Here is the claim:

Given a polytope $K$ in a unit ball in $\mathbb{R}^d$, there exists a universal constant $C(d)>0$ depending only on $d$ and a countable collection of convex polytopes $\{K_i\}_{i=1}^{\infty}$ such that $$K \subset \cup_{i=1}^{\infty} K_i$$and $$\sum_{i=1}^{\infty}S(K_i) \le C(d)S(K),$$where $S$ denotes the surface area.


One of the motivation of the claim is from the link I gave below. The other motivation is from my research. In my research I use again and again polytopes to approximate sets of finite perimeter (this is a term in geometric measure theory, basically means the finite surface area in a distributional sense), and in some cases, if I can use convex sets to approximate sets of finite perimeter, then it makes a difference, and some results can be refined.

In 2D the claim is straightforward, because the surface area is the perimeter, and taking a convex hull decreases the perimeter, thus the convex hull of $K$ is the natural cover, and $C(2)=1$.

In 3D, the argument in 2D is never true, see here the perimeter of a non-convex set.

I cannot prove or disprove this claim. I don't know the right tools to describe nonconvex polytope, because it can be composed of a lot of thick parts and thin parts, and their numbers cannot be controlled. The polytope can be as crazy as possible, but I guess there should be a way to cover it in a way as the claim says.

I haven't found any references dealing with this problem. I'm a beginner in convex geometry. Can anyone given me some references related to this problem? Thanks very much!

  • $\begingroup$ The complexity of the problem strongly depends on what exactly is meant by "polytope" and "covers". For example, if we interpret $K$ as the union of its facets (i.e. the boundary, or "area") then there is a straightforward solution: just cover each facet with a thin parallel slab, without increasing the area too much. However, if you want to cover the "volume", then very intricate phenomena can happen. I can elaborate on some of it if needed. $\endgroup$ – Victor Protsak Jul 3 '15 at 15:18
  • 1
    $\begingroup$ @Victor Protsak, I'm considering covering the volumn while at the same time without increasing surface area too much. Can you elaborate on some of the intricate phenomena? I know the "polytope" can be as wild as possible, but as far as I can imagine, there exists such a cover. $\endgroup$ – student Jul 3 '15 at 19:29
  • $\begingroup$ Please clarify what you mean by polytope. In the usage I am familiar with, it means the same thing as "convex polytope", but that can't be what you mean here. $\endgroup$ – Hugh Thomas Jul 6 '15 at 16:02
  • $\begingroup$ @Hugh Thomas, I used the definition from en.wikipedia.org/wiki/Polytope , basically means open set with piecewise affine hyperplanes. $\endgroup$ – student Jul 7 '15 at 15:29
  • $\begingroup$ The wikipedia page says that there are many different definitions, and does not settle on one. The following is one of the definitions. Does it capture the polytopes that you want to ask about? "A polytope [is] a set of points that admits a simplicial decomposition. In this definition, a polytope is the union of finitely many simplices, with the additional property that, for any two simplices that have a nonempty intersection, their intersection is a vertex, edge, or higher dimensional face of the two." $\endgroup$ – Hugh Thomas Jul 8 '15 at 4:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.