Given a compact and sufficiently regular set $\Omega\subset\mathbb{R^n}$ and its characteristic function $\chi=\chi_\Omega$, I would like to conclude that the (inverse) Fourier transform $\check{\chi}$ reproduces itself under convolution, $\check{\chi}*\check{\chi}=\check{\chi}$ since $\chi^2=\chi$ and formally convolution corresponds to multiplication. I try to apply the formula \begin{align} \widehat{f*g}=\hat f \hat g \end{align} with $f=g=\check \chi$, but I only know of its validity for $(f,g)$ in $(L^1,L^1)$, $(D',D)$, or $(S',S)$. However, in my situation $\check \chi$ is not integrable, much less a Schwartz function, so it does not fit into these cases.
Can one nevertheless show that $\check{\chi}*\check{\chi}=\check{\chi}$ under appropriate conditions on $\Omega$?
$\check\chi$ is after all in $L^2$ (and, in the case that I need, in $O(\|x\|^{-\frac{n+1}{2}})\in L^{\frac{2n}{n+1}}_w$), so the convolution makes sense.
As a special case, we have in one dimension that $\check\chi_{[-1,1]}(x)=C\frac{\sin x}{x}=:C\operatorname{sinc} x$. Is there an abstract argument which proves that $C \operatorname{sinc}*\operatorname{sinc}=\operatorname{sinc}$, or can one only do the convolution integral explicitly?