A strong convergence theorem of common elements in Hilbert spaces

Abstract

The purpose of this article is to investigate the convergence of an iterative process for equilibrium problems, fixed point problems and variational inequalities. Strong convergence of the purposed iterative process is obtained in the framework of Hilbert spaces.

MSC:47H05, 47H09, 47J25.

1 Introduction

Nonlinear analysis plays an important role in optimization problems, economics and transportation. The theory of variational inequalities has emerged as a rapidly growing area of research because of its applications; see [1–17] fore more details and the references therein. To study variational inequalities based on iterative methods has been attracting many authors’ attention. For the iterative methods, the most popular method is the Mann iterative method which was introduced by Mann in 1953; see [18] and the references therein. The Mann iterative process has been proved to be weak convergence for nonexpansive mappings in infinite dimension spaces; see [19] and the reference therein. Recently, many authors studied the modification of Mann iterative methods. The most popular one is to use projections. We call the method a hybrid projection method; see [20] and the reference therein. In this paper, we study equilibrium problems, fixed point problems and variational inequalities based on the hybrid projection method. Strong convergence theorems for common solutions of the problems are established in infinite dimension Hilbert spaces.

2 Preliminaries

Throughout this paper, we always assume that H is a real Hilbert space with the inner product $〈\cdot ,\cdot 〉$ and the norm $\parallel \cdot \parallel$. Let C be a nonempty closed convex subset of H. Let $S:C\to C$ be a mapping. In this paper, we use $F(S)$ to denote the fixed point set of S.

Recall that the mapping 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 quasi-nonexpansive if $F(S)\ne \mathrm{\varnothing }$ and

$$\parallel Sx-y\parallel \le \parallel x-y\parallel ,\mathrm{\forall }x\in C,y\in F(S).$$

Let $A:C\to H$ be a mapping. 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 strongly monotone if there exists a constant $\alpha >0$ such that

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

For such a case, A is also said to be α-strongly monotone. A is said to be inverse-strongly monotone if there exists a constant $\alpha >0$ such that

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

For such a case, A is also said to be α-inverse-strongly monotone. A is said to be Lipschitz if there exits a constant $L>0$ such that

$$\parallel Ax-Ay\parallel \le L{\parallel x-y\parallel }^2,\mathrm{\forall }x,y\in C.$$

For such a case, A is also said to be L-Lipschitz. A set-valued mapping $T:H\to 2^H$ 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 mapping $T:H\to 2^H$ is maximal if the graph $G(T)$ of T is not properly contained in the graph of any other monotone mapping. It is known that a monotone mapping 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 $f\in Tx$.

Let F be a bifunction of $C\times C$ into ℝ, where ℝ denotes the set of real numbers and $A:C\to H$ is an inverse-strongly monotone mapping. In this paper, we consider the following generalized equilibrium problem:

$$\text{Find }x\in C\text{ such that }F(x,y)+〈Ax,y-x〉\ge 0,\mathrm{\forall }y\in C.$$

(2.1)

In this paper, the set of such an $x\in C$ is denoted by $EP(F,A)$, i.e.,

$$EP(F,A)=\{x\in C:F(x,y)+〈Ax,y-x〉\ge 0,\mathrm{\forall }y\in C\}.$$

To study the generalized equilibrium problems (2.1), we may assume that F satisfies the following conditions:

• (A1) $F(x,x)=0$ for all $x\in C$;

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

• (A3) for each $x,y,z\in C$,

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

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

Next, we give two special cases of the problem (2.1).

• (I) If $A\equiv 0$, then the generalized equilibrium problem (2.1) is reduced to the following equilibrium problem:

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

  (2.2)

• (II) If $F\equiv 0$, then the problem (2.1) is reduced to the following classical variational inequality:

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

  (2.3)

It is known that $x\in C$ is a solution to (2.3) if and only if x is a fixed point of the mapping $P_C(I-\lambda A)$, where $\lambda >0$ is a constant and I is the identity mapping.

Recently, many authors studied the problems (2.1), (2.2) and (2.3) based on hybrid projection methods; see, for example, [21–36] and the references therein. Motivated by these results, we investigated the common element problems of the generalized equilibrium problem (2.1) and quasi-nonexpansive mappings based on the shrinking projection algorithm. A strong convergence theorem of common elements is established in the framework of Hilbert spaces.

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

The following lemma can be found in [4] and [9].

Lemma 2.1 Let C be a nonempty closed convex subset of H and let $F:C\times C\to \mathbb{R}$ be 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 2.2 [37]

Let B be a monotone mapping 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 mapping M on C by

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

Then M is maximal monotone and $0\in Mv$ if and only if $〈Bv,u-v〉\ge 0$ for all $u\in C$.

3 Main results

Theorem 3.1 Let C be a nonempty, closed and convex subset of a real Hilbert space H. Let $F_m$ be a bifunction from $C\times C$ to ℝ which satisfies (A1)-(A4) and $A_m:C\to H$ be a ${\kappa }_m$-inverse-strongly monotone mapping for every $1\le m\le N$, where N denotes some positive integer. Let $S:C\to C$ be a continuous quasi-nonexpansive mapping which is assumed to be demiclosed at zero and let $B:C\to H$ be a β-inverse-strongly monotone mapping. Assume that $\mathcal{F}:={\bigcap }_{m=1}^{N}EP(F_m,A_m)\cap VI(C,B)\cap F(S)\ne \mathrm{\varnothing }$. Let $\{{\lambda }_n\}$ be a positive sequence in $[0,2\beta ]$ and $\{r_{n,m}\}$ be a positive sequence in $[0,2{\kappa }_m]$ for every $1\le m\le N$. Let $\{{\alpha }_n\},\{{\beta }_{n,1}\},\dots$ and $\{{\beta }_{n,N}\}$ be sequences in $[0,1]$. Let $\{x_n\}$ be a sequence generated in the following iterative process:

$$\{\begin{array}{c}{x}_1\in C,\hfill \\ C_1=C,\hfill \\ z_n={Proj}_C({\sum }_{m=1}^N{\beta }_{n,m}{u}_{n,m}-{\lambda }_{n}B{\sum }_{m=1}^N{\beta }_{n,m}{u}_{n,m}),\hfill \\ y_n={\alpha }_n{x}_n+(1-{\alpha }_n)S{z}_n,\hfill \\ C_{n+1}=\{v\in C_n:\parallel y_n-v\parallel \le \parallel x_n-v\parallel \},\hfill \\ x_{n+1}=P_{C_{n+1}}{x}_1,n\ge 1,\hfill \end{array}$$

where $\{u_{n,m}\}$ is such that

$$F_m(u_{n,m},u_m)+〈A_m{x}_n,u_m-u_{n,m}〉+\frac{1}{r_{n,m}}〈u_m-u_{n,m},u_{n,m}-x_n〉\ge 0,\mathrm{\forall }{u}_m\in C$$

for each $1\le m\le N$. Assume that the above sequence also satisfies the following restrictions:

• (a) ${\alpha }_n\le a<1$;

• (b) ${\sum }_{m=1}^N{\beta }_{n,m}=1$ and $0\le b\le {\beta }_{n,m}<1$ for each $1\le m\le N$;

• (c) $0<c\le {\lambda }_n\le d<2\beta$ and $0<e\le r_{n,m}\le f<2{\kappa }_m$ for each $1\le m\le N$,

where a, b, c, d, e and f are real numbers. Then the sequence $\{x_n\}$ strongly converges to ${Proj}_{\mathcal{F}}{x}_1$.

Proof

In view of Lemma 2.1, we see that

$$u_{n,m}=T_{r_{n,m}}(x_n-r_{n,m}{A}_m{x}_n),\mathrm{\forall }1\le m\le N.$$

Letting $p\in \mathcal{F}$, we obtain that

$$p=Sp={Proj}_C(I-{\lambda }_{n}B)p=T_{r_{n,m}}(p-r_{n,m}{A}_{m}p),\mathrm{\forall }m\in \{1,2,\dots ,N\}.$$

In view of the restriction (c), we obtain that

This shows that $I-r_{n,m}{A}_m$ is nonexpansive for every $m\in \{1,2,\dots ,N\}$. In view of the restriction (c), we also see that $I-{\lambda }_{n}B$ is nonexpansive.

Next, we show that $C_n$ is closed and convex. In view of the assumption in the main body of the theorem, we see that $C_1=C$ is closed and convex. Suppose that $C_i$ is closed and convex for some $i\ge 1$. We show that $C_{i+1}$ is closed and convex for the same i. Indeed, for any $v\in C_i$, we see that

$$\parallel y_i-v\parallel \le \parallel x_i-v\parallel$$

is equivalent to

$${\parallel y_i\parallel }^2-{\parallel x_i\parallel }^2-2〈v,y_i-x_i〉\ge 0.$$

Thus $C_{i+1}$ is closed and convex. This shows that $C_n$ is closed and convex.

Next, we show that $\mathcal{F}\subset C_n$ for each $n\ge 1$. From the assumption, we see that $\mathcal{F}\subset C=C_1$. Assume that $\mathcal{F}\subset C_i$ for some $i\ge 1$. For any $v\in \mathcal{F}\subset C_i$, we see that

$$\begin{array}{rl}\parallel y_i-v\parallel & =\parallel {\alpha }_i{x}_i+(1-{\alpha }_i)S{z}_i-v\parallel \\ \le {\alpha }_i\parallel x_i-v\parallel +(1-{\alpha }_i)\parallel z_i-v\parallel \\ \le {\alpha }_i\parallel x_i-v\parallel +(1-{\alpha }_i)\sum _{m=1}^N{\beta }_{i,m}\parallel u_{i,m}-v\parallel \\ \le {\alpha }_i\parallel x_i-v\parallel +(1-{\alpha }_i)\sum _{m=1}^N{\beta }_{i,m}\parallel T_{r_{i,m}}(I-r_{i,m}{A}_m)x_i-v\parallel \\ \le {\alpha }_i\parallel x_i-v\parallel +(1-{\alpha }_i)\sum _{m=1}^N{\beta }_{i,m}\parallel (I-r_{i,m}{A}_m)x_i-v\parallel \\ \le \parallel x_i-v\parallel .\end{array}$$

This shows that $v\in C_{i+1}$. This proves that $\mathcal{F}\subset C_n$. Notice that $x_n={Proj}_{C_n}{x}_1$. For each $v\in \mathcal{F}\subset C_n$, we have

$$\parallel x_1-x_n\parallel \le \parallel x_1-v\parallel .$$

In particular, we have

$$\parallel x_1-x_n\parallel \le \parallel x_1-{Proj}_{\mathcal{F}}{x}_1\parallel .$$

This implies that $\{x_n\}$ is bounded. Since $x_n={Proj}_{C_n}{x}_1$ and $x_{n+1}={Proj}_{C_{n+1}}{x}_1\in C_{n+1}\subset C_n$, we arrive at

$$0\le 〈x_1-x_n,x_n-x_{n+1}〉\le -{\parallel x_1-x_n\parallel }^2+\parallel x_1-x_n\parallel \parallel x_1-x_{n+1}\parallel .$$

It follows that

$$\parallel x_n-x_1\parallel \le \parallel x_{n+1}-x_1\parallel .$$

This implies that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-x_1\parallel$ exists. On the other hand, we have

It follows that

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

(3.1)

Notice that $x_{n+1}={Proj}_{C_{n+1}}{x}_1\in C_{n+1}$. It follows that

$$\parallel y_n-x_{n+1}\parallel \le \parallel x_n-x_{n+1}\parallel .$$

This in turn implies that

$$\parallel y_n-x_n\parallel \le \parallel y_n-x_{n+1}\parallel +\parallel x_n-x_{n+1}\parallel \le 2\parallel x_n-x_{n+1}\parallel .$$

In view of (3.1), we obtain that

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

(3.2)

On the other hand, we have

$$\parallel x_n-y_n\parallel =(1-{\alpha }_n)\parallel x_n-S{z}_n\parallel .$$

It follows from (3.2) that

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

(3.3)

For any $p\in \mathcal{F}$, we have from Lemma 2.1 that

$$\begin{array}{rl}{\parallel u_{n,m}-p\parallel }^2& ={\parallel T_{r_{n,m}}(I-r_{n,m}{A}_m)x_n-T_{r_{n,m}}(I-r_{n,m}{A}_m)p\parallel }^2\\ \le {\parallel (x_n-p)-r_{n,m}(A_m{x}_n-A_{m}p)\parallel }^2\\ ={\parallel x_n-p\parallel }^2-2r_{n,m}〈x_n-p,A_m{x}_n-A_{m}p〉+r_{n,m}^2{\parallel A_m{x}_n-A_{m}p\parallel }^2\\ \le {\parallel x_n-p\parallel }^2-r_{n,m}(2{\kappa }_m-r_{n,m}){\parallel A_m{x}_n-A_{m}p\parallel }^2,\\ \phantom{\le }\mathrm{\forall }m\in \{1,2,\dots ,N\}.\end{array}$$

(3.4)

On the other hand, we have

$$\begin{array}{rl}{\parallel y_n-p\parallel }^2& ={\parallel {\alpha }_n{x}_n+(1-{\alpha }_n)S{z}_n-p\parallel }^2\\ \le {\alpha }_n{\parallel x_n-p\parallel }^2+(1-{\alpha }_n){\parallel S{z}_n-p\parallel }^2\\ \le {\alpha }_n{\parallel x_n-p\parallel }^2+(1-{\alpha }_n){\parallel z_n-p\parallel }^2\\ \le {\alpha }_n{\parallel x_n-p\parallel }^2+(1-{\alpha }_n)\sum _{m=1}^N{\beta }_{n,m}{\parallel u_{n,m}-p\parallel }^2.\end{array}$$

(3.5)

Substituting (3.4) into (3.5), we arrive at

$${\parallel y_n-p\parallel }^2\le {\parallel x_n-p\parallel }^2-(1-{\alpha }_n)\sum _{m=1}^N{\beta }_{n,m}{r}_{n,m}(2{\kappa }_m-r_{n,m}){\parallel A_m{x}_n-A_{m}p\parallel }^2.$$

(3.6)

This in turn implies that

In view of the restrictions (a)-(c), we obtain from (3.2) that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel A_m{x}_n-A_{m}p\parallel =0,\mathrm{\forall }m\in \{1,2,\dots ,N\}.$$

(3.7)

On the other hand, we have from Lemma 2.1 that

$$\begin{array}{rcl}{\parallel u_{n,m}-p\parallel }^2& =& {\parallel T_{r_{n,m}}(I-r_{n,m}{A}_m)x_n-T_{r_{n,m}}(I-r_{n,m}{A}_m)p\parallel }^2\\ \le & 〈(I-r_{n,m}{A}_m)x_n-(I-r_{n,m}{A}_m)p,u_{n,m}-p〉\\ =& \frac{1}{2}({\parallel (I-r_{n,m}{A}_m)x_n-(I-r_{n,m}{A}_m)p\parallel }^2+{\parallel u_{n,m}-p\parallel }^2\\ -{\parallel (I-r_{n,m}{A}_m)x_n-(I-r_{n,m}{A}_m)p-(u_{n,m}-p)\parallel }^2)\\ \le & \frac{1}{2}({\parallel x_n-p\parallel }^2+{\parallel u_{n,m}-p\parallel }^2-{\parallel x_n-u_{n,m}-r_{n,m}(A_m{x}_n-A_{m}p)\parallel }^2)\\ =& \frac{1}{2}({\parallel x_n-p\parallel }^2+{\parallel u_{n,m}-p\parallel }^2-({\parallel x_n-u_{n,m}\parallel }^2\\ -2r_{n,m}〈x_n-u_{n,m},A_m{x}_n-A_{m}p〉+r_{n,m}^2{\parallel A_m{x}_n-A_{m}p\parallel }^2\left)\right).\end{array}$$

This implies that

$${\parallel u_{n,m}-p\parallel }^2\le {\parallel x_n-p\parallel }^2-{\parallel x_n-u_{n,m}\parallel }^2+2r_{n,m}\parallel x_n-u_{n,m}\parallel \parallel A_m{x}_n-A_{m}p\parallel .$$

(3.8)

Notice that

$$\begin{array}{rl}{\parallel y_n-p\parallel }^2& \le {\alpha }_n{\parallel x_n-p\parallel }^2+(1-{\alpha }_n){\parallel S{z}_n-p\parallel }^2\\ \le {\alpha }_n{\parallel x_n-p\parallel }^2+(1-{\alpha }_n){\parallel z_n-p\parallel }^2\\ \le {\alpha }_n{\parallel x_n-p\parallel }^2+(1-{\alpha }_n)\sum _{m=1}^N{\beta }_{n,m}{\parallel u_{n,m}-p\parallel }^2.\end{array}$$

(3.9)

Substituting (3.8) into (3.9), we see that

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& \le & {\parallel x_n-p\parallel }^2+(1-{\alpha }_n)\sum _{m=1}^N{\beta }_{n,m}2r_{n,m}\parallel x_n-u_{n,m}\parallel \parallel A_m{x}_n-A_{m}p\parallel \\ -(1-{\alpha }_n)\sum _{m=1}^N{\beta }_{n,m}{\parallel x_n-u_{n,m}\parallel }^2\\ \le & {\parallel x_n-p\parallel }^2+\sum _{m=1}^{N}2r_{n,m}\parallel x_n-u_{n,m}\parallel \parallel A_m{x}_n-A_{m}p\parallel \\ -(1-{\alpha }_n)\sum _{m=1}^N{\beta }_{n,m}{\parallel x_n-u_{n,m}\parallel }^2,\mathrm{\forall }1\le m\le N.\end{array}$$

(3.10)

It follows that

(3.11)

In view of the restrictions (a) and (b), we obtain from (3.2) and (3.7) that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-u_{n,m}\parallel =0,\mathrm{\forall }1\le m\le N.$$

(3.12)

Since $\{x_n\}$ is bounded, we may assume that there is a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ converging weakly to some point x. It follows from (3.12) that $u_{n_i,m}$ converges weakly to x for every $m\in \{1,2,\dots ,N\}$.

Next, we show that $x\in EP(F_m,A_m)$ for every $m\in \{1,2,\dots ,N\}$. Since $u_{n,m}=T_{r_{n,m}}(x_n-r_{n,m}{A}_m{x}_n)$ for any $u\in C$, we have

$$F_m(u_{n,m},u_m)+〈A_m{x}_n,u_m-u_{n,m}〉+\frac{1}{r_{n,m}}〈u_m-u_{n,m},u_{n,m}-x_n〉\ge 0.$$

From the condition (A2), we see that

$$〈A_m{x}_n,u_m-u_{n,m}〉+\frac{1}{r_{n,m}}〈u_m-u_{n,m},u_{n,m}-x_n〉\ge F_m(u_m,u_{n,m}).$$

(3.13)

Replacing n by $n_i$, we arrive at

$$〈A_m{x}_{n_i},u_m-u_{n_i,m}〉+〈u_m-u_{n_i,m},\frac{u_{n_i,m}-x_{n_i}}{r_{n_i,m}}〉\ge F_m(u_m,u_{n_i,m}).$$

(3.14)

For $t_m$ with $0<t_m\le 1$ and $u_m\in C$, let $u_{t_m}=t_m{u}_m+(1-t_m)x$. Since $u_m\in C$ and $x\in C$, we have $u_{t_m}\in C$ for every $1\le m\le N$. It follows from (3.14) that

(3.15)

From (3.12), we have $A_m{u}_{n_i,m}-A_m{x}_{n_i}\to 0$ as $i\to \mathrm{\infty }$ for every $1\le m\le N$. On the other hand, we obtain from the monotonicity of $A_m$ that $〈u_{t_m}-u_{n_i,m},A_m{u}_{t_m}-A_m{u}_{n_i,m}〉\ge 0$. It follows from (A4) that

$$〈u_{t_m}-x,A_m{u}_{t_m}〉\ge F_m(u_{t_m},x),\mathrm{\forall }1\le m\le N.$$

(3.16)

From (A1) and (A4), we obtain from (3.16) that

$$\begin{array}{rl}0& =F_m(u_{t_m},u_{t_m})\le t_m{F}_m(u_{t_m},u_m)+(1-t_m)F_m(u_{t_m},x)\\ \le t_m{F}_m(u_{t_m},u_m)+(1-t_m)〈u_{t_m}-x,A_m{u}_{t_m}〉\\ =t_m{F}_m(u_{t_m},u_m)+(1-t_m)t_m〈u_m-x,A_m{u}_{t_m}〉,\end{array}$$

which yields that

$$F_m(u_{t_m},u_m)+(1-t_m)〈u_m-x,A_m{u}_{t_m}〉\ge 0,\mathrm{\forall }1\le m\le N.$$

Letting $t_m\to 0$ in the above inequality for every $1\le m\le N$, we arrive at

$$F_m(x,u_m)+〈u_m-x,A_m\xi 〉\ge 0,\mathrm{\forall }1\le m\le N.$$

This shows that $x\in EP(F_m,A_m)$ for every $1\le m\le N$, that is, $x\in {\bigcap }_{m=1}^{N}EP(F_m,A_m)$. Putting $w_n={\sum }_{m=1}^N{\beta }_{n,m}{u}_{n,m}$, we see that

$$\parallel w_n-p\parallel \le \parallel x_n-p\parallel$$

and

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& =& {\parallel {\alpha }_n{x}_n+(1-{\alpha }_n)S{z}_n-p\parallel }^2\\ \le & {\alpha }_n{\parallel x_n-p\parallel }^2+(1-{\alpha }_n){\parallel S{z}_n-p\parallel }^2\\ \le & {\alpha }_n{\parallel x_n-p\parallel }^2+(1-{\alpha }_n){\parallel z_n-p\parallel }^2\\ \le & {\alpha }_n{\parallel x_n-p\parallel }^2+(1-{\alpha }_n){\parallel (I-{\lambda }_{n}B)w_n-p\parallel }^2\\ \le & {\parallel x_n-p\parallel }^2-(1-{\alpha }_n){\lambda }_n(2\beta -{\lambda }_n){\parallel B{w}_n-Bp\parallel }^2.\end{array}$$

(3.17)

This in turn implies that

$$\begin{array}{rl}(1-{\alpha }_n){\lambda }_n(2\beta -{\lambda }_n){\parallel B{w}_n-Bp\parallel }^2& \le {\parallel x_n-p\parallel }^2-{\parallel y_n-p\parallel }^2\\ \le (\parallel x_n-p\parallel +\parallel y_n-p\parallel )\parallel x_n-y_n\parallel .\end{array}$$

(3.18)

In view of the restriction (a)-(c), we obtain from (3.2) that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel B{w}_n-Bp\parallel =0.$$

(3.19)

On the other hand, we have from the firm nonexpansivity of ${Proj}_C$ that

$$\begin{array}{rcl}{\parallel z_n-p\parallel }^2& =& {\parallel {Proj}_C(I-{\lambda }_{n}B)w_n-{Proj}_C(I-{\lambda }_{n}B)p\parallel }^2\\ \le & 〈(I-{\lambda }_{n}B)w_n-(I-{\lambda }_{n}B)p,z_n-p〉\\ =& \frac{1}{2}({\parallel (I-{\lambda }_{n}B)w_n-(I-{\lambda }_{n}B)p\parallel }^2+{\parallel z_n-p\parallel }^2\\ -{\parallel (I-{\lambda }_{n}B)w_n-(I-{\lambda }_{n}B)p-(z_n-p)\parallel }^2)\\ \le & \frac{1}{2}({\parallel w_n-p\parallel }^2+{\parallel z_n-p\parallel }^2-{\parallel w_n-z_n-{\lambda }_n(B{w}_n-Bp)\parallel }^2)\\ \le & \frac{1}{2}({\parallel x_n-p\parallel }^2+{\parallel z_n-p\parallel }^2-{\parallel w_n-z_n\parallel }^2\\ +2{\lambda }_n〈w_n-z_n,B{w}_n-Bp〉-{\lambda }_n^2{\parallel B{w}_n-Bp\parallel }^2).\end{array}$$

This implies that

$${\parallel z_n-p\parallel }^2\le {\parallel x_n-p\parallel }^2-{\parallel w_n-z_n\parallel }^2+2{\lambda }_n\parallel w_n-z_n\parallel \parallel B{w}_n-Bp\parallel ,$$

from which it follows that

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& \le & {\alpha }_n{\parallel x_n-p\parallel }^2+(1-{\alpha }_n){\parallel S{z}_n-p\parallel }^2\\ \le & {\alpha }_n{\parallel x_n-p\parallel }^2+(1-{\alpha }_n){\parallel z_n-p\parallel }^2\\ \le & {\parallel x_n-p\parallel }^2-(1-{\alpha }_n){\parallel w_n-z_n\parallel }^2\\ +2{\lambda }_n\parallel w_n-z_n\parallel \parallel B{w}_n-Bp\parallel .\end{array}$$

Hence, we get that

In view of the restriction (a), we obtain from (3.2) and (3.19) that

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

(3.20)

Note that

$$\parallel z_n-x_n\parallel \le \parallel z_n-w_n\parallel +\parallel w_n-x_n\parallel \le \parallel z_n-w_n\parallel +\sum _{m=1}^N{\beta }_{n,m}\parallel u_{n,m}-x_n\parallel .$$

In view of (3.12) and (3.20), we get that

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

(3.21)

Next, we prove that $x\in VI(C,B)$. In fact, let M be the maximal monotone mapping defined by

$$My=\{\begin{array}{cc}By+N_{C}y,\hfill & y\in C,\hfill \\ \mathrm{\varnothing },\hfill & y\notin C.\hfill \end{array}$$

For any given $(s,t)\in G(T)$, we have $t-Bs\in N_{C}s$. Since $z_n\in C$, by the definition of $N_C$, we have

$$〈s-z_n,t-Bs〉\ge 0.$$

(3.22)

In view of the algorithm, we obtain that

$$〈s-z_n,z_n-(I-{\lambda }_{n}B)w_n〉\ge 0$$

and hence

$$〈s-z_n,\frac{z_n-w_n}{{\lambda }_n}+B{w}_n〉\ge 0.$$

(3.23)

Since B is monotone, we obtain from (3.23) that

$$\begin{array}{rcl}〈s-z_{n_i},t〉& \ge & 〈s-z_{n_i},Bs〉\\ \ge & 〈s-z_{n_i},Bs〉-〈s-z_{n_i},\frac{z_{n_i}-w_{n_i}}{{\lambda }_{n_i}}+B{w}_{n_i}〉\\ =& 〈s-z_{n_i},Bs-B{z}_{n_i}〉+〈s-z_{n_i},B{z}_{n_i}-B{w}_{n_i}〉\\ -〈s-z_{n_i},\frac{z_{n_i}-w_{n_i}}{{\lambda }_{n_i}}〉\\ \ge & 〈s-z_{n_i},B{z}_{n_i}-B{w}_{n_i}〉-〈s-z_{n_i},\frac{z_{n_i}-w_{n_i}}{{\lambda }_{n_i}}〉.\end{array}$$

It follows from (3.21) that $z_{n_i}\rightharpoonup x$. On the other hand, we have that B is $\frac{1}{\beta }$-Lipschitz continuous. It follows from (3.20) that

$$〈s-x,t〉\ge 0.$$

Notice that M is maximal monotone and hence $0\in Mx$. This shows that $x\in VI(C,B)$. Notice that

$$\parallel x_n-S{x}_n\parallel \le \parallel x_n-S{z}_n\parallel +\parallel S{z}_n-S{x}_n\parallel .$$

We find from (3.3) and (3.21) that

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

(3.24)

Next, we prove that $x\in F(S)$. Since S is demiclosed at zero, we see that $x\in F(S)$. This proves that $x\in \mathcal{F}$. Notice that ${Proj}_{\mathcal{F}}{x}_1\subset C_{n+1}$ and $x_{n+1}={Proj}_{C_{n+1}}{x}_1$, we have

$$\parallel x_1-x_{n+1}\parallel \le \parallel x_1-{Proj}_{\mathcal{F}}{x}_1\parallel .$$

On the other hand, we have

$$\begin{array}{rl}\parallel x_1-{Proj}_{\mathcal{F}}{x}_1\parallel & \le \parallel x_1-x\parallel \\ \le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_1-x_{n_i}\parallel \\ \le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_1-x_{n_i}\parallel \\ \le \parallel x_1-{Proj}_{\mathcal{F}}{x}_1\parallel .\end{array}$$

We, therefore, obtain that

$$\parallel x_1-x\parallel =\underset{i\to \mathrm{\infty }}{lim}\parallel x_1-x_{n_i}\parallel =\parallel x_1-{Proj}_{\mathcal{F}}{x}_1\parallel .$$

This implies $x_{n_i}\to x={Proj}_{\mathcal{F}}{x}_1$. Since $\{x_{n_i}\}$ is an arbitrary subsequence of $\{x_n\}$, we obtain that $x_n\to {Proj}_{\mathcal{F}}{x}_1$ as $n\to \mathrm{\infty }$. This completes the proof. □

If S is an identity mapping, we obtain from Theorem 3.1 the following.

Corollary 3.2 Let C be a nonempty, closed and convex subset of a real Hilbert space H. Let $F_m$ be a bifunction from $C\times C$ to ℝ which satisfies (A1)-(A4) and $A_m:C\to H$ be a ${\kappa }_m$-inverse-strongly monotone mapping for every $1\le m\le N$, where N denotes some positive integer. Let $B:C\to H$ be a β-inverse-strongly monotone mapping. Assume that $\mathcal{F}:={\bigcap }_{m=1}^{N}EP(F_m,A_m)\cap VI(C,B)\ne \mathrm{\varnothing }$. Let $\{{\lambda }_n\}$ be a positive sequence in $[0,2\beta ]$ and $\{r_{n,m}\}$ be a positive sequence in $[0,2{\kappa }_m]$ for every $1\le m\le N$. Let $\{{\alpha }_n\},\{{\beta }_{n,1}\},\dots$ and $\{{\beta }_{n,N}\}$ be sequences in $[0,1]$. Let $\{x_n\}$ be a sequence generated in the following iterative process:

$$\{\begin{array}{c}{x}_1\in C,\hfill \\ C_1=C,\hfill \\ y_n={\alpha }_n{x}_n+(1-{\alpha }_n){Proj}_C({\sum }_{m=1}^N{\beta }_{n,m}{u}_{n,m}-{\lambda }_{n}B{\sum }_{m=1}^N{\beta }_{n,m}{u}_{n,m}),\hfill \\ C_{n+1}=\{v\in C_n:\parallel y_n-v\parallel \le \parallel x_n-v\parallel \},\hfill \\ x_{n+1}=P_{C_{n+1}}{x}_1,n\ge 1,\hfill \end{array}$$

where $\{u_{n,m}\}$ is such that

$$F_m(u_{n,m},u_m)+〈A_m{x}_n,u_m-u_{n,m}〉+\frac{1}{r_{n,m}}〈u_m-u_{n,m},u_{n,m}-x_n〉\ge 0,\mathrm{\forall }{u}_m\in C$$

for each $1\le m\le N$. Assume that the above sequence also satisfies the following restrictions:

• (a) ${\alpha }_n\le a<1$;

• (b) ${\sum }_{m=1}^N{\beta }_{n,m}=1$ and $0\le b\le {\beta }_{n,m}<1$ for each $1\le m\le N$;

• (c) $0<c\le {\lambda }_n\le d<2\beta$ and $0<e\le r_{n,m}\le f<2{\kappa }_m$ for each $1\le m\le N$,

where a, b, c, d, e and f are real numbers. Then the sequence $\{x_n\}$ strongly converges to ${Proj}_{\mathcal{F}}{x}_1$.

If $N=1$, we obtain from Theorem 3.1 the following.

Corollary 3.3 Let C be a nonempty, closed and convex subset of a real Hilbert space H. Let F be a bifunction from $C\times C$ to ℝ which satisfies (A1)-(A4) and $A:C\to H$ be a κ-inverse-strongly monotone mapping. Let $S:C\to C$ be a continuous quasi-nonexpansive mapping which is assumed to be demiclosed at zero and let $B:C\to H$ be a β-inverse-strongly monotone mapping. Assume that $\mathcal{F}:=EP(F,A)\cap VI(C,B)\cap F(S)\ne \mathrm{\varnothing }$. Let $\{{\lambda }_n\}$ be a positive sequence in $[0,2\beta ]$ and $\{r_n\}$ be a positive sequence in $[0,2\kappa ]$. Let $\{{\alpha }_n\}$ be a sequence in $[0,1]$. Let $\{x_n\}$ be a sequence generated in the following iterative process:

$$\{\begin{array}{c}{x}_1\in C,\hfill \\ C_1=C,\hfill \\ y_n={\alpha }_n{x}_n+(1-{\alpha }_n)S{Proj}_C(u_n-{\lambda }_{n}B{u}_n),\hfill \\ C_{n+1}=\{v\in C_n:\parallel y_n-v\parallel \le \parallel x_n-v\parallel \},\hfill \\ x_{n+1}=P_{C_{n+1}}{x}_1,n\ge 1,\hfill \end{array}$$

where $\{u_n\}$ is such that

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

Assume that the above sequence also satisfies the following restrictions:

• (a) ${\alpha }_n\le a<1$;

• (b) $0<b\le {\lambda }_n\le c<2\beta$ and $0<d\le r_n\le e<2\kappa$ for each $1\le m\le N$,

where a, b, c, d and e are real numbers. Then the sequence $\{x_n\}$ strongly converges to ${Proj}_{\mathcal{F}}{x}_1$.

If B is a zero operator, we obtain from Theorem 3.1 the following.

Corollary 3.4 Let C be a nonempty, closed and convex subset of a real Hilbert space H. Let $F_m$ be a bifunction from $C\times C$ to ℝ which satisfies (A1)-(A4) and $A_m:C\to H$ be a ${\kappa }_m$-inverse-strongly monotone mapping for every $1\le m\le N$, where N denotes some positive integer. Let $S:C\to C$ be a continuous quasi-nonexpansive mapping which is assumed to be demiclosed at zero. Assume that $\mathcal{F}:={\bigcap }_{m=1}^{N}EP(F_m,A_m)\cap F(S)\ne \mathrm{\varnothing }$. Let $\{r_{n,m}\}$ be a positive sequence in $[0,2{\kappa }_m]$ for every $1\le m\le N$. Let $\{{\alpha }_n\},\{{\beta }_{n,1}\},\dots$ and $\{{\beta }_{n,N}\}$ be sequences in $[0,1]$. Let $\{x_n\}$ be a sequence generated in the following iterative process:

$$\{\begin{array}{c}{x}_1\in C,\hfill \\ C_1=C,\hfill \\ y_n={\alpha }_n{x}_n+(1-{\alpha }_n)S{\sum }_{m=1}^N{\beta }_{n,m}{u}_{n,m},\hfill \\ C_{n+1}=\{v\in C_n:\parallel y_n-v\parallel \le \parallel x_n-v\parallel \},\hfill \\ x_{n+1}=P_{C_{n+1}}{x}_1,n\ge 1,\hfill \end{array}$$

where $\{u_{n,m}\}$ is such that

$$F_m(u_{n,m},u_m)+〈A_m{x}_n,u_m-u_{n,m}〉+\frac{1}{r_{n,m}}〈u_m-u_{n,m},u_{n,m}-x_n〉\ge 0,\mathrm{\forall }{u}_m\in C$$

for each $1\le m\le N$. Assume that the above sequence also satisfies the following restrictions:

• (a) ${\alpha }_n\le a<1$;

• (b) ${\sum }_{m=1}^N{\beta }_{n,m}=1$ and $0\le b\le {\beta }_{n,m}<1$ for each $1\le m\le N$;

• (c) $0<c\le r_{n,m}\le d<2{\kappa }_m$ for each $1\le m\le N$,

where a, b, c and d are real numbers. Then the sequence $\{x_n\}$ strongly converges to ${Proj}_{\mathcal{F}}{x}_1$.

Finally, we consider the following optimization problem: Find an $x^{\ast }$ such that

$$\{\begin{array}{c}{\phi }_1(x^{\ast })={min}_{x\in C}{\phi }_1(x),\hfill \\ {\phi }_2(x^{\ast })={min}_{x\in C}{\phi }_2(x),\hfill \\ ⋮\hfill \\ {\phi }_N(x^{\ast })={min}_{x\in C}{\phi }_N(x),\hfill \end{array}$$

where ${\phi }_m:C\to \mathbb{R}$ is a convex and lower semicontinuous function for each $1\le m\le N$, where $N\ge 1$ is some positive integer.

Theorem 3.5 Let C be a nonempty, closed and convex subset of a real Hilbert space H. Let ${\phi }_m$ be a proper convex and lower semicontinuous function for every $1\le m\le N$, where N denotes some positive integer. Assume that $\mathcal{F}:=OP(\phi )\cap VI(C,B)\cap F(S)\ne \mathrm{\varnothing }$, $OP(\phi )$ denotes the solution set of the above optimization problem. Let $\{{\alpha }_n\},\{{\beta }_{n,1}\},\dots$ and $\{{\beta }_{n,N}\}$ be sequences in $[0,1]$. Let $\{x_n\}$ be a sequence generated in the following iterative process:

$$\{\begin{array}{c}{x}_1\in C,\hfill \\ C_1=C,\hfill \\ y_n={\alpha }_n{x}_n+(1-{\alpha }_n){\sum }_{m=1}^N{\beta }_{n,m}{u}_{n,m},\hfill \\ C_{n+1}=\{v\in C_n:\parallel y_n-v\parallel \le \parallel x_n-v\parallel \},\hfill \\ x_{n+1}=P_{C_{n+1}}{x}_1,n\ge 1,\hfill \end{array}$$

where $\{u_{n,m}\}$ is such that

$${\phi }_m(u_m)-{\phi }_m(u_{n,m})+\frac{1}{r_{n,m}}〈u_m-u_{n,m},u_{n,m}-x_n〉\ge 0,\mathrm{\forall }{u}_m\in C$$

for each $1\le m\le N$. Assume that the above sequence also satisfies the following restrictions:

• (a) ${\alpha }_n\le a<1$;

• (b) ${\sum }_{m=1}^N{\beta }_{n,m}=1$ and $0\le b\le {\beta }_{n,m}<1$ for each $1\le m\le N$;

• (c) $0<c\le r_{n,m}\le d<\mathrm{\infty }$ for each $1\le m\le N$,

where a, b, c and d are real numbers. Then the sequence $\{x_n\}$ strongly converges to ${Proj}_{\mathcal{F}}{x}_1$.

Proof Putting $S=I$, $A_m=B=0$ and $F_m(x,y)=\phi (y)-\phi (x)$, we find from Theorem 3.1 the desired conclusion immediately. □