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I read in a paper by E. A. Rakhmanov "Orthogonal Polynomials and S-curves" the following statement, in Lemma 9.9 therein, which I state here in my words.

Under the standard hypotheses for the Riemann mapping theorem, if $\Omega =\widehat{ \mathbb C}\setminus \mathcal K$ is a simply connected domain (of the Riemann sphere), then the Green's potential $G(z)$ of $\Omega$ (i.e. $G(z)=0$ on $\mathcal K$ and $G(z) \simeq \ln |z|$ at infinity) satisfies $$ G(z) \leq \sqrt{ \frac{\operatorname{dist}(z,\mathcal K)}{\operatorname{Cap}(\mathcal K)}} $$ where $\operatorname{Cap}(\mathcal K)$ is the capacity. I tried to follow the proof but it invokes a Theorem 1 Ch. IV of Goluzin's book, which seemingly has nothing to do with the property where it is used.

Question: is such an inequality proven somewhere else (or provable)? I am thinking of Köbe 1/4 theorem….

Note that the statement should not rely on any regularity whatsoever of $\Omega$. It would seem to me that this should be some very classical result (and if true, a quite nice one at that).

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  • $\begingroup$ I can now prove H\"older continuity for simply connected domains of $\mathbb C$. Will report when I get the full picture! $\endgroup$
    – user528012
    Commented Sep 6 at 14:04

2 Answers 2

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Scaling $\mathcal K$ by a factor $1/Cap(\mathcal K)$, we can assume $Cap(\mathcal{K})=1$. In this case the estimate $$ G(z) \leq C \sqrt{ {\rm dist}(z,\mathcal K)} $$ indeed follows from Koebe-type inequalities. This is explained for instance in the book Complex Dynamics by Carleson and Gamelin, see the second paragraph of p.139.

The claim is that one can take $C=1$, however. For that I am not sure, I am a bit skeptical.

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    $\begingroup$ I claim nothing, I am reporting others' claim! Thanks. $\endgroup$
    – user528012
    Commented Sep 6 at 21:59
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    $\begingroup$ Right, sorry about that. I edited my answer accordingly. It may be true that $C=1$ but I don't think the method I outlined can give that. For your purposes do you need $C=1$? $\endgroup$
    – Malik Younsi
    Commented Sep 6 at 22:17
  • $\begingroup$ I need an estimate where the constant C is either absolute or depend explicitly on the capacity. $\endgroup$
    – user528012
    Commented Sep 6 at 22:36
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    $\begingroup$ I managed to reconstruct the proof (see below). Thanks for your comment! $\endgroup$
    – user528012
    Commented Sep 7 at 9:23
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The inequality is correct and here is the proof, reconstructed after the quoted paper. You need the inequality here: $$ {\rm dist}(z,\mathcal K) = \min_{|\rho|=1} |F(\zeta)-F(\rho)| \geq C \frac{(1-|\zeta|)^2}{|\zeta|} $$ Here $z=F(\zeta) = C \zeta +a_0 + \mathcal O(\zeta^{-1})$ is the uniformizing map of the complement of $\mathcal K$ and $C$ its capacity. The Green’s potential is related to $\zeta$ as $$ {\rm e}^{G(z)} = |\zeta|$$. So the inequality reads, setting $\delta = {\rm dist} (z,\mathcal K)$, $$ \frac {\delta}{{\rm Cap}(\mathcal K)}\geq 4\sinh(G/2)^2\geq G^2. $$ Done!. Thanks to all.

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  • $\begingroup$ I don't understand what you mean by $\min_{|\rho|=1} |F(\zeta)-F(\rho)|$. What are your assumptions on $\mathcal{K}$? Note that in general the uniformizing map of the complement of $\mathcal{K}$ may not extend to the boundary. $\endgroup$
    – Malik Younsi
    Commented Sep 9 at 18:37
  • $\begingroup$ It means the minimum of the set of numbers $\{ |F(\zeta)-F(\rho)|: \rho\in S^1\}$. This is the same as the distance of $z = F(\zeta)$ from the boundary of the image of $F$. $\mathcal K$ (the complement in the $z$--plane of the image of $F$) is compact, because the complement of an unbounded open domain that contains a full neighbourhood of $\infty$. $\endgroup$
    – user528012
    Commented Sep 10 at 2:17
  • $\begingroup$ $F(\rho)$ may not even be defined...! $\endgroup$
    – Malik Younsi
    Commented Sep 10 at 7:10
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    $\begingroup$ This is true if and only if $\mathcal{K}$ is locally connected! If $\mathcal{K}$ is not locally connected then the argument you provide does not work. $\endgroup$
    – Malik Younsi
    Commented Sep 10 at 18:51
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    $\begingroup$ See en.wikipedia.org/wiki/… $\endgroup$
    – Malik Younsi
    Commented Sep 10 at 19:10

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