Home Mathematics A class of strongly convergent subgradient extragradient methods for solving quasimonotone variational inequalities
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A class of strongly convergent subgradient extragradient methods for solving quasimonotone variational inequalities

  • Habib ur Rehman , Poom Kumam EMAIL logo , Murat Ozdemir , Isa Yildirim and Wiyada Kumam EMAIL logo
Published/Copyright: July 29, 2023
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Demonstratio Mathematica
From the journal Demonstratio Mathematica Volume 56 Issue 1

Abstract

The primary goal of this research is to investigate the approximate numerical solution of variational inequalities using quasimonotone operators in infinite-dimensional real Hilbert spaces. In this study, the sequence obtained by the proposed iterative technique for solving quasimonotone variational inequalities converges strongly toward a solution due to the viscosity-type iterative scheme. Furthermore, a new technique is proposed that uses an inertial mechanism to obtain strong convergence iteratively without the requirement for a hybrid version. The fundamental benefit of the suggested iterative strategy is that it substitutes a monotone and non-monotone step size rule based on mapping (operator) information for its Lipschitz constant or another line search method. This article also provides a numerical example to demonstrate how each method works.

Keywords: variational inequality problem; subgradient extragradient method; strong convergence results; quasimonotone operator; Lipschitz continuity
MSC 2010: 65Y05; 65K15; 68W10; 47H05; 47H10

1 Introduction

The primary goal of this article is to investigate the iterative methods used to approximate the solution of the variational inequality problem (VIP) using quasimonotone operators in any real Hilbert space. Let ℰ be a real Hilbert space and ℳ be a nonempty convex, closed subset of ℰ . Let K : ℰ → ℰ be an operator. A problem (VIP) for K on ℳ is described as follows [28]:

(VIP) Find r ∗ ∈ ℳ such that ⟨ K ( r ∗ ) , y − r ∗ ⟩ ≥ 0 , ∀ y ∈ ℳ .

To validate the strong convergence, it is assumed that the following requirements are met:

( K 1) The solution set for problem (VIP) is denoted by VI ( ℳ , K ) and it is nonempty.

( K 2) An operator K : ℰ → ℰ is said to be quasimonotone such that

⟨ K ( y 1 ) , y 2 − y 1 ⟩ > 0 ⇒ ⟨ K ( y 2 ) , y 2 − y 1 ⟩ ≥ 0 , ∀ y 1 , y 2 ∈ ℳ .

( K 3) An operator K : ℰ → ℰ is said to be Lipschitz continuous with constant L > 0 such that

‖ K ( y 1 ) − K ( y 2 ) ‖ ≤ L ‖ y 1 − y 2 ‖ , ∀ y 1 , y 2 ∈ ℳ .

( K 4) An operator K : ℰ → ℰ is said to be sequentially weakly continuous, i.e., { K ( u n ) } weakly converges to K ( u ) because each sequence { u n } weakly converges to u .

It is well recognized that the problem (VIP) is a crucial problem in non-linear analysis. It is a key mathematical model that integrates a number of important concepts in applied mathematics, such as a non-linear system of equations, optimization conditions for problems with the optimization process, complementarity problems, network equilibrium problems, and finance (see for more details [15,18–21, 26,30]). As a result, this concept has several applications in mathematical programming, engineering, transportation analysis, network economics, game theory, and computer science.

The regularized approach and the projection method are two popular and generic methods for solving variational inequalities. It should also be mentioned that the first technique is most usually used to deal with variational inequalities accompanied by the monotone operator class. In this method, the regularized subproblem is strongly monotone, and its unique solution is obtained more conveniently than the initial problem. In this article, we will look into projection methods that are well known for their ease of numerical computation.

Furthermore, projection techniques can be used to find a numerical solution to variational inequalities. Many researchers have created original projection methods to solve various types of variational inequalities (for more details, see [5–8,12,14,17,22,23,25,27,29,31,38] and others in [3,4,9–11,16,32–36]). All techniques for resolving the (VIP) problem are focused on computing a projection on the appropriate set ℳ . Korpelevich [22] and Antipin [1] introduced the equivalent extragradient method. Their method is as follows:

(1) u 1 ∈ ℳ , y n = P ℳ [ u n − ρ K ( u n ) ] , u n + 1 = P ℳ [ u n − ρ K ( y n ) ] ,

where 0 < ρ < 1 L . In keeping with the previous technique, we used two projections on the underlying set ℳ for each iteration. Indeed, if the feasible set ℳ has a complicated structure, the method’s computing efficiency may decrease. This section will go through various methods to obtain through this limitation. In the study by Censor et al. [12], the subgradient extragradient technique was first used. The following strategy is used in this technique:

(2) u 1 ∈ ℳ , y n = P ℳ [ u n − ρ K ( u n ) ] , u n + 1 = P ℰ n [ u n − ρ K ( y n ) ] ,

where 0 < ρ < 1 L through

ℰ n = { z ∈ ℰ : ⟨ u n − ρ K ( u n ) − y n , z − y n ⟩ ≤ 0 } .

Tseng’s extragradient technique [31], which uses only one projection every iteration, is another notable method that does not require two projections. The following strategy is used in this technique:

(3) u 1 ∈ ℳ , y n = P ℳ [ u n − ρ K ( u n ) ] , u n + 1 = y n + ρ [ K ( u n ) − K ( y n ) ] ,

where 0 < ρ < 1 L . It is important to note that the previous techniques have two main flaws: a fixed step size rule that is reliant on the Lipschitz modulus of the cost operator and a weakly converging iterative procedure. The Lipschitz modulus is frequently uncertain or difficult to calculate. A fixed step size limitation that impacts the method’s efficacy and speed of convergence may be difficult to explain theoretically. Additionally, in the situation of an infinite-dimensional Hilbert space, the investigation of a strongly convergent iterative sequence is crucial.

The gradient projection technique was the first well-established projection method for determining variational inequalities, and it was followed by numerous additional projection methods, including the well-known extragradient approach [22], the subgradient extragradient methods [12,13], and others [14,17,25, 31,39]. The aforementioned methods are used to solve variational inequalities using monotone, strongly monotone, or inverse monotone. Furthermore, while generating approximation solutions and determining their convergence, fixed or variable step sizes frequently depend on the Lipschitz constants of the operators. This can limit implementations since, in some cases, some parameters are unknown or impossible to estimate.

The purpose of this research is to look at variational inequalities using quasimonotone operators in infinite-dimensional Hilbert spaces. Furthermore, this study shows that the iterative sequences generated by all four subgradient extragradient algorithms strongly converge to a solution. Subgradient extragradient methods use both monotone and non-monotone variable step size rules. The study of inertial algorithms is also presented, which typically enhances the efficiency of the iterative sequence. The article’s main contribution is that it investigates explicit monotone and non-monotone step size rules using inertial schemes and achieves strong convergence.

This article is written as follows. Section 2 provides preliminary results. Section 3 describes four novel methods and their convergence analysis. Finally, Section 4 provides some numerical findings to explain the practical efficiency of the proposed methods.

2 Preliminaries

This section contains various important identities as well as significant lemmas. Let us define the following set:

VI ( ℳ , K ) + = { r ∗ ∈ ℳ : ⟨ ℳ ( r ∗ ) , y − r ∗ ⟩ > 0 , ∀ y ∈ ℳ } .

For any u , y ∈ ℰ , we have

‖ u + y ‖ 2 = ‖ u ‖ 2 + 2 ⟨ u , y ⟩ + ‖ y ‖ 2 .

A metric projection P ℳ ( y 1 ) of y 1 ∈ ℰ is described by:

P ℳ ( y 1 ) = arg min { ‖ y 1 − y 2 ‖ : y 2 ∈ ℳ } .

Lemma 2.1

[2] Suppose that P ℳ : ℰ → ℳ is a metric projection. Then, the following conditions are satisfied:

  1. e 3 = P ℳ ( e 1 ) if and only if

    ⟨ e 1 − e 3 , e 2 − e 3 ⟩ ≤ 0 , ∀ e 2 ∈ ℳ ,

  2. ‖ e 1 − P ℳ ( e 2 ) ‖ 2 + ‖ P ℳ ( e 2 ) − e 2 ‖ 2 ≤ ‖ e 1 − e 2 ‖ 2 , e 1 ∈ ℳ , e 2 ∈ Σ ,

  3. ‖ e 1 − P ℳ ( e 1 ) ‖ ≤ ‖ e 1 − e 2 ‖ , e 2 ∈ ℳ , e 1 ∈ Σ .

Lemma 2.2

[37] Let { p n } ⊂ [ 0 , + ∞ ) be a sequence such that

p n + 1 ≤ ( 1 − q n ) p n + q n r n , ∀ n ∈ N .

Moreover, two sequences { q n } ⊂ ( 0 , 1 ) and { r n } ⊂ R such that

lim n → + ∞ q n = 0 , ∑ n = 1 + ∞ q n = + ∞ and limsup n → + ∞ r n ≤ 0 .

Then, lim n → + ∞ p n = 0 .

Lemma 2.3

[24] Let { p n } be a real sequence, and there exists a subsequence { n i } of { n } such that

p n i < p n i + 1 for a l l i ∈ N .

Then, there exists a non-decreasing sequence m k ⊂ N such that m k → + ∞ as k → + ∞ , and satisfying the following inequality for k ∈ N :

p m k ≤ p m k + 1 and p k ≤ p m k + 1 .

Indeed, m k = max { j ≤ k : p j ≤ p j + 1 } .

3 Main results

In this section, we propose four new methods to solve quasimonotone variational inequalities in a real Hilbert space and prove strong convergence results for the proposed method. The first and second methods involve a monotonic self-adaptive step rule to make the algorithm independent of the Lipschitz constant. Let g : ℰ → ℰ be a strict contraction function through constant ξ ∈ [ 0 , 1 ) . The main algorithm is as follows:

Algorithm 1 (Explicit monotonic viscosity-type subgradient extragradient method)
Step 0: Let u 1 ∈ ℳ , μ ∈ ( 0 , 1 ) and ρ 1 > 0 . Moreover, sequence { γ n } ⊂ ( 0 , 1 ) such that
lim n → + ∞ γ n = 0 and ∑ n = 1 + ∞ γ n = + ∞ .
Step 1: Compute
y n = P ℳ ( u n − ρ n K ( u n ) ) .
If u n = y n , then STOP. Otherwise, go to Step 2.
Step 2: Construct a set ℰ n = { z ∈ ℰ : ⟨ u n − ρ n K ( u n ) − y n , z − y n ⟩ ≤ 0 } and evaluate
t n = P ℰ n ( u n − ρ n K ( y n ) ) .
Step 3: Calculate
u n + 1 = γ n g ( u n ) + ( 1 − γ n ) t n .
Step 4: Calculate
ρ n + 1 = min ρ n , μ ‖ u n − y n ‖ ‖ K ( u n ) − K ( y n ) ‖ if K ( u n ) ≠ K ( y n ) , ρ n else. ( 4 )
Set n ≔ n + 1s and go back to Step 1.

Lemma 3.1

A step size sequence { ρ n } generated in (4) is decreasing monotonically with a lower bound min { μ L , ρ 0 } and converges to a fixed ρ > 0 .

Proof

It is obvious that { ρ n } is a monotone and non-increasing sequence. It is given that operator K is Lipschitz continuous with a constant L > 0 such that

‖ K ( u n ) − K ( y n ) ‖ ≤ L ‖ u n − y n ‖ .

Let K ( u n ) ≠ K ( y n ) such that

(7) μ ‖ u n − y n ‖ ‖ K ( u n ) − K ( y n ) ‖ ≥ μ ‖ u n − y n ‖ L ‖ u n − y n ‖ ≥ μ L .

As a result of the aforementioned expression, { ρ n } has a lower bound of min { μ L , ρ 0 } . Moreover, there exists ρ > 0 such that lim n → ∞ ρ n = ρ . □

Algorithm 2 (Inertial monotonic explicit subgradient extragradient method)
Step 0: Let u 0 , u 1 ∈ ℳ , μ ∈ ( 0 , 1 ) and ρ 1 > 0 . Moreover, { γ n } ⊂ ( 0 , 1 ) such that
lim n → + ∞ γ n = 0 and ∑ n = 1 + ∞ γ n = + ∞ .
Step 1: Evaluate s n = u n + χ n ( u n − u n − 1 ) − γ n [ u n + χ n ( u n − u n − 1 ) ] , where χ n such that
0 ≤ χ n ≤ χ n ˆ and χ n ˆ = min χ 2 , ε n ‖ u n − u n − 1 ‖ if u n ≠ u n − 1 , χ 2 else , ( 5 )
with positive sequence ε n = ∘ ( γ n ) such that lim n → + ∞ ε n γ n = 0 .
Step 2: Compute
y n = P ℳ ( s n − ρ n K ( s n ) ) .
If u n = y n , then STOP. Otherwise, go to Step 3.
Step 3: Construct a set ℰ n = { z ∈ ℰ : ⟨ s n − ρ n K ( s n ) − y n , z − y n ⟩ ≤ 0 } and evaluate
u n + 1 = P ℰ n ( s n − ρ n K ( y n ) ) .
Step 4: Calculate
ρ n + 1 = min ρ n , μ ‖ s n − y n ‖ ‖ K ( s n ) − K ( y n ) ‖ if K ( s n ) ≠ K ( y n ) , ρ n else. ( 6 )
Set n ≔ n + 1 and go back to Step 1.
Algorithm 3 (Non-monotonic explicit viscosity-type subgradient extragradient method)
Step 0: Let u 1 ∈ ℳ , ρ 1 > 0 , μ ∈ ( 0 , 1 ) and choose a non-negative real sequence { φ n } such that ∑ n ∞ φ n < + ∞ . Moreover, { γ n } ⊂ ( 0 , 1 ) such that
lim n → + ∞ γ n = 0 and ∑ n = 1 + ∞ γ n = + ∞ .
Step 1: Compute
y n = P ℳ ( u n − ρ n K ( u n ) ) .
If u n = y n , then STOP. Otherwise, go to Step 2.
Step 2: Construct a set ℰ n = { z ∈ ℰ : ⟨ u n − ρ n K ( u n ) − y n , z − y n ⟩ ≤ 0 } and evaluate
t n = P ℰ n ( u n − ρ n K ( y n ) ) .
Step 3: Calculate u n + 1 = γ n g ( u n ) + ( 1 − γ n ) t n .
Step 4: Calculate
ρ n + 1 = min ρ n + φ n , μ ‖ u n − y n ‖ ‖ K ( u n ) − K ( y n ) ‖ if K ( u n ) ≠ K ( y n ) , ρ n + φ n else. ( 8 )
Set n ≔ n + 1 and go back to Step 1.
Algorithm 4 (Inertial non-monotonic explicit subgradient extragradient method)
Step 0: Let u 0 , u 1 ∈ ℳ , μ ∈ ( 0 , 1 ) , ρ 1 > 0 and choose a non-negative real sequence { φ n } such that ∑ n ∞ φ n < + ∞ . Moreover, { γ n } ⊂ ( 0 , 1 ) such that
lim n → + ∞ γ n = 0 and ∑ n = 1 + ∞ γ n = + ∞ .
Step 1: Evaluate s n = u n + χ n ( u n − u n − 1 ) − γ n [ u n + χ n ( u n − u n − 1 ) ] , where χ n such that
0 ≤ χ n ≤ χ ˆ n and χ ˆ n = min χ 2 , ε n ‖ u n − u n − 1 ‖ if u n ≠ u n − 1 , χ 2 else , (9)
with positive sequence ε n = ∘ ( γ n ) such that lim n → + ∞ ε n γ n = 0 .
Step 2: Compute
y n = P ℳ ( s n − ρ n K ( s n ) ) .
If u n = y n , then STOP. Otherwise, go to Step 3.
Step 3: Construct a set ℰ n = { z ∈ ℰ : ⟨ s n − ρ n K ( s n ) − y n , z − y n ⟩ ≤ 0 } and evaluate
u n + 1 = P ℰ n ( s n − ρ n K ( y n ) ) .
Step 4: Calculate
ρ n + 1 = min ρ n + φ n , μ ‖ s n − y n ‖ ‖ K ( s n ) − K ( y n ) ‖ if K ( s n ) ≠ K ( y n ) , ρ n + φ n else. ( 10 )
Set n ≔ n + 1 and go back to Step 1.

Lemma 3.2

A sequence { ρ n } generated by expression (8) is convergent to ρ and satisfying the following inequality:

min μ L , ρ 1 ≤ ρ ≤ ρ 1 + P where P = ∑ n = 1 + ∞ φ n .

Proof

Due to the Lipschitz continuity of a mapping K , there exists a fixed number L > 0 . Let K ( u n ) ≠ K ( y n ) such that

(11) μ ‖ u n − y n ‖ ‖ K ( u n ) − K ( y n ) ‖ ≥ μ ‖ u n − y n ‖ L ‖ u n − y n ‖ ≥ μ L .

By using mathematical induction on the definition of ρ n + 1 , we have

min μ L , ρ 1 ≤ ρ n ≤ ρ 1 + P .

Let

[ ρ n + 1 − ρ n ] + = max { 0 , ρ n + 1 − ρ n }

and

[ ρ n + 1 − ρ n ] − = max { 0 , − ( ρ n + 1 − ρ n ) } .

From the definition of { ρ n } , we have

(12) ∑ n = 1 + ∞ ( ρ n + 1 − ρ n ) + = ∑ n = 1 + ∞ max { 0 , ρ n + 1 − ρ n } ≤ P < + ∞ .

That is, the series ∑ n = 1 + ∞ ( ρ n + 1 − ρ n ) + is convergent. Next, we need to prove the convergence of ∑ n = 1 + ∞ ( ρ n + 1 − ρ n ) − . Let ∑ n = 1 + ∞ ( ρ n + 1 − ρ n ) − = + ∞ , due to the reason that

ρ n + 1 − ρ n = ( ρ n + 1 − ρ n ) + − ( ρ n + 1 − ρ n ) − .

Thus, we have

(13) ρ k + 1 − ρ 1 = ∑ n = 0 k ( ρ n + 1 − ρ n ) = ∑ n = 0 k ( ρ n + 1 − ρ n ) + − ∑ n = 0 k ( ρ n + 1 − ρ n ) − .

By allowing k → + ∞ in expression (13), we have ρ k → − ∞ as k → ∞ . This is a contradiction. Due to the convergence of ∑ n = 0 k ( ρ n + 1 − ρ n ) + and ∑ n = 0 k ( ρ n + 1 − ρ n ) − as k → + ∞ in expression (13), we obtain lim n → ∞ ρ n = ρ . This completes the proof.□

Lemma 3.3

Let the mapping K : ℰ → ℰ satisfy conditions ( K 1)–( K 4). For any r ∗ ∈ VI ( ℳ , K ) + , we have

‖ t n − r ∗ ‖ 2 ≤ ‖ u n − r ∗ ‖ 2 − 1 − μ ρ n ρ n + 1 ‖ u n − y n ‖ 2 − 1 − μ ρ n ρ n + 1 ‖ t n − y n ‖ 2 .

Proof

Let us consider that

(14) ‖ t n − r ∗ ‖ 2 = ‖ P ℰ n [ u n − ρ n K ( y n ) ] − r ∗ ‖ 2 = ‖ P ℰ n [ u n − ρ n K ( y n ) ] + [ u n − ρ n K ( y n ) ] − [ u n − ρ n K ( y n ) ] − r ∗ ‖ 2 = ‖ [ u n − ρ n K ( y n ) ] − r ∗ ‖ 2 + ‖ P ℰ n [ u n − ρ n K ( y n ) ] − [ u n − ρ n K ( y n ) ] ‖ 2 + 2 ⟨ P ℰ n [ u n − ρ n K ( y n ) ] − [ u n − ρ n K ( y n ) ] , [ u n − ρ n K ( y n ) ] − r ∗ ⟩ .

By using r ∗ ∈ VI ( ℳ , K ) + ⊂ ℰ n , we obtain

(15) ‖ P ℰ n [ u n − ρ n K ( y n ) ] − [ u n − ρ n K ( y n ) ] ‖ 2 + ⟨ P ℰ n [ u n − ρ n K ( y n ) ] − [ u n − ρ n K ( y n ) ] , [ u n − ρ n K ( y n ) ] − r ∗ ⟩ = ⟨ [ u n − ρ n K ( y n ) ] − P ℰ n [ u n − ρ n K ( y n ) ] , r ∗ − P ℰ n [ u n − ρ n K ( y n ) ] ⟩ ≤ 0 ,

which implies that

(16) ⟨ P ℰ n [ u n − ρ n K ( y n ) ] − [ u n − ρ n K ( y n ) ] , [ u n − ρ n K ( y n ) ] − r ∗ ⟩ ≤ − ‖ P ℰ n [ u n − ρ n K ( y n ) ] − [ u n − ρ n K ( y n ) ] ‖ 2 .

Combining (14) and (16), we obtain

(17) ‖ t n − r ∗ ‖ 2 ≤ ‖ u n − ρ n K ( y n ) − r ∗ ‖ 2 − ‖ P ℰ n [ u n − ρ n K ( y n ) ] − [ u n − ρ n K ( y n ) ] ‖ 2 ≤ ‖ u n − r ∗ ‖ 2 − ‖ u n − t n ‖ 2 + 2 ρ n ⟨ K ( y n ) , r ∗ − t n ⟩ .

Since r ∗ ∈ VI ( ℳ , K ) + , we have

⟨ K ( r ∗ ) , y − r ∗ ⟩ > 0 , for all y ∈ ℳ .

Thus, the aforementioned expression implies that

⟨ K ( y ) , y − r ∗ ⟩ ≥ 0 , for all y ∈ ℳ .

By using y = y n ∈ ℳ , we obtain

⟨ K ( y n ) , y n − r ∗ ⟩ ≥ 0 .

Thus, we have

(18) ⟨ K ( y n ) , r ∗ − t n ⟩ = ⟨ K ( y n ) , r ∗ − y n ⟩ + ⟨ K ( y n ) , y n − t n ⟩ ≤ ⟨ K ( y n ) , y n − t n ⟩ .

Combining expressions (17) and (18), we obtain

(19) ‖ t n − r ∗ ‖ 2 ≤ ‖ u n − r ∗ ‖ 2 − ‖ u n − t n ‖ 2 + 2 ρ n ⟨ K ( y n ) , y n − t n ⟩ ≤ ‖ u n − r ∗ ‖ 2 − ‖ u n − y n + y n − t n ‖ 2 + 2 ρ n ⟨ K ( y n ) , y n − t n ⟩ ≤ ‖ u n − r ∗ ‖ 2 − ‖ u n − y n ‖ 2 − ‖ y n − t n ‖ 2 + 2 ⟨ u n − ρ n K ( y n ) − y n , t n − y n ⟩ .

Note that t n = P ℰ n [ u n − ρ n K ( y n ) ] implies that

(20) 2 ⟨ u n − ρ n K ( y n ) − y n , t n − y n ⟩ = 2 ⟨ u n − ρ n K ( u n ) − y n , t n − y n ⟩ + 2 ρ n ⟨ K ( u n ) − K ( y n ) , t n − y n ⟩ ≤ ρ n ρ n + 1 2 ρ n + 1 ‖ K ( u n ) − K ( y n ) ‖ ‖ t n − y n ‖ ≤ μ ρ n ρ n + 1 ‖ u n − y n ‖ 2 + μ ρ n ρ n + 1 ‖ t n − y n ‖ 2 .

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

(21)□ ‖ t n − r ∗ ‖ 2 ≤ ‖ u n − r ∗ ‖ 2 − ‖ u n − y n ‖ 2 − ‖ y n − t n ‖ 2 + ρ n ρ n + 1 [ μ ‖ u n − y n ‖ 2 + μ ‖ t n − y n ‖ 2 ] ≤ ‖ u n − r ∗ ‖ 2 − 1 − μ ρ n ρ n + 1 ‖ u n − y n ‖ 2 − 1 − μ ρ n ρ n + 1 ‖ t n − y n ‖ 2 .

Lemma 3.4

Let the mapping K : ℰ → ℰ satisfy conditions ( K 1)–( K 4). If there exists a subsequence { u n k } weakly convergent to u ˆ and lim k → ∞ ‖ u n k − y n k ‖ = 0 , then u ˆ ∈ VI ( ℳ , K ) .

Proof

Since { u n k } is weakly convergent to u ˆ and lim k → ∞ ‖ u n k − y n k ‖ = 0 , the sequence { y n k } is weakly convergent to u ˆ . Next, we need to show that u ˆ ∈ VI ( ℳ , K ) . Thus, we have

y n k = P ℳ [ u n k − ρ n k K ( u n k ) ] ,

which is equivalent to

(22) ⟨ u n k − ρ n k K ( u n k ) − y n k , y − y n k ⟩ ≤ 0 , ∀ y ∈ ℳ .

The aforementioned inequality implies that

(23) ⟨ u n k − y n k , y − y n k ⟩ ≤ ρ n k ⟨ K ( u n k ) , y − y n k ⟩ , ∀ y ∈ ℳ .

Thus, we obtain

(24) 1 ρ n k ⟨ u n k − y n k , y − y n k ⟩ + ⟨ K ( u n k ) , y n k − u n k ⟩ ≤ ⟨ K ( u n k ) , y − u n k ⟩ , ∀ y ∈ ℳ .

By the use of min { μ L , ρ 1 } ≤ ρ ≤ ρ 1 and { u n k } is a bounded sequence. It is given that lim k → ∞ ‖ u n k − y n k ‖ = 0 and k → ∞ in (24), we obtain

(25) liminf k → ∞ ⟨ K ( u n k ) , y − u n k ⟩ ≥ 0 , ∀ y ∈ ℳ .

Moreover, we have

(26) ⟨ K ( y n k ) , y − y n k ⟩ = ⟨ K ( y n k ) − K ( u n k ) , y − u n k ⟩ + ⟨ K ( u n k ) , y − u n k ⟩ + ⟨ K ( y n k ) , u n k − y n k ⟩ .

Since lim k → ∞ ‖ u n k − y n k ‖ = 0 and K is L -Lipschitz continuity on ℰ , we obtain

(27) lim k → ∞ ‖ K ( u n k ) − K ( y n k ) ‖ = 0 .

From expressions (26) and (27), we obtain

(28) liminf k → ∞ ⟨ K ( y n k ) , y − y n k ⟩ ≥ 0 , ∀ y ∈ ℳ .

For the continuity of demonstration, let us take a positive sequence { ε k } that is decreasing to zero. For every { ε k } , we denote by m k the positive smallest integer in order that

(29) ⟨ K ( u n i ) , y − u n i ⟩ + ε k > 0 , ∀ i ≥ m k .

Since { ε k } is decreasing, it is easy to observe that the sequence { m k } is increasing.

Case I: If there exists subsequence { u n m k j } of { u n m k } such that K ( u n m k j ) = 0 ( ∀ j ). Let j → ∞ , we obtain

(30) ⟨ K ( u ˆ ) , y − u ˆ ⟩ = lim j → ∞ ⟨ K ( u n m k j ) , y − u ˆ ⟩ = 0 .

Thus, u ˆ ∈ ℳ and imply that u ˆ ∈ VI ( ℳ , K ) .

Case II: If there exists a fixed number N 0 ∈ N such that for all n m k ≥ N 0 , K ( u n m k ) ≠ 0 . Consider that

(31) ϒ n m k = K ( u n m k ) ‖ K ( u n m k ) ‖ 2 , ∀ n m k ≥ N 0 .

Due to the aforementioned definition, we obtain

(32) ⟨ K ( u n m k ) , ϒ n m k ⟩ = 1 , ∀ n m k ≥ N 0 .

Moreover, using expressions (29) and (32), for all n m k ≥ N 0 , we have

(33) ⟨ K ( u n m k ) , y + ε k ϒ n m k − u n m k ⟩ > 0 .

Since K is quasimonotone, then

(34) ⟨ K ( y + ε k ϒ n m k ) , y + ε k ϒ n m k − u n m k ⟩ > 0 .

For all n m k ≥ N 0 , we have

(35) ⟨ K ( y ) , y − u n m k ⟩ ≥ ⟨ K ( y ) − K ( y + ε k ϒ n m k ) , y + ε k ϒ n m k − u n m k ⟩ − ε k ⟨ K ( y ) , ϒ n m k ⟩ .

Since { u n k } converges weakly to u ˆ ∈ ℳ through K is weakly continuous on the set ℳ , we obtain { K ( u n k ) } converges weakly to K ( u ˆ ) . Let K ( u ˆ ) ≠ 0 , we have

(36) ‖ K ( u ˆ ) ‖ ≤ liminf k → ∞ ‖ K ( u n k ) ‖ .

Since { u n m k } ⊂ { u n k } and lim k → ∞ ε k = 0 , we have

(37) 0 ≤ lim k → ∞ ‖ ε k ϒ n m k ‖ = lim k → ∞ ε k ‖ K ( u n m k ) ‖ ≤ 0 ‖ K ( u ˆ ) ‖ = 0 .

Next, letting k → ∞ in expression (35), we obtain

(38) ⟨ K ( y ) , y − u ˆ ⟩ ≥ 0 , ∀ y ∈ ℳ .

Consider the case when u ∈ ℳ is an arbitrary element and 0 < ρ ≤ 1 . Thus, we have

(39) u ˆ ρ = ρ u + ( 1 − ρ ) u ˆ .

Then, u ˆ ρ ∈ ℳ , and from expression (38), we have

(40) ρ ⟨ K ( u ˆ ρ ) , u − u ˆ ⟩ ≥ 0 .

Hence, we have

(41) ⟨ K ( u ˆ ρ ) , u − u ˆ ⟩ ≥ 0 .

It is clear from equation (41) that

(42) ⟨ K ( u ˆ ) , u − u ˆ ⟩ ≥ 0 .

Hence, u ˆ ∈ VI ( ℳ , K ) . This completes the proof of lemma.□

Theorem 3.5

Let the mapping K : ℰ → ℰ satisfy conditions ( K 1)–( K 4). Then, { u n } generated by the Algorithm 1 strongly converges to r ∗ = P VI ( ℳ , K ) ∘ g ( r ∗ ) .

Proof

Since ρ n → ρ , there exists a fixed number ε ∈ ( 0 , 1 − μ ) such that

lim n → ∞ 1 − μ ρ n ρ n + 1 = 1 − μ > ε > 0 .

Therefore, there exists a fixed number M 1 ∈ N such that

(43) 1 − μ ρ n ρ n + 1 > ε > 0 , ∀ n ≥ M 1 .

From expression (21), we obtain

(44) ‖ t n − r ∗ ‖ 2 ≤ ‖ u n − r ∗ ‖ 2 , ∀ n ≥ M 1 .

It is given that r ∗ ∈ Ω and due to the fact that g is a contraction with ξ ∈ [ 0 , 1 ) , we have

(45) ‖ u n + 1 − r ∗ ‖ = ‖ γ n g ( u n ) + ( 1 − γ n ) t n − r ∗ ‖ = ‖ γ n [ g ( u n ) − r ∗ ] + ( 1 − γ n ) [ t n − r ∗ ] ‖ = ‖ γ n [ g ( u n ) + g ( r ∗ ) − g ( r ∗ ) − r ∗ ] + ( 1 − γ n ) [ t n − r ∗ ] ‖ ≤ γ n ‖ g ( u n ) − g ( r ∗ ) ‖ + γ n ‖ g ( r ∗ ) − r ∗ ‖ + ( 1 − γ n ) ‖ t n − r ∗ ‖ ≤ γ n ξ ‖ u n − r ∗ ‖ + γ n ‖ g ( r ∗ ) − r ∗ ‖ + ( 1 − γ n ) ‖ t n − r ∗ ‖ .

Combining expressions (44) and (45) and γ n ⊂ ( 0 , 1 ) , we obtain

(46) ‖ u n + 1 − r ∗ ‖ ≤ γ n ξ ‖ u n − r ∗ ‖ + γ n ‖ g ( r ∗ ) − r ∗ ‖ + ( 1 − γ n ) ‖ u n − r ∗ ‖ = [ 1 − γ n + ξ γ n ] ‖ u n − r ∗ ‖ + γ n ( 1 − ξ ) ‖ g ( r ∗ ) − r ∗ ‖ ( 1 − ξ ) ≤ max ‖ u n − r ∗ ‖ , ‖ g ( r ∗ ) − r ∗ ‖ ( 1 − ξ ) ≤ max ‖ u M 1 − r ∗ ‖ , ‖ g ( r ∗ ) − r ∗ ‖ ( 1 − ξ ) .

Therefore, we conclude that { u n } is a bounded sequence. By using expression (21), we have

(47) ‖ u n + 1 − r ∗ ‖ 2 = ‖ γ n g ( u n ) + ( 1 − γ n ) t n − r ∗ ‖ 2 = ‖ γ n [ g ( u n ) − r ∗ ] + ( 1 − γ n ) [ t n − r ∗ ] ‖ 2 = γ n ‖ g ( u n ) − r ∗ ‖ 2 + ( 1 − γ n ) ‖ t n − r ∗ ‖ 2 − γ n ( 1 − γ n ) ‖ g ( u n ) − t n ‖ 2 ≤ γ n ‖ g ( u n ) − r ∗ ‖ 2 + ( 1 − γ n ) ‖ u n − r ∗ ‖ 2 − 1 − μ ρ n ρ n + 1 ‖ u n − y n ‖ 2 − 1 − μ ρ n ρ n + 1 ‖ t n − y n ‖ 2 − γ n ( 1 − γ n ) ‖ g ( u n ) − t n ‖ 2 ≤ γ n ‖ g ( u n ) − r ∗ ‖ 2 + ‖ u n − r ∗ ‖ 2 − ( 1 − γ n ) 1 − μ ρ n ρ n + 1 ‖ u n − y n ‖ 2 − ( 1 − γ n ) 1 − μ ρ n ρ n + 1 ‖ t n − y n ‖ 2 .

The aforementioned expression implies that

(48) ( 1 − γ n ) 1 − μ ρ n ρ n + 1 ‖ u n − y n ‖ 2 + ( 1 − γ n ) 1 − μ ρ n ρ n + 1 ‖ t n − y n ‖ 2 ≤ γ n ‖ g ( u n ) − r ∗ ‖ 2 + ‖ u n − r ∗ ‖ 2 − ‖ u n + 1 − r ∗ ‖ 2 .

By using expression (44), we obtain

(49) ‖ u n + 1 − r ∗ ‖ 2 = ‖ γ n g ( u n ) + ( 1 − γ n ) t n − r ∗ ‖ 2 = ‖ γ n [ g ( u n ) − r ∗ ] + ( 1 − γ n ) [ t n − r ∗ ] ‖ 2 ≤ ( 1 − γ n ) 2 ‖ t n − r ∗ ‖ 2 + 2 γ n ⟨ g ( u n ) − r ∗ , ( 1 − γ n ) [ t n − r ∗ ] + γ n [ g ( u n ) − r ∗ ] ⟩ = ( 1 − γ n ) 2 ‖ t n − r ∗ ‖ 2 + 2 γ n ⟨ g ( u n ) − g ( r ∗ ) + g ( r ∗ ) − r ∗ , u n + 1 − r ∗ ⟩ = ( 1 − γ n ) 2 ‖ t n − r ∗ ‖ 2 + 2 γ n ⟨ g ( u n ) − g ( r ∗ ) , u n + 1 − r ∗ ⟩ + 2 γ n ⟨ g ( r ∗ ) − r ∗ , u n + 1 − r ∗ ⟩ ≤ ( 1 − γ n ) 2 ‖ t n − r ∗ ‖ 2 + 2 γ n ξ ‖ u n − r ∗ ‖ ‖ u n + 1 − r ∗ ‖ + 2 γ n ⟨ g ( r ∗ ) − r ∗ , u n + 1 − r ∗ ⟩ ≤ ( 1 + γ n 2 − 2 γ n ) ‖ u n − r ∗ ‖ 2 + 2 γ n ξ ‖ u n − r ∗ ‖ 2 + 2 γ n ⟨ g ( r ∗ ) − r ∗ , u n + 1 − r ∗ ⟩ = ( 1 − 2 γ n ) ‖ u n − r ∗ ‖ 2 + γ n 2 ‖ u n − r ∗ ‖ 2 + 2 γ n ξ ‖ u n − r ∗ ‖ 2 + 2 γ n ⟨ g ( r ∗ ) − r ∗ , u n + 1 − r ∗ ⟩ = [ 1 − 2 γ n ( 1 − ξ ) ] ‖ u n − r ∗ ‖ 2 + 2 γ n ( 1 − ξ ) γ n ‖ u n − r ∗ ‖ 2 2 ( 1 − ξ ) + ⟨ g ( r ∗ ) − r ∗ , u n + 1 − r ∗ ⟩ 1 − ξ .

Case 1: Let us consider M 2 ∈ N ( M 2 ≥ M 1 ) such that

(50) ‖ u n + 1 − r ∗ ‖ ≤ ‖ u n − r ∗ ‖ , ∀ n ≥ M 2 .

Then, lim n → ∞ ‖ u n − r ∗ ‖ exists. Let lim n → ∞ ‖ u n − r ∗ ‖ = l . By the use of expression (48), we obtain

(51) ( 1 − γ n ) 1 − μ ρ n ρ n + 1 ‖ u n − y n ‖ 2 + ( 1 − γ n ) 1 − μ ρ n ρ n + 1 ‖ t n − y n ‖ 2 ≤ γ n ‖ g ( u n ) − r ∗ ‖ 2 + ‖ u n − r ∗ ‖ 2 − ‖ u n + 1 − r ∗ ‖ 2 .

Since lim n → ∞ ‖ u n − r ∗ ‖ exists and γ n → 0 , we obtain

(52) lim n → ∞ ‖ u n − y n ‖ = lim n → ∞ ‖ t n − y n ‖ = 0 .

By using the aforementioned results, we obtain

(53) lim n → ∞ ‖ u n − t n ‖ ≤ lim n → ∞ ‖ u n − y n ‖ + lim n → ∞ ‖ y n − t n ‖ = 0 .

It further implies that

(54) ‖ u n + 1 − u n ‖ = ‖ γ n g ( u n ) + ( 1 − γ n ) t n − u n ‖ = ‖ γ n [ g ( u n ) − u n ] + ( 1 − γ n ) [ t n − u n ] ‖ ≤ γ n ‖ g ( u n ) − u n ‖ + ( 1 − γ n ) ‖ t n − u n ‖ ⟶ 0 .

The aforementioned term specifically includes that

(55) lim n → ∞ ‖ u n + 1 − u n ‖ = 0 .

By using Lemma 3.4, we obtain

(56) limsup n → ∞ ⟨ g ( r ∗ ) − r ∗ , u n − r ∗ ⟩ = limsup k → ∞ ⟨ g ( r ∗ ) − r ∗ , u n k − r ∗ ⟩ = ⟨ g ( r ∗ ) − r ∗ , u ˆ − r ∗ ⟩ ≤ 0 .

By the use of lim n → ∞ ‖ u n + 1 − u n ‖ = 0 , we may deduce that

(57) limsup n → ∞ ⟨ g ( r ∗ ) − r ∗ , u n + 1 − r ∗ ⟩ ≤ limsup n → ∞ ⟨ g ( r ∗ ) − r ∗ , u n + 1 − u n ⟩ + limsup n → ∞ ⟨ g ( r ∗ ) − r ∗ , u n − r ∗ ⟩ ≤ 0 .

It is evident from expressions (49) and (57) that we are going to have it

(58) limsup n → ∞ γ n ‖ u n − r ∗ ‖ 2 2 ( 1 − ξ ) + ⟨ g ( r ∗ ) − r ∗ , u n + 1 − r ∗ ⟩ 1 − ξ ≤ 0 .

By choosing n ≥ M 3 ∈ N ( M 3 ≥ M 2 ) large enough such that 2 γ n ( 1 − ξ ) < 1 , and by the use of (49), (58), and through Lemma 2.2, we conclude that ‖ u n − r ∗ ‖ → 0 as n → ∞ .

Case 2: Suppose that there exists a subsequence { n i } of { n } such that

‖ u n i − r ∗ ‖ ≤ ‖ u n i + 1 − r ∗ ‖ , ∀ i ∈ N .

From Lemma 2.3, there exists a sequence { m k } ⊂ N as { m k } → ∞ , such that

(59) ‖ u m k − r ∗ ‖ ≤ ‖ u m k + 1 − r ∗ ‖ and ‖ u k − r ∗ ‖ ≤ ‖ u m k + 1 − r ∗ ‖ , ∀ k ∈ N .

As in case 1, expression (47) implies that:

(60) ( 1 − γ m k ) 1 − μ ρ m k ρ m k + 1 ‖ u m k − y m k ‖ 2 + ( 1 − γ m k ) 1 − μ ρ m k ρ m k + 1 ‖ t m k − y m k ‖ 2 ≤ γ m k ‖ g ( u m k ) − r ∗ ‖ 2 + ‖ u m k − r ∗ ‖ 2 − ‖ u m k + 1 − r ∗ ‖ 2 .

By using γ m k → 0 , we obtain the following result:

(61) lim k → ∞ ‖ u m k − y m k ‖ = lim k → ∞ ‖ t m k − y m k ‖ = 0 .

Next, we are going to obtain the following:

(62) ‖ u m k + 1 − u m k ‖ = ‖ γ m k g ( u m k ) + ( 1 − γ m k ) t m k − u m k ‖ = ‖ γ m k [ g ( u m k ) − u m k ] + ( 1 − γ m k ) [ t m k − u m k ] ‖ ≤ γ m k ‖ g ( u m k ) − u m k ‖ + ( 1 − γ m k ) ‖ t m k − u m k ‖ ⟶ 0 .

We use the same argument as in Case 1 such that

(63) limsup k → ∞ ⟨ g ( r ∗ ) − r ∗ , u m k + 1 − r ∗ ⟩ ≤ 0 .

By using expressions (49) and (59), we have

(64) ‖ u m k + 1 − r ∗ ‖ 2 ≤ [ 1 − 2 γ m k ( 1 − ξ ) ] ‖ u m k − r ∗ ‖ 2 + 2 γ m k ( 1 − ξ ) γ m k ‖ u m k − r ∗ ‖ 2 2 ( 1 − ξ ) + ⟨ g ( r ∗ ) − r ∗ , u m k + 1 − r ∗ ⟩ 1 − ξ ≤ [ 1 − 2 γ m k ( 1 − ξ ) ] ‖ u m k + 1 − r ∗ ‖ 2 + 2 γ m k ( 1 − ξ ) γ m k ‖ u m k − r ∗ ‖ 2 2 ( 1 − ξ ) + ⟨ g ( r ∗ ) − r ∗ , u m k + 1 − r ∗ ⟩ 1 − ξ .

It continues on from that

(65) ‖ u m k + 1 − r ∗ ‖ 2 ≤ γ m k ‖ u m k − r ∗ ‖ 2 2 ( 1 − ξ ) + ⟨ g ( r ∗ ) − r ∗ , u m k + 1 − r ∗ ⟩ 1 − ξ .

Since γ m k → 0 and ‖ u m k − r ∗ ‖ is a bounded sequence, then, expressions (63) and (65) indicate that

(66) ‖ u m k + 1 − r ∗ ‖ 2 → 0 , as k → ∞ .

The aforementioned equation means that

(67) lim k → ∞ ‖ u k − r ∗ ‖ 2 ≤ lim k → ∞ ‖ u m k + 1 − r ∗ ‖ 2 ≤ 0 .

Consequently, u n → r ∗ . This completes the proof of the theorem.□

Theorem 3.6

Let the mapping K : ℰ → ℰ satisfy conditions ( K 1)–( K 4). Then, { u n } generated by the Algorithm 2 converges strongly to a solution r ∗ = P VI ( ℳ , K ) ( 0 ) .

Proof

By using the definition of { s n } , we obtain

(68) ‖ s n − r ∗ ‖ = ‖ u n + χ n ( u n − u n − 1 ) − γ n u n − χ n γ n ( u n − u n − 1 ) − r ∗ ‖ = ‖ ( 1 − γ n ) ( u n − r ∗ ) + ( 1 − γ n ) χ n ( u n − u n − 1 ) − γ n r ∗ ‖

(69) ≤ ( 1 − γ n ) ‖ u n − r ∗ ‖ + ( 1 − γ n ) χ n ‖ u n − u n − 1 ‖ + γ n ‖ r ∗ ‖ ≤ ( 1 − γ n ) ‖ u n − r ∗ ‖ + γ n K 1 ,

where

( 1 − γ n ) χ n γ n ‖ u n − u n − 1 ‖ + ‖ r ∗ ‖ ≤ K 1 .

It is given that ρ n → ρ such that there exists a finite number ℑ ∈ ( 0 , 1 − μ ) such that

lim n → ∞ 1 − μ ρ n ρ n + 1 = 1 − μ > ℑ > 0 .

Thus, there is N 1 ∈ N such that

(70) 1 − μ ρ n ρ n + 1 > ℑ > 0 , ∀ n ≥ N 1 .

From Lemma 3.3, we may rewrite

(71) ‖ u n + 1 − r ∗ ‖ 2 ≤ ‖ s n − r ∗ ‖ 2 , ∀ n ≥ N 1 .

By the use of expressions (69) and (71), we obtain

(72) ‖ u n + 1 − r ∗ ‖ ≤ ( 1 − γ n ) ‖ u n − r ∗ ‖ + γ n K 1 ≤ max { ‖ u n − r ∗ ‖ , K 1 } ⋯ ≤ max { ‖ u N 1 − r ∗ ‖ , K 1 } .

As a result, we can conclude that { u n } is a bounded sequence. Indeed, by expression (69), we have

(73) ‖ s n − r ∗ ‖ 2 ≤ ( 1 − γ n ) 2 ‖ u n − r ∗ ‖ 2 + γ n 2 K 1 2 + 2 K 1 γ n ( 1 − γ n ) ‖ u n − r ∗ ‖ ≤ ‖ u n − r ∗ ‖ 2 + γ n [ γ n K 1 2 + 2 K 1 ( 1 − γ n ) ‖ u n − r ∗ ‖ ] ≤ ‖ u n − r ∗ ‖ 2 + γ n K 2 ,

for some K 2 > 0 . Both expressions (21) and (73) imply that

(74) ‖ u n + 1 − r ∗ ‖ 2 ≤ ‖ u n − r ∗ ‖ 2 + γ n K 2 − 1 − μ ρ n ρ n + 1 ‖ s n − y n ‖ 2 − 1 − μ ρ n ρ n + 1 ‖ u n + 1 − y n ‖ 2 .

From expression (68), we can write

(75) ‖ s n − r ∗ ‖ 2 = ‖ u n + χ n ( u n − u n − 1 ) − γ n u n − χ n γ n ( u n − u n − 1 ) − r ∗ ‖ 2 = ‖ ( 1 − γ n ) ( u n − r ∗ ) + ( 1 − γ n ) χ n ( u n − u n − 1 ) − γ n r ∗ ‖ 2 ≤ ‖ ( 1 − γ n ) ( u n − r ∗ ) + ( 1 − γ n ) χ n ( u n − u n − 1 ) ‖ 2 + 2 γ n ⟨ − r ∗ , s n − r ∗ ⟩ = ( 1 − γ n ) 2 ‖ u n − r ∗ ‖ 2 + ( 1 − γ n ) 2 χ n 2 ‖ u n − u n − 1 ‖ 2 + 2 χ n ( 1 − γ n ) 2 ‖ u n − r ∗ ‖ ‖ u n − u n − 1 ‖ + 2 γ n ⟨ − r ∗ , s n − u n + 1 ⟩ + 2 γ n ⟨ − r ∗ , u n + 1 − r ∗ ⟩ ≤ ( 1 − γ n ) ‖ u n − r ∗ ‖ 2 + χ n 2 ‖ u n − u n − 1 ‖ 2 + 2 χ n ( 1 − γ n ) ‖ u n − r ∗ ‖ ‖ u n − u n − 1 ‖ + 2 γ n ‖ r ∗ ‖ ‖ s n − u n + 1 ‖ + 2 γ n ⟨ − r ∗ , u n + 1 − r ∗ ⟩ = ( 1 − γ n ) ‖ u n − r ∗ ‖ 2 + γ n χ n ‖ u n − u n − 1 ‖ χ n γ n ‖ u n − u n − 1 ‖ + 2 ( 1 − γ n ) ‖ u n − r ∗ ‖ χ n γ n ‖ u n − u n − 1 ‖ + 2 ‖ r ∗ ‖ ‖ s n − u n + 1 ‖ + 2 ⟨ r ∗ , r ∗ − u n + 1 ⟩ .

From expressions (71) and (75), we obtain

(76) ‖ u n + 1 − r ∗ ‖ 2 ≤ ( 1 − γ n ) ‖ u n − r ∗ ‖ 2 + γ n χ n ‖ u n − u n − 1 ‖ χ n γ n ‖ u n − u n − 1 ‖ + 2 ( 1 − γ n ) ‖ u n − r ∗ ‖ χ n γ n ‖ u n − u n − 1 ‖ + 2 ‖ r ∗ ‖ ‖ s n − u n + 1 ‖ + 2 ⟨ r ∗ , r ∗ − u n + 1 ⟩ .

Case 1: Consider that there exists a finite number N 2 ∈ N ( N 2 ≥ N 1 ) such that

(77) ‖ u n + 1 − r ∗ ‖ ≤ ‖ u n − r ∗ ‖ , ∀ n ≥ N 2 .

Thus, the aforementioned relation implies that lim n → ∞ ‖ u n − r ∗ ‖ exists and let lim n → ∞ ‖ u n − r ∗ ‖ = l , for l ≥ 0 . By using expression (74), we can rewrite

(78) 1 − μ ρ n ρ n + 1 ‖ s n − y n ‖ 2 + 1 − μ ρ n ρ n + 1 ‖ u n + 1 − y n ‖ 2 ≤ ‖ u n − r ∗ ‖ 2 + γ n K 2 − ‖ u n + 1 − r ∗ ‖ 2 .

Since limit of ‖ u n − r ∗ ‖ exists and γ n → 0 , we can deduce that

(79) ‖ s n − y n ‖ → 0 and ‖ u n + 1 − y n ‖ → 0 as n → ∞ .

It continues from expression (79) that

(80) lim n → ∞ ‖ s n − u n + 1 ‖ ≤ lim n → ∞ ‖ s n − y n ‖ + lim n → ∞ ‖ y n − u n + 1 ‖ = 0 .

Next, we need to evaluate

(81) ‖ s n − u n ‖ = ‖ u n + χ n ( u n − u n − 1 ) − γ n [ u n + χ n ( u n − u n − 1 ) ] − u n ‖ ≤ χ n ‖ u n − u n − 1 ‖ + γ n ‖ u n ‖ + χ n γ n ‖ u n − u n − 1 ‖ = γ n χ n γ n ‖ u n − u n − 1 ‖ + γ n ‖ u n ‖ + γ n 2 χ n γ n ‖ u n − u n − 1 ‖ ⟶ 0 .

Thus, the aforementioned expression implies that

(82) lim n → ∞ ‖ u n − u n + 1 ‖ ≤ lim n → ∞ ‖ u n − s n ‖ + lim n → ∞ ‖ s n − u n + 1 ‖ = 0 .

From given r ∗ = P VI ( ℳ , K ) ( 0 ) , we have

(83) ⟨ 0 − r ∗ , y − r ∗ ⟩ ≤ 0 , ∀ y ∈ VI ( ℳ , K ) .

Moreover, it is considered that

(84) limsup n → ∞ ⟨ r ∗ , r ∗ − u n ⟩ = lim k → ∞ ⟨ r ∗ , r ∗ − u n k ⟩ = ⟨ r ∗ , r ∗ − u ˆ ⟩ ≤ 0 .

By using the fact, lim n → ∞ ‖ u n + 1 − u n ‖ = 0 . Thus, using expression (84), we can deduce that

(85) limsup n → ∞ ⟨ r ∗ , r ∗ − u n + 1 ⟩ ≤ limsup n → ∞ ⟨ r ∗ , r ∗ − u n ⟩ + limsup n → ∞ ⟨ r ∗ , u n − u n + 1 ⟩ ≤ 0 .

By using expressions (76) and (85) and taking Lemma 2.2 imply that ‖ u n − r ∗ ‖ → 0 as n → ∞ .

Case 2: Let there exists a subsequence { n i } of { n } such that

‖ u n i − r ∗ ‖ ≤ ‖ u n i + 1 − r ∗ ‖ , ∀ i ∈ N .

Thus, by using Lemma 2.3, there exists a sequence { m k } ⊂ N as { m k } → ∞ such that

(86) ‖ u m k − r ∗ ‖ ≤ ‖ u m k + 1 − r ∗ ‖ and ‖ u k − r ∗ ‖ ≤ ‖ u m k + 1 − r ∗ ‖ , for all k ∈ N .

Expression (78) implies that

(87) 1 − μ ρ m k ρ m k + 1 ‖ s m k − y m k ‖ 2 + 1 − μ ρ m k ρ m k + 1 ‖ u m k + 1 − y m k ‖ 2 ≤ ‖ u m k − r ∗ ‖ 2 + γ m k K 2 − ‖ u m k + 1 − r ∗ ‖ 2 .

Due to sequence γ m k → 0 , we deduce the following:

(88) lim k → ∞ ‖ s m k − y m k ‖ = lim k → ∞ ‖ u m k + 1 − y m k ‖ = 0 .

It follows that

(89) lim k → ∞ ‖ u m k + 1 − s m k ‖ ≤ lim k → ∞ ‖ u m k + 1 − y m k ‖ + lim k → ∞ ‖ y m k − s m k ‖ = 0 .

Next, we have to evaluate

(90) ‖ s m k − u m k ‖ = ‖ u m k + α m k ( u m k − u m k − 1 ) − γ m k [ u m k + α m k ( u m k − u m k − 1 ) ] − u m k ‖ ≤ α m k ‖ u m k − u m k − 1 ‖ + γ m k ‖ u m k ‖ + α m k γ m k ‖ u m k − u m k − 1 ‖ = γ m k α m k γ m k ‖ u m k − u m k − 1 ‖ + γ m k ‖ u m k ‖ + γ m k 2 α m k γ m k ‖ u m k − u m k − 1 ‖ ⟶ 0 .

It follows that

(91) lim k → ∞ ‖ u m k − u m k + 1 ‖ ≤ lim k → ∞ ‖ u m k − s m k ‖ + lim k → ∞ ‖ s m k − u m k + 1 ‖ = 0 .

By using the same argument as in Case 1, such that

(92) limsup k → ∞ ⟨ r ∗ , r ∗ − u m k + 1 ⟩ ≤ 0 .

Now, using expressions (76) and (86), we have

(93) ‖ u m k + 1 − r ∗ ‖ 2 ≤ ( 1 − γ m k ) ‖ u m k − r ∗ ‖ 2 + γ m k α m k ‖ u m k − u m k − 1 ‖ α m k γ m k ‖ u m k − u m k − 1 ‖ + 2 ( 1 − γ m k ) ‖ u m k − r ∗ ‖ α m k γ m k ‖ u m k − u m k − 1 ‖ + 2 ‖ r ∗ ‖ ‖ s m k − u m k + 1 ‖ + 2 ⟨ r ∗ , r ∗ − u m k + 1 ⟩ ≤ ( 1 − γ m k ) ‖ u m k + 1 − r ∗ ‖ 2 + γ m k α m k ‖ u m k − u m k − 1 ‖ α m k γ m k ‖ u m k − u m k − 1 ‖ + 2 ( 1 − γ m k ) ‖ u m k − r ∗ ‖ α m k γ m k ‖ u m k − u m k − 1 ‖ + 2 ‖ r ∗ ‖ ‖ s m k − u m k + 1 ‖ + 2 ⟨ r ∗ , r ∗ − u m k + 1 ⟩ .

Thus, we obtain

(94) ‖ u m k + 1 − r ∗ ‖ 2 ≤ α m k ‖ u m k − u m k − 1 ‖ α m k γ m k ‖ u m k − u m k − 1 ‖ + 2 ( 1 − γ m k ) ‖ u m k − r ∗ ‖ α m k γ m k ‖ u m k − u m k − 1 ‖ + 2 ‖ r ∗ ‖ ‖ s m k − u m k + 1 ‖ + 2 ⟨ r ∗ , r ∗ − u m k + 1 ⟩ .

Since γ m k → 0 and ‖ u m k − r ∗ ‖ is a bounded sequence, then expressions (92) and (94) imply that

(95) ‖ u m k + 1 − r ∗ ‖ 2 → 0 , as k → ∞ .

It implies that

(96) lim n → ∞ ‖ u k − r ∗ ‖ 2 ≤ lim n → ∞ ‖ u m k + 1 − r ∗ ‖ 2 ≤ 0 .

As a consequence, u n → r ∗ . This will complete the proof of the theorem.□

Theorem 3.7

Let the mapping K : ℰ → ℰ satisfy conditions ( K 1)–( K 4). Then, { u n } generated by Algorithm 3 converges strongly to a solution of the problem VIP.

Proof

The proof is the same as the proof of Theorem 3.5.□

Theorem 3.8

Let the mapping K : ℰ → ℰ meet conditions ( K 1)–( K 4). Then, { u n } generated by Algorithm 4 converges strongly to a solution of the problem VIP.

Proof

The proof is the same as the proof of Theorem 3.6.□

4 Numerical illustrations

The numerical results of the proposed iterative schemes are given in this section, in contrast to some related work in the literature and also in the analysis of how variations in control parameters affect the numerical effectiveness of the proposed algorithms. All computations are done in MATLAB R2018b and run on an HP i5-6200 8.00 GB (7.78 GB usable) RAM laptop.

Example 4.1

Let H = l 2 be a real Hilbert space with sequences of real numbers satisfying the following condition:

(97) ‖ u 1 ‖ 2 + ‖ u 2 ‖ 2 + ⋯ + ‖ u n ‖ 2 + ⋯ < + ∞ .

Assume that K : ℳ → ℳ is defined by:

G ( u ) = ( 5 − ‖ u ‖ ) u , ∀ u ∈ H ,

where C = { u ∈ H : ‖ u ‖ ≤ 3 } . It is to note that K is sequentially weakly continuous on ℰ and VI ( ℳ , K ) = { 0 } . For each u , y ∈ ℰ , we have

(98) ∥ K ( u ) − K ( y ) ∥ = ∥ ( 5 − ‖ u ‖ ) u − ( 5 − ‖ y ‖ ) y ∥ = ∥ 5 ( u − y ) − ‖ u ‖ ( u − y ) − ( ‖ u ‖ − ‖ y ‖ ) y ∥ ≤ 5 ‖ u − y ‖ + ‖ u ‖ ‖ u − y ‖ + ∣ ‖ u ‖ − ‖ y ‖ ∣ ‖ y ‖ ≤ 5 ‖ u − y ‖ + 3 ‖ u − y ‖ + 3 ‖ u − y ‖ ≤ 11 ‖ u − y ‖ .

Hence, K is L -Lipschitz continuous with L = 11 . For any u , y ∈ ℰ , let ⟨ K ( u ) , y − u ⟩ > 0 such that

( 5 − ‖ u ‖ ) ⟨ u , y − u ⟩ > 0 .

Since ‖ u ‖ ≤ 3 , it implies that

⟨ u , y − u ⟩ > 0 .

Thus, it implies that

(99) ⟨ K ( y ) , y − u ⟩ = ( 5 − ‖ y ‖ ) ⟨ y , y − u ⟩ ≥ ( 5 − ‖ y ‖ ) ⟨ y , y − u ⟩ − ( 5 − ‖ y ‖ ) ⟨ u , y − u ⟩ ≥ 2 ‖ u − y ‖ 2 ≥ 0 .

Thus, we show that K is quasimonotone on ℳ . Let u = ( 5 2 , 0 , 0 , … , 0 , … ) and y = ( 3 , 0 , 0 , … , 0 , … ) such that

⟨ K ( u ) − K ( y ) , u − y ⟩ = ( 2.5 − 3 ) 2 < 0 .

The formula for a projection on C is given in the following manner:

P C ( u ) = u if ‖ u ‖ ≤ 3 , 3 u ‖ u ‖ , otherwise .

The following conditions have been taken for numerical study. (i) Algorithm 1 (Alg1): ρ 1 = 0.22 , μ = 0.44 , γ n = 1 ( n + 2 ) ; (ii) Algorithm 2 (Alg2): ρ 1 = 0.22 , μ = 0.44 , γ n = 1 ( n + 2 ) , p n = 100 ( n + 1 ) 2 ; (iii) Algorithm 3 (Alg3): ρ 1 = 0.22 , μ = 0.44 , χ = 0.50 , γ n = 1 ( n + 2 ) , ε n = 1 ( n + 1 ) 2 ; (iv) Algorithm 4 (Alg4): ρ 1 = 0.22 , μ = 0.44 , χ = 0.50 , γ n = 1 ( n + 2 ) , ε n = 1 ( n + 1 ) 2 , p n = 100 ( n + 1 ) 2 (Tables 1 and 2).

Table 1

Numerical values for Example 4.1

Number of iterations Execution time in seconds
u 1 Alg1 Alg3 Alg1 Alg3
( 2 , 2 , … , 2 5,000 , 0 , 0 , … ) 41 36 3.93847480000000 3.33763350000000
( 1 , 2 , … , 5,000 , 0 , 0 , … ) 57 48 4.87583390000000 4.24728740000000
( 5 , 5 , … , 5 10,000 , 0 , 0 , … ) 48 39 4.37341940000000 3.82418350000000
( 50 , 50 , … , 5 0 10,000 , 0 , 0 , … ) 61 49 5.84570940000000 4.57478350000000
( 500 , 500 , … , 50 0 10,000 , 0 , 0 , … ) 89 67 8.34746348000000 6.92528350000000
Table 2

Numerical values for Example 4.1

Number of iterations Execution time in seconds
u 1 Alg2 Alg4 Alg2 Alg4
( 2 , 2 , … , 2 5,000 , 0 , 0 , … ) 29 21 2.14638330000000 2.00018330000000
( 1 , 2 , … , 5,000 , 0 , 0 , … ) 33 27 2.87463850000000 2.14639210000000
( 5 , 5 , … , 5 10,000 , 0 , 0 , … ) 30 21 2.72444340000000 1.92738850000000
( 50 , 50 , … , 5 0 10,000 , 0 , 0 , … ) 41 34 4.56044940000000 3.93847350000000
( 500 , 500 , … , 50 0 10,000 , 0 , 0 , … ) 47 38 7.46292840000000 5.46293350000000

5 Conclusion

We developed various types of explicit extragradient-type methods for finding a numerical solution to quasimonotone variational inequalities in a real Hilbert space. This approach is seen as a variant of the two-step extragradient method. Two strongly convergent findings are well proven and correspond to the suggested methods. The numerical findings were analyzed to demonstrate the numerical performance of the suggested methods. These computational results show that the non-monotone variable step size rule continues to improve the effectiveness of the iterative sequence in this scenario.

  1. Funding information: This research was supported by The Science, Research and Innovation Promotion Funding (TSRI) (Grant No. FRB660012/0168). This research block grants were managed under Rajamangala University of Technology Thanyaburi (FRB66E0653S.2).

  2. Author contributions: All the authors contributed equally to this manuscript.

  3. Conflict of interest: No potential conflict of interest was reported by the authors.

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Received: 2021-06-29
Revised: 2022-11-16
Accepted: 2023-01-20
Published Online: 2023-07-29

© 2023 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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https://doi.org/10.1515/dema-2022-0202
Keywords for this article
variational inequality problem; subgradient extragradient method; strong convergence results; quasimonotone operator; Lipschitz continuity
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  2. A novel class of bipolar soft separation axioms concerning crisp points
  3. Duality for convolution on subclasses of analytic functions and weighted integral operators
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