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Suppose $f$ is entire and $\lim_{z\to\infty}f(z)=\infty$. Show that $f(\mathbb{C})=\mathbb{C}$.

First of all I don't really understand this question. I know $z\to\infty$ means $|z|\to\infty$, but what does $f(z)\to\infty$ means? Does it mean $|f(z)|\to \infty$? Also, I just learnt about the one point compactification $\infty$ to the complex plane. So the reason we write $z\to\infty$ instead of $|z|\to \infty$ is because we are referring to $\infty$ as a point in the extended complex plane $\bar{\mathbb{C}}$? So $f(z)\to\infty$ is also referring to $\infty$ in $\bar{\mathbb{C}}$?

Asaf Karagila
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Fluffy Skye
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3 Answers3

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The niceties of the one-point compactification of $\Bbb C$ aside, consider:

$f(z) \ne w, \; \forall z \in \Bbb C; \tag 1$

$f(z) - w \ne 0, \; \forall z \in \Bbb C; \tag 2$

$(f(z) - w)^{-1}$ is then entire; since

$f(z) \to \infty \; \text{as} \; z \to \infty \tag 3$

$(f(z) - w)^{-1}$ is bounded; hence by Liouville's theorem, constant; but then we cannot have (3); thus (1) fails.

Robert Lewis
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    Nice answer! +1 vote. – Kavi Rama Murthy Feb 25 '19 at 07:33
  • @KaviRamaMurthy: quite a complement, considering the source! Cheers! – Robert Lewis Feb 25 '19 at 07:34
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    Why is it bounded? At first I thought you can separate it in to a compact set and outside the compact set we can bound it. But I just noticed that a punctured disc is not compact. – Fluffy Skye Feb 28 '19 at 02:20
  • Roughly speaking, $(f(z) - w)^{-1}$ is bounded because $f(z)$ can be made arbitrarily large by taking $z$ large enough. So for fixed $w$, we also have $f(z) - w \to \infty$ as $z \to \infty$. So picking $R$ large enough, we can ensure $(f(z) - w)^{-1}$ is bounded for $\vert z \vert \ge R$; but for $\vert z \vert \le R$, the function is bounded since it is a continuous function on the compact set $z \le R$. Is that clear? – Robert Lewis Feb 28 '19 at 02:30
  • The disk $\vert z \vert \le R$ isn't punctured, since $\forall z, f(z) \ne w$; we don't need to remove $w$ to ensure $(f(z) - w)^{-1}$ is holomorphic. – Robert Lewis Feb 28 '19 at 02:35
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    Ok, your argument sounds right. But I just don't feel very confident. It is possible for f(z) to go arbitrarily close to w right? How did we rule that out? – Fluffy Skye Feb 28 '19 at 04:25
  • We want to show that $f(\Bbb C) = \Bbb C$; so my proof is by contradiction, right? So I look at what must happen if $\forall z, ; f(z) \ne w$; then we can't have $f(\Bbb C) = \Bbb C$; If fact, inside the disk of radius $R$, $\vert f(z) - w \vert$ will be bounded below, by compactness; and outside obviously we can make $f(z) - w$ get as big as we want. So there is some real $\epsilon$ with $\forall z, ; \vert f(z) - w \vert > \epsilon$, which may be small but it is fixed. So I would say $f(z)$ *can't* get arbitrarily close to $w$. – Robert Lewis Feb 28 '19 at 04:33
  • You're right. What troubles me is $e^z$. I know it doesn't satisfy $\lim_{z\to \infty}e^z=\infty$. But it can get arbitrarily close but can never get to $0$. By Picard Theorem we know non-constant entire function achieve all but maybe except one value in $\mathbb{C}$. This problem is saying $\lim_{z\to \infty}f(z)=\infty$ will rule out cases like $e^z$. Is my understanding correct? – Fluffy Skye Feb 28 '19 at 04:46
  • The condition $f(z) \to \infty$ as $z \to \infty$ is pretty strong, as the other answers illustrate; $f(z)$ has to be a polynomial. In the present case this condition on $f$ implies $f(z) - w$ *can't* approach $0$ on a compact set. – Robert Lewis Feb 28 '19 at 04:51
  • I see. Thanks for your explanation! – Fluffy Skye Feb 28 '19 at 04:52
  • No problem. And thanks for re-accepting my answer! Cheers! – Robert Lewis Feb 28 '19 at 04:53
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The statement $\lim_{z\to\infty}f(z)=\infty$ means $|f(z)|\to \infty$ as $|z|\to\infty$. It can be seen as convergence to $\infty \in\Bbb C_\infty$ where $\Bbb C_\infty$ is a one-point compactification of $\Bbb C$. In fact, this property implies that $f(z)=p(z)$ for some non-constant polynomial $p$. By fundamental theorem of algebra, it follows $f(\Bbb C)=\Bbb C$. To see this, let $ g(z)=f(\frac1{z}) $ for $z\ne 0$. Since $\lim_{z\to 0}|g(z)|=\infty$ (which is true because as $z\to 0$, $1/z \to \infty$ and $g(z)=f(1/z)\to\infty$ by the assumption), it follows that $g$ has an $n$-th pole at $0$, i.e. for some $n\ge 1$, $a_{-n}\ne 0$ and for all $z\ne 0$, $$ g(z)=\sum_{k=-n}^\infty a_kz^k. $$ (Here, $g$ having a pole is a classical result. Since $1/g$ tends to $0$ as $z\to 0$, it has a removable singularity at $0$, which is also an $n$-th zero. It follows $1/g(z)=z^nh(z)$ for some $h\ne 0$ on a neighborhood of $0$. So $g(z) = z^{-n}\frac1{h(z)}$ has an $n$-th pole.) Thus $f(z)=g(\frac1{z})=\sum_{k=0}^n a_{-k}z^k +\sum_{i>0}a_{i}z^{-i}$, and since $f$ is entire, it follows $$f(z)=\sum_{k=0}^n a_{-k}z^k=p(z).$$

Myunghyun Song
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  • The statement right after $\lim_{z \to 0} |g(z)| = \infty$ is not that clear to me. It would require more arguments. – nicomezi Feb 25 '19 at 07:39
  • Thank you but I was rather talking about what is after that limit (the fact that $g$ has an $n$-th pole). I am not saying it is false. Since we have the limit on $f$ at $\infty$ it cannot be an essential singularity. But it seems quite hard (compared to others answers) to write this rigorously. – nicomezi Feb 25 '19 at 07:44
  • @nicomezi I've added some explanation except for the result about removable singularity (Riemann's theorem is referred to.) – Myunghyun Song Feb 25 '19 at 07:55
  • Thank you for the edit. – nicomezi Feb 25 '19 at 08:00
  • Question about the expansion of $g$. I know if $g$ has an order $n$ pole at $w$. Then in a small neighborhood of $w$ we can write $g$ as a series $\sum_{k=-n}^{\infty}a_k (z-w)^k$. But in this problem how do we know that this series agrees with $g$ everywhere? – Fluffy Skye Mar 01 '19 at 01:03
  • @FluffySkye To add some more details, it can be seen this way; we have that $g(x)-\sum_{k<0}a_k z^k =\sum_{k\ge 0}a_k z^k$ on a neighborhood of $0$. The LHS is an entire function since we removed the singularity ($n$-th pole) from $g$ and the RHS is the power series representation of this entire function. Now, we can observe that the $=$ holds on $\Bbb C$. – Myunghyun Song Mar 01 '19 at 17:04
  • @Song I see! Thanks! – Fluffy Skye Mar 01 '19 at 20:31
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Let $g(z):=f(1/z)$ for $z \ne 0$. Then $g$ has at $0$ an isolated singularity. Since $g(z) \to \infty$ as $z \to 0$, $g$ has a pole at $0$.

Let $f(z)=\sum_{n=0}^{\infty}a_nz^n$, then the Laurent expansion of $g$ at $0$ is given by

$g(z)=\sum_{n=0}^{\infty}a_n \frac{1}{z^n}$, which shows that there is $m$ such that $a_n=0$ for $n>m$.

Hence $f$ is a non- constant poynomial. Now use the fundamental theorem of algebra.

Fred
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