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What is the the rate of decrease of $|\widehat{f_p}(t)|$ (as $t\to\infty$), where $p\in(0,\infty)$, $$\widehat{f_p}(t):=\int_{\mathbb R} e^{itx}f_p(x)\,dx,$$ and $$f_p(x):=e^{-1/(1-x^2/p)^p}1(|x|<\sqrt p),$$ so that $f_p\in C^\infty$, $f_p'(0)=0$, and $f_p''(0)=-2/e$ for all $p$.

Here are the graphs $\{(x,f_p(x))\colon|x|<\sqrt p\}$ for $p=1$ (red), $p=2$ (green), and $p=3$ (blue):

enter image description here

The decrease seems to be very fast, almost to $0$ in a finite time $t>0$. Below are parts of the graphs of the functions $\ln\ln\dfrac1{|\widehat{f_p}|}$ for $p=1$ (red), $p=2$ (green), and $p=3$ (blue):

enter image description here

This seems to almost contradict Hardy's uncertainty principle -- which implies, as noted in Dmitry Krachun's comment, that $\widehat{f_p}(t)$ cannot decay faster than $e^{-ct^2}$ for any $c>0$.

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  • $\begingroup$ I think the function $f_{p}$ never vanishes, as it is defined as an exponential of something. $\endgroup$ Dec 16, 2021 at 20:07
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    $\begingroup$ If I'm not mistaken, since $f$ is compactly supported, Hardy's uncertainty principle implies that $\hat{f}$ cannot decay faster than $e^{-\epsilon t^2}$ for any $\epsilon>0$. $\endgroup$ Dec 16, 2021 at 20:12
  • $\begingroup$ @DmitryKrachun : Good point. $\endgroup$ Dec 16, 2021 at 20:21
  • $\begingroup$ The Fourier transform of the Lévy distribution, $a x^{-3/2} \exp(-1/x)$, decays as $\exp(-b\sqrt{|t|})$ as $|t| \to \infty$. This suggests a similar decay rate for the Fourier transform of $f_p$ with $p = 1$. Other one-sided stable distributions may give a hint what to expect for other values of $p$, but I do not quite remember their behaviour at $0$, and unfortunately I have no time to look it up now. $\endgroup$ Dec 16, 2021 at 21:37
  • $\begingroup$ @MateuszKwaśnicki : Thank you for your comment. $\endgroup$ Dec 16, 2021 at 21:41

1 Answer 1

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So, morally, the only real way to bound the decay of $\hat{f}$ is to obtain some bounds on the derivatives of $f$. Let me do a little dilation and consider $f(x) = \exp(-1/(1-x)^p - 1/(1+x)^p)$, it does a well-known thing on the Fourier side and doesn't affect general decay properties. And fortunately for us, bounds for the derivatives of this function were written out by Arie Israel in this beautiful paper (I think it's a third time I've given this reference to someone, see also this MO question for related matters). Specifically, Lemma 2 of his paper says that

$$|f^{(k)}(x)| \le (16p)^k k^{(1+1/p)k}.$$ (he considers the case $p\in \mathbb{N}$ but his methods should work for any $p > 0$).

We have $$|\hat{f}(\xi)| \le 2\frac{||f^{(k)}||_{L^\infty}}{|\xi|^k}.$$ Factor of $2$ here is because $\text{supp} f = [-1, 1]$ has length $2$ and there also might be some powers of $2\pi$ thrown here and there depending on the normalization of the Fourier transform. Neither of those will affect our argument of course.

Combining this with the above estimate we get

$$|\hat{f}(\xi)| \le 2 \left(\frac{16p k^{(1+1/p)}}{|\xi|}\right)^k.$$

We want to choose $k\in \mathbb{N}$ so that the number in the brackets is some constant, say it's at most $\frac{1}{2}$ and at least $\frac{1}{4}$ (this can always be done for $|\xi|$ big enough). This gives us $$|\hat{f}(\xi)| \le 2^{1-k}.$$ On the other hand we have $$k \ge \left(\frac{|\xi|}{64p}\right)^{p/(p+1)},$$ which gives us the desired decay in terms of $|\xi|$ of the form $$|\hat{f}(\xi)|\le A\exp(-B|\xi|^{p/(p+1)})$$ for some $A, B>0$.

Note that $\hat{f}$ can't have exponential decay because otherwise $f$ would be analytic in a strip and thus not compactly supported; one can actuallly deduce even more from the general results, see Denjoy-Carleman Theorem and Beurling-Malliavin Theorem in my answer to the above MO question. But here I think it is probable that the above decay is the correct one up to the values of $A$ and $B$: indeed, if $\hat{f}$ decays very fast then we can obtain better pointwise bounds on the derivatives of $f$ and the Israel's bounds are probably sharp, although I'm not sure how to prove it.

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  • $\begingroup$ Apparently, I can’t do simple arithmetic and the function I considered is not directly related to the one from the OP. I will write an edit later today $\endgroup$ Dec 18, 2021 at 12:26

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