## 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 78

Page 2325

This , however , follows from Lemma 3.5 and from

This , however , follows from Lemma 3.5 and from

**formulas**( 16 ) and ( 14 ) . It also follows , from Lemma 3.5 ,**formula**( 16 ) , and**formula**( 14 ) , that the zero Šn of M ( u ) in R7 has the asymptotic representation ξ , ο 2πη + α + ...Page 2341

We wish to show , using

We wish to show , using

**formula**( 58 ) , that the family of all sums Σ Eml : me ) J ranging over all finite sets of integers , is uniformly bounded . This follows from ( 58 ) by an argument using Lemma 7 , which is similar to the ...Page 2347

It follows from

It follows from

**formula**( 28 ) ( in Case 1A ) and from the corresponding**formula**( 57 ) ( in Case 2 ) that Ang Y m m ... is represented asymptotically by a**formula**of exactly the same sort , ( 28 ) or ( 57 ) , as ( E ( Am ) f ) ( t ) .### 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 | |

28 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 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