Vanishing Bergman Kernels on the Disk

Equivalent Bergman Spaces with Inequivalent Weights

“…It is known [10] that the kernel B D,µ ( · , w) of an integrable radial weight µ on the unit disk D ⊂ C cannot have infinitely many zeroes for a fixed w ∈ D. In this section we show that the analogous result fails when D is replaced with the complex plane. In fact, we exhibit a family of radial weights W ⊂ L 1 (C) such that for every W ∈ W, the associated weighted Bergman kernel B C,W ( · , w) has infinitely many zeroes for each fixed nonzero w in the plane.…”

“…Following the convention of A. Perälä [10], we say that two admissible weights µ 1 and µ 2 are equivalent, or µ 1 ∼ µ 2 , if L 2 H (D, µ 1 ) = L 2 H (D, µ 2 ) as sets. For example, if g is a positive measurable function on D with the property that ess inf z∈D g(z) > 0 and ess sup z∈D g(z) < ∞, then g · µ ∼ µ for any weight µ on D.…”

“…The purpose of this note is to answer two questions [10]. The firsts asks if every space L 2 H (D, µ) can be equipped with an equivalent weight µ * so that the Bergman kernel K D,µ * has zeroes.…”

Hankel operators induced by radial Bekollé–Bonami weights on Bergman spaces

“…In the case of a standard weight, the Bergman reproducing kernels are given by the neat formula p1´zζq´p 2`ηq . However, for a general radial weight ν the Bergman reproducing kernels B ν z may have zeros [19] and such explicit formulas for the kernels do not necessarily exist. This is one of the main obstacles in dealing with P ν [9,17].…”

Bergman projection induced by radial weight

“…This is in stark contrast with the neat expression p1 ´zζq ´p2`αq of the standard Bergman kernel B α z which is easy to work with as it is zero-free and its modulus is essentially constant in hyperbolically bounded regions. In general, the Bergman reproducing kernel induced by a radial weight may have a wild behavior in the sense that for a given radial weight ω there exists another radial weight ν such that A 2 ω " A 2 ν , but B ν z have zeros, see [8, Proof of Theorem 2] and also [30]. The proof of Theorem 1 draws strongly on [25, Theorem 1], which says, in particular, that › › ›pB ω z q pN q › › ›…”