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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 $U$.

This point of view is important in the more advanced theory of such functions (e.g. in the theory of hyperfunctions and microlocal analysis).

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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 $\mathbb{R}$.

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