Dirac Comb is Distribution

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.

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

By definition, $\sequence {\phi_n}$ is 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:

By definition, the Dirac comb is a distribution.

Also see

 * Closed Real Interval is Compact