# Analytic functions of a real variable which do not extend to an open complex neighborhood

Do such functions exist? If not, is it appropriate to think of real analytic functions as "slices" of holomorphic functions?

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If $f(x)$ is analytic at a point $a \in \mathbb R$, then its Taylor series has a non-zero radius of convergence, say $R$, at $a$, and so it actually converges in the disk of radius $R$ around $a$ in $\mathbb C$. Thus $f(x)$ always extends analytically to some complex n.h. of $a$. Hence we can find some complex open n.h. of the domain of $f$ over which it extends.
So yes, it is reasonable to think of a real analytic function as a restriction to $\mathbb R \cap U$ of a real analytic function on a complex open set $U$.
If $f$ is real analytic on an open interval $(a,b)$. Then at every point $x_0\in (a,b)$, there is a power series $P_{x_0}(x)=\sum_{n=0}^\infty a_n(x-x_0)^n$ with radius of convergence $r(x_0)>0$ such that $f(x)=P_{x_0}(x)$ for all $x$ in $(a,b)\cap \{x:|x-x_0|<r\}$. Then $f$ can be extended to an open neighborhood $B(x_0,r(x_0))$ of $x_0$ in $\mathbb{C}$ by the power series $P_{x_0}(z)$. Now let $O\subset \mathbb{C}$ be the union of these open balls $B(x_0,r(x_0), x_0\in (a,b)$. Define $F(z), z\in O$ such that $F(z)=P_{x_0}(z)$ if $z\in B(x_0,r(x_0))$ (for some $x_0\in (a,b)$). This is well-defined since any two analytic functions agreeing on a set with accumulation points in a connected open set must be identically equal. So $F$ is an extension of $f$ to an open set in $\mathbb{C}$ containing $(a,b)$. Since every open set in $\mathbb{R}$ is a countable union of disjoint open intervals, $f$ can be so extended if its domain is open in $\mathbb{R}$.