# Extension of Fourier transform to $L^2$ by density of Schwartz functions

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 $$n$$?

• @KaviRamaMurthy why are you invoking the derivatives of $f$? I did not assume differentiability of that function. Oct 13, 2020 at 9:32
• @KaviRamaMurthy Okay, but I still don't understand this tip, since showing that the functions $\hat{f}_n$ are Schwartz is precisely what I want to do. Oct 13, 2020 at 9:39
• Yes, but here I still don't have a definition of the FT for an $L^2$ function. Oct 13, 2020 at 9:43
• @DavidC.Ullrich I did not read the question properly. Thank you for your comment. Oct 13, 2020 at 11:50
• You mention the "density of Schwartz functions in $L^2$". Doesn't this mean that $f_n\in\mathcal{S}$ so that $\hat f_n\in\mathcal{S}$? Or is your problem understanding that the Fourier transform maps $\mathcal{S}$ to $\mathcal{S}$? Oct 13, 2020 at 12:37

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 $$||f-f_n||_2\to0$$.
• Thanks for clarifying that. Yes, $f_n$ cannot be Schwartz, of course, but what about $\hat{f}_n(\nu)$ defined above? Oct 13, 2020 at 11:55
• My point is that I did not understand your $f_\nu$ notation. Oct 13, 2020 at 12:18