Volume 2008, Article ID 581917,9pages 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,[email protected] 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 functionsin the sense of Baire category, we establish that the corresponding equilibrium problem possesses a unique solution and is well-posed.
Copyrightq2008 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,1–3and the references mentioned therein.
LetX, ρbe a complete metric space. In this paper, we consider the following equilib- rium problem:
To findx∈X such thatfx, y≥0 ∀y∈X, P
wheref belongs to a complete metric space of functionsAdefined 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 Ppossesses a unique solution. In other words, the problem P possesses a unique solution for a generictypicalelement ofA4–6.
Set
ρ1 x1, y1
, x2, y2
ρ x1, x2
ρ y1, y2
, x1, x2, y1, y2∈X. 1.1 Clearly,X×X, ρ1is a complete metric space.
Denote byA0the set of all continuous functionsf:X×X→R1such that
fx, x 0 ∀x∈X. 1.2
We equip the setA0with the uniformity determined by the base
U
f, g∈ A0× A0: fz−gz≤∀z∈X×X
, 1.3
where >0. It is clear that the spaceA0with this uniformity is metrizableby a metricdand complete.
Denote byAthe set of allf∈ A0for which the following properties hold.
P1For each >0, there existsx∈Xsuch thatfx, y≥ −for allx∈X.
P2For each >0, there existsδ > 0 such that|fx, y| ≤for allx, y∈X satisfying ρx, y≤δ.
Clearly,Ais a closed subset ofX. We equip the spaceAwith the metricdand consider the topological subspaceA ⊂ A0with the relative topology.
For eachx∈Xand each subsetD⊂X, put ρx, D inf
ρx, y: y∈D
. 1.4
For eachx∈Xand eachr >0, set
Bx, r
y∈X:ρx, y≤r , Box, r
y∈X:ρx, y< r
. 1.5
Assume that the following property holds.
P3There exists a positive number Δsuch that for eachy ∈ X and each pair of real numberst1, t2satisfying 0< t1< t2<Δ, there isz∈Xsuch thatρz, y∈t1, t2.
In this paper, we will establish the following result.
Theorem 1.1. There exists a setF ⊂ Awhich is a countable intersection of open everywhere dense subsets ofAsuch that for eachf ∈ F, the following properties hold:
ithere exists a uniquexf ∈Xsuch that
f xf, y
≥0 ∀x, y∈X; 1.6
iifor each >0, there areδ >0 and a neighborhoodV offinAsuch that for eachh∈V and eachx∈Xsatisfying inf{hx, y: y∈X}>−δ, the inequalityρxf, x< holds.
In other words, for a generictypicalf∈ A, the problemPis well-posed7–9.
2. An auxiliary density result
Lemma 2.1. Letf∈ Aand∈0,1. Then there existf0∈ Aandx0∈Xsuch thatf, f0∈U andfx0, y≥0 for ally∈X.
Proof. ByP1there isx0∈Xsuch that f
x0, y
≥ −
16 ∀y∈X. 2.1
Set
E1
x, y∈X×X: fx, y≥ − 16
, E2
x, y∈X×X\E1: fx, y≥ − 8
, E3 X×X\
E1∪E2 .
2.2
For eachy1, y2∈E1, there isr1y1, y2∈0,1such that f
z1, z2
>−
14 ∀z1, z2∈X satisfyingρ zi, yi
≤r1 y1, y2
, i1,2. 2.3
For eachy1, y2∈E2, there isr1y1, y2∈0,1such that f
z1, z2
>−
6 ∀z1, z2∈X satisfyingρ zi, yi
≤r1 y1, y2
, i1,2. 2.4
For eachy1, y2∈E3, there isr1y1, y2∈0,1such that f
z1, z2
<−
8 ∀z1, z2∈X satisyingρ zi, yi
≤r1
y1, y2
, i1,2. 2.5
For eachy1, y2∈X×X, set U
y1, y2 Bo
y1, r1 y1, y2
×Bo y2, r1
y1, y2
. 2.6
For anyy1, y2∈E1∪E2, put
gy1,y2z max
fz,0
, z∈X×X 2.7
and for anyy1, y2∈E3, put
gy1,y2z fz, z∈X×X. 2.8
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{φβ :β∈ B}subordinated to the covering{Uy1, y2:y1, y2∈X}. Namely, for anyβ∈ B,φβ :X×X→0,1is a continuous function and there existy1β, y2β∈Xsuch that suppφβ⊂Uy1β, y2βand that
β∈B
φβz 1 ∀z∈X×X. 2.9
Define
f0z
β∈B
φβzgy1β,y2βz, z∈X×X. 2.10
Clearly,f0is well defined, continuous, and satisfies
f0z≥fz ∀z∈X×X. 2.11
Letz1, z2∈E1. Then
f z1, z2
≥ −
16. 2.12
Assume thatβ∈ Band thatφβz1, z2>0. Then z1, z2
∈supp φβ
⊂U
y1β, y2β
. 2.13
Ify1β, y2β∈E3, then in view of2.5,2.6, and2.13,fz1, z2<−/8, a contradiction see2.12. Theny1β, y2β∈E1∪E2, and by2.7,
gy1β,y2β z1, z2
max f
z1, z2 ,0
. 2.14
Since this equality holds for anyβ∈ Bsatisfyingφβz1, z2>0, it follows from2.10that f0
z1, z2
max f
z1, z2 ,0
2.15 for allz1, z2∈E1.
Relations2.1,2.2, and2.15imply that f0
x0, y
≥0, y∈X. 2.16
By1.2,2.7,2.8, and2.10
f0x, x 0, x∈X. 2.17 Assume that
z1, z2
∈E2. 2.18
Then in view of2.2and2.18,fz1, z2≥ −/8.Together with2.7and2.10, this implies that
f0
z1, z2
≤
β∈B
φβ
z1, z2
f z1, z2
8
f
z1, z2
8. 2.19
Combined with2.11, this implies that f
z1, z2
≤f0 z1, z2
≤f z1, z2
8 2.20
for allz1, z2∈E2.
Let
z1, z2
∈E3 2.21
and assume that
β∈ B, φβ
z1, z2
>0. 2.22
Then in view of2.22,
z1, z2
∈supp φβ
⊂U
y1β, y2β
. 2.23
By 2.23 and the choice ofUy1β, y2β see 2.3–2.6,y1β, y2β/∈E1 and by2.4, 2.6,2.7, and2.8,
gy1β,y2β z1, z2
≤f z1, z2
6. 2.24
Since the inequality above holds for anyβ∈ Bsatisfying2.22, the relation2.10implies that f0
z1, z2
≤f z1, z2
6. 2.25
Together with2.11,2.12, and2.15, this implies that for allz1, z2∈X×X f
z1, z2
≤f0 z1, z2
≤f z1, z2
6. 2.26
By2.17,f0 ∈ A0. In view of2.16,f0possessesP1. SincefpossessesP2, it follows from 2.7,2.8, and2.10thatf0 possessesP2. Thereforef0 ∈ AandLemma 2.1now follows from2.16and2.26.
3. A perturbation lemma
Lemma 3.1. Let∈0,1,f∈ A, and letx0∈Xsatisfy f
x0, y
≥0 ∀y∈X. 3.1
Then there existg∈ Aandδ >0 such that g
x0, y
≥0 ∀y∈X, g−fx, y≤
4 ∀x, y∈X 3.2 and ifx∈Xsatisfies inf
gx, y:y∈X
>−δ,thenρx0, x< /8.
Proof. ByP2there is a positive number δ0<min
16−1,16−1Δ
3.3 such that
fy, z≤
16 ∀y, z∈Xsatisfyingρy, z≤4δ0. 3.4
Set
δ2−1δ0. 3.5
Define
φt 1, t∈ 0, δ0
, φt 0, t∈
2δ0,∞ , φt 2−tδ0−1, t∈
δ0,2δ0 ,
3.6
f1x, y −φ
ρx, y
ρx, y 1−φ
ρx, y
fx, y, x, y∈X. 3.7 Clearly,f1is continuous and
f1x, x 0 ∀x∈X. 3.8 By3.6and3.7,
f1x, y −ρx, y ∀x, y∈X satisfyingρx, y≤δ0. 3.9 Letx, y∈X. We estimate|fx, y−f1x, y|. Ifρx, y≥2δ0, then by3.6and3.7,
f1x, y−fx, y0. 3.10
Assume that
ρx, y≤2δ0. 3.11
By3.3and3.11,
fx, y≤
16. 3.12
By3.5,3.6,3.7,3.11, and3.12,
f1x, y−fx, y≤ρx, y fx, y≤2δ0 16<
4. 3.13
Together with3.10this implies that
f1x, y−fx, y<
4 ∀x, y∈X. 3.14
Assume thatx∈X. In view ofP3and3.3, there isy∈Xsuch that ρy, x∈
2−1δ0, δ0
. 3.15
It follows from3.15and3.9that
f1x, y −ρy, x≤ −2−1δ0, 3.16 inf
f1x, z:z∈X
≤ −2−1δ0 3.17
for allx∈X. Set
gx, y φ ρ
x, x0
fx, y 1−φ
ρ
x, x0
f1x, y, x, y∈X. 3.18 Clearly, the functiongis continuous and
gx, x 0 ∀x∈X. 3.19
In view of3.1,3.18, and3.6, g
x0, y f
x0, y
≥0 ∀y∈X. 3.20
Since the functionfpossessesP2, it follows from3.9,3.20, and3.18thatgpossesses the propertyP2. Thusg∈ A.
By3.6,3.14, and3.18for allx, y∈X
f−gx, y≤f1x, y−fx, y≤
4. 3.21
Assume that
x∈X, inf
gx, y: y∈X
>−2−1δ0−δ. 3.22
Ifρx0, x≥2δ0, then by3.6and3.18,
gx, y f1x, y ∀y∈Y 3.23
and together with3.17, this implies that inf
gx, y:y∈X
≤ −2−1δ0. 3.24
This inequality contradicts3.22. The contradiction we have reached proves that ρ
x0, x
<2δ0<
8. 3.25
This completes the proof of the lemma.
4. Proof ofTheorem 1.1
Denote byEthe set of allf∈ Afor which there existsx∈Xsuch thatfx, y≥0 for ally∈X.
ByLemma 2.1,Eis an everywhere dense subset ofA.
Letf∈Eandnbe a natural number. There existsxf ∈Xsuch that f
xf, y
≥0 ∀y∈X. 4.1
ByLemma 3.1, there existgf,n∈ Aandδf,n>0 such that gf,n
xf, y
≥0 ∀y∈X, gf,n−f
x, y≤4n−1 ∀x, y∈X, 4.2 and the following property holds.
P4For eachx∈Xsatisfying inf{gf,nx, y: y∈X}>−δf,n,the inequalityρxf, x<
4n−1holds.
Denote byVf, nthe open neighborhood ofgf,ninAsuch that Vf, n⊂
h∈ A: h, gf,n
∈U
4−1δf,n
. 4.3
Assume that
x∈X, h∈Vf, n, inf
hx, y:y∈X
>−2−1δf,n. 4.4 By1.3,4.3, and4.4,
inf
gf,nx, y:y∈X
≥inf
hx, y:y∈X
−4−1δf,n>−δf,n. 4.5 In view of4.5andP4,
ρ xf, x
<4n−1. 4.6
Thus we have shown that the following property holds.
P5For eachx∈Xand eachh∈Vf, nsatisfying4.4, the inequalityρxf, x<4n−1 holds.
Set
F∞
k1
∪
Vf, n:f∈Eand an integern≥k
. 4.7
Clearly,Fis a countable intersection of open everywhere dense subset ofA. Let
ξ∈ F, >0. 4.8
Choose a natural numberk >8−11. There existf∈Eand an integern≥ksuch that ξ∈Vf, n. 4.9 The propertyP4,4.3, and4.9imply that for eachx∈Xsatisfying
inf
ξx, y:y∈X
>−2−1δf,n, 4.10
we have
inf
gf,nx, y:y∈X
>−2−1δf,n−4−1δf,n>−δf,n, ρ
xf, x
<4n−1<
8. 4.11
Thus we have shown that the following property holds.
P6For eachx∈Xsatisfying4.10, the inequalityρxf, x< /8 holds.
ByP1there is a sequence{xi}∞i1⊂Xsuch that lim inf
i→∞
inf
ξxi, y:y∈X
≥0. 4.12
In view of4.12andP6for all large enough natural numbersi, j, we have ρ
xi, xj
≤ρ xi, xf
ρ xf, xj
<
4. 4.13
Sinceis any positive number, we conclude that{xi}∞i1is a Cauchy sequence and there exists xξ lim
i→∞xi. 4.14
Relations4.12and4.14imply that for ally∈X ξ
xξ, y lim
i→∞ξ xi, y
≥0. 4.15
We have also shown that any sequence{xi}∞i1 ⊂ X satisfying4.12converges. This implies that ifx∈Xsatisfiesξx, y≥0 for ally∈X, thenxxξ. ByP6and4.15,
ρ xξ, xf
≤
8. 4.16
Letx ∈ Xandh∈ Vf, nsatisfy4.4. ByP5,ρxf, x <4n−1. Together with4.16, this implies that
ρ x, xξ
≤ρ x, xf
ρ xf, xξ
<4n−1
8 < . 4.17
Theorem 1.1is proved.
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