Smoothness of Fourier transform of $\frac{1}{|x|^p}$

Question

Consider the "function" (more precisely it is a tempered distribution) given by $f : \mathbb{R}^n \to \mathbb{R}$, $f(x) = \frac{1}{|x|^p}$, where $0 < p < n$. It can be calculated that the Fourier transform of $f$ is given (upto a constant) by $\hat{f}(\xi) = \frac{1}{|\xi|^{n - p}}$. Now, I am trying to prove that the Fourier transform of $f$ is smooth eveywhere, except at the origin, which is clear from above. But I am unclear how to prove this without calculating the Fourier transform explicitly first. One should be able to exchange derivatives and decay, but the singularity at $0$ is throwing me off. Any hints would be highly appreciated.

Answer 1

Let $\chi$ be a compactly supported smooth function that is equal to $1$ near the origin. Write $f_1=f\chi$ and $f_2=f(1-\chi)$. Note that $f_1$ is compactly supported, therefore its Fourier transform is real-analytic. The function $f_2$ is smooth, but has heavy tail at infinity, which is the main issue.

Fix $\xi_0\ne 0$. Let $\phi$ be a bump function that is equal to $1$ near $\xi_0$, and is zero in a neighborhood of the origin. The goal is to prove that $\widehat {f_2} \phi$ is smooth. Equivalently, we must prove that $f_2*\check \phi$ vanishes at infinity faster than any power of $|x|$.

Given a positive integer $k$, let $\psi(\xi) = |\xi|^{-2k} \phi(\xi)$. Note that $\psi$ is still smooth and compactly supported. Going back to space variable, observe that $\check \phi = \Delta^k \check \psi$ (Laplacian applied $k$ times, ignoring some multiplicative constants.)

So, $f_2*\check \phi = f_2* \Delta^k \check \psi = (\Delta^k f_2 ) * \check \psi$. The benefit of all this juggling of Laplacians is that $\Delta^k f_2 $ decays at infinity faster than $f_2$: namely, like $|x|^{-p-2k}$. Since $\check \psi$ is a Schwartz function, it follows that $(\Delta^k f_2 ) * \check \psi$ is $O(|x|^{-p-2k})$ as $|x|\to\infty$. This completes the proof.