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An operator inequality related to Jensen's inequality

Author(s): Mitsuru Uchiyama
Journal: Proc. Amer. Math. Soc. 129 (2001), 3339-3344.
MSC (2000): Primary 47A63, 15A48
Posted: April 9, 2001
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Abstract:

For bounded non-negative operators $A$ and $B$, Furuta showed

\begin{displaymath}0\leq A \leq B {\rm implies } A^{\frac{r}{2}}B^sA^{\frac{r}{2... ... A^{\frac{r}{2}})^{\frac{s+r}{t+r}} (0\leq r, 0\leq s \leq t).\end{displaymath}

We will extend this as follows: $0\leq A\leq B \underset{\lambda}{!}C $ $(0<\lambda <1)$ implies

\begin{displaymath}A^{\frac{r}{2}}(\lambda B^s+ (1-\lambda)C^s)A^{\frac{r}{2}} \... ...bda B^t+ (1- \lambda)C^t) A^{\frac{r}{2}}\}^{\frac{s+r}{t+r}} ,\end{displaymath}

where $B \underset{\lambda}{!}C$ is a harmonic mean of $B$ and $C$. The idea of the proof comes from Jensen's inequality for an operator convex function by Hansen-Pedersen.

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Additional Information:

Mitsuru Uchiyama
Affiliation: Department of Mathematics, Fukuoka University of Education, Munakata, Fukuoka, 811-4192, Japan
Email: uchiyama@fukuoka-edu.ac.jp

DOI: 10.1090/S0002-9939-01-06130-5
PII: S 0002-9939(01)06130-5
Keywords: Order of selfadjoint operators, Jensen inequality, Furuta inequality
Received by editor(s): March 21, 2000
Posted: April 9, 2001
Communicated by: Joseph A. Ball
Copyright of article: Copyright 2001, American Mathematical Society

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