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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.

Varun Vejalla
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Xoff
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    $\displaystyle f(x)=\frac{1}{\Gamma!\left(-\log_2!x\right),\sqrt{x^{\log_2!x+1}}}\int_0^1 \frac{1-F(t)}{t^{\log_2!x+1}},dt$ looks like a plausible candidate. It is real-analytic and we can show that the equality $f(x)=F(x)$ holds exactly on an infinite set of points in any neighborhood of $x=0$. Its graph suggests that the required limit holds, but I still have to fill some gaps in my proof. – Vladimir Reshetnikov Mar 29 '18 at 02:53
  • @VladimirReshetnikov If you are using F itself, it's not really useful, because you can just say $f(x)=F(x)$ or $f(x)=2\int F(2t)dt$. But your expressions are really complex and puzzling – Xoff Mar 31 '18 at 14:55
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    The question ask for a "usual" function asymptotic to $F(x)$. It is not really clear what "usual" means, so I interpreted it as "analytic". Although my example is defined in terms of an integral of $F(x)$, it is an analytic function (in some vicinity of $x=0$, but excluding that point itself; see https://math.stackexchange.com/questions/2495466/an-integral-involving-a-smooth-function). Of course, it would be much more interesting to find an elementary(-ish) asymptotic function, not defined in terms of $F(x)$; I tried to do that for quite some time, but I do not know if it is possible. – Vladimir Reshetnikov Mar 31 '18 at 17:31

1 Answers1

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Based oh this answer, the asymptotic of the Fabius function for $x\to0^+$: $$\small\begin{align}\log F(x)&=-\frac{\log^2x}{2\log2}+\frac{\log x\cdot\log(-\log x)}{\log2}-\left(\frac{1+\log\log2}{\log2}+\frac12\right)\log x-\frac{\log^2(-\log x)}{2\log2}\\&+\frac{\log\log 2\cdot\log(-\log x)}{\log2}+\left(\frac{6\gamma_0^2+12\gamma_1-\pi^2-6\log^2\log2}{12\log 2}-\frac{7\log 2}{12}-\frac{\log\pi}2\right)\\&+\frac{\log^2(-\log x)}{2\log2\cdot\log x}-\frac{\log\log2\cdot\log(-\log x)}{\log2\cdot\log x}+\mathcal O\!\left(\frac1{\log x}\right),\end{align}$$ where $\gamma_n$ are the Stieltjes constants (in particular, $\gamma_0$ is the Euler–Mascheroni constant $\gamma$). If I'm not mistaken—series manipulations become cumbersome there—the next term should be $\left(\frac{\log^2\log2}{2\log2}-\frac{\log2}{24}-\frac12\right)\frac1{\log x}.$ Although $\frac{\log^2\log2}{4\log2}\frac1{\log x}$ is almost perfect fit numerically. Anyways, more careful analysis is needed for next terms.

Vladimir Reshetnikov
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