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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5$\begingroup$ Obviously $\delta *\delta=\delta$. $\endgroup$– Hui YuDec 3, 2012 at 14:53
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$\begingroup$ This is a related question (on MathOverflow). $\endgroup$– Giuseppe NegroAug 30, 2019 at 6:25
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1 Answer
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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:
- $A=\mathbb{R}$, $f_A=\delta$.
- $A=\emptyset$, $f_A=-\delta$.
- $A=[0,\infty)$, $f_A=\dfrac{i}{\pi}\operatorname{Principal Value}\dfrac1x$.
answered Dec 3, 2012 at 15:55
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$\begingroup$ Thank you Julián. But how can I be sure that all solutions are of that form? $\endgroup$– flavioDec 4, 2012 at 8:26
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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, 2012 at 10:13