Let $A(1,r) = \{z \in \mathbb{C} : 1 < |z| < r\}$. I would like to prove the standard result that $A(1,r)$ and $A(1,r')$ are conformally equivalent iff $r = r'$. To prove the nontrivial direction, suppose that we have an analytic isomorphism $f \colon A(1,r) \to A(1,r')$. I would like to iteratively use the Schwarz reflection principle to extend this isomorphism to an automorphism of the punctured plane with a removable discontinuity at the origin (I've seen this idea suggested in various sources). Then, I can use the fact that the only automorphisms of the plane are of the form $z \mapsto az + b$ to deduce the desired result. However, to use Schwarz, I seem to require that $f$ have a continuous extension to the boundary of the annulus $A(1,r)$. Is there any way to show that $f$ has such a continuous extension, or is there a version of Schwarz that doesn't put this requirement on $f$?

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    $\begingroup$ Yes, the continuity assumption in the reflection principle can be relaxed. See The continuity assumption in Schwarz's reflection principle $\endgroup$
    – user357151
    Mar 30, 2018 at 22:57
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    $\begingroup$ If the domain is simply-connected, the continuity assumption can be relaxed to just stipulating that $\Im(f)$ is continuous because one can apply the Schwarz reflection principle for harmonic functions and then use the fact that there is a unique analytic function on a simply connected with given harmonic function as imaginary part. But in this case, the domain is not simply-connected, and I don't even know that $\Im(f)$ is continuous anyway. So this strategy appears to fail here. $\endgroup$ Mar 30, 2018 at 23:21
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    $\begingroup$ It's more about $\operatorname{Re}\log f$ in case of reflection in a circle. But leaving that aside, one can use Carathéodory's theorem to show that a conformal map between annuli has a continuous extension to the boundary. (Again, the annuli are not simply connected but the theorem applies after making a cut.) $\endgroup$
    – user357151
    Mar 31, 2018 at 0:55

1 Answer 1


Some initial observations:

  • Riemann's theorem on removable singularities requires only boundedness, not continuity.
  • It is certainly correct, for Schwarz reflection on the half-plane, that one only need assume that the imaginary part is continuous up to the boundary.
  • There are of course standard results that say that a conformal map of annuli must be smooth up to the boundary. Carathéodory's result does not apply just to simply connected domains, because you can localize the arguments near a boundary point.

Is there any way to show that $f$ has such a continuous extension, or is there a version of Schwarz that doesn't put this requirement on $f$?

The answer to both questions is yes.

First question. The Osgood-Carathéodory Theorem states that if $f$ is a conformal mapping of a Jordan domain $U$ onto a Jordan domain $V$, then $f$ extends to a homeomorphism of the closure of $U$ onto the closure of $V$. This theorem has been extended to domains bounded by finitely many disjoint Jordan curves (more precisely to domains which are a Jordan domain with finitely disjoint closed Jordan domains removed from the interior). See Theorem 5.3 in the following paper.


In particular domains bounded by two disjoint Jordan domains (such as annuli). Getting back to your first question, this shows that your $f$ does have a continuous extension.

Second question. Your question was answered in the discussion in the comments. You are concerned that the annuli are not simply connected, but the extension problem is local, as noticed by taking a small disk in the answer to the question linked to in the comments.

The continuity assumption in Schwarz's reflection principle

If you take a small disk about a boundary point of an annulus, its intersection with the annulus is simply connected.

In looking at the linked paper, Riemann surfaces can be replaced by plane domains. Every domain (open connected set) in the plane is a Riemann surface.


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