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Let $U\subseteq \mathbb{C}$ be a connected open set, and let $f,g,h:U\to\mathbb{C}$ be holomorphic functions such that $$|f(z)|+|g(z)|+|h(z)|=1$$ for all $z\in U$. How does one prove that $f,g,h$ are constant functions?

Any hints would be appreciated.

Cloudscape
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felipeuni
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    Are there any conditions on $U$? Does it hold for only two functions satisfying an analogous condition? and what about only one function? – MASL Sep 19 '15 at 02:29
  • of course, $U$ should be open (for holomorphicity) and then connected, otherwise this is not true. – Groups Sep 19 '15 at 04:26

1 Answers1

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We recall the following identity for harmonic functions $$\Delta (u^2 + v^2) = 2(|\nabla u|^2 + |\nabla v|^2).$$ Considering a more general problem, suppose $$\sum_{i=1}^n |f_i(z)| =1$$ for all $z\in U$ for some set of holomorphic functions $f_i$. Working locally, we may assume that $f_i= g_i^2$ for some holomorphic functions $g_i=u_i+iv_i$. Then taking the Laplacian of both sides yields $$\sum_{i=1}^n |\nabla u_i|^2 + |\nabla v_i|^2 = 0.$$ Hence the gradients of the real and imaginary parts of $u_i$ and $v_i$ vanish for all $i$, so each $f_i$ must be constant.

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Potato
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  • But $|f_i(z)|$ is not differentiable. – felipeuni Sep 19 '15 at 05:01
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    @felipeuni |g^2|=|g|^2=u^2+v^2, which is. – Potato Sep 19 '15 at 05:01
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    So $g$ is the square root of $f$ but if $f$ is zero anywhere the square root is not differentiable in which case $g$ would not be holomorphic. So your answer assumes that $f,g,h$ are non-zero on $U$. Or am I missing something here? – Rudy the Reindeer Sep 19 '15 at 13:36
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    @RudytheReindeer You only need to show they're zero on some open disk in $U$, and it's not hard to see there's one that will avoid all the zeros of the functions. – Potato Sep 19 '15 at 15:26
  • @Potato Ok, right, of course it's enough to show they are constant on an open subset. I believe you but I don't yet see how to get the open subset. Concretely, if they are not constant each one has some open subset on which they are non zero, say $U_f, U_g$ and $U_h$. How do we guarantee that their intersection is non empty? – Rudy the Reindeer Sep 19 '15 at 23:53
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    @RudytheReindeer For each function, its set of zeros is discrete in $U$. So the union of these sets is too. – Potato Sep 20 '15 at 00:31