How to show that the fabius function is nowhere analytic? Consider the fabius function
https://en.m.wikipedia.org/wiki/Fabius_function
https://people.math.osu.edu/edgar.2/selfdiff/
How does one show that this function is nowhere analytic ?
Probably related , Maybe even a step in the answer : how to evaluate this function for nonreals ? Is it defined there ? Can it be extended to the complex ? 
Is the best strategy to show it does not equal its Taylor series anywhere ?
Also : is it sufficient to show it is not analytic at rational $x$ ? Is there Some theorem that says nowhere analytic for rationals implies nowhere analytic for reals ?

How about other self-differentiating functions ; are all of them either constant or nowhere analytic ??
 A: Note $F(x)=0\iff x=0$. From
$$F'(x)=\begin{cases}2F(2x)&x\in[0,1/2]\\ 2F(2(1-x))& x\in[1/2,1]\end{cases}$$
It follows $F'(x)=0\iff x\in\{0,1\}$. Differentiating again gives $F''(x)=0\iff x\in\{0,1/2,1\}$. One can see that this pattern continues (do it with induction if necessary) so that:
$$F^{(n)}(x)=0\iff x\in\left\{\frac k{2^n}\mid k\in\{0,...,2^n\}\right\}$$
The important consequence is this: If $x = k/2^n$ for some $k\in\mathbb N$ then only finitely many derivatives of $F$ are non-zero at $x$. This means the taylor series of $F$ at this point is a polynomial.
For $F$ to be analytic at $x$ it is necessary and sufficient that there exist an open neighbourhood of $x$ in which $F$ is equal to its taylor series, here a polynomial. There cannot be any such neighbourhood however, since if $y$ is not of the form $k/2^{n}$ then no derivative of $F$ vanishes at $y$ and $F$ cannot be a polynomial in a neighbourhood of $y$. Since every neighbourhood of $x$ contains irrational points it follows $F$ is not analytic at $x$.
The set $\left\{\frac k{2^n}\mid n\in\mathbb N, k\in\{0,...,2^n\}\right\}$ is dense in $[0,1]$ and the set of non-analyticities is always closed ($*$), so $F$ is not analytic anywhere on $[0,1]$.
( ($*$) follows from power series being analytic, if $F$ is analytic at some $y$ it must be equal to a power series on some open neighbourhood of $y$ and thus analytic on this entire neighbourhood)
This was the step in the original paper by Fabius:
J. Fabius, "A probabilistic example of a nowhere analytic
$C^\infty$-function".  Zeitschrift für Wahrscheinlichkeitstheorie und Verwandte Gebiete 5 (1966) 173--174

As to other self differentiating functions (I'm not entirely sure what this means?): $\exp'(x)=\exp(x)$ and $\exp$ is analytic.
