0

Let G be a domain, and let $f_1 \ldots f_n$ analytics in G such that $$ h = |f_1|^2 + \ldots + |f_n|^2$$ is constant

prove that every $f_i$ is also constant in G

the question has a hint to calculate the laplacian

I'm not sure how to do it,

I tried to do: $|f_i|^2 = f\bar{f} \to (|f_i|^2)'' = f_i''\bar{f_i} + 2f_i'\bar{f_i}' + f_i\bar{f_i}'' = f_i''\bar{f_i} + |f_i'|^2 + f_i\bar{f_i}''$

I think I should somehow get rid of the $f_i''\bar{f_i} + f_i\bar{f_i}''$ part

and I would get something of the form $$|f_1'|^2 + \ldots + |f_n'|^2 = 0 $$

so I can conclude every $f_i' = 0$ so every $f_i =constant$

but not sure how to do it.. thx

Daniel Katzan
  • 782

2 Answers2

3

You seem to have misunderstood how the Laplacian works in this case. Generally, for functions $g$ defined on open subsets of $\mathbb{C}$, the Laplacian of $g$ is

$$\Delta g = \frac{\partial^2 g}{\partial x^2} + \frac{\partial^2 g}{\partial y^2}.$$

One can also express the Laplacian in terms of the Wirtinger derivatives

$$\frac{\partial}{\partial z} = \frac{1}{2}\biggl(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y}\biggr) \quad \text{and}\quad \frac{\partial}{\partial \overline{z}} = \frac{1}{2}\biggl(\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}\biggr),$$

namely

$$\Delta = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} = 4 \frac{\partial}{\partial z}\frac{\partial}{\partial \overline{z}}.$$

This representation is particularly useful when holomorphic and antiholomorphic functions occur, for we have $\frac{\partial h}{\partial \overline{z}} \equiv 0$ for holomorphic $h$ and $\frac{\partial a}{\partial z} \equiv 0$ for antiholomorphic $a$.

Since the Wirtinger derivatives satisfy the relation

$$\frac{\partial \overline{g}}{\partial \overline{z}} = \overline{\frac{\partial g}{\partial z}}$$

and those that can be derived from it like

$$\frac{\partial g}{\partial \overline{z}} = \overline{\frac{\partial \overline{g}}{\partial z}},$$

the Laplacian of the square of the modulus of a holomorphic function is easily computed:

$$\frac{\partial}{\partial \overline{z}}\lvert f\rvert^2 = \frac{\partial}{\partial\overline{z}}(f\cdot \overline{f}) = \frac{\partial f}{\partial \overline{z}}\cdot \overline{f} + f\cdot \frac{\partial \overline{f}}{\partial \overline{z}} = 0\cdot \overline{f} + f\cdot \overline{\frac{\partial f}{\partial z}} = f\cdot \overline{f'},$$

and then, since $f'$ is again holomorphic, so $\overline{f'}$ antiholomorphic,

$$\frac{\partial}{\partial z}\frac{\partial}{\partial\overline{z}}\lvert f\rvert^2 = \frac{\partial}{\partial z}(f\cdot \overline{f'}) = \frac{\partial f}{\partial z}\cdot \overline{f'} + f\cdot \frac{\partial \overline{f'}}{\partial z} = f'\cdot \overline{f'} + f\cdot 0 = f'\cdot \overline{f'} = \lvert f'\rvert^2.$$

No second derivative actually occurs, since each factor of $\lvert f\rvert^2 = f\cdot \overline{f}$ is holomorphic or antiholomorphic, so one of the two Wirtinger derivatives annihilates that factor.

For

$$h = \sum_{k = 1}^n \lvert f_k\rvert^2,$$

the constantness of $h$ immediately implies $\Delta h \equiv 0$, so

$$0 \equiv \Delta h = \sum_{k = 1}^n \Delta \lvert f_k\rvert^2 = 4\sum_{k = 1}^n \frac{\partial}{\partial z}\frac{\partial}{\partial\overline{z}} f_k\cdot \overline{f_k} = 4\sum_{k = 1}^n f_k'\cdot \overline{f_k'} = 4\sum_{k = 1}^n \lvert f_k'\rvert^2.$$

Since each term in the sum is non-negative, the sum can only be $0$ if each term is $0$, i.e. $f'_k \equiv 0$ for all $k$.

Daniel Fischer
  • 206,697
0

I'm sort of amazed that nobody has used Parseval's identity for this.

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

user387394
  • 594