Newtonian potential of smooth compact data is smooth and solves the Poisson equation pointwise

Substituting $z=x-y$ gives $Nf(x)={\int }_{{\mathbb{R}}^3}\mathrm{\Phi }(z)f(x-z)dz$, an absolutely convergent integral for every $x$: $f$ is bounded with support in some closed ball ${\bar{B}}_R(0)$, so the integrand is supported in $x-{\bar{B}}_R(0)$ and bounded there by $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }\mathrm{\Phi }(z)\mathrm{\mid }$ with $\mathrm{\Phi }\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1({\mathbb{R}}^3)$. The same substitution exhibits $Nf$ as the Newtonian potential of [L1], so the distributional identity $-\mathrm{\Delta }{T}_{Nf}=T_f$ of [L1] holds.

Fix a multi-index $\alpha =({\alpha }_1,{\alpha }_2,{\alpha }_3)$ and a point $x$, and let $K\subseteq {\mathbb{R}}^3$ be a compact set containing $x$ in its interior. For $z$ in the fixed compact set $K-\mathrm{supp}f$ and $x^{\mathrm{\prime }}\in K$, the function $z\mapsto \mathrm{\Phi }(z)D^{\alpha }f(x^{\mathrm{\prime }}-z)$ is bounded by $\mathrm{\parallel }{D}^{\alpha }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }\mathrm{\Phi }(z)\mathrm{\mid }$, integrable on that compact set, and $z\mapsto \mathrm{\Phi }(z)D^{\alpha +e_j}f(x^{\mathrm{\prime }}-z)$ likewise; moreover $D^{\alpha }f$ is uniformly continuous on the compact set $K-\mathrm{supp}f$ shifted appropriately. Applying the differentiation-under-the-integral-sign theorem [L2] successively in the three coordinates (the parameter being one coordinate of $x$ and the measure space ${\mathbb{R}}^3$ with the dominated bounds above) gives $D^{\alpha }Nf(x)=\int \mathrm{\Phi }(z)D^{\alpha }f(x-z)dz=N(D^{\alpha }f)(x)$ at every $x$, and the resulting function is continuous in $x$ because the bound $\mathrm{\parallel }{D}^{\alpha }f(x-z)-D^{\alpha }f(x^{\mathrm{\prime }}-z){\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }\mathrm{\Phi }(z)\mathrm{\mid }$ on the compact $z$-set is uniform in $x,x^{\mathrm{\prime }}$ near $x$ and tends to $0$. Hence $Nf$ has continuous derivatives of every order, so $Nf\in C^{\mathrm{\infty }}({\mathbb{R}}^3)$, and the identity $D^{\alpha }Nf=N(D^{\alpha }f)$ holds for every multi-index $\alpha$; in particular the classical Laplacian $\mathrm{\Delta }Nf={\sum }_{j=1}^{3}N({\mathrm{\partial }}_j^{2}f)$ is a continuous function, and it is the classical Laplacian of the smooth function $Nf$.

By [L3] the distributional Laplacian of the smooth function $Nf$ is the regular distribution of the classical Laplacian, so the identity $-\mathrm{\Delta }{T}_{Nf}=T_f$ of step 1.1 says that ${\int }_{{\mathbb{R}}^3}(f(y)+\mathrm{\Delta }Nf(y))\phi (y)dy=0$ for every test function $\phi$. The function $g:=f+\mathrm{\Delta }Nf$ is continuous by step 2.1, so it is locally integrable and the fundamental lemma [L5] gives $g=0$ almost everywhere. A continuous function vanishing almost everywhere vanishes everywhere: if $g(x_0)\mathrm{\ne }0$, continuity gives $r>0$ with $\mathrm{\mid }g\mathrm{\mid }>\mathrm{\mid }g(x_0)\mathrm{\mid }\mathrm{/}2>0$ on the nonempty open ball $B_r(x_0)$, which has positive Lebesgue measure by [L4], contradicting $g=0$ a.e.; hence $g(x_0)=0$ for every $x_0$. Therefore $-\mathrm{\Delta }Nf=f$ pointwise on ${\mathbb{R}}^3$.

For smooth compactly supported $\rho$, the derivative identity of step 2.1 with $\alpha =e_j$ shows that $\mathrm{\nabla }N\rho$ is a smooth field, and $E=-{\epsilon }_0^{-1}\mathrm{\nabla }N\rho$ is smooth because ${\epsilon }_0^{-1}$ is a constant. Countable Choice is inherited from the exact statement of the Newtonian-potential theorem [L1] and from the positive-density-measure corollary [L4]; the differentiation and upgrade arguments use no further choice. This proves the smoothness of $Nf$, the pointwise Poisson identity and the derivative identity $D^{\alpha }Nf=N(D^{\alpha }f)$, and hence the smoothness of $E$.