Theorem 16.2.3 (Bauer-Namioka).label Let $E$ be an ordered vector space over $K \in \RC$ with positive cone $C$, $\topo$ be a vector space topology on $E$[1], $F \subset E$ be a subspace, and $\phi \in (F, \topo)^{*}$, then the following are equivalent:
- (1)
There exists a continuous positive linear functional $\Phi \in (E, \topo)^{*}$ such that $\Phi|_{F} = \phi$.
- (2)
There exists $U \in \cn_{\topo}(0)$ convex such that
\[\sup\bracs{\text{Re}\dpn{x, \phi}{F}|x \in F \cap (U - C)}< \infty\]
Proof, [V.5.4, SW99]. (1) $\Rightarrow$ (2): Let $U = \bracs{\text{Re}(\Phi) < 1}$, then $U \in \cn_{\topo}(0)$ is convex. Since $\Phi$ is positive, for any $x \in U$ and $y \in C$, $\text{Re}\dpn{x - y, \Phi}{E}\le \text{Re}\dpn{x, \Phi}{E}< 1$.
(2) $\Rightarrow$ (1): Assume without loss of generality that $K = \real$. Let $\alpha > 0$ such that $F \cap (U - C) \subset \bracs{\phi < \alpha}$. Since $U$ is convex and open, and $C$ is convex, $U - C \subset E$ is an open convex set, disjoint from the convex set $\bracs{\phi = \alpha}$. By the Hahn-Banach Theorem, there exists $\Phi \in E^{*}$ such that $U - C \subset \bracs{\Phi < \alpha}$, and $\bracs{\phi = \alpha}\subset \bracs{\Phi \ge \alpha}$.
After rescaling, assume without loss of generality that $\bracs{\phi = \alpha}\subset \bracs{\Phi = \alpha}$, then $\Phi \in E^{*}$ is an extension of $\phi$. For each $x \in C$ and $\lambda > 0$, $-\lambda x \in U - C$, and $-\lambda\dpn{x, \Phi}{E}< \alpha$. As this holds for all $\lambda > 0$, $\dpn{x, \Phi}{E}\ge 0$, so $\Phi$ is the desired extension.$\square$
- The order and the topology need not to be compatible. keyboard_return
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