Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
De Branges's theorem
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
==Schlicht functions== The normalizations :<math>a_0=0 \ \text{and}\ a_1=1</math> mean that :<math> f(0)=0\ \text{and}\ f'(0)=1 . </math> This can always be obtained by an [[affine transformation]]: starting with an arbitrary injective holomorphic function <math>g</math> defined on the open unit disk and setting :<math>f(z)=\frac{g(z)-g(0)}{g'(0)}.</math> Such functions <math>g</math> are of interest because they appear in the [[Riemann mapping theorem]]. A '''schlicht function''' is defined as an analytic function <math>f</math> that is one-to-one and satisfies <math>f(0)=0</math> and <math>f'(0)=1</math>. A family of schlicht functions are the [[rotated Koebe function]]s :<math>f_\alpha(z)=\frac{z}{(1-\alpha z)^2}=\sum_{n=1}^\infty n\alpha^{n-1} z^n</math> with <math>\alpha</math> a [[complex number]] of [[absolute value]] <math>1</math>. If <math>f</math> is a schlicht function and <math>|a_n|=n</math> for some <math>n\geq 2</math>, then <math>f</math> is a rotated Koebe function. The condition of de Branges' theorem is not sufficient to show the function is schlicht, as the function :<math>f(z)=z+z^2 = (z+1/2)^2 - 1/4</math> shows: it is holomorphic on the unit disc and satisfies <math>|a_n|\leq n</math> for all <math>n</math>, but it is not injective since <math>f(-1/2+z) = f(-1/2-z)</math>.
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)