Hardy space

Remarks

The integral means increase with $r$, because $\abs{f}^p$ is subharmonic, so the supremum is a limit as $r \to 1$. For $f(z) = \sum_{n \ge 0} a_n z^n$, a power series on $\mathbb{D}$, Parseval's identity gives $\frac{1}{2\pi}\int_0^{2\pi} \abs{f(re^{i\theta})}^2 d\theta = \sum \abs{a_n}^2 r^{2n}$, so $\norm{f}_{H^2}^2 = \sum_{n \ge 0} \abs{a_n}^2$. Example: $f(z) = \log(1 - z) = -\sum_{n \ge 1} z^n/n$ has $\sum 1/n^2 < \infty$, so $f \in H^2$. Non-example: $g(z) = 1/(1 - z) = \sum_{n \ge 0} z^n$ has all coefficients $1$, so $g \notin H^2$; in fact $g \in H^p$ exactly for $p < 1$. For $1 \le p \le \infty$, $H^p$ is a Banach space; for $0 < p < 1$ it is a complete metric space but not a normed space. There are also Hardy spaces of the upper half-plane and of domains in $\C^n$ and $\R^n$.