Internat. J. Math. & Math. Sci.
VOL. Ii NO. 4 (1988) 735-742
735
MAXIMAL SUBALGEBRA OF DOUGLAS ALGEBRA
CARROLL J. GULLORY
Department of Mathematics
University of Southwestern Louisiana
Lafayette, Louisiana 70504
(Received April i, 1987)
ABSTRACT.
When q is an interpolating Blaschke product, we find necessary and sufficient
conditions for a subalgebra B of
H[q]
to be a maximal subalgebra in terms of the
nonanalytic points of the noninvertible interpolating Blaschke products in B.
set
M(B) N Z(q)
Z(q),
is not open in
such that B is maximal in
xistence of a factor q0 of q in H
If the
we also find a condition that guarantees the
H[].
We also give
conditions that show when two arbitrary Douglas algebras A and B, with A c B have
property that A is maximal in B.
KEY WORDS AND PHRASES.
Maximal subalgebra, Douglas algebra, interpolating sequence,
sparse sequence, Blaschke product, inner functions, open and closed subset, nonanalytic
points, support set, Q-C level sets.
1980 AMS SUBJECT CLASSIFICATION CODE.
I.
46J15, 46J20.
INTRODUCTION.
Let D be the open unit disk in the complex plane and T be its boundary.
Let L
be the space of essentially measurable functions on T with respect to the Lebesgue
By H
measure.
we mean the fmmily of all bounded analytic functions in D.
fication with boundary functions, H
of L
Via identi-
can be considered as a uniformly closed subalgebra
A uniformly closed subalgebra B between H
and L
is called a Douglas algebra.
If we let C be the family of continuous functions on T, then it is well known that
is the smallest Douglas algebra containing
H properly.
H+C
For any Douglas algebra B, we
denote by M(B) the space of nonzero multiplicative linear functionals on B, that is, the
set of all maximal ideals in B.
if B
An algebra B 0 is said to be a maximal subalgebra of B,
BI
B0
B, then either B
B 0 or B I B.
is another algebra with the property that
I
An interpolating sequence
{Zn}=l
bounded sequence of complex numbers {
for all n.
D with the property that for any
there exists f in H such that f(z
n
n
is a sequence in
n
}n=l’
A well-known condition states that a sequence
{Zn}n=l
is interpolating if
and only if
inf
H
n
nn
m
n
-z z
n m
0.
(i.i)
C.J. GUILLORY
736
A Blaschke product
q(z)
H
n=l
(1.2)
Blaschke product if its zero set {Zn}n__1 is an interpolating
0). A sequence __(Zn}n= is said to be
is called an interpolating
sequence
-,,([Zn[/Zn
is understood whenever z
n
sparse if it is an interpolatin sequence and
z
lim
n
H
nm
n
I.
(1.3)
z z
n m
H+C,
For a function q in
M(H+C).
z
m
we let Z(q)=
{m6M(H+C):qkm)=
O} be the zero set of
almost everywhere on
H of modulus
An
# in
T. We denote by H [b] the Douglas algebra generated by H and the complex conjugate of
inner function is a function in
the inner function b.
M(L). Then X
M(H), we denote by
We put X
point in
support set for
x"
eM(H+C)
and
points of q.
f e OC}.
Qxo
For a
x
x the
the representing measure on X for x and by supp
For a function q in L
Blaschke product), we put
x
is the Shilow boundary for every Douglas algebra.
H
+
C fl H
+ C
an interpolating
x
N(q) the closure of the
Isuppx HIsupp x"
Set QC
(in particular if q is
such that
union set of supp
Roughly speaking, N(q) is the set of nonanalytic
and for x
is called the QC-level set for x
0
0
in X, let
[9].
Qxo
{xe X: f(x)=f(x
O)
for
For an inner function q, K. Izuchi
has shown the following [5, Theorem l(i)].
THEOREM i.
If q is an inner function that is not a finite Blaschke product,
then,
N(q)
U
{Ox;
x
(1.4)
Z(q)}.
In particular, the right side of I..4 is a closed set. Now assume that q is an
interpolating Blaschke product, and let B be a Douglas algebra contained in H[]. We
will always assume that M(B) 8 Z(q) is not an open set in Z(q), for Izuchi has shown [6]
that if B is a maximal subalgebra of H[], then M{B) fl Z(q) is not open in Z(q). We
will give answers to the following two questions.
H[]
or when is there a factor
qo
of q in H
When is B a maximal subalgebra of
such that B is maximal in
H[q-O]?
These
answers will be in terms of the nonanalytic points of q and the invertible inner functions
of
H[q]
that are not invertible in B.
x 6M(H+C)/M(B)}.
For a Douglas algebra B, we denote by N(B) the closure of U {supp
In particular N(H[])
N(). In general if A and B are Douglas algebras such that
x;
MAXIMAL SUBALGEBRA OF DOUGLAS ALGEBRA
x:
737
M(A)/M(B)} and for any inner function
M(A),
i}.
Ib(x)l
hA(b)
It is shown in [7, Corollary 2.5] that if B c H[], then N(B) c N(), and it is
not hard to show that N(q)/N(B) 2 NB(q) (in a sense the set NB(q) is generated by the
A
B, we put
hA(B)
x:
the closure of U {supp
b,
nonanalytic points
2.
the closure of U {supp
M(B)/M(H[]) e
with b
gH[q]},
C)/M(H[]))
It shows how small
H[q].
{bOg :
and n(B)
g
M( H +
OUR MAIN RESULT.
We’ll need the following lemma.
be a maximal subalgebra of
x
xg
Let
b
M(B)/M(H[])
must be if B is to
b is an interpolating Blaschke product
{b
B}.
0
LEMMA i.
Let q be an interpolating Blaschke product and B be a Douglas algebra
contained in H[]. Suppose for all bog (B), we have that NB(q)
Then B
NB(b0).
H[q].
is a maximal subalgebra of
H[q]
Douglas algebra between B and
M(H[q]) 5
M(H[q])
Then
bl supp Um HIsupp
M(B[b])
c
qlsupp
M(H[q]),
m# M(H[]),
__HIsupp
m m
Ib(m)
Thus
m"
Hence the only
that contains B properly is H [q].
M(H[q]).
We show that M(B[B])
M(B[b]).
It suffices to show that if
m
H[q].
It suffices to show that if be (B), then B[b]
PROOF.
m#M(B[]).
NB(q)
then
and B is maximal in
Let me M(B) such that
NB(b),
and since
< 1 and we
get m
It is clear that
{me M(B)
Ib(m)l=l }.
Now M(B[D])
M(B[D]).
we have that
This shows that
H[q].
Using Theorem i above, it is not hard to show directly that N(B[b])
N(q).
However, by Proposition 4.1 of [7], this condition is not sufficient
We let E
NB(q). This can be a very complicated set. For example, it can contain
x where x belongs to a trivial Gleason part or a Gleason part where
supp
0 on this part [see 3].
yet q
So for B to be maximal in
as possible.
To see how simple, we set A(B)
{ag (B): a
A(B)}.
{b n(B)
N(b), E
Now let E
be A(B)
E0
a
1
E
in
E.
(B) such that
NB(q)
n N(a 0)
see this, just note that both
belong to B (B).
a01supp
Xl
Obviously, B
,
Note that if E
0
H[q],
B
.
B[a0]
PROPOSITION i.
LetB
e 0
and
NB(q)
NB[0](q)
N(q),
N(a 1)
DouNlas
N(B)
(iii)
E0
E0
(iv)
E0
NB(q).
E;
E and
since a
subalKebra
of
.
H[q];
0
and a
Xlg N(B) such that
H[q] unless E 0
algebra properly contained in
N(q);
B is a maximal
e
which implies that B[a
Then the following statements are equivalent:
(i)
(ii)
E
B[a0]
so B cannot be maximal in
be a
E0
then there are interpolating Blaschke products a and
0
c ’[q].
To
a N(a
Thus we get B
1)
N(a
NB(q)
0)
Thus
and A (B)
N(b0),
Since their intersection is empty, there is an
HIsupp x
i, but
E must be as simple
H[]}
bog
lql
0]
H[q].
c
We_now
H[q],
state.
and suppose
C.J. GUILLORY
738
We will show that
NB(b)
NB(q)
N(B) U
and
: H’Isupp x
-qlsupp
NB(q)
NB(b).
implies that supp
It is clear that E
EIE
0e
-x E0"
B[b0].
It is clear by Theorem 1 that supp
bog A(B),
Since
Ib0(x)
that
B[b0].
B
So E
Since supp
Again this implies that
E0
c
E01E0
blsupp ye
E 0.
H
A(B and
b01supp x
N N(
0)
x 5 N(q)
.
,
[H
B0
Isupp
H’[q]
N(B0)).
B0 (because N(B)
H[ q],
When is there a factor
H[q0
(B
0 (B)
q0e0 (B)
B c_
B
{q0 g 0
H[q0 ].
So
q0 of q
possible)?
is not
Set F
and N(B
which is a contradiction.
&](B), then by (iv)
H[q].
N(q).
H
N(q).
Since
IIQe HIQ].
N(B)
,
0 Q
By
we
Thus N(B)
(E 0
is
N(q).
possible).
such that B is a maximal subalgebra of
To answer this question, let
0
H[q0
{q0:qq0
e H }, and
in H
Then
H[q0]}.
N(q0).
q0
in
E
0)
.
Suppose not.
Put
This implies that B is not a maximal sub-
which is a contradiction
Now suppose we have that E 0 5 E 0
y
M(B)
N(b).
B[b] (since
Thus we have that B
there is a Q-C level set Q with N(B) N Q
Proposition 4.1 of [7], we have B 0
algebra of
have
y
I; I is an interpolating Blaschke product with le H [q] and
also have B
N(b0)’ we
that B[b 0]
Hence there is a y
We are going to show that N(B)
Finally, suppose (ii) holds.
By Theorem
x
There is a b A(b) such that supp
so by Lemma i, B is a maximal subalgebra of
N(q).
Consider the algebra
This implies
If (iv) holds and b is any interpolating Blaschke product in
Then N(B) c
Isupp x and
and supp
y N(q)/NB[](q)),
NB(q) c_ NB(b
Suppose not.
is empty.
H[q] (since supp
B[b]
(iii).
E. This shows that (ii)
So we get E
0
It is trivial that if (iii) holds, E 0
NB(q).
b
We must show that
E.
e H
Again suppose not.
is empty.
0
E, but supp by
y
H’Isupp x
blsupp
Thus
E 0.
0
Now we show that E/E
such that supp
x
0
x N(q)/NB[0](q)"
I, so we have supp
which is a contradiction.
Thus, if x e M(B)
N(B[b]).
NB(q)
xN(B[b])
are empty sets. First we show that
and
0
A (B) such that
Then there is an xeM(B) and a b
supp
Let b e (B) and consider the
N(q). Now N(q)
N(B[b])
(ii).
We have (i)
Next suppose that (ii) holds.
E01E0
Using Lemma i,
hence N(B)
so by the above equality we have that
NB(q),
such that
H’[q],
B[b]
We have B
Douglas algebra B[b].
for all b e (B).
H[q].
this will prove that B is a maximal subalgebra of
(i).
(ii)
(iv)
(iii)
(ii)
(i)
We prove the following:
PROOF.
Suppose (i) holds.
Suppose F
N(q0)
for some factor
is our possible candidate.
Blaschke product with c e
H[q0]},
Next, let
q0 of q
q0
{c:c is an interpolating
MAXIMAL SUBALGEBRA OF DOUGLAS ALGEBRA
fl
q0
(B)
ri
F0
E fl
F0
F
fl(B)
f
q0
N(qo),
A
q0
(B)
riq0
E 0 N F, F
F
A*
c
q0
A * (B),
q0(B) flq0 (B)
A(B)
(B)
739
(B)
N(c), F 0
F
n F 0,
and finally
N(q0)
and assume F
n F0
.
We have the following.
COROLLARY I.
If any of the
Let q0 be a factor of q i__qn H such that F
followin
(i) F
conditions hold:
F0
0
F0
NB(H0),
(ii) F 0
where H
N(q0)
H
0
for some
where
fl0 (B)
q0
[qo ].
qoEri(B)
qo rio(B).
H[q0
Then B is a maxima subalgebra of H
0
The fact that F
0
H[0
implies that H 0
and our
corollary follows from Proposition i.
We now consider this question for the genral Douglas algebras.
Douglas algebras such that A
c
Let A and B be
B and there is an inner function q with B
It is well known that if B
When this occurs we say that A is near B.
L
c
A[q].
and A is
any Douglas algebra properly contained in B, then A is not near B, that is, B S A[q]
In fact L
for any inner function q.
algebra A [I0].
is not countably generated over any Douglas
Sundberg [i0] any Douglas algebra B which is
So by the results of C.
countably generated over A is also near it.
The following result comes from [2, Lemma 5] and gives equivalent conditions for
two Douglas algebras to be near each other [see II, Theorem
THEOREM 2.
function.
Le___!tA
and B be
(ii) # B
ZA(q)
PROOF.
ZA(q
Z(q)
M(A).
Assume (i) holds; we show that
is in B.
B c_ A.
If x is in
Let b be any interpolating
ZA(b),
ZA(q).
ZA(b)
c_
[A] we have b is in A. Now let f be any function in B.
Theorem [1,8] there is a sequence of functions {hn} in H
polating Blaschke products
i
{bn}
we show that x is also in
0 implies that x is not in M(B), since b is in B.
So by (i) we have that x must be in
n
be an inner
U M(B)
Now x is in M(A) and b(x)
But h (b)
B and
A.
c
Blaschke product for which
ZA(q).
for a similar result].
A
Then the following statements are equivalent.
(i) M(A)
where
DouRlas alRebras with H +C
with
f belongs to A since b
n
bn
s
Thus
ZA(q),
of
By the Chang Marshall
and a sequence of inter-
B for all n, such that
in A for all n.
and by Theorem
hnbn
f"
This proves (ii)
C.J. GUILLORY
740
Let x be in M(A) bat not in M(B).
Assume (ii) holds.
function b which is invertible in B such that
n, the function f
qn
n
Set
(x)
we get
ZB(q)
M(B)
Then there is an inner
For any positive integer
=< [b(x)[ n.
This proves (i).
0.
ZA(q)
0
i.
so
[b(x)In[fn(X)[
[g(x)[
Letting n
A,
is in
Ib(x)l
ZA(q)/zB(q);
ZB(q)
and
then
M(A)/M(B)
,U
Px,
XeZB(q)
since
M(A)
ZA(q).
M(B) U
As we have previously done, let (B,A) be the set of interpolating Blaschke
products b such that b eB but bA and set W
NA(b). We assume W #
b(B,A)
.
Using Proposition 1, Theorem 2 and Lemma 1, we have the following result.
Le__!tA
PROPOSITION 2.
and B be arbitrary Douglas algebras such that A is near B.
Then the following statements are equivalent:
NA(B
(ii)
A is a maximal subalgebra of B;
(iii)
PROOF.
(B,A);
(i)
NA(b
for all b
NA(B)-
W
Assume that (i) holds.
such that M(A)
M(B) U
Since
A
is near to
B, there is an inner function
If we set A
P }.
X
U
U
xeZ
XeZB(#)
B
(@)
PX
then it is
immediate that
NA(B)
closure of U {supp
Let b be any element in (B,A).
Lemma 1 we have that A[b]
B.
Px:
x A }.
By (i) we have that
NA(B) S NA(b)-
As in proof of
Thus is a maximal in B.
Since A is near B, we have that M(A)
Assume that (ii) holds, and let x eA
B, which implies that if
M(B) U A
If b e fl(B,A), then by our hypothesis A[b]
M(B)).
i}
{g M(A): Ib(g)l
i, then y M(B) (since M(A[b])
y eM(A) and Ib(y)
c
c
b n(B,A).
for
all
Thus
the
H
So, if supp
NA(B) NA(b
NA(B),
x
blsupp
This implies that
To show what
NA(B) W
W _c NA(B),
NA(b)
x
let b e
Isuppx
(B,A).
Hence bE B; therefore we have
x; xM(A), Ib(x)l < i}
{supp x; x e M(A)/M(B), Ib(x)
{supp x; x eM(A)/M(B)}
closure of U{supp
closure of U
c_
closure of U
NA(B).
i}
MAXIMAL SUBALGEBRA OF DOUGLAS ALGEBRA
Since this is true for any b
e(B,A),
we have
NA(B)
_D W
Thus W
741
NA(B)
if A is
maximal in B.
It is trivial that if (iii) holds,
NA(B) c_ NA(b
for all
We are done.
In Proposltion 4.1 of [7] Izuchi constructed a family of Douglas algebras B
H’[q] with the property that N(B) N(). By Proposition I, we have that
this family is a family of maximal subalgebras of H[q].
contained in
Finally we close this paper with the following question that I have been uable to
answer.
QUESTION i.
N(q)
N(B) U
Recall that if q is an interpolating Blaschke product, then
for any Douglas algebra with B
H[q]. Does there exist a
-c H[q] with N (q) N(q).
NB(q)
Douglas algebra
Bo
B
O
REFERENCES
2.
CHANG, S.Y. A Characterization of Douglas Subalgebras, Acta. Math. 137 (1976) 81-89.
GUILLORY, C.J. Lemmas on Thin Blaschke Products and Nearness of Douglas Algebras,
3.
GUILLORY, C.J.
i.
Preprint.
Douglas Algebras of the form
H[q],
to appear in J. Math Anal. &
Appl., 1987.
4.
GUILLORY, C.J., IZUCHI, K., and SARASON, D.
5.
IZUCHI, K.
6.
IZUCHI, K.
Interpolating Blaschke Products
and Division in Douglas Algebras, Proc. Royal Irish Acad. Sect., A84 (1984) 1-7.
QC-Level Sets and Quotients of Douglas Algebras, J. Funct. Anal.,
65 (1986), 293-308.
19
Zero sets of Interpolating Blaschke Products, Pacific J. Math.,
(1985), 337-342.
,
7.
IZUCHI, K.
8.
MARSHALL, D. Subalgebras of L containing H
Acta Math. 137 (1976), 91-98.
SARASON, D. The Shilov and Bishop decomposition of H+C, Conference on Harmonic
Analysis in Honor of A. Zygmund, in Wadsworth Math Series, pp. 461-474,
9.
Countably Generated Douglas Algebras, To appear in Trans. Amer. Math. Soc.
California, 1981.
A Note on Algebras between L
2 (1981), 333-335.
I0.
SUNDBERG, C.
ii.
YOUNIS, R.
and H
Rocky Mountain J. Math. Ii
Division in Douglas Algebras and Some Applications, Arch. Math.,
Vo__l1. 4_5, 555,560.