Unbounded Entire Function must be Polynomial

1. Feb 9, 2012

Poopsilon

1. The problem statement, all variables and given/known data

Let $f$ be entire. Then if $lim_{z\rightarrow \infty}|f(z)|=\infty$ then $f$ must be a non-constant polynomial.

2. Relevant equations

3. The attempt at a solution

So we know f is entire. Thus I suppose it makes sense to go ahead and expand it as a power series centered at zero. Thus what it seems to come down to is showing that if this power series is infinite, then there exists some path to infinity which z can travel such that |f(z)| remains bounded. And then I would take the contrapositive of this statement to prove the claim.

I looked at e^z for a bit of intuition and it's along the complex axis and negative real axis that this function stays bounded. But I just can't see how to generalize this observation.

I was thinking maybe it had something to do with having non-zero derivatives of all orders, but I can't see how to use that either. All in all I'm really stumped.

Last edited: Feb 9, 2012
2. Feb 9, 2012

Dick

Suppose the power series expansion doesn't terminate. Now think about f(1/z).

3. Feb 9, 2012

Poopsilon

I'm really trying here but I just can't figure out how to proceed with your clue. Is there some theorem I need? I mean I feel like I somehow have to use the fact that f is entire in order to come up with some clever path to infinity that will cause some major cancellation or telescoping process in the infinite series.

4. Feb 9, 2012

Dick

What do you know about essential singularities? Because f(1/z) has an essential singularity at z=0 if the power series doesn't terminate. That would mean f(z) has an essential singularity at infinity.

Last edited: Feb 9, 2012
5. Feb 9, 2012

Poopsilon

As best I can figure, this would mean that f maps the exterior of an arbitrarily large disk centered at 0 to ℂ minus some point, which I think would violate the condition that
|f(z)|→∞ as z→∞.

I'm a bit troubled by the difference between a function growing without bound, and a function converging to the point 'at infinity'.

With e^z I have a path to infinity along which e^z neither grows without bound nor converges to the point at infinity. What you have just shown me is a way to prove that |f(z)| does not converge to the point at infinity, since it keeps constellating almost the entire positive real axis no matter how far out one goes. Nevertheless it still grows without bound.

The Riemann sphere has not been formally introduced to us yet, but I should probably interpret |f(z)|→∞ as z→∞ as a statement about convergence.

Contradict me if I have not understood this correctly, thanks.

6. Feb 9, 2012

Dick

I'm talking about the behavior of functions near essential singularities. Stuff like the Picard theorems, you don't know them? If there's an essential singularity at infinity, it can't satisfy |f(z)|->infinity as |z|->infinity.

Last edited: Feb 9, 2012
7. Feb 9, 2012

Poopsilon

ya, Picard's big theorem was what I was using. The concept that a function can have a singularity at infinity I think is what threw me on this problem.

8. Feb 10, 2013

eulerfan

Trying to solve this in the most elementary way:
If f(z) has infinitely many zeroes inside a disc, it will be identically zero which is impossible. If it had a sequence of zeroes going to infinity it would have limit infinity at infinity, so f has finitely many zeroes.
write f(z) = g(z) (z-z1)^n1(z-z2)^n2..(Z-zk)^nk, where g(z) now has no zeroes and is entire.
Since lim f(z) = infinity, |g(z)| must eventually (i.e. for |z| > R) be >= 1/|z|^m, where m = sum of the multiplicities n1, n2... nk.
So 1/g(z) is entire and <= |z|^m for |z|>R. Then by the Cauchy integral formula for the mth derivative, we can see that the (m+1)st derivative and higher of 1/g(z) at zero are zero. Thus 1/g(z) is a polynomial of degree at most m. But 1/g(z) has no roots so it is actually a (nonzero) constant.
Hence g(z) is itself a constant and f(z) is a polynomial of degree m.

Last edited: Feb 10, 2013