6
$\begingroup$

Here is my problem:

Let $f(z)$ be an entire function such that |f '(z)| < |f(z)| for all $z \in \mathbb{C}$, Show that there exists a constant A such that $|f(z)| < A*e^{|z|}$ for all $z \in \mathbb{C}$. I am trying to use Liouville's theorem to prove and try to set g(z)=|f '(z)|/|f(z)| < 1 and then g(z) is constant. I am not sure if my thinking is right and how to prove this problem? Thanks

  • 2
    The reason why Willie suggests $f'/f$ and not your $g$ is that your function is not analytic, so Liouville' stheorem doesn't apply.2011-04-18

1 Answers 1

5

Consider g(z) = \frac{f'(z)}{f(z)}. Since |f'(z)| \lt |f(z)| we see that $f$ has no zeros, so $g$ is entire. Thus, $g$ is an entire function satisfying $|g(z)| \lt 1$, so by Liouville it is constant. This means that f'(z) = C f(z) for some $C \lt 1$.

Since $f$ is entire we can write $f(z) = \sum_{n=0}^{\infty} a_n z^n$ for all $z \in \mathbb{C}$ and the equation f'(z) = C f(z) leads to f'(z) = \sum_{n=0}^{\infty} (n +1) a_{n+1} z^{n} = C \sum_{n=0}^{\infty} a_n z^n. By comparing coefficients we see that $a_1 = C a_0$, $2 a_2 = C a_1$, $3a_3 = Ca_2$, etc, so that $a_n = \frac{C^{n}}{n!} a_0$. In other words, $f(z) = \sum_{n=0}^{\infty} \frac{a_{0}}{n!} (Cz)^n = a_{0} e^{Cz}$.

Now let us prove the estimate. Since $C \lt 1$ we have for all $z \in \mathbb{C}$ that $|f(z)| \leq \sum_{n=0}^{\infty} \frac{|a_0|}{n!}\,|Cz|^n \leq \sum_{n=0}^{\infty} \frac{|a_0|}{n!} |z|^n = |a_0| e^{|z|},$ almost as we wanted. By choosing $A \gt |a_0|$ we get the desired strict inequality $|f(z)| \lt A e^{|z|}$.