Let $e_1=(0,1)^T$, $$ S=\left\{x\in \mathbb{R}^2\Big| \frac{|\langle x, e_1\rangle|}{|x|}>\delta>0\right\}, $$ is a cone in $\mathbb{R}^2$.

I want to find a non-trivial smooth function which satisfies: $$ f\geq 0; \ \ f\equiv0 \ \mbox{on} \ S^c; \ \ \mbox{supp}\ \hat{f}\subset B_1(0). $$ Here $\hat{f}$ is the Fourier transform of $f$. I found using so called "real paley wiener theorem", which was proved in "Andersen, Nils, and Marcel de Jeu. Real Paley-Wiener theorems and local spectral radius formulas. Transactions of the American Mathematical Society 362.7 (2010): 3613-3640," the last condition can be rewrote as:
$$ |\nabla^k f(x)|\leq Ck^N(1+|x|)^N. $$

Thanks a lot if one can give me any comment or reference.

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    $\begingroup$ What is $\hat f$? Fourier transform of $f$? If so, such a function does not exist. If the Fourier transform has compact support, the function is analytic, so it cannot vanish on the complement of $S$ unless it is identically zero. $\endgroup$ – Michael Renardy May 11 '17 at 15:50
  • $\begingroup$ Yes, $\hat{f}$ is the Fourier transform of $f$. You mean a real analytic function cannot vanish on an open set in $\mathbb{R}^2$? $\endgroup$ – John Zhao May 11 '17 at 22:36
  • $\begingroup$ @JohnZhao: yes, as explained here $\endgroup$ – Nik Weaver May 12 '17 at 2:46
  • $\begingroup$ @MichaelRenardy, Oh, I see. My question is really trivial! Thank you so much for your help. $\endgroup$ – John Zhao May 12 '17 at 13:47

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