 Solution

Write $({\varphi }_{\epsilon }\ast f)(x)=\int \varphi (y)f(x-\epsilon y)dy$. Since $\int \varphi =1$,A Schwartz function is bounded and uniformly continuous. Given a tolerance, first choose $R$ so the tail integral of $|\varphi |$ is small. On $|y|\le R$, uniform continuity makes the difference uniformly small for every $x$ when $\epsilon R$ is small. On the tail it is bounded by $2\Vert f{\Vert }_{\infty }$. This proves uniform convergence, without assuming $\varphi \ge 0$, and is the approximate identity argument.

For the radial regularization of the three-dimensional fundamental solution of the Laplace equation, direct differentiation givesThe radial integral is $\int \varphi =3{\int }_0^{\infty }{r}^2(1+r^2{)}^{-5/2}dr=1$. This even approximate identity therefore gives $\int (-\Delta N_{\epsilon })f\to f(0)$ for $f\in \mathcal{S}({\mathbb{R}}^3)$.

The function $N=1/(4\pi |x|)$ is locally integrable and determines a tempered distribution. We have $0<N_{\epsilon }\le N$, and $N|\Delta f|$ is integrable near zero and infinity for a Schwartz test function. Integration by parts and the dominated convergence theorem giveThus $-\Delta N={\delta }_0$, proving that $N$ is a fundamental solution of a linear differential operator. In this last part the test-function space is $\mathcal{S}({\mathbb{R}}^3)$; the printed ${\mathbb{R}}^n$ in the three-dimensional integral is understood with $n=3$.