 Solution

The function $U(x)=1/|x|$ is locally integrable in ${\mathbb{R}}^3$, so it defines a distribution. The claimed distributional identity means that for every $\phi \in C_c^{\infty }({\mathbb{R}}^3)$,Choose a ball containing the support and remove $B_{\epsilon }(0)$. On the punctured region $\Delta U=0$. Green's second identity givesThe outer boundary contributes zero. On the inner boundary, the outward normal for the punctured region is $-\hat{r}$, so ${\partial }_{n}U=+1/{\epsilon }^2$, fixing the sign.

The first term is $O(\epsilon )$; the second equals $-{\int }_{S^2}\phi (\epsilon \omega )d\omega \to -4\pi \phi (0)$. Local integrability allows the left side to converge to its full-space integral. Therefore $\boxed{-\Delta (1/|x|)=4\pi {\delta }_0}$. In this sign convention, $1/(4\pi |x|)$ is the fundamental solution of the Laplace equation for the operator $-\Delta$.