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Entire Functions and Lacunary Values.

  1. Aug 9, 2015 #1

    WWGD

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    #Hi All,
    Let ## f: \mathbb C \rightarrow \mathbb C ## be entire, i.e., analytic in the whole Complex plane. By one of Picard's theorems, ##f ## must be onto , except possibly for one value, called the lacunary value.
    Question: say ##0## is the lacunary value of ##f ##. Must ## f ## be of the form ##e^{g(z)}## , with ##g(z)## analytic?.
    Clearly if ##g(z)## has no lacunary values, then ## e^{g(z)}## will have only ##0## as its lacunary value,
    and if ##g(z) ## has only ##w## as its lacunary value, then this value will be assumed in ## w+i2\pi n ##, so ##0## will still be the lacunary value of ## e^{g(z)}##. Maybe we can consider composing functions with known lacunary values, but I don't see offhand how, since I don't know the lacunary values of general entire functions.
    I considered using Weirstrass factorization thm, but it seems overkill and has led nowhere.
    Thanks.
     
    Last edited: Aug 9, 2015
  2. jcsd
  3. Aug 9, 2015 #2

    micromass

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    Theorem: Let ##D## be any elementary domain (for example ##D = \mathbb{C}##), let ##f:D\rightarrow \mathbb{C}## be an analytic function such that ##f(z) \neq 0## for each ##z\in D##, then there exist an analytic function ##h## such that ##f(z) = e^{h(z)}## for all ##z\in D##.

    Proof: Let ##F## be a primitive of ##f^\prime / f##. Then put ##G(z) = \frac{e^{F(z)}}{f(z)}##. One checks easily that ##G^\prime (z) = 0## for all ##z\in D##. Therefore, ##e^{F(z)} = Cf(z)## for some nonzero constant ##C##. Since ##e^z## is surjective, we can write ##C = e^c##. Then ##h(z) = F(z) - c## does the trick.
     
  4. Aug 9, 2015 #3

    WWGD

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    Thanks, that shows the side I am aware of, but does it show that all entire functions that do not hit zero are of this form? Sorry, I know I am being lazy and using composition of entire functions would give an answer, but I wondered if someone knew the answer offhand.
     
  5. Aug 9, 2015 #4

    mfb

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    They all can be expressed in that way. That's what the theorem says.
     
  6. Aug 9, 2015 #5

    micromass

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    I don't get it, how does my answer not answer your question?
     
  7. Aug 9, 2015 #6

    WWGD

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    Sorry for not being clear: I am looking for a common way of generating/describing all such functions, I would say with an iff rule. I know that every function _can_ be expressed that way. What I meant to say is: does that form exhaust all possible functions with lacunary value 1?. I am looking for something along the lines of ##f: \mathbb C \rightarrow \mathbb C ## is an automorphism (global diffeomorphism) iff ##f(z)= az+b ##. Yes, you are right that any function ##f(z)## with ##f(z)## entire and never ##0## does have a logarithm ## g(z) ## in a simply-connected region so that ## e^{g(z)}=f(z) ## , but what I should have asked is whether this exhausts all possible such functions. Hope I was clearer this time
     
  8. Aug 9, 2015 #7

    micromass

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    Sorry, I really don't get it.
     
  9. Aug 9, 2015 #8

    WWGD

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    Let me think it through and I will try to be more clear. Thanks.
     
  10. Aug 10, 2015 #9

    mfb

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    "For every entire nonzero non-constant function f there exists g with [property]" is as exhaustive as it can get.
    The other direction is also true:
    For every non-constant entire g there is a non-constant nonzero entire f with f(z)=eg(z).
     
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