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(1)

FIXED POINTS AND CONTINUITY FOR MULTIVALUED MAPPINGS

TROY HICKSandB.E. RHOADES

Department

of Mathematics University ofMissouriatRolla

Rolla,

Missouri 65401

Department

ofMathematics Indiana University

Bloomington,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 AND

PHRASES.

Continuity, fixed point, multivaluedmapping.

1980AMS

SUB:IECT CLASSIFICATION

CODE.

54

H

25

In

arecent paper, Rhoades

[1],

it wasverified

that,

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. Let

T

bea selfmapofacompletemetric space with afixedpointp. Then thefollowingareequivalent:

(a)

T

is continuousatp.

(b)

If

{y,,}

is anysequencecontained in

X

with y,,

-

p,then

limd(y.,Ty.)

O.

PROOF. To

show that

(a)

and

(b)

are equivalent, suppose that

(a)

is satisfied. Then Yr,

--

Pimplies that

Tyn

Tp

p,andthus lim

d(yn,Tyn)

O.

Conversely,

d(Tyn,Tp)

<

d(Ty,,y,)

+

d(yn,p),

sothat

limsupd(Tv.,Tp)

<_

limsup[d(Tv.,V,)

+

d(y,,Tp)]

0. Then

T

is continuousatp.

In

themultivaluedcase,amultivaluedmap

T

hasafixedpointp ifp E

Tp.

Let

CL(X)

denote thecollectionofnonempty closed subsets of

X,D(A,B):= inf{d(a,b):a e

A,b

e

B}.

A

multivalued map

T

(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
(2)

(a)

T

is continuous at p,

(b)

If

{y,,}

C

X

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,then

D(Ty.,Tp)

<

D(Ty.,y,,)

+

D(y,,,p)

-

0, and

(a)is

satisfied.

In

Proposition2,weobtainthesameconclusionsbyreplacing

CL(X)

with

B(X),

the setof allnonemptybounded subsets of

X,

andreplacing

D

by

,(A,B),

where

,(A,B)

:=

sup{d(a,b):

a E A,b

B}.

Wecanalso replace

CL(X)

with

CB(X),

thecollection ofnonempty closed,

boundedsubsets of

X,

andreplace

D

with

H,

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. If

F

and

I

commute and

I(X)

D_

F(X),

then

F

and

I

have a uniquecommonfixed point z,

Fz

{z},

and

F

iscontinuous atz.

The fact that

F

and

I

haveauniquecommonfixed pointcomesfromFisher

[2 ].

Although

not mentioned inthestatement of Theorem 1,Fisher

[2]

has shown that

Iz

{z}.

Toshow that

F

is continuousat z, let

{y,,}

C

X,

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, and

F

is continuousat z.

By

setting

I

equal to the identity map, Theorem 1 is ageneralization of the result in Fisher

[3].

Foraninteger n,x

.

X,F"

isdefined inductivelyby

F"(x)

F(F"-I(x)).

THEOREM2. Let

F:X

B(X)

satisfying

6(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. If

F

alsomaps

B(X)

intoitself, then

F

hasaunique fixed pointz,

Fz

{z},

and

F

is continuousatz.

PROOF.

The fact that

F

hasaunique fixed pointzand that

Fz

{z}

comesfromFisher

[41.

Let {y,}

C

X,

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 aboveinequalitybecomes

6(z,Fy.)

<_ c[6(z,y.)

+

6(z,Fy.)],

which implies that

6(z,Fy.)

<

c[6(z,y.)]/(1- c)

0 as n cx. Thus

F

is continuousat z.

Theorem2 alsogeneralizesthecorrespondingresult inFisher

[4].

For

ametricspace

X,N(e,A)

:=

{x

6

X"

d(x,a) <

efor some a 6

A

6

CL(X),e >

0}.

TheHansdorffmetric

H

isdefinedby

H(A,B)

inf{e >

O"

A

C_

N(e,B)

and

B

C_

N(e,A)},

ifthe infimumexists, and

H(A,B)

cotherwise.

An

equivalent definitionof

H

is
(3)

THEOREM3. Let F1,

F2

X

CB(X)

satisfying

H(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 and

a

a2 or a3 a4. Ifz is acommon fixedpoint of

F1

and

F2

such that

F1

z

{ z}

or

F2

z

{ z},

thenz isthe uniquecommon fixed point of

F1

and

F2.

Moreover,

F

and

F2

arecontinuousatz.

PROOF.

Theconclusionsabout the fixed point zcomefromMendaglioandDube

[5].

We

first observethat the hypotheses of the theorem implies that both

F1

z

F2z

{z}.

For,

supposethat

F2z

{z}.

Then,in

(3)

withx y z,wehave

H(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).

Therefore

Fz

{z}.

Similarly,

F1

z

{z }

implies

F

z

{ z}.

Toprovecontinuity, let

{y.}

C

X,y.

---}z.

In

(3)

setx y.,y ztoget

H(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, and

F1

is continuousatz.

Settingx zand y y. in

(3)

leads tothe resultthat

F2

is continuous atz.

THEOREM

4. Let

F,

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 zin

X

satisfying

Fz

Gz

{z},

F

and

G

have aunique commonfixed point in

X,

and

F

and

G

arecontinuousatz.

Thatz isthe uniquecommonfixedpoint of

F

andGfollows from SarnantaandBaisnab

[6].

Toprove continuity, let

{y,

}

C

X,

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);

(4)

whichtends to 0 asn o0, and

F

is continuousat z.

A

similar argumentshows that, Gis continuousatz.

THEOREM

5. Let

F,

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. Then

F

andG

haveacommonfixed point. If,

further,

a3

+

aa

+

as

<

1, then

Fz

Gz

{z}

and zis the uniquecommonfixed point of

F

and G.

Moreover,

if

F

and

G

have aunique common fixed pointz,then

F

andGarecontinuousatz.

The fixed point properties comefrom Theorem 2 of Dixit

[7].

To prove continuity, let

{y,,}

C

X,

y,,

-

zand set x y., y zin

(5)

toget

Thus

(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-a3

d(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. Then

F

andGhaveauniquecommon fixedpoint z,

Fz

Gz

{z},

and

F

and

G

arecontinuousatz.

Theexistenceand uniqueness of the fixed pointcomefrom Fisher

[8].

Toprove continuity, let

{y,,}

C

X,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, and

F

is continuousatz.

A

similarproofshow that

G

is continuous atz.

THEOREM

7. Let

F,

G

:X

B(X),

X bounded, F

continuous,

F

commutes with

G,

andsatisfying

,(Ft’Gt’x,Gy)

<_

cmax{$(F

G

x,a

y),,5(F G

x,F

a’x),,5(y,Gy)

0

<_

r,r

,s,s

<_

p;i

0,1}

(7)

(5)

Theexistence and uniqueness of thefixedpointcome from Theorem 2 of Fisher

[9].

To

provethe continuity of

G,

let

{y,}

C

X,

y. z,and setx z,y y.in

(7)

toget

5(r’G’

z,Gy.)

<_ cmax{,(r"G"

z,Giy.),5(F"G’z,F

G

z), (y.,Gy.)

0

<_

r,r,s,s

<_

p;i

O,

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 that

6(z,Gy.)

<_

c[d(y,.,,z)/(1

c)]

0 asn-.oo,and

G

iscontinuous atz.

THEOREM

8. Let

F,

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. Then

F

and

G

haveauniquecommonfixedpoint z,

Fz

Gz

{z},

and

F

and

G

arecontinuousatz.

The existence and uniqueness of the fixed point follow from Fisher

[10].

To prove the continuity of

G,

let

{y.}

C

X,

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)

<_

c

d(y.,z)

0asn.--.oo, and

F

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

satisfying

6(Fx,Gy)

<_

cmax{d(Ix,Jy), 5(Ix,Gy), 5(Jy,Fx)}

(9)

for all x,V in X,0

_<

c

<

1. If

F

commutes with

I

and

G

commutes with

J,F(X)

C_

I(X),

G(X)

c_

J(X)

and, if

F

or

I

and G or

J

are continuous, then

F,

G,I,

and

J,

have auniquecommonfixed point z.

Further, Fz

Gz

{z},

and z isthe uniquecommonfixed point of

F

and

I

and

G

and

J.

If

I

and

J

arecontinuous,then

F

and

G

arecontinuous atz. If

F

andGarecontinuous, then

I

and

J

arecontinuous atz.

PROOF.Theexistenceand uniqueness ofzfollow from Theorem1ofFisher

[11].

Suppose

that

I

and

J

arecontinuous. Let

{y.}

C

X,

yn z, andsetx y.,y zin

(9)

toget

5(Fy.,Gz)

<_

cmax{d(Iy.,Jz),

5(Iy.,Gz), (Jz,Fy.)}.

Thus

,(Fy,,z)

<_

c

d(Iy,.,,z)

0asn oo, and

F

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.
(6)

The specialcaseofTheorem 9 in which

I

is the identitymap on

X

yields the result in Fisher

[12].

THEOREM

10. Let

F,

G"

X

--,

B(X), I,

d,"

X

--,

X

satisfying

6(Fx,Gy)

<

cmax{d(Ix,Jy),6(Ix,Fx),*(Jy,Gy)}

(10)

for all x,V in X,0

_<

c

<

1. If

F

commutes with

I

and G commutes with

J, G(X)

C_

I(X),F(X)

C__

J(X)

and, if

I

or

J

iscontinuous, then

F,

G,I,

and

J

have aunique common fixed point z.

Further, Fz

Gz

{z},

and z is the unique common fixed point of

F,

G,

I,

and

J. Further,

the continuity of

I

implies that

F

is continuousatz,and the continuityof

J

implies thatGis continuousatz.

Theexistenceand uniqueness ofzfollowsfrom Theorem1ofFisher

[13].

Suppose

that

I

is continuous. Let

{y,)

C

X,

y,---,z, and setx y,, y zin

(10)

toget

6(Fy,,,Gz)

<

cmax{d(Iy,,,Jz),

6(Iy,,Fy,,),

6(Jz,Gz)};

6(Fy.,z)

<_

cmax{d(Iy.,z),6(Iy.,Fy.)},

which,since

6(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 that

J

is continuousleads to the continuity of

G

at z.

Thespecialcaseof Theorem 10with

I

J

Ix

yields the result ofFisher

[10],

and the continuityof both

F

and

G

at z.

THEOREM

11. Let

F,

G X

-

B(X),I,

J,:

X

-,

X

satisfying

6(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. If

F

also maps

B(X)

intoitself, then

F

andGhave auniquecommonfixed point z. Further,z isthe unique fixed point of

F

and

G,

Fz

Gz

{z},

andGis continuous atz.

The existence and uniqueness of z comefrom Theorem 2 in Fisher

[14].

To

prove the

continuityof

F,

let

{y.

}

C

X,

y,, z,and set x z,y y, in

(11)

toget

6(

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,)

<_

c

6(z,y,)

-

0asn--, oo, and

G

is continuousat z.

A

similar calculation verifiesthat

G

is continuousat z.

We

now establish continuity for multivalued mappingswith metric defined by the Hausdorff

metric.

THEOREM

12. Let

T X

CB(X)

satisfying

H(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)

(7)

limsup(a

+a’

+b+b’

+c)(x,y)

<

1, d(,)o

then foreach X in

X

thereexists a sequence ofiteratesconverging to afixed point z of

T,

and

T

is continuousatz.

The existence ofafixed point isa consequence of Garegnani and Massa

[15].

Toprove continuity, let

{y,,}

C

X,y.

z, and set x y.,y zin

(12)

toget

Thus,

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

a

b’) >

0, and lim,,

D(Ty.,z)

0,whichimplies that

lima

D(ya,Ty.)

O. Takingthe limit as n oo of the inequality involving

H

yields

lima

H(Tya,Tz)

O,

and

T

is continuousat z.

THEOREM 13. Let T:

X

--.

CL(X),f

X

X

such that

TX

C

fX,

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. Then

T

and

f

haveacoincidence point;i.e.,thereexists azin

X

such that s

tz

ETz.

COROLLARY

1. Let

f

be the identitymaponX. Then,under thehypothesesof Theorem

13,

T

hasafixed point z, and

T

is continuousatz.

The fact that

T

has afixed point comes from Singh and Kulshrestha

[16].

To prove continuity, let

{ya

}

C

X,

ya z, and setx y,,, y zin

(13)

toget

Thus

H(Tya,Tz)

<_

qmax{d(ya,z),D(ya,Tya),D(z,Tz),[D(y.,Tz)

+

D(z,Tya)]/2}.

D(Tya,z)

<

q

max{d(y.,z),

D(y.,Tya), O,

[D(y.,Tz)

+

D(z,Tya)]/2}

q

D(ya,z),

q

qD(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,
(8)

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 that

X

be a reflexive space and the range of

T

is the family of allnonempty weakly compact subsets ofX. Heis

apparentlyunawarethat the twostandard definitions of the Hausdorffmetric areequivalent. Theorems1 and 2in Czerwiek

[20]

are specialeasesofCorollary 1, as areTheorem 1of Iseki

[21]

and Theorem1of

Ray

[22].

THEOREM

14. Let

(X,d)

beacomplete metricallyconvexmetric space,

K

anonempty

closedsubset of

X,

T

X

CB(X)

such that there exist

a,/3,

7

>-

0,a

+

2/3

+

27

<

1 such that for all x, yin

X,

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

C

K

and

(a

+/3

+

-),)(1

+

13

+

/)/(1 -/3

3,) <

1,then thereexistsaz in

K

with z ETz.

Moreover T

is continuousat z.

Theexistence ofz is Theorem 1 of Itoh

[23].

To

show continuity, let

{y,}

C

X,

y,, z, and setx y,, y z in

(14)

toget

Then

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 for

H

to getlim,

H(Tz,Tyn)

O,

and

T

is continuousatz.

THEOREM

15.

Let

(X,d)

beacomplete boundedmetric space,F,

X

---,

CL(X),

1,2

satisfying

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 and

aa

a2 or

as

a4. Then

F1

and

F2

have acommon fixed point z.

Moreover

F

and

F2

arecontinuousat z.

Theexistenceofz comesfrom Theorem1 ofBoseandMukherjee

[24].

To

prove continuity of

F,

let

{Yn}

C

X,

y,

--

z, and setx Yn,Y zin

(15)

toget

H(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 that

FlZ

Fz.

Thus
(9)

and

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 and

lira. D(y.,Fiy.)

0. Taking thelimit ofthe inequalityfor

H,

and using the fact that

Fz

Fz,

yields

lim. H(Fy.,Fz)

0, and

F,

is continuousatz.

Similarly,

F

is continuous at z.

THEOREM

16.

Let

(X,d)

beacompletemetricallyconvexmetric space,

K

anonempty closed subset ofX. Let

S,

T" K

--..

CB(X)

satisfying

H(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)

C

K, T(x)

C

K

and

(a+fl+7)(1

+fl+7)/(1-fl-"r)

<

1, then thereexistsazin

K

withz6_

Tz

andz 6-Sz.

Also

S

and

T

arecontinuousat z.

The properties ofz comefrom Theorem 3.1 of Khan

[24].

To

establishthe continuity of

S,

let

{y,,

}

C

X,

y,, z,and set x y,,,y z in

(16)

toget

Thus

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,.,,z

which implies that

lim. D(Sy.,z)

0,and thus that

lim.

D(y,,,Sy.)

0. Settingx y z in

(16)

yields

Sz

Tz. Substitutingintotheinequalityfor

H

yields

H(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,and

S

is continuousatz.

A

similarcalculationverifiesthat

T

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 and

F(T)

F(T.),

n

>

1,where

F(T)

denotesthe fixed point set ofT.

Moreover

the

{T.}

arecontinuous ateachfixed point.
(10)

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)

yields

Tlz

Tr,

z. Substituting in the above inequality yields

limsup{H(Tr,

z,T,y,)}2

0 and each

T,

is continuousat z.

THEOREM

18.

Let

(X,d)

beacompletemetricspace,T1,

T2:

X

-,

CB(X)

satisfying

d’(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. Then

T

and

T2

havecommonfixed points.

Moreover

T

and

T2

arecontinuousateach fixed point.

Thefactthat

T1

and

T2

have fixed pointsisTheorem2of

Popa

[27].

To

establish continu-ity, letzbeacommonfixedpoint

ofT1

and

T2

andlet

{y,}

C

X,y,

z,and setx z,y y, in

(18)

toget

H’n(Tlz,Ty,)

<_

c

d.(/,,,T/,,)

c

<_

c

d’(y.,Tyn),

--(u,Tu)

since

d(y,,T2y,,)

<_

(y,,T2yr,).

Therefore

D(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,

th

T

T

{}.

f

,-’(,T)+-"(z,T)

#

0,

then,

from

(18),

Tz

Tz.

Substituting in the inequality for

H

yields

H(T2z,T2yn)

H(Tz,Ty)

c/D(y,,Ty)

0 n

,

d

T

iscontinuous at z.

Simflly,

T

is continuo atz.

Let

C(X)

denote thenonempty compactsubsets of

X.

A

spe

X

issd to be xo-jointly orbitMlycompleteif

eve

Cauchyquenceofeh orbit atXo isconvergentin

X.

THEOM

19.

Let

Fi

X

C(X),i

1,2,X

xo-jointly completefor some Xo

X.

Suppose

thereexists afunction

(R

+)s

R

+,

upper secontinous d nondecreing ineach

vable,

such that
(11)

satisfies

-),(t)

<

foreach

>

0. Supposethat theF,satisfy

H(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

and

F2

have a common fixed point. Moreover

F1

md

F2

are

continuousat eachfixedpoint.

The existence of a fixed point is Theorem 2.1 of

Guay

et al

[28].

To prove that

F1

is continuousat afixedpoint z, let

{y,}

C

X,y,

--

z,andset x yn,y zin

(19)

toget

H(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 limsup

D(Fly,,z) >

0. Since

D(y,Fly,)

_

D(y,z)

+

D(z,Fy,),

limsup

D(y,

,F1

y,)

_<

limsup

D(z,F1

y,,).

SincezE

F2

z,lim sup

D(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

z

F2z.

Substitutinginthe inequality for

H

wehave

H(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, and

F1

is continuousat z.

A

similarargumentverifiesthat

F2

iscontinuous atz.

THEOREM20. Let

X

beacompletemetric space,

F,

:X

--

C(X).

Suppose

that there existsafunction satisfyingtheconditionsof Theorem 19 and such that

H(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 the

Fi

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 satisfying
(12)

The proofthat there isa commonfixed point zisTheorem3.1of Mishra

[29].

Toprove that

F,

is continuousat z, let

{y.}

C

X,y.

-,z,and set x y.,y zin

(21)

toget

Hx(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)

yields

Fz

F2z.

Substitutingintheinequality for

H

gives

H(r.,rz)

< d(.,r.)

+

d(.,Fz)

+

d(,F.)

+

Ad(.,z).

Takingthelimit asn ooweobtainlim,

Hx(Fy,,FI

z)

O,

and

F

iscontinuousatz. Similarly,

Fz

is continuousat z.

Theresult inMishraandSingh

[30]

isaspecialcaseof Theorem3.1 ofMishra

[29].

THEOREM

22. Let

X

beareflexiveBanachspace,

K

anonempty closed boundedconvex subset of

X.

Let

T

beamappingof

K

intothe family ofnonempty weakly compact convex subsets of

K

satisfying

H(Tx,Ty)

< (max{D(x,Tx),

D(y,Ty)}

(22)

for each x,y in

X,

where

4:

[0,oo)

-,

[0,oo),

nondecreasing, fight continuous, such that

(t) <

for each

>

0. Then thereexists anonemptysubset

M

of

K

such that

Tx

M

for eachx6

M.

Moreover,

T

is continuousat each point of

M.

Thefactthat asubset

M

existswiththe stated propertiesisTheorem 1 of Kaneko

[31].

Toprovethe continuity of

T,

letz 6 M. Let

{y.}

C

X,y.

--,z,and set x y.,y zin

(22)

toget

H(Ty.,Tz)

<_ (max{D(y.,Ty.),D(z,Tz)}).

Thus

D(Ty.,z)

<_

H(Ty.,Tz)

<_ (D(y.,Ty.)).

Assume

that6 limsup

D(Ty.,z) >

0. Then wehave 6

_<

(limsup[D(y.,z)+

D(Ty.,z)])

(6)

<

6,

acontradiction. Therefore6 0 and

T

is continuousatz.

THEOREM

23. Let

(X,d)

be acomplete metricspace,

{S.},

{T.}

sequences of maps from

X

--+

CB(X).

Suppose

that thereexistsan

h,

0

<

h

<

1, such

that,

for each m, n, and eachx,y in

X,

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 thateach

S.,

iscontinuous, let

{y.

}

C

X,

y.--,z,and set x yk,y zin

(23)

toget
(13)

Since

D(SmV,,z) <

H(Smyt:,Tnz),

the aboveinequaltiy yields

D(S.,y,,z)

<_

h

d(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,,,z

T,,z.

Substitutinginthe inequality for

H

yields

H(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 each

S.,

iscontinuous atz.

A

similarargumentverifiesthat each

T.

is continuous at z.

Specialcasesof Theorem 1 ofgubiak

[32]

appear in

Avram

[33],

Iseki

[21],

Popa

[34-

35],

Ray

[36],

aus

[37],

and

Tong

[38].

Achara

[39]

has thesame resultasKubiak

[32],

butwith

CB(X)

replacedby

C(X).

THEOREM24. Let

(X,d)

beacomplete metric space,T,., :X

CB(X).

Supposethat there exist a,

>_

0, 1, 5 such that

min{al

+

a2

+

as

+

2a4,al

+

a2

+

a3-t-

2as}

<

1

andm

#

n imphes

H(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,,}

C

X,

yn

-

z.

Suppose

thata4

<

as.

Then

nfin{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),

weobtain

D(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)

verifiesthat

T,z

T.z.

It then follows that

H(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)

yields

D(T,y,,z)

<_

(a,

+

a2)d(z,y,)

+

a.tD(y,,T,z)

0as k oo. 1-as-as
(14)

Althoughwehave experiencedsomesuccesswithestablishingthe continuity of multivalued maps atafixedpoint,thereaxesomedefinitionsthatdo notlendthemselvesto suchananalysis.

We

citeherethreeexamples.

THEOREM K.

(Khan

and Kubiaczyk

[41],

Theorem

1)

Let

S, T,

X

-,

B(X)

besuchthat,

forsome E :=

{b:

(R+)5

--,

R+

isupper semicontinuous fromthe right, nondecreasing

ineach variable coordinate such that

(t,t,t,

at,

bt) <

foreacht

>

O,a,b

>

O,

with

a+b

<

2},

,5( Sx,Ty) <_

(d(x,y),

,5(x,Sx),

5(y,Ty),

D(x,Ty),

D(y,Tx)

foreachz,

l/in X.

Then

S

and

T

haveauniquecommonfixed pointusuchthatuE

Su

N

Tu.

In

the abovetheoremthe difficultyarisesfromthe fact that

5(u, Tu)

neednot be zero.

THEOREM

M.

(Mukherjee and

Som

[42],

Theorem

1)

Let

(X, d)

beacomplete metric space,T1,

T2

X

--

CB(X)

satisfyinganyoneof the following conditions for eachz, I/in

X:

(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. Then

T

and

T2

haveacommonfixed point.

THEOREM S. (Singh

etal

[43],

Theorem

2.1)

Let

S, T,

be multivalued mappings froma metricspace

X

--.

CL(X).

If thereexists afunction

f

X

--

X

suchthat

SX

U

TX

C_

f(X)

and,foreach

,

yin

X,

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 semicontinuousandnondecreasingwith

b(t) <

tforeach

>

0,

there exists a point

xo

in

X

such that

(S,T)

is asymptotically

regular

at Xo and

f(x)

is

(S,T;f,

Xo)-orbitally complete,then

f,S,

and

T

haveacoincidence point. Further,ifz isa coincidencepoint of

f,

S, T,

and

fz

is afixed point of

f,

then

(a) fz

isafixed point of

S (resp

T)

provided

f

commutes weaklywith

S (resp

T)

at z,

and

(b)

fz

isacommonfixed point of

S

and

T

at z.

ADDED IN PROOF.

1.

A

closerexamination ofthe proofof Theorem

K

shows that

5(u,

Su)

(u, Tu)

0. Thereforeanargumentsimilar to theonealreadyusedrepeatedlyin this paperyieldsthat

S

and

T

axecontinuousat the fixed point.

2. Theorem

M

containsanerrorintheproof.

Moreover,

conditions

(i)

and

(ii)

imply that

Tz

T2z

{z}

foreachz inX. Theparametervaluesonconditions

(iii)

and

(iv)

makeit impossible tousestandardprooftechniquestoobtainafixed point.
(15)

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References

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