tailieunhanh - Báo cáo nghiên cứu khoa học: "Connections between an operator and a compact operator that yield hyperinvariant subspaces "

Tuyển tập các báo cáo nghiên cứu khoa học ngành toán học tạp chí Journal of Operator Theory đề tài: Các kết nối giữa một nhà điều hành và điều hành một nhỏ gọn năng suất subspaces hyperinvariant. | J. OPERATOR THEORY 1 1979 117-121 Copyright by INCREST 1979 CONNECTIONS BETWEEN AN OPERATOR AND A COMPACT OPERATOR THAT YIELD HYPERINVARIANT SUBSPACES SCOTT BROWN The algebra of all bounded linear operators on a complex Banach space X will be denoted by f y . Lomonosov s theorem 4 states that if an element SetX X is not a multiple of the identity operator on X and if s commutes with a nonzero compact operator K e Z then s has a nontrivial hyperinvariant subspace. The theorem can be easily modified to include the case where s and K anticommute SK KS 0 . This and other variations on the hypotheses of the theorem are discussed in this note. Definition. For s K elX X and ẦeC s -commutes with K if SK ẦKS. A subalgebra si .Y is called transitive if no nontrivial closed subspaces of X are invariant under si. The following is extracted from 4 in the Pearcy-Shields article 5 Theorem 2 p. 222 . Theorem 1. Lomonosov . If si is a transitive subalgebra affix and if nonzero K eXf X is compact then there exists Ae sđ such that 1 is an eigenvalue of AK. This can be used to prove a slight generalization of Lomonosov s theorem. Theorem 2. Suppose s e C X is not a multiple of the identity of ãfX . Let si denote the commutant of s. If s Ằ-commutes with a nonzero compact KeS X then st has a nontrivial invariant subspace. Proof. Case I A A 1- By Theorem 1 if si is transitive then there exists A e s such that AKS has eigenvalue 1. It is known . see 8 problem 2 b p. 259 that for any two elements a and b of a Banach algebra ST a ab u 0 a bd u 0 where a x denotes the spectrum of X e ST. Hence rtfAK S u 0 ơ S AK u 0 and so o AKS u 0 ẰơịAKS u 0 since SAK ẰAKS. This can only happen if a AKS 0 . This contradicts the fact that AKS has eigenvalue 1. 118 SCOTT BROWN Case II A 1. If sđ is transitive then there exists A 6 si such that the set 3 x e X AKx fix for some p e c ỈS 1 contains more than just Oe X. Let be the subspace of X spanned by 3. Then HF 0 and is finite dimensional. If ee .@