{"title":"Berezin number inequalities for operators","authors":"M. Bakherad, M. Garayev","doi":"10.1515/conop-2019-0003","DOIUrl":null,"url":null,"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) $${\\bf{ber}}(AB){\\bf{ber}}(A){\\bf{ber}}(B)$$","PeriodicalId":53800,"journal":{"name":"Concrete Operators","volume":null,"pages":null},"PeriodicalIF":0.3000,"publicationDate":"2019-01-01","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"https://sci-hub-pdf.com/10.1515/conop-2019-0003","citationCount":"38","resultStr":null,"platform":"Semanticscholar","paperid":null,"PeriodicalName":"Concrete Operators","FirstCategoryId":"1085","ListUrlMain":"https://doi.org/10.1515/conop-2019-0003","RegionNum":0,"RegionCategory":null,"ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"Q4","JCRName":"MATHEMATICS","Score":null,"Total":0}
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) $${\bf{ber}}(AB){\bf{ber}}(A){\bf{ber}}(B)$$