New estimations for the Berezin number inequality

Abstract

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 $\frac{1}{p}+\frac{1}{q}=1$ .

Keywords

• Inequalities
• Berezin number
• Berezin symbol
• Geometric mean