Davenport's proof of the Hadamard theorem In Multiplicative Number Theory, H. Davenport proves the following theorem in the paragraph §11 (page 74) :

Theorem : (Hadamard)
Let $f:\mathbb{C}\to\mathbb{C}$ be entire of order $\leqslant1$ (that is, $f(z)\in O(\exp[s^{1+\varepsilon}])$ for all $\varepsilon>0$) with $f(0)\neq0$. Then, there exists constants $A,B$ such that $$f(z)=e^{A+Bz}\prod_{\rho}\left(1-\frac{z}{\rho}\right)e^{z/\rho},$$
where the product ranges over all zeroes of $f$.

This theorem makes use of Jensen's formula. In its proof, he tries to show that the product part, defined as $P(z)$, is such that $f/P$ is entire of order $\leqslant1$. Since it has no zeroes (those of $f$ are cancelled by the product), it will turn out that $f/P$ must have the form $e^{A+Bz}$. However, there are two things I'm concerned about :

*

*He has shown that $|P(z)|>\exp(R^{-1+3\varepsilon})$ on $|z|=R_n$, for infintely many $R_n$ where $R_n\underset{n\to+\infty}{\longrightarrow}+\infty$. He then concludes that $f/P$ must have order $\leqslant1$. But for this, we would need to prove that the $\limsup$ of $|\exp(-z^{1+4\varepsilon})f(z)/P(z)|$ is finite. This only proves that one accumulation point is finite, doesn't it ? How to prove that the $\limsup$ is finite ?*


*When handling the case for $P_3(s)$, he basically writes that $(1-u)e^{-u}\geqslant e^{-cu^2}$ for some constant $c$, where $u\in]0,1/2[$. (There used to be a typo here in my original question, which I commented about in Daniel's answer...) However, this seems to be impossible ; I tried computing $\displaystyle\inf_{0<u<1/2}(1-u)e^{-u+cu^2}$ for arbitrarily large values of $c$ with Wolfram, and it seems like it's always $<1$ for some $u$ close enough to zero (corresponding to zeroes of $f$ infinitely far away to the origin). Am I getting something wrong, or is this really a mistake ?
I have uploaded the extracted pages from the book so people don't owning a digital/printed copy of the book can still follow along. Is this against the rules ? If so, I'll just remove it.
 A: I haven't looked at how Davenport argues, but if I interpret rain1's comment correctly, he earlier remarked that it suffices to have the estimate on some unbounded family of radii. There are many ways to see that. One can use Hadamard's three circle theorem to interpolate between $R_n$ and $R_{n+1}$. Alternatively, with
$$\frac{f(z)}{P(z)} = e^{g(z)}$$
the estimates on the circles $\lvert z\rvert = R_n$ give upper bounds for $\operatorname{Re} g(z)$ on these circles. Then for example the Borel–Carathéodory theorem gives you upper bounds for $\lvert g(z)\rvert$ on discs, say with radius $R_n/2$, from which you can obtain the desired bounds. Another method uses properties of (real-valued) harmonic functions. The upper bounds for $\operatorname{Re} g(z)$ on the circles $\lvert z\rvert = R_n$ then show that that is a polynomial (in $x,y$) whose degree is in this case at most $1$. From that it follows that $g$ is a polynomial (in $z$) of the same degree. (Pedantic note: If $\operatorname{Re} g(z) \equiv 0$, then $g$ may have degree $0$, which is larger than the degree of the zero polynomial.)
Concerning the second point, it is useful to take the logarithm. We have
\begin{align}
\log \bigl((1-u)e^{u}\bigr)
&= u + \log (1-u) \\
&= u - \sum_{k = 1}^{\infty} \frac{u^k}{k} \\
&= -\sum_{k = 2}^{\infty} \frac{u^k}{k} \\
&> -\frac{u^2}{2} \sum_{m = 0}^{\infty} u^m \\
&= -\frac{u^2}{2(1-u)} \\
&> -u^2
\end{align}
for $u \in \mathopen{]}0, 1/2\mathclose{[}$, thus every $c \geqslant 1$ works.
Regarding the edit, the minus sign in the exponent is a typo. For $(1-u)e^{-u} = 1 - 2u + O(u^2)$ we cannot have a lower bound of the form $e^{-cu^2} = 1 - cu^2 + O(u^4)$, and since there may be infinitely many factors to estimate, a bound $(1 - u)e^{-u} \geqslant ae^{-cu^2}$ with an $a \in (0,1)$ doesn't help.
The factors in the product have the form $(1 - u)e^{u}$ with $u = \frac{z}{\rho}$. The factor $e^{z/\rho}$ in that guarantees convergence of the product by ensuring that the full factors have an asymptotic expansion of $1 - O\bigl(\frac{z^2}{\rho^2}\bigr)$, and $\sum \frac{1}{\lvert \rho\rvert^2}$ converges for an entire function of order $\leqslant 1$, while $\sum \frac{1}{\lvert \rho\rvert}$ may diverge. For the estimate of the tail product
$$\prod_{\lvert \rho\rvert > 2R}\biggl(1 - \frac{z}{\rho}\biggr)e^{z/\rho}$$
we must not take the modulus of $z/\rho$ too early, we must estimate the factors as they stand and only replace $z/\rho$ with its modulus later.
The first three lines of the above computation are valid for complex $u$ with $\lvert u\rvert < 1$ (using the principal branch of the logarithm), thus we have
$$\bigl\lvert \log \bigl((1 - u)e^{u}\bigr)\bigr\rvert = \Biggl\lvert \sum_{k = 2}^{\infty} \frac{u^k}{k}\Biggr\rvert \leqslant \sum_{k = 2}^{\infty} \frac{\lvert u\rvert^k}{k} \leqslant \lvert u\rvert^2$$
for all complex $u$ with $\lvert u\rvert \leqslant \frac{1}{2}$. From that it follows that
$$- \operatorname{Re} \log \bigl((1-u)e^{u}\bigr) \leqslant \lvert u\rvert^2$$
and therefore
$$\bigl\lvert (1-u)e^{u}\bigr\rvert \geqslant e^{- \lvert u\rvert^2}$$
for these $u$. Thus
$$\Biggl\lvert \prod_{\lvert \rho\rvert > 2R} \biggl(1 - \frac{z}{\rho}\biggr)e^{z/\rho}\Biggr\rvert \geqslant \exp \Biggl(- \sum_{\lvert \rho\rvert > 2R} \frac{\lvert z\rvert^2}{\lvert \rho\rvert^2}\Biggr)\,.$$
In that, we have $\lvert z\rvert = R$, and then we use bounds for the tail sum
$$\sum_{\lvert \rho \rvert > 2R} \frac{1}{\lvert \rho\rvert^2}$$
to obtain the desired estimate.
