Difference quotient for functions of bounded variation

Question

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 $0$?

Answer 1

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.