Dirac Comb is Distribution

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Theorem

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

Suppose $\map {\operatorname {III} } 0$ is a Dirac comb such that:

$\ds \map {\map {\operatorname {III} } 0} \phi := \sum_{n \mathop \in \Z} \map \phi n$


Then $\map {\operatorname {III} } 0$ is a distribution.


Proof

By definition of test function, $\phi$ is supported on a compact subset of $\R$.

Hence:

$\exists N \in \N : \forall x \in \R \setminus \closedint {-N} N : \map \phi x = 0$

Therefore:

$\ds \sum_{n \mathop \in \Z} \map \phi n = \sum_{n \mathop = - N}^N \map \phi n$

This is a finite sequence of real numbers.

Thus, the sequence converges, and $\map {\operatorname {III} } 0$ is a mapping such that $\map {\operatorname {III} } 0 : \map \DD \R \to \R$.


Linearity

Follows from Summation is Linear.

$\Box$


Convergence in $\map \DD \R$

Let $\sequence {\phi_n}$ be a convergent sequence in $\map \DD \R$ with the limit $\mathbf 0$.

By definition of convergence, all $\sequence {\phi_n}$ are supported on a compact subset of $\R$, say, $\closedint {-K} K$ with $K \in \N$.

By definition of convergence, $\sequence {\phi_n}$ converges to $\mathbf 0$ uniformly.

Hence:

$\ds \forall k \in \R : \size k < K : \lim_{n \mathop \to \infty} \map {\phi_n} k = 0$

Therefore:

\(\ds \map {\map {\operatorname {III} } 0} {\phi_n}\) \(=\) \(\ds \sum_{k \mathop \in \Z} \map {\phi_n} k\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = -K}^K \map {\phi_n} k\)
\(\ds \leadsto \ \ \) \(\ds \lim_{n \mathop \to \infty} \map {\map {\operatorname {III} } 0} {\phi_n}\) \(=\) \(\ds \lim_{n \mathop \to \infty} \sum_{k \mathop = -K}^K \map {\phi_n} k\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = -K}^K 0\) Sum Rule for Real Sequences
\(\ds \) \(=\) \(\ds 0\)

$\Box$

By definition, the Dirac comb is a distribution.

$\blacksquare$


Also see


Sources

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