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Scharwz lemma

  1. May 9, 2006 #1
    so the way book states it, for the sc. lemma to work, |f(z)| has to be less than or equal to 1 and and z has to be less than 1. However, the book seems to use the lemma in some problems even if one of the conditions is not satisfied ... any help with gretaly appreciated

    also ...if you could rephrase the theorem so that it makes more practical sense, then that would be great too ... like so I know when I should apply the theorem to achieve the desired result.
     
  2. jcsd
  3. May 10, 2006 #2

    matt grime

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    A function f that maps the unit complex disc inside itself and fixes the origin satisfies |f(z)|<=|z| and |f'(0)|<=1. there is also the fact that if there is any point where |f(z)|=|z| or if |f'(0)|=1 then f is a rotation.

    Can't this just be seen by imagining a taylor series?

    unless you post the other problems then no one can properly help you (ie how do they fail to meet the hypotheses). this is a purely local statement, so you can shift g(z) by a constant so that g(z)-c satisfies the hypotheses and apply to g(z)-c and get info about g, or you can scale g so that it maps the disc into the disc by replacing g by a contsnat times g, or you can restrict g to some region where you can apply the lemma.

    Eg if g mapped the disc of radius 2 about the origin into itself and g(0)=0 set f(z)=g(2z)/2 to get something you can apply schwarz's lemma too.
     
  4. May 11, 2006 #3
    f (z) = lamdba (a - z)/(1 - abar*z) |lambda| = 1; |a| <1
    where f is any one-to-one analytic function mapping delta = {z:|z| < 1} onto itself
    set g(z) = (a - f(z))/(1-abar*f(z))
    now they say that g is one-to-one analytic function mapping delta onto delta (why?) and g(0) = 0
    and then they applied schwarz lemma to the function g(z) (why??)

    i might be missing something fundamental about properties of conjugate of a complex number (oh i know what conjugate is though ... so dont restate the obvious lol) that I dont see how |g(z)| <=1

    and also why is g(z) is one-to-one analytic function mapping delta onto delta?

    thanks
     
  5. May 13, 2006 #4

    matt grime

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    Work it out: f is a mobius transformation and g(z) is f(f(z).

    presumably because it shows them something useful. as we have no idea what they want to prove we cannot explain it.

    Whic part don't you see? Which part have you tried to prove? It is clearly analytic, and injective: it is a mobius map. g(0) is obviously 0, so all it remains is to show that it transforms delta inside itself. so what is the |g(z)| if |z|<=1?
     
  6. May 13, 2006 #5

    matt grime

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    Work it out: f is a multiple of mobius transformation and g(z) is easy to work out.

    Are you claiming that f is any analytic bijection? Or have you proved that all analytic bijections of the disc are of that form? What exactly is it you are trying to prove? It is very unclear from what you have written what is given and what is to be proven, if anything.

    presumably because it shows them something useful. as we have no idea what they want to prove we cannot explain it.

    Whic part don't you see? Which part have you tried to prove?
     
    Last edited: May 13, 2006
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