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 $n$?

  • $\begingroup$ @KaviRamaMurthy why are you invoking the derivatives of $f$? I did not assume differentiability of that function. $\endgroup$
    – user809418
    Oct 13, 2020 at 9:32
  • $\begingroup$ @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. $\endgroup$
    – user809418
    Oct 13, 2020 at 9:39
  • $\begingroup$ Yes, but here I still don't have a definition of the FT for an $L^2$ function. $\endgroup$
    – user809418
    Oct 13, 2020 at 9:43
  • $\begingroup$ @DavidC.Ullrich I did not read the question properly. Thank you for your comment. $\endgroup$ Oct 13, 2020 at 11:50
  • $\begingroup$ 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 $\mathcal{S}$? $\endgroup$
    – md2perpe
    Oct 13, 2020 at 12:37

1 Answer 1


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 $||f-f_n||_2\to0$.

  • $\begingroup$ Thanks for clarifying that. Yes, $f_n$ cannot be Schwartz, of course, but what about $\hat{f}_n(\nu)$ defined above? $\endgroup$
    – user809418
    Oct 13, 2020 at 11:55
  • $\begingroup$ @thetouristbr ??? All this is towards the Plancherel theorem, right? If we using the Schwarz space to prove Plancherel surely we've already shown that the FT is an isomorphism on the Schwarz space... $\endgroup$ Oct 13, 2020 at 12:14
  • $\begingroup$ My point is that I did not understand your $f_\nu$ notation. $\endgroup$
    – user809418
    Oct 13, 2020 at 12:18
  • $\begingroup$ (It seems that Duoandikoetxea's book has both approaches; Theorem 1.18, p15) $\endgroup$ Jun 4, 2021 at 6:28

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