Let $f \in H^{2}(\mathbb D)$. I'd like to show that $ f\in L^{2}(\mathbb T)$, with: $\Vert f \Vert_{L^{2}(\mathbb T)} \le \Vert f \Vert_{H^{2}(\mathbb D)}$, maybe up to a constant.
$f \in H^{2}(\mathbb D)$ means $f$ is analytic, i.e. there exists $(a_{n})_{n} \in \mathbb C$ s.t. $f$ can be written as: $f(z) = \sum_{n \ge 0} a_{n}z^{n}$, for $z \in \mathbb D$.
Let $M = \Vert f \Vert_{H^{2}(\mathbb D)} = sup_{0 \le r \lt 1} \Vert f \Vert _{L^{2}(0,2\pi)}$. By some easy computation, setting $z = re^{i \theta}$, on my teacher's proof we end up having $ \forall r \in [0,1[$:
$M^{2} \ge \int_{0}^{2\pi} \sum_{n,m \ge 0} a_{n} \overline a_{m}r^{n+m}e^{i(n-m) \theta} d\theta$
Since the series are uniformly convergent, one can switch the sum and the integral to get:
$M^{2} \ge 2\pi \sum_{n \ge0} \vert a_{n}\vert^{2}r^{2k}$
Here my problem starts. It is about a huge confusion I have over the initial hypothesis and a need of remind about convergence theorems. To eventually end up having the $L^{2}(\mathbb T)$ norm of $f$ (:= $ \sum_{n \ge 0 } \vert a_{n}\vert ^{2}$), one need to let $r \to 1^{-}$ and justify we can do so. I'll present two ways to get this, one is my teacher's which I don't understand, the other one is mine. Teacher's way:
$M^{2} \ge 2\pi \sum_{n \ge0} \vert a_{n}\vert^{2}r^{2k}$.
For some fixed $N \in \mathbb N$, the following is still true:
$M^{2} \ge 2\pi \sum_{n \ge 0}^{N} \vert a_{n}\vert^{2}r^{2k}$, $ 0 \le r \lt 1$.
Since the boundary does not depend on $r$, one has: $M^{2} \ge 2\pi \lim_{r \to 1^{-}} \sum_{n \ge 0}^{N} \vert a_{n}\vert^{2}r^{2k}$, and by Fatou's lemma (my guess), one has:
$M^{2} \ge 2\pi \sum_{n \ge 0}^{N} \vert a_{n}\vert^{2} \lim_{r \to 1^{-}} r^{2k} = 2\pi \sum_{n \ge 0}^{N} \vert a_{n}\vert^{2}$
Since $N$ was arbitrary fixed and again, the boundary is independent to $N$, by the Dominated Convergence Theorem (my guess again), one end up having:
$M^{2} \ge = 2\pi \lim_{N \to \infty} \sum_{n \ge 0}^{N} \vert a_{n}\vert^{2} = \Vert f \Vert_{L^{2}(\mathbb T)}$
; which lead us to the expected inequality.
Here is the way I've gotten to it:
Starting to $M^{2} \ge 2\pi \sum_{n \ge0} \vert a_{n}\vert^{2}r^{2k}$;
I denote $g_{k}(r) = \vert a_{n}\vert r^{k}$. Obviously, $\forall$ fixed $r \lt 1$, the sequence $(g_{k}(r))_{k} \in l^{2}(\mathbb N)$. By arguing that the boundary does not depend on $r$, I then affirm that:
$M^{2} \ge 2\pi \lim_{r \to 1^{-}} \sum_{n \ge0} \vert a_{n}\vert^{2}r^{2k}$
Thanks to Fatou's lemma:
$M^{2} \ge = 2\pi \lim_{N \to \infty} \sum_{n \ge 0}^{N} \vert a_{n}\vert^{2} = \Vert f \Vert_{L^{2}(\mathbb T)}$
Moreover, $(a_{n})_{n} \in l^{2}(\mathbb N)$.
Questions:
Where is the difference between my teacher's way and mine ?
Why is she shrinking the sum up to $N$ since I think Fatou's lemma can be enough to explain the switch between the lim and the sum ?
What hypothesis would be missing then ? And to finish, are my guesses right ?
I think both proofs are OK, but the argument can be streamlined:
Owing to the isometry $\ell^2(\mathbb Z)\leftrightarrow L^2(\mathbb T),$ it is enough to prove that $\sum |a_n|^2$ converges, and for this, it is enough to prove that the sum is bounded. We have $\sum^N |a_n|^2r^{2n}\le M^2$ for all integers, $N$, so the monotone convergence theorem (with counting measure) implies that $\sum |a_n|^2\le M$.