The far-field gradient of a compactly supported Newtonian potential

Fix $x_0$ with $\mathrm{\mid }{x}_0\mathrm{\mid }>2R$, choose $0<\delta <\mathrm{\mid }{x}_0\mathrm{\mid }-2R$, put $U:=B_{\delta }(x_0)$ and $c:=\mathrm{\mid }{x}_0\mathrm{\mid }-\delta -R>0$. For measurable real $a$, the function $a^2$ is measurable because $\{a^2>b\}$ is the whole space for $b<0$ and $\{\mathrm{\mid }a\mathrm{\mid }>\sqrt{b}\}$ for $b\ge 0$; for measurable real $a,b$, the identity $ab=((a+b{)}^2-(a-b{)}^2)\mathrm{/}4$ therefore gives measurable products by [F6]. Since $f$ vanishes almost everywhere on the complement of ${\bar{B}}_R$, the function $f_0:=f{\mathbf{1}}_{{\bar{B}}_R}$ is measurable, integrable, agrees with $f$ almost everywhere and is pointwise zero outside ${\bar{B}}_R$; replacing $f$ by $f_0$ changes no integral of the form $\int \mathrm{\Phi }(x-y)f(y)dy$ or $\int {\mathrm{\partial }}_i\mathrm{\Phi }(x-y)f(y)dy$ ([F6]), so we may assume $f$ vanishes pointwise off ${\bar{B}}_R$. Differentiating the displayed formula of [F1] with the chain and algebra rules gives, for every $z\mathrm{\ne }0$, $\mathrm{\nabla }\mathrm{\Phi }(z)=\frac{1}{(n-2){\omega }_{n-1}}(2-n)\mathrm{\mid }z{\mathrm{\mid }}^{-n}z=-\frac{z}{{\omega }_{n-1}\mathrm{\mid }z{\mathrm{\mid }}^n}$ and hence $\mathrm{\mid }\mathrm{\nabla }\mathrm{\Phi }(z)\mathrm{\mid }=\frac{1}{{\omega }_{n-1}\mathrm{\mid }z{\mathrm{\mid }}^{n-1}}$ ([F2]).

For $x\in U$ and $y\in {\bar{B}}_R$ the triangle inequality of [F3] gives $\mathrm{\mid }x-y\mathrm{\mid }\ge \mathrm{\mid }{x}_0-y\mathrm{\mid }-\mathrm{\mid }x-x_0\mathrm{\mid }\ge \mathrm{\mid }{x}_0\mathrm{\mid }-R-\delta =c,$ so with $f$ vanishing off ${\bar{B}}_R$ the bounds $\mathrm{\mid }\mathrm{\Phi }(x-y)f(y)\mathrm{\mid }\le \frac{c^{2-n}\mathrm{\mid }f(y)\mathrm{\mid }}{(n-2){\omega }_{n-1}},\mathrm{\mid }{\mathrm{\partial }}_i\mathrm{\Phi }(x-y)f(y)\mathrm{\mid }\le \frac{c^{1-n}\mathrm{\mid }f(y)\mathrm{\mid }}{{\omega }_{n-1}}$ hold for all $x\in U$ and $y\in {\mathbb{R}}^n$, the exponents being nonpositive because $n\ge 3$ and $c>0$ ([F3]). Both majorants are integrable: $f\in L^1$ and the constants are finite. Hence $Nf(x)$ and each integral $\int {\mathrm{\partial }}_i\mathrm{\Phi }(x-y)f(y)d{\lambda }_n(y)$ exist absolutely for every $x\in U$, and $g:=c^{1-n}\mathrm{\mid }f\mathrm{\mid }\mathrm{/}{\omega }_{n-1}$ is a single integrable majorant for all partial-derivative integrands on $U$. Moreover each integrand $y\mapsto {\mathrm{\partial }}_i\mathrm{\Phi }(x-y)f(y)$ is measurable, as follows: the kernel restricted to ${\bar{B}}_R$ is continuous and hence Borel measurable there; its extension by zero outside that closed ball is Borel (preimages of open sets are a relative Borel preimage, possibly with the complement of the ball added). Its product with $f$ is measurable by step 1.1. The same reasoning applies to the potential integrand.

Fix $i\in \{0,\dots ,n-1\}$ and $x\in U$; since $U$ is open there is $\epsilon >0$ with $x+t{e}_i\in U$ for $\mathrm{\mid }t\mathrm{\mid }<\epsilon$. Put $J:=(-\epsilon ,\epsilon )$ and $h(y,t):=\mathrm{\Phi }(x+t{e}_i-y)f(y)$ on ${\mathbb{R}}^n\times J$. For every $t\in J$ the section $y\mapsto h(y,t)$ is integrable by step 2.1; for $y\in {\bar{B}}_R$ the map $t\mapsto h(y,t)$ is differentiable on $J$ because $\mathrm{\mid }x+t{e}_i-y\mathrm{\mid }\ge c$ there, with derivative ${\mathrm{\partial }}_i\mathrm{\Phi }(x+t{e}_i-y)f(y)$; outside ${\bar{B}}_R$ set $h(y,t)=0$, whose derivative is zero; this derivative is measurable in $y$ by step 2.1; and it is dominated uniformly on $J$ by the integrable $g$ of step 2.1. The differentiation-under-the-integral-sign theorem [F4] therefore gives that $t\mapsto Nf(x+t{e}_i)$ is differentiable at $t=0$ with derivative $\int {\mathrm{\partial }}_i\mathrm{\Phi }(x-y)f(y)d{\lambda }_n(y)$; by the definition of the partial derivative [F2] this says that ${\mathrm{\partial }}_{i}Nf(x)$ exists and equals that integral.

Let $x_m\to x$ be a sequence in $U$. For every $y$ with $f(y)\mathrm{\ne }0$ (so $y\in {\bar{B}}_R$) the points $x_m-y$ stay at distance at least $c$ from the origin by step 2.1, so ${\mathrm{\partial }}_i\mathrm{\Phi }(x_m-y)f(y)\to {\mathrm{\partial }}_i\mathrm{\Phi }(x-y)f(y)$ by continuity of the kernel off the origin ([F1], [F2]); the convergence also holds where $f(y)=0$, both sides being $0$. All these integrands are dominated by the integrable $g$ of step 2.1, so dominated convergence [F5] gives $\int {\mathrm{\partial }}_i\mathrm{\Phi }(x_m-y)f(y)d{\lambda }_n(y)\to \int {\mathrm{\partial }}_i\mathrm{\Phi }(x-y)f(y)d{\lambda }_n(y)\mathrm{.}$ By step 3.1 the left-hand side is ${\mathrm{\partial }}_{i}Nf(x_m)$, so each partial derivative of $Nf$ is sequentially continuous on $U$, hence continuous by Metric continuity characterisations, with countable choice for the sequential converse under the assumed ${\mathrm{A}\mathrm{C}}_{\omega }$.

By steps 3.1 and 4.1 the function $Nf$ has all first partial derivatives at every point of $U$ and each of them is continuous on $U$; the characterization of total differentiability by continuous partial derivatives [F2] therefore makes $Nf$ differentiable at every $x\in U$, with $\mathrm{\nabla }(Nf)(x)$ the vector of the partial derivatives, namely $\mathrm{\nabla }(Nf)(x)={\int }_{{\mathbb{R}}^n}\mathrm{\nabla }\mathrm{\Phi }(x-y)f(y)d{\lambda }_n(y),$ the vector integral being absolutely convergent by step 2.1. As $x_0$ with $\mathrm{\mid }{x}_0\mathrm{\mid }>2R$ was arbitrary, this proves (a) throughout the open far field $\{\mathrm{\mid }x\mathrm{\mid }>2R\}$.

For (b), let $\mathrm{\mid }x\mathrm{\mid }>2R$ and let $y$ satisfy $f(y)\mathrm{\ne }0$, so $\mathrm{\mid }y\mathrm{\mid }\le R$; the reverse triangle inequality gives $\mathrm{\mid }x-y\mathrm{\mid }\ge \mathrm{\mid }x\mathrm{\mid }-R>\mathrm{\mid }x\mathrm{\mid }\mathrm{/}2$ ([F3]), hence by step 1.1 $\mathrm{\mid }\mathrm{\nabla }\mathrm{\Phi }(x-y)\mathrm{\mid }=\frac{1}{{\omega }_{n-1}\mathrm{\mid }x-y{\mathrm{\mid }}^{n-1}}\le \frac{2^{n-1}}{{\omega }_{n-1}\mathrm{\mid }x{\mathrm{\mid }}^{n-1}}\mathrm{.}$ Integrating against $\mathrm{\mid }f\mathrm{\mid }$ and using the estimate of step 5.1 gives $\mathrm{\mid }\mathrm{\nabla }(Nf)(x)\mathrm{\mid }\le \int \mathrm{\mid }\mathrm{\nabla }\mathrm{\Phi }(x-y)f(y)\mathrm{\mid }d{\lambda }_n(y)\le \frac{2^{n-1}\mathrm{\parallel }f{\mathrm{\parallel }}_1}{{\omega }_{n-1}\mathrm{\mid }x{\mathrm{\mid }}^{n-1}}$, which is (b). The argument used only the $L^1$ hypothesis and the a.e. compact support, never boundedness of $f$, and it makes no assertion at points with $\mathrm{\mid }x\mathrm{\mid }\le 2R$, where the singularity of $\mathrm{\Phi }$ can meet the support of $f$; this is exactly the scope recorded in (c).