## Linear Operators: Part III: Spectral Operators [by] Nelson Dunford and Jacob T. Schwartz, with the Assistance of William G. Bade and Robert G. Bartle, Volume 1 |

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Page 2038

... is

: — Т ( t ) q ( 0 ) , O < t < 00 , and by letting t = 0 + we see that Ś = g ( 0 ) — 0 0 ) =

0 . This shows that Pe = T ( t ) o ( 0 ) = o ( t ) and proves the uniqueness of y ( t ) .

... is

**independent**of t and thus also**independent**of u . By letting u + t + we have =: — Т ( t ) q ( 0 ) , O < t < 00 , and by letting t = 0 + we see that Ś = g ( 0 ) — 0 0 ) =

0 . This shows that Pe = T ( t ) o ( 0 ) = o ( t ) and proves the uniqueness of y ( t ) .

Page 2327

it follows that there exist at least two linearly

equation to = doq such that Bi ( 91 ) = B ; ( 42 ) = 0 , i = 1 , . . . , k . Then 21 may be

represented uniquely as P1 = { } = 1 C1 ; 0 ; ( u ( ) ) ) and , similarly , 92 = { } = 1 ...

it follows that there exist at least two linearly

**independent**solutions 91 , 92 of theequation to = doq such that Bi ( 91 ) = B ; ( 42 ) = 0 , i = 1 , . . . , k . Then 21 may be

represented uniquely as P1 = { } = 1 C1 ; 0 ; ( u ( ) ) ) and , similarly , 92 = { } = 1 ...

Page 2441

This shows that there exists a finite absolute constant c ' such that 1 ( Q ) < c ' ( 1 +

101 ) - n + 1 . Using this inequality and using ( 37 ) , we see that there exists a

finite constant M "

...

This shows that there exists a finite absolute constant c ' such that 1 ( Q ) < c ' ( 1 +

101 ) - n + 1 . Using this inequality and using ( 37 ) , we see that there exists a

finite constant M "

**independent**of ε such that 22V ( 43 ) 17 6 , wy + s ( 0 , 09 + ok...

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### Contents

SPECTRAL OPERATORS | 1924 |

Introduction | 1927 |

Terminology and Preliminary Notions | 1929 |

Copyright | |

28 other sections not shown

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