4.4.24 The Wiener criterion and planar continuaED2AAB8DC124

Proof. An interior point of a segment is regular for a disk slit along the segment by the explicit exterior-segment barrier. The boundary nonthinness theorem therefore makes the segment nonthin at that point. If it were thin at an endpoint, reflect an amplified barrier across the endpoint and add the two barriers. The doubled segment would then be thin at its interior point, a contradiction. The radial projection $y\mapsto a+|y-a|e_1$ is a contraction preserving distances from $a$. If every sufficiently small circle met a thin complement, its projection would contain a nondegenerate segment ending at $a$, contradicting contraction invariance of thinness. Hence the required circles exist. An exterior continuum containing $a$ and another point $b$ meets every circle of radius less than $|a-b|$, by connectedness of its distance image. This contradicts the circle conclusion if the complement were thin. Nonthinness gives regularity. For the final assertion use the standard planar topological characterization of a simply connected domain: its complement in the sphere is connected. For a bounded domain this complement is a nondegenerate compact continuum containing infinity; its radial distance image at each finite boundary point meets every sufficiently small circle. The same argument applies directly on the sphere, without assuming a finite exterior arc. This planar topological characterization is an explicit topological prerequisite, not a new potential-theoretic premise. $\square$

Proof. We give the kernel and measure details used in both directions. If $U\Subset B$ is open, compact exhaustion and the equilibrium measures give a measure $\mu _ U$ on $\partial U$ with mass $C(U)$ and $G_ B\mu _ U=1$ on $U$. Indeed their masses are bounded by $C(\overline U)$; weak compactness gives a limit supported on $\partial U$, and for each interior $x$ the eventual supports avoid a neighborhood of $x$, so the continuous kernel can be tested there. A cutoff equal to one around $\overline U$ gives the limiting mass. Also, if $\nu $ is supported on a closed $F$ and $G_ B\nu \ge b$ quasi-everywhere on $F$, then $\nu (F)\ge b C_*(F)$. To see this, integrate against each compact equilibrium measure $\mu _ K$, $K\subset F$, use reciprocity and $G_ B\mu _ K\le 1$, and then take the compact supremum.

Here is the uniform separated-shell estimate. For $\Phi (y-a)\in [\lambda ^ j,\lambda ^{j+1}]$ and $\Phi (z-a)\notin (\lambda ^{j-1},\lambda ^{j+2})$,

In $d\ge 3$, the radii of the two shells differ by at least the fixed factor $\lambda ^{1/(d-2)}$; the reverse triangle inequality bounds $|y-z|$ below by a fixed positive multiple of $|z-a|$. In the plane, write the radii as $e^{-t}$. In the outer case their ratio is at most $q=e^{-(\lambda -1)}<1$, hence $\log (1/|y-z|)\le \Phi (z-a)+\log (1/(1-q))$. In the inner case the same estimate with the variables reversed gives $\Phi (y-z)\le \lambda ^{-1}\Phi (z-a)+\log (1/(1-q))$. Since $z\in B$ gives $\Phi (z-a)\ge \log 2$, these are uniform ratio bounds. The same argument applies to shells of width three whose indices are separated by six. If $G_ B\nu (a)<\infty $ and $\nu $ is finite, then $\int \Phi (z-a)d\nu =G_ B\nu (a)+\Phi (1/2)\nu (B)<\infty $. Thus the potential contributed by all sufficiently separated shells is bounded by a constant independent of their index and target point.

Suppose first the series converges. Choose open $U_ j\Supset E_ j$ with compact closure in $B$, avoiding $\{ \Phi >\lambda ^{j+2}\} $, and $C(U_ j)\le C^*(E_ j)+\varepsilon _ j$, where $\sum \lambda ^ j\varepsilon _ j<\infty $. Finite initial indices outside $B$ may be discarded. Their open equilibrium potentials $p_ j$ satisfy $p_ j=1$ on $U_ j$ and $p_ j(a)\le \lambda ^{j+2}C(U_ j)$. Hence $v_ k=\sum _{j\ge k}p_ j$ is superharmonic, finite at $a$, with $v_ k(a)\to 0$, while $v_ k\ge 1$ on all sufficiently near points of $E\setminus \{ a\} $. For large $k$ this is the strict-gap criterion.

Conversely use an amplified thinness potential $v=G_ B\nu $, with $\nu $ finite and $v(a)<\infty $. To obtain it on this fixed ball, first extend the local amplified barrier exactly to an entire superharmonic function, then restrict its Riesz measure to $B$. That measure is finite because the function is defined on a neighborhood of $\overline B$. The discarded harmonic function is bounded near $a$ and has no effect on the infinite approach. Remove $a$ from $E$ and discard empty shells. Set $b_ j=\min (j,\inf _{E_ j}v)$, so $b_ j$ is finite and tends to infinity. The Borel sets

contain $E_ j$ and are capacitable. Choose compact $K_ j\subset U_ j$ with $C(K_ j)>C(U_ j)-\varepsilon _ j$, with summable weighted errors. Fix a residue class modulo six and put $K=\{ a\} \cup \bigcup K_{6j+p}$. This is compact. Its reduction $w=\widehat R_ Kv=G_ B\eta $ has $w(a)\le v(a)<\infty $, is supported on $K$, and $\eta (\{ a\} )=0$. The separated-shell estimate bounds the contribution of all the other $K_{6k+p}$ at points of $K_{6j+p}$ by a fixed $\beta $. Since $w=v>b_{6j+p}-1$ quasi-everywhere on that compact set, for all sufficiently large $j$ its own contribution is at least one quasi-everywhere. The preceding mass-capacity inequality gives $\eta (K_{6j+p})\ge C(K_{6j+p})$. Finally $\int \Phi (z-a)d\eta <\infty $, so

Apply this to all six classes, restore the summable errors, and use $C^*(E_ j)\le C(U_ j)$. This proves necessity. The last assertion is exactly the already proved equivalence between irregularity and thinness of the complement. $\square$