Limit in the sense of distributions of $n^2\chi_{[-1/n,0[}$ - $n^2\chi_{]0,1/n]}$ 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.
 A: 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 $\delta'_n\to \delta'$.
A: 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 $C=0.$
A: 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> $
