# schwarz lemma, a simple example

Let $f$ be a holomorphic function on $B(0,R)$ , R>0. Assume that there exist an $M>0$ such that $| f(z) | \le M$ $\forall z\in B(0,R)$. , and a natural number, such that : $$0 = f(0)=f'(0) = ... = f^{(n)}(0)$$

$1)$ Prove that $|f(z)| \le M (\frac{|z|}{R})^{n+1}$, $\forall z\in B(0,R)$, with equality iff there exist an $\alpha \in \mathbb{C}$ such that $|\alpha|=1$ , such that $f(z)=\alpha M (\frac{|z|}{R})^{n+1}$.

$2)$ Assume that either $f(z_0)=M (\frac{|z_0|}{R})^{n+1}$ for some $0 < |z_0| < R$ or $|f^{n+1}(0)|=(n+1)!M$ , then $f(z)=\alpha M (\frac{|z|}{R})^{n+1}$.

Sorry for asking this problem, but I want to see some examples , of Schwarz lemma. Thanks!

-
Oh I did this problem :D! –  Daniel Oct 8 '12 at 15:30
if you solved it, you can post an answer. –  Davide Giraudo Oct 10 '12 at 14:50