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Suppose I have a function $f:\mathbb{C}\rightarrow \mathbb{C}$, holomorphic on some neighborhood of an annulus $r\le|z|\le R$, $r<R$. If, for $z\in\{|z|=r\text{ or }|z|=R\}$, $|f(z)|=C$ for some constant C, does it follow that $f(z)$ is a constant function?

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To make the question interesting you should add the assumption that $r<R$. – Colin McQuillan Jan 1 '12 at 12:28
Looking forward to answers. I can't do better than saying "no" at the moment with no proof. Such a function has at least 2 zeros in the annulus counting multiplicity. See the related… and…. I think the Blaschke products referred to in might be examples, involving Green's function. But I'm guessing there are more topological ways to approach it. – Jonas Meyer Jan 1 '12 at 13:12
I don't understand Ahlfors's paper, Open Riemann Surfaces and Extremal Problems on Compact Subregions, Commentarii Mathematici Helvetici 24 (1950), 100-134, well enough to make a definitive statement, but it seems to me that the answer to your question is no. If I understand correctly, Ahlfors considers an open Riemann surface of $W$ of genus $p$ with $q \geq 1$ boundary arcs and establishes that there are analytic functions from $\overline{W}$ onto the unit disk with constant modulus $1$ on the boundary in section 4.2 p124ff. – t.b. Jan 1 '12 at 15:41
I hope somebody more qualified can expand on this. See also the remarks at the beginning of section I in E.L. Stout, On Some Algebras of Analytic Functions on Finite Open Riemann Surfaces., Math. Z. 92 (5), 1966, 366-379. – t.b. Jan 1 '12 at 15:45
I feel like I am missing something. Why cannot you use the Schwarz reflection principle repeatedly across the arcs $|z|= r(R/r)^n$, $n\in\mathbb Z$, to get an analytic function on $\mathbb C \setminus\{0\}$ with a removable singularity at $0$? – Stephen Montgomery-Smith Jan 1 '12 at 23:56
up vote 16 down vote accepted

The answer is no. An example of such a function is the Ahlfors function, which in the case of the annulus is a 2-to-1 mapping of the annulus onto the unit disk, which extends to be holomorphic on a neighborhood of the annulus and which maps each boundary circle to the unit circle.

More generally, let $\Omega$ be a finitely connected domain , whose boundary consists of $n$ disjoint analytic Jordan curves. Then there is a function $F$, called the Ahlfors function for $\Omega$, which has the following properties :

  1. $F$ is a $n$-to-$1$ holomorphic mapping of $\Omega$ onto the unit disk.
  2. $F$ extends analytically across each boundary curve of $\Omega$, and maps each of theses curves homeomorphically onto the unit circle.

The Ahlfors function is the solution to an extremal problem involving a quantity called analytic capacity. When $\Omega$ is simply connected, the Ahlfors function is simply the Riemann map. A good reference is the book "analytic capacity and measure" by Garnett. See also "Geometric function theory: explorations in complex analysis", by Krantz, theorem 4.5.9.

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No, it does not follow. Look at $$f(z)=z+1/z$$

As $z\rightarrow 0$ the function $|f|$ tends to $\infty$, similary when $z\rightarrow \infty$ On the other hand, $|f| |_{S^1}$ is bounded by some constant $C$ (e.g. $C =2 $ :-)

Choose a regular value $R$ of $|f|$ bigger than $C=|f| |_{S^1}$. $f^{-1}(R)$ will be a smooth $1$- dimensional submanifold of $\mathbb{C}$ with at least one component interior to the unit ball and one to the exterior of the unit ball. To see that (one of) the curves outside the unit ball will actually surround the unit ball at least once just look at the definition of $f$, which for large values of $|z|$ is just a distortion of $z$ (choose $R$ bigger if necessary). The same reasoning shows that the curve in the interior may be chosen so it surrounds the origin (i.e. the origin is contained in the disc bounded by the curve), because $f$ is basically a distortion of $1/z$ near $z=0$.

That is, you can choose two components of $f^{-1}(R)$ which form the boundary of a (topological) annulus $G$ containing $S^1$ as a homotopically nontrivial curve, on which $f$ is defined and holomorphic.

Any topological annulus is conformally equivalent to a standard annulus $A(r) := \{z: 1 < |z| < r\}$ - this is not trivial, but well known. Let $\phi: A(r) \rightarrow G$ denote a conformal equilvalence. Now look at $f\circ\phi$. This is holomorphic and it's norm equals $R$ at the boundary.

(This, admittedly does not answer the question completely, since it is not clear whether $f\circ\phi$ can be extended holomorphically beyond the annulus. But it's continuous up to the boundary, so I think it's close enough to an answer to at least state the result.)

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