## 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 2017

2 and the fraction in the

2 and the fraction in the

**inequality**( i ) of Theorem 6 is Isi - sl2 + 2 | 8182 / ? { ( si – $ 3 ) * + 1681 s * } 1/2 which is bounded on all of R " . Thus As has a resolution of the identity . Furthermore , the set S , is the null set ...Page 2190

Thus for some constant K we have | S ( f ) SK ||| and to prove the final

Thus for some constant K we have | S ( f ) SK ||| and to prove the final

**inequality**of the present theorem it will ... this**inequality**are continuous functions off , it will suffice to prove it for every function f in a set dense in EB ...Page 2403

First we shall prove an

First we shall prove an

**inequality**for integral operators ( Lemma 5 below ) which is elementary in the sense that it relates only to the norms of the integral kernels involved . We then use this**inequality**to apply Theorem 1 in an ...### 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 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