New estimations for the Berezin number inequality

Autor: Mojtaba Bakherad, Ulas Yamancı
Jazyk: angličtina
Rok vydání: 2020
Předmět:
Zdroj: Journal of Inequalities and Applications, Vol 2020, Iss 1, Pp 1-9 (2020)
Druh dokumentu: article
ISSN: 1029-242X
DOI: 10.1186/s13660-020-2307-0
Popis: Abstract 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 ∈ L ( H ) $X, Y, Z\in {\mathcal{L}}(\mathcal{H})$ such that X and Y are positive operators, then ber r ( ( X ♯ Y ) Z ) ≤ ber ( ( Z ⋆ Y Z ) r q 2 q + X r p 2 p ) − 1 p inf λ ∈ Ω ( [ X ˜ ( λ ) ] r p 4 − [ ( Z ⋆ Y Z ) ˜ ( λ ) ] r q 4 ) 2 , $$\begin{aligned} \operatorname{ber}^{r} \bigl( ( X\mathbin{\sharp} Y ) Z \bigr) &\leq \operatorname{ber} \biggl(\frac{ ( Z^{\star }YZ ) ^{\frac{rq}{2}}}{q}+ \frac{X^{ \frac{rp}{2}}}{p} \biggr) -\frac{1}{p}\inf_{\lambda \in \varOmega } \bigl( \bigl[ \widetilde{X} ( \lambda ) \bigr] ^{\frac{rp}{4}}- \bigl[ \widetilde{ \bigl( Z^{\star }YZ \bigr) } ( \lambda ) \bigr] ^{ \frac{rq}{4}} \bigr) ^{2}, \end{aligned}$$ in which X ♯ Y = X 1 2 ( X − 1 2 Y X − 1 2 ) 1 2 X 1 2 $X\mathbin{\sharp} Y=X^{\frac{1}{2}} ( X^{-\frac{1}{2}}YX^{- \frac{1}{2}} ) ^{\frac{1}{2}}X^{\frac{1}{2}}$ , p ≥ q > 1 $p\geq q>1$ such that r ≥ 2 q $r\geq \frac{2}{q}$ and 1 p + 1 q = 1 $\frac{1}{p}+\frac{1}{q}=1$ .
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