Hadamard factorization of a function in the Fock space An entire function $F: \mathbb C \to \mathbb C$ belongs to the Fock space $\mathcal F^2$ if
$$
\int_{\mathbb C} |F(z)|^2e^{-|z|^2} \, dA(z) < \infty.
$$
It is well-known that every $F \in \mathcal F^2$ has order $\rho$ which satisfies $\rho \leq 2$, that is,
$$
\rho = \limsup_{r \to \infty} \frac{\log \log M(r)}{\log r} \leq 2, \quad M(r) = \sup_{\theta \in \mathbb R} |F(re^{i\theta})|.
$$
Moreover, if $\rho = 2$ then its type $\lambda$ satisfies $\lambda \leq 1/2$ (see e.g. here) where
$$
\lambda = \limsup_{r \to \infty} \frac{\log M(r)}{r^\rho}.
$$
Since $F$ is of finite order $\rho \leq 2$, we can look at its Hadamard factorization
$$
F(z)=z^m e^{az^2+bz+c} P(z), \quad a,b,c \in \mathbb C,
$$
with $P$ the canonical product corresponding of the zeros of $F$.
Questions: Is it true that the constant $a$ must satisfy $|a| < 1/2$? Or can $a$ be equal to $1/2$ and $F$ is still in the Fock space, i.e. it satisfies the integrability condition above?
 A: Yes, one can have $a=1/2$.
Let $P$ be an entire function of order $3/2$, normal type, whose indicator
$h$ has the properties $h(0)<0$, $h(\pi)<0$. Such a function exists, see for example
B. Ya. Levin, Distribution of zeros of entire functions, AMS, 1980, Chap II, section 4.
Then by continuity, the indicator is negative for $|\theta|<\delta$ and for $|\theta-\pi|<\delta$, with some positive $\delta$, which implies exponential decrease in these sectors. Since the exponential growth on any ray is at most $\exp|z|^{3/2}$,
we conclude that $\exp(z^2/2)P(z)$ is in the Fock class.
Remark. Actually $a$ can be any complex number.
I will assume wlog that $a$ is positive.
Then, $f(z)=e^{az^2}P(z)$ will be in the Fock space if the indicator $h$ of $P$
satisfies
$$h(\theta)<1-a\cos2\theta.$$
For example, one can take $h(\theta)=-a\cos2\theta$.
To construct a canonical product $P$ with such indicator, one chooses zeros $a_n$ satisfying
the following conditions:
$$\lim_{r\to\infty}\sum_{n: |a_n|\leq r}a_n^{-2}=-2a,$$
$$\sum |a_n|^{-2}=\infty,$$
and $\sqrt{n}=o(|a_n|)$. Such a sequence is easy to construct. In view of the second condition, the canonical product with zeros at $a_n$ will be of genus $2$, and the third condition assures that zeros have zero density. Now the first condition
implies that $P$ has the required indicator
by Theorem 2 in section 1, Chap. II of Levin's book.
