Can every real-analytic function $f : \mathbb{R} \rightarrow \mathbb{R}$
be extended to a holomorphic function $\tilde{f} : \mathbb{C}\setminus A \rightarrow \mathbb{C}$ where $A \subseteq \mathbb{C}$ is some discrete subset, with $A \cap \mathbb{R} = \emptyset$?
If yes, what are the conditions on $f$ for $\tilde{f}$ to be furthermore meromorphic (i.e. such that the points where it is not defined are poles)?
More generally does it hold that every real-analytic function $\mathbb{R}\setminus A \rightarrow \mathbb{R}$, where $A \subset \mathbb{R}$ is discrete,
can be extended to a complex function holomorphic except on a discrete subset? And what would be the conditions for meromorphicity?
Best Answer
No, this is not possible.
To begin, recall the standard example of an analytic function that cannot be extended to $\mathbb{C}$: $$\phi(z)=\sum_{j=0}^{\infty}{z^{2^j}}$$ The standard proof of this notes that analytic functions cannot have accumulating poles, but $\phi$ has poles at every binary root of unity.
Now, $\phi$ is not real-analytic (although it has real coefficients), because it "blows up" outside $(-1,1)$. But that difficulty is easy enough to fix: I claim $$z\mapsto\phi\left(\frac{\sin{z}}{2}\right)$$ is well-defined on all of $\mathbb{R}$, analytic, and has maximal domain of extension that is much smaller than $\mathbb{C}$.
For analyticity, note $\sin$ is entire, and $\sin{0}=0$. $\phi$ has a Taylor expansion at $0$, and so the power series compose term-by-term (even as formal series). For maximal domain of extension, note that $$|\sin{\!(u+vi)}|=\sqrt{\frac{\cosh{\!(2v)}-\cos{2u}}{2}}$$