schwarz class and $L^2(\mathbb{R})$

Question

Schwarz and $L^2$ both have the property that the Fourier transform is defined and bijective as a self-map of these spaces. Are they related in anyway or is this coincidence? (i.e. dual in some sense, or perhaps both arising from some more general construction.)

Answer 1

$L^2$ is where it all starts; from it new such spaces (including $\mathcal S$, for Schwartz with the t) can be constructed in a routine way. Let $w$ be a locally integrable function on $\mathbb R$ such that $\inf w>0$. The space $$S_w=\{f: wf\in L^2 \ \text{ and } \ w\hat f\in L^2\}$$ is a subset of $L^2$ on which the Fourier transform $\mathcal F$ is a bijection.

If $w\equiv 1$, we have $L^2$ itself. Another example is given by $w(x)=(1+x^2)^{p/2}$, for any $p>0$.

The property of being mapped onto itself by $\mathcal F$ is preserved under intersection of function spaces. Taking the intersection of $S_w$ with $w(x)=(1+x^2)^{p/2}$ over all $p>0$, we get $\mathcal S$. (To see that this agrees with the usual definition, observe that derivatives remain in this intersection, and that the integral norm of a sufficiently high derivative controls the supremum of a function.)

We get something smaller than $\mathcal S$ with $w=\exp{|x|}$; this space is still nontrivial (contains the Gaussian function). I suspect that beyond $w=\exp(x^2)$ the space $S_w$ becomes trivial, but don't have a proof.