Abstract m-space
In mathematics, specifically in order theory and functional analysis, an abstract m-space or an AM-space is a Banach lattice {\displaystyle (X,\|\cdot \|)} whose norm satisfies {\displaystyle \left\|\sup\{x,y\}\right\|=\sup \left\{\|x\|,\|y\|\right\}} for all x and y in the positive cone of X.
We say that an AM-space X is an AM-space with unit if in addition there exists some u ≥ 0 in X such that the interval [−u, u] := { z ∈ X : −u ≤ z and z ≤ u } is equal to the unit ball of X; such an element u is unique and an order unit of X.[1]
Examples
[edit ]The strong dual of an AL-space is an AM-space with unit.[1]
If X is an Archimedean ordered vector lattice, u is an order unit of X, and pu is the Minkowski functional of {\displaystyle [u,-u]:=\{x\in X:-u\leq x{\text{ and }}x\leq x\},} then the complete of the semi-normed space (X, pu) is an AM-space with unit u.[1]
Properties
[edit ]Every AM-space is isomorphic (as a Banach lattice) with some closed vector sublattice of some suitable {\displaystyle C_{\mathbb {R} }\left(X\right)}.[1] The strong dual of an AM-space with unit is an AL-space.[1]
If X ≠ { 0 } is an AM-space with unit then the set K of all extreme points of the positive face of the dual unit ball is a non-empty and weakly compact (i.e. {\displaystyle \sigma \left(X^{\prime },X\right)}-compact) subset of {\displaystyle X^{\prime }} and furthermore, the evaluation map {\displaystyle I:X\to C_{\mathbb {R} }\left(K\right)} defined by {\displaystyle I(x):=I_{x}} (where {\displaystyle I_{x}:K\to \mathbb {R} } is defined by {\displaystyle I_{x}(t)=\langle x,t\rangle }) is an isomorphism.[1]
See also
[edit ]References
[edit ]Bibliography
[edit ]- Schaefer, Helmut H.; Wolff, Manfred P. (1999). Topological Vector Spaces. GTM. Vol. 8 (Second ed.). New York, NY: Springer New York Imprint Springer. ISBN 978-1-4612-7155-0. OCLC 840278135.