If I have an analytic function in plane $F(x,y)$ that is zero on a curve $y=f(x)$, is it true that $F=(y-f(x))^n h$, where $h$ is nonzero on the curve? More general, can be somethink said about factorisation of analytic functions? How much is it determined by its zero set? Thx
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$\begingroup$ You are using $f$ for two different things... $\endgroup$– Mariano Suárez-ÁlvarezOct 14, 2010 at 14:58
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$\begingroup$ Thank you very much. Does somethink like that hold also over the reals? Peter $\endgroup$– Peter FranekOct 14, 2010 at 20:38
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$\begingroup$ Yes it does.... $\endgroup$– Thierry ZellOct 14, 2010 at 20:45
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2 Answers
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The answer is Weirstrass preparation Theorem.
answered Oct 14, 2010 at 15:20
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You need a combination of Weierstrass preparation and Puiseux series expansion to factor the analytic function, but it is indeed possible. Keep in mind that this is a local factorization near a point of your choice, that the factors may be complex valued and singular (=Holder continuous) at the point, but they are analytic outside the point. Better than writing here a lengthy explanation let me point you at a paper where I wrote all the details since I could not find them in the literature, although this stuff must be well known. See Section 2 of this paper.
answered Oct 14, 2010 at 18:08
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$\begingroup$ sorry, but the link to the paper doesn't work. $\endgroup$– user111Feb 26, 2023 at 8:53
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$\begingroup$ It is on Communications in PDEs, Volume 26, Issue 5-6, Pages 779-811 $\endgroup$ Feb 26, 2023 at 8:58