FIXED POINTS AND CONTINUITY FOR MULTIVALUED MAPPINGS
TROY HICKSandB.E. RHOADES
Department
of Mathematics University ofMissouriatRollaRolla,
Missouri 65401Department
ofMathematics Indiana UniversityBloomington,Indiana47405
(Received July 31, 1990 and in revised form January II, 1991)
ABSTRACT.
Many
ofthecontractive definitionsdonot require continuity of themap.How-ever,inapreviouspaper the second author has shownin
[1]
that,
inmost cases, thefunction is continuousat afixed point.In
this paperweshow that thesame behavior isexhibited for manymultivalued mappings.KEY
WORDS ANDPHRASES.
Continuity, fixed point, multivaluedmapping.1980AMS
SUB:IECT CLASSIFICATION
CODE.
54H
25In
arecent paper, Rhoades[1],
it wasverifiedthat,
for most contractivedefinitions, the contractivedefinition wasstrongenough toforce continuity ofthe function at afixed point, even thoughcontinuitywasneitherassumednorimpliedbythecontractive definition.The purpose ofthispaperis to prove that the situation isthe same in the multivalued arena.
Before examiningthemultivaluedcases,weshall establisharesult,whichisofinterestin itsownright.
PROPOSITION
1. LetT
bea selfmapofacompletemetric space with afixedpointp. Then thefollowingareequivalent:(a)
T
is continuousatp.(b)
If{y,,}
is anysequencecontained inX
with y,,-
p,thenlimd(y.,Ty.)
O.PROOF. To
show that(a)
and(b)
are equivalent, suppose that(a)
is satisfied. Then Yr,--
Pimplies thatTyn
Tp
p,andthus limd(yn,Tyn)
O.Conversely,
d(Tyn,Tp)
<
d(Ty,,y,)
+
d(yn,p),
sothatlimsupd(Tv.,Tp)
<_
limsup[d(Tv.,V,)
+
d(y,,Tp)]
0. ThenT
is continuousatp.In
themultivaluedcase,amultivaluedmapT
hasafixedpointp ifp ETp.
LetCL(X)
denote thecollectionofnonempty closed subsets of
X,D(A,B):= inf{d(a,b):a e
A,b
e
B}.
A
multivalued mapT
(X,d)
-
(CL(X),D)
will be said to be continuous at apoint p if lira,d(z,,p)
0 implies that lira,D(Tx,,Tp)
O.PROPOSITION
2.Let T
(X,d)
(CL(X),D),
withp afixed point ofT. Then the(a)
T
is continuous at p,(b)
If{y,,}
CX
withlimy,, p, thenlimD(y.,Ty.) O.Suppose
that(a)
is satisfied,D(y.,Ty,,)
<
D(y.,p)+
D(p, Ty.)
0, and(b)
issatisfied.Conversely,if
(b)
holds,thenD(Ty.,Tp)
<
D(Ty.,y,,)
+
D(y,,,p)
-
0, and(a)is
satisfied.In
Proposition2,weobtainthesameconclusionsbyreplacingCL(X)
withB(X),
the setof allnonemptybounded subsets ofX,
andreplacingD
by,(A,B),
where,(A,B)
:=sup{d(a,b):
a E A,bB}.
Wecanalso replaceCL(X)
withCB(X),
thecollection ofnonempty closed,boundedsubsets of
X,
andreplaceD
withH,
theHausdorffmetric.THEOREM1. Let
F X
B(X),
I X
X,
I
continuous,F,I
satisfying$(Fx,Fy)
<_ cmax{6(Ix,Iy),(Ix,Fx),$(Iy,Fy),6(Ix,Fy),6(Iy,Fx)}
(1)
for all x,yin
X,
0_<
c<
1. IfF
andI
commute andI(X)
D_F(X),
thenF
andI
have a uniquecommonfixed point z,Fz
{z},
andF
iscontinuous atz.The fact that
F
andI
haveauniquecommonfixed pointcomesfromFisher[2 ].
Althoughnot mentioned inthestatement of Theorem 1,Fisher
[2]
has shown thatIz
{z}.
Toshow thatF
is continuousat z, let{y,,}
CX,
y.-
z.From
(1),
$(Fy,,,z)
<_
cmax{$(Iy,,,Iz),
$(Iy,,,Fy,,),$(Iz,Fz), (Iy.,Fz),
(Iz,
Fy.)}.
Since y,, zand
I
is continuous,Iy,,
Iz
z.Also, g(Iy.,fy,,)
<_
(Iy,,,z)
+
(z,
Fy.).
Therefore
$(Fy,,,z)
<_
c[(Iy.,z)
+
g(z,Fy.)],
whichimpliesthat(Fy,,,z)
<
c[$(Iy,,,z)]](1
-c)
--,0asn-
oo, andF
is continuousat z.By
settingI
equal to the identity map, Theorem 1 is ageneralization of the result in Fisher[3].
Foraninteger n,x
.
X,F"
isdefined inductivelybyF"(x)
F(F"-I(x)).
THEOREM2. Let
F:X
B(X)
satisfying6(FPx,Fy) <
cmax{6(Frx,F’y),(Frx,Fr’z),6(y,Fy)
0<_
r,r’
<_
p,s=0,1}
(2)
for all x,yin
X,
0<
c<
1 forsomefixed integerp. IfF
alsomapsB(X)
intoitself, thenF
hasaunique fixed pointz,
Fz
{z},
andF
is continuousatz.PROOF.
The fact thatF
hasaunique fixed pointzand thatFz
{z}
comesfromFisher[41.
Let {y,}
CX,
yn z.In
(2)
setx z,y y, toget(FPz,Fy.)
<_
cmax{$(Frz,FSy,,),$(Frz,Fr’z),$(y,,,Fy,,)
0<_
r,r’
<_
p,s 0,I}
cmax{g(z,y.), (z,Fy.), (y.,Fy,,)}.
Since
6(y.,Fy.) <_ 6(z,y.)
+
6(z,Fy.),
the aboveinequalitybecomes6(z,Fy.)
<_ c[6(z,y.)
+
6(z,Fy.)],
which implies that6(z,Fy.)
<
c[6(z,y.)]/(1- c)
0 as n cx. ThusF
is continuousat z.Theorem2 alsogeneralizesthecorrespondingresult inFisher
[4].
For
ametricspaceX,N(e,A)
:={x
6X"
d(x,a) <
efor some a 6A
6CL(X),e >
0}.
TheHansdorffmetricH
isdefinedbyH(A,B)
inf{e >
O"A
C_N(e,B)
andB
C_N(e,A)},
ifthe infimumexists, and
H(A,B)
cotherwise.An
equivalent definitionofH
isTHEOREM3. Let F1,
F2
X
CB(X)
satisfyingH(Flx,F2y)
<_ al6(x,Fx)
+
a26(y, F2y)
+
a:6(x,F2y)
+
a46(y,Fx)
+
ar,d(x,y)
(3)
5
for all x,y in
X,
wherethe al>
0,E,=I
ai<
1 anda
a2 or a3 a4. Ifz is acommon fixedpoint ofF1
andF2
such thatF1
z{ z}
orF2
z{ z},
thenz isthe uniquecommon fixed point ofF1
andF2.
Moreover,
F
andF2
arecontinuousatz.PROOF.
Theconclusionsabout the fixed point zcomefromMendaglioandDube[5].
Wefirst observethat the hypotheses of the theorem implies that both
F1
zF2z
{z}.
For,
supposethatF2z
{z}.
Then,in(3)
withx y z,wehaveH(Fz,F2z) <_ al(Z,flz)
4-a2(z,f2z)
4-a3(z,f2z
4-a4(z,Flz
4-and
H(Flz,{z}) < (al
+
a,)6(z,Fz).
But
H(Fz,{z})
6(z,Fz).
ThereforeFz
{z}.
Similarly,
F1
z{z }
impliesF
z{ z}.
Toprovecontinuity, let
{y.}
CX,y.
---}z.In
(3)
setx y.,y ztogetH(Fy.,F2z)
<_
a(y.,Fly.)
+
a2(z,F2z)
+
a:$(y.,F2z)
+
a4(z,Fy.)
+
ahd(y.,z).
Thus
H(fltn,Z)
(flln,Z)
<
(hi
4-a24-a3)
d(In,Z),
1
a
-a4 whichtends to 0as n-}oo, andF1
is continuousatz.Settingx zand y y. in
(3)
leads tothe resultthatF2
is continuous atz.THEOREM
4. LetF,
G
X
B(X)
(Fx,Gy)
<_
a,(p)(x,Fx)
+
a(p)(y,Gy)
+
a3(p)(x,Gy)
+
a4(p)(y,Fx)
+
ah(p)d(x,y)
(4)
for allx,yinX, where p
:=8(Fx,Gy)
>
0,and where eachai:(0, oo)
--,[0,1)
isadecreasing functionsuch that(a
+a
+2a3
+
2a4
+a)(t) <
I
foreach>
0.Then
thereexistsaunique point zinX
satisfyingFz
Gz
{z},
F
andG
have aunique commonfixed point inX,
and
F
andG
arecontinuousatz.Thatz isthe uniquecommonfixedpoint of
F
andGfollows from SarnantaandBaisnab[6].
Toprove continuity, let{y,
}
CX,
y, z.From
(4),
withx It,,y z,wehave(Fy.,Gz)
<
al(p)$(y.,Fy.)
+
a2(p)$(z,Gz)
+
a:(p)(yn,Gz)
+
o,(p)$(z,Fy.)
+
a(p)d(y.,z),
or
(Fy.,z)
<_ (a(p)
+
a4(p))(z,Fy.)
+
(I(P)
4-c3(p)
4-os(p))d(yn,z);
whichtends to 0 asn o0, and
F
is continuousat z.A
similar argumentshows that, Gis continuousatz.THEOREM
5. LetF,
G"X
,
B(X)
satisfying,(Fx,Gy)
<_ ol,(x,Fx)
+
a2,(y,Gy)
+
a3,(x,Gy)
+
a4,5(y,Fx)
+
o,d(x,y)
(5)
for all x,y in
X,
where theai>_
0,al+
tr2+
2a4
+
as
<
1 and ct+
a
<
1. ThenF
andGhaveacommonfixed point. If,
further,
a3+
aa
+
as
<
1, thenFz
Gz{z}
and zis the uniquecommonfixed point ofF
and G.Moreover,
ifF
andG
have aunique common fixed pointz,thenF
andGarecontinuousatz.The fixed point properties comefrom Theorem 2 of Dixit
[7].
To prove continuity, let{y,,}
CX,
y,,-
zand set x y., y zin(5)
togetThus
(Fy,,,Gz)
<_
al,(y,,,Fy,,)
4-a2,5(zGz)
4-a3,5(y,,,Gz)
4-a4(z,Fy,.,)
4-ar,d(y,.,,z).
,5(Fy,,,z)
< (al
4-a),5(z,Fy,,)
4-(ct
4-a34-a,)d(y.,z),
or
,(Fy,,,z)
<
al 4-a:
4-a3d(y,,,z),
1-a-a4
whichtends to 0 as n oo, and
F
is continuousat z.A
similar argument shows that G is continuousat z.THEOREM6. Let
F,
G"X
.-B(X)
satisfying,(Fx,Cy) <_ cmax{,(x,Fx), ,(y,Cy),
,5(x,Cy), ,(y,Fx), d(x,y)}
(6)
for all x,yinX,0
<
c<
1. ThenF
andGhaveauniquecommon fixedpoint z,Fz
Gz{z},
and
F
andG
arecontinuousatz.Theexistenceand uniqueness of the fixed pointcomefrom Fisher
[8].
Toprove continuity, let{y,,}
CX,y,,
---,zand setx y.,y zin(6)
toget,(Fy.,Gz)
<_
cmax{,(y.,Fy,.,), ,(z,Gz),
6(y.,Gz),
6(z,Fy.), d(y.,z)},
or,since
6(y.,Fy.)
<
6(y.,z) +,5(z,Fy.),6(Fy.,z)
<_ c[6(y.,z)
+
6(z,Fy.)],
whichimplies that,(Fy.,z)
<
c[,5(y.,z)/(1
-c)]
--.0asn oo, andF
is continuousatz.A
similarproofshow thatG
is continuous atz.THEOREM
7. LetF,
G:X
B(X),
X bounded, F
continuous,F
commutes withG,
andsatisfying
,(Ft’Gt’x,Gy)
<_
cmax{$(F
Gx,a
y),,5(F G
x,F
a’x),,5(y,Gy)
0
<_
r,r,s,s
<_
p;i0,1}
(7)
Theexistence and uniqueness of thefixedpointcome from Theorem 2 of Fisher
[9].
Toprovethe continuity of
G,
let{y,}
CX,
y. z,and setx z,y y.in(7)
toget5(r’G’
z,Gy.)
<_ cmax{,(r"G"
z,Giy.),5(F"G’z,F
Gz), (y.,Gy.)
0
<_
r,r,s,s
<_
p;iO,
1},
or,
(z,Gv,) _<
cmax{d(z,vn),5(v.,Gy.)}.
Since
6(y.,ay.)
<
6(z,Gy.)
+
d(z,yn),6(z,Gy.)
< c[6(z,ay.)
+
d(z,y.)],
which implies that6(z,Gy.)
<_
c[d(y,.,,z)/(1
c)]
0 asn-.oo,andG
iscontinuous atz.THEOREM
8. LetF,
G" X
--,B(X)
satisfying,(Fx,Gy)
<_
cmax{d(x,y), ,5(x,Gy), (y,Fx)}
(8)
for each x,y in
X,
0_<
c<
1. ThenF
andG
haveauniquecommonfixedpoint z,Fz
Gz{z},
andF
andG
arecontinuousatz.The existence and uniqueness of the fixed point follow from Fisher
[10].
To prove the continuity ofG,
let{y.}
CX,
y. z,and set x y., y zin(8)
toget,5(Fy,.,,Gz)
<_
cmax{d(y.,z),$(y,.,,Gz),5(z,Fy.)}
Thus
,5(Fy.,z)
<_
cd(y.,z)
0asn.--.oo, andF
iscontinuousatz. Settingx z,y y. in(8)
leds tothe fact thatGis continuousat z.THEOREM
9.Let
F,
G"
X
B(X),
I, J,"
X ---. X
satisfying6(Fx,Gy)
<_
cmax{d(Ix,Jy), 5(Ix,Gy), 5(Jy,Fx)}
(9)
for all x,V in X,0
_<
c<
1. IfF
commutes withI
andG
commutes withJ,F(X)
C_I(X),
G(X)
c_
J(X)
and, ifF
orI
and G orJ
are continuous, thenF,
G,I,
andJ,
have auniquecommonfixed point z.Further, Fz
Gz
{z},
and z isthe uniquecommonfixed point ofF
andI
andG
andJ.
IfI
andJ
arecontinuous,thenF
andG
arecontinuous atz. IfF
andGarecontinuous, thenI
andJ
arecontinuous atz.PROOF.Theexistenceand uniqueness ofzfollow from Theorem1ofFisher
[11].
Suppose
that
I
andJ
arecontinuous. Let{y.}
CX,
yn z, andsetx y.,y zin(9)
toget5(Fy.,Gz)
<_
cmax{d(Iy.,Jz),
5(Iy.,Gz), (Jz,Fy.)}.
Thus
,(Fy,,z)
<_
cd(Iy,.,,z)
0asn oo, andF
is continuousat z.Nowsetx z, y y,, in
(9)
toget(
Fz,Gy.
<_
cmax{
d(
Iz,Jy.
),
5(
Iz,Gy,
),
5(
Jy,
,Fz },
or
5(z,Gy.) <_
cd(z,Jy,.,)
0 asn oo, andGis continuousat z.The specialcaseofTheorem 9 in which
I
is the identitymap onX
yields the result in Fisher[12].
THEOREM
10. LetF,
G"X
--,B(X), I,
d,"X
--,X
satisfying6(Fx,Gy)
<
cmax{d(Ix,Jy),6(Ix,Fx),*(Jy,Gy)}
(10)
for all x,V in X,0
_<
c<
1. IfF
commutes withI
and G commutes withJ, G(X)
C_I(X),F(X)
C__J(X)
and, ifI
orJ
iscontinuous, thenF,
G,I,
andJ
have aunique common fixed point z.Further, Fz
Gz{z},
and z is the unique common fixed point ofF,
G,
I,
and
J. Further,
the continuity ofI
implies thatF
is continuousatz,and the continuityofJ
implies thatGis continuousatz.
Theexistenceand uniqueness ofzfollowsfrom Theorem1ofFisher
[13].
Suppose
thatI
is continuous. Let{y,)
CX,
y,---,z, and setx y,, y zin(10)
toget6(Fy,,,Gz)
<
cmax{d(Iy,,,Jz),
6(Iy,,Fy,,),
6(Jz,Gz)};
6(Fy.,z)
<_
cmax{d(Iy.,z),6(Iy.,Fy.)},
which,since6(Iy.,Fy.)
<_
6(Iy.,z)
+
6(z,Fy.),
implies that
6(fy.,z)
<_
c[6(Iy.,z)
+
6(z,Fy.)]
-
0asn oo, and fis continuousat z. The assumption thatJ
is continuousleads to the continuity ofG
at z.Thespecialcaseof Theorem 10with
I
J
Ix
yields the result ofFisher[10],
and the continuityof bothF
andG
at z.THEOREM
11. LetF,
G X
-
B(X),I,
J,:
X
-,X
satisfying6(F"x,Gy)
<_ cmax{(Frx,Gy),6(Frx,y)
0<_
r<
p}
(II)
for all x,y in
X,
0_<
c<
1, pafixed positive integer. IfF
also mapsB(X)
intoitself, thenF
andGhave auniquecommonfixed point z. Further,z isthe unique fixed point ofF
andG,
Fz
Gz
{z},
andGis continuous atz.The existence and uniqueness of z comefrom Theorem 2 in Fisher
[14].
To
prove thecontinuityof
F,
let{y.
}
CX,
y,, z,and set x z,y y, in(11)
toget6(
F
pz,Gy.
<_
cmax{
6(F
z,Gy.
), 6(
F
z,y.)"
0<
r<_
p};
6(z,Gy,)
<_
cmax{6(z,Gy,,),
6(z,yn)},
whichimpliesthat
6(z,Gy,)
<_
c6(z,y,)
-
0asn--, oo, andG
is continuousat z.A
similar calculation verifiesthatG
is continuousat z.We
now establish continuity for multivalued mappingswith metric defined by the Hausdorffmetric.
THEOREM
12. LetT X
CB(X)
satisfyingH(Tx,Ty)
<
a(x,y)D(x,Tx)
+
a’(x,y)D(y,Ty)
+
b(x,y)D(x,Ty)+
b’(x,y)D(y,Tx)
+
c(x,y)d(x,y)
(12)
limsup(a
+a’
+b+b’
+c)(x,y)
<
1, d(,)othen foreach X in
X
thereexists a sequence ofiteratesconverging to afixed point z ofT,
and
T
is continuousatz.The existence ofafixed point isa consequence of Garegnani and Massa
[15].
Toprove continuity, let{y,,}
CX,y.
z, and set x y.,y zin(12)
togetThus,
H(Ty.,Tz)
<_ a(y.,z)D(y.,Ty.)
+
a’(y.,z)D(z,Tz)
+
b(y.,z)D(y.,Tz)
+
b’(y.,z)D(z,Ty.)
+
c(y.,z)d(y.,z).
D(Ty.,z)
<_
H(Ty.,Tz)
<_ a(d(u.,z)
+
D(z,Ty.))
+
bD(y.,Tz)
+
b’D(z,Tya)
+
cd(y,,,z),
or
D(Ty.,z)
<
(a
+
c)d(y.,z)
+
bD(u.,Tz)
1-a-b’whichtends to 0asn---,oo,since
lim(1
ab’) >
0, and lim,,D(Ty.,z)
0,whichimplies thatlima
D(ya,Ty.)
O. Takingthe limit as n oo of the inequality involvingH
yieldslima
H(Tya,Tz)
O,
andT
is continuousat z.THEOREM 13. Let T:
X
--.CL(X),f
X
X
such thatTX
CfX,
fX
is(T,f)-orbitallycompleteand
H(Tx,Ty)
<_
qmax{d(fx,fy),D(fx,Tx),D(fu,Tu), [D(fx,Ty)
+
D(fy,Tx)]/2}
(13)
for allx,y in
X,
0<
q<
1. ThenT
andf
haveacoincidence point;i.e.,thereexists azinX
such that s
tz
ETz.COROLLARY
1. Letf
be the identitymaponX. Then,under thehypothesesof Theorem13,
T
hasafixed point z, andT
is continuousatz.The fact that
T
has afixed point comes from Singh and Kulshrestha[16].
To prove continuity, let{ya
}
CX,
ya z, and setx y,,, y zin(13)
togetThus
H(Tya,Tz)
<_
qmax{d(ya,z),D(ya,Tya),D(z,Tz),[D(y.,Tz)
+
D(z,Tya)]/2}.
D(Tya,z)
<
qmax{d(y.,z),
D(y.,Tya), O,
[D(y.,Tz)
+
D(z,Tya)]/2}
q
D(ya,z),
qqD(ya,Tz)}
--.
0as n--.oo.<
max{qd(y.,z),
1 q 2
and lim,,
D(Tya,z)
0,whichimpliesthat lim,,D(y.,Ty.)
0. Takingthe limitasn --,oo,The fixed point portionofCorollary 1 isessentiallydue to Ciric
[17].
The theoremof Ciric alsocontainsaresult ofReich[18]
asaspecial ease.Kaneko
[19]
proves the Cirie result under the weaker conditions thatX
be a reflexive space and the range ofT
is the family of allnonempty weakly compact subsets ofX. Heisapparentlyunawarethat the twostandard definitions of the Hausdorffmetric areequivalent. Theorems1 and 2in Czerwiek
[20]
are specialeasesofCorollary 1, as areTheorem 1of Iseki[21]
and Theorem1ofRay
[22].
THEOREM
14. Let(X,d)
beacomplete metricallyconvexmetric space,K
anonemptyclosedsubset of
X,
T
X
CB(X)
such that there exista,/3,
7>-
0,a+
2/3
+
27
<
1 such that for all x, yinX,
H(Tx,Ty)
<_
od(x,y)
+
{D(x,Tx)
+
D(y,Ty)}
+
"7{D(x,Ty)
+
D(y,Tx)}.
(14)
If for eachx
c:: OK,
Tz
CK
and(a
+/3
+
-),)(1
+
13
+
/)/(1 -/3
3,) <
1,then thereexistsaz inK
with z ETz.Moreover T
is continuousat z.Theexistence ofz is Theorem 1 of Itoh
[23].
To
show continuity, let{y,}
CX,
y,, z, and setx y,, y z in(14)
togetThen
H(Ty,.,,Tz)
<_
od(y,,,z)
+
3{D(y,,Ty,)
+
D(z,Tz)}
+
"7{D(y,.,,Tz)
+
D(z,Ty,.,)}.
D(z,Ty,.,)
<_ H(Ty,.,,Tz) <_
od(y,,z)
+
fl{D(y,,,z)
+
D(z,Ty,)}
+
"),{D(y,,Tz)
+
D(z,Ty,,)},
or,
D(z,Ty,,)
<_
(a
+
3)d(y,.,,z)
+
"TD(y,,,Tz)
0asn c,and lim,
D(z,Ty,.,)
0,whichimpliesthat lim,D(y,,Ty,)
0.Now
take the limitasn -inthe inequlality forH
to getlim,H(Tz,Tyn)
O,
andT
is continuousatz.THEOREM
15.Let
(X,d)
beacomplete boundedmetric space,F,X
---,CL(X),
1,2satisfying
H(Fax,F:y)
<
aaD(x,Fax)
+
a2D(y,F:y)
+
a:D(y,Fax)
+
a,D(x,Fy)
+
asd(x,y)
(15)
5
for allx,y in
X,
ai>_
O,
Ei=aai
<
1 andaa
a2 oras
a4. ThenF1
andF2
have acommon fixed point z.Moreover
F
andF2
arecontinuousat z.Theexistenceofz comesfrom Theorem1 ofBoseandMukherjee
[24].
To
prove continuity ofF,
let{Yn}
CX,
y,--
z, and setx Yn,Y zin(15)
togetH(Fay,,,Fz)
<_ aaD(y,,Fay,,)
+
a:D(z,Fz)
+
a:D(z,Fy,.,)
+
a,D(yr,,F:z)
+
ar,d(y,.,,z).
Settingx y zin
(15)
yields the fact thatFlZ
Fz.
Thusand
D(z,Fly.) <_
a,tD(y.,F:z)
+
(al
+
as)d(y.,z)
1-al
whichleads to the facts that
lim.
D(z,Fy,,)
0 andlira. D(y.,Fiy.)
0. Taking thelimit ofthe inequalityforH,
and using the fact thatFz
Fz,
yieldslim. H(Fy.,Fz)
0, andF,
is continuousatz.Similarly,
F
is continuous at z.THEOREM
16.Let
(X,d)
beacompletemetricallyconvexmetric space,K
anonempty closed subset ofX. LetS,
T" K
--..CB(X)
satisfyingH(Sx,Ty)
<_
ad(x,y)
+
${D(x,Sx)
+
D(y,Ty)}
+
7{D(x,Ty)
+
D(y,Sx)}
(16)
for all x,y in
X,a,
fl,7>
0witha+
2fl
+
27
<
1. If for each x 6-OK, S(x)
CK, T(x)
CK
and(a+fl+7)(1
+fl+7)/(1-fl-"r)
<
1, then thereexistsazinK
withz6_Tz
andz 6-Sz.Also
S
andT
arecontinuousat z.The properties ofz comefrom Theorem 3.1 of Khan
[24].
To
establishthe continuity ofS,
let{y,,
}
CX,
y,, z,and set x y,,,y z in(16)
togetThus
H(Sy.,Tz)
<
od(y.,z)
+
fl{D(y.,Sy.)
+
D(z,Tz)}
+
7{D(y.,Tz)
+
D(z,Sy.)}.
D(Sy.,z) <_ H(Sy.,Tz)
ad(y.,z)
+
flD(y.,Sy.)
+
7{D(y.,Tz)
+
D(z,Sy.)},
or,
(a
+
fl)d(y,,,z)
+
7D(y.,Tz),
D(
Sy,.,,zwhich implies that
lim. D(Sy.,z)
0,and thus thatlim.
D(y,,,Sy.)
0. Settingx y z in(16)
yieldsSz
Tz. SubstitutingintotheinequalityforH
yieldsH(Sy,,,Tz)
H(Sy,,,Sz) <_
ad(y,,,z)
+
D(y,,,Sy,.,)
+
7{D(y.,Sz)}
+
D(z,Sy.),
which impies that lim,,
H(Sy.,Sz)
0,andS
is continuousatz.A
similarcalculationverifiesthatT
is continuous atz.THEOREM
17.Let T,., X
---,CB(X)
satisfying{H(Tl
x,T.y)}
<
kmax{
D(x,Tx)D(y,T.y),D(x,T.y)D(y,Tx),
D(x,T
x)D(x,T.y),
D(y,T
x)D(y,T.y),
d(x,y)}
(17)
for all x,y in
x,n
>
2,0<
k<
1/2.
Then{T.}
has a coon fixed point andF(T)
F(T.),
n>
1,whereF(T)
denotesthe fixed point set ofT.Moreover
the{T.}
arecontinuous ateachfixed point.or
{H(,T)} _<
={D(,T)O(,T),O(,)O(,), O(,)O(,),
D(y,Tlz)D(y,,T.yk),
d2(z,y,)},
{H(Tz,Tr,
y,)}
2<_ kmax{0, D(z,Tr,
y,)D(y,Tz),O, D(ye,Tz)D(y.,T,,y,),
d2(z,y)}.
Setting z y z in
(17)
yieldsTlz
Tr,
z. Substituting in the above inequality yieldslimsup{H(Tr,
z,T,y,)}2
0 and eachT,
is continuousat z.THEOREM
18.Let
(X,d)
beacompletemetricspace,T1,T2:
X
-,CB(X)
satisfyingd’(x,T,
x)
+
d’(y,T2y)
(18)
H’(T,
,T/)
<
_,,,(,T,
z)
+
-,,,(/,T/)
foreach x,y in
X
such that’-"n(x,Tx)
+
$’-’n(y,T2y)#
0,0<
c<
1,rn>_
1,p>_
2,rn<
p. ThenT
andT2
havecommonfixed points.Moreover
T
andT2
arecontinuousateach fixed point.Thefactthat
T1
andT2
have fixed pointsisTheorem2ofPopa
[27].
To
establish continu-ity, letzbeacommonfixedpointofT1
andT2
andlet{y,}
CX,y,
z,and setx z,y y, in(18)
togetH’n(Tlz,Ty,)
<_
cd.(/,,,T/,,)
c
<_
cd’(y.,Tyn),
--(u,Tu)
since
d(y,,T2y,,)
<_
(y,,T2yr,).
ThereforeD(z,T2y,)
<_
H(Tz,T2yr,)
<_
c/rnd(y,,T2yr,)
c/nD(y,,T2y,)
<_
c/’[DO/,,,z) +
D(z,T/,,)],
or,
D(z,T2y,,) <_
cl/’n[D(y,,z)]/(1-
c)
---, 0 as n oo, and limr,D(z,T2y,)
O. Also,lim,
D(yr,,Tyr,)
O.f
,-’(,T)+,-’(z,Tz)
O,
thT
T
{}.
f,-’(,T)+-"(z,T)
#
0,
then,
from(18),
Tz
Tz.
Substituting in the inequality forH
yieldsH(T2z,T2yn)
H(Tz,Ty)
c/D(y,,Ty)
0 n,
dT
iscontinuous at z.Simflly,
T
is continuo atz.Let
C(X)
denote thenonempty compactsubsets ofX.
A
speX
issd to be xo-jointly orbitMlycompleteifeve
Cauchyquenceofeh orbit atXo isconvergentinX.
THEOM
19.Let
Fi
X
C(X),i
1,2,X
xo-jointly completefor some XoX.
Suppose
thereexists afunction(R
+)s
R
+,
upper secontinous d nondecreing ineachvable,
such thatsatisfies
-),(t)
<
foreach>
0. Supposethat theF,satisfyH(Flx,F2y)
<_
{D(x,Flx),D(y,F2y),D(x,F2y),D(y,Flx),d(x,y)}
(19)
for all x,y in X. Then
F1
andF2
have a common fixed point. MoreoverF1
mdF2
arecontinuousat eachfixedpoint.
The existence of a fixed point is Theorem 2.1 of
Guay
et al[28].
To prove thatF1
is continuousat afixedpoint z, let{y,}
CX,y,
--
z,andset x yn,y zin(19)
togetH(Fyn,Fz)
<_
{D(yn,Fly),D(z,F=z),D(y,F=z), D(z,Fly,), d(yn,z)}.
Thus
D(Fly,z)
<_
{D(yn,Flyn),
O,D(ynF=z),D(z,Fly,),d(yz)}.
Suppose
that 5 limsupD(Fly,,z) >
0. SinceD(y,Fly,)
_
D(y,z)
+
D(z,Fy,),
limsup
D(y,
,F1y,)
_<
limsupD(z,F1
y,,).
SincezEF2
z,lim supD(y,
F2
z)
0. Thereforewe have5_
(5, 0,0,5,0)
_
(5,
5, 5,
5,5) <
5,
acontradiction. Consequentlylim,D(Fy,z)=
O,and
lim D(y,,Fly,)
O.From
(19)
with x y z,H(Fz,F=z) <_
{D(z,Fz),D(z,F=z),D(z,F.z),D(z,Fz),d(z,z)}
(0,0,0,0,0)
0,and
F
zF2z.
Substitutinginthe inequality forH
wehaveH(Fly,,F
z) <_
y{D(y,,Fly), O,D(y,Fl z),D(z,Fyn), d(y,,z)}.
Taking the limsup as n c, yields limsup,
H(Fly,,Fz)
_
(0,0,0,0,0)
0, andF1
is continuousat z.A
similarargumentverifiesthatF2
iscontinuous atz.THEOREM20. Let
X
beacompletemetric space,F,
:X--
C(X).
Suppose
that there existsafunction satisfyingtheconditionsof Theorem 19 and such thatH(Fix,Fjy)
_
{D(x,Fix),D(y,Fjy),D(x,Fy),D(y,Fix),
d(x,y)}
(20)
for eachx,y in
X,
foreach i, j,:/:
j. Then{F,
}
hasa commonfixedpoint, and each of theFi
is continuous at this fixedpoint.Theexistenceofacommonfixedpointis Theorem 2.5 of
Guay
et al[28].
The continuity isprovedinthesamewayasinTheorem 19.Theorem 4 of Kaneko
[19]
isaspecialcaseof Theorem 20.THEOREM21. Let
(X,d)
beacompleteHausdorffuniform space definedby{dx
A
e
I}.
Let
Fi X
2x,i
1,2 satisfyingThe proofthat there isa commonfixed point zisTheorem3.1of Mishra
[29].
Toprove thatF,
is continuousat z, let{y.}
CX,y.
-,z,and set x y.,y zin(21)
togetHx(Flyn,F2z)
<_
axdx(y.,F,y.)
+
bxdx(z,F2z)
+
cxdx(y.,Fz)
+
edx(z,Fly.)
+
fdx(y.,z).
Since
dx(Fly.,z)
<_ Ha(FIv.,F2z),
the above inequality implies that(
+
A)d(.,z)
+
d(.,)
dx
Fa
y,.,,z)
<
0 asn cx.1
-ax-bx
It then follows that limndx(y,,Fy,) 0. Setting x y z in
(21)
yieldsFz
F2z.
Substitutingintheinequality forH
givesH(r.,rz)
< d(.,r.)
+
d(.,Fz)
+
d(,F.)
+
Ad(.,z).
Takingthelimit asn ooweobtainlim,
Hx(Fy,,FI
z)
O,
andF
iscontinuousatz. Similarly,Fz
is continuousat z.Theresult inMishraandSingh
[30]
isaspecialcaseof Theorem3.1 ofMishra[29].
THEOREM
22. LetX
beareflexiveBanachspace,K
anonempty closed boundedconvex subset ofX.
LetT
beamappingofK
intothe family ofnonempty weakly compact convex subsets ofK
satisfyingH(Tx,Ty)
< (max{D(x,Tx),
D(y,Ty)}
(22)
for each x,y in
X,
where4:
[0,oo)
-,[0,oo),
nondecreasing, fight continuous, such that(t) <
for each>
0. Then thereexists anonemptysubsetM
ofK
such thatTx
M
for eachx6M.
Moreover,
T
is continuousat each point ofM.
Thefactthat asubset
M
existswiththe stated propertiesisTheorem 1 of Kaneko[31].
Toprovethe continuity of
T,
letz 6 M. Let{y.}
CX,y.
--,z,and set x y.,y zin(22)
togetH(Ty.,Tz)
<_ (max{D(y.,Ty.),D(z,Tz)}).
Thus
D(Ty.,z)
<_
H(Ty.,Tz)
<_ (D(y.,Ty.)).
Assume
that6 limsupD(Ty.,z) >
0. Then wehave 6_<
(limsup[D(y.,z)+
D(Ty.,z)])
(6)
<
6,
acontradiction. Therefore6 0 andT
is continuousatz.THEOREM
23. Let(X,d)
be acomplete metricspace,{S.},
{T.}
sequences of maps fromX
--+CB(X).
Suppose
that thereexistsanh,
0<
h<
1, suchthat,
for each m, n, and eachx,y inX,
H(Smz,T.y)
<
hmax{d(x,y),D(x,Smz),D(y,T.y), [(Dx,T.y)
+
D(y,S.x)]/2}.
(23)
Then
{S,,
}
and{T,
}
haveacommonfixed pointz.Moreover,
{
S,
}
and{T,
}
arecontinuous atz.The existence ofacommonfixed pointz isTheorem1 of Kubiak
[32].
To
prove thateachS.,
iscontinuous, let{y.
}
CX,
y.--,z,and set x yk,y zin(23)
togetSince
D(SmV,,z) <
H(Smyt:,Tnz),
the aboveinequaltiy yieldsD(S.,y,,z)
<_
hd(y,,z)/(1
h)
0 as k-
oo.Itthen follows that
limk D(y,,S.y,)
O.Settingx y zin
(23)
gives the result that S,,,zT,,z.
Substitutinginthe inequality forH
yieldsH(S.,y,,S..,z)
<_
hmax{d(y,,z),D(y,,S,,,y,),O,[(D(y,,S,-,,z)
+
D(z,S.,yk)]/2},
and, takingthelimit ask oogives lim,
H(S,y,,
S,,,z)
0,and eachS.,
iscontinuous atz.A
similarargumentverifiesthat eachT.
is continuous at z.Specialcasesof Theorem 1 ofgubiak
[32]
appear inAvram
[33],
Iseki[21],
Popa
[34-
35],
Ray
[36],
aus
[37],
andTong
[38].
Achara[39]
has thesame resultasKubiak[32],
butwithCB(X)
replacedbyC(X).
THEOREM24. Let
(X,d)
beacomplete metric space,T,., :XCB(X).
Supposethat there exist a,>_
0, 1, 5 such thatmin{al
+
a2+
as
+
2a4,al+
a2+
a3-t-2as}
<
1andm
#
n imphesH(T,,,x,T.y)
<
ald(x,y)
+
a2D(x,T,-.x)
+
a3D(y,T,.,y)
+
a4D(x,T,y)
+
asD(y,T.,x)
(24)
for all x, yinX. Then
{T,,
hasa commonfixedpoint.Theproofthat the
{T,,}
hasa commonfixed point isTheorem 1 ofKith[40].
Toprove continuity, let zbeacomonfixedpoint of{T.},
andlet{y,,}
CX,
yn-
z.Suppose
thata4<
as.
Thennfin{al
+a2+as+2a4,al+a2+a3+2as}
a
+a2+a3+2a4 <
1.Set x z,y y,. From
(24),
H(T,..z,T,,y,)
<_
a,d(z,y,)
+
a2D(z,T..,z)
+
a:D(y,,T,,y,)
+
a,tD(z,T,,y,)
+
asD(yr,,T,z),
or, since
D(z,T,.,V,) < H(T..,z,T.y),
weobtainD(z,T.y,)
<
(01
q-a3)d(z,y,)
+
asD(y,,T,.,,z)
0 ask oo.1 a3
a
Thus lim,
D(z,T,,y,)
0,whichimplies that lim,D(y,,T,.,y,)
O.Substitutingx y zinto
(24)
verifiesthatT,z
T.z.
It then follows thatH(T,.,z,T.y,)
H(Tmz,T.y,)
<_
ald(z,y.)
+
a3D(y,,Tnyt:)
+
a4D(z,T,.,y,)
+
asD(y,,Tmz).
Taking thelimitas k ooyields the result that each
T.
is continuousat z. If a,>
as,then,
settingx y,, y zin(24)
yieldsD(T,y,,z)
<_
(a,
+
a2)d(z,y,)
+
a.tD(y,,T,z)
0as k oo. 1-as-asAlthoughwehave experiencedsomesuccesswithestablishingthe continuity of multivalued maps atafixedpoint,thereaxesomedefinitionsthatdo notlendthemselvesto suchananalysis.
We
citeherethreeexamples.THEOREM K.
(Khan
and Kubiaczyk[41],
Theorem1)
Let
S, T,
X
-,B(X)
besuchthat,forsome E :=
{b:
(R+)5
--,R+
isupper semicontinuous fromthe right, nondecreasingineach variable coordinate such that
(t,t,t,
at,
bt) <
foreacht>
O,a,b>
O,
witha+b
<
2},
,5( Sx,Ty) <_
(d(x,y),
,5(x,Sx),
5(y,Ty),
D(x,Ty),
D(y,Tx)
foreachz,
l/in X.
ThenS
andT
haveauniquecommonfixed pointusuchthatuESu
NTu.
In
the abovetheoremthe difficultyarisesfromthe fact that5(u, Tu)
neednot be zero.THEOREM
M.
(Mukherjee andSom
[42],
Theorem1)
Let(X, d)
beacomplete metric space,T1,T2
X
--
CB(X)
satisfyinganyoneof the following conditions for eachz, I/inX:
(i) (z,T] x,)
4-(I/,T2y)
_<
ad(x,,y),
1_<
et<
2,(ii)
(z,T,=)
+
(y,T2y) <_ {(z,T2y)
q-H(y,T2z)
+
d(z,y)},
1/2
< /<
2/3,
(iii) &(X,Tlz)
+
&(y,T2y)
+
&(Tlx,
T2F)
<_
7{H(x.,T2F)
+
H(y,Tx)},
1_<
7<
3/2,
(iv) &(Tz,T2!t)
<
lmax{d(z,y),H(z,T=),U(!t,T2y),[H(=,Ty)
+
H(y,Tz)]/2},O <_
1<
1. ThenT
andT2
haveacommonfixed point.THEOREM S. (Singh
etal[43],
Theorem2.1)
Let
S, T,
be multivalued mappings froma metricspaceX
--.CL(X).
If thereexists afunctionf
X
--
X
suchthatSX
UTX
C_f(X)
and,foreach,
yinX,
H(Sx,Ty) <_
(max{
D(fx,Sx),
D(fy,Ty),
D(fz,Tll),
D(fy,Sx), d(fx,fy)
})
where
b
R
+ --,R
+,
b
upper semicontinuousandnondecreasingwithb(t) <
tforeach>
0,there exists a point
xo
inX
such that(S,T)
is asymptoticallyregular
at Xo andf(x)
is(S,T;f,
Xo)-orbitally complete,thenf,S,
andT
haveacoincidence point. Further,ifz isa coincidencepoint off,
S, T,
andfz
is afixed point off,
then(a) fz
isafixed point ofS (resp
T)
providedf
commutes weaklywithS (resp
T)
at z,and
(b)
fz
isacommonfixed point ofS
andT
at z.ADDED IN PROOF.
1.A
closerexamination ofthe proofof TheoremK
shows that5(u,
Su)
(u, Tu)
0. Thereforeanargumentsimilar to theonealreadyusedrepeatedlyin this paperyieldsthatS
andT
axecontinuousat the fixed point.2. Theorem
M
containsanerrorintheproof.Moreover,
conditions(i)
and(ii)
imply thatTz
T2z
{z}
foreachz inX. Theparametervaluesonconditions(iii)
and(iv)
makeit impossible tousestandardprooftechniquestoobtainafixed point.REFERENCES
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