$A\geq B\geq 0$ assures $(B^ rA^ pB^ r)^ {1/q}\geq B^ {(p+2r)/q}$ for $r\geq 0$, $p\geq 0$, $q\geq 1$ with $(1+2r)q\geq p+2r$
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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.References
- N. N. Chan and Man Kam Kwong, Hermitian matrix inequalities and a conjecture, Amer. Math. Monthly 92 (1985), no. 8, 533–541. MR 812096, DOI 10.2307/2323157
- Frank Hansen, An operator inequality, Math. Ann. 246 (1979/80), no. 3, 249–250. MR 563403, DOI 10.1007/BF01371046
- Gert K. Pedersen, Some operator monotone functions, Proc. Amer. Math. Soc. 36 (1972), 309–310. MR 306957, DOI 10.1090/S0002-9939-1972-0306957-4
Additional Information
- © Copyright 1987 American Mathematical Society
- Journal: Proc. Amer. Math. Soc. 101 (1987), 85-88
- MSC: Primary 47A60; Secondary 15A45, 47B15
- DOI: https://doi.org/10.1090/S0002-9939-1987-0897075-6
- MathSciNet review: 897075