What’s going on again

In this post we discussed the topological properties of the zero points of an entire nonzero function, or roughly, how those points look like. The set of zero points contains no limit point, and at most countable (countable or finite). So if it’s finite, then we can find them out one by one. For example, the function $f(z)=z$ has simply one zero point. But what if it’s just countable? How fast the number grows?

Another question. Suppose we have an entire function $f$, and the zeros of $f$, namely $z_1,z_2,\cdots,z_n$, are ordered increasingly by moduli:

$|z_1| \leq |z_2| \leq \cdots \leq |z_n| \leq \cdots$

Is it possible to get a fine enough estimation of $|z_n|$? Interesting enough, we can get there with the help of Jensen’s formula.

Jensen’s formula

Suppose $\Omega=D(0;R)$, $f \in H(\Omega)$, $f(0) \neq 0$, $0<r<R$, and $z_1,z_2,\cdots,z_{n(r)}$ are the zeros of $f$ in $\overline{D}(0;R)$, then
$|f(0)|\prod_{n=1}^{n(r)}\frac{r}{|z_n|}=\exp\left[\frac{1}{2\pi}\int_{-\pi}^{\pi}\log|f(re^{i\theta})|d\theta\right]$

There is no need to worry about the assumption $f(0) \neq 0$. Take another look at this proof. Every zero point $a$ has a unique positive number $m$ such that $f(z)=(z-a)^mg(z)$ and $g \in H(\Omega)$ but $g(a) \neq 0$. The number $m$ is called the order of the zero at $a$. Therefore if we have $f(0)=0$ we can simply consider another function, namely $\frac{f}{z^m}$ where $m$ is the order of zero at $0$.

We are not proving this identity at this point. But it can be done by considering the following function

$g(z)=f(z)\prod_{n=1}^{m}\frac{r^2-\overline{z}_nz}{r(z_n-z)}\prod_{n=m+1}^{n(r)}\frac{z_n}{z_n-z}$

where $m$ is found by ordering $z_j$ in such a way that $z_1,\cdots,z_m \in D(0;r)$ and $|z_{m+1}|=\cdots=|z_{n}|$. One can prove this identity by considering $|g(0)|$ as well as $\log|g(re^{i\theta})|$.

Several applications

The number of zeros of $f$ in $\overline{D}(0;r)$

For simplicity we shall assume $f(0)=1$ which has no loss of generality. Let

$M(r)=\sup_{\theta}|f(re^{i\theta})|\quad 0<r<\infty$

and $n(r)$ be the number of zeros of $f$ in $\overline{D}(0;r)$. By the maximum modulus theorem, we have

$\log|f(2re^{i\theta})| \leq |f(2re^{i\theta})| \leq M(2r)$

If we insert Jensen’s formula into this inequality and order $|z_n|$ by increasing moduli, we get

$\log M(2r) \geq \frac{1}{2\pi}\int_{-\pi}^{\pi}\log|f(2re^{i\theta})|d\theta=\sum_{n=1}^{n(2r)}\log\frac{2r}{|z_n|}\geq\sum_{n=1}^{n(r)}\log\frac{2r}{|z_n|}\geq n(r)\log2$

Which implies

$n(r)\leq\log_2M(2r)$

So $n(r)$ is controlled by $M(2r)$. The second and third inequalities look tricky, which require more explanation.

First we should notice the fact that $z_n \in \overline{D}(0;R)$ for all $R \in \mathbb{R}$. Hence we have $\log\frac{2r}{|z_n|} \geq \log1=0$ for all $z_n \in \overline{D}(0;R)$. Hence the second inequality follows. For the third one, we simply have

$\sum_{n=1}^{n(r)}\log\frac{2r}{|z_n|}=\sum_{n=1}^{n(r)}(\log2+\log\frac{r}{|z_n|}) \geq n(r)\log2.$

So this is it, the rapidity with which $n(r)$ can grow is dominated by $M(r)$. Namely, the number of zeros of $f$ in the closed disc with radius $r$ is controlled by the maximum modulus of $f$ on a circle with bigger radius.

Examples based on different $M(r)$

Let’s begin with a simple example. Let $f(z)=1$, we have $M(r)=1$ for all $r$, but also we have $n(r)=0$, in which sense this estimation does nothing. Indeed, as long as $M(r)$ is bounded by a constant, which implies $f(z)$ is bounded, then by Liouville’s theorem, $f(z)$ is constant and this estimation is not available.

But if $M(r)$ grows properly, things become interesting. For example, if we have

$M(r) \leq \exp(Ar^k)$

where $A$ and $k$ are given positive numbers, we have a good enough estimation by

$n(r) \leq \frac{A+(2r)^k}{\log2}$

This estimation becomes interesting if we consider the logarithm of $n(r)$ and $r$, that is

$\begin{aligned}\limsup_{r\to\infty}\frac{\log{n(r)}}{\log{r}} &\leq \lim_{r\to\infty} \frac{\log(A+(2r)^k)-\log{2}}{\log{r}} \\ & =k\end{aligned}$

If we have $f(z)=1-\exp(z^k)$ where $k$ is a positive integer, we have $n(r) \sim \frac{kr^k}{\pi}$, also

$\lim_{r\to\infty}\frac{\log{n(r)}}{\log r}=k$

Lower bound of $|z_{n(r)}|$

We’ll see here, how to evaluate the lower bound of $|z_{n(r)}|$ using Jensen’s formula, provided that $M(r)$, or simply the upper bound of $f(z)$ is properly described. Without loss of generality we shall assume that $f(0)=1$. Also, we assume that the zero points of $f(z)$ are ordered by increasing moduli.

First we still consider

$M(r) \leq \exp(Ar^k)$

and see what will happen.

By Jensen’s, we have

$\prod_{n=1}^{n(r)}\frac{r}{|z_n|}=\exp\left[\frac{1}{2\pi}\int_{-\pi}^{\pi}\log|f(re^{i\theta})|d\theta\right] \leq \exp{Ar^k}$

This gives

$\prod_{n=1}^{n(r)}|z_n| \geq r^{n(r)}\exp(-Ar^k)$

By the arrangement of ${z_n}$, we have

$|z_{n(r)}| \geq \sqrt[n(r)]{\prod_{n=1}^{n(r)}|z_n|}\geq r\exp(-Ar^{k-n(r)})$

Another example is when we have

$|f(z)| \leq \exp(A|\Im{z}|)$

where $\Im{z}$ means the imagine part of $z$.

We shall notice that in this case,

$\begin{aligned}\frac{1}{2\pi}\int_{-\pi}^{\pi}\log|f(re^{i\theta})|d\theta &\leq \frac{1}{2\pi}\int_{-\pi}^{\pi}A|r\sin\theta|d\theta=\frac{2Ar}{\pi}\end{aligned}$

Following Jensen’s formula, we therefore have

$|z_{n(r)}| \geq \exp(\frac{2A}{\pi}r^{1-n(r)})$