Embedding and Volterra integral operators on a class of Dirichlet-Morrey spaces

1.   Introduction

Let ${\mathbb D}$ and $\partial {\mathbb D}$ denote the unit disc of complex plane ${\mathbb{C}}$ and its boundary, respectively. Let $H({\mathbb D})$ be the class of holomorphic functions on ${\mathbb D}$. For $0 < p < \infty$, the Hardy space $H^p$ consists of those functions $f\in H({\mathbb D})$ satisfying

Let $H^\infty$ denote the space of bounded analytic functions with the supremum norm $\|f\|_{H^\infty} = \sup_{z\in {\mathbb D}}|f(z)|.$

For $\alpha > -1$ and $0 < p < \infty$, the weight Dirichlet space $D_\alpha^p$ consists of those functions $f\in H({\mathbb D})$ satisfying

where $dA$ denotes the normalized area measure on ${\mathbb D}$. When $\alpha = 1$ and $p = 2$, the space $D_\alpha^p$ is the Hardy space $H^2$. When $\alpha = p$, $D_\alpha^p$ is just the Bergman space $A^p$.

Let $f\in H({\mathbb D})$, $0 < p < \infty$, $-2 < q < \infty$ and $0\leq s < \infty$. We say that $f\in F(p, q, s)$ if

where $\sigma_a = {a-z \over 1-\bar{a}z}$ is a M$\ddot{{\rm o}}$bius map that interchanges 0 and $a$. The space $F(p, q, s)$ was introduced by Zhao in ^[25]. For $q+s > -1$, the space $F(p, q, s)$ is nontrivial. When $q = p-2$, $F(p, p-2, s)$ are Möbius invariant spaces that contain some classical spaces. For instance, when $s > 1$, $F(p, p-2, s)$ is the classical Bloch space ${\mathcal{B}}$. When $p = 2$, $F(p, p-2, s)$ is the $Q_s$ space. If $p = 2$ and $s = 1$, $F(p, p-2, s)$ is the $BMOA$ space.

Let $g \in H({\mathbb D})$. The Volterra integral operator $T_g$ is defined by

The operator $T_g$ has been investigated by many researchers. Pommerenke ^[13] showed that $T_g$ is bounded on $H^2$ if and only if $g\in BMOA$. Aleman and Siskakis ^[2] proved that $T_g$ is bounded on $H^p$ if and only if $g\in BMOA$ when $p\geq 1$. See ^{[1,2,3,6,8,9,14,15,17,18,19]} and the references therein for more information of the operator $T_g$.

For any arc $I\subset \partial {\mathbb D}$, let $|I| = \int_I{|d\xi|\over 2\pi }$ be the normalized arc length of $I$ and

be the Carleson box based on $I$. Let $0 < p, s < \infty$ and $\mu$ be a positive Borel measure on ${\mathbb D}$. The tent space $\mathcal{T}_s^p(\mu)$ consists of all $\mu$-measure functions $f$ satisfying

It was first introduced by Pau and Zhao in ^[12]. They also showed that $\mathcal{T}_s^p(\mu)$ is a Banach space for $p\ge 1$. In ^[24], Xiao showed that the $Q_p$ $(0 < p < 1)$ space is continuously contained in $\mathcal{T}_s^2(\mu)$ if and only if

Let $0\leq \lambda\leq 1$. The analytic Morrey space $\mathcal L^{2, \lambda}({\mathbb D})$, which introduced by Wu and Xie in ^[22], consists of all functions $f\in H^2({\mathbb D})$ such that

where $f_I = {1\over |I|}\int_If(\xi){|d\xi|\over 2\pi }$. From ^[8], the equivalent norm of $f\in\mathcal{L}^{2, \lambda}({\mathbb D})$ can be defined as

It is obvious that $\mathcal L^{2, 1}({\mathbb D}) = BMOA$, $\mathcal L^{2, 0}({\mathbb D}) = H^2$. Moreover,

See ^[23] for the generalization of the Morrey space.

Recently, Galanopoulos, Merch$\acute{{\rm a}}$n and Siskakis ^[6] defined the Dirichlet-Morrey space $D_p^{2, \lambda}$, which consisting of all functions $f\in D_p^{2 }$ such that

where $0\leq p, \lambda\leq1$. It is easy to check that $D_1^{2, \lambda} = \mathcal L^{2, \lambda}, D_p^{2, 1} = Q_p, D_p^{2, 0} = D^2_p$ and

Recently, Morrey-type spaces have received a lot of attention and many results have been obtained. For example, Li, Liu and Lou proved that $T_g$ is bounded on $\mathcal L^{2, \lambda}({\mathbb D})$ if and only if $g\in BMOA$ when $0 < \lambda < 1$ in ^[8]. In ^[6], Galanopoulos, Merch$\acute{{\rm a}}$n and Siskakis proved that if $T_g$ is bounded on $D_p^{2, \lambda}$, then $g\in Q_p$, while if $g\in W_p$, then $T_g$ is bounded on $D_p^{2, \lambda}$. Here the space $W_p$ is the space consisting of all functions $g\in H({\mathbb D})$ such that

Clearly, the necessary and sufficient condition for the boundedness of $T_g$ on $D_p^{2, \lambda}$ are not obtained. See ^[4,10,16,27] and references therein for more information on other Morrey-type spaces.

Motivated by the definitions of the Morrey space $\mathcal L^{2, \lambda}$ and the Dirichlet-Morrey space $D_p^{2, \lambda}$, we introduce a class of Dirichlet-Morrey spaces as follows. Assume that $-1 < \beta < 0$, $0\leq \lambda \leq 1$ and $f\in D_\beta^1$. We say that $f$ belongs to the Dirichlet-Morrey space $D_{\beta, \lambda}$ if

It is obvious that $D_{\beta, \lambda}$ is a linear space. Under the above norm, it is easy to check that $D_{\beta, \lambda}$ is a Banach space. By a simple calculation, we have that $D_{\beta, 0} = D_\beta^1, \; \; D_{\beta, 1} = F(1, -1, \beta+1)$ and

In this paper, we first state some basic properties for the Dirichlet-Morrey space $D_{\beta, \lambda}$ and then investigate the boundedness and compactness of the identity operator $I_d : D_{\beta, \lambda} \to \mathcal{T}_{s}^1(\mu)$. Using the embedding theorem, we give a necessary and sufficient condition for the boundedness of the operator $T_g: D_{\beta, \lambda} \to F(1, \beta-s, s)$ when $-1 < \beta < 0$, $0 < \lambda, s < 1$ such that $s\ge\lambda(\beta +1)$. Moreover, the essential norm and compactness of $T_g: D_{\beta, \lambda} \to F(1, \beta-s, s)$ are also investigated. In particular, we will prove that $T_g: D_{\beta, \lambda} \to D_{\beta, \lambda}$ is bounded (compact) if and only if $g\in F(1, -1, \beta+1)$ ($g\in F_0(1, -1, \beta+1)$).

In this paper, we say that $f\preceq g$ if there exists a constant $C$ such that $f\leq Cg$. If both $f\preceq g$ and $g\preceq f$ are valid, we write $f\approx g$.

2.   Some basic properties

In this section, we characterize some basic properties of the space $D_{\beta, \lambda}$. These properties play an important role in the proof of our main results. We first recall the definition of $\alpha$-Carleson measure.

Suppose that $0 < \alpha < \infty$ and $\mu$ is a positive Borel measure on ${\mathbb D}$. We say that $\mu$ is a $\alpha$-Carleson measure if (see ^[12])

When $\alpha = 1$, $\mu$ is called the Carleson measure. We say that $\mu$ is a vanishing $\alpha$-Carleson measure if

The Carleson measure is a very important tool in the theory of function spaces and operator theory (see ^[5,7,11,24]).

Lemma 1. ^[12] Let $\alpha, q > 0$ and $\mu$ be a positive Borel measure on ${\mathbb D}$. Then $\mu$ is $\alpha$-Carleson measure if and only if

Proposition 1. Let $-1 < \beta < 0$, $0 < \lambda < 1$ and $f\in H({\mathbb D})$. Then $f\in D_{\beta, \lambda}$ if and only if

Proof. First, suppose that $f\in D_{\beta, \lambda}$. For any arc $I\subset \partial {\mathbb D}$, let $a = (1-|I|)\xi$, where $\xi$ is the center of arc $I$. Then

Changing the variable $z = \sigma_a(w)$, we have

which implies the desired result by the arbitrariness of $I$.

Conversely, assume that (2.1) holds. Let $d\mu_f(z) = |f'(z)|(1-|z|^2)^\beta dA(z)$. Then

So $\mu_f$ is a $\lambda(\beta +1)$-Carleson measure. Then for each $a\in {\mathbb D}$,

Therefore, by Lemma 1 we have

where $\alpha = \lambda(\beta +1) > 0$, $q = (\beta+1)(2-\lambda) > 0$. The proof is complete.

Proposition 2. Let $-1 < \beta < 0$, $0 < \lambda < 1$. Then the following statements hold.

(i) For any $f\in D_{\beta, \lambda}$,

(ii) The function $f_{\beta, \lambda}(z) = {1\over (1-z)^{(\beta+1)(1-\lambda)}} \ \ \ $ belongs to $D_{\beta, \lambda}$.

Proof. $(i)$ Suppose that $f\in D_{\beta, \lambda}$. For each $a\in {\mathbb D}$, applying the Lemma 4.12 in ^[26], we get

So

Since $f(z)-f(0) = \int_0^z f'(w)dw$, by integrating both sides of the last inequality, we obtain the desired result.

$(ii)$ By Proposition 1, it suffices to show that

Set

Then the inequality (2) is equivalent to

Since

we see that the inequality (2.3) holds. The proof is complete.

3.   Embedding theorem from $D_{\beta, \lambda}$ to tent spaces

In this section, we study the boundedness and compactness of the identity operator $I_d: D_{\beta, \lambda} \to \mathcal{T}_s^1(\mu)$. We say that $I_d$ is compact if

where $I\subset \partial {\mathbb D}$, $\{f_n\}$ is a bounded sequence in $D_{\beta, \lambda}$ and converges to zero uniformly on every compact subset of ${\mathbb D}$.

We begin this section with several lemmas.

Lemma 2. [12,Corollary 2.5] Let $a, b\in {\mathbb D}$ and $r > -1, s, t > 0$ such that $0 < s+t-r-2 < s$. Then

Lemma 3. Let $-1 < \beta < 0$, $0 < \lambda < 1$, $q\geq\lambda(\beta+1)$ and $f\in D_{\beta, \lambda}$. Then

Proof. From Lemma 1 in ^[11], for any $a\in {\mathbb D}$ and $f\in H({\mathbb D})$,

where $h_z(t) = 1-(1-\bar{z}t)^3$ is uniformly bounded on ${\mathbb D}$ and satisfies $h_z(0) = 0$. Employing Schwarz's Lemma, we have $|h_z(t)|\preceq |t|$. Using this, we deduce that

According to the fact that $1-|t| \leq |1-\bar{a}t|$, Lemma 2 and Fubini's Theorem, we get

The proof is complete.

Lemma 4. ^[21] Let $0 < \alpha < 1$ and $\mu$ be a positive Borel measure on ${\mathbb D}$. Then the identity operator $I_d:D_{\alpha-1}^1 \to L^1(\mu)$ is bounded if and only if $\mu$ is $\alpha$-Carleson measure.

The following theorem is the main result in this section.

Theorem 1. Let $-1 < \beta < 0$, $0 < \lambda < 1$, $s\geq\lambda(\beta+1)$ and $\mu$ be a positive Borel measure on ${\mathbb D}$. Then the identity operator $I_d : D_{\beta, \lambda} \to \mathcal{T}_{s}^1(\mu)$ is bounded if and only if the measure $\mu$ is a $s+(\beta+1)(1-\lambda)$-Carleson measure.

Proof. First, assume that $I_d : D_{\beta, \lambda} \to \mathcal{T}_{s}^1(\mu)$ is bounded. For any arc $I\subset \partial {\mathbb D}$, let $a = (1-|I|)\xi$, where $\xi$ is the center of arc $I$. Then

Set

Using Proposition 2, we obtain that $f_a \in D_{\beta, \lambda}$ with $\|f_a\|_{D_{\beta, \lambda}}\preceq 1$. Moreover $|f_a(z)|\approx {1\over|I|^{(\beta+1)(1-\lambda)}}, \, \, z\in S(I).$ Hence,

which implies that $\mu$ is a $s+(\beta+1)(1-\lambda)$-Carleson measure.

Conversely, suppose that $\mu$ is a $s+(\beta+1)(1-\lambda)$-Carleson measure. Let $f\in D_{\beta, \lambda}$. For any arc $I\subset \partial {\mathbb D}$, let $a = (1-|I|)\xi$, where $\xi$ is the center of arc $I$.

From Proposition 2, we have that

Since

by Lemma 4 we obtain

where

and

Changing the variable $z = \sigma_a(w)$, we have

Changing the variable $z = \sigma_a(w)$ and using Lemma 3, we obtain

So the identity operator $I_d : D_{\beta, \lambda} \to \mathcal{T}_s^1(\mu)$ is bounded. The proof is complete.

Theorem 2. Let $-1 < \beta < 0$, $0 < \lambda < 1$, $s\geq\lambda(\beta+1)$ and $\mu$ be a positive Borel measure on ${\mathbb D}$ such that point evaluation is a bounded functional on $\mathcal{T}_{s}^1(\mu)$. Then the identity operator $I_d : D_{\beta, \lambda} \to \mathcal{T}_s^1(\mu)$ is compact if and only if the measure $\mu$ is a vanishing $s+(\beta+1)(1-\lambda)$-Carleson measure.

Proof. First, we suppose that $I_d : D_{\beta, \lambda} \to \mathcal{T}_{s}^1(\mu)$ is compact. Let $\{I_n\}$ be a sequence of subarcs of $\partial {\mathbb D}$ with $\lim\limits_{n\to \infty}|I_n| = 0$. Set $a_n = (1-|I_n|)\xi_n$, where $\xi_n$ is the midpoint of ${I_n}$. By simple calculation we have that, for any $z\in S(I_n)$, $1-|a_n|^2\approx |1-\bar{a}_nz|\approx |I_n|$. Set

By Proposition 2, we see that $\{f_n\}$ is a bounded sequence in $D_{\beta, \lambda}$ and converges to zero uniformly on every compact subset of ${\mathbb D}$. Then

as $n\to \infty$, which implies that $\mu$ is a vanishing $s+(\beta+1)(1-\lambda)$-Carleson measure.

Conversely, suppose that $\mu$ is a vanishing $s+(\beta+1)(1-\lambda)$-Carleson measure. Then $\mu$ is a $s+(\beta+1)(1-\lambda)$-Carleson measure. Therefore, the identity operator $I_d : D_{\beta, \lambda} \to \mathcal{T}_{s}^1(\mu)$ is bounded. From ^[9] we have $\|\mu-\mu_r\|_{CM_{s+(\beta+1)(1-\lambda)}}\to 0,$ as $r\to 1$, where $\mu_r(z) = 0$ for $r\leq |z| < 1$ and $\mu_r(z) = \mu(z)$ for $|z| < r$. Let $\{f_n\}$ be a bounded sequence in $D_{\beta, \lambda}$ with $\sup_{n\in \mathbb{N } }\|f_n\|_{D_{\beta, \lambda}}\preceq 1$ and converges to zero uniformly on every compact subset of ${\mathbb D}$. We obtain

Letting $n\to \infty$ and $r\to 1$, we get $\lim_{n\to \infty}\|f_n\|_{\mathcal{T}_{s}^1} = 0$. So the identity operator $I_d : D_{\beta, \lambda} \to \mathcal{T}_{s}^1(\mu)$ is compact. The proof is complete.

4.   Boundedness of integral operator

In this section, we characterize the boundedness of the operator $T_g: D_{\beta, \lambda} \to F(1, \beta-s, s)$ when $-1 < \beta < 0$, $0 < \lambda, s < 1$ such that $s\ge\lambda(\beta +1)$.

Theorem 3. Let $g\in H({\mathbb D})$, $-1 < \beta < 0$, $0 < \lambda, s < 1$ such that $s\ge\lambda(\beta +1)$. Then $T_g: D_{\beta, \lambda} \to F(1, \beta-s, s)$ is bounded if and only if

Proof. First, suppose that $T_g: D_{\beta, \lambda} \to F(1, \beta-s, s)$ is bounded. For any fixed arc $I\subset \partial {\mathbb D}$, let $e^{i\theta}$ denote the center of $I$ and $a = (1-|I|)e^{i\theta}.$ Set

Then we get $\|T_gf_a\|_{F(1, \beta-s, s)}\preceq \|T_g\|_{D_{\beta, \lambda} \to F(1, \beta-s, s)}\|f_a\|_{D_{\beta, \lambda}} < \infty,$ by the assumption and Proposition 2. Since $(T_gf_a)'(z) = f_a(z)g'(z)$, we have

where $d\mu_g(z) = |g'(z)|(1-|z|^2)^{\beta} dA(z)$. Hence $\mu_g$ is a $s+(\beta+1)(1-\lambda)$ -Carleson measure. Employing Lemma 1, we obtain that

which implies that $g\in F(1, \beta-s -(\beta+1)(1-\lambda), s+(\beta+1)(1-\lambda))$.

Conversely, assume that $g\in F(1, \beta-s -(\beta+1)(1-\lambda), s+(\beta+1)(1-\lambda))$. Then we see that $\mu_g$ is a $s+(\beta+1)(1-\lambda)$ -Carleson measure. For each $f\in D_{\beta, \lambda}$, by Theorem 1, we have

Therefore, $T_g: D_{\beta, \lambda} \to F(1, \beta-s, s)$ is bounded. The proof is complete.

In particular, taking $s = \lambda(\beta+1)$, we get the following result.

Theorem 4. Let $-1 < \beta < 0$, $0 < \lambda < 1$ and $g\in H({\mathbb D})$. Then $T_g: D_{\beta, \lambda} \to D_{\beta, \lambda}$ is bounded if and only if $g\in F(1, -1, \beta+1).$

5.   Essential norm of integral operator

In this section, we investigate the essential norm of the operator $T_g: D_{\beta, \lambda} \to F(1, \beta-s, s)$. We first recall some definitions. The essential norm of $T:X\to Y$ is defined by

where $(X, \|\cdot\|_X)$ and $(Y, \|\cdot\|_Y)$ are Banach spaces, $\|T-K\|_{X\to Y}$ is the operator norm of the operator $T-K$ from $X$ to $Y$. It is easy to see that $T:X\to Y$ is compact if $\|T\|_{e, X\to Y} = 0$.

For a closed subspace $A$ of $X$, given $f\in X$, the distance from $f$ to $A$ is defined by ${\rm dist}_X(f, A) = \inf_{g\in A}\|f-g\|_X$.

Let $F_{0}(1, \beta-s-(\beta+1)(1-\lambda), s+(\beta+1)(1-\lambda))$ denote the space of all functions $f\in F(1, \beta-s-(\beta+1)(1-\lambda), s+(\beta+1)(1-\lambda))$ such that

The following lemma gives the distance from $F(1, \beta-s-(\beta+1)(1-\lambda), s+(\beta+1)(1-\lambda))$ to $F_0(1, \beta-s-(\beta+1)(1-\lambda), s+(\beta+1)(1-\lambda))$.

Lemma 5. Let $-1 < \beta < 0$, $0 < \lambda < 1$. If $g\in F(1, m, t)$, then

Here $m = \beta-s-(\beta+1)(1-\lambda)$, $t = s+(\beta+1)(1-\lambda)$, $g_r(z) = g(rz)$, $0 < r < 1$, $z\in {\mathbb D}$.

Proof. Given any $g \in F(1, m, t)$, then $g_r \in F_0(1, m, t)$ and $\|g_r\|_{F(1, m, t)}\preceq \|g\|_{F(1, m, t)}.$ For any $\xi\in (0, 1)$, there exists a $a\in (0, \xi)$ such that $\sigma_a(z)$ lies in a compact subset of ${\mathbb D}$. Then $\lim_{r\to 1^{-}}\sup_{z\in{\mathbb{D}}}|g'(\sigma_a(z))-rg'(r\sigma_a(z))| = 0.$ Changing the variable $z = \sigma_a(w)$, we have

By the definition of distance mentioned above, we obtain that

We write $\Phi_{r, a}(z) = \sigma_{ra}\circ r \sigma_a(z)$. Then $\Phi_{r, a}$ is an analytic self-map of ${\mathbb D}$ and $\Phi_{r, a}(0) = 0$. Changing the variable $z = \sigma_a(w)$, and using the Littlewood's Subordination Theorem (see Theorem 1.7 of ^[5]), we get

Take the supremum on the above inequality over $w\in {\mathbb D}$. Because of the arbitrariness of $\xi$, we obtain

For each $g\in F(1, m, t)$, it is easy to get

The proof is complete.

Lemma 6. Let $-1 < \beta < 0$, $0 < \lambda, s < 1$ such that $s\ge\lambda(\beta +1)$. If $f \in F(1, \beta-s-(\beta+1)(1-\lambda), s+(\beta+1)(1-\lambda))$, then

Proof. For any $a\in {\mathbb D}$, by a change of variable argument, we have

The last inequality used the Lemma 4.12 in ^[26]. The proof is complete.

Lemma 7. Let $-1 < \beta < 0$, $0 < \lambda, s < 1$ such that $s\ge\lambda(\beta +1)$. If $g \in F(1, \beta-s-(\beta+1)(1-\lambda), s+(\beta+1)(1-\lambda))$ and $0 < r < 1$, then $T_{g_r}:D_{\beta, \lambda}\to F(1, \beta-s, s)$ is compact.

Proof. Let $\{f_n\}$ be a bounded sequence in $D_{\beta, \lambda}$ such that $\{f_n\}$ converges to zero uniformly on any compact subset of ${\mathbb D}$. Changing the variable $z = \sigma_a(w)$, for any $a\in {\mathbb D}$, from Lemma 6 and Proposition 1 we have

where $m = \beta-s-(\beta+1)(1-\lambda)$, $t = s+(\beta+1)(1-\lambda)$. Using the Dominated Convergence Theorem we obtain

which implies that $T_{g_r}:D_{\beta, \lambda}\to F(1, \beta-s, s)$ is compact. The proof is complete.

The following conclusion is important for studying the essential norm of operators on some analytic function spaces, see ^[20].

Lemma 8. Let $X, Y$ be two Banach spaces of analytic functions on ${\mathbb D}$. Suppose that

(i) The point evaluation functionals on $Y$ are continuous.

(ii) The closed unit ball of $X$ is a compact subset of $X$ in the topology of uniform convergence on compact sets.

(iii) $T : X\rightarrow Y$ is continuous when $X$ and $Y$ are given the topology of uniform convergence on compact sets.

Then, $T$ is a compact operator if and only if for any bounded sequence $\{f_n\}$ in $X$ such that $\{f_n\}$ converges to zero uniformly on every compact set of ${\mathbb D}$, then the sequence $\{Tf_n\}$ converges to zero in the norm of $Y$.

Theorem 5. Let $g \in H({\mathbb D})$, $-1 < \beta < 0$, $0 < \lambda, s < 1$ such that $s\ge\lambda(\beta +1)$. If $T_g:D_{\beta, \lambda}\to F(1, \beta-s, s)$ is bounded, then

Here $m = \beta-s-(\beta+1)(1-\lambda)$, $t = s+(\beta+1)(1-\lambda)$.

Proof. Assume that $\{I_n\}$ is a sequence of subarcs of $\partial {\mathbb D}$ with $\lim\limits_{n\to \infty}|I_n| = 0$. Let $a_n = (1-|I_n|)\xi_n$, where $\xi_n$ is the center of arc $I_n$. Then $\{a_n\}$ is a bounded sequence in ${\mathbb D}$ such that $\lim_{n\to \infty}|a_n| = 1$. Set

Then $\{f_n\}$ is a bounded sequence in $D_{\beta, \lambda}$ and converges to zero uniformly on every compact subset of ${\mathbb D}$. Moreover,

For any compact operator $K:D_{\beta, \lambda}\to F(1, \beta-s, s)$, by Lemma 8, we have $\lim_{n\to \infty}\|Kf_{n}\|_{F(1, \beta-s, s)} = 0$. Hence

Then it is obvious that

Since $\{a_n\}$ is arbitrary, using Lemma 5, we have

Conversely, by Lemma 7 and Theorem 3,

Using Lemma 5 again, we obtain that

The proof is complete.

From the last theorem, we get the following corollary.

Corollary 1. Let $-1 < \beta < 0$, $0 < \lambda, s < 1$ such that $s\ge \lambda(\beta +1)$. If $g \in H({\mathbb D})$, then $T_g:D_{\beta, \lambda}\to F(1, \beta-s, s)$ is compact if and only if

In particular, when $s = \lambda(\beta +1)$, we get the following result.

Corollary 2. Let $-1 < \beta < 0$, $0 < \lambda < 1$. If $g \in H({\mathbb D})$, then $T_g:D_{\beta, \lambda}\to D_{\beta, \lambda}$ is compact if and only if $g\in F_0(1, -1, \beta+1).$

6.   Conclusions

In this paper, we mainly prove that the identity operator $I_d : D_{\beta, \lambda} \to \mathcal{T}_s^1(\mu)$ is bounded(compact) if and only if the measure $\mu$ is a $s+(\beta+1)(1-\lambda)$-Carleson measure(vanishing $s+(\beta+1)(1-\lambda)$-Carleson measure). As an application, we prove that Volterra integral operator $T_g: D_{\beta, \lambda} \to F(1, \beta-s, s)$ is bounded(compact) if and only if

In particular, $T_g:D_{\beta, \lambda}\to D_{\beta, \lambda}$ is bounded(compact) if and only if $g\in F(1, -1, \beta+1)$($g\in F_0(1, -1, \beta+1)$).

Acknowledgments

The authors thank five referees for useful remarks and comments that led to the improvement of this paper. This work was supported by NNSF of China (No.11720101003).

Conflict of interest

We declare that we have no conflict of interest.