Some Berezin number inequalities for operator matrices

Abstract: The Berezin symbol $\widetilde{A}$ of an operator $A$ acting on the reproducing kernel Hilbert space ${\mathscr H}={\mathscr H(}\Omega)$ over some (non-empty) set is defined by $\widetilde{A}(\lambda)=\langle A\hat{k}{\lambda},\hat{k}{\lambda}\rangle\,\,\,(\lambda\in\Omega)$, where $\hat{k}{\lambda}=\frac{{k}{\lambda}}{|{k}{\lambda}|}$ is the normalized reproducing kernel of ${\mathscr H}$. The Berezin number of operator $A$ is defined by $\mathbf{ber}(A) = \underset{\lambda \in \Omega}{\sup} \big|\tilde{A}(\lambda)\big|=\underset{\lambda \in \Omega }{\sup} \big|\langle A\hat{k}{\lambda}, \hat{k}_{\lambda}\rangle\big|$. Moreover $\mathbf{ber}(A)\leqslant w(A)$ (numerical radius). In this paper, we present some Berezin number inequalities. Among other inequalities, it is shown that if $\mathbf{T}=\left[\begin{array}{cc} A&B C&D \end{array}\right]\in {\mathbb B}({\mathscr H(\Omega_1)}\oplus{\mathscr H(\Omega_2)})$, then \begin{align*} \mathbf{ber}(\mathbf{T}) \leqslant\frac{1}{2}\left( \mathbf{ber}(A)+ \mathbf{ber}(D)\right)+\frac{1}{2}\sqrt{\left( \mathbf{ber}(A)- \mathbf{ber}(D)\right)^2+(|B|+|C|)^2}. \end{align*}