Carleson measures and the range of a Cesàro-like operator acting on $H^\infty$

Abstract

In this paper, we determine the range of a Cesàro-like operator acting on $H^\infty$ by describing characterizations of Carleson type measures on [0, 1). A special case of our result gives an answer to a question posed by P. Galanopoulos, D. Girela and N. Merchán recently.

1 Introduction

Let ${\mathbb {D}}$ be the open unit disk in the complex plane ${\mathbb {C}}$. Denote by $H({\mathbb {D}})$ the space of functions analytic in ${\mathbb {D}}$. For $f(z)=\sum _{n=0}^\infty a_nz^n$ in $H({\mathbb {D}})$, the Cesàro operator ${\mathcal {C}}$ is defined by

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

See [7, 12, 14, 21, 23, 24] for the investigation of the Cesàro operator acting on some analytic function spaces.

Recently, P. Galanopoulos, D. Girela and N. Merchán [16] considered a Cesàro-like operator ${\mathcal {C}}_\mu$ on $H({\mathbb {D}})$. For nonnegative integer n, let $\mu _n$ be the moment of order n of a finite positive Borel measure $\mu$ on [0, 1); that is,

$$\begin{aligned} \mu _n=\int _{[0, 1)} t^{n}d\mu (t). \end{aligned}$$

For $f(z)=\sum _{n=0}^\infty a_nz^n$ belonging to $H({\mathbb {D}})$, the Cesàro-like operator ${\mathcal {C}}_\mu$ is defined by

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

If $d\mu (t)=dt$, then ${\mathcal {C}}_\mu ={\mathcal {C}}$. In [16, 19], the authors studied the action of ${\mathcal {C}}_\mu$ on distinct spaces of analytic functions.

We also need to recall some function spaces. For $0<p<\infty$, $H^p$ denotes the classical Hardy space [13] of those functions $f\in H({\mathbb {D}})$ for which

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

where

$$\begin{aligned} M_p(r, f)= \left( \frac{1}{2\pi }\int _0^{2\pi }|f(re^{i\theta })|^p d\theta \right) ^{1/p}. \end{aligned}$$

As usual, denote by $H^\infty$ the space of bounded analytic functions in ${\mathbb {D}}$. It is well known that $H^\infty$ is a proper subset of the Bloch space ${\mathcal {B}}$ which consists of those functions $f\in H({\mathbb {D}})$ satisfying

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

Denote by $\text {Aut}({\mathbb {D}})$ the group of Möbius maps of ${\mathbb {D}}$, namely,

$$\begin{aligned} \text {Aut}({\mathbb {D}})=\{e^{i\theta }\sigma _a:\ \ a\in {\mathbb {D}}\ \text {and} \ \theta \ \ \text {is real}\}, \end{aligned}$$

where

$$\begin{aligned} \sigma _a(z)=\frac{a-z}{1-{\overline{a}}z}, \qquad z\in {\mathbb {D}}. \end{aligned}$$

In 1995 R. Aulaskari, J. Xiao and R. Zhao [2] introduced ${\mathcal {Q}}_p$ spaces. For $0\le p<\infty$, a function f analytic in ${\mathbb {D}}$ belongs to ${\mathcal {Q}}_p$ if

$$\begin{aligned} \Vert f\Vert _{{\mathcal {Q}}_p}^2=\sup _{w\in {\mathbb {D}}} \int _{\mathbb {D}}|f'(z)|^2(1-|\sigma _w(z)|^2)^p dA(z)<\infty , \end{aligned}$$

where dA is the area measure on ${\mathbb {C}}$ normalized so that $A({\mathbb {D}})=1$. ${\mathcal {Q}}_p$ spaces are Möbius invariant in the sense that

$$\begin{aligned} \Vert f\Vert _{{\mathcal {Q}}_p}=\Vert f\circ \phi \Vert _{{\mathcal {Q}}_p} \end{aligned}$$

for every $f\in {\mathcal {Q}}_p$ and $\phi \in \text {Aut}({\mathbb {D}})$. It was shown in [25] that ${\mathcal {Q}}_2$ coincides with the Bloch space ${\mathcal {B}}$. This result was extended in [1] by showing that ${\mathcal {Q}}_p={\mathcal {B}}$ for all $1<p<\infty$. The space ${\mathcal {Q}}_1$ coincides with BMOA, the set of analytic functions in ${\mathbb {D}}$ with boundary values of bounded mean oscillation (see [5, 17]). The space ${\mathcal {Q}}_0$ is the Dirichlet space ${\mathcal {D}}$. For $0<p<1$, the space ${\mathcal {Q}}_p$ is a proper subset of BMOA and has many interesting properties. See J. Xiao’s monographs [26, 27] for the theory of ${\mathcal {Q}}_p$ spaces.

For $1\le p<\infty$ and $0<\alpha \le 1$, the mean Lipschitz space $\Lambda ^p_\alpha$ is the set of those functions $f\in H({\mathbb {D}})$ with 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 well known (cf. [13, Chapter 5]) that $\Lambda ^p_\alpha$ is a subset of $H^p$ and $\Lambda ^p_\alpha$ consists of those functions $f\in H({\mathbb {D}})$ satisfying

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

Among these spaces, the spaces $\Lambda ^p_{1/p}$ are of special interest. $\Lambda ^p_{1/p}$ spaces increase with $p\in (1, \infty )$ in the sense of inclusion and they are contained in BMOA (cf. [10]). By Theorem 1.4 in [4], $\Lambda ^p_{1/p}\subseteq {\mathcal {Q}}_q$ when $1\le p<2/(1-q)$ and $0<q<1$. In particular, $\Lambda ^2_{1/2}\subseteq {\mathcal {Q}}_q \subseteq {\mathcal {B}}$ for all $0<q<\infty$.

Given an arc I of the unit circle ${\mathbb {T}}$ with arclength |I| (normalized such that $|{\mathbb {T}}|=1$), the Carleson box S(I) is given by

$$\begin{aligned} S(I)=\{r\zeta \in {\mathbb {D}}: 1-|I|<r<1, \ \zeta \in I\}. \end{aligned}$$

For $0<s<\infty$, a positive Borel measure $\nu$ on ${\mathbb {D}}$ is said to be an s-Carleson measure if

$$\begin{aligned} \sup _{I\subseteq {\mathbb {T}}}\frac{\nu (S(I))}{|I|^s}<\infty . \end{aligned}$$

If $\nu$ is a 1-Carleson measure, we write that $\nu$ is a Carleson measure characterizing $H^p\subseteq L^p(d\nu )$ for $0<p<\infty$ (cf. [13]). A positive Borel measure $\mu$ on [0, 1) can be seen as a Borel measure on ${\mathbb {D}}$ by identifying it with the measure ${\tilde{\mu }}$ defined by

$$\begin{aligned} {\tilde{\mu }}(E)=\mu (E \cap [0, 1)), \end{aligned}$$

for any Borel subset E of ${\mathbb {D}}$. Thus $\mu$ is an s-Carleson measure on [0, 1) if there is a positive constant C such that

$$\begin{aligned} \mu ([t, 1)) \le C (1-t)^s \end{aligned}$$

for all $t\in [0, 1)$. We refer to [8] for the investigation of this kind of measures associated with Hankel measures.

It is known that the Cesàro operator ${\mathcal {C}}$ is bounded on $H^p$ for all $0<p<\infty$ (cf. [21, 23, 24]) but this is not true on $H^\infty$. In fact, N. Danikas and A. Siskakis [12] gave that ${\mathcal {C}}(H^\infty )\nsubseteq H^\infty$ but ${\mathcal {C}}(H^\infty )\subseteq BMOA$. Later M. Essén and J. Xiao [14] proved that ${\mathcal {C}}(H^\infty )\subsetneqq {\mathcal {Q}}_p$ for $0<p<1$. Recently, the relation between ${\mathcal {C}}(H^\infty )$ and a class of Möbius invariant function spaces was considered in [7].

It is quite natural to study ${\mathcal {C}}_\mu (H^\infty )$. In [16] the authors characterized positive Borel measures $\mu$ such that ${\mathcal {C}}_\mu (H^\infty )\subseteq H^\infty$ and proved that ${\mathcal {C}}_\mu (H^\infty )\subseteq {\mathcal {B}}$ if and only if $\mu$ is a Carleson measure. Moreover, they showed that if ${\mathcal {C}}_\mu (H^\infty )\subseteq BMOA$, then $\mu$ is a Carleson measure. In [16, p. 20], the authors asked whether or not $\mu$ being a Carleson measure implies that ${\mathcal {C}}_\mu (H^\infty )\subseteq BMOA$. In this paper, by giving some descriptions of s-Carleson measures on [0, 1), for $0<p<2$, we show that ${\mathcal {C}}_\mu (H^\infty )\subseteq {\mathcal {Q}}_p$ if and only if $\mu$ is a Carleson measure, which gives an affirmative answer to their question. We also consider another Cesàro-like operator ${\mathcal {C}}_{\mu , s}$ and describe the embedding ${\mathcal {C}}_{\mu , s}(H^\infty )\subseteq X$ in terms of s-Carleson measures, where X is between $\Lambda ^p_{1/p}$ and ${\mathcal {B}}$ for $\max \{1, 1/s\}<p<\infty$.

Throughout this paper, the symbol $A\thickapprox B$ means that $A\lesssim B\lesssim A$. We say that $A\lesssim B$ if there exists a positive constant C such that $A\le CB$.

2 Positive Borel measures on [0, 1) as Carleson type measures

In this section, we give some characterizations of positive Borel measures on [0, 1) as Carleson type measures.

The following description of Carleson type measures (cf. [9] ) is well known.

Lemma A

Suppose $s>0$, $t>0$ and $\mu$ is a positive Borel measure on ${\mathbb {D}}$. Then $\mu$ is an s-Carleson measure if and only if

$$\begin{aligned} \sup _{a\in {\mathbb {D}}}\int _{{\mathbb {D}}} \frac{(1-|a|^2)^t}{|1-{\overline{a}}w|^{s+t}}d\mu (w)<\infty . \end{aligned}$$

(2.1)

For Carleson type measures on [0, 1), we can obtain some descriptions that are different from Lemma A. Now we give the first main result in this section.

Proposition 2.1

Suppose $0<t<\infty$, $0\le r<s<\infty$ and $\mu$ is a finite positive Borel measure on [0, 1). Then the following conditions are equivalent:

• (i) $\mu$ is an s-Carleson measure;

• (ii)
  $$\begin{aligned} \sup _{a\in {\mathbb {D}}}\int _{[0,1)}\frac{(1-|a|)^t}{(1-x)^{r}(1-|a|x)^{s+t-r}}d\mu (x)<\infty ; \end{aligned}$$

  (2.2)
• (iii)
  $$\begin{aligned} \sup _{a\in {\mathbb {D}}}\int _{[0,1)}\frac{(1-|a|)^t}{(1-x)^{r}|1-ax|^{s+t-r}}d\mu (x)<\infty . \end{aligned}$$

  (2.3)

Proof

$(i)\Rightarrow (ii)$. Let $\mu$ be an s-Carleson measure. Fix $a\in {\mathbb {D}}$ with $|a|\le 1/2$. If $r=0$, the desired result holds. For $0<r<s$, using a well-known formula about the distribution function(cf. [15, p.20 ]), we get

$$\begin{aligned} \int _{[0,1)}&\frac{(1-|a|)^t}{(1-x)^{r}(1-|a|x)^{s+t-r}}d\mu (x)\nonumber \\ \thickapprox&\quad \int _{[0,1)}\left( \frac{1}{1-x}\right) ^rd\mu (x) \nonumber \\ \thickapprox&\quad r \int _0^\infty \lambda ^{r-1} \mu (\{x\in [0, 1): 1-\frac{1}{\lambda }<x \})d\lambda \nonumber \\ \lesssim&\quad \int _0^1 \lambda ^{r-1} \mu ([0, 1))d\lambda + \int _1^\infty \lambda ^{r-1} \mu ([1-\frac{1}{\lambda }, 1))d\lambda \nonumber \\ \lesssim&\quad 1+ \int _1^\infty \lambda ^{r-s-1} d\lambda \lesssim 1. \end{aligned}$$

(2.4)

Fix $a\in {\mathbb {D}}$ with $|a|>1/2$ and let

$$\begin{aligned} S_n(a)=\{x\in [0,1): 1-2^n(1-|a|)\le x<1\}, \ \ n=1, 2, \cdots . \end{aligned}$$

Let $n_a$ be the minimal integer such that $1-2^{n_a}(1-|a|)\le 0$. Then $S_n(a)=[0, 1)$ when $n\ge n_a$. If $x\in S_1(a)$, then

$$\begin{aligned} 1-|a| \le 1-|a|x. \end{aligned}$$

(2.5)

Also, for $2\le n\le n_a$ and $x\in S_n(a)\backslash S_{n-1}(a)$, we have

$$\begin{aligned} 1-|a|x \ge |a|-x \ge |a|-(1-2^{n-1}(1-|a|))=(2^{n-1}-1)(1-|a|). \end{aligned}$$

(2.6)

We write

$$\begin{aligned} \int _{[0,1)}&\frac{(1-|a|)^t}{(1-x)^r(1-|a|x)^{s+t-r}}d\mu (x)\\ =&\int _{S_1(a)}\frac{(1-|a|)^t}{(1-x)^r(1-|a|x)^{s+t-r}}d\mu (x)\\&+\sum ^{n_a}_{n=2}\int _{S_n(a)\backslash S_{n-1}(a)}\frac{(1-|a|)^t}{(1-x)^r(1-|a|x)^{s+t-r}}d\mu (x)\\ =:&J_1(a)+J_2(a). \end{aligned}$$

If $r=0$, bearing in mind (2.5), (2.6) and that $\mu$ is an s-Carleson measure, it is easy to check that $J_i(a)\lesssim 1$ for $i=1, 2$. Now consider $0<t<\infty$ and $0< r<s<\infty$. Using (2.5) and some estimates similar to (2.4), we have

$$\begin{aligned} J_1(a) \lesssim (1-|a|)^{r-s}\int _{S_1(a)}\left( \frac{1}{1-x}\right) ^rd\mu (x) \lesssim 1. \end{aligned}$$

Note that (2.6) holds, $0<t<\infty$, $0< r<s<\infty$ and $\mu$ is an s-Carleson measure. Then

$$\begin{aligned} J_2(a) \lesssim&\sum ^{n_a}_{n=2}\frac{(1-|a|)^{r-s}}{2^{n(s+t-r)}}\int _{S_n(a)\backslash S_{n-1}(a)}\left( \frac{1}{1-x}\right) ^rd\mu (x)\\ \lesssim&\sum ^{n_a}_{n=2}\frac{(1-|a|)^{r-s}}{2^{n(s+t-r)}}\int _0^\infty \lambda ^{r-1}\mu \Big (\Big \{x\in [1-2^n(1-|a|),1): 1-\frac{1}{\lambda }<x\Big \}\Big )d\lambda \\ \thickapprox&\sum ^{n_a}_{n=2}\frac{(1-|a|)^{r-s}}{2^{n(s+t-r)}} \bigg (\int _0^{\frac{1}{2^n(1-|a|)}}\lambda ^{r-1}\mu \big ([1-2^n(1-|a|),1)\big )d\lambda \\&+\int _{\frac{1}{2^n(1-|a|)}}^\infty \lambda ^{r-1} \mu \Big (\Big [1-\frac{1}{\lambda },1\Big )\Big )d\lambda \Bigg )\\ \lesssim&\sum ^{n_a}_{n=2}\frac{(1-|a|)^{r-s}}{2^{n(s+t-r)}} \Bigg (2^{ns}(1-|a|)^s\int _0^{\frac{1}{2^n(1-|a|)}}\lambda ^{r-1}d\lambda +\int _{\frac{1}{2^n(1-|a|)}}^\infty \lambda ^{r-1-s}d\lambda \Bigg )\\ \thickapprox&\sum ^{n_a}_{n=2} \frac{1}{2^{tn}}<\infty . \end{aligned}$$

Consequently,

$$\begin{aligned} \sup _{a\in {\mathbb {D}}}\int _{[0,1)}\frac{(1-|a|)^t}{(1-x)^r(1-|a|x)^{s+t-r}}d\mu (x)<\infty . \end{aligned}$$

The implication of $(ii)\Rightarrow (iii)$ is clear.

$(iii)\Rightarrow (i)$. For $r\ge 0$, it is clear that

$$\begin{aligned} \int _{[0,1)}\frac{(1-|a|)^t}{(1-x)^{r}|1-ax|^{s+t-r}}d\mu (x)\ge \int _{[0,1)}\frac{(1-|a|)^t}{|1-ax|^{s+t}}d\mu (x) \end{aligned}$$

for all $a\in {\mathbb {D}}$. Combining this with Lemma A, we see that if (2.3) holds, then $\mu$ is an s-Carleson measure. $\square$

Remark 1

The condition $0\le r<s<\infty$ in Proposition 2.1 can not be changed to $r\ge s>0$. For example, let $d\mu _1(x)=(1-x)^{s-1}dx$, $x\in [0, 1)$. Then $\mu _1$ is an s-Carleson measure but for $r\ge s>0$,

$$\begin{aligned} \sup _{a\in {\mathbb {D}}}\int _{[0,1)}&\frac{(1-|a|)^t}{(1-x)^{r}|1-ax|^{s+t-r}}d\mu _1(x)\\ \ge&\int _0^1 (1-x)^{s-1-r}dx=+\infty . \end{aligned}$$

Remark 2

$\mu$ supported on [0, 1) is essential in Proposition 2.1. For example, consider $0<t<1$, $0< r<s<1$ and $s=r+t$. Set $d\mu _2(w)=|f'(w)|^2(1-|w|^2)^sdA(w)$, $w\in {\mathbb {D}}$, where $f\in {\mathcal {Q}}_s\setminus {\mathcal {Q}}_t$. Note that for $0<p<\infty$ and $g\in H({\mathbb {D}})$, $|g'(w)|^2(1-|w|^2)^pdA(w)$ is a p-Carleson measure if and only if $g\in {\mathcal {Q}}_p$ (cf. [26]). Hence $d\mu _2$ is an s-Carleson measure. But

$$\begin{aligned} \sup _{a\in {\mathbb {D}}}\int _{{\mathbb {D}}}&\frac{(1-|a|)^t}{(1-|w|)^{r}|1-a{\overline{w}}|^{s+t-r}}d\mu _2(w)\\&=\sup _{a\in {\mathbb {D}}}\int _{{\mathbb {D}}}|f'(w)|^2\frac{(1-|a|)^t(1-|w|)^{s-r}}{|1-a{\overline{w}}|^{s+t-r}}dA(w)\\&\thickapprox \sup _{a\in {\mathbb {D}}}\int _{{\mathbb {D}}}|f'(w)|^2(1-|\sigma _a(w)|^2)^tdA(w)=+\infty . \end{aligned}$$

Before giving the other characterization of Carleson type measures on [0, 1), we need to recall some results.

The following result is Lemma 1 in [20], which generalizes Lemma 3.1 in [18] from $p=2$ to $1<p<\infty$.

Lemma B

Let $f\in H({\mathbb {D}})$ with $f(z)=\sum ^\infty _{n=0}a_n z^n$. Suppose $1<p<\infty$ and the sequence $\{a_n\}$ is a decreasing sequence of nonnegative numbers. If X is a subspace of $H({\mathbb {D}})$ with $\Lambda ^p_{1/p}\subseteq X\subseteq {\mathcal {B}}$, then

$$\begin{aligned} f\in X \iff a_n=O\left( \frac{1}{n}\right) . \end{aligned}$$

We recall a characterization of s-Carleson measure $\mu$ on [0, 1) as follows (cf. [6, Theorem 2.1] or [11, Proposition1]).

Proposition C

Let $\mu$ be a finite positive Borel measure on [0, 1) and $s>0$. Then $\mu$ is an s-Carleson measure if and only if the sequence of moments $\{\mu _n\}_{n=0}^\infty$ satisfies $\sup _{n\ge 0} (1+n)^s \mu _n<\infty$.

The following characterization of functions with nonnegative Taylor coefficients in ${\mathcal {Q}}_p$ is Theorem 2.3 in [3].

Theorem D

Let $0<p<\infty$ and let $f(z)=\sum _{n=0}^\infty a_nz^n$ be an analytic function in ${\mathbb {D}}$ with $a_n\ge 0$ for all n. Then $f\in {\mathcal {Q}}_p$ if and only if

$$\begin{aligned} \sup _{0\le r<1} \sum _{n=0}^\infty \frac{(1-r)^p}{(n+1)^{p+1}}\left( \sum _{k=0}^n(k+1)a_{k+1}(n-k+1)^{p-1}r^{n-k}\right) ^2<\infty . \end{aligned}$$

We need the following well-known estimates (cf. [28, Lemma 3.10]).

Lemma E

Let $\beta$ be any real number. Then

$$\begin{aligned} \int ^{2\pi }_0\frac{d\theta }{|1-ze^{-i\theta }|^{1+\beta }}\thickapprox {\left\{ \begin{array}{ll}1 &{} \text {if} \ \ \beta <0,\\ \log \frac{2}{1-|z|^2} &{} \text {if} \ \ \beta =0,\\ \frac{1}{(1-|z|^2)^\beta } &{} \text {if}\ \ \beta >0, \end{array}\right. } \end{aligned}$$

for all $z\in {\mathbb {D}}$.

For $0<s<\infty$ and a finite positive Borel measure $\mu$ on [0, 1), set

$$\begin{aligned} f_{\mu , s}(z)=\sum _{n=0}^\infty \frac{\Gamma (n+s)}{\Gamma (s)n!} \mu _n z^n, \ \ z\in {\mathbb {D}}. \end{aligned}$$

Now we state the other main result in this section which is inspired by Lemma B and Proposition C.

Proposition 2.2

Suppose $0<s<\infty$ and $\mu$ is a finite positive Borel measure on [0, 1). Let $1<p<\infty$ and let X be a subspace of $H({\mathbb {D}})$ with $\Lambda ^p_{1/p}\subseteq X\subseteq {\mathcal {B}}$. Then $\mu$ is an s-Carleson measure if and only if $f_{\mu , s}\in X$.

Proof

Let $\mu$ be an s-Carleson measure. Clearly,

$$\begin{aligned} f_{\mu , s}(z)=\int _{[0, 1)} \frac{1}{(1-tz)^s}d\mu (t) \end{aligned}$$

for any $z\in {\mathbb {D}}$. For $p>1$, it follows from the Minkowski inequality and Lemma E that

$$\begin{aligned} M_p(r, f'_{\mu , s})\le&s \left( \frac{1}{2\pi }\int _0^{2\pi } \left( \int _{[0, 1)}\frac{1}{|1-tre^{i\theta }|^{s+1}}d\mu (t)\right) ^pd\theta \right) ^{1/p}\\ \le&s \int _{[0, 1)} \left( \frac{1}{2\pi }\int _0^{2\pi } \frac{1}{|1-tre^{i\theta }|^{(s+1)^p}}d\theta \right) ^{1/p} d\mu (t)\\ \lesssim&\int _{[0, 1)} \frac{1}{(1-tr)^{s+1-\frac{1}{p}}} d\mu (t) \end{aligned}$$

for all $0<r<1$. Combining this with Proposition 2.1, we get $f_{\mu , s}\in \Lambda ^p_{1/p}$ and hence $f_{\mu , s}\in X$.

On the other hand, let $f_{\mu , s}\in X$. Then $f_{\mu , s}\in {\mathcal {Q}}_q$ with $q>1$. By the Stirling formula,

$$\begin{aligned} \frac{\Gamma (n+s)}{\Gamma (s)n!}\thickapprox (n+1)^{s-1} \end{aligned}$$

for all nonnegative integers n. Consequently, by Theorem D we deduce

$$\begin{aligned} \infty> & {} \sum _{n=0}^\infty \frac{(1-r)^q}{(n+1)^{q+1}}\left( \sum _{k=0}^n(k+2)^{s}\mu _{k+1}(n-k+1)^{q-1}r^{n-k}\right) ^2\\ > rsim & {} \sum _{n=0}^\infty \frac{(1-r)^q}{(4n+1)^{q+1}}\left( \sum _{k=0}^{4n}(k+2)^{s}\mu _{k+1}(4n-k+1)^{q-1}r^{4n-k}\right) ^2\\ > rsim & {} \sum _{n=0}^\infty \frac{(1-r)^q}{(4n+1)^{q+1}}\left( \sum _{k=n}^{2n}(k+2)^{s}\int _r^1 t^{k+1}d\mu (t) (4n-k+1)^{q-1}r^{4n-k}\right) ^2\\ > rsim & {} \mu ^2([r, 1)) (1-r)^q\sum _{n=0}^\infty \frac{r^{8n+2}}{(4n+1)^{q+1}}\left( \sum _{k=n}^{2n}(k+2)^{s}(4n-k+1)^{q-1} \right) ^2\\ > rsim & {} \mu ^2([r, 1)) (1-r)^q \sum _{n=0}^\infty (4n+2)^{2s+q-1} r^{8n+2}\\&\thickapprox&\frac{\mu ^2([r, 1))}{(1-r)^{2s}} \end{aligned}$$

for all $r\in [0, 1)$ which yields that $\mu$ is an s-Carleson measure. The proof is complete. $\square$

3 ${\mathcal {Q}}_p$ spaces and the range of ${\mathcal {C}}_\mu$ acting on $H^\infty$

In this section, we characterize finite positive Borel measures $\mu$ on [0, 1) such that ${\mathcal {C}}_\mu (H^\infty )\subseteq {\mathcal {Q}}_p$ for $0<p<2$. Descriptions of Carleson measures in Proposition 2.1 play a key role in our proof.

The following lemma is from [22].

Lemma F

Suppose $s>-1$, $r>0$, $t>0$ with $r+t-s-2>0$. If r, $t<2+s$, then

$$\begin{aligned} \int _{\mathbb {D}}\frac{(1-|z|^2)^s}{|1-{\overline{a}}z|^r|1-{\overline{b}}z|^t}dA(z)\lesssim \frac{1}{|1-{\overline{a}}b|^{r+t-s-2}} \end{aligned}$$

for all a, $b\in {\mathbb {D}}$. If $t<2+s<r$, then

$$\begin{aligned} \int _{\mathbb {D}}\frac{(1-|z|^2)^s}{|1-{\overline{a}}z|^r|1-{\overline{b}}z|^t}dA(z)\lesssim \frac{(1-|a|^2)^{2+s-r}}{|1-{\overline{a}}b|^{t}} \end{aligned}$$

for all a, $b\in {\mathbb {D}}$.

We give our result as follows.

Theorem 3.1

Suppose $0<p<2$ and $\mu$ is a finite positive Borel measure on [0, 1). Then ${\mathcal {C}}_\mu (H^\infty )\subseteq {\mathcal {Q}}_p$ if and only if $\mu$ is a Carleson measure.

Proof

Suppose ${\mathcal {C}}_\mu (H^\infty )\subseteq {\mathcal {Q}}_p$. Then ${\mathcal {C}}_\mu (H^\infty )$ is a subset of the Bloch space. By [16, Theorem 5], $\mu$ is a Carleson measure.

Conversely, suppose $\mu$ is a Carleson measure and $f\in H^\infty$. Then f is also in the Bloch space ${\mathcal {B}}$. From Proposition 1 in [16],

$$\begin{aligned} {\mathcal {C}}_{\mu }(f)(z)=\int _{[0, 1)} \frac{f(tz)}{1-tz}d\mu (t), \ \ z\in {\mathbb {D}}. \end{aligned}$$

Hence for any $z\in {\mathbb {D}}$,

$$\begin{aligned} \Vert {\mathcal {C}}_\mu (f)\Vert _{{\mathcal {Q}}_p}&\lesssim \sup _{a\in {\mathbb {D}}} \left( \int _{{\mathbb {D}}}\left( \int _{[0,1)}\frac{|tf'(tz)|}{|1-tz|}d\mu (t)\right) ^2(1-|\sigma _a(z)|^2)^p dA(z)\right) ^{\frac{1}{2}} \nonumber \\&\quad +\sup _{a\in {\mathbb {D}}}\left( \int _{{\mathbb {D}}}\left( \int _{[0,1)}\frac{|tf(tz)|}{|1-tz|^2}d\mu (t)\right) ^2(1-|\sigma _a(z)|^2)^p dA(z)\right) ^{\frac{1}{2}} \nonumber \\&\lesssim \Vert f\Vert _{\mathcal {B}}\sup _{a\in {\mathbb {D}}} \left( \int _{{\mathbb {D}}}\left( \int _{[0,1)}\frac{1}{(1-|tz|)|1-tz|}d\mu (t)\right) ^2(1-|\sigma _a(z)|^2)^p dA(z)\right) ^{\frac{1}{2}} \nonumber \\&\quad +\Vert f\Vert _{H^\infty }\sup _{a\in {\mathbb {D}}}\left( \int _{{\mathbb {D}}}\left( \int _{[0,1)}\frac{1}{|1-tz|^2}d\mu (t)\right) ^2(1-|\sigma _a(z)|^2)^p dA(z)\right) ^{\frac{1}{2}}.\nonumber \\ \end{aligned}$$

(3.1)

Let c be a positive constant such that $2c<\min \{2-p, p\}$. Then

$$\begin{aligned} (1-|tz|)^2\ge (1-t)^{2-2c} (1-|z|)^{2c} \end{aligned}$$

(3.2)

for all $t\in [0, 1)$ and all $z\in {\mathbb {D}}$. By the Minkowski inequality, (3.2), Lemma F and Proposition 2.1, we get

$$\begin{aligned}&\sup _{a\in {\mathbb {D}}} \left( \int _{{\mathbb {D}}}\left( \int _{[0,1)}\frac{1}{(1-|tz|)|1-tz|}d\mu (t)\right) ^2(1-|\sigma _a(z)|^2)^p dA(z)\right) ^{\frac{1}{2}} \nonumber \\&\quad \le \sup _{a\in {\mathbb {D}}} \int _{[0,1)} \left( \int _{{\mathbb {D}}} \frac{1}{(1-|tz|)^2|1-tz|^2} (1-|\sigma _a(z)|^2)^p dA(z) \right) ^{\frac{1}{2}} d\mu (t)\nonumber \\&\quad \lesssim \sup _{a\in {\mathbb {D}}}(1-|a|^2)^{\frac{p}{2}}\int _{[0,1)}\frac{1}{(1-t)^{1-c}}d\mu (t)\big (\int _{{\mathbb {D}}}\frac{(1-|z|^2)^{p-2c}}{|1-tz|^2|1-{\bar{a}}z|^{2p}} dA(z)\big )^{\frac{1}{2}} \nonumber \\&\quad \lesssim \sup _{a\in {\mathbb {D}}}\int _{[0,1)}\frac{(1-|a|^2)^{\frac{p}{2}}}{(1-t)^{1-c}|1-ta|^{\frac{p}{2}+c}}d\mu (t)<\infty . \end{aligned}$$

(3.3)

Similarly, it follows from Lemma F and Proposition 2.1 that

$$\begin{aligned} \sup _{a\in {\mathbb {D}}}&\left( \int _{{\mathbb {D}}}\left( \int _{[0,1)}\frac{1}{|1-tz|^2}d\mu (t)\right) ^2(1-|\sigma _a(z)|^2)^p dA(z)\right) ^{\frac{1}{2}} \nonumber \\&\quad \le \sup _{a\in {\mathbb {D}}} \int _{[0,1)} \left( \int _{{\mathbb {D}}} \frac{1}{|1-tz|^4} (1-|\sigma _a(z)|^2)^p dA(z)\right) ^{\frac{1}{2}} d\mu (t)\nonumber \\&\quad \lesssim \sup _{a\in {\mathbb {D}}}\int _{[0,1)}\frac{(1-|a|^2)^{\frac{p}{2}}}{(1-t^2)^{1-\frac{p}{2}}|1-at|^p}d\mu (t)<\infty . \end{aligned}$$

(3.4)

From (3.1), (3.3) and (3.4), we get that ${\mathcal {C}}_\mu (f)\in {\mathcal {Q}}_p$. The proof is complete. $\square$

Remark 3

Set $d\mu _0(x)=dx$ on [0, 1). Then $d\mu _0$ is a Carleson measure and ${\mathcal {C}}_{\mu _0}(1)(z)=\frac{1}{z}\log \frac{1}{1-z}$. Clearly, the function ${\mathcal {C}}_{\mu _0}(1)$ is not in the Dirichlet space. Thus Theorem 3.1 does not hold when $p=0$.

Note that ${\mathcal {Q}}_p={\mathcal {B}}$ for any $p>1$. Theorem 3.1 generalizes Theorem 5 in [16] from the Bloch space ${\mathcal {B}}$ to all ${\mathcal {Q}}_p$ spaces. For $p=1$, Theorem 3.1 gives an answer to a question raised in [16, p. 20]. The proof given here highlights the role of Proposition 2.1. In the next section, we give a more general result where an alternative proof of Theorem 3.1 will be provided.

4 s-Carleson measures and the range of another Cesàro-like operator acting on $H^\infty$

It is also natural to consider how the characterization of s-Carleson measures in Proposition 2.2 can play a role in the investigation of the range of Cesàro-like operators acting on $H^\infty$. We consider this topic by another kind of Cesàro-like operators.

Suppose $0<s<\infty$ and $\mu$ is a finite positive Borel measure on [0, 1). For $f(z)=\sum _{n=0}^\infty a_nz^n$ in $H({\mathbb {D}})$, we define

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

Clearly, ${\mathcal {C}}_{\mu , 1}$ is equal to ${\mathcal {C}}_{\mu }$.

Lemma 4.1

Suppose $0<s<\infty$ and $\mu$ is a finite positive Borel measure on [0, 1). Then

$$\begin{aligned} {\mathcal {C}}_{\mu , s} (f)(z)=\int _{[0,1)}\frac{f(tz)}{(1-tz)^s}d\mu (t) \end{aligned}$$

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

Proof

The proof follows from a simple calculation with power series. We omit it. $\square$

We have the following result.

Theorem 4.2

Suppose $0<s<\infty$ and $\mu$ is a finite positive Borel measure on [0, 1). Let $\max \{1, \frac{1}{s}\}<p<\infty$ and let X be a subspace of $H({\mathbb {D}})$ with $\Lambda ^p_{1/p}\subseteq X\subseteq {\mathcal {B}}$. Then ${\mathcal {C}}_{\mu , s}(H^\infty )\subseteq X$ if and only if $\mu$ is an s-Carleson measure.

Proof

Let ${\mathcal {C}}_{\mu , s}(H^\infty )\subseteq X$. Then ${\mathcal {C}}_{\mu , s}(1)\in X$; that is, $f_{\mu , s}\in X$. It follows from Proposition 2.2 that $\mu$ is an s-Carleson measure.

On the other hand, let $\mu$ be an s-Carleson measure and $f\in H^\infty$. By Lemma 4.1, we see

$$\begin{aligned} {\mathcal {C}}_{\mu , s}(f)'(z)=\int _{[0,1)}\frac{tf'(tz)}{(1-tz)^s}d\mu (t)+ \int _{[0,1)}\frac{stf(tz)}{(1-tz)^{s+1}}d\mu (t), \quad z\in {\mathbb {D}}. \end{aligned}$$

Then

$$\begin{aligned}&\sup _{0<r<1}(1-r)^{1-\frac{1}{p}}\left( \frac{1}{2\pi }\int _0^{2\pi }|{\mathcal {C}}_{\mu , s}(f)'(re^{i\theta })|^p d\theta \right) ^{\frac{1}{p}}\nonumber \\&\quad \lesssim \Vert f\Vert _{\mathcal {B}}\sup _{0<r<1}(1-r)^{1-\frac{1}{p}} \left( \frac{1}{2\pi }\int _0^{2\pi }\left( \int _{[0,1)} \frac{1}{|1-tre^{i\theta }|^{s}(1-tr)} d\mu (t)\right) ^p d\theta \right) ^{\frac{1}{p}} \nonumber \\&\qquad +\Vert f\Vert _{H^\infty }\sup _{0<r<1}(1-r)^{1-\frac{1}{p}} \left( \frac{1}{2\pi }\int _0^{2\pi }\left( \int _{[0,1)} \frac{1}{|1-tre^{i\theta }|^{s+1}} d\mu (t)\right) ^p d\theta \right) ^{\frac{1}{p}}. \nonumber \\ \end{aligned}$$

(4.1)

Note that $ps>1$. By the Minkowski inequality, Lemma E and Lemma A, we deduce

$$\begin{aligned} \sup _{0<r<1}&(1-r)^{1-\frac{1}{p}} \left( \frac{1}{2\pi }\int _0^{2\pi }\left( \int _{[0,1)} \frac{1}{|1-tre^{i\theta }|^{s}(1-tr)} d\mu (t)\right) ^p d\theta \right) ^{\frac{1}{p}} \nonumber \\&\quad \le \sup _{0<r<1}(1-r)^{1-\frac{1}{p}} \int _{[0,1)} \left( \frac{1}{2\pi }\int _0^{2\pi } \frac{1}{|1-tre^{i\theta }|^{sp}(1-tr)^p} d\theta \right) ^{\frac{1}{p}} d\mu (t) \nonumber \\&\quad \lesssim \sup _{0<r<1}(1-r)^{1-\frac{1}{p}} \int _{[0,1)} \frac{1}{(1-tr)^{s+1-\frac{1}{p}}} d\mu (t)<\infty , \nonumber \\ \end{aligned}$$

(4.2)

and

$$\begin{aligned} \sup _{0<r<1}&(1-r)^{1-\frac{1}{p}} \left( \frac{1}{2\pi }\int _0^{2\pi }\left( \int _{[0,1)} \frac{1}{|1-tre^{i\theta }|^{s+1}} d\mu (t)\right) ^p d\theta \right) ^{\frac{1}{p}}\nonumber \\&\quad \lesssim \sup _{0<r<1}(1-r)^{1-\frac{1}{p}} \int _{[0,1)} \left( \frac{1}{2\pi }\int _0^{2\pi }\frac{1}{|1-tre^{i\theta }|^{(s+1)p} } d\theta \right) ^{\frac{1}{p}} d\mu (t) \nonumber \\&\quad \lesssim \sup _{0<r<1}(1-r)^{1-\frac{1}{p}} \int _{[0,1)} \frac{1}{(1-tr)^{s+1-\frac{1}{p}}} d\mu (t)<\infty . \end{aligned}$$

(4.3)

From (4.1), (4.2) and (4.3), ${\mathcal {C}}_{\mu , s}(f)\in \Lambda ^p_{1/p}$. Note that $\Lambda ^p_{1/p}\subseteq X$. The desired result follows. $\square$