Some Berezin number inequalities for operator matrices

Mojtaba Bakherad

Czechoslovak Mathematical Journal (2018)

  • Volume: 68, Issue: 4, page 997-1009
  • ISSN: 0011-4642

Abstract

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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 $$ {\bf ber}({\bf T}) \leq \frac {1}{2}({\bf ber}(A)+{\bf ber}(D))+\frac {1}{2}\sqrt {({\bf ber}(A)- {\bf ber}(D))^2+(\|B\|+\|C\|)^2}. $$

How to cite

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Bakherad, Mojtaba. "Some Berezin number inequalities for operator matrices." Czechoslovak Mathematical Journal 68.4 (2018): 997-1009. <http://eudml.org/doc/294711>.

@article{Bakherad2018,
abstract = {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 $$ \{\bf ber\}(\{\bf T\}) \leq \frac \{1\}\{2\}(\{\bf ber\}(A)+\{\bf ber\}(D))+\frac \{1\}\{2\}\sqrt \{(\{\bf ber\}(A)- \{\bf ber\}(D))^2+(\|B\|+\|C\|)^2\}. $$},
author = {Bakherad, Mojtaba},
journal = {Czechoslovak Mathematical Journal},
language = {eng},
number = {4},
pages = {997-1009},
publisher = {Institute of Mathematics, Academy of Sciences of the Czech Republic},
title = {Some Berezin number inequalities for operator matrices},
url = {http://eudml.org/doc/294711},
volume = {68},
year = {2018},
}

TY - JOUR
AU - Bakherad, Mojtaba
TI - Some Berezin number inequalities for operator matrices
JO - Czechoslovak Mathematical Journal
PY - 2018
PB - Institute of Mathematics, Academy of Sciences of the Czech Republic
VL - 68
IS - 4
SP - 997
EP - 1009
AB - 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 $$ {\bf ber}({\bf T}) \leq \frac {1}{2}({\bf ber}(A)+{\bf ber}(D))+\frac {1}{2}\sqrt {({\bf ber}(A)- {\bf ber}(D))^2+(\|B\|+\|C\|)^2}. $$
LA - eng
UR - http://eudml.org/doc/294711
ER -

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