Inertial Modification Using Self-Adaptive Subgradient Extragradient Techniques for Equilibrium Programming Applied to Variational Inequalities and Fixed-Point Problems

Abstract

Equilibrium problems are articulated in a variety of mathematical computing applications, including minimax and numerical programming, saddle-point problems, fixed-point problems, and variational inequalities. In this paper, we introduce improved iterative techniques for evaluating the numerical solution of an equilibrium problem in a Hilbert space with a pseudomonotone and a Lipschitz-type bifunction. These techniques are based on two computing steps of a proximal-like mapping with inertial terms. We investigated two simplified stepsize rules that do not require a line search, allowing the technique to be carried out more successfully without knowledge of the Lipschitz-type constant of the cost bifunction. Once control parameter constraints are put in place, the iterative sequences converge on a particular solution to the problem. We prove strong convergence theorems without knowing the Lipschitz-type bifunction constants. A sequence of numerical tests was performed, and the results confirmed the correctness and speedy convergence of the new techniques over the traditional ones.

1. Introduction

Let $\Delta$ be a nonempty closed convex subset of a real Hilbert space $\Sigma .$ The study focuses on new iterative techniques for solving an equilibrium problem. Assume that a bifunction $ϝ:\Delta \times \Delta \to \mathbb{R}$ is satisfying $ϝ(q_1,q_1)=0,$ for each $q_1\in \Delta .$ An equilibrium problem for a bifunction $ϝ$ on $\Delta$ is described in the following manner: Find ${\nu }^{*}\in \Delta$ such that

$$ϝ({\nu }^{*},q_1)\ge 0,\forall q_1\in \Delta .$$

(1)

Let us denote the solution set of a problem (1) as $Sol(ϝ,\Delta ).$ Furthermore, we will assume in the following text that this solution set is nonempty. This study investigates the numerical evaluation of the equilibrium problem under the following conditions. We assume the following conditions are satisfied:

($ϝ$1)

The solution set $Sol(ϝ,\Delta )$ is nonempty for the problem (1);

($ϝ$2)

$ϝ$ is said to be pseudomonotone [1,2], i.e.,

$$ϝ(q_1,q_2)\ge 0⟹ϝ(q_2,q_1)\le 0,\forall q_1,q_2\in \Delta ;$$

($ϝ$3)

$ϝ$ is said to be Lipschitz-type continuous [3] on set $\Delta$ if there exists two constants $c_1,c_2>0,$ such that

$$ϝ(q_1,q_3)\le ϝ(q_1,q_2)+ϝ(q_2,q_3)+c_1\parallel q_1-q_2{\parallel }^2+c_2{\parallel q_2-q_3\parallel }^2,\forall q_1,q_2,q_3\in \Delta ;$$

($ϝ$4)

For any sequence $\left\{q_k\right\}\subset \Delta$ satisfying $q_k\rightharpoonup q^{*},$ then the following inequality holds

$$\underset{k\to +\infty }{lim\; sup}ϝ(q_k,q_1)\le ϝ(q^{*},q_1),\forall q_1\in \Delta ;$$

($ϝ$5)

$ϝ(q_1,·)$ is convex and subdifferentiable on $\Sigma$ for each fixed $q_1\in \Sigma .$

Researchers have concentrated on the equilibrium problem because it encompasses numerous additional mathematical problems in the literature (check for more details [1,4,5,6,7]). The term “equilibrium problem” was originally proposed in the literature in 1992 by Muu and Oettli [4] and was further examined by Blum and Oettli [1]. More precisely, we consider two applications for the problem (1). (i) A variational inequality problem for $\mathcal{E}:\Delta \to \Sigma$ is stated in the following manner: Find ${\nu }^{*}\in \Delta$ such that

$$〈\mathcal{E}\left({\nu }^{*}\right),q_1-{\nu }^{*}〉\ge 0,\forall q_1\in \Delta .$$

(2)

Let us define a bifunction $ϝ$ as

$$ϝ(q_1,q_2):=〈\mathcal{E}\left(q_1\right),q_2-q_1〉,\forall q_1,q_2\in \Delta .$$

(3)

The equilibrium problem is then transformed into the variational inequalities problem described in (2) and Lipschitz constants of a mapping $\mathcal{E}$ is $L=2c_1=2c_2.$ (ii) Let a mapping $\mathcal{D}:\Delta \to \Delta$ is said to $\kappa$-strict pseudocontraction [8] if there is a constant $\kappa \in (0,1)$ such that

$$\parallel \mathcal{D}{q}_1-\mathcal{D}{q}_2{\parallel }^2\le \parallel q_1-q_2{\parallel }^2+\kappa {\parallel (q_1-\mathcal{D}{q}_1)-(q_2-\mathcal{D}{q}_2)\parallel }^2,\forall q_1,q_2\in \Delta .$$

(4)

A fixed-point problem (FPP) for $\mathcal{D}:\Delta \to \Delta$ is to evaluate ${\nu }^{*}\in \Delta$ such that $\mathcal{D}\left({\nu }^{*}\right)={\nu }^{*}.$ Let us define a bifunction $ϝ$ as follows:

$$ϝ(q_1,q_2)=\langle q_1-\mathcal{D}{q}_1,q_2-q_1\rangle ,\forall q_1,q_2\in \Delta .$$

(5)

It can be easily seen in [9] that the expression (5) satisfies the conditions ($ϝ$1)–($ϝ$5) as well as the value of Lipschitz constants which are $c_1=c_2=\frac{3-2\kappa }{2-2\kappa }.$

The extragradient method developed by Tran et al. [10] is one useful method. This technique was built as follows. Take an arbitrary starting point $s_0\in \Sigma ;$ using the current iteration $s_k,$ take the next iteration as follows:

$$\left\{\begin{array}{c}{s}_0\in \Delta ,\hfill \\ \\ q_k=\underset{q\in \Delta }{argmin}\{\sigma ϝ(s_k,q)+\frac{1}{2}\parallel s_k-q{\parallel }^2\},\hfill \\ s_{k+1}=\underset{q\in \Delta }{argmin}\{\sigma ϝ(q_k,q)+\frac{1}{2}\parallel s_k-q{\parallel }^2\},\hfill \end{array}\right.$$

(6)

where $0<\sigma <min\left\{\frac{1}{2c_1},\frac{1}{2c_2}\right\}$ and $c_1,c_2$ are two Lipschitz-type constants. The main objective is to create an inertial-type technique in the case of [10] that will be designed to increase the convergence rate of the iterative sequence. Such techniques were previously developed as a result of the oscillator equation with damping and conservative force restoration. This second-order dynamical system is known as a “heavy friction ball”, and it was first proposed by Polyak in [11]. The important characteristic of this method is that the next iteration is built up of two prior iterations. In this context, numerical findings show that inertial terms improve the performance of the techniques in terms of the number of iterations and the elapsed time. Inertial-type iterative methods have been intensively investigated in recent years for certain classes of equilibrium problems [12,13,14,15,16,17,18], and others in [19,20,21,22,23,24].

As a result, the following obvious question arises:

Is it possible to construct new inertial-type strongly convergent extragradient-type techniques for solving equilibrium problems using monotone and nonmonotone stepsize rules?

In this work, we present a positive response to this question, namely that when solving equilibrium problems involving pseudomonotone bifunctions using a new monotone and nonmonotone variable stepsize rule, the gradient technique still yields a strong convergence sequence. Motivated by the work of Censor et al. [25] and Tran et al. [10], we present a new inertial extragradient-type technique for solving problem (1) in the setting of an infinite-dimensional real Hilbert space. Our main factors that contributed to this work were as follows:

(i)

To solve equilibrium problems in a real Hilbert space, we develop an inertial subgradient extragradient approach with a new monotone variable stepsize rule and show that the resulting sequence is strongly convergent.

(ii)

To solve equilibrium problems, we developed another inertial subgradient extragradient technique based on a novel variable nonmonotone stepsize rule independent of the Lipschitz constants.

(iii)

In order to solve various types of equilibrium issues in a real Hilbert space, several conclusions are derived.

(iv)

We show illustrative computations of the proposed methods for confirming theoretical conclusions and comparing them to earlier results [26,27,28]. Our numerical findings show that the new methods are useful and outperform the existing ones. A variety of effective techniques were also evaluated in the recent work by Rehman et al. [29].

The following is how this paper is organized: Preliminary results were reported in Section 2. All new methods and their convergence analysis are presented in Section 3. Finally, Section 5 provides numerical findings to demonstrate the practical efficacy of the proposed methodologies.

2. Preliminaries

In this section, we will go over some elementary identities as well as key definitions and lemmas. A metric projection $P_{\Delta }\left(q_1\right)$ of $q_1\in \Sigma$ is formulated as having:

$$P_{\Delta }\left(q_1\right)=\underset{}{argmin}\{\parallel q_1-q_2\parallel :q_2\in \Delta \}.$$

 The key characteristics of projection mapping are described below.

Lemma 1

([30]). Suppose that a metric projection $P_{\Delta }:\Sigma \to \Delta .$ Then, we have the following characteristics:

(i)

$$\parallel q_1-P_{\Delta }\left(q_2\right){\parallel }^2+\parallel P_{\Delta }\left(q_2\right)-q_2{\parallel }^2\le {\parallel q_1-q_2\parallel }^2,\forall q_1\in \Delta ,q_2\in \Sigma ;$$

(ii)  $q_3=P_{\Delta }\left(q_1\right)$ if and only if

$$\langle q_1-q_3,q_2-q_3\rangle \le 0,\forall q_2\in \Delta ;$$

(iii)

$$\parallel q_1-P_{\Delta }\left(q_1\right)\parallel \le \parallel q_1-q_2\parallel ,q_2\in \Delta ,q_1\in \Sigma .$$

Lemma 2

([30]). For any $q_1,q_2\in \Sigma$ and $\ell \in \mathbb{R}$. Then, the respective conditions are satisfied:

(i)

$$\parallel \ell q_1+(1-\ell )q_2{\parallel }^2=\ell \parallel q_1{\parallel }^2+(1-\ell )\parallel q_2{\parallel }^2-\ell (1-\ell ){\parallel q_1-q_2\parallel }^2.$$

(ii)

$$\parallel q_1+q_2{\parallel }^2\le {\parallel q_1\parallel }^2+2\langle q_2,q_1+q_2\rangle .$$

 A normal cone of Δ at $q_1\in \Delta$ is characterized with

$$N_{\Delta }\left(q_1\right)=\{q_3\in \Sigma :\langle q_3,q_2-q_1\rangle \le 0,\forall q_2\in \Delta \}.$$

 Suppose that $\mho :\Delta \to \mathbb{R}$ is a convex function and subdifferential of ℧ at $q_1\in \Delta$ is characterized with

$$\partial \mho \left(q_1\right)=\{q_3\in \Sigma :\mho \left(q_2\right)-\mho \left(q_1\right)\ge \langle q_3,q_2-q_1\rangle ,\forall q_2\in \Delta \}.$$

Lemma 3

([31]). Suppose that $\mho :\Delta \to \mathbb{R}$ is a convex, subdifferentiable, and lower semicontinuous function. An element $p\in \Delta$ is a minimizer of a function ℧ if and only if

$$0\in \partial \mho \left(p\right)+N_{\Delta }\left(p\right),$$

where $\partial \mho \left(p\right)$ denoted subdifferential of ℧ at $p\in \Delta$ and $N_{\Delta }\left(p\right)$ the normal cone of Δ at $p.$

Lemma 4

([32]). Let $\left\{l_k\right\}\subset [0,+\infty ),$$\left\{m_k\right\}\subset (0,1)$ and $\left\{n_k\right\}\subset \mathbb{R}$ be three sequences subsequent the following conditions:

$$l_{k+1}\le (1-m_k)l_k+m_k{n}_k,\forall k\in \mathbb{N}\mathrm{and}\sum _{k=1}^{+\infty }{m}_k=+\infty .$$

If ${lim\; sup}_{j\to +\infty }{n}_{k_j}\le 0$ for each subsequence $\left\{l_{k_j}\right\}$ of a sequence $\left\{l_k\right\}$ satisfies

$$\underset{j\to +\infty }{lim\; inf}(l_{k_j+1}-l_{k_j})\ge 0.$$

Then, ${lim}_{k\to +\infty }{l}_k=0.$

3. Main Results

In this section, we describe a numerical iterative method for increasing the rate of convergence of an iterative sequence by combining two strong convex optimization problems with an inertial term. We provide the following techniques for resolving equilibrium problems.

Lemma 5.

A sequence $\left\{{\sigma }_k\right\}$ is convergent to σ and $min\left\{\frac{\mu (2-\sqrt{2}-\rho )}{max\{2c_1,2c_2\}},{\sigma }_1\right\}\le \sigma \le {\sigma }_1.$

Proof. 

Due to the Lipschitz-type continuity of a bifunction, there exists two constants $c_1>0$ and $c_2>0.$ Let $ϝ(r_k,s_{k+1})-ϝ(r_k,q_k)-ϝ(q_k,s_{k+1})>0.$ Thus, we have

$$\begin{array}{cc}\hfill \frac{\mu (2-\sqrt{2}-\rho )(\parallel r_k-q_k{\parallel }^2+\parallel s_{k+1}-q_k{\parallel }^2)}{2[ϝ(r_k,s_{k+1})-ϝ(r_k,q_k)-ϝ(q_k,s_{k+1})]}& \ge \frac{\mu (2-\sqrt{2}-\rho )(\parallel r_k-q_k{\parallel }^2+\parallel s_{k+1}-q_k{\parallel }^2)}{2[c_1\parallel r_k-q_k{\parallel }^2+c_2\parallel s_{k+1}-q_k{\parallel }^2]}\hfill \\ & \ge \frac{\mu (2-\sqrt{2}-\rho )}{2max\{c_1,c_2\}}.\hfill \end{array}$$

(7)

Thus, we obtain ${lim}_{k\to +\infty }{\sigma }_k=\sigma .$ This completes the proof of the lemma.    □

Lemma 6.

A sequence $\left\{{\sigma }_k\right\}$ is convergent to σ and $min\left\{\frac{\mu (2-\sqrt{2}-\rho )}{max\{2c_1,2c_2\}},{\sigma }_1\right\}\le \sigma \le {\sigma }_1+P,$ where ${\displaystyle P=\sum _{k=1}^{+\infty }{\phi }_k.}$

Proof. 

Due to the Lipschitz-type continuity of a bifunction, there exists two constants $c_1>0$ and $c_2>0.$ Let $ϝ(r_k,s_{k+1})-ϝ(r_k,q_k)-ϝ(q_k,s_{k+1})>0.$ Thus, we have

$$\begin{array}{cc}\hfill \frac{\mu (2-\sqrt{2}-\rho )(\parallel r_k-q_k{\parallel }^2+\parallel s_{k+1}-q_k{\parallel }^2)}{2[ϝ(r_k,s_{k+1})-ϝ(r_k,q_k)-ϝ(q_k,s_{k+1})]}& \ge \frac{\mu (2-\sqrt{2}-\rho )(\parallel r_k-q_k{\parallel }^2+\parallel s_{k+1}-q_k{\parallel }^2)}{2[c_1\parallel r_k-q_k{\parallel }^2+c_2\parallel s_{k+1}-q_k{\parallel }^2]}\hfill \\ & \ge \frac{\mu (2-\sqrt{2}-\rho )}{2max\{c_1,c_2\}}.\hfill \end{array}$$

(8)

By definition of ${\sigma }_{k+1},$ we have

$$min\left\{\frac{\mu (2-\sqrt{2}-\rho )}{max\{2c_1,2c_2\}},{\sigma }_1\right\}\le {\sigma }_k\le {\sigma }_1+P.$$

Suppose that ${[{\sigma }_{k+1}-{\sigma }_k]}^{+}=max\left\{0,{\sigma }_{k+1}-{\sigma }_k\right\}$ and

$${[{\sigma }_{k+1}-{\sigma }_k]}^{-}=max\left\{0,-({\sigma }_{k+1}-{\sigma }_k)\right\}.$$

Due to the definition of $\left\{{\sigma }_k\right\},$ we obtain

$$\sum _{k=1}^{+\infty }{({\sigma }_{k+1}-{\sigma }_k)}^{+}=\sum _{k=1}^{+\infty }max\left\{0,{\sigma }_{k+1}-{\sigma }_k\right\}\le P<+\infty .$$

(9)

That is to say, the series ${\sum }_{k=1}^{+\infty }{({\sigma }_{k+1}-{\sigma }_k)}^{+}$ is convergent. The convergence must now be established

$$\sum _{k=1}^{+\infty }{({\sigma }_{k+1}-{\sigma }_k)}^{-}.$$

Let ${\sum }_{k=1}^{+\infty }{({\sigma }_{k+1}-{\sigma }_k)}^{-}=+\infty$. Due to the fact that

$${\sigma }_{k+1}-{\sigma }_k={({\sigma }_{k+1}-{\sigma }_k)}^{+}-{({\sigma }_{k+1}-{\sigma }_k)}^{-}.$$

We obtain the given equations:

$${\displaystyle {\sigma }_{k+1}-{\sigma }_1=\sum _{k=0}^k({\sigma }_{k+1}-{\sigma }_k)=\sum _{k=0}^k{({\sigma }_{k+1}-{\sigma }_k)}^{+}-\sum _{k=0}^k{({\sigma }_{k+1}-{\sigma }_k)}^{-}.}$$

(10)

Letting $k\to +\infty$ in expression (10), we have ${\sigma }_k\to -\infty$ as $k\to +\infty .$ This is an absurdity. As a result of the series convergence ${\sum }_{k=0}^k{({\sigma }_{k+1}-{\sigma }_k)}^{+}$ and ${\sum }_{k=0}^k{({\sigma }_{k+1}-{\sigma }_k)}^{-}$ taking $k\to +\infty$ in expression (10), we obtain ${lim}_{k\to +\infty }{\sigma }_k=\sigma .$ This completes the proof of the theorem.    □

Lemma 7.

From Algorithm 1 we have the following useful inequality

$${\sigma }_kϝ(q_k,q)-{\sigma }_kϝ(q_k,s_{k+1})\ge \langle r_k-s_{k+1},q-s_{k+1}\rangle ,\forall q\in {\Sigma }_k.$$

Algorithm 1: Explicit Subgradient Extragradient Method With Monotone Stepsize Rule

STEP 0: Choose ${\sigma }_1>0,$$s_{-1},s_0\in \Sigma ,$$\varkappa \in (0,1),$$\mu \in (0,1),$$\rho \in (0,2-\sqrt{2})$ with $\left\{{\chi }_k\right\}\subset (0,1)$ satisfy the following conditions: $$\underset{k\to +\infty }{lim}{\chi }_k=0\mathrm{and}\sum _{k=1}^{+\infty }{\chi }_k=+\infty .$$ STEP 1: Compute $$r_k=s_k+{\varkappa }_k(s_k-s_{k-1})-{\chi }_k\left[s_k+{\varkappa }_k(s_k-s_{k-1})\right],$$ where choose ${\varkappa }_k$ such that $$0\le {\varkappa }_k\le \widehat{{\varkappa }_k}\mathrm{and}\widehat{{\varkappa }_k}=\left\{\begin{array}{cc}min\left\{\varkappa ,\frac{{ϵ}_k}{\parallel s_k-s_{k-1}\parallel }\right\}\hfill & \mathrm{if}{s}_k\ne s_{k-1},\hfill \\ \varkappa \hfill & \mathrm{otherwise},\hfill \end{array}\right.$$ (11) where ${ϵ}_k=\circ \left({\chi }_k\right)$ a positive sequence such that ${lim}_{k\to +\infty }\frac{{ϵ}_k}{{\chi }_k}=0.$ STEP 2: Compute $$q_k=\underset{q\in \Delta }{argmin}\{{\sigma }_kϝ(r_k,q)+\frac{1}{2}\parallel r_k-q{\parallel }^2\}.$$ STEP 3: Given the current iterates $s_{k-1},$$s_k,$$q_k.$ Firstly, choose ${\omega }_k\in {\partial }_2ϝ(r_k,q_k)$ satisfying $r_k-{\sigma }_k{\omega }_k-q_k\in N_{\Delta }\left(q_k\right)$ and generate a half-space $${\Sigma }_k=\{z\in \Sigma :\langle r_k-{\sigma }_k{\omega }_k-q_k,z-q_k\rangle \le 0\}.$$ Compute $$s_{k+1}=\underset{q\in {\Sigma }_k}{argmin}\{{\sigma }_kϝ(q_k,q)+\frac{1}{2}\parallel r_k-q{\parallel }^2\}.$$ STEP 4: Compute $${\sigma }_{k+1}=\left\{\begin{array}{c}min\left\{{\sigma }_k,\frac{(2-\sqrt{2}-\rho )\mu \parallel r_k-q_k{\parallel }^2+(2-\sqrt{2}-\rho )\mu {\parallel s_{k+1}-q_k\parallel }^2}{2[ϝ(r_k,s_{k+1})-ϝ(r_k,q_k)-ϝ(q_k,s_{k+1})]}\right\}\hfill \\ \mathrm{if}ϝ(r_k,s_{k+1})-ϝ(r_k,q_k)-ϝ(q_k,s_{k+1})>0,\hfill \\ {\sigma }_k,\mathrm{otherwise}.\hfill \end{array}\right.$$ (12) STEP 5: If $q_k=r_k,$ then complete the computation. Otherwise, set $k:=k+1$ and go back STEP 1.

Proof. 

By the use of Lemma 3, we have

$$0\in {\partial }_2\left\{{\sigma }_kϝ(q_k,·)+\frac{1}{2}{\parallel r_k-·\parallel }^2\right\}\left(s_{k+1}\right)+N_{{\Sigma }_k}\left(s_{k+1}\right).$$

Thus, $\upsilon \in \partial ϝ(q_k,s_{k+1})$ there exists a vector $\overline{\upsilon }\in N_{{\Sigma }_k(}{s}_{k+1)}$ such that

$${\sigma }_k\upsilon +s_{k+1}-r_k+\overline{\upsilon }=0.$$

As a result of this, we now have

$$\langle r_k-s_{k+1},q-s_{k+1}\rangle ={\sigma }_k\langle \upsilon ,q-s_{k+1}\rangle +\langle \overline{\upsilon },q-s_{k+1}\rangle ,\forall q\in {\Sigma }_k.$$

Because $\overline{\upsilon }\in N_{{\Sigma }_k}\left(s_{k+1}\right)$, thus there exists a vector $\langle \overline{\upsilon },q-s_{k+1}\rangle \le 0$ for all $q\in {\Sigma }_k.$ It implies that

$$\langle r_k-s_{k+1},q-s_{k+1}\rangle \le {\sigma }_k\langle \upsilon ,q-s_{k+1}\rangle ,\forall q\in {\Sigma }_k.$$

(13)

Because $\upsilon \in \partial ϝ(q_k,s_{k+1}),$ we have

$$ϝ(q_k,q)-ϝ(q_k,s_{k+1})\ge \langle \upsilon ,q-s_{k+1}\rangle ,\forall q\in \Sigma .$$

(14)

We obtain by combining the Formulas (13) and (14)

$${\sigma }_kϝ(q_k,q)-{\sigma }_kϝ(q_k,s_{k+1})\ge \langle r_k-s_{k+1},q-s_{k+1}\rangle ,\forall q\in {\Sigma }_k.$$

   □

Lemma 8.

The following important inequality is obtained from Algorithm 1

$${\sigma }_kϝ(r_k,q)-{\sigma }_kϝ(r_k,q_k)\ge \langle r_k-q_k,q-q_k\rangle ,\forall q\in \Delta .$$

Proof. 

The proof is analogous to the proof of Lemma 7. Next, substitute $q=s_{k+1},$ we have

$${\sigma }_k\left\{ϝ(r_k,s_{k+1})-ϝ(r_k,q_k)\right\}\ge \langle r_k-q_k,s_{k+1}-q_k\rangle .$$

(15)

   □

Theorem 1.

Let $\left\{s_k\right\}$ be a sequence generated by Algorithm 1 that meets the conditions ($ϝ$1)–($ϝ$5). The sequence $\left\{s_k\right\}$ then strongly converges to ${\nu }^{*}.$

Proof. 

By using $q={\nu }^{*}$ in Lemma 7, we obtain

$${\sigma }_kϝ(q_k,{\nu }^{*})-{\sigma }_kϝ(q_k,s_{k+1})\ge \langle r_k-s_{k+1},{\nu }^{*}-s_{k+1}\rangle .$$

(16)

By the use of condition ($ϝ$2), we obtain

$$\langle r_k-s_{k+1},s_{k+1}-{\nu }^{*}\rangle \ge {\sigma }_kϝ(q_k,s_{k+1}).$$

(17)

From expression (12), we obtain

$$ϝ(r_k,s_{k+1})-ϝ(r_k,q_k)-ϝ(q_k,s_{k+1})\le \frac{(2-\sqrt{2}-\rho )\mu \left(\parallel r_k-q_k{\parallel }^2+{\parallel s_{k+1}-q_k\parallel }^2\right)}{2{\sigma }_{k+1}}.$$

After multiplying both sides by ${\sigma }_k>0,$ we have

$$\begin{array}{cc}\hfill {\sigma }_kϝ(q_k,s_{k+1})& \ge {\sigma }_kϝ(r_k,s_{k+1})-{\sigma }_kϝ(r_k,q_k)\hfill \\ & -\frac{(2-\sqrt{2}-\rho ){\sigma }_k\mu \left(\parallel r_k-q_k{\parallel }^2+{\parallel s_{k+1}-q_k\parallel }^2\right)}{2{\sigma }_{k+1}}.\hfill \end{array}$$

(18)

Integrating Equations (17) and (18) yields

$$\begin{array}{cc}\hfill \langle r_k-s_{k+1},s_{k+1}-{\nu }^{*}\rangle & \ge {\sigma }_k\{ϝ(r_k,s_{k+1})-ϝ(r_k,q_k)\}\hfill \\ & -\frac{(2-\sqrt{2}-\rho ){\sigma }_k\mu \left(\parallel r_k-q_k{\parallel }^2+{\parallel s_{k+1}-q_k\parallel }^2\right)}{2{\sigma }_{k+1}}.\hfill \end{array}$$

(19)

By using expression (15), we have

$${\sigma }_k\left\{ϝ(r_k,s_{k+1})-ϝ(r_k,q_k)\right\}\ge \langle r_k-q_k,s_{k+1}-q_k\rangle .$$

(20)

Combining expressions (19) and (20), we have

$$\begin{array}{cc}\hfill \langle r_k-s_{k+1},s_{k+1}-{\nu }^{*}\rangle & \ge \langle r_k-q_k,s_{k+1}-q_k\rangle \hfill \\ & -\frac{(2-\sqrt{2}-\rho ){\sigma }_k\mu \left(\parallel r_k-q_k{\parallel }^2+{\parallel s_{k+1}-q_k\parallel }^2\right)}{2{\sigma }_{k+1}}.\hfill \end{array}$$

(21)

The following facts are available to us:

$$2\langle r_k-s_{k+1},s_{k+1}-{\nu }^{*}\rangle =\parallel r_k-{\nu }^{*}{\parallel }^2-\parallel s_{k+1}-r_k{\parallel }^2-{\parallel s_{k+1}-{\nu }^{*}\parallel }^2,$$

$$2\langle q_k-r_k,q_k-s_{k+1}\rangle =\parallel r_k-q_k{\parallel }^2+\parallel s_{k+1}-q_k{\parallel }^2-{\parallel r_k-s_{k+1}\parallel }^2.$$

As a result of this, we now have

$$\begin{array}{cc}\hfill \parallel s_{k+1}-{\nu }^{*}{\parallel }^2& \le \parallel r_k-{\nu }^{*}{\parallel }^2-\parallel r_k-q_k{\parallel }^2-{\parallel s_{k+1}-q_k\parallel }^2\hfill \\ & +\frac{(2-\sqrt{2}-\rho ){\sigma }_k\mu \left(\parallel r_k-q_k{\parallel }^2+{\parallel s_{k+1}-q_k\parallel }^2\right)}{{\sigma }_{k+1}}.\hfill \end{array}$$

(22)

Due to ${\sigma }_k\to \sigma ,$ there is a specific natural number $N_1\in \mathbb{N}$ such that

$$\underset{k\to +\infty }{lim}\frac{\mu {\sigma }_k}{{\sigma }_{k+1}}\le 1.$$

Thus, we have

$$\begin{array}{cc}\hfill \parallel s_{k+1}-{\nu }^{*}{\parallel }^2& \le \parallel r_k-{\nu }^{*}{\parallel }^2-\parallel r_k-q_k{\parallel }^2-{\parallel s_{k+1}-q_k\parallel }^2\hfill \\ & +(2-\sqrt{2}-\rho )\left(\parallel r_k-q_k{\parallel }^2+{\parallel s_{k+1}-q_k\parallel }^2\right).\hfill \end{array}$$

(23)

Furthermore, it implies that

$$\begin{array}{cc}\hfill \parallel s_{k+1}-{\nu }^{*}{\parallel }^2& \le \parallel r_k-{\nu }^{*}{\parallel }^2-(\sqrt{2}-1)\parallel r_k-q_k{\parallel }^2-(\sqrt{2}-1){\parallel s_{k+1}-q_k\parallel }^2\hfill \\ & -\rho \left(\parallel r_k-q_k{\parallel }^2+{\parallel s_{k+1}-q_k\parallel }^2\right).\hfill \end{array}$$

(24)

From expression (24), we obtain

$$\parallel s_{k+1}-{\nu }^{*}{\parallel }^2\le {\parallel r_k-{\nu }^{*}\parallel }^2,\forall k\ge N_1.$$

(25)

To use the formula of $\left\{r_k\right\},$ we obtain

$$\begin{array}{}\hfill & \begin{array}{cc}\hfill ∥r_k-{\nu }^{*}∥& =∥s_k+{\varkappa }_k(s_k-s_{k-1})-{\chi }_k{s}_k-{\varkappa }_k{\chi }_k(s_k-s_{k-1})-{\nu }^{*}∥\hfill \end{array}\hfill \hfill \mathrm{(26)}& \begin{array}{cc}\hfill & =∥(1-{\chi }_k)(s_k-{\nu }^{*})+(1-{\chi }_k){\varkappa }_k(s_k-s_{k-1})-{\chi }_k{\nu }^{*}∥\hfill \end{array}\hfill \hfill & \begin{array}{cc}\hfill & \le (1-{\chi }_k)∥s_k-{\nu }^{*}∥+(1-{\chi }_k){\varkappa }_k∥s_k-s_{k-1}∥+{\chi }_k∥{\nu }^{*}∥\hfill \end{array}\hfill \hfill \mathrm{(27)}& \begin{array}{cc}\hfill & \le (1-{\chi }_k)\parallel s_k-{\nu }^{*}\parallel +{\chi }_k{K}_1,\hfill \end{array}\hfill \end{array}$$

where

$$(1-{\chi }_k)\frac{{\varkappa }_k}{{\chi }_k}∥s_k-s_{k-1}∥+∥{\nu }^{*}∥\le K_1.$$

Combining the expressions (25) and (27), we obtain

$$\begin{array}{cc}\hfill \parallel s_{k+1}-{\nu }^{*}\parallel & \le (1-{\chi }_k)\parallel s_k-{\nu }^{*}\parallel +{\chi }_k{K}_1\hfill \\ \hfill & \le max\left\{\parallel s_k-{\nu }^{*}\parallel ,K_1\right\}\hfill \\ \hfill ⋮\\ \hfill & \le max\left\{\parallel s_{N_1}-{\nu }^{*}\parallel ,K_1\right\}.\hfill \end{array}$$

(28)

As a result, we may conclude that $\left\{s_k\right\}$ is a bounded sequence. By using expression (22), we have

$$\begin{array}{cc}\hfill \parallel s_{k+1}-{\nu }^{*}{\parallel }^2& \le \parallel r_k-{\nu }^{*}{\parallel }^2-\parallel r_k-q_k{\parallel }^2-{\parallel s_{k+1}-q_k\parallel }^2\hfill \\ \hfill & +\frac{(2-\sqrt{2}-\rho ){\sigma }_k\mu \left(\parallel r_k-q_k{\parallel }^2+{\parallel s_{k+1}-q_k\parallel }^2\right)}{{\sigma }_{k+1}}.\hfill \end{array}$$

(29)

Indeed, by expression (27), we have

$$\begin{array}{cc}\hfill {∥r_k-{\nu }^{*}∥}^2& \le {(1-{\chi }_k)}^2\parallel s_k-{\nu }^{*}{\parallel }^2+{\chi }_k^2{K}_1^2+2K_1{\chi }_k(1-{\chi }_k)\parallel s_k-{\nu }^{*}\parallel \hfill \\ \hfill & \le \parallel s_k-{\nu }^{*}{\parallel }^2+{\chi }_k\left[{\chi }_k{K}_1^2+2K_1(1-{\chi }_k)\parallel s_k-{\nu }^{*}\parallel \right]\hfill \\ \hfill & \le \parallel s_k-{\nu }^{*}{\parallel }^2+{\chi }_k{K}_2,\hfill \end{array}$$

(30)

for some $K_2>0.$ By using expressions (29) and (30), we have

$$\begin{array}{cc}\hfill & \parallel s_{k+1}-{\nu }^{*}{\parallel }^2\hfill \\ \hfill & \le \parallel s_k-{\nu }^{*}{\parallel }^2+{\chi }_k{K}_2\hfill \\ \hfill & -\left(1-\frac{(2-\sqrt{2}-\rho )\mu {\sigma }_k}{{\sigma }_{k+1}}\right)\parallel r_k-q_k{\parallel }^2-\left(1-\frac{(2-\sqrt{2}-\rho )\mu {\sigma }_k}{{\sigma }_{k+1}}\right){\parallel q_k-s_{k+1}\parallel }^2.\hfill \end{array}$$

(31)

Consequently, this implies that

$$\begin{array}{cc}\hfill & \left(1-\frac{(2-\sqrt{2}-\rho )\mu {\sigma }_k}{{\sigma }_{k+1}}\right)\parallel r_k-q_k{\parallel }^2+\left(1-\frac{(2-\sqrt{2}-\rho )\mu {\sigma }_k}{{\sigma }_{k+1}}\right){\parallel q_k-s_{k+1}\parallel }^2\hfill \\ \hfill & \le \parallel s_k-{\nu }^{*}{\parallel }^2+{\chi }_k{K}_2-{\parallel s_{k+1}-{\nu }^{*}\parallel }^2.\hfill \end{array}$$

(32)

From the definition of $\left\{r_k\right\}$, we can rewrite

$$\begin{array}{cc}\hfill & {∥r_k-{\nu }^{*}∥}^2\hfill \\ \hfill & ={∥s_k+{\varkappa }_k(s_k-s_{k-1})-{\chi }_k{s}_k-{\varkappa }_k{\chi }_k(s_k-s_{k-1})-{\nu }^{*}∥}^2\hfill \\ \hfill & ={∥(1-{\chi }_k)(s_k-{\nu }^{*})+(1-{\chi }_k){\varkappa }_k(s_k-s_{k-1})-{\chi }_k{\nu }^{*}∥}^2\hfill \\ \hfill & \le {∥(1-{\chi }_k)(s_k-{\nu }^{*})+(1-{\chi }_k){\varkappa }_k(s_k-s_{k-1})∥}^2+2{\chi }_k\langle -{\nu }^{*},r_k-{\nu }^{*}\rangle \hfill \\ \hfill & ={(1-{\chi }_k)}^2{∥s_k-{\nu }^{*}∥}^2+{(1-{\chi }_k)}^2{\varkappa }_k^2{∥s_k-s_{k-1}∥}^2\hfill \\ \hfill & +2{\varkappa }_k{(1-{\chi }_k)}^2∥s_k-{\nu }^{*}∥∥s_k-s_{k-1}∥+2{\chi }_k\langle -{\nu }^{*},r_k-s_{k+1}\rangle +2{\chi }_k\langle -{\nu }^{*},s_{k+1}-{\nu }^{*}\rangle \hfill \\ \hfill & \le (1-{\chi }_k){∥s_k-{\nu }^{*}∥}^2+{\varkappa }_k^2{∥s_k-s_{k-1}∥}^2+2{\varkappa }_k(1-{\chi }_k)∥s_k-{\nu }^{*}∥∥s_k-s_{k-1}∥\hfill \\ \hfill & +2{\chi }_k∥{\nu }^{*}∥∥r_k-s_{k+1}∥+2{\chi }_k\langle -{\nu }^{*},s_{k+1}-{\nu }^{*}\rangle \hfill \\ \hfill & =(1-{\chi }_k){∥s_k-{\nu }^{*}∥}^2+{\chi }_k[{\varkappa }_k∥s_k-s_{k-1}∥\frac{{\varkappa }_k}{{\chi }_k}∥s_k-s_{k-1}∥\hfill \\ \hfill & +2(1-{\chi }_k)∥s_k-{\nu }^{*}∥\frac{{\varkappa }_k}{{\chi }_k}∥s_k-s_{k-1}∥+2∥{\nu }^{*}∥∥r_k-s_{k+1}∥+2\langle {\nu }^{*},{\nu }^{*}-s_{k+1}\rangle ].\hfill \end{array}$$

(33)

Combining expressions (25) and (33), we obtain

$$\begin{array}{cc}\hfill & {∥s_{k+1}-{\nu }^{*}∥}^2\hfill \\ \hfill & \le (1-{\chi }_k){∥s_k-{\nu }^{*}∥}^2+{\chi }_k[{\varkappa }_k∥s_k-s_{k-1}∥\frac{{\varkappa }_k}{{\chi }_k}∥s_k-s_{k-1}∥\hfill \\ \hfill & +2(1-{\chi }_k)∥s_k-{\nu }^{*}∥\frac{{\varkappa }_k}{{\chi }_k}∥s_k-s_{k-1}∥+2∥{\nu }^{*}∥∥r_k-s_{k+1}∥+2\langle {\nu }^{*},{\nu }^{*}-s_{k+1}\rangle ].\hfill \end{array}$$

(34)

Next, we need to prove that the sequence ${∥s_k-{\nu }^{*}∥}^2$ converges to zero. Set

$$l_k:={\parallel s_k-{\nu }^{*}\parallel }^2$$

and

$$\begin{array}{cc}\hfill m_k& :={\varkappa }_k∥s_k-s_{k-1}∥\frac{{\varkappa }_k}{{\chi }_k}∥s_k-s_{k-1}∥+2(1-{\chi }_k)∥s_k-{\nu }^{*}∥\frac{{\varkappa }_k}{{\chi }_k}∥s_k-s_{k-1}∥\hfill \\ \hfill & +2∥{\nu }^{*}∥∥r_k-s_{k+1}∥+2\langle {\nu }^{*},{\nu }^{*}-s_{k+1}\rangle .\hfill \end{array}$$

Then, next expression can be rewritten as follows:

$$l_{k+1}\le (1-{\chi }_k)l_k+{\chi }_k{m}_k.$$

Indeed, from Lemma 4, it is adequate to demonstrate that ${lim\; sup}_{j\to +\infty }{m}_{k_j}\le 0$ such that for every subsequence $\left\{l_{k_j}\right\}$ of $\left\{l_k\right\}$ satisfying

$$\underset{j\to +\infty }{lim\; inf}(l_{k_j+1}-l_{k_j})\ge 0.$$

This is proportionately needed to show that

$$\underset{j\to +\infty }{lim\; sup}\langle {\nu }^{*},{\nu }^{*}-s_{k_j+1}\rangle \le 0$$

and

$$\underset{j\to +\infty }{lim\; sup}∥r_{k_j}-s_{k_j+1}∥\le 0,$$

for each subsequence $\{\parallel s_{k_j}-{\nu }^{*}\parallel \}$ of $\{\parallel s_k-{\nu }^{*}\parallel \}$ satisfying

$$\underset{j\to +\infty }{lim\; inf}\left(\parallel s_{k_j+1}-{\nu }^{*}\parallel -\parallel s_{k_j}-{\nu }^{*}\parallel \right)\ge 0.$$

Consider that $\{\parallel s_{k_j}-{\nu }^{*}\parallel \}$ is a subsequence of $\{\parallel s_k-{\nu }^{*}\parallel \}$ satisfying

$$\underset{j\to +\infty }{lim\; inf}\left(\parallel s_{k_j+1}-{\nu }^{*}\parallel -\parallel s_{k_j}-{\nu }^{*}\parallel \right)\ge 0.$$

Then, we have

$$\begin{array}{cc}\hfill & \underset{j\to +\infty }{lim\; inf}\left(\parallel s_{k_j+1}-{\nu }^{*}{\parallel }^2-{\parallel s_{k_j}-{\nu }^{*}\parallel }^2\right)\hfill \\ \hfill & =\underset{j\to +\infty }{lim\; inf}\left(\parallel s_{k_j+1}-{\nu }^{*}\parallel -\parallel s_{k_j}-{\nu }^{*}\parallel \right)\left(\parallel s_{k_j+1}-{\nu }^{*}\parallel +\parallel s_{k_j}-{\nu }^{*}\parallel \right)\ge 0.\hfill \end{array}$$

(35)

It derives from expression (32) that

$$\begin{array}{cc}\hfill & \underset{j\to +\infty }{lim\; sup}\left(1-\frac{(2-\sqrt{2}-\rho )\mu {\sigma }_{k_j}}{{\sigma }_{k_j+1}}\right)\parallel r_{k_j}-q_{k_j}{\parallel }^2+\left(1-\frac{(2-\sqrt{2}-\rho )\mu {\sigma }_{k_j}}{{\sigma }_{k_j+1}}\right){\parallel q_{k_j}-s_{k_j+1}\parallel }^2\hfill \\ \hfill & \le \underset{j\to +\infty }{lim\; sup}\left[\parallel s_{k_j}-{\nu }^{*}{\parallel }^2-{\parallel s_{k_j+1}-{\nu }^{*}\parallel }^2\right]+\underset{j\to +\infty }{lim\; sup}{\chi }_{k_j}{K}_2\hfill \\ \hfill & =-\underset{j\to +\infty }{lim\; inf}\left[\parallel s_{k_j+1}-{\nu }^{*}{\parallel }^2-{\parallel s_{k_j}-{\nu }^{*}\parallel }^2\right]\hfill \\ \hfill & \le 0.\hfill \end{array}$$

(36)

The preceding relationship suggests that

$$\underset{j\to +\infty }{lim}\parallel r_{k_j}-q_{k_j}\parallel =0,\underset{j\to +\infty }{lim}\parallel s_{k_j+1}-q_{k_j}\parallel =0.$$

(37)

Therefore, we obtain

$$\underset{j\to +\infty }{lim}\parallel s_{k_j+1}-r_{k_j}\parallel =0.$$

(38)

Next, we have to compute

$$\begin{array}{cc}\hfill \parallel r_{k_j}-s_{k_j}\parallel & =\parallel s_{k_j}+{\varkappa }_{k_j}(s_{k_j}-s_{k_j-1})-{\chi }_{k_j}\left[s_{k_j}+{\varkappa }_{k_j}(s_{k_j}-s_{k_j-1})\right]-s_{k_j}\parallel \hfill \\ \hfill & \le {\varkappa }_{k_j}\parallel s_{k_j}-s_{k_j-1}\parallel +{\chi }_{k_j}\parallel s_{k_j}\parallel +{\varkappa }_{k_j}{\chi }_{k_j}\parallel s_{k_j}-s_{k_j-1}\parallel \hfill \\ \hfill & ={\chi }_{k_j}\frac{{\varkappa }_{k_j}}{{\chi }_{k_j}}\parallel s_{k_j}-s_{k_j-1}\parallel +{\chi }_{k_j}\parallel s_{k_j}\parallel +{\chi }_{k_j}^2\frac{{\varkappa }_{k_j}}{{\chi }_{k_j}}\parallel s_{k_j}-s_{k_j-1}\parallel ⟶0.\hfill \end{array}$$

(39)

This, together with ${lim}_{j\to +\infty }\parallel s_{k_j+1}-r_{k_j}\parallel =0,$ yields that

$$\underset{j\to +\infty }{lim}\parallel s_{k_j+1}-s_{k_j}\parallel =0.$$

(40)

Because $\left\{s_{k_j}\right\}$ is a bounded sequence, without loss of generality, we can consider that $\left\{s_{k_j}\right\}$ converges weakly to some $\widehat{p}\in \Sigma .$ Thus, we have

$$\langle 0-{\nu }^{*},q-{\nu }^{*}\rangle \le 0,\forall q\in Sol(ϝ,\Delta ).$$

(41)

Next, let us demonstrate that $\widehat{p}\in Sol(ϝ,\Delta ).$ We have obtained as follows by combining Lemma 7 with expressions (18) and (15), and we have

$$\begin{array}{cc}\hfill {\sigma }_{k_j}ϝ(q_{k_j},q)& \ge {\sigma }_{k_j}ϝ(q_{k_j},s_{k_j+1})+\langle r_{k_j}-s_{k_j+1},q-s_{k_j+1}\rangle \hfill \\ \hfill & \ge {\sigma }_{k_j}ϝ(r_{k_j},s_{k_{j+1}})-{\sigma }_{k_j}ϝ(r_{k_j},q_{k_j})-\frac{(2-\sqrt{2}-\rho )\mu {\sigma }_{k_j}}{2{\sigma }_{k_j+1}}{\parallel q_{k_j}-r_{k_j}\parallel }^2\hfill \\ \hfill & -\frac{(2-\sqrt{2}-\rho )\mu {\sigma }_{k_j}}{2{\sigma }_{k_j+1}}{\parallel q_{k_j}-s_{k_j+1}\parallel }^2+\langle r_{k_j}-s_{k_j+1},q-s_{k_j+1}\rangle \hfill \\ \hfill & \ge \langle r_{k_j}-q_{k_j},s_{k_j+1}-q_{k_j}\rangle -\frac{(2-\sqrt{2}-\rho )\mu {\sigma }_{k_j}}{2{\sigma }_{k_j+1}}{\parallel q_{k_j}-r_{k_j}\parallel }^2\hfill \\ \hfill & -\frac{(2-\sqrt{2}-\rho )\mu {\sigma }_{k_j}}{2{\sigma }_{k_j+1}}{\parallel q_{k_j}-s_{k_j+1}\parallel }^2+\langle r_{k_j}-s_{k_j+1},q-s_{k_j+1}\rangle ,\hfill \end{array}$$

(42)

where q is any member of ${\Sigma }_k.$ The right-hand side of the last inequality equals zero due to formula (37) and the boundedness of sequence $\left\{s_k\right\}.$ By making the use of the condition ($ϝ$4) and $q_{k_j}\rightharpoonup \widehat{p},$ we have ${\sigma }_{k_j}\ge \sigma >0$ such that

$$0\le \underset{j\to +\infty }{lim\; sup}ϝ(q_{k_j},q)\le ϝ(\widehat{p},q),\forall q\in {\Sigma }_k.$$

Because $\Delta$ is a subset of ${\Sigma }_k$ and $ϝ(\widehat{p},q)\ge 0,\forall q\in \Delta .$ This demonstrates that $\widehat{p}\in Sol(ϝ,\Delta ).$ Thus, we have

$$\begin{array}{cc}\hfill & \underset{k\to +\infty }{lim\; sup}\langle {\nu }^{*},{\nu }^{*}-s_k\rangle \hfill \\ \hfill & =\underset{j\to +\infty }{lim}\langle {\nu }^{*},{\nu }^{*}-s_{k_j}\rangle =\langle {\nu }^{*},{\nu }^{*}-\widehat{p}\rangle \le 0.\hfill \end{array}$$

(43)

By using the fact ${lim}_{j\to +\infty }∥s_{k_j+1}-s_{k_j}∥=0$. Thus, we have

$$\begin{array}{cc}\hfill & \underset{j\to +\infty }{lim\; sup}\langle {\nu }^{*},{\nu }^{*}-s_{k_j+1}\rangle \hfill \\ \hfill & \le \underset{j\to +\infty }{lim\; sup}\langle {\nu }^{*},{\nu }^{*}-s_{k_j}\rangle +\underset{j\to +\infty }{lim\; sup}\langle {\nu }^{*},s_{k_j}-s_{k_j+1}\rangle \le 0.\hfill \end{array}$$

(44)

By considering expression (34) with Lemma 4, we infer that $s_k\to {\nu }^{*}$ is the same as $k\to +\infty .$ This concludes the proof of Theorem 1.    □

4. Results to Solve Fixed-Point Problem and Variational Inequalities

In this section, we apply our primary results to solve fixed-point problems and variational inequalities. The Formulas (3) and (5) are being used to retrieve relevant conclusions. All the techniques are based on our primary results, which are discussed in further detail below.

Corollary 2.

Assume that $\mathcal{E}:\Delta \to \Sigma$ is a weakly continuous, L-Lipschitz continuous, and pseudomonotone operator with solution set $Sol(\mathcal{E},\Delta )\ne \varnothing .$ Choose ${\sigma }_1>0,$$s_{-1},s_0\in \Sigma ,$$\varkappa \in (0,1),$$\mu \in (0,1),$$\rho \in (0,2-\sqrt{2})$ with a sequence $\left\{{\chi }_k\right\}\subset (0,1)$ satisfy the following conditions:

$$\underset{k\to +\infty }{lim}{\chi }_k=0\mathrm{and}\sum _{k=1}^{+\infty }{\chi }_k=+\infty .$$

Compute

$$r_k=s_k+{\varkappa }_k(s_k-s_{k-1})-{\chi }_k\left[s_k+{\varkappa }_k(s_k-s_{k-1})\right],$$

and choose ${\varkappa }_k$ such that

$$0\le {\varkappa }_k\le \widehat{{\varkappa }_k}\mathrm{and}\widehat{{\varkappa }_k}=\left\{\begin{array}{cc}min\left\{\varkappa ,\frac{{ϵ}_k}{\parallel s_k-s_{k-1}\parallel }\right\}\hfill & \mathrm{if}{s}_k\ne s_{k-1},\hfill \\ \varkappa \hfill & \mathrm{otherwise},\hfill \end{array}\right.$$

where ${ϵ}_k=\circ \left({\chi }_k\right)$ a positive sequence such that ${lim}_{k\to +\infty }\frac{{ϵ}_k}{{\chi }_k}=0.$ First, we have to compute

$$q_k=P_{\Delta }(r_k-{\sigma }_k\mathcal{E}\left(r_k\right)).$$

Given the current iterates $s_{k-1},$$s_k,$$q_k$ and construct a half-space

$${\Sigma }_k=\{z\in \Sigma :\langle r_k-{\sigma }_k\mathcal{E}\left(r_k\right)-q_k,z-q_k\rangle \le 0\}\mathrm{for}\mathrm{each}k\ge 0.$$

Compute

$$s_{k+1}=P_{{\Sigma }_k}(r_k-{\sigma }_k\mathcal{E}\left(q_k\right)).$$

The stepsize should be updated as follows:

$${\sigma }_{k+1}=\left\{\begin{array}{c}min\left\{{\sigma }_k,\frac{(2-\sqrt{2}-\rho )\mu \parallel r_k-q_k{\parallel }^2+(2-\sqrt{2}-\rho )\mu {\parallel s_{k+1}-q_k\parallel }^2}{2〈\mathcal{E}\left(r_k\right)-\mathcal{E}\left(q_k\right),s_{k+1}-q_k〉}\right\}\hfill \\ \mathrm{if}〈\mathcal{E}\left(r_k\right)-\mathcal{E}\left(q_k\right),s_{k+1}-q_k〉>0,\hfill \\ {\sigma }_k,\mathrm{otherwise}.\hfill \end{array}\right.$$

Then, the sequences $\left\{s_k\right\}$ converge strongly to ${\nu }^{*}\in Sol(\mathcal{E},\Delta ).$

Corollary 3.

Assume that $\mathcal{E}:\Delta \to \Sigma$ is a weakly continuous, L-Lipschitz continuous, and pseudomonotone operator with solution set $Sol(\mathcal{E},\Delta )\ne \varnothing .$ Choose ${\sigma }_1>0,$$s_{-1},s_0\in \Sigma ,$$\varkappa \in (0,1),$$\mu \in (0,1),$$\rho \in (0,2-\sqrt{2})$ with a sequence $\left\{{\chi }_k\right\}\subset (0,1)$ satisfy the following conditions:

$$\underset{k\to +\infty }{lim}{\chi }_k=0\mathrm{and}\sum _{k=1}^{+\infty }{\chi }_k=+\infty .$$

Compute

$$r_k=s_k+{\varkappa }_k(s_k-s_{k-1})-{\chi }_k\left[s_k+{\varkappa }_k(s_k-s_{k-1})\right],$$

and choose ${\varkappa }_k$ such that

$$0\le {\varkappa }_k\le \widehat{{\varkappa }_k}\mathrm{and}\widehat{{\varkappa }_k}=\left\{\begin{array}{cc}min\left\{\varkappa ,\frac{{ϵ}_k}{\parallel s_k-s_{k-1}\parallel }\right\}\hfill & \mathrm{if}{s}_k\ne s_{k-1},\hfill \\ \varkappa \hfill & \mathrm{otherwise},\hfill \end{array}\right.$$

where ${ϵ}_k=\circ \left({\chi }_k\right)$ a positive sequence such that ${lim}_{k\to +\infty }\frac{{ϵ}_k}{{\chi }_k}=0.$ First, we have to compute

$$q_k=P_{\Delta }(r_k-{\sigma }_k\mathcal{E}\left(r_k\right)).$$

Given the current iterates $s_{k-1},$$s_k,$$q_k$ and construct a half-space

$${\Sigma }_k=\{z\in \Sigma :\langle r_k-{\sigma }_k\mathcal{E}\left(r_k\right)-q_k,z-q_k\rangle \le 0\}\mathrm{for}\mathrm{each}k\ge 0.$$

Compute

$$s_{k+1}=P_{{\Sigma }_k}(r_k-{\sigma }_k\mathcal{E}\left(q_k\right)).$$

Select a non-negative real sequence $\left\{{\phi }_k\right\}$ such that ${\sum }_{k=1}^{+\infty }{\phi }_k<+\infty .$ Update

$${\sigma }_{k+1}=\left\{\begin{array}{c}min\left\{{\sigma }_k+{\phi }_k,\frac{(2-\sqrt{2}-\rho )\mu \parallel r_k-q_k{\parallel }^2+(2-\sqrt{2}-\rho )\mu {\parallel s_{k+1}-q_k\parallel }^2}{2〈\mathcal{E}\left(r_k\right)-\mathcal{E}\left(q_k\right),s_{k+1}-q_k〉}\right\}\hfill \\ \mathrm{if}〈\mathcal{E}\left(r_k\right)-\mathcal{E}\left(q_k\right),s_{k+1}-q_k〉>0,\hfill \\ {\sigma }_k+{\phi }_k,\mathrm{otherwise}.\hfill \end{array}\right.$$

Then, the sequences $\left\{s_k\right\}$ converge strongly to ${\nu }^{*}\in Sol(\mathcal{E},\Delta ).$

Corollary 4.

Assume that $\mathcal{E}:\Delta \to \Sigma$ is a weakly continuous, L-Lipschitz continuous, and pseudomonotone operator with solution set $Sol(\mathcal{E},\Delta )\ne \varnothing .$ Choose ${\sigma }_1>0,$$s_{-1},s_0\in \Sigma ,$$\varkappa \in (0,1),$$\mu \in (0,1),$$\rho \in (0,2-\sqrt{2})$ with a sequence $\left\{{\chi }_k\right\}\subset (0,1)$ satisfy the following conditions:

$$\underset{k\to +\infty }{lim}{\chi }_k=0\mathrm{and}\sum _{k=1}^{+\infty }{\chi }_k=+\infty .$$

Compute

$$r_k=s_k+{\varkappa }_k(s_k-s_{k-1})-{\chi }_k\left[s_k+{\varkappa }_k(s_k-s_{k-1})\right],$$

and choose ${\varkappa }_k$ such that

$$0\le {\varkappa }_k\le \widehat{{\varkappa }_k}\mathrm{and}\widehat{{\varkappa }_k}=\left\{\begin{array}{cc}min\left\{\varkappa ,\frac{{ϵ}_k}{\parallel s_k-s_{k-1}\parallel }\right\}\hfill & \mathrm{if}{s}_k\ne s_{k-1},\hfill \\ \varkappa \hfill & \mathrm{otherwise},\hfill \end{array}\right.$$

where ${ϵ}_k=\circ \left({\chi }_k\right)$ a positive sequence such that ${lim}_{k\to +\infty }\frac{{ϵ}_k}{{\chi }_k}=0.$ First, we have to compute

$$\left\{\begin{array}{c}{q}_k=P_{\Delta }(r_k-{\sigma }_k\mathcal{E}\left(r_k\right)),\hfill \\ s_{k+1}=P_{\Delta }(r_k-{\sigma }_k\mathcal{E}\left(q_k\right)).\hfill \end{array}\right.$$

Update the stepsize in the following way:

$${\sigma }_{k+1}=\left\{\begin{array}{c}min\left\{{\sigma }_k,\frac{(2-\sqrt{2}-\rho )\mu \parallel r_k-q_k{\parallel }^2+(2-\sqrt{2}-\rho )\mu {\parallel s_{k+1}-q_k\parallel }^2}{2〈\mathcal{E}\left(r_k\right)-\mathcal{E}\left(q_k\right),s_{k+1}-q_k〉}\right\}\hfill \\ \mathrm{if}〈\mathcal{E}\left(r_k\right)-\mathcal{E}\left(q_k\right),s_{k+1}-q_k〉>0,\hfill \\ {\sigma }_k,\mathrm{otherwise}.\hfill \end{array}\right.$$

Then, the sequences $\left\{s_k\right\}$ converge strongly to ${\nu }^{*}\in Sol(\mathcal{E},\Delta ).$

Corollary 5.

Assume that $\mathcal{E}:\Delta \to \Sigma$ is a weakly continuous, L-Lipschitz continuous, and pseudomonotone operator with solution set $Sol(\mathcal{E},\Delta )\ne \varnothing .$ Choose ${\sigma }_1>0,$$s_{-1},s_0\in \Sigma ,$$\varkappa \in (0,1),$$\mu \in (0,1),$$\rho \in (0,2-\sqrt{2})$ with a sequence $\left\{{\chi }_k\right\}\subset (0,1)$ satisfy the following conditions:

$$\underset{k\to +\infty }{lim}{\chi }_k=0\mathrm{and}\sum _{k=1}^{+\infty }{\chi }_k=+\infty .$$

Compute

$$r_k=s_k+{\varkappa }_k(s_k-s_{k-1})-{\chi }_k\left[s_k+{\varkappa }_k(s_k-s_{k-1})\right],$$

and choose ${\varkappa }_k$ such that

$$0\le {\varkappa }_k\le \widehat{{\varkappa }_k}\mathrm{and}\widehat{{\varkappa }_k}=\left\{\begin{array}{cc}min\left\{\varkappa ,\frac{{ϵ}_k}{\parallel s_k-s_{k-1}\parallel }\right\}\hfill & \mathrm{if}{s}_k\ne s_{k-1},\hfill \\ \varkappa \hfill & \mathrm{otherwise},\hfill \end{array}\right.$$

where ${ϵ}_k=\circ \left({\chi }_k\right)$ a positive sequence such that ${lim}_{k\to +\infty }\frac{{ϵ}_k}{{\chi }_k}=0.$ First, we have to compute

$$\left\{\begin{array}{c}{q}_k=P_{\Delta }(r_k-{\sigma }_k\mathcal{E}\left(r_k\right)),\hfill \\ s_{k+1}=P_{\Delta }(r_k-{\sigma }_k\mathcal{E}\left(q_k\right)).\hfill \end{array}\right.$$

Furthermore, select a non-negative real sequence $\left\{{\phi }_k\right\}$ such that ${\sum }_{k=1}^{+\infty }{\phi }_k<+\infty .$ Update

$${\sigma }_{k+1}=\left\{\begin{array}{c}min\left\{{\sigma }_k+{\phi }_k,\frac{(2-\sqrt{2}-\rho )\mu \parallel r_k-q_k{\parallel }^2+(2-\sqrt{2}-\rho )\mu {\parallel s_{k+1}-q_k\parallel }^2}{2〈\mathcal{E}\left(r_k\right)-\mathcal{E}\left(q_k\right),s_{k+1}-q_k〉}\right\}\hfill \\ \mathrm{if}〈\mathcal{E}\left(r_k\right)-\mathcal{E}\left(q_k\right),s_{k+1}-q_k〉>0,\hfill \\ {\sigma }_k+{\phi }_k,\mathrm{otherwise}.\hfill \end{array}\right.$$

Then, the sequences $\left\{s_k\right\}$ converge strongly to ${\nu }^{*}\in Sol(\mathcal{E},\Delta ).$

Corollary 6.

Suppose that $\mathcal{D}:\Delta \to \Sigma$ is weakly continuous, L-Lipschitz continuous, and κ-strict pseudocontraction operator with $Sol(\mathcal{D},\Delta )\ne \varnothing .$ Select ${\sigma }_1>0,$$s_{-1},s_0\in \Sigma ,$$\varkappa \in (0,1),$$\mu \in (0,1),$$\rho \in (0,2-\sqrt{2})$ with a sequence $\left\{{\chi }_k\right\}\subset (0,1)$ satisfy the conditions:

$$\underset{k\to +\infty }{lim}{\chi }_k=0\mathrm{and}\sum _{k=1}^{+\infty }{\chi }_k=+\infty .$$

Compute

$$r_k=s_k+{\varkappa }_k(s_k-s_{k-1})-{\chi }_k\left[s_k+{\varkappa }_k(s_k-s_{k-1})\right],$$

and choose ${\varkappa }_k$ such that

$$0\le {\varkappa }_k\le \widehat{{\varkappa }_k}\mathrm{and}\widehat{{\varkappa }_k}=\left\{\begin{array}{cc}min\left\{\varkappa ,\frac{{ϵ}_k}{\parallel s_k-s_{k-1}\parallel }\right\}\hfill & \mathrm{if}{s}_k\ne s_{k-1},\hfill \\ \varkappa \hfill & \mathrm{otherwise},\hfill \end{array}\right.$$

where ${ϵ}_k=\circ \left({\chi }_k\right)$ a positive sequence such that ${lim}_{k\to +\infty }\frac{{ϵ}_k}{{\chi }_k}=0.$

Compute

$$q_k=P_{\Delta }\left[r_k-{\sigma }_k(r_k-\mathcal{D}\left(r_k\right))\right].$$

Given $s_{k-1},s_k,$$q_k,$ and construct a half-space

$${\Sigma }_k=\{z\in \Sigma :\langle (1-{\sigma }_k)r_k+{\sigma }_k\mathcal{D}\left(r_k\right)-q_k,z-q_k\rangle \le 0\}.$$

Compute

$$s_{k+1}=P_{{\Sigma }_k}\left[r_k-{\sigma }_k(q_k-\mathcal{D}\left(q_k\right))\right].$$

Evaluate stepsize rule for next iteration, which is evaluated as follows:

$${\sigma }_{k+1}=\left\{\begin{array}{c}min\left\{{\sigma }_k,\frac{(2-\sqrt{2}-\rho )\mu \parallel r_k-q_k{\parallel }^2+(2-\sqrt{2}-\rho )\mu {\parallel s_{k+1}-q_k\parallel }^2}{2〈(r_k-q_k)-[\mathcal{D}\left(r_k\right)-\mathcal{D}\left(q_k\right)],s_{k+1}-q_k〉}\right\}\hfill \\ \mathrm{if}〈(r_k-q_k)-[\mathcal{D}\left(r_k\right)-\mathcal{D}\left(q_k\right)],s_{k+1}-q_k〉>0,\hfill \\ {\sigma }_k\mathrm{otherwise}.\hfill \end{array}\right.$$

Then, the sequence $\left\{s_k\right\}$ converges strongly to ${\nu }^{*}\in Sol(\mathcal{D},\Delta ).$

Corollary 7.

Suppose that $\mathcal{D}:\Delta \to \Sigma$ is weakly continuous, L-Lipschitz continuous, and κ-strict pseudocontraction operator with $Sol(\mathcal{D},\Delta )\ne \varnothing .$ Select ${\sigma }_1>0,$$s_{-1},s_0\in \Sigma ,$$\varkappa \in (0,1),$$\mu \in (0,1),$$\rho \in (0,2-\sqrt{2})$ with a sequence $\left\{{\chi }_k\right\}\subset (0,1)$ satisfy the conditions:

$$\underset{k\to +\infty }{lim}{\chi }_k=0\mathrm{and}\sum _{k=1}^{+\infty }{\chi }_k=+\infty .$$

Compute

$$r_k=s_k+{\varkappa }_k(s_k-s_{k-1})-{\chi }_k\left[s_k+{\varkappa }_k(s_k-s_{k-1})\right],$$

and choose ${\varkappa }_k$ such that

$$0\le {\varkappa }_k\le \widehat{{\varkappa }_k}\mathrm{and}\widehat{{\varkappa }_k}=\left\{\begin{array}{cc}min\left\{\varkappa ,\frac{{ϵ}_k}{\parallel s_k-s_{k-1}\parallel }\right\}\hfill & \mathrm{if}{s}_k\ne s_{k-1},\hfill \\ \varkappa \hfill & \mathrm{otherwise},\hfill \end{array}\right.$$

where ${ϵ}_k=\circ \left({\chi }_k\right)$ a positive sequence such that ${lim}_{k\to +\infty }\frac{{ϵ}_k}{{\chi }_k}=0.$

Compute

$$q_k=P_{\Delta }\left[r_k-{\sigma }_k(r_k-\mathcal{D}\left(r_k\right))\right].$$

Given $s_{k-1},s_k,$$q_k,$ and construct a half-space

$${\Sigma }_k=\{z\in \Sigma :\langle (1-{\sigma }_k)r_k+{\sigma }_k\mathcal{D}\left(r_k\right)-q_k,z-q_k\rangle \le 0\}.$$

Compute

$$s_{k+1}=P_{{\Sigma }_k}\left[r_k-{\sigma }_k(q_k-\mathcal{D}\left(q_k\right))\right].$$

Select a non-negative real sequence $\left\{{\phi }_k\right\}$ such that ${\sum }_{k=1}^{+\infty }{\phi }_k<+\infty .$ Update

$${\sigma }_{k+1}=\left\{\begin{array}{c}min\left\{{\sigma }_k+{\phi }_k,\frac{(2-\sqrt{2}-\rho )\mu \parallel r_k-q_k{\parallel }^2+(2-\sqrt{2}-\rho )\mu {\parallel s_{k+1}-q_k\parallel }^2}{2〈(r_k-q_k)-[\mathcal{D}\left(r_k\right)-\mathcal{D}\left(q_k\right)],s_{k+1}-q_k〉}\right\}\hfill \\ \mathrm{if}〈(r_k-q_k)-[\mathcal{D}\left(r_k\right)-\mathcal{D}\left(q_k\right)],s_{k+1}-q_k〉>0,\hfill \\ {\sigma }_k+{\phi }_k\mathrm{otherwise}.\hfill \end{array}\right.$$

Then, the sequence $\left\{s_k\right\}$ converges strongly to ${\nu }^{*}\in Sol(\mathcal{D},\Delta ).$

Corollary 8.

Suppose that $\mathcal{D}:\Delta \to \Sigma$ is weakly continuous, L-Lipschitz continuous, and κ-strict pseudocontraction operator with $Sol(\mathcal{D},\Delta )\ne \varnothing .$ Select ${\sigma }_1>0,$$s_{-1},s_0\in \Sigma ,$$\varkappa \in (0,1),$$\mu \in (0,1),$$\rho \in (0,2-\sqrt{2})$ with a sequence $\left\{{\chi }_k\right\}\subset (0,1)$ satisfy the conditions:

$$\underset{k\to +\infty }{lim}{\chi }_k=0\mathrm{and}\sum _{k=1}^{+\infty }{\chi }_k=+\infty .$$

Compute

$$r_k=s_k+{\varkappa }_k(s_k-s_{k-1})-{\chi }_k\left[s_k+{\varkappa }_k(s_k-s_{k-1})\right],$$

and choose ${\varkappa }_k$ such that

$$0\le {\varkappa }_k\le \widehat{{\varkappa }_k}\mathrm{and}\widehat{{\varkappa }_k}=\left\{\begin{array}{cc}min\left\{\varkappa ,\frac{{ϵ}_k}{\parallel s_k-s_{k-1}\parallel }\right\}\hfill & \mathrm{if}{s}_k\ne s_{k-1},\hfill \\ \varkappa \hfill & \mathrm{otherwise},\hfill \end{array}\right.$$

where ${ϵ}_k=\circ \left({\chi }_k\right)$ a positive sequence such that ${lim}_{k\to +\infty }\frac{{ϵ}_k}{{\chi }_k}=0.$ Compute

$$\left\{\begin{array}{c}{q}_k=P_{\Delta }\left[r_k-{\sigma }_k(r_k-\mathcal{D}\left(r_k\right))\right],\hfill \\ s_{k+1}=P_{\Delta }\left[r_k-{\sigma }_k(q_k-\mathcal{D}\left(q_k\right))\right].\hfill \end{array}\right.$$

Evaluate stepsize rule for next iteration, which is evaluated as follows:

$${\sigma }_{k+1}=\left\{\begin{array}{c}min\left\{{\sigma }_k,\frac{(2-\sqrt{2}-\rho )\mu \parallel r_k-q_k{\parallel }^2+(2-\sqrt{2}-\rho )\mu {\parallel s_{k+1}-q_k\parallel }^2}{2〈(r_k-q_k)-[\mathcal{D}\left(r_k\right)-\mathcal{D}\left(q_k\right)],s_{k+1}-q_k〉}\right\}\hfill \\ \mathrm{if}〈(r_k-q_k)-[\mathcal{D}\left(r_k\right)-\mathcal{D}\left(q_k\right)],s_{k+1}-q_k〉>0,\hfill \\ {\sigma }_k\mathrm{otherwise}.\hfill \end{array}\right.$$

Then, the sequence $\left\{s_k\right\}$ converges strongly to ${\nu }^{*}\in Sol(\mathcal{D},\Delta ).$

Corollary 9.

Suppose that $\mathcal{D}:\Delta \to \Sigma$ is weakly continuous, L-Lipschitz continuous, and κ-strict pseudocontraction operator with $Sol(\mathcal{D},\Delta )\ne \varnothing .$ Select ${\sigma }_1>0,$$s_{-1},s_0\in \Sigma ,$$\varkappa \in (0,1),$$\mu \in (0,1),$$\rho \in (0,2-\sqrt{2})$ with a sequence $\left\{{\chi }_k\right\}\subset (0,1)$ satisfy the conditions:

$$\underset{k\to +\infty }{lim}{\chi }_k=0\mathrm{and}\sum _{k=1}^{+\infty }{\chi }_k=+\infty .$$

Compute

$$r_k=s_k+{\varkappa }_k(s_k-s_{k-1})-{\chi }_k\left[s_k+{\varkappa }_k(s_k-s_{k-1})\right],$$

and choose ${\varkappa }_k$ such that

$$0\le {\varkappa }_k\le \widehat{{\varkappa }_k}\mathrm{and}\widehat{{\varkappa }_k}=\left\{\begin{array}{cc}min\left\{\varkappa ,\frac{{ϵ}_k}{\parallel s_k-s_{k-1}\parallel }\right\}\hfill & \mathrm{if}{s}_k\ne s_{k-1},\hfill \\ \varkappa \hfill & \mathrm{otherwise},\hfill \end{array}\right.$$

where ${ϵ}_k=\circ \left({\chi }_k\right)$ a positive sequence such that ${lim}_{k\to +\infty }\frac{{ϵ}_k}{{\chi }_k}=0.$

Compute

$$\left\{\begin{array}{c}{q}_k=P_{\Delta }\left[r_k-{\sigma }_k(r_k-\mathcal{D}\left(r_k\right))\right],\hfill \\ s_{k+1}=P_{\Delta }\left[r_k-{\sigma }_k(q_k-\mathcal{D}\left(q_k\right))\right].\hfill \end{array}\right.$$

Furthermore, select a non-negative real sequence $\left\{{\phi }_k\right\}$ such that ${\sum }_{k=1}^{+\infty }{\phi }_k<+\infty .$ Update

$${\sigma }_{k+1}=\left\{\begin{array}{c}min\left\{{\sigma }_k+{\phi }_k,\frac{(2-\sqrt{2}-\rho )\mu \parallel r_k-q_k{\parallel }^2+(2-\sqrt{2}-\rho )\mu {\parallel s_{k+1}-q_k\parallel }^2}{2〈(r_k-q_k)-[\mathcal{D}\left(r_k\right)-\mathcal{D}\left(q_k\right)],s_{k+1}-q_k〉}\right\}\hfill \\ \mathrm{if}〈(r_k-q_k)-[\mathcal{D}\left(r_k\right)-\mathcal{D}\left(q_k\right)],s_{k+1}-q_k〉>0,\hfill \\ {\sigma }_k+{\phi }_k\mathrm{otherwise}.\hfill \end{array}\right.$$

Then, the sequence $\left\{s_k\right\}$ converges strongly to ${\nu }^{*}\in Sol(\mathcal{D},\Delta ).$

5. Numerical Illustrations

This section describes a number of numerical experiments conducted to demonstrate the efficacy of the suggested techniques. Some of these numerical experiments provide a thorough understanding of how to select optimal control parameters. Some of them show how the suggested strategies outperform current ones in the literature. All MATLAB codes were executed in MATLAB 9.5 (R2018b) on an Intel(R) Core(TM) i5-6200 Processor CPU @ 2.30–2.40 GHz, with 8.00 GB RAM.

Example 1.

The first sample problem given is taken from the Nash–Cournot Oligopolistic Equilibrium model in [10]. In this case, the bifunction ϝ could be written as follows:

$$ϝ(p,q)=\langle Pp+Qq+c,q-p\rangle ,$$

where matrices $P,Q$ and vector c are defined by

$$P=\left(\begin{array}{ccccc}3.1& 2& 0& 0& 0\\ 2& 3.6& 0& 0& 0\\ 0& 0& 3.5& 2& 0\\ 0& 0& 2& 3.3& 0\\ 0& 0& 0& 0& 3\end{array}\right)Q=\left(\begin{array}{ccccc}1.6& 1& 0& 0& 0\\ 1& 1.6& 0& 0& 0\\ 0& 0& 1.5& 1& 0\\ 0& 0& 1& 1.5& 0\\ 0& 0& 0& 0& 2\end{array}\right)c=\left(\begin{array}{c}1\\ -2\\ -1\\ 2\\ -1\end{array}\right).$$

The matrix $Q-P$ has eigenvalues $-2.9050,-2.7808,-1.0000,-0.8950,-0.7192.$ As a result, the matrices $Q-P$ and Q are symmetrically negative semidefinite and symmetrically positive semidefinite, respectively. In addition, the Lipschitz-type constants are $c_1=c_2=\frac{1}{2}\parallel P-Q\parallel =1.4525.$ The constraint set $\Delta \subset {\mathbb{R}}^M$ is defined by

$$\Delta :=\{p\in {\mathbb{R}}^5:-2\le p_i\le 5;i=1,2,3,4,5\}.$$

Consider the following details regarding control settings: (1) Algorithm 3.2 in [27] (to make it short,Iter.Mthd.1): $\sigma =\frac{1}{max\left\{3c_1,3c_2\right\}},{\varkappa }_k=\frac{1}{5(k+2)},Error-Term={\parallel s_k-q_k\parallel }^2;$ (2) Algorithm 4.1 in [26] (to make it short,Iter.Mthd.2): ${\sigma }_1=0.25,\mu =0.33,{\varkappa }_k=\frac{1}{{(k+1)}^{0.5}},Error-Term=max\left\{\parallel s_{k+1}-q_k{\parallel }^2,{\parallel s_k-q_k\parallel }^2\right\};$ (3) Algorithm 3 in [28] (to make it short,Iter.Mthd.3): $\sigma =\frac{1}{max\left\{3c_1,3c_2\right\}},\rho =0.50,{ϵ}_k=\frac{1}{{(k+1)}^2},{\gamma }_k=\frac{1}{5(k+2)},{\chi }_k=\frac{5}{10}(1-{\gamma }_k),Error-Term={\parallel r_k-q_k\parallel }^2;$ (4) For Algorithm 1 (to make it short,Iter.Mthd.4): ${\sigma }_1=0.50,\varkappa =0.50,\mu =0.55,\rho =0.05,{ϵ}_k=\frac{1}{k^2},Error-Term={\parallel r_k-q_k\parallel }^2;$ (5) For Algorithm 2 (to make it short,Iter.Mthd.5): ${\sigma }_1=0.50,\varkappa =0.50,\mu =0.55,\rho =0.05,{ϵ}_k=\frac{1}{k^2},{\phi }_k=\frac{100}{{(1+k)}^2},Error-Term={\parallel r_k-q_k\parallel }^2.$ Figure 1, Figure 2, Figure 3, Figure 4, Figure 5, Figure 6, Figure 7 and Figure 8 and

Table 1. The first three techniques have numerical data for Figure 1, Figure 2, Figure 3, Figure 4, Figure 5, Figure 6, Figure 7 and Figure 8 by using Example 1.

	Number of Iterations			Execution Time in Seconds
${\mathbf{s}}_{\mathbf{0}}$	Iter.Mthd.1	Iter.Mthd.2	Iter.Mthd.3	Iter.Mthd.1	Iter.Mthd.2	Iter.Mthd.3
${(1,1,1,1,1)}^T$	40	28	24	0.344801	0.26301	0.24393
${(1,2,1,2,3)}^T$	36	30	25	0.358533	0.39007	0.25731
${(2,2,3,-4,4)}^T$	50	35	27	0.467655	0.38310	0.29807
${(2,-2,3,-4,6)}^T$	44	41	27	0.448825	0.40952	0.27013

and

Table 2. The proposed two techniques have numerical data for Figure 1, Figure 2, Figure 3, Figure 4, Figure 5, Figure 6, Figure 7 and Figure 8 by using Example 1.

	Number of Iterations		Execution Time in Seconds
${\mathbf{s}}_{\mathbf{0}}$	Iter.Mthd.4	Iter.Mthd.5	Iter.Mthd.4	Iter.Mthd.5
${(1,1,1,1,1)}^T$	13	8	0.1261769	0.0854048
${(1,2,1,2,3)}^T$	18	8	0.1857996	0.0932000
${(2,2,3,-4,4)}^T$	19	11	0.2098914	0.1436989
${(2,-2,3,-4,6)}^T$	20	10	0.2199371	0.1250085

are shown numerical results for Example 1.

We present two iterative methods Algorithms 3 and 4 based on a monotone and nonmonotone variable stepsize rule, as well as two strongly convex minimization problems that do not require subgradient methods. The second most significant methods are described below.

Example 2.

Let $ϝ:\Delta \times \Delta \to ϝ$ be a bifunction defined by

$$ϝ(p,q)=\sum _{i=2}^5(q_i-p_i)\parallel p\parallel ,\forall p,q\in {ϝ}^5.$$

Furthermore, the set Δ is taken into account as follows:

$$\Delta =\left\{(s_1,\cdots ,s_5):s_1\ge -1,s_i\ge 1,i=2,\cdots ,5\right\}.$$

As a result, the bifunction ϝ is Lipschitz-type continuous with $c_1=c_2=2$ and meets the criterion ($ϝ$1)–($ϝ$5). The solution set of an equilibrium problem is $Sol(ϝ,\Delta )=\{(s_1,1,1,1,1):s_1>1\};$ for details, see [9]. Consider the following information regarding control settings: (1) Algorithm 3.2 in [27] (to make it short,Iter.Mthd.1): $\sigma =\frac{1}{max\left\{4c_1,4c_2\right\}},{\varkappa }_k=\frac{1}{4(k+2)},Error-Term={\parallel s_k-q_k\parallel }^2;$ (2) Algorithm 4.1 in [26] (to make it short,Iter.Mthd.2): ${\sigma }_1=0.33,\mu =0.53,$${\varkappa }_k=\frac{1}{{(k+1)}^{0.4}},Error-Term=max\left\{\parallel s_{k+1}-q_k{\parallel }^2,{\parallel s_k-q_k\parallel }^2\right\};$ (3) Algorithm 3 in [28] (to make it short,Iter.Mthd.3): $\sigma =\frac{1}{max\left\{4c_1,4c_2\right\}},\rho =0.55,{ϵ}_k=\frac{1}{{(k+1)}^2},{\gamma }_k=\frac{1}{4(k+2)},{\chi }_k=\frac{7}{10}(1-{\gamma }_k),Error-Term={\parallel r_k-q_k\parallel }^2;$ (4) For Algorithm 1 (to make it short,Iter.Mthd.4): ${\sigma }_1=0.33,\varkappa =0.76,\mu =0.53,\rho =0.045,{ϵ}_k=\frac{1}{k^2},Error-Term={\parallel r_k-q_k\parallel }^2;$ (5) For Algorithm 2 (to make it short,Iter.Mthd.5): ${\sigma }_1=0.33,\varkappa =0.76,\mu =0.53,\rho =0.045,{ϵ}_k=\frac{1}{k^2},{\phi }_k=\frac{100}{{(1+k)}^2},Error-Term={\parallel r_k-q_k\parallel }^2.$ Figure 9, Figure 10, Figure 11, Figure 12, Figure 13, Figure 14, Figure 15 and Figure 16 and

Table 3. The first three techniques have numerical data for Figure 9, Figure 10, Figure 11, Figure 12, Figure 13, Figure 14, Figure 15 and Figure 16 by using Example 2.

	Number of Iterations			Execution Time in Seconds
${\mathbf{s}}_{\mathbf{0}}$	Iter.Mthd.1	Iter.Mthd.2	Iter.Mthd.3	Iter.Mthd.1	Iter.Mthd.2	Iter.Mthd.3
${(1,-1,2,0,3)}^T$	44	33	22	0.3408147	0.3129066	0.236492066
${(2,-1,3,2,5)}^T$	54	35	23	0.6523779	0.3518180	0.256393849
${(-1,6,3,-4,4)}^T$	56	35	25	0.5266949	0.3325744	0.259483922
${(6,-1,5,-4,1)}^T$	57	40	25	0.4948373	0.3590396	0.258739392

and

Table 4. The proposed two techniques have numerical data for Figure 9, Figure 10, Figure 11, Figure 12, Figure 13, Figure 14, Figure 15 and Figure 16 by using Example 2.

	Number of Iterations		Execution Time in Seconds
${\mathbf{s}}_{\mathbf{0}}$	Iter.Mthd.4	Iter.Mthd.5	Iter.Mthd.4	Iter.Mthd.5
${(1,-1,2,0,3)}^T$	14	10	0.1276095	0.1807799
${(2,-1,3,2,5)}^T$	16	11	0.1522214	0.1913611
${(-1,6,3,-4,4)}^T$	16	13	0.1542963	0.1918947
${(6,-1,5,-4,1)}^T$	16	13	0.1445121	0.1881485

are shown numerical results for Example 2.

Algorithm 2: Explicit Subgradient Extragradient Method With Non-Monotone Stepsize Rule

STEP 0: Choose ${\sigma }_1>0,$$s_{-1},s_0\in \Sigma ,$$\varkappa \in (0,1),$$\mu \in (0,1),$$\rho \in (0,2-\sqrt{2})$ with $\left\{{\chi }_k\right\}\subset (0,1)$ satisfy the following conditions: $$\underset{k\to +\infty }{lim}{\chi }_k=0\mathrm{and}\sum _{k=1}^{+\infty }{\chi }_k=+\infty .$$ STEP 1: Compute $$r_k=s_k+{\varkappa }_k(s_k-s_{k-1})-{\chi }_k\left[s_k+{\varkappa }_k(s_k-s_{k-1})\right],$$ where choose ${\varkappa }_k$ such that $$0\le {\varkappa }_k\le \widehat{{\varkappa }_k}\mathrm{and}\widehat{{\varkappa }_k}=\left\{\begin{array}{cc}min\left\{\varkappa ,\frac{{ϵ}_k}{\parallel s_k-s_{k-1}\parallel }\right\}\hfill & \mathrm{if}{s}_k\ne s_{k-1},\hfill \\ \varkappa \hfill & \mathrm{otherwise},\hfill \end{array}\right.$$ (45) where ${ϵ}_k=\circ \left({\chi }_k\right)$ a positive sequence such that ${lim}_{k\to +\infty }\frac{{ϵ}_k}{{\chi }_k}=0.$ STEP 2: Compute $$q_k=\underset{q\in \Delta }{argmin}\{{\sigma }_kϝ(r_k,q)+\frac{1}{2}\parallel r_k-q{\parallel }^2\}.$$ STEP 3: Given the current iterates $s_{k-1},$$s_k,$$q_k.$ Firstly choose ${\omega }_k\in {\partial }_2ϝ(r_k,q_k)$ satisfying $r_k-{\sigma }_k{\omega }_k-q_k\in N_{\Delta }\left(q_k\right)$ and generate a half-space $${\Sigma }_k=\{z\in \Sigma :\langle r_k-{\sigma }_k{\omega }_k-q_k,z-q_k\rangle \le 0\}.$$ Compute $$s_{k+1}=\underset{q\in {\Sigma }_k}{argmin}\{{\sigma }_kϝ(q_k,q)+\frac{1}{2}\parallel r_k-q{\parallel }^2\}.$$ STEP 4: Select a non-negative real sequence $\left\{{\phi }_k\right\}$ such that ${\sum }_{k=1}^{+\infty }{\phi }_k<+\infty .$ Compute $${\sigma }_{k+1}=\left\{\begin{array}{c}min\left\{{\sigma }_k+{\phi }_k,\frac{(2-\sqrt{2}-\rho )\mu \parallel r_k-q_k{\parallel }^2+(2-\sqrt{2}-\rho )\mu {\parallel s_{k+1}-q_k\parallel }^2}{2[ϝ(r_k,s_{k+1})-ϝ(r_k,q_k)-ϝ(q_k,s_{k+1})]}\right\}\hfill \\ \mathrm{if}ϝ(r_k,s_{k+1})-ϝ(r_k,q_k)-ϝ(q_k,s_{k+1})>0,\hfill \\ {\sigma }_k+{\phi }_k,\mathrm{otherwise}.\hfill \end{array}\right.$$ (46) STEP 5: If $q_k=r_k,$ then complete the computation. Otherwise, set $k:=k+1$ and go back STEP 1.

Algorithm 3: Explicit Extragradient Method With Monotone Stepsize Rule

STEP 0: Choose ${\sigma }_1>0,$$s_{-1},s_0\in \Sigma ,$$\varkappa \in (0,1),$$\mu \in (0,1),$$\rho \in (0,2-\sqrt{2})$ with $\left\{{\chi }_k\right\}\subset (0,1)$ satisfy the following conditions: $$\underset{k\to +\infty }{lim}{\chi }_k=0\mathrm{and}\sum _{k=1}^{+\infty }{\chi }_k=+\infty .$$ STEP 1: Compute $$r_k=s_k+{\varkappa }_k(s_k-s_{k-1})-{\chi }_k\left[s_k+{\varkappa }_k(s_k-s_{k-1})\right],$$ where choose ${\varkappa }_k$ such that $$0\le {\varkappa }_k\le \widehat{{\varkappa }_k}\mathrm{and}\widehat{{\varkappa }_k}=\left\{\begin{array}{cc}min\left\{\varkappa ,\frac{{ϵ}_k}{\parallel s_k-s_{k-1}\parallel }\right\}\hfill & \mathrm{if}{s}_k\ne s_{k-1},\hfill \\ \varkappa \hfill & \mathrm{otherwise},\hfill \end{array}\right.$$ (47) where ${ϵ}_k=\circ \left({\chi }_k\right)$ a sequence such that ${lim}_{k\to +\infty }\frac{{ϵ}_k}{{\chi }_k}=0.$ STEP 2: Compute $$q_k=\underset{q\in \Delta }{argmin}\{{\sigma }_kϝ(r_k,q)+\frac{1}{2}\parallel r_k-q{\parallel }^2\}.$$ STEP 3: Compute $$s_{k+1}=\underset{q\in \Delta }{argmin}\{{\sigma }_kϝ(q_k,q)+\frac{1}{2}\parallel r_k-q{\parallel }^2\}.$$ STEP 4: Compute $${\sigma }_{k+1}=\left\{\begin{array}{c}min\left\{{\sigma }_k,\frac{(2-\sqrt{2}-\rho )\mu \parallel r_k-q_k{\parallel }^2+(2-\sqrt{2}-\rho )\mu {\parallel s_{k+1}-q_k\parallel }^2}{2[ϝ(r_k,s_{k+1})-ϝ(r_k,q_k)-ϝ(q_k,s_{k+1})]}\right\}\hfill \\ \mathrm{if}ϝ(r_k,s_{k+1})-ϝ(r_k,q_k)-ϝ(q_k,s_{k+1})>0,\hfill \\ {\sigma }_k,\mathrm{otherwise}.\hfill \end{array}\right.$$ (48) STEP 5: If $q_k=r_k,$ then complete the computation. Otherwise, set $k:=k+1$ and go back STEP 1.

Algorithm 4: Explicit Extragradient Method With Non-Monotone Stepsize Rule

STEP 0: Choose ${\sigma }_1>0,$$s_{-1},s_0\in \Sigma ,$$\varkappa \in (0,1),$$\mu \in (0,1),$$\rho \in (0,2-\sqrt{2})$ with a sequence $\left\{{\chi }_k\right\}\subset (0,1)$ satisfy the following conditions: $$\underset{k\to +\infty }{lim}{\chi }_k=0\mathrm{and}\sum _{k=1}^{+\infty }{\chi }_k=+\infty .$$ STEP 1: Compute $$r_k=s_k+{\varkappa }_k(s_k-s_{k-1})-{\chi }_k\left[s_k+{\varkappa }_k(s_k-s_{k-1})\right],$$ where choose ${\varkappa }_k$ such that $$0\le {\varkappa }_k\le \widehat{{\varkappa }_k}\mathrm{and}\widehat{{\varkappa }_k}=\left\{\begin{array}{cc}min\left\{\varkappa ,\frac{{ϵ}_k}{\parallel s_k-s_{k-1}\parallel }\right\}\hfill & \mathrm{if}{s}_k\ne s_{k-1},\hfill \\ \varkappa \hfill & \mathrm{otherwise},\hfill \end{array}\right.$$ (49) where ${ϵ}_k=\circ \left({\chi }_k\right)$ a positive sequence such that ${lim}_{k\to +\infty }\frac{{ϵ}_k}{{\chi }_k}=0.$ STEP 2: Compute $$q_k=\underset{q\in \Delta }{argmin}\{{\sigma }_kϝ(r_k,q)+\frac{1}{2}\parallel r_k-q{\parallel }^2\}.$$ STEP 3: Compute $$s_{k+1}=\underset{q\in \Delta }{argmin}\{{\sigma }_kϝ(q_k,q)+\frac{1}{2}\parallel r_k-q{\parallel }^2\}.$$ STEP 4: Furthermore, select a non-negative real sequence $\left\{{\phi }_k\right\}$ as well as ${\sum }_{k=1}^{+\infty }{\phi }_k<+\infty .$ Compute $${\sigma }_{k+1}=\left\{\begin{array}{c}min\left\{{\sigma }_k+{\phi }_k,\frac{(2-\sqrt{2}-\rho )\mu \parallel r_k-q_k{\parallel }^2+(2-\sqrt{2}-\rho )\mu {\parallel s_{k+1}-q_k\parallel }^2}{2[ϝ(r_k,s_{k+1})-ϝ(r_k,q_k)-ϝ(q_k,s_{k+1})]}\right\}\hfill \\ \mathrm{if}ϝ(r_k,s_{k+1})-ϝ(r_k,q_k)-ϝ(q_k,s_{k+1})>0,\hfill \\ {\sigma }_k+{\phi }_k,\mathrm{otherwise}.\hfill \end{array}\right.$$ (50) STEP 5: If $q_k=r_k,$ then complete the computation. Otherwise, set $k:=k+1$ and go back STEP 1.

Some Observations through Numerical Illustrations: The following conclusions can be drawn from the above-mentioned numerical examples: (1) Numerical findings for multiple methods in finite-dimensional domains have been presented for Examples 1 and 2. It is apparent that the supplied techniques are efficacious in terms of the number of iterations and the lapse of time in practically all scenarios. All trials demonstrate that the presented algorithms outperform the techniques available beforehand. (2) In both situations, the existence of an insufficient stepsize results in a hump in the graph of the methods. It has no effect on the overall effectiveness of the techniques. (3) Examples 1 and 2 have provided data for a variety of approaches in both finite-dimensional and infinite-dimensional domains. In more situations, we could see that the computation development is enhanced by the problem’s complexity and the tolerance value employed. (4) It has also been revealed that adopting a specific formula for a stepsize evaluation enhances the efficiency and speed of convergence of the algorithm. In other words, rather than having a fixed stepsize, a variable stepsize improves algorithm efficiency. (5) In Examples 1 and 2, it is also proved that the beginning point selection and the complexity of the operators have an influence on the effectiveness of techniques in proportion to the number of iterations and the time of operation in seconds.

6. Conclusions

The paper proposes four different explicit extragradient-like techniques for solving an equilibrium problem in a real Hilbert space by combining a pseudomonotone with a Lipschitz-type bifunction. An innovative stepsize rule has been presented that would not rely on Lipschitz-type constant information. The algorithm’s convergence has been verified. Several experiments are constructed to demonstrate the numerical behavior of our two algorithms and to compare them to others widely known in the literature.