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Let $u \in BV(\Omega \subset \mathbb R^N, \mathbb{R}^N)$. Is it true that there exists a function $f$ in the weak $L^1$ space such that $$|u(y)-u(x)| \le |x-y|\big|f(y) - f(x)\big|$$ holds for a.e. $x,y$?

This question is motivated by Convergence of the difference quotient of a BV function.

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  • $\begingroup$ The estimate looks strange to me for the following reasons. First, interchanging $x,y$ and multiplying by $-1$ we get $u(y)-u(x)\ge |x-y| (f(y)-f(x))$, so the inequality becomes an equality. Second, the function $u$ can always be perturbed on negligible sets, so probably the estimate cannot hold for all $x,y$. $\endgroup$
    – Skeeve
    Commented Apr 20, 2019 at 8:14
  • $\begingroup$ @Skeeve You're right. I forgot to put norms and it should holds for a.e. $x,y$. $\endgroup$
    – Riku
    Commented Apr 20, 2019 at 10:37

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A result of this kind can be found in [1] (Lemma A.3):

If $u\in BV(\mathbb R^N)$ then there exists a Lebesgue negligible set $F \subset \mathbb R^N$ such that $$ |u(x) - u(y)| \le c_N |x-y| (M_R Du(x) + M_R Du(y)) $$ for $x,y\in \mathbb R^N \setminus F$ with $|x-y|\le R$.

Here $$ M_R Du(x) = \sup_{r\in(0,R)} \frac{|Du|(B_r(x))}{|B_r(x)|}, $$ where $|B_r(x)|$ denotes the Lebesgue measure of $B_r(x)$.

From Lemma A.2 (which is stated for $L^1$ functions but holds for measures as well) if follows that $x\mapsto M_R Du(x)$ belongs to the weak $L^1$ space.

References

[1] De Lellis C., Crippa G. Estimates And Regularity Results For The Diperna–Lions Flow. J. Reine Angew. Math. 616 (2008), 15–46.

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  • $\begingroup$ That's great. I wonder what's the role of weak $L^1$ vs $L^1$ to estimate $M_R Du$. That is, I'd like to ask: $\forall \epsilon>0$ is it true that $M_R Du = F_1 + F_2$, where $\| F_1\|_{w-L^1} \le \epsilon$ and $\|F_2\|_{L^1} \le C_\epsilon$? $\endgroup$
    – Riku
    Commented Apr 20, 2019 at 11:17
  • $\begingroup$ I doubt that $M_R Du$ can be written this way. Suppose for instance that $u$ is the Heaviside step function, $N=1$. Then $M_R Du(x) = \frac{1}{2|x|}$ if $|x|\le R$ and zero otherwise. $\endgroup$
    – Skeeve
    Commented Apr 20, 2019 at 11:53
  • $\begingroup$ I see that this example is not in $L^1_{loc}$, but can't it be decomposed in the sum of two functions $F_1,F_2$ such that $\|F_1\|_{w-L^1} \le \epsilon$ and $\|F_2\|_{L^1} \le C_\epsilon$? $\endgroup$
    – Riku
    Commented Apr 20, 2019 at 12:24
  • $\begingroup$ I've asked the follow up question in a separate post: mathoverflow.net/questions/328571/…. $\endgroup$
    – Riku
    Commented Apr 21, 2019 at 10:32

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