Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2013, Article ID 487545, 6 pages http://dx.doi.org/10.1155/2013/487545 Research Article Multivalued Variational Inequalities with π·π½-Pseudomonotone Mappings in Reflexive Banach Spaces A. M. Saddeek1 and S. A. Ahmed1,2 1 2 Department of Mathematics, Faculty of Science, Assiut University, Assiut, Egypt Department of Mathematics, University College, Umm Al-Qura University, Saudi Arabia Correspondence should be addressed to A. M. Saddeek; a m saddeek@yahoo.com Received 28 August 2012; Accepted 31 January 2013 Academic Editor: Feyzi BasΜ§ar Copyright © 2013 A. M. Saddeek and S. A. Ahmed. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. This paper is concerned with the study of a class of variational inequalities with multivalued Dπ½ -pseudomonotone mappings in reflexive Banach spaces by using the Dπ½ -antiresolvent technique. An application to the multivalued nonlinear Dπ½ -complementarity problem is also presented. The results coincide with the corresponding results announced by many others for the gradient state. 1. Introduction and Preliminaries Variational inequalities give a convenient mathematical framework for discussing a large variety of interesting problems appearing in pure and applied sciences. It is well known that the theory of pseudomonotone mappings plays an important part in the study of the above-mentioned variational inequalities. In recent years, pseudomonotone theory has become an attractive field for many mathematicians (see [1–8]). In a very recent paper [9], by using the π·π½ -antiresolvent technique (where π½ is the duality mapping) devised by the first author, the author introduced a new concept of monotonicity, which is called the π·π½ -pseudomonotone type. In the present paper, the concept of multivalued π·π½ pseudomonotone mappings in reflexive Banach spaces is used to study a wide class of variational inequalities, called the multivalued π·π½ -pseudomonotone variational inequalities. Moreover, the results obtained in this paper can be applied to the multivalued nonlinear π·π½ -complementarity problem. This problem contains, in particular, a mathematical model arising in the study of the postcritical equilibrium state of a thin plate resting, without friction on a flat rigid support (see [10–12]). The results coincide with the corresponding results (see [2, 13–15]) in the case of gradient mappings. Unless otherwise stated, π stands for a real reflexive Banach space with norm β⋅βπ and πβ stands for the uniformly convex dual of π with the dual norm β ⋅ βπβ . The duality pairing between π and πβ is denoted by β¨⋅, ⋅β©. The set of all nonnegative integers is denoted by N. The field of real (resp., positive real) numbers is denoted by R (resp., R+ ). Notation “ → ” stands for strong convergence and “β” for weak convergence. A mapping π½ : π → πβ is said to be a duality mapping (see, e.g., [16]) with gauge function Φ (i.e., Φ is continuous strictly increasing real-valued function satisfying Φ(0) = 0 and limπ‘ → +∞ Φ(π‘) = +∞) if for every π’ ∈ π, β¨π½π’, π’β© = βπ½π’βπβ βπ’βπ = Φ(βπ’βπ)(βπ’βπ). If π = π» is a Hilbert space, then π½ ≡ πΌ, the identity mapping. Assume that πβ has a weakly sequentially continuous duality mapping π½ (i.e., if {π’π }π∈N is a sequence in π which weakly convergent to a point π’, then the sequence {π½(π’π )}π∈N converges to π½(π’) (see, e.g., [17])). Let π : π → R ∪ {+∞} be a function. The domain of π is dom π = {π’ ∈ π : π(π’) < +∞}. When dom π =ΜΈ π, π is called proper (see, e.g., [18]). The interior of the domain of π is denoted by int dom π. The function π is said to be GΜateaux differentiable at π’ ∈ int dom π (see, e.g., [18]), if πσΈ (π’, π) = lim π‘→0 exists for all π ∈ π. π (π’ + π‘π) − π (π’) π‘ (1) 2 Abstract and Applied Analysis Let π be proper, convex, lower semicontinuous, and GΜateaux differentiable at π’ ∈ int dom π; then the gradient of π at π’ is the function ∇π(π’) which is defined by β¨∇π(π’), πβ© = πσΈ (π’, π) for any π ∈ π. It is known (see, e.g., [19]) that the conjugate πβ : πβ → R ∪ {+∞} is also proper, convex, and lower semicontinuous. The convex function π is said to be of Legendre type (see, e.g., [20]) if the following conditions hold: (πΏ 1 ) int dom(π) =ΜΈ π, π is GΜateaux differentiable on int dom(π) and dom ∇π = int dom π; β β (πΏ 2 ) int dom(π ) =ΜΈ π, π is GΜateaux differentiable on int dom(πβ ) and dom ∇πβ = int dom(πβ ). It is well known (see, e.g., [21]) that if π is a proper, convex, lower semicontinuous, and Legendre type, then ∇πβ = (∇π)−1 and range ∇πβ = dom ∇π. Throughout this paper, the function π : π → R+ ∪ {+∞} is proper, convex, and lower semicontinuous which is also Legendre on int dom(π). The Bregman distance (see, e.g., [22]) is the function π·π : π × int dom(π) → R ∪ {+∞} defined by π·π (], π’) = π (]) − π (π’) − β¨∇π (π’) , ] − π’β© , (2) dom π·π = (dom π) × int dom π. (3) with It should be pointed out that if π = π» is a Hilbert space and π = (1/2)β ⋅ β2π», then ∇π = πΌ (the identity mapping) and π·π½ (], π’) = (1/2)β] − π’β2π». β For a multivalued mapping π : π → 2π , the associated π·π½ -antiresolvent (where π½ : π → πβ is the duality mapping) of π½ − π (see [9]) is the mapping ππ½ : π → 2π, defined by ππ½ = ∇πβ (π½ − π) . (4) Such a mapping is known as (see [23]) a π·π -antiresolvent mapping of π when π½ = ∇π (in this case, the mapping ππ½ is denoted by ππ ). In light of the above-mentioned discussion, we note that if π½ − π = ∇π, then ππ½ is the identity mapping πΌ. β Following [9], the mapping π½ − π : π → 2π is said to be π·π½ -pseudomonotone, if for every π’, π ∈ dom π½ ∩ dom π, ] ∈ (π½ − π)(π’), π ∈ (π½ − π)(π), and every sequence {π’π }π∈N ⊂ dom π½ ∩ dom π and ]π ∈ (π½ − π)(π’π ) the conditions ∇πβ (]π ) β ∇πβ (]) , lim sup β¨]π , ∇πβ (]π ) − ∇πβ (])β© ≤ 0 (5) π→∞ π·π -pseudomonotonicity in the sense of Bregman distance π·π ) of the mapping ∇π, if π½ − π = ∇π. The multivalued variational inequality defined by the π·π½ mapping (or multivalued π·π½ -variational inequality) π½ − π : β π → 2π and the set πΎ ⊂ π is to find π’ ∈ πΎ such that ∃] ∈ (π½ − π) (π’) , π ∈ (π½ − π) (π) : β¨] − π, ∇πβ (π) − ∇πβ (])β© ≥ 0, (7) ∀π ∈ πΎ, β where π ∈ 2π . The multivalued nonlinear complementarity problem defined by the π·π½ -mapping (or multivalued nonlinear π·π½ β complementarity problem) π½ − π : π → 2π and the set πΎ is to find π’ ∈ πΎ such that β¨] − π, ∇πβ (])β© = 0, β¨] − π, ∇πβ (π) − ∇πβ (])β© ≥ 0, ∀π ∈ πΎ, (8) β where π ∈ 2π , ] ∈ (π½ − π)(π’), and π ∈ (π½ − π)(π). The multivalued π·π½ -variational inequality and multivalued nonlinear π·π½ -complementarity problem are very general in the sense that they include, as special cases, multivalued variational inequality and multivalued nonlinear complementarity problem. The following definition and results will be used in the sequel. Definition 1 (see, e.g., [15, p. 84]). The mapping π΄ : π → β 2π is continuous on finite dimensional subspaces if for any finite dimensional subspace π0 ⊂ π, the restriction of π΄ to π0 ∩ dom(π΄) is weakly continuous. Corollary 2 (see [24]). Let π : π0 ⊂ π → π be the injection mapping. Let πβ : πβ → π0β be its dual mapping. Then, πβ π΄π : β dom(π΄) ∩ ππ0 → 2π0 is continuous. Corollary 3 (see [25]). Let πΎ be a nonempty compact convex set of Rπ and let π : πΎ → 2πΎ be continuous. Then π admits a fixed point. 2. Main Results Theorem 4. Let πΎ be a closed convex set in π» and let π : πΎ → 2π» be a multivalued mapping. Then the following are equivalent: (1) ∇πβ (]σΈ ) ∈ πππΎ (π₯) = arg min {π·πΌ (∇πβ (πσΈ ) , π₯) imply that lim inf β¨]π , ∇πβ (]π ) − ∇πβ (π)β© ≥ β¨π, ∇πβ (]) − ∇πβ (π)β© . π→∞ (6) As remarked in [9], the π·π½ -pseudomonotonicity of the mapping π½−π coincides with the pseudomonotonicity (or the 1σ΅© σ΅©2 = σ΅©σ΅©σ΅©σ΅©∇πβ (πσΈ ) − π₯σ΅©σ΅©σ΅©σ΅©π» : ∇πβ (πσΈ ) 2 ∈ πΎ, πσΈ ∈ (πΌ − π) (π) , π ∈ πΎ} , (9) Abstract and Applied Analysis 3 the multivalued projection for πΎ; Then there exists a solution to the multivalued π·π½ -variational inequality (7). (2) ∇πβ (]σΈ ) ∈ πΎ : β¨∇πβ (]σΈ ) − π₯, ∇πβ (πσΈ ) −∇πβ (]σΈ )β© ≥ 0 ∀∇πβ (πσΈ ) ∈ πΎ, (10) where πσΈ ∈ (πΌ − π) (π) , ]σΈ ∈ (πΌ − π) (π¦) , π¦, π ∈ πΎ. Proof. Assume that (1) holds. Let π₯ ∈ 2π» and ∇πβ (]σΈ ) ∈ πππΎ (π₯) ⊂ πΎ. For every ∇πβ (πσΈ ) ∈ πΎ and π‘ ∈ (0, 1], we have π·πΌ (π₯, ∇πβ (]σΈ )) β σΈ β β β¨π π − ππ π, ∇πβ (π) − ∇πβ (]π )β© ≥ 0. σΈ ≤ π·πΌ (π₯, (1 − π‘) ∇π (] ) + π‘∇π (π )) = π·πΌ (π₯, ∇πβ (]σΈ )) (11) − π‘ β¨π₯ − ∇πβ (]σΈ ) , ∇πβ (πσΈ ) − ∇πβ (]σΈ )β© 2 β σΈ β + π‘ π·πΌ (∇π (π ) , ∇π (] )) . β π − π π , ∇πβ (π) − ∇πβ (]π )β© . ≥ β¨∇πβ (]π ) + ππ This implies β¨π₯ − ∇πβ (]σΈ ) , ∇πβ (πσΈ ) − ∇πβ (]σΈ )β© (12) Hence, π‘ → 0+ implies (2). On the other hand, assume that (2) holds. For every ∇πβ (πσΈ ) ∈ πΎ, we have π·πΌ (∇πβ (πσΈ ) , π₯) β σΈ β σΈ = π·πΌ (∇π (π ) , ∇π (] )) + β¨∇πβ (πσΈ ) − π₯, ∇πβ (πσΈ ) − ∇πβ (]σΈ )β© (13) + π·πΌ (∇πβ (]σΈ ) , π₯) ≥ π·πΌ (∇πβ (]σΈ ) , π₯) . This implies (1). Remark 5. In the particular situation when πΌ − π = ∇π Theorem 4 coincides (in gradient setting) with Theorem 2.3 in [15] and also with Proposition 2.1 (1) and (2) in [2]. Theorem 6. In addition to conditions on π, πβ , πβ , and π½, one assumes that π is separable, πΎ ⊂ π is nonempty closed and β convex, π : πΎ → 2π , ∇πβ are weakly continuous mappings, and either π or ∇πβ is continuous. Moreover, assume that the β mapping π½ − π : πΎ → 2π is a bounded π·π½ -pseudomonotone mapping and that, for each π ∈ πΎ, there exist π0 ∈ πΎ, π0 ∈ (π½ − π)(π0 ), π ∈ (π½ − π)(π), and π > 0 such that β¨π, ∇πβ (π) − ∇πβ (π0 )β© > β¨π, ∇πβ (π) − ∇πβ (π0 )β© , β π ∈ 2π , σ΅© σ΅©σ΅© β σ΅©σ΅©∇π (π)σ΅©σ΅©σ΅©π ≥ π. (15) The equivalent form of problem (15) is to find π’π ∈ πΎπ such β (π½ − π)ππ (π’π ), ]π ∈ that for each π ∈ πΎπ , there exist π π ∈ ππ (π½ − π)(π’π ), and π ∈ (π½ − π)(π) such that β¨∇πβ (]π ) , ∇πβ (π) − ∇πβ (]π )β© σΈ ≤ π‘π·πΌ (∇πβ (πσΈ ) , ∇πβ (]σΈ )) . Proof. Suppose that {π₯1 , π₯2 , π₯3 , . . .} is an infinite dense set in πΎ and ππ , π ∈ N, is the linear span of {π₯1 , π₯2 , π₯3 , . . . , π₯π }. Let πΎπ = con{π₯1 , π₯2 , π₯3 , . . . , π₯π } = {∑π π=1 π π π₯π , π π ≥ 0, ∑π π=1 π π = 1}. Let ππ : ππ → π be the injection mapping β : πβ → ππβ be its restriction dual. Observe that and let ππ ∪πΎπ is dense in πΎ, for π ∈ N. Now, fix an integer π ≥ 1 and consider the finite dimensional problem. Find π’π ∈ πΎπ such that for each π ∈ πΎπ , there exist β (π½ − π)ππ (π’π ), ]π ∈ (π½ − π)(π’π ), and π ∈ (π½ − π)(π) π π ∈ ππ such that (16) Using the identification of ππ with Rπ and ππβ and Theorem 4 (with Rπ = π» and π½ = πΌ), we see that (16) is equivalent β π − π π ). to ∇πβ (]π ) ∈ πππΎπ (∇πβ (]π ) + ππ Let π΅π (0) be any closed ball containing πΎπ . It is well known (see, e.g., [26, p. 54, 224]) that π΅π (0) is compact and convex in πΎπ ; thus it is weakly closed. From Corollary 2, πβ (π½ − π)π : π΅π (0) → 2π΅π (0) is continuous; hence the function ∇πβ (]) σ³¨→ πππ΅π (0) (∇πβ (]) + β ππ π − π π ) is continuous from π΅π (0) into 2π΅π (0) . Hence, by Corollary 3, this equation admits a solution. If the closed convex set πΎ is assumed to be bounded, then by the reflexivity of π it is weakly compact (by employing the Banach -Alaoglu theorem (see, e.g., [16, p. 3]). Then we have a subsequence denoted by {π’π }π∈N such that π’π β π’ ∈ πΎ. Since π½ − π is bounded, we have β]π βπβ ≤ π for all π ∈ N. Since π½ − π is weakly continuous and since either π½ − π or ∇πβ is continuous by hypothesis, it follows that ππ½ is weakly continuous by [27, Lemma 1]. So, we have ∇πβ (]π ) β ∇πβ (]). Now, we prove that lim sup β¨]π , ∇πβ (]π ) − ∇πβ (])β© ≤ 0. π→∞ (17) Μ ∈ πΎπ such that For any π > 0, choose π so large and π’ σ΅© σ΅©σ΅© β β π’) . σ΅©σ΅©∇π (]) − ∇π (Μ])σ΅©σ΅©σ΅©π < π for Μ] ∈ (π½ − π) (Μ (18) Therefore, we have (14) β¨]π − π, ∇πβ (]π ) − ∇πβ (Μ])β© ≤ 0 for π ≥ π. (19) 4 Abstract and Applied Analysis Since πΎπ ⊂ πΎπ , we have β Theorem 8. Let all assumptions of Theorem 6 hold, except for condition (14) let it be replaced by the π·π½ -coercive condition: for π ∈ (π½ − π)(π), β lim sup β¨]π , ∇π (]π ) − ∇π (])β© π→∞ = lim sup [β¨]π , ∇πβ (]π ) − ∇πβ (Μ])β© π→∞ lim β (20) β∇π (π)βπ → ∞ + β¨]π , ∇πβ (Μ]) − ∇πβ (])β©] Since π is arbitrary, this shows the desired inequality. By the π·π½ -pseudomonotonicity of π½ − π, it follows that lim inf β¨]π , ∇πβ (]π ) − ∇πβ (π)β© ≥ β¨], ∇πβ (]) − πβ (π)β© π→∞ (21) for all π ∈ dom π½ ∩ dom π and π ∈ (π½ − π)(π). If π ∈ πΎπ , π ≥ π, we have Hence (23) for every π in πΎπ , π ∈ N, π ∈ (π½ − π)(π). Since ∪π πΎπ is dense in πΎπ , so we have that π’ is a solution to (7). Now, to complete the proof, we consider the case when πΎ is unbounded. In this case we consider the set πΎπ = {π ∈ πΎ : β∇πβ (π)βπ ≤ π}, where π = max{β∇πβ (π0 )βπ, π}. Since πΎπ is bounded, there exists at least one π’π ∈ πΎπ : β β¨]π − π, ∇π (π) − ∇π (]π )β© ≥ 0 π, π0 ∈ πΎ. Suppose further that πΎ has the following property (W): πΌ∇πβ (π) ∈ πΎ for all ∇πβ (π) ∈ πΎ and πΌ ≥ 0. β Then for every π ∈ 2π there exist π’ ∈ πΎ, ] ∈ (π½ − π)(π’) such that β¨] − π, ∇πβ (])β© = 0, β¨] − π, ∇πβ (π) − ∇πβ (])β© ≥ 0 (28) β Proof. Let π ∈ 2π satisfy βπβπβ < π and σ΅© σ΅© 2π − σ΅©σ΅©σ΅©πσ΅©σ΅©σ΅©πβ σ΅©σ΅© β σ΅©σ΅© ∇π (π ) < π. σ΅©π σ΅©σ΅© σ΅©σ΅© 0 σ΅© σ΅©σ΅© σ΅©σ΅©πσ΅©σ΅©πβ (29) The π·π½ -coercivity of π½ − π implies that there exists π > 0 such that σ΅© σ΅© β¨π, ∇πβ (π) − ∇πβ (π0 )β© ≥ 2πσ΅©σ΅©σ΅©∇πβ (π)σ΅©σ΅©σ΅©π (30) for π ∈ (π½ − π)(π), π ∈ πΎ with β∇πβ (π)βπ ≥ π. So we conclude β¨π − π, ∇πβ (π) − ∇πβ (π0 )β© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© ≥ 2πσ΅©σ΅©σ΅©∇πβ (π)σ΅©σ΅©σ΅©π − σ΅©σ΅©σ΅©πσ΅©σ΅©σ΅©πβ σ΅©σ΅©σ΅©∇πβ (π)σ΅©σ΅©σ΅©π − σ΅©σ΅©σ΅©πσ΅©σ΅©σ΅©πβ σ΅©σ΅©σ΅©∇πβ (π0 )σ΅©σ΅©σ΅©π σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© ≥ (2π − σ΅©σ΅©σ΅©πσ΅©σ΅©σ΅©πβ ) π − σ΅©σ΅©σ΅©πσ΅©σ΅©σ΅©πβ σ΅©σ΅©σ΅©∇πβ (π0 )σ΅©σ΅©σ΅©π (24) for ]π ∈ (π½ − π)(π’π ) and π ∈ πΎπ . Since π0 ∈ πΎπ , we have β (27) for all π ∈ πΎ, π ∈ (π½ − π)(π). β¨]π , ∇πβ (]π ) − ∇πβ (π)β© ≤ β¨π, ∇πβ (]π ) − ∇πβ (π)β© . (22) β β¨π, ∇πβ (π) − ∇πβ (π0 )β© ] = +∞, σ΅© σ΅©σ΅© β σ΅©σ΅©∇π (π)σ΅©σ΅©σ΅©π π0 ∈ (π½ − π) (π0 ) , σ΅© σ΅© ≤ (σ΅©σ΅©σ΅©πσ΅©σ΅©σ΅©πβ + π) π. β¨], ∇πβ (]) − ∇πβ (π)β© ≤ β¨π, ∇πβ (]) − ∇πβ (π)β© [ σ΅© σ΅© > 0 for π ∈ πΎ, π ∈ (π½ − π) (π) π€ππ‘β σ΅©σ΅©σ΅©∇πβ (π)σ΅©σ΅©σ΅©π ≥ π. (31) β β¨]π − π, ∇π (]π ) − ∇π (π0 )β© ≤ 0 for π0 ∈ (π½ − π) (π0 ) . (25) This, together with (14), implies that β∇πβ (]π )βπ < π. To clarify that π’π is also a solution to original problem on πΎ, for any π ∈ πΎ, set ∇πβ (ππ‘ ) = (1 − π‘)∇πβ (]π ) + π‘∇πβ (π) for π‘ > 0 is sufficiently small, where ππ‘ ∈ (π½ − π)(ππ‘ ) and ππ‘ ∈ πΎπ . Consequently π’π ∈ πΎπ ⊂ πΎ : 0 ≤ β¨]π − π, ∇πβ (ππ‘ ) − ∇πβ (]π )β© = π‘ β¨]π − π, ∇πβ (π) − ∇πβ (]π )β© (26) for π ∈ πΎ. The second part of (28) thus follows from Theorem 6. To prove the first part of (28), observe that we can choose a point π in πΎ and π ∈ (π½−π)(π) and assume that ∇πβ (π) = 0. Therefore, from Theorem 6, we have β¨] − π, ∇πβ (])β© ≤ 0 for all ] ∈ (π½ − π)(π’), π’ ∈ πΎ. On the other hand, setting ∇πβ (π) = πΌ∇πβ (]), where πΌ > 1, we get 0 ≤ β¨] − π, ∇πβ (π) − ∇πβ (])β© = (πΌ − 1) β¨] − π, ∇πβ (])β© . (33) This implies This completes the proof. Remark 7. In the particular situation when π½ − π = ∇π, Theorem 6 coincides with the Brezis Theorem (see, e.g., [13, 14]) for the case of gradient mapping. We are now in a position to state and prove the following theorem. (32) β¨] − π, ∇πβ (])β© ≥ 0. (34) β¨] − π, ∇πβ (])β© = 0. (35) So, This completes the proof of (28). Abstract and Applied Analysis 5 The following proposition gives a characterization of the sum of two π·π½ -Pseudomonotone mappings. 3. Application to Multivalued Nonlinear π·π½ Complementarity Problem Proposition 9. Let π, πβ , and π½ be as above and let ππ : π → β 2π , π = 1, 2, and ∇πβ be weakly continuous mappings. If π½ − β ππ : π → 2π , π = 1, 2, are π·π½ -pseudomonotone mappings such that dom π½ ∩ dom ππ =ΜΈ π, π = 1, 2, then ∑2π=1 (π½ − ππ ) is π·π½ -pseudomonotone. As applications of Theorem 8 we consider the multivalued nonlinear π·π½ -complementarity problem (8) with π = ππ1 + π2 − ππ½, where ππ , π = 1, 2 are two nonlinear multivalued β mappings from πΎ to 2π , and π ∈ (0, ∞). Μ π ∈ ∑2π=1 (π½ − ππ )(ππ ), π¦ Μ ∈ ∑2π=1 (π½ − ππ )(π), ππ , π ∈ Proof. Let π¦ π¦π ) β ∇πβ (Μ π¦) and dom π½ ∩ dom ππ , π = 1, 2, with ∇πβ (Μ π¦π , ∇πβ (Μ π¦π ) − ∇πβ (Μ π¦)β© ≤ 0. lim sup β¨Μ π→∞ (36) Now, we prove for π¦π(π) ∈ (π½ − ππ )(ππ ), π¦(π) ∈ (π½ − ππ )(π), π = 1, 2, that lim sup β¨π¦π(1) , ∇πβ (π¦π(1) ) − ∇πβ (π¦(1) )β© ≤ 0, π→∞ lim sup β¨π¦π(2) , ∇πβ (π¦π(2) ) − ∇πβ (π¦(2) )β© ≤ 0. Theorem 10. Let π, πβ , πβ , π½, and πΎ be the same as in Theorem 6, and suppose that πΎ has the property (W). Let the mappings ∇πβ and ππ , π = 1, 2 be weakly continuous and let either ∇πβ or both ππ , π = 1, 2 be continuous. Let π½ − ππ : πΎ → β 2π , π = 1, 2 be two bounded π·π½ -pseudomonotone mappings. Let 1 = π1 (37) inf π∈πΎ β∇πβ (ππ€1 +π€2 )βπ =ΜΈ 0 ( β¨π€1 , ∇πβ (ππ€1 + π€2 ) −∇πβ (ππ€0(1) + π€0(2) )β© π→∞ If lim sup β¨π¦π(2) , ∇πβ (π¦π(2) ) − ∇πβ (π¦(2) )β© = π > 0, π→∞ (38) (note that otherwise, by symmetry), then there exists a subsequence {π¦π(2) } ⊂ {π¦π(2) }π∈N such that π π∈N lim sup β¨π¦π(2) , ∇πβ (π¦π(2) ) − ∇πβ (π¦(2) )β© = π. π π π→∞ σ΅©2 −1 σ΅© ×(σ΅©σ΅©σ΅©∇πβ (ππ€1 + π€2 )σ΅©σ΅©σ΅©π) ) , π= lim inf ( β¨π€2 , ∇πβ (ππ€1 + π€2 ) π∈πΎ β∇πβ (ππ€1 +π€2 )βπ → ∞ −∇πβ (ππ€0(1) + π€0(2) )β© (39) σ΅©2 −1 σ΅© ×(σ΅©σ΅©σ΅©∇πβ (ππ€1 + π€2 )σ΅©σ΅©σ΅©π) ) , This implies that lim sup β¨π¦π(1) , ∇πβ π π→∞ (π¦π(1) ) π β (1) − ∇π (π¦ )β© β π = lim inf π‘→∞ β = lim sup [ β¨Μ π¦ππ , ∇π (Μ π¦ππ ) − ∇π (Μ π¦)β© π→∞ − β¨π¦π(2) , ∇πβ π (π¦π(2) ) π β (40) (2) − ∇π (π¦ )β©] ≤ 0 − π. From the π·π½ -pseudomonotonicity of π½−π1 , we get for all π¦σΈ ∈ (π½ − π1 )(πσΈ ), πσΈ ∈ dom π½ ∩ dom π1 (1) β (1) β σΈ β¨π¦ , ∇π (π¦ ) − ∇π (π¦ )β© , ∇πβ (π¦π(1) ) − ∇πβ (π¦σΈ )β© . ≤ lim inf β¨π¦π(1) π π (41) π→∞ Letting π¦ = π¦(1) , we obtain 0 ≤ lim inf π→∞ (π¦π(1) ) π β (1) − ∇π (π¦ )β© ≤ 0 − π, Proof. By Proposition 9, π½ − π = π(π½ − π1 ) + (π½ − π2 ) is π·π½ pseudomonotone for every π ≥ 0. Set π > π(π − π). Then lim inf σ΅©2 −1 σ΅© ×(σ΅©σ΅©σ΅©∇πβ (ππ€1 + π€2 )σ΅©σ΅©σ΅©π) ) a contradiction. Hence, π→∞ lim sup β¨π¦π(2) , ∇πβ (π¦π(2) ) − ∇πβ (π¦(2) )β© ≤ 0. π π ≥ (43) π→∞ This holds for any subsequence, so (37) holds and the proof follows immediately by the superadditivity of the lim inf. ( β¨ππ€1 + π€2 , ∇πβ (ππ€1 + π€2 ) −∇πβ (ππ€0(1) + π€0(2) )β© (42) , ∇πβ (π¦π(1) ) − ∇πβ (π¦(1) )β© ≤ 0, lim sup β¨π¦π(1) π π Φ (π‘) π‘ Be such that π < π, where π€π ∈ (π½ − ππ )(π), π€0(π) ∈ (π½ − ππ )(π0 ), π = 1, 2, π, π0 ∈ πΎ, π1 > 0 and Φ is the gauge function. 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