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?

  • 5
    $\begingroup$ how about $e^{ix}$? It doesn't look very rational. $\endgroup$ – user8268 May 2 '11 at 16:01
  • $\begingroup$ e^{ix} is not holomorphic on the upper half plane. $\endgroup$ – ergo May 2 '11 at 16:35
  • $\begingroup$ but it's the composition of two holomorphic functions? $\endgroup$ – quanta May 2 '11 at 16:37
  • 2
    $\begingroup$ @ergo: user8268 means $f(z) = e^{iz}$. $\endgroup$ – Robert Israel May 2 '11 at 16:42
  • $\begingroup$ 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. $\endgroup$ – Plop May 2 '11 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.

  • $\begingroup$ Isn't it necessary that $f(r)\in\mathbb{R}$ for all $r\in\mathbb{R}$ in order to apply the Schwarz Reflection principle? $\endgroup$ – ergo May 2 '11 at 16:46
  • 1
    $\begingroup$ @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. $\endgroup$ – t.b. May 2 '11 at 16:51
  • $\begingroup$ @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). $\endgroup$ – Plop May 2 '11 at 19:06
  • $\begingroup$ @Robert why don't we need the stronger condition that $\lim_{|z| \rightarrow + \infty} f(z)$ exists? $\endgroup$ – Plop May 2 '11 at 19:09
  • $\begingroup$ @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 $U \cap \mathbb{R}$. $\endgroup$ – t.b. May 2 '11 at 19:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.