Lemma 9.6.3 (Separation of Point and Convex Set). Let $E$ be a TVS over $\real$, $A \subset E$ be a non-empty open convex set, and $x \in E \setminus A$, then there exists $\phi \in E^{*}$ and $\alpha \in \real$ such that $\dpb{x, \phi}{E}= \alpha$ and $A \subset \bracs{\phi < \alpha}$
Proof. By translation, assume without loss of generality that $0 \in A$. In which case, $A \in \cn^{o}(0)$ is convex.
Let $[\cdot]_{A}: E \to [0, \infty)$ be the gaugeg of $A$, then $[\cdot]_{A}$ is a sublinear functional on $E$. For any $y, z \in E$ and $t > 0$ with $y, z \in tA$,
Hence $[\cdot]_{A}$ is continuous on $E$.
Let $\phi_{0}: \real x \to \real$ be defined by $\lambda x \mapsto \lambda$. Since $x \not\in A$, $[x]_{A} > 1$. Hence $\phi_{0}|_{Kx}\le [\cdot]_{A}|_{Kx}$. By the Hahn-Banach Theorem, there exists $\phi \in \hom(E; \real)$ such that $\phi|_{\real x}= \phi_{0}$ and $\phi(y) \le [y]_{A}$ for all $y \in E$.
For any $y \in A \cap (-A)$,
so $\phi \in E^{*}$.
Finally, let $y \in A$, then $\dpb{y, \phi}{E}\le [y]_{A} < 1 = \dpb{x, \phi}{E}$.$\square$