Let $Psubseteq mathbb{R}^n$ be a full dimensional polytope. Let us assume that $P$ has a facet description with the following inequalities:

$$ left<x,u_Fright> geq -a_F$$

where $u_Fin mathbb{R}^n$ and $a_Fin mathbb{R}$ for each facet $F$ of the polytope.

In other words, $P$ is the bounded intersection of the halfspaces $H_{u_F,-a_F}^+$. For each $Q$ face of $P$ we define the cone

$$sigma_Q := operatorname{Cone}(u_F: F text{ facet, } Fsupseteq Q)$$

What I want is to prove that $$sigma_Q = {uin mathbb{R}^n :

left<x,uright> leq left<y,uright> text{ for all } xin Q, yin P}$$

My thoughts: It is easy to prove one of the inclusions. For the other, I wanted to use (one of the many versions of) Farkas Lemma.

If $uin mathbb{R}^n$ satisfies $left<x,uright>leq left<y,uright>$ for each $xin Q$ and $yin P$, in particular, there exists a face $Q’$ of $P$ such that $Q’ = P cap H_{u,a}$ for certain $a=left<x,uright>$ for any $xin Q$ (this value is independent of $xin Q$), also $Qsubseteq Q’$ and $Psubseteq H_{u,a}^+$. I tried to use Farkas Lemma II (see Ziegler’s book). So, to conclude that $uin sigma_Q$, I need to prove that there cannot exist a vector $min mathbb{R}^n$ with the property that:

$$left<m,u_Fright> geq 0 text{ for all facets } Fsupseteq Q’$$

and simultaneously

$$left<m,uright> < 0.$$

To get a contradiction I think it suffices to show that one can find a point $xin Psmallsetminus Q’$ such that $x = q – lambda m$ for a $qin Q’$ and $lambda > 0$.

In that case, $xin Psmallsetminus Q’$ implies that for some facet $F$ with $Fsupseteq Q’$ it has to be: $left<x,u_Fright> > -a_F$, which yields $left<q,u_Fright> – lambdaleft<m,u_Fright> > -a_F$. The fact that $qin Q’$ implies that $left<q,u_Fright> = -a_F$, from where it follows $lambdaleft<m,u_Fright> < 0$ and $lambda > 0$ yields $left<m,u_Fright> < 0$ which is the desired contradiction. I am having trouble to prove the existence of such an $x$.

Of course, a different approach to prove this is also welcome!