I have the following true or false question:
If $f$ is a non-constant, entire function that satisfies $|f(z)| ≤ 2|ze^z
|$, then f has an
essential singularity at $\infty$
I believe the answer is true. If $f$ had a removable singularity at $\infty$, then $f$ would be bounded in a neighborhood of $\infty$. That is, there is $M$ such that $|f(z)| \leq M$ for $|z| > N$ for some sufficiently large $N$. Seeing that $|f|$ is bounded in the compact set $|z|\leq N$ we we would have that $|f|$ is bounded which by Liouville's Theorem would imply that $f$ is constant… a contradiction.
If $f(1/z)$ had a pole at $\infty$, then $f$ would be a polynomial (this can be seen by looking at the laurent series of $f(1/z)$ at $z = 0$. Say $f(z) = a_0 + a_1z+ \cdot + z^n$. For large enough $|z|$ this would imply $|f(z)| \geq |z|^n – 1$. But, for $z = iy$ where $y > 0$ this would mean
$$|y|^n – 1 \leq |f(z)| \leq 2|iye^{iy}| = 2|y|$$
Which is not true if $y$ is large. All told, $f$ has an essential singularity at $\infty$.
Does the following reasoning/answer seem correct?
Best Answer
The hypothesis implies that $f(0)=0$. Thus $h(z)=\frac{1}{z}e^{-z}f(z)$ (once the singularity at $0$ is removed) is entire and bounded. Thus $f(z)=Cze^{z}$ for some constant $C\neq0$. The rest should be straight forward.