Distributional Derivative on Distributions is Continuous Operator

Theorem
The distributional derivative on distributions is a continuous operator.

Proof
Let $\mathbf 0$ be the zero mapping.

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

Let $\sequence {\phi_n}_{n \mathop \in \N}$ converge to $\mathbf 0$ in the test function space:


 * $\phi_n \stackrel \DD {\longrightarrow} \mathbf 0$

By definition, a test function is a smooth real function with a compact support $K$.

By differentiating a test function we get another smooth real function with a compact support not larger than $K$.

Thus, $\phi_n'$ is also a test function.

Since $\phi_n \stackrel \DD {\longrightarrow} \mathbf 0$, we have that for every multiindex $k$ the sequence $\sequence {D^k \phi_n}_{n \mathop \in \N}$ converges uniformly to $\mathbf 0$.

Obviously, all the sequences of $\sequence {D^k \phi_n}_{n \mathop \in \N}$ with $k \ge 1$ converge uniformly to $\mathbf 0$.

By definition, $\sequence {\phi_n'}_{n \mathop \in \N}$ converges to $\mathbf 0$ in the test function space:


 * $\phi_n' \stackrel \DD {\longrightarrow} \mathbf 0$

Let $T \in \map {\DD'} \R$ be a distribution.

Then:


 * $\paren {\phi_n' \stackrel \DD {\longrightarrow} \mathbf 0} \implies \paren {\map T {\phi_n'} \to \map T {\mathbf 0} = 0}$

By definition of the distributional derivative:


 * $\map T {\phi_n'} = - \map {T'} {\phi_n}$

Therefore:


 * $\paren {\phi_n' \stackrel \DD {\longrightarrow} \mathbf 0} \implies \paren {\map {T'} {\phi_n} \to 0 }$

By definition, the distributional derivative on distributions is a continuous operator.