Let $V$ be an arbitrary set. Given that $\sup_{x\in V} \langle x,u\rangle \ge 1$ for all unit $u\in \mathbb{R}^n,$ show that convex hull of $V$ contains oepn unit ball centered at 0.
This looks easy at first but I have spent hours thinking about it to no avail.
On
(Separating Hyperplane Theorem) Let $A$ and $B$ be two disjoint nonempty convex subsets of $\mathbb{R}^n$. Then there exist a nonzero vector $v$ and a real number $c$ such that :$\langle x, v \rangle \ge c \, \text{ and } \langle y, v \rangle \le c$ for all $x$ in $A$ and $y$ in $B$; i.e., the hyperplane $\langle \cdot, v \rangle = c$, $v$ the normal vector, separates $A$ and $B$.
Suppose that $u \in B(0,1)$ but $u \notin conv(V)$. Then let $A:=\{u\},B:=conv(V)$ in the theorem. Then there exist a nonzero vector $v$ and a real number $c$ such that:
$$\langle u, v \rangle \ge c \implies \|v\|_2> c$$
and $$ \sup_{y \in conv(V)}\langle y, v \rangle \le c \implies \|v\|_2\le \sup_{y \in V}\langle y, v \rangle \le c$$
The former implication is by Cauchy Schwarz inequality and that the $B(0,1)$ is open; the latter implication is by our assumption. This gives the desired contradiction.
Suppose $u$ lies in the open unit ball but $u \notin V$. The Hahn Banach shows that there is some unit vector $h$ such that $\langle h, x \rangle \le \langle h, u \rangle $ for all $x \in V$. Hence $1 \le \langle h, u \rangle \le \|u\| < 1$ which is a contradiction.