# Multi-step implicit iterative methods with regularization for minimization problems and fixed point problems

## Abstract

In this paper we introduce a multi-step implicit iterative scheme with regularization for finding a common solution of the minimization problem (MP) for a convex and continuously Fréchet differentiable functional and the common fixed point problem of an infinite family of nonexpansive mappings in the setting of Hilbert spaces. The multi-step implicit iterative method with regularization is based on three well-known methods: the extragradient method, approximate proximal method and gradient projection algorithm with regularization. We derive a weak convergence theorem for the sequences generated by the proposed scheme. On the other hand, we also establish a strong convergence result via an implicit hybrid method with regularization for solving these two problems. This implicit hybrid method with regularization is based on the CQ method, extragradient method and gradient projection algorithm with regularization.

MSC:49J30, 47H09, 47J20.

## 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 ${P}_{C}$ be the metric projection of H onto C. Let $S:C\to C$ be a self-mapping on C. We denote by $Fix\left(S\right)$ the set of fixed points of S and by R the set of all real numbers. A mapping $A:C\to H$ is called L-Lipschitz continuous if there exists a constant $L\ge 0$ such that $\parallel Ax-Ay\parallel \le L\parallel x-y\parallel$ for all $x,y\in C$. In particular, if $L=1$, then A is called a nonexpansive mapping [1]; if $L\in \left[0,1\right)$ then A is called a contraction.

Let $f:C\to \mathbf{R}$ be a convex and continuously Fréchet differentiable functional. Consider the minimization problem (MP) of minimizing f over the constraint set C

$\underset{x\in C}{min}f\left(x\right).$
(1.1)

We denote by Γ the set of minimizers of MP (1.1) which are assumed to be nonempty.

On the other hand, consider the following variational inequality problem (VIP): find a $\overline{x}\in C$ such that

$〈A\overline{x},y-\overline{x}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in C.$
(1.2)

The solution set of VIP (1.2) is denoted by $VI\left(C,A\right)$.

We remark that VIP (1.2) was first discussed by Lions [2] and now is well known. There are a lot of different approaches towards solving VIP (1.2) in finite-dimensional and infinite-dimensional spaces, and the research is intensively investigated. VIP (1.2) has many applications in computational mathematics, mathematical physics, operations research, mathematical economics, optimization theory, and other fields; see, e.g., [36] and the references therein.

Recently, motivated by the work of Takahashi and Zembayashi [7], Cholamjiak [8] introduced a new hybrid projection algorithm for finding a common element of the set of solutions of the equilibrium problem and the set of solutions of the variational inequality problem and the set of fixed points of relatively quasi-nonexpansive mappings in a Banach space. Here, the involved operator in [8] is an inverse-strongly monotone operator. Furthermore, Nadezhkina and Takahashi [9] introduced an iterative process for finding an element of $Fix\left(S\right)\cap VI\left(C,A\right)$ and obtained a strong convergence theorem.

Theorem NT (see [[9], Theorem 3.1])

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $A:C\to H$ be a monotone and L-Lipschitz-continuous mapping and $S:C\to C$ be a nonexpansive mapping such that $Fix\left(S\right)\cap VI\left(C,A\right)\ne \mathrm{\varnothing }$. Let $\left\{{x}_{n}\right\}$, $\left\{{y}_{n}\right\}$ and $\left\{{z}_{n}\right\}$ be the sequences generated by

$\left\{\begin{array}{c}{x}_{0}=x\in C\phantom{\rule{1em}{0ex}}\mathit{\text{chosen arbitrarily}},\hfill \\ {y}_{n}={P}_{C}\left({x}_{n}-{\lambda }_{n}A{x}_{n}\right),\hfill \\ {z}_{n}={\alpha }_{n}{x}_{n}+\left(1-{\alpha }_{n}\right){P}_{C}\left({x}_{n}-{\lambda }_{n}A{y}_{n}\right),\hfill \\ {C}_{n}=\left\{z\in C:\parallel {z}_{n}-z\parallel \le \parallel {x}_{n}-z\parallel \right\},\hfill \\ {Q}_{n}=\left\{z\in C:〈{x}_{n}-z,x-{x}_{n}〉\ge 0\right\},\hfill \\ {x}_{n+1}={P}_{{C}_{n}\cap {Q}_{n}}x,\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 0,\hfill \end{array}$

where $\left\{{\lambda }_{n}\right\}\subset \left[a,b\right]$ for some $a,b\in \left(0,1/L\right)$ and $\left\{{\alpha }_{n}\right\}\subset \left[0,c\right]$ for some $c\in \left[0,1\right)$. Then the sequences $\left\{{x}_{n}\right\}$, $\left\{{y}_{n}\right\}$ and $\left\{{z}_{n}\right\}$ converge strongly to ${P}_{Fix\left(S\right)\cap VI\left(C,A\right)}x$.

Also, it is remarkable that the joint work of Nadezhkina and Takahashi [9], which introduced a new iterative method, combines Korpelevich’s extragradient method and the so-called CQ method. We note that Nadezhkina and Takahashi employed the monotonicity and Lipschitz-continuity of A to define a maximal monotone operator T [10]. However, if the mapping A is pseudomonotone Lipschitz-continuous, then T is not necessarily a maximal monotone operator. To overcome this difficulty, Ceng et al. [11] suggested another iterative method. They established necessary and sufficient mild conditions such that the sequences generated by their proposed method converge weakly to some common solution of VIP (1.2) and the common fixed point problem of a finite family of nonexpansive mappings.

Theorem CTY ([[11], Theorem 3.1])

Let C be a nonempty closed convex subset of a real Hilbert space H. Let A be a pseudomonotone, k-Lipschitz-continuous and $\left(w,s\right)$-sequentially-continuous mapping of C into H, and let ${\left\{{S}_{n}\right\}}_{i=1}^{N}$ be N nonexpansive mappings of C into itself such that ${\bigcap }_{i=1}^{N}Fix\left({S}_{i}\right)\cap VI\left(C,A\right)\ne \mathrm{\varnothing }$. Let $\left\{{x}_{n}\right\}$, $\left\{{y}_{n}\right\}$, $\left\{{z}_{n}\right\}$ be the sequences generated by

$\left\{\begin{array}{c}{x}_{1}=x\in C\phantom{\rule{1em}{0ex}}\mathit{\text{chosen arbitrarily}},\hfill \\ {y}_{n}={P}_{C}\left({x}_{n}-{\lambda }_{n}A{x}_{n}\right),\hfill \\ {z}_{n}={\alpha }_{n}{x}_{n}+\left(1-{\alpha }_{n}\right){S}_{n}{P}_{C}\left({x}_{n}-{\lambda }_{n}A{y}_{n}\right),\hfill \\ {C}_{n}=\left\{z\in C:\parallel {z}_{n}-z\parallel \le \parallel {x}_{n}-z\parallel \right\},\hfill \\ \phantom{\rule{1em}{0ex}}\mathit{\text{find}}{x}_{n+1}\in {C}_{n}\mathit{\text{such that}}〈{x}_{n}-{x}_{n+1}+{e}_{n}-{\sigma }_{n}A{x}_{n+1},{x}_{n+1}-x〉\ge -{\epsilon }_{n},\phantom{\rule{0.25em}{0ex}}\mathrm{\forall }x\in {C}_{n},\hfill \end{array}$

for every $n=1,2,\dots$ , where ${S}_{n}={S}_{nmodN}$, $\left\{{e}_{n}\right\}$ is an error sequence in H such that ${\sum }_{n=1}^{\mathrm{\infty }}\parallel {e}_{n}\parallel <\mathrm{\infty }$ and the following conditions hold:

1. (i)

$\left\{{\sigma }_{n}\right\}\subset \left(0,1/k\right)$, $\left\{{\epsilon }_{n}\right\}\subset \left[0,\mathrm{\infty }\right)$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\epsilon }_{n}<\mathrm{\infty }$;

2. (ii)

$\left\{{\lambda }_{n}\right\}\subset \left[a,b\right]$ for some $a,b\in \left(0,1/k\right)$;

3. (iii)

$\left\{{\alpha }_{n}\right\}\subset \left[0,c\right]$ for some $c\in \left[0,1\right)$.

Then the sequences $\left\{{x}_{n}\right\}$, $\left\{{y}_{n}\right\}$, $\left\{{z}_{n}\right\}$ converge weakly to the same element of ${\bigcap }_{i=1}^{N}Fix\left({S}_{i}\right)\cap VI\left(C,A\right)$ if and only if ${lim inf}_{n\to \mathrm{\infty }}〈A{x}_{n},x-{x}_{n}〉\ge 0$, $\mathrm{\forall }x\in C$.

In this paper, we aim to find a common solution of the minimization problem (MP) for a convex and continuously Fréchet differentiable functional and the common fixed point problem of an infinite family of nonexpansive mappings in the setting of Hilbert spaces. Motivated and inspired by the research going on in this area, we propose two iterative schemes for this purpose. One is called a multi-step implicit iterative method with regularization which is based on three well-known methods: extragradient method, approximate proximal method and gradient projection algorithm with regularization. Another is an implicit hybrid method with regularization which is based on the CQ method, extragradient method and gradient projection algorithm with regularization. Weak and strong convergence results for these two schemes are established, respectively. Recent results in this direction can be found, e.g., in [732].

## 2 Preliminaries

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 ${x}_{n}⇀x$ to indicate that the sequence $\left\{{x}_{n}\right\}$ converges weakly to x and ${x}_{n}\to x$ to indicate that the sequence $\left\{{x}_{n}\right\}$ converges strongly to x. Moreover, we use ${\omega }_{w}\left({x}_{n}\right)$ to denote the weak ω-limit set of the sequence $\left\{{x}_{n}\right\}$, i.e.,

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

$\parallel x-{P}_{C}x\parallel =\underset{y\in C}{inf}\parallel x-y\parallel =:d\left(x,C\right).$

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

Proposition 2.1 For given $x\in H$ and $z\in C$:

1. (i)

$z={P}_{C}x⇔〈x-z,y-z〉\le 0$, $\mathrm{\forall }y\in C$;

2. (ii)

$z={P}_{C}x⇔{\parallel x-z\parallel }^{2}\le {\parallel x-y\parallel }^{2}-{\parallel y-z\parallel }^{2}$, $\mathrm{\forall }y\in C$;

3. (iii)

$〈{P}_{C}x-{P}_{C}y,x-y〉\ge {\parallel {P}_{C}x-{P}_{C}y\parallel }^{2}$, $\mathrm{\forall }y\in H$.

Consequently, ${P}_{C}$ is nonexpansive and monotone.

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

1. (a)

pseudomonotone if for all $x,y\in C$

$〈Ay-Ay,x-y〉\ge 0\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}〈Ax,x-y〉\ge 0;$
2. (b)

monotone if

$〈Ax-Ay,x-y〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C;$
3. (c)

η-strongly monotone if there exists a constant $\eta >0$ such that

$〈Ax-Ay,x-y〉\ge \eta {\parallel x-y\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C;$
4. (d)

α-inverse-strongly monotone (α-ism) if there exists a constant $\alpha >0$ such that

$〈Ax-Ay,x-y〉\ge \alpha {\parallel Ax-Ay\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$

It is obvious that if A is α-inverse-strongly monotone, then A is monotone and $\frac{1}{\alpha }$-Lipschitz continuous.

Recall that a mapping $S:C\to C$ is said to be nonexpansive [1] if

$\parallel Sx-Sy\parallel \le \parallel x-y\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$

Denote by $Fix\left(S\right)$ the set of fixed points of S; that is, $Fix\left(S\right)=\left\{x\in C:Sx=x\right\}$. It can be easily seen that if $S:C\to C$ is nonexpansive, then $I-S$ is monotone. It is also easy to see that a projection ${P}_{C}$ is 1-ism. Inverse strongly monotone (also referred to as co-coercive) operators have been applied widely in solving practical problems in various fields.

We need some facts and tools which are listed as lemmas below.

Lemma 2.1 Let X be a real inner product space. Then the following inequality holds:

${\parallel x+y\parallel }^{2}\le {\parallel x\parallel }^{2}+2〈y,x+y〉,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in X.$

Lemma 2.2 Let $\left\{{x}_{n}\right\}$ be a bounded sequence in a reflexive Banach space X. If ${\omega }_{w}\left(\left\{{x}_{n}\right\}\right)=\left\{x\right\}$, then ${x}_{n}⇀x$.

Lemma 2.3 Let $A:C\to H$ be a monotone mapping. In the context of the variational inequality problem, the characterization of the projection (see Proposition  2.1(i)) implies

$u\in VI\left(C,A\right)\phantom{\rule{1em}{0ex}}⇔\phantom{\rule{1em}{0ex}}u={P}_{C}\left(u-\lambda Au\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }\lambda >0.$

Lemma 2.4 Let H be a real Hilbert space. Then the following hold:

1. (a)

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

2. (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 $\lambda ,\mu \in \left[0,1\right]$ with $\lambda +\mu =1$;

3. (c)

If $\left\{{x}_{n}\right\}$ is a sequence in H such that ${x}_{n}⇀x$, it follows that

$\underset{n\to \mathrm{\infty }}{lim sup}{\parallel {x}_{n}-y\parallel }^{2}=\underset{n\to \mathrm{\infty }}{lim sup}{\parallel {x}_{n}-x\parallel }^{2}+{\parallel x-y\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in H.$

Lemma 2.5 ([[33], Lemma 2.5])

Let H be a real Hilbert space. Given a nonempty closed convex subset of H and points $x,y,z\in H$ and given also a real number $a\in \mathbf{R}$, the set

$\left\{v\in C:{\parallel y-v\parallel }^{2}\le {\parallel x-v\parallel }^{2}+〈z,v〉+a\right\}$

is convex (and closed).

Let C be a nonempty closed convex subset of a real Hilbert space H. Let ${\left\{{S}_{i}\right\}}_{i=1}^{\mathrm{\infty }}$ be an infinite family of nonexpansive mappings of C into itself and let ${\left\{{\xi }_{i}\right\}}_{i=1}^{\mathrm{\infty }}$ be a sequence in $\left[0,1\right]$. For any $n\ge 1$, define a mapping ${W}_{n}$ of C into itself as follows:

$\left\{\begin{array}{c}{U}_{n,n+1}=I,\hfill \\ {U}_{n,n}={\xi }_{n}{S}_{n}{U}_{n,n+1}+\left(1-{\xi }_{n}\right)I,\hfill \\ {U}_{n,n-1}={\xi }_{n-1}{S}_{n-1}{U}_{n,n}+\left(1-{\xi }_{n-1}\right)I,\hfill \\ \cdots ,\hfill \\ {U}_{n,k}={\xi }_{k}{S}_{k}{U}_{n,k+1}+\left(1-{\xi }_{k}\right)I,\hfill \\ {U}_{n,k-1}={\xi }_{k-1}{S}_{k-1}{U}_{n,k}+\left(1-x{i}_{k-1}\right)I,\hfill \\ \cdots ,\hfill \\ {U}_{n,2}={\xi }_{2}{S}_{2}{U}_{n,3}+\left(1-{\xi }_{2}\right)I,\hfill \\ {W}_{n}={U}_{n,1}={\xi }_{1}{S}_{1}{U}_{n,2}+\left(1-{\xi }_{1}\right)I.\hfill \end{array}$
(2.1)

Such ${W}_{n}$ is called a W-mapping generated by ${\left\{{S}_{i}\right\}}_{i=1}^{\mathrm{\infty }}$ and ${\left\{{\xi }_{i}\right\}}_{i=1}^{\mathrm{\infty }}$. We need the following lemmas for proving our main results.

Lemma 2.6 [34]

Let C be a nonempty, closed and convex subset of a real Hilbert space H. Let ${S}_{1},{S}_{2},\dots$ be nonexpansive mappings of C into itself such that ${\bigcap }_{n=1}^{\mathrm{\infty }}{S}_{n}$ is nonempty, and let ${\xi }_{1},{\xi }_{2},\dots$ be real numbers such that $0<{\xi }_{i}\le b<1$ for all $i\ge 1$. Then, for every $x\in C$ and $k\ge 1$, the limit ${lim}_{n\to \mathrm{\infty }}{U}_{n,k}x$ exists.

Using Lemma 2.6, one can define a mapping W of C into itself as follows:

$Wx=\underset{n\to \mathrm{\infty }}{lim}{W}_{n}x=\underset{n\to \mathrm{\infty }}{lim}{U}_{n,1}x,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in C.$

Lemma 2.7 ([34])

Let C be a nonempty, closed and convex subset of a real Hilbert space H. Let ${S}_{1},{S}_{2},\dots$ be nonexpansive mappings of C into itself such that ${\bigcap }_{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)$ is nonempty, and let ${\xi }_{1},{\xi }_{2},\dots$ be real numbers such that $0<{\xi }_{i}\le b<1$ for all $i\ge 1$. Then

$Fix\left(W\right)=\bigcap _{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right).$

Lemma 2.8 ([35])

If $\left\{{x}_{n}\right\}$ is a bounded sequence in C, then

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

Lemma 2.9 ([[36], Demiclosedness principle])

Let C be a nonempty, closed and convex subset of a real Hilbert space H. Let $S:C\to C$ be a nonexpansive mapping such that $Fix\left(S\right)\ne \mathrm{\varnothing }$. Then S is demiclosed on C, i.e., if ${y}_{n}⇀z\in C$ and ${y}_{n}-S{y}_{n}\to y$, then $\left(I-S\right)z=y$.

To prove a weak convergence theorem by the multi-step implicit iterative method with regularization for MP (1.1) and infinitely many nonexpansive mappings ${\left\{{S}_{n}\right\}}_{n=1}^{\mathrm{\infty }}$, we need the following lemma due to Osilike et al. [37].

Lemma 2.10 ([[37], p.80])

Let ${\left\{{a}_{n}\right\}}_{n=1}^{\mathrm{\infty }}$, ${\left\{{b}_{n}\right\}}_{n=1}^{\mathrm{\infty }}$ and ${\left\{{\delta }_{n}\right\}}_{n=1}^{\mathrm{\infty }}$ be sequences of nonnegative real numbers satisfying the inequality

${a}_{n+1}\le \left(1+{\delta }_{n}\right){a}_{n}+{b}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1.$

If ${\sum }_{n=1}^{\mathrm{\infty }}{\delta }_{n}<\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}{b}_{n}<\mathrm{\infty }$, then ${lim}_{n\to \mathrm{\infty }}{a}_{n}$ exists. If, in addition, ${\left\{{a}_{n}\right\}}_{n=1}^{\mathrm{\infty }}$ has a subsequence which converges to zero, then ${lim}_{n\to \mathrm{\infty }}{a}_{n}=0$.

Corollary 2.1 ([[38], p.303])

Let ${\left\{{a}_{n}\right\}}_{n=0}^{\mathrm{\infty }}$ and ${\left\{{b}_{n}\right\}}_{n=0}^{\mathrm{\infty }}$ be two sequences of nonnegative real numbers satisfying the inequality

${a}_{n+1}\le {a}_{n}+{b}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 0.$

If ${\sum }_{n=0}^{\mathrm{\infty }}{b}_{n}$ converges, then ${lim}_{n\to \mathrm{\infty }}{a}_{n}$ exists.

Lemma 2.11 ([36])

Every Hilbert space H has the Kadec-Klee property; that is, given a sequence $\left\{{x}_{n}\right\}\subset H$ and a point $x\in H$, we have

$\begin{array}{r}\parallel {x}_{n}\parallel \to \parallel x\parallel \\ {x}_{n}⇀x\end{array}\right\}\phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}{x}_{n}\to x.$

It is well known that every Hilbert space H satisfies Opial’s condition [39], i.e., for any sequence $\left\{{x}_{n}\right\}$ with ${x}_{n}⇀x$, the inequality

$\underset{n\to \mathrm{\infty }}{lim inf}\parallel {x}_{n}-x\parallel <\underset{n\to \mathrm{\infty }}{lim inf}\parallel {x}_{n}-y\parallel$

holds for every $y\in H$ with $y\ne x$.

A set-valued mapping $T:H\to {2}^{H}$ is called monotone if for all $x,y\in H$, $f\in Tx$ and $g\in Ty$ imply $〈x-y,f-g〉\ge 0$. A monotone mapping $T:H\to {2}^{H}$ is maximal if its graph $G\left(T\right)$ is not properly contained in the graph of any other monotone mapping. It is known that a monotone mapping T is maximal if and only if for $\left(x,f\right)\in H×H$, $〈x-y,f-g〉\ge 0$ for all $\left(y,g\right)\in G\left(T\right)$ implies $f\in Tx$. Let $A:C\to H$ be a monotone, L-Lipschitz continuous mapping and let ${N}_{C}v$ be the normal cone to C at $v\in C$, i.e., ${N}_{C}v=\left\{w\in H:〈v-u,w〉\ge 0,\mathrm{\forall }u\in C\right\}$. Define

It is known that in this case T is maximal monotone, and $0\in Tv$ if and only if $v\in \Omega$; see [10].

## 3 Weak convergence theorem

In this section, we derive weak convergence criteria for a multi-step implicit iterative method with regularization for finding a common solution of the common fixed point problem of infinitely many nonexpansive mappings ${\left\{{S}_{n}\right\}}_{n=1}^{\mathrm{\infty }}$ and MP (1.1) for a convex functional $f:C\to \mathbf{R}$ with an L-Lipschitz continuous gradient f. This implicit iterative method with regularization is based on the extragradient method, approximate proximal method and gradient projection algorithm (GPA) with regularization.

Theorem 3.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let ${W}_{n}$ be a W-mapping defined by (2.1), let $\mathrm{\nabla }f:C\to H$ be an L-Lipschitz continuous mapping with $L>0$, and let ${\left\{{S}_{n}\right\}}_{n=1}^{\mathrm{\infty }}$ be an infinite family of nonexpansive mappings of C into itself such that ${\bigcap }_{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma$ is nonempty and bounded. Let $\left\{{x}_{n}\right\}$, $\left\{{y}_{n}\right\}$ and $\left\{{z}_{n}\right\}$ be the sequences generated by

$\left\{\begin{array}{c}{x}_{1}=x\in C\phantom{\rule{1em}{0ex}}\mathit{\text{chosen arbitrarily}},\hfill \\ {y}_{n}={P}_{C}\left({x}_{n}-{\lambda }_{n}{\mu }_{n}\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({x}_{n}\right)-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)\right),\hfill \\ {t}_{n}={P}_{C}\left({x}_{n}-{\lambda }_{n}\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({t}_{n}\right)\right),\hfill \\ {z}_{n}={\gamma }_{n}{x}_{n}+\left(1-{\gamma }_{n}\right){W}_{n}{t}_{n},\hfill \\ {C}_{n}=\left\{z\in C:{\parallel {z}_{n}-z\parallel }^{2}\le {\parallel {x}_{n}-z\parallel }^{2}+{\theta }_{n}\right\},\hfill \\ {x}_{n+1}={P}_{{C}_{n}}\left(\left(1-{\beta }_{n}\right){x}_{n}+{e}_{n}-{\sigma }_{n}\mathrm{\nabla }f\left({x}_{n+1}\right)\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,\hfill \end{array}$
(3.1)

where $\left\{{e}_{n}\right\}\subset H$ an error sequence with ${\sum }_{n=1}^{\mathrm{\infty }}\parallel {e}_{n}\parallel <\mathrm{\infty }$, and ${\theta }_{n}={\alpha }_{n}\frac{18}{{\mu }_{n}^{4}}{\Delta }_{n}$ with

${\Delta }_{n}=sup\left\{{\parallel {x}_{n}-z\parallel }^{2}+\left(1+\frac{1}{18{L}^{2}}\right){\parallel z\parallel }^{2}:z\in \bigcap _{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma \right\}<\mathrm{\infty }.$

Assume the following conditions hold:

1. (i)

$\left\{{\alpha }_{n}\right\}\subset \left(0,\mathrm{\infty }\right)$ and ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$;

2. (ii)

$\left\{{\gamma }_{n}\right\}\subset \left[0,c\right]$ for some $c\in \left[0,1\right)$;

3. (iii)

$\left\{{\mu }_{n}\right\}\subset \left(0,1\right]$ and ${lim}_{n\to \mathrm{\infty }}{\mu }_{n}=1$;

4. (iv)

${\lambda }_{n}\left({\alpha }_{n}+L\right)<1$, $\mathrm{\forall }n\ge 1$ and $\left\{{\lambda }_{n}\right\}\subset \left[a,b\right]$ for some $a,b\in \left(0,1/L\right)$;

5. (v)

$\left\{{\sigma }_{n}\right\}\subset \left(0,1/L\right)$ and $\left\{{\beta }_{n}\right\}\subset \left[0,1\right]$ satisfy ${\sum }_{n=1}^{\mathrm{\infty }}{\beta }_{n}<\mathrm{\infty }$.

Then the sequences $\left\{{x}_{n}\right\}$, $\left\{{y}_{n}\right\}$ and $\left\{{z}_{n}\right\}$ generated by (3.1) converge weakly to some $u\in {\bigcap }_{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma$.

Remark 3.1 In the proof of Theorem 3.1 below, we show that every ${C}_{n}$ is closed and convex and that ${\bigcap }_{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma \subset {C}_{n}$, $\mathrm{\forall }n\ge 1$.

Now, we observe that for all $x,y\in C$ and all $n\ge 1$,

$\begin{array}{r}\parallel {P}_{C}\left({x}_{n}-{\lambda }_{n}{\mu }_{n}\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({x}_{n}\right)-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left(x\right)\right)\\ \phantom{\rule{2em}{0ex}}-{P}_{C}\left({x}_{n}-{\lambda }_{n}{\mu }_{n}\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({x}_{n}\right)-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left(y\right)\right)\parallel \\ \phantom{\rule{1em}{0ex}}\le \parallel \left({x}_{n}-{\lambda }_{n}{\mu }_{n}\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({x}_{n}\right)-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left(x\right)\right)\\ \phantom{\rule{2em}{0ex}}-\left({x}_{n}-{\lambda }_{n}{\mu }_{n}\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({x}_{n}\right)-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left(y\right)\right)\parallel \\ \phantom{\rule{1em}{0ex}}={\lambda }_{n}\left(1-{\mu }_{n}\right)\parallel \mathrm{\nabla }{f}_{{\alpha }_{n}}\left(x\right)-\mathrm{\nabla }{f}_{{\alpha }_{n}}\left(y\right)\parallel \\ \phantom{\rule{1em}{0ex}}\le {\lambda }_{n}\left({\alpha }_{n}+L\right)\parallel x-y\parallel .\end{array}$

Hence, by the Banach contraction principle, we know that for each $n\ge 1$ there exists a unique ${y}_{n}\in C$ such that

${y}_{n}={P}_{C}\left({x}_{n}-{\lambda }_{n}{\mu }_{n}\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({x}_{n}\right)-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)\right).$
(3.2)

Also, observe that for all $x,y\in C$ and all $n\ge 1$,

$\begin{array}{r}\parallel {P}_{C}\left({x}_{n}-{\lambda }_{n}\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left(x\right)\right)\\ \phantom{\rule{2em}{0ex}}-{P}_{C}\left({x}_{n}-{\lambda }_{n}\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left(y\right)\right)\parallel \\ \phantom{\rule{1em}{0ex}}\le \parallel \left({x}_{n}-{\lambda }_{n}\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left(x\right)\right)\\ \phantom{\rule{2em}{0ex}}-\left({x}_{n}-{\lambda }_{n}\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left(y\right)\right)\parallel \\ \phantom{\rule{1em}{0ex}}={\lambda }_{n}\left(1-{\mu }_{n}\right)\parallel \mathrm{\nabla }{f}_{{\alpha }_{n}}\left(x\right)-\mathrm{\nabla }{f}_{{\alpha }_{n}}\left(y\right)\parallel \\ \phantom{\rule{1em}{0ex}}\le {\lambda }_{n}\left({\alpha }_{n}+L\right)\parallel x-y\parallel .\end{array}$

So, by the Banach contraction principle, we know that for each $n\ge 1$ there exists a unique ${z}_{n}\in C$ such that

${t}_{n}={P}_{C}\left({x}_{n}-{\lambda }_{n}\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({t}_{n}\right)\right).$
(3.3)

In addition, observe that for all $x,y\in C$ and all $n\ge 1$,

$\begin{array}{r}\parallel {P}_{{C}_{n}}\left(\left(1-{\beta }_{n}\right){x}_{n}+{e}_{n}-{\sigma }_{n}\mathrm{\nabla }f\left(x\right)\right)-{P}_{{C}_{n}}\left(\left(1-{\beta }_{n}\right){x}_{n}+{e}_{n}-{\sigma }_{n}\mathrm{\nabla }f\left(y\right)\right)\parallel \\ \phantom{\rule{1em}{0ex}}\le \parallel \left(\left(1-{\beta }_{n}\right){x}_{n}+{e}_{n}-{\sigma }_{n}\mathrm{\nabla }f\left(x\right)\right)-\left(\left(1-{\beta }_{n}\right){x}_{n}+{e}_{n}-{\sigma }_{n}\mathrm{\nabla }f\left(y\right)\right)\parallel \\ \phantom{\rule{1em}{0ex}}={\sigma }_{n}\parallel \mathrm{\nabla }f\left(x\right)-\mathrm{\nabla }f\left(y\right)\parallel \\ \phantom{\rule{1em}{0ex}}\le {\sigma }_{n}L\parallel x-y\parallel .\end{array}$

Thus, by the Banach contraction principle, we know that for each $n\ge 1$ there exists a unique ${x}_{n+1}\in {C}_{n}$ such that

${x}_{n+1}={P}_{{C}_{n}}\left(\left(1-{\beta }_{n}\right){x}_{n}+{e}_{n}-{\sigma }_{n}\mathrm{\nabla }f\left({x}_{n+1}\right)\right).$
(3.4)

Therefore, the sequences $\left\{{x}_{n}\right\}$, $\left\{{y}_{n}\right\}$ and $\left\{{z}_{n}\right\}$ generated by (3.1) are well defined.

Next, we divide our detailed proof into several propositions. For this purpose, in the sequel, we assume that all our assumptions are satisfied.

Proposition 3.1 ${\bigcap }_{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma \subset {C}_{n}$, $\mathrm{\forall }n\ge 1$.

Proof First we note that every set ${C}_{n}$ is closed and convex. As a matter of fact, since the defining inequality in ${C}_{n}$ is equivalent to the inequality

$〈2\left({x}_{n}-{z}_{n}\right),z〉\le {\parallel {x}_{n}\parallel }^{2}-{\parallel {z}_{n}\parallel }^{2}+{\theta }_{n},$

by Lemma 2.5 we also have that ${C}_{n}$ is convex and closed for every $n=1,2,\dots$ . Also, note that the L-Lipschitz continuity of the gradient f implies that f is $1/L$-ism [31], that is,

$〈\mathrm{\nabla }f\left(x\right)-\mathrm{\nabla }f\left(y\right),x-y〉\ge \frac{1}{L}{\parallel \mathrm{\nabla }f\left(x\right)-\mathrm{\nabla }f\left(y\right)\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$

Observe that

$\begin{array}{r}\left(\alpha +L\right)〈\mathrm{\nabla }{f}_{\alpha }\left(x\right)-\mathrm{\nabla }{f}_{\alpha }\left(y\right),x-y〉\\ \phantom{\rule{1em}{0ex}}=\left(\alpha +L\right)\left[\alpha {\parallel x-y\parallel }^{2}+〈\mathrm{\nabla }f\left(x\right)-\mathrm{\nabla }f\left(y\right),x-y〉\right]\\ \phantom{\rule{1em}{0ex}}={\alpha }^{2}{\parallel x-y\parallel }^{2}+\alpha 〈\mathrm{\nabla }f\left(x\right)-\mathrm{\nabla }f\left(y\right),x-y〉+\alpha L{\parallel x-y\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}+L〈\mathrm{\nabla }f\left(x\right)-\mathrm{\nabla }f\left(y\right),x-y〉\\ \phantom{\rule{1em}{0ex}}\ge {\alpha }^{2}{\parallel x-y\parallel }^{2}+2\alpha 〈\mathrm{\nabla }f\left(x\right)-\mathrm{\nabla }f\left(y\right),x-y〉+{\parallel \mathrm{\nabla }f\left(x\right)-\mathrm{\nabla }f\left(y\right)\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}={\parallel \alpha \left(x-y\right)+\mathrm{\nabla }f\left(x\right)-\mathrm{\nabla }f\left(y\right)\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}={\parallel \mathrm{\nabla }{f}_{\alpha }\left(x\right)-\mathrm{\nabla }{f}_{\alpha }\left(y\right)\parallel }^{2}.\end{array}$

Hence, it follows that $\mathrm{\nabla }{f}_{\alpha }=\alpha I+\mathrm{\nabla }f$ is $1/\left(\alpha +L\right)$-ism. Now, take $u\in {\bigcap }_{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma$ arbitrarily. Taking into account ${\lambda }_{n}\left({\alpha }_{n}+L\right)<1$, $\mathrm{\forall }n\ge 1$, we deduce that

$\begin{array}{r}{\parallel {x}_{n}-u-{\lambda }_{n}{\mu }_{n}\left(\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({x}_{n}\right)-\mathrm{\nabla }{f}_{{\alpha }_{n}}\left(u\right)\right)\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-u\parallel }^{2}+{\lambda }_{n}{\mu }_{n}\left({\lambda }_{n}{\mu }_{n}-\frac{2}{{\alpha }_{n}+L}\right){\parallel \mathrm{\nabla }{f}_{{\alpha }_{n}}\left({x}_{n}\right)-\mathrm{\nabla }{f}_{{\alpha }_{n}}\left(u\right)\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-u\parallel }^{2}+{\lambda }_{n}\left({\lambda }_{n}-\frac{2}{{\alpha }_{n}+L}\right){\parallel \mathrm{\nabla }{f}_{{\alpha }_{n}}\left({x}_{n}\right)-\mathrm{\nabla }{f}_{{\alpha }_{n}}\left(u\right)\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-u\parallel }^{2},\end{array}$

and

$\begin{array}{rcl}\parallel {y}_{n}-u\parallel & =& \parallel {P}_{C}\left({x}_{n}-{\lambda }_{n}{\mu }_{n}\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({x}_{n}\right)-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)\right)\\ -{P}_{C}\left(u-{\lambda }_{n}{\mu }_{n}\mathrm{\nabla }f\left(u\right)-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }f\left(u\right)\right)\parallel \\ \le & \parallel \left({x}_{n}-{\lambda }_{n}{\mu }_{n}\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({x}_{n}\right)-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)\right)\\ -\left(u-{\lambda }_{n}{\mu }_{n}\mathrm{\nabla }f\left(u\right)-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }f\left(u\right)\right)\parallel \\ \le & \parallel {x}_{n}-u-{\lambda }_{n}{\mu }_{n}\left(\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({x}_{n}\right)-\mathrm{\nabla }f\left(u\right)\right)\parallel +{\lambda }_{n}\left(1-{\mu }_{n}\right)\parallel \mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)-\mathrm{\nabla }f\left(u\right)\parallel \\ \le & \parallel {x}_{n}-u-{\lambda }_{n}{\mu }_{n}\left(\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({x}_{n}\right)-\mathrm{\nabla }{f}_{{\alpha }_{n}}\left(u\right)\right)\parallel +{\lambda }_{n}{\mu }_{n}\parallel \mathrm{\nabla }{f}_{{\alpha }_{n}}\left(u\right)-\mathrm{\nabla }f\left(u\right)\parallel \\ +{\lambda }_{n}\left(1-{\mu }_{n}\right)\left[\parallel \mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)-\mathrm{\nabla }{f}_{{\alpha }_{n}}\left(u\right)\parallel +\parallel \mathrm{\nabla }{f}_{{\alpha }_{n}}\left(u\right)-\mathrm{\nabla }f\left(u\right)\parallel \right]\\ \le & \parallel {x}_{n}-u\parallel +{\lambda }_{n}{\mu }_{n}{\alpha }_{n}\parallel u\parallel +{\lambda }_{n}\left(1-{\mu }_{n}\right)\left[\left({\alpha }_{n}+L\right)\parallel {y}_{n}-u\parallel +{\alpha }_{n}\parallel u\parallel \right]\\ =& \parallel {x}_{n}-u\parallel +{\lambda }_{n}\left(1-{\mu }_{n}\right)\left({\alpha }_{n}+L\right)\parallel {y}_{n}-u\parallel +{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \\ \le & \parallel {x}_{n}-u\parallel +\left(1-{\mu }_{n}\right)\parallel {y}_{n}-u\parallel +{\lambda }_{n}{\alpha }_{n}\parallel u\parallel ,\end{array}$

which implies that

$\parallel {y}_{n}-u\parallel \le \frac{1}{{\mu }_{n}}\left(\parallel {x}_{n}-u\parallel +{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \right).$
(3.5)

Meantime, we also have

$\begin{array}{r}\parallel {t}_{n}-u\parallel \\ \phantom{\rule{1em}{0ex}}=\parallel {P}_{C}\left({x}_{n}-{\lambda }_{n}\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({t}_{n}\right)\right)\\ \phantom{\rule{2em}{0ex}}-{P}_{C}\left(u-{\lambda }_{n}\mathrm{\nabla }f\left(u\right)-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }f\left(u\right)\right)\parallel \\ \phantom{\rule{1em}{0ex}}\le \parallel \left({x}_{n}-{\lambda }_{n}\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({t}_{n}\right)\right)-\left(u-{\lambda }_{n}\mathrm{\nabla }f\left(u\right)-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }f\left(u\right)\right)\parallel \\ \phantom{\rule{1em}{0ex}}\le \parallel {x}_{n}-u\parallel +{\lambda }_{n}\parallel \mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)-\mathrm{\nabla }f\left(u\right)\parallel +{\lambda }_{n}\left(1-{\mu }_{n}\right)\parallel \mathrm{\nabla }{f}_{{\alpha }_{n}}\left({t}_{n}\right)-\mathrm{\nabla }f\left(u\right)\parallel \\ \phantom{\rule{1em}{0ex}}\le \parallel {x}_{n}-u\parallel +{\lambda }_{n}\left[\parallel \mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)-\mathrm{\nabla }{f}_{{\alpha }_{n}}\left(u\right)\parallel +\parallel \mathrm{\nabla }{f}_{{\alpha }_{n}}\left(u\right)-\mathrm{\nabla }f\left(u\right)\parallel \right]\\ \phantom{\rule{2em}{0ex}}+{\lambda }_{n}\left(1-{\mu }_{n}\right)\left[\parallel \mathrm{\nabla }{f}_{{\alpha }_{n}}\left({t}_{n}\right)-\mathrm{\nabla }{f}_{{\alpha }_{n}}\left(u\right)\parallel +\parallel \mathrm{\nabla }{f}_{{\alpha }_{n}}\left(u\right)-\mathrm{\nabla }f\left(u\right)\parallel \right]\\ \phantom{\rule{1em}{0ex}}\le \parallel {x}_{n}-u\parallel +{\lambda }_{n}\left[\left({\alpha }_{n}+L\right)\parallel {y}_{n}-u\parallel +{\alpha }_{n}\parallel u\parallel \right]+{\lambda }_{n}\left(1-{\mu }_{n}\right)\left[\left({\alpha }_{n}+L\right)\parallel {t}_{n}-u\parallel +{\alpha }_{n}\parallel u\parallel \right]\\ \phantom{\rule{1em}{0ex}}\le \parallel {x}_{n}-u\parallel +\parallel {y}_{n}-u\parallel +{\lambda }_{n}{\alpha }_{n}\parallel u\parallel +\left(1-{\mu }_{n}\right)\parallel {t}_{n}-u\parallel +{\lambda }_{n}\left(1-{\mu }_{n}\right){\alpha }_{n}\parallel u\parallel \\ \phantom{\rule{1em}{0ex}}\le \parallel {x}_{n}-u\parallel +\parallel {y}_{n}-u\parallel +\left(2-{\mu }_{n}\right){\lambda }_{n}{\alpha }_{n}\parallel u\parallel +\left(1-{\mu }_{n}\right)\parallel {t}_{n}-u\parallel ,\end{array}$

which hence implies that

$\begin{array}{rl}\parallel {t}_{n}-u\parallel & \le \frac{1}{{\mu }_{n}}\left[\parallel {x}_{n}-u\parallel +\parallel {y}_{n}-u\parallel +\left(2-{\mu }_{n}\right){\lambda }_{n}{\alpha }_{n}\parallel u\parallel \right]\\ \le \frac{1}{{\mu }_{n}}\left\{\parallel {x}_{n}-u\parallel +\frac{1}{{\mu }_{n}}\left(\parallel {x}_{n}-u\parallel +{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \right)+\left(2-{\mu }_{n}\right){\lambda }_{n}{\alpha }_{n}\parallel u\parallel \right\}\\ =\left(\frac{1}{{\mu }_{n}}+\frac{1}{{\mu }_{n}^{2}}\right)\parallel {x}_{n}-u\parallel +\left(\frac{1}{{\mu }_{n}^{2}}+\frac{2-{\mu }_{n}}{{\mu }_{n}}\right){\lambda }_{n}{\alpha }_{n}\parallel u\parallel \\ \le \frac{1+2{\mu }_{n}-{\mu }_{n}^{2}}{{\mu }_{n}^{2}}\left(\parallel {x}_{n}-u\parallel +{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \right)\\ \le \frac{1+2{\mu }_{n}}{{\mu }_{n}^{2}}\left(\parallel {x}_{n}-u\parallel +{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \right).\end{array}$
(3.6)

Thus, from (3.5) and (3.6) it follows that

$\begin{array}{r}\parallel {y}_{n}-u\parallel +\left(1-{\mu }_{n}\right)\parallel {t}_{n}-u\parallel \\ \phantom{\rule{1em}{0ex}}\le \frac{1}{{\mu }_{n}}\left(\parallel {x}_{n}-u\parallel +{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \right)+\left(1-{\mu }_{n}\right)\frac{1+2{\mu }_{n}}{{\mu }_{n}^{2}}\left(\parallel {x}_{n}-u\parallel +{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \right)\\ \phantom{\rule{1em}{0ex}}=\frac{1+2{\mu }_{n}\left(1-{\mu }_{n}\right)}{{\mu }_{n}^{2}}\left(\parallel {x}_{n}-u\parallel +{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \right)\\ \phantom{\rule{1em}{0ex}}\le \frac{3}{{\mu }_{n}^{2}}\left(\parallel {x}_{n}-u\parallel +{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \right),\end{array}$

which together with ${\lambda }_{n}\left({\alpha }_{n}+L\right)<1$ implies that

$\begin{array}{rl}{\left[\parallel {y}_{n}-u\parallel +\left(1-{\mu }_{n}\right)\parallel {t}_{n}-u\parallel \right]}^{2}& \le \frac{9}{{\mu }_{n}^{4}}{\left(\parallel {x}_{n}-u\parallel +{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \right)}^{2}\\ \le \frac{9}{{\mu }_{n}^{4}}\left(2{\parallel {x}_{n}-u\parallel }^{2}+2{\lambda }_{n}^{2}{\alpha }_{n}^{2}{\parallel u\parallel }^{2}\right)\\ \le \frac{9}{{\mu }_{n}^{4}}\left(2{\parallel {x}_{n}-u\parallel }^{2}+2{\parallel u\parallel }^{2}\right)\\ =\frac{18}{{\mu }_{n}^{4}}{\parallel {x}_{n}-u\parallel }^{2}+\frac{18}{{\mu }_{n}^{4}}{\parallel u\parallel }^{2}.\end{array}$
(3.7)

Furthermore, from Proposition 2.1(ii), the monotonicity of f, and $u\in \Gamma$, we have

$\begin{array}{rcl}{\parallel {t}_{n}-u\parallel }^{2}& \le & {\parallel \left({x}_{n}-{\lambda }_{n}\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({t}_{n}\right)\right)-u\parallel }^{2}\\ -{\parallel \left({x}_{n}-{\lambda }_{n}\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({t}_{n}\right)\right)-{t}_{n}\parallel }^{2}\\ =& {\parallel {x}_{n}-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({t}_{n}\right)-u\parallel }^{2}\\ -{\parallel {x}_{n}-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({t}_{n}\right)-{t}_{n}\parallel }^{2}+2{\lambda }_{n}〈\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right),u-{t}_{n}〉\\ =& {\parallel {x}_{n}-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({t}_{n}\right)-u\parallel }^{2}-{\parallel {x}_{n}-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({t}_{n}\right)-{t}_{n}\parallel }^{2}\\ +2{\lambda }_{n}\left(〈\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right),u-{y}_{n}〉+〈\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right),{y}_{n}-{t}_{n}〉\right)\\ =& {\parallel {x}_{n}-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({t}_{n}\right)-u\parallel }^{2}-{\parallel {x}_{n}-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({t}_{n}\right)-{t}_{n}\parallel }^{2}\\ +2{\lambda }_{n}\left(〈\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)-\mathrm{\nabla }{f}_{{\alpha }_{n}}\left(u\right),u-{y}_{n}〉+〈\mathrm{\nabla }{f}_{{\alpha }_{n}}\left(u\right),u-{y}_{n}〉\\ +〈\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right),{y}_{n}-{t}_{n}〉\right)\\ \le & {\parallel {x}_{n}-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({t}_{n}\right)-u\parallel }^{2}-{\parallel {x}_{n}-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({t}_{n}\right)-{t}_{n}\parallel }^{2}\\ +2{\lambda }_{n}\left({\alpha }_{n}〈u,u-{y}_{n}〉+〈\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right),{y}_{n}-{t}_{n}〉\right)\\ =& {\parallel {x}_{n}-u\parallel }^{2}-{\parallel {x}_{n}-{t}_{n}\parallel }^{2}-2{\lambda }_{n}\left(1-{\mu }_{n}\right)〈\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({t}_{n}\right),{t}_{n}-u〉\\ +2{\lambda }_{n}\left({\alpha }_{n}〈u,u-{y}_{n}〉+〈\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right),{y}_{n}-{t}_{n}〉\right)\\ =& {\parallel {x}_{n}-u\parallel }^{2}-{\parallel {x}_{n}-{y}_{n}\parallel }^{2}-2〈{x}_{n}-{y}_{n},{y}_{n}-{t}_{n}〉-{\parallel {y}_{n}-{t}_{n}\parallel }^{2}\\ +2{\lambda }_{n}\left({\alpha }_{n}〈u,u-{y}_{n}〉+〈\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right),{y}_{n}-{t}_{n}〉\right)\\ -2{\lambda }_{n}\left(1-{\mu }_{n}\right)\left(〈\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({t}_{n}\right)-\mathrm{\nabla }{f}_{{\alpha }_{n}}\left(u\right),{t}_{n}-u〉+〈\mathrm{\nabla }{f}_{{\alpha }_{n}}\left(u\right),{t}_{n}-u〉\right)\\ \le & {\parallel {x}_{n}-u\parallel }^{2}-{\parallel {x}_{n}-{y}_{n}\parallel }^{2}-{\parallel {y}_{n}-{t}_{n}\parallel }^{2}+2〈{x}_{n}-{\lambda }_{n}\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)-{y}_{n},{t}_{n}-{y}_{n}〉\\ +2{\lambda }_{n}{\alpha }_{n}\left(〈u,u-{y}_{n}〉+\left(1-{\mu }_{n}\right)〈u,u-{t}_{n}〉\right)\\ \le & {\parallel {x}_{n}-u\parallel }^{2}-{\parallel {x}_{n}-{y}_{n}\parallel }^{2}-{\parallel {y}_{n}-{t}_{n}\parallel }^{2}+2〈{x}_{n}-{\lambda }_{n}\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)-{y}_{n},{t}_{n}-{y}_{n}〉\\ +2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\left(1-{\mu }_{n}\right)\parallel {t}_{n}-u\parallel \right].\end{array}$

Since ${y}_{n}={P}_{C}\left({x}_{n}-{\lambda }_{n}{\mu }_{n}\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({x}_{n}\right)-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)\right)$ and $\mathrm{\nabla }{f}_{{\alpha }_{n}}$ $\left({\alpha }_{n}+L\right)$-Lipschitz continuous, by Proposition 2.1(i) we have

$\begin{array}{rl}〈{x}_{n}-{\lambda }_{n}\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)-{y}_{n},{t}_{n}-{y}_{n}〉=& 〈{x}_{n}-{\lambda }_{n}{\mu }_{n}\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({x}_{n}\right)-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)-{y}_{n},{t}_{n}-{y}_{n}〉\\ +{\lambda }_{n}{\mu }_{n}〈\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({x}_{n}\right)-\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right),{t}_{n}-{y}_{n}〉\\ \le & {\lambda }_{n}{\mu }_{n}〈\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({x}_{n}\right)-\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right),{t}_{n}-{y}_{n}〉\\ \le & {\lambda }_{n}{\mu }_{n}\parallel \mathrm{\nabla }{f}_{{\alpha }_{n}}\left({x}_{n}\right)-\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)\parallel \parallel {t}_{n}-{y}_{n}\parallel \\ \le & {\lambda }_{n}\left({\alpha }_{n}+L\right)\parallel {x}_{n}-{y}_{n}\parallel \parallel {t}_{n}-{y}_{n}\parallel .\end{array}$

So, we have

$\begin{array}{r}{\parallel {t}_{n}-u\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-u\parallel }^{2}-{\parallel {x}_{n}-{y}_{n}\parallel }^{2}-{\parallel {y}_{n}-{t}_{n}\parallel }^{2}+2{\lambda }_{n}\left({\alpha }_{n}+L\right)\parallel {x}_{n}-{y}_{n}\parallel \parallel {t}_{n}-{y}_{n}\parallel \\ \phantom{\rule{2em}{0ex}}+2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\left(1-{\mu }_{n}\right)\parallel {t}_{n}-u\parallel \right]\\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-u\parallel }^{2}-{\parallel {x}_{n}-{y}_{n}\parallel }^{2}-{\parallel {y}_{n}-{t}_{n}\parallel }^{2}+{\lambda }_{n}^{2}{\left({\alpha }_{n}+L\right)}^{2}{\parallel {x}_{n}-{y}_{n}\parallel }^{2}+{\parallel {t}_{n}-{y}_{n}\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}+2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\left(1-{\mu }_{n}\right)\parallel {t}_{n}-u\parallel \right]\\ \phantom{\rule{1em}{0ex}}={\parallel {x}_{n}-u\parallel }^{2}+\left({\lambda }_{n}^{2}{\left({\alpha }_{n}+L\right)}^{2}-1\right){\parallel {x}_{n}-{y}_{n}\parallel }^{2}+2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\left(1-{\mu }_{n}\right)\parallel {t}_{n}-u\parallel \right]\\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-u\parallel }^{2}+2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\left(1-{\mu }_{n}\right)\parallel {t}_{n}-u\parallel \right].\end{array}$
(3.8)

Therefore, from (3.7) and (3.8), together with ${z}_{n}={\gamma }_{n}{x}_{n}+\left(1-{\gamma }_{n}\right){W}_{n}{t}_{n}$ and $u={W}_{n}u$, by Lemma 2.4(b) we have

$\begin{array}{rcl}{\parallel {z}_{n}-u\parallel }^{2}& =& {\parallel {\gamma }_{n}\left({x}_{n}-u\right)+\left(1-{\gamma }_{n}\right)\left({W}_{n}{t}_{n}-u\right)\parallel }^{2}\\ \le & {\gamma }_{n}{\parallel {x}_{n}-u\parallel }^{2}+\left(1-{\gamma }_{n}\right){\parallel {W}_{n}{t}_{n}-u\parallel }^{2}\\ \le & {\gamma }_{n}{\parallel {x}_{n}-u\parallel }^{2}+\left(1-{\gamma }_{n}\right){\parallel {t}_{n}-u\parallel }^{2}\\ \le & {\gamma }_{n}{\parallel {x}_{n}-u\parallel }^{2}+\left(1-{\gamma }_{n}\right)\left\{{\parallel {x}_{n}-u\parallel }^{2}+\left({\lambda }_{n}^{2}{\left({\alpha }_{n}+L\right)}^{2}-1\right){\parallel {x}_{n}-{y}_{n}\parallel }^{2}\\ +2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\left(1-{\mu }_{n}\right)\parallel {t}_{n}-u\parallel \right]\right\}\\ \le & {\parallel {x}_{n}-u\parallel }^{2}+\left(1-{\gamma }_{n}\right)\left({\lambda }_{n}^{2}{\left({\alpha }_{n}+L\right)}^{2}-1\right){\parallel {x}_{n}-{y}_{n}\parallel }^{2}\\ +2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\left(1-{\mu }_{n}\right)\parallel {t}_{n}-u\parallel \right]\\ \le & {\parallel {x}_{n}-u\parallel }^{2}+2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\left(1-{\mu }_{n}\right)\parallel {t}_{n}-u\parallel \right]\\ \le & {\parallel {x}_{n}-u\parallel }^{2}+{\alpha }_{n}\left[{\lambda }_{n}^{2}{\parallel u\parallel }^{2}+{\left(\parallel {y}_{n}-u\parallel +\left(1-{\mu }_{n}\right)\parallel {t}_{n}-u\parallel \right)}^{2}\right]\\ \le & {\parallel {x}_{n}-u\parallel }^{2}+{\alpha }_{n}\left[\frac{1}{{L}^{2}}{\parallel u\parallel }^{2}+\frac{18}{{\mu }_{n}^{4}}{\parallel {x}_{n}-u\parallel }^{2}+\frac{18}{{\mu }_{n}^{4}}{\parallel u\parallel }^{2}\right]\\ =& {\parallel {x}_{n}-u\parallel }^{2}+{\alpha }_{n}\frac{18}{{\mu }_{n}^{4}}\left[{\parallel {x}_{n}-u\parallel }^{2}+\left(1+\frac{{\mu }_{n}^{4}}{18{L}^{2}}\right){\parallel u\parallel }^{2}\right]\\ \le & {\parallel {x}_{n}-u\parallel }^{2}+{\alpha }_{n}\frac{18}{{\mu }_{n}^{4}}\left[{\parallel {x}_{n}-u\parallel }^{2}+\left(1+\frac{1}{18{L}^{2}}\right){\parallel u\parallel }^{2}\right]\\ \le & {\parallel {x}_{n}-u\parallel }^{2}+{\alpha }_{n}\frac{18}{{\mu }_{n}^{4}}{\Delta }_{n}\\ =& {\parallel {x}_{n}-u\parallel }^{2}+{\theta }_{n},\end{array}$
(3.9)

which implies that $u\in {C}_{n}$. Therefore,

$\bigcap _{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma \subset {C}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,$

and this completes the proof. □

Proposition 3.2 The sequences $\left\{{x}_{n}\right\}$, $\left\{{y}_{n}\right\}$, $\left\{{t}_{n}\right\}$ and $\left\{{z}_{n}\right\}$ are all bounded.

Proof Since $u\in \Gamma$ and ${x}_{n}\in C$ for all $n\ge 1$, from the monotonicity of f, we have

$〈\mathrm{\nabla }f\left(u\right),{x}_{n}-u〉\ge 0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}〈\mathrm{\nabla }f\left({x}_{n}\right)-\mathrm{\nabla }f\left(u\right),{x}_{n}-u〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,$

which hence implies that

$〈\mathrm{\nabla }f\left({x}_{n}\right),{x}_{n}-u〉\ge 〈\mathrm{\nabla }f\left(u\right),{x}_{n}-u〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1.$
(3.10)

Note that ${x}_{n+1}={P}_{{C}_{n}}\left(\left(1-{\beta }_{n}\right){x}_{n}+{e}_{n}-{\sigma }_{n}\mathrm{\nabla }f\left({x}_{n+1}\right)\right)$ is equivalent to the inequality

$〈\left(1-{\beta }_{n}\right){x}_{n}-{x}_{n+1}+{e}_{n}-{\sigma }_{n}\mathrm{\nabla }f\left({x}_{n+1}\right),{x}_{n+1}-x〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in {C}_{n}.$

Taking $x=u$ in the last inequality, we deduce

$〈\left(1-{\beta }_{n}\right){x}_{n}-{x}_{n+1}+{e}_{n}-{\sigma }_{n}\mathrm{\nabla }f\left({x}_{n+1}\right),{x}_{n+1}-u〉\ge 0,$

which implies that

$〈\left(1-{\beta }_{n}\right){x}_{n}-{x}_{n+1}+{e}_{n},{x}_{n+1}-u〉\ge {\sigma }_{n}〈\mathrm{\nabla }f\left({x}_{n+1}\right),{x}_{n+1}-u〉.$
(3.11)

From (3.10) and (3.11) we get

$〈\left(1-{\beta }_{n}\right){x}_{n}-{x}_{n+1}+{e}_{n},{x}_{n+1}-u〉\ge 0,$

which can be rewritten as

$〈\left(1-{\beta }_{n}\right)\left({x}_{n}-u\right)-\left({x}_{n+1}-u\right)-{\beta }_{n}u+{e}_{n},{x}_{n+1}-u〉\ge 0.$
(3.12)

It follows that

$\begin{array}{rl}{\parallel {x}_{n+1}-u\parallel }^{2}& \le 〈\left(1-{\beta }_{n}\right)\left({x}_{n}-u\right)-{\beta }_{n}u+{e}_{n},{x}_{n+1}-u〉\\ \le \left(1-{\beta }_{n}\right)\parallel {x}_{n}-u\parallel \parallel {x}_{n+1}-u\parallel +{\beta }_{n}\parallel u\parallel \parallel {x}_{n+1}-u\parallel +\parallel {e}_{n}\parallel \parallel {x}_{n+1}-u\parallel .\end{array}$

Hence,

$\begin{array}{rl}\parallel {x}_{n+1}-u\parallel & \le \left(1-{\beta }_{n}\right)\parallel {x}_{n}-u\parallel +{\beta }_{n}\parallel u\parallel +\parallel {e}_{n}\parallel \\ \le max\left\{\parallel {x}_{n}-u\parallel ,\parallel u\parallel \right\}+\parallel {e}_{n}\parallel .\end{array}$
(3.13)

By induction, we can obtain

$\parallel {x}_{n+1}-u\parallel \le max\left\{\parallel {x}_{1}-u\parallel ,\parallel u\parallel \right\}+\sum _{i=1}^{n}\parallel {e}_{i}\parallel .$

Since ${\sum }_{n=1}^{\mathrm{\infty }}\parallel {e}_{n}\parallel <\mathrm{\infty }$, we immediately conclude that the sequence $\left\{{x}_{n}\right\}$ is bounded. Thus, from ${\alpha }_{n}\to 0$, ${\mu }_{n}\to 1$, ${\lambda }_{n}\left({\alpha }_{n}+L\right)<1$, (3.5), (3.6) and (3.9) it follows that $\left\{{y}_{n}\right\}$, $\left\{{t}_{n}\right\}$ and $\left\{{z}_{n}\right\}$ are bounded. This completes the proof. □

Proposition 3.3 The following statements hold:

1. (i)

${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-u\parallel$ exists for each $u\in {\bigcap }_{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma$;

2. (ii)

${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-{x}_{n+1}\parallel ={lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-{y}_{n}\parallel ={lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-{z}_{n}\parallel ={lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-{t}_{n}\parallel =0$;

3. (iii)

${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-{W}_{n}{x}_{n}\parallel ={lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-W{x}_{n}\parallel =0$.

Proof For each $u\in {\bigcap }_{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma$, we get from (3.13)

$\begin{array}{rl}\parallel {x}_{n+1}-u\parallel & \le \left(1-{\beta }_{n}\right)\parallel {x}_{n}-u\parallel +{\beta }_{n}\parallel u\parallel +\parallel {e}_{n}\parallel \\ \le \parallel {x}_{n}-u\parallel +{\beta }_{n}\parallel u\parallel +\parallel {e}_{n}\parallel .\end{array}$

Since the conditions ${\sum }_{n=1}^{\mathrm{\infty }}{\beta }_{n}<\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}{e}_{n}<\mathrm{\infty }$ lead to ${\sum }_{n=1}^{\mathrm{\infty }}\left({\beta }_{n}\parallel u\parallel +\parallel {e}_{n}\parallel \right)<\mathrm{\infty }$, by Corollary 2.1, we know that ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-u\parallel$ exists for each $u\in {\bigcap }_{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma$. Note that by Lemma 2.4(a) we have from (3.12)

$\begin{array}{rl}{\parallel {x}_{n}-{x}_{n+1}\parallel }^{2}& ={\parallel {x}_{n}-u\parallel }^{2}-{\parallel {x}_{n+1}-u\parallel }^{2}+2〈{x}_{n+1}-{x}_{n},{x}_{n+1}-u〉\\ \le {\parallel {x}_{n}-u\parallel }^{2}-{\parallel {x}_{n+1}-u\parallel }^{2}+2〈-{\beta }_{n}{x}_{n}+{e}_{n},{x}_{n+1}-u〉\\ \le {\parallel {x}_{n}-u\parallel }^{2}-{\parallel {x}_{n+1}-u\parallel }^{2}+2\left({\beta }_{n}\parallel {x}_{n}\parallel +\parallel {e}_{n}\parallel \right)\parallel {x}_{n+1}-u\parallel .\end{array}$

Since ${\beta }_{n}\to 0$ and $\parallel {e}_{n}\parallel \to 0$ as $n\to \mathrm{\infty }$, from the existence of ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-u\parallel$ and the boundedness of $\left\{{x}_{n}\right\}$, we obtain that

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

Since ${x}_{n+1}\in {C}_{n}$, it follows that

${\parallel {z}_{n}-{x}_{n+1}\parallel }^{2}\le {\parallel {x}_{n}-{x}_{n+1}\parallel }^{2}+{\theta }_{n},$

which implies that

$\parallel {z}_{n}-{x}_{n+1}\parallel \le \parallel {x}_{n}-{x}_{n+1}\parallel +\sqrt{{\theta }_{n}},$

and hence

$\begin{array}{rl}\parallel {x}_{n}-{z}_{n}\parallel & \le \parallel {x}_{n}-{x}_{n+1}\parallel +\parallel {x}_{n+1}-{z}_{n}\parallel \\ \le 2\parallel {x}_{n+1}-{x}_{n}\parallel +\sqrt{{\theta }_{n}}\to 0.\end{array}$

For each $u\in {\bigcap }_{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma$, from (3.9) we have

$\begin{array}{r}\left(1-{\gamma }_{n}\right)\left(1-{\lambda }_{n}^{2}{\left({\alpha }_{n}+L\right)}^{2}\right){\parallel {x}_{n}-{y}_{n}\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-u\parallel }^{2}-{\parallel {z}_{n}-u\parallel }^{2}+2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\left(1-{\mu }_{n}\right)\parallel {t}_{n}-u\parallel \right]\\ \phantom{\rule{1em}{0ex}}\le \left(\parallel {x}_{n}-u\parallel +\parallel {z}_{n}-u\parallel \right)\parallel {x}_{n}-{z}_{n}\parallel +2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\parallel {t}_{n}-u\parallel \right].\end{array}$

Since $\left\{{\gamma }_{n}\right\}\subset \left[0,c\right]$, $\left\{{\lambda }_{n}\right\}\subset \left[a,b\right]$, $1-{b}^{2}{L}^{2}>0$, ${\alpha }_{n}\to 0$ and $\parallel {x}_{n}-{z}_{n}\parallel \to 0$ as $n\to \mathrm{\infty }$, from the boundedness of $\left\{{x}_{n}\right\}$, $\left\{{y}_{n}\right\}$, $\left\{{t}_{n}\right\}$ and $\left\{{z}_{n}\right\}$ we conclude that

$\begin{array}{r}\parallel {x}_{n}-{y}_{n}\parallel \\ \phantom{\rule{1em}{0ex}}\le \frac{1}{\left(1-{\gamma }_{n}\right)\left(1-{\lambda }_{n}^{2}{\left({\alpha }_{n}+L\right)}^{2}\right)}\left\{\left(\parallel {x}_{n}-u\parallel +\parallel {z}_{n}-u\parallel \right)\parallel {x}_{n}-{z}_{n}\parallel \\ \phantom{\rule{2em}{0ex}}+2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\parallel {t}_{n}-u\parallel \right]\right\}\\ \phantom{\rule{1em}{0ex}}\le \frac{1}{\left(1-c\right)\left(1-{b}^{2}{\left({\alpha }_{n}+L\right)}^{2}\right)}\left\{\left(\parallel {x}_{n}-u\parallel +\parallel {z}_{n}-u\parallel \right)\parallel {x}_{n}-{z}_{n}\parallel \\ \phantom{\rule{2em}{0ex}}+2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\parallel {t}_{n}-u\parallel \right]\right\}\\ \phantom{\rule{1em}{0ex}}\to 0.\end{array}$

Utilizing the arguments similar to those in (3.8),

$\begin{array}{r}{\parallel {t}_{n}-u\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-u\parallel }^{2}-{\parallel {x}_{n}-{y}_{n}\parallel }^{2}-{\parallel {y}_{n}-{t}_{n}\parallel }^{2}+2{\lambda }_{n}\left({\alpha }_{n}+L\right)\parallel {x}_{n}-{y}_{n}\parallel \parallel {t}_{n}-{y}_{n}\parallel \\ \phantom{\rule{2em}{0ex}}+2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\left(1-{\mu }_{n}\right)\parallel {t}_{n}-u\parallel \right]\\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-u\parallel }^{2}-{\parallel {x}_{n}-{y}_{n}\parallel }^{2}-{\parallel {y}_{n}-{t}_{n}\parallel }^{2}+{\lambda }_{n}^{2}{\left({\alpha }_{n}+L\right)}^{2}{\parallel {t}_{n}-{y}_{n}\parallel }^{2}+{\parallel {x}_{n}-{y}_{n}\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}+2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\left(1-{\mu }_{n}\right)\parallel {t}_{n}-u\parallel \right]\\ \phantom{\rule{1em}{0ex}}={\parallel {x}_{n}-u\parallel }^{2}+\left({\lambda }_{n}^{2}{\left({\alpha }_{n}+L\right)}^{2}-1\right){\parallel {t}_{n}-{y}_{n}\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}+2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\left(1-{\mu }_{n}\right)\parallel {t}_{n}-u\parallel \right].\end{array}$

Hence,

$\begin{array}{rcl}{\parallel {z}_{n}-u\parallel }^{2}& \le & {\gamma }_{n}{\parallel {x}_{n}-u\parallel }^{2}+\left(1-{\gamma }_{n}\right){\parallel {t}_{n}-u\parallel }^{2}\\ \le & {\gamma }_{n}{\parallel {x}_{n}-u\parallel }^{2}+\left(1-{\gamma }_{n}\right)\left\{{\parallel {x}_{n}-u\parallel }^{2}+\left({\lambda }_{n}^{2}{\left({\alpha }_{n}+L\right)}^{2}-1\right){\parallel {t}_{n}-{y}_{n}\parallel }^{2}\\ +2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\left(1-{\mu }_{n}\right)\parallel {t}_{n}-u\parallel \right]\right\}\\ \le & {\parallel {x}_{n}-u\parallel }^{2}+\left(1-{\gamma }_{n}\right)\left({\lambda }_{n}^{2}{\left({\alpha }_{n}+L\right)}^{2}-1\right){\parallel {t}_{n}-{y}_{n}\parallel }^{2}\\ +2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\parallel {t}_{n}-u\parallel \right].\end{array}$

It follows that

$\begin{array}{r}\left(1-{\gamma }_{n}\right)\left(1-{\lambda }_{n}^{2}{\left({\alpha }_{n}+L\right)}^{2}\right){\parallel {t}_{n}-{y}_{n}\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-u\parallel }^{2}-{\parallel {z}_{n}-u\parallel }^{2}+2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\parallel {t}_{n}-u\parallel \right]\\ \phantom{\rule{1em}{0ex}}\le \left(\parallel {x}_{n}-u\parallel +\parallel {z}_{n}-u\parallel \right)\parallel {x}_{n}-{z}_{n}\parallel +2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\parallel {t}_{n}-u\parallel \right].\end{array}$

Since $\left\{{\gamma }_{n}\right\}\subset \left[0,c\right]$, $\left\{{\lambda }_{n}\right\}\subset \left[a,b\right]$, $1-{b}^{2}{L}^{2}>0$, ${\alpha }_{n}\to 0$ and $\parallel {x}_{n}-{z}_{n}\parallel \to 0$ as $n\to \mathrm{\infty }$, from the boundedness of $\left\{{x}_{n}\right\}$, $\left\{{y}_{n}\right\}$, $\left\{{t}_{n}\right\}$ and $\left\{{z}_{n}\right\}$ we deduce that

$\begin{array}{r}{\parallel {t}_{n}-{y}_{n}\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le \frac{1}{\left(1-{\gamma }_{n}\right)\left(1-{\lambda }_{n}^{2}{\left({\alpha }_{n}+L\right)}^{2}\right)}\left\{\left(\parallel {x}_{n}-u\parallel +\parallel {z}_{n}-u\parallel \right)\parallel {x}_{n}-{z}_{n}\parallel \\ \phantom{\rule{2em}{0ex}}+2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\parallel {t}_{n}-u\parallel \right]\right\}\\ \phantom{\rule{1em}{0ex}}\le \frac{1}{\left(1-c\right)\left(1-{b}^{2}{\left({\alpha }_{n}+L\right)}^{2}\right)}\left\{\left(\parallel {x}_{n}-u\parallel +\parallel {z}_{n}-u\parallel \right)\parallel {x}_{n}-{z}_{n}\parallel \\ \phantom{\rule{2em}{0ex}}+2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\parallel {t}_{n}-u\parallel \right]\right\}\\ \phantom{\rule{1em}{0ex}}\to 0.\end{array}$

Taking into consideration that

$\parallel {x}_{n}-{t}_{n}\parallel \le \parallel {x}_{n}-{y}_{n}\parallel +\parallel {y}_{n}-{t}_{n}\parallel ,$

we also have

$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-{t}_{n}\parallel =0.$

Since ${z}_{n}={\gamma }_{n}{x}_{n}+\left(1-{\gamma }_{n}\right){W}_{n}{t}_{n}$, we have

$\left(1-{\gamma }_{n}\right)\left({W}_{n}{t}_{n}-{t}_{n}\right)={\gamma }_{n}\left({t}_{n}-{x}_{n}\right)+{z}_{n}-{t}_{n}.$

Then

$\begin{array}{rl}\left(1-c\right)\parallel {W}_{n}{t}_{n}-{t}_{n}\parallel & \le \left(1-{\gamma }_{n}\right)\parallel {W}_{n}{t}_{n}-{t}_{n}\parallel \le {\gamma }_{n}\parallel {t}_{n}-{x}_{n}\parallel +\parallel {z}_{n}-{t}_{n}\parallel \\ \le \left(1+{\gamma }_{n}\right)\parallel {t}_{n}-{x}_{n}\parallel +\parallel {z}_{n}-{x}_{n}\parallel ,\end{array}$

and hence $\parallel {t}_{n}-{W}_{n}{t}_{n}\parallel \to 0$. Observe also that

$\begin{array}{rl}\parallel {x}_{n}-{W}_{n}{x}_{n}\parallel & \le \parallel {x}_{n}-{t}_{n}\parallel +\parallel {t}_{n}-{W}_{n}{t}_{n}\parallel +\parallel {W}_{n}{t}_{n}-{W}_{n}{x}_{n}\parallel \\ \le \parallel {x}_{n}-{t}_{n}\parallel +\parallel {t}_{n}-{W}_{n}{t}_{n}\parallel +\parallel {t}_{n}-{x}_{n}\parallel \\ \le 2\parallel {x}_{n}-{t}_{n}\parallel +\parallel {t}_{n}-{W}_{n}{t}_{n}\parallel .\end{array}$

So, we have $\parallel {x}_{n}-{W}_{n}{x}_{n}\parallel \to 0$. On the other hand, since $\left\{{x}_{n}\right\}$ is bounded, from Lemma 2.8, we have ${lim}_{n\to \mathrm{\infty }}\parallel {W}_{n}{x}_{n}-W{x}_{n}\parallel =0$. Therefore, we obtain

$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-W{x}_{n}\parallel =0,$

and this completes the proof. □

Proposition 3.4 ${\omega }_{w}\left({x}_{n}\right)\subset {\bigcap }_{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma$.

Proof By Proposition 3.3(iii), we know that

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

Take $\stackrel{ˆ}{u}\in {\omega }_{w}\left({x}_{n}\right)$ arbitrarily. Then there exists a subsequence $\left\{{x}_{{n}_{i}}\right\}$ of $\left\{{x}_{n}\right\}$ such that ${x}_{{n}_{i}}⇀\stackrel{ˆ}{u}$; hence, we have ${lim}_{i\to \mathrm{\infty }}\parallel {x}_{{n}_{i}}-W{x}_{{n}_{i}}\parallel =0$. Note that from Lemma 2.9 it follows that $I-W$ is demiclosed at zero. Thus, $\stackrel{ˆ}{u}\in Fix\left(W\right)$. Now, let us show $\stackrel{ˆ}{u}\in \Gamma$. Since ${x}_{n}-{t}_{n}\to 0$ and ${x}_{n}-{y}_{n}\to 0$, we have ${t}_{{n}_{i}}⇀\stackrel{ˆ}{u}$ and ${y}_{{n}_{i}}⇀\stackrel{ˆ}{u}$. Let

where ${N}_{C}v$ is the normal cone to C at $v\in C$. We have already mentioned that in this case the mapping T is maximal monotone, and $0\in Tv$ if and only if $v\in VI\left(C,\mathrm{\nabla }f\right)$; see [10] for more details. Let $G\left(T\right)$ be the graph of T and let $\left(v,w\right)\in G\left(T\right)$. Then we have $w\in Tv=\mathrm{\nabla }f\left(v\right)+{N}_{C}v$ and hence $w-\mathrm{\nabla }f\left(v\right)\in {N}_{C}v$. So, we have $〈v-t,w-\mathrm{\nabla }f\left(v\right)〉\ge 0$ for all $t\in C$. On the other hand, from ${t}_{n}={P}_{C}\left({x}_{n}-{\lambda }_{n}\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({t}_{n}\right)\right)$ and $v\in C$, we have

$〈{x}_{n}-{\lambda }_{n}\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({t}_{n}\right)-{t}_{n},{t}_{n}-v〉\ge 0$

and hence

$〈v-{t}_{n},\frac{{t}_{n}-{x}_{n}}{{\lambda }_{n}}+\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)+\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({t}_{n}\right)〉\ge 0.$

Therefore, from $〈v-t,w-\mathrm{\nabla }f\left(v\right)〉\ge 0$ for all $t\in C$ and ${t}_{{n}_{i}}\in C$, we have

$\begin{array}{rcl}〈v-{t}_{{n}_{i}},w〉& \ge & 〈v-{t}_{{n}_{i}},\mathrm{\nabla }f\left(v\right)〉\\ \ge & 〈v-{t}_{{n}_{i}},\mathrm{\nabla }f\left(v\right)〉-〈v-{t}_{{n}_{i}},\frac{{t}_{{n}_{i}}-{x}_{{n}_{i}}}{{\lambda }_{{n}_{i}}}+\mathrm{\nabla }{f}_{{\alpha }_{{n}_{i}}}\left({y}_{{n}_{i}}\right)+\left(1-{\mu }_{{n}_{i}}\right)\mathrm{\nabla }{f}_{{\alpha }_{{n}_{i}}}\left({t}_{{n}_{i}}\right)〉\\ =& 〈v-{t}_{{n}_{i}},\mathrm{\nabla }f\left(v\right)〉-〈v-{t}_{{n}_{i}},\frac{{t}_{{n}_{i}}-{x}_{{n}_{i}}}{{\lambda }_{{n}_{i}}}+\mathrm{\nabla }f\left({y}_{{n}_{i}}\right)〉\\ -{\alpha }_{{n}_{i}}〈v-{t}_{{n}_{i}},{y}_{{n}_{i}}〉-\left(1-{\mu }_{{n}_{i}}\right)〈v-{t}_{{n}_{i}},\mathrm{\nabla }{f}_{{\alpha }_{{n}_{i}}}\left({t}_{{n}_{i}}\right)〉\\ =& 〈v-{t}_{{n}_{i}},\mathrm{\nabla }f\left(v\right)-\mathrm{\nabla }f\left({t}_{{n}_{i}}\right)〉+〈v-{t}_{{n}_{i}},\mathrm{\nabla }f\left({t}_{{n}_{i}}\right)-\mathrm{\nabla }f\left({y}_{{n}_{i}}\right)〉\\ -〈v-{t}_{{n}_{i}},\frac{{t}_{{n}_{i}}-{x}_{{n}_{i}}}{{\lambda }_{{n}_{i}}}〉-{\alpha }_{{n}_{i}}〈v-{t}_{{n}_{i}},{y}_{{n}_{i}}〉-\left(1-{\mu }_{{n}_{i}}\right)〈v-{t}_{{n}_{i}},\mathrm{\nabla }{f}_{{\alpha }_{{n}_{i}}}\left({t}_{{n}_{i}}\right)〉\\ \ge & 〈v-{t}_{{n}_{i}},\mathrm{\nabla }f\left({t}_{{n}_{i}}\right)-\mathrm{\nabla }f\left({y}_{{n}_{i}}\right)〉-〈v-{t}_{{n}_{i}},\frac{{t}_{{n}_{i}}-{x}_{{n}_{i}}}{{\lambda }_{{n}_{i}}}〉\\ -{\alpha }_{{n}_{i}}〈v-{t}_{{n}_{i}},{y}_{{n}_{i}}〉-\left(1-{\mu }_{{n}_{i}}\right)〈v-{t}_{{n}_{i}},\mathrm{\nabla }{f}_{{\alpha }_{{n}_{i}}}\left({t}_{{n}_{i}}\right)〉.\end{array}$

Since $\parallel \mathrm{\nabla }f\left({t}_{n}\right)-\mathrm{\nabla }f\left({y}_{n}\right)\parallel \to 0$ (due to the Lipschitz continuity of f), $\frac{{t}_{n}-{x}_{n}}{{\lambda }_{n}}\to 0$ (due to $\left\{{\lambda }_{n}\right\}\subset \left[a,b\right]$), ${\alpha }_{n}\to 0$ and ${\mu }_{n}\to 1$, we obtain $〈v-\stackrel{ˆ}{u},w〉\ge 0$ as $i\to \mathrm{\infty }$. Since T is maximal monotone, we have $\stackrel{ˆ}{u}\in {T}^{-1}0$ and hence $\stackrel{ˆ}{u}\in VI\left(C,\mathrm{\nabla }f\right)$. Clearly, $\stackrel{ˆ}{u}\in \Gamma$. Consequently, $\stackrel{ˆ}{u}\in {\bigcap }_{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma$. This implies that ${\omega }_{w}\left({x}_{n}\right)\subset {\bigcap }_{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma$. □

Finally, according to Propositions 3.1-3.4, we prove the remainder of Theorem 3.1.

Proof It is sufficient to show that ${\omega }_{w}\left({x}_{n}\right)$ is a single-point set because ${x}_{n}-{y}_{n}\to 0$ and ${x}_{n}-{z}_{n}\to 0$ as $n\to \mathrm{\infty }$. Since ${\omega }_{w}\left({x}_{n}\right)\ne \mathrm{\varnothing }$, let us take two points $u,\stackrel{ˆ}{u}\in {\omega }_{w}\left({x}_{n}\right)$ arbitrarily. Then there exist two subsequences $\left\{{x}_{{n}_{j}}\right\}$ and $\left\{{x}_{{m}_{k}}\right\}$ of $\left\{{x}_{n}\right\}$ such that ${x}_{{n}_{j}}⇀u$ and ${x}_{{m}_{k}}⇀\stackrel{ˆ}{u}$, respectively. In terms of Proposition 3.4, we know that $u,\stackrel{ˆ}{u}\in {\bigcap }_{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma$. Meantime, according to Proposition 3.3(i), we also know that there exist both ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-u\parallel$ and ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-\stackrel{ˆ}{u}\parallel$. Let us show that $u=\stackrel{ˆ}{u}$. Assume that $u\ne \stackrel{ˆ}{u}$. From the Opial condition [39] it follows that

$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-u\parallel & =\underset{j\to \mathrm{\infty }}{lim inf}\parallel {x}_{{n}_{j}}-u\parallel <\underset{j\to \mathrm{\infty }}{lim inf}\parallel {x}_{{n}_{j}}-\stackrel{ˆ}{u}\parallel \\ =\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-\stackrel{ˆ}{u}\parallel =\underset{k\to \mathrm{\infty }}{lim inf}\parallel {x}_{{m}_{k}}-\stackrel{ˆ}{u}\parallel \\ <\underset{k\to \mathrm{\infty }}{lim inf}\parallel {x}_{{m}_{k}}-u\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-u\parallel .\end{array}$

This leads to a contradiction. Thus, we must have $u=\stackrel{ˆ}{u}$. This implies that ${\omega }_{w}\left({x}_{n}\right)$ is a singleton. Without loss of generality, we may write ${\omega }_{w}\left({x}_{n}\right)=\left\{u\right\}$. Consequently, by Lemma 2.2 we obtain that ${x}_{n}⇀u\in {\bigcap }_{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma$. Since ${x}_{n}-{y}_{n}\to 0$ and ${x}_{n}-{z}_{n}\to 0$ as $n\to \mathrm{\infty }$, we also have that ${y}_{n}⇀u$ and ${z}_{n}⇀u$. This completes the proof. □

Remark 3.2 Our Theorem 3.1 improves, extends, supplements and develops Nadezhkina and Takahashi [[9], Theorem 3.1] and Ceng et al. [[11], Theorem 3.1] in the following aspects.

1. (i)

The combination of the problem of finding an element of $Fix\left(S\right)\cap VI\left(C,A\right)$ in [[9], Theorem 3.1] and the one of finding an element of ${\bigcap }_{i=1}^{N}Fix\left({S}_{i}\right)\cap VI\left(C,A\right)$ in [[11], Theorem 3.1] is extended to develop the one of finding an element of ${\bigcap }_{i=1}^{\mathrm{\infty }}Fix\left({S}_{i}\right)\cap \Gamma$ in our Theorem 3.1.

2. (ii)

Our Theorem 3.1 drops the required condition ${lim inf}_{n\to \mathrm{\infty }}〈A{x}_{n},x-{x}_{n}〉\ge 0$, $\mathrm{\forall }x\in C$ in [[11], Theorem 3.1].

3. (iii)

The iterative scheme in [[11], Theorem 3.1] is extended to develop the iterative scheme (3.1) of our Theorem 3.1 by virtue of the iterative scheme of [[9], Theorem 3.1]. The iterative scheme (3.1) of our Theorem 3.1 is more advantageous and more flexible than the iterative scheme of [[11], Theorem 3.1] because it involves several parameter sequences $\left\{{\lambda }_{n}\right\}$, $\left\{{\mu }_{n}\right\}$, $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$, $\left\{{\gamma }_{n}\right\}$, $\left\{{\sigma }_{n}\right\}$ and $\left\{{e}_{n}\right\}$.

4. (iv)

The iterative scheme (3.1) in our Theorem 3.1 is very different from every one in [[9], Theorem 3.1] and [[11], Theorem 3.1] because the final iteration steps of computing ${x}_{n+1}$ in [[9], Theorem 3.1] and [[11], Theorem 3.1] are replaced by the implicit iteration step ${x}_{n+1}={P}_{{C}_{n}}\left(\left(1-{\beta }_{n}\right){x}_{n}+{e}_{n}-{\sigma }_{n}\mathrm{\nabla }f\left({x}_{n+1}\right)\right)$ in the iterative scheme (3.1) of our Theorem 3.1.

5. (v)

The argument technique of our Theorem 3.1 combines the argument one in [[9], Theorem 3.1] and the argument one in [[11], Theorem 3.1]. Because the problem of finding an element of ${\bigcap }_{i=1}^{\mathrm{\infty }}Fix\left({S}_{i}\right)\cap \Gamma$ in our Theorem 3.1 involves a countable family of nonexpansive mappings $\left\{{S}_{n}\right\}$, the proof of our Theorem 3.1 depends on the properties of the W-mapping (see Lemmas 2.6-2.8 of Section 2 in this paper). Therefore, the proof of our Theorem 3.1 is very different from every one in [[9], Theorem 3.1] and [[11], Theorem 3.1].

## 4 Strong convergence theorem

In this section, we prove a strong convergence theorem via an implicit hybrid method with regularization for finding a common element of the set of common fixed points of an infinite family of nonexpansive mappings ${\left\{{S}_{n}\right\}}_{n=1}^{\mathrm{\infty }}$ and the set of solutions of MP (1.1) for a convex functional $f:C\to \mathbf{R}$ with an L-Lipschitz continuous gradient f. This implicit hybrid method with regularization is based on the CQ method, extragradient method and gradient projection algorithm (GPA) with regularization.

Theorem 4.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let ${W}_{n}$ be a W-mapping defined by (2.1), let $\mathrm{\nabla }f:C\to H$ be an L-Lipschitz continuous mapping with $L>0$, and let ${\left\{{S}_{n}\right\}}_{n=1}^{\mathrm{\infty }}$ be an infinite family of nonexpansive mappings of C into itself such that ${\bigcap }_{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma$ is nonempty and bounded. Let $\left\{{x}_{n}\right\}$, $\left\{{y}_{n}\right\}$ and $\left\{{z}_{n}\right\}$ be the sequences generated by

$\left\{\begin{array}{c}{x}_{1}=x\in C\phantom{\rule{1em}{0ex}}\mathit{\text{chosen arbitrarily}},\hfill \\ {y}_{n}={P}_{C}\left({x}_{n}-{\lambda }_{n}{\mu }_{n}\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({x}_{n}\right)-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)\right),\hfill \\ {t}_{n}={P}_{C}\left({x}_{n}-{\lambda }_{n}\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({y}_{n}\right)-{\lambda }_{n}\left(1-{\mu }_{n}\right)\mathrm{\nabla }{f}_{{\alpha }_{n}}\left({t}_{n}\right)\right),\hfill \\ {z}_{n}={\gamma }_{n}{x}_{n}+\left(1-{\gamma }_{n}\right){W}_{n}{t}_{n},\hfill \\ {C}_{n}=\left\{z\in C:{\parallel {z}_{n}-z\parallel }^{2}\le {\parallel {x}_{n}-z\parallel }^{2}+{\theta }_{n}\right\},\hfill \\ {Q}_{n}=\left\{z\in C:〈{x}_{n}-z,x-{x}_{n}〉\ge 0\right\},\hfill \\ {x}_{n+1}={P}_{{C}_{n}\cap {Q}_{n}}x,\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,\hfill \end{array}$
(4.1)

where ${\theta }_{n}={\alpha }_{n}\frac{18}{{\mu }_{n}^{4}}{\Delta }_{n}$ and

${\Delta }_{n}=sup\left\{{\parallel {x}_{n}-z\parallel }^{2}+\left(1+\frac{1}{18{L}^{2}}\right){\parallel z\parallel }^{2}:z\in \bigcap _{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma \right\}<\mathrm{\infty }.$

Assume the following conditions hold:

1. (i)

$\left\{{\alpha }_{n}\right\}\subset \left(0,\mathrm{\infty }\right)$ and ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$;

2. (ii)

$\left\{{\gamma }_{n}\right\}\subset \left[0,c\right]$ for some $c\in \left[0,1\right)$;

3. (iii)

$\left\{{\mu }_{n}\right\}\subset \left(0,1\right]$ and ${lim}_{n\to \mathrm{\infty }}{\mu }_{n}=1$;

4. (iv)

${\lambda }_{n}\left({\alpha }_{n}+L\right)<1$, $\mathrm{\forall }n\ge 1$ and $\left\{{\lambda }_{n}\right\}\subset \left[a,b\right]$ for some $a,b\in \left(0,1/L\right)$.

Then the sequences $\left\{{x}_{n}\right\}$, $\left\{{y}_{n}\right\}$ and $\left\{{z}_{n}\right\}$ generated by (4.1) converge strongly to the same point ${P}_{{\bigcap }_{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma }x$.

Proof Utilizing the condition ${\lambda }_{n}\left({\alpha }_{n}+L\right)<1$, $\mathrm{\forall }n\ge 1$, and repeating the same arguments as in Remark 3.1, we can see that $\left\{{y}_{n}\right\}$, $\left\{{t}_{n}\right\}$ and $\left\{{z}_{n}\right\}$ are defined well. Note that the L-Lipschitz continuity of the gradient f implies that f is $\frac{1}{L}$-ism [31], that is,

$〈\mathrm{\nabla }f\left(x\right)-\mathrm{\nabla }f\left(y\right),x-y〉\ge \frac{1}{L}{\parallel \mathrm{\nabla }f\left(x\right)-\mathrm{\nabla }f\left(y\right)\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$

Repeating the same arguments as in the proof of Proposition 3.1, we know that $\mathrm{\nabla }{f}_{\alpha }=\alpha I+\mathrm{\nabla }f$ is $1/\left(\alpha +L\right)$-ism. It is clear that ${C}_{n}$ is closed and ${Q}_{n}$ is closed and convex for every $n=1,2,\dots$ . As the defining inequality in ${C}_{n}$ is equivalent to the inequality

$〈2\left({x}_{n}-{z}_{n}\right),z〉\le {\parallel {x}_{n}\parallel }^{2}-{\parallel {z}_{n}\parallel }^{2}+{\theta }_{n},$

by Lemma 2.5 we also have that ${C}_{n}$ is convex for every $n=1,2,\dots$ . As ${Q}_{n}=\left\{z\in C:〈{x}_{n}-z,x-{x}_{n}〉\ge 0\right\}$, we have $〈{x}_{n}-z,x-{x}_{n}〉\ge 0$ for all $z\in {Q}_{n}$, and by Proposition 2.1(i) we get ${x}_{n}={P}_{{Q}_{n}}x$.

We divide the rest of the proof into several steps.

Step 1. $\left\{{x}_{n}\right\}$, $\left\{{y}_{n}\right\}$ and $\left\{{z}_{n}\right\}$ are bounded.

Indeed, take $u\in {\bigcap }_{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma$ arbitrarily. Taking into account ${\lambda }_{n}\left({\alpha }_{n}+L\right)<1$, $\mathrm{\forall }n\ge 1$ and repeating the same arguments as in (3.5) and (3.6), we deduce that

$\parallel {y}_{n}-u\parallel \le \frac{1}{{\mu }_{n}}\left(\parallel {x}_{n}-u\parallel +{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \right)$
(4.2)

and

$\parallel {t}_{n}-u\parallel \le \frac{1+2{\mu }_{n}}{{\mu }_{n}^{2}}\left(\parallel {x}_{n}-u\parallel +{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \right).$
(4.3)

Thus, from (4.2) and (4.3) it follows that

$\parallel {y}_{n}-u\parallel +\left(1-{\mu }_{n}\right)\parallel {t}_{n}-u\parallel \le \frac{3}{{\mu }_{n}^{2}}\left(\parallel {x}_{n}-u\parallel +{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \right),$

which together with ${\lambda }_{n}\left({\alpha }_{n}+L\right)<1$ implies that

$\begin{array}{rl}{\left[\parallel {y}_{n}-u\parallel +\left(1-{\mu }_{n}\right)\parallel {t}_{n}-u\parallel \right]}^{2}& \le \frac{9}{{\mu }_{n}^{4}}{\left(\parallel {x}_{n}-u\parallel +{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \right)}^{2}\\ \le \frac{9}{{\mu }_{n}^{4}}\left(2{\parallel {x}_{n}-u\parallel }^{2}+2{\parallel u\parallel }^{2}\right)\\ =\frac{18}{{\mu }_{n}^{4}}{\parallel {x}_{n}-u\parallel }^{2}+\frac{18}{{\mu }_{n}^{4}}{\parallel u\parallel }^{2}.\end{array}$

Repeating the same arguments as in (3.8) and (3.9), we can deduce that

$\begin{array}{r}{\parallel {t}_{n}-u\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-u\parallel }^{2}-{\parallel {x}_{n}-{y}_{n}\parallel }^{2}-{\parallel {y}_{n}-{t}_{n}\parallel }^{2}+2{\lambda }_{n}\left({\alpha }_{n}+L\right)\parallel {x}_{n}-{y}_{n}\parallel \parallel {t}_{n}-{y}_{n}\parallel \\ \phantom{\rule{2em}{0ex}}+2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\left(1-{\mu }_{n}\right)\parallel {t}_{n}-u\parallel \right]\\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-u\parallel }^{2}+\left({\lambda }_{n}^{2}{\left({\alpha }_{n}+L\right)}^{2}-1\right){\parallel {x}_{n}-{y}_{n}\parallel }^{2}+2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\left(1-{\mu }_{n}\right)\parallel {t}_{n}-u\parallel \right]\\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-u\parallel }^{2}+2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\left(1-{\mu }_{n}\right)\parallel {t}_{n}-u\parallel \right]\end{array}$
(4.4)

and

$\begin{array}{rl}{\parallel {z}_{n}-u\parallel }^{2}\le & {\gamma }_{n}{\parallel {x}_{n}-u\parallel }^{2}+\left(1-{\gamma }_{n}\right){\parallel {t}_{n}-u\parallel }^{2}\\ \le & {\parallel {x}_{n}-u\parallel }^{2}+\left(1-{\gamma }_{n}\right)\left({\lambda }_{n}^{2}{\left({\alpha }_{n}+L\right)}^{2}-1\right){\parallel {x}_{n}-{y}_{n}\parallel }^{2}\\ +2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\left(1-{\mu }_{n}\right)\parallel {t}_{n}-u\parallel \right]\\ \le & {\parallel {x}_{n}-u\parallel }^{2}+{\alpha }_{n}\left[{\lambda }_{n}^{2}{\parallel u\parallel }^{2}+{\left(\parallel {y}_{n}-u\parallel +\left(1-{\mu }_{n}\right)\parallel {t}_{n}-u\parallel \right)}^{2}\right]\\ \le & {\parallel {x}_{n}-u\parallel }^{2}+{\alpha }_{n}\frac{18}{{\mu }_{n}^{4}}\left[{\parallel {x}_{n}-u\parallel }^{2}+\left(1+\frac{1}{18{L}^{2}}\right){\parallel u\parallel }^{2}\right]\\ \le & {\parallel {x}_{n}-u\parallel }^{2}+{\alpha }_{n}\frac{18}{{\mu }_{n}^{4}}{\Delta }_{n}\\ =& {\parallel {x}_{n}-u\parallel }^{2}+{\theta }_{n}\end{array}$
(4.5)

for every $n=1,2,\dots$ and hence $u\in {C}_{n}$. So,

$\bigcap _{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma \subset {C}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1.$

Next, let us show by mathematical induction that $\left\{{x}_{n}\right\}$ is well defined and ${\bigcap }_{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma \subset {C}_{n}\cap {Q}_{n}$ for every $n=1,2,\dots$ . For $n=1$ we have ${Q}_{1}=C$. Hence we obtain ${\bigcap }_{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma \subset {C}_{1}\cap {Q}_{1}$. Suppose that ${x}_{k}$ is given and ${\bigcap }_{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma \subset {C}_{k}\cap {Q}_{k}$ for some integer $k\ge 1$. Since ${\bigcap }_{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma$ is nonempty, ${C}_{k}\cap {Q}_{k}$ is a nonempty closed convex subset of C. So, there exists a unique element ${x}_{k+1}\in {C}_{k}\cap {Q}_{k}$ such that ${x}_{k+1}={P}_{{C}_{k}\cap {Q}_{k}}x$. It is also obvious that there holds $〈{x}_{k+1}-z,x-{x}_{k+1}〉\ge 0$ for every $z\in {C}_{k}\cap {Q}_{k}$. Since ${\bigcap }_{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma \subset {C}_{k}\cap {Q}_{k}$, we have $〈{x}_{k+1}-z,x-{x}_{k+1}〉\ge 0$ for every $z\in {\bigcap }_{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma$ and hence ${\bigcap }_{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma \subset {Q}_{k+1}$. Therefore, we obtain ${\bigcap }_{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma \subset {C}_{k+1}\cap {Q}_{k+1}$.

Step 2. ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-{x}_{n+1}\parallel =0$ and ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-{z}_{n}\parallel =0$.

Indeed, let $q={P}_{{\bigcap }_{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma }x$. From ${x}_{n+1}={P}_{{C}_{n}\cap {Q}_{n}}x$ and $q\in {\bigcap }_{n=1}^{\mathrm{\infty }}Fix\left({S}_{n}\right)\cap \Gamma \subset {C}_{n}\cap {Q}_{n}$, we have

$\parallel {x}_{n+1}-x\parallel \le \parallel q-x\parallel$
(4.6)

for every $n=1,2,\dots$ . Therefore, $\left\{{x}_{n}\right\}$ is bounded. From (4.2), (4.3) and (4.5) we also obtain that $\left\{{y}_{n}\right\}$, $\left\{{t}_{n}\right\}$ and $\left\{{z}_{n}\right\}$ are bounded. Since ${x}_{n+1}\in {C}_{n}\cap {Q}_{n}\subset {Q}_{n}$ and ${x}_{n}={P}_{{Q}_{n}}x$, we have

$\parallel {x}_{n}-x\parallel \le \parallel {x}_{n+1}-x\parallel$

for every $n=1,2,\dots$ . Therefore, there exists ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-x\parallel$. Since ${x}_{n}={P}_{{Q}_{n}}x$ and ${x}_{n+1}\in {Q}_{n}$, using Proposition 2.1(ii), we have

${\parallel {x}_{n+1}-{x}_{n}\parallel }^{2}\le {\parallel {x}_{n+1}-x\parallel }^{2}-{\parallel {x}_{n}-x\parallel }^{2}$

for every $n=1,2,\dots$ . This implies that

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

Since ${x}_{n+1}\in {C}_{n}$, we have

${\parallel {z}_{n}-{x}_{n+1}\parallel }^{2}\le {\parallel {x}_{n}-{x}_{n+1}\parallel }^{2}+{\theta }_{n},$

which implies that

$\parallel {z}_{n}-{x}_{n+1}\parallel \le \parallel {x}_{n}-{x}_{n+1}\parallel +\sqrt{{\theta }_{n}}.$

Hence we get

$\parallel {x}_{n}-{z}_{n}\parallel \le \parallel {x}_{n}-{x}_{n+1}\parallel +\parallel {x}_{n+1}-{z}_{n}\parallel \le 2\parallel {x}_{n+1}-{x}_{n}\parallel +\sqrt{{\theta }_{n}}$

for every $n=1,2,\dots$ . From $\parallel {x}_{n+1}-{x}_{n}\parallel \to 0$ and ${\theta }_{n}\to 0$, we conclude that $\parallel {x}_{n}-{z}_{n}\parallel \to 0$ as $n\to \mathrm{\infty }$.

Step 3. ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-{y}_{n}\parallel ={lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-{t}_{n}\parallel =0$ and ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-{W}_{n}{x}_{n}\parallel ={lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-W{x}_{n}\parallel =0$.

Indeed, since $\left\{{\gamma }_{n}\right\}\subset \left[0,c\right]$, $\left\{{\lambda }_{n}\right\}\subset \left[a,b\right]$, $1-{b}^{2}{L}^{2}>0$, ${\alpha }_{n}\to 0$ and $\parallel {x}_{n}-{z}_{n}\parallel \to 0$ as $n\to \mathrm{\infty }$, from the boundedness of $\left\{{x}_{n}\right\}$, $\left\{{y}_{n}\right\}$, $\left\{{t}_{n}\right\}$ and $\left\{{z}_{n}\right\}$ we conclude from (4.5) that

$\begin{array}{r}{\parallel {x}_{n}-{y}_{n}\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le \frac{1}{\left(1-{\gamma }_{n}\right)\left(1-{\lambda }_{n}^{2}{\left({\alpha }_{n}+L\right)}^{2}\right)}\left\{{\parallel {x}_{n}-u\parallel }^{2}-{\parallel {z}_{n}-u\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}+2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\left(1-{\mu }_{n}\right)\parallel {t}_{n}-u\parallel \right]\right\}\\ \phantom{\rule{1em}{0ex}}\le \frac{1}{\left(1-{\gamma }_{n}\right)\left(1-{\lambda }_{n}^{2}{\left({\alpha }_{n}+L\right)}^{2}\right)}\left\{\left(\parallel {x}_{n}-u\parallel +\parallel {z}_{n}-u\parallel \right)\parallel {x}_{n}-{z}_{n}\parallel \\ \phantom{\rule{2em}{0ex}}+2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\parallel {t}_{n}-u\parallel \right]\right\}\\ \phantom{\rule{1em}{0ex}}\le \frac{1}{\left(1-c\right)\left(1-{b}^{2}{\left({\alpha }_{n}+L\right)}^{2}\right)}\left\{\left(\parallel {x}_{n}-u\parallel +\parallel {z}_{n}-u\parallel \right)\parallel {x}_{n}-{z}_{n}\parallel \\ \phantom{\rule{2em}{0ex}}+2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\parallel {t}_{n}-u\parallel \right]\right\}\\ \phantom{\rule{1em}{0ex}}\to 0.\end{array}$

Utilizing the arguments similar to those in (4.4),

$\begin{array}{r}{\parallel {t}_{n}-u\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-u\parallel }^{2}-{\parallel {x}_{n}-{y}_{n}\parallel }^{2}-{\parallel {y}_{n}-{t}_{n}\parallel }^{2}+2{\lambda }_{n}\left({\alpha }_{n}+L\right)\parallel {x}_{n}-{y}_{n}\parallel \parallel {t}_{n}-{y}_{n}\parallel \\ \phantom{\rule{2em}{0ex}}+2{\lambda }_{n}{\alpha }_{n}\parallel u\parallel \left[\parallel {y}_{n}-u\parallel +\left(1-{\mu }_{n}\right)\parallel {t}_{}\end{array}$