characterizations of C (X) arllong its subalgebras
R. B. Burckel
Lecture Notes in Pure and Applied Mathematics
COORDINATOR OF THE EDITORIAL BOARD S. Kobayashi UNIVERSITY OF CALIFORNIA AT BERKELEY
1. N. Jacobson, Exceptional Lie Algebras 2. L.-l. Lindahl and F. Poulsen, Thin Sets in Harmonic Analysis 3. I. Satake, Classification Theory of Semi-Simple Algebraic Groups 4. F. Hirzebruch, W. D. Neumann, and S. S. Koh, Differentiable Manifolds and Quadratic Forms 5. I. Chavel, Riemannian Symmetric Spaces of Rank One 6. R. B. Burckel, Characterizations of C(X) among Its Subalgebras
Other volumes in preparation
Characterizations of C(X) among Its Subalgebras R. B. BURCKEL Kansas State University Manhattan. Kansas
MARCEL
DEKKER, INC.
New York
1972
COPYRIGHT © 1972 by MARCEL DEKKER, INC. ALL RIGHTS RESERVED
No part of this work may be reproduced or utilized in any form or by any means, electronic or mechanical, including xerography, photocopying, microfilm, and recording, or by any information storage and retrieval system, without the written permission of the publisher.
MARCEL DEKKER, INC. 95 Madison Avenue, New York, New York 10016
LIBRARY OF CONGRESS CATALOG CARD NUMBER 72-90373 ISBN 0-8247-6038-7
PRINTED IN THE UNITED STATES OF AMERICA
To the Memory of my Father
PREFACE These notes are a revised version of lectures I gave at the University of Oregon in the Spring of 1970.
Since that
time I have had access to several more recent papers, lecture notes and monographs and have freely used them to expand and improve my original account.
Still,some of the material here
has not appeared in monographic form before. My aim is to present a detailed account of some recent resu~ts--almost
all the material is less than a decade old--
about subalgebras of
C(X).
These algebras carry a Banach
algebra norm which is often, but not always, the uniform norm.
Accordingly, students of Banach algebra theory will
perceive that we are really concerned with commutative semisimple Banach algebras, but I do not use the language of or any non-elementary results about Banach algebras.
In fact,
nothing more recondite than the Gelfand spectral radius formula is used, and this is used only once, in one proof of a theorem for which two proofs are provided.
I have made these self-
imposed limitations because this is an evangelistic effort: I wish to communicate these beautiful results to as large a public as possible, especially students.
The prospective
reader should have had a standard graduate real-variable course and be acquainted with a few odds and ends of functional analysiS and complex-variables.
For example, the first 225
pages of Rudin's book [41] should be more than adequate
vi
equipment.
Modulo this background, all proofs are complete
and, in fact, some will find the proof detail excessive. However, my attitude here is that it is easier for the reader to skip over details of a proof which he understands than to supply these details if he does not. I thank Irving Glicksberg for making the notes [22] available to me, Stuart Sidney for the notes [46] and some helpful correspondence and errata, and Lee Stout, William Bade and Yitzhak Katznelson for helpful correspondence over their work.
Especial thanks to Kenneth A. Ross for his meticulous
reading of my account, to Carolyn Hobbs for typing it and to the original seminar members for critically listening to it.
R. B. Burckel
CONTENTS
PREFACE
v
SOME NOTATIONS AND GENERAL REMARKS
1
I.
BISHOP'S STONE-WEIERSTRASS THEOREM
5
RESTRICTION ALGEBRAS DETERMINING C(X)
16
II. III.
IV. V.
WERMER'S THEOREM ON ALGEBRAS WITH MULTIPLICATIVELY CLOSED REAL PART
29
THE WORK OF ALAIN BERNARD
40 v
THE THEOREMS OF GORIN AND CIRKA
68 ,
VI.
VII. VIII.
BOUNDED APPROXIMATE NORMALITY, THE WORK OF BADE AND CURTIS
88
KATZNELSON'S BOUNDED IDEMPOTENT THEOREM
111
CHARACTERIZATION OF C(X) BY FUNCTIONS WHICH OPERATE
126
APPENDIX:
KATZNELSON'S IDEMPOTENT MODIFICATION TECHNIQUE
135
REFERENCES
150
SYMBOL INDEX
155
SUBJECT INDEX
157
�haracterizations of
C(X)
among Its Subalgebras
SOME NOTATIONS AND GENERAL REMARKS (I)
We use
for the complexes,
G:
the integers,
lR+
lR
for the reals, Z
for the non-negative reals, z+
the non-negative integers,
IN
for for
for the positive
integers (viz. the natural numbers). (II)
If
X
is a topological space, we write
C(X)
for the
set of all bounded continuous complex-valued functions on
X, and
Co(X)
for the subset of these functions
which vanish at infinity in the sense of being arbitrarily small off compact sets:
Ifl-l[£,~)
f E C(X)
and
We write
Cn(X)
in
C(X).
If
f E Co(X)
is compact for each
Y C X
and
restrictions to
of the functions in
C(X).
Y
We wri te
also write
/If/l y
for
to
f E C(X), we write
f
C
IIfll~
Y
or
norm topology generated by X
and
Indeed if
C(X), then A E G:
and
fly
for
for the set of
for
E
whenever sup If(X) I.
We
Any otherwise unqualified C(X)
will refer to the
"/I~.
is a topological space and
lying in
ElY
/If/l x
/lfIYII~.
topological statements about
If
£ > o.
for the set of real-valued functions
the restriction of
E
(III)
if and only if
/lfll~ s IIfllA
A
a Banach algebra
for every
1>..1 > /lfl/ A , then
and so for any positive integers
m
and
f E A.
"" -lfH A = r < 1 p,
2
Since
A
converges in
A, say to
that the function It follows that f(X) c (IV)
'"~ (A-lf)n
is complete, the series
g
therefore n=l A trivial calculation shows
g.
satisfies
f(x)
F
A
g - A-lf - A-lfg
for every
x E X.
O.
Thus
fA E «:: IA I ~ !If!1A}, that is, Ilfl!", ~ IlfilA.
Our attitude toward restriction algebras and vector spaces is ambivalent: Y
C
X, and E
injected in constant
C
C(X)
Let
X
a normed linear space continuously
C (X), that is, "
M.
The set
in the norm of
be a topological space,
kY
=
"OJ
s;
Mil liE
{f E E: fey)
for some
= O}
is closed
E, because of this inequality.
[The use
.~
of "k" here is to suggest kernel, as in the hull-kernel topology in Banach algebra theory.] E/kY
and the restriction set
ElY
The coset space are clearly
algebraically isomorphic vector spaces.
This set will
always be topologized by the quotient norm in
E/kY
and we shall treat it in whichever manifestation is more convenient for the problem at hand. I!f + kYI!E/kY
=
inf{lff + gllE: g E kY)
Thus for
fEE
for
h E ElY.
and h)
(a)
When
E
is an algebra and
11
liE
is submultiplicative,
then this quotient is an algebra and this quotient norm is also submultiplicative, as is easy to see.
3
(b)
If
1/ /IE
is a complete norm, so is this quotient
norm.
The standard proof of this is as follows.
Given
thn}
Cauchy in
ElY
that some subsequence of
it suffices to show
{hn }
has a limit.
Passing to an appropriate subsequence, we may suppos~
therefore
without loss of generality that
n IIhn+1 - h n II'E Iy < 2- • with
f n Iy = h n+1 - h n
series
""
and
n IIfnllE < 2- •
therefore converges in
E fk k=l We then have
fn E E
Thus there exist
""
The
E, say to
f.
'"
'"E 2- k = 2- n +1 •
<
k=n It follows that
th n }
converges in
to
ElY
h1 + fly. (c)
If
H II""
~ Mil f1E
then
IIhll", ~ M/lh/rE\Y Indeed, for any
for all
fEE
with
h E ElY. fly = h
and the infimum on the right over such
(v)
If
X
is a topological space and
separates the points of
X
E
C
we'have
f
is
CO(X), we say
or is a point-separating
E
4 set if for each fEE (VI)
such that
x E X f(x)
and
y E X\{x}
there is an
F f(y) F O.
All other notation is either standard, self-explanatory, or temporary and defined where introduced.
Chapter I
BISHOP'S STONE-WEIERSTRASS THEOREM Definition 1.1 Let
X
be a compact Hausdorff space, A
complex) subalgebra of (real or complex) the constants.
Call
K C X
an A-antisymmetric set if f
constant on
Call
A
2!
a set
f EA
f
With
X
A
and
containing
real on
implies
an anti symmetric algebra if
set, that is, if
C(X)
antisymroetry
and
(i.e., g,g E AIK
K
a (real or
X
g
2! K
A
or
implies
constant).
is an A-antisymmetric
contains no non-constant real functions.
A
as above note the following elementary
facts: 1.
If
Kl ,K 2
Kl U K2 and
are A-antisymmetric sets and is A-antisymmetric.
flKl U K2
For if
x E Kl flKj
is real-valued then
n K2 F ¢
Kl
n K2 ,
then
f E A
are each constant
so
whence 2.
flKl U K2
If ¢FYCX and X
C
is constant. is a set of A-antisymmetric subsets of
which each contain
metric set. for each
For if
K E C, f
Y
f E A
K E C, so
and
is real on
and this constant must be for all
then
f
f(K)
UC f K
is again an A-antisymis real on
then
whence constant on
K
f(Y), i.e. is the same
is constant on
above is a special case of this.
UC
UC.
Of course 1.
Evidently then the union
6
of all the A-antisymrnetric sets which contain maximal A-antisymrnetric set which contains 3.
As each singleton set, each
4.
Let
K
x
K
is the
Y.
is trivially an A-antisymrnetric
lies in a maximal A-antisymrnetric set
be a maximal A-antisymrnetric set.
real on
then
continuity. 5.
{x}
Y
f
is constant on
So maximality forces
If
f E A
is
K
by
K, whence on
K
K.
= K.
Finally by 1., distinct maximal A-antisymmetric sets must be disjoint. This establishes the trivial parts of
Theorem 1.2 (Bishop [9])
Let
X
be a compact Hausdorff space, A
uniformly closed subalgebra of constants.
Let
may be all of
I A).
in a maximal one.
C(X)
which contains the
be a uniformly closed ideal of
f E C(X)
(iU
11K
The collection
Y
&
flK E I!K
as a functional on
C(X)
is the annihilator of
1.1
C(K)
1.1 E M(X)
satisfying:
for each C(X)*
X
K E 1(.
= M(X)
is
with the norm of
equal to its total variation. I
in
C(X)*
the extreme points of the unit ball of closed support of
on
X
K E 1(=f E I.
for each
is uniformly closed in
the finite regular Borel measures
IJ.
(which
of maximal A-antisymrnetric
For (i) we need a lemma and some notation.
1.1
A
Then every A-antisymrnetric set is contained
sets forms a pairwise disjoint closed cover of (i)
a
by
and IJ. •
(ball IJ.)e We denote the
supp 1.1, and for
f E C(X)
is
7
we write
~(f)
for
Jxfd~
and
f~
for the measure defined
= ~(fh)
(f~)(h)
(as a linear functional) by
for all
h E C(X).
Lemma 1.3 (de Branges [llJ)
If
set of antisymmetry of
A.
~:
Let
constant on
K
K
= supp
~ E (ball I~)e, then
~,
f E A, f(K)
it suffices (as
A
C
JR.
supp ~
To see
is a
f
is
is a vector space containing
1) to suppose
Then
< 1
on
(1 -
f)~
0 < f
(0) f~
and
K.
are non-zero measures so
while II
(2 )
f~II
+
II (1
- f)
~II
=
JIf Id I~ I + XJ 11 -
X
IKlfldl~1 JKfdl~1
f Id
!~ I
+ JKII - fldlul
+ IK(l -
JKId I~ I = I~ !(K)
f)dl~1
by (0)
= I~ I (X) = II ~ 1/
1.
[For the first equality the reader may wish to consult Rudin [41 J, p. 126, and for the last equality recall that extreme in the unit ball of
I~
so
II~II ~ 1
so
IIJ..1I/
of that ball, i.e. (1 -
f)~
lie in
I~
U(fg)
since for
o
~
~
is
is not in the interior 1.
J Now
f~
g E I
(fg E I ideal, ~ E I ~ )
and
8
=
[(1 - f)~J(g)
=
U«l - f)g)
«1 - f)g E I ideal, U E I~).
0
~
So by (1), (2) and the extremality of
l~l - a.e.
In particular, then, we have f
1 - f
TITu1r whence, setting
/In - f)~" c l = /lfull/H (1
-
f)~11
I~I
-
a.e.
V
=
(3)
As
f
is continuous, the set
open.
of the support of a measure, V f
has the constant value
then by Hahn-Banach
cl/c2
f
not annihilated by
*
supp u
throughout
If
f E C(X)
is
(ball I~)e.
K.
and
f ' I,
Then by Krein-Milman For
= K,
I~, equivalently
1s not annihilated by
is not annihilated by ball I~.
weak
F cl/c2}
(x E X:f(x)
1s disjoint from
Proof of (1) of Theorem 1.2:
f
and
has l~l-measure 0 and therefore by definition
By (3), V
i.e.
we get
{f}~ n ball I~
f
is
is a
compact convex set and if it conta1ns
it contains the weak
*
closed convex hull of
which by the Krein-Milman Theorem is all of ball I~~ ~ E (ball I~)e
lies in some
be such that K E
(hypothesis on
1(.
U(f)
So for some
F O.
By the lemma, supp ~
gEl, f
f) and then
This however contradicts the choice of
Let
u.
=
9
on
K
9
Corollary 1.4 (Stone-Weierstrass) and
A
If
is a compact Hausdorff space
is a uniformly closed subalgebra of
separates the points of conjugate closed, then ~:
A
is a singleton, so
Note:
which
= C(X). A
A n Cm(X)
the theorem (with
C(X)
X, contains the constants and is
Evidently, since
conjugate closed, K E ~
X
separates points and is
separates points.
f IK E A IK
I = A) yields
C(X)
for every C
But then each f E C(X)
and
A.
Lest the reader sense circularity in this deduction
of the traditional Stone-Weierstrass Theorem from Bishop's Theorem, let us point out that the Riesz Representation Theorem, the Hahn-Banach Theorem and the Krein-Milman Theorem, upon which alone Bishop's Theorem depends, are proven without recourse to the Stone-Weierstrass Theorem.
Because of the way Stone-
Weierstrass permeates analysis, such a fear would not prima
~
have been ill-founded. For (ii) of the Bishop Theorem we need some more lemmas. Call
E
C
X
a
~
E = {x E X:f(x) = 11 that i£ we set IF(x)
I =1
and
IF(x) I < 1
iff
F
for some
= ~(l F(x)
set of
+ f)
= f(x)
for each
x
this last property peaks 2!!.
~:
A
or an A-peak set i£
f EA
= 1, so
E.
IIfl/ ...
1.
F E A, !IF" ... = 1
then
t
with
E = (x E X:F(x)
Notice
and
= 11
Let us say that an
F
with
.!.
A countable intersection of A-peak sets
En
is
10
also an A-peak set.
For if
Fn E A
peaks on
En' then clearly
CD
nE
peaks on
and belongs to
A.
n=l n Lemma 1.5 If norm in
E
is an intersection of A-peak sets, then the uniform
liE
equals the quotient norm.
is a Banach space, it follows that closed in
liE
Since the quotient is complete, hence
C(E).
~:
Recall (IV (c), p. 3) that for all
/If IEI/..,
:s;
Hf + I
f
EA
n kE/I.
On the other hand, if
Ve
is the open set
{x E X: If(x) I < IIfIEI/.., + el, then some finite intersection of the A-peak sets containing
E
compactness.
As noted above, Ee
there exists
ge E A
all
On the compact set
x lEe.
supremum less than
with
lies entirely in
Ve
Ee
by
is again an A-peak set, so
=
ge(E e )
1
X\V e
1, so for large
and then
Ige(x) I < 1 ge
for
has a
n
It follows that
sup!g~f(X)!
:s;
max(sup!f(V e ) !+e,sUpI9~f(Ve) I) = suplf(V e ) I+e.
Consulting the definition of
Ve
we see
sup!f(VE)!:S; /If!EII.., + E,
so we've proved
Since
g~£
gn f _ £ = (gn _ 1)£ = 0 on E~ ~ E, we see that e e .. £ EkE. Also, I is an ideal, so £ E I implies
11
g~f - f E I.
g~f - f E I
Therefore
n kE
and our last
inequality yields
As
€
is arbitrary,we're done.
Lemma 1.6 (Bishop [lOJ)
If
A-peak sets and
L
some A-peak set
S.
then so is
L
F
C
X, F
C
is an intersection of
is an A1F-peak set, then If in particular
F
L
=S n
F
for
is an A-peak set,
L.
~:
Let
f EA
be such that
flF
peaks on
L, that
is, L = (x E F: f(x) = l} For each positive integer G
n
= (x E X:
If(x)
and
n
I
set
< 1 +
This is an open set containing of the A-peak sets containing Gn , by compactness.
F F
hn E A
As the continuous function
and so some finite number have their intersection in
Ihnl
we have
is then less than
X\Gn , we have
sup!h n (X\Gn )!
=
Choose positive integer k
....!}. 2n
This finite intersection is again an
A-peak set, so for some
compact set
If(x)1 < 1 for each x E F\L.
6n < 1. kn
so that
6n n < [2 n cHf/r ... + u r l .
1
on the
12
Then Yx E X\G • n
Consider
Note that since
series converges uniformly (so
9 E A) and
...
k
If(X) I t 2- n lh (x) InS If(x) I n=l n
Ig(x) I s
Yx E X.
...
x EnG n=l n
Therefore if
(3)
If(x) I s 1
then
and so
...
Ig(x) I s 1
...
If
nG n=l n
x,
then, defining
x E Gm_ 1
for which
while
GO = X, there exists
x , Gn
for all
n
~
m.
m ~ 1 Thus
k
Ihnn(X)f(X) I < 1 + 2m: 1 since
for
If I < 1 + _1_ 2m- 1
Hhnll ... = 1, while k
Ihnn(X)f(X) I <...l. < __ 1_ 2n 2m- 1 by (2) and the fact
x
n = 1,2, ••• ,m-1
t
Gn •
for
On
Also
n ~ m
Therefore
(4) = (1
1 ) (1 1 ) + _1_ 1 1 + 2m-1 - 2m- 1 2m- 1 • 2m- 1 =
..
V (/. n G x I'- n=l n •
From (3) and (4) we get
Therefore, if A. F
Since each
S = g-1(1), we have that
hn
is
1
on
F
S
is a peak set of
by(1), we have
9 = f
and so S
n
F
{x E F: g(x)
{x E F: f(x)
1) = L.
on
13
Lemma 1.7 Every maximal A-antisymmetric set
K
is an intersection
of A-peak sets. Proof: X.
Let
There are peak sets which contain
E
show that
denote the intersection of all of them. K = E.
If not, then by maXimality of
a set of A-antisymmetry so there eXists an flE ~,
K, for example
real but not constant. there exist
Un
f(E)
K, E
in
A
is not
with
being a compact subset of ~
open subsets of
Un = {z E~: d(z,f(E»
(E.g., let
f
We will
f(E) =
such that
< ~).)
Then
nU
•
n=l n..,
f-l(f(E»
= n Vn ,
n=l
where
Vn
is the open set
f-l(U n ).
By compactness, E CV n
implies that some finite intersection determining
..,
..,
E
already lies in
Vn •
n E e n V C f-l(f(E», and so n=l n n=l n as noted above, F Since E, f in
~
E C F =
Moreover,
is again a peak set.
f(E) = f(F)
and
f
is real and not constant on F.
But
f
K, is constant on the antisymmetry set
f(K) = c. a
Thus
of the peak sets def
f(E) = f(F).
is real and not constant on E
En
being real K, say
We therefore have
= min
f(F) < b
= max
f(F).
c E [a,b].
Pick
N
large enough that
consider the polynomial p(t)
=1
One checks that
t-C]2 - [ -W- •
[c - N, c + N] ~ [a,b]
and
14
o and that
~
~
p(t) c
=1
p(c)
Vt E [c - N, c + N]
is the unique maximizing point in this interval.
Consider now
pof E A.
We have that
f(F) c [a,b] c [c - N, c + N] and so (pof) (K)
p(f(K»
p(c)
~
Therefore if
L
we see that Moreover
L
a < b
least one point of
L
L
p(a) < p(c) F
is a peak set for
definition of and since
f(E)
p(f(F»
E
A.
f(F)
K
C
1
L.
a # c
p(b) < p(c).
or
So at
As we have now shown
So
L
pof(E)
~
K
and the
pof(L)
C
=
{ll
this gives C
{ll.
This contradicts the fact that the peak set
L
F.
Proof of (ii) of Theorem 1.2: lemmas 1.5 and 1.7 to conclude that in
.is
AIF, the last lemma insures that
EeL.
= p(f(E»
is a proper subset of
pof
and that
But then the fact
imply
=
L.
where
so either
or
is not in
is a peak set for
F
pOflF
a,b E f(F)
and both
b # c, whence either
that
max p ( f ( F) ) •
denotes the subset of is a peak set for
max p[a,b]
max p[C-N,C+N]
1
It remains only to combine 11K
is uniformly closed
C(K).
Corollary 1.8 Let
X
be a compact Hausdorff space, A
closed subalgebra of '" l1nii'ormlv
C(X)
closed ideal of
a uniformly
which contains the constants, I A.
If
f E I
and
f
E A, then
15
f
E I.
That is, uniformly closed ideals are conjugate closed
to the extent that Proof: of
f + f
X.
A ~
Let and
itself is.
be the maximal A-antisymmetric decomposition i(f - f)
are real-valued functions in
and so they are constant on each
K
~(f - f)
is constant on
f
then
= cK =
since
flK
0 E 11K, while if
f E I, fEA Remark.
K, say
and
I
E~.
cK cK
F0
is an ideal.
Then on
K.
flK
f
= ~(f If
= cK
A
+ f) _
=0 f).fIK = (-1:. c cK
K Now quote Theorem 1.2.
This proof of Bishop's Theorem and its corollaries
is from Glicksberg [20].
EIIK,
Chapter II RESTRICTION ALGEBRAS DETERMINING
C(X)
Our next result seems to be due independently to Chalice [14], Mullins [38] and Wilken & Gamelin [52]. [30].
Compare also
The proof below is adapted from [38].
Theorem 2.1 Let
X
be a compact Hausdorff space, A
closed, point-separating subalgebra of the constants.
Suppose
F l ,F 2 ,...
a uniformly
C(X)
which contains
is a sequence of closed
0>
subsets of
X
such that
U
F. = X
and
each
j.
Then
A
for
AIF. = C(F.)
J
j=l J
J
= C(X).
It is convenient to introduce a term and prove some lemmas first. Definition 2.2 Let C (X).
Call
hood of ~A
X
x
be a compact Hausdorff space, A x E X
a suba+gebra of
a strong A-boundary point if each neighbor.
contains an A-peak set which contains
x.
Let
denote the closure of the set of strong A-boundary points.
Lemma 2.3 (Runge) K
=
Let
B U {z E <1::
function
'XB
of
0 < rO < r l
Il-z I :s: l-rll. B
and
B
Then on
K
< 1
=
{z E <1::
Iz I :s: rol
the characteristic
is a uniform limit of polynomials.
17 Proof:
Clearly
'XB E C(K), since
B
is clopen in ~
By Hahn-Banach we have only to show that if ~
annihilates all polynomials then
K.
E C(K) * = M(K)
annihilates
'XB•
So we
suppose
JKSnd~(;)
(1)
=0
n /\
~
Define the function
/\
=
~(z)
(2)
If, say,
Izl
JK z-g 1
~ 2
on
= 0,1,2, •••• ~\K
d~(~)
by
t
Yz
K.
then the series
~
converges uniformly for
E K
to
I I I Z • 1 _ = z-g
I
z
IS I
g E K implies
(for
Therefore for such /\
=
~(z)
J!
s: 2-r l
implies
~ l-zS I --~ ~
< 1).
z
~ (!)ndU(~) =
K Z n=O Z
by (U. Thus we have /\
(3)
~(z)
=0
/\
Now
~
is analytic in the open set
~\K,
for it is easy
to justify differentiating under the integral (2) with respect to
Z
(in
~\K).
Moreover the set
~\K
is obviously connected.
Therefore from (3) and the uniqueness Theorem of Analytic Function Theory we conclude /\
(4)
~(z)
Let
r0 < r
Yz E ~\K.
= 0
< r 1.
Then
18
2TT
I
2TTi
I0
,it ~dt it re -z
if
Iz \
< r
(Cauchy's Integral Formula for a disk)
if
\z \
> r
(Cauchy's Theorem for a disk).
In particular XB (z)
(5 )
=
..l:. 2TT
Now the function
lTT reiitt 0
re
(~,z)
't
fre 1 : 0 s t s 2TT} X K 2TT
I K'YB(z)du(z)
'fz E K.
dt -z
~ is continuous on s-z and so the simplest form of Fubini gives
....
(5)
=
reit I KI[o,2TTJ
reit_z
it I[ O,2TT JIK re reit_z
But
dtd\J(z)
du(z)dt
{re it : 0 s t s 2TT} C ¢\K, so by (4) and (2) the inner
integral above is identically zero and we're done. Lemma 2.4 Let
X
be a compact Hausdorff space, A
closed point-separating subalgebra of C(X) the constants.
a uniformly
which contains
Then every A-peak set contains a strong A-
boundary point. Proof: set, say
e
E
Zornicate.
=
{x E X:
In more detail, let
\f(x) \
=
IIfH..,),f E A.
be an A-peak
Evidently if
is a linearly ordered (by inclusion) chain of non-void
intersections of A-peak sets lying in void (by compactness of
F
E
then
nc
is non-
E) and an intersection of A-peak sets.
Therefore Zorn's lemma implies that set
E
E
contains a non-void
which is an intersection of A-peak sets and does not
properly contain any such non-void intersection.
We contend
19
that
F
is a single point.
point, A
For if
will contain a function
F g
contains more than one which is not constant on
F, by the point-separation hypothesis. Pick h
E F
Xo
= ~(l
+
ari:r) g 0
proper (since
E A.
g
L
=
F
and consider
IIgIFII..,
{x E F: h(x)
= I}
is a
F), non-void (since
which is an AIF-peak set.
Xo
By lemma 1.6,
is an intersection of A-peak sets and the minima1ity of
is compromised. V
The set
=
is not constant on
belongs) subset of L
0 < Ig(xo ) I
such that
is any open neighborhood of
the A-peak sets containing by compactness. set, and so
x
F = {x}
We conclude that x
F
o.
glF ~
In particular
for some
x.
F If
then some finite number of
have their intersection in
V,
This finite intersection is again an A-peak is a strong A-boundary point.
Lemma 2.5 With the notation and hypotheses of the last lemma, suppose
A
is an anti symmetriC algebra and that some strong
A-boundary point has an open neighborhood Alv
= C(V).
Then
X
Xo
Let
~:
let
g E A
and
Ig(x) I < 1
such that
is a single point.
V
be a strong A-boundary point in {x E X: g(x)
be such that for
V
x
t
V.
= g(xO )
So there exists
IIgll..,
= I}
0 < rO < 1
that Ig(x\v) I < rOo Pick
rO < r 1 < 1 B
=
(z E a::
and set Iz I < r 0 } ,
D
=
By lemma 2.3 there are polynomials uniformly on
BUD.
If we set
{z E a:: Pn
I z-l
1
such that
< 1-r 1 ).
Pn
~
~B
and
cV such
20
U = {x EX: then
U
\g(x) - 1\ < l-r l ),
is an open subset of
in
V.
Let
U.
Since
f
X
be any element of
= C(V),
A\V
Xo
which contains C(X)
we can pick
and lies
which vanishes outside
h EA
with
h\V
= fiVe
Now we have g(U) CD
and
so the sequence
g(X\V) C B,
{Pnog}
converges uniformly on
the characteristic function of vanishes on But
h· Xu. on
V ::> U.
Since
sequence is in If
U
since
f
vanishes outside
finally
f E C(X)
X
U.
This
= X,
whence
U
= X.
Since
C(X)
A\V
f
X
and
h
f
=f
of this
A
would belong would be
U CV, we have
=A
and because of
the anti symmetry and point-separation properties of that
to
there would be a non-constant
vanishing off
We conclude that V
X
A.
A, as just shown, and the anti symmetry of
violated.
to
h
U
h· (Pnog) E A, the uniform limit
were not all of
real-valued to
=f
Therefore,since
converges uniformly on
V\U, (h· (Pnog) } h·Xu
U.
U U (X\V)
A
we see
must reduce to a single point.
Lemma 2.6 Under the hypotheses of Theorem 2.1 and the additional hypothesis that that
X
A
is antisymmetric and
~A
=X
it follows
is a single point.
Proof: void interior
By the Baire Category Theorem some Vj •
The hypothesis
~A
strong A-boundary points are dense in
=X X
Fj
has non-
means that the
so there is one in
21 The asserted conclusion follows then from the last lemma. Lemma 2.7 Under the hypotheses of Theorem 2.1 and the additional hypothesis that
A
is antisymmetric it follows that
X
is a
single point. Proof:
We first show that
Suppose not. real but
g
oA
is an A-antisymmetric set.
g EA
Then there exists is not constant on
oA.
As
an A-antisymmetric set,it must be that of
X.
function on
1m g(xO) =
Let, say,
= expo(ig)
f
~
< 0
belongs to
such that X
9
is
is by hypothesis
is not real on all
Xo
for some A
g(oA)
and since
E X. 9
The
is real
M i t satisfies If( M) I = 1
If(XO) I = e- Im g(xO) Therefore
Hfll"" > 1
and
E
~A.
an A-peak set disjoint from this means that
E
{x E X:
If(x) I
=
IIfH",,}
contains
By the definition of
oA
contains no strong A-boundary points, in
contradiction to lemma 2.4. Another application of lemma 2.4 shows that every function in
A
attains its maximum modulus on
f ~ floA
is an isometry.
C(X), we see that course for each
AI~A
AIFj = C(F j ) j.
Since
A
oA
and so the map
is uniformly closed in
is uniformly closed in
implies
(AlaA) I(F j
Finally it is clear that
n aA)
o(AloA)
C(oA).
n M)
C(F j aA
Of
and we
are at last in a position to apply lemma 2.6 to the restriction algebra as
AlaA.
f ~ flaA
It follows that is
oA
is a single point.
1-1, it must be that
A
Then
consists of constants.
22 Since
A
separates the points of
X, this means that
X
is
a single point. Proof of Theorem 2.1:
~
If
symmetric decomposition of
is Bishop's maximal A-anti-
X, then for each
K f ~, AIK
is
closed (Bishop's Theorem) and satisfies the hypotheses of lemma 2.7 with
n F.
K
J
in the role of K E
is a single point for each A
=
~,
Conclude that
K
hence by Bishop's Theorem
C(X).
Corollary 2.8 If
Y
is a compact, countable subset of
continuous complex function on
Y
~
then every
is a uniform limit of
polynomials.
= {y.}, let A J J of the polynomials (in the F.
~:
be the uniform closure in
C(Y)
z) and apply Theorem 2.1 to conclude
complex variable
A
C(Y) •
Of course Corollary 2.8 is just the special case of Corollary 2.9 below needed to prove the latter: Corollary 2.9 (Rudin [48J)
Let
X
be a compact Hausdorff space without
non-void perfect subsets, A subalgebra of Proof:
C(X)
Let
f
1 E A.
with
sequence of polynomials
I'pn I
0
f
f(X).
E A, Y
compact set is countable.
follows that
a uniformly closed, point-separating
-
Then
A
= C(X).
We will show that this
The last corollary then provides a
Pn
fl'I ...
-+
such that 0
and so
sup\p (z) zEy n f E A. Thus
z\ A
-+
O.
It
is
conjugate closed and so the conclusion follows from Stone-Weierstr 8
23
Let
Vl 'V 2 '...
be an enumeration of the open disks with
rational radii and centers having rational coordinates which intersect
Y
'" Ll V
C = Y n
Then
in finite or countable sets.
n=l n is countable and For if
x
~
P
P = Y\C
is closed and has no isolated points.
is isolated, there is a disc
V
of rational
radius and center having rational coordinates such that V
nP
= {xl.
But then
V
nY
= (V
countable and so by definition x E
vn
Y
C
V
C, a contradiction.
we show that
P
showing that if
n p) u
n C)
(V
is among the
c
fx} U C
is
Vn ' whence
We are therefore finished
~f
is void and we argue this by contradiction, P
F¢
then
this end we want a minimal
P f
has an isolated point. pre-image for
P
To
and, of course,
if
e
is a linearly ordered (by inclusion)
set of compact subsets
K
of
Zorn provides one:
for each
X
yEp, f- l (y) n K F ¢
such that
f{K) = P, then
K E e
for each
so by
compactness (and the linearity of the inclusion order on
~ f-l{y) KEC
nK
= f-l{y)
n r:
K. KEC
That is,
f(
P
a minimal one
E.
By hypothesis on
not perfect and so has an isolated point is closed and so by minima1ity of
E
f{x O )
is compact,so is is isolated in
f
pre-images
X, the set
xO.
Then
we must have
P = f{E), that is, P\{f{XO)} = f{E)\{f{X O )} E\{xO }
n K) = P.
KEC
There is therefore in the family of all compact of
e),
=
E
is
E\{XO} f{E\{X O })
f{E\{X O}).
F
As
f{E\(X O }) = P\{f{XO)}' and so
P, our final contradiction.
We proceed next to locally compact generalizations of Theorem 2.1.
The first result in this direction, Theorem 2.10,
is from the paper of Bade and Curtis in [8J, pp. 90-92: they
24
attribute the proof to Katznelson.
A generalization of Theorem
2.1 of a different kind will appear in chapter six (Corollary
6.15). Theorem 2.10 Let
Y
be a locally compact Hausdorff space, A a
uniformly closed subalgebra of for every compact Proof: f
on
Y
FeY.
CO(Y)
Then
AIF = C(F)
such that
A = CO(Y).
It suffices to show that every continuous function
which has compact support belongs to
A.
For
A
is
uniformly closed and (as an easy consequence of Urysohn's lemma) every function in
Co(Y)
is a uniform limit of compactly
supported functions. Let
K
provides a
be a compact set off which Ifl E CO(Y)
of Ifllv.
S
s::::> K.
Let
V
with compact closure and pick Let
Igl-l[~,,,,) compact and
F = Igl-l[~,~)\v. is compact, so S
=V n
F
Xs
hypothesis there is then an
such that
Now
is open therein. S
h E A
Now for each positive integer
and
cp
glV =
9 E CO(Y)' the set
is compact.
of
Urysohn
be an open neighborhood 9 E A
Since
(S U F)
characteristic function
vanishes.
Ifl(Y) c [0,1], Ifl(K) = 1
with
supported in a compact
f
belongs to such that
n, g~ E A
S UF
is
Therefore the C(S U F).
By
=
Xsf.
his U F
and we have
1/ 9 n h - f ll y \v U F = Ilg~ I'y\v U F since f is supported in K c V, ~ 2-nnhVy since
(1) IIh-f/lF
/lg~/IF
since
f
V U F ::::> Igl-l[~,,,,).
is supported in
Kc V
which is disjoint from F (definition of F),
25
o
( 2)
since
Hg~Hv\s
IIgnf-fHs = 0
(4)
since
since
It follows from (1)-(4) that g~ E A
is supported in
since
h = f
g =
in
(!'
the support of
since each
f
since g = !P in in S.
o
(3)
= O.
h(F)
and
A
V
and
in
K
S
C
is supported
(!'
S,
V::> Sand
(!'
=
1
on
f.
IIg~-fHy:S: 2- n /lhH y
0, so
-t
f EA
is closed.
Lemma 2.11 Let
X
be a locally compact Hausdorff space, A
uniformly closed point-separating subalgebra of
Co (X) ,
X
a sequence of compact subsets of
Fl ,F 2 ,···
a
such that
."
U F. = X j=l J
and
Proof:
AIFj = C(Fj)
Let
X
the point adjoined, FO
in
A
A of
j.
Then
A=CO(X).
be the one point compactification of
xo
extended to
for each
X
{xol.
as
0
at
Let
AO
x O' and set
X,
be the functions
A =
~
+ AO.
Then
is evidently a uniformly closed, point-separating subalgebra C(X)
which contains the constants. C
C(FO )
C + AIFj and the cover
F j (j = 0,1,2, ••• ) X.
Moreover
= C(F j )
AO
which vanish at
and so
xo
= 1,2, •••
are compact subsets of
Therefore Theorem 2.1 ensures that
follows easily that
j
X
A = C(X).
comprises all the functions in A = Co(X).
which It C(X)
26
Corollary 2.12 Let
Y
be a locally compact Hausdorff space, A
uniformly closed, point-separating subalgebra of {Ya )
Let
be a family of closed subsets of
A\Ya = Co(Ya )
for every
a.
Y
C(F)
If each point of
Y
some function in there exist
Each
A
Fey.
Ya , then
is not zero.
fl, ••• ,fn E A
F
U
X =
(2)
Ya •
F
is covered by the
fk E Co(Y).
Set
The hypothesis and
a covering argument show that every o-compact subset of covered by countably many
F
Then by compactness of
such that
is a-compact, since
A\F =
At every point of such an
Vk = \fk\-l(o,m), k = 1,2, ••• ,n.
Vk
has a
By Theorem 2.10 it suffices to show that
for each compact
open sets
CO(Y). such that
neighborhood which is covered by countably many
~:
a
Y
It follows that we can write
F
m=l m where each
for some
F
m is a compact set such that
k
and some
a.
In particular, we then have
(4)
A\F = (A\Ya ) IF = Co(Ya ) IF = C(F ). m m m m
Given
cp E C(F )
m
Vk ::> Fm· f E A fkf
we have, wi th
k
as in (3),
If k \ > 0
on
cw'fk IFm E C(Fm) and by (4) there exists f IFm = cw'fk IFm• Therefore fkflFm = cpo Now
Thus
with
vanishes outside
Vk
C
X.
If therefore we set
{f E A: f(Y\X) = oJ, a closed ideal in
I
=
A, we shall have
is
27
IIFm
(5 )
Evidently
= C(Fm)
Ilx
m
is a uniformly closed subalgebra of
will show that
I
separates the points of
light of (2) and (5), Ilx
Ifk (x)
I
> 0
fk E I.
and
else
=
fk(X)
y
y E Vk
If
fk(y) 1= fk (x).
= ffk
E I
and
t
and
then
Vk
fk(y)
then either
separates points, there is an g
y EX
for some
f E A
k, so
y 1= x, there
= 0,
so
fk(x) 1= fk(y)
In the latter case, because
fk(y) 1= o.
We
and then, in the
x E Vk
then
If also
If
are two possibilities.
X
Co(X).
will satisfy the hypotheses of
x E X
Indeed, if
lemma 2.11.
= 1,2, ••••
or A
f(x) 1= fey).
with
Then
g(x) 1= g(y).
Apply lemma 2.11 to learn that is a compact subset of
X
= Co(X).
Ilx
Since
F
it follows that
Corollary 2.13 Let
X
be a compact Hausdorff space, A
closed, point-separating subalgebra of the constants.
Suppose that each
C(X)
= C(F). x ~:
Then
A
x
Fx
such that
= C(X).
There are (distinct) points
each point
which contains
x E X, with at most finitely
many exceptions, has a compact neighborhood AIFx
a uniformly
xl, ••• ,xn
in the locally compact set
Xo
= x\{xl, ••• ,xn }
has a compact neighborhood of the indicated type. the ideal of functions in Evidently B and
I
A
which vanish on
is uniformly closed.
Let
is a uniformly closed subalgebra of x 1= y, set
Xo
=x
and
xn+l
= y.
B
=
such that
Let
I
be
{xl, ••• ,xn }. IlxO.
Co(Xo ).
If
Evidently x,y E Xo
Use the fact that
28
1 E A and
A separates points to find functions
= 1,2, ••• ,n+l) with gj E A (j = O,l, ••• ,n) f = fl ••• f n + l , 9 = go (j
function a
at
af + ag
y.
Thus
with
••• gn.
belongs to B
BIK
=0
fj(X j )
= 1,
gj(Y)
x =C(K). x
= O.
and has value
E Xo
Let
a,~ E ~
Then for any I
x
and functions
gj(X j )
separates the points of
We will show that each such that
= 1,
fj(X)
fj E A
a
the
at
x
and
XO.
has a compact neighborhood
Consider the
FX
provided by
the hypothesis and use the result of the first paragraph to fx E I
find a function
compact neighborhood of (Note that then g/fxlKx.
Let (Tietze)
There is then an
flKx
= g/fxlKx
follows that B 1Kx
x
Kx C XO.)
Fx.
h
inside If
Fx
Let
be a
Ifxl > O.
on which
belongs to
9
Kx
C(Kx )'
so does
be an extension of this function to
f EA
and so IIKx
Ifx(X) I > O.
with
such that
ffxlKx
= C(Kx )
= g.
flFx
= h.
ffx E I.
Moreover
or, since
Then It
CXO ' IlxolKx
Kx
= C(Kx ). We now apply Corollary 2.12 to assert that
It follows at once that which vanish on
I
B
= CO(XO).
contains all continuous functions
{xl, ••• ,xnl.
For each
j
=
1,2, ••• ,n
there
is, by a construction like that in the first paragraph, a fj E A
function for any
f
fj(~)
with
E C(X), the function
{Xl, ••• ,x ), hence belongs to n
f
= t
j=l
=
n
= 1,2, ••• ,n).
6' k J
(j,k
f
I: f(x.)f. j=l J J
-
leA.
n
vanishes on
It follows that
n
f(Xj)f. + [f J
I:f(x.)f.J j=l J J
belongs to
Then
A.
Chapter III WERMER'S THEOREM ON ALGEBRAS WITH MULTIPLICATIVELY CLOSED REAL PART The principal result of this chapter, Corollary 3.6, is an easy consequence of the more general results of the next chapter, but we offer here the original proof of i t because it is such an architectonic display of the tools of the trade in operation. Theorem 3.1 Let
X
be a compact Hausdorff space, A
closed subalgebra of
C(X)
a uniformly
which separates the points of
and contains the constants.
Suppose also that
A
X
is an anti-
symmetric algebra (i.e. contains no non-constant real functions) and that
Re A is closed under multiplication.
Then
X
is
a single point. Preliminaries In lemmas 3.2, 3.3 and 3.5 to follow the notation and hypotheses will be that of theorem 3.1. u E part
Fix
Xo
Re A, the difference of any two functions in u
E X. A
If with real
is a pure imaginary-valued function, hence is constant
by the antisymmetry of there is a unique Re f
= u,
We denote this
f
A.
E A
1m f(X O ) f
by
Therefore, recalling that satisfying
= O.
H(u):
lEA,
30
o.
H ( u ) E A, Re H ( u ) = u, [ 1m H ( u) ] ( Xo ) Define a norm in
=
I/ull
Re A
by
IIH(u) II",.
This is evidently a real linear space norm.
We show in the
next lemma that it is (essentially) an algebra norm. H(Re A)
Since
is the uniformly closed real subspace of all 1m f(x O)
such that
= 0,
H
onto a complete space, so
effects an isometry of ""
f
E A
(Re A,II II>
is a complete norm.
Lemma 3.2 There exists a constant
K
lIuv II ~ K/lul/ IIvll ~:
map
u
~
uv
f
= H(u),
Yu,v E Re A.
We first show that for each fixed
Indeed suppose
un,u,w ERe A. gn
Re A
is continuous in
Graph Theorem. for some
such that
= H(UnV),
Therefore for every
g
This means that for we have
x E X
whence
= Re fn(x) ~ Re (u v)(x) = Re g (x) ~ n n
un(x)
whence
and so u(x)v(x)
w(x) ,
= u(x) and g(x) = w(x)
f(x) Re
the
by the (real) Closed
lIu n - ull ~ 0
= H(w)
v E Re A
and fn
lIu n v - wll ~ 0
= H(u n ),
31
that is,
= w.
uv
Hence the graph of
so the map is continuous. fixed
u
Let
K(v)
u
~
uv
is closed and
be its norm.
Then for
the family
{uv:v E Re A, /lv/l
I}
S
is bounded: /luv/l
S
/lv/lK(U)
S
K(u).
Hence by the (real) Uniform Boundedness Principle the family of bounds
{K(v): v E Re A, /lv/l
Then for any
So
u,v E Re A
Huvll
S
lIullK(v)
lIuvll
S
Kllu/l IIv/l
Next define
B
S
with
K(v)
S
l}
S
lIu/l
is bounded, say by S
and
I
to be the set of all functions
Re
is closed under multiplication) of
Then
B
.e
/le i6 f/l
which is
Yf E B.
is a complex linear space norm on iv
I/(au-bv) + S
C(X)
Yu, v E Re A
/If/l B = supllel. fl/ 6
f
u + iv
Define
A.
/lu + iv/l = Hull + "v/l
that if
we have
is a complex subalgebra (since
conjugate Closed and contains
/I /l B =u +
1
Yu,v E Re A.
u,v E Re A.
Evidently
S
K.
with A
IIvll
K.
e ie
then for any i ( av+bu)
/I =
=a
+ ib
B.
Notice
we have
/lau-bvU + /lav+bull
laillull + Iblllvl/ + lall/v/l + \bl/lul/ = (lal+lbl)(l/ull+/lv/l)
= (Ial + Ibl)l/flr
S
Taking the supremum over all
2l1fll.
e,
we find
32
Lemma 3.3 B
is complete in
1/ /l B
and this norm satisfies Yf,g E B.
Hence
B
is a complex Banach algebra under a norm equivalent
~:
II
/l B ~
where
II 1/,
fn E Band
= 1m
f n , Vn
fn.
u,v E Re A with
f = u + iv E B.
Finally for
IIfn - fmllB .... 0
then, since
IIfn - fmll = /lun - umll + II v n - vmH .... 0,
we have
= Re
un
there exist Let
If
As
II 1/,
Re A is complete in
/lun - un .... 0
and
/lvn - vII .... O.
~en
f,g E B with,say, u = Re f, v = 1m f, u· = Re g,
v· = 1m g, we have /lfg/l
=
lI(uu' - vv' > + i{uv' + u'v>/I
=
lIuu' - vv'lI + /luv' + u'v/l
~K"u/l I/u'll +Kllv/l/lv'/I +K/lu/l /lv'/I +K/lU'/I /lvll = K(/lull+llv/l> = K/lfll IIgll.
Therefore for all real
e
and taking the supremum on
e
Next we present a working lemma about analytic functions which we need here and also later on. Lemma 3.4 Given
A > 0
there exist a polynomial
p
and an
£
> 0
33
such that
p
the corridor ~:
maps the open disk IRe zl < 1
and satisfies
r > 0
For each
Iz I < 1 + E:}
{z E G::
let
into
p(O) = 0, Im pel) > A.
Dr = {z E G:: Izl < r).
First
note that the series
.., ,f,(z) = - t 1. zn n=l n converges absolutely in
Dl
and so defines an analytic function
there which may be differentiated termwise to give
.., ,f,o(z) = - t zn-l = n=l
(2 )
-(1 -
z ) -1 •
It follows that the analytic function (1 - z)-le,f,(z) in Dl (1 )[e,f,(z)]. + e,f,(z) (l_z),f,'(z)et(z) + e,f,(z) has derivative -z 2 = ~~~~~~~2~~~~--- = 0 (l-z) (l-z) and so this function is constant in the disk
et(z) = 1 _ z It follows that
and therefore
Yz E Dl •
= Re[.....!=.L] = _1_ (l-Re z) > 0 Il-z I Il-z I
cos(Im ,f,(z»
Im ,f,(z) E 2nZ + (-n/2,n/2).
the image of the connected set Im,f,
and so is a connected set.
1m ,f,(D l ) c lnn + (-n/2,n/2) ,f,(0) = 0
shows that
n
Dl :
Dl
But
(5)
This implies that
for a single integer
= 0:
r3 > r 2 > r l > 1
1 + r- l lo9[----~3":""] n A• - l > 2' 1 - r 3
is
under the continuous map
(4)
Next we choose
1m t(D l )
so that
n.
And
34
Now for
2 Re[l+w] = ![~ + l+W] = 1_lw1 2 > 1-w 2 1-w 1-
we have
Iwl < 1
-w
)-1 z E -D } r -1 3 z ·• r 2
{w E
(1::
Iw-tl < t). 2'
f(z) = :[~(1 -
(6)
(-1,1)
t > 1
We can therefore form -1
1
1+r3 z
t .
-1) + log t] ,
1-r 3 z
We have from (4) and (6) that the compact set in
o.
is a compact
subset of the open right half plane and so for some lies in
II-wi
Re f(Dr
)
lies
2
and so
( 7)
sup IRe f(Dr
) I = ~ < 1. 2
Also from (6) -1
1m f(1) =~[Re ~(1_!.1 +r3) + log t] n t 1 -1 - r3 2 1 1 + r- 1 = -[log(-. --";;':'="1) + log t] n t 1 - r3
by (3)
r- 1
(8)
=
4" 10g[ 11 -+ r;31 ]
> A
by (5).
We set
(9)
~
= 1m f(l) - A > 0,
and then choose a partial sum of
f
about
o in
p
in the power series expansion
such that in the compact subset
we have
(10)
Ip(z) - f(z) I < min(~,l-~)
Then ( 11)
and for all
p(O) = f(O)
o
by (6) and (3)
35
(12)
(10) (7) IRe p(z) I ~ IRe f(z) 1+lp(z)-f(z) I < IRe f(z) 1+1-11 s 11+(1-11)=1
(13)
1m p(l)
~ 1m f(l)-IIm p(l)-Im f(l) I ~ 1m f(l)-Ip(l)-f(l)
I
(10) (9) > 1m f(l) - 6 = A. Lemma 3.5 If
u E Re A
Proof: ~
If
u > 6
and
By compactness
u
on ~
X
then
5 > 0
on
logou E Re A. X
5 > O.
for some
is any non-zero complex homomorphism of
~
A, then
is a continuous functional of norm 1 (see III, p. 1) and so lifts to a functional of norm 1 on Theorem.
C(X)
by the Hahn-Banach
This extension is represented by a (complex, regular,
Borel) measure
U on
X
so is a positive measure.
IIull
which satisfies Then if
f E A
= 1
= ,..I< 1)
is such that
and Re f
5 > O.
W is
Now if 9 ~ W(g)
a non-zero complex homomorphism of 9 ~
and
the subalgebra With
t 1
~
f
~l(f) Re
A
i~
~2(f)
> 0
right half plane.
v
=
1 -
[z E
r -1 u.
B.
w(u)
Call them
= W(~(f+f»
and therefore
u
¢:
then
are non-zero complex homomorphisms on ~1'~2
respectively.
= ~ ~(f)
+ ~ W(f)
=~
Re W(u)
by two applications of (*).
the spectrum of
lies in
of
as before +
W(q)
B
in the Banach algebra
Re
~l(f)
+
This says that B
lies in the open
As the spectrum is compact it therefore Iz - r
I
< r}
for some
r > O.
Consider
Its B-spectrum lies in the open unit disk so
its B-spectral radius is less than
1:
that is, by the Gelfand
=u
36
spectral radius formula, liml/vn,,~/n p < 1. Therefore by '" n-+"" the root test E n-l/lvnl/ B converges and so the series n=l
- t n-lv n
converges in
B, say to
w.
x E X
for each we get
evaluation at
x
Therefore
wERe B = Re A
log r = w + log r
a contradiction. I/gl/ ... = 1.
p
Let
corridor
Then
We suppose
a E X K
q
and
1m q(g(a»
=
qOg
belongs to
~lm
A.
(1)
o
(2)
1m f(x O )
(3 )
/I f II... ."
~ (xO)
X
g E A
with
\g( a) \ = 1.
A
Moreover
p(l) >
~A.
Call it
f.
< Re f < 1
and deduce g(xO) = 0
The magic number
and define \zl ~ 1
1m q(O) = And as
q
~lm
into the p(O) = 0
is a polynomial,
Thus on
X
0
1m f ( a )
with
Set
V
expo~F
F
= 10goRe f. E A
Let
q(z)
1m q(g(a»
>
~A.
Now by lemma 3.5, 10goRe f E Re A, i.e. there exists
Re
In
is the constant of lemma 3.2.
maps the disk
0 < Re z < 1.
and
to
logou = logo (I-v) +
and finally
be as in Lemma 3.4 for this
~ + ~p(grarz).
10gO(1-v) = w.
Then there exists
A = 16K + 5, where
is
Band
E Re A.
Proof of Theorem 3.1:
and
so
[recall equations (1) and (3) in
the proof of Lemma 3.4J. particular
"liB
and this
\v(x) \ ~ p < 1)
numerical series converges (recall = log(r-lu(x»
s;
is continuous in
"" "" w(x) = _ t n-lvn(x) = _ t n-l(v(x»n n=l n=l
log(l-v(x»
1/""
1/
Now
n-l
and get
F E A
37
(4)
Iv 12 = Re f
(5)
/lvl/..,
= /IRe f/l~
by (1).
< 1
z
NoW for any complex number
and applying this to various
we have
V(X)
we get by (4).
(6)
Observe that for any h - i 1m h(xO)
h
we clearly have that
E A
H(Re h)
=
and so
Therefore from (6) /I(Re v)211 = ~IIRe(v2+f)1I ~ ~[lIv2+f/l.., -
IIm(V£(Xo)+f(Xo » IJ
~ ~[lIf".., - /lvlI! - 11m V2 (xo) IJ
by (2)
by (3) and (5). On the other hand by lemma 3.2 (8 )
h E A, H(Re h)
=h
h)/I..,
s;
From (8) and (5) we get
" (Re V)
£"
since for any liRe hll
= I/H(Re
2I1hll..,.
- i 1m h(X O)
and so
s 4K
which with (7) leads to 4K
~ ~A
16K + 4
- 1 ~
A,
manifestly contrary to the definition of
A.
Corollary 3.6 (Wermer [50])
Let
X
be a compact Hausdorff space,
A
38
a uniformly closed subalgebra of points of
X
ring, then
X.
K E~
which separates the
and contains the constants.
A
Then
If
Re A
~
AIK
be the maximal A-antisymmetric decomposition
is a closed subalgebra of
C(K)
by Bishop's Theorem (1.2) and evidently
is a ring.
Moreover, AIK
functions.
By Theorem 3.1 each
then
is a
= C(X).
Let
~:
of
C(X)
f IK E A IK
for each
so by Bishop's Theorem
for each
= Re
Re(AIK)
contains no non-constant real K
K E ~
f E A
is a single point.
and every
for every
f E C (X)
But and
f E C(X).
Corollary 3.7 (Kahane 1961, unpublished) real function conjugate
u
There exists a continuous
on the unit circle having a continuous real u + iv
v, in the sense that
into the disk, but such that
u
2
extends analytically
does not have such a
conjugate. For
~:
A
in Corollary 3.6 take the disk algebra:
continuous complex functions on the circle admitting analytic extensions into the disk.
Re A
is then the class of real
continuous functions having real continuous conjugates. Evidently
A
does not comprise all the continuous functions
on the unit circle. to
A:
zF(z)
if
F
zz
=1
For example, fez)
=z
does not belong
were an analytic extension of
f
then
would hold on the boundary of the unit disk,
hence throughout the disk by the Uniqueness Theorem for Analytic Functions. z
=0
But the validity of this equality for
is a manifest absurdity.
By Corollary 3.6 then
Re A
AIK
39
is not closed under multiplication, hence not closed under 222 squaring either because of the identity 2ab = (a+b) - a - b • Remarks Harmonic analyzers may wish to translate the corollary into the language of conjugate functions in the sense of Fourier Theory.
We also remark that it is easy to adapt the
techniques of lemma 3.2 to prove some interesting relatives of the last corollary. these.
We refer the reader to Brown [12] for
Chapter IV THE WORK OF ALAIN BERNARD
In this chapter we present some exciting recent work of Alain Bernard.
This work is scattered among Comptes Rendus
notes [4J, [5], [6] but will probably be given a unified exposition in
[7].
The account here follows some Grenoble
seminar notes of Bernard and the recent lectures [22] of Glicksberg.
We thank both of them for making this material
available to us. Lemma 4.1 Let C(X)
X
be a compact Hausdorff space, A
a subalgebra of
which contains the constants and separates the points
of
X, K
If
K
that
a maximal set of antisymmetry of
is not a singleton, there exist
A
n C JR(X),
x l ,x 2 E K, f
A > O.
E A
such
f(X l ) ~:
There is
9 E A
K.
which is non-constant on
Pulling off an appropriate constant and taking an appropriate scalar multiple, we can have g(x l ) 1
=0
= g(x 2 )
some xl E K sup{ Ig(x)
I:
By lemma 3.4 there exists a polynomial that
p
maps the open disk
the corridor
IRe zl < 1
D
=
some x 2 E K.
x E K},
(z E~:
and satisfies
p
and an
€
Izi < 1 + €) p(O)
= 0,
> 0
such
into
1m pel) > A.
41 Set
H = {x E X:
disjoint from of
K.
n C~(X),
A
hx E A
I
\g(x)
d.
~ 1 +
Because
K
is a maximal set of antisymmetry
there exists for each
which is real on all of
{x} UK
x, K
Ihx (x)
hx (K) = 1, For example, if t<
=
1\r(x) F- O.
[
ql
9:C
lI~cll""
For
E A
<-nd-- , II hx /I "" s 1.
I
"g" ..
C lR(X •
=c
cp( K)
and
]2 E n )
hx
K): say, without loss
"gil .. ;;., \g(x 2 ) I = 1)
n C lR (X) A
a function
but not constant on
X
(though of course constant on
of generality (recalling that
first
This set is compact and
Then
0
F-
slits
cp( x)
consider
1, !JI (K) = 0,
1 - -1\1.
take a sufficiently high power of
There is a neighborhood
Nx
of
x
in which the middle
inequality above persists: sup{ Ihx(Y) I: yEN }
x
and set
h
= h x3 ·hx 4 • ••• ·hxn h(K)
For the function g' (x 2 )
=
= 1, g'
<-nd--. "g" ..
and get
sup{lh(x) I:x E H}
= g·h
in
s-nd-ligll..,
A we have
•
g'(x l ) - 0,
1,
Ig(x) I < 1 + Ig(x) I Ih(x) I <
Ig' (x) I •
{ "gil"" •
and so since
X
The function
f
IIgt
is compact
= pog'
in
A
then satisfies
€
x'H
= 1
x E H
42
f(X l )
= p(g'(x l » = p(O) = 0
Im f(x 2 ) IRe f(x) and so
= Im p(g'(x 2 » = Im p(l) > A I = IRe p(g'(x» I < 1 for x
EX,
liRe fll.., < 1.
Lemma 4.2 Let lying in
X
be a compact Hausdorff space, A
C(X)
which separates the points of
the constants.
If
Re A
separates the points of ~:
II
Let
f
X
and contains
is uniformly closed in
Cm(X), then
X.
IIA
p. 1) that for all
a Banach algebra
denote the norm in
A.
Recall (III,
E A
Therefore .... Re f
f
is a norm decreasing map of uniform norm.
As
Re A
A
onto
Re A, the latter in its
is by hypothesis a Banach space in
this norm, the Open Map Theorem (for real Banach spaces) M> 0
provides an (1)
u ERe A
&
such that
/lull..,
:s;
1 ~:ilv E Cm(X) such that u+iv E A
&
/Iu+ivHA < M. If we suppose, contrary to what is claimed, that not separate the points of symmetry set x l ,x 2 E K
and
K
Am
does
X, then some (maximal) Am-anti-
is non-degenerate and lemma 4.1 then provides f
such that
43
(2)
APply (1) to Evidently u(X l )
so
With this
u = Re f.
h E A
= Re
and
h
v
consider
is real-valued on
f(X l ) = 0
and
X.
h = f - (u+iv) i
But
f(X l ) = 0
h(X l ) = -v(X l ),
Ih(X l ) I = Iv(x l ) I s: /lvl/.., s: /lu + iv/l.., s: IIu + ivl/ A s: M Ih(X 2 ) I = Ii 1m f(x 2 )-iv(x 2 ) I ~ 11m f(x 2 ) 1-lv(x 2 ) 1~2M+l-lv(X2)1 ~ 2M+l - /lvl/.., ~ 2M+l - /lu + ivl/.., ~ 2M+l - /lu + ivl/ A ~
Therefore
by (1).
M+ 1
h E A n C ]R (X)
the fact that
x l ,x 2
but
h(x l
) -/r. h(x 2 ), contrary to
both belong to the set
set of antisymmetry for
K
which is a
A n C ]R (X).
Corollary 4.3 (Hoffman & Wermer [29]) space and
If
X is a compact Hausdorff
A is a uniformly closed subalgebra of
separates points and contains the constants, then uniformly closed implies ~:
which
Re A
C(X) •
A
The last lemma allows real Stone-Weierstrass to
be applied to the real algebra A]R
C(X)
is uniformly dense in
of
C ]R (X)
in
C(X), and so
A]R = A nC]R(X).
C]R(X).
But this real subalgebra
is uniformly closed since A]R = C ]R(X).
Therefore
A
Hence
is uniformly closed
A::> A]R + iA]R
C]R(X) +iC]R(X) =C(X).
A. Browder gives another nice proof of this on pp. 88-89 of [8].
Also Arenson [1] gives a proof.
44 More generally Corollary 4.4 If
X
lying in
is a compact Hausdorff space, A
C(X)
which contains the constants and separates
the points of implies
~ A~
must have
Re A
C~(X)
uniformly closed in
C ~(X).
By lemma 4.2 and real Stone-Weierstrass, the
real algebra Re A
X, then
Re A
E!22!:
a Banach algebra
A~ and
C ~ (X).
is uniformly dense in Re A
Since
is assumed to be uniformly closed, we
Re A = C ~(X).
Lemma 4.5 (Bernard's Lemma) normed linear spaces, E Let to
E E
= t=(
Suppose C
F
E
E!22!:
and the injection is continuous.
-=
t=( IN,F).
F
is complete and
E
C
F
E C F. F
If
then
E
M such that
is bounded in
F
so
and evidently
Now suppose
E
is complete and
show first that for some (2)
is dense in
E
= {1,2, •••• }
Then
By hypothesis there is a constant
Therefore any bounded sequence in E
are (real or complex)
IN,E), the bounded functions from IN
normed as usual, and
in addition
E,F
x E F
&
E
is dense in
F.
We will
r > 0
/lx/l F < 1 => !!ly E E, lIyl/E < r
Indeed, if this fails for every
&
II x -yllF
< 1/2.
r > 0, we can select a
= F.
45
x
sequence of points
EF
n
with
(3)
such that I\xn - yI1 F ;;, ~
(4.n) Now
(x n } E F
[Y n } E E
with
(5)
But
for every E
and
ly n } E E
~ ~
for each
n
= 1,2, •••
means that for some constant n
= 1,2, •••
Finally we use to show that each Y
(2 )
x EF
to prove that with
.
Yl'···'Y n E E
n > K.
FeE.
It suffices
belongs to
IlxllF < 1
E.
Suppose
be chosen according to (2).
= Yl
.
K
(5 ) and (6) contradict (4.n) when
Then
Let
{yn}\lF' ~ ~, that is,
llYn liE ~ K
(6)
with l'yll E < n.
is dense therein, so there exists
lI{x n )
I\xn - YnllF
Y EE
chosen so that
(7.n)
(B.n) Apply (2) (appropriately scaled) to find
Yn+l E E
x - (Yl + ••• + Yn )
to
with
(B.n+l) by (7.n) such that \Ix - (Y l + ••• + Yn ) - Yn+lll F s: ~llx-(Yl+ ••• +Yn)nF s: 2- n - l
This completes the inductive construction.
Since
by (7.n). E
is
46 eo
complete, the series
L y n=l n
converges in
to
y.
F.
But by (7.n) the series converges to
E
by (8.n), say
Therefore by (1) the series also converges to
x
in
F.
y
in
It
x = y E E.
follows that Theorem 4.6 (Bernard)
Let
X
be a compact Hausdorff space, A
Banach algebra lying in
C(X)
contains the constants.
If
A
a
which separates points and Re A
is uniformly closed,. then
= C(X). Proof:
Let bar denote uniform closure.
uniformly closed, Re
Re A = Re A
so
Re
A=
Re A
Re A
is
is
The Hoffman-Wermer result (4.3) is therefore
closed.
A
applicable to
A
Re A = Re
( 1)
AC
Since
A
and we conclude
= C(X).
Whence
= C lR (X) •
Now, as observed before (III, p. 1), we have
1\ l\x s: 11 IlA
and therefore if we define !lull = inft \1fIlA : f then
n 1\
E A, Re f = u}
u E Re A = C lR (X)
is a complete (quotient) norm in
C lR(X)
which
dominates the uniform norm.
By the Open Map Theorem there
is a constant
II II
such that
K
s: K II
IIx.
\I II
Since
is a
quotient norm, this implies (2)
Yu E C lR(X) :!Iv
+
(K
Since (3)
1\
1)
IIx
such that
u + iv E A
and
I\u + ivllA s:
I\u 11x-
s:
11
HA
A = .t eo ( IN,A)
is a suba1gebra of
.teo ( IN,C(X))
C(X)
47 whence Re ACRe .to'( IN,C(X)). But (2) implies that
Re .t..,( IN,C(X))
Re A, for if
C
(un} E Re .t..,( IN,C(X)) [= .tQ)( IN,C JR(X» note], say n, then by (2) there exist
for all
un + i v n E A
and so
We have therefore
Re A = Re .teo ( IN,C(X)).
( 4)
Now if IN X X IN
such that
nUn + i v n nA ~ (K + l) M
and
{un + iVn } E A.
vn
llunllx ~ M
X
X, then
C( IN X X)
v
denotes the Stone-Cech compactification of .teo ( IN,C(X))
and
is naturally identified with
with a subalgebra thereof and (4) translates
A
into Re A = CJR(lN X X). In particular
separates the points of IN X X
A
bar denoting uniform closure in CJR( IN X X) = Re
A.
C( IN
X
and, with
X), Re ACRe A
C
Hoffman-Wermer (4.3) therefore applies
10
to
A
and says that A
= C(
IN X X),
that is,
(6)
A
is dense in
.t..,( IN,C(X)) = C(X).
It follows from Lemma 4.5 that
A
= C(X).
Corollary 4.7 (Sidney & Stout [47])
If
X
is a compact Hausdorff
space, A
is a uniformly closed point-separating subalgebra
of
C(X)
which contains the constants, Y
of
X
and
Re AIY
is uniformly closed in
is a closed subset C]R(Y)' then
48
~:
Apply Theorem 4.6 to the algebra
AIY
with its
quotient norm (IV, p. 2). Our next corollary (and the proof given) is valid in any
compact abelian group with totally ordered dual, as the reader will at once perceive.
The idea of deducing this result
from Bernard's theorem is due to J.-P. Kahane. Corollary 4.8 (Wik [51]).
For a closed subset
E
T,
of the circle
the following are equivalent C (E) = .tl (Z) " IE
C(E) • (i) •
For the converse assume that the closed set Let
where
A
= .(..1 (Z+)" IE
E .tl(Z+):W(E)
Banach algebra lying in ~
there is a f
(2 )
we have
E .tl(Z)
= o}.
C(E).
Thus
Given
A
.tl(Z+)/kE
is a complex
f E C m(E), by (i)
such that
= ~(~(n) "A + ~)
~(~
+ ~(-nl)
= Re "~
and therefore, as
f
is real-
valued, (1) gives ( 3)
Since
f
w< -n)
satisfies
=" ~IE.
*(n)
"W=
with the quotient norm of
"
= (w
kE
E
=
"WIE.
=
lTiiT
by (2), we have for every
z E T
49 co
co
~(z) = I:
ao
w(n)zn = 1: w(n)zn + I: $(_n)z-n = 1: $(n) zn + I: Tril)zn n=_ao n=O n=l n=O n=l co
ao
co
= 1: w(n)zn + t $(n)zn n=O n=O
$(0) = Re [- W(0) + t 2 $ (n) zn] n=O
( 4)
It follows from (3) and (4), since
f
E
C~(E)
is
arbi trary, that C ~(E)
!
fortiori then
the constants and
Re A. A
separates the points of Re A
is closed in
E
C~(E).
and contains So by Bernard's
A = C(E).
Theorem (4.6) Definition 4.9 Let
X
be a compact Hausdorff space, A
a (real
C C(X)
or complex) normed linear space continuously injected in A
C(X).
is called ultraseparating if
[cf. lemma 4.5] as a subset of
t
A = .tao ( IN,A), regarded
ao ( IN,C(X))
= C( IN x X),
separates the points of IN x X, the latter space being the Stone-Cech compactification of IN x X. Lemma 4.10 Let
X
be a compact Hausdorff space, A
a (real
C C(X)
or complex) normed linear space continuously injected in C(X). If
A
Space in
is ultraseparating then for each compact AIK
C(K)
K
C
X
the
with the quotient norm is continuously injected and is ultraseparating.
50
Proof:
n llx flK
:s;
IIA•
Ml\
=g
By hypothesis there is a constant Thus for any
g E AIK
and any
M such that f E A
with
we have
Therefore
As the latter infimum is the definition of
I\gnA IK' we have
and so
C(K).
AIK
is continuously injected into v
Clearly we may regard the Stone-Cech compactification of IN X K
as a closed subset of IN X X.
u1traseparating, i.e. that
(AIK)~
To show that
A
that if
f
n
AllN X K
separates the points of IN x X
the points of the subset IN X K. E A
and
co
,
(A IK)~,
and so surely
then
s~pllfnlKIIAIK
is the extension of
f
to a continuous function on lNX X is the extension of
C
But this amounts to showing
s~pllfnllA <
(and this is obvious) , for if
is
separates the points of
IN X K, it is then enough to show that because
A IK
then evidently
<
co
(f n }
fllNXK
(fnIK}.
Lemma 4.11 Let
X
be a compact Hausdorff space, A
closed suba1gebra of X in
C(X)
which separates the points of
and contains the constants. C~(X),
then
A
a uniformly
If
Re A
is ultraseparating.
is uniformly dense
51 ~:
If
U = 10g o f E
fEe
Given
C ]R(X).
f > 0
and
]R(X)
€ >
on
then
X,
there is then a g E A
0
with lIu - Re gllx < €. Then a simple application of the Mean Value Theorem (to eU(x) _ eRe g(x)
for each
x E X) gives
lieU - eRe gllx s lIu _ Re gllx. ellu - Re gl\x +
lI u llx
s €e €l1fllx' that is,
If
x,y
with
E
IN
and
X X
cp(x) = 4, cp(y)
x -F y, Urysohn provides a
cp
o.
as
1
q; >
and
g
n
EA
cp
IN
X X)
CPn > 0
Thus each
to produce
Regard
E C(
and we may apply (1) gn such that the element h n = e of A
satisfies (2)
It follows that, (hn } belongs to to IN
X
X.
t~(
IN,A).
By
(2)
is bounded (since Let
h
[ tn} ....n
is) and so
be its continuous extension
and the densensss of IN
X X
in IN
X X
have I\cp -
Ih I n
s 1
JNXX
and so
It follows that
A
14 -
lhex) II
Icp(x) -
Ih(X) I l S I
11 -
lhey) II
Icp(y) -
Ih(y) II s 1.
lhex) I -F lhey) I.
As
separates the points of
h E t ... ( IN,A) IN
x
X.
A, we see
we
52
Lemma 4.12 (Bernard [5])
Let
X
be a compact Hausdorff space, A
and B Banach algebras lying in B
is conjugate closed and Proof:
and
If
so does
is u1traseparating,
1 E A C B, then
B
= C(X). II II X s: II II A
liB' so it is easy to apply the Closed Graph
Theorem to see that the inclusion of Therefore
A
As noted before (III, p. 1) we have
"x s: II
II
C(X).
A
C
B.
B.
Since
A
A
into
B
is continuous.
separates the points of IN x X,
Moreover an application of the (real) Closed
Graph Theorem in
B, using as before
conjugation is continuous in is conjugate closed:
B.
II II X s: II \I B' shows that
This clearly implies
II ffl B s: MllfliB
if
for all
implies M S~Plfn"B' i.e. with
with
II (fnlllB' s: MII{fnlllB'.
B
fEB, then supllf liB s:
Thus
n
n
B
is
(naturally identified with) a conjugate Closed point-separating suba1gebra of
C( IN x X)
Stone-Weierstrass C(X)
B
and therefore
which contains the constants.
is dense in B = C(X)
C( IN x X)
= .t..,(
By
IN,C(X))
=
by Bernard's lemma (4.5).
Theorem 4.13 (Bernard [5])
Let
X
a Banach algebra lying in and is ultraseparating. tion, then ~:
A
=
C(X) If
which contains the constants
Re A
is closed under multiplica-
= C(X).
Let
B = Re A + iRe A.
under multiplication, B E
be a compact Hausdorff space, A
{f E A: Re f
= 0).
Since
is a subalgebra of Then
E
Re A C(X).
is closed Let
is a real linear space and
53
is closed in
A
because
liRe fll ...
II fll ...
:s;
:s;
II filA
(III, p. 1).
Therefore u E Re A is a quotient norm in II
IIx
:s;
II II.
Re A, hence is complete.
(Cf. IV(c), p. 3) o~
easily mimic the proof
The reader can therefore
lemma 3.3 (i.e. a Closed Graph
argument) to see that multiplication in
II II-continuous.
Re A
lIu + iVlI = lIuli +
Then
Moreover
complete real linear space norm in
B
II VII
1S
jointly
defines a
II fll B = sUPII e i Sfll
and
e
is an equivalent complex linear space norm.
Again an
imitation of the proof of lemma 3.3 shows that multiplication in
B
is jointly
a Banach algebra.
II II B -continuous.
Therefore all the hypotheses
lemma are met and i t follows that Re A
= Re
B
= Re
It fOJ.lows that
C(X)
= Cm(X).
C(X).
B
As
A
it certainly separates the points of Theorem (4.6) applies and gives
A
o~
B
is
the last
In particular
is ultraseparating,
X, so Bernard's
= C(X).
Corollary 4.14 (Wermer)
Let
X
be a compact Hausdorff space, A
unitormly Closed point-separating subalgebra of Contains the constants. plication, then ~:
A
If
Re A
C(X)
which
is Closed under multi-
= C(X).
By the real Stone-Weierstrass Theorem
uniformly dense in
a
Cm(X).
Therefore
A
Re A
is
is ultraseparating
by lemma 4.11 and the conclusion follows from the last theorem.
54
Defini tion 4.15 Let
X
cp: S -+ JR (1)
(ii)
be a topological space, A
C
C(X), S
and
JR
C
a function.
Say
cp
and
f(X) C S.
If
A
operates in
A
such that
f E A, f(X)
if
cpof E A
whenever
cp
is a normed linear space say
boundedly in M(e) > 0
A
S, and
C
if for every
cpof E A IIfllA
and
€
> 0
IIcpOfli A
f E A
operates
there is an S
M( €)
whenever
€.
S
Lemma 4.16 Let
X
be a compact Hausdorff space, E
closed vector subspace of constants.
Let
supported.
If
~:
CJR(X)
cp,h: JR -+ JR h
operates in
E
then so does
cp
compactly
~h.
It is a routine exercise in epsilonics to move
thereto on compact subsets of JR translates of operates in
h.
Since
E
fEE,
(~h)Of
(~h)
cp.h
to uniform approximations
by linear combinations of
contains the constants and
E, any translate of
h
also operates.
h O(f - t)
for various
h
So for
is uniformly approximable on
by linear combinations of hence
which contains the
be continuous with
from the integral definition of
a given
a uniformly
t
X E JR,
Of E E.
A less direct proof (Hahn-Banach, Riesz Representation and Fubinito all intervene) but one of a type which occurs frequently elsewhere in this monograph (and so would be pointless to eschew here) is as follows. show that
JX( cp.h)
0
fdIJ = 0
for every
It suffices to
fEE
and every
55 ~
finite regular Borel measure For any
t
E JR
we have
on
X
which annihilates
E.
ho (f - t) E E, as noted above, and
~
so for such a
o
Yt E JR.
The most elemental form of Fubini's Theorem (finite regular measures, continuous functions) gives
J JRCil(t)[Ixh
(f-t)dU]dt
by (*).
= 0
Lemma 4.17 Let
X
be a compact Hausdorff space, E
closed subset of
C JR (X).
If
Ciln : JR'"
JR
a uniformly
operate in
Ciln ... Cil uniformly on compact subsets of JR, then in
Cil
E
and
operates
E. ~:
Ciln ... Cil
Let
fEE.
uniformly on
We have
Cilnof E E
Closed.
So
f(X)
is a compact subset of JR
f(X), i.e.
Cilnof'" Cilof
by assumption on
Ciln
and
uniformly on E
Cilof E E.
a < b
are real numbers, there is a C'" function
which is supported in ~:
For
(a,b)
c E JR
and strictly positive there.
define
X.
is uniformly
Lemma 4.18 If
so
56 1
e t-c
o
t <
C
t
c.
~
The reader may confirm by induction that for each non-negative integer
n
there is a polynomial
( 2)
such that Yt < c.
P n (t:c)
lim xme- x = 0
Since
Pn
for every integer
m, it follows from
x-+ co
(2) that (3 )
lim
Therefore
(_co,c)
=
0
n = 0,1,2, •••
Note that
is supported in
and is strictly positive there.
function
Therefore the
defined by
t
E lR
has the desired properties. Lemma 4.19 Let
h,
be continuous,
and continuously differentiable. and
> 0
Let
be given.
the compact set (1)
is differentiable
(!p*h) , =
£
Then
compactly supported
be supported in
[a,b], let
The continuous function
h
Xo E lR
is bounded on
[XO-b-l,xo-a+l], say
Ih(t)1 <M
The function
is uniformly continuous on lR
supported in the compact set o < 6(£) < 1
such that
(being
[a,b]) and so there exists
and
57
lu-v I < 6( E:) ~ Ic:pl (U)_c:p' (v) I
(2 )
If we consider only
x
with
Ix-xol < 1, then
..,
= S_cJ c:p' (xt-t)_c:pl (xo-t) ]h( t)dt
( 3)
between
for some Since
c:pl
and
is supported in
we get from I
x
x o ' by the Mean Value Theorem.
[a,b]
Ixt-XOI ~ Ix-xol < 1,
and
(3)
(cpA'h) (x) - (cplrh) (x O) x-xO - (c:pl *h) (X O )
xO-a+l
1= IS xO-b-l [c:pl (xt-t)_c:p' (xo-t) ]h( t)dt I (1) xO-a+l ~ M] x -b-ll c:p' (xt-t)_c:pl (xo-t) Idt.
( 4)
o
But E lR.
yt
Therefore if
Ix-xol < 6(E:)
it follows from (2) that Yt E
lR
and so (4) gives I ( cpA' h) (x) - ( cpk h) (x O) €,
x-xO holding whenever
Ix-xol < 6(€).
Lemma 4.20 Let
c:p,h: lR .... lR
supported in integer
[-1,1]
n, set
c:pn(t)
be continuous, with 1
and
]_1C:P
= nc:p(nt).
on compact subsets of lR.
= 1.
Then
c:p
non-negative,
For each positive c:pn*h .... h
uniformly
58
Proof:
Note that t:P.n
is supported in
[- 1. 1.J n'n
and (2)
Yn.
1
Therefore
'"
'"
'"
= S_",[h(x-t) - h(x)J~n(t)dt 1
= Sn
( 3)
1 [h(x-t) - h(X)J~n(t)dt. n
€ > 0
Given
[-M-l,M+1J
M> 0
and
the uniform continuity of
h
implies that there is a positive integer
on n(M,€)
such that (4)
\h(x-t)-h(X)
I
< €
Yx E [-M,MJ
It follows from (4) and (3) that for
viti < l/n(M,€). n > n(M,€)
and all
x E [-M,MJ 1
1(~n*h)(x)-h(x)1 s
sn -
1 Ih(x-t)-h(X) I~n(t)dt
-n
1 n
s
s_
1.€~n(t)dt
by
€
(l)
and (2).
n
Theorem 4.21 (deLeeuw & Katznelson [31])
Let
X
be a compact
Hausdorff space, E
a uniformly closed point-separating
vector subspace of
C~(X)
which contains the constants.
If any continuous non-affine function on then
E
=
C ~ (X) •
~
operates in
E,
59
It suffices to show that
~:
E
contains the square
of each of its elements. For then E is an algebra: 222 fg = (f+9) -2 f -g ,and we may cite real Stone-Weierstrass. Let
h: JR... JR
operate in
non-affine, i.e. for some
and be continuous and
E
a,b E JR
and some
0 < ~ < 1
(1) C""
Use lemmas 4.18 and 4.20 to produce a compactly supported function
~
with
I~h - hi
so small at
a,b
and
Xa + (l-~)b
that the inequality (1) holds as well for the function H = c:pIrh, i.e.
H(Aa +
(2)
(I-~)b)
I:
~H(a)
By lemma 4.19 the function affine. with
Therefore
H"(t ) I: 0
o
H"
+ (1-) )H(b).
H
is
C""
and by (2) is not
is not identically
Pick
O.
and consider t E JR.
( 3)
Then
is
t
(4)
C"" f( 0)
but not affine. " (0) = 0
By lemma 4.16 the function h
to E JR
does and so (since
too.
and H =
Moreover f" ( 0) = 2 • ~h
operates in
t
1 E E) the function
Taylor I s Theorem provides for each
t
E
because
operates in
E JR
a
e( t)
E wi th
la(t) I ~ It\
(5) such that
(6)
where
dt) = !t(''''oa)(t).
Notice that by (5)
e
t
E JR
t
E lR
is continuous
60
at
0
€
SO
is also
lim d t) t-+O
( 7)
=
O.
If for each positive integer
= n 2 t(*),
'n (t) then
'n
(8)
=t2
'n (t)
+ t
€n (t)
=
t d-) n
t E JR
and from (6)
E
operates in
we set
n
2
€n (t).
Notice that by ( 7) we have
o uniformly on compact subsets of JR. In particular if
fEE
then
€nof -+ 0
uniformly on
X, and
so
Since
'nof E E
and
E
is closed, it follows that
f2 E E.
Lemma 4.22 Let
X
be a compact Hausdorff space, E C CJR(X)
Banach space continuously injected in
CJR(X)
which contains
the constants, separates points and is a lattice. x E X
E\Kx
F
E. C
X
Let let
Kx
such that
~(t) = It I
operates boundedly
with its quotient norm.
~:
in
Then each
with at most finitely many exceptions has a compact
neighborhood in
a
Since N NF
If I
=
fVO - fAa
denote the norm in
we see that E
~
and for each compact
denote the quotient norm induced in
N:
f}
operates
ElF
by
61
Notice first that if
E
is in
a function
o.
u(X\U)
Indeed
that for each
u
K with
1 E E
(2 )
x E K, Y E X\U
x E
Cover compact
f;~y(l,CD)
u(X) C [0,1], u(K) = 1,
and
fx,y(X) > 1, fx,y(Y) <
with
compact c U open c X, then there
E
point separating means
there is a function
o.
f
E E
x,y
Thus
open, y E
f;~y(_CD,O)
open.
K
and
consider f
Y
= f
fEE. x 1 ,y v ••• v xn'y
This function satisfies K C f;l(l,CD), y E f;l(_CD,O) open. Cover the compact set
X\U
with, say, f y- 1 (_CD,0), ••• ,f;1(_CD,0) 1 m
and set f
f
Y1
/\... /\ f E E . Ym
This function satisfies
Finally let
u =
(f /\ 1)
v 0 E E.
This function satisfies
It follows that any which vanishes on and
-(g /\ 0)
for
g
X\U.
has an extension
By considering separately
gEE g v 0
and scaling, i t suffices to demonstrate this
satisfying
extension and
gEE IK
u
g(K)
C
[0,1].
If then
is as above, the function
f
is any g
=
(f v 0) /\ u
62
has the desired properties. We now use the hypothesis that into
evaluations at the points of E
X\U
and so the set
k(X\U)
E. X
This evidently makes continuous linear functiona1s
of
fEE
is a closed vector subspace of g ~ glK
paragraph, the map course continuously. is a constant (3)
is continuously injected
Cm(X), that is, the supremum norm is dominated by a
multiple of the given norm in
on
E
maps
which vanish on
E.
As noted in the last
k(X\U)
onto
ElK, and of
Therefore by the Open Map Theorem, there
M(K,U)
such that
Yg E ElK :!!g E k(X\U) with glK = g
&
N(g) :s: M(K,U)NK(g) •.
We now prove the assertion of the lemma by reductio ad absurdum. ~
Suppose
Xo
is an infinite subset of
X
such that
operates bounded1y in no neighborhood of any point of
Let
x
be a limit point of
XO' xl
V1
disjoint open neighborhoods of
If
x 1 , ••• ,xn E Xo
U1 ' ••• 'Un 'V n
of
xn+1 E Vn
n
hoods of
x n +1 'x
any point of xl
and
x
XO.
XO\(x}, U1 '
respectively.
and disjoint open neighborhoods x 1 ' ••• ,xn ,x
(X\(x})
and
have been selected, let
Un +1 'Vn+1
respectively in
be disjoint open neighborVn •
We have inductively
constructed
Let ~
Kn
C
be a compact neighborhood of
does not operate bounded1y in
E) there exists (5)
Un
gn E EIKn
EIKn
such that
xn•
Since
(but does operate in
63
and NK (\gn \) ~ n.
(6)
n
According to (3) then there is
gn E k(X\Un ), that is with
( 7)
such that
(8) and
It follows from (9) and (5) that g
=
.9
converges in L n=l n
E.
As
N(gn) E
::s;
2- n
and therefore
\91
is a lattice,
E E.
Moreover (10)
= 1,2, ••• ,
n
because if by (7).
k ~ n
we have
Kn C Un C X\Uk
whence
gk(K n ) = 0
From (10) and the definition of the quotient norm
we have n = 1,2, ••••
The validi ty of this inequaU ty for
n > N(
\9 \ )
is in
contradiction with (6). Lemma 4.23 Let
X
be a compact Hausdorff space, E C C JR(X)
Banach space continuously injected into the constants and is ultraseparating. affine function on JR ~:
Let
cp
CJR(X), which contains If any continuous non-
operates boundedly in
E, then
be a continuous function on
operates boundedly in
E.
a real
That is, for each
€
JR ~
0
E
which there is
64
M(€) > 0
a
such that
(u ) E E
If
n
~
(2 )
Consider
E
= t m(
IN,E), it follows at once from (1) that
operates in
E.
as a subspace of
C lR ( IN x X),
'¥'
Cech compactification of IN x X.
IN x X
the Stone-
If bar denotes uniform
closure, then it follows from (2) that operates in E. ,... fk E E and converge to ~
(3)
Indeed if
then there is a compact subset fk ( IN X X)
C
K
for all
fk
converges to
converges to
~ofO
uniformly on IN x X,
of lR
fO
such that Since
k = 0,1,2, •••
continuous on the compact set that
K
fO
~
is uniformly
K, it follows from the fact
uniformly on lNXX
uniformly on lNX
that
~ofk
,...
X.
As each
~ofk
E E
7V
by (2) we see that
~ofO
E E.
The ultraseparation hypothesis means that E, separates the points of IN X X.
If
~
E, and so also
is not affine it
follows from (3) and the Katznelson-deLeeuw Theorem (4.21) that t
m(
-
E = C lR ( IN X X).
IN,ClR(X)) = ClR(X).
tha t
Hence
E
is dense in
C lR ( IN X X)
It follows from Bernard's lemma (4.5)
E = C lR (X) •
Corollary 4.24 (Bernard [6])
Let
X
be a compact Hausdorff space, A
a uniformly Closed point-separating subalgebra of contains the constants.
C(X)
which
If some non-affine continuous function
65 operates boundedly in
Re A
with its quotient norm, then
A=C{X). Proof:
Let
lR
ql:
operate boundedly in
lR
-t
Re A
be continuous, non-affine and with its quotient norm.
shown in the course of the last proof, uniform closure
Re A
of
Re A.
operates in the
ql
Since
Then, as
Re A
separates
points it follows from the Katznelson-deLeeuw Theorem (4.2l)
ReA
that
= C lR{X).
Then by lemma 4.11, A
is ultraseparating.
We next deduce from this last fact that quotient norm is also ultraseparating. notion makes sense because for any liRe £Ilx :s: inf{llgllx:g E A, Re 9
(*)
f
E A
= Re
is continuously injected into
definition 4.9. Since A
A
Now let
in its
Notice first that this
(by definition of the quotient norm in
Re A
Re A
f} = liRe fli Re A Re A) and therefore
ClR{X)
as required in
x,y E IN x X, x ,t. Y
is ultraseparating there is
'" E A
f
be given. with
f{x),t. f{y).
being a complex linear space, we may suppose without loss
of generality that sion to IN x X
Re f{x) ,t. Re f{y).
Now
f
is the exten-
of some
Since the norm in
Re A
is the quotient norm,we have
liRe fn"Re A:S:
(* )
II fnll A. h h
Therefore
(Re fnl E t..,{ IN,Re A)
is the continuous extension to IN x X
= Re
h{x)
f
= Re
on the dense subset IN x X f(x)
Now that
F Re Re A
fey)
C
.t..,{ IN,ClR{X)). of
{Re fn 1, then
of IN x X
and so
= hey).
is ultraseparating we can apply the
previous lemma to conclude that
Re A
= C~(X).
It follows
If
66 then from anyone of several previous results (for example from the Hoffman-Wermer Theorem) that
A
= C(X).
Theorem 4.25 (Bernard [6])
Let
X
be a compact Hausdorff space, A
a uniformly closed point-separating subalgebra of contains the constants.
ClR(X).
Re A
If
~:
Indeed i f
regulari ty of
Re A
is a lattice, then
A
X
is a real Borel measure on X
a compact subset of
provides open U=>K
IJ
which
= C(X).
is a lattice it is uniformly dense in
IJ
Re A, K
annihilates
If
C(X)
and
which
> 0,
€
IIJI (u\J<) <
so that
€.
At the beginning of the proof of lemma 4.22 (before the completeness hypothesis on the there exists
E
u EEl: Re A
there was used) i t was shown that wi th
It follows that
o = IxudIJ = IuudIJ = IJ(K) IIJ(K) I s JU\J
> O.
holding for all
€
compact
Since
K C X.
+ IU\J
IIJI(u\1<) <
Therefore IJ
IJ(K)
€,
= 0,
for every
is regular i t follows
IJ
= O.
As shown in the second paragraph of the proof of Corollary 4.24, i t follows from the uniform denseness of Re A
in
ClR(X)
ultraseparating.
that
Re A
with its quotient norm is
For any compact
therefore continuously injected in separating by lemma 4.10. C lR(K)
K
C
X, (Re A) IK
C lR (K)
is
and is ultra-
Therefore by lemma 4.23
(Re A) IK
whenever the absolute value operates boundedly in
=
67
E = (Re A) IK.
Consider such a
K.
We have
(Re A)IK
= Re(AIK),
a set theoretic banality, and in fact the two norms are also equal:
if
u E Re(AIK)
the norm of
u
in either case is
inf{lIfIlX:f E A, Re f IK = u}. From
Re(AIK) = C lR(K)
AIK
C(K).
and Bernard's Theorem 4.6 we get
Finally lemma 4.22 ensures that every point
x EX
with
at most finitely many exceptions has a compact neighborhood KX
such that the absolute value operates boundedly in
(Re A) IKx.
By the result of the last paragraph
AIKx
= C(Kx ).
It remains only to cite Corollary 2.13 to conclude that A=C(X).
Chapter V \I
THE THEOREMS OF GORIN AND CIRKA This chapter is devoted to some recent work of two Soviet mathematicians.
Here topological hypotheses are imposed on
the underlying compact Hausdorff space.
The proof of the
major theorem is quite intricate and the more eclectic reader is reassured that this result is not needed in later chapters. For our first result we begin with a lemma which is elementary but of sufficient intrinsic interest to be dignified with the ti tIe "theorem". Theorem 5.1 If
X
is a compact Hausdorff space, then
separable in the uniform norm if and only if ~:
(~)
If
d
radius
n
cover compact
balls for each
on
X
defined by
the functions
fn
X.
X.
If we take the centers of such
Let
fn
{x l ,x 2 ' ••• }
fn(X) = d(X,xn ).
It is easy to see that
separate the points of
X
and so the algebra
C(X)
by Stone-
Evidently this algebra is separable.
i t is the countable set of multinomials in the rational complex coefficients.)
which
be the continuous function
which they generate is uniformly dense in Weierstrass.
is metrizable.
n, finitely many d-balls of
n, we have a countable set
is clearly dense in
is
is a metric giving the topology of
X, then for each positive integer 1
X
C(X)
fn
(Dense in with
69 (=> )
If
{f:n = 1,2, ••• ) n
is q countable subset of C(X)
which is dense in the uniform norm, then these functions must separate the points of
X
d:X X X .... lR
and so the function
defined by
r'"
1 n n=l 2
d(x,y) =
is a metric on
X.
Ifn (X)-f n (y)
I
l+llfnll",
The Hausdorff topology determined by
is weaker than the given compact topology on
X
d
and consequently
coincides with it.
If
A
is a commutative Banach algebra with unit 1, then
the (multiplicative) group A
E
of exponentials of elements of
is open. ~:
show that
t
Letting
B = {a E A: II a-III < I}, it suffices to
BeE, for then
E
converges in
A
= U aB
is open.
aEE
to an element
The series
b, say, and by
n=l checking the coefficients in the various power series the reader may confirm that p. 7 ff.J
In case
A
[For details see Stout [49J, is an algebra of functions, as in the
application below, this latter tedious procedure can be circumvented as was done in proof of lemma 3.5. Theorem 5.3 (Gorin [25])
Let
X
be a compact metric space, A
uniformly closed subalgebra of
C(X)
which contains
that every strictly positive function in modulus of an invertible element of
A.
ClR(X) Then
1
is the
A = C(X).
a such
70 ~:
Let E
A-I
Consider first the case where
A
is antisymmetric.
denote the group of invertible elements of
the group of exponentials of elements of
is then that
= Cm(X).
10g1A-ll
A.
A
and
The hypothesis
If we can show that
E
= A-I.
then we shall have
and then
A = C(X)
follows from anyone of several previous
results. for example from Corollary 4.3. To see that
E
case the subgroup
= A-I
E
Since by lemma 5.2
we shall show that in the contrary
of
E
A-I
has uncountable index in
is open. the cosets of
E
in
A-I.
A-I
will constitute uncountably many disjoint open subsets of Therefore
A
and hence
C(X)
A.
is surely not separable.
contradicting Theorem 5.1. E r~ A-I
So assume f E A-l\E
such that
(1)
f(x O)
For each
loglftl
and some
= 1.
t E m. t loglfl E Cm(X)
which equals
x0 E X
and fOx 1 some
(since
10gIA-ll. so there is an
ft
f
is never
E A-I
0)
such that
t loglfl. that is.
(2)
Iftl
=
Iflt.
Moreover we may scale
ft
to achieve in addition to (2)
(3 )
and then Igli
=
ft
Ig 2 1
is uniquely determined. and
gl(x O )
= g2(xO )'
For if
gl.g2 E A-I.
then consider
h
= gl/g2
E A -1 •
71
We have
Re h
then
=
Ihl
1
h =
so
and
By antisymmetry of
are each constant.
1m h
these constants are
llh E A.
and
1
A
h(x O)
Since
0, respectively, so
h
1
= Re
h
1.
From the uniqueness thus established follows at once that f t +s
= ftf s
and
fO
(the additive group) fs
t -+ f t
and so
lR
into
A-I.
is a homomorphism of
Finally if
belong to different cosets of f t _ s -_ ftf-s l E E , say, f t-s
. otherw1se eh(XO)
=f
=1
Ifl t - s
and
may that
h(xO)
agree at
Xo
= 0,
=
t,t. s
then For
E
= eh ,
h E A•
Th en
lehl, that is, assuming as we thus
the invertible functions
and 'e h/t - s
f
and have identical moduli, hence are identical
by the uniqueness result above.
But this puts
fEE, a
contradiction. It remains to reduce the general situation to the antisymmetric case.
But it is obvious that if
A-antisymmetric decomposition of algebra
AIK
~
is the maximal
X, then for each
K E
~
the
(which by part (ii) of Bishop's Theorem (1.2) is
uniformly closed in this theorem.
C(K»
Since
satisfies all the hypotheses of
AIK
established shows that
is antisymmetric, the result already
AIK
= C(K)
and then
A
= C(X)
from
part (i) of Bishop's Theorem. Remarks Gorin shows by an example in [25J that in general the metrizability hypothesis above cannot be dropped. it is not enough that every positive function in the modulus of some function in
A:
Moreover ClR(X)
be
In [28J and [39J the
authors construct a proper, uniformly closed, point-separating subalgebra
A
of
cC1N)
..,.
("IN'
the Stone-Cech compactification
72
m) which contains the constants
of the positive integers
C~(lN)
and is such that each non-negative function in modulus of some element of
A.
is the
The proof above is essentially
Gorin's, simplified and made more elementary by Sidney and Stout (cf. Stout [49]).
Gorin's example may also be found
in [49], p. 130. v
The next theorem is due to Cirka [15], the (somewhat intimidating) proof below being an elaboration of that in Stout [49J. Theorem 5.4 (Cirka)
Let
X
be a locally connected, compact Hausdorff
space, A a point-separating, uniformly closed subalgebra of C(X)
which contains the constants.
is the square of another, then
A
If each function in
A
= C(X).
For the proof we assemble a phalanx of lemmas. Lemma 5.5 Let
Y
be a connected topological space, fj: Y
continuous functions such that
=1
lim f.(yO) j~CD
)
If there exist positive integers n. lim f.) = 1 uniformly on Y, then
nl
Yo E Y.
that
j~CD
on
< n2 < 1
such uniformly
Y.
n.
for all large that nOn fj
then
¢
for some
)
Proof:
have
~
gjj ~ 1
From
f.)~ 1 )
n.
uniformly on
Y
we have
If.)1 )
> 0
j, so without loss of generality we can suppose is ever zero. uniformly on
Then setting Y.
gj
=
fj/lfjl, we
Without loss of generality
73
n. Ig.) - 11 < 1
Yj.
)
gj(Y) n {e
and so
ni/n .
2 nik/n .
): k E z} =
) e
¢.
Igjl = 1, it follows that
Since
nj min . g.(Y) c U e j{em.8: .Ji < 8 < k~~), J k=l nj) a disjoint union of open subsets of the unit circle. gj(Y)
Since
is connected, it follows that g.(Y) c e )
for some
TTi/n.
k
nj
k E Z.
Igj(y)-ll s
.8
){eTTl.: -
k 1
< 8 <...±....} nj
y E Y
Therefore for all
Igj(y)-gj(Yo) 1+lgj(YO)-ll s
gj(Y O) = fj(Yo)/lfj(yO) I ~ 1
Since
Ie
TTi/n. )-ll+lgj(YO)-ll.
by hypothesis, we thus
have ~
gj
o
Now given
uniformly on
1
Y.
there exists
< E: < 1
I nj fj - 11 S
Yj
E:
such that
jo OO!
joe
Then
n. 1 -
E: S
If.)1 S 1 +
1 - £ < ( 1 - £)
)
lin. )
S
Ifj I
£
S
(1 + d
lin. )
< 1 +
£
that is Ifj I ~ 1
(2)
uniformly on
Y.
fj = gjlfj \, it follows from (1) and (2) that
Since
uniformly on
fj ~ 1
Y.
Lemma 5.6 If
Y
is a connected topological space, f,g
continuous
74 which are never zero and m 2m for some Yo E Y and if f = g2
complex-valued functions on f(yO)
satisfy
= g(yo)
for some positive integer proof: G(yO)
Setting F 2 __ G2 •
and
as soon as the case f 2 = g2.
F
Then
=f
Y
m, then
2m- l
, G
f
=g
= g. 2m- l
we have
Therefore the result follows by induction m = 1
is established.
( f -) g ( f +) g = 0
Thus suppose
and so
Both the sets on the right hand of (*) are closed since and
f + g
are continuous functions on
f - g
Y, and they are
disjoint because at any common point we would have f(y) - g(y), whence
=
F ( yO)
f(y) + g(y)
g(y) = 0, contrary to hypothesis.
Y, since
these sets must be void, and so the other is all of otherwise (*) would constitute a disconnection of Yo E (f - g)-leo}, it follows that
One of
Y.
As
(f _ g)-leO} = Y.
Lemma 5.7 r > 0
There exists an For any disk
a = lale ie E ~
D(a,r) = {z E~:
such that
1m L
~:
Let
with
lzl < I}
i,
Recall
and satisfies
(1)
et(z) = 1 _ z
(2 )
t(O)
= o.
(e -
an analytic logarithm e +
L
i ).
be the function defined at the beginning
of the proof of lemma 3.4. Dl = {z E~:
lal ~ 1, there is in the open
lz-al < r}
maps into t
with the following property.
t
is analytic in
75
From (2) and continuity of
t
0 < r < 1
there exists
such
that (-~/8,
1m {(Dr) c
( 3)
{z E
Iz I
< r}.
~/8)
From (1) we see
ae{(z) = all - z) and therefore, setting L(w) = loglal + is + t(~) a
(4 )
w E D(a, \a\),
we have
I
Yw E D ( a, a \ )
(5 )
and
L
is analytic in
I< \~ a
then (6)
1m L(w)
r TaT = e
s;
r
D(a, \al) ~ D(a,r).
If
w E D(a,r)
and so (3) and ( 4 ) show
+ 1m t(a;w) E S + (-~/8, ~/8).
Lemma 5.8 t > 0, then for all sufficiently large positive
If integers
n
We have [1 -
1
n lin
(I+t)]
•
o <
1 -
1 n (l+t) < 1
and so
( __ l_)n-l ~ 0 l+t
the assertion follows.
Proof of Theorem 5.4 We first show that if (1)
E
n
<
It follows that
[(l+t)n_l]l/n = (l+t)[l_(l;t)n]l/n > (l+t)[l-(l;t)n] Since
1
o < 1 - (l+t)
f
E A,
compact, connected non-void
£ > C
0
and
Ifl-l(l,~)
1 )n-l l+t- ( I+t •
76
(2)
F
compact, connected non-void C Ifl-l[o,l)
then there exists
Let
Xo E F
g E A
with
m = 1,2, ••• let
For each
be fixed.
fm E A
satisfy f
( 4)
o <
e
< 1
If I < 1
is compact and
F
Because
2m 2m f + 1. m =
such that
If I <
e
on
F, there is a
on
It follows from this
F.
and (4) that
f2
(5) Now if
m
m
t ~ 1
~
1
then
1 - t n ~ 1 - t > O.
uniformly on
F
as
tn - 1 ~ t - 1 ~ 0
m
~
00.
0 ~ t < 1
and if
It follows that Yt
~
0
and positive integers
n,
and so
It follows from (5) then that (6)
Ifml ~ 1
uniformly on
F.
It is clear from (6) that if we multiply each appropriate
by the m 2mth root of unity (so as to rotate the
argument of
fm(x O)
actually converge to
into 1.
[0
.!!.»
' 2m
f
then the values at
Xo
As this does not vitiate (4), we
may suppose without loss of generality that in addition to (4) we have ( 7)
then
77 It follows from (5), (7) and lemma 5.5 that
F.
uniformly on
(8) If I > 1
Because
E
on
E
and
is compact, there is
1 > 5 > 0
such that
(9) From (4) and the definition of If m - 11 ~ If m I
~
S
we have
m
-
1
112m + 11
If2
- 1
[If\2 m _ IJ l / 2m _ 1
s.
on It follows from lemma 5.8 that there is an
Let
r > 0
be as provided by lemma 5.7.
locally connected and
f
such that
is continuous, each point of
E
V
such that V C S
E
(12)
Vj open connected, diam f(V j ) < r.
VI U ••• U Vq
C
as well.
S
In addition we may require that (13)
E
n Vj
~
¢
for each
j = l,2, ••• ,q.
i6. Select
Xj E Vj , write
open disk of radius (12 )
(14)
r
f(x.)
= r.e J
J centered at
J
and let
f(X j ).
is
diam f(V) < r.
with finitely many such neighborhoods:
(11)
C
2
X
Of course because of (9) we may require
E
~
Because
lies in a connected open neighborhood
Cover compact
mO
D. be the J Note that by
78
and by (9) and (11), r. > 1 + 6. J
D.
that there is in
It follows from lemma 5.7
a continuous logarithm
J
L.
with
J
(15 ) Since
is a logarithm we get from (14)
Lj
=
f(x)
(16) Now let
L.(f(x» e J
Yx E V .•
J
be the continuous function defined at the begin-
t
ning of the proof of lemma 3.4.
(17)
- z 1m Hz) E (_n/2,n/2)
(18) Since
Recall that
viz I
< 1
vlzl
< 1.
we get from (17) m 2 (x» et(-C 2 (x) = 1 + C
If I > 1
S
on
m
(19 )
Yx E S.
If we form (20)
.t.
then
t.
J ,m
m (x) = 2- m.t(_f- 2 (x»
+ Lj(f(X»
Yx E V .
J
is a continuous complex function on
J ,m
(16) and (19) satisfies m ( 21) e 2 tj,m = [1 + f- 2m ).f 2m
= f2
m
+ 1
in
V.
J
and by
Vj
and by (14) , (15) and (18) satisfies (22)
1m .t.
J ,m
c
TT
e.J
+ (- 'B' -
TT n n 2 m+l ' 'B' + 2m+l )
in
Vr
From (21) and (4)
Since
is connected it follows from this and lemma 5.6 that t.
e J,m for some integer
k
+ 2TTik2- m
= k(j,m).
It follows from this expression
79 for (23)
fm
and from (22) that
Re[e
_(2T1k.2- m+9. ) i J f] > 0
is disjoint from (_CIO,O) J according as the exponential there has non-
It follows from (23) that or from
Vm ;", 2.
m
(O,CIO)
f
(V.)
m
negative or negative real part. disjoint from (24)
fm-l
[-1,1]
Moreover by (10), fm(S) m ~ mO.
for all
is
Therefore
is disjoint from either (_CIO,OJ or from [O,CIO) on Vj Ym
~
mQ.
For the next phase of the proof select for each positive integer
N
and each positive integer
m
a function
f
N,m
E A
such that (25)
f
2N N,m =
We have from (4) IIfmHim :S; IIfilim + 1 Therefore
:s;
IIfmHx:S; 2/1f/l x
m 211 f /li
:s;
m m 22 /l f /li •
and from (25)
N
/l f N , m/li (26)
:S; 2/1f/l x + 1
/lfN m/lx :S; (2/1f/l x + 1)1/2N ,
YN,m.
Also (25) and (10) give (27)
IfN,m l ~ (~)1/2N ,L
on
S
for all
It follows from (26) and (27) (recall:
(28)
limlfN I = 1 N-+CIO ,m
uniformly on
Because of (24) and (25), fN rays
[0,CIO)e2ITik2-N
,m (V.) J
S
N
and all
m'"" mO·
0 < 6 < 1) that and uniformly in
m ~ mO.
is disjoint from all the
or from all of the rays
[o,CIO)e~ik2-N+ni2-N
80
(k E Z)
and therefore, as in the proof of lemma 5.5,
(29)
Apply the diagonal process to produce a sequence such that for each
= 1,2, ••• ,q
j
and all
Nl < N2 < •••
m
lim fN (x.) = A(j,m) k....... k,m J
(30) exists.
It follows from (28) that IA(j,m) I = 1
( 31)
j
= 1,2, ••• ,q:
m ~ mO.
As noted before in a similar circumstance, we may, in view of N
(30) and (31), multiply the
f
by appropriate
2 kth
Nk,m roots of unity so as to have, without disturbing (25) lim fN (xl) k .... '" k,m (32)
=1
for every
(29),
m ~ mO' that is,
A{l,m) = 1
It follows from (28), (29), and (30) that (33)
lim f k....... Nk,m
= A(j,m) m
~
uniformly on
mO
for each
Recall now (11), (12) and (13). that
VI
meets one of
j
and uniformly in
Vj
= 1,2, ••• ,q.
Connectedness of
v 2 , ••• ,Vq •
Say
V2 •
E
means
It follows then
from (32) and (33) that (34) Similarly
A(2,m)
= 1.
VI U V2
meets one of
Say
V3' ••• ' Vq.
follows then from (32), (34) and (33) that
A(3,m) = 1.
Proceeding in this way we learn finally that A(
1 ,m) =
...
= A(q,m) = 1
Vr
Vm
~
mO·
It
81
It follows from this and (33) that VI U ••• U V and uniformly q in m ~ mO.
lim f = 1 uniformly on k-+CX) Nk,m
(35)
e > 0
Finally if pick
k
is given we use (35) [and (ll)J to
so that
(36 ) Because of (26) we can also require that
(37)
/l f N
k'
m/lx < 1 +
Then choose for this
k
an
Ym.
f:
m
~
mO
according to (8) so that
and i t will follow from (25) that
(38)
/l f N
k'
m/lF < e.
The reader may now confirm that, in view of (37), (38) and fNk,m (36), the function g = 1 + e does the job promised in (3).
The next phase of the proof is to show that separation like (3) can be aChieved for any two disjoint compact subsets
E
and
F
of
X
with no additional hypotheses.
a pair be given and of
X, for each
with
e > O.
x E E
fx,y(X) f. fx,y(Y).
(1 E A)
we may suppose
and
Since
A
y E F
there is a
on
of
Vx,y.
sets
Y
such that
separates the points f
x,y
E A
By scaling and adding constants
o = IfX,y(X) I <
there are connected open neighborhoods Vx,y
So let such
Ifx,yl < 1
on
1 <
I fx,y(Y) I.
Ux,y
of
Ux,y
and
x
Then
and Ifx,yl > 1
By a finite covering argument we find open connected and functions
such that
82
(39)
E
(40)
Ui
If
Ul U ••• U Up' F
C
0 < c < 1
Ifijl-l[O,l), Vj
C
C
Ifijl-l(l,~).
the function
z-c () cp z = l-cz
(41)
Vl U ••• U Vq
C
is a conformal map of
Iz I ~ 1 Dl = (z E <1::' Izl < l}
onto itself and
satisfies
(42) and
cp(l) = 1
11
Icp(O) +
= l-c.
Therefore we may choose
0 < c ~ 1
so
that Icp(O) + Let
0 <
~
< 1
< 2£/q.
be so small that
Icp( Z )
(43)
11
+ 1
I
< 2 £/ q
for all
Izl < ~
and
(44)
\cp(z) -
11
< 2£
for all
z E Dl with IZ-l\ < p~.
Use the result of the first part of the proof to find
gij E A
with 1,2, ••• ,p; j = 1,2, •• "
If we set g
j
p n g .. - i=l l.J
then we clearly have
IIg j ll x
(46) and on
6
i=l
< 1
we have
U. l.
(47)
On the set
V. J
we have
83
P P P P Ig.-ll s: In gi'- n g .. 1 + I n g .. - n g .. 1 + ••• + IgpJ.-ll J i=l J i=2 1J i=2 1J i=3 1J
s: Iglj-ll + Ig 2j -ll + ••• + Igpj - 11
ll
I/gij x < 1)
(recalling that all
< P11 (48)
by (45)
< P11
I/g. - ll/v J j
= l, 2, ••• , q.
j
Because the power series of
= .qn1 J=
converges uniformly in the
IIgjl/x < 1, the function
closed disk of radius h
~
71(1 -
~ogj
) E A
~og.
J
Form
E A.
•
We clearly have = l,2, ••• ,q
j
so I/hl/ x < 1.
(50)
If
x E F
then
Igj(x) - 11 < P11
x E Vj
by (48).
All the other factors in
for some
j
Then by (44), h
and so 11 - ~(gj(x» I < 2e.
are of modulus less than
1
by
Ih(x) I < e, that is,
(49), so this gives IIhllF < e.
(51)
If for each
x E E
then
Igj(x) I < 11
j = 1,2, ••• ,q
have modulus less than
by (43).
by (47) so
Since all the factors in
So:
t
. 1 J=
14.(1 _ ,£
h
1, i t follows just as in the derivation
of (48) that Ih(x)-ll
1~(gj(X»+ll < 2e/q
~og.) (x) J
- 11
84
that is, (52)
Finally, with (50) - (52) in hand, the proof that is concluded by an application of Hahn-Banach. ~
that if
~(E)
show
We are to show
is a finite regular Borel measure on
A, then
annihilates
= 0
= O.
~
F
n
C
x\E
X
which
By regularity it suffices to
for each compact
choose compact
A = e(x)
E C X.
Use regularity to
with
1~I(X\Fn U E) < lin.
(53)
h n E A with
Use (50) - (52) to choose (54)
/lhnll x < 1, /lhn/lF
n
< lin, /lh n - l//E < lin.
It follows from (54) and (53) that
hn
characteristic function
almost everywhere with
respect to
I~I.
annihilates
A
XE
of
E
converges to the
By Dominated Convergence and the fact
~
~(E).
Corollary 5.9 (Cirka) space, A C(X)
X
be a locally connected compact Hausdorff
a uniformly closed, point-separating subalgebra of
which contains the constants.
of squares in ~:
show that if f = g2 of
Let
{gn}
A
is dense in
If the set
A, then
(g2: g E A)
A = C(X).
To reduce this to the previous result we must gn,f E A
for some
g E A.
and
2 gn -+ f
uni formly on
X, then
Since the limit of any subsequence
which happens to converge uniformly on
X
will
85
belong to
A
subsequence.
and have square We have
~ Re( If 12 + f2)
f, we shall look for such a
(Re g )2 = ~ Re( Ig 12+g2) ~ n ~ n n 1 Re( Ig 12-g) 2 ~ and (Im gn)2 = ~
Ignl2
(Re f)2
~
If\
so
n
~
~ Re( Ifl2 _ f2) = (Im f)2.
n
Therefore to produce the desired
uniformly convergent subsequence of
{gn)
i t suffices to make
two successive applications of Lemma 5.10 Let
X be a locally connected compact Hausdorff space
~n ' W E C lR ( X ) {~n)
2
~n
and
w.
uniformly convergent to
Then
has a uniformly convergent subsequence. Proof:
Notice that
(From [49])
ETKT
For each positive integer
compact c E(2K) open.
connectedness and compactness, E(K) many connected subsets of
E(2K).
finitely many components of them be
K
set
Therefore by local
may be covered by finitely Thus there are only
E(2K)
which meet
E(K).
Let
C(K,l), ••• ,C(K,qK): qK
U C (K , j)
E ( K) c
(2)
C
E ( 2K ) •
j=l For each
K
and
j
pick
~,j
E C(K,j)
n E(K).
diagonal argument to produce a subsequence of converges at each
~,j'
say to
on
~,j' we will show that
{~n}
{~n}
lim ~(~,j) n~CD
which
itself converges
is uniformly Cauchy
X. Notice that
{~n)
Assuming, as we thus
may without loss of generality, that at each
Use a
and
86
2 AK ,J.
1jr('1<,j) > 1/1<, since
=
(3 )
IAK ,J.1 >1/1<
Given
K
K
Therefore
= 1,2, ••••
so large that
1 - AK ,J.1 < '2R'
1qln ('1<, j )
(4)
= 1,2,···,QK:
j n(K)
choose
E E(K).
'1<,j
'q"n
j = 1,2,···,QK
~
n(K)
and such that
IIcp~
(5)
qI~lIx
-
<
1
ZK
Yn,m
~
n(K)
Yn
n(K).
and therefore
11q1~ - ~lIx sZK1
(6)
We claim that for each the function
CPn
j
= 1,2, ••• ,~
For suppose
and for each
has throughout the set AK .
(Recall that the
sign as
~
AK . > ,J
o.
C(K,j)
n
~
the same
are not zero by (3).)
,J
Then
(4) 1(3)1 1 ~ AK . -I ~ (~ .) -A . 1 ~ AK ,J.- 2K > 1< - ZK > CPn ('1<, j ) ,J , J ,KJ
( 7)
Since
C(K,j)
of lR
which, according to (7), meets
it also meets for some 1!r(x) >
zk
is connected, qln(C(K,j»
and so
As
C(K,j)
Icp~(x) - 1Ir(x)
C
I=
If therefore
(0, "') •
(-"',oJ, then it contains
x E C(K,j).
0: that is, cpn(x)
x E E(K),
Sign for all Consider definition of
m,n
~
x E X\E(K).
~(x)
and
I\r(x) >~, in contradiction
cpm(x)
have the same
n(K). Then
= 1Ir(x)
11Ir(x) I
E(K), so from (6) for
0
E(2K), it follows that
In particular we have, recalling (2), for each
o.
is a connected subset
to (6).
(8) {
n(K)
n
~
n(K)
21112 CPn (x) S 1Ir(x) + 2K S K + '2K < K·
s 1/1<
by
87
Therefore
(9)
ICPo (x)-cp, (x) I n m
s;
ICPo (x) 1+ICPo (x) I <2.fi K n m
~
n(K),x E X\E(K).
n,m ~ nOn
If on the other hand, x E E(K), then for
(8)
Vn,m
(6)
J
1
I~n(x)+~m{x) I = I~n{x) I + I~m(x) I ~ 2 w(x) - 2K >
2J ft - ~
by definition of
2 =-.(1.K
(10)
n,m
~
x E E(K)
n(K), x E E{K).
It follows that
Iq£(x)-~;(x)
I
I~n (x) +CIlm (x) I
s;.If . ~ = -1-
( 11)
2J'lK
by (5) Yn,m
~
n{K), x E E{K).
Yn,m
~
n(K).
It follows from (9) and (11) that
As this is the case for any
K, the sequence
(M) Tn
is
uniformly Cauchy, as desired. Remarks As to the status of the local connectedness hypothesis in Theorem 5.4 we remark that some hypothesis beyond the basic compact Hausdorff is needed. compact metric spaces
X
proper subalgebras
of
function in
A
A
For Cole [16] constructs
and uniformly closed, point-separating C{X)
with
the square of another.
1 E A
and every
For the ingenious
details the reader may wish to consult the exposition in
[49], pp. 194 - 201.
Chapter VI
BOUNDED APPROXIMATE NORMALITY, THE WORK OF BADE & CURTIS Definition 6.1 Let
X
be a locally compact Hausdorff space, Co(X)
the continuous complex functions vanishing at infinity on (i)
If
F l ,F 2
are disjoint compact subsets of
e ~ 0, then a function and
(ii)
!1-f(F2 )!
respect
.!2.
Call
C
E
~
Co(X)
S
€
f E CO(X)
(iii)
x
Call
t E
such that
and !f(Fl )!
pair
(iv)
If E
E
regular if
E
contains a O-idempotent
({x},F)
with
F
compact
F. C
Co(X)
€-normal if
E
contains an €-idempotent
E
X
simply normal if it is O-normal.
is a normed linear space lying in
Co(X), call
boundedly €-normal if there is a constant
K
such
that the €-idempotents required in (iii) can be found with E-norm less than (v)
If E
E
€
(F l ,F 2 ).
with respect to each pair of disjoint compacta in and call
S
is called an €-idempotent with
with respect to each pair and
X
X.
K.
is a normed linear space lying in
locally boundedly €-normal if
X
Co(X), call
is a union of open
sets
U
for which there exist constants
that
E
contains an €-idempotent of norm less than
for every pair of disjoint compacta in
U.
KU
such KU
89
Under certain conditions regular implies normal (Theorem 6.4
below).
We shall show next that for normed algebras, (iv)
€ <
for some
~
implies (iv) for every € > O.
Lemma 6.2 Let
X
be a locally compact Hausdorff space, A
algebra lying in some
<~,
€O
CO(X).
A
then
If
a normed
A is boundedly eO-normal for € > O.
is boundedly €-normal for every
We are asserting only that there is a bound, as per (iv) of the above definition, for each that a single bound independent of ~:
€ > O. €
We are not asserting
exists.
€ > 0, lemma 2.3 provides a polynomial
Given
q€
such that
< e/2 Then
Iq€(O) I < €/2
if
Iz -
if
Iz I
11
€O
eO.
s;
P€ = qe - q€(O)
and
s;
is a polynomial
without constant term satisfying
11
(1)
IPe(z) -
(2)
\P€(z) I < € n
< €
i f
Iz
if
Iz
.
t c zJ j=l j
- 1
I
P€(z) =
s;
€O
eO.
s;
n
let
I .
t i c . IzJ. j=l J
Let
K
be a
finite constant such that for any two disjoint compact subsets E,F
of
the pair
X, the algebra (E,F)
with
zero constant term) and
A IIfllA
contains an eo-idempotent s; K.
Then
p Of E A €
( as
f
for
p
has
90
(3) Since
If(E) I s
€o
and
11 - f(F)1 s
€o
i t follows from (1)
and (2) that
that is, Peof
is an e-idempotent for the pair
(E,F).
Remark 6.3 We remark for later use that the version of the last lemma in which all reference to a norm in
A
is suppressed
read the above proof with its 9 th through
is also valid:
14th lines suppressed. In connection with normality hypotheses which will appear in several subsequent theorems, we note that normality is implied by certain other natural hypotheses: Theorem 6.4
(i)
Let
X
be a compact Hausdorff space.
If
A
is a Banach algebra lying in
regular, contains 1 and for which ideal space, then (ii)
If
A
A
C(X) X
which is
is the maximal
is normal.
is a Banach algebra lying in
C(X)
which is
normal and conjugate closed, then the separation in the definition of normality can be achieved by functions with range in
[0,1].
~:
(i)
Let I
E,F
= kF,
be disjoint closed subsets of J
= kE
X
and let
be the ideals of functions in
A
91
vanishing on
F,E
respectively.
f(x) F- 0
then either
Note that if f E I
for some
or
g(x) F- 0
Indeed if
x E F
then
x t. E
so by
regularity there exists
g EA
with
g(E)
0, g(x)
g E J.
some
g EJ
i.e.
g(x) F- O.
and
Similarly if
I + J
follows that the ideal
the maximal ideals of
= 0)
f(x)
for some f + g
say, 1
vanishes on
F
In this case
x E X.
1
0
¢
on
F
and
1
on
[f E A:
=1
f
Let, - g
and
1
X
and
A
on
contains
X.
f EA
If
E,F
is
0
E, then consider
- 4l[ expo(i
TT -
~ff)
which belongs to
A
into
= sin 2 (0) = 0,
[O,lJ, h(F)
is
E.
are disjoint closed subsets of on
Then
1 E A, because by normality
a function which is
I UJ
1 E I + J.
Therefore
on
F- 0,
A, since by hypothesis
f E I, g E J.
and is
for
It
F.
are all of the form
A
where
t
x
generated by
contained in no maximal ideal of
(ii)
x EX
- expo(-i Then
(power series).
2,.,. ff- ) J2 h
maps
X
sin 2 (¥) = 1.
h(E)
Theorem 6.5 (Bade & Curtis [3J) and
c > 0
Let
finite constants, E
equilibrated convex hull of the closed unit disk in (i)
(ii)
X,Y
coe E
~
T:X
Y
~
for each
~).
E
C
be Banach spaces, k < 1
Y
(i.e.
and
coeE
D·co E
the Closed where
Suppose
B, the closed unit ball of
Y, and let
be a bounded linear operator such that y E E, there exists
x E X
such that
D
is
92
Then
'iTx - yll
~ k, )IxI' ~ c.
TX = Y.
I f moreover
T
is 1-1 then
HT-ll/ s: c(l-k)-l,
We need the following simple variant of the Separation Theorem. Lemma 6.6
Y is a Banach space, E
If of
Y, then
coe E
~
B
Yy* E y*.
yEE
Note that
coe E
J=l
consists of all finite sums
n
n
.r
where
c j Yj
then for every
r
E c&,
cj
the closed unit ball
if and only if
I/y*l' s: suply*(y) I
~:
Y, B
C
j=l
y* E y*
Ic j l
~
1, Yj E E •
If
coe E
~
B
n
and
e > 0
we can find
Yo =
r
c.y. E coe j=l J J
such that
But then for at least one IY*(Yj) (since
tlcjl ~ 1)
I
j
> I/y*/I -
we must have €
and we see that
suply*(y) I ~ Hy*/I.
yEE
Conversely, suppose
yO E B\coe E.
By the Separation
Theorem (viz. the Hahn-Banach Theorem) there exists
Yo E y*
such that Re Yo(Yo) > sup{Re Yo(y): y E Since
y E
coe
E
implies
cy E
coe
'COe" E
E).
for every unimodular
E
93
complex
c, the right side above equals sup{ !yo (y) !: y E 'C'Oe E)
and so we get, since
I!yo II s: 1,
Proof of Theorem 6.5: select
For any
y* E y*
and
y E E
x E X by (ii) and note that !y*(y)! s: !y*(y _ Tx)! + !y*(Tx)! s: IIy*/I /Iy - Txl/ + !T*(y*)(x)! s: /Iy*H.k + HT*(y*)II
"xII
s: /ly*lI·k + HT*(y*) /I·c.
Take the supremum over
y EE
and get via lemma 6.6
IIy*/I s: k/ly*/1 + c/lT*(y*)H (2)
/IT*(y*)1! > (1 - k)C-lrry*/I.
It follows at once from this inequality that range.
has a closed
But it is well known and not too difficult that
closed implies [35J.)
T*
T(X)
closed.
(Por a brief proof, see Kaufman
However, it follows from (2) that
if not, there exists
o
~
T*(Y*)
y* E y*
with
T(X)
is dense:
y*(TX) = T*(y*) (X) = 0,
so from (2) 0 < lIy*n s: (1 _ k)-lcIIT*(y*)1I = 0,
summarily, TX = Y.
T
is an onto
Banach space isomorphism and so its adjoint
T*
is also,
and (as shown by simple calculations) (T- I )*
= (T*)-l,
II(T- I )*II = liT-III.
If now
T
is 1-1, then
It follows then from (2) that
94
IIT-lil = II(T-l)*11 = II(T*)-lll(t)(l _ k)-lc.
Bade and Curtis [3] make an interesting application of Theorem 6.5 to Helson sets.
It depends upon the following
fact, of quite independent interest. Theorem 6.7 (Phelps [40])
Let
X
be a Hausdorff space, C(X)
bounded continuous complex functions on (uniformly) closed unit ball convex hull of the set
U
B
of
X.
C(X)
the
Then the is the closed
of functions in
B
of constant
unit modulus. Remarks There is a far-reaching non-commutative version of Theorem 6.7, due to Russo and Dye [44], which is central to the modern theory of Banach *-algebras.
One avatar of i t is
that in any norm closed, adjoint Closed algebra of (bounded) linear operators on a complex Hilbert space which contains the identity operator, the closed unit ball is the closed convex hull of the unitary operators in the algebra.
The
proof by Russo and Dye is neither elementary nor transparent. But recently an exceedingly simple and natural proof was found by L. Harris [26], [27].
As is to be expected, his
idea simplifies still further in a commutative setting and there emerges the following proof of Theorem 6.7. Harris' Proof of Theorem 6.7: f
E C (X)
the number
wi th
"f II co < 1.
~
1-zu
u
For any
It suffices to consider z ,u E
(J;
has modulus 1.
z-u
wi th
Iz I
< 1
Therefore the
Iu I
95
function f
belongs to fu E U.
f-u u =-l-uf'
C(X)
and has constant unit modulus, that is, 1f'1 < 1, we can expand
Since
'" un~ = ncO
= f + (lfl 2 _ 1)
thus:
r uI1fO- l
r.
fu = (f-u)
fu
n=l
r'"
unf'n-l.
n=l Take for fr
(*)
u
for
various roots of unity and average.
If we write
(r positive rational) , this gives
f e 2nir
1 m I: f m k=l kim
-
(lfI 2 - 1 )
f
1 m r
2nik m-(e m )n:wn- l • n=l m k=l '"
r. [1 r
2nin
-(e m)k
Now the average is 1 or o according as m k=l 2nin e m is 1 or not, that is, according as m divides n or not.
Thus in particular the first
right of (*) are
0
m-l
terms on the
and we get the following crude but quite
adequate estimate = (1_lfI2) Ifl m- l
r'"
If In
n=l =
(l-If 12) If Im-l 1 -
As
II~I",
< 1, it follows that
convex sums
If I f
is the uniform limit of the
m
1 r:
m k~l
f
kim
and the proof is complete.
Remarks Because fu(X) c
~\[ru:
II~I",
r
~
< 1, a simple calculation shows that
OJ.
Now there exists an analytic logarithm
96
LU
in this latter region and so fu -1m gu -iL of E C (X) and in fact e
u
so
u
1m gu
=0
and
gu E C JR (X) •
=e
=
igu
=
where g Re(igu ) ! i9 e = e
U! =
Therefore we have proved
more than was claimed, namely that the unit ball of is the closed convex hull of the functions
e ig
C(X)
with
g E C m(X), Bade &
In fact Phelps' original proof also establishes this.
Curtis [3] give a proof of this improved form of Theorem 6.7 too.
For yet another proof, see Sine [48]. Recalling that a point in a convex set is extreme if it
is not the midpoint of two distinct points of the set, and checking the elementary fact that the set of extreme points of
B
B
= co
is just U.
U, we see the geometric significance of
(Compare also the Krein-Milman and Choquet Theorems.)
It is interesting that the corresponding result for does not hold for all compact Hausdorff
X.
CJR(X)
Here again a
function is extreme in the unit ball if and only if i t has constant unit modulus, that is, takes only the values 1 and -1. of
The existence of such a function means a disconnection X
and such functions will be abundant enough to recover
the whole unit ball by (limits of) their convex combinations only if
X
has lots of disconnection.
The theorem, due to
Bade (see [2]), is that the closed unit ball of
CJR(X)
(X compact Hausdorff) is the Closed convex hull of its extreme points if and only if
X
is totally disconnected (that is,
the only connected subsets of Goodner [23]. n ~ 2
X
are single points).
See
However Cantwell [13] has shown that for every
the unit ball in the space of bounded continuous JRn_
valued functions on
X
is the closed convex hull of its
97
extreme points. proof below.
We present the case
n
=
2
of his clever
In this same vein Fisher [18] (cf. also Rudin
[42]) shows that the closed unit ball in the disk algebra of continuous functions on the unit circle admitting analytic extensions into the unit disk is the closed convex hull of its unimodular functions: the latter, it is not hard to show, are the finite Blaschke products and each is an extreme point in the unit ball. Let for each
D = {z E
¢:
x E
E JR.
¢,
€
\z \ s 1}, B(x,
= {z
e)
\z - x \ < e}
E ¢:
Lemma 6.8 Let on
X
X
be a Hausdorff space and
such that
f(X)
D\{y)
C
f
~:
EVidently 9
and
h
D\B(y,l-\y\)
A is continuous. X
f A
f
x E f-l(B(y,l_\yl»
h).
If
on
X
on
with X
\y\ < 1.
with
+ h). by
if
x E f- 1 (B(y,1_\y\»
if
x E X\f- 1 (B(y, l-\Y
\».
So also then are the functions
by
=
f(x) _ iA(x) f(x)-y • If(x)-y I
x E X\f- 1 (B(y,l_lyl»
E f(X)\B(y,l-lyl) then
g,h
= ~(g
f(x) + iA(x) f(x)-y , h(x) If(x)-y I
= ~(g + = h(x) = f(x)
Clearly
and
Define a function
defined on
g(x)
g(x)
C
y E¢
for some
Then there exist continuous functions g(X) U h(X)
a continuous function
C
then
D\B(y,l-\y\), while if
98
Ig(x)-y I
11
= If(x)-y I
and similarly for D\B(y,l-ly\)
~£!
namely that if Given such an
Then
g,h
map all of
X
into
f (X) c.
Theorem 6.7 establishes a little more,
f E C(X)
and
IIfll.., < 1
then
let
= 1 -
IIfll ... > 0
so that
f
If< 0 , 1- E:)
Pick positive integer
(2 )
Therefore
= l-Iyl
[recall, this is an open ballJ.
Cantwell's
( 1)
h(x).
12
+ iJ(l-\y \)2 - \f(x)-y If(x) - yl
€
D \B (l , E:) •
C
m > 2/£
k { Yl = 1 Yk = 1 - m
2
Y2 E B(Yl'£)
f E co U.
S
k
and set
m.
S
so by (1)
Y2
t
f(X)
and by Lemma 6.8
(3.2)
j
= 1,2
such that
(4.2) Proceed inductively.
Suppose for some
2 S k < m j
= 1,2, ••• ,2
such that
(4.k)
2k - l
1
f=-kl 2 -
Then
\Yk+l - ykl
and by (3.k) then
=~
I: fk '. j=l 'J <
~ =
1 -
1Yk\
Yk+l t fk,j(X).
applications of lemma 6.8 we see
so
Yk+l E B(yk,l-Iyk \) Therefore by 2k - l
k-l
99
such that J" = 1 2 ,
,
0
0
0
2k - l
,
and consequently
(4 o k+l)
_ _l_
f
2k - l
r
- 2 k - l j=l
2k - l
f
= -1 r 2k j=l
k,j
and so m =0 D\B(O,l) is the unit circle. Now
f
" m,J
(f
)
Therefore for the functions If
0
k+l,2j
and
y
we have by (3 m) that
+ f
k+l,2j-l
"I m,J
=1
so
f
"E U m,J
and
2 m- l
= __ 1_ t f "E co U. 2m- l j=l m,J
f
(4.m) gives
Corollary 6.9 If
G
is a locally compact abelian group and
X
a
A A
compact subset of
G, then
Ll(G)
Ix = C(X) k < 1
a Helson set) if there exist constants that for every wi th
II fill
s;
K
F E C(X)
with
(that is, X
IF l E I
and
K
there is an
is
such f
"
E Ll(G)
and A
suplf(x) - F(x) I
xEX ~:
S;
k. A
Apply Theorem 6.5 to the map
T:Ll(G) ~ C(X)
A
given by
T(f) = fix, taking
E
equal to the
U
of Theorem 6.7.
Next consider a locally compact Hausdorff space
X,CO(X)
the continuous complex functions vanishing at infinity on X, B
the closed unit ball in
functions in
B.
CO(X)
and
B+
the non-negative
100
Lemma 6.10 If
E
contains
CO(X)
then
continuous
It suffices to show f:X ~ [0,1)
be given.
E
contains every
with compact support.
Ilfll.., < 1.
Then
co
Let
U
Let such an
be an open set with
compact closure such that the Closed support of
(1)
co E
B+.
~:
f
is a bounded normal family in
f-l(O,l) C U open c
and consider any
f
lies in
U compact
n > [1 - IIfll..,]-l, so that
f(X) c [O,n~l).
(2 )
For each positive integer
k
1,2, ••• ,n-l
Uk = (x E X:f(x) ~
Vk
= (x
E X:f(x)
set
kin) n U.
S
(k-l)/n)
k
= 1,2, ••• ,n-l.
Notice that
U}
(3)
Uk c U, V k c,
(4)
Uk
Thus
E, being normal, contains an
n Vk = ¢
for
fl
such that
(5 )
Set
(6)
WI = {x E X: If 1 (x) I ~ I/n2) n (X\U),
a compact set since by (3).
Therefore
there exists
Let
fl E CO(X).
U2 'V 2 U WI
f2 E E
such that
It is disjoint from
U2
are disjoint compacta and
U:
101
etc.
Wl 'W 2 ' ••• 'W n _ l
associated sets
x E Vk U WI U ••• U Wk _ 1 }
Wk = {x E X: Ifk(X) I
(8)
Notice that Wk
is
fk
in
0
and
such that
x E Uk
{:
fk(X) =
( 7)
in
f l ,f 2 , ••• ,fn _ 1 E E
Continuing we finally obtain
k
= 2, ••• ,n-l.
~ 1/n2} n (X\U) • j < k
for
Wj
Ifkl ~ 1/n2
while
so
(9)
Wj
n Wk
=
¢
if
j
~
k.
Define 1 n-l E n-l g = L f. - n co E.
(10)
. 1 J J=
n
We estimate IIf - gllo>: (I)
If
x E U. Then (recalling (2»
lS=! n
such that x E Ul
n
S f(x) s~ n·
pick
k E {I, 2, ••• , n-l }
It follows that
n ••• n Uk _ l n Vk+l n Vk+2 n ••• nvn-l
U2
so from (7) flex) = f 2 (x) = ••• = fk_1(x) = 1 fk+l(x) = f k + 2 (x)
= ••• = fn_l(x)
and = O.
We therefore have lk-l 1 1 n-l If(x)-g(x) ISlf(x)- nj~lfj(X) 1+ n1fk(x) 1+lnj=~+lfj(x)1 = If(x) _ k~ll + ilfk(x) 1 S
i
+
~Ifkllo>
102
where (II)
M is some bound on
E.
n-l x E (X\U)\ UWk • k=1
If
Then by (8) we have Ifk(X) I < l/n2 for k = 1,2, ••• ,n_l 1 n-1 1 1 and so Ig(x) I
is supported in
U.
Thus
If(X) - g(x) I <~.
(12)
n
(III)
x E (x\U) n
If
n-1 ~
Wk.
k=l Say (8)
x E Wk.
Then
x E X\U\W j
Ifj (x) I < l/n 2 •
Again
for
j,l: k
f(x) = 0
so
If(x)-g(x) I = Ig(x) I s: -n1Ifk(x) I
<.k n
( 13)
whence by
n-l
+! ~ If, (x) I n j,l:k
J
+ 2· 1 n
Combining (11), (12), (13), we get If(x) - g(x) I s: (14) IIf - gil"" s:
i(M
i(M +
+ 1)
'Ix E X.
1).
Now IIf -
n~lgll""
s: IIf - gil"" + n:lllgll..,
1 n
S:-(M
+
1)
1 1 n-1 + n-l n,~ IIfjll"" J=1
s: ~(M + 1) 1 1 n-1 n + n-l n ,1: M
J=1
1
= il(2M + 1).
by
(10)
&
(14)
103 n
As
n-l g E co E
(10) and
large, we see that
f E
-Theorem(Bade6.11& Curtis [3J) space, Y
Let
co E. X
be a locally compact Hausdorff
a Banach space continuously embedded in
Y is boundedly €-normal for some moreover then
K
~:
€
<~,
then
Y
Co(X).
If
= Co(X).
If
is a bound as in part (iv) of Definition 6.1,
Ily ~ 4K(l - 4E:)-1
II
may be chosen arbitrarily
n
Let
B
\I \I"".
denote the closed unit ball in
Fl and F2 are disjoint compact subsets of X pick h
Co(X).
If
=h F E ,F 2 y l
with Ilh I\y ~ K,
(1)
and an
f = fF
Ih
FEB
€,
Ih
€
wi th
l' 2
(2)
f(F l )
= 1,
f(F 2 )
= O.
Then
Let (Tietze)
9
gF
F
E CO(X)
be an extension to
X
of
l' 2
(f - h) IFI U F2
119 II ...
(3)
9 + h
We have
€.
=f
(g + h)(F l )
(4) Let
s
with
A
on
= 1,
Fl U F2
so by (2)
(g + h)(F 2 )
= O.
be the set consisting of all such functions
9 F ,F + h F ,F • 2 l l 2
1\ 9
+ h
1\...
Then
~
\I gil...
where we have taken for
A +
is uniformly bounded:
II b \I...
~
1\ 9 1\...
+ M
II h II y
~ e:+MK < ~+MK,
M the norm of the injection of
Y
104
into
Co(X).
Moreover from (4) we see that
co A ~
From Lemma 6.10 then
family.
(5)
coe 4A
~
B+
A
is a normal
and so obviously
B.
Since by (3) 114(g + h) - 4hll"" = 411gll""
:!: 4€,
(6)
we are in a position to apply Theorem 6.5 with injection map of
Y
into
CO(X), E
= 4A,
T
= 4€
k
the and
= 4K.
c
Because of (6), (1) and (5) the hypotheses of Theorem 6.5 are met and we conclude that the range
Y
of
T
is all of
CO(X)
and the norm inequality holds.
If
Y
is a Banach algebra, lemma 6.2 allows the hypothesis
in the last theorem and in Corollary 6.13 below to be relaxed to
€
<~.
Moreover the continuity of the embedding need not
be hypothesized because
II II",:!: II IIy is automatic (III, p. 1>.
Corollary 6.12 If
X
is a compact Hausdorff space and
algebra lying in
C(X)
exists a constant quotient algebra then
A
=
C(X)
E!22!:
and
then normality of (1)
a Banach
which is normal and for which there
M such that the idempotents in each A/kF
If
A
(F closed
are bounded by
M,
are disjoint closed subsets of
X,
II II A
E l ,E 2 A
~
C
X)
4MII II",.
provides a function
f
E A
such that
105 So
U E2 )
f + k(E l
is an idempotent in
hence has (quotient) norm no greater than
M.
of the quotient norm this means that, given f E A
U E2 )
Alk(E l
and
By definition 6 > 0, the
can be chosen to satisfy in addition to (1) IIfliA < M
(2)
+ 6. 6 > 0, A
We have shown that for each M+ 6
with bound
e = O.
for
is boundedly e-normal
All conclusions therefore
follow from the last theorem. Corollary 6.13 Let
X
be a compact Hausdorff space, Y
continuously embedded in
C(X).
If for some
locally boundedly e-normal, then finitely many open sets
Uj
X
a Banach space e
<~,
Y
is
can be covered by
=
ylUj = C(Uj ), j
such that
1,2, ••• ,n. ~:
From the open sets
6.l(v) of locally boundedly Ul' ••• 'Un
of
X.
in
Y
zero on
such that
II II...
closed in
Y
implies that
:s;
so
in the quotient
the vector subspace of functions
By hypothesis there is a constant
Mil lIy. Yj
II II ...
extract a finite cover
Yj _ ylUj
Y/kUj , kUj
Uj.
provided by the definition
~-normal,
Consider
(pseudo-) norm of
U
This implies that each
is a Banach space.
:s;
Mil lIy. J
in
Yj •
that
C(Uj ).
is
This inequality also
See (:[V), p. 2.
also clear upon a little reflection that e-normal family in
kUj
M
Yj
It is
is a boundedly
It follows from Theorem 6.11 then
Yj = C(Uj ).
When
Y
is an algebra we can make a (not entirely routine)
106
partition of unity argument and conclude
Y
= C(X)
(Corollary
6.15 below). Lemma 6.14 Let
X be a compact Hausdorff space, A C(X)
lying in X
which is eO-normal for some
is covered by open sets
Then for each
€ > 0
k
€O <
such that
ul,···,Un
there exists
a Banach algebra
fk E A
= c
A !Uk
before
and
II II",
the quotient norm in
norm in
S
A !Uk = C
= 1,2, ••• ,n
an €-partition of unity subordinate to the cover Since
Suppose
such that
Let us call such a set of functions
~:
~.
{fl,···,f n } {Ul ,···, Un}·
II II A, whence, as noted
A!Uk
dominates the uniform
C(Uk ), the Open Map Theorem provides a constant
Ck
such that f!Uk for some f E A with IIfliA
( 1)
Ckllfll",.
S
We set (2 )
and let
U be some
Closed S
C
A
is
6/2C+l
{
(3)
Let have
U.
T
Uk.
Let
0 < 6 <
~
be given and any
According to Remark 6.3 and our hypothesis, normal, so there exists
11 -
h(S) I
Ih(X\U) I
$
S
h E A with
6/2C+l
6/2C+l.
(x E X: Ih(x) I ~ ~l.
Because of (3) and
6 <
~
we
107
seT c U.
(4)
~IT E C(T)
Tietze insures that the function to
9 E C(U)
has an extension
which satisfies \\gl\.., = sup I~I s 2. T
According to (1) (and (2» therefore there is a g E A ,., (5) Iigl\A S Cligl\.., S 2C
~Iu We set
= g.
(6)
C
with
and have
(since
9
extends
kiT)
while ( 7)
and by definition of (8)
!
T
I\gll.., Ih(U\T) I
S
(5)
IIgIlA·~ s: C.
S
C ~ lJ
In particular from (6), (7), (8) [upon recalling that we have
\\ crIl..., s:
C.
Summarily
Now choose covers
X.
Vk open
X.
Vk
Uk
C
so that
(Vl, ••• ,V n )
This can be done successively thus:
is a closed subset of with
C
Ul •
X\U2 U ••• U UncVlc~ With open sets
So there exists an open set
eli.
Vl' ••• 'Vk
Vl' ••• 'Vk'Uk+l' ••• 'Un
cover
X\U2 U ••• U Un
Then
Vl 'U 2 ' ••• 'Un
VI
cover
chosen (1 s: k < n) so that X
and
Vj
C
Uj
(1 s: j
S
k),
108
X\V l U ••• U Vk U Uk +2 U ••• U Un
the set Uk +l •
in
!
(11)
Vk +l
is now clear.
~n' ••• '~
E A such that
The choice of
choose successively
n
s e[ II,
n
J=k+l
where a void product means (13)
=
fk
is closed and lies
1.
n
n
j=k+l
~j)lI
(1 -
... + 1 r l
Set ~,)
(1 -
Use (10) to
J
We have from (12) and the fact
X
E A.
= VI
U ••• U Vn
that
n
n (1
(14)
-
~) =
O.
k=l
But for any complex numbers
~l'
••• ')n
a trivial induction
establishes that (15) Therefore from (14) 1 = ~n
(16)
n-l
n
k=l
j=k+l
+ r CIk
This is desideratum (ii).
n (1 -
~j) = f
n-l
n
+ t
k=l
f k•
Finally from (11) and (13) follow
Corollary 6.15 Let lying in X
X
be a compact Hausdorff space, A
C(X)
which is eO_normal for some
is covered by open sets
C(Uk ).
Then
A = C(X).
Ul, ••• ,Un
a Banach algebra" eO
such that
<~.
Suppose
A!Uk =
109 ~:
Let
proof and set
C
be the constant (2) defined in the last
e = (2n(C+l»-1.
For this
be as provided by the last lemma. Theorem 6.5 with
E
e
let
fl, ••• ,f n
We have in mind applying
the unit ball of
C(X), k
=~,
c =
n
C !: Ilf .II A , and T injection. For if gEE is given we j=l J may, because of how C is defined, find gl, ••• ,gn E A satisfying
and
Set (2)
n
f
I: g.f. = j=l J J
E A.
Then (3)
On the other hand gf j = gjfj
in
Uj
and
\gfj-gjf j I ~ (II gil ",,+lIgjll ",,) Ifj I ~ (l+C) Ifj I ~ (l+C) e by property (i) of the
fj•
Thus
(4)
It follows (recalling (ii) and (2» (5)
n
that
n
IIg-fll"" = II I: gf. - I: g.f·ll"" j=l J j=l J J ~
n
t II gf. - gJ.fJ·11 co ~ n(l+C) e
j=l
=~
J
by definition of
e.
in X\U j
110
Since (3) and (5) hold for each
gEE, the unit ball of
Theorem 6.5 applies and we conclude
A
C(X),
= C(X).
Corollary 6.16 (Bade & Curtis [3J) A
Let
a Banach algebra lying in
e-normal for some Proof:
e
<~.
X
be a compact Hausdorff space,
C(X)
Then
A
which is locally boundedly
= C(X).
Quote Corollary 6.15 and Corollary 6.13 (plus the
note to Theorem 6.11). Remarks In connection with Corollary 6.16, it should be noted that, even for uniformly closed subalgebras
A, normality
without some boundedness hypothesis is not enough to conclude A
= C(X).
subset A
of
X C(X)
McKissick [36J and [37J has constructed a compact of
~
and a uniformly closed proper subalgebra
which is normal.
is the uniform closure in with poles outside
X.
C(X)
In fact, in his example
A
of the rational functions
There is a detailed exposition of
McKissick's example in [49J, pp. 346-355.
Chapter VII KATZNELSON'S BOUNDED IDEMPOTENT THEOREM Our goal here is to relax the requirement in Corollary 6.12 that the bound on the idempotents in the quotient algebras be uniform.
We need some definitions and lemmas.
The setting
for the next three lemmas (from Katznelson [32]) is a compact Hausdorff space
X
and a Banach algebra
A
lying in
C(X)
which is normal and for which the idempotents in each quotient algebra
A/kF
are bounded.
(Theorem 7.8 below) that
A
We will eventually prove must be all of
C(X).
Defini tion 7.1 Let us say
A
exists a constant
is bounded on a subset
Bv
V.
Say
A
of
X
such that the idempotents in
all bounded (quotient norm) by subset of
V
!.!
Bv
bounded
A/kF
are
whenever
F
is a closed
.! point
x
of
~
is bounded on some neighborhood of
if there
X
if
x.
Lemma 7.2 If A
A
is bounded on each of the open sets
is bounded on every closed subset
E!.2.2!.:
Since
there exists open
F\V 2 Wl
F
of
Vl 'V 2
Vl U V2 •
is a closed subset of open
such that
F\V 2 C Wl CWlcV 1 • Then similarly there exists open
W2
then
such that
Vl
A
112
As
F\W l , F\W 2
are disjoint closed subsets of ~
normal, there is a
X, and
A
is
E A with
(1) Now let
P
be any closed subset of
be any idempotent in
A/kP.
at every point of
and so
of
P
n Wi (i
P
f
0
is
is bounded and
Therefore, as the
P
n Wi
IIf + k(P
f
f + k(P
Vi
S
or
0
1
at every point
n Wi)
(i = 1,2)
are sets on which
is a closed subset of
n Wi) II
is
1
or
f + kP
and let
This means that
= 1,2), which means that
is an idempotent.
F
A
Vi'
IV • i
By definition of the quotient norm then there exist
fi E f+k(Pnii )
such that (2 )
Now
n Wl ) means fl
fl E f + k(P
= f
on
P
n Wl.
Therefore
(3 )
Of course since
and so by (4) From
(1)
P
C
is
~
on
p\(P
on
p.
on
0
p\(P
n Wl)
n Wl).
and (4) qxf = (fXf 1
Similarly
we have
we have that
(fXf = ¢ l (3 )
F
(1-
~)f =
f = ¢
+
u-
~)f2
U-~)f =
on
¢l +
P
and so
(1-~)f2
on
P,
so
113
that is,
~l
+
(1-~)f2
E f + kP.
s \I~IA(BV
(5 )
It follows that
+ 1) + 111-cIiI A(Bv
1
2
+ 1)
by (2).
Therefore the right side of (5) may be taken as a constant BF
which bounds all idempotents in
P
F.
C
A/kP
for every closed
Lemma 7.3 If then
A
A
is bounded on each of the open sets
true for
By induction on
n = k
then
F\Vk +l
open
U
and and
Vl'···'Vk'Vk +l
U
F
Trivial for
n = 1.
If
is bounded on each of the open sets
A F
n.
Vl U
is a closed subset of
is a closed subset of
Vl U ••• U Vk •
U Vk U Vk +l ' Pick
such that
By the induction hypothesis, A on
Vl U ••• U Vn •
is bounded on every closed subset of
~:
Vl' ••• 'Vn
is bounded on
U, consequently
also, and therefore by lemma 7.2 on the closed subset U U Vk +l •
of
Corollary 7.4 If then
A
F
is closed and
A
is bounded at every point of
is bounded on some open neighborhood
V
of
Immediate from the definition of
A
bounded at
~:
a point, the last lemma, and the compactness of where each open
V
V. J
so that
is an open set in which F eve V
C
A
Vl U ••• U Vn •
F:F
F,
F.
C
is bounded.
Vl U••• UV n
Pick
By lemma 7.3,
114 A is bounded on
V
and so on
V.
Lemma 7.5 There are at most finitely many
x EX
at which
A
is
not bounded. ~:
As in the proof of lemma 4.22, if the conclusion
fails there exist mutually disjoint open A
is not bounded. F c n
closed (1)
U
not bounded on
and idempotent
n
Ilfn + kF n "
U
n
n on which A means there exist U
in
fn + kFn
A/kF n
with
n.
O!:
co
F = ( U F )-. n=l n
Now let F
F
n
u ( U F )m,t.n m
there exist (2 )
Since
U
n
n Fm
=~
and
if
n ,t. m,
Thus by normality
Cl>n E A with Cl>n (F n) = 1,
Cl>n (F\Fn) =
f n + kF n is idempotent we know so from (2) we get
As
o. fn
is
0
or
on
1
F
n'
(3)
Yet
F
~
Fn
clearly implies
(4)
kF
C
kF n
so that VCI> E A
and we see (1) (2) (4) n ~ II fn + kFnil = II Cl>nfn + kF nil ~ II Cl>nfn + kF11 , which says (recalling (3»
that the idempotents in
not bounded, contrary to the basic hypothesis about
A/kF A.
are
115
-Lemma
7.6 (Katznelson [33J)
and
X
be a compact Hausdorff space
a Banach algebra lying in
B
for some Xo
Let
Xo E X
BO (H)
~
and some base
there is a constant
\x\v
K
each
= 0).
Choose
I EB
Since
g E C(X) VI E '2r
Suppose
of open neighborhoods of V E
~
such that
Then
B = C(X).
it suffices to show that
g(x O)
with
Kllf \X\VII",
:5:
B
= O.
We may also assume
\x\v I
~ 0
g
(1)
g
contains ~
O.
and Yx E VI.
By (i) there exists
= O.
hl(X\V I )
Choose
I E B.
= C(X\V)
BO = (f E B:f(xO) Proof:
with
such that for every
Ilf + k(X\V) "B/k(X\V)
where
C(X)
WI
(3)
hI E C(X)
such that
By (ii) we can even choose
g + hI E BO hI
and
so that
such that
{
and
Xo E WI open c WI c VI
ligil",
\h l (x\wl)1 < - -
•
4
This is possible since
hl(X\V I )
=
0
implies
hI
some neighborhood of the compact set
X\V I •
at
vanishes at
and
Therefore
g + hI E BO
hl(xO)
so
= 0 = g(xO)
g + hI
and we may choose
such that (4)
J Xo
E U1 open
C
WI
Now
and
llh1(X)1 + Ig(x)1 <4(R;3) 1Ig11",
is small on g
vanishes
116 Take
CPl E C (X)
wi th
o.
CPl(X) c [0,1],
(5)
Then use (i) and (ii) to pick CPl + hi E BO
hi E C(X)
and
IICP1+hiIiB < (K+1>!!CPllx\U11I." s: K+l
(6)
such that
hi(X\U l ) = O.
and
Then
Set
Claim:
(8) If and
x E X\V l
then
CPl(x) = 1
hl(x) = 0
(by (5».
(by (2», hi(x) = 0
(by (6»
Therefore
gl(x) - g(x) = O. If
x E Vl \W l
then
Igl(x)-g(x) 1 s: Ig(x) 1 + [1~(x)I+lhi(x)I][lg(x) 1+lhl(X) IJ
<~!gll ... + [ICP1(x)I+lhi(x)IJ[~lglI.,,+lhl(x)I] by (1) s: ~lglI." + [l+O][~!gll.,,+lhl (x) 1J s:
~!gll."
by (6) and (5)
by (3)
+ [ftllglI.., + ii!gll."J
= ~I gil.".
If
x E Wl\Ul
(by (6».
then
CPl(x)
=0
(by (5»
and
hi(x)
Therefore
Ig 1 (x) - g(x) 1
I-g(x) 1 < ~I 911..,
by
(1).
=0
117
If
x E Ul
then
\gl (x)-g(x) \ s: \g(x) \+\gl (x) Is: ~lIgll..,+[ \CPl (x)+hi (x) \][ \g(x) \+ \hl(X)\] s: illgll.., + [0+2+K][ \g(x) \+\h l (x) \]
by (5) and (7)
s: illglI", + [2+K] 4(2;K) Ilgll",
by (4)
<
i IIgll",.
Thus (8) is established. definition of
gl
Finally from (2) and (6) and the
we get
(9)
We are now in a position to quote Theorem 6.5. we can iterate the construction with etc.
We get
gl,g2,g3' ••• E BO
g-gl
Alternatively
in the role of
g
such that 1
(10)
IIg - (gl + ••• + gn)lI", s: '2llg - (gl +
(11 )
Ilgnll B s: (K+l) IIg - (gl + ••• + gn-l)II",.
+ gn-l)lI",
2
Induction on (10) gives
and then (11) yields
The series
L g n=l n
therefore converges in the norm of
B
to
Lemma 7.7 Let
X
be a compact Hausdorff space, A
lying in
C(X).
the space
A\V
each Closed
F
Let
V
be a closed subset of
with the quotient norm of C
V
a Banach space X
A/kV.
the two quotient norms in
AIF
and
B
Then for are equal:
g.
118
IIf Iv + kFIiB/kF = IIf + kFIIA/kF
Yf E A
where (suffering a little abuse of language)
kF
vector space of functions in which vanish on Proof:
g E A with
II f Iv + g IvII B
B = Alv
II f Iv + kE11 B/kF €
> 0
II f Iv + g IViI B
S
S
S
IIf + kFIiA/kF
and as
E:
> 0
S
g(F) = 0
with
pick
h EA
€.
h E f + kF
we have
IIhllA < IIf Iv + gila +
Taking the infimum on such (2)
II f + 911 A.
S
and
hlF = (flv) IF + glF = flF
II f + kFIiA/kF
Therefore
g
g E B
and
II hli A < II f Iv + ejl B + As
gil A
II f + kFIl A/kF.
h Iv = f Iv + g
such that
we have
is the quotient norm.
Taking the infimum over such
Next given
g(F) = 0
= II (f + g) IVII B s II f +
II f Iv + kFIl B/kF
(1)
B, as the case may be,
F.
Given
as the norm in
A or
denotes the
g
so
€.
in this last inequality gives
IIf Iv + kFIiB/kF +
E:
is arbitrary, (1) and (2) give the desired
equality. Theorem 7.8 (Katznelson [32] and [34]) space and
A
Let
X
a Banach algebra lying in
be a compact Hausdorff c(X)
which is normal
and for which the idempotents in each quotient algebra (F closed
C
X) are bounded.
Then
A
= C(X).
A/kF
119 ~:
point of
It suffices to show that
X.
A
is bounded at each
For then by Corollary 7.4
A
is bounded on
X
and the desired conclusion is guaranteed by Corollary 6.12. Toward this end we notice that as a consequence of lemma 7.5 each point that
A
x
of
has a neighborhood
is bounded at every point of
Next we notice that if AlE
X
E
V
such
v\{x}.
X
is a closed subset of
is considered in the quotient norm of
satisfies the same hypotheses that
A
A/kE
does:
then
and
AlE
normality is
trivial and for the boundedness of idempotents in quotients we just quote lemma 7.7. Because of the last paragraph it suffices to deal in
V = X.
the second paragraph only with the case
Summarily
then the theorem will be proved if, under the assumption that Xo E X, A
for some
we can prove that
is bounded at every point of
A
must also be bounded at
x\{xO },
xO.
With a view to using lemma 7.6 we will next show that there is a closed neighborhood
V
of
Xo
and a constant
with the property that the idempotents in bounded by
M
for all compact
Ajk(F
F c V\{xoJ.
U (xO})
are
The argument
for this is like that in the proof of lemma 7.5:
we suppose
it false and construct inductively closed neighborhoods of
M
V
n
and compact
contains an idempotent, say norm greater than
n+l
gn + k(F n U (xo })
and such that for
diSjoint from
F l ,F 2 , ••• ,F n _ l •
o
Fn U [xO}
and
1
on
and so
Now
gn
l-gn
(gn E A). of
n > 1, Vn
is
takes only the values also has this property,
120
that is, 1 - gn + k(F n U {Xo)
is also an idempotent.
Moreover its norm is greater than
II
IIl-gn+k(Fn U (Xo )
n:
11- II l +k (Fn ( Xo }) II - 1
~ IIgn +k (Fn U (Xo )
~
II gn
+ k ( Fn U
U (Xo }) \I
> n+l - 1 = n. (Note that we have used the fact
1 E A, an immediate consequence
of normality.)
and
Xo
Since one of
A/k(F n U (xo })
we see that
{xo})
fn + k(Fn U
gn
(fn
= gn
l-gn
must be
0
at
contains an idempotent
or
l-gn
as the case may be)
with
(1)
IIfn + k(F n U {Xo})
(2 )
fn(x o )
II
> n
= O. CD
As in the proof of lemma 7.S we set have that
Fm C Vn+l
for
m
~
n+l
F
= (xo )
U ( U F )-. n=l n
We
so
CD
( U
F )m=n+l m
and as the
V +1 n
C
x\Fl U F2 U ••• U Fn
C
are all disjoint we see that CD
F
C
n
X\[F I U F2 U ••• U F n _ l U (
CD
U
F )-]=X\( U F )-. m=n+l m m~n m
CD
Therefore
F
= Fn
U (
U
F
m~n m
U (xo )-
we can appeal to the normality of
is a disjoint union and A
for a function
o
at
~n
EA
with (3 )
As that
o
or
1
on
F
n
and
we see from (3)
121
and also from (3) that
"" 't'n f
o
is
n
or
F·,
throughout
1
so ~nfn
(5)
+ kF
is an idempotent.
Then, as before, (1)
n < Ilf n +k(Fn which says that the idempotents in
A/kF
contrary to the basic hypothesis on
A.
With the neighborhood
V
assert that for every compact
are not bounded,
and constant F
M in hand we
v\{xo ), AIF U (xO)
C
(quotient norm) satisfies the hypotheses of Corollary 6.12. For if If
P closed
Xo
E P
then
P
compact subset of A/k(F O U {xc})
two cases must be looked at:
F U (xc)
C
=
FO U {xO }
v\(x O }·
= A/kP
where
V\{x O} M.
M.
are bounded by
xo
If
'
P
then
so the idempotents in
By normality
A
= p\{xO}
is a
Then the idempotents in M and so (lemma 7.7)
those of the corresponding quotient of bounded by
FO
P
AIF U {xc}
is a closed subset of
A/k(P U (xc)~
contains an
h
are also
are bounded by
such that
h(P) = 1, h(x O ) = O. Then for any idempotent hf + kP
=f
+ kP
and
f + kP
in
A/kP
hf + k(P U (xc)~
we shall have is also an idempotent.
It follows that
11£
+ kPII = Ilhf + kPII s IIhf + k(P U (xO }) II s M.
That is, the idempotents in
A/kP
are bounded by
M.
By
lemma 7.7 the idempotents in the corresponding quotient of AIF U {xo)
are also bounded by
M.
122
Because of the result of the last paragraph we can use Corollary 6.12 on
AIF U (Xo )
for each closed
F ~ V\(xo )
and get
I
(6 )
A F U {xo }
(7)
IIf + ldF U {xo ) l~k(FU(X )
= C (F U {xo })
o
s: 4MllfiF U (x o ) II", 'If E A.
Now we have in mind applying lemma 7.6 to X and the algebra
Setting
= {f
AO
Alv
E
V
in the role of
(quotient norm) in the role of
A: f(X O)
= 0)
and
BO
= Aolv
B.
it is
obvious from (6) that
= C(F)
BolF
~
that is, if Which lie in (i)
for all compact Fe V\{X O)
is the set of all open neighborhoods of V,
BO Iv\u = C(V\U)
for all
U E ~.
But from (7) we have for any closed F
IIf + kFII A/kF s: IIf + keF U {Xo }) s: 4MII flF U {Xo ) = 4M 1\ f IF
That is, for every (8)
U
II '"
V\{xo )
C
II
A/k(F U {Xo )
II", Vi E AO.
E: 'Ii
IIf + k(V\U) II A/k(V\U) s: 4M IIf Iv\u II",
From this and lemma 7.7 it follows at once that (ii)
Ilf + k(V\U)
holding for every
1\ B/k(V\U) U
E~.
B = C(V), that is, (9)
Alv=c(v).
s: 4M II f Iv\u II",
It follows from lemma 7.6 that
Xo
123
This implies, as we have noted several times before in similar circumstances (via the Open Map Theorem), that the quotient norm in
Alv
evidently
is equivalent to the uniform norm and so
A
is bounded on the neighborhood
V
of
xO.
Which we noted in the fourth paragraph of the proof, was all that had to be demonstrated. Corollary 7.9 (Glicksberg [21]) A
Let
X
be a compact Hausdorff space,
a point-separating subalgebra of
the constants and for which for each closed F Proof:
X.
Then
f EA
F,K
of
X
which contains·
is uniformly closed in
A
is normal.
be given.
If
Let disjoint
x E F, Y E K, there
f(x) = 0, f(y) = 1 (1 E A, A
with
separates
points) and so there eXist disjoint closed neighborhoods of
x
and
Wx
of
y
which converges uniformly on ~:
Il-z I s 1/4)
{z E
~:
Iz I s 1/4).
Vx U Wx
Alvx U Wx exists
(z E~:
Consequently
is closed in
function of Keeping
Pn
Izl s 1/4) U
{(Pn of) Ivx U Wx )
such that
is a
which converges uniformly
Alvx U Wx
to the characteristic function of
e EA
C(V x U Wx )
Vx •
Since
by hypothesis, there
elv x u Wx
is the characteristic
vx • y EK
x
to the characteristic function of
sequence of elements of on
V
such that
By lemma 2.2 there exists a sequence of polynomials
{z E
C(F)
A = C(X).
We first show that
closed subsets exists
C
AIF
C(X)
fixed, cover
F
with finitely many
124
and let e. V , ••• ,Vx J xl n above for the pair xj,y
so that
W.
for some neighborhood I -
be the element of
of
J
e j (W j )
and
ej(v x .) = 1 J
Then the function
y.
belongs to
(1-e l )(1-e 2 )···(1-e n )
constructed
A
A
and is
f
1
Y
=
F
on
U Cover K with y of y. and take f = f Y ••• f y to get a function in Uy , ••• , Uy 1 m I m which is I on F and 0 on K. and
=0
on a neighborhood
0
A
It is an immediate consequence of the Open Map Theorem and the fact (hypothesis) that
AIF
C(F)
AIF, which is always dominated
that the uniform norm in
is uniformly Closed in
by the quotient norm, is in fact equivalent to it. that the idempotents in
A/kF
norm as the idempotents in uniform norm.
It follows
are bounded in the quotient
AIF
are bounded (by 1) in the
All the hypotheses of Theorem 7.8 are therefore
satisfied by
A
example, taking
in its uniform norm. F
=X
is uniformly Closed in
(Note that, for
in the hypothesis shows that C(X).)
We conclude that
A
A
itself
= C(X).
Corollary 7.10 Let
Y
be a locally compact Hausdorff space, A
uniformly Closed point-separating subalgebra of the property that
AIF
each compact Fey. Proof:
Let
closed subset ¢ + AIF
F
of
Then
A
X
and
CO(Y) with C(F)
for
= Co(Y).
be a compact subset of we have
is also closed in
fn E A IF, f E C(F) {cn )
X
is uniformly closed in
a
C(F).
AIF
Y.
Then for any
closed in
C(F).
To see this let
c n + fn ... f
is bounded then some subsequence
uniformly on {c n )
"
Then
c n E ¢, F.
If
converges to
125
some
c E~, so
f - c E AIF.
(f n .) converges to f - c and therefore J It follows that f = c + (f - c) E ~ + AIF. If
on the other hand ( Ic n . I) J
(c n )
diverges to
converges to
-1
CD
is not bounded, then some subsequence and
c
uniformly on
f
1
nj
nj
F, so
= __ l_(c + f ) _ 1 c n. n. nj J J -1 E AIF.
in this case that ~+A IF = A I~ The algebra
It follows
~ + A Ix
thus meets
the hypotheses of Corollary 7.9 and so we conclude that ~ + Alx
C(X)
=
But then
Alx
is clearly a closed ideal in
and so by Corollary 1.8 is conjugate closed.
fact that to find zero on Since
C(X).
A
separates points and a finite covering argument
f. E A J X. As
9 = Ifll2 + ••• + If n 12 is never is conjugate closed, h = glx E Alx.
such that Alx
h > 0, l/h E C(X)
is an ideal.
Use the
It follows
so
=t
·h E A Ix, as the latter
Alx = ~ + Alx = C(X).
an arbitrary compact subset of follows from Theorem 2.9.
1
Y, the conclusion
Since
X
is
A = Co(Y)
Chapter VIII cHARACTERIZATION OF
C(X)
BY FUNCTIONS WHICH OPERATE
Our goal is a theorem of Katznelson (8.6) asserting that if the square root function on JR+
operates (Definition 4.15)
in a point-separating and conjugate-closed Banach subalgebra of
C(X), then that algebra is all of
C(X).
Lemma 8.1 Let
A
be a commutative Banach algebra with unit.
the set of idempotents in
A
is not bounded, then
If
A contains
an unbounded sequence of mutually orthogonal idempotents. ~:
N(h)
(1)
For each idempotent
= sup{ Ilhxlli
The hypothesis is that For idempotents
If
f.g
= 0,
then
h E A
set
x idempotent) E [0,"'].
N(l) = ... , 1
being the unit of
A.
f,g,x
f + g
is idempotent and taking the
supremum on the left yields N(f + g) S N(f) + N(g). In particular (2)
f,g orthogonal idempotents & N(f+g) = '"' ~ N(f)='"' or N(g)='"'
Now choose idempotent
gl
with
127
Then let (recalling (2 as
N( I-gl) =
(3)
IIhl
CD
IIA >
»
.
or
N(gl) =
'"
&
N(l-h l )
=
1
or
hl = gl
hl
= I-gl
according
It follows that
0).
If mutually orthogonal idempotents
hl, ••• ,hn _ l
have been
chosen with n-l
II ~ 1\ A > k
(4)
( k = 1, 2 , ••• , n-1>
& N(l -
=
r.~)
k=l then pick idempotent
x
CD
such that
n-l n-l 1: ~)xIlA> n + \\1 - r. ~IIK k=l k=l n-l Then let (recalling (2» h n = (1 r. ~)x or h n = k=l n-l n-l (1 - r. ~)( l-x) according as N«l - r. ~)(l-X» k=l k=l n-l N«l - t ~)x) = "'. It follows from (5) then that k=l 11(1 -
(5)
CD
or
(6) while N(1
( 7)
Of course
h j (1
j = 1,2, ••• ,n-l
(8)
n
r hk ) - k=l -
=
'"
.
n-l
n-l
k=l
J k=l J
r.~) = h.-
r.
h.~
= hj-h j = 0
so whichever choice is made for
h .·h = 0 J n
j
for hn
we have
= 1,2, ••• ,n-l.
Now (6), (7) and (8) complete the inductive construction. Lemma 8.2 Let lying in
X
be a compact Hausdorff space, A
C(X), (hnJ:=l
idempotents of
A,
~n
~
a Banach algebra
a sequence of mutually orthogonal 1 (n
= 1,2, ••• )
real numbers.
128
c n (n = 1,2, ••• )
Suppose that for every sequence
of non-
negative numbers such that CD
A
X which takes the value
contains the function on
throughout the support of
X.
elsewhere in
hn
for every
n
cn
0
and is
Then there exists a finite constant
K
such that
= 1,2, ••• ).
(n ~:
~
Let
1: IJ E [0, CD] nEB n
be the measure on IN
for every
usual Ll-space on IN ~
IJn
1
8
C
Let
IN.
defined by
tl (IJ)
denote the
with respect to this measure.
i t is clear that
tl(lJ)
=
1.1(8)
As
is an algebra under pointwise
multiplication and that the tl-norm is submultiplicative.
For
CD
~ f(n)h n , a function on X. Because n=l of the mutual orthogonality of the h n , the supports of the
f
E "'1 (u)
hn
let
t(f) =
are disjoint so there is no convergence question here.
The hypothesis is that
t(f) E A
so by linearity for every morphism of
tl(lJ)
into
f E tl(IJ). A.
(1)
E tl(\J)
and
F E
A
Ilfn - flit (IJ) .... 0 1 then since \I 110) S II IIA
hence
Thus
t
t
E tr(u)
and
is a homo-
is continuous.
satisfy &
f
An easy application of the
Closed Graph Theorem shows that fn, f
for every
II t( fn) - FilA .... 0
(III, p.
1)
we get
For if
129 I Hf n ) (x) - F(x) I -+ 0
(2)
for every
x E X.
Also from (1) follows Ifn(k) - f(k) I -+ 0
(3)
Now each the
t(f n )
~
~
F
of
(k 1.
t(f n )
k Em
takes only the value
takes only the value
F
It follows that
Now for each 'Xk
also has this property, while in
each
and so again from (2) there.
k Em.
vanishes outside the union of the supports of
and so by (2)
the support of
for each
lim f n-+ CD
F
fn(k) n
(k)(~)f(k)
= t(f).
consider the characteristic function
\'1e have evidently
H 'Xk )
=~
II 'Xk III = ~k.
and
So
So take for
K
t.
the norm of the bounded linear operator
Lemma 8.3 Let
X be an infinite compact Hausdorff space, A
Banach algebra lying in closed.
If
t
on
C(X)
[0,1)
a
which is normal and conjugate
operates in
A, then
t
is
continuous. Proof:
As in the proof of lemma 4.22, because
X
is
infinite there is a sequence of disjoint non-void open neighborhoods
in
X.
the closure of the remaining Theorem 6.4(ii) provide
Now suppose that
t
E U
If
n
~,
hn E A
then
is not in
so the normality of such that
is not continuous at some point
A
plus
130 6 > 0
and
an E [0,1>
a O E [0,1>, so that there exist
such
that (2)
lim an = a O n .... ..,
(3)
\t(an ) -
t(a O) \
Choose positive reals ( 4)
L bk
k=l Then select
II ~ \I A
~
6
.
n = 1,2, •••
b l ,b 2 ,....
so that
< 1 - a O•
n l < n 2 < n3 < •••
so that
\ a~ - a O \ < b k •
(5)
Define then (recalling that
1 EA
by normality)
The series converges in A-norm by (4) and (5). (5) and the fact (III, p.l) that
\I 1\..,
has range in
h
Since
t
[0,1], we see that
operates we then have
cluster point of the set
s
\I \I A
and each
has range in
to h E A.
(Xk);=l.
Also from (4),
Let
Then since
hn
[0,1).
Xo toh
be a is
continuous (tOh)(X O)
(6)
is a cluster point of the numerical sequence {( tOh)"{"k) );=1
and
h
being continuous
(7) h(X O) is a cluster point of the numerical sequence (h("k»);=l· But
h.
h("k)
And
= a~
is immediate from (1) and the definition of so from (7)
(8)
Thus (6) becomes the following contradiction to (3):
131
(9) t(a O ) is a cluster point of the numerical sequence [~(~)~=l. Remark For a related but deeper result see Corollary 9.2 of [17J. Lemma 8.4 Let
X
lying in
be a compact Hausdorff space, A
C(X)
F:[O,l) .... lR
(*) If
which is normal and conjugate closed.
Let
satisfy
F(O) = 0 F
a Banach algebra
lim IFit ) I = t-+O+
and
operates in
A
m.
then the set of idempotents in
A
is
bounded. Proof:
Let (lemma 8.1)
Suppose not.
hn
sequence of mutually orthogonal idempotents.
be an unbounded Passing to a
subsequence we may in view of (*) suppose without loss of generality that (1)
Let
IF( t) I
~
nt
for
n- l IIh IIA n
Iln
o
s t s IIh n lIi l < n- l
and let
dn
for all
n.
be such that
m
(2 )
!: d IJ < 1 n=l n n
We have
0
S
dn/n
and s;
d
n
;?;
l/(nl-ln) =
for every
0
n.
IIhn lIil, since by (2), dnl-ln < 1.
So by (1) d F(..1!.)
n
Now if
X
is finite our lemma is trivial.
lemma 8.3
F
is continuous and so there exist d
(3)
0
s;
an
Otherwise by
s;
~
and
F(a n )
=
dn
an
such that
132
(by the Intermediate Value Theorem for continuous functions). But
... I:dU
n=l n n
<1
and consequently
... f= and
r:ah EA n=l n n
f(X) C [0,1).
Therefore
Fof(x) = F(f(x»
...
Fof E A.
Now Yx E X
= F( 1: anhn(X» n=l
and so we see that the function
FOf
of
A
takes the value
F(a n ) = d throughout the support of h n and the value n F(O) = 0 elsewhere in X. We conclude from lemma 8.2 that
II h n II AIJ~ 1
is bounded, contrary to the definition of
IJn •
Corollary 8.5 Let lying in
X
be a compact Hausdorff space, A
C(X)
F: (-1,1) .... lR
a Banach algebra
which is normal and conjugate closed.
Let
satisfy
F(O) = 0, limIF(t) I = .... t .... O t If
F
operates in
~:
If
A, then
E
is a closed subset of
given the quotient norm of algebra lying in conjugate closed.
A = C(X).
C(E)
A/kE, then
X
AlE
and
AlE
is
is a Banach
and it is evidently normal and
We will show that
F
operates in
AlE
and then conClude from the last lemma that in each such algebra the idempotents are bounded. Theorem 7.8 that
A = C(X).
It will follow from
133
So let glE
= f.
assume
f E AlE
Then g
A
Evidently
Then
~(g +
g) IE = f
g-l(_l,l)
g E A
is an open set
1
on
E
and
cp(X)
= f. and
(-1,1).
According to Theorem 6.4 (ii )
[O,lJ.
Consider
cp.g E A.
Moreover if
cp(x) t 0
then
C
Icp(x)g(x) I
As
F
cp
X\g-l(_l,l)=
on the complement
0
with
so we may
is normal there is a function
cp.glE
Fo(cp.g) E A
Pick
A
of this open set.
Ig(x) I < 1 C
and
(-1,1).
As
we may suppose that
cp.g(X)
E A
C
E.
which is
Igl-l[l,co)
so
g)
~(g +
f(E)
is real-valued.
which contains in
and
$
Ig(x) I < 1.
operates in
A
x E g-l(_l,l),
Therefore
we have that
and so
that is, Fo(cp·gIE) = Fof E AlE. Theorem 8.6 (Katznelson [33J) and
A
Let
X
be a compact Hausdorff space
a Banach algebra lying in
C(X)
which is conjugate
closed, contains the constants and separates the points of If the function Proof:
~
operates in
A
then
First notice that for any
= Ji! belongs JTtT operates in If I
A
f
= C(X).
E A
the function
Therefore the function
F(t)
=
to
A.
A.
This function obviously satisfies the
conditions of the last corollary and so the desired result will follow from that corollary as soon as we show that
A
is normal. If
f
is any real-valued function in
then the function
f v c
belongs to
A
A
since
and
c E JR,
X.
134
f v c = If-CJ+f-C + C.
/I 1)
Cl(l , "') x.
with
f E A
valued g = (f
are distinct points of
x,y
Next i f
of
V 0
y
f(x) < 0, f(y) > 1.
belongs to and
X, there is a real-
0
A
and is
Then the function 1
in the neighborhood
in the neighborhood
f-l(_""O)
The argument in the second paragraph of the proof of
Corollary 7.9 then establishes the normality of
A.
of
Appendix KATZNELSON'S IDEMPOTENT MODIFICATION TECHNIQUE Prior to Bade and Curtis' proof of Corollary 6.16 in 1966, Katznelson [32J and [34J had proved a somewhat weaker version of it (the theorem below).
Compare also Gorin [24J.
Because
Katznelson's technique in [34J involves a clever and intricate construction and because [34J was never published in a journal (and consequently has not perhaps reached as large an audience as i t deserves) we are going to present the construction here. From now on A
X
shall be a compact Hausdorff space and
a Banach algebra lying in
for some
€
<
~
C(X)
which is bounded1y €-norma1
€ > a
(hence for all
by lemma 6.2).
Definition A.1 Define a function 1 + inf(K: If A
F 1 ,F 2
K1 : (a,m)
~
[l,m)
A-norm not greater than
K}.
By its definition then
K1 (€)
A
€ > a
K1 (€)
are disjoint closed subsets of
contains an €-idempotent with respect to
For every
by
(F 1 ,F 2 )
X
then
of
satisfies
and disjoint closed subsets
contains an €-idempotent with respect to
A-norm less than
=
F 1 ,F 2
of
(F 1 ,F 2 )
of
X
K1 (€).
The advantage of the function
Kl
is that it is non-increasing.
But in the next lemma,which is preparatory to the idempotent
136
modification technique,it is advantageous to consider any
such that
K
is non-increasing and
~
K(t)
Kl(t)
for all
t > O. Lemma A.2 Fix disjoint non-void closed subsets £1 > 0
of
X.
Let
be an £l-idempotent with respect to
with
(E l ,E 2 )
(0.1)
hl E A
and
E l ,E 2
max(lm h l ) > 8£1 > O.
Then there exists h2 E A such that def def 6 max(lm h l ) £1 0.2) () = mine ,~l, £2 = £1 + 4K( 6) 2K( £1) (2.2)
IIh2 IIA
s;
IIhl IIA + !max(lm h l )
(3.2)
Ih21
(4.2)
Ih2 - 11
(5.2)
max(Im h 2 )
s;
[1 -
(6.2)
min(Im h 2 )
~
min(lm h l ) -
s;
~:
on
£2 s;
El on
£2
E2
~]max(lm h l )
P l = ex E
Set
X.
disjoint closed subsets of
( 7)
Ilh~
(8)
Ih~1
(9)
Ih~
Such
* hl
(P l ,P 2 )
IIA
s;
K( 6)
s;
6
on
Pl
6
on
- 11
s;
1m hl(x)
(x E X: lm hl(x)
P2
with respect to
x:
6 max(Im h l ) 4K( 6)
Let
* hl
s;
~ max(Im h l )}
~ ~
max(Im h l ) ),
be a 6-idempotent
with
P2 •
exist by definition of
Kl
and the fact
Kl
s;
K.
137
Set (10)
h2 = hI -
i max(Im h l ) * 4K( 6~ hI E A.
Then (2.2) is clear from (7) [and (O.l)J. hI
is an £l-idempotent with respect to
Notice that since (E l
,E 2 )
we have on
El U E2 by (0.1). Therefore
Therefore (3.2) and (4.2) follow from the definition of (1.2), the fact
Ih~ I s 6 in
hI
£2
is an £l-idempotent and the fact
PI::> El ::> E 2 •
Now from (10) (12)
We analyze (5.2) and (6.2) in cases. (I)
In IRe
PI:
h~ I s 6 by (8) and
1m hI s:
l
max(Im hI) by definition of PI'
so from (12) 1m h2 s:
~
6
max(Im hI) + 4K1"'6T max( 1m hI)
[~ + ~Jmax(Im
hI)
s:
[3 1 Jmax (Im hI) 8 + 8KT6T
s:
[~ +
lJmax
~ax(Im
s: [1 -
since
hI)
~Jmax(Im hI)
which gives (5.2) in this case.
since K
6 s: ~
by (1.2)
maps into
[l,co)
138
On the other hand from (12) and 1m h2 ? 1m h1 (13)
1m h2 In
(II)
?
6 4KT6T
IRe h~1 ~ 6
max(Im h 1 )
~ max(Im h 1 ).
min(Im h 1 ) -
P2 :
IRe h~-ll s 6 1m h2
~
and (12) gives
by (9) so
1m h1 -
s [1 -
( 1-6) 4KT6T
max(Im h 1 )
4~(~)Jmax(Im
h1 )
from which (5.2) follows in this case, upon recalling (1.2) that
6 s
~.
On the other hand in (by definition of
Since
h1
P2
we have
1m h1 ?
~ax(Im
P 2 ) and so (12) and
(9)
is an €l-idempotent with respect to
(E 1 ,E 2 )
have
and as
E1
is non-void by hypothesis this shows
min(Im h 1 )
:s;
S
€1 1
'S" max(Im hI)
by (0.1> since
and so
Putting this into (14) gives
K? 1
h1 ) give
we
139
(III)
In
x\p 1 U P 2 :
Here we have IRe h~ I
while
$
~ax(Im
Im hI <
IIh~ II",
IIh~ IIA s
$
hI)
(definition of
P2)
by (7), so from (12)
K( 6)
max(Im hI) Im h2
$
~ax(Im hI) + [~
$
4K(6)
oK(6)
+ ~]max(Im hI)
[1 -
4Kt6) Jmax(Im hI)
since
K
~ 1,
giving (5.2) in this case. Im hI > ~ max(Im hI)
On the other hand we also have (defini tion of Im h2
~
IRe h~1 $ K( 6) PI) so (12) and max(Im hI) • K( 6) 83 max(Im hI) 4K( 6)
give
by (O.l> ~
min(Im hI)
as noted in the analysis of the previous case
(16) Summarily (13), (15), (16) establish (6.2) and complete the proof of the lemma.
Now we iterate this construction: (0.2)
max(Im h 2 ) > 8€2
then we apply the construction to (hl'€l). ~
> 0
If
Suppose for and functions
=
k ~
(h 2 '€2)
2,3, ••• ,n
E A
we have constructed
satisfying
(O.k-l>
max(Im ~-l) > 8~k_l > 0
(l.k)
€k = €k-l +
6 max( Im h k _ l ) 4K( 6)
in the role of
here
140
II~ IIA ~ II~-l IIA + ~ax(Im ~-l)
(2.k) (3.k)
I~I ~ ~
(4.k)
I~-ll ~ ~
on
El
on
E2
(S.k)
max(Im
~) ~ [1 - ~]max(Im ~-l)
(6.k)
min(Im
~) ~
Induction on (S.k)
[k
min(Im
~-l)
= 2,3, ••• ,n]
-
~
max(Im
~-l).
gives
max(Im ~) ~ [1 - ~]k-lmax(Im hI)
~ [1- md-orJk-lllhlllA
( 7.k)
It follows by induction from (7.k) \lhn IIA ~ Ilhl IIA +
~
and~.k)
CD
II A[ 1
I: ~(1 - SKt6»k- l J k=l
+ 2K( 6) ]
= Ilhl IIA[l +
2K(min{~, 2K~~
It follows from (7.k) and (1.k) that
By induction on this
whence (l.n)
I
that
n-l 1 k 1 I: ~[l - "SiITIT] - Ilhl IIA k=l
IIhlllA[l +
= Ilhl
(S.n)
k = 1,2, ••• ,n
by (1.2).
»)) J. 1
141
From (7.k) and (6.k) [also (O.k) is involved] we get
By induction on this min(Im h n }
:?
min(Im hI} -
~ min(Im hI} -
n-l 6 1 ,k 1 1: 4KTIT[! - 8iIT6T r - IlhlllA k=l 6 IIhl IIA 4K(6)
'"
1
k 1
L [1 - 8iIT6T J -
k=l
= mine 1m hI} - 26 IIhl IIA ~ min(Im hI} -
( ll.n)
Now if in addition
hI
el IIhl IIA K(€l}
by (1.2).
satisfies
(17) as will be the case in the applications to follow, then (9.n), (lO.n) and (ll.n) give (12.n)
Ihnl s 2€1
on
El
(13.n)
Ihn-ll s 2€1
on
E2
(!4.n)
min(Im h n }
~
min(Im hI}
Now while we do not claim that decreasing function of fact (l.k) that the
-
el •
max(Im~}
is a
k, it does follow from (7.k) and the
€k
do not decrease with increasing
that (O.k-l) cannot hold for all positive integers us take for
n
the last
k
k.
for which (O.k-l) holds.
k
Let Then
142 (O.n) fails for
=
en
en _ l +
6 max(Im h n _ l ) 4K(6)
that is,
by (1.n)
I
by (17).
(15.n)
e > 0
Let us define for all
= Kl (e/16)[1
+ 2K1(min{~, 32K 1 (e/16)})]
(18)
K 2 (e)
(19)
K3 (e) = 2K 2 (e)[1 + 4K 2 (min{2, 4K (e»))]' 2
formidable functions to be sure, hut. they may be ignored for the present. We now apply (twice) the idempotent modification technique just developed to prove Lemma A.3 For any two disjoint non-void closed subsets of
X
e > 0, A
and
Proof:
Set
(20) ( 21) (22)
=
e/16
(F 2 ,F l )
with respect to If
el
contains an e-idempotent
max(Im hI) :s: 8e l
IIh IIA Ih I
s;;
s;;
on
max(Im h) :s: e/2.
F2 ,
wi th
and pick an el-idempotent
having set
IIhl IIA :s: Kl ( e l ).
= hI
h
and have
Kl (e/16) < K2 ( e)
e/16
h
F l ,F 2
Ih-ll
s;;
e/16
on
Fl
hI
143
Otherwise we can apply the above construction with E2
= Fl
and
K
= Kl
to the pair
(hl,f. l ).
El = F 2 ,
Then set
= hn
h
and get by (8.n) and (18)
(22)'
max(Im h)
~
f.
by (15.n) and (17).
Now the modification technique truncates the imaginary part of the function from above.
We would like to effect a
truncation of the imaginary part of
h
consider applying this modification to
from below so we I-h, since
Im(l-h)
-1m h.
However we must contend with the possibility that
1 'A.
But this is an inconsequential annoyance since
boundedly f.-normal for every function
g
f. > 0
and so
A
A is
contains a
such that
(23) (that is, g
is an ../8-idempotent for the pair
(r,X»
and
(24) Set (25) By (21) and (21)' we have (26)
Ihll
(27)
Ihl-ll s
so that
hI
~
Ig-ll + Il-h I s £/8 + f./8 = ../4 Ig-ll + Ihl s .. /8 + €/8
f./4
on
Fl
on
F2
is an f./4-idempotent with respect to
which by (20) , (20)' and (24) satisfies
(F 1 ,F 2 )
144 using the fact that
Kl
is nonincreasing, by (18).
Note
...
min(Im(g - h»
min(Im hI) ~
min(Im(- h» - max(Im g)
= min(Im(- h» - max(Im(g - 1» ~
min(Im(- h» -max(Im h) -
(29)
~
- €
~
-€ -
-
IIg - 1 II"" Ilg - 1
IIg - 1 II ...
*
> -
lz.
II."
by (22) and (22)' by (23).
If max(Im hI)
(30)
h = hI
then we set
2€
:s;
and have from (28)
( 31)
and from (26) and (27) (32)
on
on
while from (29) ( 33)
min(Im h)
(34)
max(Im h)
by (30),
and the lemma is proven.
Otherwise
(35 )
and we can apply the above construction with K(€) = 2K 2 (4€), hI = h1 nonincreasing (since while tion.]
K(€) Set
~
Kl
2K l (€/4) h = hn
and
~
€l = €/4.
is) and so this Kl(e)
and have
El = F l , E2
[Note that K
K2
= F2 , is
is nonincreasing,
as required for the construc-
145
IIh IIA s Ilhl "A[l + 2K(min(~, 8K(
~/4) }) ]
s Ilhl IIA[l + 4K 2 (min(2, 2K(
by (8.n)
~/4) }) ]
by definition of K
s 2K 2 (e)[1 + 4K 2 (min(2, 4K e (e)})] by (28) and 2 definition of K (31)' On
by (19).
= K3 ( e)
Fl
we have Ih-I s €/4[1 +
Ilhl IIA] R(e/4) 2K 2 ( e) s e/4[1 + K(e/4)]
by (9.n) by (28)
~/2
by definition of
and similarly with
h-l
on
min(Im h)
<':
-
min(Im hI) -
e/4
..... 3€ e > 2" ""---Z-4' - .. (33)'
F2 •
by (14.n) and (17) by (29)
min(Im h) > -2e.
Finally we have by definition of (34)'
here;
F2 : on
Next
K
max(Im h)
From (31) see that
h
S
l&i
= 4e
(34) or (31)' -
.... h
as
hn :
by (15.n) and (17). (34)' as the case may be, we
does the required jobs.
Theorem A.4 (Katznelson [34J and Gorin [24J) Hausdorff space and
A
is a compact
a Banach algebra lying in
which is boundedly €-normal for some Proof:
X
If
It suffices to show
e
C +JR{X)
<~,
C
A.
then
C(X)
A
= c(X).
For then
lEA
146
C lR(X) = C~(X) + lR
and so
C
A
and then of course
C(X) =
ClR(X) + iClR(X) cA. Let
c
be any number satisfying
(1)
where
is the function defined in (19) above.
K3
f EC~(X).
then any (2 )
PI
(3)
P2
We may suppose
facts
f
~
PI 0, f
~
o.
E X: f(x) s 1/2 Ilf II ... ).
II ...
IIf
> 0) closed subsets of
is non-void by definition of ~
oJ.
(4)
IIh IIA s K3 (1/B)
(5 )
IIIm h II ... s
(6)
IRe hi
( 7)
IRe h-l I
S
~
liB S
P2
on
liB
on
Pl·
Then define fl = c Ilf II ...h E A.
We have from this and (4) (9)
IIfl
(10) On
II ...
S
IIfl IIA
S
CK 3 (l/B) IIf
< K3(1/B) IIf II ....
PI
we have
IIf II...
X. [and the
We consider the two possibilities for
Then lemma A.3 is applicable and provides an
(B)
Define
E X: f(x) ~ 3/4 IIf II ... )
= {x = {x
These are disjoint (as Note that
f
Consider
II",
h E A
such that
147
f-Re f1 (11)
( 7)
~
f-Re f1
IIf II ..-c Ilf II ..Re h
P2
~
[1 -
~]
f-Re f1
~ ~
f-Re f 1
~
~
[! + ~j IIf II ..
(6)
-Re f 1 = -c II f II ..Re h
~
[1 -
~]
~
-
~
(1)
~
[1-
(1)
~
II f II a>
~J
(3)
~ ~
II f
CK3(~)]
~]
II f II ..
[1-
~J
IIf IIa>
-[1-
~J
IIf 1Ia>
- [1-
(1)
~
(12),
[1 -
~J
~
IIf II ..
(9)
1I ..-lIf111 .. ~ [~-CK3(1/8)]
IIf
If - Re f11
From (11),
(1) IIf 1Ia>
~
IIf 1Ia>.
(13)
~
- Re f I l l '"
[1 -
~ ] 1\ f II a>.
Then II f -
f I l l a>
~ II 1m f I l l a> + II f - Re f I l l ..
II f -
f I l l a>
~
.z
~ ~
=
(15 )
P2 =
¢.
Then since
f
(II)
(16)
~
Ilf II ..
II f II ...
(2)3 (9) 3 ~ '4llf1 1Ia>+ IIf111 a> ~ ['4 +
f-Re f1
(14)
(6) h
U P2
f-Re f1
(13)
Ilf II ..
II f II ...
II ..Re
IIf II ..-c Ilf
If - Re f 1 I
x\p 1
On
~J
Ilf 110'> < [1-
we have
(3)
(12 )
~J
[1-
(2)3 (7)3 9 3 9 ~ 4""llf 1Ia>-c Ilf IIa>Re h ~ 4"lIf II..,-c Ilf 11a>· S=['4-SC] IIf 1Ia»o
If - Re f 1 I
On
~
IIf
~
II f
II..
+ II f - Re f I l l a>
IIf 1Ia> + [1 -
(1 -
~)
~J
IIf 11a>
by (5) and (8) by (14)
IIf 11a>.
0
1Ia> :;;
f
~
IIf II ...
Arguing as in lines 19 through 24 of the proof of lemma A.3,
148
we see
A
contains an
(17)
IIh - 1 II ... s ~
(18)
Ilh IIA s K1 (~).
h
with and
Define then
and have from this and (18)
since
K1
is nonincreasing
From (17) and (8)'
It follows from this and (16) that IIf II.., - ~ IIf
f - Re f1 s
~
=
f - Re f 1
IIf - Re f1
II..,
Ilf - f1 II.., s
IIf II..,
!
IIf II",
s
!
!)
IIf II..,.
IIf - Re f1 II", + IIIm f1 II",
=
II f - Re f 1
s
II f
s (15)'
!)
~ ~ II f II.., - ~ II f II.., (1 + = -
(19)
11..,(1 -
- Re f 1
II '" II '"
~ II f II '" + ~ II f II
+
-d IIf II..,
CD
II Im h II '" by (1 7) by (19)
< (1 - ~) IIf II",.
With (10), (10)' and (15), (15)' in hand, we could wrapup the proof by appealing to Theorem 6.5.
Alternatively we
149
can apply this construction to
f-f l
in the role of
f, etc.
and produce
such that for every
n > 1
~) IIf - (f l + ••• + fn_l)
(20)
Ilf - (f l + ••• + fn II .. s
(21)
IIfn IIA s K 3 (l/S) IIf - (f l + ••• + fn_l)
Induction on
(20)
IIf - (f l + ••• + fn) II .. s
(23)
Ilfn IIA s K 3 (l/S) (1 0 < 1 - ~ < 1 f
=
.
I: f
n=l n
II ...
gives
(22)
As
(1 -
_
(1 -
~)n-l IIf
~)n Ilf II..,
II ...
it follows from (22) and (23) that E A.
II ..
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SYMBOL INDEX
closure of set S
A-1 ~,
invertible elements of A 42
f
Fourier transform of f ( L l(G)
G
dual of locally compact abelian group G
aA., 16 ~
H( ), 29
= complex numbers, 1
C(X), v, 1
1m f = imaginary part of f
Co(X), Ck(X), 1
key), 2
CooQR) = infinitely differentiable
K, K1 • K2 • K3 = special functions, 135, 136, 142
functions onlR L
COO-function = element of cooQR)
1
(G) = functions integrable with
co(E) = convex hull of E
respect to Haar measure on the
coe(E) = equilibrated convex hull
locally compact abe1ain group G. £l(Z) = L 1 (Z) = absolutely conver-
of E, 92 d (z , f (E» = inf { Iz-w I :w
E
gent two-sided sequences
f (E) }, 13
£l(Z+) = absolutely convergent
0jk = Kronecker delta (equals 1
sequences
if k=j, 0 otherwise)
log = natural logarithm on (0,00)
)e, 6
Ely,
M(X), 6, 17
1
~
fly,
(Cauchy transform), 17
I~I = total variation measure
1
z=
complex conjugate of z EO:
f
complex conjugate of function
determined by f~, ~
f
155
7
(f), 7
~E
M(X)
156 supp
II 11
~,
II Ilx' liliA'
00 ,
= natural
IN
y* = space of continuous linear
6
numbers, 1
= characteristic
XB
1
function of set
B real part of f
Re A
={Re f
: f" A}
IR, IR+, 1 non-negative square root
defined on [0,(0) T* = adjoint of the bounded operator T
Z, z+, 1 ~,
6
V= universal quantifier
Re f
;-=
functiona1s on the Banach space Y
3 = existential quantifier
,,= set theoretic difference 0
= functional composition
*
convolution :
SUBJECT INDEX A
C
A-antisymmetric set, 5
Cantwell, 96, 98
A-peak set, 5
Cauchy's integral formula, 18
Antisymmetric algebra, 5
Cauchy's theorem, 18
Antisymmetric set, 5
Chalice, 16
Arenson, 43
Choquet, 96 B
Bad~, vi, 23, 88, 91, 94, 96,
103, 110, 135
...
Cirka, 68, 72, 84 Closed Graph Theorem, 30, 52, 53, 128
Baire, 20
Cole, 87
Ball, 6
Conformal, 82
Bernard, 40, 44, 46, 48, 49, 52,
Conjugate function, 38, 39
53, 64, 66 Bishop, 6, 9, 11, 15, 22, 38, 71
Curtis, 23, 88, 91, 94, 96, 103, 110, 131, 135
Blaschke product, 97
D
Borel measure, 6, 66, 84
Disk algebra, 38, 97
Bounded at a point, 111
Dye, 94
Bounded on a set, 111
E
Bounded1y £-norma1, 88
£-idempotent, 88
de Branges, 9
£-norma1, 88
Browder, 43
£-partition of unity, 106
Brown, 39
Exponentials, group of, 69, 70 Extreme point, 96
158 F
Kaufman, 93
Fisher, 97
Kernel, 2
Fubini, 18, 54, 55
Krein-Milman, 8, 9, 96
G
L
Gamelin, 16
Lattice, 60, 63, 66
Gelfand, v, 5
de Leeuw, 58, 64, 65
Glicksberg, vi, 15, 40, 123
Locally boundedly £-normal, 88
Goodner, 96
Locally compact, 23-27, 103, 124
Gorin, 68, 69, 71, 72, 135, 145
Locally connected, 72, 77, 84,
Grenoble, 40
85, 87
Group, compact abelian, 48 Group, locally compact abelian, 99 H Hahn-Banach, 8, 9, 17, 35, 54, 84,
M
McKissick, 110 Mean Value Theorem, 51, 57 Mullins, 16
92
N
Harris, 94
Negrepontis, 71
Helson, 94, 99
Non-affine, 59
Hobbs, vi
Normal, 88
o
Hoffman, 43, 46, 47, 66, 71 Hull-kernel, 2 I Intermediate Value Theorem, 132 Ishikawa, 16 K
Kahane, 38, 48 Katznelson, vi, 24, 58, 64, 65, 111, 115, 118, 126, 133, 135, 145
One-point compactification, 25 Open Map Theorem, 42, 46, 62, 106, 123, 124 Operates in, 54 Operates boundedly in, 54 Orthogonal idempotents, 126, 127, 131 P
Partition of unity, 106
159
Peaks (on a set), 9, 11 Peak set, 9
Stone-Weierstrass, 9, 22, 43, 44, 52, 53, 59, 68
Perfect set, 22, 23
Strong A-boundary point, 16 T
Phelps, 94, 96 Point-separating, 3 Q
Quotient norm, 2 R
Taylor's theorem, 59 Tietze, 28, 103, 107 Tomiyama, 16 Totally ordered dual, 48 Total variation, 6
Ramsey, 71
U
Regular, 88 Restriction, 1
Ultraseparating, 49
Riesz, 9, 54
Uniform Boundedness Principle, 31
Ross, vi
Uniform norm (= II
Rudin, v, 7, 22, 97
Uniqueness Theorem, 17, 38
Runge, 16
Urysohn, 24, 51
I t),
1
v
Russo, 94 S
Vanish at infinity, 1 W
Separates points, 3 Separation Theorem, 92
Wada, 16
Sidney, vi, 47, 72
Wermer, 29, 37, 43, 46, 47, 53,
Sine, 96
66
a-compact, 26
Wik, 48
Spectral radius formula, v, 36
Wilken, 16 Z
*-algebra, 94
Stone-~ech compactification, 47, 49, 50, 64, 71
Zorn, 18, 23
about the book . . . This work presents a detailed account of some recent results about subalgebras of C(X). These algebras carry a Banach algebra norm which is often, but not always, the uniform norm. Conditions on the algebra A (and occasionally on the compact space X) which ensure that A contains every continuous function on X are examined. For example, generalizations of the prototypical Weierstrass Approximation Theory are proved. The prospective reader should have taken a graduate course in real-variables and should have some acquaintance with functional analysis and complex-variables. With this background in mind, the author proves all results completely and in detail. The methods are generally elementary in order to make the results comprehensible to a broad range of readers. Characterization of C( X) among Its Subalgebras will be of interest to advanced undergraduates, graduate students, and professional mathematicians alike.
about the author . . . R. B. BURCKEL is Associate Professor of Mathematics at Kansas State University. He previously taught at the University of Oregon (I 966-71). His research interests primarily . concern functional analysis. He received his B.S. (1961) from the University of Notre Dame and his M.S. (1963) and Ph.D. (1968) from Yale University. Professor Burckel is the author of the book, Weakly Almost Periodic Functions on Semigroups and is a member of the American Mathematical Society. Cover Design by Mary Ann Rosenfeld Liebert Printed in the United States of America
ISBN: 0-8247-6038-7
MARCEL DEKKER, INC., NEW YORK
