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
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 $\,|h(z)\leq 1\,$ ?