Zeroes of Functions of Finite Order

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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:

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


Proof

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

By Jensen's Formula:

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


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

Then:

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

and

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


Moreover, by hypothesis we have that:

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


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:

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

$\blacksquare$