Analytic/meromorphic continuation is a difficult problem in general. For "motivic L-functions", the idea of proving their analytic continuations by first proving their modularity goes back, I guess, to Riemann.

Here I just want to ask a purely complex-analytic question. Let's restrict ourselves to the case of one variable functions. Let $U$ be a region in the complex plane, and let $f$ be a holomorphic function on $U.$ Is there any criterion for $f$ to have analytic continuation to a larger region? And what is this "maximal domain of regularity"?

Feel free to assume $U$ and $f$ to have the shape you like, e.g. a power series on an open disk or a Dirichlet series on some half plane. I guess even if $U$ is an annulus or a punctured disk, where one can compute (theoretically or numerically) the value of the extended function (if exists) at the points inside the inner loop by Cauchy's formula, it is still difficult to decide if this extension is continuous or analytic.

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    $\begingroup$ I doubt that there is or can be a criterion (for the existence of morphic continuation) of the generality you are looking for. Experience teaches us that one usually needs additional pieces of information like functional equation or integral representation 'ready' for regularization, or like in the case of the monodromy theorem, that there exists analytic continuation along the paths of curves in a simply connected domain containing $U$. $\endgroup$
    – M.G.
    May 27, 2010 at 1:40
  • $\begingroup$ Hi Shenghao, for functions defined on open simple connected subset of $\mathbb C$, I heard once, that the analytic continuation can be made until the domain get surrounded by a dense set o singularities. (Perhaps someone else could give us a precise statement regarding to this fact) Even supposing this fact is true, for some class of holomorphic functions, It sounds for you as an acceptable criterion ? I am asking because frequently will be very hard to verify it. But if you have any interest on this kind of statement, I can look for you the details. $\endgroup$
    – Leandro
    May 27, 2010 at 3:07
  • $\begingroup$ To Leandro: Thanks, and yes please, I'm interested in this. Thank you in advance. $\endgroup$
    – shenghao
    May 27, 2010 at 4:38

3 Answers 3


Well, in case of power series some criterions do exist. Roughly speaking, one can take the element $$ \sum\limits_{n=1}^{\infty}c_nz^n,\quad z < 1,\label{1}\tag{$\ast$}$$ and consider an analytic function $\phi$, such that $\phi(n)=c_n$ for every $n$. The element \eqref{1} can be analytically extended onto some angular domain iff $\phi(z)$ has finite exponential growth.

Let $E\subset \mathbb C$ be a closed unbounded domain and let $H(E)$ denote the set of functions such that each of them is analytic in a neighborhood of $E$. For a function $\phi\in H(E)$, the exponential type of $\phi$ on $E$ is defined as $$\sigma_\phi(E)=\limsup\limits_{z\to\infty,\\ z\in E}\frac{\log^+|\phi(z)|}{|z|}.$$ The following result is due to LeRoy and Lindelöf.

Theorem 1. Let $\Pi=\{z\in\mathbb C|\ \Re z \geq 0\}$. Assume that $\phi\in H(\Pi)$ is of finite exponential type $\sigma<\pi$. Then the series $$f(z)=\sum\limits_{n=1}^{\infty}\phi(n)z^n$$ can be analytically extended onto the angular domain $\{z\in\mathbb C| \ |\arg z|>\sigma\}$.

The LeRoy-Lindelöf theorem gives only a sufficient condition. A criterion can be obtained if we relax a bit the condition that $\phi$ is of finite exponential type.

Let $\Omega=\{z\in\mathbb C|\ \Re z > 0\}$ be the interior of $\Pi$. An analytic function $\phi\in H(\Omega)$ is said to be of (finite) interior exponential type iff $$\sigma_\phi^\Omega=\sup\limits_{\{\Delta\}}\sigma_\phi(\Delta)< \infty,$$ where $\{\Delta\}$ is the set of all closed angular domains such that $\Delta\subset \Omega\cup \{0\}.$

Theorem 2. The element $$\sum\limits_{n=1}^{\infty}c_nz^n,\quad z < 1,$$ can be analytically extended onto the angular domain ${{{{}}{}}}\{z\in\mathbb C| \ |\arg z|>\sigma\}$ for some $\sigma\in[0,\pi)$ iff there is a function $\phi\in H(\Pi)$ of interior exponential type less or equal to $\sigma$ such that $$c_n=\phi(n),\quad n=0,1,2,\dots.$$

You might be interested in this article.



I would say analytic continuation is, in many cases, almost like black magic!

These problems are very, very hard. There is probably no sensible general answer. Not surprisingly, since solving the Riemann Hypothesis is trivially equivalent to the question of whether $1/\zeta$ has a continuation to $\{ \mathrm{Re}(z) > 1/2 \}$ or not!

Analytic continuation is very ill-posed: approximating a function $F$ by a sequence $F_n$ usually is of no use at all in determining the domain you can extend $F$ to. So you usually need exact formulae for your functions, expressed as infinite series, products, integrals or something else; numerical computations are almost certainly useless for these problems.

Often, existence of a continuation depends on cancellation in very complicated oscillating series or integrals, so you rarely have nice things like absolute convergence (or even conditional convergence!) on the true domain of the function. It can be very difficult to analyse the formulae you get.

Unfortunately you probably will have to do lots of detailed algebraic calculations with the specific functions you are considering. Cauchy's theorem for functions expressed using contour integration is about the only general method I can think of, although it's not always possible.

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    $\begingroup$ +1 for "Often, existence of a continuation depends on cancellation in very complicated oscillating series or integrals..." ; it cannot be emphasized enough. $\endgroup$ Aug 5, 2010 at 1:12

To answer your specific questions, a) a maximal (by inclusion) domain does not have to exist. Consider $\sqrt{1-z}$ in the unit disk. That is actually the reason why Riemann introduced Riemann surfaces.

But there are maximal domains in various other senses. Suppose you have an analytic germ at infinity. Then there is a unique (!) set $K$ in the plane of of minimal logarithmic capacity, such that our germ has a (single-valued) analytic continuation to $C\backslash K$. This is due to H. Stahl.

b) The answer depends on what you exactly mean by a criterion. There are necessary and sufficient conditions for existence of an analytic continuation (multiple valued) to a given region. Not surprisingly they are difficult to check. See, for example MR1711032 Atzmon, A., Eremenko, A., Sodin, M. Spectral inclusion and analytic continuation. Bull. London Math. Soc. 31 (1999), no. 6, 722–728.

There is a necessary and sufficient condition for a function defined in the unit disk to have an analytic continuation to a given point on the circumference (Euler).

And there are many separate necessary (but not sufficient) and sufficient (but not necessary) conditions in terms of coefficients in various special cases. This is a very vast subject. A good survey of these conditions can be found in Bieberbach, Analytische Fortsetzung, Springer, 1955, and V. Bernstein, Lecons sur les progres recents de la théorie des séries de Dirichlet, Paris: Gauthier- Villars. XIV, 320 S. (1933).


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