# Does the derivative of a BV function with no jump part vanish on level sets?

Let $$u: \mathbb R^n \to \mathbb R$$ be a $$BV$$ function with no jump part, i.e., writing $$Du = D^a u + D^s u + D^j u$$ for the decomposition of $$Du$$ into absolutely continuous, Cantor, and jump part respectively, we have $$D^j u = 0$$.

Since $$u$$ is a priori an equivalence class of functions $$L_{loc}^1$$, in the following question which is sensitive to variation on Lebesgue null sets, we use the precise representative of $$u$$, which is defined to be equal to the approximate limit of $$u$$, wherever it exists. This defines the function $$u$$ up to a set of $$\mathcal H^{n-1}$$ finite measure, which is enough for the purposes of the following:

Question: Does the weak derivative of $$u$$ vanish on level sets? In other words, is it true that $$|Du|(u^{-1}(p)) = 0$$ for all $$p \in \mathbb R$$?

• Can you confirm that this is what you intended to write? When $u$ is of class $C^1$ for example then $Du = D^a u = \nabla u \, \mathrm{d} \mathcal{H}^n$ with no jump or Cantor part, no? – Leo Moos Apr 5 at 23:40
• Yes that’s right. For such a function, if the hypothesis is true we should have for each $r$ that $\ ∇ u = 0$ $\mathcal, H^n$-a.e. on the set $\{x \in \mathbb R \ | \ u(x) = r \}$. Although I think I want to change $Du(f^{-1} (p))$ to $|Du|(f^{-1} (p))$, which I will do so now. – Nate River Apr 5 at 23:50
• How do you ensure that $|Du|(u^{-1}(p))$ is well defined? The standard-definition of $BV$ in $\mathbb{R}^n$ is as a subset of $L^1$, so a function is only defined up to Lebesgue zero-sets. This is enough to define $|D^au|(u^{-1}(p))$, but for the singular part one could take a representative with either $u = p$ on the support of $D^su$ or one with $u \neq p$, which would change the answer. – mlk Apr 6 at 12:54
• Yes, but as stated in my previous comment, as long as there is any Cantor-part, there is always a representative for which the statement is false. I.e. you can set $u=p$ on the zero-set $\mathrm{supp} D^su$, and then $|Du|(u^{-1}(p)) \geq |D^su|(u^{-1}(p)) = |D^su|(\mathbb{R}^n) > 0$. – mlk Apr 10 at 9:13
• Ah you’re right. The proper move here is to use its “precise representative” which is equal to its approximate limit everywhere except the jump part (which doesn’t exist in this case). I will update the post, thanks for pointing this out. – Nate River Apr 10 at 10:17

## 1 Answer

This is about the univariate case ($$n = 1$$) only, but way too long for a comment.

Step 1: notation. Let $$u$$ be a continuous function of bounded variation, let $$\mu = Du$$ be the corresponding Radon measure, let $$|\mu|$$ denote the variation measure of $$\mu$$, and let $$U$$ be the distribution function of $$|\mu|$$: $$U(x) = \begin{cases} |\mu|([0,x)) & \text{if } x \geqslant 0 \\ |\mu|([x, 0)) & \text{if } x < 0 \end{cases}$$ (so that $$|\mu| = D U$$). Let $$S$$ be the support of $$\mu$$, and let $$S'$$ be equal to $$S$$ right endpoints of every connected component of $$S$$ removed. Then $$|\mu|$$ assigns no measure to the complement of $$S'$$ (because it assigns no measure both to the complement of $$S$$ and to the countable set $$S \setminus S'$$), and $$U$$ is a one-to-one mapping of $$S'$$ onto some interval $$I$$. The notation would have been much simpler if we assumed that $$S$$ is an interval, but let us deal with the general case.

Define a measure $$\nu$$ on $$I$$ by the formula $$\nu(B) = \mu(U^{-1}(B)) = \mu(U^{-1}(B) \cap S') .$$ Since $$U : S' \to I$$ is a bijection and $$\mu(A) = \mu(A \cap S') = \nu(U(A))$$, we have $$\nu_\pm(B) = \mu_\pm(U^{-1}(B)) .$$ Therefore, $$|\nu|(B) = |\mu|(U^{-1}(B)) = |B|$$ is the Lebesgue measure of $$B$$ (as long as $$B$$ is contained in $$I$$, of course). It follows that $$\nu$$ is absolutely continuous on $$I$$, with density function equal to $$\pm 1$$ almost everywhere.

Let $$w(y) = u(U^{-1}(y))$$, where $$U^{-1}(y)$$ is the unique point $$x$$ in $$S'$$ such that $$U(x) = y$$ (note, however, that since $$v$$ is constant on every connected component of the complement of $$S$$, any other choice of $$x$$ would work equally well). It is easy to check that $$w$$ is the distribution function of $$\nu$$ (that is, $$\nu = D w$$), so that $$w'$$ exists almost everywhere, and $$|w'| = 1$$ almost everywhere.

Clearly, for any $$p$$ we have $$|\mu|(u^{-1}(\{p\})) = |\mu|(u^{-1}(\{p\}) \cap S') = |\nu|(U(u^{-1}(\{p\}))) = |\nu|(w^{-1}(\{p\})) = |w^{-1}(\{p\})|.$$ Note that $$w' = 0$$ almost everywhere on $$w^{-1}(\{p\})$$ (indeed: if $$w'(y)$$ exists for some non-isolated point $$y$$ in $$w^{-1}(\{p\})$$, then $$w'(y) = 0$$). However, $$|w'| = 1$$ almost everywhere. This means that $$w^{-1}(\{p\})$$ has zero Lebesgue measure, and consequently $$|\mu|(u^{-1}(\{p\})) = 0 ,$$ as desired.

I have no experience with multivariate functions of bounded variation, so I will stop here. Let me remark, however, that the multivariate case might follow from the univariate one if, for example, there is a bounded variation counterpart of the "absolutely continuous on lines" (ACL) characterisation of weakly differentiable functions.

• That’s a good idea on the ACL part! I believe there is a “BVL” (bounded variation on lines) characterisation of BV functions, provided in Ambrosio’s book, Functions of Bounded Variation and Free Discontinuity Problems. I know where to take a look now. Thanks! – Nate River Apr 6 at 0:30