Set of Distributional Derivatives of Dirac Delta Distribution is Linearly Independent

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

Then for any $n \in \N$ the set of distributional derivatives $\set {\delta, \delta' \ldots, \delta^{\paren n} }$ is linearly independent in $\map {\DD'} \R$.

Proof
there exist scalars $c_0, c_1, \ldots, c_n \in \R$ such that:


 * $\ds \sum_{i \mathop = 0}^n c_i \delta^{\paren i} = \mathbf 0$

where $\mathbf 0 : \map \DD \R \to 0$ is the zero distribution.

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

Let $\lambda \in \R_{\mathop > 0}$

Let $\map {\phi_\lambda} x := \map \phi {\lambda x}$.

We have that $\phi_\lambda$ is a test function.

Then:

This has to hold for any $\lambda > 0$.

Hence $\paren \star$:


 * $c_0 \map \phi 0 = \ldots = c_n \map {\phi^{\paren n} } 0 = 0$

We have that:


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

By Leibniz's Rule in One Variable:


 * $\ds \map {\paren {x^n \phi}^{\paren n} } 0 = \paren {-1}^n \binom n n n! \map \phi 0 \ne 0$

Using $x^n \phi$ as our test function in $\paren \star$ yields:


 * $c_0 = \ldots = c_n = 0$

and we have a contradiction.