## 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 86

Page 2149

Condition ( A ) of Section 2 is satisfied if the spectrum of T is nowhere

Condition ( A ) of Section 2 is satisfied if the spectrum of T is nowhere

**dense**in the complex plane . PROOF . If the resolvent set is**dense**, then any two analytic , or even continuous , extensions of R ( ; T ' ) x must coincide on ...Page 2156

are both

are both

**dense**in X. Since M , is**dense**in X , the manifold ( 1,1 – T ) ^ M2 + { x | ( 9,1 – T ) Nx = 0 } is**dense**in X , so that ( 1,1 – T ) ^ ( 121 – T ) * X + { x | ( 9,1 – T ) Nx = 0 } + { x | ( 121 – T ) ^ x = 0 } is also**dense**in ...Page 2159

The union of all intervals of constancy relative to T is an open set

The union of all intervals of constancy relative to T is an open set

**dense**in I .. PROOF . It is clear that the union of intervals of constancy is open . To see that it is**dense**, let y be a closed subarc of T ' , having positive length ...### 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 domain elements equation equivalent established exists extension 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