Iterative algorithms with regularization for hierarchical variational inequality problems and convex minimization problems

Abstract

In this paper, we consider a variational inequality problem which is defined over the set of intersections of the set of fixed points of a ζ-strictly pseudocontractive mapping, the set of fixed points of a nonexpansive mapping and the set of solutions of a minimization problem. We propose an iterative algorithm with regularization to solve such a variational inequality problem and study the strong convergence of the sequence generated by the proposed algorithm. The results of this paper improve and extend several known results in the literature.

1 Introduction

Let H be a real Hilbert space with the inner product $〈\cdot ,\cdot 〉$ and the norm $\parallel \cdot \parallel$, let C be a nonempty closed convex subset of H, and let $f:C\to \mathbf{R}$ be a convex and continuously Fréchet differentiable functional. We consider the following minimization problem (MP):

$$\underset{x\in C}{min}f(x).$$

(1.1)

We denote by Ξ the set of minimizers of problem (1.1), and we assume that $\Xi \ne \mathrm{\varnothing }$. The gradient-projection algorithm (GPA) is one of the most elegant methods to solve the minimization problem (1.1). The convergence of the sequence generated by the GPA depends on the behavior of the gradient ∇f. If ∇f is strongly monotone and Lipschitz continuous, then we get the strong convergence of the sequence generated by the GPA to a unique solution of MP (1.1). However, if the gradient ∇f is assumed to be only Lipschitz continuous, then the sequence generated by the GPA converges weakly if H is infinite-dimensional (a counterexample is given in [1]). Since the Lipschitz continuity of the gradient ∇f implies that it is actually inverse strongly monotone (ism) [2], its complement can be an averaged mapping (that is, it can be expressed as a proper convex combination of the identity mapping and a nonexpansive mapping) [1]. Consequently, the GPA can be rewritten as the composite of a projection and an averaged mapping, which is again an averaged mapping. This shows that averaged mappings play an important role in the GPA. Very recently, Xu [1] used averaged mappings to study the convergence analysis of the GPA, which is an operator-oriented approach. He showed that the sequence generated by the GPA converges in norm to a minimizer of MP (1.1), which is also a unique solution of a particular type of variational inequality problem (VIP). It is worth to emphasize that the regularization, in particular the traditional Tikhonov regularization, is usually used to solve ill-posed optimization problems. The advantage of a regularization method is its possible strong convergence to the minimum-norm solution of the optimization problem. In [1], Xu introduced a hybrid gradient-projection algorithm with regularization and proved the strong convergence of the sequence to the minimum-norm solution of MP (1.1). Some iterative algorithms with or without regularization for MP (1.1) are proposed and analyzed in [3–5] for finding a common solution of MP (1.1) and the set of solutions of a nonexpansive mapping.

On the other hand, the theory of variational inequalities [6, 7] has emerged as an important tool to study a wide class of problems from science, engineering, social sciences. If the underlying set in the formulation of a variational inequality problem is a set of fixed points of a mapping or, more precisely, of a nonexpansive mapping, then the variational inequality problem is called hierarchical variational problem. For further details on hierarchical variational inequalities, we refer to [8–11] and the references therein.

In this paper, we consider a variational inequality problem which is defined over the set of intersections of the set of fixed points of a ζ-strictly pseudocontractive mapping, the set of fixed points of a nonexpansive mapping and the set of solutions of MP (1.1). We propose an iterative algorithm with regularization to solve such a variational inequality problem and study the strong convergence of the sequence generated by the proposed algorithm. The results of this paper improve and extend several known results in the literature.

2 Preliminaries and formulations

Throughout the paper, unless otherwise specified, we use the following assumptions and notations. Let H be a real Hilbert space whose inner product and norm are denoted by $〈\cdot ,\cdot 〉$ and $\parallel \cdot \parallel$, respectively. Let C be a nonempty closed convex subset of H. We write $x_n\to x$ (respectively, $x_n\rightharpoonup x$) to indicate that the sequence $\{x_n\}$ converges strongly (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,

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

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

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

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

Proposition 2.1 For given $x\in H$ and $z\in C$, we have

• (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) $〈P_{C}x-P_{C}y,x-y〉\ge {\parallel P_{C}x-P_{C}y\parallel }^2$, $\mathrm{\forall }y\in H$, which concludes that $P_C$ is nonexpansive and monotone.

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

• (a) ζ-strictly pseudocontractive if there exists a constant $\zeta \in [0,1)$ such that

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

  If $\zeta =0$, then it is called nonexpansive;

• (b) firmly nonexpansive if $2T-I$ is nonexpansive, or equivalently,

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

  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 a nonexpansive mapping.

It can be easily seen that the projection mappings are firmly nonexpansive. It is clear that $T:C\subseteq H\to C$ is ζ-strictly pseudocontractive if and only if

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

Definition 2.2 Let T be a nonlinear operator with domain $D(T)\subseteq H$ and range $R(T)\subseteq H$.

• (a) T is said to be monotone if

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

• (b) Given a number $\beta >0$, T is said to be β-strongly monotone if

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

• (c) Given a number $\nu >0$, T is said to be ν-inverse strongly monotone (ν-ism) if

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

Clearly,

• if T is nonexpansive, then $I-T$ is monotone;

• a projection $P_K$ is 1-ism;

• if T is a ζ-strictly pseudocontractive mapping, then $I-T$ is $\frac{1-\zeta }{2}$-inverse strongly monotone.

Definition 2.3 [1]

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 a nonexpansive mapping. 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 maps.

Proposition 2.2 [12]

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 2.3 [12, 13]

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 $\alpha ={\alpha }_1+{\alpha }_2-{\alpha }_1{\alpha }_2$.

Lemma 2.1 [[14], Proposition 2.1]

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

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

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

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

• (c) If T is a ζ-(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.

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

Lemma 2.2 In a real Hilbert space H, we have

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

The following elementary result on real sequences is quite well known.

Lemma 2.3 [15]

Let $\{a_n\}$ be a sequence of nonnegative real numbers such that

$$a_{n+1}\le (1-s_n)a_n+s_n{t}_n+{ϵ}_n,\mathrm{\forall }n\ge 0,$$

where $\{s_n\}\subset (0,1]$ and $\{t_n\}$ satisfy the following conditions:

• (i) ${\sum }_{n=0}^{\mathrm{\infty }}{s}_n=\mathrm{\infty }$;

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

• (iii) ${\sum }_{n=0}^{\mathrm{\infty }}{ϵ}_n<\mathrm{\infty }$, where ${ϵ}_n\ge 0$, $\mathrm{\forall }n\ge 0$.

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

Lemma 2.4 [10]

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

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

The following lemma appeared implicitly in the paper of Reineermann [16].

Lemma 2.5 [16]

Let H be a real Hilbert space. Then, for all $x,y\in H$ and $\lambda \in [0,1]$,

$${\parallel \lambda x+(1-\lambda )y\parallel }^2=\lambda {\parallel x\parallel }^2+(1-\lambda ){\parallel y\parallel }^2-\lambda (1-\lambda ){\parallel x-y\parallel }^2.$$

Let C be a nonempty closed convex subset of a real Hilbert space H, and let $A:C\to H$ be a monotone mapping. The variational inequality problem (VIP) is to find $x\in C$ such that

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

The solution set of the VIP is denoted by $VI(C,A)$. It is well known that

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

A set-valued mapping $V:H\to 2^H$ is called monotone if for all $x,y\in H$, $f\in Vx$ and $g\in Vy$ imply that $〈x-y,f-g〉\ge 0$. A monotone set-valued mapping $V:H\to 2^H$ is called maximal if its graph $Gph(V)$ is not properly contained in the graph of any other monotone set-valued mapping. It is known that a monotone set-valued mapping $V:H\to 2^H$ is maximal if and only if for $(x,f)\in H\times H$, $〈x-y,f-g〉\ge 0$ for every $(y,g)\in Gph(V)$ implies that $f\in Vx$. Let $A:C\to H$ be a monotone and Lipschitz continuous mapping and $N_{C}v$ be the normal cone to C at $v\in C$, that is,

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

Define

$$Vv=\{\begin{array}{cc}Av+N_{C}v\hfill & \text{if }v\in C,\hfill \\ \mathrm{\varnothing }\hfill & \text{if }v\notin C.\hfill \end{array}$$

Lemma 2.6 [17]

Let $A:C\to H$ be a monotone mapping. Then

• (i) V is maximal monotone;

• (ii) $v\in V^{-1}0\iff v\in VI(C,A)$.

Throughout the paper, we denote by $Fix(T)$ and $Fix(\Gamma )$ the set of fixed points of T and Γ, respectively. We also assume that the set $Fix(T)\cap Fix(\Gamma )\cap \Xi$ is nonempty closed and convex.

Let $S,T:C\to C$ be nonexpansive mappings and $\Gamma :C\to C$ be a ζ-strictly pseudocontractive mapping with $\zeta \in [0,1)$. In this paper, we consider and study the following hierarchical variational inequality problem which is defined on $Fix(T)\cap Fix(\Gamma )\cap \Xi$.

$$\begin{array}{r}\text{Find }\tilde{x}\in Fix(T)\cap Fix(\Gamma )\cap \Xi \text{such that}\\ 〈\tilde{x}-S\tilde{x},\tilde{x}-x〉\le 0,\mathrm{\forall }x\in Fix(T)\cap Fix(\Gamma )\cap \Xi .\end{array}$$

(2.1)

We denote by Ω the solution set of problem (2.1). It is not difficult to verify that solving (2.1) is equivalent to the fixed point problem of finding $\tilde{x}\in C$ such that

$$\tilde{x}=P_{Fix(T)\cap Fix(\Gamma )\cap \Xi }S\tilde{x},$$

where $P_{Fix(T)\cap Fix(\Gamma )\cap \Xi }$ stands for the metric projection onto the closed convex set $Fix(T)\cap Fix(\Gamma )\cap \Xi$.

Problem (2.1) contains the hierarchical variational inequality problems considered and studied in [8, 18, 19] and the references therein.

By using the definition of the normal cone to $Fix(T)\cap Fix(\Gamma )\cap \Xi$, we have the mapping $N_{Fix(T)\cap Fix(\Gamma )\cap \Xi }:H\to 2^H$:

$$x\mapsto \{\begin{array}{c}\{u\in H:(\mathrm{\forall }y\in Fix(T)\cap Fix(\Gamma )\cap \Xi )〈y-x,u〉\le 0\},\hfill \\ \text{if }x\in Fix(T)\cap Fix(\Gamma )\cap \Xi ;\hfill \\ \mathrm{\varnothing },\text{otherwise},\hfill \end{array}$$

and we readily prove that (2.1) is equivalent to the variational inequality

$$0\in (I-S)\tilde{x}+N_{Fix(T)\cap Fix(\Gamma )\cap \Xi }\tilde{x}.$$

By combining the hybrid gradient-projection method of Xu [1] and a two-step method of Yao et al. [11], we introduce the following three-step iterative algorithm:

$$\{\begin{array}{c}{y}_n={\theta }_{n}S{x}_n+(1-{\theta }_n)x_n,\hfill \\ z_n={\beta }_{n}Q{y}_n+(1-{\beta }_n)T{P}_C(y_n-\lambda \mathrm{\nabla }{f}_{{\alpha }_n}(y_n)),\hfill \\ x_{n+1}={\sigma }_n{z}_n+{\gamma }_n{P}_C(z_n-\lambda \mathrm{\nabla }{f}_{{\alpha }_n}(z_n))+{\delta }_n\Gamma P_C(z_n-\lambda \mathrm{\nabla }{f}_{{\alpha }_n}(z_n)),\mathrm{\forall }n\ge 0,\hfill \end{array}$$

(2.2)

where $Q:C\to C$ is a ρ-contraction mapping, $\{{\alpha }_n\}\subset (0,\mathrm{\infty })$, $\{{\beta }_n\},\{{\theta }_n\},\{{\sigma }_n\}\subset (0,1)$ and $\{{\gamma }_n\},\{{\delta }_n\}\subset [0,1]$ with ${\sigma }_n+{\gamma }_n+{\delta }_n=1$, $\mathrm{\forall }n\ge 0$. It is proven that under appropriate assumptions, the above iterative sequence $\{x_n\}$ converges strongly to an element $\tilde{x}\in Fix(T)\cap Fix(\Gamma )\cap \Xi$.

3 Main results

Let us consider the following assumptions:

• the mapping $Q:C\to C$ is a ρ-contraction;

• the mapping $\Gamma :C\to C$ is a ζ-strict pseudocontraction;

• $S,T:C\to C$ are two nonexpansive mappings;

• $\mathrm{\nabla }f:C\to H$ is Lipschitz continuous with $0<\lambda <\frac{2}{L}$;

• $\{{\alpha }_n\}$ is a sequence in $(0,\mathrm{\infty })$ with ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n<\mathrm{\infty }$;

• $\{{\beta }_n\}$, $\{{\theta }_n\}$, $\{{\sigma }_n\}$ are sequences in $(0,1)$ with $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\sigma }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\sigma }_n<1$;

• $\{{\gamma }_n\}$, $\{{\delta }_n\}$ are sequences in $[0,1]$ with ${\sigma }_n+{\gamma }_n+{\delta }_n=1$, $\mathrm{\forall }n\ge 0$;

• ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\delta }_n>0$ and $({\gamma }_n+{\delta }_n)\zeta \le {\gamma }_n$, $\mathrm{\forall }n\ge 0$.

Theorem 3.1 Let $\{x_n\}$ be a bounded sequence generated from any given $x_0\in C$ by (2.2). Assume that the following conditions hold:

• (H1) ${\sum }_{n=0}^{\mathrm{\infty }}{\beta }_n=\mathrm{\infty }$, ${lim}_{n\to \mathrm{\infty }}\frac{1}{{\beta }_n}|1-\frac{{\theta }_{n-1}}{{\theta }_n}|=0$;

• (H2) ${lim}_{n\to \mathrm{\infty }}\frac{1}{{\beta }_n}|\frac{1}{{\theta }_n}-\frac{1}{{\theta }_{n-1}}|=0$, ${lim}_{n\to \mathrm{\infty }}\frac{1}{{\theta }_n}|1-\frac{{\beta }_{n-1}}{{\beta }_n}|=0$;

• (H3) ${lim}_{n\to \mathrm{\infty }}{\theta }_n=0$ and ${lim}_{n\to \mathrm{\infty }}\frac{{\alpha }_n+{\beta }_n}{{\theta }_n}=0$;

• (H4) ${lim}_{n\to \mathrm{\infty }}\frac{|{\alpha }_n-{\alpha }_{n-1}|}{{\beta }_n{\theta }_n}=0$, ${lim}_{n\to \mathrm{\infty }}\frac{|{\sigma }_n-{\sigma }_{n-1}|}{{\beta }_n{\theta }_n}=0$;

• (H5) ${lim}_{n\to \mathrm{\infty }}\frac{1}{{\beta }_n{\theta }_n}|\frac{{\gamma }_n}{1-{\sigma }_n}-\frac{{\gamma }_{n-1}}{1-{\sigma }_{n-1}}|=0$.

Then the following assertions hold:

• (i) ${lim}_{n\to \mathrm{\infty }}\frac{\parallel x_{n+1}-x_n\parallel }{{\theta }_n}=0$;

• (ii) ${\omega }_w(x_n)\subset \Omega$.

Proof First of all, we show that $P_C(I-\lambda \mathrm{\nabla }{f}_{\alpha })$ is ξ-averaged for each $\lambda \in (0,\frac{2}{\alpha +L})$, where

$$\xi =\frac{2+\lambda (\alpha +L)}{4}\in (0,1).$$

Indeed, the Lipschitz continuity of ∇f implies that ∇f is $\frac{1}{L}$-ism [2], that is,

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

Observe that

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

Therefore, it follows that $\mathrm{\nabla }{f}_{\alpha }=\alpha I+\mathrm{\nabla }f$ is $\frac{1}{\alpha +L}$-ism. Thus, by Proposition 2.2(b), $\lambda \mathrm{\nabla }{f}_{\alpha }$ is $\frac{1}{\lambda (\alpha +L)}$-ism. From Proposition 2.2(c), the complement $I-\lambda \mathrm{\nabla }{f}_{\alpha }$ is $\frac{\lambda (\alpha +L)}{2}$-averaged. Therefore, noting that $P_C$ is $\frac{1}{2}$-averaged and utilizing Proposition 2.3(d), we obtain that for each $\lambda \in (0,\frac{2}{\alpha +L})$, $P_C(I-\lambda \mathrm{\nabla }{f}_{\alpha })$ is ξ-averaged with

$$\xi =\frac{1}{2}+\frac{\lambda (\alpha +L)}{2}-\frac{1}{2}\cdot \frac{\lambda (\alpha +L)}{2}=\frac{2+\lambda (\alpha +L)}{4}\in (0,1).$$

This shows that $P_C(I-\lambda \mathrm{\nabla }{f}_{\alpha })$ is nonexpansive. For $\lambda \in (0,\frac{2}{L})$, utilizing the fact that ${lim}_{n\to \mathrm{\infty }}\frac{2}{{\alpha }_n+L}=\frac{2}{L}$, we may assume that

$$0<\lambda <\frac{2}{{\alpha }_n+L},\mathrm{\forall }n\ge 0.$$

Consequently, it follows that for each integer $n\ge 0$, $P_C(I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})$ is ${\xi }_n$-averaged with

$${\xi }_n=\frac{1}{2}+\frac{\lambda ({\alpha }_n+L)}{2}-\frac{1}{2}\cdot \frac{\lambda ({\alpha }_n+L)}{2}=\frac{2+\lambda ({\alpha }_n+L)}{4}\in (0,1).$$

This implies that $P_C(I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})$ is nonexpansive for all $n\ge 0$.

The rest of the proof is divided into several steps.

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

For simplicity, we put ${\tilde{y}}_n=P_C(y_n-\lambda \mathrm{\nabla }{f}_{{\alpha }_n}(y_n))$ and ${\tilde{z}}_n=P_C(z_n-\lambda \mathrm{\nabla }{f}_{{\alpha }_n}(z_n))$ for every $n\ge 0$. Then $z_n={\beta }_{n}Q{y}_n+(1-{\beta }_n)T{\tilde{y}}_n$ and $x_{n+1}={\sigma }_n{z}_n+{\gamma }_n{\tilde{z}}_n+{\delta }_n\Gamma {\tilde{z}}_n$ for every $n\ge 0$.

Taking into account $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\sigma }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\sigma }_n<1$, without loss of generality, we may assume that $\{{\sigma }_n\}\subset [c,d]$ for some $c,d\in (0,1)$. We write $x_n={\sigma }_{n-1}{z}_{n-1}+(1-{\sigma }_{n-1})v_{n-1}$, $\mathrm{\forall }n\ge 1$, where $v_{n-1}=\frac{x_n-{\sigma }_{n-1}{z}_{n-1}}{1-{\sigma }_{n-1}}$. It follows that for all $n\ge 1$,

$$\begin{array}{rl}{v}_n-v_{n-1}=& \frac{x_{n+1}-{\sigma }_n{z}_n}{1-{\sigma }_n}-\frac{x_n-{\sigma }_{n-1}{z}_{n-1}}{1-{\sigma }_{n-1}}\\ =& \frac{{\gamma }_n{\tilde{z}}_n+{\delta }_n\Gamma {\tilde{z}}_n}{1-{\sigma }_n}-\frac{{\gamma }_{n-1}{\tilde{z}}_{n-1}+{\delta }_{n-1}\Gamma {\tilde{z}}_{n-1}}{1-{\sigma }_{n-1}}\\ =& \frac{{\gamma }_n({\tilde{z}}_n-{\tilde{z}}_{n-1})+{\delta }_n(\Gamma {\tilde{z}}_n-\Gamma {\tilde{z}}_{n-1})}{1-{\sigma }_n}\\ +(\frac{{\gamma }_n}{1-{\sigma }_n}-\frac{{\gamma }_{n-1}}{1-{\sigma }_{n-1}}){\tilde{z}}_{n-1}+(\frac{{\delta }_n}{1-{\sigma }_n}-\frac{{\delta }_{n-1}}{1-{\sigma }_{n-1}})\Gamma {\tilde{z}}_{n-1}.\end{array}$$

(3.1)

Since $({\gamma }_n+{\delta }_n)\zeta \le {\gamma }_n$ for all $n\ge 0$, by Lemma 2.4, we have

$$\parallel {\gamma }_n({\tilde{z}}_n-{\tilde{z}}_{n-1})+{\delta }_n(\Gamma {\tilde{z}}_n-\Gamma {\tilde{z}}_{n-1})\parallel \le ({\gamma }_n+{\delta }_n)\parallel {\tilde{z}}_n-{\tilde{z}}_{n-1}\parallel .$$

(3.2)

Now, we estimate $\parallel z_n-z_{n-1}\parallel$. Observe that for every $n\ge 1$,

$$\begin{array}{rl}\parallel {\tilde{y}}_n-{\tilde{y}}_{n-1}\parallel \le & \parallel P_C(I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})y_n-P_C(I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})y_{n-1}\parallel \\ +\parallel P_C(I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})y_{n-1}-P_C(I-\lambda \mathrm{\nabla }{f}_{{\alpha }_{n-1}})y_{n-1}\parallel \\ \le & \parallel y_n-y_{n-1}\parallel +\parallel P_C(I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})y_{n-1}-P_C(I-\lambda \mathrm{\nabla }{f}_{{\alpha }_{n-1}})y_{n-1}\parallel \\ \le & \parallel y_n-y_{n-1}\parallel +\parallel (I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})y_{n-1}-(I-\lambda \mathrm{\nabla }{f}_{{\alpha }_{n-1}})y_{n-1}\parallel \\ =& \parallel y_n-y_{n-1}\parallel +\parallel \lambda \mathrm{\nabla }{f}_{{\alpha }_n}(y_{n-1})-\lambda \mathrm{\nabla }{f}_{{\alpha }_{n-1}}(y_{n-1})\parallel \\ =& \parallel y_n-y_{n-1}\parallel +\lambda |{\alpha }_n-{\alpha }_{n-1}|\parallel y_{n-1}\parallel .\end{array}$$

(3.3)

Similarly, for all $n\ge 1$, we have

$$\parallel {\tilde{z}}_n-{\tilde{z}}_{n-1}\parallel \le \parallel z_n-z_{n-1}\parallel +\lambda |{\alpha }_n-{\alpha }_{n-1}|\parallel z_{n-1}\parallel .$$

From (2.2), we have

$$\{\begin{array}{c}{y}_n={\theta }_{n}S{x}_n+(1-{\theta }_n)x_n,\hfill \\ y_{n-1}={\theta }_{n-1}S{x}_{n-1}+(1-{\theta }_{n-1})x_{n-1},\mathrm{\forall }n\ge 1,\hfill \end{array}$$

and therefore

$$y_n-y_{n-1}={\theta }_n(S{x}_n-S{x}_{n-1})+({\theta }_n-{\theta }_{n-1})(S{x}_{n-1}-x_{n-1})+(1-{\theta }_n)(x_n-x_{n-1}),$$

which implies that

$$\begin{array}{rl}\parallel y_n-y_{n-1}\parallel & \le {\theta }_n\parallel S{x}_n-S{x}_{n-1}\parallel +|{\theta }_n-{\theta }_{n-1}|\parallel S{x}_{n-1}-x_{n-1}\parallel +(1-{\theta }_n)\parallel x_n-x_{n-1}\parallel \\ \le \parallel x_n-x_{n-1}\parallel +|{\theta }_n-{\theta }_{n-1}|\parallel S{x}_{n-1}-x_{n-1}\parallel .\end{array}$$

(3.4)

Also, from (2.2) we have

$$\{\begin{array}{c}{z}_n={\beta }_{n}Q{y}_n+(1-{\beta }_n)T{\tilde{y}}_n,\hfill \\ z_{n-1}={\beta }_{n-1}Q{y}_{n-1}+(1-{\beta }_{n-1})T{\tilde{y}}_{n-1},\mathrm{\forall }n\ge 1,\hfill \end{array}$$

then simple calculations show that

$$z_n-z_{n-1}=(1-{\beta }_n)(T{\tilde{y}}_n-T{\tilde{y}}_{n-1})+({\beta }_n-{\beta }_{n-1})(Q{y}_{n-1}-T{\tilde{y}}_{n-1})+{\beta }_n(Q{y}_n-Q{y}_{n-1}),$$

and thus, from (3.3)-(3.4), we have

$$\begin{array}{r}\parallel z_n-z_{n-1}\parallel \\ \le (1-{\beta }_n)\parallel T{\tilde{y}}_n-T{\tilde{y}}_{n-1}\parallel +|{\beta }_n-{\beta }_{n-1}|\parallel Q{y}_{n-1}-T{\tilde{y}}_{n-1}\parallel +{\beta }_n\parallel Q{y}_n-Q{y}_{n-1}\parallel \\ \le (1-{\beta }_n)\parallel {\tilde{y}}_n-{\tilde{y}}_{n-1}\parallel +|{\beta }_n-{\beta }_{n-1}|\parallel Q{y}_{n-1}-T{\tilde{y}}_{n-1}\parallel +{\beta }_n\parallel Q{y}_n-Q{y}_{n-1}\parallel \\ \le (1-{\beta }_n)(\parallel y_n-y_{n-1}\parallel +\lambda |{\alpha }_n-{\alpha }_{n-1}|\parallel y_{n-1}\parallel )+|{\beta }_n-{\beta }_{n-1}|\parallel Q{y}_{n-1}-T{\tilde{y}}_{n-1}\parallel \\ +{\beta }_n\rho \parallel y_n-y_{n-1}\parallel \\ \le (1-(1-\rho ){\beta }_n)\parallel y_n-y_{n-1}\parallel +\lambda |{\alpha }_n-{\alpha }_{n-1}|\parallel y_{n-1}\parallel +|{\beta }_n-{\beta }_{n-1}|\parallel Q{y}_{n-1}-T{\tilde{y}}_{n-1}\parallel \\ \le (1-(1-\rho ){\beta }_n)[\parallel x_n-x_{n-1}\parallel +|{\theta }_n-{\theta }_{n-1}|\parallel S{x}_{n-1}-x_{n-1}\parallel ]\\ +\lambda |{\alpha }_n-{\alpha }_{n-1}|\parallel y_{n-1}\parallel +|{\beta }_n-{\beta }_{n-1}|\parallel Q{y}_{n-1}-T{\tilde{y}}_{n-1}\parallel \\ \le (1-(1-\rho ){\beta }_n)\parallel x_n-x_{n-1}\parallel +|{\theta }_n-{\theta }_{n-1}|\parallel S{x}_{n-1}-x_{n-1}\parallel \\ +\lambda |{\alpha }_n-{\alpha }_{n-1}|\parallel y_{n-1}\parallel +|{\beta }_n-{\beta }_{n-1}|\parallel Q{y}_{n-1}-T{\tilde{y}}_{n-1}\parallel \\ \le (1-(1-\rho ){\beta }_n)\parallel x_n-x_{n-1}\parallel +M_1[|{\theta }_n-{\theta }_{n-1}|+|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|],\end{array}$$

(3.5)

where $\parallel S{x}_n-x_n\parallel +\lambda \parallel y_n\parallel +\parallel Q{y}_n-T{\tilde{y}}_n\parallel \le M_1$, $\mathrm{\forall }n\ge 0$ for some $M_1>0$. This together with (3.1)-(3.3) implies that

$$\begin{array}{r}\parallel v_n-v_{n-1}\parallel \\ \le \frac{\parallel {\gamma }_n({\tilde{z}}_n-{\tilde{z}}_{n-1})+{\delta }_n(\Gamma {\tilde{z}}_n-\Gamma {\tilde{z}}_{n-1})\parallel }{1-{\sigma }_n}+|\frac{{\gamma }_n}{1-{\sigma }_n}-\frac{{\gamma }_{n-1}}{1-{\sigma }_{n-1}}|\parallel {\tilde{z}}_{n-1}\parallel \\ +|\frac{{\delta }_n}{1-{\sigma }_n}-\frac{{\delta }_{n-1}}{1-{\sigma }_{n-1}}|\parallel \Gamma {\tilde{z}}_{n-1}\parallel \\ \le \frac{({\gamma }_n+{\delta }_n)\parallel {\tilde{z}}_n-{\tilde{z}}_{n-1}\parallel }{1-{\sigma }_n}+|\frac{{\gamma }_n}{1-{\sigma }_n}-\frac{{\gamma }_{n-1}}{1-{\sigma }_{n-1}}|\parallel {\tilde{z}}_{n-1}\parallel +|\frac{{\gamma }_n}{1-{\sigma }_n}-\frac{{\gamma }_{n-1}}{1-{\sigma }_{n-1}}|\parallel \Gamma {\tilde{z}}_{n-1}\parallel \\ =\parallel {\tilde{z}}_n-{\tilde{z}}_{n-1}\parallel +|\frac{{\gamma }_n}{1-{\sigma }_n}-\frac{{\gamma }_{n-1}}{1-{\sigma }_{n-1}}|(\parallel {\tilde{z}}_{n-1}\parallel +\parallel \Gamma {\tilde{y}}_{n-1}\parallel )\\ \le \parallel z_n-z_{n-1}\parallel +\lambda |{\alpha }_n-{\alpha }_{n-1}|\parallel z_{n-1}\parallel +|\frac{{\gamma }_n}{1-{\sigma }_n}-\frac{{\gamma }_{n-1}}{1-{\sigma }_{n-1}}|(\parallel {\tilde{z}}_{n-1}\parallel +\parallel \Gamma {\tilde{z}}_{n-1}\parallel )\\ \le (1-(1-\rho ){\beta }_n)\parallel x_n-x_{n-1}\parallel +M_1[|{\theta }_n-{\theta }_{n-1}|+|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|]\\ +\lambda |{\alpha }_n-{\alpha }_{n-1}|\parallel z_{n-1}\parallel +|\frac{{\gamma }_n}{1-{\sigma }_n}-\frac{{\gamma }_{n-1}}{1-{\sigma }_{n-1}}|(\parallel {\tilde{z}}_{n-1}\parallel +\parallel \Gamma {\tilde{z}}_{n-1}\parallel )\\ \le (1-(1-\rho ){\beta }_n)\parallel x_n-x_{n-1}\parallel +M_2[|{\theta }_n-{\theta }_{n-1}|+2|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|\\ +|\frac{{\gamma }_n}{1-{\sigma }_n}-\frac{{\gamma }_{n-1}}{1-{\sigma }_{n-1}}|],\end{array}$$

(3.6)

where $M_1+\lambda \parallel z_n\parallel +\parallel {\tilde{z}}_n\parallel +\parallel \Gamma {\tilde{z}}_n\parallel \le M_2$, $\mathrm{\forall }n\ge 0$ for some $M_2>0$.

Further, we observe that

$$\{\begin{array}{c}{x}_{n+1}={\sigma }_n{z}_n+(1-{\sigma }_n)v_n,\hfill \\ x_n={\sigma }_{n-1}{z}_{n-1}+(1-{\beta }_{n-1})v_{n-1},\mathrm{\forall }n\ge 1,\hfill \end{array}$$

and then by simple calculations, we have

$$x_{n+1}-x_n=(1-{\sigma }_n)(v_n-v_{n-1})+({\sigma }_n-{\sigma }_{n-1})(z_{n-1}-v_{n-1})+{\sigma }_n(z_n-z_{n-1}).$$

By taking norm and using (3.5)-(3.6), we get

$$\begin{array}{r}\parallel x_{n+1}-x_n\parallel \\ \le (1-{\sigma }_n)\parallel v_n-v_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel z_{n-1}-v_{n-1}\parallel +{\sigma }_n\parallel z_n-z_{n-1}\parallel \\ \le (1-{\sigma }_n)\{(1-(1-\rho ){\beta }_n)\parallel x_n-x_{n-1}\parallel +M_2[|{\theta }_n-{\theta }_{n-1}|+2|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|\\ +|\frac{{\gamma }_n}{1-{\sigma }_n}-\frac{{\gamma }_{n-1}}{1-{\sigma }_{n-1}}|\left]\right\}+|{\sigma }_n-{\sigma }_{n-1}|\parallel z_{n-1}-v_{n-1}\parallel \\ +{\sigma }_n\{(1-(1-\rho ){\beta }_n)\parallel x_n-x_{n-1}\parallel +M_1[|{\theta }_n-{\theta }_{n-1}|+|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|]\}\\ \le (1-(1-\rho ){\beta }_n)\parallel x_n-x_{n-1}\parallel +M_2[|{\theta }_n-{\theta }_{n-1}|+2|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|\\ +|\frac{{\gamma }_n}{1-{\sigma }_n}-\frac{{\gamma }_{n-1}}{1-{\sigma }_{n-1}}|]+|{\sigma }_n-{\sigma }_{n-1}|\parallel z_{n-1}-v_{n-1}\parallel \\ \le (1-(1-\rho ){\beta }_n)\parallel x_n-x_{n-1}\parallel +M_3[|{\theta }_n-{\theta }_{n-1}|+2|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|\\ +|\frac{{\gamma }_n}{1-{\sigma }_n}-\frac{{\gamma }_{n-1}}{1-{\sigma }_{n-1}}|+|{\sigma }_n-{\sigma }_{n-1}|],\end{array}$$

where $M_2+\parallel z_n-v_n\parallel \le M_3$, $\mathrm{\forall }n\ge 0$ for some $M_3\ge 0$. Therefore,

$$\begin{array}{r}\frac{\parallel x_{n+1}-x_n\parallel }{{\theta }_n}\\ \le (1-(1-\rho ){\beta }_n)\frac{\parallel x_n-x_{n-1}\parallel }{{\theta }_n}+M_3[\frac{|{\theta }_n-{\theta }_{n-1}|}{{\theta }_n}+2\frac{|{\alpha }_n-{\alpha }_{n-1}|}{{\theta }_n}+\frac{|{\beta }_n-{\beta }_{n-1}|}{{\theta }_n}\\ +\frac{1}{{\theta }_n}|\frac{{\gamma }_n}{1-{\sigma }_n}-\frac{{\gamma }_{n-1}}{1-{\sigma }_{n-1}}|+\frac{|{\sigma }_n-{\sigma }_{n-1}|}{{\theta }_n}]\\ =(1-(1-\rho ){\beta }_n)\frac{\parallel x_n-x_{n-1}\parallel }{{\theta }_{n-1}}+(1-(1-\rho ){\beta }_n)(\frac{\parallel x_n-x_{n-1}\parallel }{{\theta }_n}-\frac{\parallel x_n-x_{n-1}\parallel }{{\theta }_{n-1}})\\ +M_3[\frac{|{\theta }_n-{\theta }_{n-1}|}{{\theta }_n}+2\frac{|{\alpha }_n-{\alpha }_{n-1}|}{{\theta }_n}+\frac{|{\beta }_n-{\beta }_{n-1}|}{{\theta }_n}\\ +\frac{1}{{\theta }_n}|\frac{{\gamma }_n}{1-{\sigma }_n}-\frac{{\gamma }_{n-1}}{1-{\sigma }_{n-1}}|+\frac{|{\sigma }_n-{\sigma }_{n-1}|}{{\theta }_n}]\\ \le (1-(1-\rho ){\beta }_n)\frac{\parallel x_n-x_{n-1}\parallel }{{\theta }_{n-1}}+M[|\frac{1}{{\theta }_n}-\frac{1}{{\theta }_{n-1}}|\\ +\frac{|{\theta }_n-{\theta }_{n-1}|}{{\theta }_n}+2\frac{|{\alpha }_n-{\alpha }_{n-1}|}{{\theta }_n}+\frac{|{\beta }_n-{\beta }_{n-1}|}{{\theta }_n}+\frac{1}{{\theta }_n}|\frac{{\gamma }_n}{1-{\sigma }_n}-\frac{{\gamma }_{n-1}}{1-{\sigma }_{n-1}}|+\frac{|{\sigma }_n-{\sigma }_{n-1}|}{{\theta }_n}]\\ =(1-(1-\rho ){\beta }_n)\frac{\parallel x_n-x_{n-1}\parallel }{{\theta }_{n-1}}+(1-\rho ){\beta }_n\cdot \frac{M}{1-\rho }\{\frac{1}{{\beta }_n}|\frac{1}{{\theta }_n}-\frac{1}{{\theta }_{n-1}}|\\ +\frac{1}{{\beta }_n}|1-\frac{{\theta }_{n-1}}{{\theta }_n}|+2\frac{|{\alpha }_n-{\alpha }_{n-1}|}{{\beta }_n{\theta }_n}+\frac{1}{{\theta }_n}|1-\frac{{\beta }_{n-1}}{{\beta }_n}|\\ +\frac{1}{{\beta }_n{\theta }_n}|\frac{{\gamma }_n}{1-{\sigma }_n}-\frac{{\gamma }_{n-1}}{1-{\sigma }_{n-1}}|+\frac{|{\sigma }_n-{\sigma }_{n-1}|}{{\beta }_n{\theta }_n}\},\end{array}$$

(3.7)

where $M_3+\parallel x_n-x_{n-1}\parallel \le M$, $\mathrm{\forall }n\ge 1$ for some $M\ge 0$. From (H1)-(H5), it follows that ${\sum }_{n=0}^{\mathrm{\infty }}(1-\rho ){\beta }_n=\mathrm{\infty }$ and

$$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim}\frac{M}{1-\rho }\left\{\frac{1}{{\beta }_n}\right|\frac{1}{{\theta }_n}-\frac{1}{{\theta }_{n-1}}|+\frac{1}{{\beta }_n}|1-\frac{{\theta }_{n-1}}{{\theta }_n}|+2\frac{|{\alpha }_n-{\alpha }_{n-1}|}{{\beta }_n{\theta }_n}\\ +\frac{1}{{\theta }_n}|1-\frac{{\beta }_{n-1}}{{\beta }_n}|+\frac{1}{{\beta }_n{\theta }_n}|\frac{{\gamma }_n}{1-{\sigma }_n}-\frac{{\gamma }_{n-1}}{1-{\sigma }_{n-1}}|+\frac{|{\sigma }_n-{\sigma }_{n-1}|}{{\beta }_n{\theta }_n}\}=0.\end{array}$$

Thus, by applying Lemma 2.3 to (3.7), we conclude that

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

which implies that

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

(3.8)

Step 2. ${lim}_{n\to \mathrm{\infty }}\parallel x_n-z_n\parallel =0$.

Indeed, let $p\in Fix(T)\cap Fix(\Gamma )\cap \Xi$. Then we have

$$\begin{array}{rl}\parallel {\tilde{y}}_n-p\parallel =& \parallel P_C(I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})y_n-P_C(I-\lambda \mathrm{\nabla }f)p\parallel \\ \le & \parallel P_C(I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})y_n-P_C(I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})p\parallel \\ +\parallel P_C(I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})p-P_C(I-\lambda \mathrm{\nabla }f)p\parallel \\ \le & \parallel y_n-p\parallel +\parallel P_C(I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})p-P_C(I-\lambda \mathrm{\nabla }f)p\parallel \\ \le & \parallel y_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel .\end{array}$$

(3.9)

Similarly, we get

$$\parallel {\tilde{z}}_n-p\parallel \le \parallel z_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel .$$

By Lemma 2.5 and (3.9), we have

$$\begin{array}{r}{\parallel z_n-p\parallel }^2\\ ={\parallel {\beta }_n(Q{y}_n-p)+(1-{\beta }_n)(T{\tilde{y}}_n-p)\parallel }^2\\ \le {\beta }_n{\parallel Q{y}_n-p\parallel }^2+(1-{\beta }_n){\parallel {\tilde{y}}_n-p\parallel }^2\\ \le {\beta }_n{\parallel Q{y}_n-p\parallel }^2+{\parallel {\tilde{y}}_n-p\parallel }^2\\ \le {\beta }_n{\parallel Q{y}_n-p\parallel }^2+{(\parallel y_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel )}^2\\ ={\beta }_n{\parallel Q{y}_n-p\parallel }^2+{\parallel y_n-p\parallel }^2+\lambda {\alpha }_n\parallel p\parallel (2\parallel y_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel )\\ \le {\beta }_n{\parallel Q{y}_n-p\parallel }^2+{\theta }_n{\parallel S{x}_n-p\parallel }^2+(1-{\theta }_n){\parallel x_n-p\parallel }^2+\lambda {\alpha }_n\parallel p\parallel (2\parallel y_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel )\\ \le {\beta }_n{\parallel Q{y}_n-p\parallel }^2+{\theta }_n{\parallel S{x}_n-p\parallel }^2+{\parallel x_n-p\parallel }^2+\lambda {\alpha }_n\parallel p\parallel (2\parallel y_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel ).\end{array}$$

Since $({\gamma }_n+{\delta }_n)\zeta \le {\gamma }_n$ for all $n\ge 0$, utilizing Lemma 2.4, we obtain

$$\begin{array}{r}{\parallel x_{n+1}-p\parallel }^2\\ ={\parallel {\sigma }_n(z_n-p)+{\gamma }_n({\tilde{z}}_n-p)+{\delta }_n(\Gamma {\tilde{z}}_n-p)\parallel }^2\\ ={\parallel {\sigma }_n(z_n-p)+({\gamma }_n+{\delta }_n)\frac{1}{{\gamma }_n+{\delta }_n}[{\gamma }_n({\tilde{z}}_n-p)+{\delta }_n(\Gamma {\tilde{z}}_n-p)]\parallel }^2\\ ={\sigma }_n{\parallel z_n-p\parallel }^2+({\gamma }_n+{\delta }_n){\parallel \frac{1}{{\gamma }_n+{\delta }_n}[{\gamma }_n({\tilde{z}}_n-p)+{\delta }_n(\Gamma {\tilde{z}}_n-p)]\parallel }^2\\ -{\sigma }_n({\gamma }_n+{\delta }_n){\parallel (z_n-p)-\frac{1}{{\gamma }_n+{\delta }_n}[{\gamma }_n({\tilde{z}}_n-p)+{\delta }_n(\Gamma {\tilde{z}}_n-p)]\parallel }^2\\ ={\sigma }_n{\parallel z_n-p\parallel }^2+({\gamma }_n+{\delta }_n){\parallel \frac{1}{{\gamma }_n+{\delta }_n}[{\gamma }_n({\tilde{z}}_n-p)+{\delta }_n(\Gamma {\tilde{z}}_n-p)]\parallel }^2\\ -{\sigma }_n({\gamma }_n+{\delta }_n){\parallel \frac{1}{{\gamma }_n+{\delta }_n}[{\gamma }_n({\tilde{z}}_n-z_n)+{\delta }_n(\Gamma {\tilde{z}}_n-z_n)]\parallel }^2\\ ={\sigma }_n{\parallel z_n-p\parallel }^2+({\gamma }_n+{\delta }_n){\parallel \frac{1}{{\gamma }_n+{\delta }_n}[{\gamma }_n({\tilde{z}}_n-p)+{\delta }_n(\Gamma {\tilde{z}}_n-p)]\parallel }^2\\ -\frac{{\sigma }_n}{{\gamma }_n+{\delta }_n}{\parallel x_{n+1}-z_n\parallel }^2\\ \le {\sigma }_n{\parallel z_n-p\parallel }^2+({\gamma }_n+{\delta }_n){\parallel {\tilde{z}}_n-p\parallel }^2-\frac{{\sigma }_n}{{\gamma }_n+{\delta }_n}{\parallel x_{n+1}-z_n\parallel }^2\\ ={\sigma }_n{\parallel z_n-p\parallel }^2+(1-{\sigma }_n){\parallel {\tilde{z}}_n-p\parallel }^2-\frac{{\sigma }_n}{1-{\sigma }_n}{\parallel x_{n+1}-z_n\parallel }^2\\ \le {\sigma }_n{\parallel z_n-p\parallel }^2+(1-{\sigma }_n)[{\parallel z_n-p\parallel }^2+\lambda {\alpha }_n\parallel p\parallel (2\parallel z_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel )]\\ -\frac{{\sigma }_n}{1-{\sigma }_n}{\parallel x_{n+1}-z_n\parallel }^2\\ \le {\parallel z_n-p\parallel }^2+\lambda {\alpha }_n\parallel p\parallel (2\parallel z_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel )-\frac{{\sigma }_n}{1-{\sigma }_n}{\parallel x_{n+1}-z_n\parallel }^2\\ \le {\beta }_n{\parallel Q{y}_n-p\parallel }^2+{\theta }_n{\parallel S{x}_n-p\parallel }^2+{\parallel x_n-p\parallel }^2+\lambda {\alpha }_n\parallel p\parallel (2\parallel y_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel )\\ +\lambda {\alpha }_n\parallel p\parallel (2\parallel z_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel )-\frac{{\sigma }_n}{1-{\sigma }_n}{\parallel x_{n+1}-z_n\parallel }^2\\ ={\parallel x_n-p\parallel }^2+{\beta }_n{\parallel Q{y}_n-p\parallel }^2+{\theta }_n{\parallel S{x}_n-p\parallel }^2\\ +2\lambda {\alpha }_n\parallel p\parallel (\parallel y_n-p\parallel +\parallel z_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel )\\ -\frac{{\sigma }_n}{1-{\sigma }_n}{\parallel x_{n+1}-z_n\parallel }^2.\end{array}$$

Since $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\sigma }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\sigma }_n<1$, we may assume that $\{{\sigma }_n\}\subset [c,d]$ for some $c,d\in (0,1)$. Therefore, we deduce

$$\begin{array}{r}\frac{c}{1-c}{\parallel x_{n+1}-z_n\parallel }^2\\ \le \frac{{\sigma }_n}{1-{\sigma }_n}{\parallel x_{n+1}-z_n\parallel }^2\\ \le {\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+{\beta }_n{\parallel Q{y}_n-p\parallel }^2+{\theta }_n{\parallel S{x}_n-p\parallel }^2\\ +2\lambda {\alpha }_n\parallel p\parallel (\parallel y_n-p\parallel +\parallel z_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel )\\ \le (\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel +{\beta }_n{\parallel Q{y}_n-p\parallel }^2+{\theta }_n{\parallel S{x}_n-p\parallel }^2\\ +2\lambda {\alpha }_n\parallel p\parallel (\parallel y_n-p\parallel +\parallel z_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel ).\end{array}$$

Since ${\alpha }_n\to 0$, ${\beta }_n\to 0$, ${\theta }_n\to 0$ and $\parallel x_n-x_{n+1}\parallel \to 0$ as $n\to \mathrm{\infty }$, we conclude from the boundedness of $\{x_n\}$, $\{y_n\}$ and $\{z_n\}$ that $\parallel x_{n+1}-z_n\parallel \to 0$ as $n\to \mathrm{\infty }$. This together with $\parallel x_n-x_{n+1}\parallel \to 0$ implies that

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

(3.10)

Step 3. ${lim}_{n\to \mathrm{\infty }}\parallel y_n-{\tilde{y}}_n\parallel =0$ and ${lim}_{n\to \mathrm{\infty }}\parallel z_n-{\tilde{z}}_n\parallel =0$.

Let $p\in Fix(T)\cap Fix(\Gamma )\cap \Xi$. Then, by Lemmas 2.2 and 2.5, we have

$$\begin{array}{r}{\parallel z_n-p\parallel }^2\\ ={\parallel {\beta }_n(Q{y}_n-p)+(1-{\beta }_n)(T{\tilde{y}}_n-p)\parallel }^2\\ \le {\beta }_n{\parallel Q{y}_n-p\parallel }^2+(1-{\beta }_n){\parallel {\tilde{y}}_n-p\parallel }^2\\ ={\beta }_n{\parallel Q{y}_n-p\parallel }^2+(1-{\beta }_n){\parallel P_C(I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})y_n-P_C(I-\lambda \mathrm{\nabla }f)p\parallel }^2\\ \le {\beta }_n{\parallel Q{y}_n-p\parallel }^2+(1-{\beta }_n){\parallel (I-\lambda \mathrm{\nabla }f)y_n-(I-\lambda \mathrm{\nabla }f)p-\lambda {\alpha }_n{y}_n\parallel }^2\\ \le {\beta }_n{\parallel Q{y}_n-p\parallel }^2+(1-{\beta }_n)[{\parallel (I-\lambda \mathrm{\nabla }f)y_n-(I-\lambda \mathrm{\nabla }f)p\parallel }^2\\ -2\lambda {\alpha }_n〈y_n,(I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})y_n-(I-\lambda \mathrm{\nabla }f)p〉]\\ \le {\beta }_n{\parallel Q{y}_n-p\parallel }^2+(1-{\beta }_n)[{\parallel y_n-p\parallel }^2+\lambda (\lambda -\frac{2}{L}){\parallel \mathrm{\nabla }f(y_n)-\mathrm{\nabla }f(p)\parallel }^2\\ +2\lambda {\alpha }_n\parallel y_n\parallel \parallel (I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})y_n-(I-\lambda \mathrm{\nabla }f)p\parallel ]\\ \le {\beta }_n{\parallel Q{y}_n-p\parallel }^2+(1-{\beta }_n)[{\theta }_n{\parallel S{x}_n-p\parallel }^2+(1-{\theta }_n){\parallel x_n-p\parallel }^2\\ +\lambda (\lambda -\frac{2}{L}){\parallel \mathrm{\nabla }f(y_n)-\mathrm{\nabla }f(p)\parallel }^2+2\lambda {\alpha }_n\parallel y_n\parallel \parallel (I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})y_n-(I-\lambda \mathrm{\nabla }f)p\parallel ]\\ \le {\beta }_n{\parallel Q{y}_n-p\parallel }^2+{\theta }_n{\parallel S{x}_n-p\parallel }^2+{\parallel x_n-p\parallel }^2+(1-{\beta }_n)\lambda (\lambda -\frac{2}{L}){\parallel \mathrm{\nabla }f(y_n)-\mathrm{\nabla }f(p)\parallel }^2\\ +2\lambda {\alpha }_n\parallel y_n\parallel \parallel (I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})y_n-(I-\lambda \mathrm{\nabla }f)p\parallel .\end{array}$$

Therefore, we obtain

$$\begin{array}{r}(1-{\beta }_n)\lambda (\frac{2}{L}-\lambda ){\parallel \mathrm{\nabla }f(y_n)-\mathrm{\nabla }f(p)\parallel }^2\\ \le {\beta }_n{\parallel Q{y}_n-p\parallel }^2+{\theta }_n{\parallel S{x}_n-p\parallel }^2+{\parallel x_n-p\parallel }^2-{\parallel z_n-p\parallel }^2\\ +2\lambda {\alpha }_n\parallel y_n\parallel \parallel (I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})y_n-(I-\lambda \mathrm{\nabla }f)p\parallel \\ \le {\beta }_n{\parallel Q{y}_n-p\parallel }^2+{\theta }_n{\parallel S{x}_n-p\parallel }^2+(\parallel x_n-p\parallel +\parallel z_n-p\parallel )(\parallel x_n-p\parallel -\parallel z_n-p\parallel )\\ +2\lambda {\alpha }_n\parallel y_n\parallel \parallel (I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})y_n-(I-\lambda \mathrm{\nabla }f)p\parallel \\ \le {\beta }_n{\parallel Q{y}_n-p\parallel }^2+{\theta }_n{\parallel S{x}_n-p\parallel }^2+(\parallel x_n-p\parallel +\parallel z_n-p\parallel )\parallel x_n-z_n\parallel \\ +2\lambda {\alpha }_n\parallel y_n\parallel \parallel (I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})y_n-(I-\lambda \mathrm{\nabla }f)p\parallel .\end{array}$$

Since ${\alpha }_n\to 0$, ${\beta }_n\to 0$, ${\theta }_n\to 0$, $\parallel x_n-z_n\parallel \to 0$ and $0<\lambda <\frac{2}{L}$, from the boundedness of $\{x_n\}$, $\{y_n\}$ and $\{z_n\}$, we obtain ${lim}_{n\to \mathrm{\infty }}\parallel \mathrm{\nabla }f(y_n)-\mathrm{\nabla }f(p)\parallel =0$, and hence

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

Also, since

$$\parallel y_n-z_n\parallel \le \parallel y_n-x_n\parallel +\parallel x_n-z_n\parallel ={\theta }_n\parallel S{x}_n-x_n\parallel +\parallel x_n-z_n\parallel ,$$

from ${\theta }_n\to 0$ and $\parallel x_n-z_n\parallel \to 0$, it follows that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel y_n-z_n\parallel =0\text{and}\underset{n\to \mathrm{\infty }}{lim}\parallel \mathrm{\nabla }{f}_{{\alpha }_n}(z_n)-\mathrm{\nabla }f(p)\parallel =0.$$

(3.11)

Furthermore, from the firm nonexpansiveness of $P_C$, we obtain

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

and so,

$$\begin{array}{r}{\parallel {\tilde{y}}_n-p\parallel }^2\\ \le {\parallel y_n-p\parallel }^2-{\parallel y_n-{\tilde{y}}_n\parallel }^2\\ +2\lambda \parallel \mathrm{\nabla }{f}_{{\alpha }_n}(y_n)-\mathrm{\nabla }f(p)\parallel \parallel (I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})y_n-(I-\lambda \mathrm{\nabla }f)p\parallel \\ +2\lambda 〈y_n-{\tilde{y}}_n,\mathrm{\nabla }{f}_{{\alpha }_n}(y_n)-\mathrm{\nabla }f(p)〉-{\lambda }^2{\parallel \mathrm{\nabla }{f}_{{\alpha }_n}(y_n)-\mathrm{\nabla }f(p)\parallel }^2.\end{array}$$

Similarly, we have

$$\begin{array}{r}{\parallel {\tilde{z}}_n-p\parallel }^2\\ \le {\parallel z_n-p\parallel }^2-{\parallel z_n-{\tilde{z}}_n\parallel }^2+2\lambda \parallel \mathrm{\nabla }{f}_{{\alpha }_n}(z_n)-\mathrm{\nabla }f(p)\parallel \parallel (I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})z_n-(I-\lambda \mathrm{\nabla }f)p\parallel \\ +2\lambda 〈z_n-{\tilde{z}}_n,\mathrm{\nabla }{f}_{{\alpha }_n}(z_n)-\mathrm{\nabla }f(p)〉-{\lambda }^2{\parallel \mathrm{\nabla }{f}_{{\alpha }_n}(z_n)-\mathrm{\nabla }f(p)\parallel }^2.\end{array}$$

(3.12)

Thus, we have

$$\begin{array}{rl}{\parallel z_n-p\parallel }^2\le & {\beta }_n{\parallel Q{y}_n-p\parallel }^2+(1-{\beta }_n){\parallel {\tilde{y}}_n-p\parallel }^2\\ \le & {\beta }_n{\parallel Q{y}_n-p\parallel }^2+{\parallel {\tilde{y}}_n-p\parallel }^2\\ \le & {\beta }_n{\parallel Q{y}_n-p\parallel }^2+{\parallel y_n-p\parallel }^2-{\parallel y_n-{\tilde{y}}_n\parallel }^2\\ +2\lambda \parallel \mathrm{\nabla }{f}_{{\alpha }_n}(y_n)-\mathrm{\nabla }f(p)\parallel \parallel (I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})y_n-(I-\lambda \mathrm{\nabla }f)p\parallel \\ +2\lambda 〈y_n-{\tilde{y}}_n,\mathrm{\nabla }{f}_{{\alpha }_n}(y_n)-\mathrm{\nabla }f(p)〉-{\lambda }^2{\parallel \mathrm{\nabla }{f}_{{\alpha }_n}(y_n)-\mathrm{\nabla }f(p)\parallel }^2\\ \le & {\beta }_n{\parallel Q{y}_n-p\parallel }^2+{\parallel y_n-p\parallel }^2-{\parallel y_n-{\tilde{y}}_n\parallel }^2\\ +2\lambda \parallel \mathrm{\nabla }{f}_{{\alpha }_n}(y_n)-\mathrm{\nabla }f(p)\parallel \parallel (I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})y_n-(I-\lambda \mathrm{\nabla }f)p\parallel \\ +2\lambda 〈y_n-{\tilde{y}}_n,\mathrm{\nabla }{f}_{{\alpha }_n}(y_n)-\mathrm{\nabla }f(p)〉\\ \le & {\beta }_n{\parallel Q{y}_n-p\parallel }^2+{\parallel y_n-p\parallel }^2-{\parallel y_n-{\tilde{y}}_n\parallel }^2\\ +2\lambda \parallel \mathrm{\nabla }{f}_{{\alpha }_n}(y_n)-\mathrm{\nabla }f(p)\parallel (\parallel (I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})y_n-(I-\lambda \mathrm{\nabla }f)p\parallel +\parallel y_n-{\tilde{y}}_n\parallel ),\end{array}$$

which implies that

$$\begin{array}{r}{\parallel y_n-{\tilde{y}}_n\parallel }^2\\ \le {\beta }_n{\parallel Q{y}_n-p\parallel }^2+{\parallel y_n-p\parallel }^2-{\parallel z_n-p\parallel }^2\\ +2\lambda \parallel \mathrm{\nabla }{f}_{{\alpha }_n}(y_n)-\mathrm{\nabla }f(p)\parallel (\parallel (I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})y_n-(I-\lambda \mathrm{\nabla }f)p\parallel +\parallel y_n-{\tilde{y}}_n\parallel )\\ \le {\beta }_n{\parallel Q{y}_n-p\parallel }^2+(\parallel y_n-p\parallel +\parallel z_n-p\parallel )\parallel y_n-z_n\parallel \\ +2\lambda \parallel \mathrm{\nabla }{f}_{{\alpha }_n}(y_n)-\mathrm{\nabla }f(p)\parallel (\parallel (I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})y_n-(I-\lambda \mathrm{\nabla }f)p\parallel +\parallel y_n-{\tilde{y}}_n\parallel ).\end{array}$$

Since ${\beta }_n\to 0$, $\parallel y_n-z_n\parallel \to 0$ and $\parallel \mathrm{\nabla }{f}_{{\alpha }_n}(y_n)-\mathrm{\nabla }f(p)\parallel \to 0$, from the boundedness of $\{x_n\}$, $\{y_n\}$, $\{z_n\}$ and $\{{\tilde{y}}_n\}$, it follows that

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

In addition, since $({\gamma }_n+{\delta }_n)\zeta \le {\gamma }_n$ for all $n\ge 0$, utilizing Lemma 2.4, we get from (3.12)

$$\begin{array}{rl}{\parallel x_{n+1}-p\parallel }^2\le & {\sigma }_n{\parallel z_n-p\parallel }^2+({\gamma }_n+{\delta }_n){\parallel {\tilde{z}}_n-p\parallel }^2\\ =& {\sigma }_n{\parallel z_n-p\parallel }^2+(1-{\sigma }_n){\parallel {\tilde{z}}_n-p\parallel }^2\\ \le & {\sigma }_n{\parallel z_n-p\parallel }^2+(1-{\sigma }_n)\{{\parallel z_n-p\parallel }^2-{\parallel z_n-{\tilde{z}}_n\parallel }^2\\ +2\lambda \parallel \mathrm{\nabla }{f}_{{\alpha }_n}(z_n)-\mathrm{\nabla }f(p)\parallel \parallel (I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})z_n-(I-\lambda \mathrm{\nabla }f)p\parallel \\ +2\lambda 〈z_n-{\tilde{z}}_n,\mathrm{\nabla }{f}_{{\alpha }_n}(z_n)-\mathrm{\nabla }f(p)〉-{\lambda }^2{\parallel \mathrm{\nabla }{f}_{{\alpha }_n}(z_n)-\mathrm{\nabla }f(p)\parallel }^2\}\\ \le & {\sigma }_n{\parallel z_n-p\parallel }^2+(1-{\sigma }_n)\{{\parallel z_n-p\parallel }^2-{\parallel z_n-{\tilde{z}}_n\parallel }^2\\ +2\lambda \parallel \mathrm{\nabla }{f}_{{\alpha }_n}(z_n)-\mathrm{\nabla }f(p)\parallel \parallel (I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})z_n-(I-\lambda \mathrm{\nabla }f)p\parallel \\ +2\lambda \parallel z_n-{\tilde{z}}_n\parallel \parallel \mathrm{\nabla }{f}_{{\alpha }_n}(z_n)-\mathrm{\nabla }f(p)\parallel \}\\ \le & {\parallel z_n-p\parallel }^2-(1-{\sigma }_n){\parallel z_n-{\tilde{z}}_n\parallel }^2\\ +2\lambda \parallel \mathrm{\nabla }{f}_{{\alpha }_n}(z_n)-\mathrm{\nabla }f(p)\parallel (\parallel (I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})z_n-(I-\lambda \mathrm{\nabla }f)p\parallel +\parallel z_n-{\tilde{z}}_n\parallel ),\end{array}$$

which implies that

$$\begin{array}{r}(1-{\sigma }_n){\parallel z_n-{\tilde{z}}_n\parallel }^2\\ \le {\parallel z_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2\\ +2\lambda \parallel \mathrm{\nabla }{f}_{{\alpha }_n}(z_n)-\mathrm{\nabla }f(p)\parallel (\parallel (I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})z_n-(I-\lambda \mathrm{\nabla }f)p\parallel +\parallel z_n-{\tilde{z}}_n\parallel )\\ \le (\parallel z_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel z_n-x_{n+1}\parallel \\ +2\lambda \parallel \mathrm{\nabla }{f}_{{\alpha }_n}(z_n)-\mathrm{\nabla }f(p)\parallel (\parallel (I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})z_n-(I-\lambda \mathrm{\nabla }f)p\parallel +\parallel z_n-{\tilde{z}}_n\parallel ).\end{array}$$

Since $\{{\sigma }_n\}\subset [c,d]$, $\parallel z_n-x_{n+1}\parallel \to 0$ and $\parallel \mathrm{\nabla }{f}_{{\alpha }_n}(z_n)-\mathrm{\nabla }f(p)\parallel \to 0$, from the boundedness of $\{x_n\}$, $\{z_n\}$ and $\{{\tilde{z}}_n\}$, it follows that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel z_n-{\tilde{z}}_n\parallel =0.$$

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

Let $p^{\ast }\in {\omega }_w(x_n)$. Then there exists a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ such that $x_{n_i}\rightharpoonup p^{\ast }$. Since

$$\begin{array}{rl}{z}_n-y_n& ={\beta }_n(Q{y}_n-y_n)+(1-{\beta }_n)(T{\tilde{y}}_n-y_n)\\ ={\beta }_n(Q{y}_n-y_n)+(1-{\beta }_n)(T{\tilde{y}}_n-{\tilde{y}}_n)+(1-{\beta }_n)({\tilde{y}}_n-y_n),\end{array}$$

we have

$$\begin{array}{rl}(1-{\beta }_n)\parallel T{\tilde{y}}_n-{\tilde{y}}_n\parallel & =\parallel z_n-y_n-{\beta }_n(Q{y}_n-y_n)-(1-{\beta }_n)({\tilde{y}}_n-y_n)\parallel \\ \le \parallel z_n-y_n\parallel +{\beta }_n\parallel Q{y}_n-y_n\parallel +(1-{\beta }_n)\parallel {\tilde{y}}_n-y_n\parallel \\ \le \parallel z_n-y_n\parallel +{\beta }_n\parallel Q{y}_n-y_n\parallel +\parallel {\tilde{y}}_n-y_n\parallel .\end{array}$$

Hence from $\parallel z_n-y_n\parallel \to 0$, ${\beta }_n\to 0$ and $\parallel {\tilde{y}}_n-y_n\parallel \to 0$, we get ${lim}_{n\to \mathrm{\infty }}\parallel T{\tilde{y}}_n-{\tilde{y}}_n\parallel =0$. Since $\parallel x_n-y_n\parallel \to 0$ and $\parallel y_n-{\tilde{y}}_n\parallel \to 0$, we have ${\tilde{y}}_{n_i}\rightharpoonup p^{\ast }$. By Lemma 2.1(b) (demiclosedness principle), we obtain $p^{\ast }\in Fix(T)$.

Meanwhile, observe that

$$\begin{array}{rl}{x}_{n+1}-z_n& ={\gamma }_n({\tilde{z}}_n-z_n)+{\delta }_n(\Gamma {\tilde{z}}_n-{\tilde{z}}_n)+{\delta }_n({\tilde{z}}_n-z_n)\\ =({\gamma }_n+{\delta }_n)({\tilde{z}}_n-z_n)+{\delta }_n(\Gamma {\tilde{z}}_n-{\tilde{z}}_n)\\ =(1-{\sigma }_n)({\tilde{z}}_n-z_n)+{\delta }_n(\Gamma {\tilde{z}}_n-{\tilde{z}}_n).\end{array}$$

Thus,

$$\begin{array}{rl}{\delta }_n\parallel \Gamma {\tilde{z}}_n-{\tilde{z}}_n\parallel & =\parallel x_{n+1}-z_n-(1-{\sigma }_n)({\tilde{z}}_n-z_n)\parallel \\ \le \parallel x_{n+1}-z_n\parallel +(1-{\sigma }_n)\parallel {\tilde{z}}_n-z_n\parallel \\ \le \parallel x_{n+1}-z_n\parallel +\parallel {\tilde{z}}_n-z_n\parallel \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

This together with ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\delta }_n>0$ yields ${lim}_{n\to \mathrm{\infty }}\parallel \Gamma {\tilde{z}}_n-{\tilde{z}}_n\parallel =0$. Since $\parallel x_n-z_n\parallel \to 0$ and $\parallel z_n-{\tilde{z}}_n\parallel \to 0$, we have ${\tilde{z}}_{n_i}\rightharpoonup p^{\ast }$. By Lemma 2.1(b) (demiclosedness principle), we have $p^{\ast }\in Fix(\Gamma )$.

Further, let us show $p^{\ast }\in \Xi$. Indeed, from $\parallel x_n-y_n\parallel \to 0$ and $\parallel {\tilde{y}}_n-y_n\parallel \to 0$, we have $y_{n_i}\rightharpoonup p^{\ast }$ and ${\tilde{y}}_{n_i}\rightharpoonup p^{\ast }$. Define

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

where $N_{C}v=\{w\in H:〈v-u,w〉\ge 0,\mathrm{\forall }u\in C\}$. Then V is maximal monotone and $0\in Vv$ if and only if $v\in VI(C,\mathrm{\nabla }f)$ (see [17]). Let $(v,w)\in graph(V)$. Then we have

$$w\in Vv=\mathrm{\nabla }f(v)+N_{C}v,$$

and hence

$$w-\mathrm{\nabla }f(v)\in N_{C}v.$$

Therefore, we have

$$〈v-u,w-\mathrm{\nabla }f(v)〉\ge 0,\mathrm{\forall }u\in C.$$

On the other hand, from

$${\tilde{y}}_n=P_C(y_n-\lambda \mathrm{\nabla }{f}_{{\alpha }_n}(y_n))\text{and}v\in C,$$

we have

$$〈y_n-\lambda \mathrm{\nabla }{f}_{{\alpha }_n}(y_n)-{\tilde{y}}_n,{\tilde{y}}_n-v〉\ge 0,$$

and hence

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

Therefore, from

$$w-\mathrm{\nabla }f(v)\in N_C(v)\text{and}{\tilde{y}}_{n_i}\in C,$$

we have

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

Hence, we obtain

$$〈v-p^{\ast },w〉\ge 0\text{as }i\to \mathrm{\infty }.$$

Since V is maximal monotone, we have $p^{\ast }\in V^{-1}0$, and hence $p^{\ast }\in VI(C,\mathrm{\nabla }f)$, which leads to $p\in \Xi$. Consequently, $p^{\ast }\in Fix(T)\cap Fix(\Gamma )\cap \Xi$. This shows that ${\omega }_w(x_n)\subset Fix(T)\cap Fix(\Gamma )\cap \Xi$.

Finally, let us show $p^{\ast }\in \Omega$. Indeed, it follows from (2.2) that for every $p\in Fix(T)\cap Fix(\Gamma )\cap \Xi$,

$$\begin{array}{rl}{\parallel y_n-p\parallel }^2& ={\parallel (1-{\theta }_n)(x_n-p)+{\theta }_n(S{x}_n-Sp)+{\theta }_n(Sp-p)\parallel }^2\\ \le {\parallel (1-{\theta }_n)(x_n-p)+{\theta }_n(S{x}_n-Sp)\parallel }^2+2{\theta }_n〈Sp-p,y_n-p〉\\ \le (1-{\theta }_n){\parallel x_n-p\parallel }^2+{\theta }_n{\parallel S{x}_n-Sp\parallel }^2+2{\theta }_n〈Sp-p,y_n-p〉\\ \le {\parallel x_n-p\parallel }^2+2{\theta }_n〈Sp-p,y_n-p〉,\end{array}$$

and hence

$$\begin{array}{r}{\parallel z_n-p\parallel }^2\\ \le {\beta }_n{\parallel Q{y}_n-p\parallel }^2+(1-{\beta }_n){\parallel {\tilde{y}}_n-p\parallel }^2\\ \le {\beta }_n{\parallel Q{y}_n-p\parallel }^2+{\parallel {\tilde{y}}_n-p\parallel }^2\\ \le {\beta }_n{\parallel Q{y}_n-p\parallel }^2+{(\parallel y_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel )}^2\\ \le {\beta }_n{\parallel Q{y}_n-p\parallel }^2+{\parallel y_n-p\parallel }^2+\lambda {\alpha }_n\parallel p\parallel (2\parallel y_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel )\\ \le {\beta }_n{\parallel Q{y}_n-p\parallel }^2+{\parallel x_n-p\parallel }^2+2{\theta }_n〈Sp-p,y_n-p〉\\ +\lambda {\alpha }_n\parallel p\parallel (2\parallel y_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel ).\end{array}$$

Since $({\gamma }_n+{\delta }_n)\zeta \le {\gamma }_n$ for all $n\ge 0$, by Lemma 2.4, we have

$$\begin{array}{r}{\parallel x_{n+1}-p\parallel }^2\\ \le {\sigma }_n{\parallel z_n-p\parallel }^2+({\gamma }_n+{\delta }_n){\parallel {\tilde{z}}_n-p\parallel }^2\\ \le {\sigma }_n{\parallel z_n-p\parallel }^2+(1-{\sigma }_n){(\parallel z_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel )}^2\\ \le {\parallel z_n-p\parallel }^2+\lambda {\alpha }_n\parallel p\parallel (2\parallel z_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel )\\ \le {\beta }_n{\parallel Q{y}_n-p\parallel }^2+{\parallel x_n-p\parallel }^2+2{\theta }_n〈Sp-p,y_n-p〉\\ +\lambda {\alpha }_n\parallel p\parallel (2\parallel y_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel )+\lambda {\alpha }_n\parallel p\parallel (2\parallel z_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel )\\ ={\parallel x_n-p\parallel }^2+{\beta }_n{\parallel Q{y}_n-p\parallel }^2+2{\theta }_n〈Sp-p,y_n-p〉\\ +2\lambda {\alpha }_n\parallel p\parallel (\parallel y_n-p\parallel +\parallel z_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel ),\end{array}$$

which implies that

$$\begin{array}{rl}2〈p-Sp,y_n-p〉\le & \frac{1}{{\theta }_n}({\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2)+\frac{{\beta }_n}{{\theta }_n}{\parallel Q{y}_n-p\parallel }^2\\ +\frac{{\alpha }_n}{{\theta }_n}2\lambda \parallel p\parallel (\parallel y_n-p\parallel +\parallel z_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel )\\ \le & \frac{\parallel x_n-x_{n+1}\parallel }{{\theta }_n}(\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )+\frac{{\beta }_n}{{\theta }_n}{\parallel Q{y}_n-p\parallel }^2\\ +\frac{{\alpha }_n}{{\theta }_n}2\lambda \parallel p\parallel (\parallel y_n-p\parallel +\parallel z_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel ).\end{array}$$

Since $\frac{{\alpha }_n+{\beta }_n}{{\theta }_n}\to 0$ and $\frac{\parallel x_n-x_{n+1}\parallel }{{\theta }_n}\to 0$ as $n\to \mathrm{\infty }$, from the boundedness of $\{x_n\}$, $\{y_n\}$ and $\{z_n\}$, we deduce that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈p-Sp,y_n-p〉\le 0,\mathrm{\forall }p\in Fix(T)\cap Fix(\Gamma )\cap \Xi .$$

So, from $y_{n_i}\rightharpoonup p^{\ast }$, we get

$$〈p-Sp,p^{\ast }-p〉\le 0,\mathrm{\forall }p\in Fix(T)\cap Fix(\Gamma )\cap \Xi .$$

Taking into consideration that $I-S$ is monotone and continuous, utilizing Minty’s lemma [7], we have

$$〈p^{\ast }-S{p}^{\ast },p^{\ast }-p〉\le 0,\mathrm{\forall }p\in Fix(T)\cap Fix(\Gamma )\cap \Xi .$$

Therefore, $p^{\ast }=P_{Fix(T)\cap Fix(\Gamma )\cap \Xi }S{p}^{\ast }$; that is, $p^{\ast }\in \Omega$. □

Remark 3.1 Iterative algorithm (2.2) is different from the algorithms in [1, 11]. The two-step iterative scheme in [11] for two nonexpansive mappings and the gradient-projection iterative schemes in [1] for MP (1.1) are extended to develop three-step iterative scheme (2.2) with regularization for MP (1.1), two nonexpansive mappings and a strictly pseudocontractive mapping.

Remark 3.2 The following sequences satisfy the hypotheses on the parameter in Theorem 3.1.

• (a) ${\alpha }_n=\frac{1}{n^{1+s+t}}$, ${\beta }_n=\frac{1}{n^s}$ and ${\theta }_n=\frac{1}{n^t}$, where $t\in (0,\frac{1}{2})$ and $s\in (t,1-t)$;

• (b) ${\sigma }_n=\frac{1}{2}+\frac{2}{n}$ and ${\gamma }_n={\delta }_n=\frac{1}{4}-\frac{1}{n}$ for all $n>4$.

Theorem 3.2 Let $\{x_n\}$ be the bounded sequence generated from any given $x_0\in C$ by (2.2). Assume that hypotheses (H1)-(H5) of Theorem  3.1 hold and

• (H6) ${lim}_{n\to \mathrm{\infty }}\frac{{\theta }_n^2}{{\beta }_n}=0$;

• (H7) There is a constant $k>0$ such that $\parallel x-T{P}_C(I-\lambda \mathrm{\nabla }f)x\parallel \ge kdist(x,Fix(T)\cap Fix(\Gamma )\cap \Xi )$ for each $x\in C$, where $dist(x,Fix(T)\cap Fix(\Gamma )\cap \Xi )={inf}_{y\in Fix(T)\cap Fix(\Gamma )\cap \Xi }\parallel x-y\parallel$.

Then the sequences $\{x_n\}$, $\{y_n\}$ and $\{z_n\}$ converge strongly to $x^{\ast }=P_{\Omega }Q{x}^{\ast }$ provided $\parallel x_n-z_n\parallel =o({\theta }_n)$, where $x^{\ast }$ solves the following variational inequality:

$$〈x^{\ast }-S{x}^{\ast },x^{\ast }-x〉\le 0,\mathrm{\forall }x\in Fix(T)\cap Fix(\Gamma )\cap \Xi .$$

Proof Let $p\in Fix(T)\cap Fix(\Gamma )\cap \Xi$. From (2.2), we have

$$z_n-p={\beta }_n(Q{y}_n-Qp)+{\beta }_n(Qp-p)+(1-{\beta }_n)(T{\tilde{y}}_n-p),$$

and therefore,

$$\begin{array}{r}{\parallel z_n-p\parallel }^2\\ \le {\parallel {\beta }_n(Q{y}_n-Qp)+(1-{\beta }_n)(T{\tilde{y}}_n-p)\parallel }^2+2{\beta }_n〈Qp-p,z_n-p〉\\ \le (1-{\beta }_n){\parallel T{\tilde{y}}_n-p\parallel }^2+{\beta }_n{\parallel Q{y}_n-Qp\parallel }^2+2{\beta }_n〈Qp-p,z_n-p〉\\ \le (1-{\beta }_n){\parallel {\tilde{y}}_n-p\parallel }^2+{\beta }_n{\rho }^2{\parallel y_n-p\parallel }^2+2{\beta }_n〈Qp-p,z_n-p〉\\ \le (1-{\beta }_n){(\parallel y_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel )}^2+{\beta }_n\rho {\parallel y_n-p\parallel }^2+2{\beta }_n〈Qp-p,z_n-p〉\\ \le (1-(1-\rho ){\beta }_n){\parallel y_n-p\parallel }^2+\lambda {\alpha }_n\parallel p\parallel (2\parallel y_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel )\\ +2{\beta }_n〈Qp-p,z_n-p〉.\end{array}$$

(3.13)

Again from (2.2), we obtain

$$\begin{array}{rl}{\parallel y_n-p\parallel }^2& ={\parallel (1-{\theta }_n)(x_n-p)+{\theta }_n(S{x}_n-Sp)+{\theta }_n(Sp-p)\parallel }^2\\ \le {\parallel (1-{\theta }_n)(x_n-p)+{\theta }_n(S{x}_n-Sp)\parallel }^2+2{\theta }_n〈Sp-p,y_n-p〉\\ \le (1-{\theta }_n){\parallel x_n-p\parallel }^2+{\theta }_n{\parallel S{x}_n-Sp\parallel }^2+2{\theta }_n〈Sp-p,y_n-p〉\\ \le {\parallel x_n-p\parallel }^2+2{\theta }_n〈Sp-p,y_n-p〉.\end{array}$$

(3.14)

Substituting (3.14) into (3.13), we get

$$\begin{array}{rl}{\parallel z_n-p\parallel }^2\le & (1-(1-\rho ){\beta }_n)({\parallel x_n-p\parallel }^2+2{\theta }_n〈Sp-p,y_n-p〉)\\ +\lambda {\alpha }_n\parallel p\parallel (2\parallel y_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel )+2{\beta }_n〈Qp-p,z_n-p〉\\ =& (1-(1-\rho ){\beta }_n){\parallel x_n-p\parallel }^2+2(1-(1-\rho ){\beta }_n){\theta }_n〈Sp-p,y_n-p〉\\ +2{\beta }_n〈Qp-p,z_n-p〉+\lambda {\alpha }_n\parallel p\parallel (2\parallel y_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel ).\end{array}$$

(3.15)

Since $({\gamma }_n+{\delta }_n)\zeta \le {\gamma }_n$ for all $n\ge 0$, utilizing Lemma 2.4, we get from (2.2) and (3.15)

$$\begin{array}{rl}{\parallel x_{n+1}-p\parallel }^2\le & {\sigma }_n{\parallel z_n-p\parallel }^2+({\gamma }_n+{\delta }_n){\parallel {\tilde{z}}_n-p\parallel }^2\\ \le & {\sigma }_n{\parallel z_n-p\parallel }^2+(1-{\sigma }_n){(\parallel z_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel )}^2\\ \le & {\parallel z_n-p\parallel }^2+\lambda {\alpha }_n\parallel p\parallel (2\parallel z_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel )\\ \le & (1-(1-\rho ){\beta }_n){\parallel x_n-p\parallel }^2+2(1-(1-\rho ){\beta }_n){\theta }_n〈Sp-p,y_n-p〉\\ +2{\beta }_n〈Qp-p,z_n-p〉+\lambda {\alpha }_n\parallel p\parallel (2\parallel y_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel )\\ +\lambda {\alpha }_n\parallel p\parallel (2\parallel z_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel )\\ =& (1-(1-\rho ){\beta }_n){\parallel x_n-p\parallel }^2+2(1-(1-\rho ){\beta }_n){\theta }_n〈Sp-p,y_n-p〉\\ +2{\beta }_n〈Qp-p,z_n-p〉+2\lambda {\alpha }_n\parallel p\parallel (\parallel y_n-p\parallel +\parallel z_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel )\\ \le & (1-(1-\rho ){\beta }_n){\parallel x_n-p\parallel }^2+2(1-(1-\rho ){\beta }_n){\theta }_n〈Sp-p,y_n-p〉\\ +2{\beta }_n〈Qp-p,z_n-p〉+\tilde{M}{\alpha }_n,\end{array}$$

(3.16)

where $\tilde{M}={sup}_{n\ge 0}\{2\lambda \parallel p\parallel (\parallel y_n-p\parallel +\parallel z_n-p\parallel +\lambda {\alpha }_n\parallel p\parallel )\}<\mathrm{\infty }$.

Taking into consideration that $P_{\Omega }\circ Q$ is a contractive mapping, we know that $P_{\Omega }\circ Q$ has a unique fixed point $x^{\ast }\in \Omega$. That is, there is a unique solution $x^{\ast }\in \Omega$ of the following variational inequality problem (VIP):

$$〈Q{x}^{\ast }-x^{\ast },q-x^{\ast }〉\le 0,\mathrm{\forall }q\in \Omega .$$

(3.17)

Since $x^{\ast }\in \Omega$, it is clear that $x^{\ast }=P_{Fix(T)\cap Fix(\Gamma )\cap \Xi }S{x}^{\ast }$, and hence $x^{\ast }\in Fix(T)\cap Fix(\Gamma )\cap \Xi$. Thus, from (3.16), we conclude that

$$\begin{array}{r}{\parallel x_{n+1}-x^{\ast }\parallel }^2\\ \le (1-(1-\rho ){\beta }_n){\parallel x_n-x^{\ast }\parallel }^2+2(1-(1-\rho ){\beta }_n){\theta }_n〈S{x}^{\ast }-x^{\ast },y_n-x^{\ast }〉\\ +2{\beta }_n〈Q{x}^{\ast }-x^{\ast },z_n-x^{\ast }〉+\tilde{M}{\alpha }_n\\ =(1-(1-\rho ){\beta }_n){\parallel x_n-x^{\ast }\parallel }^2+(1-\rho ){\beta }_n\{\frac{2(1-(1-\rho ){\beta }_n)}{1-\rho }\frac{{\theta }_n}{{\beta }_n}〈S{x}^{\ast }-x^{\ast },y_n-x^{\ast }〉\\ +\frac{2}{1-\rho }〈Q{x}^{\ast }-x^{\ast },z_n-x^{\ast }〉\}+\tilde{M}{\alpha }_n.\end{array}$$

(3.18)

Consider a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈Q{x}^{\ast }-x^{\ast },x_n-x^{\ast }〉=\underset{i\to \mathrm{\infty }}{lim}〈Q{x}^{\ast }-x^{\ast },x_{n_i}-x^{\ast }〉.$$

Without loss of generality, we may further assume that $x_{n_i}\rightharpoonup \tilde{x}$. Then, in view of Theorem 3.1, $\tilde{x}\in \Omega$. Since $x^{\ast }$ is a unique solution of VIP (3.17) and $\parallel x_n-z_n\parallel \to 0$, we have

$$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈Q{x}^{\ast }-x^{\ast },z_n-x^{\ast }〉& =\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(〈Q{x}^{\ast }-x^{\ast },z_n-x_n〉+〈Q{x}^{\ast }-x^{\ast },x_n-x^{\ast }〉)\\ =\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈Q{x}^{\ast }-x^{\ast },x_n-x^{\ast }〉=\underset{i\to \mathrm{\infty }}{lim}〈Q{x}^{\ast }-x^{\ast },x_{n_i}-x^{\ast }〉\\ =〈Q{x}^{\ast }-x^{\ast },\tilde{x}-x^{\ast }〉\le 0,\end{array}$$

which implies that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{2}{1-\rho }〈Q{x}^{\ast }-x^{\ast },z_n-x^{\ast }〉\le 0.$$

(3.19)

Meanwhile, from $x^{\ast }\in \Omega$ and (H7), we infer that

$$\begin{array}{r}〈S{x}^{\ast }-x^{\ast },y_n-x^{\ast }〉\\ =〈S{x}^{\ast }-x^{\ast },y_n-P_{Fix(T)\cap Fix(\Gamma )\cap \Xi }{y}_n〉+〈S{x}^{\ast }-x^{\ast },P_{Fix(T)\cap Fix(\Gamma )\cap \Xi }{y}_n-x^{\ast }〉\\ \le 〈S{x}^{\ast }-x^{\ast },y_n-P_{Fix(T)\cap Fix(\Gamma )\cap \Xi }{y}_n〉\\ \le \parallel S{x}^{\ast }-x^{\ast }\parallel \parallel y_n-P_{Fix(T)\cap Fix(\Gamma )\cap \Xi }{y}_n\parallel \\ =dist(y_n,Fix(T)\cap Fix(\Gamma )\cap \Xi )\parallel S{x}^{\ast }-x^{\ast }\parallel \\ \le \frac{1}{k}\parallel S{x}^{\ast }-x^{\ast }\parallel \parallel y_n-T{P}_C(I-\lambda \mathrm{\nabla }f)y_n\parallel .\end{array}$$

From (2.2), we have

$$\frac{z_n-x_n}{{\theta }_n}=\frac{{\beta }_n}{{\theta }_n}(Q{y}_n-x_n)+\frac{1-{\beta }_n}{{\theta }_n}(T{\tilde{y}}_n-x_n).$$

This together with ${lim}_{n\to \mathrm{\infty }}\frac{\parallel z_n-x_n\parallel }{{\theta }_n}=0$ and $\frac{{\beta }_n}{{\theta }_n}=0$ implies that

$$\underset{n\to \mathrm{\infty }}{lim}\frac{\parallel T{\tilde{y}}_n-x_n\parallel }{{\theta }_n}=0.$$

Hence,

$$\underset{n\to \mathrm{\infty }}{lim}\frac{{\theta }_n\parallel T{\tilde{y}}_n-x_n\parallel }{{\beta }_n}=\underset{n\to \mathrm{\infty }}{lim}\frac{\parallel T{\tilde{y}}_n-x_n\parallel }{{\theta }_n}\frac{{\theta }_n^2}{{\beta }_n}=0.$$

Observe that

$$y_n-x_n={\theta }_n(S{x}_n-x_n).$$

Therefore, we get

$$\underset{n\to \mathrm{\infty }}{lim}\frac{{\theta }_n}{{\beta }_n}\parallel y_n-x_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\frac{{\theta }_n^2}{{\beta }_n}\parallel S{x}_n-x_n\parallel =0,$$

and hence

$$\begin{array}{r}\frac{{\theta }_n}{{\beta }_n}\parallel y_n-T{P}_C(I-\lambda \mathrm{\nabla }f)y_n\parallel \\ \le \frac{{\theta }_n}{{\beta }_n}(\parallel y_n-x_n\parallel +\parallel x_n-T{P}_C(I-\lambda \mathrm{\nabla }f)y_n\parallel )\\ \le \frac{{\theta }_n}{{\beta }_n}(\parallel y_n-x_n\parallel +\parallel x_n-T{P}_C(I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})y_n\parallel \\ +\parallel T{P}_C(I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})y_n-T{P}_C(I-\lambda \mathrm{\nabla }f)y_n\parallel )\\ \le \frac{{\theta }_n}{{\beta }_n}(\parallel y_n-x_n\parallel +\parallel x_n-T{\tilde{y}}_n\parallel +\parallel (I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})y_n-(I-\lambda \mathrm{\nabla }f)y_n\parallel )\\ =\frac{{\theta }_n}{{\beta }_n}\parallel y_n-x_n\parallel +\frac{{\theta }_n}{{\beta }_n}\parallel x_n-T{\tilde{y}}_n\parallel +\frac{{\theta }_n}{{\beta }_n}\lambda {\alpha }_n\parallel y_n\parallel \\ =\frac{{\theta }_n}{{\beta }_n}\parallel y_n-x_n\parallel +\frac{{\theta }_n}{{\beta }_n}\parallel x_n-T{\tilde{y}}_n\parallel +\frac{{\theta }_n^2}{{\beta }_n}\frac{{\alpha }_n}{{\theta }_n}\lambda \parallel y_n\parallel \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

Thus, it follows that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{{\theta }_n}{{\beta }_n}〈S{x}^{\ast }-x^{\ast },y_n-x^{\ast }〉\le 0,$$

and hence

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{2(1-(1-\rho ){\beta }_n)}{1-\rho }\frac{{\theta }_n}{{\beta }_n}〈S{x}^{\ast }-x^{\ast },y_n-x^{\ast }〉\le 0.$$

(3.20)

Utilizing Lemma 2.3, from ${\sum }_{n=0}^{\mathrm{\infty }}\tilde{M}{\alpha }_n<\mathrm{\infty }$ and (3.18)-(3.20), we conclude that the sequence $\{x_n\}$ converges strongly to $x^{\ast }$. Taking into consideration that $\parallel x_n-y_n\parallel \to 0$ and $\parallel x_n-z_n\parallel \to 0$, we obtain that $\parallel y_n-x^{\ast }\parallel \to 0$ and $\parallel z_n-x^{\ast }\parallel \to 0$ as $n\to \mathrm{\infty }$. This completes the proof. □

Remark 3.3 The following parametric sequences satisfy the hypotheses of Theorem 3.2.

• (a) ${\alpha }_n=\frac{1}{n^{1+s+t}}$, ${\beta }_n=\frac{1}{n^s}$ and ${\theta }_n=\frac{1}{n^t}$, where $t\in (0,\frac{1}{3}]$ and $s\in (t,2t)$ or $t\in (\frac{1}{3},\frac{1}{2})$, $s\in (t,1-t)$;

• (b) ${\sigma }_n=\frac{1}{2}+\frac{2}{n}$, ${\gamma }_n={\delta }_n=\frac{1}{4}-\frac{1}{n}$, $\mathrm{\forall }n>4$.

Remark 3.4 Theorems 3.1 and 3.2 improve, extend, supplement and develop [[11], Theorems 3.1 and 3.2] and [[1], Theorems 5.2 and 6.1] in the following aspects:

• (a) Three-step iterative algorithm (2.2) with regularization for MP (1.1), two nonexpansive mappings and a strictly pseudocontractive mapping are more flexible and more subtle than the algorithms in [1, 11].

• (b) The argument techniques in Theorems 3.1 and 3.2 are different from the ones in [[11], Theorems 3.1 and 3.2] and the ones in [[1], Theorems 5.2 and 6.1] because we use the properties of strict pseudocontractive mappings and maximal monotone mappings (see, for example, Lemmas 2.1, 2.4 and 2.6).

• (c) Compared with the proof of Theorems 5.2 and 6.1 in [1], the proof of Theorems 3.1 and 3.2 shows ${lim}_{n\to \mathrm{\infty }}\parallel y_n-P_C(I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})y_n\parallel ={lim}_{n\to \mathrm{\infty }}\parallel z_n-P_C(I-\lambda \mathrm{\nabla }{f}_{{\alpha }_n})z_n\parallel =0$ via the argument of ${lim}_{n\to \mathrm{\infty }}\parallel \mathrm{\nabla }{f}_{{\alpha }_n}(y_n)-\mathrm{\nabla }f(p)\parallel ={lim}_{n\to \mathrm{\infty }}\parallel \mathrm{\nabla }{f}_{{\alpha }_n}(z_n)-\mathrm{\nabla }f(p)\parallel =0$, $\mathrm{\forall }p\in Fix(T)\cap Fix(\Gamma )\cap \Xi$ (see Step 3 in the proof of Theorem 3.1).

• (e) Theorems 3.1 and 3.2 remove the condition $Fix(T)\cap intC\ne \mathrm{\varnothing }$ in [[11], Theorems 3.1 and 3.2].