## 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 1934

Since , by VII.3.4 , lim x * x ( ) = lim x * R ( $ ; T ' ) x = 0 , & → & + 00 it is

Since , by VII.3.4 , lim x * x ( ) = lim x * R ( $ ; T ' ) x = 0 , & → & + 00 it is

**seen**that x * x ( ) = 0 for all & and all x * e X * . Hence , by Corollary II.3.14 , x ( ) = 0 and thus x = ( I - T ' ) x ( 8 ) = 0 .Page 2163

It is

It is

**seen**from Corollary XV.3.7 that F ( $ ) also commutes with the projections in the range of E , that is , ( iii ) F ( Ė ) E ( 0 ) = E ( 0 ) F ( K ) ...Page 2185

Since every function f in C ( 1 ) is bounded and Borel measurable , the integral sf ( ) A ( d ) ) exists and it is

Since every function f in C ( 1 ) is bounded and Borel measurable , the integral sf ( ) A ( d ) ) exists and it is

**seen**from ( i ) that ( ii ) S * ( S ) = S 5 ( A ) A ( da ) , feC ( 4 ) . It will now be shown that A is a spectral ...### 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