## Linear Operators: Spectral theory |

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Page 993

Then it

Then it

**follows**from what has just been demonstrated that ay , = Apuv , = Qy , i.e. , Qy is independent of V. Q.E.D. 16 THEOREM . If the bounded measurable function q has its spectral set consisting of the single point m then , for some ...Page 996

Since f * q is continuous by Lerma 3.1 ( d ) it

Since f * q is continuous by Lerma 3.1 ( d ) it

**follows**from the above equation that f * ° +0 . From Lemma 12 ( b ) it is seen that olf * 9 ) Colp ) and from Lemma 12 ( c ) and the equation of = Tf it**follows**that o ( f * ) contains no ...Page 1708

Since se is in H ( m ) , it

Since se is in H ( m ) , it

**follows**that there exists some F in Hm + p ) such that ( 11 +0 ) F = ĝe . However , since teq is in Hm + p - 1 ) , and since by ( 5 ) , ( Tito ) teq = ge , it**follows**that feq = F is in Hm + P ) ( C ) so that ...### What people are saying - Write a review

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### Contents

BAlgebras | 861 |

Commutative BAlgebras | 868 |

Commutative BAlgebras | 874 |

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 complete Consequently constant contains continuous converges Corollary corresponding defined Definition denote dense derivatives 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 Nauk 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