Theorem: Conditions for Preservation of Covergence in Test Function Space under Differentiation

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:
{{begin-eqn}}
{{eqn | l = \size {\map {\Phi_n} x} 
      | r = \size {\int_{-\infty}^x \map {\phi_n} x \rd x}
}}
{{eqn | o = \le 
      | r = 2a \sup_{x \mathop \in \R} \size {\map {\phi_n} x}
}}
{{eqn | r = 2 a \norm {\phi_n}_\infty
      | c = {{Defof|Supremum Norm|subdef = Continuous on Closed Interval Real-Valued Function}}
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | q = \forall x \in \R
      | l = \lim_{n \mathop \to \infty} \size {\map {\Phi_n} x}
      | o = \le
      | r = \lim_{n \mathop \to \infty} 2 a \norm {\phi_n}_\infty
}}
{{eqn | r = 0
}}
{{end-eqn}} 
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:
{{begin-eqn}}
{{eqn | l = \size {\map {\Phi^{\paren k}_n} x}
      | r = \size {\map {\phi^{\paren {k - 1} } } x}
}}
{{eqn | o = \le
      | r = \sup_{x \mathop \in \R} \size {\map {\phi^{\paren {k - 1} } } x}
}}
{{eqn | r = \norm {\phi_n^{\paren {k - 1} } }_\infty
      | c = {{Defof|Supremum Norm|subdef = Continuous on Closed Interval Real-Valued Function}}
}} 
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | q = \forall x \in \R : \forall k \in \N_{\mathop > 0}
      | l = \lim_{n \mathop \to \infty} \size {\map {\Phi_n^{\paren k } } x}
      | r = \lim_{n \mathop \to \infty} \norm {\phi_n^{\paren {k - 1} } }_\infty
}}
{{eqn | r = 0
}}
{{end-eqn}}
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 $\map \DD \R$.
{{qed}}