Prove that the sequence $f_{n}(x)=\arctan(x^{2n})$ converges in $\mathcal{D'}(\mathbb{R})$.

Question

I have to prove this fact. $$a_{n}(x)=\arctan(x^{2n})$$ converges on $\mathcal{D}'(\mathbb{R})$ to $\pi/2-\chi_{(-1,1)}$ as $n\rightarrow \infty$. ($\chi_{(-1,1)}=1$in such interval, $0$ otherwise). Any idea to start? Thank you.

Answer 1

The limit is $a_\infty=\frac{\pi}{2} (1-\chi_{(-1,1)})=\frac{\pi}{2}\chi_{\mathbb R\setminus (-1,1)} $. By definition of convergence of distributions you have to show $$\int_{\mathbb R} a_n(t) \varphi(t)dt \to \int_{\mathbb R} a_\infty(t) \varphi(t)dt.$$ This follows from $a_n(t)\to \frac{\pi}{2}$ for $|t|> 1$ and $a_n(t)\to 0$ for $|t|<1$. To interchange limits and integrals you can use Lebegue's theorem on dominated convergence.