Unbounded Bergman operators

Let G be an open subset of the plane, and denote by L2aG the Bergman space of all square integrable analytic functions with respect to the Lebesgue area measure. If dom SG = f∈L2aG :zf∈L2a G , define the Bergman operator SG: dom SG → L2aG , by SG: f(z) zf(z). We show that the problem regarding the density of dom SG in L2aG is equivalent to the problem regarding the density of the range of the operator of multiplication by z on some open subset of the unit disc D . If U is a open subset of D containing 0 in its topological boundary, using Wiener capacity we present two sufficient conditions such that zL2aU is dense in L2aU . As a consequence, it follows that if G has finite area, or the component of the complement of G with respect to the extended plane containing infinity does not equal the singleton {infinity}, then the Bergman operator SG is densely defined. Furthermore, we prove that if G is a simply connected region of finite area, or a half plane, then the self-commutator of S G is densely defined. It is also shown that for a Bergman operator SG, the spectrum equals the closure of G, and its point spectrum is the empty set. Finally, we show that if the Bergman operator SG is densely defined, then the set of all bounded operators in L2aG that commute with SG equals { Mϕ: ϕ ∈ Hinfinity (G)} where Mϕ denotes the operator of multiplication by ϕ on L2aG .