Proving that $\frac{x}{x^2+\varepsilon^2}\to \mathsf{p.v.}\left(\frac1x\right)$ in $\mathscr{S}'$

Question

Let $\mathscr{S}'$ be the space of tempered distributions. I've to prove that:

$$\frac{x}{x^2+\varepsilon^2}\to \mathsf{p.v.}\left(\frac1x\right) \ \ \text{ in $\mathscr{S}'$}$$

My book says that it's obvious (by dominated convergence) that

$$\frac12 \log(x^2+\varepsilon^2)\to \log|x| \ \ \text{ in $\mathscr{S}'$}$$

and then uses the fact that the derivative is a continous automorphism of $\mathscr{S'}$. I'm trying to see why this is obvious. Let $\psi$ be a Schwartz function. I need to prove that:

$$\int \frac12 \log(x^2+\varepsilon^2)\psi \to \int \log|x|\psi \ \ \ \text{ as $\varepsilon\to 0^+$} $$

Basically I need to bound $|\log(x^2+\varepsilon^2)\psi|$ with an $L^1$ function, but I don't know how since $|\log(\varepsilon^2)|$ keeps growing as $\varepsilon\to 0^+$.

Answer 1

Let $f(x;\varepsilon)=\frac{x}{x^2+\varepsilon^2}$ and $\phi\in C_C^\infty$. Then, we have

$$\begin{align} \lim_{\varepsilon\to 0}\langle f, \phi\rangle&=\lim_{\varepsilon\to 0}\int_{-\infty}^\infty f(x;\varepsilon)\phi(x)\,dx\\\\ &=\lim_{\varepsilon\to 0}\int_{-\infty}^\infty \frac{x}{x^2+\varepsilon^2}\phi(x)\,dx\\\\ &=\lim_{\varepsilon\to 0}\left(\int_{|x|\le \delta} \frac{x}{x^2+\varepsilon^2}\phi(x)\,dx+\int_{|x|\ge \delta} \frac{x}{x^2+\varepsilon^2}\phi(x)\,dx\right)\\\\ &=\lim_{\varepsilon\to 0}\left(\int_{|x|\le \delta} \frac{x}{x^2+\varepsilon^2}\phi(x)\,dx\right)+\int_{|x|\ge \delta} \frac{1}{x}\phi(x)\,dx\\\\ \end{align}$$

Now, note that

$$\begin{align} \left|\int_{|x|\le \delta} \frac{x}{x^2+\varepsilon^2}\phi(x)\,dx\right|&=\left|\int_{|x|\le \delta} \frac{x(\phi(x)-\phi(0))}{x^2+\varepsilon^2}\,dx\right|\\\\ &\le ||\phi'||_{\infty}\int_{|x|\le \delta} \frac{x^2}{x^2+\varepsilon^2}\,dx\\\\ &=||\phi'||_{\infty}O(\delta) \end{align}$$

Hence, we see that

$$\begin{align} \lim_{\varepsilon\to 0}\langle f, \phi\rangle&=\int_{|x|\ge\delta}\frac{\phi(x)}{x}\,dx+O(\delta) \end{align}$$

Letting $\delta\to 0$ yields the result

$$\lim_{\varepsilon\to 0}\langle f, \phi\rangle=\text{PV}\int_{-\infty}^\infty \frac{\phi(x)}{x}\,dx$$

which means that in distribution $\lim_{\varepsilon\to 0} \frac{x}{x^2+\varepsilon^2}=\text{PV}\left(\frac1x\right)$, as was to be shown!

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Now, let $g(x;\varepsilon)=\frac12\log(x^2+\varepsilon^2)$. Then, clearly $\frac{dg(x;\varepsilon)}{dx}=f(x;\varepsilon)$ so that we have

$$\begin{align} \langle f,\phi\rangle &=-\langle g,\phi'\rangle\\\\ &=-\frac12\int_{-\infty}^\infty \log(x^2+\varepsilon^2)\phi'(x)\,dx\\\\ \end{align}$$

Let $\varepsilon$ be restricted such that $\varepsilon\in (0,1/\sqrt2]$. Then, when $x^2+\varepsilon^2\le 1$, $|\log(x^2+\varepsilon^2)|\le |\log(x^2)|$ and when $x^2+\varepsilon^2\ge 1$. $|\log(x^2+\varepsilon^2)|\le \log(2x^2)$. Thus, the Dominated convergence theorem guarantees that

$$\begin{align} \lim_{\varepsilon\to0^+}\langle f,\phi\rangle &=-\frac12\int_{-\infty}^\infty \lim_{\varepsilon\to0^+}\log(x^2+\varepsilon^2)\phi'(x)\,dx\\\\ &=-\frac12\int_{-\infty}^\infty \log(x^2)\phi'(x)\,dx\\\\ &=-\int_{-\infty}^\infty \log(|x|)\phi'(x)\,dx\\\\ &=\text{PV} \int_{-\infty}^\infty \frac{\phi(x)}{x}\,dx \end{align}$$

And we are done!

Answer 2

You can also use the monotone convergence theorem on intervals.

Note that $$ \frac{\partial}{\partial\epsilon} \frac12\log(x^2+\epsilon^2) = \frac{\epsilon}{x^2+\epsilon^2} > 0 $$ when $\epsilon>0.$ That implies that $\frac12\log(x^2+\epsilon^2)$ decreases pointwise as $\epsilon\to 0^+.$

For a fixed $\varphi\in C^\infty_c(\mathbb R)$ you then have that $\frac12\log(x^2+\epsilon^2)\,\varphi(x)$ decreases pointwise on the intervals where $\varphi>0$ and increases pointwise on the intervals where $\varphi<0.$ On each of these types of intervals you can apply the monotone convergence theorem to finally get $$ \int \frac12\log(x^2+\epsilon^2)\,\varphi(x) \, dx \to \log|x|\,\varphi(x)\,dx, $$ i.e. $\frac12\log(x^2+\epsilon^2)\,\varphi(x) \to \log|x|$ in $\mathscr{D}'(\mathbb R).$