Invariance of Schwartz space under unitary group of Laplacian.

Question

Let $\mathcal S(\mathbb R^n)$ denote the Schwartz space of rapidly decreasing functions and let $H = -\Delta$ denote the free Schrödinger operator on $L^2(\mathbb R^n)$ and $U(t) = e^{-itH}$ the associated unitary group. How does one show that $\mathcal S(\mathbb R^n)$ is invariant under $U(t)$?

Answer 1

Assuming you know how the Fourier transform plays with the Schwarz space and these operators:

First you want to show this: If $m\in C^\infty$ and $D^\alpha m$ has polynomial growth for every $\alpha$ then $f\mapsto mf$ takes the Schwarz space to itself.

Now if you take the Fourier transform of your problem you see that you just have to show that if $m(\xi)=e^{it|\xi|^2}$ then every partial of $m$ has polynomial growth. This is easy; compute a few partials to see why.

Background: Say $S$ is the Schwarz space and $F$ is the Fourier transform. Then $F$ is an isomorphism of $S$ (bounded, with a bounded inverse). So a given operator $T$ maps $S$ to $S$ if and only if $F^{-1}TF$ does so.

Now $F^{-1}HF$ is just multiplication by $|\xi|^2$. Hence if $P$ is any polynomial, $F^{-1}P(H)F=P(F^{-1}HF)$ is multiplication by $P(|\xi|^2)$. Taking limits with an appropriate sequence $P_n$ shows that $F^{-1}U(t)F$ is multiplication by $m$.