$C_0^\infty(\mathbb{R}^k)$ functions can be approximated via polynomials

Question

Let $f:\mathbb{R}^k\rightarrow \mathbb{R}$ be a vanishing at infinity function, also infinitely differentiable, i.e. $f\in C_0^\infty(\mathbb{R}^k,\mathbb{R})$. Is it true that I can always approximate $f$ with some polynomial $p$? Can I use here directly the right version of the Stone-Weierstrass theorem considering the subalgebra of polynomials $P:=\lbrace p:\mathbb{R}^k\rightarrow \mathbb{R} \mid p \text{ polynomial} \rbrace$?

Answer 1

Yes and no.

If you restrain $f$ to take a compact domain $C \subseteq \mathbb{R}^k$, then the answer is an unambiguous yes. For the Stone-Weirstrass theorem only requires an algebra of functions which can discriminate between points, and this clearly applies to polynomials in $k$ variables.

If you wish to distinguish between points $(x_1, ..., x_k)$ and $(y_1, ..., y_k)$, just take the polynomial $P(z_1, ..., z_n) = \sum\limits_{i = 1}^n (z_i - x_i)^2$.

If you do not restrict $f$ to have a compact domain, you can still construct a sequence of polynomials which converges pointwise to $f$ (but not necessarily uniformly). This is true even if $k = 1$; take $f(x) = e^x$, for example.

Answer 2

If you're interested in uniform convergence of polynomials on all of $\mathbb R^k,$ there's not much good news: The only $f\in C_0^\infty$ that can be uniformly approximated by polynomials is the zero function. That's because if $P$ is a polynomial, then either $P$ is a constant or $P$ is unbounded.