# Do real-analytic functions always extend uniquely to complex-analytic functions on $\mathbb{C}$?

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.

• Analytic continuations are unique if they exist. – Rammus Nov 23 '14 at 6:20
• Look at the identity theorem for holomorphic functions. – Jeremy Daniel Nov 23 '14 at 6:21
• @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. – 346699 Nov 23 '14 at 6:38
• @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. – 346699 Nov 23 '14 at 6:39
• Any two analytic functions on the same domain that agree on a set that has a limit point in the domain are equal. – Robert Israel Nov 23 '14 at 8:10

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 $-i$.