Schwartz impossibility result 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?
 A: 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 $0$.
