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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
DOI: https://doi.org/10.1090/S0002-9939-96-03055-9
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 $(A^r B^p A^r)^{1/q}\nleq A^{(p+2r)/q}$.


References [Enhancements On Off] (What's this?)

  • 1 T. Furuta, $A\ge B\ge 0$ assures $(B^r A^p B^r)^{1/q}\ge B^{(p+2r)/q}$ for $r\ge 0$, $p\ge 0$, $q\ge 1$ with $(1+2r)q\ge (p+2r)$ Proc. Amer. Math. Soc. 101 (1987), 85--88. MR 89b:47028
  • 2 E. Heinz, Beiträge zur Störungstheorie der Spektralzerlegung, Math. Ann. 123 (1951), 415--438. MR 13:471f
  • 3 K. Löwner, Über monotone Matrixfunktionen, Math. Z. 38 (1934), 177--216.

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Additional Information

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

DOI: https://doi.org/10.1090/S0002-9939-96-03055-9
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

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