Berezin number inequalities for operators

IF 0.3 Q4 MATHEMATICS
M. Bakherad, M. Garayev
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引用次数: 38

Abstract

Abstract The Berezin transform à of an operator A, acting on the reproducing kernel Hilbert space ℋ = ℋ (Ω) over some (non-empty) set Ω, is defined by Ã(λ) = 〉Aǩ λ, ǩ λ〈 (λ ∈ Ω), where k⌢λ=kλ‖ kλ ‖ ${\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}}\over k} _\lambda } = {{{k_\lambda }} \over {\left\| {{k_\lambda }} \right\|}}$ is the normalized reproducing kernel of ℋ. The Berezin number of an operator A is defined by ber(A)=supλ∈Ω| A˜(λ) |=supλ∈Ω| 〈 Ak⌢λ,k⌢λ 〉 | ${\bf{ber}}{\rm{(}}A) = \mathop {\sup }\limits_{\lambda \in \Omega } \left| {\tilde A(\lambda )} \right| = \mathop {\sup }\limits_{\lambda \in \Omega } \left| {\left\langle {A{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}}\over k} }_\lambda },{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}}\over k} }_\lambda }} \right\rangle } \right|$ . In this paper, we prove some Berezin number inequalities. Among other inequalities, it is shown that if A, B, X are bounded linear operators on a Hilbert space ℋ, then ber(AX±XA)⩽ber12(A*A+AA*)ber12(X*X+XX*) $${\bf{ber}}(AX \pm XA) \leqslant {\bf{be}}{{\bf{r}}^{{1 \over 2}}}\left( {A*A + AA*} \right){\bf{be}}{{\bf{r}}^{{1 \over 2}}}\left( {X*X + XX*} \right)$$ and ber2(A*XB)⩽‖ X ‖2ber(A*A)ber(B*B). $${\bf{be}}{{\bf{r}}^2}({A^*}XB) \leqslant {\left\| X \right\|^2}{\bf{ber}}({A^*}A){\bf{ber}}({B^*}B).$$ We also prove the multiplicative inequality ber(AB)⩽ber(A)ber(B) $${\bf{ber}}(AB){\bf{ber}}(A){\bf{ber}}(B)$$
算子的Berezin数不等式
算子A的Berezin变换Ã作用于某个(非空)集Ω上的再现核希尔伯特空间h = h (Ω),定义为Ã(λ) = > a λ, λ < (λ∈Ω),其中kλ =kλ‖kλ‖ ${\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}}\over k} _\lambda } = {{{k_\lambda }} \over {\left\| {{k_\lambda }} \right\|}}$ 是h的归一化再现核。算子A的Berezin数定义为ber(A)=supλ∈Ω| A ~ (λ) |=supλ∈Ω| < Ak λ,k λ > | ${\bf{ber}}{\rm{(}}A) = \mathop {\sup }\limits_{\lambda \in \Omega } \left| {\tilde A(\lambda )} \right| = \mathop {\sup }\limits_{\lambda \in \Omega } \left| {\left\langle {A{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}}\over k} }_\lambda },{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}}\over k} }_\lambda }} \right\rangle } \right|$ 。本文证明了一些Berezin数不等式。在其他不等式中,证明了如果A, B, X是Hilbert空间h上的有界线性算子,则ber(AX±XA)≤ber12(A*A+AA*)ber12(X*X+XX*) $${\bf{ber}}(AX \pm XA) \leqslant {\bf{be}}{{\bf{r}}^{{1 \over 2}}}\left( {A*A + AA*} \right){\bf{be}}{{\bf{r}}^{{1 \over 2}}}\left( {X*X + XX*} \right)$$ ber2(A*XB)≥‖X‖2ber(A*A)ber(B*B)。 $${\bf{be}}{{\bf{r}}^2}({A^*}XB) \leqslant {\left\| X \right\|^2}{\bf{ber}}({A^*}A){\bf{ber}}({B^*}B).$$ 我们还证明了乘法不等式ber(AB)≤ber(A)ber(B) $${\bf{ber}}(AB){\bf{ber}}(A){\bf{ber}}(B)$$
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来源期刊
Concrete Operators
Concrete Operators MATHEMATICS-
CiteScore
1.00
自引率
16.70%
发文量
10
审稿时长
22 weeks
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