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

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 p_u 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, p_u) is an AM-space with unit u.[1]

Properties

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

• Vector lattice
• AL-space

References

1. Schaefer & Wolff 1999, pp. 242–250.

• Schaefer, Helmut H.; Wolff, Manfred P. (1999). Topological Vector Spaces. GTM. 3. New York, NY: Springer New York Imprint Springer. ISBN 978-1-4612-7155-0. OCLC 840278135.CS1 maint: ref=harv (link)