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Let $u:\mathbb{R}^N \to \mathbb{R}^N$, $u \in BV(\mathbb{R}^N)$, be a function of bounded variation. We have that the following holds

$$(\ast) \qquad \frac{1}{|B_r(0)|}\int_{B_r(0)} \frac{|u(x+z)-u(x)-Az|}{|z|} dz =0$$ for the points where $Du$ is not singular, where $A$ is the approximate differential of $u$.

What happens in the points where $Du$ is singular? What should replace $A$ to still get convergence of that integral to $0$?

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  • $\begingroup$ I don't think that your question is quite correct. For example if $u$ is linear (at least in a small neighborhood of $z$), then $[u(x+z)-u(x)]/|z| = A z / |z|$ has integral zero over $B_r (0)$, simply because the integrand is odd. But the differential would be $A$, which does not have to be zero. $\endgroup$ – PhoemueX Dec 29 '18 at 23:37
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In fact the following is true:

Theorem. If $u\in BV(\mathbb{R}^N)$, then for almost all $x\in\mathbb{R}^n$: $$ \frac{1}{|B_r(0)|}\int_{B_r(0)}\left|\frac{u(x+z)-u(x)-[Du(x)]_{ac}z}{r}\right|^{\frac{N}{N-1}}\, > dz\to 0 \quad \text{as $r\to 0$.} $$

Here $[Du(x)]_{ac}$ is the density of the absolutely continuous part of the distributional derivative $Du$ which, by the definition of $BV$ is a Radon measure.

In fact the above convergence holds for all $x$ such that:

  1. $x$ is a Lebesgue point of $u$,
  2. Lebesgue point of $Du$,
  3. $$\lim_{r\to 0}\frac{[Du]_s(B(x,r))}{r^n}=0$$

where $[Du]_s$ is the singular part of $Du$.

For other points you need not have concergence no matter what linear operator $A$ you take.

Moreover, $[Du]_{ac}$ equals almost everywhere to the approximate derivative of $u$ at all points at which the above convergence to zero holds.

The above results is proven in Evans and Gariepy, Measure theory and fine properties of functions., Theorems 1 and 4 in Chapter 6.

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  • $\begingroup$ Thank you for your answer. "For other points you need not have concergence no matter what linear operator $A$ you take." --- that's the key point for me. What about if we take a nonlinear operator. Is there one such that we have convergence to $0$? $\endgroup$ – Dal Jan 5 at 22:13
  • $\begingroup$ @Dal That is not a well stated problem. If you take a nonlinear operator $Az=u(x+z)-u(x)$, then clearly you have convergence to zero so you need to be more specific what nonlinear operators you want to consider. $\endgroup$ – Piotr Hajlasz Jan 5 at 22:17
  • $\begingroup$ A kind of approximate derivative for non-Lebesgue singular points. $\endgroup$ – Dal Jan 5 at 22:17
  • $\begingroup$ @Dal Derivative, approximate or not should be linear in $z$. $\endgroup$ – Piotr Hajlasz Jan 5 at 22:21
  • $\begingroup$ So having convergence to $0$ in the case of singular point is hopeless? $\endgroup$ – Dal Jan 5 at 22:26

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