VOL. 16 NO. 4 (1993) 669-674
A COMMON FIXED POINT THEOREM OF MEIR AND KEELER TYPE
Y.J. CHO P.P.MURTHY G.JUNGCK
DepartmentofMathematics GyeongsangNational University
Jinju660-701, KOREA
Balajee GuestHouse Street-37, Sector-5 Bhilai(M.P.)490006,INDIA
DepartmentofMathematics BradleyUniversity Peoria, Illinois61625,U.S.A.
(Received
June 10, 1992andin revisedform September 22,1992))
ABSTRACT. In this paper, weintroduce theconcept of compatiblemappings oftype (A)on a metricspace, which isequivalent to theconcept ofcompatiblemappings undersomeconditions, and giveacommonfixed point theoremofMeirandKeeler type.
Our
resultextends,
generalized and improvessomeresults ofMeir-Keeler, Park-Bae,Park-Rhoades,
PantandRao-Rao,
etc.1991AMS SUBJECT
CLASSIFICATION
CODE. 54H25.KEY
WORDS AND PHRASES. Common fixed points, compatible mappings of type (A), generalized(e,6) {S,T}-contractions, and {S, T}-iterations.1.
INTRODUCTION.
In [6],
Jungck proved a common fixed point theorem of commuting mappings on ametric space. Sincethen, he andmanyauthors extended, generalizedand unifiedthistheoremin many ways([2], [4], [5], [7]-[10], [15]-[20], [22], [23]).
For example, Sessa([22])introduced
the concept of weakly commuting mappings,which isageneralization of the conceptofcommuting mappings, and heand others provedsomefixedpoint theorems for weakly commuting mappings([20]-[23]).
Recently, Jungck
([8])
proposed a generalization of the concept of weakly commuting mappings, whichis calledcompatible mappings, and hegeneralizedsomefixed point theorems of Meir-Keeler type, especially, atheorem of Park-Bae([16]),
andin[11], Jungck,
Murthy and Cho introduced the concept of compatible mappings oftype (A) on metric spacesandobtained some fixedpoint theorems for these mappings.On the other hand, in
[14],
Meir and Keeler established a fixed point theorem for a self- mappingfofametric space(X,d) satisfying thefollowingcondition:Forevery >0, thereexistsa6>0suchthat
< d(z,V) < +6 implies d(fz, fv) <
. (1.1) In [13],
Maiti and Pal also proved a fixed point theorem for a self-mappingI
of a metricspae(X,d)satisfyingthefollowingcondition,whichisageneralizationof
(1.1):
For
everye>0, thereexistsa 6>0such thate<_rnaz{d(z,v),d(z,fz),d(y,fv)}<e+6 implies d(fz, fv)<
. (1.2)
In [17]
and[18],
Park-Rhoades andRaRao
proved some fixed point theorems for self- mappings fand gofametric space (X,d)satisfying the following condition, respectively,which is ageneralization of(1.2):
Foreverye>0, thereexistsa6>0suchthat
<_maz{d(fz,fy),d(fz, gz),d(fy,
gy),d(fz,
gy)+d(fy,gx))}<+
6 implies d(gz,gy)<e(1.3)
Manyother fixed point theoremsof Meir-Keeler typearegivenin
[1], [3], IS], [12], [15], [16], [19]
and[21].
In
this paper, we introduce the concept of compatible mappings of type (A), which is equivalent to the concept ofcompatible mappings under some conditions, and give a common fixed point theorem for compatible mappings of type (A). which extends, generalizes and improvessomecommonfixedpoint theorems ofMeir-Keelertype.2.
COMPATIBLE
MAPPINGS OFTYPE (A).
In
this section, weshow that two pairs of compatiblemappingsandcompatible mappings of type (A) axeequivalent undersomeconditionsand give several properties of compatible mappings of type (A) forourmainresults. Throughoutthispaper,(X,d)denotesametric spce.DEFINITION
2.1.Let
S,T:(X,d)-.(X,d) be mappings. $and Taxesaid tobecompatibleifimood(ST(n),TS(zn)
0whenever{z,} isasequence in Xsuch that
ImooS(z,) T(z,,)
forsome in X.DEFINITION 2.2. LetS,T:(X,d)---,(X,d) bemappings. Sand Taxesaid to be compatible of type(A) if
lnimood(TS(zn),SS(zn)
0andldrnood(ST(zn),
rT(zn) 0whenever {z,}isasequencein Xsuch that
id._mooS(z,, IooT(z,,
forsome inX.’In [11],
the following propositions show that Definitions 2.1 and 2.2 are equivalent under someconditions:PROPOSITION
2.1. Let S,T X,d)-.(X,d) be continuous mappings. If S and T axe compatible, then theyaxecompatibleof type (A).PROPOSITION2.2. Let$,T (X,d)-,(X,d) becompatiblemappings oftype(A). IfoneofS andT iscontinuous, then$andTaxecompatible.
Thefollowingisadirectconsequence of Propositions2.1 and 2.2:
PROPOSITION 2.3. Let $,T: X,d)-,(X,d) be continuous mappings. Then S and T axe
compatibleifandonlyiftheyaxecompatibleoftype(A)
Thefollowing examples showthat Proposition 2.3 isnot true ifSandT axediscontinuousin somepoint ofX.
EXAMPLE
2.1. Let X R, theset of real numbers, with the usual metric d(z,y)=Define S,T:(X, d)-(X, d)asfollows:
[r
ifr#0,$(z)
ifz=0
and Then S and T axe not continuous at =0.
n2,n
1,2,--.. Thenwehave,
as n-oo,and
but and
T(z)={z--2
ifift=0.z’#0,
Consider a sequence
{r,,}
inx
defined byso:.)
0,T(tn 1 O,
n4
Id.rnd(ST(xn),
TS(xn)i.rnood(n4,
n4)
0md(ST(zn)’TT(zn)) mood(ns’n) moo ns- n41
ooIrnd(SS(zn),TS(zn) =/n/mcd(n:, n4) -/n/moo In n4l
co.Therefore,SandTarecompatible butarenot compatibleof type(A).
EXAMPLE
2.2. LetX=[0,1]with the usualmetric d(x,u)= Ix-yl- Define$,T:[0,1 ]--.[0,1 by{
if[0,1/2),
and T(z){:-r if[0,!)
S(z) ifz
[!
2’1] ifz[1/2,1].
Then Sand Taxe not continuous at t=
1/2. Now,
weassert that S and Taxe not compatible butaxe compatible of type (A).
To
see this, suppose that {r,}_c
[0,1] and that T(z,,),S(r,,)--.t.By
definition ofSand T,
{-,2
1}. Since $and Tagreeon[1/2,1],
we needonly consider1/2. So
wecan suppose that
r,,--
and that r, <1/2
for all n. Then T(r,)=l-z,,--
from the right ands(r,)
r,--
fromthe left. Thus,since r, >,
for allnand,sincez,<
1/2,
Consequently, but
and
ST(xn) S(1 xn) TS(zn) T(zn)
r,t--"
d(ST(zn),TS(zn)
)-’
d(ST(z,,),TT(r.n)=
IST(r.)-TT(:,.,)I I1-T(1-r.)lI1-
I--,0 d(TS(rn),SS(zn) ITS(z,,)- SS(:,,) I(1 :,,)-r,, 2:,,I-’0
as
r,,-.
Therefore,SandTarecompatible mappings of type (A) butarenotcompatible.Next,
wegive several properties of compatible of type(A)forourmain theorems([11]):
PROPOSITION 2.4. Let S, T: X,d)(X,d) be mappings. IfSand Tare compatibleof type (A) and S(t) T(t) forsome X, then ST(t) TT(t) TS(t) SS(t).
PROPOSITION 2.5. Let S,T: X,d)--,(X, d)be mappings. Let $andTbe compatible oftype (A) and let S(r,,), T(r,,)-,t forsome eX. Thenwehave thefollowing:
(1) /,,imooTS(r,,
S(t)ifSis continuous att.(2)
ST(t)=TS(t) andS(t)= T(t) ifSandTaxecontinuous att.PROOF. Immediate,from Proposition2.2and Proposition2.2
(2)
of[81
3.
A COMMON FIXED
POINT THEOREM.Before stating and provingourmaintheorem,wegivesomedefinitionsandlemmas:
DEFINITION
3.1([8]). Let
A,B,S and T be mappings of a metric space (X,d) intoitself such that A(X) T(X) andB(X)C$(X). Foro
X, anysequence{v,,}definedbyY2n
Tz2n Az2n
2’1 (3.1)
Y2n
SZ2n BZ2n-
for n 1,2,..-, iscalledan{S,T}-iterationofz underAandB.
Note
that Definition 3.1 assures us that {S,T}-iterations will exist since A(X)CT(X) and B(X)CS(X),althoughthe sequence{Yn}
certainlyneed not be unique.DEFINITION 3.2. Let A,B,S and T be mappings ofa metric space (X,d)intoitself. The pair {A,B}iscalledageneralized(,$)-{S,T}-contractionif
A(X)CT(X) and B(X)CS(X),
(3.2)
thereexistsafunction&(0,oo)-(0,oo) suchthat,foranye>0and (e) <
,
<_ maz{d(Sr,Ty),d(Sz, Az),d(Ty,
By),(d(Sr,
By)+d(Ty,Az))} < ()(3.3)
impliesd(Az,By)< for allz,y X.
Forourmaintheorem,firstwegive the following:
LEMMA
3.1. LetSandT bemappings ofametric space (X,d) intoitself and the pair {A,B}beageneralized (,6)-{S,T}-contraction. If Zo Xand {y.} is an{$,T}-iteration of Zo under Aand B, thenwehave thefollowing:
(1)
for every >0, <d(y,,yq)
< 6(e) impliesd(y,+
l,Yq+i)<,
where,
andqareofopposite parity.(2) _m(R)d(n,y.
+) O.(3) {.}
isaCauchysequencein X.PROOF.
(1)
Since the pair {A,B}isageneralized(,)-{S,T}-contraction,forevery >0,<_maz{d(Sz,Ty),d(Sz, Az),d(Ty,
By),(d(Sz,
By)+d(Ty,Az))}< 6(f) implies d(Az,By)< for allz,y.
X.Suppose
that <d(yp, y)
<6(e). Putting p 2n and q 2m- in the above inequality, we haveand
d(yp+
1,Yq+1) d(Y2.+1,Y2m) d(A::tn,Bz2m-
1)<_
d(yp, yq)
d(Yan,Y2m_l) d(Sz2n’Tzam- 1)<- maz(d{Sz2a, Tz2m-
1)’d(Sz2n’Az2n)’d(Tz2m 1’Bz2m l)’(d(Sz2n’Bz2m-
1)+
d(Tz:m-1’Az2-))},
<6(), whichimpliesthat
d(y,+
1,Yq+l) d(Az2n, Bz:tm-l)<"
(2)
ForZo X, by(3.3),
wehave d(u.,.
+ d(Az.,
Bz2,,<maz{d(Sz2n,
Tza.-
)’d(Sz2n’Az2n)’d(Tz2.-l’Bz2. l)’d(Sz2.’Bzan-
1)+
d(Tzan-, Aza.))}
{d(n,
yah l),d(y2n,yn+ 1), d(y.1,yn),d(y2n,
y2.)+
d(y2._I’Y2.+1))}
d(2. 1,Yah)"
Sillily,wehaved(u.+l,Ua. +) < d(ua.,
a.
+).Thus the
uen {d(y.,y.+
1)} isnon-increing dnvergto theeatt
lower d =0of its rge rE0.
In
ft, other,(1)
impH that d(y+,+)<t whenev Sd(y,u+)<6(t). But since {d(y,y+)} converg to t, there ests a t such that d(t, ut+)<
6(t) d d(yt+l,Yk+2)
< t, whichcontracts
thedesiation
of t. Therefore, wehave
d(y.,
y. +)
0.Theprf of
(3)
followsom
the linmofthe prfofmma
3.1(c) ([8]). Ts
complet theprof.Nowwe ereadyto prove o
mn
threm:THEOM
3.2. t A,B,S d T mappings ofacomplete metc spe (X,d)intoit satisngtheconditions(3.2),
(3.4)
oneofA,B,S, dT iscontinuous,(3.5)
theprs
A,Sd B,Tecompatible of ty (A)onX,(3.6)
the pair {A,B} is a generalized (e,6)-{S,T}-contraction such that $ is lower semi- continuous.ThenA,B,SandThaveauniquecommonfixedpoint in X.
PROOF. By
Lemma 3.1(3),
the {S,T}-iteration ofx under A and B, {y,}, is a Cauchy sequenceinx.
Since (X,d) iscomplete, {y,} converges toapoint in X. Since{Sr,,} and {Tr2,_t} aresubsequencesof{y,}, they also convergetoz.
Suppose that S is continuous. Then wehave SS:2,,SAz,Sz asnoo. Since A and Sare
compatible of type (A), by Proposition 2.5
(1), ASz,-,Sz
as n-.oo.Now,
we claim that Sz z.Suppose Sz:/:z and let M(x,V) ma{d(Sz,Tv),d(Sz,Ar),d(Tv,
Bv),(d(Sx,
Bv)+d(Tv, A:))}. If M,=M(Sz,,:2,_,), then we have M,,--..d(Sz,z)#0 asn---,oo. Let e d(Sz,z) and remember that () > bydefinition. Since:(0,o)-.(0,o0)is lower semi-continuous, thereexists an a (0,)such that (t)>e for t(-a,+a). Choose to(-a,). Then wehave O<to<<,(to). But since M,--,as n--,o, thereexists an integer n such that M, e. (to,,(to))for n>_no.
Therefore, by(3.3),
wehave
d(ASz2,,Bz2,,_)< <efor n>
no.
But d(ASz2n,Bz2n_)d(Sz,z)asn-oandsowehave d(Sz, z) < <
,
which isacontradiction.Tus
wehaveSz z.We
also claim Az z.Suppose
not and let d(Az, z)="
andM,[=
M(z,z2,_). ThenwehaveMn’’as
n---,o. Now duplicatetheargument usingthe lower semi-continuity of6toproducethe contradiction d(Az,z) <’.
Thus we obtain Sz Az z. Since A(X)CT(X), there exists a pointw Xsuch thatz Sz Az Tw. Further,weclaimthatBw z. IfBw z, thenwehave d(z,Bw) d(Az, Bw)
<maz{d(Sz, Tw),d(Sz, Az),d(Tw,
Bw),1/2(d(Sz,
Bw)+
d(Tw,Az))}d(z,Bw),
which is a contradiction. Hence Bw z Tw. Since B and T are compatible of type (A), by Proposition2.4, Bz BTw TTw Tz, thatis,Bz Tz.
Finally, weshallprovethatBz z. IfBz z, thenwehave d(z, Bz) d(Az, Bz)
< maz{d(Sz, Tz),d(Sz, Az),d(Tz,
Bz),d(Sz,
Bz) +d(Tz,Az))}=d(z,Bz),
which is a contradiction and so Bz z. Thus, zis acommon fixed point of A,B,S and T. The uniqueness of thecommonfixed pointzfollows easily from
(3.6).
Similarly, we can also completethe proofwhen A or Bor T is continuous. This completes theproof.
REMARK.
Theorem 3.2 extends, generalizes and improves some results of Chung[3],
Jungck[8],
Maiti-Pal[13],
Meir-Keeler[14],
Pant[15],
Park-Bae[16],
Park-Rhoades[17],
aao-Rao[lS],
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