For $n \in \mathbb{N}$ let $f_n(x) := x^n \exp\left(-\frac{x^2}{2}\right)$. I want to prove that $A := \mathrm{span} \lbrace f_n: n \in \mathbb{N} \rbrace$ is dense in $L^2(\mathbb{R}^n)$.
Clearly, it would be enough to approximate charcteristic functions, but I am not quite sure how to go about this. Maybe $\displaystyle \sum_{n = 1}^\infty \frac{(-1)^n(x)^{2n}}{2 ^nn!} \rightarrow \exp\left(-\frac{x^2}{2}\right)$ pointwise helps...
Take $f \in L^2(\mathbb{R})$ such that : $$\forall n \geq 0, \quad \displaystyle\int_{\mathbb{R}} f(x) x^n e^{-\frac{x^2}2} dx = 0$$
We want to prove that $f = 0$ everywhere. Define :
$$\forall z \in \mathbb{C},\quad F(z) = \displaystyle\int_{\mathbb{R}} f(x) e^{izx} e^{-\frac{x^2}2} dx $$
We can prove that $F$ is an entire function and that $F^{(n)}(0) = i^n\displaystyle\int_{\mathbb{R}} f(x) x^n e^{-\frac{x^2}2} dx = 0$ for all $n \geq 0$. Thus, $F = 0$ on all $\mathbb{C}$ (by analyticity). But, $F$ restrited to $\mathbb{R}$ is the fourier transform of $x \mapsto f(x) e^{-x^2/2}$. By injectivity of the Fourier transform, $f(x) e^{-x^2/2} = 0$ for almost all $x \in \mathbb{R}$. Finally $f = 0$, and the subset is dense in $L^2(\mathbb{R})$.