Research Article Coupled Fixed Point Theorems with New Implicit Relations and an Application

17 Pages·2015·1.96 MB·English

Hindawi Publishing Corporation
Journal of Operators
Volume 2014, Article ID 389646, 16 pages
http://dx.doi.org/10.1155/2014/389646
Research Article
Coupled Fixed Point Theorems with New Implicit
Relations and an Application
G.V.R.Babu1andP.D.Sailaja1,2
1DepartmentofMathematics,AndhraUniversity,Visakhapatnam530003,India
2DepartmentofMathematics,LendiInstituteofEngineeringandTechnology,Vizianagaram535005,India
CorrespondenceshouldbeaddressedtoG.V.R.Babu;gvr [email protected]
Received28June2014;Accepted1October2014;Published11November2014
AcademicEditor:HagenNeidhardt
Copyright©2014G.V.R.BabuandP.D.Sailaja. This is an open access article distributed under the Creative Commons
AttributionLicense,whichpermitsunrestricteduse,distribution,andreproductioninanymedium,providedtheoriginalworkis
properlycited.
WeintroducetwonewclassesofimplicitrelationsSandS󸀠whereS󸀠isapropersubsetofS,andtheseclassesaremoregeneral
thantheclassofimplicitrelationsdefinedbyAltunandSimsek(2010).Weprovetheexistenceofcoupledfixedpointsforthemaps
satisfyinganimplicitrelationinS.Thesecoupledfixedpointsneednotbeunique.Inordertoestablishtheuniquenessofcoupled
fixedpointsweuseanimplicitrelationS󸀠,whereS󸀠⊂S.Ourresultsextendthefixedpointtheoremsonorderedmetricspacesof
AltunandSimsek(2010)tocoupledfixedpointtheoremsandgeneralizetheresultsofGnanaBhaskarandLakshimantham(2006).
Asanapplicationofourresults,wediscusstheexistenceanduniquenessofsolutionofFredholmintegralequation.
1.Introduction monotonepropertyif𝐹(𝑥,𝑦)ismonotonenondecreasingin𝑥
andmonotonenonincreasingin𝑦;thatis,forany𝑥,𝑦∈𝑋:
Existenceoffixedpointtheoremsinpartiallyorderedmetric
𝑥 ,𝑥 ∈𝑋, 𝑥 ⪯𝑥 󳨐⇒𝐹(𝑥 ,𝑦)⪯𝐹(𝑥 ,𝑦),
spaces with a contractive condition has been considered by 1 2 1 2 1 2
several authors (see [1–6]). Guo and Lakshmikantham [7] (1)
𝑦 ,𝑦 ∈𝑋, 𝑦 ⪯𝑦 󳨐⇒𝐹(𝑥,𝑦 )⪰𝐹(𝑥,𝑦 ).
introducedmixedmonotoneoperators.GnanaBhaskarand 1 2 1 2 1 2
Lakshmikantham [8] established the existence of coupled Theorem3(see[8]). Let(𝑋,⪯)beapartiallyorderedsetand
fixed points of mappings satisfying mixed monotone prop- supposethat𝑑isametricon𝑋suchthat(𝑋,𝑑)isacomplete
erty in partially ordered metric spaces. Later, Lakshmikan- metric space. Let 𝐹 : 𝑋×𝑋 → 𝑋 be a mapping satisfying
tham and Ciric [9] extended this property to two maps by mixedmonotoneproperty.Assumethatthereexistsa𝑘∈[0,1)
introducingmixed𝑔-monotonepropertyandestablishedthe with
existenceofcoupledcoincidencepointandcoupledcommon 𝑘
fixedpointsforapairofcommutingmaps. 𝑑(𝐹(𝑥,𝑦),𝐹(𝑢,V))≤ [𝑑(𝑥,𝑢)+𝑑(𝑦,V)]
2
Choudhury and Kundu [10] extended the existence of (2)
𝑓𝑜𝑟 𝑒𝑎𝑐ℎ 𝑥⪰𝑢, 𝑦⪯V.
coupled coincidence and coupled common fixed points for
apairofnoncommutingmaps,particularlyforapairofcom-
Supposethateither𝐹iscontinuousorthefollowingconditions
patiblemaps.
holdin𝑋:
Definition1(see[7]). Let𝑋beanonemptyset.Anelement (i)ifanondecreasingsequence{𝑥𝑛} ⊆ 𝑋with𝑥𝑛 → 𝑥,
(𝑥,𝑦)in𝑋×𝑋iscalledacoupledfixedpointofthemapping then𝑥𝑛 ⪯𝑥forall𝑛and
𝑥𝐹∈: 𝑋𝑋i×sc𝑋alle→da𝑋fixiefd𝑥po=int𝐹o(𝑥f,𝐹𝑦i)fa𝐹n(d𝑥,𝑦𝑥)==𝐹𝑥(.𝑦,𝑥).Apoint (ii)itfheann𝑦o𝑛ni⪰nc𝑦refaosrinaglls𝑛e.quence{𝑦𝑛} ⊆ 𝑋with𝑦𝑛 → 𝑦,
Definition2(see[7]). Let(𝑋,⪯)beapartiallyorderedsetand If there exist 𝑥0,𝑦0 ∈ 𝑋 such that 𝑥0 ⪯ 𝐹(𝑥0,𝑦0) and 𝑦0 ⪰
𝐹:𝑋×𝑋 → 𝑋beamapping.Wesaythat𝐹satisfiesmixed 𝐹(𝑦0,𝑥0)then𝐹hasacoupledfixedpoint.
2 JournalofOperators
In2011,LuongandThuan[11]provedthefollowingcou- Manyauthors,namely,AltunandSimsek[12],Altunand
pledfixedpointtheorem. Turkoglu [14, 15], Imdad et al. [16], Popa [17], Popa and
Mocanu[18],SharmaandDeshpande[19],andTurkogluand
Theorem4(see[11]). Let(𝑋,⪯)beapartiallyorderedsetand Altun [20], continued the study of the existence of fixed
supposethat𝑑isametricon𝑋suchthat(𝑋,𝑑)isacomplete points in this direction. The purpose of using an implicit
metric space. Let 𝐹 : 𝑋×𝑋 → 𝑋 be a mapping satisfying relationisthattheconditioninvolvingimplicitrelationgener-
mixedmonotonepropertyon𝑋andthereexist𝑥0,𝑦0 ∈𝑋such alizesmanycontractionconditions,whichinturngeneralizes
that 𝑥0 ⪯ 𝐹(𝑥0,𝑦0) and 𝑦0 ⪰ 𝐹(𝑦0,𝑥0). Suppose that there manyexistingresults.In2010,AltunandSimsek[12]defined
existsa𝜓 : [0,∞) → [0,∞)withlim𝑡→𝑟+𝜓(𝑡) > 0forall anewclassofimplicitrelationsasfollowsandestablisheda
𝑟>0andlim𝑡→0+𝜓(𝑡)=0suchthat fixedpointtheoreminorderedmetricspaces.
𝑑(𝑥,𝑢)+𝑑(𝑦,V) LetTbethesetofallcontinuousfunctions𝑇:(𝑅+)6 →
𝑑(𝐹(𝑥,𝑦),𝐹(𝑢,V))≤ 𝑅satisfyingthefollowingconditions:
2
𝑑(𝑥,𝑢)+𝑑(𝑦,V) (3) (T1):𝑇(𝑡1,...,𝑡6)isnonincreasinginvariables𝑡2,𝑡3,...,𝑡6;
−𝜓( ),
2
(T2):thereexistsarightcontinuousfunction𝑓:𝑅+ → 𝑅+,
foreach𝑥,𝑦,𝑢,V∈𝑋,𝑥⪰𝑢and𝑦⪯V.Supposethateither 𝑓(0)=0,𝑓(𝑡)<𝑡for𝑡>0,suchthatfor𝑢,V≥0
(a)𝐹iscontinuousor
𝑇(𝑢,V,𝑢,V,0,𝑢+V)≤0 or 𝑇(𝑢,V,0,0,V,V)≤0 (4)
(b)𝑋hasthefollowingproperty:
(i)ifanondecreasingsequence{𝑥𝑛}⊆𝑋with𝑥𝑛 → implies𝑢≤𝑓(V);
𝑥,then𝑥𝑛 ⪯𝑥forall𝑛and (T3):𝑇(𝑢,0,𝑢,0,0,𝑢)>0for𝑢>0.
(ii)ifanonincreasingsequence{𝑦𝑛}⊆𝑋with𝑦𝑛 →
𝑦,then𝑦𝑛 ⪰𝑦forall𝑛. Theorem5(see[12]). Let(𝑋,⪯)beapartiallyorderedsetand
supposethat𝑑isametricon𝑋suchthat(𝑋,𝑑)isacomplete
Then𝐹hasacoupledfixedpoint. metric space. Let 𝐹 : 𝑋 → 𝑋 be a nondecreasing mapping
suchthat,forall𝑥,𝑦∈𝑋with𝑦⪯𝑥,
InSection2ofthispaper,weintroducetwonewclassesof
implicitrelationsSandS󸀠whereS󸀠isapropersubsetofS,
𝑇(𝑑(𝐹𝑥,𝐹𝑦),𝑑(𝑥,𝑦),𝑑(𝑥,𝐹𝑥),
andtheseclassesaremoregeneralthantheclassofimplicit
(5)
relationsdefinedbyAltunandSimsek[12].InSection3,we
𝑑(𝑦,𝐹𝑦),𝑑(𝑥,𝐹𝑦),𝑑(𝑦,𝐹𝑥))≤0,
prove the existence of coupled fixed points for the maps
satisfyinganimplicitrelationinS.Thesecoupledfixedpoints
need not be unique (Example17). In order to establish the where𝑇 ∈ T.Also,supposethateither𝐹iscontinuousorthe
uniquenessofcoupledfixedpointsweuseanimplicitrelation followingconditionholds:“ifanondecreasingsequence{𝑥𝑛} →
S󸀠,whereS󸀠 ⊂ S.Ourresultsextendthefixedpointtheo- 𝑥,then𝑥𝑛 ⪯𝑥forall𝑛”.
rems on ordered metric spaces of Altun and Simsek [12] to Ifthereexistsan𝑥0 ∈ 𝑋with𝑥0 ⪯ 𝐹(𝑥0),then𝐹hasa
coupled fixed point theorems and generalize the results of fixedpoint.
GnanaBhaskarandLakshimantham[8].InSection4,asan
application to results of Section3, we discuss the existence Wenowdefineamoregeneralclassofimplicitrelations
anduniquenessofsolutionofFredholmintegralequation. thanthatof[12]asfollows.
LetSbethesetofallcontinuousfunctions𝑇: (𝑅+)6 →
𝑅satisfyingthefollowingconditions:
2.ImplicitRelations
In1999,Popa[13]introducedtheideaoftheclassFofimplicit (𝑇1):𝑇(𝑡1,...,𝑡6)isnonincreasinginvariables𝑡5,𝑡6;
functionstoprovetheexistenceoffixedpointsasfollows.
Let𝑅+denotethesetofallnonnegativerealnumbers.Let (𝑇2):t𝑡h>er0e,esxuicshtsthaamtfaoprp𝑢in,Vg≥𝑓0: 𝑅+ → 𝑅+, 𝑓(𝑡) < 𝑡 for
Fbethefamilyofreallowersemicontinuousfunctions𝐹 :
(𝑅+)6 → 𝑅satisfyingthefollowingconditions: 𝑇(𝑢,V,𝑢,V,0,𝑢+V)≤0 or 𝑇(𝑢,V,0,0,V,V)≤0, (6)
(𝐹1):𝐹isnonincreasinginfifthandsixthvariables;
(𝐹2): there exists ℎ ∈ (0,1) such that for every 𝑢,V ≥ 0 implies𝑢≤𝑓(V);
with
(𝑇3):𝑇(𝑢,0,𝑢,0,0,𝑢)>0for𝑢>0.
𝐹2𝑎 :𝐹(𝑢,V,V,𝑢,𝑢+V,0)≤0or
𝐹2𝑏 : 𝐹(𝑢,V,𝑢,V,0,𝑢+V)≤0 HereweobservethatT⊆S.
wehave𝑢≤ℎV;
Example6. 𝑇(𝑡1,...,𝑡6) = 𝑡1 −[𝑎𝑡2 +𝑏𝑡3 +𝑐𝑡4 +𝑑𝑡5 +𝑒𝑡6],
(𝐹3): 𝐹(𝑢,𝑢,0,0,𝑢,𝑢)>0,∀𝑢>0. where𝑎,𝑏,𝑐,𝑑,𝑒≥0and𝑎+𝑏+𝑐+𝑑+2𝑒<1.
JournalofOperators 3
Clearly(𝑇1)holds.Let𝑇(𝑢,V,𝑢,V,0,𝑢+V) = 𝑢−(𝑎V+ In this paper, we prove a coupled fixed point theorem
𝑏𝑢 + 𝑐V + 𝑒(𝑢 + V)) = (1 − 𝑏 − 𝑒)𝑢 − (𝑎 + 𝑐 + 𝑒)V ≤ 0. (Theorem16) using an implicit relation in S. In order to
Thus 𝑢 ≤ ((𝑎+𝑐+𝑒)/(1−𝑏−𝑒))V. Let 𝑇(𝑢,V,0,0,V,V) = obtain the uniqueness of coupled fixed point we define
𝑢 − (𝑎V + 𝑑V + 𝑒V) = 𝑢 − (𝑎 + 𝑑 + 𝑒)V ≤ 0. Thus 𝑢 ≤ anotherclassS󸀠ofimplicitrelationsasfollows.
(𝑎+𝑑+𝑒)V.Nowdefine𝑓: 𝑅+ → 𝑅+ by𝑓(𝑡) = 𝑘𝑡,where LetS󸀠bethesetofallcontinuousfunctions𝑇:(𝑅+)6 →
𝑘=max{((𝑎+𝑐+𝑒)/(1−𝑏−𝑒)),𝑎+𝑑+𝑒}<1,𝑓(𝑡)<𝑡for 𝑅satisfyingthefollowingconditions:
𝑡>0.Hence𝑢≤𝑓(V).Also,𝑇(𝑢,0,𝑢,0,0,𝑢)=𝑢−(𝑏𝑢+𝑒𝑢)=
(1−𝑏−𝑒)𝑢>0for𝑢>0.Therefore𝑇∈S. (𝑇1󸀠):𝑇(𝑡1,...,𝑡6)isnonincreasinginvariables𝑡3,𝑡5,𝑡6;
wExhaemrep𝜑le7:. 𝑅𝑇+(𝑡1,→...,𝑅𝑡+6)is=a𝑡1c−o𝜑n(tminauxo{u𝑡2s,m𝑡3a,𝑡p4,w(1it/h2)𝜑((𝑡05)+𝑡=6)}0),, (𝑇2󸀠):𝑡t>he0r,esuexchisttshaatm,foaprpeiancgh𝑓𝑢,:V𝑅≥+0,→ 𝑅+,𝑓(𝑡) < 𝑡for
𝜑(𝑡)<𝑡for𝑡>0. 𝑇(𝑢,V,𝑢,V,0,𝑢+V)≤0 or 𝑇(𝑢,V,0,0,V,V)≤0
Clearly(𝑇1)holds.Let𝑢>0and𝑇(𝑢,V,𝑢,V,0,𝑢+V)=𝑢− (8)
𝜑(max{V,𝑢,V,(1/2)(𝑢+V)})=𝑢−𝜑(max{𝑢,V})≤0.If𝑢≥V or 𝑇(𝑢,V,𝑢+V,0,V,𝑢)≤0,
then𝑢−𝜑(𝑢)≤0,whichisacontradiction.Hence𝑢<Vand
hence𝑢≤𝜑(V).Let𝑢>0and𝑇(𝑢,V,0,0,V,V)=𝑢−𝜑(V)≤0; implies𝑢≤𝑓(V);
sthataitsfiise,d𝑢w≤ith𝜑(𝑓V)=.I𝜑f.𝑢A=lso0,t𝑇h(e𝑢n,0c,le𝑢a,r0ly,0𝑢,𝑢≤)=𝜑(𝑢V)−.𝜑Th(𝑢u)s>(𝑇02)foirs (𝑇3󸀠):𝑇(𝑢,0,𝑢,0,0,𝑢)>0for𝑢>0.
𝑢>0.Therefore𝑇∈S. HereweobservethatS󸀠 ⊆S.
Example8. 𝑇(𝑡1,...,𝑡6) = 𝑡1−(𝑡2−𝜑(𝑡2)),where𝜑 : 𝑅+ → All functions in Examples 8–10 are in S󸀠. Some more
𝑅+ is a continuousmap with 𝜑(𝑡) = 0 if and only if 𝑡 = 0, examplesinthisdirectionareasfollows.
𝜑(𝑡)C<le𝑡arfolyr(𝑡𝑇>1)0h.olds. Let 𝑇(𝑢,V,𝑢,V,0,𝑢 + V) = 𝑢 − (V − wExhaemrep𝑎le,𝑏12,𝑐.,𝑇𝑑,(𝑡𝑒1,≥..0.a,𝑡n6d)𝑎=+𝑡1𝑏−+[𝑐𝑎𝑡+2𝑑++𝑏𝑡23𝑒+<𝑐𝑡14a+nd𝑑𝑡𝑏5≤+𝑐𝑒.𝑡6],
𝜑(V)) ≤ 0.Let𝑇(𝑢,V,0,0,V,V) = 𝑢−(V−𝜑(V)) ≤ 0;thatis, Clearly(𝑇󸀠)holds.Let𝑇(𝑢,V,𝑢,V,0,𝑢+V) = 𝑢−(𝑎V+
𝑢 ≤ V−𝜑(V).Thus𝑢 ≤ 𝑓(V)with𝑓 : 𝑅+ → 𝑅+ by𝑓(𝑡) = 𝑏𝑢+𝑐V+𝑒(𝑢1+V)) = (1−𝑏−𝑒)𝑢−(𝑎+𝑐+𝑒)V ≤ 0.Thus
𝑡−𝜑(𝑡).Hence(𝑇2)issatisfied.Also,𝑇(𝑢,0,𝑢,0,0,𝑢) = 𝑢− 𝑢≤((𝑎+𝑐+𝑒)/(1−𝑏−𝑒))V.Let𝑇(𝑢,V,0,0,V,V)=𝑢−(𝑎V+
(𝑢−𝜑(𝑢))=𝜑(𝑢)>0for𝑢>0.Hence𝑇∈S. 𝑑V+𝑒V)=𝑢−(𝑎+𝑑+𝑒)V≤0.Thus𝑢≤(𝑎+𝑑+𝑒)V.Nowlet
tEhxaatCmislpe,lae𝑢r9ly≤.(𝑇𝑇𝑘(1V𝑡)1.h,L.oe.lt.d,s𝑇𝑡.6(L)𝑢e,=tV𝑇,𝑡01(,𝑢−0,,𝑘VV𝑡,,2𝑢V,,)wV,h=0e,r𝑢𝑢e0+−≤V𝑘)V𝑘=<≤𝑢10−.;𝑘thVa≤t i0s;, 𝑅𝑇im+(𝑢pb,lyiVe,𝑓s𝑢(𝑢𝑡+)≤V=,((0𝑘𝑎,𝑡V,+w,𝑢𝑐h)+e=re𝑑𝑢)𝑘/−=(1[m𝑎−Vax𝑏+{−(𝑏((𝑎𝑒𝑢)+)+V𝑐.VN+)+o𝑒w)𝑑/Vd(e+1fi−𝑒n𝑢e𝑏]𝑓−≤:𝑒0𝑅).)+Th,𝑎→i+s
A𝑢l≤so𝑘,V𝑇.(Th𝑢,0u,s𝑢,𝑢,0≤,0𝑓,𝑢()V)=w𝑢it>h𝑓0(f𝑡o)r=𝑢𝑘>𝑡.0H.Thenuces,𝑇𝑇2∈iSss.atisfied. 𝑑0𝑢+f≤o𝑒r𝑓,𝑢(((V>𝑎).+0A.𝑐lThs+o,e𝑑𝑇r)e(f/𝑢o(,r10e,−𝑇𝑢,𝑏∈0−,S0𝑒,󸀠)𝑢.)})<=1𝑢,−𝑓((𝑏𝑡𝑢)+<𝑒𝑢𝑡)fo=r𝑡(1>−0𝑏.−H𝑒e)𝑢nc>e
aExcaomntpinleu1o0u.s𝑇m(𝑡a1p,.w..it,h𝑡6𝜑)(=0)𝑡1=−0𝜑,𝜑(𝑡(2𝑡)),<wh𝑡eforer𝜑𝑡>:𝑅0+. → 𝑅+is Example13. 𝑇(𝑡1,...,𝑡6)=𝑡1−𝜑(max{𝑡2,𝑡3/2,𝑡4,(1/2)(𝑡5+
thatCisle,a𝑢rl≤y(𝜑𝑇(1V))h.oLledts𝑇.L(𝑢et,V𝑇,(0𝑢,,0V,,V𝑢,,VV),=0,𝑢𝑢+−V𝜑)(=V)𝑢≤−0𝜑;(tVh)a≤ti0s;, 𝑡6)})C,lweahrelrye(𝜑𝑇1󸀠:)𝑅h+ol→ds.𝑅Le+tw𝑢it>h𝜑0(a0n)d=𝑇0(,𝑢𝜑,V(𝑡,)𝑢<,V,𝑡0fo,𝑢r𝑡+>V)0.=
A𝑢l≤so,𝜑𝑇(V()𝑢.,Th0,u𝑢s,0𝑢,0≤,𝑢𝑓)(V=)𝑢wi−th𝜑𝑓(0)==𝜑.𝑢H>en0cefo(r𝑇𝑢2)>is0sa.tHisefinecde. 𝑢If−𝑢𝜑≥(mV,atxh{eVn,𝑢𝑢/2−,𝜑V,((𝑢1)/2≤)0(𝑢,w+hVic)}h)i=sa𝑢co−n𝜑tr(amdaicxt{i𝑢o,nV.}H)e≤nc0e.
𝑇∈S. 𝑢 < Vandhence𝑢 ≤ 𝜑(V).Let𝑢 > 0and𝑇(𝑢,V,0,0,V,V) =
𝑢 − 𝜑(V) ≤ 0; that is, 𝑢 ≤ 𝜑(V). Let 𝑢 > 0 and 𝑇(𝑢,V,𝑢 +
Example11. V,0,V,𝑢)=𝑢−𝜑(max{V,𝑢,(𝑢+V)/2,0,(1/2)(𝑢+V)})=𝑢−
𝜑(max{𝑢,V}) ≤ 0. If 𝑢 ≥ V, then 𝑢 − 𝜑(𝑢) ≤ 0, which is a
𝑡 𝑡 𝑡 contradiction. Hence 𝑢 < V and hence 𝑢 ≤ 𝜑(V). If 𝑢 = 0,
𝑇(𝑡1,...,𝑡6)=𝑡1−𝛼𝑡 +4𝑡5 6+1, thenclearly𝑢≤𝜑(V).Thus(𝑇󸀠)issatisfiedwith𝑓=𝜑.Also,
2 3 (7) 𝑇(𝑢,0,𝑢,0,0,𝑢)=𝑢−𝜑(𝑢)>20for𝑢>0.Therefore,𝑇∈S󸀠.
where 0≤𝛼<1.
ThefollowingexamplessuggestthatS󸀠isapropersubset
𝑇isnonincreasingin𝑡5and𝑡6butnotinanyof𝑡2,𝑡3.Let ofS.
𝑢>0,𝑇(𝑢,V,𝑢,V,0,𝑢+V)=𝑢−𝛼(0)=𝑢<0,acontradiction.
Nowlet𝑇(𝑢,V,0,0,V,V)=𝑢−𝛼(0)=𝑢<0,acontradiction. Example 14. 𝑇(𝑡1,...,𝑡6) = 𝑡1 − 𝜑(max{𝑡2,𝑡3,𝑡4,(1/2)(𝑡5 +
If 𝑢 = 0 then clearly 𝑢 ≤ 𝛼V. Thus (𝑇2) is satisfied. Also, 𝑡6)}),where𝜑 : 𝑅+ → 𝑅+ definedby𝜑(𝑡) = 𝑡/2.Inviewof
𝑇(𝑢,0,𝑢,0,0,𝑢) = 𝑢 > 0for𝑢 > 0.Thus𝑇 ∈ S.But𝑇does Example7,𝑇∈S.But𝑇isnotinS󸀠forlet𝑢,V>0,V≤𝑢and
notbelongtoT. supposethat𝑇(𝑢,V,𝑢+V,0,V,𝑢)=𝑢−𝜑(max{V,𝑢+V,0,(𝑢+
InviewofExample11,theclassofimplicitfunctionsSis V)/2}) = 𝑢 − 𝜑(𝑢 + V) = 𝑢 − (𝑢 + V)/2 ≤ 0. This implies
larger than that of T introduced by Altun and Simsek [12] 𝑢 ≤ V.Therefore𝑢 = V.Hence,therecannotexistafunction
and we use the implicit relation involving implicit function 𝑓 : 𝑅+ → 𝑅+ with𝑓(𝑡) < 𝑡for𝑡 > 0suchthat𝑢 ≤ 𝑓(V).
ofStoproveourmainresults. HenceS󸀠 ⊂S.
4 JournalofOperators
Example15. Letusconsider𝑇asinExample11.Then𝑇∈S. If𝑥𝑛 = 𝑥𝑛+1 and𝑦𝑛 = 𝑦𝑛+1 forsome𝑛,then(𝑥𝑛,𝑦𝑛)is
But𝑇doesnotbelongtoS󸀠,since𝑇isnotnonincreasingin acoupledfixedpointof𝐹.So,withoutlossofgenerality,we
𝑡3,sothatS󸀠isapropersubsetofS. assumethat𝑥𝑛 ≠𝑥𝑛+1or𝑦𝑛 ≠𝑦𝑛+1forall𝑛.
Nowusing(9),wehave
3.ExistenceofCoupledFixedPoints
𝑑(𝐹(𝑥 ,𝑦 ),𝐹(𝑥 ,𝑦 ))+𝑑(𝐹(𝑦 ,𝑥 ),𝐹(𝑦 ,𝑥 ))
UsingaNewImplicitRelation 𝑇( 𝑛+1 𝑛+1 𝑛 𝑛 𝑛+1 𝑛+1 𝑛 𝑛 ,
2
Theorem16. Let(𝑋,⪯)beapartiallyorderedsetandsuppose 𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 )
that𝑑is ametricon 𝑋suchthat (𝑋,𝑑)is acompletemetric 𝑛+1 𝑛 2 𝑛+1 𝑛 ,
space. Let 𝐹 : 𝑋 × 𝑋 → 𝑋 be a mapping satisfying mixed
monotonepropertyandsupposethatthereexists𝑇 ∈ Ssuch 𝑑(𝑥𝑛+1,𝐹(𝑥𝑛+1,𝑦𝑛+1))+𝑑(𝑦𝑛+1,𝐹(𝑦𝑛+1,𝑥𝑛+1)),
that 2
𝑇(𝑑(𝐹(𝑥,𝑦),𝐹(𝑢,V))+𝑑(𝐹(𝑦,𝑥),𝐹(V,𝑢)), 𝑑(𝑥𝑛,𝐹(𝑥𝑛,𝑦𝑛))+𝑑(𝑦𝑛,𝐹(𝑦𝑛,𝑥𝑛)),
2
2
𝑑(𝑥 ,𝐹(𝑥 ,𝑦 ))+𝑑(𝑦 ,𝐹(𝑦 ,𝑥 ))
𝑑(𝑥,𝑢)+𝑑(𝑦,V) 𝑑(𝑥,𝐹(𝑥,𝑦))+𝑑(𝑦,𝐹(𝑦,𝑥)) 𝑛+1 𝑛 𝑛 𝑛+1 𝑛 𝑛 ,
, , 2
2 2
𝑑(𝑥 ,𝐹(𝑥 ,𝑦 ))+𝑑(𝑦 ,𝐹(𝑦 ,𝑥 ))
𝑑(𝑢,𝐹(𝑢,V))+𝑑(V,𝐹(V,𝑢)) 𝑛 𝑛+1 𝑛+1 𝑛 𝑛+1 𝑛+1 )≤0,
, 2
2
𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 )
𝑑(𝑥,𝐹(𝑢,V))+𝑑(𝑦,𝐹(V,𝑢)) 𝑇( 𝑛+2 𝑛+1 𝑛+2 𝑛+1 ,
, 2
2
𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 )
𝑑(𝑢,𝐹(𝑥,𝑦))+𝑑(V,𝐹(𝑦,𝑥)) 𝑛+1 𝑛 𝑛+1 𝑛 ,
)≤0, 2
2
𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 )
(9) 𝑛+1 𝑛+2 𝑛+1 𝑛+2 ,
2
foreach𝑥,𝑦,𝑢,V∈𝑋with𝑥⪰𝑢and𝑦⪯V.
𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 )
Supposethateither 𝑛 𝑛+1 𝑛 𝑛+1 ,
2
(a)𝐹iscontinuousor
𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 )
(b)𝑋hasthefollowingproperty: 𝑛+1 𝑛+1 2 𝑛+1 𝑛+1 ,
(i)ifanondecreasingsequence{𝑥𝑛}⊆𝑋with𝑥𝑛 → 𝑑(𝑥𝑛,𝑥𝑛+2)+𝑑(𝑦𝑛,𝑦𝑛+2))≤0.
𝑥,then𝑥𝑛 ⪯𝑥forall𝑛and 2
(ii)ifanonincreasingsequence{𝑦𝑛}⊆𝑋with𝑦𝑛 → (12)
𝑦,then𝑦𝑛 ⪰𝑦forall𝑛.
Since 𝑇 is nonincreasing in 6th variable, using triangle
If there exist 𝑥0,𝑦0 ∈ 𝑋 such that 𝑥0 ⪯ 𝐹(𝑥0,𝑦0) and 𝑦0 ⪰ inequality,weget
𝐹(𝑦0,𝑥0)then𝐹hasacoupledfixedpoint.
𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 )
Proof. Supposethatthereexists(𝑥0,𝑦0) ∈ 𝑋×𝑋suchthat 𝑇( 𝑛+2 𝑛+1 2 𝑛+2 𝑛+1 ,
𝑥0 ⪯ 𝐹(𝑥0,𝑦0)and𝑦0 ⪰ 𝐹(𝑦0,𝑥0).Choose𝑥1,𝑦1 ∈ 𝑋such
that 𝑥1 = 𝐹(𝑥0,𝑦0) and 𝑦1 = 𝐹(𝑦0,𝑥0). Then 𝑥0 ⪯ 𝑥1 and 𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 )
𝑦0 ⪰𝑦1.Nowchoose𝑥2,𝑦2 ∈𝑋suchthat𝑥2 =𝐹(𝑥1,𝑦1)and 𝑛+1 𝑛 2 𝑛+1 𝑛 ,
𝑦2 =𝐹(𝑦1,𝑥1).Since𝐹hasmixedmonotoneproperty,weget
𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 )
𝑥 =𝐹(𝑥 ,𝑦 )⪯𝐹(𝑥 ,𝑦 )=𝑥 , 𝑛+1 𝑛+2 𝑛+1 𝑛+2 ,
1 0 0 1 1 2 2
(10)
𝑦 =𝐹(𝑦 ,𝑥 )⪰𝐹(𝑦 ,𝑥 )=𝑦 .
1 0 0 1 1 2 𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 )
𝑛 𝑛+1 𝑛 𝑛+1 ,0,
Continuingthisprocess,weobtainsequences{𝑥𝑛}and{𝑦𝑛} 2
in 𝑋 such that 𝑥𝑛+1 = 𝐹(𝑥𝑛,𝑦𝑛), 𝑦𝑛+1 = 𝐹(𝑦𝑛,𝑥𝑛) for 𝑛 = (𝑑(𝑥 ,𝑥 )+𝑑(𝑥 ,𝑥 )
0,1,2,...and 𝑛 𝑛+1 𝑛+1 𝑛+2
+𝑑(𝑦 ,𝑦 )+𝑑(𝑦 ,𝑦 ))
𝑥0 ⪯𝑥1 ⪯𝑥2 ⪯⋅⋅⋅ , 𝑦0 ⪰𝑦1 ⪰𝑦2 ⪰⋅⋅⋅ . (11) 𝑛 𝑛+1 𝑛+1 𝑛+2
Let𝑥𝑛+1 =𝐹(𝑥𝑛,𝑦𝑛)=𝐹𝑛(𝑥0,𝑦0)=𝐹(𝐹𝑛−1(𝑥0,𝑦0),𝐹𝑛−1(𝑦0, × (2)−1)≤0.
𝑥0)) and 𝑦𝑛+1 = 𝐹(𝑦𝑛,𝑥𝑛) = 𝐹𝑛(𝑦0,𝑥0) = 𝐹(𝐹𝑛−1(𝑦0,𝑥0),
𝐹𝑛−1(𝑥0,𝑦0))foreach𝑛=0,1,2,.... (13)
JournalofOperators 5
From(𝑇2),thereexists𝑓: 𝑅+ → 𝑅+,𝑓(𝑡) < 𝑡for𝑡 > 0such Similarly,wecanshowthat
that
𝛿 = 𝑑(𝑥𝑛+1,𝑥𝑛+2)+𝑑(𝑦𝑛+1,𝑦𝑛+2) li𝑘m→i∞nf𝑟𝑘 =li𝑘m→i∞nf𝑑(𝑥𝑚(𝑘),𝑥𝑛(𝑘))+𝑑(𝑦𝑚(𝑘),𝑦𝑛(𝑘))=𝜖.
𝑛+1 2 (24)
(14)
𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 )
≤𝑓( 𝑛+1 𝑛 𝑛+1 𝑛 )=𝑓(𝛿 ).
2 𝑛 Hence
Then𝛿𝑛 ≠0forall𝑛.Hence,from(14),wehave lim 𝑟𝑘 = lim 𝑑(𝑥𝑚(𝑘),𝑥𝑛(𝑘))+𝑑(𝑦𝑚(𝑘),𝑦𝑛(𝑘))=𝜖. (25)
𝑘→∞ 𝑘→∞
𝛿𝑛+1 ≤𝑓(𝛿𝑛)<𝛿𝑛, ∀𝑛. (15)
Now
Therefore{𝛿𝑛}isstrictlydecreasingsequenceofpositivereals.
Hencethereexists𝛿≥0suchthat 𝜖≤ 𝑟 =𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 )
𝑘 𝑚(𝑘) 𝑛(𝑘) 𝑚(𝑘) 𝑛(𝑘)
𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 )
𝑛l→im∞𝛿𝑛 =𝑛l→im∞ 𝑛+1 𝑛+2 2 𝑛+1 𝑛+2 =𝛿. (16) ≤ 𝑑(𝑥𝑚(𝑘),𝑥𝑚(𝑘)+1)+𝑑(𝑥𝑚(𝑘)+1,𝑥𝑛(𝑘)+1) (26)
+𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 )
𝑛(𝑘)+1 𝑛(𝑘) 𝑚(𝑘) 𝑚(𝑘)+1
Weshowthat𝛿 = 0.Supposethat𝛿 > 0.Onletting𝑛 → ∞
in(13),weget +𝑑(𝑦 ,𝑦 )+𝑑(𝑦 ,𝑦 ).
𝑚(𝑘)+1 𝑛(𝑘)+1 𝑛(𝑘)+1 𝑛(𝑘)
𝑇(𝛿,𝛿,𝛿,𝛿,0,2𝛿)≤0. (17)
Since𝑚(𝑘)≥𝑛(𝑘),wehave𝑥𝑚(𝑘) ⪰𝑥𝑛(𝑘)and𝑦𝑚(𝑘) ⪯𝑦𝑛(𝑘)for
Then from (𝑇2), we have 𝛿 ≤ 𝑓(𝛿), a contradiction. Hence each𝑘.Hence,from(9),weget
𝛿=0.
𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 ) 𝑇((𝑑(𝐹(𝑥 ,𝑦 ),𝐹(𝑥 ,𝑦 ))
That is, lim 𝑛+1 𝑛+2 𝑛+1 𝑛+2 =0. (18) 𝑚(𝑘) 𝑚(𝑘) 𝑛(𝑘) 𝑛(𝑘)
𝑛→∞ 2
+𝑑(𝐹(𝑦 ,𝑥 ),𝐹(𝑦 ,𝑥 )))
Nowweshowthat{𝑥𝑛}and{𝑦𝑛}areCauchysequencesin𝑋. 𝑚(𝑘) 𝑚(𝑘) 𝑛(𝑘) 𝑛(𝑘)
Supposethatatleastoneof{𝑥𝑛}or{𝑦𝑛}isnotCauchy.Then ×(2)−1,
there exist 𝜖 > 0 and sequences of positive integers {𝑚(𝑘)}
and{𝑛(𝑘)}with𝑚(𝑘)≥𝑛(𝑘)>𝑘suchthat
𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 )
𝑚(𝑘) 𝑛(𝑘) 𝑚(𝑘) 𝑛(𝑘) ,
𝑟𝑘 =𝑑(𝑥𝑚(𝑘),𝑥𝑛(𝑘))+𝑑(𝑦𝑚(𝑘),𝑦𝑛(𝑘))≥𝜖. (19) 2
𝑑(𝑥 ,𝐹(𝑥 ,𝑦 ))+𝑑(𝑦 ,𝐹(𝑦 ,𝑥 ))
Now choose 𝑚(𝑘) the least positive integer such that (19) 𝑚(𝑘) 𝑚(𝑘) 𝑚(𝑘) 𝑚(𝑘) 𝑚(𝑘) 𝑚(𝑘) ,
2
holds.Then
𝑑(𝑥 ,𝐹(𝑥 ,𝑦 ))+𝑑(𝑦 ,𝐹(𝑦 ,𝑥 ))
𝑑(𝑥𝑚(𝑘)−1,𝑥𝑛(𝑘))+𝑑(𝑦𝑚(𝑘)−1,𝑦𝑛(𝑘))<𝜖. (20) 𝑛(𝑘) 𝑛(𝑘) 𝑛(𝑘) 𝑛(𝑘) 𝑛(𝑘) 𝑛(𝑘) ,
2
Now 𝑑(𝑥 ,𝐹(𝑥 ,𝑦 ))+𝑑(𝑦 ,𝐹(𝑦 ,𝑥 ))
𝑚(𝑘) 𝑛(𝑘) 𝑛(𝑘) 𝑚(𝑘) 𝑛(𝑘) 𝑛(𝑘) ,
𝜖≤ 𝑟 =𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 ) 2
𝑘 𝑚(𝑘) 𝑛(𝑘) 𝑚(𝑘) 𝑛(𝑘)
𝑑(𝑥 ,𝐹(𝑥 ,𝑦 ))+𝑑(𝑦 ,𝐹(𝑦 ,𝑥 ))
≤ 𝑑(𝑥𝑚(𝑘),𝑥𝑚(𝑘)−1)+𝑑(𝑥𝑚(𝑘)−1,𝑥𝑛(𝑘)) 𝑛(𝑘) 𝑚(𝑘) 𝑚(𝑘) 𝑛(𝑘) 𝑚(𝑘) 𝑚(𝑘) )
(21) 2
+𝑑(𝑦 ,𝑦 )+𝑑(𝑦 ,𝑦 )
𝑚(𝑘) 𝑚(𝑘)−1 𝑚(𝑘)−1 𝑛(𝑘)
≤0,
< 𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 )+𝜖.
𝑚(𝑘) 𝑚(𝑘)−1 𝑚(𝑘) 𝑚(𝑘)−1
𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 )
Ontakinglimitsupremumas𝑘 → ∞,weget 𝑇( 𝑚(𝑘)+1 𝑛(𝑘)+1 2 𝑚(𝑘)+1 𝑛(𝑘)+1 ,
𝜖≤limsup𝑟𝑘 =limsup𝑑(𝑥𝑚(𝑘),𝑥𝑛(𝑘))+𝑑(𝑦𝑚(𝑘),𝑦𝑛(𝑘))≤𝜖. 𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 )
𝑘→∞ 𝑘→∞ 𝑚(𝑘) 𝑛(𝑘) 𝑚(𝑘) 𝑛(𝑘) ,
2
(22)
𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 )
Hence 𝑚(𝑘) 𝑚(𝑘)+1 𝑚(𝑘) 𝑚(𝑘)+1 ,
2
limsup𝑟𝑘 =limsup𝑑(𝑥𝑚(𝑘),𝑥𝑛(𝑘))+𝑑(𝑦𝑚(𝑘),𝑦𝑛(𝑘))=𝜖. 𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 )
𝑘→∞ 𝑘→∞ 𝑛(𝑘) 𝑛(𝑘)+1 𝑛(𝑘) 𝑛(𝑘)+1 ,
(23) 2
6 JournalofOperators
𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 )
𝑚(𝑘) 𝑛(𝑘)+1 𝑚(𝑘) 𝑛(𝑘)+1 , Hence
2
𝜖 𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 )
𝑑(𝑥𝑛(𝑘),𝑥𝑚(𝑘)+1)+𝑑(𝑦𝑛(𝑘),𝑦𝑚(𝑘)+1))≤0. 2 ≤𝑘l→im∞ 𝑚(𝑘)+1 𝑛(𝑘)+1 2 𝑚(𝑘)+1 𝑛(𝑘)+1 (32)
2 𝜖
≤𝑓( ), acontradiction.
(27) 2
Since 𝑇 is nonincreasing in 5th and 6th variable, using tri- Therefore{𝑥𝑛}and{𝑦𝑛}areCauchysequencesin𝑋.Since𝑋
angleinequality,weget is complete thereexists 𝑥,𝑦 ∈ 𝑋 withlim𝑛→∞𝑥𝑛 = 𝑥and
lim𝑛→∞𝑦𝑛 =𝑦.
𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 ) Supposethat𝐹iscontinuous.Then
𝑇( 𝑚(𝑘)+1 𝑛(𝑘)+1 𝑚(𝑘)+1 𝑛(𝑘)+1 ,
2 𝑥=𝑛l→im∞𝑥𝑛+1 =𝑛l→im∞𝐹(𝑥𝑛,𝑦𝑛)=𝐹(𝑥,𝑦),
(33)
𝑑(𝑥𝑚(𝑘),𝑥𝑛(𝑘))+𝑑(𝑦𝑚(𝑘),𝑦𝑛(𝑘)), 𝑦=𝑛l→im∞𝑦𝑛+1 =𝑛l→im∞𝐹(𝑦𝑛,𝑥𝑛)=𝐹(𝑦,𝑥).
2
Therefore(𝑥,𝑦)isacoupledfixedpointof𝐹.
𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 )
𝑚(𝑘) 𝑚(𝑘)+1 𝑚(𝑘) 𝑚(𝑘)+1 , Now suppose that conditions (b) hold. Hence, we have
2 𝑥𝑛 ⪯𝑥forall𝑛and𝑦⪯𝑦𝑛forall𝑛.
𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 ) Nowfrom(9),wehave
𝑛(𝑘) 𝑛(𝑘)+1 𝑛(𝑘) 𝑛(𝑘)+1 ,
2 𝑑(𝐹(𝑥,𝑦),𝐹(𝑥 ,𝑦 ))+𝑑(𝐹(𝑦,𝑥),𝐹(𝑦 ,𝑥 ))
𝑇( 𝑛 𝑛 𝑛 𝑛 ,
2
(𝑑(𝑥 ,𝑥 )+𝑑(𝑥 ,𝑥 )
𝑚(𝑘) 𝑚(𝑘)+1 𝑚(𝑘)+1 𝑛(𝑘)+1
𝑑(𝑥,𝑥 )+𝑑(𝑦,𝑦 ) 𝑑(𝑥,𝐹(𝑥,𝑦))+𝑑(𝑦,𝐹(𝑦,𝑥))
+𝑑(𝑦𝑚(𝑘),𝑦𝑚(𝑘)+1)+𝑑(𝑦𝑚(𝑘)+1,𝑦𝑛(𝑘)+1)) 𝑛 𝑛 , ,
2 2
×(2)−1,
𝑑(𝑥 ,𝐹(𝑥 ,𝑦 ))+𝑑(𝑦 ,𝐹(𝑦 ,𝑥 ))
𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 ,
(𝑑(𝑥 ,𝑥 )+𝑑(𝑥 ,𝑥 ) 2
𝑛(𝑘) 𝑛(𝑘)+1 𝑛(𝑘)+1 𝑚(𝑘)+1
𝑑(𝑥,𝐹(𝑥 ,𝑦 ))+𝑑(𝑦,𝐹(𝑦 ,𝑥 ))
+𝑑(𝑦 ,𝑦 )+𝑑(𝑦 ,𝑦 )) 𝑛 𝑛 𝑛 𝑛 ,
𝑛(𝑘) 𝑛(𝑘)+1 𝑛(𝑘)+1 𝑚(𝑘)+1
2
×(2)−1)≤0. 𝑑(𝑥 ,𝐹(𝑥,𝑦))+𝑑(𝑦 ,𝐹(𝑦,𝑥))
𝑛 𝑛 )≤0.
2
(28) (34)
Onletting𝑘 → ∞,weget Onletting𝑛 → ∞,weget
𝑑(𝐹(𝑥,𝑦),𝑥)+𝑑(𝐹(𝑦,𝑥),𝑦)
𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 )
𝑇( lim 𝑚(𝑘)+1 𝑛(𝑘)+1 𝑚(𝑘)+1 𝑛(𝑘)+1 , 𝑇( 2 ,0,
𝑘→∞ 2
(29) 𝑑(𝑥,𝐹(𝑥,𝑦))+𝑑(𝑦,𝐹(𝑦,𝑥))
𝜖 𝜖 𝜖 ,0,0, (35)
,0,0, , )≤0. 2
2 2 2
𝑑(𝑥,𝐹(𝑥,𝑦))+𝑑(𝑦,𝐹(𝑦,𝑥))
)≤0.
From(𝑇2),weget 2
𝑘l→im∞𝑑(𝑥𝑚(𝑘)+1,𝑥𝑛(𝑘)+1)+2𝑑(𝑦𝑚(𝑘)+1,𝑦𝑛(𝑘)+1) ≤𝑓(2𝜖). Ififx𝑑ed(𝐹p(o𝑥i,n𝑦t)o,f𝑥𝐹).+𝑑(𝐹(𝑦,𝑥),𝑦) = 0,then(𝑥,𝑦)isacoupled
(30) If 𝑑(𝐹(𝑥,𝑦),𝑥) + 𝑑(𝐹(𝑦,𝑥),𝑦) > 0, then (35) yields a
contradiction to (𝑇3). Hence 𝐹(𝑥,𝑦) = 𝑥 and 𝐹(𝑦,𝑥) = 𝑦.
Nowonletting𝑘 → ∞,in(26),weget Therefore(𝑥,𝑦)isacoupledfixedpointof𝐹.
𝜖≤ 𝑘l→im∞[𝑑(𝑥𝑚(𝑘),𝑥𝑚(𝑘)+1) +𝑑(𝑥𝑚(𝑘)+1,𝑥𝑛(𝑘)+1) fixedThpeooirnetmof16𝐹d.oThesenfootllgouwairnagnteexeatmhepluensiuqpupenoertsssothficsoauspsleerd-
tion.
+𝑑(𝑥 ,𝑥 )+𝑑(𝑦 ,𝑦 )
𝑛(𝑘)+1 𝑛(𝑘) 𝑚(𝑘) 𝑚(𝑘)+1
(31) Example 17. Let 𝑋 = {2,4,6,8} with the usual metric. We
+𝑑(𝑦 ,𝑦 )+𝑑(𝑦 ,𝑦 )]
𝑚(𝑘)+1 𝑛(𝑘)+1 𝑛(𝑘)+1 𝑛(𝑘) defineapartialorder⪯on𝑋asfollows:
= lim [𝑑(𝑥𝑚(𝑘)+1,𝑥𝑛(𝑘)+1)+𝑑(𝑦𝑚(𝑘)+1,𝑦𝑛(𝑘)+1)]. ⪯={(2,2),(4,4),(6,6),(8,8),(2,8)}. (36)
𝑘→∞
JournalofOperators 7
We define 𝑇 : (𝑅+)6 → 𝑅 by 𝑇(𝑡1,...,𝑡6) = 𝑡1 − 𝛼(𝑡4𝑡5𝑡6/ Hence inequality (9) holds. Therefore all the hypotheses of
(𝑡2 +𝑡3+1)),𝑡1,𝑡2,...,𝑡6 ≥ 0and𝛼 = 15/16.Then𝑇 ∈ S Theorem16holdand(8,4)and(4,4)aretwocoupledfixed
(Example 11). pointsof𝐹.
Let
In the following theorem, we prove the uniqueness of
𝐴= {(2,2),(2,8),(4,2),(4,4),(4,8), coupledfixedpointsbyusinganimplicitrelationinS󸀠.
(6,2),(6,8),(8,2),(8,8)} Wedefineapartialorder⪯1on𝑋×𝑋by(𝑥,𝑦)⪯1(𝑢,V)if
𝑥⪯𝑢,𝑦⪰V.
(37)
𝐵= {(2,4),(8,4),(8,6)},
Theorem18. InadditiontothehypothesesofTheorem16with
𝐶= {(2,6),(4,6),(6,4),(6,6)}. 𝑇∈S󸀠supposethatthefollowingconditionholds:
Wedefine𝐹:𝑋×𝑋 → 𝑋by for any (𝑥,𝑦) and (𝑢,V) in 𝑋×𝑋
there exists (𝑧,𝑡)∈𝑋×𝑋 that is comparable with
{{4, if (𝑥,𝑦)∈𝐴
𝐹(𝑥,𝑦)={{8, if (𝑥,𝑦)∈𝐵 (38) (𝑥,𝑦) and (𝑢,V).
{2, if (𝑥,𝑦)∈𝐶. (40)
Clearly𝐹satisfiesthemixedmonotoneproperty.Wechoose Then𝐹hasauniquecoupledfixedpoint.
(t(h𝑢𝑥a,0tV,𝑦)𝑥0i0n)⪯𝑋=𝐹×((2𝑥𝑋,04,)s𝑦.u0Thc)haetnnhda𝐹t𝑦(𝑥0𝑥0⪰⪰,𝑦𝐹𝑢0()a𝑦n0=d,𝑥𝑦80)a⪯.nThdVae𝐹re(e𝑦let0hm,e𝑥e0fno)tlsl=o(w𝑥4,in𝑦sg)o,: Pfixroeodf.pWoinitth(t𝑥h,e𝑦h),ypthoathteisse,s𝑥of=Th𝐹eo(𝑥re,m𝑦)16an,d𝐹𝑦has=a c𝐹o(u𝑦p,l𝑥e)d.
(2,2), (2,2); (2,2), (2,8); (2,4), (2,4); (2,6), (2,6); (2,8), Supposethatifpossiblethereexists(𝑢,V) ∈ 𝑋×𝑋suchthat
(2,8); (4,2), (4,2); (4,2), (4,8); (4,4), (4,4); (4,6), (4,6); 𝑢=𝐹(𝑢,V)andV=𝐹(V,𝑢).Weshowthat𝑥=𝑢and𝑦=V.
(4,8), (4,8); (6,2), (6,2); (6,2), (6,8); (6,4), (6,4); (6,6), Casei.(𝑥,𝑦)and(𝑢,V)arecomparable.
((68,,62));, ((68,,88));, ((68,,84));, ((82,,42));, ((28,,24));, ((88,,24)),; ((28,,86));, ((28,,62));, ((88,,62)),; ThatWisi,th𝑥o⪰ut𝑢loasnsdof𝑦g⪯enVe.rality,weassumethat(𝑢,V)⪯1(𝑥,𝑦).
(8,6);(8,8),(2,8);(8,8),(8,8).Wenowverifyinequality(9) Supposethateither𝑥≠𝑢or𝑦≠V.Then
fortheabovepairs.
𝑑(𝐹(𝑥,𝑦),𝐹(𝑢,V))+𝑑(𝐹(𝑦,𝑥),𝐹(V,𝑢))
Case i. (𝑥,𝑦), (𝑢,V) ∈ 𝑋×𝑋 such that(𝑥,𝑦), (𝑢,V), (𝑦,𝑥), 𝑇( ,
2
(V,𝑢)areeitherin𝐴or𝐵or𝐶or(𝑥,𝑦)=(𝑢,V).
Inthiscase(𝑑(𝐹(𝑥,𝑦),𝐹(𝑢,V)) + 𝑑(𝐹(𝑦,𝑥),𝐹(V,𝑢)))/2=0 𝑑(𝑥,𝑢)+𝑑(𝑦,V)
sothatinequality(9)holdstrivially. ,
2
Caseii.(𝑥,𝑦)=(6,2),(𝑢,V)=(6,8)or(𝑥,𝑦)=(6,8),(𝑢,V)= 𝑑(𝑥,𝐹(𝑥,𝑦))+𝑑(𝑦,𝐹(𝑦,𝑥))
(2,6). 2 ,
Inthiscase
𝑑(𝑢,𝐹(𝑢,V))+𝑑(V,𝐹(V,𝑢))
,
𝑑(𝐹(𝑥,𝑦),𝐹(𝑢,V))+𝑑(𝐹(𝑦,𝑥),𝐹(V,𝑢)) 2
2 𝑑(𝑥,𝐹(𝑢,V))+𝑑(𝑦,𝐹(V,𝑢)) (41)
,
1516 15 2
=3= =
16 5 16 𝑑(𝑢,𝐹(𝑥,𝑦))+𝑑(V,𝐹(𝑦,𝑥))
)≤0,
𝑑(𝑢,𝐹(𝑢,V))+𝑑(V,𝐹(V,𝑢)) 2
×( )
2
𝑑(𝑥,𝑢)+𝑑(𝑦,V) 𝑑(𝑥,𝑢)+𝑑(𝑦,V)
𝑇( , ,0,0,
𝑑(𝑥,𝐹(𝑢,V))+𝑑(𝑦,𝐹(V,𝑢)) 2 2
×( )
2 (39) 𝑑(𝑥,𝑢)+𝑑(𝑦,V) 𝑑(𝑢,𝑥)+𝑑(V,𝑦)
, )≤0.
𝑑(𝑢,𝐹(𝑥,𝑦))+𝑑(V,𝐹(𝑦,𝑥)) 2 2
×( )
2
Then,from(𝑇2󸀠),thereexists𝑓:𝑅+ → 𝑅+,𝑓(𝑡)<𝑡for𝑡>0
𝑑(𝑥,𝑢)+𝑑(𝑦,V) suchthat
×(
2
𝑑(𝑥,𝑢)+𝑑(𝑦,V) 𝑑(𝑥,𝑢)+𝑑(𝑦,V)
≤𝑓( ), (42)
𝑑(𝑥,𝐹(𝑥,𝑦))+𝑑(𝑦,𝐹(𝑦,𝑥)) −1 2 2
+ +1) .
2
acontradiction.Hence𝑥=𝑢and𝑦=V.
8 JournalofOperators
Caseii.Thereexists(𝑧,𝑡)∈𝑋×𝑋whichiscomparablewith 𝑑(𝑥,𝐹(𝑥,𝑦))+𝑑(𝑦,𝐹(𝑦,𝑥)),
(𝑥,𝑦)and(𝑢,V). 2
Withoutlossofgenerality,weassumethat(𝑥,𝑦)⪯1(𝑧,𝑡) 𝑑(𝐹𝑛(𝑧,𝑡),𝑥)+𝑑(𝐹𝑛(𝑡,𝑧),𝑦)
and(𝑢,V)⪯1(𝑧,𝑡),thatis,𝑧⪰𝑥,𝑡⪯𝑦and𝑧⪰𝑢,𝑡⪯V. ,
We denote 𝐹𝑛+1(𝑥,𝑦) = 𝐹(𝐹𝑛(𝑥,𝑦),𝐹𝑛(𝑦,𝑥)) for each 2
𝑥,𝑦∈𝑋. (𝑑(𝑥,𝐹(𝐹𝑛(𝑧,𝑡)),𝐹𝑛(𝑡,𝑧))
Since𝐹hasmixedmonotoneproperty,wehave
+𝑑(𝑦,𝐹(𝐹𝑛(𝑡,𝑧),𝐹𝑛(𝑧,𝑡))))
𝐹(𝑥,𝑦)⪯𝐹(𝑧,𝑡), 𝐹(𝑢,V)⪯𝐹(𝑧,𝑡),
(43) ×(2)−1)≤0.
𝐹(𝑦,𝑥)⪰𝐹(𝑡,𝑧), 𝐹(V,𝑢)⪰𝐹(𝑡,𝑧).
(47)
Nowsupposethat
Thisimplies
𝐹𝑛(𝑥,𝑦)⪯𝐹𝑛(𝑧,𝑡), 𝐹𝑛(𝑢,V)⪯𝐹𝑛(𝑧,𝑡),
𝑑(𝐹𝑛+1(𝑧,𝑡),𝑥)+𝑑(𝐹𝑛+1(𝑡,𝑧),𝑦)
𝐹𝑛(𝑦,𝑥)⪰𝐹𝑛(𝑡,𝑧), 𝐹𝑛(V,𝑢)⪰𝐹𝑛(𝑡,𝑧) 𝑇( 2 ,
for some 𝑛. 𝑑(𝐹𝑛+1(𝑡,𝑧),𝑥)+𝑑(𝐹𝑛+1(𝑧,𝑡),𝑦)
,
𝐹𝑛+1(𝑥,𝑦)=𝐹(𝐹𝑛(𝑥,𝑦),𝐹𝑛(𝑦,𝑥)) 2
⪯𝐹(𝐹𝑛(𝑧,𝑡),𝐹𝑛(𝑡,𝑧))=𝐹𝑛+1(𝑧,𝑡), 𝑑(𝐹𝑛(𝑧,𝑡),𝐹𝑛+1(𝑧,𝑡))+𝑑(𝐹𝑛(𝑡,𝑧),𝐹𝑛+1(𝑡,𝑧))
,0,
2
𝐹𝑛+1(𝑢,V)=𝐹(𝐹𝑛(𝑢,V),𝐹𝑛(V,𝑢))
(44) 𝑑(𝐹𝑛(𝑧,𝑡),𝑥)+𝑑(𝐹𝑛(𝑡,𝑧),𝑦)
,
⪯𝐹(𝐹𝑛(𝑧,𝑡),𝐹𝑛(𝑡,𝑧))=𝐹𝑛+1(𝑧,𝑡), 2
𝐹𝑛+1(𝑦,𝑥)=𝐹(𝐹𝑛(𝑦,𝑥),𝐹𝑛(𝑥,𝑦)) 𝑑(𝑥,𝐹𝑛+1(𝑧,𝑡))+𝑑(𝑦,𝐹𝑛+1(𝑡,𝑧))
)≤0.
2
⪰𝐹(𝐹𝑛(𝑡,𝑧),𝐹𝑛(𝑧,𝑡))=𝐹𝑛+1(𝑡,𝑧),
(48)
𝐹𝑛+1(V,𝑢)=𝐹(𝐹𝑛(V,𝑢),𝐹𝑛(𝑢,V)) Since 𝑇 is nonincreasing in 3rd variable, using triangle
inequality,weget
⪰𝐹(𝐹𝑛(𝑡,𝑧),𝐹𝑛(𝑧,𝑡))=𝐹𝑛+1(𝑡,𝑧).
𝑑(𝐹𝑛+1(𝑧,𝑡),𝑥)+𝑑(𝐹𝑛+1(𝑡,𝑧),𝑦)
Hencebymathematicalinduction, 𝑇( ,
2
𝐹𝑛(𝑥,𝑦)⪯𝐹𝑛(𝑧,𝑡), 𝐹𝑛(𝑢,V)⪯𝐹𝑛(𝑧,𝑡),
𝑑(𝐹𝑛+1(𝑡,𝑧),𝑥)+𝑑(𝐹𝑛+1(𝑧,𝑡),𝑦)
𝐹𝑛(𝑦,𝑥)⪰𝐹𝑛(𝑡,𝑧), 𝐹𝑛(V,𝑢)⪰𝐹𝑛(𝑡,𝑧) (45) ,
2
for each 𝑛=0,1,2,.... (𝑑(𝐹𝑛(𝑧,𝑡),𝑥)+𝑑(𝑥,𝐹𝑛+1(𝑧,𝑡))
Wenowshowthat
+𝑑(𝐹𝑛(𝑡,𝑧),𝑦)+𝑑(𝑦,𝐹𝑛+1(𝑡,𝑧))) (49)
lim 𝐹𝑛(𝑧,𝑡)=𝑥, lim 𝐹𝑛(𝑡,𝑧)=𝑦. (46)
𝑛→∞ 𝑛→∞ ×(2)−1,0,
Using(9),weget
𝑑(𝐹𝑛(𝑧,𝑡),𝑥)+𝑑(𝐹𝑛(𝑡,𝑧),𝑦)
𝑇((𝑑(𝐹(𝐹𝑛(𝑧,𝑡),𝐹𝑛(𝑡,𝑧)),𝐹(𝑥,𝑦)) 2 ,
+𝑑(𝐹(𝐹𝑛(𝑡,𝑧),𝐹𝑛(𝑧,𝑡)),𝐹(𝑦,𝑥))) 𝑑(𝑥,𝐹𝑛+1(𝑧,𝑡))+𝑑(𝑦,𝐹𝑛+1(𝑡,𝑧))
)≤0.
2
×(2)−1,
𝑑(𝐹𝑛(𝑧,𝑡),𝑥)+𝑑(𝐹𝑛(𝑡,𝑧),𝑦) Therefore,from(𝑇2󸀠),thereexists𝑓: 𝑅+ → 𝑅+,𝑓(𝑡) < 𝑡for
, 𝑡>0suchthat
2
𝑑(𝐹𝑛+1(𝑧,𝑡),𝑥)+𝑑(𝐹𝑛+1(𝑡,𝑧),𝑦)
(𝑑(𝐹𝑛(𝑧,𝑡),𝐹(𝐹𝑛(𝑧,𝑡)),𝐹𝑛(𝑡,𝑧))
2
+𝑑(𝐹𝑛(𝑡,𝑧),𝐹(𝐹𝑛(𝑡,𝑧),𝐹𝑛(𝑧,𝑡)))) 𝑑(𝐹𝑛(𝑧,𝑡),𝑥)+𝑑(𝐹𝑛(𝑡,𝑧),𝑦) (50)
≤𝑓( ).
×(2)−1, 2
JournalofOperators 9
If𝑑(𝐹𝑛(𝑧,𝑡),𝑥)+𝑑(𝐹𝑛(𝑡,𝑧),𝑦)=0forsome𝑛then That is, lim 𝐹𝑛(𝑧,𝑡)=𝑥, lim 𝐹𝑛(𝑡,𝑧)=𝑦.
𝑛→∞ 𝑛→∞
(53)
𝐹𝑛+1(𝑧,𝑡)=𝐹(𝐹𝑛(𝑧,𝑡),𝐹𝑛(𝑡,𝑧))=𝐹(𝑥,𝑦)=𝑥,
(51) Since(𝑢,V)isalsoacoupledfixedpointof𝐹,wehave
𝐹𝑛+1(𝑡,𝑧)=𝐹(𝐹𝑛(𝑡,𝑧),𝐹𝑛(𝑧,𝑡))=𝐹(𝑦,𝑥)=𝑦.
lim 𝐹𝑛(𝑧,𝑡)=𝑢, lim 𝐹𝑛(𝑡,𝑧)=V. (54)
𝑛→∞ 𝑛→∞
Hence,𝐹𝑛+𝑘(𝑧,𝑡) = 𝑥and𝐹𝑛+𝑘(𝑡,𝑧) = 𝑦for𝑘 = 0,1,2,..., Hence, by theuniquenessoflimit,we have𝑥 = 𝑢 and𝑦 =
thatis,lim𝑛→∞𝐹𝑛(𝑧,𝑡)=𝑥andlim𝑛→∞𝐹𝑛(𝑡,𝑧)=𝑦,sothat V.
(46)holds.
Supposethat𝑑(𝐹𝑛(𝑧,𝑡),𝑥)+𝑑(𝐹𝑛(𝑡,𝑧),𝑦) > 0forall𝑛; Remark 19. In Example17, 𝑇 does not belong to 𝑆󸀠 (Exam-
ple 15) and condition (40) does not hold and 𝑇 has two
then,from(49),weget
coupledfixedpoints.
𝑑(𝐹𝑛+1(𝑧,𝑡),𝑥)+𝑑(𝐹𝑛+1(𝑡,𝑧),𝑦) Theorem20. InadditiontothehypothesesofTheorem16with
2 𝑇∈S󸀠supposethat𝑥0and𝑦0arecomparable.Further,assume
that(40)holds.Then𝐹hasauniquecoupledfixedpoint(𝑥,𝑦)
𝑑(𝐹𝑛(𝑧,𝑡),𝑥)+𝑑(𝐹𝑛(𝑡,𝑧),𝑦) and𝑥=𝑦.
≤𝑓( ) (52)
2
Proof. From the proof of Theorem16, we have that 𝐹 has a
𝑑(𝐹𝑛(𝑧,𝑡),𝑥)+𝑑(𝐹𝑛(𝑡,𝑧),𝑦) coupledfixedpointsay(𝑥,𝑦).Thatis,𝑥 = 𝐹(𝑥,𝑦)and𝑦 =
< , ∀𝑛.
2 𝐹(𝑦,𝑥).Supposethat𝑥0 ⪯𝑦0.Weshowthat
Therefore{(𝑑(𝐹𝑛(𝑧,𝑡),𝑥)+𝑑(𝐹𝑛(𝑡,𝑧),𝑦))/2}isadecreasing 𝑥𝑛 ⪯𝑦𝑛, ∀𝑛, (55)
sequenceofnonnegativerealnumbers.Thus,thereexists𝛿≥ usingprincipleofmathematicalinductionon𝑛,where𝑥𝑛 =
0suchthatlim𝑛→∞(𝑑(𝐹𝑛(𝑧,𝑡),𝑥)+𝑑(𝐹𝑛(𝑡,𝑧),𝑦))/2=𝛿. 𝐹(𝑥𝑛−1,𝑦𝑛−1) and 𝑦𝑛 = 𝐹(𝑦𝑛−1,𝑥𝑛−1), 𝑛 = 0,1,2,... are the
If𝛿 > 0,then,onletting𝑛 → ∞in(49),weget𝑇(𝛿,𝛿, sequencesdefinedintheproofofTheorem16.Supposethat
2𝛿,0,𝛿,𝛿)≤0. (55) is true for some 𝑛 ≥ 0. Then, by mixed monotone
Then, from (𝑇󸀠), we have 𝛿 ≤ 𝑓(𝛿), a contradiction. propertyof𝐹,
2
Hence𝛿=0.
𝑥𝑛+1 =𝐹(𝑥𝑛,𝑦𝑛)≤𝐹(𝑦𝑛,𝑥𝑛)=𝑦𝑛+1. (56)
𝑑(𝐹𝑛(𝑧,𝑡),𝑥)+𝑑(𝐹𝑛(𝑡,𝑧),𝑦) Thus(55)istrueforeach𝑛 = 1,2,3,....Now,from(9),we
That is, lim =0.
𝑛→∞ 2 have
𝑑(𝐹(𝐹𝑛(𝑥,𝑦),𝐹𝑛(𝑦,𝑥)),𝐹(𝐹𝑛(𝑦,𝑥),𝐹𝑛(𝑥,𝑦)))+𝑑(𝐹(𝐹𝑛(𝑦,𝑥),𝐹𝑛(𝑥,𝑦)),𝐹(𝐹𝑛(𝑥,𝑦),𝐹𝑛(𝑦,𝑥)))
𝑇( ,
2
𝑑(𝐹𝑛(𝑥,𝑦),𝐹𝑛(𝑦,𝑥))+𝑑(𝐹𝑛(𝑦,𝑥),𝐹𝑛(𝑥,𝑦))
,
2
𝑑(𝐹𝑛(𝑥,𝑦),𝐹(𝐹𝑛(𝑥,𝑦),𝐹𝑛(𝑦,𝑥)))+𝑑(𝐹𝑛(𝑦,𝑥),𝐹(𝐹𝑛(𝑦,𝑥),𝐹𝑛(𝑥,𝑦)))
,
2
(57)
𝑑(𝐹𝑛(𝑦,𝑥),𝐹(𝐹𝑛(𝑦,𝑥),𝐹𝑛(𝑥,𝑦)))+𝑑(𝐹𝑛(𝑥,𝑦),𝐹(𝐹𝑛(𝑥,𝑦),𝐹𝑛(𝑦,𝑥)))
2
𝑑(𝐹𝑛(𝑥,𝑦),𝐹(𝐹𝑛(𝑦,𝑥),𝐹𝑛(𝑥,𝑦)))+𝑑(𝐹𝑛(𝑦,𝑥),𝐹(𝐹𝑛(𝑥,𝑦),𝐹𝑛(𝑦,𝑥)))
,
2
𝑑(𝐹𝑛(𝑦,𝑥),𝐹(𝐹𝑛(𝑥,𝑦),𝐹𝑛(𝑦,𝑥)))+𝑑(𝐹𝑛(𝑥,𝑦),𝐹(𝐹𝑛(𝑦,𝑥),𝐹𝑛(𝑥,𝑦)))
)≤0.
2
10 JournalofOperators
Thisimplies map𝜓 : 𝑅+ → 𝑅+,𝜓(𝑡) < 𝑡for𝑡 > 0,𝜓(𝑡) = 0ifandonlyif
𝑡=0suchthat
𝑑(𝐹𝑛+1(𝑥,𝑦),𝐹𝑛+1(𝑦,𝑥))+𝑑(𝐹𝑛+1(𝑦,𝑥),𝐹𝑛+1(𝑥,𝑦))
𝑇( , 𝑑(𝑥,𝑢)+𝑑(𝑦,V)
2 𝑑(𝐹(𝑥,𝑦),𝐹(𝑢,V))≤
2
𝑑(𝐹𝑛(𝑥,𝑦),𝐹𝑛(𝑦,𝑥))+𝑑(𝐹𝑛(𝑦,𝑥),𝐹𝑛(𝑥,𝑦)) (61)
𝑑(𝑥,𝑢)+𝑑(𝑦,V)
,
2 −𝜓( ),
2
𝑑(𝐹𝑛(𝑥,𝑦),𝐹𝑛+1(𝑥,𝑦))+𝑑(𝐹𝑛(𝑦,𝑥),𝐹𝑛+1(𝑦,𝑥))
2 , foreach𝑥,𝑦,𝑢,V∈𝑋with𝑥⪰𝑢and𝑦⪯V.
Supposethateither(a)or(b)ofTheorem16holds.
𝑑(𝐹𝑛(𝑦,𝑥),𝐹𝑛+1(𝑦,𝑥))+𝑑(𝐹𝑛(𝑥,𝑦),𝐹𝑛+1(𝑥,𝑦))
2 If there exist 𝑥0,𝑦0 ∈ 𝑋 such that 𝑥0 ⪯ 𝐹(𝑥0,𝑦0) and 𝑦0 ⪰
𝐹(𝑦0,𝑥0)then𝐹hasacoupledfixedpoint.
𝑑(𝐹𝑛(𝑥,𝑦),𝐹𝑛+1(𝑦,𝑥))+𝑑(𝐹𝑛(𝑦,𝑥),𝐹𝑛+1(𝑥,𝑦))
,
2 Proof. From(61),weobtain
𝑑(𝐹𝑛(𝑦,𝑥),𝐹𝑛+1(𝑥,𝑦))+𝑑(𝐹𝑛(𝑥,𝑦),𝐹𝑛+1(𝑦,𝑥)) 𝑑(𝑦,V)+𝑑(𝑥,𝑢)
) 𝑑(𝐹(V,𝑢),𝐹(𝑦,𝑥))≤
2 2
(62)
≤0. 𝑑(𝑦,V)+𝑑(𝑥,𝑢)
−𝜓( ).
(58)
2
Onletting𝑛 → ∞,weget
Adding (61) and (62), we obtain inequality (60). Hence the
conclusionfollowsfromCorollary22.
𝑇(𝑑(𝑥,𝑦),𝑑(𝑥,𝑦),0,0,𝑑(𝑥,𝑦),𝑑(𝑥,𝑦))≤0, (59)
Remark 24. Corollary23 is another version of Theorem4
acontradictionto(𝑇󸀠),if𝑑(𝑥,𝑦)>0.Hence,𝑑(𝑥,𝑦)=0;that withdifferent𝜓.
3
is,𝑥=𝑦.
Corollary25. Let(𝑋,⪯)beapartiallyorderedsetandsuppose
that𝑑is a metricon𝑋suchthat (𝑋,𝑑)is acompletemetric
Remark 21. Theorem16 is an extension of Theorem5 to
coupledfixedpointswithanimplicitrelationof𝑆. space. Let 𝐹 : 𝑋 × 𝑋 → 𝑋 be a mapping satisfying mixed
monotonepropertyandsupposethatthereexistsacontinuous
Corollary22. Let(𝑋,⪯)beapartiallyorderedsetandsuppose map𝜑:𝑅+ → 𝑅+,𝜑(𝑡)<𝑡for𝑡>0,𝜑(0)=0suchthat
that𝑑is ametricon 𝑋suchthat (𝑋,𝑑)is acompletemetric 𝑑(𝐹(𝑥,𝑦),𝐹(𝑢,V))+𝑑(𝐹(𝑦,𝑥),𝐹(V,𝑢))
space. Let 𝐹 : 𝑋 × 𝑋 → 𝑋 be a mapping satisfying mixed
2
monotonepropertyandsupposethatthereexistsacontinuous
(63)
map𝜓 : 𝑅+ → 𝑅+,𝜓(𝑡) < 𝑡for𝑡 > 0,𝜓(𝑡) = 0ifandonlyif 𝑑(𝑥,𝑢)+𝑑(𝑦,V)
𝑡=0suchthat ≤𝜑( 2 ),
𝑑(𝐹(𝑥,𝑦),𝐹(𝑢,V))+𝑑(𝐹(𝑦,𝑥),𝐹(V,𝑢)) foreach𝑥,𝑦,𝑢,V∈𝑋with𝑥⪰𝑢and𝑦⪯V.
2
Supposethateither(a)or(b)ofTheorem16holds.
(60)
𝑑(𝑥,𝑢)+𝑑(𝑦,V) 𝑑(𝑥,𝑢)+𝑑(𝑦,V) If there exist 𝑥0,𝑦0 ∈ 𝑋 such that 𝑥0 ⪯ 𝐹(𝑥0,𝑦0) and 𝑦0 ⪰
≤ 2 −𝜓( 2 ), 𝐹(𝑦0,𝑥0)then𝐹hasacoupledfixedpoint.
foreach𝑥,𝑦,𝑢,V∈𝑋with𝑥⪰𝑢and𝑦⪯V. P𝜑r(o𝑡2o)f,.𝑡B1,y𝑡d2,e.fi.n.,in𝑡6g≥𝑇0:,(𝑅w+e)o6b→tain𝑅(i6n3()9.)Haesn𝑇c(e𝑡1th,.e..c,o𝑡n6c)lu=s𝑡io1−n
Supposethateither(a)or(b)ofTheorem16holds.
followsfromTheorem16.
If there exist 𝑥0,𝑦0 ∈ 𝑋 such that 𝑥0 ⪯ 𝐹(𝑥0,𝑦0) and 𝑦0 ⪰
𝐹(𝑦0,𝑥0)then𝐹hasacoupledfixedpoint. Corollary26. Let(𝑋,⪯)beapartiallyorderedsetandsuppose
that𝑑is a metricon𝑋suchthat (𝑋,𝑑)is acompletemetric
Proof. Bydefining𝑇 : (𝑅+)6 → 𝑅in(9)as𝑇(𝑡1,...,𝑡6) = space. Let 𝐹 : 𝑋 × 𝑋 → 𝑋 be a mapping satisfying mixed
𝑡1 −(𝑡2 −𝜓(𝑡2)),𝑡1,𝑡2,...,𝑡6 ≥ 0,weobtain(60).Hencethe monotonepropertyandsupposethatthereexistsacontinuous
conclusionfollowsfromTheorem16. map𝜑:𝑅+ → 𝑅+,𝜑(𝑡)<𝑡for𝑡>0,𝜑(0)=0suchthat
Corollary23. Let(𝑋,⪯)beapartiallyorderedsetandsuppose 𝑑(𝑥,𝑢)+𝑑(𝑦,V)
that𝑑is ametricon 𝑋suchthat (𝑋,𝑑)is acompletemetric 𝑑(𝐹(𝑥,𝑦),𝐹(𝑢,V))≤𝜑( 2 ), (64)
space. Let 𝐹 : 𝑋 × 𝑋 → 𝑋 be a mapping satisfying mixed
monotonepropertyandsupposethatthereexistsacontinuous foreach𝑥,𝑦,𝑢,V∈𝑋with𝑥⪰𝑢and𝑦⪯V.