Products of distributions ill-defined

Solution 1:

The famous "Impossibility Result" of Laurent Schwartz precisely says:

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.

Basically the proof shows the impossibility of the coexistence of singular elements like the $\delta$-distribution, multiply continuous functions as usual, and diffentiate them like distributions.

A nice example shows that even the very simple multiplication $C^\infty \cdot \mathcal{D}', (f\cdot S)(\varphi):=S(f\varphi)$, is not associative: You have the obvious formulas $$ 1\cdot\delta = \delta, \,\,\,\,\,\,\,\,\,\, \delta\cdot x = 0, \,\,\,\,\,\,\,\,\,\, x\cdot pv(\frac{1}{x}) = 1, $$ where $pv(\frac{1}{x})(\varphi)=\lim_{\varepsilon\to 0}\int_{|x|>\varepsilon}\frac{\varphi(x)}{x}\, dx$ denotes the principal value distribution. This implies $$ 0 = (\delta\cdot x)\cdot pv(\frac{1}{x}) \neq \delta\cdot( x\cdot pv(\frac{1}{x})) = \delta\cdot 1 = \delta. $$ So if you want to embed $\mathcal{D}'$ into an associative algebra, you necessarily need to change even the product $C^\infty\cdot\mathcal{D}'$.

This shows that you cannot define a useful product for all kind of distributions, only for some of them - basically the more singular one factor, the less must be the other ...