Distributional Derivatives of Dirac Delta Distribution do not Vanish

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Theorem

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


Then for any $n \in \N$ the distributional derivative $\delta^{\paren n}$ does not vanish.


Proof

Let $\phi \in \map \DD \R$ be a test function such that $\map \phi 0 \ne 0$.

Then:

$\forall n \in \N : \forall x \in \R : x^n \phi \in \map \DD \R$

By the definition of the distributional derivative:

\(\ds \map {\delta^{\paren n} } {x^n \phi}\) \(=\) \(\ds \paren {-1}^n \map \delta {\paren {x^n \phi}^{\paren n} }\)
\(\ds \) \(=\) \(\ds \paren {-1}^n \map \delta {\sum_{k \mathop = 0}^n \binom n k \paren {\dfrac {\d^k} {\d x^k} x^n} \phi^{\paren {n - k} } }\) Leibniz's Rule for One Variable
\(\ds \) \(=\) \(\ds \paren {-1}^n \binom n n n! \map \phi 0\) Nth Derivative of Nth Power, Definition of Dirac Delta Distribution
\(\ds \) \(\ne\) \(\ds 0\)


$\blacksquare$


Sources