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Proceedings of the American Mathematical Society
Proceedings of the American Mathematical Society
ISSN 1088-6826(online) ISSN 0002-9939(print)

$ A\geq B\geq 0$ assures $ (B\sp rA\sp pB\sp r)\sp {1/q}\geq B\sp {(p+2r)/q}$ for $ r\geq 0$, $ p\geq 0$, $ q\geq 1$ with $ (1+2r)q\geq p+2r$


Author: Takayuki Furuta
Journal: Proc. Amer. Math. Soc. 101 (1987), 85-88
MSC: Primary 47A60; Secondary 15A45, 47B15
MathSciNet review: 897075
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Abstract: An operator means a bounded linear operator on a Hilbert space. This paper proves the assertion made in its title. Theorem 1 yields the famous result that $ A \geq B \geq 0$ assures $ {A^\alpha } \geq {B^\alpha }$ for each $ \alpha \in [0,1]$ when we put $ r = 0$ in Theorem 1. Also Corollary 1 implies that $ A \geq B \geq 0$ assures $ {(B{A^p}B)^{1/p}} \geq {B^{(p + 2)/p}}$ for each $ p \geq 1$ and this inequality for $ p = 2$ is just an affirmative answer to a conjecture posed by Chan and Kwong. We cite three counterexamples related to Theorem 1 and Corollary 1.


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DOI: http://dx.doi.org/10.1090/S0002-9939-1987-0897075-6
PII: S 0002-9939(1987)0897075-6
Keywords: Positive operator, operator monotone function
Article copyright: © Copyright 1987 American Mathematical Society