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Does Hartogs's extension theorem hold if one replaces the word holomorphic by analytic (of course still in several variables)? For Hartogs's Extension Theorem see here:'_extension_theorem

If yes, is there any reference ?

Thanks bernard

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In several complex variables, as in a single variable, all the various possible definitions of "holomorphic", including "analytic", are proven equivalent. There is even the further theorem of Hartogs, that separate analyticity (expandability in convergent power series) in each variable implies analyticity in several variables. – paul garrett May 12 '13 at 14:05
is there any reference that it actually works? – bernard May 12 '13 at 14:09
The argument that holomorphic = analytic is standard--you use Cauchy's Integral formula one variable at a time. The only difference between the one- and multi-variable cases is that you have more complicated polynomials showing up. (Products of the polynomials you see in the 1-D case.) A standard reference is Griffiths and Harris. – Hiro Lee Tanaka May 12 '13 at 14:21

When you use the word analytic, do you mean (1) meromorphic, (2) holomorphic, (3) real analytic, or (4) something else? If (1), this is known as the Levi extension theorem, (2) the Hartogs extension theorem, (3) this is false, (4) you need to be more specific. If you want proof that analytic functions are holomorphic, that depends on how you define analytic. If you define analytic as being solutions to the Cauchy--Riemann equations, and holomorphic as being given locally by convergent complex Taylor series, every book on several complex variables (for example, Krantz, Function Theory of Several Complex Variables, section 1.2, p. 29) proves the equivalence. Krantz also considers several other possible definitions of the word holomorphic and proves equivalence of all of them.

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are there any theorems which treat the extension of some real-analytic function (of several variables) in a single point ? – bernard May 13 '13 at 5:50
No. Think about $f(x_1^2+x_2^2+\dots+x_n^2)$ where, for example, $f(t)=e^{1/t}$. The singularity at a single point can be as bad as you like. Maybe if your function satisfies a differential equation, you might have a chance to use some regularity theorem. – Ben McKay May 13 '13 at 6:35

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