Property of dual generalized inequalities

Question

For $K$ a proper cone we define $x<_K y$ if and only if $y-x\in\text{int}K$. Denote $K^*$ as the dual cone of $K$ (also a proper cone). I would like to prove that $$x<_Ky\iff \lambda^T x<\lambda^T y \text{ for all } \lambda\in K^*-\{0\}.$$ This result is geometrically intuitive. I am able to rigorously prove the first direction:

Proof: Let $y-x\in\text{int}K$ and $\lambda\in K^*-\{0\}$. By definition of $K^*$, we have $\lambda^T(y-x)\ge 0$, so all we must show is that $\lambda^T(y-x)\ne 0$. Suppose the previous expression is $0$. As $y-x\in\text{int}K$, there exists $\epsilon>0$ such that $B_{\epsilon}(y-x)\subseteq K$. We claim that $\exists v\in B_{\epsilon}(y-x)$ such that $\lambda^T v<0$, contradicting $\lambda\in K^*$. Indeed, $y-x\in\{v\vert \lambda^T v=0\}=\text{bd}(\{v\vert\lambda^Tv\ge 0\})$, thus $B_{\epsilon}(y-x)$ must contain points outside $\{v\vert\lambda^Tv\ge 0\}$ (since $\epsilon>0$. So the claim is proven and the result follows.

However, while geometrically intuitive, I am not able to rigorously prove the reverse direction. As an attempt, I proceeded by contradiction that $y-x\in K-\text{int}K\subseteq \text{bd}K$. To complete the proof is would be enough to show there is a nonzero $\lambda\in K^*$ such that $\lambda^T(y-x)=0$. Basically I want to prove (the forward direction of) the following: $$z\in\text{bd}K\iff \exists \lambda\in K^*-\{0\} \text{ s.t. } \lambda^T z=0.$$

Again, the picture in $\mathbb{R}^2$ is telling of why this is true, but a formal proof would be appreciated!

Answer 1

Now let assume $K$ is a convex closed cone. To prove right to left:

Let $a=y-x \in R^n$ such that $\lambda^t.a > 0$ for all $\lambda \in K^* - \{0\}$. Thus $a \in K^{**} = K$ (Since $K$ is convex and closed cone). We only need to show that $a \in \text{int}K$. To this end, let $\delta=\inf\{\lambda^t.a ~; \quad \lambda \in K^*,~ \| \lambda \| =1 \}$; note that $\delta > 0$. We show that $B(a;\varepsilon) \subseteq K^{**}=K$ where $\varepsilon < \delta$. Pick $x=a+\varepsilon b \in B(a;\varepsilon)$ where $\|b\| \leq 1.$ Now let $\lambda \in K^{*} $ with $\| \lambda \| =1$. Then we have $$ \langle \lambda , x \rangle =\langle \lambda , a + \varepsilon b \rangle=\langle \lambda , a \rangle + \varepsilon \langle \lambda , b \rangle \ge \delta - \varepsilon > 0 $$

Therefore $x \in K^{**} = K.$

Answer 2

If $K$ is not convex then Right to Left is wrong. Take $$K= \text{graph of} ~y=|x| ~\cup ~X-\text{axis}$$ Then $K^* = \{(x,y) : x=0,~ y\ge 0 \}$

Now take $x=0 , y=(1,1)$ Then Right side is satisfied but $y-x=y \notin int K$