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