# Liouville's Theorem and constant functions

Homework Statement
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? Thanks :)

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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?

Dick
Homework Helper
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?

Dick
Homework Helper
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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But aren't you dividing by 0?

Dick
Homework Helper
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.

Dick
Homework Helper
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.

Dick
Homework Helper
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

Dick
Homework Helper
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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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
Homework Helper
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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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

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Dick
Homework Helper
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.

Dick