I wan't understand the proof of a version of the maximum principle (complex analysis).
The "normal" version is: If $f$ is holomorphic and not constant on a domain $U$, then $|f|$ does not attain a maximum on $U$.
The version I'm considering states: If $f$ is holomorphic on an open set $U$ and $K$ is a compact set of $U$. Then $|f|_K|$ attains it's maximum on the boundary of K.
Given proof: Since $|f|$ is continuous it attains a maximum on the compact set K. Suppose the maximum is in the point $z_0 \in \mathring{K}$ in the interior of $K$. If $C_{z_0}$ is the connected component of $z_0$ in $\mathring{K}$, then by contraposition of the "normal" version it follows that $f$ is constant on $C_{z_0}$. Hence $|f|$ attains it's maximum also on the boundary of $K$.
My Question: Why is the last sentence of the proof true? So why can we follow from the fact that $f$ is constant on $C_{z_0}$ that $f$ attains it's max on the boundary of $K$?
Best Answer
Let $M$ be the maximum of $|f|$ on $K$. As you said, if $|f(z_0)| = M$ for some $z_0 \in \mathring{K}$ then $f$ is constant on the connected component $C_{z_0}$ of $\mathring{K}$ containing $z_0$.
So $|f(z)| = M$ for all $z \in C_{z_0}$, and therefore $|f(z)| = M$ for all $z \in \partial C_{z_0}$. But every boundary point of $C_{z_0}$ is in $K$, but not in $\mathring{K}$, so that $\partial C_{z_0} \subset \partial K$.
Therefore $|f|$ attains the maximum $M$ at all points of the boundary of $ C_{z_0}$, and those points are also on the boundary of $K$.