Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Can someone help me solve the following question please?

Let v be a vertex of a d-polytope P such that $ 0 \in intP $ . Prove that $ P^{*} \cap \{ y \in \mathbb{R}^d \mid\left < y, v\right>=1\ \} $ is a facet of $P^{*} $.

The definitions are: $P^*=\{ y\in\mathbb{R}^{d}\mid\left < x, y\right>\leq 1\ \forall x\in P\} $ and a face of P is the empty set, P itself, or an intersection of P with a supporting hyperplane (i.e.- a hyperplane, such that P is located in one of the halfspaces it determines). A facet is a face of maximal degree

I tried showing that if there exists a vertex v such that this isn't a facet, then P is a convex hull of a finite set not containing v, which is a contradiction, but without success.

HOpe you'll be able to help me

share|improve this question
Have you tried any other proof directions that do not involve contradiction? –  Samuel Reid Jan 14 '12 at 17:38
This analogy may help: Suppose $P$ is determined by a minimal set of inequalities $f_i(x) \leq c_i$. Then each set $P \cap \{ x \in \mathbb{R}^d \;|\; f_i(x) = c_i\}$ is a facet of $P$. –  Shaun Ault Jan 14 '12 at 18:12
Actually, I've already thought of this analogy but without success... I've also tried using the fact that $ P^* = \cap_{v \in V} D_0 (v) ^{-} $ where $D_0(v)^-= \{y \in \mathbb{R}^d | <y,v> \leq 1 \} $ . –  joshua Jan 15 '12 at 6:55
Hope you'll be able to help me continue... Thanks a lot! –  joshua Jan 15 '12 at 6:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.