A Corona Theorem for Multipliers on Dirichlet Space - Internet ...

Integr. equ. oper. theory 49 (2004), 123–139
0378-620X/010123-17, DOI 10.1007/s00020-002-1196-6
c○ 2004 Birkhäuser Verlag Basel/Switzerland
Integral Equations
and Operator Theory
A <strong>Corona</strong> <strong>Theorem</strong> <strong>for</strong> <strong>Multipliers</strong> on Dirichlet
Space
Tavan T. Trent
Abstract. We prove a corona theorem <strong>for</strong> infinitely many functions from the
multiplier algebra on Dirichlet space.
Mathematics Subject Classification (2000). 30H05, 46E22, 46J15.
Keywords. <strong>Corona</strong> theorem, multipliers, Dirichlet space, reproducing kernels.
In this paper we wish to extend the corona theorem on the multiplier algebra
of Dirichlet space to infinitely many functions. For a finite number of functions,
the corresponding theorem is due to Tolokonnikov [T]. (Another source <strong>for</strong> this is
Nikolskii [N].) For infinitely many functions in H ∞ (D), the corona theorem is due
to Rosenblum [R] and to Tolokonnikov [T]. Our methods are in principle close to
those of Rosenblum [R]. All of these ef<strong>for</strong>ts were made possible by Wolff’s beautiful
proof of Carleson’s original corona theorem. (See [G].)
Our proof is based on three parts. First, since the reproducing kernel of
Dirichlet space has one positive square, the commutant lifting theorem comes into
play. This reduces the general M(D 2 (D))-corona problem to the D 2 (D)-corona
problem and we may employ Hilbert space methods. Next, we have a series of tedious
lemmas that basically say that multipliers on Dirichlet space can be naturally
extended to multipliers on (boundary values of) Harmonic Dirichlet space. Third,
a linear algebra result allows us to explicitly write down proposed solutions <strong>for</strong>
the D(D)-corona problem in the smooth case. This is our key innovation. Finally,
simple estimates and a compactness argument allow us to remove the smoothness
condition.
Partially supported by NSF Grant DMS-0100294.

124 Trent IEOT
We will establish our notation. D2 (D) orjustDwill denote the Dirichlet
space on the unit disk, D. Thatis
∞
D = { f : D → C |f is analytic on D and <strong>for</strong> f(z) = anz n ,
f 2 D =
∞
(n +1)|an| 2 < ∞}.
n=0
For a nice account of many interesting properties of Dirichlet space see the survey
article of Wu [W].
We will use two other equivalent norms <strong>for</strong> smooth functions in D. Namely,
and
f 2 π
D = |f| 2
dσ +
−π
f 2 π
D =
−π
D
n=0
|f ′ (z)| 2 dA(z), where dA(z) = dm2(z)
π
|f| 2 π π
|f(e
dσ +
−π −π
it ) − f(eiθ )| 2
|eit − eiθ | 2 dσdσ. (1)
Also, we will consider D 2 l 2(D), or ∞
⊕ 1 D, which can be considered as l 2 -valued
Dirichlet space. The norms in this case are as above, but with absolute values
replaced by l 2 -norms in the appropriate spots. In addition, we will need Harmonic
Dirichlet space, HD (restricted to the boundary of D), and its vector analog. But
again we will only use this norm <strong>for</strong> smooth functions in this space. So if f is
smooth on ∂D, the boundary of the unit disk, then
f 2 π
HD = |f|
−π
2 π π
|f(e
dσ +
−π −π
it ) − f(eiθ )| 2
|eit − eiθ | 2 dσdσ.
Of course, this is <strong>for</strong>mally the same as (1), but the functions in HD may have
nonvanishing negative fourier coefficients.
The algebra of operators which we consider is the multiplier algebra on Dirichlet
space, M(D) ={ φ ∈D: φf ∈D ∀f ∈D}, and the multiplier algebra
on Harmonic Dirichlet space, M(HD), defined similarly (but only on ∂D). We
will use Mφ to denote the operator multiplication by φ <strong>for</strong> φ ∈ M(D) (and <strong>for</strong>
φ ∈ M(HD)).
Given {fj} ∞ j=1 ⊂ M(D), we let F (z) =(f1(z),f2(z),...). We use M R F to
denote the (row) operator from ∞
⊕ D to D defined by
1
Similarly, M C F
M R F ({hj} ∞ j=1) =
∞
j=1
fjhj <strong>for</strong> {hj} ∞ j=1 ∈ ∞
⊕ D.
1
will denote the (column) operator from D to ∞
⊕ 1 D defined by
M C F (h) =
∞
fjh <strong>for</strong> h ∈D.
j=1

Vol. 49 (2004) A <strong>Corona</strong> <strong>Theorem</strong> <strong>for</strong> <strong>Multipliers</strong> on Dirichlet Space 125
The corona theorem <strong>for</strong> H∞ (D) is due to Carleson [C]. The infinite version,
given by Rosenblum [R] and Tolokonnikov [T], can be <strong>for</strong>mulated as follows:
<strong>Theorem</strong> C. Let {fj} ∞ j=1 ⊂ H∞ (D) with 0 0 <strong>for</strong> all n.
This property of the reproducing kernel referred to as “one-positive square”
or “Nevanlinna–Pick” (N-P) has been much studied recently in, <strong>for</strong> example, [A],
[M], [AM], [BT], [BLTT], [BTV], [CM], and [GRS]. This property will be exploited
<strong>for</strong> “ 1
2 ” of the corona theorem in Dirichlet space. The useful relationship between
multiplier spaces and reproducing kernels is that <strong>for</strong> φ ∈ M(D) andz∈D M ∗ φkz = φ(z)kz. (2)
It immediately follows from this that φ∞ ≤Mφ, soM(D) ⊂ H ∞ (D).
we have
Similarly, if φij ∈ M(D) andM [φij] ∞ j=1
It again follows from this that
and
sup
z∈D
M ∗ [φij] (xkz) =[φij(z)] ∗ xkz.
[φij(z)]B(l2 ) ≤M [φij] ∞
B( ⊕ D)
1
M( ∞
⊕ 1 D) ⊂ H ∞ B(l 2 ) (D).
∈ B(∞⊕
D), then <strong>for</strong> x ∈ l
1 2 and z ∈ D,
For part of the pointwise hypothesis of <strong>Theorem</strong> C (that ∞
j=1 |fj(z)| 2 ≤ 1),
we note that if T R F and T C ∞
F are defined on ⊕ H
1
2 (D) andH2 (D) (Hardy spaces) in
analogy to that of M R F and M C F ,wehave
T R F = T C F =sup
z∈D
∞
(
|fj(z)|
j=1
2 ) 1
2 .
Thus one part of the pointwise hypothesis of <strong>Theorem</strong> C gives the boundedness
of the operators T R F and T C F . We will need a similar hypothesis <strong>for</strong> our version

126 Trent IEOT
on Dirichlet space. But since M(D) H∞ (D), (e.g., ∞ n=1 zn3
n2 is in H∞ (D) but
is not in D), a pointwise upper bound hypothesis will not suffice. Moreover, it is
not hard to check that I − 1 ⊗ 1= ∞ n=1 cnM n z M ∗n
z and ∞ n=1 cn =1,where
1 − 1
kw(z) = ∞ n=1 cnznw n .Thus,<strong>for</strong>fj = √ cjMz j with j ≥ 1andf0 =1,wehave
M R F
=(Mf0 ,Mf1 ,...) ∈ B(∞⊕
D, D) andM
1
R F =1.
Now
M C F (1)2 D
=
∞
n=1
√ cnz n 2 D =
∞
∞
(n +1)cn≥ ncn.
But
∞
2 ncnu 2n−1 = d
du (−(ku(u)) −1 ) ≈ (1 − u) −1 1
(ln (
(1 − u) ))−2 →∞as u ↗ 1.
n=1
Thus M C F
M R F ≤√ 18 M C F .
Lemma 1. Let M C F
n=1
∞
/∈ B(D, ⊕ D). However, as the next lemma shows, we always have
1
∈B(D, ∞
⊕ 1 D). ThenM R F
n=1
∈ B(∞⊕
D, D) and M
1 R F ≤√18 M C F .
Proof. First note that from our earlier discussion, M C F ≤1 implies that F (z)F (z)∗
≤ 1 <strong>for</strong> all z in D. Let{uk} ∞ k=1
So
M R F ({uk} ∞ k=1) 2 =
∂D
|
∈ ∞
⊕ 1 D.Then
k=1
≤u 2 σ +
≤u 2 σ +2
∞
fkuk| 2 ∞
dσ + | (fkuk)
D
k=1
′ | 2 dA
∞
| fku ′ ∞
k + f ′ kuk| 2 dA
D
≤ 2 u 2
D +2
≤ 2 u 2
D +4
≤ 2 u 2 D +4 2
≤ 18 u 2 D.
k=1
D
k=1
k=1
∞
|u ′ k| 2
dA +2
∞
∞
D
k=1 j=1
∞
D
k=1 j=1
∞
j=1
D
f ′ kukf ′
juj dA
∞
|f ′ kuj| 2 dA
M C F (uj) 2 D
|
∞
f ′ kuk| 2 dA
k=1
M R F ≤ √ 18 M C F .

Vol. 49 (2004) A <strong>Corona</strong> <strong>Theorem</strong> <strong>for</strong> <strong>Multipliers</strong> on Dirichlet Space 127
Thus our replacement <strong>for</strong> “ ∞
j=1 |fj(z)| 2 ≤ 1” <strong>for</strong> z ∈ D will be M C F ≤1.
We plan to prove
<strong>Theorem</strong> 1. Let {fj} ∞ j=1 ⊂ M(D). Assume that
M C F ≤1 and 0

128 Trent IEOT
Of course, <strong>Theorem</strong>s A and B with δ = 1,500
ɛ3 −1
complete the proof of
<strong>Theorem</strong> 1.
To prove <strong>Theorem</strong> B, we use the commutant lifting theorem <strong>for</strong> multipliers
on Dirichlet space from [CM]. But we state it in our context using the version
from [BTV]. The reader should note that the second part of the corona theorem
<strong>for</strong> multiplier algebras on reproducing Hilbert space with complete N-P kernels
holds as in the following argument.
<strong>Theorem</strong> 2 (CLT). Let M∗, N∗ be invariant subspaces <strong>for</strong> M ∗ z on M
⊕ D and
1 N
⊕ D,
1
respectively, where 1 ≤ M, N ≤ ∞. Assume that X∗ ∈ B(M∗, N∗) satisfies
X∗M ∗ z |M∗ = M ∗ z |N∗ X∗ . Then there exists a Y ∗ ∈ B( M
⊕ D,
1
N
⊕ D) so that
1
(i) Y ∗ |M∗ = X∗
(ii) Y ∗ M ∗ z = M ∗ z Y ∗
(iii) Y ∗ = X ∗ .
Note that if we view Y as an M by N matrix, then the entries of Y are in
M(D) by (ii). This follows since analytic polynomials are dense in D.
Proof of <strong>Theorem</strong> B. Define M = ∞, N =1,M∗ = range MR ∗
F , N∗ = D. Let
Then
and
So
X ∗ =[M R F (M R F ) ∗ ] −1 (M R F ).
X ∗ ≤ 1
δ
X ∗ M ∗ z ((M R F ) ∗ u)=X ∗ (M R F ) ∗ M ∗ z u = M ∗ z u
= M ∗ z (X ∗ (M R F ) ∗ u).
X ∗ M ∗ z |M∗ = M ∗ z X ∗ .
Thus, by CLT there exists a Y ∗ ∈ B( ∞
⊕ D, D) satisfying (i), (ii), and (iii).
1
By (ii), Y has entries in M(D), say gj, soY = M C G .By(i),Y∗ (M R F )∗ = I, so
M R F M C G = I. Finally, (iii) gives us that M C 1
G = Y = X ≤ δ .
Now we proceed with our proof of <strong>Theorem</strong> A. First we note that it is a
simple operator theoretic fact that <strong>Theorem</strong> A is equivalent to <strong>Theorem</strong> A ′ .
<strong>Theorem</strong> A ′ . Let {fj} ∞ j=1 ⊂ M(D). Assume that M C F ≤ 1 and 0 < ɛ2 ≤
∞ j=1 |fj(z)| 2 <strong>for</strong> all z ∈ D. Then <strong>for</strong> every h ∈D,thereexistsuh∈ ∞
⊕ D with
1
(i) M R F (u h )=h

Vol. 49 (2004) A <strong>Corona</strong> <strong>Theorem</strong> <strong>for</strong> <strong>Multipliers</strong> on Dirichlet Space 129
and
(ii) u hD ≤
1, 500
hD.
ɛ3 Actually, it suffices to prove <strong>Theorem</strong> A ′ <strong>for</strong> any dense set of functions in D.
We take functions of D smooth across ∂D. The general plan is as follows. Assume
F is analytic on D1+ɛ(0). Given h ∈Danalytic on D1+ɛ(0), write the most general
solution of the pointwise problem on D. Thatis
v h (z) =F (z) ∗ (F (z)F (z) ∗ ) −1 h − Q(z)k(z),
where range Q(z) =kernel F(z), Q(z) isanalytic,andk(z) ∈ l 2 <strong>for</strong> z ∈ D. In
fact, we will show that <strong>for</strong> each z ∈ D,
(F (z)F (z) ∗ )I − F (z) ∗ F (z) =Q(z)Q(z) ∗ .
We must find k(z) sothatv h ∈ ∞
⊕ 1 D.Thuswewant
∂zv h =0 inD.
So take u h = F ∗ (FF ∗ ) −1 h − Q Q∗ F ′∗ h
(FF ∗ ) 2 ,where k is the Cauchy trans<strong>for</strong>mation
of k on D. Thatis,<strong>for</strong>ksmooth on D and z ∈ D,
k(z) def
k(w)
=
z − w dA(w)
D
and we have ∂z k = k in D.
Clearly, Fu h = h and u h is analytic. Thus we are done (in the smooth case
of F )if
u hD ≤
1, 500
hD.
ɛ3 Let k = Q∗ F ′∗ h
(FF ∗ ) 2 . Our procedure is to show that
and the main estimate,
F ∗ (FF ∗ ) −1 hHD ≤ C1hD,
Q kHD ≤ C2 kHD,
kHD ≤ C3hD.
The next three lemmas involve extending multipliers on D to multipliers on
HD, where we are just considering HD on ∂D.
Lemma 2. (a) Let α ∈ M(D), then α ∈ M(HD) and α B(HD) ≤ √ 20 α B(D).
(b) Let {fi} ∞ i=1 ⊂ M(D). Then M C F B(HD, ∞
⊕ 1
HD) ≤ √ 20 M C F B(D, ∞
⊕ D)
1
.

130 Trent IEOT
Proof. We will only show (a), since (b) follows by summing the results of (a). Let
def
α ∈ M(D). As we have observed, α∞ ≤αB(D) = C. Letp, q0 be analytic
polynomials with q0(0) = 0. We need only estimate α(p + q0)HD, since{p + q0} is dense in HD.
α(p + q0) 2 π
HD = |α(p + q0)| 2 dσ
−π
π π
+
−π
π
−π
≤ 2 |αp| 2 π
dσ +2 |αq0| 2 dσ
−π
π π
+2
−π
π π
+2
−π
|(α(p + q 0))(e it ) − (α(p + q 0))(e iθ )| 2
|e it − e iθ | 2
−π
|(αp)(e
−π
it ) − (αp)(eiθ )| 2
|eit − eiθ | 2
|(αq0)(eit ) − (αq0)(eiθ )| 2
−π
≤ 2C 2 p 2 D +2C 2
π
+4
−π
|e it − e iθ | 2
π
|q0| 2 dσ
dσdσ
dσdσ
−π
|α(e
−π
it ) − α(eiθ )| 2
|eit − eiθ | 2 |q0(e it )| 2 dσdσ
|α(e
−π
it )| 2 |q0(eit ) − q0(eiθ )| 2
|eit − eiθ | 2 dσdσ
π
π π
+4
−π
≤ 2C 2 p 2 D +2C 2
+4C 2
π
−π
π
|q0| 2 dσ +16C 2 q0 2 D
−π
|q0(e
−π
it ) − q0(eiθ )| 2
|eit − eiθ | 2
π
≤ 2C 2 p 2 D +(4C 2 +16C 2 )q0 2 D
≤ 20C 2 (p 2 D + q0 2 D)
=20C 2 p + q 0 2 HD.
dσdσ
dσ(t)dσ(θ)
Hereweareusingthefactthatp⊥q0in HD and q02 HD = q02 D .
Lemma 3. Assume that M C F ≤1. Then
and
M (FF ∗ ) ∈ B(HD)
MFF ∗≤86.

Vol. 49 (2004) A <strong>Corona</strong> <strong>Theorem</strong> <strong>for</strong> <strong>Multipliers</strong> on Dirichlet Space 131
Proof. We show that M (FF ∗ ) 1 2 ∈ B(HD) withM (FF ∗ ) 1 2 ≤ √ 86. This will com-
plete the proof. Let p and q0 be analytic polynomials with q0(0) = 0. We let u
denote p + q 0.Then
M (FF ∗ ) 1 2 (u) 2 HD =
So
∂D
+
≤u 2 σ +2
|FF ∗ ||u| 2 dσ
∂D
∂D
|(FF ∗ ) 1
2 (e it )u(e it ) − (FF ∗ ) 1
2 (e iθ )u(e iθ )| 2
|e it − e iθ | 2
dσdσ
|(FF
∂D ∂D
∗ ) 1
2 (eit ) − (FF∗ ) 1
2 (eiθ )| 2 |u(eit )| 2
|eit − eiθ | 2
dσdσ
a
+2 (FF
∂D ∂D
∗ )(e it ) |u(eit ) − u(eiθ )| 2
|eit − eiθ | 2 dσdσ.
|(FF ∗ ) 1
2 (e it ) − (FF ∗ ) 1
(a) ≤ 2
≤ 4
∂D
∂D
by Lemma 2. Thus
+4
∂D
∂D
∂D
k
k
∂D
2 (e iθ )| 2 ≤
k
|Fk(e it ) − Fk(e iθ )| 2 .
|Fk(eit ) − Fk(eiθ )| 2
|eit − eiθ | 2 |u(e it )| 2 dσdσ
|Fk(eit )u(eit ) − Fk(eiθ )u(eiθ )| 2
|eit − eiθ | 2 |u(e it )| 2 dσdσ
≤ 4 M C F (u) 2 HD +4u 2 HD
≤ (4 · 20 + 4)k 2 HD
k
|Fk(e iθ ) u(eit ) − u(e iθ )| 2
|e it − e iθ | 2
dσdσ
M (FF ∗ ) 1 2 (u) 2 HD ≤ 2 u 2 HD +84u 2 HD =86u 2 HD.
Lemma 4. Let H ∈ Mult(HD) with 1 ≥|H(e it )|≥ɛ>0 <strong>for</strong> σ-a.e. t ∈ [−π, π].
Then 1
H
∈ Mult(HD) and 1
H B(HD) ≤ √ 10
ɛ 2 H B(D).

132 Trent IEOT
Proof. Let r ∈ HD be a rational polynomial on ∂D and let H B(HD) = C. Then
1
H r2 π
HD =
−π
| r
H |2 π π
dσ +
−π −π
≤ 1
ɛ2 π
|r|
−π
2 dσ + 1
ɛ4 π π
−π −π
≤ 1
ɛ2 π
|r|
−π
2 dσ + 2
ɛ4 π π
−π −π
+ 2
ɛ4 π π
|( r
H )(eit ) − ( r
H )(eiθ )| 2
|eit − eiθ | 2
dσdσ
|H(e iθ )r(e it ) − H(e it )r(e iθ )| 2
|e it − e iθ | 2
|H(e iθ )| 2 | r(eit ) − r(eiθ )
eit − eiθ | 2 dσdσ
|H(e
−π −π
iθ ) − H(eit )| 2
|eit − eiθ |r(e
|
it )| 2 dσdσ
≤ 2C2
ɛ4 r2HD + 2
ɛ4 (4C2r 2 HD)
= 10C2
ɛ 4 r2 HD.
dσdσ
Lemma 5. Let {fj} ∞ j=1 ⊂ M(D). Assume that M C F ≤1 and 0

Vol. 49 (2004) A <strong>Corona</strong> <strong>Theorem</strong> <strong>for</strong> <strong>Multipliers</strong> on Dirichlet Space 133
Proof. For k =1, 2,...,let
⎡
0 0 0
⎤
...
⎢ .
⎢ .
⎢ck+1
⎢
Ak = ⎢−ck
⎢ 0
⎢
⎣ 0
.
.
ck+2
0
−ck
0
.
.
ck+3
0
0
−ck
.
. ..
⎥
... ⎥
... ⎥
... ⎥
... ⎥
⎦
. ..
where the first k rows of Ak have only 0 entries.
Then
AkA ∗ ⎡
0 ... 0 0 0 0 ...
⎢
.
⎢
⎢0
k = ⎢
⎢0
⎢
⎣
0
...
...
.
0
0
.
0
.
0
.
0
...
...
∞ j=k+1 |cj| 2 0 ... 0 −ckck+2
−ckck+2
|ck|
−ckck+3 ...
2 0 ... 0 −ckck+3 0
0
|ck|
...
2 ⎤
.
.
.
.
.
.
⎥
... ⎥
⎦
. ..
Thus
So let
∞
k=1
AkA ∗ k =
⎡
⎢
⎣
∞ k=1 |ck| 2 −c1c2 −c1c3 ...
∞ −c2c1 k=2 |ck| 2 −c2c3 ...
∞ −c3c1 −c3c2 k=3 |ck| 2 ...
.
.
.
.
= CC ∗ I − C ∗ C.
.
.
⎤
⎥
⎦
. ..
Q =[A1,A2,...] ∈ B( ∞
⊕ 1 l 2 ,l 2 ).
Lemma 7. Let {fj} ∞ j=1 ⊂ M(D). Assume that <strong>for</strong> each j, fj is analytic on D1+ɛ(0)
and M C F B(D) ≤ 1. Associate Q(z) to F (z) <strong>for</strong> each |z| =1.Then
Q ∞
B( ⊕ HD)
1
≤ √ 86.
Proof. Since M C F B(D) ≤ 1, we have F (z) 2 l 2 ≤ 1. By Lemma 6, <strong>for</strong> z ∈
D, Q(z)Q(z) ∗ ≤ (F (z)F (z) ∗ ) I l 2,soQ(z) B(l 2 ) ≤ 1. First, note that if r ∈ HD
is a rational polynomial in z, then

134 Trent IEOT
π π
F (eit ) − F (eiθ )2 l2 −π
−π
by Lemma 2.
Now <strong>for</strong> r ∈ ∞
⊕ 1 HD
π
QrHD =
so
−π
π
≤
−π
|e it − e iθ | 2
π π
≤ 2
|r(e it )| 2 dσdσ
(F r)(e
−π −π
it ) − (F r)(eiθ )2 |eit − eiθ | 2
π π
|r(e
+2
−π −π
it ) − r(eiθ )| 2
|eit − eiθ | 2 dσdσ
≤ 2 M C F 2
B(HD, ∞
⊕HD) 1
r2HD +2r 2 HD
≤ 42 r 2 HD
(Qr)(e it ) 2 π
l2 dσ +
−π
(r)(e it ) 2 π
l2dσ +2
−π
π
π
−π
−π
π π
+2 Q(e
−π −π
iθ ) 2 B(l2 )
dσdσ
(Qr)(e it ) − (Qr)(e iθ ) 2 l 2
|e it − e iθ | 2
dσdσ
Q(eit ) − Q(eiθ )2 B(l2 )
|eit − eiθ | 2 r(e it ) 2 l2dσdσ r(e it ) − r(e iθ ) 2
|e it − e iθ | 2
dσdσ.
But using Lemma 6 pointwise with cj = fj(e it ) − fj(e iθ ), we get
(Q(e it ) − Q(e iθ ))(Q(e it ) − Q(e iθ )) ∗ ≤ (F (e it ) − F (e iθ ))(F (e it ) − F (e iθ )) ∗ I l 2
Q(e it ) − Q(e iθ ) 2 B(l 2 ) ≤F (eit ) − F (e iθ ) 2 .
Combining the two estimates above, we get that
Qr 2 HD ≤ 2 r 2 HD +2· 42 r 2 HD
≤ 86 r 2 HD.
We need one more lemma to handle Cauchy trans<strong>for</strong>ms.
Lemma 8. Let k be smooth and l 2 -valued on ∂D. Then
Proof.
k 2 HD = k 2 π
σ +
k 2 HD ≤k 2 A + k 2 σ.
π
−π −π
k(e it ) − k(e iθ )2 l2 |eit − eiθ | 2 dσdσ
a
Since all entries in k(e it ) involve only negative Fourier coefficients, we see that

Vol. 49 (2004) A <strong>Corona</strong> <strong>Theorem</strong> <strong>for</strong> <strong>Multipliers</strong> on Dirichlet Space 135
π π
(a) =sup{| 〈
−π −π
k(e it ) − k(e iθ )
eit − eiθ , p 0 (eit ) − p (e
0 iθ )
eit − eiθ 〉 dσdσ
b
analytic polynomial entries that vanish at 0 and p D ≤ 1}.
0
| 2 : p 0 has
But
π π
1 1
(b) =−
〈k(z)[ −
D −π −π z − eit z − eiθ ](eit − e iθ ) −1 , p 0 (eit ) − p (e
0 iθ )
eit − eiθ 〉dσdσdA
π
π
1 p0 (e
=
〈k(z)
it ) − p (e
0 iθ
)
〉e it e iθ dσ(t)dσ(θ)dA
=
=
=
D
D
D
D
−π
1
2πi
1
2πi
−π
∂D
∂D
(e it − z)(e iθ − z) ,
e it − e iθ
1
1
〈k(z)
2πi ∂D (u − z)(v − z) ,
p0 (u) − p (v)
0 〉 du dv dA(z)
u − v
1
〈k(z)
(v − z) ,
p0 (z) − p (v)
0 〉 dv dA(z)
(z − v)
〈k(z), p ′ (z)〉 dA(z)
0
by two applications of Cauchy’s theorem.
Now
| 〈k(z), p
D
′
(z)〉 dA| ≤(
0
D
≤
So
k 2 dA) 1
2 (
k
D
2 1
2
dA
D
p ′
0 2 dA) 1
2
(a) ≤k 2 A.
We are now ready to proceed with a proof of <strong>Theorem</strong> A ′ in the smooth case.
Proof. Assume that {fj} ∞ j=1 are analytic in |z| < 1+δ <strong>for</strong> all j, M C F ≤1, and
0

136 Trent IEOT
Thus we need only show that
u h 2 D ≤
1, 500
Combining Lemmas 5, 7, and 8, we get
u hD = u hHD = F ∗ h
FF
≤ F ∗h
+ Q
FF∗ ≤ 86 · √ 200
ɛ 2
But since Q(z)
√ FF ∗ B(l 2 ) ≤ 1,
ɛ 3
2
h 2 D.
Q − Q ∗ ∗F ′∗h (FF∗ ) 2
z
HD
Q∗F ′∗h (FF∗ ) 2
z
HD
hD + √
86
Q∗ F ′∗h (FF∗ ) 2 2A +
a
Q∗F ′∗h (FF∗ ) 2
z
2 σ.
b
(a) ≤ 1
ɛ 6 M C F (h) 2 HD ≤ 20
ɛ 6 h2 D.
To estimate (b), we need the corona estimates <strong>for</strong> the H ∞ (D) corona theorem.
Using the Wolff procedure (see Garnett [G]) of Paley-Littlewood estimates, we get
that
8 1
(b) ≤ ln
ɛ2 ɛ2 2 h 2 σ
8
≤
ɛ3 2 h 2 D.
See Trent [Tr2] <strong>for</strong> more details.
Combining these estimates we see that in the smooth case,
u h≤
1, 500
ɛ 3 hD.
We show that the same estimate holds <strong>for</strong> the general case. The following
two lemmas hold <strong>for</strong> any N-P r.k. kernel on the ball or polydisk in C n .
Lemma 9. Let {fj} ∞ j=1 ⊂ M(D) with M C F =1.For0≤ r ≤ 1, letFr(z) =F (rz).
Then M C Fr≤MC F and thus Fr ∈ M(D, ∞
⊕ D).
1
Proof. We claim that
I − M C Fr (M C Fr )∗ ≥ 0.
That is, <strong>for</strong> any {c j } n j=1 ⊂ l2 and {zj} n j=1
⊂ D,
0 ≤ 〈(I − F (rzk)F (rzj) ∗ )c j ,c k 〉kzj (rzk). (3)

Vol. 49 (2004) A <strong>Corona</strong> <strong>Theorem</strong> <strong>for</strong> <strong>Multipliers</strong> on Dirichlet Space 137
But
(3) =
n
n
j=1 k=1
〈(I − F (rzk)F (rzj) ∗ ) c j ,c k 〉krzj (rzk) · [
kzj (zk)
] (4)
krzj (rzk)
The expression (4) without the “boxed terms” is positive since I −M C F
M C∗
F ≥
0. We need only note that the matrix whose ij-th entry is the boxed term is
positive. Then Schur’s lemma gives us that (4) is positive.
Now kw(z) isanN-Pkernel,infact
Thus
1 − 1
kw(z) =
kzj (zk)
kzjr(zkr) =(1−
=(1−
=1+
∞
n=1
∞
n=1
∞
n=1
∞
cnzw n and cn > 0 <strong>for</strong> all n.
n=1
cnr 2n z n j z n k )kzj (zk)
cnz n j z n k +
∞
n=1
cn(1 − r 2n )z n j z n kkzj (zk).
(1 − r 2n )cnz n k z n j )kzj (zk)
Thus, [ kz (zk) j
krz (rzk) j ]n j,k=1 is positive and we are done.
Lemma 10. Let F∈M( ∞
⊕ D). Thens-limr→1− M
1
∗ Fr = M ∗ F .
Proof. Assume that MF B( ∞
⊕ 1
D) ≤ 1. By Lemma 8, MFr B( ∞
⊕ D)
1
≤ 1 <strong>for</strong> all
0 ≤ r ≤ 1. Thus we need only show that limr→1− (M ∗ Fr − M ∗ F )x =0<strong>for</strong>xin a dense subset of ∞
⊕ D. By considering finite sums of the <strong>for</strong>m
1
N j=1 c kzj j ,with
{cj } N j=1 ⊂ l2 and {zj} N j=1 ⊂ D, we need only show that <strong>for</strong> e ∈ l2 and z ∈ D,
limr→1− (M ∗ Fr − M ∗ F ) ekzD =0.
Now
Thus
(M ∗ Fr − M ∗ F )(ekz) =F(rz) ∗ ekz −F(z) ∗ ekz
= F(rz) ∗ ekrz
= M ∗ F(ekrz) kz
kz
krz
krz
(M ∗ Fr − M ∗ F)(ekz) ≤krz − kz + krz
−F(z) ∗ ekz
− M ∗ F(ekz).
sup |
w∈D
kz(w)
− 1| . (5)
krz(w)

138 Trent IEOT
But
| kz(w)
krz(w) − 1| = |kz(w) − kz(rw)
|≤
krz(w)
|kz(w) − kz(rw)|
. (6)
kz(1)
Since kz is uni<strong>for</strong>mly continuous on D, we see that combining (5) and (6) completes
the proof.
Proof of <strong>Theorem</strong> 1. Let {fj} ∞ j=1 ⊂ M(D), M C F ≤1andɛ2 ≤ F (z)F (z) ∗ <strong>for</strong> all
|z| < 1. By Lemma 8 <strong>for</strong> 0 ≤ r

Vol. 49 (2004) A <strong>Corona</strong> <strong>Theorem</strong> <strong>for</strong> <strong>Multipliers</strong> on Dirichlet Space 139
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Tavan T. Trent
Department of Mathematics
The University of Alabama
Box 870350
Tuscaloosa, AL 35487-0350
USA
e-mail: ttrent@gp.as.ua.edu
Submitted: May 10, 2002
Revised: August 30, 2002