Half-space comparison of perimeter - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T06:09:32Z http://mathoverflow.net/feeds/question/56176 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56176/half-space-comparison-of-perimeter Half-space comparison of perimeter Martijn 2011-02-21T13:51:37Z 2011-02-23T09:42:11Z <p>Claim: suppose that $E$ is a set of finite perimeter, and $H$ is a half space. Then $P(F\cap H)\le P(F)$. In words: restricting a Caccioppoli set to a half-space will not increase the perimeter.</p> <p>My question: how to prove this? </p> <p>I have a feeling it should be very simple, as in Almgren, Taylor, Wang's paper "Curvature-driven flows: a variational appraoch", it is claimed that it follows from "Stokes' theorem and Jensen's inequality". However, I'm having trouble to apply Jensen's inequality to this problem, except in trivial cases that boil down to linearity of the integral.</p> <p>EDIT: Although no satisfying answer has been given as to what Jensen has to do with this theorem, a complete proof has been given. I've accepted this as an answer.</p> http://mathoverflow.net/questions/56176/half-space-comparison-of-perimeter/56341#56341 Answer by Alexander Volkmann for Half-space comparison of perimeter Alexander Volkmann 2011-02-23T00:05:05Z 2011-02-23T00:05:05Z <p>Hello Martijn,</p> <p>i guess the following argument should work:</p> <p>For simplicity I shall assume that $E$ is bounded, i.e. there exists some ball $B_R(0)\subset \mathbb R^n$ strictly containing $E$. W.l.o.g. we may also assume that $H =$ {$x\in \mathbb R^n:x_n &lt; r$} for some $r$. It holds the following (see the book of Giusti for details) $$P(E \cap H)=P(E,H)+\int_{\partial H}\phi_E^+ d\mathcal H^{n-1},$$ where $\phi_E^+$ denotes the inner trace of the characteristic function $\chi_E$ of $E$ on $\partial H$. Now by definition of the trace operator we have (inserting the vector field $X=-\eta e_n$, where $\eta$ is some smooth cutoff function that equals $1$ on $B_R(0)$) \begin{align} 0&amp;=\int_{E\setminus \overline H}div X d\mathcal L^n=-\int_{\mathbb R^n \setminus \overline H} \langle e_n,\nu_E \rangle d\mu_E - \int_{\partial H}\phi_E^{-}\langle e_n,\nu_H\rangle d\mathcal H^{n-1}, \end{align} where $\nu_H$ denotes the outer unit normal to $H$, $\nu_E$ denotes the generalized outer unit normal of $E$ and $\phi_E^-$ is the outer trace of $E$ on $\partial H$. But since $\nu_H=e_n$ we get, using Cauchy-Schwarz' inequality: $$\int_{\partial H}\phi_E^- d\mathcal H^{n-1} \leq P(E,\mathbb R^n \setminus \overline H),$$ and so for a.e. value of $r$, $\phi_E^+=\phi_E^-$ $\mathcal H^{n-1}$-a.e. on $\partial H$, which yields $$P(E \cap H)\leq P(E,H)+P(E,\mathbb R^n \setminus \overline H)=P(E).$$ For an arbitrary value of $r$ just choose a sequence of values $r_k\to r$ for which this is satisfies. Then $E\cap H_{r_k} \to E \cap H$ in $L_{loc}^1(\mathbb R^n)$. Using the lower semi-continuity of the perimeter you get the claim for all values of $r$. </p>