Prove that if $F$ is in the Bargmann-Fock space. Then so is $\exp\cdot F$ The Bargmann-Fock $\mathcal{F}^2(\mathbb{C})$ is the Hilbert space of entire functions $F:\mathbb{C}\to\mathbb{C}$ such that $\int_{\mathbb{C}}|F(z)|^2e^{-\pi|z|^2}\,dz<\infty$.
I now want to prove that if $F\in\mathcal{F}^2(\mathbb{C})$ then it holds that $\exp\cdot F\in\mathcal{F}^2(\mathbb{C})$.
For this let $F\in\mathcal{F}^2(\mathbb{C})$. Then
\begin{align*}
  \|\exp\cdot F\|^2_{\mathcal{F}^2}=&\int_{\mathbb{C}}|e^zF(z)|^2e^{-\pi|z|^2}\,dz\\
  =&\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}|e^{x+iy}F(x+iy)|^2e^{-\pi(x^2+y^2)}\,dxdy\\
  =&\int_{-\infty}^{\infty}e^{-\pi y^2}\int_{-\infty}^{\infty}e^{2x}|F(x+iy)|^2e^{-\pi x^2}\,dxdy\\
 \end{align*}
I then tried to show that the inner integral converges and is uniformly bounded (since it technically depends on y), which I failed to do.
Showing this would prove the statement since $\int_{-\infty}^{\infty}e^{-\pi y^2}dy=1\\$.
 A: The claim in the OP is not true. Note that there are functions $f$ in the Fock space for which $zf(z)$ is not in the Fock space; eg $$f(z)=\frac{\sin \pi z^2/2}{z^2}$$ and $e^z$ is much bigger than $z$ from that point of view.
So for example for $f$ above, choosing a domain $D$ given by  $z=Re^{i\theta}, \pi/4-\delta \le \theta \le \pi/4 +\delta, \delta \le 1/R$ for $R >1$ say, one easily sees that $|e^zf(z)|$ grows like $C\frac{e^{\pi R^2 (\sin 2\theta)+aR}}{R^4}, C, a>0$ for $z \in D$ so $$\int_{\mathbb{C}}|e^zF(z)|^2e^{-\pi|z|^2}dA \ge \int_1^{\infty}\int_{\pi/4-\delta(R) }^{\pi/4+\delta(R)}C\frac{e^{-\pi R^2 (1-\sin 2\theta)+aR}}{R^3}d\theta dR$$
Since on $D$ we have $e^{-\pi R^2 (1-\sin 2\theta)} \ge k$ as $1-\sin 2\theta \le C/R^2$ by our choices, we get that
$$\int_{\mathbb{C}}|e^z f(z)|^2e^{-\pi|z|^2}dA \ge \int_1^{\infty}\int_{\pi/4-\delta(R)}^{\pi/4+\delta(R)}C_2\frac{e^{aR}}{R^3}d\theta dR \ge $$ $$\ge C_3\int_1^{\infty}\frac{e^{aR}}{R^4}dR =\infty$$ which proves our claim that $e^zf(z)$ is not in the Fock space, while the proof that $f$ is the Fock space follows from the fact that $$\int_1^{\infty}\int_{0}^{2\pi}\frac{1}{R^3}d\theta dR < \infty$$ and that $|f(z)|^2e^{-\pi|z|^2} \le C/|z|^4, |z| \ge 1$ say.
Note that if $f$ is in the Fock space, so it must have order at most $2$ and type at most $\pi/2$, is not precisely on the extremal order, type combination of $2, \pi/2$ the result clearly holds by simple majorization as indeed $e^z$ is then small enough, not to increase order or type, so we are by default in the Fock space. But for the extremal $2, \pi/2$ combination of order, type, some functions belong and some do not in the Fock space and there $e^z$ counts as we saw above.
The paper Extremal Problems in the Fock Space, by Beneteau, Carswell, Kouchekian, p193 at the start of part $3$ gives references for the more powerful claim that if $f M \subset M$ for a fixed $M \ne 0$ closed subspace of the Fock space, then $f$ constant.
