# Article

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Keywords:
reproducing kernel; Berezin number; numerical radius; operator matrix
Summary:
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 $${\bf ber}({\bf T}) \leq \frac {1}{2}({\bf ber}(A)+{\bf ber}(D))+\frac {1}{2}\sqrt {({\bf ber}(A)- {\bf ber}(D))^2+(\|B\|+\|C\|)^2}.$$
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