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Suppose I have a closed convex cone $C\subseteq \mathbb R^n$ and suppose that for every $x$ in the non-negative orthant $\mathbb R_{0+}^n$ there is a $y\in C$ such that $x\cdot y>0$ (with the standard scalar product). Does it follow that intersection of $C$ with the non-negative orthant contains more than just the origin?

This is true for $n=2$.

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Assume $C$ and $\mathbb{R}_{\ge 0}^n$ can be (non-strictly) separated by a subspace of dimension $n-1$.
Then a normal vector $x$ to that subspace lies in $\mathbb{R}_{\ge 0}^n$; see e.g. here for a proof.

But then by your assumption, there exists $y \in C$ on the same side of this subspace as $\mathbb{R}_{\ge 0}^n$, contradiction.

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  • $\begingroup$ Thanks! I haven't worked much with convex sets and didn't know about (obvious as it is geometrically) the hyperplane separation theorem. $\endgroup$ Commented Jan 29, 2021 at 23:54

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