Let $\DeclareMathOperator{\Conv}{\mathrm{Conv}} C=\Conv(v_1,v_2,...,v_m) $, where $ v_1,v_2,...,v_m $ are $ m $ points in $ \mathbb{R}^n $ and 'Conv' means the convex hull. Please prove
$$
\partial C=\cup_{D\in S}\Conv(v_i:i\in D)
$$
where
- $\partial C $ denotes the boundary of $ C $ and
- $S$ is defined as the set $$ S=\left\{D\subset\{1,2,\dots,m\} : \exists d\in\mathbb{R}^n \text{ such that } \begin{split} \langle v_i,d\rangle=1 &\;\forall i\in D \\ \langle v_i,d\rangle<1 &\:\forall i \notin D \end{split}\right\} $$
I have tried to prove it by using the fact that the boundary of $ C $ is the union of all edges. However I cannot give an explicit proof. Can you give me some concrete hints and references?
As Eric said, the statement should assume that origin is in the interior of $C$.
Hint: Consider that, there is $d$ such that $\langle d, v_i \rangle =1$ for all $v_i\in D$ iff (when thinking elements of $D$ as points instead of vectors) $D\subset H$ for some $H$, such that $H$ is a hyperplane not through origin and $dim(H)<n$. When this condition is met, defining $T$ to be the hyperplane orthogonal to $H$ and through origin, $d$ is a vector on $T$ (the hyperplane orthogonal to $H$ and through origin) such that $\langle d, proj_{T}H\rangle=1$.
Partial solution: Because the convex object $C$ contains the origin and it is suppsed that $H$ does not, the condition that $\partial conv(D)\subset\partial C$ is equivalent to $|proj_{T}(D)|>|proj_{T}(v_j)|$ for every $v_j$ not included in $D$. Divide the inequality by $|proj_{T}(D)|$ on both sides and the latter condition recovers definition $\langle d, v_j\rangle<1$ for $v_j\notin D$.
I think of the last paragraph as a criterion of deciding whether $D$ contains all the element (so the equality is straight) of the "most outward" hyperplane in any direction/dimension.