Iterative methods for hierarchical common fixed point problems and variationalinequalities

Abstract

The purpose of this paper is to deal with the problem of finding hierarchically acommon fixed point of a sequence of nearly nonexpansive self-mappings defined ona closed convex subset of a real Hilbert space which is also a solution of someparticular variational inequality problem. We introduce two explicit iterativeschemes and establish strong convergence results for sequences generatediteratively by the explicit schemes under suitable conditions. Our strongconvergence results include the previous results as special cases, and can beviewed as an improvement and refinement of several corresponding known resultsfor hierarchical variational inequality problems.

MSC: 47J20, 49J40.

1 Introduction

Variational inequality problems were initially studied by Stampacchia [1] in 1964. Since then, many kinds of variational inequalities have beenextended and generalized in several directions using novel and innovative techniques(see [2–5] and the references therein).

The classical variational inequality problem or the Stampacchia variationalinequality problem in a Hilbert space is defined as follows:

Let C be a nonempty closed convex subset of a real Hilbert space H,and let $T:C\to H$ be a nonlinear mapping. Then the classicalvariational inequality problem is a problem of finding $x^{\ast }\in C$ such that

$$〈T{x}^{\ast },y-x^{\ast }〉\ge 0\text{for all }y\in C.$$

(1.1)

Problem (1.1) is denoted by $VI(C,T)$. A point $x^{\ast }\in C$ is a solution of $VI(C,T)$ if and only if $x^{\ast }$ is a fixed point of $P_C(I-\lambda T)$, where $\lambda >0$ is a constant, I is the identity mappingfrom C into itself, and $P_C$ is the metric projection from H onto aclosed convex subset C of H. The set of solutions of (1.1) isdenoted by $\mathrm{\Omega }(C,T)$, that is,

$$\mathrm{\Omega }(C,T)=\{x^{\ast }\in C:〈T{x}^{\ast },y-x^{\ast }〉\ge 0\text{ for all }y\in C\}.$$

The variational inequality $VI(C,T)$ is called a monotone variational inequality if theoperator T is a monotone operator. The problem of kind (1.1) is connectedwith the convex minimization problem, complementarity problem, saddle point problem,fixed point problem, Nash equilibrium problem, the problem of finding a point$x\in H$ satisfying $Tx=0$ and so on, and it has several applications indifferent branches of natural sciences, social sciences, management and engineering(see [6–11] and the references therein). In this context, we discuss the variationalinequality problem over the set of fixed points of a mapping which is known as ahierarchical variational inequality problem or a hierarchical fixed point problem.

The hierarchical variational inequality problem over the set of fixed points of anonexpansive mapping is defined as follows:

Let C be a nonempty closed convex subset of a real Hilbert space H,and let $T,S:C\to C$ be two nonexpansive mappings. Then the hierarchicalvariational inequality problem is given as follows:

$$\text{Find }{x}^{\ast }\in F(S)\text{ such that }〈(I-T)x^{\ast },y-x^{\ast }〉\ge 0\text{ for all }y\in F(S).$$

(1.2)

Problem (1.2) is denoted by ${VI}_{F(S)}(C,T)$. The set of solutions of (1.2) is denoted by${\mathrm{\Omega }}_{F(S)}(C,T)$, that is,

$${\mathrm{\Omega }}_{F(S)}(C,T)=\{x^{\ast }\in F(S):〈(I-T)x^{\ast },y-x^{\ast }〉\ge 0\text{ for all }y\in F(S)\}.$$

It is easy to observe that ${VI}_{F(S)}(C,T)$ is equivalent to the fixed point problem$x^{\ast }=P_{F(S)}T{x}^{\ast }$, that is, $x^{\ast }$ is a fixed point of the nonexpansive mapping$P_{F(S)}T$. Of course, if $T=I$, then the set of solutions of ${VI}_{F(S)}(C,T)$ is ${\mathrm{\Omega }}_{F(S)}(C,T)=F(S)$.

Firstly, Moudafi and Maingé [12] introduced an implicit iterative algorithm to solve problem (1.2) andproved weak and strong convergence results, and after that, many iterative methodshave been developed for solving hierarchical problem (1.2) by several authors (see,e.g., [13–17]).

Very recently, Gu et al.[18] motivated and inspired by the results of Marino and Xu [15], Yao et al.[17] introduced and studied two iterative schemes for solving hierarchicalvariational inequality problem and proved the corresponding strong convergenceresults for the generated sequences in the context of a countable family ofnonexpansive mappings under suitable conditions on parameters.

In this paper, inspired by Gu et al.[18] and Sahu et al.[19], we introduce two explicit iterative schemes which generate sequences viaiterative algorithms. We prove that the generated sequences converge strongly to theunique solutions of particular variational inequality problems defined over the setof common fixed points of a sequence of nearly nonexpansive mappings. Our results,in one sense, extend the results of Gu et al.[18] to the sequence of nonexpansive mappings and, in another sense, to thesequence of nearly nonexpansive mappings, which is a wider class of sequence ofnonexpansive mappings. Our results also generalize the results of Cianciaruso etal.[13], Yao et al.[17], Moudafi [20], Xu [21] and many other related works.

2 Preliminaries

Let C be a nonempty subset of a real Hilbert space H with the innerproduct $〈\cdot ,\cdot 〉$ and the norm $\parallel \cdot \parallel$, respectively. A mapping $T:C\to H$ is called

• (1) monotone if

  $$〈Tx-Ty,x-y〉\ge 0\text{for all }x,y\in C,$$

• (2) η-strongly monotone if there exists a positive real number η such that

  $$〈Tx-Ty,x-y〉\ge \eta {\parallel x-y\parallel }^2\text{for all }x,y\in C,$$

• (3) k-Lipschitzian if there exists a constant $k>0$ such that

  $$\parallel Tx-Ty\parallel \le k\parallel x-y\parallel \text{for all }x,y\in C,$$

• (4) ρ-contraction if there exists a constant $\rho \in (0,1)$ such that

  $$\parallel Tx-Ty\parallel \le \rho \parallel x-y\parallel \text{for all }x,y\in C,$$

• (5) nonexpansive if

  $$\parallel Tx-Ty\parallel \le \parallel x-y\parallel \text{for all }x,y\in C,$$

• (6) nearly nonexpansive [22, 23] with respect to a fixed sequence $\{a_n\}$ in $[0,\mathrm{\infty })$ with $a_n\to 0$ if

  $$\parallel T^{n}x-T^{n}y\parallel \le \parallel x-y\parallel +a_n\text{for all }x,y\in C\text{ and }n\in \mathbb{N}.$$

Throughout this paper, we denote by I the identity mapping of H.Also, we denote by → and ⇀ the strong convergence and weak convergence,respectively. The symbol ℕ stands for the set of all natural numbers and${\omega }_w(\{x_n\})$ denotes the set of all weak subsequential limits of$\{x_n\}$.

Let C be a nonempty closed convex subset of H. Then, for any$x\in H$, there exists a unique nearest point in C,denoted by $P_C(x)$, such that

$$\parallel x-P_C(x)\parallel =inf\parallel x-y\parallel \text{for all }y\in C.$$

The mapping $P_C$ is called the metric projection fromH onto C.

It is observed that $P_C$ is a nonexpansive and monotone mapping fromH onto C (see Agarwal et al.[22] for other properties of projection operators).

Let C be a nonempty subset of a real Hilbert space H, and let$S_1,S_2:C\to H$ be two mappings. We denote by $\mathcal{B}(C)$ the collection of all bounded subsets of C.The deviation between $S_1$ and $S_2$ on $B\in \mathcal{B}(C)$[24], denoted by ${\parallel S_1-S_2\parallel }_{\mathrm{\infty },B}$, is defined by

$${\parallel S_1-S_2\parallel }_{\mathrm{\infty },B}=sup\{\parallel S_1(x)-S_2(x)\parallel :x\in B\}.$$

In what follows, we shall make use of the following lemmas and proposition.

Lemma 2.1 ([11])

Let C be a nonempty closed convex subset of a real Hilbert space H. If$x\in H$and$y\in C$, then$y=P_C(x)$if and only if the following inequality holds:

$$〈x-y,z-y〉\le 0\mathit{\text{for all}}z\in C.$$

Lemma 2.2 ([25])

Let$f:C\to H$be a λ-contraction mapping and$T:C\to C$be nonexpansive. Then the following hold:

• (a) The mapping $I-f$ is $(1-\lambda )$-strongly monotone, i.e.,

  $$〈x-y,(I-f)x-(I-f)y〉\ge (1-\lambda ){\parallel x-y\parallel }^2\mathit{\text{for all}}x,y\in C.$$

• (b) The mapping $I-T$ is monotone, i.e.,

  $$〈x-y,(I-T)x-(I-T)y〉\ge 0\mathit{\text{for all}}x,y\in C.$$

Lemma 2.3 ([22])

Let T be a nonexpansive self-mapping of a nonempty closed convex subset C of a real Hilbert space H. Then$I-T$is demiclosed at zero, i.e., if$\{x_n\}$is a sequence in C weakly converging to some$x\in C$and the sequence$\{(I-T)x_n\}$strongly converges to 0, then$x\in F(T)$.

Lemma 2.4 ([26])

Let $\{t_n\}$ and $\{d_n\}$ be the sequences of nonnegative real numbers such that

$$t_{n+1}\le (1-b_n)t_n+c_n+d_n\mathit{\text{for all}}n\in \mathbb{N},$$

where$\{b_n\}$is a real number sequence in$(0,1)$and$\{c_n\}$is a real number sequence. Assume that the following conditionshold:

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

• (ii) ${\sum }_{n=1}^{\mathrm{\infty }}{b}_n=\mathrm{\infty }$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\frac{c_n}{b_n}\le 0$.

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

Lemma 2.5 ([24, 27])

Let C be a nonempty closed convex subset of a real Hilbert space H and$t_i>0$ ($i=1,2,3,\dots ,N$) such that${\sum }_{i=1}^N{t}_i=1$. Let$T_1,T_2,T_3,\dots ,T_N:C\to C$be nonexpansive mappings with${\bigcap }_{i=1}^{N}F(T_i)\ne \mathrm{\varnothing }$and let$T={\sum }_{i=1}^N{t}_i{T}_i$. Then T is nonexpansive from C into itself and$F(T)={\bigcap }_{i=1}^{N}F(T_i)$.

Let C be a nonempty subset of a real Hilbert space H. Let$\mathcal{T}:={\{T_n\}}_{n=1}^{\mathrm{\infty }}$ be a sequence of mappings from C intoitself. We denote by $F(\mathcal{T})$ the set of common fixed points of the sequence, that is,$F(\mathcal{T})={\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)$. Fix a sequence $\{a_n\}$ in $[0,\mathrm{\infty })$ with $a_n\to 0$, and let $\{T_n\}$ be a sequence of mappings from C intoH. Then the sequence $\{T_n\}$ is called a sequence of nearly nonexpansivemappings[19] with respect to a sequence $\{a_n\}$ if

$$\parallel T_{n}x-T_{n}y\parallel \le \parallel x-y\parallel +a_n\text{for all }x,y\in C\text{ and }n\in \mathbb{N}.$$

One can observe that the sequence of nonexpansive mappings is essentially a sequenceof nearly nonexpansive mappings.

We now introduce the following:

Let C be a nonempty closed convex subset of a Hilbert space H. Let$\mathcal{T}={\{T_n\}}_{n=1}^{\mathrm{\infty }}$ be a sequence of nearly nonexpansive mappings fromC into itself with sequence $\{a_n\}$ such that ${\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)\ne \mathrm{\varnothing }$. Let $T:C\to C$ be a mapping such that $Tx={lim}_{n\to \mathrm{\infty }}{T}_{n}x$ for all $x\in C$ with $F(T)={\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)$. Then $\{T_n\}$ is said to satisfy condition (G) iffor each sequence $\{x_n\}$ in C with $x_n\rightharpoonup w$ and $x_n-T_i{x}_n\to 0$ for all $i\in \mathbb{N}$ imply $w\in F(T)$.

Proposition 2.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let$\{T_n\}$be a sequence of nonexpansive mappings from C into itself. Then$\{T_n\}$satisfies condition (G).

Proposition 2.2 Let C be a nonempty closed convex subset of a real Hilbert space H. Let$\mathcal{T}={\{T_n\}}_{n=1}^{\mathrm{\infty }}$be a sequence of nearly nonexpansive mappings from C into itself with sequence$\{a_n\}$such that${\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)\ne \mathrm{\varnothing }$. Then

$${\parallel T_{n}x-x\parallel }^2\le 2〈x-T_{n}x,x-\tilde{x}〉+(a_n+2\parallel x-\tilde{x}\parallel )a_n$$

for all$x\in C$and$\tilde{x}\in {\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)$.

Proof Let $x\in C$ and $\tilde{x}\in {\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)$. Then

$${\parallel T_{n}x-T_n\tilde{x}\parallel }^2\le {(\parallel x-\tilde{x}\parallel +a_n)}^2={\parallel x-\tilde{x}\parallel }^2+(a_n+2\parallel x-\tilde{x}\parallel )a_n.$$

Since $\tilde{x}\in {\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)$, we have

$$\begin{array}{r}{\parallel x-\tilde{x}\parallel }^2+(a_n+2\parallel x-\tilde{x}\parallel )a_n\\ \ge {\parallel T_{n}x-\tilde{x}\parallel }^2={\parallel (T_{n}x-x)+(x-\tilde{x})\parallel }^2\\ ={\parallel T_{n}x-x\parallel }^2+{\parallel x-\tilde{x}\parallel }^2+2〈T_{n}x-x,x-\tilde{x}〉.\end{array}$$

Therefore,

$${\parallel T_{n}x-x\parallel }^2\le 2〈x-T_{n}x,x-\tilde{x}〉+(a_n+2\parallel x-\tilde{x}\parallel )a_n.$$

 □

Proposition 2.3 Let C be a nonempty closed convex subset of a real Hilbert space H. Let$\mathcal{T}={\{T_n\}}_{n=1}^{\mathrm{\infty }}$be a sequence of nearly nonexpansive mappings from C into itself with sequence$\{a_n\}$. Then

$$〈(I-T_n)x-(I-T_n)y,x-y〉\ge -a_n\parallel x-y\parallel \mathit{\text{for all}}x,y\in C\mathit{\text{and}}n\in \mathbb{N}.$$

Proof Let $x,y\in C$. Then

$$\begin{array}{r}〈(I-T_n)x-(I-T_n)y,x-y〉\\ ={\parallel x-y\parallel }^2-〈T_{n}x-T_{n}y,x-y〉\\ \ge {\parallel x-y\parallel }^2-\parallel T_{n}x-T_{n}y\parallel \parallel x-y\parallel \\ \ge {\parallel x-y\parallel }^2-(\parallel x-y\parallel +a_n)\parallel x-y\parallel =-a_n\parallel x-y\parallel .\end{array}$$

 □

3 Main result

Theorem 3.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let$f:C\to H$be a λ-contraction, and let$\{S_n\}$be a sequence of nonexpansive mappings from C into itself. Let S be a nonexpansive mapping from C into itself such that${lim}_{n\to \mathrm{\infty }}{S}_{n}x=Sx$for all$x\in C$. Let$\mathcal{T}={\{T_n\}}_{n=1}^{\mathrm{\infty }}$be a sequence of uniformly continuous nearly nonexpansive mappings from C into itself with sequence$\{a_n\}$such that$F(\mathcal{T})\ne \mathrm{\varnothing }$. Let T be a mapping from C into itself defined by$Tx={lim}_{n\to \mathrm{\infty }}{T}_{n}x$for all$x\in C$. Suppose that$F(T)={\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)$. For arbitrary$x_1\in C$, consider the sequence$\{x_n\}$generated by the following iterative process:

$$\{\begin{array}{c}{x}_1\in C,\hfill \\ y_n=(1-{\beta }_n)x_n+{\beta }_n{S}_n{x}_n,\hfill \\ x_{n+1}=P_C[{\alpha }_{n}f(x_n)+{\sum }_{i=1}^n({\alpha }_{i-1}-{\alpha }_i)T_i{y}_n]\hfill \end{array}$$

(3.1)

for all$n\in \mathbb{N}$, where${\alpha }_0=1$, $\{{\alpha }_n\}$is a strictly decreasing sequence in$(0,1)$and$\{{\beta }_n\}$is a sequence in$(0,1)$satisfying the conditions:

• (i) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (ii) ${\sum }_{n=1}^{\mathrm{\infty }}({\alpha }_{n-1}-{\alpha }_n)<\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}|{\beta }_{n-1}-{\beta }_n|<\mathrm{\infty }$ and ${lim}_{n\to \mathrm{\infty }}\frac{{\beta }_n}{{\alpha }_n}=\tau \in (0,\mathrm{\infty })$;

• (iii) ${lim}_{n\to \mathrm{\infty }}\frac{(({\alpha }_{n-1}-{\alpha }_n)+|{\beta }_n-{\beta }_{n-1}|)}{{\alpha }_n{\beta }_n}=0$;

• (iv) there exists a constant $N>0$ such that $\frac{1}{{\alpha }_n}|\frac{1}{{\beta }_n}-\frac{1}{{\beta }_{n-1}}|\le N$;

• (v) ${lim}_{n\to \mathrm{\infty }}\frac{{\gamma }_n}{{\beta }_n}=0$, where ${\gamma }_n:={\sum }_{i=1}^n({\alpha }_{i-1}-{\alpha }_i)a_i$, and either ${\sum }_{n=1}^{\mathrm{\infty }}{\parallel S_{n+1}-S_n\parallel }_{\mathrm{\infty },B}<\mathrm{\infty }$ or ${lim}_{n\to \mathrm{\infty }}\frac{{\parallel S_{n+1}-S_n\parallel }_{\mathrm{\infty },B}}{{\alpha }_n}=0$ for each $B\in \mathcal{B}(C)$.

Then the sequence$\{x_n\}$converges strongly to a point$x^{\ast }\in {\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)$, which is the unique solution of the followingvariational inequality:

$$〈\frac{1}{\tau }(I-f)x^{\ast }+(I-S)x^{\ast },x-x^{\ast }〉\ge 0\mathit{\text{for all}}x\in \bigcap _{n=1}^{\mathrm{\infty }}F(T_n).$$

(3.2)

Proof It was proved in [17] that variational inequality problem (3.2) has the unique solution. Let$p\in {\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)$. We now break the proof into the following steps.

Step 1. $\{x_n\}$ is bounded.

From (3.1), we have

$$\begin{array}{rl}\parallel y_n-p\parallel & =\parallel (1-{\beta }_n)x_n+{\beta }_n{S}_n{x}_n-p\parallel \\ \le (1-{\beta }_n)\parallel x_n-p\parallel +{\beta }_n(\parallel S_n{x}_n-S_{n}p\parallel +\parallel S_{n}p-p\parallel )\\ \le \parallel x_n-p\parallel +{\beta }_n\parallel S_{n}p-p\parallel .\end{array}$$

It follows that

$$\begin{array}{r}\parallel x_{n+1}-p\parallel \\ =\parallel P_C[{\alpha }_{n}f(x_n)+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)T_i{y}_n]-P_C(p)\parallel \\ \le \parallel {\alpha }_{n}f(x_n)+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)T_i{y}_n-p\parallel \\ =\parallel {\alpha }_n(f(x_n)-p)+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)(T_i{y}_n-p)\parallel \\ \le {\alpha }_n(\parallel f(x_n)-f(p)\parallel +\parallel f(p)-p\parallel )+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)\parallel T_i{y}_n-T_{i}p\parallel \\ \le {\alpha }_n\lambda \parallel x_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel +\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)(\parallel y_n-p\parallel +a_i)\\ \le {\alpha }_n\lambda \parallel x_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel \\ +(1-{\alpha }_n)(\parallel x_n-p\parallel +{\beta }_n\parallel S_{n}p-p\parallel )+{\gamma }_n\\ \le [1-{\alpha }_n(1-\lambda )]\parallel x_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel +{\beta }_n\parallel S_{n}p-p\parallel +{\gamma }_n.\end{array}$$

Note that ${lim}_{n\to \mathrm{\infty }}\frac{{\beta }_n}{{\alpha }_n}=\tau \in (0,\mathrm{\infty })$ and ${lim}_{n\to \mathrm{\infty }}\frac{{\gamma }_n}{{\beta }_n}=0$, so there exists a constant $K>0$ such that

$$\frac{{\alpha }_n\parallel f(p)-p\parallel +{\beta }_n\parallel S_{n}p-p\parallel +{\gamma }_n}{{\alpha }_n}\le K\text{for all }n\in \mathbb{N}.$$

Thus, we have

$$\begin{array}{rl}\parallel x_{n+1}-p\parallel & \le (1-{\alpha }_n(1-\lambda ))\parallel x_n-p\parallel +{\alpha }_{n}K\\ \le max\{\parallel x_n-p\parallel ,\frac{K}{1-\lambda }\}\text{for all }n\in \mathbb{N}.\end{array}$$

Hence $\{x_n\}$ is bounded. So, $\{f(x_n)\}$, $\{y_n\}$, $\{T_i{x}_n\}$ and $\{T_i{y}_n\}$ are bounded.

Step 2. $\parallel x_{n+1}-x_n\parallel \to 0$ as $n\to \mathrm{\infty }$.

Set $u_n:={\alpha }_{n}f(x_n)+{\sum }_{i=1}^n({\alpha }_{i-1}-{\alpha }_i)T_i{y}_n$ for all $n\in \mathbb{N}$. Set $M:={sup}_{n>1}\{\parallel f(x_{n-1})\parallel +\parallel T_n{y}_{n-1}\parallel +\parallel S_{n-1}{x}_{n-1}\parallel +\parallel x_{n-1}\parallel \}$. From (3.1) we have

$$\begin{array}{r}\parallel x_{n+1}-x_n\parallel \\ =\parallel P_C(u_n)-P_C(u_{n-1})\parallel \\ \le \parallel u_n-u_{n-1}\parallel \\ =\parallel {\alpha }_n(f(x_n)-f(x_{n-1}))+({\alpha }_n-{\alpha }_{n-1})f(x_{n-1})\\ +\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)(T_i{y}_n-T_i{y}_{n-1})+({\alpha }_{n-1}-{\alpha }_n)T_n{y}_{n-1}\parallel \\ \le {\alpha }_n\parallel f(x_n)-f(x_{n-1})\parallel +\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)(\parallel y_n-y_{n-1}\parallel +a_i)\\ +({\alpha }_{n-1}-{\alpha }_n)(\parallel f(x_{n-1})\parallel +\parallel T_n{y}_{n-1}\parallel )\\ \le {\alpha }_n\lambda \parallel x_n-x_{n-1}\parallel +\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)\parallel y_n-y_{n-1}\parallel \\ +({\alpha }_{n-1}-{\alpha }_n)(\parallel f(x_{n-1})\parallel +\parallel T_n{y}_{n-1}\parallel )+{\gamma }_n\\ \le {\alpha }_n\lambda \parallel x_n-x_{n-1}\parallel +(1-{\alpha }_n)\parallel y_n-y_{n-1}\parallel \\ +({\alpha }_{n-1}-{\alpha }_n)M+{\gamma }_n.\end{array}$$

(3.3)

Set $B:=\{x_n\}$. Now, from (3.1) we have

$$\begin{array}{r}\parallel y_n-y_{n-1}\parallel \\ =\parallel ((1-{\beta }_n)x_n+{\beta }_n{S}_n{x}_n)-((1-{\beta }_{n-1})x_{n-1}+{\beta }_{n-1}{S}_{n-1}{x}_{n-1})\parallel \\ =\parallel (1-{\beta }_n)(x_n-x_{n-1})+({\beta }_{n-1}-{\beta }_n)x_{n-1}+{\beta }_n(S_n{x}_n-S_n{x}_{n-1})\\ +{\beta }_n(S_n{x}_{n-1}-S_{n-1}{x}_{n-1})+({\beta }_n-{\beta }_{n-1})S_{n-1}{x}_{n-1}\parallel \\ \le \parallel (1-{\beta }_n)(x_n-x_{n-1})+({\beta }_{n-1}-{\beta }_n)x_{n-1}+{\beta }_n(S_n{x}_n-S_n{x}_{n-1})\\ +({\beta }_n-{\beta }_{n-1})S_{n-1}{x}_{n-1}\parallel +{\beta }_n{\parallel S_n-S_{n-1}\parallel }_{\mathrm{\infty },B}\\ \le \parallel x_n-x_{n-1}\parallel +|{\beta }_n-{\beta }_{n-1}|M+{\beta }_n{\parallel S_n-S_{n-1}\parallel }_{\mathrm{\infty },B}.\end{array}$$

(3.4)

Now, using (3.4) in (3.3), we obtain that

$$\begin{array}{r}\parallel x_{n+1}-x_n\parallel \\ \le \parallel u_n-u_{n-1}\parallel \\ \le {\alpha }_n\lambda \parallel x_n-x_{n-1}\parallel +(1-{\alpha }_n)[\parallel x_n-x_{n-1}\parallel \\ +|{\beta }_n-{\beta }_{n-1}|M+{\beta }_n{\parallel S_n-S_{n-1}\parallel }_{\mathrm{\infty },B}]+({\alpha }_{n-1}-{\alpha }_n)M+{\gamma }_n\\ \le (1-{\alpha }_n(1-\lambda ))\parallel x_n-x_{n-1}\parallel +M[({\alpha }_{n-1}-{\alpha }_n)+|{\beta }_n-{\beta }_{n-1}|]\\ +(1-{\alpha }_n){\beta }_n{\parallel S_n-S_{n-1}\parallel }_{\mathrm{\infty },B}+{\gamma }_n\\ \le (1-{\alpha }_n(1-\lambda ))\parallel x_n-x_{n-1}\parallel +M[({\alpha }_{n-1}-{\alpha }_n)+|{\beta }_n-{\beta }_{n-1}|]\\ +{\beta }_n{\parallel S_n-S_{n-1}\parallel }_{\mathrm{\infty },B}+{\gamma }_n.\end{array}$$

(3.5)

Thus, by using conditions (i), (v), ${\sum }_{n=1}^{\mathrm{\infty }}({\alpha }_{n-1}-{\alpha }_n)<\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}|{\beta }_{n-1}-{\beta }_n|<\mathrm{\infty }$ and applying Lemma 2.4, we conclude that

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

(3.6)

Step 3. We claim ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T_i{x}_n\parallel =0$ for all $i\in \mathbb{N}$.

Since $T_i{x}_n\in C$ for all $i\in \mathbb{N}$ and ${\sum }_{i=1}^n({\alpha }_{i-1}-{\alpha }_i)+{\alpha }_n=1$, we get

$$\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)T_i{x}_n+{\alpha }_{n}z\in C\text{for all }z\in C.$$

Noticing $x_{n+1}=P_C(u_n)$ and fixing $z\in {\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)$, from (3.1) we have

$$\begin{array}{r}\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)(x_n-T_i{x}_n)\\ =P_C(u_n)+(1-{\alpha }_n)x_n-(\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)T_i{x}_n+{\alpha }_{n}z)\\ +{\alpha }_{n}z-x_{n+1}\\ =P_C(u_n)-P_C(\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)T_i{x}_n+{\alpha }_{n}z)\\ +(1-{\alpha }_n)(x_n-x_{n+1})+{\alpha }_n(z-x_{n+1}).\end{array}$$

Hence,

$$\begin{array}{r}\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)〈x_n-T_i{x}_n,x_n-p〉\\ =〈P_C(u_n)-P_C(\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)T_i{x}_n+{\alpha }_{n}z),x_n-p〉\\ +〈(1-{\alpha }_n)(x_n-x_{n+1}),x_n-p〉+{\alpha }_n〈z-x_{n+1},x_n-p〉\\ \le \parallel u_n-\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)T_i{x}_n-{\alpha }_{n}z\parallel \parallel x_n-p\parallel \\ +(1-{\alpha }_n)\parallel x_n-x_{n+1}\parallel \parallel x_n-p\parallel +{\alpha }_n\parallel z-x_{n+1}\parallel \parallel x_n-p\parallel \\ =\parallel {\alpha }_n(f(x_n)-z)+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)(T_i{y}_n-T_i{x}_n)\parallel \parallel x_n-p\parallel \\ +(1-{\alpha }_n)\parallel x_n-x_{n+1}\parallel \parallel x_n-p\parallel +{\alpha }_n\parallel z-x_{n+1}\parallel \parallel x_n-p\parallel \\ \le {\alpha }_n\parallel f(x_n)-z\parallel \parallel x_n-p\parallel +\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)(\parallel y_n-x_n\parallel +a_i)\parallel x_n-p\parallel \\ +(1-{\alpha }_n)\parallel x_n-x_{n+1}\parallel \parallel x_n-p\parallel +{\alpha }_n\parallel z-x_{n+1}\parallel \parallel x_n-p\parallel \\ \le {\alpha }_n\parallel f(x_n)-z\parallel \parallel x_n-p\parallel +(1-{\alpha }_n)\parallel y_n-x_n\parallel \parallel x_n-p\parallel \\ +(1-{\alpha }_n)\parallel x_n-x_{n+1}\parallel \parallel x_n-p\parallel \\ +{\alpha }_n\parallel z-x_{n+1}\parallel \parallel x_n-p\parallel +{\gamma }_n\parallel x_n-p\parallel \\ \le {\alpha }_n\parallel f(x_n)-z\parallel \parallel x_n-p\parallel +(1-{\alpha }_n){\beta }_n\parallel S_n{x}_n-x_n\parallel \parallel x_n-p\parallel \\ +(1-{\alpha }_n)\parallel x_n-x_{n+1}\parallel \parallel x_n-p\parallel +{\alpha }_n\parallel z-x_{n+1}\parallel \parallel x_n-p\parallel +{\gamma }_{n}R\\ \le (2{\alpha }_n+{\beta }_n)M^{\prime }+(1-{\alpha }_n)\parallel x_n-x_{n+1}\parallel R+{\gamma }_{n}R,\end{array}$$

(3.7)

where R is a positive constant such that $\parallel x_n-p\parallel \le R$ for all $n\in \mathbb{N}$ and

$$M^{\prime }=\underset{n\in \mathbb{N}}{sup}\{\parallel f(x_n)-z\parallel \parallel x_n-p\parallel ,\parallel S_n{x}_n-x_n\parallel \parallel x_n-p\parallel ,\parallel z-x_{n+1}\parallel \parallel x_n-p\parallel \}.$$

Set ${\xi }_n:=\frac{1}{2}(a_n+2\parallel x_n-p\parallel )a_n$. From Proposition 2.2, using (3.7), we obtainthat

$$\begin{array}{r}\frac{1}{2}\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i){\parallel x_n-T_i{x}_n\parallel }^2\\ \le \sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)〈x_n-T_i{x}_n,x_n-p〉+{\xi }_n\\ \le (2{\alpha }_n+{\beta }_n)M^{\prime }+(1-{\alpha }_n)\parallel x_n-x_{n+1}\parallel R+{\gamma }_{n}R.\end{array}$$

Using (3.6), condition (i), ${lim}_{n\to \mathrm{\infty }}\frac{{\gamma }_n}{{\beta }_n}=0$ and ${lim}_{n\to \mathrm{\infty }}\frac{{\beta }_n}{{\alpha }_n}=\tau \in (0,\mathrm{\infty })$, we have

$$\underset{n\to \mathrm{\infty }}{lim}\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)\parallel x_n-T_i{x}_n\parallel =0.$$

Since $({\alpha }_{i-1}-{\alpha }_i)\parallel x_n-T_i{x}_n\parallel \le {\sum }_{i=1}^n({\alpha }_{i-1}-{\alpha }_i)\parallel x_n-T_i{x}_n\parallel$ for all $i\in \mathbb{N}$ and $\{{\alpha }_n\}$ is strictly decreasing, we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T_i{x}_n\parallel =0\text{for all }i\in \mathbb{N}.$$

Step 4. $\parallel y_n-T_i{y}_n\parallel \to 0$ as $n\to \mathrm{\infty }$ for all $i\in \mathbb{N}$.

Noticing that ${lim}_{n\to \mathrm{\infty }}\frac{{\beta }_n}{{\alpha }_n}=\tau \in (0,\mathrm{\infty })$ and using condition (i), we have${\beta }_n\to 0$ as $n\to \mathrm{\infty }$. Therefore, we obtain that

$$\parallel y_n-x_n\parallel ={\beta }_n\parallel S_n{x}_n-x_n\parallel \to 0\text{as }n\to \mathrm{\infty }.$$

(3.8)

So that for all $i\in \mathbb{N}$, we have

$$\parallel y_n-T_i{x}_n\parallel \le \parallel y_n-x_n\parallel +\parallel x_n-T_i{x}_n\parallel \to 0\text{as }n\to \mathrm{\infty }.$$

(3.9)

Since each $T_i$ is uniformly continuous, from (3.8) and (3.9), wehave

$$\parallel y_n-T_i{y}_n\parallel \le \parallel y_n-T_i{x}_n\parallel +\parallel T_i{x}_n-T_i{y}_n\parallel \to 0\text{as }n\to \mathrm{\infty }$$

for all $i\in \mathbb{N}$.

Step 5. ${lim}_{n\to \mathrm{\infty }}\frac{\parallel x_{n+1}-x_n\parallel }{{\beta }_n}=0$ and ${lim}_{n\to \mathrm{\infty }}\frac{\parallel u_n-u_{n-1}\parallel }{{\beta }_n}=\frac{\parallel u_n-u_{n-1}\parallel }{{\alpha }_n}=0$.

From (3.5) we obtain that

$$\begin{array}{r}\frac{\parallel x_{n+1}-x_n\parallel }{{\beta }_n}\\ \le \frac{\parallel u_n-u_{n-1}\parallel }{{\beta }_n}\\ \le (1-{\alpha }_n(1-\lambda ))\frac{\parallel x_n-x_{n-1}\parallel }{{\beta }_n}\\ +M(\frac{({\alpha }_{n-1}-{\alpha }_n)}{{\beta }_n}+\frac{|{\beta }_n-{\beta }_{n-1}|}{{\beta }_n})+{\parallel S_n-S_{n-1}\parallel }_{\mathrm{\infty },B}+\frac{{\gamma }_n}{{\beta }_n}\\ =(1-{\alpha }_n(1-\lambda ))\frac{\parallel x_n-x_{n-1}\parallel }{{\beta }_{n-1}}+(1-{\alpha }_n(1-\lambda ))\parallel x_n-x_{n-1}\parallel (\frac{1}{{\beta }_n}-\frac{1}{{\beta }_{n-1}})\\ +M(\frac{({\alpha }_{n-1}-{\alpha }_n)}{{\beta }_n}+\frac{|{\beta }_n-{\beta }_{n-1}|}{{\beta }_n})+{\parallel S_n-S_{n-1}\parallel }_{\mathrm{\infty },B}+\frac{{\gamma }_n}{{\beta }_n}.\end{array}$$

(3.10)

We observe that

$$(1-{\alpha }_n(1-\lambda ))(\frac{1}{{\beta }_n}-\frac{1}{{\beta }_{n-1}})\le {\alpha }_n\frac{1}{{\alpha }_n}|\frac{1}{{\beta }_n}-\frac{1}{{\beta }_{n-1}}|\le {\alpha }_{n}N.$$

Set

$$\begin{array}{r}{\mu }_n={\alpha }_n(1-\lambda ),\\ {\phi }_n={\alpha }_{n}N\parallel x_n-x_{n-1}\parallel +M(\frac{({\alpha }_{n-1}-{\alpha }_n)}{{\beta }_n}+\frac{|{\beta }_n-{\beta }_{n-1}|}{{\beta }_n}),\\ {\nu }_n={\parallel S_n-S_{n-1}\parallel }_{\mathrm{\infty },B}+\frac{{\gamma }_n}{{\beta }_n}.\end{array}$$

From (3.10) we obtain that

$$\begin{array}{rl}\frac{\parallel x_{n+1}-x_n\parallel }{{\beta }_n}\le & \frac{\parallel u_n-u_{n-1}\parallel }{{\beta }_n}\\ \le & (1-{\alpha }_n(1-\lambda ))\frac{\parallel x_n-x_{n-1}\parallel }{{\beta }_{n-1}}+{\alpha }_{n}N\parallel x_n-x_{n-1}\parallel \\ +M(\frac{({\alpha }_{n-1}-{\alpha }_n)}{{\beta }_n}+\frac{|{\beta }_n-{\beta }_{n-1}|}{{\beta }_n})+{\parallel S_n-S_{n-1}\parallel }_{\mathrm{\infty },B}+\frac{{\gamma }_n}{{\beta }_n}\\ \le & (1-{\alpha }_n(1-\lambda ))\frac{\parallel u_{n-1}-u_{n-2}\parallel }{{\beta }_{n-1}}+{\alpha }_{n}N\parallel x_n-x_{n-1}\parallel \\ +M(\frac{({\alpha }_{n-1}-{\alpha }_n)}{{\beta }_n}+\frac{|{\beta }_n-{\beta }_{n-1}|}{{\beta }_n})+{\parallel S_n-S_{n-1}\parallel }_{\mathrm{\infty },B}+\frac{{\gamma }_n}{{\beta }_n}\\ \le & (1-{\mu }_n)\frac{\parallel u_{n-1}-u_{n-2}\parallel }{{\beta }_{n-1}}+{\phi }_n+{\nu }_n.\end{array}$$

Using conditions (i), (iii), (v) and applying Lemma 2.4, we have

$$\underset{n\to \mathrm{\infty }}{lim}\frac{\parallel x_{n+1}-x_n\parallel }{{\beta }_n}=0\text{and}\underset{n\to \mathrm{\infty }}{lim}\frac{\parallel u_n-u_{n-1}\parallel }{{\beta }_n}=\frac{\parallel u_n-u_{n-1}\parallel }{{\alpha }_n}=0.$$

Step 6. Set $v_n:=\frac{x_n-x_{n+1}}{(1-{\alpha }_n){\beta }_n}$ and $u_n:={\alpha }_{n}f(x_n)+{\sum }_{i=1}^n({\alpha }_{i-1}-{\alpha }_i)T_i{y}_n$ for all $n\in \mathbb{N}$. Then, for any $z\in {\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)$, we can calculate $〈v_n,x_n-z〉$.

From (3.1) we have

$$x_{n+1}=P_C(u_n)-u_n+{\alpha }_{n}f(x_n)+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)(T_i{y}_n-y_n)+(1-{\alpha }_n)y_n,$$

which gives that

$$\begin{array}{r}{x}_n-x_{n+1}\\ =(1-{\alpha }_n)x_n+{\alpha }_n{x}_n\\ -(P_C(u_n)-u_n+{\alpha }_{n}f(x_n)+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)(T_i{y}_n-y_n)+(1-{\alpha }_n)y_n)\\ =(1-{\alpha }_n){\beta }_n(x_n-S_n{x}_n)+(u_n-P_C(u_n))\\ +\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)(y_n-T_i{y}_n)+{\alpha }_n(x_n-f(x_n)).\end{array}$$

Then we conclude that

$$\begin{array}{r}\frac{x_n-x_{n+1}}{(1-{\alpha }_n){\beta }_n}\\ =x_n-S_n{x}_n+\frac{1}{(1-{\alpha }_n){\beta }_n}(u_n-P_C(u_n))\\ +\frac{1}{(1-{\alpha }_n){\beta }_n}\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)(y_n-T_i{y}_n)+\frac{{\alpha }_n}{(1-{\alpha }_n){\beta }_n}(x_n-f(x_n)).\end{array}$$

Noticing that $x_{n+1}=P_C(u_n)$. For any $z\in {\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)$, we have

$$\begin{array}{r}〈v_n,x_n-z〉\\ =\frac{1}{(1-{\alpha }_n){\beta }_n}〈u_n-P_C(u_n),P_C(u_{n-1})-z〉+〈x_n-S_n{x}_n,x_n-z〉\\ +\frac{1}{(1-{\alpha }_n){\beta }_n}\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)〈y_n-T_i{y}_n,x_n-z〉\\ +\frac{{\alpha }_n}{(1-{\alpha }_n){\beta }_n}〈(I-f)x_n,x_n-z〉.\end{array}$$

(3.11)

By using Lemma 2.2, we obtain that

$$\begin{array}{r}〈x_n-S_n{x}_n,x_n-z〉\\ =〈(I-S_n)x_n-(I-S_n)z,x_n-z〉+〈(I-S_n)z,x_n-z〉\\ \ge 〈(I-S_n)z,x_n-z〉,\end{array}$$

(3.12)

$$\begin{array}{r}〈(I-f)x_n,x_n-z〉\\ =〈(I-f)x_n-(I-f)z,x_n-z〉+〈(I-f)z,x_n-z〉\\ \ge (1-\lambda ){\parallel x_n-z\parallel }^2+〈(I-f)z,x_n-z〉,\end{array}$$

(3.13)

$$\begin{array}{r}〈y_n-T_i{y}_n,x_n-z〉\\ =〈(I-T_i)y_n-(I-T_i)z,x_n-y_n〉+〈(I-T_i)y_n-(I-T_i)z,y_n-z〉\\ \ge 〈(I-T_i)y_n-(I-T_i)z,x_n-y_n〉-a_i\parallel y_n-z\parallel \\ ={\beta }_n〈(I-T_i)y_n,x_n-S_n{x}_n〉-a_i\parallel y_n-z\parallel \text{for all }i\in \mathbb{N}.\end{array}$$

(3.14)

Also, from Lemma 2.1 we get that

$$\begin{array}{rl}〈u_n-P_C(u_n),P_C(u_{n-1})-z〉=& 〈u_n-P_C(u_n),P_C(u_{n-1})-P_C(u_n)〉\\ +〈u_n-P_C(u_n),P_C(u_n)-z〉\\ \ge & 〈u_n-P_C(u_n),P_C(u_{n-1})-P_C(u_n)〉.\end{array}$$

(3.15)

Substituting (3.12), (3.13), (3.14) and (3.15) into (3.11), we have

$$\begin{array}{r}〈v_n,x_n-z〉\\ \ge \frac{1}{(1-{\alpha }_n){\beta }_n}〈u_n-P_C(u_n),P_C(u_{n-1})-P_C(u_n)〉+〈(I-S_n)z,x_n-z〉\\ +\frac{1}{(1-{\alpha }_n)}\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)[〈(I-T_i)y_n,x_n-S_n{x}_n〉-a_i\parallel y_n-z\parallel ]\\ +\frac{{\alpha }_n}{(1-{\alpha }_n){\beta }_n}〈(I-f)z,x_n-z〉+\frac{{\alpha }_n(1-\lambda )}{(1-{\alpha }_n){\beta }_n}{\parallel x_n-z\parallel }^2.\end{array}$$

(3.16)

Step 7. $x_n\to \tilde{x}$, where $\tilde{x}$ is a strong cluster point of the sequence$\{x_n\}$.

From (3.16) we observe that

$$\begin{array}{rl}{\parallel x_n-z\parallel }^2\le & \frac{(1-{\alpha }_n){\beta }_n}{{\alpha }_n(1-\lambda )}[〈v_n,x_n-z〉-〈(I-S_n)z,x_n-z〉]\\ -\frac{1}{1-\lambda }〈(I-f)z,x_n-z〉\\ -\frac{{\beta }_n}{{\alpha }_n(1-\lambda )}\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)[〈(I-T_i)y_n,x_n-S_n{x}_n〉-a_i\parallel y_n-z\parallel ]\\ +\frac{\parallel u_n-u_{n-1}\parallel }{{\alpha }_n(1-\lambda )}\parallel u_n-P_C(u_n)\parallel .\end{array}$$

Noticing that $v_n=\frac{x_n-x_{n+1}}{(1-{\alpha }_n){\beta }_n}\to 0$, $\frac{\parallel u_n-u_{n-1}\parallel }{{\alpha }_n}\to 0$, and for all $i\in \mathbb{N}$, $(I-T_i)y_n\to 0$ as $n\to \mathrm{\infty }$. It is easy to see that a weak cluster point of$\{x_n\}$ is also a strong cluster point. Note that thesequence $\{x_n\}$ is bounded, then there exists a subsequence$\{x_{n_k}\}$ of $\{x_n\}$ which converges to a point $x^{\ast }\in C$. We also have, for all $i\in \mathbb{N}$, $(I-T_i)x_n\to 0$ as $n\to \mathrm{\infty }$. By using condition (G), we get that$x^{\ast }\in F(T)$. Thus, for all $z\in {\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)$, we obtain that

$$\begin{array}{r}〈(I-f)x_{n_k},x_{n_k}-z〉\\ \le \frac{(1-{\alpha }_{n_k}){\beta }_{n_k}}{{\alpha }_{n_k}}〈v_{n_k},x_{n_k}-z〉-\frac{1}{{\alpha }_{n_k}}\parallel u_{n_k}-P_C(u_{n_k})\parallel \parallel u_{n_k-1}-u_{n_k}\parallel \\ -\frac{(1-{\alpha }_{n_k}){\beta }_{n_k}}{{\alpha }_{n_k}}〈(I-S_{n_k})z,x_{n_k}-z〉\\ -\frac{{\beta }_{n_k}}{{\alpha }_{n_k}}\sum _{i=1}^{n_k}({\alpha }_{i-1}-{\alpha }_i)〈(I-T_i)y_{n_k},x_{n_k}-S_{n_k}{x}_{n_k}〉+\frac{{\beta }_{n_k}{\gamma }_{n_k}}{{\alpha }_{n_k}}\parallel y_{n_k}-z\parallel .\end{array}$$

Now, letting $k\to \mathrm{\infty }$, we have

$$〈(I-f)x^{\ast },x^{\ast }-z〉\le -\tau 〈(I-S)z,x^{\ast }-z〉\text{for all }z\in \bigcap _{n=1}^{\mathrm{\infty }}F(T_n).$$

Now, since variational inequality problem (3.2) has the unique solution, then we getthat ${\omega }_w(x_n)=\{\tilde{x}\}$. Note that every weak cluster point of the sequence$\{x_n\}$ is also a strong cluster point. Then we have${lim}_{n\to \mathrm{\infty }}{x}_n=\tilde{x}$. □

Recently, Marino et al.[28] used a different approach to obtain the convergence of a more generaliterative method that involves an equilibrium problem. We now present the result ofGu et al. [[18], Theorem 3.5] as a corollary.

Corollary 3.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let$f:C\to H$be a λ-contraction. Let$S:C\to C$be a nonexpansive mapping, and let$\mathcal{T}={\{T_n\}}_{n=1}^{\mathrm{\infty }}$be a countable family of nonexpansive mappings from C into itself such that$F(\mathcal{T})={\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)\ne \mathrm{\varnothing }$. For arbitrary$x_1\in C$, consider the sequence$\{x_n\}$generated by the following iterative process:

$$\{\begin{array}{c}{x}_1\in C,\hfill \\ y_n={\beta }_{n}S{x}_n+(1-{\beta }_n)x_n,\hfill \\ x_{n+1}=P_C[{\alpha }_{n}f(x_n)+{\sum }_{i=1}^n({\alpha }_{i-1}-{\alpha }_i)T_i{y}_n]\hfill \end{array}$$

for all$n\in \mathbb{N}$, where${\alpha }_0=1$, $\{{\alpha }_n\}$is a strictly decreasing sequence in$(0,1)$and$\{{\beta }_n\}$is a sequence in$(0,1)$satisfying conditions (i)-(iv) of Theorem  3.1. Then thesequence$\{x_n\}$converges strongly to a point$x^{\ast }\in {\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)$, which is the unique solution of the followingvariational inequality:

$$〈\frac{1}{\tau }(I-f)x^{\ast }+(I-S)x^{\ast },x-x^{\ast }〉\ge 0\mathit{\text{for all}}x\in \bigcap _{n=1}^{\mathrm{\infty }}F(T_n).$$

Again, we present the result of Yao et al. [[17], Theorem 3.2] as a corollary.

Corollary 3.2 Let C be a nonempty closed convex subset of a real Hilbert space H. Let$f:C\to H$be a λ-contraction. Let$S:C\to C$be a nonexpansive mapping, and let$T:C\to C$be a nonexpansive mapping such that$F(T)\ne \mathrm{\varnothing }$. For arbitrary$x_1\in C$, consider the sequence$\{x_n\}$generated by the following iterative process:

$$\{\begin{array}{c}{x}_1\in C,\hfill \\ y_n=(1-{\beta }_n)x_n+{\beta }_{n}S{x}_n,\hfill \\ x_{n+1}=P_C[{\alpha }_{n}f(x_n)+(1-{\alpha }_n)T{y}_n]\hfill \end{array}$$

for all$n\in \mathbb{N}$, where$\{{\alpha }_n\}$and$\{{\beta }_n\}$are two sequences in$(0,1)$satisfying conditions (i)-(iv) of Theorem  3.1. Then thesequence$\{x_n\}$converges strongly to a point$x^{\ast }\in F(T)$, which is the unique solution of the followingvariational inequality:

$$〈\frac{1}{\tau }(I-f)x^{\ast }+(I-S)x^{\ast },x-x^{\ast }〉\ge 0\mathit{\text{for all}}x\in F(T).$$

Theorem 3.2 Let C be a nonempty closed convex subset of a real Hilbert space H. Let$f:C\to H$be a λ-contraction, and let$\{S_n\}$be a sequence of nonexpansive mappings from C into itself. Let S be a nonexpansive mapping from C into itself such that${lim}_{n\to \mathrm{\infty }}{S}_{n}x=Sx$for all$x\in C$. Let$t_1,t_2,t_3,\dots ,t_N>0$such that${\sum }_{i=1}^N{t}_i=1$. Let$T_1,T_2,T_3,\dots ,T_N:C\to C$be nonexpansive mappings such that${\bigcap }_{i=1}^{N}F(T_i)\ne \mathrm{\varnothing }$, and assume that$T={\sum }_{i=1}^N{t}_i{T}_i$. For arbitrary$x_1\in C$, consider the sequence$\{x_n\}$generated by the following iterative process:

$$\{\begin{array}{c}{x}_1\in C,\hfill \\ y_n={\beta }_n{S}_n{x}_n+(1-{\beta }_n)x_n,\hfill \\ x_{n+1}=P_C[{\alpha }_{n}f(x_n)+(1-{\alpha }_n){\sum }_{i=1}^N{t}_i{T}_i{y}_n]\hfill \end{array}$$

for all$n\in \mathbb{N}$, where$\{{\alpha }_n\}$and$\{{\beta }_n\}$are two sequences in$(0,1)$satisfying conditions (i)-(iv) of Theorem  3.1. Then thesequence$\{x_n\}$converges strongly to a point$x^{\ast }\in {\bigcap }_{i=1}^{N}F(T_i)$, which is the unique solution of the followingvariational inequality:

$$〈\frac{1}{\tau }(I-f)x^{\ast }+(I-S)x^{\ast },x-x^{\ast }〉\ge 0\mathit{\text{for all}}x\in \bigcap _{i=1}^{N}F(T_i).$$

Proof The proof follows from Lemma 2.5 andCorollary 3.2. □

4 Numerical example

We present an example to show the effectiveness and convergence of the sequencegenerated by the considered iterative scheme.

Example 4.1 Let $H=\mathbb{R}$ and $C=[0,1]$. Let T be a self-mapping defined by$Tx=1-x$ for all $x\in C$. Let $f:C\to H$ be a contraction mapping defined by$f(x)=\frac{2}{3}x-\frac{1}{6}$ for all $x\in C$, and let $\{S_n\}$ be a sequence of nonexpansive mappings fromC into itself defined by $S_{n}x=x+\frac{1}{2n}$ such that $Sx={lim}_{n\to \mathrm{\infty }}{S}_{n}x$ for all $x\in C$, where S is a nonexpansive mapping definedby $Sx=x$ for all $x\in C$. Define sequences $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ in $(0,1)$ by ${\alpha }_n={\beta }_n=\frac{1}{{(n+1)}^{1/2}}$. Without loss of generality, we may assume that$a_n=\frac{1}{n^{3/2}}$ for all $n\in \mathbb{N}$. For each $n\in \mathbb{N}$, define $T_n:C\to C$ by

$$T_{n}x=\{\begin{array}{cc}1-x\hfill & \text{if }x\in [0,1),\hfill \\ a_n\hfill & \text{if }x=1.\hfill \end{array}$$

It was shown in [19] that $\mathcal{T}=\{T_n\}$ is a sequence of nearly nonexpansive mappings fromC into itself such that $F(\mathcal{T})=\{\frac{1}{2}\}$ and $Tx={lim}_{n\to \mathrm{\infty }}{T}_{n}x$ for all $x\in C$, where T is a nonexpansive mapping.

One can observe that all the assumptions of Theorem 3.1 are satisfied, and thesequence $\{x_n\}$ generated by (3.1) converges to a unique solution$\frac{1}{2}$ of variational inequality (3.2) over$F(\mathcal{T})$.