Let $f\in C_c^\infty$ be a compactly supported smooth function function with support say in $(0,1)$. It is clear that when I multiply $\tau$ with $\hat{f}(\tau)$ (Fourier transform of $f$), then inverse Fourier transform of $\tau\hat{f}(\tau)$ is equivalent to $g(x)=id/dx(f(x))$ which is clearly compactly supported with the same support at the largest. How about other powers of $\tau$. Let's say $F(\tau)=\tau^r\hat{f}(\tau)$ where $0<r<1$. Is it true that inverse Fourier transform of $F$ has compact support. If so, how big would be the support? Many thanks!


For $k\ge 0$, the inverse Fourier transform of $ (i\tau)^k \hat{f}$ is $ f^{(k)}$ which is $C^\infty_c$ with a support at most as large as $f$.

This is because $(i\tau)^k$ is the Fourier transform of the distribution $ \delta^{(k)}$ with support $\{0\}$ so the support of $f\ast \delta^{(k)}=f^{(k)}$ is included in that of $f$.

The Fourier transform of a smooth compactly supported function is entire, so $\hat{f}(\tau)$ is entire. For $f\ne 0$ and $k\ge 0$ then $\hat{f}(\tau)\tau^k$ is entire iff $k\in \Bbb{Z}_{\ge 0}$, ie. its inverse Fourier transform is compactly supported iff $k\in \Bbb{Z}_{\ge 0}$.

  • $\begingroup$ I think you wrote this for integer $k$. My question was for fractional $r$. $\endgroup$ – Carl Lincoln Mar 10 at 15:52
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    $\begingroup$ For $k$ not an integer the inverse Fourier transform is not compactly supported. $\endgroup$ – reuns Mar 10 at 15:53
  • $\begingroup$ That was my question thanks. What is the reason? $\endgroup$ – Carl Lincoln Mar 10 at 15:53

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