Internat. J. Math. & Math. Sci. VOL. 19 NO. (1996) 103-110
103
FIXED
POINT
THEOREMS FOR NON-SELF MAPS IN
d-COMPLETE
TOPOLOGICAL SPACES
LINDAMARIESAUGA
Dcpartncnt
of Mathematical Sciences TheUnivcrsityof AkronAkron,
Ohio 44325-4002USA
(Received July 7, 1994 and in revised form August 8, 1994)
ABSTRACT.
Fixed point theorems are given for non-self maps and pairs of non-self maps definedond-complete topologicalspaces.KEY
WORDS AND PHRASES. d-complete
topological spaces, fixed points,non-selfmaps,pairs of mappings.1991
SUBJECT CLASSIFICATION CODE. 47H10,
54H25. 1.INTRODUCTION.
Let
(X, t)
beatopological
spaceand d"X
xX
[0, cx)
such thatd(x,
y)
0ifand onlyifx=y.
X
is said to be d-complete if n=ld(xn,
xn+I<
c implies that the sequence{xn}n=l
isconvergent
in(X, t).
Complete
metric spacesandcomplete
quasi-metric spacesareexamples of d-complete topological spaces. The d-complete semi-metric spaces form an
importantclass ofexamplesof
d-complete
topologicalspaces.Let
X
be an infinite set and anyT
non-discrete first countabletopology
forX.
There exists acompletemetricdforX
suchthat<
dandthe metrictopology
dis non-discrete.Now
o is
Cauchy
in d.(X
d)
isd-complete
since-,oo
n=ld(xn,
xn
+
1)
<
c implies that{xn}
nThus,
xn--*x,asn o,in dand thereforeinthetopology
t. Theconstructionof disgivenby
T. L.
HicksandW.
R.
Crisler in[1].
Recently, T. L.
Hicks in[2]
andT. L.
Hicks adB.
E.
Rhoadesin[3]
and[4]
proved
several metric space fixed point theorems ind-complete
topological spaces.We
shall prove additional theorems in thissetting.Let
T X
X
beamapping.T
is w-continuous at xifXn ximpliesTxn
xA
real-valued functionG"
X
[0, c)
is lower semi-continuous if and only ifn}n=O
sequence inX
andlim Xn pimpliesG(p) <
lim infG(xn).
2.
RESULTS.
In
[2],
Hicksgave thefollowing
result.THEOREM
([2],
Theorem2)"
Suppose X
is ad-complete
Hausdorff topological space,T" X
X
isw-continuousand satisfiesd(Wx,
W2x)
_<
k(d(x,
Wx))
for allxEX,
whereTnx
Tp.
[k
is not t.h(,r(, exists x inX
with xkn((l(x, Tx))<
oc.In
this case,x,,
assumed to 1,c (’onti,mousdud
k2(d)
k(k(a)).]
The following conditi(ms are exa,nincd.
Let
T:C
X
withC
a closi’d subs(’t of thed-,’omph’t("
topological spa,’,;X
dudC
CT(C). Let
k:[0,
cxz)
[0,
exz)
1)(" su(h thatk(0)
0, kis,o,-(lccr(’asing.au(l
k(d(Tx,
Ty))_>
d(x,
y)(2.1)
for ,illx,y
C,
orfor all x,y G
C,
orforall x,y
C,
ord(Tx,
Ty)
>_ k(d(x, y))
(2.2)
d(x,
y)>_
k(d(Tx,
Ty))
(2.3)
k(d(x,
y))
_> d(Tx,
Ty)
(2.4)
for all x,y
C.
Itwill be shown that condition
(2.1)
leadstoafixed point, but that the other three conditions do notguarantee
afixed point.THEOREM
1.Suppose X
is ad-complete
Hausdorff topological space,C
is aclosed subset ofX,
andT" C
X
is anopenmappingwithC
CT(C)
which satisfiesd(x,
y)
_<
k(d(Tx,
Ty))
for all x, yC
where k"[0, oo)--,
[0, oc), k(0)=
0, and k is non-decreasing. ThenT
has a fixed pointif andonlyif there existsx0C
with-oOn=l
kn(d(Tx0’
x0))
<
PROOF.
Noticethat theconditiond(x,
y)
_< k(d(Tx,
Ty))
forcesT
tobe one-to-one.Hence
T-
exists.Also,
T
is openimplies thatT-
is continuous, and thusw-continuous.Ifp
Tp
thenSuppose
there exists x0C
such that-OOn=l
kn(d(Tx0
x0))
<
oo.We
know thatT-1
exists,soletT
beT-
restricted toC.
ThenT :C
---+C
andd(Tlx
TlY
_< k(d(x,
y))
for allx,
y
C. Let
yTlX.
Thend(TlX
W21
x) < k(d(x,
TlX))
for allxC.
In
particular,d(TlX0,
Tx0)
<_
k(d(x0,
TlX0)
<_
k2(d(Wx0,
x0)
).
By
induction,d(T-
lx0,
Wx0)
_<
kn(d(Tx0,
x0)
).
Thus,
E
d(T-lx0,
T?x0)
_<
E
kn(d(Tx0 x0))
<
oo.k=l k=l
Since
X
is d-complete,Tx
0 converges, say to p.Note
that p is inC
sinceC
is closed.Now
TI(Tx0)
TlP
as n400 sinceT
is w-continuous.But
T
+
lx
0 p as n-oo, and sincelimits are unique in
X,
Tlp=p.
Now
T(Tlp
)=T(p)
andT(TlP
)=p
soTp=p
andT
has afixed point.
COROLLARY
1.Suppose T"
C
X
whereC
isaclosed subset ofad-complete
Hausdorff symmetrizabletopological
spacewithC
CT(C).
Suppose
d(x,
y)
< [d(Tx,
Wy)]
pwhere p>
1for all x,yC.
If thereexists x0C
suchthatd(Tx0, x0)
<
1, thenT
has afixedpoint.PROOF.
Ifx#
y, 0<
d(x,
y) _<
[d(Tx,
Ty)]
pandTx
Ty.
ThusT
isone-to-one andFIXED POINT THEOREMS IN A TOPOLOGICAL SPACE 105
Ol)(,n. L,t x
0lwa
point in C.lullthatd(Tx
0.x0)<l.
Ifd(Tx
0,x0)=0.
thcnxllisafixcdpoint
fT.
Sqposc O<d(Tx
O.x0)<l.
L,tk(t)=t
p.
and=d(Tx
O.x0).
Note
that(ot)t’<o,t
pif 0<
o<
1. Since p<
t. thc. i an (! G(0. 1)
smh that p (1t. Now(t/’)
1’<
1’and there is’p p p 2p p
an
o.G(0.
1)
such that t"=.et
But
o.et
=t=(t
p)
=(olt)P<oltt’
Hence
o.<o
1.Now 2p
o.2tP
o.2<
(Ol)2t.
Assunm
"p<
]t.
Thent(,+
l)t,=
(t,,p)p
<
(,],t)t,
(],pt
p(,],pnl
,,],,
+
l)Pt.
Hen,,’. l,vindu,tion, ’’t’<
(]’t
for allk"(d(Tx
o,xo)):
[d(Txo,xo)
’’p< o ,< since0<
o<
1. Applying Theorem 1, wegetthatT
hasafixed point.If
T
isnot openonecould check thefollowing condition.THEOREM
2.Let X
1)ca d-conpletc Hausdorff topological space,C
be aclosed subsetofX,
T" C
X
withC
CT(C). Suppose
thercexistsk’[0, o)
[0, c)
such thatk(d(Wx,
Wy))
_> d(x,
y)
for all x,yGC,
kisnon-decreasing,k(0)
0, and thereexists x0C
such thaty
kn(d(Tx0 x0)
<
cxz. IfG(x)=
d(Tx,
x)is
lower semi-continuousonC
thenT
hasafixed point.
PROOF.
Ifx y, 0< d(x,
y)
< k(d(Tx, Ty))
sothatd(Tx,
Ty)
7
0.Hence
T
is one-to-one andT-
exists.Let
T
beT-
restricted toC. Now
T
C
C
and forxC,
d(x,
WlX
<
k(d(Tx, x)),
d(TlX,
Tx,)
_<
k(d(x,
TlX))
<_
k2(d(Tx,
x)).
By
induction,d(W?-lx,
W?x)
_<
kn(d(Tx, x)).
There exists x0e
C
withyn=l kn(d(Tx0
x0))
<
cx impliesn=l d(W?-lx0,
Tx0)
<
cx. SinceX
is d-completethereexists pe
X
such thatTx
0 p as n cx.Note
that pC
sinceTx
0C
for all n andC
isclosed.Now
G(x)
d(Wx,
x)
islower sexni-continuous onC
givesG(p) <
lim infG(T?x0)or
d(Wp,
p) <
lim infd(W-lx0
Wx0,
0. ThusTp
p.In
[5],
Hicksgivesseveral examplesof functions kwhichsatisfy theconditionof theorem of thatpaper. Theseexamples,
withaslight modification,carryovertothenon-selfmap case. The non-selfmap version ofExample
isgivcn for completeness. The otherexamples
carryoverinasimilarmazmer.
EXAMPLE
1.Suppose
0<,<1.Let
k(t)=At
for t>0. Ifd(x, y)<Ad(Tx, Ty),
T
is open sinceT-
existsand iscontinuous.Let
xC.
There exists yC
such thatTy
x.Now
d(x,
y)
d(Ty,
y)
<
Ad(W2y,
Ty)
andy
kn(d(Ty
y))
<
yo
And(W2y,
Wy) <
cx.n=l n=l
Applying Theorem 1 we
get
afixed point forT.
(Note"
d(x,
y)
_<
A
d(Tx,
Ty)
for 0<
,
<
1 is equivalenttod(Tx,
Ty)
>
ad(x,
y)
foro> 1.)
The following
examples
show that conditions(2.2),
(2.3)
and(2.4)
do notguarantee
fixed points.EXAMPLE
2.Let
R
denote the real numbers andCB(IR,
I)
denote the collection of all bounded and continuous functions which map1into1.Let
Define
T"
C
CB(,
)
byTf(t)=
1/2f(t
+
1)
and1,,tk(t)=
-.
Thend(Tf,
Tg)=
1/2d(f,
g)
_>
l:(d(i’,
g)). ksatisfies (ondition(2.2)but,
as shownin[6],
W
does not haveafixed point.
EXAMPLE
3. LetT’[1,
cx3)-
[0,
cx3)
1,c d,’fined byTx
X-x
1-
and h’tt:(t)=,.
Th,’nd(Tx,
Ty)
<
2d(x,
y)
ord(x, y)
_> t:(d(Tx,
Ty)).
ksatisficscondition(2.3)
butT
do,,snot hav,,afixedpoint.
EXAMPLE
4.Let
c0 denote the collection of all sequences that converge to zero.Let
C={xc
0:llxll=l
andx0=1}.
DcfincT:Cc0byTx=ywhcrcyn=x1+l,n=0,
1,2,..., and letk(t)
2t. Thend(Tx,
Ty)
d(x, y)
_<
2d(x,
y)
k(d(x,
y))
for all x, yC.
ksatisfies condition(2.4)
but,
asshovnin[6],
T
doesnothaveafixed point.Thefollowing theoremsweremotivatedbythe work of Hicks andRhoades
[3].
THEOREM
3.Let C
be a compact subset of a Hausdorff topological space(X, t)
and dX
xX
[0, co)
such thatd(x,
y)
0ifand only if x y.Suppose
T C
X
withC
CT(C),
T
andG(x)=
d(x,
Tx)
areboth continuous, andd(Tx,
T2x)>
d(x, Tx)
for allxT-I(C)
withx
Tx.
ThenT
hasafixedpoint inC.
PROOF.
C
is a compact subset of a Hausdorff space so it is closed.T
is continuous soT-I(C)
is closedand hence is compact sinceT-I(C)C
C.
G(x)
is continuous so it attains itsminimumon
T-
I(C),
say at z.Now
zC
CT(C)
sothere exists yT-
I(C)
such thatTy
z.Ify z then
d(z, Tz)
d(Ty,
T2y)
> d(y, Ty),
acontradiction. Thusy zTy
isafixedpoint ofT.
THEOREM
4. LetC
be a compact subset of a Hausdorff topological space(X,
t)
and dX
X
[0, co)
suchthatd(x,
y)
0 if andonly
if x y.Suppose
T C
X
withC
CT(C),
T
andG(x)=
d(x, Tx)
are both continuous, f:[0, co) [0, co)
is continuous andf(t)>
0 for 0. Ifweknow thatd(Tx,
W2x)
_<
f(d(x, Tx))
for allxT-
(C)
impliesW
has afixedpoint where 0<
<
1, thend(Wx,
T2x)<
f(d(x, Wx))
for all xT-I(C)
such thatf(d(x,
Tx))
0 givesafixedpoint.PROOF. C
is acompact
subset ofaHausdorffspace soit is closed.T
is continuousgivesthat
T-(C)
isclosed,
andT-I(c)C
C
soT-(C)
is compact.Suppose
xTx
for allxW-(C).
Thend(x, Wx)>0
so thatf(d(x, Wx))>0
for allxW-l(c).
DefineP(x)
ond(Tx
T2x)
T-I(C)
byP(x)=
f(d(x, Tx))"
P
is continuous sinceT,
f andG(x)
arecontinuous. ThereforeP
attainsits maximumon
T-
I(C),
say atz.P(x) < P(z)
<
sod(Wx,
W2x)
_< P(z)f(d(x, Ix))
andT
must haveafixedpoint.THEOREM
5.Let C
be a compact subset of a Hausdorff topological space(X, t)
and dX
xX
[0, co)
suchthatd(x,
y)
0 ifandonlyif x y.Suppose
T C
X
withC
CT(C),
W
andG(x)---d(x, Ix)
are both continuous, f:[0, co) [0,
co)
is continuous andf(t)>
0 for 0. Ifweknowthatd(Wx,
W2x)
_>
A
f(d(x, Tx))
for all xT-
(C)
impliesW
hasafixedpoint wherA
>
1, thend(Wx,
T2x)
>
f(d(x, Tx))
for all xT-
I(C)
such thatf(d(x, Tx))
0gives aFIXED POINT THEOREMS IN A TOPOLOGICAL SPACE 107
PROOF.
C
is acompact subset ofa Hausdorff spaceso it is closed.T
is continuousgiw,sthat
T-(C)
is closed and hence compact, sinceT-(C)
c
C.
Suppose
x#
Tx
hr alld(Tx,
T2x)
P
is continuousx
T-(C).
Thend(x,
Tx)
>
0andf(d(x, Tx))
>
0. DefineP(x)
f(d(x Tx))"
sin"
T,
andG
arccontimmus.P
attains it minimumonT-
I(C),
sayat z.P(x)
P(z)
>
so,l(Tx,
T2x)
P(z)f(d(x, Tx))
andT
musthaveafixed l,oint.Theorems fi, 7 and 8 arcgeneralizations oftheorems by
Kang
[7].
The following family of real functionswas originally introducedbyM. A.
Khan,M. S. Khan,
andS. Sessa
in[8].
Let
denotethefamily of allreal fimctions
6"(R
+
)3
R
+
satisfying the followingconditions:(C
l)
islower-scmicontinuous ineachcoordinatevariable,(C2)
Let
v,wR +
besuchthat eithervd(v,
w,w)
orv(w,
v,w).
Thenvhw,
where(1,
1,1)=h>1.
THEOREM
6.Let
(X,
t,d)
be a d-complete topological space where d is a continuous symmetric.Let
A
andB
mapC,
a closed subset ofX,
into(onto)
X
such thatC
CA(C),
C
CB(C),
andd(Ax,
By)
>
$(d(Ax,
x),
d(By,
y), d(x,
y))
for all x,y inC
where E.
ThenA
andB
have acommonfixedpointinC.
PROOF.
Fix x0EC.
SinceCcA(C)
there exists xIC
such thatAx
l=x
0.Now
x oo by
C
CB(C)
so there existsx2C
such thatBx2=x
1. Build the sequencen}n=O
Ax2n
+
x2n,Bx2n
+
2X2n
+
l"Now
ifX2n
+
x2n
forsomen, theX2n
+
isafixedpoint ofA.
Thend(x2n
+
1,X2n
+
1)
d(x2n,
X2n
+
1)
d(Ax2n
+
1,Bx2n
+
2)
->
(d(Ax2n
+
1,X2n
+
1), d(Bx2n
+
2,X2n
+
2), d(x2n
+
1’X2n
+
2))
(0,
d(x2n
+
1,X2n
+
2), d(x2n
+
1,X2n
+
2))
By
property(C2)
d(x2n
X2n+
1)
>-
hd(x2n+
1,X2n+2)"
Hence, X2n+
X2n+2
andBx2n
+
Bx2n
+
2X2n
+
1" ThereforeX2n
+
is a common fixed point ofA
andB.
Now
ifX2n
+
X2n
+
2for somen, thenBx2n
+
2Bx2n
+
X2n
+
2" Thend(x2n
+
2,X2n
+
I)
d(Ax2n
+
3,Bx2-
+
2)
->
(d(Ax2n
+
3,x2n
+
3),
d(Bx2n
+
2,X2n
+
2),
d(x2n
+
3’X2n
+
2))
(d(x2n
+
2,X2n
+
3), d(x2n
+
1,X2n +
2), d(x2n
+
3,X2n
+
2))"
By
property(C2)
d(x2n+l, x2n+2)>-hd(x2n+2, X2n+3)r
X2n+2=x2n+3.
Ax2n
+
2Ax2n
+
3x2n
+
2andx2n
+
2isafixedpoint ofA
also.Suppose
xn #
xn+
foralln. ThenThus
d(x2n,
X2n
+
1)=
d(Ax2n
+
1,Bx2n
+
2)
>--
(d(Ax2n
+
1,X2n
+
1), d(Bx2n
+
2,X2n
+
2), d(x2n
+
1,X2n
+
2))
Again by
(C2) d(x2n
x2,,+1)
->
hd(x2n
+
1’x2,,+’2)
ord(x,2n
+
1,x2n+
2)
-<
d(x2n,
x2n+
1)"
Also,
d(x2n
+
1’X2n
+
2)
d(Ax2n
+
3’Bx2n
+
2)
->
$(d(Ax2n
+
3,x2.
+
3), d(Bx2n
+
2,x2.
+2), d(x’2n
+
3’x2n +
2))
(d(x2n
+
2,x2n +
3), d(x2n
+
1,x2n +
2), d(x2n
+
3’X2n +
2))"
By
(C2)
wegetd(x2n
+
2,x2n
+
3)
-<
d(x2n
+
1,x2n
+
2)"
Inductiongives()n
+
n
,.
d(Xn+l,
xn+2)
d(x0, xl).
ThusE
d(Xn+l’Xn+2)
[)
d(x0, x,)
<.X
n=l --1
is d-complete so xnp as n where
p,
since is closed.We
also hve x2np dxn+lP
as n. This givesAx2n+lp
andBx2n+2p
n. Sincep,
p
A(C)
andpB(C),
sothere existv,wC
such thatAv
pandBw
p.Now
d(x2n,
P)
d(Ax2n
+
1,Bw)
>
(d(Ax2n
+
1,x2n
+
1),
d(Bw,
w),
d(x2n
+
1,w)).
Since is
lower-semicontinuous, letting
n oogivesd(p, p)
>_ (0,
d(p,
w),
d(p,
w))
andby
(C2)
wehave0
>
hd(p,
w).
Hence
p w.Also,
d(p,
x2n
+
1)
d(Av,
Bx2n
+
1)
> (d(Av, v),
d(Bx2n
+
2,x2n
+
2),
d(v,
X2n
+
1))"
Letting n-oo gives
d(p,
p)_> (d(p,
v),
0,d(v,
p))
or, by(C2)
0_>
hd(p,
v).
Hence
p v.Therefore,
Ap
Av
pBw
Bp.
COROLLARY
2.Let
A
andB
mapC,
a closed subset ofX,
into(onto)
X
such thatC
CA(C),
C
CB(C),
andd(Ax,
By)
_>
ad(Ax,
x)+
bd(By,
y)
+
cd(x,
y)
for all x, y6C,
where a,b,
and c are non-negative real numbers with a<l,b<l,
anda+b+c>l.
ThenA
andB
haveacommonfixedpointinC.
The
proof
ofCorollary
2isidenticaltotheproof
of Corollary 2.3in[9].
In
[7],
Kang
defined q* to be the family of all real functions o--,(R+)3--.
R
+
satisfying
condition
(C1)
and the followingcondition:(C3)
Let
v, w 6R +
{0}
be such thateither v> qo(v,
w,w)
or v>
qo(w,
v,w).
Then v>
hw,
where
p(1,
1,1)
h>
1.Kang
showedthat thefamilyq*isstrictlylarger
than thefamily.
THEOREM
7.Let
(X,
t,
d)
be ad-complete
Hausdorfftopological
space where d is acontinuoussymmetric. If
A
andB
arecontinuousmappings fromC,
aclosed subset ofX,
intoX
such thatC
CA(C),
C
CB(C),
andd(Ax,
By)
> (d(Ax, x),
d(Sy, y),
d(x,
y))
for all x, ye
C
such that x
#-y
whereo
q*,
thenA
orB
has afixed point orA
andB
have a commonfixed point.PROOF. Let
{Xn}
n=0
bedefined asintheproof
of Theorem 6. Ifxn
xn+
for some nthen
A
orB
has a fixed point.Suppose
xn#
xn+
for all n.As
in theproof
of Theorem6,
oo and{X2n
}
x oo andhencexnp as n--,oo.
Now
{X2n}n=0
+
n=0 are subsequences ofn}n=l
each
converges to p. SinceA
andB
are continuous,Ax2n
+
X2n
Ap
andFIXED POINT THEOREMS IN A TOPOLOGICAL SPACE 109 C()R()LLARY3. L(’t
A
andB
bc continuousmappings fromC,
aclosedsubset ofX,intoX
sat,sfyingC
CA(C),
C
CB(C),timd(Ax, By)
hlnin{d(Ax, x),
d(By,y), d(x, y)}
for 11 x,yC
with x#y whereh>1. TlwnA
orB
has afixcdl)oint orA
andB
have a commonfixed point.PROOF.
Note
that(tl,
t2,
ta)
hmin{t,
2.t3},
h>
isin’I’*.
Apply Thcol,’m7. IfA
B
inCorollary 3w.get ageneralizationofTheorem 3 in[9].
Boyd and
Wong
[10]
(’all the collection of all real functions:+
+
vhich satisfy the follmvngonditaons:
(C4)
isut)I)er-semicontinuousandnon-decreasing,(C5)
(t)<t
for eacht>0.THEOREM
8.Let
(X,
t,d)
be ad-complete
symmetric Hausdorff topological space. IfA
and
B
are continuous mappings fromC,
a closed subset ofX,
intoX
such thatC
CA(C),
C
CB(C),
and(d(Ax,
By))
>_ ,nin{d(Ax, x),
d(By, y),
d(x, y)}
for all x,yGC
where,
e
andcn
0l’)n(t)
<
oo for each>
0, then eitherA
orB
has a fixed point orA
andB
have a comnonfixed point.PROOF. Let
{xn}
n=0
be defined as in theproof
ofTheorem 6.Ifxn=Xn+l
for some nthen
A
orB
hasafixed point.Suppose
xn
xn+
forall n. Then’(d(x2n,
X2n +
1))
(d(Ax2n
+
1,Bx2n
+
2))
_>
mind(Ax2n
+
1,x2n
+
1), d(Bx2n
+
2,X2n
+
2), d(x2n
+
1,x2n +
2)
min{d(x2n,
X2n+
1), d(x2n
+
1,X2n+2),
d(x2n
+
1,X2n
+2)}
d(x2n
+
1,X2n +
2)
since
(t)
<
for all>
0.Similarly,
d(x2n
+
2,X2n
+
3)
-<
(d(x2n
+
1,X2n
+
2))
and henced(x
+
1,Xn
+
2)
-<
(d(xn,
Xn +
1))
for eachn. Since isnon-decreasing,
d(x
+
1,Xn
+
2)
-<
n(d(x0, x)).
NowE
d(xn,
Xn +
1)
--<
E
n(d(x0 Xl))
<
cxz.n=0 n=0
The space
X
is d-complete so there exists pC
such that Xn p as n- cx. The mappingsA
andB
arecontinuous soAx2n
+
X2n
Ap
andBx2n
+
2X2n
+
Bp.
Limits areuniquesoAp
pBp.
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L.
HicksandW. R.
Crisler,A
noteonT
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L.
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