1
$\begingroup$

Let $f\in L^1(\mathbb{R})$. One may easily check that $$(*)~~~f', f''\in L^1(\mathbb{R})\Rightarrow \int_\mathbb{R}|\hat{f}| ~\text{is finite} \Rightarrow \int_\mathbb{R}\hat{f}(s)e^{2\pi is x}ds ~\text{is finite where} ~ x\in\mathbb{R} $$ Where $\hat{f}$ is the Fourier transform of $f$. Any aforesaid conditions in $*$ guarantees $f(x)=\int_\mathbb{R}\hat{f}(s)e^{2\pi is x}ds$ almost everywhere.

Q. Any other one(s) that expands the chain $(*)$ and guarantees the Fourier inversion formula.

$\endgroup$
6
  • $\begingroup$ What about integrability of both $f$ and $\hat{f}$ ? $\endgroup$ Commented Dec 2, 2020 at 16:23
  • $\begingroup$ He assumes the function in $L^1(\mathbf{R})$. $\endgroup$ Commented Dec 2, 2020 at 17:01
  • 5
    $\begingroup$ $f\in L^1\cap C^2$ doesn't imply that $\hat{f}\in L^1$ for the reason that you mentioned: because it would imply that $f$ is bounded whereas for $\phi \in C^\infty_c$, $f(x)=\sum_{n\ge 1} n \phi(n^3 (x-n))$ is unbounded. What implies $\hat{f}\in L^1$ is $f,f''\in L^1$ or $f,f'\in L^2$. $\endgroup$
    – reuns
    Commented Dec 2, 2020 at 20:20
  • $\begingroup$ As pointed out, your given assumption is not quite enough. The optimal condition, in a certain sense, is that $f$ should belong to $L^1(R)$ and to the Fourier algebra $A(R)$ (note that since functions in $A(R)$ are continuous and vanish at infinity, if they belong to $L^1(R)$ then they belong to every other $L^p(R)$). For general groups $G$ it is not always easy to find "simple" criteria that ensure a function belongs to $A(G)$, but if $G=R^d$ then Sobolev-type conditions will suffice as pointed out by @reuns $\endgroup$
    – Yemon Choi
    Commented Dec 2, 2020 at 22:13
  • $\begingroup$ See also the comments and answer to this older MO question mathoverflow.net/questions/159388/fourier-inversion $\endgroup$
    – Yemon Choi
    Commented Dec 2, 2020 at 22:15

1 Answer 1

1
$\begingroup$

Abstract nonsense: the Fourier inversion formula is valid on tempered distributions. Let $T\in \mathscr S'(\mathbb R^n)$ and define the Fourier transform $\hat T$ by the bracket of duality $$ \langle \hat T,\phi\rangle_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)}= \langle T,\hat\phi\rangle_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)}. $$ Let $\mathcal C$ be the map $T(x)\mapsto T(-x)$ (which makes sense weakly) and let $\mathcal F$ be the Fourier transformation. Then the Fourier transformation is an isomorphism of $\mathscr S'(\mathbb R^n)$ and we have $$ \mathcal C \mathcal F^2=Id, \quad [\mathcal C ,\mathcal F]=0, $$ and as a result $ T=\mathcal F \mathcal C \hat T $ for any tempered distribution. If $\hat T$ happens to belong to $L^1(\mathbb R^n)$, it is enough to get $T(x)=\int e^{2iπ x\cdot \xi} \hat T(\xi) d\xi$ and $T$ is a continuous function with limit 0 at infinity: note that there is something to prove, but it is easy by using test functions in the Schwartz class. To sum-up, Fourier inversion formula is always true for tempered distributions and when the integrals make sense they coincide with the weak definition. Note also that the proof of the abstract nonsense argument is trivial and follows from the inversion formula in $\mathscr S(\mathbb R^n)$ since $$ \langle \mathcal C\mathcal F^2 T,\phi \rangle_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)}= \langle \mathcal F^2 T,\mathcal C\phi \rangle_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)} = \langle T,\underbrace{\mathcal F^2\mathcal C\phi}_{=\phi} \rangle_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)}. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.