## Linear Operators, Part 2 |

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Page 893

Let E be a

Let E be a

**bounded**self adjoint spectral measure in Hilbert space defined on a field & of subsets of a set S. Then the map f T ( f ) defined by the equation ( f ) T ( f ) = $$ + ( s ) E ( ds ) , fe B ( S , E ) , is a continuous ...Page 900

and thus there is a

and thus there is a

**bounded**function to on S with f ( s ) = fo ( s ) except for s in a set having E measure zero . If f is E - measurable then to is a**bounded**E - measurable function , i.e. , an element of the B * -algebra BS , E ) .Page 1452

0 x ) X E D ( T ) , so that if | x is

0 x ) X E D ( T ) , so that if | x is

**bounded**, ( Tr , x ) is**bounded**below . Conversely , suppose that for each n , en = ( -0 , -n ) n o ( T ) is : , T non - void . By Theorem XII.2.9 ( b ) , Elen ) # 0 for any n .### 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