Volume 2010, Article ID 745769,13pages doi:10.1155/2010/745769
Research Article
Fixed Points of Discontinuous Multivalued
Operators in Ordered Spaces with Applications
Shihuang Hong and Zheyong Qiu
Department of Mathematics, Hangzhou Dianzi University, Hangzhou 310018, China
Correspondence should be addressed to Shihuang Hong,[email protected] Received 24 September 2009; Accepted 3 March 2010
Academic Editor: T. Dominguez Benavides
Copyrightq2010 S. Hong and Z. Qiu. 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.
Existence theorems of fixed points for multivalued increasing operators in partially ordered spaces are presented. Here neither the continuity nor compactness is assumed for multivalued operators.
As an application, we lead to the existence principles for integral inclusions of Hammerstein type multivalued maps.
1. Introduction
The influence of fixed point theorems for contractive and nonexpansive mappingssee1,2 on fixed point theory is so huge that there are many results dealing with fixed points of mappings satisfying various types of contractive and nonexpansive conditions. On the other hand, it is also huge that well-known Brouwer’s and Schauder’s fixed point theorems for set-contractive mappings exert an influence on this theory. However, if a mapping is not completely continuous, in general, it is difficult to verify that the mapping satisfies the set- contractive condition. In 1980, M ¨onch3has obtained the following important fixed point theorem which avoids the above mentioned difficulty.
Theorem 1.1. LetEbe a Banach space,K⊂Ea closed convex subset. Suppose that (single) operator F:K → Kis continuous and satisfies that
ithere existsx∈Ksuch that ifC⊂K∩co{x} ∪FCis countable, thenCis relatively compact,
thenFhas a fixed point inK.
It has been observed that continuity is an ideal and important property in the above cited works, while in some applications the mapping under consideration may not be
continuous, yet at the same time it may be “not very discontinuous”. this idea has motivated many authors to study corresponding problems, for instance, the stability of Brouwer’s fixed point theorem 4, similar result for nonexpansive mappings 5, and existence and approximation of the synthetic approaches to fixed point theorems6. Recently, fixed point theory for discontinuous multivalued mappings has attracted much attention and many authors studied the existence of fixed points for such mappings. We refer to 7–11. For example, Hong 8 has extended M ¨onch 3 to discontinuous multivalued operators in ordered Banach spaces by using a quite weak compactness condition; that is, assuming the following condition is satisfied.
HIfC{xn}is a countable totally ordered set andC⊂ wcl{x1} ∪AC, thenCis weakly relatively compact. HereAis a multivalued operator and wclBdenotes the weak closure of the setB.
The purpose of this paper is to present some results on fixed point theorems of M ¨onch type of multivalued increasing operators for which neither the continuity nor the compactness is assumed in ordered topological spaces. However, we will use the following hypothesis.
H1IfC{xn} ⊂Kis a countable totally ordered set andC⊂cl{x1} ∪AC, thenC has a supremum.
Eis a topological vector space endowed with partial ordering “≤”, clBstands for the closure of the setB, andK{x∈E|x≥u0}withu0∈Eis a given ordered set ofE.
This paper is organized as follows. InSection 2, we introduce some definitions and preliminary facts from partially ordered theory and multivalued analysis which are used later. In especial, we introduce a new partial ordering of sets which forms a basis to our main results. InSection 3, we state and prove existence of fixed points, also, maximal and minimal fixed point theorem is presented for discontinuous multivalued increasing operators which are our main results. To illustrate the applicability of our theory, inSection 4, we discuss the existence of solutions to the Hammerstein integral inclusions of the form
ut∈ T
0
kt, sGs, usds a.e.on0, T. 1.1
2. Preliminaries
LetE,≤be a partially ordered topological vector space. By the notation “x < y” we always mean thatx≤yandx /y. Let 2Estand for the collection of all nonempty subsets ofE. Take u0∈Eand letKu0{x∈E|x≥u0}be a given ordered set ofE. The ordered interval ofEis written asu, v {x∈E:u≤x≤v}.
For two subsetsP, QofE, we writeP≤QorQ≥Pif
∀p∈P, ∃q∈Q such thatp≤q. 2.1
Given a nonempty subsetsΩofEwe say thatA : Ω 2E is increasing upwards if u, v∈Ω,u≤v, andx∈Auimply that there existsy∈Avsuch thatx≤y.Ais increasing downwards ifu, v∈Ω,u≤v, andy∈Avimply an existence ofx∈Ausuch thatx≤y.
IfAis increasing upwards and downwards we say thatAis increasing.
LetΓ ⊂ Ebe nonempty. The elementy ∈ Eis called an upperlowerbound ofΓif x≤yx≥ywheneverx∈Γ.Γis called upperlowerbounded with respect to the ordering if its upperlowerbounds exist. The elementz ∈ Eis called a supremum ofΓ, written as zsupΓ, ifzis an upper bound andz≤yas long asyis another upper bound ofΓ. Similarly, we can define the infimum infΓofΓ.
Throughout this paper, unless otherwise mentioned, the partial ordering ofEalways introduced by a closed cone ifEis a Banach space. The following lemmas will be used in after sections.
Lemma 2.1see12. LetEbe an ordered Banach space andBa totally ordered and weakly relatively compact subset ofE, then there existsx∗∈wclBsuch thatx≤x∗foe allx∈B.
An ordered topological vector space E is said to have the limit ordinal property if xn, yn∈Ewithxn≤ynforn1,2, . . ., andxn → x∗, yn → y∗forn → ∞implyx∗≤y∗. By an analogy of the proof of Lemma 1.1.2 in12, we have the following.
Lemma 2.2. IfEhas the limit ordinal property and{xn}is a relatively compact monotone sequence of E, then{xn}is convergent. Moreover,xn≤x∗if{xn}is increasing andx∗≤xnif{xn}is decreasing forn1,2, . . . .Here limn→ ∞xnx∗.
Remark 2.3. Under the assumptions of Lemma 2.2, it is evident that x∗ is the supremum infimumof increasingdecreasingsequence{xn}.
Lemma 2.4. Let the increasing sequence{xn}have the supremumz. If{xni}is a infinity subsequence of{xn}, then{xni}has the supremumz, too.
Proof. Evidently,zis an upper bound of{xni}. Letybe the other one, thenxni ≤ yfori 1,2, . . . .For any givenn, since{xni}is infinity, there existsi0such thatxn≤xni0, which implies thatxn ≤yfor alln≥1. From the definition of supremums it follows thatz≤y, that is,zis the supremum of{xni}.
Lemma 2.5. Suppose that every countable totally ordered subset of the partially ordered setY has a supremum inY. Let the operatorF:Y → Y satisfyFx≥xfor allx∈Y, then there existsx0 ∈Y such thatFx0 x0.
Proof. Takez0 ∈Y any fixed and letzi1 Fzifori0,1, . . ., thenzi1 ≥zithat is,{zi}is increasing. From our assumption it follows that{zi}has a supremum denoted byz10supzi. Let
Γ1{z0, z1, . . .} ∪ z10
. 2.2
Ifz10 Fz10, then the conclusion of the lemma is proved. Otherwise, takez1i Fz1i−1for i1,2, . . . .Again, the set{z10, z11, . . .}has the supremumz20supz1i. DenoteΓ2{z10, z11, . . .}∪
{z20}. Ifz20Fz20, then the conclusion of the lemma is proved. Otherwise, takez2i Fz2i−1 fori 1,2, . . ., and letΓ3 {z20, z21, . . .} ∪ {z30}withz30 supz2i. In general, having defined Γk {zk−10 , zk−11 , . . .} ∪ {zk0}with zk−1i Fzk−1i andzk0 supzk−1i , wherez0i zi andk, i 1,2, . . ., ifzk0 Fzk0, which completes the proof. Otherwise, repeating this process, either the conclusion of the lemma is proved, or we can obtain a set sequenceΓ1,Γ2, . . .satisfying
i Γk{zk−11 , zk−12 , . . .} ∪ {zk0}withzk0 supzk−1i andzki Fzki−1,i, k1,2, . . .;
iizki−1≤zki fori, k1,2, . . .;
iiizk−1j ≤zkt j, t0,1,2, . . ., andz0i zi. LetΓ ∞
k1Γk, thenΓis a countable subset and
z0≤x ∀x∈Γ. 2.3
We claim that
FΓ⊂Γ. 2.4
In fact, for anyy ∈ FΓ, there existsx ∈ Γsuch thaty Fx. There existsΓk such that x∈Γk. Ifx zk−1i for some nature numberi, thenyFx zk−1i1 ∈Γkwhich yieldsy ∈Γ.
Otherwise, we havexsupzk−1i zk0 ∈Γk1. This implies thatyFx Fk0 zk1 ∈Γk1. Consequently,y∈Γ. From the arbitrariness ofyit follows that2.4is satisfied.
Finally, combiningiiandiiiwe see easily thatΓis totally ordered. Our hypothesis guarantees thatΓhas a supremum, written asx∗supΓ. Note that2.4guaranteesFx∗∈ Γ, we haveFx∗≤x∗. On the other hand, the definition ofFensures thatFx∗≥x∗. Hence Fx∗ x∗. This proof is completed.
LetΩbe a nonempty subset ofKu0. In this section we impose the following hypotheses on the increasing upwards multivalued operatorA:Ω2E. Set
R{x∈Ω|there exists u∈Axsuch thatx≤u} 2.5
and for anyx∈ Rdefine that
Cx {x, u1, u2, . . . , un, . . .}, Dx Cx∪ {wx}, 2.6 where,wx supCxandui i 1,2, . . . is given as follows: sincex ∈ R, there exists u1 ∈ Ax such thatx ≤ u1. In virtue of the fact thatAis increasing upwards, there exists u2 ∈ Au1 such thatu1 ≤ u2. On the analogy of this process, there existsun ∈ Aun−1 such thatun−1 ≤ un forn 2,3, . . . ,Obviously,Cx ⊂ cl{x} ∪AC, thus, the conditionH1 guarantees that the supremumwtofCxexists.
Remark 2.6. In general, the sequences of these kinds,{un}, may not be unique, that is, every {un}corresponds toCx, moreover, corresponds toDx. For givenx∈ R, we denote with CxandDxthe families ofCxandDxas above, respectively.
In addition, ifEhas the limit ordinal property,Dxis a closed set for anyx∈ R. In fact, let{uni}be any infinity subsequence ofDfor which
uni−→x∗ fori−→ ∞. 2.7
observing that{uni}is increasing, byLemma 2.2we get thatx∗is a supremum of{uni}and byLemma 2.4we getwx x∗.
Definition 2.7. A setΓis said to be sup-closed if the supremum of each countable subset of Γ provided that it existsbelongs toΓ. A multivalued operatorA:Ω 2E is said to have sup-closed values ifAxis sup-closed for eachx∈Ω.
Defining
Xx {u: there exists Dx∈ Dxsuch thatu∈Dx}. 2.8
Lemma 2.8. LetEbe an ordered topological space,Ω a nonempty subset ofKu0 with u0 ∈ E; let A:Ω2Ehave sup-closed values and satisfy hypothesis (H1). Moreover, assume that
H2Ais increasing upwards and satisfiesu0≤Au0,
then for anyCx∈ Cx,Cxhas the supremumwxwhich belongs toR,that is,
wx≤x∗ for some x∗∈Awx. 2.9
Proof. It is clear that Cx has the supremum wx ∈ E. For any ui ∈ Cx/{x}, from ui ∈ Aui−1and ui−1 ≤ wxthere exists xi ∈ Awxsuch thatui ≤ xi. We can assume that the sequence {xi} is increasing. Indeed, if xi ≤ xi1 for i 1,2, . . ., our purpose is reached. Otherwise, there existsi0 such thatxi0/≤xi01, then we takexi01 instead of xi0. Let M{wx, x1, x2, . . . , xn, . . .}, thenM⊂cl{wx} ∪AM. ConditionH1guarantees that Mhas a supremumx∗supM. Clearly,wx≤x∗. By virtue of the fact thatAhas sup-closed values, we havex∗∈Awx. This proof is complete.
For the sake of convenience, in this paper, bywxwe always stand for the supremum of Cx. For given x ∈ R, letWxbe a set consisting of all wxgiven as inLemma 2.8, thenWis an increasing map. Now for anyun ∈CxLemma 2.4showsDun⊂Dx, thus, Wun⊂ Wx. Define
Z{Dx:x∈ R}. 2.10
It is obvious thatDu0∈Z. Hence,Zis nonempty. A relation “≤1on Z is defined as follows it is easy to see thatZ,≤1is a partially ordered set:
Dx Dy⇔xy, wx wy;
Dx<1Dy⇔x < yandwx≤wy.
Remark 2.9. It is clear we may assume that, for anyu∈Dx, there existsv∈Dysuch that u≤vifDx<1Dy.
Let us assume that there exists someu0∈ Rsuch that H3Wu0⊂clAXu0.
Define
S{Dx:x∈ R, Wx⊂clAXx}. 2.11
Obviously,Du0∈S, that is,Sis nonempty ifAis increasing upwards. Now we denote≤2
as a relation onSdefined by, for anyDx,Dy∈S, IDx Dy⇔xy;
IIDx<2Dy ⇔ afor all Dx ∈ Dx, there exists Dy ∈ Dy such that Dx<1Dyand
bthere exists a countable at most and totally ordered subsetQ⊂ Rsuch that b1x < q < yfor anyq∈Q.
b2There existsDx∈ Dxsuch that
y∈cl
⎛
⎝{wx} ∪
q∈Q
W q⎞
⎠, y
≥ W q
, wx≤q ∀q∈Q
; 2.12
b3
q∈QXq is a totally ordered set and satisfies
q∈QXq ⊂ clWx∪
A
q∈QXq.Qis called a link of linkingDxwithDy.
Remark 2.10. b2may be satisfied. In fact, we can take empty set as a link of linkingDx andDy. Thus,Dx<2Dyimplies that for anyDx∈ Dxwe can findDy∈ Dysuch thatDx<1Dy. In this case, we takewx y. Besides,Qcan be a finite set, for example, Q{q1, q2, . . . , qm}withq1 < q2<· · ·< qm,thenq1 infWx, ysupWqm.b3and the conditionH1ensure
q∈QXqto exist the supremum, so, fromLemma 2.2the elementy satisfyingb2exists.
Lemma 2.11. The relation “≤2” satisfies that iDx≤2Dx;
iiDx≤2DyandDy≤2DximpliesDx Dy;
iiiDx≤2Dy, Dy≤2DzimpliesDx≤2Dz.
Therefore,S, ≤2is a partially ordered set.
Proof. iandiiare satisfied. Trivial byIandIIa. To proveiii, for any givenDx∈ Dxwe takeDy ∈ Dysuch thatDx≤1Dyand we can find Dz ∈ Dzsuch that Dy≤1Dz. It is sufficient to assume that at least one of the above equalities does not hold.
The definition of<1guarantees that
x≤y≤z, 2.13
wx≤w y
≤wz, 2.14
and at least one strictly inequality in 2.13 holds. The links linking Dx with Dyand linkingDywithDzare written, respectively, asQandQ. LetQQ∪Q∪ {y}, for any q∈Q, q∈Q, if none of equalities in2.13holds, then byb1we have
x < q< y, y < q< z. 2.15
If at least one equality in2.13holds, for instance,xy, thenb1andb2show that wx≤q≤w y
, w y
≤q< z. 2.16
Hence,Q⊂ Ris a countable totally ordered subset and satisfiesb1.
Next we will prove thatQsatisfiesb2. It is clear that the following consequences are true, that is,z≥wy≥wxandz∈cl{wy} ∪
q∈QWq⊂cl{wx} ∪
q∈QWq. It is easy to see that{z} ≥ Wqfor allq∈Qby{z} ≥ WqandWq≥ Wqfor allq∈Q. Also,wx≤qfor allq∈Q.
Finally, we prove thatQsatisfiesb3. For allx1, x2∈
q∈QXq, there existq, q∈Q with q ≤ q because Q is totally ordered and Dq ∈ Dq, Dq ∈ Dq such that x1 ∈ Dq, x2 ∈ Dq. Ifq, q ∈ Qor q, q ∈ Q, thenx1 andx2 are ordered by b3. Ifq ∈Q, q ∈Q, fromb1andb2it follows thatDq<1Dq, which shows thatx1 ≤ wq≤q ≤x2. To conclude,
q∈QXqis totally ordered. Noting that bothQandQhave supremums, by the definition ofS, we have
W y
⊂cl A X y
. 2.17
Therefore,
q∈Q
X q
⎛
⎝
q∈Q
X q⎞
⎠⎛
⎝
q∈Q
X q⎞
⎠ X y
⊂cl
⎛
⎝Wx∪A
⎛
⎝
q∈Q
X q⎞
⎠
⎞
⎠ cl
⎛
⎝w y
∪A
⎛
⎝
q∈Q
X q⎞
⎠
⎞
⎠ X y
⊂cl
⎛
⎝Wx∪A
⎛
⎝
q∈Q
X q⎞
⎠
⎞
⎠⎛
⎝W y
∪A
⎛
⎝
q∈Q
X q⎞
⎠
⎞
⎠
cl A X y
⊂
⎛
⎝Wx∪A
⎛
⎝
q∈Q
X q⎞
⎠
⎞
⎠.
2.18
This shows thatQsatisfiesb3. Consequently,Dx<2Dz,which completes this proof.
3. Main Results
Now we can state and prove our main results.
Definition 3.1. u∈Eis said to be a fixed point of the multivalued operatorAifu∈Au. The fixed pointx∗ofAis said to be a maximal fixed point ofAifux∗wheneveru∈Auand x∗ ≤u. Ifx∗is a fixed point and ifx∗ uwheneveru∈Auandu≤x∗, we say thatx∗is a minimal fixed point ofA.
Theorem 3.2. Assume thatEis an ordered topological space. Letu0∈E,Ω⊂Ku0be nonempty and the multivalued operatorA :Ω2E have sup-closed values such that hypotheses (H1)–(H3) hold.
ThenAadmits at least one fixed point inKu0.
Proof. IfShas a maximal elementDx∗, thenx∗is a fixed point ofA. In fact, sincex∗ ∈ R, we can findu∈Ax∗such thatx∗ ≤u. From the definition ofCx∗we can letu∈Dx∗∈ Dx∗. This impliesu≤ wx∗. We claim thatx∗ u. Suppose thatx∗ < u, thenx∗ < wx∗ and Dx∗<1Dwx∗. Take empty set as a link of linking Dx∗ withDwx∗, we have Dx∗<2Dwx∗, which contradicts the definition of maximal element.
To prove the existence of maximal element ofS, by Zorn’s lemma, is thus sufficient to show that every totally ordered subset ofS has an upper bound. LetMbe any such a subset ofS. To this purpose, we consider the setN
Dx∈MXx. Obviously,N⊂Ku0. We claim thatNis totally ordered. Indeed, for anyy1, y2∈N,there existDx1,Dx2∈Mand Dx1 ∈ Dx1, Dx2 ∈ Dx2such thaty1 ∈ Dx1, y2 ∈ Dx2. IfDx1 Dx2, then y1, y2 is ordered. Otherwise, we can assume thatDx1<1Dx2, thus, fromLemma 2.8and b2it follows thaty1≤wx1≤x2≤y2. Conclusively,Nis a totally ordered subset.
We will prove that any countable totally ordered subset ofN has a supremum. It is enough to prove that any given strictly monotone sequence{yn}ofNthere is a supremum.
From the definition ofN, there existDxn∈MandDxn∈ Dxnsuch thatyn∈Dxnfor n 1,2, . . . .For anyx ∈ R, from the definition ofDx, it followsDxhas a supremum.
Moreover, Lemma 2.4 guarantees that {yn} has a supremum if yn ∈ Dxm with n ≥ m for some givenm. It is suffices to consider the fact that there exists a subsequence of{yn} without loss of generality, we may assume that it is{yn}itselfsuch thatyn/∈Dxm n /m.
Case 1. If{yn}is strictly increasing, thenyi< yi1 i1,2, . . .. We claim that
Dx1<2Dx2<2· · ·<2Dxn<2· · · . 3.1
If it is contrary, there exists some i such that Dxi1≤2Dxi. It is easy to know that Dxi1/Dxi. b2 implies thatwxi1 ≤ xi, therefore, yi1 ≤ wxi1 ≤ xi ≤ yi. This contradicts{yn}increasing. The claim follows.
TakingQias the link of linkingDxi1withDxifori1,2, . . . .Let
C∞
i1
⎛
⎝Xxi∪
⎛
⎝
q∈Qi
X q⎞
⎠
⎞
⎠, 3.2
b3shows thatCis countable totally ordered. For anyz∈C\ {x1}, there existsj such that z∈Xxj∪
q∈QjXq. Ifzxj, by means ofb2and
W xj−1
∈cl A X xj−1
, 3.3
we have
z∈cl
⎛
⎝W xj−1
∪A
⎛
⎝
q∈Qj−1
X q⎞
⎠
⎞
⎠⊂cl
⎛
⎝A X xj−1
∪A
⎛
⎝
q∈Qj−1
X q⎞
⎠
⎞
⎠
cl
⎛
⎝A
⎛
⎝X xj−1
∪
⎛
⎝
q∈Qj−1
X q⎞
⎠
⎞
⎠
⎞
⎠⊂clAC.
3.4
Ifz ∈ Xj withz /xj, then, by 3.3,z ∈ clAXxj ⊂ clAC. If z ∈
q∈QjXqwith z /∈Xxj, then from conditionb3it follows thatz∈clA
q∈QjXq⊂clAC. To sum up,C⊂ {x1} ∪clAC⊂cl{x1} ∪AC, which, combining the conditionH1, yields that Chas a supremum. Hence, byLemma 2.4,{yn}has a supremum.
Case 2. It is clear that{yn}has a supremum when{yn}is decreasing.
Now, we prove that N has a maximal element. Suppose, on the contrary, for any y ∈ N, that there existsy1 ∈ N such thaty ≤ y1 and y1/y. LetFy y1, thenF is an operator mappingNintoNand satisfiesFy≥yandFy/yfor everyy∈N. In virtue of Lemma 2.5, there existsy∗∈Nsuch thatFy∗ y∗. On the other hand, by the definition of F, we haveFy∗≥y∗andFy∗/y∗, a contradiction. Therefore,Nhas a maximal element, that is, there existsx∗∈Nsuch thatx≤x∗for allx∈N.
Finally, we shall prove thatDx∗is an upper bound ofM. Sincex∗ ∈N, there exists Dx ∈ Mand Dx ∈ Dxsuch that x∗ ∈ Dx, which implies thatx∗ ≤ wx.On the other hand, sincewx ∈ N, we havewx ≤ x∗. This compels x∗ wx. Taking empty set as a link of linkingDx with Dwx, we have that Dx≤2Dwx Dx∗. Given Du∈M, in virtue ofMbeing totally ordered, orDu≤2Dxwhich impliesDu≤2Dx∗; orDx<2Du, which, applyingb2, yieldswx≤u. Therefore,x∗wx≤u. Noting that u ∈ N, we have thatu ≤ x∗. Conclusively,u x∗, so, byawe haveDx∗ Du. This shows thatDx∗is an upper bound ofM. This proof is completed.
Remark 3.3. We observe that the result of Theorem 3.2 is true under assumptions of Theorem 3.2if all “ clare written as “wcl.The following corollary shows thatTheorem 3.2 extends and improves the results of8.
Corollary 3.4. LetEbe an ordered Banach space,A:Ω⊂E → 2Ebe a multivalued operator having nonempty and weakly closed values. Assume that there existsu0 ∈Esuch that conditions (H2), (H3) and (H) hold, thenAhas at least a fixed point.
Proof. Lemma 2.1shows that there existsx∗∈wclCsuch thatx≤x∗for allx∈C. By means of Eberlein’s theorem andLemma 2.2we have thatx∗is the supremum ofC, that is,H1is satisfied. Moreover, this implies thatwx∈wclAD, that is,Ais upper sequentially order closed in the sense of “weak.” SinceAhas weakly closed values,Ahas sup-closed values.
FromRemark 3.3Ahas a fixed point.
Corollary 3.5. LetEbe a weakly sequently completed ordered Banach space,P a normal cone. If the operatorAis bounded and satisfies conditions (H2) and (H3), thenAhas at least one fixed point in Ku0.
Proof. It is suffice to prove that condition H1 holds. Under these hypotheses, every bounded subset is weakly relatively compactsee4, which implies thatH1is true.
In what follows, we shall consider the existence of maximal and minimal fixed points.
Theorem 3.6. Under assumptions ofTheorem 3.2,Ahas a minimal fixed point inKu0.
Proof. LetFixAdenote the set consisting of fixed points ofA. FromTheorem 3.2it follows thatFixAis nonempty. SetS1 {Dx:x∈ R, x≤yfory∈FixA}. Clearly,S1 ⊂Sand S1,≤2is a partially ordered set. By the same methods as to proveTheorem 3.2, we can prove thatS1has a maximal elementDx∗andx∗is a fixed point ofAinS1. It is easy to see thatx∗ is minimal fixed point ofA. This completes the proof ofTheorem 3.6.
The next result is dual to that ofTheorem 3.6.
Theorem 3.7. Assume thatEis an ordered topological space. Letv0 ∈ E,Ω ⊂ Kv0 : {x ∈ E : x≤v0}be nonempty and the multivalued operatorA:Ω2Ehave inf-closed values such that the following hypotheses are satisfied.
h1IfC{xn} ⊂Kv0 is a countable totally ordered set andC⊂cl{x1} ∪AC, thenChas a infimum.
h2Ais increasing downwards andAv0≤v0.
h3Wv0 ⊂ clAXv0, whereWxstands for a set which consists of all infimums of Cx(its definition is similar toWx).
ThenAadmits at least one fixed point inKv0.
Theorem 3.8. Assume that the operatorAis increasing and satisfies conditions (H1)–(H3) and (h1)–
(h3), thenAhas maximal and minimal fixed points onu0, v0.
Remark 3.9. If E has the limit ordinal property, A is increasing and has nonempty closed values. Assume thatAu0, v0is relatively sequentially compact and conditionsH3and h3hold, thenAhas sup-closed and inf-closed values and satisfies conditionsH1andh1 onu0, v0. Thereby,Ahas maximal and minimal fixed points onu0, v0. In this sense, we extend and improve the corresponding results of Theorem 2.1 in10.
Corollary 3.10. LetEbe a partially ordered Banach space. If there existu0, v0 ∈Ewithu0≤v0such thatu0≤Au0, Av0≤v0. Assume thatAis increasing, has nonempty closed values, and satisfies one of the following hypotheses, thenAhas maximal and minimal fixed points onu0, v0.
s1Pis a regular cone.
s2IfC{xn} ⊂u0, v0is countable totally ordered subset andC⊂cl{x1} ∪AC,then Cis relatively compact subset.
s3Au0, v0is a weakly relatively compact set.
s4 u0, v0 is a bounded ordered interval, and for any countable noncompact subset C ⊂ u0, v0 with αC/0, one has αAC < αC, where α· denotes Kuratowskii’s noncompactness measure.
Proof. s2impliesH1andh1holds. The rest is clear.
Remark 3.11. s2is main condition of12for single-valued operators,s4is main condition of13. Hence the results presented here extend and improve the corresponding results of the above mentioned papers.
4. Application
In this section we assume thatE, · is a Banach space with partial ordering derived by the continuous bounded functionϕ:E → R as followssee14:
x≤y iffx−y≤ϕx−ϕ y
. 4.1
To illustrate the ideas involved in Theorem 3.8 we discuss the Hammerstein integral inclusions of the form
ut∈ T
0
kt, sGs, usds on0, T. 4.2
Herekis a real single-valued function, whileG:0, T×E → 2Eis a multivalued map with nonempty closed values.
Let 0 < T < ∞, I 0, T, p ∈ 1,∞, q ∈ 0,∞andr ∈ 1,∞be the conjugate exponent ofq, that is, 1/q1/r1. Letup T
0uspds1/pdenote the norm of the space LpI, E. Foru, v∈LpI, Estipulate thatu≤vif and only ifut≤vtwith allt∈I.
In order to prove the existence of solutions to 4.2 in LpI, E we assume the following.
S1The functionk :I2 → Rsatisfies thatkt,·∈LrIandt → kt,·r belongs to LpI.
S2Gt, uis increasing with regard toufor fixedt∈0, T.
S3There exist u0, v0 ∈ CI, E with u0 ≤ v0 such that u0t ≤ Gt, u0t and Gt, v0t≤v0tfor everyt∈I.
S4G·, xhas a strongly measurable selection onIfor eachx∈E.
S5sup{ut:ut∈Gt, x} ≤hta.e. onIfor allx∈E. Hereh∈LqI,R.
Theorem 4.1. Assume that conditions (S1)–(S5) hold, then4.2has maximal and minimal solutions inu0, v0.
Proof. Define a multivalued operatorAas follows:
Axt T
0
kt, sGs, xsds. 4.3
S4guarantees thatAmakes sense. For anyv∈Axwithx∈LpI, E, there existsu∈G·, x
such thatvt T
0kt, susds. FromS5and H ¨older inequality, it follows that vt ≤
T
0
kt, susds≤ T
0
hskt, sds≤ hqkt,·r :at. 4.4
This implies that v ∈ LpI, E, that is, A maps LpI, E into itself. We seek to apply Theorem 3.8. Note thatS1andS3guarantee thatu0≤Au0, Av0≤v0. For every given t ∈ I and anyCu0 ∈ Cu0, setCu0t {xnt} with xnt ≤ xn1tfor n 1,2, . . . . Thus,{ϕxnt}is a decreasing sequence. Note thatϕis a bounded function, we obtain that the sequence{ϕxnt}is convergent. Hence, for anyε >0, there exists a natural numbern0
such that
xmt−xnt ≤ϕxnt−ϕxmt< ε, 4.5 wheneverm > n ≥ n0. This shows that{xnt} is a Cauchy sequence, thereby {xnt} is convergent. Lemma 2.2guarantees supCu0t wu0t limn→ ∞xnt, which yields wu0t ∈ clACu0t ⊂ clAXu0t. From the arbitrariness of t it follows that wu0is supremum ofCu0. From4.4and the dominated convergence theorem, it follows thatwu0 ∈ LpI, E. Moreover, xn−wu0p → 0 forn → ∞. Consequently, Wu0 ⊂ clAXu0. Similarly, we haveWv0 ⊂ clAXv0. This shows thatH3andh3are satisfied fort ∈ I.S2 guarantees thatAis increasing. It is easy to see that Ahas closed values. This yields thatAhas sup-closed and inf-closed values.
Finally, we check conditionsH1and h1. Suppose that the setC {xn} ⊂ Ku0 is countable, totally ordered, and satisfiesCt ⊂ cl{x1t} ∪ACtfor all t ∈ I. We have to prove that the setCthas a supremum. SinceCtis countable totally ordered, we can assume Ct {xnt : n ≥ 1}with xnt ≤ xn1t forn 1,2, . . . .This implies that the sequence{ϕxnt}is decreasing. In the same way, we can prove that the sequence{xnt}
is convergent. Again,Lemma 2.2guarantees thatCthas a supremum, which implies that conditionH1is satisfied. Similarly, we can prove that conditionh1holds. All conditions ofTheorem 3.8are satisfied, consequently, the operatorAhas minimum and maximum fixed points inu0, v0and this proof is completed.
Remark 4.2. By comparing the results of Theorem 4.2 in 15 in which Couchouron and Precup have proved that4.2has at least one solution, we omit the conditions thatGt, xis continuous and has compact values inTheorem 4.1.
Acknowledgments
This work was supported by the Natural Science Foundation of Zhejiang ProvinceY607178 and the Natural Science Foundation of China10771048.
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