Berezin number of operators and related questions (Q2850447)

Let \(\mathcal H=\mathcal H(\Omega )\) be a reproducing kernel Hilbert space of complex-valued functions defined on a set \(\Omega\). For any linear bounded operator \(A\) on \(\mathcal H\), its Berezin symbol \(\tilde{A}(\lambda )\), \(\lambda \in \Omega\), is defined. The number \(\operatorname{ber}(A)=\sup_{\lambda \in \Omega}| \tilde{A}(\lambda )| \) is called the Berezin number of the operator \(A\). The space \(\mathcal H\) is called possessing the (Ber) property if, for any two operators \(A_1,A_2\), the equality \(\tilde{A_1}(\lambda)=\tilde{A_2}(\lambda)\) for all \(\lambda \in \Omega\) implies \(A_1=A_2\). The authors consider invertible operators on a space with the (Ber) property and find necessary and sufficient conditions of their unitarity in terms of the Berezin numbers. For an operator \(A\) on the Hardy space \(H^2(\mathbb D)\), a lower estimate of \(\operatorname{ber}(A)\) is found.