- Research Article
- Open access
- Published:
Solvability Criteria for Some Set-Valued Inequality Systems
Journal of Inequalities and Applications volume 2010, Article number: 543061 (2010)
Abstract
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 the
-dimensional Euclidean space,
its dual, and
the duality pairing on
; moreover, we denote that
and int 
is its interior. We also define
(or
) in
by
(or by
int 
).
It is known that the generalized (linear or nonlinear) von Neumann model, which is composed of an inequality system and a growth factor problem described by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ1_HTML.gif)
is one of the most important issues in the input-output analysis [1–3], where ,
(
may not be equal to
), and
are two nonnegative or positive maps from
to
. A series of researches on (1.1) have been made by the authors of [1–5] for the linear case (i.e.,
are
matrices) and by the authors of [6, 7] for the nonlinear case (i.e.,
are some types of nonlinear maps). Since (a) or (b) of (1.1) is precisely a special example of the inequality
if we restrict
or
, it is enough for (1.1) to consider the inequality system. This idea can be extended to the set-valued version. Indeed, if
and
are replaced by set-valued maps
and
, 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 [8] (as a set-valued extension to [6, 7]) we have considered the following multivalued inequality system:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ2_HTML.gif)
and obtained several necessary and sufficient conditions for its solvability, where and
is a class of set-valued maps from
to
. Along the way, three further set-valued inequality systems that we will study in the sequel can be stated as follows.
Let ,
be as above, and let
be set-valued maps from
to
, then we try to find the solvability criteria (i.e., the necessary and sufficient conditions) that
solves
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ3_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ4_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ5_HTML.gif)
When and
are single-valued maps, then (1.3)–(1.5) return to the models of [6, 7]. When
and
,
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
. Indeed, if
is compact and
is continuous, compact valued with
, then
is compact and there is
with
for
. Hence
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
. However, this type of result is trivial because it depends only on the property of
but not on the respective information of
and
. This opinion is also applicable to (1.3) and (1.5).
-
(3)
Clearly, (1.3) (or (1.5)) is more fine and more useful than (1.2) (or (1.4)). However, the method used for
in [8] (or the possible idea for (1.4)) to obtain solvability criteria fails to be applied to find the similar characteristic results for (1.3) (or (1.5)) because there are some examples (see Examples 3.5 and 4.4) to show that, without any additional restrictions, no necessary and sufficient conditions concerning existence for them can be obtained. This is also a main cause that the author did not consider
and
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 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 and
?
Assumption.
is a fixed point and
is a convex compact subset.
Assumption.
Consider the following: and
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 ,
, and
(
). Let
,
, and
be functions and
a set-valued map. We need some concepts concerning
,
and
and
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
,
, and
as follows.
Definition 2.1.
(1) is said to be closed if its graph defined by graph 
is closed in
. Moreover,
is said to be upper semicontinuous (in short, u.s.c.) if, for each
and each neighborhood
of
, there exists a neighborhood
of
such that
.
-
(2)
Assume that
(
), and define
,
(
). Then
is said to be upper hemicontinuous (in short, u.h.c.) if
is u.s.c. on
for any
.
(3) is said to be convex if
is convex and
for any
and
.
(4)
(a)If , then one claims that the minimax equality of
holds. Denoting by
the value of the preceding equality, one also says that the minimax value
of
exists. If
such that
, then one calls
a saddle point of
. Denote by
the set of all saddle points of
(i.e.,
), and define
, the restriction of
to
if
is nonempty.
(b)Replacing by
and
by
, with the similar method one can also define
(the minimax value of
),
(the saddle-point set of
), and
(the restriction of
to
).
-
(5)
If
is a convex set and
a subset of
, one claims that
is an extremal subset of
if
and
for some
entails
.
is an extremal point of
if
is an extremal subset of
, and the set of all extremal points of
is denoted by ext
.
Remark 2.2.
-
(1)
Since
and
, we can see that
is u.h.c. if and only if
is l.s.c. on
for any
.
-
(2)
For the function
on
,
exists if and only if
, and
if and only if
if and only if
for any
. If
, then
exists, and
for any
. The same properties are also true for
on
. Moreover, we have
(2.1)
We also need three known results as follows.
Lemma 2.3.
-
(1)
(see [9]) If
is u.s.c., then
is u.h.c.
-
(2)
(see [9]) If
is u.s.c. with closed values, then
is closed.
-
(3)
(see [9]) If
(the closure of
) is compact and
is closed, then
is u.s.c.
-
(4)
If
is compact and
is u.s.c. with compact values, then
is compact in
.
-
(5)
If
is convex (or compact) and
are convex (or u.s.c. with compact values), then
are also convex (or u.s.c.) for all
.
Proof.
We only need to prove (5).
-
(a)
If
are convex,
,
, and
, then
(2.2)
Hence is convex.
-
(b)
Now we assume that
is compact.
In case is u.s.c. with compact values and
, then by (2), (4),
is closed and the range
of
is compact. If
, then
for any
; hence,
is u.s.c. If
, supposing that
graph
with
, then
graph 
such that
as
, which implies that
. Hence,
is closed and also u.s.c. because of (3).
In case are u.s.c. with compact values, if
graph
with
, then
and there exist
,
such that
for all
By (4),
and
are compact, so we can suppose
and
as
. By (2), both
are closed, this implies that
, and thus
is closed. Hence by (3),
is u.s.c. because
and
is compact.
Lemma 2.4 (see [8, Theorems and
]).
Let be convex compact with
,
, and
. Assume that
is convex and u.s.c. with nonempty convex compact values, and define
on
by
for
. Then
(1) exists and
is a convex compact subset of
,
(2) solves (1.2)
for
.
In particular, both (1) and (2) are also true if .
Lemma 2.5.
-
(1)
(see [10, 11]) If
is convex or l.s.c. (resp., concave or u.s.c.) on
for
and
(resp.,
) is finite for
, then
(resp.,
) is also convex or l.s.c. (resp., concave or u.s.c.) on
.
-
(2)
(see [11]) If
is l.s.c. (or u.s.c.) and
is compact, then
defined by
(or
defined by
) is also l.s.c. (or u.s.c.).
-
(3)
(see [9–11], Minimax Theorem) Let
,
be convex compact, and let
be defined on
. If, for each
is convex and l.s.c. and, for each
is concave and u.s.c., then
and there exists
such that
.
3. Solvability Theorem to (1.3)
Let be introduced as in Lemma 2.4, and define the functions
on
and
on
by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ8_HTML.gif)
Remark 3.1.
By both Assumptions in Section 1, Definition 2.1, and Lemmas 2.4 and 2.5, we can see that
(1) for all
,
(2) and
exist, and
and
are nonempty,
(3) solves (1.2) if and only if
if and only if
with
.
Hence, and
are nonempty. Moreover, we have the following.
Theorem 3.2.
For (1.3), the following three statements are equivalent to each other:
(1),
(2)for all , for all
,
,
,
(3),
,
,
.
Remark 3.3.
Clearly, each of (2) and (3) implies that solves (1.3) because
and
. So we conclude from Theorem 3.2 that
solves (1.3) in the way of (2) or in the way of (3) if and only if
.
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ9_HTML.gif)
Then for each and each
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ10_HTML.gif)
By taking , it follows that
. Hence,
. On the other hand, it is easy to verify that
is a nonempty extremal subset of
. The Crain-Milmann Theorem (see [12]) shows that ext 
is nonempty with ext 
ext 
. So there exists
such that
ext 
. This implies by (3.3) that
, and therefore (2) follows.
(3)(1). Let
,
, and let
be presented in (3). Since
solves (1.3) and also solves (1.2), by (3.1) and Remarks 2.2 and 3.1, we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ11_HTML.gif)
where . Hence,
and the theorem follows.
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_IEq329_HTML.gif)
Remark 3.4.
From the Theorem, we know that implies that
solves (1.3). However, without any additional restricting conditions, the inverse may not be true.
Example 3.5.
Let , and let
be defined by
for
Then
is an u.s.c. and convex set-valued map with convex compact values, and for each
,
and
,
. Hence,
for all
and therefore,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ12_HTML.gif)
This implies that (or
) may not be the necessary (or the sufficient) condition that
solves (1.3).
4. Solvability Theorems to (1.4) and (1.5)
Taking , then from both Assumptions, Lemma 2.4, and Theorem 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
. To get some further solvability criteria to (1.4) and (1.5) depending on the respective information of
and
, we define the functions
on
and
on
by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ13_HTML.gif)
By both Assumptions, we know that and
are finite with
and
for
and
, so the functions
and
defined by (4.1) are well defined.
In view of Definition 2.1, we denote by (or
) the minimax value of
(or
) if it exists,
(or
) the saddle point set if it is nonempty, and
(or
) the restriction of
to
(or
to
). Then we have the solvability result to (1.4) and (1.5) as follows.
Theorem 4.1.
(i) exists if and only if
.
(ii)
(1) solves (1.4) if and only if
exists with
if and only if
with
for
.
(2)In particular, if exists with
, then for each
, there exists
such that
Theorem 4.2.
For (1.5), the following three statements are equivalent to each other:
(1),
(2),
, and for all
, for all
,
,
,
(3),
, and
,
,
,
That is, solves (1.5) in the way of (2) or in the way of (3) if and only if
.
Remark 4.3.
It is also needed to point out that is not the necessary condition of
making (1.5) solvable without any other restricting conditions.
Example 4.4.
Let ,
, and
be defined by
,
for
Then both
and
are u.s.c. convex set-valued maps with convex compact values, and for any
,
, and
, we have
,
,
, and
Therefore,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ14_HTML.gif)
This implies that, for each ,
solves (1.5) but
.
The proof of both Theorems 4.1 and 4.2 can be divided into eight lemmas.
Let ,
, and
. Consider the auxiliary inequality system
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ15_HTML.gif)
Then solves (4.3) if and only if
solves (1.2) for
, and in particular,
solves (4.3) if and only if
solves (1.4). Define
on
by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ16_HTML.gif)
denote by the minimax value of
if it exists, and denote by
the saddle point set if it is nonempty. Then we have the following.
Lemma 4.5.
-
(1)
For each
,
exists and
is nonempty. Moreover,
solves (4.3) if and only if
if and only if
for
.
-
(2)
The function
is continuous and strictly decreasing on
with
.
Proof.
-
(1)
By both Assumptions and Lemmas 2.3(4) and 2.3(5),
is convex and u.s.c. with nonempty convex compact values for each
. Since
for
, applying Lemma 2.4 to
and substituting
for
, we know that (1) is true.
-
(2)
We prove (2) in three steps as follows.
(a)By Lemma 2.3(1), and
are u.h.c., which implies by Definition 2.1(2) and Remark 2.2(1) that for each
,
is u.s.c. on
. Then from Lemma 2.5(1)(2), we know that both functions
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ17_HTML.gif)
Since and
are compact by both Assumptions and Lemma 2.3(4),
and
are finite. Then for any
,
and
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ18_HTML.gif)
This implies that, for each ,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ19_HTML.gif)
Hence for each ,
is continuous on
. Also from Lemmas 2.5(1) and 2.5(2), it follows that both functions
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ20_HTML.gif)
So we conclude from (4.5), (4.8), and Statement (1) that is continuous on
.
(b)Assume that . Since
int 
is compact, it is easy to see that
. Thus for any
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ21_HTML.gif)
which implies that , and hence
is strict decreasing on
.
(c)Let and
. By both Assumptions,
and
are finite. Thus for any
and
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ22_HTML.gif)
This implies that . Therefore,
. This completes the proof.
Lemma 4.6.
(1) and
are l.s.c. on
.
(2)) and
are u.s.c. on
.
(3) exists if and only if
is nonempty.
Proof.
-
(1)
Since, for each
and
, the function
is continuous on
, from Lemma 2.5(1), we can see that
and
are l.s.c., hence (1) is true.
-
(2)
Assume that
is a sequence with
, then for each
, there exist
and
such that
,
. Since
are compact and
,
, we may choose
and
such that
(4.11)
By Lemma 2.3(2), both and
are closed. Hence,
and
, which in turn imply that
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ24_HTML.gif)
Combining this with for
, it follows that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ25_HTML.gif)
Hence by (4.1), is u.s.c. on
, so is
on
thanks to Lemma 2.5(1).
-
(3)
Assume that
exists. By (1) and (2), there exist
and
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ26_HTML.gif)
By Remark 2.2(2), . Hence
is nonempty. The inverse is obvious. This completes the proof.
Lemma 4.7.
-
(1)
If
solves(1.4), then
exists with
.
(2)If exists with
, then
and
for
.
Proof.
-
(1)
If
solves (1.4), then
solves (4.3). From Lemma 4.5, we know that
, and there is a unique
such that
. Moreover, also from Lemma 4.5,
is the biggest number that makes (4.3) solvable, and thus
solves (4.3) if and only if
. We will prove that
exists with
. Let
(4.15)
then . It is needed to show that
.
Since solves (4.3), there exist
,
, and
such that
. Hence for each
As
for
, it follows from (4.1) that
and thus
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ28_HTML.gif)
On the other hand, by (4.15), for each we have
. By (4.1) and Lemma 4.6(2), there exists
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ29_HTML.gif)
It deduces from (4.4) that for each
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ30_HTML.gif)
Hence by Lemma 4.5(1), , and
solves (4.3). Since
is the biggest number that makes (4.3) solvable, we have
. Combining this with (4.16), we obtain
. Therefore,
exists and
-
(2)
follows immediately from Lemma 4.6(3) and Remark 2.2(2). The third lemma follows.
Lemma 4.8.
If with
for
, then
solves (1.4). Moreover, for each
, there exists
such that
Proof.
By (4.1) and Remark 2.2(2), we know that, for each ,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ31_HTML.gif)
Combining this with the definition of (i.e., (4.4) for
), it follows that, for each
and each
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ32_HTML.gif)
Hence by Definition 2.1(4) and Remark 2.2(2),
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ33_HTML.gif)
It follows that implies that
with
and so
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ34_HTML.gif)
Applying Lemma 4.5(1) to , we then conclude that
solves (4.3). So there exist
,
, and
such that
. Hence
solves (1.4) because
,
, and
.
For each . Since
by (4.4), applying Lemma 2.5(3) to the function
on
and associating with (4.22), we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ35_HTML.gif)
Since is u.s.c. on
, from (4.23) there exist
,
such that
satisfies
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ36_HTML.gif)
By taking , we get
, and therefore
satisfies
because
. This completes the proof.
Proof of Theorem 4.1.
By Lemmas 4.6(3), 4.7, and 4.8, we know that Theorem 4.1 is true.
To prove Theorem 4.2, besides using Lemmas 4.5–4.8, for and
, we also need to study the condition that
solves
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ37_HTML.gif)
Define on
by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ38_HTML.gif)
We denote by the minimax values of
if it exists,
the saddle point set if it is nonempty, and
its restriction to
.
Lemma 4.9.
Let and
be fixed. Then one has the following.
(1)For each exists and
is nonempty.
(2) solves (4.25) if and only if
if and only if
implies that
(3) is continuous and strict decreasing on
with
Proof.
Define from
to
by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ39_HTML.gif)
Then is a single-valued continuous map with the convex condition defined by Definition 2.1(3) because
for all
and
.
Since is convex and compact in
, replacing
by
,
by
, and
by
, from Lemma 2.4, we know that both (1) and (2) are true. Moreover, with the same method as in proving Lemma 4.5(2), we can show that (3) is also true. (In fact, since
is continuous on
and
and
are compact, by (4.26) and Lemmas 2.5(1) and 2.5(2), we can see that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ40_HTML.gif)
Hence by (1), is continuous on
.
Let ,
, and
be defined as in the proof of Lemma 4.5(2).
If , also by (4.26), we can see that
for
. It follows that
and thus
is strict decreasing on
.
If , then
for
. This implies that
with
.) Hence the fifth lemma follows.
Lemma 4.10.
(1) exists if and only if
is nonempty, where
is defined by (4.1)(b).
-
(2)
If
solves (4.25) for
and
, then
exists with
,
is nonempty with
for
, and
for
.
Proof.
Since is continuous on
, by Lemma 2.5(1), it is easy to see that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ41_HTML.gif)
(1)By (4.29) and with the same method as in proving Lemma 4.6(3), we can show that (1) is true. (Indeed, we only need to prove the necessary part. If exists, then by (4.29), there exists
such that
. Hence
is nonempty.)
(2)If solves (4.25) for
and
, then from Lemma 4.9 we know that
exists with
, and there is a unique
such that
. In particular,
is the biggest number that makes (4.25) solvable for
and
.
Applying the same method as in proving Lemma 4.7(1), we can show that exists with
. (In fact, let
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ42_HTML.gif)
Then . We need to show that
.
Since solves (4.25) for
and
, there exist
and
such that
. It follows that
for any
, hence
On the other hand, by the definition of
, we have
for any
. By (4.29)(a) and (4.1)(b), there exists
such that
which implies by (4.26) that
for any
. Hence from Lemma 4.9,
solves (4.25), and
. Therefore,
exist with
.) So we conclude from (1) and Remark 2.2 that (2) is true. This completes the proof.
Lemma 4.11.
If , then Theorem 4.2(2) is true.
Proof.
-
(i)
If
, then by Lemma 4.6(3) and Remark 2.2,
and
(4.31)
By the same proof of (4.21) we can show that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ44_HTML.gif)
Combining this with Lemma 4.5(1) and using Remark 2.2(2), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ45_HTML.gif)
As by (4.4), applying Lemma 2.5(3) to the function
on
, we obtain that, for each
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ46_HTML.gif)
Since is u.s.c. on
, from (4.34) there exists
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ47_HTML.gif)
Hence, This implies that
solves (4.25) for
and any
. So we conclude from Lemma 4.10 and Remark 2.2(2) that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ48_HTML.gif)
On the other hand, by (4.1), . Combining this with (4.36), it follows that, for each
and each
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ49_HTML.gif)
Hence, also by (4.1), . This implies that, for each
and there exists
such that
So we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ50_HTML.gif)
By using the same method as in proving (1)(2) of Theorem 3.2, we conclude that
and there exists
such that
. Hence Theorem 4.2(2) is true.
Lemma 4.12.
If Theorem 4.2(3) holds,then .
Proof.
If Theorem 4.2(3) holds, then solves both (1.4) and (1.5), and by Lemma 4.7(1),
exists with
.
Now we let ,
, and
satisfy
and
, then we have
for
and
(where
This implies by (4.1) that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ51_HTML.gif)
Combining this with the fact that and using Remark 2.2 and (4.1), we obtain that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F543061/MediaObjects/13660_2010_Article_2183_Equ52_HTML.gif)
Hence, .
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, and 4.2) composed of necessary and sufficient conditions regarding these inequality systems have been obtained.
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Liu, Y. Solvability Criteria for Some Set-Valued Inequality Systems. J Inequal Appl 2010, 543061 (2010). https://doi.org/10.1155/2010/543061
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DOI: https://doi.org/10.1155/2010/543061