Relative isoperimetric inequality on bounded connected Lipschitz domains

The empty domain gives the inequality with both sides zero, so assume $\mathrm{\Omega }\mathrm{\ne }\mathrm{\varnothing }$. Its openness and nonemptiness give an open box with positive side lengths contained in $\mathrm{\Omega }$, and its boundedness puts $\mathrm{\Omega }$ inside a bounded box; [F5] therefore gives $0<\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }<\mathrm{\infty }$. For $m=\min (\mathrm{\mid }E\mathrm{\mid },\mathrm{\mid }\mathrm{\Omega }\setminus E\mathrm{\mid })$, take $A$ to be the smaller side. Its indicator is BV: the distributional derivative of the constant function $1$ in $\mathrm{\Omega }$ is zero, so $D{\chi }_{\mathrm{\Omega }\setminus E}=-D{\chi }_E$ and both sides have the same relative perimeter $P$. Its zero extension has $L^{n\mathrm{/}(n-1)}$ norm $m^a$. By [F1] and [F4], $m^a\le C(m+P)$ for a fixed $C\ge 1$ depending only on $\mathrm{\Omega }$. If $0<m\le \delta =(2C{)}^{-n}$, then $Cm\le m^a\mathrm{/}2$, so $m^a\le 2CP$. For $m=0$ the conclusion is immediate.

Among sets with $\delta \le \mathrm{\mid }E\mathrm{\mid }\le \mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }-\delta$, the relative perimeters have a positive lower bound, unless that family is empty. Otherwise, if the infimum were zero, Countable Choice would select a sequence $E_j$ in this family with $P(E_j;\mathrm{\Omega })<1\mathrm{/}j$. Its indicators are uniformly BV-bounded. By [F2], a subsequence converges in $L^1(\mathrm{\Omega })$ to $v\in BV(\mathrm{\Omega })$ with $TV(v;\mathrm{\Omega })=0$. The distance from $v(x)$ to $\{0,1\}$ is bounded by $\mathrm{\mid }v(x)-{\chi }_{E_j}(x)\mathrm{\mid }$ for every $j$, so its integral is zero by $L^1$ convergence; hence $v$ is an indicator almost everywhere. Its volume lies in $[\delta ,\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }-\delta ]$ by $L^1$ convergence. But its zero variation makes every coordinate distributional derivative zero, so [F3] makes $v$ constant almost everywhere. An indicator constant must be zero or one, contradicting the volume bounds. This proves a positive lower bound $b_{\mathrm{\Omega }}$ whenever the family is nonempty.

In the remaining case $m\ge \delta$, both side volumes meet the bounds in step 2.1, so $P\ge b_{\mathrm{\Omega }}$ and $m^a\le (\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }\mathrm{/}2{)}^a\le (\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }\mathrm{/}2{)}^a{b}_{\mathrm{\Omega }}^{-1}P$. Combine this constant with $2C$ from step 1.1. If the intermediate-volume family is empty, step 1.1 already covers every set. Thus a finite domain constant gives the asserted inequality for every finite-perimeter $E$.