Some refinements of Berezin number inequalities via convex functions

IF 0.7 Q2 MATHEMATICS
S. Saltan, Nazlı Baskan
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引用次数: 0

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 \[ \mathrm{ber}^{2r}\left( A\right) \leq\frac{1}{4}\left( \left\Vert A^{\ast }A+AA^{\ast}\right\Vert _{\mathrm{ber}}^{r}+\left\Vert A^{\ast}A-AA^{\ast }\right\Vert _{\mathrm{ber}}^{r}\right) +\frac{1}{2}\mathrm{ber}^{r}\left( A^{2}\right) . \]
通过凸函数对Berezin数不等式的若干改进
一个具有标准化再生核的集合$\Omega$上再生核Hilbert空间上算子$A$的Berezin变换$\widetilde{A}$和Berezin数{k}_{\lambda}$分别由$\widetilde{A}(\lambda)=\left\langle{A}\widehat定义{k}_{\lambda},\widehat{k}_{\lambda}\right\rangle,\\lambda\in\Omega$和$\mathrm{ber}%(A):=\sup_{\lamba\in\Omega}\left \vert\widetilde{A}{(\lambda)}\right \vert$这些特征之间的直接比较产生等式$\mathrm{ber}\left(A\right)\leq\frac{1}{2}\lift(\left\VertA\right\Vert_。在本文中,我们进一步研究了与它们相关的不等式。也就是说,我们得到了包含凸函数的Berezin数不等式的一些精化。特别是,对于$A\In\mathcal{B}\left(\mathcal{H}%\right)$和$r\geq1$,我们证明了\[\mathrm{ber}^{2r}\left(A\right)\leq\frac{1}{4}\left right)+\frac{1}{2}\mathrm{ber}^{r}\left(A^{2}\right)。\]
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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