Operators Induced by Radial Measures Acting on the Dirichlet Space

Abstract

Let ${\mathbb {D}}$ be the unit disc in the complex plane. Given a positive finite Borel measure $\mu$ on the radius [0, 1), we let $\mu _n$ denote the n-th moment of $\mu$ and we deal with the action on spaces of analytic functions in ${\mathbb {D}}$ of the operator of Hibert-type ${\mathcal {H}}_\mu$ and the operator of Cesàro-type ${\mathcal {C}}_\mu$ which are defined as follows: If f is holomorphic in ${\mathbb {D}}$, $f(z)=\sum _{n=0}^\infty a_nz^n$ ($z\in {\mathbb {D}})$, then ${\mathcal {H}}_\mu (f)$ is formally defined by ${\mathcal {H}}_\mu (f)(z) = \sum _{n=0}^\infty \left( \sum _{k=0}^\infty \mu _{n+k}a_k\right) z^n$ ($z\in {\mathbb {D}}$) and ${\mathcal {C}}_\mu (f)$ is defined by $\mathcal C_\mu (f)(z) = \sum _{n=0}^\infty \mu _n\left( \sum _{k=0}^na_k\right) z^n$ ($z\in {\mathbb {D}}$). These are natural generalizations of the classical Hilbert and Cesàro operators. A good amount of work has been devoted recently to study the action of these operators on distinct spaces of analytic functions in ${\mathbb {D}}$. In this paper we study the action of the operators ${\mathcal {H}}_\mu$ and ${\mathcal {C}}_\mu$ on the Dirichlet space ${\mathcal {D}}$ and, more generally, on the analytic Besov spaces $B^p$ ($1\le p<\infty$).

1 Introduction

The open unit disc in the complex plane ${\mathbb {C}}$ will be denoted by ${\mathbb {D}}$ and $\textrm{Hol}({\mathbb {D}})$ will stand for the space of all analytic functions in ${\mathbb {D}}$. Also, dA will 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 $0\,\le \,r\,<\,1$, $0<p\le \infty$, and f analytic in ${\mathbb {D}}$, the integral means $M_p(r, f)$ of f are defined by

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

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

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

We refer to [20] for the theory of Hardy spaces.

For $0<p<\infty$ and $\alpha >-1$ the weighted Bergman space $A^p_\alpha$ consists of those $f\in \textrm{Hol}({\mathbb {D}})$ such that

$$\begin{aligned} \Vert f\Vert _{A^p_\alpha }\,{\mathop {=}\limits ^{\text {def}}}\, \left( (\alpha +1)\int _{\mathbb {D}}(1-\vert z\vert ^2)^{\alpha }\vert f (z)\vert ^p\,dA(z)\right) ^{1/p}\,<\,\infty . \end{aligned}$$

The unweighted Bergman space $A^p_0$ is simply denoted by $A^p$. We refer to [21, 31, 48] for the notation and results about Bergman spaces.

The space of Dirichlet type ${\mathcal {D}}^p_\alpha$ ($0<p<\infty$, $\alpha >-1$) is the space of those $f\in \textrm{Hol}({\mathbb {D}})$ such that $f^\prime \in A^p_{\alpha }$. Thus, a function $f\in \textrm{Hol}({\mathbb {D}})$ belongs to ${\mathcal {D}}^p_\alpha$ if and only if

$$\begin{aligned} \Vert f\Vert _{{\mathcal {D}}^p_\alpha }\,{\mathop {=}\limits ^{\text {def}}}\, \vert f(0)\vert +\left( (\alpha +1)\int _{\mathbb {D}}(1-\vert z\vert ^2)^{\alpha }\vert f^\prime (z)\vert ^p\,dA(z)\right) ^{1/p}\,<\,\infty . \end{aligned}$$

In this paper we shall be mainly concerned with the Dirichlet space ${\mathcal {D}}={\mathcal {D}}^2_0$ which consists of those $f\in \textrm{Hol}({\mathbb {D}})$ whose image Riemann surface has a finite area. We recall that if $f\in {\mathcal {D}}$, $f(z)=\sum _{n=0}^\infty a_nz^n$ ($z\in {\mathcal {D}}$), then

$$\begin{aligned} \Vert f\Vert _{{\mathcal {D}}}\,{\mathop {=}\limits ^{\text {def}}}\Vert f\Vert _{\mathcal D^2_0}=\vert f(0)\vert + \!\left( \int _{{\mathbb {D}}}\vert f^\prime (z)\vert ^2\,dA(z)\right) ^{1/2}=\,\vert a_0\vert \!+\!\left( \sum _{k=1}^\infty k\vert a_k\vert ^2\right) ^{1/2}.\qquad \quad \end{aligned}$$

(1.1)

Throughout the paper $\mu$ will be a positive finite Borel measure on the radius [0, 1) and, for $n=0, 1, 2, \dots$, we shall let $\mu _n$ denote the moment of order n of $\mu$, that is, $\mu _n=\int _{[0, 1)}t^n\,d\mu (t)$. The matrices ${\mathcal {H}}_\mu$ and ${\mathcal {C}}_\mu$ are defined as follows

$$\begin{aligned}{\mathcal {H}}_\mu =\,\left( \begin{array}{ccccc} \mu _0 &{} \mu _1 &{} \mu _2 &{} . &{} . \\ \mu _1 &{} \mu _2 &{} \mu _3 &{} . &{} . \\ \mu _2 &{} \mu _3 &{} \mu _4 &{} . &{} . \\ . &{} . &{} . &{} . &{} . \\ . &{} . &{} . &{} . &{} . \\ \end{array} \right) ;&\,\,\,\,\,\, {\mathcal {C}}_{\mu }\,=\,\left( \begin{array}{cccccc} \mu _0 &{} 0 &{} 0 &{} 0 &{} . &{} . \\ \mu _1 &{} \mu _1 &{} 0 &{} 0 &{} . &{} . \\ \mu _2 &{} \mu _2 &{} \mu _2 &{} 0 &{} . &{} . \\ . &{} . &{} . &{} . &{} . &{} . \\ . &{} . &{} . &{} . &{} . &{} . \\ \end{array}\right) . \end{aligned}$$

As we shall see in Sects. 2 and  3, these matrices induce operators acting on spaces of analytic functions which are natural generalizations of the classical Hilbert and Cesàro operators. Recently a good amount of work has been devoted to study the action of these operators of Hilbert type and of Cesàro type on distinct subspaces of $\textrm{Hol}({\mathbb {D}})$. Carleson-type measures play a basic role in this work.

Let us recall that if $\mu$ is a positive finite Borel measure on [0, 1) then:

• If $s>0$, then $\mu$ is said to be an s-Carleson measure if there exists a positive constant C such that

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

• If $0\le \alpha <\infty$, and $0<s<\infty$ we say that $\mu$ is an $\alpha$-logarithmic s-Carleson measure if there exists a positive constant C such that

  $$\begin{aligned} \mu \left( [t, 1)\right) \le C(1-t)^s\left( \log \frac{2}{1-t}\right) ^{-\alpha }, \quad 0\le t<1. \end{aligned}$$

Let us close this section by saying that, as usual, 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 $K_1\asymp K_2$.

2 Hilbert-Type Operators

The matrix ${\mathcal {H}}_\mu$ induces formally an operator, which will be also called ${\mathcal {H}}_\mu$, on spaces of analytic functions by its action on the Taylor coefficients:

$$\begin{aligned} a_n\mapsto \sum _{k=0}^{\infty } \mu _{n+k}{a_k}, \quad n=0,1,2, \dots . \end{aligned}$$

To be precise, if  $f(z)=\sum _{k=0}^\infty a_kz^k\in \textrm{Hol}({\mathbb {D}})$ we define

$$\begin{aligned} {\mathcal {H}}_\mu (f)(z)= \sum _{n=0}^{\infty }\left( \sum _{k=0}^{\infty } \mu _{n+k}{a_k}\right) z^n, \end{aligned}$$

(2.1)

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

If $\mu$ is the Lebesgue measure on [0, 1) the matrix ${\mathcal {H}}_\mu$ reduces to the classical Hilbert matrix   ${\mathcal {H}}= \left( {(n+k+1)^{-1}}\right) _{n,k\ge 0}$, which induces the classical Hilbert operator ${\mathcal {H}}$ which has extensively studied recently (see [1, 16, 17, 19, 32,33,34]).

The finite positive Borel measures $\mu$ for which ${\mathcal {H}}_\mu$ is a bounded operator on distinct spaces of analytic functions in ${\mathbb {D}}$ have been characterized in a number of papers such as [9, 14, 25, 27,28,29, 35, 37, 38, 45]. Obtaining an integral representation of ${\mathcal {H}}_\mu$ plays a basic role in these works. If $\mu$ is as above, we shall write throughout the paper

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

(2.2)

whenever the right hand side makes sense and defines an analytic function in ${\mathbb {D}}$. It turns out that the operators ${\mathcal {H}}_\mu$ and ${\mathcal {I}}_\mu$ are very closely related.

Let us mention the following results.

Theorem A

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

• (i) The operator ${\mathcal {H}}_\mu$ is bounded from $H^1$ into itself if and only if $\mu$ is a 1-logarithmic 1-Carleson measure. In such a case ${\mathcal {H}}_\mu$ and ${\mathcal {I}}_\mu$ coincide on $H^1$.

• (ii) If $1<p<\infty$, then ${\mathcal {H}}_\mu$ is a bounded operator from $H^p$ into itself if and only if $\mu$ is a 1-Carleson measure. In such a case ${\mathcal {H}}_\mu$ and ${\mathcal {I}}_\mu$ coincide on $H^p$.

• (iii) If $p>1$ and $-1<\alpha <\,p-2$ then the operator $\mathcal H_\mu$ is well defined on $A^p_\alpha$ and it is bounded from $A^p_\alpha$ into itself if and only if $\mu$ is a 1-Carleson measure. In such a case ${\mathcal {H}}_\mu$ and ${\mathcal {I}}_\mu$ coincide on $A^p_\alpha$.

• (iv) If $p>1$ and $p-2<\alpha \le p-1$, then $\mathcal H_\mu$ is well defined on ${\mathcal {D}}^p_\alpha$ and it is bounded from ${\mathcal {D}}^p_\alpha$ into itself if and only if $\mu$ is a 1-Carleson measure. In such a case ${\mathcal {H}}_\mu$ and ${\mathcal {I}}_\mu$ coincide on ${\mathcal {D}}^p_\alpha$.

• (v) If $0<\alpha <2$, ${\mathcal {H}}_\mu$ is a bounded operator from ${\mathcal {D}}^2_\alpha$ into itself if and only if $\mu$ is a 1-Carleson measure. In such a case ${\mathcal {H}}_\mu$ and ${\mathcal {I}}_\mu$ coincide on $\mathcal D^2_\alpha$.

The questions of characterizing those $\mu$ for which $\mathcal H_\mu$ is bounded on either the Dirichlet space ${\mathcal {D}}$ or on the Bergman space $A^2$ are more delicate and remain open. Regarding the Dirichlet space, the following results are proved in [28].

Theorem B

• (i) Let $\mu$ be a positive and finite Borel measure on [0, 1). If $\gamma >1$ and $\mu$ is a $\gamma$-logarithmic 1-Carleson measure, then ${\mathcal {H}}_\mu$ is bounded from ${\mathcal {D}}$ into itself.

• (ii) If $0<\beta \le \frac{1}{2}$, then there exists a positive and finite Borel measure $\mu$ on [0, 1) which is a $\beta$-logarithmic 1-Carleson measure but such that ${\mathcal {H}}_\mu (\mathcal D)\not \subset {\mathcal {D}}$.

We improve this result showing that being a 1-logarithmic 1-Carleson measure is enough to insure that ${\mathcal {H}}_\mu$ is bounded from ${\mathcal {D}}$ into itself and closing the gap between (i) and (ii). Indeed, we shall prove the following result.

Theorem 1

• (i) Let $\mu$ be a positive and finite Borel measure on [0, 1). If $\mu$ is a 1-logarithmic 1-Carleson measure, then ${\mathcal {H}}_\mu$ is bounded from ${\mathcal {D}}$ into itself.

• (ii) If $0<\beta <1$, then there exists a positive and finite Borel measure $\mu$ on [0, 1) which is a $\beta$-logarithmic 1-Carleson measure but such that ${\mathcal {H}}_\mu (\mathcal D)\not \subset {\mathcal {D}}$.

As a corollary of part (i) we obtain the following.

Corollary 2

• (a) Let $\mu$ be a positive and finite Borel measure on [0, 1) and suppose that $\mu$ is a 1-logarithmic 1-Carleson measure. Then there exists a positive constant C such that

  $$\begin{aligned} \int _{[0, 1)}\vert tf(t)f^\prime (t)\vert \,d\mu (t)\,\le C\Vert f\Vert _{{\mathcal {D}}}^2,\quad f\in \mathcal D.\end{aligned}$$

  (2.3)

• (b) There exists a positive constant C such that

  $$\begin{aligned} \int _0^1\vert tf(t)f^\prime (t)\vert \,\log \frac{2}{1-t}\,dt\,\le C\Vert f\Vert _{{\mathcal {D}}}^2,\quad f\in {\mathcal {D}}. \end{aligned}$$

  (2.4)

Regarding the Bergman space $A^2$, Theorem 1.5 of [25] asserts the following.

Theorem C

Let $\mu$ be a positive and finite Borel measure on [0, 1) and let $h_\mu$ be defined by $h_\mu (z)=\sum _{n=0}^\infty \mu _nz^n$  ($z\in {\mathbb {D}}.$) If $\mu$ satisfies the condition

$$\begin{aligned} \int _{[0, 1)} \frac{\mu \left( [t, 1)\right) }{(1-t)^2}\,d\mu (t)<\infty ,\end{aligned}$$

(2.5)

then ${\mathcal {H}}_\mu$ is bounded from $A^2$ into itself if and only if the measure $\vert h_\mu ^\prime (z)\vert ^2dA(z)$ is a Dirichlet-Carleson measure.

We recall that a finite positive Borel measure $\nu$ on ${\mathbb {D}}$ is said to be a Dirichlet-Carleson messure if ${\mathcal {D}}$ is continuously embedded in $L^2(d\nu )$. Stegenga [43] gave a characterization of these measures involving the logarithmic capacity of a finite union of intervals of $\partial {\mathbb {D}}$. Shields [39] obtained a simpler characterization when dealing with measures supported on [0, 1). This result of Shields will be used below.

Using Theorem 1 we shall prove the following result.

Theorem 3

• (i) Let $\mu$ be a positive and finite Borel measure on [0, 1). If $\mu$ is a 1-logarithmic 1-Carleson measure, then ${\mathcal {H}}_\mu$ is bounded from $A^2$ into itself.

• (ii) If $0<\beta <1$, then there exists a positive and finite Borel measure $\mu$ on [0, 1) which is a $\beta$-logarithmic 1-Carleson measure but such that ${\mathcal {H}}_\mu (A^2)\not \subset {\mathcal {A}}^2$.

In order to prove our results we start using the above mentioned result of Shields [39] to find a weak condition which insures that ${\mathcal {H}}_\mu$ and ${\mathcal {I}}_\mu$ are well defined in ${\mathcal {D}}$ and that ${\mathcal {H}}_\mu (f)=\mathcal I_\mu (f)$ for all $f\in {\mathcal {D}}$.

Proposition 4

Let $\mu$ be a positive and finite Borel measure on [0, 1). If there exists a positive constant C such that

$$\begin{aligned} \mu \left( [t, 1)\right) \le C\left( \log \frac{2}{1-t}\right) ^{-1},\quad 0<t<1, \end{aligned}$$

(2.6)

then $\mathcal H_\mu$ and ${\mathcal {I}}_\mu$ are well defined in ${\mathcal {D}}$ and, furthermore, ${\mathcal {H}}_\mu (f)={\mathcal {I}}_\mu (f)$ for all $f\in {\mathcal {D}}$.

Proof

Suppose that $\mu$ satisfies (2.6). Shields proved in [39, Theorem 2] that this is equivalent to saying that there exists a positive constant A such that

$$\begin{aligned} \int _{[0, 1)}\vert f(t)\vert ^2 \,d\mu (t)\,\le \,A\Vert f\Vert _{{\mathcal {D}}}^2,\quad f\in \mathcal D. \end{aligned}$$

(2.7)

We can express (2.7) simply by saying that $\mu$ is a radial Carleson-Dirichlet measure. Also, it is easy to see that (2.6) implies that there exists $B>0$ such that

$$\begin{aligned} \mu _n\,\le \,\frac{B}{\log (n+2)},\quad n=0, 1, 2, \dots . \end{aligned}$$

(2.8)

Take $f\in {\mathcal {D}}$, $f(z)=\sum _{n=0}^\infty a_nz^n$ ($z\in {\mathbb {D}}$).

Let us prove that ${\mathcal {I}}_\mu (f)$ is well defined.

Using (2.7) and (2.8), we see that

$$\begin{aligned} \int _{[0, 1)}t^n\vert f(t)\vert \,d\mu (t)&\le \left( \int _{[0, 1)}t^{2n}\,d\mu (t)\right) ^{1/2}\left( \int _{[0, 1)}\vert f(t)\vert ^2\,d\mu (t)\right) ^{1/2}\\&\le A^{1/2}\mu _{2n}^{1/2}\Vert f\Vert _{\mathcal D}\\&\le \frac{A^{1/2}B^{1/2}\Vert f\Vert _{{\mathcal {D}}}}{\left( \log (2n+2)\right) ^{1/2}}, \end{aligned}$$

for all n. Then we have

$$\begin{aligned} \sum _{n=0}^\infty \left( \int _{[0, 1)}t^n\vert f(t)\vert \,d\mu (t)\right) \vert z\vert ^n\lesssim \sum _{n=0}^\infty \frac{\vert z\vert ^n}{\left( \log (2n+2)\right) ^{1/2}},\quad z\in {\mathbb {D}}. \end{aligned}$$

This implies that, for all $z\in {\mathbb {D}}$, the integral

$$\begin{aligned} \int _{[0, 1)}\frac{f(t)}{1-tz}\,d\mu (t)\,=\,\int _{[0, 1)}f(t)\left( \sum _{n=0}^\infty t^nz^n\right) \,d\mu (t) \end{aligned}$$

converges and that

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

So $\mathcal I_\mu (f)$ is a well defined analytic function in ${\mathbb {D}}$ and

$$\begin{aligned} {\mathcal {I}}_\mu (f)(z)=\sum _{n=0}^\infty \left( \int _{[0, 1)}t^nf(t)\,d\mu (t)\right) z^n,\quad z\in \mathbb D.\end{aligned}$$

(2.9)

Let us see now that ${\mathcal {H}}_\mu (f)$ is also well defined and that ${\mathcal {H}}_\mu (f)={\mathcal {I}}_\mu (f)$. Using (2.8), for all n, we have

$$\begin{aligned} \sum _{k=0}^\infty \vert \mu _{n+k}a_k\vert&\lesssim \mu _n\vert a_0\vert +\sum _{k=1}^\infty \frac{k^{1/2}\vert a_k\vert }{k^{1/2}\log (n+k+2)}\\ {}&\lesssim \mu _0\vert a_0\vert +\left( \sum _{k=1}^\infty k\vert a_k\vert ^2\right) ^{1/2}\left( \sum _{k=1}^\infty \frac{1}{k\left( \log (k+1)\right) ^2}\right) ^{1/2}\\&\lesssim \Vert f\Vert _{{\mathcal {D}}}. \end{aligned}$$

Clearly, this implies that ${\mathcal {H}}_\mu$ is a well defined analytic function in ${\mathbb {D}}$. Also,

$$\begin{aligned} \int _{[0, 1)}t^nf(t)\,d\mu (t)\,=\,\int _{[0, 1)}t^n\left( \sum _{k=0}^\infty a_kt^k\right) \,d\mu (t)\,=\,\sum _{k=0}^\infty \mu _{n+k}a_k \end{aligned}$$

for all k. Then (2.9) yields that ${\mathcal {H}}_\mu (f)={\mathcal {I}}_\mu (f)$. $\square$

Let us turn now to prove Theorem 1

Proof of Theorem 1 (i)

Suppose that $\mu$ is a 1-logarithmic 1-Carleson measure. Take $f\in {\mathcal {D}}$, $f(z)=\sum _{k=0}^\infty a_kz^k$ ($z\in {\mathbb {D}}$). Proposition 4 implies that ${\mathcal {H}}_\mu (f)$ and ${\mathcal {I}}_\mu (f)$ are well defined and that ${\mathcal {H}}_\mu (f)={\mathcal {I}}_\mu (f)$. The above mentioned result of Shields yields that

$$\begin{aligned} \vert {\mathcal {H}}_{\mu }(f)(0)\vert \,=\,&\vert {\mathcal {I}}_{\mu }(f)(0)\vert \,=\,\left| \int _{[0, 1)}f(t)\,d\mu (t)\right| \nonumber \\ \,\lesssim \,&\left( \int _{[0, 1)}\vert f(t)\vert ^2\,d\mu (t)\right) ^{1/2} \,\lesssim \, \Vert f\Vert _{\mathcal D}. \end{aligned}$$

(2.10)

Since $\mu$ is a 1-logarithmic 1-Carleson measure,

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

(2.11)

(see e. g. [28, pp. 380-381]). Using (2.10) and (2.11), we obtain

$$\begin{aligned} \Vert {\mathcal {H}}_{\mu }(f)\Vert _{{\mathcal {D}}}^2\,\lesssim \,&\vert {\mathcal {H}}_{\mu }(f)(0)\vert ^2\,+\,\sum _{n=1}^\infty n\left( \sum _{k=0}^\infty \mu _{n+k}\vert a_k\vert \right) ^2 \\ \,\lesssim \,&\Vert f\Vert _{{\mathcal {D}}}^2\,+\,\sum _{n=1}^\infty n\left( \sum _{k=0}^\infty \frac{\vert a_k\vert }{(n+k)\log (n+k+1)}\right) ^2 \\ \,\lesssim \,&\Vert f\Vert _{{\mathcal {D}}}^2\,+\,I\,+\,II, \end{aligned}$$

where

$$\begin{aligned}I\,=\,&\sum _{n=1}^\infty n\left( \sum _{k=0}^n \frac{\vert a_k\vert }{(n+k)\log (n+k+1)}\right) ^2,\\II\,=\,&\sum _{n=1}^\infty n\left( \sum _{k=n+1}^\infty \frac{\vert a_k\vert }{(n+k)\log (n+k)}\right) ^2. \end{aligned}$$

Now, using a result of Holland and Walsh [30, Theorem 7] and simple estimates we deduce that

$$\begin{aligned} I\,=\,&\sum _{n=1}^\infty n\left( \sum _{k=0}^n \frac{\vert a_k\vert }{(n+k)\log (n+k+1)}\right) ^2\\ \,\le \,&\sum _{n=1}^\infty \frac{1}{n\left( \log (n+1)\right) ^2}\left( \sum _{k=0}^n\vert a_k\vert \right) ^2\,\lesssim \Vert f\Vert _{\mathcal D}^2. \end{aligned}$$

Also, since, for every n,

$$\begin{aligned} \sum _{k=n+1}^\infty \frac{\vert a_k\vert }{(n+k)\log (n+k)}\,\le \,&\frac{1}{\log (n+1)}\sum _{k=n+1}^\infty \frac{k^{1/2}\vert a_k\vert }{k^{1/2}(n+k)} \\ \,\le \,&\frac{1}{\log (n+1)}\left( \sum _{k=n+1}^\infty k\vert a_k\vert ^2\right) ^{1/2}\left( \sum _{k=n+1}^\infty \frac{1}{k(n+k)^2}\right) ^{1/2} \\ \,\le \,&\frac{\Vert f\Vert _{{\mathcal {D}}}}{\log (n+1)}\left( \sum _{k=n+1}^\infty \frac{1}{k(n+k)^2}\right) ^{1/2}\\ \,\le \,&\frac{\Vert f\Vert _{{\mathcal {D}}}}{n^{1/2}\log (n+1)}\left( \sum _{k=n+1}^\infty \frac{1}{(n+k)^2}\right) ^{1/2} \\ \,\lesssim \,&\frac{\Vert f\Vert _{{\mathcal {D}}}}{n\log (n+1)}, \end{aligned}$$

it follows that

$$\begin{aligned} II\,=\,&\sum _{n=1}^\infty n\left( \sum _{k=n+1}^\infty \frac{\vert a_k\vert }{(n+k)\log (n+k)}\right) ^2 \\ \,\lesssim \,&\Vert f\Vert _{{\mathcal {D}}}^2 \sum _{n=1}^\infty \frac{1}{n\left( \log (n+1)\right) ^2}\\\,\lesssim \,&\Vert f\Vert _{{\mathcal {D}}}^2. \end{aligned}$$

Putting everything together, we obtain $\Vert {\mathcal {H}}_\mu (f)\Vert _{{\mathcal {D}}}^2\lesssim \Vert f\Vert _{{\mathcal {D}}}^2$. $\square$

Proof of Theorem 1 (ii)

Suppose that $0<\beta <1$. Take $\alpha \in {\mathbb {R}}$ with

$$\begin{aligned} \frac{1}{2}\,<\alpha \,<\min \left( 1, \frac{3-2\beta }{2}\right) . \end{aligned}$$

Let $\mu$ be the Borel measure on [0, 1) defined by $d\mu (t)=\left( \log \frac{2}{1-t}\right) ^{-\beta }\,dt$. Then (see [28, p. 392]) $\mu$ is a $\beta$-logarithmic 1-Carleson measure and

$$\begin{aligned} \mu _n\,\asymp \,\frac{1}{n\left[ \log (n+1)\right] ^\beta }. \end{aligned}$$

Set $a_n=\frac{1}{(n+1)\left[ \log (n+1)\right] ^\alpha }$ ($n=1, 2, \dots$) and $g(z)=\sum _{n=1}^\infty a_nz^n$ ($z\in {\mathbb {D}}$).

The condition $\alpha >\frac{1}{2}$ implies that $g\in \mathcal D$. We are going to see that ${\mathcal {H}}_\mu (g)\notin \mathcal D$, this will finish the proof.

We have

$$\begin{aligned} \Vert {\mathcal {H}}_\mu (g)\Vert ^2_{{\mathcal {D}}}\,\gtrsim \,&\sum _{n=2}^\infty n\left( \sum _{k=2}^n\mu _{n+k}a_k\right) ^2\\ \,\asymp \,&\sum _{n=2}^\infty n\left( \sum _{k=2}^n\frac{1}{(n+k)\left[ \log (n+k)\right] ^\beta k\left[ \log k\right] ^\alpha }\right) ^2\\ \,\gtrsim \,&\sum _{n=2}^\infty \frac{n}{n^2\left[ \log n\right] ^{2\beta }}\left( \sum _{k=2}^n\frac{1}{k\left[ \log k\right] ^\alpha }\right) ^2 \\ \,=\,&\sum _{n=2}^\infty \frac{1}{n\left[ \log n\right] ^{2\beta }}\left( \sum _{k=2}^n\frac{1}{k\left[ \log k\right] ^\alpha }\right) ^2 \\ \,\gtrsim \,&\sum _{n=2}^\infty \frac{1}{n\left[ \log n\right] ^{2\beta +2\alpha -2}}. \end{aligned}$$

Since $2\alpha +2\beta -2<1$, $\sum _{n=2}^\infty \frac{1}{n\left[ \log n\right] ^{2\beta +2\alpha -2}}=\infty$ and, hence, $\mathcal H_\mu (g)\notin {\mathcal {D}}$ as desired. $\square$

Proof of Corollary 2

The Dirichlet space is a Hilbert space with the inner product

$$\begin{aligned} <f, g>= f(0)\overline{g(0)}\,+\,\int _{{\mathbb {D}}}f^\prime (z)\overline{g^\prime (z)}\,dA(z),\quad f, g\in {\mathcal {D}}. \end{aligned}$$

Hence, ${\mathcal {D}}$ is identifiable with its dual with this pairing.

Assume that $\mu$ is a finite Borel measure on [0, 1) which is a 1-logarithmic 1-Carleson measure. If $f\in {\mathcal {D}}$, using Theorem 1, we see that ${\mathcal {H}}_\mu (f)\in {\mathcal {D}}$ and $\Vert {\mathcal {H}}_\mu (f)\Vert _{{\mathcal {D}}}\lesssim \Vert f\Vert _{{\mathcal {D}}}$. Then ${\mathcal {H}}_\mu (f)$ induces a bounded linear functional on ${\mathcal {D}}$ with norm controlled by $\Vert f\Vert _{\mathcal D}$. Thus

$$\begin{aligned} \left| \int _{\mathbb D}{\mathcal {H}}_\mu (f)^\prime (z)\overline{g^\prime (z)}\,dA(z)\right| \,\lesssim \Vert f\Vert _{{\mathcal {D}}}\Vert g\Vert _{{\mathcal {D}}},\quad f, g\in {\mathcal {D}}. \end{aligned}$$

(2.12)

Now, using the definitions, Fubini’s theorem, and the reproducing formula for the Bergman space $A^2$, we have

$$\begin{aligned} \int _{{\mathbb {D}}}{\mathcal {H}}_\mu (f)^\prime (z)\overline{g^\prime (z)}\,dA(z)\,= \,&\int _{{\mathbb {D}}}\left( \int _{[0, 1)}\frac{tf(t)}{(1-tz)^2}\,d\mu (t)\right) \overline{g^\prime (z)}\,dA(z) \\ \,=\,&\int _{[0, 1)}tf(t)\left( \int _{{\mathbb {D}}}\frac{\overline{g^\prime (z)}}{(1-tz)^2}\,dA(z)\right) \,d\mu (t) \\ \,=\,&\int _{[0, 1)}tf(t)\overline{g^\prime (t)}\,d\mu (t). \end{aligned}$$

Using (2.12), we obtain

$$\begin{aligned} \left| \int _{[0, 1)}tf(t)\overline{g^\prime (t)}\,d\mu (t)\right| \lesssim \Vert f\Vert _{{\mathcal {D}}}\Vert g\Vert _{{\mathcal {D}}},\quad f, g\in {\mathcal {D}}. \end{aligned}$$

(2.13)

Take $f, g\in {\mathcal {D}}$, $f(z)=\sum _{n=0}^\infty a_nz^n$, $g(z)=\sum _{n=0}^\infty b_nz^n$ ($z\in {\mathbb {D}}$). Set

$$\begin{aligned} f_1(z)=\sum _{n=0}^\infty \vert a_n\vert z^n,\quad g_1(z)=\sum _{n=0}^\infty \vert b_n\vert z^n\quad (z\in {\mathbb {D}}). \end{aligned}$$

Then $f_1, g_1\in {\mathcal {D}}$, $\Vert f_1\Vert _{{\mathcal {D}}}=\Vert f\Vert _{{\mathcal {D}}}$, and $\Vert g_1\Vert _{{\mathcal {D}}}=\Vert g\Vert _{{\mathcal {D}}}$. Using (2.13) with $f_1$ and $g_1$ in the places of f and g, we obtain

$$\begin{aligned} \int _{[0, 1)}\left| tf(t)\overline{g^\prime (t)}\right| \,d\mu (t)\,\le \,&\int _{[0, 1)}\left| tf_1(t)\overline{g_1^\prime (t)}\right| \,d\mu (t) \\ \,\lesssim&\Vert f_1\Vert _{{\mathcal {D}}}\Vert g_1\Vert _{{\mathcal {D}}}\\ \,=\,&\Vert f\Vert _{{\mathcal {D}}}\Vert g\Vert _{{\mathcal {D}}}. \end{aligned}$$

Taking $f=g$, (2.3) follows.

Part (b) follows taking $d\mu (t)=\log \frac{2}{1-t}\,dt$ in part (a). $\square$

Proof of Theorem 3

Our proof of Theorem 3 is based on the fact that the pairing

$$\begin{aligned} <f, g>\,=\,f(0)\overline{g(0)}\,+\,\int _{{\mathbb {D}}}f^\prime (z)\overline{\left( \frac{g(z)-g(0)}{z}\right) }\,dA(z),\quad f\in {\mathcal {D}},\,\,g\in A^2 \end{aligned}$$

is a “duality paring” between the Dirichlet space ${\mathcal {D}}$ and the Bergman space $A^2$. Notice that if $f(z)=\sum _{n=0}^\infty a_nz^n$ and $g(z)=\sum _{n=0}^\infty b_nz^n$ ($z\in {\mathbb {D}}$), then

$$\begin{aligned} <f, g>\,=\,\sum _{n=0}^\infty a_n\overline{b_n}. \end{aligned}$$

It is a simple exercise to show that $<{\mathcal {H}}_\mu (P), Q>=<P, {\mathcal {H}}_\mu (Q)>$ if P and Q are polynomials. Then it follows that if ${\mathcal {H}}_\mu$ is a bounded operator from ${\mathcal {D}}$ into itself then its adjoint (via this pairing) is ${\mathcal {H}}_\mu$, and then we see that ${\mathcal {H}}_\mu$ is a bounded operator from $A^2$ into itself. Using this and Theorem 1 (i) we obtain part (a) of Theorem 3.

Similarly, if ${\mathcal {H}}_\mu$ is a bounded operator from $A^2$ into itself, then ${\mathcal {H}}_\mu$ is also a bounded operator from ${\mathcal {D}}$ into itself and then part (b) of Theorem 3 follows using Theorem 1 (ii). $\square$

3 Cesàro-Type Operators

For $\mu$ a finite positive Borel measure on [0, 1) as above, the matrix ${\mathcal {C}}_\mu$ induces a linear operator, also called ${\mathcal {C}}_\mu$, from $\textrm{Hol}({\mathbb {D}})$ into itself as follows: If $f\in \textrm{Hol}({\mathbb {D}})$, $f(z)=\sum _{n=0}^\infty a_nz^n$ ($z\in {\mathbb {D}}$),

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

Let us remark that the operator ${\mathcal {C}}_\mu$ has the following integral representation: If $f\in \textrm{Hol}({\mathbb {D}})$ then

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

(3.1)

When $\mu$ is the Lebesgue measure on [0, 1), the operator ${\mathcal {C}}_\mu$ reduces to the classical Cesàro operator ${\mathcal {C}}$.

The Cesàro operator ${\mathcal {C}}$ acting on distinct subspaces of $\textrm{Hol}({\mathbb {D}})$ has been extensively studied in a good number of articles such as [2, 10, 12, 15, 23, 36, 40,41,42, 44]. Let us recall that it is bounded on $H^p$ ($0<p<\infty$) and on $A^p_\alpha$ ($0<p<\infty$, $\alpha >\,-1$).

The operators ${\mathcal {C}}_\mu$ were introduced in [23] where, among other results, it was proved that the following conditions are equivalent:

• (i) $\mu$ is a Carleson measure, that is, $\mu ([t, 1))\le C(1-t)$ ($0<t<1$).

• (ii) $\mu _n={{\,\textrm{O}\,}}\left( \frac{1}{n}\right)$.

• (iii) $1\le p<\infty$ and ${\mathcal {C}}_\mu$ is bounded from $H^p$ into itself.

• (iv) $1<p<\infty$, $\alpha >-1$, and ${\mathcal {C}}_\mu$ is bounded from $A^p_\alpha$ into itself.

Blasco [12] has generalized the definition of the operators ${\mathcal {C}}_\mu$ by dealing with complex Borel measures on [0, 1) and he has extended results of [23] to this more general setting.

A further generalization has been given in [24] by working with the operators ${\mathcal {C}}_\mu$ associated to arbitrary complex Borel measures on ${\mathbb {D}}$, not necessarily supported on a radius. The complex Borel measures on ${\mathbb {D}}$ for which the operator ${\mathcal {C}}_\mu$ is bounded or Hilbert-Schmidt on $H^2$ or on $A^2_\alpha$ ($\alpha >-1$) are characterized in the mentioned paper [24].

We devote this section to study the operators ${\mathcal {C}}_\mu$ on the Dirichlet space, a question which has not been considered in the just mentioned papers. Our main results are contained in the following two theorems.

Theorem 5

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

• (i) If $\mu$ is a 1-logarithmic 1-Carleson measure, then ${\mathcal {C}}_\mu$ is a bounded operator from the Dirichlet space ${\mathcal {D}}$ into itself.

• (ii) If ${\mathcal {C}}_\mu$ is a bounded operator from ${\mathcal {D}}$ into itself then $\mu$ is a 1/2-logarithmic 1-Carleson measure.

Theorem 6

Suppose that $\frac{1}{2}<\beta <1$. Then there exists a finite positive Borel measure $\mu$ on [0, 1) which is $\beta$-logarithmic 1-Carleson measure for which ${\mathcal {C}}_\mu (\mathcal D)\not \subset {\mathcal {D}}$.

Proof of Theorem 5 (i)

Since $\mu$ is a 1-logarithmic 1-Carleson measure, we have that

$$\begin{aligned} \mu _n={{\,\textrm{O}\,}}\left( \frac{1}{(n+1)\log (n+2)}\right) . \end{aligned}$$

(3.2)

Take $f\in {\mathcal {D}}$, $f(z)=\sum _{n=0}^\infty a_nz^n$ ($z\in {\mathbb {D}}$). Using (3.2) and Theorem 7 of [30], we obtain

$$\begin{aligned} \Vert {\mathcal {C}}_\mu (f)\Vert _{\mathcal D}^2\,\le \,&\sum _{n=0}^\infty (n+1)\mu _n^2\left( \sum _{k=0}^n\vert a_k\vert \right) ^2\\ \,\lesssim \,&\sum _{n=0}^\infty \frac{\left( \sum _{k=0}^n\vert a_k\vert \right) ^2}{(n+1)[\log (n+2)]^2} \\ \,\lesssim \,&\Vert f\Vert _{{\mathcal {D}}}^2. \end{aligned}$$

$\square$

Proof of Theorem 5 (ii)

Suppose that $\mathcal C_\mu$ is a bounded operator from ${\mathcal {D}}$ into itself. For $N\in {\mathbb {N}}$, set

$$\begin{aligned} f_N(z)\,=\,\sum _{n=1}^N\frac{z^n}{n},\quad z\in {\mathbb {D}}. \end{aligned}$$

Then,

$$\begin{aligned} \Vert f_N\Vert _{{\mathcal {D}}}^2\,=\,\sum _{n=1}^N\frac{1}{n}\,\asymp \,\log (N+1). \end{aligned}$$

Since ${\mathcal {C}}_\mu$ is bounded on ${\mathcal {D}}$, bearing in mind that the sequence of moments $\{ \mu _n\}$ is decreasing, we have

$$\begin{aligned} \log (N+1)\,\asymp \,&\Vert f_N\Vert _{{\mathcal {D}}}^2\,\gtrsim \,\sum _{n=1}^\infty n\mu _n^2\left( \sum _{k=1}^n\frac{1}{k}\right) ^2 \\ \,\gtrsim \,&\mu _N^2\sum _{n=1}^Nn[\log (n+1)]^2 \,\asymp \,\mu _N^2N^2[\log (N+1)]^2. \end{aligned}$$

Then it follows that $\mu _N={{\,\textrm{O}\,}}\left( \frac{1}{N[\log (N+1)]^{1/2}}\right)$. This implies that $\mu$ is a 1/2-logarithmic 1-Carleson measure. $\square$

Proof of Theorem 6

Assume that $1/2<\beta <1$. Let $\mu$ be the Borel measure on [0, 1) defined by $d\mu (t)=\left( \log \frac{2}{1-t}\right) ^{-\beta }\,dt$. Then, as mentioned before, $\mu$ is a $\beta$-logarithmic 1-Carleson measure and $\mu _n\asymp \frac{1}{n[\log (n+1)]^\beta }$.

Set $\alpha =\beta -\frac{1}{2}$. Then $0<\alpha <\frac{1}{2}$. Define

$$\begin{aligned} g(z)=\left( \log \frac{2}{1-z}\right) ^\alpha =\sum _{n=0}^\infty A_nz^n,\quad z\in {\mathbb {D}}. \end{aligned}$$

We have that

$$\begin{aligned} A_n\asymp \frac{1}{(n+1)[\log (n+2)]^{1-\alpha }}. \end{aligned}$$

Since $\alpha <\frac{1}{2}$, we have that $g\in {\mathcal {D}}$. Also

$$\begin{aligned} \Vert {\mathcal {C}}_\mu (g)\Vert _{{\mathcal {D}}}^2\,\ge \,&\sum _{n=2}^\infty n\mu _n^2\left( \sum _{k=2}^nA_k\right) ^2\,\gtrsim \,\sum _{n=2}^\infty \frac{n}{n^2[\log n]^{2\beta }[\log n]^{-2\alpha }}\\ \,=&\sum _{n=2}^\infty \frac{1}{n[\log n]^{2(\beta -\alpha )}} \,=\,\sum _{n=2}^\infty \frac{1}{n[\log n]}\,=\,\infty . \end{aligned}$$

$\square$

Danikas and Siskakis [15] proved that ${\mathcal {C}}(H^\infty )\not \subset H^\infty$ and that $\mathcal C(H^\infty )\subset BMOA$. This was improved by Essén and Xiao who proved in [22] that ${\mathcal {C}}(H^\infty )\subset Q^p$ for $0<p<\infty$. This result has been sharpened in [10].

We recall that BMOA is the space of those functions $f\in H^1$ whose boundary values have bounded mean oscillation. Alternatively, a function $f\in \textrm{Hol}({\mathbb {D}})$ belongs to BMOA if and only if

$$\begin{aligned} \sup _{T\in \textrm{Aut}({\mathbb {D}})}\Vert f\circ T\,-\,f(T(0))\Vert _{H^2}<\infty , \end{aligned}$$

where $\textrm{Aut}({\mathbb {D}})$ denotes the set of all Möbius transformations from ${\mathbb {D}}$ onto itself. We refer to [26] for the theory of BMOA-functions.

For $0<s<\infty$ the space $Q_s$ consists of those $f\in \textrm{Hol}({\mathbb {D}})$ such that

$$\begin{aligned} \sup _{T\in \textrm{Aut}({\mathbb {D}})}\int _{{\mathbb {D}}}\vert f^\prime (z)\vert ^2(1-\vert T(z)\vert ^2)^s\,dA(z)<\infty . \end{aligned}$$

The spaces $Q_s$ were introduced in [6] and [7]. We refer to [46] for the theory of $Q_s$ spaces. Let us recall that

$$\begin{aligned} \mathcal D\subsetneq Q_{s_1}\subsetneq Q_{s_2}\subsetneq Q_1=BMOA,\quad 0<s_1<s_2<1. \end{aligned}$$

For $s>1$ the space $Q_s$ coincides with the Bloch space ${\mathcal {B}}$ of those functions $f\in \textrm{Hol}({\mathbb {D}})$ for which

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

The paper [3] is an excellent reference for the theory of Bloch functions. Let us recall that $BMOA\subsetneq {\mathcal {B}}$.

Blasco [12] has proved that

$$\begin{aligned} {\mathcal {C}}(H^\infty )\subset \bigcap _{1<p<\infty }\Lambda ^p_{1/p}. \end{aligned}$$

(3.3)

Here, for $p\ge 1$, $\Lambda ^p_{1/p}$ is the space of those functions $f\in \textrm{Hol}({\mathbb {D}})$ having a non-tangential limit at almost every point of $\partial {\mathbb {D}}$ and so that $\omega _ p(\cdot , f)$, the integral modulus of continuity of order p of the boundary values $f(e^{i\theta })$ of f, satisfies $\omega _ p(\delta , f)={{\,\textrm{O}\,}}(\delta ^{1/p})$, as $\delta \rightarrow 0$. Classical results of Hardy and Littlewood (see [13] and [20, Chapter 5]) show that $\Lambda ^p_{1/p}\subset H^p$ and that

$$\begin{aligned} \Lambda ^p_{1/p}=\left\{ f \hbox { analytic in } {\mathbb {D}}: M_ p(r, f^\prime )= {{\,\textrm{O}\,}}\left( \frac{1}{(1-r)^{1-\frac{1}{p}}}\right) , \quad \hbox { as } r\rightarrow 1 \right\} . \end{aligned}$$

In particular, $\Lambda ^1_1$ is the space of those $f\in \textrm{Hol}({\mathbb {D}})$ such that $f^\prime \in H^1$. The spaces $\Lambda ^p_{1/p}$ increase with p and they are all contained in BMOA [13]. Since $\Lambda ^2_{1/2}\subset Q_s$ for all $s>0$ (see [5, p. 427]), (3.3) improves the mentioned result in [22].

Bao, Sun and Wulan [8, Theorem 3.1] have proved that for any given $s>0$, ${\mathcal {C}}_\mu (H^\infty )\subset Q_s$ if and only if $\mu$ is a Carleson measure.

It is natural to look for a result like (3.3) with ${\mathcal {D}}$ in the place of $H^\infty$. It is easy to see that

$$\begin{aligned} {\mathcal {C}}({\mathcal {D}})\not \subset {\mathcal {B}}. \end{aligned}$$

(3.4)

Indeed, set $a_n=\frac{1}{(n+1)\log (n+1)}$ ($n\ge 1$) and $f(z)=\sum _{n=1}^\infty a_nz^n$ ($z\in \mathbb D$). Then $f\in {\mathcal {D}}$ and, setting $A_n=\sum _{k=1}^na_k$, we have, for $0<r<1$,

$$\begin{aligned} (1-r){\mathcal {C}}(f)^\prime (r)\,=\,&(1-r)\sum _{n=1}^\infty \frac{n}{n+1}A_nr^{n-1}\,\ge \,\frac{1}{2}(1-r)\sum _{n=1}^\infty A_nr^{n-1}\\ \,=\,&\frac{1}{2}\left[ A_1\,+\,\sum _{n=2}^\infty (A_n\,-\,A_{n-1})r^{n-1}\right] \,=\,\frac{1}{2}\left[ A_1\,+\,\sum _{n=2}^\infty a_nr^{n-1}\right] \\ \,\asymp \,&\log \log \frac{2}{1-r}. \end{aligned}$$

Hence, $C(f)\not \in {\mathcal {B}}$.

The next natural step is trying to characterize the measures $\mu$ such that ${\mathcal {C}}_\mu ({\mathcal {D}})\subset \mathcal B$. We have the following result.

Theorem 7

Let X be a Banach space of analytic functions in ${\mathbb {D}}$ with $\Lambda ^2_{1/2}\subset X\subset {\mathcal {B}}$ and let $\mu$ be a positive finite Borel measure on [0, 1).

• (i) If $\mu$ is a $\frac{1}{2}$-logarithmic 1-Carleson measure, then ${\mathcal {C}}_\mu$ is a bounded operator from ${\mathcal {D}}$ into X.

• (ii) If $\mathcal C_\mu$ is a bounded operator from ${\mathcal {D}}$ into X and $0<\beta <\frac{1}{2}$, then $\mu$ is a $\beta$-logarithmic 1-Carleson measure.

Proof

Suppose that $\mu$ is a $\frac{1}{2}$-logarithmic 1-Carleson measure. Then

$$\begin{aligned} \mu _n\lesssim \frac{1}{n[\log (n+1)]^{1/2}}. \end{aligned}$$

(3.5)

Take $f\in {\mathcal {D}}$, $f(z)=\sum _{n=0}^\infty a_nz^n$ ($z\in {\mathbb {D}}$). We have

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

where $A_n=\mu _n\left( \sum _{k=0}^na_k\right)$. We have,

$$\begin{aligned} \left| \sum _{k=0}^na_k\right|&\le \vert a_0\vert \,+\,\sum _{k=1}^n\frac{k^{1/2}\vert a_k\vert }{k^{1/2}}\\&\le \vert a_0\vert \,+\,\left( \sum _{k=1}^nk\vert a_k\vert ^2\right) ^{1/2}\left( \sum _{k=1}^n\frac{1}{k}\right) ^{1/2}\,\lesssim \Vert f\Vert _{{\mathcal {D}}}[\log (n+1)]^{1/2}. \end{aligned}$$

This and (3.5) imply that $\vert A_n\vert \lesssim \frac{\Vert f\Vert _{{\mathcal {D}}}}{n}$ a fact which easily yields that ${\mathcal {C}}_\mu (f)\in \Lambda ^2_{1/2}$. This finishes the proof of (i).

Let us turn to prove (ii). Assume that $0<\beta <\frac{1}{2}$ and that ${\mathcal {C}}_\mu$ is a bounded operator from ${\mathcal {D}}$ into X.

Since $X\subset {\mathcal {B}}$, ${\mathcal {C}}_\mu$ is a bounded operator from ${\mathcal {D}}$ into ${\mathcal {B}}$.

Set $\alpha =1-\beta$, and $f(z)=\sum _{n=0}^\infty \frac{z^n}{(n+1)[\log (n+2)]^\alpha }$ ($z\in {\mathbb {D}}$).

Notice that $\frac{1}{2}<\alpha <1$. This implies that $f\in \mathcal D$ and, hence, ${\mathcal {C}}_\mu (f)\in {\mathcal {B}}$. Then, bearing in mind that the sequence $\{ \mu _n\}$ is decreasing, we see that, for $0<r<1$ and $N\in {\mathbb {N}}$,

$$\begin{aligned} \frac{1}{1-r}&\gtrsim \,\sum _{n=1}^\infty n\mu _n\left( \sum _{k=1}^n\frac{1}{(k+1)[\log (k+2)]^\alpha }\right) r^{n-1} \\&\ge \sum _{n=1}^N n \mu _n\left( \sum _{k=1}^n\frac{1}{(k+1)[\log (k+2)]^\alpha }\right) r^n \\&\gtrsim \,\mu _N\sum _{n=1}^Nn[\log (n+2)]^{1-\alpha }r^n. \end{aligned}$$

Taking $r=1-\frac{1}{N}$, we obtain

$$\begin{aligned} N\,\gtrsim \,\mu _NN^2[\log (N+2)]^{1-\alpha }\,=\,\mu _NN^2[\log (N+2)]^{\beta } \end{aligned}$$

and, hence, $\mu _N\lesssim \frac{1}{N[\log (N+2)]^\beta }$. This implies that $\mu$ is a $\beta$-logarithmic 1-Carleson measure. $\square$

4 Extensions to Besov Spaces

The Dirichlet space is one among the analytic Besov spaces $B^p$. For $1<p<\infty$, the analytic Besov space $B^p$ is the space ${\mathcal {D}}^p_{p-2}$. Thus $B^2\,=\,{\mathcal {D}}$.

The minimal Besov space $B^1$ requires a special definition. It is the space of all $f\in \textrm{Hol}({\mathbb {D}})$ such that $f^{\prime \prime }\in A^1$. It is a Banach space with the norm $\Vert \cdot \Vert _{B^1}$ defined by $\Vert f\Vert _{B^1}\,=\,\vert f(0)\vert \,+\,\vert f^\prime (0)\vert \,+\,\Vert f^{\prime \prime }\Vert _{A^1}$.

The Besov spaces $B^p$ form a nested scale of conformally invariant spaces and they are all contained in BMOA:

$$\begin{aligned} B^p\,\subsetneq B^q\,\subsetneq BMOA\subsetneq {\mathcal {B}},\quad 1\le p<q<\infty . \end{aligned}$$

Also $B^p\,\subsetneq \,\Lambda ^p_{1/p}$ for all $p\in [1,\infty )$. We mention [4, 11, 18, 30, 47, 48] for information on Besov spaces. Let us remark that, letting $d\lambda$ be the Möbius invariant measure on ${\mathbb {D}}$ defined by $d\lambda (z)=\frac{dA(z)}{(1-\vert z\vert ^2)^2}$, we have:

• (a) The Bergman projection P is a continuous linear operator from $L^\infty ({\mathbb {D}})$ onto the Bloch space ${\mathcal {B}}$,

• (b) For $1<p<\infty$, the Bergman projection P is a continuous linear operator from $L^p(d\lambda )$ onto $B^p$

(see [48, Chapter 5]).

Our aim in this section is trying to extend to the spaces $B^p$ some of the results obtained in the preceding ones for the Dirichlet space.

For the space $B^1$ we have the following result.

Theorem 8

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

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

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

• (iii) The operator ${\mathcal {H}}_\mu$ is a bounded operator from $B^1$ into itself.

• (iv) The operator ${\mathcal {C}}_\mu$ is a bounded operator from $B^1$ into itself.

Proof

The equivalence (i)  $\Leftrightarrow$  (ii) is clear.

Suppose that (iii) holds. Let f be the constant function $f(z)=1$, for all $z\in {\mathbb {D}}$. Then ${\mathcal {H}}_\mu (f)={\mathcal {I}}_\mu (f)\in B^1\subset H^\infty$ and then

$$\begin{aligned} \int _{[0, 1)}\frac{d\mu (t)}{1-t}\,=\,\lim _{r\rightarrow 1^-}{\mathcal {I}}_\mu (f)(r)\,\le \,\Vert {\mathcal {I}}_\mu (f)\Vert _{H^\infty }\,<\infty . \end{aligned}$$

Thus (i) holds.

Conversely, suppose that (i) holds. Take $f\in B^1$. We have

$$\begin{aligned} {\mathcal {H}}_\mu (f)^{\prime \prime }(z)\,=\,\int _{[0, 1)}\frac{2t^2f(t)}{(1-tz)^3}\,d\mu (t),\quad z\in {\mathbb {D}}. \end{aligned}$$

Then using Fubini’s theorem, [48, Lemma 3.10], and the fact that $B^1\subset H^\infty$, we obtain

$$\begin{aligned} \int _{{\mathbb {D}}}\left| {\mathcal {H}}_\mu (f)^{\prime \prime }(z)\right| \,dA(z)\,&\lesssim \,\int _{{\mathbb {D}}}\int _{[0, 1)}\frac{\vert f(t)\vert }{\vert 1-tz\vert ^3}\,d\mu (t)\,dA(z)\\&= \int _{[0, 1)}\vert f(t)\vert \int _{{\mathbb {D}}}\frac{dA(z)}{\vert 1-tz\vert ^3} \,d\mu (t)\\&\lesssim \, \Vert f\Vert _{H^\infty }\int _{[0, 1)}\frac{d\mu (t)}{1-t}\\&\lesssim \,\Vert f\Vert _{B^1}\int _{[0, 1)}\frac{d\mu (t)}{1-t}. \end{aligned}$$

Thus, (iii) follows.

Let us prove next the equivalence (i)  $\Leftrightarrow$  (iv).

Suppose (i). Take $f\in B^1$. Bearing in mind (3.1) and using Fubini’s theorem, we see that

$$\begin{aligned}&\int _{{\mathbb {D}}}\left| {\mathcal {C}}_\mu (f)^{\prime \prime }(z)\right| dA(z)\\&\quad \lesssim \int _{[0, 1)}\int _{{\mathbb {D}}}\frac{\vert f^{\prime \prime }(tz)\vert dA(z)}{\vert 1-tz\vert }d\mu (t)\,+\, \int _{[0, 1)}\int _{{\mathbb {D}}}\frac{\vert f^{\prime }(tz)\vert dA(z) }{\vert 1-tz\vert ^2}d\mu (t)\\&\qquad +\,\int _{[0,1)}\int _{\mathbb D}\frac{\vert f(tz)\vert dA(z)}{\vert 1-tz\vert ^3}d\mu (t). \end{aligned}$$

We now estimate each of the three terms in the last formula separately. For the first one we have

$$\begin{aligned} \int _{[0,1)} \int _{{\mathbb {D}}} \frac{\vert f^{\prime \prime }(tz)\vert }{\vert 1-tz\vert }\,dA(z)\,d\mu (t)&\le \int _{[0,1)}\frac{1}{1-t} \int _{{\mathbb {D}}} \vert f^{\prime \prime }(tz)\vert \,dA(z)\,d\mu (t) \\&\lesssim \Vert f\Vert _{B^1} \int _{[0,1)}\frac{d\mu (t)}{1-t}. \end{aligned}$$

For the second one, we use the fact that $B^1\subset \Lambda ^1_1$ to obtain

$$\begin{aligned} \int _{[0,1)} \int _{{\mathbb {D}}} \frac{\vert f^{\prime }(tz)\vert }{\vert 1-tz\vert ^2} \,dA(z)\,d\mu (t)&\lesssim \int _{[0,1)} \int _0^1 \frac{M_1(tr,f^\prime )}{(1-tr)^2} \,dr\,d\mu (t) \\&\le \Vert f\Vert _{\Lambda ^1_1} \int _{[0,1)} \frac{d\mu (t)}{1-t} \lesssim \Vert f\Vert _{B^1} \int _{[0,1)}\frac{d\mu (t)}{1-t}. \end{aligned}$$

For the last integral, we use that $B^1\subset H^\infty$ and Lemma 3.10 of [48] to see that

$$\begin{aligned} \int _{[0,1)} \int _{{\mathbb {D}}} \frac{\vert f(tz)\vert }{\vert 1-tz\vert ^3} \,dA(z)\,d\mu (t) \le \,&\Vert f\Vert _{H^\infty } \int _{[0,1)} \int _{{\mathbb {D}}} \frac{dA(z)}{\vert 1-tz\vert ^3} \,d\mu (t)\\ \, \lesssim \,&\Vert f\Vert _{B^1} \int _{[0,1)}\frac{d\mu (t)}{1-t}. \end{aligned}$$

Putting everything together we obtain (iv).

Suppose now that (iv) holds. Let f be the constant function given by $f(z)=1$, for all $z\in {\mathbb {D}}$. Then ${\mathcal {C}}_\mu (f)\in B^1\subset H^\infty$. Using the integral representation of ${\mathcal {C}}_\mu$ we see that

$$\begin{aligned} \int _{[0, 1)}\frac{d\mu (t)}{1-t}\,=\,\lim _{r\rightarrow 1^-}{\mathcal {C}}_\mu (f)(r)\,\le \,\Vert {\mathcal {C}}_\mu (f)\Vert _{H^\infty }. \end{aligned}$$

Thus, $\int _{[0, 1)}\frac{d\mu (t)}{1-t}<\infty$. This is (i). $\square$

Let us turn now to deal with the possible extensions in the range $1<p<\infty$. The following result comes from [28, Theorem 2.4] and [23, Theorem 7].

Theorem D

Let $\mu$ be a positive finite Borel measure on [0, 1). If $\mu$ is a 1-logarithmic 1-Carleson measure then the operators ${\mathcal {H}}_\mu$ and ${\mathcal {C}}_\mu$ are bounded from the Bloch space ${\mathcal {B}}$ into itself.

Using this result and those obtained in Sects. 2 and  3 we will prove the following.

Theorem 9

Suppose that $2<p<\infty$ and let $\mu$ be a positive finite Borel measure on [0, 1). If $\mu$ is a 1-logarithmic 1-Carleson measure then the operators ${\mathcal {H}}_\mu$ and ${\mathcal {C}}_\mu$ are bounded from the Besov space $B^p$ into itself.

Proof

We shall use complex interpolation in the proof. Let us refer to [48, 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. As a consequence of the above mentioned results characterizing the spaces $B^p$ as the image of $L^p(d\lambda )$ under the Bergman projection and the Bloch space as the image of $L^\infty (d\lambda )$ under the Bergman projection, Zhu proves in [48, Theorem 5.25] that if $1<p_0<\infty$, $0<\theta <1$, and $1/p=(1-\theta )/p_0$, then

$$\begin{aligned}{}[B^{p_0}, {\mathcal {B}}]_{\theta }=B^p. \end{aligned}$$

(4.1)

In particular,

$$\begin{aligned} B^p=[{\mathcal {D}}, {\mathcal {B}}]_\theta , \quad \text {if } 2<p<\infty \text { and } \theta =1-\frac{2}{p}. \end{aligned}$$

(4.2)

Theorem 9 follows using (4.2), Theorem 1 (i), Theorem 5 (i), and the interpolation theorem of operators [48, Theorem 2.4]. $\square$

Regarding the sharpness of Theorem 9, we have the following result.

Theorem 10

Suppose that $0<\beta <1$.

• (i) If $1<p<\infty$ then there exists a positive Borel measure $\mu$ on [0, 1) which is a $\beta$-logarithmic 1-Carleson measure with the property that ${\mathcal {H}}_\mu (B^p)\not \subset B^p$.

• (ii) If $1<p\le 2$ then there exists a positive Borel measure $\mu$ on [0, 1) which is a $\beta$-logarithmic 1-Carleson measure with the property that ${\mathcal {C}}_\mu (B^p)\not \subset B^p$.

Proof

Assume that $1<p<\infty$ and $0<\beta <1$. Take $\alpha \in {\mathbb {R}}$ with

$$\begin{aligned} \frac{1}{p}\,<\alpha \,<\,\min \left( 1, 1+\frac{1}{p}-\beta \right) . \end{aligned}$$

Let $\mu$ be the Borel measure on [0, 1) defined by $d\mu (t)=\left( \log \frac{2}{1-t}\right) ^{-\beta } \,dt$. We know that $\mu$ is a $\beta$-logarithmic 1-Carleson measure and that $\mu _n\asymp \frac{1}{(n+1)[\log (n+2)]^\beta }.$

For $n\ge 1$, set $a_n=\frac{1}{n[\log (n+1)]^\alpha }$ and $g(z)=\sum _{n=1}^\infty a_nz^n$ ($z\in {\mathbb {D}}$).

Since the sequence $\{ a_n\}$ is decreasing and $\sum _{n=1}^\infty n^{p-1}\vert a_n\vert ^p<\infty$, using [28, Theorem 3.10] we see that $g\in B^p$.

We have that ${\mathcal {H}}_\mu (g)(z)=\sum _{n=0}^\infty \left( \sum _{k=0}^\infty \mu _{n+k}a_k\right) z^n$ ($z\in {\mathbb {D}}$). Since the $a_k$’s are positive and the sequence of moments $\{ \mu _n\}$ is decreasing, it follows that the sequence $\left\{ \sum _{k=0}^\infty \mu _{n+k}a_k\right\}$ is also decreasing. Then using again [28, Theorem 3.10] we see that

$$\begin{aligned} H_\mu (g)\in B^p\,\,\,\,\Leftrightarrow \,\,\,\,\,\sum _{n=1}^\infty n^{p-1}\left( \sum _{k=0}^\infty \mu _{n+k}a_k\right) ^p<\infty . \end{aligned}$$

(4.3)

Now,

$$\begin{aligned}\sum _{n=1}^\infty n^{p-1}\left( \sum _{k=0}^\infty \mu _{n+k}a_k\right) ^p\,\gtrsim \,&\sum _{n=2}^\infty n^{p-1}\left( \sum _{k=2}^\infty \frac{1}{(n+k)[\log (n+k)]^\beta k(\log k)^\alpha }\right) ^p\\ \,\ge \,&\sum _{n=2}^\infty n^{p-1}\left( \sum _{k=2}^n \frac{1}{(n+k)[\log (n+k)]^\beta k(\log k)^\alpha }\right) ^p \\ \,\gtrsim \,&\sum _{n=2}^\infty \frac{n^{p-1}}{n^p(\log n)^{p\beta }}\left( \sum _{k=2}^n\frac{1}{k(\log k)^\alpha }\right) ^p \\ \,\asymp \,&\sum _{n=2}^\infty \frac{1}{n(\log n)^{p\beta }(\log n)^{p(\alpha -1)}} \\ \,= \,&\sum _{n=2}^\infty \frac{1}{n(\log n)^{p(\beta +\alpha -1)}}. \end{aligned}$$

Since $p(\beta +\alpha -1)<1$, it follows that $\sum _{n=1}^\infty n^{p-1}\left( \sum _{k=0}^\infty \mu _{n+k}a_k\right) ^p=\infty$ and then (4.3) gives that $H_\mu (g)\not \in B^p$.

Assume now that $1<p\le 2$. We have

$$\begin{aligned} {\mathcal {C}}_\mu (g)(z)\,=\,\sum _{n=0}^\infty \mu _n\left( \sum _{k=0}^na_k\right) z^n. \end{aligned}$$

Using the fact that $1<p\le 2$ and [20, Theorem 6.s2] it readily follows that

$$\begin{aligned} {\mathcal {C}}_\mu (g)\in B^p\,\,\,\,\Rightarrow \,\,\,\,\,\sum _{n=1}^\infty n^{p-1}\mu _n^p\left( \sum _{k=1}^na_k\right) ^p<\infty .\end{aligned}$$

(4.4)

But,

$$\begin{aligned}\sum _{n=1}^\infty n^{p-1}\mu _n^p\left( \sum _{k=1}^na_k\right) ^p\,\gtrsim \,&\sum _{n=1}^\infty \frac{1}{n[\log (n+1)]^{\beta p}}\left( \sum _{k=2}^n\frac{1}{k(\log k)^\alpha }\right) ^p \\ \,\gtrsim \,&\sum _{n=1}^\infty \frac{1}{n[\log (n+1)]^{p(\beta +\alpha -1)}}\\ \,=\,&\infty . \end{aligned}$$

Using (4.4) we obtain that ${\mathcal {C}}_\mu (g)\not \in B^p$. $\square$