Annals of Functional Analysis

An operator inequality implying the usual and chaotic orders

Jun Ichi Fujii, Masatoshi Fujii, and Ritsuo Nakamoto

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We prove that if positive invertible operators $A$ and $B$ satisfy an operator inequality $(B^{s/2}A^{(s-t)/2}B^tA^{(s-t)/2}B^{s/2})^{1\over{2s}}\geq B$ for some $t > s > 0$, then

(1) If $t \ge 3s-2 \ge 0$, then $\log B \geq \log A$, and if $t \ge s+2$ is additionally assumed, then $B \ge A$.

(2) If $s \in (0, 1/2)$, then $\log B \geq \log A$, and if $t \ge s+2$ is additionally assumed, then $B \ge A$.

It is an interesting application of the Furuta inequality. Furthermore we consider some related results.

Article information

Ann. Funct. Anal., Volume 5, Number 1 (2014), 24-29.

First available in Project Euclid: 5 February 2014

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Mathematical Reviews number (MathSciNet)

Zentralblatt MATH identifier

Primary: 47A63: Operator inequalities
Secondary: 47A56: Functions whose values are linear operators (operator and matrix valued functions, etc., including analytic and meromorphic ones)

Operator inequality chaotic order Furuta inequality


Fujii, Jun Ichi; Fujii, Masatoshi; Nakamoto, Ritsuo. An operator inequality implying the usual and chaotic orders. Ann. Funct. Anal. 5 (2014), no. 1, 24--29. doi:10.15352/afa/1391614565.

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