# Convex Hull

• Aug 30th 2010, 01:22 AM
qspeechc
Convex Hull
Hello Everyone.

While I was reading along in a book I came across this: It says the convex hull is
$\displaystyle co(\{ x_n : n\in \mathbb{N}\}) = \{ \sum _{i=1}^m \alpha _i x_i : \sum |\alpha _i|\leq 1\}$
The $\displaystyle x_n$ are vectors in a vector space and the $\displaystyle \alpha _n$ complex scalars. But from my understanding (and wikipedia verifies this) the definition of a convex hull requires $\displaystyle \sum |\alpha _i|=1$. Normally I'd just think this was a typo, but the text goes on to use the fact that $\displaystyle \sum |\alpha _i|\leq 1$. So are the two things equivalent or what? I can't see how they are. (Crying). Help please!
• Aug 30th 2010, 05:19 AM
Iondor
The definitions are only equivalent, if 0 is one of the $\displaystyle x_{n}$.
I guess one could prove the following:
$\displaystyle co(\{x_{n}-x_{0} : n \in \mathbb{N}\})+x_{0}={ \{ \sum _{i=0}^m \alpha _i x_i : \sum |\alpha _i|= 1\}$
• Aug 31st 2010, 08:23 AM
qspeechc
No 0 is not one of them. I do not see how $\displaystyle co(\{x_{n}-x_{0} : n \in \mathbb{N}\})+x_{0}={ \{ \sum _{i=0}^m \alpha _i x_i : \sum |\alpha _i|= 1\}$ helps...?
• Aug 31st 2010, 11:17 AM
Iondor
Quote:

Originally Posted by qspeechc
No 0 is not one of them. I do not see how $\displaystyle co(\{x_{n}-x_{0} : n \in \mathbb{N}\})+x_{0}={ \{ \sum _{i=0}^m \alpha _i x_i : \sum |\alpha _i|= 1\}$ helps...?