The Aluthge transform of unilateral weighted shifts and the square root problem for finitely atomic measures.

*(English)*Zbl 1447.47029Let \(T=V|T|\) 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.

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.

Reviewer: Abdellatif Bourhim (Syracuse)

##### MSC:

47B20 | Subnormal operators, hyponormal operators, etc. |

47B37 | Linear operators on special spaces (weighted shifts, operators on sequence spaces, etc.) |