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Homework Help: Liouville's Theorem and constant functions

  1. Apr 1, 2008 #1
    The problem statement, all variables and given/known data
    Suppose that f is entire and [tex]\lim_{z \to \infty}\frac{f(z)}{z} = 0[/tex]. Prove that f is constant.

    z, and f(z) are in the complex plane

    The attempt at a solution
    I've tried to find out how the condition [tex]\lim_{z \to \infty}\frac{f(z)}{z} = 0[/tex] implies the function itself is bounded, but I've not been successful in doing so.

    Any hints?:confused:

    Thanks :)
     
    Last edited: Apr 1, 2008
  2. jcsd
  3. Apr 1, 2008 #2
    If a function is entire, then it has no poles for finite z. The limit condition means that f(z) is finite at infinity --- so the entire function is bounded.
     
  4. Apr 1, 2008 #3
    Hi, does this mean that the lack of singularities for finite z and it being finite at infinity imply that it's bounded on the complex plane?

    Thanks

    (Just looking for a clarification here)
     
  5. Apr 1, 2008 #4
    Suppose f(z) = 0 if z = 0 and log z otherwise. f is entire right? So

    [tex]\lim_{z \to \infty} \frac{f(z)}{z} = \lim_{z \to \infty} \frac{f'(z)}{1} = \lim_{z \to \infty} \frac{1}{z} = 0[/tex]

    by l'Hopital's rule. However f is not bounded.
     
  6. Apr 1, 2008 #5
    log z is not entire -- there is a branch cut.
     
  7. Apr 1, 2008 #6
    So genneth, is my previous statement true? Thanks
     
  8. Apr 1, 2008 #7
    There needn't be a branch cut, but then f wouldn't be a function any more. Nevermind

    I'm having a hard time showing that f is bounded from the definition of the limit: for any e > 0 there is a d such that |f(z)/z -0| < e for all z satisfying |z| >= d. In other words, |f(z)| < ed for any e > 0, for all z, |z| >= d. This doesn't imply that f is bounded though. And if it does, how?
     
  9. Apr 1, 2008 #8

    Dick

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    Let a0+a1*z+a2*z^2+... be the power series expansion of f(z). Let g(z)=(f(z)-a0)/z. Limit of |g(z)| as z->infinity is also 0. And g(z) is bounded (use the maximum modulus principle) and entire. So g(z) is a constant. What constant?
     
  10. Apr 1, 2008 #9
    I've never heard of the maximum modulus principle. And how can g be entire if it isn't even defined for z = 0?
     
  11. Apr 1, 2008 #10

    Dick

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    You MIGHT look it up. It says an analytic function takes on it's maximum modulus on the boundary of a domain, not the interior. f(z)-a0 has a power series with a common factor of z. Divide it out. g(0)=a1.
     
    Last edited: Apr 1, 2008
  12. Apr 1, 2008 #11
    But aren't you dividing by 0?
     
  13. Apr 1, 2008 #12

    Dick

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    g(z)=(a1*z+a2*z^2+....)/z=a1+a2*z+....
     
  14. Apr 1, 2008 #13
    Yes. If you do that, you're assuming z is not zero so g(0) is undefined.
     
  15. Apr 1, 2008 #14

    Dick

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    Fine. Then DEFINE g(0)=a1, if you wish. Or just define g(z) by the power series after z is factored out.
     
  16. Apr 1, 2008 #15
    You could have just stated that g has a removable singularity at 0.

    How does the maximum modulus principle apply, i.e. what domain are you using? It certainly can't be the whole complex plane because it doesn't have a boundary.
     
  17. Apr 1, 2008 #16

    Dick

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    Maybe we should let the OP get a question or two in here. Since |g(z)| goes to 0 as z goes to infinity, use a circle of large enough radius that the values at |z|=R are less than epsilon.
     
  18. Apr 1, 2008 #17
    Would this solution be sufficient?

    f is entire, then it has a Taylor representation about 0:

    f(z) = a0 + a1*z + a2*z^2 + ...

    Also, we have lim z-> infinity f(z)/z = 0, i.e. we must have a1, a2, .... = 0

    thus f(z) = a0, a constant.

    I used some reasoning from real analysis, can this be transferred for the complex plane though?

    I still find it hard to see how the condition that lim z -> infinity f(z)/z = 0 and f is entire implies f is bounded
     
  19. Apr 1, 2008 #18

    Dick

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    That's REALLY unconvincing because you just stated the conclusion with no reason. It's not even true for real variables. Take f(z)=exp(-z^2). The theorem isn't true for real functions. You need to use complex methods.
     
    Last edited: Apr 1, 2008
  20. Apr 1, 2008 #19
    Whoops, my mistake then

    But really, can someone explain to me how does the condition that f is entire and lim z -> infinity f(z)/z = 0 imply that f is bounded?

    Thanks :)

    [edit: It might be helpful to note that in the course I'm taking, we're doing really basic complex analysis, nothing on the maximum modulus principle, just things such as Cauchy's Theorem/Cauchy's Integral Theorem/Liouville's Theorem/Taylor, Laurent Series, classification of singular points and Cauchy's Residue Theorem]
     
    Last edited: Apr 1, 2008
  21. Apr 1, 2008 #20

    Dick

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    The maximum modulus principle follows pretty directly from Cauchy's theorem. So you can also prove that g(x)=0 using Cauchy's theorem directly. But you do need to use complex methods. If somebody else can come up with a simpler method, that would, of course, be nice.
     
    Last edited: Apr 1, 2008
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