Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Show the function doesnt exist

  1. Mar 19, 2012 #1
    Show that there is no holomorphic function f in the unit disc D that extends continuously to |z|=1 such that f(z) =1/z for |z|=1

    Some thoughts that might not be relevant:
    If such f existed then, I can see that f maps the unit circle to the unit circle and the unit disc onto the unit disc.
    On |z|=1 f would be equal to the conjugate function which is not differentiable anywhere.

    I'm kind of stuck. I appreciate any suggestions.
     
  2. jcsd
  3. Mar 19, 2012 #2

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Edit: wow it's way too late. Think integration
     
  4. Mar 20, 2012 #3
    Do you mean that if I integrate f on circles with radii approaching 1, I get
    0 = int f approaches int 1/z =2ipi ?

    but does the continuity of f imply the continuity of the integral like that?
     
  5. Mar 20, 2012 #4

    jgens

    User Avatar
    Gold Member

    Does integration work with this problem? I assume you had an argument like the following in mind:
    But since the extension of [itex]f[/itex] to [itex]\partial D[/itex] is only continuous, none of the integration theorems for analytic functions will work here. The notion of what it means for [itex]f[/itex] to be analytic on [itex]D \cup \partial D[/itex] is ill-defined as well, since [itex]D \cup \partial D[/itex] is not open. Maybe I am just missing something here and you found a way to do this with integration (which would be pretty neat).

    I am pretty sure that you can do this without integration though. Notice that the maximum/minimum modulus principle apply since [itex]f[/itex] is holomorphic on [itex]D[/itex] and continuous on [itex]\partial D[/itex]. This means [itex]|f|[/itex] assumes its maximum and minimum values on [itex]\partial D[/itex]; in particular, it follows that [itex]|f|=1[/itex] on [itex]D[/itex]. From here you should be able to derive a contradiction.
     
  6. Mar 20, 2012 #5
    Thank you jgens.
    About your approach, you say f doesnt have a min in D.
    How do we know 0 is not a min ?
     
  7. Mar 20, 2012 #6

    jgens

    User Avatar
    Gold Member

    I said [itex]|f|=1[/itex] on [itex]D[/itex] which means that every point of [itex]D[/itex] is a minimum for [itex]|f|[/itex].
     
  8. Mar 20, 2012 #7
    I want to understand how do you conclude |f| = 1 on D.
    I know f cannot have a maximum in D. So |f|<1 on D.
    ....what else?
     
  9. Mar 20, 2012 #8

    jgens

    User Avatar
    Gold Member

    The maximum modulus principle says that [itex]|f| \leq 1[/itex] on [itex]D[/itex]. The minimum modulus principle says that [itex]1 \leq |f|[/itex]. This means that [itex]|f|=1[/itex]. There is not much to it.
     
  10. Mar 20, 2012 #9

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    To get the minimum you can take 1/f as long as f us non-zero.

    If you integrate around circles of radius r and let r approach one you can get a contradiction can't you?
     
  11. Mar 20, 2012 #10
    To jgens: you say 1 leq |f|. How do we know f is not a function like f(z)=z which has a minimum in D ?

    To office shredder: to get that contradiction don't we need to know that integrals are continuous ? is that a fact?
     
  12. Mar 21, 2012 #11

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper

    wouldn't the winding number around the origin be positive? whereas it seems to be -1.
     
  13. Mar 21, 2012 #12
    I don't follow you mathwonk.
    If the curve is going counterclockwise, the winding number is greater or equal to 0.
    And how is that related to the question?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Show the function doesnt exist
Loading...