A novel accelerated extragradient algorithm to solve pseudomonotone variational inequalities

In this paper, we propose a new inertial iterative method to solve classical variational inequalities with pseudomonotone and Lipschitz continuous operators in the setting of a real Hilbert space. The proposed iterative scheme is basically analogous to the extragradient method used to solve the problems of variational inequalities in real Hilbert spaces. The strong convergence of the proposed algorithm is set up with the prior knowledge of Lipschitz’s constant of an operator. Finally, several computational experiments are listed to show the applicability and efficiency of the proposed algorithm.


Introduction
This article studies the iterative method that is used to estimate the solution of variational inequality problem (shortly, VIP) in the setting of a real Hilbert space. Let X be a real Hilbert space and D be a non-empty, closed, and convex subset of X . Let S : X → X to be an operator. The problem (VIP) for S on D is given as follows [15,22]: Find u * ∈ D such that S(u * ), y − u * ≥ 0, ∀ y ∈ D. (VIP) Let us consider that is the solution set of the problem (VIP). This idea of variational inequalities includes different disciplines such as partial differential equations, optimization, optimal control, mechanics, mathematical programming, and finance (see [6,[11][12][13][14]18,24]). This problem is an important topic in the physical sciences, and a considerable amount of discussion has been given to many authors who have dedicated themselves to studying not only the theory of existence and the stability of solutions but also the iterative method used to solve the problem. Korpelevich [16] and Antipin [2] have established the following extragradient method. Their method consists of the following: Let u 0 ∈ D and 0 < τ < 1 L such that On the other hand, many projection methods are used to figure out the numerical solution of variational inequalities. Many researchers have suggested various forms of projection techniques to solve the problem (VIP) (see for details [10,16,19,25,26,[28][29][30][31][32]34,36]). Almost all the methods for solving the problem (VIP) are based on the projection method, which is computed on the feasible set D. It is important to note that the above well-established method has two significant flaws, the first being the fixed constant step size, which involves the knowledge or approximation of the Lipschitz constant of the respective operator and is only weakly convergent in the Hilbert spaces. From the computational point of view, it might be problematic to use a fixed step size, and hence the convergence rate and appropriateness of the method could be affected.
The main contribution of this study is to develop an inertial-type method used to improve the convergence rate of the sequence. Previously, such approaches have been developed based on the oscillator equation with damping and conservative force restoration. This second-order dynamical structure is called a strong friction ball and was originally studied by Polyak in [20]. Primarily, the functionality of the inertial-type method is that it will use the two prior iterations to execute the next iteration. Therefore, a natural question is raised: "Is it possible to introduce new strongly convergent inertial extragradient-like method to solve the problem (VIP)"?
In this study, we give a positive answer to the above question, i.e., the gradient method indeed generates a strongly convergent iterative sequence by letting a fixed and variable step size rule. In this paper, we study a different method to obtain strong convergence and introduce a new iterative method for solving variational inequalities involving pseudomonotone and the Lipschitz operator in a real Hilbert space. Our method is inspired by one projection method [16] and the inertial technique in [20]. At each iteration, the method only needs to compute one projection onto the feasible set. Under some suitable conditions imposed on control parameters, the iterative sequences generated by our method converge strongly to some solution of the considered problem. We also provide numerical examples to illustrate the computational effectiveness of the new method over some existing methods.
The paper is organized in the following way. In Sect. 2, we review some concepts and preliminary results used in the paper. Section 3 deals with the description of the method and proves its convergence theorems. Finally, Sect. 4 presents some numerical results to illustrate the convergence and effectiveness of the proposed method.

Background
In this section of the manuscript, we have written a number of important identities and relevant lemmas and definitions. For all u, y ∈ X , we have u + y 2 = u 2 + 2 u, y + y 2 .
A metric projection P D (y 1 ) of y 1 ∈ X is defined by First, we list some of the important features of projection operator. Lemma 2.1 [3] Suppose that P D : X → D is a metric projection. Then, the following conditions were satisfied.
Then, lim n→+∞ p n = 0. [17] Assume that { p n } is a sequence of real numbers such that there exists a subsequence {n i } of {n} such that

Lemma 2.3
Then, there is a non decreasing sequence m k ⊂ N such that m k → +∞ as k → +∞, and meet the following requirements for numbers k ∈ N: Indeed, m k = max{ j ≤ k : p j ≤ p j+1 }.
Next, we list some of the important identities that were used to prove the convergence analysis.

Lemma 2.4 [3]
For any y 1 , y 2 ∈ X and ∈ R. Then, the following inequalities hold: (ii) Lemma 2.5 [23] Assume that S : D → X is a pseudomonotone and continuous operator. Then, u * is a solution to the problem (VIP) if and only if u * is a solution to the following problem: Find u ∈ D such that S(y), y − u ≥ 0, ∀ y ∈ D.
Step 2: Compute y n = P D (t n − τ S(t n )). If t n = y n , then STOP and y n is a solution. Otherwise, go to next step.

Main results
To investigate the convergence analysis, it is considered that the following requirements are satisfied: (S1) A solution set of problem (VIP) is denoted by and it is non-empty; (S2) An operator S : X → X is said to be pseudomonotone if (S3) An operator S : X → X is said to be Lipschitz continuous with constant L > 0 if there exists L > 0 such that (S4) An operator S : X → X is said to be sequentially weakly continuous if {S(u n )} converges weakly to S(u) for every sequence {u n } converges weakly to u. The main algorithm is given the following form: Lemma 3.1 Assume that S : X → X satisfies the conditions (S1)-(S4). For a given u * ∈ = ∅, we have Proof First consider the following It is given that u * ∈ ⊂ D such that which implies that By the use of expressions (3) and (5), we obtain By given that u * is the solution of problem (VIP), we have Due to the pseudomonotonicity of S on D, we get By substituting y = y n ∈ D, we get By the use of expressions (6) and (7), we obtain It is given that Combining expressions (8) and (9), we obtain Then, {u n } strongly converges to u * ∈ . Moreover, P (0) = u * .
Proof It is given in expression (2) that By the use of definition of {t n } and inequality (11), we obtain where By the use of Lemma 3.1, we obtain Combining (13) with (14), we obtain Thus, we conclude that the {u n } is bounded sequence. Indeed, by (13) we have for some M 2 > 0. Combining the expressions (10) with (16), we have Due to the Lipschitz-continuity and pseudomonotonicity of S implies that the solution set is a closed and convex set. It is given that u * = P (0) and using Lemma 2.1(ii), we have The remainder of the facts shall be split into the following two parts: Thus, above implies that lim n→+∞ u n − u * exists and let lim n→+∞ u n − u * = l, for some l ≥ 0. From the expression (17), we have Due to existence of a limit of sequence u n − u * and ψ n → 0, we infer that t n − y n → 0 and u n+1 − y n → 0 as n → +∞.
By the use of expression (21), we have Next, we will evaluate The above provides that The above explanation guarantees that the sequences {t n } and {y n } are also bounded. By the use of reflexivity of X and the boundedness of {u n } guarantees that there exits a subsequence {u n k } in order that {u n k } û ∈ X as k → +∞. Next, we have to prove thatû ∈ . This is provided that The above inequality implies that Thus, we shall obtain Due to boundedness of the sequence {t n k } implies that {S(t n k )} is also bounded. By the use of lim k→∞ t n k − y n k = 0 and k → ∞ in (27), we obtain Moreover, we have Since lim k→∞ t n k − y n k = 0 and S is L-Lipschitz continuous on X implies that which together with (29) and (30), we obtain Let consider a sequence of positive numbers { k } that is decreasing and converge to zero. For each k, we denote m k by the smallest positive integer such that Due to { k } is decreasing and {m k } is increasing. Case I: If there is a t n m k j subsequence of t n m k such that S(t n m k j ) = 0 (∀ j). Let j → ∞, we obtain Henceû ∈ D, therefore we obtainû ∈ .
Case II: If there exits N 0 ∈ N such that for all n m k ≥ N 0 , S(t n m k ) = 0. Consider that Due to the above definition, we obtain Moreover, expressions (32) and (35), for all n m k ≥ N 0 , we have Due to the pseudomonotonicity of S for n m k ≥ N 0 , For all n m k ≥ N 0 , we have Due to {t n k } weakly converges toû ∈ D through S is sequentially weakly continuous on the set D, we get {S(t n k )} weakly converges to S(û). Suppose that S(û) = 0, we have Since {t n m k } ⊂ {t n k } and lim k→∞ k = 0, we have Next, consider k → ∞ in (38), we obtain By the use of Minty Lemma 2.5, we inferû ∈ . Next, we have lim sup By the use of lim n→+∞ u n+1 − u n = 0. Therefore, (42) implies that lim sup Consider the expression (12), we have From expressions (14) and (44), we obtain By the use of (22), (43), (45) and applying Lemma 2.2, conclude that lim n→+∞ u n − u * = 0.
Using Lemma 2.3, there exists a sequence {m k } ⊂ N as {m k } → +∞ such that As in Case 1, the relation (20) gives that Due to ψ m k → 0, we deduce the following: It follows that Next, evaluate This follows that Using the same explanation as in the Case 1, such that lim sup k→+∞ u * , u * − u m k +1 ≤ 0.
Using the expressions (45) and (46), we get Thus, above implies that Since ψ m k → 0, and u m k − u * is a bounded. Therefore, expressions (52) and (54) implies that This means that As a consequence u n → u * . This is going to conclude the proof of the theorem.
The second projection in Algorithm 1 is substituted with half-space to minimize computing cost, as inspired by Algorithm 4.1 in the paper [4]. The second main algorithm is written as follows:
Step 2: Compute y n = P D (t n − τ S(t n )). If t n = y n , then STOP and y n is a solution.

Numerical illustrations
This section discusses two numerical tests to explain the efficacy of the proposed algorithms. All these numerical studies give a detailed understanding of how better control parameters can be chosen. Each of them shows the advantages of the proposed methods relative to the existing ones in the literature.
Example 4.1 First consider the HpHard problem which is taken from [7]. This example has been considered by many people for experimental test (see, [5,8,21]). A operator S : R m → R m is defined by In above definition, we take N = rand(m) to be a random matrix and B = 0.5K − 0.5K T to be a skewsymmetric matrix with K = rand(m) and D = diag(rand(m, 1)) is a diagonal matrix. The set D is taken as follows: where Q = rand(100, m) and b = rand(100, 1). It is obvious that S is monotone and that Lipschitz is continuous by L = M . The starting point for this experiment are u 0 = u 1 = (1, 1, . . . , 1). The numerical consequences of these methods are seen in Figs Moreover, the feasible set D is defined by D = {u ∈ R 4 : 1 ≤ u i ≤ 5, i = 1, 2, 3, 4}.  It is clear to see that S is not monotone on the set D. By the use of Monte-Carlo technique [9], it can be shown that S is pseudo-monotone on D. There exits a unique solution u * = (5, 5, 5, 5) T for given problem. The starting point for this experiment are u 0 = u 1 = (1, 1, . . . , 1) and D n = t n − y n ≤ 10 −3 . The numerical consequences of these methods are seen in Tables 3, 4  3 Let X = l 2 be a real Hilbert space with the sequences of real numbers satisfying the following condition u 1 2 + u 2 2 + · · · + u n 2 + · · · < +∞.
Assume that a mapping S : D → D is defined by where D = {u ∈ X : u ≤ 3}. We can easily see that S is weakly sequentially continuous on X and the solution set is V I (D, S) = {0}. Moreover, S is L-Lipschitz continuous with L = 11. The mapping S is pseudomonotone on D but not monotone (see for more details [33]). Let considered the following projection  formula: The numerical consequences of these methods are seen in Figs. 5, 6, 7 and 8 and Table 7. The control requirement shall be taken as follows: (1) Algorithm 1 in [16] (shortly, Algorithm A): τ = 0.8 L , D n = u n − y n ≤ 10 −4 ;   1, 2, . . . , 5000, 0, 0 (Table 8).