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Best possibility of the Furuta inequality

Author: Kôtarô Tanahashi
Journal: Proc. Amer. Math. Soc. 124 (1996), 141-146
MSC (1991): Primary 47B15
MathSciNet review: 1291794
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Abstract: Let $0\le p,q,r\in\mathbb R, p+2r\le(1+2r)q$, and $1\le q$. Furuta (1987) proved that if bounded linear operators $A,B\in B(H)$ on a Hilbert space $H$ $(\dim(H)\ge 2)$ satisfy $0\le B\le A$, then $(A^r B^p A^r)^{1/q} \le A^{(p+2r)/q}$. In this paper, we prove that the range $p+2r\le (1+2r)q$ and $1\le q$ is best possible with respect to the Furuta inequality, that is, if $(1+2r) q<p+2r$ or $0<q<1$, then there exist $A,B\in B(\mathbb R^2)$ which satisfy $0\le B\le A$ but $(A^r B^p A^r)^{1/q}\nleq A^{(p+2r)/q}$.

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Kôtarô Tanahashi
Affiliation: Department of Mathematics, Tohoku College of Pharmacy, Komatsushima, Aoba-ku, Sendai 981, Japan

Keywords: The L\"owner-Heinz inequality, the Furuta inequality, positive operator
Received by editor(s): February 25, 1994
Received by editor(s) in revised form: July 7, 1994
Communicated by: Palle E. T. Jorgensen
Article copyright: © Copyright 1996 American Mathematical Society