# Condition on the support of $f$ which ensure that $\widehat{f}$ has a zero-measure nodal region

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?

• Of course you want to assume $f$ is not a.e. $0$. Commented Dec 16, 2021 at 18:06
• 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 Commented Dec 16, 2021 at 18:12
• 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/… Commented Dec 16, 2021 at 18:23
• @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? Commented Dec 17, 2021 at 12:16
• @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. Commented Dec 17, 2021 at 15:36