Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have another question that is really catching me off guard, but it looks very promising and wholesome as it combines complex analysis theory and algebra. The question is as below:

Let $f$ be entire and have the property that if $B \subset \mathbb{C}$ is any bounded set,$\hspace{1.7in}$ then $f^{-1}(B)$ is bounded (or perhaps empty). Could it be shown that for any $\omega \in \mathbb{C}$, there $\hspace{0.7in}$ exists $z \in \mathbb{C}$ such that $f(z)=\omega$. Also, can one show that $f(\mathbb{C})$ is both open and $\hspace{1.1in}$ closed and deduce that $f(\mathbb{C})=\mathbb{C}$. Apply this result to polynomials to deduce $\hspace{1.4in}$ yet another proof of the Fundamental Theorem of Algebra.

share|cite|improve this question
Couldn't we user Picard's theorem to solve this one? Since $f$ is entire, if $f$ is not onto, and not constant, there exists precisely one element in $\Bbb{C} \setminus f(\Bbb{C})$. How does this contradict the assumption that the preimage of any bounded set is bounded? – Beni Bogosel Jun 2 '11 at 10:35
up vote 4 down vote accepted

I'm not sure what theorems you've seen, but if you know the open mapping theorem, then you know that $f(\mathbb C)$ is open. Let $(z_n)$ be a sequence in $f(\mathbb C)$ converging to $z\in\mathbb C$. Then there is a sequence $(w_n)$ with $f(w_n)=z_n$. Since convergent sequences are bounded, $(z_n)$ is a bounded sequence. Since the sequence $(w_n)$ lies in $f^{-1}(\{z_n\})$, $(w_n)$ is also a bounded sequence, and therefore there is a subsequence $(w_{n_k})_k$ converging to some $w\in\mathbb C$. By continuity of $f$, $f(w)=\lim_k f(w_{n_k})=\lim_k z_{n_k}=z$. Therefore $f(\mathbb C)$ is closed. Since $\mathbb C$ is connected, it follows that $f$ is onto.

If $f$ is a nonconstant polynomial, then $\lim_{z\to\infty}f(z)=\infty$, which implies that $f$ satisfies the hypothesis. Therefore $f$ is onto, and in particular $0$ is in its image, which is the fundamental theorem of algebra.

An alternative solution to the original problem that is probably overkill and actually uses the fundamental theorem of algebra goes as follows. If $f$ is a nonconstant polynomial, then it is onto by the fundamental theorem of algebra. If $f$ is an entire function that is not a polynomial, then it has an essential singularity at $\infty$, and hence $f(\{z:|z|>M\})$ is dense in $\mathbb C$ for all $M>0$ by the Casorati-Weierstrass theorem. In particular, $f^{-1}(\{z:|z|<1\})$ intersects $\{z:|z|>M\}$ for all $M>0$, and therefore it is unbounded. (I guess this is a good approach if for some reason you want to avoid using the open mapping theorem.)

Edit: Removed remark about assuming that $f$ is nonconstant, since that is implied by the hypothesis as Theo pointed out.

share|cite|improve this answer
$f$ is of course not constant because $f^{-1}(c)$ would be unbounded. – t.b. Jun 2 '11 at 9:01
@Theo: Yes, of course, thanks! – Jonas Meyer Jun 2 '11 at 9:05
+1$\ldots$ Very nice reasoning and like the approach you taken. The open mapping theorem article was very good. Thanks. – night owl Jun 2 '11 at 9:17
@night owl: Glad you like it. Note that the first approach I gave is similar to Theo's (and his answer was posted a half a minute before mine). – Jonas Meyer Jun 2 '11 at 9:24
We seem to say the exact same things in different words at more or less the same time today... (see my comment to my answer) – t.b. Jun 2 '11 at 9:28

Such a map $f$ is called proper. It is easy to show (using local compactness of $\mathbb{C}$, continuity of $f$ and compactness of $f^{-1}(K)$ for each compact $K$) that $f$ maps closed sets to closed sets and since $f$ is not constant it maps open sets to open sets by the open mapping theorem. Hence it does what you want: the image of $f$ is both open and closed and of course non-empty in $\mathbb{C}$, hence it is all of $\mathbb{C}$ by connectedness.

Using the growth lemma for polynomials you can easily show that non-constant polynomials are proper, hence are onto. I leave the details for you as they are rather straightforward.

share|cite|improve this answer
Thanks, I like the article on the proper map. The proof is very nice and understandable. I am back to read over it after this comment. Good approach. – night owl Jun 2 '11 at 9:21
@night owl. If you work out the details, you'll see that it is actually the same argument as Jonas's in the end. I'm just using words instead of (more explicit) formulas :). On the other hand, I give an explicit reason why polynomials are proper by referring to the growth lemma. I guess it's a matter of taste and I like both ways of putting it. Jonas's answer is really good and to the point, as usual. – t.b. Jun 2 '11 at 9:24
Yea I looked at your work for the growth lemma. It was nicely constructed. Good work. – night owl Jun 3 '11 at 7:14

Building on my comment:

Picard's theorem states that an entire function which misses more than two values is constant. Our function is clearly not constant, as mentioned before. Suppose it is not onto. Then there exists only one element $z \notin f(\Bbb{C})$. Pick a disk $D=D(z,r)$ centred in $z$. Since $D$ is bounded, it follows that $f^{-1}(D)$ is bounded. There exists $z_n \in D,\ z_n \to z$ for which there exist $w_n$ with the property that $f(w_n)=z_n$. The sequence $w_n$ is bounded, therefore it has a convergent subsequence $(w_{n_k})\to w$ and finally $f(w)=z$. Contradiction. This means that $f$ is onto.

This is a valid argument, but clearly, not as simple as the first one.

share|cite|improve this answer
Yes, a valid argument, but "clearly, not as simple" is clearly an euphemism :) To get a bit more feeling what is going on (even if it doesn't seem to yield an argument for answering the question), I think it is good to remember that an entire map is proper iff it is a polynomial (by Casorati-Weierstrass and the other two answers). Since non-constant polynomials are onto, an $f$ leaving out one value of $\mathbb{C}$ must be (either constant or) non-proper. An example is $z \mapsto e^z$ for which non-properness is clear. (I'm sure that all this is clear to you, I'm adding it for night owl). – t.b. Jun 2 '11 at 16:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.