Analytic Functions

Question

Prove or give a counter-example:

If $f_j(j=1,2,...,n)$ is analytic on the domain $D$ such that $\sum_{j=1}^n |f_j(z)|^2$ is constant on $D$. Then each $f_j$ is a constant function.

Inputs:

We know that, if $f$ is analytic on a domain D and if $|f|$ is a constant then $f$ must also be a constant.

I tried to prove the problem, but I am having a real difficult time. Starting with

$\sum_{j=1}^n |f_j(z)|^2=k $, where $k\in\Bbb C$ is a constant

$\sum_{j=1}^n (u_j^2 +v_j^2)=k$, (As $f_j=u_j+iv_j$)

Then I differentiated the above equation partially w.r.t $x$ as well as $y$ and tried a lot of manipulations and used CR equations in order to prove it. But I have had no success.

Now I am wondering whether the above result is at all true? Is there a counter-example for the problem?

Answer 1

Here's another proof of the more general statement mentioned by @MartinR. The domain $D$ is assumed to be connected and the constant equal to $1$. Take a fixed $z_0\in D$ such that $$\sum_{k=1}^n\lvert f_k(z_0) \rvert^2 =1$$ and define the function $f(z)$ by

$$f(z)=\sum_{k=1}^nf_k(z)\overline{f_k(z_0)}.$$

Then

1. $f$ is holomorphic on $D$
2. $f(z_0) = 1$
3. $\lvert f(z) \rvert \leq 1$ by Cauchy-Schwarz.

Then $f(z) = 1$ for all $z\in D$ by the maximum modulus principle. Again by Cauchy-Schwarz this means that $(f_1(z), \ldots, f_n(z))$ is a scalar multiple of $(f_1(z_0), \ldots, f_n(z_0))$ for all $z\in D$. Since $f(z)=1$ it follows that this scalar must be $1$ and therefore $f_k(z)=f_k(z_0)$ for $k\in\{1, \ldots, n\}$.

Alternatively, a bit more verbose (using $f(z)=1$ in the second equality):

$$\begin{eqnarray} 1 \geq \sum_{k=1}^n\lvert f_k(z) \rvert^2 &=& \sum_{k=1}^n\lvert f_k(z) - f_k(z_0) + f_k(z_0) \rvert^2 \\ &=& \sum_{k=1}^n\lvert f_k(z) - f_k(z_0)\rvert^2 + \sum_{k=1}^n \lvert f_k(z_0) \rvert^2\\ &=& 1+\sum_{k=1}^n\lvert f_k(z) - f_k(z_0)\rvert^2 \end{eqnarray}$$ and so $f_k(z) - f_k(z_0)=0$ for $k\in\{1, \ldots, n\}$.

Answer 2

We may suppose wlog that $0\in D$, and $B(0,R)\subset D$. Put $f_j(z)=\sum_{n\geq 0} a_{n,j}z^n$, that is convergent in $B(0,R)$. Let $0<r<R$. We have $$\int_0^{2\pi}|f_j(r\exp(i\theta)|^2d\theta=\int_0^{2\pi}\sum_{n,m}a_{n,j}\overline{a_{m,j}}r^{n+m}\exp(i(n-m)\theta)d\theta=2\pi\sum_{n\geq 0}|a_{n,j}|^2 r^{2n}$$

Now if $\sum_{j=1}^m|f_j(z)|^2=c$ is constant, we get that $$2\pi c=2\pi\sum_{n\geq 0}(\sum_{j=1}^m|a_{n,j}|^2)r^{2n}$$

As this is true for all $r$, $0\leq r<R$, we get that $\sum_{j=1}^m|a_{n,j}|^2=0$, and hence $a_{n,j}=0$, for all $j$ and $n\geq 1$. So each $f_j$ is constant on $B(0,R)$, and it is easy to finish.