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Let $f$ be a nonvanishing entire function of finite order. It follows immediately from Hadamard's factorization theorem that $f(z)=e^{P(z)}$ for some polynomial $P(z)$ whose degree is at most the order of $f$. However, I'm wondering if I can prove this in a more elementary way.

Let's say for simplicity that I know $|f(z)|\leq e^{|z|^5}$ for all $z\in\mathbb C$. Then I know I can write $f(z)=e^{g(z)}$ for some entire function $g$. The inequality tells me that $\Re(g(z))\leq |z|^5$ for all $z$. At this point I'm not sure how I can deduce that $g$ must be a polynomial of degree at most $5$.

justintomb
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  • Do you know Liouville’s Theorem? – Clayton Apr 16 '18 at 21:07
  • I do know Liouville's theorem. – justintomb Apr 16 '18 at 21:08
  • No Liouville. Bring infinity to the origin. $|z|^d\Re(g(1/z))\leq 1$. Therefore, the singularity of $z^dg(1/z)$ is removable: Casorati-Weierstrass. Therefore $g$ is a polynomial of degree at most $d$. –  Apr 16 '18 at 21:25
  • Would you mind giving a few more details as to how you are using Casorati-Weierstrass? I'm not sure how you are going from saying that $|z|^d\Re(g(1/z))\leq 1$ to saying that $z^dg(1/z)$ has a removable singularity. – justintomb Apr 16 '18 at 21:49
  • I know a bound on $\Re(g(z))$, but I do not have a bound on $|g(z)|$. If I had a bound on $|g(z)|$ I would be done. – justintomb Apr 16 '18 at 21:58
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    Here is one such example. I think you can also use Liouville since $\frac{\Re g}{|z|^5}$ is bounded (it would require a “special” kind of argument, but I think it isn’t too hard). – Clayton Apr 16 '18 at 23:28

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