Extension of real analytic function to a complex analytic function

Question

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?

Answer 1

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.

Answer 2

We indeed have a local extension, given by the Taylor expansion. That is, given $x_0\in\mathbb R$ there exists $\epsilon>0$ such that $$\sum_{n=0}^\infty\frac{f^{(n)}(x_0)}{n!}(z-x_0)^n$$ converges for $|z-x_0|<\epsilon$, and we call the limit $g(z)$. This defines an analytic function $g$ on a neighborhood of $\mathbb R$ which clearly extends $f$. (We do need to verify that if $z$ is in the disk of convergence of two points $x_0$ and $\tilde x_0$ then the series above is equal to the same series with $x_0$ replaced by $\tilde x_0$, but this follows from the identity theorem.)

It is worth pointing out that the $\epsilon>0$ described above depends on $x_0$, and there does not necessarily exist such $\epsilon$ independent of $x_0$. So in general we cannot expect the neighborhood to be of the form $\mathbb R\times(-\epsilon,\epsilon)$.

Answer 3

As David C. Ullrich and Jason pointed out, the answer is not always affirmative. Let me quote a Theorem that gives a partial affirmative answer to your question, with somewhat different assumptions.

Theorem: Assume that $f$ satisfies the Fourier inversion formula, i.e. $f(x) = \int_{\mathbb{R}}\hat{f}(\xi)e^{2\pi i x\xi}\,d\xi$ and that $$|\hat{f}(\xi)| \le Ae^{-2\pi a|\xi|}$$ for some constants $a,A > 0$. Then $f$ is the restriction to $\mathbb{R}$ of a function $f(z)$ holomorphic in the strip $S_b = \{z \in \mathbb{C}: |\text{Im}(z)| < b\}$, for any $0 < b < a$.

I think this is an interesting result since it also tells how much you can push the $\epsilon$ is you question. You can find a proof of this theorem and further discussions on the possibility of extending functions on the real line to complex-analytic functions in Complex Analysis by E.M. Stein and R. Shakarchi.