Approximation of a compactly supported function by Gaussians Let $f:\mathbb{R}\to\mathbb{R}$ be a smooth function whose support is a closed interval, e.g. $\text{supp}(f)=[a,b]$. Then $f$ can be approximated (e.g. in $L^2$) by a linear combination of Gaussian densities, i.e.
$$
f(x)
\approx 
G_n(x)
:=\sum_{i=1}^n a_{i,n} g(x;\mu_{i,n},\sigma_{i,n}^2),
\quad
a_{i,n}\in\mathbb{R},
\quad
g(x;\mu,\sigma^2) 
=\frac1{\sqrt{2\pi\sigma^2}}e^{-\frac1{2\sigma^2}(x-\mu)^2}.
$$
Suppose that for sufficiently large $n$, a best approximation to $f$ exists in some norm (again, e.g. $L^2$, but it doesn't really matter), i.e.
$$
\Vert f-G_n^*\Vert
\le \inf\big\{\Vert f-G_n\Vert : a_{i,n}\in\mathbb{R},\, \mu_{i,n}\in\mathbb{R},\, \sigma_{i,n}^2>0 \big\}
$$
and furthermore $G_n^*\to f$.
I'd like to show that as long as $n$ is sufficiently large, then $\mu_{i,n}^*\in [a,b]$ for all $i=1,\ldots,n$. In other words, if a best approximation is arbitrarily close to $f$, it can't do something stupid like use a Gaussian component whose mass is mostly concentrated off of $\text{supp}(f)$.
The difficulty as I see it is that the weights $a_i$ can be arbitrarily close to 0, so there exist (arbitrarily) good approximators $G_n$ with "rogue" components (i.e. $\mu_i\not\in\text{supp}(f)$), but whose weight $a_i$ is very small. the challenge is showing somehow that "best" approximators avoid this kind of pathology.
Edit: Changed the assumptions so that the support of $f$ is necessarily an interval---this may not be true for arbitrary compact sets.
 A: Case 1: Let $f\geq 0$. We consider the following easier problem :$$ \|f-G_n^*\|_{L^1}\leq \inf \{\|f-G_n\|_{L^1}:a_{i,n}\geq 0, \mu_{i,n}\in \mathbb{R}, \sigma_{i,n}>0 \}. $$
Since $f\geq 0$, we have $G_n^*\rightarrow f $. In this case, for any $n$ we have $\mu_{i,n}^*\in [a,b]$. Intuitively, otherwise more than half of the mass of a gaussian would be outside $[a,b]$ and gives a strictly positive contribution to the norm. 
Detailed proof:
We have
$$ \int_\mathbb{R} |G_n - f|dt=\int_a^b |G_n - f|dt + \sum_{i=1}^n a_i \left(\int_{-\infty}^a g_i(t)+ \int^{-\infty}_b g_i(t)dt\right)$$where $g_i(t)=g(t;\mu_i,\sigma_i)$. Let $G_n^*$ a best approximation and suppose that there exists $\mu_i^*<a$ (resp $>b$). For $s\in [0,1]$, define $$G_n(s)=G_n^* - (1-s)a_i g_i. $$ $G_n(s)$ is the same as $G_n^*(s)$ with $a_i$ replaced by $s a_i^*$.
Therefore
$$\begin{align}
\int_{\mathbb{R}} |G_n(s) - f|dt & = \int_\mathbb{R} |G_n^* -(1-s)a_ig_i- f|dt\\
&\leq \int_a^b \Big\{|G_n^* - f|+(1-s)|a_i g_i|\Big\}dt + \int_{t\notin [a,b]} |G_n^*-(1-s)a_i g_i| \\
&= \int_a^b \Big\{|G_n^* - f|+(1-s)|a_i g_i|\Big\}dt + \int_{t\notin [a,b]} G_n^*-(1-s)a_i g_i \\
&= \int_a^b |G_n^* - f|dt+\int_{t\notin [a,b]} G_n^* dt  \\ &\quad+(1-s)a_i \left(\int_a^b g_idt-\int_{t\notin[a,b]} g_i(t)dt\right)\\
\end{align}$$
But because $\int_{-\infty}^a g_i(t)dt> \frac{1}{2}$ and $\int_a^b g_i(t)dt< \frac{1}{2}$ the last term is stictly negative and we have 
$$\|G_n(s)-f\|=\int_{\mathbb{R}} |G_n(s) - f|dt < \int_{\mathbb{R}} |G_n^* - f|dt = \|G_n^* - f\|$$
for all $s\in (0,1)$ which is in contradiction to the minimisation of $G_n^*$. $\square$
Case 2: We don't make any assumption in $f$, the norm $\|.\|$ or the positivity of $a_i$. Here I don't have a rigorous proof but I argue that your claim is not true. I try to construct the following counter example:  Let $$f(t)=1_{[-1,1]}(t)g(t)$$ and  $$h(t)=f(t)-g(t)$$ where $g(t)=g(0,1,t)$.
I make the following natural generalisation of your claim: if $\text{supp}(f) = (-\infty,a]$ or $\text{supp}(f) = \mathbb{R}-[a,b]$ and $f$ decay to 0 fast enough at $\pm \infty$ then your claim is also true for $f$. 
We note $F_n^*$ be the best approximation of $f$ and $H_n^*$ the best approximation of $h$. Now remark that 
$$\|h - H_{n+1}^*\| \leq \| h - F_n^* +g\|= \|f-F_n^*\|$$ and 
$$\|f - F_{n+1}^*\| \leq \| f - H_n^* -g\|= \|h-H_n^*\|$$
So 
$$\|f - F_{n+2}^*\| \leq  \|h-H_{n+1}^*\| \leq \|f - F_{n}^*\|$$
Therefore $H_{n+2}^* + g $ is also a very good approximation of $f$ and is least better than $F_n^*$. 
But if both yours claim and the natural generalisation I proposed were true then as $n\rightarrow \infty$ all the gaussian of $H_n^*$ should be centered in $\mathbb{R}-[-1,1]$ and all the gaussiaan of $F_n^*$ should be centered in $[-1,1]$.
But this seems to me very hard to believe that $H_n^*-g$ and $F_n^*$ are so different one to another and yet gives almost as good approximation of $f$.
