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Math Help - Family of Entire Functions

  1. #1
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    Family of Entire Functions

    Problem statement:

    Let f be an entire function and set \mathscr{F}(f) = \{f(2^n z) : n \in \{0, 1, 2, \dots \} \}. Determine all entire functions f such that \mathscr{F}(f) is a compact normal family on \mathbb{C}.

    Defs: I'm including these as I'm clear on the definition of normal and compact, but unsure of the definition for "compact normal". Please comment if you think I have a definition in error.

    Compact: I'm assuming the problem's author implies that we have sequential compactness, i.e., every sequence from \mathscr{F}(f) has a limit in the set.

    Normal: Every sequence from \mathscr{F}(f) has a subsequence that converges uniformly on any compact subset of \mathbb{C}.

    So compact normal means that any sequence from \mathscr{F}(f) has a subsequence that converges uniformly on any compact subset of \mathbb{C} to a function in \mathscr{F}(f).

    Ideas:

    Any function f_n \in \mathscr{F}(f) is entire since it is the composition of two entire functions, f and 2^n z.

    Examples: The constant functions work trivially.

    In the class of non-constant entire functions, polynomials fail, e^z fails.

    I don't see how to make the general argument. Liouville's theorem identifies that any non-constant functions are unbounded. So if we can prove f must be bounded we're done. Montel's theorem gives us \mathscr{F}(f) a normal family if we have \mathscr{F}(f) locally uniformly bounded. This doesn't imply f bounded though.

    I'm sure that I'm overlooking something ... thank you for your help.
    Last edited by huram2215; July 5th 2010 at 12:26 PM.
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