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Title:
Applications of degree theories to nonlinear operator equations in Banach spaces
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Adhikari, Dhruba R
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University of South Florida
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Maximal monotone operators
M-accretive operators
Mappings of type (S+)
Excision property
Compact resolvents
Nonzero solutions
Dissertations, Academic -- Mathematics -- Doctoral -- USF   ( lcsh )
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ABSTRACT: Let X be a real Banach space and G1; G2 two nonempty, open and bounded subsets of X such that 0 \in G2 and G2 \subset G1. The problem (*) Tx + Cx = 0 is considered, where T : X \supset D(T) \rightarrow X is an accretive or monotone operator with 0 \in D(T) and T(0) = 0, while C : X \supset D(C) \rightarrow X can be, e.g., one of the following types: (a) compact; (b) continuous and bounded with the resolvents of T compact; (c) demicontinuous, bounded and of type (S+) with T positively homogeneous of degree one; (d) quasi-bounded and satisfies a generalized (S+)-condition w.r.t. the operator T; while T is positively homogeneous of degree one. Solutions are sought for the problem (*) lying in the set D(T + C) \cap (G1 \ G2). Nontrivial solutions of (*) exist even when C(0) = 0. The degree theories of Leray and Schauder, Browder, and Skrypnik as well as the degree theory by Kartsatos and Skrypnik for densely defined operators T; C are used.The last three degree theories do not assume any compactness conditions on the operator C. The excision and additivity properties of these degree theories are employed, and the main results are significant extensions or generalizations of previous results by Krasnoselskii, Guo, Ding and Kartsatos involving the relaxation of compactness conditions and/or conditions on the boundedness of the operator T. Moreover, a new degree theory developed by Kartsatos and Skrypnik has been used to prove a similar result for operators of type T + C, where T : X \supset D(T) \rightarrow 2X* is a multi-valued maximal monotone operator, with 0 \in D(T) and 0 \in T(0), and C : X \supset D(C) \rightarrow X* is a densely defined quasi-bounded and finitely continuous operator of type (~S+). The problem of existence of nonzero solutions for Tx + Cx + Gx \ni 0 is also considered. Here, T is maximal monotone, C is bounded demicontinuous of type (S+), and G is of class (P).Eigenvalue and invariance of domain results have also been established for the sum L + T + C : G \cap D(L) \rightarrow 2X* , where G \subset X is open and bounded, L : X \supset D(L) \rightarrow X* densely defined linear maximal monotone, T : X \rightarrow 2X* bounded maximal monotone, and C : G \rightarrow X* bounded demicontinuous of type (S+) w. r. t. D(L).
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Dissertation (Ph.D.)--University of South Florida, 2007.
Bibliography:
Includes bibliographical references.
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by Dhruba R. Adhikari.
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Document formatted into pages; contains 94 pages.
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Applications of degree theories to nonlinear operator equations in Banach spaces
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ABSTRACT: Let X be a real Banach space and G1; G2 two nonempty, open and bounded subsets of X such that 0 \in G2 and G2 \subset G1. The problem (*) Tx + Cx = 0 is considered, where T : X \supset D(T) \rightarrow X is an accretive or monotone operator with 0 \in D(T) and T(0) = 0, while C : X \supset D(C) \rightarrow X can be, e.g., one of the following types: (a) compact; (b) continuous and bounded with the resolvents of T compact; (c) demicontinuous, bounded and of type (S+) with T positively homogeneous of degree one; (d) quasi-bounded and satisfies a generalized (S+)-condition w.r.t. the operator T; while T is positively homogeneous of degree one. Solutions are sought for the problem (*) lying in the set D(T + C) \cap (G1 \ G2). Nontrivial solutions of (*) exist even when C(0) = 0. The degree theories of Leray and Schauder, Browder, and Skrypnik as well as the degree theory by Kartsatos and Skrypnik for densely defined operators T; C are used.The last three degree theories do not assume any compactness conditions on the operator C. The excision and additivity properties of these degree theories are employed, and the main results are significant extensions or generalizations of previous results by Krasnoselskii, Guo, Ding and Kartsatos involving the relaxation of compactness conditions and/or conditions on the boundedness of the operator T. Moreover, a new degree theory developed by Kartsatos and Skrypnik has been used to prove a similar result for operators of type T + C, where T : X \supset D(T) \rightarrow 2X* is a multi-valued maximal monotone operator, with 0 \in D(T) and 0 \in T(0), and C : X \supset D(C) \rightarrow X* is a densely defined quasi-bounded and finitely continuous operator of type (~S+). The problem of existence of nonzero solutions for Tx + Cx + Gx \ni 0 is also considered. Here, T is maximal monotone, C is bounded demicontinuous of type (S+), and G is of class (P).Eigenvalue and invariance of domain results have also been established for the sum L + T + C : G \cap D(L) \rightarrow 2X* where G \subset X is open and bounded, L : X \supset D(L) \rightarrow X* densely defined linear maximal monotone, T : X \rightarrow 2X* bounded maximal monotone, and C : G \rightarrow X* bounded demicontinuous of type (S+) w. r. t. D(L).
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Dissertation (Ph.D.)--University of South Florida, 2007.
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Advisor: Athanassios G. Kartsatos, Ph.D.
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Maximal monotone operators.
M-accretive operators.
Mappings of type (S+).
Excision property.
Compact resolvents.
Nonzero solutions.
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by DhrubaR.Adhikari Adissertationsubmittedinpartialfulllment oftherequirementsforthedegreeof DoctorofPhilosophy DepartmentofMathematicsandStatistics CollegeofArtsandSciences UniversityofSouthFlorida MajorProfessor:AthanassiosG.Kartsatos,Ph.D. Wen-XiuMa,Ph.D. MarcusMcWaters,Ph.D. ArunavaMukherjea,Ph.D. BorisShekhtman,Ph.D. YunchengYou,Ph.D. DateofApproval: April26,2007 Keywords:Maximalmonotoneoperators,m-accretiveoperators,mappingsoftype(S+),excisionproperty,compactresolvents,nonzerosolutions cCopyright2007,DhrubaR.Adhikari

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Iwouldalsoliketoexpressmysincerethankstotheothermembersoftheexaminingcommittee:Dr.Wen-XiuMa,Dr.MarcusMcWaters,Dr.ArunavaMukherjea,Dr.BorisShekhtman,andDr.YunchengYou. Finally,IexpressmygratitudetomyparentsandwifeBhagabatifortheirencour-agementandunderstanding.

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1Introduction-Preliminaries1 1.1BanachSpacesandOperatorsofMonotoneType............1 1.2TopologicalDegreeTheories........................4 2AccretiveandMonotoneOperatorEquations8 2.1AccretiveOperatorsTandLeray-SchauderDegree............9 2.2MonotoneOperatorsandBrowderandSkrypnikDegrees........22 2.3OperatorsoftheformT+C+G.....................36 2.4Application.................................46 3DenselyDenedandPerturbedMaximalMonotoneOperators49 3.1DenselyDenedOperatorsT,C......................49 3.2(~S+){PerturbationsofMaximalMonotoneOperators..........63 4NonlinearPerturbationsofLinearMaximalMonotoneOperators71 4.1Introduction.................................72 4.2TheEigenvalueProblem..........................75 4.3InvarianceofDomain............................83 References91 AbouttheAuthorEndPagei

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DhrubaR.Adhikari ABSTRACT

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ThegraphG(T)ofamonotoneoperatorTissaidtobemonotonesetinXX.IntermsofthegraphofT,TismaximalmonotoneifandonlyifG(T)isamaximalmonotonesetinXX,whereXXispartiallyorderedbyinclusion. IfXisreexive,amonotoneoperatorT:XD(T)!2XismaximalmonotoneifandonlyifR(T+J)=Xforevery>0. Forfactsinvolvingmonotoneandaccretiveoperators,andotherrelatedconcepts,thereaderisreferredtoBarbu[3],Browder[7],Cioranescu[11],andZeidler[35].Forasurveypaperoncompactnessandaccretivity,wecitethepaperbyKartsatos[19]. ThefollowinglemmacanbefoundinZeidler([35],p.915).Lemma1.4

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Theinteger-valuedfunctiondissaidtobeaclassicaltopologicaldegreeifthefol-lowingconditionsaresatised:(a)

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LetX=YbearealinnitedimensionalBanachspace,Oallopenboundedsets,FGallcontinuousmapsf: LetXbearealreexiveBanachspace,Y=X,Oallopenboundedsets,FGallboundeddemicontinuousmappingsoftype(S+)from

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Thefollowingexcisionpropertyoftopologicaldegree(cf.Brown[10],Theorem11.5),whichalsoholdsforBrouwerdegree,willbeusedinChapter2andChapter3.Theorem1.12(Excision) InChapter3,weconsidertheproblemoftheexistenceofnonzerosolutionsofTx+Cx=0fordenselydenedoperatorsT;C.DegreetheoriesdevelopedbyKartsatosandSkrypnik([22],[24])havebeenemployed. Finally,inChapter4,weconsidereigenvalueproblemsandinvarianceofdomainresultsfortheoperatorsoftheformL+T+C,whereL:XD(L)!Xisdenselydenedlinearmaximalmonotone,T:X!2Xisboundedmaximalmonotone,andCisoperatorofmonotonetypew.r.t.D(L).TheoperatorsoftheformL+Cwere6

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InSection2.1weapplytheLeray-Schauderdegreetheorytogeneralizeresultsin[12]foroperatorsoftheformT+C,whereTisaccretiveormaximalmonotoneandCiscompact.WedonotassumetheboundednessofT. InSection2.2,weapplytheBrowderandSkrypnikdegreetheorytogiveexten-sionsoftheresultsinSection2.1tomaximalmonotoneoperatorsTandboundeddemicontinuousoperatorCoftype(S+).Thepropertiessuchasexistenceofsolution,invarianceunderhomotopy,normalizationandexcisionofunderlyingdegreetheorieshavebeenemployed.Theorem2.1(Krasnoselskii)

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ForG1;G2opensubsetsofX,G1boundedand Condition(A1)wasrstusedbyKartsatosandSkrypnikin[25]inordertoshowtheexistenceofeigenvaluesforvariousproblemsinvolvingperturbationsofmonotoneandaccretiveoperators.Closedpositively-homogeneousoperators(e.g.,closedlinearoperators),satisfycondition(A1)oneveryboundedsubsetoftheirdomains.Indeed,letFXbeboundedandletfxngD(T)\FbesuchthatkTxnk!1and(2.1.1)holds.ThenTxn

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ThenTx+Cx=0hasanonzerosolutionx2D(T)\(G1nG2). Wenowshowtheexistenceofa0>0suchthattheequationTx+Cx=v0(2.1.3) hasnosolutionx2G1for0.Thevectorv0istheonegivenin(H3).Supposethatthisisnottrue.Then,foreveryn=1;2;:::,thereexistsn>0suchthatn!110

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SinceCxnisbounded,v06=0andn!1,wemusthavekTxnk!1.Consequently,inviewof(2.1.4),wehave whichimpliesnkv0k kTxnk!1andn From(2.1.5)and(2.1.6),weobtainTxn ThiscontradictsthefactthatTsatisescondition(A1)onG1.Now,wexsuchanumber0andconsiderthehomotopyfunctionH1(t;u)=(I+CT1)ut0v0(2.1.8) denedontheset[0;1]

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WenextshowthatH1(t;u)=0hasnosolutionontheboundary@U1foranyt2[0;1].Infact,ifthisisnottruethentherearet02[0;1]andu02@U1=@(T(D(T)\G1)\Bs(0))@(T(D(T)\G1))[@Bs(0)T(D(T)\@G1)[@Bs(0) suchthat(I+CT1)u0=t00v0: Thus,theLeray-Schauderdegreefunctiond(H1(t;);U1;0)iswell-denedforallt2[0;1].Thenwehaved(I+CT1;U1;0)=d(I+CT10v0;U1;0):

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Now,weconsideranotherhomotopyfunctionH2(t;u)=u+tCT1u(2.1.11) denedontheset[0;1]

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By(2.1.10)and(2.1.12),wehaved(I+CT1;U1;0)6=d(I+CT1;U2;0);(2.1.13) withU2=T(D(T)\G2)\Bs(0)T(D(T)\G1)\Bs(0)=U1: suchthatu+CT1u=0.Letx=T1u.Thenx2T1[(T(D(T)\G1)\Bs(0))n(T(D(T)\G2)\Bs(0))]=(D(T)\G1)\T1(Bs(0))n(D(T)\G2)\T1(Bs(0))(2.1.14)=D(T)\T1(Bs(0))\(G1nG2):

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Then(a)inCase(i),foreveryopenMX,thesetT(D(T)\M)isopen.IfMXisclosed,thenT(D(T)\M)isclosed.(b)InCase(ii),(iii),foreveryopenMGthesetTMisopenandforeveryclosedMG,thesetTMisclosed. Nowweturnourattentiontotheresultsinvolvingm-accretiveoperatorsT.ItisastandardargumentthatthedesiredinclusionmaybeachievedbysolvingtheapproximateproblemsofthetypeTx+Cx+1 Thefollowingtheoremreectsasituationofthistype.Theorem2.4

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forsomeh2Xwithkhk=1.Itisthenclearthatlimj!1Txj Now,by(2.1.17),weobtainTxj kTxjkTxj AsintheproofofTheorem2.3,itfollowsthatthecondition(A1)impliesthebound-ednessofthesequencefng.Wemayassumethatn!02R+.Sincefxngisbounded,wemayassume,forasubsequenceoffxngagaindenotedbyfxng,thatCxn!yforsomey2X.Since1

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Inordertoshowthat(H4)holdsforlargenwithTninplaceofT,wesupposethatitdoesn't.Thenwemayassumethatthereexistsequencesftng[0;1]withtn!t02[0;1]andfxngD(T)\@G2suchthatTnxn+tnCxn=0;n=1;2;::::(2.1.20) SinceCiscompactandfxngisbounded,wemayassume,asusual,thatCxn!yforsomey2X.ThenthestrongaccretivityofTandarelationlike(2.1.19)implythatfxngisCauchyandsoitconvergestosomepointx02 BeforeweapplyTheorem2.3,weneedtoshowthatthenumber0inthatproof,whichisdenedsothattheequationTx+Cx+1 hasnosolutionx2G1forall0,isactuallyindependentofnforallsucientlylargen.Weneedthisfactbecausewewilluseauniformbounds>0forthesolutionsofthetwohomotopyequationsH1(t;u)=0andH2(t;u)=0,whereH1(t;u)=u+CT+1 ThesesolutionsinitiallylieintheclosuresofthesetsT+1

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Toshowthat0isindependentofnforallsucientlylargen,letusassumethatthereexistsequencesfng(0;1)andfxngG1suchthatn!1andTxn+Cxn+1 Thus,byTheorem2.3,thereexistn01ands>0independentofn,suchthattheequation(2.1.15)holdsforeverynn0withasolutionxn2D(T)\G1\T+1 Sincefxngisbounded,CiscompactandTism-accretiveandstronglyaccretiveonD(T)\ Notethatx0=2@G1[@G2.Therefore,x02D(T)\(G1nG2)andwehaveprovedthetheoremforthiscase. (b)WenowassumethatCisbounded,continuousandtheresolvent(T+I)1iscompact.Wenotethatthecompactnessoftheresolvent(T+I)1impliesthecom-pactnessofanyresolvent(T+I)1,>0,>0,bythewell-knownresolventidentityform-accretiveoperators.WeconsiderthesamehomotopymappingsH1,H2andmaintainthesameconstantss,0asintherstcase. Toprovethat(H3)holdsfortheoperatorsTninplaceofT,foralllargen,weassumethatitdoesn't.Thenthereexistsequencesfng(0;1),fxngD(T)\@G118

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Ifn!1,asintheproofofTheorem2.3,weobtainacontradictiontothecondition(A1).Thus,thesequencesfngandfTxngarebounded.Wemaynowassumethatn!02R+.Also,from(2.1.25),weseethatTxn+xn=Cxn+xn1 Since(T+I)1iscompactandfCxngisbounded,weseefrom(2.1.26)thatfxngliesinarelativelycompactsetandsofxnghasaconvergentsubsequencewhichweagaindenotebyitself.Lettingxn!x02 Wenowshowthat(H4)holdsforTninplaceofTforalllargen.Assumethatitdoesn't.Wemaythenchoosesequencesftng[0;1],fxngD(T)\@G2suchthatTnxn+tnCxn=0;n=1;2;:::;(2.1.27) andtn!t02[0;1].From(2.1.27),weseethatxn=(T+I)1tnCxn+11 Since(T+I)1iscompactandCisbounded,itfollowsthatfxngliesinarelativelycompactset.Wemayassume,withoutlossofgenerality,thatxn!x02

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whichisinitiallydenedontheset[0;1] wheres1=supx2D(T)\ iswell-denedandconstanton[0;1].Consequently,wehaved(I+CT1;U1;0)=d(Iv0;U1;0)=d(I;U1;v0)=0;

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Infact,let(B1)betrueandxt2[0;1],x2D(T)\@G2suchthatTx+tCx=0.Wepickx2Jxandconsidertwocases:(a)hCx;xi<0;(b)hCx;xi0.Incase(a),weget00.ItisalsoknownthatthedualitymappingJisabounded,single-valuedandbicontinuousoperator.OurrsttworesultsinthissectiongeneralizeTheorem2.3andTheorem2.5tomaximalmonotoneoperatorsT:XD(T)!XandoperatorsC:

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Thenthereexistsanonzerosolutionx2D(T)\(G1nG2)oftheequationTx+Cx=0. arewell-denedcompactdisplacementsofidentityontheclosureofopenandboundedsetsU1=T(D(T)\G1)\Bs(0)andU2=T(D(T)\G2)\Bs(0)respectively.Thenumbers0andsarechoseninamannersimilartothatofTheorem2.3.Theorem2.8

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respectively.Therefore,weomitthedetailsoftheproof. IfT:XD(T)!Xismaximalmonotone,then,fort>0,theoperatorTt:=(T1+tJ1)1:X!Xcalledthe\YosidaApproximant"ofTissingle-valued,boundeddemicontinuousandmaximalmonotone.KartsatosandSkrypnikin([22],Lemma3.1,p.127)showedthatthemapping(t;x)7!Ttxiscontinuouson(0;1)X.Inaddition,Ttx*Txast!0+foreveryx2D(T).Also,Ttx=TJtxwhereJt:=ItJ1Tt:X!Xandsatiseslimt!0+Jtx=xforallx2 Browderdevelopedin[6]adegreetheoryforoperatorsofthetypeT+C,where24

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Inthefollowingresult,weshowtheexistenceofanonzerosolutiontotheproblemTx+Cx=0byusingtheBrowderandSkrypnikdegrees.OuroperatorTismaximalmonotoneandhomogeneousofdegree1.Inparticular,Tcouldbealinearmaximalmonotoneoperator.Weneedthefollowinglemma.Lemma2.9 Letvn=Ttnxn2T(xntnJ1vn).Thenhvn;xni=hvn;xntnJ1vn+tnJ1vni=hvn;xntnJ1vni+hvn;tnJ1vni(2.2.35)tnkvnk2:25

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Next,wex[x;x]2G(T+S).Thenx=x1+x2withx12Txandx22Sx.Thus,hvnx1;xntnJ1vnxi0 giveshvnx1;xnxihvnx1;tnJ1vni=tnhvn;J1vnihx1;tnJ1vni(2.2.36)tnkvnkkx1k

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SinceT+Sismaximalmonotone,weconcludethatx02D(T+S)andy02(T+S)x0.This,however,contradicts(2.3.62)becausewemaynowtakex=x0. If(2.2.33)istrue,thenwecanrepeattheaboveargumenttoarriveat(2.3.62),butwith\>"replacedby\".Inthiscaseweconcludeagainthatx02D(T+S)andy02(T+S)x0andwearedone.

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Fix[x;x]2G(T+S).Thenx=x1+x2withx12Txandx22Sx.UsingthemonotonicityofT+S,wehavehyn+wnx;xnxi0: SinceT+Sismaximalmonotone,wehavethatx02D(T+S)andy02(T+S)x0.Thisisacontradictionto(2.2.42)becausewemaytakex=x0.If(2.2.40)istrue,wecanthenrepeatalltheargumentabovetoarriveat(2.2.42)with\>"replacedby\".Thus,weconcludethatx02D(T+S)andy02(T+S).Theorem2.11

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Thenthereexistsx2D(T)\(G1nG2)suchthatTx+Cx=0. andtheassociatedequationTtx+Cx=0;t2(0;1);x2 Weshowthattheequation(2.2.44)hasasolutionxt2G1nG2forallsucientlysmallt.Tothisend,weshowrstthatthereexist0>0,t0>0suchthattheequationTtx+Cx=v0;(2.2.45) hasnosolutioninG1forevery0,t2(0;t0].Assumethatthisisnotthecase,andletfng(0;1),ftng(0;1],fxngG1besuchthatn!1,tn#0andTtnxn+Cxn=nv0:(2.2.46) Sinceknv0k!1andfCxngisbounded,wemusthavekTtnxnk!1.WealsonotethatTtispositivelyhomogeneousofdegree1foreacht>0.Infact,lety=Tt(sx)=(T1+tJ1)1(sx);

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sJ1y+x=sTt sJ1y+x=sTtJ1y s+x; s+tJ1y s Itcanbeeasilyseenfrom(2.2.47)thatn

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Now,weconsiderthehomotopyfunctionH1(s;x)=Ttx+Cxs0v0;(s;x)2[0;1] wheret2(0;t0]isxed.Itiseasytoseethat,foreverys2[0;1],theoperatorx7!Cxs0v0isdemicontinuousandboundedon Iflimsupn!1hCxn;xnx0i>0;31

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forsomeconstantq>0.Thiscombinedwith(2.2.49)giveslimn!1hTtnxn;xnx0i=q<0: Wemaynowchoosethenumbert0furthersmallerifnecessarysothattheequa-tionH1(s;x)=0hasnosolutionon[0;1]@G1forallt2(0;t0].ThemappingH1(s;x)isanadmissiblehomotopyfortheSkrypnikdegree.Thus,theSkrypnikde-greedS(H1(s;);G1;0)iswell-denedandconstantfors2[0;1].Here,wenotethattheBrowderdegreedB(T+C0v0;G1;0)satisesdB(T+C0v0;G1;0)=dS(Tt+C0v0;G1;0);(2.2.51) forallsucientlysmallt2(0;t0].AssumethatdS(H1(1;);G1;0)6=0forsomet12(0;t0].Weseefrom(2.2.51)andthebasicpropertyoftheSkrypnikdegreethatthereexistsx2G1suchthatTt1x+Cx=0v0;

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WenextconsiderthehomotopyfunctionH2(s;x)=s(Tt+C)x+(1s)Jx;(s;x)2[0;1] Werstshowthatthereexistst12(0;t0]suchthattheequationH2(s;x)=0hasnosolutionon[0;1]@G2foranyt2(0;t1].Ifweassumethecontrary,thenthereexistsequencesftng(0;t0],fsng[0;1],fxng@G2suchthattn#0,sn!s02[0;1],xn*x02X,Cxn*c02X,Jxn*z02X,andsn(Ttnxn+Cxn)+(1sn)Jxn=0:(2.2.54) Sincesn=0isimpossibleasJ(0)=0andJisinjective,wemayassumethatsn>0foralln.Ifsn!0,thenhTtnxn+Cxn;xni=1 becausefkxnkgisboundedawayfromzero.SincehTtnxn;xni0 andfhCxn;xnigisbounded,weseethat(2.2.55)isimpossible.Thismeansthats02(0;1]andthen(2.2.54)impliesTtnxn*c01

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Then,forsomesubsequenceoffng,denotedbyfngagain,wehavelimn!1hCxn;xnx0i=q>0: wherean=hTtnxn+Cxn;xnx0i:

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Finally,sincex0=2@G1[@G2inviewoftheconditions(H3)and(H4)andx02

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andthentheassociatedequation Here,>0andg:

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Weshowthatequation(2.3.58)hasasolutionxt;forallsucientlysmalltand.Tothisend,werstshowthatthereexist0>0,t0>0and0>0suchthattheequationTtx+Cx+gx=v0(2.3.59) hasnosolutioninG1forevery0,t2(0;t0]and2(0;0]. Assumingthecontrary,letfng(0;1),ftng(0;1),fng(0;1)andfxngG1besuchthatn!1,tn#0,n#0andTtnxn+Cxn+gnxn=nv0:(2.3.60) Wemayassumethatgnxn!g2XinviewofthepropertiesofG.So,kTtnxnk!1asknv0k!1andfCxngisbounded. Thus,from(2.3.60),wegetTtnxn WeclaimthatTt(sx)=sTts1(x)forallt;s>0:(2.3.62) Infact,lety=Tt(sx)=(T1+tJ1)1(sx); sJ1y+x=sTt sJ1y+x=sTt s1J1y s+x: s+ts1J1y s38

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Inviewof(2.3.62),weobtain,Ttnxn wheren=kTtnxnk(1)=: kTtnxnk1+Cxn kTtnxnk!1andn Letun=xn WenowconsiderthehomotopymappingH1(s;x;t;)=Ttx+Cx+gxs0v0;s2[0;1];x2

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Letusassumethatlimsupn!1hCxn;xnx0i>0:(2.3.67) Thenthereexistsasubsequenceoffxng,whichwestilldenotebyfxng,suchthatlimn!1hCxn;xnx0i=q;(2.3.68)40

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fort2(0;t0];2(0;0]. AssumethatdS(H1(1;;t1;1);G1;0)6=0;41

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Werstshowthatthereexistt12(0;t0],12(0;0]suchthattheequationH2(s;x;t;)=0hasnosolutionon@G2foranys2[0;1],anyt2(0;t1]andany2(0;1]. Letusassumethecontrary.Thenthereexistsequencestn2(0;t0],n2(0;1],sn2[0;1],andxn2@G2suchthattn#0,n#0,sn!s02[0;1],xn*x02X,Cxn*y02X,gnxn!g2X,Jxn*z02X,andsn(Ttnxn+Cxn+gnxn)+(1sn)Jxn=0:(2.3.71)sn=0isimpossiblebecauseJ(0)=0andJisinjective,wemayassumethatsn>0,foralln.Ifsn!0,hTtnxn+Cxn;xni=1 becausefkxnkgisboundedbelowawayfromzero.SincehTxnxn;xni0andfhCxn;xnigisbounded,weseethat(2.3.72)isimpossible.Thuss02(0;1]and(2.3.71)impliesthatTtn*y0g1

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Then,forsomesubsequenceoffngdenotedbyfngagain,wehave limn!1hCxn;xnx0i=q>0:(2.3.75) FromhTtnxn;xnx0i=qnhCxn;xnx0i;

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ItisthereforeobviousthatthemappingH2(s;x;t;)isanadmissiblehomotopyforSkrypnik'sdegreeandsotheSkrypnikdegreedS(H2(s;;t;);G2;0)iswell-denedandconstantforalls2[0;1],allt2(0;t0]andall2(0;0].BytheinvarianceoftheSkrypnikdegree,forallt2(0;t0],2(0;0],wehavedS(H2(1;;t;);G2;0)=dS(Tt+C+g;G2;0)=dS(H2(0;;t;);G2;0)=dS(J;G2;0)=1:

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@xii: Wenowconsiderthepartialdierentialoperatorindivergenceform(Au)(x)=Xjjm(1)jjDA(x;u(x);:::;Dmu(x));x2: Thereexistp2(1;1),c1>0and12Lq()suchthatjA(x;)jc1jjp1+1(x);x2;2RN0;jjm:(A2) TheLeray-LionsConditionXjj=m(A(x;;1)A(x;;2)(12))>0 issatisedforeveryx2,2RN1,1;22RN2with16=2.46

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issatisedforeveryx2,1;22RN0.(A4) Thereexistc2>0,22L1()suchthatXjjmA(x;)c2jjp2(x);x2;2RN0: thenconditions(A1),(A3)implythatitisbounded,continuousandmonotone(cf.e.g.Kittila[27,pp.25-26],PascaliandSburlan[30,pp.274-275]).Sinceitiscontinuous,itismaximalmonotone.Similarly,condition(A1),withAreplacedbyB,impliesthattheoperatorhCu;vi=ZXjjmB(x;(u))Dv;u;v2Wm;p0();(2.4.77) isaboundedcontinuousmapping.Wealsoknowthatconditions(A1),(A2)and(A4),withBinplaceofAeverywhere,implythattheoperatorCisoftype(S+)(cf.Kittila[27,p.27]).Wethenhavethefollowingtheorem.Theorem2.16

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InSection3.1weapplythedegreetheoryin[24]togivenewresultsaboutexistenceofnonzerosolutionsofoperatorequationsoftheform()Tx+Cx=0withdenselydenedoperatorsT;C. InSection3.2weapplythedegreetheoryin[22]togivesimilarresultsfortheopera-torequationsoftheform()fordenselydenedquasi-boundedandnitelycontinuous(~S+){perturbationCofmultivaluedmaximalmonotoneoperatorT.ThedomainofTneednothavenowadenselinearsubspaceinit.Sincethisnewdegreeisalsobasedonthedegreetheorydevelopedbytheaboveauthorsin[24],theexcisionpropertyofthisdegreestillholds.3.1DenselyDenedOperatorsT,C

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foreveryu;v2D(T).Moreover,LD(T)and ifforevery(u0;h0)2XXwithhTuh0;uu0i0;foru2L;(3.1.3) thenwehaveu02D(T)andTu0=h0;(T3) foranyu02D(T)wehaveinffhTuTu0;uu0i:u2Lg=0;(3.1.4)(T4) foreveryF2F(L),v2Lthemappingt(F;v):F!R,denedbyt(F;v)u=hTu;vi; FortheoperatorC,weconsiderthefollowingassumptions:(C1) andCisquasi-boundedwithrespecttoT,i.e.,foreverynumberS>0thereexistsanumberK(S)>0suchthatfromtheinequalitieshTu+Cu;ui0;kukS;u2L;(3.1.6) wehavekCukK(S);(C2) theoperatorCsatisesthefollowinggeneralized(S+)conditionwithrespect50

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forsomeu02X,h02X,wehaveun!u0,u02D(C)andCu0=h0;(C3) foreveryF2F(L),v2Lthemappingt(F;v):F!R,denedbyt(F;v)u=hCu;vi; Wenotethattheconditions(T2)and(T3)aresatisedforamaximalmonotoneoperatorTwhenD(T)isasubspaceofX.Inthiscase,wemaytakeD(T)=L. ThefollowingtheoremdenesthedegreeforthemappingT+Casin([24],p.427).

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Next,weconsideradmissiblehomotopyfortheabovedegree(cf.[24],p.432).LetXbearealreexiveBanachspace,LbeasubspaceofXsuchthat Considertheone-parameterfamilyofoperatorsTt:XD(Tt)!X,t2[0;1],satisfyingthefollowingconditions:t(1): foreveryt2[0;1],theoperatorTtsatises(T1){(T3)withthespaceLindependentoft;t(2): foreveryv2L,themapping(v):[0;1]!Xdenedby(v)(t)=Tt(v)iscontinuous;t(3): foreveryF2F(L)andv2L,themapping~m(F;v):F[0;1]!Rdenedby~m(F;v)(x;t)=hTt(x);viiscontinuous. Considerthesecondone-parameterfamilyofoperatorsCt:XD(Ct)!X,t2[0;1]satisfyingthefollowingconditions:c(1): thefamilyCtisuniformlyquasi-boundedw.r.t.Tt,i.e.,foreveryS>0thereexistsK(S)>0suchthathTtx+Ctx;xiS;kukS;x2L;t2[0;1](3.1.8) impliesthatkCtukK(S)forallt2[0;1];52

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foreverypairofsequencesftng[0;1],fxngLsuchthatxn*x0,Ctnxn*h0,tn!t0andlimsupn!1hCtnxn;xnx0i0;h(Ttn+Ctn)xn;xni0;(3.1.9) forsomex02X,h02X,t02[0;1],wehavexn!x0,x02D(Ct0)andCt0x0=h0;c(3): foreveryF2F(L)andv2L,themapping~c(F;v):F[0;1]!Rdenedby~c(F;v)(x;t)=hCt(x);viiscontinuous.Denition3.2 WealsoneedthefollowingconditionontheoperatorCwhichisstrongerthan(C2)meaningthattheclassofoperatorssatisfyingthecondition(~C2)belowissmallerthantheclassofoperatorssatisfyingthecondition(C2).53

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BeforeweusetheabovedegreetheorybyKartsatosandSkrypnik(cf.[24])toestablisharesultonthenonzerosolution,weneedthefollowinglemma.Lemma3.4 SincefxngandfJxngareboundedandhTxn+Cxn;xni=nhJxn;xni+hp;xnikpkkxnk;

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assumethatthisisnottrue.Thenforasuitablesubsequenceoffng,denotedbyfngagain,wehavelimn!1hCxn;xnx0i=q; SinceTxn*c0+p,(3.1.14)implieslimsupn!1hTxn;xni
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Let~x2Lbearbitrary.ThemonotonicityofTimplieshT~x;~xxni+hTxn;xnihTxn;~xi=hT~xTxn;~xxni0: sothathT~x+Cx0p;~xx0i0: ThenthereisasolutionoftheproblemTx+Cx=0inL\(G1nG2).56

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Therstinequalityof(3.1.17)implieslimsupn!1hCxn;xnx0i0(3.1.18) andthesecondinequalityof(3.1.17)implieshTxn+Cxn;xniS;(3.1.19) whereS=supf0kv0kkxnk:n1g.Usingcondition(~C2),weobtainxn!x02D(C)andCxn*Cx0.Thusx02D(Ct0)=D(C)andCtnxn=Cxn+Jxntn0v0*Cx0+Jx0t00v0=Ct0x0: WenextconsiderthehomotopyfunctionH2(t;x)=t(Tx+Cx+J)+(1t)Jx;(t;x)2[0;1](L\

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Werstshowthat0=2H2(t;x)forall(t;x)2[0;1](L\@G2)foreach2(0;0].Otherwise,thereexistsapoint(t;x)2[0;1](L\@G2)forsome2(0;0],i.e.,t(Tx+Cx+J)+(1t)Jx=0: tJ=0; implieshTx+Cx;xi+1 Toverifyconditionc(2),letusconsiderthesequencesftng[0;1],fxngLsuchthattn!t02[0;1],xn*x02X,Ctnxn*c0andlimsupn!1hCtnxn;xnx0i0;hTtnxn+Ctnxn;xni0:(3.1.21) Assumethatt0=0.Ifthereisasubsequenceofftkgoftnsuchthattk=0,k=1;2;:::;thenCtkxk=tk[Cxk+Jxk]+(1tk)Jxk=Jxk59

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becausefortn=0,wehavexn=0asabove.Fortn>0,theinequalityfollowsinanobviousmanner.Bythequasi-boundednessofCw.r.t.T,itfollowsthatfCxngisbounded.SincefJxngisalsoboundedandt0=0,therstinequalityinc(2)nowimplieslimsupn!1hJxn;xnx0i0:(3.1.23) The(S+)-propertyofJimpliesthatxn!x0=0.Also,x0=02X=D(J)=D(C0)=D(Ct0)andCt0(x0)=h=0.Thus,c(2)issatisedinthiscase.Wenowassumethattn>0foralln.Thentheinequalitiesin(3.1.22)and(3.1.23)aretrueforthesamereasonasaboveandweobtainxn!x0.Also,sincet0=0,wehavelimn!1Ctnxn=limn!1[tn(Cxn+Jxn)+(1tn)Jxn]=Jx0=c0:

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forsomepositivenumberq>0.Thenfrom(3.1.24)weobtainlimsupn!1h(tn+1tn)Jxn;xnx0ilimn!1tnhCxn;xnx0i=t0q<0: Itnowfollowsthatd(H2(t;);G2;0)=d(H(1;);G2;0)=d(T+C+J;G2;0)=1 forsucientlysmall>0.Thus,thereexists0suchthat0=d(T+C+J;G1;0)6=d(T+C+J;G2;0)=161

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forevery2(0;0].Lettingn=1 SincefxngisboundedandhTxn+Cxn;xni=1 bycondition(C1)onCweobtainfCxngisalsobounded.Wemaynowassumethatxn*x0,Cxn*c0.Inviewof(3.2.36),wealsoobtainhTxn+Cxn;xnx0i=1 leadstoacontradiction.Wethenhavelimsupn!1hCxn;xnx0i0:

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LetXbearealreexiveinnitedimensionalBanachspacewithbothXandXlocallyuniformlyconvexandLadenselinearsubspaceofX.LetF(L)bethesetofallnite-dimensionalsubspacesofL.Weassumethatt1: theoperatorCsatisescondition(~S+);c3: foreveryF2F(L),v2Lthemappingc(F;v):F!R,denedbyc(F;v)(u)=hCu;vi,iscontinuous. Conditionc3hereisthesameasconditionC3.WeneedthefollowingtheoremwhichisacombinationTheorem1andTheorem2in[22].63

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foreachtheoperatorTismaximalmonotone,02D(T),02T(0);t(2): givensequencesfng[0;1],fung,fvngsuchthatun2D(Tn),vn2Tnun,n!0,un*u0,vn*v0andlimsupn!1hvn;unihv0;u0i; thefamilyCis\uniformlyquasibounded",i.e.,foreveryS>0thereexists64

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foreverypairofsequencesfng[0;1],fungLsuchthatun*u0,Cnun*h,n!0andlimsupn!1hCnun;unu0i0;hCnun;uni0; foreveryF2F(L),v2L,themapping~c(F;v):F[0;1]!R,denedby~c(F;v)(u;)=hCu;vi,iscontinuous. WhentheoperatorsTandCareasabove,wesaythatthemappingH(;x)=(T+C)xisan\admissiblehomotopy"forthedegreeinDenition3.7.Lemma3.8

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assumethatthisisnottrue.Then,forasuitablesubsequenceoffngdenotedbyfngagain,wehavelimn!1hCxn;xnx0i=q; Sinceyn*c0+p,(3.2.32)implieslimsupn!1hyn;xni
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Combiningthiswith(3.2.33),wegethc0+p;x0i>hc0+p;~xi+hy;x0~xi; SinceTismaximalmonotone,itfollowsthatx02D(T).Thisisacontradictionto(3.2.35)becausewecannowtake~x=x0.Thus,(3.2.31)istrue.SinceCsatisesconditionc2,wehavexn!x02D(C)\

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ThenthereisasolutionoftheproblemTx+Cx30inD(T)\D(C)\(G1nG2). WenowconsiderthehomotopyfunctionH1(;x)=Tx+Cx+Jx0v0;(;x)2[0;1](D(T)\D(C)\ WenextconsiderthehomotopyfunctionH2(;x)=(Tx+Cx+J)+(1)Jx;(;x)2[0;1](D(T)\D(C)\ Itnowfollowsthatd(H2(;);G2;0)=d(H(1;);G2;0)=d(T+C+J;G2;0)=168

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forevery2(0;0].Hence,bytheexcisionpropertyofthedegree,forevery2(0;0]thereexistsasolutionx2(G1nG2)oftheequationTx+Cx+Jx30: Letyn2Txn.Sincefxngisbounded,02T(0)andhCxn;xnihyn+Cxn;xni=1 byconditionc1onCweobtainfCxngisalsobounded.Wemaynowassumethatxn*x0,Cxn*c0.Inviewof(3.2.36),wealsoobtainhyn+Cxn;xnx0i=1 leadstoacontradiction.Wethenhavelimsupn!1hCxn;xnx0i0:69

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InSection4.2weextendtheeigenvalueproblemin[21]totheoperatorsoftheformL+T+C,whereL:XD(L)!Xisdenselydenedlinearmaximalmonotone,T:X!2Xboundedmaximalmonotone,andC:XD(C)!Xisboundeddemicontinuousoftype(S+)w.r.t.D(L).71

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LetL:XD(L)!XbeadenselydenedlinearmaximalmonotoneoperatorandC:X!Xaboundeddemicontinuousoperator.WesaythatC:X!Xisoftype(S+)w.r.t.toD(L)ifforeverysequencefxngD(L)withxn*x0inX,Lxn*Lx0inXandlimsupn!1hCxn;xnx0i0; SincethegraphG(L)=f[x;Lx]:x2D(L)gofLisclosedinXX,wecanequipY=D(L)withthegraphnormkxkY=kxkX+kLxkX;x2Y Letj:Y!Xbethenaturalembeddingandj:X!Yitsadjoint.LetGXbeanopenboundedset.LetFG(L:T:S+)denotetheclassofoperatorsoftheformL+T+C:

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Here,thedualitypair(;)isinYYandJ1istheinverseofthedualitymapJ:X!XandisidentiedwiththedualitymapfromXtoX.Infact,foreveryx2YsuchthatMx2j(X),wehaveJ1(Lx)2D(L)andby(4.1.1)Mx=jLJ1(Lx):Denition4.3

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Obviously,pointwisegraph-continuityimpliesgraph-continuity. ForamaximalmonotoneoperatorT:XD(T)!2XwithYosidaapproximantsTs,s>0,thefollowingistrue.Proposition4.6 IfTnG!T,thenforeverysequencefxngXsuchthatxn!xinX,wehaveTn;s(xn)!Ts(x)inXforeverys>0.Moreover,ifthesequencefTngisbounded,wehaveTn;sn(xn)*w2T(x)forasubsequencefxngandanysequencefsng(0;1)withsn!0;(b) IfthesequenceTnispointwisegraph-continuoustoT,thenwehaveTn;sn(x)!T(x)foreverysequencefsng(0;1)withsn!0,whereT(x)istheelementofminimumnormoftheclosedconvexsubsetTxofX.74

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(i)thereexists(0;x0)2(0;](D(L)\@G)suchthatLx0+Tx0+C(0;x0)+Jx030;(4.2.2) hasnosolutiononD(L)\@Gforevery2[0;].Here,L+T+Jisstrictlymonotoneandsotheassumptionisobviouslytruefor=0.Thus,H(;D(L)\@G)630;2[0;]:(4.2.3) LetY=D(L)beequippedwiththegraphnorm.Wearenowgoingtoshowthatthereexists0>0,02(0;]suchthatforeverys2(0;s0]and2(0;0],theequation hasnosolutionx2@GR(Y),whereGR(Y)=j1(G)\BY(0;R).Here,BY(0;R)=fy2Y:kykY
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SinceTnwithTnTconvergestoTinthegraphsenseandTisbounded,byLemma4.7(a)wehaveTsnxn*w2Tx0forasubsequenceoffxngwhichweagaindenotebyfxng.Then(4.2.5)implies^Lxn+^Tsnxn+^Jxn*^Lx0+jw+^Jx0=0. Lety2Y.Then(^Lx0+jw+^Jx0;y)=0; Now,wexs2(0;s0]and2(0;0]andconsiderthehomotopyfunctionH2(t;x)^Lx+^Tsx+^C(t;x)+^Jx+sMx;(t;x)2[0;1]

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SetS(t)=C(t;)+JandTt=Ts.Then^S(t)=^C(t;)+^Jand^Tt=^Ts.InordertoshowthatH2(t;x)isanadmissiblehomotopyfortheBrowderandSkrypnikdegree,inviewofLemma4.8,itsucestoshowthatS(t)isaboundedhomotopyoftype(S+)withrespecttoD(L)andTtisaboundedpointwisegraph-continuoushomotopyofmaximalmonotoneoperators.ThelatterfollowsimmediatelybecauseTsisboundedwheneverTisbounded.So,letfxngD(L)besuchthatxn*x0inX,Lxn*Lx0inX,tn!t2[0;1]andlimsupn!1hS(tn)xn;xnx0i0:(4.2.8) WeobservethathS(tn)xn;xnx0i=hC(tn;xn);xnx0i+hJxn;xnx0i=hC(tn;xn);xnx0i+hJxnJx0;xnx0i(4.2.9)+hJx0;xnx0ihC(tn;xn);xnx0i+hJx0;xnx0i: Ift=0,thenC(tn;xn)!0andlimn!1hC(tn;xn);xnx0i=0:

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Wenextconsiderthecaset>0andobservethathC(tn;xn);xnx0i=hC(tn;xn)C(t;xn);xnx0i+hC(t;xn);xnx0i: ConsideranotherhomotopyfunctionH0(t;x)=t(^L+^Ts+^J+sM)+(1t)(^L+^J+sM);(t;x)2[0;1]

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whichimpliesx0=0.Butx02@(j1(G))whichimpliesx02G.Thisisacontradic-tion.Thus,bytheinvarianceunderhomotopyofthedegree,wehavedS+(^L+^Ts+^J+sM;GR(Y);0)=dS+(^L+^J+sM;GR(Y);0):(4.2.13) Thetopologicaldegreedevelopedin[2]isbasedonthemethodologyofdegreedevelopedin[4]byBerkovitsandMustonenandthedegreeisthelimitd(H(;);G;0)=lims#0dS(H1(s;;);GR(Y);0)=lims#0dS(H2(1;);GR(Y);0)=lims#0dS(^L+^Ts+^J+sM;GR(Y);0)=lims#0dS(^L+^J+sM;GR(Y);0)=d(L+J;G;0)=1: (ii)Inviewof(i),foreachpositiveintegern,thereexistfxngG\D(L),xn2Txn,n2(0;]suchthatLxn+xn+C(n;xn)+1

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Wenowconsidertwocases:(a)0=0;(b)0>0. (a)SinceLxn+xn=C(n;xn)1 andL+Tsatises(Sq),wehavexn!x02@GinX.SincethesumL+Tismaximalmonotone,itsclosednessimpliesx02D(L)and02Lx0+Tx0,whichcontradict0=2(L+T)(@G\D(L)). (b)Werstassertthatlimsupn!1hC(n;xn);xnx0i0:(4.2.15) Assumethatitisnottrue.Thenthereisasubsequenceoffxng,whichweagaindenotebyfxng,suchthatlimn!1hC(n;xn);xnx0i=q>0:(4.2.16) SinceLxn+xn*c,weinvoke(4.2.14)and(4.2.16)toobtainlimn!1hLxn+xn;xnx0i<0;

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Weclaimthatthereisr>0suchthat(L+T+C)(D(L)\@Bq(0))\Br(0)=;.Supposethatthecontraryistrue.Thenthereexistsasequencefrng,rn#0and83

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LetY=D(L)withthegraphnorm.SinceTandCarebounded,itfollowsthatfkLxnkgisboundedandhencefkxnkYgisbounded.SinceYisreexive,wemayassumethatxn*x0inXandLxn*Lx0inX.Wearenowgoingtoshowthatlimsupn!1hCxn;xnx0i0:(4.3.18) If(4.3.18)isnottrue,wemayassumethatlimn!1hCxn;xnx0i>0:(4.3.19) Inviewof(4.3.17)and(4.3.19),weobtainlimn!1hLxn+vn;xnx0i<0; Wenowxp2Br(0)anddenef(t)=tp,t2[0;1].Clearly,f(t)liesinBr(0)forallt2[0;1].Wenextclaimthatthereexistanintegern0>0ands0>0suchthat^Lx+^Tsx+^Cx+sMx+1 hasnosolutionx2@GR(Y),whereGR(Y)=j1(Bq(0))\BY(0;R).Here,BY(0;R)=fy2Y:kykY
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Sincexn*x0inYimpliesxn*x0inXandLxn*Lx0inx,itfollowsthatfMxngisboundedandhence,from(4.3.21),f^Tsnxngisbounded.Wearegoingtoshowthat limsupn!1hCxn;xnx0i0:(4.3.22) Supposethatthisisnottrue.Thenwemayassumethat limn!1hCxn;xnx0i>0:(4.3.23) WeobservethathLxn+Tsnxn;xnx0i=(^Lxn+^Tsnxn;xnx0)=(^Cxn;xnx0)sn(Mxn;xnx0)1 SinceTnwithTnTconvergestoTinthegraphsenseandTisbounded,byLemma4.7(a)wehaveTsnxn*w2Tx0forasubsequenceoffxngwhichweagaindenotebyfxng.Then(4.3.21)implies^Lxn+^Tsnxn+^Cxn*^Lx0+jw+^Cx0=j(f(t0)).85

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WenowconsiderthehomotopyfunctionH(s;t;x;n)=t^Lx+^Tsx+^Cx+1 where(t;x)2[0;1]j1( Wearegoingtoshowthatthereexistanintegern1>0andanumbers1>0suchthat(4.3.25)hasnosolutionx2@GR(Y)foranys2(0;s1],nn1andt2[0;1].Assumingthatthecontraryistrue,lettherebesequencesfxng@GR(Y),fsng(0;1),andftng[0;1]suchthatxn*x0inY,sn!0,tn!t0andtn^Lxn+^Tsnxn+^Cxn+1 Iftn=0foralln,thensnMxn+^Jxn=0; Case(a):Sincexn*x0inY,itfollowsthatxn*x0inXandLxn*Lx0inX.Inparticular,fkxnkgisbounded.BytheboundednessofC,fCxngisalsobounded.Now,tn^Lxn+tn^Tsnxn+^Jxn=tn^Cxntn1

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Case(b):Ift0=1,letdn=1 Wearenowgoingtoshowthat(4.3.22)istrue.Assumingthecontrary,supposethat(4.3.23)holdstrue.WeobservethathLxn+Tsnxn;xnx0i=(^Lxn+^Tsnxn;xnx0)=(^Cxn;xnx0)sn

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SinceTnwithTnTconvergestoTinthegraphsenseandTisbounded,byLemma4.7(a)wehaveTsnxn*w2Tx0forasubsequenceoffxngwhichweagaindenotebyfxng.Then(4.3.27)implies^Lxn+^Tsnxn+^Cxn*^Lx0+jw+^Cx0=1t0 whichimplies02L+T+C+1t0 SinceH(s;t;x;n)isananehomotopiesofboundeddemicontinuousoperatorsoftype(S+)fromj1(

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sothat,foreachnn0,thereexistsxn2Bq(0)\D(L)suchthatp=Lxn+wn+Cxn+1 wheneveritexistsbyLemma2.10(i),weconcludethatlimsupn!1hCxn;xnx0i0:

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