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We say that an ordered vector space $(V, \ge)$ (over $\mathbb{R}$) is "bidual embeddable" (I made up this name, not sure whether this concept already exists) if for every $x \in V$, if $x$ satisfies the property "$f(x) \ge 0$ for all order-preserving linear functional $f:V \to \mathbb{R}$" (recall that $f$ is order-preserving if $f(y)\ge 0$ for all $y \ge 0$), then we must have $x \ge 0$. Loosely speaking, this means that the evaluation map from $(V, \ge)$ to its bidual is an order isomorphism between $(V, \ge)$ and the image of the evaluation map.

Question: What is known about this "bidual embeddable" property? Is it equivalent to regularly ordered vector space?

I believe this is true for the finite dimensional case, though I am unsure about the infinite dimensional case.

(I noticed that there is a similar result in Schaefer & Wolff, Topological Vector Spaces, Theorem 1.6, though it is about vector lattices. I am more interested in the non-lattice case.)

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  • $\begingroup$ If $V$ is a normed space (or more generally a locally convex topological vector space, I think) and you only consider continuous linear functionals, then this is true if and only if the cone in $V$ is closed. That's a consequence of the Hahn-Banach separation theorem. $\endgroup$
    – Jochen Glueck
    Commented May 22, 2023 at 20:48
  • $\begingroup$ (By the way, the same is also true if $V$ is only pre-ordered. This indicates that the property you're interested in might not be directly related to the regularity property that you linked on Wikipedia, since the order dual can be very small of $V$ is only pre-ordered.) $\endgroup$
    – Jochen Glueck
    Commented May 22, 2023 at 20:50
  • $\begingroup$ @JochenGlueck Thanks for the comment. Unfortunately, I cannot assume continuity. $\endgroup$
    – Cheuk Ting Li
    Commented May 22, 2023 at 21:32
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    $\begingroup$ Every real vector space $V$ has a finest locally convex topology, which is such that all linear functionals are continuous. This is helpful because elements outside of the positive cone of $V$ can be separated by monotone functionals if and only if the cone is closed in that topology. I think this means that your bidual embeddability holds iff the cone is closed in this sense. For certain kinds of cones this can be translated into a simpler condition, e.g. if the cone has an order unit. Let me know if you need more details. $\endgroup$
    – Tobias Fritz
    Commented May 23, 2023 at 9:26
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    $\begingroup$ Oh, I realize that we've recently communicated independently of this, and now I wonder whether this question has to do with our email exchange ;) $\endgroup$
    – Tobias Fritz
    Commented May 23, 2023 at 9:28

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