Compact operators via the Berezin transform

Abstract

In this paper we prove that if S equals a finite sum of finite products of Toeplitz operators on the Bergman space of the unit disk, then S is compact if and only if the Berezin transform of S equals 0 on ∂D. This result is new even when S equals a single Toeplitz operator. Our main result can be used to prove, via a unified approach, several previously known results about compact Toeplitz operators, compact Hankel operators, and appropriate products of these operators.

Compact operators on the weighted Bergman space A1(ψ)

“…If S is a compact operator on L 2 a (D), then Sk z → 0 as z → ∂D. Examples in [1] and [4] show that the converse does not hold. However, in A 1 (ψ), Sk z → 0 as z → ∂D is a sufficient condition for the compactness of S. Moreover, the following example shows that A 0 (ϕ) being an invariant subspace of S * in Theorem 1 is necessary.…”

“…The function k z (w) := K z (w)/ K z ψ will be called the normalized reproducing kernel of A 1 (ψ). In the Bergman space L 2 a (D) setting, Axler and Zheng [1] proved that an operator S which is a finite sum of finite products of Toeplitz operators, is compact if and only if Sk z → 0 as |z| → 1 − . This result also holds for the spaces L p a (D) (1 < p < ∞) (see [7]), A 2 v (Ω) with Ω a regular bounded symmetric domain in C n (see [2]), and H 2 (Ω, dv) with Ω a smoothly bounded multiply connected domain in the complex plane (see [5]).…”

Compact Toeplitz Operators for Weighted Bergman Spaces on Bounded Symmetric Domains

“…[2][3][4][5]7,8,14,15]). For example, it was shown in [8] that under some condition on the weight ν the Toeplitz operator T ν g with bounded symbol g is compact on H 2 ν if and only if g ν ∈ C 0 (Ω).…”

Compact Toeplitz operators with continuous symbols on weighted Bergman spaces

“…For the Bergman space (that is, d' 2rdrY the Theorem is well known; see [5, p. 107] and [1]. When d' 1 À r 2 drÀ1``I, the Theorem is also true; see [3] and [4].…”