The Schwartz space $\mathcal{S}(\Bbb{R})$ consists of all indefinitely differentiable functions $f$ such that for all $\ell, k \geq 0$, we have
$$
\sup_{x \in \Bbb{R}} |x|^k |f^{(\ell)}(x)| \lt \infty
$$
i.e. $f$ and all of its derivatives are rapidly decreasing in that sense.
How do I show that this is true of $f(x) = e^{-x^2}$?
I know that $\mathcal{S}(\Bbb{R})$ is closed under multiplication by polynomials in $x$ over $\Bbb{C}$ and that it is also a vector space over $\Bbb{C}$. Obviously it's closed under taking derivatives too.
We have that $f^{(\ell)}(x) = P(x) e^{-x^2}$ for some polynomial $P$.
Best Answer
$f^{(\ell)}(x) = P_{\ell}(x) e^{-x^2}$ and $|P_{\ell}(x)|$ is bounded by $C_{\ell}|x|^d$, where $C_{\ell}$ is some constant and $d \geq$ degree of $P_{\ell}$. Let $d$ be even for convenience (so we can divide by $2$).
We can write $e^{x^2} = 1 + x^2 + (x^2)^2/2! + \dots$ so that $e^{x^2} \geq 1 + (x^2)^{d/2}/(d/2)!$ (all terms are positive) and $e^{-x^2} \leq \frac{1}{1+x^d/(d/2)!} \implies |e^{-x^2}| \leq |\frac{1}{1+x^d/(d/2)!}|$.
So we have $\sup_{x \in \Bbb{R}}|x|^k|f^{\ell}(x)| \leq \sup C_{\ell}\left | \frac{x^d}{1 + \frac{x^{d}}{(d/2)!}} \right | \leq C_{\ell}\left | \frac{x^d}{x^d / (d/2)!} \right | \leq C_{\ell}(d/2)!$
Thus $f(x) = e^{-x^2}$ and all of its derivatives are rapidly decreasing in the Schwartz space sense and so $f$ belongs to $\mathcal{S}(\Bbb{R})$.