Would any two consecutive extreme points on a convex hull can be linearly projected still being extreme? 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
 A: 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.
