# If $f$ is holomorphic in a compact Riemann surface, it is constant. Why doesn't this work for compact subsets of $\mathbb{C}$?

Let $$X$$ be a compact Riemann surface. Suppose that $$f$$ is holomorphic over all of $$X$$. Then $$f$$ is constant

I proved this in the following way:

The function $$f$$ is continuous and hence $$|f|$$ attains a maximum value $$M$$. Let $$p$$ be a point of $$X$$ such that $$|f(p)|=M$$. By the Maximum Modulus Principle, $$|f|$$ is constant (equal to $$M$$) in a neighborhood of $$p$$. That is, the set of all $$x$$ that $$|f(x)|=M$$ is open. However such set is also closed since $$f$$ is continuous. Connectedness implies that $$|f|$$ is constant. Then $$f(X)$$ is contained in a circle of radius $$M$$. This contradicts the open mapping theorem.

I'm fairly confident that this proof is correct. What I don't understand is why the same proof does not apply to proving that a holomorphic function defined on a compact subset of $$\mathbb{C}$$ is constant, which is not true.

• How do you define $f$ being holomorphic at a point $p$ if $p$ is a boundary point of $M$? – Wojowu Oct 11 '18 at 19:42
• @Wojowu What is $M$? A compact subset of $\mathbb{C}$ or a Riemann surface? – Gabriel Oct 11 '18 at 19:44

I would point out two reasons for why the proof does not work for compact subsets of $$\mathbb{C}$$.
1) In general a compact subspace $$K \subset \mathbb{C}$$ is not connected. Of course this is not the real problem, because there are non-constant functions on connected and compact subspaces, so you can ask the question "why doesn't the proof work for compact and connected subsets of $$\mathbb{C}$$?"
2) The real problem I think is that for a compact subspace $$K \subset \mathbb{C}$$, you cannot guarantee that the set of points where the function attains its maximum modulus is open in $$K$$, think about the identity function restricted to the unit disc, for example. In the case of a Riemann surface, the existence of charts around any point and the usual Maximum Modulus Theorem for holomorphic functions on $$\mathbb{C}$$ guarantee that this set is open.