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 that fx, y≥0 ∀y∈X, P
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 generictypicalelement ofA4–6. Set
ρ1
x1, y1
,x2, y2
ρx1, x2
ρy1, 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
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 ∈ Xsatisfying 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 infρx, 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 each y ∈ 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 ⊂ A which is a countable intersection of open everywhere dense
subsets ofAsuch that for eachf ∈ F, the following properties hold:
ithere exists a uniquexf ∈Xsuch that
fxf, y≥0 ∀x, y ∈X; 1.6
iifor each >0, there areδ >0and a neighborhoodV off inAsuch that for eachh∈V and eachx∈Xsatisfyinginf{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 ∈ A and∈0,1. Then there existf0 ∈ Aandx0 ∈Xsuch thatf, f0∈U
Proof. ByP1there isx0∈Xsuch that
fx0, 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
fz1, 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
fz1, z2
>−
6 ∀z1, z2∈X satisfying ρ
zi, yi
≤r1
y1, y2
, i1,2. 2.4
For eachy1, y2∈E3, there isr1y1, y2∈0,1such that
fz1, z2
<−
8 ∀z1, z2∈X satisyingρ
zi, yi≤r1
y1, y2
, i1,2. 2.5
For eachy1, y2∈X×X, set
Uy1, y2
Boy1, r1
y1, y2
×Boy2, 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
Clearly,f0is well defined, continuous, and satisfies
f0z≥fz ∀z∈X×X. 2.11
Letz1, z2∈E1. Then
fz1, z2
≥ −
16. 2.12
Assume thatβ∈ Band thatφβz1, z2>0. Then
z1, z2
∈suppφβ⊂Uy1β, 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
maxfz1, z2
,0. 2.14
Since this equality holds for anyβ∈ Bsatisfyingφβz1, z2>0, it follows from2.10that
f0
z1, z2
maxfz1, 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
φβz1, z2
fz1, z2
8
fz1, z2
8. 2.19
Combined with2.11, this implies that
fz1, z2
≤f0
z1, z2
≤fz1, z2
8 2.20
Let
z1, z2
∈E3 2.21
and assume that
β∈ B, φβz1, z2
>0. 2.22
Then in view of2.22,
z1, z2
∈suppφβ⊂Uy1β, y2β
. 2.23
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, and2.8,
gy1β,y2β
z1, z2
≤fz1, z2
6. 2.24
Since the inequality above holds for anyβ∈ Bsatisfying2.22, the relation2.10implies that
f0
z1, z2
≤fz1, z2
6. 2.25
Together with2.11,2.12, and2.15, this implies that for allz1, z2∈X×X
fz1, z2
≤f0
z1, z2
≤fz1, z2
6. 2.26
By2.17,f0 ∈ A0. In view of2.16,f0possessesP1. Sincef possessesP2, it follows from
2.7,2.8, and 2.10thatf0possesses P2. Thereforef0 ∈ A andLemma 2.1now follows from2.16and2.26.
3. A perturbation lemma
Lemma 3.1. Let∈0,1,f ∈ A, and letx0∈Xsatisfy
fx0, y
≥0 ∀y∈X. 3.1
Then there existg∈ Aandδ >0such that
gx0, y
≥0 ∀y∈X, g−fx, y≤
4 ∀x, y∈X 3.2 and ifx∈Xsatisfiesinfgx, 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≤
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, yρx, y 1−φρx, yfx, 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 satisfying ρx, 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
inff1x, z:z∈X
≤ −2−1δ
for allx∈X. Set
gx, y φρx, x0
fx, y 1−φρx, x0
f1x, y, x, y ∈X. 3.18
Clearly, the functiong is continuous and
gx, x 0 ∀x∈X. 3.19
In view of3.1,3.18, and3.6,
gx0, y
fx0, 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, infgx, 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
infgx, y:y∈X≤ −2−1δ0. 3.24
This inequality contradicts3.22. The contradiction we have reached proves that
ρx0, 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
fxf, y≥0 ∀y∈X. 4.1
ByLemma 3.1, there existgf,n∈ Aandδf,n>0 such that
gf,nxf, y≥0 ∀y∈X, gf,n−fx, y≤4n−1 ∀x, y ∈X, 4.2
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∈U4−1δf,n. 4.3
Assume that
x∈X, h∈Vf, n, infhx, y:y∈X>−2−1δf,n. 4.4
By1.3,4.3, and4.4,
infgf,nx, y:y∈X≥infhx, y:y∈X−4−1δf,n >−δf,n. 4.5
In view of4.5andP4,
ρxf, 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 ∈E and 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
infξx, y:y∈X>−2−1δf,n, 4.10
we have
infgf,nx, y:y∈X>−2−1δf,n−4−1δf,n>−δf,n,
ρxf, x<4n−1< 8.
4.11
Thus we have shown that the following property holds.
ByP1there is a sequence{xi}∞i1⊂Xsuch that
lim inf i→∞
infξxi, y:y∈X≥0. 4.12
In view of4.12andP6for all large enough natural numbersi, j, we have
ρxi, xj≤ρxi, xfρxf, 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
ξxξ, y
lim
i→∞ξ
xi, y
≥0. 4.15
We have also shown that any sequence {xi}∞i1 ⊂ X satisfying 4.12converges. This implies that ifx∈Xsatisfiesξx, y≥0 for ally∈X, thenxxξ. ByP6and4.15,
ρxξ, xf≤
8. 4.16
Let x ∈ X andh ∈ Vf, nsatisfy4.4. ByP5,ρxf, x < 4n−1. Together with4.16, this implies that
ρx, xξ≤ρx, xfρxf, xξ<4n−1
8 < . 4.17
Theorem 1.1is proved.
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