proving extreme points of closed convex sets don't have nontrivial convex combinaton

Question

I have to show that if $x$ is an extreme point of a closed convex set $C$ then their is no convex combination $x=\sum_{i=1}^k a_ix_i$ other than $x_1=x_2=...=x_k=x$.
I tried like this: let $x\in C$ is an extreme point and it has convex combination as above then $x=\sum_{i=1}^{k-1} a_ix_i+a_kx_k$ then I used the extreme point property (if $x=ax_1+(1-a)x_2,a\in (0,1)$ then $x=x_1=x_2$) to get $x=x_k$.
can it be proved in this manner? can anybody help?

Answer 1

Use induction. Supose the result is true for less than $k$ vectors $x_i$. Write $x=t\sum\limits_{i=1}^{k-1}(a_i/t)x_i+(1-t)x_k$ where $t=\sum\limits_{i=1}^{k-1}a_i=1-a_k$. It follows that $x=x_k$. Hence, $x_k$ is an exterme point and $x_k=t\sum\limits_{i=1}^{k-1}(a_i/t)x_i+(1-t)x_k$ or $x_k=\sum\limits_{i=1}^{k-1}(a_i/t)x_i$. By induction hypothesis this implies $x_k=x_1=x_2=\cdots=x_{k-1}$.