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For a set $S\subseteq\mathbb R^m$ we denote with $\text{pos}(S)$ the set \begin{equation} \{\alpha_1x_1+\cdots+\alpha_nx_n:\alpha_i\geq 0,x_i\in S, n\in\mathbb N\}. \end{equation} How to prove that $\text{conv}(S)=\text{aff}(S)\cap \text{pos}(S)$ if $0\notin\text{aff}(S)$, where $\text{aff}(S)$ is afine hull, and $\text{conv}(S)$ is convex hull of $S$?

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  • $\begingroup$ You should try showing $\mathrm{conv}(S) \subseteq \mathrm{aff}(S) \cap \mathrm{pos}(S)$ and $\mathrm{aff}(S) \cap \mathrm{pos}(S) \subseteq \mathrm{conv}(S)$ . Start with an arbitrary element of the LHS side and show it satisfies the conditions of the RHS set, then $\mathrm{LHS} \subseteq \mathrm{RHS}$. $\endgroup$
    – Rammus
    May 12, 2020 at 17:38
  • $\begingroup$ Well, first inclusion is obvious. $\endgroup$ May 12, 2020 at 17:52

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Let $C = \operatorname{co} S$, $A= \operatorname{aff} S$ and $P= \operatorname{pos} S$. It is clear that $C \subset A$, $C \subset P$ so $C \subset A \cap P$.

If it is assumed that $0 \notin A$ then we have $A \cap P \subset C$.

To see this, suppose $x \in A \cap P$ and let $x = \sum_k \alpha_k x_n$ with $\alpha_k \ge 0, x_k \in S$. If $\sum_l \alpha_k =1$ we are finished, since $x \in C$, otherwise note that $x'= {1 \over \sum_k \alpha_k} \sum_k \alpha_k x_k \in A$ and the line through $x,x'$ passes through the origin and so $0 \in A$ a contradiction. Hence $0 \in C$.

If the condition $0 \notin A$ is removed then the other inclusion is not true. Take $S= \{ e_1,e_2,e_2+e_2 \} \subset \mathbb{R}^2$. Then $A=\mathbb{R}^2, P = \{x | x \ge 0 \}$ and so $0 \in A \cap P$ but clearly $0 \notin C$ (for example, $\phi(x)= x_1+x_2 \ge 1$ for $x \in C$) so the other inclusion is false in general.

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  • $\begingroup$ What if $0\notin aff(S)$? Then, geometric intuition says it should be true, but how to prove it. $\endgroup$ May 12, 2020 at 22:46
  • $\begingroup$ Are you changing your question? $\endgroup$
    – copper.hat
    May 12, 2020 at 22:51
  • $\begingroup$ Yes, should i edit it? $\endgroup$ May 12, 2020 at 22:52
  • $\begingroup$ Your call. I will add an answer in to your question in any event. $\endgroup$
    – copper.hat
    May 12, 2020 at 22:57
  • $\begingroup$ Ok, i have edited it. $\endgroup$ May 12, 2020 at 22:59

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