Internal. J. Math. & Math. Sci.
VOL. 18 NO. 4 (1995) 677-680
677
A
NOTE ON FINITE CODIMENSIONAL LINEAR ISOMETRIES OFC(X)INTO C(Y)
SIN-ElTAKAHASI
Departmentof BasicTechnology AppliedMathematicsand Physics
Yamagata
University, Yomezawa 992, JAPANand
TAKATERU OKAYASU
Department
of Mathematics FacultyofScienceYamagata
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 ontoX
is equivalent to the non-existence oflinear isometryofC(X)
intoC(Y)
whoserangehas finite codimension>
0.KEY WORDSANDPIRASES. CompactHausdorff space,
C(X),
linearisometry,finitecodimension 1991AMSSUBJECTCLASSIFICATION CODES. 46B04,46J101. INTRODUCTION
In ], A. Gutek, D.
Hart,
J. JamisonandM. Rajagopalan provedthat there arenoisometric shift operators onC([a,b]),
aresultfirstproved inthe real scalarsbyHolub[3].
Here[a,b]
is any closed interval inthe reallineandC([a, b])
isthe Banach space of all continuouscomplex-valuedfunctions on[a,b].
By observing carefully the proofgiven in [1], one can note thatC([a,b])
does not admit anisometricshittoperator 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 apairof
compactHausdorff
spaces. Then thefollowingtwo conditionsareequivalent:(i)
If
thereisa continuousmapof Y
ontoXwhichisnotinjective, then theset{(, ) e r r: () (), # }
is
infinite.
(i
0 If
there isalinear isometryof G(X)
intoC(Y)
which hasafinite
codtmension, thenit is surjective.Since both
([0, 1], [0,1])
and(T1,T1)
satisfy the condition(i), whereTlis
the unit circle inthe complexplane,weget fromthis678 S-E TAKAttASIAN[)T()KAYASU
COROLLARY 1. "lheonlypossiblecodtmenston
of
hneartsometrtes(7([0, 1] C([0, 1])
andC(T C(T
are zeroormfimte.
Moreover,ifV isthe canonical linearmap
ofC(T
intoC([0, 1])
definedby(Vf)(t)- f(e ’’) (f EC(T 1),O<_t <
1),thenV isansometry and therangeofVisthesetofall9 E
C([0, 1])
such that9(0) 9(1)
HenceV has codimension 1, and if there isafinite codimensional linearisometryofC([0, 1])
intoC(T
), sayT, then VT is a linear isometry ofC([0,1])
into itself such thatVT(C([O, 1]))c C([0,1])
andcodim(T) +
From Corollary itfollows that VTmustbesurjective, acontradiction hencewehave alsoprovedCOROLLARY2. /heretsno
fimte
codtmenstonallineartsometryof C([O, 1])
intoC(T
2. LEMMAS
Inordertoprove themaintheorem,wehavetopreparesomelemmas
LEMMA 1. LetXbeacompact
Hausdorff
space,M
asubspaceof C(X)
whosecodtmenstonts n< +
oo, and tfaclosedboundaryof
Xwith respecttoM O.e.,for
anyf
Mthereexistsa point:cm tf wuh
f(z)t If t,,,
thesupremumnormoff
onX). 7henthesetX\K
hasat most n points.PROOF. Assumethat
X\K
hasatleastn+
points, say:rl :r,,+ Foreach_< <
n+
1,chooseafunction
f,
nC(X)
suchthatf,(z)
andf,(a:)
0forz K{a:, z,+}\{a:,}
since Ksclosed Inthiscase,{fl +
M f,,++ M}
is linearlyindependentinC(X)/M
since ifc(ft +
M)+ +
c,-I(f,,+l +
M) 0for 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
Thent9 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 thatcodim(M) >
rz+
LEMMA2. LetXandYbecompacl
Hausdorff
.spacesand acontinuousmapof
YontoX.If
g s afimcton
mC(Y)
such thatg(yl)=9(y) for
all pars(Yl,Ye)
Y Y sansfymg(Yl) (Ye),
thentheres afimcton f
mC(X)
such thatf((y)) g(y)for
all y Y.PROOF. Let g be a function in
C(Y)
such that9(Yl) g(Y2)
for all pairs (y,y.,.) Yx Y satisf3,ing(y) (y.:)
LetY/
bethe quotientspaceofY
definedby,
7r thecanonicalmap ofYonto
Y/,
and r the canonical map ofY/
onto X Then the complex-valued function.
onY/
definedby
.()
g(y)foreach9 Y/
iscontinuous,sosetnngf .
---1 it iseasytoseethatf
isafunctionwiththe desiredproperties
Finally,wewillneedthefollowingresult whoseproofisstraightforward
LEMMA 3. Let X be a compact
Hausdorff
space, K a compact subsetof
X, andAI.
theBanach subspace
of C(X)
conststmgof
allf C(X)whtch
are constant on K. Then the Banach spaceC(X)/Aa-
tstsomorphtctoaquotwntspaceof 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 toTiC(X)), acontinuousmaphofKontoX,andacontinuous unimodular functionuonYsuchthat
(T f )()
FINITE CODIMENSIONAL LINEAR ISOMETRIES 6 9
for all
.f
EC(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,...
inY
such that(a,) z(/3,,)
for allpositiveintegers n, andwherewe canassumewithoutlossof generality that al,/31, a2,/32,
CK. Letnbeanypositive integer, and for each 1_< <
n choose a function 9, inC(Y)
such that9,(az)=
1 and9,(Y)=
0for all y
Y\U,,
whereU,
is a sufficiently small neighborhood of cq. In this case{91 + T(C(X))
9,+ T(C(X))}
islinearly independentinC(Y)/T(C(X)),
since ifCl(g -- T(C(X))) + + c,(9, + T(C(X)))
0forsomecomplexnumberscl, c,thereexists
f C(X)
such thatc191+ +
Cn9 Tf, implyingq
(,) + +
(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 ofY
onto X. If for any gC(Y),
we set1 g(h_l(x))
y(z) (h_()
for eachx E
X,
thenwe obtainthatf C(X)
andTf
g, sothatT
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 yY.
ThenT
is a linearisometry ofC(X)
intoC(Y)
and since isnot injective,itfollowsthatT
isnotsurjective. PutA, {g C(Y)"
g is constanton-1(x,)} (i
1,n)
and
A {g C(Y)
g is constant on i=lU -(,) }
n
Then
A
C_I"1
Ai, and henceC(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 inclusioni=1 i=1
680 S-ETAKAtlASIANDT OKAYASU
TO(C(X))
C_I’]
A, istrivial, and the reverse inclusionfollows immediatelyfromLemma2 Also by Lemma3,C(Y)/A
isisomorphictoaquotient ofC(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 isprovedBut 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.