Distributional limit of functions

Question

Consider the sequence $\{g_n\}_{n\in\mathbb{N}}\subset\mathcal{C}^\infty(\mathbb{R})$, $g_n(x)=\frac{x}{x^2+(1/n)^2}$. Prove that $g_n\to P.V.\frac{1}{x}$, where $P.V.\frac{1}{x}$ is the Cauchy Principal Value of $1/x$, defined by $$ \left<P.V.\frac{1}{x},\phi\right>=\lim_{\epsilon\to0}\int_{\mathbb{R}\setminus[-\epsilon,\epsilon]}\frac{\phi(x)}{x}dx. $$ Any help is appreciate.

Answer 1

I assume that $g_n \to PV \frac{1}{x}$ means that for any $φ\in C^\infty_c(\mathbb{R})$, we have $$\langle g_n, φ\rangle \to \langle PV \frac{1}{x}, φ\rangle$$ or equivalently $$a_n := \langle g_n - PV\frac{1}{x}, φ\rangle \to 0.$$ Writing out this term gives $$\begin{align*} a_n &= \lim_{ε\to 0} \left[ \int_{[-ε;ε]} \frac{x}{x^2 + 1/n^2} φ(x)dx + \int_{\mathbb{R}\setminus [-ε;ε]}\left(\frac{x}{x^2 + 1/n^2} - \frac{1}{x}\right)φ(x)dx \right] \\ &= \lim_{ε\to 0} \left[ \int_{[-ε;ε]} \frac{x}{x^2 + 1/n^2} φ(x)dx - \int_{\mathbb{R}\setminus [-ε;ε]}\frac{1}{x(n^2x^2 + 1)}φ(x)dx \right] \end{align*}$$ For simpler notation, let $M = ||φ||_\infty$, i.e. the sup of $φ$. In absolute value we can hence bound $$\begin{align*} |a_n| &\leq \limsup_{ε\to 0} \left[ \int_{[-ε;ε]} \frac{|x|}{x^2 + 1/n^2} |φ(x)|dx + \left|\int_{\mathbb{R}\setminus [-ε;ε]}\frac{1}{n^2x^2 + 1}\frac{φ(x)}{x}dx \right|\right] \\ &\leq \limsup_{ε\to 0} \left[ εM\int_{[-ε;ε]} \frac{1}{x^2 + 1/n^2} dx + \left|\int_{\mathbb{R}\setminus [-ε;ε]}\frac{1}{n^2x^2 + 1}\frac{φ(x)}{x}dx \right|\right] \\ &= \limsup_{ε\to 0} \left[ 2εMn^2\arctan(n^2ε) + \left|\int_{\mathbb{R}\setminus [-ε;ε]}\frac{1}{n^2x^2 + 1}\frac{φ(x)}{x}dx \right|\right] \\ &\leq \limsup_{ε\to 0} \left|\int_{\mathbb{R}\setminus [-ε;ε]}\frac{1}{n^2x^2 + 1}\frac{φ(x)}{x}dx \right| \end{align*}$$ and using that $PV\frac{1}{x}$ is well-defined, you can conclude that the last term $\to 0$ as $n\to\infty$.