Let $f$ be an entire function such that $\int_{\mathbb{C}}|f(z)|^2e^{-|z|^2} <\infty$ (with Lebesgue measure on $\mathbb{C}$). I need to prove that $f(z)$ has order $\le 2$.
My ideas:
- Try to find bounds on coefficients and derive information about order from this bound.
- Try to rewrite integral as integral over integrals over circles and with some tricks like moving constant from the right under the integral by using $1 = \int_0^\infty (-a)e^{(-a)|z|}$ and connecting result that some integral is zero with order of function
- Try to find some information about areas where function is "bad" and if they are not empty find some contradiction.
None of them were successful, so i'm asking for hints/help.
Best Answer
Combine idea 1 with the start of idea 2.
On the circle $\lvert z\rvert = r$, writing $z = re^{i\varphi}$ yields $$\lvert f(z)\rvert^2 = \sum_{m,n = 0}^{\infty} a_n\overline{a_m} r^{n+m} e^{i\varphi(n-m)}\,.$$
Plugging this into the integral and using polar coordinates gives \begin{align} \int_{\mathbb{C}} \lvert f(z)\rvert^2 e^{-\lvert z\rvert^2}\,d\lambda &= \int_0^{\infty} \int_0^{2\pi} \sum_{n,m = 0}^{\infty} a_n\overline{a_m} r^{n+m} e^{i\varphi(n-m)}\,d\varphi\; e^{-r^2} r\,dr \\ &= \pi \int_0^{\infty} \sum_{n = 0}^{\infty} \lvert a_n\rvert^2 r^{2n} e^{-r^2}\: 2r\,dr \\ &= \pi \sum_{n = 0}^{\infty} \lvert a_n\rvert^2 \int_0^{\infty} u^n e^{-u}\,du \\ &= \pi \sum_{n = 0}^{\infty} \lvert a_n\rvert^2 \cdot n!\,. \end{align}
In particular, $\sqrt{n!}\,\lvert a_n\rvert$ is bounded. From this you can deduce that the order of $f$ is at most $2$ (e.g. using the argument here).