General iterative methods for generalized equilibrium problems and fixed point problems of k-strict pseudo-contractions

Abstract

In this paper, we modify the general iterative method to approximate a common element of the set of solutions of generalized equilibrium problems and the set of common fixed points of a finite family of k-strictly pseudo-contractive nonself mappings. Strong convergence theorems are established under some suitable conditions in a real Hilbert space, which also solves some variation inequality problems. Results presented in this paper may be viewed as a refinement and important generalizations of the previously known results announced by many other authors.

MSC:47H05, 47H09, 47H10.

1 Introduction

Let H be a real Hilbert space with inner product $〈\cdot ,\cdot 〉$ and norm $\parallel \cdot \parallel$, respectively. Let K be a nonempty closed convex subset of H. Let $A:K\to H$ be a nonlinear mapping and $F:K\times K\to \mathbb{R}$ be a bi-function, where $\mathbb{R}$ denotes the set of real numbers. We consider the following generalized equilibrium problem: Find $x\in K$ such that

$$F(x,y)+〈Ax,y-x〉\ge 0,\mathrm{\forall }y\in K.$$

(1.1)

We use $\mathit{EP}(F,A)$ to denote the solution set of the problem (1.1). If $A\equiv 0$, the zero mapping, then the problem (1.1) is reduced to the normal equilibrium problem: Find $x\in K$ such that

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

(1.2)

We use $\mathit{EP}(F)$ to denote the solution set of the problem (1.2). If $F\equiv 0$, then the problem (1.1) is reduced to the classical variational inequality problem: Find $x\in K$ such that

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

The generalized equilibrium problem (1.1) is very general in the sense that it includes, as special cases, optimization problems, variational inequalities, mini-max problems, the Nash equilibrium problem in noncooperative games and others (see, e.g., [1–3]).

Recall that a nonself mapping $T:K\to H$ is called a k-strict pseudo-contraction if there exists a constant $k\in [0,1)$ such that

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

(1.3)

We use $F(T)$ to denote the fixed point set of T, i.e., $F(T):=\{x\in K:Tx=x\}$. As $k=0$, T is said to be nonexpansive, i.e.,

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

T is said to be pseudo-contractive if $k=1$, and is also said to be strongly pseudo-contractive if there exists a positive constant $\lambda \in (0,1)$ such that $T+\lambda I$ is pseudo-contractive. Clearly, the class of k-strict pseudo-contractions falls into the one between classes of nonexpansive mappings and pseudo-contractions. We remark also that the class of strongly pseudo-contractive mappings is independent of the class of k-strict pseudo-contractions (see, e.g., [4, 5]).

Iterative methods for equilibrium problems and fixed point problems of nonexpansive mappings have been extensively investigated. However, iterative schemes for strict pseudo-contractions are far less developed than those for nonexpansive mappings though Browder and Petryshyn [5] initiated their work in 1967; the reason is probably that the second term appearing in the right-hand side of (1.3) impedes the convergence analysis for iterative algorithms used to find a fixed point of the strict pseudo-contraction. On the other hand, strict pseudo-contractions have more powerful applications than nonexpansive mappings do in solving inverse problems; see, e.g., [6–18, 20–27] and the references therein. Therefore it is interesting to develop the effective iterative methods for equilibrium problems and fixed point problems of strict pseudo-contractions.

In 2006, Marino and Xu [8] introduced a general iterative method and proved that for a given $x_0\in H$, the sequence $\{x_n\}$ generated by

$$x_{n+1}={\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}B)T{x}_n,\mathrm{\forall }n\in N,$$

where T is a self-nonexpansive mapping on H, f is a contraction of H into itself and $\{{\alpha }_n\}\subseteq (0,1)$ satisfies certain conditions, B is a strongly positive bounded linear operator on H, converges strongly to $x^{\ast }\in F(T)$, which is the unique solution of the following variational inequality:

$$〈(B-\gamma f)x^{\ast },x^{\ast }-x〉\le 0,\mathrm{\forall }x\in F(T),$$

and is also the optimality condition for some minimization problem.

Recently, Takahashi and Takahashi [12] considered the equilibrium problem and nonexpansive mapping by viscosity approximation methods. To be more precise, they proved the following theorem.

Theorem of TT Let K be a nonempty closed convex subset of H. Let F be a bi-function from $K\times K$ to $\mathbb{R}$ satisfying (A 1)-(A 4) and let $T:K\to H$ be a nonexpansive mapping such that $F(T)\cap \mathit{EP}(F)\ne \varphi$. Let $f:H\to H$ be a contraction and let $\{x_n\}$ and $\{y_n\}$ be sequences generated by $x_1\in H$ and

$$\{\begin{array}{c}F(y_n,z)+\frac{1}{r_n}〈z-y_n,y_n-x_n〉\ge 0,\mathrm{\forall }z\in K,\hfill \\ x_{n+1}={\alpha }_{n}f(x_n)+(1-{\alpha }_n)T{y}_n,n\ge 1,\hfill \end{array}$$

where $\{{\alpha }_n\}\subset [0,1]$ and $\{r_n\}$ satisfy

Then $\{x_n\}$ and $\{y_n\}$ converge strongly to $q\in F(T)\cap \mathit{EP}(F)$, where $q=P_{F(T)\cap \mathit{EP}(F)}f(q)$.

In 2009, Ceng et al. [15] further studied the equilibrium problem and fixed point problems of strict pseudo-contraction mappings T by an iterative scheme for finding an element of $\mathit{EP}(F)\cap F(T)$. Very recently, by using the general iterative method Liu [16] proposed the implicit and explicit iterative processes for finding an element of $\mathit{EP}(F)\cap F(T)$ and then obtained some strong convergence theorems, respectively. On the other hand, Takahashi and Takahashi [18] considered the generalized equilibrium problem and nonexpansive mapping in a Hilbert space. Moreover, they constructed an iterative scheme for finding an element of $\mathit{EP}(F,A)\cap F(T)$ and then proved a strong convergence of the iterative sequence under some suitable conditions.

In this paper, inspired and motivated by research going on in this area, we introduce a general iterative method for generalized equilibrium problems and strict pseudo-contractive nonself mappings, which is defined in the following way:

$$\{\begin{array}{c}F(u_n,y)+〈A{x}_n,y-u_n〉+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in K,\hfill \\ y_n={\beta }_n{u}_n+(1-{\beta }_n){\sum }_{i=1}^N{\eta }_i^{(n)}{T}_i{u}_n,\hfill \\ x_{n+1}={\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}B)y_n,n\ge 1,\hfill \end{array}$$

(1.4)

where constant $\gamma >0$, f is a contraction and A, B are two operators, ${\{T_i\}}_{i=i}^N:K\to H$ is a finite family of $k_i$-strict pseudo-contractions, ${\{{\eta }_i^{(n)}\}}_{i=1}^N$ is a finite sequence of positive numbers, $\{{\alpha }_n\}$, $\{{\beta }_n\}$ and $\{r_n\}$ are some sequences with certain conditions.

Our purpose is not only to modify the general iterative method to the case of a finite family of $k_i$-strictly pseudo-contractive nonself mappings, but also to establish strong convergence theorems for a generalized equilibrium problem and $k_i$-strict pseudo-contractions in a real Hilbert space, which also solves some variation inequality problems. Our theorems presented in this paper improve and extend the corresponding results of [12, 15, 16, 18, 20, 21, 25].

2 Preliminaries

Let K be a nonempty closed convex subset of a real Hilbert H space with inner product $〈\cdot ,\cdot 〉$ and norm $\parallel \cdot \parallel$, respectively. Recall that a mapping $f:K\to K$ is a contraction, if there exists a constant $\rho \in (0,1)$ such that

$$\parallel f(x)-f(y)\parallel \le \rho \parallel x-y\parallel ,\mathrm{\forall }x,y\in K.$$

We use ${\mathrm{\Pi }}_K$ to denote the collection of all contractions on K. The operator $A:K\to H$ is said to be monotone if

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

$A:K\to H$ is said to be r-strongly monotone if there exists a constant $r>0$ such that

$$〈Ax-Ay,x-y〉\ge r{\parallel x-y\parallel }^2,\mathrm{\forall }x,y\in K.$$

$A:K\to H$ is said to be α-inverse strongly monotone if there exists a constant $\alpha >0$ such that

$$〈Ax-Ay,x-y〉\ge \alpha {\parallel Ax-Ay\parallel }^2,\mathrm{\forall }x,y\in K.$$

Recall that an operator B is strongly positive if there exists a constant $\overline{\gamma }>0$ with the property

$$〈Bx,x〉\ge \overline{\gamma }{\parallel x\parallel }^2,\mathrm{\forall }x\in H.$$

To study the generalized equilibrium problem (1.1), we may assume that the bi-function $F:K\times K\to \mathbb{R}$ satisfies the following conditions:

(A1) $F(x,x)=0$ for all $x\in K$;

(A2) F is monotone, i.e., $F(x,y)+F(y,x)\le 0$ for all $x,y\in K$;

(A3) for each $x,y,z\in K$, ${lim}_{t\to 0}F(tz+(1-t)x,y)\le F(x,y)$;

(A4) for each $x\in K$, $y\mapsto F(x,y)$ is convex and lower semi-continuous.

In order to prove our main results, we need the following lemmas and propositions.

Lemma 2.1 [1, 3]

Let $F:K\times K\to \mathbb{R}$ be a bi-function satisfying (A 1)-(A 4). Then, for any $r>0$ and $x\in H$, there exists $z\in K$ such that

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

Further, if $T_{r}x=\{z\in K:F(z,y)+\frac{1}{r}〈y-z,z-x〉\ge 0,\mathrm{\forall }y\in K\}$, then the following hold:

• (1) $T_r$ is single-valued;

• (2) $T_r$ is firmly nonexpansive, i.e, ${\parallel T_{r}x-T_{r}y\parallel }^2\le 〈T_{r}x-T_{r}y,x-y〉$ for all $x,y\in H$;

• (3) $F(T_r)=\mathit{EP}(F)$;

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

Lemma 2.2 [8]

In the Hilbert space H, there hold the following identities:

• (i) ${\parallel x+y\parallel }^2={\parallel x\parallel }^2+2〈x,y〉+{\parallel y\parallel }^2\le {\parallel x\parallel }^2+2〈y,(x+y)〉$, $\mathrm{\forall }x,y\in H$;

• (ii) ${\parallel tx+(1-t)y\parallel }^2=t{\parallel x\parallel }^2+(1-t){\parallel y\parallel }^2-t(1-t){\parallel x-y\parallel }^2$, $\mathrm{\forall }t\in [0,1]$, $\mathrm{\forall }x,y\in H$.

Lemma 2.3 [8]

Assume that B is a strongly positive linear bounded operator on the Hilbert space H with a coefficient $\overline{\gamma }>0$ and $0<\varrho <{\parallel B\parallel }^{-1}$. Then $\parallel I-\varrho B\parallel \le 1-\varrho \overline{\gamma }$.

Lemma 2.4 [10]

If $T:K\to H$ is a k-strict pseudo-contraction, then the fixed point set $F(T)$ is closed convex so that the projection $P_{F(T)}$ is well defined.

Lemma 2.5 [2, 10]

Let $T:K\to H$ be a k-strict pseudo-contraction. For $\lambda \in [k,1)$, define $S:K\to H$ by $Sx=\lambda x+(1-\lambda )Tx$ for each $x\in K$. Then S is a nonexpansive mapping such that $F(S)=F(T)$.

Lemma 2.6 [19]

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

$$a_{n+1}\le (1-{\gamma }_n)a_n+{\gamma }_n{\delta }_n,n\ge 0,$$

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

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

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

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

Proposition 2.1 (See, e.g., Acedo and Xu [20])

Let K be a nonempty closed convex subset of the Hilbert space H. Given an integer $N\ge 1$, assume that ${\{T_i\}}_{i=1}^N:K\to H$ is a finite family of $k_i$-strict pseudo-contractions. Suppose that ${\{{\lambda }_i\}}_{i=1}^N$ is a positive sequence such that ${\sum }_{i=1}^N{\lambda }_i=1$. Then ${\sum }_{i=1}^N{\lambda }_i{T}_i$ is a k-strict pseudo-contraction with $k=max\{k_i:1\le i\le N\}$.

Proposition 2.2 (See, e.g., Acedo and Xu [20])

Let ${\{T_i\}}_{i=1}^N$ and ${\{{\lambda }_i\}}_{i=1}^N$ be given as in Proposition 2.1 above. Then $F({\sum }_{i=1}^N{\lambda }_i{T}_i)={\bigcap }_{i=1}^{N}F(T_i)$.

3 Main results

Theorem 3.1 Let K be a nonempty closed convex subset of the Hilbert space H and $F:K\times K\to \mathbb{R}$ be a bi-function satisfying (A 1)-(A 4). Let A be an α-inverse strongly monotone mapping and B be a strongly positive bounded linear operator on H with $\overline{\gamma }>0$. Assume that ${\{T_i\}}_{i=1}^N:K\to H$ be a finite family of $k_i$-strict pseudo-contractions such that $\mathcal{F}={\bigcap }_{i=1}^{N}F(T_i)\cap \mathit{EP}(F,A)\ne \varphi$. Suppose $f\in {\mathrm{\Pi }}_K$ with a coefficient $\rho \in (0,1)$ and ${\{{\eta }_i^{(n)}\}}_{i=1}^N$ are finite sequences of positive numbers such that ${\sum }_{i=1}^N{\eta }_i^{(n)}=1$ for all $n\ge 0$, for a given point $x_0\in K$, ${\alpha }_n,{\beta }_n\in (0,1)$, $r_n\in (0,2\alpha )$ and $0<\gamma <\frac{\overline{\gamma }}{\rho }$, the following control conditions are satisfied:

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

• (ii) $k_i\le {\beta }_n\le \lambda <1$, ${lim}_{n\to \mathrm{\infty }}{\beta }_n=\lambda$ and ${\sum }_{n=1}^{\mathrm{\infty }}|{\beta }_n-{\beta }_{n-1}|<\mathrm{\infty }$;

• (iii) ${\sum }_{n=1}^{\mathrm{\infty }}{\sum }_{i=1}^N|{\eta }_i^{(n)}-{\eta }_i^{(n-1)}|<\mathrm{\infty }$;

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

Then the sequence $\{x_n\}$ generated by (1.4) converges strongly to $q\in \mathcal{F}$, which solves the variational inequality

$$〈(B-\gamma f)q,q-p〉\le 0,\mathrm{\forall }p\in \mathcal{F}.$$

Proof Putting $W_n={\sum }_{i=1}^N{\eta }_i^{(n)}{T}_i$, we have $W_n:K\to H$ is a k-strict pseudo-contraction and $F(W_n)={\bigcap }_{i=1}^{N}F(T_i)$ by Proposition 2.1 and 2.2, where $k=max\{k_i:1\le i\le N\}$.

First, we show that the mapping $I-r_{n}A$ is nonexpansive. Indeed, for each $x,y\in K$, we have

$$\begin{array}{rcl}{\parallel (I-r_{n}A)x-(I-r_{n}A)y\parallel }^2& =& {\parallel x-y\parallel }^2-2r_n〈x-y,Ax-Ay〉+r_n^2{\parallel Ax-Ay\parallel }^2\\ \le & {\parallel x-y\parallel }^2-2\alpha r_n{\parallel Ax-Ay\parallel }^2+r_n^2{\parallel Ax-Ay\parallel }^2\\ =& {\parallel x-y\parallel }^2-r_n(2\alpha -r_n){\parallel Ax-Ay\parallel }^2.\end{array}$$

It follows from the condition $r_n\in (0,2\alpha )$ that the mapping $I-r_{n}A$ is nonexpansive. From Lemma 2.1, we see that $\mathit{EP}(F,A)=F(T_{r_n}(I-r_{n}A))$. Note that $u_n$ can be rewritten as $u_n=T_{r_n}(I-r_{n}A)x_n$ and $p=T_{r_n}(I-r_{n}A)p$ for each $n\ge 1$ as $p\in \mathcal{F}$.

From (1.4), condition (ii) and Lemma 2.2, we have

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& =& {\parallel {\beta }_n(u_n-p)+(1-{\beta }_n)(W_n{u}_n-p)\parallel }^2\\ =& {\beta }_n{\parallel u_n-p\parallel }^2+(1-{\beta }_n){\parallel W_n{u}_n-p\parallel }^2-{\beta }_n(1-{\beta }_n){\parallel u_n-W_n{u}_n\parallel }^2\\ \le & {\beta }_n{\parallel u_n-p\parallel }^2+(1-{\beta }_n)[{\parallel u_n-p\parallel }^2+k{\parallel u_n-W_n{u}_n\parallel }^2]\\ -{\beta }_n(1-{\beta }_n){\parallel u_n-W_n{u}_n\parallel }^2\\ =& {\parallel u_n-p\parallel }^2-(1-{\beta }_n)({\beta }_n-k){\parallel u_n-W_n{u}_n\parallel }^2\\ \le & {\parallel u_n-p\parallel }^2.\end{array}$$

(3.1)

By $u_n=T_{r_n}(I-r_{n}A)x_n$, we obtain

$$\parallel u_n-p\parallel =\parallel T_{r_n}(I-r_{n}A)x_n-p\parallel \le \parallel x_n-p\parallel .$$

This together with (3.1), we see that

$$\parallel y_n-p\parallel \le \parallel u_n-p\parallel \le \parallel x_n-p\parallel .$$

(3.2)

Furthermore, by Lemma 2.3, we have

$$\begin{array}{rcl}\parallel x_{n+1}-p\parallel & =& \parallel {\alpha }_n[\gamma f(x_n)-Bp]+(I-{\alpha }_{n}B)(y_n-p)\parallel \\ \le & (1-{\alpha }_n\overline{\gamma })\parallel y_n-p\parallel +{\alpha }_n\parallel \gamma f(x_n)-Bp\parallel \\ \le & (1-{\alpha }_n\overline{\gamma })\parallel y_n-p\parallel +{\alpha }_n[\parallel \gamma f(x_n)-\gamma f(p)\parallel +\parallel \gamma f(p)-Bp\parallel ]\\ \le & [1-(\overline{\gamma }-\gamma \rho ){\alpha }_n]\parallel x_n-p\parallel +{\alpha }_n\parallel \gamma f(p)-Bp\parallel .\end{array}$$

It follows from induction that

$$\parallel x_n-p\parallel \le max\{\parallel x_0-p\parallel ,\frac{1}{\overline{\gamma }-\gamma \rho }\parallel \gamma f(p)-Bp\parallel \},n\ge 1,$$

(3.3)

which gives that sequence $\{x_n\}$ is bounded, and so are $\{u_n\}$ and $\{y_n\}$.

Define a mapping $S_{n}x:={\beta }_{n}x+(1-{\beta }_n)W_{n}x$ for each $x\in K$. Then $S_n:K\to H$ is nonexpansive. Indeed, by using (1.3), Lemma 2.2 and condition (ii), we have for all $x,y\in K$ that

$$\begin{array}{rcl}{\parallel S_{n}x-S_{n}y\parallel }^2& =& {\parallel {\beta }_n(x-y)+(1-{\beta }_n)(W_{n}x-W_{n}y)\parallel }^2\\ =& {\beta }_n{\parallel x-y\parallel }^2+(1-{\beta }_n){\parallel W_{n}x-W_{n}y\parallel }^2\\ -{\beta }_n(1-{\beta }_n){\parallel x-W_{n}x-(y-W_{n}y)\parallel }^2\\ \le & {\beta }_n{\parallel x-y\parallel }^2+(1-{\beta }_n)[{\parallel x-y\parallel }^2+k{\parallel x-W_{n}x-(y-W_{n}y)\parallel }^2]\\ -{\beta }_n(1-{\beta }_n){\parallel x-W_{n}x-(y-W_{n}y)\parallel }^2\\ =& {\parallel x-y\parallel }^2-(1-{\beta }_n)({\beta }_n-k){\parallel x-W_{n}x-(y-W_{n}y)\parallel }^2\\ \le & {\parallel x-y\parallel }^2,\end{array}$$

which shows that $S_n:K\to H$ is nonexpansive.

Next, we show that ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$. From (1.4) and Lemma 2.3, we have

$$\begin{array}{rcl}\parallel x_{n+1}-x_n\parallel & =& \parallel {\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}B)y_n-[{\alpha }_{n-1}\gamma f(x_{n-1})+(I-{\alpha }_{n-1}B)y_{n-1}]\parallel \\ \le & {\alpha }_n\gamma \parallel f(x_n)-f(x_{n-1})\parallel +|{\alpha }_n-{\alpha }_{n-1}|[\gamma \parallel f(x_{n-1})\parallel +\parallel B{y}_{n-1}\parallel ]\\ +\parallel (I-{\alpha }_{n}B)(y_n-y_{n-1})\parallel \\ \le & {\alpha }_n\gamma \rho \parallel x_n-x_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|M_1+(1-{\alpha }_n\overline{\gamma })\parallel y_n-y_{n-1}\parallel ,\end{array}$$

(3.4)

where $M_1={sup}_{n\ge 1}\{\gamma \parallel f(x_n)\parallel +\parallel B{y}_n\parallel \}<\mathrm{\infty }$. Moreover, we note that $y_n=S_n{u}_n$ and

$$\begin{array}{rcl}\parallel y_n-y_{n-1}\parallel & \le & \parallel S_n{u}_n-S_n{u}_{n-1}\parallel +\parallel S_n{u}_{n-1}-S_{n-1}{u}_{n-1}\parallel \\ \le & \parallel u_n-u_{n-1}\parallel +\parallel {\beta }_n{u}_{n-1}+(1-{\beta }_n)W_n{u}_{n-1}\\ -[{\beta }_{n-1}{u}_{n-1}+(1-{\beta }_{n-1})W_{n-1}{u}_{n-1}]\parallel \\ \le & \parallel u_n-u_{n-1}\parallel +|{\beta }_n-{\beta }_{n-1}|\parallel u_{n-1}-W_{n-1}{u}_{n-1}\parallel \\ +(1-{\beta }_n)\parallel W_n{u}_{n-1}-W_{n-1}{u}_{n-1}\parallel \\ \le & \parallel u_n-u_{n-1}\parallel +|{\beta }_n-{\beta }_{n-1}|M_2+(1-{\beta }_n)\sum _{i=1}^N|{\eta }_i^{(n)}-{\eta }_i^{(n-1)}|\parallel T_i{u}_{n-1}\parallel ,\end{array}$$

(3.5)

where $M_2={sup}_{n\ge 1}\{\parallel u_{n-1}-W_{n-1}{u}_{n-1}\parallel \}$. On the other hand, we note that

$$\{\begin{array}{c}F(u_n,y)+〈A{x}_n,y-u_n〉+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\hfill \\ F(u_{n-1},y)+〈A{x}_{n-1},y-u_{n-1}〉+\frac{1}{r_{n-1}}〈y-u_{n-1},u_{n-1}-x_{n-1}〉\ge 0.\hfill \end{array}$$

(3.6)

Putting $y=u_{n-1}$ and $y=u_n$ in (3.6) respectively, we have

$$\{\begin{array}{c}F(u_n,u_{n-1})+〈A{x}_n,u_{n-1}-u_n〉+\frac{1}{r_n}〈u_{n-1}-u_n,u_n-x_n〉\ge 0,\hfill \\ F(u_{n-1},u_n)+〈A{x}_{n-1},u_n-u_{n-1}〉+\frac{1}{r_{n-1}}〈u_n-u_{n-1},u_{n-1}-x_{n-1}〉\ge 0.\hfill \end{array}$$

(3.7)

It follows from (A2) that

$$〈u_n-u_{n-1},\frac{u_{n-1}-(I-r_{n-1}A)x_{n-1}}{r_{n-1}}-\frac{u_n-(I-r_{n}A)x_n}{r_n}〉\ge 0,$$

and hence

$$〈u_n-u_{n-1},u_{n-1}-u_n+u_n-(I-r_{n-1}A)x_{n-1}-\frac{r_{n-1}}{r_n}[u_n-(I-r_{n}A)x_n]〉\ge 0.$$

Since ${lim}_{n\to \mathrm{\infty }}{r}_n>0$, we assume that there exists a real number μ such that $r_n>\mu >0$ for all $n\in N$. Consequently, we have

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

and hence

$$\begin{array}{rcl}\parallel u_n-u_{n-1}\parallel & \le & \parallel x_n-x_{n-1}\parallel +|r_n-r_{n-1}|\parallel A{x}_{n-1}\parallel +\frac{r_n-r_{n-1}}{r_n}\parallel u_n-(I-r_{n}A)x_n\parallel \\ \le & \parallel x_n-x_{n-1}\parallel +|r_n-r_{n-1}|[\parallel A{x}_{n-1}\parallel +\frac{1}{\mu }\parallel u_n-(I-r_{n}A)x_n\parallel ]\\ \le & \parallel x_n-x_{n-1}\parallel +|r_n-r_{n-1}|M_3,\end{array}$$

(3.8)

where $M_3=sup\{\parallel A{x}_{n-1}\parallel +\frac{1}{\mu }\parallel u_n-(I-r_{n}A)x_n\parallel ,n\in N\}$. Combining (3.4), (3.5) and (3.8), we have

$$\begin{array}{rcl}\parallel x_{n+1}-x_n\parallel & \le & {\alpha }_n\gamma \rho \parallel x_n-x_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|M_1+(1-{\alpha }_n\overline{\gamma })[\parallel x_n-x_{n-1}\parallel \\ +|{\beta }_n-{\beta }_{n-1}|M_2+|r_n-r_{n-1}|M_3+(1-{\beta }_n)\sum _{i=1}^N|{\eta }_i^{(n)}-{\eta }_i^{(n-1)}|\parallel T_i{u}_{n-1}\parallel ]\\ \le & [1-(\overline{\gamma }-\gamma \rho ){\alpha }_n]\parallel x_n-x_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|M_1+|{\beta }_n-{\beta }_{n-1}|M_2\\ +|r_n-r_{n-1}|M_3+\sum _{i=1}^N|{\eta }_i^{(n)}-{\eta }_i^{(n-1)}|\parallel T_i{u}_{n-1}\parallel .\end{array}$$

It follows from $0<\gamma <\frac{\overline{\gamma }}{\rho }$ and Lemma 2.6 that

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

(3.9)

Moreover, we observe that

$$\parallel x_n-y_n\parallel \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-y_n\parallel \le \parallel x_n-x_{n+1}\parallel +{\alpha }_n\parallel \gamma f(x_n)-B{y}_n\parallel .$$

It follows from ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and (3.9) that

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

(3.10)

For $p\in F(S_n)\cap \mathit{EP}(F,A)$, we note that $u_n=T_{r_n}(I-r_{n}A)x_n$ and

$$\begin{array}{rcl}{\parallel u_n-p\parallel }^2& =& {\parallel T_{r_n}(I-r_{n}A)x_n-T_{r_n}(I-r_{n}A)p\parallel }^2\le 〈x_n-p,u_n-p〉\\ =& \frac{1}{2}({\parallel x_n-p\parallel }^2+{\parallel u_n-p\parallel }^2-{\parallel x_n-u_n\parallel }^2),\end{array}$$

which implies that

$${\parallel u_n-p\parallel }^2\le {\parallel x_n-p\parallel }^2-{\parallel x_n-u_n\parallel }^2.$$

(3.11)

From (1.4), (3.2) and (3.11), we have

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& =& {\parallel {\alpha }_n[\gamma f(x_n)-Bp]+(I-{\alpha }_{n}B)(y_n-p)\parallel }^2\\ \le & {(1-{\alpha }_n\overline{\gamma })}^2{\parallel y_n-p\parallel }^2+{\alpha }_n^2{\parallel \gamma f(x_n)-Bp\parallel }^2\\ +2{\alpha }_n(1-{\alpha }_n\overline{\gamma })\parallel \gamma f(x_n)-Bp\parallel \parallel y_n-p\parallel \\ \le & {\parallel u_n-p\parallel }^2+{\alpha }_n^2{\parallel \gamma f(x_n)-Bp\parallel }^2+2{\alpha }_n\parallel \gamma f(x_n)-Bp\parallel \parallel y_n-p\parallel \\ \le & {\parallel x_n-p\parallel }^2-{\parallel x_n-u_n\parallel }^2+{\alpha }_n^2{\parallel \gamma f(x_n)-Bp\parallel }^2\\ +2{\alpha }_n\parallel \gamma f(x_n)-Bp\parallel \parallel y_n-p\parallel .\end{array}$$

Using ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and (3.9) again, we obtain

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-u_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T_{r_n}(I-r_{n}A)x_n\parallel =0.$$

(3.12)

By the nonexpansion of $S_n$, we have

$$\begin{array}{rcl}\parallel x_n-S_n{x}_n\parallel & \le & \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-S_n{x}_n\parallel \\ \le & \parallel x_n-x_{n+1}\parallel +\parallel {\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}B)y_n-S_n{x}_n\parallel \\ \le & \parallel x_n-x_{n+1}\parallel +{\alpha }_n[\parallel \gamma f(x_n)\parallel +\parallel B{y}_n\parallel ]+\parallel S_n{u}_n-S_n{x}_n\parallel \\ \le & \parallel x_n-x_{n+1}\parallel +{\alpha }_n[\parallel \gamma f(x_n)\parallel +\parallel B{y}_n\parallel ]+\parallel u_n-x_n\parallel .\end{array}$$

This together with (3.9) and (3.12), we obtain

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

(3.13)

Furthermore, we note that

$$\parallel x_n-S_n{x}_n\parallel =(1-{\beta }_n)\parallel x_n-W_n{x}_n\parallel .$$

It follows from condition (ii) that

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

(3.14)

On the other hand, by condition (iii), we may assume that ${\eta }_i^{(n)}\to {\eta }_i$ as $n\to \mathrm{\infty }$ for every $1\le i\le N$. It is easily seen that each ${\eta }_i>0$ and ${\sum }_{i=1}^N{\eta }_i=1$. Define $W={\sum }_{i=1}^N{\eta }_i{T}_i$, then $W:K\to H$ is a k-strict pseudo-contraction such that $F(W)=F(W_n)={\bigcap }_{i=1}^{N}F(T_i)$ by Proposition 2.1 and 2.2. Consequently,

$$\begin{array}{rcl}\parallel x_n-W{x}_n\parallel & \le & \parallel x_n-W_n{x}_n\parallel +\parallel W_n{x}_n-W{x}_n\parallel \\ \le & \parallel x_n-W_n{x}_n\parallel +\sum _{i=1}^N|{\eta }_i^{(n)}-{\eta }_i|\parallel T_i{x}_n\parallel ,\end{array}$$

which implies that

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

(3.15)

Combining (3.14) and (3.15), we obtain

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

(3.16)

Define $S:K\to H$ by $Sx=\lambda x+(1-\lambda )Wx$. By condition (ii) again, we have ${lim}_{n\to \mathrm{\infty }}{\beta }_n=\lambda \in [k,1)$. Then, S is nonexpansive with $F(S)=F(W)$ by Lemma 2.5. Notice that

$$\begin{array}{rcl}\parallel x_n-S{x}_n\parallel & \le & \parallel x_n-S_n{x}_n\parallel +\parallel S_n{x}_n-S{x}_n\parallel \\ =& \parallel x_n-S_n{x}_n\parallel +\parallel {\beta }_n{x}_n+(1-{\beta }_n)W_n{x}_n-\lambda x_n-(1-\lambda )W{x}_n\parallel \\ \le & \parallel x_n-S_n{x}_n\parallel +|{\beta }_n-\lambda |\parallel x_n-W{x}_n\parallel +(1-{\beta }_n)\parallel W_n{x}_n-W{x}_n\parallel .\end{array}$$

It follows from (3.13), (3.15) and (3.16) that

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

(3.17)

Now we claim that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈(B-\gamma f)q,q-x_n〉\le 0$, where $q={lim}_{t\to 0}{x}_t$ with $x_t$ being the fixed point of the contraction ${\mathrm{\Psi }}_n$ on H defined by

$${\mathrm{\Psi }}_{n}x=t\gamma f(x)+(I-tB)S_n{T}_{r_n}(I-r_{n}A)x,\mathrm{\forall }x\in H,n\in N,$$

where $t\in (0,1)$. Indeed, by Lemma 2.1 and 2.3, we have

$$\begin{array}{rcl}\parallel {\mathrm{\Psi }}_{n}x-{\mathrm{\Psi }}_{n}y\parallel & \le & t\gamma \parallel f(x)-f(y)\parallel +(1-t\overline{\gamma })\parallel S_n{T}_{r_n}(I-r_{n}A)x-S_n{T}_{r_n}(I-r_{n}A)y\parallel \\ \le & t\gamma \rho \parallel x-y\parallel +(1-t\overline{\gamma })\parallel T_{r_n}(I-r_{n}A)x-T_{r_n}(I-r_{n}A)y\parallel \\ \le & t\gamma \rho \parallel x-y\parallel +(1-t\overline{\gamma })\parallel x-y\parallel \\ =& [1-(\overline{\gamma }-\gamma \rho )t]\parallel x-y\parallel ,\end{array}$$

for all $x,y\in H$. Since $0<1-(\overline{\gamma }-\gamma \rho )t<1$, it follows that ${\mathrm{\Psi }}_n$ is a contraction. Therefore, by the Banach contraction principle, ${\mathrm{\Psi }}_n$ has a unique fixed point $x_t\in H$ such that

$$x_t=t\gamma f(x_t)+(I-tB)S_n{T}_{r_n}(I-r_{n}A)x_t.$$

By Lemma 2.2 and (3.10), we have

$$\begin{array}{rcl}{\parallel x_t-x_n\parallel }^2& =& {\parallel (I-tB)[S_n{T}_{r_n}(I-r_{n}A)x_t-x_n]+t[\gamma f(x_t)-B{x}_n]\parallel }^2\\ \le & {(1-\overline{\gamma }t)}^2{\parallel S_n{T}_{r_n}(I-r_{n}A)x_t-x_n\parallel }^2+2t〈\gamma f(x_t)-B{x}_n,x_t-x_n〉\\ =& {(1-\overline{\gamma }t)}^2{\parallel S_n{T}_{r_n}(I-r_{n}A)x_t-S_n{T}_{r_n}(I-r_{n}A)x_n+S_n{T}_{r_n}(I-r_{n}A)x_n-x_n\parallel }^2\\ +2t〈\gamma f(x_t)-B{x}_n,x_t-x_n〉\\ \le & {(1-\overline{\gamma }t)}^2{[\parallel x_t-x_n\parallel +\parallel y_n-x_n\parallel ]}^2+2t〈\gamma f(x_t)-B{x}_n,x_t-x_n〉\\ \le & {(1-\overline{\gamma }t)}^2{\parallel x_t-x_n\parallel }^2+{\psi }_n(t)+2t〈\gamma f(x_t)-B{x}_t,x_t-x_n〉\\ +2t〈B{x}_t-B{x}_n,x_t-x_n〉,\end{array}$$

(3.18)

where ${\psi }_n(t)={(1-\overline{\gamma }t)}^2(2\parallel x_t-x_n\parallel +\parallel y_n-x_n\parallel )\parallel y_n-x_n\parallel \to 0$ as $n\to \mathrm{\infty }$. Observe B is strongly positive, we obtain

$$〈B{x}_t-B{x}_n,x_t-x_n〉=〈B(x_t-x_n),x_t-x_n〉\ge \overline{\gamma }{\parallel x_t-x_n\parallel }^2.$$

(3.19)

Combining (3.18) and (3.19), we have

$$\begin{array}{rcl}2t〈B{x}_t-\gamma f(x_t),x_t-x_n〉& \le & ({\overline{\gamma }}^2{t}^2-2\overline{\gamma }t){\parallel x_t-x_n\parallel }^2+{\psi }_n(t)+2t〈B{x}_t-B{x}_n,x_t-x_n〉\\ \le & (\overline{\gamma }{t}^2-2t)〈B(x_t-x_n),x_t-x_n〉+{\psi }_n(t)\\ +2t〈B{x}_t-B{x}_n,x_t-x_n〉\\ =& \overline{\gamma }{t}^2〈B{x}_t-B{x}_n,x_t-x_n〉+{\psi }_n(t).\end{array}$$

It follows that

$$〈B{x}_t-\gamma f(x_t),x_t-x_n〉\le \frac{\overline{\gamma }t}{2}〈B{x}_t-B{x}_n,x_t-x_n〉+\frac{1}{2t}{\psi }_n(t).$$

(3.20)

Let $n\to \mathrm{\infty }$ in (3.20) and note that ${\psi }_n(t)\to 0$ as $n\to \mathrm{\infty }$ yields

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈B{x}_t-\gamma f(x_t),x_t-x_n〉\le \frac{t}{2}{M}_4,$$

(3.21)

where $M_4$ is an appropriate positive constant such that $M_4\ge \overline{\gamma }〈B{x}_t-B{x}_n,x_t-x_n〉$ for all $t\in (0,1)$ and $n\ge 1$. Taking $t\to 0$ from (3.21), we have

$$\underset{t\to 0}{lim\hspace{0.17em}sup}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈B{x}_t-\gamma f(x_t),x_t-x_n〉\le 0.$$

(3.22)

On the other hand, we have

$$\begin{array}{rcl}〈\gamma f(q)-Bq,x_n-q〉& =& 〈\gamma f(q)-Bq,x_n-q〉-〈\gamma f(q)-Bq,x_n-x_t〉+〈\gamma f(q)-Bq,x_n-x_t〉\\ -〈\gamma f(q)-B{x}_t,x_n-x_t〉+〈\gamma f(q)-B{x}_t,x_n-x_t〉\\ -〈\gamma f(x_t)-B{x}_t,x_n-x_t〉+〈\gamma f(x_t)-B{x}_t,x_n-x_t〉.\end{array}$$

It follows that

Therefore, from (3.22) and ${lim}_{t\to 0}{x}_t=q$, we have

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈\gamma f(q)-Bq,x_n-q〉& =& \underset{t\to 0}{lim\hspace{0.17em}sup}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈\gamma f(q)-Bq,x_n-q〉\\ \le & \underset{t\to 0}{lim\hspace{0.17em}sup}\parallel \gamma f(q)-Bq\parallel \parallel x_t-q\parallel \\ +\underset{t\to 0}{lim\hspace{0.17em}sup}\parallel B\parallel \parallel x_t-q\parallel \underset{n\to \mathrm{\infty }}{lim}\parallel x_n-x_t\parallel \\ +\underset{t\to 0}{lim\hspace{0.17em}sup}\gamma \rho \parallel x_t-q\parallel \underset{n\to \mathrm{\infty }}{lim}\parallel x_n-x_t\parallel \\ +\underset{t\to 0}{lim\hspace{0.17em}sup}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈\gamma f(x_t)-B{x}_t,x_n-x_t〉\\ \le & 0.\end{array}$$

(3.23)

Finally, we prove that $x_n\to q$ as $n\to \mathrm{\infty }$. From (1.4) and (3.2) again, we have

$$\begin{array}{rcl}{\parallel x_{n+1}-q\parallel }^2& =& 〈{\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}B)y_n-q,x_{n+1}-q〉\\ =& {\alpha }_n〈\gamma f(x_n)-Bq,x_{n+1}-q〉+〈(I-{\alpha }_{n}B)(y_n-q),x_{n+1}-q〉\\ \le & {\alpha }_n\gamma 〈f(x_n)-f(q),x_{n+1}-q〉+{\alpha }_n〈\gamma f(q)-Bq,x_{n+1}-q〉\\ +(1-{\alpha }_n\overline{\gamma })\parallel y_n-q\parallel \parallel x_{n+1}-q\parallel \\ \le & {\alpha }_n\gamma \rho \parallel x_n-q\parallel \parallel x_{n+1}-q\parallel +{\alpha }_n〈\gamma f(q)-Bq,x_{n+1}-q〉\\ +(1-{\alpha }_n\overline{\gamma })\parallel x_n-q\parallel \parallel x_{n+1}-q\parallel \\ =& [1-(\overline{\gamma }-\gamma \rho ){\alpha }_n]\parallel x_n-q\parallel \parallel x_{n+1}-q\parallel +{\alpha }_n〈\gamma f(q)-Bq,x_{n+1}-q〉\\ \le & \frac{1-(\overline{\gamma }-\gamma \rho ){\alpha }_n}{2}({\parallel x_n-q\parallel }^2+{\parallel x_{n+1}-q\parallel }^2)+{\alpha }_n〈\gamma f(q)-Bq,x_{n+1}-q〉\\ \le & \frac{1-(\overline{\gamma }-\gamma \rho ){\alpha }_n}{2}{\parallel x_n-q\parallel }^2+\frac{1}{2}{\parallel x_{n+1}-q\parallel }^2+{\alpha }_n〈\gamma f(q)-Bq,x_{n+1}-q〉.\end{array}$$

It follows that

$${\parallel x_{n+1}-q\parallel }^2\le [1-(\overline{\gamma }-\gamma \rho ){\alpha }_n]{\parallel x_n-q\parallel }^2+2{\alpha }_n〈\gamma f(q)-Bq,x_{n+1}-q〉.$$

(3.24)

From $0<\gamma <\frac{\overline{\gamma }}{\rho }$, condition (i) and (3.23), we can arrive at the desired conclusion ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q\parallel =0$ by Lemma 2.6. This completes the proof. □

Theorem 3.2 Let K be a nonempty closed convex subset of the Hilbert space H and $F:K\times K\to \mathbb{R}$ be a bi-function satisfying (A 1)-(A 4). Let A be an α-inverse strongly monotone mapping, $f\in {\mathrm{\Pi }}_K$ with a coefficient $\rho \in (0,1)$ and B be a strongly positive bounded linear operator on H with $\overline{\gamma }>0$ and $0<\gamma <\frac{\overline{\gamma }}{\rho }$. Let $T:K\to H$ be a k-strict pseudo-contraction such that $\mathcal{F}=F(T)\cap \mathit{EP}(F,A)\ne \varphi$. Let $\{x_n\}$ be a sequence generated by $x_0\in K$ in the following manner:

$$\{\begin{array}{c}F(u_n,y)+〈A{x}_n,y-u_n〉+\frac{1}{r}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in K,\hfill \\ y_n={\beta }_n{u}_n+(1-{\beta }_n)T{u}_n,\hfill \\ x_{n+1}={\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}B)y_n,n\ge 1,\hfill \end{array}$$

where $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ are two sequences in $(0,1)$, constant $r\in (0,2\alpha )$. If the following control conditions are satisfied:

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

• (ii) $k\le {\beta }_n\le \lambda <1$, ${lim}_{n\to \mathrm{\infty }}{\beta }_n=\lambda$ and ${\sum }_{n=1}^{\mathrm{\infty }}|{\beta }_n-{\beta }_{n-1}|<\mathrm{\infty }$.

Then the sequence $\{x_n\}$ converges strongly to $q\in \mathcal{F}$, which solves the variational inequality

$$〈(B-\gamma f)q,q-p〉\le 0,\mathrm{\forall }p\in \mathcal{F}.$$

Proof Putting $r_n=r$ and $N=1$, i.e., $W_n=T$, the desired conclusion follows immediately from Theorem 3.1. This completes the proof. □

Theorem 3.3 Let K be a nonempty closed convex subset of the Hilbert space H and $F:K\times K\to \mathbb{R}$ be a bi-function satisfying (A 1)-(A 4). Let $f\in {\mathrm{\Pi }}_K$ with a coefficient $\rho \in (0,1)$ and B be a strongly positive bounded linear operator on H with $\overline{\gamma }>0$ and $0<\gamma <\frac{\overline{\gamma }}{\rho }$. Let $T:K\to H$ be a k-strict pseudo-contraction such that $\mathcal{F}=F(T)\cap \mathit{EP}(F)\ne \varphi$. Let $\{x_n\}$ be a sequence generated by $x_0\in K$ in the following manner:

$$\{\begin{array}{c}F(u_n,y)+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in K,\hfill \\ y_n={\beta }_n{u}_n+(1-{\beta }_n)T{u}_n,\hfill \\ x_{n+1}={\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}B)y_n,n\ge 1,\hfill \end{array}$$

where $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ are two sequences in $(0,1)$, sequence $\{r_n\}\subset (0,2\alpha )$. If the following control conditions are satisfied:

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

• (ii) $k\le {\beta }_n\le \lambda <1$, ${lim}_{n\to \mathrm{\infty }}{\beta }_n=\lambda$ and ${\sum }_{n=1}^{\mathrm{\infty }}|{\beta }_n-{\beta }_{n-1}|<\mathrm{\infty }$;

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

Then the sequence $\{x_n\}$ converges strongly to $q\in \mathcal{F}$, which solves the variational inequality

$$〈(B-\gamma f)q,q-p〉\le 0,\mathrm{\forall }p\in \mathcal{F}.$$

Proof Putting $N=1$ and $A=0$, i.e., the generalized equilibrium problem (1.1) reduces to the normal equilibrium problem (1.2), the desired conclusion follows immediately from Theorem 3.1. This completes the proof. □

Remark 3.1 Theorem 3.1 and 3.2 improve and extend the main results of Takahashi and Takahashi [18] and Qin et al. [21] in different directions.

Remark 3.2 Theorem 3.3 is mainly due to Liu [16], which improves and extends the main results of Takahashi and Takahashi [12].

Remark 3.3 If $F=A=0$ and $u_n=x_n$, then the algorithm (1.4) reduces to approximate the fixed point of k-strict pseudo-contractions, which includes the general iterative method of Marino and Xu [8] and the parallel algorithm of Acedo and Xu [20] as special cases.