5
$\begingroup$

The question is: does the sequence of characteristic functions $f_k(x) := \chi_{[-\frac{1}{k}, \frac{1}{k}]}(x)$ converge in distributional sense to the Dirac delta?

In order to answer I followed this approach, but I fear I'm neglecting something important in my lines:

Firstable, $f_k\in L^1_{loc}(\mathbb{R})$, so we can write the action of the associated distribution as $$\langle T_k(x),\psi \rangle= \int_\mathbb{R}\chi_{[-\frac{1}{k}, \frac{1}{k}]}(x)\cdot\psi(x)dx=\int_{-\frac{1}{k}}^\frac{1}{k}\psi(x)dx$$ for every test function $\psi \in C^\infty_c(\mathbb{R})$. Then I computed the limit as: $$\displaystyle{\lim_{k\to \infty}\langle T_k(x),\psi\rangle =\lim_{k\to \infty} \int_{-\frac{1}{k}}^\frac{1}{k}{\psi(x)dx} =\lim_{k\to \infty} \int_{-1}^1{\frac{1}{k}\cdot\psi(\frac{y}{k})dy}}$$ and applied the Lebesgue dominated convergence theorem, saying that $\frac{|\psi(\frac{y}{k})|}{|k|} \le \sup_{x\in [-1,1]}|\psi(x)| <\infty$ and $\displaystyle{\lim_{k\to \infty}{\frac{\psi(\frac{y}{k})}{k}} = 0 }$. Then I deduced that the sequnce $T_k$ converges to the distribution associated to the zero function.

Is my proof correct? Any check or observation would be really appreciated.

$\endgroup$

2 Answers 2

6
$\begingroup$

Your argument is correct. Indeed, characteristic functions $\chi_\varepsilon$ of smaller and smaller intervals $[-\varepsilon,\varepsilon]$ do not converge to $\delta$ (in a distributional sense), but, rather, to $0$. Still, the renormalizations to have "total mass" $1$, namely, ${1\over 2\varepsilon}\chi_{\varepsilon}$, do converge to $\delta$, by a similar computation.

$\endgroup$
2
$\begingroup$

You are making things too commplicated. $\psi$ is a bounded function and if $|\psi| \leq M$ we get $|\int_{-1/k}^{1/k} \psi (x)dx|\leq \frac M {2k} \to 0$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .