# Zeros of Functions of Finite Order It has been suggested that this page or section be merged into Exponent of Convergence is Less Than Order. (Discuss)

## Theorem

Let $\map f z$ be an entire function which satisfies:

$\map f 0 \ne 0$
$\cmod {\map f z} \ll \map \exp {\map \alpha {\cmod z} }$

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

For $T \ge 1$, let:

$\map N T = \# \set {\rho \in \C: \map f r = 0, \ \cmod \rho < T}$

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

Then:

$\map N T \ll \map \alpha {2 T}$

### Corollary

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

$\displaystyle \sum_{k \mathop \ge 1} \frac 1 {\size {\rho_k}^{1 + \epsilon} }$

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

## Proof

Fix $T \ge 1$ 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.

$\displaystyle \frac 1 {2 \pi} \int_0^{2 \pi} \ln \size {\map f {T e^{i \theta} } } \rd \theta = \ln \cmod {\map f 0} + \sum_{k \mathop = 1}^n \paren {\ln T - \ln \size {\rho_k} }$

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

Then:

 $\displaystyle \int_0^T \map N t \frac {\d t} t$ $=$ $\displaystyle \sum_{k \mathop = 0}^n \int_{r_k}^{r_{k + 1} } \map N t \frac {\d t} t$ $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = 0}^n k \, \map \ln {\frac{r_{k + 1} } {r_k} }$ as by the definition of $N$, it is constant value $k$ on each interval $\openint {\size {\rho_k} } {\size {\rho_{k + 1} } }$ $\displaystyle$ $=$ $\displaystyle \map \ln {\frac {T^n} {r_1 \dotsm r_n} }$ $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = 1}^n \paren {\ln T - \ln r_k}$

and

 $\displaystyle \int_0^T \map N t \frac {\d t} t$ $=$ $\displaystyle \int_0^2 \map N {\frac {T \theta} 2} \frac {\d \theta} \theta$ $\displaystyle$ $\ge$ $\displaystyle \map N {\frac T 2} \int_1^2 \frac {\d \theta} \theta$ Integration by Substitution $\displaystyle$ $=$ $\displaystyle \map N {\frac T 2} \ln 2$ Definition of Logarithm

Moreover, by hypothesis we have that:

$\displaystyle \frac 1 {2 \pi} \int_0^{2 \pi} \ln \size {\map f {T e^{i \theta} } } \rd \theta \le \sup_{\theta \mathop \in \closedint 0 {2 \pi} } \ln \size {\map f {T e^{i \theta} } } \ll \map \alpha T$

Putting these facts into Jensen's Formula we have:

$\map N {\dfrac T 2} \ln 2 + \cmod {\map f 0} \ll \map \alpha T$

which implies:

$\map N T \ll \map \alpha {2 T}$

$\blacksquare$

### Proof of Corollary

Let $\epsilon > 0$, $\map N 0 = 0$, so that:

$\displaystyle \sum_{k \mathop \ge 1} \size {\rho_k}^{-1 - \epsilon} \le \sum_{T \mathop \ge 1} \paren {\map N T - \map N {T - 1} } T^{-1 - \epsilon}$

We have $\map N T \ll 2 T$, so $\map N T - \map N {T - 1}$ is bounded in $T$, say by $C > 0$.

Therefore:

$\displaystyle \sum_{k \mathop \ge 1} \size {\rho_k}^{-1 - \epsilon} \le C \ \sum_{T \mathop \ge 1} \frac 1 {T^{1 + \epsilon} }$

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

$\blacksquare$