Does regularity of distributions have anything to do with definiteness of their product? Recently I've gone through some literature concerning causal perturbation theory (CPT). As is well known, it deals with UV divergences in QFT by defining products of (operator-valued) distributions rigorously.
Now I'm confused whether regularity of two distributions would be sufficient to define their product globally. Two remarks:


*

*in the paper http://arxiv.org/abs/1404.1778, pg. 4, there is a theorem, that given two distributions with disjoint singular supports, their product is well-defined; clearly it would be defined for all regular distributions since their singular supp's are empty;

*however, an example of $\frac{1}{\sqrt{x}}$ being regular does not define it's square $\frac{1}{x}$ as a distribution on all test functions.


What is going on here?
 A: What is going on here is that the example you give of $\frac{1}{\sqrt{x}}$ is not regular. The singular support is not empty, it is equal to $\{0\}$. So the theorem you mentioned does not apply. You trivially get an element of $\mathcal{D}'(\mathbb{R}\backslash\{0\})$ but you still have to work harder in order to get a distribution on the whole real line.
A: I think you are referring to the definition of product of distributions due to Hoermander based on the notion of wavefront set. The corollary you mention has this precise form: if the sigular supports of a pair of distributions have empty intersection, then their (Hoermander) product is well defined.
The singular support of the distribution $u$ is the complement of the union of the open sets $U$ such that $u(f) = \int g_U f dx$ for some $C^\infty$ function $g_U$ and every test function supported in $U$.
With the said definition, the singular support of $u= 1/\sqrt{x}$ is $0$ (though I do not understand well how you define $u$ for $x<0$). So the corollary does not apply. 
