1

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?

Kurome
  • 1,121

1 Answers1

1

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 $m = r - s \in \mathbb Z_{\geq 0}$.

Pratyush Sarkar
  • 3,734
  • I was not aware of Schwarz lemma. However it's proof is kind of the direct solution to this problem. – Kurome Dec 06 '19 at 07:09
  • Question: How do I show that $F(z)$ satisfies $|F(z)|=|z|$ for some $z$ in $D$? Or that $F'(0)=1$? – Kurome Dec 06 '19 at 10:21
  • @Kurome Yes, it essentially boils down to maximum/minimum modulus principle. It might be better to show that directly instead because the usual formulation of Schwarz lemma does miss some things that you need. Let me modify the answer accordingly. – Pratyush Sarkar Dec 11 '19 at 21:52