Normality can be relaxed in the asymptotic Fuglede-Putnam theorem
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- by Takayuki Furuta
- Proc. Amer. Math. Soc. 79 (1980), 593-596
- DOI: https://doi.org/10.1090/S0002-9939-1980-0572310-6
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Abstract:
The original form of the Fuglede-Putnam theorem states that the operator equation $AX = XB$ implies ${A^ \ast }X = X{B^ \ast }$ when A and B are normal. In our previous paper we have relaxed the normality in the hypotheses on A and B as follows: if A and ${B^ \ast }$ are subnormal and if X is an operator such that $AX = XB$, then ${A^ \ast }X = X{B^ \ast }$. We shall show asymptotic versions of this generalized Fuglede-Putnam theorem; these results are also extensions of results of Moore and Rogers.References
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Bibliographic Information
- © Copyright 1980 American Mathematical Society
- Journal: Proc. Amer. Math. Soc. 79 (1980), 593-596
- MSC: Primary 47B20
- DOI: https://doi.org/10.1090/S0002-9939-1980-0572310-6
- MathSciNet review: 572310