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Complex Analysis: Liouville's Theorem

  1. Nov 2, 2011 #1
    1. The problem statement, all variables and given/known data
    Let [tex]f=u+iv[/tex] be an entire function. Prove that if [tex]u[/tex] is bounded, then [tex]f[/tex] is constant.


    2. Relevant equations
    Liouville's Theorem states that the only bounded entire functions are the constant functions on [tex]\mathbb{C}[/tex]


    3. The attempt at a solution
    I know that if [tex]u[/tex] is bounded, then the real part of the function is bounded, obviously. I need a function that is bounded for both the real and imaginary parts when just the real part is bounded and I am not sure how to find that...
     
  2. jcsd
  3. Nov 2, 2011 #2

    Dick

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    Think about g=1/(M-f) where M is real. How can you pick a value for M that will make g bounded?
     
  4. Nov 2, 2011 #3
    I don't see how to do that.
     
  5. Nov 2, 2011 #4

    Dick

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    What is |g| in terms of u, v and M?
     
  6. Nov 2, 2011 #5
    Is this what you mean?

    [tex]|g|=\frac{(M-u)^2}{((M-u)^2+v^2)^2}+\frac{v^2}{((M-u)^2+v^2)^2}[/tex]
     
  7. Nov 2, 2011 #6

    Dick

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    That's actually |g|^2. So simplify it and take the square root. Yes, that's basically what I mean.
     
  8. Nov 2, 2011 #7
    Oops, got carried away in my LaTeX! Yes, I did mean the square root of that... Let me see if I can get anywhere from there! Thank you.
     
  9. Nov 2, 2011 #8
    OK, so now that I have g bounded, it follows that f is bounded since [tex]\frac{1}{M-f}\neq 0.[/tex] Then it follows from Liouville's Theorem that f is constant, right?
     
  10. Nov 2, 2011 #9

    Dick

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    No, you are jumping to conclusions. |g|=1/sqrt((M-u)^2+v^2) is only going to be large if (M-u)^2 and v^2 are small. You can't do much about v^2. But you can make sure (M-u)^2 isn't small. Use that u is bounded. Tell me what 'bounded' means.
     
  11. Nov 2, 2011 #10
    OK... Bounded just means that it has a limit, right?
     
  12. Nov 2, 2011 #11

    Dick

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    A 'limit' in what sense? u is bounded if there is a constant N such that -N<u<N.
     
  13. Nov 2, 2011 #12
    Well, to make sure (M-u)^2 isn't small, could we let [tex](m-u)^2 \geq max (M,1)[/tex] where [tex]M[/tex] is the bound for [tex]u?[/tex] This is remotely, and I mean VERY remotely, similar to an example we did in class, so I think it might be in the right direction.
     
  14. Nov 2, 2011 #13
    Oops, that should be [tex](M-u)^2 \geq max(M,1) ...[/tex]
     
  15. Nov 2, 2011 #14

    Dick

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    I'm not sure you are seeing how simple it is. If -N<u<N then what happens if you pick M=N+1??
     
  16. Nov 2, 2011 #15
    I actually ended up starting over and going a completely different route, as follows:

    Let [tex]g(z) = e^{f(z)},[/tex] which is also entire, by assumption. We will apply Liouville’s theorem to [tex]g.[/tex]
    [tex]|g(z)| = e^{u+iv} = e^u*|e^{iv}|= e^u[/tex]
    because [tex]|e^{iv}| = 1[/tex] for all [tex]z = x+iy.[/tex] Since [tex]u[/tex] is bounded, say [tex]u ≤ M,[/tex] [tex]M \in \mathbb{R},[/tex] and [tex]e^u[/tex] is strictly increasing, we have [tex]|g(z)|\leq e^M[/tex] which implies that [tex]g[/tex] is constant. It follows that [tex]f[/tex] is also constant. [tex]\blacksquare[/tex]
     
  17. Nov 2, 2011 #16

    Dick

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    That works too. You do need to deal with the minor detail of showing that e^f(z) is constant implies f(z) is constant, since the exponential function isn't 1-1. The other approach would have shown for g=1/((N+1)-f) that |g|<=1 so g is constant and nonzero. Then you can just solve for f.
     
    Last edited: Nov 2, 2011
  18. Nov 2, 2011 #17
    Thank you so much! Your direction caused me to think enough to go in the other direction. I really appreciate that!
     
  19. Nov 2, 2011 #18

    Dick

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    Very welcome.
     
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