analytic self-map of unit disc

**cribby** · June 2nd 2010, 03:21 PM

$f<img src=$ (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?

**xxp9** · June 2nd 2010, 07:00 PM

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

**Bruno J.** · June 2nd 2010, 10:27 PM

Originally Posted by cribby

$f<img src=$ (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. $|f(z)/z| \leq |z|^{k-1}$ .

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