3
$\begingroup$

A problem I'm working on requires the application of Cauchy's estimate for the modulus of the coefficients of a holomorphic function's power series representation, but the original functions with which I'm working are real analytic (sometimes in one variable, sometimes in several). I've located in-several-variables analogues of Cauchy's estimates, but I'm searching for theorems that address the extension of real analytic functions in several variables to holomorphic functions.

I've come across the following notes on Wikipedia and the Encyclopedia of Mathematics that address the univariate and multivariate cases, respectively:

If ƒ is an infinitely differentiable function defined on an open set D ⊂ R, then the following conditions are equivalent.

1) ƒ is real analytic.

2) There is a complex analytic extension of ƒ to an open set G ⊂ C which contains D.

(Wiki: Analytic Function)

Similarly, for any real analytic function f:U→ℝ and any $x_0 ∈ U$ there is an open neighborhood W in $ℂ^n$ of $x_0+0i$ and an holomorphic map $g:W→ℂ$ such that g coincides with f on $W \cap \{z:Im(z)=0\}$.

(Encyclopedia of Mathematics: Real Analytic Functions)

Unfortunately, searching through the references given at both pages hasn't turned up a name or theorem-to-cite for their claims, nor has digging through a number of texts/articles on real analytic functions or complex analysis in several variables. Is anyone familiar with an authoritative reference for these facts, or a name associated with either of them?

$\endgroup$
  • 5
    $\begingroup$ Isn't this just "positive radius of convergence at each point", or am I missing something? $\endgroup$ – Yemon Choi Aug 14 '15 at 6:16
  • 3
    $\begingroup$ I think that's the case in the one-variable setting, but in the multivariate setting most of the references I'm seeing are to Reinhardt domains rather than high-dimensional spheres. If there's an analogue for the root test (or some similarly general convergence result) in $\mathbb{C}^n$ (?), I think that would work though. $\endgroup$ – NeverConvex Aug 14 '15 at 6:32
  • 1
    $\begingroup$ This note seems to address my question exactly in Proposition 1.19, and supplies enough exposition that I think I can follow through his argument for its proof. $\endgroup$ – NeverConvex Aug 14 '15 at 6:54
  • 1
    $\begingroup$ Surely polydisks are enough, even if they aren't the largest possible domain? $\endgroup$ – David Roberts Aug 14 '15 at 7:13
  • $\begingroup$ One thing that distinguishes functions of a single complex variable from those of several is that the region in which a power series converges has no simple canonical form in more than one variable, and certainly the region need not be a ball or polydisk, with the usual example $\sum_n (zw)^n$ in two variables. If one tries to address this issue, one will indeed run across Reinhardt domains and such, but that's not really necessary for basic convergence properties: a polydisk will suffice, I gather. $\endgroup$ – paul garrett Aug 15 '15 at 14:51
3
$\begingroup$

You have to state a specific problem to get a reasonable answer. The proof of the statements you cite is trivial. A function is called analytic at $x_0$ if in some neighborhood of $x_0$ it is represented by its Taylor series at $x_0$. Here is does not matter whether $x_0$ and the neighborhood are real or complex.

But if a power series at $x_0$ is convergent at ANY point other than $x_0$, then it is automatically convergent is some COMPLEX disk around $x_0$.

This proves both statements that you cite. Now you are probably interested in the radius of the disk (or polydisk) in which it converges. One way to say something about it is to look how large the derivarives of $f$ at $x_0$ are. But I guess this you do not now. Then you need some information on how large is the complex neighborhood to which $f$ extends. For this you have to know something about your function.

So be more specific: what information about $f$ is available, and what you want to prove.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.