# Regarding Schwarz lemma

It is said that the following result is true using Schwarz lemma:

Suppose $g$ is a holomorphic map from the unit disc to the half plane $Re(z)\leq\beta$ such that $g(0)=0$ and $G$ is a conformal mapping from the unit disc onto the same half plane with $G(0)=0$ as well. Then $|g’(0)|\leq|G’(0)|.$

Can anyone tell me how this is true or give a reference for the same?

• Have you forgotten a hypothesis? As written, I don't see any difference between $g$ and $G$. – Alfred Yerger Dec 5 '17 at 2:44
• No. I mean $g$ can be any polynomial with real part say$\leq \alpha$. And $G$ is say $\frac {2\alpha z}{1+z}.$ – user510271 Dec 5 '17 at 3:08
• @AlfredYerger "holomorphic" \ne "conformal". Also "to" \ne "onto". – David C. Ullrich Dec 5 '17 at 3:24

Apply the Schwarz Lemma to $f=G^{-1}\circ g$.