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Given a function $f \in L^1 (\mathbb R)$, define the roughness $R_f$ of $f$ at $x \in \mathbb R$ by

$$\DeclareMathOperator{\esssup}{\operatorname{esssup}} R_f (x) := \limsup_{r \to 0+}\dfrac{r \esssup_{y \in B_r (x)} |f(y) - f(x)|}{\displaystyle\int\limits_{B_r (x)} |f(s) - f(x)| ds} $$ where $\esssup$ denotes the essential supremum, and by convention we take $\frac{0}{0} = 1$.

Question: Let $f$ be continuous. Is it true that $f$ is differentiable almost everywhere if and only if $R_f = 1$ almost everywhere?

Remark: The “only if” direction is relatively straightforward, the “if” direction is the issue.

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    $\begingroup$ Is the "only if" direction relatively straightforward? What if $f(x) = x^2$? I get esssup$_{y \in B_r(x)} |f(y)-f(x)| = xr+\frac{r^2}{4}$ and $\int_{x-r/2}^{x+r/2} |s^2-x^2|ds = \frac{1}{2}xr^2+O(r^3)$. What's your definition of $B_r(x)$? $\endgroup$ Commented Jul 21, 2021 at 17:46
  • $\begingroup$ Oh it’s the integral over the ball of radius $r$ around $x$. So, the integral in your sentence would be the integral from $x-r$ to $x + r$. $\endgroup$
    – Nate River
    Commented Jul 22, 2021 at 2:42
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    $\begingroup$ @mathworker21 : The limit is $1$ for all $x$ with $f'(x)\ne0$. $\endgroup$ Commented Jul 22, 2021 at 3:56
  • $\begingroup$ Can you check to see if you got the integral part correct? @mathworker21 $\endgroup$
    – Nate River
    Commented Jul 22, 2021 at 5:21
  • $\begingroup$ @NateRiver sorry about the noise. deleted my second comment. thanks for the clarifications. $\endgroup$ Commented Aug 18, 2021 at 6:58

1 Answer 1

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The answer is negative: If $f'(x) = 0$ "too often", then $R_f$ may fail to be equal to one almost everywhere.


Let $C$ be a fat Cantor set, let $I_n = (a_n, b_n)$ ($n \geqslant 2$) be the sequence of all finite components of the complement of $C$, and let $f$ be a differentiable function with the following properties:

  • $f(x) = 0$ for $x \in C$;

  • on $I_n$, $f$ is a smooth bump of a fixed shape, supported in $\tfrac1n I_n$ (the middle $n$th part of $I_n$), and with maximum equal to $|I_n|^2$ (and hence the integral of $f$ over $I$ is equal to $\tfrac1n |I_n|^3$).

Then $f'(x) = 0$ for $x \in C$ by the first property, so $f$ is everywhere differentiable. On the other hand, it is rather straigthforward to see that there is a constant $C$ such that for each $n$ and $t \in I_n = (a_n, b_n)$ we have $$ \int_{a_n}^t f(y) dy \leqslant \frac{C}{n} (t - a_n) \sup_{y \in [a_n, t]} f(y) $$ and $$ \int_t^{b_n} f(y) dy \leqslant \frac{C}{n} (b_n - t) \sup_{y \in [t, b_n]} f(y) . $$ It follows that if $x \in C$ and $r$ is small enough, so that $B_r(x)$ is disjoint with $\tfrac12 I_2 \cup \tfrac13 I_3 \cup \ldots \cup \tfrac1{n-1} I_{n-1}$, then $$ \int_{B_r(x)} f(y) dy \leqslant \frac{C}{n} |B_r(x)| \sup_{y \in B_r(x)} f(y) . $$ This, in turn, implies that $R_f(x) = \infty$ for every $x \in C$.

Thus, $R_f(x) \ne 1$ on a set of positive Lebesgue measure.

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    $\begingroup$ Unless I made a mistake, this shows that the ("relatively straightforward") "only if" part fails. I believe the "if" part is false, too, but I do not have a proof. A counterexample could be of the form $f(t) = \int_0^1 s^{-1/2} (\log s)^{-2} B(t+s) ds$ where $B(t)$ is the Brownian motion: this should be nowhere differentiable with probability one, but nevertheless we should have $R_f(t) = 1$ for all $t$ with probability one. (This is very similar to the fracftional Brownian motion with Hurst parameter approaching $1$.) $\endgroup$ Commented Sep 19, 2021 at 1:27
  • $\begingroup$ Ah wow I made a mistake it seems - I did think that $f’ = 0$ would be a problem, but I thought I had overcome the difficulty by some argument or other saying that “$f$ is locally constant most places where $f’$ is $0$“. But, I forgot about the existence of positive measure nowhere dense sets like $C$. Nicely done! $\endgroup$
    – Nate River
    Commented Sep 19, 2021 at 5:37

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