Normability of $\mathscr{S}(\mathbb{R}^n)$ Let $\mathscr{S}(\mathbb{R}^n)$ be the space of Schwartz functions. In this post https://mathoverflow.net/questions/218023/the-schwartz-space-is-not-normable we are provided with an example of a Schwartz function which converges in some of the semi-norms, but not all. The function is $$f(x) = \frac{e^{-\left| x \right|^2} \sin (Nx_1)}{N^k}.$$ The semi-norms on $\mathscr{S}(\mathbb{R}^n)$ are given by $$\rho_{\alpha \beta} = \sup_{x \in \mathbb{R}^n} \left| x^{\alpha} \partial^{\beta}(f)(x) \right|,$$ where $\alpha$ and $\beta$ are multi-indices. 
I've attempted to go through this counter example explicitly, but do not understand how to compute the $\rho_{\alpha\beta}$ quantities explicitly. Could someone assist me in the explicit computations of $\rho_{\alpha \beta}(f)$?
 A: You only need to differentiate with respect to the variable $x_1$. If you have two functions $g$ and $h$ of one variable, then the $n$th derivative of the product is given by
$$\frac{d^n (gh)}{dt^n}(t)=\sum_{i=0}^n\binom{n}{i}\frac{d^i g}{dt^i}(t)\frac{d^{n-i} h}{dt^{n-i}}(t).$$
So if you differentiate the function $f$ $n$ times with respect to $x_1$ you are going to get $$\frac{\partial^n f}{\partial x^n_1}(x)=\sum_{i=0}^n\binom{n}{i}\frac1{N^k}\frac{\partial^i }{\partial x^i_1}(\sin(Nx_1))\frac{\partial^{n-i}}{\partial x_1^{n-i}}(e^{-|x|^2}).$$
Note that $|\frac{\partial^i }{\partial x^i_1}(\sin(Nx_1))|=N^i|\sin(Nx_1)|$ if $i$ is odd and $|\frac{\partial^i }{\partial x^i_1}(\sin (Nx_1))|=N^i|\cos(Nx_1)|$ if $i$ is even. Assume that $n> k$ is even. Then isolating the term $i=n$ we get 
$$\left\vert\frac{\partial^n f}{\partial x^n_1}(x)\right\vert\ge N^{n-k}|\cos(Nx_1)|e^{-|x|^2}-\sum_{i=0}^{n-1}\binom{n}{i}N^{i-k}\left\vert\frac{\partial^{n-i}}{\partial x_1^{n-i}}(e^{-|x|^2})\right\vert.$$
Since $e^{-|x|^2}$ decays faster than any polynomial, you will have that 
 $\left\vert\frac{\partial^{n-i}}{\partial x_1^{n-i}}(e^{-|x|^2})\right\vert\le C_n$ for all $x$ and so
$$\left\vert\frac{\partial^n f}{\partial x^n_1}(x)\right\vert\ge N^{n-k}|\cos(Nx_1)|e^{-|x|^2}-d_nN^{n-1-k}.$$
Taking $x=0$ gives 
$$\left\vert\frac{\partial^n f}{\partial x^n_1}(0)\right\vert\ge N^{n-k}1-d_nN^{n-1-k}\to\infty $$ as $N\to\infty$.
To prove that the space is not normable you just need to show that some of the norms do not tend to zero.
A: $$\dfrac{d^k}{d x^k} e^{-x^2} = (-1)^k H_k(x) e^{-x^2}$$
where $H_k$ is the $k$'th Hermite polynomial.  Now 
$$\sup_{x \in \mathbb R} \left|x^j \dfrac{d^k}{dx^k} e^{-x^2} \right| = \left| t^j H_k(t) e^{-t^2}\right|$$
where $t$ is a zero of
$$ \eqalign{\dfrac{d}{dx} \left(x^j \dfrac{d^k}{dx^k} e^{-x^2}\right) &= j x^{j-1} \dfrac{d^k}{dx^k} e^{-x^2} + x^j \dfrac{d^{k+1}}{dx^{k+1}} e^{-x^2}
\cr &= (-1)^k j x^{j-1} H_k(x) e^{-x^2} + (-1)^{k+1} x^j H_{k+1}(x) e^{-x^2}\cr}$$
and thus either $0$ or a zero of $j H_k(x) - x H_{k+1}(x)$.  In general, I don't think you'll get a useful explicit solution, but there may be bounds
or asymptotic approximations.
