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Showing a Holomorphic Function is Constant

  1. Jul 26, 2010 #1
    1. The problem statement, all variables and given/known data
    Let [tex]f[/tex] be a holomorphic function defined for [tex]\vert z \vert < 1[/tex]. If there is a [tex]t \in (0, 1)[/tex] such that

    [tex]\left|f\left(\frac{1}{n}\right)\right| \leq t^n[/tex]

    for all [tex]n \in \mathbb{N}[/tex] show that [tex]f[/tex] is constant on [tex]\vert z \vert < 1[/tex]

    2. Relevant equations
    Hearing the word constant, I immediately think of the Identity Theorem, Open Mapping Theorem, and Maximum Modulus Principle

    3. The attempt at a solution
    The function [tex]f[/tex] has a power series representation in the unit circle about [tex]0[/tex]; that is

    [tex]f(z) = \sum_{k = 0}^{\infty} a_{k}z^{k}[/tex].

    From this we have that

    [tex]f\left(\frac{1}{n}\right) = \sum_{k = 0}^{\infty} \frac{a_{k}}{n^{k}} [/tex]

    and by the condition that

    [tex]\left|f\left(\frac{1}{n}\right)\right| \leq t^n[/tex]

    we see that [tex]f(0) = 0[/tex]. Suppose now that [tex]a_{0} = a_{1} = ... = a_{m - 1} = 0[/tex] holds for some [tex]m[/tex]. Then we have that

    [tex]\frac{1}{n^{m}}\left|a_{m} + \sum_{k = m + 1}^{\infty} \frac{a_{k}}{n^{m - k}}\right| = \left|\sum_{k = 0}^{\infty} \frac{a_{k}}{n^{k}}\right| = \left|f\left(\frac{1}{n}\right) \right| \leq t^{n} [/tex]

    so that

    [tex]\left|a_{m} + \sum_{k = m + 1}^{\infty} \frac{a_{k}}{n^{m - k}}\right| \leq n^{m}t^{n}[/tex].

    Letting [tex]n \rightarrow \infty[/tex] we see that [tex]a_{m} = 0[/tex] (note that exponentials overtake polynomials). The result follows.
    Last edited: Jul 26, 2010
  2. jcsd
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