This is an exercise from a book called "theory of complex functions" I want to solve:
$f\colon B_1(0) \Rightarrow$ holomorphic, then $M\colon[0,1)\rightarrow \mathbb{R}$ given by $M(r ) := \max_{|z|=r} |f|$ is monotonically increasing and continuous. $M (r)$ strictly monotonically increasing $\Leftrightarrow f$ is not constant.
For $m\in \mathbb{N}$ let $f(z) = \sum_{-m}^{\infty} a_{n}z^{n}$be holomorphic in $B_{1}(0)$. Then for every $r$ with $0
Proof: If $\epsilon > 0$ with $k=0$, then one can choose $n\in \mathbb{N}$ so large that the remaining series $g(z) := \sum_{n+1}^{\infty} a_{n}z^{n}$ satisfies $\max_{|z|=r}|g(z)| \le \epsilon$. Then also: $q(z) := f(z) - g(z) = \sum_{-m}^{n} a_{n}z^{n}$ satisfies$\max_{|z|=r}|q(z)| \le M( r)+ \epsilon.$
Now a lemma has been shown before that says that $|a_{0}| \le M( r) + \epsilon ; \epsilon >0$, thus $|a_{0}| \le M( r)$. Now if $\displaystyle k\ge -m$, then $z^{-k}f(z)= \sum_{-(m+k)}^{\infty} a_{k+n}z^{n}$ is holomorphic in $B_{1}(0)$. Its constant term is $a_{n}$ and $\max_{|z|=r} |z^{-k}f(z)| = r^{-k}M( r)$. So$|a_{k}| \le r^{-k} M( r),$which means that $M( r)$ is monotonically increasing and continuous.
If $f$ were constant, then $a_{k}$ is $0$, and one gets $0 \le M( r)/r^{k}$, so this series is monotonically decreasing. So $f$ can't be constant if $M(r)$ is strictly monotonically increasing.
The book doesn't provide any solutions I can compare to. Does anybody see if my arguing is correct? Please, do tell. Merci.