Hilbert type operators acting from the Bloch space into Bergman spaces

Abstract

Let µ be a finite positive Borelmeasure on $[0,1)$ and $\alpha \gt -1$. The generalized integral operator of Hilbert type $\mathcal {I}_{\mu_{\alpha+1}}$ is defined on the spaces $H(\mathbb{D})$ of analytic functions in the unit disc $\mathbb{D}$ as follows:

\begin{equation*}\mathcal {I}_{\mu_{\alpha+1}}(f)(z)=\int_{0}^{1} \frac{f(t)}{(1-tz)^{\alpha+1}}d\mu(t),\ \ f\in H(\mathbb{D}),\ \ z\in \mathbb{D} .\end{equation*}

In this paper, we give a unified characterization of the measures µ for which the operator $\mathcal {I}_{\mu_{\alpha+1}}$ is bounded from the Bloch space to a Bergman space for all $\alpha \gt -1$. Additionally, we also investigate the action of $\mathcal {I}_{\mu_{\alpha+1}}$ from the Bloch space to the Hardy spaces and the Besov spaces.

1. Introduction

Let $\mathbb{D}=\{z\in \mathbb{C}:\vert z\vert \lt 1\}$ denote the open unit disk of the complex plane $\mathbb{C}$ and $H(\mathbb{D})$ denote the space of all analytic functions in $\mathbb{D}$.

The Bloch space $\mathcal {B}$ consists of those functions $f\in H(\mathbb{D}) $ for which

\begin{equation*} \vert \vert f\vert \vert _{\mathcal {B}}=\vert f(0)\vert +\sup_{z\in \mathbb{D}}(1-\vert z\vert ^{2})\vert f'(z)\vert \lt \infty. \end{equation*}

The little Bloch space $\mathcal {B}_{0}$ is the closed subspace of $\mathcal {B}$ consisting of those functions $f\in H(\mathbb{D}) $ such that

\begin{equation*} \lim_{\vert z\vert \rightarrow 1^{-}}(1-\vert z\vert ^{2})\vert f'(z)\vert =0. \end{equation*}

Let $0 \lt p\leq\infty$, and the classical Hardy space H^p consists of those functions $f\in H(\mathbb{D})$ for which

\begin{equation*} ||f||_{p}=\sup_{0\leq r \lt 1} M_p(r, f) \lt \infty, \end{equation*}

where

\begin{equation*} M_p(r, f)= \left(\frac{1}{2\pi}\int_0^{2\pi}|f(re^{i\theta})|^p \text{d}\theta \right)^{1/p}, \ 0 \lt p \lt \infty, \end{equation*}

\begin{equation*}M_{\infty}(r, f)=\sup_{|z|=r}|f(z)|.\end{equation*}

For $0 \lt p \lt \infty$, the Bergman space A ^p consists of those $f\in H(\mathbb{D})$ such that

\begin{equation*}||f||^{p}_{A^{p}}=\int_{\mathbb{D}}|f(z)|^{p}\text{d}A(z) \lt \infty,\end{equation*}

where $\text{d}A(z)=\frac{\text{d}x\text{d}y}{\pi}$ is the normalized area measure on $\mathbb{D}$. The reader is referred to [8] for Hardy spaces and [25] for Bergman spaces and Bloch spaces.

Let $f(z)=\sum_{n=0}^{\infty} a_{n}z^{n}\in H(\mathbb{D})$. For any complex parameters t and s with the property that neither $1+t$ nor $1+t+s$ is a negative integer, the fractional differential operator $R^{t,s}$ and the fractional integral operator $R_{t,s}$ are defined as follows:

\begin{equation*} R^{t,s}f(z)=\sum_{n=0}^{\infty}\frac{\Gamma(2+t)\Gamma(n+2+t+s)}{\Gamma(2+t+s)\Gamma(n+2+t)}a_{n}z^{n}, \end{equation*}

\begin{equation*} R_{t,s}f(z)=\sum_{n=0}^{\infty} \frac{\Gamma(2+t+s)\Gamma(n+2+t)}{\Gamma(2+t)\Gamma(n+2+s+t)}a_{n}z^{n}. \end{equation*}

Let µ be a finite positive Borel measure on $[0,1)$ and $n\in \mathbb{N}$. We use µ_n to denote the sequence of order n of µ, that is, $\mu_{n}=\int_{[0,1)}t^{n}\text{d}\mu(t)$. Let $\mathcal {H}_{\mu}$ be the Hankel matrix $(\mu_{n,k})_{n,k\geq 0}$ with entries $\mu_{n,k}=\mu_{n+k}$. The matrix $\mathcal {H}_{\mu}$ induces an operator on $ H(\mathbb{D}) $ by its action on the Taylor coefficients: $a_{n}\rightarrow\displaystyle{\sum_{k=0}^{\infty}}\mu_{n,k}a_{k},\ n\in \mathbb{N}\cup \{0\}.$ The generalized Hilbert operator $\mathcal {H}_{\mu}$ defined on the spaces $H(\mathbb{D})$ of analytic functions in the unit disc $\mathbb{D}$ is as follows:

If $f\in H(\mathbb{D})$, $f(z)=\displaystyle{\sum_{n=0}^{\infty}}a_{n}z^{n}$, then

\begin{equation*} \mathcal {H}_{\mu}(f)(z)=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{\infty}\mu_{n,k}a_{k}\right)z^{n}, \ \ z\in \mathbb{D}, \end{equation*}

whenever the right hand side makes sense and defines an analytic function in $\mathbb{D}$.

If µ is the Lebesgue measure on $[0, 1)$, then the matrix $\mathcal {H}_{\mu}$ reduces to the classical Hilbert matrix $\mathcal {H}=(\frac{1}{n+k+1})_{n,k\geq 0}$, which induces the classical Hilbert operator $\mathcal {H}$. The classic Hilbert operator $\mathcal {H}$ has been extensively studied on distinct analytic function spaces recently (see [2, 5–7, 15]).

It turns out that the integral operator

\begin{equation*}\mathcal {I}_{\mu}(f)(z)=\int_{0}^{1} \frac{f(t)}{1-tz}\text{d}\mu(t)\end{equation*}

plays a key role when we study the generalized Hilbert operator $\mathcal {H}_{\mu}$. Galanopoulos and Peláez [10] characterized the measures µ for which the operator $\mathcal {I}_{\mu}$ is well defined in H ¹ and proved that $\mathcal {H}_{\mu}(f)=\mathcal {I}_{\mu}(f)$ for such measures. Chatzifountas, Girela and Peláez [4] extended this work to H ^p for all $0 \lt p \lt \infty$. Girela and Merchán [12] studied these operators acting on certain conformally invariant spaces of analytic functions. The reader is referred to [11, 13, 14, 18, 22] for more on this topic.

In this paper, we consider the generalized integral type Hilbert operator defined as follows:

\begin{equation*}\mathcal {I}_{\mu_{\alpha+1}}(f)(z)=\int_{0}^{1} \frac{f(t)}{(1-tz)^{\alpha+1}}\text{d}\mu(t),\ \ \ (\alpha \gt -1).\end{equation*}

If α = 0, then $\mathcal {I}_{\mu_{\alpha+1}}$ is just the integral operator $\mathcal {I}_{\mu}$. It is easy to see that $R^{-1,\alpha}\mathcal {I}_{\mu}(f)=\mathcal {I}_{\mu_{\alpha+1}}(f)$ whenever $\mathcal {I}_{\mu}(f)$ makes sense and defines an analytic function in $\mathbb{D}$. Moreover, the operator $\mathcal {I}_{\mu_{\alpha+1}}$ is also closely related to the following generalized Hilbert operator:

\begin{equation*} \mathcal{H}^{\alpha}_{\mu}(f)(z)= \sum_{n=0}^{\infty}\frac{\Gamma(n+1+\alpha)}{\Gamma(n+1)\Gamma(\alpha+1)}\left(\sum_{k=0}^{\infty}\mu_{n,k}a_{k}\right)z^{n}, \ \ (\alpha \gt -1). \end{equation*}

When α = 1, the operators $\mathcal {I}_{\mu_{2}}$ and $\mathcal {H}^{1}_{\mu}$ have been studied in [16, 23, 24] recently.

Let us mention the following results which were obtained by Ye and Zhou [23].

Theorem A Let µ be a positive Borel measure on $[0, 1)$ with $\int_{0}^{1} \log\frac{e}{1-t}d\mu(t) \lt \infty$. Then, the following statements are equivalent:

1. (1) $\mathcal {I}_{\mu_{2}}$ is bounded from $\mathcal {B}$ to A ¹.

2. (2) $\mathcal {I}_{\mu_{2}}$ is compact from $\mathcal {B}$ to A ¹.

3. (3) The measure µ satisfies $\int_{0}^{1} \log^{2}\frac{e}{1-t}d\mu(t) \lt \infty.$

Theorem B Let $1 \lt p \lt \infty$ and $\frac{1}{p}+\frac{1}{q}=1$. Suppose µ is a positive Borel measure on $[0, 1)$ with $\int_{0}^{1} \log\frac{e}{1-t}d\mu(t) \lt \infty$. Let ν be the positive Borel measure on $[0, 1) $ defined by $d\nu(t)=\log\frac{e}{1-t}d\mu(t)$. Then, the following statements are equivalent:

1. (1) $\mathcal {I}_{\mu_{2}}$ is bounded from $\mathcal {B}$ to A ^p.

2. (2) $\mathcal {I}_{\mu_{2}}$ is compact from $\mathcal {B}$ to A ^p.

3. (3) The measure ν satisfies

   \begin{equation*}\left|\int_{0}^{1} g(t)d\nu(t)\right|\leq C ||g||_{A^{q}}, \ \ \mbox{for every}\ g\in A^{q}. \end{equation*}

The condition (3) in Theorem B is rather abstract and difficult to verify in general. In this paper, we shall extend the results for α = 1 to all $\alpha \gt -1$, and we shall give a new characterization that makes the results more natural and practical. This also allows us to investigate the action of $\mathcal {I}_{\mu_{\alpha+1}}$ from the Bloch space to Hardy spaces and Besov spaces. To state the main results, let us introduce some notions. We let ν be the positive Borel measure on $[0, 1) $ defined by $\text{d}\nu(t)=\log\frac{e}{1-t}\text{d}\mu(t)$ and $\nu_{n}=\int_{0}^{1} t^{n}\text{d}\nu(t)$. For $1 \lt p \lt \infty$, we set $X_{p}=A^{\frac{p}{p-1}}$ and $X_{1}=\mathcal {B}_{0}$ for p = 1. Throughout the paper, we always assume that $\alpha \gt -1$ if there are no additional statements.

Our main results of this paper are stated as follows.

Theorem 1.1 Let $1\leq p \lt \infty$, and let µ be a positive Borel measure on $[0, 1)$ with $\int_{0}^{1} \log\frac{e}{1-t}d\mu(t) \lt \infty$. Then, the following statements are equivalent:

1. (1) The operator $\mathcal {I}_{\mu_{\alpha+1}}$ is bounded from $\mathcal {B}$ to A ^p.

2. (2) The operator $\mathcal {I}_{\mu_{\alpha+1}}$ is compact from $\mathcal {B}$ to A ^p.

3. (3) The measure ν satisfies $\sum_{n=0}^{\infty}(n+1)^{p(\alpha+1)-3}\nu_{n}^{p} \lt \infty.$

4. (4) The measure ν satisfies $\left|\int_{0}^{1} R^{0,\alpha-1}g(t)d\nu(t)\right|\lesssim ||g||_{X_{p}}$ for all $g\in X_{p}$.

Here, we point out that the condition $\int_{0}^{1} \log\frac{e}{1-t}d\mu(t) \lt \infty$ in Theorems A, B and 1.1 is necessary. The operator $\mathcal {I}_{\mu_{\alpha+1}}$ is well defined in the Bloch space $\mathcal {B}$ if and only if $\int_{0}^{1} \log\frac{e}{1-t}\text{d}\mu(t) \lt \infty$. In fact, if $\mathcal {I}_{\mu_{\alpha+1}}$ is well defined in the Bloch space $\mathcal {B}$, take $f(z)=\log\frac{e}{1-z}\in \mathcal {B}$ and then $|\mathcal {I}_{\mu_{\alpha+1}}(f)(0)| \lt \infty$. Since µ is a positive measure and $\log\frac{e}{1-t} \gt 0$ for all $t\in [0,1)$, we have that

\begin{equation*}|\mathcal {I}_{\mu_{\alpha+1}}(f)(0)|=\int_{0}^{1}\log\frac{e}{1-t}\text{d}\mu(t) \lt \infty.\end{equation*}

On the other hand, if $\int_{0}^{1} \log\frac{e}{1-t}\text{d}\mu(t) \lt \infty$. Then for each $f\in \mathcal {B}$, $0 \lt r \lt 1$ and all $z\in \mathbb{D}$ with $|z|\leq r$, we have

\begin{equation*} \begin{split} |\mathcal {I}_{\mu_{\alpha+1}}(f)(z)| &\leq \int_{0}^{1}\frac{|f(t)|}{|1-tz|^{\alpha+1}}\text{d}\mu(t)\leq \frac{1}{(1-|z|)^{\alpha+1}}\int_{0}^{1}|f(t)|\text{d}\mu(t)\\ & \lesssim \frac{||f||_{\mathcal {B}}}{(1-|z|)^{\alpha+1}}\int_{0}^{1}\log\frac{e}{1-t}\text{d}\mu(t)\lesssim \frac{||f||_{\mathcal {B}}}{(1-|z|)^{\alpha+1}}. \end{split} \end{equation*}

This implies that $\mathcal {I}_{\mu_{\alpha+1}}(f)(z)$ uniformly converges on any compact subset of $\mathbb{D}$ and hence is analytic in $\mathbb{D}$.

Throughout the paper, the letter C will denote an absolute constant whose value depends on the parameters indicated in the parentheses and may change from one occurrence to another. We will use the notation ‘$P\lesssim Q$’ if there exists a constant $C=C(\cdot) $ such that ‘$P \leq CQ$’, and ‘$P \gt rsim Q$’ is understood in an analogous manner. In particular, if ‘$P\lesssim Q$’ and ‘$P \gt rsim Q$’ , then we will write ‘$P\asymp Q$’.

2. Proofs and some related results

We start recalling that for a function $f(z)=\sum_{n=0}^\infty a_nz^n$ analytic in $\mathbb{D}$, the polynomials $\Delta_jf$ are defined as follows:

\begin{equation*} \Delta_jf(z)=\sum_{k=2^j}^{2^{j+1}-1} a_kz^k, \quad\text{for}~j\geq 1, \end{equation*}

\begin{equation*}\Delta_0f(z)=a_0+a_1z.\end{equation*}

The polynomials $\Delta_j=\Delta(\frac{1}{1-z})=\sum^{2^{j+1}-1}_{k=2^{j}}z^{j}$ have the following property (see, e.g. [20])

(1)\begin{equation} ||\Delta||_{p}\asymp 2^{j(1-\frac{1}{p})}, \ \ j\in \mathbb{N}, \ 1 \lt p \lt \infty. \end{equation}

Mateljević and Pavlović [17] proved the following result.

Lemma 2.1. Let $1 \lt p \lt \infty$. For a function $f\in H(\mathbb{D})$, we define:

\begin{equation*}K_{1}(f):=\int_{\mathbb{D}} |f(z)|^{p}dA(z),\ \ \ K_{2}(f):=\sum_{n=0}^{\infty} 2^{-n}||\Delta_{n}f||^{p}_{p}.\end{equation*}

Then, $K_{1}(f)\asymp K_{2}(f)$.

Lemma 2.2. [19, Theorem 1]

Let $0 \lt \beta,p \lt \infty$, and let $\{\lambda_{n}\}_{n=0}^{\infty}$ be a sequence of positive numbers. Then,

\begin{equation*}\int_{0}^{1}(1-r)^{p\beta-1}\left(\sum_{n=0}^{\infty} \lambda_{n}r^{n}\right)^{p}dr\asymp \sum_{n=0}^{\infty} 2^{-np\beta}\left(\sum_{k\in I_{n}}\lambda_{k}\right)^{p}\end{equation*}

where $I_{0}=\{0\}$, $I_{n}=[2^{n-1},2^{n})\cap \mathbb{N}$ for $n\in \mathbb{N}$.

The proof of the next result can be found in [15, Lemma 3.4].

Lemma 2.3. Let $1 \lt p \lt \infty$, and let $\gamma=\{\gamma_{k}\}_{k=0}^{\infty}$ be a monotone sequence of positive numbers. Let $g(z)=\sum_{k=0}^{\infty}b_{k}z^{k}$ and $(\gamma g)(z)=\sum_{k=0}^{\infty}b_{k}\gamma_{k}z^{k}$.

1. (1) If $ \{\gamma_{k}\}_{k=0}^{\infty}$ is nondecreasing, then

   \begin{equation*}\gamma_{2^{n}}||\Delta_{n} g||_{p}\lesssim||\Delta_{n}(\gamma g)||_{p}\lesssim \gamma_{2^{n+1}}||\Delta_{n} g||_{p}.\end{equation*}

2. (2) If $ \{\gamma_{k}\}_{k=0}^{\infty}$ is nonincreasing, then

   \begin{equation*}\gamma_{2^{n+1}}||\Delta_{n} g||_{p}\lesssim||\Delta_{n}(\gamma g)||_{p}\lesssim \gamma_{2^{n}}||\Delta_{n} g||_{p}.\end{equation*}

We also need the following estimates (see, e.g. Proposition 1.4.10 in [21]).

Lemma 2.4. Let α be any real number and $z\in \mathbb{D}$. Then

\begin{equation*} \int^{2\pi}_0\frac{d\theta}{|1-ze^{-i\theta}|^{\alpha}}\asymp \begin{cases}1 & \quad \text{if} \ \ \alpha \lt 1,\\ \log\frac{2}{1-|z|^2} & \quad \text{if} \ \ \alpha=1,\\ \frac{1}{(1-|z|^2)^{\alpha-1}} & \quad \text{if}\ \ \alpha \gt 1, \end{cases} \end{equation*}

Proof. $(1)\Rightarrow(3)$

Suppose that $\mathcal {I}_{\mu_{\alpha+1}}$ is a bounded operator from $ \mathcal {B}$ to A ^p. Take $h(z)=\log\frac{e}{1-z}\in \mathcal {B}$, then $\mathcal {I}_{\mu_{\alpha+1}} (h)\in A^{p}$ and $||\mathcal {I}_{\mu_{\alpha+1}} (h)||_{A^{p}}\lesssim ||h||_{\mathcal {B}}$.

For p = 1, by Hardy’s inequality and Stirling’s formula

\begin{equation*} \begin{split} 1& \gt rsim||h||_{\mathcal {B}} \gt rsim ||\mathcal {I}_{\mu_{\alpha+1}}(h)||_{A^{1}}=2\int_{0}^{1}M_{1}(r,\mathcal {I}_{\mu_{\alpha+1}}(h))r\text{d}r\\ & =\frac{1}{\pi} \int_{0}^{1} \int_{0}^{2\pi}\left|\int_{0}^{1} \frac{\log\frac{e}{1-t}}{(1-tr\text{e}^{i\theta})^{\alpha+1}}\text{d}\mu(t)\right|\text{d}\theta r\text{d}r\\ & =\frac{1}{\pi}\int_{0}^{1} \int_{0}^{2\pi}\left|\sum_{n=0}^{\infty} \frac{\Gamma(\alpha+1+n)\nu_{n}}{\Gamma(\alpha+1)\Gamma(n+1)}r^{n}\text{e}^{in\theta}\right|\text{d}\theta r \text{d}r\\ & \gt rsim \int_{0}^{1} \sum_{n=0}^{\infty} \frac{\Gamma(\alpha+1+n)\nu_{n}}{(n+1)\Gamma(\alpha+1)\Gamma(n+1)}r^{n+1} \text{d}r\\ & \asymp \sum_{n=0}^{\infty} (n+1)^{\alpha-2}\nu_{n}. \end{split} \end{equation*}

This implies that $ \sum_{n=0}^{\infty} (n+1)^{\alpha-2}\nu_{n} \lt \infty$.

For p > 1, by Lemma 2.1, we have

\begin{equation*}||h||_{\mathcal {B}} \gt rsim||\mathcal {I}_{\mu_{\alpha+1}} (h)||_{A^{p}}\asymp \sum_{n=0}^{\infty}2^{-n}||\Delta_{n}(\mathcal {I}_{\mu_{\alpha+1}} (h))||^{p}_{p}.\end{equation*}

It is obvious that

(2)\begin{equation} \mathcal {I}_{\mu_{\alpha+1}}(h)(z)=\sum_{n=0}^{\infty} \frac{\Gamma(n+1+\alpha)}{\Gamma(n+1)\Gamma(\alpha+1)}\left(\int_{0}^{1} t^{n}\log\frac{e}{1-t}\text{d}\mu(t)\right)z^{n}. \end{equation}

Hence, by (2), Lemma 2.3 and (1) we have

\begin{equation*} \begin{split} 1 & \gt rsim||h||_{\mathcal {B}} \gt rsim \sum_{n=0}^{\infty}2^{-n}||\Delta_{n}(\mathcal {I}_{\mu_{\alpha+1}} (h))||^{p}_{p}\\ &= \sum_{n=0}^{\infty}2^{-n}\left\|\sum^{2^{n+1}-1}_{k=2^{n}}\frac{\Gamma(\alpha+1+k)}{\Gamma(\alpha+1)\Gamma(k+1)}\nu_{k} z^{k}\right\|_{p}^{p}\\ & \gt rsim \sum_{n=0}^{\infty}2^{-n}\nu_{2^{n+1}}^{p}\left\|\Delta_{n}\left(\frac{1}{(1-z)^{\alpha+1}}\right)\right\|^{p}_{p}\\ & \gt rsim \sum_{n=0}^{\infty}2^{n(p\alpha-1)}\nu_{2^{n+1}}^{p} ||\Delta_{n}||_{p}^{p}\\ & \asymp \sum_{n=0}^{\infty}2^{n(p\alpha+p-2)}\nu_{2^{n+1}}^{p}\\ & \asymp \sum_{k=1}^{\infty}k^{(p\alpha+p-3)}\nu_{k}^{p}. \end{split} \end{equation*}

This gives $ \sum_{n=0}^{\infty} (n+1)^{p(\alpha+1)-3}\nu^{p}_{n} \lt \infty$.

Proof. $(3)\Rightarrow(2)$

If $\sum_{n=0}^{\infty}(n+1)^{p(\alpha+1)-3}\nu_{n}^{p} \lt \infty$, then

\begin{equation*} \begin{split} \sum_{n=1}^{\infty}(n+1)^{p(\alpha+1)-3}\nu_{n}^{p}&= \sum_{n=1}^{\infty}\left(\sum_{k=2^{n-1}}^{2^{n}-1}(k+1)^{p(\alpha+1)-3}\nu_{k}^{p}\right)\\ & \gt rsim \sum_{n=1}^{\infty}2^{n(p(\alpha+1)-2)}\nu_{2^{n}}^{p}\\ & \gt rsim \sum_{n=1}^{\infty}2^{-n}\left(\sum_{k=2^{n}}^{2^{n+1}-1}(k+1)^{\alpha-\frac{1}{p}}\nu_{k}\right)^{p}. \end{split} \end{equation*}

This shows that

\begin{equation*} \sum_{n=1}^{\infty}2^{-n}\left(\sum_{k=2^{n}}^{2^{n+1}-1}(k+1)^{\alpha-\frac{1}{p}}\nu_{k}\right)^{p} \lt \infty.\end{equation*}

By Lemma 2.2, we have that

\begin{equation*}\int_{0}^{1}\left(\sum_{n=0}^{\infty} (n+1)^{\alpha-\frac{1}{p}}\nu_{n}r^{n}\right)^{p}\text{d}r \asymp \sum_{n=0}^{\infty}2^{-n}\left(\sum_{k=2^{n}}^{2^{n+1}-1}(k+1)^{\alpha-\frac{1}{p}}\nu_{k}\right)^{p} \lt \infty.\end{equation*}

Therefore, for any ɛ > 0, there exists $0 \lt r_{0} \lt 1$ such that

\begin{equation*}\int_{r_{0}}^{1}\left(\sum_{n=0}^{\infty} (n+1)^{\alpha-\frac{1}{p}}\nu_{n}r^{n}\right)^{p}\text{d}r \lt \varepsilon.\end{equation*}

By the assumption of µ, for the above ɛ > 0, there exists t ₀ with $0 \lt t_{0} \lt 1$ such that

\begin{equation*}\int_{t_{0}}^{1}\log\frac{e}{1-t}\text{d}\mu(t) \lt \varepsilon.\end{equation*}

Let $\{f_{k}\}_{k=1}^{\infty}$ be a bounded sequence in $\mathcal {B}$ which converges to 0 uniformly on every compact subset of $\mathbb{D}$. It is clear that

\begin{equation*} \begin{split} ||\mathcal {I}_{\mu_{\alpha+1}}(f_{k})||^{p}_{A^{p}}&=\int_{|z|\leq r_{0}}|\mathcal {I}_{\mu_{\alpha+1}}(f_{k})(z)|^{p}\text{d}A(z)+ \int_{r_{0} \lt |z| \lt 1}|\mathcal {I}_{\mu_{\alpha+1}}(f_{k})(z)|^{p}dA(z)\\ & := J_{1,k}+J_{2,k}. \end{split} \end{equation*}

When $|z|\leq r_{0}$, since $\{f_{k}\}_{k=1}^{\infty}$ converges to 0 uniformly on every compact subset of $\mathbb{D}$, we have

\begin{equation*} \begin{split} |\mathcal {I}_{\mu_{\alpha+1}}(f_{k})(z)|& \leq \int_{0}^{1}\frac{|f_{k}(t)|}{|1-tz|^{1+\alpha}}\text{d}\mu(t)\\ & \lesssim \sup_{|t| \lt t_{0}}|f_{k}(t)|+\int_{t_{0}}^{1}\log\frac{e}{1-t}\text{d}\mu(t)\\ & \lesssim \sup_{|t| \lt t_{0}}|f_{k}(t)|+\varepsilon. \end{split} \end{equation*}

It follows that

\begin{equation*}J_{1,k}\rightarrow 0, \ (k\rightarrow \infty).\end{equation*}

Next, we estimate $J_{2,k}$.

If $p(\alpha+1)\geq 1$, then by Minkowski inequality and Lemma 2.4, we get

\begin{align*} \begin{split} M_{p}(r,\mathcal {I}_{\mu_{\alpha+1}}(f_{k})) &\leq \left\{\int_{0}^{2\pi}\left(\int_{0}^{1}\frac{|f_{k}(t)|}{|1-rte^{i\theta}|^{1+\alpha}}{\textrm{d}} \mu(t)\right)^{p} {\textrm{d}} \theta\right\}^{\frac{1}{p}}\\ &\lesssim \int_{0}^{1}\log\frac{e}{1-t}\left(\int_{0}^{2\pi}\frac{{\textrm{d}} \theta}{|1-tre^{i\theta}|^{p(\alpha+1)}}\right)^{\frac{1}{p}} {\textrm{d}} \mu(t)\\ & \lesssim\int_{0}^{1}F(t,r)\log\frac{e}{1-t}{\textrm{d}} \mu(t)\\ & \asymp \sum_{n=0}^{\infty} (n+1)^{\alpha-\frac{1}{p}}\nu_{n}r^{n}, \end{split} \end{align*}

where

\begin{equation*}F(t,r)=\left\{ \begin{array}{cc} \displaystyle{\frac{1}{(1-tr)^{\alpha+1-\frac{1}{p}}},} & \displaystyle{p(\alpha+1) \gt 1}\\ \displaystyle{\log\frac{e}{1-tr}, } & \displaystyle{p(\alpha+1)=1} \end{array}\right.\end{equation*}

By the polar coordinate formula, we have

\begin{equation*} \begin{split} J_{2,k}& =\int_{r_{0} \lt |z| \lt 1}|\mathcal {I}_{\mu_{\alpha+1}}(f_{k})(z)|^{p}\text{d}A(z)\\ & \lesssim \int_{r_{0}}^{1}M^{p}_{p}(r,\mathcal {I}_{\mu_{\alpha+1}}(f_{k}))\text{d}r\\ & \lesssim \int_{r_{0}}^{1}\left(\sum_{n=0}^{\infty} (n+1)^{\alpha-\frac{1}{p}}\nu_{n}r^{n}\right)^{p}\text{d}r\\ & \lesssim \varepsilon. \end{split} \end{equation*}

If $0 \lt p(\alpha+1) \lt 1$, a similar discussion to the previous one leads to

\begin{equation*}M_{p}(r,\mathcal {I}_{\mu_{\alpha+1}}(f_{k}))\lesssim \int_{0}^{1}\log\frac{e}{1-t}\text{d}\mu(t)\lesssim 1.\end{equation*}

Note that

\begin{equation*}\int_{r_{0}}^{1}dr \asymp \int_{r_{0}}^{1}\left(\sum_{n=0}^{\infty} (n+1)^{\alpha-\frac{1}{p}}\nu_{n}r^{n}\right)^{p}\text{d}r \lt \varepsilon.\end{equation*}

This also yields that $J_{2}\lesssim \varepsilon.$ Therefore,

\begin{equation*}\lim_{k\rightarrow\infty}||\mathcal {I}_{\mu_{\alpha+1}}(f_{k})||^{p}_{A^{p}}=0.\end{equation*}

This shows that $\mathcal {I}_{\mu_{\alpha+1}}: \mathcal {B}\rightarrow A^{p}$ is compact.

Let us remark that if $g(z)=\sum_{n=0}^{\infty} \widehat{g}(n)z^{n}\in H(\mathbb{D})$ and the sequence $\{\widehat{g}(n)\}^{\infty}_{n=0}$ is non-negatively decreasing, then $g\in H^{p}$ $(p\geq 1)$ if and only if $\sum_{n=0}^{\infty} (n+1)^{p-2}\widehat{g}(n)^{p} \lt \infty$ and $g\in \mathcal {B}_{0}$ $\Leftrightarrow$ $g\in VMOA$ $\Leftrightarrow$ $\widehat{g}(n)=o(\frac{1}{n})$ (see, e.g. [20]). Also, $g\in A^{p}$ if and only if $\sum_{n=0}^{\infty} (n+1)^{p-3}\widehat{g}(n)^{p} \lt \infty$ (see, e.g. [3], Proposition 2.4).

The equivalence $(3)\Leftrightarrow (4)$ is a consequence of the following proposition.

Proposition 2.5. Let $1\leq p \lt \infty$,$\frac{1}{p}+\frac{1}{q}=1$, and let µ be a positive Borel measure on $[0, 1)$ with $\int_{0}^{1} \log\frac{e}{1-t}d\mu(t) \lt \infty$. Then, the following statements hold:

1. (a) If p > 1, then $\sum_{n=0}^{\infty}(n+1)^{p(\alpha+1)-3}\nu^{p}_{n} \lt \infty$ if and only if

   \begin{equation*}\left|\int_{0}^{1} R^{0,\alpha-1}g(t)d\nu(t)\right|\lesssim ||g||_{A^{q}} \ \ \mbox{for all}\ g\in A^{q}.\end{equation*}

2. (b) If p = 1, then $\sum_{n=0}^{\infty}(n+1)^{p(\alpha+1)-3}\nu_{n} \lt \infty$ if and only if

   \begin{equation*}\left|\int_{0}^{1} R^{0,\alpha-1}g(t)d\nu(t)\right|\lesssim ||g||_{\mathcal {B}} \ \ \mbox{for all}\ g\in \mathcal {B}_{0}.\end{equation*}

Proof. (a). Suppose that $\sum_{n=0}^{\infty}(n+1)^{p(\alpha+1)-3}\nu^{p}_{n} \lt \infty$. As we have proved previously, $\mathcal {I}_{\mu_{\alpha+1}}$ is a bounded operator from $ \mathcal {B}$ to A ^p. Since $\int_{0}^{1}\log\frac{e}{1-s}d\mu(s) \lt \infty$, it is easy to see that

(3)\begin{equation} \int_{0}^{1}\vert f(s)\vert \text{d}\mu(s)\lesssim \vert \vert f\vert \vert _{\mathcal {B}} \ \ \ \mbox{for all} \ \ f\in \mathcal {B}. \end{equation}

For each $0\leq r \lt 1$, $f\in \mathcal {B}$ and $g\in A^{q}$, by (3) and Hölder inequality, we have that

\begin{equation*} \begin{split} & \ \ \ \ \ \int_{\mathbb{D}}\int_{0}^{1}\left| \frac{f(s)g(rz)}{(1-rsz)^{\alpha+1}} \right| \text{d}\mu(s)dA(z)\\ &\leq \frac{1}{(1-r)^{\alpha+1}}\int_{0}^{1}| f(s)| \text{d}\mu(s)\int_{\mathbb{D}}|g(rz)| dA(z)\\ &\lesssim \frac{|| f||_{\mathcal {B}}}{(1-r)^{\alpha+1}}||g_{r}||_{A^{q}}\lesssim \frac{|| f||_{\mathcal {B}}}{(1-r)^{\alpha+1}}||g||_{A^{q}} \lt \infty,\\ \end{split} \end{equation*}

where $g_{r}(z)=g(rz)$. Let $g(z)=\sum_{n=0}^{\infty}b_{n}z^{n}\in A^{q}$. By Fubini’s theorem and a simple calculation through polar coordinate, we have that

\begin{equation*} \begin{split} & \ \ \ \ \int_{\mathbb{D}} \overline{\mathcal {I}_{\mu_{\alpha+1}}(f)(rz)}g(rz)\text{d}A(z)\\ & = \int_{0}^{1} \sum_{n=0}^{\infty} \frac{\Gamma(n+1+\alpha)}{\Gamma(n+2)\Gamma(\alpha+1)}b_{n}(r^{2}t)^{n}\overline{f(t)}\text{d}\mu(t)\\ & =\int_{0}^{1} R^{0,\alpha-1} g(r^{2}t) \overline{f(t)}d\mu(t).\\ \end{split} \end{equation*}

for all $0\leq r \lt 1$, $f\in \mathcal {B}$, $g\in A^{q}$. Recall that the duality $(A^{q})^{\ast}\simeq A^{p}$ under the pairing

\begin{equation*}\langle F, G\rangle=\lim_{r\rightarrow 1^{-}} \int_{\mathbb{D}} F(rz)\overline{G(rz)}\text{d}A(z), \ \ F\in A^{q}, G\in A^{p} .\end{equation*}

Therefore, the operator $\mathcal {I}_{\mu_{\alpha+1}}: \mathcal {B}\rightarrow A^{p}$ is bounded if and only if

\begin{equation*}\left|\int_{0}^{1} R^{0,\alpha-1} g(t) \overline{f(t)}\text{d}\mu(t)\right|\lesssim ||f||_{\mathcal {B}}||g||_{A^{q}}\ \mbox{for all}\ f \in \mathcal {B},\ g\in A^{q}. \end{equation*}

Now, take $f(z)=\log\frac{e}{1-z}\in \mathcal {B}$ in above inequality, and we get

\begin{equation*}\left|\int_{0}^{1} R^{0,\alpha-1}g(t)\text{d}\nu(t)\right|\lesssim ||g||_{A^{q}} \ \ \mbox{for all}\ g\in A^{q}.\end{equation*}

Conversely, assume $h(z)=\sum_{n=0}^{\infty} \widehat{h}(n) z^{n}\in A^{q}$ and the sequence $\{\widehat{h}(n)\}_{n=0}^{\infty}$ is a decreasing sequence of non-negative real numbers. Since the measure ν satisfies

\begin{equation*}\int_{0}^{1} R^{0,\alpha-1}h(t)\text{d}\nu(t)\lesssim ||h||_{A^{q}},\end{equation*}

it follows that

\begin{equation*} \begin{split} ||h||_{A^{q}} & \gt rsim\int_{0}^{1} R^{0,\alpha-1}h(t)\text{d}\nu(t)\\ & = \int_{0}^{1} \sum_{n=0}^{\infty} \frac{\Gamma(n+1+\alpha)}{\Gamma(n+2)\Gamma(\alpha+1)}\widehat{h}(n)t^{n}\text{d}\nu(t)\\ & = \sum_{n=0}^{\infty} \frac{\Gamma(n+1+\alpha)}{\Gamma(n+2)\Gamma(\alpha+1)}\widehat{h}(n)\nu_{n}\\ & \asymp \sum_{n=0}^{\infty} (n+1)^{\frac{q-3}{q}}\widehat{h}(n)(n+1)^{\alpha-1-\frac{q-3}{q}}\nu_{n}. \end{split} \end{equation*}

As we remarked before, the sequence $\{\widehat{h}(n)(n+1)^{\frac{q-3}{q}}\}_{n=0}^{\infty}\in \ell^{q}$. The well-known duality $(\ell^{q})^{*}=\ell^{p}$ implies that $\{(n+1)^{\alpha-1-\frac{q-3}{q}}\nu_{n}\}_{n=0}^{\infty}\in l^{p}$. Therefore,

\begin{equation*}\sum_{n=0}^{\infty} (n+1)^{p\alpha+p-3}\nu^{p}_{n} \lt \infty.\end{equation*}

(b). The proof of the necessity is similar to (a). It is enough to notice that $(\mathcal {B}_{0})^{\ast}\simeq A_{1}$ under the pairing

\begin{equation*}\langle F, G\rangle=\lim_{r\rightarrow 1^{-}} \int_{\mathbb{D}} F(rz)\overline{G(rz)}\text{d}A(z), \ \ F\in \mathcal {B}_{0}, G\in A_{1} .\end{equation*}

For the ‘if’ part, let $h(z)=\sum_{n=0}^{\infty} \widehat{h}(n)z^{n}\in \mathcal {B}_{0}$ and the sequence $\{\widehat{h}(n)\}_{n=0}^{\infty}$ is a decreasing sequence of non-negative real numbers, then $\widehat{h}(n)=o(\frac{1}{n}).$ This implies that the sequence $\{(n+1)\widehat{h}(n)\}_{n=0}^{\infty}\in c_{0}$. Arguing as the proof of (a), we have that

\begin{equation*}||h||_{\mathcal {B}} \gt rsim \sum_{n=0}^{\infty} (n+1)^{\alpha-1}\widehat{h}(n)\nu_{n}.\end{equation*}

Since $(c_{0})^{*}=\ell^{1}$, it follows that $\{(n+1)^{\alpha-2}\nu_{n}\}_{n=0}^{\infty}\in \ell^{1}$.

Corollary 2.6. Let α > 1, and let µ be a positive Borel measure on $[0, 1)$ with $\int_{0}^{1} \log\frac{e}{1-t}\text{d}\mu(t) \lt \infty$. Then the following statements are equivalent:

1. (1) The operator $\mathcal {I}_{\mu_{\alpha+1}}$ is bounded from $\mathcal {B}$ to A ¹.

2. (2) The operator $\mathcal {I}_{\mu_{\alpha+1}}$ is compact from $\mathcal {B}$ to A ¹.

3. (3) The measure µ satisfies $\sum_{n=0}^{\infty}(n+1)^{\alpha-2}\nu_{n} \lt \infty.$

4. (4) The measure µ satisfies $\int_{0}^{1} \frac{\log\frac{e}{1-t}}{(1-t)^{\alpha-1}}d\mu(t) \lt \infty. $

Corollary 2.7. Let µ be a positive Borel measure on $[0, 1)$ with $\int_{0}^{1} \log\frac{e}{1-t}\text{d}\mu(t) \lt \infty$. Then, the following statements are equivalent:

1. (1) The operator $\mathcal {I}_{\mu_{2}}$ is bounded from $\mathcal {B}$ to A ¹.

2. (2) The operator $\mathcal {I}_{\mu_{2}}$ is compact from $\mathcal {B}$ to A ¹.

3. (3) The measure µ satisfies $\sum_{n=0}^{\infty}(n+1)^{-1}\nu_{n} \lt \infty.$

4. (4) The measure µ satisfies $\int_{0}^{1} \log^{2}\frac{e}{1-t}d\mu(t) \lt \infty.$

Corollary 2.8. Let $\alpha \gt -1$, $1\leq p \lt \infty$, and let µ be a positive Borel measure on $[0, 1)$ with $\int_{0}^{1} \log\frac{e}{1-t}d\mu(t) \lt \infty$. If $0 \lt p(\alpha+1) \lt 2$, then the operator $\mathcal {I}_{\mu_{\alpha+1}}$ is compact from $ \mathcal {B}$ to A ^p.

Proof. Since $\int_{0}^{1}\log\frac{e}{1-t}\text{d}\mu(t) \lt \infty$, we have that $\nu_{n}\lesssim 1$ for all $n \in \mathbb{N}\cup \{0\}$. If $0 \lt p(\alpha+1) \lt 2$, then the sum $\sum_{n=0}^{\infty}(n+1)^{p(\alpha+1)-3}\nu_{n}^{p}$ is always finite. This implies that $\mathcal {I}_{\mu_{\alpha+1}}: \mathcal {B}\rightarrow A^{p}$ is compact.

Corollary 2.9. Let $1\leq p \lt \infty$, and let µ be a finite positive Borel measure on $[0, 1)$. If $\mathcal {I}_{\mu_{\alpha+1}}: \mathcal {B}\rightarrow A^{p}$ is bounded (equivalent to compact) for some $\alpha \gt -1$, then $\mathcal {I}_{\mu_{\alpha^{'}+1}}$ is bounded (equivalent to compact) from $ \mathcal {B}$ to A ^p for any $-1 \lt \alpha^{'} \lt \alpha$.

Carleson measures play a key role when we study the generalized Hilbert operators. Recall that if µ is a positive Borel measure on $[0, 1)$, $0\leq\gamma \lt \infty$ and $0 \lt s \lt \infty$, then µ is a γ-logarithmic s-Carleson measure if there exists a positive constant C such that

\begin{equation*}\mu([t,1))\log^{\gamma}\frac{e}{1-t} \leq C (1-t)^{s}, \ \ \mbox{for all}\ 0\leq t \lt 1 .\end{equation*}

In particular, µ is an s-Carleson measure if γ = 0.

In the following, we provide a sufficient condition and a necessary condition by using the Carleson-type measure.

Corollary 2.10. Suppose $1 \lt p \lt \infty$, $\alpha+1-\frac{2}{p} \gt 0$, $\gamma \gt 1+\frac{1}{p}$ and µ is a finite positive Borel measure on $[0, 1)$, then the following statements hold:

1. (1) If µ is a γ-logarithmic $\alpha+1-\frac{2}{p}$-Carleson measure, then $\mathcal {I}_{\mu_{\alpha+1}}: \mathcal {B}\rightarrow A^{p}$ is bounded (equivalent to compact).

2. (2) If $\mathcal {I}_{\mu_{\alpha+1}}: \mathcal {B}\rightarrow A^{p}$ is bounded, then µ is a 1-logarithmic $\alpha+1-\frac{2}{p}$-Carleson measure.

Proof. (1) Assuming µ is a γ-logarithmic $\alpha+1-\frac{2}{p}$-Carleson measure and then integrating by parts, we have that

\begin{equation*} \begin{split} &\ \ \ \ \ \ \int_{0}^{1} t^{n}\log\frac{2}{1-t}\text{d}\mu(t)\\ &=n \int_{0}^{1} t^{n-1}\mu([t,1))\log\frac{2}{1-t}\text{d}t+\int_{0}^{1} t^{n}\mu([t,1))\frac{\text{d}t}{1-t}\\ & \lesssim n \int_{0}^{1} t^{n-1}(1-t)^{\alpha+1-\frac{2}{p}}\log^{1-\gamma}\frac{e}{1-t}dt+ \int_{0}^{1} t^{n}(1-t)^{\alpha-\frac{2}{p}}\log^{-\gamma}\frac{e}{1-t}\text{d}t. \end{split} \end{equation*}

It is easy to estimate that

\begin{equation*} n \int_{0}^{1} t^{n-1}(1-t)^{\alpha+1-\frac{2}{p}}\log^{1-\gamma}\frac{e}{1-t}\text{d}t \asymp \frac{\log^{1-\gamma}(n+1)}{n^{\alpha+1-\frac{2}{p}}} \end{equation*}

and

\begin{equation*}\int_{0}^{1} t^{n}(1-t)^{\alpha-\frac{2}{p}}\log^{-\gamma}\frac{e}{1-t}\text{d}t\asymp \frac{\log^{-\gamma}(n+1)}{n^{\alpha+1-\frac{2}{p}}}.\end{equation*}

These inequalities imply that

\begin{equation*} \nu_{n}= \int_{0}^{1} t^{n}\log\frac{2}{1-t}\text{d}\mu(t)\lesssim \frac{\log^{1-\gamma}(n+1)}{n^{\alpha+1-\frac{2}{p}}}. \end{equation*}

It follows that

\begin{equation*} \begin{split} \sum_{n=0}^{\infty}(n+1)^{p(\alpha+1)-3}\nu_{n}^{p} \lesssim \sum_{n=0}^{\infty}\frac{(\log(n+1))^{p(1-\gamma)}}{n+1}\lesssim 1. \end{split} \end{equation*}

Theorem 1.1 shows that $\mathcal {I}_{\mu_{\alpha+1}}: \mathcal {B}\rightarrow A^{p}$ is bounded (equivalent to compact).

(2) If $\mathcal {I}_{\mu_{\alpha+1}}: \mathcal {B}\rightarrow A^{p}$ is bounded, then for any $N\geq 2$, we have

\begin{equation*} \begin{split} 1 & \gt rsim \sum_{n=0}^{\infty}(n+1)^{p(\alpha+1)-3}\nu_{n}^{p}\\ & \gt rsim \sum_{n=0}^{N}(n+1)^{p(\alpha+1)-3}\nu_{n}^{p}\\ & \gt rsim \nu^{p}_{N} \sum_{n=0}^{N}(n+1)^{p(\alpha+1)-3}\\ & \asymp \nu^{p}_{N}N^{p(\alpha+1)-2}. \end{split} \end{equation*}

This implies that ν is an $\alpha+1-\frac{2}{p}$-Carleson measure which is equivalent to saying that µ is a 1-logarithmic $\alpha+1-\frac{2}{p}$-Carleson measure.

In fact, the exponent $1+\frac{1}{p}$ in Corollary 2.10 is sharp.

Proposition 2.11. Let $1 \lt p \lt \infty$, $\frac{1}{p}+\frac{1}{q}=1$ and $\alpha+1-\frac{2}{p} \gt 0$. Then there exist $1+\frac{1}{p}$-logarithmic $\alpha+1-\frac{2}{p}$-Carleson measure µ and $ g\in A^{q}$ such that

\begin{equation*}\left|\int_{0}^{1} R^{0,\alpha-1}g(t)d\nu(t)\right|=\infty.\end{equation*}

Proof. Let $\text{d}\mu(s)=(1-s)^{\alpha-\frac{2}{p}}\log^{-(1+\frac{1}{p})}\frac{e}{1-s}\text{d}s$. Since $\alpha-\frac{2}{p} \gt -1$, we have that

\begin{equation*}\lim_{t\rightarrow 1}\frac{\int_{t}^{1}(1-s)^{\alpha-\frac{2}{p}}\log^{-(1+\frac{1}{p})}\frac{e}{1-s}\text{d}s}{(1-t)^{\alpha+1-\frac{2}{p}}\log^{-(1+\frac{1}{p})}\frac{e}{1-t}}=\frac{p}{p(\alpha+1)-2} \gt 0.\end{equation*}

This implies that $\mu([t,1))\lesssim (1-t)^{\alpha+1-\frac{2}{p}}\log^{-(1+\frac{1}{p})}\frac{e}{1-t}$ for all $0 \lt t \lt 1$, and hence µ is a $1+\frac{1}{p}$-logarithmic $\alpha+1-\frac{2}{p}$-Carleson measure. Choosing a positive real number δ satisfies $\frac{1}{q} \lt \delta \lt 1$. Let

\begin{equation*}g(z)=\frac{1}{(1-z)^{\frac{2}{q}}}\log^{-\frac{1}{q}}\frac{e}{1-z}(\log\log\frac{e^{2}}{1-z})^{-\delta}, \ \ z\in \mathbb{D}.\end{equation*}

Arguing as the proof of Lemma 1 in [1], we have that

\begin{equation*} \begin{split} M^{q}_{q}(r,g)&=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{1}{|1-re^{i\theta}|^{2}}|\log^{-1}\frac{e}{1-re^{i\theta}}||\log\log\frac{e^{2}}{1-re^{i\theta}}|^{-q\delta}\text{d}\theta\\ & \lesssim \frac{1}{1-r}\log^{-1}\frac{e}{1-r}\left(\log\log\frac{e^{2}}{1-r}\right)^{-q\delta}. \end{split} \end{equation*}

Therefore,

\begin{equation*} \begin{split} ||g||_{A^{q}}^{q} &= 2\int_{0}^{1} M^{q}_{q}(r,g)r\text{d}r\\ & \lesssim \int_{0}^{1} \frac{\text{d}r}{(1-r)\log\frac{e}{1-r}\left(\log\log\frac{e^{2}}{1-r}\right)^{q\delta}}\\ & \lesssim 1. \end{split} \end{equation*}

This shows that $g\in A^{q}$.

We now need the fact that the Taylor coefficients $\widehat{h}(n)$ of the function

\begin{equation*}h(z)=\frac{1}{(1-z)^{a}}\log^{b}\frac{e}{1-z}\left(\log\log\frac{e^{2}}{1-z}\right)^{c}, \ \ a \gt 0, b,c \in \mathbb{R},\end{equation*}

have the property $\widehat{h}(n)\asymp (n+1)^{a-1}\log^{b}(n+2)\left(\log\log(n+4)\right)^{c}$. The proof of this result is similar to the proof of Theorem 2.31 on pages 192–194 in [26], so we omit its proof.

It follows that

\begin{equation*} \begin{split} R^{0,\alpha-1}g(t)&=\sum_{n=0}^{\infty} \frac{\Gamma(n+1+\alpha)}{\Gamma(\alpha+1)\Gamma(n+2)}\widehat{g}(n)t^{n}\\ & \asymp \sum_{n=0}^{\infty} (n+1)^{\alpha+\frac{2}{q}-2}\log^{-\frac{1}{q}}(n+2)\left(\log\log(n+4)\right)^{-\delta}t^{n}\\ & \asymp \frac{1}{(1-t)^{\alpha+1-\frac{2}{p}}}\log^{-\frac{1}{q}}\frac{e}{1-t}\left(\log\log\frac{e^{2}}{1-t}\right)^{-\delta}. \end{split} \end{equation*}

Note that $\text{d}\nu(t)=\log\frac{e}{1-t}\text{d}\mu(t)=(1-t)^{\alpha-\frac{2}{p}}\log^{-\frac{1}{p}}\frac{e}{1-t}\text{d}t$. Hence,

\begin{equation*} \left|\int_{0}^{1} R^{0,\alpha-1}g(t)\text{d}d\nu(t)\right|\asymp \int_{0}^{1} \frac{1}{(1-t)\log\frac{e}{1-t}\left(\log\log\frac{e^{2}}{1-t}\right)^{\delta}}\text{d}t =\infty. \end{equation*}

The proof is complete.

The following Proposition gives a sufficient condition for the operator $\mathcal {H}^{\alpha}_{\mu}$ to be well defined on $\mathcal {B}$ and $\mathcal {H}^{\alpha}_{\mu}(f)=\mathcal {I}_{\mu_{\alpha+1}}(f)$.

Proposition 2.12. Let µ be a positive Borel measure on $[0, 1)$. For any s > 0, if µ is an s-Carleson measure, then the operator $\mathcal {H}^{\alpha}_{\mu}$ is well defined on $\mathcal {B}$ and $\mathcal {H}^{\alpha}_{\mu}(f)=\mathcal {I}_{\mu_{\alpha+1}}(f)$ for all $f\in \mathcal {B}$.

Proof. If s > 0 and µ is an s-Carleson measure, then it is easy to check that $\int_{0}^{1} \log\frac{e}{1-t}\text{d}\mu(t) \lt \infty$, and hence the operator $\mathcal {I}_{\mu_{\alpha+1}}$ is well defined on Bloch space.

Now, let $f(z)=\sum_{n=0}^{\infty} a_{n}z^{n}\in \mathcal {B}$ and $f_{N}(z)=\sum_{n=0}^{N}a_{n}z^{n}$, then we have that

\begin{equation*} \begin{split} \mathcal{H}^{\alpha}_{\mu}(f_{N})(z)&= \sum_{n=0}^{\infty}\frac{\Gamma(n+1+\alpha)}{\Gamma(n+1)\Gamma(\alpha+1)}\left(\sum_{k=0}^{N}\mu_{n,k}a_{k}\right)z^{n}\\ & =\sum_{n=0}^{\infty}\frac{\Gamma(n+1+\alpha)}{\Gamma(n+1)\Gamma(\alpha+1)}\sum_{k=0}^{N}\int_{0}^{1} t^{n+k}\text{d}\mu(t)a_{k}z^{n}\\ &=\sum_{n=0}^{\infty}\frac{\Gamma(n+1+\alpha)}{\Gamma(n+1)\Gamma(\alpha+1)}\int_{0}^{1} f_{N}(t) (tz)^{n}\text{d}\mu(t)\\ &=\mathcal {I}_{\mu_{\alpha+1}}(f_{N})(z), \end{split} \end{equation*}

which implies that $\mathcal{H}^{\alpha}_{\mu}$ is well defined on polynomials. Take $p \gt \frac{2}{s}$ and then integrating by parts and using the fact that µ is an s-Carleson measure, we have that

\begin{equation*} \begin{split} \int_{0}^{1} \frac{1}{(1-t)^{\frac{2}{p}}}d\mu(t)&= \mu([0,1))-\lim_{t\rightarrow 1}\frac{\mu([t,1))}{(1-t)^{\frac{2}{p}}}+\frac{2}{p}\int_{0}^{1} \frac{\mu([t,1))}{(1-t)^{\frac{2}{p}+1}}\text{d}t\\ & \lesssim \mu([0,1))+\int_{0}^{1} (1-t)^{s-\frac{2}{p}-1}\text{d}t\\ & \lesssim \mu([0,1))+ 1 \lesssim1. \end{split} \end{equation*}

Since $\mathcal {B}\subsetneq A^{p}$, we have that

\begin{equation*} \begin{split} \int_{0}^{1} |f(t)-f_{N}(t)|\text{d}\mu(t)&\lesssim ||f-f_{N}||_{A^{p}} \int_{0}^{1} \frac{1}{(1-t)^{\frac{2}{p}}}\text{d}\mu(t)\\ & \lesssim ||f-f_{N}||_{A^{p}} \lesssim ||f-f_{N}||_{\mathcal {B}}. \end{split} \end{equation*}

This implies that

\begin{equation*} \begin{split} &\ \ \ \ \left|\mathcal {I}_{\mu_{\alpha+1}}(f)(z)-\sum_{n=0}^{\infty} \frac{\Gamma(n+1+\alpha)}{\Gamma(n+1)\Gamma(\alpha+1)}\left(\sum_{k=0}^{N}\mu_{n,k}a_{k}\right)z^{n}\right|\\ & \lesssim \int_{0}^{1} \frac{|f(t)-f_{N}(t)|}{(1-|z|)^{\alpha+1}}\text{d}\mu(t) \lesssim \frac{1}{(1-|z|)^{\alpha+1}} ||f-f_{N}||_{\mathcal {B}}. \end{split} \end{equation*}

Thus, as $N\rightarrow \infty$, the series

\begin{equation*}\sum_{n=0}^{\infty}\frac{\Gamma(n+1+\alpha)}{\Gamma(n+1)\Gamma(\alpha+1)}\left(\sum_{k=0}^{N}\mu_{n,k}a_{k}\right)z^{n}\end{equation*}

converges and defines an analytic function

\begin{equation*}\mathcal {H}^{\alpha}_{\mu}(f)=\mathcal {I}_{\mu_{\alpha+1}}(f)=\int_{0}^{1}\frac{f(t)}{(1-tz)^{1+\alpha}}\text{d}\mu(t).\end{equation*}

The proof is complete.

Corollary 2.13. Suppose $1 \lt p \lt \infty$, $\frac{1}{p}+\frac{1}{q}=1$, $\gamma \gt 1+\frac{1}{p}$ and $\alpha+1-\frac{2}{p} \gt 0$, and let µ be a positive Borel measure on $[0, 1) $ which is a γ-logarithmic $\alpha+\frac{2}{q}-1$-Carleson measure. Then $\mathcal {H}_{\mu}^{\alpha}$ is a compact operator from $ \mathcal {B}$ into A ^p.

The Dirichlet type space $D^{p}_{p-1} $ consists of those functions $f\in H(\mathbb{D})$ such that

\begin{equation*}||f||^{p}_{B^{p}}= |f(0)|^{p}+\int_{\mathbb{D}}|f'(z)|^{p}(1-|z|^{2})^{p-1}{\text{d}}A(z) \lt \infty.\end{equation*}

It is known that $D^{p}_{p-1}\subseteq H^{p}$ for $1\leq p\leq2$. Now, we characterize the measures µ for which $\mathcal {I}_{\mu_{\alpha+1}}$ is a bounded (compact) operator from the Bloch space to Hardy spaces.

Theorem 2.14 Let $1\leq p \lt \infty$, and let µ be a positive Borel measure on $[0, 1)$ with $\int_{0}^{1} \log\frac{e}{1-t}d\mu(t) \lt \infty$. Then, the following statements are equivalent:

1. (1) The operator $\mathcal {I}_{\mu_{\alpha+1}}$ is bounded from $\mathcal {B}$ to H ^p.

2. (2) The operator $\mathcal {I}_{\mu_{\alpha+1}}$ is compact from $\mathcal {B}$ to H ^p.

3. (3) The measure ν satisfies $\sum_{n=0}^{\infty}(n+1)^{p(\alpha+1)-2}\nu_{n}^{p} \lt \infty.$

4. (4) The measure ν satisfies

   \begin{equation*}\left|\int_{0}^{1} R^{-1,\alpha}g(t)d\nu(t)\right|\lesssim ||g||_{Y_{p}}, \ \ \mbox{for all} \ g\in Y_{p},\end{equation*}

   where $ Y_{p}=H^{\frac{p}{p-1}}$ for p > 1 and $Y_{1}= VMOA$ for p = 1.

Proof. The proof is similar to the proof of Theorem 1.1, and we omit some specifics. Using the well-known duality $(H^{\frac{p}{p-1}})^{*}=H^{p}$ $ (1 \lt p \lt \infty)$ and $(\text{VMOA})^{*}=H^{1}$ under the pairing

\begin{equation*}\langle F,G\rangle_{H^{2}}=\int_{0}^{2\pi}F(e^{i\theta})\overline{G(e^{i\theta})}\text{d}\theta,\end{equation*}

then arguing as the proof of Proposition 2.5, one can obtain the implications $(1)\Rightarrow (4)$ and $(4)\Rightarrow (3)$. The proof of $(3)\Rightarrow (2)$ is split into $1\leq p \leq 2$ and p > 2. For $1\leq p\leq 2$, bearing in mind that $D_{p-1}^{p}\subseteq H^{p}$ and then adapting the proof of the implication $(3)\Rightarrow(2)$ in Theorem 1.1, one can deduce that $\mathcal {I}_{\mu_{\alpha+1}}$ is compact from $\mathcal {B}$ to $D_{p-1}^{p}$ and hence compact from $\mathcal {B}$ to H ^p. For $2 \lt p \lt \infty$, the Hardy–Littlewood Theorem (see [8, Theorem 6.3]) asserts that if $f(z)=\sum_{n=0}^{\infty}\widehat{f}(n)z^{n}\in H(\mathbb{D})$ and $\sum_{n=0}^{\infty} (n+1)^{p-2} |\widehat{f}(n)|^{p} \lt \infty$, then

(4)\begin{equation} ||f||^{p}_{p}\lesssim \sum_{n=0}^{\infty} (n+1)^{p-2} |\widehat{f}(n)|^{p}\end{equation}

Suppose that $\sum_{n=0}^{\infty}(n+1)^{p(\alpha+1)-2}\nu_{n}^{p} \lt \infty$, then for any ɛ > 0, there exists a positive integer N such that

(5)\begin{equation} \sum_{n=N+1}^{\infty}(n+1)^{p(\alpha+1)-2}\nu_{n}^{p} \lt \varepsilon.\end{equation}

Since $\int_{0}^{1} \log\frac{e}{1-t}d\mu(t) \lt \infty$, there exists t ₀ with $0 \lt t_{0} \lt 1$ such that

(6)\begin{equation} \int_{t_{0}}^{1}\log\frac{e}{1-t}\text{d}\mu(t) \lt \varepsilon.\end{equation}

Let $\{f_{k}\}_{k=1}^{\infty}$ be a bounded sequence in $\mathcal {B}$, which converges to 0 uniformly on every compact subset of $\mathbb{D}$. We have to prove that $\lim_{k\rightarrow0}||\mathcal {I}_{\mu_{\alpha+1}}(f_{k})||_{p}^{p}\rightarrow 0$. By (4) and Stirling’s formula, (5)–(6) we have that

\begin{equation*} \begin{split} ||\mathcal {I}_{\mu_{\alpha+1}}(f_{k})||_{p}^{p} &\lesssim \sum_{n=0}^{\infty} (n+1)^{p-2}\left|\frac{\Gamma(n+1+\alpha)}{\Gamma(\alpha+1)\Gamma(n+1)}\int_{0}^{1} t^{n}f_{k}(t){\text{d}}\mu(t)\right|^{p}\\ &\lesssim (\sup_{|w|\leq t_{0}}|f_{k}(w)|)^{p}\sum_{n=0}^{N}(n+1)^{p\alpha+p-2}\left (\int_{0}^{t_{0}}t^{n}{\text{d}}\mu(t)\right)^{p} \\ &+ \sum_{n=0}^{N}(n+1)^{p\alpha+p-2}\left(\int_{t_{0}}^{1}\log\frac{e}{1-t}{\text{d}}\mu(t)\right)^{p}\\ &+ \sum_{n=N+1}^{\infty}(n+1)^{p\alpha+p-2}\left(\int_{0}^{1} t^{n}|f_{k}(t)|{\text{d}}\mu(t)\right)^{p}\\ & \lesssim (\sup_{|w|\leq t_{0}}|f_{k}(w)|)^{p}+ \varepsilon^{p} +\sum_{n=N+1}^{\infty}(n+1)^{p\alpha+p-2}\nu_{n}^{p}\\ & \lesssim (\sup_{|w|\leq t_{0}}|f_{k}(w)|)^{p} + \varepsilon^{p}+\varepsilon.\\ \end{split} \end{equation*}

This implies that $\lim_{k\rightarrow0}||\mathcal {I}_{\mu_{\alpha+1}}(f_{k})||_{p}^{p}= 0$.

The analytic Besov space B ^p $(1 \lt p \lt \infty)$ consists of those functions $f\in H(\mathbb{D})$ such that

\begin{equation*}||f||^{p}_{B^{p}}= |f(0)|^{p}+\int_{\mathbb{D}}|f'(z)|^{p}(1-|z|^{2})^{p-2}{\text{d}}A(z) \lt \infty.\end{equation*}

Lemma 2.1 implies that

(7)\begin{equation} f\in B^{p} \Leftrightarrow \sum_{n=0}^{\infty} 2^{-n(p-1)}||\Delta_{n}f'||^{p}_{p} \lt \infty.\end{equation}

The space B ₁ consists of those functions $f\in H(\mathbb{D})$ such that

\begin{equation*}\int_{\mathbb{D}}|f''(z)|{\text{d}}A(z) \lt \infty.\end{equation*}

Based on (7), we can obtain the analogue of Theorem 1.1 for Besove spaces B ^p $(1\leq p \lt \infty)$.

Theorem 2.15 Let $1\leq p \lt \infty$ and µ be a positive Borel measure on $[0, 1)$ with $\int_{0}^{1} \log\frac{e}{1-t}d\mu(t) \lt \infty$. Then, the following statements are equivalent:

1. (1) The operator $\mathcal {I}_{\mu_{\alpha+1}}$ is bounded from $\mathcal {B}$ to B ^p.

2. (2) The operator $\mathcal {I}_{\mu_{\alpha+1}}$ is compact from $\mathcal {B}$ to B ^p.

3. (3) The measure µ satisfies $\sum_{n=0}^{\infty}(n+1)^{p(\alpha+1)-1}\nu_{n}^{p} \lt \infty.$

The proof of this theorem is analogous to Theorem 1.1, and we left it to the interested readers.

Remark 2.16. Let µ be a finite positive Borel measure on $[0, 1)$ and $f\in H(\mathbb{D})$, and Galanopoulos, Girela and Merchán [9] introduced a Cesàro-like operator $\mathcal {C}_{\mu}$ which has been widely studied recently. The operator $\mathcal {C}_{\mu}$ is defined by

\begin{equation*}\mathcal {C}_{\mu}(f)(z)=\sum^\infty_{n=0}\left(\mu_n\sum^n_{k=0}a_k\right)z^n=\int_{0}^{1}\frac{f(tz)}{1-tz}{\text{d}}\mu(t), \ z\in\mathbb{D}.\end{equation*}

We remarked that the proof of Theorem 1.1 can be used to study the action of $\mathcal{C}_\mu$ from the Bloch space to Bergman spaces, Hardy spaces and Besov spaces. We leave it to the interested readers.