TITLE: $f(z)$ 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 [6 upvotes]: 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 $f(a)$, $a \neq 0$.

REPLY [3 votes]: 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.