A counter example on common periodic points of functions

VOL. 21 NO. 4 (1998) 823-827

A

COUNTER EXAMPLE

ON

COMMON

PERIODIC POINTS OF

FUNCTIONS

ALIASGHARALIKHANI-KOOPAEI

Mathematics

Department

UniversityofIsfahan

Isfahan, Iran

(Received July 19, 1996)

ABSTRACT.

By

acounterexampleweshow that two continuous functions defined on a compactmetric spacesatisfyingacertain semi metricneednothaveacommonperiodicpoint.

KEY WORDS AND PHRASES"

Fixed points,periodicpoints. 1992 AMS

SUBJECT CLASSIFICATION CODES: 26A16,

47H10.

1.

INTRODUCTION

In

[1]

we definedthe notion ofasemi-metric and used it inacontractivetypeinequalityto obtain someresults regarding commonfixed points of two functions.

We

provedTheorem 1.1 andgaveacounterexample illustrating thatwecannot

replace

thecontractive coefficientawith 1.

However,

it isnaturaltoask

(see

[2])

ifit ispossibleto proveaversionof Theorem 1.1 with

(1.1)

amended toread strict inequality, a

replaced

by 1, and with theadditional requirement thatx y,for the situation in whichthefunctionsaredefinedon acompactmetric space

X

Theorem 1.2providesapartialanswer to this question

Here

weshow that ingeneral we can ot expect to provesucharesult.

We

begin withTheorem 1.1 and Theorem 1.2as wellassome preliminariesfrom

[1].

THEOREM

1.1.

Let

f

andg be selfmaps

of

the unitintervalandleth"

I

x

I

[0, oo)

bca

function

having

property

P1.

Suppose

g is continuous on

I

and

A

is a nonempty closed g-znvariantsubset

of

F(f).

If

thereexists arealnumber a, 0

<

a

<

suchthat

for

allx andy

,,,

F(.f),

f

andg satisfy the following inequality:

h(fz,

fy)

<_

a.

max{h(gx, gy), h(gx,

fx),

h(gy, fy),

h(gy,

Ix), h(f

x,

gy)

},

(1.1)

THEOREM

1.2.

Suppose

f

and garetwo

selfmaps

of

acompactmetricspace

X

with continuous, and let h

X

X

[0, c)

be a

function

_{having property}

P1.

If

for

all z 1 in

X. ./"

and#satisfythefollowing inequality:

h(fx,

fy)

<

max{h(gx,gy),h(gx,

fx),h(gy, fy),

h(gy,

f

z),

h(f

z,

gy)

},

(1.2)

then one

of

the following holds:

(z)

either

f

and g have acommon

fized

point.

(zz)

o; every nonempty closedg-invariant subset

of F(f)

contains a perfect minimal set

B

.uch that the

functions

1(x)

h(gx,

x)

and

2(x)

h(x,gx),

do not attain theirminimum or mazimumon

B.

Throughout g"

denotes thenfold compositionof g with itself and

X

is a compactmetric space. Theorbitofz underthehomeomorphismg aonetoonefunction g

),

O(g,

x)

istheset

{gk(x)

-oo

<

k

<

oo}.

A

subset

Y

of

X

iscalled invariantunderg if

g(Y)

C_

Y.

A

closed,

invariant, nonempty subset of

X

is called minimal if it contains no proper subset that is also closed, invariant and nonempty. The sets

P(f)

and

F(f)

arethe sets ofperiodic points and tlie fixed points of

f,

respectively. Thespace

2

{s

(soss...)

sj 0or

1}

is called the sequencespaceonthe twosymbols0and 1.

For

two sequencess

(soslsz...)

and

(totlt2...),

their distance is definedby

dis,

t]

E,__

o z,-t,

I/2’.

It

isclearthat

(E,

d)

isacompactmetric spa.ce.

Let

C

bethe

Cantor

Middle-Third set obtainedasfollows.

Let

Ao

(1/3,

2/3)

bethe middle third ofthe unit interval

I

and

10

1

A0.

Let

A

(1/9,2/9)

U

(7/9,8/9)

be the middle third ofthe two intervals in

lo

and

1

10-

A.

Inductively, let

A,

denote themiddle third of the intervals in 1,_ and let 1,, 1,_

A,,

and

C

I%,>oI,.

For

each x q

C,

weattach

a.n infinite sequenceof

O’s

and

l’s,

S(z)

(0ss2...),

accordingto

the’rule:

so

if x

belongs

to the left component of

I0;

so

0 if z

belongs

tothe right component of

10.

Since x

belongs

tosome component of

1,_,

and 1, is obtained by removing the middlethird of this interval. Thereforewe mayset

s

if x

belongs

to the left hand interval ands,, 0 otherwise.

By

thisway we canthinkofelements of the

Cantor

set

C

aselements of

:E

andviceversa. Define

A

2

by

A(soSS...)

(sossz...)

+

(100...)

mod 2, i.e.,

A

isobtained by adding1 mod 2 to .so and carryingthe result tothe right.

For

example

A(0000...)

(1000...),A(1000...)

(0100...),A(0100...)

(1100...),A(110]’1-0...)= (0011-i’0...),A(111...)

(000...).

The map

A

is knownastheddingmachine

(see

[4]).

2.

RESULTS

LEMMA

2.1

A

is ahomeomorphism

from

C

to

itself.

PROOF. To

see this we show that

A

is continuous, one one, onto on

C

with

A

-

also continuous.

To

see that

A

is continuous, let x be an arbitrary point of

C

and

>

0.

Let

N

be a positiveintegersuchthat

1/2

N

<

_{e. Choose 8}

1/2/v.

_If

d(,x)

<

,

the sequencesx nd y have identical first

N elements,

hence

A(z)

and

A(y)

have also identical first

N

terms. Thus

d(A(x),A(y)) <

1/2

N

<

_{e,implyingthe continuity of}

A

_{at x.}

To

seethat

A

isone one on

C,

let x

(xox...),y

(yoy...)

betwo points of

C

withz y, thenthere existsaleastnonnegativeinteger

N

suchthatXN YlV. Obviously the corresponding elements ofthesequences

A(x)

and

A(y)

aredifferent, hence

A

isone one.

To

seethat

A

isonto, lety

(yoyy...)

and

N

bethe smallest nonnegative integer such that YN 1. Then forx

(ll...0yN+yN+...)

wehave

A(x)

y.

Since

(Ez, d)

isacompactmetric spaceand

A

is continuouson

C

E,

the image ofevery closed subsetof

C

under

A

isaclosedset,implying thecontinuity of

A

-.

LEMMA

2.2. Theorbit

of

everypoint

of

C

under

A

is dense in

C.

PROOF.

Let

x

(xoxx...)

and y

(yoyy...)

be tworbitrarypoints of

C

E.

For

>

0, chooseapositive integer

N

so that

1/2

iv

<

_e.

Let

N

_{be the least positive integer such}

that the sequences x and y haveidentical first

N

elements. Then for

k

2N’- thetwo sequences

A"(x)

and y have at least

N

identicalfirst elements. Similarly suppose

N

is the lea,st positiveinteger such that thefirst

N

elements of thetwo sequences

Aa(x)

andyare identical. Then

N2

>

N

+

and for

k

2N- the two sequences

Ak’+(x)

and y have identical first

N

elements.

By

repeating this processweobtain apositiveintegerm

k

+

k

+

+

k

sch that

A(x)

and y have identical first

N elements,

implying

d(A’(x),y)

<

1/2

g

<

e. Since

:r and ywerearbitrarywemayinterchange the role ofxwith y. Thus the resultis established.

DEFINITION

2.1.

Let

X

beacompact metric space. The functionh

X

x

X

[0,

oo)

issaid tohave property

P

if itsatisfiesthe followingconditions:

(i)"

h(z,y)

0 ifand onlyifx y,

(ii)"

if

lim-oo

x

xo,lirn,,_.oo y yo,and lirr_.oo

h(x,,, y)

O,

the, Xo y0.

Thefollowingtheoremisbasedon anexamplewhich illustrates that assertion

(ii)

of Theorem 1.2 ma.y occur.

THEOREM

2.1. There exist two continuous

functions

f

andg

selfmaps

of

a compact metric space

(X,d),

ag-minimal

perfect

set

B

C_

F(f)

and a

function

h

X

x

Z

--,

[0, oo)

having

property

P,

such that

for

allx

#

y bothin

B,

f

andg satisfy the following:

h(fx,

fy)

< max{h(gx,gy),h(gx,

fx),h(gy, fy),

h(gy,

f

x), h(f

x,

gy)

},

(2.1)

yet

.f

and g do not have a commonperiodicpoint.

g(:r)

A(z),

where

A

isthe addingmachine.

It

isclear that

B

E2

isag-invariantperfect subset of

F(f).

Suppose g

{O(g,x):

x

B}.

By

the axiomofchoicethereisaset

E

such tha.t

E

has exactly oneelementfromeach element of

H.

Choose an arbitrary point

xo

E.

Definethe function h:B

B

[0,

x)

asfollows:

(a): h(t,s)

0ifs t.

(b):

Suppose

M

O(g,

xo)

O(g,

zo).

We

definehon

M

as

(i):

for n,m

>

O,n

:fl

m,

h(g"xo,

g’xo)

h(g’xo,

g"xo)

7-

1/2

(’+").

(ii): For

m

<

0,n 0,

h(g"xo,

g’Xo)

h(g’xo,g’xo)

3

1/2-’.

(iii): For

m

<

0

<

n,

h(g=zo,

g’zo)

h(g’zo, g=zo)

5

1/2

(-’+").

(iv):

For

m

#

n,m

<

0,n

<

0,

h(g’‘xo,

g"xo)- h(g’xo, g=xo)

+

1/2

-(’+’).

(c):

Let

z

E

andz

--

x0,

(i):

foreachintegersmand n, define

h(g’‘z,g’z)

h(g"z,g"z)

h(g’‘xo,g’zo)

h(g’xo, g"zo)

(ii)- For

each integers m and n, m

#

nand each

E,

z

E,

#

z, define

h(g’z,g’t)

h(g’t, g=z)

h(g’‘xo,

g’xo)

(iii):

For

each integersm

-

n, define

h(g’‘z,g’xo)

h(g’xo,

g’z)

h(g"xo, gmxo).

(iv)-

Foreach integernandeach

#

zboth differentfromx0,define

h(g"(z),g"(t))

h(g’*(t),g(z)):

4-

1/2

(’‘+2) ₀

<

_n,

h(g’‘z, g’‘xo)

h(g’‘xo, g"z)

3+1/2-"

n<0.

Since g isonetooneon

B,

thefunction hiswelldefinedon

B

x

B.

Thefunctionhsatisfies tleproperty

P1

since,forx y,

h(z,

y)

0and foreach

(x,

y)

B

x

B,

h(x,

y)

>

1.

It

remains to showthat for every

#

sin

B

the inequality

(2.1)

issatisfied.

To

show this lett,s

B,

and

#

s.

We

distinguish severaldifferentcases.

(:ase 1" Thereexistsx0

E

suchthat

g’‘(xo),

s

g’(xo)

forsomeintegersrnandn.

(i)"

Ifrn nthen

h(t,s)

0, but

h(gt,

s) >

O.

(ii)

If n,m

>_

0 and m

-

n, then

h(t,s)

h(g’‘xo,

g’xo)

7-

1/2

(’’+") _and

h(gt,gs)

h(.q(=+)xo,g(’+)Xo)

7-

1/2

(’++). Hence

h(t,s)

< h(gt,gs).

(iii)

Ifm

< O,

n

O,

h(t,s)

h(xo,g’xo)

3-

1/2

-’.

Suppose

m

<

-1. Then

h(gt,gs)

h(gxo,g(’+)Xo)

5

1/2

-’. Ifm -1,then

h(t,s)

5/2,

but

h(gt,gs)

h(gxo,

g(’+)Xo)

h(gz,xo)

13/2.

Hence

in thiscase wehave

h(t,s)

< h(gt,gs).

(iv)

If m

<

0

<

n, then

h(t,s)

h(gzo,g’xo)

5-

1/2

(-’+’}.

If m -1, we have

h(gt,gs)

h(g("+)xo,g(’+)zo)

h(g("+}zo, Xo)

7

1/2

(’*+).

If m

<

-1, then

h(gt,s)

h(g(+)xo,g’xo)

5-

1/2

(-’+’‘+) _and

h(gt,

t)

h(g{"+)xo,g"xo)

_7-

1/2

("++’)

7-1/2

(’‘+).

Hence

wehave

h(t,s)

< max{h(gt, g),h(gt,s)}

and

h(t,s)

< max{h(gt,gs),h(gt,

t)}.

(v)

Ifm

-7(:

n,m

<

0,n

<

0, then

h(t,s)

h(g"xo,g"xo)

1+1/2

-(’+’).

Suppose

eitherm -1 or n -1. Without loss of generalitywemay assumethat m -1. Then wehave

h(t,s)

nd

h(t,s) < max{h(gt,gs),h(gt,

t)}.

Case 2" Thereexistxl E

E,

x2 E

E

such that

g’(xl),

s

g"(x2)

for someintegers m a.nd n. Ifm

=fi

n, then

h(t,s)

h(g’xl,g’x)

h(g"xo, g’xo)

< max{h(gt,gs),h(gt,

s)}

and

h(t,s) < max{h(gt, gs),h(gt,t)}.

Ifn m

_>

0, then

h(t,s)

h(g"xl,g"x2)

4-

1/2

("+2) and

h(gt, gs)=

h(g(+l)xl,

g("+)x2)

4-

1/2(’+3),

thus

h(t,s) <

h(gt,gs).

Ifn m -1, then

h(t,s)

h(g-x,g-x2)

7/2,

and

h(gt,

gs)

h(x,x2)

15/4.

Ifn m

<

-1, then

h(t,s)

3

+

1/2

and

h(gt,gs)

3

+

1/2

-("+).

Hence

for this case we also have

h(t,)

<

max{h(gt,gs),h(gt,

s)}

and

h(t,s) < max{h(gt,gs),h(gt,

t)}.

We see that inequality

(2.1)

is satisfied for every t,s E

B,

yet

f

and g do not have any common periodic point.

In

fact the set offixed points of

f

is identicalto

B,

while g does not have anyperiodic point in

B.

Remark2.1.

We

maychoose thefunctions

f

and g to be continuousselfmaps ofthe unit interval.

For

examplelet

(am, b,)

},=

be thecomplementaryintervalsofthe middle third

Cantor

set

C.

Define

x

xEC,

f(x)

a,, a,,

<

x

<_

(a,

+

b,)/2,

2(x

b,)

+

b,

(a,

+

b,)/2

<

x

<_

b,,

(2.2)

for n 1,2,

3,...,

g a continuous extension of

A,

and the semi metric h on

C

as in the above theorem.

It

is clear that

F(f)

C

and the orbit of each point outside

C

under

.f (z.e:.{x,f(x),f(x),f3(x),...})is

attracted toapoint in

C,

also

f

andgsatisfy theinequality

(2.1)

on

F(f),

yetthey donothave any common periodic points.

ACKNOWLEDGMENT.

wouldliketothankprofessor

B.E.

Rhoadesfor his comments

(ref-erence[3])

whichled to thiswork. Alsowishtothankprofessor

A. M.

Brucknerfor hisvaluable sggestions.

AUTHOR’S PRESENT ADDRESS.

Department

ofMathematics, The Universityof

Ten-nesseeat

Chattanooga, Chattanooga, TN

37403- 2598.

REFERENCES

1.

ALIKHANI-KOOPAEI, A. A.,

On

common

fixed

and periodicpoints

of

commuting

functions,

InternationalJournal ofMathematicsand MathematicalSciences,to appear. 2.

R

HOADES,

B.E., A

comparison

of

various

definitions

of

contractivemappings,

Trans.

Amer.

Math.

Soc.

226(1977),

257-290. 3.

RHOADES,

B.E.,

Private Communication.