Convergence of sequence of functions in a the sense of distribution.

Question

Consider the sequence of functions on $\Bbb R$ defined by $$ f_n(x)=\frac{n}{\pi(1+n^2x^2)},\quad n=1,2,\ldots $$ Show that $\{f_n\}$ converges to $\delta$ in the sense of distribution.

Answer 1

We have $$\lim_{n\to\infty}<f_n,\phi>=\lim_{n\to\infty}\int_{-\infty}^\infty\frac{n\phi(x)}{\pi(1+n^2x^2)}dx=\lim_{n\to\infty}\int_{-M}^M\frac{n\phi(x)}{\pi(1+n^2x^2)}dx,$$ since $\phi$ is test function (and therefore $\phi(x)=0$ for $x\notin [-M,M]$).

By substituting $t=nx$, we have $$\lim_{n\to\infty}<f_n,\phi>=\lim_{n\to\infty}\int_{-nM}^{nM}\frac{\phi(\frac{t}{n})}{\pi(1+t^2)}dt=\lim_{n\to\infty}\int_{-\infty}^{\infty}\frac{\chi_{[-nM,nM]}(t)\phi(\frac{t}{n})}{\pi(1+t^2)}dt,$$ from where we get (by Lebesgue dominated theorem), $$\lim_{n\to\infty}<f_n,\phi>=\int_{-\infty}^{\infty}\lim_{n\to\infty}\frac{\chi_{[-nM,nM]}(t)\phi(\frac{t}{n})}{\pi(1+t^2)}dt=\int_{-\infty}^{\infty}\frac{\phi(0)}{\pi(1+t^2)}dt=\frac{\phi(0)}{\pi}\arctan{t}|_{-\infty}^{\infty},$$ $$\lim_{n\to\infty}<f_n,\phi>=\phi(0)=<\delta,\phi>$$

Answer 2

Another simpler idea (depending on what information you are allowed to use) is to show that $f_n(x)$ is a delta sequence (all of which converge to $\delta$ in terms of distributions). As all the terms are nonnegative, it suffices to show that

1. $\int_\mathbb{R} f_n(x)dx = 1$ for all $n$.
2. For $\eta >0$, $\int_{|x|> \eta} f_n(x)dx \to 0$ as $n \to \infty$.

Both are routine if you use the same $t=nx$ substitution and evaluate the integrals directly.