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
Aiming for a contradiction, suppose 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:
\(\ds 0\) | \(=\) | \(\ds \map {\mathbf 0} {\phi_\lambda}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {\paren{\sum_{i \mathop = 0}^n c_i \delta^{\paren i} } } {\phi_\lambda}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 0}^n c_i \map {\delta^i} {\phi_\lambda}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 0}^n \paren {-1}^i c_i \lambda^i \map {\phi^{\paren i} } 0\) | Definition of Distributional Derivative |
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.
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 6.2$: A glimpse of distribution theory. Derivatives in the distributional sense