Bounds on the behaviour of |E_p|

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.

$${\displaystyle \ln |E_{n}(z)|=Re\left(-z^{n+1}\sum _{k=0}^{\infty }{\frac {z^{k}}{k+n+1}}\right)\leq |z|^{n+1}|g(z)|}$$

where ${\displaystyle g(z):=\sum _{k=0}^{\infty }{\frac {z^{k}}{k+n+1}}}$ is an entire function. Since it is entire, for any ${\displaystyle R>0}$, it is bounded in ${\displaystyle B(0,R)}$. So ${\displaystyle \ln |E_{n}(z)|=O(|z|^{n+1})}$ inside ${\displaystyle B(0,R)}$. Outside ${\displaystyle B(0,R)}$, we have

$${\displaystyle \ln |E_{n}(z)|=\ln |1-z|+Re(Q(z))=o(|z|)+O(|z|^{n})<O(|z|^{n+1})}$$

g is well-defined

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:

$${\displaystyle \sum _{k=1}^{\infty }\ln |E_{p}(z/a_{k})|=\sum _{|a_{k}|<KR}\ln |E_{p}(z/a_{k})|+\sum _{|a_{k}|\geq KR}\ln |E_{p}(z/a_{k})|}$$

The bulk term is a finite sum, so it converges uniformly. It remains to bound the tail term. 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]}

$${\displaystyle \sum _{|a_{k}|\geq KR}\ln |1-(1-E_{p}(z/a_{k}))|\leq \sum _{|a_{k}|\geq KR}B'B|z/a_{k}|^{p+1}<|z|^{p+1}B'B\sum _{k}{\frac {1}{|a_{k}|^{p+1}}}}$$

Since ${\displaystyle p+1={\text{floor}}(\alpha )+1>\alpha }$, the last sum is finite.

g ≤ floor(ρ)

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

$${\displaystyle \sum _{k=1}^{\infty }\ln |E_{p}(z/a_{k})|\geq -B|z|^{\rho +\epsilon }}$$

for all large ${\displaystyle z}$ that avoids these forbidden disks. 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:

$${\displaystyle \sum _{|a_{k}|<|z_{k}|/2}\ln |E_{p}(z/a_{k})|+\sum _{|a_{k}|\geq |z_{k}|/2}\ln |E_{p}(z/a_{k})|}$$

The upper-bounding can be accomplished by the bounds (2), (3) on ${\displaystyle |E_{p}|}$, and the assumption that ${\displaystyle z}$ is outside every forbidden disk. Details are found in.^[2]

ρ ≤ g + 1

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:

$${\displaystyle \ln |f(z)|=Re(Q(z))+m\,Re(\ln z)+\sum _{k}\ln |E_{p}(z/a_{k})|.}$$

The first two terms are ${\displaystyle O(|z|^{p})=o(|z|^{g+1})}$. The third term is bounded by bound (4) of ${\displaystyle |E_{p}|}$:

$${\displaystyle \sum _{k}\ln |E_{p}(z/a_{k})|\leq B|z|^{q+1}\sum _{k}{\frac {1}{|a_{k}|^{q+1}}}.}$$

By assumption, ${\displaystyle p={\text{floor}}(\alpha )}$, so ${\displaystyle \sum _{k}{\frac {1}{|a_{k}|^{q+1}}}<\infty }$. Hence the above sum is ${\displaystyle O(|z|^{q+1})=O(|z|^{g+1}).}$