On variational inequality, fixed point and generalized mixed equilibrium problems

Abstract

In this article, variational inequality, fixed point, and generalized mixed equilibrium problems are investigated based on an extragradient iterative algorithm. Weak convergence of the extragradient iterative algorithm is obtained in Hilbert spaces.

1 Introduction

In this paper, we always assume that H is a real Hilbert space with the inner product $〈\cdot ,\cdot 〉$ and the norm $\parallel \cdot \parallel$, and C is a nonempty, closed, and convex subset of H. ℝ is denoted by the set of real numbers. Let F be a bifunction of $C\times C$ into ℝ. Consider the problem: find a p such that

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

(1.1)

In this paper, the solution set of the problem is denoted by $EP(F)$, i.e.,

$$EP(F)=\{p\in C:F(p,y)\ge 0,\mathrm{\forall }y\in C\}.$$

The above problem is first introduced by Ky Fan [1]. In the sense of Blum and Oettli [2], the Ky Fan problem is also called an equilibrium problem.

Recently, the ‘so-called’ generalized mixed equilibrium problem has been investigated by many authors: The generalized mixed equilibrium problem is to find $p\in C$ such that

$$F(p,y)+〈Ap,y-p〉+\phi (y)-\phi (p)\ge 0,\mathrm{\forall }y\in C,$$

(1.2)

where $\phi :C\to \mathbb{R}$ is a real valued function and $A:C\to H$ is map**. We use $GMEP(F,A,\phi )$ to denote the solution set of the equilibrium problem. That is,

$$GMEP(F,A,\phi ):=\{p\in C:F(p,y)+〈Ap,y-p〉+\phi (y)-\phi (z)\ge 0,\mathrm{\forall }y\in C\}.$$

Next, we give some special cases.

If $A=0$, then the problem (1.2) is equivalent to find $p\in C$ such that

$$F(p,y)+\phi (y)-\phi (z)\ge 0,\mathrm{\forall }y\in C,$$

(1.3)

which is called the mixed equilibrium problem.

If $F=0$, then the problem (1.2) is equivalent to find $p\in C$ such that

$$〈Ap,y-p〉+\phi (y)-\phi (z)\ge 0,\mathrm{\forall }y\in C,$$

(1.4)

which is called the mixed variational inequality of Browder type.

If $\phi =0$, then the problem (1.2) is equivalent to find $p\in C$ such that

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

(1.5)

which is called the generalized equilibrium problem.

If $A=0$ and $\phi =0$, then the problem (1.2) is equivalent to (1.1).

For solving the above equilibrium problems, let us assume that the bifunction $F:C\times C\to \mathbb{R}$ satisfies the following conditions:

(A1) $F(x,x)=0$, $\mathrm{\forall }x\in C$;

(A2) F is monotone, i.e., $F(x,y)+F(y,x)\le 0$, $\mathrm{\forall }x,y\in C$;

(A3)

$$\underset{t\downarrow 0}{lim\hspace{0.17em}sup}F(tz+(1-t)x,y)\le F(x,y),\mathrm{\forall }x,y,z\in C;$$

(A4) for each $x\in C$, $y\mapsto F(x,y)$ is convex and weakly lower semicontinuous.

Equilibrium problems have intensively been studied. It has been shown that equilibrium problems cover fixed point problems, variational inequality problems, inclusion problems, saddle problems, complementarity problem, minimization problem, and Nash equilibrium problem; see [1–20] and the references therein.

Let $S:C\to C$ be a map**. In this paper, we use $F(S)$ to stand for the set of fixed points. Recall that the map** S is said to be nonexpansive if

$$\parallel Sx-Sy\parallel \le \parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

S is said to be κ-strictly pseudocontractive if there exists a constant $\kappa \in [0,1)$ such that

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

It is clear that the class of κ-strictly pseudocontractive includes the class of nonexpansive map**s as a special case. The class of κ-strictly pseudocontractive map**s was introduced by Browder and Petryshyn [21]; for existence and approximation of fixed points of the class of map**s, see [22–29] and the references therein.

Let $A:C\to H$ be a map**. Recall that A is said to be monotone if

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

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

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

It is clear that the κ-inverse being strongly monotone is monotone and Lipschitz continuous.

A set-valued map** $T:H\to 2^H$ is said to be monotone if, for all $x,y\in H$, $f\in Tx$ and $g\in Ty$ imply $〈x-y,f-g〉>0$. A monotone map** $T:H\to 2^H$ is maximal if the graph $G(T)$ of T is not properly contained in the graph of any other monotone map**. It is well known that a monotone map** T is maximal if and only if, for any $(x,f)\in H\times H$, $〈x-y,f-g〉\ge 0$ for all $(y,g)\in G(T)$ implies $f\in Tx$. The class of monotone operators is one of the most important classes of operators. Within the past several decades, many authors have been devoting their efforts to the studies of the existence and convergence of zero points for maximal monotone operators.

Let $F(x,y)=〈Ax,y-x〉$, $\mathrm{\forall }x,y\in C$. We see that the problem (1.1) is reduced to the following classical variational inequality. Find $x\in C$ such that

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

(1.6)

It is well known that $x\in C$ is a solution to (1.6) if and only if x is a fixed point of the map** $P_C(I-\rho A)$, where $\rho >0$ is a constant, and I is the identity map**. If C is bounded, closed, and convex, then the solution set of the variational inequality (1.6) is nonempty.

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

Lemma 1.1 [21]

Let $S:C\to C$ be a κ-strictly pseudocontractive map**. Define $S_t:C\to C$ by $S_{t}x=tx+(1-t)Sx$ for each $x\in C$. Then, as $t\in [\kappa ,1)$, $S_t$ is nonexpansive such that $F(S_t)=F(S)$.

Lemma 1.2 [2]

Let C be a nonempty, closed, and convex subset of H, and $F:C\times C\to \mathbb{R}$ a bifunction satisfying (A1)-(A4). Then, for any $r>0$ and $x\in H$, 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, define

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

for all $r>0$ and $x\in H$. Then the following hold:

• (a) $T_r$ is single-valued;

• (b) $T_r$ is firmly nonexpansive, i.e., for any $x,y\in H$,

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

• (c) $F(T_r)=EP(F)$;

• (d) $EP(F)$ is closed and convex.

Lemma 1.3 [30]

Let A be a monotone map** of C into H and $N_{C}v$ the normal cone to C at $v\in C$, i.e.,

$$N_{C}v=\{w\in H:〈v-u,w〉\ge 0,\mathrm{\forall }u\in C\}$$

and define a map** T on C by

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

Then T is maximal monotone and $0\in Tv$ if and only if $〈Av,u-v〉\ge 0$ for all $u\in C$.

Lemma 1.4 [31]

Let ${\{a_n\}}_{n=1}^{\mathrm{\infty }}$ be real numbers in $[0,1]$ such that ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n=1$. Then we have the following:

$${\parallel \sum _{i=1}^{\mathrm{\infty }}{a}_i{x}_i\parallel }^2\le \sum _{i=1}^{\mathrm{\infty }}{a}_i{\parallel x_i\parallel }^2$$

for any given bounded sequence ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ in H.

Lemma 1.5 [32]

Let $0<p\le t_n\le q<1$ for all $n\ge 1$. Suppose that $\{x_n\}$ and $\{y_n\}$ are sequences in H such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_n\parallel \le d,\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel y_n\parallel \le d$$

and

$$\underset{n\to \mathrm{\infty }}{lim}\parallel t_n{x}_n+(1-t_n)y_n\parallel =d$$

hold for some $r\ge 0$. Then ${lim}_{n\to \mathrm{\infty }}\parallel x_n-y_n\parallel =0$.

Lemma 1.6 [21]

Let C be a nonempty, closed, and convex subset of H, and $S:C\to C$ a strictly pseudocontractive map**. If $\{x_n\}$ is a sequence in C such that $x_n\rightharpoonup x$ and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-S{x}_n\parallel =0$, then $x=Sx$.

Lemma 1.7 [33]

Let $\{a_n\}$, $\{b_n\}$, and $\{c_n\}$ be three nonnegative sequences satisfying the following condition:

$$a_{n+1}\le (1+b_n)a_n+c_n,\mathrm{\forall }n\ge n_0,$$

where $n_0$ is some nonnegative integer, ${\sum }_{n=1}^{\mathrm{\infty }}{b}_n<\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}{c}_n<\mathrm{\infty }$. Then the limit ${lim}_{n\to \mathrm{\infty }}{a}_n$ exists.

2 Main results

Theorem 2.1 Let C be a nonempty, closed, and convex subset of H, $S:C\to C$ a κ-strictly pseudocontractive map** with a nonempty fixed point set, and $A:C\to H$ an L-Lipschitz continuous and monotone map**. Let $F_m$ be a bifunction from $C\times C$ to ℝ which satisfies (A1)-(A4), $B_m:C\to H$ a continuous and monotone map**, ${\phi }_m:C\to \mathbb{R}$ a lower semicontinuous and convex function for each $m\ge 1$. Assume that $\mathcal{F}:={\bigcap }_{m=1}^{\mathrm{\infty }}GMEP(F_m,B_m,{\phi }_m)\cap VI(C,A)\cap F(S)$ is not empty. Let $\{{\alpha }_n\}$, $\{{\beta }_n\}$, and $\{{\delta }_{n,m}\}$ be real number sequences in $(0,1)$. Let $\{{\lambda }_n\}$, $\{r_{n,m}\}$ be positive real number sequences. Let $\{x_n\}$ be a sequence generated in the following manner:

$$\{\begin{array}{c}{x}_1\in H,\hfill \\ x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)({\beta }_{n}I+(1-{\beta }_n)S){Proj}_C({\sum }_{m=1}^{\mathrm{\infty }}{\delta }_{n,m}{z}_{n,m}-{\lambda }_{n}A{y}_n),n\ge 1,\hfill \\ y_n={Proj}_C({\sum }_{m=1}^{\mathrm{\infty }}{\delta }_{n,m}{z}_{n,m}-{\lambda }_{n}A{\sum }_{m=1}^{\mathrm{\infty }}{\delta }_{n,m}{z}_{n,m}),\hfill \end{array}$$

where $z_{n,m}$ is such that

$$F_m(z_{n,m},z)+〈B_m{z}_{n,m},z-z_{n,m}〉+{\phi }_m(z)-{\phi }_m(z_{n,m})+\frac{1}{r_{n,m}}〈z-z_{n,m},z_{n,m}-x_n〉\ge 0,\mathrm{\forall }z\in C.$$

Assume that $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\delta }_{n,m}\}$, $\{{\lambda }_n\}$, $\{r_{n,m}\}$ satisfy the following restrictions:

• (a) $0<a\le {\alpha }_n\le b<1$;

• (b) $\kappa \le {\beta }_n\le c<1$;

• (c) ${\sum }_{m=1}^{\mathrm{\infty }}{\delta }_{n,m}=1$, and $0<d\le {\delta }_{n,m}\le 1$;

• (d) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_{n,m}>0$ and $e\le {\lambda }_n\le f$, where $e,f\in (0,1/L)$.

Then the sequence $\{x_n\}$ weakly converges to some point $\overline{x}\in \mathcal{F}$.

Proof The proof is split into five steps.

Step 1. Show that the sequence $\{x_n\}$ is bounded.

Define $G_m(p,y)=F_m(p,y)+〈B_{m}p,y-p〉+{\phi }_m(y)-{\phi }_m(p)$, $\mathrm{\forall }p,y\in C$. Next, we prove that the bifunction $G_m$ satisfies the conditions (A1)-(A4). Therefore, generalized mixed equilibrium problem is equivalent to the following equilibrium problem: find $p\in C$ such that $G_m(p,y)\ge 0$, $\mathrm{\forall }y\in C$. It is clear that $G_m$ satisfies (A1). Next, we prove $G_m$ is monotone. Since $B_m$ is a continuous and monotone operator, we find from the definition of G that

$$\begin{array}{rcl}{G}_m(y,z)+G_m(z,y)& =& F_m(y,z)+〈B_{m}y,z-y〉+{\phi }_m(z)-{\phi }_m(y)+F_m(z,y)\\ +〈B_{m}z,y-z〉+{\phi }_m(y)-{\phi }_m(z)\\ =& F_m(z,y)+F_m(y,z)+〈B_{m}z,y-z〉+〈B_{m}y,z-y〉\\ \le & 〈B_{m}z-B_{m}y,y-z〉\\ \le & 0.\end{array}$$

Next, we show $G_m$ satisfies (A3), that is,

$$\underset{t\downarrow 0}{lim\hspace{0.17em}sup}{G}_m(tz+(1-t)x,y)\le G_m(x,y),\mathrm{\forall }x,y,z\in C.$$

Since $B_m$ is continuous and ${\phi }_m$ is lower semicontinuous, we have

$$\begin{array}{rcl}\underset{t\downarrow 0}{lim\hspace{0.17em}sup}{G}_m(tz+(1-t)x,y)& =& \underset{t\downarrow 0}{lim\hspace{0.17em}sup}{F}_m(tz+(1-t)x,y)\\ +\underset{t\downarrow 0}{lim\hspace{0.17em}sup}〈B_m(tz+(1-t)x),y-(tz+(1-t)x)〉\\ +\underset{t\downarrow 0}{lim\hspace{0.17em}sup}({\phi }_m(y)-{\phi }_m(tz+(1-t)x))\\ \le & F_m(x,y)+〈B_{m}x,y-x〉+{\phi }_m(y)-{\phi }_m(x)\\ =& G_m(x,y).\end{array}$$

Next, we show that, for each $x\in C$, $y\mapsto G_m(x,y)$ is a convex and lower semicontinuous. For each $x\in C$, for all $t\in (0,1)$ and for all $y,z\in C$, since $F_m$ satisfies (A4) and ${\phi }_m$ is convex, we have

$$\begin{array}{c}{G}_m(x,ty+(1-t)z)\hfill \\ =F_m(x,ty+(1-t)z)+〈B_{m}x,ty+(1-t)z-x〉+{\phi }_m(ty+(1-t)z)-{\phi }_m(x)\hfill \\ \le t(F_m(x,y)+〈B_{m}x,y-x〉+{\phi }_m(y)-{\phi }_m(x))\hfill \\ +(1-t)(F_m(x,z)+〈B_{m}x,z-x〉+{\phi }_m(z)-{\phi }_m(x))\hfill \\ =t{G}_m(x,y)+(1-t)G_m(x,z).\hfill \end{array}$$

Thus, $y\mapsto G_m(x,y)$ is convex. Similarly, we find that $y\mapsto G_m(x,y)$ is also lower semicontinuous. Put $u_n={Proj}_C({\sum }_{m=1}^N{\delta }_{n,m}{z}_{n,m}-{\lambda }_{n}A{y}_n)$ and $v_n={\sum }_{m=1}^N{\delta }_{n,m}{z}_{n,m}$. Letting $p\in \mathcal{F}$, we see that

$$\begin{array}{rcl}{\parallel u_n-p\parallel }^2& \le & {\parallel v_n-{\lambda }_{n}A{y}_n-p\parallel }^2-{\parallel v_n-{\lambda }_{n}A{y}_n-u_n\parallel }^2\\ =& {\parallel v_n-p\parallel }^2-{\parallel v_n-u_n\parallel }^2+2{\lambda }_n(〈A{y}_n-Ap,p-y_n〉+〈Ap,p-y_n〉\\ +〈A{y}_n,y_n-u_n〉)\\ \le & {\parallel v_n-p\parallel }^2-{\parallel v_n-y_n\parallel }^2-{\parallel y_n-u_n\parallel }^2+2〈v_n-{\lambda }_{n}A{y}_n-y_n,u_n-y_n〉.\end{array}$$

Notice that A is L-Lipschitz continuous and $y_n={Proj}_C(v_n-{\lambda }_{n}A{v}_n)$. It follows that

$$〈v_n-{\lambda }_{n}A{y}_n-y_n,u_n-y_n〉\le {\lambda }_{n}L\parallel v_n-y_n\parallel \parallel u_n-y_n\parallel .$$

It follows that

$${\parallel u_n-p\parallel }^2\le {\parallel v_n-p\parallel }^2+({\lambda }_n^2{L}^2-1){\parallel v_n-y_n\parallel }^2.$$

(2.1)

On the other hand, we have

$$\begin{array}{rcl}{\parallel v_n-p\parallel }^2& \le & {\parallel \sum _{m=1}^{\mathrm{\infty }}{\delta }_{n,m}{z}_{n,m}-p\parallel }^2\\ \le & \sum _{m=1}^{\mathrm{\infty }}{\delta }_{n,m}{\parallel T_{r_{n,m}}{x}_n-p\parallel }^2\\ \le & {\parallel x_n-p\parallel }^2,\end{array}$$

(2.2)

where $T_{r_{n,m}}=\{z\in C:G_m(z,y)+\frac{1}{r}〈y-z,z-x〉\ge 0,\mathrm{\forall }y\in C\}$. Substituting (2.2) into (2.1), we obtain

$${\parallel u_n-p\parallel }^2\le {\parallel x_n-p\parallel }^2+({\lambda }_n^2{L}^2-1){\parallel v_n-y_n\parallel }^2.$$

Putting $S_n={\beta }_{n}I+(1-{\beta }_n)S$, we find from Lemma 1.1 that $S_n$ is nonexpansive and $F(S_n)=F(S)$. It follows that

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& \le & {\alpha }_n{\parallel x_n-p\parallel }^2+(1-{\alpha }_n){\parallel S_n{u}_n-p\parallel }^2\\ \le & {\alpha }_n{\parallel x_n-p\parallel }^2+(1-{\alpha }_n){\parallel u_n-p\parallel }^2\\ \le & {\alpha }_n{\parallel x_n-p\parallel }^2+(1-{\alpha }_n)({\parallel x_n-p\parallel }^2+({\lambda }_n^2{L}^2-1){\parallel v_n-y_n\parallel }^2)\\ \le & {\parallel x_n-p\parallel }^2+(1-{\alpha }_n)({\lambda }_n^2{L}^2-1){\parallel v_n-y_n\parallel }^2\\ \le & {\parallel x_n-p\parallel }^2.\end{array}$$

(2.3)

It follows from Lemma 1.7 that the ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ exists. This shows that $\{x_n\}$ is bounded. Since $\{x_n\}$ is bounded, we may assume that a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ converges weakly to ξ.

Step 2. Show that $\xi \in VI(C,A)$

From (2.3), we find that ${\beta }_n(1-{\lambda }_n^2{L}^2){\parallel v_n-y_n\parallel }^2\le {\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2$. In view of the restrictions (b) and (d), we see that ${lim}_{n\to \mathrm{\infty }}\parallel v_n-y_n\parallel =0$. Since $\parallel y_n-u_n\parallel \le \lambda L\parallel v_n-y_n\parallel$, we have that ${lim}_{n\to \mathrm{\infty }}\parallel y_n-u_n\parallel =0$. It follows that

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

(2.4)

Notice that

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

This implies that ${\parallel z_{n,m}-p\parallel }^2\le {\parallel x_n-p\parallel }^2-{\parallel z_{n,m}-x_n\parallel }^2$. Since $v_n={\sum }_{m=1}^{\mathrm{\infty }}{\delta }_{n,m}{z}_{n,m}$, where ${\sum }_{m=1}^{\mathrm{\infty }}{\delta }_{n,m}=1$, we find that

$$\begin{array}{rcl}{\parallel v_n-p\parallel }^2& \le & \sum _{m=1}^{\mathrm{\infty }}{\delta }_{n,m}{\parallel z_{n,m}-p\parallel }^2\\ \le & {\parallel x_n-p\parallel }^2-\sum _{m=1}^{\mathrm{\infty }}{\delta }_{n,m}{\parallel z_{n,m}-x_n\parallel }^2.\end{array}$$

It follows that

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& \le & {\alpha }_n{\parallel x_n-p\parallel }^2+(1-{\alpha }_n){\parallel S_n{u}_n-p\parallel }^2\\ \le & {\alpha }_n{\parallel x_n-p\parallel }^2+(1-{\alpha }_n){\parallel u_n-p\parallel }^2\\ \le & {\alpha }_n{\parallel x_n-p\parallel }^2+(1-{\alpha }_n){\parallel v_n-p\parallel }^2\\ \le & {\parallel x_n-p\parallel }^2-(1-{\alpha }_n)\sum _{m=1}^{\mathrm{\infty }}{\delta }_{n,m}{\parallel z_{n,m}-x_n\parallel }^2.\end{array}$$

This implies that $(1-{\alpha }_n){\delta }_{n,m}{\parallel z_{n,m}-x_n\parallel }^2\le {\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2$. In view of the restrictions (a) and (c), we find that

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

(2.5)

Let T be the maximal monotone map** defined by

$$Tx=\{\begin{array}{cc}Ax+N_{C}x,\hfill & x\in C,\hfill \\ \mathrm{\varnothing },\hfill & x\notin C.\hfill \end{array}$$

For any given $(x,y)\in G(T)$, we have $y-Ax\in N_{C}x$. So, we have $〈x-m,y-Ax〉\ge 0$, for all $m\in C$. On the other hand, we have $u_n={Proj}_C(v_n-{\lambda }_{n}A{y}_n)$. We obtain

$$〈x-u_n,\frac{u_n-v_n}{{\lambda }_n}+A{y}_n〉\ge 0.$$

In view of the monotonicity of A, we see that

$$\begin{array}{rcl}〈x-u_{n_i},y〉& \ge & 〈x-u_{n_i},Ax〉\\ \ge & 〈x-u_{n_i},Ax〉-〈x-u_{n_i},\frac{u_{n_i}-v_{n_i}}{{\lambda }_{n_i}}+A{y}_{n_i}〉\\ =& 〈x-u_{n_i},Ax-A{u}_{n_i}〉+〈x-u_{n_i},A{u}_{n_i}-A{y}_{n_i}〉\\ -〈x-u_{n_i},\frac{u_{n_i}-v_{n_i}}{{\lambda }_{n_i}}〉\\ \ge & 〈x-u_{n_i},A{u}_{n_i}-A{y}_{n_i}〉-〈x-u_{n_i},\frac{u_{n_i}-v_{n_i}}{{\lambda }_{n_i}}〉\end{array}$$

in view of $\parallel v_n-x_n\parallel \le {\sum }_{m=1}^{\mathrm{\infty }}{\delta }_{n,m}\parallel z_{n,m}-x_n\parallel$. It follows from (2.5) that ${lim}_{n\to \mathrm{\infty }}\parallel v_n-x_n\parallel =0$. Notice that $\parallel u_n-x_n\parallel \le \parallel u_n-v_n\parallel +\parallel v_n-x_n\parallel$. It follows that

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

(2.6)

This in turn implies that $u_{n_i}\rightharpoonup \xi$. It follows that $〈x-\xi ,y〉\ge 0$. Notice that T is maximal monotone and hence $0\in T\xi$. This shows from Lemma 1.3 that $\xi \in VI(C,A)$.

Step 3. Show that $\xi \in GMEP(F_m,B_m,{\phi }_m)$.

It follows from (2.5) that $\{z_{n_i,m}\}$ converges weakly to ξ for each $m\ge 1$. Since $z_{n,m}=T_{r_{n,m}}{x}_n$, we have

$$G_m(z_{n,m},z)+\frac{1}{r_{n,m}}〈z-z_{n,m},z_{n,m}-x_n〉\ge 0,\mathrm{\forall }z\in C.$$

From the assumption (A2), we see that

$$〈z-z_{n_i,m},\frac{z_{n_i,m}-x_{n_i}}{r_{n_i,m}}〉\ge G_m(z,z_{n_i,m}),\mathrm{\forall }z\in C.$$

In view of the assumption (A4), we find from (2.5) that $G_m(z,\xi )\le 0$, $\mathrm{\forall }z\in C$. For $t_m$ with $0<t_m\le 1$ and $z\in C$, let $z_{t_m}=t_{m}z+(1-t_m)\xi$, for each $1\le m\le N$. Since $z\in C$ and $\xi \in C$, we have $z_{t_m}\in C$. It follows that $G_m(z_{t_m},\xi )\le 0$. Notice that

$$0=G_m(z_{t_m},z_{t_m})\le t_m{G}_m(z_{t_m},z)+(1-t_m)G_m(z_{t_m},\xi )\le t_m{G}_m(z_{t_m},z),$$

which yields $G_m(z_{t_m},z)\ge 0$, $\mathrm{\forall }z\in C$. Letting $t_m\downarrow 0$, one sees that $G_m(\xi ,z)\ge 0$, $\mathrm{\forall }z\in C$. This implies that $\xi \in GMEP(F_m,B_m,{\phi }_m)$ for each $m\ge 1$. This proves that $\xi \in {\bigcap }_{m=1}^{\mathrm{\infty }}GMEP(F_m,B_m,{\phi }_m)$.

Step 4. Show that $\xi \in F(S)$.

Since ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ exists, we put ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel =d>0$. It follows that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-p\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel {\alpha }_n(x_n-p)+(1-{\alpha }_n)(S_n{u}_n-p)\parallel =d.$$

Notice that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\parallel S_n{u}_n-p\parallel \le d$. From Lemma 1.5, we see that

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

(2.7)

Since

$$\begin{array}{rcl}\parallel S_n{x}_n-x_n\parallel & \le & \parallel S_n{x}_n-S_n{u}_n\parallel +\parallel S_n{u}_n-x_n\parallel \\ \le & \parallel x_n-u_n\parallel +\parallel S_n{u}_n-x_n\parallel ,\end{array}$$

we find from (2.6) and (2.7) that

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

(2.8)

In view of $\parallel S{x}_n-x_n\parallel \le \parallel S{x}_n-S_n{x}_n\parallel +\parallel S_n{x}_n-x_n\parallel$, we find from (2.8) that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-S{x}_n\parallel =0$. This implies from Lemma 1.6 that $\xi \in F(S)$. This completes the proof that $\xi \in \mathcal{F}$.

Step 5. Show that the whole sequence $\{x_n\}$ weakly converges to ξ.

Let $\{x_{n_j}\}$ be another subsequence of $\{x_n\}$ converging weakly to ${\xi }^{\prime }$, where ${\xi }^{\prime }\ne \xi$. In the same way, we can show that ${\xi }^{\prime }\in \mathcal{F}$. Since the space H enjoys Opial’s condition, we, therefore, obtain

$$\begin{array}{rcl}d& =& \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_{n_i}-\xi \parallel <\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_{n_i}-{\xi }^{\prime }\parallel \\ =& \underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_j-{\xi }^{\prime }\parallel <\underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_j-\xi \parallel =d.\end{array}$$

This is a contradiction. Hence $\xi ={\xi }^{\prime }$. This completes the proof. □

3 Applications

In this section, we consider solutions of the mixed equilibrium problem (1.3), which includes the Ky Fan inequality as a special case.

The so-called mixed equilibrium problem is to find $p\in C$ such that

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

The mixed equilibrium problem includes the Ky Fan inequality, fixed point problems, saddle problems, and complementary problems as special cases.

Theorem 3.1 Let C be a nonempty, closed, and convex subset of H, $S:C\to C$ a κ-strictly pseudocontractive map** with a nonempty fixed point set, and $A:C\to H$ a L-Lipschitz continuous and monotone map**. Let $F_m$ be a bifunction from $C\times C$ to ℝ which satisfies (A1)-(A4), and ${\phi }_m:C\to \mathbb{R}$ a lower semicontinuous and convex function for each $m\ge 1$. Assume that $\mathcal{F}:={\bigcap }_{m=1}^{\mathrm{\infty }}MEP(F_m,{\phi }_m)\cap VI(C,A)\cap F(S)$ is not empty. Let $\{{\alpha }_n\}$, $\{{\beta }_n\}$ and $\{{\delta }_{n,m}\}$ be real number sequences in $(0,1)$. Let $\{{\lambda }_n\}$, $\{r_{n,m}\}$ be positive real number sequences. Let $\{x_n\}$ be a sequence generated in the following manner:

$$\{\begin{array}{c}{x}_1\in H,\hfill \\ x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)({\beta }_{n}I+(1-{\beta }_n)S){Proj}_C({\sum }_{m=1}^{\mathrm{\infty }}{\delta }_{n,m}{z}_{n,m}-{\lambda }_{n}A{y}_n),n\ge 1,\hfill \\ y_n={Proj}_C({\sum }_{m=1}^{\mathrm{\infty }}{\delta }_{n,m}{z}_{n,m}-{\lambda }_{n}A{\sum }_{m=1}^{\mathrm{\infty }}{\delta }_{n,m}{z}_{n,m}),\hfill \end{array}$$

where $z_{n,m}$ is such that

$$F_m(z_{n,m},z)+{\phi }_m(z)-{\phi }_m(z_{n,m})+\frac{1}{r_{n,m}}〈z-z_{n,m},z_{n,m}-x_n〉\ge 0,\mathrm{\forall }z\in C.$$

Assume that $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\delta }_{n,m}\}$, $\{{\lambda }_n\}$, $\{r_{n,m}\}$ satisfy the following restrictions:

• (a) $0<a\le {\alpha }_n\le b<1$;

• (b) $\kappa \le {\beta }_n\le c<1$;

• (c) ${\sum }_{m=1}^{\mathrm{\infty }}{\delta }_{n,m}=1$, and $0<d\le {\delta }_{n,m}\le 1$;

• (d) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_{n,m}>0$ and $e\le {\lambda }_n\le f$, where $e,f\in (0,1/L)$.

Then the sequence $\{x_n\}$ weakly converges to some point $\overline{x}\in \mathcal{F}$.

Proof If $B_m=0$, we draw the desired conclusion immediately from Theorem 2.1. □

Further, if S is nonexpansive, we find from Theorem 3.1 the following result.

Corollary 3.2 Let C be a nonempty, closed, and convex subset of H, $S:C\to C$ a nonexpansive map** with a nonempty fixed point set, and $A:C\to H$ an L-Lipschitz continuous and monotone map**. Let $F_m$ be a bifunction from $C\times C$ to ℝ which satisfies (A1)-(A4), and ${\phi }_m:C\to \mathbb{R}$ a lower semicontinuous and convex function for each $m\ge 1$. Assume that $\mathcal{F}:={\bigcap }_{m=1}^{\mathrm{\infty }}MEP(F_m,{\phi }_m)\cap VI(C,A)\cap F(S)$ is not empty. Let $\{{\alpha }_n\}$, $\{{\beta }_n\}$, and $\{{\delta }_{n,m}\}$ be real number sequences in $(0,1)$. Let $\{{\lambda }_n\}$, $\{r_{n,m}\}$ be positive real number sequences. Let $\{x_n\}$ be a sequence generated in the following manner:

$$\{\begin{array}{c}{x}_1\in H,\hfill \\ x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)S{Proj}_C({\sum }_{m=1}^{\mathrm{\infty }}{\delta }_{n,m}{z}_{n,m}-{\lambda }_{n}A{y}_n),n\ge 1,\hfill \\ y_n={Proj}_C({\sum }_{m=1}^{\mathrm{\infty }}{\delta }_{n,m}{z}_{n,m}-{\lambda }_{n}A{\sum }_{m=1}^{\mathrm{\infty }}{\delta }_{n,m}{z}_{n,m}),\hfill \end{array}$$

where $z_{n,m}$ is such that

$$F_m(z_{n,m},z)+{\phi }_m(z)-{\phi }_m(z_{n,m})+\frac{1}{r_{n,m}}〈z-z_{n,m},z_{n,m}-x_n〉\ge 0,\mathrm{\forall }z\in C.$$

Assume that $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\delta }_{n,m}\}$, $\{{\lambda }_n\}$, $\{r_{n,m}\}$ satisfy the following restrictions:

• (a) $0<a\le {\alpha }_n\le b<1$;

• (b) ${\sum }_{m=1}^{\mathrm{\infty }}{\delta }_{n,m}=1$, and $0<d\le {\delta }_{n,m}\le 1$;

• (c) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_{n,m}>0$ and $e\le {\lambda }_n\le f$, where $e,f\in (0,1/L)$.

Then the sequence $\{x_n\}$ weakly converges to some point $\overline{x}\in \mathcal{F}$.

If $A=0$, we find from Theorem 2.1 the following result.

Theorem 3.3 Let C be a nonempty, closed, and convex subset of H, $S:C\to C$ a κ-strictly pseudocontractive map** with a nonempty fixed point set. Let $F_m$ be a bifunction from $C\times C$ to ℝ which satisfies (A1)-(A4), $B_m:C\to H$ a continuous and monotone map**, ${\phi }_m:C\to \mathbb{R}$ a lower semicontinuous and convex function for each $m\ge 1$. Assume that $\mathcal{F}:={\bigcap }_{m=1}^{\mathrm{\infty }}GMEP(F_m,B_m,{\phi }_m)\cap F(S)$ is not empty. Let $\{{\alpha }_n\}$, $\{{\beta }_n\}$, and $\{{\delta }_{n,m}\}$ be real number sequences in $(0,1)$. Let $\{r_{n,m}\}$ be a positive real number sequence. Let $\{x_n\}$ be a sequence generated in the following manner:

$$x_1\in H,x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)({\beta }_{n}I+(1-{\beta }_n)S)\sum _{m=1}^{\mathrm{\infty }}{\delta }_{n,m}{z}_{n,m},n\ge 1,$$

where $z_{n,m}$ is such that

$$F_m(z_{n,m},z)+〈B_m{z}_{n,m},z-z_{n,m}〉+{\phi }_m(z)-{\phi }_m(z_{n,m})+\frac{1}{r_{n,m}}〈z-z_{n,m},z_{n,m}-x_n〉\ge 0,\mathrm{\forall }z\in C.$$

Assume that $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\delta }_{n,m}\}$, and $\{r_{n,m}\}$ satisfy the following restrictions:

• (a) $0<a\le {\alpha }_n\le b<1$;

• (b) $\kappa \le {\beta }_n\le c<1$;

• (c) ${\sum }_{m=1}^{\mathrm{\infty }}{\delta }_{n,m}=1$, and $0<d\le {\delta }_{n,m}\le 1$;

• (d) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_{n,m}>0$.

Then the sequence $\{x_n\}$ weakly converges to some point $\overline{x}\in \mathcal{F}$.