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A function $f(x)$ is an real function and analytic in an open interval of $x$-axis or the whole $x$-axis. Is there only unique way to analytically extend it to the whole complex plane?

I know identity theorem for holomorphic functions, but it requires that two functions equal in an open and connected set, while here I only need they equal in an open interval of x-axis which is a closed set of complex plane.

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  • $\begingroup$ Analytic continuations are unique if they exist. $\endgroup$
    – Rammus
    Nov 23, 2014 at 6:20
  • $\begingroup$ Look at the identity theorem for holomorphic functions. $\endgroup$ Nov 23, 2014 at 6:21
  • $\begingroup$ @PeterBrown I know this theorem. But this need two functions equal in an connected open set. While a open interval of x-axis is not an open set of complex plane. $\endgroup$
    – 346699
    Nov 23, 2014 at 6:38
  • $\begingroup$ @JeremyDaniel I know this theorem. But this needs two functions equal in an connected open set. While an open interval of x-axis is not an open set of complex plane. $\endgroup$
    – 346699
    Nov 23, 2014 at 6:39
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    $\begingroup$ Any two analytic functions on the same domain that agree on a set that has a limit point in the domain are equal. $\endgroup$ Nov 23, 2014 at 8:10

1 Answer 1

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Given any real-analytic function $f: (a, b) \to \mathbb{R}$ (allowing $(a, b)$ to be half-infinite or infinite) and a point $x_0 \in (a, b)$, one can compute the unique extension of $f|_{(x_0 - r, x_0 + r)}$ to the ball $B_r(x_0) \subset \mathbb{C}$ for any $r$ no larger than the radius of convergence of the Taylor series of $f$ at $x_0$ (and small enough that $(x_0 - r, x_0 + r) \subseteq (a, b)$) using the Taylor series of $f$. So, given any two complex-analytic extensions of $f$ to $\mathbb{C} \to \mathbb{C}$, they agree on such a ball and by the Identity Principle must agree everywhere. In this sense, the answer to your question is yes.

In general, however, a real-analytic function $\mathbb{R} \to \mathbb{R}$ need not admit an analytic extension to a map $\mathbb{C} \to \mathbb{C}$. For example, consider $$ f(x) = \frac{1}{1 + x^2} ; $$ it is certainly a real-analytic map $\mathbb{R} \to \mathbb{R}$, but it admits no complex-analytic extension to any domain containing $+i$ or $-i$.

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  • $\begingroup$ Is the class of real-valued functions extensible to entire functions a good class to "approximate" the class of analytic functions? And for infinite differentiable functions? $\endgroup$
    – Canjioh
    Dec 9, 2020 at 21:24
  • $\begingroup$ What sense of "approximation" do you have in mind? $\endgroup$ Dec 16, 2020 at 2:58
  • $\begingroup$ I mean, for instance, something which could be linked to density arguments. $\endgroup$
    – Canjioh
    Dec 24, 2020 at 16:10
  • $\begingroup$ I'm not sure. But there are strong density results for even smaller classes of functions, e.g., the Stone-Weierstrass Theorem, which applies to the class of polynomials. $\endgroup$ Dec 26, 2020 at 2:23

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