ZASLAVSKIReceived 17 November 2005; Revised 19 March 2006; Accepted 24 March 2006 We study a class of vector minimization problems on a complete metric space such that all its bounded cl
Trang 1ALEXANDER J ZASLAVSKI
Received 17 November 2005; Revised 19 March 2006; Accepted 24 March 2006
We study a class of vector minimization problems on a complete metric space such that all its bounded closed subsets are compact We show that for most (in the sense of Baire category) problems in the class the sets of minimal values are infinite
Copyright © 2006 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 vector optimization problems has recently been a rapidly growing area of research See, for example, [1–5] and the references mentioned therein In this paper, we study a class of vector minimization problems on a complete metric space such that all its bounded closed subsets are compact This class of problems is associated with a complete metric space of continuous vector functionsᏭ defined below For each F from Ꮽ, we
denote byv(F) the set of all minimal elements of the image F(X) = { F(x) : x ∈ X } In this paper, we will study the setsv(F) with F ∈Ꮽ It is clear that for a minimization problem with only one criterion the set of minimal values is a singleton In the present paper, we will show that for mostF ∈ Ꮽ (in the sense of Baire category) the sets v(F) are infinite.
Such approach is often used in many situations when a certain property is studied for the whole space rather than for a single element of the space See, for example, [7,8] and the references mentioned there Our results show that in general the setsv(F), F ∈Ꮽ, are rather complicated Note that in our paper as in many other works on optimization theory [1–6] inequalities are of great use
In this paper, we use the convention that∞ / ∞ =1 and denote by Card(E) the
cardi-nality of the setE.
LetRbe the set of real numbers and letn be a natural number Consider the
finite-dimensional spaceRnwith the Chebyshev norm
x =x1, , x n =maxx i: =1, , n
, x =x1, , x n
∈ R n (1.1)
Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2006, Article ID 54027, Pages 1 14
DOI 10.1155/JIA/2006/54027
Trang 22 A generic result
Let{ e1, , e n }be the standard basis inR n:
e1=(1, 0, , 0), , e n =(0, , 0, 1). (1.2) Letx =(x1, , x n),y =(y1, , y n)∈ R n We equip the spaceR nwith the natural order and say that
x ≥ y ifx i ≥ y i ∀ i ∈ {1, , n },
x > y ifx ≥ y, x = y,
x y ifx i > y i ∀ i ∈ {1, , n }
(1.3)
We say thatx y (resp., x < y, x ≤ y) if y x (resp., y > x, y ≥ x).
Let (X, ρ) be a complete metric space such that each of its bounded closed subsets is
compact Fixθ ∈ X.
Denote byᏭ the set of all continuous mappings F =(f1, , f n) :X → R nsuch that for alli ∈ {1, , n },
lim
For eachF =(f1, , f n),G =( 1, , g n)∈Ꮽ, set
d(F, G) =supf i(x) − g i(x):x ∈ X, i =1, , n
,
d(F, G) = d(F, G)
1 +d(F, G) −1
.
(1.5)
Clearly, the metric space (Ꮽ,d) is complete.
Note thatd(F, G) =sup{ F(x) − G(x) :x ∈ X }for allF, G ∈Ꮽ
LetA ⊂ R nbe a nonempty set An elementx ∈ A is called a minimal element of A if
there is noy ∈ A for which y < x.
Let F ∈ Ꮽ A point x ∈ X is called a point of minimum of F if F(x) is a minimal
element ofF(X) If x ∈ X is a point of minimum of F, then F(x) is called a minimal value
ofF Denote by M(F) the set of all points of minimum of F and put v(F) = F(M(F)).
The following proposition will be proved inSection 2
Proposition 1.1 Let F =(f1, , f n)∈ Ꮽ Then M(F) is a nonempty bounded subset of
(X, ρ) and for each z ∈ F(X) there is y ∈ v(F) such that y ≤ z.
In the sequel we assume thatn ≥2 and that the space (X, ρ) has no isolated points.
The following theorem is our main result It will be proved inSection 4
Theorem 1.2 There exists a setᏲ⊂ Ꮽ which is a countable intersection of open everywhere dense subsets of Ꮽ such that for each F ∈ Ᏺ the set v(F) is infinite.
It is clear that ifX is a finite-dimensional Euclidean space, then X is a complete metric
space such that all its bounded closed subsets are compact andTheorem 1.2holds It is also clear thatTheorem 1.2holds ifX is a convex compact subset of a Banach space or if
X is a convex closed cone generated by a convex compact subset of a Banach space which
does not contain zero
Trang 32 Proof of Proposition 1.1
Letz ∈ F(X) We show that there is y ∈ v(F) such that y ≤ z Set
Ω0=h ∈ F(X) : h ≤ z
We consider the setΩ0with the natural order and show thatΩ0has a minimal element
by using Zorn’s lemma
Assume thatD is a nonempty subset ofΩ0such that for eachh1,h2∈ D either h1≥ h2
orh1≤ h2 Since all bounded closed subsets ofX are compact, it follows from (1.4) that the setF(X) is bounded from below Together with (2.1) this implies that the setD is
bounded For each integeri ∈ {1, , n }, set
¯h i =inf
λ ∈ R: there isx =x1, , x n
∈ D for which x i = λ
(2.2) and set
¯h =¯h1, , ¯h n
Clearly, the vector ¯h is well defined.
Let p be a natural number By (2.2) and (2.3) for each natural number j ∈ {1, , n }
there exists
z(p, j) = z1(p, j), , z n(p, j) ∈ D (2.4) such that
¯h j ≥ z(j p, j) −1
It is clear that there is
z(p) ∈z(p, j): =1, , n
(2.6) such that
It follows from (2.5), (2.7), (2.2), (2.6), and (2.4) that for eachj =1, , n,
¯h j ≤ z(j p) ≤ ¯h j+ 1
By (2.6), (2.4), and (2.1) for each integerp ≥1, there isx p ∈ X such that
F
x p
If the sequence { x p } ∞
p=1 is unbounded, then in view of (2.9) and (1.4) the sequence
{ z(p) } ∞
p=1is also unbounded and this contradicts (2.8) Therefore the sequence{ x p } ∞
p=1
Trang 44 A generic result
is bounded Since any bounded closed set in (X, ρ) is compact, there is a subsequence
{ x p i } ∞
i=1of the sequence{ x p } ∞
p=1which converges to some point ¯x ∈ X In view of (2.8) and (2.9)
F( ¯x) =lim
i→∞ F
x p i
=lim
and ¯h ∈ F(X) Together with (2.1) and (2.2) this implies that ¯h ∈Ω0 Definition (2.2) implies that ¯h ≤ h for all h ∈ D By Zorn’s lemma there is a minimal element y ∈ F(X)
such thaty ≤ z This completes the proof ofProposition 1.1
3 Auxiliary results
Proposition 3.1 Let F =(f1, , f n)∈ Ꮽ and let Card(v(F)) = p, where p is a natural number Then there is a neighborhood W of F in ( Ꮽ,d) such that Card(v(G)) ≥ p for each
G =( 1, , g n)∈ W.
Proof Let
y i = y j for each (i, j) ∈Ω := {1, , p } × {1, , p } \(i, i) : i =1, , p
For eachi ∈ {1, , p }, there isx i ∈ X such that
F
x i
By (3.2) and (3.3) for each (i, j) ∈ Ω, there is p(i, j) ∈ {1, , n }such that
f p(i, j)
x i
> f p(i, j)
x j
Choose > 0 such that
f p(i, j)
x i
> f p(i, j)
x j
for all (i, j) ∈Ω Set
Let
G =g1, , g n
For eachi ∈ {1, , p }, we haveG(x i)∈ G(X) and it follows fromProposition 1.1 that there is
such that
¯y i ≤ G
x i
Trang 5Leti ∈ {1, , p } By (3.8) there is ¯x i ∈ X such that
G
¯x i
In view of (3.7) and (3.6)
G
¯x i
− F
It follows from (3.1), (3.2), the equality Card(v(F)) = p, andProposition 1.1that there is
k(i) ∈ {1, , p }such that
F
¯x i
By (3.3), (3.12), (3.11), (3.10), and (3.9)
F
x k(i)
= y k(i) ≤ F
¯x i
≤ G
¯x i
+(1, 1, , 1)
≤ G
x i
+(1, 1, , 1) ≤ F
x i
+ 2(1, 1, , 1). (3.13)
Together with (3.5) this implies thatk(i) = i Combined with (3.13), (3.10), and (3.3) this equality implies that
y i ≤ ¯y i+(1, 1, , 1) ≤ y i+ 2(1, 1, , 1). (3.14)
It follows from this inequality, (3.5), and (3.3) that
¯y i = ¯y j ifi, j ∈ {1, , p }satisfyi = j. (3.15)
Proposition 3.2 Assume that F =(f1, , f n)∈ Ꮽ, p is a natural number, Card(v(F)) =
p and that
v(F) =y1, , y p
, x i ∈ X, F
x i
= y i, i =1, , p,
y i = y j ∀ i, j ∈ {1, , p } satisfying i = j. (3.16) Then for each i =1, , p the inequality F(x i)≤ F(x) holds for all x belonging to a neighbor-hood of x i
Proof It is su fficient to consider the case with i =1 Clearly, for eachj ∈ {2, , n }, there
iss( j) ∈ {1, , n }such thatf s( j)(x1)< f s( j)(x j) Choose > 0 such that
f s( j)
x1
< f s( j)
x j
−2 ∀ j ∈ {2, , n } (3.17) There isδ > 0 such that for each x ∈ X satisfying ρ(x, x1)≤ δ we have
F(x) − F
x1 ≤
Trang 66 A generic result
Letx ∈ X satisfy ρ(x, x1)≤ δ Then (3.18) is true ByProposition 1.1there existsy ∈ v(F)
such that
In order to complete the proof it is sufficient to show that y= F(x1).
Let us assume the converse Then there is j ∈ {2, , n }such that y = y j = F(x j) By this relation, (3.18), and (3.19)
F
x j
= y j = y ≤ F(x) ≤ F
x1
+
2 (1, 1, , 1),
f s( j)
x j
≤ f s( j)
x1
+
2.
(3.20)
This contradicts (3.17) The contradiction we have reached provesProposition 3.2
Proposition 3.3 Assume that F =(f1, , f n)∈ Ꮽ, > 0, p is a natural number and that
Card
v(F)
= p, x1, , x p ∈ X, y i = F
x i
, i =1, , p, v(F) =y i: =1, , p
Then there exists G =( 1, , g n)∈ Ꮽ such that
f i(x) ≤ g i(x), x ∈ X, i =1, , n, g i
x j
= f i
x j
, i =1, , n, j =1, , p,
(3.22)
v(G) =G
x j
: =1, , p
(3.24)
and that for each x ∈ X \ { x1, , x p } there is j ∈ {1, , p } for which
G(x) ≥ G
x j
+min
1,ρ
x, x i
: =1, , p
(1, 1, , 1). (3.25)
Proof For each x ∈ X and i =1, , n, set
g i(x) = f i(x) + min
1,ρ
x, x j
: =1, , p
(3.26) and setG =( 1, , g n) Clearly,G ∈Ꮽ,
g i(x) ≥ f i(x), x ∈ X, i =1, , n,
g i
x j
= f i
x j
for eachi ∈ {1, , n }and eachj ∈ {1, , p } (3.27)
andd(F, G) ≤ Therefore (3.22) and (3.23) hold.
Let j ∈ {1, , p } We will show thatG(x j)∈ v(G) Assume that x ∈ X and
G(x) ≤ G
x j
Trang 7By (3.22), (3.26), and (3.28)
F(x) ≤ F(x) + min
1,ρ
x, x i
: =1, , p
(1, 1, , 1) = G(x) ≤ G
x j
= F
x j
.
(3.29) Together with (3.21) this relation implies that
F(x) = F
x j
, x ∈x i: =1, , p
Thus
G
x j
: =1, , p
Assume that
x ∈ X \x1, , x p
ByProposition 1.1and (3.21) there is j ∈ {1, , p }such that
F
x j
Relations (3.22), (3.33), (3.26), and (3.32) imply that
G
x j
= F
x j
≤ F(x) < F(x) + min
1,ρ
x, x i
: =1, , p
(1, , 1) ≤ G(x) (3.34)
This relation implies that
G(x) > G
x j
+min
1,ρ
x, x i
: =1, , p
(1, 1, , 1) (3.35) andG(x) ∈ v(G) Together with (3.31) this relation implies (3.24) This completes the
Proposition 3.4 Assume that F =(f1, , f n)∈ Ꮽ, p is a natural number,
Card
v(F)
and that > 0 Then there exists G ∈ Ꮽ such that d(F, G) ≤ and Card(v(G)) = p + 1.
Proof Let
v(F) =y1, , y p
wherey1, , y p ∈ R n Clearly,
y i = y j for eachi, j ∈ {1, , p }such thati = j. (3.38) For eachi ∈ {1, , p }, there isx i ∈ X such that
F
x i
Trang 88 A generic result
ByProposition 3.3there existsF(1)=(f1(1), , f1(n))∈Ꮽ such that
f i(1)(x) ≥ f i(x) ∀ x ∈ X, i =1, , n, (3.40)
f i(1)
x j
= f i
x j
, i =1, , n, j =1, , p, (3.41)
d
F, F(1)
≤
v
F(1)
=F(1)
x j
: =1, , p
(3.43) and that for eachx ∈ X \ { x1, , x p }there isj ∈ {1, , p }such that
F(1)(x) ≥ F(1)
x j
+min
1,ρ
x, x i
: =1, , p
(1, 1, , 1). (3.44)
It is clear that there exists a positive number
0< min {1,}
and that for eachi, j ∈ {1, , p }satisfyingi = j there exists s(i, j) ∈ {1, , n }such that
f s(i, j)(1)
x i
< f s(i, j)(1)
x j
Chooseδ0∈(0, 1/8) such that
ρ
x i,x j
≥8 0 for eachi, j ∈ {1, , p }satisfyingi = j. (3.47) There isδ1∈(0,δ0/2) such that for each x ∈ X satisfying ρ(x1,x) ≤2 1
F(1)
x1
− F(1)(x) ≤ 0
Put
1= 0
There isx0∈ X such that
0< ρ
x0,x1
By (3.50) and (3.47)
x0∈x i: =1, , p
Choose a positive number
2< min 0ρ
x0,x1
Trang 9
Choose a positive numberδ2such that
4 2< ρ
x0,x1
f i(1)
x0
− f i(1)(x)≤ 2
4 for eachi ∈{1, , n }and eachx ∈ X satisfying ρ
x, x0
≤4 2.
(3.54) Choose a positive numberλ such that
Set
g1(x) = f1(1)(x) for eachx ∈ X satisfying ρ
x, x0
> 2δ2, (3.56)
g1(x) =min
f1(1)(x), f1(1)
x0
− 1+λρ
x, x0
for eachx ∈ X satisfying ρ
x, x0
≤2 2.
(3.57) Fori ∈ {2, , n }, set
g i(x) = f i(1)(x) for eachx ∈ X satisfying ρ
x, x0
> 2δ2, (3.58)
g i(x) =min f i(1)(x), f i(1)
x0
− 2+λρ
x, x0
for eachx ∈ X satisfying ρ
x, x0
≤2 2.
(3.59) SetG =( 1, , g n) By (3.54) and (3.55) for eachi ∈ {1, , n }and eachx ∈ X satisfying
δ2≤ ρ(x, x0)≤2 2,
f i(1)
x0
− 1+λρ
x, x0
≥ f1
i
x0
− 1+λδ2
≥ f i1
x0
+1+ 22≥ f i(1)(x) − 2/4 + 1+ 22.
(3.60)
In view of (3.60), (3.57), and (3.59) for eachi ∈ {1, , n }and eachx ∈ X satisfying δ2≤
ρ(x, x0)≤2 2,
Together with (3.56)–(3.59) this implies thatG is continuous By (3.56) and (3.58)G ∈Ꮽ Relations (3.61), (3.56), and (3.58) imply that for eachx ∈ X satisfying ρ(x0,x) ≥ δ2we have
By (3.57) and (3.59) for eachx ∈ X satisfying ρ(x0,x) ≤2 2, we have
Letx ∈ X satisfy
ρ
x, x
Trang 1010 A generic result
By (3.56)–(3.59), (3.52), (3.64), and (3.54) for eachi ∈ {1, , n },
f i(1)(x) ≥ g i(x) ≥min
f i(1)(x), f i(1)
x0
− 1
≥min f i(1)(x), f i(1)(x) − 2
4 − 1
≥ f i(1)(x) −
5
4
1,
F(1)(x) ≥ G(x) ≥ F(1)(x) −
2
(1, 1, , 1).
(3.65)
Together with (3.62) this inequality implies thatd(F (1),G) ≤ /2 Combined with (3.42) this implies that
d(F, G) ≤ d
F, F(1)
+dF(1),G
<
4+
Letx ∈ X We show that there exists j ∈ {0, , p }such thatG(x) ≥ G(x j) There are two cases:
ρ
x, x0
ρ
x, x0
Assume that (3.67) holds Then by (3.62)G(x) = F(1)(x) In view ofProposition 1.1and (3.43) there is j ∈ {1, , p }such that
F(1)
x j
Ifj =1, then (3.53) implies that
ρ
x j,x0
= ρ
x0,x1
Ifj =1, then by (3.47) and (3.50)
ρ
x j,x0
≥ ρ
x j,x1
− ρ
x1,x0
≥8 0− δ1≥7 0> 4δ2. (3.71) Thus in both casesρ(x j,x0)> 4δ2 In view of this inequality and (3.62),
F(1)
x j
= G
x j
Together with (3.69) this equality implies thatG(x j)≤ G(x) Assume that (3.68) holds
We will show thatG(x0)≤ G(x) Relations (3.57) and (3.59) imply that
G
x0
= f1(1)
x0
− 1, 2(1)
x0
− 2, , f n(1)
x0
− 2 = F(1)
x0
−1,2, , 2
.
(3.73)
By (3.68)
F(1)(x) ≥ F(1)
x0
− 2
Trang 11By (3.68), (3.57), (3.59), (3.74), and (3.52)
g1(x) ≥min f1(1)
x0
− 2
4,
(1) 1
x0
− 1
= f1(1)
x0
and fori ∈ {1, , p } \ {1},
g i(x) ≥min f i(1)
x0
− 2
4,
(1)
i
x0
− 2
= f i(1)
x0
Together with (3.73) these inequalities imply thatG(x) ≥ G(x0) Thus we have shown that for eachx ∈ X there is j ∈ {0, , p }such thatG(x) ≥ G(x j)
Now assume that j1, 2∈ {0, , p }satisfy
G
x j1
≤ G
x j2
We will show that j1= j2 In view of (3.47) and (3.50) for eachi ∈ {2, , p },
ρ
x i,x0
≥ ρ
x i,x1
− ρ
x0,x1
≥8 0− δ1> 7δ0> 4δ2. (3.78)
By (3.53)ρ(x0,x1)> 4δ2 Therefore, for eachi ∈ {1, , p },
ρ
x i,x0
Together with (3.56) and (3.58)
G
x i
= F(1)
x i
If j1, 2∈ {1, , p }, then in view of (3.77), (3.80), and (3.46)F(1)(x j1)= F(1)(x j2) and
j1= j2 Therefore we may consider only the case with 0∈ { j1, 2} Leti ∈ {1, , p } \ {1}
By (3.80)
G
x i
= F(1)
x i
By (3.46)
f s(i,1)(1) (x i)< f s(i,1)(1)
x1
−80, f s(i,1)(1)
x1
< f s(1,i)(1)
x i
−80. (3.82)
It follows from (3.81), (3.82), (3.48), (3.50), (3.49), (3.57), and (3.59) that
g s(i,1)
x i
= f s(i,1)(1)
x i
< f s(i,1)(1)
x1
−80
≤ f s(i,1)(1) (x0) +0
4 −80< f s(i,1)(1)
x0
−21≤ g s(i,1)
x0
− 1,
g s(i,1)
x i
= f s(i,1)(1)
x i
> f s(1,i)(1)
x1
+ 80≥ f s(i,1)(1)
x0
− 0
4 + 80> g s(1,i)
x0
+ 70.
(3.83)
These inequalities imply that the inequalityG(x i)≤ G(x0) does not hold and that the inequalityG(x0)≤ G(x i) does not hold too Together with (3.77) and the inclusion 0∈ { j1, 2}this implies that
j1,
... Trang 1010 A generic result< /p>
By (3.56)–(3.59), (3.52), (3.64), and (3.54) for eachi ∈... 4δ2. (3.71) Thus in both casesρ(x j,x0)> 4δ2 In view of this inequality and (3.62),
F(1)...
(3.83)
These inequalities imply that the inequalityG(x i)≤ G(x0) does not hold and that the inequalityG(x0)≤