Linearspan of Gaussians dense in Schwartz space as the title already says I am trying to show that the linear span "A" of the gaussians $e^{\frac{-|x|^2}{2}}$ and their translations/ dilations are dense in the Schwartzspace.
This is the space of rapidly decreasing smooth functions, i.e. $\mathcal{S}(\mathbb{R}^n) =\{f\in C^\infty(\mathbb{R}^n) | \forall_{\alpha, \beta \in \mathbb{N}_0^n} \sup_{x\in\mathbb{R}^n}|x^\alpha \partial^\beta u(x)| < \infty \}$ which is equipped the metric $$\rho(u,v)= \sum _{\alpha, \beta \in\mathbb{N}_0^n} 2^{-|\alpha|-|\beta|} \frac{\|u-v\|_{\alpha, \beta}}{1+\|u-v\|_{\alpha, \beta}},$$ 
where $\alpha$ and $\beta$ are multiindices and $\|u\|_{\alpha,\beta}= \sup_{x\in\mathbb{R}^n}|x^\alpha \partial^\beta u(x)|$
Thus I need to show that for every $\varphi \in \mathcal{S}(\mathbb{R}^n)$ there is a sequence $(u_k)_{k\geq 1}$ such that for all multiindices $\alpha$ and $\beta$ as above $\|\varphi-u_k\|_{\alpha, \beta}$ converges to $0$ as  $k\rightarrow \infty$. 
Does anyone have a hint/source on how to construct such a sequence? I'd really like to figure this out but as it stands I'm stuck at the beginning. My first guess was to use that $C_0^\infty(\mathbb{R}^n)$ is dense in $\mathcal{S}(\mathbb{R}^n)$, then show that the linear span of the Gaussians is dense in $C_0^\infty(\mathbb{R}^n)$ with respect to $\rho$ and afterwards apply some diagonal sequence argument. But so far I haven't succeeded in finding such an approximation.  
EDIT: 
So I have thought about the suggestion and this is what I understand so far: 
I know that the convolution for $\varphi \in\mathcal{D}(\mathbb{R}^n)$ and $T \in \mathcal{D}'(\mathbb{R}^n)$ is defined for all  distributions $\psi$ via 
$$< \varphi *T,\psi> := <T, \mathcal{R}(\varphi) * \psi> $$
where $\mathcal{R}f(x) = f(-x)$ is the reflexion operator. But this definition should also translate to a tempered distribution $\phi$ convoluted with a gaussian of the type $G=ae^{\frac{|x-x_0|^2}{2b}}$ for $a\in \mathbb{R}$,$b>0$ and $x\in\mathbb{R}^n$. 
Therefore if I choose $\phi$ in $\mathcal{S}'(\mathbb{R}^n)$such that $<\phi, G> =0 $ for all G and define $g_\epsilon = e^{\frac{-|x|^2}{2\epsilon^2}}$ as below the convolution $g_\epsilon*\phi$ is well-defined. Furthermore by the choice of $\phi$ I know 
$$ <g_\epsilon * \phi, G> = <\phi, \mathcal{R}(g_\epsilon)*G> = 0$$
as $\mathcal{R}(g_\epsilon)*G $ is again of type G. 
On the other hand $\lim_{\epsilon \rightarrow 0} \mathcal{R}(g_\epsilon)*G = G$ so also $$\lim_{\epsilon \rightarrow 0} <g_\epsilon * \phi, G> = <\phi,\mathcal{R}(g_\epsilon)*G>  =<\phi, g_\epsilon *G> =  <\phi, G> = 0$$ by the symmetry of $g_\epsilon$ and the definition of $\phi$. 
Maybe my thoughts are flawed but so far I don't know how to conclude that $\phi =0$. (This should be equivalent to the support of $\phi$ being the empty set, but I only see that $<\phi, \psi>$ vanishes for $\psi \in G$ and not for all $\psi \in \mathcal{S}(\mathbb{R}^n))$
 A: HINT: Take $\phi\in \mathcal S'$ (a continuous linear functional) with the property that 
$$\tag{1}\langle \phi, G\rangle = 0\qquad \forall G(x)=Ae^{\frac{|x-x_0|^2}{2\sigma^2}}, $$ 
where $x_0\in\mathbb{R}^n, A\in\mathbb{R}, \sigma >0$ are arbitrary. Claim (to be proved): $\phi=0$. Once this is proved, the exercise is done by standard duality arguments. 
EDIT 
Let us prove the Claim in the special case $\phi\in L^1(\mathbb{R}^n)$. We define 
$$
G_\epsilon(x)=\frac{1}{\epsilon^n} e^\frac{|x|^2}{2\epsilon^2}.$$ 
The property (1) now reads 
$$
\int_{\mathbb R^n} \phi(x) Ae^\frac{|x-x_0|^2}{2\sigma^2}\, dx = 0, $$
so it implies that the convolution product $\phi\ast G_\epsilon$ vanishes identically:
$$
\tag{2} \phi\ast G_\epsilon(x)=0.$$
Since we know that $\phi\ast G_\epsilon \to \phi$ (in $L^1$) as $\epsilon \to 0$, we conclude that $\phi=0$. 
TO DO 
We need to remove the assumption $\phi\in L^1$. One possibility is to use the distributional definition of the convolution. The one you used above, however, is too complicated. I think that, if $\phi\in \mathcal S '$ and $\psi\in \mathcal S$, the convolution $\phi\ast \psi$ can be defined as 
$$\tag{3}
\phi\ast \psi (x)= \left\langle \phi, \psi(x- \cdot) \right\rangle.$$ 
Note that this definition makes sense: it is a pairing of an element of $\mathcal S '$ with an element of $\mathcal S$. Using formula (3), you can carry over the $L^1$ proof to the general case. 
