## Linear Operators: Spectral theory |

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Page 984

The set of functions f in L ( R ) for which f

The set of functions f in L ( R ) for which f

**vanishes**in a neighborhood of infinity is dense in Ly ( R ) . PROOF . It follows from Lemma 3,6 that the set of all functions in L2 ( R , B , u ) which**vanish**outside of compact sets is ...Page 1650

If F

If F

**vanishes**in each set Iq , it**vanishes**in UI , PROOF . The proofs of the first four parts of this lemma are left to the reader as an exercise . To prove ( v ) , we must show from our hypothesis that F ( q ) = 0 if is in CO ( UqIą ) ...Page 1651

A set 1 , 1 - ܢܐܝܐ ] , 0 bear and g

A set 1 , 1 - ܢܐܝܐ ] , 0 bear and g

**vanishes**outside K , then Yk9-9**vanishes**outside a compact subset of 1 - CF , so that G ( 9 ) F ( Yko ) = F ( g ) . Thus G | I = F. If KCp = 0 and the function q in C ( IUI )**vanishes**outside K ...### What people are saying - Write a review

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### Contents

BAlgebras | 859 |

Miscellaneous Applications | 937 |

Compact Groups | 945 |

Copyright | |

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additive adjoint operator 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 satisfies seen sequence singular solution spectral square-integrable statement subset subspace sufficiently Suppose symmetric Theorem theory topology transform unique vanishes vector zero