I do not understand the proof for the maximum modulus theorem done with the open mapping theorem. Unfortunately my notes are a little bit cryptic.
What I understand: Let $z_0$ be the maximum that is not on the boundary of a holomorphic non-constant function $f$ If we choose an open ball $D_r(z_0)$, then the image $f(D_r(z_0))$ is also an open ball around $z_0$ by the OMT.
Then my notes say $f(D_r(z_0))\subset\overline{D_{|f(z_0)|}(z_0)}$ and that $\overline{D_{|f(z_0)|}(z_0)}$ is not neighbourhood of $f(z_0)$ and therefore $f(z_0)$ must be constant by the OMT, but I struggle heavily to understand this. How do I get a maximum that is bigger than $z_0$? Why do I take a closed ball with radius $|f(z_0)|$? Can someone explain this to me?
Unfortunately most proofs I found are not using the OMT and the identity theorem, but I have to do it with atleast the OMT.
Thank you very much.
Actually it should be $$ \tag{*} f(D_r(z_0))\subset\overline{D_{|f(z_0)|}(0)} $$ which is true because $|f(z)| \le |f(z_0)|$ for all $z \in D_r(z_0)$. Also note that $f(z_0)$ is on the boundary of $\overline{D_{|f(z_0)|}(0)}$ .
The open mapping theorem states that $f(D_r(z_0))$ is an open set, in particular $f(z_0)$ must have a open neighbourhood $U$ such that $f(z_0) \in U \subset f(D_r(z_0))$.
On the other hand, every open neighbourhood $U$ of $f(z_0)$ contains points outside of $\overline{D_{|f(z_0)|}(0)}$. This gives a contradiction to $(*)$.