# Complex Analysis: Liouville's Theorem

1. Nov 2, 2011

### tarheelborn

1. The problem statement, all variables and given/known data
Let $$f=u+iv$$ be an entire function. Prove that if $$u$$ is bounded, then $$f$$ is constant.

2. Relevant equations
Liouville's Theorem states that the only bounded entire functions are the constant functions on $$\mathbb{C}$$

3. The attempt at a solution
I know that if $$u$$ is bounded, then the real part of the function is bounded, obviously. I need a function that is bounded for both the real and imaginary parts when just the real part is bounded and I am not sure how to find that...

2. Nov 2, 2011

### Dick

Think about g=1/(M-f) where M is real. How can you pick a value for M that will make g bounded?

3. Nov 2, 2011

### tarheelborn

I don't see how to do that.

4. Nov 2, 2011

### Dick

What is |g| in terms of u, v and M?

5. Nov 2, 2011

### tarheelborn

Is this what you mean?

$$|g|=\frac{(M-u)^2}{((M-u)^2+v^2)^2}+\frac{v^2}{((M-u)^2+v^2)^2}$$

6. Nov 2, 2011

### Dick

That's actually |g|^2. So simplify it and take the square root. Yes, that's basically what I mean.

7. Nov 2, 2011

### tarheelborn

Oops, got carried away in my LaTeX! Yes, I did mean the square root of that... Let me see if I can get anywhere from there! Thank you.

8. Nov 2, 2011

### tarheelborn

OK, so now that I have g bounded, it follows that f is bounded since $$\frac{1}{M-f}\neq 0.$$ Then it follows from Liouville's Theorem that f is constant, right?

9. Nov 2, 2011

### Dick

No, you are jumping to conclusions. |g|=1/sqrt((M-u)^2+v^2) is only going to be large if (M-u)^2 and v^2 are small. You can't do much about v^2. But you can make sure (M-u)^2 isn't small. Use that u is bounded. Tell me what 'bounded' means.

10. Nov 2, 2011

### tarheelborn

OK... Bounded just means that it has a limit, right?

11. Nov 2, 2011

### Dick

A 'limit' in what sense? u is bounded if there is a constant N such that -N<u<N.

12. Nov 2, 2011

### tarheelborn

Well, to make sure (M-u)^2 isn't small, could we let $$(m-u)^2 \geq max (M,1)$$ where $$M$$ is the bound for $$u?$$ This is remotely, and I mean VERY remotely, similar to an example we did in class, so I think it might be in the right direction.

13. Nov 2, 2011

### tarheelborn

Oops, that should be $$(M-u)^2 \geq max(M,1) ...$$

14. Nov 2, 2011

### Dick

I'm not sure you are seeing how simple it is. If -N<u<N then what happens if you pick M=N+1??

15. Nov 2, 2011

### tarheelborn

I actually ended up starting over and going a completely different route, as follows:

Let $$g(z) = e^{f(z)},$$ which is also entire, by assumption. We will apply Liouville’s theorem to $$g.$$
$$|g(z)| = e^{u+iv} = e^u*|e^{iv}|= e^u$$
because $$|e^{iv}| = 1$$ for all $$z = x+iy.$$ Since $$u$$ is bounded, say $$u ≤ M,$$ $$M \in \mathbb{R},$$ and $$e^u$$ is strictly increasing, we have $$|g(z)|\leq e^M$$ which implies that $$g$$ is constant. It follows that $$f$$ is also constant. $$\blacksquare$$

16. Nov 2, 2011

### Dick

That works too. You do need to deal with the minor detail of showing that e^f(z) is constant implies f(z) is constant, since the exponential function isn't 1-1. The other approach would have shown for g=1/((N+1)-f) that |g|<=1 so g is constant and nonzero. Then you can just solve for f.

Last edited: Nov 2, 2011
17. Nov 2, 2011

### tarheelborn

Thank you so much! Your direction caused me to think enough to go in the other direction. I really appreciate that!

18. Nov 2, 2011

### Dick

Very welcome.