Can a monotone exponentially decreasing function be uniformely approximated bt Gaussians?

Question

This question originates an engineering application.

There is a certain process that is presumed to be a sequence of diffusions and is usually modelled as a sum of Gaussians:

$$\Sigma_n w_ne^{-\frac{x^2}{\sigma_n^2}}$$

It is known that the function $f(x)$ that is being approximated is monotone exponentially decreasing on $[0,\infty)$. The question is: can any such function be uniformly approximated with a convergent series of Gaussians?

The preferred metrics for convergence would be $L^\infty$, although any $L^p$ answer would be satisfactory.

Answer 1

You want the $w_n$ to be nonnegative, right?

With that interpretation, the answer is no. Just substitute $x^2=y$, and the problem becomes one of approximating by a sum of decaying exponentials. Such functions are well studied; they are called completely monotone.

Answer 2

The collection of functions, $\newcommand{\bR}{\mathbb{R}}$

$$f_r:\bR\to\bR,\;\;f_r(x)= e^{-rx^2}\;\;r>0, $$

spans a vector space dense in the space of even Schwartz functions on $\bR$. However, if you want the coefficients $w_n$ to be $\geq 0$ you need to first observe that the functions

$$ g_r(t)= e^{-rt}, \;\; r>0\, \;\; t\geq 0 $$

are completely monotone, i.e.,

$$ (-1)^ng_r^{(n)}(t)\geq 0,\;\;\forall t>0$. $$

In particular, any finite superposition of functions $w_rg_t(t)$, $w_r>0$ will be completely monotone. More generally for any finite positive measure $\mu$ on $(0,\infty)$ the function

$$ F(t)=\int_0^\infty g_r(t) \mu(dr) $$

is completely monotone. Observe that $F(t)$ is an infinite superposition of $g_r$'s. Conversely, Bernstein's theorem states that any completely monotone function $F(t)$ admits an integral description as above.