Fourier Transform of a harmonic polynomial

Question

Let $x \in S^n$ and $P_d$ be a homogenous harmonic polynomial of degree $d$ in $n+1$ variables. I would like to find Fourier transform of the following function: $$ \frac{1}{|x|^{n+d}}P_d(x). $$

Answer 1

Of course we treat this as a tempered distribution, since it's not in $L^1$.

As your notation already maybe suggests, writing the function as ${1\over |x|^s}P\Big({x\over |x|}\Big)$ usefully separates the radial and spherical coordinates.

Fourier transform behaves well with respect to homogeneity, sending a tempered distribution of degree $-s$ (thinking of $1/|x|^s$ to normalize) to a tempered distribution of degree $-(\hbox{dim}-s)$. Fourier transform commutes with rotations. We do somehow know that the space of harmonic degree $d$ polynomials (with or without dividing by $|x|^d$) is an irreducible representation of the rotation group, so (Schur's lemma) the only automorphisms of that space, commuting with rotations, are scalars.

Thus, up to some constant $C=C(s,n,d)$, the Fourier transform of the (tempered distribution, possibly requiring meromorphic continuation for some values of $s\in\mathbb C$), is $$ C\cdot {1\over |x|^{(n+1)-s}}\cdot P\bigg({x\over |x|}\bigg) \;=\; C \cdot {1\over |x|^{(n+1)+d-s}} \cdot P(x) $$ At $s=n$, this is $$ C\cdot {1\over |x|^{1+d}}\cdot P(x) $$ The constant(s) can be determined by evaluation at aptly chosen Schwartz functions, such as $P(x)e^{-\pi |x|^2}$.

EDIT: in fact, since Fourier transform is an isomorphism on spaces of harmonic functions, and these are irreducible, it suffices to take $P(x_1,\ldots,x_n)=(x_1+ix_2)^k$, which reduces the computation of the constant to a quite tractable elementary issue.