# Convergence of Sequence of Test Functions in Test Function Space implies Convergence in Schwartz Space

## Theorem

Let $\map \DD \R$ be the test function space.

Let $\map \SS \R$ be the Schwartz space.

Let $\sequence {\phi_n}_{n \mathop \in \N}$ be a sequence of test functions in $\map \DD \R$.

Let $\mathbf 0 : \R \to 0$ be the zero mapping.

Suppose $\sequence {\phi_n}$ converges to $\mathbf 0$ in $\map \DD \R$:

$\phi_n \stackrel \DD \longrightarrow \mathbf 0$

Then $\sequence {\phi_n}$ converges to $\mathbf 0$ in $\map \SS \R$:

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

## Proof

For all $n \in \N$ let $\phi_n$ be a test function.

By definition, $\phi_n$ has a compact support $I_n \subset \R$:

$\forall x \notin I_n \implies \map {\phi_n} x = 0$

Let:

$a \in \R : a > 0 : \forall n \in \N : I_n \subseteq \closedint {-a} a$

Then:

 $\ds \forall m,k \in \N : \sup_{x \mathop \in \R} \size {x^k \map {\phi^{\paren m}_n} x}$ $=$ $\ds \map {\sup_{\size x \mathop \le a} } {\size x^k \size {\map {\phi^{\paren m}_n} x} }$ Absolute Value of Product $\ds$ $\le$ $\ds a^k \sup_{\size x \mathop \le a} \size {\map {\phi^{\paren m}_n} x}$

Suppose:

$\phi_n \stackrel \DD \longrightarrow \mathbf 0$

Then all derivatives of $\phi_n$ on $\closedint {-a} a$ converge uniformly to $\mathbf 0$.

Hence:

 $\ds \lim_{n \mathop \to \infty} \sup_{x \mathop \in \R} \size {x^k \map {\phi^{\paren m}_n} x}$ $\le$ $\ds a^k \lim_{n \mathop \to \infty} \sup_{\size x \mathop \le a} \size {\map {\phi^{\paren m}_n} x}$ $\ds$ $=$ $\ds a^k \sup_{\size x \mathop \le a} \size {\mathbf 0}$ $\phi_n \stackrel \DD \longrightarrow \mathbf 0$ $\ds$ $=$ $\ds a^k \cdot 0$ $\ds$ $=$ $\ds 0$

In other words:

$\ds \forall m,k \in \N : \lim_{n \mathop \to \infty} \sup_{x \mathop \in \R} \size {x^k \map {\phi^{\paren m}_n} x} = 0$

By definition:

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

$\blacksquare$