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Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2010, Article ID 543061, 17 pages
doi:10.1155/2010/543061
Research Article
Solvability Criteria for Some Set-Valued
Inequality Systems
Yingfan Liu
Department of Mathematics, College of Science, Nanjing University of Posts and Telecommunications,
Nanjing 210009, China
Correspondence should be addressed to Yingfan Liu,
Received 23 May 2010; Accepted 9 July 2010
Academic Editor: Qamrul Hasan Ansari
Copyright q 2010 Yingfan Liu. 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.
Arising from studying some multivalued von Neumann model, three set-valued inequality
systems are introduced, and two solvability questions are considered. By constructing some
auxiliary functions and studying their minimax and saddle-point properties, solvability criteria
composed of necessary and sufficient conditions regarding these inequality systems are obtained.
1. Introduction
Arising from considering some multivalued von Neumann model, this paper aims to study
three set-valued inequality systems and try to find their solvability criteria. Before starting
with this subject, we need to review some necessary backgrounds as follows.
We denote by R
k
R
k
, · the k-dimensional Euclidean space, R
k∗
R
k
its dual, and
·, · the duality pairing on R
k∗
,R
k
; moreover, we denote that R
k
{x ∈ R
k
: x
i
≥ 0 ∀i} and
int R
k
is its interior. We also define ≥ or > in R
k
by x ≥ y ⇔ x −y ∈ R
k
or by x>y⇔ x−y ∈
int R
k
.
It is known that the generalized linear or nonlinear von Neumann model, which is
composed of an inequality system and a growth f actor problem described by
a
∃x ∈ X ⇒ Bx − Ax ≥ c,
b
λ>1s.t. ∃x ∈ X ⇒ Bx ≥ λAx c,
respectively, 1.1
is one of the most important issues in the input-output analysis 1–3, where c ∈ R
m
, X ⊆ R
n
m may not be equal to n,andB, A are two nonnegative or positive maps from X to R
m
.
2 Journal of Inequalities and Applications
A series of researches on 1.1 have been made by the authors of 1–5 for the linear case
i.e., B, A are m × n matrices and by the authors of 6, 7 for the nonlinear case i.e., B, A are
some types of nonlinear maps. Since a or b of 1.1 is precisely a special example of the
inequality λ ∈ 1, ∞ s.t. ∃x ∈ X ⇒ S
λ
xB − λAx Bx − λAx ≥ c if we restrict λ 1or
λ>1, it is enough for 1.1 to consider the inequality system. This idea can be extended to the
set-valued version. Indeed, if B and A are replaced by set-valued maps G and F, respectively,
then 1.1 yields a class of multivalued von Neumann model, and it solves a proper set-valued
inequality system to study. With this idea, by 8as a set-valued extension to 6, 7 we have
considered the following multivalued inequality system:
c ∈ R
m
s.t.
∃x ∈ X, ∃y ∈ Tx ⇒ y ≥ c
1.2
and obtained several necessary and sufficient conditions for its solvability, where X ⊂ R
n
and T : X → 2
R
m
is a class of set-valued maps from X to R
m
. Along the way, three further
set-valued inequality systems that we will study in the sequel can be stated as follows.
Let X, T be as above, and let G, F : X → 2
R
m
be set-valued maps from X to R
m
, then
we try to find the solvability criteria i.e., the necessary and sufficient conditions that c ∈ R
m
solves
∃x ∈ X, ∃y ∈ Tx,
∃i
0
∈
{
1, 2, ,m
}
⇒ y ≥ c, y
i
0
c
i
0
,
1.3
∃x ∈ X,
∃y ∈
G − F
x ⇒ y ≥ c,
1.4
or
∃x ∈ X, ∃y ∈
G − F
x,
∃i
0
∈
{
1, 2, ,m
}
⇒ y ≥ c, y
i
0
c
i
0
.
1.5
When T and G, F are single-valued maps, then 1.3–1.5 return to the models of 6,
7. When T and G, F are set-valued maps, there are three troubles if we try to obtain some
meaningful solvability criteria regarding 1.3–1.5 just like what we did in 8.
1 For 1.2 and 1.3, it is possible that only 1.2 has solution for some c ∈ R
m
.
Indeed, if X is compact and T is continuous, compact valued with TX ⊂ int R
m
, then TX is
compact and there is c ∈ R
m
with c<yfor y ∈ TX. Hence c solves 1.2 but does not solve
1.3.
2 It seems that the solvability criteria namely, necessary and sufficient results
concerning existence to 1.4 can be obtained immediately by 8 with the replacement
T G − F. However, this type of result is trivial because it depends only on the property
of G − F but not on the respective information of G and F. This opinion is also applicable to
1.3 and 1.5.
3 Clearly, 1.3or 1.5 is more fine and more useful than 1.2or 1.4. However,
the method used for 1.2 in 8or the possible idea for 1.4 to obtain solvability criteria
fails to be applied to find the similar characteristic results for 1.3or 1.5
because there are
some examples see Examples 3.5 and 4.4 to show that, without any additional restrictions,
Journal of Inequalities and Applications 3
no necessary and sufficient conditions concerning existence for them can be obtained. This is
also a main cause that the author did not consider 1.3 and 1.5 in 8.
So some new methods should be introduced if we want to search out the solvability
criteria to 1.3–1.5. In the sections below, we are devoted to study 1.3–1.5 by considering
two questions under two assumptions as follows:
Question 1. Whether there exist any criteria that c solves 1.3 in some proper way?
Question 2. Like Question 1, whether there exist any solvability criteria to 1.4 or 1.5 that
depend on the respective information of G and F?
Assumption 1. c ∈ R
m
is a fixed point and X ⊂ R
n
is a convex compact subset.
Assumption 2. Consider the following: T : X → 2
R
m
,G : X → 2
R
m
, and F : X → 2
int R
m
are
upper semicontinuous and convex set-valued maps with nonempty convex compact values.
By constructing some functions and studying their minimax properties, some progress
concerning both questions has been made. The paper is arranged as follows. We review some
concepts and known results in Section 2 and prove three Theorems composed of necessary
and sufficient conditions regarding the solvability of 1.3–1.5 in Sections 3 and 4. Then we
present the conclusion in Section 5.
2. Terminology
Let P ⊆ R
m
, X ⊆ R
n
,andY
i
⊆ R
m
i
i 1, 2.Letf,f
α
: X → R α ∈ Λ, ϕ ϕp, x : P × X →
R,andψ ψp, u, v : P × Y
1
× Y
2
→ R be functions and T : X → 2
R
m
a set-valued map.
We need some concepts concerning f, f
α
α ∈ Λ and ϕ and ψ such as convex or concave and
upper or lower semicontinuous in short, u.s.c. or l.s.c. and continuous i.e., both u.s.c. and
l.s.c., whose definitions can be found in 9–11, therefore, the details are omitted here. We
also need some further concepts to T, ϕ,andψ as follows.
Definition 2.1. 1 T is said to be closed if its graph defined by graph T {u, v ∈ X × R
m
:
u ∈ X, v ∈ Tx} is closed in R
n
× R
m
. Moreover, T is said to be upper semicontinuous in short,
u.s.c. if, for each x ∈ X and each neighborhood V Tx of Tx, there exists a neighborhood
Ux of x such that TUx ∩ X ⊆ V Tx.
2 Assume that Y ⊆ R
m
Y
/
∅,anddefineσ
#
Y, psup
y∈Y
p, y, σ
Y, p
inf
y∈Y
p, y p ∈ R
m
. Then T is said to be upper hemicontinuous in short, u.h.c.
if x → σ
#
Tx,p is u.s.c. on X for any p ∈ R
n
.
3 T is said to be convex if X ⊆ R
n
is convex and αTx
1
1 − αTx
2
⊆ Tαx
1
1 −
αx
2
for any α ∈ 0, 1 and x
i
∈ X i 1, 2.
4a If inf
p∈P
sup
x∈X
ϕp, xsup
x∈X
inf
p∈P
ϕp, x, then one claims that the minimax
equality of ϕ holds. Denoting by vϕ the value of the preceding equality,
one also says that the minimax value vϕ of ϕ exists. If
p, x ∈ P × X
such that sup
x∈X
ϕp, xinf
p∈P
ϕp, x, then one calls p, x a saddle point
of ϕ. Denote by Sϕ the set of all saddle points of ϕ i.e., Sϕ{
p, x ∈
P × X :sup
x∈X
ϕp, xinf
p∈P
ϕp, x},anddefineSϕ|
X
{x ∈ X : ∃ p ∈
P s.t.
p, x ∈ Sϕ}, the restriction of Sϕ to X if Sϕ is nonempty.
4 Journal of Inequalities and Applications
b Replacing X by Y
1
× Y
2
and ϕp, x by ψp, u, v, with the similar method one
can also define vψthe minimax value of ψ, Sψthe saddle-point set of
ψ,andSψ|
Y
1
×Y
2
the restriction of Sψ to Y
1
× Y
2
.
5 If Y is a convex set and A asubsetofY, one claims that A is an extremal subset of
Y if x,y ∈ Y and tx 1 − ty ∈ A for some t ∈ 0, 1 entails x,y ∈ A. x
0
∈ Y is an
extremal point of Y if A {x
0
} is an extremal subset of Y, and the set of all extremal
points of Y is denoted by ext Y.
Remark 2.2. 1 Since p ∈ R
m
⇔−p ∈ R
m
and σ
#
Tx,−p−σ
Tx,p, we can see that
T : X ⊂ R
n
→ 2
R
m
is u.h.c. if and only if x → σ
Tx,p is l.s.c. on X for any p ∈ R
m
.
2 For the function ϕ ϕp, x on P ×X, vϕ exists if and only if inf
p∈P
sup
x∈X
ϕp, x ≤
sup
x∈X
inf
p∈P
ϕp, x,andp, x ∈ Sϕ if and only if sup
x∈X
ϕp, x ≤ inf
p∈P
ϕp, x if and
only if ϕ
p, x ≤ ϕp, x ≤ ϕp, x for any p, x ∈ P × X.IfSϕ
/
∅, then vϕ exists, and
vϕϕ
p, xsup
x∈X
ϕp, xinf
p∈P
ϕp, x for any p, x ∈ Sϕ. The same properties are
also true for ψ ψp, u, v on P × Y
1
× Y
2
. Moreover, we have
∀
x ∈ S
ϕ
|
X
, inf
p∈P
ϕ
p, x
v
ϕ
,
∀
u, v
∈ S
ψ
|
Y
1
×Y
2
, inf
p∈P
ψ
p,
u, v
v
ψ
.
2.1
We also need three known results as follows.
Lemma 2.3. 1 (see [9]) If T is u.s.c., then T is u.h.c.
2 (see [9]) If T is u.s.c. with closed values, then T is closed.
3 (see [9]) If
TX (the closure of TX) is compact and T is closed, then T is u.s.c.
4 If X ⊂ R
n
is compact and T : X → 2
R
m
is u.s.c. with compact values, then TX is compact
in R
m
.
5 If X is convex (or compact) and T
1
,T
2
: X ⊂ R
n
→ 2
R
m
are convex (or u.s.c. with compact
values), then αT
1
βT
2
are also convex (or u.s.c.) for all α, β ∈ R.
Proof. We only need to prove 5.
a If T
i
i 1, 2 are convex, α, β ∈ R, x
i
i 1, 2 ∈ X,andt ∈ 0, 1, then
αT
1
βT
2
tx
1
1 − t
x
2
αT
1
tx
1
1 − t
x
2
βT
2
tx
1
1 − t
x
2
⊇ α
tT
1
x
1
1 − t
T
1
x
2
β
tT
2
x
1
1 − t
T
2
x
2
t
αT
1
βT
2
x
1
1 − t
αT
1
βT
2
x
2
.
2.2
Hence αT
1
βT
2
is convex.
b Now we assume that X is compact.
In case T : X → 2
R
m
is u.s.c. with compact values and α ∈ R, then by 2, 4, T is
closed and the range αTX of αT is compact. If α 0, then αTx 0 for any x ∈ X; hence,
αT is u.s.c. If α
/
0, supposing that x
j
,y
j
∈ graphαT with x
j
,y
j
→ x
0
,y
0
j →∞,
then x
j
,y
j
/α ∈ graph T such that x
j
,y
j
/α → x
0
,y
0
/α as j →∞, which implies t hat
y
0
∈ αTx
0
. Hence, αT is closed and also u.s.c. because of 3.
Journal of Inequalities and Applications 5
In case T
i
i 1, 2 : X → 2
R
m
are u.s.c. with compact values, if x
k
,y
k
∈ graphT
1
T
2
with x
k
,y
k
→ x
0
,y
0
k →∞, then x
0
∈ X and there exist u
k
∈ T
1
x
k
, v
k
∈ T
2
x
k
such
that y
k
u
k
v
k
for all k 1, 2, By 4, T
1
X and T
2
X are compact, so we can suppose
u
k
→ u
0
and v
k
→ v
0
as k →∞.By2,bothT
i
i 1, 2 are closed, this implies that
y
0
u
0
v
0
∈ T
1
T
2
x
0
,andthusT
1
T
2
is closed. Hence by 3, T
1
T
2
is u.s.c. because
T
1
T
2
X
x∈X
T
1
T
2
x ⊆ T
1
X T
2
X and T
1
X T
2
X is compact.
Lemma 2.4 see 8, Theorems 4.1and4.2. Let X ⊂ R
n
,P ⊂ R
m
be convex compact with R
P
R
m
, Σ
m−1
{p ∈ R
m
: Σ
m
i1
p
i
1}, and c ∈ R
m
. Assume that T : X → 2
R
m
is convex and
u.s.c. with nonempty convex compact values, and define ϕp, xφ
c
p, x on P × X by φ
c
p, x
sup
y∈Tx
p, y − c for p, x ∈ P × X.Then
1 vφ
c
exists and Sφ
c
is a convex compact subset of P × X,
2 c solves 1.2 ⇔ vφ
c
≥ 0 ⇔ φ
c
p, x ≥ 0 for p, x ∈ Sφ
c
.
In particular, both (1) and (2) are also true if P Σ
m−1
.
Lemma 2.5. (1) (see [10, 11]) If x → f
α
x is convex or l.s.c. (resp., concave or u.s.c.) on X for
α ∈ Λ and sup
α∈Λ
f
α
x (resp., inf
α∈Λ
f
α
x) is finite for x ∈ X,thenx → sup
α∈Λ
f
α
x (resp.,
x → inf
α∈Λ
f
α
x) is also convex or l.s.c. (resp., concave or u.s.c.) on X.
2 (see [11]) If g : X×Y ⊂ R
n
×R
m
→ R is l.s.c. (or u.s.c.) and Y is compact, then h : U → R
defined by hxinf
y∈Y
gx, y (or k : U → R defined by kxsup
y∈Y
gx, y)isalso
l.s.c. (or u.s.c.).
3 (see [9–11], Minimax Theorem) Let P ⊂ R
m
, X ⊂ R
n
be convex compact, and let ϕp, x
be defined on P × X. If, for each x ∈ X, p → ϕp, x is convex and l.s.c. and, for each p ∈
P, x → ϕp, x is concave and u.s.c., then inf
p∈P
sup
x∈X
ϕp, xsup
x∈X
inf
p∈P
ϕp, x
and there exists
p, x ∈ P × X such that sup
x∈X
ϕp, xinf
p∈P
ϕp, x.
3. Solvability Theorem to 1.3
Let Σ
m−1
be introduced as in Lemma 2.4, and define the functions φ
c
p, x on Σ
m−1
× X and
φ
c,x
p, y on Σ
m−1
× Tx x ∈ X by
a
φ
c
p, x
σ
#
Tx − c, p
sup
y∈Tx
p, y − c
for
p, x
∈ Σ
m−1
× X,
b
φ
c,x
p, y
p, y − c
for
p, y
∈ Σ
m−1
× Tx, x ∈ X.
3.1
Remark 3.1. By both Assumptions in Section 1, Definition 2.1, and Lemmas 2.4 and 2.5,we
can see that
1 φ
c
p, xsup
y∈Tx
φ
c,x
p, y for all p, x ∈ P × X,
2 vφ
c
and vφ
c,x
exist, and Sφ
c
and Sφ
c,x
are nonempty,
3 c solves 1.2 if and only if vφ
c
≥ 0 if and only if p, x ∈ Sφ
c
with φ
c
p, x ≥ 0.
Hence, Sφ
c
|
X
and Sφ
c,x
|
Tx
x ∈ X are nonempty. Moreover, we have the following.
6 Journal of Inequalities and Applications
Theorem 3.2. For 1.3, the following three statements are equivalent to each other:
1 vφ
c
0,
2 for all
x ∈ Sφ
c
|
X
, for all y ∈ Sφ
c,x
|
Tx
, ∃i
0
∈{1, 2, ,m}⇒y ≥ c, y
i
0
c
i
0
,
3 ∃
x ∈ Sφ
c
|
X
, ∃y ∈ Sφ
c,x
|
Tx
, ∃i
0
∈{1, 2, ,m}⇒y ≥ c, y
i
0
c
i
0
.
Remark 3.3. Clearly, each of 2 and 3 implies that c solves 1.3 because
x ∈ X and y ∈ Tx.
So we conclude from Theorem 3.2 that c solves 1.3 in the way of 2 or in the way of 3 if
and only if vφ
c
0.
Proof of Theorem 3.2. We only need to prove 1⇒2 and 3⇒1.
1⇒2. Assume that 1 holds. By 3.1 and Remarks 2.2 and 3.1, it is easy to see that
∀
x ∈ S
φ
c
|
X
, inf
p∈Σ
m−1
sup
y∈T x
p, y − c inf
p∈Σ
m−1
φ
c
p,
x
v
φ
c
0,
∀
y ∈ S
φ
c,x
|
Tx
, inf
p∈Σ
m−1
p, y − c v
φ
c,x
inf
p∈Σ
m−1
sup
y∈T x
p, y − c.
3.2
Then for each
x ∈ Sφ
c
|
X
and each y ∈ Sφ
c,x
|
Tx
, we have
inf
p∈Σ
m−1
p, y − c v
φ
c,x
v
φ
c
0.
3.3
By taking p
i
e
i
i
0, ,0, 1, 0, ,0 ∈ Σ
m−1
i 1, 2, ,m, it follows that y
i
≥ c
i
i
1, 2, ,m. Hence,
y ≥ c. On the other hand, it is easy to verify that Σ
∗
{p ∈ Σ
m−1
:
p,
y − c 0} is a nonempty extremal subset of Σ
m−1
. The Crain-Milmann Theorem see
12 shows that ext Σ
∗
is nonempty with ext Σ
∗
⊂ ext Σ
m−1
{e
1
,e
2
, ,e
m
}. So there exists
i
0
∈{1, 2, ,m} such that p e
i
0
∈ ext Σ
∗
. This implies by 3.3 that y
i
0
c
i
0
, and therefore
2 follows.
3⇒1.Let
x ∈ Sφ
c
|
X
, y ∈ Sφ
c,x
|
Tx
,andleti
0
be presented in 3. Since c solves
1.3 and also solves 1.2,by3.1 and Remarks 2.2 and 3.1,weobtain
0 e
i
0
, y − c≥ inf
p∈Σ
m−1
p, y − c
v
φ
c,
x
inf
p∈Σ
m−1
sup
y∈T x
φ
c,
x
p, y
inf
p∈Σ
m−1
φ
c
p,
x
v
φ
c
≥ 0,
3.4
where e
i
0
i
0
0, ,0, 1, 0, ,0 ∈ Σ
m−1
. Hence, vφ
c
0 and the theorem follows.
Remark 3.4. From the Theorem, we know that vφ
c
0 implies that c solves 1.3. However,
without any additional restricting conditions, the inverse may not be true.
Journal of Inequalities and Applications 7
Example 3.5. Let X 0, 1
2
,cs, ss ∈ 0, 1,andletT : X → 2
R
2
be defined by
Tx 1/2, 1
2
for x x
1
,x
2
∈ X. Then T is an u.s.c. and convex set-valued map with convex
compact values, and for each p p
1
,p
2
∈ Σ
1
, x x
1
,x
2
∈ X and c s, ss ∈ 0, 1,
φ
c
p, xσ
#
Tx − c, pp
1
p
2
− sp
1
p
2
1 − s. Hence, vφ
c
1 − s for all s ∈ 0, 1
and therefore,
v
φ
c
> 0andc solves
1.3
for s ∈
1
2
, 1
,
v
φ
c
> 0andc does not solve
1.3
for s ∈
0,
1
2
.
3.5
This implies that vφ
c
0 or vφ
c
> 0 may not be the necessary or the sufficient
condition that c solves 1.3.
4. Solvability Theorems to 1.4 and 1.5
Taking T G−F, then from both Assumptions, Lemma 2.4,andTheorem 3.2, we immediately
obtain the necessary and sufficient conditions to the solvability of 1.4 and 1.5. However,
just as indicated in Section 1, this type of result is only concerned with G − F. To get some
further solvability criteria to 1.4 and 1.5 depending on the respective information of G and
F, we define the functions H
c
p, x on Σ
m−1
×X and H
c,x
p, u, v on Σ
m−1
×Gx×Fxx ∈ X
by
a
H
c
p, x
σ
#
Gx, p
−
p, c
σ
Fx,p
for
p, x
∈ Σ
m−1
× X,
b
H
c,x
p,
u, v
p, u − c
p, v
for
p,
u, v
∈ Σ
m−1
×
Gx × Fx
,x∈ X.
4.1
By both Assumptions, we know that σ
#
Gx, psup
u∈Gx
p, u and σ
Fx,p
inf
v∈Fx
p, v are finite with σ
#
Gx, p ≥ 0andp, v≥σ
Fx,p > 0 v ∈ Fx for x ∈ X
and p ∈ Σ
m−1
, so the functions H
c
p, x and H
c,x
p, u, v x ∈ X defined by 4.1 are well
defined.
In view of Definition 2.1, we denote by vH
c
or vH
c,x
the minimax value of
ϕp, xH
c
p, xor ψp, u, v H
c,x
p, u, v if it exists, SH
c
or SH
c,x
the saddle
point set if it is nonempty, and SH
c
|
X
or SH
c,x
|
Gx×Fx
x ∈ X the restriction of SH
c
to
X or SH
c,x
to Gx × Fx. Then we have the solvability result to 1.4 and 1.5 as follows.
Theorem 4.1. i vH
c
exists if and only if SH
c
/
∅.
ii1 c solves 1.4 if and only if vH
c
exists with vH
c
≥ 1 if and only if SH
c
/
∅
with H
c
p, x ≥ 1 for p, x ∈ SH
c
.
2 In particular, if vH
c
exists with vH
c
≥ 1, then for each x ∈ SH
c
|
X
,thereexists
y ∈ G − Fx such that y ≥ c.
8 Journal of Inequalities and Applications
Theorem 4.2. For 1.5, the following three statements are equivalent to each other:
1 vH
c
1,
2 SH
c
/
∅, SH
c,x
/
∅ x ∈ SH
c
|
X
, and for all x ∈ SH
c
|
X
, for all u, v ∈
SH
c,x
|
Gx×Fx
, ∃i
0
∈{1, ,m}⇒u − v ≥ c, u
i
0
− v
i
0
c
i
0
,
3 SH
c
/
∅, SH
c,x
/
∅ x ∈ SH
c
|
X
, and ∃x ∈ SH
c
|
X
, ∃u, v ∈ SH
c,x
|
Gx×F x
,
∃i
0
∈{1, ,m}⇒u − v ≥ c, u
i
0
− v
i
0
c
i
0
.
That is, c solves 1.5 in the way of (2) or in the way of (3) if and only if vH
c
1.
Remark 4.3. It is also needed to point out that vH
c
1 is not the necessary condition of c
making 1.5 solvable without any other restricting conditions.
Example 4.4. Let X 0, 1
2
, c s, s ∈ R
2
s ∈ 5/12, 3/4,andG, F : X → 2
R
m
be defined
by Gx ≡ 3/4, 1
2
, Fx ≡ 1/4, 1/3
2
for x x
1
,x
2
∈ X. Then both G and F are u.s.c. convex
set-valued maps with convex compact values, and for any p p
1
,p
2
∈ Σ
1
, x x
1
,x
2
∈ X,
and c s, s ∈ R
2
, we have G − Fx Gx − Fx 3/4, 1
2
− 1/4, 1/3
2
5/12, 3/4
2
,
σ
#
Gx, p1, σ
Fx,p1/4, and p, c s. Therefore,
H
c
p, x
σ
#
Gx, p
−
p, c
σ
Fx,p
4
1 − s
for p ∈ Σ
1
,x∈ X,
v
H
c
exists with v
H
c
4
1 − s
> 1for
5
12
<s<
3
4
.
4.2
This implies that, for each c s, ss ∈ 5/12, 3/4, c solves 1.5 but vH
c
> 1.
The proof of both Theorems 4.1 and 4.2 can be divided into eight lemmas.
Let t ∈ R
, T
t
G − tF,andc ∈ R
m
. Consider the auxiliary inequality system
∃x ∈ X,
∃y ∈ T
t
x Gx − tFx ⇒ y ≥ c.
4.3
Then t ∈ R
solves 4.3 if and only if c solves 1.2 for T T
t
, and in particular, t 1 solves
4.3 if and only if c solves 1.4. Define ϕp, xK
t
p, x on Σ
m−1
× X by
K
t
p, x
σ
#
T
t
x − c, p
σ
#
Gx, p
− tσ
Fx,p
−p, c,
p, x
∈ Σ
m−1
× X,
4.4
denote by vK
t
the minimax value of ϕ K
t
if it exists, and denote by SK
t
the saddle point
set if it is nonempty. Then we have the following.
Lemma 4.5. 1 For each t ∈ R
0, ∞, vK
t
exists and SK
t
is nonempty. Moreover, t ∈ R
solves 4.3 if and only if vK
t
≥ 0 if and only if K
t
p, x ≥ 0 for p, x ∈ SK
t
.
2 The function t → vK
t
is continuous and strictly decreasing on R
with vK
∞
lim
t → ∞
vK
t
−∞.
Proof. 1 By both Assumptions and Lemmas 2.34 and 2.35, T
t
G−tF is convex and u.s.c.
with nonempty convex compact values for each t ∈ R
. Since K
t
p, xsup
y∈Gx−tFx
p, y − c
Journal of Inequalities and Applications 9
for p, x ∈ Σ
m−1
× X, applying Lemma 2.4 to T T
t
and substituting K
t
p, x for φ
c
p, x,we
know that 1 is true.
2 We prove 2 in three steps as follows.
a By Lemma 2.31, G and F are u.h.c., which implies by Definition 2.12 and
Remark 2.21 that for each p ∈ Σ
m−1
, t, x → K
t
p, xσ
#
Gx, p −
tσ
Fx,p −p, c is u.s.c. on R
× X. Then from Lemma 2.512,weknow
that both functions
t, x
−→ inf
p∈Σ
m−1
K
t
p, x
on R
× X,
and t −→ sup
x∈X
inf
p∈Σ
m−1
K
t
p, x
on R
are u.s.c.
4.5
Since GX and FX are compact by both Assumptions and Lemma 2.34, C
GX
sup
u∈GX
u and C
FX
sup
v∈FX
v are finite. Then for any x ∈ X, p, p
0
∈ Σ
m−1
,
and t, t
0
∈ R
, we have
σ
#
Gx, p
sup
u∈Gx
p
0
,u p − p
0
,u
≤ σ
#
Gx, p
0
p − p
0
C
GX
,
σ
Fx,p
inf
v∈Fx
p
0
,v p − p
0
,v
≥ σ
Fx,p
0
−p − p
0
C
FX
.
4.6
This implies that, for each x ∈ X,
σ
#
Gx, p
− σ
#
Gx, p
0
≤p − p
0
C
GX
,
σ
Fx,p
− σ
Fx,p
0
≤p − p
0
C
FX
.
4.7
Hence for each x ∈ X, t, p → K
t
p, xσ
#
Gx, p − tσ
Fx,p −p, c is
continuous on R
× Σ
m−1
. Also from Lemmas 2.51 and 2.52, it f ollows that
both functions
t, p
−→ sup
x∈X
K
t
p, x
on R
× Σ
m−1
,
and t −→ inf
p∈Σ
m−1
sup
x∈X
K
t
p, x
on R
are l.s.c.
4.8
So we conclude from 4.5, 4.8, and Statement 1 that t → vK
t
is
continuous on R
.
b Assume that t
2
>t
1
≥ 0. Since FX ⊂ int R
m
is compact, it is easy to see that
ε
0
inf{p, v : p, v ∈ Σ
m−1
× FX} > 0. Thus for any p, x ∈ Σ
m−1
× X,we
have
K
t
1
p, x
σ
#
Gx, p
− t
2
σ
Fx,p
−p, c
t
2
− t
1
σ
Fx,p
≥ K
t
2
p, x
t
2
− t
1
ε
0
,
4.9
10 Journal of Inequalities and Applications
which implies that vK
t
1
>vK
t
2
, and hence t → vK
t
is strict decreasing
on R
.
c Let ε
1
sup{p, u : p ∈ Σ
m−1
,u∈ GX} and ε
2
inf{p, c : p ∈ Σ
m−1
}.Byboth
Assumptions, ε
1
and ε
2
are finite. Thus for any t>0andp, x ∈ Σ
m−1
× X,we
have
K
t
p, x
σ
#
Gx, p
− tσ
Fx,p
−p, c
≤ ε
1
− tε
0
− ε
2
.
4.10
This implies that vK
t
≤ ε
1
−tε
0
−ε
2
. Therefore, vK
∞
lim
t → ∞
vK
t
−∞.
This completes the proof.
Lemma 4.6. 1 p → H
c
p, xx ∈ X and p → sup
x∈X
H
c
p, x are l.s.c. on Σ
m−1
.
2 x → H
c
p, xp ∈ Σ
m−1
) and x → inf
p∈P
H
c
p, x are u.s.c. on X.
3 vH
c
exists if and only if SH
c
is nonempty.
Proof. 1 Since, for each x ∈ X and u, v ∈ Gx × Fx, the function p →p, u − c/p, v is
continuous on Σ
m−1
,fromLemma 2.51, we can see that p → H
c
p, xsup
u,v∈Gx×Fx
p, u−
c/p, vx ∈ X and p → sup
x∈X
H
c
p, x are l.s.c., hence 1 is true.
2 Assume that {p
k
,x
k
}⊂Σ
m−1
× X is a sequence with p
k
,x
k
→ p
0
,x
0
k →∞,
then for each k, there exist u
k
∈ Gx
k
and v
k
∈ Fx
k
such that σ
#
Gx
k
,p
k
p
k
,u
k
,
σ
Fx
k
,p
k
p
k
,v
k
. Since GX, FX are compact and Gx
k
⊂ GX, Fx
k
⊂ FX k ≥ 1,
we may choose {u
k
j
}⊂{u
k
} and {v
k
j
}⊂{v
k
} such that
u
k
j
−→ u
0
,v
k
j
−→ v
0
k →∞
,
lim sup
k →∞
p
k
,u
k
lim
j →∞
p
k
j
,u
k
j
, lim inf
k →∞
p
k
,v
k
lim
j →∞
p
k
j
,v
k
j
.
4.11
By Lemma 2.32,bothG and F are closed. Hence, x
k
j
,u
k
j
→ x
0
,u
0
∈ graph G
and x
k
j
,v
k
j
→ x
0
,v
0
∈ graph F, which in turn imply that u
0
∈ Gx
0
,v
0
∈ Fx
0
and
lim sup
k →∞
σ
#
Gx
k
,p
k
lim
j →∞
p
k
j
,u
k
j
p
0
,u
0
≤σ
#
Gx
0
,p
0
,
lim inf
k →∞
σ
Fx
k
,p
k
lim
j →∞
p
k
j
,v
k
j
p
0
,v
0
≥σ
Fx
0
,p
0
.
4.12
Combining this with σ
Fx,p > 0forp, x ∈ Σ
m−1
× X, it follows that
lim sup
k →∞
σ
#
Gx
k
,p
k
−p
k
,c
σ
Fx
k
,p
k
≤
lim sup
k →∞
σ
#
Gx
k
,p
k
−p
k
,c
lim inf
k →∞
σ
Fx
k
,p
k
≤
σ
#
Gx
0
,p
0
−p
0
,c
σ
Fx
0
,p
0
.
4.13
Journal of Inequalities and Applications 11
Hence by 4.1, p, x → H
c
p, x is u.s.c. on Σ
m−1
× X,soisx → inf
p∈Σ
m−1
H
c
p, x
on X thanks to Lemma 2.51.
3 Assume that vH
c
exists. By 1 and 2, there exist p ∈ Σ
m−1
and x ∈ X such that
sup
x∈X
H
c
p, x
inf
p∈Σ
m−1
sup
x∈X
H
c
p, x
v
H
c
sup
x∈X
inf
p∈Σ
m−1
H
c
p, x
inf
p∈Σ
m−1
H
c
p,
x
.
4.14
By Remark 2.22,
p, x ∈ SH
c
. Hence SH
c
is nonempty. The inverse is obvious. This
completes the proof.
Lemma 4.7. 1 If c solves1.4,thenvH
c
exists with vH
c
≥ 1.
2 If vH
c
exists with vH
c
≥ 1,thenSH
c
/
∅ and H
c
p, x ≥ 1 for p, x ∈ SH
c
.
Proof. 1 If c solves 1.4, then t 1 solves 4.3.FromLemma 4.5, we know that vK
1
≥ 0,
and there is a unique t
0
≥ 1 such that vK
t
0
0. Moreover, also from Lemma 4.5, t
0
is the
biggest number that makes 4.3 solvable, and thus t ∈ R
solves 4.3 if and only if t ∈ 0,t
0
.
We will prove that vH
c
exists with vH
c
t
0
.Let
v
∗
sup
x∈X
inf
p∈Σ
m−1
H
c
p, x
,v
∗
inf
p∈Σ
m−1
sup
x∈X
H
c
p, x
,
4.15
then v
∗
≤ v
∗
. It is needed to show that v
∗
≤ t
0
≤ v
∗
.
Since t
0
solves 4.3, there exist x
0
∈ X, u
0
∈ Gx
0
,andv
0
∈ Fx
0
such that u
0
− t
0
v
0
≥ c.
Hence for each p ∈ Σ
m−1
,σ
#
Gx
0
,p−t
0
σ
Fx
0
,p−p, c≥p, u
0
−t
0
v
0
−c≥0. As σ
Fx
0
,p >
0forp ∈ Σ
m−1
, it follows from 4.1 that H
c
p, x
0
σ
#
Gx
0
,p −p, c/σ
Fx
0
,p ≥ t
0
p ∈
Σ
m−1
and thus
v
∗
≥ inf
p∈Σ
m−1
H
c
p, x
0
≥ t
0
.
4.16
On the other hand, by 4.15, for each p ∈ Σ
m−1
we have sup
x∈X
H
c
p, x ≥ v
∗
.By4.1
and Lemma 4.62, there exists x
p
∈ X such that
σ
#
Gx
p
,p
−p, c
σ
Fx
p
,p
H
c
p, x
p
sup
x∈X
H
c
p, x
≥ v
∗
.
4.17
It deduces from 4.4 that for each p ∈ Σ
m−1
sup
x∈X
K
v
∗
p, x
≥ K
v
∗
p, x
p
σ
#
Gx
p
,p
− v
∗
σ
Fx
p
,p
−p, c≥0.
4.18
Hence by Lemma 4.51, vK
v
∗
inf
p∈Σ
m−1
sup
x∈X
K
v
∗
p, x ≥ 0, and t v
∗
solves 4.3. Since
t
0
is the biggest number that makes 4.3 solvable, we have v
∗
≤ t
0
. Combining this with
4.16,weobtainv
∗
v
∗
t
0
. Therefore, vH
c
exists and vH
c
t
0
≥ 1.
2 follows immediately from Lemma 4.63 and Remark 2.22. The third lemma
follows.
12 Journal of Inequalities and Applications
Lemma 4.8. If SH
c
/
∅ with H
c
p, x ≥ 1 for p, x ∈ SH
c
,thenc solves 1.4. Moreover, for
each
x ∈ SH
c
|
X
,thereexistsy ∈ G − Fx such that y ≥ c.
Proof. By 4.1 and Remark 2.22, we know that, for each
p, x ∈ SH
c
,
σ
#
Gx,
p
−p, c
σ
Fx,
p
H
c
p, x
≤ v
H
c
≤ H
c
p,
x
σ
#
G
x, p
−p, c
σ
F
x, p
,
p, x
∈ Σ
m−1
× X.
4.19
Combining this with the definition of K
vH
c
p, xi.e., 4.4 for t vH
c
, it follows that, for
each
p, x ∈ SH
c
and each p, x ∈ Σ
m−1
× X,
K
vH
c
p, x
σ
#
Gx,
p
− v
H
c
σ
Fx,
p
−p, c
≤ 0 ≤ σ
#
G
x, p
− v
H
c
σ
F
x, p
−p, c
K
vH
c
p,
x
.
4.20
Hence by Definition 2.14 and Remark 2.22,
∀
p, x
∈ S
H
c
, sup
x∈X
K
vH
c
p, x
0 inf
p∈Σ
m−1
K
vH
c
p,
x
.
4.21
It follows that
p, x ∈ SH
c
implies that p, x ∈ SK
vH
c
with vK
vH
c
K
vH
c
p, x0,
and so
S
H
c
|
X
⊂ S
K
vH
c
|
X
,
∀
x ∈ S
H
c
|
X
, inf
p∈Σ
m−1
K
vH
c
p,
x
v
K
vH
c
0.
4.22
Applying Lemma 4.51 to T
vH
c
G − vH
c
F, we then conclude that t vH
c
solves 4.3.
So there exist
x ∈ X, u ∈ Gx,andv ∈ Fx such that u − vH
c
v ≥ c. Hence c solves 1.4
because Gx ⊂ R
m
, Fx ⊂ int R
m
x ∈ X,andvH
c
≥ 1.
For each
x ∈ SH
c
|
X
. Since K
vH
c
p, xsup
y∈G−vH
c
Fx
p, y − c by 4.4, applying
Lemma 2.53 to the function p, y →p, y− c on Σ
m−1
×G−vH
c
Fx and associating with
4.22,weobtain
sup
y∈G−vH
c
Fx
inf
p∈Σ
m−1
p, y − c inf
p∈Σ
m−1
sup
y∈G−vH
c
Fx
p, y − c
inf
p∈Σ
m−1
K
vH
c
p,
x
v
K
vH
c
0.
4.23
Journal of Inequalities and Applications 13
Since y → inf
p∈Σ
m−1
p, y − c is u.s.c. on G − vH
c
Fx,from4.23 there exist u ∈ Gx, v ∈ Fx
such that y
u − vH
c
v ∈ G − vH
c
Fx satisfies
inf
p∈Σ
m−1
p, y − c sup
y∈G−vH
c
Fx
inf
p∈Σ
m−1
p, y − c 0.
4.24
By taking p
i
e
i
∈ Σ
m−1
i 1, 2, ,m,wegety u − vH
c
v ≥ c, and therefore y u − v ∈
G − F
x satisfies y ≥ c because vH
c
≥ 1. This completes the proof.
Proof of Theorem 4.1. By Lemmas 4.63, 4.7,and4.8, we know that Theorem 4.1 is true.
To prove Theorem 4.2, besides using Lemmas 4.5–4.8,forc ∈ R
m
and x ∈ X,wealso
need to study the condition that t ∈ R
solves
∃u ∈ Gx, ∃v ∈ Fx
⇒ u − tv ≥ c.
4.25
Define ψp, u, v L
t,c,x
p, u, v on P × Y
1
× Y
2
Σ
m−1
× Gx × Fx by
L
t,c,x
p,
u, v
p, u − tv − c
for
p,
u, v
∈ Σ
m−1
×
Gx × Fx
.
4.26
We denote by vL
t,c,x
the minimax values of L
t,c,x
if it exists, SL
t,c,x
the saddle point set if it
is nonempty, and SL
t,c,x
|
Gx×Fx
its restriction to Gx × Fx.
Lemma 4.9. Let c ∈ R
m
and x ∈ X be fixed. Then one has the following.
1 For each t ∈ R
,vL
t,c,x
exists and SL
t,c,x
is nonempty.
2 t ∈ R
solves 4.25 if and only if vL
t,c,x
≥ 0 if and only if p, u, v ∈ SL
t,c,x
implies
that L
t,c,x
p, u, v ≥ 0.
3 t → vL
t,c,x
is continuous and strict decreasing on R
with vL
∞,c,x
−∞.
Proof. Define T
t,x
from Gx × Fx ⊂ R
2m
to R
m
by
T
t,x
u, v
u − tv,
u, v
∈ Gx × Fx. 4.27
Then T
t,x
is a single-valued continuous map with the convex condition defined by
Definition 2.13 because T
t,x
αu
1
,v
1
1 − αu
2
,v
2
αT
t,x
u
1
,v
1
1 − αT
t,x
u
2
,v
2
for all α ∈ 0, 1 and u
i
,v
i
∈ Gx × Fx i 1, 2.
14 Journal of Inequalities and Applications
Since Gx × Fx is convex and compact in R
2m
, replacing x ∈ X⊂ R
n
by u, v ∈ Gx ×
Fx⊂ R
2m
, Tx by T
t,x
u, v,andφ
c
p, xsup
y∈Tx
p, y −c by L
t,c,x
p, u, v p, T
t,x
u, v−
c,fromLemma 2.4, we know that both 1 and 2 are true. Moreover, with the same method
as in proving Lemma 4.52, we can show that 3 is also true. In fact, since t, p, u, v →
p, u − tv − c is continuous on R
× Σ
m−1
× Gx × Fx and Gx × Fx and Σ
m−1
are compact, by
4.26 and Lemmas 2.51 and 2.52, we can see that
t −→ sup
u,v∈Gx×Fx
inf
p∈Σ
m−1
L
t,c,x
p,
u, v
is u.s.c.,
t −→ inf
p∈Σ
m−1
sup
u,v
∈Gx×Fx
L
t,c,x
p,
u, v
is l.s.c.
4.28
Hence by 1, t → vL
t,c,x
is continuous on R
.
Let ε
0
, ε
1
,andε
2
be defined as in the proof of Lemma 4.52.
If t
2
>t
1
≥ 0, also by 4.26, we can see that L
t
1
,c,x
p, u, v ≥ L
t
2
,c,x
p, u, vt
2
−t
1
ε
0
for p, u, v ∈ Σ
m−1
× Gx × Fx. It follows that vL
t
1
,c,x
≥ vL
t
2
,c,x
t
2
− t
1
ε
0
and thus
t → vL
t,c,x
is strict decreasing on R
.
If t>0, then L
t,c,x
p, u, v p, u − tv − c≤ε
1
− tε
0
− ε
2
for p, u, v ∈ Σ
m−1
× Gx ×
Fx. This implies that vL
t,c,x
≤ ε
1
− tε
0
− ε
2
with vL
∞,c,x
−∞. Hence the fifth lemma
follows.
Lemma 4.10. 1 vH
c,x
exists if and only if SH
c,x
is nonempty, where H
c,x
is defined by 4.1(b).
2 If t 1 solves 4.25 for c ∈ R
m
and x ∈ X,thenvH
c,x
exists with vH
c,x
≥
1, SH
c,x
is nonempty with H
c,x
p, u, v ≥ 1 for p, u, v ∈ SH
c,x
, and
inf
p∈Σ
m−1
H
c,x
p, u, v vH
c,x
for u, v ∈ SH
c,x
|
Gx×Fx
.
Proof. Since p, u, v → H
c,x
p, u, v p, u − c/p, v is continuous on Σ
m−1
× Gx × Fx,
by Lemma 2.51,itiseasytoseethat
a
u, v
−→ H
c,x
p,
u, v
p ∈ Σ
m−1
,
u, v
−→ inf
p∈Σ
m−1
H
c,x
p,
u, v
are u.s.c. on Gx × Fx,
b
p −→ H
c,x
p,
u, v
u, v
∈ Gx × Fx
,
p −→ sup
u,v
∈Gx×Fx
H
c,x
p,
u, v
are l.s.c. on Σ
m−1
.
4.29
1 By 4.29 and with the same method as in proving Lemma 4.63, we can show
that 1 is true. Indeed, we only need to prove the necessary part. If vH
c,x
exists, then by 4.29, there exists
p, u, v ∈ Σ
m−1
× Gx × Fx such that
sup
u,v∈Gx×Fx
H
c,x
p, u, v vH
c,x
inf
p∈Σ
m−1
H
c,x
p, u, v. Hence SH
c,x
is
nonempty.
2 If t 1 solves 4.25 for c ∈ R
m
and x ∈ X, then from Lemma 4.9 we know that
vL
1,c,x
exists with vL
1,c,x
≥ 0, and there is a unique
t
0
≥ 1 such that vL
t
0
,c,x
0.
In particular,
t
0
is the biggest number that makes 4.25 solvable for c and x.
Journal of Inequalities and Applications 15
Applying the same method as in proving Lemma 4.71, we can show that vH
c,x
exists with vH
c,x
t
0
≥ 1. In fact, let
v
∗
sup
u,v∈Gx×Fx
inf
p∈Σ
m−1
H
c,x
p,
u, v
, v
∗
inf
p∈Σ
m−1
sup
u,v∈Gx×Fx
H
c,x
p,
u, v
.
4.30
Then v
∗
≤ v
∗
. We need to show that v
∗
≥
t
0
≥ v
∗
.
Since
t
0
solves 4.25 for c and x, there exist u
x
∈ Gx and v
x
∈ Fx such that
u
x
−
t
0
v
x
≥ c. It follows that H
c,x
p, u
x
,v
x
p, u
x
− c/p, v
x
≥
t
0
for any p ∈
Σ
m−1
, hence v
∗
≥ inf
p∈Σ
m−1
H
c,x
p, u
x
,v
x
≥
t
0
. On the other hand, by the definition
of v
∗
, we have sup
u,v∈Gx×Fx
H
c,x
p, u, v ≥ v
∗
for any p ∈ Σ
m−1
.By4.29a and
4.1b, there exists u
p
,v
p
∈ Gx × Fx such that p, u
p
− c/p, v
p
H
c,x
p, u
p
,v
p
sup
u,v∈Gx×Fx
H
c,x
p, u, v ≥ v
∗
, which implies by 4.26 that sup
u,v∈Gx×Fx
L
v
∗
,c,x
p, u, v ≥
p, u
p
− v
∗
v
p
− c≥0 for any p ∈ Σ
m−1
. Hence from Lemma 4.9, vL
v
∗
,c,x
≥ 0,t v
∗
solves
4.25,and
t
0
≥ v
∗
. Therefore, vH
c,x
exist with vH
c,x
t
0
≥ 1. So we conclude from 1
and Remark 2.2 that 2 is true. This completes the proof.
Lemma 4.11. If vH
c
1,thenTheorem 4.2(2) is true.
Proof. i If vH
c
1, then by Lemma 4.63 and Remark 2.2, SH
c
/
∅ and
∀
p
,
x
∈ S
H
c
,
σ
#
Gx,
p
−p, c
σ
Fx,
p
≤ 1 ≤
σ
#
G
x, p
−p, c
σ
F
x, p
for
p, x
∈ Σ
m−1
× X.
4.31
By the same proof of 4.21 we can show that
∀
p, x
∈ S
H
c
, sup
x∈X
K
1
p, x
0 inf
p∈Σ
m−1
K
1
p,
x
.
4.32
Combining this with Lemma 4.51 and using Remark 2.22, we have
S
H
c
|
X
⊆ S
K
1
|
X
, ∀ x ∈ S
H
c
|
X
, inf
p∈Σ
m−1
K
1
p,
x
v
K
1
0.
4.33
As K
1
p, xsup
y∈G−Fx−c
p, y by 4.4, applying Lemma 2.53 to the function
p, y →p, y on Σ
m−1
× G − Fx − c, we obtain that, for each x ∈ SH
c
|
X
,
sup
y∈G−Fx−c
inf
p∈Σ
m−1
p, y inf
p∈Σ
m−1
sup
y∈G−Fx−c
p, y inf
p∈Σ
m−1
K
1
p,
x
0.
4.34
Since y → inf
p∈Σ
m−1
p, y is u.s.c. on G−Fx−c,from4.34 there exists y ∈ G−Fx−c
such that
inf
p∈Σ
m−1
p, y sup
y∈
G−F
x−c
inf
p∈Σ
m−1
p, y 0.
4.35
16 Journal of Inequalities and Applications
Hence,
y ≥ 0. This implies that t 1 solves 4.25 for c and any x ∈ SH
c
|
X
. So we conclude
from Lemma 4.10 and Remark 2.22 that
∀
x ∈ S
H
c
|
X
,v
H
c,x
exists and S
H
c,x
is nonempty,
∀
u, v
∈ S
H
c,x
|
Gx×Fx
, inf
p∈Σ
m−1
H
c,x
p,
u, v
v
H
c,x
≥ 1.
4.36
On the other hand, by 4.1, H
c
p, xsup
u,v∈Gx×Fx
H
c,x
p, u, v. Combining this
with 4.36, it f ollows that, for each
x ∈ SH
c
|
X
and each u, v ∈ SH
c,x
|
Gx×Fx
,
1 ≤ inf
p∈Σ
m−1
H
c,x
p,
u, v
v
H
c,x
inf
p∈Σ
m−1
sup
u,v∈Gx×Fx
H
c,x
p,
u, v
inf
p∈Σ
m−1
H
c
p,
x
v
H
c
1.
4.37
Hence, also by 4.1,inf
p∈Σ
m−1
p, u−c/p, vinf
p∈Σ
m−1
H
c,x
p, u, v 1. This implies that,
for each p ∈ Σ
m−1
, p, u−c/p, v≥1 and there exists p ∈ Σ
m−1
such that p, u−c/p, v 1.
So we obtain
∀
x ∈ S
H
c
|
X
, ∀
u, v
∈ S
H
c,x
|
Gx×Fx
, inf
p∈Σ
m−1
p, u − v − c 0.
4.38
By using the same method as in proving 1⇒2 of Theorem 3.2, we conclude that
u − v ≥ c
and there exists i
0
∈{1, 2, ,m} such that u
i
0
− v
i
0
c
i
0
. Hence Theorem 4.22 is true.
Lemma 4.12. If Theorem 4.2(3) holds,then vH
c
1.
Proof. If Theorem 4.23 holds, then c solves both 1.4 and 1.5,andbyLemma 4.71, vH
c
exists with vH
c
≥ 1.
Nowweletx ∈ SH
c
|
X
, u, v ∈ SH
c,x
|
Gx×F x
,andi
0
∈{1, 2, ,m} satisfy u − v ≥ c
and u
i
0
− v
i
0
c
i
0
, then we have p, u − v − c≥0forp ∈ Σ
m−1
and e
i
0
, u − v − c 0 where
e
i
0
i
0
0, ,0, 1, 0, ,0 ∈ Σ
m−1
. This implies by 4.1 that
inf
p∈Σ
m−1
H
c,x
p,
u, v
inf
p∈Σ
m−1
p, u − c
p, v
1.
4.39
Combining this with t he fact that vH
c
≥ 1andusingRemark 2.2 and 4.1,weobtainthat
1 inf
p∈Σ
m−1
H
c,x
p,
u, v
v
H
c,x
inf
p∈Σ
m−1
sup
u,v∈G x×F x
H
c,x
p,
u, v
inf
p∈Σ
m−1
H
c
p, x
v
H
c
≥ 1.
4.40
Hence, vH
c
1.
Journal of Inequalities and Applications 17
Proof of Theorem 4.2. Since 2⇒3 of is clear, Theorem 4.2 follows immediately from Lemmas
4.11 and 4.12.
5. Conclusion
Based on the generalized and multivalued input-output inequality models, in this paper we
have considered three types of set-valued inequality systems namely, 1.3–1.5 and two
corresponding solvability questions. By constructing some auxiliary functions and studying
their minimax and saddle point properties with the nonlinear analysis approaches, three
solvability theorems i.e., Theorems 3.2, 4.1,and4.2 composed of necessary and sufficient
conditions regarding these inequality systems have been obtained.
References
1 E. W. Leontief, Input-Output Analysis, Pergamon Press, Oxford, UK, 1985.
2 E. W. Leontief, Input-Output Analysis, Oxford University Press, New York, NY, USA, 2nd edition, 1986.
3 R. E. Miller and P. D. Blair, Input-Output Analysis: Foundations and Extension, Prentice-Hall, Englewood
Cliffs, NJ, USA, 1985.
4 P. Medvegyev, “A general existence theorem for von Neumann economic growth models,”
Econometrica, vol. 52, no. 4, pp. 963–974, 1984.
5 C. Bidard and E. Hosoda, “On consumption baskets in a generalized von Neumann model,”
International Economic Review, vol. 28, no. 2, pp. 509–519, 1987.
6 Y. F. Liu, “Some results on a class of generalized Leontief conditional input output inequalities,” Acta
Mathematicae Applicatae Sinica, vol. 28, no. 3, pp. 506–516, 2005 Chinese.
7 Y. Liu and Q. Zhang, “Generalized input-output inequality systems,” Applied Mathematics and
Optimization, vol. 54, no. 2, pp. 189–204, 2006.
8 Y. Liu, “A set-valued type inequality system,” Nonlinear Analysis: Theory, Methods & Applications,vol.
69, no. 11, pp. 4131–4142, 2008.
9 J P. Aubin and I. Ekeland, Applied Nonlinear Analysis, Pure and Applied Mathematics, John Wiley &
Sons, New York, NY, USA, 1984.
10 J P. Aubin, Mathematical Methods of Game and Economic Theory, vol. 7 of Studies in Mathematics and Its
Applications, North-Holland, Amsterdam, The Netherlands, 1979.
11 J P. Aubin, Optima and Equilibria: An Introduction to Nonlinear Analysis, vol. 140 of Graduate Texts in
Mathematics, Springer, Berlin, Germany, 2nd edition, 1998.
12 R. B. Holmes, Geometric Functional Analysis and Applications, Springer, Berlin, Germany, 1974.
