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Math Help - analytic self-map of unit disc

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    analytic self-map of unit disc

    (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?
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    consider \frac{f(z)}{z^{k-1}}?
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by cribby View Post
    (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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