Fabius function and equivalent

The Fabius function $F$ can be defined on $[0,1]$ by

• $F(0)=0$
• $F(1)=1$
• on $[0,\frac{1}{2}]$ $F'(x)=2F(2x)$
• on $[\frac{1}{2},1]$ $F'(x)=2F(2(1-x))$

It's a known example of a not analytic $C^\infty$ function.

The Fabius function can also be defined as the CDF of the random variable $X$ such that $$X=\sum_{i=1}^\infty 2^{-i}U_i$$ where $U_i$ are independent random variables uniform on $[0,1]$.

Is it possible to find a usual function $f$ such that $$\lim_{x\rightarrow 0}\frac{F(x)}{f(x)}=1$$

Some functions like $f(x)=e^{-\frac{1}{x^2}}$ could be good candidates, but I don't really know how to find such a function, or given a function $f$, how to compute the limit.

Solution 1:

Based oh this answer, the asymptotic of the Fabius function for $x\to0^+$: $$\small F(x)\sim \frac{1}{2^{7/12} \sqrt{\pi \, x}}\, \exp \left(\frac{1}{\log 2} \left(\gamma _1+\frac{\gamma_0^2}{2}-\frac{\pi ^2}{12}-q-\frac{q^2}{2}\right)\right),\quad q=W_{-1}(-x \log2),$$ where $W_{-1}(x)$ is the non-principal real-valued branch of the Lambert W-function, and $\gamma_n$ are the Stieltjes constants (in particular, $\gamma_0$ is the Euler–Mascheroni constant $\gamma$).