Product of Smooth Function and Dirac Delta Distribution

Theorem
Let $a \in \R^d$ be a vector in Euclidean space.

Let $\delta_a \in \map {\DD'} {\R^d}$ be the Dirac delta distribution.

Let $\alpha \in \map {C^\infty} {\R^d}$ be a smooth real function.

Then in the distributional sense we have that:


 * $\alpha \delta_a = \map \alpha a \delta_a$

Proof
Let $\phi \in \map \DD {\R^d}$ be a test function.

Then: