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I was reading the answer to this question:

on the boundary of analytic functions

This answer makes sense to me up until the last line. What does having isolated zeros have to do with $f$ being identically $0$?

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user235654
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1 Answers1

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Since $g(z)$ is identically zero, at each $w$ at least one of the $f(e^{\frac{2\pi ij}{n}}w)$-s must also be zero. Suppose $g(w) = h(w)\ell(w)$, where $h(w)$ is the product of the $f_j(w) = f(e^{\frac{2\pi ij}{n}}w)$ that are zero at $w$, and $\ell(w)$ the product of the $f_j$ that are nonzero at $w$. Then by continuity, $\ell(w) \neq 0$ in a neighborhood of $w$, so $h(w)$ must be identically zero in this neighborhood. But then $h(w)$ is analytic with nonisolated zeroes, so it must be identically zero.

Now assuming $h$ is a product of strictly fewer factors $f_j$ than $g$, you can repeat this argument on $h$, and inductively continue until you conclude that one of the factors $f(e^{\frac{2\pi ij}{n}}z)$ is analytic with nonisolated zeroes, which is to say $f(z)$ has nonisolated zeroes.

If $h$ is a product of the same number of $f_j$ as $g$, then move to a different $w$ to get strictly fewer factors; if there is no such $w$, then every $f_j$ is identically zero.

Gyu Eun Lee
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  • This is overly complicated. You have infinitely many zeros of a product. And finitely many factors. At least one factor has infinitely many zeros. That factor is identically zero by the identity principle. And that factor is a rotation of $f$. So $f$ is identically zero. That argument with the induction on the number of factors is unnecessary. – Alamos Apr 30 '15 at 04:11
  • I don't think my argument is particularly complicated...but I do like yours, I hadn't thought of it. If you don't like my argument that's ok, but I suggest you post your proof as an answer rather than using it as a belittling comment on my own. – Gyu Eun Lee Apr 30 '15 at 04:13
  • @neuguy : It is alleged on meta that you cannot receive any notification of my comments at the URL below. Is that true? http://math.stackexchange.com/questions/1258577/error-function-etymology-why-the-name ${}\qquad{}$ – Michael Hardy Apr 30 '15 at 22:34
  • @MichaelHardy I don't see any of your comments on that question...is there a reason you're asking me in particular? This seems rather random. – Gyu Eun Lee May 01 '15 at 05:56
  • @neuguy : I addressed each person who participated in closing the question. You were one of those. ${}\qquad{}$ – Michael Hardy May 01 '15 at 17:38