## 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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Page 1930

However , in Corollary X.2.4 we have seen that a bounded normal operator in Hilbert space always has a uniquely defined bounded and

However , in Corollary X.2.4 we have seen that a bounded normal operator in Hilbert space always has a uniquely defined bounded and

**countably additive**resolution of the identity defined on the field of all Borel subsets of the plane .Page 2143

This spectral measure is bounded , is

This spectral measure is bounded , is

**countably additive**on S ( T ) , and commutes with T. > PROOF . From Definitions 1 , 4 , and 7 it is seen that S ( T ) = 41 ( T ) , and thus for each S in S ( T ) there is , by Lemma 2 , one and only ...Page 2144

It clearly preserves finite disjoint unions , takes complements into complements , is

It clearly preserves finite disjoint unions , takes complements into complements , is

**countably additive**in the X topology of X * , and is bounded . It remains only to show that A ( 0 ) A ( S ) = A ( od ) .### 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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### 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