Fall 2016 - Problem 8

Solution.

1. For such an $r$, Jensen's formula applies. Thus, of $\set{a_n}_n$ denotes the zeroes of $f$ on the disk (since $f\p{0} \neq 0$, $f$ is not identically zero, so there are only countably many zeroes), we have

   $$\frac{1}{2\pi} \int_0^{2\pi} \log\,\abs{f\p{re^{i\theta}}} \,\diff\theta = \log\,\abs{f\p{0}} - \sum_{\abs{a_n}<r} \log\frac{\abs{a_n}}{r} \geq \log\,\abs{f\p{0}}.$$

2. Since $f$ has at most countably many zeroes in the disk, there exists a sequence $\set{r_n}_n \subseteq \p{0,1}$ of radii increasing to $1$ such that $\inf_{\abs{z}=r_n} \abs{f\p{z}} > 0$ for all $n \geq 1$. Since $f$ is continuous and not identically zero on $\cl{\D}$, we have $0 < \norm{f}_{L^\infty} < \infty$, which gives

   $$\begin{aligned} \abs{\log\,\abs{f\p{r_ne^{i\theta}}}} &= \abs{\log\frac{\abs{f\p{r_ne^{i\theta}}}}{\norm{f}_{L^\infty}} + \log\,\norm{f}_{L^\infty}} \\ &\leq \abs{\log\frac{\abs{f\p{r_ne^{i\theta}}}}{\norm{f}_{L^\infty}}} + \abs{\log\,\norm{f}_{L^\infty}} \\ &= \log\,\norm{f}_{L^\infty} - \log\,\abs{f\p{r_ne^{i\theta}}} + \abs{\log\,\norm{f}_{L^\infty}} \\ &\leq 2\abs{\log\,\norm{f}_{L^\infty}} - \log\,\abs{f\p{r_ne^{i\theta}}}. \end{aligned}$$

   Averaging, we obtain

   $$\frac{1}{2\pi} \int_0^{2\pi} \abs{\log\,\abs{f\p{r_ne^{i\theta}}}} \,\diff\theta \leq 2\abs{\log\,\norm{f}_{L^\infty}} - \log\,\abs{f\p{0}}.$$

   By Fatou's lemma,

   $$\begin{aligned} \frac{1}{2\pi} \int_0^{2\pi} \abs{\log\,\abs{f\p{e^{i\theta}}}} \,\diff\theta &= \frac{1}{2\pi} \int_0^{2\pi} \liminf_{n\to\infty}\,\abs{\log\,\abs{f\p{r_ne^{i\theta}}}} \,\diff\theta \\ &\leq \liminf_{n\to\infty} \frac{1}{2\pi} \int_0^{2\pi} \abs{\log\,\abs{f\p{r_ne^{i\theta}}}} \,\diff\theta \\ &\leq 2\abs{\log\,\norm{f}_{L^\infty}} - \log\,\abs{f\p{0}} \\ &< \infty, \end{aligned}$$

   since $f\p{0} \neq 0$. Thus, $\log\,\abs{f\p{e^{i\theta}}} \neq -\infty$ for almost every $\theta$, hence $f\p{e^{i\theta}} \neq 0$ for almost every $\theta$.