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Let $f_1,f_2,\ldots,f_n $ be analytic complex functions in domain $D$. and $f = \sum_{k=1}^n|f_k|$ is not constant.

Can I show the maximum of $f$ only appears on boundary of $D\,$?

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Related: – Willie Wong Jan 31 '12 at 10:14
up vote 11 down vote accepted

Suppose by contradiction that the maximum of $f$ is an interior point $z_0$.


$$f_i(z_0)= |f_i(z_0)| \omega_i \,,$$

with $\omega_i$ unit.

Let $g(z):= \sum_i \overline{\omega_i} f_i(z) \,.$

Then, for all $z \in D$ you have

$$| g(z)| = \left| \sum_i \overline{\omega_i} f_i(z) \right| \leq \sum_i \left| \omega_i f_i(z) \right| =f(z) \leq f(z_0)= g(z_0) = |g(z_0)|\,.$$

Now apply the maximum modulus principle to $g(z)$, and use the fact that if $g$ is constant then $|g(z)| \leq f(z) \leq g(z_0)$ implies $f$ is constant.

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$f(z_0)=\sum |f_i(z_0)|$ , $g(z_0)=\sum \omega_i f_i(z_0)$, why they are equal? I think $g(z)= \sum \bar{\omega_i} f_i(z)$ – Leitingok Jan 4 '12 at 5:36
Yup, forgot the conjugates, fixed. thank you. – N. S. Jan 4 '12 at 20:59
I am stupid, I don't know why $g(z)$ is constant.? – Laura May 15 '13 at 9:51
@Rush $g$ is constant, lets say $g(z)=C$ for all $z$... What does that inequality yield? – N. S. May 15 '13 at 13:02
@N.S.thank you. I think $g(z)$ is actually $\sum|f_k|$,it maps $\Bbb{C}$ to $\Bbb{R}$, so I think it isn't a holomorphic function. why we can use maximum modulus principle to it ? – Laura May 15 '13 at 13:59

Yes: $|f_k|$ is a subharmonic function in $\Omega$, a sum of subharmonic functions is subharmonic, and a subharmonic function can't have a local maximum in a connected open set without being constant.

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