$f(z)$ is holomorphic on $|z| < 1$, with $|f(z)| ≤ 3$ and $f(\frac{1}{2}) = 2$. Show that $f(z) \neq 0$ when $|z| < \frac{1}{8}.$

Question

This is an old Schwarz lemma problem from the August 2020 UMD qualifying exam for analysis, which is posted here. The precise wording from the test is:

Suppose $f(z)$ is a holomorphic function on the unit disk with $|f(z)| \le 3$ for all $|z| < 1$, and $f(1/2) = 2$. Show that $f(z)$ has no zeros in the disk $|z| < 1/8$. (Hint: first show $f(0) \neq 0$).

There is another kind of standard problem that assumes $|f(0)| \ge r$ and asks to prove $|f(z)| \ge \frac{r - |z|}{1 - r|z|}$ on the disk $|z| < r$; see problem 5 from chapter IX.1 of Gamelin. This problem instead gives you a value for $f(a)$, $a \neq 0$.

Answer 1

We know that the hyperbolic unit disk metric, $d(z, w) := |\frac{z - w}{1 - \bar{w}z}|$, satisfies $d(g(z), g(w)) \le d(z, w)$ for analytic maps $g : \mathbb{D} \to \mathbb{D}$. Note that $|\frac{f(z)}{3}| \le 1$ when $|z| < 1$, and the maximum principle implies we can assume $|\frac{f(z)}{3}| < 1$, otherwise $|\frac{f(z)}{3}|$ is constantly $1$, contradicting $f(\frac{1}{2}) = 2$. Hence

\begin{align*} \left|\frac{\frac{2}{3} - \frac{f(0)}{3}}{1 - \overline{\frac{f(0)}{3}}\cdot \frac{2}{3}}\right| = d(\frac{2}{3}, \frac{f(0)}{3}) = d(\frac{f(\frac{1}{2})}{3}, \frac{f(0)}{3}) \le d(\frac{1}{2}, 0) = \frac{1}{2}. \end{align*} If we assume that $|\frac{f(0)}{3}| \le \frac{2}{3}$, then the triangle inequality applied twice implies

\begin{align*} \frac{2}{3} - \frac{|f(0)|}{3} \le \frac{1}{2} + \frac{|f(0)|}{9}. \end{align*} Simplifying this implies $|\frac{f(0)}{3}| \ge \frac{1}{8}$. This inequality also holds with the negation of our assumption, and so it holds generally.

Recall that $\phi_a (z) := \frac{z - a}{1 - \overline{a}z}$ is a conformal map $\mathbb{D} \to \mathbb{D}$ which swaps $a$ and $0$. So $\phi_{\frac{f(0)}{3}} \circ (\frac{f}{3})$ maps $0$ to $0$. The Schwarz lemma implies \begin{align*} \frac{|\frac{f(0)}{3}| - |\frac{f(z)}{3}|}{1 + |\frac{f(0)}{3}||\frac{f(z)}{3}|} \le \frac{\left||\frac{f(0)}{3}| - |\frac{f(z)}{3}|\right|}{1 + |\frac{f(0)}{3}||\frac{f(z)}{3}|} \le \left| \frac{\frac{f(z)}{3} - \frac{f(0)}{3}}{1 - \overline{\frac{f(0)}{3}} \cdot \frac{f(z)}{3}} \right|= |\phi_{\frac{f(0)}{3}}(\frac{f(z)}{3})| \le |z|. \end{align*}

Rearranging this inequality implies

\begin{align*} \frac{|\frac{f(0)}{3}| - |z|}{1 + |\frac{f(0)}{3}||z|} \le |\frac{f(z)}{3}|. \end{align*} If $|z| < \frac{1}{8}$, then $|\frac{f(0)}{3}| \ge \frac{1}{8}$ implies the first term in the above inequality is nonzero. And so we finally conclude that $f(z) \neq 0$ if $|z| < \frac{1}{8}$.

Answer 2

Another, perhaps more direct approach, but which also doesn't really use the hint: Let $|a|<\frac{1}{8}$. Consider $g_a = f\circ \phi_a$ where $\phi_a(z)=\frac{z-a}{1-\overline{a}z}$ as in your answer. If $f(-a)=0$, i.e. $g(0)=0$, the Schwarz lemma would imply that $|g(z_a)|=|f(\frac{1}{2})|\leq 3|z_a|$ where $z_a = \phi_a^{-1}(\frac{1}{2})=\frac{1+2a}{2+\overline{a}}$. But we have $f(\frac{1}{2})=2$ and $3|z_a|=3\frac{|1+2a|}{|2+\overline{a}|}<3\cdot\frac{1+\frac{2}{8}}{2-\frac{1}{8}}=2$, which would contradict the above inequality.