If $\mathbb f$ is analytic and bounded on the unit disc with zeros $a_n$ then $\sum_{n=1}^\infty \left(1-\lvert a_n\rvert\right) \lt \infty$

Question

I'm going over old exam problems and I got stuck on this one. Suppose that $\mathbb{f}\colon \mathbb{D} \to \mathbb{C}$ is analytic and bounded. Let $\{a_n\}_{n=1}^\infty$ be the non-zero zeros of $\mathbb{f}$ in $\mathbb{D}$ counted according to multiplicity. Show that $$ \sum_{n=1}^\infty \left( 1 - \left|a_n \right|\right)\lt\infty $$ I can understand that $\left|a_n \right|$ goes to $1$ since zeros are isolated, but it doesn't help showing the series is convergent. Any help will be appreciated!

Answer 1

This is known as Blaschke's condition and is in fact also true for functions in the so called Nevanlinna class. The simplest way to prove this is using Jensen's formula.

Assume $f \in H^\infty$. You may as well assume that $f(0) \neq 0$ and that $f$ has infinitely many zeros. Let $n(r)$ be the number of zeros in the disc $D_r$. Fix any integer $k$ and choose $r < 1$ so large that $n(r) > k$. By Jensen's formula, $$ |f(0)| \prod_{n=1}^{n(r)} \frac{r}{|a_n|} = \exp\left( \frac{1}{2\pi} \int_0^{2\pi} \log|f(re^{i\theta})|\,d\theta \right). $$

Hence (if $|f(z)| < M$ on $D$): $$ |f(0)| \prod_{n=1}^{k} \frac{r}{|a_n|} \le |M|. $$

In other words $$ \prod_{n=1}^{k} |a_n| \ge \frac{|f(0)|}{|M|} r^k $$ for every $k$. Letting $r \to 1$ and $k\to\infty$ it follows that $$ \prod_{n=1}^{\infty} |a_n| \ge \frac{|f(0)|}{|M|} > 0, $$ which implies that $\sum_{n=1}^\infty (1-|a_n|) < \infty$.

Answer 2

First, assume $f(0) \ne 0$. Let $r < 1$ and $n(r)$ be the number of zeros of $f$ inside $\overline{D}(0, r)$. Let $k$ be a positive integer so that $k < n(r)$. By Jensen's formula: $$ \left|f(0)\right| \prod_{n=1}^{n(r)}\frac{r}{\left|\alpha_n\right|} = \exp \left\{\frac{1}{2\pi}\int_{-\pi}^\pi \log\left|f(re^{i\theta})\right|\right\} $$

Since $f$ is bounded (say by $M > 0$), we have: $$ \prod_{n=1}^{k}\left|\alpha_n\right| \ge \frac{\left|f(0)\right|r^k}{M} $$

Now, by letting $r \to 1$, we get: $$ \prod_{n=1}^{\infty}\left|\alpha_n\right| \ge \frac{\left|f(0)\right|}{M} > 0 $$

Which implies that: $$ \sum_{n=1}^\infty \left(1 - \left|\alpha_n\right|\right) < \infty $$

Finally, if $f(0) = 0$. Put $f(z) = z^m g(z)$ where $g(0) \ne 0$ and apply the reasoning above to $g$. Since $g$ and $f$ have the same zeros apart from $z = 0$, the result follows for $f$.