show that $f(z)\equiv z$.

Question

Let $f:D\to D$ be analytic where $D$ is the open unit disc such that $f(0)=0;f(\frac{1}{2})=\frac{1}{2}$ then show that $f(z)\equiv z$.

By Schwartz Lemma; $|f(z)|\le |z|$ and $|f^{'}(0)|\le 1$

Also since $f(0)=0$ we have $f(z)=z^p g(z) ;g(0)\neq 0$

Also since $f:D\to D$ then $|f(z)|\le 1$

How to prove from here that $f(z)\equiv z$? Please give some hints

Answer 1

The Schwartz Lemma contains this addendum:

If $|f(z)| = |z|$ for some non-zero $z$ or $|f'(0)| = 1$, then $f(z) = az$ for some $a \in \mathbb{C}$ with $|a| = 1$.