# $f:B_{1}(0) \Rightarrow$ holomorphic, $\max_{|z|=r} |f|$ monotonically and continuous

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 $0<r<s$ defining $M( r) := \max_{|z|=r} |f(z)|$ gives :$$|a_{k}|\le \frac{M (r )}{r^{k}},\qquad \forall k\ge -m .$$

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.

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I don't know how do you deduce from $|a_k|\leq r^{-k}M(r)$ the fact that $M$ is strictly increasing. Is the maximum principle already shown in the book? – Davide Giraudo Dec 6 '11 at 18:00
Yes, the maximum principle is shown before this exercise… – VVV Dec 6 '11 at 18:16

Consider restriction of $f\colon B_1(0)\to\mathbb{C}$ on $B_r(0)$ where $r\in(0,1)$. Since $f\in\mathcal{O}(B_1(0))$ then $f\in\mathcal{O}(B_r(0))$ By maximum principle $|f|$ accepts its maximum on $S_r(0)$, i.e. on the boundary of $B_r(0)$. As the consequence for all $\rho\in(0,r)$ we have $$M(\rho)=\max\limits_{|z|=\rho}|f(z)|\leq\max\limits_{z\in B_r(0)}|f(z)|=\max\limits_{|z|=r}|f(z)|=M(r)$$ Since $r\in(0,1)$ and $\rho\in(0,r)$ are arbitrary $M(r)$ is monotone. Assume that for some $0<r_1<r_2<1$ we have $M(r_1)=M(r_2)$, i.e $M$ is not strictly monotone. Denote $m=M(r_1).$Since $M$ is monotone we have $M(\rho)=m$ for all $\rho\in[r_1,r_2]$. Hence there are infinitely many points $z\in B_{r_2}(0)\setminus B_{r_1}(0)$ such that $f(z)=m$. Since $f\in\mathcal{O}(B_1(0))$. By uniqueness principle $f(z)=m$ for all $z\in B_1(0)$.