Fuglede-Putnam theorem for class A operators

Salah Mecheri. "Fuglede-Putnam theorem for class A operators." Colloquium Mathematicae 138.2 (2015): 183-191. <http://eudml.org/doc/283796>.

@article{SalahMecheri2015,
abstract = {Let A ∈ B(H) and B ∈ B(K). We say that A and B satisfy the Fuglede-Putnam theorem if AX = XB for some X ∈ 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 ∈ B(H) is a class A operator with reducing kernel and B* ∈ B(K) is a class 𝓨 operator, and AX = XB for some X ∈ B(K,H), then A*X = XB*.},
author = {Salah Mecheri},
journal = {Colloquium Mathematicae},
keywords = {class operator; Fuglede-Putnam's theorem},
language = {eng},
number = {2},
pages = {183-191},
title = {Fuglede-Putnam theorem for class A operators},
url = {http://eudml.org/doc/283796},
volume = {138},
year = {2015},
}

TY - JOUR
AU - Salah Mecheri
TI - Fuglede-Putnam theorem for class A operators
JO - Colloquium Mathematicae
PY - 2015
VL - 138
IS - 2
SP - 183
EP - 191
AB - Let A ∈ B(H) and B ∈ B(K). We say that A and B satisfy the Fuglede-Putnam theorem if AX = XB for some X ∈ 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 ∈ B(H) is a class A operator with reducing kernel and B* ∈ B(K) is a class 𝓨 operator, and AX = XB for some X ∈ B(K,H), then A*X = XB*.
LA - eng
KW - class operator; Fuglede-Putnam's theorem
UR - http://eudml.org/doc/283796
ER -