In this paper,the following main resvlts have been proved:1.Let T∈B(X) be a super-decomposable operator,then for any hyperinvariant subspace Y of T,the operator T/Y is super-decomposable.2.The operator T∈B(X) is sup...In this paper,the following main resvlts have been proved:1.Let T∈B(X) be a super-decomposable operator,then for any hyperinvariant subspace Y of T,the operator T/Y is super-decomposable.2.The operator T∈B(X) is super-decomposable if and only if T/Y is super-decomposable for any hyperinvariant subspace Y of T.3. If T∈B(X) has on open spectral resolvent,then T is superdecomposable.展开更多
文摘In this paper,the following main resvlts have been proved:1.Let T∈B(X) be a super-decomposable operator,then for any hyperinvariant subspace Y of T,the operator T/Y is super-decomposable.2.The operator T∈B(X) is super-decomposable if and only if T/Y is super-decomposable for any hyperinvariant subspace Y of T.3. If T∈B(X) has on open spectral resolvent,then T is superdecomposable.