Smooth Real Function times Derivative of Dirac Delta Distribution

From ProofWiki
Jump to navigation Jump to search

Theorem

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

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


Then in the distributional sense it holds that:

$\alpha \cdot \delta' = \map \alpha 0 \delta' - \map {\alpha'} 0 \delta$


Proof

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

\(\ds \map {\alpha \cdot \delta'} \phi\) \(=\) \(\ds \map {\delta'} {\alpha \phi}\) Definition of Multiplication of Distribution by Smooth Function
\(\ds \) \(=\) \(\ds - \map \delta {\paren {\alpha \phi}'}\) Definition of Distributional Derivative
\(\ds \) \(=\) \(\ds - \map \delta {\alpha' \phi + \alpha \phi'}\)
\(\ds \) \(=\) \(\ds - \map {\alpha'} 0 \map \phi 0 - \map \alpha 0 \map {\phi'} 0\) Definition of Dirac Delta Distribution
\(\ds \) \(=\) \(\ds -\map {\alpha'} 0 \map \delta \phi - \map \alpha 0 \map \delta {\phi'}\) Definition of Dirac Delta Distribution
\(\ds \) \(=\) \(\ds -\map {\alpha'} 0 \map \delta \phi + \map \alpha 0 \map {\delta'} \phi\) Definition of Distributional Derivative

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Smooth_Real_Function_times_Derivative_of_Dirac_Delta_Distribution&oldid=567240"