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.

enter image description here

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.

enter image description here

Finally, is the function $f$ one-to-one despite the appearance?


1 Answer 1


$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 $c_0 = 0$.


$\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.


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