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If $f(z)$ is an entire function of a single complex variable, then the following are indirect methods for recognizing that $f$ is a polynomial.

1) Show that $f^{(n)}\equiv0$ for some $n\geq0$.

2) Show that $\displaystyle\lim_{z\to\infty}f(z)$ exists.

I suppose that the first one could be adopted to several variables, but I do not think the second one can be. Does anyone know other similar methods?

EDIT: I would also be interested in an answer to the same question with "rational function" subbed in for "polynomial".

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  • $\begingroup$ Aside: $(2)$ should be simply that the limit exists: if $f$ is a constant polynomial, the limit won't be $\infty$. $\endgroup$
    – user14972
    Sep 5, 2014 at 17:45
  • $\begingroup$ @Hurkyl Thanks, correction made. $\endgroup$ Sep 5, 2014 at 19:12

1 Answer 1

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You can use the fact that an entire function in several complex variables is a polynomial if and only if it is separately polynomial in each coordinate direction (i.e. when restricting to the coordinate complex lines with all coordinates but one fixed). (A similar fact also holds for rational functions.)

Then your condition 2 (or 1) can be restated to hold along each coordinate line.

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  • $\begingroup$ Actaully, remarkably, condition 2 cannot be so restated. In an answer here :(math.stackexchange.com/questions/1468743/…), a paper (American Mathematical Monthly 114 (2007), David Armitage: "Entire functions that tend to zero on every line") is referenced with an entire function which approaches zero along every line. I suspect there might similarly be an entire non-polynomial function which approaches $\infty$ along every line. $\endgroup$ Feb 7, 2017 at 19:45
  • $\begingroup$ @TrevorRichards That answer deals with real lines, not complex. $\endgroup$ Feb 15, 2017 at 16:46
  • $\begingroup$ Do you have a reference about the fact that a function is rational if and only if it is rational separately on each variable? I have tried to prove it but failed. $\endgroup$ Oct 6, 2023 at 19:19

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