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Vanishing of the product of a function and its own Fourier transform
I have found the following question to be surprisingly hard:
Is there a non-zero $f\in L^1(\mathbb R)$ or $f\in L^2(\mathbb R)$ such that
$$
f\cdot\hat f=0 \qquad \text{Lebesgue-almost everywhere},
$$
...
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