báo cáo hóa học:" Research Article Solving the Set Equilibrium Problems Yen-Cherng Lin and Hsin-Jung Chen" pot

We study the weak solutions and strong solutions of set equilibrium problems in real Hausdorff topological vector space settings.. Several new results of existence for the weak solutions

Volume 2011, Article ID 945413, 13 pages

doi:10.1155/2011/945413

Research Article

Solving the Set Equilibrium Problems

Yen-Cherng Lin and Hsin-Jung Chen

Department of Occupational Safety and Health, China Medical University, Taichung 40421, Taiwan

Correspondence should be addressed to Yen-Cherng Lin,yclin@mail.cmu.edu.tw

Received 17 September 2010; Accepted 21 November 2010

Academic Editor: Qamrul Hasan Ansari

Copyrightq 2011 Y.-C Lin and H.-J Chen 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

We study the weak solutions and strong solutions of set equilibrium problems in real Hausdorff topological vector space settings Several new results of existence for the weak solutions and strong solutions of set equilibrium problems are derived The new results extend and modify various existence theorems for similar problems

1 Introduction and Preliminaries

LetX, Y, Z be arbitrary real Hausdorff topological vector spaces, let K be a nonempty closed

convex set ofX, and let C ⊂ Y be a proper closed convex and pointed cone with apex at

the origin and intC / ∅, that is, C is proper closed with int C / ∅ and satisfies the following

conditions:

1 λC ⊆ C, for all λ > 0;

2 C  C ⊆ C;

3 C ∩ −C  {0}.

LettingA, B be two sets of Y, we can define relations “≤ C” and “/≤ C” as follows:

1 A≤ C B ⇔ B − A ⊂ C;

2 A/≤ C B ⇔ B − A /⊂ C.

Similarly, we can define the relations “≤intC” and “/≤intC” if we replace the setC by int C.

The trimappingf : Z × K × K → 2 Y and mappingT : K → 2 Z are given The set equilibrium problemSEPIis to find anx ∈ K such that

for ally ∈ K and for some s ∈ Tx Such solution is called a weak solution for SEP I We note that1.1 is equivalent to the following one:

for ally ∈ K and for some s ∈ Tx.

For the case whens does not depend on y, that is, to find an x ∈ K with some s ∈ Tx

such that

for ally ∈ K, we will call this solution a strong solution of SEP I We also note that1.3 is equivalent to the following one:

for ally ∈ K.

We note that iff is a vector-valued function and the mapping s → fs, x, y is constant

for eachx, y ∈ K, then SEP I reduces to the vector equilibrium problemVEP, which is to findx ∈ K such that

fx, y/≤intC0 1.4

for ally ∈ K Existence of a solution of this problem is investigated by Ansari et al 1,2

Iff is a vector-valued function and Z  LX, Y which is denoted the space of all

continuous linear mappings fromX to Y and fs, x, y  s, y − x, where s, y denotes the

evaluation of the linear mappings at y, then SEP I reduces toGVVIP: to find x ∈ K and

s ∈ Tx such that



for ally ∈ K It has been studied by Chen and Craven 3

If we considerF : K → K, Z  LX, Y, A : LX, Y → LX, Y, and fs, x, y 

As, y − x  Fy − Fx, where s, y denotes the evaluation of the linear mapping s at y,

thenSEPIreduces to theGVVIP which is discussed by Huang and Fang 4 and Zeng and Yao5: to find a vector x ∈ K and s ∈ Tx such that



As, y − x Fy− Fx/≤intC0, ∀y ∈ K. 1.6

IfZ  LX, Y, T : K → LX, Y is a single-valued mapping, fs, x, y  Tx, y − x, then

SEPI reduces to theweak vector variational inequalities problem which is considered by Fang and Huang6, Chiang and Yao 7, and Chiang 8 as follows: to find a vector x ∈ K

such that



for ally ∈ K The vector variational inequalities problem was first introduced by Giannessi

9 in finite-dimensional Euclidean space

Summing up the above arguments, they show that for a suitable choice of the mapping

T and the spaces X, Y, and Z, we can obtain a number of known classes of vector equilibrium

problems, vector variational inequalities, and implicit generalized variational inequalities

It is also well known that variational inequality and its variants enable us to study many important problems arising in mathematical, mechanics, operations research, engineering sciences, and so forth

In this paper we aim to derive some solvabilities for the set equilibrium problems

We also study some results of existence for the weak solutions and strong solutions of set equilibrium problems LetK be a nonempty subset of a topological vector space X A

set-valued function Φ from K into the family of subsets of X is a KKM mapping if for any

nonempty finite setA ⊂ K, the convex hull of A is contained inx∈A Φx Let us first recall

the following results

Fan’s Lemma see 10 Let K be a nonempty subset of Hausdorff topological vector space X Let

G : K → 2 X be a KKM mapping such that for any y ∈ K, Gy is closed and Gy∗ is compact for

some y∗∈ K Then there exists x∗ ∈ K such that x∗∈ Gy for all y ∈ K.

Definition 1.1see 11 Let Ω be a vector space, let Σ be a topological vector space, let K be

a nonempty convex subset ofΩ, and let C ⊂ Σ be a proper closed convex and pointed cone

with apex at the origin and intC / ∅, and ϕ : K → 2Σis said to be

1 C-convex if tϕx1  1 − tϕx2 ⊂ ϕtx1 1 − tx2  C for every x1, x2 ∈ K and

t ∈ 0, 1;

2 naturally quasi -C-convex if ϕtx1  1 − tx2 ⊂ co{ϕx1 ∪ ϕx2} − C for every

x1, x2∈ K and t ∈ 0, 1.

The following definition can also be found in11

Definition 1.2 Let Y be a Hausdorff topological vector space, let C ⊂ Y be a proper closed

convex and pointed cone with apex at the origin and intC / ∅, and let A be a nonempty

subset ofY Then

1 a point z ∈ A is called a minimal point of A if A ∩ z − C  {z}; Min A is the set of

all minimal points ofA;

2 a point z ∈ A is called a maximal point of A if A ∩ z  C  {z}; Max A is the set of

all maximal points ofA;

3 a point z ∈ A is called a weakly minimal point of A if A ∩ z − int C  ∅; Min w A is

the set of all weakly minimal points ofA;

4 a point z ∈ A is called a weakly maximal point of A if A ∩ z  int C  ∅; Max w A is

the set of all weakly maximal points ofA.

Definition 1.3 Let X, Y be two topological spaces A mapping T : X → 2 Y is said to be

1 upper semicontinuous if for every x ∈ X and every open set V in Y with Tx ⊂ V ,

there exists a neighborhoodWx of x such that TWx ⊂ V ;

2 lower semicontinuous if for every x ∈ X and every open neighborhood V y of

everyy ∈ Tx, there exists a neighborhood Wx of x such that Tu ∩ V y / ∅ for

allu ∈ Wx;

3 continuous if it is both upper semicontinuous and lower semicontinuous

We note thatT is lower semicontinuous at x0if for any net{x ν } ⊂ X, x ν → x0,y0 ∈ Tx0 implies that there exists net y ν ∈ Tx ν  such that y ν → y0 For other net-terminology properties about these two mappings, one can refer to12

Lemma 1.4 see 13 Let X, Y, and Z be real topological vector spaces, and let K and C be

nonempty subsets of X and Y, respectively Let F : K × C → 2 Z , S : K → 2 C be set-valued mappings If both F and S are upper semicontinuous with nonempty compact values, then the set-valued mapping G : K → 2 Z defined by

Gx  

y∈Sx

Fx, y Fx, Sx, ∀x ∈ K 1.8

is upper semicontinuous with nonempty compact values.

By using similar technique of 11, Proposition 2.1, we can deduce the following lemma that slight-generalized the original one

Lemma 1.5 Let L, K be two Hausdorff topological vector spaces, and let L, K be nonempty compact

convex subsets of L and K, respectively Let G : L × K → 2Ê

be continuous mapping with nonempty compact valued on L × K; the mapping s → −Gs, x is naturally quasiÊ -convex on L for each

x ∈ K, and the mapping x → Gs, x isÊ -convex on K for each s ∈ L Assume that for each x ∈ K, there exists s x ∈ L such that

MinGs x , x≥Ê Min 

x∈K

Maxw

Then, one has

Min

x∈K

Maxw

s∈L Gs, x  Max

s∈L

Minw

2 Existence Theorems for Set Equilibrium Problems

Now, we state and show our main results of solvabilities for set equilibrium problems

Theorem 2.1 Let X, Y, Z be real Hausdorff topological vector spaces, let K be a nonempty closed

convex subset of X, and let C ⊂ Y be a proper closed convex and pointed cone with apex at the origin and int C / ∅ Given mappings f : Z × K × K → 2 Y , T : K → 2 Z , and ν : K × K → 2 Y , suppose that

1 {0}≤C νx, x for all x ∈ K;

2 for each x ∈ K, there is an s ∈ Tx such that for all y ∈ K,

νx, y≤C fs, x, y, 2.1

3 for each x ∈ K, the set {y ∈ K : {0}/≤ C νx, y} is convex;

4 there is a nonempty compact convex subset D of K, such that for every x ∈ K\D, there is

a y ∈ D such that for all s ∈ Tx,

5 for each y ∈ K, the set {x ∈ K : fs, x, y≤intC {0} for all s ∈ Tx} is open in K.

Then there exists an x ∈ K which is a weak solution of (SEP) I That is, there is an x ∈ K such that

for all y ∈ K and for some s ∈ Tx.

Proof Define Ω : K → 2 Dby

Ωyx ∈ D : fs, x, y/≤intC {0} for some s ∈ Tx 2.4

for ally ∈ K From condition 5 we know that for each y ∈ K, the set Ωy is closed in K,

and hence it is compact inD because of the compactness of D.

Next, we claim that the family{Ωy : y ∈ K} has the finite intersection property, and

then the whole intersection 

y∈K Ωy is nonempty and any element in the intersection



y∈K Ωy is a solution of SEP I, for any given nonempty finite subsetN of K Let D N  co{D ∪ N}, the convex hull of D ∪ N Then DNis a compact convex subset ofK Define the

mappingsS, R : D N → 2DN, respectively, by

Syx ∈ D N:fs, x, y/≤intC {0} for some s ∈ Tx,

Ryx ∈ D N :{0}≤C νx, y, 2.5

for eachy ∈ D N From conditions1 and 2, we have

{0}≤C νy, y ∀y ∈ D N , 2.6 and for eachy ∈ K, there is an s ∈ Ty such that

νy, y− fs, y, y≤C {0}. 2.7

Hence{0}≤C fs, y, y, and then y ∈ Sy for all y ∈ D N

We can easily see thatS has closed values in D N Since, for eachy ∈ D N,Ωy  Sy∩

D, if we prove that the whole intersection of the family {Sy : y ∈ D N} is nonempty, we can deduce that the family{Ωy : y ∈ K} has finite intersection property because N ⊂ D Nand due to condition4 In order to deduce the conclusion of our theorem, we can apply Fan’s

lemma if we claim thatS is a KKM mapping Indeed, if S is not a KKM mapping, neither is R

sinceRy ⊂ Sy for each y ∈ D N Then there is a nonempty finite subsetM of D Nsuch that

coM /⊂ 

Thus there is an elementu ∈ co M ⊂ D Nsuch thatu /∈ Ru for all u ∈ M, that is, {0}/≤ C νu, u

for allu ∈ M By 3, we have

u ∈ co M ⊂y ∈ K : {0}/≤ C νu, y, 2.9

and hence{0}/≤ C νu, u which contradicts 2.6 Hence R is a KKM mapping, and so is S.

Therefore, there exists anx ∈ K which is a solution of SEP I This completes the proof

Theorem 2.2 Let X, Y, Z be real Hausdorff topological vector spaces, let K be a nonempty closed

convex subset of X, and let C ⊂ Y be a proper closed convex and pointed cone with apex at the origin and int C / ∅ Let the mapping f : Z × K × K → 2 Y be such that for each y ∈ K, the mappings

s, x → fs, x, y and T : K → 2 Z are upper semicontinuous with nonempty compact values and

ν : K × K → 2 Y Suppose that conditions (1)–(4) of Theorem 2.1 hold Then there exists an x ∈ K which is a solution of (SEP) I That is, there is an x ∈ K such that

for all y ∈ K and for some s ∈ Tx.

Proof For any fixed y ∈ K, we define the mapping G : K → 2 Y by

Gx  

s∈Tx

for alls ∈ Z and x ∈ K Since the mappings s, x → fs, x, y and T : K → 2 Zare upper semicontinuous with nonempty compact values, byLemma 1.4, we know that G is upper

semicontinuous onK with nonempty compact values Hence, for each y ∈ K, the set



x ∈ K : fs, x, y≤intC {0} ∀s ∈ Tx {x ∈ K : Gx ⊂ − int C} 2.12

is open in K Then all conditions of Theorem 2.1 hold From Theorem 2.1, SEPI has a solution

In order to discuss the results of existence for the strong solution of SEPI, we introduce the condition  It is obviously fulfilled that if Y Ê,f is single-valued function.

Theorem 2.3 Under the framework of Theorem 2.2 , one has a weak solution x of (SEP) I with s ∈ Tx In addition, if Y  Ê, C  Ê , and K is compact, Tx is convex, the mapping s, x → fs, x, x is continuous with nonempty compact valued on Tx × K, the mapping s → −fs, x, x

is naturally quasiÊ -convex on Tx for each x ∈ K, and the mapping x → fs, x, x isÊ -convex

on K for each s ∈ Tx Assuming that for each x ∈ K, there exists t x ∈ Tx such that

Minft x , x, x≥ CMin

x∈K

Maxw 

s∈Tx

then x is a strong solution of (SEP) I ; that is, there exists s ∈ Tx such that

for all x ∈ K Furthermore, the set of all strong solutions of (SEP) I is compact.

Proof FromTheorem 2.2, we know thatx ∈ K such that 1.1 holds for all x ∈ K and for some

s ∈ Tx Then we have Minx∈KMax

s∈Tx fs, x, x≥ C0

From condition   and the convexity of Tx, Lemma 1.5 tells us that Max

s∈TxMinw

x∈K fs, x, x≥ C0 Then there is ans ∈ Tx such that Min wx∈K fs, x, x≥ C0 Thus for allρ ∈x∈K fs, x, x, we have ρ ≥ C0 Hence there existss ∈ Tx such that

for allx ∈ K Such an x is a strong solution of SEP I

Finally, to see that the solution set ofSEPI is compact, it is sufficient to show that the solution set is closed due to the coercivity condition4 ofTheorem 2.2 To this end, letΓ denote the solution set ofSEPI Suppose that net{x α } ⊂ Γ which converges to some p Fix

anyy ∈ K For each α, there is an s α ∈ Tx α such that

fs α , x α , y/≤intC {0}. 2.15

SinceT is upper semicontinuous with compact values and the set {x α } ∪ {p} is compact, it

follows thatT{x α } ∪ {p} is compact Therefore without loss of generality, we may assume

that the sequence{s α } converges to some s Then s ∈ Tp and fs α , x α , y /⊂ − int C Let

Ω  {s, x ∈ z∈K Tz × K : fs, x, y ⊂ − int C} Since the mapping s, x → fs, x, y is

upper semicontinuous with nonempty compact values, the setΩ is open in z∈K Tz × K.

Hencez∈K Tz×K\Ω is closed in z∈K Tz×K By the facts s α , x α ∈ z∈K Tz×K\Ω

ands α , x α →α s, p, we have s, p ∈ z∈K Tz×K\Ω This implies that fs, p, y /⊂ −int C.

We then obtain

Hencep ∈ Γ and Γ is closed.

We would like to point out that condition  is fulfilled if we take Y  Êandf is a

single-valued function The following is a concrete example for both Theorems2.1and2.3

Example 2.4 Let X  Y  Ê, Z  LX, Y, K  1, 2, C  Ê , andD  1, 2 Choose T :

K → 2 LX,Y to be defined byTx  {ax : a ∈ 1, 2} ∈ 2 LX,Y for everyx ∈ K and f :

TK × K × K → 2 Y is defined byfs, x, y  {a  δxy − x : δ ∈ 0, 1}, where x ∈ K,

s ∈ Tx with s  ax, for some a ∈ 1, 2, y ∈ K, and ν : K × K → 2 Y is defined by

νx, y

⎧

⎨

⎩



xy − x, y ≥ x,



Then all conditions of Theorems 2.1 and 2.3 are satisfied By Theorems 2.1 and 2.3, respectively, theSEPInot only has a weak solution, but also has a strong solution A simple geometric discussion tells us thatx  1 is a strong solution for SEP I

Corollary 2.5 Under the framework of Theorem 2.1 , one has a weak solution x of (SEP) I with s ∈ Tx In addition, if Y Êand C Ê , K is compact, Tx is convex,Ê -convex on Tx for each

x ∈ K and the mapping x → fs, x, x isÊ -convex on K for each s ∈ Tx, f : Z × K × K → 2 Y

such that s, x → fs, x, y is continuous with nonempty compact values for each y ∈ K, and

T : K → 2 Z is upper semicontinuous with nonempty compact values Assume that condition 

holds, then x is a strong solution of (SEP) I ; that is, there exists s ∈ Tx such that

for all x ∈ K Furthermore, the set of all strong solutions of (SEP) I is compact.

Theorem 2.6 Let X, Y, Z, K, C, T, f be as in Theorem 2.1 Assume that the mapping y → fs, x, y is C-convex on K for each x ∈ K and s ∈ Tx such that

1 for each x ∈ K, there is an s ∈ Tx such that fs, x, x/≤intC {0};

2 there is a nonempty compact convex subset D of K, such that for every x ∈ K \ D, there is

a y ∈ D such that for all s ∈ Tx,

3 for each y ∈ K, the set {x ∈ K : fs, x, y≤intC {0} for all s ∈ Tx} is open in K.

Then there is an x ∈ K which is a weak solution of (SEP) I

Proof For any given nonempty finite subset N of K Letting D N  coD ∪ N, then D Nis a nonempty compact convex subset ofK Define S : D N → 2DN as in the proof ofTheorem 2.1, and for eachy ∈ K, let

Ωyx ∈ D : fs, x, y/≤intC {0} for some s ∈ Tx. 2.20

We note that for eachx ∈ D N,Sx is nonempty and closed since x ∈ Sx by conditions

1 and 3 For each y ∈ K, Ωy is compact in D Next, we claim that the mapping S

is a KKM mapping Indeed, if not, there is a nonempty finite subset M of D N, such that

coM /⊂ x∈M Sx Then there is an x∗∈ co M ⊂ D Nsuch that

for allx ∈ M and s ∈ Tx∗ Since the mapping

isC-convex on D N, we can deduce that

for all s ∈ Tx∗ This contradicts condition 1 Therefore, S is a KKM mapping, and by

Fan’s lemma, we have

x∈DN Sx / ∅ Note that for any u ∈ x∈DN Sx, we have u ∈ D by

condition2 Hence, we have

y∈N

Ωy 

y∈N

Sy∩ D / ∅, 2.24

for each nonempty finite subset N of K Therefore, the whole intersection y∈K Ωy is

nonempty Letx ∈y∈K Ωy Then x is a solution of SEP I

Corollary 2.7 Let X, Y, Z, K, C, T, f be as in Theorem 2.1 Assume that the mapping y → fs, x, y is C-convex on K for each x ∈ K and s ∈ Tx, f : Z × K × K → 2 Y such that

s, x → fs, x, y is continuous with nonempty compact values for each y ∈ K, and T : K → 2 Z

is upper semicontinuous with nonempty compact values Suppose that

1 for each x ∈ K, there is an s ∈ Tx such that fs, x, x/≤intC {0};

2 there is a nonempty compact convex subset D of K, such that for every x ∈ K\D, there is

a y ∈ D such that for all s ∈ Tx,

Then there is an x ∈ K which is a weak solution of (SEP) I

Proof Using the technique of the proof inTheorem 2.2and applyingTheorem 2.6, we have the conclusion

The following result is another existence theorem for the strong solutions ofSEPI

We need to combineTheorem 2.6and use the technique of the proof inTheorem 2.3

Theorem 2.8 Under the framework of Theorem 2.6 , on has a weak solution x of (SEP) I with s ∈ Tx.

In addition, if Y Êand C Ê , K is compact, Tx is convex and the mapping s → −fs, x, x

is naturally quasi C-convex on Tx for each x ∈ K, f : Z × K × K → 2 Y such that s, x →

fs, x, y is continuous with nonempty compact values for each y ∈ K, and T : K → 2 Z is upper semicontinuous with nonempty compact values Assuming that condition   holds, then x is a strong

solution of (SEP) I ; that is, there exists s ∈ Tx such that

for all x ∈ K Furthermore, the set of all strong solutions of (SEP) I is compact.

Using the technique of the proof inTheorem 2.3, we have the following result

Corollary 2.9 Under the framework of Corollary 2.7 , one has a weak solution x of (SEP) I with

s ∈ Tx In addition, if Y  Ê and C  Ê , K is compact, Tx is convex, and the mapping

s → −fs, x, x is naturally quasi C-convex on Tx for each x ∈ K Assuming that condition

  holds, then x is a strong solution of (SEP) I ; that is, there exists s ∈ Tx such that

for all x ∈ K Furthermore, the set of all strong solutions of (SEP) I is compact.

Next, we discuss the existence results of the strong solutions forSEPIwith the setK

without compactness setting from Theorems2.10to2.14below

Theorem 2.10 Letting X be a finite-dimensional real Banach space, under the framework of

Theorem 2.1 , one has a weak solution x of (SEP) I with s ∈ Tx In addition, if Y Êand C Ê , Tx is convex, fs, x, x  {0} for all s ∈ Tx and for all x ∈ K, the mapping y → fs, x, y is C-convex on K for each x ∈ K and s ∈ Tx and the mapping s → −fs, x, x is naturally quasi C-convex on Tx for each x ∈ K, f : Z × K × K → 2 Y such that s, x → fs, x, y is continuous

for each y ∈ K, and T : K → 2 Z is upper semicontinuous with nonempty compact values Assume that for some r > x, such that for each x ∈ K r , there is a t x ∈ Tx such that the condition

Minft x , x, x≥ CMin 

x∈Kr

Maxw 

s∈Tx

is satisfied, where K r  B0,r∩K.Then x is a strong solution of (SEP) I ; that is, there exists s ∈ Tx such that

for all x ∈ K Furthermore, the set of all strong solutions of (SEP) I is compact.

Proof Let us choose r > x such that condition   holds Letting B0, r  {x ∈ X : x ≤

r}, then the set K r is nonempty and compact inX We replace K by K r inTheorem 2.3; all conditions ofTheorem 2.3hold Hence byTheorem 2.3, we haves ∈ Tx such that

for allz ∈ K r For anyx ∈ K, choose t ∈ 0, 1 small enough such that 1 − tx  tx ∈ K r Puttingz  1 − tx  tx in 2.29, we have

fs, x, 1 − tx  tx/≤intC {0}. 2.30

Proof Using the technique of the proof inTheorem 2. 2and applyingTheorem 2.6, we have the conclusion

The following result is another existence theorem for the strong solutions... ofSEPI

We need to combineTheorem 2. 6and use the technique of the proof inTheorem 2.3

Theorem 2.8 Under the framework of Theorem 2.6 , on has a weak solution x of... we discuss the existence results of the strong solutions forSEPIwith the set< i>K

without compactness setting from Theorems2.10to2.14below

Theorem 2.10