I am trying rigorously to prove that $e^{-|x|^2}$ on $\mathbb{R}^n$ is in the Schwartz space but not in the space of infinitely differentiable functions with compact support.
The function $e^{-|x|^2}$ is in $\mathcal{S}$.
Proof: From the fact that for any polynomial $p$, $p(x)e^{-|x|^2}$ is bounded and for any multi index $\beta$, $D^{\beta}(e^{-|x|^2})=p_{\beta}(x)e^{-|x|^2}$ for some polynomial $p_{\beta}$ we have to $\sup_{x\in \mathbb{R}^n} |x^{\alpha}D^{\beta}(e^{-|x|^2})|<\infty$.
Affirmation 1. $(\forall \beta$ multi index)($\exists p_{\beta}$ polynomial)($D^{\beta}(e^{-|x|^2})=p_{\beta}(x)e^{-|x|^2})$
Proof: It is true for $|\beta|=0$ and $|\beta|=1$.
Suppose true for any multiindex with large $l$. Let $\gamma=\beta+\epsilon$ with $\epsilon=(0,\ldots, 1,0,\ldots )$ with 1 in the $j$-esime position.
\begin{eqnarray} D^{\gamma}(e^{-|x|^2})&=&D^{\epsilon} D^{\beta}(e^{-|x|^2}\\ &=&\partial_{x_j}(p_{\beta}(x)e^{-|x|^2})\\ &=&[\partial_{x_j}(p_{\beta}(x)]e^{-|x|^2}+p_{\beta}(x)[\partial_{x_j}(e^{-|x|^2})\\ &=&q_{\beta}(x)e^{-|x|^2}+p_{\beta}(x)(-2x_j)e^{-|x|^2} \end{eqnarray}
Affirmation 2. For any $p$ polynomial, $p(x)e^{-|x|^2}$ is bounded.
Proof: $\lim_{|x|\to\infty} \frac{p(x)}{e^{|x|^2}}=0$ then for any $M>0$ exists $R>0$: $|x|>R$ implies $|p(x)e^{-|x|^2}|<M$.
Therefore, in $\left\{x\in\mathbb{R}^n:|x|>R\right\}$, $p(x)e^{-|x|^2}$ is bounded.
From the fact that $\left\{x\in\mathbb{R}^n:|x|\leq R\right\}$ is compact and $p(x)e^{-|x|^2}$ continuous we have that $p(x)e^{-|x|^2} $is bounded for $|x|\leq R.$
Now I would like to prove that the function $e^{-|x|^2}$ is not in $\mathcal{C}_{0}^{\infty}(\mathbb{R}^n)$. Is it because the function is not differentiable at the origin?
Thanks.
