_n{\alpha }_n|\parallel v_n\parallel \hfill \\ +|{\lambda }_{n+1}-{\lambda }_n|\parallel \mathrm{\nabla }f(v_n)\parallel ]+|{\beta }_{n+1}-{\beta }_n|\parallel x_{n+1}-w_{n+1}\parallel \hfill \\ \le \parallel x_{n+1}-x_n\parallel +{\tilde{M}}_0\sum _{i=1}^N|{\lambda }_{i,n+1}-{\lambda }_{i,n}|+{\tilde{M}}_1\sum _{k=1}^M|r_{k,n+1}-r_{k,n}|\hfill \\ +\frac{|{\gamma }_{n+1}-{\gamma }_n|(1-{\beta }_n)+{\gamma }_n|{\beta }_{n+1}-{\beta }_n|}{1-{\beta }_{n+1}}(\parallel t_n\parallel +\parallel T{t}_n\parallel )\hfill \\ +|{\lambda }_{n+1}{\alpha }_{n+1}-{\lambda }_n{\alpha }_n|\parallel v_n\parallel +|{\lambda }_{n+1}-{\lambda }_n|\parallel \mathrm{\nabla }f(v_n)\parallel +|{\beta }_{n+1}-{\beta }_n|\parallel x_{n+1}-w_{n+1}\parallel \hfill \\ \le \parallel x_{n+1}-x_n\parallel +{\tilde{M}}_0\sum _{i=1}^N|{\lambda }_{i,n+1}-{\lambda }_{i,n}|+{\tilde{M}}_1\sum _{k=1}^M|r_{k,n+1}-r_{k,n}|\hfill \\ +|{\gamma }_{n+1}-{\gamma }_n|\frac{\parallel t_n\parallel +\parallel T{t}_n\parallel }{1-d}+|{\beta }_{n+1}-{\beta }_n|(\parallel x_{n+1}-w_{n+1}\parallel +\frac{\parallel t_n\parallel +\parallel T{t}_n\parallel }{1-d})\hfill \\ +|{\lambda }_{n+1}{\alpha }_{n+1}-{\lambda }_n{\alpha }_n|\parallel v_n\parallel +|{\lambda }_{n+1}-{\lambda }_n|\parallel \mathrm{\nabla }f(v_n)\parallel \hfill \\ \le \parallel x_{n+1}-x_n\parallel +{\tilde{M}}_2(\sum _{i=1}^N|{\lambda }_{i,n+1}-{\lambda }_{i,n}|+\sum _{k=1}^M|r_{k,n+1}-r_{k,n}|\hfill \\ +|{\gamma }_{n+1}-{\gamma }_n|+|{\beta }_{n+1}-{\beta }_n|+|{\lambda }_{n+1}{\alpha }_{n+1}-{\lambda }_n{\alpha }_n|+|{\lambda }_{n+1}-{\lambda }_n|),\hfill \end{array}$$

(3.13)

where ${sup}_{n\ge 0}\{\parallel x_{n+1}-w_{n+1}\parallel +\frac{\parallel t_n\parallel +\parallel T{t}_n\parallel }{1-d}+\parallel v_n\parallel +\parallel \mathrm{\nabla }f(v_n)\parallel +{\tilde{M}}_0+{\tilde{M}}_1\}\le {\tilde{M}}_2$ for some ${\tilde{M}}_2>0$.

On the other hand, we define $z_n:={\delta }_{n}V{x}_n+(1-{\delta }_n)S{x}_n$, for all $n\ge 0$. Then it is well known that $x_{n+1}={ϵ}_n\gamma z_n+(I-{ϵ}_n\mu F)y_n$, for all $n\ge 0$. Simple calculations show that

$$\{\begin{array}{l}{z}_{n+1}-z_n=({\delta }_{n+1}-{\delta }_n)(V{x}_n-S{x}_n)+{\delta }_{n+1}(V{x}_{n+1}-V{x}_n)\\ \phantom{z_{n+1}-z_n=}+(1-{\delta }_{n+1})(S{x}_{n+1}-S{x}_n),\\ x_{n+2}-x_{n+1}=({ϵ}_{n+1}-{ϵ}_n)(\gamma z_n-\mu F{y}_n)+{ϵ}_{n+1}\gamma (z_{n+1}-z_n)\\ \phantom{x_{n+2}-x_{n+1}=}+(I-{\lambda }_{n+1}\mu F)y_{n+1}-(I-{\lambda }_{n+1}\mu F)y_n.\end{array}$$

Since V is a ρ-contraction with coefficient $\rho \in [0,1)$ and S is a nonexpansive mapping, we conclude that

$$\begin{array}{rcl}\parallel z_{n+1}-z_n\parallel & \le & |{\delta }_{n+1}-{\delta }_n|\parallel V{x}_n-S{x}_n\parallel +{\delta }_{n+1}\parallel V{x}_{n+1}-V{x}_n\parallel \\ +(1-{\delta }_{n+1})\parallel S{x}_{n+1}-S{x}_n\parallel \\ \le & |{\delta }_{n+1}-{\delta }_n|\parallel V{x}_n-S{x}_n\parallel +{\delta }_{n+1}\rho \parallel x_{n+1}-x_n\parallel \\ +(1-{\delta }_{n+1})\parallel x_{n+1}-x_n\parallel \\ =& (1-{\delta }_{n+1}(1-\rho ))\parallel x_{n+1}-x_n\parallel +|{\delta }_{n+1}-{\delta }_n|\parallel V{x}_n-S{x}_n\parallel ,\end{array}$$

which together with (3.13) and $0<\gamma \le \tau$ implies that

$$\begin{array}{c}\parallel x_{n+2}-x_{n+1}\parallel \hfill \\ \le |{ϵ}_{n+1}-{ϵ}_n|\parallel \gamma z_n-\mu F{y}_n\parallel +{ϵ}_{n+1}\gamma \parallel z_{n+1}-z_n\parallel \hfill \\ +\parallel (I-{ϵ}_{n+1}\mu F)y_{n+1}-(I-{ϵ}_{n+1}\mu F)y_n\parallel \hfill \\ \le |{ϵ}_{n+1}-{ϵ}_n|\parallel \gamma z_n-\mu F{y}_n\parallel +{ϵ}_{n+1}\gamma \parallel z_{n+1}-z_n\parallel +(1-{ϵ}_{n+1}\tau )\parallel y_{n+1}-y_n\parallel \hfill \\ \le |{ϵ}_{n+1}-{ϵ}_n|\parallel \gamma z_n-\mu F{y}_n\parallel +{ϵ}_{n+1}\gamma [(1-{\delta }_{n+1}(1-\rho ))\parallel x_{n+1}-x_n\parallel \hfill \\ +|{\delta }_{n+1}-{\delta }_n|\parallel V{x}_n-S{x}_n\parallel ]+(1-{ϵ}_{n+1}\tau )[\parallel x_{n+1}-x_n\parallel \hfill \\ +{\tilde{M}}_2(\sum _{i=1}^N|{\lambda }_{i,n+1}-{\lambda }_{i,n}|+\sum _{k=1}^M|r_{k,n+1}-r_{k,n}|+|{\gamma }_{n+1}-{\gamma }_n|+|{\beta }_{n+1}-{\beta }_n|\hfill \\ +|{\lambda }_{n+1}{\alpha }_{n+1}-{\lambda }_n{\alpha }_n|+|{\lambda }_{n+1}-{\lambda }_n|\left)\right]\hfill \\ \le (1-{ϵ}_{n+1}(\tau -\gamma )-{ϵ}_{n+1}{\delta }_{n+1}(1-\rho )\gamma )\parallel x_{n+1}-x_n\parallel +|{ϵ}_{n+1}-{ϵ}_n|\parallel \gamma z_n-\mu F{y}_n\parallel \hfill \\ +{ϵ}_{n+1}|{\delta }_{n+1}-{\delta }_n|\parallel V{x}_n-S{x}_n\parallel +{\tilde{M}}_2(\sum _{i=1}^N|{\lambda }_{i,n+1}-{\lambda }_{i,n}|+\sum _{k=1}^M|r_{k,n+1}-r_{k,n}|\hfill \\ +|{\gamma }_{n+1}-{\gamma }_n|+|{\beta }_{n+1}-{\beta }_n|+|{\lambda }_{n+1}{\alpha }_{n+1}-{\lambda }_n{\alpha }_n|+|{\lambda }_{n+1}-{\lambda }_n|)\hfill \\ \le (1-{ϵ}_{n+1}{\delta }_{n+1}(1-\rho )\gamma )\parallel x_{n+1}-x_n\parallel +{\tilde{M}}_3\{\sum _{i=1}^N|{\lambda }_{i,n+1}-{\lambda }_{i,n}|\hfill \\ +\sum _{k=1}^M|r_{k,n+1}-r_{k,n}|+|{ϵ}_{n+1}-{ϵ}_n|+{ϵ}_{n+1}|{\delta }_{n+1}-{\delta }_n|+|{\beta }_{n+1}-{\beta }_n|\hfill \\ +|{\gamma }_{n+1}-{\gamma }_n|+|{\lambda }_{n+1}{\alpha }_{n+1}-{\lambda }_n{\alpha }_n|+|{\lambda }_{n+1}-{\lambda }_n|\},\hfill \end{array}$$

where ${sup}_{n\ge 0}\{\parallel \gamma z_n-\mu F{y}_n\parallel +\parallel V{x}_n-S{x}_n\parallel +{\tilde{M}}_2\}\le {\tilde{M}}_3$ for some ${\tilde{M}}_3>0$. Consequently,

$$\begin{array}{c}\frac{\parallel x_{n+1}-x_n\parallel }{{ϵ}_n}\hfill \\ \le (1-{ϵ}_n{\delta }_n(1-\rho )\gamma )\frac{\parallel x_n-x_{n-1}\parallel }{{ϵ}_n}+{\tilde{M}}_3\{\sum _{i=1}^N\frac{|{\lambda }_{i,n}-{\lambda }_{i,n-1}|}{{ϵ}_n}+\sum _{k=1}^M\frac{|r_{k,n}-r_{k,n-1}|}{{ϵ}_n}\hfill \\ +\frac{|{ϵ}_n-{ϵ}_{n-1}|}{{ϵ}_n}+|{\delta }_n-{\delta }_{n-1}|+\frac{|{\beta }_n-{\beta }_{n-1}|}{{ϵ}_n}+\frac{|{\gamma }_n-{\gamma }_{n-1}|}{{ϵ}_n}\hfill \\ +\frac{|{\lambda }_n{\alpha }_n-{\lambda }_{n-1}{\alpha }_{n-1}|}{{ϵ}_n}+\frac{|{\lambda }_n-{\lambda }_{n-1}|}{{ϵ}_n}\}\hfill \\ =(1-{ϵ}_n{\delta }_n(1-\rho )\gamma )\frac{\parallel x_n-x_{n-1}\parallel }{{ϵ}_{n-1}}+(1-{ϵ}_n{\delta }_n(1-\rho )\gamma )\parallel x_n-x_{n-1}\parallel (\frac{1}{{ϵ}_n}-\frac{1}{{ϵ}_{n-1}})\hfill \\ +{\tilde{M}}_3\{\sum _{i=1}^N\frac{|{\lambda }_{i,n}-{\lambda }_{i,n-1}|}{{ϵ}_n}+\sum _{k=1}^M\frac{|r_{k,n}-r_{k,n-1}|}{{ϵ}_n}+\frac{|{ϵ}_n-{ϵ}_{n-1}|}{{ϵ}_n}+|{\delta }_n-{\delta }_{n-1}|\hfill \\ +\frac{|{\beta }_n-{\beta }_{n-1}|}{{ϵ}_n}+\frac{|{\gamma }_n-{\gamma }_{n-1}|}{{ϵ}_n}+\frac{|{\lambda }_n{\alpha }_n-{\lambda }_{n-1}{\alpha }_{n-1}|}{{ϵ}_n}+\frac{|{\lambda }_n-{\lambda }_{n-1}|}{{ϵ}_n}\}\hfill \\ \le (1-{ϵ}_n{\delta }_n(1-\rho )\gamma )\frac{\parallel x_n-x_{n-1}\parallel }{{ϵ}_{n-1}}+{ϵ}_n{\delta }_n(1-\rho )\gamma \cdot \frac{{\tilde{M}}_4}{(1-\rho )\gamma }\{\frac{|{ϵ}_n-{ϵ}_{n-1}|}{{\delta }_n{ϵ}_n^2{ϵ}_{n-1}}\hfill \\ +\sum _{i=1}^N\frac{|{\lambda }_{i,n}-{\lambda }_{i,n-1}|}{{\delta }_n{ϵ}_n^2}+\sum _{k=1}^M\frac{|r_{k,n}-r_{k,n-1}|}{{\delta }_n{ϵ}_n^2}+\frac{|{ϵ}_n-{ϵ}_{n-1}|}{{\delta }_n{ϵ}_n^2}+\frac{|{\delta }_n-{\delta }_{n-1}|}{{\delta }_n{ϵ}_n}\hfill \\ +\frac{|{\beta }_n-{\beta }_{n-1}|}{{\delta }_n{ϵ}_n^2}+\frac{|{\gamma }_n-{\gamma }_{n-1}|}{{\delta }_n{ϵ}_n^2}+\frac{|{\lambda }_n{\alpha }_n-{\lambda }_{n-1}{\alpha }_{n-1}|}{{\delta }_n{ϵ}_n^2}+\frac{|{\lambda }_n-{\lambda }_{n-1}|}{{\delta }_n{ϵ}_n^2}\},\hfill \end{array}$$

(3.14)

where ${sup}_{n\ge 1}\{\parallel x_n-x_{n-1}\parallel +{\tilde{M}}_3\}\le {\tilde{M}}_4$ for some ${\tilde{M}}_4>0$. Utilizing Lemma 2.8, we conclude from conditions (C1)-(C6) and (C8)-(C9) that ${\sum }_{n=0}^{\mathrm{\infty }}{ϵ}_n{\delta }_n(1-\rho )\gamma =\mathrm{\infty }$ and

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

So, as ${ϵ}_n\to 0$, it follows that

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

Step 3. We prove that ${lim}_{n\to \mathrm{\infty }}\frac{\parallel x_n-u_n\parallel }{{ϵ}_n}=0$, ${lim}_{n\to \mathrm{\infty }}\frac{\parallel x_n-v_n\parallel }{{ϵ}_n}=0$, ${lim}_{n\to \mathrm{\infty }}\frac{\parallel v_n-t_n\parallel }{{ϵ}_n}=0$ and ${lim}_{n\to \mathrm{\infty }}\frac{\parallel t_n-T{t}_n\parallel }{{ϵ}_n}=0$.

Indeed, utilizing Lemmas 2.2 and 2.7(b), from (3.1), (3.4)-(3.5) and $({\gamma }_n+{\sigma }_n)\xi \le {\gamma }_n$, we deduce that

$$\begin{array}{c}{\parallel y_n-p\parallel }^2\hfill \\ ={\parallel {\beta }_n{x}_n+{\gamma }_n{t}_n+{\sigma }_{n}T{t}_n-p\parallel }^2\hfill \\ ={\parallel {\beta }_n(x_n-p)+(1-{\beta }_n)(\frac{{\gamma }_n{t}_n+{\sigma }_{n}T{t}_n}{1-{\beta }_n}-p)\parallel }^2\hfill \\ ={\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n){\parallel \frac{{\gamma }_n{t}_n+{\sigma }_{n}T{t}_n}{1-{\beta }_n}-p\parallel }^2-{\beta }_n(1-{\beta }_n){\parallel \frac{{\gamma }_n{t}_n+{\sigma }_{n}T{t}_n}{1-{\beta }_n}-x_n\parallel }^2\hfill \\ ={\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n){\parallel \frac{{\gamma }_n(t_n-p)+{\sigma }_n(T{t}_n-p)}{1-{\beta }_n}\parallel }^2-{\beta }_n(1-{\beta }_n){\parallel \frac{y_n-x_n}{1-{\beta }_n}\parallel }^2\hfill \\ \le {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n)\frac{{({\gamma }_n+{\sigma }_n)}^2{\parallel t_n-p\parallel }^2}{{(1-{\beta }_n)}^2}-\frac{{\beta }_n}{1-{\beta }_n}{\parallel y_n-x_n\parallel }^2\hfill \\ ={\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n){\parallel t_n-p\parallel }^2-\frac{{\beta }_n}{1-{\beta }_n}{\parallel y_n-x_n\parallel }^2\hfill \\ \le {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n){(\parallel x_n-p\parallel +{\lambda }_n{\alpha }_n\parallel p\parallel )}^2-\frac{{\beta }_n}{1-{\beta }_n}{\parallel y_n-x_n\parallel }^2\hfill \\ \le {\beta }_n{(\parallel x_n-p\parallel +{\lambda }_n{\alpha }_n\parallel p\parallel )}^2+(1-{\beta }_n){(\parallel x_n-p\parallel +{\lambda }_n{\alpha }_n\parallel p\parallel )}^2-\frac{{\beta }_n}{1-{\beta }_n}{\parallel y_n-x_n\parallel }^2\hfill \\ ={(\parallel x_n-p\parallel +{\lambda }_n{\alpha }_n\parallel p\parallel )}^2-\frac{{\beta }_n}{1-{\beta }_n}{\parallel y_n-x_n\parallel }^2\hfill \\ \le {(\parallel x_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )}^2-\frac{{\beta }_n}{1-{\beta }_n}{\parallel y_n-x_n\parallel }^2.\hfill \end{array}$$

(3.15)

Observe that

$$\begin{array}{rcl}{\parallel {\Delta }_n^k{x}_n-p\parallel }^2& =& {\parallel T_{r_{k,n}}^{({\Theta }_k,{\phi }_k)}(I-r_{k,n}{A}_k){\Delta }_n^{k-1}{x}_n-T_{r_{k,n}}^{({\Theta }_k,{\phi }_k)}(I-r_{k,n}{A}_k)p\parallel }^2\\ \le & {\parallel (I-r_{k,n}{A}_k){\Delta }_n^{k-1}{x}_n-(I-r_{k,n}{A}_k)p\parallel }^2\\ \le & {\parallel {\Delta }_n^{k-1}{x}_n-p\parallel }^2+r_{k,n}(r_{k,n}-2{\mu }_k){\parallel A_k{\Delta }_n^{k-1}{x}_n-A_{k}p\parallel }^2\\ \le & {\parallel x_n-p\parallel }^2+r_{k,n}(r_{k,n}-2{\mu }_k){\parallel A_k{\Delta }_n^{k-1}{x}_n-A_{k}p\parallel }^2\end{array}$$

(3.16)

and

$$\begin{array}{rcl}{\parallel {\Lambda }_n^i{u}_n-p\parallel }^2& =& {\parallel J_{R_i,{\lambda }_{i,n}}(I-{\lambda }_{i,n}{B}_i){\Lambda }_n^{i-1}{u}_n-J_{R_i,{\lambda }_{i,n}}(I-{\lambda }_{i,n}{B}_i)p\parallel }^2\\ \le & {\parallel (I-{\lambda }_{i,n}{B}_i){\Lambda }_n^{i-1}{u}_n-(I-{\lambda }_{i,n}{B}_i)p\parallel }^2\\ \le & {\parallel {\Lambda }_n^{i-1}{u}_n-p\parallel }^2+{\lambda }_{i,n}({\lambda }_{i,n}-2{\eta }_i){\parallel B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p\parallel }^2\\ \le & {\parallel u_n-p\parallel }^2+{\lambda }_{i,n}({\lambda }_{i,n}-2{\eta }_i){\parallel B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p\parallel }^2\\ \le & {\parallel x_n-p\parallel }^2+{\lambda }_{i,n}({\lambda }_{i,n}-2{\eta }_i){\parallel B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p\parallel }^2\end{array}$$

(3.17)

for $i\in \{1,2,\dots ,N\}$ and $k\in \{1,2,\dots ,M\}$. Combining (3.5), (3.15)-(3.17), we get

$$\begin{array}{c}{\parallel y_n-p\parallel }^2\hfill \\ \le {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n){\parallel t_n-p\parallel }^2-\frac{{\beta }_n}{1-{\beta }_n}{\parallel y_n-x_n\parallel }^2\hfill \\ \le {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n){\parallel t_n-p\parallel }^2\hfill \\ \le {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n){(\parallel v_n-p\parallel +{\lambda }_n{\alpha }_n\parallel p\parallel )}^2\hfill \\ \le {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n){(\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )}^2\hfill \\ ={\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n)[{\parallel v_n-p\parallel }^2+{\alpha }_{n}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )]\hfill \\ \le {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n){\parallel v_n-p\parallel }^2+{\alpha }_{n}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )\hfill \\ \le {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n){\parallel {\Lambda }_n^i{u}_n-p\parallel }^2+{\alpha }_{n}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )\hfill \\ \le {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n)[{\parallel u_n-p\parallel }^2+{\lambda }_{i,n}({\lambda }_{i,n}-2{\eta }_i){\parallel B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p\parallel }^2]\hfill \\ +{\alpha }_{n}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )\hfill \\ \le {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n)[{\parallel {\Delta }_n^k{x}_n-p\parallel }^2+{\lambda }_{i,n}({\lambda }_{i,n}-2{\eta }_i){\parallel B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p\parallel }^2]\hfill \\ +{\alpha }_{n}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )\hfill \\ \le {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n)[{\parallel x_n-p\parallel }^2+r_{k,n}(r_{k,n}-2{\mu }_k){\parallel A_k{\Delta }_n^{k-1}{x}_n-A_{k}p\parallel }^2\hfill \\ +{\lambda }_{i,n}({\lambda }_{i,n}-2{\eta }_i){\parallel B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p\parallel }^2]+{\alpha }_{n}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )\hfill \\ ={\parallel x_n-p\parallel }^2+(1-{\beta }_n)[r_{k,n}(r_{k,n}-2{\mu }_k){\parallel A_k{\Delta }_n^{k-1}{x}_n-A_{k}p\parallel }^2\hfill \\ +{\lambda }_{i,n}({\lambda }_{i,n}-2{\eta }_i){\parallel B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p\parallel }^2]\hfill \\ +{\alpha }_{n}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel ),\hfill \end{array}$$

(3.18)

which immediately leads to

$$\begin{array}{c}(1-d)[r_{k,n}(2{\mu }_k-r_{k,n})\frac{{\parallel A_k{\Delta }_n^{k-1}{x}_n-A_{k}p\parallel }^2}{{ϵ}_n^2}+{\lambda }_{i,n}(2{\eta }_i-{\lambda }_{i,n})\frac{{\parallel B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p\parallel }^2}{{ϵ}_n^2}]\hfill \\ \le (1-{\beta }_n)[r_{k,n}(2{\mu }_k-r_{k,n})\frac{{\parallel A_k{\Delta }_n^{k-1}{x}_n-A_{k}p\parallel }^2}{{ϵ}_n^2}+{\lambda }_{i,n}(2{\eta }_i-{\lambda }_{i,n})\frac{{\parallel B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p\parallel }^2}{{ϵ}_n^2}]\hfill \\ \le \frac{{\parallel x_n-p\parallel }^2-{\parallel y_n-p\parallel }^2}{{ϵ}_n^2}+\frac{{\alpha }_n}{{ϵ}_n^2}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )\hfill \\ \le \frac{\parallel x_n-y_n\parallel }{{ϵ}_n^2}(\parallel x_n-p\parallel +\parallel y_n-p\parallel )+\frac{{\alpha }_n}{{ϵ}_n^2}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel ).\hfill \end{array}$$

Since $\{{\lambda }_{i,n}\}\subset [a_i,b_i]\subset (0,2{\eta }_i)$, $\{r_{k,n}\}\subset [c_k,d_k]\subset (0,2{\mu }_k)$, $i\in \{1,2,\dots ,N\}$, $k\in \{1,2,\dots ,M\}$ and $\{v_n\}$, $\{x_n\}$, $\{y_n\}$ are bounded sequences, we obtain from $\parallel x_n-y_n\parallel +{\alpha }_n=o({ϵ}_n^2)$,

$$\underset{n\to \mathrm{\infty }}{lim}\frac{\parallel A_k{\Delta }_n^{k-1}{x}_n-A_{k}p\parallel }{{ϵ}_n}=0\text{and}\underset{n\to \mathrm{\infty }}{lim}\frac{\parallel B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p\parallel }{{ϵ}_n}=0$$

(3.19)

for all $k\in \{1,2,\dots ,M\}$ and $i\in \{1,2,\dots ,N\}$.

Furthermore, by Proposition 2.4(b) and Lemma 2.7(a), we have

$$\begin{array}{c}{\parallel {\Delta }_n^k{x}_n-p\parallel }^2\hfill \\ ={\parallel T_{r_{k,n}}^{({\Theta }_k,{\phi }_k)}(I-r_{k,n}{A}_k){\Delta }_n^{k-1}{x}_n-T_{r_{k,n}}^{({\Theta }_k,{\phi }_k)}(I-r_{k,n}{A}_k)p\parallel }^2\hfill \\ \le 〈(I-r_{k,n}{A}_k){\Delta }_n^{k-1}{x}_n-(I-r_{k,n}{A}_k)p,{\Delta }_n^k{x}_n-p〉\hfill \\ =\frac{1}{2}({\parallel (I-r_{k,n}{A}_k){\Delta }_n^{k-1}{x}_n-(I-r_{k,n}{A}_k)p\parallel }^2+{\parallel {\Delta }_n^k{x}_n-p\parallel }^2\hfill \\ -{\parallel (I-r_{k,n}{A}_k){\Delta }_n^{k-1}{x}_n-(I-r_{k,n}{A}_k)p-({\Delta }_n^k{x}_n-p)\parallel }^2)\hfill \\ \le \frac{1}{2}({\parallel {\Delta }_n^{k-1}{x}_n-p\parallel }^2+{\parallel {\Delta }_n^k{x}_n-p\parallel }^2\hfill \\ -{\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n-r_{k,n}(A_k{\Delta }_n^{k-1}{x}_n-A_{k}p)\parallel }^2),\hfill \end{array}$$

which implies that

$$\begin{array}{c}{\parallel {\Delta }_n^k{x}_n-p\parallel }^2\hfill \\ \le {\parallel {\Delta }_n^{k-1}{x}_n-p\parallel }^2-{\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n-r_{k,n}(A_k{\Delta }_n^{k-1}{x}_n-A_{k}p)\parallel }^2\hfill \\ ={\parallel {\Delta }_n^{k-1}{x}_n-p\parallel }^2-{\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel }^2-r_{k,n}^2{\parallel A_k{\Delta }_n^{k-1}{x}_n-A_{k}p\parallel }^2\hfill \\ +2r_{k,n}〈{\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n,A_k{\Delta }_n^{k-1}{x}_n-A_{k}p〉\hfill \\ \le {\parallel {\Delta }_n^{k-1}{x}_n-p\parallel }^2-{\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel }^2+2r_{k,n}\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel \parallel A_k{\Delta }_n^{k-1}{x}_n-A_{k}p\parallel \hfill \\ \le {\parallel x_n-p\parallel }^2-{\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel }^2\hfill \\ +2r_{k,n}\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel \parallel A_k{\Delta }_n^{k-1}{x}_n-A_{k}p\parallel .\hfill \end{array}$$

(3.20)

By Lemma 2.7(a) and Lemma 2.10, we obtain

$$\begin{array}{c}{\parallel {\Lambda }_n^i{u}_n-p\parallel }^2\hfill \\ ={\parallel J_{R_i,{\lambda }_{i,n}}(I-{\lambda }_{i,n}{B}_i){\Lambda }_n^{i-1}{u}_n-J_{R_i,{\lambda }_{i,n}}(I-{\lambda }_{i,n}{B}_i)p\parallel }^2\hfill \\ \le 〈(I-{\lambda }_{i,n}{B}_i){\Lambda }_n^{i-1}{u}_n-(I-{\lambda }_{i,n}{B}_i)p,{\Lambda }_n^i{u}_n-p〉\hfill \\ =\frac{1}{2}({\parallel (I-{\lambda }_{i,n}{B}_i){\Lambda }_n^{i-1}{u}_n-(I-{\lambda }_{i,n}{B}_i)p\parallel }^2+{\parallel {\Lambda }_n^i{u}_n-p\parallel }^2\hfill \\ -{\parallel (I-{\lambda }_{i,n}{B}_i){\Lambda }_n^{i-1}{u}_n-(I-{\lambda }_{i,n}{B}_i)p-({\Lambda }_n^i{u}_n-p)\parallel }^2)\hfill \\ \le \frac{1}{2}({\parallel {\Lambda }_n^{i-1}{u}_n-p\parallel }^2+{\parallel {\Lambda }_n^i{u}_n-p\parallel }^2\hfill \\ -{\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n-{\lambda }_{i,n}(B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p)\parallel }^2)\hfill \\ \le \frac{1}{2}({\parallel u_n-p\parallel }^2+{\parallel {\Lambda }_n^i{u}_n-p\parallel }^2-{\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n-{\lambda }_{i,n}(B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p)\parallel }^2)\hfill \\ \le \frac{1}{2}({\parallel x_n-p\parallel }^2+{\parallel {\Lambda }_n^i{u}_n-p\parallel }^2-{\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n-{\lambda }_{i,n}(B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p)\parallel }^2),\hfill \end{array}$$

which immediately leads to

$$\begin{array}{c}{\parallel {\Lambda }_n^i{u}_n-p\parallel }^2\hfill \\ \le {\parallel x_n-p\parallel }^2-{\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n-{\lambda }_{i,n}(B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p)\parallel }^2\hfill \\ ={\parallel x_n-p\parallel }^2-{\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^k{u}_n\parallel }^2-{\lambda }_{i,n}^2{\parallel B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p\parallel }^2\hfill \\ +2{\lambda }_{i,n}〈{\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n,B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p〉\hfill \\ \le {\parallel x_n-p\parallel }^2-{\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n\parallel }^2\hfill \\ +2{\lambda }_{i,n}\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n\parallel \parallel B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p\parallel .\hfill \end{array}$$

(3.21)

Combining (3.15) and (3.21), we conclude that

$$\begin{array}{c}{\parallel y_n-p\parallel }^2\hfill \\ \le {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n){\parallel v_n-p\parallel }^2+{\alpha }_{n}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )\hfill \\ \le {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n){\parallel {\Lambda }_n^i{u}_n-p\parallel }^2+{\alpha }_{n}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )\hfill \\ \le {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n)[{\parallel x_n-p\parallel }^2-{\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n\parallel }^2\hfill \\ +2{\lambda }_{i,n}\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n\parallel \parallel B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p\parallel ]+{\alpha }_{n}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )\hfill \\ \le {\parallel x_n-p\parallel }^2-(1-{\beta }_n){\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n\parallel }^2\hfill \\ +2{\lambda }_{i,n}\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n\parallel \parallel B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p\parallel +{\alpha }_{n}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel ),\hfill \end{array}$$

which yields

$$\begin{array}{c}(1-d)\frac{{\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n\parallel }^2}{{ϵ}_n^2}\hfill \\ \le (1-{\beta }_n){\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n\parallel }^2\hfill \\ \le \frac{{\parallel x_n-p\parallel }^2-{\parallel y_n-p\parallel }^2}{{ϵ}_n^2}+2{\lambda }_{i,n}\frac{\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n\parallel \parallel B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p\parallel }{{ϵ}_n^2}\hfill \\ +{\alpha }_{n}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )\hfill \\ \le \frac{\parallel x_n-y_n\parallel }{{ϵ}_n^2}(\parallel x_n-p\parallel +\parallel y_n-p\parallel )+2{\lambda }_{i,n}\frac{\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n\parallel }{{ϵ}_n}\frac{\parallel B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p\parallel }{{ϵ}_n}\hfill \\ +\frac{{\alpha }_n}{{ϵ}_n^2}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel ).\hfill \end{array}$$

So, it follows from $\{{\lambda }_{i,n}\}\subset [a_i,b_i]\subset (0,2{\eta }_i)$, $i=1,2,\dots ,N$, that

$$\begin{array}{c}(1-d)\frac{{\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n\parallel }^2}{{ϵ}_n^2}\hfill \\ \le \frac{\parallel x_n-y_n\parallel }{{ϵ}_n^2}(\parallel x_n-p\parallel +\parallel y_n-p\parallel )\hfill \\ +2b_i\frac{\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n\parallel }{{ϵ}_n}\frac{\parallel B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p\parallel }{{ϵ}_n}\hfill \\ +\frac{{\alpha }_n}{{ϵ}_n^2}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel ).\hfill \end{array}$$

(3.22)

Now we claim that

$$\underset{n\to \mathrm{\infty }}{lim}\frac{\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n\parallel }{{ϵ}_n}=0,\mathrm{\forall }i\in \{1,2,\dots ,N\}.$$

(3.23)

As a matter of fact, it is easy to see that, for each $i\in \{1,2,\dots ,N\}$,

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n\parallel }{{ϵ}_n}\le \mathrm{\infty }.$$

If ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\frac{\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n\parallel }{{ϵ}_n}<\mathrm{\infty }$, then from (3.22) and ${lim}_{n\to \mathrm{\infty }}\frac{\parallel B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p\parallel }{{ϵ}_n}=0$ (due to (3.9)), we have

$$\begin{array}{c}(1-d)\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{{\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n\parallel }^2}{{ϵ}_n^2}\hfill \\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{\parallel x_n-y_n\parallel }{{ϵ}_n^2}(\parallel x_n-p\parallel +\parallel y_n-p\parallel )\hfill \\ +2b_i\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n\parallel }{{ϵ}_n}\frac{\parallel B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p\parallel }{{ϵ}_n}\hfill \\ +\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{{\alpha }_n}{{ϵ}_n^2}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )\hfill \\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{\parallel x_n-y_n\parallel }{{ϵ}_n^2}(\parallel x_n-p\parallel +\parallel y_n-p\parallel )\hfill \\ +2b_i\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n\parallel }{{ϵ}_n}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{\parallel B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p\parallel }{{ϵ}_n}\hfill \\ +\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{{\alpha }_n}{{ϵ}_n^2}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )\hfill \\ =0.\hfill \end{array}$$

That is, ${lim}_{n\to \mathrm{\infty }}\frac{\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n\parallel }{{ϵ}_n}=0$. If ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\frac{\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n\parallel }{{ϵ}_n}=\mathrm{\infty }$, then from (3.22), we have

$$\begin{array}{c}(1-d)\frac{\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n\parallel }{{ϵ}_n}[\frac{\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n\parallel }{{ϵ}_n}-\frac{2b_i}{1-d}\frac{\parallel B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p\parallel }{{ϵ}_n}]\hfill \\ \le \frac{\parallel x_n-y_n\parallel }{{ϵ}_n^2}(\parallel x_n-p\parallel +\parallel y_n-p\parallel )+\frac{{\alpha }_n}{{ϵ}_n^2}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel ).\hfill \end{array}$$

(3.24)

Since ${lim}_{n\to \mathrm{\infty }}\frac{\parallel B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p\parallel }{{ϵ}_n}=0$ (due to (3.19)), it is easy to see that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n\parallel }{{ϵ}_n}[\frac{\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n\parallel }{{ϵ}_n}-\frac{2b_i}{1-d}\frac{\parallel B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p\parallel }{{ϵ}_n}]=\mathrm{\infty }.$$

Thus, from (3.24), it follows that

$$\begin{array}{rcl}\mathrm{\infty }& =& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(1-d)\frac{\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n\parallel }{{ϵ}_n}[\frac{\parallel {\Lambda }_n^{i-1}{u}_n-{\Lambda }_n^i{u}_n\parallel }{{ϵ}_n}-\frac{2b_i}{1-d}\frac{\parallel B_i{\Lambda }_n^{i-1}{u}_n-B_{i}p\parallel }{{ϵ}_n}]\\ \le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{\parallel x_n-y_n\parallel }{{ϵ}_n^2}(\parallel x_n-p\parallel +\parallel y_n-p\parallel )\\ +\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{{\alpha }_n}{{ϵ}_n^2}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )\\ =& 0,\end{array}$$

which leads to a contradiction. This shows that (3.23) holds.

Also, combining (3.3), (3.15), and (3.20), we deduce that

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& \le & {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n){\parallel v_n-p\parallel }^2+{\alpha }_{n}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )\\ \le & {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n){\parallel u_n-p\parallel }^2+{\alpha }_{n}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )\\ \le & {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n){\parallel {\Delta }_n^k{x}_n-p\parallel }^2+{\alpha }_{n}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )\\ \le & {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n)[{\parallel x_n-p\parallel }^2-{\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel }^2\\ +2r_{k,n}\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel \parallel A_k{\Delta }_n^{k-1}{x}_n-A_{k}p\parallel ]\\ +{\alpha }_{n}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )\\ \le & {\parallel x_n-p\parallel }^2-(1-{\beta }_n){\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel }^2\\ +2r_{k,n}\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel \parallel A_k{\Delta }_n^{k-1}{x}_n-A_{k}p\parallel \\ +{\alpha }_{n}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel ),\end{array}$$

which yields

$$\begin{array}{c}(1-d)\frac{{\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel }^2}{{ϵ}_n^2}\hfill \\ \le (1-{\beta }_n)\frac{{\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel }^2}{{ϵ}_n^2}\hfill \\ \le \frac{{\parallel x_n-p\parallel }^2-{\parallel y_n-p\parallel }^2}{{ϵ}_n^2}+2r_{k,n}\frac{\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel \parallel A_k{\Delta }_n^{k-1}{x}_n-A_{k}p\parallel }{{ϵ}_n^2}\hfill \\ +\frac{{\alpha }_n}{{ϵ}_n^2}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )\hfill \\ \le \frac{\parallel x_n-y_n\parallel }{{ϵ}_n^2}(\parallel x_n-p\parallel +\parallel y_n-p\parallel )+2r_{k,n}\frac{\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel }{{ϵ}_n}\frac{\parallel A_k{\Delta }_n^{k-1}{x}_n-A_{k}p\parallel }{{ϵ}_n}\hfill \\ +\frac{{\alpha }_n}{{ϵ}_n^2}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel ).\hfill \end{array}$$

So, it follows from $\{r_{k,n}\}\subset [c_k,d_k]\subset (0,2{\mu }_k)$, $k=1,2,\dots ,M$, that

$$\begin{array}{c}(1-d)\frac{{\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel }^2}{{ϵ}_n^2}\hfill \\ \le \frac{\parallel x_n-y_n\parallel }{{ϵ}_n^2}(\parallel x_n-p\parallel +\parallel y_n-p\parallel )\hfill \\ +2d_k\frac{\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel }{{ϵ}_n}\frac{\parallel A_k{\Delta }_n^{k-1}{x}_n-A_{k}p\parallel }{{ϵ}_n}\hfill \\ +\frac{{\alpha }_n}{{ϵ}_n^2}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel ).\hfill \end{array}$$

(3.25)

Next, we claim that

$$\underset{n\to \mathrm{\infty }}{lim}\frac{\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel }{{ϵ}_n}=0,\mathrm{\forall }k\in \{1,2,\dots ,M\}.$$

(3.26)

As a matter of fact, it is easy to see that, for each $k\in \{1,2,\dots ,M\}$,

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel }{{ϵ}_n}\le \mathrm{\infty }.$$

If ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\frac{\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel }{{ϵ}_n}<\mathrm{\infty }$, then from (3.25) and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\frac{\parallel A_k{\Delta }_n^{k-1}{x}_n-A_{k}p\parallel }{{ϵ}_n}=0$ (due to (3.19)), we have

$$\begin{array}{c}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(1-d)\frac{{\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel }^2}{{ϵ}_n^2}\hfill \\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{\parallel x_n-y_n\parallel }{{ϵ}_n^2}(\parallel x_n-p\parallel +\parallel y_n-p\parallel )\hfill \\ +2d_k\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel }{{ϵ}_n}\frac{\parallel A_k{\Delta }_n^{k-1}{x}_n-A_{k}p\parallel }{{ϵ}_n}\hfill \\ +\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{{\alpha }_n}{{ϵ}_n^2}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )\hfill \\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{\parallel x_n-y_n\parallel }{{ϵ}_n^2}(\parallel x_n-p\parallel +\parallel y_n-p\parallel )\hfill \\ +2d_k\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel }{{ϵ}_n}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{\parallel A_k{\Delta }_n^{k-1}{x}_n-A_{k}p\parallel }{{ϵ}_n}\hfill \\ +\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{{\alpha }_n}{{ϵ}_n^2}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )\hfill \\ =0.\hfill \end{array}$$

That is, ${lim}_{n\to \mathrm{\infty }}\frac{\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel }{{ϵ}_n}=0$. If ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\frac{\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel }{{ϵ}_n}=\mathrm{\infty }$, then from (3.25), we have

$$\begin{array}{c}(1-d)\frac{\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel }{{ϵ}_n}[\frac{\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel }{{ϵ}_n}-\frac{\parallel A_k{\Delta }_n^{k-1}{x}_n-A_{k}p\parallel }{{ϵ}_n}]\hfill \\ \le \frac{\parallel x_n-y_n\parallel }{{ϵ}_n^2}(\parallel x_n-p\parallel +\parallel y_n-p\parallel )+\frac{{\alpha }_n}{{ϵ}_n^2}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel ).\hfill \end{array}$$

(3.27)

Since ${lim}_{n\to \mathrm{\infty }}\frac{\parallel A_k{\Delta }_n^{k-1}{x}_n-A_{k}p\parallel }{{ϵ}_n}=0$ (due to (3.19)), it is easy to see that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel }{{ϵ}_n}[\frac{\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel }{{ϵ}_n}-\frac{\parallel A_k{\Delta }_n^{k-1}{x}_n-A_{k}p\parallel }{{ϵ}_n}]=\mathrm{\infty }.$$

Consequently, from (3.27), it follows that

$$\begin{array}{rcl}\mathrm{\infty }& =& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(1-d)\frac{\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel }{{ϵ}_n}[\frac{\parallel {\Delta }_n^{k-1}{x}_n-{\Delta }_n^k{x}_n\parallel }{{ϵ}_n}-\frac{\parallel A_k{\Delta }_n^{k-1}{x}_n-A_{k}p\parallel }{{ϵ}_n}]\\ \le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{\parallel x_n-y_n\parallel }{{ϵ}_n^2}(\parallel x_n-p\parallel +\parallel y_n-p\parallel )\\ +\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{{\alpha }_n}{{ϵ}_n^2}b\parallel p\parallel (2\parallel v_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )\\ =& 0,\end{array}$$

which leads to a contradiction. This shows that (3.26) is valid. Therefore, from (3.23) and (3.26), we get

$$\begin{array}{rcl}\frac{\parallel x_n-u_n\parallel }{{ϵ}_n}& =& \frac{\parallel {\Delta }_n^0{x}_n-{\Delta }_n^M{x}_n\parallel }{{ϵ}_n}\\ \le & \frac{\parallel {\Delta }_n^0{x}_n-{\Delta }_n^1{x}_n\parallel }{{ϵ}_n}+\frac{\parallel {\Delta }_n^1{x}_n-{\Delta }_n^2{x}_n\parallel }{{ϵ}_n}\\ +\cdots +\frac{\parallel {\Delta }_n^{M-1}{x}_n-{\Delta }_n^M{x}_n\parallel }{{ϵ}_n}\\ \to & 0\text{as }n\to \mathrm{\infty }\end{array}$$

(3.28)

and

$$\begin{array}{rcl}\frac{\parallel u_n-v_n\parallel }{{ϵ}_n}& =& \frac{\parallel {\Lambda }_n^0{u}_n-{\Lambda }_n^N{u}_n\parallel }{{ϵ}_n}\\ \le & \frac{\parallel {\Lambda }_n^0{u}_n-{\Lambda }_n^1{u}_n\parallel }{{ϵ}_n}+\frac{\parallel {\Lambda }_n^1{u}_n-{\Lambda }_n^2{u}_n\parallel }{{ϵ}_n}\\ +\cdots +\frac{\parallel {\Lambda }_n^{N-1}{u}_n-{\Lambda }_n^N{u}_n\parallel }{{ϵ}_n}\\ \to & 0\text{as }n\to \mathrm{\infty },\end{array}$$

(3.29)

respectively. Thus, from (3.28) and (3.29), we obtain

$$\frac{\parallel x_n-v_n\parallel }{{ϵ}_n}\le \frac{\parallel x_n-u_n\parallel }{{ϵ}_n}+\frac{\parallel u_n-v_n\parallel }{{ϵ}_n}\to 0\text{as }n\to \mathrm{\infty }.$$

(3.30)

On the other hand, note that $\Gamma =VI(C,\mathrm{\nabla }f)$. Then, utilizing Lemma 2.6 and the $\frac{1}{{\parallel A\parallel }^2}$-inverse strong monotonicity of ∇f, we deduce from (2.1) that

$$\begin{array}{rcl}{\parallel t_n-p\parallel }^2& \le & {\parallel (I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n-(I-{\lambda }_n\mathrm{\nabla }f)p\parallel }^2\\ =& {\parallel v_n-p-{\lambda }_n(\mathrm{\nabla }f(v_n)-\mathrm{\nabla }f(p))-{\lambda }_n{\alpha }_n{v}_n\parallel }^2\\ \le & {\parallel v_n-p-{\lambda }_n(\mathrm{\nabla }f(v_n)-\mathrm{\nabla }f(p))\parallel }^2\\ -2{\lambda }_n{\alpha }_n〈v_n,(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n-(I-{\lambda }_n\mathrm{\nabla }f)p〉\\ \le & {\parallel v_n-p\parallel }^2+{\lambda }_n({\lambda }_n-\frac{2}{{\parallel A\parallel }^2}){\parallel \mathrm{\nabla }f(v_n)-\mathrm{\nabla }f(p)\parallel }^2\\ +2{\alpha }_{n}b\parallel v_n\parallel \parallel v_n-p-{\lambda }_n(\mathrm{\nabla }{f}_{{\alpha }_n}(v_n)-\mathrm{\nabla }f(p))\parallel .\end{array}$$

(3.31)

Combining (3.4), (3.15), and (3.31), we obtain

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& \le & {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n){\parallel t_n-p\parallel }^2\\ \le & {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n)[{\parallel v_n-p\parallel }^2+{\lambda }_n({\lambda }_n-\frac{2}{{\parallel A\parallel }^2}){\parallel \mathrm{\nabla }f(v_n)-\mathrm{\nabla }f(p)\parallel }^2\\ +2{\alpha }_{n}b\parallel v_n\parallel \parallel v_n-p-{\lambda }_n(\mathrm{\nabla }{f}_{{\alpha }_n}(v_n)-\mathrm{\nabla }f(p))\parallel ]\\ \le & {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n)[{\parallel x_n-p\parallel }^2+{\lambda }_n({\lambda }_n-\frac{2}{{\parallel A\parallel }^2}){\parallel \mathrm{\nabla }f(v_n)-\mathrm{\nabla }f(p)\parallel }^2\\ +2{\alpha }_{n}b\parallel v_n\parallel \parallel v_n-p-{\lambda }_n(\mathrm{\nabla }{f}_{{\alpha }_n}(v_n)-\mathrm{\nabla }f(p))\parallel ]\\ \le & {\parallel x_n-p\parallel }^2+(1-{\beta }_n){\lambda }_n({\lambda }_n-\frac{2}{{\parallel A\parallel }^2}){\parallel \mathrm{\nabla }f(v_n)-\mathrm{\nabla }f(p)\parallel }^2\\ +2{\alpha }_{n}b\parallel v_n\parallel \parallel v_n-p-{\lambda }_n(\mathrm{\nabla }{f}_{{\alpha }_n}(v_n)-\mathrm{\nabla }f(p))\parallel ,\end{array}$$

which together with $\{{\lambda }_n\}\subset [a,b]\subset (0,\frac{2}{{\parallel A\parallel }^2})$ and $\{{\beta }_n\}\subset [c,d]\subset (0,1)$ leads to

$$\begin{array}{c}(1-d)a(\frac{2}{{\parallel A\parallel }^2}-b)\frac{{\parallel \mathrm{\nabla }f(v_n)-\mathrm{\nabla }f(p)\parallel }^2}{{ϵ}_n^2}\hfill \\ \le (1-{\beta }_n){\lambda }_n(\frac{2}{{\parallel A\parallel }^2}-{\lambda }_n)\frac{{\parallel \mathrm{\nabla }f(v_n)-\mathrm{\nabla }f(p)\parallel }^2}{{ϵ}_n^2}\hfill \\ \le \frac{{\parallel x_n-p\parallel }^2-{\parallel y_n-p\parallel }^2}{{ϵ}_n^2}+2\frac{{\alpha }_n}{{ϵ}_n^2}b\parallel v_n\parallel \parallel v_n-p-{\lambda }_n(\mathrm{\nabla }{f}_{{\alpha }_n}(v_n)-\mathrm{\nabla }f(p))\parallel \hfill \\ \le \frac{\parallel x_n-y_n\parallel }{{ϵ}_n^2}(\parallel x_n-p\parallel +\parallel y_n-p\parallel )\hfill \\ +2\frac{{\alpha }_n}{{ϵ}_n^2}b\parallel v_n\parallel \parallel v_n-p-{\lambda }_n(\mathrm{\nabla }{f}_{{\alpha }_n}(v_n)-\mathrm{\nabla }f(p))\parallel .\hfill \end{array}$$

Since $\{v_n\}$, $\{x_n\}$, and $\{y_n\}$ are bounded sequences, we deduce from $\parallel x_n-y_n\parallel +{\alpha }_n=o({ϵ}_n^2)$ that

$$\underset{n\to \mathrm{\infty }}{lim}\frac{\parallel \mathrm{\nabla }f(v_n)-\mathrm{\nabla }f(p)\parallel }{{ϵ}_n}=0.$$

So, it is clear that

$$\underset{n\to \mathrm{\infty }}{lim}\frac{\parallel \mathrm{\nabla }{f}_{{\alpha }_n}(v_n)-\mathrm{\nabla }f(p)\parallel }{{ϵ}_n}=0.$$

(3.32)

Again, utilizing Proposition 2.3(c), from $t_n=P_C(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n$ and $p=P_C(I-{\lambda }_n\mathrm{\nabla }f)p$, we get

$$\begin{array}{rcl}{\parallel t_n-p\parallel }^2& =& {\parallel P_C(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n-P_C(I-{\lambda }_n\mathrm{\nabla }f)p\parallel }^2\\ \le & 〈(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n-(I-{\lambda }_n\mathrm{\nabla }f)p,t_n-p〉\\ =& \frac{1}{2}({\parallel (I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n-(I-{\lambda }_n\mathrm{\nabla }f)p\parallel }^2+{\parallel t_n-p\parallel }^2\\ -{\parallel (I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n-(I-{\lambda }_n\mathrm{\nabla }f)p-(t_n-p)\parallel }^2)\\ =& \frac{1}{2}({\parallel (I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n-(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})p-{\lambda }_n{\alpha }_{n}p\parallel }^2+{\parallel t_n-p\parallel }^2\\ -{\parallel (I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n-(I-{\lambda }_n\mathrm{\nabla }f)p-(t_n-p)\parallel }^2)\\ =& \frac{1}{2}({\parallel (I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n-(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})p\parallel }^2-2{\lambda }_n{\alpha }_n〈p,(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n\\ -(I-{\lambda }_n\mathrm{\nabla }f)p〉\\ +{\parallel t_n-p\parallel }^2-{\parallel (I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n-(I-{\lambda }_n\mathrm{\nabla }f)p-(t_n-p)\parallel }^2)\\ \le & \frac{1}{2}({\parallel v_n-p\parallel }^2-2{\lambda }_n{\alpha }_n〈p,(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n-(I-{\lambda }_n\mathrm{\nabla }f)p〉+{\parallel t_n-p\parallel }^2\\ -{\parallel (I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n-(I-{\lambda }_n\mathrm{\nabla }f)p-(t_n-p)\parallel }^2)\\ \le & \frac{1}{2}({\parallel v_n-p\parallel }^2+2{\lambda }_n{\alpha }_n\parallel p\parallel \parallel (I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n-(I-{\lambda }_n\mathrm{\nabla }f)p\parallel +{\parallel t_n-p\parallel }^2\\ -{\parallel v_n-t_n-{\lambda }_n(\mathrm{\nabla }{f}_{{\alpha }_n}(v_n)-\mathrm{\nabla }f(p))\parallel }^2),\end{array}$$

which immediately leads to

$$\begin{array}{rcl}{\parallel t_n-p\parallel }^2& \le & {\parallel v_n-p\parallel }^2+2{\lambda }_n{\alpha }_n\parallel p\parallel \parallel (I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n-(I-{\lambda }_n\mathrm{\nabla }f)p\parallel \\ -{\parallel v_n-t_n-{\lambda }_n(\mathrm{\nabla }{f}_{{\alpha }_n}(v_n)-\mathrm{\nabla }f(p))\parallel }^2.\end{array}$$

(3.33)

Combining (3.4), (3.15), and (3.33), we obtain

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& \le & {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n){\parallel t_n-p\parallel }^2\\ \le & {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n)[{\parallel v_n-p\parallel }^2\\ +2{\lambda }_n{\alpha }_n\parallel p\parallel \parallel (I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n-(I-{\lambda }_n\mathrm{\nabla }f)p\parallel \\ -{\parallel v_n-t_n-{\lambda }_n(\mathrm{\nabla }{f}_{{\alpha }_n}(v_n)-\mathrm{\nabla }f(p))\parallel }^2]\\ \le & {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n)[{\parallel x_n-p\parallel }^2\\ +2{\lambda }_n{\alpha }_n\parallel p\parallel \parallel (I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n-(I-{\lambda }_n\mathrm{\nabla }f)p\parallel \\ -{\parallel v_n-t_n-{\lambda }_n(\mathrm{\nabla }{f}_{{\alpha }_n}(v_n)-\mathrm{\nabla }f(p))\parallel }^2]\\ \le & {\parallel x_n-p\parallel }^2+2{\lambda }_n{\alpha }_n\parallel p\parallel \parallel (I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n-(I-{\lambda }_n\mathrm{\nabla }f)p\parallel \\ -(1-{\beta }_n){\parallel v_n-t_n-{\lambda }_n(\mathrm{\nabla }{f}_{{\alpha }_n}(v_n)-\mathrm{\nabla }f(p))\parallel }^2,\end{array}$$

which immediately yields

$$\begin{array}{c}(1-d)\frac{{\parallel v_n-t_n-{\lambda }_n(\mathrm{\nabla }{f}_{{\alpha }_n}(v_n)-\mathrm{\nabla }f(p))\parallel }^2}{{ϵ}_n^2}\hfill \\ \le (1-{\beta }_n)\frac{{\parallel v_n-t_n-{\lambda }_n(\mathrm{\nabla }{f}_{{\alpha }_n}(v_n)-\mathrm{\nabla }f(p))\parallel }^2}{{ϵ}_n^2}\hfill \\ \le \frac{{\parallel x_n-p\parallel }^2-{\parallel y_n-p\parallel }^2}{{ϵ}_n^2}+2{\lambda }_n\frac{{\alpha }_n}{{ϵ}_n^2}\parallel p\parallel \parallel (I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n-(I-{\lambda }_n\mathrm{\nabla }f)p\parallel \hfill \\ \le \frac{\parallel x_n-y_n\parallel }{{ϵ}_n^2}(\parallel x_n-p\parallel +\parallel y_n-p\parallel )+2\frac{{\alpha }_n}{{ϵ}_n^2}b\parallel p\parallel \parallel (I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})v_n-(I-{\lambda }_n\mathrm{\nabla }f)p\parallel .\hfill \end{array}$$

Since $\{v_n\}$, $\{x_n\}$, and $\{y_n\}$ are bounded sequences, we deduce from $\parallel x_n-y_n\parallel +{\alpha }_n=o({ϵ}_n^2)$ that

$$\underset{n\to \mathrm{\infty }}{lim}\frac{\parallel v_n-t_n-{\lambda }_n(\mathrm{\nabla }{f}_{{\alpha }_n}(v_n)-\mathrm{\nabla }f(p))\parallel }{{ϵ}_n}=0.$$

(3.34)

Observe that

$$\frac{\parallel v_n-t_n\parallel }{{ϵ}_n}\le \frac{\parallel v_n-t_n-{\lambda }_n(\mathrm{\nabla }{f}_{{\alpha }_n}(v_n)-\mathrm{\nabla }f(p))\parallel }{{ϵ}_n}+\frac{{\lambda }_n\parallel \mathrm{\nabla }{f}_{{\alpha }_n}(v_n)-\mathrm{\nabla }f(p)\parallel }{{ϵ}_n}.$$

Thus, from (3.32) and (3.34), we have

$$\underset{n\to \mathrm{\infty }}{lim}\frac{\parallel v_n-t_n\parallel }{{ϵ}_n}=0.$$

(3.35)

Taking into account that $\frac{\parallel x_n-t_n\parallel }{{ϵ}_n}\le \frac{\parallel x_n-v_n\parallel }{{ϵ}_n}+\frac{\parallel v_n-t_n\parallel }{{ϵ}_n}$, from (3.30) and (3.35), we get

$$\underset{n\to \mathrm{\infty }}{lim}\frac{\parallel x_n-t_n\parallel }{{ϵ}_n}=0.$$

(3.36)

Utilizing the relation $y_n-x_n={\gamma }_n(t_n-x_n)+{\sigma }_n(T{t}_n-x_n)$, we have

$$\begin{array}{rl}\frac{\parallel {\sigma }_n(T{t}_n-t_n)\parallel }{{ϵ}_n}& =\frac{\parallel {\sigma }_n(T{t}_n-x_n)-{\sigma }_n(t_n-x_n)\parallel }{{ϵ}_n}\\ =\frac{\parallel y_n-x_n-{\gamma }_n(t_n-x_n)-{\sigma }_n(t_n-x_n)\parallel }{{ϵ}_n}\\ =\frac{\parallel y_n-x_n-(1-{\beta }_n)(t_n-x_n)\parallel }{{ϵ}_n}\\ \le \frac{\parallel y_n-x_n\parallel }{{ϵ}_n}+(1-{\beta }_n)\frac{\parallel t_n-x_n\parallel }{{ϵ}_n}\\ \le \frac{\parallel y_n-x_n\parallel }{{ϵ}_n}+\frac{\parallel t_n-x_n\parallel }{{ϵ}_n},\end{array}$$

which together with (3.36) and $\parallel x_n-y_n\parallel =o({ϵ}_n^2)$, implies that

$$\underset{n\to \mathrm{\infty }}{lim}\frac{\parallel {\sigma }_n(T{t}_n-t_n)\parallel }{{ϵ}_n}=0.$$

Since ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\sigma }_n>0$, we obtain

$$\underset{n\to \mathrm{\infty }}{lim}\frac{\parallel t_n-T{t}_n\parallel }{{ϵ}_n}=0.$$

(3.37)

Step 4. We prove that ${\omega }_w(x_n)\subset \Omega$.

Indeed, since H is reflexive and $\{x_n\}$ is bounded, there exists at least a weak convergence subsequence of $\{x_n\}$. Hence, ${\omega }_w(x_n)\ne \mathrm{\varnothing }$. Now, take an arbitrary $w\in {\omega }_w(x_n)$. Then there exists a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ such that $x_{n_i}\rightharpoonup w$. From (3.23), (3.26), (3.28), (3.30) and (3.36), we have $u_{n_i}\rightharpoonup w$, $v_{n_i}\rightharpoonup w$, $t_{n_i}\rightharpoonup w$, ${\Lambda }_{n_i}^m{u}_{n_i}\rightharpoonup w$ and ${\Delta }_{n_i}^k{x}_{n_i}\rightharpoonup w$, where $m\in \{1,2,\dots ,N\}$ and $k\in \{1,2,\dots ,M\}$. Utilizing Lemma 2.1(b), we deduce from $t_{n_i}\rightharpoonup w$ and (3.37) that $w\in Fix(T)$.

Next, we prove that $w\in {\bigcap }_{m=1}^N\mathrm{I}(B_m,R_m)$. As a matter of fact, since $B_m$ is ${\eta }_m$-inverse strongly monotone, $B_m$ is a monotone and Lipschitz-continuous mapping. It follows from Lemma 2.13 that $R_m+B_m$ is maximal monotone. Let $(v,g)\in G(R_m+B_m)$, that is, $g-B_{m}v\in R_{m}v$. Again, since ${\Lambda }_n^m{u}_n=J_{R_m,{\lambda }_{m,n}}(I-{\lambda }_{m,n}{B}_m){\Lambda }_n^{m-1}{u}_n$, $n\ge 0$, $m\in \{1,2,\dots ,N\}$, we have

$${\Lambda }_n^{m-1}{u}_n-{\lambda }_{m,n}{B}_m{\Lambda }_n^{m-1}{u}_n\in (I+{\lambda }_{m,n}{R}_m){\Lambda }_n^m{u}_n,$$

that is,

$$\frac{1}{{\lambda }_{m,n}}({\Lambda }_n^{m-1}{u}_n-{\Lambda }_n^m{u}_n-{\lambda }_{m,n}{B}_m{\Lambda }_n^{m-1}{u}_n)\in R_m{\Lambda }_n^m{u}_n.$$

In terms of the monotonicity of $R_m$, we get

$$〈v-{\Lambda }_n^m{u}_n,g-B_{m}v-\frac{1}{{\lambda }_{m,n}}({\Lambda }_n^{m-1}{u}_n-{\Lambda }_n^m{u}_n-{\lambda }_{m,n}{B}_m{\Lambda }_n^{m-1}{u}_n)〉\ge 0,$$

and hence

$$\begin{array}{c}〈v-{\Lambda }_n^m{u}_n,g〉\hfill \\ \ge 〈v-{\Lambda }_n^m{u}_n,B_{m}v+\frac{1}{{\lambda }_{m,n}}({\Lambda }_n^{m-1}{u}_n-{\Lambda }_n^m{u}_n-{\lambda }_{m,n}{B}_m{\Lambda }_n^{m-1}{u}_n)〉\hfill \\ =〈v-{\Lambda }_n^m{u}_n,B_{m}v-B_m{\Lambda }_n^m{u}_n+B_m{\Lambda }_n^m{u}_n-B_m{\Lambda }_n^{m-1}{u}_n+\frac{1}{{\lambda }_{m,n}}({\Lambda }_n^{m-1}{u}_n-{\Lambda }_n^m{u}_n)〉\hfill \\ \ge 〈v-{\Lambda }_n^m{u}_n,B_m{\Lambda }_n^m{u}_n-B_m{\Lambda }_n^{m-1}{u}_n〉+〈v-{\Lambda }_n^m{u}_n,\frac{1}{{\lambda }_{m,n}}({\Lambda }_n^{m-1}{u}_n-{\Lambda }_n^m{u}_n)〉.\hfill \end{array}$$

In particular,

$$\begin{array}{rl}〈v-{\Lambda }_{n_i}^m{u}_{n_i},g〉\ge & 〈v-{\Lambda }_{n_i}^m{u}_{n_i},B_m{\Lambda }_{n_i}^m{u}_{n_i}-B_m{\Lambda }_{n_i}^{m-1}{u}_{n_i}〉\\ +〈v-{\Lambda }_{n_i}^m{u}_{n_i},\frac{1}{{\lambda }_{m,n_i}}({\Lambda }_{n_i}^{m-1}{u}_{n_i}-{\Lambda }_{n_i}^m{u}_{n_i})〉.\end{array}$$

Since $\parallel {\Lambda }_n^m{u}_n-{\Lambda }_n^{m-1}{u}_n\parallel \to 0$ (due to (3.23)) and $\parallel B_m{\Lambda }_n^m{u}_n-B_m{\Lambda }_n^{m-1}{u}_n\parallel \to 0$ (due to the Lipschitz continuity of $B_m$), we conclude from ${\Lambda }_{n_i}^m{u}_{n_i}\rightharpoonup w$ and $\{{\lambda }_{i,n}\}\subset [a_i,b_i]\subset (0,2{\eta }_i)$ that

$$\underset{i\to \mathrm{\infty }}{lim}〈v-{\Lambda }_{n_i}^m{u}_{n_i},g〉=〈v-w,g〉\ge 0.$$

It follows from the maximal monotonicity of $B_m+R_m$ that $0\in (R_m+B_m)w$, that is, $w\in \mathrm{I}(B_m,R_m)$. Therefore, $w\in {\bigcap }_{m=1}^N\mathrm{I}(B_m,R_m)$.

Next we prove that $w\in {\bigcap }_{k=1}^{M}GMEP({\Theta }_k,{\phi }_k,A_k)$. Since ${\Delta }_n^k{x}_n=T_{r_{k,n}}^{({\Theta }_k,{\phi }_k)}(I-r_{k,n}{A}_k){\Delta }_n^{k-1}{x}_n$, $n\ge 0$, $k\in \{1,2,\dots ,M\}$, we have

$$\begin{array}{c}{\Theta }_k({\Delta }_n^k{x}_n,y)+{\phi }_k(y)-{\phi }_k\left({\Delta }_n^k{x}_n\right)+〈A_k{\Delta }_n^{k-1}{x}_n,y-{\Delta }_n^k{x}_n〉\hfill \\ +\frac{1}{r_{k,n}}〈y-{\Delta }_n^k{x}_n,{\Delta }_n^k{x}_n-{\Delta }_n^{k-1}{x}_n〉\ge 0.\hfill \end{array}$$

By (A2), we have

$$\begin{array}{c}{\phi }_k(y)-{\phi }_k\left({\Delta }_n^k{x}_n\right)+〈A_k{\Delta }_n^{k-1}{x}_n,y-{\Delta }_n^k{x}_n〉+\frac{1}{r_{k,n}}〈y-{\Delta }_n^k{x}_n,{\Delta }_n^k{x}_n-{\Delta }_n^{k-1}{x}_n〉\hfill \\ \ge {\Theta }_k(y,{\Delta }_n^k{x}_n).\hfill \end{array}$$

Let $z_t=ty+(1-t)w$, for all $t\in (0,1]$ and $y\in C$. This implies that $z_t\in C$. Then we have

$$\begin{array}{c}〈z_t-{\Delta }_n^k{x}_n,A_k{z}_t〉\hfill \\ \ge {\phi }_k\left({\Delta }_n^k{x}_n\right)-{\phi }_k(z_t)+〈z_t-{\Delta }_n^k{x}_n,A_k{z}_t〉-〈z_t-{\Delta }_n^k{x}_n,A_k{\Delta }_n^{k-1}{x}_n〉\hfill \\ -〈z_t-{\Delta }_n^k{x}_n,\frac{{\Delta }_n^k{x}_n-{\Delta }_n^{k-1}{x}_n}{r_{k,n}}〉+{\Theta }_k(z_t,{\Delta }_n^k{x}_n)\hfill \\ ={\phi }_k\left({\Delta }_n^k{x}_n\right)-{\phi }_k(z_t)+〈z_t-{\Delta }_n^k{x}_n,A_k{z}_t-A_k{\Delta }_n^k{x}_n〉\hfill \\ +〈z_t-{\Delta }_n^k{x}_n,A_k{\Delta }_n^k{x}_n-A_k{\Delta }_n^{k-1}{x}_n〉-〈z_t-{\Delta }_n^k{x}_n,\frac{{\Delta }_n^k{x}_n-{\Delta }_n^{k-1}{x}_n}{r_{k,n}}〉\hfill \\ +{\Theta }_k(z_t,{\Delta }_n^k{x}_n).\hfill \end{array}$$

(3.38)

By (3.26), we have $\parallel A_k{\Delta }_n^k{x}_n-A_k{\Delta }_n^{k-1}{x}_n\parallel \to 0$ as $n\to \mathrm{\infty }$. Furthermore, by the monotonicity of $A_k$, we obtain $〈z_t-{\Delta }_n^k{x}_n,A_k{z}_t-A_k{\Delta }_n^k{x}_n〉\ge 0$. Then, by (A4), we obtain

$$〈z_t-w,A_k{z}_t〉\ge {\phi }_k(w)-{\phi }_k(z_t)+{\Theta }_k(z_t,w).$$

(3.39)

Utilizing (A1), (A4), and (3.39), we obtain

$$\begin{array}{rl}0& ={\Theta }_k(z_t,z_t)+{\phi }_k(z_t)-{\phi }_k(z_t)\\ \le t{\Theta }_k(z_t,y)+(1-t){\Theta }_k(z_t,w)+t{\phi }_k(y)+(1-t){\phi }_k(w)-{\phi }_k(z_t)\\ \le t[{\Theta }_k(z_t,y)+{\phi }_k(y)-{\phi }_k(z_t)]+(1-t)〈z_t-w,A_k{z}_t〉\\ =t[{\Theta }_k(z_t,y)+{\phi }_k(y)-{\phi }_k(z_t)]+(1-t)t〈y-w,A_k{z}_t〉,\end{array}$$

and hence

$$0\le {\Theta }_k(z_t,y)+{\phi }_k(y)-{\phi }_k(z_t)+(1-t)〈y-w,A_k{z}_t〉.$$

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

$$0\le {\Theta }_k(w,y)+{\phi }_k(y)-{\phi }_k(w)+〈y-w,A_{k}w〉.$$

This implies that $w\in GMEP({\Theta }_k,{\phi }_k,A_k)$, and hence, $w\in {\bigcap }_{k=1}^{M}GMEP({\Theta }_k,{\phi }_k,A_k)$. Thus, $w\in Fix(T)\cap {\bigcap }_{k=1}^{M}GMEP({\Theta }_k,{\phi }_k,A_k)\cap {\bigcap }_{m=1}^N\mathrm{I}(B_m,R_m)$.

Furthermore, let us show that $w\in \Gamma$. In fact, define

$$\tilde{T}v=\{\begin{array}{ll}\mathrm{\nabla }f(v)+N_{C}v,& \text{if }v\in C,\\ \mathrm{\varnothing },& \text{if }v\notin C,\end{array}$$

where $N_{C}v=\{u\in H:〈v-x,u〉\ge 0,\mathrm{\forall }x\in C\}$. Then $\tilde{T}$ is maximal monotone and $0\in \tilde{T}v$ if and only if $v\in VI(C,\mathrm{\nabla }f)$; see [16]. Let $(v,\tilde{v})\in G(\tilde{T})$. Then we have $\tilde{v}\in \tilde{T}v=\mathrm{\nabla }f(v)+N_{C}v$, and hence, $\tilde{v}-\mathrm{\nabla }f(v)\in N_{C}v$. So, we have $〈v-x,\tilde{v}-\mathrm{\nabla }f(v)〉\ge 0$, for all $x\in C$.

On the other hand, from $t_n=P_C(v_n-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n}(v_n))$ and $v\in C$, we get $〈v_n-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n}(v_n)-t_n,t_n-v〉\ge 0$, and hence,

$$〈v-t_n,\frac{t_n-v_n}{{\lambda }_n}+\mathrm{\nabla }{f}_{{\alpha }_n}(v_n)〉\ge 0.$$

Therefore, from $\tilde{v}-\mathrm{\nabla }f(v)\in N_{C}v$ and $t_{n_i}\in C$, we have

$$\begin{array}{rcl}〈v-t_{n_i},\tilde{v}〉& \ge & 〈v-t_{n_i},\mathrm{\nabla }f(v)〉\\ \ge & 〈v-t_{n_i},\mathrm{\nabla }f(v)〉-〈v-t_{n_i},\frac{t_{n_i}-v_{n_i}}{{\lambda }_{n_i}}+\mathrm{\nabla }{f}_{{\alpha }_{n_i}}(v_{n_i})〉\\ =& 〈v-t_{n_i},\mathrm{\nabla }f(v)〉-〈v-t_{n_i},\frac{t_{n_i}-v_{n_i}}{{\lambda }_{n_i}}+\mathrm{\nabla }f(v_{n_i})〉-{\alpha }_{n_i}〈v-t_{n_i},v_{n_i}〉\\ =& 〈v-t_{n_i},\mathrm{\nabla }f(v)-\mathrm{\nabla }f(t_{n_i})〉+〈v-t_{n_i},\mathrm{\nabla }f(t_{n_i})-\mathrm{\nabla }f(v_{n_i})〉\\ -〈v-t_{n_i},\frac{t_{n_i}-v_{n_i}}{{\lambda }_{n_i}}〉-{\alpha }_{n_i}〈v-t_{n_i},v_{n_i}〉\\ \ge & 〈v-t_{n_i},\mathrm{\nabla }f(t_{n_i})-\mathrm{\nabla }f(v_{n_i})〉-〈v-t_{n_i},\frac{t_{n_i}-v_{n_i}}{{\lambda }_{n_i}}〉-{\alpha }_{n_i}〈v-t_{n_i},v_{n_i}〉.\end{array}$$

Hence, it is easy to see that $〈v-w,\tilde{v}〉\ge 0$ as $i\to \mathrm{\infty }$. Since $\tilde{T}$ is maximal monotone, we have $w\in {\tilde{T}}^{-1}0$, and hence, $w\in VI(C,\mathrm{\nabla }f)=\Gamma$. Consequently, $w\in {\bigcap }_{k=1}^{M}GMEP({\Theta }_k,{\phi }_k,A_k)\cap {\bigcap }_{i=1}^N\mathrm{I}(B_i,R_i)\cap Fix(T)\cap \Gamma =:\Omega$. This shows that ${\omega }_w(x_n)\subset \Omega$.

Step 5. We prove that ${\omega }_w(x_n)\subset \Xi$.

Indeed, utilizing Lemmas 2.6 and 2.4, from (3.1) and (3.6), we find that, for all $p\in \Omega$,

$$\begin{array}{c}{\parallel x_{n+1}-p\parallel }^2\hfill \\ ={\parallel {ϵ}_n\gamma ({\delta }_{n}V{x}_n+(1-{\delta }_n)S{x}_n)+(I-{ϵ}_n\mu F)y_n-p\parallel }^2\hfill \\ ={\parallel {ϵ}_n\gamma ({\delta }_{n}V{x}_n+(1-{\delta }_n)S{x}_n)-{ϵ}_n\mu Fp+(I-{ϵ}_n\mu F)y_n-(I-{ϵ}_n\mu F)p\parallel }^2\hfill \\ ={\parallel {ϵ}_n[{\delta }_n(\gamma V{x}_n-\mu Fp)+(1-{\delta }_n)(\gamma S{x}_n-\mu Fp)]+(I-{ϵ}_n\mu F)y_n-(I-{ϵ}_n\mu F)p\parallel }^2\hfill \\ =\parallel {ϵ}_n[{\delta }_n(\gamma V{x}_n-\gamma Vp)+(1-{\delta }_n)(\gamma S{x}_n-\gamma Sp)]+(I-{ϵ}_n\mu F)y_n-(I-{ϵ}_n\mu F)p\hfill \\ {+{ϵ}_n[{\delta }_n(\gamma Vp-\mu Fp)+(1-{\delta }_n)(\gamma Sp-\mu Fp)]\parallel }^2\hfill \\ \le {\parallel {ϵ}_n[{\delta }_n(\gamma V{x}_n-\gamma Vp)+(1-{\delta }_n)(\gamma S{x}_n-\gamma Sp)]+(I-{ϵ}_n\mu F)y_n-(I-{ϵ}_n\mu F)p\parallel }^2\hfill \\ +2{ϵ}_n{\delta }_n〈(\gamma Vp-\mu Fp),x_{n+1}-p〉+2{ϵ}_n(1-{\delta }_n)〈(\gamma Sp-\mu Fp),x_{n+1}-p〉\hfill \\ \le {[{ϵ}_n\parallel {\delta }_n(\gamma V{x}_n-\gamma Vp)+(1-{\delta }_n)(\gamma S{x}_n-\gamma Sp)\parallel +\parallel (I-{ϵ}_n\mu F)y_n-(I-{ϵ}_n\mu F)p\parallel ]}^2\hfill \\ +2{ϵ}_n{\delta }_n〈(\gamma Vp-\mu Fp),x_{n+1}-p〉+2{ϵ}_n(1-{\delta }_n)〈(\gamma Sp-\mu Fp),x_{n+1}-p〉\hfill \\ \le {[{ϵ}_n({\delta }_n\gamma \rho \parallel x_n-p\parallel +(1-{\delta }_n)\gamma \parallel x_n-p\parallel )+(1-{ϵ}_n\tau )\parallel y_n-p\parallel ]}^2\hfill \\ +2{ϵ}_n{\delta }_n〈(\gamma Vp-\mu Fp),x_{n+1}-p〉+2(1-{\delta }_n){ϵ}_n〈(\gamma Sp-\mu Fp),x_{n+1}-p〉\hfill \\ ={[{ϵ}_n(1-{\delta }_n(1-\rho ))\gamma \parallel x_n-p\parallel +(1-{ϵ}_n\tau )\parallel y_n-p\parallel ]}^2\hfill \\ +2{ϵ}_n{\delta }_n〈(\gamma Vp-\mu Fp),x_{n+1}-p〉+2{ϵ}_n(1-{\delta }_n)〈(\gamma Sp-\mu Fp),x_{n+1}-p〉\hfill \\ \le {[{ϵ}_n(1-{\delta }_n(1-\rho ))\gamma \parallel x_n-p\parallel +(1-{ϵ}_n\tau )(\parallel x_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )]}^2\hfill \\ +2{ϵ}_n{\delta }_n〈(\gamma Vp-\mu Fp),x_{n+1}-p〉+2{ϵ}_n(1-{\delta }_n)〈(\gamma Sp-\mu Fp),x_{n+1}-p〉\hfill \\ \le {[{ϵ}_n(1-{\delta }_n(1-\rho ))\gamma (\parallel x_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )+(1-{ϵ}_n\tau )(\parallel x_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )]}^2\hfill \\ +2{ϵ}_n{\delta }_n〈(\gamma Vp-\mu Fp),x_{n+1}-p〉+2{ϵ}_n(1-{\delta }_n)〈(\gamma Sp-\mu Fp),x_{n+1}-p〉\hfill \\ ={(1-{ϵ}_n(\tau -\gamma )-{ϵ}_n{\delta }_n(1-\rho )\gamma )}^2{(\parallel x_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )}^2\hfill \\ +2{ϵ}_n{\delta }_n〈(\gamma Vp-\mu Fp),x_{n+1}-p〉+2{ϵ}_n(1-{\delta }_n)〈(\gamma Sp-\mu Fp),x_{n+1}-p〉\hfill \\ \le (1-{ϵ}_n(\tau -\gamma )-{ϵ}_n{\delta }_n(1-\rho )\gamma ){(\parallel x_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )}^2\hfill \\ +2{ϵ}_n{\delta }_n〈(\gamma Vp-\mu Fp),x_{n+1}-p〉+2{ϵ}_n(1-{\delta }_n)〈(\gamma Sp-\mu Fp),x_{n+1}-p〉\hfill \\ \le (1-{ϵ}_n{\delta }_n(1-\rho )\gamma )[{\parallel x_n-p\parallel }^2+{\alpha }_{n}b\parallel p\parallel (2\parallel x_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )]\hfill \\ +2{ϵ}_n{\delta }_n〈(\gamma Vp-\mu Fp),x_{n+1}-p〉+2{ϵ}_n(1-{\delta }_n)〈(\gamma Sp-\mu Fp),x_{n+1}-p〉\hfill \\ \le (1-{ϵ}_n{\delta }_n(1-\rho )\gamma ){\parallel x_n-p\parallel }^2+2{ϵ}_n{\delta }_n〈(\gamma Vp-\mu Fp),x_{n+1}-p〉\hfill \\ +2{ϵ}_n(1-{\delta }_n)〈(\gamma Sp-\mu Fp),x_{n+1}-p〉+{\alpha }_{n}b\parallel p\parallel (2\parallel x_n-p\parallel +{\alpha }_{n}b\parallel p\parallel ).\hfill \end{array}$$

(3.40)

Take an arbitrary $w\in {\omega }_w(x_n)$. Then there exists a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ such that $x_{n_i}\rightharpoonup w$. Utilizing (3.40), we obtain, for all $p\in \Omega$,

$$\begin{array}{c}{\parallel x_{n+1}-p\parallel }^2\hfill \\ \le (1-{ϵ}_n(\tau -\gamma )-{ϵ}_n{\delta }_n(1-\rho )\gamma ){(\parallel x_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )}^2\hfill \\ +2{ϵ}_n{\delta }_n〈(\gamma Vp-\mu Fp),x_{n+1}-p〉+2{ϵ}_n(1-{\delta }_n)〈(\gamma Sp-\mu Fp),x_{n+1}-p〉\hfill \\ \le {(\parallel x_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )}^2+2{ϵ}_n{\delta }_n〈(\gamma Vp-\mu Fp),x_{n+1}-p〉\hfill \\ +2{ϵ}_n(1-{\delta }_n)〈(\gamma Sp-\mu Fp),x_{n+1}-p〉,\hfill \end{array}$$

which implies that

$$\begin{array}{c}〈(\mu F-\gamma S)p,x_n-p〉\hfill \\ \le 〈(\mu F-\gamma S)p,x_n-x_{n+1}〉+〈(\mu F-\gamma S)p,x_{n+1}-p〉\hfill \\ \le \parallel (\mu F-\gamma S)p\parallel \parallel x_n-x_{n+1}\parallel +\frac{{(\parallel x_n-p\parallel +{\alpha }_{n}b\parallel p\parallel )}^2-{\parallel x_{n+1}-p\parallel }^2}{2{ϵ}_n(1-{\delta }_n)}\hfill \\ +\frac{{\delta }_n}{1-{\delta }_n}〈(\gamma V-\mu F)p,x_{n+1}-p〉\hfill \\ \le \parallel (\mu F-\gamma S)p\parallel \parallel x_n-x_{n+1}\parallel \hfill \\ +\frac{(\parallel x_n-x_{n+1}\parallel +{\alpha }_{n}b\parallel p\parallel )(\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel +{\alpha }_{n}b\parallel p\parallel )}{2{ϵ}_n(1-{\delta }_n)}\hfill \\ +\frac{{\delta }_n}{1-{\delta }_n}\parallel (\gamma V-\mu F)p\parallel \parallel x_{n+1}-p\parallel .\hfill \end{array}$$

(3.41)

Since the combination of the boundedness of $\{x_n\}$, ${\delta }_n\to 0$, ${\alpha }_n=o({ϵ}_n^2)$, and $\parallel x_n-x_{n+1}\parallel =o({ϵ}_n)$ (due to Step 2) implies that

$$\underset{n\to \mathrm{\infty }}{lim}\frac{(\parallel x_n-x_{n+1}\parallel +{\alpha }_{n}b\parallel p\parallel )(\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel +{\alpha }_{n}b\parallel p\parallel )}{2{ϵ}_n(1-{\delta }_n)}=0,$$

from (3.41), we conclude that

$$\begin{array}{rl}〈(\mu F-\gamma S)p,w-p〉& =\underset{i\to \mathrm{\infty }}{lim}〈(\mu F-\gamma S)p,x_{n_i}-p〉\\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(\mu F-\gamma S)p,x_n-p〉\\ \le 0,\mathrm{\forall }p\in \Omega ,\end{array}$$

that is,

$$〈(\mu F-\gamma S)p,w-p〉\le 0,\mathrm{\forall }p\in \Omega .$$

(3.42)

Since $\mu F-\gamma S$ is $(\mu \eta -\gamma )$-strongly monotone and $(\mu \kappa +\gamma )$-Lipschitz continuous, by Minty’s lemma [41] we know that (3.42) is equivalent to the VIP

$$〈(\mu F-\gamma S)w,p-w〉\ge 0,\mathrm{\forall }p\in \Omega .$$

(3.43)

So, it follows that $w\in VI(\Omega ,\mu F-\gamma S)=:\Xi$. This shows that ${\omega }_w(x_n)\subset \Xi$.

Step 6. We prove that $x_n\to x^{\ast }$ where $\{x^{\ast }\}=VI(\Xi ,\mu F-\gamma V)$.

Indeed, note that $\{x^{\ast }\}=VI(\Xi ,\mu F-\gamma V)$. Since $\{x_n\}$ is bounded and H is reflexive, there exists a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ such that $x_{n_i}\rightharpoonup w$ and

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(\gamma V-\mu F)x^{\ast },x_n-x^{\ast }〉=\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(\gamma V-\mu F)x^{\ast },x_{n_i}-x^{\ast }〉=〈(\gamma V-\mu F)x^{\ast },w-x^{\ast }〉.$$

According to Step 5, we get $w\in \Xi$. So, it follows from $\{x^{\ast }\}=VI(\Xi ,\mu F-\gamma V)$ that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(\gamma V-\mu F)x^{\ast },x_n-x^{\ast }〉=〈(\gamma V-\mu F)x^{\ast },w-x^{\ast }〉\le 0.$$

However, from $x^{\ast }\in \Xi \subset \Omega$ and condition (C10), we deduce that, for sufficiently large $n\ge 0$,

$$\begin{array}{c}〈(\gamma S-\mu F)x^{\ast },x_{n+1}-x^{\ast }〉\hfill \\ =〈(\gamma S-\mu F)x^{\ast },x_{n+1}-P_{\Omega }{x}_{n+1}〉+〈(\gamma S-\mu F)x^{\ast },P_{\Omega }{x}_{n+1}-x^{\ast }〉\hfill \\ \le 〈(\gamma S-\mu F)x^{\ast },x_{n+1}-P_{\Omega }{x}_{n+1}〉\hfill \\ \le \parallel (\gamma S-\mu F)x^{\ast }\parallel d(x_{n+1},\Omega )\hfill \\ \le \parallel (\gamma S-\mu F)x^{\ast }\parallel {(\frac{1}{\overline{k}}\parallel x_{n+1}-T{x}_{n+1}\parallel )}^{1/\theta }.\hfill \end{array}$$

(3.44)

Utilizing Lemma 2.1(a), we have, for sufficiently large $n\ge 0$,

$$\begin{array}{c}\parallel x_{n+1}-T{x}_{n+1}\parallel \hfill \\ \le \parallel x_{n+1}-T{x}_n\parallel +\parallel T{x}_n-T{x}_{n+1}\parallel \hfill \\ \le \frac{1+\xi }{1-\xi }\parallel x_n-x_{n+1}\parallel +\parallel {ϵ}_n\gamma ({\delta }_{n}V{x}_n+(1-{\delta }_n)S{x}_n)+(I-{ϵ}_n\mu F)y_n-T{x}_n\parallel \hfill \\ \le \frac{1+\xi }{1-\xi }\parallel x_n-x_{n+1}\parallel +\parallel y_n-T{x}_n\parallel +{ϵ}_n\parallel \gamma ({\delta }_{n}V{x}_n+(1-{\delta }_n)S{x}_n)-\mu F{y}_n\parallel \hfill \\ \le \frac{1+\xi }{1-\xi }\parallel x_n-x_{n+1}\parallel +\parallel y_n-x_n\parallel +\parallel x_n-T{x}_n\parallel +{ϵ}_n\parallel \gamma ({\delta }_{n}V{x}_n+(1-{\delta }_n)S{x}_n)-\mu F{y}_n\parallel \hfill \\ \le \frac{1+\xi }{1-\xi }\parallel x_n-x_{n+1}\parallel +\parallel y_n-x_n\parallel +\parallel x_n-T{t}_n\parallel +\parallel T{t}_n-T{x}_n\parallel \hfill \\ +{ϵ}_n\parallel \gamma {\delta }_n(V{x}_n-S{x}_n)+\gamma S{x}_n-\mu F{y}_n\parallel \hfill \\ \le \frac{1+\xi }{1-\xi }\parallel x_n-x_{n+1}\parallel +\parallel y_n-x_n\parallel +\parallel x_n-t_n\parallel +\parallel t_n-T{t}_n\parallel \hfill \\ +\parallel T{t}_n-T{x}_n\parallel +{ϵ}_n\parallel \gamma {\delta }_n(V{x}_n-S{x}_n)+\gamma S{x}_n-\mu F{y}_n\parallel \hfill \\ \le \frac{1+\xi }{1-\xi }\parallel x_n-x_{n+1}\parallel +\parallel y_n-x_n\parallel +(1+\frac{1+\xi }{1-\xi })\parallel t_n-x_n\parallel +\parallel t_n-T{t}_n\parallel +{ϵ}_n{\tilde{M}}_4,\hfill \end{array}$$

(3.45)

where ${\tilde{M}}_4={sup}_{n\ge 0}\parallel \gamma {\delta }_n(V{x}_n-S{x}_n)+\gamma S{x}_n-\mu F{y}_n\parallel <\mathrm{\infty }$. Hence, for a large enough constant ${\overline{k}}_1>0$, from (3.44) and (3.45), we have, for sufficiently large $n\ge 0$,

$$\begin{array}{c}〈(\gamma S-\mu F)x^{\ast },x_{n+1}-x^{\ast }〉\hfill \\ \le \parallel (\gamma S-\mu F)x^{\ast }\parallel {(\frac{1}{\overline{k}}\parallel x_{n+1}-T{x}_{n+1}\parallel )}^{1/\theta }\hfill \\ \le \parallel (\gamma S-\mu F)x^{\ast }\parallel \left\{\frac{1}{\overline{k}}\right[\frac{1+\xi }{1-\xi }\parallel x_n-x_{n+1}\parallel +\parallel y_n-x_n\parallel +(1+\frac{1+\xi }{1-\xi })\parallel t_n-x_n\parallel \hfill \\ {+\parallel t_n-T{t}_n\parallel +{ϵ}_n{\tilde{M}}_4]\}}^{1/\theta }\hfill \\ \le {\overline{k}}_1{({ϵ}_n+\parallel x_n-x_{n+1}\parallel +\parallel y_n-x_n\parallel +\parallel x_n-t_n\parallel +\parallel t_n-T{t}_n\parallel )}^{1/\theta }\hfill \\ \le {\overline{k}}_1{ϵ}_n^{1/\theta }{(1+\frac{\parallel x_n-x_{n+1}\parallel +\parallel y_n-x_n\parallel +\parallel x_n-t_n\parallel +\parallel t_n-T{t}_n\parallel }{{ϵ}_n})}^{1/\theta }.\hfill \end{array}$$

(3.46)

Next we prove that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-x^{\ast }\parallel =0$. As a matter of fact, putting $p=x^{\ast }$ in (3.40), we obtain from (3.46) that

$$\begin{array}{c}{\parallel x_{n+1}-x^{\ast }\parallel }^2\hfill \\ \le (1-{ϵ}_n{\delta }_n(1-\rho )\gamma ){\parallel x_n-x^{\ast }\parallel }^2+2{ϵ}_n{\delta }_n〈(\gamma V-\mu F)x^{\ast },x_{n+1}-x^{\ast }〉\hfill \\ +2{ϵ}_n(1-{\delta }_n)〈(\gamma S-\mu F)x^{\ast },x_{n+1}-x^{\ast }〉+{\alpha }_{n}b\parallel x^{\ast }\parallel (2\parallel x_n-x^{\ast }\parallel +{\alpha }_{n}b\parallel x^{\ast }\parallel )\hfill \\ \le (1-{ϵ}_n{\delta }_n(1-\rho )\gamma ){\parallel x_n-x^{\ast }\parallel }^2+{ϵ}_n{\delta }_n(1-\rho )\gamma \cdot \frac{2}{(1-\rho )\gamma }[〈(\gamma V-\mu F)x^{\ast },x_{n+1}-x^{\ast }〉\hfill \\ +(1-{\delta }_n)\frac{〈(\gamma S-\mu F)x^{\ast },x_{n+1}-x^{\ast }〉}{{\delta }_n}]+{\alpha }_{n}b\parallel x^{\ast }\parallel (2\parallel x_n-x^{\ast }\parallel +{\alpha }_{n}b\parallel x^{\ast }\parallel )\hfill \\ \le (1-{ϵ}_n{\delta }_n(1-\rho )\gamma ){\parallel x_n-x^{\ast }\parallel }^2+{ϵ}_n{\delta }_n(1-\rho )\gamma \cdot \frac{2}{(1-\rho )\gamma }[〈(\gamma V-\mu F)x^{\ast },x_{n+1}-x^{\ast }〉\hfill \\ +{\overline{k}}_1\frac{{ϵ}_n^{1/\theta }}{{\delta }_n}{(1+\frac{\parallel x_n-x_{n+1}\parallel +\parallel y_n-x_n\parallel +\parallel x_n-t_n\parallel +\parallel t_n-T{t}_n\parallel }{{ϵ}_n})}^{1/\theta }]\hfill \\ +{\alpha }_{n}b\parallel x^{\ast }\parallel (2\parallel x_n-x^{\ast }\parallel +{\alpha }_{n}b\parallel x^{\ast }\parallel ).\hfill \end{array}$$

(3.47)

Since ${\sum }_{n=1}^{\mathrm{\infty }}{ϵ}_n{\delta }_n=\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n<\mathrm{\infty }$, $\frac{{ϵ}_n^{1/\theta }}{{\delta }_n}\to 0$, $\parallel x_n-y_n\parallel =o({ϵ}_n^2)$, and $\parallel x_n-x_{n+1}\parallel =o({ϵ}_n)$, we conclude from (3.36), (3.37), and $x_n\rightharpoonup x^{\ast }$ that ${\sum }_{n=1}^{\mathrm{\infty }}{ϵ}_n{\delta }_n(1-\rho )\gamma =\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}b\parallel x^{\ast }\parallel (2\parallel x_n-x^{\ast }\parallel +{\alpha }_{n}b\parallel x^{\ast }\parallel )\le \mathrm{\infty }$, and

$$\begin{array}{c}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{2}{(1-\rho )\gamma }[〈(\gamma V-\mu F)x^{\ast },x_{n+1}-x^{\ast }〉\hfill \\ +{\overline{k}}_1\frac{{ϵ}_n^{1/\theta }}{{\delta }_n}{(1+\frac{\parallel x_n-x_{n+1}\parallel +\parallel y_n-x_n\parallel +\parallel x_n-t_n\parallel +\parallel t_n-T{t}_n\parallel }{{ϵ}_n})}^{1/\theta }]\le 0.\hfill \end{array}$$

Therefore, applying Lemma 2.8 to (3.47), we infer that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-x^{\ast }\parallel =0$. This completes the proof. □

Remark 3.1 Algorithm 3.1 and Theorem 3.1 extend and generalize algorithms and convergence results in [28, 30].