Derivatives of the Berezin Transform

For a rotation invariant domain Ω, we consider A2 Ω, μ the Bergman space and we investigate some properties of the rank one projection A z : 〈·, kz〉kz. We prove that the trace of all the strong derivatives of A z is zero. We also focus on the generalized Fock space A2 μm , where μm is the measure with weight e−|z| m ,m > 0, with respect to the Lebesgue measure on C and establish estimations of derivatives of the Berezin transform of a bounded operator T on A2 μm .


Introduction and Statement of the Main Results
We consider a rotation invariant open set Ω in C n and μ a positive rotation invariant measure on Ω; we suppose that μ has moments of every order.Let L 2 Ω, μ be the Hilbert space of square integrable complex-valued functions on Ω and A 2 Ω, μ its subspace consisting of holomorphic elements.We assume that for each compact set K ⊂ Ω there exists C C K such that for all f ∈ A 2 Ω, μ 1.1 It is known that A 2 Ω, μ is a closed space of L 2 Ω, μ and possesses a reproducing kernel K ω, z : K z ω : we have for all f ∈ A 2 Ω, μ and z ∈ Ω.

Journal of Function Spaces and Applications
For a bounded linear operator T on A 2 Ω, μ , the Berezin transform of T is the function T defined on Ω by where k z is the normalized reproducing kernel The case of the Fock space, where Ω C n and μ 2 is the Gaussian measure, was considered by Coburn, Englis, and Zhang.Coburn 1 has shown that T is a Lipschitz function.Namely, for x, y ∈ C n , T x − T y ≤ 2 T x − y , 1.5 the constant 2 being sharp see 2 .
Englis and Zhang 3 have shown that T has bounded derivatives of all orders.Namely, for any multi-indices β, γ, there exists a constant c β,γ , depending on β, γ, and n only, such that where ∂ β will stand for , where ∂ ∂z j 1.9 In this paper we will investigate some properties of the derivatives of the Berezin transform on A 2 Ω, μ .For z in Ω, if A z is the rank one projection We first fix some notations.Let N n 0 denote the set of all n-tuples with components in the set N 0 of all nonnegative integers.If β β 1 , . . ., β n ∈ N n 0 , we let |β| : β 1 • • • β n denote the length of β and β! stands for n j 1 β j !.If γ γ 1 , . . ., γ n ∈ N n 0 satisfies γ j ≤ β j for all j 1, . . ., n, then we write γ ≤ β.Our first main result is about the strong derivatives of A z .
Theorem 1.1.Let Ω be a rotation invariant open set in C n and μ a rotation invariant positive measure on Ω that satisfies 1.1 and has moments of every order.Moreover one assumes that, for any multiindex α and any compact set K of Ω, there exists G α,K ∈ L 2 Ω, μ such that Then for all β, γ multi-indices, the operators ∂ β ∂ γ A z and ∂ β ∂ γ A z are adjoint to each other; their rank is smaller or equal to the infimum of #{δ ∈ N n 0 , δ ≤ γ} and #{δ ∈ N n 0 , δ ≤ β}.Moreover, if at least β or γ is different from 0, one has Our second main result generalizes the estimates of Englis and Zhang for the strong derivatives of the Berezin transform on weighted Fock spaces.
Theorem 1.2.For a bounded linear operator T on A 2 C n , μ m , the Berezin transform T has derivatives of all orders.In addition to any multi-indices β, γ, there exist positive constants c β,γ and A, depending on β, γ, and n only such that 1.14

Preliminaries
We recall some properties of the Bergman kernel K, when μ is a positive rotation invariant measure on Ω.The kernel K is given by for z, ω in Ω, with the usual convention that, for z z 1 , . . ., z n and β β 1 , . . ., β n , z β stands for z n and where By Lemma 2.1 in 4 , we know that Thus we can write where is a holomorphic function of one complex variable.
For a bounded operator T on A 2 Ω, μ , the Berezin transform can be written in the form where A z is the rank one projection A z •, k z k z .Recall 3 that a mapping h from a domain in C n into a Banach space possesses a strong holomorphic derivative ∂h/∂z 1 z at a point z z similarly one can define the antiholomorphic derivative ∂.Englis and Zhang showed that the mapping z → A z has strong derivatives.
Lemma 2.1.Let Ω be a domain in C n , μ a measure on Ω that satisfies 1.1 .Then the function z → k z has strong derivatives of all orders.

Lemma 2.2. Under the same hypothesis, the trace-class-operator-valued function
from Ω into the space of trace-class operators on A 2 Ω, μ has strong derivatives of all orders.
The mapping z → A z and its derivatives can be expressed in terms of the function F. It is easy to see that, for z, u in Ω, 2.9

Proof of Theorem 1.1
We write, for f ∈ A 2 Ω, μ and z, u in Ω, Since it is possible to differentiate under the integral sign at any order with respect to z, for β in N n 0 , we have The Leibnitz rule leads to and then

3.8
It follows that the rank of the operator 3.9 Now let β and γ be multi-indices.Due to Lemma 2.2 and the continuity of the linear form Tr X , we have 3.10

Proof of Theorem 1.2
When Ω C n and dμ m z e −|z| m dv z , for brevity we set A 2 μ m for A 2 C n , μ m .The Bergman kernel K m of A 2 μ m can be expressed in terms of the Mittag-Leffler function see 5 .Putting α 2/m, we have being the n − 1-th derivative of the Mittag-Leffler function E α,α , entire on C and defined by In what follows, we will use some asymptotic properties of this function see 6 near the positive real axis.
Lemma 4.1.Let p a nonnegative integer.There exists a constant > 0, such that, for any complex number z with | arg z| < , one has where and Z s z |z| 1/α exp i/α arg z 2πs , for s integer.The polynomials P p p∈N are defined by d p /du p e u 1/α e u 1/α /u p P p u 1/α , P p is of degree p, and Z 0 for any nonnegative integers r and q.
We also need asymptotic estimates for some auxiliaries functions.
Lemma 4.2.Let p and d be nonnegative integers.When t is real and tends to ∞,

4.5
For T a bounded linear operator on A 2 μ m , the Berezin transform is given by see 3 T z tr TA z .

4.6
Let β and γ be some multi-indices.By differentiation, Lemma 2.2 gives where To estimate the eigenvalues of the finite rank positive operator SS * , we compute its trace:

4.12
With the operator being of finite rank, we observe that Tr SS * Tr S * S .We now compute and then estimate each diagonal term a δ .Fix some multi-indices δ and δ such that δ ≤ γ and δ ≤ γ.To simplify the notation we set L δ and M δ instead of L z,δ and M z,δ .We obtain

4.13
For the second term, since

4.16
By the Leibnitz rule, we see that

4.25
Using the Leibnitz rule

4.26
An induction process shows that

4.29
On the other hand

4.33
Notice that, for p and q integers, we have, when t is real and t → ∞,

4.34
Hence by Lemmas 4.1 and 4.2 we have the following estimates:

4 . 7
Due to the continuity of the bilinear form X, Y → tr XY , we have |tr XY | ≤ X Y tr , 4.8 for all bounded operators X and trace class operators Y .Therefore, we obtain Like in Section 3, we set S : ∂ β ∂ γ A z .Next we shall estimate tr SS * 1/2 .Fix f in A 2 μ m and z in C n .We recall that Sf δ≤γ f, L z,δ M z,δ , 4.10