# Let $f$ be an entire function which takes every value no more than three times. What can it be?

Let $f$ be an entire function which takes every value no more than three times. What can it be?

Consider the singularity at infinity. If it is removable then $f$ is constant. If it is a pole then $f$ is a polynomial, and it's clear that degree of $f$ is less than or equal to three. Suppose $f$ has an essential singularity at infinity. The Big Picard theorem says that it takes every value except for possibly one (call it $\omega$) in a neighborhood of $\infty$. Let $w\neq \omega$ and take some value $z_1$ where $f(z_1)=w$. Then there is a neighborhood of infinity which does not include $z_1$, and so there must exist another point $z_2$ where $|z_2|>|z_1|$ and $f(z_2)=2$. By induction there must exist a sequence $z_n$ of distinct points with strictly increasing moduli where $f(z_n)=w$, for all $w\neq \omega$.

Edit: I suppose I should clarify why the degree of $f$ must be at most three. This should follow from the density of separable polynomials...suppose $f(z)$ is a polynomial of degree greater than three. If $f(z)$ is separable, then $f$ takes the value zero at $>3$ distinct points. If $f$ is not separable, then I want to say that there is a constant $b$ such that $f+b$ is separable, and then $f$ takes the value $-b$ at $>3$ distinct points. But I should make this rigorous...after all, separable polys may be dense, but how do we know that a set of inseparable polynomials is not parallel to the subspace spanned by the constant polynomials (so $f+b$ is inseparable for all $b\in \mathbb{C}$)? Any tips for this?

• You have basically proved your $f$ cannot have an essential singularity at $\infty$ and hence is a polynomial of degree one to three. What is your question???? – achille hui Jan 2 '14 at 1:40
• @achillehui Take a look now...what do you think? – Eric Auld Jan 2 '14 at 1:46
• Are you asking whether your proof is valid or how can it be improved? Betty's answer speak my mind and in a much better and coherent way. – achille hui Jan 2 '14 at 1:59
• About the last part, for any polynomial $g$ of degree $n > 0$ and $\omega \in \mathbb{C}$. If $$\omega \not\in \bigg\{ g(z) : z \in \mathbb{C},\;s.t.\;g'(z) = 0 \bigg\}$$ then $|g^{-1}(\omega)| = n$. Since the set of exceptional $\omega$ is finite. One can be sure if $\deg f > 3$, then it will take most values more than three times. – achille hui Jan 2 '14 at 2:25
• @achillehui Thank you, that is a better way to see it. And yes, the question was a "proof verification", as the tag indicates. – Eric Auld Jan 2 '14 at 2:41

First, did you mean there must exist another point $z_2$ ... such that $f(z_2)$ = w? You said 2.
You said if the singularity is a pole then f is a polynomial of degree 3. Can you explain why it couldn't be a polynomial like p(z) = $(x-1)^3(x-2)? Is 1 counted as 3 separate values? Could you reference a theorem? You took a w for$z_1$; I would have found it easier to follow if you could stated more clearly why w can't be omitted from any closer neighborhood of$\infty$. (I know big Picard does say that, but I have trouble visualizing neighborhoods of$\infty$-- maybe others do too?) Obviously if you can find a sequence of longer than 3 such that$f(z_n)$= w you have a contradiction; so f cannot have an essential singularity at$\infty$. At that point it helps if you specifically state that you have a contradiction and what it is; as well as stating that f must look like what? Do you mean a 3rd degree polynomial? Any old 3rd degree polynomial? Or as I asked above, could it be a special polynomial of higher degree? • If I may suggest, a polynomial$p$of degree$n\leq 3$cannot take any value$w_0$at more than three points simply because$p(z)-w_0$is another polynomial of degree$n$, which has at most$n\$ distinct roots. – Jonathan Y. Jan 2 '14 at 2:03