# Convergence of a sequence of functions in sense of distributions

I'm trying to determine the limit in sense of distributions of the sequence $f_{n}=\frac{n}{1+n^2x^2}$. First I wrote this:

\begin{align}\int_{\mathbb{R}}f_n(x) \phi(x) dx&=\arctan(nx)\phi(x)|_{-\infty}^{+\infty} -\int_{\mathbb{R}}\arctan(nx) \phi'(x) dx\\\\ &=-\int_{-\infty}^{0}\arctan(nx) \phi'(x) dx-\int_{0}^{+\infty}\arctan(nx) \phi'(x) dx \end{align}

I tried proving that the sequence converges to $\pi\delta_{0}$, since $\arctan(nx)\rightarrow \text{sgn(x)}\frac{\pi}{2}$, as $n\rightarrow +\infty$, but without success. Could someone hint what the limit could be so I can try to prove it?

Note that

$$\int_{-\infty}^\infty \frac{n}{1+n^2x^2}\phi(x)\,dx\overbrace{=}^{x\mapsto x/n}\int_{-\infty}^\infty \frac1{1+x^2}\phi(x/n)\,dx$$

Now, since $\phi(x)$ is a test function, it is bounded and continuous. Hence, applying the Dominated Convergence Theorem yields

\begin{align} \lim_{n\to \infty}\int_{-\infty}^\infty \frac{n}{1+n^2x^2}\phi(x)\,dx&=\lim_{n\to\infty}\int_{-\infty}^\infty \frac1{1+x^2}\phi(x/n)\,dx\\\\ &=\int_{-\infty}^\infty \frac1{1+x^2}\lim_{n\to\infty}\phi(x/n)\,dx\\\\ &=\phi(0)\int_{-\infty}^\infty \frac{1}{1+x^2}\,dx\\\\ &=\pi\phi(0) \end{align}

If we wish to proceed as in the OP, then using the fact that $\phi$ is of compact support, we have

$$\int_{-\infty}^\infty \frac{n}{1+n^2x^2}\phi(x)=-\int_{-\infty}^\infty \arctan(nx)\phi'(x)\,dx$$

Since $|\arctan(nx)|\le \pi/2$, then $|\arctan(nx)\phi'(x)|\le \frac\pi2 |\phi'(x)|$. Inasmuch as $\phi'(x)$ is continuous and of compact support, then $\int_{-\infty}^\infty|\phi'(x)|\,dx<\infty$. Again, we appeal to the Dominated Convergence Theorem to find

\begin{align} \lim_{n\to \infty}\int_{-\infty}^\infty \arctan(nx)\phi'(x)\,dx&=-\int_{-\infty}^\infty \lim_{n\to\infty}\arctan(nx)\phi'(x)\,dx\\\\ &=-\int_{-\infty}^\infty \phi'(x)\frac\pi 2\text{sgn}(x)\,dx\\\\ &=-\frac\pi2 \left(\int_0^\infty \phi'(x)\,dx-\int_{-\infty}^0 \phi'(x)\,dx\right)\\\\ &=-\frac\pi2 \left(-\phi(0)-\phi(0)\right)\\\\ &=\pi\phi(0) \end{align}

• The first argument is definitely the "right" one, because it applies in much greater generality - proceeding as in the OP gives the idea that there's something special about $1/(1+x^2)$. In fact if $f\in L^1$, $\int f=1$, and $f_n(x)=nf(nx)$ then $\int f_n\phi\to\phi(0)$ for every bounded continuous $\phi$, exactly as above. – David C. Ullrich Apr 7 '18 at 14:12
• @DavidC.Ullrich Indeed. The specific function is only an example within the general class that you identified. – Mark Viola Apr 7 '18 at 14:16

Hint: $$H(x)=\lim_{n\to\infty}\left(\dfrac{1}{2}+ \dfrac{1}{\pi} arctan(nx) \right)$$ where $H(x)$ is heaviside hunction defined by $H'=\delta_0$ or $H(x)=\int_{- \infty}^{x} \delta(s) ds$