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Find the distributional limit of the sequence of distributions

$$F_n=n\delta_{-\frac{1}{n}}-n\delta_{\frac{1}{n}}$$

Hint: $F_n$ is the distributional derivative of some function

So far I've tried kind of working backwards towards the definition of the distributional derivative to make use of the hint:

$$F_n=n\delta_{-\frac{1}{n}}-n\delta_{\frac{1}{n}}$$ $$F_n=<n\delta_{-\frac{1}{n}},\phi>-<n\delta_{\frac{1}{n}},\phi>$$ $$n\int_{-1/n}^{1/n}\phi'(x) \ dx$$ $$-\int_{\mathbb{R}}f(x)\phi'(x) \ dx$$ $$-<f,\phi'> \ = \ <f',\phi>$$ where
$$f(x) = \begin{cases} -n & \text{$-\frac{1}{n}\le x\le\frac{1}{n}$} \\ 0 & \text{otherwise} \end{cases}$$

But I'm not sure how to proceed from here and how following this hint got me closer to the answer.

Any help is greatly appreciated!

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Take some test function $\phi$, then

$$\langle F_n, \phi\rangle = \int n\big(\delta(x+\tfrac 1n) - \delta(x-\tfrac 1n)\big)\phi(x) d x = n\big(\phi(-\tfrac 1n)-\phi(+\tfrac 1n)\big) \overset{n\to\infty}{\longrightarrow} -2\phi'(0) $$

So the limit distribution $F$ is the linear functional $F(\phi) = -2\phi'(0)$. In particular $F=2\delta'$.

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    $\begingroup$ how did you get to $-2\phi'(0)$ for when $n\rightarrow\infty$? $\endgroup$ – nickoba Sep 29 '19 at 8:55
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    $\begingroup$ @nickoba en.wikipedia.org/wiki/Symmetric_derivative Note that $\phi$ is smooth by assumption. $\endgroup$ – Hyperplane Sep 29 '19 at 8:57
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    $\begingroup$ the -2 is still throwing me off, could u care to explain that part of it? because wouln't it be $\infty$ instead of -2? $\endgroup$ – nickoba Sep 30 '19 at 2:17
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    $\begingroup$ @nickoba It is $-2\phi'(0)$ but there was a small sign mistake, which I fixed now. We have $\lim _{h \rightarrow 0} \frac{f(x+h)-f(x-h)}{2 h} = f'(x)$. Substitute $x=0$, $h=1/n$ and $f=\phi$. $\endgroup$ – Hyperplane Sep 30 '19 at 9:59

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