The Corona Problem
Brett D. Wick
Brett D. Wick, School of Mathematics, Georgia Institute of Technology, 686 Cherry Street, Atlanta, GA USA 30332-0160
E-mail address: [email protected]
Research supported in part by a National Science Foundation DMS grant.
Abstract. These are the lecture notes generated for the CIMPA Summer School: Real and
Complex Analysis with Applications to other Sciences that took place in Buea, Cameroon
May 2 - 13, 2011.
The topic of the series of lectures that were given focused on the Corona Problem in
complex analysis. This problem can serve as a point of entry to numerous areas of analysis:
complex analysis, harmonic analysis, operator theory, and real analysis. In fact, mastering
many of the ideas that appear in the proof of the Corona Theorem will be immensely
beneficial for students of analysis.
There were three lectures given during the CIMPA school. Additionally, there were two
problem sessions associated with the lectures so that the students could gain some mastery
of the material being presented. The course notes first introduce most of the topics necessary
for the Hardy space and then turned to the necessary background for the Corona problem.
The topics covered in the course include the following
• Definitions of these Spaces;
• Computations of their Reproducing Kernels;
• Definitions of their Carleson measures and Geometric Characterizations;
• Corona Theorems.
Based on my personal interests, we focused much more on the function theory and
harmonic analysis aspects associated with these spaces.
Contents
Notation
5
Chapter 1. The Hardy Space on the Disc
1. Basic Definitions
1.1. The Reproducing Kernel for H 2 (D)
2. Littlewood-Paley Identities and H 2 (D)
2.1. Besov-Sobolev Spaces
3. Carleson Measures for H 2 (D)
3.1. Carleson Measures via Reproducing Kernels
7
7
9
10
13
13
14
Chapter 2. The Maximal Ideal Space of H ∞ (D), ∂-equations and Wolff’s Theorem
1. The Maximal Ideal Space of H ∞ (D)
2. Solving ∂-Equations
3. Wolff’s Theorem
21
21
24
26
Chapter 3. The Corona Theorem
1. Wolff’s Proof of the Corona Theorem
2. Peter Jones’ Constructive Solution to ∂
29
29
33
Chapter 4. Corona Theorems and Complete Nevanlinna-Pick Kernels
1. Calculus of kernel functions and proof of the Toeplitz Corona Theorem
37
38
Bibliography
45
3
Notation
:=
equal by definition;
C
the complex plane, C;
D
the unit disc in the complex plane, D = {z ∈ C : |z| < 1};
H
the upper half plane, H = {z = x + iy ∈ C : y > 0};
T
the unit circle in the complex plane, T = {z ∈ C : |z| = 1};
dm(θ)
normalized Lebesgue measure on the unit circle T, dm =
dA(z)
normalized Lebesgue measure on the unit disc D, dA(z) = π1 dxdy;
∂
derivative with respect to z, ∂ = 21 (∂x − i∂y );
∂
derivative with respect to z, ∂ = 12 (∂x + i∂y );
∆
Laplacian operator, ∆ = ∂x2 + ∂y = 4∂∂ = 4∂∂;
L2 (T)
the standard Lebesgue space on the unit circle with respect to dm
Z
2
2
iθ 2
L (T) = f : |f (e )| dm(θ) := kf kL2 < ∞ ;
1
dθ;
2π
T
H 2 (D)
the Hardy space on the unit disc,
Z
2
2
iθ 2
H (D) = f ∈ Hol(D) : sup
|f (re )| dm(θ) := kf kH 2 < ∞ ;
0<r<1
H ∞ (D)
T
the Hardy algebra on the unit disc,
∞
H (D) = f ∈ Hol(D) : sup |f (z)| := kf kH ∞ < ∞ ;
z∈D
5
CHAPTER 1
The Hardy Space on the Disc
In this first lecture we will focus on the Hardy space H 2 (D). We begin with a “crash
course” on the necessary theory for the Hardy space.
1. Basic Definitions
We now introduce the space H 2 (D). Let f ∈ Hol(D), then we say that f ∈ H 2 (D) if
Z
(1.1)
sup
|f (reiθ )|2 dm(θ) := kf k2H 2 (D) < ∞.
0<r<1
T
A related space that will play a distinguished role in our space is the Hardy space H ∞ (D)
(1.2)
sup |f (z)| := kf kH ∞ (D) < ∞.
z∈D
We will see that with the norms we have introduced, the space H 2 (D) is a Hilbert space,
while the space H ∞ (D) is a Banach space.
Exercise 1.1. Show that it is possible to replace the sup0<r<1 by limr→1 in the definition
of H 2 (D).
Exercise 1.2. Show that H ∞ (D) is a Banach algebra.
We now show other norms that can be used to study the functions in H 2 (D). First, recall
that the Fourier transform of a function f ∈ L2 (T) is given by
Z
fˆ(n) = f (eiθ )e−inθ dm(θ).
T
Then, a simple computation shows that
Z
1 : n=m
i(n−m)θ
e
dm(θ) =
0 : n 6= m.
T
P∞
Using this, we see that for f (z) = n=0 an z n that
Z
2
kf kH 2 (D) = sup
|f (reiθ )|2 dm(θ)
0<r<1 T
2
Z X
∞
n inθ a
r
e
= sup
dm(θ)
n
0<r<1 T n=0
Z
∞
X
n m
= sup
an am r r
ei(n−m)θ dm(θ)
0<r<1
=
∞
X
T
n,m=0
|an |2 = kf k2H 2 (D) .
n=0
7
8
1. THE HARDY SPACE ON THE DISC
Note that this norm says that it is possible to study the behavior of the functions in
H (D) via their Fourier coefficients. For 0 < r < 1 and z ∈ D let fr (z) = f (rz). Then the
computations done above, prove that the following proposition.
2
Proposition 1.3. Suppose that f ∈ H 2 (D). Then, the sequence {fr } is Cauchy in
L (T).
2
Proof. Using the computations from
Z
2
kfr − fs kL2 (T) =
above, and obvious modifications, we see
2
∞
X
n
n inθ an (r − s )e dm(θ)
T
n=0
∞
X
=
|(rn − sn )|2 |an |2 .
n=0
But, as r, s → 1 and the dominated convergence theorem, since
conclude that this last summand goes to zero.
P∞
n=0
|an |2 < ∞, we can
Now note that since L2 (T) is a complete space, then we have an element f ∗ ∈ L2 (T)
given by f ∗ = limr→1 fr is also in L2 (T). Since f ∗ ∈ L2 (T) we can compute the Fourier
coefficients to be
Z
∗
fb (n) =
f ∗ (eiθ )e−inθ dm(θ)
T
Z
= lim fr (eiθ )e−inθ dm(θ)
r→1
T
an : n ≥ 0
=
0 : n < 0.
Note that the computations we have done thus far proves the following proposition.
Proposition 1.4. Suppose that f ∈ H 2 (D) and f ∗ (eiθ ) = limr→1 f (reiθ ) then
kf k2H 2 (D)
=
∞
X
|an |2 = kf ∗ k2L2 (T) .
n=0
The only fact that remains to complete the proof of this proposition is that
∞
X
|an |2 = kf ∗ k2L2 (T)
n=0
which holds by Parseval’s Theorem. This also shows that the inner product on H 2 (D) will
satisfy
Z
∞
X
∗ iθ ∗ iθ
hf, giH 2 (D) = f (e )g (e )dm(θ) =
an b n
T
where we have f (z) =
P∞
n=0
an z n and g(z) =
n=0
P∞
n=0 bn z
n
.
Exercise 1.5. Let H02 denote the collection of functions in H 2 (D) such that f (0) = 0.
Show that L2 (T) = H 2 (D) ⊕ H02 (D). Hint: This follows easily from the Fourier coefficients.
1. BASIC DEFINITIONS
9
1.1. The Reproducing Kernel for H 2 (D). The Hardy space H 2 (D) also has an additional property of being a reproducing kernel Hilbert space. This means that for each point
z ∈ D there is a special function kz ∈ D such that
hf, kz iH 2 (D) = f (z)
We now turn to determining what this function is.
Proposition 1.6. Let z ∈ D, then
1
|f (z)| ≤ kf kH 2 (D) p
.
1 − |z|2
Proof. Note that we have via taking limits in Cauchy’s formula that
f (z) = lim fr (z)
r→1
Z ∗
f (w)
1
dw
=
2πi T w − z
Z ∗ iθ
f (e ) iθ
1
ie dθ
=
2πi T eiθ − z
Z
f ∗ (eiθ )
=
dm(θ).
−iθ
T 1 − ze
This then implies by an application of Cauchy-Schwarz and then computing that
Z
1
1
dm(θ) =
.
−iθ
2
|
1 − |z|2
T |1 − ze
1
1
|f (z)| ≤ kf ∗ kL2 (T) p
= kf kH 2 (D) p
.
2
1 − |z|
1 − |z|2
Exercise 1.7. Give an alternate proof of Proposition 1.6 by using the Fourier coefficient
representation of functions in H 2 (D).
With this proposition proved, we see that pointwise evaluation of functions f ∈ H 2 (D)
is a bounded operator. So, by the Riesz representation theorem, we know that there is a
unique function kz ∈ H 2 (D) such that
f (z) = hf, kz iH 2 (D) .
We now turn to determining what this function in fact is. Observe that the following
property holds since the functions kz ∈ H 2 (D)
kλ (ξ) = hkλ , kξ iH 2 (D) .
We now collect a general fact about reproducing kernel Hilbert spaces. Good references
for the facts presented here are [1, 9, 10]. These are Hilbert spaces H of functions over some
domain X. For each point λ ∈ X we have that point evaluation of the functions f ∈ H is a
continuous operation. Therefore, we have a special function kλ ∈ H such that
f (λ) = hf, kλ i
∀f ∈ H.
10
1. THE HARDY SPACE ON THE DISC
This vector kλ is called the reproducing kernel for the space H. We also have the following
property holding:
kλ (ξ) = k(λ, ξ) = hkλ , kξ i .
This function is called the kernel function for the Hilbert space H. A simple fact is then the
following.
Proposition 1.8. Let H be a Hilbert function space on X and let {ei } be an orthonormal
basis for H. Then
X
kλ (ξ) =
ei (λ)ei (ξ)
i
Exercise 1.9. Prove Proposition 1.8.
The proof of this fact is just an application of Parseval’s Identity in a general Hilbert
space, and two applications of the reproducing kernel property for H. When we specialize
to the case H 2 (D), then one can show that the functions {z k } are an orthonormal system of
functions in H 2 (D). Thus, appropriately normalizing we have that
kλ (ξ) =
∞
X
k
λ ξk.
k=0
Using this formula, we see that the function
kz (w) =
1
1 − zw
is the reproducing kernel for the Hardy space H 2 (D).
Exercise 1.10. Determine the norm of the function kz ∈ H 2 (D).
Exercise 1.11. Using Proposition 1.6, determine the reproducing kernel for H 2 (D).
Exercise 1.12. Show that
Z
T
1
1
dm(θ) =
.
−iθ
2
|1 − ze |
1 − |z|2
2
Exercise 1.13. Let A (D) denote the Bergman space of functions,
Z
2
2
2
A (D) := f ∈ Hol(D) : kf kA2 (D) :=
|f (z)| dA(z) < ∞ .
D
Show that the reproducing kernel for the Bergman space is given by
1
kλ (z) =
.
(1 − λz)2
2. Littlewood-Paley Identities and H 2 (D)
We now show how it is possible to obtain another norm on H 2 (D) using information
about the function on the disc D. This equivalent norm will prove useful when we study the
space of Carleson measures for H 2 (D) since it will allow us to generate a natural family of
examples of functions which generate Carleson measures. Also, this new norm will allow us
to place the Hardy space in a scale of Besov-Sobolev spaces.
2. LITTLEWOOD-PALEY IDENTITIES AND H 2 (D)
11
First, we begin by recalling Green’s formula in the case of the unit disc D and its boundary
T. Then Green’s formula
Z
Z takes the form:
1
u(ξ)dm(ξ) − u(0) =
∆u(z) log dA(z)
|z|
T
D
Note that we can move the point 0 to any other point z ∈ D by a Möbius map of the form
w−z
ϕz (w) = 1−zw
.
Exercise 2.1. Work out Green’s formula for an arbitrary point z ∈ D.
We will begin with a function g ∈ L1 (T) and, as usual, let g(z) denote the Poisson
extension of the function g. The gradient of a function g is given by ∇g = (∂x g, ∂y g) and we
have
|∇g(z)|2 = |∂x g(x, y)|2 + |∂y g(x, y)|2 .
In the case when g happens to be an analytic function we have that
2
|∇g(z)|2 = |∂g(z)|2 = |g 0 (z)| .
Lemma 2.2 (Littlewood-Paley Identity). Suppose that g ∈ L1 (T) and if g(0) =
then
Z
Z
Z
1
2
2
2 |∇g(z)| log dA(z) = |g − g(0)| dm = |g|2 dm − |g(0)|2 .
|z|
T
T
D
R
T
gdm
Proof. With out loss of generality we may assume that g(0) = 0, since we can reduce
to this case by considering the function g̃ = g − g(0). We will apply Green’s Theorem with
the function u = |g|2 . Since g(0) = 0 we have that u(0) = |g(0)|2 = 0. Now observe that
∂∂ |g(z)|2 = ∂ ∂gg + g∂g
= ∂∂gg + g∂∂g + ∂g∂g + ∂g∂g
2
= ∂g∂g + ∂g∂g = |∂g|2 + ∂g 1
|∇g(z)|2 .
2
Here the last equality follows from the definitions of the operators ∂ and ∂. Using this we
see that
∆ |g(z)|2 = 2 |∇g(z)|2 .
Substituting into Green’s formula we have
Z
Z
Z
1
1
2
2
|g(ξ)| dm =
∆(|g(z)| ) log dA(z) = 2 |∇g(z)|2 log dA(z)
|z|
|z|
T
D
D
proving the Lemma.
=
Using this lemma, we have another way to compute the norm of a function in H 2 (D).
Proposition 2.3. Suppose that g ∈ H 2 (D) then we have
Z
1
2
2
2
kgkH 2 (D) = |g(0)| + 2 |g 0 (z)| log dA(z).
|z|
D
The proof of this follows by simple rearrangement of the above Lemma. We give a slightly
different way to see the resulting norm that in some cases is easier to use. More importantly
for us, it will allow us to place the Hardy space in a scale of analytic function spaces that
are very interesting.
12
1. THE HARDY SPACE ON THE DISC
Lemma 2.4. If g ∈ L1 (T) then
Z
Z
Z
1
2
2
2
|∇g(z)| (1 − |z| )dA(z) ≤ 2 |∇g(z)| log dA(z) ≤ C |∇g(z)|2 (1 − |z|2 )dA(z).
|z|
D
D
D
Proof. First note that 1 − t ≤ 2 log 1t if 0 ≤ t < 1. So we have that
Z
Z
1
2
2
|∇g(z)| (1 − |z| )dA(z) ≤ 2 |∇g(z)|2 log dA(z),
|z|
D
D
which is the first inequality in the Lemma.
To prove the second inequality, first, suppose that the integral on the right hand side is
finite and then normalize it so that
Z
|∇g(z)|2 (1 − |z|2 )dA(z) = 1.
D
Now, if |z| >
1
4
1
then we have that log |z|
≤ C(1 − |z|2 ), and so we then have that
Z
Z
1
2
|∇g(z)| log dA(z) ≤ C |∇g(z)|2 (1 − |z|2 )dA(z).
1
|z|
D
≤|z|≤1
4
In the case when |z| <
have that
1
4
we exploit the subharmonicity of |∇g(z)|. By subharmonicity we
2
|∇g(z)|
Z
|∇g(ξ)|2 dA(ξ)
≤ 16
|ξ−z|< 14
Z
≤ 32
|∇g(ξ)|2 (1 − |ξ|2 )dA(ξ) = 32.
|ξ|< 12
For the last inequality, we have used that for |z| < 41 and |ξ − z| < 14 that |ξ| < 21 . We then
use the fact that when |ξ| < 21 that 1 − |ξ|2 ≥ 34 ≥ 12 . Using this, we see that
Z
Z
1
1
2
|∇g(z)| log dA(z) ≤ C
log dA(z) = C.
1
1
|z|
|z|
|z|< 4
|z|< 4
Combining the estimates we have obtained when |z| ≥ 41 and when |z| ≤ 14 then gives that
Z
Z
1
2
2 |∇g(z)| log dA(z) ≤ C |∇g(z)|2 (1 − |z|2 )dA(z).
|z|
D
D
Again, by rearrangement of the above Lemma we have another equivalent norm on the
space H 2 (D)
Z
Z
2
2
2
2
2
2
2
0
0
|g(0)| + |g (z)| (1 − |z| )dA(z) ≤ kgkH 2 (D) ≤ C |g(0)| + |g (z)| (1 − |z| )dA(z) .
D
D
Exercise 2.5. Give an alternate proof of the above equivalent norm on H 2 (D) using
Fourier series. Doing this, you can obtain a better (in fact sharp) estimate of the constant
C.
3. CARLESON MEASURES FOR H 2 (D)
13
2.1. Besov-Sobolev Spaces. We now (briefly) introduce the Besov-Sobolev spaces on
the unit disc D. We fix a parameter 0 ≤ σ ≤ 21 and define the Besov-Sobolev space Bσ2 (D)
as the collection of analytic functions on the disc such that
Z
2
2
2
kf kBσ2 (D) = |f (0)| + |f 0 (z)| (1 − |z|2 )2σ dA(z) < ∞.
D
Based on the Lemmas above, we have that when σ = 21 that H 2 (D) = B 21 (D), with equivalent
2
norms. When σ = 0, then we are looking at the functions that are analytic on D and such
that its derivative is square integrable.
Exercise 2.6. Show that an equivalent norm on the space Bσ2 (D) is given by
∞
X
n1−2σ |an |2
n=0
where f (z) =
P∞
n=0
n
an z .
3. Carleson Measures for H 2 (D)
Our next goal is to prove the following important theorem about certain measures for
H 2 (D).
Theorem 3.1 (Carleson Embedding Theorem). Let µ be a non-negative Borel measure
in D. Then the following are equivalent:
2
2
(i) The embedding operator
2 J : L (T) → L (D, µ), with J (f )(z) = f (z), is bounded.
2 1/2
)
(ii) C(µ)2 := supz∈D J k̃z = supz∈D kPz kL1 (µ) < ∞, where k̃z (ξ) = (1−|z|
, the
(1−ξz)
L2 (µ)
reproducingkernel T
for the Hardy space H 2 (D).
1
(iii) I(µ) = sup r µ(D Q(ξ, r)) : r > 0, ξ ∈ T < ∞, where Q(ξ, r) is a ball measured
centered at ξ with radius r.
Moreover, the following inequalities hold
C(µ) ≤ kJ k ≤ 4C(µ)
and
32−1 I(µ) ≤ C(µ)2 ≤ 32I(µ)
Here the embedding operator J is given by
Z
J (f )(z) = Pz (ξ)f (ξ)dm(ξ)
T
where
Pz (ξ) =
1 − |z|2
|1 − ξz|2
is the Poisson kernel.
We will give one proof of this fact, one that will take advantage of the special structure
of the reproducing kernel of the space H 2 (D). An alternate proof can be based on the tools
of harmonic analysis in particular maximal functions. The interested reader can see the
books [5, 11] to learn more about this connection between Carleson measures and harmonic
analysis.
First, we begin with a definiton.
14
1. THE HARDY SPACE ON THE DISC
Definition 3.2. A non-negative Borel measure µ on D is a H 2 (D)-Carleson measure if
Z
|f (z)|2 dµ(z) ≤ C(µ)2 kf k2H 2 (D) ∀f ∈ H 2 (D).
D
Observe that this is saying that H 2 (D) continuously embeds into L2 (D; µ). This is nothing
other than condition (i) in Theorem 3.1 when applied to the boundary values of f ∈ H 2 (D)
and then carried back to the unit disc via harmonic extension.
Note that since we have a natural set of functions by which to test this inequality, namely
the reproducing kernels, {kz }z∈D , we see that a necessary condition is that
Z
|kz (w)|2 dµ(w) ≤ C(µ)2 kkz k2H 2 (D) .
D
Upon rearrangement, this gives condition (ii) from Theorem 3.1.
Finally, observe that if we restrict the integration to a natural set associated to the kernel
kz then we see that condition (iii) arises naturally. Namely, note that to each point z ∈ D
z
we can associate an interval in T with center |z|
and total length (1 − |z|). We let T (I)
denote the “tent” over an interval I,
T (I) = {z = reiθ ∈ D : 0 < 1 − r ≤ |I|, θ ∈ I}.
Then it is possible to phrase condition (iii) as
µ (T (I)) ≤ T (µ)|I| ∀I ⊂ T.
Exercise 3.3. Show that the above inequality is equivalent to condition (iii).
3.1. Carleson Measures via Reproducing Kernels. We first give a proof of Theorem 3.1 that uses the reproducing kernel directly. This proof can be found as an exercise in
the book [7].
3.1.1. (i) ⇐⇒ (ii). One direction of this equivalence is immediate, namely (i)⇒(ii). For
if we know that the embedding operator J is bounded then we have
= kJ k
≤ kJ k k̃z J k̃z L2 (µ)
L2 (T)
since the reproducing kernel k̃z is normalized to have L2 (T) norm one. This also proves that
C(µ) ≤ kJ k. Finally, we should indicate why the equality
2
sup J k̃z z∈D
L2 (µ)
= sup kPz kL1 (µ)
z∈D
holds. Since k̃z ∈ H 2 (D) then the Poisson kernel Pw reproduces the function value at w,
namely
Z
k̃z (w) =
k̃z (ξ)Pw (ξ)dm(ξ).
T
This then implies that J (k̃z )(w) = kz (w), and using this we have
3. CARLESON MEASURES FOR H 2 (D)
2
J k̃z =
L2 (µ)
=
=
=
=
15
Z 2
J (k̃z )(w) dµ(w)
ZD 2
k̃z (w) dµ(w)
D
Z 2 1/2 2
(1
−
|z|
)
dµ(w)
(1 − zw) D
Z
(1 − |z|2 )
2 dµ(w)
D |1 − zw|
Z
Pz (w)dµ(w).
D
2
This then shows that J k̃z L2 (µ)
= kPz kL1 (µ) , and then these suprema over z ∈ D are of
course equal.
It only remains to prove that (ii)⇒(i). To prove this direction it is enough to show that
the formal adjoint J ∗ : L2 (D, µ) → L2 (T) with
Z
∗
Pz (ξ)f (z)dµ(z),
J (f )(ξ) =
D
is bounded. With this in mind we have,
kJ
∗
(f )k2L2 (T)
Z
=
|J ∗ (f )(ξ)|2 dm(ξ)
ZT
J ∗ (f )(ξ)J ∗ (f )(ξ)dm(ξ)
Z
ZT Z
0
Pz0 (ξ)f (z 0 )dµ(z ) dm(ξ)
Pz (ξ)f (z)dµ(z)
=
D
T
D
Z Z Z
Pz (ξ)Pz0 (ξ)dm(ξ) f (z)f (z 0 )dµ(z)dµ(z 0 ).
=
=
D
D
T
The following lemma will be used to prove that J ∗ is bounded.
Lemma 3.4 (Vinogradov-Senichkin Test). Let Z be a measurable space and k a nonnegative measurable function on Z × Z. If
Z
k(s, t)k(s, x)ds ≤ C(k(t, x) + k(x, t))
Z
for a.e. (t, x) ∈ Z × Z, then
ZZ
k(s, t)g(s)g(t)dsdt ≤ 2C
Q :=
Z×Z
for any non-negative function g with kgkL2 (Z) ≤ 1 and Q < ∞.
Exercise 3.5. Prove the Vinogradov-Senichkin Test.
16
1. THE HARDY SPACE ON THE DISC
We want to apply the Vinogradov-Senichkin Test, so we define the integral operator
Tk : L2 (D, µ) → L2 (D, µ) with kernel k(z 0 , z) = Pz (z 0 ) by
Z
k(z 0 , z)g(z 0 )dµ(z 0 ) ∀g ∈ L2 (D, µ).
Tk (g)(z) :=
D
Then we have that
kJ
∗
(f )k2L2 (T)
Z Z
≤
D
Pz (z 0 ) |f (z)| |f (z 0 )| dµ(z)dµ(z 0 )
D
= (Tk |f | , |f |).
Now, we make the following claim:
Proposition 3.6. Let z, z 0 , w ∈ D then
Pz (z 0 )Pw (z 0 ) ≤ 8 (Pz (w)Pw (z 0 ) + Pw (z)Pz (z 0 ))
Proof. Begin by noting that the following inequality holds
1/2
1/2
a−1 := |1 − zw|1/2 ≤ 1 − zz 0 + |1 − z 0 w| := b−1 + c−1 .
Exercise 3.7. Prove this. Hint: This is the triangle inequality for a certain metric.
Using this inequality we have that
bc ≤ a(b + c),
which in turn implies that
b4 c4 ≤ a4 (b + c)4 ≤ 23 a4 (b4 + c4 ).
Using this inequality, but with the appropriate substitutions for a,b, and c and using the
numerators for Pz (z 0 ) and Pw (z 0 ) we find that
!
3
1
1
2
2
2
(1 − |z| )(1 − |w| ) Pz (z 0 )Pw (z 0 ) ≤
+
1 − zz 0 2 |1 − z 0 w|2
|1 − zw|2
#
"
2
2
2
2
(1
−
|w|
)
(1
−
|z|
)
(1
−
|w|
)
(1
−
|z|
)
= 23
+
|1 − zw|2 1 − zz 0 2
|1 − z 0 w|2 |1 − zw|2
= 23 (Pw (z)Pz (z 0 ) + Pz (w)Pw (z 0 )).
Since we want to apply the Vinogradov-Senichkin Test we need to know that the kernel
k(z 0 , z) = Pz (z 0 ) satisfies the hypothesis of the lemma. To this end we need to estimate
Z
k(z 0 , z)k(z 0 , w)dµ(z 0 ).
D
Using the estimate we found for the product of two Poisson kernels that we illustrated
above, we have
3. CARLESON MEASURES FOR H 2 (D)
Z
0
0
0
17
Z
Pz (z 0 )Pw (z 0 )dµ(z 0 )
D
Z
3
≤ 2
Pz (w)Pw (z 0 ) + Pw (z)Pz (z 0 )dµ(z 0 )
D
Z
Z
0
0
0
0
3
= 2 Pz (w) Pw (z )dµ(z ) + Pw (z) Pz (z )dµ(z )
k(z , z)k(z , w)dµ(z ) =
D
D
3
D
2
≤ 2 C(µ) (Pz (w) + Pw (z)).
The last inequality holds because we are trying to prove that (ii)⇒(i). So the kernel
satisfies the Vinogradov-Senichkin Test, thus we have
kJ ∗ (f )k2L2 (T) ≤ (Tk |f | , |f |)
≤ 2 · (23 C(µ)2 ) kf k2L2 (µ)
= 24 C(µ)2 kf k2L2 (µ) .
A duality argument then gives that kJ k ≤ 4C(µ), proving that J is bounded and giving
the relationship between kJ k and C(µ).
3.1.2. (ii) ⇐⇒ (iii). To finish the proof of this theorem we only need to dispose of the
final equivalence. Again, we have an easy implication and a harder implication. We begin
by showing that (ii)⇒(iii). Under the hypothesis of (ii) we have
Z
Z
(1 − |z|2 )
2
0
0
C(µ) ≥
Pz (z )dµ(z ) =
2 dµ(z 0 ).
0
D
D 1 − zz Now take ξ ∈ T and 0 < r < 2, and set z = (1 − 2r )ξ. Now consider the non-isotropic
ball Q(ξ, r). A simple calculation shows that z ∈ Q(ξ, r). Then we have
2
C(µ)
Z
≥
D
(1 − |z|2 )
dµ(z 0 )
1 − zz 0 2
(1 − |z|2 )
≥
2 dµ(z 0 )
T
D Q(ξ,r) 1 − zz 0 Z
−1 −2
≥ 16 r (1 − |z|) T
dµ(z 0 )
D Q(ξ,r)
\
= 16−1 r−2 (1 − |z|)µ D Q(ξ, r) .
Z
The last inequality follows since z ∈ D and we are considering z, z 0 ∈ Q(ξ, r). Because if
we have two points z, z 0 ∈ Q(ξ, r) then by the triangle inequality for the non-isotropic metric
we have
1 − zz 0 1/2 ≤ 1 − zξ 1/2 + 1 − z 0 ξ 1/2 .
Squaring this last inequality gives
1 − zz 0 ≤ 2 1 − zξ + 1 − ξz 0 ≤ 4r.
18
1. THE HARDY SPACE ON THE DISC
−2
This then gives 1 − zz 0 ≥ 2−4 r−2 . We also have that
r
1 − |z| = .
2
Combining these estimates gives,
\
C(µ)2 ≥ 16−1 r−2 (1 − |z|)µ D Q(ξ, r)
\
≥ 16−1 2−1 rr−2 µ D Q(ξ, r) .
Taking the supremum over 0 < r < 2 then gives
32−1 I(µ) ≤ C(µ)2 .
It only remains to prove that (iii)⇒(ii). We will break this part up into two different cases.
First, consider the case where |z| ≤ 43 . Now we have the following inequality holding for the
Poisson kernel.
(1 − |z|2 )
Pz (ξ) = 1 − zξ 2
(1 − |z|2 )
≤
(1 − |z|)2
(1 + |z|)2
=
(1 − |z|2 )
22
≤
.
(1 − |z|2 )
Then if |z| ≤
3
4
we have that
Z
16
µ(D)
7
\
26
=
2µ(D Q(ξ, 2))
7
27
≤
I(µ)
7
≤ 20I(µ).
Pz (w)dµ(w) ≤ 4
D
z
So we only need to deal with the case when |z| > 43 . Let z̃ = |z|
and define the following
sets
\
Qk := D Q z̃, 2k+1 (1 − |z|2 )
∀k ∈ N.
Then for w ∈ Qk+1 \ Qk we have
1 − wz̃ ≥ 2k+1 (1 − |z|2 ).
By the triangle inequality for the non-isotropic metric we have
3. CARLESON MEASURES FOR H 2 (D)
19
1 − wz̃ 1/2 ≤ |1 − wz|1/2 + 1 − z z̃ 1/2
= |1 − wz|1/2 + (1 − |z|)1/2
≤ |1 − wz|1/2 + (1 − |z|2 )1/2 ,
with the last inequality following since z ∈ D. Squaring this last inequality gives,
1 − wz̃ ≤ 2 |1 − wz| + (1 − |z|2 ) .
Now using this we can conclude that
|1 − wz| ≥ 2−1 1 − wz̃ − (1 − |z|2 )
≥ 2k (1 − |z|2 ) − (1 − |z|2 )
≥ 2k−1 (1 − |z|2 ).
But this last inequality implies that
|1 − wz|−2 ≤ 2−2(k−1) (1 − |z|2 )−2
when w ∈ Qk+1 \ Qk . Now using this we have
Z
Z
Pz (w)dµ(w) =
Pz (w)dµ(w) +
Q1
D
∞ Z
X
k=1
Pz (w)dµ(w)
Qk+1 \Qk
∞ Z
X
(1 − |z|2 )
2
dµ(w)
≤
2 dµ(w) +
4k−1 (1 − |z|2 )2
Q1 (1 − |z| )
k=1 Qk+1 \Qk
∞ k−1
X
1
µ(Qk+1 )
µ(Q1 )
.
≤ 4
2 +
(1 − |z| ) k=1 4
(1 − |z|2 )
2
Z
Now we need to recall how I(µ) was defined and how each of the Qk was defined. Doing
this we have
∞
Z
X
µ(Q1 )
µ(Qk+1 )
Pz (w)dµ(w) ≤ 16
+
(4−1 )k−1 (2k+2 )
2
2
k+2
(2 (1 − |z| )) k=1
(2 (1 − |z|2 ))
D
4
≤ 16I(µ) + 2 I(µ)
∞
X
(2−1 )k
k=1
5
≤ 2 I(µ).
Combining the estimates if |z| ≤ 43 and if |z| > 43 , we have that
Z
Pz (w)dµ(w) ≤ 25 I(µ),
D
then taking the supremum over z ∈ D proves the theorem.
The careful reader will have noticed that this proof can actually be used to show the
following theorem.
Theorem 3.8 (Carleson Embedding Theorem). Let µ be a non-negative Borel measure
in D. Then the following are equivalent:
20
1. THE HARDY SPACE ON THE DISC
2
2
(i) The embedding operator
J2: L (T) → L (D, µ), with J (f )(z) = f (z), is bounded.
2 1/2
)
(ii) C(µ)2 := supz∈suppµ J k̃z = supz∈suppµ kPz kL1 (µ) < ∞, where k̃z (ξ) = (1−|z|
,
(1−ξz)
2
L (µ)
the reproducing kernel for the Hardy space H 2 (D).
CHAPTER 2
The Maximal Ideal Space of H ∞ (D), ∂-equations and Wolff ’s
Theorem
1. The Maximal Ideal Space of H ∞ (D)
We now wish to study the maximal ideals space associated to the algebra H ∞ (D). But,
before we can do that, we need to review a little about the maximal ideal spaces associated
to a Banach algebra.
Recall that a (commutative) Banach algebra A is a complex (commutative) algebra A
that is also a Banach space under a norm that satisfies
kf gk ≤ kf k kgk
f, g ∈ A.
We will also assume that there is an identity element 1 ∈ A and that our algebra is commutative. An element f ∈ A is invertible if there exists an element g ∈ A such that f g = 1.
When this happens, we will simply write f −1 for g. We let
A−1 = {f ∈ A : f −1 exists}.
Finally, we need to consider the multiplicative linear functionals on the algebra A. These are
simply the complex homomorphisms m : A → C. Note that we trivially have that m(1) = 1.
The first observation is that the complex homomorphisms are continuous and bounded
with norm at most 1.
Lemma 1.1. Every complex homomorphism from A to C is a continuous linear functional
with norm at most 1. Namely,
kmk =
sup
|m(f )| ≤ 1.
f ∈A,kf k≤1
Proof. If m is unbounded or if kmk > 1P
then we can find an element f ∈ A with
n
kf k < 1 but m(f ) = 1. Consider the element ∞
n=0 f ∈ A. This element exists since we
have kf k < 1 and so
∞
∞
∞
X X
X
1
n
n
f ≤
kf k ≤
kf kn =
.
1
−
kf
k
n=0
n=0
n=0
P∞ n
Note that we have (1 − f ) n=0 f = 1. So 1 − f ∈ A−1 . However,
1 = m(1) = m((1 − f )(1 − f )−1 ) = m((1 − f )−1 )(m(1) − m(f )) = 0
which is a contradiction.
Exercise 1.2. Prove that mz (f ) = f (z) is a multiplicative linear functional on H ∞ (D).
Next, we will connect the algebraic property of maximal ideals and the function analytic
property of the multiplicative linear functionals.
21
22
2. THE MAXIMAL IDEAL SPACE OF H ∞ (D), ∂-EQUATIONS AND WOLFF’S THEOREM
Lemma 1.3. Suppose that M is a maximal ideal of A. Then M is the kernel of a
multiplicative linear functional m : A → C. Conversely, suppose that m : A → C is a
multiplicative linear functional. Then ker m is a maximal ideal.
Proof. We first show that the maximal ideal M is closed. Note that M ⊂ M . If M is
proper, i.e, M 6= A, then M is also an ideal. However, since M is maximal, if M 6= A then
we must have M = M and so M is closed. If g ∈ M then g ∈
/ A−1 (otherwise we would have
1 ∈ M and so M = A). Consider the element f = 1 − g, then we have that k1 − gk ≥ 1,
and so 1 ∈
/ M and so M is closed.
Now we next show that the quotient algebra B = A/M satisfies
B = C1
where 1 = 1 + M denotes the unit in the quotient algebra. It is obvious that C1 ⊂ B,
and so we need to handle the other inclusion. Once we have shown this, then the quotient
mapping will then define the multiplicative linear functional, and the kernel of this mapping
with then be M .
Since we have M maximal, then we have that B = A/M is a field. Moreover, since M is
closed we have that B is complete in the quotient norm
kf + M k = inf kf + gkA .
g∈M
It is also the case that the norm makes B into a Banach algebra (use the ideal property of
M ).
Suppose that f ∈ B \ C1. Then, we have that f − λ ∈ B −1 for all λ ∈ C since B is a
field. Choose λ0 and note that on the disc centered at λ0 of radius k(f − λ0 )−1 k the series,
∞
X
(λ − λ0 )n ((f − λ0 )−1 )n+1
n=0
converges in norm to the element (f − λ)−1 because we have the following identity holding
1
1
1
=
.
0
f −λ
f − λ0 1 − (fλ−λ
−λ )
0
−1
Now f 6= 0 and by the Hahn-Banach Theorem, there is a bounded linear functional L on
B such that kLk = 1 and L(f −1 ) 6= 0.
Define the function on the disc centered at λ0 of radius k(f − λ0 )−1 k
∞
X
F (λ) = L((f − λ)−1 ) =
(λ − λ0 )n L((f − λ0 )−1 )n+1 ).
n=0
Since the kLk = 1 and the series defining (f − λ)−1 is norm convergent this function makes
sense. Because λ0 is arbitrary, we have that F (λ) is an entire function. Suppose that |λ| is
large, then we have that
∞
(1 − f )−1 1 X kf kn
λ
(f − λ)−1 =
.
≤
|λ|
|λ| n=0 |λ|n
Note that this implies for |λ| large that we have
|F (λ)| ≤
C
|λ|
1. THE MAXIMAL IDEAL SPACE OF H ∞ (D)
23
which by Louiville’s Theorem implies that F = 0. But, this contradict the existence of the
functional L and hence gives us that B = C1.
For the converse, it is immediate that the ker m is an ideal. The maximality follows since
the codimension of any linear functional is 1. Namely, dim(A \ ker m) = 1.
Exercise 1.4. Verify that if A is a Banach algebra, and if M is a closed proper maximal
ideal, then A/M is a commutative Banach algebra.
Exercise 1.5. Show that for a linear functional that dim(A \ ker m) = 1.
Given a Banach algebra A, we let MA denote the set of complex homorphisms of A.
This is called the maximal ideal space of the Banach algebra. By the above, we have that
MA is contained in the unit ball of the dual Banach algebra A∗ . We now endow MA with
the weak-* topology of A∗ . Namely, the basic neighborhood of a m0 ∈ MA is determined by
> 0 and by elements f1 , . . . , fn ∈ A such that
V = {m ∈ A∗ : kmk ≤ 1, |m(fj ) − m0 (fj )| < ,
1 ≤ j ≤ n}.
Also, note that we have that MA is a weak-* closed subset of the unit ball of A∗ . This is
because
MA = {m ∈ A∗ : kmk ≤ 1, m(f g) = m(f )m(g), f, g ∈ A}.
This topology on MA is called the Gelfand topology. In this topology we have that MA is
a weak-* closed subset of the unit ball of A∗ . Now by the Banach-Alaoglu Theorem, we have
that the ball of A∗ is weak-* compact and so we can have that MA is compact Hausdorff
space.
We now turn from these abstractions and focus on a particular case of interest A =
∞
H (D). By the exercise above we see that D ⊂ MH ∞ . Now, we know that the disc D is
open, and that the space M∞
H is compact, so we can not have them being equal. However,
it is conceivable that by taking the closure of the disc D is the Gelfand topology we could
have
Gelfand
MH ∞ = D
.
Gelfand
We then define the “Corona” of the algebra H ∞ (D) to be MH ∞ \ D
this question of density to a question about analytic functions.
. We now translate
Theorem 1.6. The open disc D is dense in MH ∞ if and only if the following condition
holds: If f1 , . . . , fn ∈ H ∞ (D) and if
max |fj (z)| ≥ δ > 0
1≤j≤n
then there exists g1 , · · · , gn ∈ H ∞ (D) such that
f1 g1 + · · · + fn gn = 1.
Proof. Suppose that D is dense in MH ∞ . Then, by continuity we have that
max |m(fj )| ≥ δ
1≤j≤n
for all m ∈ MH ∞ . This implies that {f1 , . . . , fn } is in no proper ideal of H ∞ (D). Hence
the ideal generated by {f1 , . . . , fn } must contain the constant function 1 and so there exists
g1 , . . . , gn ∈ H ∞ (D) such that
1 = f1 g1 + · · · + fn gn .
2. THE MAXIMAL IDEAL SPACE OF H ∞ (D), ∂-EQUATIONS AND WOLFF’S THEOREM
24
Conversely, suppose that D is not dense in MA . Then for some m0 ∈ MA has a neighborhood
disjoint from D and this neighborhood has the form
V = ∩nj=1 {m : |m(jj )| < δ}
where δ > 0 and f1 , . . . , fn ∈ H ∞ (D) with m0 (fj ) = 0. Since D ∩ V = ∅ we have that
max |fj (z)| ≥ δ > 0.
1≤j≤n
But, it is not possible that we have
1 = f1 g1 + · · · + fn gn
since we have that m0 (fj ) = 0 for all 1 ≤ j ≤ n.
Exercise 1.7. Let A(D) = {f ∈ Hol(D) ∩ C(D) : supz∈D |f (z)| < ∞} denote the disc
algebra. Show that
MA(D) = D.
Our goal in later lectures is to prove that a collection of functions f1 , . . . , fn ∈ H ∞ (D)
that don’t simultaneously vanish as in Theorem 1.6 generate all of H ∞ (D) as an ideal. To
accomplish this, we first need to introduce a little more machinery and tools that are very
valuable in their own right.
2. Solving ∂-Equations
Recall that a function h is analytic if
∂h = 0
where ∂ = 12 (∂x + i∂y ). We first begin showing how to solve equations of the form
∂F = G.
Equations of this type will arise naturally in the proof of the Corona Theorem. Suppose
that G is smooth with compact support. Then we have the following Theorem
Theorem 2.1. Suppose that G is a smooth compactly supported function in D. Then
Z
1
1
F (z) =
G(ξ)
dξ ∧ dξ
2πi D
z−ξ
solves
∂F = G.
Before we prove this theorem, since we will apply Stokes Theorem, we briefly recall the
theory of differential forms in this context. Recall that Stokes Theorem can (roughly) be
stated as
Z
Z
ω=
∂Ω
dω.
Ω
Here Ω is a nice domain, ∂Ω is the boundary of Ω, ω is a differential form, and d is the
exterior differential.
In two variables, for a smooth function we have that
df = ∂x dx + ∂y dy
2. SOLVING ∂-EQUATIONS
25
Recall also that dx ∧ dy = −dy ∧ dx and that dx ∧ dx = dy ∧ dy = 0. Since we are working
with complex variables it is more conducive to write this in the variables z and z. In this
notation we have
df = ∂f dz + ∂f dz
where
1
1
∂ = (∂x + i∂y ) ∂ = (∂x − i∂y ) dz ∧ dz = −2idx ∧ dy.
2
2
To apply Stokes Theorem, we will have to integrate 2-forms, i.e. expressions like ω(ξ)dξ ∧
dξ over the domain Ω. And we will have to integrate 1-forms, i.e., expressions of the form
ω(z)dz + σ(z)dz. With these notions out of the way, we can turn to the proof
dz = dx + idy
dz = dx − idy
Proof of Theorem 2.1. Fix > 0 and z ∈ D let D (z) = {ξ ∈ D : |z − ξ| ≥ }. Note
that ∂D = T ∪ {ξ : |ξ − z| = }. Suppose that ϕ is a smooth compactly supported function
in D. Then, we have
Z
Z
1
1
ϕ(ξ)
1
dξ ∧ dξ = −
∂ϕ(ξ)
∂
dξ ∧ dξ
2πi D
ξ−z
2πi D
ξ−z
Z
Z
1
ϕ(ξ)
ϕ(ξ)
=
dξ +
dξ
2πi |ξ−z|= ξ − z
T ξ −z
Z
1
ϕ(ξ)
=
dξ.
2πi |ξ−z|= ξ − z
Here we have used the fact that the support of ϕ ⊂ D to conclude that the last integral is 0.
We have also used the following computation
ϕ(ξ)
ϕ(ξ)
ϕ(ξ)
dξ = ∂
dξ ∧ dξ + ∂
dξ ∧ dξ
d
ξ−z
ξ−z
ξ−z
Now note that as → 0 we have that
Z
ϕ(ξ)
1
dξ → ϕ(z).
2πi |ξ−z|= ξ − z
This says that
1
2πi
Z
∂ϕ(ξ)
D
1
dξ ∧ dξ = ϕ(z).
ξ−z
So, if we have a solution to the problem ∂F = G then one solution should be given by
Z
1
1
F (z) =
G(ξ)
dξ ∧ dξ.
2πi D
ξ−z
Note that this solution is continuous in the complex plane and smooth in the disc since it is
the convolution of a continuous function and a bounded function. We now show that we do
indeed have ∂F = G. First, note that
Z
Z
Z
F ∂ϕdz ∧ dz + ∂F ϕdz ∧ dz =
∂(F ϕ)dz ∧ dz
D
D
D
Z
=
F ϕdz = 0.
T
26
2. THE MAXIMAL IDEAL SPACE OF H ∞ (D), ∂-EQUATIONS AND WOLFF’S THEOREM
Here again we have used the support of ϕ and the analogous computations from above. This
then implies
Z
Z
F ∂ϕdz ∧ dz = −
D
∂F ϕdz ∧ dz.
D
Using these computations we see that
Z
Z
∂F ϕdz ∧ dz = − F ∂ϕdz ∧ dz
D
ZD Z
1
1
dξ ∧ dξ ∂ϕdz ∧ dz
= −
G(ξ)
2πi D
ξ−z
D
Z
Z
1
1
= − G
∂ϕ
dz ∧ dz dξ ∧ dξ
2πi D
ξ−z
D
Z
=
Gϕdξ ∧ dξ.
D
Since this is true for all smooth compactly supported ϕ in D we have that
∂F = G
as claimed.
3. Wolff ’s Theorem
The above Theorem demonstrates that it is possible to solve equations for the form
∂F = G. However, we will want to solve the equation with some norm control, in particular
we want to solve the equation and obtain estimates on kF k∞ in terms of information from
G. To accomplish this, we will assume the the function G “generates” Carleson measures
for H 2 (D). We now prove a result of Wolff that gives the desired estimates.
Theorem 3.1 (Wolff, [5]). Suppose that G(z) is bounded and smooth on the disc D.
Further, assume that the measures
1
1
|G|2 log dA(z) and |∂G| log dA(z)
|z|
|z|
are H 2 (D)-Carleson measures. Then there exists a continuous function b(z) on D, smooth
on D such that
∂b = G
and there exists constants C1 and C2 such that
2
2
1
1
kbkL∞ (D) ≤ C1 |G| log dA(z)
+ C2 |∂G| log dA(z)
2
|z|
|z|
H 2 (D)−Carl
H (D)−Carl
The proof of this Theorem is a clever application of Green’s Theorem and using the
conditions on the measures appropriately.
Proof. By Theorem 2.1 above, we clearly have one solution to the problem
∂b = G.
Note now that we can obtain lots of solutions by adding functions that are in the kernel of
the operator ∂ and any function h in the disc algebra A(D) allows us to have that b + h also
3. WOLFF’S THEOREM
27
satisfies that ∂(b + h) = G. The goal is to select a good choice of the function h that allows
us to obtain the estimates we seek.
A duality argument show that
Z
2
2
inf kbk∞ : ∂b = G = sup F k1 k2 dm : k1 ∈ H (D), k2 ∈ H0 (D) .
T
Here we have that the function F is defined as in the Theorem 2.1. Now since we are
supposing that G is bounded and smooth, we have that F is smooth on the D and continuous
on D. A density argument lets us further assume that the functions k1 and k2 are smooth
across the boundary of D (just consider dilates of the functions fr (z) = f (rz) and apply a
normal family argument). Without loss of generality, we can also assume that kk1 k2 ≤ 1
and kk1 k2 ≤ 1.
Now, we apply Green’s Theorem to the function F k1 k2 . First, we compute the Laplacian
of the function F k1 k2 since it will appear in Green’s Theorem. Doing so, we find
∆(F k1 k2 ) = 4∂ ∂F k1 k2 + F ∂(k1 k2 )
= 4 ∂Gk1 k2 + G k10 k2 + k1 k20 .
Here we have used that ∂F = G and that k1 k2 is anti-holomorphic.
Substituting in this information we find:
Z
Z
1
F k1 k2 dm = F (0)k1 (0)k2 (0) + ∆(F k1 k2 ) log dA
|z|
T
D
Z
Z
1
1
= 2 ∂Gk1 k2 log dA(z) + 2 G k10 k2 + k1 k20 log dA(z)
|z|
|z|
D
D
= I + II.
We estimate each of these integrals separately. First, consider the integral corresponding to
I. Making obvious estimates, we have
Z
Z
1
1
∂Gk1 k2 log dA(z) ≤
|k1 | |k2 | |∂G| log dA(z)
|z|
|z|
D
D
Z
21 Z
12
1
1
2
2
≤
|k1 | |∂G| log dA(z)
|k2 | |∂G| log dA(z)
|z|
|z|
D
D
2
1
kk1 kH 2 (D) kk2 kH 2 (D) .
≤ |∂G| log |z| dA(z) 2
H (D)−Carl
Next, turning to term II, one easily sees that it suffices to handle the term k10 k2 since
the other will follow by symmetry. So, consider the following,
Z
Z
Gk10 k2 log 1 dA(z) ≤
Gk10 k2 log 1 dA(z)
|z|
|z|
D
D
Z
21 Z
21
1
1
2
2
0 2
≤
|k1 | log dA(z)
|k2 | |G| log dA(z)
|z|
|z|
D
D
2
1
≤ |G|
log
kk1 kH 2 (D) kk2 kH 2 (D)
|z| H 2 (D)−Carl
28
2. THE MAXIMAL IDEAL SPACE OF H ∞ (D), ∂-EQUATIONS AND WOLFF’S THEOREM
Thus, we see that we have the following estimate for term II
Z
2
1
1
0
0
G k1 k2 + k1 k2 log dA(z) ≤ C |G| log kk1 kH 2 (D) kk2 kH 2 (D) .
|z|
|z| H 2 (D)−Carl
D
Putting the estimates for terms I and II together gives us that
Z
F k1 k2 dm ≤ C1 |G|2 log 1 kk1 kH 2 (D) kk2 kH 2 (D)
|z| H 2 (D)−Carl
T
2
1
+C2 |∂G|
log
dA(z)
kk1 kH 2 (D) kk2 kH 2 (D) ,
2
|z|
H (D)−Carl
which then proves the claim.
CHAPTER 3
The Corona Theorem
In the last lecture we studied the problem
∂u = v
and gave a solution operator of the form
1
u(z) =
2πi
Z
D
v(w)
dA(w).
w−z
We then wanted to study the bounded of the solution u in terms of boundedness properties
of the right hand side v. We finished the last lecture by showing that if the right hand
side has some Carleson measure estimates, then we always have a bounded solution to this
problem.
In this lecture we will return to studying the ∂-problem when the right hand side is
a Carleson measure. We will show that in this case, it is possible to give a constructive
solution to this problem. This is a very clever construction due to Peter Jones. Finally, in
this lecture we will use the results from the last lecture to complete the proof of Carleson’s
Corona Theorem.
1. Wolff ’s Proof of the Corona Theorem
Recall that in the last lecture we finished by proving the following Theorem of Wolff.
Theorem 1.1 (Wolff, [5]). Suppose that G(z) is bounded and smooth on the disc D.
Further, assume that the measures
1
1
|G|2 log dA(z) and |∂G| log dA(z)
|z|
|z|
are H 2 (D)-Carleson measures. Then there exists a continuous function b(z) on D, smooth
on D such that
∂b = G
and there exists constants C1 and C2 such that
2
2
1
1
kbkL∞ (D) ≤ C1 + C2 |G| log |z| dA(z) 2
|∂G| log |z| dA(z) 2
H (D)−Carl
H (D)−Carl
With this tool, we are now is a place to prove the following important Theorem due to
Carleson. The proof we give, will exploit the Theorem above due to Wolff.
Theorem 1.2 (Carleson, [4]). Suppose that f1 , . . . , fn ∈ H ∞ (D) and there exists a δ > 0
such that
1 ≥ max {|fj (z)|} ≥ δ > 0.
1≤j≤n
29
30
3. THE CORONA THEOREM
Then there exists g1 , . . . , gn ∈ H ∞ (D) such that
1 = f1 (z)g1 (z) + · · · + fn (z)gn (z) ∀z ∈ D
and
kgj kH ∞ (D) ≤ C(δ, n)
∀j = 1, . . . , n.
An obvious remark is that the condition on the functions f is clearly necessary. We can’t
have all the functions simultaneously vanish if they can generate the function 1.
Before we prove this Theorem, we first consider the case of two functions so that we can
see the connections between this problem and the ∂-problem we initially studied. Suppose we
have two functions f1 , f2 ∈ H ∞ (D) such that max(|f1 (z)| , |f2 (z)|) ≥ δ. Define the following
functions
f1 (z)
f2 (z)
ϕ1 (z) =
ϕ2 (z) =
.
2
2
|f1 (z)| + |f2 (z)|
|f1 (z)|2 + |f2 (z)|2
The hypotheses on f1 and f2 imply that the functions ϕ1 and ϕ2 are in fact bounded and
smooth on D. Note that we have that
1 = f1 (z)ϕ1 (z) + f2 (z)ϕ2 (z) ∀z ∈ D
but the functions ϕ1 and ϕ2 are in general not analytic. Now, observe for any function r we
have that the functions
g1 = ϕ1 + rf2 g2 = ϕ2 − rf1
also solve the problem
f1 g1 + f2 g2 = 1.
Our goal is to select a good choice of function r so that the resulting choice will make g1 and
g2 be analytic and bounded. Now, we have that g1 is analytic if and only if
0 = ∂g1 = ∂ϕ1 + f2 ∂r.
Similarly, g2 is analytic if and only if
0 = ∂g2 = ∂ϕ2 − f1 ∂r.
Using these two equations and the condition that f1 ϕ1 + f2 ϕ2 = 1 gives that the function r
must satisfy the equation
∂r = ϕ1 ∂ϕ2 − ϕ2 ∂ϕ1 .
Thus, we need to solve for the choice of r that will give a bounded solution. Here we will use
the fact that when we express what appears on the right hand side of the above equation,
we have certain Carleson measures appearing.
Proof. With out lose of generality, we may assume that the functions are analytic in a
neighborhood of the closed disc D. Define the functions
fj (z)
2
j=1 |fj (z)|
ϕj (z) = Pn
∀z ∈ D.
Then by the hypotheses on the functions fj we clearly have that |ϕj (z)| ≤ C(n, δ). These
functions are in general not analytic, and so we must correct them to be so. We thus, will
set
n
X
gj (z) = ϕj (z) +
aj,k (z)fk (z)
k=1
1. WOLFF’S PROOF OF THE CORONA THEOREM
31
where the functions aj,k (z) are to be determined. However, we will require that aj,k (z) =
−ak,j (z). Note that this alternating condition implies that
n
X
fj (z)gj (z) =
j=1
n
X
fj (z)ϕj (z) +
j=1
n X
n
X
aj,k (z)fj (z)fk (z) = 1.
j=1 k=1
To have the alternating characterisitic of aj,k we set aj,k = bj,k (z) − bk,j (z) for some yet to
be determined functions. We will chose the functions bj,k to be solutions to the follow ∂
problem:
∂bj,k = ϕj ∂ϕk := Gj,k .
Using this, we see that
∂gj = ∂ϕj +
= ∂ϕj +
= ∂ϕj +
n
X
k=1
n
X
k=1
n
X
fk ∂aj,k
fk ∂bj,k − ∂bk,j
fk ϕj ∂ϕk − ϕk ∂ϕj
k=1
= ∂ϕj + ϕj ∂
n
X
!
fk ϕk
− ∂ϕj
k=1
n
X
fk ϕk
k=1
= ∂ϕj + ϕj ∂1 − ∂ϕj 1 = 0.
So the functions gj are analytic. Suppose that we prove the functions bj,k are bounded by
C(n, δ), then similarly we have that aj,k are bounded, and so
|gj (z)| ≤ C(n, δ).
We are thus left with proving that the functions bj,k are bounded. With this in mind, and
having the result of Wolff at our disposal, we must show that the measures
|Gj,k |2 log
1
1
dA(z) and |∂Gj,k | log dA(z)
|z|
|z|
are H 2 (D)-Carleson measures. We claim that each of these measures can be dominated (up
to a constant C(n, δ)) by the following measure
n
X
0 2
fj (z) log 1 dA(z).
|z|
j=1
To see that this is a H 2 (D)-Carleson measure, it suffices to show that for f ∈ H ∞ (D) that
1
we have |f 0 (z)|2 log |z|
dA(z) is a H 2 (D)-Carleson measure, which follows easily from the
alternate norm on H 2 (D) we previous defined (see the exercise below).
Consider the expression |Gj,k |2 . Note that by the hypotheses on fj we have that |ϕj | ≤
C(n, δ) and so,
2
|Gj,k |2 ≤ C(n, δ) ∂ϕk .
32
3. THE CORONA THEOREM
Now, if we compute we see that
∂ϕk
P
fk nj=1 fj fj0
fk0
= Pn
2
2 − P
n
2
j=1 |fj |
|f |
j=1
j
Pn
=
fj (fj fk0 − fk fj0 )
.
P
2
n
2
j=1 |fj |
j=1
Using this, we see that
∂ϕk 2 ≤ C
Pn
0 2
n
X
0 2
j=1 fj
fj ≤
C(n,
δ)
2
2
j=1
j=1 |fj |
2
j=1 |fj |
P
n
Pn
which proves that
n
X
0 2
1
fj log 1 dA(z).
≤ C(n, δ)
|Gj,k | log
|z|
|z|
j=1
2
1
We now turn to showing that |∂Gj,k | log |z|
dA(z) is dominated appropriately. First, observe that
∂Gj,k = ∂ϕj ∂ϕk + ϕj ∂∂ϕk
By the computations above, we have that
Pn
∂ϕk =
fj (fj fk0 − fk fj0 )
.
2
P
2
n
|f
|
j=1 j
j=1
Direct computation also gives, that
P
fj nl=1 fl0 fl
∂ϕj = − P
2 .
n
2
|f
|
j=1 j
Finally, we have that
Pn
Pn 0 Pn
0
0
fj0 (fj fk0 − fk fj0 )
l=1 fl fl
l=1 fl fk − fk fl
−2
.
∂∂ϕk =
2
3
P
P
n
2
n
2
|f
|
|f
|
j=1 j
j=1 j
Now consider the term ∂ϕj ∂ϕk . It is obvious that we can dominate this expression by
j=1
n
X X
2
0
0
C(n, δ)
fj |fk | ≤ C(n, δ)
|fk0 | .
j,k
k=1
Similarly, we have that ϕj ∂∂ϕk can be dominated by an identical expression. Altogether
this then gives that
|∂Gj,k | log
n
X
1
1
2
dA(z) ≤ C(n, δ)
|fk0 | log dA(z).
|z|
|z|
k=1
2. PETER JONES’ CONSTRUCTIVE SOLUTION TO ∂
33
1
Exercise 1.3. Show that for f ∈ H ∞ (D) that we have |f 0 (z)|2 log |z|
dA(z) is a H 2 (D)Carleson measure. Hint: Use the alternate norm for H 2 (D) and think about the product rule
for derivatives.
Exercise 1.4. Suppose that f1 , . . . , fn , g ∈ H ∞ (D) such that
|g(z)| ≤
n
X
|fj (z)| .
j=1
∞
Show that there exists g1 , . . . , gn ∈ H (D) such that
3
g =
n
X
f j gj .
j=1
Hint: Mimic Wolff ’s proof of the Corona Theorem but start with ψj = gϕj , with ϕj as defined
above.
We state one more theorem, closely related to Carleson’s Corona Theorem.
Theorem 1.5. Suppose that f1 , . . . , fn ∈ H ∞ (D) and there exists a δ > 0 such that
1 ≥ max {|fj (z)|} ≥ δ > 0.
1≤j≤n
Let h ∈ H 2 (D). Then there exists g1 , . . . , gn ∈ H 2 (D) such that
h(z) = f1 (z)g1 (z) + · · · + fn (z)gn (z)
∀z ∈ D
and
kgj kH 2 (D) ≤ C(δ, n) khkH 2 (D)
∀j = 1, . . . , n.
It is clear that Carleson’s Corona Theorem implies this result. What isn’t immediately
obvious though, is that knowing this result, one can deduce Carleson’s Corona Theorem.
This is connected to deeper facts about operator theory and will play a role in later lectures.
Exercise 1.6. Give a proof of the above Theorem without appealing to Carleson’s Theorem directly, but instead appealing to the proof of the Theorem.
2. Peter Jones’ Constructive Solution to ∂
We now give another method to solve the ∂ problem that is more constructive. The proof
this time is based on the extremely clever solution operator constructed by Peter Jones, [6].
The starting point of the construction is the following observation. We have seen from
above, that to solve the equation ∂F = G we can set
Z
1
1
F =
G(ξ)
dξ ∧ dξ.
2πi D
z−ξ
However, we can also use another kernel to accomplish this. Choose a function K(z, ξ) that
is analytic in z, K(z, z) = 1 and that is smooth, then we will also have that
Z
1
1
F =
K(z, ξ)G(ξ)
dξ ∧ dξ.
2πi D
z−ξ
solves ∂F = G.
34
3. THE CORONA THEOREM
Exercise 2.1. Show that this is true. Hint: K(z, ξ) = 1 + (z − ξ)K̃(z, ξ) and argue as
before.
This then gives us lots of freedom by which kernel we choose to solve the problem. The
construction by Jones makes a very clever and judicious choice of the kernel so that one can
obtain the solutions directly.
Theorem 2.2 (Jones, [6]). Let µ be a complex H 2 (D) Carleson measure on D. Then
with S (µ) (z) given by
Z
(2.1)
S (µ) (z) =
K (σ, z, ζ) dµ (ζ)
D
where σ =
|µ|
kµkCM (H 2 )
and
2i
1 − |ζ|2
exp
K (σ, z, ζ) ≡
π (z − ζ) 1 − ζz
1 + ωz 1 + ωζ
−
+
dσ (ω) ,
1 − ωz 1 − ωζ
|ω|≥|ζ|
Z
we have that:
(1) S (µ) ∈ L1loc (D).
(2) ∂S(µ)= µ in the sense
of distributions.
R |µ|
(3) D K kµk
, x, ζ d |µ| (ζ) . kµkCM (H 2 ) for all x ∈ T = ∂D,
2
CM (H )
so kS (µ)kL∞ (T) . kµkCM (H 2 ) .
Note that the kernel K(σ, z, ξ) is analytic in z,
K(σ, z, ξ) =
2i 1
K̃(σ, z, ξ)
π z−ξ
and K̃(σ, z, ξ) is smooth with
1 − |z|2
K̃(σ, z, z) =
exp
1 − |z|2
Z
1 + ωz 1 + ωz
−
+
dσ(ω) = 1.
1 − ωz 1 − ωz
|ω|≥|z|
So (2) follows the argument above. While (1) follows from (3). We now turn to proving that
(3) holds (though in the course of the proof we will address (1) and (3) at the same time).
Proof. Observe that if we prove
x ∈ T = ∂D, then we have
Z |S(µ)(z)| ≤
K
D
R K kµk |µ|
D
CM (H 2 )
, x, ζ d |µ| (ζ) . kµkCM (H 2 ) for all
!
, z, ζ d |µ| (ζ) . kµkCM (H 2 ) .
kµkCM (H 2 )
|µ|
2. PETER JONES’ CONSTRUCTIVE SOLUTION TO ∂
We turn now to the proof of this remaining fact. Note that for the measure σ =
we have
Z
Z
1 + wζ
1 + wζ
Re
dσ(w)
=
dσ(w)
Re
1 − wζ
1 − wζ
|w|≥|ζ|
|w|≥|ζ|
Z
1 − |ζ|2
≤ 2
2 dσ(w)
D |1 − wζ|
2
= 2,
≤ 2 k̃ζ 35
µ
kµkCM (H 2 )
H 2 (D)
1
2 2
)
where k̃ζ (z) = (1−|ζ|
is the normalized reproducing kernel and clearly has H 2 (D) norm
1−ζz
equal to 1. Next, observe that it will suffice to control the boundary values of the function
S(µ), and so we can take z ∈ T. We then see, using the estimate from above and that z ∈ T
that we have
Z
2 1 − |ξ|2 1 + ωz 1 + ωξ
|K(σ, z, ξ)| =
−
+
dσ(ω) 2 exp
π 1 − ξz 1 − ωz 1 − ωξ
|ω|≥|ξ|
Z
Z
1 + ωz
2 1 − |ξ|2
1 + ωz
dσ(ω) exp Re
dσ(ω)
=
−
exp Re
π 1 − ξz 2
1 − ωz
|ω|≥|ξ| 1 − ωz
|ω|≥|ξ|
)
( Z
2 2 1 − |ξ|2
1 − |ω|2
≤
e exp −
2 dσ(ω)
π 1 − ξz 2
|ω|≥|ξ| |1 − ωz|
We can then use this estimate on the kernel K(σ, z, ζ) to give
Z
|S (µ) (z)| = K (σ, z, ζ) dµ (ζ)
D
Z
≤ kµkCM (H 2 ) |K (σ, z, ζ)| dσ(ζ)
D
)
( Z
Z
2
2
2
1
−
|ζ|
1
−
|ω|
≤ e2 kµkCM (H 2 )
exp −
2 dσ(ω) dσ(ζ).
π D 1 − ζz 2
|ω|≥|ζ| |1 − ωz|
It remains to show that
( Z
)
Z
1 − |ζ|2
1 − |ω|2
2 exp −
2 dσ(ω) dσ(ζ) ≤ 1.
|ω|≥|ζ| |1 − ωz|
D 1 − ζz P
First, suppose that we have dσ = N
j=1 aj δζj with |ζj | ≤ |ζj+1 |. Further, set βj = aj
1−|ζj |2
2
|1−ζj z|
and
tj =
N
X
βk , and so βj = tj − tj−1 .
k=j
If we evaluate the above integral for this measure and use the resulting notation, we see that
the integral becomes
Z ∞
N
X
−tj
(tj − tj−1 )e ≤
e−t dt = 1.
j=1
0
36
3. THE CORONA THEOREM
A standard measure theory argument then finishes the proof that
( Z
)
Z
1 − |ζ|2
1 − |ω|2
2 exp −
2 dσ(ω) dσ(ζ) ≤ 1.
D 1 − ζz |ω|≥|ζ| |1 − ωz|
This then completes the proof of the Theorem.
The original proof that Jones gave was for the upper half-plane H. The idea is identical,
but requires certain modifications.
Exercise 2.3. State and give the proof of Jones’ Theorem in the case of the upper half
plane H.
CHAPTER 4
Corona Theorems and Complete Nevanlinna-Pick Kernels
The Hardy space has a special property that allows one to solve the Pick interpolation
problem. Namely, given a finite collection of points zj ∈ D and a finite collection of targets
wj it is possible to find a function f ∈ H ∞ (D) such that f (zj ) = wj with kf k∞ if and only if
a certain matrix formed from the points zj and wj is positive semi-definite. It turns out that
the Hardy space has a slightly stronger property of being able to solve the Nevanlinna-Pick
interpolation problem, but with matrix targets or arbitrary size. When a reproducing kernel
Hilbert space has this property, then we will say that it has the complete Nevanlinna-Pick
property.
It turns out that for certain function spaces that have a complete Nevanlinna-Pick kernel,
it is possible to solve the Corona problem for the multiplier algebra by solving a relatively
easier Corona question for the space of analytic functions itself.
Let X be a Hilbert space of holomorphic functions in an open set Ω in Cn that is a
reproducing kernel Hilbert space with a complete irreducible Nevanlinna-Pick kernel (see [1]
for the definition). The following Toeplitz corona theorem is due to Ball, Trent and Vinnikov
[3] (see also Ambrozie and Timotin [2] and Theorem 8.57 in [1]).
N
N
For f = (fα )N
α=1 ∈ ⊕ X and h ∈ X, define Mf h = (fα h)α=1 and
kf kM ult(X,⊕N X) = kMf kX→⊕N X = sup kMf hk⊕N X .
khkX ≤1
Note that max1≤α≤N kMfα kMX ≤ kf kM ult(X,⊕N X) ≤
qP
N
α=1
kMfα k2MX .
Theorem 0.4 (Toeplitz Corona Theorem). Let X be a Hilbert function space in an open
set Ω in Cn with an irreducible complete Nevanlinna-Pick kernel. Let δ > 0 and N ∈ N.
Then g1 , . . . , gN ∈ MX satisfy the following “baby corona property”; for every h ∈ X, there
are f1 , . . . , fN ∈ X such that
(0.2)
1
khk2X ,
δ
g1 (z) f1 (z) + · · · + gN (z) fN (z) = h (z) ,
z ∈ Ω,
kf1 k2X + · · · + kfN k2X ≤
if and only if g1 , . . . , gN ∈ MX satisfy the following “multiplier corona property”; there are
ϕ1 , . . . , ϕN ∈ MX such that
(0.3)
kϕkM ult(X,⊕N X) ≤ 1,
√
g1 (z) ϕ1 (z) + · · · + gN (z) ϕN (z) =
δ,
z ∈ Ω.
The baby corona theorem is said to hold for X if whenever g1 , · · · , gN ∈ MX satisfy
(0.4)
|g1 (z)|2 + · · · + |gN (z)|2 ≥ c > 0,
then g1 , . . . , gN satisfy the baby corona property (0.2).
37
z ∈ Ω,
38
4. CORONA THEOREMS AND COMPLETE NEVANLINNA-PICK KERNELS
More succinctly, (0.2) is equivalent to the operator lower bound
Mg M∗g − δIX ≥ 0,
N
P
∗
∗
where g ≡ (g1 , ...gN ), Mg : ⊕N X → X by Mg f = N
α=1 gα fα , and Mg h = Mgα f α=1 . We
note that (0.4) with c = δ is necessary for (0.5) as can be seen by testing on reproducing
kernels kz .
(0.5)
Remark 0.5. A standard abstract argument applies to show that the absence of a corona
for the multiplier algebra MX , i.e. the density of the linear span of point evaluations in the
maximal ideal space of MX , is equivalent to the following assertion: for each finite set
N
{gj }N
j=1 ⊂ MX such that (0.4) holds for some c > 0, there are {ϕj }j=1 ⊂ MX and δ > 0 such
that condition (0.3) holds. See for example Lemma 9.2.6 in [8] or the proof of Criterion 3.5
on page 39 of [9].
Here we recall the proof of the Toeplitz Corona Theorem 0.4 for holomorphic Hilbert
function spaces with a complete Nevanlinna-Pick kernel. First we note the equivalence of
(0.2) and (0.5). To see this note that (0.5) is equivalent to
(0.6)
δ hh, hiX ≤ h, Mϕ M∗ϕ h X = M∗ϕ h, M∗ϕ h ⊕N X .
From functional analysis, we obtain that the bounded map Mϕ : ⊕N X → X is onto. If
dϕ : N ⊥ → X is invertible. Now (0.6) implies that M
dϕ ∗ : X → N ⊥
N = ker Mϕ , then M
−1 ∗ −1 dϕ
dϕ
≤ √1 . By duality we then have M
≤ √1 . Thus
is invertible and that M
δ
δ
given h ∈ X, there is f ∈ N ⊥ satisfying Mϕ f = h and
−1 2
1
2
2
d
kf k⊕N X = Mϕ
h
N ≤ δ khkX ,
⊕ X
which is (0.2). Conversely, using (0.2)
∗ Mϕ h N
=
sup
(0.7)
⊕ X
we compute that
g, M∗ϕ h ⊕N X =
kgk⊕N X ≤1
sup
hMϕ g, hi kgk⊕N X ≤1
X
√
khk2X
f
,h =
≥ δ khkX ,
≥ Mϕ
kf k⊕N X
kf k⊕N X
X
which is (0.6), and hence (0.5).
Next we note that (0.4) with c = δ is necessary for (0.5) as can be seen by testing (0.6)
on reproducing kernels kz :
δ hkz , kz i ≤ M∗ϕ kz , M∗ϕ kz ⊕N X = |ϕ (z)|2 hkz , kz i
N
since M∗ϕ kz = ϕα (z)kz
.
α=1
1. Calculus of kernel functions and proof of the Toeplitz Corona Theorem
A crucial theme for the proof of the Toeplitz Corona Theorem is that operator bounds
for Hilbert function spaces, such as Mϕ M∗ϕ − δIX ≥ 0 for X in (0.5), can be recast in terms
of kernel functions, namely
(1.1)
{hϕ (ζ) , ϕ (λ)iCN − δ} k (ζ, λ) 0.
1. CALCULUS OF KERNEL FUNCTIONS AND PROOF OF THE TOEPLITZ CORONA THEOREM 39
Indeed, if we let h =
δ
J
X
PJ
i=1 ξi kxi
in (0.6) we obtain
J
X
ξi ξj k (xj , xi ) = δ
i,j=1
ξi ξj kxi , kxj
i,j=1
* J
N
X
X
≤
α=1
J
X
=
ξi ϕα (xi )kxi ,
J
X
i=1
ξi ξj
+
ξj ϕα (xj )kxj
j=1
( N
X
X
)
ϕα (xi )ϕα (xj ) k (xj , xi ) ,
α=1
i,j=1
which is (1.1). A similar calculation shows that the operator upper bound IX − Mf M∗f ≥ 0,
which is equivalent to kMf k⊕N X→X ≤ 1, is recast in terms of kernel functions as
{1 − hf (ζ) , f (λ)iCN } k (ζ, λ) 0.
(1.2)
In order to recast the multiplier bound in the first line of (0.3) in terms of kernel functions,
we must consider N × N matrix-valued kernel functions. Recall that Mf : X → ⊕N X by
PN
N
Mf h = (fα h)N
α=1 fα gα ). Then for
α=1 (compare with Mf : ⊕ X → X by Mf g =
g ∈ ⊕N MX ,
+
* N
N
X
X
∗
Mf α g α
,
hMf h, gi⊕N X =
hfα h, gα iX = h,
α=1
M∗f g
and so
hence to
(1.3)
=
PN
∗
α=1 Mfα gα .
0 ≤
N
X
=
X
2
Thus the first line in (0.3) is equivalent to M∗f ⊕N X→X ≤ 1,
2
N
X
∗
−
Mfα gα = hg, gi⊕N X − M∗f g, M∗f g X
α=1
X
− Mf M∗f g, g ⊕N X ,
kgα k2X
α=1
α=1
I⊕N X
which is the operator bound
I⊕N X − Mf M∗f ≥ 0.
(1.4)
To obtain an equivalent kernel estimate, let
J
X
g = (gα )N
α=1 =
!N
ξiα kxαi
i=1
,
α=1
so that
M∗f g =
N
X
α=1
M∗fα gα =
N X
J
X
ξiα fα (xαi )kxαi .
α=1 i=1
If we substitute this in (1.3) we obtain
N
J
X
X
ξiα ξjβ fα (xαi )fβ xβj k xβj , xαi
= M∗f g, M∗f g X
α,β=1 i,j=1
≤ hg, gi⊕N X =
N
X
α,β=1
δβα
J
X
i,j=1
ξiα ξjβ k
xβj , xαi
,
40
4. CORONA THEOREMS AND COMPLETE NEVANLINNA-PICK KERNELS
or
N
J
X
X
ξiα ξjβ
hn
δβα
−
fα (xαi )fβ
xβj
o i
β
α
k xj , xi
≥ 0.
α,β=1 i,j=1
If we view f (ζ) ∈ B C, CN we can rewrite this last expression as
{ICN − f (ζ) f (λ)∗ } k (ζ, λ) 0,
(1.5)
which is the required matrix-valued kernel equivalence of the multiplier bound in (0.3).
Now √
we turn to the proof of Theorem 0.4. To see that (0.3) implies (0.2) just multiply
fh
= h where
Mϕ f = δ by √hδ to get Mϕ √
δ
f h 2
1
√ kf1 hk2X + ... + kfN hk2X
=
δ N
δ
⊕ X
1
1
≤
kf1 k2MX + ... + kfN k2MX khk2X ≤ khk2X .
δ
δ
The computation (0.7) above then shows that (0.5) holds for the same δ > 0.
However, we can give another short proof, but using the language of positive semidefinite kernel functions to characterize operator boundedness. This will afford us our first
opportunity to use the ”calculus” of positive semidefinite forms. If (0.3) holds then
√
∗
ϕ (ζ) f (ζ) = Mϕ f = δ,
and (1.5) holds:
{ICN − f (ζ) f (λ)∗ } k (ζ, λ) 0.
These two relations imply the positivity of the kernel function,
{hϕ (ζ) , ϕ (λ)iCN − δ} k (ζ, λ)
n
√ √ o
=
hϕ (ζ) , ϕ (λ)iCN − δ δ k (ζ, λ)
n
o
∗
=
hϕ (ζ) , ϕ (λ)iCN − ϕ (ζ) f (ζ) f (λ)∗ ϕ (λ) k (ζ, λ)
∗
= ϕ (ζ) [{ICN − f (ζ) f (λ)∗ } k (ζ, λ)] ϕ (λ) 0.
By (1.1) this is equivalent to (0.6), and hence to (0.5).
Conversely, normalize k at a fixed point λ0 ∈ Ω so that kλ0 ≡ 1. Since k is an irreducible
complete Nevanlinna-Pick kernel, we can find a Hilbert space K and a map b : Ω → K with
b (λ0 ) = 0 and such that
(1.6)
k (ζ, λ) =
1
.
1 − hb (ζ) , b (λ)iK
This theorem has a long history and is not easy to prove (Theorem 7.31 in [1]). In fact, one
2
can take K to be the Drury-Arveson Hardy space Hm
for some cardinal number m (Theorem
8.2 in [1]), but we will not need this. From (0.5) we now obtain (1.1):
K (ζ, λ) ≡ {hϕ (ζ) , ϕ (λ)iCN − δ} k (ζ, λ) 0.
1. CALCULUS OF KERNEL FUNCTIONS AND PROOF OF THE TOEPLITZ CORONA THEOREM 41
By a kernel-valued version of the Lax-Milgram Theorem, we can factor the left hand side
K (ζ, λ) as hG (ζ) , G (λ)iH where G : Ω → H for some auxiliary space H. Indeed, define
F : Ω → XK by F (ζ) = Kζ so that
K (ζ, λ) = hKλ , Kζ iXK = hF (λ) , F (ζ)iXK .
Now fix an orthonormal basis {eα }α for XK and define a conjugate linear operator Γ by
!
X
X
Γ
cα e α =
cα e α .
α
α
Then G = Γ ◦ F satisfies
K (ζ, λ) = hF (λ) , F (ζ)iXK = hΓ ◦ F (ζ) , Γ ◦ F (λ)iXK = hG (ζ) , G (λ)iXK ,
with H = XK as required. Hence
hϕ (ζ) , ϕ (λ)iCN − δ = [1 − hb (ζ) , b (λ)iK ] hG (ζ) , G (λ)iH ,
or equivalently,
(1.7)
hϕ (ζ) , ϕ (λ)iCN + hb (ζ) , b (λ)iK hG (ζ) , G (λ)iH = δ + hG (ζ) , G (λ)iH .
Now we rewrite (1.7) in terms of inner products of direct sums of Hilbert spaces,
hϕ (ζ) , ϕ (λ)iCN + hb (ζ) ⊗ G (ζ) , b (λ) ⊗ G (λ)iK⊗H
D√ √ E
δ, δ + hG (ζ) , G (λ)iH ,
=
C
so that it can be interpreted as saying that the map from N1 to N2 that sends the element
(1.8)
(ϕ (λ) u, b (λ) ⊗ G (λ) u) ∈ CN ⊕ (K ⊗ H)
with u ∈ C to the element
(1.9)
√
δu, G (λ) u ∈ C ⊕ H
is an isometry! Here the spaces N1 and N2 are given by
ϕ (λ)
N1 = Span
u : u ∈ C, λ ∈ Ω ⊂ CN ⊕ (K ⊗ H) ,
b (λ) ⊗ G (λ)
√ δ
N2 = Span
u : u ∈ C, λ ∈ Ω ⊂ C ⊕ H.
G (λ)
Thus using (0.5) we have obtained (1.7) that defines a linear isometry V 0 from the linear
span N1 of the elements ϕ (λ) u ⊕ (b (λ) ⊗ G (λ)) u in the direct sum CN ⊕ (K ⊗ H) onto a
subspace N2 of the direct sum C ⊕ H: the element in (1.8) goes to the element in (1.9). Now
extend this isometry V 0 to an isometry V from all of CN ⊕ (K ⊗ H) onto C ⊕ H, where we
add an infinite-dimensional summand to H if necessary. Indeed, V 0 extends by continuity to
an isometry from N1 onto N2 , and provided the orthogonal complements of N1 and N2 have
the same dimension, we can then trivially extend the isometry from all of CN ⊕ (K ⊗ H)
onto C ⊕ H. But the dimensions of the complements can be made equal by adding an
infinite-dimensional summand to H.
Decompose the extended isometry V as a block matrix
A B
C
CN
:
.
(1.10)
V =
→
C D
H
K⊗H
42
4. CORONA THEOREMS AND COMPLETE NEVANLINNA-PICK KERNELS
Since V is an onto isometry we obtain the formulas,
∗
∗
A B
A A + C ∗ C A∗ B + C ∗ D
A C∗
(1.11)
=
B ∗ D∗
C D
B ∗ A + D∗ C B ∗ B + D∗ D
ICN
0
= V ∗ V = ICN ⊕(K⊗H) =
.
0 IK⊗H
Then (1.10) on the subspace N1 becomes
√
Aϕ (λ) + B [b (λ) ⊗ G (λ)] =
δ,
Cϕ (λ) + D [b (λ) ⊗ G (λ)] = G (λ) .
Now define f : Ω → B C, CN (which is of course isomorphic to CN ) by
(1.12)
(1.13)
n
−1 o
∗
f (λ) = A + B b (λ) ⊗ I − DEb(λ)
C ,
where Eb is the map Eb : H → K ⊗ H given by
Eb v = b ⊗ v,
(1.14)
v ∈ H.
∗
Note that this formula for f (λ) is obtained by solving the second line in (1.12) for G (λ) =
−1
I − DEb(λ)
Cϕ (λ), and then substituting this in the first line and dropping ϕ (λ). Ob∗
serve that Eb (c ⊗ w) = hc, biK w, so that
Eb∗ Ec = hc, biK IH .
(1.15)
From this we conclude that I − DEb(λ) is invertible. Indeed, (1.6) shows that hb, biK < 1 and
(1.15) then implies that Eb(λ) is a strict contraction. From the equation B ∗ B + D∗ D = IK⊗H
∗
in (1.11) we see that D is a contraction, which altogether implies DEb(λ) < 1. Thus f (λ)
satisfies
h
i
−1
∗
(1.16)
f (λ) ϕ (λ) = Aϕ (λ) + B b (λ) ⊗ I − DEb(λ)
Cϕ (λ)
= Aϕ (λ) + B [b (λ) ⊗ G (λ)]
√
=
δ,
which is the second line in (0.3).
To see that the first line in (0.3) holds, we must show that f (λ) is a contractive multiplier,
i.e. that (1.5) holds:
{ICN − f (ζ) f (λ)∗ } k (ζ, λ) 0.
(1.17)
For this we use (1.11). We compute with
∗
−1
= A + BEb(λ) I − DEb(λ)
C,
−1
∗
∗
D∗
Eb(ζ)
B∗,
f (ζ) = A∗ + C ∗ I − Eb(ζ)
f (λ)
1. CALCULUS OF KERNEL FUNCTIONS AND PROOF OF THE TOEPLITZ CORONA THEOREM 43
that
i
h
−1 ∗
∗
Eb(ζ) B ∗
D∗
ICN − f (ζ) f (λ)∗ = I − A∗ + C ∗ I − Eb(ζ)
h
−1 i
× A + BEb(λ) I − DEb(λ)
C
−1
= I − A∗ A − A∗ BEb(λ) I − DEb(λ)
C
−1
∗
∗
Eb(ζ)
B∗A
D∗
−C ∗ I − Eb(ζ)
−1
−1 ∗
∗
C,
Eb(ζ) B ∗ BEb(λ) I − DEb(λ)
D∗
−C ∗ I − Eb(ζ)
and then using (1.11) we obtain
(1.18)
ICN − f (ζ) f (λ)∗
−1
= C ∗ C + C ∗ DEb(λ) I − DEb(λ)
C
−1
∗
∗
Eb(ζ)
D∗ C
D∗
+C ∗ I − Eb(ζ)
−1 ∗
−1
∗
+C ∗ I − Eb(ζ)
Eb(ζ) (D∗ D − I) Eb(λ) I − DEb(λ)
C
D∗
−1
∗
= C ∗ I − Eb(ζ)
D∗
∗
∗
× I − Eb(ζ)
D∗ I − DEb(λ) + I − Eb(ζ)
D∗ DEb(λ)
∗
∗
∗
+Eb(ζ)
D∗ I − DEb(λ) − Eb(ζ)
Eb(λ) + Eb(ζ)
D∗ DEb(λ)
−1
C
× I − DEb(λ)
−1
−1
∗
∗
C
= C ∗ I − Eb(ζ)
D∗
I − Eb(ζ)
Eb(λ) I − DEb(λ)
−1
−1
∗
D∗
I − DEb(λ)
= 1 − hb (ζ) , b (λ)iK C ∗ I − Eb(ζ)
C,
where the last line follows from (1.15). Thus using (1.6) the left side of (1.17), which is an
N × N matix-valued kernel function, has its complex conjugate equal to
−1
−1
∗
C ∗ I − Eb(ζ)
D∗
I − DEb(λ)
C
D
−1
−1 E
= I − DEb(λ)
C, I − DEb(ζ)
C ,
H
which is an N × N matix-valued Grammian, hence a positive kernel as required.
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