# Bound for analytic function on a disk given values at 0

I am trying to prove the following problem from An Introduction to Complex Function Theory by Palka.

If $f$ is analytic in the unit disk $D(0,1)$, and if $f(0)=f'(0)=\ldots= f^{(k)}(0)=0$, and $|f^{(k)}(z)|\leq 1$ for every $z\in D(0,1)$, then show that $|f(z)|\leq |z|^{k+1}/(k+1)!$ in the disk.

I tried to use Schwarz lemma or maximum modulus principle, but what I get is something like $|f(z)|\leq |z|^{k+1}$, and I can't seem to use all the information given.

## 1 Answer

Hint:

1. By Schwarz Lemma, $|f^{(k)}(z)|\le|z|$.
2. Using $g(z)-g(0)= \int_0^zg'(w)dw$ for any holomorphic $g$ defined on $D(0,1)$ or not, prove that $|f^{(k-j)}(z)|\le \frac{|z|^{j+1}}{(j+1)!}$ for $j\le k$ by induction on $j$.