# Holomorphic function on Upper Half Plane must be rational

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?

• how about $e^{ix}$? It doesn't look very rational. Commented May 2, 2011 at 16:01
• e^{ix} is not holomorphic on the upper half plane.
– ergo
Commented May 2, 2011 at 16:35
• but it's the composition of two holomorphic functions? Commented May 2, 2011 at 16:37
• @ergo: user8268 means $f(z) = e^{iz}$. Commented May 2, 2011 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
Commented May 2, 2011 at 19:33

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.

• 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
Commented May 2, 2011 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.
Commented May 2, 2011 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
Commented May 2, 2011 at 19:06
• @Robert why don't we need the stronger condition that $\lim_{|z| \rightarrow + \infty} f(z)$ exists?
– Plop
Commented May 2, 2011 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.
Commented May 2, 2011 at 19:19