Conditions for Preservation of Covergence in Test Function Space under Differentiation

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Theorem

For all $n \in \N$ let $\Phi_n, \phi_n \in \map \DD \R$ be test functions.

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

Let $\phi_n$ be such that:

$\ds \int_{-\infty}^\infty \map {\phi_n} x \rd x = 0$

Let $\Phi_n$ be such that $\Phi_n' = \phi_n$.

Let $\sequence {\Phi_n}_{n \mathop \in \N}$ and $\sequence {\phi_n}_{n \mathop \in \N}$ be sequences in $\map \DD \R$.

Suppose $\sequence {\phi_n}_{n \mathop \in \N}$ converges to $\mathbf 0$ in $\map \DD \R$.


Then $\sequence {\Phi_n}_{n \mathop \in \N}$ converges to $\mathbf 0$ in $\map \DD \R$ as well.


Proof

By Characterization of Derivative of Test Function we have that for every $\phi_n$ there is a unique $\Phi_n$ such that:

$\ds \map {\Phi_n} x = \int_{-\infty}^x \map {\phi_n} x \rd x$

Let $K = \closedint {-a} a$ be a closed real interval.

Suppose that $\sequence {\phi_n}_{n \mathop \in \N}$ is supported on $K$.

Hence:

$\forall n \in \N : \forall x \in \R \setminus K : \map {\phi_n} x = 0$

From assumption about $\phi_n$ it follows that:

$\forall n \in \N : \forall x \in \R \setminus K : \map {\Phi_n} x = 0$

Furthermore, for all $x \in \R$ we have that:

\(\ds \size {\map {\Phi_n} x}\) \(=\) \(\ds \size {\int_{-\infty}^x \map {\phi_n} x \rd x}\)
\(\ds \) \(\le\) \(\ds 2a \sup_{x \mathop \in \R} \size {\map {\phi_n} x}\)
\(\ds \) \(=\) \(\ds 2 a \norm {\phi_n}_\infty\) Definition of Supremum Norm

By assumption, $\sequence {\phi_n}$ converges to $\mathbf 0$ in $\map \DD \R$.

Hence, $\sequence {\phi_n}$ converges uniformly to $\mathbf 0$ on $\R$:

$\ds \forall \epsilon \in \R_{>0}: \exists N \in \R: \forall n \ge N: \sup_{x \mathop \in \R} \size {\map {\phi_n} x} < \epsilon$

Therefore:

\(\ds \forall x \in \R : \ \ \) \(\ds \lim_{n \mathop \to \infty} \size {\map {\Phi_n} x}\) \(\le\) \(\ds \lim_{n \mathop \to \infty} 2 a \norm {\phi_n}_\infty\)
\(\ds \) \(=\) \(\ds 0\)

Thus, $\sequence {\map {\Phi_n} x}$ converges uniformly on $\R$ to $\mathbf 0$.

Since $\Phi'_n = \phi_n$, it follows that $\Phi_n^{\paren k} = \phi_n^{\paren {k - 1} }$ with $k \in \N_{\mathop > 0}$.

Hence, for all $k \in \N_{> 0}$ we have that:

\(\ds \size {\map {\Phi^{\paren k}_n} x}\) \(=\) \(\ds \size {\map {\phi^{\paren {k - 1} } } x}\)
\(\ds \) \(\le\) \(\ds \sup_{x \mathop \in \R} \size {\map {\phi^{\paren {k - 1} } } x}\)
\(\ds \) \(=\) \(\ds \norm {\phi_n^{\paren {k - 1} } }_\infty\) Definition of Supremum Norm

By assumption, $\sequence {\phi_n}$ converges to $\mathbf 0$ in $\map \DD \R$.

Hence, for all $k \in \N_{\mathop > 0}$ we have that $\sequence {\phi_n^{\paren {k - 1} } }$ converges uniformly to $\mathbf 0$ on $\R$:

$\ds \forall k \in \N_{\mathop > 0} : \forall \epsilon \in \R_{>0}: \exists N \in \R: \forall n \ge N: \sup_{x \mathop \in \R} \size {\map {\phi_n^{\paren k} } x} < \epsilon$

Then:

\(\ds \forall x \in \R : \forall k \in \N_{\mathop > 0} \ \ \) \(\ds \lim_{n \mathop \to \infty} \size {\map {\Phi_n^{\paren k } } x}\) \(=\) \(\ds \lim_{n \mathop \to \infty} \norm {\phi_n^{\paren {k - 1} } }_\infty\)
\(\ds \) \(=\) \(\ds 0\)

Hence, for all $k \in \N_{\mathop > 0}$ we have that $\sequence {\Phi_n^{\paren k} }$ converges uniformly to $\mathbf 0$ on $\R$.

By definition, $\sequence {\Phi_n}$ converges to $\mathbf 0$ in $\map \DD \R$.

$\blacksquare$


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