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Elliptic function - does limit at infinity exist?

  1. Mar 26, 2017 #1
    1. The problem statement, all variables and given/known data

    I am wanting to show that

    ##lim_{z\to\infty} f(z)=c## does not exist for ##c \in C##, ##C## the complex plane, where ##f## is non-constant periodic meromorphic function. (elliptic)

    2. Relevant equations


    3. The attempt at a solution

    So I want to proove this is not true
    .
    Beginning with the definition:

    For any ##\epsilon >0## can find a ##R## s.t ##|z|>R \implies |f(z)-c|<\epsilon## (1)

    Now let ##w## be a period.

    Then we also have

    ##|f(z+mw)-c|<\epsilon##(2), for ##m## some integer

    Now I attach the solution:

    yoyo.png

    Questions:

    - I thought that (2) would hold automatically, from (1) by periodicty, without additionally needing to have ##|z+mw |>R ##

    - Also this would hold following from (1), if not using periodicity, but simply looking at ##|z+mw |>R ##, for ##z## large since a shift by some constant is insignificant? Therefore in the solution, I do not understand the need to specify both, by periodicity and include this condition ##|z+mw |>R ##

    - Finally I do not understand the conclusion itself, perhaps due to being unclear on the above points- I understand the goal is to show that if this limit exists it implies that ##f## is holomoprhic (since I then have the theorem that an elliptic function holomorphic is constant, and so proof by contradiction).

    I do not understand the conclusion that

    ##|f(z)-c|=|f(z+mw)-c|<1## (3) implies the ##f(z)## is a bounded holomorphic function.

    My interpretation is that since we can vary ##m## to cover a range of values of ##f(z)## which are bounded this conclusion is made? So ##m## generating alot of points? However isn't this only for large ##z## ? We have only showed bounded for large ##z##? (When ##|z|>R## is satisfied)? Or am I barking up the wrong tree as to why we conclude ##f(z)## is constant from (3)?


    Many thanks in advance
     
  2. jcsd
  3. Mar 26, 2017 #2

    mfb

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    Your (2) is valid for |z|>R, but that is not really what you want to show.

    You want to show that |f(z)-c| < ##\epsilon## for all z, including those not larger than R in magnitude. So you need the reverse direction. Start with those z, then find (using periodicity) a complex value with a magnitude larger than R.

    If f(z) does not differ from c by more than 1 for all z, it is bounded. A bounded meromorphic function is constant.
     
  4. Mar 26, 2017 #3
    So this is ##|z+mw|## but then how does this show that |f(z)-c| < ##\epsilon## for all z, including those not larger than R in magnitude?

    Because we can choose any ##m## integer to make this small?
     
  5. Mar 26, 2017 #4

    mfb

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    We can choose an m such that |z+mw|>R, and there we know that |f(z+mw)-c|< ϵ.
     
    Last edited: Mar 27, 2017
  6. Mar 27, 2017 #5
    but that's large ##m## isn't it? so again only large ##z##, not all ##z##?


    I'm not understanding, what's happened to the ##c##?
     
  7. Mar 27, 2017 #6

    mfb

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    It is sufficient that some m exists, it does not matter how large it is. z can be anything.
    I forgot it. Should be in there of course.
     
  8. Mar 28, 2017 #7
    Ok, think I may understand at last, so whilst ##|f(z+mw)-c|<\epsilon## could simply follow from a condition only on ##z## , ##|z|>R## , by periodicity, this is still only considering large ##z##.

    However if we instead have ##|z+mw|>R##, than the periodicity can be used to make |z+mw| large through ##m## and not ##z##, and so it holds for small ##z## too?
     
  9. Mar 28, 2017 #8

    mfb

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    I don't understand your last post.

    Let's make an example, maybe that is easier to understand.

    ##\epsilon=0.01##, R=1, w=0.5, c=3.
    For |z|>1, we know that |f(z)-3|<0.01, or, in words, f(z) is close to 3. We want to show that f(z) is close to 3 everywhere.

    What about z=0.2? We don't have direct condition on that, because 0.2<R. But we know f(0.2)=f(0.7)=f(1.2), and |f(1.2)-3|<0.01, i. e. f(1.2) is close to 3. Therefore, |f(0.2)-3|<0.01, which means f(0.2) is close to 3 as well.
    z=-0.4+0.2i? Same idea: f(-0.4+0.2i) = f(1.1+0.2i), and |1.1+0.2i|>1. Again we can conclude that |f(-0.4+0.2i)-3|<0.01.

    This approach works for all z with |z|<1. Therefore, |f(z)-3|<0.01 for all z.
    We can repeat this with every ##\epsilon>0##, and conclude that |f(z)-3|<##\epsilon## for all ##\epsilon##. Therefore, f(z)=3 everywhere.
     
  10. Apr 10, 2017 #9
    This doesn't need to be ##1## right? The more general definition is |f(z)-c| < ##\epsilon## for ##0< \epsilon <1 ##?
     
  11. Apr 10, 2017 #10
    Ahh okay thanks, I think this answers my next question: ##m## is integer so for a fixed ##\epsilon## with the procedure you descried you could only cover a discrete set of points in the complex plane but then by varying ##\epsilon## you can cover continuous points so the whole complex plane? thanks.
     
  12. Apr 10, 2017 #11

    mfb

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    It works with every positive real number, yes.
    No, you can cover the whole plane with a single value of ##\epsilon##.
    Check the example in post 8. I picked z=0.2 and z=-0.4+0.2i as examples, but it works for every z.
     
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