$|f(z)|^{2}\leq \frac{1}{\pi r^{2}}\iint_{D(z,r)} |f(\theta)|^{2}dm(\theta)$ for $f \in H(\Omega)$

Question

Let $\Omega $ be a domain ,$\overline{D(z,r)} \subset \Omega $, $f$ holomorphic in $\Omega$.

a) Show that $$|f(z)|^{2}\leq \frac{1}{\pi r^{2}}\iint_{D(z,r)} |f(\theta)|^{2}dm(\theta)$$ where $dm$ denotes the lebesgue measure in $\mathbb{C} \equiv \mathbb{R^{2}} $.

b) For $M\geq 0 $ let $$ F = \left\{f \in H(\Omega)| \ \iint_{\Omega} |f(\theta)|^{2}dm(\theta) \leq M\right\}$$. Then show that $F$ is normal.

Any hint ?

Answer 1

Since $g(z)=f^2(z)$ is also analytic, Mean Value Theorem for Analytic Functions provides that $$ f^2(z)=\frac{1}{2\pi}\int_0^{2\pi} f^2(z+\varrho\mathrm{e}^{i\vartheta})d\vartheta, $$ which implies using polar coordinates \begin{align} \int_{D_r(z)} f^2(x+iy)\,dx\,dy&= \int_0^r\int_0^{2\pi}f^2\big(z+\varrho\mathrm{e}^{i\vartheta}\big)\,d\vartheta\,\varrho\,d\varrho=\int_0^r 2\pi\, f^2(z)\,\varrho\,d\varrho=\pi r^2 f(z), \end{align} and thus $$ \int_{D_r(z)} |f(x+iy)|^2\,dx\,dy \ge \left| \int_{D_r(z)} f^2(x+iy)\,dx\,dy \right| =\pi r^2\,|f^2(z)|. $$

Answer 2

Hint for a)

1. WLOG assume $z=0$, by subharmonicity we have $$|f(0)|^2\leq \frac{1}{2\pi}\int_0^{2\pi}|f(\rho e^{i\phi})|^2d\phi$$

2. Now, $d\phi$ is not the area measure - what is missing?

Answer 3

As an alternative to the subharmonicity argument, you can also explicitly (well, sort of) evaluate

$$\int_{D(z;r)} \lvert f(\theta)\rvert^2\,dm(\theta)$$

using the Taylor expansion of $f$ about $z$,

$$f(\theta) = \sum_{n=0}^\infty a_n(\theta-z)^n \Rightarrow \lvert f(\theta)\rvert^2 = \sum_{m,n=0}^\infty a_n\overline{a_m}(\theta-z)^n\overline{(\theta-z)^m}.$$

Write $\theta = z + \rho e^{i\varphi}$ and integrate in polar coordinates to get a nice expression for the integral. Use that $f(z) = a_0$.

For b), the condition implies that

$$\int_{D(z;r)} \lvert f(\theta)\rvert^2 \,dm(\theta) \leqslant M$$

for all $z \in \Omega$ and $r > 0$ such that $D(z;r) \subset \Omega$.

Then use a) to obtain the premises of the little Montel theorem.