The Fourier transform is usually extended to the $L^2(\mathbb{R})$ space by invoking an argument that relies on the density of Schwartz functions in $L^2$.
Often, this extension is explicitly written as $$ \hat{f} = \lim_{n \rightarrow \infty} \hat{f}_n \quad \text{(in } L^2),$$ where $$\hat f_n(\nu) = \int_{[-n;+n]} f(t) e^{-2i\pi \nu t} \, dt,$$ with $f \in L^2$.
My question is: are the functions $\hat f_n(\nu)$ in that construction Schwartz? They are clearly $C^\infty$, but can one show that $\nu^k \hat f_n(\nu)$ vanishes at infinity for all $n$?
No, $f_n$ is certainly not a Schwarz function, even if $f$ is. (Regardless of whether it's been proved yet, the inversion theorem shows that ) $f_n=f\chi_{[-n,n]}$, not even continuous. (Hence $\hat f_n\notin L^1$.)
I don't see why it matters. I see people on MSE state that Plancherel is proved by using the Schwarz space this way, but I've never seen a book that actually takes that approach. Instead one just shows directly that $$||\hat f||_2=||f||_2\quad(f\in L^2\cap L^1),$$and then one notes that if $f\in L^2(\Bbb R)$ and $f_n=f\chi_{[-n,n]}$ then $f_n\in L^2\cap L^1$ and $||f-f_n||_2\to0$.