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 $f$, $OOOf = OOf$?

  • $\begingroup$ Here B_d (x) is the ball of radius d around x. $\endgroup$ – James Baxter Dec 13 '18 at 13:58
  • $\begingroup$ This was essentially proved by Sierpinski in 1910 (reprinted on pp. 42-43 of Volume 2 of his Oeuvres Choisies), then generalized by Henry Blumberg 1 2 3 to cases in which the oscillation is computed by allowing for the neglect of countable, measure zero, and first category sets. Although the result appears in a few books (e.g. Hahn's 1921 treatise on real functions) and papers, (continued) $\endgroup$ – Dave L Renfro Dec 15 '18 at 12:44
  • $\begingroup$ it was not sufficiently well known to avoid rediscovery, which Donald C. Benson did in 1960. In recent years there has been some research done on certain properties of the oscillation function --- see the references at the end of my answer to Oscillation of a Function. $\endgroup$ – Dave L Renfro Dec 15 '18 at 12:53

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.

  • $\begingroup$ I think you meant $g:=Of$ (and then everything seems to work), rather than $g:=f$. $\endgroup$ – Iosif Pinelis Dec 14 '18 at 15:51
  • $\begingroup$ @IosifPinelis of course, thank you. Fixed. $\endgroup$ – Fedor Petrov Dec 14 '18 at 21:06

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