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Solution of the quadratically hyponormal completion problem

Authors:	Raú l E Curto Woo Young Lee

Affiliation:	Department of Mathematics, University of Iowa, Iowa City, Iowa 52242 ; Department of Mathematics, SungKyunKwan University, Suwon 440-746, Korea

Abstract:	For $m\ge 1$ , let $\alpha : \alpha _{0}<\cdots <\alpha _{m}$ be a collection of () positive weights. The Quadratically Hyponormal Completion Problem seeks necessary and sufficient conditions on $\alpha$ to guarantee the existence of a quadratically hyponormal unilateral weighted shift with $\alpha$ as the initial segment of weights. We prove that $\alpha$ admits a quadratically hyponormal completion if and only if the self-adjoint $m\times m$ matrix $\begin{displaymath}D_{m-1}(s):= \begin{pmatrix}q_{0}&\bar r_{0}&0&\hdots &0&0 ... ...m-2}&\bar r_{m-2} 0&0&0&\hdots &r_{m-2}&q_{m-1}\end{pmatrix}\end{displaymath}$ is positive and invertible, where $q_{k}:=u_{k}+\vert s\vert^{2} v_{k}$ , $r_{k}:=s\sqrt {w_{k}}$ , $u_{k}:=\alpha _{k}^{2}-\alpha _{k-1}^{2}$ , $v_{k}:=\alpha _{k}^{2}\alpha _{k+1}^{2}-\alpha _{k-1}^{2}\alpha _{k-2}^{2}$ , $w_{k}:=\alpha _{k}^{2}(\alpha _{k+1}^{2}-\alpha _{k-1}^{2})^{2}$ , and, for notational convenience, $\alpha _{-2}=\alpha _{-1}=0$ . As a particular case, this result shows that a collection of four positive numbers $\alpha _{0}<\alpha _{1}<\alpha _{2}<\alpha _{3}$ always admits a quadratically hyponormal completion. This provides a new qualitative criterion to distinguish quadratic hyponormality from 2-hyponormality.

Keywords:	Weighted shifts propagation subnormal $k$-hyponormal quadratically hyponormal completions

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