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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.
MSC:
 47B20 Subnormal operators, hyponormal operators, etc. 47B37 Linear operators on special spaces (weighted shifts, operators on sequence spaces, etc.)
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