= 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 $\varphi\in C_c^\infty(\mathbb R^3)$,
$$
-\int_{\mathbb R^3}\frac1{|x|}\Delta\varphi(x)\,dx=4\pi\varphi(0).
$$
Choose a ball containing the support and remove $B_\varepsilon(0)$. On the punctured region $\Delta U=0$. <Green's second identity> gives
$$
\int_{|x|>\varepsilon}U\Delta\varphi\,dx
=\int_{|x|=\varepsilon}
\left(-\frac1\varepsilon\partial_r\varphi-\frac1{\varepsilon^2}\varphi\right)\,dS.
$$
The outer boundary contributes zero. On the inner boundary, the outward normal for the punctured region is $-\widehat r$, so $\partial_nU=+1/\varepsilon^2$, fixing the sign.
The first term is $O(\varepsilon)$; the second equals $-\int_{S^2}\varphi(\varepsilon\omega)\,d\omega\to-4\pi\varphi(0)$. 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 $-\Delta$.
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