Show that, $ f(x)= e ^{-\pi|x|^2}\in\mathcal S (\mathbb R^d)$.

Question

I got this question while reading about Fourier Inversion Formula.

Can anyone help me to prove this:

Let $ f(x)= e ^{-\pi|x|^2}$. Show that $f\in \mathcal S (\mathbb R^d)$.

where $\mathcal S (\mathbb R^d)$ is a Schwartz Space.

Answer 1

Hint: show that for any multiindex $\alpha\in\mathbb N^d$

$$\partial_\alpha f (x) = P_\alpha(x)f(x)$$ Where $P_\alpha(x)$ is some polynomial . To this scope observe that
$$f(x) =\prod_{i=1}^d e^{-\pi x_i^2}$$ and Show that for any multiindex $\beta\in\mathbb N^d$

$$\lim_{|x|\to\infty}x^\beta f(x) = 0$$ conclude that for all multiindex $\alpha,\beta\in\mathbb N^d$

$$\lim_{|x|\to\infty}x^\beta \partial_\alpha f (x) = 0$$ that is $f\in\mathcal S(\mathbb R^d)$ since $f$ is $C^\infty$