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axioms Article Inertial Iterative Self-Adaptive Step Size Extragradient-Like Method for Solving Equilibrium Problems in Real Hilbert Space with Applications 1 2, Wiyada Kumam and Kanikar Muangchoo * Program in Applied Statistics, Department of Mathematics and Computer Science, Faculty of Science and Technology, Rajamangala University of Technology Thanyaburi, Thanyaburi, Pathumthani 12110, Thailand; wiyada.kum@rmutt.ac.th Faculty of Science and Technology, Rajamangala University of Technology Phra Nakhon (RMUTP), 1381 Pracharat 1 Road, Wongsawang, Bang Sue, Bangkok 10800, Thailand * Correspondence: kanikar.m@rmutp.ac.th; Tel.: +66-2-836-3000 (ext. 4193) Received: 2 September 2020; Accepted: 22 October 2020; Published: 31 October 2020 Abstract: A number of applications from mathematical programmings, such as minimization problems, variational inequality problems and ﬁxed point problems, can be written as equilibrium problems. Most of the schemes being used to solve this problem involve iterative methods, and for that reason, in this paper, we introduce a modiﬁed iterative method to solve equilibrium problems in real Hilbert space. This method can be seen as a modiﬁcation of the paper titled “A new two-step proximal algorithm of solving the problem of equilibrium programming” by Lyashko et al. (Optimization and its applications in control and data sciences, Springer book pp. 315–325, 2016). A weak convergence result has been proven by considering the mild conditions on the cost bifunction. We have given the application of our results to solve variational inequality problems. A detailed numerical study on the Nash–Cournot electricity equilibrium model and other test problems is considered to verify the convergence result and its performance. Keywords: pseudomonotone bifunction; Lipschitz-type conditions; equilibrium problem; variational inequalities 1. Introduction An equilibrium problem (EP) is a generalized concept that uniﬁes several mathematical problems, such as the variational inequality problems, minimization problems, complementarity problems, the ﬁxed point problems, non-cooperative games of Nash equilibrium, the saddle point problems and scalar and vector minimization problems (see e.g., [1–3]). The particular form of an equilibrium problem was ﬁrstly established in 1992 by Muu and Oettli [4] and then further elaborated by Blum and Oettli [1]. Next, we consider the concept of an equilibrium problem introduced by Blum and Oettli in [1]. Let C be a non-empty, closed and convex subset H of a real Hilbert space and f : HH ! R is bifunction with f (v, v) = 0, for each v 2 C. A equilibrium problem regarding f on the set C is deﬁned in the following way: Find p 2 C such that f ( p, v) 0, for all v 2 C. (1) Many methods have been already established over the past couple of years to ﬁgure out the equilibrium problem in Hilbert spaces [5–15], the inertial methods [11,16–18] and others in [18–24]. In particular, Tran et al. introduced an iterative scheme in [8], in that a sequence fu g was generated in the following way: Axioms 2020, 9, 127; doi:10.3390/axioms9040127 www.mdpi.com/journal/axioms Axioms 2020, 9, 127 2 of 21 u 2 C, > 0 1 2 v = arg minfl f (u , y) + ku yk : y 2 Cg, n n n (2) 1 2 u = arg minfl f (v , y) + ku yk : y 2 Cg, n n n+1 1 1 where 0 < l < min , and c , c are Lipschitz constants. Lyashko et al. [25] in 2016 introduced 1 2 2c 2c 1 2 an improvement of the method (2) to solve equilibrium problem and sequence fu g was generated in the following way: u , v 2 C, > 0 0 1 2 u = arg minfl f (v , y) + ku yk : y 2 Cg, n n n+1 (3) > 1 v = arg minfl f (v , y) + ku yk : y 2 Cg, n+1 n n+1 where 0 < l < and c , c are Lipschitz constants. 1 2 2c +4c 2 1 In this paper, we consider the extragradient method in (3) and to provide its improvement by using the inertial scheme [26] and continue to improve the step size rule for its second step. The step size is not ﬁxed in our case, but it is dependent on a particular formula by using prior information of the bifunction values. A weak convergence theorem dealing with the suggested technique is presented by having the speciﬁc bi-functional condition. We have also considered how our results are presented to the problems of a variational inequality. A few other formulations of the problem of variational inequalities are discussed, and many computational examples in ﬁnite and inﬁnite dimensions spaces are also presented to demonstrate the applicability of our proposed results. In this study, we study the equilibrium problem through the following assumptions: ( f ) A bifunction f : HH ! R is said to be (see [1,27]) pseudomonotone on C if f (v , v ) 0 =) f (v , v ) 0, for all v , v 2 C. 1 2 2 1 1 2 ( f ) A bifunction f : H H ! R is said to be Lipschitz-type continuous [28] on C if there exist c , c > 0 such that 1 2 2 2 f (v , v ) f (v , v ) + f (v , v ) + c kv v k + c kv v k , for all v , v , v 2 C. 1 3 1 2 2 3 1 1 2 2 2 3 1 2 3 ( f ) lim sup f (v , z) f (v , z) for each z 2 C and fv g C satisfying v * v ; 3 n n n n!+¥ ( f ) f (u,) is convex and subdifferentiable on H for each u 2 H. The rest of this paper will be organized as follows: In Section 2, we give a few deﬁnitions and important lemmas to be used in this paper. Section 3 includes the main algorithm involving pseudomonotone bifunction and provides a weak convergence theorem. Section 4 describes some applications in the variational inequality problems. Section 5 sets out the numerical studies to demonstrate the algorithmic efﬁciency. 2. Preliminaries In this section, some important lemmas and basic deﬁnitions are provided. Moreover, EP( f , C) denotes the solution set of an equilibrium problem on the set C and p is any arbitrary element of EP( f , C). A metric projection P (u) of u onto a closed, convex subset C of H is deﬁned by P (u) = arg minfkv ukg. v2C Lemma 1. [29] Let P : H ! C be a metric projection from H onto C. Then C Axioms 2020, 9, 127 3 of 21 (i) For all u 2 C, v 2 H and 2 2 2 ku P (v)k +kP (v) vk ku vk . C C (ii) w = P (u) if and only if hu w, v wi 0, for all v 2 C. Lemma 2. [29] For all u, v 2 H with k 2 R. Then, the following relationship is holds. 2 2 2 2 kku + (1k)vk = kkuk + (1k)kvk k(1k)ku vk . Assume that g : C ! R be a convex function and subdifferential of g at u 2 C is deﬁned by ¶g(u) = fw 2 C : g(v) g(u) hw, v ui, for all v 2 Cg. Given that f (u, .) is convex and subdifferentiable on H for each ﬁxed u 2 H and subdifferential of f (u, .) at x 2 H deﬁned by ¶ f (u, .)(x) = ¶ f (u, x) = fz 2 H : f (u, v) f (u, x) hz, v xi, for all v 2 Hg. 2 2 A normal cone of C at u 2 C is deﬁned by N (u) = fw 2 H : hw, v ui 0, for all v 2 Cg. Lemma 3. [30] Assume that C is a nonempty, closed and convex subset of a real Hilbert spaceH and h : C ! R be a convex, lower semi-continuous and subdifferentiable function on C. Then, u 2 C is a minimizer of a function h if and only if 0 2 ¶h(u) + N (u) where ¶h(u) and N (u) denotes the subdifferential of h at u and the normal C C cone of C at u, respectively. Lemma 4. [31] Let a , b and c are non-negative real sequences such that n n n +¥ a a + b (a a ) + c , for all n 1, with c < +¥, n+1 n n n n1 n å n n=1 where b > 0 such that 0 b b < 1 for all n 2 N. Then, the following relations are true. +¥ (i) [a a ] < +¥, with [s] := maxfs, 0g; n + + å n1 n=1 (ii) lim a = a 2 [0, +¥). n!+¥ n Lemma 5. [32] Let a sequence fa g in H and C H and the following conditions have been met: (i) for each a 2 C, lim ka ak exists; n!+¥ n (ii) each weak sequentially limit point of fa g belongs to set C. Then, fa g weakly converges to an element in C. 3. Main Results In this section, we present our main algorithm and provide a weak convergence theorem for our proposed method. The detailed method is given below. Remark 1. By Expression (5), we obtain h i 2 2 l f (v , u ) f (v , v ) c kv v k c kv u k m f (v , u ). (4) n+1 n1 n+1 n1 n 1 n1 n 2 n n+1 n n+1 Axioms 2020, 9, 127 4 of 21 Lemma 6. Let fu g be a sequence generated by Algorithm 1. Then, the following inequality holds. ml f (v , y) ml f (v , u ) hr u , y u i, for all y 2 C. n n n n n+1 n n+1 n+1 Algorithm 1 Modiﬁed Popov’s subgradient extragradient-like iterative scheme. Step 1: Choose u , v , u , v 2 H and a sequence } is non-decreasing such that 0 } } < , 0 0 n n 1 1 n o 13} l > 0 and 0 < s < min , and m 2 (0, s). 0 2 2 (1}) +4c (}+} ) 2c +4c (1+}) 1 2 1 Step 2: Evaluate u = arg minfml f (v , y) + kr yk : y 2 Cg, n+1 n n n where r = u + } (u u ). n n n n n1 Step 3: Updated the step size in the following order: n o m f (v ,u ) n n+1 min s, , > 2 2 f (v ,u ) f (v ,v )c kv v k c ku v k +1 n1 n+1 n1 n 1 n1 n 2 n+1 n m f (v ,u ) n n+1 l = (5) n+1 if > 0, 2 2 > f (v ,u ) f (v ,v )c kv v k c ku v k +1 n1 n+1 n1 n 1 n1 n 2 n+1 n l else. Step 4: Evaluate v = arg minfl f (v , y) + kr yk : y 2 Cg, n+1 n+1 n n+1 where r = u + } (u u ). If u = v = r or r = v = v then Stop. n n n n n+1 n+1 n+1 n+1 n+1 n+1 n+1 Else, take n := n + 1 and go back to Step 2. Proof. By the use of Lemma 3, we get n o 0 2 ¶ ml f (v , y) + kr yk (u ) + N (u ). 2 n n n n+1 C n+1 From above there is a w 2 ¶ f (v , u ) and w 2 N (u ) such that 2 n n+1 C n+1 ml w + u r + w = 0. n n+1 n Therefore, we obtain hr u , y u i = ml hw, y u i +hw, y u i, for all y 2 C. n n n+1 n+1 n+1 n+1 Due to w 2 N (u ) then hw, y u i 0, for each y 2 C. It implies that C n+1 n+1 ml hw, y u i hr u , y u i, for all y 2 C. (6) n n+1 n n+1 n+1 Given that w 2 ¶ f (v , u ), we have 2 n n+1 f (v , y) f (v , u ) hw, y u i, for all y 2 H. (7) n n n+1 n+1 By combining Expressions (6) and (7), we obtain ml f (v , y) ml f (v , u ) hr u , y u i, for all y 2 C. n n n n n+1 n n+1 n+1 Axioms 2020, 9, 127 5 of 21 Lemma 7. Let fv g be a sequence generated by Algorithm 1. Then, the following inequality holds. l f (v , y) l f (v , v ) hr v , y v i, for all y 2 C. n+1 n n+1 n n+1 n+1 n+1 n+1 Proof. The proof is same as the proof of Lemma 6. Lemma 8. If u = v = r and r = v = v in Algorithm 1, then v 2 EP( f , C). n+1 n n n+1 n+1 n n Proof. The proof of this can easily be seen from Lemmas 6 and 7. Lemma 9. Let f : H H ! R be a bifunction and satisﬁes the conditions ( f )–( f ). Then, for each p 2 1 4 EP( f , C) 6= Æ, we have ku pk n+1 2 2 2 kr pk (1 l )ku r k + 4c l l kr v k n n+1 n+1 n 1 n+1 n n n1 2 2 l (1 4c l )kr v k l (1 2c l )ku v k . n+1 1 n n n n+1 2 n n+1 n Proof. By Lemma 6, we obtain ml f (v , p) ml f (v , u ) hr u , p u i. (8) n n n n n+1 n n+1 n+1 Thus, p 2 EP( f , C) and the condition ( f ) implies that f (v , p) 0. From (8), we have hr u , u pi ml f (v , u ). (9) n n+1 n+1 n n n+1 From Expression (4), we obtain 2 2 m f (v , u ) l f (v , u ) f (v , v ) c kv v k c kv u k . (10) n n+1 n+1 n1 n+1 n1 n 1 n1 n 2 n n+1 Combining expression (9) and (10), implies that hr u , u pi l l f (v , u ) f (v , v ) n n n n+1 n+1 n+1 n1 n+1 n1 i (11) 2 2 c l kv v k c l ku v k . 1 n n1 n 2 n n+1 n By Lemma 7, we have l f (v , u ) f (v , v ) hr v , u v i. (12) n n1 n+1 n1 n n n n+1 n Thus, combining (11) and (12) we get hr u , u pi l hr v , u v i n n+1 n+1 n+1 n n n+1 n i (13) 2 2 c l kv v k c l ku v k . n n n n 1 n1 2 n+1 We have the following mathematical expressions: 2 2 2 2hr u , u pi = kr pk ku r k ku pk . n n+1 n+1 n n+1 n n+1 2 2 2 2hr v , u v i = kr v k +ku v k kr u k . n n n n n n n n+1 n+1 n+1 Axioms 2020, 9, 127 6 of 21 From the above equation and (13), we have ku pk n+1 2 2 2 kr pk (1 l )ku r k l (1 2c l )ku v k n n+1 n+1 n n+1 2 n n+1 n 2 2 l kr v k + l (2c l )kv v k n n n n n+1 n+1 1 n1 We also have 2 2 2 kv v k kv r k +kr v k 2kv r k + 2kr v k . n1 n n1 n n n n1 n n n Finally, we get ku pk n+1 2 2 2 kr pk (1 l )ku r k + 4c l l kr v k n n+1 n+1 n 1 n n+1 n n1 2 2 l (1 4c l )kr v k l (1 2c l )ku v k . n+1 1 n n n n+1 2 n n+1 n Theorem 1. Assume that f : HH ! R satisﬁes the conditions ( f )–( f ). Then, for some p 2 EP( f , C) 6= Æ, 1 4 the sequence fr g, fu g and fv g generated by Algorithm 1, weakly converge to p 2 EP( f , C). n n n Proof. By Lemma 9, we obtain ku pk n+1 2 2 2 kr pk (1 l )ku r k + 4c l l kr v k n n+1 n+1 n 1 n n+1 n n1 2 2 l (1 4c l )kr v k l (1 2c l )ku v k . (14) n n n n n n+1 1 n+1 2 n+1 By deﬁnition of r in the Algorithm 1, we have 2 2 kr v k = ku + } (u u ) v k n n1 n n n n1 n1 = k(1 + } )(u v ) } (u v )k n n n n1 n1 n1 2 2 2 = (1 + } )ku v k } ku v k + } (1 + } )ku u k n n n1 n n1 n1 n n n n1 2 2 (1 + })ku v k + }(1 + })ku u k . (15) n n n1 n1 Axioms 2020, 9, 127 7 of 21 Adding the term 4c sl (1 + })ku v k on both sides in (14) with (15) for n 1, we have 1 n+1 n+1 n 2 2 ku pk + 4c sl (1 + })ku v k n+1 1 n+1 n+1 n 2 2 2 kr pk (1 s)ku r k + 4c sl (1 + })ku v k n n n n+1 1 n+1 n+1 2 2 + 4c sl (1 + })ku v k + }(1 + })ku u k 1 n n n1 n n1 2 2 l (1 4c s)kr v k l (1 2c s)ku v k (16) n n 2 n n+1 1 n+1 n+1 2 2 2 kr pk (1 s)ku r k + 4c sl (1 + })ku v k n n+1 n 1 n n n1 2 2 2 + 4c s(} + } )ku u k l (1 4c s)kr v k n n n 1 n1 n+1 1 l (1 2c s 4c s(1 + }))ku v k (17) n+1 2 1 n+1 n 2 2 2 kr pk (1 s)ku r k + 4c sl (1 + })ku v k n n n n n+1 1 n1 2 2 + 4c s(} + } )ku u k 1 n n1 n+1 2 2 (1 2c s 4c s(1 + })) 2ku v k + 2kr v k (18) 2 1 n+1 n n n 2 2 2 kr pk (1 s)ku r k + 4c sl (1 + })ku v k n n+1 n 1 n n n1 2 2 + 4c s(} + } )ku u k 1 n1 n+1 (1 2c s 4c s(1 + }))ku r k . (19) 2 1 n+1 Given that 0 < l s < , then the last inequality turns into 2c +4c (1+}) 2 1 2 2 ku pk + 4c sl (1 + })ku v k n+1 1 n+1 n+1 2 2 2 kr pk (1 s)ku r k + 4c sl (1 + })ku v k n n+1 n 1 n n n1 2 2 + 4c s(} + } )ku u k . (20) 1 n1 From the deﬁnition of r , we have 2 2 kr pk = ku + } (u u ) pk n n n n n1 = k(1 + } )(u p) } (u p)k n n n n1 2 2 2 = (1 + } )ku pk } ku pk + } (1 + } )ku u k . (21) n n n n1 n n n n1 From r , we obtain n+1 2 2 ku r k = ku u } (u u )k n n n n n+1 n+1 n1 2 2 2 = ku u k + } ku u k 2} hu u , u u i (22) n+1 n n n1 n n+1 n n n1 2 2 2 ku u k + } ku u k 2} ku u kku u k n n n n n n+1 n n1 n+1 n1 2 2 2 2 2 ku u k + } ku u k } ku u k } ku u k n+1 n n n1 n n+1 n n n n1 2 2 2 = (1 } )ku u k + (} } )ku u k . (23) n n n n n+1 n n1 Axioms 2020, 9, 127 8 of 21 Combining the Expressions (20), (21) and (23) we have 2 2 ku pk + 4c sl (1 + })ku v k n+1 1 n+1 n+1 n 2 2 2 (1 + } )ku pk } ku pk + } (1 + } )ku u k n n n n n n n1 n1 2 2 2 (1 s) (1 } )ku u k + (} } )ku u k n n+1 n n n n1 2 2 2 + 4c sl (1 + })ku v k + 4c s(} + } )ku u k (24) n n n 1 n1 1 n1 2 2 2 (1 + } )ku pk } ku pk + 4c sl (1 + })ku v k n n n n1 1 n n n1 h i 2 2 2 + }(1 + }) (1 s)(} } ) + 4c s(} + } ) ku u k n 1 n n1 (1 s)(1 } )ku u k (25) n n+1 n 2 2 2 (1 + } )ku pk } ku pk + 4c sl (1 + })ku v k n n n n n n1 1 n1 2 2 + r ku u k q ku u k , (26) n n n1 n n+1 n where h i 2 2 r = }(1 + }) (1 s)(} } ) + 4c s(} + } ) ; n n n 1 q = (1 s)(1 } ). n n Assume that G = Y + r ku u k , n n n n n1 2 2 2 where Y = ku pk } ku pk + 4c sl (1 + })ku v k . Next, (26) implies that n n n n1 1 n n n1 G G n+1 2 2 2 2 = ku pk } ku pk + 4c sl (1 + })ku v k + r ku u k n+1 n+1 n 1 n+1 n+1 n n+1 n+1 n 2 2 2 2 ku pk + } ku pk 4c sl (1 + })ku v k r ku u k n n n n n n n1 1 n1 n1 2 2 2 2 ku pk (1 + } )ku pk + } ku pk + 4c sl (1 + })ku v k n+1 n n n n1 1 n+1 n+1 n 2 2 2 + r ku u k 4c sl (1 + })ku v k r ku u k n n n n n n+1 n+1 1 n1 n1 (q r )ku u k . (27) n n+1 n+1 n Next, we have to compute 2 2 (q r ) = (1 s)(1 } ) }(1 + }) + (1 s)(} } ) 4c s(} + } ) n n n n+1 n 1 2 2 (1 s)(1 }) }(1 + }) 4c s(} + } ) 2 2 2 = (1 }) }(1 + }) s(1 }) 4c s(} + } ) 2 2 = 1 3} s (1 }) + 4c (} + } ) 0. (28) By the use of (27) and (28) for some d 0 implies that 2 2 G G (q r )ku u k dku u k 0. (29) n+1 n n n+1 n+1 n n+1 n The relation (29) implies that fG g is non-increasing. From G we have n n+1 2 2 2 2 G = ku pk } ku pk + r ku u k + 4c sl (1 + })ku v k n n n n+1 n+1 n+1 n+1 n+1 1 n+1 n+1 (30) } ku pk . n+1 n Axioms 2020, 9, 127 9 of 21 By deﬁnition G , we have 2 2 ku pk G + } ku pk n n n n1 G + }ku pk 1 n1 n1 n 2 G (} + + 1) + } ku pk 1 0 n 2 + } ku pk . (31) 1 } From Equations (30) and (31), we obtain G } ku pk n+1 n+1 }ku pk n+1 2 } + } ku pk . (32) 1 } It follows (29) and (32) that d ku u k G G å n+1 n 1 k+1 n=1 1 k+1 2 G + } + } ku pk 1 0 1 } +ku pk . (33) 1 } By letting k ! +¥ in (33), we obtain +¥ ku u k < +¥ implies that lim ku u k = 0. (34) n+1 n n+1 n n!+¥ n=1 From Expressions (22) with (34), we obtain ku r k ! 0 as n ! +¥. (35) n+1 From (32), we have 1 n+1 2 2 Y } + } ku pk + r ku u k . (36) n+1 0 n+1 n+1 n 1 } From Expression (18) and using (21), we have h i 2 2 l (1 2c s 4c s(1 + })) ku v k +kr v k n+1 2 1 n+1 n n n (37) 2 2 Y Y + }(1 + })ku u k + 4c s}(1 + })ku u k . n n n n+1 n1 1 n1 Axioms 2020, 9, 127 10 of 21 Fix k 2 N and using above expression for n = 1, 2, , k. Summing them up, we obtain k h i 2 2 l (1 2c s 4c s(1 + })) ku v k +kr v k n n n n+1 2 1 å n+1 n=1 k k 2 2 Y Y + }(1 + }) ku u k + 4c s}(1 + }) ku u k n n 0 k+1 å n1 1 å n1 n=1 n=1 k+1 2 2 Y + } + } ku pk + r ku u k 0 0 k+1 k+1 k 1 } k k 2 2 + }(1 + }) ku u k + 4c s}(1 + }) ku u k , (38) å n n1 1 å n n1 n=1 n=1 letting k ! +¥, and due to sum of the positive terms series, we obtain +¥ +¥ 2 2 ku v k < +¥ and kr v k < +¥. (39) å n+1 n å n n n=1 n=1 Moreover, we obtain lim ku v k = lim kr v k = 0. (40) n n n n+1 n!+¥ n!+¥ By using the triangular inequality, we get lim ku v k = lim ku r k = lim kv v k = 0. (41) n n n n n1 n n!+¥ n!+¥ n!+¥ It is follow from the relation (24), we obtain ku pk n+1 2 2 2 (1 + } )ku pk } ku pk + }(1 + })ku u k n n n n1 n n1 2 2 2 + 4c s(1 + })ku v k + 4c s(} + } )ku u k , (42) n n 1 n1 1 n1 with (34), (39) and Lemma 4 imply that the sequences ku pk, kr pk and kv pk limits exist for n n n every p 2 EP( f , C). It means that fu g, fr g and fv g are bounded sequences. Take z an arbitrary n n n sequential cluster point of the sequence fu g. Now our aim to prove that z 2 EP( f , C). By Lemma 6 with Expressions (10) and (12), we write ml f (v , y) ml f (v , u ) +hr u , y u i n n n n n n +1 n +1 n +1 k k k k k k k k l l f (v , u ) l l f (v , v ) c l l kv v k n n n 1 n +1 n n n 1 n 1 n n n 1 n k k+1 k k k k+1 k k k k+1 k k c l l kv u k +hr u , y u i n n n n 2 n +1 n +1 n +1 k k+1 k k k k k l hr v , u v i c l l kv v k n n n n +1 n 1 n n n 1 n k+1 k k k k k k+1 k k c l l kv u k +hr u , y u i (43) n n n n 2 n +1 n +1 n +1 k k+1 k k k k k where y in C. Next, from (35), (40), (41) and due to boundedness of fu g gives that the right hand side reaches to zero. Due to m, l > 0 and v * z, we have n n k k 0 lim sup f (v , y) f (z, y), for all y 2 C. (44) k!+¥ Thus, z 2 C implies that f (z, y) 0, for all y 2 C. It proves that z 2 EP( f , C). By Lemma 5, the sequence fu g converges weakly to p 2 EP( f , C). If } = 0 in Algorithm 1, we have a better version of Lyashko et al. [25] extragradient method in terms of step size improvement. Axioms 2020, 9, 127 11 of 21 Corollary 1. Let f : HH ! R satisfy the conditions ( f )-( f ). For some p 2 EP( f , C) 6= Æ, the sequence 1 4 fu g and fv g generated in the following way: n n (i) Given u , v , v 2 H, 0 < s < min 1, , m 2 (0, s) and l > 0. 0 1 0 0 2c +4c 2 1 (ii) Compute 1 2 u = arg minfml f (v , y) + ku yk g, n+1 n n n < 2 y2C 1 2 v = arg minfl f (v , y) + ku yk g, : n+1 n+1 n+1 y2C where n o m f (v , u ) n+1 l = min s, . n+1 2 2 f (v , u ) f (v , v ) c kv v k c ku v k + 1 n1 n+1 n1 n 1 n1 n 2 n+1 n Then, the sequences fu g and fv g converge weakly to p 2 EP( f , C). n n 4. Applications Now, we consider the applications of Theorem 1 to solve the variational inequality problems involving pseudomonotone and Lipschitz continuous operator. A variational inequality problem is deﬁned in the following way: Find p 2 C such that F( p ), v p 0, for all v 2 C. We consider that F meets the following conditions. (F ) Solution set V I(F, C) is non-empty and F is pseudomonotone on C, i.e., F(u), v u 0 implies that F(v), u v 0, for all u, v 2 C; (F ) F is L-Lipschitz continuous on C if there exists a positive constants L > 0 such that kF(u) F(v)k Lku vk, for all u, v 2 C. (F ) lim suphF(u ), v u i hF( p ), v p i for every v 2 C and fu g C satisfying u * p . 3 n n n n n!+¥ Corollary 2. Assume that F : C ! H meet the conditions (F )–(F ). Letfr g,fu g andfv g be the sequences 1 3 n n n are generated in the following way: (i) Choose u , v , u , v 2 H and a sequence } is non-decreasing such that 0 } } < , l > 0, 1 1 0 0 n n 0 n o 13} 1 0 < s < min , and m 2 (0, s). 2 2 3L+2}L) (1}) +2L(}+} ) (ii) Compute u = P (r ml F(v )), where r = u + } (u u ), n+1 n n n n n n n n1 v = P (r l F(v )), where r = u + } (u u ), n+1 C n+1 n+1 n n+1 n+1 n+1 n+1 n while mhFv , u v i n n+1 n l = min s, . n+1 L L 2 2 hFv , u v i kv v k ku v k + 1 n1 n+1 n n1 n n+1 n 2 2 Then, the sequence fr g, fu g and fv g converge weakly to p. n n n Axioms 2020, 9, 127 12 of 21 Corollary 3. Assume that F : C ! H meets the conditions (F )-(F ). Let fu g and fv g be the sequences are 1 3 n n generated in the following way: (i) Choose v , u , v 2 H, 0 < s < min 1, and l > 0. 1 0 0 0 3L (ii) Compute u = P (u ml F(v )), n n n n+1 C v = P (u l F(v )), n+1 C n+1 n+1 n while mhFv , u v i n n+1 n l = min s, . n+1 L L 2 2 hFv , u v i kv v k ku v k + 1 n1 n+1 n n1 n n+1 n 2 2 Then, the sequence fu g and fv g converge weakly to p. n n 5. Computational Illustration Numerical ﬁndings are discussed in this section to show the efﬁciency of our suggested method. Moreover, for Lyashko et al.’s [25] method (L.EgA), our proposed algorithm (Algo.1) and we use error term D = ku u k. n n+1 n Example 1. Consider the Nash–Cournot equilibrium of electricity markets as in [7]. In this problem, there are total three electricity producing ﬁrms: i (i = 1, 2, 3). The ﬁrm’s 1,2,3 have generating units named as I = f1g, I = f2, 3g and I = f4, 5, 6g, respectively. Assume that u denote the producing power of the unit 1 2 3 j for i = f1, 2, 3, 4, 5, 6g. Suppose that the value p of electricity can be taken as p = 378.4 2 u . The cost j=1 of the manufacture j unit follows: c (u ) := maxfc (u ), c (u )g, j j j j j j (b +1) a b j j b b j j where c (u ) := u + b u + g and c (u ) := a u + g (u ) . The values are provided in a , b , j j j j j j j j j j j j j 2 j b +1 g , a , b and g in Table 1. Proﬁt of the ﬁrm i is j j j j f (u) := p u c (u ) = 378.4 2 u u c (u ), i å j å j j å l å j å j j j2 I j2 I l=1 j2 I j2 I i i i i T 6 min max min max where u = (u , , u ) with reference to set u 2 C := fu 2 R : u u u g, with u and u 1 6 j j j j j give in Table 2. Deﬁne the equilibrium bifunction f in the following way: f (u, v) := f (u, u) f (u, v) , å i i i=1 where f (u, v) := 378.4 2 u + v v c (v ). i å j å j å j å j j j2 I j2 I j2 I j62 I i i i i This model of electricity markets can be viewed as an equilibrium problem Find u 2 C such that f (u , v) 0, for all v 2 C. Numerical conclusions have shown in Figures 1–4 and Table 3. For these numerical experiments we take u = v = u = v = (48, 48, 30, 27, 18, 24) and l = 0.01, s = 0.026, m = 0.024, } = 0.20, l = 0.1. 0 0 n 0 1 1 Axioms 2020, 9, 127 13 of 21 Table 1. Parameters for cost bi-function. Unit j a b g a b g j j j j j j 1 0.04 2 0 2 1 25 2 0.035 1.75 0 1.75 1 28.5714 3 0.125 1 0 1 1 8 4 0.0116 3.25 0 3.25 1 86.2069 5 0.05 3 0 3 1 20 6 0.05 3 0 3 1 20 Table 2. Values used for constraint set. min max j u u j j 1 0 80 2 0 80 3 0 50 4 0 55 5 0 30 6 0 40 1 1 10 10 0 0 10 10 -1 -1 10 10 -2 -2 10 10 0 20 40 60 80 100 120 140 0 1 2 3 4 5 6 7 8 Number of iterartions Elapsed time [sec] Figure 1. Example 1 while tolerance is 0.01. 1 1 10 10 0 0 10 10 -1 -1 10 10 -2 -2 10 10 -3 -3 10 10 0 500 1000 1500 2000 2500 3000 0 20 40 60 80 100 120 140 160 180 200 Number of iterartions Elapsed time [sec] Figure 2. Example 1 while tolerance is 0.001. Axioms 2020, 9, 127 14 of 21 1 1 10 10 0 0 10 10 -1 -1 10 10 -2 -2 10 10 -3 -3 10 10 -4 -4 10 10 0 2000 4000 6000 8000 10000 12000 0 100 200 300 400 500 600 700 800 900 Number of iterartions Elapsed time [sec] Figure 3. Example 1 while tolerance is 0.0001. 1 1 10 10 0 0 10 10 -1 -1 10 10 -2 -2 10 10 -3 -3 10 10 -4 -4 10 10 -5 -5 10 10 0 0.5 1 1.5 2 2.5 0 500 1000 1500 Number of iterartions Elapsed time [sec] Figure 4. Example 1 while tolerance is 0.00001. Table 3. Figures 1–4 numerical values. L.EgA Algo.1 TOL Iter. time (s) Iter. time (s) 0.01 125 7.3692 61 3.4055 u = (47.3245, 47.3245, 47.3245, 47.3245, 47.3245, 47.3245) L.EgA u = (47.3245, 47.3245, 47.3245, 47.3245, 47.3245, 47.3245) Algo.1 0.001 2761 193.3939 2063 150.6757 u = (47.3245, 47.3245, 47.3245, 47.3245, 47.3245, 47.3245) L.EgA u = (47.3245, 47.3245, 47.3245, 47.3245, 47.3245, 47.3245) Algo.1 0.0001 11,526 818.7184 4687 324.3571 u = (47.3245, 47.3245, 47.3245, 47.3245, 47.3245, 47.3245) L.EgA u = (47.3245, 47.3245, 47.3245, 47.3245, 47.3245, 47.3245) Algo.1 0.00001 20,946 1449.3959 7307 526.9766 u = (47.3245, 47.3245, 47.3245, 47.3245, 47.3245, 47.3245) L.EgA u = (47.3245, 47.3245, 47.3245, 47.3245, 47.3245, 47.3245) Algo.1 Example 2. Assume that the following cost bifunction f deﬁned by f (u, v) = ( A A + B + C)u, v u , on the convex set C = fu 2 R : Du dg while D is an 100 n matrix and d is a non-negative vector. In the above bifunction deﬁnition we take A is an n n matrix, B is an n n skew-symmetric matrix, C is an n n diagonal matrix having diagonal entries are non-negative. The matrices are generated as; A = rand(n), K = rand(n), B = 0.5K 0.5K and C = diag(rand(n,1)). The bifunction f is monotone and having Lipschitz-type Axioms 2020, 9, 127 15 of 21 1 T constants are c = c = k A A + B + Ck. Numerical results are presented in the Figures 5–8 and Table 4. 1 2 T 1 1 For these numerical experiments we take u = v = u = v = (1, 1, , 1) and l = , s = , 1 1 0 0 10c 8c 1 1 1 1 m = , } = , l = 1/4c . n 0 1 8.2c 5 1 1 10 10 -1 -1 -2 -2 -3 -3 -4 10 10 0 5 10 15 20 25 30 35 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 Number of iterartions Number of iterartions Figure 5. Example 2 for average number of iterations while n = 5. 1 1 10 10 -1 -1 -2 -2 -3 -3 -4 10 10 0 10 20 30 40 50 60 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 Number of iterartions Number of iterartions Figure 6. Example 2 for average number of iterations while n = 10. 1 1 10 10 0 0 10 10 -1 -1 10 10 -2 -2 10 10 -3 -3 10 10 -4 -4 10 10 0 10 20 30 40 50 60 70 80 90 1 2 3 4 5 6 7 Number of iterartions Number of iterartions Figure 7. Example 2 for average number of iterations while n = 20. Axioms 2020, 9, 127 16 of 21 1 1 10 10 -1 -1 -2 10 10 -3 -2 -4 -3 -5 10 10 0 5 10 15 20 25 30 1 2 3 4 5 6 7 8 Number of iterartions Number of iterartions Figure 8. Example 2 for average number of iterations while n = 40. Table 4. Numerical results for Figures 5–8. L.EgA Algo.1 n T. Samples Avg Iter. Avg time(s) Avg Iter. Avg time(s) 5 10 35 0.8066 6 0.1438 10 10 51 1.1779 6 0.1302 20 10 84 1.7441 7 0.1801 40 10 30 0.6859 8 0.1999 2 2 Example 3. Assume that F : R ! R is deﬁned by 0.5u u 2u 10 1 2 2 F(u) = 2 7 4u 0.1u 10 2 2 2 with C = fu 2 R : (u 2) + (u 2) 1g. It is not hard to check that F is Lipschitz continuous 1 2 with L = 5 and pseudomonotone. The step size l = 10 for Lyashko et al. [25] and l = 0.1, s = 0.129, } = 0.20 and m = 0.119. Computational results are shown in the Table 5 and in Figures 9–12. -2 -4 -6 -8 -10 0 2 4 6 8 10 12 14 16 18 20 Number of iterartions Figure 9. Example 3 while u = (1.5, 1.7). 0 Axioms 2020, 9, 127 17 of 21 -2 -4 -6 -8 -10 0 5 10 15 20 25 Number of iterartions Figure 10. Example 3 while u = (2, 3). -2 -4 -6 -8 -10 0 5 10 15 20 25 Number of iterartions Figure 11. Example 3 while u = (1, 2). -2 -4 -6 -8 -10 0 2 4 6 8 10 12 14 16 18 20 Number of iterartions Figure 12. Example 3 while u = (2.7, 2.6). 0 Axioms 2020, 9, 127 18 of 21 Table 5. Numerical results for Figures 9–12. L.EgA Algo.1 u Iter. time(s) Iter. time(s) (1.5, 1.7) 20 0.7506 8 0.5316 (2.0, 3.0) 21 0.7879 8 0.6484 (1.0, 2.0) 23 1.1450 14 0.9730 (2.7, 2.6) 19 0.7254 7 0.5835 2 2 Example 4. Let F : R ! R is deﬁned by 2 2 (u + (u 1) )(1 + u ) 2 2 F(u) = 3 2 u u (u 1) 1 2 2 2 2 and C = fu 2 R : (u 2) + (u 2) 1g. Here, F is not monotone but pseudomonotone on C and 1 2 L-Lipschitz continuous through L = 5 (see, e.g., [33]). The stepsize l = 10 for Lyashko et al. [25] and l = 0.01, s = 0.129, } = 0.15 and m = 0.119. The computational experimental ﬁndings are written in 0 n Table 6 and in Figures 13–15. -1 -2 -3 -4 -5 0 10 20 30 40 50 60 70 Number of iterartions Figure 13. Example 4 while u = (10, 10). -1 -2 -3 -4 -5 0 10 20 30 40 50 60 70 80 90 100 Number of iterartions Figure 14. Example 4 while u = (10,10). 0 Axioms 2020, 9, 127 19 of 21 -1 -2 -3 -4 -5 0 10 20 30 40 50 60 Number of iterartions Figure 15. Example 4 while u = (10, 20). Table 6. Figures 13–15 numerical values. L.EgA Algo.1 u Iter. time(s) Iter. time(s) (10, 10) 67 1.9151 31 1.0752 (10,10) 92 2.5721 71 2.0469 (10, 20) 60 1.7689 41 1.1864 6. Conclusions We have established an extragradient-like method to solve pseudomonotone equilibrium problems in real Hilbert space. The main advantage of the proposed method is that an iterative sequence has been incorporated with a certain step size evaluation formula. The step size formula is updated for each iteration based on the previous iterations. Numerical ﬁndings were presented to show our algorithm’s numerical efﬁciency compared with other methods. Such numerical investigations indicate that inertial effects often generally improve the effectiveness of the iterative sequence in this context. Author Contributions: Conceptualization, W.K. and K.M.; methodology, W.K. and K.M.; writing–original draft preparation, W.K. and K.M.; writing–review and editing, W.K. and K.M.; software, W.K. and K.M.; supervision, W.K.; project administration and funding acquisition, K.M. All authors have read and agreed to the published version of the manuscript. Funding: This project was supported by Rajamangala University of Technology Phra Nakhon (RMUTP). Acknowledgments: The ﬁrst author would like to thanks the Rajamangala University of Technology Thanyaburi (RMUTT) (Grant No. NSF62D0604). The second author would like to thanks the Rajamangala University of Technology Phra Nakhon (RMUTP). Conﬂicts of Interest: The authors declare that they have no conﬂict of interest. References 1. Blum, E. From optimization and variational inequalities to equilibrium problems. Math. Stud. 1994, 63, 123–145. 2. Facchinei, F.; Pang, J.S. Finite-Dimensional Variational Inequalities and Complementarity Problems; Springer Science & Business Media: Berlin/Heidelberg, Germany, 2007. Axioms 2020, 9, 127 20 of 21 3. Konnov, I. Equilibrium Models and Variational Inequalities; Elsevier: Amsterdam, The Netherlands, 2007; Volume 210. 4. Muu, L.D.; Oettli, W. Convergence of an adaptive penalty scheme for ﬁnding constrained equilibria. Nonlinear Anal. Theory Methods Appl. 1992, 18, 1159–1166. 5. Combettes, P.L.; Hirstoaga, S.A. 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This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).
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Published: Oct 31, 2020
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