Vanishing Carleson measures and power compact weighted composition operators

In this paper, we characterize Carleson measure and vanishing Carleson measure on Bergman spaces with admissible weights in terms of {\it t-Berezin transform} and {\it averaging function} as key tools. Moreover, power bounded and power compact weighted composition operators are characterized as application of Carleson measure and vanishing Carleson measure respectively on Bergman spaces with admissible weights.


Introduction
Let H(D) denote the space of analytic functions on the unit disk D = {z∈C : |z| < 1}.Given a positive integrable function σ ∈ C 2 [0, 1], we extend it on D by defining σ(z) = σ(|z|), z ∈ D and call such σ a weight function.For 0 < p < ∞ and a positive Borel measure Ω, the space L p (Ω) consists of all measurable functions f on D for which In the case p = ∞, the space of all complex-valued measurable functions f on D is defined as where the essential supremum is taken with respect to the measure Ω.A sequence {f n } n∈N is norm bounded in L ∞ (Ω) if sup n∈N f n ∞ is finite.Let dA(z) = dxdy π be the normalized Lebesgue area measure on D, we define the weighted Bergman space as Note that A 2 σ is a closed subspace of L 2 (σdA) and hence is a Hilbert space endowed with the inner product Throughout this paper, we will consider σ as admissible weight function.Recall that if a weight function σ is non-increasing on [0, 1) and σ(r)(1 − r) −(1+δ) is non decreasing on [0, 1) for some δ > 0, then σ is called admissible weight.We refer the readers [12] for useful fact over pseudohyperbolic metric.The pseudohyperbolic metric is defined as ρ(a, z) = |φ a (z)|, where φ a (z) = a−z 1−āz is möbius transformation.For r in (0, 1) and a in D, E(a, r) = {z ∈ D : ρ(z, a) < r} = φ a (E(0, r)) = φ a ({z : |z| < r}) denote the pseudohyperbolic disk centre at a and radius r ∈ (0, 1).It turns out E(a, r) is a Euclidean disk with center 1 − r 2 1 − |z| 2 r 2 z and radius (1 − |z| 2 )r 1 − |z| 2 r 2 .For every z ∈ E(a, r), (1  2 and area of E(a, r) is denoted by |E(a, r)| and |E(a, r)| ≍ (1 − |a| 2 ) 2 are well known facts.Here the symbol " ≍ " denotes that the left hand side is bounded above and below by constant multiples of the right hand side, where the constants are positive and independent of variables.Given r in (0, 1), a sequence {z k } ∞ k=1 ⊂ D is said to be an r-lattice if the disk {E(z k , r)} ∞ k=1 cover D and there is some integer M > 0 such that each z in D belongs to at most M of the disks Recall that, where For a finite positive Borel measure Ω on D, the t-Berezin transform is defined to be Note that for t = 2, the classical Berezin transform is denoted by Ω 2 .Given r in (0, 1), the Averaging function of Ω is defined to be If we set dΩ = f dA, for a Lebesgue measurable function f , then we can write f t = Ω t and f r = Ω r for simplicity.
Motivated by [21] and [9], in this article we study the power bounded and power compact weighted composition operator on A p σ by using Carleson measure characterization in terms of t-Berezin transform and averaging function as a key tool.An operator T on a normed linear space (X, .X ) is called power bounded if {T n } is a bounded sequence in the space of all bounded operators from X to itself.Also, recall that an operator T on Banach space (X, .X ) is said to be power compact, see [2] if there exist some integer m > 0 such that T m is compact from X to itself.
Denote by Λ 2 (C), the linear space of all double sequences with complex entries.A double sequence {γ j,k } j,k∈N of complex numbers is bounded if there exists some M > 0 such that sup j,k |γ j,k | ≤ M .The space Λ 2 ∞ of all bounded double sequences is defined as Let C ψ,φ denoted the well known weighted composition operator on the space H(D) is defined as where ψ ∈ H(D) and φ is an analytic self map of D. If φ(z) = z and ψ = 1, then C ψ,φ becomes the multiplication operator M ψ and the composition operator C φ respectively.Denote by φ n the nth iteration of φ, that is Note that any power of C ψ,φ on H(D) is a weighted composition operators which is defined as For the sake of simplicity, we set Thus, We define dΩ n = | ψ, φ, n |dΩ • φ −1 n , one can easily see that Ω n is a measure and therefore .For t > 0, the t-Berezin transform and for 0 < r < 1, the averaging function of Ω n are defined as respectively.For more about weighted composition operators, Carleson measures and vanishing Carleson measures, we refer to [16]- [12].Throughout the paper, the expression E F means that there exists a constant C such that E ≤ CF .

Preliminaries
In this section, we prove and collect some useful facts and lemmas that are required for the proof of our main results.The next lemma gives a growth estimate for functions in A p σ and an asymptotic estimate for norm of K α (z, •) already proved in [1].
Lemma 1.Let p > 0, σ be an admissible weight and α > −1.Then (1) for each z ∈ D, we have that (2) for each z ∈ D, we have that The boundedness of the operator f −→ f r is trivially holds for p = ∞ , that is f r ∞ ≤ f ∞ and also holds for 1 < p < ∞, by complex interpolation.
Lemma 3. Suppose 0 < p < ∞, r ∈ (0, 1) and Ω be a finite positive Borel measure.Then there exists some constant C such that for all non-negative subharmonic function h : Proof.Since h be non negative subharmonic function h : D −→ [0, ∞).Then, we have for all z ∈ D, see page no.125 of [12].Using above inequality, Fubini's theorem and (1.3), we have that This accomplished the result.
Note that lemma 4 and lemma 5 already prove in [21] and [22] in more general sense.For the sake of completeness, we include the proofs.Note that lemma 4 reveals that the function Ω(E(z, r)) behaves like a subharmonic function.For r = R, the following lemma, lemma 4 was first obtained in [?].
Lemma 4. Suppose r, R > 0 and Ω be a positive borel measure on D, then there exists a constant C such that Proof.For r, R > 0, we have For w ∈ (E(a, R), then there exists a Euclidean disk with diameter 1 2 min{r, R} contained in (E(a, R) ∩ E(w, r)).Therefore (2.3) holds.
Moreover, Ω r (z) Proof.For r and R, by lemma 1 there exists a constant C such that for all z ∈ D. Above inequality and (1.3) implies that, where 1) .Combining this with Lemma 3, we find that Ω R (z) Interchanging the role of r and R, we get the desired result.
Lemma 6. Suppose 1 < p < ∞ and t > 0. Then the integral operator p and V = s p .Then claim that the intervals (P, Q) and (U, V ) are non-empty.By using hypothesis one can easily find that Implies that the intervals (P, Q) and (U, V ) are non-empty.Also, z ∈ D. By Schur's test, boundedness of the operator T t,s on L p (D) holds. (2.4) The proof is an easy modification of arguments in Theorem 4.1 in [1].We omit the details.
belongs to l p for some (or Proof.We will prove the result in the order: (a) ⇔ (b) and (b) ⇔ (c).
(a) ⇒ (b).For any R ∈ (0, 1), there exist positive constant C R such that for any z ∈ D kernel estimate holds.Thus for s ∈ R , we have Above implies that, M R,s p ≤ C M t,s p .
(b) ⇒ (a).By Lemma 2 and Lemma 5, there is a positive constant C such that for any z ∈ D and s ∈ R, we have k=1 be any r-lattice.By Lemma 5, we may assume R < r.By triangle inequality, we have E(z k , r) ⊂ E(z, 2r) for z ∈ E(z k , r) and for all k.Thus, we have (2.5) whenever z ∈ E(z k , r).Therefore, we have Thus by Lemma 5, we have ∈ l p for some r-lattice Above inequality and Lemma 5 implies that for any R ∈ (0, 1).

Carleson measure characterizations
In this section, we are using averaging function and t-Berezin transform as our main tools to characterize the (p, q, σ)−Bergman Carleson measure for 0 < p, q < ∞ and t > 0. Let Ω be a finite positive Borel measure.Recall that, • Ω is a (p, q, σ)−Bergman Carleson measure if the embedding i : A p σ −→ L q (Ω) is bounded.In other words, we can say Ω is a (p, q, σ)−Bergman Carleson measure if there exists a finite constant C > 0 such that as n −→ ∞ whenever {f n } is a bounded sequence in A p σ which converges to 0 uniformly on any compact subset of D. Note that, by taking p = q and σ(z) = 1, Ω becomes a Bergman Carleson measure and vanishing Carleson measure.We divide our result into two cases: 0 < p ≤ q < ∞ and 0 < q < p < ∞.
(c) The function Ω R (z) is bounded on D for some (or any) R ∈ (0, 1).
is bounded for some (or any) Proof.(d) ⇒ (a) We assume that {z k } ∞ k=1 be an r-lattice.We use the elementary inequality , by taking l = p q ≥ 1, using Lemma 1 and (1.1), we obtain Above inequality reveals that .
(a) ⇒ (c).Set f z (w) = K α (z, w), w ∈ D. By Lemma 1 and statement(a), we have Ω R (z) Above inequality reveals that The equivalence of (a),(c) and (d) follows from above proof of implications.Moreover, .
(b) For t > 2q p(α+2) , we have (d) ⇒ (a).Suppose (d) holds.For any ǫ > 0, there exists a positive integer k 0 such that Ω r (z k ) is relatively compact in D. Let us consider a bounded sequence {f j } ∞ j=1 in A p σ such that f j −→ 0 uniformly on any compact subset of D as j → ∞.Similar to the proof of (3.2) and if j is large enough, we have that where C is independent of ǫ.
(a) ⇒ (b) The equivalence of (a), (c) and (d) shows that the measure Ω is a vanishing (N p, N q, σ)-Bergman Carleson measure if Ω is a vanishing (p, q, σ)- One can easily find that f z ∈ A p σ , f z p,σ ≤ C and f z → 0 uniformly on any compact subset of D as z → ∂D.Since N p > 4 (α+2) and it follows from (3.7), we have that Statement (a) yields that Ω is a vanishing ( tp N q , t N , σ)−Bergman Carleson measure.Therefore lim z→∂D Ω N q (z) The proof is completed.

Moreover, we have
Proof.Since Lemma 8 implies that the statements (c), (d) and (e) are equivalent with the corresponding norm estimate (3.9) and the implication (b) ⇒ (a) is trivially true, it is sufficient to prove that (d) ⇒ (a), (a) ⇒ (e) and (a) ⇒ (b).(d) ⇒ (a).Since 0 < q < p < ∞ implies p q > 1 ,therefore the conjugate exponent of p q is p p−q .For f ∈ A p σ , we have Above inequality follows from Lemma 3 and Holder's inequality shows that Ω is a (p, q, σ)−Bergman Carleson measure and i q Recall that Rademacher functions ψ k are defined by , where [t] denotes the greatest integer less than or equal to t.For 0 < q < ∞, Khinchine's inequality is given as , which holds for all m ≥ 1 and all complex numbers b 1 , b Thus, we have (3.11) Since ∪ k0 k=1 E(z k , R) is relatively compact in D, so Hence by duality argument,we have Ω r (z k ) Therefore, |f n (z)| q dΩ(z) = 0

FinallyF
, we will prove the implication (a) ⇒ (b).Let us consider a bounded sequence {f n } ∞ n=1 in A p σ such that f n −→ 0 uniformly on each compact subset of D. Let F be any compact subset of D and Ω F be the restriction of Ω to F .Then we haveD |f n (z)| q dΩ(z) = F + D\F |f n (z)| q dΩ(z) = I 1 + I 2 .(3.12)Since f n → 0 uniformly on F as n → ∞, we haveI 1 = F |f n (z)| q dΩ(z) ≤ C sup z∈F |f n (z)| q −→ 0.Fix R ∈ (0, 1) and z ∈ D, and we obtain that (Ω F ) R (z) extended to D. By the equivalence of (a) and (d), we have (Ω F ) as F extended to D, (3.13) above follows from (1.1) and the dominated convergence theorem.Hence lim n→∞ D 2 , • • • b m .Let ψ k (t) be the kth Rademacher function on [0, 1].Replacing λ k with ψ k (t)λ k , integrating w.r.t t from 0 to 1 and applying Khinchine's inequality in [3.10], we see that