# Convergence of a distribution to delta function

I was reading a book on distributions, and the following is classified as a $$\delta$$-convergent sequence. $$u_n(x)=\frac{n}{\pi(1+n^2x^2)}$$

This was my attempt at understanding why it converges to $$\delta(x)$$.

$$\displaystyle \lim_{n \to \infty}\left\langle u_n, \phi\right\rangle = \displaystyle \lim_{n \to \infty}\int_{-\infty}^{\infty}\frac{n}{\pi(1+n^2x^2)} \ \phi(x)\ dx$$

Then by the Lebesgue dominated convergence theorem,

$$\displaystyle \lim_{n \to \infty}\left\langle u_n, \phi\right\rangle = \displaystyle \int_{-\infty}^{\infty}\lim_{n \to \infty}\frac{n}{\pi(1+n^2x^2)} \ \phi(x)\ dx$$

$$=\int_{-\infty}^{\infty}\delta(x)\ \phi(x) \ dx\quad\quad (1)$$ $$=\left\langle u, \phi\right\rangle$$ Thus, $$\displaystyle \lim_{n \to \infty}u_n=u=\delta(x)$$

First of all, I'm not sure if my process is correct. Second of all, I'm not certain why I would be able to make the claim in step $$(1)$$.

Thank you in advance.

• It's not hard to show that $f_n\to\delta$, but your application of DCT is absurd. When you apply a theorem you need to verify the hypotheses - that's clearly impossible here since $\delta$ is not even a function. – David C. Ullrich Jun 30 at 23:14
• That's my question. Why does $f_n→\delta$? – Salinas Jun 30 at 23:19

I suppose $$u_n$$ is same as $$f_n$$. We have $$\int f_n(x)\phi(x)dx=\int \frac 1 {\pi (1+y^{2})} \phi(\frac y n)dy$$ (by the substitution $$y=nx$$). By DCT we get the limit as $$\phi(0)$$ (because $$\frac 1 {\pi (1+y^{2})}$$ is integrable). Since $$\int \phi d\delta=\delta( \phi)$$ the result follows.
Hint: Choose $$A>0$$ so $$|\phi(x)-\phi(0)|<\epsilon$$ if $$|x|. Then $$\phi(0)-\int\phi(x)f_n(x)=\int(\phi(0)-\phi(x))f_n(x)=\int_{|x|A}.$$
Now $$\left|\int_{|x|while $$\left|\int_{|x|>A}\right|<2||\phi||_\infty\int_{|x|>A}f_n.$$Show that $$\int_{|x|>A}f_n\to0$$ as $$n\to\infty$$.