As you could see in Wikipedia the Fabius function is a known "example of an infinitely differentiable function that is nowhere analytic".
In the following answer where it is explained how is nowhere analytic its shown that (also it is Wikipedia): $$F'(x)=\begin{cases}2F(2x)&x\in[0,1/2]\\ 2F(2(1-x))& x\in[1/2,1]\end{cases}$$
And here is where something I don't understand: the Fabius function showed a behavior of flat function near zero $x=0$ and at the top near $x=1$, so since its derivative matches a scaled version of the function, this means that this "flat top" will be also in its derivative $F'(x)$, implying that there is a neighborhood near $x=\frac12$ where the function have constant slope, so in this neighborhood the function should be analytical since is going to be fitted perfectly by polynomial of order 1 (its just a straight line!).
So in order to be nowhere analytical near $x=\frac12$, it will imply that the Fabius function cannot be a flat function near $x=0$ neither $x=1$, but from the plots through polynomial approximations it really looks like is flat at these points. So I don't know what is happening: It is possible to prove that the Fabius function is not a flat function?
These doubt rises in this other question's answers by @Artes from where I am taking the following plot of $F'(x)$:
Even if I plot a polynomial approximation of $F(x+1)$ obtained as is shown in this answer (considering $m=3$) with function that do have a flat top like $g(x)=\left(1+\exp\left(\frac{1-2|x|}{x^2-|x|}\right)\right)^{-1}$ as you could see here is hard to believe it is not a flat function.
Maybe I have misunderstood what means to be nowhere analytical.
Added later
Using what is stated in the mentioned answer that the Fabius function fulfill: $$ \frac{d^n}{dx^n}F(x)\biggr|_{x=\frac12} = 0,\quad\forall n>1,\,n\in\mathbb{Z}$$ I should have that the Taylor Series at $x=\frac{1}{2}$ will be: $$T(x) = \sum\limits_{n=0}^{\infty}\frac{F^{(n)}(a)}{n!}(x-a)^n \Biggr|_{a=\frac12} = F\left(\frac12\right)+F'\left(\frac12\right)\left(x-\frac12\right)$$ and from the Wikipedia I know that $ F\left(\frac12\right) = \frac12$ and $F(1)=1$, and also from the formulas shown at the beginning one can see that: $$\lim\limits_{x\to \frac12^{\pm}}F'(x) = 2F(1)=2 \Rightarrow F'\left(\frac12\right) = 2$$ With this is possible to find the Taylor series at $x=1/2$ as: $$T(x) = 2x -\frac12$$
So for me the Fabius function do "looks analytical" at $x = 1/2$.
Since I cannot plot exactly the Fabius function, if I compare it with a polynomial approximation (as shown here for $m=3$) which should look analytical as is polynomial, but also with another approximation: $$q(x) = \left(1+\exp\left(\frac{1-2|x-1|}{(x-1)^2-|x-1|}\right)\right)^{-1}$$ which has flat top but it don't have a "straight-line slope" near $x=\frac12$ (also fulfill $F(0)=q(0^+)=0$, $F(1)=q(1)=1$, and $F(1/2)=q(1/2)=1/2$), so I could think as the real Fabius function being something "close" to both of them.
Seeing them plotted against $T(x)$ make hard to believe it is not analytical at $x = 1/2$ since the straight line is quite a good approximation to both functions (and the fact it have a defined Taylor series at $x=\frac12$):
Hope you could explain in an intuitive way How it is nowhere analytical when it have a defined Taylor series expansion at $x=\frac12$.
Added after the question was answered
I have realized later that the following function: $$q(x)=\frac{1}{1+\exp\left(\frac{1-2|x|}{x^2-|x|}\right)}$$ Matches "almost perfectly" the comstruction of a "smooth" transition curve show in Wikipedia Non-analytic smooth function:
- Define $$f(x)=\begin{cases}e^{-\frac{1}{x}},\quad x>0\\ 0,\quad x\leq 0\end{cases}$$
- Define $$p(x)=\frac{f(x)}{f(x)+f(1-x)}$$
- Then for some real valued constants $a<b<c<d$ one could built a smooth bump function as: $$r(x)=p\left(\frac{x-a}{b-a}\right)\cdot p\left(\frac{d-x}{d-c}\right)$$ which is non-zero in $[a,\ d]$ such is flat on $[b,\ c]$
One can "experimentally" see that for $[a,\ d]\equiv [-1,\ 1]$, by chosing $b\to 0,\ c\to 0$ then $q(x)$ matches $r(x)$ on $(-1,\ 1)\setminus\{0\}$.
Then, by playing with $q(x)$ I realized two interesting things about the function: $$q(x,a)=\frac{1}{1+\exp\left(\frac{a(1-2|x|)}{x^2-|x|}\right)}$$
- The function $q(x,4)$ fits "really good" the function $r(x)$ if I change the definition of $f(x)$ to $$f(x)=\begin{cases}e^{-\frac{1}{x^2}},\quad x>0\\ 0,\quad x\leq 0\end{cases}$$
- The function $q(x,\frac{\sqrt{3}}{2})$ "looks like" a perfect smooth transition function with straight-line edges, fulfilling having a derivative with flat tops - but unfortunately it don't fulfill solving a DDE like $y'(x)=k\left(y(2x+1)-y(2x-1)\right)$ for some constant $k$.
You could see them on Desmos:
Jointly with what was noted by comments and answers, this kind of discard the flat slope on an open interval of the Fabius function.
A function $f(x)$ is real-analytic$^{[1]}$ at a given point if it is infinitely differentiable at that point (and so, it has the Taylor series at that point) and the Taylor series converges exactly to $f(x)$ in some open interval (however small) containing that point.
How does it apply to the Fabius function? The Fabius function is infinitely differentiable everywhere. For $x<0$, the Fabius function is exactly $0$, so its Taylor series is also $0$, so it coincides with the function in any open interval that is small enough to not include any point $x\ge0$, so the Fabius function is real-analytic there. At $x=0$, its Taylor series is also $0$, but any open interval containing $x=0$ necessarily contains some point $0<x<1$, where the Fabius function is not $0$, so its Taylor series does not converge to the Fabius function in any open interval. So, the Fabius function is not real-analytic at $x=0$. Now, what about $x>0$? At dyadic rational points, the Taylor series of the Fabius function has only a finite number of non-zero coefficients, so it's just a polynomial and has an infinite radius of convergence. But the Fabius function is not a polynomial, so, although the Taylor series converges, it doesn't converge to the Fabius function, and hence, it's not real-analytic there. At all the other points (that are not dyadic rationals), the Taylor series has an infinite number of terms, but they grow too fast and its radius of convergence is $0$, so it doesn't converge in any open interval.
So, the Fabius function is real-analytic for all $x<0$, and is not real-analytic for all $x\ge0$.