Flux normalization on every centered sphere

For $n\ge 3$, put $q_n(s)=s^{2-n}\mathrm{/}((n-2){\omega }_{n-1})$ for $s>0$. By [F6], $q_n^{\mathrm{\prime }}(s)=\frac{2-n}{(n-2){\omega }_{n-1}}{s}^{1-n}=-\frac{1}{{\omega }_{n-1}{s}^{n-1}}\mathrm{.}$ For $n=2$, put $q_2(s)=-(2\pi {)}^{-1}\log s$ for $s>0$. By [F2], [F8] and [F9], ${\omega }_1=\mathrm{\mid }{S}^1\mathrm{\mid }=2\mathrm{\mid }{B}_1\mathrm{\mid }=2V_2(1)=2\pi$, and [F7] gives $q_2^{\mathrm{\prime }}(s)=-\frac{1}{2\pi s}=-\frac{1}{{\omega }_{1}s}\mathrm{.}$ Thus in both cases $q_n^{\mathrm{\prime }}(s)=-\frac{1}{{\omega }_{n-1}{s}^{n-1}}(s>0)\mathrm{.}$

The sphere $S_r=\mathrm{\partial }{B}_r(0)$ is nonempty because $(r,0,\dots ,0)\in S_r$. It is closed and bounded, hence compact by [F14]. At each $x\in S_r$, some coordinate is nonzero; take the least index $j$ with $x_j\mathrm{\ne }0$. Near $x$, the sphere is the graph of $x_j=\mathrm{sgn}(x_j)\sqrt{r^2-{\sum }_{k\mathrm{\ne }j}{x}_k^2}\mathrm{.}$ Its radicand is positive near the projected point; the inner polynomial and positive-base square-root are $C^1$ by [F6], [F12] and [F13]. These graph charts cover $S_r$, have rank $n-1$, and their coordinate transitions are $C^1$ by [F13], giving $S_r$ the compact embedded $C^1$ hypersurface structure used by [F3]. Every tangent vector is the velocity of a differentiable curve in the sphere. Differentiating ${\sum }_k{\gamma }_k(t{)}^2=r^2$ at $x$ by [F11] shows the tangent plane is contained in $x^{\perp }$; its dimension is $n-1$ by the chart rank, so it equals $x^{\perp }$. The unit normal candidates are $\pm \omega$ for $x=r\omega$, $\mathrm{\mid }\omega \mathrm{\mid }=1$. Since $x+t\omega =(r+t)\omega$ is outside the ball and $x-t\omega =(r-t)\omega$ is inside for $0<t<r$, the outward unit normal is $\nu (x)=\omega$.

By [F2] and [F3], ${\int }_{S_r}1dS=r^{n-1}{\int }_{S^{n-1}}1dS={\omega }_{n-1}{r}^{n-1}\mathrm{.}$ This is finite: [F2] gives ${\omega }_{n-1}=n\mathrm{\mid }{B}_1\mathrm{\mid }$, and [F10] gives $\mathrm{\mid }{B}_1\mathrm{\mid }<\mathrm{\infty }$. Thus constant functions are integrable on $S_r$.

For $x=r\omega \in S_r$, the line in direction $\nu (x)=\omega$ satisfies $x+t\omega =(r+t)\omega$ near $t=0$. By [F5] and step 1.1, ${\mathrm{\partial }}_{\nu }\mathrm{\Phi }(x)=D_{\omega }\mathrm{\Phi }(x)={\lim }_{t\to 0}\frac{q_n(r+t)-q_n(r)}{t}=q_n^{\mathrm{\prime }}(r)=-\frac{1}{{\omega }_{n-1}{r}^{n-1}}\mathrm{.}$ The derivative is taken on a neighborhood of each sphere point and does not evaluate the singular kernel at the pole.

By [F4], steps 1.3 and 2.1 give $-{\int }_{S_r}{\mathrm{\partial }}_{\nu }\mathrm{\Phi }dS=-q_n^{\mathrm{\prime }}(r){\int }_{S_r}1dS=\frac{1}{{\omega }_{n-1}{r}^{n-1}}{\omega }_{n-1}{r}^{n-1}=1.$ This uses only the stated ${\mathrm{A}\mathrm{C}}_{\omega }$ convention in the surface-integration interface [F3].

For $0<r<R$, take $0<t<\min (r,R-r)$. At $x=r\omega$, $x-t\omega =(r-t)\omega$ lies in the excised hole, while $x+t\omega =(r+t)\omega$ lies in the annulus. Thus its outward normal at the inner boundary is $-\omega$. Hence ${\mathrm{\partial }}_{{\nu }_{\mathrm{i}\mathrm{n}\mathrm{n}\mathrm{e}\mathrm{r}}}\mathrm{\Phi }(r\omega )=D_{-\omega }\mathrm{\Phi }(r\omega )=-q_n^{\mathrm{\prime }}(r)=1\mathrm{/}({\omega }_{n-1}{r}^{n-1})$. Using the same area and integrability calculation as in step 1.3 and linearity [F4], this contextual inner-boundary calculation gives negative flux $-1$.