Newton shell theorem from harmonic mean values

Let $S_R=\mathrm{\partial }{B}_R(0)$. By [F3] and [F7], its chart surface measure is $\mathrm{\mid }{S}_R\mathrm{\mid }=R^{n-1}{\omega }_{n-1}=R^{n-1}n\mathrm{\mid }{B}_1\mathrm{\mid },$ which is finite and strictly positive. The sphere is a compact embedded $C^1$ hypersurface by [F8], so [F6] defines the integral. The same scaling relation gives, for every continuous $h$ on $S_R$, $\frac{1}{\mathrm{\mid }{S}_R\mathrm{\mid }}{\int }_{S_R}h(y)d{S}_y=\frac{1}{{\omega }_{n-1}}{\int }_{S^{n-1}}h(R\theta )d\sigma (\theta )\mathrm{.}$ If $\mathrm{\mid }x\mathrm{\mid }\mathrm{\ne }R$, the distance from $x$ to $S_R$ is positive by [F15], so the integrand in the definition of $U(x)$ is continuous and bounded on the compact shell; its integral is finite.

Suppose $\mathrm{\mid }x\mathrm{\mid }>R$. Choose $s$ with $R<s<\mathrm{\mid }x\mathrm{\mid }$ and put $g_x(y)=\mathrm{\Phi }(x-y)$ on $\mathrm{\Omega }=B_s(0)$. The pole $x$ lies outside $\overline{B_s(0)}$, so [F1]–[F2] show that $g_x\in C^2(\mathrm{\Omega })$ and ${\mathrm{\Delta }}_y{g}_x=0$ there. The closed ball $\overline{B_R(0)}$ is compact and lies inside $\mathrm{\Omega }$ by [F8]. Applying [F5] at its center and then the scaling identity of step 1.1 gives $\frac{1}{\mathrm{\mid }{S}_R\mathrm{\mid }}{\int }_{S_R}\mathrm{\Phi }(x-y)d{S}_y=\frac{1}{{\omega }_{n-1}}{\int }_{S^{n-1}}\mathrm{\Phi }(x-R\theta )d\sigma (\theta )=g_x(0)=\mathrm{\Phi }(x)\mathrm{.}$ Multiplication by $M$ proves the exterior formula, including $M=0$.

Put $V(x)=\mathrm{\mid }{S}_R{\mathrm{\mid }}^{-1}{\int }_{S_R}\mathrm{\Phi }(x-y)d{S}_y$ for $\mathrm{\mid }x\mathrm{\mid }<R$. Fix $r<R$. For $\mathrm{\mid }x\mathrm{\mid }\le r$ and $y\in S_R$, $R-r\le \mathrm{\mid }x-y\mathrm{\mid }\le R+r$ by [F15]. The closed annulus $K=\{z:R-r\le \mathrm{\mid }z\mathrm{\mid }\le R+r\}$ is bounded and closed by [F18], hence compact; it is nonempty since it contains $R{e}_1$. All derivatives of $\mathrm{\Phi }$ through order two are continuous and bounded on $K$ by [F2] and [F18], and are uniformly continuous there by [F19]. As $S_R$ has finite measure by step 1.1, constant bounds on these derivatives are integrable majorants. The derivative integrands are Borel measurable because they are continuous on the shell. For each $x\in B_r(0)$, choose a parameter interval small enough that $x+t{e}_i$ stays in $B_r(0)$. Apply [F9, F13] to each coordinate parameter, first to $\mathrm{\Phi }(x-y)$ and then to each first derivative. This gives ${\mathrm{\partial }}_{i}V(x)=\frac{1}{\mathrm{\mid }{S}_R\mathrm{\mid }}{\int }_{S_R}{\mathrm{\partial }}_i\mathrm{\Phi }(x-y)d{S}_y,{\mathrm{\partial }}_j{\mathrm{\partial }}_{i}V(x)=\frac{1}{\mathrm{\mid }{S}_R\mathrm{\mid }}{\int }_{S_R}{\mathrm{\partial }}_j{\mathrm{\partial }}_i\mathrm{\Phi }(x-y)d{S}_y\mathrm{.}$ For $x,x^{\mathrm{\prime }}\in B_r(0)$ and $y\in S_R$, the points $x-y,x^{\mathrm{\prime }}-y$ lie in $K$ and have distance $\mathrm{\mid }x-x^{\mathrm{\prime }}\mathrm{\mid }$. Uniform continuity [F19] therefore bounds each derivative-integrand difference uniformly in $y$ by a modulus tending to zero with $\mathrm{\mid }x-x^{\mathrm{\prime }}\mathrm{\mid }$. Applying [F20] and dividing by $\mathrm{\mid }{S}_R\mathrm{\mid }$ shows that each derivative integral is continuous in $x$. Since $r<R$ was arbitrary, $V\in C^2(B_R(0))$ by [F11]. Each $y\in S_R$ is different from $x$; [F2] and [F10] now give $\mathrm{\Delta }V(x)=\frac{1}{\mathrm{\mid }{S}_R\mathrm{\mid }}{\int }_{S_R}\mathrm{\Delta }\mathrm{\Phi }(x-y)d{S}_y=0.$ Thus $V$ is harmonic inside the shell.

The function $V$ is radial. Indeed, take $x,x^{\mathrm{\prime }}$ with $\mathrm{\mid }x\mathrm{\mid }=\mathrm{\mid }{x}^{\mathrm{\prime }}\mathrm{\mid }<R$. If $x=x^{\mathrm{\prime }}$, use the identity map. Otherwise set $v=x-x^{\mathrm{\prime }}$ and $Qz=z-2\frac{⟨z,v⟩}{⟨v,v⟩}v\mathrm{.}$ Bilinearity of the inner product [F12] gives $⟨Qz,Qw⟩=⟨z,w⟩$. Since $⟨Qz,v⟩=-⟨z,v⟩$, the formula gives $Q^{2}z=z$; hence $Q$ is an orthogonal involution. Also $2⟨x,v⟩=⟨v,v⟩$, hence $Qx=x^{\mathrm{\prime }}$. Thus $Q$ maps $S_R$ onto itself. It is an isometry, hence continuous [F24], and [F13] makes it a Borel-measurable self-map. The unit-sphere invariance and radius-scaling clauses of [F3] imply that $Q$ preserves the surface measure on $S_R$, so it preserves surface integrals by [F22]. Using $y=Qz$ and the radial formula [F1], $V(x^{\mathrm{\prime }})=\frac{1}{\mathrm{\mid }{S}_R\mathrm{\mid }}{\int }_{S_R}\mathrm{\Phi }(Qx-y)d{S}_y=\frac{1}{\mathrm{\mid }{S}_R\mathrm{\mid }}{\int }_{S_R}\mathrm{\Phi }(Qx-Qz)d{S}_z=\frac{1}{\mathrm{\mid }{S}_R\mathrm{\mid }}{\int }_{S_R}\mathrm{\Phi }(x-z)d{S}_z=V(x)\mathrm{.}$

Write $V(x)=v(\mathrm{\mid }x\mathrm{\mid })$ on $B_R(0)$ and set $v(r)=V(r{e}_1)$ for $0<r<R$. By the chain and product rules [F14, F16], for $\mathrm{\mid }x\mathrm{\mid }=r>0$, ${\mathrm{\partial }}_{i}V(x)=v^{\mathrm{\prime }}(r)\frac{x_i}{r},{\mathrm{\partial }}_{ii}V(x)=v^{\mathrm{\prime }\mathrm{\prime }}(r)\frac{x_i^2}{r^2}+v^{\mathrm{\prime }}(r)(\frac{1}{r}-\frac{x_i^2}{r^3})\mathrm{.}$ Summing over $i$ and using [F11] yields $0=\mathrm{\Delta }V(x)=v^{\mathrm{\prime }\mathrm{\prime }}(r)+\frac{n-1}{r}{v}^{\mathrm{\prime }}(r)\mathrm{.}$ Consequently $(r^{n-1}{v}^{\mathrm{\prime }}(r){)}^{\mathrm{\prime }}=0$, so [F17] makes $r^{n-1}{v}^{\mathrm{\prime }}(r)=c$ on $(0,R)$ for one constant $c$. By [F16], the derivative of $v(r)-\frac{c}{2-n}{r}^{2-n}$ is zero; another application of [F17] gives $v(r)=a+b{r}^{2-n}(0<r<R)$ for constants $a,b$. Since $V$ is continuous at the origin by [F11, step 2.2], there are $\rho >0$ and $C>0$ such that $\mathrm{\mid }v(r)\mathrm{\mid }\le C$ for $0<r<\rho$. Set $m=n-2\ge 1$. For any $L>0$ and $0<r<\min \{1,L^{-1}\}$, induction on $k\in \mathbb{N}$ proves $0<r^{k+1}\le r$: the base $k=0$ is $r^1=r$ by [F25]; if $r^{k+1}\le r$, then multiplying by $r>0$ and using $r<1$ gives $r^{k+2}=r^{k+1}r\le r^2\le r$ by [F27]. Since $m\ge 1$, it is a nonzero natural; [F29] gives $k\in \mathbb{N}$ with $m=\sigma (k)$, whose real exponent is $k+1$ by [F26]. Thus $0<r^m\le r<L^{-1}$. By [F28], $r^{2-n}=r^{-m}=(r^m{)}^{-1}>L$. This is the quantified limit $r^{2-n}\to +\mathrm{\infty }$ as $r\downarrow 0$. If $b\mathrm{\ne }0$, choose $L>(C+\mathrm{\mid }a\mathrm{\mid })\mathrm{/}\mathrm{\mid }b\mathrm{\mid }$ and take $r$ small enough to satisfy the preceding bound and $r<\min \{R,\rho \}$. Reverse triangle inequality [F15] gives $\mathrm{\mid }v(r)\mathrm{\mid }=\mathrm{\mid }a+b{r}^{-m}\mathrm{\mid }\ge \mathrm{\mid }b\mathrm{\mid }{r}^{-m}-\mathrm{\mid }a\mathrm{\mid }>C$, a contradiction. Thus $b=0$, and $V$ is constant throughout $B_R(0)$.

At $x=0$, $\mathrm{\mid }y\mathrm{\mid }=R$ on $S_R$, so [F1] gives $V(0)=\frac{1}{\mathrm{\mid }{S}_R\mathrm{\mid }}{\int }_{S_R}\mathrm{\Phi }(-y)d{S}_y=\mathrm{\Phi }(R)\mathrm{.}$ Step 4.1 makes this the value of $V$ at every interior point. Since $U=MV$, the interior formula follows for every real $M$, including zero.

The outside and inside cases are disjoint and cover exactly the points $\mathrm{\mid }x\mathrm{\mid }\mathrm{\ne }R$. At $\mathrm{\mid }x\mathrm{\mid }=R$ the pole lies on the shell, and the statement makes no claim there. The $n\ge 3$ assumption is used when the radial ODE produces $r^{2-n}$ and continuity at zero removes its singular term; dimensions one and two are outside this statement. Countable Choice is the sole set-theoretic assumption, carried by the cited kernel, surface/polar measure, spherical-mean and ball-measure interfaces [A1, F1, F2, F3, F4, F5, F6, F7]; the explicit differentiation and reflection calculations require no full Axiom of Choice. ∎