# Convex hull as intersection of affine hull and positive hull

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

• 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 $\mathrm{LHS} \subseteq \mathrm{RHS}$. – Rammus May 12 at 17:38
• Well, first inclusion is obvious. – Madara Uchiha May 12 at 17:52

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.

• What if $0\notin aff(S)$? Then, geometric intuition says it should be true, but how to prove it. – Madara Uchiha May 12 at 22:46
• Are you changing your question? – copper.hat May 12 at 22:51
• Yes, should i edit it? – Madara Uchiha May 12 at 22:52