Spring 2009 - Problem 9

Solution.

1. Let $R > 0$ with the property stated and consider $h\p{z} = f\p{Rz}$. Thus, $\abs{h\p{z}} > 0$ when $\abs{z} = 1$ and $h$ has zeroes $\set{\frac{a_n}{R}}_n$. Consider the Blaschke product

   $$\phi_a\p{z} = \frac{z - a}{1 - \conj{a}z}$$

   for $a \in B\p{0, 1}$. Notice that if $\abs{z} = 1$, then $\abs{\phi_a\p{z}} = 1$. Next, consider

   $$g\p{z} = \prod_{\abs{a_n} < R} \phi_{a_n/R}\p{z},$$

   which is an analytic function on $B\p{0, 1}$ with the same zeroes and with the same order as $h$. Since $a_n \neq 0$ for any $n \geq 1$, we see that $g\p{0} \neq 0$ as well. Thus, $h/g$ is holomorphic and non-zero on $B\p{0, 1}$.

   When $\abs{z} = R$, $f\p{z} \neq 0$ which implies that $h\p{z} \neq 0$ on $\abs{z} = 1$. $h$ is holomorphic near the boundary and the Blaschke products are continuous up to the boundary, so we see that $\log\abs{h/g}$ is a harmonic function on $B\p{0, 1}$ and continuous up to the boundary, so we apply the mean value property:

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

   On the other hand, $\abs{g\p{e^{i\theta}}} = 1$ for all $0 \leq \theta \leq 2\pi$, so

   $$\frac{1}{2\pi} \int_0^{2\pi} \log\abs{\frac{h\p{e^{i\theta}}}{g\p{e^{i\theta}}}} \,\diff\theta = \frac{1}{2\pi} \int_0^{2\pi} h\p{e^{i\theta}} \,\diff\theta = \frac{1}{2\pi} \int_0^{2\pi} f\p{Re^{i\theta}} \,\diff\theta.$$

   Combining these together, we obtain

   $$\frac{1}{2\pi} \int_0^{2\pi} f\p{Re^{i\theta}} \,\diff\theta = \log\abs{f\p{0}} + \sum_{\abs{a_n} < R} \log\frac{R}{\abs{a_n}},$$

   as required.

2. Let $N\p{R} = \set{n \mid \abs{a_n} < R}$, i.e., the number of zeroes of $f$ on the disk $B\p{0, R}$. Then

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

   On the other hand, by assumption on $f$,

   $$\frac{1}{2\pi} \int_0^{2\pi} \log\abs{f\p{2Re^{i\theta}}} \,\diff\theta \leq \frac{1}{2\pi} \int_0^{2\pi} \log\abs{C} + \abs{2R}^\lambda \,\diff\theta = \log\abs{C} + \p{2R}^\lambda.$$

   Thus, combining these, we obtain the estimate

   $$N\p{R} \leq \frac{\p{2R}^\lambda + \log\abs{C} - \log\abs{f\p{0}}}{\log{2}} \leq A\p{2R}^\lambda$$

   for $A, R$ large enough. Let $M$ be large enough so that the inequality holds for $R \geq 2^{M-1}$. Let $\epsilon > 0$ and observe that

   $$\sum_n \frac{1}{\abs{a_n}^{\lambda+\epsilon}} = \sum_{\abs{a_n} < 2^{M-1}} \frac{1}{\abs{a_n}^{\lambda+\epsilon}} + \sum_{\abs{a_n} \geq 2^{M-1}} \frac{1}{\abs{a_n}^{\lambda+\epsilon}}.$$

   Notice $f$ can only have finitely many zeroes in $\cl{B\p{0, 2^{M-1}}}$, or else they accumulate and $f$ is identically zero, which is impossible. Moreover, the zeroes avoid $0$, since $f\p{0} \neq 0$, so the first sum is finite and bounded. Hence, it remains to bound the other sum:

   $$\begin{aligned} \sum_{\abs{a_n} \geq 2^{M-1}} \frac{1}{\abs{a_n}^{\lambda+\epsilon}} &= \sum_{r=M}^\infty \sum_{2^{r-1} \leq \abs{a_n} < 2^r} \frac{1}{\abs{a_n}^{\lambda+\epsilon}} \\ &= \sum_{r=M}^\infty \sum_{2^{r-1} \leq \abs{a_n} < 2^r} \frac{1}{2^{\p{r-1}\p{\lambda+\epsilon}}} \\ &= \sum_{r=M}^\infty \frac{N\p{2^r} - N\p{2^{r-1}}}{2^{\p{r-1}\p{\lambda+\epsilon}}} \\ &\leq \sum_{r=M}^\infty \frac{A\p{2R}^\lambda}{2^{\p{r-1}\p{\lambda+\epsilon}}} \\ &= \frac{AR^\lambda}{2} \sum_{r=M}^\infty \frac{1}{\p{2^\epsilon}^r} \\ &< \infty. \end{aligned}$$

   Indeed, this is a geometric series with common ratio $2^{-\epsilon} < 1$, since $\epsilon > 0$, which completes the proof.