I want to find all $\varphi\in\mathcal S(\mathbb R)$ for which the sequence $\varphi_n(x):=\frac{\varphi(nx)}{n}$ converges in $\mathcal S(\mathbb R)$.
The first step, I have already managed to do by myself: $\forall n\in\mathbb N, \varphi_n\in\mathcal S(\mathbb R)$. I used the substitution $x'=nx, \frac{\partial }{\partial x'}=n\frac{\partial }{\partial x}$: Then we have
$|\varphi_n|_{l,m}=\sup_{x\in\mathbb R}{|x^l\partial^m\frac{\varphi(nx)}{n}|}\\=\sup_{x'\in\mathbb R}{|\frac{1}{n}(\frac{x'}{n})^l (n\frac{\partial }{\partial x'})^m \varphi(x')}|\\ =\sup_{x'\in\mathbb R}{|n^{m-l-1}x'^l (\frac{\partial }{\partial x'})^m\varphi(x')}|<\infty,\ \forall n\in\mathbb N$
But now I'm stuck on showing the convergence: My guess is, that above statement holds for all $\varphi\in\mathcal S(\mathbb R)$. Has anyone an idea, how to show that?
Any help is appreciated.
Assume that the sequence $(\varphi_n,n\geqslant 1)$ converges in the Schwartz space. Since $\varphi$ is bounded, the limit function is necessarily $0$. In particular, we have $\sup_{x\in\mathbb R}|\varphi''_n(x)|\leqslant M$ for some $M$ independent on $n$, hence $\varphi''=0$ and $\varphi(x)=ax+b$. Since $\varphi\in\mathcal S(\mathbb R)$, we should have $a=b=0$.