Let $R$ be a Riemann surface that admits a non-constant bounded holomorphic function. Then is it true that any two points of $R$ can be separated by a bounded holomorphic function? This is easy to see when $R$ is a planar domain.

  • $\begingroup$ What do you mean by "separated" ? $f(z_1)=1$ and $f(z_2)=-1$? $\endgroup$ Dec 18, 2018 at 15:26
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    $\begingroup$ @LoïcTeyssier We say $f$ separates $a$ and $b$ if $f(a) \neq f(b)$. $\endgroup$ Dec 18, 2018 at 16:53
  • $\begingroup$ If R has finite type, can't you embed R into a ball in C^n and then project to a line through the two points? $\endgroup$ Dec 18, 2018 at 19:22
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    $\begingroup$ @AutumnKent What is your definition of finite type? If $R$ can be embedded into any bounded domain in $\mathbb{C}^n$ then the claim is certainly true. $\endgroup$ Dec 18, 2018 at 19:36
  • $\begingroup$ If it has finitely generated fundamental group. $\endgroup$ Dec 18, 2018 at 21:54

1 Answer 1


The answer is no.


Stanton, Charles M., Bounded analytic functions on a class of open Riemann surfaces. Pacific J. Math. 59 (1975), no. 2, 557–565.

Stanton shows that a branched cover $R$ of the disk is separated by $H^\infty(R)$ if and only if the branch points lie over a set that is the zero set of a Blaschke product. So you can build an example like this:

Take a $2$-fold branched cover $\pi:\Delta_2 \to \Delta$ of the unit disk $\Delta$ with branched points above a sequence $x_1, x_2, \ldots$ with $0 <x_1 < x_2 < \cdots < 1$ tending to $1$ and with $\sum (1-x_n) = \infty$.

If $f$ is a bounded analytic function on $\Delta_2$ define $h(z) = (f(z_1) - f(z_2))^2$, where $\pi^{-1}(z) = \{z_1,z_2\}$. The function $h$ is an analytic function on the disk, and the Blaschke theorem and the choice of $x_n$ forces $h$ to be identically zero.

It seems like the characterization of Riemann surfaces $R$ separated by $H^\infty(R)$ is still an open problem.

(My first answer cited example 2 on page 241 of

Nakai, Mitsuru, Valuations on meromorphic functions of bounded type. Trans. Amer. Math. Soc. 309 (1988), no. 1, 231–252.

This is a variation of the above that is not a branched disk and yet $H^\infty(R) \cong H^\infty(\Delta)$.)

  • $\begingroup$ I know this is an old answer, but I am struggling a little bit with understanding it. Surely a branched cover of the disc with infinitely many critical values has infinite degree, so how could you choose a degree 2 cover branched over the points $x_j$? $\endgroup$ Sep 15, 2022 at 22:05
  • $\begingroup$ @LasseRempe For example, consider an infinite genus surface with one end, such as the picture of the standard genus g surface but that continues forever to the right. There is an involution of this surface that rotates it by 180 degrees. The fixed point set of this is an infinite discrete set, and the quotient is homeomorphic to the disk. $\endgroup$ Sep 30, 2022 at 16:29

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