# Liouville's Theorem and constant functions

1. Apr 1, 2008

### brian87

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?

Thanks :)

Last edited: Apr 1, 2008
2. Apr 1, 2008

### genneth

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.

3. Apr 1, 2008

### brian87

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)

4. Apr 1, 2008

### e(ho0n3

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.

5. Apr 1, 2008

### genneth

log z is not entire -- there is a branch cut.

6. Apr 1, 2008

### brian87

So genneth, is my previous statement true? Thanks

7. Apr 1, 2008

### e(ho0n3

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?

8. Apr 1, 2008

### Dick

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?

9. Apr 1, 2008

### e(ho0n3

I've never heard of the maximum modulus principle. And how can g be entire if it isn't even defined for z = 0?

10. Apr 1, 2008

### Dick

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.

Last edited: Apr 1, 2008
11. Apr 1, 2008

### e(ho0n3

But aren't you dividing by 0?

12. Apr 1, 2008

### Dick

g(z)=(a1*z+a2*z^2+....)/z=a1+a2*z+....

13. Apr 1, 2008

### e(ho0n3

Yes. If you do that, you're assuming z is not zero so g(0) is undefined.

14. Apr 1, 2008

### Dick

Fine. Then DEFINE g(0)=a1, if you wish. Or just define g(z) by the power series after z is factored out.

15. Apr 1, 2008

### e(ho0n3

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.

16. Apr 1, 2008

### Dick

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.

17. Apr 1, 2008

### brian87

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

18. Apr 1, 2008

### Dick

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.

Last edited: Apr 1, 2008
19. Apr 1, 2008

### brian87

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]

Last edited: Apr 1, 2008
20. Apr 1, 2008

### Dick

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.

Last edited: Apr 1, 2008