Iterative methods for mixed equilibrium problems and strictly pseudocontractive mappings

Abstract

In this paper, we introduce new implicit and explicit iterative schemes for finding a common element of the set of solutions of the mixed equilibrium problem and the set of fixed points of a k-strictly pseudocontractive non-self mapping in Hilbert spaces. We establish results of the strong convergence of the sequences generated by the proposed schemes to a common point of two sets, which is a solution of a certain variational inequality. Our results extend and improve the corresponding results given by many authors recently in this area.

MSC:47H05, 47H06, 47H09, 47H10, 47J25, 47J05, 49M05.

1 Introduction

Let H be a real Hilbert space with inner product $〈\cdot ,\cdot 〉$ and induced norm $\parallel \cdot \parallel$. Let C be a nonempty closed convex subset of H and $S:C\to C$ be a self-mapping on C. We denote by $F(S)$ the set of fixed points of S, that is, $F(S):=\{x\in C:Sx=x\}$.

Let Θ be a bifunction of $C\times C$ into ℝ and $\phi :C\to \mathbb{R}$ be a function, where ℝ is the set of real numbers. Then we consider the following mixed equilibrium problem (for short, MEP): finding $x\in C$ such that

$$\mathrm{\Theta }(x,y)+\phi (y)-\phi (x)\ge 0,\mathrm{\forall }y\in C,$$

(1.1)

which was studied by Ceng and Yao [1] (see also [2]). The set of solutions of the MEP (1.1) is denoted by $MEP(\mathrm{\Theta },\phi )$. We see that x being a solution of the problem (1.1) implies that $x\in dom\phi =\{\phi (x)<\mathrm{\infty }\}$.

If $\phi =0$, then the MEP (1.1) becomes the following equilibrium problem (for short, EP): finding $x\in C$ such that

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

(1.2)

The set of solutions of the EP (1.2) is denoted by $EP(\mathrm{\Theta })$.

The MEP (1.1) is very general in the sense that it includes, as special cases, fixed point problems, optimization problems, variational inequality problems, minmax problems, Nash equilibrium problems in noncooperative games and others; see, e.g., [1, 3–5].

The class of pseudocontractive mappings is one of the most important classes of mappings among nonlinear mappings. Recently, many authors have devoted their studies to the problems of finding fixed points for pseudocontractive mappings; see, for example, [6–9] and the references therein. We recall that a mapping $S:C\to H$ is said to be k-strictly pseudocontractive if there exists a constant $k\in [0,1)$ such that

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

Note that the class of k-strictly pseudocontractive mappings includes the class of nonexpansive mappings as a subclass. That is, S is nonexpansive (i.e., $\parallel Sx-Sy\parallel \le \parallel x-y\parallel$, $\mathrm{\forall }x,y\in C$) if and only if S is 0-strictly pseudocontractive. The mapping S is also said to be pseudocontractive if $k=1$, and S is said to be strongly pseudocontractive if there exists a constant $\lambda \in (0,1)$ such that $S-\lambda I$ is pseudocontractive. Clearly, the class of k-strictly pseudocontractive mappings falls into the one between classes of nonexpansive mappings and pseudocontractive mappings. Also, we remark that the class of strongly pseudocontractive mappings is independent of the class of k-strictly pseudocontractive mappings (see [10, 11]).

Recently, in order to study the EP (1.2) coupled with the fixed point problem, many authors have introduced some iterative schemes for finding a common element of the set of solutions of the EP (1.2) and the set of fixed points of a countable family of nonexpansive mappings or strictly pseudocontractive mappings; see [12–14] and the references therein.

On the other hand, in 2001 Yamada [15] introduced the hybrid iterative method for the nonexpansive mapping to solve a variational inequality related to a Lipschitzian and strongly monotone operator. Since then, by using the ideas of Marino and Xu [16], Tien [17, 18] and Ceng et al. [19] provided the general iterative schemes for finding a fixed point of the nonexpansive mapping, which is a solution of a certain variational inequality related to a Lipschitzian and strongly monotone operator. Cho et al. [7] and Jung [8, 20] gave the general iterative schemes for finding a fixed point of the k-strictly pseudocontractive mapping, which is a solution of a certain variational inequality.

Inspired and motivated by the above mentioned recent works, in this paper, we introduce new implicit and explicit iterative schemes for finding a common element of the set of the solutions of the MEP (1.1) and the set of fixed points of a k-strictly pseudocontractive mapping. Then we establish results of the strong convergence of the sequences generated by the proposed schemes to a common point of two sets, which is a solution of a certain variational inequality. Our results extend and improve the recent well-known results in this area.

2 Preliminaries and lemmas

Let H be a real Hilbert space, and let C be a nonempty closed convex subset of H. In the following, we write $x_n\rightharpoonup x$ to indicate that the sequence $\{x_n\}$ converges weakly to x. $x_n\to x$ implies that $\{x_n\}$ converges strongly to x.

Recall that the mapping $V:H\to H$ is said to be l-Lipschitzian if

$$\parallel Vx-Vy\parallel \le l\parallel x-y\parallel ,\mathrm{\forall }x,y\in H,$$

and that the nonlinear operator $F:H\to H$ is said to be ρ-Lipschitzian and η-strongly monotone, where $\rho >0$ and $\eta >0$ are constants, if

$$\parallel Fx-Fy\parallel \le \rho \parallel x-y\parallel$$

and

$$〈Fx-Fy,x-y〉\ge \eta {\parallel x-y\parallel }^2,\mathrm{\forall }x,y\in H.$$

In a real Hilbert space H, we have

$${\parallel x-y\parallel }^2={\parallel x\parallel }^2+{\parallel y\parallel }^2-2〈x,y〉,$$

(2.1)

for all $x,y\in H$ and $\lambda \in \mathbb{R}$. For every point $x\in H$, there exists a unique nearest point in C, denoted by $P_{C}x$, such that

$$\parallel x-P_{C}x\parallel \le \parallel x-y\parallel$$

for all $y\in C$. $P_C$ is called the metric projection of H onto C. It is well known that $P_C$ is nonexpansive and $P_C$ is characterized by the property

$$u=P_{C}x\iff 〈x-u,u-y〉\ge 0,\mathrm{\forall }x\in H,y\in C.$$

(2.2)

It is also well known that H satisfies the Opial condition; that is, for any sequence $\{x_n\}$ with $x_n\rightharpoonup x$, the inequality

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_n-x\parallel <\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_n-y\parallel$$

holds for every $y\in H$ with $y\ne x$.

For solving the equilibrium problem for a bifunction $\mathrm{\Theta }:C\times C\to \mathbb{R}$, let us assume that Θ and φ satisfy the following conditions:

(A1) $\mathrm{\Theta }(x,x)=0$ for all $x\in C$;

(A2) Θ is monotone, that is, $\mathrm{\Theta }(x,y)+\mathrm{\Theta }(y,x)\le 0$ for all $x,y\in C$;

(A3) for each $x,y,z\in C$,

$$\underset{t\downarrow 0}{lim}\mathrm{\Theta }(tz+(1-t)x,y)\le \mathrm{\Theta }(x,y);$$

(A4) for each $x\in C$, $y\mapsto \mathrm{\Theta }(x,y)$ is convex and lower semicontinuous;

(A5) for each $y\in C$, $x\mapsto \mathrm{\Theta }(x,y)$ is weakly upper semicontinuous;

(B1) for each $x\in H$ and $r>0$, there exist a bounded subset $D_x\subseteq C$ and $y_x\in C$ such that for any $z\in C\setminus D_x$,

$$\mathrm{\Theta }(z,y_x)+\phi (y_x)-\phi (z)+\frac{1}{r}〈y_x-z,z-x〉<0;$$

(B2) C is a bounded set.

The following lemmas were given in [3, 21].

Lemma 2.1 [3]

Let C be a nonempty closed convex subset of H and Θ be a bifunction of $C\times C$ into ℝ satisfying (A1)-(A4). Let $r>0$ and $x\in H$. Then there exists $z\in C$ such that

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

Lemma 2.2 [21]

Let C be a nonempty closed convex subset of H. Let Θ be a bifunction from $C\times C$ to ℝ satisfying (A1)-(A5) and $\phi :C\to \mathbb{R}$ be a proper lower semicontinuous and convex function. For $r>0$ and $x\in H$, define a mapping $T_r:H\to C$ as follows:

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

for all $x\in H$. Assume that either (B1) or (B2) holds. Then the following hold:

• (1) for each $x\in H$, $T_{r}x\ne \mathrm{\varnothing }$;

• (2) $T_r$ is single-valued;

• (3) $T_r$ is firmly nonexpansive; that is, for any $x,y\in H$,

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

• (4) $F(T_r)=MEP(\mathrm{\Theta },\phi )$;

• (5) $MEP(\mathrm{\Theta },\phi )$ is closed and convex.

We need the following lemmas for the proof of our main results.

Lemma 2.3 [22]

Let H be a Hilbert space, C be a closed convex subset of H. If S is a k-strictly pseudocontractive mapping on C, then the fixed point set $F(S)$ is closed convex, so that the projection $P_{F(S)}$ is well defined.

Lemma 2.4 [22]

Let H be a real Hilbert space and C be a closed convex subset of H. Let $S:C\to H$ be a k-strictly pseudocontractive mapping with $F(S)\ne \mathrm{\varnothing }$. Then $F(P_{C}S)=F(S)$.

Lemma 2.5 [22]

Let H be a real Hilbert space, C be a closed convex subset of H, and $S:C\to H$ be a k-strictly pseudocontractive mapping. Define a mapping $T:C\to H$ by $Tx=\lambda x+(1-\lambda )Sx$ for all $x\in C$. Then as $\lambda \in [k,1)$, T is a nonexpansive mapping such that $F(T)=F(S)$.

Lemma 2.6 [23]

Let $\{s_n\}$ be a sequence of non-negative real numbers satisfying

$$s_{n+1}\le (1-{\xi }_n)s_n+{\xi }_n{\delta }_n,\mathrm{\forall }n\ge 1,$$

where $\{{\xi }_n\}$ and $\{{\delta }_n\}$ satisfy the following conditions:

• (i) $\{{\xi }_n\}\subset [0,1]$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\xi }_n=\mathrm{\infty }$,

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

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

Lemma 2.7 [24]

Let $\{x_n\}$ and $\{z_n\}$ be bounded sequences in a real Banach space E and $\{{\gamma }_n\}$ be a sequence in $[0,1]$ which satisfies the following condition:

$$0<\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\gamma }_n\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\gamma }_n<1.$$

Suppose that $x_{n+1}={\gamma }_n{x}_n+(1-{\gamma }_n)z_n$ for all $n\ge 1$ and

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

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

Lemma 2.8 In a real Hilbert space H, the following inequality holds:

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

The following lemma can be easily proven, and therefore, we omit the proof.

Lemma 2.9 Let $V:H\to H$ be an l-Lipschitzian mapping with a constant $l\ge 0$, and $F:H\to H$ be a ρ-Lipschitzian and η-strongly monotone operator with constants $\rho ,\eta >0$. Then for $0\le \gamma l<\mu \eta$,

$$〈(\mu F-\gamma V)x-(\mu F-\gamma V)y,x-y〉\ge (\mu \eta -\gamma l){\parallel x-y\parallel }^2,\mathrm{\forall }x,y\in C.$$

That is, $\mu F-\gamma V$ is strongly monotone with a constant $\mu \eta -\gamma l$.

Finally, the following lemma is an improvement of Lemma 2.9 in [20] (see also [15]).

Lemma 2.10 Let H be a real Hilbert space H. Let $F:H\to H$ be a ρ-Lipschizian and η-strongly monotone operator with $0<\eta \le \rho$. Let $0<\mu <\frac{2\eta }{{\rho }^2}$ and $0<t<\varsigma \le 1$. Then $S:=\varsigma I-t\mu F:H\to H$ is a contraction with a contractive constant $\varsigma -t\tau$, where $\tau =1-\sqrt{1-\mu (2\eta -\mu {\rho }^2)}$.

Proof First, we show that $I-\mu F$ is strictly contractive. In fact, by applying the ρ-Lipschitz continuity and η-strongly monotonicity of F, we obtain for $x,y\in H$,

$$\begin{array}{c}{\parallel (I-\mu F)x-(I-\mu F)y\parallel }^2\hfill \\ ={\parallel (x-y)-\mu (Fx-Fy)\parallel }^2\hfill \\ ={\parallel x-y\parallel }^2-2\mu 〈Fx-Fy,x-y〉+{\mu }^2{\parallel Fx-Fy\parallel }^2\hfill \\ \le {\parallel x-y\parallel }^2-2\mu \eta {\parallel x-y\parallel }^2+{\mu }^2{\rho }^2{\parallel x-y\parallel }^2\hfill \\ =(1-\mu (2\eta -\mu {\rho }^2)){\parallel x-y\parallel }^2,\hfill \end{array}$$

and so

$$\parallel (I-\mu F)x-(I-\mu F)y\parallel \le \sqrt{1-\mu (2\eta -\mu {\rho }^2)}\parallel x-y\parallel .$$

(2.3)

Now, noting that $S:=\varsigma I-t\mu F=(\varsigma -t)I-t(\mu F-I)$, by (2.3) we have for $x,y\in H$,

$$\begin{array}{rcl}\parallel Sx-Sy\parallel & =& \parallel (\varsigma -t)(x-y)-t((\mu F-I)x-(\mu F-I)y)\parallel \\ \le & (\varsigma -t)\parallel x-y\parallel +t\parallel (\mu F-I)x-(\mu F-I)y\parallel \\ \le & (\varsigma -t)\parallel x-y\parallel +t\sqrt{1-\mu (2\eta -\mu {\rho }^2)}\parallel x-y\parallel \\ =& (\varsigma -t(1-\sqrt{1-\mu (2\eta -\mu {\rho }^2)}))\parallel x-y\parallel \\ =& (\varsigma -t\tau )\parallel x-y\parallel .\end{array}$$

Hence, S is a contraction with a contractive constant $\varsigma -t\tau$. □

3 Main results

Let H be a real Hilbert space, and let C be a nonempty closed convex subset of H. Let $V:H\to H$ be an l-Lipschitzian mapping with a constant $l>0$, and $F:H\to H$ be a ρ-Lipschitzian and η-strongly monotone operator with $0<\eta \le \rho$. Let $0<\mu <\frac{2\eta }{{\rho }^2}$ and $0<\gamma l<\tau$, where $\tau =1-\sqrt{1-\mu (2\eta -\mu {\rho }^2)}$. Let $\{T_{r_n}\}$ be a sequence of mappings defined as in Lemma 2.2 and $S:C\to H$ be a k-strictly pseudocontractive mapping. Define a mapping $S_n:C\to H$ by $S_{n}x={\beta }_{n}x+(1-{\beta }_n)Sx$, $\mathrm{\forall }x\in C$, where ${\beta }_n\in [k,1)$. Then, by Lemma 2.5, $S_n$ is nonexpansive.

Consider the following mapping $Q_n$ on H defined by

$$Q_{n}x={\alpha }_n\gamma Vx+(I-{\alpha }_n\mu F)S_n{T}_{r_n}x,\mathrm{\forall }x\in H,n\ge 1,$$

where ${\alpha }_n\in (0,1)$. By Lemmas 2.2 and 2.10, we have

$$\begin{array}{rcl}\parallel Q_{n}x-Q_{n}y\parallel & \le & {\alpha }_n\gamma \parallel Vx-Vy\parallel +\parallel (I-{\alpha }_n\mu F)S_n{T}_{r_n}x-(I-{\alpha }_n\mu F)S_n{T}_{r_n}y\parallel \\ \le & {\alpha }_n\gamma l\parallel x-y\parallel +(1-{\alpha }_n\tau )\parallel x-y\parallel \\ =& (1-{\alpha }_n(\tau -\gamma l))\parallel x-y\parallel .\end{array}$$

Since $0<1-{\alpha }_n(\tau -\gamma l)<1$, $Q_n$ is a contraction. Therefore, by the Banach contraction principle, $Q_n$ has a unique fixed point $x_n\in H$, which uniquely solves the fixed point equation

$$x_n={\alpha }_n\gamma V{x}_n+(I-{\alpha }_n\mu F)S_n{T}_{r_n}{x}_n.$$

Now, we prove the convergence of the sequence $\{x_n\}$ and show the existence of the $q\in MEP(\mathrm{\Theta },\phi )\cap F(S)$, which solves the variational inequality

$$〈(\mu F-\gamma V)q,p-q〉\ge 0,\mathrm{\forall }p\in MEP(\mathrm{\Theta },\phi )\cap F(S).$$

(3.1)

Equivalently, $q=P_{MEP(\mathrm{\Theta },\phi )\cap F(S)}(I-\mu F+\gamma V)q$.

Theorem 3.1 Let C be a nonempty closed convex subset of a real Hilbert space H and Θ be a bifunction from $C\times C\to \mathbb{R}$ satisfying (A1)-(A5). Let $S:C\to H$ be a k-strictly pseudocontractive non-self mapping such that $F(S)\cap MEP(\mathrm{\Theta },\phi )\ne \mathrm{\varnothing }$. Let $F:H\to H$ be a ρ-Lipschitzian and η-strongly monotone operator with $0<\eta \le \rho$. Let $V:H\to H$ be an l-Lipschitzian mapping with a constant $l>0$. Let $0<\mu <\frac{2\eta }{{\rho }^2}$ and $0<\gamma l<\tau$, where $\tau =1-\sqrt{1-\mu (2\eta -\mu {\rho }^2)}$. Assume that either (B1) or (B2) holds. Let $\{x_n\}$ be a sequence generated by

$$\{\begin{array}{c}\mathrm{\Theta }(u_n,y)+\phi (y)-\phi (u_n)+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ y_n={\beta }_n{u}_n+(1-{\beta }_n)S{u}_n,\hfill \\ x_n={\alpha }_n\gamma V{x}_n+(I-{\alpha }_n\mu F)y_n,\mathrm{\forall }n\ge 1,\hfill \end{array}$$

(3.2)

where $u_n=T_{r_n}{x}_n$, $y_n=S_n{u}_n$, and $\{r_n\}\subset (0,\mathrm{\infty })$ satisfying ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$. If $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ satisfy the following conditions:

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

• (ii) $0\le k\le {\beta }_n\le \lambda <1$ and ${lim}_{n\to \mathrm{\infty }}{\beta }_n=\lambda$,

then $\{x_n\}$ converges strongly to a point $q\in F(S)\cap MEP(\mathrm{\Theta },\phi )$, which solves the variational inequality (3.1).

Proof Note that from the condition (i), without loss of generality, we assume that ${\alpha }_n\tau <1$ for $n\ge 1$.

First, we can show easily the uniqueness of a solution of the variational inequality (3.1). In fact, noting that $0\le \gamma l<\tau$ and $\mu \eta \ge \tau \iff \rho \ge \eta$, it follows from Lemma 2.9 that

$$〈(\mu F-\gamma V)x-(\mu F-\gamma V)y,x-y〉\ge (\mu \eta -\gamma l){\parallel x-y\parallel }^2.$$

That is, $\mu F-\gamma V$ is strongly monotone for $0\le \gamma l<\tau \le \mu \eta$. So, the variational inequality (3.1) has only one solution. In what follows, we use $q\in F(S)\cap MEP(\mathrm{\Theta },\phi )$ to denote the unique solution of the variational inequality (3.1).

Now, take $p\in F(S)\cap MEP(\mathrm{\Theta },\phi )$. Since $u_n=T_{r_n}{x}_n$ and $p=T_{r_n}p$, from Lemma 2.2, we know that

$$\parallel u_n-p\parallel \le \parallel x_n-p\parallel ,\mathrm{\forall }n\ge 1.$$

Moreover, from $S_{n}p=p$, it follows that

$$\parallel y_n-p\parallel =\parallel S_n{u}_n-S_{n}p\parallel \le \parallel u_n-p\parallel \le \parallel x_n-p\parallel .$$

(3.3)

Thus, we have

$$\begin{array}{rcl}\parallel x_n-p\parallel & =& \parallel {\alpha }_n(\gamma V{x}_n-\mu Fp)+(I-{\alpha }_n\mu F)y_n-(I-{\alpha }_n\mu F)p\parallel \\ \le & (1-{\alpha }_n\tau )\parallel y_n-p\parallel +{\alpha }_n(\gamma l\parallel x_n-p\parallel +\parallel \gamma Vp-\mu Fp\parallel )\\ \le & (1-{\alpha }_n(\tau -\gamma l))\parallel x_n-p\parallel +{\alpha }_n\parallel \gamma Vp-\mu Fp\parallel .\end{array}$$

This implies that $\parallel x_n-p\parallel \le \frac{\parallel \gamma Vp-\mu Fp\parallel }{\tau -\gamma l}$. Hence, $\{x_n\}$ is bounded, and we also obtain that $\{u_n\}$, $\{y_n\}$ and $\{V{x}_n\}$ are bounded. We note that

$$\begin{array}{rcl}\parallel u_n-y_n\parallel & \le & \parallel u_n-x_n\parallel +\parallel x_n-y_n\parallel \\ =& \parallel u_n-x_n\parallel +{\alpha }_n\parallel \gamma V{x}_n-\mu F{y}_n\parallel .\end{array}$$

(3.4)

Using Lemma 2.2, we obtain

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

and so

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

(3.5)

Then, from Lemma 2.8, (3.3) and (3.5), we have

$$\begin{array}{rcl}{\parallel x_n-p\parallel }^2& =& {\parallel {\alpha }_n(\gamma V{x}_n-\mu Fp)+(I-{\alpha }_n\mu F)y_n-(I-{\alpha }_n\mu F)p\parallel }^2\\ \le & {(1-{\alpha }_n\tau )}^2{\parallel y_n-p\parallel }^2+2{\alpha }_n〈\gamma V{x}_n-\mu Fp,x_n-p〉\\ \le & {(1-{\alpha }_n\tau )}^2{\parallel u_n-p\parallel }^2+2{\alpha }_n\gamma 〈V{x}_n-Vp,x_n-p〉\\ +2{\alpha }_n\parallel \gamma Vp-\mu Fp\parallel \parallel x_n-p\parallel \\ \le & {(1-{\alpha }_n\tau )}^2({\parallel x_n-p\parallel }^2-{\parallel x_n-u_n\parallel }^2)+2{\alpha }_n\gamma l{\parallel x_n-p\parallel }^2\\ +2{\alpha }_n\parallel \gamma Vp-\mu Fp\parallel \parallel x_n-p\parallel \\ =& (1-2{\alpha }_n(\tau -\gamma l)+{({\alpha }_n\tau )}^2){\parallel x_n-p\parallel }^2-{(1-{\alpha }_n\tau )}^2{\parallel x_n-u_n\parallel }^2\\ +2{\alpha }_n\parallel \gamma Vp-\mu Fp\parallel \parallel x_n-p\parallel \\ \le & {\parallel x_n-p\parallel }^2+{\alpha }_n^2{\tau }^2{\parallel x_n-p\parallel }^2-{(1-{\alpha }_n\tau )}^2{\parallel x_n-u_n\parallel }^2\\ +2{\alpha }_n\parallel \gamma Vp-\mu Fp\parallel \parallel x_n-p\parallel ,\end{array}$$

and hence

$${(1-{\alpha }_n\tau )}^2{\parallel x_n-u_n\parallel }^2\le {\alpha }_n^2{\tau }^2{\parallel x_n-p\parallel }^2+2{\alpha }_n\parallel \gamma Vp-\mu Fp\parallel \parallel x_n-p\parallel .$$

Since ${\alpha }_n\to 0$, it follows that

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

From (3.4), we know that

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

(3.6)

Define $T:C\to H$ by $Tx=\lambda x+(1-\lambda )Sx$. Then by Lemma 2.5, T is nonexpansive with $F(T)=F(S)$. Notice that

$$\begin{array}{rcl}\parallel T{u}_n-u_n\parallel & \le & \parallel T{u}_n-y_n\parallel +\parallel y_n-u_n\parallel \\ \le & |\lambda -{\beta }_n|\parallel u_n-S{u}_n\parallel +\parallel y_n-u_n\parallel .\end{array}$$

By (3.6) and ${\beta }_n\to \lambda$, we obtain

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

Consider a subsequence $\{u_{n_i}\}$ of $\{u_n\}$. Since $\{u_n\}$ is bounded, there exists a subsequence $\{u_{n_{i_j}}\}$ of $\{u_{n_i}\}$ which converges weakly to q.

Next, we show that $q\in F(S)\cap MEP(\mathrm{\Theta },\phi )$. Without loss of generality, we can assume that $u_{n_i}\rightharpoonup q$. Since C is closed and convex, C is weakly closed. So, we have $q\in C$. Let us show $q\in F(T)$. Assume that $q\notin F(T)$. Since $u_{n_i}\rightharpoonup q$ and $q\ne Tq$, it follows from the Opial condition that

$$\begin{array}{rcl}\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel u_{n_i}-q\parallel & <& \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel u_{n_i}-Tq\parallel \\ \le & \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}(\parallel u_{n_i}-T{u}_{n_i}\parallel +\parallel T{u}_{n_i}-Tq\parallel )\\ \le & \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel u_{n_i}-q\parallel ,\end{array}$$

which is a contradiction. So, we get $q\in F(T)$, and hence $q\in F(S)$.

We shall show that $q\in MEP(\mathrm{\Theta },\phi )$. Since $u_n=T_{r_n}{x}_n$, for any $y\in C$, we have

$$\mathrm{\Theta }(u_n,y)+\phi (y)-\phi (u_n)+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0.$$

It follows from (A2) that

$$\phi (y)-\phi (u_n)+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge \mathrm{\Theta }(y,u_n).$$

Replacing n by $n_i$, we have

$$\phi (y)-\phi (u_{n_i})+\frac{1}{r_{n_i}}〈y-u_{n_i},u_{n_i}-x_{n_i}〉\ge \mathrm{\Theta }(y,u_{n_i}).$$

Since $\frac{u_{n_i}-x_{n_i}}{r_{n_i}}\to 0$ and $u_{n_i}\rightharpoonup q$, it follows from (A4) that

$$0\ge -\phi (y)+\phi (q)+\mathrm{\Theta }(y,q),\mathrm{\forall }y\in C.$$

Put $z_t=ty+(1-t)q$ for all $t\in (0,1]$ and $y\in C$. Then we have $z_t\in C$ and

$$-\phi (z_t)+\phi (q)+\mathrm{\Theta }(z_t,q)\le 0.$$

(3.7)

By (A1), (A4) and (3.7), we have

$$\begin{array}{rcl}0& =& \mathrm{\Theta }(z_t,z_t)+\phi (z_t)-\phi (z_t)\\ \le & t\mathrm{\Theta }(z_t,y)+(1-t)\mathrm{\Theta }(z_t,q)+t\phi (y)+(1-t)\phi (q)-\phi (z_t)\\ \le & t(\mathrm{\Theta }(z_t,y)+\phi (y)-\phi (z_t))+(1-t)(\mathrm{\Theta }(z_t,q)+\phi (q)-\phi (z_t))\\ \le & t(\mathrm{\Theta }(z_t,y)+\phi (y)-\phi (z_t)),\end{array}$$

and hence

$$0\le \mathrm{\Theta }(z_t,y)+\phi (y)-\phi (z_t).$$

(3.8)

Letting $t\to 0$, by (A3) we have for each $y\in C$,

$$0\le \mathrm{\Theta }(q,y)+\phi (y)-\phi (q).$$

This implies that $q\in MEP(\mathrm{\Theta },\phi )$. Therefore, $q\in F(S)\cap MEP(\mathrm{\Theta },\phi )$.

On the other hand, we note that

$$x_n-q={\alpha }_n(\gamma V{x}_n-\mu Fq)+(I-{\alpha }_n\mu F)y_n-(I-{\alpha }_n\mu F)q.$$

It follows that

$$\begin{array}{c}{\parallel x_n-q\parallel }^2\hfill \\ ={\alpha }_n〈\gamma V{x}_n-\mu Fq,x_n-q〉+〈(I-{\alpha }_n\mu F)y_n-(I-{\alpha }_n\mu F)q,x_n-q〉\hfill \\ \le {\alpha }_n〈\gamma V{x}_n-\mu Fq,x_n-q〉+\parallel (I-{\alpha }_n\mu F)y_n-(I-{\alpha }_n\mu F)q\parallel \parallel x_n-q\parallel \hfill \\ \le {\alpha }_n〈\gamma V{x}_n-\mu Fq,x_n-q〉+(1-{\alpha }_n\tau )\parallel y_n-q\parallel \parallel x_n-q\parallel \hfill \\ \le {\alpha }_n〈\gamma V{x}_n-\mu Fq,x_n-q〉+(1-{\alpha }_n\tau ){\parallel x_n-q\parallel }^2.\hfill \end{array}$$

Hence, we obtain

$$\begin{array}{rcl}{\parallel x_n-q\parallel }^2& \le & \frac{1}{\tau }〈\gamma V{x}_n-\mu Fq,x_n-q〉\\ =& \frac{1}{\tau }(\gamma 〈V{x}_n-Vq,x_n-q〉+〈\gamma Vq-\mu Fq,x_n-q〉)\\ \le & \frac{1}{\tau }(\gamma l{\parallel x_n-q\parallel }^2+〈\gamma Vq-\mu Fq,x_n-q〉).\end{array}$$

This implies that

$${\parallel x_n-q\parallel }^2\le \frac{〈\gamma Vq-\mu Fq,x_n-q〉}{\tau -\gamma l}.$$

In particular, we have

$${\parallel x_{n_i}-q\parallel }^2\le \frac{〈\gamma Vq-\mu Fq,x_{n_i}-q〉}{\tau -\gamma l}.$$

(3.9)

Since $x_{n_i}\rightharpoonup q$, it follows that $x_{n_i}\to q$ as $i\to \mathrm{\infty }$.

Now, we show that q solves the variational inequality (3.1). Since $x_n={\alpha }_n\gamma V{x}_n+(I-{\alpha }_n\mu F)S_n{T}_{r_n}{x}_n$, we have

$$(\mu F-\gamma V)x_n=-\frac{1}{{\alpha }_n}((I-{\alpha }_n\mu F)x_n-(I-{\alpha }_n\mu F)S_n{T}_{r_n}{x}_n).$$

It follows that for $p\in F(S)\cap MEP(\mathrm{\Theta },\phi )$,

$$\begin{array}{rcl}〈(\mu F-\gamma V)x_n,x_n-p〉& =& -\frac{1}{{\alpha }_n}〈(I-{\alpha }_n\mu F)x_n-(I-{\alpha }_n\mu F)S_n{T}_{r_n}{x}_n,x_n-p〉\\ =& -\frac{1}{{\alpha }_n}〈(I-S_n{T}_{r_n})x_n-(I-S_n{T}_{r_n})p,x_n-p〉\\ +〈\mu F{x}_n-\mu F{S}_n{T}_{r_n}{x}_n,x_n-p〉\\ \le & 〈\mu F{x}_n-\mu F{S}_n{T}_{r_n}{x}_n,x_n-p〉\end{array}$$

(3.10)

since $I-S_n{T}_{r_n}$ is monotone (i.e., $〈x-y,(I-S_n{T}_{r_n})x-(I-S_n{T}_{r_n})y〉\ge 0$ for all $x,y\in H$. This is due to the nonexpansivity of $S_n{T}_{r_n}$). Since $\parallel x_n-y_n\parallel ={\alpha }_n\parallel \gamma V{x}_n-\mu F{y}_n\parallel \to 0$ as $n\to \mathrm{\infty }$, by replacing n in (3.10) with $n_i$ and letting $i\to \mathrm{\infty }$, we obtain

$$\begin{array}{rcl}〈(\mu F-\gamma V)q,q-p〉& =& \underset{i\to \mathrm{\infty }}{lim}〈(\mu F-\gamma V)x_{n_i},x_{n_i}-p〉\\ \le & \underset{i\to \mathrm{\infty }}{lim}〈\mu F{x}_{n_i}-\mu F{y}_{n_i},x_{n_i}-p〉=0.\end{array}$$

(3.11)

That is, $q\in F(S)\cap MEP(\mathrm{\Theta },\phi )$ is a solution of the variational inequality (3.1).

Finally, we show that the sequence $\{x_n\}$ converges strongly to q. To this end, let $\{x_{n_k}\}$ be another subsequence of $\{x_n\}$ and assume $x_{n_k}\to \hat{q}$. By the same proof as the one above, we have $\hat{q}\in F(S)\cap MEP(\mathrm{\Theta },\phi )$. Moreover, it follows from (3.10) that

$$〈(\mu F-\gamma V)q,q-\hat{q}〉\le 0.$$

(3.12)

Interchanging q and $\hat{q}$, we obtain

$$〈(\mu F-\gamma V)\hat{q},\hat{q}-q〉\le 0.$$

(3.13)

Lemma 2.9 and adding these two inequalities (3.12) and (3.13) yield

$$(\mu \eta -\gamma l){\parallel q-\hat{q}\parallel }^2\le 〈(\mu F-\gamma V)q-(\mu F-\gamma V)\hat{q},q-\hat{q}〉\le 0.$$

Hence, $q=\hat{q}$. Therefore, we conclude that $x_n\to q$ as $n\to \mathrm{\infty }$.

The variational inequality (3.1) can be rewritten as

$$〈(I-\mu F+\gamma V)q-q,q-p〉\ge 0,\mathrm{\forall }p\in F(S)\cap MEP(\mathrm{\Theta },\phi ).$$

By (2.2), this is equivalent to the fixed point equation

$$P_{F(S)\cap MEP(\mathrm{\Theta },\phi )}(I-\mu F+\gamma V)q=q.$$

 □

Now, we establish the strong convergence of an explicit iterative scheme for finding a common element of the set of solutions of a mixed equilibrium problem and the set of fixed points of a k-strictly pseudocontractive non-self mapping.

Theorem 3.2 Let C be a nonempty closed convex subset of a real Hilbert space H and Θ be a bifunction from $C\times C\to \mathbb{R}$ satisfying (A1)-(A5). Let $S:C\to H$ be a k-strictly pseudocontractive non-self mapping such that $F(S)\cap MEP(\mathrm{\Theta },\phi )\ne \mathrm{\varnothing }$. Let $F:H\to H$ be a ρ-Lipschitzian and η-strongly monotone operator with $0<\eta \le \rho$. Let $V:H\to H$ be an l-Lipschitzian mapping with a constant $l>0$. Let $0<\mu <\frac{2\eta }{{\rho }^2}$ and $0<\gamma l<\tau$, where $\tau =1-\sqrt{1-\mu (2\eta -\mu {\rho }^2)}$. Assume that either (B1) or (B2) holds. Let $\{x_n\}$ and $\{u_n\}$ be sequences generated by

$$\{\begin{array}{c}\mathrm{\Theta }(u_n,y)+\phi (y)-\phi (u_n)+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ y_n={\beta }_n{u}_n+(1-{\beta }_n)S{u}_n,\hfill \\ x_{n+1}={\alpha }_n\gamma V{x}_n+{\lambda }_n{x}_n+((1-{\lambda }_n)I-{\alpha }_n\mu F)y_n,\mathrm{\forall }n\ge 1,\hfill \end{array}$$

(3.14)

where $u_n=T_{r_n}{x}_n$ and $y_n=S_n{u}_n$. If $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{r_n\}$ and $\{{\lambda }_n\}$ satisfy the following conditions:

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

• (ii) $0\le k\le {\beta }_n\le \lambda <1$ and ${lim}_{n\to \mathrm{\infty }}{\beta }_n=\lambda$, ${lim}_{n\to \mathrm{\infty }}|{\beta }_{n+1}-{\beta }_n|=0$;

• (iii) $\{r_n\}\subset (0,\mathrm{\infty })$, ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$, ${lim}_{n\to \mathrm{\infty }}|r_{n+1}-r_n|=0$;

• (iv) $\{{\lambda }_n\}\subset (0,1)$ and $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\lambda }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\lambda }_n<1$,

then $\{x_n\}$ and $\{u_n\}$ converge strongly to a point $q\in F(S)\cap MEP(\mathrm{\Theta },\phi )$, which solves the variational inequality (3.1).

Proof First, from the condition (i), without loss of generality, we assume that ${\alpha }_n\tau <1$, $\frac{2{\alpha }_n(\tau -\gamma l)}{1-{\alpha }_n\gamma l}<1$ and ${\alpha }_n(1-{\lambda }_n)<1$ for $n\ge 1$.

We divide the proof into several steps as follows.

Step 1. We show that $\parallel x_n-p\parallel \le max\{\parallel x_0-p\parallel ,\frac{\parallel \gamma Vp-\mu Fp\parallel }{\tau -\gamma l}\}$ for all $n\ge 0$ and all $p\in F(S)\cap MEP(\mathrm{\Theta },\phi )$. Indeed, let $p\in F(S)\cap MEP(\mathrm{\Theta },\phi )$. Then from Lemma 2.10, we have

From induction, we have

$$\parallel x_n-p\parallel \le max\{\parallel x_1-p\parallel ,\frac{\parallel \gamma Vp-\mu Fp\parallel }{\tau -\gamma l}\},\mathrm{\forall }n\ge 1.$$

Hence, $\{x_n\}$ is bounded. From (3.3), $\{u_n\}$, $\{y_n\}$, $\{V{x}_n\}$, $\{S{u}_n\}$ and $\{F{y}_n\}$ are also bounded.

Step 2. We show that ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$ and ${lim}_{n\to \mathrm{\infty }}\parallel u_{n+1}-u_n\parallel =0$. To show this, define

$$x_{n+1}={\lambda }_n{x}_n+(1-{\lambda }_n)z_n,\mathrm{\forall }n\ge 1.$$

Observe that from the definition of $z_n$,

$$\begin{array}{rcl}{z}_{n+1}-z_n& =& \frac{x_{n+2}-{\lambda }_{n+1}{x}_{n+1}}{1-{\lambda }_{n+1}}-\frac{x_{n+1}-{\lambda }_n{x}_n}{1-{\lambda }_n}\\ =& \frac{{\alpha }_{n+1}\gamma V{x}_{n+1}+((1-{\lambda }_{n+1})I-{\alpha }_{n+1}\mu F)y_{n+1}}{1-{\lambda }_{n+1}}\\ -\frac{{\alpha }_n\gamma V{x}_n+((1-{\lambda }_n)I-{\alpha }_n\mu F)y_n}{1-{\lambda }_n}\\ =& \frac{{\alpha }_{n+1}}{1-{\lambda }_{n+1}}\gamma V{x}_{n+1}-\frac{{\alpha }_n}{1-{\lambda }_n}\gamma V{x}_n\\ +y_{n+1}-y_n+\frac{{\alpha }_n}{1-{\lambda }_n}\mu F{y}_n-\frac{{\alpha }_{n+1}}{1-{\lambda }_{n+1}}\mu F{y}_{n+1}\\ =& \frac{{\alpha }_{n+1}}{1-{\lambda }_{n+1}}(\gamma V{x}_{n+1}-\mu F{y}_{n+1})+\frac{{\alpha }_n}{1-{\lambda }_n}(\mu F{y}_n-\gamma V{x}_n)+y_{n+1}-y_n.\end{array}$$

Thus, it follows that

$$\begin{array}{rcl}\parallel z_{n+1}-z_n\parallel & \le & \frac{{\alpha }_{n+1}}{1-{\lambda }_{n+1}}(\gamma \parallel V{x}_{n+1}\parallel +\mu \parallel F{y}_{n+1}\parallel )\\ +\frac{{\alpha }_n}{1-{\lambda }_n}(\mu \parallel F{y}_n\parallel +\gamma \parallel V{x}_n\parallel )+\parallel y_{n+1}-y_n\parallel .\end{array}$$

(3.15)

On the one hand, we note that

$$\begin{array}{rcl}\parallel y_{n+1}-y_n\parallel & =& \parallel S_{n+1}{u}_{n+1}-S_n{u}_n\parallel \\ \le & \parallel S_{n+1}{u}_{n+1}-S_{n+1}{u}_n\parallel +\parallel S_{n+1}{u}_n-S_n{u}_n\parallel \\ \le & \parallel u_{n+1}-u_n\parallel +\parallel S_{n+1}{u}_n-S_n{u}_n\parallel .\end{array}$$

(3.16)

Noticing that

$$\begin{array}{rcl}\parallel s_{n+1}{u}_n-S_n{u}_n\parallel & =& \parallel {\beta }_{n+1}{u}_n+(1-{\beta }_{n+1})S{u}_n-({\beta }_n{u}_n+(1-{\beta }_n)S{u}_n)\parallel \\ \le & |{\beta }_{n+1}-{\beta }_n|\parallel u_n-S{u}_n\parallel ,\end{array}$$

from (3.16) we have

$$\parallel y_{n+1}-y_n\parallel \le \parallel u_{n+1}-u_n\parallel +|{\beta }_{n+1}-{\beta }_n|\parallel u_n-S{u}_n\parallel .$$

(3.17)

On the other hand, from $u_{n+1}=T_{r_{n+1}}{x}_{n+1}$ and $u_n=T_{r_n}{x}_n$, we have

$$\mathrm{\Theta }(u_{n+1},y)+\phi (y)-\phi (u_{n+1})+\frac{1}{r_{n+1}}〈y-u_{n+1},u_{n+1}-x_{n+1}〉\ge 0,\mathrm{\forall }y\in C$$

(3.18)

and

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

(3.19)

Putting $y=u_n$ in (3.18) and $y=u_{n+1}$ in (3.19), we obtain

$$\mathrm{\Theta }(u_{n+1},u_n)+\phi (u_n)-\phi (u_{n+1})+\frac{1}{r_{n+1}}〈u_n-u_{n+1},u_{n+1}-x_{n+1}〉\ge 0$$

and

$$\mathrm{\Theta }(u_n,u_{n+1})+\phi (u_{n+1})-\phi (u_n)+\frac{1}{r_n}〈u_{n+1}-u_n,u_n-x_n〉\ge 0.$$

By (A2), we have

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

and hence

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

Since ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$, we assume that there exists a real number c such that $r_n>c>0$ for all $n\ge 1$. Thus, we have

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

and hence

$$\begin{array}{rcl}\parallel u_{n+1}-u_n\parallel & \le & \parallel x_{n+1}-x_n\parallel +\frac{1}{r_{n+1}}|r_{n+1}-r_n|\parallel u_{n+1}-x_{n+1}\parallel \\ \le & \parallel x_{n+1}-x_n\parallel +\frac{1}{c}|r_{n+1}-r_n|L,\end{array}$$

(3.20)

where $L=sup\{\parallel u_n-x_n\parallel :n\ge 1\}$. Therefore, from (3.15), (3.17) and (3.20), we obtain

$$\begin{array}{c}\parallel z_{n+1}-z_n\parallel -\parallel x_{n+1}-x_n\parallel \hfill \\ \le \frac{{\alpha }_{n+1}}{1-{\lambda }_{n+1}}(\gamma \parallel V{x}_{n+1}\parallel +\mu \parallel F{y}_{n+1}\parallel )+\frac{{\alpha }_n}{1-{\lambda }_n}(\mu \parallel F{y}_n\parallel +\gamma \parallel V{x}_n\parallel )\hfill \\ +|{\beta }_{n+1}-{\beta }_n|\parallel u_n-S{u}_n\parallel +\frac{1}{c}|r_{n+1}-r_n|L\hfill \\ \le \frac{{\alpha }_{n+1}}{1-{\lambda }_{n+1}}(\gamma \parallel V{x}_{n+1}\parallel +\mu \parallel F{y}_{n+1}\parallel )+\frac{{\alpha }_n}{1-{\lambda }_n}(\mu \parallel F{y}_n\parallel +\gamma \parallel V{x}_n\parallel )\hfill \\ +(|{\beta }_{n+1}-{\beta }_n|+|r_{n+1}-r_n|)M_1,\hfill \end{array}$$

where $M_1$ is an appropriate constant such that $M_1=\frac{L}{c}+sup\{\parallel u_n-S{u}_n\parallel :n\ge 1\}$. Thus, from conditions (i)-(iv), it follows that

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

Hence, by Lemma 2.7, we have

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

Consequently,

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-x_n\parallel =\underset{n\to \mathrm{\infty }}{lim}(1-{\lambda }_n)\parallel z_n-x_n\parallel =0,$$

and by (3.17) and (3.20),

$$\underset{n\to \mathrm{\infty }}{lim}\parallel u_{n+1}-u_n\parallel =0,\text{and}\underset{n\to \mathrm{\infty }}{lim}\parallel y_{n+1}-y_n\parallel =0.$$

Step 3. We show that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-y_n\parallel =0$. Indeed, since

$$x_{n+1}={\alpha }_n\gamma V{x}_n+{\lambda }_n{x}_n+((1-{\lambda }_n)I-{\alpha }_n\mu F)y_n,$$

we have

$$\begin{array}{rcl}\parallel x_n-y_n\parallel & \le & \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-y_n\parallel \\ \le & \parallel x_n-x_{n+1}\parallel +{\alpha }_n\parallel \gamma V{x}_n-\mu F{y}_n\parallel +{\lambda }_n\parallel x_n-y_n\parallel ,\end{array}$$

that is,

$$\parallel x_n-y_n\parallel \le \frac{1}{1-{\lambda }_n}\parallel x_n-x_{n+1}\parallel +\frac{{\alpha }_n}{1-{\lambda }_n}(\gamma \parallel V{x}_n\parallel +\mu \parallel F{y}_n\parallel ).$$

So, from the conditions ${\alpha }_n\to 0$ and (iv) and Step 2, it follows that

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

Step 4. We show that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-u_n\parallel =0$ and ${lim}_{n\to \mathrm{\infty }}\parallel u_n-y_n\parallel =0$. Indeed, since $T_{r_n}$ is firmly nonexpansive, for $p\in F(S)\cap MEP(\mathrm{\Theta },\phi )$, we have

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

and hence

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

(3.21)

Then, by using the convexity of ${\parallel \cdot \parallel }^2$, we have from (3.14) and (3.21),

(3.22)

where

$$M_n={\lambda }_n\parallel x_n-y_n\parallel [2{\alpha }_n(\gamma \parallel V{x}_n\parallel +\mu \parallel F{y}_n\parallel +(2-{\alpha }_n)\parallel y_n-p\parallel )+{\lambda }_n\parallel x_n-y_n\parallel ].$$

By Step 3, we know that ${lim}_{n\to \mathrm{\infty }}{M}_n=0$. Then from (3.22), we have

$$\begin{array}{rcl}(1-{\alpha }_n){\parallel x_n-u_n\parallel }^2& \le & {\alpha }_n{\parallel \gamma V{x}_n+(I-\mu F)y_n-p\parallel }^2\\ +{\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+M_n\\ \le & {\alpha }_n{(\gamma \parallel V{x}_n\parallel +\mu \parallel F{y}_n\parallel +\parallel y_n-p\parallel )}^2\\ +\parallel x_n-x_{n+1}\parallel (\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )+M_n.\end{array}$$

Since ${\alpha }_n\to 0$ and $\parallel x_n-x_{n+1}\parallel \to 0$, we obtain

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

(3.23)

From Step 3, we also have

$$\parallel u_n-y_n\parallel \le \parallel u_n-x_n\parallel +\parallel x_n-y_n\parallel \to 0,\text{as }n\to \mathrm{\infty }.$$

(3.24)

Step 5. We show that ${lim}_{n\to \mathrm{\infty }}\parallel T{u}_n-u_n\parallel =0$, where $T:C\to H$ is defined by $Tx=\lambda x+(1-\lambda )Sx$. We know that T is nonexpansive with $F(T)=F(S)$ by Lemma 2.5. Notice that

$$\begin{array}{rcl}\parallel T{u}_n-u_n\parallel & \le & \parallel T{u}_n-y_n\parallel +\parallel y_n-u_n\parallel \\ \le & |\lambda -{\beta }_n|\parallel u_n-S{u}_n\parallel +\parallel y_n-u_n\parallel .\end{array}$$

By (3.24) and ${\beta }_n\to \lambda$, we obtain

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

Step 6. We show that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(\mu F-\gamma V)q,q-x_n〉\le 0,$$

where $q=P_{F(S)\cap MEP(\mathrm{\Theta },\phi )}(I-\mu F+\gamma V)q$ is a unique solution of the variational inequality (3.1). To show this inequality, we choose a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ such that

$$\underset{i\to \mathrm{\infty }}{lim}〈(\mu F-\gamma V)q,q-x_{n_i}〉=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(\mu F-\gamma V)q,q-x_n〉.$$

Since $\{u_{n_i}\}$ is bounded, there exists a subsequence $\{u_{n_{i_j}}\}$ which converges weakly to w. Without loss of generality, we can assume that $u_{n_i}\rightharpoonup w$. From Step 4 and Step 5, we obtain $x_{n_i}\rightharpoonup w$ and $T{u}_{n_i}\rightharpoonup w$. By the same argument as in the proof of Theorem 3.1, we have $w\in F(S)\cap MEP(\mathrm{\Theta },\phi )$. Since $q=P_{F(S)\cap MEP(\mathrm{\Theta },\phi )}(I-\mu F+\gamma V)q$, it follows that

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(\mu F-\gamma V)q,q-x_n〉& =& \underset{i\to \mathrm{\infty }}{lim}〈(\mu F-\gamma V)q,q-x_{n_i}〉\\ =& 〈(\mu F-\gamma V)q,q-w〉\le 0.\end{array}$$

Step 7. We show that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q\parallel =0$, where $q=P_{F(S)\cap MEP(\mathrm{\Theta },\phi )}(I-\mu F+\gamma V)q$ is a unique solution of the variational inequality (3.1). From (3.14), we know that

$$x_{n+1}-q={\alpha }_n(\gamma V{x}_n-\mu Fq)+{\lambda }_n(y_n-q)+((1-{\lambda }_n)I-{\alpha }_n\mu F)y_n-((1-{\lambda }_n)I-{\alpha }_n\mu F)q.$$

Applying Lemma 2.8 and Lemma 2.10, we have

$$\begin{array}{rcl}{\parallel x_{n+1}-q\parallel }^2& \le & {\parallel {\lambda }_n(x_n-q)+((1-{\lambda }_n)I-{\alpha }_n\mu F)y_n-((1-{\lambda }_n)I-{\alpha }_n\mu F)q\parallel }^2\\ +2{\alpha }_n〈\gamma V{x}_n-\mu Fq,x_{n+1}-q〉\\ \le & {((1-{\lambda }_n-{\alpha }_n\tau )\parallel y_n-q\parallel +{\lambda }_n\parallel x_n-q\parallel )}^2\\ +2{\alpha }_n\gamma 〈V{x}_n-Vq,x_{n+1}-q〉+2{\alpha }_n〈\gamma Vq-\mu Fq,x_{n+1}-q〉\\ \le & {(1-\tau {\alpha }_n)}^2{\parallel x_n-q\parallel }^2+2{\alpha }_n\gamma l\parallel x_n-q\parallel \parallel x_{n+1}-q\parallel \\ +2{\alpha }_n〈\gamma Vq-\mu Fq,x_{n+1}-q〉\\ \le & {(1-\tau {\alpha }_n)}^2{\parallel x_n-q\parallel }^2+{\alpha }_n\gamma l({\parallel x_n-q\parallel }^2+{\parallel x_{n+1}-q\parallel }^2)\\ +2{\alpha }_n〈\gamma Vq-\mu Fq,x_{n+1}-q〉.\end{array}$$

This implies that

$$\begin{array}{rcl}{\parallel x_{n+1}-q\parallel }^2& \le & \frac{1-2\tau {\alpha }_n+{\tau }^2{\alpha }_n^2+{\alpha }_n\gamma l}{1-{\alpha }_n\gamma l}{\parallel x_n-q\parallel }^2\\ +\frac{2{\alpha }_n}{1-{\alpha }_n\gamma l}〈\gamma Vq-\mu Fq,x_{n+1}-q〉\\ =& (1-\frac{2(\tau -\gamma l){\alpha }_n}{1-{\alpha }_n\gamma l}){\parallel x_n-q\parallel }^2+\frac{{\tau }^2{\alpha }_n^2}{1-{\alpha }_n\gamma \alpha }{\parallel x_n-q\parallel }^2\\ +\frac{2{\alpha }_n}{1-{\alpha }_n\gamma l}〈\gamma Vq-\mu Fq,x_{n+1}-q〉\\ \le & (1-\frac{2(\tau -\gamma l)}{1-{\alpha }_n\gamma l}{\alpha }_n){\parallel x_n-q\parallel }^2\\ +\frac{2(\tau -\gamma l){\alpha }_n}{1-{\alpha }_n\gamma l}(\frac{{\tau }^2{\alpha }_n}{2(\tau -\gamma l)}{M}_2+\frac{1}{\tau -\gamma l}〈\mu Fq-\gamma Vq,q-x_{n+1}〉)\\ =& (1-{\xi }_n){\parallel x_n-q\parallel }^2+{\xi }_n{\delta }_n,\end{array}$$

where $M_2=sup\{{\parallel x_n-q\parallel }^2:n\ge 1\}$, ${\xi }_n=\frac{2(\tau -\gamma l)}{1-{\alpha }_n\gamma l}{\alpha }_n$ and

$${\delta }_n=\frac{{\tau }^2{\alpha }_n}{2(\tau -\gamma l)}{M}_2+\frac{1}{\tau -\gamma l}〈\mu Fq-\gamma Vq,q-x_{n+1}〉.$$

From the condition (i) and Step 6, it is easy to see that ${\xi }_n\to 0$, ${\sum }_{n=0}^{\mathrm{\infty }}{\xi }_n=\mathrm{\infty }$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\delta }_n\le 0$. Hence, by Lemma 2.6, we conclude $x_n\to q$ as $n\to \mathrm{\infty }$. This completes the proof. □

Remark 3.1

• (1) Theorem 3.1 and Theorem 3.2 extend and develop Theorem 3.1 and Theorem 3.2 of Liu [12], respectively, in the following ways:

• (a) The EP (1.2) in Theorem 3.1 and Theorem 3.2 of [12] is extended to the case of the MEP (1.1).

• (b) The strongly positive bounded linear operator A in Theorem 3.1 and Theorem 3.2 of [12] is extended to the case of the ρ-Lipschitzian and η-strongly monotone operator F. In fact, from the definitions, a strongly positive bounded linear operator A with a constant $\overline{\gamma }>0$ is a $\parallel A\parallel$-Lipschitzian and $\overline{\gamma }$-strongly monotone operator.

• (c) The contractive mapping $f:H\to H$ with contractive coefficient $\alpha \in (0,1)$ in Theorem 3.1 and Theorem 3.2 of [12] is extended to the case of a Lipschitzian mapping $V:H\to H$ with a constant $l\in [0,\mathrm{\infty })$.

• (d) The condition ${\sum }_{n=1}^{\mathrm{\infty }}|{\alpha }_{n+1}-{\alpha }_n|<\mathrm{\infty }$ in Theorem 3.2 of [12] is removed.

• (e) The conditions ${\sum }_{n=1}^{\mathrm{\infty }}|{\beta }_{n+1}-{\beta }_n|<\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}|r_{n+1}-r_n|<\mathrm{\infty }$ in Theorem 3.2 of [12] are also relaxed by the conditions ${lim}_{n\to \mathrm{\infty }}|{\beta }_{n+1}-{\beta }_n|=0$ and ${lim}_{n\to \mathrm{\infty }}|r_{n+1}-r_n|=0$, respectively.

• (2) Even if $C=H$, S is nonexpansive, $\{{\beta }_n\}=\{0\}$, $r_n=1$, $u_n=x_n$, $\mathrm{\Theta }(x,y)=0$ and $\phi (x)=0$, $\mathrm{\forall }x,y\in C$, Theorem 3.1 and Theorem 3.2 improve Theorem 3.1 and Theorem 3.2 of Tian [18] and Theorem 3.1 and Theorem 3.2 of Ceng et al. [19] from the class of nonexpansive mappings to the class of k-strictly pseudocontractive mappings. In particular, Theorem 3.2 develops Theorem 3.2 of Tian [18] and Theorem 3.2 of Ceng et al. [19] by removing the condition either ${\sum }_{n=1}^{\mathrm{\infty }}|{\alpha }_{n+1}-{\alpha }_n|<\mathrm{\infty }$ or ${lim}_{n\to \mathrm{\infty }}{\alpha }_n/{\alpha }_{n+1}=1$.

• (3) Theorem 3.1 also contains Theorem 3.1 of Plubtieng and Pungaeng [13] as a special case with the nonexpansive mapping S, and $\{{\beta }_n\}=\{0\}$, and $\phi \equiv 0$.

• (4) Theorem 3.2 also includes and improves Theorem 3.3 of Plubtieng and Punpaeng [13], Theorem 3.2 of Takahashi and Takahashi [14] as well as Theorem 3.2 of Tian [17], Theorem 2.1 of Jung [8], Theorem 3.1 of Jung [20] Theorem 2.1 of Cho et al. [7] and Theorem 3.4 of Marino and Xu [16] as some special cases.