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This question concerns distributions, as often encountered in PDE theory, which are defined as continuous linear functionals on the space $C_0^{\infty}(\Omega)$ of test functions.

The product of two distributions, and the convolution of two distributions, is not defined in general. Are there some easy/well-known counterexamples which show why these operations cannot make sense?

user142700
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  • Consider the following: \begin{equation} P_1(x)=P_2(x)=\begin{cases} 0.1, & \text{if $0\leq x \leq 10$}.\ 0, & \text{otherwise}. \end{cases} \end{equation}

    Then, \begin{equation} (P_1*P_2)(x)=\begin{cases} .01, & \text{if $0\leq x \leq 10$}.\ 0, & \text{otherwise}. \end{cases} \end{equation}

    Thus, the integral of $P_1*P_2$ over their domain does not equal 1, and it is therefore not a probability distribution function.

     – Austin D Aug 17 '14 at 07:57

1 Answers1

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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 ...

Vobo
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