Spring 2017 - Problem 9

Solution.

1. Let $R > 0$ be as in the problem statement. Observe that $f\p{Rz}$ has roots $\frac{a_n}{R}$ for $\abs{a_n} < R$, so let

   $$g\p{z} = \prod_{\abs{a_n} < R} \frac{z - \p{a_n/R}}{1 - \p{\conj{a_n}/R}z}$$

   be the associated Blaschke product. $g$ is a product of automorphisms of the disk, hence holomorphic, and it has the exact same zeroes as $f\p{Rz}$ with the same multiplicities. Moreover, since Blaschke factors have norm $1$ whenever $\abs{z} = 1$, we have $\abs{g\p{z}} = 1$ on $\partial\D$. Thus, $\frac{f\p{Rz}}{g\p{z}}$ is a holomorphic function which does not vanish on $\cl{\D}$, which means that $\log\,\abs{\frac{f\p{Rz}}{g\p{z}}}$ is harmonic as the real part of a logarithm. Hence, by the mean value property,

   $$\begin{aligned} \frac{1}{2\pi} \int_0^{2\pi} \log\,\abs{f\p{Re^{i\theta}}} \,\diff\theta &= \frac{1}{2\pi} \int_0^{2\pi} \log\,\abs{\frac{f\p{Re^{i\theta}}}{g\p{e^{i\theta}}}} \,\diff\theta \\ &= \log\,\abs{\frac{f\p{0}}{g\p{0}}} \\ &= \log\,\abs{f\p{0}} - \sum_{\abs{a_n} < R} \log\,\abs{\frac{a_n}{R}} \\ &= \log\,\abs{f\p{0}} + \sum_{\abs{a_n} < R} \log\,\abs{\frac{R}{a_n}}. \end{aligned}$$

2. Let $N\p{R} = \set{n \mid \abs{a_n} < R}$. Then

   $$\begin{aligned} \frac{1}{2\pi} \int_0^{2\pi} \log\,\abs{f\p{2Re^{i\theta}}} \,\diff\theta &= \log\,\abs{f\p{0}} + \sum_{\abs{a_n} < 2R} \log\,\abs{\frac{2R}{a_n}} \\ &\geq \log\,\abs{f\p{0}} + \sum_{\abs{a_n} < R} \log\,\abs{\frac{2R}{R}} \\ &\geq \log\,\abs{f\p{0}} + N\p{R} \log{2}. \end{aligned}$$

   Thus, applying the estimate on $f$,

   $$\begin{aligned} \log\,\abs{f\p{0}} + N\p{R} \log{2} &\leq \frac{1}{2\pi} \int_0^{2\pi} \log{Ce^{\p{2R}^\lambda}} \,\diff\theta \\ &= \log{C} + \p{2R}^\lambda \\ \implies N\p{R} &\leq \frac{\p{2R}^\lambda + \log{C} - \log\,\abs{f\p{0}}}{\log{2}} \\ &\leq A\p{2R}^\lambda \end{aligned}$$

   for $R$ large enough, say, for $R \geq 2^N$ where $N \in \N$. Next, observe that

   $$\sum_n \frac{1}{\abs{a_n}^{\lambda+\epsilon}} = \sum_{\abs{a_n} < 2^N} \frac{1}{\abs{a_n}^{\lambda+\epsilon}} + \sum_{n=N}^\infty \sum_{2^n \leq \abs{a_n} < 2^{n+1}} \frac{1}{\abs{a_n}^{\lambda+\epsilon}}.$$

   Since $f$ is entire and non-zero at the origin, there are only finitely many $n$ such that $\abs{a_n} < 2^N$, so it remains to estimate the second term:

   $$\begin{aligned} \sum_{n=N}^\infty \sum_{2^n \leq \abs{a_n} < 2^{n+1}} \frac{1}{\abs{a_n}^{\lambda+\epsilon}} &\leq \sum_{n=N}^\infty \sum_{2^n \leq \abs{a_n} < 2^{n+1}} \frac{1}{2^{n\p{\lambda+\epsilon}}} \\ &\leq \sum_{n=N}^\infty \frac{N\p{2^{n+1}}}{2^{n\p{\lambda+\epsilon}}} \\ &\leq \sum_{n=N}^\infty \frac{A\p{2 \cdot 2^{n+1}}^\lambda}{2^{n\p{\lambda+\epsilon}}} \\ &= A \cdot 4^\lambda \sum_{n=N}^\infty \frac{1}{\p{2^\epsilon}^n} \\ &< \infty, \end{aligned}$$

   since $\epsilon > 0 \implies 2^\epsilon > 1$, and this completes the proof.