TITLE: Convolution square root of $\delta $
QUESTION [8 upvotes]: 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?

REPLY [6 votes]: 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:

$A=\mathbb{R}$, $f_A=\delta$.
$A=\emptyset$, $f_A=-\delta$.
$A=[0,\infty)$, $f_A=\dfrac{i}{\pi}\operatorname{Principal Value}\dfrac1x$.