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Bergman spaces with exponential type weights

Abstract

For \(1\le p<\infty \), let \(A^{p}_{\omega }\) be the weighted Bergman space associated with an exponential type weight ω satisfying

$$ \int _{{\mathbb{D}}} \bigl\vert K_{z}(\xi ) \bigr\vert \omega (\xi )^{1/2} \,dA(\xi ) \le C \omega (z)^{-1/2}, \quad z\in {\mathbb{D}}, $$

where \(K_{z}\) is the reproducing kernel of \(A^{2}_{\omega }\). This condition allows us to obtain some interesting reproducing kernel estimates and more estimates on the solutions of the ∂̅-equation (Theorem 2.5) for more general weight \(\omega _{*}\). As an application, we prove the boundedness of the Bergman projection on \(L^{p}_{\omega }\), identify the dual space of \(A^{p}_{\omega }\), and establish an atomic decomposition for it. Further, we give necessary and sufficient conditions for the boundedness and compactness of some operators acting from \(A^{p}_{\omega }\) into \(A^{q}_{\omega }\), \(1\le p,q<\infty \), such as Toeplitz and (big) Hankel operators.

1 Introduction and results

Let \(H({\mathbb{D}})\) denote the space of all analytic functions on \({\mathbb{D}}\), where \({\mathbb{D}}\) is the open unit disk in the complex plane \(\mathbb{C}\). A weight is a positive function \(\omega \in L^{1}(\mathbb{D}, dA)\), where \(dA(z) = \frac{dx\,dy}{\pi }\) is the normalized area measure on \({\mathbb{D}}\). For \(0< p < \infty \), the weighted Bergman space \(A^{p}(\omega )\) consists of those functions \(f\in H({\mathbb{D}})\) for which

$$ \Vert f \Vert _{A^{p}(\omega )} = \biggl( \int _{{\mathbb{D}}} \bigl\vert f(z) \bigr\vert ^{p} \omega (z)\,dA(z) \biggr)^{1/p} < \infty . $$

For \(p=\infty \), we introduce the growth space \(L^{\infty }(\omega ^{1/2})\) of all measurable functions f on \({\mathbb{D}}\) with

$$ \Vert f \Vert _{L^{\infty }(\omega ^{1/2})}: = \operatorname{ess} \sup _{ \substack{z\in \mathbb{D}}} \bigl\vert f(z) \bigr\vert \omega (z)^{1/2} < \infty , $$

and we let

$$ A^{\infty }\bigl(\omega ^{1/2}\bigr):= L^{\infty }\bigl(\omega ^{1/2}\bigr)\cap H( \mathbb{D}). $$

In this paper we study Bergman spaces with weights belonging to a certain class \(\mathcal{W}\), which we describe now. Decreasing weights ω satisfying conditions will be specified in what follows. The class \(\mathcal{W}\), considered previously in [24] and [12], consists of the radial decreasing weights of the form \(\omega (z)=e^{-2\varphi (z)}\), where \(\varphi \in C^{2}({\mathbb{D}})\) is a radial function such that \((\Delta \varphi (z) )^{-1/2}\asymp \tau (z)\) for some positive radial function \(\tau (z)\) that decreases to 0 as \(|z|\rightarrow 1^{-}\) and satisfies \(\lim_{r\to 1^{-}}\tau '(r)=0\). Here Δ denotes the standard Laplace operator. Furthermore, we assume that either there exists a constant \(C>0\) such that \(\tau (r)(1-r)^{-C}\) increases for r close to 1 or

$$ \lim_{r\to 1^{-}}\tau '(r)\log \frac{1}{\tau (r)}=0. $$

The prototype is the exponential weight

$$ \omega _{\sigma }(z) = \exp \biggl(\frac{-b}{(1- \vert z \vert ^{2})^{\sigma }} \biggr) ,\quad b,\sigma > 0 . $$
(1.1)

For the weights ω in our class, the point evaluations \(L_{z}: f\longmapsto f(z)\) are bounded linear functionals on \(A^{2}(\omega )\) for each \(z\in {\mathbb{D}}\). In particular, the space \(A^{2}(\omega )\) is a reproducing kernel Hilbert space: for each \(z\in {\mathbb{D}}\), there are functions \(K_{z}\in A^{2}(\omega )\) with \(\|L_{z}\|=\|K_{z}\|_{A^{2}(\omega )}\) such that \(L_{z} f=f(z)=\langle f, K_{z} \rangle _{\omega }\), where

$$ \langle f,g \rangle _{\omega }= \int _{{\mathbb{D}}} f(z) \overline{g(z)} \omega (z)\,dA(z). $$

The function \(K_{z}\) is called the reproducing kernel for the Bergman space \(A^{2}(\omega )\) and has the property that \(K_{z}(\xi )=\overline{K_{\xi }(z)}\). The Bergman spaces with exponential type weights have attracted a lot of attention in recent years [9, 12, 16, 17, 24, 25] since new techniques, different from those used for standard Bergman spaces, are required. Consider the class \(\mathcal{E}\) that consists of the weights \(\omega \in \mathcal{W}\) satisfying

$$ \int _{{\mathbb{D}}} \bigl\vert K_{z}(\xi ) \bigr\vert \omega (\xi )^{1/2}\,dA(\xi ) \le C \omega (z)^{-1/2},\quad z\in { \mathbb{D}}. $$
(1.2)

It has been proved in [7] that the exponential type weights \(\omega _{\sigma }\) given by (1.1) with \(\sigma =1\) satisfy the previous condition and, therefore, they are in the class \(\mathcal{E}\). Note that following the proof given in [7] with nontrivial modifications, we are able to show that every weight of the form (1.1), with \(0<\sigma <\infty \), is in the class \(\mathcal{E}\). Recently, Hu, Xiaofen, and Schuster proved that the prototype weights, considered in (1.1), satisfy (1.2) (see [14, Corollary 3.2]). This integral estimate allows to study other properties and operators, such as the Bergman projection which is given by

$$ P_{\omega }(f) (z)= \int _{{\mathbb{D}}}f(\xi ) \overline{K_{z}(\xi )} \omega ( \xi )\,dA(\xi ) ,\quad z\in {\mathbb{D}}. $$

The boundedness of the Bergman projection \(P_{\omega }\) on \(L^{2}(\omega )\) is trivial from the general theory of Hilbert spaces. In contrast with the case of standard Bergman spaces (where the Bergman projection is bounded for \(1< p<\infty \)) in the case of exponential type weights, it turns out that the natural Bergman projection is not bounded on \(L^{p}(\omega )\) unless \(p=2\) (see [8] and [35]). At first glance, this may look surprising, but when one takes into account the similarities with Fock spaces, this seems to be more natural. It turns out that, similarly to the Fock space setting, when studying problems where reproducing kernels are involved, the most convenient setting is provided by the spaces \(A^{p}(\omega ^{p/2})\). As a consequence of condition (1.2), we get the right estimates for the norm of reproducing kernels in \(A^{p}(\omega ^{p/2})\) for \(1\le p\le \infty \). Also, we prove in Theorem 3.2 that, for weights in the class \(\mathcal{E}\), the Bergman projection \(P_{\omega }: L^{p}(\omega ^{p/2})\rightarrow A^{p}(\omega ^{p/2})\) is bounded for \(1\le p\le \infty \). A consequence of that result will be the identification of the dual space of \(A^{p}(\omega ^{p/2})\) with the space \(A^{p'}(\omega ^{{p'}/2})\) and \(A^{1}(\omega ^{1/2})\) with \(A^{\infty }(\omega ^{1/2})\) under the natural integral pairing \(\langle \cdot , \cdot \rangle _{\omega }\), where \(p'\) denotes the conjugate exponent of p. Afterwards, by using the duality and the estimates for the p-norms of reproducing kernels, we are going to obtain an atomic decomposition for Bergman spaces with exponential type weights: for weights \(\omega \in \mathcal{E}\), every function in the weighted Bergman space \(A^{p}(\omega ^{p/2})\), \(1\le p<\infty \), can be decomposed into a series of very nice functions (called atoms). These atoms are defined in terms of kernel functions and in some sense act as a basis for the space \(A^{p}(\omega ^{p/2})\). The atomic decomposition for Bergman space with standard weights was obtained by Coifman and Rochberg [6], and it has become a powerful tool in the study of weighted Bergman spaces. We refer to the books [26, 37, 38] for a modern proof of these results. The norm estimates for the reproducing kernels in \(A^{p}(\omega ^{p/2})\) for \(1\le p<\infty \) permit to extend the results on the boundedness and compactness of Toeplitz operators \(T_{\mu }\), to consider the action of \(T_{\mu }\) between different large weighted Bergman spaces, and to find a general description of when \(T_{\mu }: A^{p}(\omega ^{p/2})\longrightarrow A^{q}(\omega ^{q/2})\) is bounded or compact for all values of \(1\le p,q<\infty \). Furthermore, we also generalize the results obtained in [12] on the boundedness and compactness of big Hankel operators \(H_{\overline{g}}\) with conjugate analytic symbols to the non-Hilbert space setting, characterizing for all \(1< p,q<\infty \) the operators

$$ H_{\overline{g}}: A^{p}\bigl(\omega ^{p/2}\bigr) \longrightarrow L^{q}\bigl(\omega ^{q/2}\bigr) $$

that are bounded or compact. As mentioned earlier, one of the key tools consists in using the estimates for the ∂̅-equation obtained in Sect. 2.

In what follows we use the notation \(a \lesssim b\) to indicate that there is a constant \(C > 0\) with \(a \le C b\); and the notation \(a \asymp b\) means that \(a \lesssim b\) and \(b \lesssim a\). Also, respectively the expressions \(L^{p}_{\omega }\) and \(A^{p}_{\omega }\) mean \(L^{p}({\mathbb{D}},\omega ^{p/2}\,dA)\) and \(A^{p}(\omega ^{p/2})\) for \(1 le p<\infty \).

2 Preliminaries and basic properties

A positive function τ on \({\mathbb{D}}\) is said to be of class \(\mathcal{L}\) if it satisfies the following two properties:

  1. (A)

    There is a constant \(c_{1}\) such that \(\tau (z)\le c_{1} (1-|z|)\) for all \(z\in {\mathbb{D}}\);

  2. (B)

    There is a constant \(c_{2}\) such that \(|\tau (z)-\tau (\zeta )|\le c_{2} |z-\zeta |\) for all \(z,\zeta \in {\mathbb{D}}\).

We also use the notation

$$ m_{\tau }: = \frac{\min (1, c_{1}^{-1}, c_{2}^{-1})}{4}, $$

where \(c_{1}\) and \(c_{2}\) are the constants appearing in the previous definition. For \(a\in {\mathbb{D}}\) and \(\delta >0\), we use \(D(\delta \tau (a))\) to denote the Euclidean disc centered at a and having radius \(\delta \tau (a)\). It is easy to see from conditions (A) and (B) (see [24, Lemma 2.1]) that if \(\tau \in \mathcal{L}\) and \(z \in D(\delta \tau ((a)) \), then

$$ \frac{1}{2} \tau (a)\leq \tau (z) \leq 2 \tau (a) $$
(2.1)

for sufficiently small \(\delta > 0 \), that is, for \(\delta \in (0, m_{\tau })\). This fact will be used many times in this paper.

Definition 2.1

We say that a weight ω is of class \(\mathcal{L}^{*}\) if it is of the form \(\omega =e^{-2\varphi }\), where \(\varphi \in C^{2}({\mathbb{D}})\) with \(\Delta \varphi >0\) and \((\Delta \varphi (z) )^{-1/2}\asymp \tau (z)\) with \(\tau (z)\) is a function in the class \(\mathcal{L}\).

It is straightforward to see that \(\mathcal{W}\subset \mathcal{L}^{*}\). The following result is from [24, Lemma 2.2] and gives the boundedness of the point evaluation functional on \(A^{2}_{\omega }\).

Lemma A

Let \(\omega \in \mathcal{L}^{*}\), \(0< p<\infty \), and \(z\in {\mathbb{D}}\). If \(\beta \in \mathbb{R}\), there exists \(M \geq 1\) such that

$$ \bigl\vert f(z) \bigr\vert ^{p} \omega (z)^{\beta } \leq \frac{M}{\delta ^{2}\tau (z)^{2}} \int _{D(\delta \tau (z))}{ \bigl\vert f(\xi ) \bigr\vert ^{p} \omega (\xi )^{\beta }\,dA( \xi )} $$

for all \(f \in H(\mathbb{D})\) and all sufficiently small \(\delta > 0 \).

We also need a similar estimate for the gradient of \(|f| \omega ^{1/2}\).

Lemma 2.2

Let \(\omega \in \mathcal{L}^{*}\) and \(0< p < \infty \). For any \(\delta _{0} > 0\) sufficiently small, there exists a constant \(C(\delta _{0})> 0\) such that

$$ \bigl\vert \nabla \bigl( \vert f \vert \omega ^{1/2}\bigr) (z) \bigr\vert \leq \frac{C(\delta _{0})}{\tau (z)^{1+\frac{2}{p}}} \biggl( \int _{D(\delta _{0} \tau (z)/2)}{ \bigl\vert f(\xi ) \bigr\vert ^{p} \omega (\xi )^{p/2}\,dA(\xi )} \biggr)^{1/p} $$

for all \(f \in H(\mathbb{D})\).

Proof

We follow the method used in [22]. Without loss of generality we can assume \(z= 0\). Then, applying the Riesz decomposition (see for example [29]) of the subharmonic function φ in \(D(0,\frac{\delta _{0}}{2}\tau (0))\), we obtain

$$ \varphi (\xi ) = u(\xi ) + \int _{D(\frac{r}{2})}{G(\xi ,\eta ) \Delta \varphi (\eta )\,dA(\eta )}, $$
(2.2)

where \(r = \delta _{0} \tau (0)\), u is the least harmonic majorant of φ in \(D(0,\frac{r}{2})\), and G is the Green function defined for every \(\xi ,\eta \in D(0,r)\), \(\xi \neq \eta \), by

$$ G(\xi ,\eta ): = \log \biggl\vert \frac{r(\xi - \eta )}{r^{2} - \bar{\eta }\xi } \biggr\vert ^{2}. $$

For \(\xi ,\eta \in D(0,\frac{r}{2})\), we have

$$\begin{aligned} \begin{aligned} \biggl\vert \frac{\partial G}{\partial \xi } (\xi ,\eta ) \biggr\vert &= \frac{r^{2} - \vert \eta \vert ^{2}}{ \vert \xi - \eta \vert \vert r^{2} - \eta \bar{\xi } \vert } \\ &\leq \frac{r^{2}}{ \vert \xi - \eta \vert \cdot \vert r^{2} - \vert \eta \vert \vert \xi \vert \vert } \leq \frac{4}{3 \vert \xi - \eta \vert } . \end{aligned} \end{aligned}$$

Then

$$ \begin{aligned} \biggl\vert \frac{\partial \varphi (0)}{\partial \xi } - \frac{\partial u(0)}{\partial \xi } \biggr\vert &\leq \int _{D(\frac{r}{2})}{ \biggl\vert \frac{\partial G}{\partial \xi } (0,\eta ) \biggr\vert \Delta \varphi ( \eta )\,dA(\eta )} \\ &\lesssim \frac{1}{\tau (0)^{2}} \int _{D(\frac{r}{2})}{ \frac{dA(\eta )}{ \vert \eta \vert }}= \frac{\delta _{0}}{\tau (0)}. \end{aligned} $$
(2.3)

We pick a function \(h \in H(\mathbb{D})\) such that \(\operatorname{Re}(h) = u\). Also,

$$\begin{aligned} \begin{aligned} \bigl\vert \nabla \bigl( \vert f \vert e^{-\varphi }\bigr) (\xi ) \bigr\vert &= 2\cdot \biggl\vert \frac{1}{2}\frac{f'(\xi )\overline{f(\xi )}}{ \vert f(\xi ) \vert }- \frac{\partial \varphi }{\partial \xi }(\xi ) \bigl\vert f(\xi ) \bigr\vert \biggr\vert e^{- \varphi (\xi )} \\ &= \biggl\vert f'(\xi )-2\frac{ \vert f(\xi ) \vert ^{2}}{\overline{f(\xi )}} \frac{\partial \varphi }{\partial \xi }( \xi ) \biggr\vert e^{-\varphi (\xi )}. \end{aligned} \end{aligned}$$

Therefore, since \(h'(0) = 2 \frac{\partial u}{\partial \xi }(0)\), we get

$$\begin{aligned} \begin{aligned} \bigl\vert \nabla \bigl( \vert f \vert e^{-\varphi }\bigr) (0) \bigr\vert &= \biggl\vert f'(0)-2 f(0) \frac{\partial \varphi }{\partial \xi }(0) \biggr\vert e^{-\varphi (0)} \\ &\leq \biggl\vert f'(0)-2 f(0) \frac{\partial u}{\partial \xi }(0) \biggr\vert e^{- \varphi (0)}+ 2 \biggl\vert \frac{\partial u}{\partial \xi }(0) - \frac{\partial \varphi }{\partial \xi }(0) \biggr\vert \bigl\vert f(0) \bigr\vert e^{-\varphi (0)} \\ &\lesssim \biggl\vert \frac{\partial (f e^{-h})(0)}{\partial \xi } \biggr\vert e^{u(0)- \varphi (0)}+ \biggl\vert \frac{\partial u}{\partial \xi }(0) - \frac{\partial \varphi }{\partial \xi }(0) \biggr\vert \bigl\vert f(0) \bigr\vert e^{-\varphi (0)}. \end{aligned} \end{aligned}$$

By (2.3) we have

$$ \biggl\vert \frac{\partial u}{\partial \xi }(0) - \frac{\partial \varphi }{\partial \xi }(0) \biggr\vert \bigl\vert f(0) \bigr\vert e^{-\varphi (0)} \lesssim \frac{\delta _{0}}{\tau (0)} \bigl\vert f(0) \bigr\vert e^{-\varphi (0)}. $$

This gives

$$ \bigl\vert \nabla \bigl( \vert f \vert e^{-\varphi } \bigr) (0) \bigr\vert \lesssim \biggl\vert \frac{\partial (f e^{-h})(0)}{\partial \xi } \biggr\vert e^{u(0)-\varphi (0)}+ \frac{ \vert f(0) \vert }{\tau (0)} e^{-\varphi (0)}. $$
(2.4)

It follows from Lemma A that

$$ \begin{aligned} \frac{ \vert f(0) \vert }{\tau (0)} e^{-\varphi (0)} & \lesssim \frac{1}{\tau (0)^{1+\frac{2}{p}}} \biggl( \int _{D(\delta _{0}\tau (0)/2)}{ \bigl\vert f(z) \bigr\vert ^{p} e^{-p\varphi (z)}\,dA(z)} \biggr)^{1/p}. \end{aligned} $$
(2.5)

To deal with the other term appearing in (2.4), notice that if we use identity (2.2) with the function \(\phi (\xi ) = |\xi |^{2} -(r/2)^{2}\) (since \(\Delta \phi (\xi )= 4\) and its least harmonic majorant is \(u_{\phi } = 0\)), we obtain

$$ \int _{D(\frac{r}{2})}{G(\xi ,\eta )\,dA(\eta )}= \frac{1}{4} \bigl( \vert \xi \vert ^{2} - (r/2)^{2} \bigr). $$

Therefore, since \(\Delta \varphi (\eta )= \frac{1}{\tau (\eta )^{2}}\lesssim \frac{1}{\tau (0)^{2}}= \Delta \varphi (0) \) and the Green function \(G\leq 0\), we obtain, for every \(\xi \in D(0,\frac{r}{2})\),

$$ \begin{aligned} u(\xi ) - \varphi (\xi )&=- \int _{D(\frac{r}{2})}{G( \xi ,\eta ) \Delta \varphi (\eta )\,dA(\eta )} \\ &\lesssim \frac{\Delta \varphi (0)}{4} \bigl((r/2)^{2} - \vert \xi \vert ^{2} \bigr)=\frac{1}{4\tau (0)^{2}} \bigl((r/2)^{2} - \vert \xi \vert ^{2} \bigr). \end{aligned} $$

This gives

$$ e^{u(0)-\varphi (0)} \le e^{C\delta _{0}^{2}}. $$

Therefore

$$ \biggl\vert \frac{\partial (f e^{-h})(0)}{\partial \xi } \biggr\vert e^{u(0)- \varphi (0)} \lesssim \biggl\vert \frac{\partial (f e^{-h})(0)}{\partial \xi } \biggr\vert . $$
(2.6)

On the other hand, using Cauchy’s inequality, the fact that \(\varphi -u\le 0\), and Lemma A, we get

$$\begin{aligned} \begin{aligned} \biggl\vert \frac{\partial (fe^{-h})}{\partial \xi }(0) \biggr\vert & \lesssim \biggl\vert \int _{ \vert \eta \vert =\frac{ \delta _{0}\tau (0)}{4}}{ \frac{f(\eta )e^{-h(\eta )}}{ \eta ^{2}}\,d\eta } \biggr\vert \\ &\lesssim \frac{1}{\delta _{0}^{2}\tau (0)^{2}} \int _{ \vert \eta \vert = \frac{ \delta _{0}\tau (0)}{4}}{ \bigl\vert f(\eta ) \bigl\vert e^{-\varphi (\eta )} e^{ \varphi (\eta )-u(\eta )} \bigr\vert \,d\eta } \bigr\vert \\ &\lesssim \frac{1}{\tau (0)^{2}} \int _{ \vert \eta \vert = \frac{ \delta _{0}\tau (0)}{4}}{ \biggl(\frac{1}{\tau (\eta )^{2}} \int _{D( \delta _{0}\tau (\eta )/4)}{ \bigl\vert f(z) \bigr\vert ^{p} e^{-p\varphi (z)}\,dA(z)} \biggr)^{1/p} \vert \,d\eta \vert }. \end{aligned} \end{aligned}$$

Finally, using \(\tau (\eta ) \asymp \tau (0)\), we obtain

$$ \begin{aligned} \biggl\vert \frac{\partial fe^{-h}}{\partial \xi }(0) \biggr\vert & \lesssim \frac{1}{\tau (0)^{2}} \int _{ \vert \eta \vert = \frac{ \delta _{0}\tau (0)}{4}}{ \biggl(\frac{1}{\tau (0)^{2}} \int _{D( \delta _{0}\tau (0)/2)}{ \bigl\vert f(z) \bigr\vert ^{p} e^{-p\varphi (z)}\,dA(z)} \biggr)^{1/p} \vert \,d\eta \vert } \\ &\lesssim \frac{1}{\tau (0)^{1+\frac{2}{p}}} \biggl( \int _{D(\delta _{0} \tau (0)/2)}{ \bigl\vert f(z) \bigr\vert ^{p} e^{-p\varphi (z)}\,dA(z)} \biggr)^{1/p}. \end{aligned} $$

Bearing in mind (2.6) this gives

$$ \biggl\vert \frac{\partial (f e^{-h})(0)}{\partial \xi } \biggr\vert e^{u(0)- \varphi (0)}\lesssim \frac{1}{\tau (0)^{1+\frac{2}{p}}} \biggl( \int _{D( \delta _{0}\tau (0)/2)}{ \bigl\vert f(z) \bigr\vert ^{p} e^{-p\varphi (z)}\,dA(z)} \biggr)^{1/p}. $$

Plugging this and (2.5) into (2.4), we get the desired result. □

The following lemma on coverings is due to Oleinik, see [21].

Lemma B

Let τ be a positive function on \({\mathbb{D}}\) of class \(\mathcal{L}\), and let \(\delta \in (0,m_{\tau })\). Then there exists a sequence of points \(\{z_{n}\}\subset {\mathbb{D}}\) such that the following conditions are satisfied:

  1. (i)

    \(z_{n} \notin D(\delta \tau (z_{k})) \), \(n \neq k \);

  2. (ii)

    \(\bigcup_{\substack{n}} D(\delta \tau (z_{n})) = \mathbb{D}\);

  3. (iii)

    \(\widetilde{D}(\delta \tau (z_{n}))\subset D(3\delta \tau (z_{n})) \), where \(\widetilde{D}(\delta \tau (z_{n})) = \bigcup_{ \substack{z\in D(\delta \tau (z_{n}))}} D(\delta \tau (z)) \), \(n = 1, 2 ,3 ,\ldots\) ;

  4. (iv)

    \(\lbrace D(3\delta \tau (z_{n}))\rbrace \) is a covering of \(\mathbb{D}\) of finite multiplicity N.

The multiplicity N in Lemma B is independent of δ, and it is easy to see that one can take, for example, \(N=256\). Any sequence satisfying the conditions in Lemma B will be called a \((\delta ,\tau )\)-lattice. Note that \(|z_{n}|\to 1^{-}\) as \(n\to \infty \). In what follows, the sequence \(\{z_{n}\}\) will always refer to the sequence chosen in Lemma B.

2.1 Integral estimates for reproducing kernels

We use the notation \(k_{z}\) for the normalized reproducing kernels in \(A^{p}_{\omega }\), that is,

$$ k_{z}=\frac{K_{z}}{ \Vert K_{z} \Vert _{A^{p}_{\omega }}}. $$

The next result (see [4, 17, 24] for (a) when \(p=2\) and [18, Lemma 3.6] for part (b)) provides useful estimates involving reproducing kernels.

Theorem A

Let \(K_{z} \) be the reproducing kernel of \(A^{2}_{\omega }\). Then

  1. (a)

    For \(\omega \in \mathcal{W}\), one has

    $$ \Vert K_{z} \Vert _{A^{2}_{\omega }} \asymp \omega (z)^{-1/2} \tau (z)^{-1}, \quad z\in {\mathbb{D}}. $$
    (2.7)
  2. (b)

    For all sufficiently small \(\delta \in (0,m_{\tau })\) and \(\omega \in \mathcal{W}\), one has

    $$ \bigl\vert K_{z}(\zeta ) \bigr\vert \asymp \Vert K_{z} \Vert _{A^{2}_{\omega }}\cdot \Vert K_{\zeta } \Vert _{A^{2}_{\omega }} ,\quad \zeta \in D_{\delta }(z). $$
    (2.8)

Lemma 2.3

Let \(\omega \in \mathcal{E}\). For each \(z\in {\mathbb{D}}\), we have

$$ \Vert K_{z} \Vert _{A^{\infty }(\omega ^{1/2})} \lesssim \omega (z)^{-1/2} \tau (z)^{-2}. $$

Proof

By Lemma A and condition (1.2), we have

$$ \begin{aligned} \omega (\xi )^{1/2} \bigl\vert K_{z}(\xi ) \bigr\vert &=\omega (\xi )^{1/2} \bigl\vert K_{\xi }(z) \bigr\vert \\ &\lesssim \frac{\omega (\xi )^{1/2}}{\omega (z)^{1/2} \tau (z)^{2}} \int _{D( \delta \tau (z)) } \bigl\vert K_{\xi }(s) \bigr\vert \omega (s)^{1/2}\,dA(s) \\ & \lesssim \omega (z)^{-1/2} \tau (z)^{-2}, \end{aligned} $$

which finishes the proof. □

Lemma 2.3 together with condition (1.2) allows us to obtain the following estimate for the norm of the reproducing kernel in \(A^{p}_{\omega }\).

Lemma 2.4

Let \(1\le p<\infty \), \(\omega \in \mathcal{E}\), and \(z\in {\mathbb{D}}\). Then

$$ \Vert K_{z} \Vert _{A^{p}_{\omega }}\asymp \omega (z)^{-1/2} \tau (z)^{-2(p-1)/p}. $$

Proof

By (1.2) and Lemma 2.3, we have

$$ \begin{aligned} \Vert K_{z} \Vert ^{p}_{A^{p}_{\omega }} &= \int _{{\mathbb{D}}} \bigl\vert K_{z}(\xi ) \bigr\vert ^{p} \omega (\xi )^{p/2}\,dA(\xi ) = \int _{{\mathbb{D}}} \bigl\vert K_{z}(\xi ) \bigr\vert \omega (\xi )^{1/2} \bigl( \bigl\vert K_{z}(\xi ) \bigr\vert \omega (\xi )^{1/2} \bigr)^{p-1}\,dA(\xi ) \\ & \le \Vert K_{z} \Vert _{A^{\infty }(\omega ^{1/2})}^{p-1} \int _{{\mathbb{D}}} \bigl\vert K_{z}(\xi ) \bigr\vert \omega (\xi )^{1/2}\,dA(\xi ) \lesssim \Vert K_{z} \Vert _{A^{ \infty }(\omega ^{1/2})}^{p-1} \omega (z)^{-1/2} \\ &\lesssim \omega (z)^{-p/2} \tau (z)^{-2(p-1)}. \end{aligned} $$

On the other hand, by using statement (b) of Theorem A, we have

$$ \begin{aligned} \Vert K_{z} \Vert ^{p}_{A^{p}_{\omega }} &\ge \int _{D(\delta \tau (z))} \bigl\vert K_{z}(\xi ) \bigr\vert ^{p} \omega (\xi )^{p/2}\,dA(\xi ) \\ & \ge C \Vert K_{z} \Vert _{A^{2}_{\omega }}^{p} \int _{D(\delta \tau (z))} \Vert K_{\xi } \Vert ^{p}_{A^{2}_{\omega }} \omega (\xi )^{p/2}\,dA(\xi ) \\ &\ge C \omega (z)^{-p/2} \tau (z)^{-2(p-1)}. \end{aligned} $$

This completes the proof. □

2.2 Estimates for the ∂̅-equation

The following result, which provides more estimates on the solutions of the ∂̅-equation, will play a crucial role in describing the bounded Hankel operators acting from \(A^{p}_{\omega }\) to \(A^{q}_{\omega }\) when \(1\le p\le q <\infty \). Also, it can be of independent interest.

Theorem 2.5

Let \(\omega \in \mathcal{W}\), and consider the associated weight \(\omega _{*}(z) : = \omega (z) \tau (z)^{\alpha }\), \(z\in {\mathbb{D}}\), and \(\alpha \in \mathbb{R}\). Then there exists a solution u of the equation \(\overline{\partial }u= f\) such that

$$ \int _{\mathbb{D}} \bigl\vert u(z) \bigr\vert ^{p} \omega _{*}(z)^{p/2}\,dA(z) \leq C \int _{\mathbb{D}} \bigl\vert f(z) \bigr\vert ^{p} \omega _{*}(z)^{p/2} \tau (z)^{p}\,dA(z) $$

for all \(1\le p<\infty \), provided the right-hand side is finite. Moreover, one also has the \(L^{\infty }\)-estimate

$$ \sup_{z\in {\mathbb{D}}} \bigl\vert u(z) \bigr\vert \omega _{*} (z)^{1/2}\le C \sup_{z \in {\mathbb{D}}} \bigl\vert f(z) \bigr\vert \omega _{*} (z)^{1/2} \tau (z). $$

Proof

We follow the method used in [5] where the case \(\alpha =0\) was proved. By Lemma 3.1 in [24], there are holomorphic functions \(F_{a}\) and some \(\delta _{0} \in (0, m_{\tau })\) such that

$$ \begin{aligned} \text{(i)}\quad & \bigl\vert F_{a}(\xi ) \bigr\vert \asymp \omega (\xi )^{-1/2},\quad \xi \in D\bigl(\delta _{0}\tau (a)\bigr). \\ \text{(ii)}\quad & \bigl\vert F_{a}(\xi ) \bigr\vert \leq C \omega (\xi )^{-1/2} \biggl( \frac{\min (\tau (\xi ),\tau (a) )}{ \vert a-\xi \vert } \biggr)^{M},\quad (a, \xi )\in \mathbb{D}\times {\mathbb{D}}. \end{aligned} $$
(2.9)

Let \(\delta _{1}< \delta _{0}\). Then there is a sequence \(\lbrace z_{n}\rbrace _{n\geq 1}\) such that \(\lbrace D(\delta _{1}\tau (z_{n}))\rbrace \) is a covering of \(\mathbb{D}\) of finite multiplicity N and satisfies the other statements of Lemma B. Let \(\chi _{n}\) be a partition of unity subordinate to the covering \(D(\delta _{1}\tau (z_{n}))\). Consider

$$ S_{n} f (z) = F_{z_{n}}(z) \int _{\mathbb{D}} \frac{f(\xi ) \chi _{n} (\xi )}{(\xi -z) F_{z_{n}}(\xi )}\,dA( \xi ). $$

Since \(F_{z_{n}}\) are holomorphic functions on \(\mathbb{D}\), by the Cauchy–Pompeiu formula, we have

$$ \overline{\partial }S_{n} f(z) = f(z) \chi _{n}(z),\quad n= 1,2.... $$

Then

$$\begin{aligned} \begin{aligned} Sf(z) &= \sum_{n = 1 }^{\infty }S_{n}f(z) = \omega _{*}(z)^{-1/2} \int _{\mathbb{D}} G(z,\xi ) f(\xi ) \omega _{*}(\xi )^{1/2}\,dA( \xi ), \end{aligned} \end{aligned}$$

where

$$ G(z,\xi ) = \sum_{n = 1 }^{\infty } \frac{ F_{z_{n}}(z)}{\xi -z} \frac{\chi _{n}(\xi )}{ F_{z_{n}}(\xi )} \omega _{*}(\xi )^{-1/2} \omega _{*}(z)^{1/2}. $$

Since \(\chi _{n}\) is a partition of the unity, we have

$$ \overline{\partial }(Sf) = \sum_{n=1}^{\infty } \overline{\partial }(S_{n} f) = f \sum_{n=1}^{\infty } \chi _{n} = f $$

on \({\mathbb{D}}\), so that Sf solves the equation \(\overline{\partial }S f(z) = f(z)\).

On the other hand, assume that

$$ \begin{aligned} \int _{\mathbb{D}} \bigl\vert G(z,\xi ) \bigr\vert \frac{dA(\xi )}{\tau (\xi )}\lesssim 1 \end{aligned} $$
(2.10)

and

$$ \begin{aligned} \int _{\mathbb{D}} \frac{ \vert F_{z_{n}}(z) \vert }{ \vert \xi -z \vert } \omega _{*}(z)^{1/2}\, dA(z) \lesssim \tau (\xi )^{1+\alpha /2},\quad \xi \in D\bigl(\delta _{1} \tau (z_{n})\bigr). \end{aligned} $$
(2.11)

Then, by (2.13), it is straightforward that the \(L^{\infty }\)-estimate holds. Our next goal is to prove the inequality

$$ \int _{\mathbb{D}} \bigl\vert Sf(z) \bigr\vert ^{p} \omega _{*}(z)^{p/2}\,dA(z) \lesssim \int _{\mathbb{D}} \bigl\vert f(z) \bigr\vert ^{p} \omega _{*}(z)^{p/2} \tau (z) ^{p}\,dA(z) . $$

Consider \(g(\xi ): = f(\xi ) \omega _{*}(\xi )^{1/2}\) and \(Tg(z): = \int _{\mathbb{D}}G(z,\xi ) g(\xi )\,dA(\xi )\). Then the last inequality takes the form

$$ \int _{\mathbb{D}} \bigl\vert Tg(z) \bigr\vert ^{p}\,dA(z) \lesssim \int _{\mathbb{D}} \bigl\vert g(z) \bigr\vert ^{p} \tau (z) ^{p}\,dA(z) . $$

Therefore, using Hölder’s inequality and (2.13), we have

$$\begin{aligned} \begin{aligned} \bigl\vert Tg(z) \bigr\vert ^{p} &\leq \int _{\mathbb{D}} \bigl\vert g(\xi ) \bigr\vert ^{p} \tau (\xi )^{p-1} \bigl\vert G(z,\xi ) \bigr\vert dA(\xi ) \biggl( \int _{\mathbb{D}} \bigl\vert G(z, \xi ) \bigr\vert \frac{dA(\xi )}{\tau (\xi )} \biggr)^{p-1} \\ &\lesssim \int _{\mathbb{D}} \bigl\vert g(\xi ) \bigr\vert ^{p} \tau (\xi )^{p-1} \bigl\vert G(z, \xi ) \bigr\vert dA(\xi ). \end{aligned} \end{aligned}$$

These estimates and Fubini’s theorem give

$$\begin{aligned} \begin{aligned} \int _{\mathbb{D}} \bigl\vert Tg(z) \bigr\vert ^{p}\,dA(z) &\lesssim \int _{ \mathbb{D}} \biggl( \int _{\mathbb{D}} \bigl\vert g(\xi ) \bigr\vert ^{p} \tau (\xi )^{p-1} \bigl\vert G(z, \xi ) \bigr\vert dA(\xi ) \biggr)\,dA(z) \\ &\lesssim \int _{\mathbb{D}} \bigl\vert g(\xi ) \bigr\vert ^{p} \tau (\xi )^{p-1} \biggl( \int _{\mathbb{D}} \bigl\vert G(z,\xi ) \bigr\vert dA(z) \biggr)\,dA(\xi ). \end{aligned} \end{aligned}$$

Now, using the expression of the kernel \(G(z,\xi )\) and the fact that \(\chi _{n}\) are supported in \(D(\delta _{1}\tau (z_{n}))\), we obtain

$$\begin{aligned} \begin{aligned} &\int _{\mathbb{D}} \bigl\vert Tg(z) \bigr\vert ^{p}\,dA(z)\\ &\quad \lesssim \int _{ \mathbb{D}} \bigl\vert g(\xi ) \bigr\vert ^{p} \tau (\xi )^{p-1} \Biggl( \int _{\mathbb{D}} \sum_{n = 1 }^{\infty } \frac{ \vert F_{z_{n}}(z) \vert }{ \vert \xi -z \vert } \frac{\chi _{n}(\xi )}{ \vert F_{z_{n}}(\xi ) \vert } \frac{\omega _{*}(z)^{1/2}}{\omega _{*}(\xi )^{1/2}}\,dA(z) \Biggr)\,dA( \xi ) \\ &\quad \lesssim \sum_{n = 1 }^{\infty } \int _{{\mathbb{D}}} \bigl\vert g(\xi ) \bigr\vert ^{p} \tau (\xi )^{p-1} \biggl( \frac{\omega _{*}(\xi )^{-1/2}}{ \vert F_{z_{n}}(\xi ) \vert } \int _{\mathbb{D}} \frac{ \vert F_{z_{n}}(z) \vert }{ \vert \xi -z \vert } \omega _{*}(z)^{1/2}\,dA(z) \biggr)\chi _{n}(\xi )\,dA(\xi ) \\ &\quad \lesssim \sum_{n = 1 }^{\infty } \int _{D(\delta _{1}\tau (z_{n}))} \bigl\vert g( \xi ) \bigr\vert ^{p} \tau (\xi )^{p-1} \biggl( \frac{\omega _{*}(\xi )^{-1/2}}{ \vert F_{z_{n}}(\xi ) \vert } \int _{\mathbb{D}} \frac{ \vert F_{z_{n}}(z) \vert }{ \vert \xi -z \vert } \omega _{*}(z)^{1/2}\,dA(z) \biggr)\,dA(\xi ). \end{aligned} \end{aligned}$$

By (2.11) and using the fact that \(| F_{z_{n}}(\xi )| \asymp \omega (\xi )^{-1/2}\), \(\xi \in D( \delta _{1}\tau (z_{n})) \), it follows that

$$ \begin{aligned} \int _{\mathbb{D}}& \bigl\vert Tg(z) \bigr\vert ^{p}\,dA(z) \\ &\lesssim \sum_{n = 1 }^{\infty } \int _{D(\delta _{1}\tau (z_{n}))} \bigl\vert g( \xi ) \bigr\vert ^{p} \tau (\xi )^{p-1-\alpha /2} \biggl( \int _{\mathbb{D}} \frac{ \vert F_{z_{n}}(z) \vert }{ \vert \xi -z \vert } \omega _{*}(z)^{1/2}\,dA(z) \biggr)\,dA(\xi ) \\ &\lesssim \sum_{n = 1 }^{\infty } \int _{D(\delta _{1}\tau (z_{n}))} \bigl\vert g( \xi ) \bigr\vert ^{p} \tau (\xi )^{p}\,dA(\xi ) \\ &\lesssim \int _{\mathbb{D}} \bigl\vert g(\xi ) \bigr\vert ^{p} \tau (\xi )^{p}\,dA(\xi ), \end{aligned} $$
(2.12)

where the last inequality above follows from the fact that \(\{D(\delta _{\tau }(z_{n}))\}\) is a covering of \({\mathbb{D}}\) of finite multiplicity.

Now we are going to prove that

$$ \begin{aligned} \int _{\mathbb{D}} \bigl\vert G(z,\xi ) \bigr\vert \frac{dA(\xi )}{\tau (\xi )}\lesssim 1. \end{aligned} $$
(2.13)

First we consider the covering of \(\lbrace \xi \in \mathbb{D} : |z-\xi | > \delta _{2} \tau (z)\rbrace \) given by

$$ R_{k}(z) = \bigl\lbrace \xi \in \mathbb{D} : 2^{k-1}\delta _{2} \tau (z) < \vert z-\xi \vert \leq 2^{k}\delta _{2}\tau (z)\bigr\rbrace ,\quad k=1,2, \dots $$

Let \(4 \delta _{1} <\delta _{2} < \frac{\delta _{0}}{5}\) and \(z\in \mathbb{D}\) be fixed. If \(\xi \in D(\delta _{2}\tau (z))\cap D(\delta _{1}\tau (z_{n}))\), using (2.1), we have

$$\begin{aligned} \begin{aligned} \vert z-z_{n} \vert &\leq \vert z-\xi \vert + \vert \xi -z_{n} \vert \leq \delta _{2} \tau (z)+ \delta _{1}\tau (z_{n}) \\ &\leq 4\delta _{2} \tau (z_{n})+ \delta _{1}\tau (z_{n}) < \delta _{0} \tau (z_{n}), \end{aligned} \end{aligned}$$

that implies \(z \in D(\delta _{0}\tau (z_{n}))\). Using (2.1) and property (i) of (2.9), it follows

$$ \begin{aligned} \bigl\vert G(z,\xi ) \bigr\vert \lesssim \frac{\omega (z)^{-1/2}}{ \vert \xi -z \vert } \frac{\omega _{*}(\xi )^{-1/2}\omega _{*}(z)^{1/2}}{ \omega (\xi )^{-1/2}} \sum_{n=1}^{\infty } \chi _{n}(\xi ) \lesssim \frac{1}{ \vert \xi -z \vert }. \end{aligned} $$

Therefore, using again (2.1) and polar coordinates, we get

$$ \int _{D(\delta _{2}\tau (z))} \bigl\vert G(z,\xi ) \bigr\vert \frac{dA(\xi )}{\tau (\xi )} \lesssim \frac{1}{\tau (z)} \int _{D( \delta _{2}\tau (z))}\frac{1}{ \vert \xi -z \vert }\,dA(\xi )\lesssim 1. $$
(2.14)

If \(\xi \in ({\mathbb{D}}\smallsetminus D(\delta _{2}\tau (z)) ) \cap D(\delta _{1}\tau (z_{n}))\), we show that \(z\notin D(\delta _{1}\tau (z_{n}))\). In fact, if not, \(z\in D(\delta _{1}\tau (z_{n}))\), then using (2.1) we have

$$ \vert z-z_{n} \vert > \vert z-\xi \vert - \vert \xi -z_{n} \vert > \delta _{2}\tau (z)-\delta _{1} \tau (z_{n})\geq (\delta _{2}/2 -\delta _{1})\tau (z_{n})\geq \delta _{1} \tau (z_{n}), $$

this implies a contradiction with our assumption. Thus,

$$\begin{aligned} \begin{aligned} \vert z-\xi \vert &\leq \vert z-z_{n} \vert + \vert z_{n}-\xi \vert \leq \vert z-z_{n} \vert + \delta _{1}\tau (z_{n}) \\ &\leq 2 \vert z-z_{n} \vert . \end{aligned} \end{aligned}$$

Also, using \(\tau (\xi )\asymp \tau (z_{n})\), we get

$$ \frac{ \vert z-z_{n} \vert }{\min (\tau (z),\tau (z_{n}) )}\geq C \frac{ \vert z-\xi \vert }{\min (\tau (z),\tau (\xi ) )} . $$

Then, again using \(\tau (\xi )\asymp \tau (z_{n})\) and property (ii) of (2.9) with

$$ M > \max (1;-\alpha /2;1+\alpha /2), $$

we have

$$ \begin{aligned} \bigl\vert G(z,\xi ) \bigr\vert &\leq C \sum_{n = 1 }^{\infty } \frac{ \vert F_{z_{n}}(z) \vert }{ \vert \xi -z \vert } \frac{\chi _{n}(\xi )}{ \vert F_{z_{n}}(\xi ) \vert } \omega _{*}(\xi )^{-1/2} \omega _{*}(z)^{1/2} \\ &\leq C \frac{\omega (z)^{-1/2}}{ \vert \xi -z \vert } \frac{\omega _{*}(\xi )^{-1/2}\omega _{*}(z)^{1/2}}{ \omega (\xi )^{-1/2}} \biggl(\frac{\min (\tau (z),\tau (\xi ) )}{ \vert \xi -z \vert } \biggr)^{M} \sum_{n=1}^{\infty }\chi _{n}(\xi ) \\ &\leq C \frac{\tau (z)^{\frac{\alpha }{2}}}{\tau (\xi )^{\frac{\alpha }{2}} \vert \xi -z \vert } \biggl(\frac{\min (\tau (z),\tau (\xi ) )}{ \vert \xi -z \vert } \biggr)^{M}. \end{aligned} $$
(2.15)

Then

$$\begin{aligned} \begin{aligned} &\int _{\mathbb{D}\smallsetminus D(\delta _{2}\tau (z))} \bigl\vert G(z, \xi ) \bigr\vert \frac{dA(\xi )}{\tau (\xi )} \\ &\quad \leq C \tau (z)^{\alpha /2} \int _{\mathbb{D}\smallsetminus D(\delta _{2} \tau (z))}\frac{\tau (\xi )^{\frac{-(2+\alpha )}{2}}}{ \vert \xi -z \vert } \biggl(\frac{\min (\tau (z),\tau (\xi )) )}{ \vert \xi -z \vert } \biggr)^{M}\,dA( \xi ). \end{aligned} \end{aligned}$$

• If \(2+\alpha > 0\).

$$\begin{aligned} \begin{aligned} \int _{\mathbb{D}\smallsetminus D(\delta _{2}\tau (z))} \bigl\vert G(z, \xi ) \bigr\vert \frac{dA(\xi )}{\tau (\xi )} &\leq C \tau (z)^{M-1}\sum _{k=1}^{ \infty } \int _{R_{k}(z)}\frac{1}{ \vert \xi -z \vert ^{M+1}}\,dA(\xi ) \\ &\leq C\tau (z)^{M-1}\sum_{k=1}^{\infty } \int _{R_{k}(z)} \frac{1}{ (2^{k}\tau (z) )^{M+1}}\,dA(\xi ) \\ &\leq C \sum_{k=1}^{\infty } \frac{1}{2^{k(M-1)}} \lesssim 1. \end{aligned} \end{aligned}$$

• If \(2+\alpha \leq 0\). Using condition \((B)\) in the definition of the class \(\mathcal{L}\), it follows that

$$ \tau (\xi )\leq C 2^{k}\delta _{2}\tau (z),\quad \xi \in R_{k}(z), k=0,1,2,\dots. $$

So,

$$\begin{aligned} \begin{aligned} \int _{\mathbb{D}\smallsetminus D(\delta _{2}\tau (z))} \bigl\vert G(z, \xi ) \bigr\vert \frac{dA(\xi )}{\tau (\xi )}&\leq C \tau (z)^{\alpha /2} \int _{ \mathbb{D}\smallsetminus D(\delta _{2}\tau (z))} \frac{\tau (\xi )^{-(2+\alpha )/2}\tau (z)^{M}}{ \vert \xi -z \vert ^{M+1}}\,dA( \xi ) \\ &\leq C\tau (z)^{M+\alpha /2}\sum_{k=1}^{\infty } \int _{R_{k}(z)} \frac{ (2^{k} \tau (z) )^{-(2+\alpha )/2}}{ (2^{k}\tau (z) )^{M+1}}\,dA(\xi ) \\ &\leq C \sum_{k=1}^{\infty } \frac{1}{2^{k(M+\alpha /2)}} \lesssim 1. \end{aligned} \end{aligned}$$

This together with (2.14) establishes (2.13).

Finally, it remains to prove inequality (2.11). We split this integral in two parts: one integrating over the disk \(D(\delta _{2}\tau (\xi ))\) and the other one over \(\mathbb{D}\smallsetminus D(\delta _{2}\tau (\xi ))\). We compute the first integral using (i) of (2.9), \(\tau (\xi )\asymp \tau (z_{n})\asymp \tau (z)\), \(z\in D(\delta _{2} \tau (\xi ))\), and by using polar coordinates, we obtain

$$ \int _{D(\delta _{2}\tau (\xi ))} \frac{ \vert F_{z_{n}}(z) \vert }{ \vert \xi -z \vert } \omega _{*}(z)^{1/2}\,dA(z)\lesssim \tau (\xi )^{\alpha /2} \int _{D( \delta _{2}\tau (\xi ))}\frac{dA(z)}{ \vert z-\xi \vert } \lesssim \tau (\xi )^{1+ \alpha /2}. $$
(2.16)

Now we consider

$$ I(\xi ) : = \int _{\mathbb{D}\smallsetminus D(\delta _{2}\tau (\xi ))} \frac{ \vert F_{z_{n}}(z) \vert }{ \vert \xi -z \vert } \omega _{*}(z)^{1/2}\,dA(z),\quad \xi \in D\bigl(\delta _{1}\tau (z_{n})\bigr). $$

For \(z\notin D(\delta _{2}\tau (\xi ))\),

$$\begin{aligned} \begin{aligned} \vert z-\xi \vert &\leq \vert z-z_{n} \vert + \vert z_{n}- \xi \vert \\ &\leq \vert z-z_{n} \vert + \delta _{1} \tau (z_{n}) \\ &\leq \vert z-z_{n} \vert +\frac{2\delta _{1}}{\delta _{2}} \vert z-z_{n} \vert \\ & \leq \biggl(1+ \frac{2\delta _{1}}{\delta _{2}} \biggr) \vert z-z_{n} \vert . \end{aligned} \end{aligned}$$

Then, again by using \(\tau (\xi )\asymp \tau (z_{n})\), we obtain

$$ \frac{\min (\tau (z_{n}),\tau (z) )}{ \vert z-z_{n} \vert }\leq C \frac{\min (\tau (\xi ),\tau (z) )}{ \vert z-\xi \vert }. $$

This together with (2.9) taking \(M> \max (1,2+\alpha /2)\) gives

$$\begin{aligned} \begin{aligned} I(\xi ) &\lesssim \int _{\mathbb{D}\smallsetminus D( \delta _{2}\tau (\xi ))} \frac{\tau (z)^{\alpha /2}}{ \vert \xi -z \vert } \frac{ (\min (\tau (z_{n}),\tau (z) ) )^{M}}{ \vert z-z_{n} \vert ^{M}}\,dA(z) \\ &\lesssim \int _{\mathbb{D}\smallsetminus D(\delta _{2}\tau (\xi ))} \tau (z)^{\alpha /2} \frac{ (\min (\tau (\xi ),\tau (z) ) )^{M}}{ \vert z-\xi \vert ^{M+1}}\,dA(z). \end{aligned} \end{aligned}$$

• Suppose first that \(\alpha \geq 0\). By \(\tau (z)\leq C 2^{k}\tau (\xi )\), for every \(z\in R_{k}(\xi )\), \(k=1,2,\dots \), we get

$$\begin{aligned} \begin{aligned} I(\xi ) &\lesssim \sum_{k=1}^{\infty } \int _{R_{k}(\xi )} \frac{ (2^{k} \tau (\xi ) )^{\alpha /2} \tau (\xi )^{M}}{ (2^{k} \tau (\xi ) )^{M+1}}\,dA(z) \\ &\lesssim \tau (\xi )^{1+\alpha /2} \sum_{k=1}^{\infty } \frac{1}{2^{k(M-2-\alpha /2)}}\lesssim \tau (\xi )^{1+\alpha /2}. \end{aligned} \end{aligned}$$

• If \(\alpha < 0\), then

$$\begin{aligned} \begin{aligned} I(\xi ) &\lesssim \int _{\mathbb{D}\smallsetminus D( \delta _{2}\tau (\xi ))} \tau (z)^{\alpha /2} \frac{ (\min (\tau (\xi ),\tau (z) ) )^{M}}{ \vert z-\xi \vert ^{M+1}}\,dA(z) \\ &\lesssim \sum_{k=1}^{\infty } \int _{R_{k}(\xi )} \frac{\tau (\xi )^{M+\alpha /2}}{ \vert z-\xi \vert ^{M+1}}\,dA(z) \\ &\lesssim \tau (\xi )^{1+\alpha /2} \sum_{k=1}^{\infty } \frac{1}{2^{k(M-1)}} \lesssim \tau (\xi )^{1+\alpha /2}. \end{aligned} \end{aligned}$$

This together with (2.16) establishes (2.11). The proof is complete. □

2.3 Carleson type measures

We are going to define (vanishing) q-Carleson measures for \(A^{p}_{\omega }\), \(0< p, q < \infty \), for weights ω in the class \(\mathcal{W}\) and give some essential theorems.

Definition 2.6

Given \(\omega \in \mathcal{W}\) and \(0< p,q\le \infty \), let μ be a positive measure on \({\mathbb{D}}\). We say that μ is a q-Carleson measure for \(A^{p}_{\omega }\) if there exists a positive constant C such that

$$ \int _{\mathbb{D}}{ \bigl\vert f(z) \bigr\vert ^{q}\,d\mu (z)}\leq C \Vert f \Vert ^{q}_{A^{p}_{\omega }} $$

for all \(f \in A^{p}_{\omega }\). Thus, by the definition, μ is q-Carleson for \(A^{p}_{\omega }\) when the inclusion \(I_{\mu }:A^{p}_{\omega }\longrightarrow L^{q}(\mathbb{D},d\mu )\) is bounded.

Next, the following theorems were essentially proved in [24, Theorem 1]. They established necessary and sufficient conditions for \(I_{\mu }:A^{p}_{\omega }\longrightarrow L^{q}(\mathbb{D},d\mu )\) to be bounded (compact) when \(0< p,q<\infty \).

Theorem B

Given \(\omega \in \mathcal{W}\) and \(0 < p \leq q < \infty \), let μ be a finite positive Borel measure on \(\mathbb{D}\). Then \(I_{\mu }:A^{p}_{\omega }\longrightarrow L^{q}(\mathbb{D},d\mu )\) is bounded if and only if, for each sufficiently small \(\delta > 0\),

$$ K_{\mu ,\omega } : = \sup_{\substack{a\in \mathbb{D}}} \frac{1}{\tau (a)^{2q/p}} \int _{D(\delta \tau (a))}{\omega (\xi )^{-q/2}\,d\mu (\xi )} < \infty . $$
(2.17)

Moreover, in that case, \(K_{\mu ,\omega } \asymp \|I_{\mu }\|^{q}_{A^{p}_{\omega }\rightarrow L^{q}( \mathbb{D},d\mu )}\).

Theorem C

Given \(\omega \in \mathcal{W}\) and \(0 < p \leq q < \infty \), let μ be a finite positive Borel measure on \(\mathbb{D}\). Then \(I_{\mu }:A^{p}_{\omega }\longrightarrow L^{q}(\mathbb{D},d\mu )\) is compact if and only if, for each sufficiently small \(\delta > 0\),

$$ \lim_{r\rightarrow 1^{-}}\sup_{\substack{|a| > r}} \frac{1}{\tau (a)^{2q/p}} \int _{D(\delta \tau (a))}{\omega (\xi )^{-q/2}\,d\mu (\xi )} = 0. $$
(2.18)

Theorem D

Given \(\omega \in \mathcal{W}\) and \(0 < q < p < \infty \), let μ be a finite positive Borel measure on \(\mathbb{D}\). The following conditions are equivalent:

  1. (a)

    \(I_{\mu }: A^{p}_{\omega }\longrightarrow L^{q}(\mathbb{D},d\mu )\) is compact.

  2. (b)

    \(I_{\mu }:A^{p}_{\omega }\longrightarrow L^{q}(\mathbb{D},d\mu )\) is bounded.

  3. (c)

    For each sufficiently small \(\delta > 0\), the function

    $$ F_{\mu }(z)= \frac{1}{\tau (z)^{2}} \int _{D(\delta \tau (z))}{ \omega (\xi )^{-q/2}\,d\mu (\xi )} $$

    belongs to \(L^{\frac{p}{p-q}}(\mathbb{D},dA)\).

Moreover, one has

$$ \Vert I_{\mu } \Vert ^{q}_{A^{p}_{\omega }\rightarrow L^{q}(\mu )}\asymp \Vert F_{\mu } \Vert _{L^{\frac{p}{p-q}}({\mathbb{D}})}. $$

3 Bounded projections

The boundedness of Bergman projection is a fact of fundamental importance. In the case of the unit disc, the boundedness of Bergman projections is studied in [13, 38], and it immediately gives the duality between the Bergman spaces. The natural Bergman projection is not necessarily bounded on \(L^{p}_{\omega }\) unless \(p = 2\) (see [8] and [35] for more details). However, we are going to see next that \(P_{\omega }\) is bounded on \(L^{p}_{\omega }\) for weights ω in the class \(\mathcal{E}\). We first prove the boundedness of the sublinear operator \(P^{+}_{\omega }\) defined as

$$ P^{+}_{\omega }f(z) = \int _{\mathbb{D}}{f(\xi ) \bigl\vert K_{z} (\xi ) \bigr\vert \omega (\xi )\,dA(\xi )}. $$

We mention here that, for the case of the exponential weight with \(\sigma =1\), the results of this section have been obtained recently in [7].

Theorem 3.1

Let \(1 \leq p < \infty \) and \(\omega \in \mathcal{E} \). The operator \(P^{+}_{\omega }\) is bounded on \(L^{p}_{\omega }\) and on \(L^{\infty }(\omega ^{1/2})\).

Proof

We first consider the easiest case \(p= 1\). By Fubini’s theorem and condition (1.2), we obtain

$$\begin{aligned} \begin{aligned} \bigl\Vert P^{+}_{\omega }f \bigr\Vert _{A^{1}_{\omega }} &= \int _{\mathbb{D}}{ \bigl\vert P^{+}_{\omega }f(z) \bigr\vert \omega (z)^{1/2}\,dA(z)} \\ &\leq \int _{\mathbb{D}}{ \int _{\mathbb{D}}{ \bigl\vert f(z) \bigr\vert |K_{z}( \xi })| \omega (\xi ) \omega (z)^{1/2}\,dA(\xi )\,dA(z)} \\ &= \int _{\mathbb{D}}{ \bigl\vert f(\xi ) \bigr\vert \omega (\xi ) \biggl( \int _{ \mathbb{D}}{ \bigl\vert K_{\xi }(z) \bigr\vert \omega (z)^{1/2}\,dA(z)} \biggr)\,dA(\xi )} \\ &\lesssim \int _{\mathbb{D}}{ \bigl\vert f(\xi ) \bigr\vert \omega (\xi )^{1/2}\,dA(\xi )} = \Vert f \Vert _{L^{1}_{\omega }}. \end{aligned} \end{aligned}$$

Next, we consider the case \(1 < p < \infty \). Let \(p'\) denote the conjugate exponent of p. By Hölder’s inequality and (1.2), we get

$$\begin{aligned} \begin{aligned} \bigl\vert P^{+}_{\omega }f(z) \bigr\vert ^{p} &\leq \biggl( \int _{\mathbb{D}}{ \bigl\vert f( \xi ) \bigr\vert ^{p} \bigl\vert K_{z}(\xi ) \bigr\vert \omega (\xi )^{\frac{p+1}{2}}\,dA( \xi )} \biggr) \biggl( \int _{\mathbb{D}}{ \bigl\vert K_{z}(\xi ) \bigr\vert \omega (\xi )^{1/2}\,dA( \xi )} \biggr)^{p-1} \\ &\lesssim \biggl( \int _{\mathbb{D}}{ \bigl\vert f(\xi ) \bigr\vert ^{p} \bigl\vert K_{z}(\xi ) \bigr\vert \omega ( \xi )^{\frac{p+1}{2}}\,dA(\xi )} \biggr) \omega (z)^{-\frac{(p-1)}{2}}. \end{aligned} \end{aligned}$$

This together with Fubini’s theorem and another application of (1.2) gives

$$\begin{aligned} \begin{aligned} \bigl\Vert P^{+}_{\omega }f \bigr\Vert ^{p}_{A^{p}_{\omega }} &= \int _{ \mathbb{D}}{ \bigl\vert P^{+}_{\omega }f(z) \bigr\vert ^{p} \omega (z)^{p/2}\,dA(z)} \\ &\lesssim \int _{\mathbb{D}} \biggl( \int _{\mathbb{D}}{ \bigl\vert f(\xi ) \bigr\vert ^{p} \bigl\vert K_{z}( \xi ) \bigr\vert \omega (\xi )^{\frac{p+1}{2}}\,dA(\xi )} \biggr)\omega (z)^{1/2}\,dA(z) \\ &= \int _{\mathbb{D}}{ \bigl\vert f(\xi ) \bigr\vert ^{p} \omega (\xi )^{\frac{p+1}{2}} \biggl( \int _{\mathbb{D}}{ \bigl\vert K_{\xi }(z) \bigr\vert \omega (z)^{1/2}\,dA(z)} \biggr)\,dA( \xi )} \\ &\lesssim \Vert f \Vert _{L^{p}_{\omega }}^{p}. \end{aligned} \end{aligned}$$

Finally, if \(f \in L^{\infty }(\omega ^{1/2})\), by condition (1.2) we get

$$\begin{aligned} \begin{aligned} \omega (z)^{1/2} \bigl\vert P^{+}_{\omega }(f) (z) \bigr\vert &\leq \omega (z)^{1/2} \int _{\mathbb{D}}{ \bigl\vert f(\xi ) \bigr\vert \bigl\vert K_{z}(\xi ) \bigr\vert \omega (\xi )\,dA(\xi )} \\ &\leq \Vert f \Vert _{L^{\infty }(\omega ^{1/2})} \omega (z)^{1/2} \int _{ \mathbb{D}}{ \bigl\vert K_{z}(\xi ) \bigr\vert \omega (\xi )^{1/2}\,dA(\xi )} \\ &\lesssim \Vert f \Vert _{L^{\infty }(\omega ^{1/2})}. \end{aligned} \end{aligned}$$

This shows that \(P^{+}_{\omega }\) is bounded on \(L^{\infty }(\omega ^{1/2})\). The proof is complete. □

Theorem 3.2

Let \(1 \leq p < \infty \) and \(\omega \in \mathcal{E} \). The Bergman projection \(P_{\omega }: L^{p}_{\omega }\longrightarrow A^{p}_{\omega }\) is bounded. Moreover, \(P_{\omega }: L^{\infty }(\omega ^{1/2})\rightarrow A^{\infty }(\omega ^{1/2})\) is also bounded.

Proof

In view of Theorem 3.1, it remains to see that \(P_{\omega } f\) defines an analytic function on \({\mathbb{D}}\). This follows easily by the density of the polynomials, the boundedness of \(P^{+}_{\omega }\), and the completeness of \(A^{p}_{\omega }\). □

Corollary 3.3

Let \(\omega \in \mathcal{E}\). The reproducing formula \(f=P_{\omega } f\) holds for each \(f\in A^{1}_{\omega }\).

Proof

This is an immediate consequence of the boundedness of the Bergman projection and the density of the polynomials. □

4 Duality

As in the case of the standard Bergman spaces, one can use the result just proved on the boundedness of the Bergman projection \(P_{\omega }\) in \(L^{p}_{\omega }\) to identify the dual spaces of \(A^{p}_{\omega }\). As usual, if X is a Banach space, we denote its dual by \(X^{*}\). Next two results (Theorems 4.1 and 4.2) on the duality of Bergman spaces with exponential type weights appear also on [7].

Theorem 4.1

Let \(\omega \in \mathcal{E}\) and \(1 < p < \infty \). The dual space of \(A^{p}_{\omega }\) can be identified (with equivalent norms) with \(A^{p'}_{\omega }\) under the integral pairing

$$ \langle f, g \rangle _{\omega }= \int _{\mathbb{D}}{f(z) \overline{g(z)} \omega (z)\,dA(z)}. $$

Here \(p'\) denotes the conjugate exponent of p, that is, \(p'= p/(p-1)\).

Proof

Let \(1 < p < \infty \) and let \(p' = p/(p-1)\) be its dual exponent. Given a function \(g\in A^{p'}_{\omega }\), Hölder’s inequality implies that the linear functional \(\psi _{g}: A^{p}_{\omega }\longrightarrow \mathbb{C}\) defined by

$$\begin{aligned} \begin{aligned} \psi _{g}(f):= \int _{\mathbb{D}}{f(\xi ) \overline{g(\xi )} \omega (\xi )\,dA(\xi ),}\quad f\in A^{p}_{\omega }, \end{aligned} \end{aligned}$$

is bounded with \(\|\psi _{g}\|\leq \|g\|_{A^{p'}_{\omega }}\).

Conversely, let \(T \in (A^{p}_{\omega } )^{*}\). By the Hahn–Banach theorem, we can extend T to an element \(\widetilde{T}\in ( L^{p}_{\omega } )^{*}\) such that \(\|\widetilde{T}\|= \|T\|\). By the Riesz representation theorem, there exists \(H \in L^{p'}(\mathbb{D},\omega ^{p/2}\,dA)\) with \(\|H\|_{L^{p'}(\omega ^{p/2})}=\|\widetilde{T}\|= \|T\| \) such that

$$ \widetilde{T}(f)= \int _{\mathbb{D}}{f(\xi ) \overline{H(\xi )} \omega (\xi )^{p/2}\,dA(\xi )} $$

for every \(f\in A^{p}_{\omega }\). Consider the function \(h(\xi ) = H(\xi ) \omega (\xi )^{\frac{p}{2}-1}\). Then \(h\in L^{p'}_{\omega }\) with

$$ \Vert h \Vert _{L^{p'}_{\omega }}= \Vert H \Vert _{L^{p'}(\omega ^{p/2})}= \Vert T \Vert $$

and

$$ T(f) = \widetilde{T}(f) = \int _{\mathbb{D}}{f(\xi ) \overline{h(\xi )} \omega (\xi )\,dA(\xi )},\quad f\in A^{p}_{\omega }. $$

Let \(g= P_{\omega }h\). By Theorem 3.2, the Bergman projection \(P_{\omega }: L^{p'}_{\omega }\longrightarrow A^{p'}_{\omega }\) is bounded. Thus \(g\in A^{p'}_{\omega }\) with

$$ \Vert g \Vert _{A^{p'}_{\omega }} = \Vert P_{\omega }h \Vert _{A^{p'}_{\omega }} \lesssim \Vert h \Vert _{L^{p'}_{\omega }}= \Vert T \Vert . $$

From Fubini’s theorem it is easy to see that \(P_{\omega }\) is self-adjoint. Indeed,

$$\begin{aligned} \begin{aligned} \langle P_{\omega } f, g \rangle _{\omega } &= \int _{ \mathbb{D}}{ P_{\omega }f(\xi ) \overline{g(\xi )} \omega (\xi )\,dA(\xi )} \\ &= \int _{\mathbb{D}}{ \biggl( \int _{\mathbb{D}}{ f(s) \overline{K_{\xi } (s)} \omega (s)\,dA(s)} \biggr) \overline{g(\xi )} \omega (\xi )\,dA(\xi )} \\ &= \int _{\mathbb{D}}{ \biggl( \overline{ \int _{\mathbb{D}}{g(\xi ) \overline{K_{s} (\xi )} \omega (\xi )\,dA(\xi )}} \biggr) f(s) \omega (s)\,dA(s)} \\ &= \int _{\mathbb{D}}{ \overline{ P_{\omega }g(s)} f(s) \omega (s)\,dA(s)} = \langle f, P_{\omega } g \rangle _{\omega }. \end{aligned} \end{aligned}$$

The interchange of the order of integration is well justified, because of the boundedness of the operator \(P^{+}_{\omega }\) (see Theorem 3.1) given by

$$ P^{+}_{\omega } f(z)= \int _{{\mathbb{D}}} f(\zeta ) \bigl\vert K_{z}(\zeta ) \bigr\vert \omega (\zeta )\,dA(\zeta ). $$

Therefore, since \(f = P_{\omega }f\) for every \(f\in A^{p}_{\omega }\), according to Corollary 3.3, we get

$$\begin{aligned} \begin{aligned} T(f) &= \int _{\mathbb{D}}{f(\xi ) \overline{h(\xi )} \omega (\xi )\,dA(\xi )} \\ &=\langle f, P_{\omega } h \rangle _{\omega } = \langle f, g \rangle _{ \omega } = \psi _{g} (f). \end{aligned} \end{aligned}$$

Finally, the function g is unique. Indeed, if there is another function \(\widetilde{g}\in A^{p'}_{\omega }\) with \(T(f) = \psi _{g}(f) = \psi _{\widetilde{g}}(f)\) for every \(f\in A^{p}_{\omega }\), then by taking \(f= K_{a}\) for each \(a\in \mathbb{D}\) (that belongs to \(A^{p}_{\omega }\) due to Lemma 2.4) and using the reproducing formula, we obtain

$$ g(a) = \overline{\psi _{g}(K_{a})} = \overline{\psi _{\widetilde{g}}(K_{a})} = \widetilde{g}(a),\quad a \in \mathbb{D}. $$

Thus, any bounded linear functional T is of the form \(T = \psi _{g}\) for some unique \(g\in A^{p'}_{\omega }\) and, furthermore,

$$ \Vert T \Vert \asymp \Vert g \Vert _{A^{p'}_{\omega }}. $$

The proof is complete. □

Theorem 4.2

Let \(\omega \in \mathcal{E}\). The dual space of \(A^{1}_{\omega }\) can be identified (with equivalent norms) with \(A^{\infty }(\omega ^{1/2})\) under the integral pairing \(\langle f, g \rangle _{\omega }\).

Proof

Let \(g\in A^{\infty }_{\omega }\). The linear functional \(\psi _{g}: A^{1}_{\omega }\longrightarrow \mathbb{C}\) defined by \(\psi _{g}(f):= \langle f, g \rangle _{\omega }\) is bounded with \(\|\psi _{g}\| \leq \|g\|_{L^{\infty }(\omega ^{1/2})}\) since, for every \(f\in A^{1}_{\omega }\),

$$ \bigl\vert \psi _{g}(f) \bigr\vert \leq \Vert g \Vert _{L^{\infty }(\omega ^{1/2})} \Vert f \Vert _{A^{1}_{\omega }}. $$

Conversely, let \(T\in (A^{1}_{\omega } )^{*}\). Consider the space X that consists of the functions of the form \(h = f \omega ^{1/2}\) with \(f\in A^{1}_{\omega }\). Clearly, X is a subspace of \(L^{1}(\mathbb{D}, dA)\) and \(F(h): = T(h\omega ^{-1/2}) = T(f)\) defines a bounded linear functional on X with \(\|F\|= \|T\|\). By the Hahn–Banach theorem, F has an extension \(\widetilde{F}\in (L^{1}(\mathbb{D}, dA) )^{*}\) with \(\|\widetilde{F}\|= \|F\|\). Hence, there is a function \(G\in L^{\infty }(\mathbb{D},dA)\) with \(\|G\|_{L^{\infty }(\mathbb{D},dA)}= \|F\|\) such that

$$ F(h) = \widetilde{F}(h) = \int _{\mathbb{D}}{h(\xi ) \overline{G(\xi )}\,dA(\xi )},\quad h\in X, $$

or

$$ T(f) = \int _{\mathbb{D}}{f(\xi ) \overline{G(\xi )} \omega (\xi )^{1/2}\,dA(\xi )},\quad f\in A^{1}_{\omega }. $$

Consider the function \(H(z) = \omega (z)^{-1/2} G(z)\). Then \(H \in L^{ \infty }(\omega ^{1/2})\) with

$$ \Vert H \Vert _{L^{ \infty }(\omega ^{1/2})}= \Vert G \Vert _{L^{\infty }({\mathbb{D}},dA)} = \Vert F \Vert = \Vert T \Vert . $$

By Theorem 3.2, the function \(g = P_{\omega }H\) is in \(A^{\infty }_{\omega }\) with

$$ \Vert g \Vert _{A^{\infty }(\omega ^{1/2})} \lesssim \Vert H \Vert _{L^{\infty }(\omega ^{1/2})} = \Vert T \Vert . $$

Also, for \(f \in A^{1}_{\omega }\), by the reproducing formula, we have

$$ T(f) = \int _{\mathbb{D}}{f(\xi ) \overline{H(\xi )} \omega (\xi )\,dA(\xi )} = \langle P_{\omega } f, H \rangle _{\omega }= \langle f, P_{\omega }H \rangle _{\omega } = \psi _{g}(f) . $$

Finally, as in the proof of Theorem 4.1, the function g is unique. □

Corollary 4.3

Let \(\omega \in \mathcal{E}\). The set E of finite linear combinations of reproducing kernels is dense in \(A^{p}_{\omega }\), \(1\leq p<\infty \).

Proof

Since E is a linear subspace of \(A^{p}_{\omega }\), by standard functional analysis and the duality results in Theorems 4.1 and 4.2, it is enough to prove that \(g\equiv 0\) if \(g\in A^{p'}_{\omega }\) satisfies \(\langle f,g \rangle _{\omega }=0\) for each f in E (with \(p'\) being the conjugate exponent of p, and \(g\in A^{\infty }(\omega ^{1/2})\) if \(p=1\)). But, taking \(f=K_{z}\) for each \(z\in {\mathbb{D}}\) and using the reproducing formula, we get \(g(z)=P_{\omega } g(z)=\langle g,K_{z} \rangle _{\omega }=0\) for each \(z\in {\mathbb{D}}\). This finishes the proof. □

Our next goal is to identify the predual of \(A^{1}_{\omega }\). For a given weight v, we introduce the space \(A_{0}(v)\) consisting of those functions \(f\in A^{\infty }(v)\) with \(\lim_{|z|\rightarrow 1^{-}} v(z) |f(z)|=0\). Clearly, \(A_{0}(v)\) is a closed subspace of \(A^{\infty }(v)\).

Theorem 4.4

Let \(\omega \in \mathcal{E}\). Under the integral pairing \(\langle f,g \rangle _{\omega }\), the dual space of \(A_{0}(\omega ^{1/2})\) can be identified (with equivalent norms) with \(A^{1}_{\omega }\).

Proof

If \(g\in A^{1}_{\omega }\), clearly \(\Lambda _{g}(f)= \langle f,g\rangle _{\omega }\) defines a bounded linear functional in \(A_{0}(\omega ^{1/2})\) with \(\|\Lambda _{g} \|\le \|g\|_{A^{1}_{\omega }}\). Conversely, assume that \(\Lambda \in (A_{0}(\omega ^{1/2}) )^{*}\). Consider the space X that consists of functions of the form \(h=f\omega ^{1/2}\) with \(f\in A_{0}(\omega ^{1/2})\). Clearly, X is a subspace of \(C_{0}({\mathbb{D}})\) (the space of all continuous functions vanishing at the boundary) and \(T(h)=\Lambda (\omega ^{-1/2} h)=\Lambda (f)\) defines a bounded linear functional on X with \(\|T\|=\|\Lambda \|\). By the Hahn–Banach theorem, T has an extension \(\widetilde{T}\in (C_{0}({\mathbb{D}}) )^{*}\) with \(\|\widetilde{T}\|=\|T\|\). Hence, by the Riesz representation theorem, there is a measure \(\mu \in \mathcal{M}({\mathbb{D}})\) (the Banach space of all complex Borel measures μ equipped with the variation norm \(\|\mu \|_{{\mathcal{M}}}\)) with \(\|\mu \|_{{\mathcal{M}}}=\|T\|\) such that

$$ T(h)=\widetilde{T}(h)= \int _{{\mathbb{D}}} h(\zeta )\,d\mu (\zeta ), \quad h\in X, $$

or

$$ \Lambda (f)= \int _{{\mathbb{D}}} f(\zeta ) \omega (\zeta )^{1/2}\,d\mu ( \zeta ),\quad f\in A_{0}\bigl(\omega ^{1/2}\bigr). $$

Consider the function g defined on the unit disk by

$$ \overline{g(z)}= \int _{{\mathbb{D}}} K_{z}(\zeta ) \omega (\zeta )^{1/2}\,d\mu (\zeta ), \quad z\in {\mathbb{D}}. $$

Clearly, g is analytic on \({\mathbb{D}}\) and, by Fubini’s theorem and condition (1.2), we have

$$\begin{aligned} \begin{aligned} \Vert g \Vert _{A^{1}_{\omega }} &\le \int _{{\mathbb{D}}} \biggl( \int _{{\mathbb{D}}} \bigl\vert K_{z}(\zeta ) \bigr\vert \omega (\zeta )^{1/2}\,d\vert \mu \vert ( \zeta ) \biggr) \omega (z)^{1/2}\,dA(z) \\ & = \int _{{\mathbb{D}}} \biggl( \int _{{\mathbb{D}}} \bigl\vert K_{\zeta }(z) \bigr\vert \omega (z)^{1/2}\,dA(z) \biggr) \omega (\zeta )^{1/2}\,d\vert \mu \vert ( \zeta ) \\ &\lesssim \vert \mu \vert ({\mathbb{D}})= \Vert \mu \Vert _{{\mathcal{M}}}= \Vert \Lambda \Vert , \end{aligned} \end{aligned}$$

proving that g belongs to \(A^{1}_{\omega }\). Now, since \(A_{0}(\omega ^{1/2})\subset A^{2}_{\omega }\), the reproducing formula \(f(\zeta )=\langle f, K_{\zeta } \rangle _{\omega }\) holds for all \(f\in A_{0}(\omega ^{1/2})\). This and Fubini’s theorem yield

$$\begin{aligned} \begin{aligned} \Lambda _{g}(f)=\langle f,g \rangle _{\omega }&= \int _{{ \mathbb{D}}} f(z) \biggl( \int _{{\mathbb{D}}} K_{z}(\zeta ) \omega ( \zeta )^{1/2}\,d\mu (\zeta ) \biggr) \omega (z)\,dA(z) \\ &= \int _{{\mathbb{D}}} \biggl( \int _{{\mathbb{D}}} f(z) \overline{K_{\zeta }(z)} \omega (z)\,dA(z) \biggr)\omega (\zeta )^{1/2}\,d\mu (\zeta ) \\ &= \int _{{\mathbb{D}}} f(\zeta ) \omega (\zeta )^{1/2}\,d\mu (\zeta )= \Lambda (f). \end{aligned} \end{aligned}$$

By the reproducing formula, the function g is uniquely determined by the identity \(g(z)=\overline{\Lambda (K_{z})}\). This completes the proof. □

For the case of normal weights, the analogues of Theorems 4.2 and 4.4 were obtained by Shields and Williams in [32]. They also asked what happens with the exponential weights, problem that is solved in the present paper.

5 Atomic decomposition

For \(1\le p < \infty \), in this section we are going to obtain an atomic decomposition for the large weighted Bergman space \(A^{p}_{\omega }\), that is, we show that every function in the Bergman spaces \(A^{p}_{\omega }\) with ω in the class \(\mathcal{E}\) can be decomposed into a series of kernel functions. With the help of the duality results and the estimates for the p-norm of the reproducing kernels \(K_{z}\), we can reach our goal. Before stating the main theorem of this section, we need two auxiliary lemmas as follows.

Lemma 5.1

Let \(\omega \in \mathcal{W}\). There is a sequence \(\lbrace z_{n}\rbrace \subset {\mathbb{D}}\) such that

$$ \sum_{k = 1}^{\infty } \bigl\vert f(z_{k}) \bigr\vert ^{p} \omega (z_{k})^{p/2} \tau (z_{k})^{2} \gtrsim \Vert f \Vert ^{p}_{A^{p}_{\omega }} $$

for all \(f \in A^{p}_{\omega }\) and \(1\le p <\infty \).

Proof

Let \(\{z_{k}\}\) be a \((\varepsilon , \tau )\)-lattice on \({\mathbb{D}}\) (that exists by Lemma B) with \(\varepsilon >0\) small enough to be specified later. Let \(f \in A^{p}_{\omega } \). We consider

$$ I_{f}(k) := \sum_{k = 1}^{\infty } \bigl\vert f(z_{k}) \bigr\vert ^{p} \omega (z_{k})^{p/2} \tau (z_{k})^{2} . $$

We have

$$\begin{aligned} \begin{aligned} \Vert f \Vert ^{p}_{A^{p}_{\omega }}&= \int _{\mathbb{D}}{ \bigl\vert f(z) \bigr\vert ^{p} \omega (z)^{p/2}\,dA(z)} \\ &\leq C \Biggl[\sum_{k = 1}^{\infty } \int _{D(\varepsilon \tau (z_{k}))}{ \bigl( \bigl\vert f(z) \bigr\vert \omega (z)^{1/2}- \bigl\vert f(z_{k}) \bigr\vert \omega (z_{k})^{1/2} \bigr)^{p}\,dA(z)} + C \varepsilon ^{2} I_{f}(k) \Biggr]. \end{aligned} \end{aligned}$$

For \(z\in D(\varepsilon \tau (z_{k}))\), there exists \(\xi _{k,z}\in [z,z_{k}]\) such that

$$\begin{aligned} \begin{aligned} \bigl( \bigl\vert f(z) \bigr\vert \omega (z)^{1/2}- \bigl\vert f(z_{k}) \bigr\vert \omega (z_{k})^{1/2} \bigr)^{p} & \leq \bigl\vert \nabla \bigl( \vert f \vert \omega ^{1/2}\bigr) (\xi _{k,z}) \bigr\vert ^{p} \vert z - z_{k} \vert ^{p} \\ &\leq \varepsilon ^{p} \tau (z_{k})^{p} \bigl\vert \nabla \bigl( \vert f \vert \omega ^{1/2}\bigr) ( \xi _{k,z}) \bigr\vert ^{p} . \end{aligned} \end{aligned}$$

This together with Lemma 2.2, with \(\delta _{0}\in (0,m_{\tau })\) fixed, yields

$$\begin{aligned} \begin{aligned} &\int _{D(\varepsilon \tau (z_{k}))} \bigl( \bigl\vert f(z) \bigr\vert \omega (z)^{1/2}- \bigl\vert f(z_{k}) \bigr\vert \omega (z_{k})^{1/2} \bigr)^{p}\,dA(z) \\ &\quad \leq C \varepsilon ^{p} \tau (z_{k})^{p} \int _{D(\varepsilon \tau (z_{k}))} \biggl(\frac{1}{\tau (\xi _{k,z})^{p+2}} \int _{D( \delta _{0}\tau (\xi _{k,z}))}{ \bigl\vert f(\eta ) \bigr\vert ^{p} \omega (\eta )^{p/2}\,dA( \eta )} \biggr)\,dA(z). \end{aligned} \end{aligned}$$

Using that \(\tau (\xi _{k,z}) \asymp \tau (z_{k})\) and \(D(\delta _{0}\tau (\xi _{k,z}))\subset D(3\delta _{0} \tau (z_{k}))\) for \(z\in D(\varepsilon \tau (z_{k}))\), we obtain

$$\begin{aligned} \begin{aligned} &\int _{D(\varepsilon \tau (z_{k}))} \bigl( \bigl\vert f(z) \bigr\vert \omega (z)^{1/2}- \bigl\vert f(z_{k}) \bigr\vert \omega (z_{k})^{1/2} \bigr)^{p}\,dA(z) \\ &\quad \leq C \varepsilon ^{p+2} \biggl( \int _{D(3\delta _{0}\tau (z_{k}))}{ \bigl\vert f( \eta ) \bigr\vert ^{p} \omega (\eta )^{p/2}\,dA(\eta )} \biggr). \end{aligned} \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned} \Vert f \Vert ^{p}_{A^{p}_{\omega }} &\leq C \varepsilon ^{p+2} \sum_{k = 1}^{\infty } \int _{D(3\delta _{0}\tau (z_{k}))}{ \bigl\vert f(\eta ) \bigr\vert ^{p} \omega (\eta )^{p/2}\,dA(\eta )} + C \varepsilon ^{2} I_{f}(k). \end{aligned} \end{aligned}$$

By Lemma B, every point \(z\in \mathbb{D}\) belongs to at most \(C\varepsilon ^{-2}\) of the sets \(D(3\delta _{0}\tau (z_{k}))\), and therefore

$$\begin{aligned} \begin{aligned} \Vert f \Vert ^{p}_{A^{p}_{\omega }} &\le C \varepsilon ^{p} \Vert f \Vert ^{p}_{A^{p}_{\omega }} + C \varepsilon ^{2} I_{f}(n). \end{aligned} \end{aligned}$$

Thus, taking \(\varepsilon >0\) so that \(C\varepsilon ^{p} < 1/2\), we get the desired result. □

We may rephrase Lemma 5.1 by saying that every \((\varepsilon ,\tau )\)-lattice, with \(\varepsilon >0\) small enough, is a sampling sequence for the Bergman space \(A^{p}_{\omega }\). Recall that \(\{z_{k}\}\subset {\mathbb{D}}\) is a sampling sequence for the Bergman space \(A^{p}_{\omega }\) if

$$ \Vert f \Vert ^{p}_{A^{p}_{\omega }} \asymp \sum _{k} \bigl\vert f(z_{k}) \bigr\vert ^{p} \omega (z_{k})^{p/2} \tau (z_{k})^{2} $$

for any \(f\in A^{p}_{\omega }\). Just note that Lemma 5.1 gives one inequality, and the other follows by standard methods using Lemma A and the lattice properties. Sampling sequences on the classical Bergman space were characterized by K. Seip [30] (see also the monographs [10] and [31]). For sampling sequences on large weighted Bergman spaces, we refer to [4].

Lemma 5.2

Let \(\omega \in \mathcal{E}\), \(1 \le p < \infty \), and \((z_{k})_{k\in \mathbb{N}}\subset {\mathbb{D}}\) be the sequence defined in Lemma B. The function given by

$$ F(z): = \sum_{\substack{k = 1}}^{\infty } \lambda _{k} \omega (z_{k})^{1/2} \tau (z_{k})^{2(\frac{p-1}{p})}K_{z_{k}}(z) $$
(5.1)

belongs to \(A^{p}_{\omega }\) for every sequence \(\lambda =\lbrace \lambda _{k}\rbrace \in \ell ^{p}\). Moreover,

$$ \Vert F \Vert _{A^{p}_{\omega }}\lesssim \Vert \lambda \Vert _{\ell ^{p}}. $$

Proof

By Hölder’s inequality, Lemma A, and Lemma 2.4, it is easy to see that the partial sums of the series in (5.1) converge uniformly on compact subsets of \({\mathbb{D}}\). Thus, F defines an analytic function on \({\mathbb{D}}\). Furthermore, for p = 1, using (1.2) we have

$$ \Vert F \Vert _{A^{1}_{\omega }}\le \sum_{k=1}^{\infty } \vert \lambda _{k} \vert \omega (z_{k})^{1/2} \Vert K_{z_{k}} \Vert _{A^{1}_{\omega }}\lesssim \Vert \lambda \Vert _{\ell ^{1}}. $$

For the case \(p > 1\), consider

$$ M(z): = \sum_{\substack{k = 1}}^{\infty }\tau (z_{k})^{2}\omega (z_{k})^{1/2} \bigl\vert K_{z_{k}}(z) \bigr\vert . $$

By Hölder’s inequality, we have

$$\begin{aligned} \begin{aligned} \Vert F \Vert _{A^{p}_{\omega }}^{p}&\leq \int _{{\mathbb{D}}}{ \Biggl(\sum_{\substack{k = 0}}^{\infty } \vert \lambda _{k} \vert \omega (z_{k})^{1/2} \tau (z_{k})^{2(\frac{p-1}{p})} \bigl\vert K_{z_{k}}(z) \bigr\vert \Biggr)^{p} \omega (z)^{p/2}\,dA(z)}. \\ & \lesssim \int _{{\mathbb{D}}}{ \Biggl(\sum_{ \substack{k = 1}}^{\infty } \vert \lambda _{k} \vert ^{p}\omega (z_{k})^{1/2} \bigl\vert K_{z_{k}}(z) \bigr\vert \Biggr) M(z)^{p-1} \omega (z)^{p/2}\,dA(z)}. \end{aligned} \end{aligned}$$

On the other hand, using Lemma A, Lemma B, and (1.2), we have

$$\begin{aligned} M(z): &= \sum_{\substack{k = 1}}^{\infty } \tau (z_{k})^{2} \omega (z_{k})^{1/2} \bigl\vert K_{z_{k}}(z) \bigr\vert \\ &\lesssim \sum_{\substack{k=1}}^{\infty } \int _{D(\delta \tau ((z_{k}))}{ \bigl\vert K_{z}( \xi ) \bigr\vert \omega (\xi )^{1/2}\,dA(\xi )} \\ &\lesssim \int _{{\mathbb{D}}}{ \bigl\vert K_{z}(\xi ) \bigr\vert \omega (\xi )^{1/2}\,dA( \xi )} \lesssim \omega (z)^{-1/2}. \end{aligned}$$

Therefore, applying again (1.2), we obtain

$$\begin{aligned} \begin{aligned} \Vert F \Vert _{A^{p}_{\omega }}^{p} & \lesssim \int _{{\mathbb{D}}}{ \Biggl(\sum_{\substack{k = 0}}^{\infty } \vert \lambda _{k} \vert ^{p}\omega (z_{k})^{1/2} \bigl\vert K_{z_{k}}(z) \bigr\vert \Biggr)\omega (z)^{1/2}\,dA(z)} \\ &\lesssim \sum_{\substack{k=1}}^{\infty } \vert \lambda _{k} \vert ^{p}\omega (z_{k})^{1/2} \int _{{\mathbb{D}}}{ \bigl\vert K_{z_{k}}(z) \bigr\vert \omega (z)^{1/2}\,dA(z)} \\ &\lesssim \sum_{\substack{k=1}}^{\infty } \vert \lambda _{k} \vert ^{p}, \end{aligned} \end{aligned}$$

which completes the proof. □

Now we are ready to state our main result related to an atomic decomposition of large weighted Bergman spaces \(A^{p}_{\omega }\) for \(1\le p <\infty \). Recall that \(k_{p,z}\) is the normalized reproducing kernel in \(A^{p}_{\omega }\), that is,

$$ k_{p,z}= K_{z} / \Vert K_{z} \Vert _{A^{p}_{\omega }}, \quad z\in {\mathbb{D}}. $$

Theorem 5.3

Let \(\omega \in \mathcal{E}\) and \(1 \le p < \infty \). There exists a τ-lattice \(\lbrace z_{n} \rbrace \subset \mathbb{D}\) such that:

  1. (i)

    For any \(\lambda =\{\lambda _{n}\}\in \ell ^{p}\), the function

    $$ f(z)=\sum_{n} \lambda _{n} k_{p,z_{n}}(z) $$

    is in \(A^{p}_{\omega }\) with \(\|f\|_{A^{p}_{\omega }}\le C \|\lambda \|_{\ell ^{p}}\).

  2. (ii)

    For every \(f\in A^{p}_{\omega }\), there exists \(\lambda =\lbrace \lambda _{n} \rbrace \in \ell ^{p} \) such that

    $$ f(z)=\sum_{n} \lambda _{n} k_{p,z_{n}}(z) $$

    and \(\|\lambda \|_{\ell ^{p}}\le C \|f\|_{A^{p}_{\omega }}\).

Proof

On the one hand, statement (i) is exactly Lemma 5.2. On the other hand, in order to prove (ii), we define a linear operator \(S: \ell ^{p} \longrightarrow A^{p}_{\omega }\) given by

$$ S\bigl(\lbrace \lambda _{n} \rbrace \bigr): = \sum _{n = 0}^{\infty }\lambda _{n} k_{p,z_{n}}. $$

By (i), the operator S is bounded. By the duality results obtained in the previous section, when \(1< p<\infty \), the adjoint operator \(S^{*}:A^{p'}_{\omega }\rightarrow \ell ^{p'}\), where \(p'\) is the conjugate exponent of p, is defined by

$$ \langle Sx,f\rangle _{\omega } = \bigl\langle x,S^{*}f\bigr\rangle _{\ell }= \sum_{n} x_{n} \overline{\bigl(S^{*} f\bigr)_{n}} $$

for every \(x\in \ell ^{p}\) and \(f \in A^{p'}_{\omega }\). To compute \(S^{*}\), let \(e_{n}\) denote the vector that equals 1 at the nth coordinate and equals 0 at the other coordinates. Then \(Se_{n} = k_{p,z_{n}} \), and using the reproducing formula, we get

$$\begin{aligned} \begin{aligned} \overline{\bigl(S^{*}f\bigr)_{n}} & =\bigl\langle e_{n},S^{*}f\bigr\rangle _{ \ell }= \langle Se_{n},f\rangle _{\omega } = \langle k_{p,z_{n}}, f \rangle _{\omega } \\ &= \Vert K_{z_{n}} \Vert _{A^{p}_{\omega }}^{-1}\cdot \langle K_{z_{n}}, f \rangle _{\omega } = \frac{\overline{f(z_{n})}}{ \Vert K_{z_{n}} \Vert _{A^{p}_{\omega }}}. \end{aligned} \end{aligned}$$

Hence, \(S^{*}: A^{p'}_{\omega }\longrightarrow \ell ^{p'} \) is given by

$$ S^{*}f = \bigl\lbrace \bigl(S^{*}f\bigr)_{n} \bigr\rbrace = \biggl\lbrace \frac{f(z_{n})}{ \Vert K_{z_{n}} \Vert _{A^{p}_{\omega }}} \biggr\rbrace _{n}. $$

We must prove that S is surjective in order to finish the proof of this case. By a classical result in functional analysis, it is enough to show that \(S^{*}\) is bounded below. By Lemma 5.1 and Lemma 2.4, we obtain

$$ \bigl\Vert S^{*}f \bigr\Vert ^{p'}_{\ell ^{p'}} \asymp \sum_{n = 1}^{\infty } \bigl\vert f(z_{n}) \bigr\vert ^{p'} \omega (z_{n})^{p'/2} \tau (z_{n})^{2} \gtrsim \Vert f \Vert ^{p'}_{A^{p'}_{\omega }}, $$

which shows that \(S^{*}\) is bounded below. Finally, once the surjectivity is proved, the estimate \(\|\lambda \|_{\ell ^{p}}\lesssim \|f\|^{p}_{A^{p}_{\omega }}\) is a standard application of the open mapping theorem. When \(p=1\), then \(S^{*}:A^{\infty }(\omega ^{1/2})\longrightarrow \ell ^{\infty }\) is given by

$$ \bigl\lbrace \bigl(S^{*}f\bigr)_{n} \bigr\rbrace = \biggl\lbrace \frac{f(z_{n})}{ \Vert K_{z_{n}} \Vert _{A^{1}_{\omega }}} \biggr\rbrace _{n}. $$

Hence we must show that

$$ \sup_{z\in {\mathbb{D}}} \omega (z)^{1/2} \bigl\vert f(z) \bigr\vert = \Vert f \Vert _{A^{\infty }( \omega ^{1/2})}\lesssim \bigl\Vert S^{*} f \bigr\Vert _{\ell ^{\infty }} \asymp \sup_{n} \omega (z_{n}) ^{1/2} \bigl\vert f(z_{n}) \bigr\vert $$

for \(f\in A^{\infty }(\omega ^{1/2})\). However, this can be proved with the same method as Lemma 5.1. Indeed, let \(z\in {\mathbb{D}}\). Then there is a point \(z_{n}\) with \(z\in D(\varepsilon \tau (z_{n}))\). By Lemma A, we have

$$ \begin{aligned} \omega (z)^{1/2} \bigl\vert f(z) \bigr\vert \le{}& \frac{C_{1}}{\varepsilon ^{2} \tau (z)^{ 2} } \int _{D( \varepsilon \tau (z))} \bigl( \bigl\vert f(\zeta ) \bigr\vert \omega ( \zeta )^{1/2}- \bigl\vert f(z_{n}) \bigr\vert \omega (z_{n})^{1/2} \bigr)\,dA(\zeta ) \\ & {}+ C_{1} \bigl\vert f(z_{n}) \bigr\vert \omega (z_{n})^{1/2}. \end{aligned} $$

As done in the proof of Lemma 5.1, we have

$$ \begin{aligned} \bigl\vert \bigl\vert f(\zeta ) \bigr\vert \omega (\zeta )^{1/2}- \bigr\vert f(z_{n}) \bigl\vert \omega (z_{n})^{1/2} \bigr\vert & \le C_{2} \varepsilon \frac{1}{\tau (z_{n})^{2}} \int _{D(3\delta _{0} \tau (z_{n}))} \bigl\vert f( \xi ) \bigr\vert \omega (\xi )^{1/2}\,dA(\xi ) \\ & \le C_{3} \varepsilon \Vert f \Vert _{A^{\infty }(\omega ^{1/2})}. \end{aligned} $$

Thus, putting this in the previous estimate, we obtain

$$ \omega (z)^{1/2} \bigl\vert f(z) \bigr\vert \le C_{4} \varepsilon \Vert f \Vert _{A^{\infty }( \omega ^{1/2})}+C_{1}\sup _{n} \omega (z_{n}) ^{1/2} \bigl\vert f(z_{n}) \bigr\vert . $$

Finally, taking the supremum on z and \(\varepsilon >0\) small enough so that \(C_{4} \varepsilon \le 1/2\), we have

$$ \Vert f \Vert _{A^{\infty }(\omega ^{1/2})}\lesssim \sup_{n} \omega (z_{n}) ^{1/2} \bigl\vert f(z_{n}) \bigr\vert . $$

The proof is complete. □

6 Toeplitz operators

In this section we are going to extend the results given in [3, Theorem 1.1] to the non-Hilbert space setting, when the weight ω is in the class \(\mathcal{E}\). Concretely, we characterize the bounded and compact operators \(T_{\mu }\) acting from \(A^{p}_{\omega }\) to \(A^{q}_{\omega }\) when \(1\le p,q<\infty \). Recall that the Toeplitz operator \(T_{\mu }\) is defined by

$$ T_{\mu }f(z) = \int _{\mathbb{D}}{f(\xi ) \overline{K_{z}(\xi )} \omega ( \xi )\,d\mu (\xi )}. $$

Note that \(T_{\mu }\) is very loosely defined here, because it is not clear when the integrals above will converge, even if the measure μ is finite. We suppose that μ is a finite positive Borel measure that satisfies the condition

$$ \begin{aligned} \int _{{\mathbb{D}}} \bigl\vert K_{z}(\xi ) \bigr\vert ^{2} \omega (\xi )\,d\mu (\xi ) < \infty . \end{aligned} $$
(6.1)

Then the Toeplitz operator \(T_{\mu }\) is well defined on a dense subset of \(A^{p}_{\omega }\), \(1\le p < \infty \). In fact, by Corollary 4.3 and Theorem 5.3, the set E of finite linear combinations of reproducing kernels is dense in \(A^{p}_{\omega }\). Therefore, it follows from condition (6.1) and the Cauchy–Schwarz inequality that \(T_{\mu }(f)\) is well defined for any \(f\in E\). Also, recall that, for \(\delta \in (0,m_{\tau })\), the averaging function of μ on \({\mathbb{D}}\) is given by

$$ \widehat{\mu }_{\delta }(z) : = \frac{\mu (D(\delta \tau (z))}{\tau (z)^{2}},\quad z\in {\mathbb{D}}. $$

Theorem 6.1

Let \(\omega \in \mathcal{E}\), \(1\le p\leq q < \infty \), and μ be a finite positive Borel measure on \(\mathbb{D}\) satisfying (6.1). Then \(T_{\mu }: A^{p}_{\omega }\rightarrow A^{q}_{\omega }\) is bounded if and only if, for each \(\delta \in (0,m_{\tau })\) sufficiently small,

$$ E(\mu )=\sup_{\substack{z\in {\mathbb{D}}}}\tau (z)^{2(\frac{1}{q}- \frac{1}{p})} \widehat{\mu }_{\delta }(z)< \infty . $$
(6.2)

Moreover,

$$ \Vert T_{\mu } \Vert _{A^{p}_{\omega }\rightarrow A^{q}_{\omega }}\asymp E(\mu ). $$

Proof

Since we have the estimate \(\|K_{z}\|_{A^{p}_{\omega }}\asymp \omega (z)^{-1/2} \tau (z)^{-2(p-1)/p}\), if we assume that the Toeplitz operator \(T_{\mu }: A^{p}_{\omega }\longrightarrow A^{q}_{\omega }\) is bounded, then we obtain (6.2) with the same argument as in the proof of Theorem 1.1 in [3].

Conversely, we suppose that (6.2) holds. We first prove that

$$ \int _{\mathbb{D}}{ \bigl\vert K_{z}(\xi ) \bigr\vert \omega (\xi )^{1/2} \tau (\xi )^{-2(\frac{1}{p}-\frac{1}{q})}\,d\mu (\xi ) \lesssim E( \mu ) \omega (z)^{-1/2}}. $$
(6.3)

Indeed, by Lemma A, we have

$$ \bigl\vert K_{z}(\xi ) \bigr\vert \omega (\xi )^{1/2} \lesssim \frac{1}{\tau (\xi )^{2}} \int _{D(\frac{\delta }{2}\tau (\xi ))}{ \bigl\vert K_{z}(s) \bigr\vert \omega (s)^{1/2}\,dA(s)}. $$

Then, by Fubini’s theorem, the fact that \(\tau (s)\asymp \tau (\xi )\) for \(s\in D(\delta \tau (\xi ))\), and condition (1.2), we get

$$\begin{aligned} \begin{aligned} \int _{\mathbb{D}} \bigl\vert K_{z}(\xi ) \bigr\vert \omega (\xi )^{1/2} \tau (\xi )^{-2(\frac{1}{p}-\frac{1}{q})}\,d\mu (\xi ) &\lesssim \int _{ \mathbb{D}}{ \bigl\vert K_{z}(s) \bigr\vert \omega (s)^{1/2} \tau (s)^{2(\frac{1}{q}- \frac{1}{p})} \widehat{\mu }_{\delta }(s)\,dA(s)} \\ &\lesssim E(\mu ) \int _{\mathbb{D}}{ \bigl\vert K_{z}(s) \bigr\vert \omega (s)^{1/2}\,dA(s)} \\ &\lesssim E(\mu ) \omega (z)^{-1/2}. \end{aligned} \end{aligned}$$

This establishes (6.3). Now we proceed to prove that \(T_{\mu }\) is bounded. If \(q > 1\), by Hölder’s inequality, we obtain

$$\begin{aligned} \begin{aligned} \bigl\vert T_{\mu }f(z) \bigr\vert ^{q}\leq{}& \biggl( \int _{\mathbb{D}}{ \bigl\vert f( \xi ) \bigr\vert \bigl\vert K_{z}(\xi ) \bigr\vert \omega (\xi )\,d\mu (\xi )} \biggr)^{q} \\ \leq{}& \biggl( \int _{\mathbb{D}}{ \bigl\vert f(\xi ) \bigr\vert ^{q} \omega (\xi )^{ \frac{q+1}{2}} \bigl\vert K_{z}(\xi ) \bigr\vert \frac{d\mu (\xi )}{\tau (\xi )^{2(\frac{1}{p}-\frac{1}{q})(1-q)}}} \biggr) \\ &{}\times \biggl( \int _{\mathbb{D}}{ \bigl\vert K_{z}(\xi ) \bigr\vert \omega (\xi )^{1/2} \frac{d\mu (\xi )}{\tau (\xi )^{2(\frac{1}{p}-\frac{1}{q})}}} \biggr)^{q-1} . \end{aligned} \end{aligned}$$

Using (6.3), we have

$$ \bigl\vert T_{\mu }f(z) \bigr\vert ^{q}\lesssim E(\mu )^{q-1} \biggl( \int _{\mathbb{D}}{ \bigl\vert f( \xi ) \bigr\vert ^{q} \omega (\xi )^{\frac{q+1}{2}} \bigl\vert K_{z}(\xi ) \bigr\vert \tau ( \xi )^{2(\frac{1}{p}-\frac{1}{q})(q-1)}\,d\mu (\xi )} \biggr) \omega (z)^{-\frac{(q-1)}{2}}. $$

If \(q=1\), this holds directly. By Fubini’s theorem and condition (1.2), we obtain

$$\begin{aligned} \begin{aligned} \Vert T_{\mu }f \Vert ^{q}_{A^{q}_{\omega }} &= \int _{\mathbb{D}} { \bigl\vert T_{\mu }f(z) \bigr\vert ^{q} \omega (z)^{q/2}\,dA(z)} \\ &\lesssim E(\mu )^{q-1} \int _{\mathbb{D}}{ \bigl\vert f(\xi ) \bigr\vert ^{q} \omega (\xi )^{ \frac{(q+1)}{2}} \tau (\xi )^{2(q-1)(\frac{1}{p}-\frac{1}{q})} \biggl( \int _{\mathbb{D}}{ \bigl\vert K_{\xi }(z) \bigr\vert \omega (z)^{1/2}\,dA(z)} \biggr)\,d\mu (\xi )} \\ &\lesssim E(\mu )^{q-1} \int _{\mathbb{D}} { \bigl\vert f(\xi ) \bigr\vert ^{q} \omega ( \xi )^{q/2} \tau (\xi )^{2(q-1)(\frac{1}{p}-\frac{1}{q})}\,d\mu (\xi )}. \end{aligned} \end{aligned}$$

Consider the measure ν given by

$$ d\nu (\xi ) := \omega (\xi )^{q/2} \tau (\xi )^{2(q-1)(\frac{1}{p}- \frac{1}{q})}\,d\mu (\xi ). $$

Since (6.2) holds, by Theorem B, the identity \(I_{\nu } : A^{p}_{\omega }\longrightarrow L^{q}(\mathbb{D},d\nu )\) is bounded. Moreover, \(\|I_{\nu }\|\lesssim E(\mu )^{1/q}\). Therefore,

$$ \Vert T_{\mu }f \Vert ^{q}_{A^{q}_{\omega }} \lesssim E(\mu )^{q-1} \int _{ \mathbb{D}}{ \bigl\vert f(z) \bigr\vert ^{q}\,d\nu (z)}\lesssim E(\mu )^{q}\cdot \Vert f \Vert ^{q}_{A^{p}_{\omega }}. $$
(6.4)

This finishes the proof. □

In order to describe the boundedness of \(T_{\mu }:A^{p}_{\omega }\rightarrow A^{q}_{\omega }\) when \(1\le q< p<\infty \), we need first an auxiliary result.

Proposition 6.2

Let \(\omega \in \mathcal{E}\) and \(1 < q < p < \infty \). If \(\widehat{\mu }_{\delta }\in L^{\frac{pq}{p-q}}(\mathbb{D},dA) \), then

$$ \begin{aligned}J_{\delta ,q} &: = \int _{\mathbb{D}}{ \biggl(\frac{1}{\tau (z)^{2}} \int _{D(\delta \tau (z))}{ \bigl\vert f(\xi ) \bigr\vert \omega (\xi )^{1/2}\,d\mu (\xi )} \biggr)^{q}\,dA(z)} \\ &\lesssim \Vert \widehat{\mu }_{\delta } \Vert _{L^{ \frac{pq}{p-q}}(\mathbb{D)}}^{q}\cdot \Vert f \Vert _{A^{p}_{\omega }}^{q} \end{aligned}$$

for any \(f\in A^{p}_{\omega }\).

Proof

For \(z\in {\mathbb{D}}\) and \(\xi \in D(\delta \tau (z))\), by Lemma A, Lemma B, and (2.1), we obtain

$$\begin{aligned} \begin{aligned} \bigl\vert f(\xi ) \bigr\vert \omega (\xi )^{1/2} &\lesssim \biggl( \frac{1}{\tau (\xi )^{2}} \int _{D(\frac{\delta }{3}\tau (\xi ))}{ \bigl\vert f(s) \bigr\vert ^{p} \omega (s)^{p/2}\,dA(s)} \biggr)^{1/p} \\ &\lesssim \biggl(\frac{1}{\tau (z)^{2}} \int _{D(\delta \tau (z))}{ \bigl\vert f(s) \bigr\vert ^{p} \omega (s)^{p/2}\,dA(s)} \biggr)^{1/p}. \end{aligned} \end{aligned}$$

This gives

$$ \begin{aligned}&\frac{1}{\tau (z)^{2}} \int _{D(\delta \tau (z))}{ \bigl\vert f(\xi ) \bigr\vert \omega ( \xi )^{1/2}\,d\mu (\xi )}\\ &\quad \lesssim \widehat{\mu }_{\delta }(z) \biggl( \frac{1}{\tau (z)^{2}} \int _{D(\delta \tau (z))}{ \bigl\vert f(s) \bigr\vert ^{p} \omega (s)^{p/2}\,dA(s)} \biggr)^{1/p}. \end{aligned}$$

Therefore,

$$ J_{\delta ,q}\lesssim \int _{\mathbb{D}} \biggl(\frac{1}{\tau (z)^{2}} \int _{D(\delta \tau (z))}{ \bigl\vert f(s) \bigr\vert ^{p} \omega (s)^{p/2}\,dA(s)} \biggr)^{q/p} \widehat{\mu }_{\delta }(z)^{q}\,dA(z). $$

Applying Hölder’s inequality, we get

$$ J_{\delta ,q} \lesssim \biggl( \int _{\mathbb{D}}\frac{1}{\tau (z)^{2}} \int _{D(\delta \tau (z))}{ \bigl\vert f(s) \bigr\vert ^{p} \omega (s)^{p/2}\,dA(s)}\,dA(z) \biggr)^{q/p} \Vert \widehat{\mu }_{\delta } \Vert _{L^{\frac{pq}{p-q}}( \mathbb{D)}}^{q}. $$
(6.5)

On the other hand, by Fubini’s theorem and \(\tau (z)\asymp \tau (s)\), for \(s\in D(\delta \tau (z))\), we have

$$\begin{aligned} \begin{aligned} \int _{\mathbb{D}} \biggl(\frac{1}{\tau (z)^{2}} \int _{D( \delta \tau (z))} \bigl\vert f(s) \bigr\vert ^{p} \omega (s)^{p/2}\,dA(s) \biggr)\,dA(z) \lesssim \Vert f \Vert ^{p}_{A^{p}_{\omega }}. \end{aligned} \end{aligned}$$

Combining this with (6.5), we get

$$ J_{\delta ,q} \lesssim \Vert f \Vert _{A^{p}_{\omega }}^{q} \cdot \Vert \widehat{\mu }_{\delta } \Vert _{L^{\frac{pq}{p-q}}(\mathbb{D)}}^{q} . $$

The proof is complete. □

Theorem 6.3

Let \(\omega \in \mathcal{E}\), \(1\le q< p < \infty \), and μ be a finite positive Borel measure on \(\mathbb{D}\) satisfying (6.1). The following conditions are equivalent:

  1. (i)

    The Toeplitz operator \(T_{\mu }: A^{p}_{\omega }\rightarrow A^{q}_{\omega }\) is bounded.

  2. (ii)

    For each sufficiently small \(\delta > 0\), \(\widehat{\mu }_{\delta }\in L^{\frac{pq}{p-q}}(\mathbb{D},dA) \).

Moreover,

$$ \Vert T_{\mu } \Vert _{ A^{p}_{\omega }\rightarrow A^{q}_{\omega }}\asymp \Vert \widehat{\mu }_{\delta } \Vert _{L^{\frac{pq}{p-q}}(\mathbb{D)}}. $$

Proof

(i) (ii) For an arbitrary sequence \(\lambda =\lbrace \lambda _{k}\rbrace \in \ell ^{p}\), we consider the function

$$ G_{t}(z) = \sum_{\substack{k=1}}^{\infty } \lambda _{k} r_{k}(t) \omega (z_{k})^{1/2} \tau (z_{k})^{2(\frac{p-1}{p})} K_{z_{k}}(z) , \quad 0 < t < 1, $$

where \(r_{k}(t)\) is a sequence of Rademacher functions (see [20] or Appendix A of [11] ) and \(\lbrace z_{k}\rbrace \) is the sequence given in Lemma B. Because of the norm estimate

$$ \Vert K_{z} \Vert _{A^{p}_{\omega }}\asymp \omega (z)^{-1/2} \tau (z)^{-2(p-1)/p} $$

given in Lemma 2.4, by part (i) of Theorem 5.3, we obtain

$$ \Vert G_{t} \Vert _{A^{p}_{\omega }}\lesssim \Biggl(\sum _{ \substack{k=0}}^{\infty } \vert \lambda _{k} \vert ^{p} \Biggr)^{1/p}. $$

Thus, the boundedness of \(T_{\mu }: A^{p}_{\omega }\rightarrow A^{q}_{\omega }\) gives

$$ \Vert T_{\mu }G_{t} \Vert _{A^{q}_{\omega }}^{q} \lesssim \Vert T_{\mu } \Vert ^{q}\cdot \Vert \lambda \Vert ^{q}_{\ell ^{p}}. $$

In other words, we have

$$ \int _{\mathbb{D}}{ \Biggl\vert \sum_{\substack{k=1}}^{\infty } \lambda _{k} r_{k}(t) \omega (z_{k})^{1/2} \tau (z_{k})^{2(\frac{p-1}{p})} T_{\mu }K_{z_{k}}(z) \Biggr\vert ^{q} \omega (z)^{q/2}\,dA(z)}\lesssim \Vert T_{\mu } \Vert ^{q}\cdot \Vert \lambda \Vert ^{q}_{\ell ^{p}}. $$

Integrating with respect to t from 0 to 1, applying Fubini’s theorem, and invoking Khinchine’s inequality (see [20]), we obtain

$$ B : = \int _{\mathbb{D}}{ \Biggl(\sum_{\substack{k=1}}^{ \infty } \vert \lambda _{k} \vert ^{2} \omega (z_{k}) \tau (z_{k})^{4( \frac{p-1}{p})} \bigl\vert T_{\mu }K_{z_{k}}(z) \bigr\vert ^{2} \Biggr)^{q/2} \omega (z)^{q/2}\,dA(z)}\lesssim \Vert T_{\mu } \Vert ^{q}\cdot \Vert \lambda \Vert ^{q}_{ \ell ^{p}}. $$

Let \(\chi _{k}\) denote the characteristic function of the set \(D(3\delta \tau (z_{k}))\). Since the covering \(\lbrace D(3\delta \tau (z_{k}))\rbrace \) of \({\mathbb{D}}\) has finite multiplicity N, we have

$$ \begin{aligned} &\sum_{\substack{k=1}}^{\infty } \vert \lambda _{k} \vert ^{q} \omega (z_{k})^{q/2} \tau (z_{k})^{2q(\frac{p-1}{p})} \int _{D(3 \delta \tau (z_{k}))}{ \bigl\vert T_{\mu }K_{z_{k}}(z) \bigr\vert ^{q} \omega (z)^{q/2}\,dA(z)} \\ &\quad = \int _{\mathbb{D}}{\sum_{\substack{k=1}}^{\infty } \vert \lambda _{k} \vert ^{q} \omega (z_{k})^{q/2} \tau (z_{k})^{2q(\frac{p-1}{p})} \bigl\vert T_{\mu }K_{z_{k}}(z) \bigr\vert ^{q} \chi _{{k}}(z) \omega (z)^{q/2}\,dA(z)} \\ &\quad \leq \max \bigl\lbrace 1, N^{1-\frac{q}{2}}\bigr\rbrace B. \end{aligned} $$
(6.6)

Now, using Lemma A yields

$$ \sum_{\substack{k=1}}^{\infty } \vert \lambda _{k} \vert ^{q} \omega (z_{k})^{q/2} \tau (z_{k})^{2q(\frac{p-1}{p})+2} \bigl\vert T_{\mu }K_{z_{k}}(z_{k}) \bigr\vert ^{q} \omega (z_{k})^{q/2}\lesssim \Vert T_{\mu } \Vert ^{q}\cdot \Vert \lambda \Vert ^{q}_{ \ell ^{p}}. $$

On the other hand, since for small \(\delta >0\) we have \(|K_{z_{k}}(z)|\asymp \|K_{z_{k}}\|_{A^{2}_{\omega }}\|K_{z}\|_{A^{2}_{\omega }}\) for every \(z \in D(\delta \tau (z_{k}))\), applying statement (a) of Theorem A and (2.1), we have

$$\begin{aligned} \begin{aligned} \bigl\vert T_{\mu }K_{z_{k}}(z_{k}) \bigr\vert &\geq \int _{D(\delta \tau (z_{k}))}{ \bigl\vert K_{z_{k}}(z) \bigr\vert ^{2} \omega (z)\,d\mu (z)} \\ &\asymp \Vert K_{z_{k}} \Vert ^{2}_{A^{2}_{\omega }} \int _{D(\delta \tau (z_{k}))}{ \Vert K_{z} \Vert ^{2}_{A^{2}_{\omega }} \omega (z)\,d\mu (z)} \\ &\asymp \frac{\omega (z_{k})^{-1} \widehat{\mu }_{\delta }(z_{k})}{\tau (z_{k})^{2}}. \end{aligned} \end{aligned}$$

That is,

$$ \bigl\vert T_{\mu }K_{z_{k}}(z_{k}) \bigr\vert ^{q} \omega (z_{k})^{q/2}\gtrsim \frac{\omega (z_{k})^{-\frac{q}{2}} \widehat{\mu }_{\delta }(z_{k})^{q}}{\tau (z_{k})^{2q}}. $$

Therefore,

$$ \sum_{\substack{k=1}}^{\infty } \vert \lambda _{k} \vert ^{q} \tau (z_{k})^{2q( \frac{1}{q}-\frac{1}{p})} \widehat{\mu }_{\delta }(z_{k})^{q} \lesssim \Vert T_{\mu } \Vert ^{q}\cdot \Vert \lambda \Vert ^{q}_{\ell ^{p}}. $$

Then, using the duality between \(\ell ^{p/q}\) and \(\ell ^{\frac{p}{p-q}}\), we conclude that

$$ \sum_{\substack{k=1}}^{\infty } \bigl( \tau (z_{k})^{2q(\frac{1}{q}- \frac{1}{p})} \widehat{\mu }_{\delta }(z_{k})^{q} \bigr)^{\frac{p}{p-q}} \lesssim \Vert T_{\mu } \Vert ^{\frac{pq}{p-q}}, $$

that means

$$ \sum_{\substack{k=1}}^{\infty }\tau (z_{k})^{2} \widehat{\mu }_{\delta }(z_{k})^{\frac{pq}{p-q}}\lesssim \Vert T_{\mu } \Vert ^{\frac{pq}{p-q}}. $$

This is the discrete version of our condition. To obtain the continuous version, simply note that

$$ \widehat{\mu }_{\delta }(z) \lesssim \widehat{\mu }_{4\delta }(z_{k}),\quad z\in D\bigl(\delta \tau (z_{k})\bigr). $$

Then

$$ \int _{\mathbb{D}}{\widehat{\mu }_{\delta }(z)^{\frac{pq}{p-q}}\,dA(z)} \leq \sum_{\substack{k=0}}^{\infty } \int _{D(\delta \tau (z_{k}))}{ \widehat{\mu }_{\delta }(z)^{\frac{pq}{p-q}}\,dA(z)} \lesssim \sum_{ \substack{k=0}}^{\infty } \tau (z_{k})^{2} \widehat{\mu }_{4\delta }(z_{k})^{ \frac{pq}{p-q}}. $$

This finishes the proof of this implication.

(ii) (i) First we begin with the easiest case \(q = 1 \). By Fubini’s theorem and condition (1.2), we have

$$ \begin{aligned} \Vert T_{\mu }f \Vert _{A^{1}_{\omega }}&= \int _{\mathbb{D}}{ \bigl\vert T_{\mu }f(z) \bigr\vert \omega (z)^{1/2}\,dA(z)} \\ &\leq \int _{\mathbb{D}} \biggl( \int _{\mathbb{D}}{ \bigl\vert f(\xi ) \bigr\vert \bigl\vert K_{z}( \xi ) \bigr\vert \omega (\xi )\,d\mu (\xi )} \biggr) \omega (z)^{1/2}\,dA(z) \\ &= \int _{\mathbb{D}} \bigl\vert f(\xi ) \bigr\vert \biggl( \int _{\mathbb{D}}{ \bigl\vert K_{\xi }(z) \bigr\vert \omega (z)^{1/2}\,dA(z)} \biggr) \omega (\xi )\,d\mu (\xi ) \\ &\lesssim \int _{\mathbb{D}} \bigl\vert f(\xi ) \bigr\vert \omega (\xi )^{1/2}\,d\mu ( \xi ). \end{aligned} $$
(6.7)

Now, by using Theorem D with the measure given by

$$ d\nu (\xi ): = \omega (\xi )^{1/2}\,d\mu (\xi ), $$

we obtain the desired result. Finally, we study the case \(1< q < \infty \). Let \(\lbrace z_{j}\rbrace \) be the sequence given in Lemma B. Applying Lemma A and Lemma B, we obtain

$$\begin{aligned} \begin{aligned} \bigl\vert T_{\mu }f(z) \bigr\vert &\leq \sum_{\substack{j=1}}^{\infty } \int _{D( \delta \tau (z_{j})}{ \bigl\vert f(\xi ) \bigr\vert \bigl\vert K_{z}(\xi ) \bigr\vert \omega (\xi )\,d\mu (\xi )} \\ &\lesssim \sum_{\substack{j=1}}^{\infty } \int _{D(\delta \tau (z_{j}))}{ \bigl\vert f( \xi ) \bigr\vert \omega (\xi )^{1/2} \biggl(\frac{1}{\tau (\xi )^{2}} \int _{D( \delta \tau (\xi ))}{ \bigl\vert K_{z}(s) \bigr\vert \omega (s)^{1/2}\,dA(s)} \biggr)\,d\mu ( \xi )} \\ &\lesssim \sum_{\substack{j=1}}^{\infty } \biggl( \int _{D(\delta \tau (z_{j}))} \bigl\vert f( \xi ) \bigr\vert \omega (\xi )^{1/2} \frac{d\mu (\xi )}{\tau (\xi )^{2}} \biggr) \int _{D(3\delta \tau (z_{j}))}{ \bigl\vert K_{z}(s) \bigr\vert \omega (s)^{1/2}\,dA(s)}. \end{aligned} \end{aligned}$$

Applying Hölder’s inequality, we get

$$ \bigl\vert T_{\mu }f(z) \bigr\vert ^{q} \lesssim M(z) \times N(z), $$

where

$$ M(z):= \sum_{\substack{j=1}}^{\infty } \biggl( \int _{D(\delta \tau (z_{j}))} \bigl\vert f( \xi ) \bigr\vert \omega (\xi )^{1/2} \frac{d\mu (\xi )}{\tau (\xi )^{2}} \biggr)^{q} \int _{D(3\delta \tau (z_{j}))}{ \bigl\vert K_{z}(s) \bigr\vert \omega (s)^{1/2}\,dA(s)}, $$

and

$$ N(z): = \Biggl(\sum_{\substack{j=1}}^{\infty } \int _{D(3\delta \tau (z_{j}))}{ \bigl\vert K_{z}(s) \bigr\vert \omega (s)^{1/2}\,dA(s)} \Biggr)^{q-1}. $$

Furthermore, by Lemma B and condition (1.2), we have

$$ N(z)\lesssim \biggl( \int _{{\mathbb{D}}} \bigl\vert K_{z}(s) \bigr\vert \omega (s)^{1/2}\,dA(s) \biggr)^{q-1}\lesssim \omega (z)^{\frac{1-q}{2}}. $$

Thus

$$\begin{aligned} \begin{aligned} \bigl\vert T_{\mu }f(z) \bigr\vert ^{q} \omega (z)^{q/2}&\lesssim M(z) \omega (z)^{1/2}. \end{aligned} \end{aligned}$$

This gives

$$ \Vert T_{\mu }f \Vert ^{q}_{A^{q}_{\omega }} \lesssim \sum _{\substack{j=1}}^{ \infty } \biggl( \int _{D(\delta \tau (z_{j}))} \bigl\vert f(\xi ) \bigr\vert \omega (\xi )^{1/2 }\frac{d\mu (\xi )}{\tau (\xi )^{2}} \biggr)^{q} K(j), $$

where

$$ K(j):= \int _{{\mathbb{D}}} \biggl( \int _{D(3\delta \tau (z_{j}))}{ \bigl\vert K_{z}(s) \bigr\vert \omega (s)^{1/2}\,dA(s)} \biggr) \omega (z)^{1/2}\,dA(z), $$

which, by Fubini’s theorem and condition (1.2), gives

$$ K(j)\lesssim \tau (z_{j})^{2}. $$

Combining this with using (2.1) and Proposition 6.2 shows that

$$\begin{aligned} \begin{aligned} \Vert T_{\mu }f \Vert ^{q}_{A^{q}_{\omega }} &\lesssim \sum_{ \substack{j=1}}^{\infty }\tau (z_{j})^{2} \biggl( \frac{1}{\tau (z_{j})^{2}} \int _{D(\delta \tau (z_{j}))}{ \bigl\vert f(\xi ) \bigr\vert \omega (\xi )^{1/2}\,d\mu (\xi )} \biggr)^{q} \\ &\lesssim \int _{\mathbb{D}}{ \biggl( \frac{1}{\tau (z)^{2}} \int _{D(4\delta \tau (z))}{ \bigl\vert f(\xi ) \bigr\vert \omega ( \xi )^{1/2}\,d\mu (\xi )} \biggr)^{q}\,dA(z)} \\ &\lesssim \Vert \widehat{\mu }_{4\delta } \Vert ^{q}_{L^{\frac{pq}{p-q}}( \mathbb{D)}} \cdot \Vert f \Vert ^{q}_{A^{p}_{\omega }}. \end{aligned} \end{aligned}$$

This proves the desired result. □

Next we characterize compact Toeplitz operators on weighted Bergman spaces \(A^{p}_{\omega }\) for weights ω in the class \(\mathcal{E}\). We need first a lemma.

Lemma 6.4

Let \(1< p<\infty \), and let \(k_{p,z}\) be the normalized reproducing kernels in \(A^{p}_{\omega }\), with \(\omega \in \mathcal{E}\). Then \(k_{p,z} \rightarrow 0\) weakly in \(A^{p}_{\omega }\) as \(|z|\rightarrow 1^{-}\).

Proof

By duality and the reproducing kernel properties, we must show that \(|g(z)|/\|K_{z}\|_{A^{p}_{\omega }}\) goes to zero as \(|z|\rightarrow 1^{-}\) whenever g is in \(A^{p'}_{\omega }\), where \(p'\) denotes the conjugate exponent of p, but this follows easily by the density of the polynomials and Lemma A. □

Theorem 6.5

Let \(\omega \in \mathcal{E}\), \(1< p\leq q < \infty \), and μ be a finite positive Borel measure on \(\mathbb{D}\) satisfying (6.1). Then the Toeplitz operator \(T_{\mu }: A^{p}_{\omega }\rightarrow A^{q}_{\omega }\) is compact if and only if, for each \(\delta \in (0,m_{\tau })\) small enough, one has

$$ \lim_{|a|\rightarrow 1^{-}}\tau (a)^{2(\frac{1}{q}- \frac{1}{p})} \widehat{\mu }_{\delta }(a) = 0. $$
(6.8)

Proof

First we assume that \(T_{\mu }\) is compact. Following the proof of the boundedness part and the fact that \(\|K_{a}\|^{-1}_{ A^{p}_{\omega }}\asymp \omega (a)^{1/2} \tau (a)^{ \frac{2(p-1)}{p}}\), we get the estimate

$$ \begin{aligned} \tau (a)^{2(\frac{1}{q}- \frac{1}{p})} \widehat{\mu }_{\delta }(a)\lesssim \frac{\omega (a)^{1/2}}{\tau (a)^{2(\frac{1}{p}-1)}} \Vert T_{\mu }K_{a} \Vert _{A^{q}_{\omega }}\lesssim \Vert T_{\mu }k_{p,a} \Vert _{A^{q}_{\omega }}, \end{aligned} $$
(6.9)

where \(k_{p,a}\) are the normalized reproducing kernels in \(A^{p}_{\omega }\). Since, by Lemma 6.4, \(k_{p,a}\) tends to zero weakly in \(A^{p}_{\omega }\) and \(T_{\mu }\) is compact, the result follows.

Conversely, we suppose that (6.8) holds. Let \(\lbrace f_{n}\rbrace \subset A^{p}_{\omega }\) be a bounded sequence converging to zero uniformly on compact subsets of \(\mathbb{D}\). By (6.4), we have

$$ \Vert T_{\mu }f_{n} \Vert _{ A^{q}_{\omega }}\lesssim \int _{\mathbb{D}} { \bigl\vert f_{n}(z) \bigr\vert ^{q}\,d\nu (z)}= \Vert I_{\nu }f_{n} \Vert _{L^{q}(\mathbb{D}, d\nu )}, $$
(6.10)

where \(I_{\nu }: A^{p}_{\omega }\longrightarrow L^{q}(\mathbb{D}, d\nu )\) with \(d\nu (\xi ) = \omega (\xi )^{q/2} \tau (\xi )^{2(q-1)(\frac{1}{p}- \frac{1}{q})}\,d\mu (\xi )\). By using \(\tau (a) \asymp \tau (\xi )\), for \(\xi \in D(\delta \tau (a))\), we have

$$ \sup_{\substack{|a| > r}}\frac{1}{\tau (a)^{2q/p}} \int _{D( \delta \tau (a))}\omega (\xi )^{-q/2}\,d\nu (\xi ) \lesssim \sup_{|a|> r}\tau (a)^{2(\frac{1}{q}- \frac{1}{p})} \widehat{\mu }_{\delta }(a). $$

By Theorem C, \(I_{\nu }\) is compact, and in view of (6.10), \(T_{\mu }\) is compact. □

Theorem 6.6

Let \(\omega \in \mathcal{E}\), \(1\le q< p < \infty \), and μ be a finite positive Borel measure on \(\mathbb{D}\) satisfying (6.1). The following conditions are equivalent:

  1. (i)

    The Toeplitz operator \(T_{\mu }: A^{p}_{\omega }\rightarrow A^{q}_{\omega }\) is compact.

  2. (ii)

    For each sufficiently small \(\delta > 0\),

    $$ \widehat{\mu }_{\delta }\in L^{\frac{pq}{p-q}}( \mathbb{D},dA). $$
    (6.11)

Proof

If \(T_{\mu }\) is compact, then it is bounded, and by Theorem 6.3 we get the desired result. Conversely, if (6.11) holds, then by Theorem 6.3\(T_{\mu }\) is bounded. Since by Theorem 5.3 the spaces \(A^{p}_{\omega }\) and \(A^{q}_{\omega }\) are isomorphic to \(\ell ^{p}\), the result is a consequence of a general result of Banach space theory: it is known that, for \(1 \le q< p< \infty \), every bounded operator from \(\ell ^{p}\) to \(\ell ^{q}\) is compact (see [19, Theorem I.2.7, p. 31]). □

7 Hankel operators

One of the most important classes of operators acting on spaces of analytic functions is the Hankel operators. When acting on the classical Hardy spaces, their study presents [23, 27] a broad range of applications such as to control theory, approximation theory, prediction theory, perturbation theory, and interpolation problems. Furthermore, one can find an extensive literature on Hankel operators acting on other classical function spaces in one or several complex variables, such as Bergman spaces [1, 2, 15, 36, 37], Fock spaces [28], or Dirichlet spaces [33, 34]. In this section, we are going to study big Hankel operators acting on our large weighted Bergman spaces.

Definition 7.1

Let \(M_{g}\) denote the multiplication operator induced by a function g, and \(P_{\omega }\) be the Bergman projection, where ω is a weight in the class \(\mathcal{E}\). The Hankel operator \(H_{g}\) is given by

$$ H_{g}=H_{g}^{\omega }:=(I-P_{\omega }) M_{g}. $$

We assume that the function g satisfies

$$ gK_{z} \in L^{1}_{\omega },\quad z\in { \mathbb{D}}. $$
(7.1)

Under this assumption, the Hankel operator \(H_{g}^{\omega }\) is well defined on the set E of all finite linear combinations of reproducing kernels and, therefore, is densely defined in the weighted Bergman space \(A^{p}_{\omega }\), \(1\le p<\infty \). Also, for \(f\in E\), one has

$$ H_{g}^{\omega } f(z)= \int _{{\mathbb{D}}} \bigl( g(z)-g(s) \bigr) f(s) \overline{K_{z}(s)} \omega (s)\,dA(s). $$

We are going to study the boundedness and compactness when the symbol is conjugate analytic. In the Hilbert space case \(A^{2}_{\omega }\), and for weights in the class \(\mathcal{W}\), a characterization of the boundedness, compactness, and membership in Schatten classes of the Hankel operator \(H_{\bar{g}}:A^{2}_{\omega }\rightarrow L^{2}_{\omega }\) was obtained in [12]. In order to extend such results to the non-Hilbert space setting, we need estimates for the p-norm of the reproducing kernels, and it is here when condition (1.2) and the exponential type class \(\mathcal{E}\) enters in action. Before going to study the boundedness of the Hankel operator on \(A^{p}_{\omega }\) with conjugate analytic symbols, we need the following lemma.

Lemma C

Let \(1\leq p < \infty \), \(g\in H({\mathbb{D}})\), and \(a\in {\mathbb{D}}\). Then

$$ \tau (a) \bigl\vert g'(a) \bigr\vert \leq C \biggl( \frac{1}{\tau (a)^{2}} \int _{D(\delta \tau (a))} \bigl\vert g(z)- g(a) \bigr\vert ^{p}\,dA(z) \biggr)^{1/p}. $$

Proof

See for example [12]. □

Now we are ready to characterize the boundedness of the Hankel operator with conjugate analytic symbols acting on large weighted Bergman spaces in terms of the growth of the maximum modulus of \(g'\). We begin with the case \(1\le p\le q<\infty \).

Theorem 7.2

Let \(\omega \in \mathcal{E}\), \(1 \leq p\le q< \infty \), and \(g \in H({\mathbb{D}})\) satisfying (7.1). The Hankel operator \(H_{\overline{g}}: A^{p}_{\omega }\longrightarrow L^{q}_{\omega }\) is bounded if and only if

$$ \sup_{\substack{z\in \mathbb{D}}}\tau (z)^{1+2(\frac{1}{q}- \frac{1}{p})} \bigl\vert g'(z) \bigr\vert < \infty . $$
(7.2)

Proof

Suppose first that \(H_{\overline{g}}: A^{p}_{\omega }\longrightarrow L^{q}_{\omega }\) is bounded. Thus

$$ \Vert H_{\overline{g}}K_{a} \Vert _{L^{q}(\omega ^{q/2})} \leq \Vert H_{ \overline{g}} \Vert \Vert K_{a} \Vert _{A^{{p}}(\omega ^{p/2})} . $$

For each \(z\in {\mathbb{D}}\), consider the function \(g_{z} (\xi ): = (g(z) - g(\xi ) )K_{z}(\xi )\). Condition (7.1) ensures that \(g_{z}\in A^{1}_{\omega }\), and by the reproducing formula in Corollary 3.3, one has

$$\begin{aligned} \begin{aligned} H_{\overline{g}}K_{a}(z) &= \int _{\mathbb{D}}{ \overline{ \bigl(g(z)-g(\xi ) \bigr)} K_{a}(\xi ) \overline{K_{z}(\xi )} \omega (\xi )\,dA(\xi )} \\ &=\overline{ \langle g_{z},K_{a}\rangle _{\omega }} = \overline{g_{z}(a)}. \end{aligned} \end{aligned}$$

Now, for δ small enough, we have \(|K_{z}(a)|\asymp \|K_{z}\|_{A^{2}_{\omega }}\|K_{a}\|_{A^{2}_{\omega }}\) for \(z\in D(\delta \tau (a))\). Hence, by the statement (a) of Theorem A and (2.1), we have

$$\begin{aligned} \begin{aligned} \Vert H_{\overline{g}}K_{a} \Vert _{L^{{q}}_{\omega }}^{q} &= \int _{ \mathbb{D}}{ \bigl\vert g(z)- g(a) \bigr\vert ^{q} \bigl\vert K_{z}(a) \bigr\vert ^{q} \omega (z)^{q/2}\,dA(z)} \\ &\geq \int _{D(\delta \tau (a))}{ \bigl\vert g(z)- g(a) \bigr\vert ^{q} \bigl\vert K_{z}(a) \bigr\vert ^{q} \omega (z)^{q/2}\,dA(z)} \\ &\asymp \int _{D(\delta \tau (a))}{ \bigl\vert g(z)- g(a) \bigr\vert ^{q} \Vert K_{z} \Vert _{A^{2}_{\omega }}^{q} \Vert K_{a} \Vert _{A^{2}_{\omega }}^{q} \omega (z)^{q/2}\,dA(z) } \\ &\asymp \frac{ \Vert K_{a} \Vert _{A^{2}_{\omega }}^{q}}{\tau (a)^{q}} \int _{D( \delta \tau (a))} \bigl\vert g(z)- g(a) \bigr\vert ^{q}\,dA(z). \end{aligned} \end{aligned}$$

Because of the boundedness of the Hankel operator \(H_{\overline{g}}\), we have

$$\begin{aligned} \begin{aligned} \Vert H_{\overline{g}} \Vert ^{q} &\geq \frac{ \Vert H_{\overline{g}}K_{a} \Vert _{L^{{q}}_{\omega }}^{q}}{ \Vert K_{a} \Vert _{A^{p}_{\omega }}^{q}} \\ &\gtrsim \Vert K_{a} \Vert _{A^{p}_{\omega }}^{-q} \frac{ \Vert K_{a} \Vert _{A^{2}_{\omega }}^{q}}{\tau (a)^{q}} \int _{D(\delta \tau (a))}{ \bigl\vert g(z)- g(a) \bigr\vert ^{q}\,dA(z)}. \end{aligned} \end{aligned}$$

Finally, by the estimates on the norm of \(K_{z}\) in Lemma 2.4 and Theorem A, we obtain

$$\begin{aligned} \begin{aligned} \Vert H_{\overline{g}} \Vert &\gtrsim \tau (a)^{2(\frac{1}{q}- \frac{1}{p})} \biggl(\frac{1}{\tau (a)^{2}} \int _{D(\delta \tau (a))}{ \bigl\vert g(z)- g(a) \bigr\vert ^{q}\,dA(z)} \biggr)^{1/q}. \end{aligned} \end{aligned}$$

By Lemma C, this completes the proof of this implication.

Conversely, assume that (7.2) holds, and let \(1\le p \le q<\infty \). By Theorem 2.5, there exists a solution u of the equation \(\overline{\partial }u = f \overline{g'}\) in \(L^{q}(\omega ^{q/2})\) such that

$$ \Vert u \Vert ^{q}_{L^{q}_{\omega }} \lesssim \int _{{\mathbb{D}}} \bigl\vert \overline{\partial }u(z) \bigr\vert ^{q} \omega (z)^{q/2} \tau (z)^{q}\,dA(z). $$

Since any solution v of the ∂̅-equation has the form \(v = u- h\) with \(h\in H({\mathbb{D}})\), and because \(H_{\overline{g}}f\) is also a solution of ∂̅-equation, there is a function \(h\in H({\mathbb{D}})\) such that \(H_{\overline{g}}f = u - h \). As a result of \(P_{\omega } (H_{\overline{g}}f )=0 \), we have \(H_{\overline{g}}f= (I-P_{\omega } )u\), where I is the identity operator. Therefore, by the boundedness of \(P_{\omega }\) on \(L^{q}_{\omega }\) (see Theorem 3.2), we obtain

$$ \begin{aligned} \Vert H_{\overline{g}}f \Vert ^{q}_{L^{q}_{\omega }} &\leq \Vert I-P_{\omega } \Vert ^{q} \Vert u \Vert ^{q}_{L^{q}_{\omega }} \lesssim \Vert u \Vert ^{q}_{L^{q}_{\omega }} \\ &\lesssim \int _{{\mathbb{D}}} \bigl\vert f(z) \bigr\vert ^{q} \bigl\vert g'(z) \bigr\vert ^{q} \omega (z)^{q/2} \tau (z)^{q}\,dA(z). \end{aligned} $$
(7.3)

By our assumption (7.2), we have

$$\begin{aligned} \begin{aligned} \Vert H_{\overline{g}}f \Vert ^{q}_{L^{q}_{\omega }} &\lesssim \int _{{\mathbb{D}}} \bigl\vert f(z) \bigr\vert ^{q} \omega (z)^{q/2} \tau (z)^{2q( \frac{1}{p}-\frac{1}{q})}\,dA(z). \end{aligned} \end{aligned}$$

On the other hand, by Lemma A,

$$ \bigl\vert f(z) \bigr\vert \omega (z)^{1/2}\lesssim \tau (z)^{-2/p} \Vert f \Vert _{A^{p}_{\omega }}. $$

Using the last pointwise estimate, we have

$$ \Vert H_{\overline{g}}f \Vert ^{q}_{L^{q}_{\omega }} \lesssim \Vert f \Vert ^{q-p}_{A^{p}_{\omega }} \int _{{\mathbb{D}}} \bigl\vert f(z) \bigr\vert ^{p} \omega (z)^{p/2}\,dA(z) = \Vert f \Vert ^{q}_{A^{p}_{\omega }}. $$

This completes the proof. □

Next we are going to characterize the boundedness of the Hankel operator with conjugate analytic symbols when \(1\le q < p < \infty \). Before that we prove the following lemma.

Lemma 7.3

Let \(\delta _{0} \in (0,m_{\tau })\) and \(0< r<\infty \). Then

$$ \bigl\vert f'(z) \bigr\vert ^{r} \lesssim \frac{1}{\tau (z)^{r+2}} \int _{D(\delta _{0} \tau (z))} \bigl\vert f(s) \bigr\vert ^{r}\,dA(s) $$

for \(f\in H({\mathbb{D}})\).

Proof

By Cauchy’s integral formula and Lemma A, we get

$$\begin{aligned} \begin{aligned} \bigl\vert f'(z) \bigr\vert &\lesssim \biggl\vert \int _{ \vert \eta -z \vert = \frac{ \delta _{0}\tau (z)}{4}}\frac{f(\eta )}{ (\eta -z)^{2}}\,d\eta \biggr\vert \lesssim \frac{1}{\tau (z)^{2}} \int _{ \vert \eta -z \vert = \frac{ \delta _{0}\tau (z)}{4}} \bigl\vert f(\eta ) \bigr\vert \vert d\eta \vert \\ &\lesssim \frac{1}{\tau (z)^{2}} \int _{ \vert \eta -z \vert = \frac{ \delta _{0}\tau (z)}{4}}{ \biggl(\frac{1}{\tau (\eta )^{2}} \int _{D( \delta _{0}\tau (\eta )/4)} \bigl\vert f(s) \bigr\vert ^{r}\,dA(s)} \biggr)^{1/r} \vert d\eta \vert . \end{aligned} \end{aligned}$$

An application of \(\tau (\eta ) \asymp \tau (z)\), for \(\eta \in D(\delta _{0}\tau (z)/2)\), gives

$$ \begin{aligned} \bigl\vert f'(z) \bigr\vert &\lesssim \frac{1}{\tau (z)^{2}} \int _{ \vert \eta -z \vert = \frac{ \delta _{0}\tau (z)}{4}}{ \biggl(\frac{1}{\tau (z)^{2}} \int _{D( \delta _{0}\tau (z)/2)}{ \bigl\vert f(s) \bigr\vert ^{r}\,dA(s)} \biggr)^{1/r} \vert d\eta \vert } \\ &\lesssim \frac{1}{\tau (z)^{1+\frac{2}{r}}} \biggl( \int _{D(\delta _{0} \tau (z)/2)}{ \bigl\vert f(s) \bigr\vert ^{r}\,dA(s)} \biggr)^{1/r}, \end{aligned} $$

which proves the desired result. □

The following result gives the characterization of the boundedness of the Hankel operator going from \(A^{p}_{\omega }\) into \(L^{q}_{\omega }\) when \(1\le q < p < \infty \).

Theorem 7.4

Let \(\omega \in \mathcal{E}\), \(1 \le q< p< \infty \), and let \(g \in H({\mathbb{D}})\) satisfy (7.1). Then the following statements are equivalent:

  1. (a)

    The Hankel operator \(H_{\overline{g}}: A^{p}_{\omega }\longrightarrow L^{q}_{\omega }\) is bounded.

  2. (b)

    The function \(\tau g'\) belongs to \(L^{r}({\mathbb{D}}, dA)\), where \(\frac{1}{r}= \frac{1}{q}-\frac{1}{p}\).

Proof

Suppose that \(\tau g' \in L^{r}({\mathbb{D}}, dA)\). By (7.3), since \(p/q > 1\), a simple application of Hölder’s inequality, yields

$$ \Vert H_{\overline{g}}f \Vert ^{q}_{L^{q}_{\omega }} \lesssim \int _{{\mathbb{D}}} \bigl\vert f(z) \bigr\vert ^{q} \bigl\vert g'(z) \bigr\vert ^{q} \omega (z)^{q/2} \tau (z)^{q}\,dA(z)\le \Vert f \Vert ^{q}_{A^{p}_{\omega }} \bigl\Vert \tau g' \bigr\Vert ^{q}_{ L^{r}({\mathbb{D}}, dA)}. $$

This proves the boundedness of \(H_{\overline{g}}\).

Conversely, pick \(\varepsilon >0\) and let \(\lbrace z_{k}\rbrace \) be a \((\varepsilon ,\tau )\)- lattice on \({\mathbb{D}}\). For a sequence \(\lambda =\lbrace \lambda _{k}\rbrace \in \ell ^{p}\), we consider the function

$$ G_{t}(z) = \sum_{\substack{k=1}}^{\infty } \lambda _{k} r_{k}(t) \omega (z_{k})^{1/2} \tau (z_{k})^{2(\frac{p-1}{p})} K_{z_{k}}(z) ,\quad 0 < t < 1, $$

where \(r_{k}(t)\) is a sequence of Rademacher functions. Because of the norm estimate for reproducing kernels given in Lemma 2.4, by part (i) of Theorem 5.3, we obtain

$$ \Vert G_{t} \Vert _{A^{p}_{\omega }}\lesssim \Vert \lambda \Vert _{\ell ^{p}}. $$

Thus, the boundedness of \(H_{\overline{g}}\) gives

$$ \Vert H_{\overline{g}} G_{t} \Vert _{L^{q}_{\omega }}^{q} \lesssim \Vert \lambda \Vert ^{q}_{ \ell ^{p}}. $$

Therefore,

$$ \int _{\mathbb{D}}{ \Biggl\vert \sum_{\substack{k=1}}^{\infty } \lambda _{k} r_{k}(t) \omega (z_{k})^{1/2} \tau (z_{k})^{2(\frac{p-1}{p})} H_{\overline{g}} K_{z_{k}}(z) \Biggr\vert ^{q} \omega (z)^{q/2}\,dA(z)} \lesssim \Vert \lambda \Vert ^{q}_{\ell ^{p}}. $$

Using the same method in (6.6), we obtain

$$ \int _{\mathbb{D}}{\sum_{\substack{k=1}}^{\infty } \vert \lambda _{k} \vert ^{q} \omega (z_{k})^{q/2} \tau (z_{k})^{2q(\frac{p-1}{p})} \bigl\vert H_{ \overline{g}} K_{z_{k}}(z) \bigr\vert ^{q} \chi _{{k}}(z) \omega (z)^{q/2}\,dA(z)}\lesssim \Vert \lambda \Vert ^{q}_{\ell ^{p}}, $$
(7.4)

where \(\chi _{k}\) is the characteristic function of the set \(D(3 \varepsilon \tau (z_{k}))\). Additionally, by applying both statements (a) and (b) of Theorem A and (2.1), we get

$$\begin{aligned} \bigl\vert H_{\overline{g}} K_{z_{k}}(z) \bigr\vert ^{q} \omega (z)^{q/2} = \bigl\vert g(z)-g(z_{k}) \bigr\vert ^{q} \bigl\vert K_{z_{k}}(z) \bigr\vert ^{q} \omega (z)^{q/2} \gtrsim \frac{\omega (z_{k})^{-q/2} }{\tau (z_{k})^{2q}} \bigl\vert g(z)-g(z_{k}) \bigr\vert ^{q}. \end{aligned}$$

Putting this in (7.4) gives

$$\begin{aligned} \sum_{\substack{k=1}}^{\infty } \vert \lambda _{k} \vert ^{q} \tau (z_{k})^{-2q/p} \int _{D(3 \varepsilon \tau (z_{k}))} \bigl\vert g(z)-g(z_{k}) \bigr\vert ^{q}\,dA(z) \lesssim \Vert \lambda \Vert ^{q}_{\ell ^{p}}. \end{aligned}$$

Furthermore, by Lemma C, we obtain

$$ \sum_{\substack{k=1}}^{\infty } \vert \lambda _{k} \vert ^{q} \tau (z_{k})^{-2q/p+2} \bigl\vert \tau (z_{k}) g'(z_{k}) \bigr\vert ^{q}\lesssim \Vert \lambda \Vert ^{q}_{\ell ^{p}}. $$

Moreover, by the duality between \(\ell ^{p/q}\) and \(\ell ^{\frac{p}{p-q}}\), it follows

$$ I_{g'}:=\sum_{\substack{k=1}}^{\infty }\tau (z_{k})^{2} \bigl( \tau (z_{k}) \bigl\vert g'(z_{k}) \bigr\vert \bigr)^{r} < \infty . $$

In order to finish the proof, we will justify that \(\|\tau g'\|^{r}_{L^{r}({\mathbb{D}},dA)} \lesssim I_{g'}\). For that, on the one hand, by Lemma 7.3 applied to \(g'\), we get

$$ \bigl\vert g''(\xi ) \bigr\vert ^{r}\lesssim \frac{1}{\tau (\xi )^{r+2}} \int _{D( \delta _{0}\tau (\xi ))} \bigl\vert g'(s) \bigr\vert ^{r}\,dA(s). $$
(7.5)

On the other hand, by Cauchy estimates, there exists \(\xi \in [z,z_{k}]\) such that

$$ \bigl\vert g'(z)-g'(z_{k}) \bigr\vert \le \bigl\vert g''(\xi ) \bigr\vert \vert z-z_{k} \vert . $$

Using \(\tau (\xi )\asymp \tau (z)\asymp \tau (z_{k})\), for \(\xi ,z\in D(\varepsilon \tau (z_{k}))\), we have

$$\begin{aligned} \begin{aligned} \bigl\Vert \tau g' \bigr\Vert ^{r}_{L^{r}({\mathbb{D}},dA)}&\leq \sum_{k} \int _{D(\varepsilon \tau (z_{k}))} \tau (z)^{r} \bigl\vert g'(z) \bigr\vert ^{r}\,dA(z) \\ &\leq C \sum_{k} \int _{D(\varepsilon \tau (z_{k}))}\tau (z_{k})^{r} \bigl\vert g'(z)-g'(z_{k}) \bigr\vert ^{r}\,dA(z)+ C \varepsilon ^{2}I_{g'} \\ &\leq C\varepsilon ^{r} \sum_{k} \int _{D(\varepsilon \tau (z_{k}))} \tau (z_{k})^{2r} \bigl\vert g''(\xi ) \bigr\vert ^{r}\,dA(z)+ C \varepsilon ^{2}I_{g'}. \end{aligned} \end{aligned}$$

By (7.5) and using again (2.1), we obtain

$$\begin{aligned} \begin{aligned} \bigl\Vert \tau g' \bigr\Vert ^{r}_{L^{r}({\mathbb{D}},dA)}&\leq C \varepsilon ^{r}\sum _{k} \int _{D(\varepsilon \tau (z_{k}))} \frac{\tau (z_{k})^{2r}}{\tau (\xi )^{r+2}} \int _{D(\delta _{0}\tau ( \xi ))} \bigl\vert g'(s) \bigr\vert ^{r}\,dA(s)\,dA(z) +\varepsilon ^{2} I_{g'} \\ &\le C \varepsilon ^{r+2}\sum_{k} \int _{D(3\delta _{0}\tau (z_{k}))} \tau (s)^{r} \bigl\vert g'(s) \bigr\vert ^{r}\,dA(s) +\varepsilon ^{2} I_{g'}. \end{aligned} \end{aligned}$$

By Lemma B, every point \(z\in \mathbb{D}\) belongs to at most \(C\varepsilon ^{-2}\) of the sets \(D(3\delta _{0}\tau (z_{k}))\). Hence

$$\begin{aligned} \begin{aligned} \bigl(1-C \varepsilon ^{r} \bigr) \bigl\Vert \tau g' \bigr\Vert ^{r}_{L^{r}({ \mathbb{D}},dA)}& \lesssim I_{g'}. \end{aligned} \end{aligned}$$

Thus, taking ε so that \(C \varepsilon ^{r} < 1/2\), we get the desired result. □

Next we characterize the compactness of the Hankel operator with conjugate analytic symbol acting from \(A^{p}_{\omega }\) into \(L^{q}_{\omega }\), \(1\le p,q < \infty \). This characterization will be given in two theorems depending on the order of p and q. We begin with the case \(1\le p \le q < \infty \).

Theorem 7.5

Let \(\omega \in \mathcal{E}\), \(1 < p\le q< \infty \), and \(g \in H({\mathbb{D}})\) satisfying (7.1). Then, the Hankel operator \(H_{\overline{g}}: A^{p}_{\omega }\longrightarrow L^{q}_{\omega }\) is compact if and only if

$$ \lim_{\substack{ \vert a \vert \rightarrow 1^{-}}}\tau (a)^{1+2( \frac{1}{q}- \frac{1}{p})} \bigl\vert g'(a) \bigr\vert =0. $$

Proof

Let \(a\in {\mathbb{D}}\) and \(1 < p\le q< \infty \). Recall that \(k_{a,p}\) is the normalized reproducing kernel in \(A^{p}_{\omega }\). By Lemma 6.4, \(k_{a,p}\rightarrow 0\) weakly. Thus, if \(H_{\overline{g}}\) is compact, then

$$ \lim_{\substack{ \vert a \vert \rightarrow 1^{-}}} \Vert H_{\overline{g}} k_{a,p} \Vert _{L^{q}_{\omega }}=0. $$

Let δ be small enough such that \(|K_{a}(z)|\asymp \|K_{a}\|_{A^{2}_{\omega }} \|K_{z}\|_{A^{2}_{\omega }}\) for \(z\in D(\delta \tau (a))\). By Theorem A, using (2.1) and Lemma 2.4, we have

$$\begin{aligned} \begin{aligned} \Vert H_{\overline{g}} k_{a,p} \Vert ^{q}_{L^{q}_{\omega }} & \geq \int _{ D(\delta \tau (a))} \bigl\vert g(a)- g(z) \bigr\vert ^{q} \frac{ \vert K_{a}(z) \vert ^{q} \omega (z)^{q/2}}{ \Vert K_{a} \Vert ^{q}_{A^{p}_{\omega }} }\,dA(z) \\ &\gtrsim \frac{1}{\tau (a)^{2q/p}} \int _{ D(\delta \tau (a))} \bigl\vert g(a)- g(z) \bigr\vert ^{q}\,dA(z). \end{aligned} \end{aligned}$$

It follows from Lemma C that

$$ \Vert H_{\overline{g}} k_{a,p} \Vert _{L^{q}_{\omega }} \gtrsim \tau (a)^{1+2( \frac{1}{q}-\frac{1}{p})} \bigl\vert g'(a) \bigr\vert . $$

This implies that

$$ \lim_{ \vert a \vert \rightarrow 1^{-}}\tau (a)^{1+2(\frac{1}{q}- \frac{1}{p})} \bigl\vert g'(a) \bigr\vert =0, $$

which completes the proof of this implication.

Conversely, let \(\lbrace f_{n} \rbrace \) be a bounded sequence in \(A^{p}_{\omega }\) such that \(f_{n} \rightarrow 0\) uniformly on compact subsets of \({\mathbb{D}}\). To show compactness, it is standard to see that it is enough to prove that \(\|H_{\bar{g}}f_{n}\|_{L^{q}_{\omega }}\rightarrow 0\). By the assumption, given any \(\varepsilon >0\), there is \(0 < r_{0} < 1\) such that

$$ \tau (z)^{1+2(\frac{1}{q}-\frac{1}{p})} \bigl\vert g'(z) \bigr\vert < \varepsilon ,\quad r_{0}< \vert z \vert < 1. $$

Since \(\lbrace f_{n}\rbrace \) converges to zero uniformly on compact subsets of \({\mathbb{D}}\), there exists an integer \(n_{0}\) such that

$$ \bigl\vert f_{n}(z) \bigr\vert < \varepsilon \quad \text{for } \vert z \vert \le r_{0} \text{ and } n\geq n_{0}. $$

According to (7.3), we have

$$\begin{aligned} \begin{aligned} \Vert H_{\overline{g}}f_{n} \Vert ^{q}_{L^{q}_{\omega }} &\lesssim \int _{{\mathbb{D}}} \bigl\vert f_{n}(z) \bigr\vert ^{q} \bigl\vert g'(z) \bigr\vert ^{q} \omega (z)^{q/2} \tau (z)^{q}\,dA(z) \\ &= \biggl( \int _{ \vert z \vert \leq r_{0}} + \int _{r_{0}< \vert z \vert < 1} \biggr) \bigl\vert f_{n}(z) \bigr\vert ^{q} \bigl\vert g'(z) \bigr\vert ^{q} \omega (z)^{q/2} \tau (z)^{q}\,dA(z). \end{aligned} \end{aligned}$$

On the one hand, it is easy to see that

$$ \int _{ \vert z \vert \leq r_{0}} \bigl\vert f_{n}(z) \bigr\vert ^{q} \bigl\vert g'(z) \bigr\vert ^{q} \omega (z)^{q/2} \tau (z)^{q}\,dA(z) \lesssim \varepsilon ^{q}. $$
(7.6)

On the other hand, by Lemma A, we have the pointwise estimate

$$ \bigl\vert f_{n}(z) \bigr\vert \lesssim \omega (z)^{-1/2} \tau (z)^{-2/p} \Vert f_{n} \Vert _{A^{p}_{\omega }}. $$

Applying this together with our assumption, we get

$$\begin{aligned} \begin{aligned} &\int _{r_{0}< \vert z \vert < 1} \bigl\vert f_{n}(z) \bigr\vert ^{q} \bigl\vert g'(z) \bigr\vert ^{q} \omega (z)^{q/2} \tau (z)^{q}\,dA(z) \\ &\quad < \varepsilon ^{q} \int _{{\mathbb{D}}} \bigl\vert f_{n}(z) \bigr\vert ^{p} \bigl\vert f_{n}(z) \bigr\vert ^{q-p} \omega (z)^{q/2} \tau (z)^{2q (\frac{1}{p}-\frac{1}{q} )}\,dA(z) \\ &\quad \lesssim \varepsilon ^{q} \Vert f_{n} \Vert ^{q-p}_{A^{p}_{\omega }} \int _{{ \mathbb{D}}} \bigl\vert f_{n}(z) \bigr\vert ^{p} \omega (z)^{p/2}\,dA(z) = \varepsilon ^{q} \Vert f_{n} \Vert ^{q}_{A^{p}_{\omega }}. \end{aligned} \end{aligned}$$

Combining this with (7.6) gives \(\lim_{\substack{n\rightarrow \infty }}\|H_{\overline{g}}f_{n} \|_{L^{q}_{\omega }} =0\). This shows that the Hankel operator \(H_{\overline{g}}:A^{p}_{\omega }\longrightarrow L^{q}_{\omega }\) is compact. □

The next theorem contains a compactness criterion for Hankel operators when \(1 \le q< p < \infty \).

Theorem 7.6

Let \(\omega \in \mathcal{E}\), \(1 \le q< p < \infty \), and \(g \in H({\mathbb{D}})\) satisfying (7.1). The following conditions are equivalent:

  1. (a)

    The Hankel operator \(H_{\overline{g}}: A^{p}_{\omega }\longrightarrow L^{q}_{\omega }\) is compact.

  2. (b)

    The function \(\tau g' \) belongs to \(L^{r}({\mathbb{D}}, dA)\), where \(\frac{1}{r}= \frac{1}{q}-\frac{1}{p}\).

Proof

\((a)\Longrightarrow (b)\) Assume that \(H_{\overline{g}}\) is compact. Then \(H_{\overline{g}}\) is bounded. Hence, by applying Theorem 7.4, we get the desired result.

\((b)\Longrightarrow (a)\) Suppose that \(\tau g' \) belongs to \(L^{r}({\mathbb{D}}, dA)\), where \(\frac{1}{r}= \frac{1}{q}-\frac{1}{p}\). By Theorem 7.4, the Hankel operator \(H_{\overline{g}} \) is bounded and, as a result of Theorem 5.3, the space \(A^{p}_{\omega }\) is isomorphic to \(\ell ^{p}\). In this case, \(H_{\overline{g}} \) is also compact, due to a general result of Banach space theory: For \(1 \le q< p < \infty \), every bounded operator from \(\ell ^{p}\) to \(\ell ^{q}\) is compact (see [19, Theorem I.2.7]). This finishes the proof. □

8 Concluding remarks

I believe that I have done a satisfactory work in order to get a better understanding of the function properties of large weighted Bergman spaces and the operators acting on them. I hope that this work is going to attract many other researchers to this area, and expect that the study of this function spaces is going to experience a period of intensive research in the next years. There is still plenty of work to do for a better understanding of the theory of large weighted Bergman spaces, and several natural problems are waiting for further study or a complete solution: atomic decomposition, coefficient multipliers, zero sets, etc. I hope that the methods developed here will be of some help in order to attach the previous mentioned problems.

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No data were used to support this study.

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The author thanks Pr. Jordi Pau Plana (University of Barcelona) for his support.

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Arroussi, H. Bergman spaces with exponential type weights. J Inequal Appl 2021, 193 (2021). https://doi.org/10.1186/s13660-021-02726-4

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