Journal of Operator Theory

Volume 21, Issue 1, Winter 1989 pp. 107-116.

$A \ge B \ge 0$ ensures

$B^rA^pB^r \ge \left({B^rA^{p - s}B^r}\right)^{\left({p + 2r}\right)\left({p - s + 2r}\right)}$ for

$1 \ge 2r \ge 0, p \ge s \ge 0$ with

$p + 2r \ge 2s$

Authors: Takayuki Furuta
Author institution:Department of Mathematics, Faculty of Science, Hirosaki University, Bunkyo-cho 3, Hirosaki 036, Aomori, Japan.

Summary:

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