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Transactions of the American Mathematical Society

ISSN 1088-6850(online) ISSN 0002-9947(print)



$ C\sp{\ast} $-extreme points

Authors: Alan Hopenwasser, Robert L. Moore and V. I. Paulsen
Journal: Trans. Amer. Math. Soc. 266 (1981), 291-307
MSC: Primary 46L05; Secondary 47D20
MathSciNet review: 613797
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Abstract: Let $ \mathcal{A}$ be a $ {C^ \ast }$-algebra and let $ \mathcal{S}$ be a subset of $ \mathcal{A}$. $ \mathcal{S}$ is $ {C^ \ast }$-convex if whenever $ {T_1},{T_2}, \ldots ,{T_n}$ are in $ \mathcal{S}$ and $ {A_1}, \ldots ,{A_n}$ are in $ \mathcal{A}$ with $ \sum\nolimits_{i = 1}^n {A_i^ \ast {A_i} = I} $, then $ \sum\nolimits_{i = 1}^n {A_i^ \ast {T_i}{A_i}} $ is in $ \mathcal{S}$. An element $ T$ in $ \mathcal{S}$ is called $ {C^ \ast }$-extreme in $ \mathcal{S}$ if whenever $ T = \sum\nolimits_{i = 1}^n {A_i^ \ast {T_i}{A_i}} $ with $ {T_i}$ and $ {A_i}$ as above and with $ {A_i}$ invertible, then $ {T_i}$ is unitarily equivalent to $ T$ for each $ i$. We investigate the linear extreme points and $ {C^ \ast }$-extreme points for three sets: first, the unit ball of operators in Hilbert space; next, the set of $ 2 \times 2$ matrices with numerical radius bounded by $ 1$; and last, the unit interval of positive operators on Hilbert space. In particular we find that for the second set, the linear and $ {C^ \ast }$-extreme points are different.

References [Enhancements On Off] (What's this?)

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Keywords: Numerical radius, $ {C^ \ast }$-convex set, $ {C^ \ast }$-extreme point
Article copyright: © Copyright 1981 American Mathematical Society

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