Some refinements of Berezin number inequalities via convex functions

Year 2023, Volume: 72 Issue: 1, 32 - 42, 30.03.2023

Suna Saltan Nazlı Baskan

https://doi.org/10.31801/cfsuasmas.1089790

Abstract

The Berezin transform $\widetilde{A}$ and the Berezin number of an operator
$A$ on the reproducing kernel Hilbert space over some set $\Omega$ with
normalized reproducing kernel $\widehat{k}_{\lambda}$ are defined,
respectively, by $\widetilde{A}(\lambda)=\left\langle {A}\widehat{k}_{\lambda
},\widehat{k}_{\lambda}\right\rangle ,\ \lambda\in\Omega$ and $\mathrm{ber}%
(A):=\sup_{\lambda\in\Omega}\left\vert \widetilde{A}{(\lambda)}\right\vert .$
A straightforward comparison between these characteristics yields the
inequalities $\mathrm{ber}\left( A\right) \leq\frac{1}{2}\left( \left\Vert
A\right\Vert _{\mathrm{ber}}+\left\Vert A^{2}\right\Vert _{\mathrm{ber}}%
^{1/2}\right) $. In this paper, we study further inequalities relating them.
Namely, we obtained some refinement of Berezin number inequalities involving
convex functions. In particular, for $A\in\mathcal{B}\left( \mathcal{H}%
\right) $ and $r\geq1$ we show that
\[
\mathrm{ber}^{2r}\left( A\right) \leq\frac{1}{4}\left( \left\Vert A^{\ast
}A+AA^{\ast}\right\Vert _{\mathrm{ber}}^{r}+\left\Vert A^{\ast}A-AA^{\ast
}\right\Vert _{\mathrm{ber}}^{r}\right) +\frac{1}{2}\mathrm{ber}^{r}\left(
A^{2}\right) .
\]

Keywords

Berezin number, Berezin transform, Cauchy-Schwarz inequality, Young inequality, reproducing kernel Hilbert space

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