# Density of the linear span of products of harmonic polymomials

Let $\mathcal{H}$ denote the space of all harmonic polynomials with complex coefficients in $n$ variables $x_1,\ldots, x_n$ in $\mathbb{R}^n$. I'm trying to show that the linear span of the set $\mathcal{M}=\{p\, q: p,q\in\mathcal{H}\}$ is dense in $C(K)$ under supremum norm, where $K$ is a compact set in $\mathbb{R}^n$, for $n\geq 2$. One may assume that $K$ is the closed unit ball in $\mathbb{R}^n$ since once it is proved for the unit ball, the result follows for any compact $K$. This seems to be classical and should be known but I have not found a reference.

When $n$ is even, it is not hard to prove. In fact, if $n=2m$, one may identify $\mathbb{R}^n$ with $\mathbb{C}^{m}$ and use the fact that the linear span of $\mathcal{N}=\{f\,\overline{g}: f, g \text{ are holomorphic polynomials}\}$ is dense in $C(K)$ by Stone-Weierstrass Theorem. Note that for any holomorhpic polynomials $f, g$ in $\mathbb{C}^m$, the polynomials $p=f$ and $q= \overline{g}$ are harmonic in $\mathbb{R}^n$. Therefore, $\mathcal{N}\subset\mathcal{M}$ and hence, $\mathcal{M}$ is dense in $C(K)$.

I have not been able to resolve the case $n$ is odd. In general, $\mathcal{M}$ is not a subalgebra of $C(K)$ even though it is self-adjoint, separates points and vanishes at no point so Stone-Weierstrass Theorem cannot apply directly to $\mathcal{M}$ (even when $n$ is even).

I would highly appreciate it if someone can point out a reference or offer a proof.

## 2 Answers

Induction on $n$. Base $n=2$ is clear, as you said. Let $\nu$ be a non-trivial finite compactly supported (complex) measure in $\mathbb R^n$ orthogonal to any product $uv$. Then we can smear it a bit and get a non-trivial continuous compactly supported function $f$ orthogonal to each product. Now choose any $n-1$ dimensional space and consider the harmonic functions that do not depend on the orthogonal variable. Applying the induction assumption, we see that the integral of $f$ along every line must be $0$ (otherwise we can project $f$ to get something non-trivial in $\mathbb R^{n-1}$). This is enough to conclude that the Fourier transform of $f$ is identically $0$, so $f$ must be $0$.

• The "smearing" while maintaining orthogonality worries me a bit... could you elaborate a bit, please? – paul garrett Aug 14 '16 at 21:56
• @paul garrett The space is shift invariant, so we can convolve with any mollifier for free. – fedja Aug 14 '16 at 22:03
• I'd like to point out that the following two properties of harmonic polynomials have been used in the answer: (1) shift invariance, as mentioned in the above comment about "convolving"; and (2) rotation invariance, in the induction step. – T. Le Aug 16 '16 at 13:29

Back to the drawing board. There was an error in my argument.

• Is $r^2-2n$ harmonic? – Fedor Petrov May 16 '16 at 17:49
• I noticed my error the second after I posted my answer''. – Liviu Nicolaescu May 16 '16 at 18:16