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

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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:

\(\ds \sup_{x \mathop \in \R} \size {\map {\phi_n^{\paren k} } x - 0}\) \(=\) \(\ds \sup_{x \mathop \in \R} \size {\frac 1 {n^k} \map {\phi^{\paren k} } x}\)
\(\ds \) \(\le\) \(\ds \frac 1 {n^k} \sup_{x \mathop \in \R} \size {\map {\phi^{\paren k} } x}\)
\(\ds \) \(=\) \(\ds \frac {M_k} {n^k}\) $\ds M_k := \sup_{x \mathop \in \R} \size {\map {\phi^{\paren k} } x} \in \R_{> 0}$

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:

\(\ds \map T {\phi_n}\) \(=\) \(\ds \sum_{k \mathop \in \Z} \map {\phi_n} k\) Definition of Dirac Comb
\(\ds \) \(=\) \(\ds \sum_{k \mathop \in \Z} \frac 1 n \map \phi {\frac k n}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = - \infty}^{- 1} \frac 1 n \map \phi {\frac k n} + \sum_{k \mathop = 0}^n \frac 1 n \map \phi {\frac k n} + \sum_{k \mathop = n}^\infty \frac 1 n \map \phi {\frac k n}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^n \frac 1 n \map \phi {\frac k n}\) Definition of Support of Continuous Real-Valued Function

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 $0$.

$\blacksquare$


Sources

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