Estimations of Heron means for positive operators (Q2790749)

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scientific article; zbMATH DE number 6551726
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Estimations of Heron means for positive operators
scientific article; zbMATH DE number 6551726

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    Estimations of Heron means for positive operators (English)
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    8 March 2016
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    arithmetic mean
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    geometric mean
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    arithmetic-geometric mean inequality
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    Horen mean
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    Assume that \(\mu\in[0, 1]\), \(A\) and \(B\) are positive operators on a Hilbert space, and \(r\in\mathbb{R}\). The \textit{Heron mean} for positive operators \(A\) and \(B\) is defined by NEWLINE\[NEWLINE H_r^\mu(A, B):=r(A\sharp_\mu B)+(1-r)(A\nabla_\mu B)\,, NEWLINE\]NEWLINE where \(A \sharp_\mu B=A^{1/2}\left(A^{-1/2}BA^{-1/2}\right)^\mu A^{1/2}\) and \(A\nabla_\mu B=(1-\mu)A+\mu B\) are the \(\mu\)-geometric mean and the \(\mu\)-arithmetic mean, respectively.NEWLINENEWLINEIn the paper under review, the authors estimate \(H_r^\mu(A, B)\) by the harmonic mean as follows:NEWLINENEWLINEIf \(\mu\in[\frac{1}{2}, 1)\) and \(r \geq \frac{2(2-\mu)}{3(1-\mu)}\), then NEWLINE\[NEWLINE H_r^\mu(A, B) \leq A!_\mu B\,, NEWLINE\]NEWLINE where \(A!_\mu B=\left((1-\mu)A^{-1}+\mu B^{-1}\right)^{-1}\) is the harmonic mean.NEWLINENEWLINEAlso, the authors prove that, if \(A, B>0\), \(C:=A^{-1/2}BA^{-1/2}\), \(0 \leq \mu \leq\frac{1}{2}\) and \(B-A\geq m >0\), then NEWLINE\[NEWLINE c_1 \leq A\nabla_mu B+B!_\mu A-2A\sharp_\mu B\,, NEWLINE\]NEWLINE where \(c_1=2m_A\{\|\left((1+m\|A\|^{-1})^{1/2}-(1+m\|A\|^{-1})^\mu\right)\|\} \geq 0\) such that \(m_X=\min\sigma(X)=\|X^{-1}\|^{-1}\) for \(X>0\) and, if \(\frac{1}{2} \leq \mu \leq 1\) and \(A-B \geq m >0\), then NEWLINE\[NEWLINE \min\{c_2, c_3\} \leq A\nabla_\mu B+B!_\mu A-2A\sharp_\mu B\,, NEWLINE\]NEWLINE where \(c_2=2m_A\left\{(1-m\|A\|^{-1})^{1/2}-(1-m\|A\|^{-1})^\mu\right\}\geq0\) and \(c_3=2m_A\left\{(m_C)^{1/2}-(m_C)^\mu\right\}\geq0\).
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