Title:
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Some Berezin number inequalities for operator matrices (English) |
Author:
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Bakherad, Mojtaba |
Language:
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English |
Journal:
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Czechoslovak Mathematical Journal |
ISSN:
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0011-4642 (print) |
ISSN:
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1572-9141 (online) |
Volume:
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68 |
Issue:
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4 |
Year:
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2018 |
Pages:
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997-1009 |
Summary lang:
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English |
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Category:
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math |
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Summary:
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The Berezin symbol $\tilde {A}$ of an operator $A$ acting on the reproducing kernel Hilbert space ${\mathcal H}={\mathcal H}(\Omega )$ over some (nonempty) set is defined by $\tilde {A}(\lambda )=\langle A\hat {k}_{\lambda },\hat {k}_{\lambda }\rangle ,$ $\lambda \in \Omega $, where $\hat {k}_{\lambda }={{k}_{\lambda }}/{\|{k}_{\lambda }\|}$ is the normalized reproducing kernel of ${\mathcal H}$. The Berezin number of the operator $A$ is defined by ${\bf ber}(A)=\sup _{\lambda \in \Omega }|\tilde {A}(\lambda )|=\sup _{\lambda \in \Omega }|\langle A\hat {k}_{\lambda },\hat {k}_{\lambda }\rangle |$. Moreover, ${\bf ber}(A)\leq w(A)$ (numerical radius). We present some Berezin number inequalities. Among other inequalities, it is shown that if ${\bf T}=\left [\smallmatrix A&B\\ C&D \endmatrix \right ]\in {\mathbb B}({\mathcal H(\Omega _1)}\oplus {\mathcal H(\Omega _2)})$, then $$ {\bf ber}({\bf T}) \leq \frac {1}{2}({\bf ber}(A)+{\bf ber}(D))+\frac {1}{2}\sqrt {({\bf ber}(A)- {\bf ber}(D))^2+(\|B\|+\|C\|)^2}. $$ (English) |
Keyword:
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reproducing kernel |
Keyword:
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Berezin number |
Keyword:
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numerical radius |
Keyword:
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operator matrix |
MSC:
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15A60 |
MSC:
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30E20 |
MSC:
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47A12 |
MSC:
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47A30 |
MSC:
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47B15 |
MSC:
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47B20 |
idZBL:
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Zbl 07031692 |
idMR:
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MR3881891 |
DOI:
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10.21136/CMJ.2018.0048-17 |
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Date available:
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2018-12-07T06:19:16Z |
Last updated:
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2021-01-04 |
Stable URL:
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http://hdl.handle.net/10338.dmlcz/147516 |
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Reference:
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Reference:
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