Fatou’s theorem, Bergman spaces, and Hardy spaces on the circle
1 Introduction
In this note I am writing out proofs of some facts about Fourier series, Bergman spaces, and Hardy spaces. §§1–3 follow the presentation in Stein and Shakarchi’s Real Analysis and Fourier Analysis. The questions in Halmos’s Hilbert Space Problem Book that deal with Hardy spaces are: §§24–35, 67, 116–117, 124–125, 127, 193–199, and I present solutions to some of these in §§4–5, on Bergman spaces, and §6, on Hardy spaces.
2 Poisson kernel
Let
If
Define
One checks that
For
hence by the dominated convergence theorem we have
and so
3 Harmonic functions
For
In polar coordinates, the Laplacian is
Then
Hence
4 Fatou’s theorem
Let
By the Cauchy integral formula, for
Hence, for
On the other hand, for
and because
Let
For
On the other hand,
It follows that
Define
this defines an element of
As
which means here that for almost all
Thus, for almost all
In words, we have proved that if
5 Bergman spaces
This section somewhat follows Problem 24 of Halmos.
Let
If
and let
It is apparent that
and so
Lemma 1.
If
Proof.
Put
We have
which tends to
For
Therefore
That is, for each
(1) |
Because
Thus, taking the limit as
∎
If
and for
This is the radius of the largest open disc centered at
Lemma 2.
If
Proof.
Now we prove that
Theorem 3.
If
Proof.
Suppose that
From this and the fact that
That is,
On the other hand,
∎
6 Inner products
In this section we follow Problem 25 of Halmos.
In this section we restrict our attention to the Bergman space
As
Theorem 4.
If
Then
Proof.
If
while
Therefore
For
One checks that
Therefore
As
Steven G. Krantz, Geometric Function Theory: Explorations in Complex Analysis,
p. 9, §1.2, writes about the Bergman space
7 Hardy spaces
In a Hilbert space
Let
We define the Hardy space
The following is Problem 26 of Halmos. Note
Theorem 5.
If
Proof.
If
Thus
If
Therefore if
(2) |
For
the first equality is because
If
and
and
(3) |
The following is Problem 35 of Halmos. In words, it states that a real valued
Theorem 6.
If
Proof.
Define
As
This is finite, hence
For any
Put
Thus
If
Thus
∎