Spring 2010 - Problem 4

Solution.

1. We will prove Jensen's formula: let $r$ be so that $\inf_{\abs{z}=r}\, \inf \abs{f\p{z}} > 0$. If $E_r$ is the set of roots of $f$ in the open disk $B\p{0, r}$ counting multiplicity, then

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

   For $a \in \D$, consider the Blaschke product $\func{\phi_a}{\D}{\D}$,

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

   which is a holomorphic function on the disk with the property that $\abs{\phi_a\p{z}} = 1$ if $\abs{z} = 1$ or $\abs{a} = 1$. Thus, if we set

   $$g\p{z} = \frac{f\p{z}}{\prod_{a \in E_r} \phi_{a/r}\p{z/r}},$$

   then $g$ is a holomorphic function on $\D$ which is continuous on the $\cl{\D}$ that vanishes nowhere on $\abs{z} \leq r$. Indeed, at each root $a \in E_r$ of $f$, the denominator of $g$ has a zero at $a$ with order equal to that of its numerator. Moreover, if $\abs{z} = r$, then $\abs{g\p{z}} = \abs{f\p{z}}$, since all the factors in the denominator become $1$.

   Hence, $\log\abs{g\p{z}}$ is a harmonic function which does not vanish on $\abs{z} = r$. By the mean value property, we have

   $$\frac{1}{2\pi} \int_0^{2\pi} \log\abs{g\p{re^{i\theta}}} \,\diff\theta = \log\abs{g\p{0}} = \log\abs{f\p{0}} - \sum_{a \in E_r} \log\frac{\abs{a}}{r}.$$

   On the other hand, since $\abs{re^{i\theta}} = r$,

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

   which proves Jensen's formula. Observe that $\log\frac{\abs{a}}{r} < 0$ for $a \in E_r$, so this gives the inequality

   $$\frac{1}{2\pi} \int_0^{2\pi} \log\abs{f\p{re^{i\theta}}} \,\diff\theta \geq \log\abs{f\p{0}},$$

   which was what we wanted to show.

2. Since $\cl{B\p{0, r}}$ is compact and $f$ does not vanish everywhere (since $f\p{0} \neq 0$), it follows that $f$ can only have finitely many zeroes in $\cl{B\p{0, r}}$. By writing

   $$\D = \bigcup_{n=1}^\infty B\p{0, 1 - \frac{1}{n}},$$

   we see that the zeroes of $f$ are a countable union of finite sets, so $f$ has at most countably many zeroes. In particular, we may pick a sequence $\set{r_n}_n \subseteq \p{0, 1}$ such that $r_n$ increases to $1$ and $\inf_{\abs{z}=r_n}\,\abs{f\p{z}} > 0$, since countably many zeroes almost removes countably many choices of radii. By Fatou's lemma,

   $$\begin{aligned} \log\abs{f\p{0}} \leq \limsup_{n\to\infty} \frac{1}{2\pi} \int_0^{2\pi} \log\abs{f\p{r_ne^{i\theta}}} \,\diff\theta \leq \frac{1}{2\pi} \int_0^{2\pi} \log\abs{f\p{e^{i\theta}}} \,\diff\theta. \end{aligned}$$

   Let $E = \set{\theta \in \br{0, 2\pi} \mid f\p{e^{i\theta}} = 0}$. Since $f\p{0} \neq 0$, we have $\log\abs{f\p{0}} > -\infty$, so if $\abs{E} > 0$,

   $$\begin{aligned} -\infty &< \log\abs{f\p{0}} \\ &\leq \frac{1}{2\pi} \int_0^{2\pi} \log\abs{f\p{e^{i\theta}}} \,\diff\theta \\ &= \frac{1}{2\pi} \int_E \log\abs{f\p{e^{i\theta}}} \,\diff\theta + \frac{1}{2\pi} \int_{E^\comp} \log\abs{f\p{e^{i\theta}}} \,\diff\theta \\ &= \frac{\abs{E}}{2\pi} \cdot \p{-\infty} + \frac{1}{2\pi} \int_{E^\comp} \log\abs{f\p{e^{i\theta}}} \,\diff\theta. \end{aligned}$$

   The second term is bounded, since $f$ is continuous on $\abs{z} = 1$, so the second term is bounded above by some finite number. Thus, if $\abs{E} \neq 0$, then

   $$-\infty < \log\abs{f\p{0}} \leq -\infty,$$

   which is impossible. Hence, $\abs{E} = 0$, as required.