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Let $f$ be holomorphic on the upper half plane and continuous on $\mathbb{R}$, with $|f(r)|=1$ for all $r\in\mathbb{R}$. Prove that $f$ is rational.

I was playing around with conformal maps and $\overline{f(\bar{z})}$, but I would really like a hint on how exactly "rationality" comes up. I'm guessing Schwarz Lemma is involved?

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how about $e^{ix}$? It doesn't look very rational. – user8268 May 2 '11 at 16:01
e^{ix} is not holomorphic on the upper half plane. – ergo May 2 '11 at 16:35
but it's the composition of two holomorphic functions? – quanta May 2 '11 at 16:37
@ergo: user8268 means $f(z) = e^{iz}$. – Robert Israel May 2 '11 at 16:42
Note that a rational function holomorphic on the upper half plane and such that $|f(r)|=1$ for all $r\in\mathbb{R}$ is a product of $z \mapsto (z-\alpha)/(z-\bar{\alpha})$ for $\alpha$ in the upper half plane. – Plop May 2 '11 at 19:33
up vote 10 down vote accepted

I think you also want $\lim_{r \to +\infty} f(r)$ and $\lim_{r \to -\infty} f(r)$ to exist and be equal. Schwarz Reflection principle shows $f$ is meromorphic on $\mathbb C$ with $f(\overline{z}) = 1/\overline{f(z)}$. Same applies to $f(1/z)$. So $f$ is an analytic function from the Riemann sphere to itself, and such functions are rational.

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Isn't it necessary that $f(r)\in\mathbb{R}$ for all $r\in\mathbb{R}$ in order to apply the Schwarz Reflection principle? – ergo May 2 '11 at 16:46
@ergo: you can compose $f$ with the inverse of the Cayley transform $z \mapsto \frac{z-i}{z+i}$, then apply the reflection principle you know and then transform back. – t.b. May 2 '11 at 16:51
@Theo letting $\varphi(z)=\frac{i(z+1)}{z-1}$ (the inverse of the Cayley transform), for $f \circ \varphi$ to be holomorphic we need $f(z) \neq 1$ for all $z$, and that is not true in general. I think an ad hoc version of the Schwarz reflection principle is needed here (to allow meromorphic functions). – Plop May 2 '11 at 19:06
@Robert why don't we need the stronger condition that $\lim_{|z| \rightarrow + \infty} f(z)$ exists? – Plop May 2 '11 at 19:09
@Plop: Right, I should have formulated this a bit more carefully. But you can simply exclude the discrete set of points that are mapped to one. The Schwarz reflection principle applies to all sets $U$ that are open in the closed upper half plane and only take real values on $U \cap \mathbb{R}$. – t.b. May 2 '11 at 19:19

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