Commutator of x and Distributional Derivative acting on Distribution

Theorem
Let $T \in \map {\DD'} \R$ be a distribution.

Let:


 * $\sqbrk {x, \dfrac \d {\d x} } T := x \dfrac {\d T}{\d x} - \dfrac {\d \paren {x T}}{\d x}$

where derivatives are to be understood in the distributional sense.

Then:


 * $\sqbrk {x, \dfrac \d {\d x}} T = - T$

Proof
Let $\phi \in \map \DD \R$ be a test function.