Fuglede–Putnam theorem for class $A$ operators
Colloquium Mathematicum, Tome 138 (2015) no. 2, pp. 183-191
Cet article a éte moissonné depuis la source Institute of Mathematics Polish Academy of Sciences
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 $A^{*}X=XB^{*}$.
Keywords:
say satisfy fuglede putnam theorem implies * * patel showed fuglede putnam theorem holds class operators mentioned still problem present article partial positive answer problem class operator reducing kernel * class mathcal operator * *
Affiliations des auteurs :
Salah Mecheri  1
@article{10_4064_cm138_2_3,
author = {Salah Mecheri},
title = {Fuglede{\textendash}Putnam theorem for class $A$ operators},
journal = {Colloquium Mathematicum},
pages = {183--191},
year = {2015},
volume = {138},
number = {2},
doi = {10.4064/cm138-2-3},
language = {en},
url = {http://geodesic.mathdoc.fr/articles/10.4064/cm138-2-3/}
}
Salah Mecheri. Fuglede–Putnam theorem for class $A$ operators. Colloquium Mathematicum, Tome 138 (2015) no. 2, pp. 183-191. doi: 10.4064/cm138-2-3
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