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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Looking for the proof of the lemma asserting that the conical surface (envelope) is a closed space. Thank you.

share|cite|improve this question
2  
Please give more details, so that we know what you have in mind. Also, writing complete words is usually a great idea! :) – Mariano Suárez-Alvarez Jan 8 '11 at 1:10
1  
If by "conical surface" you mean this: en.wikipedia.org/wiki/Conical_surface and if by "closed" you mean this: en.wikipedia.org/wiki/Closed_set then the lemma is false and there is no proof. You need some additional condition like having a closed directrix. – Ryan Budney Jan 8 '11 at 1:21
    
Please see my answer below. – mira Jan 8 '11 at 11:36

We are talking here about the convex cone $C:={\rm co}(A)$ spanned by a finite set $A$ of vectors $a\in {\mathbb R}^n$. Consider an arbitrary point $x\in C$. It has a representation $x=\sum_{i=1}^m \lambda_i a_i$ with $a_i\in A$, $\lambda_i>0$ and minimal $m$. I claim that the $a_i$ occurring in this representation are linearly independent. (This is essentially Caratheodory's theorem.)

Proof: If one had $\sum_{i=1}^m \mu_i a_i=0$, where not all $\mu_i$ are $0$, then after rearranging one may assume that for some $s\leq m$ one has $$\mu_i\leq 0 \quad (1\leq i\leq s-1), \qquad \mu_i > 0 \quad (s\leq i\leq m), \qquad {\lambda_i\over\mu_i}\geq {\lambda_m\over\mu_m} \quad (s\leq i\leq m).$$ It follows that $$x=\sum_{i=1}^{m-1} (\lambda_i-{\lambda_m\over\mu_m} \mu_i) a_i$$ were a representation of $x$ with nonnegative coefficients using at most $m-1$ of the $a_i$.

Now the cone spanned by a linearly independent set of $m$ vectors $a_i$ is homeomorphic to the set $$\{x\in {\mathbb R}^n\ |\ x_i\geq 0 \ (1\leq i\leq m), \ x_i=0 \ (m+1\leq i\leq n)\}$$ which is clearly a closed set. As $C$ is a union of finitely many such sets it is closed as well.

share|cite|improve this answer

Your Answer

 
discard

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