Graduate Why don't we multiply generalized functions?

[wrobel]

Because it drives to contradictions. Here is a nice example from E. Rosinger Generalized solutions of nonlinear PDE.

We can multiply generalized functions from $\mathcal D'(\mathbb{R})$ by functions from $C^\infty(\mathbb{R})$. This operation is well defined. For example $$x\delta(x)=0\in \mathcal D'(\mathbb{R}),\quad x\cdot\frac{1}{x}=1\in \mathcal D'(\mathbb{R}),\quad \frac{1}{x}\in \mathcal D'(\mathbb{R}).$$
On the other hand $C^\infty(\mathbb{R})\subset \mathcal D'(\mathbb{R})$

Ok then:)
$$\delta=\Big(x\cdot\frac{1}{x}\Big)\cdot\delta=\frac{1}{x}\cdot(x\delta)=0.$$

 

[anuttarasammyak]

I have interpreted $x \delta(x)=0$ as for support of good behavior function, i.e.
\int f(x) x \delta(x) dx= 0
for f(x) such that f(0) is finite. 1/x does not satisfy it.

 

[wrobel]

This example shows that you can not define a binary operation of $\mathcal D'(\mathbb R)\times \mathcal D'(\mathbb R)$ with values in $\mathcal D'(\mathbb R)$ such that
1) this operation acts on $C^\infty(\mathbb{R})\times \mathcal D'(\mathbb R)$ in the standard way;
2) this operation possesses the standard properties of the arithmetic multiplication