Holomorphic function on unit disk

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.

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 $$|zf(z)|=|\phi(z)|\le\int_0^1|\phi'(rz)||z|\,dr\le\int_0^1|rz||z|\,dr=\frac{|z|^2}{2}.$$

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 $z\in \mathbb{D}$.

• Is it true if we assume $f$ to be non-constant? Commented Jun 22, 2018 at 9:14
• No. Take $f(z)=\frac{1}{4}(z+1)$.
– Fred
Commented Jun 22, 2018 at 9:32
• Indeed, thanks! Commented Jun 22, 2018 at 9:42
• @AVATAR Note that $|f(z)|\le\frac{|z|}{2}$ implies $f(0)=0$. Any holomorphic $f$ with $f(0)\ne 0$ and bounded $f(z)$, $f'(z)$ in $\Bbb D$ can be scaled to make a counter-example.
– A.Γ.
Commented Jun 22, 2018 at 10:36
• @Fred Sorry. I forgot to put the condition $f(0)=0$. I think the problem is correct now. Commented Jun 22, 2018 at 14:58