Linear Operators: Spectral operators |
From inside the book
Results 1-3 of 88
Page 1954
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 – 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 sufficiently
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 ...
Page 2264
then , belongs neither to the point nor to the residual spectrum of S . Using the
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 the
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 ...
Page 2507
AND uniformly bounded by a constant K in the neighborhood of the spectrum of
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 of
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 ...
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