Bounding $L^1$ norm of the difference between a function $f:\mathbb R^n\to\mathbb R$ of bounded variation and a piecewise constant approximation

Question

As a follow up to this question, which deals with univariate functions, I assume that we are given a function $f:\mathbb R^n\to\mathbb R$ which is of bounded variation on bounded sets, meaning, according to Wikipedia, that for $\Omega\subset\mathbb R^n$ open and bounded $$V(f,\Omega)=\sup\left\{\int_\Omega f(x)\mathrm{div}\phi(x)\ \mathrm dx:\phi\in C_c^1(\Omega,\mathbb R^n),\|\phi\|_{L^\infty(\Omega)}\leq1\right\}$$ is finite. I want to take some bounded set $A\subset\mathbb R^n$; I'm willing to assume that $A$ is an hyperrectangle, i.e. $A$ is of the form $A=\prod_{i=1}^n[a_i,b_i)$. Then we partition $A$ into $d$ hyperrectangles $A_1^d,\ldots,A_d^d$, and we denote this partition by $\mathcal P^d$. We define its mesh by the largest Lebesgue measure (EDIT: it seems that we need to define the mesh as the largest diameter, see @dsh's comment and second update to their answer) among the elements from this partition. Then the piecewise constant function $f^d$ is obtained by averaging over the elements of the partition, i.e. with $|A_i^d|$ the Lebesgue measure of $A_i^d$ $$f^d|_{A_i^d}=\frac1{|A_i^d|}\int_{A_i^d}f(x)\ \mathrm dx.$$

Similarly to the univariate case, I was wondering whether we can prove that $$\|f^d-f\|_{L^1(A)}\leq V(f,A)\cdot \mathrm{mesh}(\mathcal P^d).$$

Since my multivariate calculus is not that strong, I'm having some difficulty proving this. I'm wondering whether this can be done similarly as in the univariate case, as done by Alex Ravsky in the linked question. Intuitively, the answer should definitely be yes, but I was wondering how to bound $\Delta_i:=\sup_{x\in A_i^d}f(x)-\inf_{x\in A_i^d}f(x)$ by the integrand involving the divergence operator. More specifically, ignoring the fact that we should deal with equivalence classes for the moment, some easy bounding gives \begin{align*} \|f^d-f\|_{L^1(A)}&=\int_A|f^d(x)-f(x)|\ \mathrm dx=\sum_{i=1}^d\int_{A_i}|f^d(x)-f(x)|\ \mathrm dx\leq\sum_{i=1}^d\int_{A_i}\Delta_i\ \mathrm dx\\ &\leq\mathrm{mesh}(\mathcal P^d)\sum_{i=1}^d\Delta_i, \end{align*} so that it suffices to prove that for any partition $\mathcal P^d$ of (the hyperrectangle) $A$ into $d$ subsets (hyperrectangles) $A_i$ it holds that $$\sum_{i=1}^d\Delta_i\leq V(f,A).$$

Any help is much appreciated.

Answer 1

Partial answer. Longer article on bounded variation functions contains Poincaré inequality: $$\|f|A^d_i − f^d|A^d_i\|_{L^{1∗}(A^d_i)} \le C_2 \|Df\|(\mathrm{int}(A^d_i))$$ where $1^* = n/(n-1)$ and $\|Df\|$ is a finite positive measure associated with $f.$ $\|Df\|(\Omega) = V(f, \Omega).$ This inequality is used for $n=1 (1^* = \infty, C_2 = 1)$ in the linked question. Holder inequality and Poincaré inequality give result up to constant.

UPD 1. In Holder inequality use $p=1^*$ and $q=n$ with $f =f|A^d_i − f^d|A^d_i, g = \chi_{A^d_i}.$

UPD 2. Consider $f=x_1+\ldots+x_n.$ Then $\|Df\|$ is Lebesgue measure $dx$ up to constant depending on $n.$ General formula is $$ \int_U f\mathrm{div}\phi\:dx = - \int_U (\phi, \sigma)\:d\|Df\| $$ where $|\sigma|_2 = 1,$ $\sigma$ is vector function with $\|Df\|$-almost everywhere (from [EG] p.195, see link to encyclopediaofmath.org). This is generalization of integration by parts, so for particular $f$ $\sigma = \nabla f/|\nabla f|_2 = (1, \ldots, 1)/\sqrt{n},$ $\|Df\| = \sqrt{n}\:dx$

Consider $A^d_i$ with all sizes equal to $\delta$ then $ c\delta^{n+1} \le \|f|A^d_i − f^d|A^d_i\|_{L^1(A^d_i)} \le C\delta^{n+1}$ where $c, C > 0$ constants depending on n. On the other hand $\|Df\|(A^d_i) = \sqrt{n}\delta^n.$ Summing up for $i$ one obtains $$ c'\delta \le \|f − f^d\|_{L^1}/V(f, A) \le C'\delta $$ where $\delta = \sqrt[n]{A^d_i}.$

Answer 2

I may have found a solution to this problem in the literature. I could use some feedback, though. Please let me know if you think this approach is valid.

It seems that my conjecture is correct for functions in $W^{1,1}$; the proof being an immediate corollary of Theorem 3.2 of An optimal Poincaré inequality in $L^1$ for convex domains by Acosta and Durán:

Let $\Omega\subset\mathbb R^n$ be a convex domain with diameter $d$ and $u\in W^{1,1}(\Omega)$ such that $\int_\Omega u=0$. Then $$\|u\|_{L^1(\Omega)}\leq\frac12d\|\nabla u\|_{L^1(\Omega)}.$$

The result then follows from Theorem 2 on wikipedia or $\S$13.2 from Leoni's textbook on Sobolev spaces, using $u=f-\frac1\Omega\int_\Omega f$, defined on $\Omega$. It seems possible to carry over this result to any function of bounded variation on $\Omega$, using an approach like the one from Theorem 1, $\S$5.6.1, p. 189 of the Evans & Gariepy textbook. Using Theorem 2 from $\S$5.2.2 of the same textbook, we select a sequence $(u_k)\subset C_c^\infty(\Omega)$ converging to $u$ in $L^1(\Omega)$, and such that ($\|\nabla u_k\|_{L^1(\Omega)}$)=$\|D u_k\|\to\|Du\|=V(u,\Omega)$. Then by Fatou's lemma and the stated result for $W^{1,1}$ functions, \begin{align*} \|u\|_{L^1(\Omega)}\leq\liminf_{k\to\infty}\|u_k\|_{L^1(\Omega)}\leq\lim_{k\to\infty}\frac12d\|\nabla u_k\|_{L^1(\Omega)}=\lim_{k\to\infty}\frac12d\|Du_k\|=\frac12d\|Du\|. \end{align*}