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Fall 2010 - Problem 12

mean value property

Let $\mathcal{F}$ be the family of functions $f$ holomorphic on $\D$ with

\iint_{\set{x^2 + y^2 < 1}} \abs{f\p{x + iy}}^2 \,\diff{x} \,\diff{y} < 1.

Prove that for each compact subset $K \subseteq \D$ there is a constant $A$ so that $\abs{f\p{z}} < A$ for all $z \in K$ and all $f \in \mathcal{F}$ .

Solution.

First, let $B\p{z_0, R}$ be an open ball with closure contained in $\D$ . Let $f \in \mathcal{F}$ , so that we can apply the mean value property:

f\p{z_0} = \frac{1}{2\pi} \int_0^{2\pi} f\p{z_0 + Re^{i\theta}} \,\diff\theta.

Multiplying by $2r$ , integrating from $0$ to $R$ , and applying a change-of-coordinates, we end up with the two-dimensional mean value property:

f\p{z_0} = \frac{1}{\pi R^2} \int_{B\p{z_0,R}} f\p{x + iy} \,\diff{A}.

By the triangle inequality, Hölder's inequality, and assumption on $\mathcal{F}$ ,

\begin{aligned} \abs{f\p{z_0}} &\leq \frac{1}{\pi R^2} \int_{B\p{z_0,R}} \abs{f\p{x + iy}} \,\diff{A} \\ &\leq \frac{1}{\pi R^2} \p{\int_{B\p{z_0,R}} \abs{f\p{x + iy}}^2 \,\diff{A}}^{1/2} \p{\int_{B\p{z_0,R}} \diff{A}}^{1/2} \\ &\leq \frac{m\p{B\p{z_0, R}}^{1/2}}{\pi R^2} \\ &\leq \frac{m\p{\D}^{1/2}}{\pi R^2}. \end{aligned}

where $m$ denotes the Lebesgue measure. Now let $K \subseteq \D$ be any compact set, and let $\delta = \frac{1}{2}d\p{K, \D^\comp}$ , which is positive since it's a distance between two disjoint closed sets. For any $z \in K$ , $B\p{z, \delta} \subseteq \D$ , so by our previous calculation,

\abs{f\p{z}} \leq \frac{m\p{\D}^{1/2}}{\pi\delta^2}

for all $z \in \D$ . This bound is independent of choice of $f$ , so this bound holds for all $f \in \mathcal{F}$ as well, which completes the proof.