Hindawi - Fixed Point Theory and Applications

17 Pages·2010·0.22 MB·English

HindawiPublishingCorporation
FixedPointTheoryandApplications
Volume2010,ArticleID621469,17pages
doi:10.1155/2010/621469
Research Article
Some Fixed Point Theorems on Ordered Metric
Spaces and Application
Ishak Altun and Hakan Simsek
DepartmentofMathematics,FacultyofScienceandArts,KirikkaleUniversity,Yahsihan,
Kirikkale71450,Turkey
CorrespondenceshouldbeaddressedtoIshakAltun,[email protected]
Received2July2009;Accepted13January2010
AcademicEditor:JuanJoseNieto
Copyrightq2010I.AltunandH.Simsek. This is an open access article distributed under the
Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproductioninanymedium,providedtheoriginalworkisproperlycited.
We present some fixed point results for nondecreasing and weakly increasing operators in a
partially ordered metric space using implicit relations. Also we give an existence theorem for
commonsolutionoftwointegralequations.
1. Introduction
Existence of fixed points in partially ordered sets has been considered recently in (cid:2)1(cid:3), and
some generalizations of the result of (cid:2)1(cid:3) are given in (cid:2)2–6(cid:3). Also, in (cid:2)1(cid:3) some applications
to matrix equations are presented, in (cid:2)3, 4(cid:3) some applications to periodic boundary value
problemandtosomeparticularproblemsare,respectively,given.Later,in(cid:2)6(cid:3)O’Reganand
Petrus¸el gave some existence results for Fredholm and Volterra type integral equations. In
someoftheaboveworks,thefixedpointresultsaregivenfornondecreasingmappings.
Wecanorderthepurposesofthepaperasfollows.
First,wegiveaslightgeneralizationofsomeoftheresultsoftheabovepapersusing
animplicitrelationinthefollowingway.
In (cid:2)1, 3(cid:3), the authors used the following contractive condition in their result, there
existsk ∈(cid:2)0,1(cid:4)suchthat
(cid:2) (cid:3) (cid:2) (cid:3)
d fx,fy ≤kd x,y for y (cid:4)x. (cid:5)1.1(cid:4)
Afterwards,in(cid:2)2(cid:3),theauthorsusedthenonlinearcontractivecondition,thatis,
(cid:2) (cid:3) (cid:2) (cid:2) (cid:3)(cid:3)
d fx,fy ≤ψ d x,y for y (cid:4)x, (cid:5)1.2(cid:4)
2 FixedPointTheoryandApplications
whereψ :(cid:2)0,∞(cid:4) → (cid:2)0,∞(cid:4)isanondecreasingfunctionwithlimn→∞ψn(cid:5)t(cid:4)(cid:6)0fort>0,instead
of (cid:5)1.1(cid:4). Also in (cid:2)2(cid:3), the authors proved a fixed point theorem using generalized nonlinear
contractivecondition,thatis,
(cid:4) (cid:5) (cid:8)(cid:9)
(cid:2) (cid:3) (cid:2) (cid:3) (cid:2) (cid:3) (cid:2) (cid:3) (cid:6) (cid:2) (cid:3) (cid:2) (cid:3)(cid:7)
1
d fx,fy ≤ψ max d x,y ,d x,fx ,d y,fy , d x,fy (cid:7)d y,fx (cid:5)1.3(cid:4)
2
for y (cid:4) x, where ψ is as above. In the Section3, we generalized the above contractive
conditionsusingtheimplicitrelationtechniqueinsuchawaythat
(cid:2) (cid:2) (cid:3) (cid:2) (cid:3) (cid:2) (cid:3) (cid:2) (cid:3) (cid:2) (cid:3)(cid:3)
T d Fx,Fy ,d x,y ,d(cid:5)x,Fx(cid:4),d y,Fy ,d x,Fy ,d y,Fx ≤0 (cid:5)1.4(cid:4)
for y (cid:4) x, where T : R6(cid:7) → R is a function as given in Section2. We can obtain various
contractiveconditionsfrom(cid:5)1.4(cid:4).Forexample,ifwechoose
(cid:4) (cid:5) (cid:8)(cid:9)
1
T(cid:5)t ,...,t (cid:4)(cid:6)t −ψ max t ,t ,t , (cid:2)t (cid:7)t (cid:3) (cid:5)1.5(cid:4)
1 6 1 2 3 4 5 6
2
in(cid:5)1.4(cid:4),then,wehave(cid:5)1.3(cid:4).Similarlywecanhavethecontractiveconditionsin(cid:2)7–9(cid:3)from
(cid:5)1.4(cid:4).
In some of the above mentioned theorems, the fixed point results are given for
nondecreasingmappings.Alsointhesetheoremsthefollowingconditionisused:
there exists x ∈X such that x (cid:4)fx . (cid:5)1.6(cid:4)
0 0 0
InSection4,wegivesomeexamplessuchthattwoweaklyincreasingmappingsneednotbe
nondecreasing.Therefore,wegiveacommonfixedpointtheoremfortwoweaklyincreasing
operators in partially ordered metric spaces using implicit relation technique. Also we did
notusethecondition(cid:5)1.6(cid:4)inthistheorem.Attheend,toseetheapplicabilityofourresult,
we give an existence theorem for common solution of two integral equations using a result
oftheSection4.
2. Implicit Relation
Implicitrelationsonmetricspaceshavebeenusedinmanyarticles.Seeforexamples,(cid:2)10–15(cid:3).
Let R(cid:7) denote the nonnegative real numbers, and let T be the set of all continuous
functionsT :R6(cid:7) → Rsatisfyingthefollowingconditions:
T :T(cid:5)t ,...,t (cid:4)isnonincreasinginvariablest ,...,t ;
1 1 6 2 6
T2 :there exists a right continuous function f : R(cid:7) → R(cid:7),f(cid:5)0(cid:4) (cid:6) 0, f(cid:5)t(cid:4) < t for t > 0,
suchthatforu≥0,
T(cid:5)u,v,u,v,0,u(cid:7)v(cid:4)≤0 (cid:5)2.1(cid:4)
FixedPointTheoryandApplications 3
or
T(cid:5)u,v,0,0,v,v(cid:4)≤0 (cid:5)2.2(cid:4)
impliesu≤f(cid:5)v(cid:4);
T :T(cid:5)u,0,u,0,0,u(cid:4)>0,forallu>0.
3
Example2.1. T(cid:5)t ,...,t (cid:4)(cid:6)t −αmax{t ,t ,t }−(cid:5)1−α(cid:4)(cid:2)at (cid:7)bt (cid:3),where0≤α<1,0≤a<1/2,
1 6 1 2 3 4 5 6
0≤b<1/2.
Let u > 0 and T(cid:5)u,v,u,v,0,u(cid:7)v(cid:4) (cid:6) u−αmax{u,v}−(cid:5)1−α(cid:4)b(cid:5)u(cid:7)v(cid:4) ≤ 0. If u ≥ v,
then (cid:5)1−b(cid:4)u ≤ bv which implies b ≥ 1/2, a contradiction. Thus u < v and u ≤ (cid:5)(cid:5)α(cid:7)(cid:5)1−
α(cid:4)b(cid:4)/(cid:5)1−(cid:5)1−α(cid:4)b(cid:4)(cid:4)v(cid:6)βv.Similarly,letu>0andT(cid:5)u,v,0,0,v,v(cid:4)(cid:6)u−αv−(cid:5)1−α(cid:4)(cid:5)a(cid:7)b(cid:4)v (cid:6)
u−(cid:5)α(cid:7)(cid:5)1−α(cid:4)(cid:5)a(cid:7)b(cid:4)(cid:4)v ≤0,thenu≤(cid:5)α(cid:7)(cid:5)1−α(cid:4)(cid:5)a(cid:7)b(cid:4)(cid:4)v (cid:6)γv.Ifu(cid:6)0,thenu≤γv.ThusT is
2
satisfiedwithf(cid:5)t(cid:4)(cid:6)max{β,γ}t.AlsoT(cid:5)u,0,u,0,0,u(cid:4)(cid:6)u−αu−(cid:5)1−α(cid:4)bu(cid:6)(cid:5)1−α(cid:4)(cid:5)1−b(cid:4)u>0,
forallu>0.Therefore,T ∈T.
Example2.2. T(cid:5)t ,...,t (cid:4)(cid:6)t −kmax{t ,t ,t ,(cid:5)1/2(cid:4)(cid:5)t (cid:7)t (cid:4)},wherek ∈(cid:5)0,1(cid:4).
1 6 1 2 3 4 5 6
Letu > 0andT(cid:5)u,v,u,v,0,u(cid:7)v(cid:4) (cid:6) u−kmax{u,v} ≤ 0.Ifu ≥ v,thenu ≤ ku,which
isacontradiction.Thusu<vandu≤kv.Similarly,letu>0andT(cid:5)u,v,0,0,v,v(cid:4)(cid:6)u−kv ≤
0, then we have u ≤ kv. If u (cid:6) 0, then u ≤ kv. Thus T is satisfied with f(cid:5)t(cid:4) (cid:6) kt. Also
2
T(cid:5)u,0,u,0,0,u(cid:4)(cid:6)u−ku>0,forallu>0.Therefore,T ∈T.
Example 2.3. T(cid:5)t1,...,t6(cid:4) (cid:6) t1 −φ(cid:5)max{t2,t3,t4,(cid:5)1/2(cid:4)(cid:5)t5 (cid:7)t6(cid:4)}(cid:4), where φ : R(cid:7) → R(cid:7) is right
continuousandφ(cid:5)0(cid:4)(cid:6)0,φ(cid:5)t(cid:4)<tfort>0.
Letu > 0andT(cid:5)u,v,u,v,0,u(cid:7)v(cid:4) (cid:6) u−φ(cid:5)max{u,v}(cid:4) ≤ 0.Ifu ≥ v,thenu−φ(cid:5)u(cid:4) ≤ 0,
whichisacontradiction.Thusu < vandu ≤ φ(cid:5)v(cid:4).Similarly,letu > 0andT(cid:5)u,v,0,0,v,v(cid:4) (cid:6)
u−φ(cid:5)v(cid:4) ≤ 0,thenwehaveu ≤ φ(cid:5)v(cid:4).Ifu (cid:6) 0,thenu ≤ φ(cid:5)v(cid:4).ThusT issatisfiedwithf (cid:6) φ.
2
AlsoT(cid:5)u,0,u,0,0,u(cid:4)(cid:6)u−φ(cid:5)u(cid:4)>0,forallu>0.Therefore,T ∈T.
Example2.4. T(cid:5)t ,...,t (cid:4) (cid:6) t2−t (cid:5)at (cid:7)bt (cid:7)ct (cid:4)−dt t ,wherea > 0,b,c,d ≥ 0,a(cid:7)b(cid:7)c < 1
1 6 1 1 2 3 4 5 6
anda(cid:7)d<1.
Letu > 0andT(cid:5)u,v,u,v,0,u(cid:7)v(cid:4) (cid:6) u2−u(cid:5)av(cid:7)bu(cid:7)cv(cid:4) ≤ 0.Thenu ≤ (cid:5)(cid:5)a(cid:7)c(cid:4)/(cid:5)1−
b(cid:4)(cid:4)v√(cid:6) h1v.Similarly,letu > 0andT(cid:5)u,v,0,0,v,v(cid:4) (cid:6) u2 −auv−dv2 ≤ 0,thenwehaveu ≤
(cid:5)(cid:5)a(cid:7) 4d(cid:7)a2(cid:4)/2(cid:4)v (cid:6)h v.Ifu(cid:6)0,thenu≤h v.ThusT issatisfiedwithf(cid:5)t(cid:4)(cid:6)max{h ,h }t.
2 2 2 1 2
AlsoT(cid:5)u,0,u,0,0,u(cid:4)(cid:6)(cid:5)1−b(cid:4)u2 >0,forallu>0.Therefore,T ∈T.
3. Fixed Point Theorem for Nondecreasing Mappings
Weneedthefollowinglemmafortheproofofourtheorems.
Lemma3.1(cid:5)see(cid:2)16(cid:3)(cid:4). Letf :R(cid:7) → R(cid:7)bearightcontinuousfunctionsuchthatf(cid:5)t(cid:4)<tforevery
t>0,thenlimn→∞fn(cid:5)t(cid:4)(cid:6)0,wherefndenotesthen-timesrepeatedcompositionoff withitself.
4 FixedPointTheoryandApplications
Theorem3.2. Let(cid:5)X,(cid:4)(cid:4)beapartiallyorderedsetandsupposethatthereisametricdonXsuchthat
(cid:5)X,d(cid:4)isacompletemetricspace.SupposeF : X → X isanondecreasingmappingsuchthatforall
x,y ∈Xwithy (cid:4)x,
(cid:2) (cid:2) (cid:3) (cid:2) (cid:3) (cid:2) (cid:3) (cid:2) (cid:3) (cid:2) (cid:3)(cid:3)
T d Fx,Fy ,d x,y ,d(cid:5)x,Fx(cid:4),d y,Fy ,d x,Fy ,d y,Fx ≤0, (cid:5)3.1(cid:4)
whereT ∈T.Also
F is continuous, (cid:5)3.2(cid:4)
or
if {x }⊂X is a nondecreasing sequence with x −→x in X,
n n
(cid:5)3.3(cid:4)
then x (cid:4)x ∀n
n
hold.Ifthereexistsanx ∈Xwithx (cid:4)F(cid:5)x (cid:4),thenF hasafixedpoint.
0 0 0
Proof. If Fx0 (cid:6) x0, then the proof is finished; so suppose x0/(cid:6)Fx0. Now let xn (cid:6) Fxn−1 for
n∈{1,2,...}.Noticethat,sincex (cid:4)Fx andF isnondecreasing,wehave
0 0
x0 (cid:4)x1 (cid:4)x2 (cid:4)···(cid:4)xn (cid:4)xn(cid:7)1 (cid:4)··· . (cid:5)3.4(cid:4)
Nowsincexn−1 (cid:4)xn,wecanusetheinequality(cid:5)3.1(cid:4)forthesepoints,thenwehave
T(cid:5)d(cid:5)Fxn,Fxn−1(cid:4),d(cid:5)xn,xn−1(cid:4),d(cid:5)xn,Fxn(cid:4),d(cid:5)xn−1,Fxn−1(cid:4),d(cid:5)xn,Fxn−1(cid:4),d(cid:5)xn−1,Fxn(cid:4)(cid:4)≤0
(cid:5)3.5(cid:4)
andso
T(cid:5)d(cid:5)xn(cid:7)1,xn(cid:4),d(cid:5)xn,xn−1(cid:4),d(cid:5)xn,xn(cid:7)1(cid:4),d(cid:5)xn−1,xn(cid:4),0,d(cid:5)xn−1,xn(cid:7)1(cid:4)(cid:4)≤0. (cid:5)3.6(cid:4)
NowusingT ,wehave
1
T(cid:5)d(cid:5)xn(cid:7)1,xn(cid:4),d(cid:5)xn,xn−1(cid:4),d(cid:5)xn,xn(cid:7)1(cid:4),d(cid:5)xn−1,xn(cid:4),0,d(cid:5)xn−1,xn(cid:4)(cid:7)d(cid:5)xn,xn(cid:7)1(cid:4)(cid:4)≤0, (cid:5)3.7(cid:4)
and from T2 there exists a right continuous function f : R(cid:7) → R(cid:7), f(cid:5)0(cid:4) (cid:6) 0, f(cid:5)t(cid:4) < t, for
t>0,suchthatforalln∈{1,2,...},
d(cid:5)xn(cid:7)1,xn(cid:4)≤f(cid:5)d(cid:5)xn,xn−1(cid:4)(cid:4). (cid:5)3.8(cid:4)
Ifwecontinuethisprocedure,wecanhave
d(cid:5)xn(cid:7)1,xn(cid:4)≤fn(cid:5)d(cid:5)x1,x0(cid:4)(cid:4), (cid:5)3.9(cid:4)
FixedPointTheoryandApplications 5
andsofromLemma3.1,
nl→im∞d(cid:5)xn(cid:7)1,xn(cid:4)(cid:6)0. (cid:5)3.10(cid:4)
Nextweshowthat{x }isaCauchysequence.Supposeitisnottrue.Thenwecanfindaδ >0
n
andtwosequenceofintegers{m(cid:5)k(cid:4)},{n(cid:5)k(cid:4)},m(cid:5)k(cid:4)>n(cid:5)k(cid:4)≥kwith
(cid:2) (cid:3)
rk (cid:6)d xn(cid:5)k(cid:4),xm(cid:5)k(cid:4) ≥δ for k ∈{1,2,...}. (cid:5)3.11(cid:4)
Wemayalsoassume
(cid:2) (cid:3)
d xm(cid:5)k(cid:4)−1,xn(cid:5)k(cid:4) <δ (cid:5)3.12(cid:4)
by choosing m(cid:5)k(cid:4) to be the smallest number exceeding n(cid:5)k(cid:4) for which (cid:5)3.11(cid:4) holds. Now
(cid:5)3.9(cid:4),(cid:5)3.11(cid:4),and(cid:5)3.12(cid:4)imply
(cid:2) (cid:3) (cid:2) (cid:3)
δ ≤rk ≤d xm(cid:5)k(cid:4),xm(cid:5)k(cid:4)−1 (cid:7)d xm(cid:5)k(cid:4)−1,xn(cid:5)k(cid:4) ≤fm(cid:5)k(cid:4)−1(cid:5)d(cid:5)x0,x1(cid:4)(cid:4)(cid:7)δ (cid:5)3.13(cid:4)
andso
limr (cid:6)δ. (cid:5)3.14(cid:4)
k
k→∞
Alsosince
(cid:2) (cid:3) (cid:2) (cid:3) (cid:2) (cid:3)
δ ≤rk ≤d xn(cid:5)k(cid:4),xn(cid:5)k(cid:4)(cid:7)1 (cid:7)d xm(cid:5)k(cid:4),xm(cid:5)k(cid:4)(cid:7)1 (cid:7)d xn(cid:5)k(cid:4)(cid:7)1,xm(cid:5)k(cid:4)(cid:7)1 , (cid:5)3.15(cid:4)
wehavefrom(cid:5)3.9(cid:4)that
(cid:2) (cid:3)
δ ≤rk ≤fn(cid:5)k(cid:4)(cid:5)d(cid:5)x0,x1(cid:4)(cid:4)(cid:7)fm(cid:5)k(cid:4)(cid:5)d(cid:5)x0,x1(cid:4)(cid:4)(cid:7)d xm(cid:5)k(cid:4)(cid:7)1,xn(cid:5)k(cid:4)(cid:7)1 . (cid:5)3.16(cid:4)
On the other hand, since xn(cid:5)k(cid:4) (cid:4) xm(cid:5)k(cid:4), we can use the condition (cid:5)3.1(cid:4) for these points.
Therefore,wehave
(cid:2) (cid:2) (cid:3) (cid:2) (cid:3) (cid:2) (cid:3)
T d Fxm(cid:5)k(cid:4),Fxn(cid:5)k(cid:4) ,d xm(cid:5)k(cid:4),xn(cid:5)k(cid:4) ,d xm(cid:5)k(cid:4),Fxm(cid:5)k(cid:4) ,
(cid:2) (cid:3) (cid:2) (cid:3) (cid:2) (cid:3)(cid:3) (cid:5)3.17(cid:4)
d xn(cid:5)k(cid:4),Fxn(cid:5)k(cid:4) ,d xm(cid:5)k(cid:4),Fxn(cid:5)k(cid:4) ,d xn(cid:5)k(cid:4),Fxm(cid:5)k(cid:4) ≤0
andso
(cid:10) (cid:2) (cid:3)
T d Fxm(cid:5)k(cid:4),Fxn(cid:5)k(cid:4) ,rk,fm(cid:5)k(cid:4)(cid:5)d(cid:5)x0,x1(cid:4)(cid:4),fn(cid:5)k(cid:4)(cid:5)d(cid:5)x0,x1(cid:4)(cid:4),
(cid:11) (cid:5)3.18(cid:4)
r (cid:7)fn(cid:5)k(cid:4)(cid:5)d(cid:5)x ,x (cid:4)(cid:4),r (cid:7)fm(cid:5)k(cid:4)(cid:5)d(cid:5)x ,x (cid:4)(cid:4) ≤0.
k 0 1 k 0 1
6 FixedPointTheoryandApplications
Nowlettingk → ∞andusing(cid:5)3.14(cid:4),wehave,bycontinuityofT,that
(cid:4) (cid:9)
(cid:2) (cid:3)
T limd xm(cid:5)k(cid:4)(cid:7)1,xn(cid:5)k(cid:4)(cid:7)1 ,δ,0,0,δ,δ ≤0. (cid:5)3.19(cid:4)
k→∞
From T2, we have limk→∞d(cid:5)xm(cid:5)k(cid:4)(cid:7)1,xn(cid:5)k(cid:4)(cid:7)1(cid:4) ≤ f(cid:5)δ(cid:4). Therefore, letting k → ∞ in (cid:5)3.16(cid:4), we
haveδ ≤f(cid:5)δ(cid:4).Thisisacontradictionsincef(cid:5)t(cid:4)<tfort>0.Thus{x }isaCauchysequence
n
inX,sothereexistsanx∈Xwithlimn→∞xn (cid:6)x.
If (cid:5)3.2(cid:4) holds, then clearly x (cid:6) Fx. Now suppose (cid:5)3.3(cid:4) holds. Suppose d(cid:5)x,Fx(cid:4) > 0.
Now since limn→∞xn (cid:6) x, then from (cid:5)3.3(cid:4), xn (cid:4) x for all n. Using the inequality (cid:5)3.1(cid:4), we
have
T(cid:5)d(cid:5)Fx,Fx (cid:4),d(cid:5)x,x (cid:4),d(cid:5)x,Fx(cid:4),d(cid:5)x ,Fx (cid:4),d(cid:5)x,Fx (cid:4),d(cid:5)x ,Fx(cid:4)(cid:4)≤0, (cid:5)3.20(cid:4)
n n n n n n
solettingn → ∞fromthelastinequality,wehave
T(cid:5)d(cid:5)Fx,x(cid:4),0,d(cid:5)x,Fx(cid:4),0,0,d(cid:5)x,Fx(cid:4)(cid:4)≤0, (cid:5)3.21(cid:4)
whichisacontradictiontoT .Thusd(cid:5)x,Fx(cid:4)(cid:6)0andsox(cid:6)Fx.
3
Remark3.3. Notethatifwetakethat
T4:thereexistsanondecreasingfunctionf : R(cid:7) → R(cid:7) withlimn→∞fn(cid:5)t(cid:4) (cid:6) 0foreach
t>0,suchthatforu≥0,
T(cid:5)u,v,u,v,0,u(cid:7)v(cid:4)≤0 (cid:5)3.22(cid:4)
or
T(cid:5)u,v,0,0,v,v(cid:4)≤0 (cid:5)3.23(cid:4)
impliesu≤f(cid:5)v(cid:4),
InsteadofT inTheorem3.2,againwecanhavethesameresult.
2
IfwecombineTheorem3.2withExample2.1,weobtainthefollowingresult.
Corollary 3.4. Let (cid:5)X,(cid:4)(cid:4) be a partially ordered set and suppose that there is a metric d on X such
that(cid:5)X,d(cid:4)isacompletemetricspace.SupposeF :X → Xisanondecreasingmappingsuchthatfor
allx,y ∈Xwithy (cid:4)x,
(cid:2) (cid:3) (cid:12) (cid:2) (cid:3) (cid:2) (cid:3)(cid:13) (cid:6) (cid:2) (cid:3) (cid:2) (cid:3)(cid:7)
d Fx,Fy ≤αmax d x,y ,d(cid:5)x,Fx(cid:4),d y,Fy (cid:7)(cid:5)1−α(cid:4) ad x,Fy (cid:7)bd y,Fx , (cid:5)3.24(cid:4)
where0≤α<1,0≤a<1/2,0≤b<1/2.Also
F is continuous (cid:5)3.25(cid:4)
FixedPointTheoryandApplications 7
or
if {x }⊂X is a nondecreasing sequence with x −→x in X,
n n
(cid:5)3.26(cid:4)
then x (cid:4)x ∀n
n
hold.Ifthereexistsanx ∈Xwithx (cid:4)F(cid:5)x (cid:4),thenF hasafixedpoint.
0 0 0
Remark3.5. Theorem2.2of(cid:2)2(cid:3)followsfromExample2.3,Remark3.3,andTheorem3.2.
Remark3.6. WecanhavesomenewresultsfromotherexamplesandTheorem3.2.
Remark3.7. InTheorem1(cid:2)1(cid:3),itisprovedthatif
every pair of elements has a lower bound and an upper bound, (cid:5)3.27(cid:4)
thenforeveryx∈X,
limFn(cid:5)x(cid:4)(cid:6)y, (cid:5)3.28(cid:4)
n→∞
whereyisthefixedpointofF suchthat
y (cid:6) limFn(cid:5)x (cid:4) (cid:5)3.29(cid:4)
n→∞ 0
andhenceF hasauniquefixedpoint.Ifcondition(cid:5)3.27(cid:4)fails,itispossibletofindexamples
of functions F with more than one fixed point. There exist some examples to illustrate this
factin(cid:2)3(cid:3).
4. Fixed Point Theorem for Weakly Increasing Mappings
Now we give a fixed point theorem for two weakly increasing mappings in ordered metric
spaces using an implicit relation. Before this, we will define an implicit relation for the
contractiveconditionofthetheorem.
LetT(cid:11)bethesetofallcontinuousfunctionsT :R6(cid:7) → RsatisfyingT1andthefollowing
conditions:
T(cid:11):there exists a right continuous function f : R(cid:7) → R(cid:7), f(cid:5)0(cid:4) (cid:6) 0,f(cid:5)t(cid:4) < t for t > 0,
2
suchthatforu≥0,
T(cid:5)u,v,u,v,0,u(cid:7)v(cid:4)≤0 (cid:5)4.1(cid:4)
or
T(cid:5)u,v,v,u,u(cid:7)v,0(cid:4)≤0 (cid:5)4.2(cid:4)
or
T(cid:5)u,v,0,0,v,v(cid:4)≤0 (cid:5)4.3(cid:4)
8 FixedPointTheoryandApplications
impliesu≤f(cid:5)v(cid:4);
T(cid:11):T(cid:5)u,0,u,0,0,u(cid:4)>0andT(cid:5)u,0,0,u,u,0(cid:4)>0,forallu>0.
3
Wecaneasilyshowthat,allfunctionsintheExamplesinSection2areinT(cid:11).
Definition4.1(cid:5)see(cid:2)17,18(cid:3)(cid:4). Let(cid:5)X,(cid:4)(cid:4)beapartiallyorderedset.TwomappingsF,G:X → X
aresaidtobeweaklyincreasingifFx(cid:4)GFxandGx(cid:4)FGxforallx∈X.
Notethat,twoweaklyincreasingmappingsneednotbenondecreasing.
Example4.2. LetX (cid:6)R(cid:7)endowedwithusualordering.LetF,G:X → Xdefinedby
⎧ ⎧
√
⎨x if 0≤x≤1 ⎨ x if 0≤x≤1
Fx(cid:6) Gx(cid:6) (cid:5)4.4(cid:4)
⎩ ⎩
0 if 1<x<∞, 0 if 1<x<∞,
then it is obvious that Fx ≤ GFx and Gx ≤ FGx for all x ∈ X. Thus F and G are weakly
increasingmappings.NotethatbothF andGarenotnondecreasing.
Example4.3. LetX (cid:6)(cid:2)1,∞(cid:4)×(cid:2)1,∞(cid:4)beendowedwiththecoordinateordering,thatis,(cid:5)x,y(cid:4)(cid:4)
(cid:5)z,w(cid:4) ⇔ x ≤ zandy ≤ w.LetF,G : X → X bedefinedbyF(cid:5)x,y(cid:4) (cid:6) (cid:5)2x,3y(cid:4)andG(cid:5)x,y(cid:4) (cid:6)
(cid:5)x2,y2(cid:4),thenF(cid:5)x,y(cid:4) (cid:6) (cid:5)2x,3y(cid:4) (cid:4) GF(cid:5)x,y(cid:4) (cid:6) G(cid:5)2x,3y(cid:4) (cid:6) (cid:5)4x2,9y2(cid:4)andG(cid:5)x,y(cid:4) (cid:6) (cid:5)x2,y2(cid:4) (cid:4)
FG(cid:5)x,y(cid:4)(cid:6)F(cid:5)x2,y2(cid:4)(cid:6)(cid:5)2x2,3y2(cid:4).ThusF andGareweaklyincreasingmappings.
Example 4.4. Let X (cid:6) R2 be endowed with the lexicographical ordering, that is, (cid:5)x,y(cid:4) (cid:4)
(cid:5)z,w(cid:4)⇔(cid:5)x<z(cid:4)or(cid:5)ifx(cid:6)z,theny ≤w(cid:4).LetF,G:X → Xbedefinedby
(cid:2) (cid:3) (cid:2) (cid:12) (cid:13) (cid:12) (cid:13)(cid:3)
F x,y (cid:6) max x,y ,min x,y ,
(cid:2) (cid:3) (cid:4) (cid:12) (cid:13) x(cid:7)y(cid:9) (cid:5)4.5(cid:4)
G x,y (cid:6) max x,y , ,
2
then
(cid:2) (cid:3) (cid:2) (cid:12) (cid:13) (cid:12) (cid:13)(cid:3)
F x,y (cid:6) max x,y ,min x,y
(cid:2) (cid:3)
(cid:4)GF x,y
(cid:2) (cid:12) (cid:13) (cid:12) (cid:13)(cid:3)
(cid:6)G max x,y ,min x,y
(cid:17) (cid:12) (cid:13) (cid:12) (cid:13)(cid:18)
(cid:12) (cid:12) (cid:13) (cid:12) (cid:13)(cid:13) max x,y (cid:7)min x,y
(cid:6) max max x,y ,min x,y ,
2
(cid:4) (cid:9)
(cid:12) (cid:13) x(cid:7)y
(cid:6) max x,y , ,
2
FixedPointTheoryandApplications 9
(cid:4) (cid:9)
(cid:2) (cid:3) (cid:12) (cid:13) x(cid:7)y
G x,y (cid:6) max x,y ,
2
(cid:2) (cid:3)
(cid:4)FG x,y
(cid:4) (cid:9)
(cid:12) (cid:13) x(cid:7)y
(cid:6)F max x,y ,
2
(cid:4) (cid:5) (cid:8) (cid:5) (cid:8)(cid:9)
(cid:12) (cid:13) x(cid:7)y (cid:12) (cid:13) x(cid:7)y
(cid:6) max max x,y , ,min max x,y ,
2 2
(cid:4) (cid:9)
(cid:12) (cid:13) x(cid:7)y
(cid:6) max x,y , .
2
(cid:5)4.6(cid:4)
Thus F and G are weakly increasing mappings. Note that (cid:5)1,4(cid:4) (cid:4) (cid:5)2,3(cid:4) but F(cid:5)1,4(cid:4) (cid:6)
(cid:5)4,1(cid:4)(cid:5)3,2(cid:4)(cid:6)F(cid:5)2,3(cid:4),thenF isnotnondecreasing.Similarly,Gisnotnondecreasing.
Theorem4.5. Let(cid:5)X,(cid:4)(cid:4)beapartiallyorderedsetandsupposethatthereisametricdonXsuchthat
(cid:5)X,d(cid:4)isacompletemetricspace.SupposeF,G : X → X aretwoweaklyincreasingmappingssuch
thatforallcomparablex,y ∈X,
(cid:2) (cid:2) (cid:3) (cid:2) (cid:3) (cid:2) (cid:3) (cid:2) (cid:3) (cid:2) (cid:3)(cid:3)
T d Fx,Gy ,d x,y ,d(cid:5)x,Fx(cid:4),d y,Gy ,d x,Gy ,d y,Fx ≤0, (cid:5)4.7(cid:4)
whereT ∈T(cid:11).Also
F is continuous (cid:5)4.8(cid:4)
or
G is continuous (cid:5)4.9(cid:4)
or
if {x }⊂X is a nondecreasing sequence with x −→x in X,
n n
(cid:5)4.10(cid:4)
then x (cid:4)x ∀n
n
hold,thenF andGhaveacommonfixedpoint.
Remark4.6. Notethat,inthistheoremweremovethecondition“thereexistsanx ∈ X with
0
x (cid:4)F(cid:5)x (cid:4)”ofTheorem3.2.AgainwecanconsidertheresultofRemark3.7forthistheorem.
0 0
ProofofTheorem4.5. FirstofallweshowthatifF orGhasafixedpoint,thenitisacommon
fixed point of F and G. Indeed, let z be a fixed point of F. Now assume d(cid:5)z,Gz(cid:4) > 0. If we
usetheinequality(cid:5)4.7(cid:4),forx(cid:6)y (cid:6)z,wehave
T(cid:5)d(cid:5)z,Gz(cid:4),0,0,d(cid:5)z,Gz(cid:4),d(cid:5)z,Gz(cid:4),0(cid:4)≤0, (cid:5)4.11(cid:4)
10 FixedPointTheoryandApplications
whichisacontradictiontoT(cid:11).Thusd(cid:5)z,Gz(cid:4)(cid:6)0andsozisacommonfixedpointofFandG.
3
Similarly,ifzisafixedpointofG,thenitisalsoafixedpointofF.Nowletx beanarbitrary
0
point of X. If x0 (cid:6) Fx0, the proof is finished, so assume x0/(cid:6)Fx0. We can define a sequence
{x }inXasfollows:
n
x2n(cid:7)1 (cid:6)Fx2n, x2n(cid:7)2 (cid:6)Gx2n(cid:7)1 for n∈{0,1,...}. (cid:5)4.12(cid:4)
Without loss of generality, we can suppose that the successive terms of {x } are different.
n
Otherwise,weareagainfinished.NotethatsinceF andGareweaklyincreasing,wehave
x (cid:6)Fx (cid:4)GFx (cid:6)Gx (cid:6)x ,
1 0 0 1 2
(cid:5)4.13(cid:4)
x (cid:6)Gx (cid:4)FGx (cid:6)Fx (cid:6)x
2 1 1 2 3
andcontinuingthisprocess,wehave
x1 (cid:4)x2 (cid:4)···(cid:4)xn (cid:4)xn(cid:7)1 (cid:4)··· . (cid:5)4.14(cid:4)
Nowsincex2n−1andx2narecomparablethen,wecanusetheinequality(cid:5)4.7(cid:4)forthesepoints
thenwehave
T(cid:5)d(cid:5)Fx2n,Gx2n−1(cid:4),d(cid:5)x2n,x2n−1(cid:4),d(cid:5)x2n,Fx2n(cid:4),d(cid:5)x2n−1,Gx2n−1(cid:4),
(cid:5)4.15(cid:4)
d(cid:5)x2n,Gx2n−1(cid:4),d(cid:5)x2n−1,Fx2n(cid:4)(cid:4)≤0
andso
T(cid:5)d(cid:5)x2n(cid:7)1,x2n(cid:4),d(cid:5)x2n,x2n−1(cid:4),d(cid:5)x2n,x2n(cid:7)1(cid:4),d(cid:5)x2n−1,x2n(cid:4),0,d(cid:5)x2n−1,x2n(cid:7)1(cid:4)(cid:4)≤0. (cid:5)4.16(cid:4)
NowusingT ,wehave
1
T(cid:5)d(cid:5)x2n(cid:7)1,x2n(cid:4),d(cid:5)x2n,x2n−1(cid:4),d(cid:5)x2n,x2n(cid:7)1(cid:4),d(cid:5)x2n−1,x2n(cid:4),0,d(cid:5)x2n−1,x2n(cid:4)(cid:7)d(cid:5)x2n,x2n(cid:7)1(cid:4)(cid:4)≤0,
(cid:5)4.17(cid:4)
andformT(cid:11) thereexistsarightcontinuousfunctionf :R(cid:7) → R(cid:7),f(cid:5)0(cid:4)(cid:6)0,f(cid:5)t(cid:4)<t,fort>0,
2
wehaveforalln∈{1,2,...}
d(cid:5)x2n(cid:7)1,x2n(cid:4)≤f(cid:5)d(cid:5)x2n,x2n−1(cid:4)(cid:4). (cid:5)4.18(cid:4)
Similarly, since x2n and x2n(cid:7)1 are comparable, then we can use the inequality (cid:5)4.7(cid:4) for these
pointsthenwehave
T(cid:5)d(cid:5)Fx2n,Gx2n(cid:7)1(cid:4),d(cid:5)x2n,x2n(cid:7)1(cid:4),d(cid:5)x2n,Fx2n(cid:4),d(cid:5)x2n(cid:7)1,Gx2n(cid:7)1(cid:4),
(cid:5)4.19(cid:4)
d(cid:5)x2n,Gx2n(cid:7)1(cid:4),d(cid:5)x2n(cid:7)1,Fx2n(cid:4)(cid:4)≤0