New estimates for the Berezin number of Hilbert space operators

Pub Date : 2024-03-25 DOI:10.1515/gmj-2024-2012
Parvaneh Zolfaghari
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引用次数: 0

Abstract

In this article, we improve some Berezin number inequalities concerning a Hilbert space. It is shown that if T is a bounded linear operator on a Hilbert space, then for any r 1 {r\geq 1} , 𝐛𝐞𝐫 2 r ( T ) 1 2 𝐛𝐞𝐫 r ( | T * | 2 ( 1 - t ) | T | 2 t ) + 1 4 | T | 4 r t + | T * | 4 r ( 1 - t ) 𝐛𝐞𝐫 ( 0 t 1 ) , \mathbf{ber}^{2r}(T)\leq\frac{1}{2}\mathbf{ber}^{r}({{|{{T}^{*}}|}^{2(1-t)}}{{% |T|}^{2t}})+\frac{1}{4}{{\|{{|T|}^{4rt}}+{{|{{T}^{*}}|}^{4r(1-t)}}\|}_{\mathbf% {ber}}}\quad(0\leq t\leq 1), where | T | = ( T * T ) 1 2 {|T|={{({{T}^{*}}T)}^{\frac{1}{2}}}} .
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希尔伯特空间算子贝雷津数的新估计值
在本文中,我们改进了一些关于希尔伯特空间的贝雷津数不等式。结果表明,如果 T 是希尔伯特空间上的有界线性算子,那么对于任意 r ≥ 1 {r\geq 1} ,𝐛𝐞𝐫𝐫是有界线性算子。 𝐛𝐞𝐫 2 r ( T ) ≤ 1 2 𝐛𝐞𝐫 r ( | T * | 2 ( 1 - t ) | T | 2 t ) + 1 4 ∥ | T | 4 r t + | T * | 4 r ( 1 - t ) ∥ 𝐛𝐞𝐫 ( 0 ≤ t ≤ 1 ) 、 \mathbf{ber}^{2r}(T)\leq\frac{1}{2}\mathbf{ber}^{r}({{|{{T}^{*}}|}^{2(1-t)}}{{% |T|}^{2t}})+\frac{1}{4}{{\|{{|T|}^{4rt}}+{{|{{T}^{*}}|}^{4r(1-t)}}\|}_{\mathbf% {ber}}}\quad(0\leq t\leq 1), 其中 | T | = ( T * T ) 1 2 {|T|={{({{T}^{*}}T)}^{\frac{1}{2}}}} .
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