I am trying to construct an example of a function $f : \mathbb{R} \rightarrow \mathbb{R}$ which is analytic everywhere, where the radius of convergence of the Taylor series of $f$ at any $x \in \mathbb{R}$ is some constant $C \in (0,\infty)$ independent of $x$.
I have come up with the following and would appreciate a verification that my argument is correct.
Claim: The function $f : \mathbb{R} \rightarrow \mathbb{R}$ given by $$f(x) = \sum_{n = 0}^\infty \left ( \frac{e^x-1}{e^x+1} \right)^{2^n}$$ satisfies the properties described above, with $C = \pi/2$.
My reasoning is based on the following:
- The lacunary series $h: z \mapsto \sum_{n = 0}^\infty z^{2^n}$ is analytic on $\mathbb{D} := \{ z \in \mathbb{C} : |z| < 1 \}$, and $\partial \mathbb{D}$ forms a natural boundary, past which the analytic function cannot be extended.
- The map $$g : z \mapsto \frac{\exp(z + i \pi/2) - i}{\exp(z + i\pi/2)+i}$$ maps $\mathbb{R} \times (-\pi/2,\pi/2) \rightarrow \mathbb{D}$, and is analytic on this domain.
- $f$ is the restriction to $\mathbb{R}$ of $h \circ g$, and hence is analytic on all of $\mathbb{R}$.
- The radius of convergence of an analytic function $k$ at a point $x$ in its domain is the distance from $x$ to the closest singularity of $k$.
- Thus, by construction, the radius of convergence of $f$ at any $x \in \mathbb{R}$ is the distance of $x$ to the boundary of the domain, which is $\pi/2$.
If this is all valid, I have a follow-up question. Suppose we are given the value of $f$ in some open interval. Is it possible to reconstruct $f$ on all of $\mathbb{R}$, by recursively taking Taylor series, which converge on some larger interval, and 'chaining' together Taylor series at different points? I suppose I am just describing analytic continuation, but I wonder if not having the imaginary direction to differentiate in changes things.