Convergence in $D'(\mathbb{R})$

Question

Since $\chi_{R}$ denote the characteristic function of $R \subset \mathbb{R}$, that is, $\chi_{R} = 1$, if $x \in R$ and $\chi_{R} = 0$, if $x \notin R$. I'm trying to show that for any $\phi \in D(\mathbb{R})$ and $\epsilon > 0$,

$\left\langle \frac{1}{\epsilon} \chi_{[-\frac{\epsilon}{2},\frac{\epsilon}{2}]}, \phi \right\rangle \rightarrow \left\langle \delta_0, \phi \right\rangle$ as $\epsilon \rightarrow 0$.

Notion of convergence in $D'(\mathbb{R})$ (Is the vector space of all distributions).

Can someone help me to show this convergence, I need to present a work on the subject.

Answer 1

Let $T_{\epsilon}$ denote the distribution given, for all $\phi\in\mathscr{D}(\mathbb{R})$, by $$ \langle T_{\epsilon},\phi\rangle=\int_{\mathbb{R}}\frac{1}{\epsilon}\chi_{[-\frac{\epsilon}{2},\frac{\epsilon}{2}]}(x) \phi(x)\ dx\ . $$ The question is about showing $\lim_{\epsilon\rightarrow 0}T_{\epsilon}=\delta_0$ in the topological space $\mathscr{D}'(\mathbb{R})$.

Just like in more elementary situations, this means showing that for all $U$ open in $\mathscr{D}'(\mathbb{R})$ which contains $\delta_0$, there exists $\epsilon_0>0$ such that for all $\epsilon\in(0,\epsilon_0) $, we have $T_{\epsilon}\in U$. This in turn means showing for every bounded set $B$ in $\mathscr{D}(\mathbb{R})$, that $$ \lim_{\epsilon\rightarrow 0}||T_{\epsilon}-\delta_0||_B = 0 $$ where $||T||_B$ denotes $\sup_{\phi\in B}|\langle T,\phi\rangle|$ for every distribution $T$.

Now $$ \langle T_{\epsilon}-\delta_0,\phi\rangle= \frac{1}{\epsilon}\int_{-\frac{\epsilon}{2}}^{\frac{\epsilon}{2}}(\phi(x)-\phi(0))\ dx $$

$$ =\int_{-\frac{1}{2}}^{\frac{1}{2}}(\phi(\epsilon s)-\phi(0))\ ds $$ By the mean value theorem, $|\phi(\epsilon s)-\phi(0)|\le \epsilon|s|\times \sup_{z\in\mathbb{R}}|\phi'(z)|$.

From this we immediately get $$ ||T_{\epsilon}-\delta_0||_B\le \epsilon\times\frac{1}{2}\times\sup_{\phi\in B}\sup_{z\in\mathbb{R}}|\phi'(z)|\ . $$ Finally, a subset $B$ in $\mathscr{D}(\mathbb{R})$ is bounded iff 1) there exists a compact set $K$ such that $\forall\phi\in B, {\rm supp}(\phi)\subset K$ and 2) $\forall k\ge 0, \sup_{\phi\in B}\sup_{z\in\mathbb{R}}|\phi^{(k)}(z)|$ is finite.