Prove the set ext(S)$=\{x\in\mathbb{R}^n:x_i^2=1,i=1\ldots n\}$ where $S=\{x\in\mathbb{R}^n:||x||_{\infty}\leq 1\}$

Question

Prove the set ext(S)$=\{x\in\mathbb{R}^n:x_i^2=1,i=1\ldots n\}$ where $S=\{x\in\mathbb{R}^n:||x||_{\infty}\leq 1\}$ I've got this exercise and I've never seen before how to prove that a group is an extreme points. Any guide will be helpful

Answer 1

For $x\in\mathbb{R}^n$ with $|x_i|<1$ for some $i\leq n$. Then $x\notin\{x\in\mathbb{R}^n:x_i^2=1, i=1,...,n\}$. We choose $z_i=x_i+0,5(1-|x_i|)$ and $y_i=x_i-0,5(1-|x_i|)$ and $y_j,z_j=x_j$ for the other indices. Clearly $y_i, z_i\leq 1$ so $y,z\in S$ as well as $0,5y + 0,5z=x$ so $x\notin ext(S)$. This means $ext(S)\subset\{x\in\mathbb{R}^n:x_i^2=1, i=1,...,n\}$.

To prove the equality, take $x\in\{x\in\mathbb{R}^n:x_i^2=1, i=1,...,n\}$ and assume that there exist $y,z\in\S$ with $z_j,y_j\neq x_j$ for some $j\leq n$ as well as there exists $t\in(0,1)$ such that $ty+(1-t)z=x$. Since $x_i^2=1$, for all $i\leq n$, this means that $ty_j+(1-t)z_j<1$ with the index where $x$ differs from $y,z$, as $y,z\in S$ and thus $y_i,z_i\leq 1$ for all $i\leq n$.

I think this should do it.