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Let $K_1,\ldots,K_m\subset\mathbb{R}^n$ with $m\geq n$ some convex bodies (i.e. compact with nonempty interior). I am interested in sufficient criteria for the convex hull $K=\textrm{conv}(K_1,\ldots,K_m)$ to have a face of dimension $n-1$. Are there some nice criteria known?

In particular, I am interested in the situation where for each subset $I\subset\{1,\ldots,m\}$ the convex hulls $\textrm{conv}(K_i,\,i\in I)$ and $\textrm{conv}(K_j,\,j\not\in I)$ are disjoint. This seems to be sufficient for $K$ having a face of dimension $n-1$ in small dimensions but I can't prove it in general. Is it true for all $m\geq n$?

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  • $\begingroup$ One way would be to show that for some $c\in S^{n-1}$ the maximum $\max_{x_i\in K_i}\langle c,x_i\rangle$ takes on the same value for $n$ indices $i\in \{1,...m\}$. This would give you the hyperplane which defines the face. $\endgroup$
    – M. Winter
    Commented Jan 31, 2018 at 16:13
  • $\begingroup$ Sure. But I don't really see how this makes the problem easier... $\endgroup$
    – Hans
    Commented Jan 31, 2018 at 17:00
  • $\begingroup$ By Radon's theorem, the condition in the second paragraph implies that $m\le n+1$. $\endgroup$
    – Jan Kyncl
    Commented Jan 31, 2018 at 23:28
  • $\begingroup$ Since each subset of the bodies $K_i$ can be separated by a hyperplane from the other bodies, and the function defined by M. Winter seems to be continuous in $c$, some fixed-point theorem might be useful. $\endgroup$
    – Jan Kyncl
    Commented Jan 31, 2018 at 23:45
  • $\begingroup$ Yes, $m=n+1$ is indeed the case I am interested in. I think I have an idea now how to prove it. Thanks for your input (both of you)! $\endgroup$
    – Hans
    Commented Feb 1, 2018 at 15:52

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