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The Aluthge transform of unilateral weighted shifts and the square root problem for finitely atomic measures. (English) Zbl 1447.47029
Let \(T=V|T|\) be the canonical polar decomposition of a bounded linear operator on a complex Hilbert space \(H\) with an orthonormal basis \((e_k)_{k\geq0}\), and let \(\widetilde{T}:=|T|^{\frac{1}{2}}V|T|^{\frac{1}{2}}\) be its Aluthge transform. Let \(W_\alpha\) be a unilateral weighted shift with bounded weighted sequence \((\alpha_k)_{k\geq0}\); i.e., \[ W_\alpha e_k=\alpha_ke_{k+1} \] for all \(k\geq0\). The well-known Berger theorem characterizing subnormal weighted shifts states that \(W_\alpha\) is subnormal if and only if there exists a probability measure \(\mu\) supported on \(\left[0,\|W_\alpha\|^2\right]\), and called the Berger measure associated with \(W_\alpha\), such that \[ (\alpha_0\dots\alpha_{n-1})^2=\int_0^{\|W_\alpha\|^2}t^n\,d\mu(t) \] for all \(n\geq1\).
Given a positive probability Borel measure \(\mu\) supported on an interval \([a,b] \subset\mathbb{R}+\), the Square Root Problem asks if there is a positive Borel measure \(\nu\) (called the square root of \(\mu\)) such that \(\mu=\nu*\nu\) where \(*\) denotes the multiplicative convolution. If \(W_\alpha\) is subnormal and \(\mu\) is its Berger measure, then it is observed that the Aluthge transform \(\widetilde{W}_\alpha\) is subnormal, provided that \(\mu\) has a square root. Among other things, the paper under review investigates whether the subnormality of \(\widetilde{W}_\alpha\) implies that \(\mu\) has a square root. This is positively answered when the measure \(\mu\) has at most five atoms.
47B20 Subnormal operators, hyponormal operators, etc.
47B37 Linear operators on special spaces (weighted shifts, operators on sequence spaces, etc.)
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