AMM 12481 (Neat Generalization of Maximum Modulus Principle)

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AMM 12481 (Neat Generalization of Maximum Modulus Principle)

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complex analysis

maximum modulus principle

Maximum modulus theorem

Open mapping theorem

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Source: American Mathematical Monthly Volume 131 (2024), Issue 7: https://doi.org/10.1080/00029890.2024.2351727

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#1 h Mar 23, 2025, 3:49 AM

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12481. Proposed by Bernhard Elsner, Université de Versailles Saint-Quentin-en-Yvelines, Versailles, France, and Eric Müller, Villingen-Schwenningen, Germany. Let $f_1, \ldots, f_n$ be holomorphic functions on $U$ , where $U$ is an open, connected subset of $\mathbb{C}$ . Suppose that the function $g : U \rightarrow \mathbb{R}$ given by $g(z) = |f_1(z)| + \cdots + |f_n(z)|$ takes a maximum value in $U$ . Must each function $f_k$ be constant on $U$ ?

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#2 h Sunday at 12:35 PM • 1 Y

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Let's assume $n>1$ and none of the functions the 0-function, otherwise we use induction.
Suppose $g$ is maximal in $a\in U$ and choose numbers $k_i\in C(0,1)$ such that $k_i f_i(a)\ge 0$ for $i=1\cdots n$ .
Define $h(z)=\sum k_i f_i(z)$ then $h$ is analytic and $|h(z)|\le \sum |f_i(z)|=g(z)\le g(a)= \sum |f_i(a)|=\sum |k_if_i(a)|=|\sum k_if_i(a)|=|h(a)|$ .
Hence $h$ is constant and for all $z\in U$ the inequalities above are in fact equalities.
Hence for $f_1(z)\not=0$ the values $k_1f(z),\cdots, k_sf(z)$ are positively dependent and $\alpha_s(z)={{k_sf_s(z)}\over {k_1f_1(z)}}\ge 0$ when $s=1\cdots n$ . Since the roots of $f_1$ lie isolated, it follows that each $\alpha_i$ is a constant function. Hence $f_i/f_1$ is constant for each $i$ and now we can apply the MMP for one function.

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