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Let $g(z)$ be entire analytic with the growth condition $|g(z)| \to \infty$, as $|z|$ $\to$ $\infty$. Then, how can I show that $g$ must be a polynomial?

Some thoughts:

Does the singularity at infinity have to be a pole? If so, why?

I think it does not, since there was no additional, stronger assumptions such as $g$ being one-to-one. If $g$ were 1:1, then by Big Picard's theorem, the singularity is obviously a pole.

I'm also thinking about the singularity of $g(z):=f(\frac{1}{z})$ at the origin. Is it a pole or an essential singularity ... or could it be either? (Again, since there was no one-to-one assumption of $g$, I don't see why the singularity must be a pole.)

EDIT: I am also concerned that the question is incorrect and perhaps needs additional assumptions - 1:1 or something else.

Thanks,

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    $\begingroup$ The statement is correct. $\endgroup$
    – Mariano Suárez-Álvarez
    Nov 11, 2015 at 22:39
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    $\begingroup$ How do you define poles? (There are several ways to do it.) I like the definition that a pole is an isolated singularity $a$ for which $|f(z)| \to \infty$ as $z\to a$. In that case, the answer to your question about the type of singularity is yes by definition. $\endgroup$
    – mrf
    Nov 11, 2015 at 22:45
  • $\begingroup$ Hi professor, yes, I use that definition of pole, too. But I guess what is confusing me is exactly the nice answer presented below by Mariano -- why must the singularity of a polynomial at infinity be a pole, and thus the singularity at infinity for non-polynomial entire functions must be an essential singularity? Having the additional one-to-one assumption quickly rules out the option of having an essential singularity at infinity for this entire function. But without this assumption, I don't see why it is clear that the singularity is a pole. What do you think? Thanks @mrf $\endgroup$
    – User001
    Nov 11, 2015 at 22:52
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    $\begingroup$ It is not true that |exp(1/z)| goes to infinity as z goes to zero! $\endgroup$
    – Mariano Suárez-Álvarez
    Nov 11, 2015 at 23:11
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    $\begingroup$ It is not the case that $|e^{1/z}| \to \infty$ as $z\to 0$. (What happens along the negative real axis?) $\endgroup$
    – mrf
    Nov 11, 2015 at 23:12

1 Answer 1

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Part of Picard's big theorem is that an analytic function assumes every complex value, with possibly one exception, infinitely often in any neighborhood of an essential singularity. If your function is not polynomial, then it has an essential singularity at infinity, and therefore it takes some value $w$ of absolute value less than one on infinitely many points, which must go to infinity. This is incompatible with the assumtion that $|g(z)|\to\infty$ as $|z|\to\infty$.

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  • $\begingroup$ Thanks so much, @MarianoSuarezAlvarez - I am having some trouble with the line of reasoning, "if the function is not polynomial, then it has an essential singularity at infinity." Can you please elaborate? Thanks, $\endgroup$
    – User001
    Nov 11, 2015 at 22:55
  • $\begingroup$ I suggest you work on that particular point. It is a standard exercise :-) $\endgroup$
    – Mariano Suárez-Álvarez
    Nov 11, 2015 at 23:10
  • $\begingroup$ Hi @MarianoSuarezAlvarez -- ok, I will do this now. Thanks so much. :-) $\endgroup$
    – User001
    Nov 11, 2015 at 23:16
  • $\begingroup$ I think I got it @MarianoSuarezAlvarez: the growth condition is equivalent to $|g(\frac{1}{z})|$ $\to$ $\infty$ as |z|$\to$ $0$. Now, looking at the Taylor series of this entire function, centered about $z=0$, and with the assumption that it is not polynomial, then the Taylor series has infinitely many terms. But this Taylor series shows that there are infinitely many negative-power terms in z, and so the singularity at $0$ must be an essential singularity, by definition. What do you think? Thanks, $\endgroup$
    – User001
    Nov 11, 2015 at 23:32
  • $\begingroup$ And I think this applies to non-entire functions with the growth condition satisfied, too. At no point did I use the fact that the function was entire, I think. The expansion of the Taylor series was locally around $0$. So, meromorphic functions not polynomial, that satisfy this growth condition must also have an essential singularity at $\infty$ ... $\endgroup$
    – User001
    Nov 11, 2015 at 23:50

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