Let $n\in\mathbf{N}$ be fixed, and $f$ entire and $|f^{-1}(\left\lbrace w\right\rbrace)|\leq n$ for every $w\in\mathbf{C}$. Then $f$ is a polynomial of degree at most $n$.

I try to prove this statement, and I think one can prove it as follows: consider $f(1/z)$. $0$ cannot be an essential singularity of $f(1/z)$, for the big Picard theorem would imply that on any neighborhood of $0$ $f(1/z)$ takes on all possible complex values (with at most one exception) infinitely often, but this is contrary to $|f^{-1}(\left\lbrace w\right\rbrace)|\leq n$. Then $f(1/z)$ has a pole of order $k$ say, and since $f$ is holomorphic it is a polynomial of degree $k$ (for its principal part vanishes). By the fundamental theorem of algebra, $k\leq n$.

Is it possible here to avoid using Picard's theorem?

  • 2
    $\begingroup$ Please what is $|f^{-1}(w)|$? Is it the number of elements in $f^{-1}(w)$? $\endgroup$ – Julien Jan 26 '13 at 22:58
  • $\begingroup$ @mrf It's 100% what julien says. $\endgroup$ – k.stm Jan 26 '13 at 23:02
  • $\begingroup$ Indeed, $|f^{-1}(w)|$ is the number of $z\in\mathbf{C}$ such that $f(z)=w$. I thought that writing $|X|$ for the cardinality of a set $X$ is standard notation, sorry. By the way, I think my argument is somewhat a bit flawed, but I'm not sure. $\endgroup$ – user55315 Jan 26 '13 at 23:05
  • $\begingroup$ Fairly standard indeed! I was just making sure. $\endgroup$ – Julien Jan 26 '13 at 23:07
  • $\begingroup$ Anyway, I changed the notation to $|f^{-1}(\left\lbrace w\right\rbrace)|$. I think the notation should be obvious now. $\endgroup$ – user55315 Jan 26 '13 at 23:18

Here's a solution avoiding big Picard, but using Casorati-Weierstrass and Baire instead.

Assume that $f$ is not a polynomial. Then $f$ has an essential singularity at $\infty$. Let $k \in \mathbb{N}$ and let $D_k = \{ |z| > k \}$. By Casorati-Weierstrass, $f(D_k)$ is dense in $\mathbb{C}$, and by the open mapping theorem, $f(D_k)$ is open.

Baire's category theorem shows that $\bigcap_k f(D_k)$ is non-empty, so there is some $w \in \mathbb{C}$ (in fact, an open dense set of $w$:s) such that the equation $f(z) = w$ has infinitely many solutions, which is our desired contradiction.

  • $\begingroup$ Beautifully simple proof, +1. $\endgroup$ – zhw. Jun 20 '18 at 21:17

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