The two-dimensional logarithmic kernel has unit normalized flux

The set $S_r$ is nonempty, since $(r,0)\in S_r$. By [F12] it is closed as the preimage of the closed singleton $\{r\}$ under $x\mapsto \parallel x{\parallel }_2$. By [F13], each coordinate of a point in $S_r$ has absolute value at most $r$, so $S_r$ is bounded; [F14] makes it compact. Norm continuity shows that points of norm less or greater than $r$ are not on $\mathrm{\partial }{B}_r(0)$. If $\parallel x{\parallel }_2=r$, write $\omega =x\mathrm{/}r$. For $0<t<r$, homogeneity [F9] gives $\parallel x-t\omega {\parallel }_2=r-t<r$ and $\parallel x+t\omega {\parallel }_2=r+t>r$. Thus every neighborhood of $x$ meets both the disk and its complement, proving $S_r=\mathrm{\partial }{B}_r(0)$.

At each $x=(x_1,x_2)\in S_r$, at least one coordinate is nonzero. If $x_1\mathrm{\ne }0$, a neighborhood of $x$ in the circle is the graph $x_1=\pm \sqrt{r^2-x_2^2}$ with the sign chosen to match $x_1$; if $x_2\mathrm{\ne }0$, use the corresponding graph solving for $x_2$. The radicand is positive on the relevant open interval, so [F15]–[F17] make these graph maps $C^1$. Each graph map has rank one because its free coordinate has derivative one; on overlaps the transition is a restriction of the same $C^1$ square-root formula. Thus $S_r$ is an embedded $C^1$ hypersurface, as required by [F3]. Differentiating ${\gamma }_1(t{)}^2+{\gamma }_2(t{)}^2=r^2$ along chart curves using [F18] shows every tangent vector is perpendicular to $x$; the tangent space is one-dimensional by the chart rank, hence equals $x^{\perp }$. For $\omega =x\mathrm{/}r$, $x-t\omega =(r-t)\omega$ is inside the disk and $x+t\omega =(r+t)\omega$ is outside; therefore the disk-outward unit normal is $\nu (x)=\omega$.

Scaling in [F2] gives ${\int }_{S_r}1dS=r\mathrm{\mid }{S}^1\mathrm{\mid }=2r\mathrm{\mid }{B}_1\mathrm{\mid }$. By [F7] and [F8], $\mathrm{\mid }{B}_1\mathrm{\mid }=V_2(1)=\pi \mathrm{/}\mathrm{\Gamma }(2)=\pi$, so the chart area is $2\pi r$. It is finite, and therefore every constant function on $S_r$ is integrable under [F3].

Put $q(s)=-(2\pi {)}^{-1}\log s$ for $s>0$. For $x=r\omega \in S_r$, $x+t\omega =(r+t)\omega$ for $t$ near zero. By [F1], [F5], and [F6], ${\mathrm{\partial }}_{\nu }\mathrm{\Phi }(x)=D_{\omega }\mathrm{\Phi }(x)=q^{\mathrm{\prime }}(r)=-1\mathrm{/}(2\pi r)$. This derivative is evaluated at a point away from the pole.

The derivative in step 3.2 is constant and absolutely integrable on $S_r$ by step 3.1. Linearity [F4] now gives ${\int }_{S_r}{\mathrm{\partial }}_{\nu }\mathrm{\Phi }dS=-\frac{1}{2\pi r}{\int }_{S_r}1dS=-\frac{1}{2\pi r}(2\pi r)=-1$. Thus $-{\int }_{S_r}{\mathrm{\partial }}_{\nu }\mathrm{\Phi }dS=1$ for the operator $-\mathrm{\Delta }$.

For the annulus $r<\mathrm{\mid }x\mathrm{\mid }<R$, with $R>r$, at $x=r\omega$ a small move in direction $-\omega$ enters the removed disk, while a move in direction $+\omega$ enters the annulus. Its outward normal on the inner circle is $-\omega$. Therefore [F5] and step 3.2 give ${\mathrm{\partial }}_{{\nu }_{\mathrm{i}\mathrm{n}\mathrm{n}\mathrm{e}\mathrm{r}}}\mathrm{\Phi }=+1\mathrm{/}(2\pi r)$; using the area from step 3.1, the inner-boundary flux is $+1$ and its negative is $-1$.