Proposition 40.4.2.label Let $D = B_{\complex}(0, 1)$ and $H^{2}(D)$ be the Hardy space, then:
- (1)
$H^{2}(D)$ is a Hilbert space. For each $f, g \in H^{2}(D)$,
\[\dpn{f, g}{H^2(D)}= \lim_{r \upto 1}\frac{1}{2\pi}\int_{0}^{2\pi}f(re^{i\theta})\ol{g(re^{i\theta})}d\theta\] - (2)
The mapping
\[l^{2}(\natz; \complex) \to H^{2}(D) \quad x \mapsto \sum_{n = 0}^{\infty} x_{n}z^{n}\]is unitary.
Proof. (1): For each $r \in [0, 1)$ and $f \in H^{2}(D)$, let
Then, $[\cdot]_{r}$ is induced by the pseudo inner product
By the Maximum Modulus Theorem, $\norm{f}_{H^2(D)}= \lim_{r \upto 1}[f]_{r}$, so the inner product
exists, and defines the norm on $H^{2}(D)$.
(2): For any $x \in l^{2}(\natz; \complex) \subset l^{\infty}(\natz; \complex)$, $\limsup_{n \to \infty}|x_{n}|^{1/n}\le 1$, so the given map is well-defined.
On the other hand, for any $f = \sum_{n = 0}^{\infty} x_{n}z^{n}, g = \sum_{n = 0}^{\infty} y_{n}z^{n} \in H^{2}(D)$ and $r \in (0, 1)$,
by the Dominated Convergence Theorem. Thus $\bracsn{z^n}_{0}^{\infty}$ forms an orthonormal basis for $H^{2}(D)$ and the given map is unitary.$\square$
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