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 $S$?
Convex hull as intersection of affine hull and positive hull
convex-analysisconvex-hulls
Best Answer
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.