An iterative method for approximating the common solutions of a variational inequality, a mixed equilibrium problem and a hierarchical fixed point problem

Abstract

In this paper, we suggest and analyze an iterative scheme for finding the approximate element of the common set of solutions of a generalized 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.

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 A is a mapping from C into H. A classical variational inequality problem, denoted by $VI(A,C)$, is to find a vector $u\in C$ such that

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

(1.1)

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

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

We now have a variety of techniques to suggest and analyze various iterative algorithms for solving variational inequalities and the related optimization problems, see [1–22]. The fixed-point theory has played an important role in the development of various algorithms for solving variational inequalities. Using the projection operator technique, one usually establishes an equivalence between the variational inequalities and the fixed-point problem. This alternative equivalent formulation was used by Lions and Stampacchia [8] to study the existence of a solution of the variational inequalities.

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

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

(1.2)

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

$$Tx=x.$$

(1.3)

We denote $F(T)$ the set of solutions of (1.3). 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 [22]).

The mixed equilibrium problem, denoted by MEP, is to find $x\in C$ such that

$$F(x,y)+〈Dx,y-x〉\ge 0,\mathrm{\forall }y\in C,$$

(1.4)

where $F:C\times C\to \mathbb{R}$ is a bifunction, and $D:C\to H$ is a nonlinear mapping. This problem was introduced and studied by Moudafi and Théra [13] and Moudafi [14]. The set of solutions of (1.4) is denoted by

$$\mathit{MEP}(F):=\{x\in C:F(x,y)+〈Dx,y-x〉\ge 0,\mathrm{\forall }y\in C\}.$$

(1.5)

If $D=0$, then it is reduced to the equilibrium problem, which is to find $x\in C$ such that

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

(1.6)

The solution set of (1.6) is denoted by $\mathit{EP}(F)$. Numerous problems in physics, optimization, and economics reduce to find a solution of (1.6), see [4, 7, 16, 17]. In 1997, Combettes and Hirstoaga [5] introduced an iterative scheme of finding the best approximation to the initial data when $\mathit{EP}(F)$ is nonempty. Recently Plubtieng and Punpaeng [16] introduced an iterative method for finding the common element of the set $F(T)\cap {\mathrm{\Omega }}^{\ast }\cap \mathit{EP}(F)$.

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

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

(1.7)

It is known that the hierarchical fixed point problem (1.7) links with some monotone variational inequalities and convex programming problems; see [6, 20]. Various methods have been proposed to solve the hierarchical fixed point problem; see Moudafi [15], Mainge and Moudafi in [9], Marino and Xu in [11] and Cianciaruso et al. [3]. Very recently, Yao et al. [20] introduced the following strong convergence iterative algorithm to solve problem (1.7):

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

(1.8)

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 (1.8) converges strongly to $z\in F(T)$, which is the unique solution of the following variational inequality:

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

(1.9)

By changing the restrictions on parameters, the authors obtained another result on the iterative scheme (1.8), the sequence $\{x_n\}$ generated by (1.8) converges strongly to a point $z\in F(T)$, which is the unique solution of the following variational inequality:

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

(1.10)

Let $S:C\to H$ be a nonexpansive mapping, and ${\{T_i\}}_{i=1}^{\mathrm{\infty }}:C\to C$ is a countable family of nonexpansive mappings. Very recently, Gu et al. [6] introduced the following iterative algorithm:

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

(1.11)

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 (1.11) converges strongly to $z\in {\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)$, which is a unique solution of one of the variational inequalities (1.9) and (1.10).

In this paper, motivated by the work of Yao et al. [20] and Gu et al. [6] and by the recent work going in this direction, we give an iterative method for finding the approximate element of the common set of solutions of (1.1), (1.4) and (1.7) 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., [3, 6, 9, 20] 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 projection onto C.

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

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

(2.1)

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

(2.2)

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

(2.3)

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

(2.4)

Lemma 2.2 [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.

Let $r>0$ and $x\in H$. Then there exists $z\in C$ such that

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

Lemma 2.3 [5]

Assume that $F:C\times C\to \mathbb{R}$ satisfies assumptions (i)-(iv) of Lemma 2.2. For $r>0$ and $\mathrm{\forall }x\in H$, define a mapping $T_r:H\to C$ as follows:

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

Then the following hold:

• (i) $T_r$ is single-valued;

• (ii) $T_r$ is firmly nonexpansive, i.e.,

  $${\parallel T_{r}x-T_{r}y\parallel }^2\le 〈T_{r}x-T_{r}y,x-y〉,\mathrm{\forall }x,y\in H;$$

• (iii) $F(T_r)=\mathit{EP}(F)$;

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

Lemma 2.4 [21]

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 [10]

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 [19]

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 [22]

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 [6]

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 the common solutions of the variational inequality (1.1), the mixed equilibrium problem (1.4) and the hierarchical fixed point problem (1.7).

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $D,A:C\to H$ be θ, α-inverse strongly monotone mappings, respectively. Let $F:C\times C\to \mathbb{R}$ be a bifunction satisfying assumptions (i)-(iv) of Lemma 2.2, $S:C\to H$ be a nonexpansive mapping and ${\{T_i\}}_{i=1}^{\mathrm{\infty }}:C\to C$ is a countable family of $k_i$-strict pseudo-contraction mappings such that $F(T)\cap {\mathrm{\Omega }}^{\ast }\cap \mathit{MEP}(F)\ne \mathrm{\varnothing }$, 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

$$\begin{array}{r}F(u_n,y)+〈D{x}_n,y-u_n〉+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in C;\\ z_n=P_C[u_n-{\lambda }_{n}A{u}_n];\\ y_n=P_C[{\beta }_{n}S{x}_n+(1-{\beta }_n)z_n];\\ x_{n+1}=P_C[{\alpha }_{n}f(x_n)+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)V_i{y}_n],\mathrm{\forall }n\ge 0,\end{array}$$

(3.1)

where $V_i=k_{i}I+(1-k_i)T_i$, $0\le k_i<1$, $\{{\lambda }_n\}\subset (0,2\alpha )$, $\{r_n\}\subset (0,2\theta ]$, ${\alpha }_0=1$, $\{{\alpha }_n\}$ 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 ${\sum }_{n=1}^{\mathrm{\infty }}|{\lambda }_{n-1}-{\lambda }_n|<\mathrm{\infty }$.

Remark 3.1 It is easy to verify that Algorithm 3.1 includes some existing methods (e.g., [3, 6, 9, 20]) as special cases. Therefore, the new algorithm is expected to be widely applicable.

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

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

$$\begin{array}{r}{\parallel (I-r_{n}D)x-(I-r_{n}D)y\parallel }^2\\ ={\parallel (x-y)-r_n(Dx-Dy)\parallel }^2\\ ={\parallel x-y\parallel }^2-2r_n〈x-y,Dx-Dy〉+r_n^2{\parallel Dx-Dy\parallel }^2\\ \le {\parallel x-y\parallel }^2-r_n(2\theta -r_n){\parallel Dx-Dy\parallel }^2\\ \le {\parallel x-y\parallel }^2.\end{array}$$

Similarly, we can show that the mapping $(I-{\lambda }_{n}A)$ is nonexpansive. It follows from Lemma 2.3 that $u_n=T_{r_n}(x_n-r_{n}D{x}_n)$. Let $x^{\ast }\in F(T)\cap {\mathrm{\Omega }}^{\ast }\cap \mathit{MEP}(F)$, we have $x^{\ast }=T_{r_n}(x^{\ast }-r_{n}D{x}^{\ast })$.

$$\begin{array}{r}{\parallel u_n-x^{\ast }\parallel }^2\\ ={\parallel T_{r_n}(x_n-r_{n}D{x}_n)-T_{r_n}(x^{\ast }-r_{n}D{x}^{\ast })\parallel }^2\\ \le {\parallel (x_n-r_{n}D{x}_n)-(x^{\ast }-r_{n}D{x}^{\ast })\parallel }^2\\ \le {\parallel x_n-x^{\ast }\parallel }^2-r_n(2\theta -r_n){\parallel D{x}_n-D{x}^{\ast }\parallel }^2\\ \le {\parallel x_n-x^{\ast }\parallel }^2.\end{array}$$

(3.2)

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

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

(3.3)

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 (3.1), we have

$$\begin{array}{rcl}\parallel x_{n+1}-x^{\ast }\parallel & \le & \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 & \text{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}$$

(3.4)

By induction on n, we obtain $\parallel x_n-x^{\ast }\parallel \le \text{max}\{\parallel x_0-x^{\ast }\parallel ,\frac{1}{1-\rho }(\parallel f(x^{\ast })-x^{\ast }\parallel +\parallel S{x}^{\ast }-x^{\ast }\parallel )\}$ 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 $x^{\ast }\in F(T)\cap {\mathrm{\Omega }}^{\ast }\cap \mathit{MEP}(F)$ 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 $(I-{\lambda }_{n}A)$ and $P_C$, we have

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

(3.5)

Next, we estimate that

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

(3.6)

It follows from (3.5) and (3.6) that

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

(3.7)

On the other hand, $u_n=T_{r_n}(x_n-r_{n}D{x}_n)$ and $u_{n-1}=T_{r_{n-1}}(x_{n-1}-r_{n-1}D{x}_{n-1})$, we have

$$F(u_n,y)+〈D{x}_n,y-u_n〉+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in C$$

(3.8)

and

$$F(u_{n-1},y)+〈D{x}_{n-1},y-u_{n-1}〉+\frac{1}{r_{n-1}}〈y-u_{n-1},u_{n-1}-x_{n-1}〉\ge 0,\mathrm{\forall }y\in C.$$

(3.9)

Take $y=u_{n-1}$ in (3.8) and $y=u_n$ in (3.9), we get

$$F(u_n,u_{n-1})+〈D{x}_n,u_{n-1}-u_n〉+\frac{1}{r_n}〈u_{n-1}-u_n,u_n-x_n〉\ge 0$$

(3.10)

and

$$F(u_{n-1},u_n)+〈D{x}_{n-1},u_n-u_{n-1}〉+\frac{1}{r_{n-1}}〈u_n-u_{n-1},u_{n-1}-x_{n-1}〉\ge 0.$$

(3.11)

Adding (3.10) and (3.11) and using the monotonicity of F, we have

$$〈D{x}_{n-1}-D{x}_n,u_n-u_{n-1}〉+〈u_n-u_{n-1},\frac{u_{n-1}-x_{n-1}}{r_{n-1}}-\frac{u_n-x_n}{r_n}〉\ge 0,$$

which implies that

$$\begin{array}{rcl}0& \le & 〈u_n-u_{n-1},r_n(D{x}_{n-1}-D{x}_n)+\frac{r_n}{r_{n-1}}(u_{n-1}-x_{n-1})-(u_n-x_n)〉\\ =& 〈u_{n-1}-u_n,u_n-u_{n-1}+(1-\frac{r_n}{r_{n-1}})u_{n-1}\\ +(x_{n-1}-r_{n}D{x}_{n-1})-(x_n-r_{n}D{x}_n)-x_{n-1}+\frac{r_n}{r_{n-1}}{x}_{n-1}〉\\ =& 〈u_{n-1}-u_n,(1-\frac{r_n}{r_{n-1}})u_{n-1}+(x_{n-1}-r_{n}D{x}_{n-1})-(x_n-r_{n}D{x}_n)-x_{n-1}+\frac{r_n}{r_{n-1}}{x}_{n-1}〉\\ -{\parallel u_n-u_{n-1}\parallel }^2\\ =& 〈u_{n-1}-u_n,(1-\frac{r_n}{r_{n-1}})(u_{n-1}-x_{n-1})+(x_{n-1}-r_{n}D{x}_{n-1})-(x_n-r_{n}D{x}_n)〉\\ -{\parallel u_n-u_{n-1}\parallel }^2\\ \le & \parallel u_{n-1}-u_n\parallel \{|1-\frac{r_n}{r_{n-1}}|\parallel u_{n-1}-x_{n-1}\parallel +\parallel (x_{n-1}-r_{n}D{x}_{n-1})-(x_n-r_{n}D{x}_n)\parallel \}\\ -{\parallel u_n-u_{n-1}\parallel }^2\\ \le & \parallel u_{n-1}-u_n\parallel \{|1-\frac{r_n}{r_{n-1}}|\parallel u_{n-1}-x_{n-1}\parallel +\parallel x_{n-1}-x_n\parallel \}-{\parallel u_n-u_{n-1}\parallel }^2,\end{array}$$

and then

$$\parallel u_{n-1}-u_n\parallel \le |1-\frac{r_n}{r_{n-1}}|\parallel u_{n-1}-x_{n-1}\parallel +\parallel x_{n-1}-x_n\parallel .$$

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

$$\parallel u_{n-1}-u_n\parallel \le \parallel x_{n-1}-x_n\parallel +\frac{1}{\mu }|r_{n-1}-r_n|\parallel u_{n-1}-x_{n-1}\parallel .$$

(3.12)

It follows from (3.7) and (3.12) that

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

(3.13)

Next, we estimate that

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

(3.14)

From (3.13) and (3.14), we have

$$\begin{array}{rcl}\parallel x_{n+1}-x_n\parallel & \le & {\alpha }_n\rho \parallel x_n-x_{n-1}\parallel \\ +(1-{\alpha }_n)\{\parallel x_n-x_{n-1}\parallel \\ +(1-{\beta }_n)(\frac{1}{\mu }|r_n-r_{n-1}|\parallel u_{n-1}-x_{n-1}\parallel +|{\lambda }_n-{\lambda }_{n-1}|\parallel A{u}_{n-1}\parallel )\\ +|{\beta }_n-{\beta }_{n-1}|(\parallel S{x}_{n-1}\parallel +\parallel z_{n-1}\parallel )\}+|{\alpha }_n-{\alpha }_{n-1}|(\parallel f(x_{n-1})\parallel +\parallel V_n{y}_{n-1}\parallel )\\ \le & (1-(1-\rho ){\alpha }_n)\parallel x_n-x_{n-1}\parallel +\frac{1}{\mu }|r_n-r_{n-1}|\parallel u_{n-1}-x_{n-1}\parallel \\ +|{\lambda }_n-{\lambda }_{n-1}|\parallel A{u}_{n-1}\parallel \\ +|{\beta }_n-{\beta }_{n-1}|(\parallel S{x}_{n-1}\parallel +\parallel z_{n-1}\parallel )+|{\alpha }_n-{\alpha }_{n-1}|(\parallel f(x_{n-1})\parallel +\parallel V_n{y}_{n-1}\parallel )\\ \le & (1-(1-\rho ){\alpha }_n)\parallel x_n-x_{n-1}\parallel \\ +M(\frac{1}{\mu }|r_n-r_{n-1}|+|{\lambda }_n-{\lambda }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|+|{\alpha }_n-{\alpha }_{n-1}|).\end{array}$$

(3.15)

Where

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

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 $x^{\ast }\in F(T)\cap {\mathrm{\Omega }}^{\ast }\cap \mathit{MEP}(F)$ and ${\alpha }_n+{\sum }_{i=1}^n({\alpha }_{i-1}-{\alpha }_i)=1$, by using (3.2) and (3.3), we obtain

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

(3.16)

Then from the inequality above, we get

$$\begin{array}{r}(1-{\alpha }_n)(1-{\beta }_n)\{r_n(2\theta -r_n){\parallel D{x}_n-D{x}^{\ast }\parallel }^2+{\lambda }_n(2\alpha -{\lambda }_n){\parallel A{u}_n-A{x}^{\ast }\parallel }^2\}\\ \le {\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2+{\beta }_n{\parallel S{x}_n-x^{\ast }\parallel }^2+{\parallel x_n-x^{\ast }\parallel }^2-{\parallel x_{n+1}-x^{\ast }\parallel }^2\\ \le {\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2+{\beta }_n{\parallel S{x}_n-x^{\ast }\parallel }^2+(\parallel x_n-x^{\ast }\parallel +\parallel x_{n+1}-x^{\ast }\parallel )\parallel x_{n+1}-x_n\parallel .\end{array}$$

Since ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\lambda }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\lambda }_n<2\alpha$, ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$, ${\alpha }_n\to 0$ and ${\beta }_n\to 0$, we obtain ${lim}_{n\to \mathrm{\infty }}\parallel D{x}_n-D{x}^{\ast }\parallel =0$ and ${lim}_{n\to \mathrm{\infty }}\parallel A{u}_n-A{x}^{\ast }\parallel =0$.

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

$$\begin{array}{rcl}{\parallel u_n-x^{\ast }\parallel }^2& =& {\parallel T_{r_n}(x_n-r_{n}D{x}_n)-T_{r_n}(x^{\ast }-r_{n}D{x}^{\ast })\parallel }^2\\ \le & 〈u_n-x^{\ast },(x_n-r_{n}D{x}_n)-(x^{\ast }-r_{n}D{x}^{\ast })〉\\ =& \frac{1}{2}\{{\parallel u_n-x^{\ast }\parallel }^2+{\parallel (x_n-r_{n}D{x}_n)-(x^{\ast }-r_{n}D{x}^{\ast })\parallel }^2\\ -{\parallel u_n-x^{\ast }-[(x_n-r_{n}D{x}_n)-(x^{\ast }-r_{n}D{x}^{\ast })]\parallel }^2\}.\end{array}$$

Hence,

$$\begin{array}{rcl}{\parallel u_n-x^{\ast }\parallel }^2& \le & {\parallel (x_n-r_{n}D{x}_n)-(x^{\ast }-r_{n}D{x}^{\ast })\parallel }^2-{\parallel u_n-x_n+r_n(D{x}_n-D{x}^{\ast })\parallel }^2\\ \le & {\parallel x_n-x^{\ast }\parallel }^2-{\parallel u_n-x_n+r_n(D{x}_n-D{x}^{\ast })\parallel }^2\\ \le & {\parallel x_n-x^{\ast }\parallel }^2-{\parallel u_n-x_n\parallel }^2+2r_n\parallel u_n-x_n\parallel \parallel D{x}_n-D{x}^{\ast }\parallel .\end{array}$$

From (3.16), (3.3) and the inequality above, we have

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

Hence,

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

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

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

(3.17)

From (2.2), we get

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

Hence,

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

From (3.16) and the inequality above, we have

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

Hence,

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

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

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

(3.18)

It follows from (3.17) and (3.18) that

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

(3.19)

Now, let $z\in F(T)\cap {\mathrm{\Omega }}^{\ast }\cap \mathit{MEP}(F)$, 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

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

It follows that

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

From Lemma 2.9 and the inequality above, we get

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

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 $z=P_{{\mathrm{\Omega }}^{\ast }\cap \mathit{MEP}(F)\cap F(T)}f(z)$, which is the unique solution of the variational inequality

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

(3.20)

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 $w\in \mathit{MEP}(F)$. Since $u_n=T_{r_n}(x_n-r_{n}D{x}_n)$, we have

$$F(u_n,y)+〈D{x}_n,y-u_n〉+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in C.$$

It follows from monotonicity of F that

$$〈D{x}_n,y-u_n〉+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge F(y,u_n),\mathrm{\forall }y\in C$$

and

$$〈D{x}_{n_k},y-u_{n_k}〉+〈y-u_{n_k},\frac{u_{n_k}-x_{n_k}}{r_{n_k}}〉\ge F(y,u_{n_k}),\mathrm{\forall }y\in C.$$

(3.21)

Since ${lim}_{n\to \mathrm{\infty }}\parallel u_n-x_n\parallel =0$ and $x_n\rightharpoonup w$, it easy to observe that $u_{n_k}\to w$. 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 (3.21), we obtain

$$\begin{array}{rcl}〈D{y}_t,y_t-u_{n_k}〉& \ge & 〈D{y}_t,y_t-u_{n_k}〉-〈D{x}_{n_k},y_t-u_{n_k}〉-〈y_t-u_{n_k},\frac{u_{n_k}-x_{n_k}}{r_{n_k}}〉+F(y_t,u_{n_k})\\ =& 〈D{y}_t-D{u}_{n_k},y_t-u_{n_k}〉+〈D{u}_{n_k}-D{x}_{n_k},y_t-u_{n_k}〉\\ -〈y_t-u_{n_k},\frac{u_{n_k}-x_{n_k}}{r_{n_k}}〉+F(y_t,u_{n_k}).\end{array}$$

(3.22)

Since D is Lipschitz continuous and ${lim}_{n\to \mathrm{\infty }}\parallel u_n-x_n\parallel =0$, we obtain ${lim}_{k\to \mathrm{\infty }}\parallel D{u}_{n_k}-D{x}_{n_k}\parallel =0$. From the monotonicity of D and $u_{n_k}\to w$, it follows from (3.22) that

$$〈D{y}_t,y_t-w〉\ge F(y_t,w).$$

(3.23)

Hence, from assumptions (i)-(iv) of Lemma 2.2 and (3.23), we have

$$\begin{array}{rcl}0=F(y_t,y_t)& \le & tF(y_t,y)+(1-t)F(y_t,w)\\ \le & tF(y_t,y)+(1-t)〈D{y}_t,y_t-w〉\\ \le & tF(y_t,y)+(1-t)t〈D{y}_t,y-w〉,\end{array}$$

(3.24)

which implies that $F(y_t,y)+(1-t)〈D{y}_t,y-w〉\ge 0$. Letting $t\to 0_{+}$, we have

$$F(w,y)+〈Dw,y-w〉\ge 0,\mathrm{\forall }y\in C,$$

which implies that $w\in \mathit{MEP}(F)$.

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

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

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 [18]). Let $G(T)$ denote the graph of T, and let $(v,u)\in G(T)$, since $u-Av\in N_{C}v$ and $z_n\in C$, we have

$$〈v-z_n,u-Av〉\ge 0.$$

(3.25)

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

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

and

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

Therefore, from (3.25) and inverse strongly monotonicity of A, we have

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

Since ${lim}_{n\to \mathrm{\infty }}\parallel u_n-z_n\parallel =0$ and $u_{n_k}\to w$, it 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

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

Next, we claim that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈f(z)-z,x_n-z〉\le 0$, where $z=P_{{\mathrm{\Omega }}^{\ast }\cap \mathit{MEP}(F)\cap F(T)}f(z)$.

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 $x_n\to z$.

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

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 $x_n\to z$.

Since $P_{{\mathrm{\Omega }}^{\ast }\cap \mathit{MEP}(F)\cap F(T)}f$ is a contraction, there exists a unique $z\in C$ such that $z=P_{{\mathrm{\Omega }}^{\ast }\cap \mathit{MEP}(F)\cap F(T)}f(z)$. From (2.1), it follows that z is the unique solution of problem (3.20). This completes the proof. □

Theorem 3.2 Let C be a nonempty closed convex subset of a real Hilbert space H. Let $D,A:C\to H$ be θ, α-inverse strongly monotone mappings, respectively. Let $F:C\times C\to \mathbb{R}$ be a bifunction satisfying the assumptions (i)-(iv) of Lemma 2.2, $S:C\to H$ be a nonexpansive mapping, and ${\{T_i\}}_{i=1}^{\mathrm{\infty }}:C\to C$ is a countable family of $k_i$-strict pseudo-contraction mappings such that $F(T)\cap {\mathrm{\Omega }}^{\ast }\cap \mathit{MEP}(F)\ne \mathrm{\varnothing }$, 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

$$\begin{array}{r}F(u_n,y)+〈D{x}_n,y-u_n〉+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in C;\\ z_n=P_C[u_n-{\lambda }_{n}A{u}_n];\\ y_n={\beta }_{n}S{x}_n+(1-{\beta }_n)z_n;\\ x_{n+1}=P_C[{\alpha }_{n}f(x_n)+\sum _{i=1}^n({\alpha }_{i-1}-{\alpha }_i)V_i{y}_n],\mathrm{\forall }n\ge 0,\end{array}$$

(3.26)

where $V_i=k_{i}I+(1-k_i)T_i$, $0\le k_i<1$, ${\alpha }_0=1$, $\{{\alpha }_n\}$ 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 sequence $\{x_n\}$ generated by (3.26) converges strongly to $x^{\ast }\in {\mathrm{\Omega }}^{\ast }\cap \mathit{MEP}(F)\cap F(T)$, which is the unique solution of the variational inequality

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

(3.27)

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 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 (3.19)) and $(I-V_i)x_n\to 0$. Then we have

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

(3.28)

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

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

(3.29)

From (3.28) and (3.29), we have

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

Set $w_n={\alpha }_{n}f(x_n)+{\sum }_{i=1}^n({\alpha }_{i-1}-{\alpha }_i)V_i{y}_n$. From (3.14) and (3.15), we obtain

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