Hankel matrices acting on the Hardy space $H^1$ and on Dirichlet spaces

Abstract

If $\,\mu \,$ is a finite positive Borel measure on the interval $\,[0,1)$, we let $\,\mathcal {H}_\mu \,$ be the Hankel matrix $\,(\mu _{n, k})_{n,k\ge 0}\,$ with entries $\,\mu _{n, k}=\mu _{n+k}$, where, for $\,n\,=\,0, 1, 2, \ldots$, $\mu _n\,$ denotes the moment of order $\,n\,$ of $\,\mu$. This matrix induces formally the operator $\,\mathcal {H}_\mu (f)(z)= \sum _{n=0}^{\infty }\left( \sum _{k=0}^{\infty } \mu _{n,k}{a_k}\right) z^n\,$ on the space of all analytic functions $\,f(z)=\sum _{k=0}^\infty a_kz^k\,$, in the unit disc $\,\mathbb {D}$. When $\,\mu \,$ is the Lebesgue measure on $\,[0,1)\,$ the operator $\,\mathcal {H}_\mu \,$ is the classical Hilbert operator $\,\mathcal {H}\,$ which is bounded on $\,H^p\,$ if $\,1<p<\infty$, but not on $\,H^1$. J. Cima has recently proved that $\,\mathcal {H}\,$ is an injective bounded operator from $\,H^1\,$ into the space $\,\mathscr {C}\,$ of Cauchy transforms of measures on the unit circle. The operator $\,\mathcal {H}_\mu \,$ is known to be well defined on $\,H^1\,$ if and only if $\,\mu \,$ is a Carleson measure and in such a case we have that $\mathcal {H}_\mu (H^1)\subset \,{\mathscr {C}}$. Furthermore, it is bounded from $\,H^1\,$ into itself if and only if $\,\mu \,$ is a 1-logarithmic 1-Carleson measure. In this paper we prove that when $\,\mu \,$ is a 1-logarithmic 1-Carleson measure then $\,{\mathcal {H}}_\mu \,$ actually maps $\,H^1\,$ into the space of Dirichlet type $\,{\mathcal {D}}^1_0\,$. We discuss also the range of $\,{\mathcal {H}}_\mu \,$ on $\,H^1\,$ when $\,\mu \,$ is an $\alpha$-logarithmic 1-Carleson measure ($0<\alpha <1$). We study also the action of the operators $\,{\mathcal {H}}_\mu \,$ on Bergman spaces and on Dirichlet spaces.

1 Introduction and main results

Let $\,{\mathbb {D}}=\{z\in {\mathbb {C}}:|z|<1\}\,$ denote the open unit disc in the complex plane $\,{\mathbb {C}}$, $\partial {\mathbb {D}}\,$ will be the unit circle. The space of all analytic functions in $\,{\mathbb {D}}\,$ will be denoted by $\,{\mathcal {H}}ol({\mathbb {D}})$. We also let $\,H^p\,$ ($0<p\le \infty$) be the classical Hardy spaces. We refer to [11] for the notation and results regarding Hardy spaces.

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

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

Here, $\,dA\,$ stands for the area measure on $\,{\mathbb {D}}$, normalized so that the total area of $\,{\mathbb {D}}\,$ is $\,1$. Thus $\,dA(z)\,=\,\frac{1}{\pi }\,dx\,dy\,=\,\frac{1}{\pi }\,r\,dr\,d\theta$. The unweighted Bergman space $\,A^p_0\,$ is simply denoted by $\,A^p$. We refer to [12, 18, 29] for the notation and results about Bergman spaces.

The space of Dirichlet type $\,{{\mathcal {D}}^p_{\alpha }}\,$ ($0<p<\infty \,$ and $\,\alpha >-1$) consists of those $f\,\in {\mathcal {H}}ol({\mathbb {D}})\,$ such that $\,f^\prime \in A^p_\alpha$. In other words, a function $\,f\in {\mathcal {H}}ol({\mathbb {D}})\,$ belongs to $\,{{\mathcal {D}}^p_{\alpha }}\,$ if and only if

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

The Hilbert matrix is the infinite matrix $\,{\mathcal {H}}=\left( \frac{1}{k+n+1}\right) _{k,n\ge 0}$. It induces formally an operator, called the Hilbert operator, on spaces of analytic functions as follows:

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

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

(1)

whenever the right-hand side of (1) makes sense for all $\,z\in {\mathbb {D}}\,$ and the resulting function is analytic in $\,{\mathbb {D}}$. We define also

$$\begin{aligned} \mathcal {I}f(z)=\int _0^1\frac{f(t)}{1-tz}\,dt,\quad z\in {\mathbb {D}}, \end{aligned}$$

(2)

if the integrals in the right-hand side of (2) converge for all $\,z\in {\mathbb {D}}\,$ and the resulting function $\,\mathcal {I}f\,$ is analytic in $\,{\mathbb {D}}$. It is clear that the correspondences $\,f\mapsto \mathcal {H}f\,$ and $\,f\mapsto \mathcal {I}f\,$ are linear.

If $\,f\in H^1$, $f(z)=\sum _{n=0}^\infty a_nz^z$, then by the Fejér-Riesz inequality [11, Theorem 3. 13, p. 46] and Hardy’s inequality [11, p. 48], we have

$$\begin{aligned} \int _0^1\vert f(t)\vert \,dt\le \pi \Vert f\Vert _{H^1}\quad \text {and}\quad \sum _{n=0}^\infty \frac{a_n}{n+1}\le \pi \Vert f\Vert _{H^1}. \end{aligned}$$

This immediately yields that if $\,f\in H^1\,$ then $\,\mathcal {H}f\,$ and $\mathcal {I}f\,$ are well defined analytic functions in $\,{\mathbb {D}}\,$ and that, furthermore, $\mathcal {H}f=\mathcal {I}f$.

Diamantopoulos and Siskakis [9] proved that $\,{\mathcal {H}}\,$ is a bounded operator from $\,H^p\,$ into itself if $\,1<p<\infty$, but this is not true for $\,p=1$. In fact, they proved that ${\mathcal {H}}\left( H^1\right) \nsubseteq H^1$. Cima [6] has recently proved the following result.

Theorem A

• (i) The operator $\,{\mathcal {H}} \,$ maps $\,H^1\,$ into the space $\,{\mathscr {C}}\,$ of Cauchy transforms of measures on the unit circle $\,\partial {\mathbb {D}}$.

• (ii) $\,{\mathcal {H}}:H^1\rightarrow {\mathscr {C}}\,$ is injective.

We recall that if $\,\sigma \,$ is a finite complex Borel measure on $\,\partial {\mathbb {D}}$, the Cauchy transform $\,C\sigma \,$ is defined by

$$\begin{aligned} C\sigma (z)=\int _{\partial {\mathbb {D}}}\frac{d\sigma (\xi )}{1-\overline{\xi }\,z},\quad z\in {\mathbb {D}}. \end{aligned}$$

We let $\mathscr {M}$ be the space of all finite complex Borel measure on $\,\partial {\mathbb {D}}$. It is a Banach space with the total variation norm. The space of Cauchy transforms is $\,{\mathscr {C}}=\{ C\sigma : \sigma \in \mathscr {M}\}$. It is a Banach space with the norm $\,\Vert C\sigma \Vert {\mathop {=}\limits ^{\text {def}}}\inf \{ \Vert \tau \Vert : C\tau =C\sigma \}$. We mention [7] as an excellent reference for the main results about Cauchy transforms. We let $\,\mathcal {A}\,$ denote the disc algebra, that is, the space of analytic functions in $\,{\mathbb {D}}\,$ with a continuous extension to the closed unit disc, endowed with the $\,\Vert \cdot \Vert _{H^\infty }$-norm. It turns out [7, Chapter 4] that $\,\mathcal {A}\,$ can be identified with the pre-dual of $\,{\mathscr {C}}\,$ via the pairing

$$\begin{aligned} \langle g, C\sigma \rangle \, {\mathop {=}\limits ^{\text {def}}}\,\lim _{r\rightarrow 1}\frac{1}{2\pi }\int _0^{2\pi }g(re^{i\theta })\overline{C\sigma (re^{i\theta })}\,d\theta .\end{aligned}$$

(3)

This is the basic ingredient used by Cima to prove the inclusion $\,{\mathcal {H}}(H^1)\subset {\mathscr {C}}$.

Now we turn to consider a class of operators which are natural generalizations of the operators $\,{\mathcal {H}}\,$ and $\,{\mathcal {I}}$. If $\,\mu \,$ is a finite positive Borel measure on $\,[0, 1)\,$ and $\,n\, = 0, 1, 2, \ldots$, we let $\,\mu _n\,$ denote the moment of order $\,n\,$ of $\,\mu$, that is, $\mu _n=\int _{[0,1)}t^n\,d\mu (t),$ and we define $\,{\mathcal {H}}_\mu \,$ to be the Hankel matrix $\,(\mu _{n,k})_{n,k\ge 0}\,$ with entries $\,\mu _{n,k}=\mu _{n+k}$. The measure $\,\mu \,$ induces formally the operators $\,{\mathcal {I}}_\mu \,$ and $\,{\mathcal {H}}_\mu \,$ on spaces of analytic functions as follows:

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

for $\,f(z)=\sum _{n=0}^\infty a_nz^n\in {\mathcal {H}}ol({\mathbb {D}})\,$ being such that the terms on the right-hand sides make sense for all $\,z\in {\mathbb {D}}$, and the resulting functions are analytic in $\,{\mathbb {D}}$. If $\,\mu \,$ is the Lebesgue measure on $\,[0,1)\,$ the matrix $\,{\mathcal {H}}_\mu \,$ reduces to the classical Hilbert matrix and the operators $\,{\mathcal {H}}_\mu \,$ and $\,{\mathcal {I}}_\mu \,$ are simply the operators $\,{\mathcal {H}}\,$ and $\,{\mathcal {I}}$.

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

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

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

A 1-Carleson measure will be simply called a Carleson measure. We recall that Carleson [4] proved that $\,H^p\,\subset \,L^p(d\mu )\,$ ($0<p<\infty$) if and only if $\,\mu \,$ is a Carleson measure (see also [11, Chapter 9]).

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

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

A positive Borel measure $\mu$ on [0, 1) can be seen as a Borel measure on ${\mathbb {D}}$ by identifying it with the measure $\tilde{\mu }$ defined by

$$\begin{aligned} {\tilde{\mu }} (A)\,=\,\mu \left( A\cap [0,1)\right) ,\quad \text {for any Borel subset}~ A~ \text {of}~ {\mathbb {D}}. \end{aligned}$$

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

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

We have a similar statement for $\alpha$-logarithmic s-Carleson measures.

The action of the operators $\,{\mathcal {I}}_\mu \,$ and $\,{\mathcal {H}}_\mu \,$ on distinct spaces of analytic functions have been studied in a number of articles (see, e. g., [2, 5, 14,15,16, 22, 25, 27]).

Combining results of [14] and of [16] we can state the following result.

Theorem B

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

• (i) The operator $\,{\mathcal {I}}_\mu \,$ is well defined on $\,H^1\,$ if and only if $\,\mu \,$ is a Carleson measure.

• (ii) If $\,\mu \,$ is a Carleson measure, then the operator $\,{\mathcal {H}}_\mu \,$ is also well defined on $\,H^1\,$ and $\,{\mathcal {I}}_\mu f={\mathcal {H}}_\mu f\,$ for all $\,f\in H^1$.

• (iii) The operator $\,{\mathcal {H}}_\mu \,$ is a bounded operator from $\,H^1\,$ into itself if and only if $\,\mu \,$ is a 1-logarithmic 1-Carleson measure.

Galanopoulos and Peláez [14, Theorem 2. 2] proved the following.

Theorem C

Let $\,\mu \,$ be a positive Borel measure on $\,[0,1)$. If $\,\mu \,$ is a Carleson measure then $\,{\mathcal {H}}_\mu (H^1)\subset {\mathscr {C}}$.

This result is stronger than Theorem A(i). In view of these results, the following question arises naturally.

Question 1

Suppose that $\,\mu \,$ is a 1-logarithmic 1-Carleson measure on $\,[0,1)$. What can we say about the image $\,{\mathcal {H}}_\mu (H^1)\,$ of $\,H^1\,$ under the action of the operator $\,{\mathcal {H}}_\mu$?

To answer Question 1, let us start noticing that it is known that, for $0<p\le 2$, the space of Dirichlet type $\,{{\mathcal {D}}^p_{p-1}}$ is continuously included in $H^p\,$ (see [26, Lemma 1. 4]). In particular, the space $\,{\mathcal {D}}^1_0\,$ is continuously included in $\,H^1$. In fact, the estimates obtained by Vinogradov in the proof of his lemma easily yield the inequality

$$\begin{aligned} \Vert f\Vert _{H^1}\le 2\Vert f\Vert _{{\mathcal {D}}^1_0},\quad f\in {\mathcal {D}}^1_0. \end{aligned}$$

We shall prove that if $\,\mu \,$ is a 1-logarithmic 1-Carleson measure on $\,[0,1)\,$ then $\,{\mathcal {H}}_\mu (H^1)\,$ is contained in the space $\,{\mathcal {D}}^1_0$. Actually, we have the following stronger result.

Theorem 1

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

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

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

• (iii) ${\mathcal {H}}_\mu \,$ is a bounded operator from $\,H^1\,$ into $\,{\mathcal {D}}^1_0$.

• (iv) ${\mathcal {H}}_\mu \,$ is a bounded operator from $\,{\mathcal {D}}^1_0\,$ into $\,{\mathcal {D}}^1_0$.

There is a gap between Theorem C and Theorem 1 and so it is natural to discuss the range of $\,H^1\,$ under the action of $\,{\mathcal {H}}_\mu \,$ when $\,\mu \,$ is an $\,\alpha$-logarithmic 1-Carleson measure with $\,0<\alpha <1$. We shall prove the following result.

Theorem 2

Let $\,\mu \,$ be a positive Borel measure on $\,[0,1)$. Suppose that $\,0<\alpha <1\,$ and that $\,\mu \,$ is an $\,\alpha$-logarithmic 1-Carleson measure. Then $\,{\mathcal {H}}_\mu \,$ maps $\,H^1\,$ into the space $\,{\mathcal {D}}^1(\log ^{\alpha -1})\,$ defined as follows:

$$\begin{aligned} {\mathcal {D}}^1(\log ^{\alpha -1})=\left\{ f\in {\mathcal {H}}ol({\mathbb {D}}) : \int _{{\mathbb {D}}} \vert f^\prime (z)\vert \left( \log \frac{2}{1-\vert z\vert }\right) ^{\alpha -1}\,dA(z)<\infty \right\} . \end{aligned}$$

These results will be proved in Sect. 2. Since the space of Dirichlet type $\,{\mathcal {D}}^1_0\,$ has showed up in a natural way in our work, it seems natural to study the action of the operators $\,{\mathcal {H}}_\mu \,$ and $\,{\mathcal {I}}_\mu \,$ on the Bergman spaces $\,A^p_{\alpha }\,$ and the Dirichlet spaces $\,{{\mathcal {D}}^p_{\alpha }}\,$ for general values of the parameters $\,p\,$ and $\,\alpha$. This will be done in Sect. 3.

Throughout this paper the letter C denotes a positive constant that may change from one step to the next. Moreover, for two real-valued functions $E_1, E_2$ we write $E_1\lesssim E_2$, or $E_1\gtrsim E_2$, if there exists a positive constant C independent of the arguments such that $E_1\le C E_2$, respectively $E_1\ge C E_2$. If we have $E_1\lesssim E_2$ and $E_1\gtrsim E_2$ simultaneously then we say that $E_1$ and $E_2$ are equivalent and we write $E_1\asymp E_2$.

2 Proofs of the theorems 1 and  2

Proof of Theorem 1

We already know that (i) and (ii) are equivalent by Theorem B.

To prove that (i) implies (iii) we shall use some results about the Bloch space. We recall that a function $\,f\in {\mathcal {H}}ol({\mathbb {D}})\,$ 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 will be denoted by $\,\mathcal {B}$. It is a non-separable Banach space with the norm $\,\Vert \cdot \Vert _{\mathcal {B}}\,$ just defined. A classical source for the theory of Bloch functions is [1]. The closure of the polynomials in the Bloch norm is the little Bloch space$\,\mathcal {B}_0\,$ which consists of those $\,f\in {\mathcal {H}}ol({\mathbb {D}})\,$ with the property that

$$\begin{aligned}\lim _{\vert z\vert \rightarrow 1}(1-\vert z\vert ^2)\vert f^\prime (z)\vert =0.\end{aligned}$$

It is well known that (see [1, p. 13])

$$\begin{aligned} \vert f(z)\vert \lesssim \Vert f\Vert _{\mathcal {B}}\log \frac{2}{1-\vert z\vert }.\end{aligned}$$

(4)

The basic ingredient to prove that (i) implies (iii) is the fact that the dual $\,\left( \mathcal {B}_0\right) ^*\,$ of the little Bloch space can be identified with the Bergman space $\,A^1$ via the integral pairing

$$\begin{aligned} \langle h,f\rangle \,=\,\int _{{\mathbb {D}}}\,h(z)\,\overline{f(z)}\,dA(z),\quad h\in \mathcal {B}_0, \, f\in A^1. \end{aligned}$$

(5)

(See [29, Theorem 5. 15]).

Let us proceed to prove the implication (i) $\Rightarrow$ (iii). Assume that $\,\mu \,$ is a 1-logarithmic 1-Carleson measure and take $\,f\in H^1$. We have to show that $\,{\mathcal {I}}_\mu f\in {\mathcal {D}}^1_0\,$ or, equivalently, that $\,\left( {\mathcal {I}}_\mu f\right) ^\prime \in A^1$. Since $\,\mathcal {B}_0\,$ is the closure of the polynomials in the Bloch norm, it suffices to show that

$$\begin{aligned} \left| \int _{{\mathbb {D}}}\,h(z)\,\overline{\left( {\mathcal {I}}_\mu f\right) ^\prime (z)}\,dA(z)\right| \,\lesssim \Vert h\Vert _{\mathcal {B}}\Vert f\Vert _{H^1},\quad \text {for any polynomial}~ h.\end{aligned}$$

(6)

So, let $\,h\,$ be a polynomial. We have

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

Because of the reproducing property of the Bergman kernel [29, Proposition 4. 23], $\,\int _{{\mathbb {D}}}\,\frac{h(z)}{(1-t\,\overline{z})^2}\,dA(z)\,=\,h(t)$. Then it follows that

$$\begin{aligned} \int _{{\mathbb {D}}}\,h(z)\,\overline{\left( {\mathcal {I}}_\mu f\right) ^\prime (z)}\,dA(z)\,=\,\int _{[0,1)}\,t\,\overline{f(t)}\,h(t)\,d\mu (t).\end{aligned}$$

(7)

Since $\,\mu \,$ is a 1-logarithmic 1-Carleson measure, the measure $\,\nu \,$ defined by

$$\begin{aligned} d\nu (t)\,=\,\log \frac{2}{1-t}\,d\mu (t) \end{aligned}$$

is a Carleson measure [15, Proposition 2. 5]. This implies that

$$\begin{aligned} \int _{[0,1)}\vert f(t)\vert \log \frac{2}{1-t}\,d\mu (t)\lesssim \Vert f\Vert _{H^1}. \end{aligned}$$

This and (4) yield

$$\begin{aligned} \int _{[0,1)}\left| t\,\overline{f(t)}\,h(t)\right| \,d\mu (t)\,\lesssim \,\Vert h\Vert _{\mathcal {B}}\Vert f\Vert _{H^1}. \end{aligned}$$

Using this and (7), (6) follows.

Since $\,{\mathcal {D}}^1_0\subset H^1\,$, the implication (iii)  $\Rightarrow$  (iv) is trivial. To prove that (iv) implies (i) we shall use the following result of Pavlović [23, Theorem 3. 2].

Theorem D

Let $\,f\in {\mathcal {H}}ol({\mathbb {D}})$, $\,f(z)=\sum _{n=0}^\infty a_nz^n$, and suppose that the sequence $\,\{ a_n\}$ is a decreasing sequence of non-negative real numbers. Then $\,f\in {\mathcal {D}}^1_0\,$ if and only if $\,\sum _{n=0}^\infty \frac{a_n}{n+1}<\infty$, and we have

$$\begin{aligned} \Vert f\Vert _{{\mathcal {D}}^1_0}\,\asymp \,\sum _{n=0}^\infty \frac{a_n}{n+1}. \end{aligned}$$

Now we turn to prove the implication (iv) $\Rightarrow$ (i). Assume that ${\mathcal {H}}_\mu$ is a bounded operator from $\,{\mathcal {D}}^1_0\,$ into $\,{\mathcal {D}}^1_0$. We argue as in the proof of Theorem 1. 1 of [16]. For $\,\frac{1}{2}<b<1\,$ set

$$\begin{aligned} f_b(z)=\frac{1-b^2}{(1-bz)^2},\quad z\in {\mathbb {D}}. \end{aligned}$$

We have $\,f_b^\prime (z)=\frac{2b(1-b^2)}{(1-bz)^3}\,$ ($z\in {\mathbb {D}}$). Then, using Lemma 3. 10 of [29] with $\,t=0\,$ and $\,c=1$, we see that

$$\begin{aligned} \Vert f_b\Vert _{{\mathcal {D}}^1_0}\,\asymp \,\int _{{\mathbb {D}}}\frac{1-b^2}{\vert 1-bz\vert ^3}\,dA(z)\,\asymp 1. \end{aligned}$$

Since $\,{\mathcal {H}}_\mu \,$ is bounded on $\,{\mathcal {D}}^1_0$, this implies that

$$\begin{aligned} 1\gtrsim \Vert {\mathcal {H}}_\mu (f_b)\Vert _{{\mathcal {D}}^1_0}.\end{aligned}$$

(8)

We also have,

$$\begin{aligned} f_b(z)\,=\,\sum _{k=0}^\infty a_{k,b}z^k,\quad \text {with}~ a_{k,b}=(1-b^2)(k+1)b^k. \end{aligned}$$

Since the $a_{k,b}$’s are positive, it is clear that the sequence $\{ \sum _{k=0}^\infty \mu _{n+k}a_{k,b}\} _{n=0}^\infty$ of the Taylor coefficients of ${\mathcal {H}}_\mu (f_b)$ is a decreasing sequence of non-negative real numbers. Using this, Theorem D, (8), and the definition of the $a_{k,b}$’s, we obtain

$$\begin{aligned} 1\,&\gtrsim \,\Vert {\mathcal {H}}_\mu (f_b)\Vert _{{\mathcal {D}}^1_0}\,\gtrsim \,\sum _{n=1}^\infty \frac{1}{n}\left( \sum _{k=0}^\infty \mu _{n+k}a_{k,b}\right) \\&= \sum _{n=1}^\infty \frac{1}{n}\left( \sum _{k=0}^\infty a_{k,b}\int _{[0,1)}t^{n+k}\,d\mu (t)\right) \\&\gtrsim \,(1-b^2)\sum _{n=1}^\infty \frac{1}{n}\left( \sum _{k=1}^\infty kb^k\int _{[b,1)}t^{n+k}\,d\mu (t)\right) \\&\gtrsim \,(1-b^2)\sum _{n=1}^\infty \frac{1}{n}\left( \sum _{k=1}^\infty kb^{n+2k}\,\mu \left( [b,1)\right) \right) \\&=\,(1-b^2)\mu \left( [b,1)\right) \sum _{n=1}^\infty \frac{b^n}{n}\left( \sum _{k=1}^\infty kb^{2k}\right) \\&= \,(1-b^2)\mu \left( [b,1)\right) \left( \log \frac{1}{1-b}\right) \frac{b^2}{(1-b^2)^2}. \end{aligned}$$

Then it follows that

$$\begin{aligned} \mu \left( [b,1)\right) \,=\,{{\,\mathrm{O}\,}}\left( \frac{1-b}{\log \frac{1}{1-b}}\right) ,\quad \text {as}~ b\rightarrow 1. \end{aligned}$$

Hence, $\mu$ is a 1-logarithmic 1-Carleson measure. $\square$

Before embarking on the proof of Theorem 2 we have to introduce some notation and results. Following [24], for $\,\alpha \in \mathbb {R}\,$ the weighted Bergman space $\,A^1(\log ^\alpha )\,$ consists of those $\,f\in {\mathcal {H}}ol({\mathbb {D}})\,$ such that

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

This is a Banach space with the norm $\,\Vert \cdot \Vert _{A^1(\log ^\alpha )}\,$ just defined and the polynomials are dense in $\,A^1(\log ^\alpha )$. Likewise, we define

$$\begin{aligned} {\mathcal {D}}^1(\log ^\alpha )=\{ f\in {\mathcal {H}}ol({\mathbb {D}}) : f^\prime \in A^1(\log ^\alpha )\} . \end{aligned}$$

We define also the Bloch-type space $\,\mathcal {B}(\log ^\alpha )\,$ as the space of those $\,f\in {\mathcal {H}}ol({\mathbb {D}})\,$ such that

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

and

$$\begin{aligned} {\mathcal {B}}_0(\log ^\alpha )=\left\{ f\in {\mathcal {H}}ol({\mathbb {D}}) : \vert f^\prime (z)\vert ={{\,\mathrm{o}\,}}\left( \frac{\left( \log \frac{2}{1-\vert z\vert }\right) ^\alpha }{1-\vert z\vert }\right) ,\,\,\,\text {as}~ \,\vert z\vert \rightarrow 1~ \right\} . \end{aligned}$$

The space $\,\mathcal {B}(\log ^\alpha )\,$ is a Banach space and $\,\mathcal {B}_0(\log ^\alpha )\,$ is the closure of the polynomials in $\,\mathcal {B}(\log ^\alpha )$.

We remark that the spaces $\,{\mathcal {D}}^1(\log ^\alpha )\,$, $\,\mathcal {B}(\log ^\alpha )\,$, and $\,\mathcal {B}_0(\log ^\alpha )\,$ were called $\,\mathfrak {B}^1_{\log ^\alpha }$, $\,\mathfrak {B}_{\log ^\alpha }$, and $\,\mathfrak {b}_{\log ^\alpha }\,$ in [24]. Pavlović identified in [24, Theorem 2. 4] the dual of the space $\,{\mathcal {B}}_0(\log ^\alpha )$.

Theorem E

Let $\,\alpha \in \mathbb {R}$. Then the dual of $\,{\mathcal {B}}_0(\log ^\alpha )\,$ is $\,A^1(\log ^\alpha )\,$ via the pairing

$$\begin{aligned} \langle h, g\rangle \,=\,\int _{{\mathbb {D}}}\,f(z)\,\overline{g(z)}\,dA(z),\quad h\in {\mathcal {B}}_0(\log ^\alpha ),\,\,\,g \in A^1(\log ^\alpha ). \end{aligned}$$

Actually, Pavlović formulated the duality theorem in another way but it is a simple exercise to show that his formulation is equivalent to this one which is better suited to our work.

Proof of Theorem 2

Let $\,\mu \,$ be a positive Borel measure on $\,[0,1)\,$ and $\,0<\alpha <1$. Suppose that $\,\mu \,$ is an $\,\alpha$-logarithmic 1-Carleson measure. Take $\,f\in H^1$. We have to show that $\,{\mathcal {I}}_\mu f\in {\mathcal {D}}^1(\log ^{\alpha -1})\,$ or, equivalently, that $\,\left( {\mathcal {I}}_\mu f\right) ^\prime \in A^1(\log ^{\alpha -1})$. Bearing in mind Theorem E and the fact that $\,{\mathcal {B}}_0(\log ^{\alpha -1})\,$ is the closure of the polynomials in $\,{\mathcal {B}}(\log ^{\alpha -1})\,$, it suffices to show that

$$\begin{aligned} \left| \int _{{\mathbb {D}}}\,h(z)\,\overline{\left( {\mathcal {I}}_\mu f\right) ^\prime (z)}\,dA(z)\right| \,\lesssim \Vert h\Vert _{{\mathcal {B}}(\log ^{\alpha -1})}\Vert f\Vert _{H^1},\quad \text {for any polynomial}~ h.\end{aligned}$$

(9)

So, let $\,h\,$ be a polynomial. Arguing as in the proof of the implication (i) $\Rightarrow$ (iii) in Theorem 1 we obtain

$$\begin{aligned} \int _{{\mathbb {D}}}h(z)\,\overline{\left( {\mathcal {I}}_\mu f\right) ^\prime (z)}\,dA(z)\,=\,\int _{[0,1)}t\,\overline{f(t)}\,h(t)\,d\mu (t).\end{aligned}$$

(10)

Now, it is clear that

$$\begin{aligned} \vert h(z)\vert \lesssim \Vert h\Vert _{{\mathcal {B}}(\log ^{\alpha -1})}\left( \log \frac{2}{1-\vert z\vert }\right) ^\alpha , \end{aligned}$$

and then it follows that

$$\begin{aligned} \int _{[0,1)}\left| t\,\overline{f(t)}\,h(t)\right| \,d\mu (t)\lesssim \Vert h\Vert _{{\mathcal {B}}(\log ^{\alpha -1})}\int _{[0,1)}\vert f(t)\vert \left( \log \frac{2}{1-t}\right) ^{\alpha }\,d\mu (t). \end{aligned}$$

Using the fact that the measure $\,\left( \log \frac{2}{1-t}\right) ^{\alpha }\,d\mu (t)\,$ is a Carleson measure [15, Proposition 2. 5], this implies that

$$\begin{aligned} \int _{[0,1)}\left| t\,\overline{f(t)}\,h(t)\right| \,d\mu (t)\lesssim \Vert h\Vert _{{\mathcal {B}}(\log ^{\alpha -1})}\Vert f\Vert _{H^1}. \end{aligned}$$

This and (10) give (9). $\square$

3 The operators ${\mathcal {H}}_\mu$ acting on Bergman spaces and on Dirichlet spaces

Jevtić and Karapetrović [20] have recently proved the following result.

Theorem F

The Hilbert operator $\,{\mathcal {H}}\,$ is a bounded operator from $\,{{\mathcal {D}}^p_{\alpha }}\,$ into itself if and only if $\,\max (-1, p-2)\,<\alpha \,<2p-2$.

Now, it is well known that $\,A^p_\alpha \,=\,{\mathcal {D}}^p_{\alpha +p}\,$ (see [29, Theorem 4. 28]). Hence, regarding Bergman spaces Theorem F says the following.

Corollary G

The Hilbert operator $\,{\mathcal {H}}\,$ is a bounded operator from $\,A^p_\alpha \,$ into itself if and only if $\,-1<\alpha <p-2$.

Let us recall that Diamantopoulos [8] had proved before that the Hilbert operator is bounded on $\,A^p\,$ for $\,p>2$, but not on $\,A^2$. The situation on $\,A^2\,$ is even worse. Dostanić, Jevtić, and Vukotić [10] proved that the Hilbert operator is not well defined on $\,A^2$. Indeed, they considered the function $\,f\,$ defined by

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

(11)

which belongs to $\,A^2.$ However, the series defining $\,\mathcal {H}f(0)\,$ is $\,\sum _{n=1}^\infty \frac{1}{(n+1)\log (n+1)}=\infty \,$ and the integral defining $\,\mathcal {I}f(0)\,$ is $\,\int _0^1\,f(t)\,dt\,=\,\infty$. Hence neither $\,{\mathcal {H}}\,$ nor $\,{\mathcal {I}}\,$ are defined on $\,A^2$.

This result can be extended. We can assert that $\,{\mathcal {H}}\,$ is not well defined on $\,A^p_{p-2}\,$ for any $\,p>1$. Indeed, let $\,f\,$ be the function defined in (11). Notice that the sequence $\,\{ \frac{1}{(n+1)\log (n+1)}\}$ is decreasing and that $\,\sum _{n=1}^\infty \frac{1}{n\left( \log (n+1)\right) ^p}<\infty$. Then (see Proposition 1 below) it follows that $\,f\in A^p_{p-2}$, and we have already seen that $\,\mathcal {H}f\,$ and $\,\mathcal {I}f\,$ are not defined. Since $\,\alpha \ge p-2\,\,\Rightarrow \,\, A^p_{p-2} \subset A^p_\alpha$, it follows that the Hilbert operator $\,{\mathcal {H}}\,$ is not defined on $\,A^p_\alpha \,$ if $\,\alpha \ge p-2$.

In this section we shall obtain extensions of the mentioned results of Jevtić and Karapetrović considering the generalized Hilbert operators $\,{\mathcal {H}}_\mu \,$.

Theorem 3

Suppose that $\,\max (-1, p-2)\,<\alpha \,<2p-2\,$ and let $\,\mu \,$ be a finite positive Borel measure on $\,[0,1)$. If $\,\mu \,$ is a Carleson measure then the operators $\,{\mathcal {H}}_\mu \,$ and $\,{\mathcal {I}}_\mu \,$ are well defined on $\,{{\mathcal {D}}^p_{\alpha }}$. Furthermore, ${\mathcal {I}}_\mu f={\mathcal {H}}_\mu f$, for all $\,f\in {{\mathcal {D}}^p_{\alpha }}$.

When dealing with Bergman spaces Theorem 3 reduces to the following.

Corollary 1

Suppose that $\,p>1\,$ and $\,-1\,<\alpha \,<\,p-2\,$, and let $\,\mu \,$ be a finite positive Borel measure on $\,[0,1)$. If $\,\mu \,$ is a Carleson measure then the operators $\,{\mathcal {H}}_\mu \,$ and $\,{\mathcal {I}}_\mu \,$ are well defined on $\,A^p_\alpha$. Furthermore, ${\mathcal {I}}_\mu f={\mathcal {H}}_\mu f$, for all $\,f\in A^p_\alpha$.

Proof of Theorem 3

Suppose that $\,\mu \,$ is a Carleson measure and take $\,f\in {{\mathcal {D}}^p_{\alpha }}$. Set $\,\beta =\frac{2+\alpha }{p}-1$. Observe that $\,0<\beta <1$. Using [29, Theorem 4. 14], we see that $\,\vert f^\prime (z)\vert \,\lesssim \frac{1}{(1-\vert z\vert )^{(2+\alpha )/p}}\,$ and, hence, $\,\vert f(z)\vert \,\lesssim \frac{1}{(1-\vert z\vert )^\beta }$. Then it follows that

$$\begin{aligned}\int _{[0,1)}\vert f(t)\vert \,d\mu (t)\,\lesssim \,\int _{[0,1)}\frac{d\mu (t)}{(1-t)^\beta }.\end{aligned}$$

Integrating by parts, using that $\,\mu \,$ is a Carleson measure, and that $\,0<\beta <1$, we obtain

$$\begin{aligned} \int _{[0,1)}\frac{d\mu (t)}{(1-t)^\beta }\,&=\,\mu ([0,1))\,-\,\lim _{t\rightarrow 1}\frac{\mu ([t,1))}{(1-t)^{\beta }}\,+\,\beta \int _0^1\,\frac{\mu ([t,1))}{(1-t)^{\beta +1}}\,dt \\&=\,\mu ([0,1))\,+\,\beta \int _0^1\,\frac{\mu ([t,1))}{(1-t)^{\beta +1}}\,dt \\&\lesssim \,\mu ([0,1))\,+\,\int _0^1\,\frac{1}{(1-t)^{\beta }}\,dt \\&<\,\infty . \end{aligned}$$

Consequently, we obtain that

$$\begin{aligned} \int _{[0,1)}\vert f(t)\vert \,d\mu (t)\,<\,\infty .\end{aligned}$$

(12)

Clearly, this implies that the integral

$$\begin{aligned} \int _{[0,1)}\frac{f(t)\,d\mu (t)}{1-tz}\quad \text {converges absolutely and uniformly on compact subsets of}~ {\mathbb {D}}.\end{aligned}$$

(13)

This gives that $\,{\mathcal {I}}_\mu f\,$ is a well defined analytic function in ${\mathbb {D}}$ and that

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

(14)

Using [19, Theorem 2. 1] (see also [20, Theorem 2. 1]) we see that for these values of $\,p\,$ and $\,\alpha \,$ we have that if $\,f\in A^p_\alpha$, $f(z)=\sum _{n=0}^\infty a_nz^n$, then $\,\sum _{k=0}^\infty \frac{\vert a_k\vert }{k+1}<\infty$. Now, since $\,\mu \,$ is a Carleson measure we have that $\,\vert \mu _n\vert \lesssim \frac{1}{n+1}$ ([5, Proposition 1]). Then it follows that

$$\begin{aligned} \sum _{k=0}^\infty \vert \mu _{n+k}a_k\vert \,\lesssim \,\sum _{k=0}^\infty \frac{\vert a_k\vert }{k+n+1}\lesssim \sum _{k=0}^\infty \frac{\vert a_k\vert }{k+1},\quad \text {for all}~ \,n. \end{aligned}$$

Clearly, this implies that $\,{\mathcal {H}}_\mu f\,$ is a well defined analytic function in $\,{\mathbb {D}}\,$ and that $\,\int _{[0,1)}t^n\,f(t)\,d\mu (t)\,=\,\sum _{k=0}^\infty \mu _{n+k}a_k\,$ for all $\,n$. This and (13) give that $\,{\mathcal {I}}_\mu f\,=\,{\mathcal {H}}_\mu f$. $\square$

Our next result is an extension of Corollary G.

Theorem 4

Suppose that $\,-1\,<\alpha \,<\,p-2\,$ and let $\,\mu \,$ be a finite positive Borel measure on $\,[0,1)$.

The operator $\,{\mathcal {H}}_\mu \,$ is well defined on $\,A^p_\alpha \,$ and it is a bounded operator from $\,A^p_\alpha \,$ to itself if and only if $\,\mu \,$ is a Carleson measure.

A number of results will be needed to prove this theorem. We start with a characterization of the functions $\,f\in {\mathcal {H}}ol({\mathbb {D}})\,$ whose sequence of Taylor coefficients is decreasing which belong to $\,A^p_\alpha$.

Proposition 1

Let $\,f\in {\mathcal {H}}ol({\mathbb {D}})$, $f(z)=\sum _{n=0}^\infty a_n\,z^n$$\,(z\in {\mathbb {D}})$. Suppose that $\,1<p<\infty$, $\,\alpha >-1$, and that the sequence $\{ a_n\} _{n=0}^\infty$ is a decreasing sequence of non-negative real numbers. Then

$$\begin{aligned} f\in A^p_\alpha \,\,\,\Leftrightarrow \,\,\,\sum _{n=1}^\infty n^{p-3-\alpha }a_n^p\,<\,\infty . \end{aligned}$$

Furthermore, $\Vert f\Vert _{A^p_\alpha }^p\,\asymp \,\vert a_0\vert ^p\,+\,\sum _{n=1}^\infty n^{p-3-\alpha }a_n^p\,<\,\infty$.

This result can be proved with arguments similar to those used in the proofs of [15, Theorem 3.10] and [23, Theorem 3. 1] where the analogous results for the Besov spaces $\,B^p={\mathcal {D}}^p_{p-2}$ ($p>1$) and for the spaces $\,{{\mathcal {D}}^p_{p-1}}\,$ ($p>1$) were proved. The case $\,\alpha =0\,$ is proved in [3, Proposition 2. 4]. Consequently, we omit the details.

The following lemma is a generalization of [13, Lemma 3 (ii)].

Lemma 1

Let $\,\mu \,$ be a positive Borel measure on $\,[0,1)\,$ which is a Carleson measure. Assume that $\,0<p<\infty \,$ and $\,\alpha >-1$. Then there exists a positive constant $\,C=C(p,\alpha ,\mu )\,$ such that for any $\,f\in A^p_\alpha \,$

$$\begin{aligned} \int _{[0,1)}M_\infty ^p(r,f)(1-r)^{\alpha +1}\,d\mu (r)\,\le C\Vert f\Vert _{A^p_\alpha }^p. \end{aligned}$$

Of course, $M_\infty (r,f)=\sup _{\vert z\vert =r}\vert f(z)\vert$.

Proof

Take $\,f\in A^p_\alpha \,$ and set

$$\begin{aligned}&g(r)=M_\infty ^p(r,f)(1-r)^{\alpha +1},\,\, \\&\quad F(r)=\mu ([0,r))\,-\mu ([0,1))\,=\,-\mu ([r,1)),\,\, 0<r<1. \end{aligned}$$

Integrating by parts, we have

$$\begin{aligned}&\int _{[0,1)}M_\infty ^p(r,f)(1-r)^{\alpha +1}\,d\mu (r)\,=\,\int _{[0,1)}g(r)\,d\mu (r)\nonumber \\&\quad =\,\lim _{r\rightarrow 1}g(r)F(r)\,-\,g(0)F(0)\,-\int _0^1g^\prime (r)F(r)\,dr \nonumber \\&\quad =\,\vert f(0)\vert ^p\mu ([0,1))\,-\,\lim _{r\rightarrow 1}M_\infty ^p(r,f)(1-r)^{\alpha +1}\mu ([r,1))\,\nonumber \\&\qquad +\,\int _0^1g^\prime (r)\mu ([r, 1))\,dr. \end{aligned}$$

(15)

Since $\,f\in A^p_\alpha \,$ we have that $\,M_\infty ^p(r,f)={{\,\mathrm{o}\,}}\left( (1-r)^{-2-\alpha }\right)$, as $\,r\rightarrow 1$ (see, e. g., [18, p. 54]). This and the fact that $\,\mu \,$ is a Carleson measure imply that

$$\begin{aligned} \lim _{r\rightarrow 1}M_\infty ^p(r,f)(1-r)^{\alpha +1}\mu ([r,1))\,=\,0.\end{aligned}$$

(16)

Using again that $\,\mu \,$ is a Carleson measure and integrating by parts we see that

$$\begin{aligned} \int _0^1g^\prime (r)\mu ([r, 1))\,dr\,&\lesssim \,\int _0^1g^\prime (r)(1-r)\,dr\\&=\,\lim _{r\rightarrow 1} g(r)(1-r)\,-g(0)\,+\,\int _0^1g(r)\,dr\\&\le \, \lim _{r\rightarrow 1}M_\infty ^p(r,f)(1-r)^{\alpha +2}\,+\,\int _0^1M_\infty ^p(r,f)(1-r)^{\alpha +1}\,dr \\&=\,\int _0^1M_\infty ^p(r,f)(1-r)^{\alpha +1}\,dr. \end{aligned}$$

Then, using [13, Lemma  3. (ii)], it follows that

$$\begin{aligned}\int _0^1g^\prime (r)\mu ([r, 1))\,dr\,\lesssim \,\Vert f\Vert _{A^p_\alpha }^p.\end{aligned}$$

Using this and (16) in (15) readily yields $\,\int _{[0,1)}M_\infty ^p(r,f)(1-r)^{\alpha +1}\,d\mu (r)\,\lesssim \,\Vert f\Vert _{A^p_\alpha }^p$. $\square$

We shall also need the following characterization of the dual of the spaces $\,A^q_\beta$ ($q>1$). It is a special case of [21, Theorem 2. 1].

Lemma 2

If $1<q<\infty$ and $\beta >-1$, then the dual of $A^q_\beta$ can be identified with $A^{p}_{\alpha }$ where $\frac{1}{p}+\frac{1}{q}=1$ and $\alpha$ is any number with $\alpha >-1$, under the pairing

$$\begin{aligned} \langle h, f\rangle _{A_{q,\beta ,\alpha }}=\int _{\mathbb {D}}h(z)\overline{f(z)}(1-|z|^2)^{\frac{\beta }{q}+\frac{\alpha }{p}}\,dA(z),\quad h\in A^q_\beta ,\,\,\,\,f\in A^p_\alpha . \end{aligned}$$

(17)

Finally, we recall the following result from [13, (5. 2), p. 242] which is a version of the classical Hardy’s inequality [17, pp. 244–245].

Lemma 3

Suppose that $\,k>0$, $\,q>1$, and $\,h\,$ is a non-negative function defined in $\,(0,1)$, then

$$\begin{aligned} \int _0^1\left( \int _{1-r}^1 h(t)\, dt \right) ^q (1-r)^{k-1}\,dr \le \left( \frac{q}{k}\right) ^q \int _0^1 (h(1-r))^q (1-r)^{q+k-1}\,dr. \end{aligned}$$

Proof of Theorem 4

Suppose first that $\,{\mathcal {H}}_\mu \,$ is a bounded operator from $\,A^p_\alpha \,$ into itself. For $0<b<1$, set

$$\begin{aligned} f_b(z)=\frac{(1-b^2)^{1-\frac{\alpha }{p}}}{(1-bz)^{\frac{2}{p}+1}},\quad z\in {\mathbb {D}}. \end{aligned}$$

Recall that $\,p-\alpha >2$. Then using [29, Lemma 3. 10] with $\,t=\alpha \,$ and $\,c=p-\alpha$, we obtain

$$\begin{aligned} \Vert f_b\Vert _{A^p_\alpha }^p\,=\,(1-b^2)^{p-\alpha }\int _{{\mathbb {D}}}\frac{(1-\vert z\vert ^2)^\alpha }{\vert 1-bz\vert ^{2+p}}\,dA(z)\,\asymp \,1.\end{aligned}$$

Since $\,{\mathcal {H}}_\mu \,$ is bounded on $\,A^p_\alpha$, this implies

$$\begin{aligned} 1\,\gtrsim \Vert {\mathcal {H}}\mu (f_b)\Vert _{A^p_\alpha }.\end{aligned}$$

(18)

We also have

$$\begin{aligned} f_b(z)=\sum _{k=0}^\infty a_{k,b}z^k,\,\,\,(z\in {\mathbb {D}}),\quad \text {with}~ \,a_{k,b}\asymp (1-b^2)^{1-\frac{\alpha }{p}}k^{\frac{2}{p}}b^k. \end{aligned}$$

Since the $a_{k,b}$’s are positive, it is clear that the sequence $\left\{ \sum _{k=0}^\infty \mu _{n+k}a_{k,b}\right\} _{n=0}^\infty$ of the Taylor coefficients of ${\mathcal {H}}_\mu (f_b)$ is a decreasing sequence of non-negative real numbers. Using this, Proposition 1, (18), and the definition of the $a_{k,b}$’s, we obtain

$$\begin{aligned} 1&\gtrsim \Vert {\mathcal {H}}_\mu (f_b)\Vert _{A^p_\alpha }^p\gtrsim \sum _{n=1}^\infty n^{p-\alpha -3}\left( \sum _{k=1}^\infty \mu _{n+k}a_{k,b}\right) ^p \\&= \sum _{n=1}^\infty n^{p-\alpha -3}\left( \sum _{k=1}^\infty a_{k,b}\int _{[0,1)} t^{n+k} d\mu (t)\right) ^p \\&\gtrsim (1-b^2)^{p-\alpha }\sum _{n=1}^\infty n^{p-\alpha -3}\left( \sum _{k=1}^\infty k^{\frac{2}{p}}b^{k}\int _{[b,1)} t^{n+k} d\mu (t)\right) ^p \\&\ge (1-b^2)^{p-\alpha }\sum _{n=1}^\infty n^{p-\alpha -3}\left( \sum _{k=1}^\infty k^{\frac{2}{p}}b^{n+2k}\mu ([b,1))\right) ^p \\&= (1-b^2)^{p-\alpha }\mu ([b,1))^p\sum _{n=1}^\infty n^{p-\alpha -3}b^{np}\left( \sum _{k=1}^\infty k^{\frac{2}{p}}b^{2k}\right) ^p \\&\asymp (1-b^2)^{p-\alpha }\mu ([b,1))^p\frac{1}{(1-b^2)^{2+p}}\sum _{n=1}^\infty n^{p-\alpha -3}b^{np} \\&\asymp (1-b^2)^{p-\alpha }\mu ([b,1))^p\frac{1}{(1-b^2)^{2+p}}\cdot \frac{1}{(1-b^2)^{p-\alpha -2}} \\&\asymp \mu ([b,1))^p\frac{1}{(1-b)^{p}}. \end{aligned}$$

Then it follows that

$$\begin{aligned} \mu \left( [b,1)\right) \,=\,{{\,\mathrm{O}\,}}\left( 1-b\right) ,\quad \text {as}~ b\rightarrow 1, \end{aligned}$$

and, hence, $\mu$ is a Carleson measure.

We turn to prove the other implication. So, suppose that $\,\mu \,$ is a Carleson measure and take $\,f\in A^p_\alpha$. Let $\,q\,$ be defined by the relation $\,\frac{1}{p}+\frac{1}{q}=1\,$ and take $\,\beta =\frac{-\alpha q}{p}=\frac{-\alpha }{p-1}$. Observe that $\,\beta >-1\,$ and that with this election of $\,\beta \,$ the weight in the pairing (17) is identically equal to 1. We have to show that $\,{\mathcal {H}}_\mu f\in A^p_\alpha \,$ which is equal to $\,\left( A^q_\beta \right) ^*\,$ under the pairing $\,\langle \cdot ,\cdot \rangle _{q,\beta ,\alpha }$. So take $\,h\in A^q_\beta$.

$$\begin{aligned} \langle h,{\mathcal {H}}_\mu f \rangle _{q,\beta ,\alpha }\,=&\,\int _{{\mathbb {D}}}h(z)\,\overline{{\mathcal {H}}_\mu f(z)}\,dA(z)\\ =&\, \int _{[0,1)}\overline{f(t)}\left( \int _{{\mathbb {D}}}\frac{h(z)}{1-t\,\overline{z}}\,dA(z)\right) \,d\mu (t)\\ =&\, \int _{[0,1)}\overline{f(t)}\left( \int _0^1\,\frac{r}{\pi }\,\int _0^{2\pi } \frac{h(re^{i\theta })}{1-tre^{-i\theta }}\,d\theta \,dr\right) \,d\mu (t) \\ =&\, \int _{[0,1)}\overline{f(t)}\left( \int _0^1\,\left( \frac{r}{\pi i}\,\int _{\vert \xi \vert =1} \frac{h(r\xi )}{\xi -tr}\,d\xi \right) \,dr\right) \,d\mu (t) \\ =&\, 2\int _{[0,1)}\overline{f(t)}\left( \int _0^1\,rh(r^2t)\,dr\right) \,d\mu (t). \end{aligned}$$

Thus,

$$\begin{aligned}\left| \langle h,{\mathcal {H}}_\mu f \rangle _{q,\beta ,\alpha }\right| \le 2\int _0^1\vert f(t)\vert G(t)\,d\mu (t),\end{aligned}$$

where $G(t)=\int _0^1 r\vert h(r^2t)\vert \,dr$. Using Hölder’s inequality we obtain,

$$\begin{aligned}&\int _{[0,1)} f(t)G(t)\,d\mu (t)=\int _{[0,1)} |f(t)|(1-t)^{\frac{\alpha +1}{p}} G(t)(1-t)^{-\frac{\alpha +1}{p}}\,d\mu (t) \\&\quad \le \left( \int _{[0,1)} |f(t)|^p(1-t)^{\alpha +1}\,d\mu (t) \right) ^{1/p} \cdot \left( \int _{[0,1)} G(t)^q(1-t)^{-\frac{q(\alpha +1)}{p}}\,d\mu (t)\right) ^{1/q}. \end{aligned}$$

Lemma 1 implies that

$$\begin{aligned} \left( \int _{[0,1)} |f(t)|^p(1-t)^{\alpha +1}\,d\mu (t) \right) ^{1/p}\,\lesssim \,\Vert f\Vert _{A^p_\alpha }. \end{aligned}$$

Next we will show that

$$\begin{aligned} \int _{[0,1)} G(t)^q(1-t)^{-\frac{q(\alpha +1)}{p}}\,d\mu (t)\lesssim \Vert h\Vert ^q_{A^q_\beta }.\end{aligned}$$

(19)

This will give that

$$\begin{aligned} \left| \langle h,{\mathcal {H}}_\mu f \rangle _{q,\beta ,\alpha }\right| \,\lesssim \,\Vert f\Vert _{A^p_\alpha }\cdot \Vert h\Vert ^q_{A^q_\beta }. \end{aligned}$$

By the duality theorem, this implies that $\,{\mathcal {H}}_\mu f\in A^p_\alpha$.

Let us prove (19). Observe first that if $0<t<1/2$ then $|h(r^2t)|\le M_\infty (\frac{1}{2},h)$ for each $r\in (0,1)$, thus

$$\begin{aligned} G(t)=\int _0^1 |h(r^2t)|r\,dr\,\le M_\infty \left( \frac{1}{2},h\right) , \quad 0<t<1/2. \end{aligned}$$

Clearly, this implies

$$\begin{aligned} \int _{[0,1/2)} G(t)^q(1-t)^{-\frac{q(\alpha +1)}{p}}\,d\mu (t)\,\lesssim M_\infty ^q\left( \frac{1}{2},h\right) \,\lesssim \,\Vert h\Vert ^q_{A^q_\beta }. \end{aligned}$$

(20)

Notice that $\,-\frac{q(\alpha +1)}{p}\,=\,\frac{p-2-\alpha }{p-1}-1\,>-1$. Making the change of variables $\,r^2t=s$, we obtain $\int _0^1 r|h(r^2t)|\,dr=\frac{1}{2t}\int _0^t |h(s)|\,ds$ and, hence,

$$\begin{aligned}&\int _{[1/2,1)} G(t)^q(1-t)^{-\frac{q(\alpha +1)}{p}}\,d\mu (t)\nonumber \\&\quad = \int _{[1/2,1)} \left( \int _0^1 |h(r^2t)|r\,dr\right) ^q(1-t)^{-\frac{q(\alpha +1)}{p}}\,d\mu (t)\nonumber \\&\quad =\int _{[1/2,1)} \frac{1}{(2t)^q} \left( \int _0^t |h(s)|\,ds\right) ^q(1-t)^{-\frac{q(\alpha +1)}{p}}\,d\mu (t)\nonumber \\&\quad \le \int _{[1/2,1)} \left( \int _0^t M_\infty (s,h)\,ds\right) ^q(1-t)^{-\frac{q(\alpha +1)}{p}}\,d\mu (t)\nonumber \\&\quad \le \int _{[0,1)} \left( \int _{1-t}^1 M_\infty (1-s,h)\,ds\right) ^q(1-t)^{-\frac{q(\alpha +1)}{p}}\,d\mu (t) \end{aligned}$$

(21)

Let us call $H(t)= \left( \int _{1-t}^1 M_\infty (1-s,h)\,ds\right) ^q(1-t)^{-\frac{q(\alpha +1)}{p}}$ for $0\le t <1$. Integrating by parts we obtain the following

$$\begin{aligned} \int _{[0,1)} H(t)\,d\mu (t)=H(0)\mu ([0,1))-\lim _{t\rightarrow 1^-} H(t)\mu ([t,1))+ \int _0^1 \mu ([t,1))H^\prime (t)\,dt. \end{aligned}$$

(22)

The first term is equal to 0. Using the fact that $\mu$ is a Carleson measure we have that

$$\begin{aligned} H(t)\mu ([t,1))&\lesssim (1-t)H(t) \\&= \left( \int _{1-t}^1 M_\infty (1-s,h)\,ds\right) ^q(1-t)^{1-\frac{q(\alpha +1)}{p}} \\&=\left( \int _{0}^t M_\infty (s,h)\,ds\right) ^q(1-t)^{1-\frac{q(\alpha +1)}{p}} . \end{aligned}$$

Since $\,h\in A^q_\beta \,$ we have $\,M_\infty (t,h)=o\left( (1-t)^{-\frac{\beta +2}{q}}\right)$, as $\,t\rightarrow 1$. Then, bearing in mind that $\,\frac{\beta +2}{q}>1$, it follows that

$$\begin{aligned} H(t)\mu ([t,1))\,=\,{{\,\mathrm{o}\,}}\left( (1-t)^{-\beta -2+q}\cdot (1-t)^{1-\frac{q(\alpha +1)}{p}}\right) ={{\,\mathrm{o}\,}}(1),\quad \text {as}~ \,t\rightarrow 1.\end{aligned}$$

(23)

Actually, we have also proved that

$$\begin{aligned} (1-t)H(t)\,=\,{{\,\mathrm{o}\,}}(1),\quad \text {as}~ \,t\rightarrow 1.\end{aligned}$$

(24)

Using that $\mu$ is a Carleson measure, integrating by parts, and using the definition of $\,H\,$ and (24), we obtain

$$\begin{aligned} \int _0^1 \mu ([t,1))H^\prime (t)\,dt&\lesssim \int _0^1 (1-t)H^\prime (t)\,dt\nonumber \\&= \lim _{t\rightarrow 1}(1-t)H(t)\,-\,H(0)\,+\,\int _0^1 H(t)\,dt \nonumber \\&= \int _0^1 \left( \int _{1-t}^1 M_\infty (1-s,h)\,ds\right) ^q(1-t)^{-\frac{q(\alpha +1)}{p}}\,dt. \end{aligned}$$

(25)

Now, using Lemma 3 and [13, Lemma 3], we see that

$$\begin{aligned}&\int _0^1 \left( \int _{1-t}^1 M_\infty (1-s,h)\,ds\right) ^q(1-t)^{-\frac{q(\alpha +1)}{p}}\,dt\\&\quad \lesssim \int _0^1 M^q_\infty (t,h)(1-t)^{\alpha +1}\,dt \lesssim \Vert h\Vert ^q_{A^q_\beta }. \end{aligned}$$

Using this, (25), (23), (22), and (21), it follows that

$$\begin{aligned} \int _{[1/2,1)} G(t)^q(1-t)^{-\frac{q(\alpha +1)}{p}}\,d\mu (t)\,\lesssim \Vert h\Vert ^q_{A^q_\beta }. \end{aligned}$$

This and (20) yield (19). $\square$

Our final aim in this article is to find the analogue of Theorem 4 for Dirichlet spaces. In other words, we wish give an answer to the following question.

Question 2

If $\,\max (-1, p-2)<\alpha <2p-2$, is it true that $\,{\mathcal {H}}_\mu \,$ is a bounded operator from $\,{{\mathcal {D}}^p_{\alpha }}\,$ into itself if and only if $\,\mu \,$ is a Carleson measure?

Since $\,p-1<\alpha <2p-2\,$ implies that $\,{{\mathcal {D}}^p_{\alpha }}\,=\,A^p_{\alpha -p},$    Theorem 4 answers the question affirmatively for these values of $\,p\,$ and $\,\alpha$. It remains to consider the case $\,\max (-1, p-2)<\alpha \le p-1$. We shall prove the following result which gives a positive answer to Question 2 in the case $\,p>1$.

Theorem 5

Suppose that $\,p>1\,$ and $\,p-2<\alpha \le p-1$, and let $\,\mu \,$ be a finite positive Borel measure on $\,[0,1)$.

The operator $\,{\mathcal {H}}_\mu \,$ is well defined on $\,{{\mathcal {D}}^p_{\alpha }}\,$ and it is a bounded operator from $\,{{\mathcal {D}}^p_{\alpha }}\,$ into itself if and only if $\,\mu \,$ is a Carleson measure.

The following two lemmas will be needed in the proof of Theorem 5. The first one follows trivially from Proposition 1.

Lemma 4

Let $\,f\in {\mathcal {H}}ol({\mathbb {D}})$, $f(z)=\sum _{n=0}^\infty a_nz^n\,$$\,(z\in {\mathbb {D}})$. Suppose that $1\,<p<\infty \,$ and $\,p-2<\alpha \le p-1$, and that the sequence $\,\{ a_n\} _{n=0}^\infty \,$ is a decreasing sequence of non-negative real numbers. Then

$$\begin{aligned} f\in {{\mathcal {D}}^p_{\alpha }}\,\,\,\Leftrightarrow \,\,\,\sum _{n=0}^\infty (n+1)^{2p-\alpha -3}a_n^p<\infty . \end{aligned}$$

The following lemma is a generalization of [13, Lemma 4].

Lemma 5

Let $\,\mu \,$ be a positive Borel measure on $\,[0,1)\,$ which is a Carleson measure. Assume that $\,0<p<\infty \,$ and $\,\alpha >-1$. Then there exists a positive constant $\,C=C(p,\alpha ,\mu )\,$ such that for any $\,f\in {{\mathcal {D}}^p_{\alpha }}\,$

$$\begin{aligned} \int _{[0,1)}M_\infty ^p(r,f)(1-r)^{\alpha -p+1}\,d\mu (r)\,\le C\Vert f\Vert _{{{\mathcal {D}}^p_{\alpha }}}^p. \end{aligned}$$

Proof

We argue as in the proof of Lemma 1. Take $\,f\in {{\mathcal {D}}^p_{\alpha }}\,$ and set

$$\begin{aligned}&g(r)=M_\infty ^p(r,f)(1-r)^{\alpha -p+1},\,\, \\&\quad F(r)=\mu ([0,r))\,-\mu ([0,1))\,=\,-\mu ([r,1)),\,\, 0<r<1. \end{aligned}$$

Integrating by parts, we have

$$\begin{aligned}&\int _{[0,1)}M_\infty ^p(r,f)(1-r)^{\alpha -p+1}\,d\mu (r)\,=\,\int _{[0,1)}g(r)\,d\mu (r)\nonumber \\&\quad =\,\lim _{r\rightarrow 1}g(r)F(r)\,-\,g(0)F(0)\,-\int _0^1g^\prime (r)F(r)\,dr \nonumber \\&\quad =\,\vert f(0)\vert ^p\mu ([0,1))\,-\,\lim _{r\rightarrow 1}M_\infty ^p(r,f)(1-r)^{\alpha -p+1}\mu ([r,1))\,\nonumber \\&\qquad +\,\int _0^1g^\prime (r)\mu ([r, 1))\,dr. \end{aligned}$$

(26)

Since $\,f\in {{\mathcal {D}}^p_{\alpha }}\,$ we have that $\,M_\infty ^p(r,f^\prime )={{\,\mathrm{o}\,}}\left( (1-r)^{-2-\alpha }\right)$, as $\,r\rightarrow 1$. Hence, $\,M_\infty ^p(r,f)={{\,\mathrm{o}\,}}\left( (1-r)^{-2-\alpha +p}\right)$, as $\,r\rightarrow 1$. This and the fact that $\,\mu \,$ is a Carleson measure imply that

$$\begin{aligned} \lim _{r\rightarrow 1}M_\infty ^p(r,f)(1-r)^{\alpha -p+1}\mu ([r,1))\,=\,0.\end{aligned}$$

(27)

Using again that $\,\mu \,$ is a Carleson measure and integrating by parts we see that

$$\begin{aligned} \int _0^1g^\prime (r)\mu ([r, 1))\,dr\,\lesssim&\int _0^1g^\prime (r)(1-r)\,dr\\ =&\lim _{r\rightarrow 1}g(r)(1-r)\,-g(0)\,+\,\int _0^1g(r)\,dr\\ \,\le \,&\lim _{r\rightarrow 1}M_\infty ^p(r,f)(1-r)^{\alpha -p +2}\,\\&+\int _0^1M_\infty ^p(r,f)(1-r)^{\alpha -p+1}\,dr \\ =&\int _0^1M_\infty ^p(r,f)(1-r)^{\alpha -p+1}\,dr. \end{aligned}$$

Then, using [13, Lemma  3], it follows that

$$\begin{aligned}\int _0^1g^\prime (r)\mu ([r, 1))\,dr\,\lesssim \,\Vert f\Vert _{{{\mathcal {D}}^p_{\alpha }}}^p.\end{aligned}$$

Using this and (27) in (26) readily yields $\,\int _{[0,1)}M_\infty ^p(r,f)(1-r)^{\alpha -p+1}\,d\mu (r)\,\lesssim \,\Vert f\Vert _{{{\mathcal {D}}^p_{\alpha }}}^p$. $\square$

Proof of Theorem 5

Suppose first that $\,{\mathcal {H}}_\mu \,$ is a bounded operator from $\,{{\mathcal {D}}^p_{\alpha }}\,$ into itself. For $\,1/2<b<1\,$ we set

$$\begin{aligned} f_b(z)=\frac{(1-b^2)^{1-\frac{\alpha }{p}}}{(1-bz)^{2/p}},\quad z\in {\mathbb {D}}. \end{aligned}$$

We have $\,\Vert f_b\Vert _{{{\mathcal {D}}^p_{\alpha }}}\asymp 1$. Then arguing as in the proof of the correspondent implication in Theorem 4 we obtain that $\,\mu \,$ is a Carleson measure. We omit the details.

To prove the other implication, suppose that $\,\mu \,$ is a Carleson measure and take $\,f\in {{\mathcal {D}}^p_{\alpha }}$. Since $\,{\mathcal {H}}_ \mu \,$ and $\,{\mathcal {I}}_\mu \,$ coincide on $\,{{\mathcal {D}}^p_{\alpha }}$, we have to prove that $\,{\mathcal {I}}_\mu f\in {{\mathcal {D}}^p_{\alpha }}\,$ and that $\,\Vert {\mathcal {I}}_\mu f\Vert _ {{\mathcal {D}}^p_{\alpha }}\lesssim \Vert f\Vert _ {{\mathcal {D}}^p_{\alpha }}\,$ or, equivalently, that $\,\left( {\mathcal {I}}_\mu f\right) ^\prime \in A^p_\alpha \,$ and

$$\begin{aligned} \,\Vert \left( {\mathcal {I}}_\mu f\right) ^\prime \Vert _ {A^p_\alpha }\lesssim \Vert f\Vert _ {A^p_\alpha }.\end{aligned}$$

(28)

We shall distinguish two cases.

First case: ${\varvec{\alpha <p-1}}$. Let $\,q\,$ be defined by the relation $\,\frac{1}{p}+\frac{1}{q}=1\,$ and take $\,\beta =\frac{-\alpha q}{p}$. In view of Lemma 2, (28) is equivalent to

$$\begin{aligned} \left| \int _{{\mathbb {D}}}\,h(z)\,\overline{\left( {\mathcal {I}}_\mu f\right) ^\prime (z)}\,dA(z)\right| \,\lesssim \Vert f\Vert _{{{\mathcal {D}}^p_{\alpha }}}\Vert h\Vert _{A^q_\beta },\quad h\in A^q_\beta .\end{aligned}$$

(29)

So, take $\,h\in A^q_\beta$. Just as in the proof of Theorem 1, we have

$$\begin{aligned} \int _{{\mathbb {D}}}\,h(z)\,\overline{\left( {\mathcal {I}}_\mu f\right) ^\prime (z)}\,dA(z)\,=\,\int _{[0,1)}\,t\,\overline{f(t)}\,h(t)\,d\mu (t).\end{aligned}$$

(30)

Set $\,s=-1 + \frac{\alpha +1}{p}$. Observe that $\,ps=\alpha -p+1\,$ and $\,-qs=\beta +1$. Then, using (30), Hölder’s inequality, Lemma 1, and Lemma 5, we obtain

$$\begin{aligned}&\left| \int _{{\mathbb {D}}}\,h(z)\,\overline{\left( {\mathcal {I}}_\mu f\right) ^\prime (z)}\,dA(z)\right| \,\le \, \int _{[0,1)}\vert f(t)\vert (1-t)^s\,\vert h(t)\vert (1-t)^{-s}\,d\mu (t)\\&\quad \le \,\left( \int _{{\mathbb {D}}}\vert f(t)\vert ^p(1-t)^{\alpha -p+1}\,d\mu (t)\right) ^{1/p}\left( \int _{[0,1)}\vert h(t)\vert ^q(1-t)^{\beta +1}\,d\mu (t)\right) ^{1/q} \\&\quad \le \,\left( \int _{{\mathbb {D}}}M_\infty ^p(t,f)(1-t)^{\alpha -p+1}\,d\mu (t)\right) ^{1/p}\\&\qquad \times \left( \int _{[0,1)}M_\infty ^q(t,h)(1-t)^{\beta +1}\,d\mu (t)\right) ^{1/q} \\&\quad \le \,\Vert f\Vert _{{{\mathcal {D}}^p_{\alpha }}}\Vert h\Vert _{A^q_\beta }. \end{aligned}$$

Thus, (29) holds.

Second case: ${\varvec{\alpha =p-1}}$. We let again $\,q\,$ be defined by the relation $\,\frac{1}{p}+\frac{1}{q}=1\,$ and take $\,\beta =q-1$. Using Lemma 2 and arguing as in the preceding case, we have to show that

$$\begin{aligned} \left| \int _{{\mathbb {D}}}\, (1-\vert z\vert ^2)\,h(z)\,\overline{\left( {\mathcal {I}}_\mu f\right) ^\prime (z)}\,dA(z)\right| \,\lesssim \,\Vert f\Vert _{{{\mathcal {D}}^p_{p-1}}}\Vert h\Vert _{A^q_{q-1}},\quad h\in A^q_{q-1}.\end{aligned}$$

(31)

We have

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

(32)

Now, $\int _{{\mathbb {D}}}\frac{h(z)}{(1-t\,\overline{z})^2}\,dA(z)\,=\,h(t)$ and

$$\begin{aligned}&\int _{{\mathbb {D}}}\frac{\vert z\vert ^2\,h(z)}{(1-t\,\overline{z})^2}\,dA(z)\,=\,\int _0^1\,\frac{r^3}{\pi }\int _0^{2\pi }\frac{h(re^{i\theta })\,d\theta }{(1-tre^{-i\theta })^2}\,dr\\&\quad =\, \int _0^1\,\frac{2r^3}{2\pi i}\int _0^{2\pi }\frac{e^{i\theta }h(re^{i\theta })ie^{i\theta }\,d\theta }{(e^{i\theta }-tr)^2}\,dr\,=\,\int _0^1\,\frac{2r^3}{2\pi i}\int _{\vert z\vert =1}\frac{zh(rz)}{(z-tr)^2}\,dz\,dr\\&\quad =\,\int _0^12r^3\left[ h(r^2t)\,+\,r^2th^\prime (r^2t)\right] \,dr. \end{aligned}$$

Then it is clear that $\left| \int _{{\mathbb {D}}}\frac{(1-\vert z\vert ^2)\,h(z)}{(1-t\,\overline{z})^2}\,dA(z)\right| \,\lesssim M_\infty (t,h)$. Using this, (32), Hölder’s inequality, Lemma 1, and Lemma 5, we obtain

$$\begin{aligned}&\left| \int _{{\mathbb {D}}}\, (1-\vert z\vert ^2)\,h(z)\,\overline{\left( {\mathcal {I}}_\mu f\right) ^\prime (z)}\,dA(z)\right| \,\lesssim \,\int _{[0,1)}M_\infty (t,f)\,M_\infty (t,h)\,d\mu (t)\\&\quad \le \,\left( \int _{[0,1)}M_\infty ^p(t,f)\,d\mu (t)\right) ^{1/p} \,\left( \int _{[0,1)}M_\infty ^q(t,h)\,d\mu (t)\right) ^{1/q} \,\\&\quad \le \,\Vert f\Vert _{{{\mathcal {D}}^p_{p-1}}}\Vert h\Vert _{A^q_{q-1}}. \end{aligned}$$

This is (31). $\square$

We shall close the article with some comments about the case $\,p=1\,$ in Question 2. We have the following result.

Theorem 6

Let $\,\mu \,$ be a finite positive Borel measure on $\,[0,1)\,$ and $\,-1<\alpha <0$. If $\,\mu \,$ is a Carleson measure then the operator $\,{\mathcal {H}}_\mu \,$ is a bounded operator form $\,{\mathcal {D}}^1_\alpha \,$ to itself.

Proof

Using [29, Theorem 5. 15, p. 113], we see that $\,A^1_\alpha \,$ can be identified as the dual of the little Bloch space under the pairing

$$\begin{aligned} \langle h, g\rangle \,=\,\int _{{\mathbb {D}}}(1-\vert z\vert ^2)^\alpha \,h(z)\,\overline{g(z)}\,dA(z),\quad h\in {\mathcal {B}}_0,\,\,\,g\in A^1_\alpha .\end{aligned}$$

(33)

Suppose that $\,\mu \,$ is a Carleson measure. Using this duality relation and the fact that $\,{\mathcal {H}}_\mu ={\mathcal {I}}_\mu \,$ on $\,{\mathcal {D}}^1_\alpha$, showing that $\,{\mathcal {H}}_\mu \,$ is a bounded operator from $\,{\mathcal {D}}^1_\alpha \,$ to itself is equivalent to showing that

$$\begin{aligned} \left| \int _{{\mathbb {D}}}(1-\vert z\vert ^2)^{\alpha }\,h(z)\,\overline{\left( {\mathcal {I}}_\mu f\right) ^\prime (z)}\,dA(z)\right| \,\lesssim \Vert h\Vert _{{\mathcal {B}}}\cdot \Vert f\Vert _{{\mathcal {D}}^1_\alpha },\quad h\in {\mathcal {B}}_0,\,\,\,f\in {\mathcal {D}}^1_\alpha .\end{aligned}$$

(34)

Let us prove (34). Take $\,h\in {\mathcal {B}}_0\,$ and $\,f\in {\mathcal {D}}^1_\alpha$. We have

$$\begin{aligned}&\int _{{\mathbb {D}}}(1-\vert z\vert ^2)^{\alpha }\,h(z)\,\overline{\left( {\mathcal {I}}_\mu f\right) ^\prime (z)}\,dA(z)\,\nonumber \\&\quad =\,\int _{[0,1)}t\,\overline{f(t)}\,\int _{{\mathbb {D}}}\frac{(1-\vert z\vert ^2)^\alpha h(z)}{(1-t\,\overline{z})^2}\,dA(z)\,d\mu (t). \end{aligned}$$

(35)

Using [29, Lemma 5. 14, pp. 113-114] we have that the operator $\,T\,$ defined by

$$\begin{aligned} T\phi (\xi )\,=\,(1-\vert \xi \vert ^2)^{-\alpha }\int _{{\mathbb {D}}}\frac{(1-\vert z\vert ^2)^\alpha \phi (z)}{(1-\xi \,\overline{z})^2}\,dA(z) \end{aligned}$$

is a bounded operator from $\,{\mathcal {B}}\,$ into $\,L^\infty ({\mathbb {D}})$. Then it follows that

$$\begin{aligned} \left| \int _{{\mathbb {D}}}\frac{(1-\vert z\vert ^2)^\alpha h(z)}{(1-t\,\overline{z})^2}\,dA(z)\right| \,\lesssim \,\Vert h\Vert _{{\mathcal {B}}}(1-t^2)^\alpha ,\quad t\in [0,1). \end{aligned}$$

Using this in (35), we obtain

$$\begin{aligned} \left| \int _{{\mathbb {D}}}(1-\vert z\vert ^2)^{\alpha }\,h(z)\,\overline{\left( {\mathcal {I}}_\mu f\right) ^\prime (z)}\,dA(z)\right| \,\lesssim \Vert h\Vert _{{\mathcal {B}}}\int _{{\mathbb {D}}}(1-t)^{\alpha }\vert f(t)\vert \,d\mu (t).\end{aligned}$$

(36)

The fact that $\,\mu \,$ is a Carleson measure readily implies that the measure $\,\nu \,$ defined by $\,d\nu (t)=(1-t)^\alpha \,d\mu (t)\,$ is a $\,(1-\alpha )$-Carleson measure. Using Theorem 1 of [28] we see that then $\,\nu \,$ is a Carleson measure for $\,{\mathcal {D}}^1_\alpha$, that is,

$$\begin{aligned} \int _{[0,1)}(1-t)^\alpha \vert g(t)\vert \,d\mu (t)\,\lesssim \,\Vert g\Vert _{{\mathcal {D}}^1_\alpha },\quad g\in {\mathcal {D}}^1_\alpha . \end{aligned}$$

Using this in (36), (34) follows. $\square$

We do not know whether the converse of Theorem 6 is true. This is due to the fact that we do not know whether Lemma 4 remains true for $\,p=1$. The inequality

$$\begin{aligned} \sum _{n=0}^\infty \vert a_n\vert (n+1)^{-(1+\alpha )}\lesssim \Vert f\Vert _{{\mathcal {D}}^1_\alpha }.\end{aligned}$$

(37)

is certainly true with no assumption on the sequence $\,\{ a_n\}$. Indeed, by Hardy’s inequality [11, p. 48], $\sum _{n=1}^\infty \vert a_n\vert r^{n-1}\lesssim \int _0^{2\pi }\vert f^\prime (re^{i\theta })\vert d\theta$. Hence

$$\begin{aligned}&\Vert f\Vert _{{\mathcal {D}}^1_\alpha }\asymp \int _0^1(1-r)^\alpha \int _0^{2\pi }\vert f^\prime (re^{i\theta })\vert d\theta dr \\&\quad \gtrsim \sum _{n=1}^\infty \vert a_n\vert \int _0^1(1-r)^\alpha r^{n-1}\,dr = \sum _{n=1}^\infty \vert a_n\vert B(\alpha +1, n), \end{aligned}$$

where $\,B(\cdot ,\cdot )\,$ is the Beta function. Stirling’s formula gives $\,B(\alpha +1, n)\asymp n^{-(\alpha +1)}$ and then (37) follows.

However, the proof of Theorem D in [23] does not seen to work to prove the opposite inequality when $\{a_n\}$ is decreasing.