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?

$\begingroup$ Yes, it is true. I wonder if it could be true when the boundary is piecewise Lipschitz. I think the easiest proof is to cover $\Omega$ by starshaped domains and then prove the result for stardomains. Note that the convex hull of elements of your cover will cover the convex hull of $\Omega$. However, what worries me about this adhoc argument is I can't detect where I use the notion of $C^1$ boundary. $\endgroup$ – Andrew Stout Mar 7 '13 at 19:56

$\begingroup$ *...prove the result for starshaped domains. $\endgroup$ – Andrew Stout Mar 7 '13 at 19:57

1$\begingroup$ I wonder what you mean by the perimeter? Is this the same as the surface area in $R^3$? If so, then what about the catenoid? Its convex hull has a larger surface area (the corners can be smoothed out to be $C^1$). $\endgroup$ – Yoav Kallus Mar 7 '13 at 20:07

$\begingroup$ Yes, this is true. $\endgroup$ – Alex Degtyarev May 16 '14 at 5:15

1$\begingroup$ I've merged in a similar question asking about nonconvex polytopes, together with Douglas's answer. $\endgroup$ – S. Carnahan♦ May 17 '14 at 4:00
In dimensions $3$ or greater, it is false. Take some noncoplanar 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.
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.