# Difference quotient for functions of bounded variation

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 $$0$$?

• 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. – PhoemueX Dec 29 '18 at 23:37

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.

• 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 $0$? – Dal Jan 5 at 22:13
• @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. – Piotr Hajlasz Jan 5 at 22:17
• A kind of approximate derivative for non-Lebesgue singular points. – Dal Jan 5 at 22:17
• @Dal Derivative, approximate or not should be linear in $z$. – Piotr Hajlasz Jan 5 at 22:21
• So having convergence to $0$ in the case of singular point is hopeless? – Dal Jan 5 at 22:26