## 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 1953

Then since the

Then since the

**range**of T is closed , it follows from Corollary 12 that TX = X. By Lemma 3.1 the operator T is one - to - one . By Theorem II.2.2 , T has a bounded inverse and hence 0 € p ( T ) = P ( S ) and so SX = X. Next suppose that ...Page 1954

Q.E.D. It was shown in the course of the preceding proof that for an operator T with a closed

Q.E.D. It was shown in the course of the preceding proof that for an operator T with a closed

**range**the point i = 0 is not in the spectrum of the operator V = T | E ( { 0 } ' ) X . Thus for all sufficiently small complex numbers 1 # 0 ...Page 2312

The closure of the

The closure of the

**range**of a densely defined linear operator T is the set of all x such that y * x = 0 whenever T * y * = 0 . PROOF . If T * y * = 0 , then y * y = y * Tz = ( T * y * ) z = 0 for all y = Tz in the**range**of T , and hence ...### 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 defined Definition denote dense determined differential operator discrete 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