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I am asked to show that if $f$ is entire with the property that $\lim_{z\to\infty} f(z)=\infty$, then $f$ must be a polynomial.

However, I feel as if I am missing something.

$e^z$ is entire (holomorphic at all finite $z$), and clearly $\lim_{z\to\infty} f(z)=\infty$ holds. However, $e^z$ is not a polynomial.

This seems to be a disproof via counterexample. Is there something wrong with my reasoning, or with the problem?

Dean Miller
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Tejas Rao
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3 Answers3

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It is not true that $\lim_{z\to\infty}e^z=\infty$. For instance, the sequence $(e^{-n})_{n\in\Bbb N}$ converges to $0$.

Let $g(z)=f\left(\frac1z\right)$. Then $\lim_{z\to0}g(z)=\infty$ and therefore $g$ has a pole at $0$, because otherwise $g$ would have an essential singularity at $0$, which is impossible, by the Casorati-Weierstrass theorem. But if $f(z)=a_0+a_1z+a_2z^2+\cdots$, then $g(z)=a_0+\frac{a_1}z+\frac{a_2}{z^2}+\cdots$ and therefore asserting that $g$ has a pole at $0$ means that $a_n=0$ if $n$ is large enough. So, $f$ is a polynomial function.

José Carlos Santos
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  • Hi! I tried to give an answer too, would you mind giving it a quick read to check? – Ovi Mar 11 '19 at 20:18
  • I saw another solution to this question asserting that since $g(z) = \frac{1}{f(1/z)}$ has a removable singularity at $z=0$, due to Riemann's theorem on removable singularities ($g(z)$ bounded near $z=0$). If we denote the order of the zero of $g(z)$ at $z=0$ by $N \geq 1$, we obtain, when solving for $f(z)$: $f(z) = \frac{1}{h(1/z)} = O(|z|^n)$ as $|z| \to \infty$. The solution then states that by Liouvilles theorem $f(z)$ must be a polynomial of degreee at most $N$.

    I don't understand what $O(|z|^n)$ refers to and how is it obtained? I can't connect $O(|z|^n)$ to Liouvilles theorem.

     – iaenstrom Aug 21 '19 at 10:54
  • sir, I can't understand why by " Casorati-Weierstrass theorem" this function will not have essential singularity at zero?. – TheStudent Oct 18 '19 at 07:39
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    @HoneyKumar Because if it had an essential singularity at $0$ then, by the Casorati-Weierstrass theorem, the image of any neighborhood of $0$ by $g$ would be dense in $\mathbb C$. But it is not, since $\lim_{z\to0}g(z)=\infty$. – José Carlos Santos Oct 18 '19 at 07:59
  • Isn't $\lim\limits_{z\to 0}f(z)=\infty$ the definition of a pole at $0$? For an essential singularity the limit shouldn't even exist (finitely or infinitely). – H.R Dec 09 '19 at 20:23
  • @HritRoy Actually, the definition of pole that I work with is that $0$ is a pole if the Laurent series of $f$ near $0$ is of the type $\sum_{n=-N}^\infty a_nz^n$, for some $N\in\mathbb N$. But it is equivalent to the assertion that $\lim_{z\to0}f(z)=\infty$. And, yes, if $0$ is an essential singularity, then the limit $\lim_{z\to0}f(z)$ doesn't exist in $\mathbb C\cup{\infty}$. – José Carlos Santos Dec 10 '19 at 07:29
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I believe you have mis-read or mis-interpreted the key part of the problem. For complex variables, $$ \lim_{z\to \infty} f(z) = X $$ means that the limit is $X$ (whatever $X$ is) as $z$ approaches infinity in any direction.

I'm not certain, but I believe you could replace that with $$ \forall u \in \Bbb C : (|u|=1 \implies \lim_{r\to\infty_{r\in\Bbb R}} f(r u) = X) $$ that is, the limits, taken as the limit of the real length along a line from the origin, of the function of points on that line, are all the same value $X$ (which in this case is $\infty$.

So you need to show that for any non-polynomial entire function, there is some direction along with the function does not go to infinity.

Mark Fischler
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One important feature about holomorphic functions is the number and position of their zeroes. The function going to $\infty$ implies that outside of some compact disk $D$ centered at the origin, the function is never zero.

Now to prove that our function is a polynomial, it suffices by analytic continuation to prove that our $f$ has only a finite number of zeroes in this disk $D$. What would go wrong if our function had infinitely many zeroes in $D$? Let $S$ be the set of zeroes of $f$. Since $D$ is compact, $S$ would have a limit point in $D$; but again, by analytic continuation, this would imply that $f$ is the zero function, which is a contradiction.

J. W. Tanner
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Ovi
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    Asserting that a $\lim_{z\to\infty}f(z)=\infty$ does not mean that, outside a compact disk centered ar the origin, $f$ is never zero. For instance, it is not true that $\lim_{z\to\infty}e^z=\infty$ but nevertheless the exponential function has no zero whatsoever. – José Carlos Santos Mar 11 '19 at 20:34
  • @JoséCarlosSantos Than you very much for the response! I've never seen $\lim_{z\to\infty}f(z) = \infty$, so I assumed it means $\forall (M \in \mathbb{R}^+) \exists (\delta > 0) \forall (z \in \mathbb{C}): |z|> \delta \implies |f(z)| > M$. I'm pretty sure that under this definition, outside of some compact disk $D$ centered at the origin, $f$ would not have any zeroes, no? Is there another definition of $\lim_{z\to\infty}f(z) = \infty$? – Ovi Mar 11 '19 at 20:42
  • You are right about the definition. But it doesn't mean that it has no zeros outside a compact disk. It just implies it. – José Carlos Santos Mar 11 '19 at 21:25
  • @JoséCarlosSantos Ahhh okay thank you, now I understand your example too. Yes, I used "means" to mean "imply". I fixed the answer. – Ovi Mar 11 '19 at 21:31