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Let be $f:B(0,1)\to B(0,1)$ holomorphic function such that $$f(0)=f'(0)=\cdots=f^{(n-1)}(0)=0$$ but $f^{(n)}(0)\neq 0.$ Show that $|f(z)|\le |z|^n,$ for every $z\in B(0,1)$ and $|f^{(n)}(0)|\le n!.$

I think that I will use the Schwarz Lemma, cause by the Lemma, I have that $|f(z)|\le |z|$, but I don't know how I can to use the conditions about the derivatives.

Any suggestion? Thanks.

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    $\begingroup$ You don't use the Schwarz lemma, but you mimic its proof. By the assumption, $$g(z) := \frac{f(z)}{z^n}$$ is holomorphic on the open unit disk. Use the maximum modulus principle. $\endgroup$ – Daniel Fischer Jul 6 '15 at 6:56
  • $\begingroup$ Hi @DanielFischer, I don't understand why $g$ is holomorphic in z=0 (z=0 is a polo of order n of $g $). Thanks you for your help. $\endgroup$ – Irddo Jul 6 '15 at 11:06
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    $\begingroup$ Since $0$ is a zero of order $n$ of $f$, it is not a pole of $g$ but a removable singularity (and since $n$ is the precise order of the zero, we have $g(0) \neq 0$, but that is not important). $\endgroup$ – Daniel Fischer Jul 6 '15 at 11:10
  • $\begingroup$ Yes, I get it. So, I need to show there is $z_0\in B (0, 1) $ such that $|g (z)|\le g |(z_0)|$, for every $z\in B (0, 1)$ to conclude that $g $ is constant? Is it? $\endgroup$ – Irddo Jul 6 '15 at 11:44
  • $\begingroup$ But $g$ need not be constant, consider e.g. $f(z) = z^n\cdot e^{z-1}$. You need to show that $\lvert g(z)\rvert \leqslant 1$ for all $z\in B(0,1)$. $\endgroup$ – Daniel Fischer Jul 6 '15 at 11:48
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By the assumption that $f(0) = f'(0) = \dotsc = f^{(n-1)}(0) = 0$, the Taylor series of $f$ centred at $0$ is

$$f(z) = \sum_{k = n}^\infty a_k z^k.$$

Hence the function

$$g(z) = \frac{f(z)}{z^n} = \sum_{k = n}^\infty a_k z^{k-n} = \sum_{k = 0}^\infty a_{n+m}\cdot z^m$$

is holomorphic on $B(0,1)$.

By the maximum modulus principle, for $0 < r < 1$, we have

$$\lvert g(z)\rvert \leqslant \max \{ \lvert g(\zeta) : \lvert \zeta\rvert = r\} = \max \biggl\{ \biggl\lvert \frac{f(\zeta)}{\zeta^n}\biggr\rvert : \lvert \zeta\rvert = r\biggr\} \leqslant \frac{1}{r^n},\tag{1}$$

for $\lvert z\rvert \leqslant r$, since $\lvert f(\zeta)\rvert < 1$. We thus have

$$\lvert g(z)\rvert \leqslant \inf \{ r^{-n} : \lvert z\rvert \leqslant r < 1\} = 1$$

for all $z\in B(0,1)$.

Hence, we have

$$\lvert f(z)\rvert = \lvert z^n\cdot g(z)\rvert = \lvert g(z)\rvert\cdot \lvert z\rvert^n \leqslant \lvert z\rvert^n$$

on $B(0,1)$.

The estimate for $f^{(n)}(0)$ follows similarly from the integral formula:

$$f^{(k)}(z) = \frac{k!}{2\pi i} \int_{\lvert \zeta\rvert = r} \frac{f(\zeta)}{(\zeta-z)^{k+1}}\,d\zeta$$

for $\lvert z\rvert < r < 1$, whence

$$\lvert f^{(k)}(0)\rvert \leqslant \frac{k!}{2\pi} \cdot (2\pi r)\cdot \frac{1}{r^{k+1}} = \frac{k!}{r^k}.$$

Since that inequality holds for all $r < 1$, it follows that

$$\lvert f^{(k)}(0)\rvert \leqslant k!$$

for all $k\in \mathbb{N}$. Note that for this estimate, it was immaterial that $f$ has a zero at $0$, only that $\lvert f\rvert$ is bounded by $1$ is required.

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