= Solution
Write $(\phi_\varepsilon*f)(x)=\int\phi(y)f(x-\varepsilon y)\,dy$. Since $\int\phi=1$,
$$
|(\phi_\varepsilon*f)(x)-f(x)|\leq\int|\phi(y)|\,|f(x-\varepsilon y)-f(x)|\,dy.
$$
A <Schwartz function> is bounded and <uniformly continuous>. Given a tolerance, first choose $R$ so the tail integral of $|\phi|$ is small. On $|y|\leq R$, <uniform continuity> makes the difference uniformly small for every $x$ when $\varepsilon R$ is small. On the tail it is bounded by $2\|f\|_\infty$. This proves \b[<uniform convergence>], without assuming $\phi\geq0$, and is the <approximate identity> argument.
For the radial regularization of the three-dimensional <fundamental solution of the Laplace equation>, direct differentiation gives
$$
-\Delta N_\varepsilon(x)=\frac{3\varepsilon^2}{4\pi(|x|^2+\varepsilon^2)^{5/2}}=\varepsilon^{-3}\phi(x/\varepsilon),\qquad\phi(x)=\frac3{4\pi}(1+|x|^2)^{-5/2}.
$$
The radial integral is $\int\phi=3\int_0^\infty r^2(1+r^2)^{-5/2}\,dr=1$. This even <approximate identity> therefore gives $\int(-\Delta N_\varepsilon)f\to f(0)$ 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 $0<N_\varepsilon\leq N$, and $N|\Delta f|$ is integrable near zero and infinity for a Schwartz test function. Integration by parts and the <dominated convergence theorem> give
$$
\langle-\Delta N,f\rangle=\lim_{\varepsilon\downarrow0}\int N_\varepsilon(-\Delta f)=\lim_{\varepsilon\downarrow0}\int(-\Delta N_\varepsilon)f=f(0).
$$
Thus \b[$-\Delta N=\delta_0$], 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 $n=3$.
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