An iterative approach to mixed equilibrium problems and fixed points problems

Abstract

In the present paper, an iterative algorithm for solving mixed equilibrium problems and fixed points problems has been constructed. It is shown that under some mild conditions, the sequence generated by the presented algorithm converges strongly to the common solution of mixed equilibrium problems and fixed points problems. As an application, we can find the minimum norm element without involving projection.

MSC:47J05, 47J25, 47H09.

1 Introduction

Let H be a real Hilbert space with the inner product $〈\cdot ,\cdot 〉$ and the norm $\parallel \cdot \parallel$, respectively. Let C be a nonempty closed convex subset of H. For a nonlinear mapping $A:C\to H$ and a bifunction $F:C\times C\to R$, the mixed equilibrium problem is to find $z\in C$ such that

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

(1.1)

The solution set of (1.1) is denoted by MEP. If $A=0$, then (1.1) reduces to the following equilibrium problem of finding $z\in C$ such that

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

(1.2)

The solution set of (1.2) is denoted by EP. If $F=0$, then (1.1) reduces to the variational inequality problem of finding $z\in C$ such that

$$〈Az,y-z〉\ge 0,\mathrm{\forall }y\in C.$$

(1.3)

The solution set of (1.3) is denoted by VI. Problem (1.1) is very general in the sense that it includes, as special cases, optimization problems, variational inequalities, minimax problems, Nash equilibrium problem in noncooperative games and others. See, e.g., [1–22].

For solving mixed equilibrium problem (1.1), Moudafi [9] introduced an iterative algorithm and proved a weak convergence theorem. Further, Takahashi and Takahashi [15] introduced the following iterative algorithm for finding an element of $F(S)\cap \mathit{MEP}$:

$$\{\begin{array}{c}F(z_n,y)+〈A{x}_n,y-z_n〉+\frac{1}{{\lambda }_n}〈y-z_n,z_n-x_n〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ x_{n+1}={\beta }_n{x}_n+S({\alpha }_{n}u+(1-{\beta }_n)z_n)\hfill \end{array}$$

(1.4)

for all $n\ge 0$, where $S:C\to C$ is a nonexpansive mapping. They proved that the sequence $\{x_n\}$ generated by (1.4) converges strongly to $z={Proj}_{F(S)\cap \mathit{MEP}}(u)$.

Recently, Yao and Shahzad [19] gave the following iteration process for nonexpansive mappings with perturbation: $x_1\in C$ and

$$x_{n+1}=(1-{\beta }_n)x_n+{\beta }_n{Proj}_C({\alpha }_n{u}_n+(1-{\alpha }_n)T{x}_n),n\ge 0,$$

where $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ are sequences in $[0,1]$, and the sequence $\{u_n\}$ in H is a small perturbation for the n-step iteration satisfying $\parallel u_n\parallel \to 0$ as $n\to \mathrm{\infty }$. In fact, there are perturbations always occurring in the iterative processes because the manipulations are inaccurate.

Using the ideas in [19], Chuang et al. [4] introduced the following iteration process for finding a common element of the set of solutions of the equilibrium problem and the set of fixed points for a quasi-nonexpansive mapping with perturbation: $q_1\in H$ and

$$\{\begin{array}{c}{x}_n\in C\text{ such that }F(x_n,y)+\frac{1}{{\lambda }_n}〈y-x_n,x_n-q_n〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ q_{n+1}={\alpha }_n{u}_n+(1-{\alpha }_n)({\beta }_n{x}_n+(1-{\beta }_n)S{x}_n)\hfill \end{array}$$

for all $n\ge 0$. They showed that the sequence $\{q_n\}$ converges strongly to ${Proj}_{F(S)\cap \mathit{EP}}$.

Motivated and inspired by the above works, in the present paper, we construct an iterative algorithm for solving mixed equilibrium problems and fixed points problems. It is shown that under some mild conditions the sequence $\{x_n\}$ generated by the presented algorithm converges strongly to the common solution of mixed equilibrium problems and fixed points problems. As an application, we can find the minimum norm element without involving projection.

2 Preliminaries

Let C be a nonempty closed convex subset of a real Hilbert space H. Recall that a mapping $A:C\to H$ is called α-inverse-strongly monotone if there exists a positive real number $\alpha >0$ such that

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

It is clear that any α-inverse-strongly monotone mapping is monotone and $\frac{1}{\alpha }$-Lipschitz continuous. A mapping $S:C\to C$ is said to be nonexpansive if $\parallel Sx-Sy\parallel \le \parallel x-y\parallel$ for all $x,y\in C$. And a mapping $S:C\to C$ is said to be strictly pseudo-contractive if there exists a constant $0\le \kappa <1$ such that

$${\parallel Sx-Sy\parallel }^2\le {\parallel x-y\parallel }^2+\kappa {\parallel (I-S)x-(I-S)y\parallel }^2,\mathrm{\forall }x,y\in C.$$

For such a case, we also say that S is a κ-strictly pseudo-contractive mapping.

Throughout this paper, we assume that a bifunction $F:C\times C\to R$ satisfies the following conditions:

• (H1) $F(x,x)=0$ for all $x\in C$;

• (H2) F is monotone, i.e., $F(x,y)+F(y,x)\le 0$ for all $x,y\in C$;

• (H3) for each $x,y,z\in C$, ${lim}_{t\downarrow 0}F(tz+(1-t)x,y)\le F(x,y)$;

• (H4) for each $x\in C$, $y\mapsto F(x,y)$ is convex and lower semicontinuous.

We need the following lemmas for proving our main results.

Lemma 2.1 [7]

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $F:C\times C\to R$ be a bifunction which satisfies conditions (H1)-(H4). Let $r>0$ and $x\in H$. Then there exists $z\in C$ such that

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

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

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

• (ii) EP is closed and convex and $\mathit{EP}=F(T_r)$.

Lemma 2.2 [19]

Let C, H, F and $T_{r}x$ be as in Lemma  2.1. Then we have

$${\parallel T_{s}x-T_{t}x\parallel }^2\le \frac{s-t}{s}〈T_{s}x-T_{t}x,T_{s}x-x〉$$

for all $s,t>0$ and $x\in H$.

Lemma 2.3 [19]

Let C be a nonempty closed convex subset of a real Hilbert space H. Let the mapping $A:C\to H$ be α-inverse strongly monotone and $r>0$ be a constant. Then we have

$${\parallel (I-rA)x-(I-rA)y\parallel }^2\le {\parallel x-y\parallel }^2+r(r-2\alpha ){\parallel Ax-Ay\parallel }^2,\mathrm{\forall }x,y\in C.$$

In particular, if $0\le r\le 2\alpha$, then $I-rA$ is nonexpansive.

Lemma 2.4 [23]

Let $\{x_n\}$ and $\{y_n\}$ be bounded sequences in a Banach space X and let $\{{\beta }_n\}$ be a sequence in $[0,1]$ with $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$. Suppose that $x_{n+1}=(1-{\beta }_n)y_n+{\beta }_n{x}_n$ for all $n\ge 0$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}(\parallel y_{n+1}-y_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0$. Then ${lim}_{n\to \mathrm{\infty }}\parallel y_n-x_n\parallel =0$.

Lemma 2.5 [24]

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $S:C\to C$ be a λ-strict pseudo-contraction. Then we have

• (i) $F(S)=\{x:Sx=x\}$ is closed convex;

• (ii) $\kappa I+(1-\kappa )S$ for $\kappa \in [\lambda ,1)$ is nonexpansive.

Lemma 2.6 [25]

Let C be a nonempty closed and convex of a real Hilbert space H. Let $S:C\to C$ be a κ-strictly pseudo-contractive mapping. Then $I-S$ is demi-closed at 0, i.e., if $x_n\rightharpoonup x\in C$ and $x_n-S{x}_n\to 0$, then $x=Sx$.

Lemma 2.7 [16]

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

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

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

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

3 Main results

In this section, we prove our main results.

Theorem 3.1 Let C be a nonempty closed convex subset of a real Hilbert space H and let $F:C\times C\to R$ be a bifunction satisfying conditions (H1)-(H4). Let $A:C\to H$ be an α-inverse-strongly monotone mapping and let $S:C\to C$ be a κ-strictly pseudo-contractive mapping. Suppose that $F(S)\cap \mathit{MEP}\ne \mathrm{\varnothing }$. Let $x_0\in C$, $\{z_n\}$ and $\{x_n\}$ be sequences in C generated by

$$\{\begin{array}{c}F(z_n,y)+〈A{x}_n,y-z_n〉+\frac{1}{{\lambda }_n}〈y-z_n,z_n-({\alpha }_n{u}_n+(1-{\alpha }_n)x_n)〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)\gamma z_n+(1-{\beta }_n)(1-\gamma )S{z}_n,\hfill \end{array}$$

(3.1)

for all $n\ge 0$, where $\{{\lambda }_n\}\subset (0,2\alpha )$, $\{{\alpha }_n\}\subset (0,1)$ and $\{{\beta }_n\}\subset (0,1)$ satisfy

• (r1) ${lim}_{n\to \mathrm{\infty }}{u}_n=u$ for some $u\in H$;

• (r2) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (r3) $0<c\le {\beta }_n\le d<1$ and $\gamma \in [\kappa ,1)$;

• (r4) $a(1-{\alpha }_n)\le {\lambda }_n\le b(1-{\alpha }_n)$, where $[a,b]\subset (0,2\alpha )$ and ${lim}_{n\to \mathrm{\infty }}({\lambda }_{n+1}-{\lambda }_n)=0$.

Then $\{x_n\}$ generated by (3.1) converges strongly to ${Proj}_{F(S)\cap \mathit{MEP}}(u)$.

Proof Note that $z_n$ can be rewritten as $z_n=T_{{\lambda }_n}({\alpha }_n{u}_n+(1-{\alpha }_n)x_n-{\lambda }_{n}A{x}_n)$ for each n. Take $z\in F(S)\cap \mathit{MEP}$. It is obvious that $z=T_{{\lambda }_n}(z-{\lambda }_{n}Az)=T_{{\lambda }_n}({\alpha }_{n}z+(1-{\alpha }_n)(z-\frac{{\lambda }_{n}Az}{1-{\alpha }_n}))$ for all $n\ge 0$. By using the nonexpansivity of $T_{{\lambda }_n}$ and the convexity of $\parallel \cdot \parallel$, we derive

$$\begin{array}{c}{\parallel z_n-z\parallel }^2\hfill \\ ={\parallel T_{{\lambda }_n}({\alpha }_n{u}_n+(1-{\alpha }_n)x_n-{\lambda }_{n}A{x}_n)-T_{{\lambda }_n}(z-{\lambda }_{n}Az)\parallel }^2\hfill \\ ={\parallel T_{{\lambda }_n}({\alpha }_n{u}_n+(1-{\alpha }_n)(x_n-\frac{{\lambda }_{n}A{x}_n}{1-{\alpha }_n}))-T_{{\lambda }_n}({\alpha }_{n}z+(1-{\alpha }_n)(z-\frac{{\lambda }_{n}Az}{1-{\alpha }_n}))\parallel }^2\hfill \\ \le {\parallel ({\alpha }_n{u}_n+(1-{\alpha }_n)(x_n-\frac{{\lambda }_{n}A{x}_n}{1-{\alpha }_n}))-({\alpha }_{n}z+(1-{\alpha }_n)(z-\frac{{\lambda }_{n}Az}{1-{\alpha }_n}))\parallel }^2\hfill \\ ={\parallel (1-{\alpha }_n)((x_n-\frac{{\lambda }_{n}A{x}_n}{1-{\alpha }_n})-(z-\frac{{\lambda }_{n}Az}{1-{\alpha }_n}))+{\alpha }_n(u_n-z)\parallel }^2\hfill \\ \le (1-{\alpha }_n){\parallel (x_n-\frac{{\lambda }_{n}A{x}_n}{1-{\alpha }_n})-(z-\frac{{\lambda }_{n}Az}{1-{\alpha }_n})\parallel }^2+{\alpha }_n{\parallel u_n-z\parallel }^2.\hfill \end{array}$$

Since A is α-inverse strongly monotone, we know from Lemma 2.3 that

$$\begin{array}{c}{\parallel (x_n-\frac{{\lambda }_{n}A{x}_n}{1-{\alpha }_n})-(z-\frac{{\lambda }_{n}Az}{1-{\alpha }_n})\parallel }^2\hfill \\ \le {\parallel x_n-z\parallel }^2+\frac{{\lambda }_n({\lambda }_n-2(1-{\alpha }_n)\alpha )}{{(1-{\alpha }_n)}^2}{\parallel A{x}_n-Az\parallel }^2.\hfill \end{array}$$

It follows that

$$\begin{array}{rcl}{\parallel z_n-z\parallel }^2& \le & (1-{\alpha }_n)({\parallel x_n-z\parallel }^2+\frac{{\lambda }_n({\lambda }_n-2(1-{\alpha }_n)\alpha )}{{(1-{\alpha }_n)}^2}{\parallel A{x}_n-Az\parallel }^2)+{\alpha }_n{\parallel u_n-z\parallel }^2\\ \le & (1-{\alpha }_n){\parallel x_n-z\parallel }^2+{\alpha }_n{\parallel u_n-z\parallel }^2.\end{array}$$

(3.2)

So, we have

$$\begin{array}{rcl}{\parallel x_{n+1}-z\parallel }^2& =& {\parallel {\beta }_n(x_n-z)+(1-{\beta }_n)((\gamma I+(1-\gamma )S)z_n-z)\parallel }^2\\ \le & {\beta }_n{\parallel x_n-z\parallel }^2+(1-{\beta }_n){\parallel z_n-z\parallel }^2\\ \le & {\beta }_n{\parallel x_n-z\parallel }^2+(1-{\beta }_n)((1-{\alpha }_n){\parallel x_n-z\parallel }^2+{\alpha }_n{\parallel u_n-z\parallel }^2)\\ =& [1-(1-{\beta }_n){\alpha }_n]{\parallel x_n-z\parallel }^2+(1-{\beta }_n){\alpha }_n{\parallel u_n-z\parallel }^2\\ \le & max\{{\parallel x_n-z\parallel }^2,{\parallel u_n-z\parallel }^2\}.\end{array}$$

Since ${lim}_{n\to \mathrm{\infty }}{u}_n=u$, $\{u_n\}$ is bounded. Therefore, by induction, we deduce that $\{x_n\}$ is bounded. Hence, $\{A{x}_n\}$, $\{z_n\}$ and $\{S{z}_n\}$ are also bounded.

Putting $y_n={\alpha }_n{u}_n+(1-{\alpha }_n)x_n-{\lambda }_{n}A{x}_n$ for all n, we have

$$z_{n+1}-z_n=T_{{\lambda }_{n+1}}{y}_{n+1}-T_{{\lambda }_{n+1}}{y}_n+T_{{\lambda }_{n+1}}{y}_n-T_{{\lambda }_n}{y}_n.$$

It follows that

$$\begin{array}{rcl}\parallel z_{n+1}-z_n\parallel & \le & \parallel T_{{\lambda }_{n+1}}{y}_{n+1}-T_{{\lambda }_{n+1}}{y}_n\parallel +\parallel T_{{\lambda }_{n+1}}{y}_n-T_{{\lambda }_n}{y}_n\parallel \\ \le & \parallel y_{n+1}-y_n\parallel +\parallel T_{{\lambda }_{n+1}}{y}_n-T_{{\lambda }_n}{y}_n\parallel .\end{array}$$

(3.3)

From Lemma 2.3, we know that $I-\lambda A$ is nonexpansive for all $\lambda \in (0,2\alpha )$. Thus, we have $I-\frac{{\lambda }_{n+1}}{1-{\alpha }_{n+1}}A$ is nonexpansive for all n due to the fact that $\frac{{\lambda }_{n+1}}{1-{\alpha }_{n+1}}\in (0,2\alpha )$. Then we get

$$\begin{array}{rcl}\parallel y_{n+1}-y_n\parallel & =& \parallel {\alpha }_{n+1}{u}_{n+1}+(1-{\alpha }_{n+1})x_{n+1}-{\lambda }_{n+1}A{x}_{n+1}-({\alpha }_n{u}_n+(1-{\alpha }_n)x_n-{\lambda }_{n}A{x}_n)\parallel \\ \le & \parallel (1-{\alpha }_{n+1})(x_{n+1}-\frac{{\lambda }_{n+1}}{1-{\alpha }_{n+1}}A{x}_{n+1})-(1-{\alpha }_n)(x_n-\frac{{\lambda }_n}{1-{\alpha }_n}A{x}_n)\parallel \\ +{\alpha }_{n+1}\parallel u_{n+1}\parallel +{\alpha }_n\parallel u_n\parallel \\ \le & (1-{\alpha }_{n+1})\parallel (I-\frac{{\lambda }_{n+1}}{1-{\alpha }_{n+1}}A)x_{n+1}-(I-\frac{{\lambda }_{n+1}}{1-{\alpha }_{n+1}}A)x_n\parallel \\ +\parallel (1-{\alpha }_{n+1})(x_n-\frac{{\lambda }_{n+1}}{1-{\alpha }_{n+1}}A{x}_n)-(1-{\alpha }_n)(x_n-\frac{{\lambda }_n}{1-{\alpha }_n}A{x}_n)\parallel \\ +{\alpha }_{n+1}\parallel u_{n+1}\parallel +{\alpha }_n\parallel u_n\parallel \\ \le & \parallel x_{n+1}-x_n\parallel +|{\alpha }_{n+1}-{\alpha }_n|\parallel x_n\parallel +|{\lambda }_{n+1}-{\lambda }_n|\parallel A{x}_n\parallel \\ +{\alpha }_{n+1}\parallel u_{n+1}\parallel +{\alpha }_n\parallel u_n\parallel .\end{array}$$

(3.4)

By Lemma 2.2, we have

$$\parallel T_{{\lambda }_{n+1}}{y}_n-T_{{\lambda }_n}{y}_n\parallel \le \frac{|{\lambda }_{n+1}-{\lambda }_n|}{{\lambda }_{n+1}}\parallel T_{{\lambda }_{n+1}}{y}_n-y_n\parallel .$$

(3.5)

From (3.3)-(3.5), we obtain

$$\begin{array}{rcl}\parallel z_{n+1}-z_n\parallel & \le & \parallel x_{n+1}-x_n\parallel +|{\alpha }_{n+1}-{\alpha }_n|\parallel x_n\parallel +|{\lambda }_{n+1}-{\lambda }_n|\parallel A{x}_n\parallel \\ +\frac{|{\lambda }_{n+1}-{\lambda }_n|}{{\lambda }_{n+1}}\parallel T_{{\lambda }_{n+1}}{y}_n-y_n\parallel +{\alpha }_{n+1}\parallel u_{n+1}\parallel +{\alpha }_n\parallel u_n\parallel .\end{array}$$

Then

$$\begin{array}{c}\parallel (\gamma I+(1-\gamma )S)z_{n+1}-(\gamma I+(1-\gamma )S)z_n\parallel \hfill \\ \le \parallel z_{n+1}-z_n\parallel \hfill \\ \le \parallel x_{n+1}-x_n\parallel +|{\alpha }_{n+1}-{\alpha }_n|\parallel x_n\parallel +|{\lambda }_{n+1}-{\lambda }_n|\parallel A{x}_n\parallel \hfill \\ +\frac{|{\lambda }_{n+1}-{\lambda }_n|}{{\lambda }_{n+1}}\parallel T_{{\lambda }_{n+1}}{y}_n-y_n\parallel +{\alpha }_{n+1}\parallel u_{n+1}\parallel +{\alpha }_n\parallel u_n\parallel .\hfill \end{array}$$

Therefore,

$$\begin{array}{c}\parallel (\gamma I+(1-\gamma )S)z_{n+1}-(\gamma I+(1-\gamma )S)z_n\parallel -\parallel x_{n+1}-x_n\parallel \hfill \\ \le |{\alpha }_{n+1}-{\alpha }_n|\parallel x_n\parallel +|{\lambda }_{n+1}-{\lambda }_n|\parallel A{x}_n\parallel +{\alpha }_{n+1}\parallel u_{n+1}\parallel +{\alpha }_n\parallel u_n\parallel \hfill \\ +\frac{|{\lambda }_{n+1}-{\lambda }_n|}{{\lambda }_{n+1}}\parallel T_{{\lambda }_{n+1}}{y}_n-y_n\parallel .\hfill \end{array}$$

Since ${\alpha }_n\to 0$, ${\lambda }_{n+1}-{\lambda }_n\to 0$ and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\lambda }_n>0$, we obtain

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(\parallel (\gamma I+(1-\gamma )S)z_{n+1}-(\gamma I+(1-\gamma )S)z_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0.$$

This together with Lemma 2.4 implies that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel (\gamma I+(1-\gamma )S)z_n-x_n\parallel =0.$$

(3.6)

Consequently, we obtain

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-x_n\parallel =\underset{n\to \mathrm{\infty }}{lim}(1-{\beta }_n)\parallel (\gamma I+(1-\gamma )S)z_n-x_n\parallel =0.$$

From (3.1) and (3.2), we have

$$\begin{array}{rcl}{\parallel x_{n+1}-z\parallel }^2& \le & (1-{\beta }_n){\parallel (\gamma I+(1-\gamma )S)T_{{\lambda }_n}({\alpha }_n{u}_n+(1-{\alpha }_n)x_n-{\lambda }_{n}A{x}_n)-z\parallel }^2\\ +{\beta }_n{\parallel x_n-z\parallel }^2\\ \le & (1-{\beta }_n)\{(1-{\alpha }_n)({\parallel x_n-z\parallel }^2+\frac{{\lambda }_n}{{(1-{\alpha }_n)}^2}({\lambda }_n-2(1-{\alpha }_n)\alpha ){\parallel A{x}_n-Az\parallel }^2)\\ +{\alpha }_n{\parallel u_n-z\parallel }^2\}+{\beta }_n{\parallel x_n-z\parallel }^2\\ =& (1-(1-{\beta }_n){\alpha }_n){\parallel x_n-z\parallel }^2+\frac{(1-{\beta }_n){\lambda }_n}{1-{\alpha }_n}({\lambda }_n-2(1-{\alpha }_n)\alpha ){\parallel A{x}_n-Az\parallel }^2\\ +(1-{\beta }_n){\alpha }_n{\parallel u_n-z\parallel }^2\\ \le & {\parallel x_n-z\parallel }^2+\frac{(1-{\beta }_n){\lambda }_n}{1-{\alpha }_n}({\lambda }_n-2(1-{\alpha }_n)\alpha ){\parallel A{x}_n-Az\parallel }^2\\ +(1-{\beta }_n){\alpha }_n{\parallel u_n-z\parallel }^2.\end{array}$$

Then we obtain

$$\begin{array}{c}\frac{(1-{\beta }_n){\lambda }_n}{1-{\alpha }_n}(2(1-{\alpha }_n)\alpha -{\lambda }_n){\parallel A{x}_n-Az\parallel }^2\hfill \\ \le {\parallel x_n-z\parallel }^2-{\parallel x_{n+1}-z\parallel }^2+(1-{\beta }_n){\alpha }_n{\parallel u_n-z\parallel }^2\hfill \\ \le (\parallel x_n-z\parallel -\parallel x_{n+1}-z\parallel )\parallel x_{n+1}-x_n\parallel +(1-{\beta }_n){\alpha }_n{\parallel u_n-z\parallel }^2.\hfill \end{array}$$

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

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

(3.7)

Next, we show $\parallel x_n-z_n\parallel =\parallel x_n-T_{{\lambda }_n}{y}_n\parallel \to 0$. By using the firm nonexpansivity of $T_{{\lambda }_n}$, we have

$$\begin{array}{rcl}{\parallel T_{{\lambda }_n}{y}_n-z\parallel }^2& =& {\parallel T_{{\lambda }_n}{y}_n-T_{{\lambda }_n}(z-{\lambda }_{n}Az)\parallel }^2\\ \le & 〈y_n-(z-{\lambda }_{n}Az),T_{{\lambda }_n}{y}_n-z〉\\ =& \frac{1}{2}({\parallel y_n-(z-{\lambda }_{n}Az)\parallel }^2+{\parallel T_{{\lambda }_n}{y}_n-z\parallel }^2\\ -{\parallel {\alpha }_n{u}_n+(1-{\alpha }_n)x_n-{\lambda }_n(A{x}_n-{\lambda }_{n}Az)-T_{{\lambda }_n}{y}_n\parallel }^2).\end{array}$$

We note that

$${\parallel y_n-(z-{\lambda }_{n}Az)\parallel }^2\le (1-{\alpha }_n){\parallel x_n-z\parallel }^2+{\alpha }_n{\parallel u_n-z\parallel }^2.$$

Thus,

$$\begin{array}{rcl}{\parallel T_{{\lambda }_n}{y}_n-z\parallel }^2& \le & \frac{1}{2}((1-{\alpha }_n){\parallel x_n-z\parallel }^2+{\alpha }_n{\parallel u_n-z\parallel }^2+{\parallel T_{{\lambda }_n}{y}_n-z\parallel }^2\\ -{\parallel {\alpha }_n{u}_n+(1-{\alpha }_n)x_n-T_{{\lambda }_n}{y}_n-{\lambda }_n(A{x}_n-{\lambda }_{n}Az)\parallel }^2).\end{array}$$

That is,

$$\begin{array}{rcl}{\parallel T_{{\lambda }_n}{y}_n-z\parallel }^2& \le & (1-{\alpha }_n){\parallel x_n-z\parallel }^2+{\alpha }_n{\parallel u_n-z\parallel }^2\\ -{\parallel {\alpha }_n{u}_n+(1-{\alpha }_n)x_n-T_{{\lambda }_n}{y}_n-{\lambda }_n(A{x}_n-{\lambda }_{n}Az)\parallel }^2\\ =& (1-{\alpha }_n){\parallel x_n-z\parallel }^2+{\alpha }_n{\parallel u_n-z\parallel }^2-{\parallel {\alpha }_n{u}_n+(1-{\alpha }_n)x_n-T_{{\lambda }_n}{y}_n\parallel }^2\\ +2{\lambda }_n〈{\alpha }_n{u}_n+(1-{\alpha }_n)x_n-T_{{\lambda }_n}{y}_n,A{x}_n-Az〉-{\lambda }_n^2{\parallel A{x}_n-Az\parallel }^2\\ \le & (1-{\alpha }_n){\parallel x_n-z\parallel }^2+{\alpha }_n{\parallel u_n-z\parallel }^2-{\parallel {\alpha }_n{u}_n+(1-{\alpha }_n)x_n-T_{{\lambda }_n}{y}_n\parallel }^2\\ +2{\lambda }_n\parallel {\alpha }_n{u}_n+(1-{\alpha }_n)x_n-T_{{\lambda }_n}{y}_n\parallel \parallel A{x}_n-Az\parallel .\end{array}$$

It follows that

$$\begin{array}{rcl}{\parallel x_{n+1}-z\parallel }^2& \le & {\beta }_n{\parallel x_n-z\parallel }^2+(1-{\beta }_n)(1-{\alpha }_n){\parallel x_n-z\parallel }^2+(1-{\beta }_n){\alpha }_n{\parallel u_n-z\parallel }^2\\ -(1-{\beta }_n){\parallel {\alpha }_n{u}_n+(1-{\alpha }_n)x_n-T_{{\lambda }_n}{y}_n\parallel }^2\\ +2{\lambda }_n(1-{\beta }_n)\parallel {\alpha }_n{u}_n+(1-{\alpha }_n)x_n-T_{{\lambda }_n}{y}_n\parallel \parallel A{x}_n-Az\parallel \\ =& (1-(1-{\beta }_n){\alpha }_n){\parallel x_n-z\parallel }^2+(1-{\beta }_n){\alpha }_n{\parallel u_n-z\parallel }^2\\ -(1-{\beta }_n){\parallel {\alpha }_n{u}_n+(1-{\alpha }_n)x_n-T_{{\lambda }_n}{y}_n\parallel }^2\\ +2{\lambda }_n(1-{\beta }_n)\parallel {\alpha }_n{u}_n+(1-{\alpha }_n)x_n-T_{{\lambda }_n}{y}_n\parallel \parallel A{x}_n-Az\parallel .\end{array}$$

Hence,

$$\begin{array}{c}(1-{\beta }_n){\parallel {\alpha }_n{u}_n+(1-{\alpha }_n)x_n-T_{{\lambda }_n}{y}_n\parallel }^2\hfill \\ \le {\parallel x_n-z\parallel }^2-{\parallel x_{n+1}-z\parallel }^2-(1-{\beta }_n){\alpha }_n{\parallel x_n-z\parallel }^2\hfill \\ +(1-{\beta }_n){\alpha }_n{\parallel u_n-z\parallel }^2+2{\lambda }_n(1-{\beta }_n)\parallel {\alpha }_n{u}_n+(1-{\alpha }_n)x_n-T_{{\lambda }_n}{y}_n\parallel \parallel A{x}_n-Az\parallel \hfill \\ \le (\parallel x_n-z\parallel +\parallel x_{n+1}-z\parallel )\parallel x_{n+1}-x_n\parallel +(1-{\beta }_n){\alpha }_n{\parallel u_n-z\parallel }^2\hfill \\ +2{\lambda }_n(1-{\beta }_n)\parallel {\alpha }_n{u}_n+(1-{\alpha }_n)x_n-T_{{\lambda }_n}{y}_n\parallel \parallel A{x}_n-Az\parallel .\hfill \end{array}$$

Since ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$, $\parallel x_{n+1}-x_n\parallel \to 0$, ${\alpha }_n\to 0$ and $\parallel A{x}_n-Az\parallel \to 0$, we deduce

$$\underset{n\to \mathrm{\infty }}{lim}\parallel {\alpha }_n{u}_n+(1-{\alpha }_n)x_n-T_{{\lambda }_n}{y}_n\parallel =0.$$

This implies that

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

(3.8)

Put $\tilde{x}={Proj}_{F(S)\cap \mathit{MEP}}(u)$. We will finally show that $x_n\to \tilde{x}$.

Setting $v_n=x_n-\frac{{\lambda }_n}{1-{\alpha }_n}(A{x}_n-A\tilde{x})$ for all n. Taking $z=\tilde{x}$ in (3.7) to get $\parallel A{x}_n-A\tilde{x}\parallel \to 0$. First, we prove ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈u-\tilde{x},v_n-\tilde{x}〉\le 0$. We take a subsequence $\{v_{n_i}\}$ of $\{v_n\}$ such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈u-\tilde{x},v_n-\tilde{x}〉=\underset{i\to \mathrm{\infty }}{lim}〈u-\tilde{x},v_{n_i}-\tilde{x}〉.$$

It is clear that $\{v_{n_i}\}$ is bounded due to the boundedness of $\{x_n\}$ and $\parallel A{x}_n-A\tilde{x}\parallel \to 0$. Then there exists a subsequence $\{v_{n_{i_j}}\}$ of $\{v_{n_i}\}$ which converges weakly to some point $w\in C$. Hence, $\{x_{n_{i_j}}\}$ also converges weakly to w. At the same time, from (3.6) and (3.8), we have

$$\underset{j\to \mathrm{\infty }}{lim}\parallel x_{n_{i_j}}-(\gamma I+(1-\gamma )S)x_{n_{i_j}}\parallel =0.$$

(3.9)

By the demi-closedness principle (see Lemma 2.6) and (3.9), we deduce $w\in F(S)$.

Further, we show that w is also in MEP. From (3.1), we have

$$F(z_n,y)+〈A{x}_n,y-z_n〉+\frac{1}{{\lambda }_n}〈y-z_n,z_n-({\alpha }_n{u}_n+(1-{\alpha }_n)x_n)〉\ge 0.$$

From (H2), we have

$$〈A{x}_n,y-z_n〉+\frac{1}{{\lambda }_n}〈y-z_n,z_n-({\alpha }_n{u}_n+(1-{\alpha }_n)x_n)〉\ge F(y,z_n).$$

(3.10)

Put $x_t=ty+(1-t)w$ for all $t\in (0,1-\frac{\lambda }{2\alpha })$ and $y\in C$. Then we have $x_t\in C$. So, from (3.10), we have

$$\begin{array}{rcl}〈x_t-z_n,A{x}_t〉& \ge & 〈x_t-z_n,A{x}_t〉-〈x_t-z_n,A{x}_n〉\\ -\frac{1}{{\lambda }_n}〈x_t-z_n,z_n-({\alpha }_n{u}_n+(1-{\alpha }_n)x_n)〉+F(x_t,z_n)\\ =& 〈x_t-z_n,A{x}_t-A{z}_n〉+〈x_t-z_n,A{z}_n-A{x}_n〉\\ -\frac{1}{{\lambda }_n}〈x_t-z_n,z_n-({\alpha }_n{u}_n+(1-{\alpha }_n)x_n)〉+F(x_t,z_n).\end{array}$$

Since $\parallel z_n-x_n\parallel \to 0$, we have $\parallel A{z}_n-A{x}_n\parallel \to 0$. Further, from monotonicity of A, we have $〈x_t-z_n,A{x}_t-A{z}_n〉\ge 0$. So, from (H4), we have

$$〈x_t-w,A{x}_t〉\ge F(x_t,w),\text{as }n\to \mathrm{\infty }.$$

(3.11)

From (H1), (H4) and (3.11), we also have

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

and hence

$$0\le F(x_t,y)+(1-t)〈y-w,A{x}_t〉.$$

Letting $t\to 0$, we have, for each $y\in C$,

$$0\le F(w,y)+〈y-w,Aw〉.$$

This implies $w\in \mathit{MEP}$. Hence, we have $w\in F(S)\cap \mathit{MEP}$. This implies that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈u-\tilde{x},v_n-\tilde{x}〉=\underset{j\to \mathrm{\infty }}{lim}〈u-\tilde{x},v_{n_{i_j}}-\tilde{x}〉=〈u-\tilde{x},w-\tilde{x}〉.$$

Note that $\tilde{x}={Proj}_{F(S)\cap \mathit{MEP}}(u)$. Then $〈u-\tilde{x},w-\tilde{x}〉\le 0$, $w\in F(S)\cap \mathit{MEP}$. Therefore,

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈u-\tilde{x},v_n-\tilde{x}〉\le 0.$$

Since $u_n\to u$, we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈u_n-\tilde{x},v_n-\tilde{x}〉\le 0.$$

From (3.1), we have

$$\begin{array}{c}{\parallel x_{n+1}-\tilde{x}\parallel }^2\hfill \\ \le {\beta }_n{\parallel x_n-\tilde{x}\parallel }^2+(1-{\beta }_n){\parallel (\gamma I+(1-\gamma )S)T_{{\lambda }_n}{y}_n-\tilde{x}\parallel }^2\hfill \\ \le {\beta }_n{\parallel x_n-\tilde{x}\parallel }^2+(1-{\beta }_n){\parallel T_{{\lambda }_n}{y}_n-\tilde{x}\parallel }^2\hfill \\ ={\beta }_n{\parallel x_n-\tilde{x}\parallel }^2+(1-{\beta }_n){\parallel T_{{\lambda }_n}{y}_n-T_{{\lambda }_n}(\tilde{x}-{\lambda }_{n}A\tilde{x})\parallel }^2\hfill \\ \le {\beta }_n{\parallel x_n-\tilde{x}\parallel }^2+(1-{\beta }_n){\parallel y_n-(\tilde{x}-{\lambda }_{n}A\tilde{x})\parallel }^2\hfill \\ ={\beta }_n{\parallel x_n-\tilde{x}\parallel }^2+(1-{\beta }_n){\parallel {\alpha }_n{u}_n+(1-{\alpha }_n)x_n-{\lambda }_{n}A{x}_n-(\tilde{x}-{\lambda }_{n}A\tilde{x})\parallel }^2\hfill \\ =(1-{\beta }_n){\parallel (1-{\alpha }_n)((x_n-\frac{{\lambda }_n}{1-{\alpha }_n}A{x}_n)-(\tilde{x}-\frac{{\lambda }_n}{1-{\alpha }_n}A\tilde{x}))+{\alpha }_n(u_n-\tilde{x})\parallel }^2\hfill \\ +{\beta }_n{\parallel x_n-\tilde{x}\parallel }^2\hfill \\ =(1-{\beta }_n)({(1-{\alpha }_n)}^2{\parallel (x_n-\frac{{\lambda }_n}{1-{\alpha }_n}A{x}_n)-(\tilde{x}-\frac{{\lambda }_n}{1-{\alpha }_n}A\tilde{x})\parallel }^2\hfill \\ +2{\alpha }_n(1-{\alpha }_n)〈u_n-\tilde{x},(x_n-\frac{{\lambda }_n}{1-{\alpha }_n}A{x}_n)-(\tilde{x}-\frac{{\lambda }_n}{1-{\alpha }_n}A\tilde{x})〉\hfill \\ +{\alpha }_n^2{\parallel u_n-\tilde{x}\parallel }^2)+{\beta }_n{\parallel x_n-\tilde{x}\parallel }^2\hfill \\ \le {\beta }_n{\parallel x_n-\tilde{x}\parallel }^2+(1-{\beta }_n)({(1-{\alpha }_n)}^2{\parallel x_n-\tilde{x}\parallel }^2\hfill \\ +2{\alpha }_n(1-{\alpha }_n)〈u_n-\tilde{x},x_n-\frac{{\lambda }_n}{1-{\alpha }_n}(A{x}_n-A\tilde{x})-\tilde{x}〉+{\alpha }_n^2{\parallel u_n-\tilde{x}\parallel }^2)\hfill \\ \le (1-(1-{\beta }_n){\alpha }_n){\parallel x_n-\tilde{x}\parallel }^2\hfill \\ +(1-{\beta }_n){\alpha }_n\{2(1-{\alpha }_n)〈u_n-\tilde{x},v_n-\tilde{x}〉+{\alpha }_n{\parallel u_n-\tilde{x}\parallel }^2\}.\hfill \end{array}$$

It is clear that ${\sum }_{n=1}^{\mathrm{\infty }}(1-{\beta }_n){\alpha }_n=\mathrm{\infty }$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}(2(1-{\alpha }_n)〈u_n-\tilde{x},v_n-\tilde{x}〉+{\alpha }_n{\parallel u_n-\tilde{x}\parallel }^2)\le 0$. We can therefore apply Lemma 2.7 to conclude that $x_n\to \tilde{x}$. This completes the proof. □

Corollary 3.2 Let C be a nonempty closed convex subset of a real Hilbert space H and let $F:C\times C\to R$ be a bifunction satisfying conditions (H1)-(H4). Let $A:C\to H$ be an α-inverse-strongly monotone mapping and let $S:C\to C$ be a nonexpansive mapping. Suppose that $F(S)\cap \mathit{MEP}\ne \mathrm{\varnothing }$. Let $x_0\in C$, $\{z_n\}$ and $\{x_n\}$ be sequences in C generated by

$$\{\begin{array}{c}F(z_n,y)+〈A{x}_n,y-z_n〉+\frac{1}{{\lambda }_n}〈y-z_n,z_n-({\alpha }_n{u}_n+(1-{\alpha }_n)x_n)〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)\gamma z_n+(1-{\beta }_n)(1-\gamma )S{z}_n\hfill \end{array}$$

(3.12)

for all $n\ge 0$, where $\{{\lambda }_n\}\subset (0,2\alpha )$, $\{{\alpha }_n\}\subset (0,1)$ and $\{{\beta }_n\}\subset (0,1)$ satisfy

• (r1) ${lim}_{n\to \mathrm{\infty }}{u}_n=u$ for some $u\in H$;

• (r2) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (r3) $0<c\le {\beta }_n\le d<1$ and $\gamma \in (0,1)$;

• (r4) $a(1-{\alpha }_n)\le {\lambda }_n\le b(1-{\alpha }_n)$, where $[a,b]\subset (0,2\alpha )$ and ${lim}_{n\to \mathrm{\infty }}({\lambda }_{n+1}-{\lambda }_n)=0$.

Then $\{x_n\}$ generated by (3.12) converges strongly to ${Proj}_{F(S)\cap \mathit{MEP}}(u)$.

Corollary 3.3 Let C be a nonempty closed convex subset of a real Hilbert space H and let $F:C\times C\to R$ be a bifunction satisfying conditions (H1)-(H4). Let $S:C\to C$ be a κ-strictly pseudo-contractive mapping. Suppose that $F(S)\cap \mathit{EP}\ne \mathrm{\varnothing }$. Let $x_0\in C$, $\{z_n\}$ and $\{x_n\}$ be sequences in C generated by

$$\{\begin{array}{c}F(z_n,y)+\frac{1}{{\lambda }_n}〈y-z_n,z_n-({\alpha }_n{u}_n+(1-{\alpha }_n)x_n)〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)\gamma z_n+(1-{\beta }_n)(1-\gamma )S{z}_n\hfill \end{array}$$

(3.13)

for all $n\ge 0$, where $\{{\lambda }_n\}\subset (0,1)$, $\{{\alpha }_n\}\subset (0,1)$ and $\{{\beta }_n\}\subset (0,1)$ satisfy

• (r1) ${lim}_{n\to \mathrm{\infty }}{u}_n=u$ for some $u\in H$;

• (r2) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (r3) $0<c\le {\beta }_n\le d<1$ and $\gamma \in [\kappa ,1)$;

• (r4) $a(1-{\alpha }_n)\le {\lambda }_n\le b(1-{\alpha }_n)$, where $[a,b]\subset (0,1)$ and ${lim}_{n\to \mathrm{\infty }}({\lambda }_{n+1}-{\lambda }_n)=0$.

Then $\{x_n\}$ generated by (3.13) converges strongly to ${Proj}_{F(S)\cap \mathit{EP}}(u)$.

Corollary 3.4 Let C be a nonempty closed convex subset of a real Hilbert space H and let $F:C\times C\to R$ be a bifunction satisfying conditions (H1)-(H4). Let $S:C\to C$ be a nonexpansive mapping. Suppose that $F(S)\cap \mathit{EP}\ne \mathrm{\varnothing }$. Let $x_0\in C$, $\{z_n\}$ and $\{x_n\}$ be sequences in C generated by

$$\{\begin{array}{c}F(z_n,y)+\frac{1}{{\lambda }_n}〈y-z_n,z_n-({\alpha }_n{u}_n+(1-{\alpha }_n)x_n)〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)\gamma z_n+(1-{\beta }_n)(1-\gamma )S{z}_n\hfill \end{array}$$

(3.14)

for all $n\ge 0$, where $\{{\lambda }_n\}\subset (0,1)$, $\{{\alpha }_n\}\subset (0,1)$ and $\{{\beta }_n\}\subset (0,1)$ satisfy

• (r1) ${lim}_{n\to \mathrm{\infty }}{u}_n=u$ for some $u\in H$;

• (r2) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (r3) $0<c\le {\beta }_n\le d<1$ and $\gamma \in (0,1)$;

• (r4) $a(1-{\alpha }_n)\le {\lambda }_n\le b(1-{\alpha }_n)$, where $[a,b]\subset (0,1)$ and ${lim}_{n\to \mathrm{\infty }}({\lambda }_{n+1}-{\lambda }_n)=0$.

Then $\{x_n\}$ generated by (3.14) converges strongly to ${Proj}_{F(S)\cap \mathit{EP}}(u)$.

Corollary 3.5 Let C be a nonempty closed convex subset of a real Hilbert space H and let $F:C\times C\to R$ be a bifunction satisfying conditions (H1)-(H4). Let $A:C\to H$ be an α-inverse-strongly monotone mapping and let $S:C\to C$ be a κ-strictly pseudo-contractive mapping. Suppose that $F(S)\cap \mathit{MEP}\ne \mathrm{\varnothing }$. Let $x_0\in C$, $\{z_n\}$ and $\{x_n\}$ be sequences in C generated by

$$\{\begin{array}{c}F(z_n,y)+〈A{x}_n,y-z_n〉+\frac{1}{{\lambda }_n}〈y-z_n,z_n-(1-{\alpha }_n)x_n〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)\gamma z_n+(1-{\beta }_n)(1-\gamma )S{z}_n\hfill \end{array}$$

(3.15)

for all $n\ge 0$, where $\{{\lambda }_n\}\subset (0,2\alpha )$, $\{{\alpha }_n\}\subset (0,1)$ and $\{{\beta }_n\}\subset (0,1)$ satisfy

• (r2) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (r3) $0<c\le {\beta }_n\le d<1$ and $\gamma \in [\kappa ,1)$;

• (r4) $a(1-{\alpha }_n)\le {\lambda }_n\le b(1-{\alpha }_n)$, where $[a,b]\subset (0,2\alpha )$ and ${lim}_{n\to \mathrm{\infty }}({\lambda }_{n+1}-{\lambda }_n)=0$.

Then $\{x_n\}$ generated by (3.15) converges strongly to ${Proj}_{F(S)\cap \mathit{MEP}}(0)$, which is the minimum norm element in $F(S)\cap \mathit{MEP}$.

Corollary 3.6 Let C be a nonempty closed convex subset of a real Hilbert space H and let $F:C\times C\to R$ be a bifunction satisfying conditions (H1)-(H4). Let $S:C\to C$ be a κ-strictly pseudo-contractive mapping. Suppose that $F(S)\cap \mathit{EP}\ne \mathrm{\varnothing }$. Let $x_0\in C$, $\{z_n\}$ and $\{x_n\}$ be sequences in C generated by

$$\{\begin{array}{c}F(z_n,y)+\frac{1}{{\lambda }_n}〈y-z_n,z_n-(1-{\alpha }_n)x_n〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)\gamma z_n+(1-{\beta }_n)(1-\gamma )S{z}_n\hfill \end{array}$$

(3.16)

for all $n\ge 0$, where $\{{\lambda }_n\}\subset (0,1)$, $\{{\alpha }_n\}\subset (0,1)$ and $\{{\beta }_n\}\subset (0,1)$ satisfy

• (r2) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (r3) $0<c\le {\beta }_n\le d<1$ and $\gamma \in [\kappa ,1)$;

• (r4) $a(1-{\alpha }_n)\le {\lambda }_n\le b(1-{\alpha }_n)$, where $[a,b]\subset (0,1)$ and ${lim}_{n\to \mathrm{\infty }}({\lambda }_{n+1}-{\lambda }_n)=0$.

Then $\{x_n\}$ generated by (3.16) converges strongly to ${Proj}_{F(S)\cap \mathit{EP}}(0)$, which is the minimum norm element in $F(S)\cap \mathit{EP}$.

Corollary 3.7 Let C be a nonempty closed convex subset of a real Hilbert space H and let $F:C\times C\to R$ be a bifunction satisfying conditions (H1)-(H4). Let $A:C\to H$ be an α-inverse-strongly monotone mapping. Suppose that $\mathit{MEP}\ne \mathrm{\varnothing }$. Let $x_0\in C$, $\{z_n\}$ and $\{x_n\}$ be sequences in C generated by

$$\{\begin{array}{c}F(z_n,y)+〈A{x}_n,y-z_n〉+\frac{1}{{\lambda }_n}〈y-z_n,z_n-(1-{\alpha }_n)x_n〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)z_n\hfill \end{array}$$

(3.17)

for all $n\ge 0$, where $\{{\lambda }_n\}\subset (0,2\alpha )$, $\{{\alpha }_n\}\subset (0,1)$ and $\{{\beta }_n\}\subset (0,1)$ satisfy

• (r2) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (r3) $0<c\le {\beta }_n\le d<1$;

• (r4) $a(1-{\alpha }_n)\le {\lambda }_n\le b(1-{\alpha }_n)$, where $[a,b]\subset (0,2\alpha )$ and ${lim}_{n\to \mathrm{\infty }}({\lambda }_{n+1}-{\lambda }_n)=0$.

Then $\{x_n\}$ generated by (3.17) converges strongly to ${Proj}_{\mathit{MEP}}(0)$, which is the minimum norm element in MEP.

Corollary 3.8 Let C be a nonempty closed convex subset of a real Hilbert space H and let $F:C\times C\to R$ be a bifunction satisfying conditions (H1)-(H4). Suppose $\mathit{EP}\ne \mathrm{\varnothing }$. Let $x_0\in C$, $\{z_n\}$ and $\{x_n\}$ be sequences in C generated by

$$\{\begin{array}{c}F(z_n,y)+\frac{1}{{\lambda }_n}〈y-z_n,z_n-(1-{\alpha }_n)x_n〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)z_n\hfill \end{array}$$

(3.18)

for all $n\ge 0$, where $\{{\lambda }_n\}\subset (0,1)$, $\{{\alpha }_n\}\subset (0,1)$ and $\{{\beta }_n\}\subset (0,1)$ satisfy

• (r2) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (r3) $0<c\le {\beta }_n\le d<1$;

• (r4) $a(1-{\alpha }_n)\le {\lambda }_n\le b(1-{\alpha }_n)$, where $[a,b]\subset (0,1)$ and ${lim}_{n\to \mathrm{\infty }}({\lambda }_{n+1}-{\lambda }_n)=0$.

Then $\{x_n\}$ generated by (3.18) converges strongly to ${Proj}_{\mathit{EP}}(0)$, which is the minimum norm element in EP.