Berezin number inequalities for operators

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

Keywords

• reproducing kernel
• berezin number
• berezin transform
• primary: 47a30
• secondary: 15a60
• 30e20
• 47a12
• 47b15
• 47b20