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Proving an inequality using maximum modulus

  1. Jul 25, 2013 #1
    1. The problem statement, all variables and given/known data
    Let f be an analytic function on the disc |z|<1 and satisfies |f(z)|≤M if |z|<1.
    Show that [itex] |f(z)| \le M \left| \frac{z-a}{1-a'z} \right| [/itex] when |z|<1

    where a' is the complex conjugate of a

    2. Relevant equations
    This section uses maximum modulus principle, but I really don't understand it.

    3. The attempt at a solution
    Since |z|<1 and consequently |f(z)|≤1, I rewrite the inequality to be proved as
    |f(z)|/M ≤ |[itex]\frac{z-a}{1-a'z}[/itex]|
    If M=0, the inequality holds, since it is a trivial case of 0 =< 0. If not, we can use the fraction above. Since [itex]|f(z)| \le M, \; \frac{|f(z)|}{M} \le 1[/itex]

    Then, I write z as e^iθ. Technically I should write e^(iθ) but for the sake of convenience, e^iθ means just that, not e^i * θ.

    Now, [itex] \frac{e^{i\theta} - a}{1-a'e^{i\theta}} = e^{i\theta} \frac{1-ae^{-i\theta}}{1-a'e^{i\theta}} [/itex]

    Note that the numerator is the complex conjugate of the denominator. Then, we can reduce |f(z)| as [itex]|e^{i\theta}|[/itex]

    Here's where it gets confusing. [itex]|e^{i\theta}|[/itex] was stipulated to be less than 1 in the problem, but I wish to show that it is greater than equal to one.

    What should I do?
     
  2. jcsd
  3. Jul 25, 2013 #2
    Ok, here is how I did it:

    1) Multiply with the complex conjugate of the denominator;
    2) Use the fact that |z|<1 to get of an 'a';
    3) Use |z|<1 again to get rid of the fraction.
     
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