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

Solution 1:

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 $-i$.