8
$\begingroup$

I want to somehow classify the distributional solutions of the equation $$ f \ast f = \delta $$ where $\delta = \delta _0$ is the Dirac delta distribution. Clearly, by Fourier transformation, we have $$ \widehat{f}^2 = 1, $$ but I'm wondering whether it is possible to obtain a more explicit solution?

$\endgroup$
2
  • 5
    $\begingroup$ Obviously $\delta *\delta=\delta$. $\endgroup$
    – Hui Yu
    Dec 3 '12 at 14:53
  • $\begingroup$ This is a related question (on MathOverflow). $\endgroup$ Aug 30 '19 at 6:25
6
$\begingroup$

Let $A\subset\mathbb{R}$ be a measurable set. Define $$ f_A=\mathcal{F}^{-1}\bigl(\chi_A-\chi_{\mathbb{R}\setminus A}\bigl), $$ where $\mathcal{F}$ denotes the Fourier transform and $\chi_B$ is the characteristic fnction of the set $B$. Then $$ f_A\ast f_A=\delta. $$ Some explicit examples are:

  • $A=\mathbb{R}$, $f_A=\delta$.
  • $A=\emptyset$, $f_A=-\delta$.
  • $A=[0,\infty)$, $f_A=\dfrac{i}{\pi}\operatorname{Principal Value}\dfrac1x$.
$\endgroup$
2
  • $\begingroup$ Thank you Julián. But how can I be sure that all solutions are of that form? $\endgroup$
    – flavio
    Dec 4 '12 at 8:26
  • 2
    $\begingroup$ From $(\hat f)^2=1$ it follows that $f(x)=\pm1$ for almost every $x\in\mathbb{R}$. Let $A^\pm=\{x:f(x)=\pm1\}$. Then $\mathbb{R}\setminus(A^+\cup A^-)$ is of measure zero. $\endgroup$ Dec 4 '12 at 10:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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