I have found in several books the following affirmation :

Let $f: \Delta \rightarrow \Delta$ be a non constant holomorphic function that extends continuously to $\overline{\Delta}$, $\Delta$ being the open unit disk. Then $f$ is a finite Blaschke product, i.e. of the form $$B(z) = e^{i \theta} \prod_{k=1}^d \frac{z- a_i}{1-\overline{a_i} z}$$

Now it is not hard to check that those rational fractions do indeed send $\Delta$ onto $\Delta$, but I do not know where to find a proof that they are the only ones.

Where can I find that proof in the literature ? I am not interested in the infinite product case, only finite. (Of course, if anyone is kind enough to post a proof, that would do fine as well =)

  • $\begingroup$ You need to multiply with a constant of modulus 1 as well. $\endgroup$ – Harald Hanche-Olsen Jun 7 '12 at 14:22
  • $\begingroup$ yup, i'll edit it $\endgroup$ – Glougloubarbaki Jun 7 '12 at 14:36

I think you have to assume that $f$ is non constant and that $f$ maps the unit circle into itself. In this case, the result follows by the maximum modulus principle.

Indeed, first note that $f$ has a finite number of zeros in $\mathbb{D}$, otherwise since $f$ is not identically zero the zeros would accumulate on the circle, which contradicts the fact that $|f|=1$ there.

Then, consider $B$ a finite blaschke product that has exactly the same zeros than $f$. Then $B/f$ and $f/B$ are holomorphic in $\mathbb{D}$, continuous in $\overline{\mathbb{D}}$. Furthermor, $|B/f|=1$ and $|f/B|=1$ on the unit circle. By the maximum modulus principle, we get that $|B/f| \equiv 1$ in $\mathbb{D}$ and thus $B/f$ is a unimodular constant. This gives the result.

Note that infinite blaschke products do not extend continuously to the unit circle : they have (essential) singularities where the zeros accumulate.

  • $\begingroup$ thanks ! I suspected there was an easy proof since nobody seemed to bother mentionning it, but I didn't think of it ! $\endgroup$ – Glougloubarbaki Jun 7 '12 at 14:38
  • 1
    $\begingroup$ Nit: zeros can't accumulate at a pole. You get essential singularities. $\endgroup$ – Harald Hanche-Olsen Jun 7 '12 at 14:47
  • $\begingroup$ You're right, I meant singularities. I'll edit the answer. Thank you! $\endgroup$ – Malik Younsi Jun 7 '12 at 18:00

The result can be found in section 6.12 (p. 197) of the encyclopedic An introduction to classical complex analysis, vol 1, by Robert B. Burckel. In case you like to go to original sources, the author cites Fatou, “Sur les fonctions holomorphes et bornées à l'intérieur d'un cercle”, Bull. Soc. Math. France 51 (1923), 191–202.


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