Cesàro-like operators acting on a class of analytic function spaces

Abstract

Let $\mu$ be a finite positive Borel measure on [0, 1) and let $H(\mathbb {D})$ be the space of all analytic function in the unit disc $\mathbb {D}$. The Cesàro-like operator $\mathcal {C}_\mu$ is defined in $H(\mathbb {D})$ as follows: If $f \in H(\mathbb {D})$, $f(z)=\sum _{n=0}^{\infty }a_{n}z^{n} (z\in \mathbb {D})$, then

$$\begin{aligned} \mathcal {C}_\mu (f)(z)=\sum ^\infty _{n=0}\left( \mu _n\sum ^n_{k=0}a_k\right) z^n,\quad z\in \mathbb {D}, \end{aligned}$$

where for $n\ge 0$, $\mu _n$ denotes the n-th moment of the measure $\mu$, that is, $\mu _n=\int _{0}^{1} t^{n}d\mu (t)$. For $s > 1$, let X be a Banach subspace of $H(\mathbb {D})$ lying between the mean Lipschtz space $\Lambda ^{s}_{\frac{1}{s}}$ and the Bloch space $\mathcal {B}$. In this paper we characterize the measures $\mu$ as above for which $\mathcal {C}_\mu$ is bounded (compact) from X into any of the Hardy spaces $H^{p} ~ (1\le p\le \infty )$.

1 Introduction

Let $\mathbb {D}=\{z\in \mathbb {C}:\vert z\vert <1\}$ 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

$$\begin{aligned} \vert \vert f\vert \vert _{\mathcal {B}}=\vert f(0)\vert +\sup _{z\in \mathbb {D}}(1-\vert z\vert ^{2})\vert f'(z)\vert <\infty . \end{aligned}$$

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

$$\begin{aligned} \Vert f\Vert _{p}=\sup _{0\le r<1} M_p(r, f)<\infty , \end{aligned}$$

where

$$\begin{aligned} M_p(r, f)= & {} \left( \frac{1}{2\pi }\int _0^{2\pi }|f(re^{i\theta })|^p d\theta \right) ^{1/p}, \quad 0<p<\infty ,\\ M_{\infty }(r, f)= & {} \sup _{|z|=r}|f(z)|. \end{aligned}$$

For $0<p<\infty$, the Dirichlet-type space $D^{p}_{p-1}$ is the space of $h\in H(\mathbb {D})$ such that

$$\begin{aligned} \Vert h\Vert _{D^{p}_{p-1}}^{p}=|h(0)|^{p}+ \int _{\mathbb {D}}|h'(z)|^{p}(1-|z|)^{p-1}dA(z)<\infty . \end{aligned}$$

When $p=2$, the space $D^{2}_{1}$ is just the Hardy space $H^{2}$.

The Hardy–Littlewood space HL(p) consists of those function $f\in H(\mathbb {D})$ for which

$$\begin{aligned} \Vert h\Vert ^{p}_{HL(p)}=\sum _{n=0}^{\infty }(n+1)^{p-2}|\widehat{h}(n)|^{p}<\infty . \end{aligned}$$

The space $HL(\infty )$ consist of $f(z)=\sum _{n=0}^{\infty }a_{n}z^{n}\in H(\mathbb {D})$ such that

$$\begin{aligned} \Vert f\Vert _{HL(\infty )}=\sup _{n\ge 0}(n+1)|a_{n}|<\infty . \end{aligned}$$

We also need the space $H(\infty ,p)=\{h\in H(\mathbb {D}): \Vert h\Vert ^{p}_{H(\infty ,p)}=\int _{0}^{1}M_{\infty }^{p}(r,h)dr<\infty \}$.

It is well known that

$$\begin{aligned} D^{p}_{p-1}\subset H^{p}\subset HL(p),\quad 0<p\le 2, \end{aligned}$$

(1)

$$\begin{aligned} HL(p) \subset H^{p}\subset D^{p}_{p-1},\quad 2\le p<\infty , \end{aligned}$$

(2)

and

$$\begin{aligned} H^{p}\subset H(\infty ,p),\quad D^{p}_{p-1}\subset H(\infty ,p),\quad 0< p<\infty . \end{aligned}$$

(3)

The proofs of (1) and (2) can be found in [10, 11, 18], and the proof of (3) appears in [21, p. 127] and [14, Lemma 4].

Let $1\le p<\infty$ and $0<\alpha \le 1$, the mean Lipschitz space $\Lambda ^p_\alpha$ consists of those functions $f\in H(\mathbb {D})$ having a non-tangential limit almost everywhere such that $\omega _p(t, f)=O(t^\alpha )$ as $t\rightarrow 0$. Here $\omega _p(\cdot , f)$ is the integral modulus of continuity of order p of the function $f(e^{i\theta })$. It is known (see [10]) that $\Lambda ^p_\alpha$ is a subset of $H^p$ and

$$\begin{aligned} \Lambda ^p_\alpha =\left\{ f\in H(\mathbb {D}):M_p(r, f')=O\left( \frac{1}{(1-r)^{1-\alpha }}\right) , \quad \text{ as }\quad r\rightarrow 1\right\} . \end{aligned}$$

The space $\Lambda ^p_\alpha$ is a Banach space with the norm $\Vert \cdot \Vert _{\Lambda ^p_\alpha }$ given by

$$\begin{aligned} \Vert f\Vert _{\Lambda ^p_\alpha }=|f(0)|+\sup _{0\le r<1}(1-r)^{1-\alpha }M_p(r, f'). \end{aligned}$$

It is known (see e.g. [5, Theorem 2.5]) that

$$\begin{aligned} \Lambda ^{p}_{\frac{1}{p}}\subsetneq \mathcal {B}, \ \ 1<p<\infty . \end{aligned}$$

The Cesàro operator $\mathcal {C}$ is defined in $H(\mathbb {D})$ as follows:

If $f\in H(\mathbb {D})$, $f(z)=\sum _{n=0}^\infty a_nz^n$, then

$$\begin{aligned} \mathcal {C}(f)(z)=\sum _{n=0}^\infty \left( \frac{1}{n+1}\sum _{k=0}^n a_k\right) z^n, \ z\in \mathbb {D}. \end{aligned}$$

The boundedness and compactness of the Cesàro operator $\mathcal {C}$ and its generalizations on various spaces of analytic functions such as Hardy spaces, Bergman spaces, Dirichlet spaces, Bloch space, $Q_{p}$ space, mixed norm space have been widely studied. See, e.g., [1, 2, 6, 9, 19, 23,24,25,26] and the references therein.

Recently, Galanopoulos et al. [12] introduced a Cesàro-like operator $\mathcal {C}_\mu$ on $H(\mathbb {D})$, which is a natural generalization of the classical Cesàro operator $\mathcal {C}$. For a finite positive Borel measure $\mu$ on the interval [0, 1), the Cesàro-like operator $\mathcal {C}_\mu$ is defined on $H(\mathbb {D})$ as follows:

$$\begin{aligned} \mathcal {C}_\mu (f)(z)=\sum ^\infty _{n=0}\left( \mu _n\sum ^n_{k=0}\widehat{f}(k)\right) z^n=\int _{0}^{1}\frac{f(tz)}{(1-tz)}d\mu (t), \quad z\in \mathbb {D}, \end{aligned}$$

where $\mu _{n}$ stands for the moment of order n of $\mu$, that is, $\mu _{n}=\int _{0}^{1}t^{n}d\mu (t)$. They studied the operators $\mathcal {C}_\mu$ acting on distinct spaces of analytic functions (e.g. Hardy space, Bergman space, Bloch space, etc.).

The Cesàro-like operator $\mathcal {C}_\mu$ defined above has attracted the interest of many mathematicians. For instance, Jin and Tang [16] studied the boundedness (compactness) of $\mathcal {C}_\mu$ from one Dirichlet-type space $\mathcal {D}_{\alpha }$ into another one $\mathcal {D}_{\beta }$. Bao, Sun and Wulan [3] studied the range of $\mathcal {C}_\mu$ acting on $H^{\infty }$. Blasco [4] investigated the operators $\mathcal {C}_\mu$ induced by complex Borel measures on [0, 1). Galanopoulos et al. [13] studied the behaviour of the operators $\mathcal {C}_\mu$ on the Dirichlet space and on the analytic Besov spaces. The operators $\mathcal {C}_\mu$ associated to arbitrary complex Borel measures on $\mathbb {D}$ the reader is referred to Galanopoulos et al. [15] and Zhou [27].

The following result was proved by Blasco [4, Theorem 3.17].

Theorem A

Let $\mu$ be a finite positive Borel measure on [0, 1). Then $\mathcal {C}_\mu : \mathcal {B}\rightarrow H^{2}$ is bounded if and only if

$$\begin{aligned} \sum _{n=0}^{\infty }\mu _{n}^{2}\log ^{2}(n+1)<\infty . \end{aligned}$$

It is natural to seek conditions on the measure $\mu$ which makes the operator $\mathcal {C}_\mu$ bounded (compact) from $\mathcal {B}$ to $H^{p}$ when $p\ne 2$. The result for $p\ne 2$ has not been given in the existing literature. In this paper, we extend the Bloch space $\mathcal {B}$ to a class of spaces X and consider the target space $H^{p}$ with any $1\le p\le \infty$, where X is a Banach subspace of $H(\mathbb {D})$ with $\Lambda ^{s}_{\frac{1}{s}}\subset X\subset \mathcal {B}$ ($1<s<\infty )$. We complete characterize the measure $\mu$ for which $\mathcal {C}_\mu$ is bounded (compact) from X to $H^{p}$ $(1\le p\le \infty )$. It turns out that the boundedness and compactness from X to $H^{p}$ are equivalent. This conclusion appears to be new even for $p = 2$. At the end of the article, we also consider the case $0<p<1$.

Our main results are included in the following.

Theorem 1.1

Suppose $1\le p<\infty$, $1<s<\infty$, and let $\mu$ be a finite positive Borel measure on [0, 1). Let $Y_{p}\in \{D^{p}_{p-1}, H^{p}, HL(p), H(\infty ,p)\}$ and let X be a Banach subspace of $H(\mathbb {D})$ with $\Lambda ^{s}_{\frac{1}{s}}\subset X\subset \mathcal {B}$. Then the following statements are equivalent.

• (1) The operator $\mathcal {C}_\mu$ is bounded from X to $Y_{p}$.

• (2) The operator $\mathcal {C}_\mu$ is compact from X to $Y_{p}$.

• (3) The measure $\mu$ satisfies

$$\begin{aligned} \sum _{n=0}^{\infty }(n+1)^{p-2}\mu _{n}^{p}\log ^{p}(n+1)<\infty . \end{aligned}$$

Theorem 1.2

Let $1<s<\infty$ and $\mu$ be a finite positive Borel measure on [0, 1). Let X be a Banach subspace of $H(\mathbb {D})$ with $\Lambda ^{s}_{\frac{1}{s}}\subset X\subset \mathcal {B}$. Then the operator $\mathcal {C}_\mu$ is bounded from X in $H^{\infty }$ if and only if

$$\begin{aligned} \int _{0}^{1}\frac{\log \frac{e}{1-t}}{1-t}d\mu (t)<\infty . \end{aligned}$$

Throughout the paper, the measure $\mu$ will be a positive finite Borel measure on the radius [0, 1), the letter C will denote an absolute constant whose value depends on the parameters indicated in the parenthesis, 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 $`` P \le CQ"$, and $`` P \gtrsim Q"$ is understood in an analogous manner. In particular, if $``P\lesssim Q"$ and $``P \gtrsim Q"$, then we will write $``P\asymp Q"$.

2 Proofs and some related results

We begin with some preliminary results which will be repeatedly used throughout the rest of the paper. The first lemma contains a characterization of $L^{p}$-integrability of power series with nonnegative coefficients. For a proof, see [20, Theorem 1].

Lemma 2.1

Let $0<\beta ,p<\infty$, $\{\lambda _{n}\}_{n=0}^{\infty }$ be a sequence of non-negative numbers. Then

$$\begin{aligned} \int _{0}^{1}(1-r)^{p\beta -1}\left( \sum _{n=0}^{\infty }\lambda _{n}r^{n}\right) ^{p}dr\asymp \sum _{n=0}^{\infty }2^{-np\beta }\left( \sum _{k\in I_{n}}\lambda _{k}\right) ^{p} \end{aligned}$$

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

The following lemma is a consequence of Theorem 2.31 on page 192 of the classical monograph [28].

Lemma 2.2

• (a) The Taylor coefficients $a_{n}$ of the function

  $$\begin{aligned} f(z)=\frac{1}{(1-z)^{\beta }}\log ^{\gamma }\frac{2}{1-z}, \quad \beta >0,\gamma \in \mathbb {R}, \quad z\in \mathbb {D} \end{aligned}$$

  have the property $a_{n}\asymp n^{\beta -1}(\log (n+1))^{\gamma }$.

• (b) The Taylor coefficients $a_{n}$ of the function

  $$\begin{aligned} f(z)=\log ^{\gamma }\frac{2}{1-z}, \quad \gamma >0,\quad z\in \mathbb {D} \end{aligned}$$

  have the property $a_{n}\asymp n^{-1}(\log (n+1))^{\gamma -1}$.

We also need the following estimates (see, e.g. Proposition 1.4.10 in [22]).

Lemma 2.3

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

$$\begin{aligned} \int ^{2\pi }_0\frac{d\theta }{|1-ze^{-i\theta }|^{\alpha }}\asymp {\left\{ \begin{array}{ll}1 &{} \hspace{5.0pt}\textrm{if} \quad \alpha <1,\\ \log \frac{2}{1-|z|^2} &{} \hspace{5.0pt}\textrm{if} \quad \alpha =1,\\ \frac{1}{(1-|z|^2)^{\alpha -1}} &{} \hspace{5.0pt}\textrm{if}\quad \alpha >1, \end{array}\right. } \end{aligned}$$

Proof of the implication $(1)\Rightarrow (3)$ Assume $\mathcal {C}_\mu$ is bounded from X to $Y_{p}$. Let $f(z)=\log \frac{1}{1-z}$, then it is easy to check that $f\in \Lambda ^{s}_{\frac{1}{s}}\subset X$. This implies that $\mathcal {C}_\mu (f)\in Y_{p}$.

Case $Y_{p}=H(\infty ,p)$. For $0<r<1$, we have

$$\begin{aligned} M_{\infty }(r,\mathcal {C}_\mu (f))=\sup _{|z|=r}|\mathcal {C}_\mu (f)(z)|\ge \sum _{n=1}^{\infty } \mu _{n}\left( \sum _{k=1}^{n}\frac{1}{k}\right) r^{n}\gtrsim \sum _{n=1}^{\infty } \mu _{n}\log (n+1)r^{n}. \end{aligned}$$

Hence, by Lemma 2.1 we have

$$\begin{aligned} \Vert f\Vert ^{p}_{X}{} & {} \gtrsim \Vert \mathcal {C}_\mu (f)\Vert ^{p}_{Y_{p}}=\int _{0}^{1}M^{p}_{\infty }(r,\mathcal {C}_\mu (f))dr\\{} & {} \gtrsim \int _{0}^{1}\left( \sum _{n=1}^{\infty } \mu _{n}\log (n+1)r^{n}\right) ^{p}dr\\{} & {} \asymp \sum _{n=1}^{\infty } 2^{-n}\left( \sum _{k=2^{n-1}}^{2^{n}-1}\mu _{k}\log (k+1)\right) ^{p}\\{} & {} \gtrsim \sum _{n=1}^{\infty }2^{n(p-1)}\log ^{p}(2^{n}+1)\mu ^{p}_{2^{n}}\\{} & {} \gtrsim \sum _{k=1}^{\infty }(k+1)^{p-2}\mu ^{p}_{k}\log ^{p}(k+1). \end{aligned}$$

Case $Y_{p}\ne H(\infty ,p)$. If $p\ge 2$, then (2) and (3) show that $Y_{p}\subset H(\infty ,p)$. This implies that

$$\begin{aligned} \Vert f\Vert ^{p}_{X}\gtrsim \Vert \mathcal {C}_\mu (f)\Vert ^{p}_{Y_{p}}\gtrsim \Vert \mathcal {C}_\mu (f)\Vert ^{p}_{H(\infty ,p)}. \end{aligned}$$

The desired result follows from the previous case.

If $1\le p\le 2$, then (1) shows that $Y_{p}\subset HL(p)$. Hence,

$$\begin{aligned} \Vert f\Vert ^{p}_{X}{} & {} \gtrsim \Vert \mathcal {C}_\mu (f)\Vert ^{p}_{Y_{p}}\gtrsim \Vert \mathcal {C}_\mu (f)\Vert ^{p}_{HL(p)}\\{} & {} = \sum _{n=1}^{\infty }(n+1)^{p-2}\mu _{n}^{p}\left| \sum _{k=1}^{n}\frac{1}{k}\right| ^{p}\\{} & {} \gtrsim \sum _{n=1}^{\infty }(n+1)^{p-2}\mu _{n}^{p}\log ^{p}(n+1). \end{aligned}$$

$\square$

Proof of the implication $(3)\Rightarrow (2)$ Let $\{f_{k}\}_{k=1}^{\infty }$ be a bounded sequence in X which converges to 0 uniformly on every compact subset of $\mathbb {D}$. Without loss of generality, we may assume that $f_{k}(0)=0$ for all $k\ge 1$ and $\sup _{k\ge 1}\Vert f\Vert _{X}\le 1$.

Case $1\le p\le 2$. Assume $\sum _{n=1}^{\infty }(n+1)^{p-2}\mu _{n}^{p}\log ^{p}(n+1)<\infty$, then

$$\begin{aligned} \sum _{n=1}^{\infty }(n+1)^{p-2}\mu _{n}^{p}\log ^{p}(n+1){} & {} = \sum _{n=1}^{\infty }\left( \sum _{k=2^{n-1}}^{2^{n}-1}(k+1)^{p-2}\mu _{k}^{p}\log ^{p}(k+1)\right) \\{} & {} \gtrsim \sum _{n=1}^{\infty }2^{n(p-1)}\mu _{2^{n}}^{p}\log ^{p}(2^{n}+1)\\{} & {} \gtrsim \sum _{n=1}^{\infty }2^{-np}\left( \sum _{k=2^{n}}^{2^{n+1}-1}(k+1)^{1-\frac{1}{p}}\mu _{k}\log (k+1)\right) ^{p}. \end{aligned}$$

It follows that

$$\begin{aligned} \sum _{n=1}^{\infty }2^{-np}\left( \sum _{k=2^{n}}^{2^{n+1}-1}(k+1)^{1-\frac{1}{p}}\mu _{k}\log (k+1)\right) ^{p}<\infty . \end{aligned}$$

By Lemma 2.1 we have that

$$\begin{aligned}{} & {} \int _{0}^{1}(1-r)^{p-1}\left( \sum _{n=0}^{\infty }(n+1)^{1-\frac{1}{p}}\mu _{n}\log (n+1)r^{n}\right) ^{p}dr \\{} & {} \quad \asymp \sum _{n=0}^{\infty }2^{-np}\left( \sum _{k=2^{n}}^{2^{n+1}-1}(k+1)^{1-\frac{1}{p}}\mu _{k}\log (k+1)\right) ^{p}<\infty . \end{aligned}$$

Therefore, for any $\varepsilon >0$ there exists a $0<r_{0}<1$ such that

$$\begin{aligned} \int _{r_{0}}^{1}(1-r)^{p-1}\left( \sum _{n=0}^{\infty }(n+1)^{1-\frac{1}{p}}\mu _{k}\log (k+1)r^{n}\right) ^{p}dr<\varepsilon . \end{aligned}$$

(4)

By (1) and (3), we have that $D_{p-1}^{p}\subset Y_{p}$. Hence, it is suffices to prove that

$$\begin{aligned} \lim _{k\rightarrow \infty }\Vert \mathcal {C}_\mu (f_{k})\Vert _{D_{p-1}^{p}}=0. \end{aligned}$$

It is clear that

$$\begin{aligned} \begin{aligned} \Vert \mathcal {C}_\mu (f_{k})\Vert ^{p}_{D_{p-1}^{p}}&=\int _{|z|\le r_{0}}|\mathcal {C}_\mu (f_{k})'(z)|^{p}(1-|z|)^{p-1}dA(z)\\&\quad +\, \int _{r_{0}<|z|<1}|\mathcal {C}_\mu (f_{k})'(z)|^{p}(1-|z|)^{p-1}dA(z)\\&:= J_{1,k}+J_{2,k}. \end{aligned} \end{aligned}$$

By the integral representation of $\mathcal {C}_\mu$ we get

$$\begin{aligned} \mathcal {C}_\mu (f_{k})'(z)=\int _{0}^{1}\frac{tf'_{k}(tz)}{(1-tz)}d\mu (t) +\int _{0}^{1}\frac{tf_{k}(tz)}{(1-tz)^{2}}d\mu (t). \end{aligned}$$

(5)

Since $\{f_{k}\}_{k=1}^{\infty }$ is converge to 0 uniformly on every compact subset of $\mathbb {D}$, the Cauchy’s integral theorem implies that $\{f'_{k}\}_{k=1}^{\infty }$ converges to 0 uniformly on every compact subset of $\mathbb {D}$. Thus, for $|z|\le r_{0}$, we have that

$$\begin{aligned} \begin{aligned} |\mathcal {C}_\mu (f_{k})'(z)|&\lesssim \int _{0}^{1}\frac{|f_{k}'(tz)|}{|1-tz|}+ \frac{|f_{k}(tz)|}{|1-tz|^{2}}d\mu (t)\\&\lesssim \sup _{|w|<r_{0}}\left( |f_{k}(w)|+|f_{k}'(w)|\right) \int _{0}^{1}\frac{1}{(1-tr_{0})^{2}}d\mu (t)\\&\lesssim \sup _{|w|<r_{0}}\left( |f_{k}(w)|+|f_{k}'(w)|\right) . \end{aligned} \end{aligned}$$

It follows that

$$\begin{aligned} J_{1,k} \rightarrow 0, \quad \text{ as }\quad k\rightarrow \infty . \end{aligned}$$

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

Since $X\subset \mathcal {B}$, for each $k\ge 1$, we have

$$\begin{aligned} |f_{k}(z)|\lesssim \log \frac{e}{1-|z|}\quad \text{ and }\quad |f_{k}'(z)|\lesssim \frac{1}{1-|z|}\quad \text{ for } \text{ all }\quad z\in \mathbb {D}. \end{aligned}$$

(6)

By (5) and (6), Minkowski’s inequity, Lemma 2.3 we have that

$$\begin{aligned} \begin{aligned} M_{p}(r,\mathcal {C}_\mu (f_{k})')&= \left\{ \int _{0}^{2\pi }\left| \int _{0}^{1}\frac{tf_{k}'(tre^{i\theta })}{1-rte^{i\theta }}+\frac{tf_{k}(tre^{i\theta })}{(1-tre^{i\theta })^{2}}d\mu (t)\right| ^{p}d\theta \right\} ^{\frac{1}{p}}\\&\lesssim \left\{ \int _{0}^{2\pi }\left( \int _{0}^{1}\frac{1}{(1-tr)|1 -tre^{i\theta }|}d\mu (t)\right) ^{p}d\theta \right\} ^{\frac{1}{p}}\\&\quad +\, \left\{ \int _{0}^{2\pi }\left( \int _{0}^{1}\frac{\log \frac{e}{1-tr}}{|1-tre^{i\theta }|^{2}}d\mu (t)\right) ^{p}d\theta \right\} ^{\frac{1}{p}}\\&\lesssim \int _{0}^{1}\frac{1}{1-tr}\left( \int _{0}^{2\pi }\frac{d\theta }{|1-tre^{i\theta }|^{p}}\right) ^{\frac{1}{p}}d\mu (t)\\&\quad +\, \int _{0}^{1}\log \frac{e}{1-tr}\left( \int _{0}^{2\pi } \frac{d\theta }{|1-tre^{i\theta }|^{2p}}\right) ^{\frac{1}{p}}d\mu (t)\\&\lesssim \int _{0}^{1}F(t,r)d\mu (t), \end{aligned} \end{aligned}$$

where

$$\begin{aligned} F(t,r)=\left\{ \begin{array}{ll} \displaystyle {\frac{\log \frac{e}{1-tr}}{(1-tr)^{2-\frac{1}{p}}},} &{} \displaystyle {1<p\le 2};\\ \displaystyle {\frac{\log \frac{e}{1-tr}}{1-tr}, } &{} \displaystyle {p=1}. \end{array}\right. \end{aligned}$$

Lemma 2.2 yields that

$$\begin{aligned} M_{p}(r,\mathcal {C}_\mu (f_{k})')\lesssim \int _{0}^{1}F(t,r)d\mu (t) \asymp \sum _{n=0}^{\infty }(n+1)^{1-\frac{1}{p}}\mu _{n}\log (n+1)r^{n}. \end{aligned}$$

This together with (4) imply that

$$\begin{aligned} J_{2,k}&=\int _{r_{0}<|z|<1}|\mathcal {C}_\mu (f_{k})'(z)|^{p}(1-|z|)^{p-1}dA(z)\\&\lesssim \int _{r_{0}}^{1}(1-r)^{p-1}M^{p}_{p}(r,\mathcal {C}_\mu (f_{k})')dr\\&\lesssim \int _{r_{0}}^{1}(1-r)^{p-1}\left( \sum _{n=0}^{\infty }(n+1)^{1-\frac{1}{p}}\mu _{n}\log (n+1)r^{n}\right) ^{p}dr\\&\lesssim \varepsilon . \end{aligned}$$

Consequently,

$$\begin{aligned} \lim _{k\rightarrow \infty }\Vert \mathcal {C}_\mu (f_{k})\Vert _{D_{p-1}^{p}}=0. \end{aligned}$$

Case $p>2$.

By (2) and (3) we see that $HL(p)\subset Y_{p}$. To complete the proof, we have to prove that $\lim _{k\rightarrow \infty }\Vert \mathcal {C}_\mu (f_{k})\Vert _{HL(p)}=0.$

Since $\sum _{n=1}^{\infty }(n+1)^{p-2}\mu _{n}^{p}\log ^{p}(n+1)<\infty$, for any $\varepsilon >0$ there exists a positive integer N such that

$$\begin{aligned} \sum _{n=N+1}^{\infty }(n+1)^{p-2}\mu _{n}^{p}\log ^{p}(n+1)<\varepsilon . \end{aligned}$$

(7)

Set

$$\begin{aligned} f_{k}(z)=\sum _{j=1}^{\infty }\widehat{f_{k}}(j)z^{j},\quad z\in \mathbb {D}. \end{aligned}$$

Since $\{f_{k}\}_{k=1}^{\infty }$ converges to 0 uniformly on every compact subsets of $\mathbb {D}$, it follows that $\widehat{f_{k}}(j)\longrightarrow 0$ as $k\rightarrow \infty$ for every j. This yields that there exists $K_{0}\in \mathbb {N}$ such that

$$\begin{aligned} \sum _{n=0}^{N}(n+1)^{p-2}\mu _{n}^{p} \left| \sum _{j=1}^{n}\widehat{f_{k}}(j)\right| ^{p}\lesssim \varepsilon \quad \text{ for } \text{ all } \ k>K_{0}. \end{aligned}$$

(8)

Note that $\{f_{k}\}_{k=1}^{\infty }\subset X\subset \mathcal {B}$, then it follows from Corollary D in [17] we have that

$$\begin{aligned} \sup _{k\ge 1}\left| \sum _{j=1}^{n}\widehat{f_{k}}(j)\right| \lesssim \log (n+1) \sup _{k\ge 1} \Vert f_{k}\Vert _{\mathcal {B}}\lesssim \log (n+1). \end{aligned}$$

(9)

Hence, for $k> K_{0}$, by (7)–(9) we have that

$$\begin{aligned} \begin{aligned} \Vert \mathcal {C}_\mu (f_{k})\Vert ^{p}_{HL(p)}&=\sum _{n=1}^{N}(n+1)^{p-2}\mu _{n}^{p}\left| \sum _{j=1}^{n}\widehat{f_{k}}(j)\right| ^{p}+ \sum _{n=N+1}^{\infty }(n+1)^{p-2}\mu _{n}^{p}\left| \sum _{j=1}^{n}\widehat{f_{k}}(j)\right| ^{p}\\&\lesssim \varepsilon +\sum _{n=N+1}^{\infty }(n+1)^{p-2}\mu _{n}^{p}\log ^{p}(n+1)\\&\lesssim \varepsilon +\varepsilon . \end{aligned} \end{aligned}$$

Therefore,

$$\begin{aligned} \lim _{k\rightarrow \infty } \Vert \mathcal {C}_\mu (f_{k})\Vert _{HL(p)}=0. \end{aligned}$$

The proof is complete. $\square$

Taking $p = 1$ in Theorem 1.1, we can obtain the following corollary.

Corollary 2.4

Let $1<s<\infty$, $\mu$ be a finite positive Borel measure on [0, 1). Let X be a Banach subspace of $H(\mathbb {D})$ with $\Lambda ^{s}_{\frac{1}{s}}\subset X\subset \mathcal {B}$ and $Y_{1}\in \{D, H^{1}, HL(1), H(\infty ,1)\}$. Then the following statements are equivalent.

• (1) The operator $\mathcal {C}_\mu$ is bounded from X into $Y_{1}$.

• (2) The operator $\mathcal {C}_\mu$ is compact from X in to $Y_{1}$.

• (3) The measure $\mu$ satisfies

$$\begin{aligned} \int _{0}^{1}\log ^{2}\frac{e}{1-t}d\mu (t)<\infty . \end{aligned}$$

Carleson measures play a key role when we study the Cesàro-like operators. Recall that if $\mu$ is a positive Borel measure on [0, 1), $0\le \gamma <\infty$ and $0< s < \infty$, then $\mu$ is a $\gamma$-logarithmic s-Carleson measure if there exists a positive constant C such that

$$\begin{aligned} \mu ([t,1))\log ^{\gamma }\frac{e}{1-t} \le C (1-t)^{s}, \quad \text{ for } \text{ all }\quad 0\le t<1. \end{aligned}$$

In particular, $\mu$ is an s-Carleson measure if $\gamma =0$.

In the following, we provide a sufficient condition and a necessary condition in term of Carleson-type measure.

Corollary 2.5

Suppose $1<p<\infty$, $1<s<\infty$ and $\gamma >1+\frac{1}{p}$, $\mu$ is a positive Borel measure on [0, 1). Let $Y_{p}\in \{D^{p}_{p-1}, H^{p}, HL(p), H(\infty ,p)\}$ and let X be a Banach subspace of $H(\mathbb {D})$ with $\Lambda ^{s}_{\frac{1}{s}}\subset X\subset \mathcal {B}$, then the following statements hold.

• (1) If $\mu$ is a $\gamma$-logarithmic $1-\frac{1}{p}$-Carleson measure, then $\mathcal {C}_\mu$ is bounded (equivalent to compact) from X to $Y_{p}$.

• (2) If $\mathcal {C}_\mu$ is bounded from X to $Y_{p}$, then $\mu$ is a 1-logarithmic $1-\frac{1}{p}$-Carleson measure.

Proof

• (1) Suppose $\mu$ is a $\gamma$-logarithmic $1-\frac{1}{p}$-Carleson measure. Integrating by parts we have

  $$\begin{aligned} \begin{aligned} \mu _{n}&=\int _{0}^{1}t^{n}d\mu (t)=n \int _{0}^{1}t^{n-1}\mu ([t,1))dt\\&\lesssim n \int _{0}^{1}t^{n-1}(1-t)^{1-\frac{1}{p}}\log ^{-\gamma }\frac{e}{1-t}dt.\\ \end{aligned} \end{aligned}$$

  It is easy to estimate that

  $$\begin{aligned} n \int _{0}^{1}t^{n-1}(1-t)^{1-\frac{1}{p}}\log ^{-\gamma }\frac{e}{1-t}dt \asymp \frac{\log ^{-\gamma }(n+1)}{(n+1)^{1-\frac{1}{p}}}. \end{aligned}$$

  This shows that

  $$\begin{aligned} \mu _{n}= O\left( \frac{\log ^{-\gamma }(n+1)}{(n+1)^{1-\frac{1}{p}}}\right) . \end{aligned}$$

  Hence,

  $$\begin{aligned} \sum _{n=0}^{\infty }(n+1)^{p-2}\mu _{n}^{p}\log ^{p}(n+1)\lesssim \sum _{n=0}^{\infty }\frac{1}{(n+1)\log ^{p(\gamma -1)}(n+1)}\lesssim 1. \end{aligned}$$

  Now, Theorem 1.1 implies that $\mathcal {C}_\mu$ is bounded (equivalent to compact) from X to $Y_{p}$.

• (2) If $\mathcal {C}_\mu$ is bounded from X to $Y_{p}$, then for any $N\ge 2$ we have

  $$\begin{aligned} \begin{aligned} 1&\gtrsim \sum _{n=0}^{\infty }(n+1)^{p-2}\mu _{n}^{p}\log ^{p}(n+1)\\&\gtrsim \sum _{n=0}^{N}(n+1)^{p-2}\mu _{n}^{p}\log ^{p}(n+1)\\&\gtrsim \mu ^{p}_{N} \sum _{n=0}^{N}(n+1)^{p-2}\log ^{p}(n+1)\\&\asymp \mu ^{p}_{N}N^{p-1} \log ^{p}(N+1). \end{aligned} \end{aligned}$$

  This yields that $\mu$ is a 1-logarithmic $1-\frac{1}{p}$-Carleson measure.

$\square$

Proof of Theorem 1.2 Suppose $\int _{0}^{1}\frac{\log \frac{e}{1-t}}{1-t}d\mu (t)<\infty$, then for any $f\in X$ we have

$$\begin{aligned} \Vert \mathcal {C}_\mu (f)\Vert _{\infty }=\sup _{z\in \mathbb {D}}\left| \int _{0}^{1}\frac{f(tz)}{1-tz}d\mu (t)\right| \lesssim \sup _{z\in \mathbb {D}} \int _{0}^{1}\frac{\log \frac{e}{1-t|z|}}{1-t|z|}d\mu (t)\le \int _{0}^{1}\frac{\log \frac{e}{1-t}}{1-t}d\mu (t). \end{aligned}$$

This shows that $\mathcal {C}_\mu$ is bounded from X to $H^{\infty }$.

Conversely, assume that $\mathcal {C}_\mu$ is bounded from X to $H^{\infty }$. Take $g(z)=\log \frac{e}{1-z}\in X$, then

$$\begin{aligned} \begin{aligned} \Vert g\Vert _{X}&\gtrsim \Vert \mathcal {C}_\mu (g)\Vert _{\infty }=\sup _{z\in \mathbb {D}}\left| \int _{0}^{1}\frac{\log \frac{e}{1-tz}}{1-tz}d\mu (t)\right| \\&\ge \sup _{x\in (0,1)}\int _{0}^{1}\frac{\log \frac{e}{1-tx}}{1-tx}d\mu (t). \end{aligned} \end{aligned}$$

By the Lebesgue control convergence theorem we have $\int _{0}^{1}\frac{\log \frac{e}{1-t}}{1-t}d\mu (t)<\infty$. $\square$

Remark 2.6

In contrast with what happens in the case $1\le p<\infty$, Theorem 1.2 is no longer valid for $HL(\infty )$. In fact, $\mathcal {C}_\mu$ is bounded from X to $HL(\infty )$ if and only if $\mu$ is a 1-logarithmic 1-Carleson measure. The proof is not difficult, we leave it to the interested reader. However, the condition $\int _{0}^{1}\frac{\log \frac{e}{1-t}}{1-t}d\mu (t)<\infty$ implies that $\mu$ is a 1-logarithmic 1-Carleson measure. But the reverse is not true, as the measure $d\mu (t)=\log ^{-1}\frac{e}{1-t}dt$ shows. It is easy to check that $\mu$ is a 1-logarithmic 1-Carleson measure but $\int _{0}^{1}\frac{\log \frac{e}{1-t}}{1-t}d\mu (t)=\infty$.

Recall that the mixed norm space $H(p,q,\alpha )$, $0<p,q<\infty$, $0<\alpha <\infty$, consists of those analytic functions on $\mathbb {D}$ such that

$$\begin{aligned} \Vert f\Vert ^{q}_{p,q,\alpha }= \int _{0}^{1}(1-r)^{q\alpha -1}M^{q}_{p}(r,f)dr<\infty . \end{aligned}$$

Hardy and Littlewood proved (see [10, Theorem 5.11]) that

$$\begin{aligned} H^{p}\subset H\Bigg (q,p,\frac{1}{p}-\frac{1}{q}\Bigg ),\quad 0<p<q<\infty . \end{aligned}$$

In particular, for $0<p<1$, $H^{p}\subset H(1,p,\frac{1}{p}-1)$. In the following, we show that the range of $\mathcal {C}_\mu$ acting on X is contained in $H(1,p,\frac{1}{p}-1)$ for $0<p<1$.

Theorem 2.7

Suppose $1<s<\infty$, $0<p<1$ and $\mu$ is a finite positive Borel measure on [0, 1). Let X be a Banach subspace of $H(\mathbb {D})$ with $\Lambda ^{s}_{\frac{1}{s}}\subset X\subset \mathcal {B}$. Then $\mathcal {C}_\mu (X)\subset H(1,p,\frac{1}{p}-1)$.

Proof

Let $f\in X\subset \mathcal {B}$, then it is known that

$$\begin{aligned} |f(z)|\lesssim \Vert f\Vert _{\mathcal {B}}\log \frac{e}{1-|z|} \quad \text{ for } \text{ all }\ \ z\in \mathbb {D}. \end{aligned}$$

By Fubini theorem and Lemma 2.1–2.3 we have

$$\begin{aligned} \begin{aligned} \Vert \mathcal {C}_\mu (f)\Vert ^{p}_{1,p,\frac{1}{p}-1}&=\int _{0}^{1}(1-r)^{p(\frac{1}{p}-1)-1}M^{p}_{1}(r,\mathcal {C}_\mu (f))dr\\&=\int _{0}^{1}(1-r)^{-p}\left( \int _{0}^{2\pi }\left| \int _{0}^{1}\frac{f(tre^{i\theta })}{1-tre^{i\theta }}d\mu (t)\right| d\theta \right) ^{p}dr\\&\lesssim \Vert f\Vert ^{p}_{\mathcal {B}}\int _{0}^{1}(1-r)^{-p}\left( \int _{0}^{1}\log ^{2}\frac{e}{1-tr}d\mu (t)\right) ^{p}dr\\&\asymp \Vert f\Vert ^{p}_{\mathcal {B}}\int _{0}^{1}(1-r)^{-p} \left( \sum _{n=0}^{\infty }\frac{\log (n+1)}{n+1}\mu _{n}r^{n}\right) ^{p}dr\\&\asymp \Vert f\Vert ^{p}_{\mathcal {B}}\sum _{n=0}^{\infty }2^{-n(1-p)}\left( \sum _{k=2^{n}-1}^{2^{n+1}}(k+1)^{-1}\log (k+1)\mu _{k}\right) ^{p}\\&\lesssim \Vert f\Vert ^{p}_{\mathcal {B}}\sum _{n=0}^{\infty }2^{-n(1-p)} (n+1)^{p}\mu ^{p}_{2^{n}-1}\\&\lesssim \Vert f\Vert ^{p}_{\mathcal {B}}\sum _{n=0}^{\infty }2^{-n(1-p)} (n+1)^{p}\lesssim \Vert f\Vert ^{p}_{\mathcal {B}}\lesssim \Vert f\Vert ^{p}_{X}. \end{aligned} \end{aligned}$$

This yields that $\mathcal {C}_\mu (X)\subset H\Bigg (1,p,\frac{1}{p}-1\Bigg )$. $\square$

We recall that $f\in H(\mathbb {D})$ is a Cauchy transform if it admits a representation

$$\begin{aligned} f(z)=\int _{0}^{2\pi }\frac{d\lambda (\theta )}{1-e^{i\theta }z}, \ \ z\in \mathbb {D}, \end{aligned}$$

where $\lambda$ is a finite complex Borel measure on $\partial \mathbb {D}$. The space of all Cauchy transforms is denoted by $\mathcal {K}$. We let $\mathcal {A}$ denote the disc algebra, that is, the space of analytic functions in $\mathbb {D}$ with a continuous extension to the closed unit disc, endowed with the $\Vert \cdot \Vert _{\infty }$-norm. It turns out [8] that

$$\begin{aligned} \Vert f\Vert _{\mathcal {K}}=\sup \{|\langle f, g\rangle _{H^{2}}|: g\in \mathcal {A}, \Vert g\Vert _{H^{\infty }}\le 1 \}. \end{aligned}$$

It is known (see [7]) that $H^{1}\subsetneq \mathcal {K}\subsetneq \bigcap _{0<p<1}H^{p}$. In view of Theorem 2.7, the following question arises naturally.

Question

Suppose $1<s<\infty$, $\mu$ is a finite positive Borel measure on [0, 1) and X is a Banach subspace of $H(\mathbb {D})$ with $\Lambda ^{s}_{\frac{1}{s}}\subset X\subset \mathcal {B}$. Is the operator $\mathcal {C}_\mu$ bounded from X to $\bigcap _{0<p<1}H^{p}$?

We do not know the answer to this question. However, there exists a finite positive measure $\mu$ on [0, 1) such that $\mathcal {C}_\mu (X)\nsubseteq \mathcal {K}$. Actually, we have the following result.

Proposition 2.8

Suppose $1<s<\infty$ and X is a Banach subspace of $H(\mathbb {D})$ with $\Lambda ^{s}_{\frac{1}{s}}\subset X\subset \mathcal {B}$. Then there exist $f\in X$ and a finite positive measure $\mu$ on [0, 1) such that $\mathcal {C}_\mu (f)\notin \mathcal {K}$.

Proof

For $1<\gamma <2$, let $d\mu (t)=\left( (1-t)\left( \log \frac{e}{1-t}\right) ^{3-\gamma }\right) ^{-1}dt$. Then it is clear that $\mu$ is a finite positive measure $\mu$ on [0, 1). By Lemma 2.2 we have

$$\begin{aligned} \frac{1}{1-t}\left( \log \frac{e}{1-t}\right) ^{\gamma -3}\asymp \sum _{k=0}^{\infty }\left( \log (k+2)\right) ^{\gamma -3}t^{k}. \end{aligned}$$

For $n\ge 2$, we have

$$\begin{aligned} \begin{aligned} \mu _{n}&=\int _{0}^{1}t^{n}d\mu (t)=\int _{0}^{1}t^{n}\left( (1-t) \left( \log \frac{e}{1-t}\right) ^{3-\gamma }\right) ^{-1}dt\\&\asymp \sum _{k=0}^{\infty }\left( \log (k+2)\right) ^{\gamma -3}\int _{0}^{1}t^{n+k}dt\\&=\sum _{k=0}^{n}\frac{\left( \log (k+2)\right) ^{\gamma -3}}{n+k+1}+\sum _{k=n+1}^{\infty } \frac{\left( \log (k+2)\right) ^{\gamma -3}}{n+k+1}\\&=S_{1}+S_{2}. \end{aligned} \end{aligned}$$

It is easy to see that

$$\begin{aligned} S_{1}=\sum _{k=0}^{n}\frac{\left( \log (k+2)\right) ^{\gamma -3}}{n+k+1}\asymp \frac{1}{n+1}\sum _{k=0}^{n} \left( \log (k+2)\right) ^{\gamma -3}\asymp \left( \log (n+2)\right) ^{\gamma -3}. \end{aligned}$$

Also, for every $n\ge 2$ we have that

$$\begin{aligned} S_{2}= & {} \sum _{k=n+1}^{\infty }\frac{\left( \log (k+2)\right) ^{\gamma -3}}{n+k+1}\\= & {} \sum _{k=n+1}^{\infty }\frac{1}{(n+k+1)[\log (k+n+2)]^{3-\gamma }} \left( \frac{\log (k+n+2)}{\log (k+2)}\right) ^{3-\gamma }\\= & {} \sum _{k=n+1}^{\infty }\frac{1}{(n+k+1)[\log (k+n+2)]^{3-\gamma }} \left( 1+\frac{\log (1+\frac{n}{k+2})}{\log (k+2)}\right) ^{3-\gamma }\\&\asymp&\sum _{k=n+1}^{\infty }\frac{1}{(n+k+1)[\log (k+n+2)]^{3-\gamma }}\\&\asymp&\int _{n+1}^{\infty }\frac{1}{(n+x)[\log (n+x)]^{3-\gamma }}dx \\&\asymp&(\log (n+2))^{\gamma -2}. \end{aligned}$$

Hence, we have that

$$\begin{aligned} \mu _{n}\asymp \left( \log (n+2)\right) ^{\gamma -3}+ (\log (n+2))^{\gamma -2}\asymp (\log (n+2))^{\gamma -2}. \end{aligned}$$

(10)

Now, let $f(z)=\log \frac{1}{1-z}\in X$ and $g(z)=\sum _{n=0}^{\infty }\frac{z^{n}}{(n+1)(\log (n+2))^{\gamma }}\in \mathcal {A}$. To complete the proof, it is suffices to show that

$$\begin{aligned} |\langle \mathcal {C}_\mu (f), g\rangle _{H^{2}}|=\lim _{r\rightarrow 1^{-}}\sum _{n=1}^{\infty } \frac{\mu _{n}\left( \sum ^{n}_{k=1}\frac{1}{k}\right) }{(n+1)(\log (n+2))^{\gamma }}r^{2n}=\infty . \end{aligned}$$

In fact, it follows from (10) that

$$\begin{aligned} \sum _{n=1}^{\infty } \frac{\mu _{n}\left( \sum ^{n}_{k=1}\frac{1}{k}\right) }{(n+1)(\log (n+2))^{\gamma }}r^{2n}\asymp \sum _{n=1}^{\infty }\frac{r^{2n}}{(n+1)\log (n+2)}\rightarrow \infty , \ \ (r\rightarrow 1^{-}). \end{aligned}$$

The proof is complete. $\square$