The density of polynomials in the space of continuous functions on the unit ball of $\ell^p$

Question

Let $$B = \{a : \|a\|_p \le 1\} \subset \ell^p(\mathbb{N})$$be the unit ball, endowed with the weak topology. For which $p$, where $1 < p \le \infty$, are the functions of the form$$f(a) = q(a_0, a_1, \dots, a_n),$$where $q$ is a polynomial in $a_0, \dots, a_n$, dense in $C(B)$?

Answer 1

It's certainly true for $1<p<\infty$. In that case the weak topology on $B$ is the same as the weak* topology regarding $\ell^p$ as the dual of $\ell^{q}$. So $B$ is a compact Hausdorff space, and so Stone-Weierstrass implies that those polynomials are dense in $C(B)$.

Note that you didn't specify what topology on $C(B)$ you're talking about when you ask about those polynomials being dense in $C(B)$; I'm assuming you intended the norm topology.

Now for $p=\infty$. If you gave $B$ the weak* topology, regarding $\ell^\infty$ as the dual of $\ell^1$, then the same answer works.

But you said "weak topology". In that case the question's meaningless for $p=\infty$, at least without some clarification: $B$ is not compact, so (unless I'm missing something) functions in $C(B)$ need not be bounded, so there's no such thing as the norm topology on $C(B)$. So the question is meaningless until you specify what topology on $C(B)$ you're talking about.

Answer 2

For $1 < p < \infty$, the space $\ell^p$ is reflexive, so the weak and weak-star topologies on $B$ coincide, and thus $B$ is a compact Hausdorff space by the Banach-Alaoglu theorem. The polynomials in question form a subalgebra of $C(B)$ that separates points, and by the Stone-Weierstrass theorem this algebra is dense in $C(B)$.

The same result fails for $p = \infty$. Let $\phi \in C(B)$ be a bounded linear functional such that$$\phi(a) = \lim a_n$$when $a_n$ converges. Given a polynomial $q(a_0, a_1, \dots, a_n)$, construct a point $a \in B$ by setting $a_i = 0$ for $i \le n$ and $a_i = \pm1$ for $i > n$, with sign chosen to disagree with the sign of $q(0, \dots, 0)$. Then$$|q(a) - \phi(a)| \ge 1,$$so $\phi$ is not in the closure of the polynomials in $C(B)$.