Why is there no “real analytic continuation”

Suppose you have an analytic function defined on open interval f: D → C where D,C ⊂ ℝ. Why can’t we extend f’s domain to ℝ in a similar sense to how we may extend a complex analytic function defined in an open set to ℂ? What is an example of a real valued function that could be continued in two distinct ways while remaining analytic?

• Who said that ? It is immediate from the definitions that $f$ is complex analytic on a complex open containing your real interval. Oct 21, 2019 at 18:00
• See theorem 3 here: sites.math.northwestern.edu/~scanez/courses/320/notes/… Oct 21, 2019 at 18:00

A singularity in a real analytic function disconnects the real line. Consequently, the parts on either side of the singularity need not agree anywhere. For instance, consider $$\int_1^x 1/t \,\mathrm{d}t$$, which gives the logarithm. Notice that $$C + \log |x|$$ is a solution left of the singularity at zero for any $$C \in \Bbb{R}$$, so there is no unique continuation.
There are $$\Bbb{C}$$ functions which exhibit a similar phenomenon. $$\sum_{n \geq 0} x^{2^n}$$ has a natural boundary along the unit circle -- it cannot be extended outside the circle.
On the other hand, what do you mean by “we may”? Sometimes we may and sometimes we may not. For instance, there are analytic functions from $$D(0,1)$$ into $$\mathbb C$$ which cannot be extended to an analytic function whose domain is a connected open set which strictly contains $$D(0,1)$$.