# Thread: analytic self-map of unit disc

1. ## analytic self-map of unit disc

$f(0,1) \to D(0,1)" alt="f(0,1) \to D(0,1)" /> has a zero of order k at the origin. Show that the function is bounded in magnitude by $\vert z \vert ^k$.

I want to use Schwarz Lemma because the function is analytic on the unit disc, its image is a subset of the unit disc, and f(0)=0. Then I can definitely say that $\vert f(z) \vert \leq \vert z \vert$, but I need more. How might I bring in powers of |z|? Would the other consequence of Schwarz Lemma ( $\vert f'(0) \vert \leq 1$) help here?

2. consider $\frac{f(z)}{z^{k-1}}$?

3. Originally Posted by cribby
$f(0,1) \to D(0,1)" alt="f(0,1) \to D(0,1)" /> has a zero of order k at the origin. Show that the function is bounded in magnitude by $\vert z \vert ^k$.

I want to use Schwarz Lemma because the function is analytic on the unit disc, its image is a subset of the unit disc, and f(0)=0. Then I can definitely say that $\vert f(z) \vert \leq \vert z \vert$, but I need more. How might I bring in powers of |z|? Would the other consequence of Schwarz Lemma ( $\vert f'(0) \vert \leq 1$) help here?
Use induction on $k$ : Schwarz's Lemma is the case $k=1$; now if the statement holds up to $k-1$ and $f$ has a zero of order $k$ at the origin, then by Schwarz's Lemma, $|f(z)|\leq |z|$, and from there $f(z)/z$ is a function $D(0,1)\rightarrow D(0,1)$ having a zero of order $k-1$ at the origin, i.e. $|f(z)/z| \leq |z|^{k-1}$.