# If $f,g$ are analytic in the unit disk, and $|f|^2+|g|^2=1$, then $f,g$ constant.

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.

• See this question. (Let $f_1=f^2$ and $f_2=g^2$.) – Potato Jul 21 '13 at 2:24
• Seems like this is just an application of Liouville's Theorem – Wintermute Jul 21 '13 at 2:33
• @mtiano I don't see the connection with Liouville's Theorem. – 40 votes Jul 21 '13 at 2:41

## 2 Answers

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.

• $\frac{\partial}{\partial z}(f\bar{f}')=f'\bar{f}'+f \frac{\partial ^2\bar{f}}{\partial z\partial \bar{z}}=0$ Why does $f \frac{\partial ^2\bar{f}}{\partial z\partial \bar{z}}=0$? – Sachchidanand Prasad Sep 21 '17 at 11:22

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 $$0, and similarly for $$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 $$n \geq 1$$.