Linear Operators, Part 1 |
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Page 287
Thus ( 0 * = \ gle The mapping x * → g is then a one - to - one isometric map of L *
into Lg . It is evident from the Hölder inequality that any g € L , determines an x * €
L * satisfying ( i ) , so that the mapping x * * → g is a one - to - one isometric ...
Thus ( 0 * = \ gle The mapping x * → g is then a one - to - one isometric map of L *
into Lg . It is evident from the Hölder inequality that any g € L , determines an x * €
L * satisfying ( i ) , so that the mapping x * * → g is a one - to - one isometric ...
Page 299
... it follows that all the functions G1 , ... , & y vanish outside some sufficiently large
interval [ -A0 , +4 . ] , so that s + S | ) Pdy = $ + S- * My ) –s : ( y ) pdy Sly - g ; ” SEP
for A 2 A0 , proving ( b ) . To prove ( a ) we note first that it is evident that lim ...
... it follows that all the functions G1 , ... , & y vanish outside some sufficiently large
interval [ -A0 , +4 . ] , so that s + S | ) Pdy = $ + S- * My ) –s : ( y ) pdy Sly - g ; ” SEP
for A 2 A0 , proving ( b ) . To prove ( a ) we note first that it is evident that lim ...
Page 337
It is then evident that v ( us , I ) = v ( 1 , 1 ) . Thus , ba ( S , E ) is isometrically
isomorphic with the closed subspace BV ( 1 ) of all | e BV ( 1 ) such that f ( a + ) =
0 . If N is the one - dimensional space of constant functions , it is evident that BV (
I ) ...
It is then evident that v ( us , I ) = v ( 1 , 1 ) . Thus , ba ( S , E ) is isometrically
isomorphic with the closed subspace BV ( 1 ) of all | e BV ( 1 ) such that f ( a + ) =
0 . If N is the one - dimensional space of constant functions , it is evident that BV (
I ) ...
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Contents
Preliminary Concepts | 1 |
The VitaliHahnSaks Theorem and Spaces of Measures | 7 |
B Topological Preliminaries | 10 |
Copyright | |
87 other sections not shown
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