Analytic functions on the unit ball such that $|f(z)| \leq |z+3/2|$

Question

$f \colon B_1(0) \to \mathbb{C}$ is analytic, $f(0)=0$ and $|f(z)| \leq |z+3/2|$ for all $z \in B_1(0)$. Prove that $|f(1/2)| \leq 1$. I tried to apply Schwarz's lemma to the function $\frac{f(z)}{z+3/2}$ but that gives a bound which is too big.

Answer 1

You are on the right track. As you proposed, consider $$ g(z)=\frac{f(z)}{z+3/2}, $$ which yields $g:\mathbb{D}\to\mathbb{D}$ with $g(0)=0$. Thus Schwarz's lemma applies, which gives $$ \left|\frac{f(z)}{z+3/2}\right|=\left|g(z)\right|\le\left|z\right| $$ holds for all $z\in\mathbb{D}$.

Note that the above inequality implies $$ \left|f(z)\right|\le\left|z\right|\left|z+\frac{3}{2}\right|. $$ Take $z=1/2\in\mathbb{D}$, and the last inequality leads to $$ \left|f\biggl(\frac{1}{2}\biggr)\right|\le\left|\frac{1}{2}\right|\left|\frac{1}{2}+\frac{3}{2}\right|=\left|\frac{1}{2}\right|\left|2\right|=1. $$