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Ordered topological vector space

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In mathematics, specifically infunctional analysis andorder theory, anordered topological vector space, also called anordered TVS, is atopological vector space (TVS)X that has apartial order ≤ making it into anordered vector space whose positive coneC:={xX:x0}{\displaystyle C:=\left\{x\in X:x\geq 0\right\}} is a closed subset ofX.[1] Ordered TVSes have important applications inspectral theory.

Normal cone

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Main article:Normal cone (functional analysis)

IfC is a cone in a TVSX thenC isnormal ifU=[U]C{\displaystyle {\mathcal {U}}=\left[{\mathcal {U}}\right]_{C}}, whereU{\displaystyle {\mathcal {U}}} is the neighborhood filter at the origin,[U]C={[U]:UU}{\displaystyle \left[{\mathcal {U}}\right]_{C}=\left\{\left[U\right]:U\in {\mathcal {U}}\right\}}, and[U]C:=(U+C)(UC){\displaystyle [U]_{C}:=\left(U+C\right)\cap \left(U-C\right)} is theC-saturated hull of a subsetU ofX.[2]

IfC is a cone in a TVSX (over the real or complex numbers), then the following are equivalent:[2]

  1. C is a normal cone.
  2. For every filterF{\displaystyle {\mathcal {F}}} inX, iflimF=0{\displaystyle \lim {\mathcal {F}}=0} thenlim[F]C=0{\displaystyle \lim \left[{\mathcal {F}}\right]_{C}=0}.
  3. There exists a neighborhood baseB{\displaystyle {\mathcal {B}}} inX such thatBB{\displaystyle B\in {\mathcal {B}}} implies[BC]CB{\displaystyle \left[B\cap C\right]_{C}\subseteq B}.

and ifX 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 familyP{\displaystyle {\mathcal {P}}} of semi-norms onX such thatp(x)p(x+y){\displaystyle p(x)\leq p(x+y)} for allx,yC{\displaystyle x,y\in C} andpP{\displaystyle p\in {\mathcal {P}}}.

If the topology onX is locally convex then the closure of a normal cone is a normal cone.[2]

Properties

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IfC is a normal cone inX andB is a bounded subset ofX then[B]C{\displaystyle \left[B\right]_{C}} is bounded; in particular, every interval[a,b]{\displaystyle [a,b]} is bounded.[2] IfX is Hausdorff then every normal cone inX is a proper cone.[2]

Properties

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  • LetX be anordered vector space over the reals that is finite-dimensional. Then the order ofX is Archimedean if and only if the positive cone ofX is closed for the unique topology under whichX is a Hausdorff TVS.[1]
  • LetX be an ordered vector space over the reals with positive coneC. Then the following are equivalent:[1]
  1. the order ofX is regular.
  2. C is sequentially closed for some Hausdorff locally convex TVS topology onX andX+{\displaystyle X^{+}} distinguishes points inX
  3. the order ofX is Archimedean andC is normal for some Hausdorff locally convex TVS topology onX.

See also

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References

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  1. ^abcSchaefer & Wolff 1999, pp. 222–225.
  2. ^abcdefSchaefer & Wolff 1999, pp. 215–222.
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