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Show that $$\lim_{\epsilon \to 0^{+}} \left\langle \frac{\epsilon}{x^{2}+ \epsilon} ,\phi \right\rangle =\langle \delta,\phi\rangle $$ where $\phi\in D(\mathbb{R}) $ and $ \frac{\epsilon}{x^{2}+ \epsilon} \in D' (\mathbb{R}) $

What I tried is this: $ \frac{\epsilon}{x^{2}+ \epsilon}$ is an integrable function hence $\left\langle \frac{\epsilon}{x^{2}+ \epsilon} ,\phi\right\rangle$ defines an integral on $\mathbb{R}$:

$$\left\langle\frac{\epsilon}{x^{2}+ \epsilon} ,\phi\right\rangle = \int_{- \infty}^{\infty}\frac{\epsilon}{x^{2}+ \epsilon} \phi (x) dx$$

How can I take limit of the last equation?

Thank you from now on.

Serkan Yaray
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  • did you try partial integration and splitting up the integral at 0? – user159517 Feb 03 '16 at 23:12
  • Wasn't it $\frac{\epsilon}{x^2 + \epsilon^2}$? And there's a constant factor ($\frac{1}{\pi}$) missing then. – Daniel Fischer Feb 03 '16 at 23:12
  • @DanielFischer no. It is how I wrote it in the question. – Serkan Yaray Feb 03 '16 at 23:23
  • In that case, substituting $x = \sqrt{\epsilon}, y$ will show the limit is $0$, not $\delta$. – Daniel Fischer Feb 03 '16 at 23:26
  • To prove what Daniel Fischer suggested, you can integrate by parts to see that you have $-\int_{-\infty}^\infty \phi'(x) \arctan(x/\epsilon) dx$. Now split this integral between $(-\infty,-\sqrt{\epsilon}),(-\sqrt{\epsilon},\sqrt{\epsilon})$ and $(\sqrt{\epsilon},\infty)$. The middle interval is small and everything is bounded so it is not important. What happens on the other two intervals? (Hint: how does $\arctan$ behave at large arguments?) – Ian Feb 03 '16 at 23:29
  • @Ian Is it $ arctan(x\ \sqrt \epsilon)$ in that integral? – Serkan Yaray Feb 03 '16 at 23:32
  • @SerkanYaray No, I had it right. – Ian Feb 03 '16 at 23:34
  • @Ian No need to integrate by parts: $\frac {\epsilon}{x^2+\epsilon}\le 1[x^2\le \sqrt \epsilon]+\sqrt \epsilon[x^2> \sqrt \epsilon]$. – A.S. Feb 03 '16 at 23:38
  • @A.S. I meant the first thing Daniel Fischer suggested (with the $\epsilon^2$ in the problem). – Ian Feb 03 '16 at 23:39
  • I am sorry,why do you need $ \epsilon ^{2} $ I really didn't understand why you are changing the problem? – Serkan Yaray Feb 03 '16 at 23:41
  • With the $\epsilon$, your problem statement is wrong, in that the family of distributions in question converges to $0$, not $\delta$. – Ian Feb 03 '16 at 23:51

1 Answers1

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As some people pointed to you you should consider the delta sequence $$ f_n(x) = \frac{n}{\pi}\frac{1}{1+n^2x^2}$$ Which gives

$$\lim_{n\to \infty} <f_n(x), \phi(x) > =<\delta(x), \phi(x) >$$

Note: you can put $\epsilon =\frac{1}{n}$ and consider the limit as $\epsilon$ goes to 0.

Mhenni Benghorbal
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