Zeros of Functions of Finite Order

Theorem

Let $f(z)$ be an entire function which satisfies $f(0) \neq 0$, and

$|f(z)| \ll \exp\left( \alpha(|z|) \right)$

for all $z \in \C$ and some function $\alpha$, where $\ll$ is the order notation.

For $T \geq 1$, let

$N(T) = \# \{\rho \in \C : f(\rho) = 0,\ |\rho| < T \}$

where $\#$ denotes the cardinality of a set.

Then $N(T) \ll \alpha(2T)$.

Corollary

If $f$ has order $1$, then for all $\epsilon > 0$, the sum

$\displaystyle \sum_{k \geq 1} \frac 1{|\rho_k|^{1+\epsilon}}$

converges, where $\{ \rho_k \}_{k \geq 1}$ is a non-decreasing enumeration of the zeros of $f$, counted with multiplicity.

Proof of Theorem

Fix $T \geq 1$ and let $\rho_1,\rho_2,\ldots,\rho_n$ be an enumeration of the zeros of $f$ with modulus less than $T$, counted with multiplicity.

By Jensen's Formula we have

$\displaystyle \frac 1{2\pi} \int_0^{2\pi} \log |f(Te^{i\theta})|\ d\theta = \log|f(0)| + \sum_{k=1}^n (\log T - \log |\rho_k|)$

Let $\rho_0 = 1$, $\rho_{n+1} = T$, $r_k = |\rho_k|$. Then

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \int_0^T N(t) \frac{dt}t\)	\(=\)	\(\displaystyle \sum_{k=0}^n \int_{r_k}^{r_{k+1} } N(t) \frac{dt}t\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{k=0}^n k \log \left( \frac{r_{k+1} } {r_k} \right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	As by the definition of $N$, it is constant value $k$ on each interval $(\vert\rho_k\vert,\vert\rho_{k+1}\vert)$.
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \log\left( \frac{T^n}{r_1\cdots r_n} \right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{k=1}^n (\log T - \log r_k)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

and

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \int_0^T N(t) \frac{dt}t\)	\(=\)	\(\displaystyle \int_0^2 N\left( \frac{T\theta}2 \right) \frac{d\theta}{\theta}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\geq\)	\(\displaystyle N\left(\frac T2 \right) \int_1^2 \frac{d\theta}{\theta}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By Integration by Substitution
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle N\left(\frac T2 \right) \log 2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By the definition of the logarithm

Moreover, by hypothesis we have that

$\displaystyle \frac 1{2\pi} \int_0^{2\pi} \log |f(Te^{i\theta})|\ d\theta \leq \sup_{\theta \in [0,2\pi)}\log |f(Te^{i\theta})| \ll \alpha(T)$

Putting these facts into Jensen's formula we have

$\displaystyle N\left(\frac T2 \right) \log 2 + |f(0)| \ll \alpha(T)$

Which implies

$N(T) \ll \alpha(2T)$

$\blacksquare$

Proof of Corollary

Let $\epsilon > 0$, $N(0) = 0$, so that

$\displaystyle \sum_{k\geq 1} |\rho_k|^{-1-\epsilon} \leq \sum_{T \geq 1} \left[ N(T) - N(T-1) \right] T^{-1-\epsilon}$

We have $N(T) \ll 2T$, so $N(T) - N(T-1)$ is bounded in $T$, say by $C > 0$. Therefore,

$\displaystyle \sum_{k\geq 1} |\rho_k|^{-1-\epsilon} \leq C\: \sum_{T \geq 1} \frac 1{T^{1+\epsilon}}$

and the sum on the right converges absolutely for $\epsilon > 0$.

$\blacksquare$