## 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 e p ( 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 e p ( 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 NI – T

...

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

with a closed

**range**the point . = 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 NI – T

...

Page 2312

The closure of the

such that y * x = 0 whenever T * y * = 0 . PROOF . If T * y * = 0 , then y * y = y * T2

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

of ...

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 * T2

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

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

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### Contents

SPECTRAL OPERATORS 1937 1941 1945 XV Spectral Operators | 1924 |

Introduction | 1927 |

Terminology and Preliminary Notions | 1929 |

Copyright | |

32 other sections not shown

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