From: israel@math.ubc.ca (Robert Israel) Subject: Re: Complex analysis Date: 7 Oct 1999 20:19:18 GMT Newsgroups: sci.math Keywords: Hartogs theorem (several complex variables) In article <051019991308425915%edgar@math.ohio-state.edu.nospam>, "G. A. Edgar" writes: > In article , Eric Brunet > wrote: > > I consider a function > > x -> f(x,c) > > which is analytical for any value of the real parameter c. > > Furthermore, I suppose that for any x I can make an expansion of f(x,c) > > for small c: > > f(x,c) = f_0 (x) + c f_1(x) + c^2 f_2(x) + ... > > Are the functions x -> f_k(x) analytical if the expansion is c has a > > non-zero radius of convergence ? Are these functions still analytical if > > the radius of convergence IS zero ? > Consider f(x,c) = x*c/(x^2+c^2) where f_1(x) is discontinuous at x=0. And Eric complained: > This function is not analytical: it is discontinuous at x=± i c. Perhaps what you want is the theorem of Hartogs: if f is defined in some open set G in C^n and is analytic in each variable separately, then f is continuous in G. Moreover (this is not really Hartogs but is a consequence of it), f can be expanded in a convergent multivariate Taylor series about any point of G. In particular your f_k(x), which are pieces of this multivariate Taylor series, would be analytic. Reference: L. H\"ormander, "An Introduction to Complex Analysis in Several Variables", Van Nostrand 1966. In Gerald's example, there is no neighbourhood of (0,0) in C^2 where f is defined. However, it fits one interpretation of your requirements: for each value of c (including 0), x -> f(x,c) is analytic in a neighbourhood of 0, and for each value of x (including 0), c -> f(x,c) is analytic in a neighbourhood of 0 (and thus has an expansion in powers of c with a non-zero radius of convergence; this radius, however, depends on x). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2