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Fuglede–Putnam theorem for class $A$ operators

Volume 138 / 2015

Salah Mecheri Colloquium Mathematicum 138 (2015), 183-191 MSC: Primary 47B47, 47A30, 47B20; Secondary 47B10. DOI: 10.4064/cm138-2-3

Abstract

Let $A\in B (H)$ and $B\in B(K)$. We say that $A$ and $B$ satisfy the Fuglede–Putnam theorem if $AX = XB$ for some $X\in B(K,H)$ implies $A^{*}X = XB^{*}$. Patel et al. (2006) showed that the Fuglede–Putnam theorem holds for class $A(s,t)$ operators with $s+t< 1$ and they mentioned that the case $s=t=1$ is still an open problem. In the present article we give a partial positive answer to this problem. We show that if $A\in B(H)$ is a class $A$ operator with reducing kernel and $B^{*}\in B(K)$ is a class $\mathcal {Y}$ operator, and $AX=XB$ for some $X\in B(K,H)$, then $A^{*}X=XB^{*}$.

Authors

  • Salah MecheriDepartment of Mathematics
    College of Sciences
    Taibah University
    Medina, Saudi Arabia
    e-mail

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