Showing Functions are Constant

Question

Suppose that $f$ and $g$ are holomorphic in $D(0,1)$, and that $|f(z)| + |g(z)| = 3$ for all $z \in D(0,1)$. Prove that both $f$ and $g$ are constant in $D(0,1)$.

Any hints or Theorems that are applicable would help.

Answer 1

Suppose $f$ is not constant. Then $g$ is not constant either, and the Maximum Modulus Theorem implies $0<|f|,|g|<3$. In particular, $g \neq 0$.

Fix $z_0 \in D=D(0,1)$, and choose some $\theta \in \mathbb{R}$ so that $e^{i\theta} g(z_0)$ and $f(z_0)$ are colinear, that is, so that $e^{i\theta} g(z_0)$ is a positive real multiple of $f(z_0)$. Then

$$|e^{i\theta} g(z_0)+f(z_0)| = |e^{i\theta} g(z_0)|+|f(z_0)|=|g(z_0)|+|f(z_0)|=3$$

On the other hand, $|e^{i\theta} g+f| \leq |e^{i\theta} g|+|f|=|g|+|f|=3$. In other words, $z_0$ maximizes the modulus of $e^{i\theta} g+f$. By the Maximum Modulus Theorem, it follows that $e^{i\theta} g + f$ is constant.

We claim that for each $z \in D$, $f(z)$ is a real multiple of $e^{i\theta} g(z)$. Indeed, if it weren't so for some $z \in D$, we'd have

$$|e^{i\theta} g(z)+f(z)| < |e^{i\theta} g(z)|+|f(z)|=|g(z)|+|f(z)|=3$$

which contradicts $e^{i\theta} g+f$ being constant and having modulus $3$. It follows that there is some $c:D\longrightarrow \mathbb{R}$ such that for all $z \in D$

$$f(z)=c(z)\cdot e^{i\theta} g(z)$$

Now, since $g\neq 0$ throughout $D$, we may write

$$c(z)=\frac{f(z)}{e^{i\theta} g(z)},$$

that is, $c$ is holomorphic. Since it takes only real values, $c$ must be constant. Hence, $f=c e^{i\theta} g$ and $|f|=|c||g|$. In particular:

$$3=|f|+|g|=(1+|c|)|g|\Longrightarrow |g|=\frac{3}{1+|c|}$$

so $g$ has constant modulus and is hence constant, a contradiction.