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Let $f, g: [0, 1] \to \mathbb R$ be functions that are differentiable a.e. with $f’ = g’$ almost everywhere. What is the supremal Hausdorff dimension of the set on which $f$ and $g$ are both differentiable but $f’ \neq g’$?

Comments:

It is natural to ask if $f' \neq g'$ can hold even at a single point. The following example shows that this can indeed hold. Let $r_n \to 0$ be a slowly decreasing sequence of positive numbers, and set

$$f(x) = \sum_n r_{n+1} \mathbf 1_{[r_{n+1}, r_n)} (x),$$ $$ g(x) = 2f(x).$$

Then $f' = g' = 0$ a.e., but $f$ and $g$ are additionally differentiable at $0$ with derivatives $1$ and $2$ respectively.

Replacing the piecewise constant staircase with a Cantor staircase, we can even arrange for $f$ and $g$ to be continuous.

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  • $\begingroup$ Is there an obvious reason that the 'supremal' dimension should be achieved? $\endgroup$
    – LSpice
    Commented Apr 24, 2021 at 15:47
  • $\begingroup$ Using $h = f - g,$ the question becomes: If $h'$ exists a.e. and $h' = 0$ a.e., then what is supremum of the Hausdorff dimension of the set at which $h'$ exists and differs from zero? (Several versions, actually, depending on whether none or one or both of the $h'$ exists requirements include infinite derivatives.) Off-hand I don't know (probably $1$ even for the strongest version and with "a.e." replaced with "everywhere"), but my answer to Set of zeroes of the derivative of a pathological function may be of use. (continued) $\endgroup$ Commented Apr 24, 2021 at 19:48
  • $\begingroup$ Unfortunately, I don't know what possibilities exist for the Hausdorff dimension of sets belonging to Zahorski's $M_4$ class. Probably any dimension strictly less than $1$ is possible, but I don't know about achieving dimension $1.$ $\endgroup$ Commented Apr 24, 2021 at 19:48
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    $\begingroup$ @DaveLRenfro If you can get any dimension less than one, can't you stitch these functions together to some $h: \mathbf{R} \to \mathbf{R}$ and reparametrise it via a diffeomorphism $(0,1) \to \mathbf{R}$? This would give dimension one, no? $\endgroup$
    – Leo Moos
    Commented Apr 24, 2021 at 20:00
  • $\begingroup$ @Keo Moos: Unless I'm overlooking something, what you suggest seems fine. In fact, I believe I made a similar comment about something similar (achieving maximal Hausdorff dimension for a single graph by gluing appropriate graphs) in a comment somewhere, probably MSE, within the last 2 or 3 years. $\endgroup$ Commented Apr 25, 2021 at 13:46

3 Answers 3

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I think the paper "A singular function with a non-zero finite derivative on a dense set with Hausdorff dimension one" answers exactly this question.

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  • $\begingroup$ Nice find! This answers the question fully. $\endgroup$
    – Nate River
    Commented May 20 at 12:49
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If by not agree you include that the derivative may not exist, you can get any dimension and measure that does not contradict the almost everywhere. Consider the worst case $d=1$:

Take the construction of the Cantor-set. If in the $n$-th step you remove the middle $\frac{1}{n+1}$-th part of each interval, you get a set $C$ of Hausdorff-dimension $1$ but with $\mathcal{H}^1(C) =0$. So if you construct the Cantor-function using the same intervals, you get a function $f$ with $f'=0$ a.e, so for $g(x):=0$ you get $f'=g'$ a.e. except on $C$.

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  • $\begingroup$ Right though this is the less interesting case... I would want them to both exist and not agree. $\endgroup$
    – Nate River
    Commented Apr 12, 2021 at 7:53
  • $\begingroup$ @NateRiver To be honest, I am struggling to come up with any example for that case. Allowing $\pm \infty$ as values for $f'$ you can do something with jumps, but apart from that, if there are any such functions, they probably need to be highly oscillating, i.e. not even in BV and that is outside of my expertise. $\endgroup$
    – mlk
    Commented Apr 12, 2021 at 12:58
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If the derivatives exist for $x \in [0,1],$ then $f$ and $g$ must be continuous on some open set $A \subset [0,1].$ If $f \neq g$ on $A$ but $f' = g'$ a.e on $A$, then by integrating $f_1 = g_1 + C$, for some constant $C$ locally on open set $A$. However, because all the functions are continuous (locally on A) $f = g + C$ everywhere on $A$. Moreover, it's enough to consider the problem locally, if the derivatives exist. So the answer is $0,$ if the derivatives exist.

EDIT:(I now try to use FTC locally)

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    $\begingroup$ I don't think the FTC necessarily holds... $\endgroup$
    – Nate River
    Commented Apr 24, 2021 at 15:26
  • $\begingroup$ Continuous a.e. definitely does not imply continuous everywhere. $\endgroup$
    – LSpice
    Commented Apr 24, 2021 at 15:46
  • $\begingroup$ yes I am sorry, I will remove this answer. $\endgroup$ Commented Apr 24, 2021 at 15:47
  • $\begingroup$ The derivatives are only assumed to exist almost everywhere. $\endgroup$
    – LSpice
    Commented Apr 26, 2021 at 1:25
  • $\begingroup$ @LSpice True, but OP was interested just that case in a comment to an another answer $\endgroup$ Commented Apr 26, 2021 at 1:30

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