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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?

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    $\begingroup$ 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. $\endgroup$ – Oliver Diaz Jun 22 at 0:28
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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.

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  • $\begingroup$ Please correct your first line. $\endgroup$ – Ted Shifrin Jun 22 at 0:43
  • $\begingroup$ Ah, that is good. Thank you for your response. While I like your solution better, is there anything wrong with mine? $\endgroup$ – Mike Jun 22 at 1:00
  • $\begingroup$ I don’t see a problem with your solution. I just thought that the problem was simpler at fist glance. $\endgroup$ – Oliver Diaz Jun 22 at 1:03
  • $\begingroup$ Thank you for your help $\endgroup$ – Mike Jun 22 at 1:07

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