# A generalization of maximum modulus principle

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.

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.