0

Let $f:D(0,1)\to D(0,1)$ holomorphic with $D(0,1)=\{ z\in\mathbb{C}: |z|<1\}$. In addition $f(0)=0$ and the root $0$ has multiplicity $N$. Show that $|f(z)|\leq |z|^N$ for all $z\in D(0,1)$. My idea: The definition of $f$ being of multiplicity $N$ yields $f(z)= z^N g(z)$ for $g(0)\neq 0$. It is obvious that ${}{}{}{}$

$$g(z)= \begin{cases} \frac{f(z)}{z^N} & z\neq 0 \\ \frac{f^{(N)}(z)}{N!} &z=0\\ \end{cases}$$

where $f^{(N)}(z)$ denotes the N-th derivative of $f$. We get $$|f(z)|=|z|^N |g(z)|$$ how do I show that $|g(z)|\leq 1$ or is this even true? Is there another way to show this inequality?

Arbuja
  • 1
shekh
  • 63
  • 1
    https://math.stackexchange.com/q/2092501/42969 – Martin R Jun 05 '23 at 18:26
  • it's indeed a duplicate, though the accepted solution in the linked question doesn't look satisfactory – user8268 Jun 05 '23 at 18:35
  • i agree with you – shekh Jun 05 '23 at 18:39
  • @shekh How is it not satisfactory? – donaastor Jun 05 '23 at 18:41
  • 1
    @donaastor The largest issue is that Schwarz doesn't say "$g(0)=0$ and $g$ holomorphic on $D(0,1)$ implies $|g(z)|\leq |z|$" (counterexamples are easy: $g(x)=2x$). The missing step is exactly what OP here is asking about, and only the (-1)-downvoted answer in the linked dupe actually fills in the missing assumption by using an inductive argument. – Brian Moehring Jun 05 '23 at 18:46
  • @BrianMoehring I don't see where the accepted solution says that Schwartz says that. EDIT: I see now. – donaastor Jun 05 '23 at 18:48
  • @BrianMoehring code I mean the link to schwarz lemma gave me the right idea to solve this slightly modified version. The Maximum modulus principle yields $|g(z)|\leq \max_{\zeta\in\partial D(0,1)}\frac{|f(\zeta)|}{|\zeta |^N}\leq \max_{\zeta\in\partial D(0,1)} |f(\zeta)| \leq 1$ – shekh Jun 05 '23 at 18:55
  • 1
    I suggest truncating and applying max modulus theorem on the closed ball $\overline{ B(0,1-\delta)}$ where $\delta \gt 0$ may be chosen to be arbitrarily small, that way you do not need to contemplate behavior on $S^1$. – user8675309 Jun 05 '23 at 18:58
  • 1
    This is also a special case of https://math.stackexchange.com/q/94122/42969 (which has a satisfactory answer IMO). – Martin R Jun 05 '23 at 20:24

0 Answers0