# Show that an entire function is a proper if and only if it is a nonconstant polynomial

Show that an entire function (Holomorphic on $\mathbb C$) is proper if and only if it is a non constant polynomial.

Def:A map $f:X\to Y$ is called proper if $f^{-1}(K)$ is compact for every compact set $K$ in $Y$.

Clearly every non constant polynomial is an entire proper fuction.Also if $f: \mathbb C \to \mathbb C$ is an entire proper function then $f$ will be nonconstant otherwise $C$ is bounded.I am having problem in showing that $f$ is a polynomial.Please help.

• What does being proper imply for $f(z)$ as $z\to \infty$? What does that then imply about the type of singularity that $f$ has at $\infty$? – Daniel Fischer May 25 '15 at 16:01
• essential singularity,right? – Dontknowanything May 25 '15 at 16:06
• If we want to show that $f$ is a polynomial, the singularity better be a pole. – Daniel Fischer May 25 '15 at 16:07
• @DanielFischer I am sorry,I don't see why it will have a pole at infinity. – Dontknowanything May 25 '15 at 16:10
• Do you know that an isolated singularity is a pole iff the function $\to \infty$ at that point? – zhw. May 25 '15 at 17:00

## 1 Answer

Suppose $f$ is proper. So it must not be constant. For any $a\in\mathbb{C}$, $f^{-1}(a)$ is isolated by Identity Theorem. By properness $f^{-1}(a)$ is a finite set. The cardinality $n$, counted with multiplicity, of this set (I.e. The degree of $f$) is independent of $a$ by continuity. Actually $n$ is the number of zeros (counted with multiplicity) of $f(z)-a$. So $f$ is a polynomial of degree $n$.

Edit: 1. By Rouche's theorem, \begin{eqnarray}n=\frac{1}{2\pi i}\int_{\gamma}\frac{f'(z)}{f(z)-a}dz\end{eqnarray} for large enough contour $\gamma$ which encloses all the zeros. Wiggling $a$ does not change $n$.

1. By the above, for any $a$, the order of zero of $f(z)-f(a)$ at $a$ is at most $n$. So $f(z)-f(a)=(z-a)^mh(z)$ for some $m\leq n$ and entire function $h(z)$ which is not zero at $z=a$. By residue theorem, \begin{eqnarray}\frac{1}{2\pi i}\int_{\gamma}\frac{f(z)-f(a)}{(z-a)^{n+2}}dz=0\end{eqnarray} On the other hand, the integral also equals $\displaystyle \frac{f^{(n+1)}(a)}{n!}$. So $f^{(n+1)}(a)=0$ for all $a\in\mathbb{C}$. Hence $f$ is a polynomial of degree $n$.
• Thank you for response.Can you please explain this: "The cardinality $n$, counted with multiplicity, of this set is independent of $a$ by continuity" – Dontknowanything May 25 '15 at 16:20
• $n$ is the number of zeros (counted with multiplicity) of $f(z)-a$. – Alex Fok May 25 '15 at 16:30
• why is that "independent" of a? – Dontknowanything May 25 '15 at 16:31
• Apply Rouche's theorem and continuity of the relevant integral with respect to $a$. – Alex Fok May 25 '15 at 16:35
• sorry i'm new in complex analysis therefore don't know Rouche's Theorem.Can you please add that in your answer?Sorry for troubling you for my silly doubts. – Dontknowanything May 25 '15 at 16:39