Need any kind of insights about a strange function The function in question is $f(x; b) = \sum_{k=0}^\infty \{ b^k x \}\cdot b^{-k}$ where the base $b$ is equal to $2$ here. Also, $x\in [0, 1[$. The curly brackets denote the fractional part function.
The function $f$ seems continuous almost everywhere but nowhere differentiable, and has a fractal look. It satisfies the recursion $f(x) = 2f(x/2) - x$. We also have $f(x) = -x \log_2 x + g(x)$ with $g$ a non-trivial, non-constant function satisfying $g(x) = 2 g(x/2)$. The function $g$ is pictured below. 

Any interesting results about this function is welcome. In particular, I am interested in the proportion $P(\alpha)$ of real numbers in $[0, 1[$ such that $f(x) < \alpha$ for any $\alpha \in [0, 2[$. This is the inverse of the percentile distribution, pictured below. But I'm not even sure that the set $\{ x \in [0, 1[  ; f(x) < \alpha\}$ is Lebesgue-measurable. 

Finally, is the function $f$ one-to-one despite the appearance?
 A: $f(x; b) 
= \sum_{k=0}^\infty \{ b^k x \}\cdot b^{-k}
$.
Here's a start.
I hope someone else
can continue it.
Let
$x
=\sum_{j=0}^{\infty} c_j b^{-j}
$
where
$0 \le c_j \le b-1
$
and
$c_0 = 0$.
Then
$\begin{array}\\
\{ b^k x \}
&=\{ b^k \sum_{j=0}^{\infty} c_j b^{-j} \}\\
&=\{ \sum_{j=0}^{\infty} c_j b^{-j+k} \}\\
&=\{ \sum_{j=0}^{k} c_j b^{-j+k}\\
&=\{ \sum_{j=0}^{\infty} c_{j+k} b^{-j} \}\\
&=\sum_{j=0}^{\infty} c_{j+k} b^{-j}\\
&=\sum_{j=0}^{\infty} c_{j+k} b^{-j}\\
\text{so}\\
f(x; b) 
&= \sum_{k=0}^\infty b^{-k}\{ b^k x \}\\
&= \sum_{k=0}^\infty b^{-k}\sum_{j=0}^{\infty} c_{j+k} b^{-j}\\
&= \sum_{k=0}^\infty \sum_{j=0}^{\infty} c_{j+k} b^{-j-k}\\
&= \sum_{n=0}^\infty c_nb^{-n} \sum_{j=0}^{n} 1\\
&= \sum_{n=0}^\infty (n+1)c_nb^{-n}\\
&= \sum_{n=1}^\infty (n+1)c_nb^{-n}\\
&= \sum_{n=1}^\infty nc_nb^{-n}+\sum_{n=1}^\infty c_nb^{-n}\\
&= \sum_{n=1}^\infty nc_nb^{-n}+x\\
&= g(x; b)+x\\
\end{array}
$
At this stage,
I'm not sure what to do,
so I'll
split the sum
into sets of $n$
with $m+1$ digits,
$0\to b-1,
b \to b^2-1,
...,
b^m \to b^{m+1}-1
$,
so that
$b^m \le n \le b^{m+1}-1
$.
$\begin{array}\\
g(x; b) 
&= \sum_{n=1}^\infty nc_nb^{-n}\\
&= \sum_{m=0}^{\infty} \sum_{n=b^m}^{b^{m+1}-1} nc_nb^{-n}\\
&= \sum_{m=0}^{\infty} \sum_{n=0}^{b^{m+1}-b^m-1} (n+b^m)c_{n+b^m}b^{-n-b^m}\\
&= \sum_{m=0}^{\infty} b^{-b^m}\sum_{n=0}^{b^{m+1}-b^m-1} (n+b^m)c_{n+b^m}b^{-n}\\
&= \sum_{m=0}^{\infty} b^{-b^m}\sum_{n=0}^{b^{m+1}-b^m-1} nc_{n+b^m}b^{-n}+\sum_{m=0}^{\infty} b^{-b^m}\sum_{n=0}^{b^{m+1}-b^m-1} b^mc_{n+b^m}b^{-n}\\
&= \sum_{m=0}^{\infty} b^{-b^m}\sum_{n=0}^{b^{m+1}-b^m-1} nc_{n+b^m}b^{-n}+\sum_{m=0}^{\infty} b^{-b^m+m}\sum_{n=0}^{b^{m+1}-b^m-1} c_{n+b^m}b^{-n}\\
&= g_1(x; b)+g_2(x; b)\\
\end{array}
$
I'm not sure what to do here.
The inner sums
look like they might be written
as a fractional part of $x$
scaled by a power of $b$,
but, again,
I don't see the best way to do it.
So I'll stop here
and hope this is of use
to others.
