Ordered topological vector space explained

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

C:=\left\{x\inX:x\geq0\right\}

is a closed subset of X. Ordered TVSes have important applications in spectral theory.

Normal cone

See main article: Normal cone (functional analysis).

If C is a cone in a TVS X then C is normal if

l{U}=\left[l{U}\right]C

, where

l{U}

is the neighborhood filter at the origin,

\left[l{U}\right]C=\left\{\left[U\right]:U\inl{U}\right\}

, and

[U]C:=\left(U+C\right)\cap\left(U-C\right)

is the C-saturated hull of a subset U of X.

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

  1. C is a normal cone.
  2. For every filter

l{F}

in X, if

\liml{F}=0

then

\lim\left[l{F}\right]C=0

.
  1. There exists a neighborhood base

l{B}

in X such that

B\inl{B}

implies

\left[B\capC\right]C\subseteqB

.

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

  1. There exists a neighborhood base at the origin consisting of convex, balanced, C-saturated sets.
  2. There exists a generating family

l{P}

of semi-norms on X such that

p(x)\leqp(x+y)

for all

x,y\inC

and

p\inl{P}

.

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

Properties

If C is a normal cone in X and B is a bounded subset of X then

\left[B\right]C

is bounded; in particular, every interval

[a,b]

is bounded. If X is Hausdorff then every normal cone in X is a proper cone.

Properties

  1. the order of X is regular.
  2. C is sequentially closed for some Hausdorff locally convex TVS topology on X and

X+

distinguishes points in X
  1. the order of X is Archimedean and C is normal for some Hausdorff locally convex TVS topology on X.