Analytic continuation of a conformal map across the unit circle

Question

I know that if $f$ is a conformal mapping of $\mathbb{D}$ onto some domain $D$ such that $\partial D$ is a Jordan curve, then $f$ has a continuous extension up to $\partial \mathbb{D}$ such that $f(\partial \mathbb{D}) = \partial D$. This is, as far as I know, called Caratheodory's extension theorem.

Suppose now that $f$ maps some arc of the unit circle -- call it $A$ -- to the whole unit circle. The rest of the unit circle -- call it $B$ -- is mapped to some arc.

Now I am trying to analytically continue $f$ to the exterior of $A$. I know that the function $$f^{*}(z) = \overline{f\left(\frac{1}{\bar z} \right)}$$ is analytic, but on $A$ we get $f^{*}(z) = \overline{f(z)}$, which is a problem. If we just had $f^{*}(z) = f(\frac{1}{\bar z})$, then the two functions would agree on $A$ and the Identity theorem would allow us to conclude that $f^{*}$ is the analytic extension of $f$ to the rest of the plane.

But I don't know if that function is analytic -- I doubt it, in fact I'm pretty sure it's not.

How, then, can we obtain an analytic continuation of $f$ to the outside of $A$? I'm sure that some version of the Schwarz reflection principle will allow this, but I'm not seeing a way. One problem is that the version of this principle that I'm familiar with imposes some condition on $f$ requiring it to take real values on some set -- I think on $A$ -- and that is not the case here.

What I particularly need is a continuation that will map a neighborhood of $z_0 \in B$ to a neighborhood of $f(z_0)$, such that points approaching $z_0$ from inside $\mathbb{D}$ will get mapped across as usual, but points approaching $z_0$ from outside the unit circle will get mapped to points outside $\partial D$ -- namely, the reflection across the unit circle of the image points under the original $f$.

To put it another way: $f$ maps some arc of the unit circle to some curve. Now I need an analytic continuation of $f$ which will map that arc to the original image plus the reflection across the unit circle of that original image. In geometric terms, the extended $f$ splits the arc $B$ into an inside arc (which has its original image) and an outside arc, which is still the arc $B$, but which gets mapped to the reflection of $f(B)$ under the extended $f$.

This is why I am interested in some concept of reflection across the unit circle.

Do we get an easy answer by some version of the Schwarz reflection principle? If so, what conditions must $f$ satisfy? Is an application of Caratheodory's extension theorem enough?

(Note: I've assumed that for $z\neq 0$, the reflection of $z$ across the unit circle is the point $\frac{1}{\bar z}$, the inverse of the conjugate of $z$, but I haven't found a definition in any textbook confirming this.)

I edited this question to make it more specific.

Answer 1

(Answering my own question.) We can apply the following version of the Schwarz reflection principle:

Let $\Omega$ be a domain symmetric with resect to the unit circle. Let $\Omega_{0} = \Omega \cap \mathbb{D}$ and let $L=\Omega \cap \partial \mathbb{D}$. Suppose that $f$ is holomorphic on $\Omega_{0}$ and continuous on $\Omega_{0}\cup L$. Suppose also that for each $z\in L$, we have $|f(z)|=1$. Then there is an analytic continuation of $f$ to the whole of $\Omega$ given by $$f^{*}(z) = \overline{f\left(\frac{1}{\bar z}\right)}^{\,-1}.$$

This is the standard version of the principle but the usual lines are now circles, and $f$ is altered as shown, by applying appropriate Mobius transformations. This is also what changes the requirement that $f$ takes real values on the real line to the requirement that $f$ takes unit-modulus values on the unit circle.

I believe that this extension of $f$ displays the correct behavior -- and it's definitely the case that $f$ and $f^{*}$ agree on the unit circle, and that $f^{*}$ is analytic on $\mathbb{C}\setminus\mathbb{D}$, so as far as I can work out, the Identity theorem (a.k.a. Coincidence principle) should guarantee that this is a valid analytic continuation. And of course it has the desired 'reflective' behavior by construction.

Note: Caratheodory's extension theorem gives us the boundary continuity requirement.

Answer 2

This is not possible to do in general. Here's the idea of an obstruction. Consider a conformal bijection $f : \mathbb{D} \to R$ where $R$ is the rectangle $(-1,1) \times (-1,1) \subset \mathbb{C}$. As you've mentioned, $f$ admits a continuous extension to $\overline{\mathbb{D}}$. However, no conformal extension across any of the boundary points $p_i$ of $\mathbb{D}$ which map to corners $r_i$ of $R$ under this map is possible.

To see this, a conformal extension of $f$ would map a neighborhood $U$ of $p_i$ to neighborhood of $r_i$ conformally and also take the arc $S^1 \cap U$ to two segments about $r_i$. This evidently does not preserve angles at $p$, and so $f$ admits no conformal extension there.

It's worth mentioning that in this specific example, using Mobius transformations and Schwarz reflection one can conformally extend $f$ across each of the arcs of $\partial{\mathbb{D}}$. But this demands compatible symmetries of the domain and image of $f$. In the absence of such symmetries, problems emerge.

Answer 3

The fundamental theorem here is that if $f: \mathbb D \to \Omega$ is a Riemann Map, where $\Omega$ is the inner domain of a Jordan curve $J$, then $f$ is extendable analytically beyond $C$ the unit circle to a conformal function, iff $J$ is an analytic Jordan curve (ie there is a univalent parametrization of $J$ from the unit circle that has a univalent analytic extension to an open neighborhood of the circle).

The proof follows from the general reflection principle across circles as one implication is trivial.

Conversely by the defintion of analytic Jordan curves, there is $\phi: A \to B$ analytic and univalent where $A$ is an annulus $\frac{1}{\rho} < |z| < \rho, \rho >1$ with $\phi(C)=J$, then $h(z)=\phi^{-1}(f(z))$ is analytic and univalent in $\frac{1}{R}< |z| <1$ for some $R>1$

But $\frac{1}{\rho} <|h(z)| < 1$ and $|h(z)| \to 1, |z| \to 1$, so by the reflection principle we can extend $h$ (conformally) to $\frac{1}{R}<|z| <R$, hence we can extend $f(z)=\phi(h(z))$ to $|z| <R$ so we are done!

Note that if $f'$ vanishes on $|z|=1$, $f$ may have an extension but $J$ is not analytic (see $f(z)=z^2-2z$ as a simple example, $f$ is univalent inside the unit disc, but $f'(1)=0$)