Limit of a functional

Question

I'd like to find: $$ \lim_{\varepsilon\rightarrow 0}\frac{\varepsilon}{\varepsilon^2+x^2}\qquad \mbox{ in }\mathcal D'(\mathbb{R}) $$ And I started with the definition: $$ \left\langle \frac{\varepsilon}{\varepsilon^2+x^2},\varphi\right\rangle $$ After doing the integration by parts I had some problems with the support of $\varphi$. I think the result could be $0$, but I don't know how to prove it in a proper way. (during the proof I used the substitution $x=\varepsilon y$ or $y=\varepsilon x$).

Answer 1

Define $$T_\varepsilon(\varphi):=\int_{-\infty}^{+\infty}\frac{\varepsilon}{\varepsilon^2+x^2}\varphi(x)\mathrm dx.$$ Then, after the substitution $x=\varepsilon t$, we obtain $$T_\varepsilon(\varphi)=\int_{-\infty}^{+\infty}\frac{1}{1+t^2}\varphi(t\varepsilon)\mathrm dt.$$ Using dominated convergence, we obtain that for each $\varphi\in\mathcal D(\mathbf R)$, $$\lim_{\varepsilon\to 0}T_\varepsilon(\varphi)=\varphi(0)\cdot\int_{-\infty}^{+\infty}\frac{1}{1+t^2}\mathrm dt =\pi\varphi(0).$$