## Linear Operators: Part III: Spectral Operators [by] Nelson Dunford and Jacob T. Schwartz, with the Assistance of William G. Bade and Robert G. Bartle, Volume 1 |

### From inside the book

Results 1-3 of 58

Page 2194

Strongly Closed Algebras and Complete

Strongly Closed Algebras and Complete

**Boolean Algebras**In this section an attempt will be made to characterize the strong closure of a commutative algebra of spectral operators . It has been observed ( cf. VI.1.5 ) that a convex set in ...Page 2195

A

A

**Boolean algebra**B of projections in a B - space X is said to be complete ( o - complete ) as an abstract**Boolean algebra**if each subset ( sequence ) of B has a greatest lower bound and a least upper bound in B. The**Boolean algebra**B ...Page 2217

Let B be a o - complete

Let B be a o - complete

**Boolean algebra**of projections in a B - space X , and let B , be its strong closure . By Lemma 3 , B is bounded and thus B , is also a bounded**Boolean algebra**of projections in X. Suppose that B , is not complete ...### What people are saying - Write a review

We haven't found any reviews in the usual places.

### Contents

SPECTRAL OPERATORS | 1924 |

Introduction | 1927 |

Terminology and Preliminary Notions | 1929 |

Copyright | |

47 other sections not shown

### Other editions - View all

### Common terms and phrases

adjoint operator Amer analytic apply arbitrary assumed B-space Banach space belongs Boolean algebra Borel set boundary conditions bounded bounded operator Chapter clear closed commuting compact complex constant contains continuous converges Corollary corresponding countably additive defined Definition denote dense determined differential operator domain elements equation equivalent established exists extension fact finite follows formal formula function given gives Hence Hilbert space hypothesis identity inequality integral invariant inverse Lemma limit linear operator Math Moreover multiplicity norm perturbation plane positive preceding present problem projections PROOF properties prove range resolution resolvent restriction Russian satisfies scalar type seen sequence shown shows similar solution spectral measure spectral operator spectrum subset sufficiently Suppose Theorem theory topology unbounded uniformly unique valued vector zero