# Multiplication by a simple function and compactly supported inverse Fourier transform

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 $$0. Is it true that inverse Fourier transform of $$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 $$k\in \Bbb{Z}_{\ge 0}$$.
• I think you wrote this for integer $k$. My question was for fractional $r$. – Carl Lincoln Mar 10 at 15:52
• For $k$ not an integer the inverse Fourier transform is not compactly supported. – reuns Mar 10 at 15:53