Berezin数不等式的改进

M. Gürdal, Hamdullah Basaran
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引用次数: 1

摘要

对于泛函希尔伯特空间$\mathcal{H}\left(\Omega\right) $上的有界线性算子$ a $,对于归一化的复制内核$\widehat {k}_{\eta}:=\frac{k_{\eta}}{\左\Vert k_{\eta}\右\Vert {\mathcal{H}}}, Berezin符号和Berezin数分别定义为$\ widetide {A}\左\langle A\widehat{k} {\eta}} $和$\mathrm{ber}(A):=\sup_{\eta\in\Omega}\左\Vert \ widetide {A}{(\eta)}\右\Vert。对这些性质进行简单的比较,可以得出$\mathrm{ber}%\left(A\right)\ leq \压裂{1}{2}\左左(\ \绿色一个\ \ Vert_ {\ mathrm{1}} + \ \绿色了^{2}\右\绿色_ {\ mathrm{1}} ^{5} \右)美元(见[17])。本文进一步证明了与它们相关的不等式,并建立了泛函Hilbert空间上算子的Berezin数的一些不等式
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Advanced refinements of Berezin number inequalities
For a bounded linear operator $A$ on a functional Hilbert space $\mathcal{H}\left( \Omega\right) $, with normalized reproducing kernel $\widehat {k}_{\eta}:=\frac{k_{\eta}}{\left\Vert k_{\eta}\right\Vert _{\mathcal{H}}},$ the Berezin symbol and Berezin number are defined respectively by $\widetilde{A}\left( \eta\right) :=\left\langle A\widehat{k}_{\eta},\widehat{k}_{\eta}\right\rangle _{\mathcal{H}}$ and $\mathrm{ber}(A):=\sup_{\eta\in\Omega}\left\vert \widetilde{A}{(\eta)}\right\vert .$ A simple comparison of these properties produces the inequality $\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) $ (see [17]). In this paper, we prove further inequalities relating them, and also establish some inequalities for the Berezin number of operators on functional Hilbert spaces
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