# Bounded holomorphic functions on a Riemann surface separating points

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.

• What do you mean by "separated" ? $f(z_1)=1$ and $f(z_2)=-1$? – Loïc Teyssier Dec 18 '18 at 15:26
• @LoïcTeyssier We say $f$ separates $a$ and $b$ if $f(a) \neq f(b)$. – Jaikrishnan Dec 18 '18 at 16:53
• 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? – Autumn Kent Dec 18 '18 at 19:22
• @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. – Jaikrishnan Dec 18 '18 at 19:36
• If it has finitely generated fundamental group. – Autumn Kent Dec 18 '18 at 21:54

## 1 Answer

The answer is no.

See

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 $$0 tending to $$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 $$H^\infty(R) \cong H^\infty(\Delta)$$.)