Cesàro-like operators acting on spaces of analytic functions

Abstract

Let ${\mathbb {D}}$ be the unit disc in ${\mathbb {C}}$. If $\mu$ is a finite positive Borel measure on the interval [0, 1) and f is an analytic function in ${\mathbb {D}}$, $f(z)=\sum _{n=0}^\infty a_nz^n$ ($z\in {\mathbb {D}}$), we define

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

where, for $n\ge 0$, $\mu _n$ denotes the n-th moment of the measure $\mu$, that is, $\mu _n=\int _{[0, 1)}t^nd\mu (t).$ In this way, ${\mathcal {C}}_\mu$ becomes a linear operator defined on the space ${\mathrm{Hol}}({\mathbb {D}})$ of all analytic functions in ${\mathbb {D}}$. We study the action of the operators ${\mathcal {C}}_\mu$ on distinct spaces of analytic functions in ${\mathbb {D}}$, such as the Hardy spaces $H^p$, the weighted Bergman spaces $A^p_\alpha$, BMOA, and the Bloch space ${\mathcal {B}}$.

1 Introduction and main results

Let ${{\mathbb {D}}}=\{z\in {{\mathbb {C}}}: |z|<1\}$ denote the open unit disc in the complex plane ${{\mathbb {C}}}$ and let ${\mathrm{Hol}}({{\mathbb {D}}})$ be the space of all analytic functions in ${{\mathbb {D}}}$ endowed with the topology of uniform convergence in compact subsets.

If $\,0<r<1\,$ and $\,f\in {\mathrm{Hol}}({{\mathbb {D}}})$, we set

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

For $\,0<p\le \infty$,  the Hardy space $H^p$ consists of those $f\in {\mathrm{Hol}}({\mathbb {D}})$ such that

$$\begin{aligned}\Vert f\Vert _{H^p}{\mathop {=}\limits ^{\text {def}}}\sup _{0<r<1}M_p(r,f)<\infty . \end{aligned}$$

We refer to [13] for the notation and results regarding Hardy spaces.

Let dA denote the area measure on ${\mathbb {D}}$, normalized so that the area of ${\mathbb {D}}$ is 1. Thus $dA(z)\,=\,\frac{1}{\pi }\,dx\,dy\,=\,\frac{1}{\pi }\,r\,dr\,d\theta$. For $0<p<\infty$ and $\alpha >-1$ the weighted Bergman space $A^p_\alpha$ consists of those $f\in {\mathrm{Hol}}({\mathbb {D}})$ such that

$$\begin{aligned} \Vert f\Vert _{A^p_\alpha }\,{\mathop {=}\limits ^{\text {def}}}\, \left( \int _{\mathbb {D}}\vert f (z)\vert ^p\,dA_\alpha (z)\right) ^{1/p}\,<\,\infty , \end{aligned}$$

where $dA_\alpha (z)=(\alpha +1)(1-\vert z\vert ^2)^{\alpha }dA(z)$. We refer to [14, 25, 39] for the notation and results about Bergman spaces.

The space BMOA consists of those functions $f\in H^1$ whose boundary values have bounded mean oscillation on $\partial \mathbb D$. We refer to [16] for the theory of BMOA-functions.

Finally, we recall that a function $f\in {\mathrm{Hol}}({\mathbb {D}})$ is said to be a Bloch function if

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

The space of all Bloch functions is denoted by ${\mathcal {B}}$. A classical reference for the theory of Bloch functions is [2]. Let us recall that

$$\begin{aligned}H^\infty \subsetneq BMOA\subsetneq {\mathcal {B}},\qquad BMOA\subsetneq \bigcap _{0<p<\infty }{H^p},\qquad {\mathcal {B}}\subsetneq A^p_\alpha \,\, (p>0,\,\,\alpha >-1). \end{aligned}$$

The Cesàro operator ${\mathcal {C}}$ is defined over the space of all complex sequences as follows: If $(a)=\{ a_k\} _{k=0}^\infty$ is a sequence of complex numbers then

$$\begin{aligned} {\mathcal {C}} \left( (a)\right) =\left\{ \frac{1}{n+1}\sum _{k=0}^na_k\right\} _{n=0}^\infty . \end{aligned}$$

The operator ${\mathcal {C}}$ is known to be bounded from $\ell ^p$ to $\ell ^p$ for $1<p<\infty$. In fact, the sharp inequalities

$$\begin{aligned} \Vert {\mathcal {C}}\left( (a)\right) \Vert _{p}\le \frac{p}{p-1}\Vert (a)\Vert _{p},\quad (a)\in \ell ^p,\quad 1<p<\infty , \end{aligned}$$

were proved by Hardy [21] and Landau [29] (see also [24, Theorem 326, p.239 ]).

Identifying any given function $f\in {\mathrm{Hol}}({\mathbb {D}})$ with the sequence $\{ a_k\}_{k=0}^\infty$ of its Taylor coefficients, the Cesàro operator ${\mathcal {C}}$ becomes a linear operator from ${\mathrm{Hol}}({\mathbb {D}})$ into itself as follows:

If $f\in {\mathrm{Hol}}({\mathbb {D}})$, $f(z)=\sum _{k=0}^\infty a_kz^k$  ($z\in {\mathbb {D}}$), then

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

The Cesàro operator is bounded on $H^p$ for $0<p<\infty$. For $1<p<\infty$, this follows from a result of Hardy on Fourier series [22] together with the M. Riesz’s theorem on the conjugate function [13, Theorem 4.1]. Siskakis [33] used semigroups of composition operators to give an alternative proof of this result and to extend it to $p=1$. A direct proof of the boundedness on $H^1$ was given by Siskakis in [34]. Miao [31] dealt with the case $0<p<1$. Stempak [36] gave a proof valid for $0<p\le 2$ and Andersen [1] provided another proof valid for all $p<\infty$.

In this paper we associate to every positive finite Borel measure on [0, 1) a certain operator ${\mathcal {C}}_\mu$ acting on ${\mathrm{Hol}}({\mathbb {D}})$ which is a natural generalization of the classical Cesàro operator ${\mathcal {C}}$.

If $\mu$ is a positive finite Borel measure on [0, 1) and n is a non-negative integer, we let $\mu _n$ denote the moment of order n of $\mu$, that is,

$$\begin{aligned}\mu _n=\int _{[0,1)}\,t^n\,d\mu (t),\quad n=0, 1, 2, \dots .\end{aligned}$$

If $f\in {\mathrm{Hol}}({\mathbb {D}})$, $f(z)=\sum _{n=0}^\infty a_nz^n$ ($z\in \mathbb D$), we define ${\mathcal {C}}_\mu (f)$ as follows

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

It is clear that $\mathcal C_\mu$ is a well defined linear operator ${\mathcal {C}}_\mu :{\mathrm{Hol}}({\mathbb {D}})\rightarrow {\mathrm{Hol}}({\mathbb {D}})$. When $\mu$ is the Lebesgue measure on [0, 1), the operator ${\mathcal {C}}_\mu$ reduces to the classical Cesàro operator ${\mathcal {C}}$.

Our main objective in this work is to characterize those positive finite Borel measures $\mu$ on [0, 1) such that the operator ${\mathcal {C}}_\mu$ is bounded or compact on classical subspaces of ${\mathrm{Hol}}({\mathbb {D}})$ such as the Hardy spaces $H^p$, the weighted Bergman spaces $A^p_\alpha$, and the spaces BMOA and ${\mathcal {B}}$.

Measures of Carleson type will play a basic role in the sequel. If $I\subset \partial {{\mathbb {D}}}$ is an interval, $\vert I\vert$ will denote the length of I. The Carleson square S(I) is defined as

$$\begin{aligned} S(I)=\left\{ re^{it}:\,e^{it}\in I,\quad 1-\frac{|I|}{2\pi }\le r <1\right\} . \end{aligned}$$

If $\, s>0$ and $\mu$ is a positive Borel measure on ${{\mathbb {D}}}$, we shall say that $\mu$ is an s-Carleson measure if there exists a positive constant C such that

$$\begin{aligned} \mu \left( S(I)\right) \le C{|I|^s}, \quad \hbox {for any interval} I\subset \partial {{\mathbb {D}}}. \end{aligned}$$

If $\mu$ satisfies $\mu \left( S(I)\right) \,=\,{{\,\mathrm{o}\,}}\left( |I|^s\right)$, as $\vert I\vert \rightarrow 0$, then we say that $\mu$ is a  vanishing s-Carleson measure.

A 1-Carleson measure, respectively, a vanishing 1-Carleson measure, will be simply called a Carleson measure, respectively, a vanishing Carleson measure.

We recall that Carleson [7] proved that $H^p\,\subset \,L^p(d\mu )$ ($0<p<\infty$), if and only if $\mu$ is a Carleson measure (see [13, Chapter 9]).

Following [38], if $\mu$ is a positive Borel measure on ${{\mathbb {D}}}$, $0\le \alpha <\infty$, and $0<s<\infty$, we say that $\mu$ is an $\alpha$-logarithmic s-Carleson measure if there exists a positive constant C such that

$$\begin{aligned}\frac{ \mu \left( S(I)\right) \left( \log \frac{2}{\vert I\vert }\right) ^\alpha }{|I|^s}\le C, \quad \hbox {for any interval} I\subset \partial {{\mathbb {D}}}. \end{aligned}$$

If $\mu \left( S(I)\right) \left( \log \frac{2}{\vert I\vert }\right) ^\alpha \,=\,{{\,\mathrm{o}\,}}\left( |I|^s\right)$, as $\vert I\vert \rightarrow 0$, we say that $\mu$ is a vanishing $\alpha$-logarithmic s-Carleson measure.

A measure $\mu$ on [0, 1) can be seen as a measure on ${\mathbb {D}}$ with support contained in the radius [0, 1). In this way, a positive Borel measure $\mu$ on [0, 1) is an s-Carleson measure if and only if there exists a positive constant C such that

$$\begin{aligned} \mu \left( [t,1)\right) \le C(1-t)^s, \quad 0\le t<1, \end{aligned}$$

and we have similar statements for vanishing s-Carleson measures, for $\alpha$-logarithmic s-Carleson measures, and for vanishing $\alpha$-logarithmic s-Carleson measures.

Among other, we shall prove the following results.

Theorem 1

Suppose that $1\le p<\infty$ and let $\mu$ be a positive finite Borel measure on [0, 1). Then the following conditions are equivalent.

• (i) The measure $\mu$ is a Carleson measure.

• (ii) The operator ${\mathcal {C}}_\mu$ is bounded from $H^p$ into itself.

Theorem 2

Suppose that $1\le p<\infty$ and let $\mu$ be a positive finite Borel measure on [0, 1). Then the following conditions are equivalent.

• (i) The measure $\mu$ is a vanishing Carleson measure.

• (ii) The operator ${\mathcal {C}}_\mu$ is compact from $H^p$ into itself.

Danikas and Siskakis [12] observed that ${\mathcal {C}}(H^\infty )\not \subset H^\infty$ and ${\mathcal {C}}(BMOA)\not \subset BMOA$ and studied the action of the the Cesàro operator on these spaces. We will devote Sects. 3.3 and 5 to study the Cesàro-like operators ${\mathcal {C}}_\mu$ acting on these spaces. Let us just mention here the following result.

Theorem 3

Let $\mu$ be a positive finite Borel measure on [0, 1). Then the following conditions are equivalent.

• (i) The measure $\mu$ is a 1-logarithmic 1-Carleson measure.

• (ii) The operator ${\mathcal {C}}_\mu$ is bounded from BMOA into itself.

• (iii) The operator ${\mathcal {C}}_\mu$ is bounded from the Bloch space ${\mathcal {B}}$ into itself.

Section 3 will be devoted to present the proofs of Theorem 1 and Theorem 2 as well as some further results concerning the action of the operators $\mathcal C_\mu$ on Hardy spaces. Section 4 will deal with the action of the operators ${\mathcal {C}}_\mu$ on Bergman spaces and, as we have already mentioned, Sect. 5 will be devoted to study the operators ${\mathcal {C}}_\mu$ acting on BMOA, the Bloch space, and some related spaces. In particular, Sect. 5 will include a proof of Theorem 3 and the substitute of this result concerning compactness.

In Sect. 2 we shall give two alternative representations of the operator ${\mathcal {C}}_\mu$, one of them is an integral representation and the other one involves the convolution with a fixed analytic function in ${\mathbb {D}}$. We shall also introduce a related operator which will be denoted $T_\mu$ and which will play a basic role in the proofs of some of our results.

Throughout the paper, if $\mu$ is a finite positive Borel measure on [0, 1), for $n\ge 0$, $\mu _n$ will denote the moment of order n of $\mu$. Also, we shall be using the convention that $C=C(p, \alpha ,q,\beta , \dots )$ will denote a positive constant which depends only upon the displayed parameters $p, \alpha , q, \beta \dots$ (which sometimes will be omitted) but not necessarily the same at different occurrences. Furthermore, for two real-valued functions $K_1, K_2$ we write $K_1\lesssim K_2$, or $K_1\gtrsim K_2$, if there exists a positive constant C independent of the arguments such that $K_1\le C K_2$, respectively $K_1\ge C K_2$. If we have $K_1\lesssim K_2$ and $K_1\gtrsim K_2$ simultaneously, then we say that $K_1$ and $K_2$ are equivalent and we write $K_1\asymp K_2$.

Let us close this section noticing that, since the subspaces X of ${\mathrm{Hol}}({\mathbb {D}})$ we shall be dealing with are Banach spaces continuously embedded in ${\mathrm{Hol}}({\mathbb {D}})$, to prove that the operator ${\mathcal {C}}_\mu$ (or $T_\mu$, to be defined below) is bounded on X it suffices to show that it maps X into X by appealing to the closed graph theorem.

2 Alternative representations of ${\mathcal {C}}_\mu$ and a related operator

A simple calculation with power series gives the following integral representation of the operators ${\mathcal {C}}_\mu$.

Proposition 1

If $\mu$ is a positive finite Borel measure on [0, 1) and $f\in {\mathrm{Hol}}({\mathbb {D}})$ then

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

(1)

Next we shall give another expression for $\mathcal C_\mu (f)$ involving the convolution of analytic functions. If f and g are two analytic functions in the unit disc,

$$\begin{aligned} f(z)=\sum _{n=0}^\infty a_nz^n,\quad g(z)=\sum _{n=0}^\infty b_nz^n,\quad z\in {\mathbb {D}}, \end{aligned}$$

the convolution $f\star g$ of f and g is defined by

$$\begin{aligned} f\star g(z)=\sum _{n=0}^\infty a_nb_nz^n,\quad z\in {\mathbb {D}}. \end{aligned}$$

Lemma 1

Let $\mu$ be a positive finite Borel measure on [0, 1) and set

$$\begin{aligned} F(z)=\sum _{n=0}^\infty \mu _nz^n,\quad z\in {\mathbb {D}}. \end{aligned}$$

If $f\in {\mathrm{Hol}}({\mathbb {D}})$ and

$$\begin{aligned} g(z)=\frac{f(z)}{1-z},\quad z\in {\mathbb {D}}, \end{aligned}$$

then ${\mathcal {C}}_\mu (f)=F\star g$.

The proof is elementary and will be omitted.

The following result regarding the radial measures $\mu$ we are considering will be used in our work.

Lemma 2

Let $\mu$ be a finite positive Borel measure on the interval [0, 1) and, for $n\ge 0$, let $\mu _n$ denote the moment of order n of $\mu$.

• (i) $\mu$ is a Carleson measure if and only if $\mu _n={{\,\mathrm{O}\,}}(\frac{1}{n})$.

• (ii) $\mu$ is a vanishing Carleson measure if and only if $\mu _n={{\,\mathrm{o}\,}}(\frac{1}{n}$).

• (iii) $\mu$ is a 1-logarithmic 1-Carleson measure if and only if $\mu _n={{\,\mathrm{O}\,}}(\frac{1}{n\log n})$.

• (iv) $\mu$ is a vanishing 1 logarithmic 1-Carleson measure if and only if $\mu _n={{\,\mathrm{o}\,}}(\frac{1}{n\log n})$.

Proof

(i) is Proposition 8 of [8] and (ii) follows with a similar argument. Lemma 2. 7 of [19] gives one implication of (iii) and the other one follows from the from the simple inequality

$$\begin{aligned} \mu \left( \left[ 1-\frac{1}{n}, 1\right) \right) \lesssim \int _{\left[ 1-\frac{1}{n}, 1\right) }t^nd\mu (t)\le \mu _n. \end{aligned}$$

Finally, (iv) can be proved with an argument similar the the one used to prove (iii). $\square$

Now we define a new operator operator $T_\mu$ associated to $\mu$ which will be important in our work because it will become the adjoint of ${\mathcal {C}}_\mu$ in distinct instances.

If $\mu$ is a finite positive Borel measure on [0, 1) and $f\in {\mathrm{Hol}}({\mathbb {D}})$, $f(z)=\sum _{n=0}^\infty a_nz^n$ ($z\in {\mathbb {D}}$) we set

$$\begin{aligned} T_\mu (f)(z)= \sum _{n=0}^\infty \left( \sum _{k=n}^\infty \mu _ka_k\right) z^n, \end{aligned}$$

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

Clearly, the operator $T_\mu$ is not defined over the whole space ${\mathrm{Hol}}({\mathbb {D}})$. We have the following result.

Proposition 2

Let $\mu$ is a finite positive Borel measure on [0, 1).

• (a) If P is a polynomial then $T_\mu (P)$ is well defined and it also a polynomial.

• (b) If $\mu$ is a Carleson measure then $T_\mu$ is well defined on $H^1$.

Proof

(a) is clear. To prove (b) we use the fact that if $\mu$ is a Carleson measure then $\mu _n={{\,\mathrm{O}\,}}\left( n^{-1}\right)$ (see Lemma 2). This and Hardy’s inequality [13, p. 48] shows that if $f\in H^1$, $f(z)=\sum _{k=1}^\infty a_kz^k$, then there exists $C>0$ such that

$$\begin{aligned} \sum _{k=n}^\infty \mu _k\vert a_k\vert \le C\sum _{k=n}^\infty \frac{\vert a_k\vert }{k+1}\le C\pi \Vert f\Vert _{H^1} \end{aligned}$$

for all n. Clearly, this implies (b). $\square$

It is well known that, for $1<p<\infty$, the dual of $H^p$ is identifiable with $H^q$, $\frac{1}{p}+\frac{1}{q}=1$, with the pairing

$$\begin{aligned} <f, g>_{H^p}\,=\,\frac{1}{2\pi }\int _0^{2\pi }f(e^{i\theta })\overline{g(e^{i\theta })}d\theta =\sum _{n=0}^\infty a_n\overline{b_n} \end{aligned}$$

where $f(z)=\sum _{n=0}^\infty a_nz^n\in H^p$ and $g(z)=\sum _{n=0}^\infty b_nz^n\in H^q$ (see [13, Theorem 7.3]).

Similarly, if $1<p<\infty$ and $\alpha >1$, the dual of $A^p_\alpha$ is identifiable with $A^q_{\alpha }$ with the pairing

$$\begin{aligned} <f, g>_{p,\alpha }\,=\,\int _{{\mathbb {D}}}f(z)\overline{g(z)}dA_\alpha (z)=\sum _{n=0}^\infty c_{n,\alpha }a_n\overline{b_n}, \end{aligned}$$

where

$$\begin{aligned}c_{n,\alpha }=\frac{n!\,\varGamma (2+\alpha )}{\varGamma (n+2+\alpha )},\quad n=0, 1, 2, \dots , \end{aligned}$$

and $f(z)=\sum _{n=0}^\infty a_nz^n\in A^p_\alpha$, $g(z)=\sum _{n=0}^\infty b_nz^n\in A^q_\alpha$ (see [25, Theorem 1.16 and p. 5]). A simple calculation gives the following result.

Proposition 3

Let $\mu$ be a positive finite Borel measure on [0, 1).

• (i) If  $1<p<\infty$, $f\in H^p$, and g is a polynomial then

  $$\begin{aligned}<{\mathcal {C}}_\mu (f), g>_{H^p}\,=\,<f, T_\mu (g)>_{H^p}. \end{aligned}$$

• (ii) If  $1<p<\infty$, $\alpha >-1$, $f\in A^p_\alpha$, and g is a polynomial then

  $$\begin{aligned}<{\mathcal {C}}_\mu (f), g>_{p,\alpha }\,=\,<f, T_\mu (g)>_{p,\alpha }.\end{aligned}$$

Proposition 3, together with the fact that the polynomials are dense in all the spaces $H^p$ ($p<\infty$) and $A^p_\alpha$ ($p<\infty ,\,\alpha >-1$), readily implies the following result.

Proposition 4

Suppose that $1<p<\infty$ and let $\mu$ be a positive finite Borel measure on [0, 1). Let q be the conjugate exponent of p, that is, $\frac{1}{p}+\frac{1}{q}=1$.

• (i) If ${\mathcal {C}}_\mu$ is a bounded operator from $H^p$ into itself, then there exists a positive constant C such that

  $$\begin{aligned} \Vert T_\mu (P)\Vert _{H^q}\le C\Vert P\Vert _{H^q} \end{aligned}$$

  for every polynomial P. Consequently, $T_\mu$ extends to a bounded linear operator from $H^q$ into itself. This extension, which will be also denoted by $T_\mu$, is the adjoint of ${\mathcal {C}}_\mu$.

• (ii) Suppose that $\alpha >-1$. If ${\mathcal {C}}_\mu$ is a bounded operator from $A^p_\alpha$ into itself, then there exists a positive constant C such that

  $$\begin{aligned} \Vert T_\mu (P)\Vert _{A^q_\alpha }\le C\Vert P\Vert _{A^q_\alpha } \end{aligned}$$

  for every polynomial P. Consequently, $T_\mu$ extends to a bounded linear operator from $A^q_\alpha$ into itself. This extension, which will be also denoted by $T_\mu$, is the adjoint of ${\mathcal {C}}_\mu$.

3 The operators ${\mathcal {C}}_\mu$ acting on Hardy spaces

In this section we shall study the action of the operators $\mathcal C_\mu$ on Hardy spaces.

We shall use complex interpolation to prove some of our results. Let us refer to [39, Chapter 2] for the terminology and basic results concerning complex interpolation.

If $X_0$ and $X_1$ are two compatible Banach spaces then, for $0<\theta <1$, $(X_0, X_1)_{\theta }$ stands for the space obtained by the complex method of interpolation of Calderón [5]. It is well known (see [6, 26, 32]) that if $1\le p_0, p_1\le \infty$, $0<\theta <1$, and $1/p=(1-\theta )/p_0+\theta /p_1$, then

$$\begin{aligned} (H^{p_0}, H^{p_1})_{\theta }=H^p. \end{aligned}$$

(2)

In particular,

$$\begin{aligned} H^p=(H^2, H^1)_\theta , \quad \text {if }1<p<2\text { and }\theta =\frac{2}{p}-1. \end{aligned}$$

(3)

3.1 Proof of Theorem 1

We shall split it in several cases.

Proof of the implication (i)  $\Rightarrow$  (ii) when $p=1$. Assume that $\mu$ is a Carleson measure and take $f\in H^1$. Set

$$\begin{aligned} g(z)=\frac{f(z)}{1-z},\quad z\in {\mathbb {D}}, \end{aligned}$$

and

$$\begin{aligned} t_k=1-\frac{1}{2^k},\quad k=0, 1, 2, \dots \end{aligned}$$

Using the integral representation on ${\mathcal {C}}_\mu$, we see that, for $0<r<1$,

$$\begin{aligned}M_1\left( r, {\mathcal {C}}_\mu (f)\right)&=\frac{1}{2\pi }\int _0^{2\pi }\left| \int _{[0,1)}\frac{f(rte^{i\theta })}{1-rte^{i\theta }}\,d\mu (t)\right| \,d\theta \\&\le \frac{1}{2\pi }\int _0^{2\pi }\int _{[0,1)}\left| g(rte^{i\theta })\right| \,d\mu (t)\,d\theta \\&=\frac{1}{2\pi }\int _0^{2\pi }\left( \sum _{k=1}^\infty \int _{[t_{k-1},t_k)}\left| g(rte^{i\theta })\right| \,d\mu (t)\right) \,d\theta \\&\le \frac{1}{2\pi }\int _0^{2\pi }\left( \sum _{k=1}^\infty \left[ \sup _{0\le t\le t_k}\left| g(rte^{i\theta })\right| \right] \right) \mu \left( [t_{k-1},t_k]\right) \,d\theta . \end{aligned}$$

Since $\mu$ is a Carleson measure, $\mu \left( [t_{k-1},t_k]\right) \lesssim \frac{1}{2^k}$. Using this, the Hardy-Littlewood maximal theorem [13, Theorem 1.9], the fact that integral means $M_1(s,g)$ increase with s, and the Cauchy-Schwarz inequality, we obtain

$$\begin{aligned} M_1\left( r, {\mathcal {C}}_\mu (f)\right)&\le \sum _{k=1}^\infty \frac{1}{2^k}\left( \frac{1}{2\pi }\int _0^{2\pi }\left[ \sup _{0\le t\le t_k}\left| g(rte^{i\theta })\right| \right] \,d\theta \right) \\&\lesssim \sum _{k=1}^\infty \frac{1}{2^k}M_1(rt_k,g) \\&\lesssim \sum _{k=1}^\infty \frac{1}{2^k}2^k\int _{t_k}^{t_{k+1}}M_1(rt,g)\,dt \\&\lesssim \int _0^r\frac{1}{2\pi }\int _0^{2\pi }\left| g(te^{i\theta })\right| \,d\theta \,dt \\&\lesssim \int _0^r\frac{1}{2\pi }\int _0^{2\pi }\left| \frac{f(te^{i\theta })}{1-te^{i\theta }}\right| \,d\theta \,dt \\&\lesssim \int _0^rM_2(t,f)\left( \frac{1}{2\pi }\int _0^{2\pi }\frac{d\theta }{\vert 1-te^{i\theta }\vert ^2}\,d\theta \right) ^{1/2}dt\\&\lesssim \int _0^rM_2(t,f)(1-t)^{-1/2}\,dt. \end{aligned}$$

Making the change of variables $t=rs$ in the last integral and setting $f_r(z)=f(rz)$ ($z\in {\mathbb {D}}$), it follows that

$$\begin{aligned}&M_1\left( r, {\mathcal {C}}_\mu (f)\right) \,\lesssim \,\int _0^1M_2(sr,f)(1-sr)^{-1/2}\,ds\,\\&\quad =\int _0^1M_2(s,f_r)(1-sr)^{-1/2}\,ds\, \le \int _0^1M_2(s,f_r)(1-s)^{-1/2}\,ds. \end{aligned}$$

Using a result of Hardy and Littlewood [23] (see also [34]) we see that

$$\begin{aligned} \int _0^1M_2(s,f_r)(1-s)^{-1/2}\,dt\lesssim \Vert f_r\Vert _{H^1}. \end{aligned}$$

Then it follows that

$$\begin{aligned} M_1\left( r, {\mathcal {C}}_\mu (f)\right) \lesssim M_1(r, f). \end{aligned}$$

(4)

This implies that ${\mathcal {C}}_\mu (f)\in H^1$ and that $\Vert \mathcal C_\mu (f)\Vert _{H^1}\lesssim \Vert f\Vert _{H^1}$. $\square$

Proof of the implication (i)  $\Rightarrow$  (ii) when $p=2$. Assume that $\mu$ is a Carleson measure and take $f\in H^2$, $f(z)=\sum _{n=0}^\infty a_nz^n$ ($z\in {\mathbb {D}}$). Using [8, Proposition 1] we see that $\vert \mu _n\vert \lesssim \frac{1}{n+1}$. Using this, the definition of $\mathcal C_\mu (f)$, and the fact that the Cesàro operator is bounded on $H^2$, it follows that

$$\begin{aligned} \Vert {\mathcal {C}}_\mu (f)\Vert _{H^2}^2&= \sum _{n=0}^\infty {\mu _n^2}\left| \sum _{k=0}^na_k\right| ^2 \,\lesssim \, \sum _{n=0}^\infty \frac{1}{(n+1)^2}\left| \sum _{k=0}^na_k\right| ^2 \\&= \Vert {\mathcal {C}}(f)\Vert _{H^2}^2 \,\lesssim \, \Vert f\Vert _{H^2}^2. \end{aligned}$$

$\square$

Proof of the implication (i)  $\Rightarrow$  (ii) for $1<p<2$. Since (i)  $\Rightarrow$  (ii) when $p=1$ and $p=2$, the fact that (i)  $\Rightarrow$  (ii) when $1<p<2$ follows using (3) and Theorem 2. 4 of [39]. $\square$

To prove the remaining case, that is, the implication (i)  $\Rightarrow$  (ii) for $2<p<\infty$ we shall use ideas of Andersen [1]. Actually, our next argument works for $1<p<\infty$.

Proof of the implication (i)  $\Rightarrow$  (ii) for $1<p<\infty$. Assume that $\mu$ is a Carleson measure, $1<p<\infty$, and $f\in H^p$.

For $0<r<1$, set

$$\begin{aligned}K_{r,\mu }(\theta ,\varphi )=\int _{[0,1)}\frac{(1+t)(1-t)}{\vert 1-te^{i\varphi }\vert ^2 (1-tre^{i\theta })}\,d\mu (t),\quad \theta , \varphi \in [-\pi , \pi ]. \end{aligned}$$

Arguing just as in [1, p. 621], using Fubini’s theorem, we have that

$$\begin{aligned} {\mathcal {C}}_\mu (f)(re^{i\theta })=\int _{-\pi }^{\pi }K_{r,\mu }(\theta ,\varphi )f(re^{i(\theta +\varphi )})\,d\varphi . \end{aligned}$$

(5)

Now, letting $\{ t_k\} _{k=0}^\infty$ be as above, using the fact that $\mu ([t_k,t_{k+1}))\lesssim \frac{1}{2^k}$, and simple estimates, we obatin

$$\begin{aligned} \vert K_{r,\mu }(\theta ,\varphi )\vert&\le 2\int _{[0,1)}\frac{1-t}{\vert 1-te^{i\varphi }\vert ^2 \vert 1-tre^{i\theta }\vert }\,d\mu (t) \\&\lesssim \int _{[0,1)}\frac{1-t}{[(1-t)^2+\varphi ^2][(1-t)^2+\theta ^2]^{1/2}}\,d\mu (t) \\&\lesssim \sum _{k=0}^\infty \int _{t_k}^{t_{k+1}}\frac{1-t}{[(1-t)^2+\varphi ^2][(1-t)^2+\theta ]^{1/2}}\,d\mu (t) \\&\lesssim \sum _{k=0}^\infty \frac{1}{2^k}\frac{\frac{1}{2^k}}{\left[ \left( \frac{1}{2^{k+1}}\right) ^2+\varphi ^2 \right] \left[ \left( \frac{1}{2^{k+1}}\right) ^2+\theta ^2 \right] ^{1/2}}\\&\lesssim \sum _{k=0}^\infty \int _{t_k}^{t_{k+1}}\frac{\frac{1}{2^k}}{\left[ \left( \frac{1}{2^{k+1}}\right) ^2+\varphi ^2 \right] \left[ \left( \frac{1}{2^{k+1}}\right) ^2+\theta ^2 \right] ^{1/2}}\,dt \\&\lesssim \int _0^1\frac{1-t}{[(1-t)^2+\varphi ^2][(1-t)^2+\theta ^2]^{1/2}}\,dt \\&=\int _0^1\frac{x}{[x^2+\varphi ^2][x^2+\theta ^2]^{1/2}}\,dx \end{aligned}$$

Then, using Lemma 2.1 of [1], we see that for all $\theta , \varphi \in (-\pi ,\pi )\setminus \{ 0\}$ and $r\in (0,1)$, we have

$$\begin{aligned} \vert K_{r,\mu }(\theta ,\varphi )\vert \lesssim \frac{H(\varphi /\theta )}{\vert \theta \vert }, \end{aligned}$$

where

$$\begin{aligned} H(s)=\frac{\log (2+1/\vert s\vert )}{1+\vert s\vert },\quad s\ne 0. \end{aligned}$$

Using this and (5) it follows that

$$\begin{aligned}\vert {\mathcal {C}}_\mu (f)(re^{i\theta })\vert \lesssim \int _{-\pi }^\pi \frac{H(\varphi /\theta )}{\vert \theta \vert }\vert f(re^{i(\theta +\varphi )})\vert \,d\varphi ,\quad \theta \in (-\pi ,\pi )\setminus \{ 0\} ,\,\,0<r<1. \end{aligned}$$

Then the argument in p. 622 of [1] yields that

$$\begin{aligned} M_p(r,{\mathcal {C}}_\mu (f)) \lesssim M_p(r,f) \end{aligned}$$

(6)

and, hence $\Vert {\mathcal {C}}_\mu (f)\Vert _{H^p}\lesssim \Vert f\Vert _{H^p}.$ $\square$

Proof of the implication (ii)  $\Rightarrow$  (i) for $1\le p\le 2$. Suppose that $1\le p\le 2$ and that ${\mathcal {C}}_\mu$ is bounded on $H^p$. Recall that, for $\alpha >0$,

$$\begin{aligned}\frac{1}{(1-z)^\alpha }=\sum _{n=0}^\infty a_n(\alpha )z^n,\quad z\in {\mathbb {D}} \end{aligned}$$

where

$$\begin{aligned} a_n(\alpha )\,\asymp \,n^{\alpha -1}.\end{aligned}$$

(7)

For $0<a<1$, set

$$\begin{aligned} f_a(z)=\left( \frac{1-a^2}{(1-az)^2}\right) ^{1/p}= (1-a^2)^{1/p}\sum _{n=0}^\infty a_n({2}/{p})a^nz^n,\quad z\in \mathbb D.\end{aligned}$$

We have that

$$\begin{aligned} f_a\in H^p\,\,\text{ and } \,\,\Vert f_a\Vert _{H^p}=1,\quad 0<a<1. \end{aligned}$$

Since ${\mathcal {C}}_\mu$ is bounded on $H^p$, we have

$$\begin{aligned} \Vert {\mathcal {C}}_\mu (f_a)\Vert _{H^p}^p\lesssim 1. \end{aligned}$$

(8)

Now

$$\begin{aligned}{\mathcal {C}}_\mu (f_a)(z)=(1-a^2)^{1/p}\sum _{n=0}^\infty \mu _n\left( \sum _{k=0}^na_k({2}/{p})a^k\right) z^n,\quad z\in {\mathbb {D}}. \end{aligned}$$

Using the fact that $1\le p\le 2$, [13, Theorem 6.2], (7), and the fact that the sequence $\{ \mu _n\}$ is decreasing, we obtain

$$\begin{aligned}&(1-a)\mu _N^p \sum _{n=0}^N (n+1)^{p-2}\left( \sum _{k=0}^nk^{\frac{2}{p}-1}a^k\right) ^p \\&\quad \le (1-a)\sum _{n=0}^N (n+1)^{p-2}\mu _n^p\left( \sum _{k=0}^nk^{\frac{2}{p}-1}a^k\right) ^p\\&\quad \le (1-a)\sum _{n=0}^\infty (n+1)^{p-2}\mu _n^p\left( \sum _{k=0}^nk^{\frac{2}{p}-1}a^k\right) ^p\\&\quad \asymp (1-a^2)\sum _{n=0}^\infty (n+1)^{p-2}\mu _n^p\left( \sum _{k=0}^na_k({2}/{p})a^k\right) ^p\\&\quad \lesssim \Vert {\mathcal {C}}_\mu (f_a)\Vert _{H^p}^p, \end{aligned}$$

for every positive integer N and every $a\in (0,1)$. Taking $a=1-\frac{1}{N}$ and using the fact that ${\mathcal {C}}_\mu$ is bounded on $H^p$, we obtain

$$\begin{aligned} \frac{\mu _N^p}{N}\sum _{n=0}^N(n+1)^{p-2}\left( \sum _{k=0}^nk^{\frac{2}{p}-1}\right) ^p\asymp \mu _N^pN^p\lesssim \Vert {\mathcal {C}}_\mu (f_a)\Vert _{H^p}^p\lesssim \Vert f_a\Vert _{H^p}. \end{aligned}$$

(9)

This and (8) imply that $\mu _N\lesssim \frac{1}{N}$. Using again Lemma 2, this yields that $\mu$ is a Carleson measure. $\square$

Proof of the implication (ii)  $\Rightarrow$  (i) for $2\le p<\infty$. Suppose that $2<p<\infty$ and that ${\mathcal {C}}_\mu$ is a bounded operator on $H^p$. Let q be the conjugate exponent of p, that is, $\frac{1}{p}+\frac{1}{q}=1$. Bearing in mind Proposition 2 and Proposition 3, we see that the operator $T_\mu$, initially defined over polynomials, extends to a bounded operator on $H^q$.

For $0<a<1$ and $N\in {\mathbb {N}}$, set

$$\begin{aligned} f_a(z)&=\left( \frac{1-a^2}{(1-az)^2}\right) ^{1/q}=(1-a^2)^{1/q}\sum _{n=0}^\infty a_n(2/q)a^nz^n,\quad z\in {\mathbb {D}}, \\ f_{a,N}(z)&=(1-a^2)^{1/q}\sum _{n=0}^N a_n(2/q)a^nz^n,\quad z\in \mathbb D. \end{aligned}$$

We have that for all $a\in (0,1)$, $f_a\in H^q$ and $\Vert f_a\Vert _{H^q}=1$. Since $T_\mu$ is bounded on $H^q$, it follows that

$$\begin{aligned} \Vert T_\mu (f_a)\Vert _{H^q}\lesssim 1\end{aligned}$$

(10)

Also, for every a, $\,f_{a,N}\, \rightarrow f_a$, as $N\rightarrow \infty$ in $H^q$ and uniformly on compact subsets of ${\mathbb {D}}$. Now, $T_\mu \left( f_{a,N}\right) (z)=(1-a^2)^{1/q}\sum _{n=0}^N\left( \sum _{k=n}^N\mu _ka_k(2/q)a^k\right) z^n$ ($z\in {\mathbb {D}}$) and then, using that $1<q<2$ and [13, Theorem 6.2], we have that

$$\begin{aligned} (1-a)\sum _{n=1}^N(n+1)^{q-2}\left( \sum _{k=n}^N\mu _ka_k(2/q)a^k\right) ^q\lesssim \Vert T_\mu \left( f_{a,N}\right) \Vert _{H^q}^q. \end{aligned}$$

Letting N tend to $\infty$, we obtain

$$\begin{aligned} (1-a)\sum _{n=1}^\infty (n+1)^{q-2}\left( \sum _{k=n}^\infty \mu _ka_k(2/q)a^k\right) ^q\lesssim \Vert T_\mu \left( f_{a}\right) \Vert _{H^q}^q. \end{aligned}$$

Taking $a=1-\frac{1}{N}$ and letting [N/2] denote the largest integer less than or equal to N/2, we obtain

$$\begin{aligned}&\Vert T_\mu \left( f_{a}\right) \Vert _{H^q}^q\gtrsim (1-a)\sum _{n=1}^N(n+1)^{q-2}\left( \sum _{k=n}^N\mu _ka_k(2/q)a^k\right) ^q\,\nonumber \\&\quad \gtrsim \frac{\mu _N^q}{N}\sum _{n=1}^Nn^{q-2}\left( \sum _{k=n}^Nk^{\frac{2}{q}-1}\right) ^q \,\gtrsim \, \frac{\mu _N^q}{N}\sum _{n=1}^{[N/2]}n^{q-2}\left( \sum _{k=[N/2]}^Nk^{\frac{2}{q}-1}\right) ^q \nonumber \\&\quad \asymp \frac{\mu _N^q}{N}\sum _{n=1}^{[N/2]}n^{q-2}\left( N^{2/q}\right) ^q \,\asymp \,\mu _N^qN^q. \end{aligned}$$

(11)

Using (10), it follows that $\mu _N\lesssim \frac{1}{N}$ and then Lemma 2 implies that $\mu$ is a Carleson measure. $\square$

3.2 Proof of Theorem 2

Proof

Let us start with the implication (ii)   $\Rightarrow$   (i). We shall consider the cases $1\le p\le 2$ and $2<p<\infty$ separately.

Suppose first that $1\le p\le 2$ and ${\mathcal {C}}_\mu$ is compact from $H^p$ into itself. As in the proof of Theorem 1, for $0<a<1$, set

$$\begin{aligned} f_a(z)=\left( \frac{1-a^2}{(1-az)^2}\right) ^{1/p},\quad z\in {\mathbb {D}}.\end{aligned}$$

We have that $\Vert f_a\Vert _{H^p}=1$ for all a and, also, $f_a\,\rightarrow \, 0,$ as $a\rightarrow 1$, uniformly on compact subsets of ${\mathbb {D}}$. Hence, $\Vert {\mathcal {C}}_\mu (f_a)\Vert _{H^p}\rightarrow 0$, as $a\rightarrow 1$. But in the course of the proof of the implication (ii)  $\Rightarrow$   (i) of Theorem 1, we obtained that $\mu _NN\lesssim \Vert {\mathcal {C}}_\mu (f_a)\Vert _{H^p}$ for $a=1-\frac{1}{N}$ (see (9)). Then it follows that $\mu _N={{\,\mathrm{o}\,}}\left( \frac{1}{N}\right)$ and this implies that $\mu$ is a vanishing Carleson measure.

Suppose now that $2<p<\infty$ and ${\mathcal {C}}_\mu$ is compact from $H^p$ into itself. By Theorem 1, $\mu$ is a Carleson measure and then it follows that the operator $T_\mu$ is well defined on $H^q$ ($\frac{1}{p}+\frac{1}{q}=1$) and it is the adjoint of $\mathcal C_\mu$. For $0<a<1$, set $f_a(z)=\left( \frac{1-a^2}{(1-az)^2}\right) ^{1/q}$ ,  ($z\in {\mathbb {D}}$). We have that $\Vert f_a\Vert _{H^q}=1$ for all a and, also, $f_a\,\rightarrow \, 0,$   as $a\rightarrow 1$, uniformly on compact subsets of ${\mathbb {D}}$. By Schauder’s theorem [10, p. 174], $T_\mu$ is a compact operator from $H^q$ into itself and, hence, $\Vert T_\mu (f_a)\Vert _{H^q}\rightarrow 0$. In the course of the proof of the implication (ii)  $\Rightarrow$   (i) of Theorem 1, we obtained that $\mu _NN\lesssim \Vert T_\mu (f_a)\Vert _{H^q}$ for $a=1-\frac{1}{N}$ (see (11)). Then it follows that $\mu _N={{\,\mathrm{o}\,}}\left( \frac{1}{N}\right)$ and, hence, $\mu$ is a vanishing Carleson measure.

To prove the other implication we shall consider the cases $p=2$, $p=1$, $1<p<2$, and $2<p<\infty$ separately.

Let us start with the case $p=2$. So assume that $\mu$ is a vanishing Carleson measure and let $\{ f_n\}$ be a sequence of functions in $H^2$ with $\Vert f_n\Vert _{H^2}\le 1$, for all n, and such that $\, f_n\, \rightarrow 0$, uniformly on compact subsets of ${\mathbb {D}}$.

Since $\mu$ is a vanishing Carleson measure $\mu _k={{\,\mathrm{o}\,}}\left( \frac{1}{k}\right)$, as $k\rightarrow \infty$. Say

$$\begin{aligned} \mu _k=\frac{\varepsilon _k}{k+1},\quad k=0, 1, 2, \dots . \end{aligned}$$

Then $\{ \varepsilon _k\} \rightarrow 0$. Say that, for every n,

$$\begin{aligned} f_n(z)=\sum _{k=0}^\infty a_k^{(n)}z^k,\quad z\in {\mathbb {D}}. \end{aligned}$$

Since the Cesàro operator ${\mathcal {C}}$ is bounded on $H^2$, there exists $M>0$ such that

$$\begin{aligned} \Vert {\mathcal {C}}(f_n)\Vert _{H^2}^2\le M,\quad \text {for all }n. \end{aligned}$$

(12)

Take $\varepsilon >0$ and next take a natural number N such that

$$\begin{aligned} k\ge N\,\,\,\, \Rightarrow \,\,\,\,\varepsilon _k^2<\frac{\varepsilon }{2M}.\end{aligned}$$

We have

$$\begin{aligned} \Vert {\mathcal {C}}_\mu (f_n)\Vert _{H^2}^2&=\sum _{k=0}^\infty \mu _k^2\left| \sum _{j=0}^ka_j^{(n)}\right| ^2\\&=\sum _{k=0}^N \mu _k^2\left| \sum _{j=0}^ka_j^{(n)}\right| ^2\,+\,\sum _{k=N+1}^\infty \frac{\varepsilon _k^2}{(k+1)^2}\left| \sum _{j=0}^ka_j^{(n)}\right| ^2\\&\le \sum _{k=0}^N \mu _k^2\left| \sum _{j=0}^ka_j^{(n)}\right| ^2\,+\,\frac{\varepsilon }{2M}\sum _{k=0}^\infty \frac{1}{(k+1)^2}\left| \sum _{j=0}^ka_j^{(n)}\right| ^2 \\&=\sum _{k=0}^N \mu _k^2\left| \sum _{j=0}^ka_j^{(n)}\right| ^2\,+\,\frac{\varepsilon }{2M}\Vert {\mathcal {C}}(f_n)\Vert _{H^2}^2 \\&\le \sum _{k=0}^N \mu _k^2\left| \sum _{j=0}^ka_j^{(n)}\right| ^2\,+\,\frac{\varepsilon }{2}. \end{aligned}$$

Now, since $\, f_n\, \rightarrow 0$, uniformly on compact subsets of ${\mathbb {D}}$, it follows that

$$\begin{aligned} \sum _{k=0}^N \mu _k^2\left| \sum _{j=0}^ka_j^{(n)}\right| ^2\rightarrow 0,\quad \text {as }n\rightarrow \infty . \end{aligned}$$

Then it follows that that there exist $n_0\in {\mathbb {N}}$ such that $\Vert {\mathcal {C}}_\mu (f_n)\Vert _{H^2}^2<\varepsilon$ for all $n\ge n_0$. So, we have proved that $\Vert {\mathcal {C}}_\mu (f_n)\Vert _{H^2}^2\rightarrow 0$. The compactness of ${\mathcal {C}}_\mu$ on $H^2$ follows.

Let us move to the case $p=1$. Assume that $\mu$ is a vanishing Carleson measure and let $\{ f_n\}$ be a sequence of functions in $H^1$ with $\Vert f_n\Vert _{H^1}\le 1$, for all n, and such that $\, f_n\, \rightarrow 0$, uniformly on compact subsets of ${\mathbb {D}}$.

Set

$$\begin{aligned} g_n(z)=\frac{f_n(z)}{1-z},\quad z\in {\mathbb {D}},\,\,\, n\in {\mathbb {N}}, \end{aligned}$$

and

$$\begin{aligned} t_k=1-\frac{1}{2^k},\quad k=0, 1, 2, \dots \end{aligned}$$

As in the proof of the implication (i)  $\Rightarrow$  (ii) in Theorem 1 when $p=1$ we see that, for $0<r<1$ and $n\in {\mathbb {N}}$,

$$\begin{aligned}M_1\left( r, {\mathcal {C}}_\mu (f_n)\right) \,\le \,&\frac{1}{2\pi }\int _0^{2\pi }\left( \sum _{k=1}^\infty \left[ \sup _{0\le t\le t_k}\left| g_n(rte^{i\theta })\right| \right] \right) \mu \left( [t_{k-1},t_k]\right) \,d\theta \end{aligned}$$

and, hence,

$$\begin{aligned} \Vert {\mathcal {C}}_\mu (f_n)\Vert _{H^1}\,\le \,&\frac{1}{2\pi }\int _0^{2\pi }\left( \sum _{k=1}^\infty \left[ \sup _{0\le t\le t_k}\left| g_n(te^{i\theta })\right| \right] \right) \mu \left( [t_{k-1},t_k]\right) \,d\theta . \end{aligned}$$

(13)

Since $\mu$ is a vanishing Carleson measure $\mu \left( [t_{k-1},t_k]\right) ={{\,\mathrm{o}\,}}(2^{-k})$ and, hence, we have

$$\begin{aligned} \mu \left( [t_{k-1},t_k]\right) =\frac{\varepsilon _k}{2^k},\,\,\text { where }\varepsilon _k\ge 0 \text {and }\{ \varepsilon _k\} \rightarrow 0. \end{aligned}$$

On the other hand, looking at the proof of Theorem 1, we see that there exists $C>0$ such that

$$\begin{aligned} \sum _{k=1}^\infty \int _{t_k}^{t_{k+1}}M_1(t,g_n)dt\le C\Vert f_n\Vert _{H^1}\le C,\quad n\in {\mathbb {N}}. \end{aligned}$$

(14)

Take $\varepsilon >0$ and then take $N\in {\mathbb {N}}$ so that $\varepsilon _k\le \frac{\varepsilon }{2CK}$, for all $k\ge N$, where K is the constant in the Hardy-Littlewood maximal estimate

$$\begin{aligned}\frac{1}{2\pi }\int _0^{2\pi }\left[ \sup _{0<t<1}\vert F(te^{i\theta })\right] \,d\theta \le K\Vert F\Vert _{H^1}.\end{aligned}$$

Using (13) we see that

$$\begin{aligned}\Vert {\mathcal {C}}_\mu (f_n)\Vert _{H^1}\,\le \,&I(n)\,+\,II(n),\end{aligned}$$

where

$$\begin{aligned}I(n)\,=\,&\frac{1}{2\pi }\int _0^{2\pi }\left( \sum _{k=1}^N\left[ \sup _{0\le t\le t_k}\left| g_n(te^{i\theta })\right| \right] \right) \mu \left( [t_{k-1},t_k]\right) \,d\theta ,\\ II(n)\,=\,&\frac{1}{2\pi }\int _0^{2\pi }\left( \sum _{k=N+1}^\infty \left[ \sup _{0\le t\le t_k}\left| g_n(te^{i\theta })\right| \right] \right) \mu \left( [t_{k-1},t_k]\right) \,d\theta .\end{aligned}$$

Using (14), we obtain

$$\begin{aligned}II(n)\,\le \,&\sum _{k=N+1}^\infty \frac{\varepsilon _k}{2^k}\frac{1}{2\pi }\int _0^{2\pi }\left[ \sup _{0\le t\le t_k}\left| g_n(te^{i\theta })\right| \right] d\theta \\ \,\le \,&\frac{\varepsilon }{2C}\sum _{k=1}^\infty \frac{1}{2^k}M_1(t_k, g_n)\\ \,\le \,&\frac{\varepsilon }{2C}\sum _{k=1}^\infty \frac{1}{2^k}\int _{t_k}^{t_{k+1}}M_1(t, g_n)dt\\ \,\le \,&\frac{\varepsilon }{2}. \end{aligned}$$

Since $\, f_n\, \rightarrow 0$, uniformly on compact subsets of ${\mathbb {D}}$, it is clear that $I(n)\rightarrow 0$, as $n\rightarrow \infty$. Then it follows that there exists $n_0\in {\mathbb {N}}$ such that $\Vert {\mathcal {C}}_\mu (f_n)\Vert _{H^1}<\varepsilon$ whenever $n\ge n_0$. Thus,we have shown that $\Vert {\mathcal {C}}_\mu (f_n)\Vert _{H^1}\rightarrow 0$, as $n\rightarrow \infty$ and the compactness of ${\mathcal {C}}_\mu$ on $H^1$ follows.

To deal with the cases $1<p<2$ and $2<p<\infty$, we use again complex interpolation.

Suppose first that $1<p<2$ and $\mu$ is a vanishing Carleson measure. Recall that

$$\begin{aligned} H^p=(H^2, H^1)_{\theta },\,\,\text {with}\,\,\, \theta =\frac{2}{p}-1. \end{aligned}$$

We have also that if $2<s<\infty$ then

$$\begin{aligned} H^2=\left( H^{s}, H^{1}\right) _{\alpha } \end{aligned}$$

for a certain $\alpha \in (0,1)$, namely, $\alpha =\left( \frac{1}{2}-\frac{1}{s}\right) /\left( 1-\frac{1}{s}\right)$. Since $H^2$ is reflexive, and ${\mathcal {C}}_\mu$ is compact from $H^2$ into $H^2$ and from $H^1$ into $H^1$, Theorem 10 of [11] gives that and ${\mathcal {C}}_\mu$ is compact from $H^p$ into $H^p$.

Suppose now that $2<p<\infty$ and $\mu$ is a vanishing Carleson measure. Let q be conjugate exponent of p. Take $q_1$ with $1<q_1<q<2$. We have that $T_\mu$ is compact from $H^2$ into itself and continuous from $H^{q_1}$ into $H^{q_1}$. Also, $H^q=(H^2, H^{q_1})_{\theta }$ for a certain $\theta \in (0,1)$. Then, Theorem 10 of [11] gives that and $T_\mu$ is compact from $H^q$ into $H^q$ and, hence, ${\mathcal {C}}_\mu$ is compact from $H^p$ into itself. $\square$

3.3 The operators ${\mathcal {C}}_\mu$ acting on $H^\infty$

For the constant function 1 we have

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

Consequently, ${\mathcal {C}} (H^\infty )\not \subset H^\infty$.

If $\mu$ is positive finite Borel measure on [0, 1) then

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

So, it follows that

$$\begin{aligned} {\mathcal {C}}_\mu (1)\in H^\infty \,\, \Leftrightarrow \,\,\, \int _{[0,1]}\frac{d\mu (t)}{1-t}<\infty \,\,\, \Leftrightarrow \,\,\,\sum _{n=0}^\infty \mu _n<\infty .\end{aligned}$$

This easily implies the following result.

Theorem 4

Let $\mu$ be positive finite Borel measure on [0, 1). Then the following conditions are equivalent.

• (i) ${\mathcal {C}}_\mu$ is a bounded operator from $H^\infty$ into itself.

• (ii) $\int _{[0,1]}\frac{d\mu (t)}{1-t}<\infty$.

• (iii) $\sum _{n=0}^\infty \mu _n<\infty .$

Danikas and Siskakis [12] proved that

$$\begin{aligned} {\mathcal {C}}(H^\infty ) \subset BMOA\subset {\mathcal {B}}. \end{aligned}$$

We extend this result obtaining a characterization of those positive finite Borel measure $\mu$ on [0, 1) for which ${\mathcal {C}}_\mu (H^\infty )\subset {\mathcal {B}}$.

Theorem 5

Let $\mu$ be positive finite Borel measure on [0, 1). Then the following conditions are equivalent

• (i) ${\mathcal {C}}_\mu$ is a bounded operator from $H^\infty$ into the Bloch space ${\mathcal {B}}$.

• (ii) $\mu$ is a Carleson measure.

Proof

Let us start with the implication (i)   $\Rightarrow$   (ii). So, assume that ${\mathcal {C}}_\mu (H^\infty )\subset {\mathcal {B}}$. Then ${\mathcal {C}}_\mu (1)\in {\mathcal {B}}$, but, as we have seen above

$$\begin{aligned} {\mathcal {C}}_\mu (1)(z)=\sum _{n=0}^\infty \mu _nz^n,\quad z\in {\mathbb {D}}, \end{aligned}$$

and then, using the fact that the sequence $\{ \mu _n\}$ is a decreasing sequence of nonnegative numbers and Lemma B, we see that $\mu _n={{\,\mathrm{O}\,}}\left( \frac{1}{n}\right)$ which is equivalent to saying that $\mu$ is a Carleson measure.

Let us turn now to prove the other implication. So, assume that $\mu$ is a Carleson measure and take $f\in H^\infty$. Using the integral representation of ${\mathcal {C}}_\mu$ we see that

$$\begin{aligned} {\mathcal {C}}_\mu (f)^\prime (z)\,=\,\int _{[0,1)}\frac{tf^\prime (tz)}{1-tz}\,d\mu (t)\,+\,\int _{[0,1)}\frac{tf (tz)}{(1-tz)^2}\,d\mu (t),\quad z\in {\mathbb {D}}.\end{aligned}$$

Hence, using that $f\in H^\infty \subset {\mathcal {B}}$, we obtain

$$\begin{aligned} \left| \mathcal C_\mu (f)^\prime (z)\right|&\le \int _{[0,1)}\frac{\vert f^\prime (tz)\vert }{\vert 1-tz\vert }\,d\mu (t)\,+\int _{[0,1)}\frac{\vert f(tz)\vert }{\vert 1-tz\vert ^2 }\,d\mu (t) \nonumber \\&\lesssim \int _{[0,1)}\frac{d\mu (t)}{(1-\vert tz\vert )^2 },\quad z\in {\mathbb {D}}. \end{aligned}$$

(15)

Take $z\in {\mathbb {D}}$ and set $r=\vert z\vert$. Set also

$$\begin{aligned}\phi (t)=\mu ([0,t))-\mu ([0,1))=-\mu ([t,1)),\quad 0\le t<1. \end{aligned}$$

Integrating by parts and using the fact that $\mu$ is a Carleson measure, we obtain

$$\begin{aligned} \int _{[0,1)}\frac{d\mu (t)}{(1-\vert tz\vert )^2 }&=\int _{[0,1)}\frac{d\mu (t)}{(1-tr)^2 }\,=\,\mu ([0,1))+2r\int _0^1\frac{\mu ([t,1))}{(1-tr)^3}\,dt\\ \lesssim&\,\mu ([0,1))\,+\,\int _0^1\frac{1-t}{(1-tr)^3}\,dt\\&=\mu ([0,1))\,+\,\int _0^r\frac{1-t}{(1-tr)^3}\,dt\,+\,\int _r^1\frac{1-t}{(1-tr)^3}\,dt \\&\quad \lesssim \mu ([0,1))\,+\int _0^r\frac{1}{(1-t)^2}\,dt\,+\,\frac{1}{(1-r)^3}\int _r^1(1-t)\,dt \\&\lesssim \frac{1}{1-r}. \end{aligned}$$

This and (15) yield that ${\mathcal {C}}_\mu (f)\in {\mathcal {B}}$. $\square$

It is natural to ask whether or not $\mu$ being a Carleson measure implies that ${\mathcal {C}}_\mu (H^\infty )\subset BMOA$. We do not know the answer to this question.

4 The operators ${\mathcal {C}}_\mu$ acting on Bergman spaces

The boundedness of the Cesàro operator on Bergman spaces was studied in [1] and [35] where the following result was proved.

Theorem A

If $p>0$ and $\alpha >-1$, then the Cesàro operator is bounded from $A^p_\alpha$ into itself.

In the course of our proof of Theorem 1, we proved that if $\mu$ is a Carleson measure, $1\le p<\infty$, and $f\in H^p$, then $M_p(r,{\mathcal {C}}_\mu (f)\lesssim M_p(r,f)$ (see (4) and (6)). This readily yields that that if $\mu$ is a Carleson measure, $1\le p<\infty$, and $\alpha >-1$, then ${\mathcal {C}}_\mu$ is bounded from $A^p_\alpha$ into itself.

For $p>1$ we shall give a different proof of this result and we shall also prove that the converse is true. Hence, our work in particular will lead to a new proof of the boundedness of the classical Cesàro operator on the spaces $A^p_\alpha$ ($1<p<\infty$, $\alpha >-1$).

Theorem 6

Suppose that $1<p<\infty$ and $\alpha >-1$. Let $\mu$ be a positive finite Borel measure on [0, 1).Then the following conditions are equivalent.

• (i) The measure $\mu$ is a Carleson measure.

• (ii) The operator ${\mathcal {C}}_\mu$ is bounded from $A^p_\alpha$ into itself.

Let us collect several results which will be needed in the proof of Theorem 6.

Let us start recalling the given $1\le p\le \infty$ and $0<\alpha \le 1$, the mean Lipschitz space $\varLambda _ \alpha ^p$ consists of those functions f analytic in ${{\mathbb {D}}}$ having a non-tangential limit almost everywhere for which $\omega _ p(\delta , f)=O(\delta ^\alpha )$, as $\delta \rightarrow 0$, where $\omega _ p(., f)$ is the integral modulus of continuity of order p of the boundary values $f(e^{i\theta })$ of f. A classical result of Hardy and Littlewood [23] (see also Chapter 5 of [13]) asserts that for $1\le p\le \infty$ and $0<\alpha \le 1$, we have that $\varLambda _ \alpha ^p\subset H^p$ and

$$\begin{aligned} \varLambda _ \alpha ^p=\left\{ f\text { analytic in }{{\mathbb {D}}}: M_ p(r, f^\prime )= {{\,\mathrm{O}\,}}\left( \frac{1}{(1-r)^{1-\alpha }}\right) , \quad \text { as }r\rightarrow 1 \right\} . \end{aligned}$$

The space $\varLambda _ \alpha ^p$ is a Banach space with the norm $\Vert \cdot \Vert _{p, \alpha }$ given by

$$\begin{aligned} \Vert f\Vert _{p, \alpha }=\vert f(0)\vert +\sup _{0\le r<1}(1-r)^{1-\alpha }M_ p(r, f^\prime ). \end{aligned}$$

Of special interest are the spaces $\varLambda _ {1/p}^ p$ since they lie in the border of continuity. Indeed, if $1<p<\infty$ and $\alpha >\frac{1}{p}$ then each $f\in \varLambda _\alpha ^p$ has a continuous extension to the closed unit disc. This is not true for $\alpha =\frac{1}{p}$. This follows easily noticing that the function $f(z)=\log (1-z)$ belongs to $\varLambda _ {1/p}^p$ for all $p\in (1, \infty )$. Cima and Petersen proved in [9] that $\varLambda _{1/2}^2\subset BMOA$ and this result was generalized by Bourdon, Shapiro and Sledd who proved in [4] that

$$\begin{aligned} \varLambda _{1/p}^p\subset BMOA,\quad 1<p<\infty . \end{aligned}$$

This was shown to be sharp in a very strong sense in [3].

The following result of Merchán [30, Lemma 1] (see also [18, Theorem 2] and [17, Theorem 2]) will be needed in our work.

Lemma B

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

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

We shall also use some results on pointwise multipliers and coefficient multipliers of Bergman spaces and Hardy spaces.

Let us start recalling that for $g\in {\mathrm{Hol}}({{\mathbb {D}}})$, the multiplication operator $M_g$ is defined by

$$\begin{aligned} M_g(f)(z){\mathop {=}\limits ^{\text {def}}}g(z)f(z),\quad f\in {\mathrm{Hol}}({{\mathbb {D}}}),\,\, z\in {{\mathbb {D}}}. \end{aligned}$$

If X and Y are two spaces of analytic functions in ${\mathbb {D}}$ (which will always be assumed to be Banach or F-spaces continuously embedded in ${\mathrm{Hol}}({\mathbb {D}} )$) and $g\in {\mathrm{Hol}}(\mathbb D )$ then g is said to be a pointwise multiplier from X to Y if $M_g(X)\subset Y$. The space of all multipliers from X to Y will be denoted by M(X, Y). Using the closed graph theorem we see that if $g\in M(X, Y)$ then $M_g$ is a bounded operator from X into Y. The following result is a particular case of Theorem C of [37].

Theorem C

Suppose that $1<p<\infty$ and $\alpha >-1$. Then

$$\begin{aligned} M\left( A^p_\alpha , A^{p/(p+1)}_\alpha \right) =A^1_\alpha . \end{aligned}$$

If X and Y are two spaces of analytic functions in $\mathbb D$, a function $F\in {\mathrm{Hol}}({\mathbb {D}})$ is said to be a coefficient multiplier (or a convolution multiplier) from X to Y if

$$\begin{aligned} f\in X\,\,\,\Rightarrow \,\,\, F\star f\in Y. \end{aligned}$$

The following result is due to Duren and Shields, it is a particular case of [15, Theorem 4].

Theorem D

Suppose that $1<p<\infty$ and $F\in {\mathrm{Hol}}({\mathbb {D}})$. Let m be a positive integer such that $(m +1)^{-1}\le \frac{p}{p+1}<m^{-1}$. Then F is a coefficient multiplier from $H^{p/(p+1)}$ to $H^p$ if and only if the $(m+1)$-th derivative $F^{(m+1)}$ of F satisfies

$$\begin{aligned}M_p\left( r, F^{(m+1)}\right) = {{\,\mathrm{O}\,}}\left( (1-r)^{\frac{1}{p}-1-m}\right) .\end{aligned}$$

We can now proceed to prove Theorem 6.

Proof of the implication (i)  $\Rightarrow$  (ii) in Theorem 6. Assume that $\mu$ is a Carleson measure and set

$$\begin{aligned} F(z)=\sum _{n=0}^\infty \mu _nz^n,\quad z\in {\mathbb {D}}. \end{aligned}$$

Since $\mu$ is a Carleson measure $\mu _n={{\,\mathrm{O}\,}}\left( \frac{1}{n}\right)$. This, the simple fact that $\{ \mu _n\}$ is a deceasing sequence of nonnegative numbers, and Lemma B imply that $F\in \varLambda ^p_{1/p}$ and, hence

$$\begin{aligned} M_p(r,F^\prime )={{\,\mathrm{O}\,}}\left( (1-r)^{\frac{1}{p}-1}\right) .\end{aligned}$$

Using [13, Theorem 5.5], we see that this implies

$$\begin{aligned} M_p(r,F^{(m+1)})={{\,\mathrm{O}\,}}\left( (1-r)^{\frac{1}{p}-1-m}\right) ,\quad m=1, 2, 3, \dots , \end{aligned}$$

and then Theorem D gives that F is a coefficient multiplier from $H^{p/(p+1)}$ into $H^p$. Trivially, this implies that

$$\begin{aligned} F\text { is also a coefficient multiplier from }A^{p/(p+1)}_\alpha \text {into }A^p_\alpha . \end{aligned}$$

(16)

Take $f\in A^p_\alpha$. We have to prove that ${\mathcal {C}}_\mu (f)\in A^p_\alpha$. Set $g(z)=\frac{f(z)}{1-z}$ ($z\in \mathbb D$). A simple computation shows that $\frac{1}{1-z}\in A^1_\alpha$. Then, using Theorem C we deduce that $g\in A^{p/(p+1)}_\alpha$. This and (16) imply that $F\star g\in A^p_\alpha$. By Lemma 1 this is equivalent to saying that ${\mathcal {C}}_\mu (f)\in A^p_\alpha$. $\square$

Proof of the implication (ii)  $\Rightarrow$  (i) in Theorem 6. Suppose that ${\mathcal {C}}_\mu$ is a bounded operator on $A^p_\alpha$. Let q be the exponent conjugate to p, that is, $\frac{1}{p}+\frac{1}{q}=1$. Let $T_\mu$ be the adjoint of ${\mathcal {C}}_\mu$, it is a bounded operator on $A^q_\alpha$.

For $0<b<1$, set

$$\begin{aligned}f_b(z)=\frac{(1-b)^{1-\frac{1}{q}}}{(1-bz)^{1+\frac{\alpha +1}{q}}}=\sum _{k=0}^\infty a_{k,b}z^k,\quad z\in {\mathbb {D}}. \end{aligned}$$

Using [39, Lemma 3.10], we see that

$$\begin{aligned} \Vert f_b\Vert _{A^q_\alpha }^q\asymp 1.\end{aligned}$$

(17)

Also,

$$\begin{aligned} a_{k,b}\asymp (1-b)^{1-\frac{1}{q}}k^{(\alpha +1)/q}b^k.\end{aligned}$$

For $N\in {\mathbb {N}}$, set

$$\begin{aligned} f_{b,N}(z)=\sum _{k=0}^N a_{k,b}z^k,\quad z\in {\mathbb {D}}. \end{aligned}$$

Bearing in mind Proposition 2 and Proposition 3, we see that

$$\begin{aligned} T_\mu (f_{b,N})(z)=\sum _{n=0}^N\left( \sum _{k=n}^N\mu _k a_{k,b}\right) z^n. \end{aligned}$$

Since the coefficients $a_{k,b}$ are nonnegative, it follows that the sequence of the Taylor coefficients of $T_\mu (f_{b,N})$ is a decreasing sequence of nonnegative numbers, then (see, e. g., [20, Proposition 1])

$$\begin{aligned} \Vert T_\mu (f_{b,N})\Vert _{A^q_\alpha }^q&\gtrsim \sum _{n=1}^Nn^{q-\alpha -3}\left( \sum _{k=n}^N\mu _ka_{k,b}\right) ^q \\&\gtrsim (1-b)^{q-1} \sum _{n=1}^Nn^{q-\alpha -3}\left( \sum _{k=n}^Nk^{\frac{\alpha +1}{q}}b^k\int _{[b,1)}t^k\,d\mu (t)\right) ^q \\&\gtrsim (1-b)^{q-1}\mu \left( [b,1)\right) ^q\sum _{n=1}^Nn^{q-\alpha -3}\left( \sum _{k=n}^Nk^{\frac{\alpha +1}{q}}b^{2k}\right) ^q. \end{aligned}$$

Since $\, f_{b,N}\, \rightarrow f_b$ in $A^q_\alpha$ as $N\rightarrow \infty$, using the fact that $T_\mu$ is bounded on $A^q_\alpha$, (17), and simple estimations, we deduce that

$$\begin{aligned} 1&\gtrsim (1-b)^{q-1}\mu \left( [b,1)\right) ^q\sum _{n=1}^\infty n^{q-\alpha -3}\left( \sum _{k=n}^\infty k^{\frac{\alpha +1}{q}}b^{2k}\right) ^q \\&\gtrsim (1-b)^{q-1}\mu \left( [b,1)\right) ^q\sum _{n=1}^\infty n^{q-\alpha -3}n^{\alpha +1}\left( \sum _{k=n}^\infty b^{2k}\right) ^q \\&\asymp (1-b)^{q-1}\mu \left( [b,1)\right) ^q\sum _{n=1}^\infty n^{q-2}\frac{b^{2nq}}{(1-b)^q} \\&\asymp \left( \frac{\mu \left( [b,1)\right) }{1-b}\right) ^q. \end{aligned}$$

Hence, $\mu$ is a Carleson measure. $\square$

5 The operators ${\mathcal {C}}_\mu$ acting on BMOA and on the Bloch space

Let $\lambda$ be defined by  $\lambda (z)=\log \frac{1}{1-z}$   ($z\in {\mathbb {D}}$). Then $\lambda \in BMOA$. In fact, it is true that $\lambda \in \varLambda ^p_{1/p}$ for all $p>1$. Danikas and Siskakis [12] observed that ${\mathcal {C}} (\lambda )\notin BMOA$. This implies that the Cesàro operator does not map BMOA into itself. Our Theorem 3 includes a characterization of those $\mu$ so that ${\mathcal {C}}_\mu$ maps BMOA into itself.

Since $\varLambda ^2_{1/2}\subset BMOA\subset {\mathcal {B}}$, Theorem 3 follows from the following result.

Theorem 7

Let $\mu$ be a positive finite Borel measure on [0, 1) and let X and Y be two Banach subspaces of ${\mathrm{Hol}}({\mathbb {D}})$ with $\varLambda ^2_{1/2}\subset X \subset {\mathcal {B}}$ and $\varLambda ^2_{1/2}\subset Y \subset {\mathcal {B}}$. Then the following conditions are equivalent.

• (i) The measure $\mu$ is a 1-logarithmic 1-Carleson measure.

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

Proof

Let us start showing that (i)  $\Rightarrow$  (ii). So assume that $\mu$ is a 1-logarithmic 1-Carleson measure and take $f\in X$. We recall that $\mu$ being a 1-logarithmic 1-Carleson measure is equivalent to

$$\begin{aligned} \mu _n={{\,\mathrm{O}\,}}\left( \frac{1}{n\log (n+1)}\right) . \end{aligned}$$

(18)

Take $f\in X$, $f(z)=\sum _{n=0}^\infty a_nz^n$ ($z\in {\mathbb {D}}$). Since $X\subset {\mathcal {B}}$, we have that $f\in {\mathcal {B}}$. Then, using a result of Kayumov and Wirths (see [27, Corollary 4] or [28, Corollary D]), we have

$$\begin{aligned} \left| \sum \limits _{k=0}^n a_k\right| \lesssim \Vert f\Vert _{{\mathcal {B}}}\log (n+1). \end{aligned}$$

(19)

The estimates (18) and (19) yield

$$\begin{aligned}M_2^2(r,{\mathcal {C}}_\mu (f)^\prime )=\sum \limits _{n=1}^\infty n^2 \mu _n^2 \left| \sum \limits _{k=0}^n a_k\right| ^2 r^{2n-2}\lesssim \sum \limits _{n=1}^\infty r^{2n-2} \lesssim \frac{1}{1-r}. \end{aligned}$$

Hence ${\mathcal {C}}_\mu (f)\in \varLambda ^2_{1/2}\subset Y$.

Suppose now that ${\mathcal {C}}_\mu (X)\subset Y$. As above, set $\lambda (z)=\log \frac{1}{1-z}=\sum \limits _{n=1}^\infty \frac{z^n}{n}$ ($z\in {\mathbb {D}}$). We have that $\lambda \in X$ and then ${\mathcal {C}}_\mu (\lambda )\in Y\subset {\mathcal {B}}$. Now, ${\mathcal {C}}_\mu (\lambda )(z)=\sum \limits _{n=1}^\infty \mu _n\left( \sum \limits _{k=1}^n\frac{1}{k}\right) z^n$ and then it follows that

$$\begin{aligned} \sum _{n=1}^\infty n\mu _n\left( \sum _{k=1}^n\frac{1}{k}\right) r^n\lesssim \frac{1}{1-r},\quad r\in (0,1). \end{aligned}$$

For $N\ge 2$ take $r_N=1-\frac{1}{N}$. Bearing in mind that the sequence $\{ \mu _n\}$ is decreasing, simple estimations lead us to the following

$$\begin{aligned}&N^2\mu _N\log N\,\asymp \,\mu _N\sum _{n=1}^Nn\log n\\&\quad \lesssim \sum _{n=1}^Nn\mu _n(\log n)r_N^n \\&\quad \lesssim \sum _{n=1}^Nn\mu _n\left( \sum _{k=1}^n\frac{1}{k}\right) r_N^n \\&\quad \lesssim \sum _{n=1}^\infty n\mu _n\left( \sum _{k=1}^n\frac{1}{k}\right) r_N^n \\&\quad \lesssim N. \end{aligned}$$

Hence $\mu _N\lesssim \frac{1}{N\log N}$ which implies that $\mu$ is a 1-logarithmic 1-Carleson measure. $\square$

We have the following result concerning compactness.

Theorem 8

Let $\mu$ be a positive finite Borel measure on [0, 1) and let X and Y be two Banach subspaces of  ${\mathrm{Hol}}({\mathbb {D}})$ with $\varLambda ^2_{1/2}\subset X \subset {\mathcal {B}}$ and $\varLambda ^2_{1/2}\subset Y \subset {\mathcal {B}}$. Then the following four conditions are equivalent.

• (i) $\mu$ is a vanishing 1-logarithmic 1-Carleson measure.

• (ii) The operator ${\mathcal {C}}_\mu$ is a compact operator from X into Y.

• (iii) The operator ${\mathcal {C}}_\mu$ is a compact operator from the Bloch space ${\mathcal {B}}$ into itself.

• (iv) The operator ${\mathcal {C}}_\mu$ is a compact operator from the BMOA into itself.

Proof

Clearly, it suffices to prove that (i) and (ii) are equivalent. Let us prove first that (i) implies (ii). So, assume that $\mu$ is a vanishing 1-logarithmic 1-Carleson measure and $\varLambda ^2_{1/2}\subset X, Y \subset {\mathcal {B}}$.

Take $\{ f_j\} \subset X$ with $\Vert f_j\Vert _{X}\le 1$, for all j, and $\, f_j\, \rightarrow 0$, as $j\rightarrow \infty$, uniformly on compact subsets of ${\mathbb {D}}$. Since X is continuously embedded in ${\mathcal {B}}$, $\{ f_j\} \subset {\mathcal {B}}$ and there exists $K_1>0$ such that $\Vert f\Vert _{{\mathcal {B}}}\le K_1$, for all j.

Say $f_j(z)=\sum _{k=0}^\infty a_k^{(j)}z^k$ ($z\in {\mathbb {D}}$). Using the result of Kayumov and Wirths that we have mentioned above, we see that there exists $K_2>0$ such that

$$\begin{aligned} \left| \sum _{k=0}^na_k^{(j)}\right| \le K_2\Vert f_j\Vert _{{\mathcal {B}}}\log (n+1)\le K_1K_2\log (n+1),\quad \text {for all } n\text { and }j. \end{aligned}$$

Set $K=K_1K_2$.

Since $\mu$ is a vanishing 1-logarithmic 1-Carleson measure, $\mu _n={{\,\mathrm{o}\,}}\left( \frac{1}{n\log (n+1)}\right)$. Say $\mu _n=\frac{\varepsilon _n}{n\log (n+1)}$, with $\{ \varepsilon _n\} \rightarrow 0$. Take $\varepsilon >0$. Take $N\in {\mathbb {N}}$ such that $\varepsilon _n^2K^2<\frac{\varepsilon }{2}$ if $n\ge N$. We have, for all $j\in {\mathbb {N}}$ and $0<r<1$,

$$\begin{aligned} M_2^2\left( r, {\mathcal {C}}_\mu (f_j)^\prime \right) =&\sum _{n=1}^\infty n^2\mu _n^2\left| \sum _{k=0}^na_k^{(j)}\right| ^2r^{2n-2}\\&\quad \le \sum _{n=1}^Nn^2\mu _n^2\left| \sum _{k=0}^na_k^{(j)}\right| ^2\,+\,\sum _{n=N+1}^\infty n^2\mu _n^2K^2[\log (n+1)]^2r^{2n-2}\\&\quad \le \sum _{n=1}^Nn^2\mu _n^2\left| \sum _{k=0}^na_k^{(j)}\right| ^2\,+\,\frac{\varepsilon /2}{1-r}. \end{aligned}$$

Thus,

$$\begin{aligned}\sup _{0\le r<1}(1-r) M_2^2\left( r, {\mathcal {C}}_\mu (f_j)^\prime \right) \le \frac{\varepsilon }{2}+\sum _{n=1}^Nn^2\mu _n^2\left| \sum _{k=0}^na_k^{(j)}\right| ^2,\quad j\in {\mathbb {N}}. \end{aligned}$$

Now, since $\sum _{n=1}^Nn^2\mu _n^2\left| \sum _{k=0}^na_k^{(j)}\right| ^2\rightarrow 0$ and $f_j(0)\rightarrow 0$, as $j\rightarrow \infty$, it follows that there exists $j_0\in {\mathbb {N}}$ such that

$$\begin{aligned} \vert f_j(0)\vert +\sum _{n=1}^Nn^2\mu _n^2\left| \sum _{k=0}^na_k^{(j)}\right| ^2<\frac{\varepsilon }{2} \end{aligned}$$

for all $j\ge j_0$. With this we have proved that ${\mathcal {C}}_\mu (f_j)\rightarrow 0$ in $\varLambda ^2_{1/2}$. Since $\varLambda ^2_{1/2}$ is continuously embedded in Y, it follows that ${\mathcal {C}}_\mu (f_j)\rightarrow 0$ in Y.

Let us prove now that (ii) implies (i). Assume that $\varLambda ^2_{1/2}\subset X, Y \subset {\mathcal {B}}$ and that ${\mathcal {C}}_\mu$ is compact from X into Y. For $0<a<1$, set

$$\begin{aligned} f_a(z)=\,\left( \log \frac{2}{1-a}\right) ^{-1}\left( \log \frac{2}{1-az}\right) ^2,\quad z\in {\mathbb {D}}. \end{aligned}$$

We have that

$$\begin{aligned} f_a^\prime (z)\,=\,\left( \log \frac{2}{1-a}\right) ^{-1}\left( \log \frac{2}{1-az}\right) \frac{2a}{1-az},\quad z\in \mathbb D,\,\, 0<a<1. \end{aligned}$$

Then it is clear that $f_a\in \varLambda ^2_{1/2}$ for all $a\in [0, 1)$ and that there exists a constant $M_1>0$ such that $\Vert f_a\Vert _{2, 1/2}\le M_1,$ for all $a\in (0, 1)$. Since $\varLambda ^2_{1/2}$ is continuously embedded in X, it follows that $f_a\in X$ for all $a\in [0, 1)$ and that there exists $M>0$ such that $\Vert f_a\Vert _{X}\le M,$ for all $a\in (0, 1)$. Also, $\, f_a\, \rightarrow 0$, as $a\rightarrow 1$, uniformly on compact subsets of $\mathbb D$. Since ${\mathcal {C}}_\mu$ is compact from X into Y, we have that $\Vert {\mathcal {C}}_\mu (f_a)\Vert _{Y}\rightarrow 0$, as $a\rightarrow 1$. This, together with the fact that Y is continuously embedded in ${\mathcal {B}}$, implies that

$$\begin{aligned} \Vert {\mathcal {C}}_\mu (f_a)\Vert _{{\mathcal {B}}}\rightarrow 0,\quad \text {as }a\rightarrow 1. \end{aligned}$$

(20)

A simple calculation gives that for $0<a<1$ and $z\in {\mathbb {D}}$,

$$\begin{aligned} {\mathcal {C}}_\mu (f_a)^\prime (z)=&\int _{[0,1)}\left[ \frac{tf_a^\prime (tz)}{1-tz}\,+\, \frac{tf_a(tz)}{(1-tz)^2}\right] d\mu (t). \end{aligned}$$

Then it follows that, for $0<a<1$,

$$\begin{aligned} \left| {\mathcal {C}}_\mu (f_a)^\prime (a)\right|&= {\mathcal {C}}_\mu (f_a)^\prime (a)\\&\quad \ge \int _{[0,1)}\frac{tf_a(ta)}{(1-ta)^2}d\mu (t) \\&= \left( \log \frac{2}{1-a}\right) ^{-1}\int _{[0,1)}\frac{t\left( \log \frac{2}{1-ta}\right) ^2}{(1-ta)^2}d\mu (t) \\&\ge \left( \log \frac{2}{1-a}\right) ^{-1}\int _{[a,1)}\frac{t\left( \log \frac{2}{1-ta}\right) ^2}{(1-ta)^2}d\mu (t) \\&\ge \left( \log \frac{2}{1-a}\right) ^{-1}\mu \left( [a, 1)\right) \frac{a\left( \log \frac{2}{1-a^2}\right) ^2}{(1-a^2)^2}. \end{aligned}$$

This gives that

$$\begin{aligned}\mu \left( [a, 1)\right) \lesssim (1-a)\left( \log \frac{2}{1-a}\right) ^{-1}\Vert \mathcal C_\mu (f_a)\Vert _{{\mathcal {B}}}. \end{aligned}$$

This and (20) imply that $\mu$ is a vanishing 1-logarithmic 1-Carleson measure. $\square$