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.



Approximation of a compactly supported function by Gaussians
**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 $f$.


