## Linear Operators, Part 2 |

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Page 1303

Clearly B ( A ) = 0 for those f which vanish in a

Clearly B ( A ) = 0 for those f which vanish in a

**neighborhood**of a . Thus B is a boundary value for ī at a . To prove the converse , let B be a boundary ...Page 1678

Let ý be a second function in C 0 ( I ) such that ý ( x ) for x in a

Let ý be a second function in C 0 ( I ) such that ý ( x ) for x in a

**neighborhood**of Kı . Then yo - yo vanishes in a**neighborhood**of K C ( F ) ...Page 1734

Let Uici , be a bounded

Let Uici , be a bounded

**neighborhood**of q chosen so small that BU , CE , and so that there exists a mapping 9 of U , onto the unit spherical**neighborhood**V ...### 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