## 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 1936

Let I = E + + En where Ej , En are bounded , disjoint projections in X , each commuting with the bounded operator T. Then T is a spectral operator if and only if each

Let I = E + + En where Ej , En are bounded , disjoint projections in X , each commuting with the bounded operator T. Then T is a spectral operator if and only if each

**restriction**T | E ; X is a spectral operator .Page 2094

**Restrictions**and quotients . Theorem 3.10 shows that if a spectral operator Te B ( X ) is reduced by a closed subspace Y 5 * and one of its complements ( that is , if T commutes with some projection of X onto Y ) , then the**restriction**...Page 2228

If o is a Borel set , and T is a spectral operator with resolution of the identity E , then the

If o is a Borel set , and T is a spectral operator with resolution of the identity E , then the

**restriction**T | E ( 0 ) X of T to E ( Q ) X is a spectral operator whose resolution of the identity is the**restriction**of E to E ( o ) X ...### 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