Let the zigzag function defined as
$$zz(x):=\left|\lfloor x+1/2\rfloor-x\right|,\quad x\in\Bbb R$$
Then for $k\in\Bbb Z$ we have that $zz(k)=0$, and $zz(\Bbb R)=[0,1/2]$, and $zz$ is increasing in any interval of the kind $[k,k+1/2]$, and decreasing in $[k+1/2,k+1]$ for $k\in\Bbb Z$.
Then we define the function
$$F(x):=\sum_{n=0}^\infty \frac{zz(4^nx)}{4^n},\quad x\in\Bbb R$$
Then I need to prove that $F(x)$ is not monotone in any interval of $\Bbb R$. Then I have the inequality
$$F(a\pm h)-F(a)\ge h$$
with $a:=k4^{-m}$ and $h:=4^{-2m-1}$ (I dont know exactly why $h$ is chosen like this). I can prove the inequality, but it is a bit tedious, then I want to know if there is a more easy way to
Show the above inequality truth or
Show that $F(x)$ is not monotone in any interval using a different way
The way I work the inequality to prove it: to prove the above inequality, or one similar bound, it is enough to consider $x\in (0,1)$ because the zigzag function is cyclic with period $1$, and $zz(0)=zz(1)=0$.
Using the above definitions for $a$ and $h$ we have that
$$zz(4^na)=0,\quad\text{ if }n\ge m$$
$$zz(4^n(a\pm h))=0,\quad\text{ if }n\ge 2m+1$$
Then
$$F(a\pm h)-F(a)=\sum_{n=0}^{2m}\frac{zz(4^n(a\pm h))}{4^n}-\sum_{k=0}^{m-1}\frac{zz(4^na)}{4^n}\ge 4^{-2m-1}$$
Then we can assume two cases: when the floor function have a value of zero or when it have a value of one, and depending of the chosen $x$ it is possible to approximate it choosing some $m$ such that $\lfloor 4^n(a\pm h)+1/2\rfloor=\lfloor 4^na+1/2\rfloor$.
(Indeed we must prove first that in any neighborhood of $x$ exists points of the kind $a$, $a+h$ and $a-h$.)
But, as I said, probably there is a better way to prove this. The described workaround seems too long and tedious.