Given an arbitrary set $A\subset\mathbb R^n$, why do all extreme points of the convex hull $\operatorname{conv}(A)$ lie in $A$?

(An extreme point of a convex set is defined as one that cannot be written as a strictly convex combination of two distinct points of this set.)


closed as off-topic by Claude Leibovici, Davide Giraudo, user91500, JonMark Perry, user223391 May 25 '17 at 0:51

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Claude Leibovici, Davide Giraudo, user91500, JonMark Perry, Community
If this question can be reworded to fit the rules in the help center, please edit the question.


An element $x\in\operatorname{conv}A$ if and only if there are a positive natural number $n$, elements $x_1,\cdots,x_n\in A$ and non-negative real numbers $t_1,\cdots,t_n$ such that $$\begin{cases}x=\sum\limits_{k=1}^n t_kx_k\\ \sum\limits_{k=1}^n t_k=1\end{cases}$$

We can assign to each $x\in\operatorname{conv}A$ the least natural number $n(x)$ such that some $x_1,\cdots, x_{n(x)}$ and $t_1,\cdots,t_{n(x)}$ as above exist. Notice that $x\in A\iff n(x)=1$.

Suppose $n(x)\ge 2$. Then, $$x=t_1x_1+\sum_{k=2}^{n(x)} t_kx_k=t_1x_1+(1-t_1)\sum_{k=2}^{n(x)}\frac{t_k}{1-t_1}x_k$$ and by minimality all the $t_i$-s are $> 0$. Notice that, by hypothesis $1-t_1=\sum\limits_{k=2}^{n(x)}t_k>0$ and, thus $$x':=\sum_{k=2}^{n(x)}\frac{t_k}{1-t_1}x_k\in\operatorname{conv} A$$

and $x$ is a strictly convex combination of $x_1$ and $x'$. If $x'\ne x_1$, then $x_1$ is not extremal. On the other hand, if $x'=x_1$, then $x=x_1\in A$, against the assumption $n(x)\ge2$.


Let $x\in \text{conv}(A)$ be extreme. By definition of the convex hull, $x=\sum_i \lambda_i a_i$ for $a_i\in A$, $\sum_i \lambda_i=1$, and $\lambda_i\ge 0$. Since $A\subset\text{conv}(A)$ and $x$ was extreme, only one $\lambda_i$ can be nonzero (and it must be equal to $1$), so $x=a_i$ for this $i$. Thus $x\in A$.


Not the answer you're looking for? Browse other questions tagged or ask your own question.