## 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 1983

Then A is a spectral operator of scalar type . PROOF . The argument of the

Then A is a spectral operator of scalar type . PROOF . The argument of the

**preceding**corollary shows that A is a spectral operator . Since Â ( s ) has distinct eigenvalues , it is a scalar operator , that is , its radical part is zero .Page 2396

Let o4 be as in the

Let o4 be as in the

**preceding**lemma , put A ( a ) = A ( Qil :; ( ) ) ) , and let B ( a ) = A ( 04 ( :, u ( a ) ) ) ( cf. Lemma 4 for the definition of u ( a ) ) . Then , by the**preceding**lemma , by Lemma 1 , and by formulas ( 2a ) and ...Page 2455

Therefore , if we let the three operators of the

Therefore , if we let the three operators of the

**preceding**lemma be H2 , H1 , H1 , we obtain the present corollary . Q.E.D. 5 COROLLARY . Under the hypotheses of the**preceding**corollary , U ( H1 , H2 ) is an isometric mapping of E ( H1 ...### What people are saying - Write a review

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### Contents

SPECTRAL OPERATORS | 1924 |

Introduction | 1927 |

Terminology and Preliminary Notions | 1929 |

Copyright | |

28 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 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