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Volume 2010, Article ID 543061, 17 pagesdoi:10.1155/2010/543061 Research Article Solvability Criteria for Some Set-Valued Inequality Systems Yingfan Liu Department of Mathematics, Colleg

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,yingfanliu@hotmail.com

Received 23 May 2010; Accepted 9 July 2010

Academic Editor: Qamrul Hasan Ansari

Copyrightq 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 kits 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 kis 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 generalizedlinear or nonlinear von Neumann model, which is composed of an inequality system and a growth factor problem described by

a ∃x ∈ X ⇒ Bx − Ax ≥ c,

is one of the most important issues in the input-output analysis1 3, where c ∈ R m

, X ⊆ R n



m may not be equal to n, and B, A are two nonnegative or positive maps from X to R m



A series of researches on1.1 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  1 or

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,

then1.1 yields a class of multivalued von Neumann model, and it solves a proper set-valued inequality system to study With this idea, by8 as a set-valued extension to 6,7 we have considered the following multivalued inequality system:

c ∈ R m

 s.t.

and obtained several necessary and sufficient conditions for its solvability, where X ⊂ Rn



and T : X → 2R 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 → 2R m

 be set-valued maps from X to R m, then

we try to find the solvability criteriai.e., the necessary and sufficient conditions that c ∈ R m



solves

∃x ∈ X, ∃y ∈ Tx,

∃x ∈ X,

When T and G, F are single-valued maps, then1.3–1.5 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 regarding1.3–1.5 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 < y for 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, 1.3 or 1.5 is more fine and more useful than 1.2 or 1.4 However, the method used for1.2 in 8 or the possible idea for 1.4 to obtain solvability criteria fails to be applied to find the similar characteristic results for1.3 or 1.5 because there are some examplessee Examples3.5and4.4 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 consider1.3 and 1.5 in 8

So some new methods should be introduced if we want to search out the solvability criteria to1.3–1.5 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 solves1.3 in some proper way?

Question 2 LikeQuestion 1, whether there exist any solvability criteria to1.4 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 → 2R 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 inSection 2and prove three Theorems composed of necessary and sufficient conditions regarding the solvability of 1.3–1.5 in Sections3and4 Then we present the conclusion inSection 5

2 Terminology

Let P ⊆ R m , X ⊆ R n , and Y i ⊆ R m i i  1, 2 Let f, f α : X → R α ∈ Λ, ϕ  ϕp, x : P × X →

R, and ψ  ψp, u, v : P × Y1× Y2 → 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 semicontinuousin 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, and ψ 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 semicontinuousin 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 / ∅, and define σ#Y, p  sup y∈Y p, y, σ  Y, p 

infy∈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 αTx1 1 − αTx2⊆ Tαx1 1 −

αx2 for any α ∈ 0, 1 and x i ∈ X i  1, 2.

4 a If infp∈Psupx∈X ϕp, x  sup x∈Xinfp∈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 supx∈X ϕp, x  inf 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, x  inf p∈P ϕp, x}, and define Sϕ| X {x ∈ X : ∃ p ∈

P s.t p, x ∈ Sϕ}, the restriction of Sϕ to X if Sϕ is nonempty.

b Replacing X by Y1×Y2and ϕp, x by ψp, u, v, with the similar method one can also define vψ the minimax value of ψ, Sψ the saddle-point set of

ψ, and Sψ| Y1×Y2the restriction of Sψ to Y1× Y2

5 If Y is a convex set and A a subset of Y, 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 x0 ∈ Y is an extremal point of Y if A  {x0} 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 → 2R 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∈Psupx∈X ϕp, x ≤

supx∈Xinfp∈P ϕp, x, and p, 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 If Sϕ / ∅, then vϕ exists, and

vϕ  ϕp, x  sup x∈X ϕp, x  inf p∈P ϕp, x for any p, x ∈ Sϕ The same properties are

also true for ψ  ψp, u, v on P × Y1× Y2 Moreover, we have

∀ x ∈ Sϕ

|X , inf

p∈P ϕ

p, x

 vϕ

,

∀u, v ∈ Sψ

|Y1×Y2, inf

p∈P ψ

p, u, v vψ

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 T1, T2 : X ⊂ R n → 2R m

are convex (or u.s.c with compact values), then αT1 βT2are also convex (or u.s.c.) for all α, β ∈ R.

Proof We only need to prove5

a If T i i  1, 2 are convex, α, β ∈ R, x i i  1, 2 ∈ X, and t ∈ 0, 1, then



αT1 βT2



tx1 1 − tx2  αT1tx1 1 − tx2  βT2tx1 1 − tx2

⊇ αtT1x1 1 − tT1x2  βtT2x1 1 − tT2x2

 tαT1 βT2



x1 1 − tαT1 βT2



x2.

2.2

Hence αT1 βT2is convex

b Now we assume that X is compact.

In case T : X → 2R m

is u.s.c with compact values and α ∈ R, then by 2, 4, T is

closed and the rangeαTX 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  → x0, y0 j → ∞,

thenx j , y j /α ∈ graph T such that x j , y j /α → x0, y0/α as j → ∞, which implies that

y0∈ αTx0 Hence, αT is closed and also u.s.c because of3

In case T i i  1, 2 : X → 2 R m

are u.s.c with compact values, ifx k , y k  ∈ graphT1

T2 with x k , y k  → x0, y0 k → ∞, then x0 ∈ X and there exist u k ∈ T1x k , v k ∈ T2x ksuch

that y k  u k  v k for all k  1, 2, By 4, T1X and T2X are compact, so we can suppose

u k → u0 and v k → v0 as k → ∞ By 2, both T i i  1, 2 are closed, this implies that

y0  u0 v0 ∈ T1 T2x0, and thus T1 T2 is closed Hence by3, T1 T2 is u.s.c because

T1 T2X x∈X T1 T2x ⊆ T1X  T2X and T1X  T2X is compact.

Lemma 2.4 see 8, Theorems 4.1 and 4.2 Let X ⊂ R n

, P ⊂ R m

 be convex compact with RP 

R m

, Σm−1  {p ∈ R m

 : Σm 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, x  φ c p, x on P × X by φ c p, x 

supy∈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, then x → 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 hx  inf y∈Y gx, y (or k : U → R defined by kx  sup y∈Y gx, y) is also 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∈Psupx∈X ϕp, x  sup x∈Xinfp∈P ϕp, x and there exists p, x ∈ P × X such that sup x∈X ϕp, x  inf p∈P ϕp, x.

3 Solvability Theorem to  1.3 

LetΣm−1be introduced as inLemma 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

∈ Σm−1 × X,

b φ c,x

p, y

p, y − c for

p, y

∈ Σm−1 × Tx, x ∈ X.

3.1

Remark 3.1 By both Assumptions inSection 1,Definition 2.1, and Lemmas2.4and2.5, we can see that

1 φ c p, x  sup 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.

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 , ∃i0∈ {1, 2, , m} ⇒ y ≥ c, y i0  c i0,

3 ∃x ∈ Sφ c|X , ∃y ∈ Sφ c,x|Tx , ∃i0∈ {1, 2, , m} ⇒ y ≥ c, y i0  c i0.

Remark 3.3 Clearly, each of 2 and 3 implies that c solves 1.3 because x ∈ X and y ∈ Tx.

So we conclude fromTheorem 3.2that c solves1.3 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 prove1⇒2 and 3⇒1

1⇒2 Assume that 1 holds By 3.1 and Remarks2.2and3.1, it is easy to see that

∀ x ∈ Sφ c

|X , inf

p∈Σ m−1sup

y∈Tx p, y − c  inf

p∈Σ m−1 φ c

p, x

 vφ c

 0,

∀ y ∈ Sφ c,x

|Tx , inf

p∈Σ m−1 p, y − c  vφ c,x

 inf

p∈Σ m−1sup

Then for each x ∈ Sφ c|X and each y ∈ Sφ c,x|Tx, we have

inf

p∈Σ m−1 p, y − c  vφ c,x

 vφ c

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 extΣ∗ ⊂ ext Σm−1  {e1, e2, , e m} So there exists

i0 ∈ {1, 2, , m} such that p  e i0 ∈ ext Σ∗ This implies by3.3 that y i0  c i0, and therefore

2 follows

3⇒1 Let x ∈ Sφ c|X , y ∈ Sφ c,x|Tx , and let i0be presented in3 Since c solves

1.3 and also solves 1.2, by 3.1 and Remarks2.2and3.1, we obtain

0 e i0, y − c ≥ inf

p∈Σ m−1 p, y − c

 vφ c,x

 inf

p∈Σ m−1sup

y∈Tx

φ c,x

p, y

 inf

p∈Σ m−1 φ c

p, x

 vφ c

≥ 0,

3.4

where e i0 

i0

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 v φ c   0 implies that c solves 1.3 However, without any additional restricting conditions, the inverse may not be true

Example 3.5 Let X  0, 12, c  s, s s ∈ 0, 1, and let T : X → 2 R2

be defined by

Tx  1/2, 12for x  x1, x2 ∈ X Then T is an u.s.c and convex set-valued map with convex compact values, and for each p  p1, p2 ∈ Σ1, x  x1, x2 ∈ X and c  s, s s ∈ 0, 1,

φ c p, x  σ#Tx − c, p  p1 p2 − sp1 p2  1 − s Hence, vφ c   1 − s for all s ∈ 0, 1

and therefore,

v

φ c

> 0 and c solves 1.3 for s ∈

1

2, 1 ,

v

φ c

> 0 and c does not solve 1.3 for s ∈

0,1

3.5

This implies that vφ c   0 or vφ c  > 0 may not be the necessary or the sufficient condition that c solves1.3

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 inSection 1, this type of result is only concerned with G − F To get some

further solvability criteria to1.4 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×Fx x ∈ X

by

a H c

p, x

 σ#



Gx, p

−p, c

σ 

p, x

∈ Σm−1 × X,

b H c,x

p, u, v



p, u − c



p, v for

p, u, v∈ Σm−1 × Gx × Fx, x ∈ X.

4.1

By both Assumptions, we know that σ#Gx, p  sup u∈Gx p, u and σ  Fx, p 

infv∈Fx p, v are finite with σ#Gx, p ≥ 0 and p, v ≥ σ  Fx, p > 0 v ∈ Fx for x ∈ X and p∈ Σm−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, x  H c p, x or 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  / ∅.

ii 1 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 , there exists

y ∈ G − Fx such that y ≥ c.

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 , ∃i0∈ {1, , m} ⇒ u − v ≥ c, u i0− v i0 c i0,

3 SH c  / ∅, SH c,x  / ∅ x ∈ SH c|X , and ∃x ∈ SH c|X , ∃u, v ∈ SH c,x|Gx×F x ,

∃i0∈ {1, , m} ⇒ u − v ≥ c, u i0− v i0  c i0.

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 v H c   1 is not the necessary condition of c

making1.5 solvable without any other restricting conditions

Example 4.4 Let X  0, 12, c  s, s ∈ R2

 s ∈ 5/12, 3/4, and G, F : X → 2 R m

 be defined

by Gx ≡ 3/4, 12, Fx ≡ 1/4, 1/32for x  x1, x2 ∈ X Then both G and F are u.s.c convex set-valued maps with convex compact values, and for any p  p1, p2 ∈ Σ1, x  x1, x2 ∈ X, and c  s, s ∈ R2

, we haveG − Fx  Gx − Fx  3/4, 12− 1/4, 1/32  5/12, 3/42,

σ#Gx, p  1, σ  Fx, p  1/4, and p, c  s Therefore,

H c

p, x

 σ#



Gx, p

−p, c

σ 

Fx, p  41 − s for p ∈ Σ1, x ∈ X, vH c  exists with vH c   41 − s > 1 for 5

12 < s < 3

4.

4.2

This implies that, for each c  s, s s ∈ 5/12, 3/4, c solves 1.5 but vH c  > 1.

The proof of both Theorems4.1and4.2can be divided into eight lemmas

Let t ∈ R, T t  G − tF, and c ∈ R m

 Consider the auxiliary inequality system

∃x ∈ X,

Then t ∈ Rsolves4.3 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, x  K 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

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

solves4.3 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∞  limt → ∞ vK t   −∞.

Proof. 1 By both Assumptions and Lemmas2.34 and2.35, T t  G−tF is convex and u.s.c with nonempty convex compact values for each t ∈ R Since K t p, x  sup y∈Gx−tFx p, y − c

forp, x ∈ Σ m−1 × X, applyingLemma 2.4to T  T t and substituting K t p, x for φ c p, x, we

know that1 is true

2 We prove 2 in three steps as follows

a ByLemma 2.31, G and F are u.h.c., which implies byDefinition 2.12 and

Remark 2.21 that for each p ∈ Σ m−1, t, x → K t p, x  σ#Gx, p −

tσ  Fx, p − p, c is u.s.c on R× X Then fromLemma 2.512, we know 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

Since GX and FX are compact by both Assumptions andLemma 2.34, C GX  supu∈GX u and C FX supv∈FX v are finite Then for any x ∈ X, p, p0∈ Σm−1 ,

and t, t0 ∈ R, we have

σ#

Gx, p

 sup

u∈Gx



p0, u  p − p0, u≤ σ#

Gx, p0

 p − p0C GX ,

σ 

Fx, p

 inf

v∈Fx



p0, v  p − p0, v≥ σ 

Fx, p0

This implies that, for each x ∈ X,

σ#

Gx, p

− σ#

Gx, p0 ≤ p − p0C GX ,

σ 

Fx, p

− σ 

Hence for each x ∈ X, t, p → K t p, x  σ#Gx, p − tσ  Fx, p − p, c is continuous on R× Σm−1 Also from Lemmas2.51 and2.52, it follows that both functions



t, p

−→ sup

x∈X

K t

p, x

on R× Σm−1 ,

and t−→ inf

p∈Σ m−1sup

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 t2 > t1 ≥ 0 Since FX ⊂ int R m

 is compact, it is easy to see that

ε0  inf{p, v : p, v ∈ Σ m−1 × FX} > 0 Thus for any p, x ∈ Σ m−1 × X, we

have

K t1

p, x

 σ#

Gx, p

− t2σ 

Fx, p

− p, c  t2− t1σ 

Fx, p

≥ K t

p, x

which implies that vK t1 > vK t2, and hence t → vK t is strict decreasing

on R

c Let ε1  sup{p, u : p ∈ Σ m−1 , u ∈ GX} and ε2 inf{p, c : p ∈ Σ m−1} By both

Assumptions, ε1and ε2are finite Thus for any t > 0 and p, x ∈ Σ m−1 × X, we

have

K t

p, x

 σ#

Gx, p

− tσ 

Fx, p

− p, c

This implies that vK t  ≤ ε1−tε0−ε2 Therefore, vK∞  limt → ∞ vK t  −∞ This completes the proof

Lemma 4.6 1 p → H c p, x x ∈ X and p → sup x∈X H c p, x are l.s.c on Σ m−1

2 x → H c p, x p ∈ 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 onΣm−1, fromLemma 2.51, we can see that p → H c p, x  sup u,v∈Gx×Fx p, u−

c/p, v x ∈ 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} ⊂ Σm−1 × X is a sequence with p k , x k  → p0, x0 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 −→ u0, v k j −→ v0 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

ByLemma 2.32, both G and F are closed Hence, x k j , u k j  → x0, u0 ∈ graph G

andx k j , v k j  → x0, v0 ∈ graph F, which in turn imply that u0 ∈ Gx0, v0 ∈ Fx0

and

lim sup

k → ∞ σ#

Gx k , p k

 lim

j → ∞ p k j , u k j   p0, u0 ≤ σ#

Gx0, p0

,

lim inf

k → ∞ σ 

Fx k , p k

 lim

j → ∞ p k j , v k j   p0, v0 ≥ σ 

Fx0, p0

.

4.12

Combining this with σ  Fx, p > 0 for p, 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 supk → ∞



σ#

Gx k , p k

− p k , c

lim infk → ∞ σ 

Fx k , p k ≤ σ#



Gx0, p0

− p0, c

σ 

Fx0, p0 .

4.13

Theorem 4.2 For 1.5 , the following three statements... t

p, x

which implies that vK t1... class="page_container" data-page ="9 ">

for< i>p, x ∈ Σ m−1 × X, applyingLemma 2.4to T  T t and substituting K t p, x for φ c