Lattices of invariant subspaces for sequentially subdecomposable operators (Q2749022)

The main result of this paper is a strengthening of an invariant subspace theorem for some sequentially subdecomposable operators on a Banach space obtained before by \textit{H. Mohebi} and \textit{M. Radjabalipour} [Integral Equations Oper. Theory 18, No. 2, 222-241 (1994; Zbl 0807.47020)]. This latter theorem says that if \(T\) and \(B\) are bounded linear operators on Banach spaces \(X\) and \(Z\), respectively, and \(G\) is a nonempty open subset of the complex plane \(\mathbb{C}\) satisfying the following conditions:NEWLINENEWLINENEWLINE(1) \(qB= T^*q\) for some surjective operator \(q\) from \(Z\) to \(X^*\),NEWLINENEWLINENEWLINE(2) there exist a sequence of open subsets \(\{G_n\}\) of \(\mathbb{C}\) and a sequence of invariant subspaces \(\{M_n\}\) of \(B\) such that \(\overline G_n\subseteq G_{n+1}\), \(G= \bigcup_n G_n\), \(\sigma(B|M_n)\subseteq\overline G_n\) and \(\sigma(B/M_n)\subseteq \mathbb{C}\setminus G_n\) for all \(n\geq 1\), andNEWLINENEWLINENEWLINE(3) the set \(K= \sigma(T)\setminus\{\lambda\in \mathbb{C}: \overline{(B-\lambda I)Z}\neq Z\) and \(\overline{(B-\lambda I)\ker q}\neq \ker q\}\) is dominating in \(G\),NEWLINENEWLINENEWLINEthen \(T\) has infinitely many invariant subspaces.NEWLINENEWLINENEWLINEIn the present paper, the author gives a different proof of this theorem, again based on Scott Brown's techniques, and, under the assumption that \(K\cap (\sigma_p(T)\cup \sigma_r(T))\) is a finite set (\(\sigma_p(T)\) and \(\sigma_r(T)\) are the point and residual spectra of \(T\), respectively), shows that the lattice of invariant subspaces of \(T\) has a sublattice which is order-isomorphic to the lattice of all (closed) subspaces of some Banach space.