Distribution acting on Sequence of Test Functions without common Support is not Continuous

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

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

Suppose $\sequence {\phi_n}_{n \mathop \in \N}$ does not have the common compact support.

Then $T$ acting on $\sequence {\phi_n}_{n \mathop \in \N}$ is not continuous.

Proof
Let $\operatorname {III} \in \map {\DD'} \R$ be the Dirac comb.

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

Let $\phi \in \map \DD \R$ be a test function with compact support $K = \closedint 0 1$ such that:


 * $\forall x \in K : \map \phi x > 0$

Let $\phi_n \in \map \DD \R$ be a test function sequence such that:


 * $\ds \forall x \in \R : \forall n \in \N : \map {\phi_n} x = \frac 1 n \map \phi {\frac x n}$

We have that:


 * $\ds \forall n \in \N : \map {\phi_{n + 1} } n = \frac 1 {n + 1} \map \phi {\frac n {n + 1} } > 0$

Thus, with each $n$ we generate a new test function with a larger support.

By taking the $k$th derivative of $\map {\phi_n} x$ and applying the chain rule we get that:


 * $\ds \forall k \in \N : \map {\phi_n^{\paren k} } x = \frac 1 {n^{k + 1}} \map {\phi^{\paren k} } {\frac x n}$

Hence:

Fix $k \in \N$.

Let $\epsilon \in \R_{\mathop > 0}$.

Let $N \in \R$ be such that $\ds \frac {M_k}{N^k} < \epsilon$.

Then:


 * $\ds \forall n \in \N : n \ge N \implies \frac {M_k}{n^k} \le \frac {M_k}{N^k} < \epsilon$

Altogether:


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

Hence, as $n \rightarrow \infty$ we get that $\phi_n^{\paren k}$ converges to $\mathbf 0$ uniformly.

However:

We have that Test Function is Riemann Integrable.

Let the subdivision of the interval $\closedint 0 1$ be normal:


 * $\ds \forall i \in \N_{\mathop \le n} : x_{i + 1} - x_i = \frac 1 n$

Then $\map T {\phi_n}$ is a Riemann sum with $\ds \Delta x_k = \frac 1 n$, $\ds c_k = \frac k n$, $x_0 = 0$ and $x_n = 1$.

Taking $n \to \infty$ yields:


 * $\ds \map T {\phi_n} \stackrel {n \mathop \to \infty} \longrightarrow \int_0^1 \map \phi x \rd x$

However, we assumed that:


 * $\forall x \in K : \map \phi x > 0$

Thus:


 * $\ds \int_0^1 \map \phi x \rd x > 0$

So we have that, as $n \to \infty$, $\sequence {\phi_n}$ converges to $\mathbf 0$ uniformly, but $\map T {\phi_n}$ does not converge to $0$.