Product of Smooth Function and Dirac Delta Distribution

Theorem

Let $a \in \R^d$ be a vector in real 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:

\(\ds \map {\alpha \delta_a} \phi\)

\(=\)

\(\ds \map {\delta_a} {\alpha \phi}\)

Definition of Multiplication of Distribution by Smooth Function

\(\ds \)

\(=\)

\(\ds \map {\paren {\alpha \phi} } a\)

Definition of Dirac Delta Distribution

\(\ds \)

\(=\)

\(\ds \map \alpha a \map \phi a\)

Definition of Pointwise Multiplication of Real-Valued Functions

\(\ds \)

\(=\)

\(\ds \map \alpha a \map {\delta_a} \phi\)

Definition of Dirac Delta Distribution

$\blacksquare$

Sources

• 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 6.4$: Multiplication by $C^\infty$ functions