14
$\begingroup$

In general, it is well known that, on the real line, say on $[0,1]$, if a function $f$ is of (pointwise) bounded variation, meaning that $$ \sum_{i=1}^n |f(x_i)-f(x_{i-1})| <+\infty $$ for every partition ${x_i}_{0}^n$ of $[0,1]$, then $f$ can be written as the difference of two monotone functions, hence it is differentiable a.e. w.r.t. the Lebesgue measure.

I am wondering if the same is true for $BV$ functions in $\mathbb R^d$ for $d \ge 2$.

Of course, the right definition of $BV$ in $d$-dimensional domains passes through the theory of distributions: $f \colon \Omega \subset \mathbb R^d \to \mathbb R$ is in $BV(\Omega)$ if it is an $L^1$ function whose distirbutional gradient $Df$ can be represented by a finite Radon measure (see here).

Question 1. Let $f \colon \Omega \subset \mathbb R^d \to \mathbb R$ be in $BV(\Omega)$. Is it true that $f$ is differentiable a.e. with respect to the Lebesgue measure?

What I know is that they are approximately differentiable a.e. (in view of Calderon-Zygmund Theorem) so an approximate differential exists a.e. but I am not aware of any link between the approximate differentiability and the pointwise a.e. one.


Addendum.

In view of Mizar's answer, it seems that the answer to Q1 is negative, as it has been exhibited a $BV$ function which does not have even a continuous a.e. representative (in $L^1$).

While checking the details of the answer I received, I would like to ask another version of question above (do not know if still meaningful or not).

Question 2. Let $f \colon \Omega \subset \mathbb R^d \to \mathbb R$ be in $BV(\Omega)$. Assume further $f \in C^0(\Omega)$ i.e. it is continuous. Is it true that $f$ is differentiable a.e. with respect to the Lebesgue measure?

$\endgroup$
  • 4
    $\begingroup$ How about including the definition of BV for $n$-dimensional domain? Is it this en.wikipedia.org/wiki/… having to do with the distributional derivative being a Radon measure? $\endgroup$ – Gerald Edgar Jun 15 '17 at 18:16
  • $\begingroup$ @GeraldEdgar Sure, I will edit accordingly. I did not write the definition because I thought it was something definitely well known. $\endgroup$ – user111164 Jun 16 '17 at 6:01
16
$\begingroup$

No. Take a dense countable set $\{x_1,x_2,\dots\}$ in $\mathbb{R}^d$ and a sequence $(r_i)\subseteq\mathbb{R}^+$ such that $\sum_i r_i^{d-1}<\infty$. Then the function $$f=1_{\bigcup_{i=1}^\infty B_{r_i}(x_i)}$$ is in $BV(\mathbb{R}^d)$ (since $|\bigcup B_{r_i}(x_i)|\le C\sum_i r_i^d$ and $f$ is the limit in $L^1$ of the functions $1_{\bigcup_{i=1}^k B_{r_i}(x_i)}$, whose gradients have total variation bounded by $C\sum_i r_i^{d-1}<\infty$).

Now, for any Lebesgue point $x_0$ of $f$ in the closed set $\{f=0\}$, no representative $g$ is continuous at $x_0$ (representative means a function which coincides a.e. with $f$). Indeed, $x_0$ lies in the closure of the open set $\bigcup B_{r_i}(x_i)$, so it belongs to the closure of $\{g=1\}$. On the other hand, since $x_0$ is a Lebesgue point for $f$, it must also belong to the closure of $\{g=0\}$. This shows that $g$ is not even a.e. continuous (since the set $\{f=0\}$ has positive measure).


Addendum.

The answer is still no even assuming $f$ continuous. Below I construct an example where the differentiability of $f$ fails on a Borel set of positive measure.

Choose a countable dense set $\{x_i\}$ in $B_1(0)$ and a sequence $r_i>0$ such that $\sum_i r_i^{d-1}<\infty$ and $\sum_i|B_{r_i}(x_i)|<|B_1(0)|$. In particular, $r_i\to 0$. Using Besicovitch covering theorem, up to a subsequence we can assume that the balls $B_{r_i}(x_i)$ have bounded overlapping (i.e. any point lies in at most $N$ such balls); in doing this, we could lose the density of $\{x_i\}$ but we still have $B_1(0)\subseteq\overline{\cup_i B_{r_i}(x_i)}$.

Let $S:=B_1(0)\setminus\cup_i\overline{B_{r_i}(x_i)}$. We remark that $|S|>0$ and that, for any $N\ge 1$, $S\subseteq\overline{\{x_i\mid i>N\}}$. It follows that we can find a sequence of positive radii $R_i\to 0$ such that $S\subseteq\cup_{i\ge j}B_{R_i}(x_i)$ for all $j\ge 1$: by compactness, we can find $n_1>0$ such that $$S\subseteq\overline{\{x_i\mid i>0\}}\subseteq\cup_{i=1}^{n_1}B_1(x_i),$$ then we use the above remark with $N=n_1$ and we find $n_2>n_1$ such that $$S\subseteq\overline{\{x_i\mid i>n_1\}}\subseteq\cup_{i=n_1+1}^{n_2}B_{1/2}(x_i),$$ and so on.

Now the function $f(x):=\sum_i R_i(1-r_i^{-1}|x-x_i|)^+$ (a superposition of 'traffic cones' with heights $R_i$ placed on our balls $B_{r_i}(x_i)$) is continuous, as it is a uniform limit of continuous functions, thanks to the bounded overlapping. It also lies in $BV(\mathbb{R}^d)$ thanks to the assumption $\sum_i r_i^{d-1}<\infty$.

I claim that $f$ cannot be differentiable at $x$, for any $x\in S$. Indeed, as $x$ is a minimum point for $f$, we would have $\nabla f(x)=0$. But there is a sequence $i_k\to\infty$ with $x\in B_{R_{i_k}}(x_{i_k})$, so $f(x_{i_k})\ge R_{i_k}\ge|x-x_{i_k}|$, which contradicts $\nabla f(x)=0$ (since $x_{i_k}\to x$).

$\endgroup$
  • $\begingroup$ What a nice and complete answer! Thanks a lot, I have checked it and it seems correct. +1 ;-) $\endgroup$ – user111164 Jun 17 '17 at 18:16
  • 1
    $\begingroup$ Yet one more question: what about $W^{1,1}$ functions? Could your examples be adapted to the $W^{1,1}$ setting or on the contrary is it true that in $W^{1,1}$ differentiabiliy a.e. holds true (I am assuming $d \ge 2$, of course)? Thanks! $\endgroup$ – user111164 Jun 18 '17 at 9:02
  • 1
    $\begingroup$ The series defining the function $f$ in the second example converges in $C^0\cap W^{1,1}$, so it is a counterexample for $W^{1,1}$ as well. If you impose that $\sum_i r_i^{(d-p)/p}<\infty$, you also get a counterexample in $C^0\cap W^{1,p}$, for $p<d$. As you probably know, functions in $W^{1,p}$ are continuous and differentiable a.e. if $p>d$. I expect that continuous counterexamples exist in the borderline case $p=d$, but probably it requires more work. $\endgroup$ – Mizar Jun 18 '17 at 14:45

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.