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This is the isoperimetric type question. We know that in $\mathbb{R}^d$, balls are the sets that minimize the isoperimetric ratio $\frac{S^{d}}{V^{d-1}}$, where $S$ is the surface area and $V$ is the volumn.

Now consider the polytopes with $f$ faces. Lindelof's theorem says, among all proper convex polytopes in $\mathbb{R}^d$ with given exterior normals of the facets, it is precisely the polytopes circumscribed to a ball that have minimum isoperimetric quotient. This theorem can be found in http://link.springer.com/book/10.1007%2F978-3-540-71133-9, page 308, Theorem 18.4.

However, on Page 309, the author made a Corollary 18.2 that among all proper convex polytopes in $\mathbb{R}^d$ with a given number of facets, there are polytopes with minimum isoperimetric quotient and these polytopes are circumscribed to a ball.

Now my question is, I think the two claims above are different. To prove the Corollary 18.2, one has to prove the existence of the polytopes minimizing the isoperimetric constant among all polytopes circumscribed to a unit ball. I searched a lot of references, but I didn't find any proofs of such an existence. Is this an obvious result?

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    $\begingroup$ It follows since the space of configurations of $n$ unit vectors is compact. $\endgroup$
    – Anton Petrunin
    Commented Jul 3, 2015 at 20:44
  • $\begingroup$ @Anton Petrunin, I'm confused because the $f$ unit exterior normal vectors are given in Lindelof's Theorem, but in the Corollary, only the number of faces are given. Maybe I didn't understand your comment. What's the meaning of "the space of configurations of n unit vectors "? $\endgroup$
    – student
    Commented Jul 3, 2015 at 21:25

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This is an explanation of Anton's comment. For each set of $f$ unit vectors, one finds the polytope minimizing the isoperimetric quotient. The set of configurations of $f$ (not necessarily distinct) unit vectors is compact (it is homeomorphic to $\mathbb{S}^{(d-1) \times f}$), and the isoperimetric quotient is continuous thereon (this requires an argument), therefore it achieves its minimum for some set of $f$ unit vectors. These vectors might not be distinct a priori, but it is easy to see that counting a face with multiplicity (which is what "not distinct" means) does not help you. So, the faces are distinct, and by Lindelof the minimizer is circumscribed.

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  • $\begingroup$ Could you explain "counting a face with multiplicity does not help you"? I'm sorry I didn't understand... If $\frac{S^d}{V^{d-1}}=G(v_1, \dots \, , v_f)$ for some continuous function $G$, then what does it mean if the minimum is attained at a point of which two components are the same? $\endgroup$
    – student
    Commented Jul 4, 2015 at 0:51
  • $\begingroup$ @student The volume is the same as when you would count the face one, but the area is bigger. $\endgroup$
    – Igor Rivin
    Commented Jul 4, 2015 at 1:08
  • $\begingroup$ Sorry I still didn't get it. If $\min_{(v_1, \dots \, ,v_f) \in \mathbb{S}^{(n-1) \times f}}\frac{S(v_1, \dots \, ,v_f)^d}{V(v_1, \dots \, ,v_f)^{d-1}}$ is attained at $(v_1, v_1, v_2, \dots \, ,v_{f-1})$, then this means the minimizer is polytope with $f-1$ faces with outer normals $v_1, \dots \, ,v_{f-1}$, but if then how to conclude this is not the minimizer? My main confusion is, I don't know the correspondence of the $(v_1, v_1, \dots \, ,v_{f-1})$ to the geometry of the convex body. It seems that you interpreted it in a different way I explained. $\endgroup$
    – student
    Commented Jul 4, 2015 at 1:26
  • $\begingroup$ I interpreted it as a polytope with $f-1$ faces with outer normals $v_1, \dots \, ,v_{f-1}$, but you suggested it should mean counting the multiplicity of the face with $v_1$ vector. Maybe in order to understand the argument, I need to know the expression of the isoperimetric ratio function? $\endgroup$
    – student
    Commented Jul 4, 2015 at 1:32

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