## Linear Operators, Part 2 |

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Page 925

a 20 16 Let Ni , N2 , ... be a countable

a 20 16 Let Ni , N2 , ... be a countable

**sequence**of normal operators in ý , all commuting with each other . Show that there exists a single Hermitian operator T such that each N. is a Borel function of T. ( Hint : Use Theorem 2.1 and ...Page 959

Since Ueem = e , the

Since Ueem = e , the

**sequence**{ eembn , m 2 1 } is an increasing**sequence**of sets whose union is ebm . Since Mo is countably additive on Bo , Molebn ) = limm Mo ( eembn ) 2 k , and so for some m , Mo ( eem ) 2 Mo ( eembn ) > k — ε .Page 1124

If En , E are in F and q ( En ) increases to the limit ( E ) , then it follows from what we have already proved that En is an increasing

If En , E are in F and q ( En ) increases to the limit ( E ) , then it follows from what we have already proved that En is an increasing

**sequence**of projections and E , SE . If Em is the strong limit of En , then E. SE and Q ( E ) = Q ...### 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