The Fourier transform of $|x|^{-a}$, where $0<a<n$

Question

1)When checking the calculation of the Fourier transform of $|x|^{-a}$, where $\frac{n}{2}<a<n$, it can be proved that the Fourier transform of this function is rotational invariant and is a homogeneous function of degree $a-n$, then it states that we can write it in the form $c_{a,n}|\xi|^{a-n}$, I don't know why

2) When proving the case when $0<a<\frac{n}{2}$, it uses Fourier inversion formula, but I don't know why it can be applied to $|\xi|^{-a}$

Answer 1

1. Being rotationally invariant implies being constant on the unit sphere: $\hat f(\xi)=c$ when $|\xi|=1$. Being homogeneous of segree $a-n$ means $$\hat f(\xi) = |\xi|^{a-n}\hat f(\xi/|\xi|)$$ Put the two things together, and you have the conclusion.

2. Fourier inversion property (the inverse being almost the same transform with a different sign and perhaps normalization) holds for tempered distributions. The function $|x|^{-a}$, $0<a<n$, is a tempered distribution.

Some references are in comments here.