Identity involving "the distance to the nearest integer" function It's a problem of an exercise list:
I want to prove that, if $||x||$ = the distance to the nearest integer to $x$, then:
$$\sum_{n=1}^{\infty}4^{-n}||2^nx||=||x||(1-2||x||)$$
is true for every $x \in \mathbb{R}/\mathbb{Z}$.
What I've tried so far:
We know that the set $D=\{\sum_{k=1}^T\frac{a_k}{2^k}$, where $a_i = 0, 1 \forall i \in \{1,2,\ldots,T\}; T \in \mathbb{N}\}$ is dense in $(0,1)$. So, it's enought to prove the identity for these numbers. Because then, we can use the Weierstrass M Test to guarantee that the identity is valid to any other number in $(0,1)$.
Why these specific numbers? Because then, the infinity sum $\sum_{n=1}^{\infty}4^{-n}||2^nx||$ will become finity and it will be easier to work with.
So, considering numbers of the form above, we obtain that
$$\sum_{n=1}^{\infty}4^{-n}||2^nx||=\sum_{n=1}^{\infty}4^{-n}\left|\left|2^n\sum_{k=1}^T\frac{a_k}{2^k}\right|\right| = \sum_{k=1}^{T-1}4^{-n}\left|\left|2^n\sum_{k=1}^{T}\frac{a_k}{2^k}\right|\right|. $$
Firt of all, the way I constructed that specific numbers, we can't obtain 1, but that's OK, because in $\mathbb{R}/\mathbb{Z}$ is 0.
Now, note that:
$||x||=x$, if $0 \leq x \leq 1/2$ and
$||x|| = 1- x$, if $1/2 < x <1$
(Remember we are working in $\mathbb{R}/\mathbb{Z}$).
So, if we define $S_n=2^n\sum_{k=1}^{T-1}\frac{a_k}{2^k}$, we get that $||S_n||=||2^n\sum_{k=1}^{T-1}\frac{a_k}{2^k}||=a_{n+1}-2a_{n+1}S_n+S_n$, because if $a_{n+1}=0$, $S_n \leq 1/2$ and if $a_{n+1}=1$, $S_n > 1/2$.
So, $$\sum_{n=1}^{\infty}4^{-n}\left|\left|2^n\sum_{k=1}^T\frac{a_k}{2^k}\right|\right|=\sum_{n=1}^{T-1}4^{-n}(a_{n+1}-2a_{n+1}S_n + S_n)$$
$$= \sum_{n=1}^{T-1}\frac{a_{n+1}}{4^n}-\sum_{n=1}^{T-1}\frac{1}{4^n}(2a_{n+1}S_n - S_n)$$
$$=\sum_{n=1}^{T-1}\frac{a_{n+1}}{4^n} + \sum_{n=1}^{T-1}\frac{S_n}{4^n}(1-2a_{n+1})$$
$$=\sum_{n=1}^{T-1}\frac{a_{n+1}}{4^n}+\sum_{n=1}^{T-1}\sum_{i=1}^{T}\frac{a_i}{2^i}\left(\frac{1-2a_{n+1}}{2^n}\right)$$.
On the other hand, we have that
$$||x||(1-2||x||)=\sum_{k=2}^{T}\frac{a_k}{2^k}-2\left(\sum_{k=2}^{T}\frac{a_k}{2^k}\right)^2$$.
I'm having trouble finding a way to prove that the two expressions are, indeed, equal. I tried to write the last expression as an open product of two series, but it didn't work... 
Does someone have a hint or a solution?
Thanks! 
 A: A simple, but unfortunately unrelated, argument would be to first prove that
$$f(x)=\sum_{n=1}^{\infty}4^{-n}||2^n x||$$
is a continuous function on $\mathbb R / \mathbb Z$. This is relatively easy - for instance, noting that $4^{-n}||2^n x||$ is Lipschitz continuous with Lipschitz constant $2^{-n}$ yields that $f$ is Lipschitz continuos with Lipschitz constant $1$.
Now, let's consider the metric space of continuous functions on $\mathbb R / \mathbb Z\rightarrow \mathbb R$ endowed with the metric
$$d(f,g)=\sup_{x\in\mathbb R / \mathbb Z}|f(x)-g(x)|$$
and, on this space, consider the map
$$T(g)=x\mapsto\frac{||2x||+g(2x)}{4}.$$
of which $f$ is obviously fixed point. The function $x\mapsto ||x||(1-2||x||)$ is also a fixed point, though the proof of this is more tedious. Moreover, $T$ is a contraction map (contracting by a factor of $\frac{1}4$) and therefore has at most one fixed point. Thus, as both $f$ and $x\mapsto ||x||(1-2||x||)$ are fixed points, they are equal.
(Your work looks good, but I'm not certain about how one would complete it; the two things you want to prove are equal look pretty different at this point).
