Results 1 to 7 of 7

Math Help - analytic, fixed point

  1. #1
    Junior Member
    Joined
    Mar 2009
    Posts
    29

    analytic, fixed point

    Let f(z) be an analytic function mapping the unit disc \mathbb{D} into itself. Suppose that f(z) has two different fixed points in the unit disc \mathbb{D}, show that f must be the identity function f(z)=z for all z \in \mathbb{D}.

    z_0 is called a fixed point for a function f, if f(z_0)=z_0.


    For this one, I was thinking to use Rouché's Theorem or Schwarz lemma. However, we are dealing with fixed points here. So, I don't see what to do right now. I need a few hints on this one. Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    If f(*) is analytic inside the unit disk, then it can be developped in Taylor expansion in [for example...] z_{0}=0...

    f(z)= \sum_{n=0}^{\infty} \frac{f^{(n)} (0)}{n!}\cdot z^{n} (1)

    ... where the f^{n} (0) are given by the Cauchy integrals...

    f^{n} (0) = \frac{n!}{2 \pi i} \int_{\gamma} \frac{f(z)}{z^{n+1}}\cdot dz (1)

    ... where \gamma is the unit circle. You can verify that the line integral in (2) vanishes forall n with the only exception of n=1 where is is equal to 2\pi i. So...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Alternatively, let the fixed points be z_1 and z_2, and let g be a conformal map from the disk to itself that takes z_1 to the origin. Then g f g^{-1} is an analytic map from the disk to itself that fixes the origin and also the point g(z_2). You should now be able to use Schwarz's lemma to show that g f g^{-1} is the identity and hence so is f.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by chisigma View Post
    You can verify that the line integral in (2) vanishes forall n with the only exception of n=1 where is is equal to 2\pi i.
    I may be missing something obvious, but I can't see how to prove that. How do you use the two fixed point to compute the integral?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member bkarpuz's Avatar
    Joined
    Sep 2008
    From
    Posts
    481
    Thanks
    2
    Quote Originally Posted by Laurent View Post
    Quote Originally Posted by chisigma View Post
    You can verify that the line integral in (2) vanishes forall n with the only exception of n=1 where is is equal to 2\pi i.
    I may be missing something obvious, but I can't see how to prove that. How do you use the two fixed point to compute the integral?
    Actually \chi\sigma proves that every analytic function in the unit disc is constant. Nice and simple.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    It is evident that I read hurriedly the post of eskimo 343... with a little more carefull reading it is not difficult to 'discover' that there is an evident counterexample: all the complex functions of the form f(z)= z^{2n+1} are analytic and have two fixed point in z=1 and z=-1, both on the unit disc... very sorry!...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by chisigma View Post
    It is evident that I read hurriedly the post of eskimo 343... with a little more carefull reading it is not difficult to 'discover' that there is an evident counterexample: all the complex functions of the form f(z)= z^{2n+1} are analytic and have two fixed point in z=1 and z=-1, both on the unit disc... very sorry!...
    That is correct if you take \mathbb{D} to be the closed unit disk. But I think that the original result is correct if \mathbb{D} is the open disk.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. show that f has a fixed point
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: November 19th 2011, 09:05 AM
  2. The fixed point
    Posted in the Geometry Forum
    Replies: 0
    Last Post: April 2nd 2011, 11:57 AM
  3. Fixed point iteration
    Posted in the Differential Geometry Forum
    Replies: 13
    Last Post: October 21st 2010, 04:36 AM
  4. fixed point free
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: February 19th 2009, 11:24 PM
  5. fixed point?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 1st 2006, 02:46 AM

Search Tags


/mathhelpforum @mathhelpforum