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**Boundary values of Berezin symbols.**
*(English)*
Zbl 0874.47013

Feintuch, A. (ed.) et al., Nonselfadjoint operators and related topics. Workshop on Operator theory and its applications, Beersheva, Israel, February 24-28, 1992. Basel: Birkhäuser Verlag. Oper. Theory, Adv. Appl. 73, 362-368 (1994).

Let \(\mathcal H\) be a functional Hilbert space of functions defined on a set \(S\) with reproducing kernel \(\{k_z\}_{z\in S}\), so that \(k_z\in{\mathcal H}\) and \(f(z)= (f,k_z)\) for \(f\in{\mathcal H}\) and \(z\in S\). Given a bounded linear operator \(A\) on \(\mathcal H\), the function \(\widetilde A\) defined on \(S\) by \(\widetilde A(z)= (A\widehat k_z,\widehat k_z)\), where \(\widehat k_z=|k_z|^{-1}k_z\), is called the Berezin symbol of \(A\). Also, \(\mathcal H\) is said to be standard if the underlying set \(S\) is a subset of a topological space, the boundary \(\partial S\) of \(S\) is non-empty and \(\widehat k_{z_n}\to 0\) weakly in \(\mathcal H\) whenever \(\{z_n\}\) is a sequence in \(S\) converging to a point in \(\partial S\). The common functional Hilbert spaces, such as the Hardy space \(H^2\) and the Bergman space, are standard. For a standard functional Hilbert space and a compact operator \(A\) on \(\mathcal H\), \(\widetilde A(z_n)\to 0\) as \(n\to\infty\) whenever \(\{z_n\}\) is a sequence in \(S\) converging to a point in \(\partial S\). This is described by saying that the Berezin symbol of a compact operator on a standard functional Hilbert space vanishes on the boundary.

The present note was stimulated by the following question raised by C. A. Berger and L. A. Coburn: on the Hardy and Bergman spaces, must a bounded linear operator be compact if its Berezin symbol vanishes on the boundary of \(S\)? (In this case \(S\) is the unit disc in the complex plane.)

Firstly, the authors present several counterexamples to this question. In a positive direction they show that, if the Berezin symbols of all unitary equivalents of an operator \(A\) on a standard functional Hilbert space vanish on the boundary, then \(A\) is compact. This result follows from a characterization of the essential numerical range of \(A\) as the set of all cluster values of the Berezin symbols of all operators unitarily equivalent to \(A\). In a similar vein, they show that the Berezin symbols of all operators unitarily equivalent to \(A\) extend continuously to \(S\cup\partial S\) if and only if \(A\) is a translate by a scalar multiple of the identity of a compact operator.

For the entire collection see [Zbl 0798.00021].

The present note was stimulated by the following question raised by C. A. Berger and L. A. Coburn: on the Hardy and Bergman spaces, must a bounded linear operator be compact if its Berezin symbol vanishes on the boundary of \(S\)? (In this case \(S\) is the unit disc in the complex plane.)

Firstly, the authors present several counterexamples to this question. In a positive direction they show that, if the Berezin symbols of all unitary equivalents of an operator \(A\) on a standard functional Hilbert space vanish on the boundary, then \(A\) is compact. This result follows from a characterization of the essential numerical range of \(A\) as the set of all cluster values of the Berezin symbols of all operators unitarily equivalent to \(A\). In a similar vein, they show that the Berezin symbols of all operators unitarily equivalent to \(A\) extend continuously to \(S\cup\partial S\) if and only if \(A\) is a translate by a scalar multiple of the identity of a compact operator.

For the entire collection see [Zbl 0798.00021].

Reviewer: T.A.Gillespie (Edinburgh)

### MSC:

47B38 | Linear operators on function spaces (general) |

47B07 | Linear operators defined by compactness properties |

47B35 | Toeplitz operators, Hankel operators, Wiener-Hopf operators |

47A12 | Numerical range, numerical radius |

46E22 | Hilbert spaces with reproducing kernels (= (proper) functional Hilbert spaces, including de Branges-Rovnyak and other structured spaces) |