Gaussian with zero mean dense in $L^2$ I have found in this article that linear combinations of Gaussian with fixed variance are dense in $L^2$. Can something similar be true for Gaussian of fixed mean and variable variance? Equivalently, can linear combination of this family of functions
$$
  f(x,a) = e^{-(x/a)^2}
   \qquad x \in \mathbb{R}^+_0
$$
be dense in $L^2(\mathbb{R}^+_0)$? Intuitively I think that is not possibile as it should be hard to approximate localized "high" peak far from the origin ( $\lambda \chi_{[n,n+1]}$ for large $\lambda \in \mathbb{R}$ and $n \in \mathbb{N}$). Any suggestions (or references if that is a known fact)?
[EDIT: as pointed in the comments, this is trivial in $\mathbb{R}$ as this functions all belong to the proper closed subspace of symmetric function with respect to the origin]
 A: Yes, it is dense in the space of even $L^2$ functions. To see this we can use a corollary of Hahn-Banach theorem:
The space would be dense if we have the following implication for every even $f\in L^2$:
$$ \int_\mathbb R f(x) e^{-a x^2} \, dx = 0 \quad \forall a > 0 \Longrightarrow f \equiv 0.$$
Lets take $f$ that satisfies
$$ \int_\mathbb R f(x) e^{-a x^2} \, dx = 0 \quad \forall a > 0.$$
If we differentiate with respect to $a$ we obtain (after dividing by $-a$)
$$\int_\mathbb R x^2 f(x) e^{-a x^2} \, dx = 0 \quad \forall a > 0.$$
In order to justify the differentiation under the integral sign, just notice that if $a\in (0,A)$, then
$$ |x^2 f(x) e^{-a x^2}| \leq x^2 |f(x)| e^{-A x^2} , $$
which is integrable.
We can continue this process iterativelly to obtain
$$\int_\mathbb R x^{2n} f(x) e^{-a x^2} \, dx = 0 \quad \forall a > 0, n\in \mathbb N.$$
If we choose $a = 1$ we obtain
$$\int_\mathbb R x^{2n} f(x) e^{-x^2} \, dx = 0 \quad \forall n\in \mathbb N, $$
hence
$$\int_\mathbb R H_{2n}(x) f(x) e^{-x^2} \, dx = 0 \quad \forall n\in \mathbb N, $$
where $H_n$ are the hermite polynomials that form an orthogonal base of $L^2(\mathbb R)$ with the weight $w(x) = e^{-x^2}$. Thanks to the fact that $f$ is even, we have also
$$\int_\mathbb R H_{2n + 1}(x) f(x) e^{-x^2} \, dx = 0 \quad \forall n\in \mathbb N, $$
from which we conclude that $f = 0 \in L^2(\mathbb R)$.
Remark: In fact this shows that it is enough to consider Gaussians with variance in a given open (and non empty) interval only.
