Show that holomorphic $f_1, . . . , f_n$ are constant if $\sum_{k=1}^n \left| f_k(z) \right|$ is constant.

While studying for an exam in complex analysis, I came across this problem. Unfortunately I was not able to solve it. Any help would be greatly appreciated.

Let $U ⊂ \mathbb{C}$ be a domain and $f_1, . . . , f_n : U \rightarrow \mathbb{C}$ holomorphic functions, such that $\sum_{k=1}^n \left| f_k(z) \right|$ is constant on $U$. Show that $f_1, . . . , f_n$ are constant.

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Tip: Show it's true for $n=2$ and then use induction. –  Daniel Rust Jan 28 '13 at 17:40
Thanks, that's a great idea! –  user59982 Jan 28 '13 at 17:42
@DanielRust - I'm a little frustrated that this problem has been closed as a duplicate, because the linked questions address something slightly different. The linked questions actually establish a "maximum principle" for $g(z) = \sum_{k=1}^n |f_k(z)|$. That, is the maximum of $g(z)$ occurs on the boundary of $U$. But this not the same as what we want to show here right? –  jtms88 Jun 25 '13 at 14:35
@DanielRust - Do you have any more tips for addressing the specific question here? I have been working on this question for a while. I do know that each individual $f_k$ satisfies a maximum principle--$|f_k(z)|$ having a local max in $U$ implies that $f_k(z)$ is constant. –  jtms88 Jun 25 '13 at 14:39
@DanielRust How do you transform $|f_1|+|f_2|+|f_3|$ into $|g_1|+|g_2|$ with still $g_1,g_2$ analytic? Unless I missed something, I don't think it is so obvious that induction helps. –  1015 Jun 25 '13 at 16:03

For simplicity, let's consider $n=2$. In $U$, we have $|f|+|g|=C$ for some constant. Fix a point $z_0$ in $U$. Then, for appropriately chosen unimodular constants

$$|\alpha f(z_0) +\beta g(z_0)|=|f(z_0)|+|g(z_0)|=C.$$

This means the holomorphic function $\alpha f(z) +\beta g(z)$ attains its maximum in $U$ (since the supremum in $U$ is at most $C$ by the triangle inequality, and it attains $C$), so it is a constant. So for all $z$, we have

$$|\alpha f(z) +\beta g(z)|=|f(z)|+|g(z)|.$$

Equality is only possible in the triangle inequality all the vectors point in the same direction. So $c(z)\alpha f(z) = \beta g(z)$ for some real holomorphic $c(z)$. But since real holomorphic functions are constant, $c$ is constant. Then $\alpha f(z) +\beta g(z)=(1+c)(\alpha f)$, and the latter is a holomorphic function that attains its maximum on $U$, so it is constant. So $f$ is constant, and it follows that $g$ is constant.

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Nicely done, +1. –  1015 Jun 25 '13 at 17:05
@julien Thanks! That means a lot coming from you. –  Potato Jun 25 '13 at 17:08
I suddenly realize I am not sure about the last step. Your $c$ depends on $z$. So you probably need to do like Davide Giraudo, restrict to a ball where $f,g$ don't vanish (at least $f$). Then $c$ will be holomorphic real-valued whence constant. –  1015 Jun 25 '13 at 17:55
@julien Thanks for the correction. –  Potato Jun 25 '13 at 18:10

The result would be easier to show if we would have as assumption that $\sum_{j=1}^n|f_j|^{\color{red}2}$ is constant (then we would take the derivative, and use Cauchy-Riemann equations, or as Julien suggest, the Laplacian).

To deal with this case, consider $z_0\in U$ and $r>0$ such that $B(z_0,r)$ contains none of the zeroes of the $f_k$ (if one is $0$, remove it, and in a ball there are finitely many zeroes of $f_k$). Then on this ball, using a determination of logarithm, we can find for each $k$, $g_k$ such that $f_k=g_k^2$ on $B(z_0,r)$ for some $g\in\mathcal O(B(z_0,r))$.

We get that each $g_k$ is constant on $B(z_0,r)$ (hence so is each $f_k$) and we conclude invoking again principle of isolated zeroes (and connectedness of $U$).

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For people slow like me, the first can be done taking the Laplacian $\Delta =4\frac{\partial^2}{\partial z\partial \overline{z}}$ which yields $0=\sum|f'_j|^2$ with the help of CR. Nice answer, +1. –  1015 Jun 25 '13 at 18:44
@julien Actually, the computations I suggest lead to similar as yours. –  Davide Giraudo Jun 26 '13 at 14:11
I supposed so. I just commented because it took me a while to figure it out... –  1015 Jun 26 '13 at 15:15

Here is a different approach at the solution: Let $c = \sum_{i=1}^n |f_i|$ where c is a constant. For any holomorphic function $f$ we have the mean value theorem as $$f(z_0) = \frac{1}{2\pi} \int_0 ^{2\pi} f(z_0 + re^{i\theta}) d\theta \\ \implies |f(z_0)| \leq \frac{1}{2\pi} \int_0 ^{2\pi} |f(z_0 + re^{i\theta})| d\theta$$ Hence $|f|$ is subharmonic and also $-|f|$ is superharmonic. Now we know that, $|f_1| = c - \sum_{i=2}^n |f_i|$. This means that $|f_1|$ is both subharmonic as well as superharmonic or in other words $|f_1|$ is harmonic.

Thus the problem reduces to showing that if the modulus of a holomorphic function is harmonic then it is a constant. This can be shown in several ways.

1) One way would be to note that $|f|$ would follow the mean value theorem itself. Hence $$|f(z_0)| = \frac{1}{2\pi} \int_0 ^{2\pi} |f(z_0 + re^{i\theta})| d\theta$$ Since this is true for any $r>0$ in $D(z_0,r)$, hence $|f(z_0 + re^{i\theta})| = |f|$ is independent of $r = |z-z_0|$. This signifies that on all rays emanating from $z_0$, $|f|$ would be constant. This would violate the maximum value theorem for $f$ unless it is a constant.

2) A second approach is this: If $f$ is non-constant, then the derivative $f'$ has isolated zeroes. On a small disk away from the zeroes of $f'$, the function $f$ has a holomorphic inverse. The composition of a harmonic function with a holomorphic function is again harmonic, so composing with $f^{-1}$ reduces the problem to showing that the function $|z|$ is not harmonic in a neighborhood of the origin. But this conclusion is evident, since $|z|$ attains a global minimum at the origin.

Now that we have shown $|f_1|$ is a constant, we can keep repeating the process for any $f_i$ from $\tilde{c} = \sum_{i=2}^n |f_i|$ to show that each $f_i$ is a constant.

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To complete @Potato's solution, where he/she did the $n=2$ case, here's how the induction step can be done (I would have added this as a comment, but it's too long to post it as a comment):

Let $n \geq 3$ and $|f_1(z)| + \dots + |f_n(z)|=C$. Then we can choose an arbitrary $z_0 \in U$ and choose $w_1,\dots,w_n$ such that $|f_i(z_0)| = w_i f_i (z_0)$, $|w_i| = 1$.

Then by the same reason as above (in Potato's answer), $g(z) = w_1 f_1(z) + \dots + w_n f_n(z)$ is constant on $U$ (by the maximum principle because $|g(z)|$ attains a maximum at $z_0$.) In fact $g(z)=C$ for all $z \in U$. So $$C = |g(z)| \leq |w_1 f_1(z) + w_2 f_2(z)| + |f_3(z)| + \dots + |f_n(z)| \leq |f_1(z)| + \dots + |f_n(z)| = C$$ So by the induction hypothesis, the $n-1$ functions $(w_1 f_1 + w_2 f_2) , f_3 , \dots , f_n$ are all constant and by considering that $(w_1 f_1(z_0) + w_2 f_2(z_0) ) = |f_1(z_0)| + |f_2(z_0)|$ we can apply @Potato's reasoning again to conclude that $f_1$ and $f_2$ are also constant.

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