The Wikipedia page on distributions says that multiplication of two distributions whose singular supports are disjoint is easy to define, but the page doesn't actually define it. I'm not sure how to define the multiplication. The multiplication of a distribution and a smooth function is easy to define. Let $f$ be a distribution acting on usual test functions, and let $g$ be a smooth function. Then the standard definition of $fg$ is $\left< fg, \phi \right> = \left<f, g \phi \right>$ for an arbitrary test function $\phi$. This definition makes sense because $g \phi$ is a test function. However, in the case of two general distributions, this definition does not make sense regardless of disjoint singular supports. How can I define the multiplication?
1 Answer
$\newcommand{\singsupp}{\operatorname{sing supp}}$ Let $u, v$ be two distributions with disjoint singular support. Take $\psi_u, \psi_v \in C^\infty$ such that $\psi_u = 1$ on a neighborhood of $\singsupp u$ and $\psi_u = 0$ on a neighborhood of $\singsupp v$, reversely for $\psi_v$. Then we decompose $u$ and $v$ according to $u = \psi_u u + (1-\psi_u) u$ and $v = \psi_v v + (1-\psi_v) v$. Further we have $\psi_u v, \psi_v u, (1-\psi_u) u, (1-\psi_v) v \in C^\infty$.
Formally, $$uv = \left(\psi_u u + (1-\psi_u) u \right) \left(\psi_v v + (1-\psi_v) v\right) \\ = \left(\psi_u u\right) \left(\psi_v v\right) + \left(\psi_u u\right) \left((1-\psi_v) v\right) + \left((1-\psi_u) u\right) \left(\psi_v v\right) + \left((1-\psi_u) u\right) \left((1-\psi_v) v\right) $$
The first term, $\left(\psi_u u\right) \left(\psi_v v\right)$, although not defined as it stands, can formally be rewritten as $\left(\psi_v u\right) \left(\psi_u v\right) \in C^\infty$ and thus be made defined.
The second term is a distribution $\psi_u u$ times a $C^\infty$ function $(1-\psi_v) v$ and is thus defined.
The third term is of the same form as the second term but with $u$ and $v$ swapped, so it is also defined.
The forth term is a product of two $C^\infty$ functions and thus defined.
Therefore we can take the above formula (edit: with $\left(\psi_u u\right) \left(\psi_v v\right)$ replaced with $\left(\psi_v u\right) \left(\psi_u v\right)$) as a definition of $uv$.
