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Liouville's Theorem and constant functions

  • Thread starter brian87
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Homework Statement
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 :)
 
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Answers and Replies

979
1
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.
 
8
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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)
 
1,356
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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.
 
979
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log z is not entire -- there is a branch cut.
 
8
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So genneth, is my previous statement true? Thanks
 
1,356
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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?
 
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?
 
1,356
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I've never heard of the maximum modulus principle. And how can g be entire if it isn't even defined for z = 0?
 
Dick
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I've never heard of the maximum modulus principle. And how can g be entire if it isn't even defined for z = 0?
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.
 
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1,356
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But aren't you dividing by 0?
 
Dick
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g(z)=(a1*z+a2*z^2+....)/z=a1+a2*z+....
 
1,356
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Yes. If you do that, you're assuming z is not zero so g(0) is undefined.
 
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.
 
1,356
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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.
 
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.
 
8
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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
 
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.
 
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8
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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]
 
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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.
 
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979
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http://en.wikipedia.org/wiki/Liouville's_theorem_(complex_analysis)
 
8
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Ah, would this then work?

By Cauchy's Integral Theorem, since f is entire, we have:

f(z0) = 1/(2*pi*i) [tex]\oint\frac{f(z)}{z-z0}dz[/tex]

But since we have [tex]\lim_{z \to \infty}\frac{f(z)}{z} = 0[/tex], pick any circle with R -> infinity, thus we have f(z0) to be 0 everywhere which yields f(z) = constant
 
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Dick
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No, it doesn't work. And it's not because of a specific flaw. It simply doesn't make sense. Look at the proof of the standard Liouville theorem and study it until you really understand it. Then tackle this problem.
 
1,356
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One thing that bugs me about that proof is that |f(u)|/|uk+1| <= M/rk+1. This is like saying: if a < b and c < d, then a/c < b/d. This is certainly not true.
 
Dick
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One thing that bugs me about that proof is that |f(u)|/|uk+1| <= M/rk+1. This is like saying: if a < b and c < d, then a/c < b/d. This is certainly not true.
But |u^(k+1)|=r^(k+1) along the contour. So it's a bit more like saying a<b and c=d implies a/c<b/d. Can't argue with that one, can you?
 

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