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Complex polynomial properties when bounded (Liouville theorem)

  1. Mar 26, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose [itex] f [/itex] is differentiable in [itex]\mathbb{C}[/itex] and [itex] |f(z)| \leq C|z|^m [/itex] for some [itex]m \geq 1, C > 0 [/itex] and all [itex] z \in \mathbb{C} [/itex], show that;

    [itex] f(z) = a_1z + a_2 z^2 + a_3 z^3 + ... a_m z^m [/itex]


    2. Relevant equations


    3. The attempt at a solution

    I can't seem to show this. It does the proof when [itex] m = 1 [/itex] because then;

    If f(z) is differentiable, it's analytic (possible to expand in to taylor series), so;

    [itex] f(z) = a_0 + a_1z + a_2 z^2 + a_3 z^3 + ...[/itex]

    [itex] |f(0)| = |a_0| \leq C|0| = 0 [/itex]

    So [itex] a_0 = 0 [/itex]

    Then take [itex] g(z) = f(z)/z [/itex]

    [itex] |g(z)| \leq C [/itex], by Liouville's theorem this means g(z) is a constant, i.e. [itex] g(z) = a_1 [/itex] so [itex] f(z) = g(z)z = a_1 z [/itex]


    I can't seem to prove this for [itex] m > 1 [/itex] though, because this means

    [itex] |f(z)| \leq C|z|^m [/itex]

    so

    [itex] |f(0)| \leq C|0|^m = 0 [/itex]

    So [itex] a_0 = 0 [/itex]

    Again take

    [itex] g(z) = f(z)/z = a_1 + a_2 z + a_3 z^2 + ... [/itex]

    Then [itex] |g(z)| = |f(z)|/|z| \leq C|z|^{m-1} [/itex]

    So [itex] |g(0)| \leq C|0|^{m-1} = 0 [/itex]

    Meaning [itex] a_1 = 0 [/itex]

    Keep on applying this shows that [itex] a_1 = a_2 = .... = a_{m-1} = 0 [/itex], which pretty much disproves exactly what I'm trying to prove. Any help?
     
  2. jcsd
  3. Mar 26, 2012 #2

    Office_Shredder

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    I think the problem is wrong (well the statement is technically true, but Liouville gives a much stronger result). It should be that f is some constant times zm which is what you got. For example in the m=2 case we have

    f(z) = az+bz2
    [itex] |f(z)| \geq |a||z|-|b||z|^2[/itex] for small values of z. If |z| is super small (smaller than |a|/(2|b|))
    [itex] |f(z)| \geq |a||z|/2 [/itex] and no matter what our constant C is, we can pick |z| small enough so that
    [tex] |a||z|/2 > C|z|^2[/tex]
    for example let [itex]|z| < |a|/(2C)[/itex]
     
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