# $f$ non-constant entire with $|f(z)| \leq 2|ze^z|$, then $f$ has essential singularity at $\infty$

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?

• Hope I am not embarrassing myself here, but since $f(0)=0$ and so $\frac{1}{z}f(z)e^{-z}$ is entire and bounded. The rest should be straight forward. – Oliver Diaz Jun 22 at 0:28

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.