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It was confirmed that Wold-type decomposition can be extended from von Neumann algebras to Baer*-rings (see this paper). By algebraic tools the notion of unilateral shifts is successfully transmitted to Baer*-rings (see Def. 2.3 of the paper). I have a question in this context. To state it, some notations are given from the paper.

$\bullet$ For a given $y$ in $A$, the left projection of $y$ is the smallest projection $[y]$ satisfying in $[y]y=y$. When $A$ is a von algebra contained in $B(H)$ for some Hilbert space $H$ then, $[y]$ is just the range projection of the operator $y$.

$\bullet$ Let $x$ be an isometry element in $A$ that is $x^*x=1$. A projection $p$ is called wandering for $x$ if $[x^np][x^mq]=0$ for every different pair of natural numbers $m$ and $n$.

$\bullet$ An isometry $x$ is called a unilateral shift in $A$ if there exists a wandering projection $p$ with $\sup_{n\geq0} \sum_0^n[x^np]=1$.

Wold-decomposition in Baer*-rings. Let $x$ be an isometry in $A$. There exists a commuting projection $q$ with $x$ such that $x$ is decomposed to a direct sum $x=xq+(1-q)x$ of a unitary $xq$ and unilateral shift $(1-q)x$ in the corner sub-rings $Aq$ and $A(1-q)$ respectively.

Question. Let A be a Baer*-ring. Let $x$ be a unilateral shift and $q$ be a finite projection in $A$. Can we conclude that $\inf_{n\geq1}[x^{*n}q]=0$?

The answer is clearly yes in von Neumann algebras and seems to be YES in general case.

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