## Linear Operators, Part 2 |

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Page 893

Returning now to the general integral S / ( s ) E ( ds ) where E is merely a bounded

Returning now to the general integral S / ( s ) E ( ds ) where E is merely a bounded

**additive**operator valued set function , we observe that the integral has been defined in terms of the uniform operator topology .Page 932

Let S be an abstract set and E a field ( resp . o - field ) of subsets of S. Let F be an

Let S be an abstract set and E a field ( resp . o - field ) of subsets of S. Let F be an

**additive**( resp . weakly countably**additive**) function on E to the set of positive operators on a Hilbert space H satisfying F ( $ ) = 0 and F ( S ) ...Page 958

This argument shows that the vector valued

This argument shows that the vector valued

**additive**set function y is weakly countably**additive**on the o - field consisting of all Borel subsets of e . By a theorem of Pettis ( IV.10.1 ) it is countably**additive**in the strong topology ...### What people are saying - Write a review

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### Contents

BAlgebras | 859 |

Bounded Normal Operators in Hilbert Space | 887 |

Miscellaneous Applications | 937 |

Copyright | |

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additive Akad algebra Amer analytic assume Banach spaces basis belongs Borel boundary conditions boundary values bounded called clear closed closure coefficients compact complex Consequently constant contains continuous converges Corollary corresponding defined Definition denote dense determined domain eigenvalues element equal equation essential spectrum evident Exercise exists extension finite follows formal differential operator formula function function f given Hence Hilbert space identity independent indices inequality integral interval Lemma limit linear mapping Math matrix measure multiplicity neighborhood norm obtained partial positive preceding present problem projection proof properties prove range regular remark representation respectively restriction result Russian satisfies seen sequence singular solution spectral square-integrable statement subset subspace sufficiently Suppose symmetric Theorem theory topology transform unique vanishes vector zero