Linear Operators, Part 1 |
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Page 495
We now prove that B , is an algebra under the natural product ( 1g ) ( s ) f ( s ) g (
s ) , s € S. To see this , let h be a fixed element of C ( S ) and let B ( h ) = { fe B. th
B , } . Evidently B ( h ) C Bo , and it is clear that B ( h ) satisfies properties ( i ) ...
We now prove that B , is an algebra under the natural product ( 1g ) ( s ) f ( s ) g (
s ) , s € S. To see this , let h be a fixed element of C ( S ) and let B ( h ) = { fe B. th
B , } . Evidently B ( h ) C Bo , and it is clear that B ( h ) satisfies properties ( i ) ...
Page 498
on to X * satisfies ( i ) then ( ii ) defines an operator T on X to L ( S , E , ” ) whose
norm satisfies ( iii ) . Furthermore T is weakly compact if and only if x * ( ) is
countably additive on in the strong topology of X * . Proof . If , for E in E , the
functional x ...
on to X * satisfies ( i ) then ( ii ) defines an operator T on X to L ( S , E , ” ) whose
norm satisfies ( iii ) . Furthermore T is weakly compact if and only if x * ( ) is
countably additive on in the strong topology of X * . Proof . If , for E in E , the
functional x ...
Page 557
... zero polynomial R exists such that R ( T ) Let R be factored as R ( 2 ) = B119–1
( 2–2 ; ) . If his o ( T ) , then ( T - 2 ; 1 ) x = 0 implies x = 0. Consequently , the
product R , of all the factors ( 2–2 ; ) " in R such that li € ( T ) , still satisfies R ( T ) =
0.
... zero polynomial R exists such that R ( T ) Let R be factored as R ( 2 ) = B119–1
( 2–2 ; ) . If his o ( T ) , then ( T - 2 ; 1 ) x = 0 implies x = 0. Consequently , the
product R , of all the factors ( 2–2 ; ) " in R such that li € ( T ) , still satisfies R ( T ) =
0.
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Contents
Preliminary Concepts | 1 |
The VitaliHahnSaks Theorem and Spaces of Measures | 7 |
B Topological Preliminaries | 10 |
Copyright | |
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