View Full Version : Liouville's Theorem and constant functions
The problem statement, all variables and given/known data
Suppose that f is entire and \lim_{z \to \infty}\frac{f(z)}{z} = 0. 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 \lim_{z \to \infty}\frac{f(z)}{z} = 0 implies the function itself is bounded, but I've not been successful in doing so.
Any hints?:confused:
Thanks :)
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.
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)
Suppose f(z) = 0 if z = 0 and log z otherwise. f is entire right? So
\lim_{z \to \infty} \frac{f(z)}{z} = \lim_{z \to \infty} \frac{f'(z)}{1} = \lim_{z \to \infty} \frac{1}{z} = 0
by l'Hopital's rule. However f is not bounded.
log z is not entire -- there is a branch cut.
So genneth, is my previous statement true? Thanks
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?
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?
I've never heard of the maximum modulus principle. And how can g be entire if it isn't even defined for z = 0?
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.
But aren't you dividing by 0?
g(z)=(a1*z+a2*z^2+....)/z=a1+a2*z+....
Yes. If you do that, you're assuming z is not zero so g(0) is undefined.
Fine. Then DEFINE g(0)=a1, if you wish. Or just define g(z) by the power series after z is factored out.
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.
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.
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
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.
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]
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.
http://en.wikipedia.org/wiki/Liouville's_theorem_(complex_analysis)
Ah, would this then work?
By Cauchy's Integral Theorem, since f is entire, we have:
f(z0) = 1/(2*pi*i) \oint\frac{f(z)}{z-z0}dz
But since we have \lim_{z \to \infty}\frac{f(z)}{z} = 0, pick any circle with R -> infinity, thus we have f(z0) to be 0 everywhere which yields f(z) = constant
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.
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.
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?
Aha! I overlooked that fact. Now it makes sense.
I've looked through the proof of Liouville's theorem, it's just that I can't get that implication that such a M exists that |f(z)| < M for all z. (the earlier method i tried above did not use anything from the proof of liouville's theorem)
Liouville's Theorem only applies for bounded f(z). In other words, it is assumed that f is such that |f(z)| <= M.
brian87, did you look at the rest of the thread? I thought by post #16 we had practically given the solution away...
Well, it seems that I got lost somewhere from post 4 onwards. (something about branch cuts)
Well, Dick, what's the part that does not make sense with the integral method?
[I note that although I know the series representation of f, this question was in a chapter before series were introduced, so I was trying to get a solution without using any series.]
Skip over the 'branch cut' stuff. If you know f can be represented as a convergent power series, then just use it. Once you've actually figured out how the proof works then you can go back and try to write it without the series. I already said I can't tell you which part of your 'integral method' doesn't make sense. Because I can't find any parts that do make sense.
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