Prove that $P=\text{conv}(x_1,...,x_m)\subset\mathbb R^n$ is the convex hull of its extreme points

Let $P\subseteq \mathbb{R}^{n}$ is convex hull of finite points: $P=conv(x_1,x_2,\ldots,x_m)$. I need to show that $P$ is convex hull of its extreme points.

I am thinking about such proof. Let $x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k}}$ is minimal subset of $x_1,x_2,\ldots,x_m$ for which $P=conv(x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k}})$. How to prove that each $x_{i_{j}}$ is extreme point?

If some $x_{i_{j}}$ is not extreme point, it can be represented $x_{i_{j}}=\frac{1}{2}x'+\frac{1}{2}x''$ where $x'\in P$ and $x''\in P$ and $x' \neq x''$. How to continue?

Solution 1:

Let $x_1,x_2,\ldots,x_n$ is minimal set for which $P=conv(x_1,x_2,\ldots,x_n)$ and $x_1$ is not extreme point. $x_1=\frac{1}{2}x'+\frac{1}{2}x''$ and $x'\neq x''$

$x'=\sum_{i=1}^{n}\lambda_i'x_i $ and $\sum_{i=1}^{n}\lambda_i'=1$

$x''=\sum_{i=1}^{n}\lambda_i''x_i $ and $\sum_{i=1}^{n}\lambda_i''=1$

So $x_1=\sum_{i=1}^{n}\frac{\lambda_i'+\lambda_i''}{2}x_i=\sum_{i=1}^{n}\mu_ix_i$

If $\mu_1<1$ then $x_1=\frac{1}{1-\mu_1}\sum_{i=2}^{n}\mu_ix_i$ and we would have contrary to minimality of set $x_1,x_2,\ldots,x_n$

If $\mu_1=1$ then $\mu_i=0$ for $i\neq1$ or $\frac{\lambda_i'+\lambda_i''}{2}=0$ which means $\lambda_i'=\lambda_i''=0$ for $i\neq1$, so that $x'= x_1 = x''$.