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Power Set of Some Quasinilpotent Weighted Shifts

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For a quasinilpotent bounded linear operator T, we write $$k_x=\limsup \limits _{z\rightarrow 0}\frac{\log \Vert (z-T)^{-1}x\Vert }{\log \Vert (z-T)^{-1}\Vert }$$ for each nonzero vector x. Set $$\Lambda (T)=\{k_x:x\ne 0\}$$ , and… Click to show full abstract

For a quasinilpotent bounded linear operator T, we write $$k_x=\limsup \limits _{z\rightarrow 0}\frac{\log \Vert (z-T)^{-1}x\Vert }{\log \Vert (z-T)^{-1}\Vert }$$ for each nonzero vector x. Set $$\Lambda (T)=\{k_x:x\ne 0\}$$ , and call it the power set $$\Lambda (T)$$ of T. This notation was introduced by Douglas and Yang (see Hermitian geometry on resolvent set. arXiv:1608.05990 , 2016; Oper Theory Adv Appl 267, Springer, Cham, 2018; Sci Cina Math. https://doi.org/10.1007/s11425-000-0000-0 ). They showed that for $$\tau \in \Lambda (T)$$ , $$M_\tau :=\{0, x:k_x\le \tau \}$$ is a linear subspace invariant under each A commuting with T; hence, if there are two different points $$\tau _j\in \Lambda (T)$$ such that $$M_{\tau _j}$$ ’s are closed, then T has a nontrivial hyperinvariant subspace. In this paper, we show that if a quasinilpotent unilateral weighted shift T is strongly strictly cyclic, then $$\Lambda (T)=\{1\}$$ . Moreover, we construct a quasinilpotent operator T such that $$\Lambda (T)=[0, 1]$$ and $$M_\tau $$ is not closed for all $$\tau $$ in [0, 1). Even so, we still find a subset $${\mathcal {N}}$$ of $${\text {Lat}}T$$ , the lattice of invariant subspaces of T, such that $${\mathcal {N}}$$ is order isomorphic to $$\Lambda (T)$$ .

Keywords: vert; tau; lambda; power set; quasinilpotent

Journal Title: Integral Equations and Operator Theory
Year Published: 2021

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