## Linear Operators, Volume 2 |

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Page 876

Then ( y + Nie ) ( 2 ) y ( 2 ) + Ni = i ( 1 + N ) , and and

Then ( y + Nie ) ( 2 ) y ( 2 ) + Ni = i ( 1 + N ) , and and

**hence**11 + N = \ y + Niel .**Hence**( 1 + N ) 2 = ly + Niel2 = | ( y + Nie ) ( y + Nie ) * 1 = ( y + Nie ) ( y - Nie ) \ y2 + Nae < \ y ? + N2 . Since this inequality must hold ...Page 1027

Then Tx = 2x + y , where y belongs to the subspace ( I– E ) H , and

Then Tx = 2x + y , where y belongs to the subspace ( I– E ) H , and

**hence**to the nullspace of T. Let u = 1 - y . Then T ( x + u ) = 1 ( x + u ) ,**hence**2 is an eigenvalue of T. ( b ) Suppose that 2 is in the resolvent set of T. From the ...Page 1227

**Hence**T * x = ix , or xe Dr.**Hence**D is closed . Similarly , D_ is closed . Since Dt and D_ are clearly linear subspaces of D ( T * ) , it remains to show that the spaces D ( T ) , Dr , and D_ are mutually orthogonal , and that their ...### What people are saying - Write a review

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### Contents

BAlgebras | 859 |

Commutative BAlgebras | 868 |

Commutative BAlgebras | 874 |

Copyright | |

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