9
$\begingroup$

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!

$\endgroup$

2 Answers 2

11
$\begingroup$

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 $\sum_{n=1}^\infty (1-|a_n|) < \infty$.

$\endgroup$
3
  • 1
    $\begingroup$ You beat me by a few seconds :) $\endgroup$ May 17, 2013 at 21:12
  • $\begingroup$ How do you know that the infinite product converges? $\endgroup$ Apr 21, 2016 at 16:25
  • $\begingroup$ @Rick: the partial products are decreasing (and by the argument above bounded away from $0$). $\endgroup$
    – mrf
    Apr 21, 2016 at 16:32
8
$\begingroup$

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 $f$.

$\endgroup$
3
  • $\begingroup$ Amazing timing with two almost identical answers! $\endgroup$
    – mrf
    May 17, 2013 at 21:11
  • $\begingroup$ @mrf Amazing indeed. And this isn't the first time I see it happen! $\endgroup$ May 17, 2013 at 21:15
  • $\begingroup$ Thank you very much for both of you. The answer and timing is amazing! $\endgroup$
    – BigTree
    May 17, 2013 at 21:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .