Lipschitz estimates for the Berezin transform

We consider 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 Cn. Improving upon a recent result of L. Coburn and J. Xia, we show that for any bounded operator X on A2(μm) , the Berezin transform of X satisfies Lipschitz estimates.


Introduction and statement of the main results
We consider the Fock space A 2 (μ m ) of those holomorphic functions which are square integrable on C n with respect to the measure dμ m (z) = e −|z| m dV (z), where dV (z)is the Lebesgue measure and m > 0 is a positive parameter.Let K m be the Bergman kernel of the Hilbert space A 2 (μ m ) and denote by k a the corresponding normalized kernel.It is given by k a (z) := K(z, a)/ K(a, a) and is a unit vector of A 2 (μ m ).The Berezin transform of a bounded linear operator X on A 2 (μ m ) is the function defined by X(a) = Xk a , k a .Recently, Coburn [4], [5] has shown that in the classical case m = 2, corresponding to the Segal-Bargmann space endowed with the Gaussian measure on C n , X is a Lipschitz function, satisfying where X is the norm of X .
The main goal of this paper is to investigate the general case of positive parameters m > 0 .Namely, we will show that the Berezin transform is locally Lipschitz and when n = 1 and m ≤ 1 , the Lipschitz estimate is uniform.In the proof of our results we also establish estimates of the quantity If A and B are two quantities, we use the symbol A ≈ B whenever A ≤ C 1 B and B ≤ C 2 A, where C 1 and C 2 are positive constants independent of the varying parameters.Our main result is the following Theorem 1.If X is a bounded operator on A 2 (μ m ), then its Berezin transform X is locally Lipschitz.In addition, there exist positive constants A, C , depending on n and m only, such that for any a ∈ C n with |a| > A, there is a neighbourhood U of a that satisfies The proof of Theorem 1 gives a more accurate estimate of ρ m (a, b) when n = 1.
Theorem 2. On C, the quantity ρ m (a, b) satisfies a Lipschitz estimate.More precisely, there exists a positive constant A, depending on m only, such that for any a ∈ C n with |a| > A, there is a neighbourhood U of a that satisfies When m is an even integer, we have the following consequence of Theorem 2: Corollary 3. Suppose n = 1 and m an even positive integer.There exists a positive constant A, depending on n and m only, such that, for any a ∈ C with |a| > A, there is a neighbourhood U of a that satisfies A careful study of ρ m (a, x) when m ≤ 1 shows that the Lipschitz condition is uniform.Corollary 4. Assume n = 1 and m ≤ 1 and let X be a bounded operator on A 2 (μ m ).Then X is Lipschitz.More precisely, there are positive constants A, C, δ , depending on n and m only, such that for any a ∈ C n with |a| > A , | X(x) − X(a)| ≤ C X |x − a|, for all x with |x − a| < δ.

Preparatory results
We will use a theorem shown by Coburn in [4], which is valid for any reproducing kernel.

Lemma 1. For a bounded operator
where K m is the Bergman kernel of A 2 (μ m ).It follows from [1] that the Bergman kernel K m can be expressed in terms of the generalized Mittag-Leffler function by the equation In order to study ρ m (a, a + u) 2 near a, we will use a Taylor's expansion formula.Thus we need to estimate F and its higher order derivatives.

Asymptotic expansion of the derivatives of the Mittag-Leffler function E α,α
Following the method developped by Wong and Zhao in [10], we consider several subcases, corresponding to different values of α .
Assume p is a non-negative integer.For z ∈ C \ {0} , we suppose the branch of the argument is chosen such that arg z ∈ [−π, π[.If s is an integer, let and denote by S the set of those integers , then there exists a polynomial P p+1 of degree p + 1 such that for any z = 0, Proof.We recall the Hankel representation of the entire function .Then from the formula above we have We write ,where I d,lin and I d,circ are respectively the integrals on the linear part and circular part of γ a,φ0 .
On the half rays of γ a,φ0 we have On the circular part of γ a,φ0 we have Then Thus we can write Let p be a non-negative integer.On the linear part of γ a,φ0 , t = where p+1 .Now we will get another expression of this integral by deforming the contour γ a,φ0 .Choose r, R, φ 1 such that 0 < r < a < R , r α < |z| < a α , and 0 < φ 1 < π − φ 0 .Consider the counter-clockwise loop which starts at Re i(−π+φ1) , follows the circular path |t| = R to Re −iφ0 , a straight line to ae −iφ0 , a part of the circle |t| = a to ae iφ0 , a straight line to Re iφ0 , the circle |t| = R to Re i(π−φ1) , a straight horizontal line to re i π 2 , the semicircle |t| = r to re −i π 2 , then a straight horizontal line to the starting point Re i (−π+φ1) .
For any integer s we put and Define the set S := {s ∈ Z, | arg Z s | < π} .If we choose r small enough for the poles Z s (such that | arg Z s | < π) of f p (t) to be inside γ a,φ0,R,r , the theorem of residues leads to On the big circular part of γ a,φ0,R,r , we have Since cos φ 0 < 0, we see that When R tends to +∞, γ a,φ0,R,r is deformed in two paths, namely γ a,φ0 and γ r , where γ r starts at −ir − ∞, follows a straight horizontal line to re −i π 2 , encirles the origin once counter-clockwise on the semi-circle |t| = r , and follows a straight horizontal line to ir − ∞.
On the rectilinear part of γ r we have Next, for s ∈ S , we shall compute Res(f p ; Z s ).Let C s be a circle centered at Z s of radius η and C s the image of C s by the function power α , where we choose η small enough for C s and C s to be included in C \ R − and the function power α is determined by u = t α = exp[α | ln t| + iα arg t], | arg t| < π.In a neighbourhood of C s , the function power 1 α is defined by The change of variables u = t α shows that

u=z
. Now, we have to compute the higher order derivatives of h(u) := e u 1 α .By induction we see that, for any p ∈ N 0 , there exists a polynomial P p of degree p such that h (p) (u) = e u 1 α ).By differentiating the last equality, we get It follows that the polynomials (P p ) p∈N satisfy the relation: Since P 0 (t) = t, P p has real coefficients , and the degree of P p is p.
We write Next we estimate the second term on the right-hand in the above equality.Observe that for any nonnegative integer N we have Integrating on γ r yields where Since α is a positive integer, each term in the sum above is zero.First assume that for any integer s, l,N (z) are the integrals given by which can be rewritten Since α is a positive integer, we see that We have the exact formula If there exists an integer s 0 such that | arg Z s0 | = π , the same identity holds by analytic continuation.
In both cases, there exists a polynomial P p+1 of degree p + 1 defined in Lemma 2 and a non-negative integer N such that for any z = 0, and there exists a constant C > 0 , depending on α, p, N, only such that Proof.Using the notations of Lemma 2 we obtain where First assume that for any integer s, l,N (z) are the following integrals: We distinguish the three cases corresponding to different values of arg z .First suppose −π ≤ arg z ≤ −απ − .We determine the set S .If s is an integer put The required result follows by taking the sum over l ∈ {0, • • • , p} .Now assume that there exists an integer s 0 such that | arg Z s0 | = π .We choose r sufficiently small so that the poles Z s with s ∈ S are to the right of γ r .
Recall that We deform γ r in another Hankel contour γ r,a surrounding the singularity Z s0 from above and below by circular parts of radius a such that the poles Z s with s ∈ S are to the right of γ r,a .We have Following the same method as above we get where If r tends to 0 , γ r,a is deformed to γ 0,a which consists of the following four disconnected lines segments, the first one starts below the negative real axis at −∞ to Z s0 − a , the second one joins Z s0 + a to 0 , the third joins 0 to Z s0 + a above the negative real axis and the fourth returning from Z s0 − a to −∞.Then we see that Let C s0 be a circle centered at Z s0 of radius η .Then we have Let be C s0 the image of C s0 by the function power α , where η is chosen small enough so that C s and C s0 are included in C \ R + and u = t α = |t| α exp[iα(arg t + 2πs 0 )], where 0 < arg t < 2π .In a neighbourhood of C s0 , we define u where 0 < arg u < 2π .The change of variables u = t α shows that for some polynomial Since Z s0 is a negative real number and the integral 1 2πi γ 0,a e t t (N +1−p+l)α (t α − z) l+1 dt can be handled in the same way as the integral 1 2πi γ r e t t (N +1−p+l)α (t α − z) l+1 dt previously treated, we see that for any l ∈ {0, . . ., p} , R In those cases, there exists a polynomial P p+1 of degree p + 1 defined in Lemma 2 and a non-negative N such that for any z = 0, and there exists a constant C > 0 , depending on α, p, N, only such that Proof.Arguing as in Lemma 2 we have where First assume that for any integer s, the set S is easy to determine.In each case, for a positive number v , the inequality (3.1) holds.Therefore we get , for l ∈ {0, • • • , p} and Taking the sum over l ∈ {0, • • • , p} we get the required result.Now we consider the results in the case when z is a negative real number.Notice that if arg z = −π , then arg Z s = π α (−1 + 2s) with s ∈ {−(L − 1), . . ., 0, . . ., L} and if arg z = π , then arg Z s = π α (1 + 2s) with s ∈ {−L, . . ., 0, . . ., L − 1} .It follows that {Z s , s ∈ {−(L − 1), . . ., 0, . . ., L}} = {Z s , s ∈ {−L, . . ., 0, . . ., L − 1}} .Therefore when z is a negative real number, the formulas obtained in the first and the third subcases coincide.
Finally if there exists an integer s 0 such that | arg Z s0 | = π , the proof is similar to that of Lemma 3.

Lemma 5. Suppose α ∈]2L, 2L + 1[, where L is a positive integer, and is a positive number with
In those cases, there exists a polynomial P p+1 of degree p + 1 defined in Lemma 2 and a non-negative integer N such that for any z = 0, and there exists a constant C > 0 , depending on α, p, N, only such that Proof.Proceeding as in Lemma 2 we get again where First we consider the case l,N (z), where The set S can then be determined as in Lemmas 3 and 4 and the estimate for |R Recall that We first estimate the numerator by using a Taylor expansion near a.To do so we need to study F and its higher order derivatives near the positive real axis.Lemma 6.Let p be a non-negative integer.There exists a constant > 0 such that, for any complex z with | arg z| ≤ , we have where Z s = |z| 1 α exp i α (arg z + 2πs) , for s integer, the polynomials (P p ) p∈N are defined by the relation and O |z| q e Z±1 for any integer q if α > 1 .
Proof.Follows from Lemmas 2, 3, 4, 5 and the fact that when arg z tends to 0 , with s in S such that |s| ≥ 2 and q, q , q are integers, e Zs |z| q is negligible in comparison of e Z±1 |z| q and e Z±1 |z| q is negligible in comparison of e Z0 |z| q .
We will also need some properties of the polynomials (P p ) p∈N .Lemma 7. The sequence (P p ) p∈N defined by (3.2) satisfies the recurrence relation: In particular each P p has real coefficients, and is of degree p .In addition, the polynomial P p (t) − 1 α p t p + p(p−1) Proof.The relation between the polynomials (P p ) and the fact that P p has real coefficients and degree p have been shown in the proof of the Lemma 2. We observe that the term of highest degree term of P p is 1 α p t p .
Then P p can be written in the form where P p (t) has a degree at most p − 2 .We need to compute b p .By differentiation we have P p (t) = p α p t p−1 +(p−1)b p t p−2 + P p (t), and therefore α t, we see that b 1 = 0 ; an induction process gives Using the identities we find that the term of highest degree in

Preparatory estimates
To study ρ m (a, a + u), we first focus on the numerator N (a, a + u).Lemma 8.There are positive constants A and C such that for any a, u in C n with |a| > A and |u| < 1 , we have Proof.The Taylor expansion formula gives and where We write We shall estimate the remainder R(a, u).By Lemma 6, there are positive constants A 1 , D 1 , such that for any p = 0, . . ., 3 and for any z ∈ C with | arg z| < and |z| > A 1 , we have , where Thus there exists a positive constant A 2 such that for any p = 0, . . ., 3 and for any a, u in C n with |a| > A 2 and |u| < 1, we have Proof.A Taylor's exansion formula gives By Lemma 6, there are positive constants A 4 , C 4 , C 5 such that for a, u in C n with |a| > A 4 and |u| < 1, we have and This completes the proof.

Proofs of the mains results
Proof of Theorem 1.By Lemma 1, it is sufficient to estimate ρ m (a, a+u), when a and u are in C n .Setting Z 0 = |a| 2 α and applying Lemma 6, there exist two functions 1 and 2 such that where the polynomials P p are defined by (3.2) and lim t→+∞ The properties of the polynomials P p shown in Lemma 7 imply that where 3 is a function satisfying lim In the case n = 1 , we can give a more precise estimate for ρ m (a, a + u).
Proof of Theorem 2. Assume n = 1 .With the notation of the previous proof, we get and where the functions 4 and 5 satisfy lim We have also So there exist positive numbers C 4 , C 4 and δ(|a|) such that for |u| < δ(|a|), we have which completes the proof.
We now prove Corollary 3.
Proof of Corollary 3. Assume n = 1 and 1 α = r is a positive integer; the function z → z r is analytic in the whole complex plane .The Taylor formula completes the proof.
In the special case when n = 1 and m ≤ 1 , we shall prove Corollary 4.

The range of Ber
The Berezin transform Ber : X → X , defined for an operator on A 2 (μ m ) such that both domains of X and X contain the set {k a , ∈ C n } is linear and one-to-one (see ).If Op denotes the algebra of all linear bounded operators on A 2 (μ m ), we showed that Ber(Op) is a linear subspace of locally Lipschitz functions on C n .As in the classical case (see [5]), we can show that smooth bounded Lipschitz functions can fail to be in Ber(Op).Theorem 5. Assunme m > 0 .There is no X in Op such that , where K m is the reproducing kernel of A 2 (μ m ) and Proof.We use the notation K instead of K m and dμ instead of dμ m .For a real number t, let R t be the linear operator on A 2 (μ m ) defined by R t (f )(z) := f (tz).First, we consider the boundedness of R t .Using the properties of the reproducing kernel, we compute We are going to show that there is no bounded operator X on A 2 (μ m ) such that X(a) = K(ta, a) K(a, a) .
We observe that the function g t (a) = K(ta, a) K(a, a) is real-differentiable at every order.Assume that t < 0 .We will show that g and its first order derivatives are bounded on C n .We have  .
On the other hand, we have , Suppose there exist a bounded operator X in Op with X(a) = g t (a).The operator A := X − R t is in Op and satisfies A(a) = 0 .The injectivity of the mapping A → A [7]leads to A = 0. Then Xk a = R t k a , which contradicts the boundedness of X .

Proof of Corollary 4 . 1
Assume α ≥ 2 .Then, in the proof of Theorem 2, the terms Z when |a| > A 4 .