10
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $0$.

$\endgroup$
  • $\begingroup$ The "smearing" while maintaining orthogonality worries me a bit... could you elaborate a bit, please? $\endgroup$ – paul garrett Aug 14 '16 at 21:56
  • 1
    $\begingroup$ @paul garrett The space is shift invariant, so we can convolve with any mollifier for free. $\endgroup$ – fedja Aug 14 '16 at 22:03
  • $\begingroup$ 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. $\endgroup$ – T. Le Aug 16 '16 at 13:29
0
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.