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.