{"title":"算子的Berezin数不等式","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":"{\"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}","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}
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) $${\bf{ber}}(AB){\bf{ber}}(A){\bf{ber}}(B)$$