New estimations for the Berezin number inequality

Journal of inequalities and applications

article

Mojtaba Bakherad¹  Ulas Yamancı² 

[1] Department of Mathematics, Faculty of Mathematics, University of Sistan and Baluchestan;Department of Statistics, Suleyman Demirel University

关键词: Inequalities;    Berezin number;    Berezin symbol;    Geometric mean;   

DOI  :  10.1186/s13660-020-2307-0

学科分类：电力

来源: SpringerOpen

【 摘 要 】

In this paper, by the definition of Berezin number, we present some inequalities involving the operator geometric mean. For instance, it is shown that if $X, Y, Z\in {\mathcal{L}}(\mathcal{H})$ such that X and Y are positive operators, then $$\begin{aligned} \operatorname{ber}^{r} \bigl( ( X\mathbin{\sharp} Y ) Z \bigr) &\leq \operatorname{ber} \biggl(\frac{ ( Z^{\star }YZ ) ^{\frac{rq}{2}}}{q}+ \frac{X^{ \frac{rp}{2}}}{p} \biggr) -\frac{1}{p}\inf_{\lambda \in \varOmega } \bigl( \bigl[ \widetilde{X} ( \lambda ) \bigr] ^{\frac{rp}{4}}- \bigl[ \widetilde{ \bigl( Z^{\star }YZ \bigr) } ( \lambda ) \bigr] ^{ \frac{rq}{4}} \bigr) ^{2}, \end{aligned}$$ in which $X\mathbin{\sharp} Y=X^{\frac{1}{2}} ( X^{-\frac{1}{2}}YX^{- \frac{1}{2}} ) ^{\frac{1}{2}}X^{\frac{1}{2}}$, $p\geq q>1$ such that $r\geq \frac{2}{q}$ and $\frac{1}{p}+\frac{1}{q}=1$.

【 授权许可】

CC BY