We have discussed the maximum value principle:
Let $\Omega \subset \Bbb{C}$ be an open connected subset and $f:\Omega \rightarrow \Bbb{C}$ be an analytic function. Assume that there exists $z_0\in \Omega$ such that $$|f(z_0)|=\sup_{z\in \Omega} |f(z)|$$ then $f$ is constant.
Now I asked myself what fails if $\Omega$ is not connected. Is there a nice illustrative counterexample which shows this?
Thanks for your help.
Best Answer
The simplest counter example would be a function with values that are two different constants on two disconnected domains.
If you're thinking things can be saved by weakening the conclusion to "$f$ is constant on each connected component of its domain", @MartinR 's counterexample ($f(z) = z$ on $\{z \mid |z| \lt 1\}$, $f(z) = 2$ on $\{z \mid |z-3| \lt 1\}$) shows that it is not that simple.
If you would like a general lesson to take away, it might be that many of our theorems show how the value(s) a function takes in one place determine the value(s) it takes in another place, but if the two places aren't connected to each other, the connection is broken, and you can't say so much.