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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?

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    $\begingroup$ I think $\ell^2(\mathbb{R})$ should read $\ell^2(\mathbb{Z}^n_{\ge 0})$. $\endgroup$ – Nate Eldredge Sep 15 '14 at 4:01
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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.

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  • $\begingroup$ thank you for this nice answer. I fixed the inaccuracies. $\endgroup$ – Hans Sep 15 '14 at 4:39

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