Call a function from $[0, 1]$ to itself a box function.


Given any box function $f$, define its oscillation function $Of$ as $$Of(x) = \lim \_{d \to 0} \sup \_{y, z \in B\_d (x)} |f(y) - f(z)| \, .$$ Then $Of(x)$ is itself a box function.




---


Is it true that for every box function $f$, $OOOf = OOf$?



Oscillation operator of a function
Yes, it is true. 


For a function $h(x)$ we denote by $LS(h)$, $LI(h)$ the functions defined as $$LS(h)(x)=\max(h(x),\limsup\_{y\to x} h(y)),\\ LI(h)(x)=\min(h(x),\liminf\_{y\to x} h(y)).$$ 


Then $$Og=LS(g)-LI(g).$$


Denote $g(x)=Of(x)$. Note that $g=LS(g)$, i.e. $$g(x)\geqslant \limsup\_{y\to x} g(y).$$ Indeed, for any $d>0$ and any $a$ such that $|a-x|<d$, we have $\sup\_{y,z\in B\_d(x)} |f(y)-f(z)|\geqslant g(a)$, taking limsup in $a\to x$ we get $\limsup\_{a\to x} g(a)\leqslant \sup\_{y,z\in B\_d(x)} |f(y)-f(z)|$, now take limit in $d\to 0$.


So we have $OOf=Og=g-LI(g)$, $OOOf=O(Og)=Og-LI(Og)$. Thus your relation $OOf=OOOf$ rewrites as $LI(Og)=0$. In other words, we should prove that for any $\varepsilon>0$, any $x\in [0,1]$ and and $d>0$ there exists $y\in (x-d,x+d)$ such that $Og(y)<\varepsilon$. Denote $s=\inf\_{(x-d,x+d)} g$ and choose $y\in (x-d,x+d)$ such that $g(y)<s+\varepsilon$. But $LI(g)(y)\geqslant s$ and we conclude that $Og(y)=g(y)-LI(g)(y)<\varepsilon$ as desired.


