1
$\begingroup$

Consider a positive definite locally Lipschitz function $V:\mathbb{R}^2\to\mathbb{R}_{\geq0}$.

Fix $c\in\mathbb{R}_{\geq0}$ and consider the sublevel-set $E_c=\{x:\in\mathbb{R}^2:V(x)\leq c\}$, define the function $v_c:\partial E_c\to\mathbb{R}^2$ given by \begin{equation*} v_c(x)=\dfrac{\nabla\chi_{E_c}(x)}{|\nabla\chi_{E_c}(x)|}, \end{equation*} where $\chi_{E_c}$ is the indicator function of $E_c$.[1]

I would like to define an indicator function $\xi$ for each sublevel set of $V$ whose gradient coincide the gradient of $V$. More precisely, for every $c\in\mathbb{R}_{\geq0}$, define a function $\xi_c:\mathbb{R}_{\geq0}\to[0,1]$ such that \begin{equation*} \xi_c(x)=\left\{\begin{array}{rclrcl} 1,&\text{if}&x&:V(x)\leq c\\ 0,&\text{if}&x&:V(x)\neq c, \end{array}\right. \end{equation*} and \begin{equation*} \nabla \xi_c(x)= \nabla V(x) \end{equation*}

I have no clue how to defined it. I would like to use it for sets of finite perimeter [1] satisfying Federer's definition of measure-theoretic normal to set [2].

[1] http://en.wikipedia.org/wiki/Caccioppoli_set#Notions_of_boundary

[2] Pfeffer, F. W. - The divergence theorem and sets of finite perimeter.

$\endgroup$

1 Answer 1

3
$\begingroup$

I think you misunderstand the basic concepts of the subject. The expression $\nabla \chi_{E_c}$ is not "a function $v:\partial E_c\to\mathbb R^2$", but a distribution. If your sublevel set $E_c$ happens to be a Caccioppoli set (which is not guaranteed; the sublevel set of a Lipschitz function can be any closed set) then $\nabla \chi_{E_c}$ is a vector-valued Radon measure supported on $\partial E_c$. It is still not a function; thus, it does not make sense to ask for it to agree with $\nabla V$.

Besides, $\nabla V$ itself is guaranteed to exist only a.e. (by Rademacher's theorem) and since $\partial E_c$ usually has zero Lebesgue measure, you may well have a situation where $\nabla V$ is not defined at any point of $\partial E_c$. So in general, you are trying to get a non-function to agree with a non-existent function.

$\endgroup$
5
  • $\begingroup$ Thank you very much for your answer. Indeed, I misunderstood the concepts that now became more clear after your explanation. Now, assuming that $E_c$ is a Caccioppoli set, then by De Giorgi's Theorem [1], the distribution \begin{equation*} \dfrac{\nabla \chi_{E_c}(B_{\rho}(x))}{|\nabla \chi_{E_c}|(B_{\rho}(x))} \end{equation*} converge to a vector $v(x)\in\mathbb{R}^2$, as $\rho\to0$. It is such that, for all $y\in\mathbb{R}^2$ and $x\in \partial_\ast E_c$, $v(x)\cdot y>0$. From the uniqueness of the normal vector, can I conclude that, for a.e. $x\in\partial_\ast E_c$, $v(x)=\nabla V(x)$? $\endgroup$ Mar 3, 2014 at 19:15
  • $\begingroup$ @Humberto You still need to assume that $\nabla V(x)$ exists, because this is not guaranteed a.e. on the boundary. But if it does, then I think you can prove this. With enough control on $V$, the sublevel set will look like half-space with normal vector $\nabla V(x)$ near $x$. $\endgroup$
    – user127096
    Mar 3, 2014 at 20:27
  • $\begingroup$ I still have a question concerning the existence of $\nabla V$. If $V$ is $\mathscr{H}^1$-a.e. differentiable in $E_c$ (i.e., differentiable in almost every $x\in E_c\subset\mathbb{R}^2$, in the unidimensional Hausdorff measure), then does it imply that $\nabla V(x)$ exist almost everywhere in $E_c$ in the Lebesgue measure in $\mathbb{R}^2$? $\endgroup$ Mar 3, 2014 at 21:50
  • $\begingroup$ @Humberto On $E_c$, yes. But you need it on the boundary $\partial E_c$, which is typically a set of Lebesgue measure zero. So, differentiability a.e. yields no conclusion about the existence of $\nabla E_c$ on the boundary. $\endgroup$
    – user127096
    Mar 3, 2014 at 22:27
  • $\begingroup$ Thank you very much for the discussion. I found a paper that gives a sufficient condition for the differentiation on the boundary $\partial E_c$. Thus, $\nabla V(x)$ is an outward normal to $E_c$, for $\mathscr{H}^1$-a.e. $x\in\partial E_c$. In case you're interested.\\ Alberti, G.; Bianchini, S.; Crippa, G. - Structure of level sets and Sard-type properties of Lipschitz maps, Ann. Scuola Norm. Sup. Pisa Cl. Sci., 2013. $\endgroup$ Mar 3, 2014 at 23:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .