Divergence and curl of the Newtonian potential of a smooth compact vector field

Since each $f_i$ is smooth and compactly supported, [F2] gives $A_i=N(f_i)\in C^{\mathrm{\infty }}({\mathbb{R}}^3)$ with $D^{\alpha }{A}_i=N(D^{\alpha }{f}_i)$ for every multi-index $\alpha$; in particular $A\in C^{\mathrm{\infty }}({\mathbb{R}}^3;{\mathbb{R}}^3)$ by [F3], and ${\mathrm{\partial }}_i{A}_i=N({\mathrm{\partial }}_i{f}_i)$ for each $i$. Summing over $i$ and using finite additivity of the integral [F8], $\mathrm{div}A={\sum }_i{\mathrm{\partial }}_i{A}_i=N({\sum }_i{\mathrm{\partial }}_i{f}_i)=N(\mathrm{div}f)$ pointwise on ${\mathbb{R}}^3$, which is the first displayed identity; its integral is absolutely convergent because $\mathrm{\mid }\mathrm{div}f\mathrm{\mid }\le \mathrm{\parallel }\mathrm{div}f{\mathrm{\parallel }}_{\mathrm{\infty }}$ is bounded and compactly supported and ${\int }_{B_R(x)}\mathrm{\mid }x-y{\mathrm{\mid }}^{-1}d{\lambda }_3<+\mathrm{\infty }$ by [F7].

Fix $x\in {\mathbb{R}}^3$ and $j,k\in \{1,2,3\}$, let $\mathrm{\Phi }(z)=1\mathrm{/}(4\pi \mathrm{\mid }z\mathrm{\mid })$, and choose $R>0$ with $\mathrm{supp}f\subset B_R(x)$. For $0<\epsilon <R$ let ${\mathrm{\Omega }}_{\epsilon }:=B_R(x)\setminus \overline{B_{\epsilon }(x)}$, presented as in [F5]. On a neighborhood of the compact closure the scalar product $y\mapsto f_k(y)\mathrm{\Phi }(x-y)$ is $C^1$, and by [F4] its partial derivative is ${\mathrm{\partial }}_{y_j}(f_k\mathrm{\Phi }(x-y))=({\mathrm{\partial }}_j{f}_k)(y)\mathrm{\Phi }(x-y)+f_k(y){\mathrm{\partial }}_{y_j}\mathrm{\Phi }(x-y)$. Applying the divergence theorem [F5] to the field $f_k\mathrm{\Phi }(x-\cdot )e_j$ on ${\mathrm{\Omega }}_{\epsilon }$ gives ${\int }_{{\mathrm{\Omega }}_{\epsilon }}({\mathrm{\partial }}_j{f}_k)\mathrm{\Phi }(x-y)d{\lambda }_3(y)={\int }_{\mathrm{\partial }{B}_R(x)}{f}_k\mathrm{\Phi }(x-y)({\nu }_R{)}_{j}dS+{\int }_{\mathrm{\partial }{B}_{\epsilon }(x)}{f}_k\mathrm{\Phi }(x-y)({\nu }_{\epsilon }{)}_{j}dS-{\int }_{{\mathrm{\Omega }}_{\epsilon }}{f}_k(y){\mathrm{\partial }}_{y_j}\mathrm{\Phi }(x-y)d{\lambda }_3(y)$. The outer sphere contributes $0$ because $f=0$ there; the inner sphere term is bounded in absolute value, by [F8] and the bound $\mathrm{\mid }\mathrm{\Phi }\mathrm{\mid }=1\mathrm{/}(4\pi \mathrm{\mid }y-x\mathrm{\mid })=1\mathrm{/}(4\pi \epsilon )$ there, by $\mathrm{\parallel }{f}_k{\mathrm{\parallel }}_{\mathrm{\infty }}\frac{1}{4\pi \epsilon }\mathrm{\mid }\mathrm{\partial }{B}_{\epsilon }(x)\mathrm{\mid }=\mathrm{\parallel }{f}_k{\mathrm{\parallel }}_{\mathrm{\infty }}\frac{\mathrm{\mid }{S}^2\mathrm{\mid }}{4\pi }\epsilon \to 0$ as $\epsilon \to 0$ by [F6].

Let $\epsilon \to 0$ in the identity of step 1.2. The shell integrands $({\mathrm{\partial }}_j{f}_k)\mathrm{\Phi }(x-y){\mathbf{1}}_{{\mathrm{\Omega }}_{\epsilon }}$ are dominated on $B_R(x)$ by $\mathrm{\parallel }{\mathrm{\partial }}_j{f}_k{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }\mathrm{\Phi }\mathrm{\mid }$, and the shell integrands $f_k{\mathrm{\partial }}_{y_j}\mathrm{\Phi }(x-y){\mathbf{1}}_{{\mathrm{\Omega }}_{\epsilon }}$ by $\mathrm{\parallel }{f}_k{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }{\mathrm{\partial }}_{y_j}\mathrm{\Phi }\mathrm{\mid }\le \mathrm{\parallel }{f}_k{\mathrm{\parallel }}_{\mathrm{\infty }}\frac{1}{4\pi }\mathrm{\mid }x-y{\mathrm{\mid }}^{-2}$; both majorants are integrable on $B_R(x)$ by [F7], and both families converge pointwise off the null singleton $\{x\}$ to the corresponding unrestricted integrands. Dominated convergence [F9] therefore gives ${\int }_{B_R(x)}({\mathrm{\partial }}_j{f}_k)\mathrm{\Phi }(x-y)d{\lambda }_3(y)=-{\int }_{B_R(x)}{f}_k(y){\mathrm{\partial }}_{y_j}\mathrm{\Phi }(x-y)d{\lambda }_3(y)$; the left side is $N({\mathrm{\partial }}_j{f}_k)(x)={\mathrm{\partial }}_j{A}_k(x)$ by step 1.1, and since ${\mathrm{\partial }}_{y_j}\mathrm{\Phi }(x-y)=\frac{x_j-y_j}{4\pi \mathrm{\mid }x-y{\mathrm{\mid }}^3}$ for $y\mathrm{\ne }x$, this reads ${\mathrm{\partial }}_j{A}_k(x)=-\frac{1}{4\pi }{\int }_{{\mathbb{R}}^3}\frac{f_k(y)(x_j-y_j)}{\mathrm{\mid }x-y{\mathrm{\mid }}^3}dy$, the integral being absolutely convergent because $\mathrm{\mid }{f}_k(y)(x_j-y_j)\mathrm{\mid }\le \mathrm{\parallel }{f}_k{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }x-y\mathrm{\mid }$ on $B_R(x)$ and ${\int }_{B_R(x)}\mathrm{\mid }x-y{\mathrm{\mid }}^{-2}d{\lambda }_3<+\mathrm{\infty }$ by [F7].

Summing the identity of step 2.1 with the alternating symbol, $(\mathrm{curl}A{)}_i={\epsilon }_{ijk}{\mathrm{\partial }}_j{A}_k=-\frac{1}{4\pi }{\int }_{{\mathbb{R}}^3}\frac{{\epsilon }_{ijk}{f}_k(y)(x_j-y_j)}{\mathrm{\mid }x-y{\mathrm{\mid }}^3}dy=\frac{1}{4\pi }{\int }_{{\mathbb{R}}^3}\frac{(f(y)\times (x-y){)}_i}{\mathrm{\mid }x-y{\mathrm{\mid }}^3}dy$ by the component identity of [F10] and finite linearity of the integral [F8]; as this holds for each $i$, $\mathrm{curl}A(x)=\frac{1}{4\pi }{\int }_{{\mathbb{R}}^3}\frac{f(y)\times (x-y)}{\mathrm{\mid }x-y{\mathrm{\mid }}^3}dy$, and the integral is absolutely convergent because $\mathrm{\mid }f(y)\times (x-y)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }x-y\mathrm{\mid }$ and ${\int }_{B_R(x)}\mathrm{\mid }x-y{\mathrm{\mid }}^{-2}<+\mathrm{\infty }$ by [F7]. With step 1.1 this proves both displayed identities at the arbitrary point $x$; the point was allowed to lie in $\mathrm{supp}f$, since the excision argument only removed a ball around it and used the bound on the differentiated kernel, so no smoothness of the kernel at the pole is assumed. Countable Choice is used only through the polar and surface-measure interfaces [F11] and the stated choice assumption of the smooth-newtonian supplier, and no stronger principle occurs.