I just learned that real analytic functions (by real analytic, I mean functions $f: \mathbb{R} \to \mathbb{R}$ which admit a local Taylor series expansion around any point $p \in \mathbb{R}$) cannot be extended to complex entire function always. I believe functions with this extension property are called real entire functions in some books, and a function which is real analytic, but not real entire is $f(x) = \frac{1}{1 + x^2}$. Clearly, with this example, the problem with any extension happens around $\pm i$. See also this MSE post.
My question is, do real analytic functions admit extensions to a complex analytic function even locally? That is, given a real analytic function $f : \mathbb{R} \to \mathbb{R}$, can we find a complex analytic function $g : \Omega \to C$, such that $g|_\mathbb{R} = f$, and $\Omega$ contains a strip $\mathbb{R} \times (-\varepsilon, \varepsilon)$ around the real axis?
Best Answer
Yes, a real analytic function on $\Bbb R$ extends locally to a complex analytic function, except that (in my opinion) "locally" doesn't/shouldn't mean what you say it does.
If $f$ is real analytic on $\Bbb R$ then there exists an open set $\Omega\subset\Bbb C$ with $\Bbb R\subset\Omega$, such that $f$ extends to a function complex-analytic in $\Omega$. This is easy to show - details on request. But $f$ need not extend to a set $\Omega$ that contains some strip $\Bbb R\times(-\epsilon,\epsilon)$.
For example consider $$f(t)=\sum_{n=1}^\infty a_n\frac{1}{n^2(n-t)^2+1},$$where $a_n>0$ tends to $0$ fast enough. The extension will have poles at $n+i/n$, so $\Omega$ cannot contain that horizontal strip.