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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?

ergo
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1 Answers1

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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.

Robert Israel
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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 02 '11 at 16:46
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    @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 02 '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 02 '11 at 19:06
  • @Robert why don't we need the stronger condition that $\lim_{|z| \rightarrow + \infty} f(z)$ exists? – Plop May 02 '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 02 '11 at 19:19
  • @Robert Israel the function $z \mapsto e^{i e^{-z^2}}$ is entire, and has modulus one on the real axis, and yet it is not a rational function. – Plop May 08 '11 at 16:52
  • @Plop: you're right, you do need that stronger condition. – Robert Israel May 10 '11 at 17:16