Dirac Delta Distribution is Tempered Distribution

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Theorem

Let $\delta$ be the Dirac delta distribution.

Let $\map {\SS'} \R$ be the tempered distribution space.


Then:

$\delta \in \map {\SS'} \R$


Proof

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

Consider the mapping $\phi \stackrel \delta \longrightarrow \map \phi 0 : \map \SS \R \to \C$.

We have that the Schwartz space is a vector space.

Let $\phi$ be a linear superposition of Schwartz test functions.

Then $\map \phi 0$ is also a linear supperposition evaluated at $x = 0$.

Hence, the mapping $\phi \stackrel \delta \longrightarrow \map \phi 0 : \map \SS \R \to \C$ is linear.


Suppose, $\sequence {\phi_n}_{n \mathop \in \N}$ converges in $\map \SS \R$:

$\phi_n \stackrel \SS \longrightarrow \mathbf 0$

We have that:

$\ds \forall x \in \R : \forall l, m \in \N : 0 \le \size {x^l \map {\phi_n^{\paren m} } x} \le \sup_{x \mathop \in \R} \size {x^l \map {\phi_n^{\paren m} } x}$

Specifically, for $x = 0$, $l = 0$, and $m = 0$ it holds that:

$\ds 0 \le \size {\map {\phi_n} 0} \le \sup_{x \mathop \in \R} \size {\map {\phi_n} x}$

Taking the limit $n \to \infty$ and applying the squeeze theorem yields:

$\ds \lim_{n \mathop \to \infty} \map {\phi_n} 0 = 0$.

$\blacksquare$


Sources

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