# Complex polynomial properties when bounded (Liouville theorem)

1. Mar 26, 2012

### Silversonic

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

Suppose $f$ is differentiable in $\mathbb{C}$ and $|f(z)| \leq C|z|^m$ for some $m \geq 1, C > 0$ and all $z \in \mathbb{C}$, show that;

$f(z) = a_1z + a_2 z^2 + a_3 z^3 + ... a_m z^m$

2. Relevant equations

3. The attempt at a solution

I can't seem to show this. It does the proof when $m = 1$ because then;

If f(z) is differentiable, it's analytic (possible to expand in to taylor series), so;

$f(z) = a_0 + a_1z + a_2 z^2 + a_3 z^3 + ...$

$|f(0)| = |a_0| \leq C|0| = 0$

So $a_0 = 0$

Then take $g(z) = f(z)/z$

$|g(z)| \leq C$, by Liouville's theorem this means g(z) is a constant, i.e. $g(z) = a_1$ so $f(z) = g(z)z = a_1 z$

I can't seem to prove this for $m > 1$ though, because this means

$|f(z)| \leq C|z|^m$

so

$|f(0)| \leq C|0|^m = 0$

So $a_0 = 0$

Again take

$g(z) = f(z)/z = a_1 + a_2 z + a_3 z^2 + ...$

Then $|g(z)| = |f(z)|/|z| \leq C|z|^{m-1}$

So $|g(0)| \leq C|0|^{m-1} = 0$

Meaning $a_1 = 0$

Keep on applying this shows that $a_1 = a_2 = .... = a_{m-1} = 0$, which pretty much disproves exactly what I'm trying to prove. Any help?

2. Mar 26, 2012

### Office_Shredder

Staff Emeritus
I think the problem is wrong (well the statement is technically true, but Liouville gives a much stronger result). It should be that f is some constant times zm which is what you got. For example in the m=2 case we have

f(z) = az+bz2
$|f(z)| \geq |a||z|-|b||z|^2$ for small values of z. If |z| is super small (smaller than |a|/(2|b|))
$|f(z)| \geq |a||z|/2$ and no matter what our constant C is, we can pick |z| small enough so that
$$|a||z|/2 > C|z|^2$$
for example let $|z| < |a|/(2C)$