My question is as follows: Given $t_0\in\mathbb{R}$. Does there exist a non-negative valued compactly supported function $f\in L^1(\mathbb{R})$ such that its Fourier transform, $\widehat f\left( t \right)$, only vanishes at $t_0$, $-t_0$?

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    $\begingroup$ Well, the Fourier transform of a real-valued function is hermitian, so I think you should at least have $\widehat f(-t_0)=0$ too $\endgroup$ – leonbloy Jan 28 '15 at 15:10
  • $\begingroup$ Yes. I completely agree with you. $\endgroup$ – User3101 Jan 28 '15 at 15:12
  • $\begingroup$ So are you looking for a compact-supported, real-valued function such that $\widehat{f}$ vanishes only at $-t_0,0,t_0$ or what? $\endgroup$ – Jack D'Aurizio Jan 28 '15 at 15:20
  • $\begingroup$ Yes. I wish to construct a function satisfying the given properties. $\endgroup$ – User3101 Jan 28 '15 at 15:26
  • $\begingroup$ Is $\widehat{f}$ allowed to have complex zeroes other than $t_0,-t_0$? $\endgroup$ – Jack D'Aurizio Jan 28 '15 at 15:42

Assuming that $f(x)$ is supported on $[-1,1]$ we have that $\widehat{f}(s)$ is an entire function of exponential type $1$ by Paley-Wiener theorem. If $\widehat{f}$ is not allowed to have any complex root out of $\pm t_0$, then $$g(s)=\frac{\widehat{f}(s)}{s^2-t_0^2}$$ is an entire function of exponential type $1$ with no roots, hence $g(s)=\exp(as+b)$ by Hadamard's theorem. In this case, however, by considering the inverse Fourier transform we do not get a real valued function.

  • $\begingroup$ I'm wondering whether or not the inverse Fourier transform of $g$ has compactly supported. $\endgroup$ – User3101 Jan 28 '15 at 16:17
  • $\begingroup$ @Cao: I bet not, since it is not square integrable over horizontal lines (second part of Paley-Wiener theorem). $\endgroup$ – Jack D'Aurizio Jan 28 '15 at 16:20

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