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I was wondering what made it impossible to define a product of distributions. Googling, I found two questions, one of which stated the following impossibility result:

There is no associative algebra over $\mathbb{R}$ containing $\mathcal{D}'$ as a vector subspace and the constant function 1 as unity element, having a differential operator acting like the differential operator on $\mathcal{D}'$ and the algebra multiplication of continuous functions is like their pointwise multiplication.

I Googled a bit, but this query doesn't seem to yield the desired result, and I wouldn't know how else to phrase it. So where can I find a proof of this result? Is it too long for an answer or can someone post one as an answer?

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MickG
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  • As I said in my linked answer, this is *the imposibility result* (by L. Schwartz). Sure you may define a product by defining $S \cdot T = 0$ for all distributions $S,T$, but most probably this is not what you want. You want the product to behave somehow "nice" (like the pointwise multiplication of continous functions etc.), and Schwartz stated that there is no such thing ... – Vobo Jul 13 '15 at 16:25
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    @Vobo it seems to me your comment would be better suited to my other related question rather than this one which is merely asking for a proof of that result. – MickG Jul 13 '15 at 16:36
  • As Schwartz books are not available online, I am afraid you have to go to a library to get his "Theorie des Distributions". There is also a full treatment of Michael Oberguggenberger, "Mulitplication of distributions and applications to pde's" from the early 90's (which was in fact my motivation for a thesis in this subject). And yes, I misunderstood your question at first. – Vobo Jul 13 '15 at 16:37

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The proof of the impossibility result goes along these lines: suppose that there exists a distribution $x^{-1}$ such that $x x^{-1} = 1$. Recall also that $\delta_0 x = 0$. If the product between distributions is associative, we deduce the contradiction $$ 0 = (\delta_0 x) x^{-1} = \delta_0 (x x^{-1}) = \delta_0. $$ The proof of the existence of the distribution $x^{-1}$ requires the use of a derivative satisfying the Leibnitz Rule (i.e. the chain rule for products of functions). If you want, you can read the original paper by Schwartz (written in French) here: http://sites.mathdoc.fr/OCLS/pdf/OCLS_1954__21__1_0.pdf

Notice however that, by relaxing some of the hypothesis of the theorem, there can be created algebras of functions extending the distributions. An example is given by Colombeau's algebras, where the multiplication is coherent only with the product between smooth functions (and not simply continuous ones). Another option is working in the setting of nonstandard analysis and relaxing the Leibnitz Rule: in this case, the distributions can be embedded in an algebra of functions where the product is coherent with pointwise product between continuous functions (see http://link.springer.com/article/10.1007%2Fs00605-014-0647-x).

I am also working on a similar result over a different algebra of nonstandard functions, where the derivative satisfies the Leibnitz Rule and product is an extension of the pointwise product between continuous functions. The catch here is that $x$ is not invertible, but there is a function $x^\star$ such that $x x^\star = 1$ at every point but $0$.

Emanuele Bottazzi
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  • Of course, $f(x)=x$ is not invertible at zero. I do not know who came with an idea it should be. Are there other contradictions besides this? – Anixx Oct 30 '19 at 20:28
  • Recall that in the space of distributions one already has the principal value distribution $P\frac{1}{x}$. Moreover, the product between $x$ and$P\frac{1}{x}$ can be defined, and it is the distribution corresponding to the constant function $c(x)=1$ (see e.g. http://moby.mib.infn.it/~zaffaron/distribuzioni.pdf). If you look at some algebras of generalized functions, you might get a feeling of what can be obtained by relaxing one of the conditions in Schwart's original theorem. For instance, if you only ask for coherence of the product between $C^\infty$ functions, you get Colombeau algebras. – Emanuele Bottazzi Nov 07 '19 at 17:00