Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2008, Article ID 581917, 9 pages doi:10.1155/2008/581917 ResearchArticleGenericWell-PosednessforaClassofEquilibrium Problems Alexander J. Zaslavski Department of Mathematics, The Technion-Israel Institute of Technology, 32000 Haifa, Israel Correspondence should be addressed to Alexander J. Zaslavski, ajzasl@tx.technion.ac.il Received 23 December 2007; Accepted 6 March 2008 Recommended by Simeon Reich We study aclassofequilibrium problems which is identified with a complete metric s pace of functions. For most elements of this space of functions in the sense of Baire category, we establish that the corresponding equilibrium problem possesses a unique solution and is well-posed. Copyright q 2008 Alexander J. Zaslavski. 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. 1. Introduction The study of equilibriumproblems has recently been a rapidly growing area of research. See, for example, 1–3 and the references mentioned therein. Let X, ρ be a complete metric space. In this paper, we consider the following equilib- rium problem: To find x ∈ X such that fx, y ≥ 0 ∀y ∈ X, P where f belongs to a complete metric space of functions A defined below. In this paper, we show that for most elements of this space of functions A in the sense of Baire category the equilibrium problem P possesses a unique solution. In other words, the problem P possesses a unique solution forageneric typical element ofA 4–6. Set ρ 1 x 1 ,y 1 , x 2 ,y 2 ρ x 1 ,x 2 ρ y 1 ,y 2 ,x 1 ,x 2 ,y 1 ,y 2 ∈ X. 1.1 Clearly, X × X, ρ 1 is a complete metric space. Denote by A 0 the set of all continuous functions f : X × X → R 1 such that fx, x0 ∀x ∈ X. 1.2 2 Journal of Inequalities and Applications We equip the set A 0 with the uniformity determined by the base U f, g ∈A 0 ×A 0 : fz − gz ≤ ∀z ∈ X × X , 1.3 where >0. It is clear that the space A 0 with this uniformity is metrizable by a metric d and complete. Denote by A the set of all f ∈A 0 for which the following properties hold. P1 For each >0, there exists x ∈ X such that fx ,y ≥− for all x ∈ X. P2 For each >0, there exists δ>0 such that |fx, y|≤ for all x, y ∈ X satisfying ρx, y ≤ δ. Clearly, A is a closed subset of X. We equip the space A with the metric d and consider the topological subspace A⊂A 0 with the relative topology. For each x ∈ X and each subset D ⊂ X,put ρx, Dinf ρx, y : y ∈ D . 1.4 For each x ∈ X and each r>0, set Bx, r y ∈ X : ρx, y ≤ r , B o x, r y ∈ X : ρx, y <r . 1.5 Assume that the following property holds. P3 There exists a positive number Δ such that for each y ∈ X and each pair of real numbers t 1 ,t 2 satisfying 0 <t 1 <t 2 < Δ,thereisz ∈ X such that ρz, y ∈ t 1 ,t 2 . In this paper, we will establish the following result. Theorem 1.1. There exists a set F⊂Awhich is a countable intersection of open everywhere dense subsets ofA such that for each f ∈F, the following properties hold: i there exists a unique x f ∈ X such that f x f ,y ≥ 0 ∀ x, y ∈ X; 1.6 ii for each >0,thereareδ>0 and a neighborhood V of f in A such that for each h ∈ V and each x ∈ X satisfying inf{hx, y : y ∈ X} > −δ, the inequality ρx f ,x <holds. In other words, forageneric typical f ∈A, the problem P is well-posed 7–9. 2. An auxiliary density result Lemma 2.1. Let f ∈Aand ∈ 0, 1. Then there exist f 0 ∈Aand x 0 ∈ X such that f, f 0 ∈ U and fx 0 ,y ≥ 0 for all y ∈ X. Alexander J. Zaslavski 3 Proof. By P1 there is x 0 ∈ X such that f x 0 ,y ≥− 16 ∀y ∈ X. 2.1 Set E 1 x, y ∈ X × X : fx, y ≥− 16 , E 2 x, y ∈ X × X \ E 1 : fx, y ≥− 8 , E 3 X × X \ E 1 ∪ E 2 . 2.2 For each y 1 ,y 2 ∈ E 1 ,thereisr 1 y 1 ,y 2 ∈ 0, 1 such that f z 1 ,z 2 > − 14 ∀z 1 ,z 2 ∈ X satisfying ρ z i ,y i ≤ r 1 y 1 ,y 2 ,i 1, 2. 2.3 For each y 1 ,y 2 ∈ E 2 ,thereisr 1 y 1 ,y 2 ∈ 0, 1 such that f z 1 ,z 2 > − 6 ∀z 1 ,z 2 ∈ X satisfying ρ z i ,y i ≤ r 1 y 1 ,y 2 ,i 1, 2. 2.4 For each y 1 ,y 2 ∈ E 3 ,thereisr 1 y 1 ,y 2 ∈ 0, 1 such that f z 1 ,z 2 < − 8 ∀z 1 ,z 2 ∈ X satisying ρ z i ,y i ≤ r 1 y 1 ,y 2 ,i 1, 2. 2.5 For each y 1 ,y 2 ∈ X × X, set U y 1 ,y 2 B o y 1 ,r 1 y 1 ,y 2 × B o y 2 ,r 1 y 1 ,y 2 . 2.6 For any y 1 ,y 2 ∈ E 1 ∪ E 2 ,put g y 1 ,y 2 zmax fz, 0 ,z∈ X × X 2.7 and for any y 1 ,y 2 ∈ E 3 ,put g y 1 ,y 2 zfz,z∈ X × X. 2.8 Clearly, {Uy 1 ,y 2 : y 1 ,y 2 ∈ X} is an open covering of X × X. Since any metric space is paracompact, there is a continuous locally finite partition of unity {φ β : β ∈B}subordinated to the covering {Uy 1 ,y 2 : y 1 ,y 2 ∈ X}. Namely, for any β ∈B, φ β : X × X → 0, 1 is a continuous function and there exist y 1 β,y 2 β ∈ X such that suppφ β ⊂ Uy 1 β,y 2 β and that β∈B φ β z1 ∀z ∈ X × X. 2.9 Define f 0 z β∈B φ β zg y 1 β,y 2 β z,z∈ X × X. 2.10 4 Journal of Inequalities and Applications Clearly, f 0 is well defined, continuous, and satisfies f 0 z ≥ fz ∀z ∈ X × X. 2.11 Let z 1 ,z 2 ∈ E 1 .Then f z 1 ,z 2 ≥− 16 . 2.12 Assume that β ∈Band that φ β z 1 ,z 2 > 0. Then z 1 ,z 2 ∈ supp φ β ⊂ U y 1 β,y 2 β . 2.13 If y 1 β,y 2 β ∈ E 3 ,theninviewof2.5, 2.6,and2.13, fz 1 ,z 2 < −/8, a contradiction see 2.12.Theny 1 β,y 2 β ∈ E 1 ∪ E 2 ,andby2.7, g y 1 β,y 2 β z 1 ,z 2 max f z 1 ,z 2 , 0 . 2.14 Since this equality holds for any β ∈Bsatisfying φ β z 1 ,z 2 > 0, it follows from 2.10 that f 0 z 1 ,z 2 max f z 1 ,z 2 , 0 2.15 for all z 1 ,z 2 ∈ E 1 . Relations 2.1, 2.2,and2.15 imply that f 0 x 0 ,y ≥ 0,y∈ X. 2.16 By 1.2, 2.7, 2.8,and2.10 f 0 x, x0,x∈ X. 2.17 Assume that z 1 ,z 2 ∈ E 2 . 2.18 Then in view of 2.2 and 2.18, fz 1 ,z 2 ≥−/8. Together with 2.7 and 2.10, this implies that f 0 z 1 ,z 2 ≤ β∈B φ β z 1 ,z 2 f z 1 ,z 2 8 f z 1 ,z 2 8 . 2.19 Combined with 2.11, this implies that f z 1 ,z 2 ≤ f 0 z 1 ,z 2 ≤ f z 1 ,z 2 8 2.20 for all z 1 ,z 2 ∈ E 2 . Alexander J. Zaslavski 5 Let z 1 ,z 2 ∈ E 3 2.21 and assume that β ∈B,φ β z 1 ,z 2 > 0. 2.22 Then in view of 2.22, z 1 ,z 2 ∈ supp φ β ⊂ U y 1 β,y 2 β . 2.23 By 2.23 and the choice of Uy 1 β,y 2 β see 2.3–2.6, y 1 β,y 2 β / ∈E 1 and by 2.4, 2.6, 2.7,and2.8, g y 1 β,y 2 β z 1 ,z 2 ≤ f z 1 ,z 2 6 . 2.24 Since the inequality above holds for any β ∈Bsatisfying 2.22,therelation2.10 implies that f 0 z 1 ,z 2 ≤ f z 1 ,z 2 6 . 2.25 Together with 2.11, 2.12,and2.15, this implies that for all z 1 ,z 2 ∈ X × X f z 1 ,z 2 ≤ f 0 z 1 ,z 2 ≤ f z 1 ,z 2 6 . 2.26 By 2.17, f 0 ∈A 0 .Inviewof2.16, f 0 possesses P1. Since f possesses P2, it follows from 2.7, 2.8,and2.10 that f 0 possesses P2. Therefore f 0 ∈Aand Lemma 2.1 now follows from 2.16 and 2.26. 3. A perturbation lemma Lemma 3.1. Let ∈ 0, 1, f ∈A, and let x 0 ∈ X satisfy f x 0 ,y ≥ 0 ∀y ∈ X. 3.1 Then there exist g ∈Aand δ>0 such that g x 0 ,y ≥ 0 ∀y ∈ X, g − fx, y ≤ 4 ∀x, y ∈ X 3.2 and if x ∈ X satisfies inf gx, y : y ∈ X > −δ, then ρx 0 ,x </8. Proof. By P2 there is a positive number δ 0 < min 16 −1 , 16 −1 Δ 3.3 such that fy,z ≤ 16 ∀y, z ∈ X satisfying ρy, z ≤ 4δ 0 . 3.4 6 Journal of Inequalities and Applications Set δ 2 −1 δ 0 . 3.5 Define φt1,t∈ 0,δ 0 , φt0,t∈ 2δ 0 , ∞ , φt2 − tδ −1 0 ,t∈ δ 0 , 2δ 0 , 3.6 f 1 x, y−φ ρx, y ρx, y 1 − φ ρx, y fx, y, x, y ∈ X. 3.7 Clearly, f 1 is continuous and f 1 x, x0 ∀x ∈ X. 3.8 By 3.6 and 3.7, f 1 x, y−ρx, y ∀ x, y ∈ X satisfying ρx, y ≤ δ 0 . 3.9 Let x, y ∈ X. We estimate |fx, y − f 1 x, y|.Ifρx, y ≥ 2δ 0 ,thenby3.6 and 3.7, f 1 x, y − fx, y 0. 3.10 Assume that ρx, y ≤ 2δ 0 . 3.11 By 3.3 and 3.11, fx, y ≤ 16 . 3.12 By 3.5, 3.6, 3.7, 3.11,and3.12, f 1 x, y − fx, y ≤ ρx, y fx, y ≤ 2δ 0 16 < 4 . 3.13 Together with 3.10 this implies that f 1 x, y − fx, y < 4 ∀x, y ∈ X. 3.14 Assume that x ∈ X.InviewofP3 and 3.3,thereisy ∈ X such that ρy, x ∈ 2 −1 δ 0 ,δ 0 . 3.15 It follows from 3.15 and 3.9 that f 1 x, y−ρy, x ≤−2 −1 δ 0 , 3.16 inf f 1 x, z : z ∈ X ≤−2 −1 δ 0 3.17 Alexander J. Zaslavski 7 for all x ∈ X. Set gx, yφ ρ x, x 0 fx, y 1 − φ ρ x, x 0 f 1 x, y,x,y∈ X. 3.18 Clearly, the function g is continuous and gx, x0 ∀x ∈ X. 3.19 In view of 3.1, 3.18,and3.6, g x 0 ,y f x 0 ,y ≥ 0 ∀y ∈ X. 3.20 Since the function f possesses P2, it follows from 3.9, 3.20,and3.18 that g possesses the property P2.Thusg ∈A. By 3.6, 3.14,and3.18 for all x, y ∈ X f − gx, y ≤ f 1 x, y − fx, y ≤ 4 . 3.21 Assume that x ∈ X, inf gx, y : y ∈ X > −2 −1 δ 0 −δ. 3.22 If ρx 0 ,x ≥ 2δ 0 ,thenby3.6 and 3.18, gx, yf 1 x, y ∀y ∈ Y 3.23 and together with 3.17, this implies that inf gx, y : y ∈ X ≤−2 −1 δ 0 . 3.24 This inequality contradicts 3.22. The contradiction we have reached proves that ρ x 0 ,x < 2δ 0 < 8 . 3.25 This completes the proof of the lemma. 4. Proof of Theorem 1.1 Denote by E the set of all f ∈Afor which there exists x ∈ X such that fx, y ≥ 0 for all y ∈ X. By Lemma 2.1, E is an everywhere dense subset of A. Let f ∈ E and n be a natural number. There exists x f ∈ X such that f x f ,y ≥ 0 ∀y ∈ X. 4.1 By Lemma 3.1, there exist g f,n ∈Aand δ f,n > 0 such that g f,n x f ,y ≥ 0 ∀y ∈ X, g f,n − f x, y ≤ 4n −1 ∀x, y ∈ X, 4.2 and the following property holds. 8 Journal of Inequalities and Applications P4 For each x ∈ X satisfying inf{g f,n x, y : y ∈ X} > −δ f,n , the inequality ρx f ,x < 4n −1 holds. Denote by V f,n the open neighborhood of g f,n in A such that V f, n ⊂ h ∈A: h, g f,n ∈ U 4 −1 δ f,n . 4.3 Assume that x ∈ X, h ∈ V f,n, inf hx, y : y ∈ X > −2 −1 δ f,n . 4.4 By 1.3, 4.3,and4.4, inf g f,n x, y : y ∈ X ≥ inf hx, y : y ∈ X − 4 −1 δ f,n > −δ f,n . 4.5 In view of 4.5 and P4, ρ x f ,x < 4n −1 . 4.6 Thus we have shown that the following property holds. P5 For each x ∈ X and each h ∈ V f, n satisfying 4.4, the inequality ρx f ,x < 4n −1 holds. Set F ∞ k1 ∪ V f, n : f ∈ E and an integer n ≥ k . 4.7 Clearly, F is a countable intersection of open everywhere dense subset of A.Let ξ ∈F,>0. 4.8 Choose a natural number k>8 −1 1. There exist f ∈ E and an integer n ≥ k such that ξ ∈ V f, n. 4.9 The property P4, 4.3,and4.9 imply that for each x ∈ X satisfying inf ξx, y : y ∈ X > −2 −1 δ f,n , 4.10 we have inf g f,n x, y : y ∈ X > −2 −1 δ f,n − 4 −1 δ f,n > −δ f,n , ρ x f ,x < 4n −1 < 8 . 4.11 Thus we have shown that the following property holds. P6 For each x ∈ X satisfying 4.10, the inequality ρx f ,x </8 holds. Alexander J. Zaslavski 9 By P1 there is a sequence {x i } ∞ i1 ⊂ X such that lim inf i→∞ inf ξx i ,y : y ∈ X ≥ 0. 4.12 In view of 4.12 and P6 for all large enough natural numbers i, j,wehave ρ x i ,x j ≤ ρ x i ,x f ρ x f ,x j < 4 . 4.13 Since is any positive number, we conclude that {x i } ∞ i1 is a Cauchy sequence and there exists x ξ lim i→∞ x i . 4.14 Relations 4.12 and 4.14 imply that for all y ∈ X ξ x ξ ,y lim i→∞ ξ x i ,y ≥ 0. 4.15 We have also shown that any sequence {x i } ∞ i1 ⊂ X satisfying 4.12 converges. This implies that if x ∈ X satisfies ξx, y ≥ 0 for all y ∈ X,thenx x ξ .ByP6 and 4.15, ρ x ξ ,x f ≤ 8 . 4.16 Let x ∈ X and h ∈ V f,n satisfy 4.4.ByP5, ρx f ,x < 4n −1 .Togetherwith4.16,this implies that ρ x, x ξ ≤ ρ x, x f ρ x f ,x ξ < 4n −1 8 <. 4.17 Theorem 1.1 is proved. References 1 E. Blum and W. Oettli, “From optimization and variational inequalities to equilibrium problems,” The Mathematics Student, vol. 63, no. 1–4, pp. 123–145, 1994. 2 G Y. Chen, X. Huang, and X. Yang, Vector Optimization, Set-Valued and Variational Analysis, Springer, Berlin, Germany, 2005. 3 X. J. Long, N J. Huang, and K. L. Teo, “Levitin-Polyak well-posednessforequilibrium problems with functional constraints,” Journal of Inequalities and Applications, vol. 2008, Article ID 657329, 14 pages, 2008. 4 J P. Aubin and I. Ekeland, Applied Nonlinear Analysis, John Wiley & Sons, New York, NY, USA, 1984. 5 Z. Nitecki, Differentiable Dynamics. 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Calculus of Variations and Optimal Control, Springer, New York, NY, USA, 2006. 7 A. D. Ioffe and A. J. Zaslavski, “Variational principles and well-posedness in optimization and calculus of variations,”. Problems Alexander J. Zaslavski Department of Mathematics, The Technion-Israel Institute of Technology, 32000 Haifa, Israel Correspondence should be addressed to Alexander J. Zaslavski, ajzasl@tx.technion.ac.il Received