Talk:Hartogs's theorem on separate holomorphicity
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Does the result imply that the function is actually analytic or just continuous? --ComplexZeta
- Well, it implies continuity, and from there it is rather easy to get analyticity. Oleg Alexandrov (talk) 04:41, 21 June 2006 (UTC) [reply ]
Hey, I have never edited Wikipedia, but the conditions for the theorem are stated incorrectly, an additional requirement of boundedness is required, and domain is incorrect as well, it should be the ${|x_k-(x_k)_0| /leq B_k, k=1,2, ldots ,n}$. If the function is bounded in all of these regions, it holds in $/mathbb{C}^n$ as well, because of glueing. This correct version of this theorem is proven in "theory of analytic functions in several variables".
It should be added that this version of the theorem is actually incorrect, or at least not implied by ZFC. Given a countable model of ZFC, you can list the complex numbers in this model. If a function is determined in a finite amount of points, you are able to fit a polynomial through these points. If you add the additional requirement that $f(/frac{1}{n},/frac{1}{n})=n,ドル and find a polynomial for the first and second variable, you will construct a function that is actually a polynomial in every variable, but this two-variable function is of course not bounded, and therefore not analytic, in the region around the origin. — Preceding unsigned comment added by 94.215.136.128 (talk) 13:48, 22 September 2024 (UTC) [reply ]
Incomplete definition of function
[edit ]The function used as a counterexample is not defined at 0. I would fix it but I see the mistake actually comes from PlanetMath. -Set theorist (talk) 09:10, 25 November 2010 (UTC) [reply ]
The proper function in this case is defined piecewise as f(x,y) = xy(x^2-y^2)/(x^2 + y^2) except at the origin, where f = 0. --Moly 21:24, 6 January 2011 (UTC) — Preceding unsigned comment added by Moly (talk • contribs)