Is there a complete "infinite mixture of gaussians representation" for densities? What I mean is, is there, for any reasonably big class of densities $\phi(x)$ I can come up with a function $c(\mu, \tau)$ such that:


$\displaystyle\phi(x) = \int d\mu d\tau\; c(\mu, \tau) \exp\left(-\frac{\tau (x-\mu)^2}{2}\right)$??


($\tau = \frac{1}{\sigma}$ is the inverse of the variance and the normalization of the gaussian was absorbed in $c(\mu,\tau)$)


I'm using the word "complete" in the sense theoretical physicists talk about the completeness of a basis for a vector space. I'm not sure if it's an adequate use of the term for mathematicians (probably not). 


If yes, does it generalize for multivariate distributions?
Also, is there a known "inverse integral transform" for this representation? Something like:


$\tilde{\phi}(\Sigma, \mu) = \int dx \; \tilde{\mathcal{N}}(\Sigma, \mu | x) \phi(x)$


$\phi(x) = \int d\Sigma \mathcal{N}(x | \Sigma, \mu) \tilde{\phi}(\Sigma,\mu)$


where $\Sigma$ and $\mu$ are the covariance matrix and the vector of means.



Completeness of an "infinite mixture of gaussians" representation
How about this: the set of finite mixtures of (non-degenerate) Gaussians is weakly dense in the space of probability measures on $\mathbb{R}$. Proof: the set of finite mixtures of constants is certainly weakly dense. But a constant can be approximated as a Gaussian with small variance.


