On Berezin number inequalities for operator matrices (Q2042271)

For a bounded linear operator, acting in the reproducing kernel Hilbert space \(\mathcal{H}=\mathcal{H}(\Omega)\) over some set \(\Omega\), its Berezin symbol \(\tilde{A}\) is defined by \(\tilde{A}(\lambda)=\langle A\tilde{k}_\lambda, \tilde{k}_\lambda \rangle\), where \(\tilde{k}_\lambda\) the normalized reproducing kernel of \(\mathcal{H}\). The Berezin set and the Berezin number of an operator \(A\) are defined respectively by \[\text{Ber}(A):=\{\tilde{A}(\lambda): \lambda\in\Omega \}\quad \hbox{and} \quad \text{ber}(A):= \sup \{\tilde{A}(\lambda): \lambda\in\Omega \}.\] Clearly, \(\text{Ber}(A)\) is contained in the numerical range of \(A\) and so \(\text{ber}(A)\) is no larger than the numerical radius of \(A\). Results related to the numerical range and numerical radius could be similarly considered for the Berezin set and the Berezin number. This paper presents refinements of certain earlier existing bounds for Berezin number of operator matrices and proves some new Berezin number inequalities for general \(n\times n\) operator matrices. Further, it also establishes several upper bounds for Berezin number and generalized Euclidean Berezin number (brought in by \textit{M. Bakherad} [Czech. Math. J. 68, No. 4, 997--1009 (2018; Zbl 1482.47003)] for off-diagonal operator matrices.