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Zaslavski, ajzasl@tx.technion.ac.il Received 23 December 2007; Accepted 6 March 2008 Recommended by Simeon Reich We study a class of equilibrium problems which is identified with a compl

Volume 2008, Article ID 581917, 9 pages

doi:10.1155/2008/581917

Research Article

Generic Well-Posedness for a Class of

Equilibrium 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 a class of equilibrium problems which is identified with a complete metric space 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

LetX, ρ be a complete metric space In this paper, we consider the following

equilib-rium problem:

where f belongs to a complete metric space of functionsA 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 for a generictypical element of A 4 6

Set

ρ1



x1, y1



,

x2, y2



 ρx1, x2



 ρy1, y2



, x1, x2, y1, y2∈ X. 1.1 Clearly,X × X, ρ1 is a complete metric space

Denote byA0the set of all continuous functions f : X × X → R1such that

We equip the setA0with the uniformity determined by the base

U  f, g ∈ A0× A0: f z − gz ≤  ∀z ∈ X × X, 1.3

where  > 0 It is clear that the spaceA0with this uniformity is metrizableby a metric d and

complete

Denote byA the set of all f ∈ A0for 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

ρ x, y ≤ δ.

Clearly,A is a closed subset of X We equip the space A with the metric d and consider

the topological subspaceA ⊂ A0with the relative topology

For each x ∈ X and each subset D ⊂ X, put

For each x ∈ X and each r > 0, set

B x, r y ∈ X : ρx, y ≤ r,

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 t1, t2satisfying 0 < t1< t2< Δ, there is z ∈ X such that ρz, y ∈ t1, t2.

In this paper, we will establish the following result

Theorem 1.1 There exists a set F ⊂ A which is a countable intersection of open everywhere dense

subsets of A such that for each f ∈ F, the following properties hold:

i there exists a unique x f ∈ X such that

f

x f , y

ii for each  > 0, there are δ > 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, for a generictypical f ∈ A, the problem P is well-posed 7 9

2 An auxiliary density result

Lemma 2.1 Let f ∈ A and  ∈ 0, 1 Then there exist f0 ∈ A and x0 ∈ X such that f, f0 ∈ U

and f x0, y  ≥ 0 for all y ∈ X.

Proof By P1 there is x0∈ X such that

f

x0, y

≥ − 

Set

E1



x, y ∈ X × X : fx, y ≥ − 

16



,

E2



x, y ∈ X × X \ E1: f x, y ≥ − 

8



,

E3 X × X \E1∪ E2



.

2.2

For eachy1, y2 ∈ E1, there is r1y1, y2 ∈ 0, 1 such that

f

z1, z2



>− 

14 ∀z1, z2∈ X satisfying ρz i , y i



≤ r1



y1, y2



For eachy1, y2 ∈ E2, there is r1y1, y2 ∈ 0, 1 such that

f

z1, z2

>−

6 ∀z1, z2∈ X satisfying ρz i , y i

≤ r1



y1, y2

For eachy1, y2 ∈ E3, there is r1y1, y2 ∈ 0, 1 such that

f

z1, z2



<−

8 ∀z1, z2∈ X satisying ρz i , y i



≤ r1



y1, y2



For eachy1, y2 ∈ X × X, set

U

y1, y2



 B o

y1, r1



y1, y2



× B o

y2, r1



y1, y2



For anyy1, y2 ∈ E1∪ E2, put

and for anyy1, y2 ∈ E3, put

Clearly, {Uy1, y2 : y1, y2 ∈ 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{Uy1, y2 : y1, y2∈ X} Namely, for any β ∈ B, φ β : X ×X → 0, 1 is a continuous function and there exist y1β, y2β ∈ X such that suppφ β  ⊂ Uy1β, y2β and that

β∈B

Define

f0z 

β∈B

φ β zg y1β,y2β z, z ∈ X × X. 2.10

Clearly, f0is well defined, continuous, and satisfies

Letz1, z2 ∈ E1 Then

f

z1, z2



≥ − 

Assume that β ∈ B and that φ β z1, z2 > 0 Then



z1, z2



∈ suppφ β



Ify1β, y2β ∈ E3, then in view of2.5, 2.6, and 2.13, fz1, z2 < −/8, a contradiction

see 2.12 Then y1β, y2β ∈ E1∪ E2, and by2.7,

g y1β,y2β

z1, z2



 maxf

z1, z2



, 0

Since this equality holds for any β ∈ B satisfying φ β z1, z2 > 0, it follows from 2.10 that

f0



z1, z2



 maxf

z1, z2



, 0

2.15

for allz1, z2 ∈ E1

Relations2.1, 2.2, and 2.15 imply that

f0



x0, y

By1.2, 2.7, 2.8, and 2.10

Assume that



z1, z2



Then in view of2.2 and 2.18, fz1, z2 ≥ −/8 Together with 2.7 and 2.10, this implies that

f0



z1, z2



β∈B

φ β



z1, z2 f

z1, z2



 8

 fz1, z2



 

Combined with2.11, this implies that

f

z1, z2



≤ f0



z1, z2



≤ fz1, z2



 

for allz1, z2 ∈ E2.



z1, z2



and assume that

β ∈ B, φ β



z1, z2



Then in view of2.22,



z1, z2



∈ suppφ β



By 2.23 and the choice of Uy1β, y2β see 2.3–2.6, y1β, y2β/∈E1 and by 2.4,

2.6, 2.7, and 2.8,

g y1β,y2β

z1, z2



≤ fz1, z2





Since the inequality above holds for any β∈ B satisfying 2.22, the relation 2.10 implies that

f0



z1, z2



≤ fz1, z2





Together with2.11, 2.12, and 2.15, this implies that for all z1, z2 ∈ X × X

f

z1, z2



≤ f0



z1, z2



≤ fz1, z2



 

By2.17, f0 ∈ A0 In view of2.16, f0possessesP1 Since f possesses P2, it follows from

2.7, 2.8, and 2.10 that f0possesses P2 Therefore f0 ∈ A andLemma 2.1now follows from2.16 and 2.26

3 A perturbation lemma

Lemma 3.1 Let  ∈ 0, 1, f ∈ A, and let x0∈ X satisfy

f

x0, y

Then there exist g ∈ A and δ > 0 such that

g

x0, y

≥ 0 ∀y ∈ X, g − fx,y ≤ 

and if x ∈ X satisfies infg x, y : y ∈ X> −δ, then ρx0, x  < /8.

Proof ByP2 there is a positive number

δ0< min

such that

f y, z ≤ 

Define

φ t  1, t ∈0, δ0 ,

φ t  0, t ∈2δ0,∞,

φ t  2 − tδ−1

0 , t∈δ0, 2δ0



,

3.6

f1x, y  −φρ x, yρ x, y 1− φρ x, yf x, y, x, y ∈ X. 3.7

Clearly, f1is continuous and

By3.6 and 3.7,

f1x, y  −ρx, y ∀ x, y ∈ X satisfying ρx, y ≤ δ0. 3.9

Let x, y ∈ X We estimate |fx, y − f1x, y| If ρx, y ≥ 2δ0, then by3.6 and 3.7,

Assume that

By3.3 and 3.11,

f x, y ≤ 

By3.5, 3.6, 3.7, 3.11, and 3.12,

f1x, y − fx, y ≤ ρx,y  fx,y ≤ 2δ0 

16 <



Together with3.10 this implies that

f1x, y − fx, y< 

Assume that x ∈ X In view of P3 and 3.3, there is y ∈ X such that

It follows from3.15 and 3.9 that

inf

for all x ∈ X Set

g x, y  φρ

x, x0



f x, y 1− φρ

x, x0



Clearly, the function g is continuous and

In view of3.1, 3.18, and 3.6,

g

x0, y

 fx0, y

Since the function f possessesP2, it follows from 3.9, 3.20, and 3.18 that g possesses the

propertyP2 Thus g ∈ A.

By3.6, 3.14, and 3.18 for all x, y ∈ X

f − gx,y ≤ f1x, y − fx, y ≤ 

Assume that

If ρx0, x  ≥ 2δ0, then by3.6 and 3.18,

and together with3.17, this implies that

inf

This inequality contradicts3.22 The contradiction we have reached proves that

ρ

x0, x

< 2δ0< 

This completes the proof of the lemma

4 Proof of Theorem 1.1

Denote by E the set of all f ∈ A for which there exists x ∈ X such that fx, y ≥ 0 for all y ∈ X.

ByLemma 2.1, E is an everywhere dense subset ofA

Let f ∈ E and n be a natural number There exists x f ∈ X such that

f

x f , y

ByLemma 3.1, there exist g f,n ∈ A and δ f,n > 0 such that

g f,n



x f , y

≥ 0 ∀y ∈ X, g f,n − fx, y ≤ 4n−1 ∀x, y ∈ X, 4.2 and the following property holds

P4 For each x ∈ X satisfying inf{g f,n x, y : y ∈ X} > −δ f,n , the inequality ρ x f , x  <

4n−1holds

Denote by V f, n the open neighborhood of g f,ninA such that

V f, n ⊂h∈ A :h, g f,n



∈ U4−1δ f,n



Assume that

x ∈ X, h ∈ V f, n, infh x, y : y ∈ X>−2−1δ f,n 4.4

By1.3, 4.3, and 4.4,

inf

g f,n x, y : y ∈ X≥ infh x, y : y ∈ X− 4−1δ f,n > −δ f,n 4.5

In view of4.5 and P4,

ρ

x f , x

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

Clearly,F is a countable intersection of open everywhere dense subset of A Let

Choose a natural number k > 8−1 1 There exist f ∈ E and an integer n ≥ k such that

The propertyP4, 4.3, and 4.9 imply that for each x ∈ X satisfying

inf

we have

inf

g f,n x, y : y ∈ X>−2−1δ f,n− 4−1δ 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.

ByP1 there is a sequence {x i}∞i1⊂ X such that

lim inf

i→∞

 inf

In view of4.12 and P6 for all large enough natural numbers i, j, we have

ρ

x i , x j



≤ ρx i , x f



 ρx f , x j



< 

Since  is any positive number, we conclude that {x i}∞

i1is a Cauchy sequence and there exists

x ξ  lim

Relations4.12 and 4.14 imply that for all y ∈ X

ξ

x ξ , y

 lim

i→∞ξ

x i , y

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, then x  x ξ ByP6 and 4.15,

ρ

x ξ , x f



≤ 

Let x ∈ X and h ∈ V f, n satisfy 4.4 By P5, ρx f , x  < 4n−1 Together with4.16, this implies that

ρ

x, x ξ



≤ ρx, x f



 ρx f , x ξ



< 4n−1

Theorem 1.1is proved

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P4 For each...

It follows from3.15 and 3.9 that

inf