Local spectra of unilateral operator weighted shifts (Q1766408)

For a sequence of uniformly bounded invertible operators \(\{A_n\}^\infty_{n=0}\) on a complex Hilbert space \(H\), the unilateral operator weighted shift \(S\) is defined on \(\widehat H= \sum^\infty_{n=0}\oplus H\) by \(S(x_0\oplus x_1\oplus\dots)= 0\oplus A_0x_0\oplus A_1 x_1\oplus\dots\)\ . The present paper studies and examines whether some of the results obtained before by various authors on the local spectral properties for scalar weighted shifts, the case when \(\dim H= 1\), remain valid for the operator weighted shifts. Among them are Dunford's condition (C) and Bishop's property \((\beta)\) for \(S\). Recall that for any operator \(T\) on \(H\) and any vector \(x\) in \(H\), the local spectrum \(\sigma_T(x)\) is the intersection of all closed subsets \(F\) of the plane for which there is no analytic function \(\phi\) from \(\mathbb{C}\setminus F\) to \(H\) which satisfies \((T- \lambda I)\phi(\lambda)= x\) on \(\mathbb{C}\setminus F\). \(T\) is said to satisfy Dunford's condition (C) if for every closed subset \(F\) of the plane, the hyperinvariant subspace \(\{x\in H: \sigma_T(x)\subseteq F\}\) of \(T\) is closed. It is shown (in Theorem 4.1) that if \(S\) satisfies Dunford's condition (C), then its spectral radius equals \(\sup_{x\neq 0}\limsup_{n\to\infty}\| B_nx\|^{1/n}\), where \(B_n= A_{n-1}A_{n-2}\cdots A_1 A_0\) for \(n> 0\) and \(B_0= I\). An operator \(T\) on \(H\) is said to possess Bishop's property \((\beta)\) if for any open subset \(U\) of the plane, the mapping \(f\mapsto (T-\lambda I)f(\lambda)\) on the space of all analytic \(H\)-valued functions \(f\) on \(U\) is injective with closed range. It is known that Bishop's property \((\beta)\) implies Dunford's condition (C). Here the author proves in Theorem 4.2 that if \(S\) possesses Bishop's property \((\beta)\), then the limit of \(m(S^n)^{1/n}\) as \(n\) approaches infinity equals \((\limsup_{n\to\infty}\| B^{-1}_n\|^{1/n})^{-1}\), where \(m(A)= \text{inf}\{\| Ax\|:\| x\|= 1\}\) for any operator \(A\). (That \(m(A^n)^{1/n}\) always converges, has been proved before by Makai and Zemánek).