Can the Fourier transform of a test function vanish on an interval?

Question

Suppose $f$ is a test function and thus compactly supported on some set $S$. Its Fourier transform is

\begin{equation} \widehat{f}(p) = \int_{-\infty}^{\infty} dx \, e^{ipx} f(x) \, . \end{equation}

Is it possible for this Fourier transform $\widehat{f}$ to vanish on some interval $(a,b)$?

I have tried a number of things and believe that this should not be possible, also with regard to this question. More generally, is this possible if $f$ is a Schwartz function?

I would also appreciate if anyone could direct me to a general discussion of the relation between the support of a function and that of its Fourier transform.

Regarding your last paragraph, A function and its Fourier transform cannot both be compactly supported — Sine of the Time, Sep 10 '24 at 16:50
Thank you for the link. While I had seen this result I still asked the question since compact support for f is a more stringent condition than vanishing within some interval. — Zarathustra, Sep 10 '24 at 17:11

score 9 · Accepted Answer · answered Sep 10 '24 at 16:58

9

Of course not. Since $f$ has compact support, then $\hat{f}$ is the restriction to $R$ of a function which is analytic in the whole complex plane (ie an entire function). It cannot be identically 0 on an interval (principle of isolated zeros) if $f$ is not zero almost everywhere.

answered Sep 10 '24 at 16:58

Letac Gérard

3,820

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I didnt know that. Do you have a link? (That the fourier transform of a compact support schwarz function is analytic). – lalala Sep 11 '24 at 07:06
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@lalala https://en.wikipedia.org/wiki/Paley–Wiener_theorem – Dan Romik Sep 11 '24 at 07:35
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Paley-Wiener does not seem necessary: a simple power expansion around $z$ is easily proved to have a positive radius on convergence. – Letac Gérard Sep 11 '24 at 10:23
additionaly with was already said, the reason an entire function descrived by a power series cannot become some constant value on an interval, like the value zero, is due the Identity Theorem – Joako Sep 13 '24 at 04:28

Ryszard Szwarc · Answer 2 · 2024-09-10T18:34:24.917

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Concerning the Schwartz class the Fourier transform is an isomorphism from $\cal{S}(\mathbb{R})$ onto itself. As $\cal{S}(\mathbb{R})$ contains functions with compact support, the answer is yes.

edited Sep 10 '24 at 18:34

answered Sep 10 '24 at 18:06

Ryszard Szwarc

44,675

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I don't quite follow why the isomorphism property (as inner product spaces I assume) implies the result. It could be that all images of compact support functions do not have the desired property. – Zarathustra Sep 10 '24 at 18:23
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We have $f\in S$ iff $\widehat{f}\in S.$ I have added a link in my answer. – Ryszard Szwarc Sep 10 '24 at 18:42
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The question in OP: "Does there exist a nonzero function in $S$ so that its Fourier transform vanishes on some interval ?" The answer is yes. The answer is no for functions with compact support. – Ryszard Szwarc Sep 11 '24 at 06:09
Thank you, that does indeed answer the question. – Zarathustra Sep 13 '24 at 03:12
Does not test functions always have compact-support? – Joako Sep 13 '24 at 04:26

Can the Fourier transform of a test function vanish on an interval?

2 Answers2