Let $u\in BV(\mathbb{R}^n)$ be a function of bounded variation, and let $x\in J_u$ be a point in its jump set. For $\mathcal{H}^{n-1}$-a.e. such $x$, we can define the unit normal $\nu$ to the boundary, and an upper and lower limit $u^+$ and $u^-$ such that $$ \lim_{r\to 0} \frac{1}{|B_r(0)|} \int_{B_r(0)\cap H^+_\nu} |u(x+y)-u^+|\,dy = 0 $$ and $$ \lim_{r\to 0} \frac{1}{|B_r(0)|} \int_{B_r(0)\cap H^-_\nu} |u(x+y)-u^-|\,dy = 0 $$ where $H^+_\nu$ and $H^-_\nu$ represent the upper and lower half-spaces with respect to the unit normal $\nu$. This gives us convergence of the blow-ups of $u$ to a piecewise constant function across a flat jump in $L^1$. In fact, according to the textbook by Evans and Gariepy we can obtain convergence in $L^{n/n-1}$.

My question is about the convergence of the blow-ups of $Du$, the distributional gradient of $u$. In the case $u=1_E$ that $u$ is the indicator function for a set of finite perimeter $E$, a consequence of De Giorgi's structure theorem is that the blowups $Du$ (and more interestingly, of $|Du|$) converge to the the surface measure of a hyperplane. I was wondering whether this is still true in the case of a general function of bounded variation.

Actual Question

More precisely, let $|Du_{x,r}|$ denote the blowup of the measure $|Du|$ at the point $x$ with scale $r$, defined by $|Du_{x,r}|(A) = |Du|(rA+x)$ for Borel sets $A\subset \mathbb{R}^n$. The question is then:

Does the sequence of blow-ups $|Du_{x,r}|$ converge weak-* to the surface measure of the half-plane $H_\nu$ for $\mathcal{H}^{n-1}$-a.e. $x\in J_u$?


1 Answer 1


I think you forgot to divide by $r^{n-1}$. Even for sets of finite perimeters what you are claiming is not true. If you take $E=\{(y,t)\in\mathbb{R}% ^{n-1}\times\mathbb{R}:\,t<f(y)\}$, where $f$ is a very regular function, then $\partial E$ is just the graph of $f$ and so the perimeter of $E$ is just the surface area given by $$ \Vert\partial E\Vert(G)=\int_{G\cap\partial E}1\,d\mathcal{H}^{n-1}=\int% _{F}\sqrt{1+|\nabla f(y)|^{2}}dy $$ where $F=\{y\in\mathbb{R}^{n-1}:\,(y,f(y))\in G\}$. So if you take $G=rB(0,1)+x=B(x,r)$ with $x=(y_{0},f(y_{0}))$ you get $$ \Vert\partial E\Vert(B(x,r))=\int_{F_{r}}\sqrt{1+|\nabla f(y)|^{2}}dy $$ where $F_{r}=\{y\in\mathbb{R}^{n-1}:\,(y,f(y))\in B((y_{0},f(y_{0})),r)\}$. When $r\rightarrow0$ you have that $\chi_{F_{r}}\rightarrow\chi_{\{(y_{0}% ,f(y_{0}))\}}$ and so the integral goes to zero by the Lebesgue dominated convergence theorem.

Instead if you divide by $r^{n-1}$ you get $$ \frac{1}{r^{n-1}}\int_{F_{r}}\sqrt{1+|\nabla f(y)|^{2}}dy $$ which converges.

For $BV$ functions you can prove that $Du$ can be written as the sum of three mutually singular parts, $$ Du=g\mathcal{L}^{N}+(u^{+}-u^{-})\nu_{u}\mathcal{H}^{n-1}\lfloor J_{u}+D^{c}u, $$ where $D^{c}u$ is the Cantor part. The total variation will be $$ \Vert Du\Vert(G)=\int_{G}|g|\,dx+\int_{G\cap J_{u}}|u^{+}-u^{-}|\,d\mathcal{H}% ^{n-1}+\Vert D^{c}u\Vert(G). $$ So if you take $G=rA+x$ with $x\in J_{u}$ and divide by $r^{n-1}$, by Besicovitch theorem you will have that the only relevant term is $\frac {1}{r^{n-1}}\int_{(rA+x)\cap J_{u}}|u^{+}-u^{-}|\,d\mathcal{H}^{n-1}$ and using graphs of functions as above you should prove that this goes weakly star to $|u^{+}(x)-u^{-}(x)|\mathcal{H}^{n-1}\lfloor H_{x}$, where $H_{x}$ is the hyperplane perpendicular to $v_{u}(x)$. A good reference is Ambrosio, Fusco, Pallara Theorems 3.77 and 3.78.

  • $\begingroup$ I am confused by your sentence "Even for sets of finite perimeters what you are claiming is not true." Can you state what it is I said about sets of finite perimeter that isn't true? $\endgroup$
    – felipeh
    Aug 31, 2017 at 13:31
  • $\begingroup$ The measures $|\partial E|(rA+x)$ do not converge weakly star to the surface measure on an hyperplane. They converge strongly to zero. It's the example at the beginning of my answer. $\endgroup$
    – Gio67
    Aug 31, 2017 at 13:35
  • $\begingroup$ Oh I see. I suppose I used the term "blowup" incorrectly, since of course $r^{n-1}$ is the correct scaling. $\endgroup$
    – felipeh
    Aug 31, 2017 at 13:36
  • $\begingroup$ yes, after rescaling everything is fine and what you want is in the book of Ambrosio Fusco Pallara $\endgroup$
    – Gio67
    Aug 31, 2017 at 13:37
  • 1
    $\begingroup$ Thanks! Another good book is Maggi's Sets of finite perimeter. $\endgroup$
    – Gio67
    Aug 31, 2017 at 13:39

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