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Is it true that the definition of approximate differentiability presented here of a function $f: \mathbb{R}^N \to \mathbb{R}$ is equivalent to the following one?

$$\lim_{r \to 0} \rlap{-}\!\!\int_{B_r(x)} \min \left\{\frac{f(y)-f(x) - L(y-x)}{|y-x|},1 \right\} dy = 0$$ for some linear $L:\mathbb{R}^N \to \mathbb{R}$.

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I think the answer is no.

First of all consider a related question: is the standard definition of approximate continuity of $g\colon \mathbb R \to \mathbb R$ equivalent to the following one

$$ \lim_{r \to 0} \rlap{-}\!\!\int_{B_r(x)} \min \left\{|g(y)-g(x)|,1 \right\} dy = 0. $$

This definition is in fact strictly weaker. Consider for instance the function $g(x) = \sum_{n=1}^\infty b_n \chi_{[-r_n, -r_n + a_n]}$, where $r_n = 3^{-n}$, $a_n = \frac12 4^{-n}$, $b_n = 2^n$ and $x=0$.

Then for $r=r_n$

$$ \rlap{-}\!\!\int_{B_r(x)} \min \left\{|g(y)-g(x)|,1 \right\} dy = \frac{1}{2 r_n}\sum_{k=n}^\infty a_k = \frac13 \left(\frac34\right)^{n} \to 0 $$

while

$$ \rlap{-}\!\!\int_{B_r(x)} |g(y)-g(x)| \, dy = \frac{1}{2 r_n}\sum_{k=n}^\infty a_k b_k = \frac12 \left(\frac32\right)^{n} \to \infty $$

as $n\to \infty$.

For the original question one can consider $f(x) = x g(x)$. Let me know if additional details are needed.

Update. For some closely related results see also Proposition 3.65 in Functions of Bounded Variation by L. Ambrosio, N. Fusco and D. Pallara (2000).

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  • $\begingroup$ I'm confused. Shouldn't it be true as in Definition 4.1 on page 6 of this paper: cvgmt.sns.it/paper/75 ? $\endgroup$
    – Riku
    Commented Mar 25, 2019 at 9:42
  • $\begingroup$ @Riku In that paper it is not claimed that Definition 4.1 is equivalent to the standard one. By the way, in your question only $\varepsilon=1$ (in the notation of the paper) is considered, right? See also p. 161 in the book of Ambrosio, Fusco and Pallara. I have appended a reference to this book to my answer. $\endgroup$
    – Skeeve
    Commented Mar 25, 2019 at 10:38
  • $\begingroup$ Why not? It think that is the claim. In fact, the definition in the question is stronger than the one in the paper (where only convergence in measure is required). $\endgroup$
    – Riku
    Commented Mar 30, 2019 at 17:14
  • $\begingroup$ I've reformulated the additional question here: mathoverflow.net/questions/326784/approximate-differentiability Definition 2 of the linked question should be the standard definition (which is also the one in that paper). $\endgroup$
    – Riku
    Commented Mar 30, 2019 at 21:27
  • $\begingroup$ @Riku You are right, the definition in the paper is different from the one in your question. But in my answer I consider the definition from your question, not the one from the paper. And it is not clear to me what claim do you refer to in your comment? $\endgroup$
    – Skeeve
    Commented Mar 30, 2019 at 22:06

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