# Extreme points and positive linear combinations

Let $C$ be a non-empty convex subset of $\mathbb{R}^n$. We say that $x\in C$ is a extreme point of $C$ if for every $z,y\in C$ and $t\in [0,1]$ such that $x=ty+(1-t)z$ we have $x=z$ or $x=y$. Or equivalently, if for every $z,y\in C$ and $t\in (0,1)$ such that $x=ty+(1-t)z$ we have $x=z=y$.

I want to prove the following:

If $x$ is a extreme point of $C$, then for every $x_1,...,x_m\in C$ and $\lambda _1,...,\lambda _m >0$ such that $x=\lambda _1x_1+...+\lambda _mx_m$ we have $x=x_i$ for some $i$.

Any hint? Thanks.

• That's false like that. Take $C=[0,1]$. Then $1=2\cdot\frac{1}{4}+1\cdot\frac{1}{2}$. It is true if you add the condition $\sum_{j=1}^m\lambda_j=1$. And then it follows from a straightforward induction on $m$. – Julien May 4 '13 at 22:12
• Thank you. But how to manipulate the expression $x = \displaystyle\sum_{j=1}^{m+1}\lambda _jx_j$ if $\displaystyle\sum_{j=1}^{m+1}\lambda _j = 1$ to apply induction? – user73564 May 4 '13 at 22:45
• If $\lambda_{m+1}=1$, you're done. If not $x=(1-\lambda_{m+1})y+\lambda_{m+1}x_{m+1}$ with $y=\sum_{j=1}^m\frac{\lambda_j}{1-\lambda_{m+1}}x_j$. – Julien May 4 '13 at 22:50
• You can write the answer and accept it, if you want. Note that the case $\lambda_{m+1}=1$ I considered was pointless. This can't happen, when you suppose $\lambda_j>0$ for every $j$ and $\sum \lambda_j=1$. – Julien May 4 '13 at 23:01
• Maybe it is another mistake. I think it must be $\lambda _j\ge 0$. If we put $\lambda _j > 0$, the conclusion would be $x=x_1=...=x_m$ instead $x=x_i$ for some i. What do you think? – user73564 May 4 '13 at 23:18

1. It must be assumed that $\sum_{j=1}^m \lambda_j =1$
2. The case $m=2$ is the definition of an extreme point.
3. For $m>2$ use induction: since $x=(1-\lambda_m)y+\lambda_m x_m$ with $y=\sum_{j=1}^{m-1}\frac{\lambda_j}{1-\lambda_m}x_j$, it follows that $x_m=x$ and $y=x$. Apply the inductive hypothesis to $y$.