Suppose $f$ is analytic in the unit disk $D$ and satisfies $0<|f(z)|<1$. Show that $|f(z)|\leq|f(0)|^{\frac{1-|z|}{1+|z|}}$ for all $z\in D$.
I tried to work with $\log|f|$. It seems that $\log|f|$ is harmonic but I couldn't get estimates.
Suppose $f$ is analytic in the unit disk $D$ and satisfies $0<|f(z)|<1$. Show that $|f(z)|\leq|f(0)|^{\frac{1-|z|}{1+|z|}}$ for all $z\in D$.
I tried to work with $\log|f|$. It seems that $\log|f|$ is harmonic but I couldn't get estimates.
Copyright © 2021 JogjaFile Inc.
Following your idea, let $g=-\log|f|$. Then $g$ is a positive harmonic function, so you can apply Harnack's inequality to $g$ to deduce that $g(z)\ge g(0)\frac{1-|z|}{1+|z|}$.