Algorithms of common solutions for a variational inequality, a split equilibrium problem and a hierarchical fixed point problem

Abstract

In this paper, we suggest and analyze an iterative scheme for finding an approximate element of the common set of solutions of a split equilibrium problem, a variational inequality problem and a hierarchical fixed point problem in a real Hilbert space. We also consider the strong convergence of the proposed method under some conditions. Results proved in this paper may be viewed as an improvement and refinement of the previously known results.

MSC:49J30, 47H09, 47J20.

Dedication

Dedicated to Professor Bingsheng He on the occasion of his sixty-fifth birthday

1 Introduction

Let H be a real Hilbert space, whose inner product and norm are denoted by $〈\cdot ,\cdot 〉$ and $\parallel \cdot \parallel$. Let C be a nonempty closed convex subset of H and D be a mapping from C into H. A classical variational inequality problem, denoted by $\mathit{VI}(D,C)$, is to find a vector $u\in C$ such that

$$〈v-u,Du〉\ge 0,\mathrm{\forall }v\in C.$$

(1)

The solution of $\mathit{VI}(D,C)$ is denoted by ${\mathrm{\Omega }}^{\ast }$. It is easy to observe that

$$u^{\ast }\in {\mathrm{\Omega }}^{\ast }⟺u^{\ast }=P_C[u^{\ast }-\lambda D{u}^{\ast }],\text{where }\lambda >0.$$

This alternative formulation has played a significant part in developing various projection-type methods for solving variational inequalities. We now have a variety of techniques to suggest and analyze various iterative algorithms for solving variational inequalities and the related optimization problems; see [1–29].

We introduce the following definitions which are useful in the following analysis.

Definition 1.1 The mapping $T:C\to H$ is said to be

• (a) monotone if

  $$〈Tx-Ty,x-y〉\ge 0,\mathrm{\forall }x,y\in C;$$

• (b) strongly monotone if there exists $\alpha >0$ such that

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

• (c) α-inverse strongly monotone if there exists $\alpha >0$ such that

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

• (d) nonexpansive if

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

• (e) k-Lipschitz continuous if there exists a constant $k>0$ such that

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

• (f) contraction on C if there exists a constant $0\le k<1$ such that

  $$\parallel Tx-Ty\parallel \le k\parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

It is easy to observe that every α-inverse strongly monotone T is monotone and Lipschitz continuous. It is well known that every nonexpansive operator $T:H_1\to H_1$ satisfies, for all $(x,y)\in H_1\times H_1$, the inequality

$$〈(x-T(x))-(y-T(y)),T(y)-T(x)〉\le \frac{1}{2}{\parallel (T(x)-x)-(T(y)-y)\parallel }^2$$

(2)

and therefore we get, for all $(x,y)\in H_1\times F(T)$,

$$〈x-T(x),y-T(x)〉\le \frac{1}{2}{\parallel T(x)-x\parallel }^2;$$

(3)

see, e.g., [9], Theorem 1 and [10], Theorem 3.

A mapping $T:C\to H$ is called a k-strict pseudo-contraction if there exists a constant $0\le k<1$ such that

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

(4)

The fixed point problem for the mapping T is to find $x\in C$ such that

$$Tx=x.$$

(5)

We denote by $F(T)$ the set of solutions of (5). It is well known that the class of strict pseudo-contractions strictly includes the class of nonexpansive mappings, then $F(T)$ is closed and convex and $P_{F(T)}$ is well defined (see [29]).

The equilibrium problem denoted by EP is to find $x\in C$ such that

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

(6)

The solution set of (6) is denoted by $\mathit{EP}(F)$. Numerous problems in physics, optimization and economics reduce to finding a solution of (6); see [7, 12, 23, 24]. In 1997, Combettes and Hirstoaga [8] introduced an iterative scheme of finding the best approximation to the initial data when $\mathit{EP}(F)$ is nonempty. Recently Plubtieng and Punpaeng [23] introduced an iterative method for finding the common element of the set $F(T)\cap {\mathrm{\Omega }}^{\ast }\cap \mathit{EP}(F)$.

Recently, Censor et al. [4] introduced a new variational inequality problem which we call the split variational inequality problem (SVIP). Let $H_1$ and $H_2$ be two real Hilbert spaces. Given operators $f:H_1\to H_1$ and $g:H_2\to H_2$, a bounded linear operator $A:H_1\to H_2$, and nonempty, closed and convex subsets $C\subseteq H_1$ and $Q\subseteq H_2$, the SVIP is formulated as follows: Find a point $x^{\ast }\in C$ such that

$$〈f\left(x^{\ast }\right),x-x^{\ast }〉\ge 0\text{for all }x\in C$$

(7)

and such that

$$y^{\ast }=A{x}^{\ast }\in Q\text{solves}〈g\left(y^{\ast }\right),y-y^{\ast }〉\ge 0\text{for all }y\in Q.$$

(8)

In [22], Moudafi introduced an iterative method which can be regarded as an extension of the method given by Censor et al. [4] for the following split monotone variational inclusions:

$$\text{Find }{x}^{\ast }\in H_1\text{ such that }0\in f\left(x^{\ast }\right)+B_1\left(x^{\ast }\right)$$

and such that

$$y^{\ast }=A{x}^{\ast }\in H_2\text{solves}0\in g\left(y^{\ast }\right)+B_2\left(y^{\ast }\right),$$

where $B_i:H_i\to 2^{H_i}$ is a set-valued mapping for $i=1,2$. Later Byrne et al. [3] generalized and extended the work of Censor et al. [4] and Moudafi [22].

Very recently, Kazmi and Rivzi [13] studied the following pair of equilibrium problems called a split equilibrium problem: Let $F_1:C\times C\to R$ and $F_2:Q\times Q\to R$ be nonlinear bifunctions and $A:H_1\to H_2$ be a bounded linear operator, then the split equilibrium problem (SEP) is to find $x^{\ast }\in C$ such that

$$F_1(x^{\ast },x)\ge 0,\mathrm{\forall }x\in C$$

(9)

and such that

$$y^{\ast }=A{x}^{\ast }\in Q\text{solves}{F}_2(y^{\ast },y)\ge 0,\mathrm{\forall }y\in Q.$$

(10)

The solution set of SEP (9)-(10) is denoted by $\mathrm{\Lambda }=\{p\in \mathit{EP}(F_1):Ap\in \mathit{EP}(F_2)\}$.

Let $S:C\to H$ be a nonexpansive mapping. The following problem is called a hierarchical fixed point problem: Find $x\in F(T)$ such that

$$〈x-Sx,y-x〉\ge 0,\mathrm{\forall }y\in F(T).$$

(11)

It is known that the hierarchical fixed point problem (11) links with some monotone variational inequalities and convex programming problems; see [11, 27]. Various methods have been proposed to solve the hierarchical fixed point problem; see Moudafi [21], Mainge and Moudafi in [15], Marino and Xu in [17] and Cianciaruso et al. [5]. In 2010, Yao et al. [27] introduced the following strong convergence iterative algorithm to solve problem (11):

$$\begin{array}{r}{y}_n={\beta }_{n}S{x}_n+(1-{\beta }_n)x_n,\\ x_{n+1}=P_C[{\alpha }_{n}f(x_n)+(1-{\alpha }_n)T{y}_n],\mathrm{\forall }n\ge 0,\end{array}$$

(12)

where $f:C\to H$ is a contraction mapping and $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ are two sequences in $(0,1)$. Under some certain restrictions on parameters, Yao et al. proved that the sequence $\{x_n\}$ generated by (12) converges strongly to $z\in F(T)$, which is the unique solution of the following variational inequality:

$$〈(I-f)z,y-z〉\ge 0,\mathrm{\forall }y\in F(T).$$

(13)

By changing the restrictions on parameters, the authors obtained another result on the iterative scheme (12), the sequence $\{x_n\}$ generated by (12) converges strongly to a point $z\in F(T)$, which is the unique solution of the following variational inequality:

$$〈\frac{1}{\tau }(I-f)z+(I-S)z,y-z〉\ge 0,\mathrm{\forall }y\in F(T).$$

(14)

Let $S:C\to H$ be a nonexpansive mapping and ${\{T_i\}}_{i=1}^{\mathrm{\infty }}:C\to C$ be a countable family of nonexpansive mappings. In 2011, Gu et al. [11] introduced the following iterative algorithm:

$$\begin{array}{r}{y}_n=P_C[{\beta }_{n}S{x}_n+(1-{\beta }_n)x_n],\\ x_{n+1}=P_C[{\alpha }_{n}f(x_n)+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)T_i{y}_n],\mathrm{\forall }n\ge 1,\end{array}$$

(15)

where ${\alpha }_0=1$, $\{{\alpha }_n\}$ is a strictly decreasing sequence in $(0,1)$ and $\{{\beta }_n\}$ is a sequence in $(0,1)$. Under some certain conditions on parameters, Gu et al. proved that the sequence $\{x_n\}$ generated by (15) converges strongly to $z\in {\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)$, which is the unique solution of one of variational inequalities (13) and (14).

In this paper, motivated by the work of Censor et al. [4], Moudafi [22], Byrne et al. [3] Kazmi and Rivzi [13], Yao et al. [27] and Gu et al. [11] and by the recent work going on in this direction, we give an iterative method for finding an approximate element of the common set of solutions of (1), (9)-(10) and (11) for a strictly pseudo-contraction mapping in a real Hilbert space. We establish a strong convergence theorem based on this method. The presented method improves and generalizes many known results for solving equilibrium problems, variational inequality problems and hierarchical fixed point problems; see, e.g., [5, 11, 15, 27] and relevant references cited therein.

2 Preliminaries

In this section, we list some fundamental lemmas that are useful in the consequent analysis. The first lemma provides some basic properties of the projection of H onto C.

Lemma 2.1 Let $P_C$ denote the projection of H onto C. Then we have the following inequalities,

$$〈z-P_C[z],P_C[z]-v〉\ge 0,\mathrm{\forall }z\in H,v\in C;$$

(16)

$$〈u-v,P_C[u]-P_C[v]〉\ge {\parallel P_C[u]-P_C[v]\parallel }^2,\mathrm{\forall }u,v\in H;$$

(17)

$$\parallel P_C[u]-P_C[v]\parallel \le \parallel u-v\parallel ,\mathrm{\forall }u,v\in H;$$

(18)

$${\parallel u-P_C[z]\parallel }^2\le {\parallel z-u\parallel }^2-{\parallel z-P_C[z]\parallel }^2,\mathrm{\forall }z\in H,u\in C.$$

(19)

Assumption 2.1 [2]

Let $F:C\times C\to \mathbb{R}$ be a bifunction satisfying the following assumptions:

• (i) $F(x,x)=0$, $\mathrm{\forall }x\in C$;

• (ii) F is monotone, i.e., $F(x,y)+F(y,x)\le 0$, $\mathrm{\forall }x,y\in C$;

• (iii) For each $x,y,z\in C$, ${lim}_{t\to 0}F(tz+(1-t)x,y)\le F(x,y)$;

• (iv) For each $x\in C$, $y\to F(x,y)$ is convex and lower semicontinuous;

• (v) Fixed $r>0$ and $z\in C$, there exists a bounded subset K of $H_1$ and $x\in C\cap K$ such that

  $$F(y,x)+\frac{1}{r}〈y-x,x-z〉\ge 0,\mathrm{\forall }y\in C\mathrm{\setminus }K.$$

Lemma 2.2 [8]

Assume that $F_1:C\times C\to \mathbb{R}$ satisfies Assumption  2.1. For $r>0$ and $\mathrm{\forall }x\in H_1$, define a mapping $T_r^{F_1}:H_1\to C$ as follows:

$$T_r^{F_1}(x)=\{z\in C:F_1(z,y)+\frac{1}{r}〈y-z,z-x〉\ge 0,\mathrm{\forall }y\in C\}.$$

Then the following hold:

• (i) $T_r^{F_1}$ is nonempty and single-valued;

• (ii) $T_r^{F_1}$ is firmly nonexpansive, i.e.,

  $${\parallel T_r^{F_1}(x)-T_r^{F_1}(y)\parallel }^2\le 〈T_r^{F_1}(x)-T_r^{F_1}(y),x-y〉,\mathrm{\forall }x,y\in H_1;$$

• (iii) $F(T_r^{F_1})=\mathit{EP}(F_1)$;

• (iv) $\mathit{EP}(F_1)$ is closed and convex.

Assume that $F_2:Q\times Q\to \mathbb{R}$ satisfies Assumption 2.1. For $s>0$ and $\mathrm{\forall }u\in H_2$, define a mapping $T_s^{F_2}:H_2\to Q$ as follows:

$$T_s^{F_2}(u)=\{v\in Q:F_2(v,w)+\frac{1}{s}〈w-v,v-u〉\ge 0,\mathrm{\forall }w\in Q\}.$$

Then $T_s^{F_2}$ satisfies conditions (i)-(iv) of Lemma 2.2. $F(T_s^{F_2})=\mathit{EP}(F_2,Q)$, where $\mathit{EP}(F_2,Q)$ is the solution set of the following equilibrium problem:

$$\text{Find }{y}^{\ast }\in Q\text{ such that }{F}_2(y^{\ast },y)\ge 0,\mathrm{\forall }y\in Q.$$

Lemma 2.3 [6]

Assume that $F_1:C\times C\to \mathbb{R}$ satisfies Assumption  2.1, and let $T_r^{F_1}$ be defined as in Lemma  2.2. Let $x,y\in H_1$ and $r_1,r_2>0$. Then

$$\parallel T_{r_2}^{F_1}(y)-T_{r_1}^{F_1}(x)\parallel \le \parallel y-x\parallel +\left|\frac{r_2-r_1}{r_2}\right|\parallel T_{r_2}^{F_1}(y)-y\parallel .$$

Lemma 2.4 [28]

Let C be a nonempty closed convex subset of a real Hilbert space H. If $T:C\to C$ is a k-strict pseudo-contraction, then:

• (i) The mapping $I-T$ is demiclosed at 0, i.e., if $\{x_n\}$ is a sequence in C weakly converging to x and if $\{(I-T)x_n\}$ converges strongly to 0, then $(I-T)x=0$;

• (ii) The set $F(T)$ of T is closed and convex so that the projection $P_{F(T)}$ is well defined.

Lemma 2.5 [16]

Let H be a real Hilbert space. Then the following inequality holds:

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

Lemma 2.6 [26]

Assume that $\{a_n\}$ is a sequence of nonnegative real numbers such that

$$a_{n+1}\le (1-{\gamma }_n)a_n+{\delta }_n,$$

where $\{{\gamma }_n\}$ is a sequence in $(0,1)$ and $\{{\delta }_n\}$ is a sequence such that

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

• (2) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\delta }_n/{\gamma }_n\le 0$ or ${\sum }_{n=1}^{\mathrm{\infty }}|{\delta }_n|<\mathrm{\infty }$.

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

Lemma 2.7 [1]

Let C be a closed convex subset of H. Let $\{x_n\}$ be a bounded sequence in H. Assume that

• (i) the weak w-limit set $w_w(x_n)\subset C$, where $w_w(x_n)=\{x:x_{n_i}\rightharpoonup x\}$;

• (ii) for each $z\in C$, ${lim}_{n\to \mathrm{\infty }}\parallel x_n-z\parallel$ exists.

Then $\{x_n\}$ is weakly convergent to a point in C.

Lemma 2.8 [29]

Let H be a Hilbert space, C be a closed and convex subset of H, and $T:C\to C$ be a k-strict pseudo-contraction mapping. Define a mapping $V:C\to H$ by $Vx=\lambda x+(1-\lambda )Tx$, $\mathrm{\forall }x\in C$. Then, as $k\le \lambda <1$, V is a nonexpansive mapping such that $F(V)=F(T)$.

Lemma 2.9 [11]

Let H be a Hilbert space, C be a closed and convex subset of H, and $T:C\to C$ be a nonexpansive mapping such that $F(T)\ne \mathrm{\varnothing }$. Then

$${\parallel Tx-x\parallel }^2\le 2〈x-Tx,x-x^{\prime }〉,\mathrm{\forall }{x}^{\prime }\in F(T),\mathrm{\forall }x\in C.$$

3 The proposed method and some properties

In this section, we suggest and analyze our method for finding common solutions of the variational inequality (1), the split equilibrium problem (9)-(10) and the hierarchical fixed point problem (11).

Let $H_1$ and $H_2$ be two real Hilbert spaces and $C\subseteq H_1$ and $Q\subseteq H_2$ be nonempty closed convex subsets of Hilbert spaces $H_1$ and $H_2$, respectively. Let $A:H_1\to H_2$ be a bounded linear operator. Let $D:C\to H_1$ be an α-inverse strongly monotone mapping. Assume that $F_1:C\times C\to \mathbb{R}$ and $F_2:Q\times Q\to \mathbb{R}$ are the bifunctions satisfying Assumption 2.1 and $F_2$ is upper semicontinuous in the first argument. Let $S:C\to H_1$ be a nonexpansive mapping and ${\{T_i\}}_{i=1}^{\mathrm{\infty }}:C\to C$ be a countable family of $k_i$-strict pseudo-contraction mappings such that $F(T)\cap {\mathrm{\Omega }}^{\ast }\cap \mathrm{\Lambda }\ne \mathrm{\varnothing }$, where $F(T)={\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)$. Let f be a ρ-contraction mapping.

Algorithm 3.1 For a given $x_0\in C$ arbitrarily, let the iterative sequences $\{u_n\}$, $\{x_n\}$, $\{y_n\}$ and $\{z_n\}$ be generated by

$$\begin{array}{r}{u}_n=T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n);\\ z_n=P_C[u_n-{\lambda }_{n}D{u}_n];\\ y_n=P_C[{\beta }_{n}S{x}_n+(1-{\beta }_n)z_n];\\ x_{n+1}=P_C[{\alpha }_{n}f(x_n)+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)V_i{y}_n],\mathrm{\forall }n\ge 0,\end{array}$$

(20)

where $V_i=k_{i}I+(1-k_i)T_i$, $0\le k_i<1$, $\{r_n\}\subset (0,\mathrm{\infty })$, $\{{\lambda }_n\}\subset (0,2\alpha )$ and $\gamma \in (0,1/L)$, L is the spectral radius of the operator $A^{\ast }A$ and $A^{\ast }$ is the adjoint of A and ${\alpha }_0=1$, $\{{\alpha }_n\}$ is a strictly decreasing sequence in $(0,1)$ and $\{{\beta }_n\}$ is a sequence in $(0,1)$ satisfying the following conditions:

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

• (b) ${lim}_{n\to \mathrm{\infty }}({\beta }_n/{\alpha }_n)=0$,

• (c) ${\sum }_{n=1}^{\mathrm{\infty }}|{\alpha }_{n-1}-{\alpha }_n|<\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}|{\beta }_{n-1}-{\beta }_n|<\mathrm{\infty }$,

• (d) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$ and ${\sum }_{n=1}^{\mathrm{\infty }}|r_{n-1}-r_n|<\mathrm{\infty }$,

• (e) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\lambda }_n<{lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\lambda }_n<2\alpha$ and ${\sum }_{n=1}^{\mathrm{\infty }}|{\lambda }_{n-1}-{\lambda }_n|<\mathrm{\infty }$.

Lemma 3.1 Let $x^{\ast }\in F(T)\cap {\mathrm{\Omega }}^{\ast }\cap \mathrm{\Lambda }$. Then $\{x_n\}$, $\{u_n\}$, $\{z_n\}$ and $\{y_n\}$ are bounded.

Proof First, we show that the mapping $(I-{\lambda }_{n}D)$ is nonexpansive. For any $x,y\in C$,

$$\begin{array}{rcl}{\parallel (I-{\lambda }_{n}D)x-(I-{\lambda }_{n}D)y\parallel }^2& =& {\parallel (x-y)-{\lambda }_n(Dx-Dy)\parallel }^2\\ =& {\parallel x-y\parallel }^2-2{\lambda }_n〈x-y,Dx-Dy〉+{\lambda }_n^2{\parallel Dx-Dy\parallel }^2\\ \le & {\parallel x-y\parallel }^2-{\lambda }_n(2\alpha -{\lambda }_n){\parallel Dx-Dy\parallel }^2\\ \le & {\parallel x-y\parallel }^2.\end{array}$$

Let $x^{\ast }\in F(T)\cap {\mathrm{\Omega }}^{\ast }\cap \mathrm{\Lambda }$, we have $x^{\ast }=T_{r_n}^{F_1}(x^{\ast })$ and $A{x}^{\ast }=T_{r_n}^{F_2}(A{x}^{\ast })$. Then

$$\begin{array}{rcl}{\parallel u_n-x^{\ast }\parallel }^2& =& {\parallel T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)-x^{\ast }\parallel }^2\\ =& {\parallel T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)-T_{r_n}^{F_1}\left(x^{\ast }\right)\parallel }^2\\ \le & {\parallel x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n-x^{\ast }\parallel }^2\\ =& {\parallel x_n-x^{\ast }\parallel }^2+{\gamma }^2{\parallel A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n\parallel }^2+2\gamma 〈x_n-x^{\ast },A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n〉\\ =& {\parallel x_n-x^{\ast }\parallel }^2+{\gamma }^2〈(T_{r_n}^{F_2}-I)A{x}_n,A{A}^{\ast }(T_{r_n}^{F_2}-I)A{x}_n〉\\ +2\gamma 〈x_n-x^{\ast },A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n〉.\end{array}$$

(21)

From the definition of L, it follows that

$$\begin{array}{rcl}{\gamma }^2〈(T_{r_n}^{F_2}-I)A{x}_n,A{A}^{\ast }(T_{r_n}^{F_2}-I)A{x}_n〉& \le & L{\gamma }^2〈(T_{r_n}^{F_2}-I)A{x}_n,(T_{r_n}^{F_2}-I)A{x}_n〉\\ =& L{\gamma }^2{\parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel }^2.\end{array}$$

(22)

It follows from (3) that

$$\begin{array}{c}2\gamma 〈x_n-x^{\ast },A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n〉\hfill \\ =2\gamma 〈A(x_n-x^{\ast }),(T_{r_n}^{F_2}-I)A{x}_n〉\hfill \\ =2\gamma 〈A(x_n-x^{\ast })+(T_{r_n}^{F_2}-I)A{x}_n-(T_{r_n}^{F_2}-I)A{x}_n,(T_{r_n}^{F_2}-I)A{x}_n〉\hfill \\ =2\gamma (〈T_{r_n}^{F_2}A{x}_n-A{x}^{\ast },(T_{r_n}^{F_2}-I)A{x}_n〉-{\parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel }^2)\hfill \\ \le 2\gamma (\frac{1}{2}{\parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel }^2-{\parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel }^2)\hfill \\ =-\gamma {\parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel }^2.\hfill \end{array}$$

(23)

Applying (23) and (22) to (21) and from the definition of γ, we get

$${\parallel u_n-x^{\ast }\parallel }^2\le {\parallel x_n-x^{\ast }\parallel }^2+\gamma (L\gamma -1){\parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel }^2\le {\parallel x_n-x^{\ast }\parallel }^2.$$

(24)

Since the mapping D is α-inverse strongly monotone, we have

$$\begin{array}{rcl}{\parallel z_n-x^{\ast }\parallel }^2& =& {\parallel P_C[u_n-{\lambda }_{n}D{u}_n]-P_C[x^{\ast }-{\lambda }_{n}D{x}^{\ast }]\parallel }^2\\ \le & {\parallel u_n-x^{\ast }-{\lambda }_n(D{u}_n-D{x}^{\ast })\parallel }^2\\ \le & {\parallel u_n-x^{\ast }\parallel }^2-{\lambda }_n(2\alpha -{\lambda }_n){\parallel D{u}_n-D{x}^{\ast }\parallel }^2\\ \le & {\parallel u_n-x^{\ast }\parallel }^2\\ \le & {\parallel x_n-x^{\ast }\parallel }^2.\end{array}$$

(25)

Next, we prove that the sequence $\{x_n\}$ is bounded, without loss of generality, we can assume that ${\beta }_n\le {\alpha }_n$ for all $n\ge 1$. From Lemma 2.8, we have $V_i$ is a nonexpansive mapping and $V_i{x}^{\ast }=x^{\ast }$. Since ${\sum }_{i=1}^n({\alpha }_{i-1}-{\alpha }_i)=1-{\alpha }_n$, we get

$$\begin{array}{rcl}\parallel x_{n+1}-x^{\ast }\parallel & =& \parallel P_C[{\alpha }_{n}f(x_n)+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)V_i{y}_n]-x^{\ast }\parallel \\ \le & \parallel {\alpha }_{n}f(x_n)+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)V_i{y}_n-x^{\ast }\parallel \\ =& \parallel {\alpha }_{n}f(x_n)+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)V_i{y}_n-{\alpha }_n{x}^{\ast }-\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)V_i{x}^{\ast }\parallel \\ \le & {\alpha }_n\parallel f(x_n)-f\left(x^{\ast }\right)\parallel +{\alpha }_n\parallel f\left(x^{\ast }\right)-x^{\ast }\parallel +\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)\parallel V_i{y}_n-V_i{x}^{\ast }\parallel \\ \le & {\alpha }_n\parallel f(x_n)-f\left(x^{\ast }\right)\parallel +{\alpha }_n\parallel f\left(x^{\ast }\right)-x^{\ast }\parallel +\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)\parallel y_n-x^{\ast }\parallel \\ =& {\alpha }_n\parallel f(x_n)-f\left(x^{\ast }\right)\parallel +{\alpha }_n\parallel f\left(x^{\ast }\right)-x^{\ast }\parallel +(1-{\alpha }_n)\parallel {\beta }_{n}S{x}_n+(1-{\beta }_n)z_n-x^{\ast }\parallel \\ \le & {\alpha }_n\parallel f(x_n)-f\left(x^{\ast }\right)\parallel +{\alpha }_n\parallel f\left(x^{\ast }\right)-x^{\ast }\parallel \\ +(1-{\alpha }_n)({\beta }_n\parallel S{x}_n-S{x}^{\ast }\parallel +{\beta }_n\parallel S{x}^{\ast }-x^{\ast }\parallel +(1-{\beta }_n)\parallel z_n-x^{\ast }\parallel )\\ \le & {\alpha }_n\rho \parallel x_n-x^{\ast }\parallel +{\alpha }_n\parallel f\left(x^{\ast }\right)-x^{\ast }\parallel \\ +(1-{\alpha }_n)({\beta }_n\parallel x_n-x^{\ast }\parallel +{\beta }_n\parallel S{x}^{\ast }-x^{\ast }\parallel +(1-{\beta }_n)\parallel x_n-x^{\ast }\parallel )\\ =& (1-{\alpha }_n(1-\rho ))\parallel x_n-x^{\ast }\parallel +{\alpha }_n\parallel f\left(x^{\ast }\right)-x^{\ast }\parallel +(1-{\alpha }_n){\beta }_n\parallel S{x}^{\ast }-x^{\ast }\parallel \\ \le & (1-{\alpha }_n(1-\rho ))\parallel x_n-x^{\ast }\parallel +{\alpha }_n\parallel f\left(x^{\ast }\right)-x^{\ast }\parallel +{\beta }_n\parallel S{x}^{\ast }-x^{\ast }\parallel \\ \le & (1-{\alpha }_n(1-\rho ))\parallel x_n-x^{\ast }\parallel +{\alpha }_n(\parallel f\left(x^{\ast }\right)-x^{\ast }\parallel +\parallel S{x}^{\ast }-x^{\ast }\parallel )\\ =& (1-{\alpha }_n(1-\rho ))\parallel x_n-x^{\ast }\parallel +\frac{{\alpha }_n(1-\rho )}{1-\rho }(\parallel f\left(x^{\ast }\right)-x^{\ast }\parallel +\parallel S{x}^{\ast }-x^{\ast }\parallel )\\ \le & max\{\parallel x_n-x^{\ast }\parallel ,\frac{1}{1-\rho }(\parallel f\left(x^{\ast }\right)-x^{\ast }\parallel +\parallel S{x}^{\ast }-x^{\ast }\parallel )\}.\end{array}$$

(26)

By induction on n, we obtain $\parallel x_n-x^{\ast }\parallel \le max\{\parallel x_0-x^{\ast }\parallel ,\frac{1}{1-\rho }(\parallel f(x^{\ast })-x^{\ast }\parallel +\parallel S{x}^{\ast }-x^{\ast }\parallel )\}$, for $n\ge 0$ and $x_0\in C$. Hence $\{x_n\}$ is bounded and consequently, we deduce that $\{u_n\}$, $\{z_n\}$ and $\{y_n\}$ are bounded. □

Lemma 3.2 Let $x^{\ast }\in F(T)\cap {\mathrm{\Omega }}^{\ast }\cap \mathrm{\Lambda }$ and $\{x_n\}$ be the sequence generated by Algorithm 3.1. Then we have

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

• (b) The weak w-limit set $w_w(x_n)\subset F(T)$ ($w_w(x_n)=\{x:x_{n_i}\rightharpoonup x\}$).

Proof From the nonexpansivity of the mapping $(I-{\lambda }_{n}D)$ and $P_C$, we have

$$\begin{array}{rcl}\parallel z_n-z_{n-1}\parallel & \le & \parallel (u_n-{\lambda }_{n}D{u}_n)-(u_{n-1}-{\lambda }_{n-1}D{u}_{n-1})\parallel \\ =& \parallel (u_n-u_{n-1})-{\lambda }_n(D{u}_n-D{u}_{n-1})-({\lambda }_n-{\lambda }_{n-1})D{u}_{n-1}\parallel \\ \le & \parallel (u_n-u_{n-1})-{\lambda }_n(D{u}_n-D{u}_{n-1})\parallel +|{\lambda }_n-{\lambda }_{n-1}|\parallel D{u}_{n-1}\parallel \\ \le & \parallel u_n-u_{n-1}\parallel +|{\lambda }_n-{\lambda }_{n-1}|\parallel D{u}_{n-1}\parallel .\end{array}$$

(27)

Next, we estimate

$$\begin{array}{c}\parallel y_n-y_{n-1}\parallel \hfill \\ \le \parallel {\beta }_{n}S{x}_n+(1-{\beta }_n)z_n-({\beta }_{n-1}S{x}_{n-1}+(1-{\beta }_{n-1})z_{n-1})\parallel \hfill \\ =\parallel {\beta }_n(S{x}_n-S{x}_{n-1})+({\beta }_n-{\beta }_{n-1})S{x}_{n-1}+(1-{\beta }_n)(z_n-z_{n-1})+({\beta }_{n-1}-{\beta }_n)z_{n-1}\parallel \hfill \\ \le {\beta }_n\parallel x_n-x_{n-1}\parallel +(1-{\beta }_n)\parallel z_n-z_{n-1}\parallel +|{\beta }_n-{\beta }_{n-1}|(\parallel S{x}_{n-1}\parallel +\parallel z_{n-1}\parallel ).\hfill \end{array}$$

(28)

It follows from (27) and (28) that

$$\begin{array}{rcl}\parallel y_n-y_{n-1}\parallel & \le & {\beta }_n\parallel x_n-x_{n-1}\parallel +(1-{\beta }_n)\{\parallel u_n-u_{n-1}\parallel +|{\lambda }_n-{\lambda }_{n-1}|\parallel D{u}_{n-1}\parallel \}\\ +|{\beta }_n-{\beta }_{n-1}|(\parallel S{x}_{n-1}\parallel +\parallel z_{n-1}\parallel ).\end{array}$$

(29)

On the other hand, $u_n=T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)$ and $u_{n-1}=T_{r_{n-1}}^{F_1}(x_{n-1}+\gamma A^{\ast }(T_{r_{n-1}}^{F_2}-I)A{x}_{n-1})$. It follows from Lemma 2.3 that

$$\begin{array}{c}\parallel u_n-u_{n-1}\parallel \hfill \\ \le \parallel x_n-x_{n-1}+\gamma (A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n-A^{\ast }(T_{r_{n-1}}^{F_2}-I)A{x}_{n-1})\parallel \hfill \\ +|1-\frac{r_{n-1}}{r_n}|\parallel T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)-(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)\parallel \hfill \\ \le \parallel x_n-x_{n-1}-\gamma A^{\ast }A(x_n-x_{n-1})\parallel +\gamma \parallel A\parallel \parallel T_{r_n}^{F_2}A{x}_n-T_{r_{n-1}}^{F_2}A{x}_{n-1}\parallel \hfill \\ +|1-\frac{r_{n-1}}{r_n}|\parallel T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)-(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)\parallel \hfill \\ \le {({\parallel x_n-x_{n-1}\parallel }^2-2\gamma {\parallel A(x_n-x_{n-1})\parallel }^2+{\gamma }^2{\parallel A\parallel }^4{\parallel x_n-x_{n-1}\parallel }^2)}^{\frac{1}{2}}\hfill \\ +\gamma \parallel A\parallel (\parallel A(x_n-x_{n-1})\parallel +|1-\frac{r_{n-1}}{r_n}|\parallel T_{r_n}^{F_2}A{x}_n-A{x}_n\parallel )\hfill \\ +|1-\frac{r_{n-1}}{r_n}|\parallel T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)-(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)\parallel \hfill \\ \le {(1-2\gamma {\parallel A\parallel }^2+{\gamma }^2{\parallel A\parallel }^4)}^{\frac{1}{2}}\parallel x_n-x_{n-1}\parallel +\gamma {\parallel A\parallel }^2\parallel x_n-x_{n-1}\parallel \hfill \\ +\gamma \parallel A\parallel |1-\frac{r_{n-1}}{r_n}|\parallel T_{r_n}^{F_2}A{x}_n-A{x}_n\parallel \hfill \\ +|1-\frac{r_{n-1}}{r_n}|\parallel T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)-(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)\parallel \hfill \\ =(1-\gamma {\parallel A\parallel }^2)\parallel x_n-x_{n-1}\parallel +\gamma {\parallel A\parallel }^2\parallel x_n-x_{n-1}\parallel +\gamma \parallel A\parallel |1-\frac{r_{n-1}}{r_n}|\parallel T_{r_n}^{F_2}A{x}_n-A{x}_n\parallel \hfill \\ +|1-\frac{r_{n-1}}{r_n}|\parallel T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)-(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)\parallel \hfill \\ =\parallel x_n-x_{n-1}\parallel +\gamma \parallel A\parallel |1-\frac{r_{n-1}}{r_n}|\parallel T_{r_n}^{F_2}A{x}_n-A{x}_n\parallel \hfill \\ +|1-\frac{r_{n-1}}{r_n}|\parallel T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)-(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)\parallel \hfill \\ =\parallel x_n-x_{n-1}\parallel +\left|\frac{r_n-r_{n-1}}{r_n}\right|(\gamma \parallel A\parallel {\sigma }_n+{\chi }_n),\hfill \end{array}$$

where ${\sigma }_n:=\parallel T_{r_n}^{F_2}A{x}_n-A{x}_n\parallel$ and ${\chi }_n:=\parallel T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)-(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)\parallel$. Without loss of generality, let us assume that there exists a real number μ such that $r_n>\mu >0$ for all positive integers n. Then we get

$$\parallel u_{n-1}-u_n\parallel \le \parallel x_{n-1}-x_n\parallel +\frac{1}{\mu }|r_{n-1}-r_n|(\gamma \parallel A\parallel {\sigma }_n+{\chi }_n).$$

(30)

It follows from (29) and (30) that

$$\begin{array}{c}\parallel y_n-y_{n-1}\parallel \hfill \\ \le {\beta }_n\parallel x_n-x_{n-1}\parallel +(1-{\beta }_n)\{\parallel x_n-x_{n-1}\parallel +\frac{1}{\mu }|r_{n-1}-r_n|(\gamma \parallel A\parallel {\sigma }_n+{\chi }_n)\hfill \\ +|{\lambda }_n-{\lambda }_{n-1}|\parallel D{u}_{n-1}\parallel \}+|{\beta }_n-{\beta }_{n-1}|(\parallel S{x}_{n-1}\parallel +\parallel z_{n-1}\parallel )\hfill \\ =\parallel x_n-x_{n-1}\parallel +(1-{\beta }_n)\{\frac{1}{\mu }|r_{n-1}-r_n|(\gamma \parallel A\parallel {\sigma }_n+{\chi }_n)+|{\lambda }_n-{\lambda }_{n-1}|\parallel D{u}_{n-1}\parallel \}\hfill \\ +|{\beta }_n-{\beta }_{n-1}|(\parallel S{x}_{n-1}\parallel +\parallel z_{n-1}\parallel ).\hfill \end{array}$$

(31)

Next, we estimate

$$\begin{array}{c}\parallel x_{n+1}-x_n\parallel \hfill \\ \le \parallel {\alpha }_{n}f(x_n)+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)V_i{y}_n-({\alpha }_{n-1}f(x_{n-1})+\sum _{i=1}^{n-1}({\alpha }_{i-1}-{\alpha }_i)V_i{y}_{n-1})\parallel \hfill \\ =\parallel {\alpha }_n(f(x_n)-f(x_{n-1}))+({\alpha }_n-{\alpha }_{n-1})f(x_{n-1})+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)(V_i{y}_n-V_i{y}_{n-1})\hfill \\ +({\alpha }_{n-1}-{\alpha }_n)V_n{y}_{n-1}\parallel \hfill \\ \le {\alpha }_n\parallel f(x_n)-f(x_{n-1})\parallel +\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)\parallel V_i{y}_n-V_i{y}_{n-1}\parallel \hfill \\ +|{\alpha }_n-{\alpha }_{n-1}|(\parallel f(x_{n-1})\parallel +\parallel V_n{y}_{n-1}\parallel )\hfill \\ \le {\alpha }_n\rho \parallel x_n-x_{n-1}\parallel +\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)\parallel y_n-y_{n-1}\parallel \hfill \\ +|{\alpha }_n-{\alpha }_{n-1}|(\parallel f(x_{n-1})\parallel +\parallel V_n{y}_{n-1}\parallel )\hfill \\ ={\alpha }_n\rho \parallel x_n-x_{n-1}\parallel +(1-{\alpha }_n)\parallel y_n-y_{n-1}\parallel \hfill \\ +|{\alpha }_n-{\alpha }_{n-1}|(\parallel f(x_{n-1})\parallel +\parallel V_n{y}_{n-1}\parallel ).\hfill \end{array}$$

(32)

From (31) and (32), we have

$$\begin{array}{c}\parallel x_{n+1}-x_n\parallel \hfill \\ \le {\alpha }_n\rho \parallel x_n-x_{n-1}\parallel +(1-{\alpha }_n)\{\parallel x_n-x_{n-1}\parallel \hfill \\ +(1-{\beta }_n)(\frac{1}{\mu }|r_{n-1}-r_n|(\gamma \parallel A\parallel {\sigma }_n+{\chi }_n)+|{\lambda }_n-{\lambda }_{n-1}|\parallel D{u}_{n-1}\parallel )\hfill \\ +|{\beta }_n-{\beta }_{n-1}|(\parallel S{x}_{n-1}\parallel +\parallel z_{n-1}\parallel )\}+|{\alpha }_n-{\alpha }_{n-1}|(\parallel f(x_{n-1})\parallel +\parallel V_n{y}_{n-1}\parallel )\hfill \\ \le (1-(1-\rho ){\alpha }_n)\parallel x_n-x_{n-1}\parallel +\frac{1}{\mu }|r_{n-1}-r_n|(\gamma \parallel A\parallel {\sigma }_n+{\chi }_n)\hfill \\ +|{\lambda }_n-{\lambda }_{n-1}|\parallel D{u}_{n-1}\parallel +|{\beta }_n-{\beta }_{n-1}|(\parallel S{x}_{n-1}\parallel +\parallel z_{n-1}\parallel )\hfill \\ +|{\alpha }_n-{\alpha }_{n-1}|(\parallel f(x_{n-1})\parallel +\parallel V_n{y}_{n-1}\parallel )\hfill \\ \le (1-(1-\rho ){\alpha }_n)\parallel x_n-x_{n-1}\parallel \hfill \\ +M(\frac{1}{\mu }|r_n-r_{n-1}|+|{\lambda }_n-{\lambda }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|+|{\alpha }_n-{\alpha }_{n-1}|),\hfill \end{array}$$

(33)

where

$$\begin{array}{rcl}M& =& max\{\underset{n\ge 1}{sup}(\gamma \parallel A\parallel {\sigma }_n+{\chi }_n),\underset{n\ge 1}{sup}\parallel D{u}_{n-1}\parallel ,\underset{n\ge 1}{sup}(\parallel S{x}_{n-1}\parallel +\parallel z_{n-1}\parallel ),\\ \underset{n\ge 1}{sup}(\parallel f(x_{n-1})\parallel +\parallel V_n{y}_{n-1}\parallel )\}.\end{array}$$

Since $\{x_n\}$, $\{u_n\}$, $\{z_n\}$ and $\{y_n\}$ are bounded, we deduce that $\{A{x}_n\}$, $\{D{u}_{n-1}\}$, $\{S{x}_{n-1}\}$, $\{f(x_{n-1})\}$ and $\{V_n{y}_{n-1}\}$ are bounded. We can conclude that ${sup}_{n\ge 1}(\gamma \parallel A\parallel {\sigma }_n+{\chi }_n)<\mathrm{\infty }$, ${sup}_{n\ge 1}\parallel D{u}_{n-1}\parallel <\mathrm{\infty }$, ${sup}_{n\ge 1}(\parallel S{x}_{n-1}\parallel +\parallel z_{n-1}\parallel )<\mathrm{\infty }$, ${sup}_{n\ge 1}(\parallel f(x_{n-1})\parallel +\parallel V_n{y}_{n-1}\parallel )<\mathrm{\infty }$, and $M<\mathrm{\infty }$.

It follows by conditions (a)-(e) of Algorithm 3.1 and Lemma 2.6 that

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

Next, we show that ${lim}_{n\to \mathrm{\infty }}\parallel u_n-x_n\parallel =0$. Since $x^{\ast }\in F(T)\cap {\mathrm{\Omega }}^{\ast }\cap \mathrm{\Lambda }$ and ${\alpha }_n+{\sum }_{i=1}^n({\alpha }_{i-1}-{\alpha }_i)=1$, by using (24) and (25), we obtain

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& =& {\parallel P_C[{\alpha }_{n}f(x_n)+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)V_i{y}_n]-x^{\ast }\parallel }^2\\ \le & {\parallel {\alpha }_{n}f(x_n)+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)V_i{y}_n-x^{\ast }\parallel }^2\\ =& {\parallel {\alpha }_{n}f(x_n)+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)V_i{y}_n-{\alpha }_n{x}^{\ast }-\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)V_i{x}^{\ast }\parallel }^2\\ \le & {\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i){\parallel V_i{y}_n-V_i{x}^{\ast }\parallel }^2\\ \le & {\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i){\parallel y_n-x^{\ast }\parallel }^2\\ \le & {\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2+(1-{\alpha }_n)({\beta }_n{\parallel S{x}_n-x^{\ast }\parallel }^2+(1-{\beta }_n){\parallel z_n-x^{\ast }\parallel }^2)\\ \le & {\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2+(1-{\alpha }_n){\beta }_n{\parallel S{x}_n-x^{\ast }\parallel }^2\\ +(1-{\alpha }_n)(1-{\beta }_n)\{{\parallel x_n-x^{\ast }\parallel }^2+\gamma (L\gamma -1){\parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel }^2\\ -{\lambda }_n(2\alpha -{\lambda }_n){\parallel D{u}_n-D{x}^{\ast }\parallel }^2\}\\ \le & {\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2+{\beta }_n{\parallel S{x}_n-x^{\ast }\parallel }^2+{\parallel x_n-x^{\ast }\parallel }^2\\ -(1-{\alpha }_n)(1-{\beta }_n)\{\gamma (1-L\gamma ){\parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel }^2\\ +{\lambda }_n(2\alpha -{\lambda }_n){\parallel D{u}_n-D{x}^{\ast }\parallel }^2\}.\end{array}$$

(34)

Then, from the above inequality, we get

$$\begin{array}{c}(1-{\alpha }_n)(1-{\beta }_n)\{\gamma (1-L\gamma ){\parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel }^2+{\lambda }_n(2\alpha -{\lambda }_n){\parallel D{u}_n-D{x}^{\ast }\parallel }^2\}\hfill \\ \le {\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2+{\beta }_n{\parallel S{x}_n-x^{\ast }\parallel }^2+{\parallel x_n-x^{\ast }\parallel }^2-{\parallel x_{n+1}-x^{\ast }\parallel }^2\hfill \\ \le {\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2+{\beta }_n{\parallel S{x}_n-x^{\ast }\parallel }^2+(\parallel x_n-x^{\ast }\parallel +\parallel x_{n+1}-x^{\ast }\parallel )\parallel x_{n+1}-x_n\parallel .\hfill \end{array}$$

Since $\gamma (1-L\gamma )>0$, ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\lambda }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\lambda }_n<2\alpha$, ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$, ${\alpha }_n\to 0$ and ${\beta }_n\to 0$, we obtain

$$\underset{n\to \mathrm{\infty }}{lim}\parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel =0$$

(35)

and

$$\underset{n\to \mathrm{\infty }}{lim}\parallel D{u}_n-D{x}^{\ast }\parallel =0.$$

Since $T_{r_n}^{F_1}$ is firmly nonexpansive, we have

$$\begin{array}{rcl}{\parallel u_n-x^{\ast }\parallel }^2& =& {\parallel T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)-T_{r_n}^{F_1}\left(x^{\ast }\right)\parallel }^2\\ \le & 〈u_n-x^{\ast },x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n-x^{\ast }〉\\ =& \frac{1}{2}\{{\parallel u_n-x^{\ast }\parallel }^2+{\parallel x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n-x^{\ast }\parallel }^2\\ -{\parallel u_n-x^{\ast }-[x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n-x^{\ast }]\parallel }^2\}\\ =& \frac{1}{2}\{{\parallel u_n-x^{\ast }\parallel }^2+{\parallel x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n-x^{\ast }\parallel }^2\\ -{\parallel u_n-x_n-\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n\parallel }^2\}\\ \le & \frac{1}{2}\{{\parallel u_n-x^{\ast }\parallel }^2+{\parallel x_n-x^{\ast }\parallel }^2-{\parallel u_n-x_n-\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n\parallel }^2\}\\ =& \frac{1}{2}\{{\parallel u_n-x^{\ast }\parallel }^2+{\parallel x_n-x^{\ast }\parallel }^2\\ -[{\parallel u_n-x_n\parallel }^2+{\gamma }^2{\parallel A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n\parallel }^2-2\gamma 〈u_n-x_n,A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n〉]\},\end{array}$$

where the last inequality follows from (21) and (24). Hence, we get

$${\parallel u_n-x^{\ast }\parallel }^2\le {\parallel x_n-x^{\ast }\parallel }^2-{\parallel u_n-x_n\parallel }^2+2\gamma \parallel A{u}_n-A{x}_n\parallel \parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel .$$

From (34), (25) and the above inequality, we have

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& \le & {\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2+(1-{\alpha }_n)({\beta }_n{\parallel S{x}_n-x^{\ast }\parallel }^2+(1-{\beta }_n){\parallel z_n-x^{\ast }\parallel }^2)\\ \le & {\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2+(1-{\alpha }_n)({\beta }_n{\parallel S{x}_n-x^{\ast }\parallel }^2+(1-{\beta }_n){\parallel u_n-x^{\ast }\parallel }^2)\\ \le & {\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2+(1-{\alpha }_n)\{{\beta }_n{\parallel S{x}_n-x^{\ast }\parallel }^2\\ +(1-{\beta }_n)({\parallel x_n-x^{\ast }\parallel }^2-{\parallel u_n-x_n\parallel }^2+2\gamma \parallel A{u}_n-A{x}_n\parallel \parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel )\}\\ \le & {\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2+{\beta }_n{\parallel S{x}_n-x^{\ast }\parallel }^2+{\parallel x_n-x^{\ast }\parallel }^2\\ -(1-{\alpha }_n)(1-{\beta }_n){\parallel u_n-x_n\parallel }^2+2\gamma \parallel A{u}_n-A{x}_n\parallel \parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel .\end{array}$$

Hence

$$\begin{array}{c}(1-{\alpha }_n)(1-{\beta }_n){\parallel u_n-x_n\parallel }^2\hfill \\ \le {\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2+{\beta }_n{\parallel S{x}_n-x^{\ast }\parallel }^2+{\parallel x_n-x^{\ast }\parallel }^2-{\parallel x_{n+1}-x^{\ast }\parallel }^2\hfill \\ +2\gamma \parallel A{u}_n-A{x}_n\parallel \parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel \hfill \\ \le {\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2+{\beta }_n{\parallel S{x}_n-x^{\ast }\parallel }^2+(\parallel x_n-x^{\ast }\parallel +\parallel x_{n+1}-x^{\ast }\parallel )\parallel x_{n+1}-x_n\parallel \hfill \\ +2\gamma \parallel A{u}_n-A{x}_n\parallel \parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel .\hfill \end{array}$$

Since ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$, ${\alpha }_n\to 0$, ${\beta }_n\to 0$ and ${lim}_{n\to \mathrm{\infty }}\parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel =0$, we obtain

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

(36)

From (17), we get

$$\begin{array}{rcl}{\parallel z_n-x^{\ast }\parallel }^2& =& {\parallel P_C[u_n-{\lambda }_{n}D{u}_n]-P_C[x^{\ast }-{\lambda }_{n}D{x}^{\ast }]\parallel }^2\\ \le & 〈z_n-x^{\ast },(u_n-{\lambda }_{n}D{u}_n)-(x^{\ast }-{\lambda }_{n}D{x}^{\ast })〉\\ =& \frac{1}{2}\{{\parallel z_n-x^{\ast }\parallel }^2+{\parallel u_n-x^{\ast }-{\lambda }_n(D{u}_n-D{x}^{\ast })\parallel }^2\\ -{\parallel u_n-x^{\ast }-{\lambda }_n(D{u}_n-D{x}^{\ast })-(z_n-x^{\ast })\parallel }^2\}\\ \le & \frac{1}{2}\{{\parallel z_n-x^{\ast }\parallel }^2+{\parallel u_n-x^{\ast }\parallel }^2-{\parallel u_n-z_n-{\lambda }_n(D{u}_n-D{x}^{\ast })\parallel }^2\}\\ \le & \frac{1}{2}\{{\parallel z_n-x^{\ast }\parallel }^2+{\parallel u_n-x^{\ast }\parallel }^2-{\parallel u_n-z_n\parallel }^2+2{\lambda }_n〈u_n-z_n,D{u}_n-D{x}^{\ast }〉\}\\ \le & \frac{1}{2}\{{\parallel z_n-x^{\ast }\parallel }^2+{\parallel u_n-x^{\ast }\parallel }^2-{\parallel u_n-z_n\parallel }^2+2{\lambda }_n\parallel u_n-z_n\parallel \parallel D{u}_n-D{x}^{\ast }\parallel \}.\end{array}$$

Hence

$$\begin{array}{rcl}{\parallel z_n-x^{\ast }\parallel }^2& \le & {\parallel u_n-x^{\ast }\parallel }^2-{\parallel u_n-z_n\parallel }^2+2{\lambda }_n\parallel u_n-z_n\parallel \parallel D{u}_n-D{x}^{\ast }\parallel \\ \le & {\parallel x_n-x^{\ast }\parallel }^2-{\parallel u_n-z_n\parallel }^2+2{\lambda }_n\parallel u_n-z_n\parallel \parallel D{u}_n-D{x}^{\ast }\parallel .\end{array}$$

From (34) and the above inequality, we have

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& \le & {\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2+(1-{\alpha }_n)({\beta }_n{\parallel S{x}_n-x^{\ast }\parallel }^2+(1-{\beta }_n){\parallel z_n-x^{\ast }\parallel }^2)\\ \le & {\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2+(1-{\alpha }_n)\{{\beta }_n{\parallel S{x}_n-x^{\ast }\parallel }^2\\ +(1-{\beta }_n)({\parallel x_n-x^{\ast }\parallel }^2-{\parallel u_n-z_n\parallel }^2+2{\lambda }_n\parallel u_n-z_n\parallel \parallel D{u}_n-D{x}^{\ast }\parallel )\}\\ \le & {\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2+{\beta }_n{\parallel S{x}_n-x^{\ast }\parallel }^2+{\parallel x_n-x^{\ast }\parallel }^2\\ -(1-{\alpha }_n)(1-{\beta }_n){\parallel u_n-z_n\parallel }^2+2{\lambda }_n\parallel u_n-z_n\parallel \parallel D{u}_n-D{x}^{\ast }\parallel .\end{array}$$

Hence

$$\begin{array}{c}(1-{\alpha }_n)(1-{\beta }_n){\parallel u_n-z_n\parallel }^2\hfill \\ \le {\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2+{\beta }_n{\parallel S{x}_n-x^{\ast }\parallel }^2+{\parallel x_n-x^{\ast }\parallel }^2-{\parallel x_{n+1}-x^{\ast }\parallel }^2\hfill \\ +2{\lambda }_n\parallel u_n-z_n\parallel \parallel D{u}_n-D{x}^{\ast }\parallel \hfill \\ \le {\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2+{\beta }_n{\parallel S{x}_n-x^{\ast }\parallel }^2+(\parallel x_n-x^{\ast }\parallel +\parallel x_{n+1}-x^{\ast }\parallel )\parallel x_{n+1}-x_n\parallel \hfill \\ +2{\lambda }_n\parallel u_n-z_n\parallel \parallel D{u}_n-D{x}^{\ast }\parallel .\hfill \end{array}$$

Since ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$, ${\alpha }_n\to 0$, ${\beta }_n\to 0$ and ${lim}_{n\to \mathrm{\infty }}\parallel D{u}_n-D{x}^{\ast }\parallel =0$, we obtain

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

(37)

It follows from (36) and (37) that

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

(38)

Now, let $z\in F(T)\cap {\mathrm{\Omega }}^{\ast }\cap \mathrm{\Lambda }$, since for each $i\ge 1$, $V_i{x}_n\in C$ and ${\alpha }_n+{\sum }_{i=1}^n({\alpha }_{i-1}-{\alpha }_i)=1$, we have ${\sum }_{i=1}^n({\alpha }_{i-1}-{\alpha }_i)V_i{x}_n+{\alpha }_{n}z\in C$, and

$$\begin{array}{c}\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)(x_n-V_i{x}_n)\hfill \\ =P_C[{\alpha }_{n}f(x_n)+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)V_i{y}_n]+(1-{\alpha }_n)x_n\hfill \\ -(\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)V_i{x}_n+{\alpha }_{n}z)+{\alpha }_{n}z-x_{n+1}\hfill \\ =P_C[{\alpha }_{n}f(x_n)+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)V_i{y}_n]+{\alpha }_n(z-x_{n+1})\hfill \\ -P_C[\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)V_i{x}_n+{\alpha }_{n}z]+(1-{\alpha }_n)(x_n-x_{n+1}).\hfill \end{array}$$

It follows that

$$\begin{array}{c}\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)〈x_n-V_i{x}_n,x_n-x^{\ast }〉\hfill \\ =〈P_C[{\alpha }_{n}f(x_n)+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)V_i{y}_n]-P_C[\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)V_i{x}_n+{\alpha }_{n}z],x_n-x^{\ast }〉\hfill \\ +{\alpha }_n〈z-x_{n+1},x_n-x^{\ast }〉+(1-{\alpha }_n)〈x_n-x_{n+1},x_n-x^{\ast }〉\hfill \\ \le \parallel {\alpha }_n(f(x_n)-z)+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)(V_i{y}_n-V_i{x}_n)\parallel \parallel x_n-x^{\ast }\parallel \hfill \\ +{\alpha }_n\parallel z-x_{n+1}\parallel \parallel x_n-x^{\ast }\parallel +(1-{\alpha }_n)\parallel x_n-x_{n+1}\parallel \parallel x_n-x^{\ast }\parallel \hfill \\ \le {\alpha }_n\parallel f(x_n)-z\parallel \parallel x_n-x^{\ast }\parallel +\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)\parallel y_n-x_n\parallel \parallel x_n-x^{\ast }\parallel \hfill \\ +{\alpha }_n\parallel z-x_{n+1}\parallel \parallel x_n-x^{\ast }\parallel +(1-{\alpha }_n)\parallel x_n-x_{n+1}\parallel \parallel x_n-x^{\ast }\parallel \hfill \\ ={\alpha }_n\parallel f(x_n)-z\parallel \parallel x_n-x^{\ast }\parallel +(1-{\alpha }_n)\parallel y_n-x_n\parallel \parallel x_n-x^{\ast }\parallel \hfill \\ +{\alpha }_n\parallel z-x_{n+1}\parallel \parallel x_n-x^{\ast }\parallel +(1-{\alpha }_n)\parallel x_n-x_{n+1}\parallel \parallel x_n-x^{\ast }\parallel \hfill \\ \le {\alpha }_n\parallel f(x_n)-z\parallel \parallel x_n-x^{\ast }\parallel \hfill \\ +(1-{\alpha }_n)\parallel {\beta }_{n}S{x}_n+(1-{\beta }_n)z_n-x_n\parallel \parallel x_n-x^{\ast }\parallel +{\alpha }_n\parallel z-x_{n+1}\parallel \parallel x_n-x^{\ast }\parallel \hfill \\ +(1-{\alpha }_n)\parallel x_n-x_{n+1}\parallel \parallel x_n-x^{\ast }\parallel \hfill \\ \le {\alpha }_n\parallel f(x_n)-z\parallel \parallel x_n-x^{\ast }\parallel +(1-{\alpha }_n){\beta }_n\parallel S{x}_n-x_n\parallel \parallel x_n-x^{\ast }\parallel \hfill \\ +(1-{\alpha }_n)(1-{\beta }_n)\parallel z_n-x_n\parallel \parallel x_n-x^{\ast }\parallel \hfill \\ +{\alpha }_n\parallel z-x_{n+1}\parallel \parallel x_n-x^{\ast }\parallel +(1-{\alpha }_n)\parallel x_n-x_{n+1}\parallel \parallel x_n-x^{\ast }\parallel .\hfill \end{array}$$

From Lemma 2.9 and the above inequality, we get

$$\begin{array}{c}\frac{1}{2}\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i){\parallel x_n-V_i{x}_n\parallel }^2\hfill \\ \le \sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)〈x_n-V_i{x}_n,x_n-x^{\ast }〉\hfill \\ \le {\alpha }_n\parallel f(x_n)-z\parallel \parallel x_n-x^{\ast }\parallel +(1-{\alpha }_n){\beta }_n\parallel S{x}_n-x_n\parallel \parallel x_n-x^{\ast }\parallel \hfill \\ +(1-{\alpha }_n)(1-{\beta }_n)\parallel z_n-x_n\parallel \parallel x_n-x^{\ast }\parallel +{\alpha }_n\parallel z-x_{n+1}\parallel \parallel x_n-x^{\ast }\parallel \hfill \\ +(1-{\alpha }_n)\parallel x_n-x_{n+1}\parallel \parallel x_n-x^{\ast }\parallel .\hfill \end{array}$$

Since ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$, ${\alpha }_n\to 0$, ${\beta }_n\to 0$ and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-z_n\parallel =0$, we obtain

$$\underset{n\to \mathrm{\infty }}{lim}\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i){\parallel x_n-V_i{x}_n\parallel }^2=0.$$

Since $({\alpha }_{i-1}-{\alpha }_i){\parallel x_n-V_i{x}_n\parallel }^2\le {\sum }_{i=1}^n({\alpha }_{i-1}-{\alpha }_i){\parallel x_n-V_i{x}_n\parallel }^2$ and $\{{\alpha }_n\}$ is strictly decreasing, we have

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

Hence, we obtain

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T_i{x}_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\frac{\parallel x_n-V_i{x}_n\parallel }{(1-k_i)}=0,\mathrm{\forall }i\ge 1.$$

Since $\{x_n\}$ is bounded, without loss of generality, we can assume that $x_n\rightharpoonup w\in C$. It follows from Lemma 2.4 that $w\in F(T)$. Therefore $w_w(x_n)\subset F(T)$. □

Theorem 3.1 The sequence $\{x_n\}$ generated by Algorithm 3.1 converges strongly to $z=P_{{\mathrm{\Omega }}^{\ast }\cap \mathrm{\Lambda }\cap F(T)}f(z)$, which is the unique solution of the variational inequality

$$〈(I-f)z,x-z〉\ge 0,\mathrm{\forall }x\in {\mathrm{\Omega }}^{\ast }\cap \mathrm{\Lambda }\cap F(T),$$

(39)

which is the optimality condition for a minimization problem

$$\underset{x\in \mathrm{\Upsilon }}{min}\{\frac{1}{2}{\parallel x\parallel }^2-h(x)\},$$

where h is a potential function for f (i.e., $h^{\prime }(x)=f(x)$ for $x\in H$) and $\mathrm{\Upsilon }={\mathrm{\Omega }}^{\ast }\cap \mathrm{\Lambda }\cap F(T)$.

Proof Since $\{x_n\}$ is bounded $x_n\rightharpoonup w$ and from Lemma 3.2, we have $w\in F(T)$. Next, we show that $w\in \mathit{EP}(F_1)$. Since $u_n=T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)$, we have

$$F_1(u_n,y)+\frac{1}{r_n}〈y-u_n,u_n-x_n〉-\frac{1}{r_n}〈y-u_n,\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n〉\ge 0,\mathrm{\forall }y\in C.$$

It follows from the monotonicity of $F_1$ that

$$-\frac{1}{r_n}〈y-u_n,\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n〉+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge F_1(y,u_n),\mathrm{\forall }y\in C$$

and

$$-\frac{1}{r_{n_k}}〈y-u_{n_k},\gamma A^{\ast }(T_{r_{n_k}}^{F_2}-I)A{x}_{n_k}〉+〈y-u_{n_k},\frac{u_{n_k}-x_{n_k}}{r_{n_k}}〉\ge F_1(y,u_{n_k}),\mathrm{\forall }y\in C.$$

(40)

Since ${lim}_{n\to \mathrm{\infty }}\parallel u_n-x_n\parallel =0$, ${lim}_{n\to \mathrm{\infty }}\parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel =0$ and $x_n\rightharpoonup w$, it is easy to observe that $u_{n_k}\to w$. It follows by Assumption 2.1(iv) that $F_1(y,w)\le 0$, $\mathrm{\forall }y\in C$.

For any $0<t\le 1$ and $y\in C$, let $y_t=ty+(1-t)w$, we have $y_t\in C$. Then, from Assumption 2.1(i) and (iv), we have

$$\begin{array}{rcl}0& =& F_1(y_t,y_t)\le t{F}_1(y_t,y)+(1-t)F_1(y_t,w)\\ \le & t{F}_1(y_t,y).\end{array}$$

Therefore $F_1(y_t,y)\ge 0$. From Assumption 2.1(iii), we have $F_1(w,y)\ge 0$, which implies that $w\in \mathit{EP}(F_1)$.

Next, we show that $Aw\in \mathit{EP}(F_2)$. Since $\{x_n\}$ is bounded and $x_n\rightharpoonup w$, there exists a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that $x_{n_k}\to w$ and since A is a bounded linear operator so that $A{x}_{n_k}\to Aw$. Now set $v_{n_k}=A{x}_{n_k}-T_{r_{n_k}}^{F_2}A{x}_{n_k}$. It follows from (35) that ${lim}_{k\to \mathrm{\infty }}{v}_{n_k}=0$ and $A{x}_{n_k}-v_{n_k}=T_{r_{n_k}}^{F_2}A{x}_{n_k}$. Therefore from the definition of $T_{r_{n_k}}^{F_2}$, we have

$$F_2(A{x}_{n_k}-v_{n_k},y)+\frac{1}{r_{n_k}}〈y-(A{x}_{n_k}-v_{n_k}),(A{x}_{n_k}-v_{n_k})-A{x}_{n_k}〉\ge 0,\mathrm{\forall }y\in C.$$

Since $F_2$ is upper semicontinuous in the first argument, taking lim sup to the above inequality as $k\to \mathrm{\infty }$ and using Assumption 2.1(iv), we obtain

$$F_2(Aw,y)\ge 0,\mathrm{\forall }y\in C,$$

which implies that $Aw\in \mathit{EP}(F_2)$ and hence $w\in \mathrm{\Lambda }$.

Furthermore, we show that $w\in {\mathrm{\Omega }}^{\ast }$. Let

$$Tv=\{\begin{array}{cc}Dv+N_{C}v,\hfill & \mathrm{\forall }v\in C,\hfill \\ \mathrm{\varnothing },\hfill & \text{otherwise,}\hfill \end{array}$$

where $N_{C}v:=\{w\in H:〈w,v-u〉\ge 0,\mathrm{\forall }u\in C\}$ is the normal cone to C at $v\in C$. Then T is maximal monotone and $0\in Tv$ if and only if $v\in {\mathrm{\Omega }}^{\ast }$ (see [25]). Let $G(T)$ denote the graph of T and let $(v,u)\in G(T)$. Since $u-Dv\in N_{C}v$ and $z_n\in C$, we have

$$〈v-z_n,u-Dv〉\ge 0.$$

(41)

On the other hand, it follows from $z_n=P_C[u_n-{\lambda }_{n}D{u}_n]$ and $v\in C$ that

$$〈v-z_n,z_n-(u_n-{\lambda }_{n}D{u}_n)〉\ge 0$$

and

$$〈v-z_n,\frac{z_n-u_n}{{\lambda }_n}+D{u}_n〉\ge 0.$$

Therefore, from (41) and inverse strong monotonicity of D, we have

$$\begin{array}{rcl}〈v-z_{n_k},u〉& \ge & 〈v-z_{n_k},Dv〉\\ \ge & 〈v-z_{n_k},Dv〉-〈v-z_{n_k},\frac{z_{n_k}-u_{n_k}}{{\lambda }_{n_k}}+D{u}_{n_k}〉\\ \ge & 〈v-z_{n_k},Dv-D{z}_{n_k}〉+〈v-z_{n_k},D{z}_{n_k}-D{u}_{n_k}〉-〈v-z_{n_k},\frac{z_{n_k}-u_{n_k}}{{\lambda }_{n_k}}〉\\ \ge & 〈v-z_{n_k},D{z}_{n_k}-D{u}_{n_k}〉-〈v-z_{n_k},\frac{z_{n_k}-u_{n_k}}{{\lambda }_{n_k}}〉.\end{array}$$

Since ${lim}_{n\to \mathrm{\infty }}\parallel u_n-z_n\parallel =0$ and $u_{n_k}\to w$, it is easy to observe that $z_{n_k}\to w$. Hence, we obtain $〈v-w,u〉\ge 0$. Since T is maximal monotone, we have $w\in T^{-1}0$ and hence $w\in {\mathrm{\Omega }}^{\ast }$. Thus we have

$$w\in {\mathrm{\Omega }}^{\ast }\cap \mathrm{\Lambda }\cap F(T).$$

Since ${\mathrm{\Omega }}^{\ast }$, Λ and $F(T)$ are convex, then ${\mathrm{\Omega }}^{\ast }\cap \mathrm{\Lambda }\cap F(T)$ is convex. Next, we claim that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈f(z)-z,x_n-z〉\le 0$, where $z=P_{{\mathrm{\Omega }}^{\ast }\cap \mathrm{\Lambda }\cap F(T)}f(z)$.

Since $\{x_n\}$ is bounded, there exists a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(z)-z,x_n-z〉=\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(z)-z,x_{n_k}-z〉=〈f(z)-z,w-z〉\le 0.$$

Next, we show that $x_n\to z$. From (16), we get

$$\begin{array}{rcl}{\parallel x_{n+1}-z\parallel }^2& =& 〈x_{n+1}-{\alpha }_{n}f(x_n)-\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)V_i{y}_n,x_{n+1}-z〉\\ +〈{\alpha }_{n}f(x_n)+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)V_i{y}_n-z,x_{n+1}-z〉\\ \le & 〈{\alpha }_{n}f(x_n)+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)V_i{y}_n-z,x_{n+1}-z〉\\ \le & {\alpha }_n〈f(x_n)-f(z),x_{n+1}-z〉+{\alpha }_n〈f(z)-z,x_{n+1}-z〉\\ +\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)〈V_i{y}_n-z,x_{n+1}-z〉\\ \le & {\alpha }_n\parallel f(x_n)-f(z)\parallel \parallel x_{n+1}-z\parallel +{\alpha }_n〈f(z)-z,x_{n+1}-z〉\\ +\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)\parallel V_i{y}_n-z\parallel \parallel x_{n+1}-z\parallel \\ \le & {\alpha }_n\rho \parallel x_n-z\parallel \parallel x_{n+1}-z\parallel +{\alpha }_n〈f(z)-z,x_{n+1}-z〉\\ +\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)\parallel y_n-z\parallel \parallel x_{n+1}-z\parallel \\ \le & {\alpha }_n\rho \parallel x_n-z\parallel \parallel x_{n+1}-z\parallel +{\alpha }_n〈f(z)-z,x_{n+1}-z〉\\ +(1-{\alpha }_n)\{{\beta }_n\parallel S{x}_n-Sz\parallel +{\beta }_n\parallel Sz-z\parallel +(1-{\beta }_n)\parallel z_n-z\parallel \}\parallel x_{n+1}-z\parallel \\ \le & {\alpha }_n\rho \parallel x_n-z\parallel \parallel x_{n+1}-z\parallel +{\alpha }_n〈f(z)-z,x_{n+1}-z〉\\ +(1-{\alpha }_n)\{{\beta }_n\parallel x_n-z\parallel +{\beta }_n\parallel Sz-z\parallel +(1-{\beta }_n)\parallel x_n-z\parallel \}\parallel x_{n+1}-z\parallel \\ \le & (1-{\alpha }_n(1-\rho ))\parallel x_n-z\parallel \parallel x_{n+1}-z\parallel +{\alpha }_n〈f(z)-z,x_{n+1}-z〉\\ +(1-{\alpha }_n){\beta }_n\parallel Sz-z\parallel \parallel x_{n+1}-z\parallel \\ \le & \frac{1-{\alpha }_n(1-\rho )}{2}({\parallel x_n-z\parallel }^2+{\parallel x_{n+1}-z\parallel }^2)+{\alpha }_n〈f(z)-z,x_{n+1}-z〉\\ +(1-{\alpha }_n){\beta }_n\parallel Sz-z\parallel \parallel x_{n+1}-z\parallel ,\end{array}$$

which implies that

$$\begin{array}{rcl}{\parallel x_{n+1}-z\parallel }^2& \le & (1-\frac{2{\alpha }_n(1-\rho )}{1+{\alpha }_n(1-\rho )}){\parallel x_n-z\parallel }^2+\frac{2{\alpha }_n}{1+{\alpha }_n(1-\rho )}〈f(z)-z,x_{n+1}-z〉\\ +\frac{2(1-{\alpha }_n){\beta }_n}{1+{\alpha }_n(1-\rho )}\parallel Sz-z\parallel \parallel x_{n+1}-z\parallel \\ \le & (1-\frac{2{\alpha }_n(1-\rho )}{1+{\alpha }_n(1-\rho )}){\parallel x_n-z\parallel }^2+\frac{2{\alpha }_n(1-\rho )}{1+{\alpha }_n(1-\rho )}\{\frac{1}{1-\rho }〈f(z)-z,x_{n+1}-z〉\\ +\frac{(1-{\alpha }_n){\beta }_n}{{\alpha }_n(1-\rho )}\parallel Sz-z\parallel \parallel x_{n+1}-z\parallel \}.\end{array}$$

Let ${\gamma }_n=\frac{2{\alpha }_n(1-\rho )}{1+{\alpha }_n(1-\rho )}$ and ${\delta }_n=\frac{2{\alpha }_n(1-\rho )}{1+{\alpha }_n(1-\rho )}\{\frac{1}{1-\rho }〈f(z)-z,x_{n+1}-z〉+\frac{(1-{\alpha }_n){\beta }_n}{{\alpha }_n(1-\rho )}\parallel Sz-z\parallel \parallel x_{n+1}-z\parallel \}$.

Since

$$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty },1+{\alpha }_n(1-\rho )\le 2\text{and}\hfill \\ \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\{\frac{1}{1-\rho }〈f(z)-z,x_{n+1}-z〉+\frac{(1-{\alpha }_n){\beta }_n}{{\alpha }_n(1-\rho )}\parallel Sz-z\parallel \parallel x_{n+1}-z\parallel \}\le 0,\hfill \end{array}$$

it follows that

$$\sum _{n=1}^{\mathrm{\infty }}{\gamma }_n=\mathrm{\infty }\text{and}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{{\delta }_n}{{\gamma }_n}\le 0.$$

Thus all the conditions of Lemma 2.6 are satisfied. Hence we deduce that $x_n\to z$.

$P_{{\mathrm{\Omega }}^{\ast }\cap \mathrm{\Lambda }\cap F(T)}f$ is a contraction, there exists a unique $z\in C$ such that $z=P_{{\mathrm{\Omega }}^{\ast }\cap \mathrm{\Lambda }\cap F(T)}f(z)$. From (16), it follows that z is the unique solution of problem (39). This completes the proof. □

Theorem 3.2 Let $H_1$ and $H_2$ be two real Hilbert spaces and $C\subseteq H_1$ and $Q\subseteq H_2$ be nonempty closed convex subsets of Hilbert spaces $H_1$ and $H_2$, respectively. Let $A:H_1\to H_2$ be a bounded linear operator. Let $D:C\to H_1$ be an α-inverse strongly monotone mapping. Assume that $F_1:C\times C\to \mathbb{R}$ and $F_2:Q\times Q\to \mathbb{R}$ are the bifunctions satisfying Assumption  2.1 and $F_2$ is upper semicontinuous in the first argument. Let $S:C\to H_1$ be a nonexpansive mapping and ${\{T_i\}}_{i=1}^{\mathrm{\infty }}:C\to C$ be a countable family of $k_i$-strict pseudo-contraction mappings such that $F(T)\cap {\mathrm{\Omega }}^{\ast }\cap \mathrm{\Lambda }\ne \mathrm{\varnothing }$, where $F(T)={\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)$. Let f be a ρ-contraction mapping. For a given $x_0\in C$ arbitrarily, let the iterative sequences $\{u_n\}$, $\{x_n\}$, $\{y_n\}$ and $\{z_n\}$ be generated by

$$\begin{array}{r}{u}_n=T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n);\\ z_n=P_C[u_n-{\lambda }_{n}A{u}_n];\\ y_n={\beta }_{n}S{x}_n+(1-{\beta }_n)z_n;\\ x_{n+1}=P_C[{\alpha }_{n}f(x_n)+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)V_i{y}_n],\mathrm{\forall }n\ge 0,\end{array}$$

(42)

where $V_i=k_{i}I+(1-k_i)T_i$, $0\le k_i<1$, $\{r_n\}\subset (0,\mathrm{\infty })$, $\{{\lambda }_n\}\subset (0,2\alpha )$ and $\gamma \in (0,1/L)$, L is the spectral radius of the operator $A^{\ast }A$ and $A^{\ast }$ is the adjoint of A and ${\alpha }_0=1$, $\{{\alpha }_n\}$ is a strictly decreasing sequence in $(0,1)$ and $\{{\beta }_n\}$ is a sequence in $(0,1)$ satisfying the following conditions:

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

• (b) ${lim}_{n\to \mathrm{\infty }}\frac{{\beta }_n}{{\alpha }_n}=\tau \in (0,\mathrm{\infty })$,

• (c) ${\sum }_{n=1}^{\mathrm{\infty }}({\alpha }_{n-1}-{\alpha }_n)<\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}|{\beta }_{n-1}-{\beta }_n|<\mathrm{\infty }$,

• (d) ${lim}_{n\to \mathrm{\infty }}\frac{\frac{1}{\mu }|r_n-r_{n-1}|+|{\lambda }_n-{\lambda }_{n-1}|+|{\alpha }_{n-1}-{\alpha }_n|+|{\beta }_{n-1}-{\beta }_n|}{{\alpha }_n{\beta }_n}=0$,

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

• (f) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$ and ${\sum }_{n=1}^{\mathrm{\infty }}|r_{n-1}-r_n|<\mathrm{\infty }$,

• (g) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\lambda }_n<{lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\lambda }_n<2\alpha$ and ${\sum }_{n=1}^{\mathrm{\infty }}|{\lambda }_{n-1}-{\lambda }_n|<\mathrm{\infty }$.

Then the sequence $\{x_n\}$ generated by Algorithm (42) converges strongly to $x^{\ast }\in {\mathrm{\Omega }}^{\ast }\cap \mathrm{\Lambda }\cap F(T)$, which is the unique solution of the variational inequality

$$〈\frac{1}{\tau }(I-f)x^{\ast }+(I-S)x^{\ast },x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in {\mathrm{\Omega }}^{\ast }\cap \mathrm{\Lambda }\cap F(T).$$

(43)

Proof From ${lim}_{n\to \mathrm{\infty }}({\beta }_n/{\alpha }_n)=\tau \in (0,\mathrm{\infty })$, without loss of generality, we can assume that ${\beta }_n\le (1+\tau ){\alpha }_n$ for all $n\ge 1$. Hence ${\beta }_n\to 0$. By a similar argument as that in Lemmas 3.1 and 3.2, we can deduce that $\{x_n\}$ is bounded, ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$, ${lim}_{n\to \mathrm{\infty }}\parallel x_n-z_n\parallel =0$ (see (38)) and $(I-V_i)x_n\to 0$. Then we have

$$\parallel y_n-x_n\parallel \le {\beta }_n\parallel x_n-S{x}_n\parallel +(1-{\beta }_n)\parallel x_n-z_n\parallel \to 0\text{as }n\to \mathrm{\infty }.$$

(44)

It follows that for all $i\ge 1$,

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

(45)

From (44) and (45), we have

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

Set $w_n={\alpha }_{n}f(x_n)+{\sum }_{i=1}^n({\alpha }_{i-1}-{\alpha }_i)V_i{y}_n$. From (32) and (33), we obtain

$$\begin{array}{rcl}\frac{\parallel x_{n+1}-x_n\parallel }{{\beta }_n}& \le & \frac{\parallel w_n-w_{n-1}\parallel }{{\beta }_n}\\ \le & (1-(1-\rho ){\alpha }_n)\frac{\parallel x_n-x_{n-1}\parallel }{{\beta }_n}\\ +M(\frac{1}{\mu }\frac{|r_n-r_{n-1}|}{{\beta }_n}+\frac{|{\lambda }_n-{\lambda }_{n-1}|}{{\beta }_n}+\frac{|{\beta }_n-{\beta }_{n-1}|}{{\beta }_n}+\frac{|{\alpha }_n-{\alpha }_{n-1}|}{{\beta }_n})\\ =& (1-(1-\rho ){\alpha }_n)\frac{\parallel x_n-x_{n-1}\parallel }{{\beta }_{n-1}}+(1-(1-\rho ){\alpha }_n)\parallel x_n-x_{n-1}\parallel (\frac{1}{{\beta }_n}-\frac{1}{{\beta }_{n-1}})\\ +M(\frac{1}{\mu }\frac{|r_n-r_{n-1}|}{{\beta }_n}+\frac{|{\lambda }_n-{\lambda }_{n-1}|}{{\beta }_n}+\frac{|{\beta }_n-{\beta }_{n-1}|}{{\beta }_n}+\frac{|{\alpha }_n-{\alpha }_{n-1}|}{{\beta }_n})\\ \le & (1-(1-\rho ){\alpha }_n)\frac{\parallel x_n-x_{n-1}\parallel }{{\beta }_{n-1}}+\parallel x_n-x_{n-1}\parallel |\frac{1}{{\beta }_n}-\frac{1}{{\beta }_{n-1}}|\\ +M(\frac{1}{\mu }\frac{|r_n-r_{n-1}|}{{\beta }_n}+\frac{|{\lambda }_n-{\lambda }_{n-1}|}{{\beta }_n}+\frac{|{\beta }_n-{\beta }_{n-1}|}{{\beta }_n}+\frac{|{\alpha }_n-{\alpha }_{n-1}|}{{\beta }_n})\\ \le & (1-(1-\rho ){\alpha }_n)\frac{\parallel x_n-x_{n-1}\parallel }{{\beta }_{n-1}}+{\alpha }_{n}K\parallel x_n-x_{n-1}\parallel \\ +M(\frac{1}{\mu }\frac{|r_n-r_{n-1}|}{{\beta }_n}+\frac{|{\lambda }_n-{\lambda }_{n-1}|}{{\beta }_n}+\frac{|{\beta }_n-{\beta }_{n-1}|}{{\beta }_n}+\frac{|{\alpha }_n-{\alpha }_{n-1}|}{{\beta }_n})\\ \le & (1-(1-\rho ){\alpha }_n)\frac{\parallel w_{n-1}-w_{n-2}\parallel }{{\beta }_{n-1}}+{\alpha }_{n}K\parallel x_n-x_{n-1}\parallel \\ +M(\frac{1}{\mu }\frac{|r_n-r_{n-1}|}{{\beta }_n}+\frac{|{\lambda }_n-{\lambda }_{n-1}|}{{\beta }_n}+\frac{|{\beta }_n-{\beta }_{n-1}|}{{\beta }_n}+\frac{|{\alpha }_n-{\alpha }_{n-1}|}{{\beta }_n}).\end{array}$$

Let ${\gamma }_n=(1-\rho ){\alpha }_n$ and ${\delta }_n={\alpha }_{n}K\parallel x_n-x_{n-1}\parallel +M(\frac{1}{\mu }\frac{|r_n-r_{n-1}|}{{\beta }_n}+\frac{|{\lambda }_n-{\lambda }_{n-1}|}{{\beta }_n}+\frac{|{\beta }_n-{\beta }_{n-1}|}{{\beta }_n}+\frac{|{\alpha }_n-{\alpha }_{n-1}|}{{\beta }_n})$. From conditions (a) and (d), we have

$$\sum _{n=1}^{\mathrm{\infty }}{\gamma }_n=\mathrm{\infty }\text{and}\underset{n\to \mathrm{\infty }}{lim}\frac{{\delta }_n}{{\gamma }_n}=0.$$

By Lemma 2.6, we obtain

$$\underset{n\to \mathrm{\infty }}{lim}\frac{\parallel x_{n+1}-x_n\parallel }{{\beta }_n}=0,\underset{n\to \mathrm{\infty }}{lim}\frac{\parallel w_{n+1}-w_n\parallel }{{\beta }_n}=\underset{n\to \mathrm{\infty }}{lim}\frac{\parallel w_{n+1}-w_n\parallel }{{\alpha }_n}=0.$$

From (42), we have

$$x_{n+1}=P_C[w_n]-w_n+{\alpha }_{n}f(x_n)+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)(V_i{y}_n-y_n)+(1-{\alpha }_n)y_n.$$

Hence it follows that

$$\begin{array}{rcl}{x}_n-x_{n+1}& =& (1-{\alpha }_n)x_n+{\alpha }_n{x}_n\\ -(P_C[w_n]-w_n+{\alpha }_{n}f(x_n)+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)(V_i{y}_n-y_n)+(1-{\alpha }_n)y_n)\\ =& (1-{\alpha }_n)[{\beta }_n(x_n-S{x}_n)+(1-{\beta }_n)(x_n-z_n)]+(w_n-P_C[w_n])\\ +\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)(y_n-V_i{y}_n)+{\alpha }_n(x_n-f(x_n)),\end{array}$$

and hence

$$\begin{array}{rcl}\frac{x_n-x_{n+1}}{(1-{\alpha }_n){\beta }_n}& =& x_n-S{x}_n+\frac{(1-{\beta }_n)}{{\beta }_n}(x_n-z_n)+\frac{1}{(1-{\alpha }_n){\beta }_n}(w_n-P_C[w_n])\\ +\frac{1}{(1-{\alpha }_n){\beta }_n}\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)(y_n-V_i{y}_n)+\frac{{\alpha }_n}{(1-{\alpha }_n){\beta }_n}(x_n-f(x_n)).\end{array}$$

Let $v_n=\frac{x_n-x_{n+1}}{(1-{\alpha }_n){\beta }_n}$. For any $z\in {\mathrm{\Omega }}^{\ast }\cap \mathrm{\Lambda }\cap F(T)$, we have

$$\begin{array}{rcl}〈v_n,x_n-z〉& =& \frac{1}{(1-{\alpha }_n){\beta }_n}〈w_n-P_C[w_n],P_C[w_{n-1}]-z〉+\frac{{\alpha }_n}{(1-{\alpha }_n){\beta }_n}〈(I-f)x_n,x_n-z〉\\ +〈x_n-S{x}_n,x_n-z〉+\frac{(1-{\beta }_n)}{{\beta }_n}〈x_n-z_n,x_n-z〉\\ +\frac{1}{(1-{\alpha }_n){\beta }_n}\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)〈y_n-V_i{y}_n,x_n-z〉.\end{array}$$

(46)

Since S is a nonexpansive mapping, f is a ρ-contraction mapping and $V_i$ is a $k_i$-strict pseudo-contraction mapping. Then $(I-S)$ and $(I-V_i)$ are monotone and f is strongly monotone with a coefficient $(1-\rho )$. We can deduce

$$\begin{array}{c}〈x_n-S{x}_n,x_n-z〉=〈(I-S)x_n-(I-S)z,x_n-z〉+〈(I-S)z,x_n-z〉\hfill \\ \phantom{〈x_n-S{x}_n,x_n-z〉}\ge 〈(I-S)z,x_n-z〉,\hfill \end{array}$$

(47)

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

(48)

$$\begin{array}{c}〈(I-V_i)y_n,x_n-z〉=〈(I-V_i)y_n-(I-V_i)z,x_n-y_n〉+〈(I-V_i)y_n-(I-V_i)z,y_n-z〉\hfill \\ \phantom{〈(I-V_i)y_n,x_n-z〉}\ge 〈(I-V_i)y_n-(I-V_i)z,x_n-y_n〉\hfill \\ \phantom{〈(I-V_i)y_n,x_n-z〉}=〈(I-V_i)y_n,x_n-y_n〉\hfill \\ \phantom{〈(I-V_i)y_n,x_n-z〉}=〈(I-V_i)y_n,{\beta }_n(x_n-S{x}_n)+(1-{\beta }_n)(x_n-z_n)〉.\hfill \end{array}$$

(49)

From (16), we get

$$\begin{array}{c}〈w_n-P_C[w_n],P_C[w_{n-1}]-z〉\hfill \\ =〈w_n-P_C[w_n],P_C[w_{n-1}]-P_C[w_n]〉+〈w_n-P_C[w_n],P_C[w_n]-z〉\hfill \\ \ge 〈w_n-P_C[w_n],P_C[w_{n-1}]-P_C[w_n]〉.\hfill \end{array}$$

Then, from (46)-(49), we have

$$\begin{array}{rcl}〈v_n,x_n-z〉& \ge & \frac{1}{(1-{\alpha }_n){\beta }_n}〈w_n-P_C[w_n],P_C[w_{n-1}]-P_C[w_n]〉+\frac{{\alpha }_n}{(1-{\alpha }_n){\beta }_n}〈(I-f)z,x_n-z〉\\ +〈(I-S)z,x_n-z〉+\frac{(1-{\beta }_n)}{{\beta }_n}〈x_n-z_n,x_n-z〉\\ +\frac{(1-{\beta }_n)}{(1-{\alpha }_n){\beta }_n}\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)〈(I-V_i)y_n,x_n-z_n〉\\ +\frac{1}{(1-{\alpha }_n)}\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)〈(I-V_i)y_n,x_n-S{x}_n〉+\frac{(1-\rho ){\alpha }_n}{(1-{\alpha }_n){\beta }_n}{\parallel x_n-z\parallel }^2.\end{array}$$

Then we obtain

$$\begin{array}{rcl}{\parallel x_n-z\parallel }^2& \le & \frac{1}{(1-\rho ){\alpha }_n}\parallel w_n-P_C[w_n]\parallel \parallel w_{n-1}-w_n\parallel -\frac{1}{(1-\rho )}〈(I-f)z,x_n-z〉\\ +\frac{(1-{\alpha }_n){\beta }_n}{(1-\rho ){\alpha }_n}(〈v_n,x_n-z〉-〈(I-S)z,x_n-z〉)\\ -\frac{(1-{\beta }_n)(1-{\alpha }_n)}{(1-\rho ){\alpha }_n}〈x_n-z_n,x_n-z〉\\ -\frac{(1-{\beta }_n)}{(1-\rho ){\alpha }_n}\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)〈(I-V_i)y_n,x_n-z_n〉\\ -\frac{{\beta }_n}{(1-\rho ){\alpha }_n}\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)〈(I-V_i)y_n,x_n-S{x}_n〉\\ \le & \frac{\parallel w_{n-1}-w_n\parallel }{(1-\rho ){\alpha }_n}\parallel w_n-P_C[w_n]\parallel -\frac{1}{(1-\rho )}〈(I-f)z,x_n-z〉\\ +\frac{(1-{\alpha }_n){\beta }_n}{(1-\rho ){\alpha }_n}(〈v_n,x_n-z〉-〈(I-S)z,x_n-z〉)\\ +\frac{1}{(1-\rho )}\frac{(1-{\beta }_n)}{{\beta }_n}\frac{{\beta }_n}{{\alpha }_n}\parallel x_n-z_n\parallel \parallel x_n-z\parallel \\ +\frac{1}{(1-\rho )}\frac{(1-{\beta }_n)}{{\beta }_n}\frac{{\beta }_n}{{\alpha }_n}\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)\parallel (I-V_i)y_n\parallel \parallel x_n-z_n\parallel \\ -\frac{{\beta }_n}{(1-\rho ){\alpha }_n}\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)〈(I-V_i)y_n,x_n-S{x}_n〉.\end{array}$$

By condition (e) of Theorem 3.2, there exists a constant $N>0$ such that $\frac{1-{\beta }_n}{{\beta }_n}\le N$. Since ${lim}_{n\to \mathrm{\infty }}\parallel x_n-z_n\parallel =0$, $v_n\to 0$, $(I-V_i)y_n\to 0$ and $\frac{\parallel w_{n-1}-w_n\parallel }{{\alpha }_n}\to 0$ as $n\to \mathrm{\infty }$, then every weak cluster point of $\{x_n\}$ is also a strong cluster point. Since $\{x_n\}$ is bounded, by Lemma 3.2 there exists a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ converging to a point $x^{\ast }\in F(T)$, and by some similar arguments in Theorem 3.1, we can show that $x^{\ast }\in {\mathrm{\Omega }}^{\ast }\cap \mathrm{\Lambda }\cap F(T)$.

From (46)-(49), it follows that for any $z\in {\mathrm{\Omega }}^{\ast }\cap \mathrm{\Lambda }\cap F(T)$,

$$\begin{array}{c}〈(I-f)x_{n_k},x_{n_k}-z〉\hfill \\ =\frac{(1-{\alpha }_{n_k}){\beta }_{n_k}}{{\alpha }_{n_k}}〈v_{n_k},x_{n_k}-z〉-\frac{1}{{\alpha }_{n_k}}〈w_{n_k}-P_C[w_{n_k}],P_C[w_{n_k-1}]-z〉\hfill \\ -\frac{(1-{\alpha }_{n_k}){\beta }_{n_k}}{{\alpha }_{n_k}}〈x_{n_k}-S{x}_{n_k},x_{n_k}-z〉-\frac{(1-{\alpha }_{n_k})(1-{\beta }_{n_k})}{{\alpha }_{n_k}}〈x_{n_k}-z_{n_k},x_{n_k}-z〉\hfill \\ -\frac{1}{{\alpha }_{n_k}}\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)〈y_{n_k}-V_i{y}_{n_k},x_{n_k}-z〉\hfill \\ \le \frac{(1-{\alpha }_{n_k}){\beta }_{n_k}}{{\alpha }_{n_k}}〈v_{n_k},x_{n_k}-z〉+\frac{1}{{\alpha }_{n_k}}\parallel w_{n_k}-P_C[w_{n_k}]\parallel \parallel w_{n_k-1}-w_{n_k}\parallel \hfill \\ -\frac{(1-{\alpha }_{n_k}){\beta }_{n_k}}{{\alpha }_{n_k}}〈x_{n_k}-S{x}_{n_k},x_{n_k}-z〉+\frac{(1-{\beta }_{n_k})}{{\beta }_{n_k}}\frac{{\beta }_{n_k}}{{\alpha }_{n_k}}\parallel x_{n_k}-z_{n_k}\parallel \parallel x_{n_k}-z\parallel \hfill \\ +\frac{(1-{\beta }_{n_k})}{{\beta }_{n_k}}\frac{{\beta }_{n_k}}{{\alpha }_{n_k}}\sum _{i=1}^{n_k}({\alpha }_{i-1}-{\alpha }_i)\parallel (I-V_i)y_{n_k}\parallel \parallel x_{n_k}-z_{n_k}\parallel \hfill \\ -\frac{{\beta }_{n_k}}{{\alpha }_{n_k}}\sum _{i=1}^{n_k}({\alpha }_{i-1}-{\alpha }_i)〈(I-V_i)y_{n_k},x_{n_k}-S{x}_{n_k}〉.\hfill \end{array}$$

(50)

Since ${lim}_{n\to \mathrm{\infty }}\parallel x_n-z_n\parallel =0$, $v_n\to 0$, $(I-V_i)y_n\to 0$ and $\frac{\parallel w_{n-1}-w_n\parallel }{{\alpha }_n}\to 0$, letting $k\to \mathrm{\infty }$ in (50), we obtain

$$〈(I-f)x^{\ast },x^{\ast }-z〉\le -\tau 〈x^{\ast }-S{x}^{\ast },x^{\ast }-z〉,$$

i.e.,

$$〈\frac{1}{\tau }(I-f)x^{\ast }+(I-S)x^{\ast },z-x^{\ast }〉\ge 0.$$

In the following, we show that (43) has a unique solution. Assume that $x^{\prime }$ is another solution. Then we have

$$〈(I-f)x^{\prime },x^{\prime }-x^{\ast }〉\le -\tau 〈x^{\prime }-S{x}^{\prime },x^{\prime }-x^{\ast }〉,$$

(51)

$$〈(I-f)x^{\ast },x^{\ast }-x^{\prime }〉\le -\tau 〈x^{\ast }-S{x}^{\ast },x^{\ast }-x^{\prime }〉.$$

(52)

Adding (51) and (52), we get

$$\begin{array}{rcl}(1-\rho ){\parallel x^{\prime }-x^{\ast }\parallel }^2& \le & 〈(I-f)x^{\prime }-(I-f)x^{\ast },x^{\prime }-x^{\ast }〉\\ \le & -\tau 〈(I-S)x^{\prime }-(I-S)x^{\ast },x^{\prime }-x^{\ast }〉\\ \le & 0.\end{array}$$

Then $x^{\prime }=x^{\ast }$. Since (43) has a unique solution, it follows that $w_w(x_n)=\{x^{\ast }\}$. Since every weak cluster point of $\{x_n\}$ is also a strong cluster point, we conclude that $\{x_n\}\to x^{\ast }$. This completes the proof. □

4 Applications

In this section, we obtain the following results by using a special case of the proposed method. The first result can be viewed as an extension and improvement of the method of Gu et al. [11] for finding an approximate element of the common set of solutions of a split equilibrium problem and a hierarchical fixed point problem in a real Hilbert space.

Corollary 4.1 Let $H_1$ and $H_2$ be two real Hilbert spaces and $C\subseteq H_1$ and $Q\subseteq H_2$ be nonempty closed convex subsets of Hilbert spaces $H_1$ and $H_2$, respectively. Let $A:H_1\to H_2$ be a bounded linear operator. Let $D:C\to H_1$ be an α-inverse strongly monotone mapping. Assume that $F_1:C\times C\to \mathbb{R}$ and $F_2:Q\times Q\to \mathbb{R}$ are the bifunctions satisfying Assumption  2.1 and $F_2$ is upper semicontinuous in the first argument. Let $S:C\to H_1$ be a nonexpansive mapping and ${\{T_i\}}_{i=1}^{\mathrm{\infty }}:C\to C$ be a countable family of $k_i$-strict pseudo-contraction mappings such that $F(T)\cap {\mathrm{\Omega }}^{\ast }\cap \mathrm{\Lambda }\ne \mathrm{\varnothing }$, where $F(T)={\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)$. Let f be a ρ-contraction mapping. For a given $x_0\in C$ arbitrarily, let the iterative sequences $\{u_n\}$, $\{x_n\}$, $\{y_n\}$ and $\{z_n\}$ be generated by

$$\begin{array}{c}{u}_n=T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n);\hfill \\ y_n={\beta }_{n}S{x}_n+(1-{\beta }_n)u_n;\hfill \\ x_{n+1}=P_C[{\alpha }_{n}f(x_n)+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)T_i{y}_n],\mathrm{\forall }n\ge 0,\hfill \end{array}$$

(53)

where $\{r_n\}\subset (0,\mathrm{\infty })$ and $\gamma \in (0,1/L)$, L is the spectral radius of the operator $A^{\ast }A$ and $A^{\ast }$ is the adjoint of A and ${\alpha }_0=1$, $\{{\alpha }_n\}$ is a strictly decreasing sequence in $(0,1)$ and $\{{\beta }_n\}$ is a sequence in $(0,1)$ satisfying the following conditions:

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

• (b) ${lim}_{n\to \mathrm{\infty }}\frac{{\beta }_n}{{\alpha }_n}=\tau \in (0,\mathrm{\infty })$,

• (c) ${\sum }_{n=1}^{\mathrm{\infty }}({\alpha }_{n-1}-{\alpha }_n)<\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}|{\beta }_{n-1}-{\beta }_n|<\mathrm{\infty }$,

• (d) ${lim}_{n\to \mathrm{\infty }}\frac{\frac{1}{\mu }|r_n-r_{n-1}|+|{\alpha }_{n-1}-{\alpha }_n|+|{\beta }_{n-1}-{\beta }_n|}{{\alpha }_n{\beta }_n}=0$,

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

• (f) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$ and ${\sum }_{n=1}^{\mathrm{\infty }}|r_{n-1}-r_n|<\mathrm{\infty }$.

Then the sequence $\{x_n\}$ generated by Algorithm (53) converges strongly to $x^{\ast }\in \mathrm{\Lambda }\cap F(T)$, which is the unique solution of the variational inequality

$$〈\frac{1}{\tau }(I-f)x^{\ast }+(I-S)x^{\ast },x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in \mathrm{\Lambda }\cap F(T).$$

Proof Put ${\lambda }_n=0$ and $k_i=0$, $\mathrm{\forall }i\ge 1$ in Theorem 3.2. Then conclusion of Corollary 4.1 is obtained. □

The following result can be viewed as an extension and improvement of the method of Yao et al. [27] for finding an approximate element of the common set of solutions of a split equilibrium problem and a hierarchical fixed point problem in a real Hilbert space.

Corollary 4.2 Let $H_1$ and $H_2$ be two real Hilbert spaces and $C\subseteq H_1$ and $Q\subseteq H_2$ be nonempty closed convex subsets of Hilbert spaces $H_1$ and $H_2$, respectively. Let $A:H_1\to H_2$ be a bounded linear operator. Let $D:C\to H_1$ be an α-inverse strongly monotone mapping. Assume that $F_1:C\times C\to \mathbb{R}$ and $F_2:Q\times Q\to \mathbb{R}$ are the bifunctions satisfying Assumption  2.1 and $F_2$ is upper semicontinuous in the first argument. Let $S:C\to H_1$ be a nonexpansive mapping and $T:C\to C$ be a k-strict pseudo-contraction mapping such that $F(T)\cap \mathrm{\Lambda }\ne \mathrm{\varnothing }$. Let f be a ρ-contraction mapping. For a given $x_0\in C$ arbitrarily, let the iterative sequences $\{u_n\}$, $\{x_n\}$, $\{y_n\}$ and $\{z_n\}$ be generated by

$$\begin{array}{c}{u}_n=T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n);\hfill \\ y_n={\beta }_{n}S{x}_n+(1-{\beta }_n)u_n;\hfill \\ x_{n+1}=P_C[{\alpha }_{n}f(x_n)+(1-{\alpha }_n)T{y}_n],\mathrm{\forall }n\ge 0,\hfill \end{array}$$

(54)

where $\{r_n\}\subset (0,\mathrm{\infty })$ and $\gamma \in (0,1/L)$, L is the spectral radius of the operator $A^{\ast }A$ and $A^{\ast }$ is the adjoint of A and ${\alpha }_0=1$, $\{{\alpha }_n\}$ is a strictly decreasing sequence in $(0,1)$ and $\{{\beta }_n\}$ is a sequence in $(0,1)$ satisfying the following conditions:

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

• (b) ${lim}_{n\to \mathrm{\infty }}\frac{{\beta }_n}{{\alpha }_n}=\tau \in (0,\mathrm{\infty })$,

• (c) ${\sum }_{n=1}^{\mathrm{\infty }}({\alpha }_{n-1}-{\alpha }_n)<\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}|{\beta }_{n-1}-{\beta }_n|<\mathrm{\infty }$,

• (d) ${lim}_{n\to \mathrm{\infty }}\frac{\frac{1}{\mu }|r_n-r_{n-1}|+|{\alpha }_{n-1}-{\alpha }_n|+|{\beta }_{n-1}-{\beta }_n|}{{\alpha }_n{\beta }_n}=0$,

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

• (f) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$ and ${\sum }_{n=1}^{\mathrm{\infty }}|r_{n-1}-r_n|<\mathrm{\infty }$.

Then the sequence $\{x_n\}$ generated by Algorithm (54) converges strongly to $x^{\ast }\in \mathrm{\Lambda }\cap F(T)$, which is the unique solution of the variational inequality

$$〈\frac{1}{\tau }(I-f)x^{\ast }+(I-S)x^{\ast },x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in \mathrm{\Lambda }\cap F(T).$$

Proof Put ${\lambda }_n=0$, $k_i=0$ and $T_i=T$, $\mathrm{\forall }i\ge 1$ in Theorem 3.2. Then conclusion of Corollary 4.2 is obtained. □