Convergence of a distribution to delta function

Question

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.

Answer 1

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.

Answer 2

Hint: Say $\phi$ is continuous with compact support. Choose $A>0$ so $|\phi(x)-\phi(0)|<\epsilon$ if $|x|<A$. Then $$\phi(0)-\int\phi(x)f_n(x)=\int(\phi(0)-\phi(x))f_n(x)=\int_{|x|<A}+\int_{|x|>A}.$$

Now $$\left|\int_{|x|<A}\right|<\epsilon\int_{|x|<A}f_n<\epsilon\int f_n=\epsilon,$$while $$\left|\int_{|x|>A}\right|<2||\phi||_\infty\int_{|x|>A}f_n.$$Show that $\int_{|x|>A}f_n\to0$ as $n\to\infty$. (Hint for that last bit: If $|x|>A$ then $$\left|\frac{n}{1+n^2x^2}\right|<\frac n{n^2A^2};$$note as well that $\int|\phi|<\infty$.