Local resolvents of operators with one-dimensional self-commutator
Proceedings of the American Mathematical Society
Let T = H + ¡J be an irreducible operator on a Hubert space with one-dimensional self-commutator. It is known that the selfadjoint operator H is absolutely continuous. Let EH denote the absolutely continuous support of H. In this note the following theorem is proven: Theorem. // there exists a real number p such that ess inf EH < p < ess sup EH and fE \t -p\ dt < oo, then the operator T has a nontrivial invariant subspace. Let % be a separable Hilbert space with inner product ( , ). Let F be a
... ounded linear operator on %. The operator T is called hyponormal in case its self-commutator F* T -TT* = D is nonnegative. If the adjoint F* is a hyponormal operator, then F is called cohyponormal. If the self-commutator T*T -TT* is a one-dimensional operator, then either F or F* is hyponormal. It is not known at present whether every operator with a one-dimensional self-commutator has a nontrivial invariant subspace. In this note a result is described which increases the class of operators with one-dimensional selfcommutator that are known to have nontrivial invariant subspaces. Let F be a cohyponormal operator. The cartesian decomposition of F will be expressed T = H + U, where H and J are selfadjoint. For the purposes of this note it can be assumed that T is an irreducible operator. In this case the selfadjoint operators H and J are absolutely continuous [10, Theorem 3.2.1]. Suppose H = / t dG, is the spectral resolution of H. Then there is a Borel set EH in the real line, determined up to a set of measure zero with the property $EHdGt = I and if ff dG, = I, for F C EH, then EH\Fis of Lebesgue measure zero. This set EH (really EH is an equivalence class of Borel sets) will be called the absolutely continuous support of H.