Entire function $f(z)$ that satisfies a given limit.

Question

I am given an entire function $f$ satisfies the limit $$ \lim_{z \to \infty} \frac{f(z)^2}{z^3} = 0 . $$ I'm asked to determine the most general form of $f$.

I've thought of the following ways;

1. We know $f(z)^2$ is a polynomial of degree at most 2, thus; $f(z)^2 = \alpha + \beta z + \gamma z^2$. The most general form of $f$ is therefore $$ f(z) = \pm \sqrt{\alpha + \beta z + \gamma z^2}. $$
2. We know $$ \lim_{z \to \infty} \frac{f(z)^2}{z^3} = 0 \implies \lim_{z \to \infty} \frac{f(z)}{z^{3/2}} = 0. $$ However, Liouville's generalized theorem only states that $f(z)$ is a polynomial of at most $q$ if it is a non-negative integer and the limit $f(z) / z^q$ exits. Therefore I figure this yields no result.

I applied method 1 to a similar question where $f$ satisfies $$ \lim_{z \to \infty} \frac{f(z)^2}{z} = 1\,,$$ concluding that $f(z) = \pm \sqrt{\alpha + z}$.

Can I do better than this?

Answer 1

If $z\rightarrow \infty$ (as a complex variable), then $f(z)=a+bz$ is the only possibility. Genuine square-roots are not entire functions.

Answer 2

If the limit condition in the assumption holds, then there exists $C > 0$ such that $|f(z)|^2 \le C |z|^3$ for all $|z| \ge 1$ and therefore (perhaps with a larger $C$) $|f(z)|^2 \le C(1 + |z|^3)$ for all $z$. Thus $|f(z| \le C(1 + |z|)^{3/2}$ for all $z$, for some $C$. Now apply Liouville's Theorem.

Answer 3

Liouville's generalized theorem holds for any positive power, not just integer powers. So if $f$ is entire and $|f(z)| \le C|z|^p $ for large $|z|,$ then $f$ is a polynomial of degree at most $\lfloor p\rfloor.$ In the problem at hand we have $|f(z)| \le C|z|^{3/2} $ for large $|z|.$ Therefore $f$ is a polynomial of degree at most $1.$