逢甲學報, Bände 21-22逢甲大學, 1988 |
Im Buch
Ergebnisse 1-3 von 29
Seite 413
... hence , we proved that A is convex . А Now , let A be a projection of H onto H11 . Suppose that there are B , C in A such that A = 1⁄2 ( B + C ) . If we can show that B = A = C , then by lemma 0 , we conclude that A E ext A. For h EH ...
... hence , we proved that A is convex . А Now , let A be a projection of H onto H11 . Suppose that there are B , C in A such that A = 1⁄2 ( B + C ) . If we can show that B = A = C , then by lemma 0 , we conclude that A E ext A. For h EH ...
Seite 415
... hence sup { < Ah , h > : || h || = 1 } ≤1 . That is || A || ≤1 . So , AЄball B ( H ) + . Now , it is time to prove the subject . To show this , we suppose that B ( H ) is reflexive , then by alaoglu's theorem , ball B ( H ) is weakly ...
... hence sup { < Ah , h > : || h || = 1 } ≤1 . That is || A || ≤1 . So , AЄball B ( H ) + . Now , it is time to prove the subject . To show this , we suppose that B ( H ) is reflexive , then by alaoglu's theorem , ball B ( H ) is weakly ...
Seite 296
... Hence w≤0 , and consequently there exists a first imteger R , such that P ≤ R ≤ Q - 1 , such that wR > 0 and WR + 1 ≤ 0. Let be a solution of ( 1.1 ) defined by 1 Д P1 = VRU - 1 → URVA - 1 PR = WR > 0 , PR + 1 = 0 , PR + 2 = -WR + ...
... Hence w≤0 , and consequently there exists a first imteger R , such that P ≤ R ≤ Q - 1 , such that wR > 0 and WR + 1 ≤ 0. Let be a solution of ( 1.1 ) defined by 1 Д P1 = VRU - 1 → URVA - 1 PR = WR > 0 , PR + 1 = 0 , PR + 2 = -WR + ...
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