Define the Hadamard canonical factors

Furthermore, Jensen's inequality implies that its roots are distributed sparsely, with critical exponent ${\displaystyle \alpha \leq \rho \leq g+1}$ .

For example, ${\displaystyle \sin }$ , ${\displaystyle \cos }$  and ${\displaystyle \exp }$  are entire functions of genus ${\displaystyle g=\rho =1}$ .

Define the critical exponent of the roots of ${\displaystyle f}$  as the following:

Theorem:^[2] If ${\displaystyle f}$  is an entire function with infinitely many roots, then

Proposition: ${\displaystyle \alpha (f)\leq \rho }$ ,^[3] by Jensen's formula.

Since ${\displaystyle f(z)/z^{m}}$  is also an entire function with the same order ${\displaystyle \rho }$  and genus, we can wlog assume ${\displaystyle m=1}$ .

If ${\displaystyle f}$  has only finitely many roots, then ${\displaystyle f(z)=e^{Q(z)}\prod _{k=1}^{n}E_{0}(z/a_{k})}$  with the function ${\displaystyle e^{Q(z)}}$  of order ${\displaystyle \rho }$ . Thus by an application of the Borel–Carathéodory theorem, ${\displaystyle P}$  is a polynomial of degree ${\displaystyle deg(Q)\leq {\text{floor}}(\rho )}$ , and so we have ${\displaystyle \rho =deg(Q)=g}$ .

Otherwise, ${\displaystyle f}$  has infinitely many roots. This is the tricky part and requires splitting into two cases. First show that ${\displaystyle g\leq {\text{floor}}(\rho )}$ , then show that ${\displaystyle \rho \leq g+1}$ .

Define the function ${\displaystyle g(z):=\prod _{k=1}^{\infty }E_{p}(z/a_{k})}$  where ${\displaystyle p={\text{floor}}(\alpha )}$ . We will study the behavior of ${\displaystyle f(z)/g(z)}$ .

Bounds on the behaviour of |E_p| edit

In the proof, we need four bounds on ${\displaystyle |E_{p}|}$ :

1. For any ${\displaystyle r\in (0,1)}$ , ${\displaystyle |1-E_{p}(z)|\leq O(|z|^{p+1})}$  when ${\displaystyle |z|\leq r}$ .
2. For any ${\displaystyle r\in (0,1)}$ , there exists ${\displaystyle B>0}$  such that ${\displaystyle \ln |E_{p}(z)|\geq -B|z|^{p+1}}$  when ${\displaystyle |z|\leq r}$ .
3. For any ${\displaystyle r\in (0,1)}$ , there exists ${\displaystyle B>0}$  such that ${\displaystyle |E_{p}(z)|\geq |1-z|e^{-B|z|^{p}}}$  when ${\displaystyle |z|\geq r}$ .
4. ${\displaystyle \ln |E_{p}(z)|\leq O(|z|^{p+1})}$  for all ${\displaystyle z}$ , and ${\displaystyle \ln |E_{p}(z)|\leq O(|z|^{p})}$  as ${\displaystyle |z|\to \infty }$ .

These are essentially proved in the similar way. As an example, we prove the fourth one.

Outside ${\displaystyle B(0,R)}$ , we have

g is well-defined edit

Source:^[2]

For any ${\displaystyle R>0}$ , we show that the sum ${\displaystyle \sum _{k=1}^{\infty }\ln |E_{p}(z/a_{k})|}$  converges uniformly over ${\displaystyle B(0,R)}$ .

Since only finitely many ${\displaystyle |a_{k}|<KR}$ , we can split the sum to a finite bulk and an infinite tail:

By bound (1) on ${\displaystyle |E_{p}|}$ , ${\displaystyle |1-E_{p}(z/a_{k})|\leq B|z/a_{k}|^{p+1}\leq B/K^{p+1}}$ . So if ${\displaystyle K}$  is large enough, for some ${\displaystyle B'>0}$ ,^{[nb 1]}

g ≤ floor(ρ) edit

As usual in analysis, we fix some small ${\displaystyle \epsilon >0}$ .

Then the goal is to show that ${\displaystyle f(z)/g(z)}$  is of order ${\displaystyle \leq \rho +\epsilon }$ . This does not exactly work, however, due to bad behavior of ${\displaystyle f(z)/g(z)}$  near ${\displaystyle a_{k}}$ . Consequently, we need to pepper the complex plane with "forbidden disks", one around each ${\displaystyle a_{k}}$ , each with radius ${\displaystyle {\frac {1}{|a_{k}|^{\rho +\epsilon }}}}$ . Then since ${\displaystyle \sum _{k}{\frac {1}{|a_{k}|^{\rho +\epsilon }}}<\infty }$  by the previous result on ${\displaystyle \alpha }$ , we can pick an increasing sequence of radii ${\displaystyle R_{1}<R_{2}<\cdots }$  that diverge to infinity, such that each circle ${\displaystyle \partial B(0,R_{n})}$  avoids all these forbidden disks.

Thus, if we can prove a bound of form ${\displaystyle \ln \left|{\frac {f(z)}{g(z)}}\right|=O(|z|^{\rho +\epsilon })=o(|z|^{\rho +2\epsilon })}$  for all large ${\displaystyle z}$  ^{[nb 2]} that avoids these forbidden disks, then by the same application of Borel–Carathéodory theorem, ${\displaystyle deg(Q)\leq {\text{floor}}(\rho +2\epsilon )}$  for any ${\displaystyle \epsilon >0}$ , and so as we take ${\displaystyle \epsilon \to 0}$ , we obtain ${\displaystyle g\leq {\text{floor}}(\rho )}$ .

Since ${\displaystyle \ln \left|f(z)\right|=o(|z|^{\rho +\epsilon })}$  by the definition of ${\displaystyle \rho }$ , it remains to show that ${\displaystyle -\ln \left|g(z)\right|=O(|z|^{\rho +\epsilon })}$ , that is, there exists some constant ${\displaystyle B>0}$  such that

As usual in analysis, this infinite sum can be split into two parts: a finite bulk and an infinite tail term, each of which is to be separately handled. There are finitely many ${\displaystyle a_{k}}$  with modulus ${\displaystyle <|z_{k}|/2}$  and infinitely many ${\displaystyle a_{k}}$  with modulus ${\displaystyle \geq |z_{k}|/2}$ . So we have to bound:

ρ ≤ g + 1 edit

This is a corollary of the following:

If ${\displaystyle f(z)=e^{Q(z)}z^{m}\prod _{k}E_{p}(z/a_{k})}$  has genus ${\displaystyle g}$ , then ${\displaystyle \ln |f(z)|=O(|z|^{g+1})}$ .

Split the sum to three parts:

With Hadamard factorization we can prove some special cases of Picard's little theorem.

Theorem:^[4] If ${\displaystyle f}$  is entire, nonconstant, and has finite order, then it assumes either the whole complex plane or the plane minus a single point.

Proof: If ${\displaystyle f}$  does not assume value ${\displaystyle z_{0}}$ , then by Hadamard factorization, ${\displaystyle f(z)-z_{0}=e^{Q(z)}}$  for a nonconstant polynomial ${\displaystyle Q}$ . By the fundamental theorem of algebra, ${\displaystyle Q}$  assumes all values, so ${\displaystyle f(z)-z_{0}}$  assumes all nonzero values.

Theorem:^[4] If ${\displaystyle f}$  is entire, nonconstant, and has finite, non-integer order ${\displaystyle \rho }$ , then it assumes the whole complex plane infinitely many times.

Proof: For any ${\displaystyle w\in \mathbb {C} }$ , it suffices to prove ${\displaystyle f(z)-w}$  has infinitely many roots. Expand ${\displaystyle f(z)-w}$  to its Hadamard representation ${\displaystyle f(z)-w=e^{Q(z)}z^{m}\prod _{k}E_{p}(z/a_{k})}$ . If the product is finite, then ${\displaystyle \rho =g}$  is an integer.