local constant $\Rightarrow$ global constant

Question

Let $f\colon M\to\mathbb{C}$ be a holomorphic, localy constant function, $M\subseteq\mathbb{C}$ open and connected. Show, that then $f$ is constant on whole $M$.

Isn't this an easy consequence of the identity theorem, i.e:

$f$ is constant localy, i.e. there exist $x_0\in M, r>0, c\in\mathbb{C}$, so that $$ f(x)=c~\forall~x\in B(r,x_0). $$

So $x_0$ is a limit point of $$ \left\{z\in M: f(z)=c\right\} $$ and the identity theorem says that $$ f=c~\forall~z\in M. $$

That's it already?

You don't even need holomorphicity. A locally constant function on a connected set is constant. — Michael Albanese, Aug 31 '13 at 18:19
@MichaelAlbanese: I guess that in OP's words a "locally constant" function is one that is constant in a neighborhood of (at least) a single point. — Giuseppe Negro, Aug 31 '13 at 18:31
Open and connected implies path connectedness. Any two points can be joined by some path $\gamma$. Hence $f \circ \gamma:[0,1] \to \mathbb{C}$ has derivative zero and so is constant. — copper.hat, Aug 31 '13 at 19:16

score 2 · Accepted Answer · answered Aug 31 '13 at 20:18

2

Just so this question has an answer, yes, that's it.

If by locally constant you mean each point has a neighbourhood on which the function is constant, then you don't even need the function to be holomorphic. If, however, you mean there is a point with a neighbourhood on which the function is constant, then you do need holomorphicity, in particular the Identity Theorem.

answered Aug 31 '13 at 20:18

Michael Albanese

99,526

Thank you very much for your answer. – Aug 31 '13 at 20:55

local constant $\Rightarrow$ global constant

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