## Linear Operators: Spectral theory |

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Page 897

It follows then from ( iii ) that the projections E ( d ) also commute with T ( f ) and this completes the proof of the theorem . Q.E.D. COROLLARY . The spectral measure is

It follows then from ( iii ) that the projections E ( d ) also commute with T ( f ) and this completes the proof of the theorem . Q.E.D. COROLLARY . The spectral measure is

**countably additive**in the strong operator topology . Proof .Page 932

... of S. Let F be an additive ( resp . weakly

... of S. Let F be an additive ( resp . weakly

**countably additive**) function on to the set of positive operators on a ... there exists a Hilbert space 12V and a self adjoint projection valued additive ( resp . weakly**countable additive**) ...Page 958

This argument shows that the vector valued additive set function y is weakly

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 given Hence Hilbert space identity independent indices inequality integral interval Lemma limit linear mapping Math matrix measure multiplicity Nauk 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