We were given the expression $\hat S(\xi, t) = e^{\, i|\xi|^2t}$, where $\xi \in \mathbb R^n$. If my calculation is correct, after the reverse Fourier transform, we have: $$S(x, t) = \left( \frac{i}{2t} \right)^{n/2} e^{\, \frac{-i|x|^2}{4t}}$$ Anyway, a tempered distribution is a continuous linear functional on the Schwartz space, but to me, $S(x, t)$ is just some function except when $t = 0$. So, how does $S(x, t)$ act on a test function?
Furthermore, how does this related to the $H^s$ spaces and their duals?
\begin{align} H^s & = \left\{ \, u \in L^2(\mathbb R^n) : (1 + |\xi|^2)^{s/2} \hat u(\xi) \in L^2(\mathbb R^n) \, \right\} \\ & = \left\{ \, u \in L^2(\mathbb R^n) : D^\alpha(u) \in L^2(\mathbb R^n) \text{ where } |\alpha| < s \, \right\} \end{align}
Thanks for the help. I'm lost.
Recall that every tempered distribution has a Fourier transform defined by $$ <\hat{T},\varphi> = <T,\hat{\varphi}>. $$ Similarly, we can define the inverse Fourier transform of any tempered distribution. Consequently, it suffices to prove that your "transformed" function $\hat{S}$ is a tempered distribution. Since it is actually a smooth and bounded function its action as a tempered distribution is just integration, i.e. it's the map $T_t$ given by $$ <T_t,\psi> = \int_{\mathbb{R}^n} \hat{S}(\xi,t) \psi(\xi) d\xi, $$ which it's not too hard to show is actually a tempered distribution. Once you have this then you can just do your computation to get the formula for the action of $S$, the inverse Fourier transform.