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I need to prove that if $f,g$ are analytic in the unit disk, and $|f|^2+|g|^2=1$ for all $z$ in the unit disk, then $f,g$ are constant.

This is an exercise question so it should not be very hard, but I don't know where to start. Any hint is appreciated.

NECing
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2 Answers2

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While the comment by Potato points a way to an answer, this problem is easier (which the answer by Davide Giraudo in the other thread indicates). Namely, for every holomorphic function $f$ we have $$ \frac{\partial}{\partial z}\frac{\partial}{\partial \bar z}(f\bar f) = \frac{\partial}{\partial z}(f\bar f') = f'\bar f' = |f'|^2 \tag1 $$ Apply (1) to $g$ as well, and add the results.

Incidentally, $\frac{\partial}{\partial z}\frac{\partial}{\partial \bar z}$ is $\frac14$ of the Laplacian.

40 votes
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  • $\frac{\partial}{\partial z}(f\bar{f}')=f'\bar{f}'+f \frac{\partial ^2\bar{f}}{\partial z\partial \bar{z}}=0$ Why does $f \frac{\partial ^2\bar{f}}{\partial z\partial \bar{z}}=0$? – Sachchidanand Prasad Sep 21 '17 at 11:22
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Here's different method. If $f(z) = \sum_{n=0}^\infty a_n z^n$ and $g(z) = \sum_{n=0}^\infty b_n z^n$, then Parseval's Identity says

$$ \frac{1}{2\pi} \int_0^{2\pi} |f(re^{it})|^2 dt = \sum_{n=0}^\infty |a_n|^2r^{2n} $$

for $0<r<1$, and similarly for $g$. But then

$$ 1 = \frac{1}{2\pi} \int_0^{2\pi} (|f(re^{it})|^2+|g(re^{it})|^2) dt = \sum_{n=0}^\infty (|a_n|^2+|b_n|^2) r^{2n},$$

and the only way this can happen is if $a_n=b_n= 0$ for $n \geq 1$.

user387394
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