Find the distributional limit of the sequence of distributions


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!


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 $F=2\delta'$.

  • 1
    $\begingroup$ how did you get to $-2\phi'(0)$ for when $n\rightarrow\infty$? $\endgroup$ – nickoba Sep 29 '19 at 8:55
  • 1
    $\begingroup$ @nickoba en.wikipedia.org/wiki/Symmetric_derivative Note that $\phi$ is smooth by assumption. $\endgroup$ – Hyperplane Sep 29 '19 at 8:57
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.