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Let $A$ be a unital Banach algebra. An element $a \in A$ is hermitian if $\|\mathrm{exp}(ita)\|=1$ for every $t \in \mathbb{R}$. An element $a \in A$ is Moore-Penrose invertible if there exists $b \in A$ such that $a=aba$, $b=bab$, and $ab$ and $ba$ are hermitian. One can show that the element $b$ with these properties is unique (if it exists), and then it is called the Moore-Penrose inverse of $a$, denoted by $a^t$. An element $a \in A$ is called a MP-partial isometry if it is Moore-Penrose invertible and both $a$ and $a^t$ are contractive.

In a unital C*-algebra, an element $a$ is hermitian if and only if it is self-adjoint ($a=a^*$), and an element $a$ is a MP-partial isometry if and only if it is a partial isometry ($a=aa^*a$).

Question: Let $a$ be a MP-partial isometry, and let $b$ be hermitian. Is $aba^t$ hermitian?

In C*-algebras this is true (easy), and it for operators on Lp-spaces it also seems to hold. What about general Banach algebras?

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    $\begingroup$ I assume "MP-partial isometry" = "PI-partial isometry"? $\endgroup$
    – Yemon Choi
    Commented Oct 30 at 0:08
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    $\begingroup$ Yes, that was a typo. It is "MP-partial isometry", short for "Moore-Penrose partial isometry". $\endgroup$ Commented Oct 30 at 6:40

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