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?
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$\begingroup$ Yes, it is true. I wonder if it could be true when the boundary is piece-wise Lipschitz. I think the easiest proof is to cover $\Omega$ by star-shaped domains and then prove the result for star-domains. Note that the convex hull of elements of your cover will cover the convex hull of $\Omega$. However, what worries me about this ad-hoc argument is I can't detect where I use the notion of $C^1$ boundary. $\endgroup$– Andrew StoutCommented Mar 7, 2013 at 19:56
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$\begingroup$ *...prove the result for star-shaped domains. $\endgroup$– Andrew StoutCommented Mar 7, 2013 at 19:57
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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 KallusCommented Mar 7, 2013 at 20:07
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$\begingroup$ Yes, this is true. $\endgroup$– Alex DegtyarevCommented May 16, 2014 at 5:15
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1$\begingroup$ I've merged in a similar question asking about non-convex polytopes, together with Douglas's answer. $\endgroup$– S. Carnahan ♦Commented May 17, 2014 at 4:00
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2 Answers
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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.
answered May 16, 2014 at 5:30
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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.