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I had an approach to the following problem which now I'm not sure will work:

If $f$ is entire, and $\lim\limits_{z \to \infty} f(z) = \infty$, show that $f$ is a polynomial.

Case 1: there exist $C > 0, N \in \mathbb{N}$ such that $|f(z)| \leq C \cdot |z|^N$ for all $z$. A standard argument shows that $f$ has to be a polynomial.

Case 2 is where Case 1 does not hold. Since $f(z)$ goes to infinity, $f$ can have only finitely many zeroes $c_1, ... , c_r$ with multiplicities $m_1, ... , m_r$. Then I'd like to say that $f$ grows so fast that $$g(z) := \frac{f(z)}{(z-c_1)^{m_1} \cdots (z - c_r)^{m_r}}$$ also goes to infinity as $z \to \infty$. But it's not clear that this will actually be the case. I might have to weaken the hypothesis of Case 1 to make this work, but I'm not sure how weak I can make it.

Assuming I can have $g$ go to infinity, then $g$ will be an entire function with no zeroes, so $\frac{1}{g}$ will be bounded, hence constant. Thus $f(z) = k(z-c_1)^{m_1} \cdots (z - c_r)^{m_r}$ for some $k \in \mathbb{C}$, as required.

Saurabh Rana
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D_S
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  • Turn it around. $\frac{1}{g}$ is an entire function that grows at most as fast as $\lvert z\rvert^{m_1 + \ldots + m_r}$, hence it's a polynomial. It has no zero, hence it is constant. – Daniel Fischer Nov 14 '20 at 23:19

1 Answers1

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I thought one can make this argument work like this:

I would start with the argument in case 2. If $g$ is not bounded you are done, as you explained. If $g$ is bounded, say $|g(z)| \le C$, then $|f(z)| \le C \prod_{i=1}^r |(z-c_i)^{m_i}|$.

Then remark (maybe you need to give some details there) that $\prod_{i=1}^r |(z-c_i)^{m_i}| \le C' |z|^{(m_1+1) + \dots + (m_r+1)}$ and recall the argument given in case 1. (That I add the $+1$s is not really needed, but it makes the "remarking" easier.)

Or if you prefer, modify your case 1 to show that $|f(z)| \le |P(z)|$ for some polynomial $P$ implies that $f$ is a polynomial.

But I am not sure anymore this line of argument works. Another way to approach this problem, found at Show that this entire function is polynomial. , is to consider $z\mapsto f(1/z)$ on $\mathbb{C} \setminus \{0\}$ instead. Then, at $0$ this function has a pole (and not an essential singularity) as it tends to infinity. Thus the principal part of the Laurent series must be finite, which implies that the power series expansion of $f$ was finite.

Community
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quid
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    That makes things a lot easier. I was trying to argue that $g$ had limit infinity (not exactly the negation of Case 1) so I could deal with $\frac{1}{g}$ being bounded, but for some reason I wasn't thinking of making $g$ itself bounded. Thanks. – D_S Jul 14 '15 at 23:42
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    You are welcome. But now that you say it I have a doubt in my solution! Namely I thought your argument handles $g$ not bounded, but on second thought I am not sure. – quid Jul 14 '15 at 23:56
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    I am afraid my argument has a hole. Here is a correct one http://math.stackexchange.com/questions/311892/show-that-this-entire-function-is-polynomial – quid Jul 15 '15 at 00:03
  • Aww man. Thanks for taking the time to answer. I really want to keep trying to make this approach work, since I put so much time into it. – D_S Jul 15 '15 at 02:02