# Simultaneous power factorization in modules over Banach algebras

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Positivity

, Volume 21, Issue 2, pp 633–672

First Online: 23 November 2016Received: 06 October 2016Accepted: 18 November 2016DOI: 10.1007-s11117-016-0457-6

Cite this article as: de Jeu, M. & Jiang, X. Positivity 2017 21: 633. doi:10.1007-s11117-016-0457-6

Abstract

Let A be a Banach algebra with a bounded left approximate identity \\{e \lambda \} {\lambda \in \Lambda }\, let \\pi \ be a continuous representation of A on a Banach space X, and let S be a non-empty subset of X such that \\lim {\lambda }\pi e \lambda s=s\ uniformly on S. If S is bounded, or if \\{e \lambda \} {\lambda \in \Lambda }\ is commutative, then we show that there exist \a\in A\ and maps \x n: S ightarrow X\ for \ge 1\ such that \s=\pi a^nx ns\ for all \ge 1\ and \s\in S\. The properties of \a\in A\ and the maps \x n\, as produced by the constructive proof, are studied in some detail. The results generalize previous simultaneous factorization theorems as well as Allan and Sinclair’s power factorization theorem. In an ordered context, we also consider the existence of a positive factorization for a subset of the positive cone of an ordered Banach space that is a positive module over an ordered Banach algebra with a positive bounded left approximate identity. Such factorizations are not always possible. In certain cases, including those for positive modules over ordered Banach algebras of bounded functions, such positive factorizations exist, but the general picture is still unclear. Furthermore, simultaneous pointwise power factorizations for sets of bounded maps with values in a Banach module such as sets of bounded convergent nets are obtained. A worked example for the left regular representation of \\mathrm {C} 0{\mathbb R}\ and unbounded S is included.

KeywordsBanach module Simultaneous power factorization Positive factorization Mathematics Subject ClassificationPrimary 46H25 Secondary 46B40 46B42

Author: **Marcel de Jeu - Xingni Jiang**

Source: https://link.springer.com/