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?

  • 3
    $\begingroup$ In the article you cite, there is a definition of product (page 6) that is coherent with the theorem you cite, but allows the definition of the product of some distributions with themselves. In particular, it gives the product of two $\theta$ functions, and of two $\frac{1}{x+i0^+}$. But not of two $\delta$ functions, or of $\frac{1}{x+i0^+}$ with $\frac{1}{x-i0^+}$. After it is introduced the Hormander's notion of Wavefront set that allows to define the product in another different fashion, and there are theorems and plenty of examples about it. $\endgroup$ – yuggib May 27 '15 at 8:26
  • $\begingroup$ @yuggib thanks for remarks; now it seems to me that what confused me was the phrase "well-defined", which probably should not be regarded as "defined on all test functions" - that will be true only after performing renormalization of 1/x $\endgroup$ – krzysiekb May 28 '15 at 0:16

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.

  • $\begingroup$ But $\frac{1}{\sqrt{x}}$ is locally integrable, so why is it not regular? $\endgroup$ – krzysiekb Jun 1 '15 at 0:21
  • $\begingroup$ locally integrable implies that integration against your function on the complement of the origin gives a well-defined distribution on the whole real line, namely an element of $\mathcal{D}'(\mathbb{R})$. However, you need to carefully review the definition of regularity on some open set and that of singular support. Here the distribution is not regular on any open set $U$ containing the origin. Otherwise, for test functions with support in $U$, the distribution would be given by $f\rightarrow\int f\phi$ for some $C^{\infty}$ function $\phi$ on $U$. If this was the case... $\endgroup$ – Abdelmalek Abdesselam Jun 1 '15 at 9:24
  • $\begingroup$ ...one would have a bound constant times the $L^{1}$ norm of $f$ for test functions in some open interval around the origin. It is easy to see that this is impossible by taking smooth approximations of indicator functions of smaller and smaller intervals around the origin. $\endgroup$ – Abdelmalek Abdesselam Jun 1 '15 at 9:27
  • $\begingroup$ @ AbdelmalekAbdesselam Is it the case that $\frac{1}{\sqrt{x}}$ is not regular at $0$ simply because it is not defined there? If that's not very clever, may I ask for some reference in the spirit of your comments/answer? $\endgroup$ – krzysiekb Jul 24 '15 at 10:20
  • $\begingroup$ @krzysiekb: I don't know else to say. In my previous comments I explained that $1/\sqrt{x}$ is not regular at 0 and I even sketched a proof of that. $\endgroup$ – Abdelmalek Abdesselam Aug 3 '15 at 12:20

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.

  • $\begingroup$ Yes, that is true. I had accepted these facts (also mentioned earlier by Abdelmalek Abdesselam) later but forgot to mark his answer as accepted... what I just did. Thanks! $\endgroup$ – krzysiekb Jan 4 '17 at 13:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.