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Does regularity of distributions have anything to do with definiteness of their product?

+ 6 like - 0 dislike
208 views

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?

This post imported from StackExchange Physics at 2015-05-28 17:11 (UTC), posted by SE-user krzysiekb
asked May 26, 2015 in Theoretical Physics by krzysiekb (30 points) [ no revision ]
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.

This post imported from StackExchange Physics at 2015-05-28 17:11 (UTC), posted by SE-user yuggib
@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

This post imported from StackExchange Physics at 2015-05-28 17:11 (UTC), posted by SE-user krzysiekb

The square is a product with itself, hence the support is not disjoint.

@Arnold But still $0$ is not singular support of $\frac{1}{\sqrt{x}}$ right? Since it is locally integrable, it defines a regular distribution, which by definition has empty singular support. Perpahs I am missing something important here ...

1 Answer

+ 3 like - 0 dislike

Local integrability only guarantees that you get a distribution, but says nothing about the singular support.

To make your example a locally integrable function you need to define it almost everywhere. What is your definition of the distribution $x^{-1/2}$ for $x<0$? I take it to be $|x|^{-1/2}$ for definiteness, but other extensions to all of $R$ behave similarly. None of these distributions is regular at 0 since the definition of regularity requires ''that x $\not\in$ sing supp u if and only if there is a neighborhood U of x such that the restriction of u to U is a smooth function''. But you cannot do this in a neighborhood of zero with your function. 

Thus 0 is in the singular support, and indeed, the singular support is exactly $\{0\}$. The square is a product with itself, hence the support is not disjoint. Thus the theorem is not applicable in your case.

answered Jul 24, 2015 by Arnold Neumaier (12,355 points) [ no revision ]

Now I feel enlightened, thanks! Edit: I think I should clarify one more thing: regularity (hence emptiness of the sing supp's) of two distributions does (or does not?) guarantee their product to be defined globally?
 

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