Ordered topological vector space

In mathematics, specifically in functional analysis and order theory, an ordered topological vector space, also called an ordered TVS, is a topological vector space (TVS) X that has a partial order ≤ making it into an ordered vector space whose positive cone ${\displaystyle C:=\left\{x\in X:x\geq 0\right\}}$ is a closed subset of X.¹ Ordered TVSes have important applications in spectral theory.

If C is a cone in a TVS X then C is normal if ${\displaystyle {\mathcal {U}}=\left[{\mathcal {U}}\right]_{C}}$, where ${\displaystyle {\mathcal {U}}}$ is the neighborhood filter at the origin, ${\displaystyle \left[{\mathcal {U}}\right]_{C}=\left\{\left[U\right]:U\in {\mathcal {U}}\right\}}$, and ${\displaystyle [U]_{C}:=\left(U+C\right)\cap \left(U-C\right)}$ is the C-saturated hull of a subset U of X.²

If C is a cone in a TVS X (over the real or complex numbers), then the following are equivalent:²

1. C is a normal cone.
2. For every filter ${\displaystyle {\mathcal {F}}}$ in X, if ${\displaystyle \lim {\mathcal {F}}=0}$ then ${\displaystyle \lim \left[{\mathcal {F}}\right]_{C}=0}$.
3. There exists a neighborhood base ${\displaystyle {\mathcal {B}}}$ in X such that ${\displaystyle B\in {\mathcal {B}}}$ implies ${\displaystyle \left[B\cap C\right]_{C}\subseteq B}$.

and if X is a vector space over the reals then also:²

1. There exists a neighborhood base at the origin consisting of convex, balanced, C-saturated sets.
2. There exists a generating family ${\displaystyle {\mathcal {P}}}$ of semi-norms on X such that ${\displaystyle p(x)\leq p(x+y)}$ for all ${\displaystyle x,y\in C}$ and ${\displaystyle p\in {\mathcal {P}}}$.

If the topology on X is locally convex then the closure of a normal cone is a normal cone.²

If C is a normal cone in X and B is a bounded subset of X then ${\displaystyle \left[B\right]_{C}}$ is bounded; in particular, every interval ${\displaystyle [a,b]}$ is bounded.² If X is Hausdorff then every normal cone in X is a proper cone.²

• Let X be an ordered vector space over the reals that is finite-dimensional. Then the order of X is Archimedean if and only if the positive cone of X is closed for the unique topology under which X is a Hausdorff TVS.¹
• Let X be an ordered vector space over the reals with positive cone C. Then the following are equivalent:¹

1. the order of X is regular.
2. C is sequentially closed for some Hausdorff locally convex TVS topology on X and ${\displaystyle X^{+}}$ distinguishes points in X
3. the order of X is Archimedean and C is normal for some Hausdorff locally convex TVS topology on X.