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Internal. J. Math. & Math. Sci.

VOL. 18 NO. 4 (1995) 677-680

677

A

NOTE ON FINITE CODIMENSIONAL LINEAR ISOMETRIES OF

C(X)INTO C(Y)

SIN-ElTAKAHASI

Departmentof BasicTechnology AppliedMathematicsand Physics

Yamagata

University, Yomezawa 992, JAPAN

and

TAKATERU OKAYASU

Department

of Mathematics FacultyofScience

Yamagata

University,Yamagata990,JAPAN

(Received April 24,1994and in revisedform

May

25, 1995)

ABSTRACT. Let

(X, Y)

be apair of compact Hausdorff spaces. It is shown that a certain property of the class of continuous maps ofY onto

X

is equivalent to the non-existence oflinear isometryof

C(X)

into

C(Y)

whoserangehas finite codimension

>

0.

KEY WORDSANDPIRASES. CompactHausdorff space,

C(X),

linearisometry,finitecodimension 1991AMSSUBJECTCLASSIFICATION CODES. 46B04,46J10

1. INTRODUCTION

In ], A. Gutek, D.

Hart,

J. JamisonandM. Rajagopalan provedthat there arenoisometric shift operators on

C([a,b]),

aresultfirstproved inthe real scalarsbyHolub

[3].

Here

[a,b]

is any closed interval inthe reallineand

C([a, b])

isthe Banach space of all continuouscomplex-valuedfunctions on

[a,b].

By observing carefully the proofgiven in [1], one can note that

C([a,b])

does not admit an

isometricshittoperator because thespace

[a, b]

has the property that theset

{(, u) [, b] [, hi: () (), # U}

is infinitefor everycontinuousmap of

[a, b]

ontoitself which is notinjective.

The purpose of thenoteistoprovethe following theoremwhichisbasedonthe above idea:

TI]EOREM. Let

(X, Y)

be apair

of

compact

Hausdorff

spaces. Then thefollowingtwo conditionsareequivalent:

(i)

If

thereisa continuousmap

of Y

ontoXwhichisnotinjective, then theset

{(, ) e r r: () (), # }

is

infinite.

(i

0 If

there isalinear isometry

of G(X)

into

C(Y)

which hasa

finite

codtmension, thenit is surjective.

Since both

([0, 1], [0,1])

and

(T1,T1)

satisfy the condition(i), where

Tlis

the unit circle inthe complexplane,weget fromthis

(2)

678 S-E TAKAttASIAN[)T()KAYASU

COROLLARY 1. "lheonlypossiblecodtmenston

of

hneartsometrtes

(7([0, 1] C([0, 1])

and

C(T C(T

are zeroor

mfimte.

Moreover,ifV isthe canonical linearmap

ofC(T

into

C([0, 1])

definedby

(Vf)(t)- f(e ’’) (f EC(T 1),O<_t <

1),

thenV isansometry and therangeofVisthesetofall9 E

C([0, 1])

such that

9(0) 9(1)

HenceV has codimension 1, and if there isafinite codimensional linearisometryof

C([0, 1])

into

C(T

), sayT, then VT is a linear isometry of

C([0,1])

into itself such that

VT(C([O, 1]))c C([0,1])

and

codim(T) +

From Corollary itfollows that VTmustbesurjective, acontradiction hencewehave alsoproved

COROLLARY2. /heretsno

fimte

codtmenstonallineartsometry

of C([O, 1])

into

C(T

2. LEMMAS

Inordertoprove themaintheorem,wehavetopreparesomelemmas

LEMMA 1. LetXbeacompact

Hausdorff

space,

M

asubspace

of C(X)

whosecodtmenstonts n

< +

oo, and tfaclosedboundary

of

Xwith respecttoM O.e.,

for

any

f

Mthereexistsa point:c

m tf wuh

f(z)t If t,,,

thesupremumnorm

off

onX). 7hentheset

X\K

hasat most n points.

PROOF. Assumethat

X\K

hasatleastn

+

points, say:rl :r,,+ Foreach

_< <

n

+

1,

chooseafunction

f,

n

C(X)

suchthat

f,(z)

and

f,(a:)

0forz K

{a:, z,+}\{a:,}

since Ksclosed Inthiscase,

{fl +

M f,,+

+ M}

is linearlyindependentin

C(X)/M

since if

c(ft +

M)

+ +

c,-I

(f,,+l +

M) 0

for some complex numbers cl,...,c,_i there exists a function 9 M such that C

lfl + +

c,+f,+l +

9-0 and (snce K is a boundary ofX with respect to M) a point :c0 in K such that

[19 9(Zo

Then

t9 Ix, Clfl(a:0) + + Cn+lLz+l(X0)l

0,

implying c 0, c._l 0 since

{fl,--,f,+l}

is hnearly independent, and it follows that

codim(M) >

rz

+

LEMMA2. LetXandYbecompacl

Hausdorff

.spacesand acontinuousmap

of

YontoX.

If

g s a

fimcton

m

C(Y)

such that

g(yl)=9(y) for

all pars

(Yl,Ye)

Y Y sansfymg

(Yl) (Ye),

thentheres a

fimcton f

m

C(X)

such thatf((y)) g(y)

for

all y Y.

PROOF. Let g be a function in

C(Y)

such that

9(Yl) g(Y2)

for all pairs (y,y.,.) Yx Y satisf3,ing

(y) (y.:)

Let

Y/

bethe quotientspaceof

Y

definedby

,

7r thecanonicalmap ofY

onto

Y/,

and r the canonical map of

Y/

onto X Then the complex-valued function

.

on

Y/

definedby

.()

g(y)foreach

9 Y/

iscontinuous,sosetnng

f .

---1 it iseasytoseethat

f

isa

functionwiththe desiredproperties

Finally,wewillneedthefollowingresult whoseproofisstraightforward

LEMMA 3. Let X be a compact

Hausdorff

space, K a compact subset

of

X, and

AI.

the

Banach subspace

of C(X)

conststmg

of

all

f C(X)whtch

are constant on K. Then the Banach space

C(X)/Aa-

tstsomorphtctoaquotwntspace

of C(K).

3. PROOFOF THEOREM

(1) (n) Let Tbeahnear isometryofC(Xi into

Ct,

Y) which has a fimte codlmension Bythe decomposition theorem of Holsztynskl [2], there exists a closed boundary K of Y with respect to

TiC(X)), acontinuousmaphofKontoX,andacontinuous unimodular functionuonYsuchthat

(T f )()

(3)

FINITE CODIMENSIONAL LINEAR ISOMETRIES 6 9

for all

.f

E

C(X)

and y EK SinceT hasa finitecodimension, itfollows fromLemma that Kis a closed subset ofYwhose complementis a finiteset. Then h hasacontinuousextension toY,say

.

We claim that the map

.

is injective Assume the contrary Then by the condition (i) there is a mutuallydifferentsequence

{a1,/31,a2,2,...

in

Y

such that

(a,) z(/3,,)

for allpositiveintegers n, andwherewe canassumewithoutlossof generality that a

l,/31, a2,/32,

CK. Letnbeanypositive integer, and for each 1

_< <

n choose a function 9, in

C(Y)

such that

9,(az)=

1 and

9,(Y)=

0

for all y

Y\U,,

where

U,

is a sufficiently small neighborhood of cq. In this case

{91 + T(C(X))

9,

+ T(C(X))}

islinearly independentin

C(Y)/T(C(X)),

since if

Cl(g -- T(C(X))) + + c,(9, + T(C(X)))

0

forsomecomplexnumberscl, c,thereexists

f C(X)

such thatc191

+ +

Cn9 Tf, implying

q

(,) + +

(T]) (,) (,)/(h(,)) u(,)f(h(,)) (Tf)($,) (,)

(,) {,z(,) + + (,)

=0

for each 1 n. Itfollows thatThasan infinite codimension sincenisarbitrary, acontradiction.

Consequently, must be injective, K

Y,

and h is a homeomorphism of

Y

onto X. If for any g

C(Y),

we set

1 g(h_l(x))

y(z) (h_()

for eachx E

X,

thenwe obtainthat

f C(X)

and

Tf

g, sothat

T

issurjective.

(ii)

=

(i). Let be a continuousmapofYontoXwhich isnotinjective. Thenwehave toshow thattheset

{(u, ) e r

’.

() (), u #

isinfinite under the condition(ii). Ifnot, then all

-1 (x)(x _ X)

arenon-emptyfinite sets, and also

{x e X" card(-l(x))> 2}

is a non-empty finite set, say

{xl ,x,,},

where "card" denotes the cardinalnumber. Set

(Tcf)(y) /((y))

for each

f C(X)

and y

Y.

Then

T

is a linearisometry of

C(X)

into

C(Y)

and since isnot injective,itfollowsthat

T

isnotsurjective. Put

A, {g C(Y)"

g is constanton

-1(x,)} (i

1,

n)

and

A {g C(Y)

g is constant on i=l

U -(,) }

n

Then

A

C_

I"1

Ai, and hence

C(Y)/ ["] A,

is isomorphic to

(C(Y)/A)/I,

where I=

i=1 i=1

{g+ A C(Y)/A’g ["] A,}.

On the other hand,

T(C(X))= f"l A,

since the inclusion

i=1 i=1

(4)

680 S-ETAKAtlASIANDT OKAYASU

TO(C(X))

C_

I’]

A, istrivial, and the reverse inclusionfollows immediatelyfromLemma2 Also by Lemma3,

C(Y)/A

isisomorphictoaquotient of

C(Y0),

where I/’o

[,_J -(a:,)

Consequently,

i=1

codim(T) dim(C(Y) /To(C(X)

<_ dim(C(Y)/A)

<_ dim(C(Y0))

< card(-l(x,))

Hence

To

has a finite codimension, and so must be surjective by the condition (ii) contradiction,sothe implication isproved

But this is a

ACKNOWLEDGMENT. The authors thank the referees forhelpful commentsandfor improving the paper Thesecond author waspartiallysupportedbythe Grant-in-Aid for Scientific Researchfromthe ministry of Education,Scienceand Culture inJapan

REFERENCES

[1 GUTEK, A.,

HART,

D., JAM/SON, J and RAJAGOPALAN, M., Shift operators on Banach spaces,J. Funct.Anal. 101(1991),97-119.

[2]

HOLSZTYNSKI, W., Continuous mapping induced by isometries of spaces of continuous functions,Studia Math. 124(1966), 133-136.

[3] HOLUB, J.R., Onshift operators,Can.Math. Bull. 31(1988),85-94.

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