Strong Convergence of Extragradient-Type Method to Solve Pseudomonotone Variational Inequalities Problems

Abstract

A number of applications from mathematical programmings, such as minimax problems, penalization methods and fixed-point problems can be formulated as a variational inequality model. Most of the techniques used to solve such problems involve iterative algorithms, and that is why, in this paper, we introduce a new extragradient-like method to solve the problems of variational inequalities in real Hilbert space involving pseudomonotone operators. The method has a clear advantage because of a variable stepsize formula that is revised on each iteration based on the previous iterations. The key advantage of the method is that it works without the prior knowledge of the Lipschitz constant. Strong convergence of the method is proved under mild conditions. Several numerical experiments are reported to show the numerical behaviour of the method.

1. Introduction

In this article, we consider the classic variational inequalities problems (VIPs) [1,2] for an operator $\mathcal{F}:\mathcal{E}\to \mathcal{E}$ is formulated in the following way:

$$\mathrm{Find}{u}^{*}\in \mathcal{K}\mathrm{such}\mathrm{that}〈\mathcal{F}\left(u^{*}\right),y-u^{*}〉\ge 0,\forall y\in \mathcal{K},$$

(1)

where $\mathcal{K}$ is a nonempty, convex and closed subset of a real Hilbert space $\mathcal{E}.$ The inner product and induced norm on $\mathcal{E}$ are denoted by $\langle .,.\rangle$ and $\parallel .\parallel$, respectively. Moreover, the set of real and natural numbers are denoted by $\mathcal{R}$ and $\mathcal{N}$, respectively. It is important to note that solving the problem (1) is equivalent to solving the following problem:

$$\mathrm{Find}\mathrm{an}\mathrm{element}{u}^{*}\in \mathcal{K}\mathrm{such}\mathrm{that}{u}^{*}=P_{\mathcal{K}}[u^{*}-\zeta \mathcal{F}\left(u^{*}\right)].$$

We assume that the following requirements have been fulfilled:

(B1)

The solution set of the problem (1), represented by SVIP is nonempty.

(B2)

A mapping $\mathcal{F}:\mathcal{E}\to \mathcal{E}$ is called to be pseudomonotone, i.e.,

$$〈\mathcal{F}\left(y_1\right),y_2-y_1〉\ge 0⟹〈\mathcal{F}\left(y_2\right),y_1-y_2〉\le 0,\forall y_1,y_2\in \mathcal{K}.$$

(B3)

A mapping $\mathcal{F}:\mathcal{E}\to \mathcal{E}$ is said to be Lipschitz continuous, i.e., there exists $L>0$ such that

$$\parallel \mathcal{F}\left(y_1\right)-\mathcal{F}\left(y_2\right)\parallel \le L\parallel y_1-y_2\parallel ,\forall y_1,y_2\in \mathcal{K}.$$

(B4)

A mapping $\mathcal{F}:\mathcal{E}\to \mathcal{E}$ is called to be sequentially weakly continuous, i.e., $\left\{\mathcal{F}\left(u_n\right)\right\}$ converges weakly to $\mathcal{F}\left(u\right)$, where $\left\{u_n\right\}$ weakly converges to u.

The concept of variational inequalities has been used as a powerful tool to study different subjects, i.e., physics, engineering, economics and optimization theory. The problem (1) was firstly introduced by Stampacchia [1] in 1964 and also provided that this problem (1) is a crucial problem in nonlinear analysis. This is an efficient mathematical technique that integrates several key elements of applied mathematics, i.e., the problems of network equilibrium, the necessary optimality conditions, the complementarity problems and the systems of non-linear equations (for more details [3,4,5,6,7,8,9]). On the other hand, the projection methods are important to find the numerical solution of variational inequalities. Many authors have proposed and studied different projection methods to solve the problem of variational inequalities (see for more details [10,11,12,13,14,15,16,17,18,19,20]) and others in [21,22,23,24,25,26,27,28,29,30,31,32]. In particular, Karpelevich [10] and Antipin [33] introduced the following extragradient method:

$$\left\{\begin{array}{c}{u}_n\in \mathcal{K},\hfill \\ v_n=P_{\mathcal{K}}[u_n-\zeta \mathcal{F}\left(u_n\right)],\hfill \\ u_{n+1}=P_{\mathcal{K}}[u_n-\zeta \mathcal{F}\left(v_n\right)].\hfill \end{array}\right.$$

(2)

Recently, the subgradient extragradient algorithm was established by Censor et al. [12] for solving problem (1) in real Hilbert space. Their method has the form

$$\left\{\begin{array}{c}{u}_n\in \mathcal{K},\hfill \\ v_n=P_{\mathcal{K}}[u_n-\zeta \mathcal{F}\left(u_n\right)],\hfill \\ u_{n+1}=P_{{\mathcal{E}}_n}[u_n-\zeta \mathcal{F}\left(v_n\right)].\hfill \end{array}\right.$$

(3)

where ${\mathcal{E}}_n=\{z\in \mathcal{E}:\langle u_n-\zeta \mathcal{F}\left(u_n\right)-v_n,z-v_n\rangle \le 0\}.$ Migorski et al. [34] proposed a viscosity-type subgradient extragradient method to solve monotone variational inequalities problems. The main contribution is the presence of a viscosity scheme in the algorithm that was used to improve the convergence rate of the iterative sequence and provide strong convergence theorem. The iterative sequence $\left\{u_n\right\}$ was generated in the following way: (i) Let $u_0\in \mathcal{K},$$\mu \in (0,1),$${\zeta }_0>0$ and a sequence ${\gamma }_n\subset (0,1)$ with ${\gamma }_n\to 0$ and ${\sum }_n^{\infty }{\gamma }_n=+\infty .$ (ii) Compute

$$\left\{\begin{array}{c}{v}_n=P_{\mathcal{K}}[u_n-{\zeta }_n\mathcal{F}\left(u_n\right)],\hfill \\ w_n=P_{{\mathcal{E}}_n}[u_n-{\zeta }_n\mathcal{F}\left(v_n\right)],\hfill \\ u_{n+1}={\gamma }_{n}f\left(u_n\right)+(1-{\gamma }_n)w_n,\hfill \end{array}\right.$$

(4)

where

$${\mathcal{E}}_n=\{z\in \mathcal{E}:\langle u_n-{\zeta }_n\mathcal{F}\left(u_n\right)-v_n,z-v_n\rangle \le 0\}.$$

(iii) Revised the stepsize in the following way:

$${\zeta }_{n+1}=\left\{\begin{array}{cc}min\left\{{\zeta }_n,\frac{\mu \parallel u_n-v_n\parallel }{\parallel \mathcal{F}\left(u_n\right)-\mathcal{F}\left(v_n\right)\parallel }\right\}\mathrm{if}\hfill & \mathcal{F}\left(u_n\right)\ne \mathcal{F}\left(v_n\right),\hfill \\ {\zeta }_n\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

In this paper, inspired by the iterative methods in [12,16,35,36], a modified subgradient extragradient algorithm is proposed for solving variational inequalities problems involving pseudomonotone mapping in real Hilbert space. It is important to note that our proposed scheme is effective. In particular, by comparing the results of Migorski et al. [34], our algorithm can solve pseudomonotone variational inequalities. Similar to the results of Migorski et al. [34] the proof of strong convergence of the proposed algorithm is proved without knowing the Lipschitz constant of the operator $\mathcal{F}.$ The proposed algorithm could be seen as a modification of the methods that are appeared in [10,12,34,35,36]. Under mild conditions, a strong convergence theorem is proved. Numerical experiments have been shown that the new approach tends to be more successful than the existing one [34].

The rest of this article has been arranged as follows: Section 2 contains some definitions and basic results that have been used throughout the paper. Section 3 contains our main algorithm and a strong convergence theorem. Section 4 presents the numerical results showing the algorithmic efficacy of the proposed method.

2. Preliminaries

This section contains useful lemmas and basic identities that have been used throughout the article. The metric projection $P_{\mathcal{K}}\left(u_1\right)$ for $u_1\in \mathcal{E}$ onto a closed and convex subset $\mathcal{K}$ of $\mathcal{E}$ is defined by

$$P_{\mathcal{K}}\left(u_1\right)=\underset{}{argmin}\{\parallel u_2-u_1\parallel :u_2\in \mathcal{K}\}.$$

Lemma 1.

[37,38]Assume $\mathcal{K}$ is a nonempty, convex and closed subset of a real Hilbert space $\mathcal{E}$ and $P_{\mathcal{K}}:\mathcal{E}\to \mathcal{K}$ is a metric projection from $\mathcal{E}$ onto $\mathcal{K}$.

(i) Let $u_1\in \mathcal{K}$ and $u_2\in \mathcal{E},$ we have

$$\parallel u_1-P_{\mathcal{K}}\left(u_2\right){\parallel }^2+\parallel P_{\mathcal{K}}\left(u_2\right)-u_2{\parallel }^2\le {\parallel u_1-u_2\parallel }^2.$$

(ii) $u_3=P_{\mathcal{K}}\left(u_1\right)$ if and only if

$$\langle u_1-u_3,u_2-u_3\rangle \le 0,\forall u_2\in \mathcal{K}.$$

(iii) For $u_2\in \mathcal{K}$ and $u_1\in \mathcal{E}$

$$\parallel u_1-P_{\mathcal{K}}\left(u_1\right)\parallel \le \parallel u_1-u_2\parallel .$$

Lemma 2.

[37] Let $u,v\in \mathcal{E}$ and $\varpi \in \mathcal{R}.$

(i) ${\parallel \varpi u+(1-\varpi )v\parallel }^2={\varpi \parallel u\parallel }^2+{(1-\varpi )\parallel v\parallel }^2-\varpi (1-\varpi ){\parallel u-v\parallel }^2.$

(ii) ${\parallel u+v\parallel }^2\le {\parallel u\parallel }^2+2\langle v,u+v\rangle$.

Lemma 3.

[39] Assume that $\left\{{\chi }_n\right\}$ be a sequence of non-negative real numbers satisfying

$${\chi }_{n+1}\le (1-{\tau }_n){\chi }_n+{\tau }_n{\delta }_n,\forall n\in \mathcal{N},$$

where $\left\{{\tau }_n\right\}\subset (0,1)$ and $\left\{{\delta }_n\right\}\subset \mathcal{R}$ satisfy the following conditions:

$$\underset{n\to \infty }{lim}{\tau }_n=0,\sum _{n=1}^{\infty }{\tau }_n=\infty ,\mathrm{and}\underset{n\to \infty }{lim\; sup}{\delta }_n\le 0.$$

Then, ${lim}_{n\to \infty }{\chi }_n=0.$

Lemma 4.

[40] Assume that $\left\{{\chi }_n\right\}$ is a sequence of real numbers such that there exists a subsequence $\left\{n_i\right\}$ of $\left\{n\right\}$ such that ${\chi }_{n_i}<{\chi }_{n_{i+1}}$ for all $i\in \mathcal{N}.$ Then, there exists a non decreasing sequence $m_k\subset \mathcal{N}$ such that $m_k\to \infty$ as $k\to \infty ,$ and the following conditions are fulfilled by all (sufficiently large) numbers $k\in \mathcal{N}$:

$${\chi }_{m_k}\le {\chi }_{m_{k+1}}and{\chi }_k\le {\chi }_{m_{k+1}}.$$

In fact, $m_k=max\{j\le k:{\chi }_j\le {\chi }_{j+1}\}.$

Lemma 5.

[41] Assume that $\mathcal{F}:\mathcal{K}\to \mathcal{E}$ is a pseudomonotone and continuous mapping. Then, $u^{*}$ is a solution of the problem (1) if and only if $u^{*}$ is a solution of the following problem.

$$\mathrm{Find}x\in \mathcal{K}\mathrm{such}\mathrm{that}\langle \mathcal{F}\left(y\right),y-x\rangle \ge 0,\forall y\in \mathcal{K}.$$

3. Main Results

We provide a method consisting of two convex minimization problems through a viscosity scheme and an explicit stepsize formula which is being used to improve the convergence rate of the iterative sequence and to make the method independent of the Lipschitz constants. The detailed method is provided in Algorithm 1.

Algorithm 1 (Explicit method for pseudomonotone variational inequalities problems).

Step 0: Let $u_0\in \mathcal{K},$$\mu \in (0,1),$${\zeta }_0>0$ and a sequence ${\gamma }_n\subset (0,1)$ satisfying

$$\underset{n\to \infty }{lim}{\gamma }_n=0\mathrm{and}\sum _n^{\infty }{\gamma }_n=+\infty .$$

Step 1: Evaluate $$v_n=P_{\mathcal{K}}[u_n-{\zeta }_n\mathcal{F}\left(u_n\right)].$$

If $u_n=v_n$; STOP. Otherwise, go to Step 2.

Step 2: Evaluate

$$w_n=P_{{\mathcal{E}}_n}[u_n-{\zeta }_n\mathcal{F}\left(v_n\right)],$$

where ${\mathcal{E}}_n=\{z\in \mathcal{E}:\langle u_n-{\zeta }_n\mathcal{F}\left(u_n\right)-v_n,z-v_n\rangle \le 0\}.$

Step 3: Compute

$$u_{n+1}={\gamma }_{n}f\left(u_n\right)+(1-{\gamma }_n)w_n.$$

Step 4: Evaluate

$${\zeta }_{n+1}=\left\{\begin{array}{cc}min\left\{{\zeta }_n,\frac{\mu \parallel u_n-v_n{\parallel }^2+\mu {\parallel w_n-v_n\parallel }^2}{2〈\mathcal{F}\left(u_n\right)-\mathcal{F}\left(v_n\right),w_n-v_n〉}\right\}\mathrm{if}\hfill & 〈\mathcal{F}\left(u_n\right)-\mathcal{F}\left(v_n\right),w_n-v_n〉>0,\hfill \\ {\zeta }_n\hfill & \mathrm{else}.\hfill \end{array}\right\}$$

Lemma 6.

The stepsize sequence $\left\{{\zeta }_n\right\}$ is monotonically decreasing with a lower bound $min\left\{\frac{\mu }{L},{\zeta }_0\right\}$ and converges to a fixed $\zeta >0.$

Proof. 

Let $〈\mathcal{F}\left(u_n\right)-\mathcal{F}\left(v_n\right),w_n-v_n〉>0,$ such that

$$\begin{array}{cc}\hfill \frac{\mu (\parallel u_n-v_n{\parallel }^2+\parallel w_n-v_n{\parallel }^2)}{2〈\mathcal{F}\left(u_n\right)-\mathcal{F}\left(v_n\right),w_n-v_n〉}& \ge \frac{2\mu \parallel u_n-v_n\parallel \parallel w_n-v_n\parallel }{2\parallel \mathcal{F}\left(u_n\right)-\mathcal{F}\left(v_n\right)\parallel \parallel w_n-v_n\parallel }\hfill \\ \hfill & \ge \frac{2\mu \parallel u_n-v_n\parallel \parallel w_n-v_n\parallel }{2\parallel u_n-v_n\parallel \parallel w_n-v_n\parallel }\hfill \\ \hfill & \ge \frac{\mu }{L}.\hfill \end{array}$$

(5)

Clearly, from above we can conclude that $\left\{{\zeta }_n\right\}$ has a lower bound $min\left\{\frac{\mu }{L},{\zeta }_0\right\}.$ Moreover, there exists a real number $\zeta >0,$ such that ${lim}_{n\to \infty }{\zeta }_n=\zeta .$ □

Lemma 7.

Assume that $\mathcal{F}:\mathcal{E}\to \mathcal{E}$ satisfies the conditions(B1)–(B4). For a given $u^{*}\in SVIP\ne \varnothing ,$ we have

$$\parallel w_n-u^{*}{\parallel }^2\le \parallel u_n-u^{*}{\parallel }^2-\left(1-\frac{\mu {\zeta }_n}{{\zeta }_{n+1}}\right)\parallel u_n-v_n{\parallel }^2-\left(1-\frac{\mu {\zeta }_n}{{\zeta }_{n+1}}\right){\parallel w_n-v_n\parallel }^2.$$

Proof. 

Consider that

$$\begin{array}{cc}\hfill {∥w_n-u^{*}∥}^2& ={∥P_{{\mathcal{E}}_n}[u_n-{\zeta }_n\mathcal{F}\left(v_n\right)]-u^{*}∥}^2\hfill \\ \hfill & ={∥P_{{\mathcal{E}}_n}[u_n-{\zeta }_n\mathcal{F}\left(v_n\right)]+[u_n-{\zeta }_n\mathcal{F}\left(v_n\right)]-[u_n-{\zeta }_n\mathcal{F}\left(v_n\right)]-u^{*}∥}^2\hfill \\ \hfill & ={∥[u_n-{\zeta }_n\mathcal{F}\left(v_n\right)]-u^{*}∥}^2+{∥P_{{\mathcal{E}}_n}[u_n-{\zeta }_n\mathcal{F}\left(v_n\right)]-[u_n-{\zeta }_n\mathcal{F}\left(v_n\right)]∥}^2\hfill \\ \hfill & +2〈P_{{\mathcal{E}}_n}[u_n-{\zeta }_n\mathcal{F}\left(v_n\right)]-[u_n-{\zeta }_n\mathcal{F}\left(v_n\right)],[u_n-{\zeta }_n\mathcal{F}\left(v_n\right)]-u^{*}〉.\hfill \end{array}$$

(6)

Given that $u^{*}\in SVIP\subset \mathcal{K}\subset {\mathcal{E}}_n,$ we get

$$\begin{array}{cc}\hfill & {∥P_{{\mathcal{E}}_n}[u_n-{\zeta }_n\mathcal{F}\left(v_n\right)]-[u_n-{\zeta }_n\mathcal{F}\left(v_n\right)]∥}^2\hfill \\ \hfill & +〈P_{{\mathcal{E}}_n}[u_n-{\zeta }_n\mathcal{F}\left(v_n\right)]-[u_n-{\zeta }_n\mathcal{F}\left(v_n\right)],[u_n-{\zeta }_n\mathcal{F}\left(v_n\right)]-u^{*}〉\hfill \\ \hfill & =〈[u_n-{\zeta }_n\mathcal{F}\left(v_n\right)]-P_{{\mathcal{E}}_n}[u_n-{\zeta }_n\mathcal{F}\left(v_n\right)],u^{*}-P_{{\mathcal{E}}_n}[u_n-{\zeta }_n\mathcal{F}\left(v_n\right)]〉\le 0,\hfill \end{array}$$

(7)

which implies that

$$\begin{array}{cc}\hfill & 〈P_{{\mathcal{E}}_n}[u_n-{\zeta }_n\mathcal{F}\left(v_n\right)]-[u_n-{\zeta }_n\mathcal{F}\left(v_n\right)],[u_n-{\zeta }_n\mathcal{F}\left(v_n\right)]-u^{*}〉\hfill \\ \hfill & \le -{∥P_{{\mathcal{E}}_n}[u_n-{\zeta }_n\mathcal{F}\left(v_n\right)]-[u_n-{\zeta }_n\mathcal{F}\left(v_n\right)]∥}^2.\hfill \end{array}$$

(8)

Using expressions (6) and (8), we obtain

$$\begin{array}{cc}\hfill \parallel w_n-u^{*}{\parallel }^2& \le {∥u_n-{\zeta }_n\mathcal{F}\left(v_n\right)-u^{*}∥}^2-{∥P_{{\mathcal{E}}_n}[u_n-{\zeta }_n\mathcal{F}\left(v_n\right)]-[u_n-{\zeta }_n\mathcal{F}\left(v_n\right)]∥}^2\hfill \\ & \le \parallel u_n-u^{*}{\parallel }^2-{\parallel u_n-w_n\parallel }^2+2{\zeta }_n〈\mathcal{F}\left(v_n\right),u^{*}-w_n〉.\hfill \end{array}$$

(9)

Since $u^{*}$ is the solution of problem (1), we have

$$\langle \mathcal{F}\left(u^{*}\right),y-u^{*}\rangle \ge 0,\mathrm{for}\mathrm{all}y\in \mathcal{K}.$$

Due to the pseudomonotonicity of $\mathcal{F}$ on $\mathcal{K}$, we get

$$\langle \mathcal{F}\left(y\right),y-u^{*}\rangle \ge 0,\mathrm{for}\mathrm{all}y\in \mathcal{K}.$$

By substituting $y=v_n\in \mathcal{K},$ we get

$$\langle \mathcal{F}\left(v_n\right),v_n-u^{*}\rangle \ge 0.$$

Thus, we have

$$〈\mathcal{F}\left(v_n\right),u^{*}-w_n〉=〈\mathcal{F}\left(v_n\right),u^{*}-v_n〉+〈\mathcal{F}\left(v_n\right),v_n-w_n〉\le 〈\mathcal{F}\left(v_n\right),v_n-w_n〉.$$

(10)

Combining expressions (9) and (10), we obtain

$$\begin{array}{cc}\hfill \parallel w_n-u^{*}{\parallel }^2& \le \parallel u_n-u^{*}{\parallel }^2-{\parallel u_n-w_n\parallel }^2+2{\zeta }_n〈\mathcal{F}\left(v_n\right),v_n-w_n〉\hfill \\ & \le \parallel u_n-u^{*}{\parallel }^2-{\parallel u_n-v_n+v_n-w_n\parallel }^2+2{\zeta }_n〈\mathcal{F}\left(v_n\right),v_n-w_n〉\hfill \\ & \le \parallel u_n-u^{*}{\parallel }^2-\parallel u_n-v_n{\parallel }^2-{\parallel v_n-w_n\parallel }^2+2〈u_n-{\zeta }_n\mathcal{F}\left(v_n\right)-v_n,w_n-v_n〉.\hfill \end{array}$$

(11)

Note that $w_n=P_{{\mathcal{E}}_n}[u_n-{\zeta }_n\mathcal{F}\left(v_n\right)]$ and by the definition of ${\zeta }_{n+1}$, we have

$$\begin{array}{cc}\hfill & 2〈u_n-{\zeta }_n\mathcal{F}\left(v_n\right)-v_n,w_n-v_n〉\hfill \\ \hfill & =2〈u_n-{\zeta }_n\mathcal{F}\left(u_n\right)-v_n,w_n-v_n〉+2{\zeta }_n〈\mathcal{F}\left(u_n\right)-\mathcal{F}\left(v_n\right),w_n-v_n〉\hfill \\ \hfill & \le \frac{2{\zeta }_n}{{\zeta }_{n+1}}{\zeta }_{n+1}〈\mathcal{F}\left(u_n\right)-\mathcal{F}\left(v_n\right),w_n-v_n〉\le \frac{{\zeta }_n}{{\zeta }_{n+1}}\left[\mu \parallel u_n-v_n{\parallel }^2+\mu {\parallel w_n-v_n\parallel }^2\right].\hfill \end{array}$$

(12)

Combining expressions (11) and (12), we obtain

$$\begin{array}{cc}\hfill & \parallel w_n-u^{*}{\parallel }^2\hfill \\ \hfill & \le \parallel u_n-u^{*}{\parallel }^2-\parallel u_n-v_n{\parallel }^2-{\parallel v_n-w_n\parallel }^2+\frac{{\zeta }_n}{{\zeta }_{n+1}}\left[\mu \parallel u_n-v_n{\parallel }^2+\mu {\parallel w_n-v_n\parallel }^2\right]\hfill \\ \hfill & \le \parallel u_n-u^{*}{\parallel }^2-\left(1-\frac{\mu {\zeta }_n}{{\zeta }_{n+1}}\right)\parallel u_n-v_n{\parallel }^2-\left(1-\frac{\mu {\zeta }_n}{{\zeta }_{n+1}}\right){\parallel w_n-v_n\parallel }^2.\hfill \end{array}$$

(13)

□

Lemma 8.

Suppose that conditions (B1)–(B4) hold. Let $\left\{u_n\right\}$ be a sequence generated by Algorithm 1. If there is a subsequence $\left\{u_{n_k}\right\}$ which is weakly convergent to $\widehat{u}\in \mathcal{E}$ and ${lim}_{n\to \infty }\parallel u_n-v_n\parallel =0,$ then $\widehat{u}\in SVIP.$

Proof. 

We have

$$v_{n_k}=P_{\mathcal{K}}[u_{n_k}-{\zeta }_{n_k}\mathcal{F}\left(u_{n_k}\right)],$$

(14)

which is equivalent to

$$\langle u_{n_k}-{\zeta }_{n_k}\mathcal{F}\left(u_{n_k}\right)-v_{n_k},y-v_{n_k}\rangle \le 0,\forall y\in \mathcal{K}.$$

(15)

From expression (15), we can write

$$\langle u_{n_k}-v_{n_k},y-v_{n_k}\rangle \le {\zeta }_{n_k}\langle \mathcal{F}\left(u_{n_k}\right),y-v_{n_k}\rangle ,\forall y\in \mathcal{K}.$$

(16)

Therefore, we get

$$\frac{1}{{\zeta }_{n_k}}\langle u_{n_k}-v_{n_k},y-v_{n_k}\rangle +\langle \mathcal{F}\left(u_{n_k}\right),v_{n_k}-u_{n_k}\rangle \le \langle \mathcal{F}\left(u_{n_k}\right),y-u_{n_k}\rangle ,\forall y\in \mathcal{K}.$$

(17)

Due to the boundedness of the sequence $\left\{u_{n_k}\right\}$ so does $\left\{\mathcal{F}\left(u_{n_k}\right)\right\}.$ By using the facts ${lim}_{n\to \infty }\parallel u_{n_k}-v_{n_k}\parallel =0,$ and ${lim}_{k\to \infty }{\zeta }_{n_k}=\zeta >0,$ limit as $k\to \infty$ in (17), we get

$$\underset{k\to \infty }{lim\; inf}\langle \mathcal{F}\left(u_{n_k}\right),y-u_{n_k}\rangle \ge 0,\forall y\in \mathcal{K}.$$

(18)

Moreover, we have

$$\begin{array}{cc}\hfill \langle \mathcal{F}\left(v_{n_k}\right),y-v_{n_k}\rangle & =\langle \mathcal{F}\left(v_{n_k}\right)-\mathcal{F}\left(u_{n_k}\right),y-u_{n_k}\rangle +\langle \mathcal{F}\left(u_{n_k}\right),y-u_{n_k}\rangle +\langle \mathcal{F}\left(v_{n_k}\right),u_{n_k}-v_{n_k}\rangle .\hfill \end{array}$$

(19)

Since ${lim}_{n\to \infty }\parallel u_{n_k}-v_{n_k}\parallel =0,$ and $\mathcal{F}$ is L-Lipschitz continuous on $\mathcal{E},$ we get

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}\parallel \mathcal{F}\left(u_{n_k}\right)-\mathcal{F}\left(v_{n_k}\right)\parallel =0.\end{array}$$

(20)

From (19) and (20), we obtain

$$\underset{k\to \infty }{lim\; inf}\langle \mathcal{F}\left(v_{n_k}\right),y-v_{n_k}\rangle \ge 0,\forall y\in \mathcal{K}.$$

(21)

Next, we show that $u^{*}\in SVIP$. We choose a sequence $\left\{{ϵ}_k\right\}$ of positive numbers decreasing and tending to 0. For each k, we denote by $m_k$ the smallest positive integer such that

$$\underset{k\to \infty }{lim\; inf}\langle \mathcal{F}\left(u_{n_i}\right),y-u_{n_i}\rangle +{ϵ}_k\ge 0,\forall i\ge m_k.$$

(22)

Due to $\left\{{ϵ}_k\right\}$ being decreasing, the sequence $\left\{m_k\right\}$ is increasing.

Case 1: If there is a subsequence $u_{n_{m_{k_j}}}$ of $u_{n_{m_k}}$ such that $\mathcal{F}\left(u_{n_{m_{k_j}}}\right)=0$ ($\forall j$). Letting $j\to \infty ,$ we obtain

$$\langle \mathcal{F}\left(u^{*}\right),y-u^{*}\rangle =\underset{j\to \infty }{lim}\langle \mathcal{F}(u_{n_{m_{k_j}}}),y-u^{*}\rangle =0.$$

(23)

Hence $u^{*}\in \mathcal{K}$, therefore we have $u^{*}\in SVIP$.

Case 2: If there exists $N_0$ such that for all $n_{m_k}\ge N_0$, $\mathcal{F}\left(u_{n_{m_k}}\right)\ne 0.$ Suppose that

$${\Theta }_{n_{m_k}}=\frac{\mathcal{F}(u_{n_{m_k}})}{\parallel \mathcal{F}\left(u_{n_{m_k}}\right){\parallel }^2},\forall n_{m_k}\ge N_0.$$

(24)

Due to the above definition, we obtain

$$\langle \mathcal{F}(u_{n_{m_k}}),\mathcal{F}({\Theta }_{n_{m_k}})\rangle =1,\forall n_{m_k}\ge N_0.$$

(25)

From (18) and (25), for all $n_{m_k}\ge N_0,$ we have

$$\langle \mathcal{F}\left(u_{n_{m_k}}\right),y+{ϵ}_k{\Theta }_{n_{m_k}}-u_{n_{m_k}}\rangle \ge 0.$$

(26)

Due to pseudomonotonicity of $\mathcal{F}$ for $n_{m_k}\ge N_0,$ we obtain

$$\langle \mathcal{F}(y+{ϵ}_k{\Theta }_{n_{m_k}}),y+{ϵ}_k{\Theta }_{n_{m_k}}-u_{n_{m_k}}\rangle \ge 0.$$

(27)

For all $n_{m_k}\ge N_0,$ we have

$$\langle \mathcal{F}\left(y\right),y-u_{n_{m_k}}\rangle \ge \langle \mathcal{F}\left(y\right)-\mathcal{F}(y+{ϵ}_k{\Theta }_{n_{m_k}}),y+{ϵ}_k{\Theta }_{n_{m_k}}-u_{n_{m_k}}\rangle -{ϵ}_k\langle \mathcal{F}\left(y\right),{\Theta }_{n_{m_k}}\rangle .$$

(28)

Since $\left\{u_{n_k}\right\}$ converges weakly to $u^{*}\in \mathcal{K}$ and $\mathcal{F}$ is sequentially weakly continuous on $\mathcal{K}$, we have $\left\{\mathcal{F}\left(u_{n_k}\right)\right\}$ converges weakly to $\mathcal{F}\left(u^{*}\right).$ We can suppose that $\mathcal{F}\left(u^{*}\right)\ne 0$. Since the norm mapping is sequentially weakly lower semicontinuous, we have

$$\parallel \mathcal{F}\left(u^{*}\right)\parallel \le \underset{k\to \infty }{lim\; inf}\parallel \mathcal{F}\left(u_{n_k}\right)\parallel .$$

(29)

Since $\left\{u_{n_{m_k}}\right\}\subset \left\{u_{n_k}\right\}$ and ${lim}_{k\to \infty }{ϵ}_k=0,$ we have

$$0\le \underset{k\to \infty }{lim}\parallel {ϵ}_k{\Theta }_{n_{m_k}}\parallel =\underset{k\to \infty }{lim}\frac{{ϵ}_k}{\parallel \mathcal{F}\left(u_{n_{m_k}}\right)\parallel }\le \frac{0}{\parallel \mathcal{F}\left(u^{*}\right)\parallel }=0.$$

(30)

Now, letting $k\to \infty$ in (28), we obtain

$$\langle \mathcal{F}\left(y\right),y-u^{*}\ge 0,\forall y\in \mathcal{K}.$$

(31)

Applying the well-known Lemma 5, we can deduce that $u^{*}\in SVIP.$ □

Theorem 1.

Assume that $\mathcal{F}:\mathcal{K}\to \mathcal{E}$ satisfies the conditions (B1)–(B4). Moreover, assume that $u^{*}$ belongs to the solution set $SVIP.$ Then, the sequences $\left\{u_n\right\},$$\left\{v_n\right\}$ and $\left\{w_n\right\}$ generated by Algorithm 1 converge strongly to $u^{*}.$

Proof. 

By using Lemma 7, we have

$$\begin{array}{cc}\hfill \parallel w_n-u^{*}{\parallel }^2& \le \parallel u_n-u^{*}{\parallel }^2-\left(1-\frac{\mu {\zeta }_n}{{\zeta }_{n+1}}\right)\parallel u_n-v_n{\parallel }^2-\left(1-\frac{\mu {\zeta }_n}{{\zeta }_{n+1}}\right){\parallel w_n-v_n\parallel }^2.\hfill \end{array}$$

(32)

Due to ${\zeta }_n\to \zeta ,$ there exists a fixed number $ϵ\in (0,1-\mu )$ such that

$$\underset{n\to \infty }{lim}\left(1-\frac{\mu {\zeta }_n}{{\zeta }_{n+1}}\right)=1-\mu >ϵ>0.$$

Then, there exists a finite number $N_1\in \mathcal{N}$ such that

$$\left(1-\frac{\mu {\zeta }_n}{{\zeta }_{n+1}}\right)>ϵ>0,\forall n\ge N_1.$$

(33)

Hence, we obtain

$$\begin{array}{c}\hfill \parallel w_n-u^{*}{\parallel }^2\le {\parallel u_n-u^{*}\parallel }^2,\forall n\ge N_1.\end{array}$$

(34)

From the definition of the sequence $\left\{u_{n+1}\right\}$ and the fact that f is a contraction with constant $\rho \in [0,1)$ and $n\ge N_1,$ we obtain

$$\begin{array}{cc}\hfill ∥u_{n+1}-u^{*}∥& =∥{\gamma }_{n}f\left(u_n\right)+(1-{\gamma }_n)w_n-u^{*}∥\hfill \\ \hfill & =∥{\gamma }_n[f\left(u_n\right)-u^{*}]+(1-{\gamma }_n)[w_n-u^{*}]∥\hfill \\ \hfill & =∥{\gamma }_n[f\left(u_n\right)+f\left(u^{*}\right)-f\left(u^{*}\right)-u^{*}]+(1-{\gamma }_n)[w_n-u^{*}]∥\hfill \\ \hfill & \le {\gamma }_n∥f\left(u_n\right)-f\left(u^{*}\right)∥+{\gamma }_n∥f\left(u^{*}\right)-u^{*}∥+(1-{\gamma }_n)∥w_n-u^{*}∥\hfill \\ \hfill & \le {\gamma }_n\rho ∥u_n-u^{*}∥+{\gamma }_n∥f\left(u^{*}\right)-u^{*}∥+(1-{\gamma }_n)∥w_n-u^{*}∥.\hfill \end{array}$$

(35)

From expressions (34) and (35) and ${\gamma }_n\subset (0,1)$, we obtain

$$\begin{array}{cc}\hfill ∥u_{n+1}-u^{*}∥& \le {\gamma }_n\rho ∥u_n-u^{*}∥+{\gamma }_n∥f\left(u^{*}\right)-u^{*}∥+(1-{\gamma }_n)∥u_n-u^{*}∥\hfill \\ \hfill & =[1-{\gamma }_n+\rho {\gamma }_n]∥u_n-u^{*}∥+{\gamma }_n(1-\rho )\frac{∥f\left(u^{*}\right)-u^{*}∥}{(1-\rho )}\hfill \\ \hfill & \le max\left\{∥u_n-u^{*}∥,\frac{∥f\left(u^{*}\right)-u^{*}∥}{(1-\rho )}\right\}\hfill \\ \hfill & \le max\left\{∥u_{N_1}-u^{*}∥,\frac{∥f\left(u^{*}\right)-u^{*}∥}{(1-\rho )}\right\}.\hfill \end{array}$$

(36)

Hence, we conclude that the sequence $\left\{u_n\right\}$ is bounded. Next, the reflexivity of $\mathcal{E}$ and the boundedness of the sequence $\left\{u_n\right\}$ guarantee that there exists a subsequence $\left\{u_{n_k}\right\}$ such that $\left\{u_{n_k}\right\}\rightharpoonup u^{*}\in \mathcal{E}$ as $k\to \infty .$ Now, we prove the strong convergence of the sequence iterative sequence $\left\{u_n\right\}$ generated by Algorithm 1. Due to the continuity and pseudomonotonicity of the operator $\mathcal{F}$ imply that the solution set $SVIP$ is a closed and convex set (for more details see [42,43]). Since the mapping f is a contraction, $P_{SVIP}\circ f$ is a contraction. The Banach contraction theorem guarantee the existence of a fixed point of $u^{*}\in SVIP$ such that

$$u^{*}=P_{SVIP}\left(f\left(u^{*}\right)\right).$$

By using Lemma 1 (ii), we have

$$\langle f\left(u^{*}\right)-u^{*},y-u^{*}\rangle \le 0,\forall y\in SVIP.$$

(37)

From given $u_{n+1}={\gamma }_{n}f\left(u_n\right)+(1-{\gamma }_n)w_n,$ and using Lemma 2 (i) and Lemma 7, we have

$$\begin{array}{cc}\hfill & {∥u_{n+1}-u^{*}∥}^2\hfill \\ \hfill & ={∥{\gamma }_{n}f\left(u_n\right)+(1-{\gamma }_n)w_n-u^{*}∥}^2\hfill \\ \hfill & ={∥{\gamma }_n[f\left(u_n\right)-u^{*}]+(1-{\gamma }_n)[w_n-u^{*}]∥}^2\hfill \\ \hfill & ={\gamma }_n\parallel f\left(u_n\right)-u^{*}{\parallel }^2+(1-{\gamma }_n)\parallel w_n-u^{*}{\parallel }^2-{\gamma }_n(1-{\gamma }_n){\parallel f\left(u_n\right)-w_n\parallel }^2\hfill \\ \hfill & \le {\gamma }_n\parallel f\left(u_n\right)-u^{*}{\parallel }^2+(1-{\gamma }_n)[\parallel u_n-u^{*}{\parallel }^2-\left(1-\frac{\mu {\zeta }_n}{{\zeta }_{n+1}}\right){\parallel u_n-v_n\parallel }^2\hfill \\ \hfill & -\left(1-\frac{\mu {\zeta }_n}{{\zeta }_{n+1}}\right)\parallel w_n-v_n{\parallel }^2]-{\gamma }_n(1-{\gamma }_n){\parallel f\left(u_n\right)-w_n\parallel }^2\hfill \\ \hfill & \le {\gamma }_n\parallel f\left(u_n\right)-u^{*}{\parallel }^2+{\parallel u_n-u^{*}\parallel }^2-(1-{\gamma }_n)\left(1-\frac{\mu {\zeta }_n}{{\zeta }_{n+1}}\right)\left[\parallel w_n-v_n{\parallel }^2+{\parallel u_n-v_n\parallel }^2\right].\hfill \end{array}$$

(38)

The rest of the proof shall be divided into the following two parts:

Case 1: Assume that there exists a fixed number $N_2\in \mathcal{N}$$(N_2\ge N_1)$ such that

$$\parallel u_{n+1}-u^{*}\parallel \le \parallel u_n-u^{*}\parallel ,\forall n\ge N_2.$$

(39)

Thus, ${lim}_{n\to \infty }\parallel u_n-u^{*}\parallel$ exists and let ${lim}_{n\to \infty }\parallel u_n-u^{*}\parallel =l.$ From expression (38), we have

$$\begin{array}{cc}\hfill & (1-{\gamma }_n)\left(1-\frac{\mu {\zeta }_n}{{\zeta }_{n+1}}\right)\left[\parallel w_n-v_n{\parallel }^2+{\parallel u_n-v_n\parallel }^2\right]\hfill \\ \hfill & \le {\gamma }_n\parallel f\left(u_n\right)-u^{*}{\parallel }^2+\parallel u_n-u^{*}{\parallel }^2-{\parallel u_{n+1}-u^{*}\parallel }^2.\hfill \end{array}$$

(40)

Due to the existence of ${lim}_{n\to \infty }\parallel u_n-u^{*}\parallel =l,$ and ${\gamma }_n\to 0$, we deduce that

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

(41)

From expression (41), we have

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

(42)

It follows that

$$\begin{array}{cc}\hfill ∥u_{n+1}-u_n∥& =∥{\gamma }_{n}f\left(u_n\right)+(1-{\gamma }_n)w_n-u_n∥\hfill \\ & =∥{\gamma }_n[f\left(u_n\right)-u_n]+(1-{\gamma }_n)[w_n-u_n]∥\hfill \\ & \le {\gamma }_n∥f\left(u_n\right)-u_n∥+(1-{\gamma }_n)∥w_n-u_n∥⟶0.\hfill \end{array}$$

(43)

Thus, the sequences $\left\{u_n\right\}$, $\left\{v_n\right\}$ and $\left\{w_n\right\}$ are bounded. Thus, we can take a subsequence $\left\{u_{n_k}\right\}$ of $\left\{u_n\right\}$ such that $\left\{u_{n_k}\right\}$ weakly converges to some $\widehat{u}\in \mathcal{E}.$ Moreover, due to $\parallel u_n-v_n\parallel \to 0$ and using Lemma 8, we have $\widehat{u}\in SVIP.$ By following expression (37), we consider that

$$\begin{array}{cc}\hfill & \underset{n\to \infty }{lim\; sup}\langle f\left(u^{*}\right)-u^{*},u_n-u^{*}\rangle \hfill \\ \hfill & =\underset{k\to \infty }{lim\; sup}\langle f\left(u^{*}\right)-u^{*},u_{n_k}-u^{*}\rangle =\langle f\left(u^{*}\right)-u^{*},\widehat{u}-u^{*}\rangle \le 0.\hfill \end{array}$$

(44)

We have ${lim}_{n\to \infty }∥u_{n+1}-u_n∥=0.$ It follows (44) that

$$\begin{array}{cc}\hfill & \underset{n\to \infty }{lim\; sup}\langle f\left(u^{*}\right)-u^{*},u_{n+1}-u^{*}\rangle \hfill \\ \hfill & \le \underset{k\to \infty }{lim\; sup}\langle f\left(u^{*}\right)-u^{*},u_{n+1}-u_n\rangle +\underset{k\to \infty }{lim\; sup}\langle f\left(u^{*}\right)-u^{*},u_n-u^{*}\rangle \le 0.\hfill \end{array}$$

(45)

From Lemma 2 (ii) and Lemma 7 for all $n\ge N_2$, we get

$$\begin{array}{cc}\hfill & {∥u_{n+1}-u^{*}∥}^2\hfill \\ \hfill & ={∥{\gamma }_{n}f\left(u_n\right)+(1-{\gamma }_n)w_n-u^{*}∥}^2\hfill \\ \hfill & ={∥{\gamma }_n[f\left(u_n\right)-u^{*}]+(1-{\gamma }_n)[w_n-u^{*}]∥}^2\hfill \\ \hfill & \le {(1-{\gamma }_n)}^2{∥w_n-u^{*}∥}^2+2{\gamma }_n\langle f\left(u_n\right)-u^{*},(1-{\gamma }_n)[w_n-u^{*}]+{\gamma }_n[f\left(u_n\right)-u^{*}]\rangle \hfill \\ \hfill & ={(1-{\gamma }_n)}^2{∥w_n-u^{*}∥}^2+2{\gamma }_n\langle f\left(u_n\right)-f\left(u^{*}\right)+f\left(u^{*}\right)-u^{*},u_{n+1}-u^{*}\rangle \hfill \\ \hfill & ={(1-{\gamma }_n)}^2{∥w_n-u^{*}∥}^2+2{\gamma }_n\langle f\left(u_n\right)-f\left(u^{*}\right),u_{n+1}-u^{*}\rangle +2{\gamma }_n\langle f\left(u^{*}\right)-u^{*},u_{n+1}-u^{*}\rangle \hfill \\ \hfill & \le {(1-{\gamma }_n)}^2{∥w_n-u^{*}∥}^2+2{\gamma }_n\rho ∥u_n-u^{*}∥∥u_{n+1}-u^{*}∥+2{\gamma }_n\langle f\left(u^{*}\right)-u^{*},u_{n+1}-u^{*}\rangle \hfill \\ \hfill & \le (1+{\gamma }_n^2-2{\gamma }_n){∥u_n-u^{*}∥}^2+2{\gamma }_n\rho {∥u_n-u^{*}∥}^2+2{\gamma }_n\langle f\left(u^{*}\right)-u^{*},u_{n+1}-u^{*}\rangle \hfill \\ \hfill & =(1-2{\gamma }_n){∥u_n-u^{*}∥}^2+{\gamma }_n^2{∥u_n-u^{*}∥}^2+2{\gamma }_n\rho {∥u_n-u^{*}∥}^2+2{\gamma }_n\langle f\left(u^{*}\right)-u^{*},u_{n+1}-u^{*}\rangle \hfill \\ \hfill & =\left[1-2{\gamma }_n(1-\rho )\right]{∥u_n-u^{*}∥}^2+2{\gamma }_n(1-\rho )\left[\frac{{\gamma }_n{∥u_n-u^{*}∥}^2}{2(1-\rho )}+\frac{\langle f\left(u^{*}\right)-u^{*},u_{n+1}-u^{*}\rangle }{1-\rho }\right].\hfill \end{array}$$

(46)

It follows from expressions (45) and (46), we obtain

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim\; sup}\left[\frac{{\gamma }_n{∥u_n-u^{*}∥}^2}{2(1-\rho )}+\frac{\langle f\left(u^{*}\right)-u^{*},u_{n+1}-u^{*}\rangle }{1-\rho }\right]\le 0.\end{array}$$

(47)

Choose $n\ge N_3\in \mathcal{N}$ ($N_3\ge N_2$) large enough such that $2{\gamma }_n(1-\rho )<1.$ Now, using expressions (46) and (47) and applying Lemma 3, we conclude that $∥u_n-u^{*}∥\to 0,$ as $n\to \infty .$

Case 2: Suppose that there exists a subsequence $\left\{n_i\right\}$ of $\left\{n\right\}$ such that

$$\parallel u_{n_i}-u^{*}\parallel \le \parallel u_{n_{i+1}}-u^{*}\parallel ,\forall i\in \mathcal{N}.$$

Thus, by Lemma 4, there exits a sequence $\left\{m_k\right\}\subset \mathcal{N}$ and $\left\{m_k\right\}\to \infty ,$ such that

$$\parallel u_{m_k}-u^{*}\parallel \le \parallel u_{m_{k+1}}-u^{*}\parallel \mathrm{and}\parallel u_k-u^{*}\parallel \le \parallel u_{m_{k+1}}-u^{*}\parallel ,\forall k\in \mathcal{N}.$$

(48)

Similar to Case 1, using (38), we have

$$\begin{array}{cc}\hfill & (1-{\gamma }_{m_k})\left(1-\frac{\mu {\zeta }_{m_k}}{{\zeta }_{m_k+1}}\right)\left[\parallel w_{m_k}-v_{m_k}{\parallel }^2+{\parallel u_{m_k}-v_{m_k}\parallel }^2\right]\hfill \\ \hfill & \le {\gamma }_{m_k}\parallel f\left(u_{m_k}\right)-u^{*}{\parallel }^2+\parallel u_{m_k}-u^{*}{\parallel }^2-{\parallel u_{m_k+1}-u^{*}\parallel }^2.\hfill \end{array}$$

(49)

Due to ${\gamma }_{m_k}\to 0$ and $\left(1-\frac{\mu {\zeta }_{m_k}}{{\zeta }_{m_k+1}}\right)\to 1-\mu ,$ we can deduce the following:

$$\underset{n\to \infty }{lim}\parallel u_{m_k}-v_{m_k}\parallel =\underset{k\to \infty }{lim}\parallel w_{m_k}-v_{m_k}\parallel =0.$$

(50)

From expression (50), we have

$$\underset{k\to \infty }{lim}\parallel u_{m_k}-w_{m_k}\parallel \le \underset{k\to \infty }{lim}\parallel u_{m_k}-v_{m_k}\parallel +\underset{k\to \infty }{lim}\parallel v_{m_k}-w_{m_k}\parallel =0.$$

(51)

Hence, we obtain

$$\begin{array}{cc}\hfill ∥u_{m_k+1}-u_{m_k}∥& =∥{\gamma }_{m_k}f\left(u_{m_k}\right)+(1-{\gamma }_{m_k})w_{m_k}-u_{m_k}∥\hfill \\ & =∥{\gamma }_{m_k}[f\left(u_{m_k}\right)-u_{m_k}]+(1-{\gamma }_{m_k})[w_{m_k}-u_{m_k}]∥\hfill \\ & \le {\gamma }_{m_k}∥f\left(u_{m_k}\right)-u_{m_k}∥+(1-{\gamma }_{m_k})∥w_{m_k}-u_{m_k}∥⟶0.\hfill \end{array}$$

(52)

We have to use the same justification as in the Case 1, such that

$$\begin{array}{c}\hfill \underset{k\to \infty }{lim\; sup}\langle f\left(u^{*}\right)-u^{*},u_{m_k+1}-u^{*}\rangle \le 0.\end{array}$$

(53)

Using (46) and (48), we have

$$\begin{array}{cc}\hfill {∥u_{m_k+1}-u^{*}∥}^2& \le \left[1-2{\gamma }_{m_k}(1-\rho )\right]{∥u_{m_k}-u^{*}∥}^2\hfill \\ & +2{\gamma }_{m_k}(1-\rho )\left[\frac{{\gamma }_{m_k}{∥u_{m_k}-u^{*}∥}^2}{2(1-\rho )}+\frac{\langle f\left(u^{*}\right)-u^{*},u_{m_k+1}-u^{*}\rangle }{1-\rho }\right]\hfill \\ & \le \left[1-2{\gamma }_{m_k}(1-\rho )\right]{∥u_{m_k+1}-u^{*}∥}^2\hfill \\ & +2{\gamma }_{m_k}(1-\rho )\left[\frac{{\gamma }_{m_k}{∥u_{m_k}-u^{*}∥}^2}{2(1-\rho )}+\frac{\langle f\left(u^{*}\right)-u^{*},u_{m_k+1}-u^{*}\rangle }{1-\rho }\right].\hfill \end{array}$$

(54)

It follows that

$$\begin{array}{cc}\hfill {∥u_{m_k+1}-u^{*}∥}^2& \le \frac{{\gamma }_{m_k}{∥u_{m_k}-u^{*}∥}^2}{2(1-\rho )}+\frac{\langle f\left(u^{*}\right)-u^{*},u_{m_k+1}-u^{*}\rangle }{1-\rho }.\hfill \\ \hfill \end{array}$$

(55)

Since ${\gamma }_{m_k}\to 0$ and $∥u_{m_k}-u^{*}∥$ is a bounded sequence. Thus, expressions (53) and (55) implies that

$$\parallel u_{m_k+1}-u^{*}{\parallel }^2\to 0,\mathrm{as}k\to \infty .$$

(56)

From the inequality (48), we have

$$\underset{n\to \infty }{lim}\parallel u_k-u^{*}{\parallel }^2\le \underset{n\to \infty }{lim}{\parallel u_{m_k+1}-u^{*}\parallel }^2\le 0.$$

(57)

Consequently, $u_n\to u^{*}.$ This completes the proof of the theorem. □

4. Numerical Experiments

Numerical investigations present in this section to demonstrate the efficiency of the introduced Algorithm 1 in four test problems, all of which are pseudomonotone. The MATLAB program has been performed on a PC (with Intel(R) Core(TM)i3-4010U CPU @ 1.70 GHz, RAM 4.00 GB) in MATLAB version 9.5 (R2018b). We use the built-in MATLAB Quadratic programming to solve the minimization problems.

Example 1.

Consider the non-linear complementarity problem of Kojima-–Shindo where the feasible set $\mathcal{K}$ which is defined by

$$\mathcal{K}=\{u\in {\mathcal{R}}^4:1\le u_i\le 5,i=1,2,3,4\}.$$

The mapping $\mathcal{F}:{\mathcal{R}}^4\to {\mathcal{R}}^4$ is defined by

$$\begin{array}{c}\hfill \mathcal{F}\left(u\right)=\left(\begin{array}{c}{u}_1+u_2+u_3+u_4-4u_2{u}_3{u}_4\\ u_1+u_2+u_3+u_4-4u_1{u}_3{u}_4\\ u_1+u_2+u_3+u_4-4u_1{u}_2{u}_4\\ u_1+u_2+u_3+u_4-4u_1{u}_2{u}_3\end{array}\right).\end{array}$$

It is easy to see that $\mathcal{F}$ is not monotone on the set $\mathcal{K}.$ By using the Monte Carlo approach [44], it can be shown that $\mathcal{F}$ is pseudomonotone on $\mathcal{K},$ This problem has a unique solution $u^{*}={(5,5,5,5)}^T.$ Generate many pairs of points u and v uniformly in $\mathcal{K}$ satisfying $\mathcal{F}{\left(u\right)}^T(v-u)\ge 0$ and then check if $\mathcal{F}{\left(v\right)}^T(v-u)\ge 0$. In this experiment, we take different initial points and $D_n=\parallel u_n-v_n\parallel .$ Moreover, control parameters ${\zeta }_0=0.33,$ $\mu =0.25,$${\gamma }_n=\frac{1}{100(n+2)}$ and $f\left(u\right)=\frac{u}{2}$ for Algorithm 1. Numerical investigation regarding the first example was shown in

Table 1. Numerical behaviour of Algorithm 1 using different starting points for Example 1.

TOL	${10}^{-2}$	${10}^{-3}$	${10}^{-4}$	${10}^{-5}$	${10}^{-2}$	${10}^{-3}$	${10}^{-4}$	${10}^{-5}$
${\mathit{u}}_0$	Iter.	Iter.	Iter.	Iter.	Time	Time	Time	Time
${[-2,2,8,10]}^T$	13	51	501	5001	0.079821	0.247776	3.251465	43.637834
${[-1,1,5,6]}^T$	12	51	501	5001	0.083870	0.236924	2.684370	39.651178
${[-5,2,-1,2]}^T$	9	51	501	5001	0.065422	0.235173	3.034747	43.630625
${[1,2,3,4]}^T$	6	1004	1004	5001	0.040866	8.051234	6.686632	42.431705

.

Example 2.

Consider the quadratic fractional programming problem in the following form [44]:

$$\begin{array}{c}\left\{\begin{array}{c}\begin{array}{c}minf\left(u\right)=\frac{u^{T}Qu+a^{T}u+a_0}{b^{T}u+b_0},\hfill \\ Subjecttou\in \mathcal{K}=\{u\in {\mathcal{R}}^4:b^{T}u+b_0>0\},\hfill \end{array}\hfill \end{array}\right.\hfill \end{array}$$

where

$$\begin{array}{c}\hfill Q=\left(\begin{array}{cccc}5& -1& 2& 0\\ -1& 5& -1& 3\\ 2& -1& 3& 0\\ 0& 3& 0& 5\end{array}\right),a=\left(\begin{array}{c}1\\ -2\\ -2\\ 1\end{array}\right),b=\left(\begin{array}{c}2\\ 1\\ 1\\ 0\end{array}\right),a_0=-2,and{b}_0=4.\end{array}$$

It is easy to verify that Q is symmetric and positive definite on ${\mathcal{R}}^4$ and consequently f is pseudo-convex on $\mathcal{K}$. Therefore, $\nabla f$ is pseudomonotone. Using the quotient rule, we obtain

$$\nabla f\left(u\right)=\frac{(b^{T}u+b_0)(2Qu+a)-b(u^{T}Q+a^{T}u+a_0)}{{(b^{T}u+b_0)}^2}.$$

(58)

In this point of view, we can set $\mathcal{F}=\nabla f$ in Theorem 1. We minimize f over $\mathcal{K}=\{u\in {\mathcal{R}}^4:1\le u_i\le 10,i=1,2,3,4\}$. This problem has a unique solution $u^{*}={(1,1,1,1)}^T\in \mathcal{K}.$ In this experiment, we take different initial points and $D_n=\parallel u_n-v_n\parallel .$ Moreover, control parameters ${\zeta }_0=0.33,$ $\mu =0.25,$${\gamma }_n=\frac{1}{100(n+2)}$ and $f\left(u\right)=\frac{u}{2}$ for Algorithm 1. Numerical investigation regarding the second example is shown in

Table 2. Numerical behaviour of Algorithm 1 using different starting points for Example 2.

TOL	${10}^{-2}$	${10}^{-3}$	${10}^{-4}$	${10}^{-5}$	${10}^{-2}$	${10}^{-3}$	${10}^{-4}$	${10}^{-5}$
${\mathit{u}}_0$	Iter.	Iter.	Iter.	Iter.	Time	Time	Time	Time
${[10,10,10,10]}^T$	43	46	99	989	0.289149	0.249285	0.475520	8.480530
${[10,20,30,40]}^T$	41	46	99	989	0.211707	0.187559	0.445240	6.898924
${[20,-20,20,-20]}^T$	29	32	99	989	0.138575	0.169190	0.394654	7.168460

.

Example 3.

The third example was taken from [45] where $\mathcal{F}:{\mathcal{R}}^2\to {\mathcal{R}}^2$ is defined by

$$\mathcal{F}\left(u\right)=\left(\begin{array}{c}0.5u_1{u}_2-2u_2-{10}^7\\ -4u_1-0.1u_2^2-{10}^7\end{array}\right),$$

on $\mathcal{K}=\{u\in {\mathcal{R}}^2:{(u_1-2)}^2+{(u_2-2)}^2\le 1\}.$ It can easily see that $\mathcal{F}$ is Lipschitz continuous with $L=5$ and $\mathcal{F}$ is not monotone on $\mathcal{K}$ but pseudomonotone. The above problem has a unique solution $u^{*}={(2.707,2.707)}^T.$ In this experiment, we take different initial points and $D_n=\parallel u_n-v_n\parallel .$ Moreover, control parameters ${\zeta }_0=0.33,$ $\mu =0.25,$${\gamma }_n=\frac{1}{100(n+2)}$ and $f\left(u\right)=\frac{u}{3}$ for Algorithm 1. Numerical investigations regarding the third example is shown in

Table 3. Numerical behaviour of Algorithm 1 using different starting points for Example 3.

TOL	${10}^{-2}$	${10}^{-3}$	${10}^{-4}$	${10}^{-5}$	${10}^{-2}$	${10}^{-3}$	${10}^{-4}$	${10}^{-5}$
${\mathit{u}}_0$	Iter.	Iter.	Iter.	Iter.	Time	Time	time	Time
${[0,0]}^T$	8	27	265	2566	0.606917	1.907212	14.120655	107.506926
${[10,10]}^T$	7	27	265	2591	0.286659	1.057623	10.764532	116.258335
${[-5,-5]}^T$	8	26	258	2596	0.388227	1.190191	11.424257	107.584978

.

Example 4.

The fourth example was taken from [45] where $\mathcal{F}:{\mathcal{R}}^2\to {\mathcal{R}}^2$ is defined by

$$\mathcal{F}\left(u\right)=\left(\begin{array}{c}(u_1^2+{(u_2-1)}^2)(1+u_2)\\ -u_1^3-u_1{(u_2-1)}^2\end{array}\right),$$

where $\mathcal{K}=\{u\in {\mathcal{R}}^2:-10\le u_i\le 10,i=1,2\}.$ It can easily see that $\mathcal{F}$ is Lipschitz continuous with $L=5$ and $\mathcal{F}$ is not monotone on $\mathcal{K}$ but pseudomonotone. In this experiment, we take different initial points and $D_n=\parallel u_n-v_n\parallel .$ Moreover, control parameters ${\zeta }_0=0.33,$ $\mu =0.25,$${\gamma }_n=\frac{1}{100(n+2)}$ and $f\left(u\right)=\frac{u}{4}$ for Algorithm 1. Numerical investigations regarding the fourth example is shown in

Table 4. Numerical behaviour of Algorithm 1 using different starting points for Example 4.

TOL	${10}^{-2}$	${10}^{-3}$	${10}^{-4}$	${10}^{-5}$	${10}^{-2}$	${10}^{-3}$	${10}^{-4}$	${10}^{-5}$
${\mathit{u}}_0$	Iter.	Iter.	Iter.	Iter.	Time	Time	Time	Time
${[0,0]}^T$	16	220	2231	29253	0.21543	2.35401	29.86562	224.95083
${[10,10]}^T$	27	190	2072	25762	0.25322	2.64742	26.84528	198.26446
${[-5,-5]}^T$	43	411	3801	47891	0.78262	4.77116	42.41738	427.904781

.

5. Conclusions

We have developed an extragradient-like method to solve pseudomonotone variational inequalities in real Hilbert space. The method had an explicit formula for an appropriate and effective stepsize evaluation on each step. For each iteration, the stepsize formula is modified based on the previous iterations. The numerical investigation was presented to explain the numerical effectiveness of our algorithm relative to other methods. These numerical studies suggest that viscosity schemes in this sense generally improve the effectiveness of the iterative sequence.