Question:
Let $G$ be a convex mapping from $\Omega \subseteq X$ into a normed space $Z$, and assume that $P \subseteq Z$ be a positive cone with nonempty interior. Show that the following two regularity conditions are equivalent:
(i) there exists $x_{1} \in \Omega$ such that $G(x_{1}) < \theta$,
(ii) for every nonnull $z^{\ast} \geq \theta$, there exists $x \in \Omega$ such that $\langle G(x) , z^{\ast} \rangle < 0$.
(The notation follows Optimization by Vector-Space Methods by Luenberger.)
(Attempted) Solution:
I can prove (i) implies (ii).
Suppose there exists $ x_{1} \in \Omega $ such that $ G(x_{1}) < 0 $. Then, we have that \begin{array}{rcrl} G(x_{1}) < 0 & \Rightarrow & -G(x_{1}) \geq 0 \\ & \Rightarrow & \langle -G(x_{1}) , z^{\ast} \rangle \geq 0 & \forall \, z^{\ast} \geq 0 \\ & \Rightarrow & \langle G(x_{1}) , z^{\ast} \rangle \leq 0 & \forall \, z^{\ast} \geq 0. \end{array} Suppose $ \langle G(x_{1}) , z_{0}^{\ast} \rangle = 0 $ for some $ z_{0}^{\ast} \geq 0 $. Since $ -G(x_{1}) $ is contained in the interior of $ P $, there exists an $ \varepsilon > 0 $ such that $ ||z + G(x_{1})|| < \varepsilon $ implies $ z \in P $. Choose any $ z \in Z $, and define $ \alpha = \frac{\varepsilon}{||z|| + 1}. $ Our choice of $ \alpha $ ensures that $ -G(x_{1}) + \alpha z $ and $ -G(x_{1}) - \alpha z $ are both contained in the interior of $ P $. Observe that $ -G(x_{1}) \pm \alpha z \in P \qquad \Rightarrow \qquad \langle -G(x_{1}) \pm \alpha z , z_{0}^{\ast} \rangle \geq 0. $ But we also have that $ \tfrac{1}{2} \cdot \langle -G(x_{1}) + \alpha z , z_{0}^{\ast} \rangle + \tfrac{1}{2} \cdot \langle -G(x_{1}) - \alpha z , z_{0}^{\ast} \rangle = \langle -G(x_{1}) , z_{0}^{\ast} \rangle = 0. $ Combining these results, we can conclude that $ \langle -G(x_{1}) \pm \alpha z , z_{0}^{\ast} \rangle = 0. $ In particular, we have that $ \langle -G(x_{1}) + \alpha z , z_{0}^{\ast} \rangle = \langle -G(x_{1}) , z_{0}^{\ast} \rangle + \alpha \cdot \langle z , z_{0}^{\ast} \rangle = \alpha \cdot \langle z , z_{0}^{\ast} \rangle = 0. $ Because $ \alpha $ is nonzero, this implies that $ \langle z , z_{0}^{\ast} \rangle = 0 $. Since $ z \in Z $ was arbitrary, this implies that $ z_{0}^{\ast} = \theta $. Thus, we have shown that for all $ z^{\ast} \geq \theta $, we have that $ \langle G(x_{1}) , z^{\ast} \rangle \leq 0 $ with equality if and only if $ z^{\ast} = \theta $. In other words, for every $ z^{\ast} \geq \theta $ such that $ z \neq \theta $, we have that $ \langle G(x_{1}) , z^{\ast}) < 0 $. Thus, $ x_{1} \in \Omega $ satisfies the desired condition uniformly for all nonnull $ z^{\ast} \geq 0 $.
Then, I am completely at a loss as to how I should prove (ii) implies (i). I was also wondering if there is a name for the second regularity condition (I think the second condition is called Slater's condition).