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It's easy to see that the Phargmen-Lindelöf theorem from complex analysis can be generalized to non-simply-connected regions. Namely to regions $G$ with the property that for each $z \in \partial_\infty G $ there is a sphere $V$ in $\mathbb{C}_\infty $ centered at $z$ such that $V \cap G$ is simply connected.

The problem is that I can't think about any example of non-simply connected regions that have this property and of simply-connected regions that don't have this property...

Does the exterior of a unit ball can be considered as an example of non-simply connected region that don't satisfy the property above? What about regions that do satisfy this property?

BTW - $ \partial _ \infty $ is the boundary of $G$ with the boundary at infinity (and $\mathbb{C}_\infty$ is the Riemann-Sphere ) .

Thanks in advance

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    $\begingroup$ A sphere in $C^\infty$? What's that? $\endgroup$
    – Marc Palm
    Commented Jul 18, 2012 at 17:04
  • $\begingroup$ It's $C$ with infinity... $\endgroup$
    – Jason Mraz
    Commented Jul 18, 2012 at 17:07
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    $\begingroup$ Firstly, there are a lot of results in complex analysis labeled "Phragmen-Lindelöf Theorems". It might be useful here to indicate which Theorem you are refering to.... Secondly, I'm not sure I understand the question... Doesn't an annulus, for example, satisfy the property? $\endgroup$
    – Malik Younsi
    Commented Jul 18, 2012 at 18:09
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    $\begingroup$ By "a sphere V in $\mathbb{C}_{\infty}$" did you mean a circle, i.e. the boundary of a disc? $\endgroup$
    – Kevin
    Commented Jul 18, 2012 at 20:59
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    $\begingroup$ For an example of unbounded non simply-connected region which satisfies your property, just take the complement in $\mathbb{C}_{\infty}$ of two disjoint closed disks... $\endgroup$
    – Malik Younsi
    Commented Jul 18, 2012 at 22:53

1 Answer 1

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Phragmen-Lindelof theorem has nothing to do with simple or multiple connectedness. The general formulation is the following. Let D be an region on the Riemann sphere. Let $w_0$ be a boundary point of D. Suppose that u is a subharmonic function in D which is bounded from above. Suppose that $$\limsup_{z\to w} u(z)\leq 0$$ for all boundary points $w$ of $D$, except $w_0$. Then $u\leq 0$.

One point $w_0$ can be replaced by any closed set of zero logarithmic capacity. But the original argument of Phragmen and Lindelof was for one point, and $u=\log|f|$ where $f$ is analytic.

For the proof, assume first that $D$ is contained in the unit disc, and $w_0=0$. Apply the Maximum principle to $v=u+\epsilon\log|z|$. You obtain that $v\leq 0$. Now fix $z$ and let $\epsilon\to 0$. Generalization to arbitrary $D$ is trivial.

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