Inequality derived Schwarz's Lemma Let $f$ be a holomorphic function on $U(0,R)$ with $0< R.$ Assume there exists an $M > 0$ such that $|f(z)| \leq M$ for $z \in U(0,R)$ and an $n \in \mathbb{Z}_{\geq 0}$ satisfying $$0 = f(0) = f'(0) = ... = f^{(n)}(0).$$
Show that $$|f(z)| \leq M(\frac{|z|}{R})^{n+1}$$ for all $z \in U(0,R)$, and $$\frac{|f^{(n+1)}(0)|}{(n+1)!} \leq \frac{M}{R^{n+1}}.$$
I think I should use Schwarz's Lemma to show it. Since $f$ is holomorphic on $U(0,R)$, then $$f(z) = \sum_{k=0}^{\infty} a_kz^k = \sum_{k = n+1}^\infty a_kz^k$$ for $z \in U(0,R)$ (because $f^{(j)}(0) = 0$ for $ 0 \leq j \leq n)$. 
Set $g : U(0,1) \rightarrow \mathbb{C}$, $$g(\frac{z}{R}) = \frac{R^nf(z)}{z^nM} =\frac{R^n}{M} \sum_{k= n+1}^\infty a_kz^{n-k}.$$ Notice that $g$ is holomorphic on $U(0,1)$ and $g(0) = 0.$ If I can show that $|g(\frac{z}{R})| < 1 $ for $z \in U(0,R)$, I can apply Schwarz 's Lemma and the result will be immediate. But I cannot show that $|g(\frac{z}{R})| < 1$ for $z \in U(0,R)$. Can anyone give a hint ? Is my approach suitable ?
 A: $f$ has a zero of multiplicity at least $n+1$ in $0$. Thus, $g$ holomorphic with with $f(z) = z^{n+1}g(z)$ exists. By maximum principle we have $|g(z)| = \left|\frac{f(z)}{z^{n+1}}\right| \leq \frac{M}{R^{n+1}}$ for all $z\in U(0,R)$.
Hence $|f(z)| \leq M\frac{|z^{n+1}|}{R^{n+1}}$ for all $z\in U(0,R)$.
A: Note that $f(Rz)/M$ maps the unit disc to itself, since $f(Rz) \leqslant M$ for $0<\lvert z \rvert < 1$.
Define
$$ g(z) = \begin{cases} 
\frac{1}{M} \frac{f(Rz)}{z^{n+1}} & z \neq 0 \\
\frac{R^{n+1}}{M}\frac{f^{(n+1)}(0)}{(n+1)!} & z=0
\end{cases},  $$
and now the proof goes in the same way as Schwarz's lemma: By the conditions in the question, $g$ is analytic on $U(0,1)$, (obviously you need to check this at $0$: use the Taylor series) so the maximum principle applies. On the boundary, $g(z) \leqslant 1$, so we find that for $0 < \lvert z \rvert \leqslant 1$,
$$ \frac{1}{M} \frac{f(Rz)}{z^{n+1}} \leqslant 1, $$
which gives the right answer after untangling the substitution. Further, what does this give at $z=0$? Why,
$$ \frac{R^{n+1}}{M}\frac{f^{(n+1)}(0)}{(n+1)!} \leqslant 1, $$
which is the other thing we wanted.

The second part can also be done using the Cauchy Integral formula:
$$ f^{(n+1)}(0) = \frac{(n+1)!}{2\pi i} \int_{|z|=r} \frac{f(z)}{z^{n+2}} \, dz, $$
by using the trivial inequality $\int_{\gamma} h(z) \, dz < l(\gamma) \max_{\gamma} \lvert h(z) \rvert $, where $l(\gamma)$ is the length of $\gamma$, and then taking $r \to R$ (note you have to do this because the function is not known to be holomorphic on the boundary, so you can't just integrate round that directly). However, given the form of the question, the first method is probably the one required.
