Báo cáo hóa học: " Research Article Generic Well-Posedness for a Class of Equilibrium Problems" pdf
Bạn đang xem bản rút gọn của tài liệu. Xem và tải ngay bản đầy đủ của tài liệu tại đây (479.27 KB, 9 trang )
Hindawi Publishing Corporation
Journal of Inequalities and Applications
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,
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 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 for a generic typical element of A 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 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
≥ 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, for a generic 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-posedness for equilibrium 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. An Introduction to the Orbit Structure of Diffeomorphisms, MIT Press,
Cambridge, Mass, USA, 1971.
6 A. J. Zaslavski, Turnpike Properties in the 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,” SIAM Journal on Control and Optimization, vol. 38, no. 2, pp. 566–581, 2000.
8 A. J. Zaslavski, “Generic well-posedness of optimal control problems without convexity assumptions,”
SIAM Journal on Control and Optimization, vol. 39, no. 1, pp. 250–280, 2000.
9 T. Zolezzi, “Well-posedness criteria in optimization with application to the calculus of variations,”
Nonlinear Analysis: Theory, Methods & Applications, vol. 25, no. 5, pp. 437–453, 1995.
