How to show that $e^{-ax^2}$ is an element of $\mathcal{S}(\mathbb{R})$?

Question

The space $\mathcal{S}(\mathbb{R})$ is defined as the space of all functions $f : \mathbb{R}\to \mathbb{R}$ such that

$$\sup_{x\in \mathbb{R}} |x|^k |f^{(l)}(x)| <\infty,$$

for all $k,l\in \mathbb{N}$. That is to say that if we fix any $k,l\in \mathbb{N}$, there is $A_{k,l}\in \mathbb{R}$ such that

$$|x|^k |f^{(l)}(x)|\leq A_{k,l}.$$

Now, it is usually said that $e^{-ax^2}$ is an element of $\mathcal{S}(\mathbb{R})$. But how can we prove this?

I've been thinking about this for a while and I have no idea on what has to be done. The function $e^{-ax^2}$ is bounded by $1$, but that is certainly not enough, since it covers just the case $k = 0$, $l = 1$.

How does one prove that $e^{-ax^2}$ is an element of $\mathcal{S}(\mathbb{R})$?

Answer 1

In fact, you need to show that any derivative of $e^{-ax^{2}}$ is of type $e^{-ax^{2}}P(x)$ for some polynomial $P$, now observe that $e^{-ax^{2}}$ decreases faster than any polynomial, you may use L'Hopital to deduce this fact.