Limit in the sense of distributions of $n^2\chi_{[-1/n,0[}$ - $n^2\chi_{]0,1/n]}$

Question

Limit in the sense of distributions of $n^2\chi_{[-1/n,0[} - n^2\chi_{]0,1/n]}$
Here's what I tried:

Let's set $T=n^2 \chi_{[-1/n,0[}$

We have for $ \phi \in \mathcal{D} $

\begin{align} <T,\phi>&=\int_{-1/n}^{0}n^2\phi(x)dx\\ &=\int_{-n}^{0}\phi(\dfrac{y}{n^2})dy\\ &=\int_{-\infty}^{\infty}\chi_{[-n,0[}(y)\phi(\dfrac{y}{n^2})dy \end{align}

I used the substitution $ y=n^2x $ above.

Since $ \phi \in \mathcal{D} $, we can use the dominated convergence theorem, which yields:

\begin{align} \lim_{n \rightarrow \infty}<T,\phi>&=\int_{-\infty}^{\infty}\chi_{[-\infty,0[}(y)\phi(0)dy\\ &=\phi(0)\int_{-\infty}^{0}dy \end{align}

Which is where I stopped. I have a particular reservation with the closed bracket on $-\infty$, is that allowed ?

Any further indications ? Thanks.

Answer 1

Let $\phi \in C_C^\infty$ and $\delta'_n(x)=-n^2\text{sgn}(x)\xi_{[-1/n,1/n]}$. Then, we have

$$\begin{align} \langle \delta'_n,\phi\rangle&=n^2 \int_{-1/n}^{0}\phi(x)\,dx-n^2\int_0^{1/n}\phi(x)\,dx\\\\ &=n^2 \int_{-1/n}^0 (\phi(0)+\phi'(0)x+O(x^2))\,dx-n^2 \int_0^{1/n} (\phi(0)+\phi'(0)x+O(x^2))\,dx\\\\ &=-\phi'(0)+O(1/n) \end{align}$$

Letting $n\to\infty$, we find that in distribution $\delta'_n\to \delta'$.

Answer 2

Using the tip from @DanielSchepler I did the following:

$T=n^2\chi_{[-1/n,0[} - n^2\chi_{]0,1/n]}$

We have for $ \phi \in \mathcal{D} $ and $F$ its antiderivative:

\begin{align} <T,\phi>&=n^2(F(0)-F(-\dfrac{1}{n})-F(\dfrac{1}{n})+F(0)\\ &=n^2(-F(\dfrac{1}{n})-F(-\dfrac{1}{n})+2F(0)) \end{align} using Taylor's expansion I get:

$ F(\dfrac{1}{n})=F(0)+\dfrac{1}{n}F'(0)+\dfrac{1}{2n^2}F''(0)+\dfrac{1}{n^2}\theta(1/n^2) $

and

$F(\dfrac{-1}{n})=F(0)-\dfrac{1}{n}F'(0)+\dfrac{1}{2n^2}F''(0)+\dfrac{1}{n^2}\theta(1/n^2) $

We finally get $ \lim_{n \rightarrow \infty}<T,\phi>=-F''(0)=-\phi'(0)=<\delta',\phi> $

Answer 3

Another way is to first take the derivative: $$\delta_n' = n^2 \left( \delta_{-1/n} - 2\delta_0 + \delta_{1/n} \right).$$ Thus, $$ \langle \delta_n', \varphi \rangle = n^2 \left( \varphi(-1/n) - 2\varphi(0) + \varphi(1/n) \right). $$ When $n\to\infty$ this converges to $\varphi''(0) = \langle \delta'',\varphi\rangle.$ (If you don't see this directly you can replace $n$ with $1/h$ and take the limit as $h\to 0.$)

So, $\delta_n' \to \delta'',$ i.e. $\delta_n \to \delta' + C$ for some constant $C.$ Studying the original limit with $\varphi \equiv 1$ on for example $[-1, 1]$ then shows that $C=0.$