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Let $ \emptyset \neq U \subset \mathbb{C} $ be a bounded open connected set and let $ f_1, \dots, f_n $ be analytic in $ \overline{U} $. Prove that $$ \max_{z \in \overline{U}} \sum_{j=1}^n |f_j(z) | = \max_{z \in \partial U} \sum_{j=1}^n |f_j(z) |. $$ Clearly we have "$\geq$" but I don't know how to reduce to the $ n = 1 $ case to use the usual maximum modulus principle. Help is appreciated.

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I will give a proof for $n=2$ because I have it typed out already. The same argument works for any $n$: given $f$ and $g$ analytic such that the maximum of $|f|+|g|$ on $\overline {U}$ is attained at an interior point $a$ we have $\left\vert f(a)\right\vert +\left\vert g(a)\right\vert \geq \left\vert f(z)\right\vert +\left\vert g(z)\right\vert $ $\forall z\in \Omega .$ Replace $f$ by $e^{is}f$ and $g$ by $e^{it}g$ where $s$ and $t$ are chosen such that $e^{is}f(a)$ and $e^{it}g(a)$ both belong to $[0,\infty ).$ This reduces the proof to the case when $f(a)$ and $g(a)$ both belong to $% [0,\infty ).$ We now have $$f(a)+g(a)\geq \left\vert f(z)\right\vert +\left\vert g(z)\right\vert \geq |f(z)+g(z)|$$ Maximum Modulus principle applied to $f+g$ shows that $f+g$ is a constant. Now $$f(a)+g(a)\geq \left\vert f(z)\right\vert +\left\vert g(z)\right\vert \geq \Re f(z)+\Re g(z)=\Re(f(z)+g(z))$$ $$=\Re (f(a)+g(a))$$ which implies that equality holds throughout. In particular $% \left\vert f(z)\right\vert =\Re(f(z))$ and $\left\vert g(z)\right\vert =\Re(g(z))$ for all z]. Hence $f$ and $g$ are both constants (because their imaginary parts vanish) in which case there is noting to prove.

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