Let $g$ maps from a unit disk $D$ to itself with the modulo of $g(z) \leq 1/2$. Prove that $|g(z)-g(0) \leq |z|$ for all $z \in D$.

Question

I'm still confused with this question. Let $g$ maps from a unit disk $D$ to itself with the modulo of $g(z) \leq 1/2$. Prove that $|g(z)-g(0) \leq |z|$ for all $z \in D$.

What I did is this

As $g(z)$ is analytic in $D$, $g(z)$ is continuous and differentiable in $D$. Then $\left|\frac{g(z) - g(0)}{z-0} \right| \leq \epsilon$. Choose $\epsilon = 1/2$. Then we have

$|g(z) - g(0)| \leq |z-0|\epsilon$

I don't know how to continue with this. I just had a feeling it might be of $\epsilon = 1/2$ but I am confused with the proof now. Can you help me? I would much appreciate that. Thanks.

Answer 1

Let $h(z)=g(z)-g(0)$. Then $|h(z)| \leq 1$ so $h$ maps the unit disk to its closure and vanishes at $0$. By Schwarz Lemma we get $|h(z)| \leq |z|$.