Conditions for Preservation of Covergence in Test Function Space under Differentiation

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:

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:

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 $k \in \N_{>0}$.

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

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

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


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

Then:


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

Hence, for all $k \in \N_{>0}$ 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 $\map \DD \R$.