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.

  • $\begingroup$ 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? $\endgroup$ – math12 Aug 31 '13 at 11:50
  • $\begingroup$ 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. $\endgroup$ – Davide Giraudo Aug 31 '13 at 12:03
  • $\begingroup$ Hm, okay. Thank you for answering. Neverhtless I would like to know if one can continue my proof. $\endgroup$ – math12 Aug 31 '13 at 12:42
  • $\begingroup$ Isn't it the continuation of your proof? $\endgroup$ – Davide Giraudo Aug 31 '13 at 12:44
  • $\begingroup$ 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. $\endgroup$ – math12 Aug 31 '13 at 12:53

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