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
$$
f\cdot\hat f=0 \qquad \text{Lebesgue-almost everywhere},
$$
where $\hat f$ is the Fourier transform of $f$ and $\cdot$ is the pointwise product?**


I expect the answer to be no, but it seems hard to prove. **Edit: Actually the answer is yes, and one can even choose $f$ to be a Schwartz function, see below.**


Note that if $f\in L^2(\mathbb R)$ or if $f\in L^1(\mathbb R)$ is sufficiently nice then $\cdot$ can equivalently be replaced by convolution $\ast$.


It is known (see [this MO question](https://mathoverflow.net/questions/275072/function-and-fourier-transform-vanish-on-an-interval)) that there is a Schwartz function $f$ on $\mathbb R$ such that both $f$ and $\hat f$ vanish on some open interval. This is similar to but obviously weaker than the above question.


**Edit: as pointed out in the comments, it is easy to see that the existing answers to the MO question referenced above also answer my question affirmatively.**



Vanishing of the product of a function and its own Fourier transform
I had to stratch my head a bit to decipher what we wrote 4 years ago but there it is


* Take $\hat{\phi} \in C^\infty\_c(\Bbb{R})$ supported on $(1/2,1)$
* Take $h \in C^\infty(\Bbb{R})$ and $1$-periodic supported on $[0,1/2]+\Bbb{Z}$, let $c\_n=\int\_0^1 h(x)e^{-2i\pi nx}dx$. Then


$$\hat{h}=\sum\_n c\_n \delta(y-n)$$ So that


$$\hat{h}\ast \hat{\phi}= \sum\_n c\_n \hat{\phi}(y-n)$$ It is Schwartz because $(c\_n)$ is rapidly decreasing, it is also non-zero and supported on $(1/2,1)+\Bbb{Z}$.


On the other hand its inverse Fourier transform $h\phi$ is supported on $[0,1/2]+\Bbb{Z}$.


