TITLE: Is every harmonic polynomial a linear combination of these?
QUESTION [6 upvotes]: In $N$-dimensional space, we can show by direct calculation that the polynomial
$$
 r^{2K+N-2}\nabla_a\nabla_b\nabla_c\cdots \frac{1}{r^{N-2}}
\hspace{1cm}
 \text{(with $K$ derivatives)}
$$
is harmonic (annihilated by the Laplacian $\nabla^2$), where $\nabla_n$ is the partial derivative with respect to the $n$th coordinate and $r$ is the distance from the origin.
Is every homogeneous harmonic polynomial of degree $K$ a linear combination of these? If so, how do we prove this? If not, what is a counterexample?
Motive: I'm studying physics, and this seems like a much nicer way to approach the theory of spherical harmonics (just divide this polynomial by $r^K$ to get a spherical harmonic) compared to the typical physics-textbook approach using spherical coordinates, but it's not obvious to me that all spherical harmonics are linear combinations of these for an arbitrary number of dimensions $N$.

REPLY [3 votes]: I notice that
$$
r^{2K+N−2} P(\partial) \frac{1}{r^{N−2}} = K(P(\partial) \frac{1}{r^{N−2}})
$$
Where $K(f)$ is Kelvin transform of the unit sphere and $P(\partial)$ is a term of derivatives.
By Theorem 5.25 in Harmonic Function Theory.
Your statement is true.
Prove of Shubin’s Conjecture
Using Corollary 5.20 in the same book. For dimension N>2, if $p$ is a homogeneous harmonic function of degree $m$. Then we have
$$
P= K(P(\partial) \frac{1}{r^{N−2}})/c_m=r^{2m+N−2} \frac{1}{c_m} P(\partial) \frac{1}{r^{N−2}}.
$$
Where $c_m$ is a constant define as $\prod_{k=1}^{m} (4-N-2k)$. The uniqueness is follow by theorem 5.18.