TAILIEUCHUNG - Báo cáo toán học: "Compact operators in the algebra of a partially ordered measure space "

Tuyển tập các báo cáo nghiên cứu khoa học ngành toán học tạp chí Journal of Operator Theory đề tài: Các nhà khai thác nhỏ gọn trong đại số của một không gian đo một phần thứ tự. | J. OPERATOR THEORY 10 1983 353. 355 Copyright by INCREST 1983 COMPACT OPERATORS IN THE ALGEBRA OF A PARTIALLY ORDERED MEASURE SPACE JOHN FROELICH In I p. 498 Arveson conjectured that alg 2 ntp contains no nonzero compact operators. By establishing a general result about the lattice of invariant subspaces of a compact operator we show that both alg L 2 mp 0 p 1 and algjt 2 contain no nonzero compact operators. Throughout this paper X z ỊÌ will denote a probability measure space. If B e z we shall denote by PB the projection on L2 X z fl induced by B. Lemma 1. Let B be a sequence of measurable sets such that for each sub- 00 X r u I 1. If f 6 L x z fi satisfies lim inf I dp 0 then 1 1 J Ấ f 0 . p . Proof Suppose lim inf ị f dp 0 and pick 0. Then there is . r 00 X sequence nJ such thatk dp fi 2 . Since fl I u 2 . 1 we have that J V 1 7 r r oo C 00 p l w l dp s IZ dA -A . J J i i J 2 X 00 B u A. j l Therefore d z 0 so f 0 . ậz . X Lemma 2. Let Bn be a sequence of measurable sets such that for each sub- 00 P O. If fi is a sequence in such that i 1 L2 support fi c Bn. for some subsequence and fi f then f 0 . u . 354 JOHN FROELICH Proof. Since f we can extract a subsequence also called J which we co may assume converges pointwise to f. Set AY - p B and E - From ỉ. 1 i . Ef. 0 we obtain p E -- 0. Suppose xệE. Then there is a subsequence Bn. such that X ệ Bn. for each Ỉ . This gives fi x 0. Since fix - x we i j I I conclude that x 0. So f- 0 . . Theorem. Suppose 5 is a sequence of measurable sets such that for each subsequence . 1 ai d ll I n n. J 0. If K L2 X X p - 1 VH i i 7 L2 X X is a compact operator and PB G K for each n then K 0. Proof Let f G L2 X X p and consider the sequence PB K - -f . We have I PB Ah j Kf p and since K is compact we can find a subsequence such that KP K -f converges. Since PB 6 Lat Q ĩĩì ni support Ẫ P K f s B i Hence by Lemma 2 KPg K f - 0. Therefore KP K - f f - 0 so PB K f

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