the perimeter of a non-convex set Let $\Omega$ be an open, bounded set in $\mathbb{R}^n$ with $C^1$ boundary. Is it true that the perimeter of the convex hull of $\Omega$ is smaller or equal the perimeter of $\Omega$ with equality if and only if $\Omega$ is a convex set? Can one always speak about the perimeter of the convex hull?
 A: While it is probably true in the case $n=2$ it's  false in dimension bigger than $2$.
The counterexample I have in mind  works as follows in $\mathbb{R}^3$. Take two  disks parallel with the $xy$ plane and  of the same radius centered on the $z$-axis and fatten them up just a little bit. Built a thin cylindrical tube that connects them (along the $z$-axis). The tube should be long enough, hence the disks should lie at quite a bit of distance. One can of course smooth the corners here without adding too much area. The convex hull will be a "cylinder" with  radius equal to the radius of the original disks and height  equal to the length of the tube plus two times the heights of the (fattened) disks. One can use appropriate values to check that the area of the convex hull is bigger than the area of the original set.
A: In dimensions $3$ or greater, it is false. Take some non-coplanar points. You can connect them with a nonconvex polyhedron with arbitrarily small surface area by thickening a spanning tree, for example. The convex hull has surface area at least as great as the surface area of the convex hull of the points.
In dimension $2$, it is true. In two dimensions, the convex hull operation decreases the number of times that a line intersects the boundary, so by the Cauchy surface area formula, it decreases the perimeter.
