## 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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The situation corresponding to an

The situation corresponding to an

**invariant**closed subspace of T is not so simple . However , Fixman [ 1 ] proved that the restriction of a spectral operator T to an**invariant**closed subspace y of X is spectral if and only if the ...Page 2214

18 THEOREM . Let B be a bounded Boolean algebra of projections in a weakly complete space . Then an operator is in the weakly closed algebra generated by B if and only if it leaves

18 THEOREM . Let B be a bounded Boolean algebra of projections in a weakly complete space . Then an operator is in the weakly closed algebra generated by B if and only if it leaves

**invariant**every closed linear manifold which is ...Page 2286

Similarly if K is a closed

Similarly if K is a closed

**invariant**subspace in H , Y ( K ) denotes the closure in X of A - 1Kn D ( A - 1 ) ) . = - 2 It follows from Theorem 19 ( b ) and Lemma 35 that Mn D ( A ) is dense in M. Similarly In D ( A - 1 ) is dense in K.### What people are saying - Write a review

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### Contents

SPECTRAL OPERATORS | 1924 |

Introduction | 1927 |

Terminology and Preliminary Notions | 1929 |

Copyright | |

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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