separating hyperplane theorem Let $\mathbb{R}^n_{+}:=\{x\in\mathbb{R}^n:x_i\geq 0, i=1,\dots,n\}$ and $C\subseteq\mathbb{R}^n_{+}$ be a convex cone, $0\notin C$. Let $f\in \mathbb{R}^{n}\setminus C.$ Can we conclude from some version of the separating hyperplane theorem, that there exists $g\in\mathbb{R}^{n}$ such that $g\cdot f\leq 0$ , $\forall y\in C, g\cdot y \geq 0$ and
$\exists y_0\in C, g\cdot y_0>0$?
 A: As stated, no. Your statement of the problem admits the possibility that $f$ can be chosen to be a positive multiple of some element of $C$. For example, let $C = \mathbb{R}^n_+ \cap \{ \sum x_i = 1\}$ and let $f = (1,1,1,\ldots,1) \in \mathbb{R}^n \setminus C$. Since $(1/n,\ldots, 1/n) = f/n \in C$, for $g\cdot f \leq 0$ and $g\cdot y\geq 0 \forall y\in C$ we must have $g\perp f$. But we then have that $f/n + \epsilon g \in C$ for all sufficiently small $|\epsilon|$. Chosen $\epsilon < 0$ with $|\epsilon| $ small we have that $y = f/n + \epsilon g$ is such that $y\cdot g < 0$. 

Perhaps you instead intend $C$ to be a convex cone? In this case the answer is yes. If $ \sum f_i \leq 0$, then just let $g = (1,1,1\ldots, 1)$. If $\sum f_i > 0$, WLOG we can assume $\sum f_i = 1$. On the affine plane $\Pi = \{ \sum x_i = 1\}$, and let $V$ be the linear subspace $\{\sum x_i = 0\}$. Let $\pi:\Pi \to V$ be the projection. 
Applying Hahn Banach separation theorem to $\pi(C\cap \Pi)$ and $\pi(f)$, we find and element $g'$ of $V$ and $s\in \mathbb{R}$ such that $\forall y\in C$ we have $g'\cdot y > s \geq g'\cdot f$. Let $g = g' - s\cdot(1,1,1\ldots,)$; it has the desired property. 
