For all $x \in \mathbb{R}^n$ and $\alpha \in \mathbb{Z}\_{\geq 0}^n$ let $x^\alpha=x\_1^{\alpha\_1} \cdots x\_n^{\alpha\_n}$. Let $$\ell^2=\{z=(z\_\alpha)\_{\alpha \in \mathbb{Z}\_{\geq 0}^n}:\, z\_{\alpha} \in \mathbb{R}, \,\, \|z\|^2=\sum\_{\alpha \in \mathbb{Z}\_{\geq 0}^n } z\_{\alpha}^2 < \infty\}.$$


Let $1 \geq \epsilon>0$ be arbitrary small. Is the $\mathbb{R}$-span of the set $\{(x^\alpha)\_{\alpha \in \mathbb{Z}\_{\geq 0}^n}: x \in (0,\epsilon)^n\}$ dense in $\ell^2$? I would guess that should be known, does somebody know a reference?



Is the span of those vectors dense in $\ell_2$?
I'm puzzled by the notation $l^2(\mathbb{R})$. (Also you surely mean $\|z\|^2$, not $\|z\|$, in the definition.)


One way to show that the span of a set is dense in a Hilbert space is by showing that the only vector orthogonal to the set is the zero vector. So let $(a\_\alpha)\_{\alpha \in \mathbb{Z}^n\_{\geq 0}}$ be an arbitrary element of your Hilbert space and
define $f(x) = \sum\_{\alpha \in \mathbb{Z}^n\_{\geq 0}} a\_\alpha x^\alpha$ for $x \in (-\epsilon,\epsilon)^n$. (I guess we assume $\epsilon \leq 1$ here.) This is the real part of an analytic function in several complex variables, but all that matters for us is that if $(a\_\alpha)$ is orthogonal to all the vectors $(x^\alpha)$ for $x \in (0,\epsilon)^n$ then $f$ is identically zero on $[0,\epsilon)^n$, and therefore all of its partial derivatives at zero vanish. This shows that every $a\_\alpha$ is zero, and it follows that the vectors $(x^\alpha)$ are dense in your Hilbert space.


