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Is there a theorem that assures the existence of a function $f$ (radial) in $S(\mathbb R^2)$; the Schwartz space, such that the support of its Fourier transformation is included in a crown $A$; that is, $\operatorname{supp} \hat{f} \subset A:=\left\{x \in \mathbb R^2: \, a\leq \|x\| \leq b\right\}$, with $a,b>0$.

Thank you in advance.

Davide Giraudo
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Z. Alfata
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1 Answers1

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Certainly.

In fact $f=0$ has all the properties you ask for. A less trivial example: Say $\psi\in C^\infty_c(\mathbb R)$, with support in $(a,b)$. Define $\phi\in C^\infty_c(\mathbb R^n)$ by $$\phi(x)=\psi(||x||).$$(If we assume $a>0$ then $\phi$ is smooth.) Now there exists a Schwarz function $f$ with $\hat f=\phi$.

David C. Ullrich
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