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


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, $$ |f(z)|\le|z^m|\tag{2} $$

  • $\begingroup$ Thanks for the answer @robjohn, but that's part of my dilemma: how can you show that $\,|g(z)|\leq 1\,$ on $\,\partial D\,$? Is it the same trick as with the MMP's proof:$$\left|\frac{f(z)}{z^m}\right|\leq \frac{1}{r^m}\xrightarrow [r\to 1]{} 1$$ on a disc $\,D_r:= z\;\;;\;\;|z|\leq r\,$? Well, I guess this is shorter than what I did. $\endgroup$ – DonAntonio Jun 30 '12 at 3:40
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    $\begingroup$ @DonAntonio: I see your concern. However, we know that $|g(z)|=\left|\dfrac{f(x)}{z^m}\right|\le\dfrac{1}{r^m}$ on $\partial D_r$ for any $r<1$. Therefore, $|f(z)|\le\dfrac{|z^m|}{r^m}$ for any $r<1$. Thus, $|f(z)|\le{z^m}$. I will modify my answer to cover this. $\endgroup$ – robjohn Jun 30 '12 at 5:02

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