# Maximum principle of holomorphic functions

Show: If $f$ is a non-constant holomorphic function on a domain $G$, then $\lvert f\rvert$ has no local maximum on $G$. Hint: Use the mean value property of holomorphic functions!

Suppose that $\lvert f\rvert$ has a local maximum in $z_0\in G$. Then it exists a $r>0$ so that $\lvert f(z)\rvert \leq\lvert f(z_0)\rvert~\forall~z\in B(r,z_0)$. (*)

The mean value property says $$f(z_0)=\frac{1}{2\pi}\int_0^{2\pi}f(z_0+re^{it})\, dt,$$ so it is, because of (*), $$\lvert f(z_0)\rvert\leq\frac{1}{2\pi}\int_0^{2\pi}\lvert f(z_0+re^{it})\rvert\, dt\leq\frac{1}{2\pi}\int_0^{2\pi}\lvert f(z_0)\rvert\, dt=\lvert f(z_0)\rvert$$ and therefore $$\lvert f(z_0)\rvert=\frac{1}{2\pi}\int_0^{2\pi}\lvert f(z_0+re^{it})\rvert\, dt.$$

Can I now use that result to get a contradiction?

The map $t\mapsto |f(z_0)|-|f(z_0+re^{it})|$ is non-negative, continuous, and its integral over $[0,2\pi]$ is $0$, hence for all $t\in [0,2\pi]$, we have that $|f(z_0+re^{it})|=|f(z_0)|$. It's true for $r$ small enough, hence the modulus of $f$ is constant on a ball centered on $z_0$. Using Cauchy-Riemann equations, it can be shown that $f$ is actually constant on the ball, and by connectedness on the domain.
• Why is $\lvert f(z_0+re^{it})\rvert = \lvert f(z_0)\rvert$ for all $t\in [0,2\pi]$? -- Is this a proof using my results or a different proof? – math12 Aug 31 '13 at 11:50
• It's actually a calculus result (for functions with real values): if a continuous non-negative function has a $0$ integral, it is actually the null function. – Davide Giraudo Aug 31 '13 at 12:03
• Oh, sorry. Of course. I had a wrong thought. So if I understood you correct: $\lvert f(z_0)\rvert=\frac{1}{2\pi}\int_0^{2\pi}\lvert f(z_0+re^{it})\rvert\, dt\Leftrightarrow\frac{1}{2\pi}\int_0^{2\pi}(\lvert f(z_0+re^{it})\rvert-\lvert f(z_0)\rvert)\ dt=0\Leftrightarrow\lvert f(z_0+re^{it})\rvert=\lvert f(z_0)\rvert$ Did I understand you right to this point of your answer? And what do you mean with "modulus of $f$" - I dont know that word. – math12 Aug 31 '13 at 12:53