# Convolution square root of $\delta$

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?

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Obviously $\delta *\delta=\delta$. –  Hui Yu Dec 3 '12 at 14:53

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:
• $A=\mathbb{R}$, $f_A=\delta$.
• $A=\emptyset$, $f_A=-\delta$.
• $A=[0,\infty)$, $f_A=\dfrac{i}{\pi}\operatorname{Principal Value}\dfrac1x$.
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. –  Julián Aguirre Dec 4 '12 at 10:13