1
$\begingroup$

Prove if $f : D(0,1) \rightarrow \overline{D}(0,1)$ is an holomorphic function in $D(0,1)$ and $f$ has in $z=0$ a zero of order $n$. Then, $$|f(z)| \leq |z|^n$$ for all $z\in D(0,1)$.

$\textbf{My attempt}$. If $f$ has a zero of orden $n$, then

$$f(0)=f'(0)=\ldots = f^{n-1)}(0)=0 \text{ and } f^{n)}(0)\neq 0$$.

Then, applying the Schwarz's Lemma I get that

$$|f(z)|\leq |z| \; , \; |f'(z)|\leq |z| \text{ and } |f^{n-1)}(z)|\leq |z| $$.

I don't know what to do now to get $|f(z)| \leq |z|^n$.

  • 0
    Consider the function $g(z) = \frac{f(z)}{z^n}$. You want to show that $\lvert g(z)\rvert \leqslant 1$ for $\lvert z\rvert < 1$. – 2017-01-10

3 Answers 3

1

Define a function $g$ on $D(0,1)$ by $g(0)=0$ and $$ g(z) = \frac{f(z)}{z^{n-1}}, z \in D(0,1) \setminus \{0\}.$$

Then $g$ is holomorphic in $D(0,1)$ and $g(0)=0$. Thus, by schwarz lemma, $|g(z)| \leq |z|$ for all $z \in D(0,1)$, that is, $|f(z)| \leq |z|^n$ for all $z \in D(0,1) \setminus \{0\}.$

Moreover, since $f(0)=0$, then $|f(z)| \leq |z|^n$ for all $z \in D(0,1).$

0

Let $\displaystyle f(z)=g(z) z^n$ and $\displaystyle g(z)=\frac{f(z)}{z^n}$ so $g(z):D\to D$ is holomorphic then apply schewarz lemma.

0

First of all, maximal modulus principle dictates that $f:D\to D$.

Notice that $f(z)=0\Rightarrow |f(z)|\leqslant|z|\Rightarrow \bigg|\dfrac{f(z)}{z}\bigg|\leqslant 1\Rightarrow \bigg|\dfrac{f(z)}{z}\bigg|\leqslant|z|$

This can be continue as long as $\dfrac{f(z)}{z^k}\bigg|_{z=0}$ is well-defined.