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Given two extreme points on a convex hull, if the straight line connects them is on the boundary , they are consecutive. Would there always exits a linear projection into lower dimension such that those two points are still extreme?

For example a convex hull in 2d, a rectangle, the claim is true, because line projection perpendicular to the line connect Consecutive extreme points (sides) is a valid one

peng yu
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  • Is this in a Euclidean space? – Zim Jul 03 '20 at 02:59
  • Yes, it’s in Euclidian space. – peng yu Jul 03 '20 at 03:00
  • If affine is easier, it’s would be very helpful to have :) – peng yu Jul 03 '20 at 03:22
  • Let $c_1,c_2$ be extreme points on a closed convex hull $C$. Then they share an outward normal vector $u$. Geometrically it looks like projecting onto, say, $c_1+\textrm{span}{u}^\perp$ should contain $C$ on one side of it, while keeping the projection of $[c_1,c_2]$ on the boundary. However, guaranteeing it is a lower-dimensional projection is another question. – Zim Jul 03 '20 at 03:28
  • @Zim I don't think this keeps the projection of $[c_1, c_2]$ on the boundary, even for polygons in $\mathbb{R}^2$, unless I'm misunderstanding. For example, consider two adjacent vertices of a regular hexagon. The line between them has a unique outward-pointing unit normal vector, but any projection onto the orthogonal complement of this normal vector maps both of the original vertices to interior points of the projection (which is an interval). – diracdeltafunk Jul 03 '20 at 21:08
  • @diracdeltafunk Oh I'm sorry I think I meant $c_1 + \textrm{span}{u}$. The idea is that $[c_1,c_2]$ projects onto the same point ($c_1 + 0$), and the rest of $C$ should project onto one side of that line since the hyperplane with normal $u$ at $c_1$ is tangent to $C$. – Zim Jul 03 '20 at 22:45

1 Answers1

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Yes, since $L\colon\mathbb{R}^N\to\mathbb{R}^N\colon x\mapsto 0$ is a projection

EDIT: If you allow the projection to be affine, then I believe there is also a nontrivial projection; here is a draft sketch: Let $[c_1,c_2]$ be the line segment on the boundary of the closed convex hull $C$. Then there exists an outer normal vector $u\in\mathbb{R}^N\setminus\{0\}$ shared by everything in $[c_1,c_2]$. In particular, \begin{equation} (\forall\lambda\in[0,1])(\forall c\in C)\quad \langle c- (\lambda c_1 + (1-\lambda)c_2)\, |\, u\rangle \leq 0 \tag{1} \end{equation} and \begin{equation} (\forall \lambda\in[0,1])\quad \langle (\lambda c_1 + (1-\lambda)c_2 - c_1\, | \, u\rangle = 0. \tag{2} \end{equation} Let $A=\{c_1+\lambda u\, | \, \lambda\in\mathbb{R}\}$ be the affine space, and note it has dimension $1$ (so it is lower-dimensional unless $C$ is a singleton point or a line segment, both of which are addressed by my first linear result.) Note that, for every $x\in\mathbb{R}^N$, $\textrm{Proj}_Ax=c_1+\textrm{Proj}_{\textrm{span}\{u\}}(x-c_1)$. It follows from (1) that $\textrm{Proj}_A(C)=\{c_1+\lambda u\, | \lambda \leq 0\}$. It follows from (2) that $\textrm{Proj}_{A}([c_1,c_2])=\{c_1+0\}$, i.e. it is on the boundary of the image.

Zim
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