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As the title asks. Let me elaborate; suppose $\mathcal K$ is a continuum (compact, connected) set of $\mathbb C$ (with at least two points!). Let's say that $g(z;a)$ is the green's function of the complement. In Saff-Totik's book ("Logarithmic potentials with external fields", App A.2, The 2.1) it is stated that $x\in \partial \mathcal K$ is regular (i.e. $\lim_{z\to x} g(z;a)=0$) iff the Wiener condition holds: $$ \sum_{n=1}^\infty \frac{n}{|\ln {\rm Cap}(A_n(x)|}=\infty $$ where $A_n$ are the annuli $\lambda^n<|z-x|<\lambda^{n-1}$, $0<\lambda<1$, intersected with $\mathcal K$.

Having little familiarity with the condition, I am asking: is it automatically satisfied at all points of a continuum? Reading the definition I would venture to say yes and argue the way below.

It would seem to me that each of the capacities is at least $4 \lambda^{n-1}(1-\lambda)$ (the capacity of a radius in the annulus) (eventually) so that the sum nicely diverges.

Any comment? Or even better a reference!

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Since you consider Green'a function of the complement, you evidently implicitly assume that this complement is connected. For this case, Wiener's criterion is an overshot, and a simple proof (without using Wiener's criterion) can be found in Hayman-Kennedy book, Subharmonic functions, vol. I. (It uses the conformal map of the complement onto the unit disk).

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  • $\begingroup$ Well, the continuum could be a circle so that there are two components, but in essence you are right, I should have added "... of the unbounded component of the complement". I was looking for a proof without using simple connectedness (I am interested in finite union of continua), but if it works for one, I know how it works for many by simple estimate. Thanks! $\endgroup$
    – user528012
    Commented Sep 12 at 1:17
  • $\begingroup$ In the case when there are several complementary components, you pick one of them. Then the statement is that the boundary of this component is regular. $\endgroup$ Commented Sep 12 at 12:26

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