## 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 |

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Even though ( A ) is taken as a standing assumption throughout this section , it will be indicated parenthetically in the statement of each lemma in the

Even though ( A ) is taken as a standing assumption throughout this section , it will be indicated parenthetically in the statement of each lemma in the

**proof**of which it is used . 1 LEMMA ( A ) . If a , ß are complex numbers and x ...Page 2218

**Proof**. Since a spectral operator of scalar type is clearly in the weakly closed algebra A generated by the projections in its resolution of the identity , it suffices to show that every operator in A is a spectral operator of scalar ...Page 2281

The

The

**proof**is similar to that of Lemma 17 and will be omitted . We turn now to the representation of E * X * when E * € & * and has finite uniform multiplicity n . The results are parallel to those in X. The weaker form of completeness ...### What people are saying - Write a review

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### Contents

SPECTRAL OPERATORS | 1924 |

Introduction | 1927 |

Terminology and Preliminary Notions | 1929 |

Copyright | |

47 other sections not shown

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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