Extension of Extragradient Techniques for Variational Inequalities

Abstract

An extragradient type method for finding the common solutions of two variational inequalities has been proposed. The convergence result of the algorithm is given under mild conditions on the algorithm parameters.

1. Introduction

Let H be a real Hilbert space equipped with inner product $⟨·,·⟩$ and norm $\parallel ·\parallel$. Let $\varnothing \ne C\subset H$ be a closed and convex set. Let $A:C\to H$ be a mapping. Recall that the variational inequality (VI) seeks an element $x^{\ast }\in C$ such that

$$⟨A{z}^{\ast },z-z^{\ast }⟩\ge 0,\forall z\in C.$$

(1)

The solution set of (1) is denoted by VI$(C,A)$.

The problem (1) introduced and studied by Stampacchia [1] is being applied as a useful tool and model to refine a multitude of problems. A large number of methods for solving VI (1) are projection methods that implement projections onto the feasible set of the VI (1), or onto another set in order to achieve a solution. Several iterative methods for solving the VI (1) have been proposed. See, e.g., [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30]. A basic one is the natural extension of the gradient projection algorithm for solving the optimization problem ${min}_{x\in C}f\left(x\right)$. For $x_0\in C$, calculate iteratively the sequence $\left\{x_n\right\}$ through

$$\begin{array}{c}\hfill x_{n+1}=pro{j}_C[x_n-{\alpha }_n\nabla f\left(x_n\right)],n\ge 0,\end{array}$$

(2)

where $proj$ is the metric projection and ${\alpha }_n>0$ is the step-size.

Korpelevich [31] introduced an iterative method for solving the VI (1), known as the extragradient method ([7]). In Korpelevich’s method, two projections are used for computing the next iteration. For the current iteration $x_n$, compute

$$\left\{\begin{array}{ccc}\hfill y_n& =pro{j}_C[x_n-\xi A{x}_n],\hfill & \\ \hfill x_{n+1}& =pro{j}_C[x_n-\xi A{y}_n],\hfill & n\ge 0,\hfill \end{array}\right.$$

(3)

where $\xi >0$ is a fixed number.

Korpelevich’s method has received so much attention by a range of scholars, who improved it in several ways; see, e.g., [32,33,34,35,36]. Now, we know that Korpelevich’s method (3) can only achieve weak convergence in a large dimensional spaces ([37,38]). In order to reach strong convergence, Korpelevich’s method was adapted by many mathematicians. For example, in [32], it is shown that several extragradient-type methods converge strongly to an element in VI$(C,A)$.

Very recently, Censor, Gibali and Reich [39] presented an alternating method for finding common solutions of two variational inequalities. In [40], Zaslavski studied an extragradient method for finding common solutions of a finite family of variational inequalities.

Inspired by the work given above, in this article, we present an extragradient type method for finding the common solutions of two variational inequalities. We prove the strong convergence of the proposed method under the mild assumptions on the parameters.

2. Preliminaries

Let H be a real Hilbert space. Let $C\subset H$ be a nonempty, closed and convex set.

Definition 1.

An operator $A:C\to H$ is called Lipschitz if

$$\parallel A{u}^{†}-A{v}^{†}\parallel \le L\parallel u^{†}-v^{†}\parallel ,\forall u^{†},v^{†}\in C,$$

where $L>0$ is a constant.

If $L=1$, we call A is nonexpansive.

Definition 2.

An operator $A:C\to H$ is called inverse strongly monotone if

$$\alpha \parallel A{u}^{†}-A{v}^{†}{\parallel }^2\le ⟨A{u}^{†}-A{v}^{†},u^{†}-v^{†}⟩,\forall u^{†},v^{†}\in C,$$

where $\alpha \ge 0$ is a constant.

In this case, we call A is $\alpha$-inverse-strongly-monotone.

Proposition 1

([41]). If C is a bounded closed convex subset of a real Hilbert space H and $A:C\to H$ is an inverse strongly monotone operator, then $VI(C,A)\ne \varnothing$.

For fixed $z\in H$, there exists a unique $z^{†}\in C$ satisfying

$$\parallel z-z^{†}\parallel =inf\{\parallel z-\tilde{z}\parallel :\tilde{z}\in C\}.$$

We denote $z^{†}$ by $pro{j}_{C}z$. The following inequality is an important property of projection $pro{j}_C$: for given $x\in H$,

$$\begin{array}{c}\hfill ⟨x-pro{j}_{C}x,y-pro{j}_{C}x⟩\le 0,\forall y\in C,\end{array}$$

(4)

which is equivalent to

$$⟨x-y,pro{j}_{C}x-pro{j}_{C}y⟩\ge {\parallel pro{j}_{C}x-pro{j}_{C}y\parallel }^2,\forall x,y\in H.$$

It follows that $pro{j}_C$ is nonexpansive. We also know that $2pro{j}_C-I$ is nonexpansive.

Lemma 1

([41]). If $\varnothing \ne C$ is a closed convex subset of a real Hilbert space H and $A:C\to H$ is an α-inverse strongly monotone operator, then

$$\begin{array}{c}\hfill {\parallel (I-\mu A)u-(I-\mu A)v\parallel }^2\le {\parallel u-v\parallel }^2+\mu (\mu -2\alpha ){\parallel Au-Av\parallel }^2,\forall u,v\in C.\end{array}$$

Especially, $I-\mu A$ is nonexpansive provided $0\le \mu \le 2\alpha$.

Lemma 2

([42]). Suppose that $\left\{u_n\right\}$ and $\left\{v_n\right\}$ are two bounded sequences in Banach spaces. Let $\left\{{\lambda }_n\right\}\subset [0,1]$ be a sequence satisfying $0<{lim\; inf}_{n\to \infty }{\lambda }_n\le {lim\; sup}_{n\to \infty }{\lambda }_n<1$. Suppose $u_{n+1}=(1-{\lambda }_n)v_n+{\lambda }_n{u}_n,\forall n\ge 0$ and ${lim\; sup}_{n\to \infty }(\parallel v_{n+1}-v_n\parallel -\parallel u_{n+1}-u_n\parallel )\le 0$. Then ${lim}_{n\to \infty }\parallel u_n-v_n\parallel =0$.

Lemma 3

([43]). Let $\left\{{\mu }_n\right\}\subset (0,\infty )$, $\left\{{\gamma }_n\right\}\subset (0,1)$ and $\left\{{\delta }_n\right\}$ be three real number sequences. If ${\mu }_{n+1}\le (1-{\gamma }_n){\mu }_n+{\delta }_n$ for all $n\ge 0$ with ${\sum }_{n=1}^{\infty }{\gamma }_n=\infty$ and ${lim\; sup}_{n\to \infty }{\delta }_n/{\gamma }_n\le 0$ or ${\sum }_{n=1}^{\infty }\left|{\delta }_n\right|<\infty$, then ${lim}_{n\to \infty }{\mu }_n=0$.

3. Main Results

Let $\varnothing \ne C$ be a convex and closed subset of a real Hilbert space H. Let the operators $A,B:C\to H$ be $\alpha$-inverse strongly monotone and $\beta$-inverse strongly monotone, respectively. Let $\left\{{\alpha }_n\right\}\subset [0,1]$, $\left\{{\beta }_n\right\}\subset [0,1]$, $\left\{{\lambda }_n\right\}\subset [0,2\alpha ]$ and $\left\{{\mu }_n\right\}\subset [0,2\beta ]$ be four sequences. In the sequel, assume $\Omega =$ VI$(C,A)\cap$ VI$(C,B)\ne \varnothing$.

Motivated by the algorithms presented in [31,39,40], we present the following iterative Algorithm 1 for finding the common solution of two variational inequalities.

Algorithm 1:

For $u,x_0\in C$. Assume the sequence $\left\{x_n\right\}$ has been constructed. Compute the next iteration $\left\{x_{n+1}\right\}$ by the following manner $$\left\{\begin{array}{c}{y}_n=pro{j}_C[x_n-(1-{\alpha }_n){\lambda }_{n}A{x}_n+{\alpha }_n(u-x_n)],\hfill \\ x_{n+1}=pro{j}_C[x_n-{\mu }_n{\beta }_{n}B{y}_n+{\beta }_n(y_n-x_n)],n\ge 0.\hfill \end{array}\right.$$ (5)

Suppose that the control parameters $\left\{{\alpha }_n\right\}$, $\left\{{\beta }_n\right\}$, $\left\{{\lambda }_n\right\}$ and $\left\{{\mu }_n\right\}$ satisfy the following assumptions: (C1): ${lim}_{n\to \infty }{\alpha }_n=0$ and ${\sum }_{n=1}^{\infty }{\alpha }_n=\infty$; (C2): ${lim\; inf}_{n\to \infty }{\beta }_n>0$ and ${lim}_{n\to \infty }({\beta }_{n+1}-{\beta }_n)=0$; (C3): ${\lambda }_n\in [a,b]\subset (0,2\alpha )$ and ${lim}_{n\to \infty }({\lambda }_{n+1}-{\lambda }_n)=0$; (C4): ${\mu }_n\in [c,d]\subset (0,2\beta )$ and ${lim}_{n\to \infty }({\mu }_{n+1}-{\mu }_n)=0$.

We will divide our main result into several propositions.

Proposition 2.

The sequence $\left\{x_n\right\}$ generated by (5) is bounded.

Proof. 

Choose any $x^{\ast }\in \Omega$. Note that $x^{\ast }=pro{j}_C\left[(I-\delta A)x^{\ast }\right]$ for any $\delta >0$. Hence,

$$\begin{array}{c}\hfill x^{\ast }=pro{j}_C[{\alpha }_n{x}^{\ast }+(1-{\alpha }_n)(x^{\ast }-{\lambda }_{n}A{x}^{\ast })],\forall n\ge 0.\end{array}$$

(6)

Thus, by (5) and (6), we have

$$\begin{array}{cc}\hfill \parallel y_n-x^{\ast }\parallel & =\parallel pro{j}_C[{\alpha }_{n}u+(1-{\alpha }_n)(I-{\lambda }_{n}A)x_n]-pro{j}_C[{\alpha }_n{x}^{\ast }+(1-{\alpha }_n)(I-{\lambda }_{n}A)x^{\ast }]\parallel \hfill \\ & \le \parallel [{\alpha }_{n}u+(1-{\alpha }_n)(I-{\lambda }_{n}A)x_n]-[{\alpha }_n{x}^{\ast }+(1-{\alpha }_n)(I-{\lambda }_{n}A)x^{\ast }]\parallel \hfill \\ & \le \parallel {\alpha }_n(u-x^{\ast })+(1-{\alpha }_n)[(I-{\lambda }_{n}A)x_n-(I-{\lambda }_{n}A)x^{\ast }]\parallel \hfill \\ & \le {\alpha }_n\parallel u-x^{\ast }\parallel +(1-{\alpha }_n)\parallel (I-{\lambda }_{n}A)x_n-(I-{\lambda }_{n}A)x^{\ast }\parallel .\hfill \end{array}$$

(7)

From Lemma 1, we know that $I-{\lambda }_{n}A$ and $I-{\mu }_{n}B$ are nonexpansive. Thus, from (7), we get

$$\begin{array}{c}\hfill \parallel y_n-x^{\ast }\parallel \le {\alpha }_n\parallel u-x^{\ast }\parallel +(1-{\alpha }_n)\parallel x_n-x^{\ast }\parallel .\end{array}$$

So,

$$\begin{array}{cc}\hfill \parallel x_{n+1}-x^{\ast }\parallel & =\parallel pro{j}_C[(1-{\beta }_n)x_n+{\beta }_n(y_n-{\mu }_{n}B{y}_n)]-pro{j}_C[(1-{\beta }_n)x^{\ast }+{\beta }_n(x^{\ast }-{\mu }_{n}B{x}^{\ast })]\parallel \hfill \\ & \le \parallel [(1-{\beta }_n)x_n+{\beta }_n(y_n-{\mu }_{n}B{y}_n)]-[(1-{\beta }_n)x^{\ast }+{\beta }_n(x^{\ast }-{\mu }_{n}B{x}^{\ast })]\parallel \hfill \\ & \le {\beta }_n\parallel (y_n-{\mu }_{n}B{y}_n)-(x^{\ast }-{\mu }_{n}B{x}^{\ast })\parallel +(1-{\beta }_n)\parallel x_n-x^{\ast }\parallel \hfill \\ & \le {\beta }_n\parallel y_n-x^{\ast }\parallel +(1-{\beta }_n)\parallel x_n-x^{\ast }\parallel \hfill \\ & \le (1-{\beta }_n{\alpha }_n)\parallel x_n-x^{\ast }\parallel +{\beta }_n{\alpha }_n\parallel u-x^{\ast }\parallel .\hfill \end{array}$$

(8)

It follows that

$$\parallel x_{n+1}-x^{\ast }\parallel \le max\{\parallel x_0-x^{\ast }\parallel ,\parallel u-x^{\ast }\parallel \}.$$

Then $\left\{x_n\right\}$ is bounded, and hence the sequences $\left\{y_n\right\}$, $\left\{A{x}_n\right\}$ and $\left\{B{y}_n\right\}$ are all bounded.  □

Proposition 3.

The following two conclusions hold

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}\parallel x_{n+1}-x_n\parallel =0and\underset{n\to \infty }{lim}\parallel x_n-y_n\parallel =0.\end{array}$$

Proof. 

Let $S=2pro{j}_C-I$. It is clear that S is nonexpansive. Set $v_n=(1-{\beta }_n)x_n+{\beta }_n(I-{\mu }_{n}B)y_n$ for all $n\ge 0$. Then, we can rewrite $x_{n+1}$ in (5) as

$$\begin{array}{cc}\hfill x_{n+1}& =\frac{1-{\beta }_n}{2}{x}_n+\frac{{\beta }_n}{2}(I-{\mu }_{n}B)y_n+\frac{1}{2}S{v}_n\hfill \\ & =\frac{1+{\beta }_n}{2}{z}_n+\frac{1-{\beta }_n}{2}{x}_n,\hfill \end{array}$$

where ${\displaystyle z_n=\frac{{\beta }_n}{1+{\beta }_n}(I-{\mu }_{n}B)y_n+\frac{1}{1+{\beta }_n}S{v}_n}$ for all $n\ge 0$.

Hence,

$$\begin{array}{c}\hfill z_{n+1}-z_n=\frac{1}{1+{\beta }_{n+1}}S{v}_{n+1}+\frac{{\beta }_{n+1}}{1+{\beta }_{n+1}}(I-{\mu }_{n+1}B)y_{n+1}-\frac{{\beta }_n}{1+{\beta }_n}(I-{\mu }_{n}B)y_n-\frac{1}{1+{\beta }_n}S{v}_n.\end{array}$$

Hence,

$$\begin{array}{ccc}\hfill \parallel z_{n+1}-z_n\parallel & \le & \frac{{\beta }_{n+1}}{1+{\beta }_{n+1}}\parallel (I-{\mu }_{n+1}B)y_{n+1}-(I-{\mu }_{n}B)y_n\parallel +\frac{1}{1+{\beta }_{n+1}}\parallel S{v}_{n+1}-S{v}_n\parallel \hfill \\ & & +|\frac{{\beta }_{n+1}}{1+{\beta }_{n+1}}-\frac{{\beta }_n}{1+{\beta }_n}|\parallel (I-{\mu }_{n}B)y_n\parallel +|\frac{1}{1+{\beta }_{n+1}}-\frac{1}{1+{\beta }_n}|\parallel S{v}_n\parallel \hfill \\ & \le & \frac{{\beta }_{n+1}}{1+{\beta }_{n+1}}\parallel (I-{\mu }_{n+1}B)y_{n+1}-(I-{\mu }_{n+1}B)y_n\parallel \hfill \\ & & +\frac{{\beta }_{n+1}}{1+{\beta }_{n+1}}|{\mu }_{n+1}-{\mu }_n|\parallel B{y}_n\parallel +|\frac{{\beta }_{n+1}}{1+{\beta }_{n+1}}-\frac{{\beta }_n}{1+{\beta }_n}|\parallel y_n-{\mu }_{n}B{y}_n\parallel \hfill \\ & & +\frac{1}{1+{\beta }_{n+1}}\parallel S{v}_{n+1}-S{v}_n\parallel +|\frac{1}{1+{\beta }_{n+1}}-\frac{1}{1+{\beta }_n}|\parallel S{v}_n\parallel .\hfill \end{array}$$

By using the nonexpansivity of $I-{\mu }_{n}B$ and S to deduce

$$\begin{array}{ccc}\hfill \parallel z_{n+1}-z_n\parallel & \le & \frac{{\beta }_{n+1}}{1+{\beta }_{n+1}}\parallel y_{n+1}-y_n\parallel +|\frac{{\beta }_{n+1}}{1+{\beta }_{n+1}}-\frac{{\beta }_n}{1+{\beta }_n}|\parallel y_n-{\mu }_{n}B{y}_n\parallel \hfill \\ & & +{\beta }_{n+1}[(I-{\mu }_{n+1}B)y_{n+1}-(I-{\mu }_{n+1}B)y_n]+|\frac{1}{1+{\beta }_n}-\frac{1}{1+{\beta }_n}|\parallel S{v}_n\parallel \hfill \\ & & +\frac{{\beta }_{n+1}}{1+{\beta }_{n+1}}|{\mu }_{n+1}-{\mu }_n|\parallel B{y}_n\parallel +\frac{1}{1+{\beta }_{n+1}}\parallel (1-{\beta }_{n+1})(x_{n+1}-x_n)\hfill \\ & & +({\beta }_{n+1}-{\beta }_n)(y_n-x_n)+({\beta }_n{\mu }_n-{\beta }_{n+1}{\mu }_{n+1})B{y}_n\parallel \hfill \\ & \le & \frac{|{\beta }_{n+1}-{\beta }_n|}{1+{\beta }_{n+1}}(\parallel x_n\parallel +\parallel y_n\parallel )+\frac{1-{\beta }_{n+1}}{1+{\beta }_{n+1}}\parallel x_{n+1}-x_n\parallel +|\frac{1}{1+{\beta }_n}-\frac{1}{1+{\beta }_n}|\parallel S{v}_n\parallel \hfill \\ & & +\frac{2{\beta }_{n+1}}{1+{\beta }_{n+1}}\parallel y_{n+1}-y_n\parallel +|\frac{{\beta }_{n+1}}{1+{\beta }_{n+1}}-\frac{{\beta }_n}{1+{\beta }_n}|\parallel y_n-{\mu }_{n}B{y}_n\parallel \hfill \\ & & +\frac{{\beta }_{n+1}}{1+{\beta }_{n+1}}|{\mu }_{n+1}-{\mu }_n|\parallel B{y}_n\parallel +\frac{|{\beta }_{n+1}{\mu }_{n+1}-{\beta }_n{\mu }_n|}{1+{\beta }_{n+1}}\parallel B{y}_n\parallel .\hfill \end{array}$$

(9)

Next, we estimate $\parallel y_{n+1}-y_n\parallel$. By (5), we get

$$\begin{array}{ccc}\hfill \parallel y_{n+1}-y_n\parallel & =& \parallel pro{j}_C[{\alpha }_{n+1}u+(1-{\alpha }_{n+1})(I-{\lambda }_{n+1}A)x_{n+1}]-pro{j}_C[{\alpha }_{n}u+(1-{\alpha }_n)(I-{\lambda }_{n}A)x_n]\parallel \hfill \\ & \le |{\alpha }_{n+1}-{\alpha }_n|\parallel x_n\parallel +|{\lambda }_{n+1}-{\lambda }_n|\parallel A{x}_n\parallel +{\alpha }_{n+1}\parallel u\parallel +{\alpha }_n\parallel u\parallel +\parallel x_{n+1}-x_n\parallel \\ & & +|{\lambda }_{n+1}-{\lambda }_n|\parallel A{x}_n\parallel +|{\lambda }_{n+1}{\alpha }_{n+1}-{\lambda }_n{\alpha }_n|\parallel A{x}_n\parallel .\end{array}$$

(10)

Substituting (10) into (9) to get

$$\begin{array}{ccc}\hfill \parallel z_{n+1}-z_n\parallel & \le & |\frac{{\beta }_{n+1}}{1+{\beta }_{n+1}}-\frac{{\beta }_n}{1+{\beta }_n}|\parallel y_n-{\mu }_{n}B{y}_n\parallel +\frac{{\beta }_{n+1}}{1+{\beta }_{n+1}}|{\mu }_{n+1}-{\mu }_n|\parallel B{y}_n\parallel \\ & & +\frac{|{\beta }_{n+1}-{\beta }_n|}{1+{\beta }_{n+1}}(\parallel x_n\parallel +\parallel y_n\parallel )+\frac{|{\beta }_{n+1}{\mu }_{n+1}-{\beta }_n{\mu }_n|}{1+{\beta }_{n+1}}\parallel B{y}_n\parallel +\parallel x_{n+1}-x_n\parallel \\ & & +|\frac{1}{1+{\beta }_n}-\frac{1}{1+{\beta }_n}|\parallel S{v}_n\parallel +2|{\lambda }_{n+1}-{\lambda }_n|\parallel A{x}_n\parallel +2{\alpha }_{n+1}\parallel u\parallel +2{\alpha }_n\parallel u\parallel \\ & & +2|{\alpha }_{n+1}-{\alpha }_n|\parallel x_n\parallel +2|{\lambda }_{n+1}-{\lambda }_n|\parallel A{x}_n\parallel +2|{\lambda }_{n+1}{\alpha }_{n+1}-{\lambda }_n{\alpha }_n|\parallel A{x}_n\parallel .\end{array}$$

Since ${\displaystyle \underset{n\to \infty }{lim}({\beta }_{n+1}-{\beta }_n)=0}$ and ${\displaystyle \underset{n\to \infty }{lim}({\mu }_{n+1}-{\mu }_n)=0}$, we derive that ${\displaystyle \underset{n\to \infty }{lim}|\frac{{\beta }_{n+1}}{1+{\beta }_{n+1}}-\frac{{\beta }_n}{1+{\beta }_n}|=0}$, ${\displaystyle \underset{n\to \infty }{lim}|{\beta }_{n+1}{\mu }_{n+1}-{\beta }_n{\mu }_n|=0}$ and ${\displaystyle \underset{n\to \infty }{lim}|\frac{1}{1+{\beta }_n}-\frac{1}{1+{\beta }_n}|=0.}$ At the same time, note that $\left\{x_n\right\}$, $\left\{A{x}_n\right\}$, $\left\{y_n\right\}$ and $\left\{B{y}_n\right\}$ are bounded. Therefore,

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim\; sup}(\parallel z_{n+1}-z_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0.\end{array}$$

Applying Lemma 2 to derive

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}\parallel z_n-x_n\parallel =0.\end{array}$$

Hence,

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}\parallel x_{n+1}-x_n\parallel =\underset{n\to \infty }{lim}\frac{1+{\beta }_n}{2}\parallel z_n-x_n\parallel =0.\end{array}$$

By virtue of (7), (8) and Lemma 1, we deduce

$$\begin{array}{ccc}\hfill \parallel x_{n+1}-x^{\ast }{\parallel }^2& \le & (1-{\beta }_n)\parallel x_n-x^{\ast }{\parallel }^2+{\beta }_n{\parallel (y_n-{\mu }_{n}B{y}_n)-(x^{\ast }-{\mu }_{n}B{x}^{\ast })\parallel }^2\hfill \\ & \le & (1-{\beta }_n)\parallel x_n-x^{\ast }{\parallel }^2+{\beta }_n\parallel y_n-x^{\ast }{\parallel }^2+{\beta }_n{\mu }_n({\mu }_n-2\beta ){\parallel B{y}_n-B{x}^{\ast }\parallel }^2\hfill \\ & \le & {\beta }_n{\parallel {\alpha }_n(u-x^{\ast })+(1-{\alpha }_n)[(x_n-{\lambda }_{n}A{x}_n)-(x^{\ast }-{\lambda }_{n}A{x}^{\ast })]\parallel }^2\hfill \\ & & +(1-{\beta }_n)\parallel x_n-x^{\ast }{\parallel }^2+{\beta }_n{\mu }_n({\mu }_n-2\beta ){\parallel B{y}_n-B{x}^{\ast }\parallel }^2\hfill \\ & \le & {\beta }_n[{\alpha }_n\parallel u-x^{\ast }{\parallel }^2+(1-{\alpha }_n)\parallel (I-{\lambda }_{n}A)x_n-(I-{\lambda }_{n}A)x^{\ast }{\parallel }^2]\hfill \\ & & +{\beta }_n{\mu }_n({\mu }_n-2\beta )\parallel B{y}_n-B{x}^{\ast }{\parallel }^2+(1-{\beta }_n){\parallel x_n-x^{\ast }\parallel }^2\hfill \\ & \le & (1-{\beta }_n)\parallel x_n-x^{\ast }{\parallel }^2+{\alpha }_n{\beta }_n\parallel u-x^{\ast }{\parallel }^2+{\beta }_n{\mu }_n({\mu }_n-2\beta ){\parallel B{y}_n-B{x}^{\ast }\parallel }^2\hfill \\ & & +(1-{\alpha }_n){\beta }_n[\parallel x_n-x^{\ast }{\parallel }^2+{\lambda }_n({\lambda }_n-2\alpha )\parallel A{x}_n-A{x}^{\ast }{\parallel }^2]\hfill \\ & \le & {\alpha }_n{\beta }_n\parallel u-x^{\ast }{\parallel }^2+\parallel x_n-x^{\ast }{\parallel }^2+{\beta }_n{\mu }_n({\mu }_n-2\beta ){\parallel B{y}_n-B{x}^{\ast }\parallel }^2\hfill \\ & & +(1-{\alpha }_n){\beta }_n{\lambda }_n({\lambda }_n-2\alpha ){\parallel A{x}_n-A{x}^{\ast }\parallel }^2.\hfill \end{array}$$

It follows that

$$\begin{array}{c}(1-{\alpha }_n){\beta }_n{\lambda }_n(2\alpha -{\lambda }_n)\parallel A{x}_n-A{x}^{\ast }{\parallel }^2+{\beta }_n{\mu }_n(2\beta -{\mu }_n){\parallel B{y}_n-B{x}^{\ast }\parallel }^2.\hfill \\ \le {\alpha }_n{\beta }_n\parallel u-x^{\ast }{\parallel }^2+\parallel x_n-x^{\ast }{\parallel }^2-{\parallel x_{n+1}-x^{\ast }\parallel }^2\hfill \\ \le {\alpha }_n{\beta }_n\parallel u-x^{\ast }{\parallel }^2+(\parallel x_n-x^{\ast }\parallel +\parallel x_{n+1}-x^{\ast }\parallel )\times \parallel x_n-x_{n+1}\parallel .\hfill \end{array}$$

This implies that

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}\parallel A{x}_n-A{x}^{\ast }\parallel =0and\underset{n\to \infty }{lim}\parallel B{y}_n-B{x}^{\ast }\parallel =0\end{array}$$

According to (4) and (5), we get

$$\begin{array}{ccc}\hfill \parallel y_n-x^{\ast }{\parallel }^2& =& \parallel pro{j}_C[{\alpha }_{n}u+(1-{\alpha }_n)(I-{\lambda }_{n}A)x_n]-pro{j}_C(I-{\lambda }_{n}A)x^{\ast }{\parallel }^2\hfill \\ & \le & \le ⟨{\alpha }_{n}u+(1-{\alpha }_n)(I-{\lambda }_{n}A)x_n-(I-{\lambda }_{n}A)x^{\ast },y_n-x^{\ast }⟩\hfill \\ & =& \frac{1}{2}\{\parallel (I-{\lambda }_{n}A)x_n-(I-{\lambda }_{n}A)x^{\ast }+{\alpha }_n(u-x_n+{\lambda }_{n}A{x}^{\ast }){\parallel }^2+{\parallel y_n-x^{\ast }\parallel }^2\hfill \\ & & -\parallel {\alpha }_{n}u+(1-{\alpha }_n)(I-{\lambda }_{n}A)x_n-(I-{\lambda }_{n}A)x^{\ast }-(y_n-x^{\ast }){\parallel }^2\}\hfill \\ & \le & \frac{1}{2}\{{\parallel (I-{\lambda }_{n}A{x}_n)-(I-{\lambda }_{n}A)x^{\ast }\parallel }^2\hfill \\ & & +2{\alpha }_n\parallel u-x_n+{\lambda }_{n}A{x}^{\ast }\parallel \parallel (I-{\lambda }_{n}A)x_n-(I-{\lambda }_{n}A)x^{\ast }+{\alpha }_n(u-x_n+{\lambda }_{n}A{x}^{\ast })\parallel \hfill \\ & & +\parallel y_n-x^{\ast }{\parallel }^2-{\parallel (x_n-y_n)-{\lambda }_n(A{x}_n-A{x}^{\ast })+{\alpha }_n(u-x_n+{\lambda }_{n}A{x}^{\ast })\parallel }^2\}\hfill \\ & \le & \frac{1}{2}\{\parallel (I-{\lambda }_{n}A)x_n-(I-{\lambda }_{n}A)x^{\ast }{\parallel }^2+{\alpha }_{n}M+{\parallel y_n-x^{\ast }\parallel }^2\hfill \\ & & -\parallel (x_n-y_n)-{\lambda }_n(A{x}_n-A{x}^{\ast })+{\alpha }_n(u-x_n+{\lambda }_{n}A{x}^{\ast }){\parallel }^2\}\hfill \\ & \le & \frac{1}{2}\{\parallel x_n-x^{\ast }{\parallel }^2-\parallel x_n-y_n{\parallel }^2+{\alpha }_{n}M+{\parallel y_n-x^{\ast }\parallel }^2+2{\lambda }_n⟨x_n-y_n,A{x}_n-A{x}^{\ast }⟩\hfill \\ & & -2{\alpha }_n⟨u-x_n+{\lambda }_{n}A{x}^{\ast },x_n-y_n⟩-{\parallel {\lambda }_n(A{x}_n-A{x}^{\ast })-{\alpha }_n(u-x_n+{\lambda }_{n}A{x}^{\ast })\parallel }^2\}\hfill \\ & \le & \frac{1}{2}\{\parallel x_n-x^{\ast }{\parallel }^2+{\alpha }_{n}M+\parallel y_n-x^{\ast }{\parallel }^2-{\parallel x_n-y_n\parallel }^2\hfill \\ & & +2{\alpha }_n\parallel u-x_n+{\lambda }_{n}A{x}^{\ast }\parallel \parallel x_n-y_n\parallel +2{\lambda }_n\parallel x_n-y_n\parallel \parallel A{x}_n-A{x}^{\ast }\parallel \},\hfill \end{array}$$

where $M>0$ is some constant such that

$$\underset{n}{sup}\{2\parallel u-x_n+{\lambda }_{n}A{x}^{\ast }\parallel \parallel (I-{\lambda }_{n}A)x_n-(I-{\lambda }_{n}A)x^{\ast }+{\alpha }_n(u-x_n+{\lambda }_{n}A{x}^{\ast })\parallel \}\le M.$$

Thus

$$\begin{array}{cc}\hfill \parallel y_n-x^{\ast }{\parallel }^2\le & \parallel x_n-x^{\ast }{\parallel }^2+{\alpha }_{n}M-\parallel x_n-y_n{\parallel }^2+2{\lambda }_n\parallel x_n-y_n\parallel \parallel A{x}_n-A{x}^{\ast }\parallel \hfill \\ & +2{\alpha }_n\parallel u-x_n+{\lambda }_{n}A{x}^{\ast }\parallel \parallel x_n-y_n\parallel ,\hfill \end{array}$$

and it follows that

$$\begin{array}{cc}\hfill \parallel x_{n+1}-x^{\ast }{\parallel }^2\le & \parallel x_n-x^{\ast }{\parallel }^2+{\alpha }_{n}M-{\beta }_n\parallel x_n-y_n{\parallel }^2+2{\lambda }_n\parallel x_n-y_n\parallel \parallel A{x}_n-A{x}^{\ast }\parallel \hfill \\ & +2{\alpha }_n\parallel u-x_n+{\lambda }_{n}A{x}^{\ast }\parallel \parallel x_n-y_n\parallel .\hfill \end{array}$$

Therefore,

$$\begin{array}{cc}\hfill {\beta }_n{\parallel x_n-y_n\parallel }^2\le & 2{\lambda }_n\parallel x_n-y_n\parallel \parallel A{x}_n-A{x}^{\ast }\parallel +(\parallel x_n-x^{\ast }\parallel +\parallel x_{n+1}-x^{\ast }\parallel )\parallel x_{n+1}-x_n\parallel \hfill \\ & +{\alpha }_{n}M+2{\alpha }_n\parallel u-x_n+{\lambda }_{n}A{x}^{\ast }\parallel \parallel x_n-y_n\parallel .\hfill \end{array}$$

Since ${lim}_{n\to \infty }{\alpha }_n=0,{lim}_{n\to \infty }\parallel x_n-x_{n+1}\parallel =0$ and ${lim}_{n\to \infty }\parallel A{x}_n-A{x}^{\ast }\parallel =0$, we derive

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

This concludes the proof.  □

Proposition 4.

${lim\; sup}_{n\to \infty }⟨\tilde{x}-u,\tilde{x}-y_n⟩\le 0$, where $\tilde{x}=P_{\Omega }u$.

Proof. 

Let $\left\{y_{n_i}\right\}$ be a subsequence of $\left\{y_n\right\}$ satisfying

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim\; sup}⟨\tilde{x}-u,\tilde{x}-y_n⟩=\underset{i\to \infty }{lim}⟨\tilde{x}-u,\tilde{x}-y_{n_i}⟩.\end{array}$$

By the boundedness of $\left\{y_{n_i}\right\}$, we can choose a subsequence $\left\{y_{n_{ij}}\right\}$ of $\left\{y_{n_i}\right\}$ such that $y_{n_{ij}}\rightharpoonup z$.

Next, we demonstrate that $z\in \Omega$. First, we prove that $z\in$ VI$(C,A)$. Let $N_{C}v$ be the normal cone of C at $v\in C$; i.e., $N_{C}v=\{w\in H:0\le ⟨v-u,w⟩,\forall u\in C\}.$ Define a mapping T by the formula

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

Let $(v,w)\in G\left(T\right)$. Since $y_n\in C$ and $w-Av\in N_{C}v$, we deduce $0\le ⟨v-y_n,w-Av⟩$. According to $y_n=pro{j}_C[{\alpha }_{n}u+(1-{\alpha }_n)(I-{\lambda }_{n}A)x_n]$, we obtain

$$\begin{array}{c}\hfill 0\le ⟨v-y_n,y_n-{\alpha }_{n}u-(1-{\alpha }_n)(x_n-{\lambda }_{n}A{x}_n)⟩,\end{array}$$

that is,

$$\begin{array}{c}\hfill 0\le ⟨v-y_n,\frac{y_n-x_n}{{\lambda }_n}+A{x}_n-\frac{{\alpha }_n}{{\lambda }_n}(u-x_n+{\lambda }_{n}A{x}_n)⟩.\end{array}$$

Thus,

$$\begin{array}{ccc}\hfill ⟨v-y_{n_i},w⟩& \ge & ⟨v-y_{n_i},Av⟩\hfill \\ & \ge ⟨v-y_{n_i},Av⟩-⟨v-y_{n_i},\frac{y_{n_i}-x_{n_i}}{{\lambda }_{n_i}}+A{x}_{n_i}-\frac{{\alpha }_{n_i}}{{\lambda }_{n_i}}(u-x_{n_i}+{\lambda }_{n_i}A{x}_{n_i})⟩\hfill \\ & =⟨v-y_{n_i},Av-A{x}_{n_i}-\frac{y_{n_i}-x_{n_i}}{{\lambda }_{n_i}}+\frac{{\alpha }_{n_i}}{{\lambda }_{n_i}}(u-x_{n_i}+{\lambda }_{n_i}A{x}_{n_i})⟩\hfill \\ & =⟨v-y_{n_i},Av-A{y}_{n_i}⟩+⟨v-y_{n_i},A{y}_{n_i}-A{x}_{n_i}⟩\hfill \\ & & -⟨v-y_{n_i},\frac{y_{n_i}-x_{n_i}}{{\lambda }_{n_i}}-\frac{{\alpha }_{n_i}}{{\lambda }_{n_i}}(u-x_{n_i}+{\lambda }_{n_i}A{x}_{n_i})⟩\hfill \\ & \ge ⟨v-y_{n_i},A{y}_{n_i}-A{x}_{n_i}⟩-⟨v-y_{n_i},\frac{y_{n_i}-x_{n_i}}{{\lambda }_{n_i}}-\frac{{\alpha }_{n_i}}{{\lambda }_{n_i}}(u-x_{n_i}+{\lambda }_{n_i}A{x}_{n_i})⟩.\hfill \end{array}$$

Noting that ${\alpha }_{n_i}\to 0$ and $\parallel y_{n_i}-x_{n_i}\parallel \to 0$, we deduce $0\le ⟨v-z,w⟩$. Hence, $z\in T^{-1}\left(0\right)$ and thus $z\in$ VI$(C,A)$.

Next, we show that $z\in$ VI$(C,B)$. Define a mapping $T^{†}$ as follows

$$T^{†}v=\left\{\begin{array}{cc}B{v}^{†}+N_C{v}^{†},\hfill & v^{†}\in C,\hfill \\ \varnothing ,& v^{†}\notin C.\hfill \end{array}\right.$$

Let $(v^{†},w^{†})\in G\left(T^{†}\right)$. Since $w^{†}-B{v}^{†}\in N_{C}v$ and $x_{n+1}\in C$, we obtain $0\le ⟨v^{†}-x_{n+1},w^{†}-B{v}^{†}⟩$. By virtue of $x_{n+1}=pro{j}_C[x_n-{\mu }_n{\beta }_{n}B{y}_n+{\beta }_n(y_n-x_n)]$, we obtain

$$\begin{array}{c}\hfill 0\le ⟨v^{†}-x_{n+1},{\mu }_n{\beta }_{n}B{y}_n+x_{n+1}-x_n-{\beta }_n(y_n-x_n)⟩,\end{array}$$

that is,

$$\begin{array}{c}\hfill 0\le ⟨v^{†}-x_{n+1},B{y}_n+\frac{x_{n+1}-x_n}{{\beta }_n{\mu }_n}-\frac{y_n-x_n}{{\mu }_n}⟩.\end{array}$$

Therefore, we have

$$\begin{array}{ccc}\hfill ⟨v^{†}-x_{n_i+1},w^{†}⟩& \ge & ⟨v^{†}-x_{n_i+1},B{v}^{†}⟩\hfill \\ & \ge ⟨v^{†}-x_{n_i+1},B{v}^{†}⟩-⟨v^{†}-x_{n_i+1},\frac{x_{n_i+1}-x_{n_i}}{{\beta }_{n_i}{\mu }_{n_i}}+B{y}_{n_i}-\frac{y_{n_i}-x_{n_i}}{{\mu }_{n_i}}⟩\hfill \\ & =⟨v^{†}-x_{n_i+1},B{v}^{†}-B{y}_{n_i}-\frac{x_{n_i+1}-x_{n_i}}{{\beta }_{n_i}{\mu }_{n_i}}+\frac{y_{n_i}-x_{n_i}}{{\mu }_{n_i}}⟩\hfill \\ & =⟨v^{†}-x_{n_i+1},B{v}^{†}-B{y}_{n_i}⟩+⟨v^{†}-x_{n_i+1},B{x}_{n_i+1}-B{y}_{n_i}⟩\hfill \\ & & -⟨v^{†}-x_{n_i+1},\frac{x_{n_i+1}-x_{n_i}}{{\beta }_{n_i}{\mu }_{n_i}}-\frac{y_{n_i}-x_{n_i}}{{\mu }_{n_i}}⟩\hfill \\ & \ge ⟨v^{†}-x_{n_i+1},B{x}_{n_i+1}-B{y}_{n_i}⟩-⟨v-x_{n_i+1},\frac{x_{n_i+1}-x_{n_i}}{{\beta }_{n_i}{\mu }_{n_i}}-\frac{y_{n_i}-x_{n_i}}{{\mu }_{n_i}}⟩.\hfill \end{array}$$

Noting that $\parallel y_{n_i}-x_{n_i}\parallel \to 0$ and $\parallel x_{n_i+1}-x_{n_i}\parallel \to 0$, we get $0\le ⟨v^{†}-z,w^{†}⟩$. Hence, $z\in {T^{†}}^{-1}\left(0\right)$ and $z\in$ VI$(C,B)$. Thus, $z\in \Omega$ and we have

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim\; sup}⟨\tilde{x}-u,\tilde{x}-y_n⟩=\underset{i\to \infty }{lim}⟨\tilde{x}-u,\tilde{x}-y_{n_i}⟩=⟨\tilde{x}-u,\tilde{x}-z⟩\le 0.\end{array}$$

 □

Finally, we prove our main result.

Theorem 1.

Suppose that $\Omega =$ VI$(C,A)\cap$ VI$(C,B)\ne \varnothing$. Assume that $\left\{{\alpha }_n\right\}$, $\left\{{\beta }_n\right\}$, $\left\{{\lambda }_n\right\}$ and $\left\{{\mu }_n\right\}$ satisfy the following restrictions (C1)–(C4). Then $\left\{x_n\right\}$ defined by (5) converges strongly to $\tilde{x}=pro{j}_{\Omega }\left(u\right)$.

Proof. 

First, we have Propositions 2–4 in hand. In terms of (4), we have

$$\begin{array}{ccc}\hfill \parallel y_n-\tilde{x}{\parallel }^2& =& \parallel pro{j}_C[{\alpha }_{n}u+(1-{\alpha }_n)(x_n-{\lambda }_{n}A{x}_n)]-pro{j}_C[x^{\ast }-(1-{\alpha }_n){\lambda }_{n}A{x}^{\ast }]{\parallel }^2\hfill \\ & \le ⟨{\alpha }_n(u-x^{\ast })+(1-{\alpha }_n)[(x_n-{\lambda }_{n}A{x}_n)-(x^{\ast }-{\lambda }_{n}A{x}^{\ast })],y_n-x^{\ast }⟩\hfill \\ & \le {\alpha }_n⟨\tilde{x}-u,\tilde{x}-y_n⟩+(1-{\alpha }_n)\parallel (x_n-{\lambda }_{n}A{x}_n)-(\tilde{x}-{\lambda }_{n}A\tilde{x})\parallel \parallel y_n-\tilde{x}\parallel \hfill \\ & \le {\alpha }_n⟨\tilde{x}-u,\tilde{x}-y_n⟩+(1-{\alpha }_n)\parallel x_n-\tilde{x}\parallel \parallel y_n-\tilde{x}\parallel \hfill \\ & \le {\alpha }_n⟨\tilde{x}-u,\tilde{x}-y_n⟩+\frac{1-{\alpha }_n}{2}\parallel x_n-\tilde{x}{\parallel }^2+\frac{1}{2}{\parallel y_n-\tilde{x}\parallel }^2.\hfill \end{array}$$

It follows that

$$\begin{array}{c}\hfill \parallel y_n-\tilde{x}{\parallel }^2\le (1-{\alpha }_n){\parallel x_n-\tilde{x}\parallel }^2+2{\alpha }_n⟨\tilde{x}-u,\tilde{x}-y_n⟩.\end{array}$$

Therefore,

$$\begin{array}{ccc}\hfill \parallel x_{n+1}-\tilde{x}{\parallel }^2& \le & (1-{\beta }_n)\parallel x_n-\tilde{x}{\parallel }^2+{\beta }_n{\parallel y_n-\tilde{x}\parallel }^2\hfill \\ & \le (1-{\alpha }_n{\beta }_n){\parallel x_n-\tilde{x}\parallel }^2+2{\alpha }_n{\beta }_n⟨\tilde{x}-u,\tilde{x}-y_n⟩.\hfill \end{array}$$

By Lemma 3 and the above inequality, we deduce $x_n\to \tilde{x}$. This completes the proof.  □

If we take $u=0$, then we have the following Algorithm 2.

Algorithm 2:

For initial value $x_0\in C$. Assume the sequence $\left\{x_n\right\}$ has been constructed. Compute the next iteration $\left\{x_{n+1}\right\}$ by the following manner $$\left\{\begin{array}{cc}{y}_n=pro{j}_C\left[(1-{\alpha }_n)(x_n-{\lambda }_{n}A{x}_n)\right],\hfill & \\ x_{n+1}=pro{j}_C[x_n-{\mu }_n{\beta }_{n}B{y}_n+{\beta }_n(y_n-x_n)],\hfill & n\ge 0.\hfill \end{array}\right.$$ (11)

Corollary 1.

Suppose that $\Omega =$ VI$(C,A)\cap$ VI$(C,B)\ne \varnothing$. Assume that $\left\{{\alpha }_n\right\}$, $\left\{{\beta }_n\right\}$, $\left\{{\lambda }_n\right\}$ and $\left\{{\mu }_n\right\}$ satisfy the following restrictions (C1)–(C4). Then $\left\{x_n\right\}$ defined by (11) converges strongly to the minimum norm element $\tilde{x}$ in Ω.

4. Conclusions

In this paper, we investigated the variational inequality problem. We suggest an extragradient type method for finding the common solutions of two variational inequalities. We prove the strong convergence of the method under the mild conditions. Noting that in our suggested iterative sequence (Equation (5)), the involved operators A and B require some form of strong monotonicity. A natural question arises, i.e., how to weaken these assumptions?