Statement of the question:
Let $D$ be the open disk centered at $0$. Let $\overline{D}$ and $\partial{D}$ denote the closure and boundary of $D$ respectively. Let $f$ and $g$ be functions holomorphic on $D$ and continuous on $\overline{D}$ which satisfy the following:
$$|f(z)|\leq |g(z)| \hspace{1in} (\forall z\in D)$$
$$|f(z)| = |g(z)| \hspace{1in} (\forall z\in \partial{D})$$
$$f(z)\neq 0 \hspace{1in} (\forall z\in \overline{D}\setminus \{0\})$$
Show that there exists $\alpha\in \mathbb{C}$ with $|\alpha|=1$ and $m\in \mathbb{N}\cup \{0\}$ such that $f(z)=\alpha z^m g(z).$
I think the solution is similar to this: Suppose $f$ and $g$ are entire functions, and $|f(z)| \leq |g(z)|$ for all $z \in \mathbb{C}$, Prove that $f(z)=cg(z)$.
However, we are not really dealing with entire functions, so I am not sure how to work with this. I am guessing the solution uses analytic continuations?
Best Answer
Firstly, $f$ and $g$ may have zeros only at $0$, say with orders $r$ and $s$ respectively. Then $r \geq s$ since $|f(z)| \leq |g(z)|$ (use series expansion at $0$ and divide by $z^s$). Consider the function $F(z) = \frac{f(z)}{z^{r - s - 1}g(z)}$ which is holomorphic and $F(z) = 0$. Let us now follow the usual proof of Schwarz lemma (e.g. in https://en.wikipedia.org/wiki/Schwarz_lemma). Let $$ G(z) = \begin{cases} \frac{F(z)}{z}, & z \neq 0 \\ F'(0), & z = 0 \end{cases} $$ which is holomorphic on $D$ and continuous on $\overline{D}$. Recalling that $|f(z)| = |g(z)|$ for all $z \in \partial D$, we have $|G(z)| = 1$. Using both the maximum and minimum modulus principle, we get $G(z) = \alpha$ for some $\alpha \in \mathbb C$ with $|\alpha| = 1$. Thus $f(z) = \alpha z^m g(z)$ where $m = r - s \in \mathbb Z_{\geq 0}$.