Delta derivative distribution identity?

Question

It is easy to show that the Dirac $\delta(x)$ distribution satisfies the distributional identity $$\delta(x) = - x \delta'(x).$$ Can we conclude that the following also holds $$\delta'(x) = - \frac{\delta(x)}{x}?$$ In other words, are we allowed to multiply and divide distributions with ordinary functions?

Answer 1

Multiplication of a distribution $u$ with a function $f$ is defined by $\langle fu, \phi \rangle := \langle u, f\phi \rangle.$ Since $C_c^\infty$ is closed under multiplication with $C^\infty$ functions, we are allowed to multiply distributions with $C^\infty$ functions. We are, using this definition, not allowed to multiply distributions with non-$C^\infty$ functions like $\frac{1}{x}.$

However, we can try to solve distribution equations like $fu = v,$ where $f$ is a $C^\infty$ function and $v$ is a given distribution. This equation often has solutions, but there might be an infinitude of solutions. For example, the solutions to $xu = \delta$ are $u = -\delta' + C \delta,$ where $C$ is a constant.

More generally, if $f$ has zeros of finite order then $fu = v$ has a family of solutions which differ by $\delta$:s and derivatives thereof placed at the zeros.

In the specific case $xu = \delta$ we can say that $u = -\delta'$ is the canonical solution, since $x$ being odd and $\delta$ being even should result in $u$ being odd. But in the general case we can not do this.