TITLE: Schwarz Lemma - like exercise
QUESTION [5 upvotes]: There's this exercise: let $\,f\,$ be analytic on $$D:=\{z\;\;;\;\;|z|<1\}\,\,,\,|f(z)|\leq 1\,\,,\,\,\forall\,z\in D$$  and $\,z=0\,$  a zero of order $\,m\,$ of $\,f\,$.
Prove that $$\forall z\in D\,\,,\,\,|f(z)|\leq |z|^m$$
My solution: Induction on $\,m\,$: for $\,m=1\,$ this is exactly the lemma of Schwarz, thus we can assume truth for $\,k<m\,$ and prove for $\,k=m>1\,$ . Since $\,f(z)=z^mh(z)\,\,,\,h(0)\neq 0\,$ analytic in $\,D\,$ ,  put 
$$g(z):=\frac{f(z)}{z}=z^{m-1}h(z)$$ 
Applying the inductive hypothesis and using Schwarz lemma $\,\,(***)\,\,$ we get that 
$$|g(z)|=\left|\frac{f(z)}{z}\right|=|z|^{m-1}|h(z)|\stackrel{ind. hyp.}\leq |z|^{m-1}\Longrightarrow |f(z)|\leq |z^m|$$
and we're done...almost: we still have to prove $\,|g(z)|\leq 1\,$ for all $\,z\in D$ in order to be able to use the inductive hypothesis and this is precisely the part where I have some doubts: this can be proved as follows (all the time we work with $\,z\in D\,$):
$(1)\,\,$ For $\,f(z)=z^mh(z)\,$ we apply directly Schwarz lemma and get
$$|f(z)|=|z|^m|h(z)|\leq |z|\Longrightarrow |z|^{m-1}h(z)|\leq 1$$
And since now the function $\,f_1(z)=z^{m-1}h(z)\,$ fulfills the conditions of S.L. we get
$(2)\,\,$ Applying again the lemma, 
$$|f_1(z)|=|z|^{m-1}|h(z)|\leq |z|\Longrightarrow |z^{m-2}h(z)|\leq 1$$and now the function $\,f_2(z):=z^{m-2}h(z)\,$ fulfills the conditions of them  lemma so...etc.
In the step$\,m-1\,$ we get 
$$|z||h(z)|\leq |z|\Longrightarrow {\color{red}{\mathbf{|h(z)|\leq 1}}}\,$$ 
and this is what allows us to use the inductive hypothesis in $\,\,(***)\,\,$ above.
My question: Is there any way I can't see right now to deduce directly, or in a shorter way, that $\,|h(z)\leq 1\,$ ?

REPLY [7 votes]: For $0< r<1$, let $D_r=\{z\in\mathbb{C}:|z|\le r\}$.
The function $g(z)=\dfrac{f(z)}{z^m}$ is analytic on $D$ (see Removable Singularity) and $|g(z)|\le\frac{1}{r^m}$ on $\partial D_r$. The maximum modulus principle says that
$$
|g(z)|\le\frac{1}{r^m}\text{ for }z\in D_r\tag{1}
$$
Since $(1)$ holds for all $r<1$, we have that $|g(z)|\le1$ for $z\in D$, and therefore,
$$
|f(z)|\le|z^m|\tag{2}
$$