Explore BrainMass
Share

Explore BrainMass

    Schwarz's lemma

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    Suppose that f is analytic in the disc |z|<1, that f(0) = f'(0) = 0 and that |f(z)| &#8804; 1 for all z in the disc. Show that |f(z)| &#8804; |z^2|, for |z|<1.
    Hint: Schwarz's Lemma:
    Suppose that f is analytic in the disc |z|<1, that f(0) = 0 and that |f(z)| &#8804; 1 for all z in the disc.
    Then |f(z)| &#8804; |z|, for |z|<1.
    Equality can hold for some z &#8800; 0 only if f(z) = &#955;z, where &#955; is a constant of absolute value 1.

    © BrainMass Inc. brainmass.com October 9, 2019, 9:21 pm ad1c9bdddf
    https://brainmass.com/math/analytic-geometry/schwarz-s-lemma-181457

    Solution Summary

    This uses Schwarz's Lemma to so details about functions.

    $2.19