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Suppose that $f\in L^2(\mathbb{R})$ is non-zero and compactly supported. Then its Fourier transform $\widehat{f}\neq 0$ is analytic, and in particular the nodal set $\{\xi\in\mathbb{R}\,s.t.\,\widehat{f}(\xi)=0\}$ has Lebesgue measure zero. I'm wondering whether the same conclusion can be obtained by slightly relaxing the compactly supported condition. So my (a bit vague) question is the following:

Are there some sufficient conditions (weaker than compactness) on the support of $f$ which guarantee that $\{\xi\in\mathbb{R}\,s.t.\,\widehat{f}(\xi)=0\}$ has Lebesgue measure zero?

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    $\begingroup$ Of course you want to assume $f$ is not a.e. $0$. $\endgroup$
    – Robert Israel
    Commented Dec 16, 2021 at 18:06
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    $\begingroup$ There exist compactly supported functions in the Gevrey class of order $s>1$, whose Fourier transform decays like $\exp(-|\xi|^{1/s})$. Thus if you are looking for suitable decay conditions, it seems there is not much room $\endgroup$
    – Piero D'Ancona
    Commented Dec 16, 2021 at 18:12
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    $\begingroup$ This question of mine suggests that not much is possible along these lines: there are Schwartz functions such that $f$ has a support of small measure, but $\widehat{f}$ vanishes on an interval. mathoverflow.net/questions/275072/… $\endgroup$
    – Christian Remling
    Commented Dec 16, 2021 at 18:23
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    $\begingroup$ @ChristianRemling Thank you, very nice construction. If I understand correctly, there is still a small room for some positive result. For example, one could assume that $f$ vanishes outside a ball except for a positive measures set which is nowhere dense. Do you think that something can be said under this assumption? $\endgroup$
    – RaffaeleScandone
    Commented Dec 17, 2021 at 12:16
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    $\begingroup$ @RaffaeleScandone: Yes, indeed, this doesn't literally answer the question you asked for exactly the reason you point out. I don't know what to expect in this situation. $\endgroup$
    – Christian Remling
    Commented Dec 17, 2021 at 15:36

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