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

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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:

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

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:

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

$\blacksquare$


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