## Linear Operators: Spectral operators |

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Results 1-3 of 89

Page 1953

Then since the

Lemma 3 . 1 the operator T is one - to - one . By Theorem II . 2 . 2 , T has a

bounded inverse and hence 0 ep ( T ) = P ( S ) and so SX = X . Next suppose that

E ( { 0 } ...

Then since the

**range**of T is closed , it follows from Corollary 12 that TX = X . ByLemma 3 . 1 the operator T is one - to - one . By Theorem II . 2 . 2 , T has a

bounded inverse and hence 0 ep ( T ) = P ( S ) and so SX = X . Next suppose that

E ( { 0 } ...

Page 1954

Q . E . D . It was shown in the course of the preceding proof that for an operator T

with a closed

{ 0 } ' ) X . Thus for all sufficiently small complex numbers 1 # 0 the operator XI ...

Q . E . D . It was shown in the course of the preceding proof that for an operator T

with a closed

**range**the point i = 0 is not in the spectrum of the operator V = T | E ({ 0 } ' ) X . Thus for all sufficiently small complex numbers 1 # 0 the operator XI ...

Page 2312

The closure of the

such that y * x = 0 whenever T * y * = 0 . PROOF . If T * y * = 0 , then y * y = y * Tz =

( T * y * ) 2 = 0 for all y = Tz in the

...

The closure of the

**range**of a densely defined linear operator T is the set of all xsuch that y * x = 0 whenever T * y * = 0 . PROOF . If T * y * = 0 , then y * y = y * Tz =

( T * y * ) 2 = 0 for all y = Tz in the

**range**of T , and hence for all y in the closure of...

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### Contents

SPECTRAL OPERATORS | 1924 |

An Operational Calculus for Bounded Spectral | 1941 |

Part | 1950 |

Copyright | |

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