## Linear Operators: Spectral operators |

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Results 1-3 of 88

Page 1954

0 is not in the

small complex numbers 1 + 0 the operator XI – T = E ( { 0 } ) — Txo ; + E { 0 } ' ) —

Toys has a bounded everywhere defined inverse . This means that the point d ...

0 is not in the

**spectrum**of the operator V = T | E ( { 0 } ' ) X . Thus for all sufficientlysmall complex numbers 1 + 0 the operator XI – T = E ( { 0 } ) — Txo ; + E { 0 } ' ) —

Toys has a bounded everywhere defined inverse . This means that the point d ...

Page 2264

then , belongs neither to the point nor to the residual

formula for the spectral resolution of S given by the preceding theorem , we find

that E ( { a } ) = 0 for 1 # vo , deo ; thus , each such , must belong to the

continuous ...

then , belongs neither to the point nor to the residual

**spectrum**of S . Using theformula for the spectral resolution of S given by the preceding theorem , we find

that E ( { a } ) = 0 for 1 # vo , deo ; thus , each such , must belong to the

continuous ...

Page 2507

AND uniformly bounded by a constant K in the neighborhood of the

H , while the integrals 1 Fie ) I – H ) - 1 | 2d 8 and 8 B * ( ( 1 + is ) I – H , ) - 10 | 2

dx 8 are bounded as ε →0 . In this case , and under the additional assumption

that ...

AND uniformly bounded by a constant K in the neighborhood of the

**spectrum**ofH , while the integrals 1 Fie ) I – H ) - 1 | 2d 8 and 8 B * ( ( 1 + is ) I – H , ) - 10 | 2

dx 8 are bounded as ε →0 . In this case , and under the additional assumption

that ...

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