Introduction to the Theory of Distributions, Friedlander and Joshi, Exercise 2.1

Question

The problem reads:

Show that $$\frac{1}{\pi}\frac{\epsilon}{x^2+\epsilon^2}\to\delta \ \text{in} \ \mathcal{D}'(\mathbb{R}) \ \text{as} \ \epsilon\to 0^+.$$

If I understand correctly, this means that I am supposed to show that for each $\phi\in C_C^{\infty}(\mathbb{R})$ $$\lim_{\epsilon\to 0^+}\int_{\mathbb{R}}\frac{1}{\pi}\frac{\epsilon}{x^2+\epsilon^2}\phi(x)\text{d}x=\phi(0)=\delta(\phi),$$ correct?

If so, any tips on how to calculate this integral (or otherwise, prove this result)?

Answer 1

You can rewrite the integral using a variable substitution: $$ \int \frac{1}{\pi} \frac{\epsilon}{x^2+\epsilon^2} \phi(x) \, dx = \{ x = \epsilon y \} = \frac{1}{\pi} \int \frac{\epsilon}{\epsilon^2 y^2+\epsilon^2} \phi(\epsilon y) \, \epsilon \, dy = \frac{1}{\pi} \int \frac{1}{y^2+1} \phi(\epsilon y) \, dy $$

Here, $\phi(\epsilon y) \to \phi(0)$ and the integral $\int \frac{1}{y^2+1} dy$ you should be able do.

Also, you need to fill in some details about taking the limit.

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Another solution, if you have learnt about derivatives of distributions, is to see that $\frac{\epsilon}{x^2+\epsilon^2} = \frac{1}{\epsilon} \frac{1}{(x/\epsilon)^2+1}$ is the derivative of $\arctan (x/\epsilon)$ and note that this tends to $\pi \operatorname{sign}(x)$ as $\epsilon \to 0.$

Answer 2

As I am currently reading the same textbook (and enjoy the presentation), I will post another solution even though the question is two years old. Let $$\eta_{\varepsilon}:=\frac{1}{\pi}\frac{\varepsilon}{x^2+\varepsilon^2},$$ then we have the following three properties

1. $\eta_{\varepsilon}\ge 0$.
2. $\int_{\mathbb R}\eta_{\varepsilon}(x)\,dx=1.$ As stated in the previous answer, this follows from recognizing that $\frac{\epsilon}{x^2+\epsilon^2} = \frac{1}{\epsilon} \frac{1}{(x/\epsilon)^2+1}$ is the derivative of $\arctan (x/\epsilon)$ which tends to $\pi \operatorname{sign}(x)$ as $\epsilon \to 0^+.$
3. $\int_{|x|\ge \alpha}\eta_{\varepsilon}(x)\,dx \to 0$ as $\varepsilon\to 0^+$ for any given $\alpha > 0.$

We need to show that any family $\{\eta_{\varepsilon}\}$ of $L^1$ functions satisfying properties $(1) - (3)$ converges to a $\delta$-function in $\mathcal{D}'(\mathbb R)$. In other words, we need to show that

$$\int_{\mathbb R} \phi(x)\eta_{\varepsilon}(x)\,dx \to \phi(0), \quad \text{for any $\phi\in C_c^{\infty}(\mathbb R)$}. $$ Given $\kappa > 0$, let $\alpha > 0$ be such that $|\phi(x) - \phi(0)| < \kappa$ for $|x| \le \alpha.$ Then \begin{align} \left|\int_{\mathbb R} \phi(x)\eta_{\varepsilon}(x)\,dx- \phi(0)\right| &= \left|\int_{\mathbb R} \Big(\phi(x)-\phi(0)\Big)\eta_{\varepsilon}(x)\,dx\right| \\&\le \int_{|x|\le \alpha}\left|\phi(x)-\phi(0)\right|\eta_{\varepsilon}(x)\,dx + \int_{|x|\ge \alpha}\left|\phi(x)-\phi(0)\right|\eta_{\varepsilon}(x)\,dx \\& < \kappa \int_{|x|\le \alpha}\eta_{\varepsilon}(x)\,dx + \sup|\phi(x)-\phi(0)|\int_{|x|\ge \alpha}\eta_{\varepsilon}(x)\,dx \\&\le \kappa, \end{align} provided $\varepsilon\to 0^+$. As $\kappa$ was arbitrary, we conclude that $\eta_{\varepsilon}\to \delta$ in $\mathcal{D}'(\mathbb R)$.