Hankel operators between different doubling Fock spaces

Abstract

In this paper, we study Hankel operators on the doubling Fock spaces for all possible $1\leq p,q<\infty$. We characterize those symbols f for which the Hankel operators $H_{f}$ and $H_{\bar{f}}$ are simultaneously bounded or compact from doubling Fock space $F^{p}_{\varphi}$ to Lebesgue space $L^{q}_{\varphi}$, where φ is a nonzero subharmonic function such that $\Delta \varphi \,dA$ is a doubling measure.

1 Introduction

Suppose v is a positive Borel measure on $\mathbb{C}$. We say v is doubling if there exists some constant $M>0$ such that

$$\begin{aligned} v\bigl(D(z,2r)\bigr)\leq Mv\bigl(D(z,r)\bigr), \quad \text{for } z\in \mathbb{C} \text{ and } r>0, \end{aligned}$$

where $D(z,r)=\{w\in \mathbb{C}:|w-z|< r\}$ is the Euclidean ball of radius r and center z. In the whole paper, we assume that φ is a subharmonic, real-valued function on $\mathbb{C}$, and φ is not identically zero with $v=\Delta \varphi \,dA$ doubling, where dA is the Lebesgue area measure on $\mathbb{C}$. We denote by $\rho (\cdot )$ the positive radius such that $v(D(z,\rho (z)))=1$ for any $z\in \mathbb{C}$. See [2, 7] for more information about $\rho ^{-2}$.

For $1\leq p<\infty$, the space $L^{p}_{\varphi}(\mathbb{C},dA)$ is the family of all Lebesgue measurable functions f on $\mathbb{C}$ such that

$$\begin{aligned} \Vert f \Vert _{L^{p}_{\varphi}}= \biggl( \int _{\mathbb{C}} \bigl\vert f(z)e^{-\varphi (z)} \bigr\vert ^{p}\,dA(z) \biggr)^{1/p}< \infty . \end{aligned}$$

Let $H(\mathbb{C})$ be the set of entire functions on $\mathbb{C}$. The doubling Fock space is defined as

$$\begin{aligned} F^{p}_{\varphi}=H(\mathbb{C})\cap L^{p}_{\varphi}( \mathbb{C},dA) (1 \leq p< \infty ). \end{aligned}$$

It is clear that $F^{p}_{\varphi}$ is a Banach space under the norm $\|\cdot \|_{F^{p}_{\varphi}}$ if $p\geq 1$, whereas $F^{p}_{\varphi}$ is a Fréchet space under $d(f,g)=\|f-g\|^{p}_{F^{p}_{\varphi}}$ if $0< p<1$. The doubling Fock spaces in this paper, always called generalized Fock spaces, cover many in the literature. When $\varphi (z)=\frac{\alpha}{2}|z|^{2}$, $F^{2}_{\varphi}$ is the classical Fock space $F^{2}_{\alpha}$, which has been studied systematically by many authors, see the monograph [15] and references therein. The so-called Fock–Sobolev spaces, whose basic properties were first introduced by Cho and Zhu in [1], is another special case when we choose that

$$\begin{aligned} \varphi (z)=\frac{1}{2} \vert z \vert ^{2}-\frac{m}{2} \ln\bigl(A+ \vert z \vert ^{2}\bigr), \end{aligned}$$

for some suitable constant $A>0$ and positive integer m. Schuster and Varolin studied the properties of the Bargmann–Fock spaces in [11], in which the weight function φ satisfies that $0< c\leq \Delta \varphi \leq C$ for two positive constants c and C. This infers $\Delta \varphi \,dA$ is doubling and it is also the special case under the case of a one-dimensional complex plane.

Let $K_{\varphi}(\cdot ,\cdot )$ be the Bergman kernel for $F^{2}_{\varphi}$. Its normalization in $F^{p}_{\varphi}$ is defined by

$$ k_{p,z}(w)= \frac{K_{\varphi}(w, z)}{ \Vert K_{\varphi}(\cdot , z) \Vert _{F^{p}_{\varphi}}},\quad \text{for } z,w\in \mathbb{C}. $$

The orthogonal projection $P_{\varphi}$ from $L^{2}_{\varphi}$ to $F^{2}_{\varphi}$ can be represented as

$$ P_{\varphi }f(z)= \int _{\mathbb{C}}f(w)K_{\varphi}(z,w)e^{-2\varphi (w)}\,dA(w),\quad \text{for } z\in \mathbb{C}. $$

Obviously, we have $P_{\varphi }f=f$ when $f\in F^{2}_{\varphi}$. As is known, when $1\leq p<\infty$, the projection $P_{\varphi}$ is well defined on $F^{p}_{\varphi}$ because $\operatorname{Span}\{k_{p,z}: z\in \mathbb{C}\}$ is dense in $F^{p}_{\varphi}$. Let Γ be the family of all measurable functions f on $\mathbb{C}$ satisfying $fk_{p,z}\in L^{p}_{\varphi}$ for each $z\in \mathbb{C}$. Given $f\in \Gamma$, the Hankel operator $H_{f}$ is densely defined on $F^{p}_{\varphi}$ by $H_{f}g(z)=(I-P_{\varphi})(fg)(z)$ or, in detail,

$$ H_{f}g(z)= \int _{\mathbb{C}}\bigl(f(z)-f(w)\bigr)g(w)K_{\varphi}(z,w)e^{-2 \varphi (w)}\,dA(w), $$

when I is the identity operator from $F^{p}_{\varphi}$ to $L^{p}_{\varphi}$.

Hankel operators have been studied for several decades in the setting of various analytic function spaces. Starting with Hankel matrices, which can be viewed as Hankel operators on Hardy spaces, the field has expanded to Hankel operators on Bergman spaces, Dirichlet-type spaces, Bergman and Hardy spaces of the unit ball or symmetric domains in $\mathbb{C}^{n}$, and Fock spaces. In 2012, Zhu characterized in [15] the Hankel operator $H_{f}$ with a suitable symbol function f on $\mathbb{C}$ on the classic Fock spaces $F^{2}_{\alpha}$. In the case of Bargmann–Fock spaces $F^{2}_{\varphi}$, Wang et al. showed in [14] that the Hankel operators $H_{f}$ and $H_{\bar{f}}$ are both bounded if and only if f is a BMO function. Furthermore, Wang et al. studied similar subjects on the Fock-type space $F^{2}_{\Psi}$ for another weight function Ψ. See [13] for more details. In 2014, Perälä et al. made a significant breakthrough, in [10], that Hankel operators $H_{f}$ and $H_{\bar{f}}$ on non-Hilbert Fock spaces $F^{p}_{\alpha}$ for $1\leq p<\infty$ are both bounded if and only if $f\in BMO^{p}$. Then, Hu and Lv extended the results on doubling Fock space $F^{p}_{\varphi}$ for $1\leq p<\infty$. See [3] for more information. More generally, Hu and Wang pointed out, in [5], that Hankel operators $H_{f}$ and $H_{\bar{f}}$ from a Fock space $F^{p}_{\alpha}$ to another space $L^{q}_{\alpha}$ are both bounded if and only if $f\in BMO$ when $1\leq p\leq q<\infty$, and relatively, f is a IMO function when $1\leq q< p<\infty$. In addition, they claimed that the BMO spaces is the special case of the IMO spaces. Subsequently, Lv gave similar results on Bargman–Fock spaces. See [6] for more details.

In this note, we will characterize the boundedness and compactness of a pair of Hankel operators $H_{f}$ and $H_{\bar{f}}$ from doubling Fock space $F^{p}_{\varphi}$ to another space $L^{q}_{\varphi}$ for $1\leq p,q<\infty$. Our results extend those in [3, 5, 6]. As stated in [4], the doubling Fock spaces are indeed different from classic Fock spaces and Bargmann–Fock spaces. In general, the Banach-space technique can not be applied to doubling Fock spaces, as the others rely on two points: one is the inclusion

$$ F^{p}_{\varphi}\subset F^{q}_{\varphi}, \quad \text{for } 0< p\leq q. $$

However, this is not available in the present case. For example, if taking $\varphi (z)=|z|^{4}$, $\Delta \varphi \,dA$ is indeed doubling, however,

$$ F^{p}_{\varphi}\setminus F^{q}_{\varphi}\neq \emptyset ,\qquad F^{q}_{ \varphi}\setminus F^{p}_{\varphi} \neq \emptyset , $$

for $p\neq q$. The other is the explicit representions of the Bergman kernel and Bergman distance induced by the Bergman metric. Unfortunately, the quadratic decay in the classic Fock spaces and generalized Bargmann–Fock spaces like

$$ e^{z\bar{w}}e^{-\frac{1}{2}|z|^{2}}e^{-\frac{1}{2}|w|^{2}}=e^{ \frac{1}{2}|z-w|^{2}} $$

is known not to hold in our context and is expected to be very rare. Furthermore, the translations, which used to play a very important role in Fock spaces, are currently not available, as are the applications of the Bergman distance. In order to tackle those difficulties, we should adopt the main ideas in [2–4, 7, 8].

The organization of this article is as follows. In Sect. 2, we will give the preliminaries including the geometrical properties of the BMO spaces and IMO spaces. Our main results about the bounedness and compactness of Hankel operators $H_{f}$ and $H_{\bar{f}}$, simultaneously, from a doubling Fock space $F^{p}_{\varphi}$ to another doubling Lebesgue space $L^{q}_{\varphi}$ will occur in Sect. 3 for $1\leq p\leq q<\infty$ and in the last section for $1\leq q< p<\infty$.

Throughout this paper we write $X\lesssim Y$ or $Y\gtrsim X$ for nonnegative quantities X and Y whenever there is a constant $C>0$ independent of X and Y such that $X\leq CY$. Similarly, we write $X\backsimeq Y$ if $X\lesssim Y$ and $Y\lesssim X$.

2 IMO and related spaces

In this section, we are going to characterize the function space of an Integrable Mean Oscillation. Also, we will give some basic properties, which will be used frequently in the following sections.

Given $r>0$ and any $z\in \mathbb{C}$, we denote by $D^{r}(z)=D(z,r\rho (z))$. According to [7], there exists a constant $\beta >0$ only dependent on r and a doubling constant such that

$$\begin{aligned} \frac{1}{\beta}\rho (z)\leq \rho (w)\leq \beta \rho (z), \end{aligned}$$

(1)

for any $w\in D^{r}(z)$. Also, there exists a $m>0$ such that

$$ D^{mr}(w)\subset D^{r}(z),\quad \forall w\in D^{mr}(z). $$

The following is the collection of the properties about the reproducing kernel $K_{\varphi}(\cdot , \cdot )$, which are referred to in [8, 9].

(1) For any $z,w\in \mathbb{C}$, there are constants $C>0$ and $\varepsilon >0$ only depending on the doubling constant such that

$$\begin{aligned} \bigl\vert K_{\varphi}(z,w) \bigr\vert \leq C \frac{e^{\varphi (w)}e^{\varphi (z)}}{\rho (w)\rho (z)}e^{-( \frac{ \vert z-w \vert }{\rho (z)})^{\varepsilon}}; \end{aligned}$$

(2)

(2) There is a $r_{0}>0$ such that, for any $z\in \mathbb{C}$ and any $w\in D^{r_{0}}(z)$,

$$\begin{aligned} \bigl\vert K_{\varphi}(z,w) \bigr\vert \simeq \frac{e^{\varphi (w)}e^{\varphi (z)}}{\rho (z)\rho (w)}; \end{aligned}$$

(3)

(3) For all $0< p<\infty$, we have

$$ \bigl\Vert K_{\varphi}(\cdot ,z) \bigr\Vert _{F^{p}_{\varphi}}\simeq e^{\varphi (z)} \rho (z)^{\frac{2}{p}-2},\quad z\in \mathbb{C}. $$

Given some $r>0$, we call a sequence $\{a_{k}\}_{k=1}^{\infty}$ in $\mathbb{C}$ an r-lattice if the balls $\{D^{r}(a_{k})\}_{k=1}^{\infty}$ cover $\mathbb{C}$ and $\{D^{\frac{r}{5}}(a_{k})\}_{k=1}^{\infty}$ are pairwise disjoint. For any positive constant r, the existence of some r-lattice comes from a stand covering lemma, see [7] for more details. Also, given an r-lattice $\{a_{k}\}_{k=1}^{\infty}$ and $m>0$ there exists some integer N such that each $z\in \mathbb{C}$ can be in at most N balls of $\{D^{mr}(a_{k})\}_{k=1}^{\infty}$. Equivalently,

$$ 1\leq \sum_{k=1}^{\infty}\chi _{D^{mr}(a_{k})}(z)\leq N. $$

For any $z,w\in \mathbb{C}$, we set the distance function as

$$ d_{\varphi}(z,w)=\inf_{\gamma} \int _{0}^{1} \frac{ \vert \gamma '(t) \vert }{\rho (\gamma (t))}\,dt, $$

where γ runs on the piecewise $C^{1}$ curves $\gamma :[0,1]\to \mathbb{C}$ with $\gamma (0)=z$ and $\gamma (1)=w$. The following lemma about the distance function was proven in Lemma 1 of [7].

Lemma 2.1

There exists a $0< t<1$ such that, for each $r>0$, there exists a constant $C_{r}>0$ depending on r such that

$$ C_{r}^{-1}\frac{ \vert z-w \vert }{\rho (z)}\leq d_{\varphi}(z,w)\leq C_{r} \frac{ \vert z-w \vert }{\rho (z)},\quad w\in D^{r}(z) $$

and

$$ C_{r}^{-1} \biggl(\frac{ \vert z-w \vert }{\rho (z)} \biggr)^{t} \leq d_{\varphi}(z,w) \leq C_{r} \biggl(\frac{ \vert z-w \vert }{\rho (z)} \biggr)^{2-t},\quad w\in \mathbb{C}\setminus D^{r}(z). $$

As noted in [8], Lemma 2.1 infers that the estimate of reproducing kernel (2) is equivalent to

$$\begin{aligned} \bigl\vert K_{\varphi}(z,w) \bigr\vert \leq C \frac{e^{\varphi (w)}e^{\varphi (z)}}{\rho (w)\rho (z)}e^{-d_{\varphi}(z,w)^{ \varepsilon}}. \end{aligned}$$

(4)

Here, ε may be different from that in (2). Without no confusion caused, we will simultaneously use two estimates about the reproducing kernel in the coming context.

Next, we introduce the definition of an IMO space. For $f\in L_{\mathrm{loc}}^{1}$ (the set of all Lebesgue functions f on $\mathbb{C}$ with $|f|$ integrable locally), the averaging function $\widehat{f}_{r}$ is defined as

$$ \widehat{f}_{r}(z)=\frac{1}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)}f(w)\,dA(w) \simeq \frac{1}{\rho ^{2}(z)} \int _{D^{r}(z)}f(w)\,dA(w). $$

Fix $0< r<\infty$ and $p\geq 1$, the pth mean oscillation of $f\in L^{p}_{\mathrm{loc}}$ is defined as

$$ MO_{p,r}(f) (z)= \biggl(\frac{1}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert f(w)- \widehat{f}_{r}(z) \bigr\vert ^{p}\,dA(w) \biggr)^{\frac{1}{p}}. $$

For a continuous function f on $\mathbb{C}^{n}$, let $\omega _{r}(f)(z)=\sup \{|f(z)-f(w)|: w\in D^{r}(z)\}$ be the oscillation of f at point z. Similar to [5], we introduce some spaces of integrable mean oscillation functions. Let $1\leq p,q<\infty$, $1\leq s\leq \infty$, $\gamma \in \mathbb{R}$ and $r>0$. The space of continuous functions f on $\mathbb{C}$ is denoted by $IO^{s,\gamma}_{r}$ such that

$$ \Vert f \Vert _{IO^{s,\gamma}_{r}}= \biggl( \int _{\mathbb{C}}\rho (z)^{s\gamma} \bigl\vert \omega _{r}f(z) \bigr\vert ^{s}\,dA(z) \biggr)^{\frac{1}{s}}< \infty . $$

The space $IA^{s,\gamma ,q}_{r}$ is the collection of functions $f\in L^{q}_{\mathrm{loc}}$ with

$$ \Vert f \Vert _{IA^{s,\gamma ,q}_{r}}= \biggl( \int _{\mathbb{C}} \bigl[\rho (z)^{q \gamma}\widehat{\bigl( \vert f \vert ^{q}\bigr)}_{r}(z) \bigr]^{\frac{s}{q}}\,dA(z) \biggr)^{ \frac{1}{s}}< \infty . $$

We denote by $\mathrm{IMO}_{r}^{s,\gamma ,q}$ the space of integrable mean oscillation f on $\mathbb{C}^{n}$ such that

$$ \Vert f \Vert _{\mathrm{IMO}^{s,\gamma ,q}_{r}}= \biggl( \int _{\mathbb{C}}\rho (z)^{s \gamma} \bigl\vert MO_{q,r}f(z) \bigr\vert ^{s}\,dA(z) \biggr)^{\frac{1}{s}}< \infty . $$

Obviously, when $s=\infty$, these spaces are similar to the well-known spaces introduced in [15]. Now, we list their definitions as follows:

$$\begin{aligned}& BO_{r}^{\gamma} = \Bigl\{ f\in C(\mathbb{C}): \Vert f \Vert _{BO^{\gamma}_{r}}= \sup_{z\in \mathbb{C}}\rho (z)^{\gamma}\omega _{r}f(z)< \infty \Bigr\} ; \\& BA_{r}^{\gamma ,q} = \Bigl\{ f\in L^{q}_{\mathrm{loc}}: \Vert f \Vert _{BA_{r}^{\gamma ,q}}= \sup_{z\in \mathbb{C}}\rho (z)^{\gamma}\widehat{\bigl( \vert f \vert ^{q} \bigr)}^{ \frac{1}{q}}_{r}(z)< \infty \Bigr\} ; \\& BMO_{r}^{\gamma ,q} = \Bigl\{ f\in L^{q}_{\mathrm{loc}}: \Vert f \Vert _{BMO_{r}^{ \gamma ,q}}=\sup _{z\in \mathbb{C}}\rho (z)^{\gamma}MO_{q,r}f(z)< \infty \Bigr\} . \end{aligned}$$

The vanishing spaces are shown, respectively, as follows:

$$\begin{aligned}& VO_{r}^{\gamma} = \Bigl\{ f\in BO_{r}^{\gamma}: \lim_{ \vert z \vert \to \infty} \rho (z)^{\gamma}\omega _{r}f(z)=0 \Bigr\} ; \\& VA_{r}^{\gamma ,q} = \Bigl\{ f\in BA_{r}^{\gamma ,q}: \lim_{ \vert z \vert \to \infty}\rho (z)^{\gamma}\widehat{\bigl( \vert f \vert ^{q}\bigr)}^{\frac{1}{q}}_{r}(z)=0 \Bigr\} ; \\& VMO_{r}^{\gamma ,q} = \Bigl\{ f\in BMO_{r}^{\gamma ,q}: \lim_{ \vert z \vert \to \infty}\rho (z)^{\gamma}MO_{q,r}f(z)=0 \Bigr\} . \end{aligned}$$

According to Chap. 3 in [15], we know the spaces $BO^{0}_{r}$, $BA^{0,q}_{r}$, $BMO^{0,q}_{r}$, $VO^{0}_{r}$, $VA^{0,q}_{r}$, and $VMO^{0,q}_{r}$ do not depend on r under the classical Fock space case. Fortunately, this situation can still be valid from the following considerations. We could not omit their details for the sake of completeness. Then, we will leave out the subscripts or denote like $IO^{s,\gamma ,q}_{r}$ as $IO^{s,\gamma ,q}$ for brevity.

Lemma 2.2

Suppose that $1\leq q<\infty$, $\gamma \in \mathbb{R}$ and $1\leq s\leq \infty$. Then, for $r,R>0$ and f Lebesgue measurable on $\mathbb{C}$, $f\in IO^{s,\gamma ,q}_{r}$ if and only if $f\in IO^{s,\gamma ,q}_{R}$. Furthermore, $\|f\|_{IO^{s,\gamma ,q}_{r}}\simeq \|f\|_{IO^{s,\gamma ,q}_{R}}$.

Proof

By the estimate (1) and the triangle inequality, there is a constant $0<\delta <1$, such that $D^{\delta}(z)\subset D(w)$ for any $z\in \mathbb{C}$ and $w\in D^{\delta}(z)$. Now, we fix this δ, and it suffices to obtain that $\|f\|_{IO^{s,\gamma ,q}_{R\delta}}\lesssim \|f\|_{IO^{s,\gamma ,q}_{1}}$, where R is large enough satisfying $R\delta >1$.

We chose that $\{a_{k}\}_{k\geq 1}$ is a δ-lattice. According to the property of the lattice, for any $R>1/\delta$, there exists a positive integral $N_{R}$ only dependent on R such that $D^{R\delta}(z)$ is covered by at most $N_{R}$ disks $\{D^{\delta}(a_{k_{z}})\}$ for any $z\in \mathbb{C}$. That is

$$\begin{aligned} D^{R\delta}(z)\subset \bigcup_{k_{z}=1}^{N_{R}} D^{\delta}(a_{k_{z}}) \subset \bigcup_{k=1}^{\infty }D^{\delta}(a_{k}), \end{aligned}$$

where $a_{k_{z}}$ is the lattice depending only on z. Therefore, we can see that

$$ \sup_{\xi \in D^{R\delta}(z)} \bigl\vert f(\xi )-f(z) \bigr\vert \lesssim \max_{1\leq k_{z} \leq N_{R}}\sup_{\xi \in D^{\delta}(a_{k_{z}})} \bigl\vert f(\xi )-f(z) \bigr\vert . $$

Subsequently, for a given z, there is some kth cover $D^{\delta}(a_{k})$ such that $z\in D^{\delta}(a_{k})$. Then, we denote that $\xi _{j,j+1}\in D^{\delta}(a_{j})\cap D^{\delta}(a_{j+1})$, where $a_{j}$, $a_{j+1}$ are the two conterminous lattices satisfied that

$$ D^{\delta}(a_{j})\cap D^{\delta}(a_{j+1}) \neq \emptyset ,\quad \text{and}\quad D^{\frac{\delta}{5}}(a_{j})\cap D^{\frac{\delta}{5}}(a_{j+1})= \emptyset . $$

Obviously, if $z,\xi \in D^{\delta}(a_{k})$,

$$ \bigl\vert f(\xi )-f(z) \bigr\vert \lesssim \omega _{\delta}(f) (a_{k}), $$

whereas, if $z\in D^{\delta}(a_{k})$, $\xi \in D^{\delta}(a_{j})\setminus D^{ \delta}(a_{k})$, $j\neq k$, we know that

$$\begin{aligned} \bigl\vert f(\xi )-f(z) \bigr\vert \leq & \bigl\vert f(\xi )-f(a_{j}) \bigr\vert + \bigl\vert f(a_{j})-f(\xi _{j,j+1}) \bigr\vert \\ &{}+ \bigl\vert f(\xi _{j,j+1})-f(a_{j+1}) \bigr\vert + \cdots + \bigl\vert f(a_{k})-f(z) \bigr\vert . \end{aligned}$$

Note that the above summation is finite because the cover is not infinite. Taking the supremum, we find that

$$ \sup_{\xi \in D^{\delta}(a_{j})} \bigl\vert f(\xi )-f(z) \bigr\vert \lesssim \sum_{k_{z}=1}^{N_{R}} \sup_{\xi \in D^{\delta}(a_{k_{z}})} \bigl\vert f(\xi )-f(a_{k_{z}}) \bigr\vert =\sum _{k_{z}=1}^{N_{R}} \omega _{\delta}(f) (a_{k_{z}}). $$

Combining the estimate (1) and the definition of $IO_{r}^{s,\gamma}$ spaces, we can see that

$$\begin{aligned}& \int _{\mathbb{C}}\rho (z)^{s\gamma} \bigl\vert \omega _{R\delta}(f) (z) \bigr\vert ^{s}\,dA(z) \\& \quad \lesssim \sum_{k=1}^{\infty} \int _{D^{\delta}(a_{k})}\rho (z)^{s \gamma} \bigl\vert \omega _{R\delta}(f) (z) \bigr\vert ^{s}\,dA(z) \\& \quad \lesssim \sum_{k=1}^{\infty}\sup _{z\in D^{\delta}(a_{k})}\rho (z)^{s \gamma} \bigl\vert \omega _{R\delta}(f) (z) \bigr\vert ^{s} \int _{D^{\delta}(a_{k})}\,dA(z) \\& \quad \lesssim \sum_{k=1}^{\infty}\sum _{k_{z}=1}^{N_{R}}\rho (a_{k_{z}})^{s \gamma} \bigl\vert \omega _{\delta}(f) (a_{k_{z}}) \bigr\vert ^{s} \int _{D^{\delta}(a_{k})}\,dA(z) \\& \quad \lesssim \sum_{k=1}^{\infty}\rho (a_{k})^{s\gamma} \bigl\vert \omega _{\delta}(f) (a_{k}) \bigr\vert ^{s} \int _{D^{\delta}(a_{k})}\,dA(z). \end{aligned}$$

On the other hand, it is known that $D^{\delta}(a_{k})\subset D(z)$ for any $z\in D^{\delta}(a_{k})$. Thus,

$$ \omega _{1} (f) (z)\gtrsim \bigl\vert f(a_{k})-f(z) \bigr\vert . $$

Furthermore,

$$ \omega _{1} (f) (z)\gtrsim \sup_{z\in D^{\delta}(a_{k})} \bigl\vert f(a_{k})-f(z) \bigr\vert = \omega _{\delta }(f) (a_{k}). $$

Together with the property of the lattice we find,

$$\begin{aligned}& \int _{\mathbb{C}}\rho (z)^{s\gamma} \bigl\vert \omega _{1}(f) (z) \bigr\vert ^{s}\,dA(z) \\& \quad \gtrsim \sum_{k=1}^{\infty} \int _{D^{\delta}(a_{k})}\rho (z)^{s \gamma} \bigl\vert \omega _{1}(f) (z) \bigr\vert ^{s}\,dA(z) \\& \quad \gtrsim \sum_{k=1}^{\infty}\rho (a_{k})^{s\gamma} \bigl\vert \omega _{\delta}(f) (a_{k}) \bigr\vert ^{s} \int _{D^{\delta}(a_{k})}\,dA(z). \end{aligned}$$

Now, we can obtain that

$$ \int _{\mathbb{C}}\rho (z)^{s\gamma} \bigl\vert \omega _{R\delta}(f) (z) \bigr\vert ^{s}\,dA(z) \lesssim \int _{\mathbb{C}}\rho (z)^{s\gamma} \bigl\vert \omega _{1}(f) (z) \bigr\vert ^{s}\,dA(z), $$

for any $R\delta >1$ and the proof is completed. □

Lemma 2.3

A continuous function f on $\mathbb{C}^{n}$ belongs to $BO^{\gamma}_{r}$ if and only if

$$ \rho (z)^{\gamma} \bigl\vert f(w)-f(z) \bigr\vert \lesssim \bigl(1+d_{\varphi}(z,w)\bigr)^{1+ \vert \gamma \vert }, $$

for any $z,w\in \mathbb{C}$.

Proof

In view of Lemma 2.1, $w\in \overline{D^{r}(z)}$ is equivalent to $d_{\varphi}(z,w)\leq M_{r}$, where $M_{r}>0$ is a constant only depending on r. Suppose that $\lambda (t)$ ($t\in [0,1]$) is the geodesic in the Bergman metric from z to w when $d_{\varphi}(z,w)>M_{r}$. If choosing

$$ n= \biggl[\frac{d_{\varphi}(z,w)}{M_{r}} \biggr]+1, $$

where $[x]$ ($x\in \mathbb{R}$) refers to a rounding function, we know that, by the estimate (2.6) in [3], there exist finite points $t_{i}\in [0,1]$ such that

$$ d_{\varphi}\bigl(\lambda (t_{i-1}),\lambda (t_{i}) \bigr)= \frac{d_{\varphi}(z,w)}{n}\leq M_{r},\quad i=1,2,\dots , n. $$

Together with the above case of $w\in \overline{D^{r}(z)}$, we can, therefore, see that

$$ \bigl\vert f(w)-f(z) \bigr\vert \lesssim \sum_{i=1}^{n} \bigl\vert f\bigl(\lambda (t_{i})\bigr)-f\bigl(\lambda (t_{i-1})\bigr) \bigr\vert \lesssim \sum _{i=1}^{n}\rho \bigl(\lambda (t_{i-1}) \bigr)^{-\gamma}. $$

Subsequently, we will claim that

$$ \biggl(\frac{\rho (w)}{\rho (z)} \biggr)^{\alpha}\lesssim \bigl(1+d_{ \varphi}(z,w) \bigr)^{|\alpha |}, $$

for any $z,w\in \mathbb{C}$ and $\alpha \in \mathbb{R}$. To this end, without loss of generality, we only discuss the case of $\alpha <0$. By Lemma 3.3 in [2] and Lemma 2.1, there is a constant $0<\delta <1$ only depending on a doubling constant such that

$$\begin{aligned} \biggl(\frac{\rho (z)}{\rho (w)} \biggr)^{-\alpha} \lesssim \biggl( \frac{ \vert z-w \vert }{\rho (w)} \biggr)^{-\delta \alpha} \lesssim \bigl(1+d_{ \varphi}(z,w) \bigr)^{ \vert \alpha \vert }. \end{aligned}$$

Therefore, we continue calculating that

$$\begin{aligned} \rho \bigl(\lambda (t_{i-1})\bigr)^{-\gamma} \lesssim &\rho (z)^{-\gamma}\bigl(1+d_{ \varphi}\bigl(z,\lambda (t_{i-1}) \bigr)\bigr)^{|\gamma |} \\ \lesssim &\rho (z)^{-\gamma}\bigl(1+d_{\varphi}(z,w) \bigr)^{1+|\gamma |}. \end{aligned}$$

Now the proof is completed, because the sufficiency is obvious. □

Lemma 2.4

Suppose that $1\leq q<\infty$, $\gamma \in \mathbb{R}$, $r>0$, and $1\leq s\leq \infty$. Then, $f\in \mathrm{IMO}_{r}^{s,\gamma ,q}$ if and only if f admits a decomposition $f=f_{1}+f_{2}$, where $f_{1}\in IO^{s,\gamma}_{r}$ and $f_{2}\in IA^{s,\gamma ,q}_{r}$. Furthermore,

$$ \Vert f \Vert _{\mathrm{IMO}_{r}^{s,\gamma ,q}}=\inf \bigl\{ \Vert f_{1} \Vert _{IO^{s,\gamma}_{r}}+ \Vert f \Vert _{IA^{s,\gamma ,q}_{r}}:f_{1}+f_{2}, f_{1}\in IO^{s,\gamma}_{r}, f_{2} \in IA^{s,\gamma ,q}_{r} \bigr\} . $$

Proof

If $f\in \mathrm{IMO}^{s,\gamma ,q}_{r}$, we set $f_{1}=\widehat{f}_{\frac{r}{2}}$ and $f_{2}=f-f_{1}$. For any $w\in D^{r}(z)$,

$$\begin{aligned} \bigl\vert f_{1}(w)-f_{1}(z) \bigr\vert \lesssim & \bigl\vert f_{1}(z)-\widehat{f}_{r}(z) \bigr\vert + \bigl\vert \widehat{f}_{r}(z)-f_{1}(w) \bigr\vert \\ \lesssim &\frac{1}{ \vert D^{\frac{r}{2}}(z) \vert } \int _{D^{\frac{r}{2}}(z)} \bigl\vert f(w)- \widehat{f}_{r}(z) \bigr\vert dA(w) \\ &{}+\frac{1}{ \vert D^{\frac{r}{2}}(w) \vert } \int _{D^{\frac{r}{2}}(w)} \bigl\vert f(u)- \widehat{f}_{r}(z) \bigr\vert dA(u). \end{aligned}$$

By the estimate (1), there is a constant $\beta >0$ only depending on r and a doubling constant such that $D^{\frac{r}{2}}(z)\subset D^{r\beta}(z)$ and $D^{\frac{r}{2}}(w)\subset D^{r\beta}(z)$. By Hölder’s inequality,

$$\begin{aligned} \sup_{w\in D^{r}(z)} \bigl\vert f_{1}(w)-f_{1}(z) \bigr\vert ^{s} \lesssim \biggl( \frac{1}{ \vert D^{r\beta}(z) \vert } \int _{D^{r\beta}(z)} \bigl\vert f(w)-\widehat{f}_{r}(z) \bigr\vert ^{q}\,dA(w) \biggr)^{\frac{s}{q}}. \end{aligned}$$

This infers that $\|f_{1}\|_{IO_{r}^{s,\gamma}}\lesssim \|f\|_{\mathrm{IMO}^{s,\gamma ,q}_{r \beta}}$.

Next, we consider the property of the function $f_{2}$.

$$\begin{aligned}& \rho (z)^{\gamma} \biggl(\frac{1}{ \vert D^{\frac{r}{2}}(z) \vert } \int _{D^{ \frac{r}{2}}(z)} \bigl\vert f_{2}(w) \bigr\vert ^{q}\,dA(w) \biggr)^{\frac{1}{q}} \\& \quad \lesssim \rho (z)^{\gamma} \biggl(\frac{1}{ \vert D^{\frac{r}{2}}(z) \vert } \int _{D^{\frac{r}{2}}(z)} \bigl\vert f(w)-f_{1}(z) \bigr\vert ^{q}\,dA(w) \biggr)^{ \frac{1}{q}} \\& \qquad {} +\rho (z)^{\gamma} \biggl(\frac{1}{ \vert D^{\frac{r}{2}}(z) \vert } \int _{D^{ \frac{r}{2}}(z)} \bigl\vert f_{1}(w)-f_{1}(z) \bigr\vert ^{q}\,dA(w) \biggr)^{\frac{1}{q}} \\& \quad \lesssim \rho (z)^{\gamma}MO_{q,\frac{r}{2}}(f) (z)+\rho (z)^{\gamma} \omega _{\frac{r}{2}}(f_{1}) (z). \end{aligned}$$

Together with the property of function $f_{1}$, we have $\|f_{2}\|_{IA_{r}^{s,\gamma ,q}}\lesssim \|f\|_{\mathrm{IMO}^{s,\gamma ,q}_{ \frac{r}{2}}}$.

On the other hand, we show that $f\in \mathrm{IMO}^{s,\gamma ,q}_{r}$ as soon as $f\in IO_{r}^{s,\gamma}$ or $f\in IA_{r}^{s,\gamma ,q}$,

$$\begin{aligned}& \rho (z)^{\gamma} \biggl(\frac{1}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert f(w)- \widehat{f}_{r}(z) \bigr\vert ^{q}\,dA(w) \biggr)^{\frac{1}{q}} \\& \quad \lesssim \rho (z)^{\gamma} \biggl(\frac{1}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert f(w)-f(z) \bigr\vert ^{q}\,dA(w) \biggr)^{\frac{1}{q}}+\rho (z)^{\gamma} \bigl\vert f(z)-\widehat{f}_{r}(z) \bigr\vert . \end{aligned}$$

Therefore, $f\in \mathrm{IMO}^{s,\gamma ,q}_{r}$ with $\|f\|_{\mathrm{IMO}^{s,\gamma ,q}_{r}}\lesssim \|f_{1}\|_{IO^{s,\gamma}}$. Secondly, if $f\in IA_{r}^{s,\gamma ,q}$,

$$\begin{aligned}& \rho (z)^{\gamma} \biggl(\frac{1}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert f(w)- \widehat{f}_{r}(z) \bigr\vert ^{q}\,dA(w) \biggr)^{\frac{1}{q}} \\& \quad \lesssim \rho (z)^{\gamma} \biggl(\frac{1}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert f(w) \bigr\vert ^{q}\,dA(w) \biggr)^{\frac{1}{q}}+\rho (z)^{\gamma} \bigl\vert \widehat{f}_{r}(z) \bigr\vert \\& \quad \lesssim \rho (z)^{\gamma} \biggl(\frac{1}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert f(w) \bigr\vert ^{q}\,dA(w) \biggr)^{\frac{1}{q}}. \end{aligned}$$

Therefore, $f\in \mathrm{IMO}_{r}^{s,\gamma ,q}$ with $\|f\|_{\mathrm{IMO}_{r}^{s,\gamma ,q}}\lesssim \|f_{2}\|_{IA_{r}^{s,\gamma}}$. □

The above tells us that the space $\mathrm{IMO}^{s,\gamma ,q}_{r}$ is independent of r, so we will write $\mathrm{IMO}^{s,\gamma ,q}$ for $\mathrm{IMO}^{s,\gamma ,q}_{r}$.

Lemma 2.5

Let $1\leq q<\infty$, $1\leq s\leq \infty$, $\gamma \in \mathbb{R}$ and $r>0$. Then, for any $f\in L^{p}_{\mathrm{loc}}$, there holds $f\in \mathrm{IMO}^{s,\gamma ,q}$ if and only if there exists a continuous function $c(z)$ on $\mathbb{C}$ such that

$$ \int _{\mathbb{C}} \biggl(\frac{\rho (z)^{q\gamma}}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert f(w)-c(z) \bigr\vert ^{q}\,dA(w) \biggr)^{\frac{s}{q}}\,dA(z)< \infty . $$

While $f\in VMO^{\gamma ,q}$ if and only if there is a continuous function $c(z)$ on $\mathbb{C}$ satisfying

$$ \lim_{ \vert z \vert \to \infty}\frac{\rho (z)^{q\gamma}}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert f(w)-c(z) \bigr\vert ^{q}\,dA(w)=0. $$

Proof

If $f\in \mathrm{IMO}^{s,\gamma ,q}$, then the result holds with $c(z)=\widehat{f}_{r}(z)$, which is continuous for $z\in \mathbb{C}$. Conversely, by Minkowski’s inequality and Hölder’s inequality,

$$\begin{aligned}& \rho (z)^{\gamma} \biggl(\frac{1}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert f(w)- \widehat{f}_{r}(z) \bigr\vert ^{q}\,dA(w) \biggr)^{\frac{1}{q}} \\& \quad \lesssim \rho (z)^{\gamma} \biggl(\frac{1}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert f(w)-c(z) \bigr\vert ^{q}\,dA(w) \biggr)^{\frac{1}{q}}+\rho (z)^{\gamma} \bigl\vert \widehat{f}_{r}(z)-c(z) \bigr\vert \\& \quad \lesssim \rho (z)^{\gamma} \biggl(\frac{1}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert f(w)-c(z) \bigr\vert ^{q}\,dA(w) \biggr)^{\frac{1}{q}}. \end{aligned}$$

Therefore, $f\in \mathrm{IMO}^{s,\gamma ,q}$. The last statement is similar and hence is omitted here. □

Lemma 2.6

Suppose that $1\leq q<\infty$, $\gamma \in \mathbb{R}$, $r>0$ and $1\leq s\leq \infty$. Then, for $f\in L^{q}_{\mathrm{loc}}$, there exist $h_{1},h_{2}\in H(D^{r}(z))$ satisfying

$$ \int _{\mathbb{C}} \biggl(\frac{\rho (z)^{q\gamma}}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert f(w)-h_{1}(w) \bigr\vert ^{q}\,dA(w) \biggr)^{\frac{s}{q}}\,dA(z)< \infty , $$

and

$$ \int _{\mathbb{C}} \biggl(\frac{\rho (z)^{q\gamma}}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert \bar{f}(w)-h_{2}(w) \bigr\vert ^{q}\,dA(w) \biggr)^{\frac{s}{q}}\,dA(z)< \infty , $$

for any $z\in \mathbb{C}^{n}$, then $f\in \mathrm{IMO}^{s,\gamma ,q}$.

Proof

To start, we will claim that if $v:D^{r}(z)\to \mathbb{R}$ is pluriharmonic, there exists a pluriharmonic function u such that $u+iv\in H(D^{r}(z))$ and, moreover,

$$ \int _{D^{r}(z)} \bigl\vert u(w)-u(z) \bigr\vert ^{q}\,dA(w)\lesssim \int _{D^{r}(z)} \bigl\vert v(w) \bigr\vert ^{q}\,dA(w). $$

In fact, Lemma 22 in [7] implies that both $u\circ \tau _{z}$ and $v\circ \tau _{z}$ are still pluriharmonic, where $\tau _{z}(w)=z+w\rho (z)$ is a translation for any $z,w\in \mathbb{C}$. Applying Theorem 6 in [12] to $u\circ \tau _{z}$ and $v\circ \tau _{z}$, we can see that

$$ \int _{D(0,1)} \bigl\vert u\circ \tau _{z}(rw) \bigr\vert ^{q}\,dA(w)\lesssim \int _{D(0,1)} \bigl\vert v \circ \tau _{z}(rw) \bigr\vert ^{q}\,dA(w),\quad 0< r< 1. $$

Using the ideas in the proof of Lemma 4.1 in [10], we choose a large enough $R>0$ satisfied $s=\frac{r}{R}<1$ when $r>1$, and denote by

$$ (u\circ \tau _{z})_{R}(w)=u\circ \tau _{z}(Rw)\quad \text{and}\quad (v \circ \tau _{z})_{R}(w)=v \circ \tau _{z}(Rw), $$

then we can obtain that

$$\begin{aligned} \int _{D(0,1)} \bigl\vert u\circ \tau _{z}(rw) \bigr\vert ^{q}\,dA(w) =& \int _{D(0,1)} \bigl\vert (u \circ \tau _{z})_{R}(sw) \bigr\vert ^{q}\,dA(w) \\ \lesssim & \int _{D(0,1)} \bigl\vert (v\circ \tau _{z})_{R}(sw) \bigr\vert ^{q}\,dA(w) \\ =& \int _{D(0,1)} \bigl\vert v\circ \tau _{z}(rw) \bigr\vert ^{q}\,dA(w). \end{aligned}$$

Therefore, for any $r>0$, $1\leq q<\infty$,

$$ \int _{D(0,r)} \bigl\vert u\circ \tau _{z}(w) \bigr\vert ^{q}\,dA(w)\lesssim \int _{D(0,r)} \bigl\vert v \circ \tau _{z}(w) \bigr\vert ^{q}\,dA(w). $$

Note the fact that $dA(\tau _{z}(w))=\rho ^{2}(z)\,dA(w)$,

$$\begin{aligned} \int _{D^{r}(z)} \bigl\vert u(w) \bigr\vert ^{q}\,dA(w) =& \int _{D(0,r)} \bigl\vert (u\circ \tau _{z}) (w) \bigr\vert ^{q} \rho ^{2}(z)\,dA(w) \\ \lesssim & \int _{D(0,r)} \bigl\vert (v\circ \tau _{z}) (w) \bigr\vert ^{q}\rho ^{2}(z)\,dA(w) \\ =& \int _{D^{r}(z)} \bigl\vert v(w) \bigr\vert ^{q}\,dA(w). \end{aligned}$$

Our goal is obtained as soon as we apply the triangle inequality.

For conciseness, we denote a new symbol as

$$ \Vert f \Vert _{q,*}= \biggl(\frac{\rho (z)^{q\gamma}}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert f(w) \bigr\vert ^{q}\,dA(w) \biggr)^{\frac{1}{q}}. $$

In terms of the triangle inequality, for any $f\in L^{q}_{\mathrm{loc}}$, we can see that

$$\begin{aligned}& \int _{\mathbb{C}} \biggl\Vert \frac{f(w)+\bar{f}(w)}{2}- \frac{h_{1}(w)+h_{2}(w)}{2} \biggr\Vert _{q,*}^{s}\,dA(z) \\& \quad \lesssim \int _{\mathbb{C}} \biggl\Vert \frac{f(w)-h_{1}(w)}{2} \biggr\Vert _{q,*}^{s}\,dA(z)+ \int _{\mathbb{C}} \biggl\Vert \frac{\bar{f}(w)-h_{2}(w)}{2} \biggr\Vert _{q,*}^{s}\,dA(z)< \infty . \end{aligned}$$

Since $f+\bar{f}$ is real valued, we can obtain that

$$ \int _{\mathbb{C}} \biggl\Vert \operatorname{Im}\frac{h_{1}(w)+h_{2}(w)}{2} \biggr\Vert _{q,*}^{s}\,dA(z)< \infty . $$

It follows from the above that

$$ \int _{\mathbb{C}} \biggl\Vert \operatorname{Re}\frac{h_{1}(w)+h_{2}(w)}{2}- \operatorname{Re} \frac{h_{1}(z)+h_{2}(z)}{2} \biggr\Vert _{q,*}^{s}\,dv(z)< \infty . $$

Hence, we can achieve that

$$\begin{aligned}& \int _{\mathbb{C}} \biggl\Vert \frac{f(w)+\bar{f}(w)}{2}-\operatorname{Re} \frac{h_{1}(z)+h_{2}(z)}{2} \biggr\Vert _{q,*}^{s}\,dv(z) \\& \quad \lesssim \int _{\mathbb{C}^{n}} \biggl\Vert \frac{f(w)+\bar{f}(w)}{2}-\operatorname{Re} \frac{h_{1}(w)+h_{2}(w)}{2} \biggr\Vert _{q,*}^{s}\,dv(z) \\& \qquad {}+ \int _{\mathbb{C}} \biggl\Vert \operatorname{Re}\frac{h_{1}(w)+h_{2}(w)}{2}- \operatorname{Re} \frac{h_{1}(z)+h_{2}(z)}{2} \biggr\Vert _{q,*}^{s}\,dv(z) \\& \quad \lesssim \int _{\mathbb{C}^{n}} \biggl\Vert \frac{f(w)+\bar{f}(w)}{2}- \frac{h_{1}(w)+h_{2}(w)}{2}\biggr|_{q,*}^{s}\,dv(z) \\& \qquad {}+ \int _{\mathbb{C}} \biggl\Vert \operatorname{Re}\frac{h_{1}(w)+h_{2}(w)}{2}- \operatorname{Re} \frac{h_{1}(z)+h_{2}(z)}{2}\biggr\| _{q,*}^{s}\,dv(z). \end{aligned}$$

This implies that

$$ \int _{\mathbb{C}} \biggl\Vert \frac{f(w)+\bar{f}(w)}{2}-\operatorname{Re} \frac{h_{1}(z)+h_{2}(z)}{2} \biggr\Vert _{q,*}^{s}\,dv(z)< \infty . $$

At the same time, we use a similar procedure to produce

$$ \int _{\mathbb{C}} \biggl\Vert \operatorname{Im}\frac{f(w)+\bar{f}(w)}{2}- \operatorname{Im} \frac{h_{1}(z)+h_{2}(z)}{2} \biggr\Vert _{q,*}^{s}\,dv(z)< \infty . $$

If we choose that

$$ c(z)=\operatorname{Re}\frac{h_{1}(z)+h_{2}(z)}{2}+i\operatorname{Im}\frac{h_{1}(z)+h_{2}(z)}{2}, $$

then we find that

$$ \int _{\mathbb{C}} \biggl(\frac{\rho (z)^{q\gamma}}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert f(w)-c(z) \bigr\vert ^{q}\,dA(w) \biggr)^{\frac{s}{q}}\,dA(z)< \infty . $$

Finally, the desired result follows from Lemma 2.5. □

Lemma 2.7

A function $f\in BMO^{\gamma ,q}$ belongs to $VMO^{\gamma ,q}$ if and only if

$$ \lim_{R\to \infty} \Vert f_{R}-f \Vert _{BMO^{\gamma ,q}}=0, $$

where $f_{R}(z)=f\circ \chi _{R}(z)$ for the characteristic function $\chi _{R}(z)$ of $D^{R}(0)$.

Proof

First, suppose that $f\in VMO^{\gamma ,q}$. By the definition, for $0< r< R$,

$$\begin{aligned}& \frac{\rho (z)^{q\gamma}}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert (f_{R}-f) (w)- \widehat{(f_{R}-f)_{r}}(z) \bigr\vert ^{q}\,dA(w) \\& \quad \lesssim \frac{\rho (z)^{q\gamma}}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert f_{R}(w)- \widehat{(f_{R})_{r}}(z)+\widehat{f}_{r}(z)-f(w) \bigr\vert ^{q}\,dA(w). \end{aligned}$$

The result follows from the Lebesgue Dominated Convergence Theorem.

On the other hand, we know, in view of the triangle inequality, that

$$\begin{aligned}& \frac{\rho (z)^{q\gamma}}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert f(w)- \widehat{f}_{r}(z) \bigr\vert ^{q}\,dA(w) \\& \quad \lesssim \frac{\rho (z)^{q\gamma}}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert (f_{R}-f) (w)- \widehat{(f_{R}-f)_{r}}(z) \bigr\vert ^{q}\,dA(w) \\& \qquad {}+\frac{\rho (z)^{q\gamma}}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert f_{R}(w)- \widehat{(f_{R})_{r}}(z) \bigr\vert ^{q}\,dA(w). \end{aligned}$$

For any $r>0$, $\varepsilon >0$, we can choose some positive R such that $\|f_{R}-f\|_{BMO^{\gamma ,q}_{r}}<\varepsilon$. Since $f_{R}(z)=0$ for $D^{r}(z)\subset D^{R}(0)^{c}$, it is clear that

$$ \frac{\rho (z)^{q\gamma}}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert f_{R}(w)- \widehat{(f_{R})_{r}}(z) \bigr\vert ^{q}\,dA(w)=0. $$

In the above assumption, we can see that

$$ \frac{\rho (z)^{q\gamma}}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert f(w)- \widehat{f_{r}}(z) \bigr\vert ^{q}\,dA(w)< \varepsilon . $$

The proof is completed for ε that is arbitrary. □

Lemma 2.8

For any continuous function f on $\mathbb{C}$, there holds

$$ \bigl\vert f(w)-f(z) \bigr\vert \lesssim \bigl(1+d_{\varphi}(z,w) \bigr)^{\frac{2-\delta}{\delta}} \int _{0}^{1}\omega _{1}(f)\circ \tau _{z}(t\xi )\,dt, $$

where $0<\delta <1$, $\tau _{z}(\eta )=z+\eta \rho (z)$ is a translation for any $z,\eta \in \mathbb{C}$ and ξ is chosen to satisfy that $w=\tau _{z}(\xi )$.

Proof

According to the Lemma 2.1 with the $r=1$ case, we can choose a $\xi \in \mathbb{C}$ such that $w=\tau _{z}(\xi )$ for any $z,w\in \mathbb{C}$. Moreover,

$$ \vert \xi \vert =\frac{ \vert z-w \vert }{\rho (z)}\leq \max \bigl\{ Cd_{\varphi}(z,w), \bigl(Cd_{ \varphi}(z,w)\bigr)^{\frac{1}{\delta}} \bigr\} . $$

When $w\in \overline{D(z)}$, $0< t<1$, we can see that

$$ \bigl\vert f(w)-f(z) \bigr\vert \leq \bigl\vert f(w)-f\circ \tau _{z}(t\xi ) \bigr\vert + \bigl\vert f\circ \tau _{z}(t \xi )-f(z) \bigr\vert \lesssim \omega _{1}(f) \circ \tau _{z}(t\xi ). $$

Integrating both sides with respect to t from 0 to 1,

$$ \bigl\vert f(w)-f(z) \bigr\vert \lesssim \int _{0}^{1}\omega _{1}(f)\circ \tau _{z}(t\xi )\,dt. $$

Now, for $w\notin \overline{D(z)}$, suppose that C is a positive constant only dependent on a doubling constant and, at the same time, let

$$ N=\max \bigl\{ 1+ \bigl[C^{\frac{1}{\delta}}d_{\varphi}(z,w)^{ \frac{2-\delta}{\delta}} \bigr], \bigl[C^{\frac{1}{\delta}}+d_{ \varphi}(z,w)^{\frac{2-\delta}{\delta}} \bigr] \bigr\} , $$

where $[x]$ denotes the largest integer less than or equal to x. Subsequently, we place N points $z_{0}=z, z_{1}, \ldots ,z_{N}=w$ from z to w on the geodesic in such a way that $z_{j+1}\in D(z_{j})$, $0\leq j\leq N-1$. We will claim that it is possible. To this end, if we set $z_{j}=\tau _{z}(\frac{j}{N}\xi )$, then we have

$$ \frac{ \vert z_{j+1}-z_{j} \vert }{\rho (z_{j})}=\frac{ \vert \xi \vert }{N} \frac{\rho (z)}{\rho (z_{j})}. $$

Obviously, $z_{1}\in D(z)$. Furthermore, we can induce that $z_{2}\in D(z_{1})$. Combining Lemma 2 in [7] and Lemma 2.1, we can see that

$$ \frac{ \vert \xi \vert }{N}\frac{\rho (z)}{\rho (z_{j})}\leq \frac{ \vert \xi \vert }{N}C^{ \frac{1}{\delta}}d_{\varphi}(z,z_{j})^{\frac{1-\delta}{\delta}} \leq \frac{C^{\frac{2}{\delta}}}{N}d_{\varphi}(z,w)^{ \frac{2-\delta}{\delta}}< 1. $$

Therefore, we can achieve that

$$\begin{aligned} \bigl\vert f(w)-f(z) \bigr\vert \leq &\sum_{j=0}^{N-1} \biggl\vert f\circ \tau _{z} \biggl( \frac{j+1}{N}\xi \biggr)-f\circ \tau _{z} \biggl(\frac{j}{N}\xi \biggr) \biggr\vert \\ \lesssim &\sum_{j=0}^{N-1} \int _{0}^{1}\omega _{1}(f)\circ \tau _{z} \biggl(\frac{j+t}{N}\xi \biggr)\,dt \\ \lesssim &N\sum_{j=0}^{N-1} \int _{\frac{j}{N}}^{\frac{j+1}{N}} \omega _{1}(f)\circ \tau _{z}(t\xi )\,dt \\ \lesssim &\bigl(1+d_{\varphi}(z,w)\bigr)^{\frac{2-\delta}{\delta}} \int _{0}^{1} \omega _{1}(f)\circ \tau _{z}(t\xi )\,dt. \end{aligned}$$

Hence, we complete the proof. □

3 Hankel operators from $F^{p}_{\varphi}$ to $L^{q}_{\varphi}$ for $1\leq p\leq q<\infty$

In this section, we will study the boundedness of Hankel operators from $F^{p}_{\varphi}$ to $L^{q}_{\varphi}$ for $1\leq p\leq q<\infty$. As the IMO functions have a decomposition that has been discussed above, we will characterize the Hankel operators with BO symbols in the beginning.

Lemma 3.1

If $f\in BO^{2(\frac{1}{q}-\frac{1}{p})}$, then the operator $H_{f}$ is bounded from $F^{p}_{\varphi}$ to $L^{q}_{\varphi}$ for $1\leq p\leq q<\infty$.

Proof

In order to give the complete argument, we first calculate the useful estimates as follows. For any $s,\varepsilon >0$, we know that

$$ \sup_{x>0}(1+x)^{s}e^{-\frac{1}{2}x^{\varepsilon}}< \infty . $$

Together with the Lemma 2.1, we can see that

$$\begin{aligned}& \int _{\mathbb{C}}\bigl(1+d_{\varphi}(z,w)\bigr)^{q+2q \vert \frac{1}{q}- \frac{1}{p} \vert } \bigl(e^{-\varphi (z)} \bigl\vert K_{\varphi}(z,w) \bigr\vert e^{-\varphi (w)} \bigr)\,dA(z) \\& \quad \lesssim \rho (w)^{-1} \biggl( \int _{D^{r}(z)}+ \int _{D^{r}(z)^{c}} \biggr) \bigl(1+d_{\varphi}(z,w) \bigr)^{q+2q \vert \frac{1}{q}-\frac{1}{p} \vert }e^{-( \frac{ \vert z-w \vert }{\rho (w)})^{\varepsilon}}\frac{dA(z)}{\rho (z)} \\& \quad \lesssim \rho (w)^{-1} \int _{D^{r}(z)} \biggl(1+ \frac{ \vert z-w \vert }{\rho (w)} \biggr)^{q+2q \vert \frac{1}{q}-\frac{1}{p} \vert } \rho (z)^{-1}e^{-( \frac{ \vert z-w \vert }{\rho (w)})^{\varepsilon}}\,dA(z) \\& \qquad {}+\rho (w)^{-1} \int _{D^{r}(z)^{c}} \biggl[1+ \biggl( \frac{ \vert z-w \vert }{\rho (w)} \biggr)^{2-t} \biggr]^{q+2q \vert \frac{1}{q}- \frac{1}{p} \vert }e^{-(\frac{ \vert z-w \vert }{\rho (w)})^{\varepsilon}} \frac{dA(z)}{\rho (z)} \\& \quad \lesssim \rho (w)^{-1} \int _{\mathbb{C}}\rho (z)^{-1}e^{-\frac{1}{2}( \frac{ \vert z-w \vert }{\rho (w)})^{\varepsilon}}\,dA(z)< \infty . \end{aligned}$$

Here, the last step follows from the ideas of Lemma 2.1 in [4] with the case of $k=-1$ and $p=\frac{1}{2}$.

Next, noting that, by Lemma 2.8 in [9],

$$\begin{aligned} \int _{\mathbb{C}}e^{-\varphi (z)} \bigl\vert K_{\varphi}(z,w) \bigr\vert e^{-\varphi (w)}\,dA(w)< \infty . \end{aligned}$$

(5)

In view of Hölder’s inequality, Fubini’s Theorem, Lemma 2.3, and the analysis above, we can continue to calculate that

$$\begin{aligned}& \int _{\mathbb{C}} \bigl\vert H_{f}g(z) \bigr\vert ^{q}e^{-q\varphi (z)}\,dA(z) \\& \quad \lesssim \int _{\mathbb{C}} \biggl[ \biggl( \int _{\mathbb{C}} \bigl\vert f(w)-f(z) \bigr\vert ^{q} \bigl\vert g(w) \bigr\vert ^{q}e^{-q \varphi (w)} \bigl(e^{-\varphi (z)} \bigl\vert K_{\varphi}(z,w) \bigr\vert e^{-\varphi (w)} \bigr)\,dA(w) \biggr) \\& \qquad {} \times \biggl( \int _{\mathbb{C}}e^{-\varphi (z)} \bigl\vert K_{\varphi}(z,w) \bigr\vert e^{- \varphi (w)}\,dA(w) \biggr)^{q-1} \biggr]\,dA(z) \\& \quad \lesssim \int _{\mathbb{C}} \biggl[ \int _{\mathbb{C}} \bigl\vert f(w)-f(z) \bigr\vert ^{q} \bigl\vert g(w) \bigr\vert ^{q}e^{-q \varphi (w)} \biggl( \frac{ \vert K_{\varphi}(z,w) \vert }{e^{\varphi (z)}e^{\varphi (w)}} \biggr)\,dA(w) \biggr]\,dA(z) \\& \quad \lesssim \int _{\mathbb{C}}\frac{ \vert g(w) \vert ^{q}}{e^{q\varphi (w)}} \biggl[ \int _{\mathbb{C}} \frac{(1+d_{\varphi}(z,w))^{q+2q \vert \frac{1}{q}-\frac{1}{p} \vert }}{\rho (w)^{2q(\frac{1}{q}-\frac{1}{p})}} \biggl(\frac{ \vert K_{\varphi}(z,w) \vert }{e^{\varphi (z)}e^{\varphi (w)}} \biggr)\,dA(z) \biggr]\,dA(w) \\& \quad \lesssim \int _{\mathbb{C}}\rho (w)^{\frac{2}{p}(q-p)} \bigl\vert g(w) \bigr\vert ^{p}e^{-p \varphi (w)} \bigl( \bigl\vert g(w) \bigr\vert e^{-\varphi (w)} \bigr)^{q-p}\,dA(w). \end{aligned}$$

Finally, we have

$$ \int _{\mathbb{C}} \bigl\vert H_{f}g(z) \bigr\vert ^{q}e^{-q\varphi (z)}\,dA(z)\lesssim \Vert g \Vert ^{q}_{F^{p}_{ \varphi}}, $$

if noting that $\sup_{w\in \mathbb{C}}\rho (w)^{\frac{2}{p}}|g(w)|e^{- \varphi (w)}\lesssim \|g\|_{F^{p}_{\varphi}}$. □

Lemma 3.2

If $f\in BA^{2(\frac{1}{q}-\frac{1}{p}),q}$, then the operator $H_{f}$ is bounded from $F^{p}_{\varphi}$ to $L^{q}_{\varphi}$ for $1\leq p\leq q<\infty$.

Proof

For any $g\in F^{p}_{\varphi}$, by the definition of Hankel’s operator, we see that

$$\begin{aligned}& \int _{\mathbb{C}} \bigl\vert H_{f}g(z) \bigr\vert ^{q}e^{-q\varphi (z)}\,dA(z) \\& \quad \lesssim \int _{\mathbb{C}} \bigl\vert f(z) \bigr\vert ^{q} \bigl\vert g(z) \bigr\vert ^{q}e^{-q\varphi (z)}\,dA(z) \\& \qquad {}+ \int _{\mathbb{C}} \biggl( \int _{\mathbb{C}} \bigl\vert f(w) \bigr\vert \bigl\vert g(w) \bigr\vert \bigl\vert K_{\varphi}(z,w) \bigr\vert \frac{dA(w)}{e^{2\varphi (w)}} \biggr)^{q} \frac{dA(z)}{e^{q\varphi (z)}}. \end{aligned}$$

Let $\{a_{k}\}_{k\geq 1}$ be an r-lattice in $\mathbb{C}^{n}$. Together with the point estimate (referred to in Lemma 19 in [7]), we will estimate the first integral as

$$\begin{aligned}& \int _{\mathbb{C}} \bigl\vert f(w) \bigr\vert ^{q} \bigl\vert g(w) \bigr\vert ^{q}e^{-q\varphi (w)}\,dA(w) \\& \quad \lesssim \sum_{k=1}^{\infty} \int _{D^{r}(a_{k})} \frac{ \vert f(w) \vert ^{q}}{\rho (w)^{2q(\frac{1}{q}-\frac{1}{p})}}\rho (w)^{2q( \frac{1}{q}-\frac{1}{p})} \bigl\vert g(w) \bigr\vert ^{q}e^{-q\varphi (w)}\,dA(w) \\& \quad \lesssim \sum_{k=1}^{\infty} \biggl(\rho (a_{k})^{2q(\frac{1}{q}- \frac{1}{p})}\frac{1}{\rho ^{2}(a_{k})} \int _{D^{r}(a_{k})} \bigl\vert f(w) \bigr\vert ^{q}\,dA(w) \biggr) \\& \qquad {} \times \Bigl(\sup_{w\in D^{r}(a_{k})}\rho (w)^{2\frac{q}{p}} \bigl\vert g(w) \bigr\vert ^{q}e^{-q \varphi (w)} \Bigr) \\& \quad \lesssim \biggl(\sup_{z\in \mathbb{C}}\rho (z)^{2q(\frac{1}{q}- \frac{1}{p})} \frac{1}{\rho ^{2}(z)} \int _{D^{r}(z)} \bigl\vert f(w) \bigr\vert ^{q}\,dA(w) \biggr) \\& \qquad {} \times \Biggl(\sum_{k=1}^{\infty}\sup _{w\in D^{r}(a_{k})}\rho ^{2}(w) \bigl\vert g(w) \bigr\vert ^{p}e^{-p \varphi (w)} \Biggr)^{\frac{q}{p}} \\& \quad \lesssim \Vert g \Vert ^{q}_{F^{p}_{\varphi}} \biggl(\sup _{z\in \mathbb{C}} \rho (z)^{2q(\frac{1}{q}-\frac{1}{p})}\frac{1}{\rho ^{2}(z)} \int _{D^{r}(z)} \bigl\vert f(w) \bigr\vert ^{q}\,dA(w) \biggr). \end{aligned}$$

Subsequently, the other integral comes into this, by Hölder’s inequality, Fubini’s Theorem, and the estimate (5),

$$\begin{aligned}& \int _{\mathbb{C}} \biggl( \int _{\mathbb{C}} \bigl\vert f(w) \bigr\vert \bigl\vert g(w) \bigr\vert \bigl\vert K_{\varphi}(z,w) \bigr\vert e^{-2 \varphi (w)}\,dA(w) \biggr)^{q}e^{-q\varphi (z)}\,dA(z) \\& \quad \lesssim \int _{\mathbb{C}} \biggl[ \biggl( \int _{\mathbb{C}} \bigl\vert f(w) \bigr\vert ^{q} \bigl\vert g(w) \bigr\vert ^{q}e^{-q \varphi (w)} \bigl(e^{-\varphi (z)} \bigl\vert K_{\varphi}(z,w) \bigr\vert e^{-\varphi (w)} \bigr)\,dA(w) \biggr) \\& \qquad {} \times \biggl( \int _{\mathbb{C}}e^{-\varphi (z)} \bigl\vert K_{\varphi}(z,w) \bigr\vert e^{- \varphi (w)}\,dA(w) \biggr)^{q-1} \biggr]\,dA(z) \\& \quad \lesssim \int _{\mathbb{C}} \biggl[ \int _{\mathbb{C}} \bigl\vert f(w) \bigr\vert ^{q} \bigl\vert g(w) \bigr\vert ^{q}e^{-q \varphi (w)} \bigl(e^{-\varphi (z)} \bigl\vert K_{\varphi}(z,w) \bigr\vert e^{-\varphi (w)} \bigr)\,dA(w) \biggr]\,dA(z) \\& \quad \lesssim \int _{\mathbb{C}} \bigl\vert f(w) \bigr\vert ^{q} \bigl\vert g(w) \bigr\vert ^{q}e^{-q\varphi (w)}\,dA(w). \end{aligned}$$

From now on, our proof is completed as soon as we apply the estimate of the first integral. □

Theorem 3.3

Suppose $1\leq p\leq q<\infty$, $f\in BMO^{2(\frac{1}{q}-\frac{1}{p}),q}$ if and only if the operators $H_{f}$ and $H_{\bar{f}}$ are both bounded from $F^{p}_{\varphi}$ to $L^{q}_{\varphi}$. Moreover,

$$ \Vert H_{f} \Vert _{F^{p}_{\varphi}\to L^{q}_{\varphi}}+ \Vert H_{\bar{f}} \Vert _{F^{p}_{ \varphi}\to L^{q}_{\varphi}}\simeq \Vert f \Vert _{BMO^{2(\frac{1}{q}- \frac{1}{p}),q}}. $$

Proof

If $f\in BMO^{2(\frac{1}{q}-\frac{1}{p}),q}$, then so is f̄ and it follows from the previous two lemmas that $H_{f}$ and $H_{\bar{f}}$ are bounded.

Suppose now that $H_{f}$ and $H_{\bar{f}}$ are both bounded from $F^{p}_{\varphi}$ to $L^{q}_{\varphi}$. Recall that $\{k_{2,z}\rho (z)^{1-\frac{2}{p}} \}_{z\in \mathbb{C}}$ is dense in $F^{p}_{\varphi}$, which was proven in Lemma 2.6 of [4]. Also, we have $H_{f}, H_{\bar{f}}:F_{\varphi}^{p}\to L_{\varphi}^{q}$ are bounded simultaneously. That is,

$$ \bigl\Vert H_{f} \bigl(k_{2,z}\rho (z)^{1-\frac{2}{p}} \bigr) \bigr\Vert _{L^{q}_{ \varphi}}< \infty \quad \text{and} \quad \bigl\Vert H_{\bar{f}} \bigl(k_{2,z}\rho (z)^{1- \frac{2}{p}} \bigr) \bigr\Vert _{L^{q}_{\varphi}}< \infty . $$

In detail, by the definition,

$$\begin{aligned} H_{f} \biggl(\frac{k_{2,z}}{\rho (z)^{\frac{2}{p}-1}} \biggr)=f(w) \biggl( \frac{k_{2,z}(w)}{\rho (z)^{\frac{2}{p}-1}} \biggr)- \int _{ \mathbb{C}}f(\lambda ) \biggl( \frac{k_{2,z}(\lambda )}{\rho (z)^{\frac{2}{p}-1}} \biggr)K_{\varphi}(w, \lambda )\frac{dA(\lambda )}{e^{2\varphi (\lambda )}}. \end{aligned}$$

Given a small enough $r>0$ that satisfies the estimates (3), we have

$$\begin{aligned}& \int _{\mathbb{C}} \biggl\vert f(w) \biggl( \frac{k_{2,z}(w)}{\rho (z)^{\frac{2}{p}-1}} \biggr)- \int _{\mathbb{C}}f( \lambda ) \biggl(\frac{k_{2,z}(\lambda )}{\rho (z)^{\frac{2}{p}-1}} \biggr)K_{\varphi}(w,\lambda ) \frac{dA(\lambda )}{e^{2\varphi (\lambda )}} \biggr\vert ^{q} \frac{dA(w)}{e^{q\varphi (w)}} \\& \quad \gtrsim \int _{D^{r}(z)} \biggl\vert f(w)- \frac{\rho (z)^{\frac{2}{p}-1}}{k_{2,z}(w)} \int _{\mathbb{C}}f( \lambda )k_{2,z}(\lambda )\rho (z)^{1-\frac{2}{p}}K_{\varphi}(w, \lambda )e^{-2\varphi (\lambda )}\,dA(\lambda ) \biggr\vert ^{q} \\& \qquad {} \times \biggl\vert \frac{k_{2,z}(w)}{\rho (z)^{\frac{2}{p}-1}} \biggr\vert ^{q}e^{-q \varphi (w)}\,dA(w) \\& \quad \gtrsim \frac{\rho (z)^{2q(\frac{1}{q}-\frac{1}{p})}}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \biggl\vert f(w)- \frac{\rho (z)^{\frac{2}{p}-1}}{k_{2,z}(w)} \int _{\mathbb{C}}f( \lambda )k_{2,z}(\lambda )\rho (z)^{1-\frac{2}{p}} \\& \qquad {} \times K_{\varphi}(w,\lambda )e^{-2\varphi (\lambda )}\,dA( \lambda ) \biggr\vert ^{q}\,dA(w). \end{aligned}$$

Here, the last estimate above follows from the fact that, for any $w\in D^{r}(z)$,

$$\begin{aligned} \biggl\vert \frac{k_{2,z}(w)}{\rho (z)^{\frac{2}{p}-1}} \biggr\vert ^{q}e^{-q \varphi (w)} \gtrsim \biggl\vert \frac{K_{\varphi}(z,w)\rho (z)}{ \rho (z)^{\frac{2}{p}-1}e^{\varphi (z)}} \biggr\vert ^{q}e^{-q\varphi (w)} \gtrsim \frac{\rho (z)^{2q(\frac{1}{q}-\frac{1}{p})}}{ \vert D^{r}(z) \vert }. \end{aligned}$$

This means we can have the estimate that

$$\begin{aligned} \biggl\Vert H_{f} \biggl(\frac{k_{2,z}}{\rho (z)^{\frac{2}{p}-1}} \biggr) \biggr\Vert ^{q}_{L^{q}_{\varphi}} \gtrsim & \frac{\rho (z)^{2q(\frac{1}{q}-\frac{1}{p})}}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \biggl\vert f(w)-\frac{\rho (z)^{\frac{2}{p}-1}}{k_{2,z}(w)} \\ &{}\times \int _{\mathbb{C}}f(\lambda ) \frac{k_{2,z}(\lambda )}{\rho (z)^{\frac{2}{p}-1}}K_{\varphi}(w, \lambda )\frac{dA(\lambda )}{e^{2\varphi (\lambda )}} \biggr\vert ^{q}\,dA(w). \end{aligned}$$

Similarly, the other estimate with the conjugate symbols becomes

$$\begin{aligned} \biggl\Vert H_{\bar{f}} \biggl(\frac{k_{2,z}}{\rho (z)^{\frac{2}{p}-1}} \biggr) \biggr\Vert ^{q}_{L^{q}_{\varphi}} \gtrsim & \frac{\rho (z)^{2q(\frac{1}{q}-\frac{1}{p})}}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \biggl\vert \bar{f}(w)-\frac{\rho (z)^{\frac{2}{p}-1}}{k_{2,z}(w)} \\ &{}\times \int _{\mathbb{C}}\bar{f}(\lambda ) \frac{k_{2,z}(\lambda )}{\rho (z)^{\frac{2}{p}-1}}K_{\varphi}(w, \lambda )\frac{dA(\lambda )}{e^{2\varphi (\lambda )}} \biggr\vert ^{q}\,dA(w). \end{aligned}$$

Since $k_{2,z}(w)\neq 0$ for any $w\in D^{r}(z)$, and setting

$$ h_{1}(w)=\frac{\rho (z)^{\frac{2}{p}-1}}{k_{2,z}(w)} \int _{\mathbb{C}}f( \lambda )k_{2,z}(\lambda )\rho (z)^{1-\frac{2}{p}}K_{\varphi}(w, \lambda )e^{-2\varphi (\lambda )}\,dA(\lambda ), $$

and, respectively,

$$ h_{2}(w)=\frac{\rho (z)^{\frac{2}{p}-1}}{k_{2,z}(w)} \int _{\mathbb{C}} \bar{f}(\lambda )k_{2,z}(\lambda )\rho (z)^{1-\frac{2}{p}}K_{\varphi}(w, \lambda )e^{-2\varphi (\lambda )}\,dA(\lambda ), $$

one can obtain that $h_{1}$ and $h_{2}$ are both holomorphic functions on $D^{r}(z)$ and moreover,

$$ \biggl(\frac{\rho (z)^{2q(\frac{1}{q}-\frac{1}{p})}}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert f(w)-h_{1}(w) \bigr\vert ^{q}\,dA(w) \biggr)^{\frac{1}{q}} \lesssim \biggl\Vert H_{f} \biggl( \frac{k_{2,z}}{\rho (z)^{\frac{2}{p}-1}} \biggr) \biggr\Vert _{L^{q}_{ \varphi}}< \infty , $$

and

$$ \biggl(\frac{\rho (z)^{2q(\frac{1}{q}-\frac{1}{p})}}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert \bar{f}(w)-h_{2}(w) \bigr\vert ^{q}\,dA(w) \biggr)^{\frac{1}{q}} \lesssim \biggl\Vert H_{ \bar{f}} \biggl(\frac{k_{2,z}}{\rho (z)^{\frac{2}{p}-1}} \biggr) \biggr\Vert _{L^{q}_{\varphi}}< \infty . $$

By Lemma 2.6 for $s=\infty$, we conclude that $f\in BMO^{2(\frac{1}{q}-\frac{1}{p}),q}$, moreover

$$ \Vert f \Vert _{BMO^{2(\frac{1}{q}-\frac{1}{p}),q}}\lesssim \Vert H_{f} \Vert _{F^{p}_{ \varphi}\to L^{q}_{\varphi}}+ \Vert H_{\bar{f}} \Vert _{F^{p}_{\varphi}\to L^{q}_{ \varphi}}. $$

Now, we complete the proof. □

Theorem 3.4

$f\in VMO^{2(\frac{1}{q}-\frac{1}{p}),q}$ if and only if the operators $H_{f}$ and $H_{\bar{f}}$ are both compact from $F^{p}_{\varphi}$ to $L^{q}_{\varphi}$.

Proof

Note that $\{k_{2,z}\rho (z)^{1-\frac{2}{p}} \}_{z\in \mathbb{C}}$ is bounded in $F^{p}_{\varphi}$ and $k_{2,z}\rho (z)^{1-\frac{2}{p}}\to 0$ uniformly on compact subsets of $\mathbb{C}^{n}$ as $|z|\to \infty$ by Lemma 2.6 in [4] and Lemma 4.8 in [9]. Then, if $H_{f}$ and $H_{\bar{f}}$ are both compact,

$$ \lim_{ \vert z \vert \to \infty} \bigl\Vert H_{f} \bigl(k_{2,z}\rho (z)^{1- \frac{2}{p}} \bigr) \bigr\Vert _{L^{q}_{\varphi}}=0 \quad \text{and}\quad \lim_{ \vert z \vert \to \infty} \bigl\Vert H_{\bar{f}} \bigl(k_{2,z}\rho (z)^{1- \frac{2}{p}} \bigr) \bigr\Vert _{L^{q}_{\varphi}}=0. $$

Similar to the proof in Theorem 3.3, we have two holomorphic functions $h_{1}$ and $h_{2}$ on $D^{r}(z)$ such that

$$ \lim_{ \vert z \vert \to \infty} \frac{\rho (z)^{2q(\frac{1}{q}-\frac{1}{p})}}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert f(w)-h_{1}(w) \bigr\vert ^{q}\,dA(w)=0, $$

and

$$ \lim_{ \vert z \vert \to \infty} \frac{\rho (z)^{2q(\frac{1}{q}-\frac{1}{p})}}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert \bar{f}(w)-h_{2}(w) \bigr\vert ^{q}\,dA(w)=0. $$

If we let

$$ c(z)=\operatorname{Re}\frac{h_{1}(z)+h_{2}(z)}{2}+i\operatorname{Im}\frac{h_{1}(z)+h_{2}(z)}{2}, $$

by Lemma 2.5 and Lemma 2.6 for $s=\infty$, we achieve that $f\in VMO^{2(\frac{1}{q}-\frac{1}{p}),q}$.

On the other hand, for any $\varepsilon >0$, by Lemma 2.7, we can find some function $f_{R}$ with respect to f such that $\|f_{R}-f\|_{BMO^{2(\frac{1}{q}-\frac{1}{p}),q}}<\varepsilon$. Together with Theorem 3.3, we have

$$ \Vert H_{f}-H_{f_{R}} \Vert _{F^{p}_{\varphi}\to L^{q}_{\varphi}}+ \Vert H_{\overline{f_{R}}}-H_{\bar{f}} \Vert _{F^{p}_{\varphi}\to L^{q}_{ \varphi}}\lesssim \Vert f_{R}-f \Vert _{BMO^{2(\frac{1}{q}-\frac{1}{p}),q}}< \varepsilon . $$

Finally, it is easy to see that both $H_{f_{R}}$ and $H_{\overline{f_{R}}}$ are compact from $F^{p}_{\varphi}$ to $L^{q}_{\varphi}$ because $\{k_{2,z}\rho (z)^{1-\frac{2}{p}} \}_{z\in \mathbb{C}}$ is bounded in $F^{p}_{\varphi}$ and $k_{2,z}\rho (z)^{1-\frac{2}{p}}\to 0$ uniformly on compact subsets of $\mathbb{C}^{n}$ as $|z|\to \infty$. Now, we conclude the results. □

4 Hankel operators from $F^{p}_{\varphi}$ to $L^{q}_{\varphi}$ for $1\leq q< p<\infty$

In this section, we will study the boundedness and compactness of Hankel operators from $F^{p}_{\varphi}$ to $L^{q}_{\varphi}$ for $1\leq q< p<\infty$. The following result shows the mapping properties of the defined operator from $L^{p}(\mathbb{C}, dA)$ to $L^{q}(\mathbb{C}, dA)$.

Lemma 4.1

Let $1\leq q< p<\infty$, $\varepsilon >0$. We define an integral operator $T_{f,\varepsilon}: L^{p}(\mathbb{C}, dA)\to L^{q}(\mathbb{C}, dA)$ as

$$ T_{f,\varepsilon}h(z) \triangleq \int _{\mathbb{C}} \biggl( \int _{0}^{1} \omega _{1}(f)\circ \tau _{z}(t\xi )\,dt \biggr) h\bigl(\tau _{z}(\xi )\bigr) \frac{\rho (z)}{\rho (\tau _{z}(\xi ))}e^{-\frac{1}{2}d_{\varphi}(z, \tau _{z}(\xi ))^{\varepsilon}}\,dA(\xi ). $$

Suppose that f is Lebesgue measurable such that $f\in IO^{\frac{pq}{p-q},0}$, then

(1) The integral operator $T_{f,\varepsilon}: L^{p}(\mathbb{C}, dA)\to L^{q}(\mathbb{C}, dA)$ is bounded, moreover,

$$ \Vert T_{f,\varepsilon} \Vert _{L^{p}\to L^{q}}\lesssim \Vert f \Vert _{IO^{ \frac{pq}{p-q},0}}. $$

(2) For bounded sequence $\{h_{k}\}_{k\geq 1}\in L^{p}(\mathbb{C}, dA)$ satisfying $\lim_{k\to \infty}\sup_{|z|\leq R}|h_{k}(z)|=0$, for all $R>0$, there holds

$$ \lim_{k\to \infty} \bigl\Vert T_{f,\varepsilon}(h_{k}) \bigr\Vert _{L^{q}}=0. $$

Proof

(1) We denote, in the beginning, $s=\frac{pq}{p-q}>1$ and $s'=\frac{pq}{pq-p+q}$. By Hölder’s inequality, we have

$$\begin{aligned} \bigl\vert T_{f,\varepsilon}h(z) \bigr\vert \lesssim & \int _{\mathbb{C}} \biggl[ \int _{0}^{1} \omega _{1}(f)^{s} \circ \tau _{z}(t\xi )\,dt \biggr]^{\frac{1}{s}} \frac{\rho (z)}{\rho (\tau _{z}(\xi ))} \frac{ \vert h(\tau _{z}(\xi )) \vert dA(\xi )}{e^{\frac{1}{2}d_{\varphi}(z,\tau _{z}(t\xi ))^{\varepsilon}}} \\ \lesssim & \biggl[ \int _{\mathbb{C}} \biggl( \int _{0}^{1}\omega _{1}(f)^{s} \circ \tau _{z}(t\xi )\,dt \biggr) e^{-\frac{s}{4}d_{\varphi}(z,\tau _{z}( \xi ))^{\varepsilon}}\,dA(\xi ) \biggr]^{\frac{1}{s}} \end{aligned}$$

(6)

$$\begin{aligned} &{}\times \biggl[ \int _{\mathbb{C}} \bigl\vert h^{s'}\bigl(\tau _{z}(\xi )\bigr) \bigr\vert \frac{\rho (z)^{s'}}{\rho (\tau _{z}(\xi ))^{s'}}e^{- \frac{s'}{4}d_{\varphi}(z,\tau _{z}(\xi ))^{\varepsilon}}\,dA( \xi ) \biggr]^{\frac{1}{s'}}. \end{aligned}$$

(7)

Next, the integral (6) in $L^{1}(dA)$-norm will be first calculated, if we use the change of variable $\eta =\rho (z)\xi$, as is known, $dA(\xi )=\rho ^{-2}(z)\,dA(\eta )$,

$$\begin{aligned}& \int _{\mathbb{C}} \biggl[ \int _{\mathbb{C}} \biggl( \int _{0}^{1} \omega _{1}(f)^{s} \circ \tau _{z}(t\xi )\,dt \biggr) e^{-\frac{s}{4}d_{ \varphi}(z,\tau _{z}(\xi ))^{\varepsilon}}\,dA(\xi ) \biggr]\,dA(z) \\& \quad \lesssim \int _{\mathbb{C}}\,dA(z) \int _{0}^{1}\,dt \int _{\mathbb{C}} \omega _{1}(f)^{s}\circ \tau _{z} \biggl(\frac{t\eta}{\rho (z)} \biggr) \frac{1}{\rho ^{2}(z)}e^{-\frac{s}{4}d_{\varphi} (z,\tau _{z} (\frac{\eta}{\rho (z)} ) )^{\varepsilon}}\,dA( \eta ) \\& \quad \lesssim \int _{0}^{1}\,dt \biggl[ \int _{\mathbb{C}} \biggl( \int _{ \mathbb{C}}\omega _{1}(f)^{s}(z+t\eta )\,dA(z) \biggr) \frac{1}{\rho ^{2}(z)}e^{-\frac{s}{4}d_{\varphi}(z,z+\eta )}\,dA(\eta ) \biggr]. \end{aligned}$$

(8)

Here, an integral in the last step is thought to be bounded in view of the translation and Lemma 2.1 in [4]. The latter is indeed

$$ \int _{\mathbb{C}}e^{-\frac{s}{4}d_{\varphi}(z,w)}\,dA(w)\lesssim \rho ^{2}(z). $$

Therefore, the estimate above becomes

$$ \int _{\mathbb{C}} \biggl[ \int _{\mathbb{C}} \biggl( \int _{0}^{1}\omega _{1}(f)^{s} \circ \tau _{z}(t\xi )\,dt \biggr) e^{-\frac{s}{4}d_{\varphi}(z,\tau _{z}( \xi ))^{\varepsilon}}\,dA(\xi ) \biggr]\,dA(z)\lesssim \Vert f \Vert _{IO^{s,0}}^{s}. $$

We continue to deal with the other integral (7) by Hölder’s inequality, a change of variable, and Fubini’s Theorem, as follows

$$\begin{aligned}& \int _{\mathbb{C}} \biggl[ \int _{\mathbb{C}} \bigl\vert h^{s'}\bigl(\tau _{z}( \xi )\bigr) \bigr\vert \frac{\rho (z)^{s'}}{\rho (\tau _{z}(\xi ))^{s'}}e^{- \frac{s'}{4}d_{\varphi}(z,\tau _{z}(\xi ))^{\varepsilon}}\,dA( \xi ) \biggr]^{\frac{p}{s'}}\,dA(z) \end{aligned}$$

(9)

$$\begin{aligned}& \quad \lesssim \int _{\mathbb{C}} \biggl[ \int _{\mathbb{C}} \bigl\vert h^{s'}\bigl( \tau _{z}(\xi )\bigr) \bigr\vert ^{\frac{p}{s'}} \frac{\rho (z)^{\frac{p}{2}}}{ \rho (\tau _{z}(\xi ))^{\frac{p}{2}}}e^{-\frac{p}{8}d_{\varphi}(z, \tau _{z}(\xi ))^{\varepsilon}}\,dA( \xi ) \biggr] \\& \qquad {}\times \biggl[ \int _{\mathbb{C}} \frac{\rho (z)^{\frac{1}{2}\frac{pq}{pq-p}}}{\rho (\tau _{z}(\xi ))^{\frac{1}{2}\frac{pq}{pq-p}}} e^{-\frac{1}{8}\frac{pq}{pq-p}d_{\varphi}(z,\tau _{z}(\xi ))^{ \varepsilon}}\,dA(\xi ) \biggr]^{(1-\frac{s'}{p})\frac{p}{s'}}\,dA(z) \end{aligned}$$

(10)

$$\begin{aligned}& \quad \lesssim \int _{\mathbb{C}} \biggl[ \int _{\mathbb{C}} \bigl\vert h\bigl(\tau _{z}( \xi ) \bigr) \bigr\vert ^{p} \frac{\rho (z)^{\frac{p}{2}}}{ \rho (\tau _{z}(\xi ))^{\frac{p}{2}}}e^{-\frac{p}{8}d_{\varphi}(z, \tau _{z}(\xi ))^{\varepsilon}}\,dA(\xi ) \biggr]\,dA(z) \\& \quad = \int _{\mathbb{C}} \biggl[ \int _{\mathbb{C}} \bigl\vert h(w) \bigr\vert ^{p} \frac{\rho (z)^{\frac{p}{2}-2}}{ \rho (w)^{\frac{p}{2}}}e^{-\frac{p}{8}d_{\varphi}(z,w)^{\varepsilon}}\,dA(w) \biggr]\,dA(z) \\& \quad \lesssim \Vert h \Vert _{L^{p}}^{p} \int _{\mathbb{C}} \frac{\rho (z)^{\frac{p}{2}-2}}{ \rho (w)^{\frac{p}{2}}}e^{-\frac{p}{8}d_{\varphi}(z,w)^{\varepsilon}}\,dA(z). \end{aligned}$$

(11)

Both the integrals (10) and (11) are finite, because, by Lemma 2.1 in [4], if we let $t>0$,

$$ \int _{\mathbb{C}}\frac{\rho (z)^{t-2}}{\rho (w)^{t}}e^{-\frac{t}{4}d_{ \varphi}(z,w)^{\varepsilon}}\,dA(z)< \infty $$

and by a change of variable,

$$\begin{aligned} \int _{\mathbb{C}}\frac{\rho (z)^{t}}{\rho (\tau _{z}(\xi ))^{t}}e^{- \frac{t}{4}d_{\varphi}(z,\tau _{z}(\xi ))^{\varepsilon}}\,dA(\xi ) = \int _{\mathbb{C}}\frac{\rho (z)^{t-2}}{\rho (w)^{t}}e^{-\frac{t}{4}d_{ \varphi}(z,w)^{\varepsilon}}\,dA(w) < \infty . \end{aligned}$$

(12)

Finally, our goal is obtained after applying Hölder’s inequality again,

$$\begin{aligned}& \int _{\mathbb{C}} \bigl\vert T_{f,\varepsilon}h(z) \bigr\vert ^{q}\,dA(z) \\& \quad \lesssim \biggl( \int _{\mathbb{C}} \biggl[ \int _{\mathbb{C}} \biggl( \int _{0}^{1}\omega _{1}(f)^{s} \circ \tau _{z}(t\xi )\,dt \biggr) e^{- \frac{s}{4}d_{\varphi}(z,\tau _{z}(\xi ))^{\varepsilon}}\,dA(\xi ) \biggr]\,dA(z) \biggr)^{\frac{q}{s}} \\& \qquad {}\times \biggl( \int _{\mathbb{C}} \biggl[ \int _{\mathbb{C}} \bigl\vert h^{s'}\bigl( \tau _{z}(\xi )\bigr) \bigr\vert \frac{\rho (z)^{s'}}{\rho (\tau _{z}(\xi ))^{s'}}e^{-\frac{s'}{4}d_{ \varphi}(z,\tau _{z}(\xi ))^{\varepsilon}}\,dA( \xi ) \biggr]^{ \frac{p}{s'}}\,dA(z) \biggr)^{\frac{q}{p}} \\& \quad \lesssim \Vert f \Vert _{IO^{s,0}}^{q} \Vert h \Vert _{L^{p}}^{q}. \end{aligned}$$

(2) In order to complete the proof, we will consider dividing $T_{f,\varepsilon}$ into two parts as follows:

$$\begin{aligned}& J_{f,r}(h) (z) = \int _{|\xi |\geq r} \biggl( \int _{0}^{1}\omega _{1}(f) \circ \tau _{z}(t\xi )\,dt \biggr) \frac{\rho (z)}{\rho (\tau _{z}(\xi ))} \frac{h(\tau _{z}(\xi ))\,dA(\xi )}{e^{\frac{1}{2}d_{\varphi}(z,\tau _{z}(\xi ))^{\varepsilon}}}; \\& Q_{f,r}(h) (z) = \int _{|\xi |< r} \biggl( \int _{0}^{1}\omega _{1}(f) \circ \tau _{z}(t\xi )\,dt \biggr) \frac{\rho (z)}{\rho (\tau _{z}(\xi ))} \frac{h(\tau _{z}(\xi ))\,dA(\xi )}{e^{\frac{1}{2}d_{\varphi}(z,\tau _{z}(\xi ))^{\varepsilon}}}. \end{aligned}$$

Then, by Lemma 2.1, we can see that, for $0< t<1$,

$$\begin{aligned} \bigl\vert J_{f,r}(h) (z) \bigr\vert \lesssim e^{-\frac{1}{4}r^{\frac{t\varepsilon}{2}}} \int _{ \vert \xi \vert \geq r} \biggl( \int _{0}^{1}\omega _{1}(f)\circ \tau _{z}(t \xi )\,dt \biggr) \frac{ \vert h(\tau _{z}(\xi )) \vert }{e^{\frac{1}{4}d_{\varphi}(z,\tau _{z}(\xi ))^{\frac{\varepsilon}{2}}}} \frac{\rho (z)\,dA(\xi )}{\rho (\tau _{z}(\xi ))}. \end{aligned}$$

The statement (1) tells us that

$$ \bigl\Vert J_{f,r}(h) \bigr\Vert _{L^{q}}\lesssim e^{-\frac{1}{4}r^{ \frac{t\varepsilon}{2}}} \bigl\Vert T_{f,{\frac{\varepsilon}{2}}}\bigl( \vert h \vert \bigr) \bigr\Vert _{L^{q}} \lesssim e^{-\frac{1}{4}r^{\frac{t\varepsilon}{2}}} \Vert f \Vert _{IO^{s,0}} \Vert h \Vert _{L^{p}}. $$

To estimate the second, let $\chi _{R}$ be the characteristic function of $\{z||z|< R\}$ and use the estimates (8) and (12), we will obtain that

$$\begin{aligned} \bigl\Vert Q_{f,r}(1)\chi _{R} \bigr\Vert _{L^{q}} \lesssim & \biggl( \int _{ \vert z \vert < R}1^{ \frac{p}{q}}\,dA(z) \biggr)^{\frac{1}{p}} \biggl( \int _{ \vert z \vert < R}Q_{f,r}(1)^{s}(z)\,dA(z) \biggr)^{\frac{1}{s}} \\ \lesssim &R^{\frac{2}{p}} \biggl\{ \int _{ \vert z \vert < R} \biggl[ \int _{ \vert \xi \vert < r} \biggl( \int _{0}^{1}\omega _{1}(f)^{s} \circ \tau _{z}(t\xi )\,dt \biggr) \frac{dA(\xi )}{e^{\frac{s}{4}d_{\varphi}(z,\tau _{z}(\xi ))^{\varepsilon}}} \biggr] \\ &{}\times \biggl[ \int _{ \vert \xi \vert < r} \frac{\rho ^{s'}(z)}{\rho ^{s'}(\tau _{z}(\xi ))}e^{-\frac{s'}{4}d_{ \varphi}(z,\tau _{z}(\xi ))^{\varepsilon}}\,dA(\xi ) \biggr]^{ \frac{s}{s'}}\,dA(z) \biggr\} ^{\frac{1}{s}} \\ \lesssim &R^{\frac{2}{p}} \biggl\{ \int _{\mathbb{C}} \biggl[ \int _{ \mathbb{C}} \biggl( \int _{0}^{1}\omega _{1}(f)^{s} \circ \tau _{z}(t\xi )\,dt \biggr) \frac{dA(\xi )}{e^{\frac{s}{4}d_{\varphi}(z,\tau _{z}(\xi ))^{\varepsilon}}} \biggr]\,dA(z) \biggr\} ^{\frac{1}{s}}. \end{aligned}$$

The above implies that

$$ \bigl\Vert Q_{f,r}(h)\chi _{R} \bigr\Vert _{L^{q}}\lesssim \Bigl(\sup_{ \vert \xi \vert \leq R+r} \bigl\vert h( \xi ) \bigr\vert \Bigr) \Vert f \Vert _{IO^{s,0}}. $$

On the other hand, by the property (1), there exists a constant β only dependent on r and a doubling constant such that $z\in D^{r\beta}(\tau _{z}(t\xi ))$ if $\tau _{z}(t\xi )\in D^{r}(z)$ for any $r>0$ and $0\leq t\leq 1$, or equivalently, $\beta ^{-1}\rho (\tau _{z}(t\xi ))\leq \rho (z)\leq \beta \rho ( \tau _{z}(t\xi ))$. Furthermore, for any $w\in D(\tau _{z}(t\xi ))$ if $\tau _{z}(t\xi )\in D^{r}(z)$, we have

$$ \frac{ \vert w-z \vert }{\rho (z)}\leq \frac{ \vert w-\tau _{z}(t\xi ) \vert }{\rho (\tau _{z}(t\xi ))} \frac{\rho (\tau _{z}(t\xi ))}{\rho (z)}+ \frac{ \vert t\xi \rho (z) \vert }{\rho (z)} \leq \beta +r. $$

Combining the triangle inequality and the definition of the oscillation, we can see that, for any $w\in D(\tau _{z}(t\xi ))$,

$$\begin{aligned} \bigl\vert f(w)-f\bigl(\tau _{z}(t\xi )\bigr) \bigr\vert \lesssim & \bigl\vert f(w)-f(z) \bigr\vert + \bigl\vert f\bigl(\tau _{z}(t\xi )\bigr)-f(z) \bigr\vert \\ \lesssim &\sup_{w\in D^{\beta +r}(z)} \bigl\vert f(w)-f(z) \bigr\vert + \sup_{w\in D^{r}(z)} \bigl\vert f(w)-f(z) \bigr\vert . \end{aligned}$$

This implies

$$ \omega _{1}(f)\circ \tau _{z}(t\xi )\lesssim \sup _{w\in D^{1+r+\beta}(z)} \bigl\vert f(w)-f(z) \bigr\vert = \omega _{1+r+\beta}(f) (z). $$

Now, applying Hölder’s inequality with the exponents $\frac{p}{p-q}$ and $\frac{p}{q}$,

$$\begin{aligned}& \bigl\Vert Q_{f,r}(h) (1-\chi _{R}) \bigr\Vert _{L^{q}} \\& \quad \lesssim \biggl[ \int _{ \vert z \vert \geq R}\omega _{1+r+\beta}(f)^{q}(z) \biggl( \int _{ \vert \xi \vert < r}\frac{\rho (z)}{\rho (\tau _{z}(\xi ))} \frac{ \vert h(\tau _{z}(\xi )) \vert }{e^{\frac{1}{2}d_{\varphi}(z,\tau _{z}(\xi ))^{\varepsilon}}}\,dA( \xi ) \biggr)^{q}\,dA(z) \biggr]^{\frac{1}{q}} \\& \quad \lesssim \biggl( \int _{ \vert z \vert \geq R}\omega _{1+r+\beta}(f)^{s}(z)\,dA(z) \biggr)^{\frac{1}{s}} \\& \qquad {}\times \biggl[ \int _{ \vert z \vert \geq R} \biggl( \int _{ \vert \xi \vert < r} \frac{\rho (z)}{\rho (\tau _{z}(\xi ))} \frac{ \vert h(\tau _{z}(\xi )) \vert }{e^{\frac{1}{2}d_{\varphi}(z,\tau _{z}(\xi ))^{\varepsilon}}}\,dA( \xi ) \biggr)^{p}\,dA(z) \biggr]^{\frac{1}{p}} \\& \quad \lesssim \biggl( \int _{ \vert z \vert \geq R}\omega _{1+r+\beta}(f)^{s}(z)\,dA(z) \biggr)^{\frac{1}{s}} \biggl\Vert \int _{ \vert \xi \vert < r} \frac{\rho (z)}{\rho (\tau _{z}(\xi ))} \frac{ \vert h(\tau _{z}(\xi )) \vert }{e^{\frac{1}{2}d_{\varphi}(z,\tau _{z}(\xi ))^{\varepsilon}}}\,dA( \xi ) \biggr\Vert _{L^{p}}. \end{aligned}$$

Using a similar argument as in the estimate (9), we can furthermore have

$$ \bigl\Vert Q_{f,r}(h) (1-\chi _{R}) \bigr\Vert _{L^{q}}\lesssim \biggl( \int _{ \vert z \vert \geq R} \omega _{1+r+\beta}(f)^{s}(z)\,dA(z) \biggr)^{\frac{1}{s}} \Vert h \Vert _{L^{p}}. $$

By the assumption in the statement (2), we may assume that $\|h_{k}\|_{L^{p}}\leq 1$ without loss of generality. Given any $\varepsilon >0$, choose some large enough $R>0$ such that

$$ \int _{|z|\geq R}\omega _{1+r+\beta}(f)^{s}(z)\,dA(z)< \varepsilon . $$

Combining all the analysis above,

$$\begin{aligned} \bigl\Vert T_{f,\varepsilon}(h_{k}) \bigr\Vert _{L^{q}} \lesssim & \bigl\Vert J_{f,r}(h_{k}) (z) \bigr\Vert _{L^{q}}+ \bigl\Vert Q_{f,r}(h_{k})\chi _{R} \bigr\Vert _{L^{q}}+ \bigl\Vert Q_{f,r}(h_{k}) (1-\chi _{R}) \bigr\Vert _{L^{q}} \\ \lesssim &e^{-\frac{1}{4}r^{\frac{t\varepsilon}{2}}} \Vert f \Vert _{IO^{s,0}} \Vert h_{k} \Vert _{L^{p}}+ \Bigl(\sup_{ \vert \xi \vert \leq R+r} \bigl\vert h_{k}(\xi ) \bigr\vert \Bigr) \Vert f \Vert _{IO^{s,0}} \\ &{}+ \biggl( \int _{ \vert z \vert \geq R}\omega _{1+r+\beta}(f)^{s}(z)\,dA(z) \biggr)^{ \frac{1}{s}} \Vert h_{k} \Vert _{L^{p}}. \end{aligned}$$

Therefore, as desired, we can achieve that $\lim_{k\to \infty}\|T_{f,\varepsilon}(h_{k})\|_{L^{q}}=0$. □

Something interesting in that the boundedness is equivalent to compactness will be concluded in the next theorem.

Theorem 4.2

Suppose $1\leq q< p<\infty$ and f is a symbol such that the Hankel operator makes sense. Then, the following statements are equivalent:

• (1) $H_{f}, H_{\bar{f}}: F^{p}_{\varphi}\to L^{q}_{\varphi}$ are bounded;

• (2) $H_{f}, H_{\bar{f}}: F^{p}_{\varphi}\to L^{q}_{\varphi}$ are compact;

• (3) $f\in \mathrm{IMO}^{\frac{pq}{p-q},0,q}$.

Furthermore,

$$ \Vert H_{f} \Vert _{F^{p}_{\varphi}\to L^{q}_{\varphi}}+ \Vert H_{\bar{f}} \Vert _{F^{p}_{ \varphi}\to L^{q}_{\varphi}}\simeq \Vert f \Vert _{\mathrm{IMO}^{\frac{pq}{p-q},0,q}}. $$

Proof

For the implication $(2)\Rightarrow (1)$ is trivial, we only prove the other cases.

To achieve the deduction $(1)\Rightarrow (3)$, we assume $\{a_{k}\}_{k\geq 1}$ is an r-lattice for some suitable $r>0$. For any $\{\lambda _{k}\}_{k\geq 1}\in l^{p}$, we let

$$ g_{t}(z)=\sum_{k=1}^{\infty}\lambda _{k}r_{k}(t) \frac{k_{2,a_{k}}(z)}{\rho (a_{k})^{\frac{2}{p}-1}}, $$

where $r_{k}$ is the Rademacher function. Then, $g_{t}\in F^{p}_{\varphi}$ and $\|g_{t}\|_{F^{p}_{\varphi}}\lesssim \|\{\lambda _{k}\}_{k\geq 1}\|_{l^{p}}$ by Lemma 2.6 in [4]. By Fubini’s theorem and Khinchine’s inequality,

$$\begin{aligned} \int _{0}^{1} \Vert H_{f}g_{t} \Vert ^{q}_{L^{q}_{\varphi}}\,dt =& \int _{0}^{1} \Biggl( \int _{\mathbb{C}} \Biggl\vert \sum_{k=1}^{\infty} \lambda _{k}r_{k}(t)H_{f} \biggl( \frac{k_{2,a_{k}}}{\rho (a_{k})^{\frac{2}{p}-1}} \biggr) (z) \Biggr\vert ^{q}e^{-q\varphi (z)}\,dA(z) \Biggr)\,dt \\ =& \int _{\mathbb{C}} \Biggl( \int _{0}^{1} \Biggl\vert \sum _{k=1}^{\infty} \lambda _{k}r_{k}(t)H_{f} \biggl( \frac{k_{2,a_{k}}}{\rho (a_{k})^{\frac{2}{p}-1}} \biggr) (z) \Biggr\vert ^{q}\,dt \Biggr)e^{-q\varphi (z)}\,dA(z) \\ \gtrsim & \int _{\mathbb{C}} \Biggl(\sum_{k=1}^{\infty} \vert \lambda _{k} \vert ^{2} \biggl\vert H_{f} \biggl(\frac{k_{2,a_{k}}}{\rho (a_{k})^{\frac{2}{p}-1}} \biggr) (z) \biggr\vert ^{2} \Biggr)^{\frac{q}{2}}e^{-q\varphi (z)}\,dA(z) \\ \gtrsim &\sum_{j=1}^{\infty} \int _{D^{r}(a_{j})} \Biggl(\sum_{k=1}^{ \infty} \vert \lambda _{k} \vert ^{2} \biggl\vert H_{f} \biggl( \frac{k_{2,a_{k}}}{\rho (a_{k})^{\frac{2}{p}-1}} \biggr) (z) \biggr\vert ^{2} \Biggr)^{\frac{q}{2}}\frac{dA(z)}{e^{q\varphi (z)}}. \end{aligned}$$

We continue to estimate that, by applying the definition of the Hankel operator,

$$\begin{aligned}& \sum_{j=1}^{\infty} \int _{D^{r}(a_{j})} \Biggl(\sum_{k=1}^{\infty} \vert \lambda _{k} \vert ^{2} \biggl\vert H_{f} \biggl( \frac{k_{2,a_{k}}}{\rho (a_{k})^{\frac{2}{p}-1}} \biggr) (z) \biggr\vert ^{2} \Biggr)^{\frac{q}{2}}e^{-q\varphi (z)}\,dA(z) \\& \quad \gtrsim \sum_{j=1}^{\infty} \vert \lambda _{j} \vert ^{q} \int _{D^{r}(a_{j})} \biggl\vert f(z) \biggl(\frac{k_{2,a_{j}}(z)}{\rho (a_{j})^{\frac{2}{p}-1}} \biggr)- \int _{\mathbb{C}}f(\lambda ) \biggl( \frac{k_{2,a_{j}}(\lambda )}{\rho (a_{j})^{\frac{2}{p}-1}} \biggr) \\& \qquad {}\times K_{\varphi}(z,\lambda )e^{-2\varphi (\lambda )}\,dA(\lambda ) \biggr\vert ^{q}e^{-q\varphi (z)}\,dA(z) \\& \quad \gtrsim \sum_{j=1}^{\infty} \vert \lambda _{j} \vert ^{q} \int _{D^{r}(a_{j})} \biggl\vert f(z)- \biggl[ \frac{\rho (a_{j})^{\frac{2}{p}-1}}{k_{2,a_{j}}(z)} \biggr] \int _{ \mathbb{C}}f(\lambda ) \biggl[ \frac{k_{2,a_{j}}(\lambda )}{\rho (a_{j})^{\frac{2}{p}-1}} \biggr] \\& \qquad {}\times K_{\varphi}(z,\lambda )e^{-2\varphi (\lambda )}\,dA( \lambda ) \biggr\vert ^{q} \biggl\vert \frac{k_{2,a_{j}}(z)}{\rho (a_{j})^{\frac{2}{p}-1}} \biggr\vert ^{q}e^{-q \varphi (z)}\,dA(z) \\& \quad \gtrsim \sum_{j=1}^{\infty} \vert \lambda _{j} \vert ^{q} \frac{\rho (a_{j})^{2q(\frac{1}{q}-\frac{1}{p})}}{ \vert D^{r}(a_{j}) \vert } \int _{D^{r}(a_{j})} \biggl\vert f(z)- \biggl[ \frac{\rho (a_{j})^{\frac{2}{p}-1}}{k_{2,a_{j}}(z)} \biggr] \int _{ \mathbb{C}}f(\lambda ) \\& \qquad {}\times \biggl[ \frac{k_{2,a_{j}}(\lambda )}{\rho (a_{j})^{\frac{2}{p}-1}} \biggr]K_{ \varphi}(z,\lambda )\frac{dA(\lambda )}{e^{2\varphi (\lambda )}} \biggr\vert ^{q}\,dA(z). \end{aligned}$$

Note that $1\leq q\leq p<\infty$ and, for any $z\in D^{r}(a_{j})$,

$$ e^{-\varphi (z)} \bigl\vert K_{\varphi}(z, a_{j}) \bigr\vert e^{-\varphi (a_{j})}\gtrsim \frac{1}{\rho (a_{j})^{2}}. $$

The last step above comes from

$$ \biggl\vert \frac{k_{2,a_{j}}(z)}{\rho (a_{j})^{\frac{2}{p}-1}} \biggr\vert ^{q} \gtrsim \biggl\vert \frac{K_{\varphi}(z,a_{j})\rho (a_{j})}{ \rho (a_{j})^{\frac{2}{p}-1}e^{\varphi (a_{j})}} \biggr\vert ^{q}\gtrsim e^{q \varphi (z)} \rho (a_{j})^{-2\frac{q}{p}}. $$

We denote the function $g_{a_{j}}$ by

$$ g_{a_{j}}(z)\triangleq \biggl( \frac{\rho (a_{j})^{\frac{2}{p}-1}}{k_{2,a_{j}}(z)} \biggr) \int _{ \mathbb{C}}f(\lambda ) \biggl( \frac{k_{2,a_{j}}(\lambda )}{\rho (a_{j})^{\frac{2}{p}-1}} \biggr)K_{ \varphi}(z,\lambda )e^{-2\varphi (\lambda )}\,dA(\lambda ). $$

It is easy to see that $g_{a_{j}}$ is a holomorphic function on $D^{r}(a_{j})$. After defining that

$$ G_{q,r}f(z)\triangleq \inf \biggl\{ \biggl(\frac{1}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert f(w)-h(w) \bigr\vert ^{q}\,dA(w) \biggr)^{\frac{1}{q}}: h\in H\bigl(D^{r}(z)\bigr) \biggr\} , $$

for any $f\in L^{q}_{\mathrm{loc}}$, we can see that

$$ \int _{0}^{1} \Vert H_{f}g_{t} \Vert ^{q}_{L^{q}_{\varphi}}\,dt\gtrsim \sum _{j=1}^{ \infty} \vert \lambda _{j} \vert ^{q}\rho (a_{j})^{2q(\frac{1}{q}-\frac{1}{p})}\bigl(G_{q,r}f(a_{j}) \bigr)^{q}. $$

The boundedness of $H_{f}: F^{p}_{\varphi}\to L^{q}_{\varphi}$ shows us that

$$ \Vert H_{f}g_{t} \Vert _{L^{q}_{\varphi}}\lesssim \Vert H_{f} \Vert _{F^{p}_{\varphi} \to L^{q}_{\varphi}} \Vert g_{t} \Vert _{F^{p}_{\varphi}}. $$

Therefore, we can achieve that

$$ \sum_{j=1}^{\infty} \vert \lambda _{j} \vert ^{q}\rho (a_{j})^{2q(\frac{1}{q}- \frac{1}{p})} \bigl(G_{q,r}f(a_{j})\bigr)^{q}\lesssim \Vert H_{f} \Vert ^{q}_{F^{p}_{ \varphi}\to L^{q}_{\varphi}} \bigl\Vert \bigl\{ \vert \lambda _{k} \vert ^{q}\bigr\} _{k\geq 1} \bigr\Vert ^{q}_{l^{ \frac{p}{q}}}. $$

The fact that $G_{q,t}f(w)\lesssim G_{q,r}f(z)$ when $D^{t}(w)\subset D^{r}(z)$ tells us that

$$\begin{aligned}& \int _{\mathbb{C}} \bigl(G_{q,r}f(z)\bigr)^{\frac{pq}{p-q}}\,dA(z) \\& \quad \lesssim \sum_{j=1}^{\infty} \int _{D^{r}(a_{j})}\bigl(G_{q,r}f(z)\bigr)^{ \frac{pq}{p-q}}\,dA(z) \\& \quad \lesssim \sum_{j=1}^{\infty} \bigl(\rho (a_{j})^{2(\frac{1}{q}- \frac{1}{p})}\bigl(G_{q,2r}f(a_{j}) \bigr) \bigr)^{\frac{pq}{p-q}}. \end{aligned}$$

This means $G_{q,r}f\in L^{\frac{pq}{p-q}}$. Similarly, the boundedness of $H_{\bar{f}}$ from $F^{p}_{\varphi}$ to $L^{q}_{\varphi}$ gives that $G_{q,r} \bar{f}\in L^{\frac{pq}{p-q}}$. By the definition of $G_{q,r} f$, there exist two holomorphic functions $h_{1}$ and $h_{2}$ on $D^{r}(z)$ such that

$$ \biggl(\frac{1}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert f(w)-h_{1}(w) \bigr\vert ^{q}\,dA(w) \biggr)^{\frac{1}{q}}\lesssim G_{q,r}f(z), $$

and

$$ \biggl(\frac{1}{ \vert D^{r}(z) \vert } \int _{D^{r}(z)} \bigl\vert \bar{f}(w)-h_{2}(w) \bigr\vert ^{q}\,dA(w) \biggr)^{\frac{1}{q}}\lesssim G_{q,r}\bar{f}(z). $$

From Lemma 2.6, it follows that $f\in \mathrm{IMO}^{\frac{pq}{p-q},0,q}$ and

$$ \Vert H_{f} \Vert _{F^{p}_{\varphi}\to L^{q}_{\varphi}}+ \Vert H_{\bar{f}} \Vert _{F^{p}_{ \varphi}\to L^{q}_{\varphi}}\gtrsim \Vert f \Vert _{\mathrm{IMO}^{\frac{pq}{p-q},0,q}}. $$

Now, we will prove $(3)\Rightarrow (2)$ and suppose $f\in \mathrm{IMO}^{\frac{pq}{p-q},0,q}$. By Lemma 2.4, f has a decomposition $f=f_{1}+f_{2}$ with

$$ \Vert f_{1} \Vert _{IO^{\frac{pq}{p-q},0}} + \Vert f_{2} \Vert _{IA^{\frac{pq}{p-q},0,q}} \lesssim \Vert f \Vert _{\mathrm{IMO}^{\frac{pq}{p-q},0,q}}. $$

First, we deal with $H_{f_{1}}$ by the linearity of the Hankel operator. By the calculation, we have that $\|H_{f_{1}}g\|^{q}_{L^{q}_{\varphi}}$ is not greater than a constant times

$$\begin{aligned}& \int _{\mathbb{C}} \biggl( \int _{\mathbb{C}} \bigl\vert f_{1}(z)-f_{1}(w) \bigr\vert \bigl\vert g(w) \bigr\vert \bigl\vert K_{ \varphi}(z,w) \bigr\vert e^{-2\varphi (w)}\,dA(w) \biggr)^{q}e^{-q\varphi (z)}\,dA(z) \\& \quad \lesssim \int _{\mathbb{C}} \biggl[ \int _{\mathbb{C}} \bigl\vert f_{1}(z)-f_{1}(w) \bigr\vert \biggl(\frac{ \vert g(w) \vert }{e^{\varphi (w)}} \biggr) \frac{e^{-(\frac{ \vert z-w \vert }{\rho (z)})^{\varepsilon}}}{\rho (z)\rho (w)}\,dA(w) \biggr]^{q}\,dA(z). \end{aligned}$$

(13)

According to Lemma 2.8, the integral in (13) becomes

$$\begin{aligned}& \int _{\mathbb{C}} \bigl\vert f_{1}(z)-f_{1}(w) \bigr\vert \bigl( \bigl\vert g(w) \bigr\vert e^{-\varphi (w)} \bigr) \frac{e^{-(\frac{ \vert z-w \vert }{\rho (z)})^{\varepsilon}}}{\rho (z)\rho (w)}\,dA(w) \\& \quad \lesssim \int _{\mathbb{C}} \biggl( \int _{0}^{1} \frac{\omega _{1}(f_{1})\circ \tau _{z}(t\xi )}{(1+d_{\varphi}(z,w))^{{\frac{\delta -2}{\delta}}}}\,dt \biggr) \biggl( \frac{ \vert g(w) \vert }{e^{\varphi (w)}} \biggr) \frac{e^{-(\frac{ \vert z-w \vert }{\rho (z)})^{\varepsilon}}}{\rho (z)\rho (w)}\,dA(w). \end{aligned}$$

(14)

As $\sup_{x>0}(1+x)^{t}e^{-\frac{1}{2}x^{\varepsilon}}<\infty$, for any $t>0$, $\varepsilon >0$,

$$\begin{aligned}& \int _{\mathbb{C}} \bigl\vert f_{1}(z)-f_{1}(w) \bigr\vert \bigl( \bigl\vert g(w) \bigr\vert e^{-\varphi (w)} \bigr) \frac{e^{-(\frac{ \vert z-w \vert }{\rho (z)})^{\varepsilon}}}{\rho (z)\rho (w)}\,dA(w) \\& \quad \lesssim \int _{\mathbb{C}} \biggl( \int _{0}^{1}\omega _{1}(f_{1}) \circ \tau _{z}(t\xi )\,dt \biggr) \bigl( \bigl\vert g(w) \bigr\vert e^{-\varphi (w)} \bigr) \frac{e^{-\frac{1}{2}d_{\varphi}(z,w)^{\varepsilon}}}{\rho (z)\rho (w)}\,dA(w). \end{aligned}$$

If we input the above into estimate (13), $\|H_{f_{1}}g\|^{q}_{L^{q}_{\varphi}}$ is not greater than a constant times

$$\begin{aligned}& \int _{\mathbb{C}} \biggl[ \int _{\mathbb{C}} \biggl( \int _{0}^{1} \omega _{1}(f_{1}) \circ \tau _{z}(t\xi )\,dt \biggr) \biggl( \frac{ \vert g(w) \vert }{e^{\varphi (w)}} \biggr) \frac{e^{-\frac{1}{2}d_{\varphi}(z,w)^{\varepsilon}}}{\rho (z)\rho (w)}\,dA(w) \biggr]^{q}\,dA(z) \\& \quad \lesssim \int _{\mathbb{C}} \biggl[ \int _{\mathbb{C}} \biggl( \int _{0}^{1} \omega _{1}(f_{1}) \circ \tau _{z}(t\xi )\,dt \biggr) \biggl( \frac{ \vert g(\tau _{z}(\xi )) \vert }{e^{\varphi (\tau _{z}(\xi ))}} \biggr) \frac{\rho ^{-1}(\tau _{z}(\xi ))\,dA(\xi )}{\rho ^{-1}(z)e^{\frac{1}{2}d_{\varphi}(z,\tau _{z}(\xi ))^{\varepsilon}}} \biggr]^{q}\,dA(z). \end{aligned}$$

In terms of Lemma 4.1,

$$ \bigl\Vert H_{f_{1}}g(z) \bigr\Vert ^{q}_{L^{q}_{\varphi}} \lesssim \int _{\mathbb{C}} \bigl(T_{f,\varepsilon}\bigl(ge^{-\varphi} \bigr) (z) \bigr)^{q}\,dA(z)\lesssim \Vert f_{1} \Vert ^{q}_{IO^{\frac{pq}{p-q},0}} \Vert g \Vert ^{q}_{F^{p}_{\varphi}}. $$

In order to obtain the boundedness of $H_{f_{2}}$, we will first obtain that

$$ \int _{\mathbb{C}} \biggl( \int _{\mathbb{C}} \bigl\vert f_{2}(w) \bigr\vert \bigl\vert g(w) \bigr\vert \bigl\vert K_{\varphi}(z,w) \bigr\vert e^{-2 \varphi (w)}\,dA(w) \biggr)^{q} e^{-q\varphi (z)}\,dA(z)< \infty . $$

After choosing a suitable r-lattice $\{a_{k}\}_{k\geq 1}$ and a suitable constant $m>0$, we apply the following fact

$$ \bigl\vert g(w) \bigr\vert \bigl\vert K_{\varphi}(z,w) \bigr\vert e^{-2\varphi (w)}\lesssim \frac{1}{ \vert D^{r}(w) \vert } \int _{D^{r}(w)} \bigl\vert g(\xi ) \bigr\vert \bigl\vert K_{\varphi}(z,\xi ) \bigr\vert e^{-2 \varphi (\xi )}\,dA(\xi ) $$

to achieve that

$$\begin{aligned}& \int _{\mathbb{C}} \bigl\vert f_{2}(w) \bigr\vert \bigl\vert g(w) \bigr\vert \bigl\vert K_{\varphi}(z,w) \bigr\vert e^{-2\varphi (w)}\,dA(w) \\& \quad \lesssim \sum_{k=1}^{\infty} \int _{D^{r}(a_{k})} \bigl\vert f_{2}(w) \bigr\vert \bigl\vert g(w) \bigr\vert \bigl\vert K_{ \varphi}(z,w) \bigr\vert e^{-2\varphi (w)}\,dA(w) \\& \quad \lesssim \sum_{k=1}^{\infty} \Bigl(\sup _{w\in D^{r}(a_{k})} \bigl\vert g(w) \bigr\vert \bigl\vert K_{ \varphi}(z,w) \bigr\vert e^{-2\varphi (w)} \Bigr) \int _{D^{r}(a_{k})} \bigl\vert f_{2}(w) \bigr\vert dA(w) \\& \quad \lesssim \sum_{k=1}^{\infty} \int _{D^{mr}(a_{k})} \bigl\vert g(w) \bigr\vert \bigl\vert K_{\varphi}(z,w) \bigr\vert e^{-2 \varphi (w)}\frac{dA(w)}{\rho ^{2}(a_{k})} \int _{D^{r}(a_{k})} \bigl\vert f_{2}(w) \bigr\vert dA(w) \\& \quad \lesssim \int _{\mathbb{C}} \bigl\vert g(w) \bigr\vert \bigl\vert K_{\varphi}(z,w) \bigr\vert e^{-2\varphi (w)} \widehat{\bigl( \vert f_{2} \vert ^{q}\bigr)}^{\frac{1}{q}}_{r}(w)\,dA(w). \end{aligned}$$

Taking together Hölder’s inequality with the estimate (5), we can see that

$$\begin{aligned}& \int _{\mathbb{C}} \biggl( \int _{\mathbb{C}} \bigl\vert f_{2}(w) \bigr\vert \bigl\vert g(w) \bigr\vert \bigl\vert K_{ \varphi}(z,w) \bigr\vert e^{-2\varphi (w)}e^{-\varphi (z)}\,dA(w) \biggr)^{q}\,dA(z) \\& \quad \lesssim \int _{\mathbb{C}} \biggl[ \biggl( \int _{\mathbb{C}} \widehat{\bigl( \vert f_{2} \vert ^{q}\bigr)}_{r}(w) \bigl\vert g(w) \bigr\vert ^{q}e^{-q\varphi (w)} \bigl\vert K_{\varphi}(z,w) \bigr\vert e^{- \varphi (w)}e^{-\varphi (z)}\,dA(w) \biggr) \\& \qquad {}\times \biggl( \int _{\mathbb{C}}e^{-\varphi (w)} \bigl\vert K_{\varphi}(z,w) \bigr\vert e^{- \varphi (z)}\,dA(w) \biggr)^{q-1} \biggr]\,dA(z) \\& \quad \lesssim \int _{\mathbb{C}} \biggl( \int _{\mathbb{C}} \widehat{\bigl( \vert f_{2} \vert ^{q}\bigr)}_{r}(w) \bigl\vert g(w) \bigr\vert ^{q}e^{-q\varphi (w)} \frac{ \vert K_{\varphi}(z,w) \vert }{e^{\varphi (w)}e^{\varphi (z)}}\,dA(w) \biggr)\,dA(z). \end{aligned}$$

After applying Fubini’s Theorem and Hölder’s inequality, we continue to estimate

$$\begin{aligned}& \int _{\mathbb{C}} \biggl( \int _{\mathbb{C}}\widehat{\bigl( \vert f_{2} \vert ^{q}\bigr)}_{r}(w) \bigl\vert g(w) \bigr\vert ^{q}e^{-q \varphi (w)}\frac{ \vert K_{\varphi}(z,w) \vert }{e^{\varphi (w)}e^{\varphi (z)}}\,dA(w) \biggr)\,dA(z) \\& \quad \lesssim \int _{\mathbb{C}}\widehat{\bigl( \vert f_{2} \vert ^{q}\bigr)}_{r}(w) \bigl\vert g(w) \bigr\vert ^{q}e^{-q \varphi (w)} \biggl( \int _{\mathbb{C}} \frac{ \vert K_{\varphi}(z,w) \vert }{e^{\varphi (w)}e^{\varphi (z)}}\,dA(z) \biggr)\,dA(w) \\& \quad \lesssim \int _{\mathbb{C}} \bigl\vert g(w) \bigr\vert ^{q}e^{-q\varphi (w)} \widehat{\bigl( \vert f_{2} \vert ^{q} \bigr)}_{r}(w)\,dA(w) \\& \quad \lesssim \biggl( \int _{\mathbb{C}} \bigl( \bigl\vert g(w) \bigr\vert ^{q}e^{-q\varphi (w)} \bigr)^{\frac{p}{q}}\,dA(w) \biggr)^{\frac{q}{p}} \biggl( \int _{ \mathbb{C}}\widehat{\bigl( \vert f_{2} \vert ^{q}\bigr)}^{\frac{p}{p-q}}_{r}(w)\,dA(w) \biggr)^{ \frac{p-q}{p}}. \end{aligned}$$

In addition, for the same lattice, we continue to estimate that

$$\begin{aligned}& \int _{\mathbb{C}} \bigl\vert f_{2}(z)g(z) \bigr\vert ^{q}e^{-q\varphi (z)}\,dA(z) \\& \quad \lesssim \sum_{k=1}^{\infty} \int _{D^{r}(a_{k})} \bigl\vert f_{2}(w)g(w) \bigr\vert ^{q}e^{-q \varphi (w)}\,dA(w) \\& \quad \lesssim \sum_{k=1}^{\infty}\sup _{w\in D^{r}(a_{k})} \bigl\vert g(w) \bigr\vert ^{q}e^{-q \varphi (w)} \int _{D^{r}(a_{k})} \bigl\vert f_{2}(w) \bigr\vert ^{q}\,dA(w) \\& \quad \lesssim \int _{\mathbb{C}} \bigl\vert g(w) \bigr\vert ^{q}e^{-q\varphi (w)} \widehat{\bigl( \vert f_{2} \vert ^{q} \bigr)}_{r}(w)\,dA(w) \\& \quad \lesssim \biggl( \int _{\mathbb{C}} \bigl( \bigl\vert g(w) \bigr\vert ^{q}e^{-q\varphi (w)} \bigr)^{\frac{p}{q}}\,dA(w) \biggr)^{\frac{q}{p}} \biggl( \int _{ \mathbb{C}}\widehat{\bigl( \vert f_{2} \vert ^{q}\bigr)}^{\frac{p}{p-q}}_{r}(w)\,dA(w) \biggr)^{ \frac{p-q}{p}}. \end{aligned}$$

The desired result that $\|H_{f}g\|_{L^{q}_{\varphi}}\lesssim \|g\|_{F^{p}_{\varphi}}\|f_{2} \|_{IA^{\frac{pq}{p-q},0,q}}$ has been achieved.

What we need next is to show that $H_{f_{1}}$ and $H_{f_{2}}$ are both compact from $F^{p}_{\varphi}$ to $L^{q}_{\varphi}$, where $f_{1}\in IO^{\frac{pq}{p-q},0}$ and $f_{2}\in IA^{\frac{pq}{p-q},0,q}$. Let $\{g_{k}\}_{k\geq 1}$ be any norm bounded sequence in $F^{p}_{\varphi}$ that converges to 0 on any compact subset of $\mathbb{C}$. According to the analysis above, we can show that

$$ \bigl\Vert H_{f_{1}}g_{k}(z) \bigr\Vert _{L^{q}_{\varphi}} \lesssim \bigl\Vert T_{f,\varepsilon}\bigl(g_{k}e^{- \varphi} \bigr) (z) \bigr\Vert _{L^{q}}\to 0, $$

when $k\to \infty$ by Lemma 2.8. For $f_{2}$ we let $f_{2,R}=f_{2}\circ \chi _{R}$, where $\chi _{R}$ is the characteristic function of $D^{R}(0)$. From the fact that

$$ \Vert H_{f_{2}} \Vert _{F^{p}_{\varphi}\to L^{q}_{\varphi}}\lesssim \Vert f_{2} \Vert _{IA^{ \frac{pq}{p-q},0,q}}, $$

which had been proven in the previous paragraph, it follows that

$$ \Vert H_{f_{2}}-H_{f_{2}\circ \chi _{R}} \Vert _{F^{p}_{\varphi}\to L^{q}_{ \varphi}}\lesssim \Vert f_{2}-f_{2}\circ \chi _{R} \Vert _{IA^{\frac{pq}{p-q},0,q}} \to 0, $$

when $R\to \infty$. Hence, $H_{f_{2}}$ is compact and so is $H_{f}$. Using the property of the conjugate, we know $H_{\bar{f}}$ is also compact. The proof is completed. □