I dont understand the following proof of Schwarz lemma: Schwarz Lemma In the last section it says, "Moreover, suppose that $|f(z)| = |z|$ for some non-zero $z$ in $D$, or $|f'(0)| = 1$. Then, $|g(z)| = 1$ at some point of $D$". Why does $g(z)$ have a local maximum then?
Asked
Active
Viewed 162 times
0
-
There is a proof following it, what is the question? – Lemon Dec 12 '18 at 11:52
2 Answers
1
Why does $g$ have a local maximum? We cannot even ask that question because $g$ is complex valued. Maximum Modulus Theorem says if $|g|$ attains its maximum at an interior point of the domain then $g$ is a constant.
It is already proved that $|g(z)|\leq 1$. So $|g(z)|=1$ for some $z$ implies that $|g|$ attains its maximum at that point. This implies that $g$ is a constant $c$ so $f(z)=cz$.
Kavi Rama Murthy
- 311,013
-
I wanted to make sure that that you fullfil all requirements to apply the Maximum Modulus Theorem. |g| =1 is a local maximum, because we proved $|g(z)| \leq 1$. Thank you very much :) – Steven33 Dec 12 '18 at 12:10
-
-
@Steven33 Just apply the hypothesis: either $|f(z)|=|z|$ (with $ z\neq 0$) or $|f'(0)|=1$. In either case you get $|c|=1$. – Kavi Rama Murthy Dec 12 '18 at 23:12
-
It is $ g(z)=c$ Then we get $ f'(0) =c $ and$ f(z)=cz. $. When you compare this expressions c must be 1? – Steven33 Dec 13 '18 at 08:59
0
On that page, $D$ is defined to be the open complex unit disc: $D=\{z\in {\mathbb C}: |z|<1\}$ and $g$ is defined as $$\left\{ g = \begin{eqnarray} \frac{f(z)}{z} \quad \mbox{ if } f(z) \not= 0 \\ f'(0) \quad \mbox{ otherwise } \end{eqnarray} \right. $$
So by definition, either $|g(z)| = |f(z)/z| = |f(z)|/|z| = 1$ (since $|f(z)| = |z|$) or $|g(z)| = |f'(0)| = 1$.
Note that this isn't a local maximum really -- by the maximum modulus principle $g$ is now constant on $D$.
postmortes
- 6,338