For entire funtion $h$, $zh(z)$ cannot have a nonzero limit as $z\to \infty$

Question

This is a question I encounter in learning complex variables.

Show that for any entire funtion $h$, $zh(z)$ cannot have a nonzero limit as $z\to \infty$.

Furthermore, use this statement to give an alternate proof of Liouville's theorem.

I am really having trouble proving the statement, and can't see any obvious connection between this and the Liouville's theorem. Any hints or suggestions are appreciated.

Answer 1

Let us assume the opposite, i.e. that $\lim_{z\to \infty} zh(z)=k;k\neq 0$

• Assuming Liouville theorem:

  $$\lim_{z\to \infty}zh(z)=k\neq 0\\ \lim_{z\to \infty}h(z)=0\\ h(z)=a (\text{const})\\ \lim_{z\to \infty}zh(z)=\infty$$

This contradiction proves that the limit, if it exists, must be either $0$ or $\infty$

• Without Liouville theorem: By the Cauchy integral formula:

$$|h^{(n)}(0)|=\left|\frac{n!}{2\pi i}\oint_{c_r}\frac{h(z)}{z^{n+1}}dz\right|\le \frac{n!}{r^n}\underset{|z|=r}{\text{Max}}|h(z)|$$

Letting $r\to \infty$ we have, thanks to the assumption, that $\forall_{n>1}h^{(n)}(0)=0$. This implies that $h(z)=az+b$, which contradicts

$$\lim_{z\to \infty}zh(z)=k$$

Proving Liouville with this statement:

Given a bounded entire function $h(z)$, let us define $$f(z):=f\left(\frac{1}{z}\right)$$ This function is holomorphic on $\mathbb{C}-\{0\}$, and is bounded in a neighbourhood of $0$; By Riemann's theorem on removable singularities $\lim_{z\to 0}f(z)=l\in \mathbb{C}$ exists, and thus $\lim_{z\to \infty}h(z)=l\in \mathbb{C}$ too. Defining $g(z)=\frac{h(z)-h(a)}{z-a}$, we obtain an entire function, for which $$\lim_{z\to \infty}g(z)z=l-h(a)$$ Thanks to our statement, $l-h(a)=0$. Since $a$ was generic, $h$ is constant

Answer 2

Your statement is wrong. What about $h(z)=1$ ? Then, for example, for $t>0$, $$\lim_{t\to \infty }t h(t)=+\infty.$$