Can n-th derivative of Schwartz function go to $\infty$ as $n\to\infty$?

Question

We define semi-norm (specificaly for $\varphi:{\mathbb R^M} \to {\mathbb R^N}$ if that matters) :

$$\|\varphi(x)\|_{\alpha,\beta}:={\sup}_{x\in\mathbb{R}^M}|x^{\alpha}D^{\beta}\varphi(x)|\,,$$

where $\alpha$,$\beta$ are multi-indices on ${\mathbb R^N}$ and $D^{\beta}$ is derivative with respect to multi-index $\beta$. The function $\varphi(x)$ belongs to Schwartz space only if the norm is finite.

My question is: Can we construct such a Schwartz function that $\|\varphi(x)\|_{\alpha,\beta}\to\infty$ for $\alpha\to\infty$ or $\beta\to\infty$?

Basicaly I am trying to proove that the norm is bounded for all $\alpha$, $\beta$ but not uniformly bounded.

Answer 1

Consider on the line that $\varphi_{n}(x)=x^{n}e^{-x^{2}}$, one can calculate the maximum of these (absolute value) functions are $(n/2)^{n/2}e^{-n/2}\rightarrow\infty$ as $n\rightarrow\infty$.

Answer 2

A bit of a hint that the answer is "yes" is the fact that we don't treat the Schwarz space as being a normed space topologically. Its topology is "generated by a family of seminorms" but that can't be stitched together into a norm, and this is precisely why.

One example to look at is $M=N=1,x=0,f(x)=e^{-x^2}$. The usual Maclaurin series of just $e^x$ gives terms of $\frac{(-x^2)^n}{n!}$, which means that in fact the absolute value of the $(2n)$th derivative at $x=0$ is $\frac{(2n)!}{n!}$.

A more abstract point of view is to note that a smooth function with derivatives bounded uniformly in both $n$ and $x$ is analytic on the whole space, yet non-analytic Schwarz functions are all over the place.