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Do you know an unbounded open set $A\subset \mathbb{R}^d$, $d\geq 2$ with the following property: if some integrable function $f$ on $\mathbb{R}^d$ has its Fourier transform vanishing on $A^c$ and all the derivatives of $f$ exist and vanish at some point, then $f=0$?

It works if $A$ is bounded because then $f$ is holomorphic, hence determined by its derivatives at some point.

To sum up: can I weaken the assumption that $A$ is bounded in the direct sense of the Paley–Wiener theorem if I agree that $A$ is "only" real analytic, not necessarily holomorphic? Ideally I am looking to apply it to tempered distributions.

If not possible, what decay assumption should I add on $f$ to make it possible?

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    $\begingroup$ I am pretty sure that for $A = [0, +\infty)$ it doesn't hold, the function is analytic in the lower half-plane, but not up to the boundary. If you are willing to assume additionally that $f$ is in $L^2$, say, then I can construct such unbounded sets. $\endgroup$
    – Aleksei Kulikov
    Commented Jun 12 at 20:35
  • $\begingroup$ Indded, I forgot this assumption sorry, I modified the post accordingly. $\endgroup$
    – kaleidoscop
    Commented Jun 13 at 5:15
  • $\begingroup$ I also relaxed the regularity assumption, I essentially need that f has a deep zero at some point. $\endgroup$
    – kaleidoscop
    Commented Jun 13 at 9:49
  • $\begingroup$ "On the real line, it works with $A = [0, +\infty)$ if $f$ is in $L^2$" -- again, I am pretty sure that this is just false. My idea was that if $A$ is extremely thin, and $f\in L^2(A)$ then $\hat{f}$ is still an entire function, hence you can use the uniqueness theorem, but I don't know if this is the type of result you are interested in? $\endgroup$
    – Aleksei Kulikov
    Commented Jun 13 at 15:26
  • $\begingroup$ When $A=(0,\infty)$, your assumption becomes $f\in H^2$. Such functions can indeed not be too small too frequently, but the precise condition is $\int\log |f|>-\infty$ (on the unit circle). That still seems to leave room for all derivatives vanishing at a point (for example, if $|f|\simeq e^{-|x|^{-1/2}}$), confirming Aleksei's comment. $\endgroup$
    – Christian Remling
    Commented Jun 13 at 17:44

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I am not sure if this is the sort of thing you want, but here it goes anyway.

The idea is that if $f\in L^2(\mathbb{R}^n)$ and $A$ is very thin then by the Cauchy--Schwarz inequality $f$ is an entire function, thus if the derivatives vanish at some point, then $f$ is identically $0$.

Indeed, we have $f(z) = \int_A \hat{f}(x)e^{2\pi i x\cdot z}dx$, which by the Cauchy--Schwarz inequality is bounded by $$ \|f\|_{L^2} \left(\int_A e^{-4 \pi Im(x\cdot z)}dx\right)^{1/2} \le \|f\|_{L^2} \left(\int_A e^{4 \pi |x||z|}dx\right)^{1/2}.$$

So, if the set $A$ is super-exponentially thin, meaning that $|(A\cap (B(0, 2R)\backslash B(0, R))|$ decays faster than any exponential in $R$ (here $B(0, r)$ is the ball with center at $0$ and radius $r$), then the integral converges for all $z$, you can differentiate under the integral sign, hence the function is entire and you have uniqueness if all derivatives at some point are $0$.

For example, you can consider the set $$ A = \left\{x\in \mathbb{R}^n \mid k < |x| < k + \frac{1}{k!}\,\, \text{for some}\,\, k\in\mathbb{N}\right\}. $$

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  • $\begingroup$ Thanks, it is indeed what I am looking for. I am not asking you necessarily to do it, but do you think there is a way to have bigger sets by optimizing this procedure? On the real line one can do better? And also what can we obtained if we have stronger decay assumptions on $f$? I thought there could be a line of literature on this topic. $\endgroup$
    – kaleidoscop
    Commented Jun 14 at 19:34

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