Holomorphic function on unit disk

Question

Suppose $f$ is a holomorphic function on the open unit disk $\mathbb{D}$ with $f(0)=0$ and $| f(z) + zf^{'}(z)| <1$ for all $z \in \mathbb{D}$. I have to show that $|f(z)| \leq \frac{|z|}{2}$ for all $z\in \mathbb{D}$.

I have tried to apply Schwarz Lemma but failed to obtain the inequality.

Answer 1

Let $\phi(z)=zf(z)$, then $\phi'(z)=f(z)+zf'(z)$.

1. Apply Schwarz lemma to $\phi'$ to get $|\phi'(z)|\le |z|$.
2. Fix $z\in\Bbb D$ and let $\psi(r)=\phi(rz)$, $r\in[0,1]$. Integrate $$ \phi(z)=\psi(1)=\int_0^1\psi'(r)\,dr=\int_0^1\phi'(rz)z\,dr $$ and use the estimate $$ |zf(z)|=|\phi(z)|\le\int_0^1|\phi'(rz)||z|\,dr\le\int_0^1|rz||z|\,dr=\frac{|z|^2}{2}. $$

Answer 2

You can not show that $ |f(z)| \leq \frac{|z|}{2}$ for all $z\in \mathbb{D}$ !

Take $c \in \mathbb D, c \ne 0$ and look at $f(z)=c$.

Then we have $| f(z) + zf^{'}(z)| <1$ for all $z \in \mathbb D$.

But we do not have $ |f(z)| \leq \frac{|z|}{2}$ for all $z\in \mathbb{D}$.