Stone-Weierstrass without the "subalgebra" condition

Question

Suppose I consider $C_0(\mathbb{N})$ consisting of function on the natural numbers vanishing at $\infty$. For an irrational $1<\alpha<2$, let $p_{m\alpha}(\cdot)$ be the function $p_{m\alpha}(n)= 1/n^{m\alpha}$. By the Stone-Weierstrass theorem, linear combinations of $\{p_{m\alpha}\},m\geq 1$ is dense in $C_0(\mathbb{N})$. If I only retain the fractional part $\{m\alpha\}$ of each $m\alpha$, is the span of $\{p_{\{m\alpha\}}\},m\geq 1$ still dense in $C_0(\mathbb{N})$?

Answer 1

We can use the Müntz–Szász theorem, which states that if $\Lambda = \{0 = \lambda_0 < \lambda_1 < \cdots\}$ then the span of $\{x^\lambda : \lambda \in \Lambda\}$ is dense in $C[0,1]$ if and only if $\sum_{k=1}^\infty 1/\lambda_k = \infty$. Note: Wikipedia restricts $\Lambda \subset \mathbb N$, but this is not needed. For example see [Rudin, Real and Complex Analysis, Theorem 15.26] or [Almira, Muntz type Theorems I, Theorem 1].

Here we take $\Lambda$ to be the union of $\{0\}$ and an infinite subset of $\{\{m\alpha\} : m \in \mathbb N\}$ accumulating at $1$. The theorem then tells us that $\{x^\lambda : \lambda \in \Lambda\}$ is dense in $C[0,1]$.

Let $f \in C_0(\mathbb N)$. Define $F \in C[0,1]$ by $F(0) = 0$, $F(1/n) = f(n)$ for $n \in \mathbb N$, and linearly interpolating. Then we can uniformly approximate $F(x)$ by some linear combination $\sum_{i=0}^K a_i x^{\lambda_i}$. By evaluating at $0$ we find that $a_0$ must be small, so we can throw it away. Now by restricting to the values $F(1/n)$, it follows that $f(n)$ is uniformly approximated by $\sum_{i=1}^K a_i / n^{\lambda_i}$.

Answer 2

Let's change variable and use $x=1/n$, so the question reads: when $\{1, x^{\lambda_1},x^{\lambda_2},\dots\}$ has dense linear span in the space of the continuous functions on $K$ vanishing at $0$, for $K:=\{0,1/n:n\in\mathbb N\}$ and $\lambda_m:=\{m\alpha\}$.

[useless assumption] At least for $\alpha$ a quadratic algebraic irrational, we have affirmative answer. In this case, by diophantine approximation, there is $C>0$ such that $\lvert\alpha-p/q\rvert>C/q^2$ holds for all $p\in\mathbb Z$ and $q\in\mathbb Z_+$, so that $\{q\alpha\}>C/q$ for all $q\ge1$.

Recall the full Müntz–Szász theorem: for a sequence of positive numbers $\{\lambda_q\}_{q\ge1}$ the set $\{1, x^{\lambda_1},x^{\lambda_2},\dots\}$ has dense span in $C^0[0,1]$ if ad only if $\displaystyle \sum_{q\ge1}\frac{\lambda_q}{1+\lambda_q^2}=+\infty$ (a condition that reduces to simply $ \sum_{q\ge1} {\lambda_q} =+\infty$ for bounded positive sequences, which is the case of $\lambda_q:=\{\alpha q\}$). Here, with change of variable $x=1/n$, we have in particular that $\{ n^{-\{ q\alpha\}}: q\ge1\}$ is dense in the space $c_0$.

[edit]: In fact the diophantine thing here is totally unnecessary, for of course here $\lambda_q$ are dense in $[0,1]$, which already ensures the assumption of the MS theorem, as in Sean Eberhard's answer!