Am trying to prove the following
Theorem. Let $A,B$ be closed convex cones in $\mathbb{R}^n$ with $A\cap B=\{0\}$. Suppose that $B$ contains no subspaces other than $\{0\}$. Then there exist a nonzero vector $v$ such that $$\langle x,v \rangle\leq 0< \langle y,v \rangle$$
for all $x\in A$ and nonzero $y\in B$.
Proof. Assume $B\neq\{0\}$, for otherwise the result holds vacuously. Let $C$ denote the convex hull of the set $\{y\in B: \|y\|=1\}$. Note that $C$ is nonempty, convex, and compact (the convex hull of a compact set in $\mathbb{R}^n$ is compact, see here). I claim that $A$ and $C$ are disjoints. Since $A\cap C\subset \{0\}$, assume by contradiction that $0\in C$. Then $0$ can be written as a convex combination of unit vectors from $B$:
$$0=\sum^n_{i=1}\lambda_i y_i, \quad \lambda_i\geq0 \text{ for each $i$,} \quad \sum_{i=1}^n\lambda_i=1$$
Choose any $k$ with $\lambda_k\neq 0$ and rearrange the above to get
$$\lambda_ky_k=-\sum_{i=1, i\neq k}^n\lambda_iy_i$$
Since $B$ is a convex cone, the LHS is an element of $B$ and the RHS is an element of $-B$. Therefore $\lambda_ky_k \in B\cap (-B)$. But $\lambda_ky_k$ is nonzero and $B\cap (-B)$ is a subspace contained in $B$, a contradiction. Therefore $A$ and $C$ are disjoints.
Now $A$ and $C$ satisfy the conditions of the separating hyperplane theorem (compact version), which means that there exist a nonzero vector $v$ and real numbers $c_1<c_2$ such that
$$\langle x,v \rangle<c_1<c_2< \langle y,v \rangle \quad \quad (1)$$
for all $x\in A$ and $y\in C$. Since $0 \in A$, we have $0<c_1$.
Let $x\in A$ and $\lambda>0$. Since $A$ is a cone, from $(1)$ we obtain
$$\langle \lambda x,v \rangle<c_1 \iff \langle x,v \rangle<\frac{1}{\lambda}c_1$$
Letting $\lambda\to\infty$ yields $\langle x,v \rangle\leq 0$. Hence $\langle x ,v \rangle\leq 0$ for all $x\in A$.
Let $y\in B$ be nonzero. Then $y=\lambda z$ for some $\lambda>0$ and $z\in C$, and from $(1)$ we obtain
$$0<\lambda c_2< \lambda \langle z,v \rangle =\langle y,v \rangle $$
Hence $\langle y ,v \rangle >0$ for all nonzero $y\in B$.
Am I missing something? Thanks a lot for your help.
